Elementary Differential Equations, 10th.pdf

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Elementary Differential

Equations and Boundary

Value Problems

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TENTHEDITION

Elementary Differential

Equations and

Boundary Value

Problems

William E. Boyce

Edward P. Hamilton Professor Emeritus

Richard C. DiPrima

formerly Eliza Ricketts Foundation Professor

Department of Mathematical Sciences

Rensselaer Polytechnic Institute

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To Elsa and in loving memory of Maureen

To Siobhan, James, Richard, Jr., Carolyn, and Ann

And to the next generation:

Charles, Aidan, Stephanie, Veronica, and Deirdre

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The Authors

William E. Boyce received his B.A. degree in Mathematics from Rhodes College,

and his M.S. and Ph.D. degrees in Mathematics from Carnegie-Mellon University.

He is a member of the American Mathematical Society, the Mathematical Associ-

ation of America, and the Society for Industrial and Applied Mathematics. He is

currently the Edward P. Hamilton Distinguished Professor Emeritus of Science Ed-

ucation (Department of Mathematical Sciences) at Rensselaer. He is the author

of numerous technical papers in boundary value problems and random differential

equations and their applications. He is the author of several textbooks including

two differential equations texts, and is the coauthor (with M.H. Holmes, J.G. Ecker,

and W.L. Siegmann) of a text on using Maple to explore Calculus. He is also coau-

thor (with R.L. Borrelli and C.S. Coleman) of Differential Equations Laboratory

Workbook (Wiley 1992), which received the EDUCOM Best Mathematics Curricu-

lar Innovation Award in 1993. Professor Boyce was a member of the NSF-sponsored

CODEE (Consortium for Ordinary Differential Equations Experiments) that led to

the widely-acclaimed ODE Architect. He has also been active in curriculum inno-

vation and reform. Among other things, he was the initiator of the “Computers in

Calculus” project at Rensselaer, partially supported by the NSF. In 1991 he received

the William H. Wiley Distinguished Faculty Award given by Rensselaer.

Richard C. DiPrima (deceased) received his B.S., M.S., and Ph.D. degrees in

Mathematics from Carnegie-Mellon University. He joined the faculty of Rensselaer

Polytechnic Institute after holding research positions at MIT, Harvard, and Hughes

Aircraft. He held the Eliza Ricketts Foundation Professorship of Mathematics at

Rensselaer, was a fellow of the American Society of Mechanical Engineers, the

American Academy of Mechanics, and the American Physical Society. He was also

a member of the American Mathematical Society, the Mathematical Association of

America, and the Society for Industrial and Applied Mathematics. He served as the

Chairman of the Department of Mathematical Sciences at Rensselaer, as President

of the Society for Industrial and Applied Mathematics, and as Chairman of the Ex-

ecutive Committee of the Applied Mechanics Division ofASME. In 1980, he was the

recipient of the William H. Wiley Distinguished Faculty Award given by Rensselaer.

He received Fulbright fellowships in 1964–65 and 1983 and a Guggenheim fellow-

ship in 1982–83. He was the author of numerous technical papers in hydrodynamic

stability and lubrication theory and two texts on differential equations and boundary

value problems. Professor DiPrima died on September 10, 1984.

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PREFACE

This edition, like its predecessors, is written from the viewpoint of the applied

mathematician, whose interest in differential equations may be sometimes quite

theoretical, sometimes intensely practical, and often somewhere in between. We

have sought to combine a sound and accurate (but not abstract) exposition of the

elementary theory of differential equations with considerable material on methods

of solution, analysis, and approximation that have proved useful in a wide variety of

applications.

The book is written primarily for undergraduate students of mathematics, science,

or engineering, who typically take a course on differential equations during their

ﬁrst or second year of study. The main prerequisite for reading the book is a working

knowledge of calculus,gained from a normal two- or three-semester course sequence

or its equivalent. Some familiarity with matrices will also be helpful in the chapters

on systems of differential equations.

To be widely useful, a textbook must be adaptable to a variety of instructional

strategies. This implies at least two things. First, instructors should have maximum

ﬂexibility to choose both the particular topics they wish to cover and the order in

which they want to cover them. Second, the book should be useful to students who

have access to a wide range of technological capability.

With respect to content, we provide this ﬂexibility by making sure that, so far as

possible, individual chapters are independent of each other. Thus, after the basic

parts of the ﬁrst three chapters are completed (roughly Sections 1.1 through 1.3, 2.1

through 2.5, and 3.1 through 3.5), the selection of additional topics,and the order and

depth in which they are covered, are at the discretion of the instructor. Chapters 4

through 11 are essentially independent of each other, except that Chapter 7 should

precede Chapter 9 and that Chapter 10 should precede Chapter 11. This means that

there are multiple pathways through the book, and many different combinations

have been used effectively with earlier editions.

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x Preface

With respect to technology, we note repeatedly in the text that computers are ex-

tremely useful for investigating differential equations and their solutions, and many

of the problems are best approached with computational assistance. Nevertheless,

the book is adaptable to courses having various levels of computer involvement,

ranging from little or none to intensive. The text is independent of any particular

hardware platform or software package.

Many problems are marked with the symbol

to indicate that we consider them

to be technologically intensive. Computers have at least three important uses in a

differential equations course. The ﬁrst is simply to crunch numbers, thereby gen-

erating accurate numerical approximations to solutions. The second is to carry out

symbolic manipulations that would be tedious and time-consuming to do by hand.

Finally, and perhaps most important of all, is the ability to translate the results of

numerical or symbolic computations into graphical form, so that the behavior of

solutions can be easily visualized. The marked problems typically involve one or

more of these features. Naturally, the designation of a problem as technologically

intensive is a somewhat subjective judgment, and the

is intended only as a guide.

Many of the marked problems can be solved, at least in part, without computa-

tional help, and a computer can also be used effectively on many of the unmarked

problems.

From a student’s point of view, the problems that are assigned as homework and

that appear on examinations drive the course. We believe that the most outstanding

feature of this book is the number, and above all the variety and range, of the prob-

lems that it contains. Many problems are entirely straightforward, but many others

are more challenging,and some are fairly open-ended and can even serve as the basis

for independent student projects. There are far more problems than any instructor

can use in any given course, and this provides instructors with a multitude of choices

in tailoring their course to meet their own goals and the needs of their students.

The motivation for solving many differential equations is the desire to learn some-

thing about an underlying physical process that the equation is believed to model.

It is basic to the importance of differential equations that even the simplest equa-

tions correspond to useful physical models, such as exponential growth and decay,

spring–mass systems, or electrical circuits. Gaining an understanding of a complex

natural process is usually accomplished by combining or building upon simpler and

more basic models. Thus a thorough knowledge of these basic models, the equations

that describe them, and their solutions is the ﬁrst and indispensable step toward the

solution of more complex and realistic problems. We describe the modeling process

in detail in Sections 1.1, 1.2, and 2.3. Careful constructions of models appear also in

Sections 2.5 and 3.7 and in the appendices to Chapter 10. Differential equations re-

sulting from the modeling process appear frequently throughout the book, especially

in the problem sets.

The main reason for including fairly extensive material on applications and math-

ematical modeling in a book on differential equations is to persuade students that

mathematical modeling often leads to differential equations, and that differential

equations are part of an investigation of problems in a wide variety of other ﬁelds.

We also emphasize the transportability of mathematical knowledge: once you mas-

ter a particular solution method, you can use it in any ﬁeld of application in which an

appropriate differential equation arises. Once these points are convincingly made,

we believe that it is unnecessary to provide speciﬁc applications of every method

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Preface xi

of solution or type of equation that we consider. This helps to keep this book to

a reasonable size, and in any case, there is only a limited time in most differential

equations courses to discuss modeling and applications.

Nonroutine problems often require the use of a variety of tools, both analytical

and numerical. Paper-and-pencil methods must often be combined with effective

use of a computer. Quantitative results and graphs, often produced by a computer,

serve to illustrate and clarify conclusions that may be obscured by complicated ana-

lytical expressions. On the other hand, the implementation of an efﬁcient numerical

procedure typically rests on a good deal of preliminary analysis—to determine the

qualitative features of the solution as a guide to computation, to investigate limit-

ing or special cases, or to discover which ranges of the variables or parameters may

require or merit special attention. Thus, a student should come to realize that investi-

gating a difﬁcult problem may well require both analysis and computation; that good

judgment may be required to determine which tool is best suited for a particular task;

and that results can often be presented in a variety of forms.

We believe that it is important for students to understand that (except perhaps

in courses on differential equations) the goal of solving a differential equation is

seldom simply to obtain the solution. Rather, we seek the solution in order to obtain

insight into the behavior of the process that the equation purports to model. In

other words, the solution is not an end in itself. Thus, we have included in the text

a great many problems, as well as some examples, that call for conclusions to be

drawn about the solution. Sometimes this takes the form of ﬁnding the value of the

independent variable at which the solution has a certain property, or determining

the long-term behavior ofthe solution. Otherproblemsask for the effectof variations

in a parameter, or for the determination of a critical value of a parameter at which

the solution experiences a substantial change. Such problems are typical of those

that arise in the applications of differential equations, and, depending on the goals

of the course,an instructor has the option of assigning few or many of these problems.

Readers familiar with the preceding edition will observe that the general structure

of the book is unchanged. The revisions that we have made in this edition are in

many cases the result of suggestions from users of earlier editions. The goals are

to improve the clarity and readability of our presentation of basic material about

differential equations and their applications. More speciﬁcally, the most important

revisions include the following:

1. Sections 8.5 and 8.6 have been interchanged, so that the more advanced topics appear at

the end of the chapter.

2. Derivations and proofs in several chapters have been expanded or rewritten to provide

more details.

3. The fact that the real and imaginary parts of a complex solution of a real problem are also

solutions now appears as a theorem in Sections 3.2 and 7.4.

4. The treatment of generalized eigenvectors in Section 7.8 has been expanded both in the

text and in the problems.

5. There are about twenty new or revised problems scattered throughout the book.

6. There are new examples in Sections 2.1, 3.8, and 7.5.

7. About a dozen ﬁgures have been modiﬁed, mainly by using color to make the essen-

tial feature of the ﬁgure more prominent. In addition, numerous captions have been

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xii Preface

expanded to clarify the purpose of the ﬁgure without requiring a search of the

surrounding text.

8. There are several new historical footnotes, and some others have been expanded.

The authors have found differential equations to be a never-ending source of in-

teresting, and sometimes surprising, results and phenomena. We hope that users of

this book, both students and instructors, will share our enthusiasm for the subject.

William E. Boyce

Grafton, New York

March 13, 2012

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Preface xiii

Supplemental Resources for Instructors and Students

An Instructor’s Solutions Manual, ISBN 978-0-470-45834-1, includes solutions for all

problems not contained in the Student Solutions Manual.

A Student Solutions Manual, ISBN 978-0-470-45833-4, includes solutions for se-

lected problems in the text.

A Book Companion Site, www.wiley.com/college/boyce, provides a wealth of re-

sources for students and instructors, including

•

PowerPoint slides of important deﬁnitions, examples, and theorems from the

book,as well as graphics for presentation in lectures or for study and note taking.

•

Chapter Review Sheets, which enable students to test their knowledge of key

concepts. For further review, diagnostic feedback is provided that refers to per-

tinent sections in the text.

•

Mathematica, Maple, and MATLAB data ﬁles for selected problems in the text

providing opportunities for further exploration of important concepts.

•

Projects that deal with extended problems normally not included among tradi-

tional topics in differential equations,many involving applications from a variety

of disciplines. These vary in length and complexity, and they can be assigned as

individual homework or as group assignments.

A series of supplemental guidebooks, also published by John Wiley & Sons, can be

used with Boyce/DiPrima in order to incorporate computing technologies into the

course. These books emphasize numerical methods and graphical analysis, showing

how these methods enable us to interpret solutions of ordinary differential equa-

tions (ODEs) in the real world. Separate guidebooks cover each of the three major

mathematical software formats, but the ODE subject matter is the same in each.

•

Hunt, Lipsman, Osborn, and Rosenberg, Differential Equations with MATLAB,

3rd ed., 2012, ISBN 978-1-118-37680-5

•

Hunt, Lardy, Lipsman, Osborn, and Rosenberg, Differential Equations with

Maple, 3rd ed., 2008, ISBN 978-0-471-77317-7

•

Hunt, Outing, Lipsman, Osborn, and Rosenberg, Differential Equations with

Mathematica, 3rd ed., 2009, ISBN 978-0-471-77316-0

WileyPLUS

WileyPLUS is an innovative, research-based online environment for effective teach-

ing and learning.

WileyPLUS builds students’ conﬁdence because it takes the guesswork out of

studying by providing students with a clear roadmap: what to do, how to do it, if they

did it right. Students will take more initiative so you’ll have greater impact on their

achievement in the classroom and beyond.

WileyPLUS, is loaded with all of the supplements above, and it also features

•

The E-book, which is an exact version of the print text but also features hyper-

links to questions, deﬁnitions, and supplements for quicker and easier support.

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xiv Preface

•

Guided Online (GO) Exercises, which prompt students to build solutions step-

by-step. Rather than simply grading an exercise answer as wrong, GO problems

show students precisely where they are making a mistake.

•

Homework management tools, which enable instructors easily to assign and

grade questions, as well as to gauge student comprehension.

•

QuickStart pre-designed reading and homework assignments. Use them as is,

or customize them to ﬁt the needs of your classroom.

•

Interactive Demonstrations, based on ﬁgures from the text, which help reinforce

and deepen understanding of the key concepts of differential equations. Use

them in class or assign them as homework. Worksheets are provided to help

guide and structure the experience of mastering these concepts.

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ACKNOWLEDGMENTS

It is a pleasure to express my appreciation to the many people who have generously

assisted in various ways in the preparation of this book.

To the individuals listed below, who reviewed the manuscript and/or provided

valuable suggestions for its improvement:

Vincent Bonini, California Polytechnic State University, San Luis Obispo

Fengxin Chen, University of Texas San Antonio

Carmen Chicone, University of Missouri

Matthew Fahy, Northern Arizona University

Isaac Goldbring, University of California at Los Angeles

Anton Gorodetski, University of California Irvine

Mansoor Haider, North Carolina State University

David Handron, Carnegie Mellon University

Thalia D. Jeffres,Wichita State University

Akhtar Khan, Rochester Institute of Technology

Joseph Koebbe, Utah State University

Ilya Kudish, Kettering University

Tong Li, University of Iowa

Wen-Xiu Ma, University of South Florida

Aldo Manfroi, University of Illinois Urbana-Champaign

Will Murray, California State University Long Beach

Harold R. Parks, Oregon State University

William Paulsen, Arkansas State University

Shagi-Di Shih, University of Wyoming

John Starrett, New Mexico Institute of Mining and Technology

David S. Torain II, Hampton University

George Yates, Youngstown State University

Nung Kwan (Aaron) Yip, Purdue University

Yue Zhao, University of Central Florida

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xvi Acknowledgments

To my colleagues and students at Rensselaer, whose suggestions and reactions

through the years have done much to sharpen my knowledge of differential equa-

tions, as well as my ideas on how to present the subject.

To those readers of the preceding edition who called errors or omissions to my

attention.

To Tamas Wiandt (Rochester Institute of Technology), who is primarily responsi-

ble for the revision of the Instructor’s Solutions Manual and the Student Solutions

Manual, and to Charles Haines (Rochester Institute of Technology), who assisted in

this undertaking.

To Tom Polaski (Winthrop University), who checked the answers in the back of

the text and the Instructor’s Solutions Manual for accuracy.

To David Ryeburn (Simon Fraser University), who carefully checked the entire

manuscript and pageproofs at leastfour times andis responsible formany corrections

and clariﬁcations.

To Douglas Meade (University of South Carolina), who gave indispensable assis-

tance in a variety of ways: by reading the entire manuscript at an early stage and

offering numerous suggestions; by materially assisting in expanding the historical

footnotes and updating the references; and by assuming the primary responsibility

for checking the accuracy of the page proofs.

To the editorial and production staff of John Wiley & Sons, who have always been

ready to offer assistance and have displayed the highest standards of professionalism.

Finally, and most important, to my wife Elsa for discussing questions both math-

ematical and stylistic, and above all for her unfailing support and encouragement

during the revision process. In a very real sense, this book is a joint product.

William E. Boyce

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xvii

CONTENTS

Chapter 1 Introduction 1

1.1 Some Basic Mathematical Models; Direction Fields 1

1.2 Solutions of Some Differential Equations 10

1.3 Classiﬁcation of Differential Equations 19

1.4 Historical Remarks 26

Chapter 2

First Order Differential Equations 31

2.1 Linear Equations; Method of Integrating Factors 31

2.2 Separable Equations 42

2.3 Modeling with First Order Equations 51

2.4 Differences Between Linear and Nonlinear Equations 68

2.5 Autonomous Equations and Population Dynamics 78

2.6 Exact Equations and Integrating Factors 95

2.7 Numerical Approximations: Euler’s Method 102

2.8 The Existence and Uniqueness Theorem 112

2.9 First Order Difference Equations 122

Chapter 3

Second Order Linear Equations 137

3.1 Homogeneous Equations with Constant Coefﬁcients 137

3.2 Solutions of Linear Homogeneous Equations; the Wronskian 145

3.3 Complex Roots of the Characteristic Equation 158

3.4 Repeated Roots; Reduction of Order 167

3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 175

3.6 Variation of Parameters 186

3.7 Mechanical and Electrical Vibrations 192

3.8 Forced Vibrations 207

Chapter 4

Higher Order Linear Equations 221

4.1 General Theory of nth Order Linear Equations 221

4.2 Homogeneous Equations with Constant Coefﬁcients 228

4.3 The Method of Undetermined Coefﬁcients 236

4.4 The Method of Variation of Parameters 241

Chapter 5

Series Solutions of Second Order Linear Equations 247

5.1 Review of Power Series 247

5.2 Series Solutions Near an Ordinary Point, Part I 254

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xviii Contents

5.3 Series Solutions Near an Ordinary Point, Part II 265

5.4 Euler Equations; Regular Singular Points 272

5.5 Series Solutions Near a Regular Singular Point, Part I 282

5.6 Series Solutions Near a Regular Singular Point, Part II 288

5.7 Bessel’s Equation 296

Chapter 6

The Laplace Transform 309

6.1 Deﬁnition of the Laplace Transform 309

6.2 Solution of Initial Value Problems 317

6.3 Step Functions 327

6.4 Differential Equations with Discontinuous Forcing Functions 336

6.5 Impulse Functions 343

6.6 The Convolution Integral 350

Chapter 7

Systems of First Order Linear Equations 359

7.1 Introduction 359

7.2 Review of Matrices 368

7.3 Systems of Linear Algebraic Equations; Linear Independence, Eigenvalues,

Eigenvectors 378

7.4 Basic Theory of Systems of First Order Linear Equations 390

7.5 Homogeneous Linear Systems with Constant Coefﬁcients 396

7.6 Complex Eigenvalues 408

7.7 Fundamental Matrices 421

7.8 Repeated Eigenvalues 429

7.9 Nonhomogeneous Linear Systems 440

Chapter 8

Numerical Methods 451

8.1 The Euler or Tangent Line Method 451

8.2 Improvements on the Euler Method 462

8.3 The Runge–Kutta Method 468

8.4 Multistep Methods 472

8.5 Systems of First Order Equations 478

8.6 More on Errors; Stability 482

Chapter 9

Nonlinear Differential Equations and Stability 495

9.1 The Phase Plane: Linear Systems 495

9.2 Autonomous Systems and Stability 508

9.3 Locally Linear Systems 519

9.4 Competing Species 531

9.5 Predator–Prey Equations 544

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Contents xix

9.6 Liapunov’s Second Method 554

9.7 Periodic Solutions and Limit Cycles 565

9.8 Chaos and Strange Attractors: The Lorenz Equations 577

Chapter 10

Partial Differential Equations and Fourier Series 589

10.1 Two-Point Boundary Value Problems 589

10.2 Fourier Series 596

10.3 The Fourier Convergence Theorem 607

10.4 Even and Odd Functions 614

10.5 Separation of Variables; Heat Conduction in a Rod 623

10.6 Other Heat Conduction Problems 632

10.7 The Wave Equation: Vibrations of an Elastic String 643

10.8 Laplace’s Equation 658

Appendix A Derivation of the Heat Conduction Equation 669

Appendix B Derivation of the Wave Equation 673

Chapter 11

Boundary Value Problems and Sturm–Liouville Theory 677

11.1 The Occurrence of Two-Point Boundary Value Problems 677

11.2 Sturm–Liouville Boundary Value Problems 685

11.3 Nonhomogeneous Boundary Value Problems 699

11.4 Singular Sturm–Liouville Problems 714

11.5 Further Remarks on the Method of Separation of Variables: A Bessel

Series Expansion 721

11.6 Series of Orthogonal Functions: Mean Convergence 728

Answers to Problems 739

Index 799

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August 7, 2012 21:03 c01 Sheet number 1 Page number 1 cyan black

CHAPTER

Introduction

In this chapter we give perspective to your study of differential equations in several

different ways. First,we use two problems to illustrate some of the basic ideas that we

will return to, and elaborate upon, frequently throughout the remainder of the book.

Later, to provide organizational structure for the book, we indicate several ways

of classifying differential equations. Finally, we outline some of the major trends

in the historical development of the subject and mention a few of the outstanding

mathematicians who have contributed to it. The study of differential equations has

attracted the attention of many of the world’s greatest mathematicians during the

past three centuries. Nevertheless, it remains a dynamic ﬁeld of inquiry today, with

many interesting open questions.

1.1 Some Basic Mathematical Models; Direction Fields

Before embarking on a serious study of differential equations (for example, by read-

ing this book or major portions of it), you should have some idea of the possible

beneﬁts to be gained by doing so. For some students the intrinsic interest of the

subject itself is enough motivation, but for most it is the likelihood of important

applications to other ﬁelds that makes the undertaking worthwhile.

Many of the principles, or laws, underlying the behavior of the natural world are

statements or relations involving rates at which things happen. When expressed

in mathematical terms, the relations are equations and the rates are derivatives.

Equations containing derivatives are differential equations.Therefore,to understand

and to investigate problems involving the motion of ﬂuids, the ﬂow of current in elec-

tric circuits, the dissipation of heat in solid objects, the propagation and detection of

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2 Chapter 1. Intr oduction

seismic waves, or the increase or decrease of populations, among many others, it is

necessary to know something about differential equations.

A differential equation that describes some physical process is often called a math-

ematical model of the process, and many such models are discussed throughout this

book. In this section we begin with two models leading to equations that are easy to

solve. It is noteworthy that even the simplest differential equations provide useful

models of important physical processes.

EXAMPLE

AFalling

Object

Suppose that an object is falling in the atmosphere near sea level. Formulate a differential

equation that describes the motion.

We begin by introducing letters to represent various quantities that may be of interest in this

problem. The motion takes place during a certain time interval, so let us use t to denote time.

Also, let us use v to represent the velocity of the falling object. The velocity will presumably

change with time, so we think of v as a function of t; in other words, t is the independent

variable and v is the dependent variable. The choice of units of measurement is somewhat

arbitrary, and there is nothing in the statement of the problem to suggest appropriate units,

so we are free to make any choice that seems reasonable. To be speciﬁc, let us measure time

t in seconds and velocity v in meters/second. Further, we will assume that v is positive in the

downward direction—that is, when the object is falling.

The physical law that governs the motion of objects is Newton’s second law, which states

that the mass of the object times its acceleration is equal to the net force on the object. In

mathematical terms this law is expressed by the equation

F = ma, (1)

where m is the mass of the object, a is its acceleration, and F is the net force exerted on the

object. To keep our units consistent, we will measure m in kilograms, a in meters/second

, and

F in newtons. Of course, a is related to v by a = dv/dt, so we can rewrite Eq. (1) in the form

F = m(dv/dt). (2)

Next, consider the forces that act on the object as it falls. Gravity exerts a force equal to

the weight of the object, or mg, where g is the acceleration due to gravity. In the units we have

chosen, g has been determined experimentally to be approximately equal to 9.8 m/s

near

the earth’s surface. There is also a force due to air resistance, or drag, that is more difﬁcult to

model. This is not the place for an extended discussion of the drag force; sufﬁce it to say that it

is often assumed that the drag is proportional to the velocity,and we will make that assumption

here.Thus the drag force has the magnitude γv,where γ is a constant called the drag coefﬁcient.

The numerical value of the drag coefﬁcient varies widely from one object to another; smooth

streamlined objects have much smaller drag coefﬁcients than rough blunt ones. The physical

units for γ are mass/time, or kg/s for this problem; if these units seem peculiar, remember that

γv must have the units of force, namely, kg·m/s

In writing an expression for the net force F , we need to remember that gravity always acts

in the downward (positive) direction, whereas, for a falling object, drag acts in the upward

(negative) direction, as shown in Figure 1.1.1. Thus

F = mg − γv (3)

and Eq. (2) then becomes

= mg − γv. (4)

Equation (4) is a mathematical model of an object falling in the atmosphere near sea level.

Note that the model contains the three constants m, g, and γ. The constants m and γ depend

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1.1 Some Basic Mathematical Models; Direction Fields 3

very much on the particular object that is falling, and they are usually different for different

objects. It is common to refer to them as parameters, since they may take on a range of values

during the course of an experiment. On the other hand, g is a physical constant, whose value

is the same for all objects.

γ υ

FIGURE 1.1.1 Free-body diagram of the forces on a falling object.

To solve Eq. (4), we need to ﬁnd a function v = v(t) that satisﬁes the equation. It

is not hard to do this, and we will show you how in the next section. For the present,

however,let us see what we can learn about solutions without actually ﬁnding any of

them. Our task is simpliﬁed slightly if we assign numerical values to m and γ, but the

procedure is the same regardless of which values we choose. So, let us suppose that

m = 10 kg and γ = 2 kg/s. Then Eq. (4) can be rewritten as

d v

= 9.8 −

. (5)

EXAMPLE

AFalling

Object

(continued)

Investigate the behavior of solutions of Eq. (5) without solving the differential equation.

First let us consider what information canbe obtained directlyfrom the differentialequation

itself. Suppose that the velocity v has a certain given value.Then, by evaluating the right side of

Eq. (5),we can ﬁnd the corresponding value of d v/dt. For instance,if v = 40,then dv/dt = 1.8.

This means that the slope of a solution v = v(t) has the value 1.8 at any point where v = 40.

We can display this information graphically in the tv-plane by drawing short line segments

with slope 1.8 at several points on the line v = 40. Similarly, if v = 50, then dv/dt =−0.2, so

we draw line segments with slope −0.2 at several points on the line v = 50. We obtain Figure

1.1.2 by proceeding in the same way with other values of v. Figure 1.1.2 is an example of what

is called a direction ﬁeld or sometimes a slope ﬁeld.

Remember that a solution of Eq. (5) is a function v = v(t) whose graph is a curve in the

tv-plane. The importance of Figure 1.1.2 is that each line segment is a tangent line to one

of these solution curves. Thus, even though we have not found any solutions, and no graphs of

solutions appear in the ﬁgure, we can nonetheless draw some qualitative conclusions about

the behavior of solutions. For instance, if v is less than a certain critical value, then all the line

segments have positive slopes, and the speed of the falling object increases as it falls. On the

other hand, if v is greater than the critical value, then the line segments have negative slopes,

and the falling object slows down as it falls.What is this critical value of v that separates objects

whose speed is increasing from those whose speed is decreasing? Referring again to Eq. (5),

we ask what value of v will cause dv/dt to be zero. The answer is v = (5)(9.8) = 49

m/s

In fact, the constant function v(t) = 49 is a solution of Eq. (5). To verify this statement,

substitute v(t) = 49 into Eq. (5) and observe that each side of the equation is zero. Because

it does not change with time, the solution v(t) = 49 is called an equilibrium solution.Itis

the solution that corresponds to a perfect balance between gravity and drag. In Figure 1.1.3

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4 Chapter 1. Intr oduction

we show the equilibrium solution v(t) = 49 superimposed on the direction ﬁeld. From this

ﬁgure we can draw another conclusion, namely, that all other solutions seem to be converging

to the equilibrium solution as t increases. Thus,in this context,the equilibrium solution is often

called the terminal velocity.

24 t6810

FIGURE 1.1.2 A direction ﬁeld for Eq. (5): d v/dt = 9.8 − (v/5).

24 t6810

FIGURE 1.1.3 Direction ﬁeld and equilibrium solution for Eq. (5): dv/dt = 9.8 − (v/5).

The approach illustrated in Example 2 can be applied equally well to the more

general Eq. (4), where the parameters m and γ are unspeciﬁed positive numbers.

The results are essentially identical to those of Example 2. The equilibrium solution

of Eq. (4) is v(t) = mg/γ. Solutions below the equilibriumsolution increasewith time,

those above it decrease with time, and all other solutions approach the equilibrium

solution as t becomes large.

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1.1 Some Basic Mathematical Models; Direction Fields 5

Direction Fields. Direction ﬁelds are valuable tools in studying the solutions of

differential equations of the form

= f (t, y), (6)

where f is a given function of the two variables t and y, sometimes referred to as the

rate function. A direction ﬁeld for equations of the form (6) can be constructed by

evaluating f at each point of a rectangular grid. At each point of the grid, a short line

segment is drawn whose slope is the value of f at that point. Thus each line segment

is tangent to the graph of the solution passing through that point. A direction ﬁeld

drawn on a fairly ﬁne grid gives a good picture of the overall behavior of solutions of

a differential equation. Usually a grid consisting of a few hundred points is sufﬁcient.

The construction of a direction ﬁeld is often a useful ﬁrst step in the investigation of

a differential equation.

Two observations are worth particular mention. First, in constructing a direction

ﬁeld, we do not have to solve Eq. (6); we just have to evaluate the given function

f (t, y) many times.Thus direction ﬁelds can be readily constructed even for equations

that may be quite difﬁcult to solve. Second, repeated evaluation of a given function

is a task for which a computer is well suited, and you should usually use a computer

to draw a direction ﬁeld. All the direction ﬁelds shown in this book, such as the one

in Figure 1.1.2, were computer-generated.

Field Mice and Owls. Now let us look at another, quite different example. Consider

a population of ﬁeld mice who inhabit a certain rural area. In the absence of

predators we assume that the mouse population increases at a rate proportional

to the current population. This assumption is not a well-established physical law

(as Newton’s law of motion is in Example 1), but it is a common initial hypothesis

in a study of population growth. If we denote time by t and the mouse population by

p(t),then the assumption about population growth can be expressed by the equation

= rp, (7)

where the proportionality factor r is called the rate constant or growth rate.Tobe

speciﬁc,suppose that time is measured in months and that the rate constant r has the

value 0.5/month. Then each term in Eq. (7) has the units of mice/month.

Now let us add to the problem by supposing that several owls live in the same

neighborhood and that they kill 15 ﬁeld mice per day.To incorporate this information

into the model, we must add another term to the differential equation (7), so that it

becomes

= 0.5p − 450. (8)

Observe that the predation term is −450 rather than −15 because time is measured

in months, so the monthly predation rate is needed.

A better model of population growth is discussed in Section 2.5.

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6 Chapter 1. Intr oduction

EXAMPLE

Investigate the solutions of Eq. (8) graphically.

A direction ﬁeld for Eq. (8) is shown in Figure 1.1.4. For sufﬁciently large values of p it can

be seen from the ﬁgure, or directly from Eq. (8) itself, that dp/dt is positive, so that solutions

increase. On the other hand, if p is small, then dp/dt is negative and solutions decrease. Again,

the critical value of p that separates solutions that increase from those that decrease is the

value of p for which dp/dt is zero. By setting dp/dt equal to zero in Eq. (8) and then solving

for p,we ﬁnd the equilibrium solution p(t) = 900, for which the growth term and the predation

term in Eq. (8) are exactly balanced. The equilibrium solution is also shown in Figure 1.1.4.

12 t345

900

850

800

950

1000

FIGURE 1.1.4 Direction ﬁeld and equilibrium solution for Eq. (8): dp/dt = 0.5p − 450.

Comparing Examples 2 and 3, we note that in both cases the equilibrium solution

separates increasing from decreasing solutions. In Example 2 other solutions con-

verge to, or are attracted by, the equilibrium solution, so that after the object falls far

enough, an observer will see it moving at very nearly the equilibrium velocity. On

the other hand, in Example 3 other solutions diverge from, or are repelled by, the

equilibrium solution. Solutions behave very differently depending on whether they

start above or below the equilibrium solution. As time passes, an observer might see

populations either much larger or much smaller than the equilibrium population,but

the equilibrium solution itself will not, in practice, be observed. In both problems,

however, the equilibrium solution is very important in understanding how solutions

of the given differential equation behave.

A more general version of Eq. (8) is

= rp − k, (9)

where the growth rate r and the predation rate k are unspeciﬁed. Solutions of this

more general equation are very similar to those of Eq. (8). The equilibrium solution

of Eq. (9) is p(t) = k /r. Solutions above the equilibrium solution increase, while

those below it decrease.

You should keep in mind that both of the models discussed in this section have

their limitations. The model (5) of the falling object is valid only as long as the

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1.1 Some Basic Mathematical Models; Direction Fields 7

object is falling freely, without encountering any obstacles. The population model

(8) eventually predicts negative numbers of mice (if p < 900) or enormously large

numbers (ifp > 900). Both of thesepredictions areunrealistic,so this model becomes

unacceptable after a fairly short time interval.

Constructing Mathematical Models. In applying differential equations to any of the

numerous ﬁelds in which they are useful, it is necessary ﬁrst to formulate the appro-

priate differential equation that describes,or models,the problem being investigated.

In this section we have looked at two examples of this modeling process, one drawn

from physics and the other from ecology. In constructing future mathematical mod-

els yourself, you should recognize that each problem is different, and that successful

modeling cannot be reduced to the observance of a set of prescribed rules. Indeed,

constructing a satisfactory model is sometimes the most difﬁcult part of the problem.

Nevertheless, it may be helpful to list some steps that are often part of the process:

1. Identify the independent and dependent variables and assign letters to represent them.

Often the independent variable is time.

2. Choose the units of measurement for each variable. In a sense the choice of units is

arbitrary, but some choices may be much more convenient than others. For example, we

chose to measure time in seconds for the falling-object problem and in months for the

population problem.

3. Articulate the basic principle that underlies or governs the problem you are investigating.

This may be a widely recognized physical law,such as Newton’s law of motion,or it may be

a more speculative assumption that may be based on your own experience or observations.

In any case, this step is likely not to be a purely mathematical one, but will require you to

be familiar with the ﬁeld in which the problem originates.

4. Express the principle or law in step 3 in terms of the variables you chose in step 1. This

may be easier said than done. It may require the introduction of physical constants or

parameters (such as the drag coefﬁcient in Example 1) and the determination of appro-

priate values for them. Or it may involve the use of auxiliary or intermediate variables

that must then be related to the primary variables.

5. Make sure that all terms in your equation have the same physical units. If this is not the

case, then your equation is wrong and you should seek to repair it. If the units agree, then

your equation at least is dimensionally consistent,although it may have other shortcomings

that this test does not reveal.

6. In the problems considered here,the result of step 4 is a single differential equation,which

constitutes the desired mathematical model. Keep in mind, though, that in more complex

problems the resulting mathematical model may be much more complicated, perhaps

involving a system of several differential equations, for example.

PROBLEMS In each of Problems 1 through 6, draw a direction ﬁeld for the given differential equation.

Based on the direction ﬁeld, determine the behavior of y as t →∞. If this behavior depends

on the initial value of y at t = 0, describe the dependency.

′

= 3 − 2y 2. y

′

= 2y − 3

′

= 3 + 2y 4. y

′

=−1 − 2y

′

= 1 + 2y 6. y

′

= y + 2

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8 Chapter 1. Intr oduction

In each of Problems 7 through 10, write down a differential equation of the form

dy/dt = ay + b whose solutions have the required behavior as t →∞.

7. All solutions approach y = 3. 8. All solutions approach y = 2/3.

9. All other solutions diverge from y = 2. 10. All other solutions diverge from y = 1/3.

In each of Problems 11 through 14, draw a direction ﬁeld for the given differential equation.

Based on the direction ﬁeld, determine the behavior of y as t →∞. If this behavior depends

on the initial value of y at t = 0, describe this dependency. Note that in these problems the

equations are not of the form y

′

= ay + b, and the behavior of their solutions is somewhat

more complicated than for the equations in the text.

11.

′

= y(4 − y) 12. y

′

=−y(5 − y)

13.

′

= y

14. y

′

= y(y − 2)

Consider the following list of differential equations, some of which produced the direction

ﬁelds shown in Figures 1.1.5 through 1.1.10. In each of Problems 15 through 20 identify the

differential equation that corresponds to the given direction ﬁeld.

(a) y

′

= 2y − 1 (b) y

′

= 2 + y (c) y

′

= y − 2

(d) y

′

= y(y + 3) (e) y

′

= y(y − 3) (f) y

′

= 1 + 2y

(g) y

′

=−2 − y (h) y

′

= y(3 − y) (i) y

′

= 1 − 2y

(j) y

′

= 2 − y

15. The direction ﬁeld of Figure 1.1.5.

16. The direction ﬁeld of Figure 1.1.6.

1234

FIGURE 1.1.5 Problem 15.

1234

FIGURE 1.1.6 Problem 16.

17. The direction ﬁeld of Figure 1.1.7.

18. The direction ﬁeld of Figure 1.1.8.

–4

–3

–2

–1

1234

FIGURE 1.1.7 Problem 17.

–4

–3

–2

–1

1234

FIGURE 1.1.8 Problem 18.

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1.1 Some Basic Mathematical Models; Direction Fields 9

19. The direction ﬁeld of Figure 1.1.9.

20. The direction ﬁeld of Figure 1.1.10.

–1

1234

FIGURE 1.1.9 Problem 19.

–1

1234

FIGURE 1.1.10 Problem 20.

21. A pond initially contains 1,000,000 gal of water and an unknown amount of an undesirable

chemical. Water containing 0.01 g of this chemical per gallon ﬂows into the pond at a rate

of 300 gal/h. The mixture ﬂows out at the same rate, so the amount of water in the pond

remains constant.Assume that the chemical is uniformly distributed throughout the pond.

(a) Write a differential equation for the amount of chemical in the pond at any time.

(b) How much of the chemicalwill bein the pondafter a verylong time?Does this limiting

amount depend on the amount that was present initially?

22. A spherical raindrop evaporates at a rate proportional to its surface area. Write a

differential equation for the volume of the raindrop as a function of time.

23. Newton’s law of cooling states that the temperature of an object changes at a rate propor-

tional to the difference between the temperature of the object itself and the temperature

of its surroundings (the ambient air temperature in most cases). Suppose that the ambient

temperature is 70

◦

F and that the rate constant is 0.05 (min)

−1

.Write a differential equation

for the temperature of the object at any time. Note that the differential equation is the

same whether the temperature of the object is above or below the ambient temperature.

24. A certain drug is being administered intravenously to a hospital patient. Fluid containing

5 mg/cm

of the drug enters the patient’s bloodstream at a rate of 100 cm

/h. The drug is

absorbed by body tissues or otherwise leaves the bloodstream at a rate proportional to

the amount present, with a rate constant of 0.4 (h)

−1

(a) Assuming that the drug is always uniformly distributed throughout the bloodstream,

write a differential equation for the amount of the drug that is present in the bloodstream

at any time.

(b) How much of the drug is present in the bloodstream after a long time?

25.

For small, slowly falling objects, the assumption made in the text that the drag force

is proportional to the velocity is a good one. For larger, more rapidly falling objects, it is

more accurate to assume that the drag force is proportional to the square of the velocity.

(a) Write a differential equation for the velocity of a falling object of mass m if the mag-

nitude of the drag force is proportional to the square of the velocity and its direction is

opposite to that of the velocity.

See Lyle N. Long and Howard Weiss, “The Velocity Dependence of Aerodynamic Drag: A Primer for

Mathematicians,”American Mathematical Monthly 106 (1999), 2, pp. 127–135.

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10 Chapter 1. Intr oduction

(b) Determine the limiting velocity after a long time.

(d) Using the data in part (c), draw a direction ﬁeld and compare it with Figure 1.1.3.

In each of Problems 26 through 33, draw a direction ﬁeld for the given differential equation.

Based on the direction ﬁeld, determine the behavior of y as t →∞. If this behavior depends

on the initial value of y at t = 0, describe this dependency. Note that the right sides of these

equations depend on t as well as y; therefore, their solutions can exhibit more complicated

behavior than those in the text.

26.

′

=−2 + t − y 27. y

′

= te

−2t

− 2y

28.

′

= e

−t

+ y 29. y

′

= t + 2y

30.

′

= 3 sin t + 1 + y 31. y

′

= 2t − 1 − y

32. y

′

=−(2t + y)/2y 33. y

′

− y −

1.2 Solutions of Some Differential Equations

In the preceding section we derived the differential equations

d v

= mg − γv (1)

and

= rp − k. (2)

Equation (1) models a falling object, and Eq. (2) models a population of ﬁeld mice

preyed on by owls. Both of these equations are of the general form

= ay − b, (3)

where a and b are given constants. We were able to draw some important qualitative

conclusions about the behavior of solutions of Eqs. (1) and (2) by considering the

associated direction ﬁelds. To answer questions of a quantitative nature,however,we

need to ﬁnd the solutions themselves, and we now investigate how to do that.

EXAMPLE

Fiel d Mi ce

and Owls

(continued)

Consider the equation

= 0.5p − 450, (4)

which describes the interaction of certain populations of ﬁeld mice and owls [see Eq. (8) of

Section 1.1]. Find solutions of this equation.

To solve Eq. (4), we need to ﬁnd functions p(t) that, when substituted into the equation,

reduce it to an obvious identity. Here is one way to proceed. First, rewrite Eq. (4) in the form

p − 900

, (5)

or, if p ̸= 900,

dp/dt

p − 900

. (6)

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1.2 Solutions of Some Differential Equations 11

By the chain rule the left side of Eq. (6) is the derivative of ln |p − 900| with respect to t,sowe

have

ln |p − 900|=

. (7)

Then, by integrating both sides of Eq. (7), we obtain

ln |p − 900|=

+ C, (8)

where C is an arbitrary constant of integration. Therefore, by taking the exponential of both

sides of Eq. (8), we ﬁnd that

|p − 900|=e

(t/2)+C

= e

t/2

, (9)

p − 900 =±e

t/2

, (10)

and ﬁnally

p = 900 + ce

t/2

, (11)

where c =±e

is also an arbitrary (nonzero) constant. Note that the constant function p = 900

is also a solution of Eq. (5) and that it is contained in the expression (11) if we allow c to take

the value zero. Graphs of Eq. (11) for several values of c are shown in Figure 1.2.1.

900

600

12 t345

700

800

1000

1100

1200

FIGURE 1.2.1 Graphs of p = 900 + ce

t/2

for several values of c.

These are solutions of dp/dt = 0.5p − 450.

Note that they have the character inferred from the direction ﬁeld in Figure 1.1.4. For

instance, solutions lying on either side of the equilibrium solution p = 900 tend to diverge

from that solution.

In Example 1 we found inﬁnitely many solutions of the differential equation (4),

corresponding to the inﬁnitely many values that the arbitrary constant c in Eq. (11)

might have.Thisis typicalof what happenswhen yousolve adifferential equation.The

solution process involves an integration, which brings with it an arbitrary constant,

whose possible values generate an inﬁnite family of solutions.

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12 Chapter 1. Intr oduction

Frequently,we want to focus our attention on a single member of the inﬁnite family

of solutions by specifying the value of the arbitrary constant. Most often, we do this

indirectly by specifying instead a point that must lie on the graph of the solution. For

example,to determine the constant c in Eq. (11),we could require that the population

have a given value at a certain time,such as the value 850 at time t = 0. In other words,

the graph of the solution must pass through the point (0, 850). Symbolically, we can

express this condition as

p(0) = 850. (12)

Then, substituting t = 0 and p = 850 into Eq. (11), we obtain

850 = 900 + c.

Hence c =−50, and by inserting this value into Eq. (11), we obtain the desired

solution, namely,

p = 900 − 50e

t/2

. (13)

The additional condition (12) that we used to determine c is an example of an

initial condition. The differential equation (4) together with the initial condition (12)

form an initial value problem.

Now consider the more general problem consisting of the differential equation (3)

= ay − b

and the initial condition

y(0) = y

, (14)

where y

is an arbitrary initial value. We can solve this problem by the same method

as in Example 1. If a ̸= 0 and y ̸= b/a, then we can rewrite Eq. (3) as

dy/dt

y − (b/a)

= a. (15)

By integrating both sides, we ﬁnd that

ln |y − (b/a)|=at + C, (16)

where C is arbitrary.Then,taking the exponential ofboth sidesof Eq.(16) and solving

for y, we obtain

y = (b/a) + ce

, (17)

where c =±e

is also arbitrary. Observe that c = 0 corresponds to the equilibrium

solution y = b/a. Finally, the initial condition (14) requires that c = y

− (b/a), so

the solution of the initial value problem (3), (14) is

y = (b/a) +[y

− (b/a)]e

. (18)

For a ̸= 0 the expression (17) contains all possible solutions of Eq. (3) and is called

the general solution.

The geometrical representation of the general solution (17) is

an inﬁnite family of curves called integral curves. Each integral curve is associated

with a particular value of c and is the graph of the solution corresponding to that

If a = 0,then the solution of Eq. (3) is not given by Eq. (17). We leave it to you to ﬁnd the general solution

in this case.

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1.2 Solutions of Some Differential Equations 13

value of c. Satisfying an initial condition amounts to identifying the integral curve

that passes through the given initial point.

To relate the solution (18) to Eq. (2), which models the ﬁeld mouse population, we

need only replace a by the growth rate r and replace b by the predation rate k. Then

the solution (18) becomes

p = (k/r) +[p

− (k/r)]e

, (19)

where p

is the initial population of ﬁeld mice. The solution (19) conﬁrms the conclu-

sions reached on the basis of the direction ﬁeld and Example 1. If p

= k/r,then from

Eq. (19) it follows that p = k/r for all t; this is the constant, or equilibrium, solution.

If p

̸= k /r, then the behavior of the solution depends on the sign of the coefﬁcient

− (k/r) of theexponential term in Eq. (19). If p

> k/r,then p grows exponentially

with time t;ifp

< k/r,then p decreases and eventually becomes zero, corresponding

to extinction of the ﬁeld mouse population. Negative values of p, while possible for

the expression (19), make no sense in the context of this particular problem.

To put the falling-object equation (1) in the form (3),we must identify a with −γ/m

and b with −g. Making these substitutions in the solution (18), we obtain

v = (mg/γ) +[v

− (mg/γ)]e

−γt/m

, (20)

where v

is the initial velocity. Again, this solution conﬁrms the conclusions reached

in Section 1.1 on the basis of a direction ﬁeld. There is an equilibrium, or constant,

solution v = mg/γ,and all other solutions tend to approach this equilibrium solution.

The speed of convergence to the equilibrium solution is determined by the exponent

−γ/m. Thus, for a given mass m, the velocity approaches the equilibrium value more

rapidly as the drag coefﬁcient γ increases.

EXAMPLE

AFalling

Object

(continued)

Suppose that, as in Example 2 of Section 1.1, we consider a falling object of mass m = 10 kg

and drag coefﬁcient γ = 2 kg/s. Then the equation of motion (1) becomes

= 9.8 −

. (21)

Suppose this object is dropped from a height of 300 m. Find its velocity at any time t. How

long will it take to fall to the ground, and how fast will it be moving at the time of impact?

The ﬁrst step is to state an appropriate initial condition for Eq. (21). The word “dropped” in

the statement of the problem suggests that the initial velocity is zero, so we will use the initial

condition

v(0) = 0. (22)

The solution of Eq. (21) can be found by substituting the values of the coefﬁcients into the

solution (20),but we will proceed instead to solve Eq. (21) directly. First,rewrite the equation as

dv/dt

v − 49

=−

. (23)

By integrating both sides, we obtain

ln |v − 49|=−

+ C, (24)

and then the general solution of Eq. (21) is

v = 49 + ce

−t/5

, (25)

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14 Chapter 1. Intr oduction

where c is arbitrary. To determine c, we substitute t = 0 and v = 0 from the initial condition

(22) into Eq. (25), with the result that c =−49. Then the solution of the initial value problem

(21), (22) is

v = 49(1 − e

−t/5

). (26)

Equation (26) gives the velocity of the falling object at any positive time (before it hits the

ground, of course).

Graphs of the solution (25) for several values of c are shown in Figure 1.2.2,with the solution

(26) shown by the black curve. It is evident that all solutions tend to approach the equilibrium

solution v = 49. This conﬁrms the conclusions we reached in Section 1.1 on the basis of the

direction ﬁelds in Figures 1.1.2 and 1.1.3.

100

24 t6 8 1210

v = 49 (1 – e

–t/5

)

(10.51, 43.01)

FIGURE 1.2.2 Graphs of the solution (25), v = 49 + ce

−t/5

, for several

values of c. The black curve corresponds to the initial condition v(0) = 0.

To ﬁnd the velocity of the object when it hits the ground, we need to know the time at which

impact occurs. In other words, we need to determine how long it takes the object to fall 300 m.

To do this, we note that the distance x the object has fallen is related to its velocity v by the

equation v = dx/dt, or

= 49(1 − e

−t/5

). (27)

Consequently, by integrating both sides of Eq. (27), we have

x = 49t + 245e

−t/5

+ c, (28)

where c is an arbitrary constant of integration. The object starts to fall when t = 0, so we know

that x = 0 when t = 0. From Eq. (28) it follows that c =−245, so the distance the object has

fallen at time t is given by

x = 49t + 245e

−t/5

− 245. (29)

Let T be the time at which the object hits the ground;then x = 300 when t = T. By substituting

these values in Eq. (29), we obtain the equation

49T + 245e

−T/5

− 545 = 0. (30)

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1.2 Solutions of Some Differential Equations 15

Thevalue of T satisfying Eq. (30) can be approximated by a numerical process

using a scientiﬁc

calculator or computer,withthe resultthat T

∼

10.51 s.At this time,the corresponding velocity

is found from Eq. (26) to be v

∼

43.01 m/s. The point (10.51, 43.01) is also shown in

Figure 1.2.2.

Further Remarks on Mathematical Modeling. Up to this point we have related our discus-

sion of differential equations to mathematical models of a falling object and of a

hypothetical relation between ﬁeld mice and owls. The derivation of these models

may have been plausible, and possibly even convincing, but you should remember

that the ultimate test of any mathematical model is whether its predictions agree

with observations or experimental results. We have no actual observations or exper-

imental results to use for comparison purposes here, but there are several sources of

possible discrepancies.

In the case of the falling object, the underlying physical principle (Newton’s law

of motion) is well established and widely applicable. However, the assumption that

the drag force is proportional to the velocity is less certain. Even if this assumption is

correct, the determination of the drag coefﬁcient γ by direct measurement presents

difﬁculties. Indeed, sometimes one ﬁnds the drag coefﬁcient indirectly—for example,

by measuring the time of fall from a given height and then calculating the value of γ

that predicts this observed time.

The model of the ﬁeld mouse population is subject to various uncertainties.

The determination of the growth rate r and the predation rate k depends on

observations of actual populations, which may be subject to considerable variation.

The assumption that r and k are constants may also be questionable. For example,

a constant predation rate becomes harder to sustain as the ﬁeld mouse population

becomes smaller. Further, the model predicts that a population above the equilib-

rium value will grow exponentially larger and larger. This seems at variance with the

behavior of actual populations; see the further discussion of population dynamics in

Section 2.5.

If the differences between actual observations and a mathematical model’s pre-

dictions are too great, then you need to consider reﬁning the model, making more

careful observations, or perhaps both. There is almost always a tradeoff between

accuracy and simplicity. Both are desirable, but a gain in one usually involves a loss

in the other. However,even if a mathematical model is incomplete or somewhat inac-

curate,it may nevertheless be useful in explaining qualitative features of the problem

under investigation. It may also give satisfactory results under some circumstances

but not others. Thus you should always use good judgment and common sense in

constructing mathematical models and in using their predictions.

PROBLEMS 1.

Solve each of the following initial value problems and plot the solutions for several values

of y

. Then describe in a few words how the solutions resemble, and differ from, each

other.

(a) dy/dt =−y + 5, y(0) = y

A computer algebra system provides this capability; many calculators also have built-in routines for

solving such equations.

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16 Chapter 1. Intr oduction

(b) dy/dt =−2y + 5, y(0) = y

2. Follow the instructions for Problem 1 for the following initial value problems:

(a) dy/dt = y − 5, y(0) = y

(b) dy/dt = 2y − 5, y(0) = y

3. Consider the differential equation

dy/dt =−ay + b,

where both a and b are positive numbers.

(a) Find the general solution of the differential equation.

(b) Sketch the solution for several different initial conditions.

i. a increases.

ii. b increases.

iii. Both a and b increase, but the ratio b/a remains the same.

4. Consider the differential equation dy/dt = ay − b.

(a) Find the equilibrium solution y

(b) Let Y(t) = y − y

; thus Y(t) is the deviation from the equilibrium solution. Find the

differential equation satisﬁed by Y(t).

5. Undetermined Coefﬁcients. Here is an alternative way to solve the equation

dy/dt = ay − b. (i)

(a) Solve the simpler equation

dy/dt = ay. (ii)

Call the solution y

(t).

(b) Observe that the only difference between Eqs. (i) and (ii) is the constant −b in Eq. (i).

Therefore, it may seem reasonable to assume that the solutions of these two equations

also differ only by a constant. Test this assumption by trying to ﬁnd a constant k such that

y = y

(t) + k is a solution of Eq. (i).

Note: This method can also be used in some cases in which the constant b is replaced

by a function g(t). It depends on whether you can guess the general form that the solution

is likely to take.This method is described in detail in Section 3.5 in connection with second

order equations.

6. Use the method of Problem 5 to solve the equation

dy/dt =−ay + b.

7. The ﬁeld mouse population in Example 1 satisﬁes the differential equation

dp/dt = 0.5p − 450.

(a) Find the time at which the population becomes extinct if p(0) = 850.

(b) Find the time of extinction if p(0) = p

, where 0 < p

< 900.

if the population is to become extinct in 1 year.

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1.2 Solutions of Some Differential Equations 17

8. Consider a population p of ﬁeld mice that grows at a rate proportional to the current

population, so that dp/dt = rp.

(a) Find the rate constant r if the population doubles in 30 days.

(b) Find r if the population doubles in N days.

9. The falling object in Example 2 satisﬁes the initial value problem

dv/dt = 9.8 − (v/5), v(0) = 0.

(a) Find the time that must elapse for the object to reach 98% of its limiting velocity.

(b) How far does the object fall in the time found in part (a)?

10. Modify Example 2 so that the falling object experiences no air resistance.

(a) Write down the modiﬁed initial value problem.

(b) Determine how long it takes the object to reach the ground.

11.

Consider the falling object of mass 10 kg in Example 2,but assume now that the drag force

is proportional to the square of the velocity.

(a) If the limiting velocity is 49 m/s (the same as in Example 2), show that the equation

of motion can be written as

dv/dt =[(49)

− v

]/245.

Also see Problem 25 of Section 1.1.

(b) If v(0) = 0, ﬁnd an expression for v(t) at any time.

axes.

(d) Based on your plots in part (c), compare the effect of a quadratic drag force with that

of a linear drag force.

(e) Find the distance x (t) that the object falls in time t.

(f) Find the time T it takes the object to fall 300 m.

12. A radioactive material,suchas the isotope thorium-234,disintegratesat a rate proportional

to the amount currently present.If Q(t) is the amount presentat time t,then dQ/dt =−rQ,

where r > 0 is the decay rate.

(a) If 100 mg of thorium-234 decays to 82.04 mg in 1 week, determine the decay rate r.

(b) Find an expression for the amount of thorium-234 present at any time t.

13. The half-life of a radioactive material is the time required for an amount of this material

to decay to one-half its original value. Show that for any radioactive material that decays

according to the equation Q

′

=−rQ,the half-lifeτ and the decay rate r satisfy the equation

rτ = ln 2.

14. Radium-226 has a half-life of 1620 years. Find the time period during which a given amount

of this material is reduced by one-quarter.

15. According to Newton’s law of cooling (see Problem 23 of Section 1.1), the temperature

u(t) of an object satisﬁes the differential equation

=−k(u − T),

where T is the constant ambient temperature and k is a positive constant. Suppose that

the initial temperature of the object is u(0) = u

(a) Find the temperature of the object at any time.

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18 Chapter 1. Intr oduction

(b) Let τ be the time at which the initial temperature difference u

− T has been reduced

by half. Find the relation between k and τ.

16. Supposethat a building loses heatin accordance with Newton’slaw ofcooling (see Problem

15)and that the rate constantk hasthevalue 0.15h

−1

.Assume thatthe interior temperature

is 70

◦

F when the heating system fails. If the external temperature is 10

◦

F, how long will it

take for the interior temperature to fall to 32

◦

17. Consider an electric circuit containing a capacitor, resistor, and battery; see Figure 1.2.3.

The charge Q(t) on the capacitor satisﬁes the equation

= V,

where R is the resistance, C is the capacitance, and V is the constant voltage supplied by

the battery.

(a) If Q(0) = 0, ﬁnd Q(t) at any time t, and sketch the graph of Q versus t.

(b) Find the limiting value Q

that Q(t) approaches after a long time.

) = Q

and that at time t = t

the battery is removed and the circuit

is closed again. Find Q(t) for t > t

and sketch its graph.

FIGURE 1.2.3 The electric circuit of Problem 17.

18.

A pond containing 1,000,000 gal of water is initially free of a certain undesirable chemical

(see Problem 21 of Section 1.1). Water containing 0.01 g/gal of the chemical ﬂows into the

pond at a rate of 300 gal/h, and water also ﬂows out of the pond at the same rate. Assume

that the chemical is uniformly distributed throughout the pond.

(a) Let Q(t) be the amount of the chemical in the pond at time t. Write down an initial

value problem for Q(t).

(b) Solve the problem in part (a) for Q(t). How much chemical is in the pond after 1 year?

pure water ﬂows into the pond, and the mixture ﬂows out at the same rate as before.Write

down the initial value problem that describes this new situation.

(d) Solve the initial value problem in part (c). How much chemical remains in the pond

after 1 additional year (2 years from the beginning of the problem)?

(e) How long does it take for Q(t) to be reduced to 10 g?

(f) Plot Q(t) versus t for 3 years.

19. Your swimming pool containing 60,000 gal of water has been contaminated by 5 kg of

a nontoxic dye that leaves a swimmer’s skin an unattractive green. The pool’s ﬁltering

system can take water from the pool, remove the dye, and return the water to the pool at

a ﬂow rate of 200 gal/min.

This equation results from Kirchhoff’s laws, which are discussed in Section 3.7.

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1.3 Classiﬁcation of Differential Equations 19

(a) Write down the initial value problem for the ﬁltering process; let q(t) be the amount

of dye in the pool at any time t.

(b) Solve the problem in part (a).

You have also determined that the effect of the dye is imperceptible if its concentration

is less than 0.02 g/gal. Is your ﬁltering system capable of reducing the dye concentration

to this level within 4 h?

(d) Find the time T at which the concentration of dye ﬁrst reaches the value 0.02 g/gal.

(e) Find the ﬂow rate that is sufﬁcient to achieve the concentration 0.02 g/gal within 4 h.

1.3 Classiﬁcation of Differential Equations

The main purpose of this book is to discuss some of the properties of solutions of

differential equations,and to present some of the methods that have proved effective

in ﬁnding solutions or, in some cases, approximating them. To provide a framework

for our presentation, we describe here several useful ways of classifying differential

equations.

Ordinar y and Partial Differential Equations. One important classiﬁcation is based on

whether the unknown function depends on a single independent variable or on sev-

eral independent variables. In the ﬁrst case, only ordinary derivatives appear in the

differential equation, and it is said to be an ordinary differential equation. In the sec-

ond case, the derivatives are partial derivatives, and the equation is called a partial

differential equation.

All the differential equations discussed in the preceding two sections are ordinary

differential equations. Another example of an ordinary differential equation is

Q(t)

+ R

dQ(t)

Q(t) = E(t), (1)

for the charge Q(t) on a capacitor in a circuit with capacitance C, resistance R, and

inductance L; this equation is derived in Section 3.7. Typical examples of partial

differential equations are the heat conduction equation

∂

u(x, t)

∂x

∂u(x, t)

∂t

(2)

and the wave equation

∂

u(x, t)

∂x

∂

u(x, t)

∂t

. (3)

Here, α

and a

are certain physical constants. Note that in both Eqs. (2) and (3) the

dependent variable u depends on the two independent variables x and t. The heat

conduction equation describes the conduction of heat in a solid body, and the wave

equation arises in a variety of problems involving wave motion in solids or ﬂuids.

Systems of Differential Equations. Another classiﬁcation of differential equations de-

pends on the number of unknown functions that are involved. If there is a single

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20 Chapter 1. Intr oduction

function to be determined, then one equation is sufﬁcient. However, if there are two

or more unknown functions, then a system of equations is required. For example, the

Lotka–Volterra, or predator–prey, equations are important in ecological modeling.

They have the form

dx/dt = ax − αxy

dy/dt =−cy + γxy,

(4)

where x(t) and y(t) are the respective populations of the prey and predator species.

The constants a, α, c, and γ are based on empirical observations and depend on the

particular species being studied. Systems of equations are discussed in Chapters 7

and 9; in particular, the Lotka–Volterra equations are examined in Section 9.5. In

some areas of application it is not unusual to encounter very large systems containing

hundreds, or even many thousands, of equations.

Order. The order of a differential equation is the order of the highest derivative that

appears in the equation. The equations in the preceding sections are all ﬁrst order

equations, whereas Eq. (1) is a second order equation. Equations (2) and (3) are

second order partial differential equations. More generally, the equation

F[t, u(t), u

′

(t), ..., u

(n)

(t)]=0 (5)

is an ordinary differential equation of the nth order. Equation (5) expresses a relation

between the independent variable t and the values of the function u and its ﬁrst n

derivatives u

′

, u

′′

, ..., u

(n)

. It is convenient and customary in differential equations to

write y for u(t), with y

′

, y

′′

, ..., y

(n)

standing for u

′

(t), u

′′

(t), ..., u

(n)

(t). Thus Eq. (5) is

written as

F(t, y, y

′

, ..., y

(n)

) = 0. (6)

For example,

′′′

+ 2e

′′

+ yy

′

= t

(7)

is a third order differential equation for y = u(t). Occasionally, other letters will be

used instead of t and y for the independent and dependent variables; the meaning

should be clear from the context.

We assume that it is always possible to solve a given ordinary differential equation

for the highest derivative, obtaining

(n)

= f (t, y, y

′

, y

′′

, ..., y

(n−1)

). (8)

This is mainly to avoid the ambiguity that may arise because a single equation of

the form (6) may correspond to several equations of the form (8). For example, the

equation

′

)

+ ty

′

+ 4y = 0 (9)

leads to the two equations

′

−t +

− 16y

or y

′

−t −

− 16y

. (10)

Linear and Nonlinear Equations. A crucial classiﬁcation of differential equations is

whether they are linear or nonlinear. The ordinary differential equation

F(t, y, y

′

, ..., y

(n)

) = 0

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1.3 Classiﬁcation of Differential Equations 21

is said to be linear if F is a linear function of the variables y, y

′

, ..., y

(n)

; a similar

deﬁnition applies to partial differential equations. Thus the general linear ordinary

differential equation of order n is

(t)y

(n)

+ a

(t)y

(n−1)

+···+a

(t)y = g(t). (11)

Most of the equations you have seen thus far in this book are linear; examples are

the equations in Sections 1.1 and 1.2 describing the falling object and the ﬁeld mouse

population. Similarly, in this section, Eq. (1) is a linear ordinary differential equation

and Eqs. (2) and (3) are linear partial differential equations. An equation that is not

of the form (11) is a nonlinear equation. Equation (7) is nonlinear because of the

term yy

′

. Similarly, each equation in the system (4) is nonlinear because of the terms

that involve the product xy.

A simple physical problem that leads to a nonlinear differential equation is the

oscillating pendulum. The angle θ that an oscillating pendulum of length L makes

with the vertical direction (see Figure 1.3.1) satisﬁes the equation

sin θ = 0, (12)

whose derivation is outlined in Problems 29 through 31. The presence of the term

involving sin θ makes Eq. (12) nonlinear.

FIGURE 1.3.1 An oscillating pendulum.

The mathematical theory and methods for solving linear equations are highly

developed. In contrast, for nonlinear equations the theory is more complicated, and

methods of solution are less satisfactory. In view of this, it is fortunate that many

signiﬁcant problems lead to linear ordinary differential equations or can be approx-

imated by linear equations. For example, for the pendulum, if the angle θ is small,

then sin θ

∼

θ and Eq. (12) can be approximated by the linear equation

θ = 0. (13)

This process of approximating a nonlinear equation by a linear one is called lineariza-

tion; it is an extremely valuable way to deal with nonlinear equations. Nevertheless,

there are many physical phenomena that simply cannot be represented adequately

by linear equations. To study these phenomena, it is essential to deal with nonlinear

equations.

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22 Chapter 1. Intr oduction

In an elementary text it is natural to emphasize the simpler and more straight-

forward parts of the subject. Therefore, the greater part of this book is devoted to

linear equations and various methods for solving them. However, Chapters 8 and 9,

as well as parts of Chapter 2, are concerned with nonlinear equations. Whenever it

is appropriate, we point out why nonlinear equations are, in general, more difﬁcult

and why many of the techniques that are useful in solving linear equations cannot

be applied to nonlinear equations.

Solutions. A solution of the ordinary differential equation (8) on the interval

α<t <βis a function φ such that φ

′

, φ

′′

, ..., φ

(n)

exist and satisfy

(n)

(t) = f [t, φ(t), φ

′

(t), ..., φ

(n−1)

(t)] (14)

for every t in α<t <β. Unless stated otherwise, we assume that the function f

of Eq. (8) is a real-valued function, and we are interested in obtaining real-valued

solutions y = φ(t).

Recall that in Section 1.2 we found solutions of certain equations by a process of

direct integration. For instance, we found that the equation

= 0.5p − 450 (15)

has the solution

p = 900 + ce

t/2

, (16)

where c is an arbitrary constant. It is often not so easy to ﬁnd solutions of differential

equations. However, if you ﬁnd a function that you think may be a solution of a given

equation, it is usually relatively easy to determine whether the function is actually a

solution simply by substituting the function into the equation. For example, in this

way it is easy to show that the function y

(t) = cos t is a solution of

′′

+ y = 0 (17)

for all t.To conﬁrm this, observe that y

′

(t) =−sin t and y

′′

(t) =−cos t;then it follows

that y

′′

(t) + y

(t) = 0. In the same way you can easily show that y

(t) = sin t is also

a solution of Eq. (17). Of course, this does not constitute a satisfactory way to solve

most differential equations,because there are far too many possible functions for you

to have a good chance of ﬁnding the correct one by a random choice. Nevertheless,

you should realize that you can verify whether any proposed solution is correct by

substituting it into the differential equation. This can be a very useful check; it is one

that you should make a habit of considering.

Some Important Questions. Although for the equations (15) and (17) we are able to

verify that certain simple functions are solutions, in general we do not have such

solutions readily available. Thus a fundamental question is the following: Does an

equation of the form (8) always have a solution? The answer is “No.” Merely writing

down an equation of the form (8) does not necessarily mean that there is a function

y = φ(t) that satisﬁes it. So, how can we tell whether some particular equation has a

solution?This is the question of existence of a solution,and it is answered by theorems

stating that under certain restrictions on the function f in Eq. (8),the equation always

has solutions. This is not a purely mathematical concern for at least two reasons.

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1.3 Classiﬁcation of Differential Equations 23

If a problem has no solution, we would prefer to know that fact before investing

time and effort in a vain attempt to solve the problem. Further, if a sensible physical

problem is modeled mathematically as a differential equation, then the equation

should have a solution. If it does not,then presumably there is something wrong with

the formulation. In this sense an engineer or scientist has some check on the validity

of the mathematical model.

If we assume that a given differential equation has at least one solution, then

we may need to consider how many solutions it has, and what additional conditions

must be speciﬁed to single out aparticular solution.This is the question ofuniqueness.

In general, solutions of differential equations contain one or more arbitrary con-

stants of integration, as does the solution (16) of Eq. (15). Equation (16) represents

an inﬁnity of functions corresponding to the inﬁnity of possible choices of the con-

stant c. As we saw in Section 1.2, if p is speciﬁed at some time t, this condition will

determine a value for c; even so, we have not yet ruled out the possibility that there

may be other solutions of Eq. (15) that also have the prescribed value of p at the

prescribed time t. As in the question of existence of solutions, the issue of uniqueness

has practical as well as theoretical implications. If we are fortunate enough to ﬁnd a

solution of a given problem, and if we know that the problem has a unique solution,

then we can be sure that we have completely solved the problem. If there may be

other solutions, then perhaps we should continue to search for them.

A third important question is: Given a differential equation of the form (8), can

we actually determine a solution, and if so, how? Note that if we ﬁnd a solution of

the given equation, we have at the same time answered the question of the exis-

tence of a solution. However, without knowledge of existence theory we might,

for example, use a computer to ﬁnd a numerical approximation to a “solution”

that does not exist. On the other hand, even though we may know that a solution

exists, it may be that the solution is not expressible in terms of the usual elemen-

tary functions—polynomial, trigonometric, exponential, logarithmic, and hyperbolic

functions. Unfortunately, this is the situation for most differential equations. Thus,

we discuss both elementary methods that can be used to obtain exact solutions of

certain relatively simple problems, and also methods of a more general nature that

can be used to ﬁnd approximations to solutions of more difﬁcult problems.

Computer Use in Differential Equations. A computer can be an extremely valuable tool

in the study of differential equations. For many years computers have been used

to execute numerical algorithms, such as those described in Chapter 8, to construct

numerical approximations to solutions of differential equations. These algorithms

have been reﬁned to an extremely high level of generality and efﬁciency. A few lines

of computer code,written in a high-level programming language and executed (often

within a few seconds) on a relatively inexpensive computer,sufﬁce to approximate to

a high degree of accuracy thesolutions of a wide rangeof differential equations. More

sophisticated routines are also readily available. These routines combine the ability

to handle very large and complicated systems with numerous diagnostic features that

alert the user to possible problems as they are encountered.

The usual output from a numerical algorithm is a table of numbers, listing selected

values of the independent variable and the corresponding values of the dependent

variable. With appropriate software it is easy to display the solution of a differen-

tial equation graphically, whether the solution has been obtained numerically or as

August 7, 2012 21:03 c01 Sheet number 24 Page number 24 cyan black

24 Chapter 1. Intr oduction

the result of an analytical procedure of some kind. Such a graphical display is often

much more illuminating and helpful in understanding and interpreting the solution of

a differential equation than a table of numbers or a complicated analytical formula.

There are on the market several well-crafted and relatively inexpensive special-

purpose software packages for the graphical investigation of differential equations.

The widespread availability of personal computers has brought powerful computa-

tional and graphical capability within the reach of individual students. You should

consider, in the light of your own circumstances, how best to take advantage of the

available computing resources. You will surely ﬁnd it enlightening to do so.

Another aspect of computer use that is very relevant to the study of differential

equations is the availability of extremely powerful and general software packages

that can perform a wide variety of mathematical operations.Among these are Maple,

Mathematica,and MATLAB, each of which can be used on various kinds of personal

computers or workstations.All three of these packages can execute extensive numer-

ical computationsand haveversatile graphicalfacilities. Maple and Mathematica also

have very extensive analytical capabilities. For example, they can perform the ana-

lytical steps involved in solving many differential equations, often in response to a

single command. Anyone who expects to deal with differential equations in more

than a superﬁcial way should become familiar with at least one of these products and

explore the ways in which it can be used.

For you, the student, these computing resources have an effect on how you should

study differential equations. To become conﬁdent in using differential equations, it

is essential to understand how the solution methods work, and this understanding is

achieved, in part,by working out a sufﬁcient number of examples in detail. However,

eventually you should plan to delegate as many as possible of the routine (often

repetitive) details to a computer, while you focus on the proper formulation of the

problem and on the interpretation of the solution. Our viewpoint is that you should

always try to use the best methods and tools available for each task. In particular,

you should strive to combine numerical, graphical, and analytical methods so as to

attain maximum understanding of the behavior of the solution and of the underlying

process that the problem models.You should also remember that some tasks can best

be done with pencil and paper, while others require a calculator or computer. Good

judgment is often needed in selecting an effective combination.

PROBLEMS In each of Problems 1 through 6, determine the order of the given differential equation; also

state whether the equation is linear or nonlinear.

1. t

+ t

+ 2y = sin t 2. (1 + y

)

+ t

+ y = e

+ y = 14.

+ ty

= 0

+ sin(t + y) = sin t 6.

+ t

+ (cos

t)y = t

In each of Problems 7 through 14, verify that each given function is a solution of the differ-

ential equation.

7. y

′′

− y = 0; y

(t) = e

, y

(t) = cosh t

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1.3 Classiﬁcation of Differential Equations 25

8. y

′′

+ 2y

′

− 3y = 0; y

(t) = e

−3t

, y

(t) = e

9. ty

′

− y = t

; y = 3t + t

10. y

′′′′

+ 4y

′′′

+ 3y = t; y

(t) = t/3, y

(t) = e

−t

+ t/3

11. 2t

′′

+ 3ty

′

− y = 0, t > 0; y

(t) = t

1/2

, y

(t) = t

−1

12. t

′′

+ 5ty

′

+ 4y = 0, t > 0; y

(t) = t

−2

, y

(t) = t

−2

ln t

13. y

′′

+ y = sec t,0< t <π/2; y = (cos t) ln cos t + t sin t

14. y

′

− 2ty = 1; y = e

−s

ds + e

In each of Problems 15 through 18, determine the values of r for which the given differential

equation has solutions of the form y = e

15. y

′

+ 2y = 0 16. y

′′

− y = 0

17. y

′′

+ y

′

− 6y = 0 18. y

′′′

− 3y

′′

+ 2y

′

= 0

In each of Problems 19 and 20, determine the values of r for which the given differential

equation has solutions of the form y = t

for t > 0.

19. t

′′

+ 4ty

′

+ 2y = 0 20. t

′′

− 4ty

′

+ 4y = 0

In each of Problems 21 through 24, determine the order of the given partial differential equa-

tion; also state whether the equation is linear or nonlinear. Partial derivatives are denoted by

subscripts.

21. u

+ u

= 0 22. u

+ u

+ uu

+ u = 0

23. u

xxxx

+ 2u

xxyy

+ u

yyyy

= 0 24. u

+ uu

= 1 + u

In each of Problems 25 through 28, verify that each given function is a solution of the given

partial differential equation.

25. u

+ u

= 0; u

(x, y) = cos x cosh y, u

(x, y) = ln(x

+ y

)

26. α

= u

; u

(x, t) = e

−α

sin x, u

(x, t) = e

−α

sin λx, λ a real constant

27. a

= u

; u

(x, t) = sin λx sin λat, u

(x, t) = sin(x − at), λ a real constant

28. α

= u

; u = (π/t)

1/2

−x

/4α

, t > 0

29. Follow the steps indicated here to derive the equation of motion of a pendulum, Eq. (12)

in the text. Assume that the rod is rigid and weightless, that the mass is a point mass, and

that there is no friction or drag anywhere in the system.

(a) Assume that the mass is in an arbitrary displaced position, indicated by the angle θ.

Draw a free-body diagram showing the forces acting on the mass.

(b) Apply Newton’s law of motion in the direction tangential to the circular arc on which

the mass moves. Then the tensile force in the rod does not enter the equation. Observe

that you need to ﬁnd the component of the gravitational force in the tangential direc-

tion. Observe also that the linear acceleration, as opposed to the angular acceleration, is

θ/dt

, where L is the length of the rod.

30. Another way to derive the pendulum equation (12) is based on the principle of

conservation of energy.

(a) Show that the kinetic energy T of the pendulum in motion is

T =

dθ

(b) Show that the potential energy V of the pendulum, relative to its rest position, is

V = mgL(1 − cos θ).

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26 Chapter 1. Intr oduction

Calculate dE/dt, set it equal to zero, and show that the resulting equation reduces to

Eq. (12).

31. A third derivation of the pendulum equation depends on the principle of angular

momentum: The rate of change of angular momentum about any point is equal to the

net external moment about the same point.

(a) Show that the angular momentum M, or moment of momentum, about the point of

support is given by M = mL

dθ/dt.

(b) Set dM/dt equal to the moment of the gravitational force,and show that the resulting

equation reduces to Eq. (12). Note that positive moments are counterclockwise.

1.4 Historical Remarks

Without knowing something about differential equations and methods of solving

them, it is difﬁcult to appreciate the history of this important branch of mathematics.

Further, the development of differential equations is intimately interwoven with the

general development of mathematics and cannot be separated from it. Nevertheless,

to provide some historical perspective, we indicate here some of the major trends in

the history of the subject and identify the most prominent early contributors. Other

historical information is contained in footnotes scattered throughout the book and

in the references listed at the end of the chapter.

The subject of differential equations originated in the study of calculus by Isaac

Newton (1642–1727) and Gottfried Wilhelm Leibniz (1646–1716) in the seventeenth

century. Newton grew up in the English countryside, was educated at Trinity Col-

lege, Cambridge, and became Lucasian Professor of Mathematics there in 1669. His

epochal discoveries of calculus and of the fundamental laws of mechanics date from

1665. They were circulated privately among his friends, but Newton was extremely

sensitive to criticism and did not begin to publish his results until 1687 with the

appearance of his most famous book, Philosophiae Naturalis Principia Mathemat-

ica. Although Newton did relatively little work in differential equations as such, his

development of the calculus and elucidation of the basic principles of mechanics pro-

vided a basis for their applications in the eighteenth century, most notably by Euler.

Newton classiﬁed ﬁrst order differential equations according to the three forms

dy/dx = f (x), dy/dx = f (y), and dy/dx = f (x, y). For the latter equation he devel-

oped a method of solution using inﬁnite series when f (x, y) is a polynomial in x

and y. Newton’s active research in mathematics ended in the early 1690s, except for

the solution of occasional “challenge problems” and the revision and publication of

results obtained much earlier. He was appointed Warden of the British Mint in 1696

and resigned his professorship a few years later. He was knighted in 1705 and, upon

his death, was buried in Westminster Abbey.

Leibniz was born in Leipzig and completed his doctorate in philosophy at the age

of 20 at the University of Altdorf.Throughout his life he engaged in scholarly work in

several different ﬁelds. He was mainly self-taught in mathematics, since his interest

in this subject developed when he was in his twenties. Leibniz arrived at the funda-

mental results of calculus independently, although a little later than Newton, but was

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1.4 Historical Remarks 27

the ﬁrst to publish them, in 1684. Leibniz was very conscious of the power of good

mathematical notation and was responsible for the notation dy/dx for the deriva-

tive and for the integral sign. He discovered the method of separation of variables

(Section 2.2) in 1691, the reduction of homogeneous equations to separable ones

(Section 2.2, Problem 30) in 1691, and the procedure for solving ﬁrst order linear

equations (Section 2.1) in 1694. He spent his life as ambassador and adviser to sev-

eral German royal families, which permitted him to travel widely and to carry on an

extensive correspondence with other mathematicians,especially the Bernoulli broth-

ers. In the course of this correspondence many problems in differential equations

were solved during the latter part of the seventeenth century.

The brothers Jakob (1654–1705) and Johann (1667–1748) Bernoulli of Basel did

much to develop methods of solving differential equations and to extend the range

of their applications. Jakob became professor of mathematics at Basel in 1687, and

Johann was appointed to the same position upon his brother’s death in 1705. Both

men were quarrelsome,jealous,and frequently embroiled in disputes, especially with

each other. Nevertheless, both also made signiﬁcant contributions to several areas of

mathematics.With the aid of calculus,they solved a number of problems in mechanics

by formulating them as differential equations. For example, Jakob Bernoulli solved

the differential equation y

′

=[a

/(b

y − a

)]

1/2

in 1690 and, in the same paper, ﬁrst

used the term “integral” in the modern sense. In 1694 Johann Bernoulli was able

to solve the equation dy/dx = y/ax. One problem which both brothers solved, and

which led to much friction between them, was the brachistochrone problem (see

Problem 32 of Section 2.3). The brachistochrone problem was also solved by Leibniz,

Newton, and the Marquis de L’H

opital. It is said, perhaps apocryphally, that Newton

learned of the problem late in the afternoon of a tiring day at the Mint and solved it

that evening after dinner. He published the solution anonymously, but upon seeing

it, Johann Bernoulli exclaimed, “Ah, I know the lion by his paw.”

Daniel Bernoulli (1700–1782),son of Johann,migrated to St. Petersburg as a young

man to join the newly established St. Petersburg Academy but returned to Basel

in 1733 as professor of botany and, later, of physics. His interests were primarily in

partial differential equations and their applications. For instance, it is his name that

is associated with the Bernoulli equation in ﬂuid mechanics. He was also the ﬁrst

to encounter the functions that a century later became known as Bessel functions

(Section 5.7).

The greatest mathematician of the eighteenth century, Leonhard Euler (1707–

1783), grew up near Basel and was a student of Johann Bernoulli. He followed his

friend Daniel Bernoulli to St. Petersburg in 1727. For the remainder of his life he

was associated with the St. Petersburg Academy (1727–1741 and 1766–1783) and

the Berlin Academy (1741–1766). Euler was the most proliﬁc mathematician of all

time; his collected works ﬁll more than 70 large volumes. His interests ranged over

all areas of mathematics and many ﬁelds of application. Even though he was blind

during the last 17 years of his life,his work continued undiminished until the very day

of his death. Of particular interest here is his formulation of problems in mechanics

in mathematical language and his development of methods of solving these mathe-

matical problems. Lagrange said of Euler’s work in mechanics,“The ﬁrst great work

in which analysis is applied to the science of movement.”Among other things, Euler

identiﬁed the condition for exactness of ﬁrst order differential equations (Section

2.6) in 1734–35, developed the theory of integrating factors (Section 2.6) in the same

August 7, 2012 21:03 c01 Sheet number 28 Page number 28 cyan black

28 Chapter 1. Intr oduction

paper,and gave the general solution of homogeneous linear equations with constant

coefﬁcients (Sections 3.1, 3.3, 3.4, and 4.2) in 1743. He extended the latter results to

nonhomogeneous equations in 1750–51. Beginning about 1750, Euler made frequent

use of power series (Chapter 5) in solving differential equations. He also proposed

a numerical procedure (Sections 2.7 and 8.1) in 1768–69, made important contribu-

tions in partial differential equations, and gave the ﬁrst systematic treatment of the

calculus of variations.

Joseph-Louis Lagrange(1736–1813) became professor of mathematics in his native

Turin at the age of 19. He succeeded Euler in the chair of mathematics at the Berlin

Academy in 1766 and moved on to the ParisAcademy in 1787. He is most famous for

his monumental work Mécanique analytique, published in 1788, an elegant and com-

prehensive treatise of Newtonian mechanics. With respect to elementary differential

equations, Lagrange showed in 1762–65 that the general solution of an nth order

linear homogeneous differential equation is a linear combination of n independent

solutions (Sections 3.2 and 4.1). Later, in 1774–75, he gave a complete development

of the method of variation of parameters (Sections 3.6 and 4.4). Lagrange is also

known for fundamental work in partial differential equations and the calculus of

variations.

Pierre-Simon de Laplace (1749–1827) lived in Normandy as a boy but came to

Paris in 1768 and quickly made his mark in scientiﬁc circles, winning election to the

Académie des Sciences in 1773. He was preeminent in the ﬁeld of celestial mechanics;

his greatest work,Traité de mécanique céleste, was published in ﬁve volumes between

1799 and 1825. Laplace’s equation is fundamental in many branches of mathematical

physics,and Laplacestudied it extensively in connection with gravitational attraction.

The Laplace transform (Chapter 6) is also named for him, although its usefulness in

solving differential equations was not recognized until much later.

By the end ofthe eighteenthcentury manyelementary methodsof solvingordinary

differential equations had been discovered. In the nineteenth century interest turned

more toward the investigation of theoretical questions of existence and uniqueness

and to the development of less elementary methods such as those based on power

series expansions (seeChapter 5).These methods ﬁnd theirnatural setting inthe com-

plex plane. Consequently, they beneﬁtted from, and to some extent stimulated, the

more or less simultaneous development of the theory of complex analytic functions.

Partial differential equations also began to be studied intensively, as their crucial role

in mathematical physics became clear. In this connection a number of functions, aris-

ing as solutions of certain ordinary differential equations, occurred repeatedly and

were studied exhaustively. Known collectively as higher transcendental functions,

many of them are associated with the names of mathematicians, including Bessel,

Legendre, Hermite, Chebyshev, and Hankel, among others.

The numerous differential equations that resisted solution by analytical means

led to the investigation of methods of numerical approximation (see Chapter 8).

By 1900 fairly effective numerical integration methods had been devised, but their

implementation was severely restricted by the need to execute the computations by

hand or with very primitive computing equipment. In the last 60 years the develop-

ment of increasingly powerful and versatile computers has vastly enlarged the range

of problems that can be investigated effectively by numerical methods. Extremely

reﬁned and robust numerical integrators were developed during the same period

and are readily available. Versions appropriate for personal computers have brought

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1.4 Historical Remarks 29

the ability to solve a great many signiﬁcant problems within the reach of individual

students.

Another characteristic of differential equations in the twentieth century was the

creation of geometrical or topological methods, especially for nonlinear equations.

The goal is to understand at least the qualitative behavior of solutions from a

geometrical, as well as from an analytical, point of view. If more detailed informa-

tion is needed, it can usually be obtained by using numerical approximations. An

introduction to geometrical methods appears in Chapter 9.

Within the past few years these two trends have come together. Computers, and

especially computer graphics, have given a new impetus to the study of systems of

nonlinear differential equations. Unexpected phenomena (Section 9.8), such as

strange attractors, chaos, and fractals, have been discovered, are being intensively

studied, and are leading to important new insights in a variety of applications.

Although it is an old subject about which much is known, the study of differen-

tial equations in the twenty-ﬁrst century remains a fertile source of fascinating and

important unsolved problems.

REFERENCES

Computer software for differential equations changes too fast for particulars to be given in a book such

as this. A Google search for Maple, Mathematica, Sage, or MATLAB is a good way to begin if you need

information about one of these computer algebra and numerical systems.

There are many instructional books on computer algebra systems, such as the following:

Cheung, C.-K., Keough, G. E., Gross, R. H., and Landraitis, C., Getting Started with Mathematica (3rd ed.)

(New York: Wiley, 2009).

Meade, D. B., May, M., Cheung, C.-K., and Keough,G. E., Getting Started with Maple (3rd ed.) (NewYork:

Wiley, 2009).

For further reading in the history of mathematics, see books such as those listed below:

Boyer, C. B., and Merzbach, U. C.,A History of Mathematics (2nd ed.) (New York:Wiley, 1989).

Kline, M., Mathematical Thought from Ancient to Modern Times (3 vols.) (New York: Oxford University

Press, 1990).

A useful historical appendix on the early development of differential equations appears in

Ince, E. L., Ordinary Differential Equations (London: Longmans, Green, 1927; New York: Dover, 1956).

Encyclopedic sources of information about the lives and achievements of mathematicians of the

past are

Gillespie, C. C., ed., Dictionary of Scientiﬁc Biography (15 vols.) (New York: Scribner’s, 1971).

Koertge, N., ed., New Dictionary of Scientiﬁc Biography (8 vols.) (New York: Scribner’s, 2007).

Koertge, N., ed., Complete Dictionary of Scientiﬁc Biography (New York: Scribner’s, 2007 [e-book]).

Much historical information can be found on the Internet. One excellent site is the MacTutor History

of Mathematics archive

http://www-history.mcs.st-and.ac.uk/history/

created by John J. O’Connor and Edmund F. Robertson, Department of Mathematics and Statistics,

University of St. Andrews, Scotland.

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CHAPTER

First Order

Differential Equations

This chapter deals with differential equations of ﬁrst order

= f (t, y), (1)

where f is a given function of two variables.Any differentiable function y = φ(t) that

satisﬁes this equation for all t in some interval is called a solution, and our object is

to determine whether such functions exist and, if so, to develop methods for ﬁnding

them. Unfortunately, for an arbitrary function f , there is no general method for solv-

ing the equation in terms of elementary functions. Instead, we will describe several

methods, each of which is applicable to a certain subclass of ﬁrst order equations.

The most important of these are linear equations (Section 2.1), separable equa-

tions (Section 2.2), and exact equations (Section 2.6). Other sections of this chapter

describe some ofthe important applicationsof ﬁrst orderdifferential equations,intro-

duce the idea of approximating a solution by numerical computation, and discuss

some theoretical questions related to the existence and uniqueness of solutions.

The ﬁnal section includes an example of chaotic solutions in the context of ﬁrst order

difference equations,which have some important points of similarity with differential

equations and are simpler to investigate.

2.1 Linear Equations; Method of Integrating Factors

If the function f in Eq. (1) depends linearly on the dependent variable y, then

Eq. (1) is called a ﬁrst order linear equation. In Sections 1.1 and 1.2 we discussed a

restricted type of ﬁrst order linear equation in which the coefﬁcients are constants.

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32 Chapter 2. First Order Differential Equations

A typical example is

=−ay + b, (2)

where a and b are given constants. Recall that an equation of this form describes the

motion of an object falling in the atmosphere.

Now we want to consider the most general ﬁrst order linear equation, which is

obtained by replacing the coefﬁcients a and b in Eq. (2) by arbitrary functions of t.

We will usually write the general ﬁrst order linear equation in the standard form

+ p(t)y = g(t), (3)

where p and g are given functions of the independent variable t. Sometimes it is more

convenient to write the equation in the form

P(t)

+ Q(t)y = G(t), (4)

where P, Q, and G are given. Of course, as long as P(t) ̸= 0, you can convert Eq. (4)

to Eq. (3) by dividing Eq. (4) by P(t).

In some cases it is possible to solve a ﬁrst order linear equation immediately by

integrating the equation, as in the next example.

EXAMPLE

Solve the differential equation

(4 + t

)

+ 2ty = 4t. (5)

The left side of Eq. (5) is a linear combination of dy/dt and y, a combination that also

appears in the rule from calculus for differentiating a product. In fact,

(4 + t

)

+ 2ty =

[(4 + t

)y];

it follows that Eq. (5) can be rewritten as

[(4 + t

)y]=4t. (6)

Thus,even though y is unknown,we can integrate both sides of Eq. (6) with respect to t,thereby

obtaining

(4 + t

)y = 2t

+ c, (7)

where c is an arbitrary constant of integration. By solving for y we ﬁnd that

y =

4 + t

. (8)

This is the general solution of Eq. (5).

Unfortunately, most ﬁrst order linear equations cannot be solved as illustrated in

Example 1 because their left sides are not the derivative of the product of y and

some other function. However, Leibniz discovered that if the differential equation is

multiplied by a certain function µ(t), then the equation is converted into one that is

immediately integrable by using the product rule for derivatives,just as in Example 1.

The function µ(t) is called an integrating factor and our main task is to determine

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2.1 Linear Equations; Method of Integrating Factors 33

how to ﬁnd it for a given equation. We will show how this method works ﬁrst for an

example and then for the general ﬁrst order linear equation in the standard form (3).

EXAMPLE

Find the general solution of the differential equation

y =

t/3

. (9)

Draw some representative integral curves; that is, plot solutions corresponding to several

values of the arbitrary constant c. Also ﬁnd the particular solution whose graph contains the

point (0, 1).

The ﬁrst step is to multiply Eq. (9) by a function µ(t), as yet undetermined; thus

µ(t)

µ(t)y =

µ(t)e

t/3

. (10)

The question now is whether we can choose µ(t) so that the left side of Eq. (10) is the derivative

of the product µ(t)y. For any differentiable function µ(t) we have

[µ(t)y]=µ(t)

dµ(t)

y, (11)

Thus the left side of Eq. (10) and the right side of Eq. (11) are identical, provided that we

choose µ(t) to satisfy

dµ(t)

µ(t). (12)

Our search for an integrating factor will be successful if we can ﬁnd a solution of Eq. (12).

Perhaps you can readily identify a function that satisﬁes Eq. (12): what well-known function

from calculus has a derivative that is equal to one-half times the original function? More

systematically, rewrite Eq. (12) as

dµ(t)/dt

µ(t)

which is equivalent to

ln |µ(t)|=

. (13)

Then it follows that

ln |µ(t)|=

t + C,

µ(t) = ce

t/2

. (14)

The function µ(t) given by Eq. (14) is an integrating factor for Eq. (9). Since we do not need

the most general integrating factor, we will choose c to be 1 in Eq. (14) and use µ(t) = e

t/2

Now we return to Eq. (9), multiply it by the integrating factor e

t/2

, and obtain

t/2

y =

5t/6

. (15)

By the choice we have made of the integrating factor, the left side of Eq. (15) is the derivative

of e

t/2

y, so that Eq. (15) becomes

t/2

y) =

5t/6

. (16)

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34 Chapter 2. First Order Differential Equations

By integrating both sides of Eq. (16), we obtain

t/2

y =

5t/6

+ c, (17)

where c is an arbitrary constant. Finally, on solving Eq. (17) for y,we have the general solution

of Eq. (9), namely,

y =

t/3

+ ce

−t/2

. (18)

To ﬁnd the solution passing through the point (0,1), we set t = 0 and y = 1 in Eq. (18),

obtaining 1 = (3/5) + c. Thus c = 2 /5, and the desired solution is

y =

t/3

−t/2

. (19)

Figure 2.1.1 includes the graphs of Eq. (18) for several values of c with a direction ﬁeld in

the background. The solution satisfying y(0) = 1 is shown by the black curve.

–2

–1

123456

FIGURE 2.1.1 Direction ﬁeld and integral curves of y

′

y =

t/3

;

the black curve passes through the point (0, 1).

Let us now extend the method of integrating factors to equations of the form

+ ay = g(t), (20)

where a is a given constant and g(t) is a given function. Proceeding as in Example 2,

we ﬁnd that the integrating factor µ(t) must satisfy

d µ

= aµ, (21)

rather than Eq. (12). Thus the integrating factor is µ(t) = e

. Multiplying Eq. (20)

by µ(t), we obtain

+ ae

y = e

g(t),

y) = e

g(t). (22)

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2.1 Linear Equations; Method of Integrating Factors 35

By integrating both sides of Eq. (22), we ﬁnd that

y =

g(t) dt + c, (23)

where c is an arbitrary constant. For many simple functions g(t), we can evaluate the

integral in Eq. (23) and express the solution y in terms of elementary functions, as

in Example 2. However, for more complicated functions g(t), it is necessary to leave

the solution in integral form. In this case

y = e

−at

g(s) ds + ce

−at

. (24)

Note that in Eq. (24) we have used s to denote the integration variable to distinguish

it from the independent variable t, and we have chosen some convenient value t

the lower limit of integration.

EXAMPLE

Find the general solution of the differential equation

− 2y = 4 − t (25)

and plot the graphs of several solutions. Discuss the behavior of solutions as t →∞.

Equation (25) is of the form (20) with a =−2; therefore, the integrating factor is

µ(t) = e

−2t

. Multiplying the differential equation (25) by µ(t), we obtain

−2t

− 2e

−2t

y = 4e

−2t

− te

−2t

y) = 4e

−2t

− te

−2t

. (26)

Then, by integrating both sides of this equation, we have

−2t

y =−2e

−2t

+ c,

where we have used integration by parts on the last term in Eq. (26). Thus the general solution

of Eq. (25) is

y =−

t + ce

. (27)

A direction ﬁeld and graphs of the solution (27) for several values of c are shown in Figure

2.1.2. The behavior of the solution for large values of t is determined by the term ce

. If c ̸= 0,

then the solution grows exponentially large in magnitude, with the same sign as c itself. Thus

the solutions diverge as t becomes large. The boundary between solutions that ultimately grow

positively and those that ultimately grow negatively occurs when c = 0. If we substitute c = 0

into Eq. (27) and then set t = 0, we ﬁnd that y =−7/4 is the separation point on the y-axis.

Note that for this initial value, the solution is y =−

t; it grows positively, but linearly

rather than exponentially.

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36 Chapter 2. First Order Differential Equations

0.5 1 1.5 2

–4

–3

–2

–1

FIGURE 2.1.2 Direction ﬁeld and integral curves of y

′

− 2y = 4 − t.

Now we return to the general ﬁrst order linear equation (3)

+ p(t)y = g(t),

where p and g are given functions. To determine an appropriate integrating factor,

we multiply Eq. (3) by an as yet undetermined function µ(t), obtaining

µ(t)

+ p(t)µ(t)y = µ(t)g(t). (28)

Following the same line of development as in Example 2, we see that the left side

of Eq. (28) is the derivative of the product µ(t)y, provided that µ(t) satisﬁes the

equation

d µ(t)

= p(t)µ(t). (29)

If we assume temporarily that µ(t) is positive, then we have

d µ(t)/dt

µ(t)

= p(t),

and consequently

ln µ(t) =

p(t) dt + k.

By choosing the arbitrary constant k to be zero, we obtain the simplest possible

function for µ, namely,

µ(t) = exp

p(t) dt. (30)

Note that µ(t) is positive for all t, as we assumed. Returning to Eq. (28), we have

[µ(t)y]=µ(t)g(t). (31)

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2.1 Linear Equations; Method of Integrating Factors 37

Hence

µ(t)y =

µ(t)g(t) dt + c, (32)

where c is an arbitrary constant. Sometimes the integral in Eq.(32) can be evaluated

in terms of elementary functions. However, in general this is not possible, so the

general solution of Eq. (3) is

y =

µ(t)

µ(s)g (s ) ds + c

, (33)

where again t

is some convenient lower limit of integration. Observe that Eq. (33)

involves two integrations,one to obtain µ(t) from Eq. (30) and the other to determine

y from Eq. (33).

EXAMPLE

Solve the initial value problem

′

+ 2y = 4t

, (34)

y(1) = 2. (35)

In order to determine p(t) and g(t) correctly, we must ﬁrst rewrite Eq. (34) in the standard

form (3). Thus we have

′

+ (2/t)y = 4t, (36)

so p(t) = 2/t and g (t) = 4t. To solve Eq. (36), we ﬁrst compute the integrating factor µ(t):

µ(t) = exp

dt = e

2 ln|t|

= t

On multiplying Eq. (36) by µ(t) = t

, we obtain

′

+ 2ty = (t

′

= 4t

and therefore

y = t

+ c,

where c is an arbitrary constant. It follows that

y = t

(37)

is the general solution of Eq. (34). Integral curves of Eq. (34) for several values of c are shown

in Figure 2.1.3. To satisfy the initial condition (35), it is necessary to choose c = 1; thus

y = t

, t > 0 (38)

is the solution of the initial value problem (34), (35).This solution is shown by the black curve

in Figure 2.1.3. Note that it becomes unbounded and is asymptotic to the positive y-axis as

t → 0 from the right. This is the effect of the inﬁnite discontinuity in the coefﬁcient p(t) at

the origin. The function y = t

+ (1/t

) for t < 0 is not part of the solution of this initial value

problem.

This is the ﬁrst example in which the solution fails to exist for some values of t. Again, this

is due to the inﬁnite discontinuity in p(t) at t = 0, which restricts the solution to the interval

0 < t < ∞.

Looking again at Figure 2.1.3,we see that some solutions (those for which c > 0) are asymp-

totic to the positive y-axis as t → 0 from the right, while other solutions (for which c < 0)

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38 Chapter 2. First Order Differential Equations

are asymptotic to the negative y-axis. The solution for which c = 0, namely, y = t

, remains

bounded and differentiable even at t = 0. If we generalize the initial condition (35) to

y(1) = y

, (39)

then c = y

− 1 and the solution (38) becomes

y = t

− 1

, t > 0 if y

̸= 1. (40)

As in Example 3, this is another instance where there is a critical initial value, namely, y

= 1,

that separates solutions that behave in one way from others that behave quite differently.

–1

11–

(1,

FIGURE 2.1.3 Integral curves of the differential equation ty

′

+ 2y = 4t

;

the black curve passes through the point (1, 2).

EXAMPLE

Solve the initial value problem

′

+ ty = 2, (41)

y(0) = 1. (42)

To convert the differential equation (41) to the standard form (3), we must divide by 2,

obtaining

′

+ (t/2)y = 1. (43)

Thus p(t) = t/2, and the integrating factor is µ(t) = exp(t

/4). Then multiply Eq. (43) by µ(t),

so that

′

y = e

. (44)

The left side of Eq. (44) is the derivative of e

y, so by integrating both sides of Eq. (44), we

obtain

y =

dt + c. (45)

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2.1 Linear Equations; Method of Integrating Factors 39

The integral on the right side of Eq. (45) cannot be evaluated in terms of the usual elementary

functions, so we leave the integral unevaluated. However, by choosing the lower limit of

integration as the initial point t = 0, we can replace Eq. (45) by

y =

ds + c, (46)

where c is an arbitrary constant. It then follows that the general solution y of Eq. (41) is

given by

y = e

−t

ds + ce

−t

. (47)

The initial condition (42) requires that c = 1.

The main purpose of this example is to illustrate that sometimes the solution must be left in

terms of an integral.This is usually at most a slight inconvenience,rather than a serious obstacle.

For a given value of t, the integral in Eq. (47) is a deﬁnite integral and can be approximated to

any desired degree of accuracy by using readily available numerical integrators. By repeating

this process for many values of t and plotting the results, you can obtain a graph of a solution.

Alternatively, you can use a numerical approximation method, such as those discussed in

Chapter 8, that proceed directly from the differential equation and need no expression for the

solution. Software packages such as Maple and Mathematica readily execute such procedures

and produce graphs of solutions of differential equations.

Figure 2.1.4 displays graphs of the solution (47) for several values of c. From the ﬁgure it

may be plausible to conjecture that all solutions approach a limit as t →∞. The limit can be

found analytically (see Problem 32).

–3

–2

–1

1234

FIGURE 2.1.4 Integral curves of 2y

′

+ ty = 2.

PROBLEMS In each of Problems 1 through 12:

(a) Draw a direction ﬁeld for the given differential equation.

(b) Based on an inspection of the direction ﬁeld, describe how solutions behave for large t.

solutions behave as t →∞.

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40 Chapter 2. First Order Differential Equations

1. y

′

+ 3y = t + e

−2t

2. y

′

− 2y = t

3. y

′

+ y = te

−t

+ 14.y

′

+ (1/t)y = 3 cos 2 t, t > 0

′

− 2y = 3e

6. ty

′

+ 2y = sin t, t > 0

′

+ 2ty = 2te

−t

8. (1 + t

′

+ 4ty = (1 + t

)

−2

9. 2y

′

+ y = 3t 10. ty

′

− y = t

−t

, t > 0

11.

′

+ y = 5 sin 2t 12. 2y

′

+ y = 3t

In each of Problems 13 through 20, ﬁnd the solution of the given initial value problem.

13. y

′

− y = 2te

, y(0) = 1

14. y

′

+ 2y = te

−2t

, y(1) = 0

15. ty

′

+ 2y = t

− t + 1, y(1) =

, t > 0

16. y

′

+ (2/t)y = (cos t)/t

, y(π) = 0, t > 0

17. y

′

− 2y = e

, y(0) = 2

18. ty

′

+ 2y = sin t, y(π/2) = 1, t > 0

19. t

′

+ 4t

y = e

−t

, y(−1) = 0, t < 0

20. ty

′

+ (t + 1)y = t, y(ln 2) = 1, t > 0

In each of Problems 21 through 23:

(a) Draw a direction ﬁeld for the given differential equation. How do solutions appear to

behave as t becomes large? Does the behavior depend on the choice of the initial value a?

Let a

be the value of a for which the transition from one type of behavior to another occurs.

Estimate the value of a

(b) Solve the initial value problem and ﬁnd the critical value a

exactly.

21.

′

−

y = 2 cos t, y(0) = a

22.

′

− y = e

t/3

, y(0) = a

23.

′

− 2y = e

−πt/2

, y(0) = a

In each of Problems 24 through 26:

(a) Draw a direction ﬁeld for the given differential equation. How do solutions appear to

behave as t → 0? Does the behavior depend on the choice of the initial value a? Let a

be the

value of a for which the transition from one type of behavior to another occurs. Estimate the

value of a

(b) Solve the initial value problem and ﬁnd the critical value a

exactly.

24.

′

+ (t + 1)y = 2te

−t

, y(1) = a, t > 0

25.

′

+ 2y = (sin t)/t, y(−π/2) = a, t < 0

26.

(sin t)y

′

+ (cos t)y = e

, y(1) = a,0< t <π

27.

Consider the initial value problem

′

y = 2 cos t, y(0) =−1.

Find the coordinates of the ﬁrst local maximum point of the solution for t > 0.

28.

Consider the initial value problem

′

y = 1 −

t, y(0) = y

Find the value of y

for which the solution touches, but does not cross, the t-axis.

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2.1 Linear Equations; Method of Integrating Factors 41

29. Consider the initial value problem

′

y = 3 + 2 cos 2t, y(0) = 0.

(a) Find the solution of this initial value problem and describe its behavior for large t.

(b) Determine the value of t for which the solution ﬁrst intersects the line y = 12.

30. Find the value of y

for which the solution of the initial value problem

′

− y = 1 + 3 sin t, y(0) = y

remains ﬁnite as t →∞.

31. Consider the initial value problem

′

−

y = 3t + 2e

, y(0) = y

Find the value of y

that separates solutions that grow positively as t →∞from those

that grow negatively. How does the solution that corresponds to this critical value of y

behave as t →∞?

32. Show that all solutions of 2y

′

+ ty = 2 [Eq. (41) of the text] approach a limit as t →∞,

and ﬁnd the limiting value.

Hint: Consider the general solution, Eq. (47), and use L’Hôpital’s rule on the ﬁrst term.

33. Show that if a and λ are positive constants, and b is any real number, then every solution

of the equation

′

+ ay = be

−λt

has the property that y → 0 as t →∞.

Hint: Consider the cases a = λ and a ̸= λ separately.

In each of Problems 34 through 37, construct a ﬁrst order linear differential equation whose

solutions have the required behavior as t →∞. Then solve your equation and conﬁrm that

the solutions do indeed have the speciﬁed property.

34. All solutions have the limit 3 as t →∞.

35. All solutions are asymptotic to the line y = 3 − t as t →∞.

36. All solutions are asymptotic to the line y = 2t − 5 as t →∞.

37. All solutions approach the curve y = 4 − t

as t →∞.

38. Variation of Parameters. Consider the following method of solving the general linear

equation of ﬁrst order:

′

+ p(t)y = g(t). (i)

(a) If g(t) = 0 for all t, show that the solution is

y = A exp

−

p(t) dt

, (ii)

where A is a constant.

(b) If g(t) is not everywhere zero, assume that the solution of Eq. (i) is of the form

y = A(t) exp

−

p(t) dt

, (iii)

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42 Chapter 2. First Order Differential Equations

where A is now a function of t. By substituting for y in the given differential equation,

show that A(t) must satisfy the condition

′

(t) = g(t) exp

p(t) dt

. (iv)

that the solution obtained in this manner agrees with that of Eq. (33) in the text. This

technique is known as the method of variation of parameters; it is discussed in detail in

Section 3.6 in connection with second order linear equations.

In each of Problems 39 through 42,use the method of Problem 38 to solve the given differential

equation.

39. y

′

− 2y = t

40. y

′

+ (1/t)y = 3 cos 2 t, t > 0

41. ty

′

+ 2y = sin t, t > 0 42. 2y

′

+ y = 3t

2.2 Separable Equations

In Section 1.2 we used a process of direct integration to solve ﬁrst order linear

equations of the form

= ay + b, (1)

where a and b are constants. We will now show that this process is actually applicable

to a much larger class of equations.

We will use x, rather than t, to denote the independent variable in this section for

two reasons. In the ﬁrst place,different letters are frequently used for the variables in

a differential equation, and you should not become too accustomed to using a single

pair. In particular, x often occurs as the independent variable. Further, we want to

reserve t for another purpose later in the section.

The general ﬁrst order equation is

= f (x, y). (2)

Linear equations were considered in the preceding section,but if Eq. (2) is nonlinear,

then there is no universally applicable method for solving the equation. Here, we

consider a subclass of ﬁrst order equations that can be solved by direct integration.

To identify this class of equations, we ﬁrst rewrite Eq. (2) in the form

M(x, y) + N(x, y)

= 0. (3)

It is always possible to do this by setting M(x, y) =−f (x, y) and N(x, y) = 1, but there

may be other ways as well. If it happens that M is a function of x only and N is a

function of y only, then Eq. (3) becomes

M(x) + N(y)

= 0. (4)

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2.2 Separable Equations 43

Such an equation is said to be separable, because if it is written in the differential

form

M(x) dx + N(y) dy = 0, (5)

then, if you wish, terms involving each variable may be placed on opposite sides

of the equation. The differential form (5) is also more symmetric and tends to

suppress the distinction between independent and dependent variables.

A separable equation can be solved by integrating the functions M and N.

We illustrate the process by an example and then discuss it in general for Eq. (4).

EXAMPLE

Show that the equation

1 − y

(6)

is separable, and then ﬁnd an equation for its integral curves.

If we write Eq. (6) as

−x

+ (1 − y

)

= 0, (7)

then it has the form (4) and is therefore separable. Recall from calculus that if y is a function

of x, then by the chain rule,

f (y) =

f (y)

= f

′

(y)

For example, if f (y) = y − y

/3, then

(y − y

/3) = (1 − y

)

Thus the second term in Eq. (7) is the derivative with respect to x of y − y

/3, and the ﬁrst

term is the derivative of −x

/3. Thus Eq. (7) can be written as

−

(

y −

(

= 0,

−

+ y −

(

= 0.

Therefore, by integrating, we obtain

−x

+ 3y − y

= c, (8)

where c is an arbitrary constant. Equation (8) is an equation for the integral curves of Eq. (6).

A direction ﬁeld and several integral curves are shown in Figure 2.2.1. Any differentiable

function y = φ(x) that satisﬁes Eq. (8) is a solution of Eq. (6). An equation of the integral

curve passing through a particular point (x

, y

) can be found by substituting x

and y

for x

and y, respectively, in Eq. (8) and determining the corresponding value of c.

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44 Chapter 2. First Order Differential Equations

422–4–

–2

–4

FIGURE 2.2.1 Direction ﬁeld and integral curves of y

′

= x

/(1 − y

Essentially the same procedure can be followed for any separable equation.

Returning to Eq. (4), let H

and H

be any antiderivatives of M and N, respectively.

Thus

′

(x) = M(x), H

′

(y) = N(y), (9)

and Eq. (4) becomes

′

(x) + H

′

(y)

= 0. (10)

If y is regarded as a function of x, then according to the chain rule,

′

(y)

(y). (11)

Consequently, we can write Eq. (10) as

(x) + H

(y)]=0. (12)

By integrating Eq. (12), we obtain

(x) + H

(y) = c, (13)

where c is an arbitrary constant. Any differentiable function y = φ(x) that satisﬁes

Eq. (13) is a solution of Eq. (4); in other words, Eq. (13) deﬁnes the solution implic-

itly rather than explicitly. In practice, Eq. (13) is usually obtained from Eq. (5) by

integrating the ﬁrst term with respect to x and the second term with respect to y.The

justiﬁcation for this is the argument that we have just given.

The differential equation (4), together with an initial condition

y(x

) = y

, (14)

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2.2 Separable Equations 45

forms an initial value problem.To solve this initial value problem,we must determine

the appropriate value for the constant c in Eq. (13). We do this by setting x = x

and

y = y

in Eq. (13) with the result that

c = H

) + H

). (15)

Substituting this value of c in Eq. (13) and noting that

(x) − H

) =

M(s) ds, H

(y) − H

) =

N(s) ds,

we obtain

M(s) ds +

N(s) ds = 0. (16)

Equation (16) is an implicit representation of the solution of the differential equation

(4) that also satisﬁes the initial condition (14). Bear in mind that to determine an

explicit formula for the solution, you need to solve Eq. (16) for y as a function of x.

Unfortunately, it is often impossible to do this analytically; in such cases you can

resort to numerical methods to ﬁnd approximate values of y for given values of x.

EXAMPLE

Solve the initial value problem

+ 4x + 2

2(y − 1)

, y(0) =−1, (17)

and determine the interval in which the solution exists.

The differential equation can be written as

2(y − 1) dy = (3x

+ 4x + 2) dx.

Integrating the left side with respect to y and the right side with respect to x gives

− 2y = x

+ 2x

+ 2x + c, (18)

where c is an arbitrary constant. To determine the solution satisfying the prescribed initial

condition, we substitute x = 0 and y =−1 in Eq. (18), obtaining c = 3. Hence the solution of

the initial value problem is given implicitly by

− 2y = x

+ 2x

+ 2x + 3. (19)

To obtain the solution explicitly, we must solve Eq. (19) for y in terms of x. That is a simple

matter in this case, since Eq. (19) is quadratic in y, and we obtain

y = 1 ±

+ 2x

+ 2x + 4. (20)

Equation (20) gives two solutions of the differential equation, only one of which, however,

satisﬁes the given initial condition. This is the solution corresponding to the minus sign in

Eq. (20), so we ﬁnally obtain

y = φ(x) = 1 −

+ 2x

+ 2x + 4 (21)

as the solution of the initial value problem (17). Note that if we choose the plus sign by mistake

in Eq. (20),then we obtain the solution of the same differential equation that satisﬁes the initial

condition y(0) = 3. Finally,to determine the interval in which the solution (21) is valid,we must

ﬁnd the interval in which the quantity under the radical is positive. The only real zero of this

expression is x =−2, so the desired interval is x > −2. Some integral curves of the differential

August 7, 2012 21:03 c02 Sheet number 16 Page number 46 cyan black

46 Chapter 2. First Order Differential Equations

equation are shown in Figure 2.2.2. The black curve passes through the point (0, −1) and thus

is the solution of the initial value problem (17). Observe that the boundary of the interval

of validity of the solution (21) is determined by the point (−2, 1) at which the tangent line is

vertical.

11––2

(–2, 1)

–2

–1

(0, –1)

FIGURE 2.2.2 Integral curves of y

′

= (3x

+ 4x + 2)/2(y − 1); the

solution satisfying y(0) =−1 is shown in black and is valid for x > −2.

EXAMPLE

Solve the equation

4x − x

4 + y

(22)

and draw graphs of several integral curves. Also ﬁnd the solution passing through the point

(0, 1) and determine its interval of validity.

Rewriting Eq. (22) as

(4 + y

) dy = (4x − x

) dx,

integrating each side, multiplying by 4, and rearranging the terms, we obtain

+ 16y + x

− 8x

= c, (23)

where c is an arbitrary constant. Any differentiable function y = φ(x) that satisﬁes Eq. (23)

is a solution of the differential equation (22). Graphs of Eq. (23) for several values of c are

shown in Figure 2.2.3.

To ﬁnd the particular solution passing through (0, 1),we set x = 0 and y = 1 in Eq. (23) with

the result that c = 17. Thus the solution in question is given implicitly by

+ 16y + x

− 8x

= 17. (24)

It is shown by the black curve in Figure 2.2.3. The interval of validity of this solution extends

on either side of the initial point as long as the function remains differentiable. From the

ﬁgure we see that the interval ends when we reach points where the tangent line is verti-

cal. It follows from the differential equation (22) that these are points where 4 + y

= 0, or

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2.2 Separable Equations 47

–3

–2

–1

–3 –2 –1 123

)4785.1– ,8843.3()4785.1– ,8843.3–(

FIGURE 2.2.3 Integral curves of y

′

= (4x − x

)/(4 + y

The solution passing through (0, 1) is shown by the black curve.

y = (−4)

1/3

∼

−1.5874. From Eq. (24) the corresponding values of x are x

∼

±3.3488. These

points are marked on the graph in Figure 2.2.3.

Note 1: Sometimes an equation of the form (2)

= f (x, y)

has a constant solution y = y

. Such a solution is usually easy to ﬁnd because if

f (x, y

) = 0 for some value y

and for all x, then the constant function y = y

is a

solution of the differential equation (2). For example, the equation

(y − 3) cos x

1 + 2y

(25)

has the constant solution y = 3. Other solutions of this equation can be found by

separating the variables and integrating.

Note 2: The investigation of a ﬁrst order nonlinear equation can sometimes be

facilitated by regarding both x and y as functions of a third variable t. Then

dy/dt

dx/dt

. (26)

If the differential equation is

F(x, y)

G(x, y)

, (27)

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48 Chapter 2. First Order Differential Equations

then, by comparing numerators and denominators in Eqs. (26) and (27), we obtain

the system

dx/dt = G(x, y), dy/dt = F(x, y). (28)

At ﬁrst sight it may seem unlikely that a problem will be simpliﬁed by replacing a

single equation by a pair of equations, but in fact, the system (28) may well be more

amenable to investigation than the single equation (27). Chapter 9 is devoted to

nonlinear systems of the form (28).

Note 3: In Example 2 it was not difﬁcult to solve explicitly for y as a function

of x. However, this situation is exceptional, and often it will be better to leave the

solution in implicit form, as in Examples 1 and 3. Thus, in the problems below and

in other sections where nonlinear equations appear, the words “solve the following

differential equation” mean to ﬁnd the solution explicitly if it is convenient to do so,

but otherwise to ﬁnd an equation deﬁning the solution implicitly.

PROBLEMS In each of Problems 1 through 8, solve the given differential equation.

1. y

′

= x

/y 2. y

′

= x

/y(1 + x

)

3. y

′

+ y

sin x = 04.y

′

= (3x

− 1)/(3 + 2y)

5. y

′

= (cos

x)(cos

2y) 6. xy

′

= (1 − y

)

1/2

x − e

−x

y + e

1 + y

In each of Problems 9 through 20:

(a) Find the solution of the given initial value problem in explicit form.

(b) Plot the graph of the solution.

′

= (1 − 2x)y

, y(0) =−1/6 10. y

′

= (1 − 2x)/y, y(1) =−2

11.

xdx+ ye

−x

dy = 0, y(0) = 1 12. dr/dθ = r

/θ, r(1) = 2

13.

′

= 2x/(y + x

y), y(0) =−2 14. y

′

= xy

(1 + x

)

−1/2

, y(0) = 1

15.

′

= 2x/(1 + 2y), y(2) = 0 16. y

′

= x(x

+ 1)/4y

, y(0) =−1/

√

17.

′

= (3x

− e

)/(2y − 5), y(0) = 1

18.

′

= (e

−x

− e

)/(3 + 4y), y(0) = 1

19.

sin 2xdx+cos 3ydy = 0, y(π/2) = π/3

20.

(1 − x

)

1/2

dy = arcsin xdx, y(0) = 1

Some of the results requested in Problems 21 through 28 can be obtained either by solving

the given equations analytically or by plotting numerically generated approximations to the

solutions. Try to form an opinion about the advantages and disadvantages of each approach.

21.

Solve the initial value problem

′

= (1 + 3x

)/(3y

− 6y), y(0) = 1

and determine the interval in which the solution is valid.

Hint: To ﬁnd the interval of deﬁnition, look for points where the integral curve has a

vertical tangent.

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2.2 Separable Equations 49

22. Solve the initial value problem

′

= 3x

/(3y

− 4), y(1) = 0

and determine the interval in which the solution is valid.

Hint: To ﬁnd the interval of deﬁnition, look for points where the integral curve has a

vertical tangent.

23.

Solve the initial value problem

′

= 2y

+ xy

, y(0) = 1

and determine where the solution attains its minimum value.

24.

Solve the initial value problem

′

= (2 − e

)/(3 + 2y), y(0) = 0

and determine where the solution attains its maximum value.

25.

Solve the initial value problem

′

= 2 cos 2x/(3 + 2y), y(0) =−1

and determine where the solution attains its maximum value.

26.

Solve the initial value problem

′

= 2(1 + x)(1 + y

), y(0) = 0

and determine where the solution attains its minimum value.

27.

Consider the initial value problem

′

= ty(4 − y)/3, y(0) = y

(a) Determine how the behavior of the solution as t increases depends on the initial

value y

(b) Suppose that y

= 0.5. Findthe time T at which the solution ﬁrst reaches the value 3.98.

28.

Consider the initial value problem

′

= ty(4 − y)/(1 + t), y(0) = y

> 0.

(a) Determine how the solution behaves as t →∞.

(b) If y

= 2, ﬁnd the time T at which the solution ﬁrst reaches the value 3.99.

by the time t = 2.

29. Solve the equation

ay + b

cy + d

where a, b, c, and d are constants.

Homogeneous Equations. If the right side of the equation dy/dx = f (x, y) can

be expressed as a function of the ratio y/x only, then the equation is said to be

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50 Chapter 2. First Order Differential Equations

homogeneous.

Such equations can always be transformed into separable equations

by a change of the dependent variable. Problem 30 illustrates how to solve ﬁrst order

homogeneous equations.

30.

Consider the equation

y − 4x

x − y

. (i)

(a) Show that Eq. (i) can be rewritten as

(y/x) − 4

1 − (y/x)

; (ii)

thus Eq. (i) is homogeneous.

(b) Introduce a new dependent variable v so that v = y/x, or y = xv(x). Express dy/dx in

terms of x, v, and dv/dx.

dv/dx. Show that the resulting differential equation is

v + x

v − 4

1 − v

− 4

1 − v

. (iii)

Observe that Eq. (iii) is separable.

(d) Solve Eq. (iii), obtaining v implicitly in terms of x.

(e) Find the solution of Eq. (i) by replacing v by y/x in the solution in part (d).

(f) Draw a direction ﬁeld and some integral curves for Eq. (i). Recall that the right side

of Eq. (i) actually depends only on the ratio y/x. This means that integral curves have

the same slope at all points on any given straight line through the origin, although the

slope changes from one line to another. Therefore, the direction ﬁeld and the integral

curves are symmetric with respect to the origin. Is this symmetry property evident from

your plot?

The method outlined in Problem 30 can be used for any homogeneous equation. That

is, the substitution y = xv(x) transforms a homogeneous equation into a separable

equation. The latter equation can be solved by direct integration, and then replacing v

by y/x gives the solution to the original equation. In each of Problems 31 through 38:

(a) Show that the given equation is homogeneous.

(b) Solve the differential equation.

the origin?

31.

+ xy + y

32.

+ 3y

2xy

33.

4y − 3x

2x − y

34.

=−

4x + 3y

2x + y

The word “homogeneous” has different meanings in different mathematical contexts. The homogeneous

equations considered here have nothing to do with the homogeneous equations that will occur in Chapter 3

and elsewhere.

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2.3 Modeling with First Order Equations 51

35.

x + 3y

x − y

36.

+ 3xy + y

) dx − x

dy = 0

37.

− 3y

2xy

38.

− x

2xy

2.3 Modeling with First Order Equations

Differential equations are of interest to nonmathematicians primarily because of the

possibility of using them to investigate a wide variety of problems in the physical,

biological, and social sciences. One reason for this is that mathematical models and

their solutions lead to equations relating the variables and parameters in the prob-

lem. These equations often enable you to make predictions about how the natural

process will behave in various circumstances. It is often easy to vary parameters in

the mathematical model over wide ranges, whereas this may be very time-consuming

or expensive, if not impossible, in an experimental setting. Nevertheless, mathemat-

ical modeling and experiment or observation are both critically important and have

somewhat complementary roles in scientiﬁc investigations. Mathematical models are

validated by comparison of their predictions with experimental results. On the other

hand, mathematical analyses may suggest the most promising directions to explore

experimentally, and they may indicate fairly precisely what experimental data will

be most helpful.

In Sections 1.1 and 1.2 we formulated and investigated a few simple mathemat-

ical models. We begin by recapitulating and expanding on some of the conclusions

reached in those sections. Regardless of the speciﬁc ﬁeld of application, there are

three identiﬁable steps that are always present in the process of mathematical

modeling.

Construction of the Model. In this step you translate the physical situation into math-

ematical terms, often using the steps listed at the end of Section 1.1. Perhaps most

critical at this stage is to state clearly the physical principle(s) that are believed to

govern the process. For example, it has been observed that in some circumstances

heat passes from a warmer to a cooler body at a rate proportional to the temperature

difference,that objects move about in accordance with Newton’s laws of motion, and

that isolated insect populations grow at a rate proportional to the current population.

Each of these statements involves a rate of change (derivative) and consequently,

when expressed mathematically, leads to a differential equation. The differential

equation is a mathematical model of the process.

It is important to realize that the mathematical equations are almost always only

an approximate description of the actual process. For example, bodies moving at

speeds comparable to the speed of light are not governed by Newton’s laws, insect

populations do not grow indeﬁnitely as stated because of eventual lack of food or

space, and heat transfer is affected by factors other than the temperature difference.

Thus you should always be aware of the limitations of the model so that you will use

it only when it is reasonable to believe that it is accurate.Alternatively, you can adopt

the point of view that the mathematical equations exactly describe the operation of

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52 Chapter 2. First Order Differential Equations

a simpliﬁed physical model, which has been constructed (or conceived of) so as to

embody the most important features of the actual process. Sometimes, the process

of mathematical modeling involves the conceptual replacement of a discrete process

by a continuous one. For instance, the number of members in an insect population

changes by discrete amounts; however, if the population is large, it seems reasonable

to consider it as a continuous variable and even to speak of its derivative.

Analysis of the Model. Once the problem has been formulated mathematically, you are

often faced with the problem of solving one or more differential equations or, failing

that, of ﬁnding out as much as possible about the properties of the solution. It may

happen that this mathematical problem is quite difﬁcult,and if so,further approxima-

tions may be indicated at this stage to make the problem mathematically tractable.

For example, a nonlinear equation may be approximated by a linear one, or a slowly

varying coefﬁcient may be replaced by a constant. Naturally, any such approxima-

tions must also be examined from the physical point of view to make sure that the

simpliﬁed mathematical problem still reﬂects the essential features of the physical

process under investigation. At the same time, an intimate knowledge of the physics

of the problem may suggest reasonable mathematical approximations that will make

the mathematical problem more amenable to analysis. This interplay of understand-

ing of physical phenomena and knowledge of mathematical techniques and their

limitations is characteristic of applied mathematics at its best, and it is indispens-

able in successfully constructing useful mathematical models of intricate physical

processes.

Comparison with Experiment or Obser vation. Finally, having obtained the solution (or at

least some information about it), you must interpret this information in the con-

text in which the problem arose. In particular, you should always check that the

mathematical solution appears physically reasonable. If possible,calculate the values

of the solution at selected points and compare them with experimentally observed

values. Or ask whether the behavior of the solution after a long time is consistent

withobservations. Or examine thesolutions corresponding tocertain special values of

parameters in the problem. Of course,the fact that the mathematical solution appears

to be reasonable does not guarantee that it is correct. However, if the predictions of

the mathematical model are seriously inconsistent with observations of the physical

system it purports to describe, this suggests that errors have been made in solving

the mathematical problem, that the mathematical model itself needs reﬁnement, or

that observations must be made with greater care.

The examples in this section are typical of applications in which ﬁrst order differ-

ential equations arise.

EXAMPLE

Mixing

At time t = 0 a tank contains Q

lb of salt dissolved in 100 gal of water;see Figure 2.3.1.Assume

that water containing

lb of salt/gal is entering the tank at a rate of r gal/min and that the

well-stirred mixture is draining from the tank at the same rate. Set up the initial value problem

that describes this ﬂow process. Find the amount of salt Q(t) in the tank at any time, and also

ﬁnd the limiting amount Q

that is present after a very long time. If r = 3 and Q

= 2Q

, ﬁnd

the time T after which the salt level is within 2% of Q

. Also ﬁnd the ﬂow rate that is required

if the value of T is not to exceed 45 min.

We assume that salt is neither created nor destroyed in the tank. Therefore, variations in

the amount of salt are due solely to the ﬂows in and out of the tank. More precisely, the rate

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2.3 Modeling with First Order Equations 53

of change of salt in the tank, dQ/dt, is equal to the rate at which salt is ﬂowing in minus the

rate at which it is ﬂowing out. In symbols,

= rate in − rate out. (1)

The rate at which salt enters the tank is the concentration

lb/gal times the ﬂow rate r gal/min,

or (r/4) lb/min. To ﬁnd the rate at which salt leaves the tank, we need to multiply the concen-

tration of salt in the tank by the rate of outﬂow, r gal/min. Since the rates of ﬂow in and out

are equal, the volume of water in the tank remains constant at 100 gal, and since the mixture

is “well-stirred,” the concentration throughout the tank is the same, namely, [Q(t)/100] lb/gal.

Therefore, the rate at which salt leaves the tank is [rQ(t)/100] lb/min. Thus the differential

equation governing this process is

−

100

. (2)

The initial condition is

Q(0) = Q

. (3)

r gal/min, lb/gal

r gal/min

FIGURE 2.3.1 The water tank in Example 1.

Upon thinking about the problem physically,wemight anticipate that eventually the mixture

originally in the tank will be essentially replaced by the mixture ﬂowing in,whoseconcentration

lb/gal. Consequently, we might expect that ultimately the amount of salt in the tank would

be very close to 25 lb. We can also ﬁnd the limiting amount Q

= 25 by setting dQ/dt equal

to zero in Eq. (2) and solving the resulting algebraic equation for Q.

To solve the initial value problem (2), (3) analytically, note that Eq. (2) is both linear and

separable. Rewriting it in the standard form for a linear equation, we have

100

. (4)

Thus the integrating factor is e

rt/100

and the general solution is

Q(t) = 25 + ce

−rt/100

, (5)

where c is an arbitrary constant.To satisfy the initial condition (3),wemust choose c = Q

− 25.

Therefore, the solution of the initial value problem (2), (3) is

Q(t) = 25 + (Q

− 25)e

−rt/100

, (6)

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54 Chapter 2. First Order Differential Equations

Q(t) = 25(1 − e

−rt/100

) + Q

−rt/100

. (7)

From Eq. (6) or (7), you can see that Q(t) → 25 (lb) as t →∞, so the limiting value Q

25, conﬁrming our physical intuition. Further, Q(t) approaches the limit more rapidly as r

increases. In interpreting the solution (7), note that the second term on the right side is the

portion of the original salt that remains at time t,while the ﬁrst term gives the amount of salt in

the tank as a consequence of the ﬂow processes. Plots of the solution for r = 3 and for several

values of Q

are shown in Figure 2.3.2.

20 40 60 80 100

FIGURE 2.3.2 Solutions of the initial value problem (2), (3):

dQ/dt = (r/4 ) − rQ/100, Q(0) = Q

for r = 3 and several values of Q

Now suppose that r = 3 and Q

= 2Q

= 50; then Eq. (6) becomes

Q(t) = 25 + 25e

−0.03t

. (8)

Since 2% of 25 is 0.5, we wish to ﬁnd the time T at which Q(t) has the value 25.5. Substituting

t = T and Q = 25.5 in Eq. (8) and solving for T, we obtain

T = (ln 50)/0.03

∼

130.4 (min). (9)

To determine r so that T = 45, return to Eq. (6), set t = 45, Q

= 50, Q(t) = 25.5, and solve

for r. The result is

r = (100/45) ln 50

∼

8.69 gal/min. (10)

Since this example is hypothetical, the validity of the model is not in question. If the ﬂow

rates are as stated, and if the concentration of salt in the tank is uniform, then the differential

equation (1) is an accurate description of the ﬂow process. Although this particular example

has no special signiﬁcance,models of this kind are often used in problems involving a pollutant

in a lake, or a drug in an organ of the body, for example, rather than a tank of salt water. In

such cases the ﬂow rates may not be easy to determine or may vary with time. Similarly, the

concentration may be far from uniform in some cases. Finally, the rates of inﬂow and outﬂow

may be different, which means that the variation of the amount of liquid in the problem must

also be taken into account.

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2.3 Modeling with First Order Equations 55

EXAMPLE

Compound

Interest

Suppose that a sum of money is deposited in a bank or money fund that pays interest at an

annual rate r. The value S(t) of the investment at any time t depends on the frequency with

which interest is compounded as well as on the interest rate. Financial institutions have various

policies concerning compounding: some compound monthly, some weekly, some even daily.

If we assume that compounding takes place continuously, then we can set up an initial value

problem that describes the growth of the investment.

The rate of change of the value of the investment is dS/dt, and this quantity is equal to

the rate at which interest accrues, which is the interest rate r times the current value of the

investment S(t). Thus

dS/dt = rS (11)

is the differential equation that governs the process. Suppose that we also know the value of

the investment at some particular time, say,

S(0 ) = S

. (12)

Then the solution of the initial value problem (11), (12) gives the balance S(t) in the account

at any time t. This initial value problem is readily solved, since the differential equation (11)

is both linear and separable. Consequently, by solving Eqs. (11) and (12), we ﬁnd that

S(t) = S

. (13)

Thus a bank account with continuously compounding interest grows exponentially.

Let us now compare the results from this continuous model with the situation in which

compounding occurs at ﬁnite time intervals. If interest is compounded once a year, then after

t years

S(t) = S

(1 + r)

If interest is compounded twice a year, then at the end of 6 months the value of the investment

is S

[1 + (r/2)], and at the end of 1 year it is S

[1 + (r/2)]

. Thus, after t years we have

S(t) = S

)

1 +

In general, if interest is compounded m times per year, then

S(t) = S

)

1 +

. (14)

The relation between formulas (13) and (14) is clariﬁed if we recall from calculus that

lim

m→∞

)

1 +

= S

The same model applies equally well to more general investments in which dividends and

perhaps capital gains can also accumulate, as well as interest. In recognition of this fact, we

will from now on refer to r as the rate of return.

Table 2.3.1 shows the effect of changing the frequency of compounding for a return rate

r of 8%. The second and third columns are calculated from Eq. (14) for quarterly and daily

compounding, respectively, and the fourth column is calculated from Eq. (13) for continuous

compounding. The results show that the frequency of compounding is not particularly impor-

tant in most cases. For example, during a 10-year period the difference between quarterly

and continuous compounding is $17.50 per $1000 invested, or less than $2/year. The differ-

ence would be somewhat greater for higher rates of return and less for lower rates. From the

ﬁrst row in the table, we see that for the return rate r = 8%, the annual yield for quarterly

compounding is 8.24% and for daily or continuous compounding it is 8.33%.

Returning now to the case of continuous compounding, let us suppose that there may be

deposits or withdrawals in addition to the accrual of interest, dividends, or capital gains. If

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56 Chapter 2. First Order Differential Equations

TABLE 2.3.1 Growth of Capital at a Return Rate r = 8%

for Several Modes of Compounding

S(t)/S(t

) from Eq.(14)

S(t)/S(t

)

Years m = 4 m = 365 from Eq. (13)

1 1.0824 1.0833 1.0833

2 1.1717 1.1735 1.1735

5 1.4859 1.4918 1.4918

10 2.2080 2.2253 2.2255

20 4.8754 4.9522 4.9530

30 10.7652 11.0203 11.0232

40 23.7699 24.5239 24.5325

we assume that the deposits or withdrawals take place at a constant rate k, then Eq. (11) is

replaced by

dS/dt = rS + k,

or, in standard form,

dS/dt − rS = k, (15)

where k is positive for deposits and negative for withdrawals.

Equation (15) is linear with the integrating factor e

−rt

, so its general solution is

S(t) = ce

− (k/r),

where c is an arbitrary constant. To satisfy the initial condition (12), we must choose

c = S

+ (k/r). Thus the solution of the initial value problem (15), (12) is

S(t) = S

+ (k/r)(e

− 1). (16)

The ﬁrst term in expression (16) is the part of S(t) that is due to the return accumulated on

the initial amount S

, and the second term is the part that is due to the deposit or withdrawal

rate k.

The advantage of stating the problem in this general way without speciﬁc values for S

, r,or

k lies in the generality of the resulting formula (16) for S(t). With this formula we can readily

compare the results of different investment programs or different rates of return.

For instance, suppose that one opens an individual retirement account (IRA) at age 25 and

makes annual investments of $2000 thereafter in a continuous manner. Assuming a rate of

return of 8%, what will be the balance in the IRA at age 65? We have S

= 0, r = 0.08, and

k = $2000, and we wish to determine S(40). From Eq. (16) we have

S(40 ) = (25,000)(e

3.2

− 1) = $588,313. (17)

It is interesting to note that the total amount invested is $80,000, so the remaining amount of

$508,313 results from the accumulated return on the investment. The balance after 40 years

is also fairly sensitive to the assumed rate. For instance, S(40) = $508,948 if r = 0.075 and

S(40 ) = $681,508 if r = 0.085.

Let us now examine the assumptions that have gone into the model. First,we have assumed

that the return is compounded continuously and that additional capital is invested continu-

ously. Neither of these is true in an actual ﬁnancial situation. We have also assumed that the

return rate r is constant for the entire period involved, whereas in fact it is likely to ﬂuctuate

considerably. Although we cannot reliably predict future rates, we can use expression (16) to

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2.3 Modeling with First Order Equations 57

determine the approximate effect of different rate projections. It is also possible to consider

r and k in Eq. (15) to be functions of t rather than constants;in that case,of course,the solution

may be much more complicated than Eq. (16).

The initial value problem (15), (12) and the solution (16) can also be used to analyze a

number of other ﬁnancial situations, including annuities, mortgages, and automobile loans.

EXAMPLE

Chemicals in

aPond

Consider a pond that initially contains 10 million gal of fresh water. Water containing an

undesirable chemical ﬂows into the pond at the rate of 5 million gal/yr, and the mixture in

the pond ﬂows out at the same rate. The concentration γ(t) of chemical in the incoming water

varies periodically with time according to the expression γ(t) = 2 + sin 2t g/gal. Construct a

mathematical model of this ﬂow process and determine the amount of chemical in the pond

at any time. Plot the solution and describe in words the effect of the variation in the incoming

concentration.

Since the incoming and outgoing ﬂows of water are the same, the amount of water in the

pond remains constant at 10

gal. Let us denote time by t, measured in years, and the chemical

by Q(t), measured in grams. This example is similar to Example 1, and the same inﬂow/outﬂow

principle applies. Thus

= rate in − rate out,

where “rate in” and “rate out” refer to the rates at which the chemical ﬂows into and out of

the pond, respectively. The rate at which the chemical ﬂows in is given by

rate in = (5 × 10

) gal/yr (2 + sin 2t) g/gal. (18)

The concentration of chemical in the pond is Q(t)/10

g/gal, so the rate of ﬂow out is

rate out = (5 × 10

) gal/yr [Q(t)/10

] g/gal = Q(t)/2 g/yr. (19)

Thus we obtain the differential equation

= (5 × 10

)(2 + sin 2t) −

Q(t)

, (20)

where each term has the units of g/yr.

To make the coefﬁcients more manageable, it is convenient to introduce a new dependent

variable deﬁned by q(t) = Q(t)/10

,orQ(t) = 10

q(t). This means that q(t) is measured in

millions of grams, or megagrams (metric tons). If we make this substitution in Eq. (20), then

each term contains the factor 10

,which can be canceled. If we also transpose the term involving

q(t) to the left side of the equation, we ﬁnally have

q = 10 + 5 sin 2t. (21)

Originally, there is no chemical in the pond, so the initial condition is

q(0) = 0. (22)

Equation (21) is linear, and although the right side is a function of time, the coefﬁcient

of q is a constant. Thus the integrating factor is e

t/2

. Multiplying Eq. (21) by this factor and

integrating the resulting equation, we obtain the general solution

q(t) = 20 −

cos 2t +

sin 2t + ce

−t/2

. (23)

The initial condition (22) requires that c =−300/17,so the solutionof the initial value problem

(21), (22) is

q(t) = 20 −

cos 2t +

sin 2t −

300

−t/2

. (24)

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58 Chapter 2. First Order Differential Equations

A plot of the solution (24) is shown in Figure 2.3.3,along with the line q = 20. The exponential

term in the solution is important for small t, but it diminishes rapidly as t increases. Later, the

solution consists of an oscillation, due to the sin 2t and cos 2t terms, about the constant level

q = 20. Note that if the sin 2t term were not present in Eq. (21), then q = 20 would be the

equilibrium solution of that equation.

8 1046

12 14 16 18 20

FIGURE 2.3.3 Solution of the initial value problem (21), (22):

dq/dt +

q = 10 + 5 sin 2t, q(0) = 0.

Let us now consider the adequacy of the mathematical model itself for this problem.

The model rests on several assumptions that have not yet been stated explicitly. In the ﬁrst

place, the amount of water in the pond is controlled entirely by the rates of ﬂow in and out—

none is lost by evaporation or by seepage into the ground, and none is gained by rainfall.

The same is true of the chemical; it ﬂows into and out of the pond, but none is absorbed by

ﬁsh or other organisms living in the pond. In addition, we assume that the concentration of

chemical in the pond is uniform throughout the pond. Whether the results obtained from the

model are accurate depends strongly on the validity of these simplifying assumptions.

EXAMPLE

Escape

Velocity

A body of constant mass m is projected away from the earth in a direction perpendicular to the

earth’s surface with an initial velocity v

. Assuming that there is no air resistance, but taking

into account the variation of the earth’s gravitational ﬁeld with distance, ﬁnd an expression

for the velocity during the ensuing motion. Also ﬁnd the initial velocity that is required to

lift the body to a given maximum altitude ξ above the surface of the earth, and ﬁnd the

least initial velocity for which the body will not return to the earth; the latter is the escape

velocity.

Let the positive x-axis point away from the center of the earth along the line of motion with

x = 0 lying on the earth’s surface; see Figure 2.3.4. The ﬁgure is drawn horizontally to remind

you that gravity is directed toward the center of the earth, which is not necessarily downward

from a perspective away from the earth’s surface. The gravitational force acting on the body

(that is, its weight) is inversely proportional to the square of the distance from the center of

the earth and is given by w(x) =−k/(x + R)

, where k is a constant, R is the radius of the

earth, and the minus sign signiﬁes that w(x) is directed in the negative x direction. We know

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2.3 Modeling with First Order Equations 59

mgR

(R + x)

FIGURE 2.3.4 A body in the earth’s gravitational ﬁeld.

that on the earth’s surface w(0) is given by −mg, where g is the acceleration due to gravity at

sea level. Therefore, k = mgR

and

w(x) =−

mgR

(R + x)

. (25)

Since there are no other forces acting on the body, the equation of motion is

=−

mgR

(R + x)

, (26)

and the initial condition is

v(0) = v

. (27)

Unfortunately, Eq. (26) involves too many variables since it depends on t, x, and v.To

remedy this situation, we can eliminate t from Eq. (26) by thinking of x, rather than t, as the

independent variable. Then we can express dv/dt in terms of dv/dx by using the chain rule;

hence

= v

and Eq. (26) is replaced by

=−

(R + x)

. (28)

Equation (28) is separable but not linear, so by separating the variables and integrating, we

obtain

R + x

+ c. (29)

Since x = 0 when t = 0,the initial condition (27) at t = 0 can be replaced by the condition that

v = v

when x = 0. Hence c = (v

/2) − gR and

v =±

− 2gR +

2gR

R + x

. (30)

Note that Eq. (30) gives the velocity as a function of altitude rather than as a function of time.

The plus sign must be chosen if the body is rising, and the minus sign must be chosen if it is

falling back to earth.

To determine the maximum altitude ξ that the body reaches, we set v = 0 and x = ξ in

Eq. (30) and then solve for ξ, obtaining

ξ =

2gR − v

. (31)

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60 Chapter 2. First Order Differential Equations

Solving Eq. (31) for v

, we ﬁnd the initial velocity required to lift the body to the altitude ξ,

namely,

2gR

R + ξ

. (32)

The escape velocity v

is then found by letting ξ →∞. Consequently,

2gR. (33)

The numerical value of v

is approximately 6.9 mi/s, or 11.1 km/s.

The preceding calculation of the escape velocity neglects the effect of air resistance, so the

actual escape velocity (including the effect of air resistance) is somewhat higher. On the other

hand, the effective escape velocity can be signiﬁcantly reduced if the body is transported a

considerable distance above sea level before being launched. Both gravitational and frictional

forces are thereby reduced;air resistance,in particular,diminishes quite rapidly withincreasing

altitude. You should keep in mind also that it may well be impractical to impart too large an

initial velocity instantaneously; space vehicles, for instance, receive their initial acceleration

during a period of a few minutes.

PROBLEMS 1. Consider a tank used in certain hydrodynamic experiments. After one experiment the

tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for

the next experiment, the tank is to be rinsed with fresh water ﬂowing in at a rate of

2 L/min, the well-stirred solution ﬂowing out at the same rate. Find the time that will

elapse before the concentration of dye in the tank reaches 1% of its original value.

2. A tank initially contains 120 L of pure water. A mixture containing a concentration of

γ g/L of salt enters the tank at a rate of 2 L/min, and the well-stirred mixture leaves the

tank at the same rate. Find an expression in terms of γ for the amount of salt in the tank

at any time t. Also ﬁnd the limiting amount of salt in the tank as t →∞.

3. A tank originally contains 100 gal of fresh water. Then water containing

lb of salt per

gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at

the same rate.After 10 min the process is stopped, and fresh water is poured into the tank

at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of

salt in the tank at the end of an additional 10 min.

4. A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt

in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and

the mixture is allowed to ﬂow out of the tank at a rate of 2 gal/min. Find the amount

of salt in the tank at any time prior to the instant when the solution begins to overﬂow.

Find the concentration (in pounds per gallon) of salt in the tank when it is on the point

of overﬂowing. Compare this concentration with the theoretical limiting concentration if

the tank had inﬁnite capacity.

A tank contains 100 gal of water and 50 oz of salt.Water containing a salt concentration of

(1 +

sin t) oz/gal ﬂows into the tank at a rate of 2 gal/min, and the mixture in the tank

ﬂows out at the same rate.

(a) Find the amount of salt in the tank at any time.

(b) Plot the solution for a time period long enough so that you see the ultimate behavior

of the graph.

What is this level? What is the amplitude of the oscillation?

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2.3 Modeling with First Order Equations 61

6. Suppose that a tank containing a certain liquid has an outlet near the bottom. Let h(t)

be the height of the liquid surface above the outlet at time t. Torricelli’s

principle states

that the outﬂow velocity v at the outlet is equal to the velocity of a particle falling freely

(with no drag) from the height h.

(a) Show that v =

2gh, where g is the acceleration due to gravity.

(b) By equating the rate of outﬂow to the rate of change of liquid in the tank, show that

h(t) satisﬁes the equation

A(h)

=−αa

2gh, (i)

where A(h) is the area of the cross section of the tank at height h and a is the area of the

outlet. The constant α is a contraction coefﬁcient that accounts for the observed fact that

the cross section of the (smooth) outﬂow stream is smaller than a. The value of α for water

is about 0.6.

the outlet. The radius of the tank is 1 m, and the radius of the circular outlet is 0.1 m.

If the tank is initially full of water, determine how long it takes to drain the tank down to

the level of the outlet.

7. Suppose that a sum S

is invested at an annual rate of return r compounded continuously.

(a) Find the time T required for the original sum to double in value as a function of r.

(b) Determine T if r = 7%.

8 years.

8. A young person with no initial capital invests k dollars per year at an annual rate of

return r. Assume that investments are made continuously and that the return is

compounded continuously.

(a) Determine the sum S(t) accumulated at any time t.

(b) If r = 7.5%,determine k so that $1 million will be available for retirement in 40 years.

available in 40 years.

9. A certain college graduate borrows $8000 to buy a car. The lender charges interest at

an annual rate of 10%. Assuming that interest is compounded continuously and that

the borrower makes payments continuously at a constant annual rate k, determine the

payment rate k that is required to pay off the loan in 3 years. Also determine how much

interest is paid during the 3-year period.

10. A home buyer can afford to spend no more than $1500/month on mortgage payments.

Suppose that the interest rate is 6%, that interest is compounded continuously, and that

payments are also made continuously.

(a) Determine the maximum amount that this buyer can afford to borrow on a 20-year

mortgage; on a 30-year mortgage.

(b) Determine the total interest paid during the term of the mortgage in each of the cases

in part (a).

Evangelista Torricelli (1608–1647), successor to Galileo as court mathematician in Florence, published

this result in 1644. He is also known for constructing the ﬁrst mercury barometer and for making important

contributions to geometry.

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62 Chapter 2. First Order Differential Equations

11. A home buyer wishes to borrow $250,000 at an interest rate of 6% to ﬁnance the pur-

chase.Assume that interest is compounded continuously and that payments are also made

continuously.

(a) Determine the monthly payment that is required to pay off the loan in 20 years;

in 30 years.

(b) Determine the total interest paid during the term of the loan in each of the cases in

part (a).

12.

A recent college graduate borrows $150,000 at an interest rate of 6% to purchase a con-

dominium. Anticipating steady salary increases, the buyer expects to make payments at a

monthly rate of 800 + 10t, where t is the number of months since the loan was made.

(a) Assuming that this payment schedule can be maintained, when will the loan be fully

paid?

(b) Assuming the same payment schedule, how large a loan could be paid off in exactly

20 years?

13. An important tool in archeological research is radiocarbon dating, developed by the

American chemistWillard F. Libby.

This is a means ofdetermining theage of certain wood

and plant remains, and hence of animal or human bones or artifacts found buried at the

same levels. Radiocarbon dating is based on the fact that some wood or plant remains con-

tain residual amounts of carbon-14,aradioactive isotope of carbon.This isotope is accumu-

lated during the lifetime of the plant and begins to decay at its death. Since the half-life of

carbon-14 is long (approximately 5730 years

), measurable amounts of carbon-14 remain

after many thousands of years. If even a tiny fraction of the original amount of carbon-14

is still present, then by appropriate laboratory measurements the proportion of the orig-

inal amount of carbon-14 that remains can be accurately determined. In other words, if

Q(t) is the amount of carbon-14 at time t and Q

is the original amount, then the ratio

Q(t)/Q

can be determined,as long as this quantity is not too small. Present measurement

techniques permit the use of this method for time periods of 50,000 years or more.

(a) Assuming that Q satisﬁes the differential equation Q

′

=−rQ, determine the decay

constant r for carbon-14.

(b) Find an expression for Q(t) at any time t, if Q(0) = Q

carbon-14 is 20% of the original amount. Determine the age of these remains.

14.

Suppose that a certain population has a growth rate that varies with time and that this

population satisﬁes the differential equation

dy/dt = (0.5 + sin t)y/5.

(a) If y(0) = 1, ﬁnd (or estimate) the time τ at which the population has doubled. Choose

other initial conditions and determine whether the doubling time τ depends on the initial

population.

(b) Suppose that the growth rate is replaced by its average value 1/10. Determine the

doubling time τ in this case.

Willard F. Libby (1908–1980) was born in rural Colorado and received his education at the University of

California at Berkeley. He developed the method of radiocarbon dating beginning in 1947 while he was

at the University of Chicago. For this work he was awarded the Nobel Prize in chemistry in 1960.

McGraw-Hill Encyclopedia of Science and Technology (8th ed.) (New York: McGraw-Hill, 1997), Vol. 5,

p. 48.

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2.3 Modeling with First Order Equations 63

the variation in the growth rate has a substantially higher frequency. What effect does this

have on the doubling time τ?

(d) Plot the solutions obtained in parts (a), (b), and (c) on a single set of axes.

15.

Suppose that a certain population satisﬁes the initial value problem

dy/dt = r(t)y − k, y(0) = y

where the growth rate r(t) is given by r(t) = (1 + sin t)/5, and k represents the rate of

predation.

(a) Suppose that k = 1/5. Plot y versus t for several values of y

between 1/2 and 1.

(b) Estimate the critical initial population y

below which the population will become

extinct.

for each one.

(d) Use the data you have found in parts (b) and (c) to plot y

versus k.

16. Newton’s law of cooling states that the temperature of an object changes at a rate pro-

portional to the difference between its temperature and that of its surroundings. Suppose

that the temperature of a cup of coffee obeys Newton’s law of cooling. If the coffee has a

temperature of 200

◦

F when freshly poured, and 1 min later has cooled to 190

◦

F in a room

at 70

◦

F, determine when the coffee reaches a temperature of 150

◦

17.

Heat transfer from a body to its surroundings by radiation, based on the Stefan–

Boltzmann

law, is described by the differential equation

=−α(u

− T

), (i)

where u(t) is the absolute temperature of the body at time t, T is the absolute temperature

of the surroundings, and α is a constant depending on the physical parameters of the body.

However, if u is much larger than T, then solutions of Eq. (i) are well approximated by

solutions of the simpler equation

=−αu

. (ii)

Suppose that a body with initial temperature 2000 K is surrounded by a medium with

temperature 300 K and that α = 2.0 × 10

−12

−3

/s.

(a) Determine the temperature of the body at any time by solving Eq. (ii).

(b) Plot the graph of u versus t.

time the error in using Eq. (ii) to approximate the solutions of Eq. (i) is no more than 1%.

18.

Consider an insulated box (a building,perhaps) with internal temperature u(t). According

to Newton’s law of cooling, u satisﬁes the differential equation

=−k[u − T(t)], (i)

where T(t) is the ambient (external) temperature. Suppose that T(t) varies sinusoidally;

for example, assume that T(t) = T

+ T

cos ωt.

Jozef Stefan (1835–1893), professor of physics at Vienna, stated the radiation law on empirical grounds

in 1879. His student Ludwig Boltzmann (1844–1906) derived it theoretically from the principles of

thermodynamics in 1884. Boltzmann is best known for his pioneering work in statistical mechanics.

August 7, 2012 21:03 c02 Sheet number 34 Page number 64 cyan black

64 Chapter 2. First Order Differential Equations

(a) Solve Eq. (i) and express u(t) in terms of t, k, T

, T

, and ω. Observe that part of

your solution approaches zero as t becomes large; this is called the transient part. The

remainder of the solution is called the steady state; denote it by S(t).

(b) Suppose that t is measured in hours and that ω = π/12, corresponding to a period of

24 h for T(t). Further, let T

= 60

◦

F, T

= 15

◦

F, and k = 0.2/h. Draw graphs of S(t) and

T(t) versus t on the same axes. From your graph estimate the amplitude R of the oscillatory

part of S(t). Also estimate the time lag τ between corresponding maxima of T(t) and S(t).

, T

, and ω now be unspeciﬁed. Write the oscillatory part of S(t) in the form

R cos[ω(t − τ)]. Use trigonometric identities to ﬁnd expressions for R and τ. Let T

and ω

have the values given in part (b), and plot graphs of R and τ versus k.

19. Consider a lake of constant volume V containing at time t an amount Q(t) of pollutant,

evenly distributed throughout the lake with a concentration c(t), where c(t) = Q(t)/V.

Assume that water containing a concentration k of pollutant enters the lake at a rate r,

and that water leaves the lake at the same rate. Suppose that pollutants are also added

directly to the lake at a constant rate P. Note that the given assumptions neglect a number

of factors that may, in some cases, be important—for example, the water added or lost

by precipitation, absorption, and evaporation; the stratifying effect of temperature differ-

ences in a deep lake; the tendency of irregularities in the coastline to produce sheltered

bays;and the fact that pollutants are deposited unevenly throughout the lake but (usually)

at isolated points around its periphery. The results below must be interpreted in the light

of the neglect of such factors as these.

(a) If at time t = 0 the concentration of pollutant is c

, ﬁnd an expression for the

concentration c(t) at any time. What is the limiting concentration as t →∞?

(b) If the addition of pollutants to the lake is terminated (k = 0 and P = 0 for t > 0),

determine the time interval T that must elapse before the concentration of pollutants is

reduced to 50% of its original value; to 10% of its original value.

for several of the Great Lakes. Using these data, determine

from part (b) the time T that is needed to reduce the contamination of each of these lakes

to 10% of the original value.

TABLE 2.3.2 Volume and Flow Data for the Great

Lakes

Lake V (km

× 10

) r (km

/year)

Superior 12.2 65.2

Michigan 4.9 158

Erie 0.46 175

Ontario 1.6 209

20. A ball with mass 0.15 kg is thrown upward with initial velocity 20 m/s from the roof of a

building 30 m high. Neglect air resistance.

(a) Find the maximum height above the ground that the ball reaches.

(b) Assuming that the ball misses the building on the way down, ﬁnd the time that it hits

the ground.

This problem is based on R. H. Rainey, “Natural Displacement of Pollution from the Great Lakes,”

Science 155 (1967), pp. 1242–1243; the information in the table was taken from that source.

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2.3 Modeling with First Order Equations 65

21. Assume that the conditions are as in Problem 20 except that there is a force due to air

resistance of magnitude |v|/30 directed opposite to the velocity, where the velocity v is

measured in m/s.

(a) Find the maximum height above the ground that the ball reaches.

(b) Find the time that the ball hits the ground.

corresponding ones in Problem 20.

22.

Assume that the conditions are as in Problem 20 except that there is a force due to air

resistance of magnitude v

/1325 directed opposite to the velocity, where the velocity v is

measured in m/s.

(a) Find the maximum height above the ground that the ball reaches.

(b) Find the time that the ball hits the ground.

corresponding ones in Problems 20 and 21.

23.

A skydiver weighing 180 lb (including equipment) falls vertically downward from an alti-

tude of 5000 ft and opens the parachute after 10 s of free fall. Assume that the force

of air resistance, which is directed opposite to the velocity, is of magnitude 0.75|v| when

the parachute is closed and is of magnitude 12|v| when the parachute is open, where the

velocity v is measured in ft/s.

(a) Find the speed of the skydiver when the parachute opens.

(b) Find the distance fallen before the parachute opens.

after the parachute opens?

(d) Determine how long the sky diver is in the air after the parachute opens.

(e) Plot the graph of velocity versus time from the beginning of the fall until the skydiver

reaches the ground.

24. A rocket sled having an initial speed of 150 mi/h is slowed by a channel of water. Assume

that during the braking process, the acceleration a is given by a(v) =−µv

, where v is the

velocity and µ is a constant.

(a) As in Example 4 in the text, use the relation dv/dt = v(dv/dx) to write the equation

of motion in terms of v and x.

(b) If it requires a distance of 2000 ft to slow the sled to 15 mi/h, determine the value of µ.

25. A body of constant mass m is projected vertically upward with an initial velocity v

in a medium offering a resistance k|v|, where k is a constant. Neglect changes in the

gravitational force.

(a) Find the maximum height x

attained by the body and the time t

at which this

maximum height is reached.

(b) Show that if kv

/mg < 1, then t

and x

can be expressed as

1 −

(

−···

1 −

(

−···

/mg is dimensionless.

26. A body of mass m is projected vertically upward with an initial velocity v

in a medium

offering a resistance k|v|, where k is a constant. Assume that the gravitational attraction

of the earth is constant.

(a) Find the velocity v(t) of the body at any time.

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66 Chapter 2. First Order Differential Equations

(b) Use the result of part (a) to calculate the limit of v(t) as k → 0—that is, as the resis-

tance approaches zero. Does this result agree with the velocity of a mass m projected

upward with an initial velocity v

in a vacuum?

approaches zero.

27. A body falling in a relatively dense ﬂuid, oil for example, is acted on by three forces

(see Figure 2.3.5): a resistive force R, a buoyant force B, and its weight w due to gravity.

The buoyant force is equal to the weight of the ﬂuid displaced by the object. For a slowly

moving spherical body of radius a,the resistive force is given by Stokes’s law,R = 6πµa|v|,

where v is the velocity of the body, and µ is the coefﬁcient of viscosity of the surrounding

ﬂuid.

(a) Find the limiting velocity of a solid sphere of radius a and density ρ falling freely in a

medium of density ρ

′

and coefﬁcient of viscosity µ.

(b) In 1910 R. A. Millikan

studied the motion of tiny droplets of oil falling in an electric

ﬁeld. A ﬁeld of strength E exerts a force Ee on a droplet with charge e. Assume that E

has been adjusted so the droplet is held stationary (v = 0) and that w and B are as given

above. Find an expression for e. Millikan repeated this experiment many times, and from

the data that he gathered he was able to deduce the charge on an electron.

FIGURE 2.3.5 A body falling in a dense ﬂuid.

28.

A mass of 0.25 kg is dropped from rest in a medium offering a resistance of 0.2|v|, where

v is measured in m/s.

(a) If the mass is dropped from a height of 30 m, ﬁnd its velocity when it hits the ground.

(b) If the mass is to attain a velocity of no more than 10 m/s, ﬁnd the maximum height

from which it can be dropped.

Sir George Gabriel Stokes (1819–1903) was born in Ireland but for most of his life was at Cambridge

University, ﬁrst as a student and later as a professor. Stokes was one of the foremost applied mathemati-

cians of the nineteenth century, best known for his work in ﬂuid dynamics and the wave theory of light.

The basic equations of ﬂuid mechanics (the Navier–Stokes equations) are named partly in his honor, and

one of the fundamental theorems of vector calculus bears his name. He was also one of the pioneers in

the use of divergent (asymptotic) series.

Robert A. Millikan (1868–1953) was educated at Oberlin College and Columbia University. Later he

was a professor at the University of Chicago and California Institute of Technology. His determination of

the charge on an electron was published in 1910. For this work, and for other studies of the photoelectric

effect, he was awarded the Nobel Prize for Physics in 1923.

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2.3 Modeling with First Order Equations 67

If the mass is dropped from a height of 30 m and must hit the ground with a velocity of

no more than 10 m/s, determine the coefﬁcient of resistance k that is required.

29. Suppose that a rocket is launched straight up from the surface of the earth with initial

velocity v

2gR, where R is the radius of the earth. Neglect air resistance.

(a) Find an expression for the velocity v in terms of the distance x from the surface of the

earth.

(b) Find the time required for the rocket to go 240,000 mi (the approximate distance from

the earth to the moon). Assume that R = 4000 mi.

30.

Let v(t) and w(t) be the horizontal and vertical components,respectively, of the velocity of

a batted (or thrown) baseball. In the absence of air resistance,v and w satisfy the equations

dv/dt = 0, dw/dt =−g.

(a) Show that

v = u cos A, w =−gt + u sin A,

where u is the initial speed of the ball and A is its initial angle of elevation.

(b) Let x(t) and y(t) be the horizontal and vertical coordinates, respectively, of the ball at

time t. If x(0) = 0 and y(0) = h, ﬁnd x(t) and y(t) at any time t.

, u = 125 ft/s, and h = 3 ft. Plot the trajectory of the ball for several

values of the angle A; that is, plot x(t) and y(t) parametrically.

(d) Suppose the outﬁeld wall is at a distance L and has height H. Find a relation between

u and A that must be satisﬁed if the ball is to clear the wall.

(e) Suppose that L = 350 ft and H = 10 ft. Using the relation in part (d),ﬁnd (or estimate

from a plot) the range of values of A that correspond to an initial velocity of u = 110 ft/s.

(f) For L = 350 and H = 10, ﬁnd the minimum initial velocity u and the corresponding

optimal angle A for which the ball will clear the wall.

31.

A more realistic model (than that in Problem 30) of a baseball in ﬂight includes the effect

of air resistance. In this case the equations of motion are

dv/dt =−rv, dw/dt =−g − rw,

where r is the coefﬁcient of resistance.

(a) Determine v(t) and w(t) in terms of initial speed u and initial angle of elevation A.

(b) Find x(t) and y(t) if x(0) = 0 and y(0) = h.

How do the trajectories differ from those in Problem 31 with r = 0?

(d) Assuming that r = 1/5 and h = 3, ﬁnd the minimum initial velocity u and the optimal

angle A for which the ball will clear a wall that is 350 ft distant and 10 ft high. Compare

this result with that in Problem 30(f).

32. Brachistochrone Problem. One of the famous problems in the history of mathematics is

the brachistochrone

problem: to ﬁnd the curve along which a particle will slide without

friction in the minimum time from one given point P to another Q, the second point being

lower than the ﬁrst but not directly beneath it (see Figure 2.3.6). This problem was posed

by Johann Bernoulli in 1696 as a challenge problem to the mathematicians of his day.

The word “brachistochrone” comes from the Greek words brachistos, meaning shortest, and chronos,

meaning time.

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68 Chapter 2. First Order Differential Equations

Correct solutions were found by Johann Bernoulli and his brother Jakob Bernoulli and

by Isaac Newton, Gottfried Leibniz, and the Marquis de L’Hôpital. The brachistochrone

problem is important in the development of mathematics as one of the forerunners of the

calculus of variations.

In solving this problem, it is convenient to take the origin as the upper point P and to

orient the axes as shown in Figure 2.3.6. The lower point Q has coordinates (x

, y

).Itis

then possible to show that the curve of minimum time is given by a function y = φ(x) that

satisﬁes the differential equation

(1 + y

′2

)y = k

, (i)

where k

is a certain positive constant to be determined later.

(a) Solve Eq. (i) for y

′

. Why is it necessary to choose the positive square root?

(b) Introduce the new variable t by the relation

y = k

sin

t. (ii)

Show that the equation found in part (a) then takes the form

sin

tdt = dx. (iii)

x = k

(θ − sin θ)/2, y = k

(1 − cos θ)/2. (iv)

Equations (iv) are parametric equations of the solution of Eq. (i) that passes through

(0, 0). The graph of Eqs. (iv) is called a cycloid.

(d) If we make a proper choice of the constant k, then the cycloid also passes through the

point(x

, y

) and is the solution ofthe brachistochrone problem. Findk if x

= 1and y

= 2.

Q(x

, y

)

FIGURE 2.3.6 The brachistochrone.

2.4 Differences Between Linear and Nonlinear Equations

Up to now, we have been primarily concerned with showing that ﬁrst order differ-

ential equations can be used to investigate many different kinds of problems in the

natural sciences, and with presenting methods of solving such equations if they are

either linear or separable. Now it is time to turn our attention to some more general

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2.4 Differences Between Linear and Nonlinear Equations 69

questions about differential equations and to explore in more detail some important

ways in which nonlinear equations differ from linear ones.

Existence and Uniqueness of Solutions. So far,we have discussed a number ofinitial value

problems, each of which had a solution and apparently only one solution. That raises

the question of whether this is true of all initial value problems for ﬁrst order equa-

tions. In other words, does every initial value problem have exactly one solution?

This may be an important question even for nonmathematicians. If you encounter

an initial value problem in the course of investigating some physical problem, you

might want to know that it has a solution before spending very much time and effort

in trying to ﬁnd it. Further, if you are successful in ﬁnding one solution, you might

be interested in knowing whether you should continue a search for other possible

solutions or whether you can be sure that there are no other solutions. For linear

equations, the answers to these questions are given by the following fundamental

theorem.

Theorem 2.4.1

If the functions p and g are continuous on an open interval I: α<t <βcontain-

ing the point t = t

, then there exists a unique function y = φ(t) that satisﬁes the

differential equation

′

+ p(t)y = g(t) (1)

for each t in I, and that also satisﬁes the initial condition

y(t

) = y

, (2)

where y

is an arbitrary prescribed initial value.

ObservethatTheorem 2.4.1 states that thegiven initial value problemhas asolution

and also that the problem has only one solution. In other words, the theorem asserts

both the existence and the uniqueness of the solution of the initial value problem (1),

(2). In addition, it states that the solution exists throughout any interval I containing

the initial point t

in which the coefﬁcients p and g are continuous.That is,the solution

can be discontinuous or fail to exist only at points where at least one of p and g is

discontinuous. Such points can often be identiﬁed at a glance.

The proof ofthis theorem is partly contained in the discussion in Section 2.1leading

to the formula [Eq. (32) in Section 2.1]

µ(t)y =

µ(t)g(t) dt + c, (3)

where [Eq. (30) in Section 2.1]

µ(t) = exp

p(t) dt. (4)

The derivation in Section 2.1 shows that if Eq. (1) has a solution, then it must be

given by Eq. (3). By looking slightly more closely at that derivation, we can also

conclude that the differential equation (1) must indeed have a solution. Since p is

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70 Chapter 2. First Order Differential Equations

continuous for α<t <β, it follows that µ is deﬁned in this interval and is a nonzero

differentiable function. Upon multiplying Eq. (1) by µ(t), we obtain

[µ(t)y]

′

= µ(t)g(t). (5)

Since both µ and g are continuous, the function µg is integrable, and Eq. (3) follows

from Eq. (5). Further, the integral of µg is differentiable, so y as given by Eq. (3)

exists and is differentiable throughout the interval α<t <β. By substituting the

expression for y from Eq. (3) into either Eq. (1) or Eq. (5), you can verify that

this expression satisﬁes the differential equation throughout the interval α<t <β.

Finally, the initial condition (2) determines the constant c uniquely, so there is only

one solution of the initial value problem; this completes the proof.

Equation (4) determines the integrating factor µ(t) only up to a multiplicative

factor that depends on the lower limit of integration. If we choose this lower limit to

be t

, then

µ(t) = exp

p(s) ds, (6)

and it follows that µ(t

) = 1. Using the integrating factor given by Eq. (6), and choos-

ing the lower limit of integration in Eq. (3) also to be t

,we obtain the general solution

of Eq. (1) in the form

y =

µ(t)

µ(s)g (s ) ds + c

. (7)

To satisfy the initial condition (2), we must choose c = y

. Thus the solution of the

initial value problem (1), (2) is

y =

µ(t)

µ(s)g (s ) ds + y

, (8)

where µ(t) is given by Eq. (6).

Turning now to nonlinear differential equations, we must replace Theorem 2.4.1

by a more general theorem, such as the one that follows.

Theorem 2.4.2

Let the functions f and ∂f /∂y be continuous in some rectangle α<t <β,

γ<y <δcontaining the point (t

, y

). Then, in some interval t

− h < t < t

+ h

contained in α<t <β, there is a unique solution y = φ(t) of the initial value

problem

′

= f (t, y), y(t

) = y

. (9)

Observe that the hypotheses in Theorem 2.4.2 reduce to those in Theorem 2.4.1 if

the differential equation is linear. In this case

f (t, y) =−p(t)y + g(t) and ∂f (t, y)/∂y =−p(t),

so the continuity of f and ∂f /∂y is equivalent to the continuity of p and g.

The proof of Theorem 2.4.1 was comparatively simple because it could be based

on the expression (3) that gives the solution of an arbitrary linear equation. There

is no corresponding expression for the solution of the differential equation (9), so

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2.4 Differences Between Linear and Nonlinear Equations 71

the proof of Theorem 2.4.2 is much more difﬁcult. It is discussed to some extent in

Section 2.8 and in greater depth in more advanced books on differential equations.

Here we note thatthe conditionsstated inTheorem2.4.2 aresufﬁcient toguarantee

the existence of a unique solution of the initial value problem (9) in some interval

− h < t < t

+ h, but they are not necessary. That is, the conclusion remains true

under slightly weaker hypotheses about the function f . In fact, the existence of a

solution (but not its uniqueness) can be established on the basis of the continuity of

f alone.

An important geometrical consequence of the uniqueness parts of Theorems 2.4.1

and 2.4.2 is that the graphs of two solutions cannot intersect each other. Otherwise,

there would be two solutions that satisfy the initial condition corresponding to the

point of intersection, in contradiction to Theorem 2.4.1 or 2.4.2.

We now consider some examples.

EXAMPLE

Use Theorem 2.4.1 to ﬁnd an interval in which the initial value problem

′

+ 2y = 4t

, (10)

y(1) = 2 (11)

has a unique solution.

Rewriting Eq. (10) in the standard form (1), we have

′

+ (2/t)y = 4t,

so p(t) = 2/t and g(t) = 4t.Thus,for this equation,g is continuous for all t,while p is continuous

only for t < 0 or for t > 0. The interval t > 0 contains the initial point; consequently,Theorem

2.4.1 guarantees that the problem (10), (11) has a unique solution on the interval 0 < t < ∞.

In Example 4 of Section 2.1 we found the solution of this initial value problem to be

y = t

, t > 0. (12)

Now suppose that the initial condition (11) is changed to y(−1) = 2. Then Theorem 2.4.1

asserts the existence of a unique solution for t < 0. As you can readily verify, the solution is

again given by Eq. (12), but now on the interval −∞ < t < 0.

EXAMPLE

Apply Theorem 2.4.2 to the initial value problem

+ 4x + 2

2(y − 1)

, y(0) =−1. (13)

Note that Theorem 2.4.1 is not applicable to this problem since the differential equation is

nonlinear. To apply Theorem 2.4.2, observe that

f (x, y) =

+ 4x + 2

2(y − 1)

∂f

∂y

(x, y) =−

+ 4x + 2

2(y − 1)

Thuseach of these functions is continuous everywhere except on the line y = 1. Consequently,a

rectangle can be drawn about the initial point (0, −1) in which both f and ∂f /∂y are continuous.

Therefore, Theorem 2.4.2 guarantees that the initial value problem has a unique solution in

some interval about x = 0. However, even though the rectangle can be stretched inﬁnitely

far in both the positive and the negative x directions, this does not necessarily mean that the

solution exists for all x. Indeed, the initial value problem (13) was solved in Example 2 of

Section 2.2, and the solution exists only for x > −2.

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72 Chapter 2. First Order Differential Equations

Now suppose we change the initial condition to y(0) = 1. The initial point now lies on the

line y = 1, so no rectangle can be drawn about it within which f and ∂f /∂y are continuous.

Consequently, Theorem 2.4.2 says nothing about possible solutions of this modiﬁed problem.

However, if we separate the variables and integrate, as in Section 2.2, we ﬁnd that

− 2y = x

+ 2x

+ 2x + c.

Further, if x = 0 and y = 1, then c =−1. Finally, by solving for y, we obtain

y = 1 ±

+ 2x

+ 2x. (14)

Equation (14) provides two functions that satisfy the given differential equation for x > 0 and

also satisfy the initial condition y(0) = 1.

EXAMPLE

Consider the initial value problem

′

= y

1/3

, y(0) = 0 (15)

for t ≥ 0. Apply Theorem 2.4.2 to this initial value problem and then solve the problem.

The function f (t, y) = y

1/3

is continuous everywhere, but ∂f /∂y does not exist when y = 0,

and hence it is not continuous there. Thus Theorem 2.4.2 does not apply to this problem, and

no conclusion can be drawn from it. However, by the remark following Theorem 2.4.2, the

continuity of f does ensure the existence of solutions, though not their uniqueness.

To understand the situation more clearly, we must actually solve the problem, which is easy

to do since the differential equation is separable. Thus we have

−1/3

dy = dt,

2/3

= t + c

and

y =

(t + c)

3/2

The initial condition is satisﬁed if c = 0, so

y = φ

(t) =

3/2

, t ≥ 0 (16)

satisﬁes both of Eqs. (15). On the other hand, the function

y = φ

(t) =−

3/2

, t ≥ 0 (17)

is also a solution of the initial value problem. Moreover, the function

y = ψ(t) = 0, t ≥ 0 (18)

is yet another solution. Indeed, for an arbitrary positive t

, the functions

y = χ(t) =

0, if 0 ≤ t < t

(t − t

)

3/2

,ift ≥ t

(19)

are continuous, are differentiable (in particular at t = t

), and are solutions of the initial value

problem (15). Hence this problem has an inﬁnite family of solutions; see Figure 2.4.1, where a

few of these solutions are shown.

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2.4 Differences Between Linear and Nonlinear Equations 73

As already noted,thenonuniqueness of the solutions of the problem (15)does not contradict

the existence and uniqueness theorem, since the theorem is not applicable if the initial point

lies on the t-axis. If (t

, y

) is any point not on the t-axis,however, then the theorem guarantees

that there is a unique solution of the differential equation y

′

= y

1/3

passing through (t

, y

(t)

–1

(t)

FIGURE 2.4.1 Several solutions of the initial value problem y

′

= y

1/3

, y(0) = 0.

Inter val of Definition. According toTheorem 2.4.1,the solution of a linear equation (1)

′

+ p(t)y = g(t),

subject to the initial condition y(t

) = y

, exists throughout any interval about t = t

in which the functions p and g are continuous. Thus vertical asymptotes or other

discontinuities in the solution can occur only at points of discontinuity of p or g.

For instance, the solutions in Example 1 (with one exception) are asymptotic to the

y-axis, corresponding to the discontinuity at t = 0 in the coefﬁcient p(t) = 2/t, but

none of thesolutions has anyother point whereit fails toexist and tobe differentiable.

The one exceptional solution shows that solutions may sometimes remain continuous

even at points of discontinuity of the coefﬁcients.

On the other hand, for a nonlinear initial value problem satisfying the hypotheses

ofTheorem 2.4.2,the interval in which a solution exists may be difﬁcult to determine.

The solution y = φ(t) is certain to exist as long as the point [t, φ(t)] remains within a

region in which the hypotheses ofTheorem 2.4.2 are satisﬁed.This is what determines

the value of h in that theorem. However, since φ(t) is usually not known, it may be

impossible to locate the point [t, φ(t)] with respect to this region. In any case, the

interval in which a solution exists may have no simple relationship to the function f

in the differential equation y

′

= f (t, y). This is illustrated by the following example.

EXAMPLE

Solve the initial value problem

′

= y

, y(0) = 1, (20)

and determine the interval in which the solution exists.

Theorem 2.4.2 guarantees that this problem has a unique solution since f (t, y) = y

and

∂f /∂y = 2y are continuous everywhere. To ﬁnd the solution, we separate the variables and

integrate with the result that

−2

dy = dt (21)

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74 Chapter 2. First Order Differential Equations

and

−y

−1

= t + c.

Then, solving for y, we have

y =−

t + c

. (22)

To satisfy the initial condition, we must choose c =−1, so

y =

1 − t

(23)

is the solution of the given initial value problem. Clearly, the solution becomes unbounded as

t → 1; therefore, the solution exists only in the interval −∞ < t < 1. There is no indication

from the differential equation itself, however, that the point t = 1 is in any way remarkable.

Moreover, if the initial condition is replaced by

y(0) = y

, (24)

then the constant c in Eq. (22) must be chosen to be c =−1/y

, and it follows that

y =

1 − y

(25)

is the solution of the initial value problem with the initial condition (24). Observe that the

solution (25) becomes unbounded as t → 1/y

, so the interval of existence of the solution

is −∞ < t < 1/y

if y

> 0, and is 1/y

< t < ∞ if y

< 0. This example illustrates another

feature of initial value problems for nonlinear equations: the singularities of the solution may

depend in an essential way on the initial conditions as well as on the differential equation.

General Solution. Another way in which linear and nonlinear equations differ concerns

the concept of a general solution. For a ﬁrst order linear equation it is possible to

obtain a solution containing one arbitrary constant, from which all possible solutions

follow by specifying values for this constant. For nonlinear equations this may not be

the case;even though a solution containing an arbitrary constant may be found,there

may be other solutions that cannot be obtained by giving values to this constant. For

instance,for the differential equation y

′

= y

in Example 4,the expression in Eq. (22)

contains an arbitrary constant but does not include all solutions of the differential

equation.Toshow this,observe that the functiony = 0 for allt is certainly a solutionof

the differential equation,but it cannot be obtained from Eq. (22) by assigning a value

to c. In this example we might anticipate that something of this sort might happen,

because to rewrite the original differential equation in the form (21),we must require

that y is not zero. However, the existence of “additional” solutions is not uncommon

for nonlinear equations; a less obvious example is given in Problem 22. Thus we will

use the term “general solution” only when discussing linear equations.

Implicit Solutions. Recall again that for an initial value problem for a ﬁrst order linear

equation, Eq. (8) provides an explicit formula for the solution y = φ(t). As long as

the necessary antiderivatives can be found, the value of the solution at any point can

be determined merely by substituting the appropriate value of t into the equation.

The situation for nonlinear equations is much less satisfactory. Usually, the best that

we can hope for is to ﬁnd an equation

F(t, y) = 0 (26)

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2.4 Differences Between Linear and Nonlinear Equations 75

involving t and y that is satisﬁed by the solution y = φ(t). Even this can be done

only for differential equations of certain particular types, of which separable equa-

tions are the most important. The equation (26) is called an integral, or ﬁrst integral,

of the differential equation, and (as we have already noted) its graph is an inte-

gral curve, or perhaps a family of integral curves. Equation (26), assuming it can

be found, deﬁnes the solution implicitly; that is, for each value of t we must solve

Eq. (26) to ﬁnd the corresponding value of y. If Eq. (26) is simple enough, it may

be possible to solve it for y by analytical means and thereby obtain an explicit

formula for the solution. However, more frequently this will not be possible, and

you will have to resort to a numerical calculation to determine (approximately) the

value of y for a given value of t. Once several pairs of values of t and y have been

calculated, it is often helpful to plot them and then to sketch the integral curve

that passes through them. You should arrange for a computer to do this for you, if

possible.

Examples 2, 3, and 4 are nonlinear problems in which it is easy to solve for an

explicit formula for the solution y = φ(t). On the other hand, Examples 1 and 3

in Section 2.2 are cases in which it is better to leave the solution in implicit form

and to use numerical means to evaluate it for particular values of the indepen-

dent variable. The latter situation is more typical; unless the implicit relation is

quadratic in y or has some other particularly simple form, it is unlikely that it can

be solved exactly by analytical methods. Indeed, more often than not, it is impos-

sible even to ﬁnd an implicit expression for the solution of a ﬁrst order nonlinear

equation.

Graphical or Numerical Construction of Integral Cur ves. Because ofthe difﬁculty in obtain-

ing exact analytical solutions of nonlinear differential equations, methods that yield

approximate solutions or other qualitative information about solutions are of corre-

spondingly greater importance. We have already described, in Section 1.1, how the

direction ﬁeld of a differential equation can be constructed. The direction ﬁeld can

often show the qualitative form of solutions and can also be helpful in identifying

regions of the ty-plane where solutions exhibit interesting features that merit more

detailedanalyticalor numerical investigation. Graphical methods forﬁrst order equa-

tions are discussed further in Section 2.5. An introduction to numerical methods for

ﬁrst order equations is given in Section 2.7, and a systematic discussion of numerical

methods appears in Chapter 8. However, it is not necessary to study the numerical

algorithms themselves in order to use effectively one of the many software pack-

ages that generate and plot numerical approximations to solutions of initial value

problems.

Summar y. The linear equation y

′

+ p(t)y = g(t) has several nice properties that can

be summarized in the following statements:

1. Assuming that the coefﬁcients are continuous, there is a general solution, containing an

arbitrary constant, that includes all solutions of the differential equation. A particular

solution that satisﬁes a given initial condition can be picked out by choosing the proper

value for the arbitrary constant.

2. There is an expression for the solution, namely, Eq. (7) or Eq. (8). Moreover, although it

involves two integrations, the expression is an explicit one for the solution y = φ(t) rather

than an equation that deﬁnes φ implicitly.

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76 Chapter 2. First Order Differential Equations

3. The possible points of discontinuity, or singularities, of the solution can be identiﬁed

(without solving the problem) merely by ﬁnding the points of discontinuity of the coefﬁ-

cients. Thus, if the coefﬁcients are continuous for all t, then the solution also exists and is

differentiable for all t.

None of these statements is true, in general, of nonlinear equations. Although a

nonlinear equation may well have a solution involving an arbitrary constant, there

may also be other solutions. There is no general formula for solutions of nonlinear

equations. If you are able to integrate a nonlinear equation,you are likely to obtain an

equation deﬁning solutions implicitly rather than explicitly. Finally, the singularities

of solutions of nonlinear equations can usually be found only by solving the equation

and examining the solution. It is likely that the singularities will depend on the initial

condition as well as on the differential equation.

PROBLEMS In each of Problems 1 through 6, determine (without solving the problem) an interval in which

the solution of the given initial value problem is certain to exist.

1. (t − 3)y

′

+ (ln t)y = 2t, y(1) = 2

2. t(t − 4)y

′

+ y = 0, y(2) = 1

3. y

′

+ (tan t)y = sin t, y(π) = 04.(4 − t

′

+ 2ty = 3t

, y(−3) = 1

5. (4 − t

′

+ 2ty = 3t

, y(1) =−36.(ln t)y

′

+ y = cot t, y(2) = 3

In each of Problems 7 through 12, state where in the ty-plane the hypotheses of Theorem 2.4.2

are satisﬁed.

7. y

′

t − y

2t + 5y

8. y

′

= (1 − t

− y

)

1/2

9. y

′

ln |ty|

1 − t

+ y

10. y

′

= (t

+ y

)

3/2

11.

1 + t

3y − y

12.

(cot t)y

1 + y

In each of Problems 13 through 16, solve the given initial value problem and determine how

the interval in which the solution exists depends on the initial value y

13. y

′

=−4t/y, y(0) = y

14. y

′

= 2ty

, y(0) = y

15. y

′

+ y

= 0, y(0) = y

16. y

′

= t

/y(1 + t

), y(0) = y

In each of Problems 17 through 20, draw a direction ﬁeld and plot (or sketch) several solutions

of the given differential equation. Describe how solutions appear to behave as t increases and

how their behavior depends on the initial value y

when t = 0.

17.

′

= ty(3 − y) 18. y

′

= y(3 − ty)

19.

′

=−y(3 − ty) 20. y

′

= t − 1 − y

21. Consider the initial value problem y

′

= y

1/3

, y(0) = 0 from Example 3 in the text.

(a) Is there a solution that passes through the point (1, 1)? If so, ﬁnd it.

(b) Is there a solution that passes through the point (2, 1)? If so, ﬁnd it.

of values that these solutions have at t = 2.

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2.4 Differences Between Linear and Nonlinear Equations 77

22. (a) Verify that both y

(t) = 1 − t and y

(t) =−t

/4 are solutions of the initial value

problem

′

−t +

+ 4y

, y(2) =−1.

Where are these solutions valid?

(b) Explain why the existence of two solutions of the given problem does not contradict

the uniqueness part of Theorem 2.4.2.

, where c is an arbitrary constant, satisﬁes the differential

equation in part (a) for t ≥−2c.Ifc =−1, the initial condition is also satisﬁed, and

the solution y = y

(t) is obtained. Show that there is no choice of c that gives the second

solution y = y

(t).

23. (a) Show that φ(t) = e

is a solution of y

′

− 2y = 0 and that y = cφ(t) is also a solution

of this equation for any value of the constant c.

(b) Show that φ(t ) = 1/t is a solution of y

′

+ y

= 0 for t > 0 but that y = cφ(t) is not

a solution of this equation unless c = 0orc = 1. Note that the equation of part (b) is

nonlinear, while that of part (a) is linear.

24. Show that if y = φ(t) is a solution of y

′

+ p(t)y = 0, then y = cφ(t) is also a solution for

any value of the constant c.

25. Let y = y

(t) be a solution of

′

+ p(t)y = 0, (i)

and let y = y

(t) be a solution of

′

+ p(t)y = g(t). (ii)

Show that y = y

(t) + y

(t) is also a solution of Eq. (ii).

26. (a) Show that the solution (7) of the general linear equation (1) can be written in the

form

y = cy

(t) + y

(t), (i)

where c is an arbitrary constant.

(b) Show that y

is a solution of the differential equation

′

+ p(t)y = 0, (ii)

corresponding to g(t) = 0.

is a solution of the full linear equation (1). We see later (for example,

in Section 3.5) that solutions of higher order linear equations have a pattern similar to

Eq. (i).

Bernoulli Equations. Sometimes it is possible to solve a nonlinear equation by making a

change of the dependent variable that converts it into a linear equation. The most important

such equation has the form

′

+ p(t)y = q(t)y

and is called a Bernoulli equation after Jakob Bernoulli. Problems 27 through 31 deal with

equations of this type.

27. (a) Solve Bernoulli’s equation when n = 0; when n = 1.

(b) Show that if n ̸= 0, 1, then the substitution v = y

1−n

reduces Bernoulli’s equation to a

linear equation. This method of solution was found by Leibniz in 1696.

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78 Chapter 2. First Order Differential Equations

In each of Problems 28 through 31, the given equation is a Bernoulli equation. In each case

solve it by using the substitution mentioned in Problem 27(b).

28. t

′

+ 2ty − y

= 0, t > 0

29. y

′

= ry − ky

, r > 0 and k > 0. This equation is important in population dynamics and is

discussed in detail in Section 2.5.

30. y

′

= ϵy − σy

, ϵ>0 and σ>0. This equation occurs in the study of the stability of ﬂuid

ﬂow.

31. dy/dt = (2 cos t + T)y − y

, where 2 and T are constants. This equation also occurs in the

study of the stability of ﬂuid ﬂow.

Discontinuous Coefﬁcients. Linear differential equations sometimes occur in which one or

both of the functions p and g have jump discontinuities. If t

is such a point of discontinuity,

then it is necessary to solve the equation separately for t < t

and t > t

. Afterward, the two

solutions are matched so that y is continuous at t

; this is accomplished by a proper choice of

the arbitrary constants. The following two problems illustrate this situation. Note in each case

that it is impossible also to make y

′

continuous at t

32. Solve the initial value problem

′

+ 2y = g(t), y(0) = 0,

where

g(t) =

1, 0 ≤ t ≤ 1,

0, t > 1.

33. Solve the initial value problem

′

+ p(t)y = 0, y(0) = 1,

where

p(t) =

2, 0 ≤ t ≤ 1,

1, t > 1.

2.5 Autonomous Equations and Population Dynamics

An important class of ﬁrst order equations consists of those in which the independent

variable does not appear explicitly. Such equations are called autonomous and have

the form

dy/dt = f (y). (1)

We will discuss these equations in the context of the growth or decline of the popula-

tion of a given species, an important issue in ﬁelds ranging from medicine to ecology

to global economics. A number of other applications are mentioned in some of the

problems. Recall that in Sections 1.1 and 1.2 we considered the special case of Eq. (1)

in which f (y) = ay + b.

Equation (1) is separable,so the discussion in Section 2.2 is applicable to it,but the

mainpurpose of this sectionis to showhow geometrical methodscan be usedto obtain

important qualitative information directly from the differential equation without

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2.5 Autonomous Equations and Population Dynamics 79

solving the equation. Of fundamental importance in this effort are the concepts

of stability and instability of solutions of differential equations. These ideas were

introduced informally in Chapter 1, but without using this terminology. They are

discussed further here and will be examined in greater depth and in a more general

setting in Chapter 9.

Exponential Growth. Let y = φ(t) be the population of the given species at time t.The

simplest hypothesis concerning the variation of population is that the rate of change

of y is proportional

to the current value of y; that is,

dy/dt = ry, (2)

where theconstant ofproportionality r is calledthe rate of growth or decline,depend-

ing on whether it is positive or negative. Here,we assume that r > 0,so the population

is growing.

Solving Eq. (2) subject to the initial condition

y(0) = y

, (3)

we obtain

y = y

. (4)

Thus the mathematical model consisting of the initial value problem (2), (3) with

r > 0 predicts that the population will grow exponentially for all time, as shown in

Figure 2.5.1 forseveral values of y

. Under ideal conditions,Eq. (4) has beenobserved

to be reasonably accurate for many populations, at least for limited periods of time.

However,it is clear that such idealconditions cannot continue indeﬁnitely;eventually,

limitations on space, food supply, or other resources will reduce the growth rate and

bring an end to uninhibited exponential growth.

1/r 2/r 3/r 4/r

FIGURE 2.5.1 Exponential growth: y versus t for dy/dt = ry (r > 0).

It was apparently the British economist Thomas Malthus (1766–1834) who ﬁrst observed that many

biological populations increase at a rate proportional to the population. His ﬁrst paper on populations

appeared in 1798.

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80 Chapter 2. First Order Differential Equations

Logistic Growth. To take account of the fact that the growth rate actually depends on

the population, we replace the constant r in Eq. (2) by a function h(y) and thereby

obtain the modiﬁed equation

dy/dt = h(y)y. (5)

We now want to choose h(y) so that h(y)

∼

r > 0 when y is small, h(y) decreases

as y grows larger, and h(y)<0 when y is sufﬁciently large.The simplest function that

has these properties is h(y) = r − ay, where a is also a positive constant. Using this

function in Eq. (5), we obtain

dy/dt = (r − ay)y. (6)

Equation (6) is known as the Verhulst

equation or the logistic equation. It is often

convenient to write the logistic equation in the equivalent form

= r

)

1 −

y, (7)

where K = r/a. The constant r is called the intrinsic growth rate—that is, the growth

rate in the absence of any limiting factors. The interpretation of K will become clear

shortly.

We will investigate the solutions of Eq. (7) in some detail later in this section.

Before doing that, however, we will show how you can easily draw a qualitatively

correct sketch of the solutions. The same methods also apply to the more general

Eq. (1).

We ﬁrst seek solutions of Eq. (7) of the simplest possible type—that is, constant

functions. For such a solution dy/dt = 0 for all t, so any constant solution of Eq. (7)

must satisfy the algebraic equation

r(1 − y/K)y = 0.

Thus the constant solutions are y = φ

(t) = 0 and y = φ

(t) = K. These solutions

are called equilibrium solutions of Eq. (7) because they correspond to no change

or variation in the value of y as t increases. In the same way, any equilibrium solu-

tions of the more general Eq. (1) can be found by locating the roots of f (y) = 0.

The zeros of f (y) are also called critical points.

To visualize other solutions of Eq. (7) and to sketch their graphs quickly, we start

by drawing the graph of f (y) versus y. In the case of Eq. (7), f (y) = r(1 − y/K)y,

so the graph is the parabola shown in Figure 2.5.2. The intercepts are (0, 0) and

(K,0), corresponding to the critical points of Eq. (7), and the vertex of the parabola

is (K/2, rK/4). Observe that dy/dt > 0 for 0 < y < K; therefore, y is an increasing

function of t when y is in this interval; this is indicated by the rightward-pointing

Pierre F. Verhulst (1804–1849) was a Belgian mathematician who introduced Eq. (6) as a model for

human population growth in 1838. He referred to it as logistic growth so Eq. (6) is often called the logistic

equation. He was unable to test the accuracy of his model because of inadequate census data, and it did

not receive much attention until many years later. Reasonable agreement with experimental data was

demonstrated by R. Pearl (1930) for Drosophila melanogaster (fruit ﬂy) populations and by G. F. Gause

(1935) for Paramecium and Tribolium (ﬂour beetle) populations.

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2.5 Autonomous Equations and Population Dynamics 81

rK/4

f ( y)

K/2

(K/2, rK/4)

FIGURE 2.5.2 f (y) versus y for dy/dt = r(1 − y/K)y.

arrows near the y-axis in Figure 2.5.2. Similarly, if y > K, then dy/dt < 0; hence y is

decreasing, as indicated by the leftward-pointing arrow in Figure 2.5.2.

In this context the y-axis is often called the phase line, and it is reproduced in its

more customary vertical orientation in Figure 2.5.3a. The dots at y = 0 and y = K

are the critical points, or equilibrium solutions. The arrows again indicate that y is

increasing whenever 0 < y < K and that y is decreasing whenever y > K.

Further, from Figure 2.5.2, note that if y is near zero or K, then the slope f (y) is

near zero, so the solution curves are relatively ﬂat. They become steeper as the value

of y leaves the neighborhood of zero or K.

To sketch the graphs of solutions of Eq. (7) in the ty-plane, we start with the equi-

librium solutions y = 0 and y = K; then we draw other curves that are increasing

when 0 < y < K, decreasing when y > K, and ﬂatten out as y approaches either of

the values 0 or K.Thus the graphs of solutions of Eq. (7) must have the general shape

shown in Figure 2.5.3b, regardless of the values of r and K.

(a)(b)

K/2

(t) = K

(t) = 0

FIGURE 2.5.3 Logistic growth: dy/dt = r(1 − y/K)y.

(a) The phase line. (b) Plots of y versus t.

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82 Chapter 2. First Order Differential Equations

Figure 2.5.3b may seem to show that other solutions intersect the equilibrium

solution y = K, but is this really possible? No, the uniqueness part of Theorem 2.4.2,

the fundamental existence and uniqueness theorem, states that only one solution

can pass through a given point in the ty-plane. Thus, although other solutions may

be asymptotic to the equilibrium solution as t →∞, they cannot intersect it at any

ﬁnite time. Consequently, a solution that starts in the interval 0 < y < K remains in

this interval for all time, and similarly for a solution that starts in K < y < ∞.

To carry the investigation one step further, we can determine the concavity of the

solution curves and the location of inﬂection points by ﬁnding d

y/dt

. From

the differential equation (1), we obtain (using the chain rule)

f (y) = f

′

(y)

= f

′

(y)f (y). (8)

The graph of y versus t is concave up when y

′′

> 0—that is, when f and f

′

have the

same sign. Similarly,it is concave down when y

′′

< 0,which occurs when f and f

′

have

opposite signs. The signs of f and f

′

can be easily identiﬁed from the graph of f (y)

versus y. Inﬂection points may occur when f

′

(y) = 0.

In the case of Eq. (7), solutions are concave up for 0 < y < K/2 where f is pos-

itive and increasing (see Figure 2.5.2), so that both f and f

′

are positive. Solutions

are also concave up for y > K where f is negative and decreasing (both f and f

′

are

negative). For K/2 < y < K, solutions are concave down since here f is positive and

decreasing,so f is positive but f

′

is negative.There is an inﬂection point whenever the

graph of y versus t crosses the line y = K/2. The graphs in Figure 2.5.3b exhibit these

properties.

Finally, observe that K is the upper bound that is approached, but not exceeded,

by growing populations starting below this value. Thus it is natural to refer to

K as the saturation level, or the environmental carrying capacity, for the given

species.

A comparison of Figures 2.5.1 and 2.5.3b reveals that solutions of the nonlinear

equation (7) are strikingly different from those of the linear equation (1), at least

for large values of t. Regardless of the value of K—that is, no matter how small

the nonlinear term in Eq. (7)—solutions of that equation approach a ﬁnite value as

t →∞, whereas solutions of Eq. (1) grow (exponentially) without bound as t →∞.

Thus even a tiny nonlinear term in the differential equation (7) has a decisive effect

on the solution for large t.

In many situations it is sufﬁcient to have the qualitative information about a solu-

tion y = φ(t) of Eq. (7) that is shown in Figure 2.5.3b. This information was obtained

entirely from the graph of f (y) versus y and without solving the differential equation

(7). However, if we wish to have a more detailed description of logistic growth—for

example, if we wish to know the value of the population at some particular time—

then we must solve Eq. (7) subject to the initial condition (3). Provided that y ̸= 0

and y ̸= K, we can write Eq. (7) in the form

(1 − y/K)y

= rdt.

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2.5 Autonomous Equations and Population Dynamics 83

Using a partial fraction expansion on the left side, we have

1/K

1 − y/K

(

dy = rdt.

Then, by integrating both sides, we obtain

ln |y|−ln

1 −

= rt + c, (9)

where c is an arbitrary constant of integration to be determined from the initial

condition y(0) = y

. We have already noted that if 0 < y

< K, then y remains in

this interval for all time. Thus in this case we can remove the absolute value bars

in Eq. (9), and by taking the exponential of both sides, we ﬁnd that

1 − (y/K)

= Ce

, (10)

where C = e

. In order to satisfy the initial condition y(0) = y

, we must choose

C = y

/[1 − (y

/K)]. Using this value for C in Eq. (10) and solving for y, we obtain

y =

+ (K − y

−rt

. (11)

We have derived the solution (11) under the assumption that 0 < y

< K. If

> K, then the details of dealing with Eq. (9) are only slightly different, and we

leave it to you to show that Eq. (11) is also valid in this case. Finally, note that

Eq. (11) also contains the equilibrium solutions y = φ

(t) = 0 and y = φ

(t) = K

corresponding to the initial conditions y

= 0 and y

= K, respectively.

All the qualitative conclusions that we reached earlier by geometrical reason-

ing can be conﬁrmed by examining the solution (11). In particular, if y

= 0, then

Eq. (11) requires that y(t) = 0 for all t.Ify

> 0, and if we let t →∞in Eq. (11),

then we obtain

lim

t→∞

y(t) = y

K/y

= K.

Thus,for each y

> 0,the solution approaches the equilibrium solution y = φ

(t) = K

asymptotically as t →∞. Therefore, we say that the constant solution φ

(t) = K is

an asymptotically stable solution of Eq. (7) or that the point y = K is an asymp-

totically stable equilibrium or critical point. After a long time, the population is

close to the saturation level K regardless of the initial population size, as long as

it is positive. Other solutions approach the equilibrium solution more rapidly as r

increases.

On the other hand, the situation for the equilibrium solution y = φ

(t) = 0 is quite

different. Even solutions that start very near zero grow as t increases and, as we have

seen,approach K as t →∞.We say that φ

(t) = 0isanunstable equilibrium solution

or that y = 0 is an unstable equilibrium or critical point. This means that the only

way to guarantee that the solution remains near zero is to make sure its initial value

is exactly equal to zero.

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84 Chapter 2. First Order Differential Equations

EXAMPLE

The logistic model has been applied to the natural growth of the halibut population in certain

areas of the Paciﬁc Ocean.

Let y, measured in kilograms, be the total mass, or biomass,

of the halibut population at time t. The parameters in the logistic equation are estimated to

have the values r = 0.71/year and K = 80.5 × 10

kg. If the initial biomass is y

= 0.25K, ﬁnd

the biomass 2 years later. Also ﬁnd the time τ for which y(τ) = 0.75K.

It is convenient to scale the solution (11) to the carrying capacity K; thus we write Eq. (11)

in the form

/K) +[1 − (y

/K)]e

−rt

. (12)

Using the data given in the problem, we ﬁnd that

y(2)

0.25

0.25 + 0.75e

−1.42

∼

0.5797.

Consequently, y(2)

∼

46.7 × 10

kg.

To ﬁnd τ, we can ﬁrst solve Eq. (12) for t. We obtain

−rt

/K)[1 − (y/K)]

(y/K)[1 − (y

/K)]

;

hence

t =−

/K)[1 − (y/K)]

(y/K)[1 − (y

/K)]

. (13)

Using the given values of r and y

/K and setting y/K = 0.75, we ﬁnd that

τ =−

0.71

(0.25)(0.25)

(0.75)(0.75)

0.71

ln 9

∼

3.095 years.

The graphs of y/K versus t for the given parameter values and for several initial conditions

are shown in Figure 2.5.4. The black curve corresponds to the initial condition y

= 0.25K.

1.50

1.75

1.25

1.00

0.75

0.50

0.25

231

456t

y/ K

≅

3.095

FIGURE 2.5.4 y/K versus t for population model of halibut in the Paciﬁc

Ocean. The black curve satisﬁes the initial condition y(0)/K = 0.25.

A good source of information on the population dynamics and economics involved in making efﬁcient

use of a renewable resource, with particular emphasis on ﬁsheries, is the book by Clark listed in the

references at the end of this chapter. The parameter values used here are given on page 53 of this book

and were obtained from a study by H. S. Mohring.

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2.5 Autonomous Equations and Population Dynamics 85

ACriticalThreshold. We now turn to a consideration of the equation

=−r

)

1 −

y, (14)

where r and T are given positive constants. Observe that (except for replacing the

parameter K by T) this equation differs from the logistic equation (7) only in

the presence of the minus sign on the right side. However,as we will see,the solutions

of Eq. (14) behave very differently from those of Eq. (7).

For Eq. (14) the graph of f (y) versus y is the parabola shown in Figure 2.5.5.

The intercepts on the y-axis are the critical points y = 0 and y = T, corresponding

to the equilibrium solutions φ

(t) = 0 and φ

(t) = T.If0< y < T, then dy/dt < 0,

and y decreases as t increases. On the other hand, if y > T, then dy/dt > 0, and y

grows as t increases. Thus φ

(t) = 0 is an asymptotically stable equilibrium solution

and φ

(t) = T is an unstable one. Further, f

′

(y) is negative for 0 < y < T/2 and

positive for T/2 < y < T, so the graph of y versus t is concave up and concave down,

respectively, in these intervals. Also, f

′

(y) is positive for y > T, so the graph of y

versus t is also concave up there.

f ( y)

–rT /4

(T/2, –rT /4)

T /2 T

FIGURE 2.5.5 f (y) versus y for dy/dt =−r(1 − y/T)y.

Figure 2.5.6(a) shows the phase line (the y-axis) for Eq. (14). The dots at y = 0 and

y = T are the critical points, or equilibrium solutions, and the arrows indicate where

solutions are either increasing or decreasing.

Solution curves of Eq. (14) can now be sketched quickly, as follows. First draw the

equilibrium solutions y = 0 and y = T.Then sketch curves in the strip 0 < y < T that

are decreasing as t increases and change concavity as they cross the line y = T/2.

Next draw some curves above y = T that increase more and more steeply as t and y

increase. Make sure that all curves become ﬂatter as y approaches either zero or T.

The result is Figure 2.5.6(b), which is a qualitatively accurate sketch of solutions of

Eq. (14) for any r and T. From this ﬁgure it appears that as time increases, y either

approaches zero or grows without bound, depending on whether the initial value y

is less than or greater than T. Thus T is a threshold level, below which growth does

not occur.

We can conﬁrm the conclusions that we have reached through geometrical rea-

soning by solving the differential equation (14). This can be done by separating the

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86 Chapter 2. First Order Differential Equations

(a)(b)

T /2

(t) = T

(t) = 0

FIGURE 2.5.6 Growth with a threshold: dy/dt =−r(1 − y/T)y.

(a) The phase line. (b) Plots of y versus t.

variables and integrating,just as we did for Eq. (7). However,if we note that Eq. (14)

can be obtained from Eq. (7) by replacing K by T and r by −r, then we can make

the same substitutions in the solution (11) and thereby obtain

y =

+ (T − y

, (15)

which is the solution of Eq. (14) subject to the initial condition y(0) = y

If 0 < y

< T, then it follows from Eq. (15) that y → 0ast →∞. This agrees with

our qualitative geometric analysis. If y

> T, then the denominator on the right side

of Eq. (15) is zero for a certain ﬁnite value of t. We denote this value by t

∗

and

calculate it from

− (y

− T)e

∗

= 0,

which gives

∗

− T

. (16)

Thus,if the initial population y

is above the thresholdT,the threshold model predicts

that the graph of y versus t has a vertical asymptote at t = t

∗

; in other words, the

population becomes unbounded in a ﬁnite time,whose value depends on y

,T,and r.

The existence and location of this asymptote were not apparent from the geometric

analysis, so in this case the explicit solution yields additional important qualitative,

as well as quantitative, information.

The populations of some species exhibit the threshold phenomenon. If too few

are present, then the species cannot propagate itself successfully and the popula-

tion becomes extinct. However, if the population is larger than the threshold level,

then further growth occurs. Of course, the population cannot become unbounded, so

eventually Eq. (14) must be modiﬁed to take this into account.

Critical thresholds also occur in other circumstances. For example,in ﬂuid mechan-

ics,equations of the form (7) or (14) often govern the evolution of a small disturbance

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2.5 Autonomous Equations and Population Dynamics 87

y in a laminar (or smooth) ﬂuid ﬂow. For instance, if Eq. (14) holds and y < T, then

the disturbance is damped out and the laminar ﬂow persists. However, if y > T,

then the disturbance grows larger and the laminar ﬂow breaks up into a turbulent

one. In this case T is referred to as the critical amplitude. Experimenters speak of

keeping the disturbance level in a wind tunnel low enough so that they can study

laminar ﬂow over an airfoil, for example.

Logistic Growth with a Threshold. As we mentioned in the last subsection, the threshold

model (14) may need to be modiﬁed so that unbounded growth does not occur when

y is above the threshold T. The simplest way to do this is to introduce another factor

that will have the effect of making dy/dt negative when y is large. Thus we consider

=−r

)

1 −

y, (17)

where r > 0 and 0 < T < K.

The graph of f (y) versus y is shown in Figure 2.5.7. In this problem there are three

critical points, y = 0, y = T, and y = K, corresponding to the equilibrium solutions

(t) = 0, φ

(t) = T, and φ

(t) = K, respectively. From Figure 2.5.7 we observe that

dy/dt > 0 for T < y < K, and consequently y is increasing there. Further, dy/dt < 0

for y < T and for y > K,soy is decreasing in these intervals. Consequently, the

equilibrium solutions φ

(t) and φ

(t) are asymptotically stable, and the solution φ

(t)

is unstable.

f ( y)

FIGURE 2.5.7 f (y) versus y for dy/dt =−r(1 − y/T)(1 − y/K)y.

The phase line for Eq. (17) is shown in Figure 2.5.8a, and the graphs of some

solutions are sketched in Figure 2.5.8b. You should make sure that you understand

the relation between these two ﬁgures, as well as the relation between Figures 2.5.7

and 2.5.8a. From Figure 2.5.8b we see that if y starts below the threshold T, then

y declines to ultimate extinction. On the other hand, if y starts above T, then y

eventually approaches the carrying capacity K. The inﬂection points on the graphs

of y versus t in Figure 2.5.8b correspond to the maximum and minimum points, y

and y

, respectively, on the graph of f (y) versus y in Figure 2.5.7. These values can

be obtained by differentiating the right side of Eq. (17) with respect to y, setting the

result equal to zero, and solving for y. We obtain

1, 2

= (K + T ±

− KT + T

)/3, (18)

where the plus sign yields y

and the minus sign y

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88 Chapter 2. First Order Differential Equations

(t) = K

(t) = T

(a)(b)

(t) = 0

FIGURE 2.5.8 Logistic growth with a threshold: dy/dt =−r(1 − y/T)(1 − y/K)y.

(a) The phase line. (b) Plots of y versus t.

A model of this general sort apparently describes the population of the passen-

ger pigeon,

which was present in the United States in vast numbers until late in the

nineteenth century. It was heavily hunted for food and for sport,and consequently its

numbers were drastically reduced by the 1880s. Unfortunately, the passenger pigeon

could apparently breed successfully only when present in a large concentration, cor-

responding to a relatively high threshold T. Although a reasonably large number of

individual birds remained alive in the late 1880s, there were not enough in any one

place to permit successful breeding, and the population rapidly declined to extinc-

tion. The last survivor died in 1914. The precipitous decline in the passenger pigeon

population from huge numbers to extinction in a few decades was one of the early

factors contributing to a concern for conservation in this country.

PROBLEMS Problems 1 through 6 involve equations of the form dy/dt = f (y). In each problem sketch the

graph of f (y) versus y, determine the critical (equilibrium) points, and classify each one as

asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions

in the ty-plane.

1. dy/dt = ay + by

, a > 0, b > 0, y

≥ 0

2. dy/dt = ay + by

, a > 0, b > 0, −∞ < y

< ∞

3. dy/dt = y(y − 1)(y − 2), y

≥ 0

4. dy/dt = e

− 1, −∞ < y

< ∞

5. dy/dt = e

−y

− 1, −∞ < y

< ∞

6. dy/dt =−2(arctan y)/(1 + y

), −∞ < y

< ∞

7. Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the

property that solutions lying on one side of the equilibrium solution tend to approach it,

See, for example, Oliver L. Austin, Jr., Birds of the World (New York: Golden Press, 1983), pp. 143–145.

August 7, 2012 21:03 c02 Sheet number 59 Page number 89 cyan black

2.5 Autonomous Equations and Population Dynamics 89

whereas solutions lying on the other side depart from it (see Figure 2.5.9). In this case the

equilibrium solution is said to be semistable.

(a) Consider the equation

dy/dt = k(1 − y)

, (i)

where k is a positive constant. Show that y = 1 is the only critical point, with the

corresponding equilibrium solution φ(t) = 1.

(b) Sketch f (y) versus y. Show that y is increasing as a function of t for y < 1 and also

for y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus

solutions below the equilibrium solution approach it,and thoseabove it grow farther away.

Therefore, φ(t) = 1 is semistable.

and conﬁrm the conclusions

reached in part (b).

(a)

(t) = k

(b)

(t) = k

FIGURE 2.5.9 In both cases the equilibrium solution φ(t) = k is semistable.

(a) dy/dt ≤ 0; (b) dy/dt ≥ 0.

Problems 8 through 13 involve equations of the form dy/dt = f (y). In each problem sketch

the graph of f (y) versus y, determine the critical (equilibrium) points, and classify each one

asymptotically stable, unstable,or semistable (see Problem 7). Draw the phase line, and sketch

several graphs of solutions in the ty-plane.

8. dy/dt =−k(y − 1)

, k > 0, −∞ < y

< ∞

9. dy/dt = y

− 1), −∞ < y

< ∞

10. dy/dt = y(1 − y

), −∞ < y

< ∞

11. dy/dt = ay − b

√

y, a > 0, b > 0, y

≥ 0

12. dy/dt = y

(4 − y

), −∞ < y

< ∞

13. dy/dt = y

(1 − y)

, −∞ < y

< ∞

14. Consider the equation dy/dt = f (y) and suppose that y

is a critical point—that is,

f (y

) = 0. Show that the constant equilibrium solution φ(t) = y

is asymptotically stable

if f

′

)<0 and unstable if f

′

)>0.

15. Suppose that a certain population obeys the logistic equation dy/dt = ry[1 − (y/K)].

(a) If y

= K/3,ﬁnd the time τ at which the initial population has doubled. Find the value

of τ corresponding to r = 0.025 per year.

(b) If y

/K = α, ﬁnd the time T at which y(T)/K = β, where 0 <α, β<1. Observe that

T →∞as α → 0orasβ → 1. Find the value of T for r = 0.025 per year, α = 0.1, and

β = 0.9.

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90 Chapter 2. First Order Differential Equations

16. Another equation that has been used to model population growth is the Gompertz

equation

dy/dt = ry ln(K/y),

where r and K are positive constants.

(a) Sketch the graph of f (y) versus y, ﬁnd the critical points, and determine whether each

is asymptotically stable or unstable.

(b) For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is

concave down.

less than dy/dt as given by the logistic equation.

17. (a) Solve the Gompertz equation

dy/dt = ry ln(K/y),

subject to the initial condition y(0) = y

Hint: You may wish to let u = ln(y/K).

(b) For the data given in Example 1 in the text (r = 0.71 per year, K = 80.5 × 10

kg,

/K = 0.25), use the Gompertz model to ﬁnd the predicted value of y(2).

y(τ) = 0.75K.

18. A pond formsas water collectsin a conical depressionof radius a and depth h. Supposethat

water ﬂows in at a constant rate k and is lost through evaporation at a rate proportional

to the surface area.

(a) Show that the volume V(t) of water in the pond at time t satisﬁes the differential

equation

dV/dt = k − απ(3a/πh)

2/3

where α is the coefﬁcient of evaporation.

(b) Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically

stable?

19. Consider a cylindrical water tank of constant cross section A. Water is pumped into the

tank at a constant rate k and leaks out through a small hole of area a in the bottom

of the tank. From Torricelli’s principle in hydrodynamics (see Problem 6 in Section 2.3)

it follows that the rate at which water ﬂows through the hole is αa

2gh, where h is the

current depth of water in the tank,g is the acceleration due to gravity,and α is a contraction

coefﬁcient that satisﬁes 0.5 ≤ α ≤ 1.0.

(a) Show that the depth of water in the tank at any time satisﬁes the equation

dh/dt = (k − αa

2gh )/A.

(b) Determine the equilibrium depth h

of water,and show that it is asymptotically stable.

Observe that h

does not depend on A.

Benjamin Gompertz (1779–1865) was an Englishactuary.Hedeveloped his model for population growth,

published in 1825, in the course of constructing mortality tables for his insurance company.

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2.5 Autonomous Equations and Population Dynamics 91

Harvesting a Renewable Resource. Suppose that the population y of a certain species of ﬁsh

(for example, tuna or halibut) in a given area of the ocean is described by the logistic equation

dy/dt = r(1 − y/K)y.

Although it is desirable to utilize this source of food, it is intuitively clear that if too many ﬁsh

are caught, then the ﬁsh population may be reduced below a useful level and possibly even

driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating

a rational strategy for managing the ﬁshery.

20. At a given level of effort, it is reasonable to assume that the rate at which ﬁsh are caught

depends on the population y: the more ﬁsh there are,the easier it is to catch them.Thus we

assume that the rate at which ﬁsh are caught is given by Ey, where E is a positive constant,

with units of 1/time, that measures the total effort made to harvest the given species of

ﬁsh. To include this effect, the logistic equation is replaced by

dy/dt = r(1 − y/K)y − Ey. (i)

This equation is known as the Schaefer model after the biologist M. B. Schaefer, who

applied it to ﬁsh populations.

(a) Show that if E < r, then there are two equilibrium points, y

= 0 and

= K(1 − E/r)>0.

(b) Show that y = y

is unstable and y = y

is asymptotically stable.

It is the product of the effort E and the asymptotically stable population y

. Find Y as a

function of the effort E; the graph of this function is known as the yield–effort curve.

(d) Determine E soas to maximize Y and thereby ﬁnd the maximum sustainable yield Y

21. In this problem we assume that ﬁsh are caught at a constant rate h independent of the size

of the ﬁsh population. Then y satisﬁes

dy/dt = r(1 − y/K)y − h. (i)

The assumption of a constant catch rate h may be reasonable when y is large but becomes

less so when y is small.

(a) If h < rK/4, show that Eq. (i) has two equilibrium points y

and y

with y

< y

;

determine these points.

(b) Show that y

is unstable and y

is asymptotically stable.

> y

, then y → y

as t →∞, but that if y

< y

, then y decreases as t increases. Note that y = 0 is not an

equilibrium point, so if y

< y

, then extinction will be reached in a ﬁnite time.

(d) If h > rK/4, show that y decreases to zero as t increases, regardless of the value of y

(e) If h = rK/4, show that there is a single equilibrium point y = K/2 and that this point

is semistable (see Problem 7). Thus the maximum sustainable yield is h

= rK/4, corre-

sponding to the equilibrium value y = K/2. Observe that h

has the same value as Y

in Problem 20(d). The ﬁshery is considered to be overexploited if y is reduced to a level

below K/2.

An excellent treatment of this kind of problem, which goes far beyond what is outlined here, may be

found in the book by Clark mentioned previously,especially in the ﬁrst two chapters. Numerous additional

references are given there.

August 7, 2012 21:03 c02 Sheet number 62 Page number 92 cyan black

92 Chapter 2. First Order Differential Equations

Epidemics. The use of mathematical methods to study the spread of contagious diseases

goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years

many mathematical models have been proposed and studied for many different diseases.

Problems 22 through 24 deal with a few of the simpler models and the conclusions that can be

drawn from them. Similar models have also been used to describe the spread of rumors and

of consumer products.

22. Suppose that a given population can be divided into two parts: those who have a given

disease and can infect others,and those who do not have it but are susceptible. Let x be the

proportion of susceptible individuals and y the proportion of infectious individuals; then

x + y = 1. Assume that the disease spreads by contact between sick and well members of

the population and that the rate of spread dy/dt is proportional to the number of such

contacts. Further, assume that members of both groups move about freely among each

other,so the number of contacts is proportional to the product of x and y. Since x = 1 − y,

we obtain the initial value problem

dy/dt = αy(1 − y), y(0) = y

, (i)

where α is a positive proportionality factor, and y

is the initial proportion of infectious

individuals.

(a) Find the equilibrium points for the differential equation (i) and determine whether

each is asymptotically stable, semistable, or unstable.

(b) Solve the initial value problem (i) and verify that the conclusions you reached in

part (a) are correct. Show that y(t) → 1ast →∞, which means that ultimately the disease

spreads through the entire population.

23. Some diseases (such as typhoid fever) are spread largely by carriers, individuals who can

transmit the disease but who exhibit no overt symptoms. Let x and y denote the propor-

tions of susceptibles and carriers, respectively, in the population. Suppose that carriers are

identiﬁed and removed from the population at a rate β, so

dy/dt =−βy. (i)

Suppose also that the disease spreads at a rate proportional to the product of x and y; thus

dx/dt =−αxy. (ii)

(a) Determine y at any time t by solving Eq. (i) subject to the initial condition y(0) = y

(b) Use the result of part (a) to ﬁnd x at any time t by solving Eq. (ii) subject to the initial

condition x(0) = x

value of x as t →∞.

24. Daniel Bernoulli’s work in 1760 had the goal of appraising the effectiveness of a contro-

versial inoculation program against smallpox, which at that time was a major threat to

public health. His model applies equally well to any other disease that, once contracted

and survived, confers a lifetime immunity.

Consider the cohort of individuals born in a given year (t = 0), and let n(t) be the

number of these individuals surviving t years later. Let x(t) be the number of members of

this cohort who have not had smallpox by year t and who are therefore still susceptible.

Let β be the rate at which susceptibles contract smallpox, and let ν be the rate at which

A standard source is the book by Bailey listed in the references. The models in Problems 22, 23, and

24 are discussed by Bailey in Chapters 5, 10, and 20, respectively.

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2.5 Autonomous Equations and Population Dynamics 93

people who contract smallpox die from the disease. Finally, let µ(t) be the death rate from

all causes other than smallpox. Then dx/dt, the rate at which the number of susceptibles

declines, is given by

dx/dt =−[β + µ(t)]x. (i)

The ﬁrst term on the right side of Eq. (i) is the rate at which susceptibles contract smallpox,

and the second term is the rate at which they die from all other causes. Also

dn/dt =−νβx − µ(t)n, (ii)

where dn/dt is the death rate of the entire cohort, and the two terms on the right side are

the death rates due to smallpox and to all other causes, respectively.

(a) Let z = x/n, and show that z satisﬁes the initial value problem

dz/dt =−βz(1 − νz), z(0) = 1. (iii)

Observe that the initial value problem (iii) does not depend on µ(t).

(b) Find z(t) by solving Eq. (iii).

. Using these values, determine the proportion of

20-year-olds who have not had smallpox.

Note: On the basis of the model just described and the best mortality data available at the

time, Bernoulli calculated that if deaths due to smallpox could be eliminated (ν = 0), then

approximately 3 years could be added to the average life expectancy (in 1760) of 26 years,

7 months. He therefore supported the inoculation program.

Bifurcation Points. For an equation of the form

dy/dt = f (a, y), (i)

where a is a real parameter, the critical points (equilibrium solutions) usually depend on the

value of a.Asa steadily increases or decreases,it often happens that at a certain value of a,called

a bifurcation point, critical points come together, or separate, and equilibrium solutions may

be either lost or gained. Bifurcation points are of great interest in many applications, because

near them the nature of the solution of the underlying differential equation is undergoing an

abrupt change. For example, in ﬂuid mechanics a smooth (laminar) ﬂow may break up and

become turbulent. Or an axially loaded column may suddenly buckle and exhibit a large lateral

displacement. Or,as the amount of one of the chemicals in a certain mixture is increased, spiral

wave patterns of varying color may suddenly emerge in an originally quiescent ﬂuid. Problems

25 through 27 describe three types of bifurcations that can occur in simple equations of the

form (i).

25. Consider the equation

dy/dt = a − y

. (ii)

(a) Find all of the critical points for Eq. (ii). Observe that there are no critical points if

a < 0, one critical point if a = 0, and two critical points if a > 0.

(b) Draw the phase line in each case and determine whether each critical point is

asymptotically stable, semistable, or unstable.

(d) If we plot the location of the critical points as a function of a in the ay-plane,we obtain

Figure 2.5.10. This is called the bifurcation diagram for Eq. (ii). The bifurcation at a = 0

is called a saddle–node bifurcation. This name is more natural in the context of second

order systems, which are discussed in Chapter 9.

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94 Chapter 2. First Order Differential Equations

–2

–1

–2 –1 1 2 3 4

Unstable

Asymptotically stable

FIGURE 2.5.10 Bifurcation diagram for y

′

= a − y

26. Consider the equation

dy/dt = ay − y

= y(a − y

). (iii)

(a) Again consider the cases a < 0, a = 0, and a > 0. In each case ﬁnd the critical points,

draw the phase line, and determine whether each critical point is asymptotically stable,

semistable, or unstable.

(b) In each case sketch several solutions of Eq. (iii) in the ty-plane.

points versus a. For Eq. (iii) the bifurcation point at a = 0 is called a pitchfork bifurcation.

Your diagram may suggest why this name is appropriate.

27. Consider the equation

dy/dt = ay − y

= y(a − y). (iv)

(a) Again consider the cases a < 0, a = 0, and a > 0. In each case ﬁnd the critical points,

draw the phase line, and determine whether each critical point is asymptotically stable,

semistable, or unstable.

(b) In each case sketch several solutions of Eq. (iv) in the ty-plane.

same number of critical points for a < 0 and a > 0 but that their stability has changed.

For a < 0 the equilibrium solution y = 0 is asymptotically stable and y = a is unstable,

while for a > 0 the situation is reversed. Thus there has been an exchange of sta bility as a

passes through the bifurcation point a = 0. This type of bifurcation is called a transcritical

bifurcation.

28. Chemical Reactions. A second order chemical reaction involves the interaction (colli-

sion) of one molecule of a substance P with one molecule of a substance Q to produce

one molecule of a new substance X ; this is denoted by P + Q → X. Suppose that p and

q, where p ̸= q, are the initial concentrations of P and Q, respectively, and let x(t) be the

concentration of X at time t. Then p − x(t)

and q − x(t) are

the

concentrations of P and

Q at time t, and the rate at which the reaction occurs is given by the equation

dx/dt = α(p − x )(q − x), (i)

where α is a positive constant.

(a) If x(0) = 0, determine the limiting value of x(t) as t →∞ without solving the

differential equation. Then solve the initial value problem and ﬁnd x(t) for any t.

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2.6 Exact Equations and Integrating Factors 95

(b) If the substances P and Q are the same, then p = q and Eq. (i) is replaced by

dx/dt = α(p − x )

. (ii)

If x(0) = 0, determine the limiting value of x(t) as t →∞without solving the differential

equation. Then solve the initial value problem and determine x(t) for any t.

2.6 Exact Equations and Integrating Factors

For ﬁrst order equations there are a number of integration methods that are applica-

ble to various classes of problems. The most important of these are linear equations

and separable equations, which we have discussed previously. Here, we consider a

class of equations known as exact equations for which there is also a well-deﬁned

method of solution. Keep in mind, however, that those ﬁrst order equations that

can be solved by elementary integration methods are rather special; most ﬁrst order

equations cannot be solved in this way.

EXAMPLE

Solve the differential equation

2x + y

+ 2xyy

′

= 0. (1)

The equation is neither linear nor separable, so the methods suitable for those types of

equations are not applicable here. However, observe that the function ψ(x, y) = x

+ xy

has

the property that

2x + y

∂ψ

∂x

,2xy =

∂ψ

∂y

. (2)

Therefore, the differential equation can be written as

∂ψ

∂x

∂ψ

∂y

= 0. (3)

Assuming that y is a function of x, we can use the chain rule to write the left side of Eq. (3) as

dψ(x, y)/dx. Then Eq. (3) has the form

dψ

(x, y) =

+ xy

) = 0. (4)

By integrating Eq. (4) we obtain

ψ(x, y) = x

+ xy

= c, (5)

where c is an arbitrary constant. The level curves of ψ(x, y) are the integral curves of Eq. (1).

Solutions of Eq. (1) are deﬁned implicitly by Eq. (5).

In solving Eq. (1) the key step was the recognition that there is a function ψ that

satisﬁes Eqs. (2). More generally, let the differential equation

M(x, y) + N(x, y)y

′

= 0 (6)

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96 Chapter 2. First Order Differential Equations

be given. Suppose that we can identify a function ψ(x, y) such that

∂ψ

∂x

(x, y) = M(x, y),

∂ψ

∂y

(x, y) = N(x, y), (7)

and such that ψ(x, y) = c deﬁnes y = φ(x) implicitly as a differentiable function

of x. Then

M(x, y) + N(x, y)y

′

∂ψ

∂x

∂ψ

∂y

ψ[x, φ(x)]

and the differential equation (6) becomes

ψ[x, φ(x)]=0. (8)

In this case Eq. (6) is said to be an exact differential equation. Solutions of Eq. (6),

or the equivalent Eq. (8), are given implicitly by

ψ(x, y) = c, (9)

where c is an arbitrary constant.

In Example 1 it was relatively easy to see that the differential equation was exact

and, in fact, easy to ﬁnd its solution, at least implicitly, by recognizing the required

function ψ. For more complicated equations it may not be possible to do this so

easily. How can we tell whether a given equation is exact, and if it is, how can we ﬁnd

the function ψ(x, y)? The following theorem answers the ﬁrst question, and its proof

provides a way of answering the second.

Theorem 2.6.1

Let the functions M, N, M

, and N

, where subscripts denote partial derivatives,

be continuous in the rectangular

region R: α<x <β, γ<y <δ. Then Eq. (6)

M(x, y) + N(x, y)y

′

= 0

is an exact differential equation in R if and only if

(x, y) = N

(x, y) (10)

at each point of R. That is, there exists a function ψ satisfying Eqs. (7),

(x, y) = M(x, y), ψ

(x, y) = N(x, y),

if and only if M and N satisfy Eq. (10).

It is not essential that the region be rectangular, only that it be simply connected. In two dimensions

this means that the region has no holes in its interior. Thus, for example, rectangular or circular regions

are simply connected, but an annular region is not. More details can be found in most books on advanced

calculus.

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2.6 Exact Equations and Integrating Factors 97

The proof of this theorem has two parts. First, we show that if there is a function

ψ such that Eqs. (7) are true, then it follows that Eq. (10) is satisﬁed. Computing

and N

from Eqs. (7), we obtain

(x, y) = ψ

(x, y), N

(x, y) = ψ

(x, y). (11)

Since M

and N

are continuous, it follows that ψ

and ψ

are also continuous. This

guarantees their equality, and Eq. (10) is valid.

We now show that if M and N satisfy Eq. (10), then Eq. (6) is exact. The proof

involves the construction of a function ψ satisfying Eqs. (7)

(x, y) = M(x, y), ψ

(x, y) = N(x, y).

We begin by integrating the ﬁrst of Eqs. (7) with respect to x, holding y constant. We

obtain

ψ(x, y) = Q(x, y) + h(y), (12)

where Q(x, y) is any differentiable function such that ∂Q(x, y)/∂x = M(x, y).For

example, we might choose

Q(x, y) =

M(s, y) ds, (13)

where x

is some speciﬁed constant in α<x

<β. The function h in Eq. (12) is an

arbitrary differentiable function of y, playing the role of the arbitrary constant. Now

we must show that it is always possible to choose h(y) so that the second of Eqs. (7)

is satisﬁed—that is, ψ

= N. By differentiating Eq. (12) with respect to y and setting

the result equal to N(x, y), we obtain

(x, y) =

∂Q

∂y

(x, y) + h

′

(y) = N(x, y).

Then, solving for h

′

(y), we have

′

(y) = N(x, y) −

∂Q

∂y

(x, y). (14)

In order for us to determine h(y) from Eq. (14), the right side of Eq. (14), despite

its appearance, must be a function of y only. One way to show that this is true is to

show that its derivative with respect to x is zero. Thus we differentiate the right side

of Eq. (14) with respect to x, obtaining

∂N

∂x

(x, y) −

∂

∂x

∂Q

∂y

(x, y). (15)

By interchanging the order of differentiation in the second term of Eq. (15), we have

∂N

∂x

(x, y) −

∂

∂y

∂Q

∂x

(x, y),

or, since ∂Q/∂x = M,

∂N

∂x

(x, y) −

∂M

∂y

(x, y),

which iszero onaccount ofEq. (10). Hence,despite its apparent form,the right side of

Eq. (14) does not, in fact, depend on x. Then we ﬁnd h(y) by integrating Eq. (14), and

August 7, 2012 21:03 c02 Sheet number 68 Page number 98 cyan black

98 Chapter 2. First Order Differential Equations

upon substituting this function in Eq. (12), we obtain the required function ψ(x, y).

This completes the proof of Theorem 2.6.1.

It is possible to obtain an explicit expression for ψ(x, y) in terms of integrals (see

Problem 17),but in solving speciﬁc exactequations,it is usually simpler and easier just

to repeat the procedure used in the preceding proof. That is, integrate ψ

= M with

respect to x, including an arbitrary function of h(y) instead of an arbitrary constant,

and then differentiate the result with respect to y and set it equal to N. Finally, use

this last equation to solve for h(y). The next example illustrates this procedure.

EXAMPLE

Solve the differential equation

(y cos x + 2xe

) + (sin x + x

− 1)y

′

= 0. (16)

By calculating M

and N

, we ﬁnd that

(x, y) = cos x + 2xe

= N

(x, y),

so the given equation is exact. Thus there is a ψ(x, y) such that

(x, y) = y cos x + 2xe

(x, y) = sin x + x

− 1.

Integrating the ﬁrst of these equations, we obtain

ψ(x, y) = y sin x + x

+ h(y). (17)

Setting ψ

= N gives

(x, y) = sin x + x

+ h

′

(y) = sin x + x

− 1.

Thus h

′

(y) =−1 and h(y) =−y.The constant of integration can be omitted since any solution

of the preceding differential equation is satisfactory; we do not require the most general one.

Substituting for h(y) in Eq. (17) gives

ψ(x, y) = y sin x + x

− y.

Hence solutions of Eq. (16) are given implicitly by

y sin x + x

− y = c. (18)

EXAMPLE

Solve the differential equation

(3xy + y

) + (x

+ xy)y

′

= 0. (19)

We have

(x, y) = 3x + 2y, N

(x, y) = 2 x + y;

since M

̸= N

, the given equation is not exact.To see that it cannot be solved by the procedure

described above, let us seek a function ψ such that

(x, y) = 3xy + y

, ψ

(x, y) = x

+ xy. (20)

Integrating the ﬁrst of Eqs. (20) gives

ψ(x, y) =

y + xy

+ h(y), (21)

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2.6 Exact Equations and Integrating Factors 99

where h is an arbitrary function of y only. To try to satisfy the second of Eqs. (20), we compute

from Eq. (21) and set it equal to N, obtaining

+ 2xy + h

′

(y) = x

+ xy

′

(y) =−

− xy. (22)

Since the right side of Eq. (22) depends on x as well as y, it is impossible to solve Eq. (22) for

h(y). Thus there is no ψ(x, y) satisfying both of Eqs. (20).

Integrating Factors. It is sometimes possible to convert a differential equation that is

not exact into an exact equation by multiplying the equation by a suitable integrating

factor. Recall that this is the procedure that we used in solving linear equations in

Section 2.1. To investigate the possibility of implementing this idea more generally,

let us multiply the equation

M(x, y) + N(x, y)y

′

= 0 (23)

by a function µ and then try to choose µ so that the resulting equation

µ(x, y)M(x, y) + µ(x, y)N(x, y)y

′

= 0 (24)

is exact. By Theorem 2.6.1, Eq. (24) is exact if and only if

(µM)

= (µN)

. (25)

Since M and N are given functions, Eq. (25) states that the integrating factor µ must

satisfy the ﬁrst order partial differential equation

Mµ

− Nµ

+ (M

− N

)µ = 0. (26)

If a function µ satisfying Eq. (26) can be found, then Eq. (24) will be exact. The

solution of Eq. (24) can then be obtained by the method described in the ﬁrst part of

this section.The solution found in this way also satisﬁes Eq. (23),since the integrating

factor µ can be canceled out of Eq. (24).

A partial differential equation of the form (26) may have more than one solution;

if this is the case, any such solution may be used as an integrating factor of Eq. (23).

This possible nonuniqueness of the integrating factor is illustrated in Example 4.

Unfortunately, Eq. (26), which determines the integrating factor µ, is ordinarily

at least as hard to solve as the original equation (23). Therefore, although in princi-

ple integratingfactors are powerful tools for solving differential equations,in practice

they can be found only in special cases.The most important situations in which simple

integrating factors canbe found occur when µ is a function ofonly one ofthe variables

x or y, instead of both.

Let us determine conditions on M and N so that Eq. (23) has an integrating factor

µ that depends on x only. If we assume that µ is a function of x only, then the partial

derivative µ

reduces to the ordinary derivative dµ/dx and µ

= 0. Making these

substitutions in Eq. (26), we ﬁnd that

d µ

− N

µ. (27)

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100 Chapter 2. First Order Differential Equations

If (M

− N

)/N is a function of x only, then there is an integrating factor µ that also

depends only on x;further,µ(x) can be found by solving Eq. (27),which is both linear

and separable.

A similar procedure can be used to determine a condition under which Eq. (23)

has an integrating factor depending only on y; see Problem 23.

EXAMPLE

Find an integrating factor for the equation

(3xy + y

) + (x

+ xy)y

′

= 0 (19)

and then solve the equation.

In Example 3 we showed that this equation is not exact. Let us determine whether it has an

integrating factor that depends on x only. On computing the quantity (M

− N

)/N, we ﬁnd

that

(x, y) − N

(x, y)

N(x, y)

3x + 2y − (2x + y)

+ xy

. (28)

Thus there is an integrating factor µ that is a function of x only, and it satisﬁes the differential

equation

dµ

. (29)

Hence

µ(x) = x. (30)

Multiplying Eq. (19) by this integrating factor, we obtain

(3x

y + xy

) + (x

+ x

y)y

′

= 0. (31)

Equation (31) is exact, since

∂

∂y

(3x

y + xy

) = 3x

+ 2xy =

∂

∂x

+ x

y).

Thus there is a function ψ such that

(x, y) = 3x

y + xy

, ψ

(x, y) = x

+ x

y. (32)

Integrating the ﬁrst of Eqs. (32), we obtain

ψ(x, y) = x

y +

+ h(y).

Substituting this expression for ψ(x, y) in the second of Eqs. (32), we ﬁnd that

+ x

y + h

′

(y) = x

+ x

so h

′

(y) = 0 and h(y) is a constant. Thus the solutions of Eq. (31), and hence of Eq. (19), are

given implicitly by

y +

= c. (33)

Solutions may also be found in explicit form since Eq. (33) is quadratic in y.

You may also verify that a second integrating factor for Eq. (19) is

µ(x, y) =

xy(2x + y)

and that the same solution is obtained, though with much greater difﬁculty, if this integrating

factor is used (see Problem 32).

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2.6 Exact Equations and Integrating Factors 101

PROBLEMS Determine whether each of the equations in Problems 1 through 12 is exact. If it is exact, ﬁnd

the solution.

1. (2x + 3) + (2y − 2)y

′

= 02.(2x + 4y) + (2x − 2y)y

′

= 0

3. (3x

− 2xy + 2) + (6y

− x

+ 3)y

′

= 04.(2xy

+ 2y) + (2x

y + 2x)y

′

= 0

=−

ax + by

bx + cy

=−

ax − by

bx − cy

7. (e

sin y − 2y sin x) + (e

cos y + 2 cos x)y

′

= 0

8. (e

sin y + 3y) − (3x − e

sin y)y

′

= 0

9. (ye

cos 2x − 2e

sin 2x + 2x) + (xe

cos 2x − 3)y

′

= 0

10. (y/x + 6x) + (ln x − 2)y

′

= 0, x > 0

11. (x ln y + xy) + (y ln x + xy)y

′

= 0; x > 0, y > 0

12.

+ y

)

3/2

+ y

)

3/2

= 0

In each of Problems 13 and 14, solve the given initial value problem and determine at least

approximately where the solution is valid.

13. (2x − y) + (2y − x)y

′

= 0, y(1) = 3

14. (9x

+ y − 1) − (4y − x)y

′

= 0, y(1) = 0

In each of Problems 15 and 16, ﬁnd the value of b for which the given equation is exact, and

then solve it using that value of b.

15. (xy

+ bx

y) + (x + y)x

′

= 0 16. (ye

2xy

+ x) + bxe

2xy

′

= 0

17. Assume that Eq. (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is

therefore exact. Show that a possible function ψ(x, y) is

ψ(x, y) =

M(s, y

) ds +

N(x, t) dt,

where (x

, y

) is a point in R.

18. Show that any separable equation

M(x) + N(y)y

′

= 0

is also exact.

In each of Problems 19 through 22, show that the given equation is not exact but becomes

exact when multiplied by the given integrating factor. Then solve the equation.

19. x

+ x(1 + y

′

= 0, µ(x, y) = 1/xy

20.

sin y

− 2e

−x

sin x

(

cos y + 2e

−x

cos x

(

′

= 0, µ(x, y) = ye

21. y + (2x − ye

′

= 0, µ(x, y) = y

22. (x + 2) siny + (x cos y)y

′

= 0, µ(x, y) = xe

23. Show that if (N

− M

)/M = Q, where Q is a function of y only, then the differential

equation

M + Ny

′

= 0

has an integrating factor of the form

µ(y) = exp

Q(y) dy.

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102 Chapter 2. First Order Differential Equations

24. Show that if (N

− M

)/(xM − yN) = R, where R depends on the quantity xy only, then

the differential equation

M + Ny

′

= 0

has an integrating factor of the form µ(xy). Find a general formula for this integrating

factor.

In each of Problems 25 through 31, ﬁnd an integrating factor and solve the given equation.

25. (3x

y + 2xy + y

) + (x

+ y

′

= 0 26. y

′

= e

+ y − 1

27. 1 + (x/y − sin y)y

′

= 0 28. y + (2 xy − e

−2y

′

= 0

29. e

+ (e

cot y + 2y csc y)y

′

= 0

30. [4(x

) + (3/y)]+[3(x/y

) + 4y]y

′

= 0

31.

3x +

(

+ 3

(

= 0

Hint: See Problem 24.

32. Solve the differential equation

(3xy + y

) + (x

+ xy)y

′

= 0

using the integrating factor µ(x, y) =[xy(2x + y)]

−1

. Verify that the solution is the same

as that obtained in Example 4 with a different integrating factor.

2.7 Numerical Approximations: Euler’s Method

Recall two important facts about the ﬁrst order initial value problem

= f (t, y), y(t

) = y

. (1)

First, if f and ∂f /∂y are continuous, then the initial value problem (1) has a unique

solution y = φ(t) in some interval surrounding the initial point t = t

. Second, it is

usually not possible to ﬁnd the solution φ by symbolic manipulations of the differ-

ential equation. Up to now we have considered the main exceptions to the latter

statement: differential equations that are linear, separable, or exact, or that can be

transformed into one of these types. Nevertheless, it remains true that solutions of

the vast majority of ﬁrst order initial value problems cannot be found by analytical

means, such as those considered in the ﬁrst part of this chapter.

Therefore, it is important to be able to approach the problem in other ways. As we

have already seen, one of these ways is to draw a direction ﬁeld for the differential

equation (which does not involve solving the equation) and then to visualize the

behavior of solutions from the direction ﬁeld. This has the advantage of being a

relatively simple process, even for complicated differential equations. However, it

does not lend itself to quantitative computations or comparisons, and this is often a

critical shortcoming.

For example, Figure 2.7.1 shows a direction ﬁeld for the differential equation

= 3 − 2t − 0.5y. (2)

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2.7 Numerical Approximations: Euler’s Method 103

From the direction ﬁeld you can visualize the behavior of solutions on the rectangle

shown in the ﬁgure. On this rectangle a solution starting at a point on the y-axis

initially increases with t, but it soon reaches a maximum value and then begins to

decrease as t increases further.

–1

321

FIGURE 2.7.1 A direction ﬁeld for Eq. (2): dy/dt = 3 − 2t − 0.5y.

You may also observe that in Figure 2.7.1 many tangent line segments at successive

values of t almost touch each other. It takes only a bit of imagination to consider

starting at a point on the y-axis and linking line segments for successive values of t in

the grid, thereby producing a piecewise linear graph. Such a graph would apparently

be an approximation to a solution of the differential equation. To convert this idea

into a useful method for generating approximate solutions, we must answer several

questions, including the following:

1. Can we carry out the linking of tangent lines in a systematic and straightforward

manner?

2. If so, does the resulting piecewise linear function provide an approximation to an actual

solution of the differential equation?

3. If so, can we assess the accuracy of the approximation? That is, can we estimate how far

the approximation deviates from the solution itself?

It turns out that the answer to each of these questions is afﬁrmative. The resulting

method was originated by Euler about 1768 and is referred to as the tangent line

method or the Euler method. We will deal with the ﬁrst two questions in this section

but will defer a systematic discussion of the third question until Chapter 8.

To see how the Euler method works, let us consider how we might use tangent

lines to approximate the solution y = φ(t) of Eqs. (1) near t = t

. We know that the

solution passes through the initial point (t

, y

),and from the differential equation,we

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104 Chapter 2. First Order Differential Equations

also know that its slope at this point is f (t

, y

). Thus we can write down an equation

for the line tangent to the solution curve at (t

, y

), namely,

y = y

+ f (t

, y

)(t − t

). (3)

The tangent line is a good approximation to the actual solution curve on an interval

short enough so that the slope of the solution does not change appreciably from its

value at the initial point; see Figure 2.7.2. Thus, if t

is close enough to t

, we can

approximate φ(t

) by the value y

determined by substituting t = t

into the tangent

line approximation at t = t

; thus

= y

+ f (t

, y

)(t

− t

). (4)

)

y = y

+ f(t

, y

) (t – t

)

(t)

y =

FIGURE 2.7.2 A tangent line approximation.

To proceed further,we can tryto repeat the process. Unfortunately,we do not know

the value φ(t

) of the solution at t

. The best we can do is to use the approximate

value y

instead. Thus we construct the line through (t

, y

) with the slope f (t

, y

y = y

+ f (t

, y

)(t − t

). (5)

To approximate the value of φ(t) at a nearby point t

,we use Eq. (5) instead,obtaining

= y

+ f (t

, y

)(t

− t

). (6)

Continuing in this manner, we use the value of y calculated at each step to deter-

mine the slope of the approximation for the next step. The general expression for the

tangent line starting at (t

, y

) is

y = y

+ f (t

, y

)(t − t

); (7)

hence the approximate value y

n+1

at t

n+1

in terms of t

, t

n+1

, and y

n+1

= y

+ f (t

, y

)(t

n+1

− t

), n = 0, 1, 2, .... (8)

If we introduce the notation f

= f (t

, y

), then we can rewrite Eq. (8) as

n+1

= y

+ f

· (t

n+1

− t

), n = 0, 1, 2, .... (9)

Finally,if we assume that there is a uniform step size h between the points t

, t

, ...,

then t

n+1

= t

+ h for each n, and we obtain Euler’s formula in the form

n+1

= y

+ f

h, n = 0, 1, 2, .... (10)

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2.7 Numerical Approximations: Euler’s Method 105

To use Euler’s method, you repeatedly evaluate Eq. (9) or Eq. (10), depending on

whether or not the step size is constant, using the result of each step to execute the

next step.In thisway you generate a sequence of valuesy

, y

, ...that approximate

the values of the solution φ(t) at the points t

, t

, ....If, instead of a sequence of

points, you need a function to approximate the solution φ(t), then you can use the

piecewise linear function constructed from the collection of tangent line segments.

That is, let y be given in [t

, t

] by Eq. (7) with n = 0, in [t

, t

] by Eq. (7) with n = 1,

and so on.

EXAMPLE

Consider the initial value problem

= 3 − 2t − 0.5y, y(0) = 1. (11)

UseEuler’s method with step sizeh = 0.2 to ﬁnd approximate values of the solution of Eqs. (11)

at t = 0.2,0.4,0.6,0.8,and 1. Compare them with the corresponding values of the actual solution

of the initial value problem.

Note that the differential equation in thegiven initial value problem is the same as in Eq. (2).

This equation is linear, so it can be solved as in Section 2.1, using the integrating factor e

t/2

The resulting solution of the initial value problem (11) is

y = φ(t) = 14 − 4t − 13e

−t/2

. (12)

To approximate this solution by means of Euler’s method, note that in this case

f (t, y) = 3 − 2t − 0.5y. Using the initial values t

= 0 and y

= 1, we ﬁnd that

= f (t

, y

) = f (0, 1) = 3 − 0 − 0.5 = 2.5

and then, from Eq. (3), the tangent line approximation near t = 0 is

y = 1 + 2.5(t − 0) = 1 + 2 .5t. (13)

Setting t = 0.2 in Eq. (13), we ﬁnd the approximate value y

of the solution at t = 0.2, namely,

= 1 + (2.5)(0.2) = 1.5.

At the next step we have

= f (0.2, 1.5) = 3 − 2(0.2) − (0.5)(1.5) = 3 − 0.4 − 0.75 = 1.85.

Then the tangent line approximation near t = 0.2 is

y = 1.5 + 1.85(t − 0.2) = 1.13 + 1.85t. (14)

Evaluating the expression in Eq. (14) for t = 0.4, we obtain

= 1.13 + 1.85(0.4) = 1.87.

Repeating this computational procedure three more times, we obtain the results shown in

Table 2.7.1.

The ﬁrst column contains the t-values separated by the step size h = 0.2. The third

column shows the corresponding y-values computed from Euler’s formula (10). In the

fourth column are the tangent line approximations found from Eq. (7). The second column

contains values of the solution (12) of the initial value problem (11), correct to ﬁve decimal

places. The solution (12) and the tangent line approximation are also plotted in Figure 2.7.3.

FromTable 2.7.1 and Figure2.7.3 we see that the approximations givenby Euler’smethod for

this problem are greater than the corresponding values of the actual solution. This is because

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106 Chapter 2. First Order Differential Equations

the graph of the solution is concave down and therefore the tangent line approximations lie

above the graph.

The accuracy of the approximations in this example is not good enough to be satisfactory in

a typical scientiﬁc or engineering application. For example, at t = 1 the error in the approxi-

mation is 2.32363 − 2.11510 = 0.20853, which is a percentage error of about 9.86% relative to

the exact solution. One way to achieve more accurate results is to use a smaller step size, with

a corresponding increase in the number of computational steps. We explore this possibility in

the next example.

TABLE 2.7.1 Results of Euler’s Method with h = 0.2 for

′

= 3 − 2t − 0.5y, y(0 ) = 1

Euler

t Exact with h = 0.2 Tangent line

0.01.00000 1.00000 y = 1 + 2.5t

0.21.43711 1.50000 y = 1.13 + 1.85t

0.41.75650 1.87000 y = 1.364 + 1.265t

0.61.96936 2.12300 y = 1.6799 + 0.7385t

0.82.08584 2.27070 y = 2.05898 + 0.26465t

1.02.11510 2.32363

Tangent line approximation

Solution

2.4

1.6

1.2

10.80.60.40.2

FIGURE 2.7.3 Plots of the solution and a tangent line approximation

with h = 0.2 for the initial value problem (11): dy/dt = 3 − 2t − 0.5y, y(0) = 1.

Of course, computations such as those in Example 1 and in the other examples in

this section are usually done on a computer. Some software packages include code

for the Euler method, while others do not. In any case, it is straightforward to write

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2.7 Numerical Approximations: Euler’s Method 107

a computer program that will carry out the calculations required to produce results

such as those in Table 2.7.1. Basically, what is required is a loop that will evaluate

Eq. (10) repetitively,along with suitable instructions for input and output.The output

can be a list of numbers, as in Table 2.7.1, or a plot, as in Figure 2.7.3. The speciﬁc

instructions can be written in any high-level programming language with which you

are familiar.

EXAMPLE

Consider again the initial value problem (11)

= 3 − 2t − 0.5y, y(0) = 1.

Use Euler’s method with various step sizes to calculate approximate values of the solution

for 0 ≤ t ≤ 5. Compare the calculated results with the corresponding values of the exact

solution (12)

y = φ(t) = 14 − 4t − 13e

−t/2

We used step sizes h = 0.1, 0.05, 0.025, and 0.01, corresponding respectively to 50, 100, 200,

and 500 steps, to go from t = 0tot = 5. The results of these calculations, along with the values

of the exact solution, are presented in Table 2.7.2. All computed entries are rounded to four

decimal places, although more digits were retained in the intermediate calculations.

TABLE 2.7.2 A Comparison of Exact Solution with Euler’s Method for Several

Step Sizes h for y

′

= 3 − 2t − 0.5y, y(0 ) = 1

t Exact h = 0.1 h = 0.05 h = 0.025 h = 0 .01

0.01.0000 1.0000 1.0000 1.0000 1.0000

1.02.1151 2.2164 2.1651 2.1399 2.1250

2.01.2176 1.3397 1.2780 1.2476 1.2295

3.0 −0.9007 −0.7903 −0.8459 −0.8734 −0.8898

4.0 −3.7594 −3.6707 −3.7152 −3.7373 −3.7506

5.0 −7.0671 −7.0003 −7.0337 −7.0504 −7.0604

What conclusions can we draw from the data inTable 2.7.2?The most important observation

is that, for a ﬁxed value of t, the computed approximate values become more accurate as the

step size h decreases. You can see this by reading across a particular row in the table from left

to right.This is what we would expect, of course,but it is encouraging that the data conﬁrm our

expectations. For example, for t = 2 the approximate value with h = 0.1 is too large by 0.1221

(about 10%), whereas the value with h = 0.01 is too large by only 0.0119 (about 1%). In this

case, reducing the step size by a factor of 10 (and performing 10 times as many computations)

also reduces the error by a factor of about 10. By comparing the errors for other pairs of

values in the table, you can verify that this relation between step size and error holds for them

also: reducing the step size by a given factor also reduces the error by approximately the same

factor. Does this mean that for the Euler method the error is approximately proportional to

the step size? Of course, one example does not establish such a general result, but it is at least

an interesting conjecture.

A more detailed discussion of the errors in using the Euler method appears in Chapter 8.

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108 Chapter 2. First Order Differential Equations

A second observation from Table 2.7.2 is that, for a ﬁxed step size h, the approximations

become more accurate as t increases, at least for t > 2. For instance, for h = 0.1 the error for

t = 5 is only 0.0668, which is a little more than one-half of the error at t = 2. We will return to

this matter later in this section.

All in all, Euler’s method seems to work rather well for this problem. Reasonably good

results are obtained even for a moderately large step size h = 0.1, and the approximation can

be improved by decreasing h.

Let us now look at another example.

EXAMPLE

Consider the initial value problem

= 4 − t + 2y, y(0) = 1. (15)

The general solution of this differential equation was found in Example 2 of Section 2.1, and

the solution of the initial value problem (15) is

y =−

t +

. (16)

Use Euler’s method with several step sizes to ﬁnd approximate values of the solution on the

interval 0 ≤ t ≤ 5. Compare the results with the corresponding values of the solution (16).

Using the same range of step sizes as in Example 2, we obtain the results presented in

Table 2.7.3.

TABLE 2.7.3 A Comparison of Exact Solution with Euler’s Method for Several Step Sizes h

for y

′

= 4 − t + 2y, y(0) = 1

t Exact h = 0.1 h = 0.05 h = 0.025 h = 0.01

0.01.000000 1.000000 1.000000 1.000000 1.000000

1.019.06990 15.77728 17.25062 18.10997 18.67278

2.0 149.3949 104.6784 123.7130 135.5440 143.5835

3.0 1109.179 652.5349 837.0745 959.2580 1045.395

4.0 8197.884 4042.122 5633.351 6755.175 7575.577

5.0 60573.53 25026.95 37897.43 47555.35 54881.32

The data in Table 2.7.3 again conﬁrm our expectation that, for a given value of t, accuracy

improves as the step size h is reduced. For example, for t = 1 the percentage error diminishes

from 17.3% when h = 0.1 to2.1% whenh = 0.01. However,the error increases fairly rapidly as

t increases for a ﬁxed h. Even for h = 0.01, the error at t = 5 is 9.4%, and it is much greater for

larger step sizes. Of course, the accuracy that is needed depends on the purpose for which the

results are intended,but the errors inTable 2.7.3 are too large for most scientiﬁc or engineering

applications.To improve the situation,we might either try even smaller step sizesor else restrict

the computations to a rather short interval away from the initial point. Nevertheless, it is clear

that Euler’s method is much less effective in this example than in Example 2.

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2.7 Numerical Approximations: Euler’s Method 109

To understand better what is happening in these examples, let us look again at

Euler’s method for the general initial value problem (1)

= f (t, y), y(t

) = y

whose solution we denote by φ(t). Recall that a ﬁrst order differential equation has

an inﬁnite family of solutions, indexed by an arbitrary constant c, and that the initial

condition picks out one member of this inﬁnite family by determining the value of c.

Thus in the inﬁnite family of solutions,φ(t) is the one that satisﬁes the initial condition

φ(t

) = y

At the ﬁrst step Euler’s method uses the tangent line approximation to the graph

of y = φ(t) passing through the initial point (t

, y

), and this produces the approx-

imate value y

at t

. Usually y

̸= φ(t

), so at the second step Euler’s method uses

the tangent line approximation not to y = φ(t), but to a nearby solution y = φ

(t)

that passes through the point (t

, y

). So it is at each subsequent step. Euler’s method

uses a succession of tangent line approximations to a sequence of different solutions

φ(t), φ

(t), φ

(t), ...of the differential equation. At each step the tangent line is con-

structed to the solution passing through the point determined by the result of the

preceding step,as shown in Figure 2.7.4.The quality of the approximation after many

steps depends strongly on the behavior of the set of solutions that pass through the

points (t

, y

) for n = 1, 2,3, ....

(t)y =

, y

)

, y

)

, y

)

(t)y =

FIGURE 2.7.4 The Euler method.

In Example 2 the general solution of the differential equation is

y = 14 − 4t + ce

−t/2

(17)

and the solution of the initial value problem (11) corresponds to c =−13.The family

of solutions (17) is a converging family since the term involving the arbitrary constant

c approaches zero as t →∞. It does not matter very much which solutions we are

approximating by tangent lines in the implementation of Euler’s method, since all

the solutions are getting closer and closer to each other as t increases.

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110 Chapter 2. First Order Differential Equations

On the other hand,in Example 3 the general solution of the differential equation is

y =−

t + ce

, (18)

and this is a diverging family. Note that solutions corresponding to two nearby values

of c become arbitrarily far apart as t increases. In Example 3 we are trying to follow

the solution for c = 11/4, but in the use of Euler’s method we are actually at each

step following another solution that separates from the desired one faster and faster

as t increases. This explains why the errors in Example 3 are so much larger than

those in Example 2.

In using a numerical procedure such as the Euler method, you must always keep

in mind the question of whether the results are accurate enough to be useful. In

the preceding examples, the accuracy of the numerical results could be determined

directly by a comparison with the solution obtained analytically. Of course, usually

the analytical solution is not available if a numerical procedure is to be employed, so

what we usually need are bounds for, or at least estimates of, the error that do not

require a knowledge of the exact solution. You should also keep in mind that the best

that we can expect, or hope for, from a numerical approximation is that it reﬂects

the behavior of the actual solution. Thus a member of a diverging family of solutions

will always be harder to approximate than a member of a converging family.

If you wish to read more about numerical approximations to solutions of initial

value problems, you may go directly to Chapter 8 at this point. There we present

some information on the analysis of errors and also discuss several algorithms that

are computationally much more efﬁcient than the Euler method.

PROBLEMS Most of the problems in this section call for fairly extensive numerical computations. To han-

dle these problems you need suitable computing hardware and software. Keep in mind that

numerical results may vary somewhat, depending on how your program is constructed and on

how your computer executes arithmetic steps, rounds off, and so forth. Minor variations in the

last decimal place may be due to such causes and do not necessarily indicate that something is

amiss. Answers in the back of the book are recorded to six digits in most cases, although more

digits were retained in the intermediate calculations.

In each of Problems 1 through 4:

(a) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2,

0.3, and 0.4 using the Euler method with h = 0.1.

(b) Repeat part (a) with h = 0.05. Compare the results with those found in (a).

(d) Find the solution y = φ(t) of the given problem and evaluate φ(t) at t = 0 .1, 0.2, 0.3, and

0.4. Compare these values with the results of (a), (b), and (c).

′

= 3 + t − y, y(0) = 12.y

′

= 2y − 1, y(0) = 1

′

= 0.5 − t + 2y, y(0) = 14.y

′

= 3 cos t − 2y, y(0) = 0

In each of Problems 5 through 10, draw a direction ﬁeld for the given differential equation

and state whether you think that the solutions are converging or diverging.

′

= 5 − 3

√

y 6. y

′

= y(3 − ty)

′

= (4 − ty)/(1 + y

) 8. y

′

=−ty + 0.1y

9. y

′

= t

+ y

10. y

′

= (y

+ 2ty)/(3 + t

)

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2.7 Numerical Approximations: Euler’s Method 111

In each of Problems 11 through 14, use Euler’s method to ﬁnd approximate values of the

solution of the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, and 3:

(a) With h = 0.1. (b) With h = 0.05.

11.

′

= 5 − 3

√

y, y(0) = 2 12. y

′

= y(3 − ty), y(0) = 0.5

13.

′

= (4 − ty)/(1 + y

), y(0) =−2 14. y

′

=−ty + 0.1y

, y(0) = 1

15.

Consider the initial value problem

′

= 3t

/(3y

− 4), y(1) = 0.

(a) Use Euler’s method with h = 0.1 to obtain approximate values of the solution at

t = 1.2, 1.4, 1.6, and 1.8.

(b) Repeat part (a) with h = 0.05.

t = 1.2, 1.4, and 1.6 but are quite different for t = 1.8. Also note (from the differ-

ential equation) that the line tangent to the solution is parallel to the y-axis when

y =±2/

√

∼

±1.155. Explain how this might cause such a difference in the calculated

values.

16.

Consider the initial value problem

′

= t

+ y

, y(0) = 1.

Use Euler’s method with h = 0.1, 0.05, 0.025, and 0.01 to explore the solution of this

problem for 0 ≤ t ≤ 1. What is your best estimate of the value of the solution at t = 0.8?

At t = 1? Are your results consistent with the direction ﬁeld in Problem 9?

17.

Consider the initial value problem

′

= (y

+ 2ty)/(3 + t

), y(1) = 2.

Use Euler’s method with h = 0.1, 0.05, 0.025, and 0.01 to explore the solution of this

problem for 1 ≤ t ≤ 3. What is your best estimate of the value of the solution at t = 2.5?

At t = 3? Are your results consistent with the direction ﬁeld in Problem 10?

18.

Consider the initial value problem

′

=−ty + 0.1y

, y(0) = α,

where α is a given number.

(a) Draw a direction ﬁeld for the differential equation (or reexamine the one from Prob-

lem 8). Observe that there is a critical value of α in the interval 2 ≤ α ≤ 3 that separates

converging solutions from diverging ones. Call this critical value α

(b) Use Euler’smethod with h = 0.01 to estimate α

. Do this by restricting α

to an interval

[a, b], where b − a = 0.01.

19.

Consider the initial value problem

′

= y

− t

, y(0) = α,

where α is a given number.

(a) Draw a direction ﬁeld for the differential equation. Observe that there is a critical

value of α in the interval 0 ≤ α ≤ 1 that separates converging solutions from diverging

ones. Call this critical value α

(b) Use Euler’smethod with h = 0.01 to estimate α

. Do this by restricting α

to an interval

[a, b], where b − a = 0.01.

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112 Chapter 2. First Order Differential Equations

20. Convergence of Euler’s Method. It can be shown that under suitable conditions on f ,

the numerical approximation generated by the Euler method for the initial value problem

′

= f (t, y), y(t

) = y

converges to the exact solution as the step size h decreases. This is

illustrated by the following example. Consider the initial value problem

′

= 1 − t + y, y(t

) = y

(a) Show that the exact solution is y = φ(t) = (y

− t

t−t

+ t.

(b) Using the Euler formula, show that

= (1 + h)y

k−1

+ h − ht

k−1

, k = 1, 2, ....

= (1 + h)(y

− t

) + t

, show by induction that

= (1 + h)

− t

) + t

(i)

for each positive integer n.

(d) Consider a ﬁxed point t > t

and for a given n choose h = (t − t

)/n. Then t

= t for

every n. Note also that h → 0asn →∞. By substituting for h in Eq. (i) and letting n →∞,

show that y

→ φ(t) as n →∞.

Hint: lim

n→∞

(1 + a/n)

= e

In each of Problems 21 through 23, use the technique discussed in Problem 20 to show that

the approximation obtained by the Euler method converges to the exact solution at any ﬁxed

point as h → 0.

21. y

′

= y, y(0) = 1

22. y

′

= 2y − 1, y(0) = 1 Hint: y

= (1 + 2h)/2 + 1/2

23. y

′

− t + 2y, y(0) = 1 Hint: y

= (1 + 2h) + t

2.8 The Existence and Uniqueness Theorem

In this section we discuss the proof of Theorem 2.4.2, the fundamental existence and

uniqueness theorem for ﬁrst order initial value problems. This theorem states that

under certain conditions on f (t, y), the initial value problem

′

= f (t, y), y(t

) = y

(1)

has a unique solution in some interval containing the point t

In some cases (for example, if the differential equation is linear) the existence

of a solution of the initial value problem (1) can be established directly by actually

solving the problem and exhibiting a formula for the solution. However, in general,

this approach is not feasible because there is no method of solving the differential

equation that applies in all cases. Therefore, for the general case, it is necessary to

adopt an indirect approach that demonstrates the existence of a solution of Eqs. (1)

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2.8 The Existence and Uniqueness Theorem 113

but usually does not provide a practical means of ﬁnding it. The heart of this method

is the construction of a sequence of functions that converges to a limit function sat-

isfying the initial value problem, although the members of the sequence individually

do not. As a rule, it is impossible to compute explicitly more than a few members

of the sequence; therefore, the limit function can be determined only in rare cases.

Nevertheless, under the restrictions on f (t, y) stated in Theorem 2.4.2, it is possible

to show that the sequence in question converges and that the limit function has the

desired properties.The argument is fairly intricateand depends,in part,on techniques

and results that are usually encountered for the ﬁrst time in a course on advanced

calculus. Consequently, we do not go into all the details of the proof here; we do,

however, indicate its main features and point out some of the difﬁculties that must

be overcome.

First of all, we note that it is sufﬁcient to consider the problem in which the initial

point (t

, y

) is the origin; that is, we consider the problem

′

= f (t, y), y(0) = 0. (2)

If some other initial point is given, then we can always make a preliminary change

of variables, corresponding to a translation of the coordinate axes, that will take the

given point (t

, y

) into the origin. The existence and uniqueness theorem can now

be stated in the following way.

Theorem 2.8.1

If f and ∂f /∂y are continuous in a rectangle R: |t|≤a, |y|≤b, then there is some

interval |t|≤h ≤ a in which there exists a unique solution y = φ(t) of the initial

value problem (2).

For the method of proof discussed here it is necessary to transform the initial value

problem (2) into a more convenient form. If we suppose temporarily that there is a

differentiable function y = φ(t) that satisﬁes the initial value problem, then f [t, φ(t)]

is a continuous function of t only. Hence we can integrate y

′

= f (t, y) from the initial

point t = 0 to an arbitrary value of t, obtaining

φ(t) =

f [s, φ(s)]ds, (3)

where we have made use of the initial condition φ(0) = 0.We also denote the dummy

variable of integration by s.

Since Eq. (3) contains an integral of the unknown function φ, it is called an integral

equation. This integral equation is not a formula for the solution of the initial value

problem, but it does provide another relation satisﬁed by any solution of Eqs. (2).

Conversely, suppose that there is a continuous function y = φ(t) that satisﬁes the

integral equation (3); then this function also satisﬁes the initial value problem (2). To

show this,we ﬁrst substitute zero for t in Eq. (3),which shows that the initial condition

is satisﬁed. Further, since the integrand in Eq. (3) is continuous, it follows from the

fundamental theorem of calculus that φ is differentiable and that φ

′

(t) = f [t, φ(t)].

Therefore, the initial value problem and the integral equation are equivalent in the

sense that any solution of one is also a solution of the other. It is more convenient

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114 Chapter 2. First Order Differential Equations

to show that there is a unique solution of the integral equation in a certain interval

|t|≤h. The same conclusion will then hold also for the initial value problem.

One method of showing that the integral equation (3) has a unique solution is

known as the method of successive approximations or Picard’s

iteration method.

In using this method, we start by choosing an initial function φ

, either arbitrarily or

to approximate in some way the solution of the initial value problem. The simplest

choice is

(t) = 0; (4)

then φ

at least satisﬁes the initial condition in Eqs. (2), although presumably not the

differential equation. The next approximation φ

is obtained by substituting φ

(s) for

φ(s) in the right side of Eq. (3) and calling the result of this operation φ

(t). Thus

(t) =

f [s, φ

(s)]ds. (5)

Similarly, φ

is obtained from φ

(t) =

f [s, φ

(s)]ds, (6)

and, in general,

n+1

(t) =

f [s, φ

(s)]ds. (7)

In this manner we generate the sequence of functions {φ

}={φ

, φ

, ..., φ

, ...}.

Each member of the sequence satisﬁes the initial condition, but in general none sat-

isﬁes the differential equation. However, if at some stage, say, for n = k, we ﬁnd that

k+1

(t) = φ

(t),then it follows that φ

is a solution of the integral equation (3). Hence

is also a solution of the initial value problem (2), and the sequence is terminated

at this point. In general, this does not occur, and it is necessary to consider the entire

inﬁnite sequence.

To establish Theorem 2.8.1, we must answer four principal questions:

1. Do all members of the sequence {φ

} exist, or may the process break down at some stage?

2. Does the sequence converge?

3. What are the properties of the limit function? In particular, does it satisfy the integral

equation (3) and hence the initial value problem (2)?

4. Is this the only solution, or may there be others?

We ﬁrst show how these questions can be answered in a speciﬁc and relatively simple

example and then comment on some of the difﬁculties that may be encountered in

the general case.

Charles-Émile Picard (1856–1914) was appointed professor at the Sorbonne before the age of 30. Except

for Henri Poincaré, he is perhaps the most distinguished French mathematician of his generation. He

is known for important theorems in complex variables and algebraic geometry as well as differential

equations. A special case of the method of successive approximations was ﬁrst published by Liouville

in 1838. However, the method is usually credited to Picard, who established it in a general and widely

applicable form in a series of papers beginning in 1890.

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2.8 The Existence and Uniqueness Theorem 115

EXAMPLE

Solve the initial value problem

′

= 2t(1 + y), y(0) = 0 (8)

by the method of successive approximations.

Note ﬁrst that if y = φ(t), then the corresponding integral equation is

φ(t) =

2s[1 + φ(s)]ds. (9)

If the initial approximation is φ

(t) = 0, it follows that

(t) =

2s[1 + φ

(s)]ds =

2sds= t

. (10)

Similarly,

(t) =

2s[1 + φ

(s)]ds =

2s[1 + s

]ds = t

(11)

and

(t) =

2s[1 + φ

(s)]ds =

1 + s

ds = t

2 · 3

. (12)

Equations (10), (11), and (12) suggest that

(t) = t

+···+

(13)

for each n ≥ 1, and this result can be established by mathematical induction, as follows. Equa-

tion (13) is certainly true for n = 1; see Eq. (10). We must show that if it is true for n = k, then

it also holds for n = k + 1. We have

k+1

(t) =

2s[1 + φ

(s)]ds

1 + s

+···+

(

= t

+···+

2k+2

(k + 1)!

, (14)

and the inductive proof is complete.

A plot of the ﬁrst four iterates, φ

(t), φ

(t), and φ

(t), is shown in Figure 2.8.1. As k

increases, the iterates seem to remain close over a gradually increasing interval, suggesting

eventual convergence to a limit function.

It follows from Eq. (13) that φ

(t) is the nth partial sum of the inﬁnite series

∞

k=1

; (15)

hence lim

n→∞

(t) exists if and only if the series (15) converges. Applying the ratio test, we see

that, for each t,

2k+2

(k + 1)!

k + 1

→ 0 as k →∞. (16)

Thus the series (15) converges for all t, and its sum φ(t) is the limit of the sequence {φ

(t)}.

Further, since the series (15) is a Taylor series, it can be differentiated or integrated term by

term as long as t remains within the interval of convergence, which in this case is the entire

t-axis. Therefore, we can verify by direct computation that φ(t) =

∞

k=1

/k! is a solution of the

integral equation (9).Alternatively,by substituting φ(t) for y in Eqs. (8), we can verify that this

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116 Chapter 2. First Order Differential Equations

(t)

–1.5 –1 –0.5 0.5 1 1.5

2.5

1.5

0.5

FIGURE 2.8.1 Plots of the ﬁrst four Picard iterates

(t), ..., φ

(t) for Example 1: dy/dt = 2 t(1 + y), y(0) = 0.

function satisﬁes the initial value problem. In this example it is also possible, from the series

(15), to identify φ in terms of elementary functions, namely, φ(t) = e

− 1. However, this is not

necessary for the discussion of existence and uniqueness.

Explicitknowledge of φ(t) does make itpossible to visualize the convergenceof the sequence

of iterates more clearly by plotting φ(t) − φ

(t) for various values of k. Figure 2.8.2 shows

this difference for k = 1, 2, 3, 4. This ﬁgure clearly shows the gradually increasing interval

over which successive iterates provide a good approximation to the solution of the initial value

problem.

–1.5 –1 –0.5 0.5 1 1.5

0.8

0.6

0.4

0.2

k = 1

k = 2

k = 3

k = 4

FIGURE 2.8.2 Plots of φ(t) − φ

(t) for Example 1 for k = 1, ...,4.

Finally, to deal with the question of uniqueness, let us suppose that the initial value problem

has two different solutions φ and ψ. Since φ and ψ both satisfy the integral equation (9), we

have by subtraction that

φ(t) − ψ(t) =

2s[φ(s) − ψ(s)]ds.

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2.8 The Existence and Uniqueness Theorem 117

Taking absolute values of both sides, we have, if t > 0,

|φ(t) − ψ(t)|=

2s[φ(s) − ψ(s)]ds

≤

2s|φ(s) − ψ(s)|ds.

If we restrict t to lie in the interval 0 ≤ t ≤ A/2, where A is arbitrary, then 2t ≤ A and

|φ(t) − ψ(t)|≤A

|φ(s) − ψ(s)|ds for 0 ≤ t ≤ A/2. (17)

It is now convenient to introduce the function U deﬁned by

U(t) =

|φ(s) − ψ(s)|ds. (18)

Then it follows at once that

U(0) = 0, (19)

U(t) ≥ 0, for t ≥ 0. (20)

Further, U is differentiable, and U

′

(t) =|φ(t ) − ψ(t)|. Hence, by Eq. (17),

′

(t) − AU (t) ≤ 0 for 0 ≤ t ≤ A/2. (21)

Multiplying Eq. (21) by the positive quantity e

−At

gives

−At

U(t)]

′

≤ 0 for 0 ≤ t ≤ A/2. (22)

Then, upon integrating Eq. (22) from zero to t and using Eq. (19), we obtain

−At

U(t) ≤ 0 for 0 ≤ t ≤ A/2.

Hence U(t) ≤ 0 for 0 ≤ t ≤ A/2. However, since A is arbitrary, we conclude that U(t) ≤ 0 for

all nonnegative t. This result and Eq. (20) are compatible only if U(t) = 0 for each t ≥ 0. Thus

′

(t) = 0 and therefore ψ(t) = φ(t) for all t ≥ 0. This contradicts the hypothesis that φ and ψ

are two different solutions. Consequently, there cannot be two different solutions of the initial

value problem for t ≥ 0. A slight modiﬁcation of this argument leads to the same conclusion

for t ≤ 0.

Returning now to the general problem of solving the integral equation (3), let us

consider brieﬂy each of the questions raised earlier:

1. Do all members of the sequence {φ

} exist?

In the example, f and ∂f /∂y were continuous in the whole ty-plane, and each member

of the sequence could be explicitly calculated. In contrast, in the general case, f and ∂f /∂y

are assumed to be continuous only in the rectangle R: |t|≤a, |y|≤b (see Figure 2.8.3).

Furthermore, the members of the sequence cannot as a rule be explicitly determined. The

danger is that at some stage, say, for n = k, the graph of y = φ

(t) may contain points

that lie outside the rectangle R. More precisely, in the computation of φ

k+1

(t) it would be

necessary to evaluate f (t, y) at points where it is not known to be continuous or even to

exist. Thus the calculation of φ

k+1

(t) might be impossible.

To avoid this danger, it may be necessary to restrict t to a smaller interval than |t|≤a.

To ﬁnd such an interval, we make use of the fact that a continuous function on a closed

bounded region is bounded. Hence f is bounded on R; thus there exists a positive number

M such that

|f (t, y)|≤M, (t, y) in R. (23)

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118 Chapter 2. First Order Differential Equations

We have mentioned before that

(0) = 0

for each n. Since f [t, φ

(t)] is equal to φ

′

k+1

(t), the maximum absolute slope of the graph of

the equation y = φ

k+1

(t) is M. Since this graph contains the point (0, 0), it must lie in

the wedge-shaped shaded region in Figure 2.8.4. Hence the point [t, φ

k+1

(t)] remains

in R at least as long as R contains the wedge-shaped region, which is for |t|≤b/M.We

hereafter consider only the rectangle D:|t|≤h, |y|≤b, where h is equal either to a or to

b/M, whichever is smaller. With this restriction, all members of the sequence {φ

(t)} exist.

Note that whenever b/M < a, you can try to obtain a larger value of h by ﬁnding a better

(that is, smaller) bound M for |f (t, y)|, if this is possible.

(–a, b)(a, b)

(–a, –b)(a, –b)

FIGURE 2.8.3 Region of deﬁnition for Theorem 2.8.1.

(a)

y = b

t = a

t = – a

y = –b

t =

y =

(t)

(b)

t = at = – a

y = b

y = –b

y =

(t)

FIGURE 2.8.4 Regions in which successive iterates lie. (a) b/M < a;(b) b/M > a.

2. Does the sequence {φ

(t)} converge?

We can identify φ

(t) = φ

(t) +[φ

(t) − φ

(t)]+···+[φ

(t) − φ

n−1

(t)] as the nth par-

tial sum of the series

(t) +

∞

k=1

[φ

k+1

(t) − φ

(t)]. (24)

The convergence of the sequence {φ

(t)} is established by showing that the series (24)

converges. To do this, it is necessary to estimate the magnitude |φ

k+1

(t) − φ

(t)| of the

general term. The argument by which this is done is indicated in Problems 15 through

18 and will be omitted here. Assuming that the sequence converges, we denote the limit

function by φ, so that

φ(t) = lim

n→∞

(t). (25)

3. What are the properties of the limit function φ?

In the ﬁrst place, we would like to know that φ is continuous. This is not, however, a

necessary consequence of the convergence of thesequence {φ

(t)},even though each mem-

ber of the sequence is itself continuous. Sometimes a sequence of continuous functions

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2.8 The Existence and Uniqueness Theorem 119

converges to a limit function that is discontinuous. A simple example of this phenomenon

is given in Problem 13. One way to show that φ is continuous is to show not only that the

sequence {φ

} converges, but also that it converges in a certain manner,known as uniform

convergence. We do not take up this matter here but note only that the argument referred

to in the discussion of question 2 is sufﬁcient to establish the uniform convergence of the

sequence {φ

} and, hence, the continuity of the limit function φ in the interval |t|≤h.

Now let us return to Eq. (7)

n+1

(t) =

f [s, φ

(s)]ds.

Allowing n to approach ∞ on both sides, we obtain

φ(t) = lim

n→∞

f [s, φ

(s)]ds. (26)

We would like to interchange the operations of integrating and taking the limit on the

right side of Eq. (26) so as to obtain

φ(t) =

lim

n→∞

f [s, φ

(s)]ds. (27)

In general, such an interchange is not permissible (see Problem 14, for example),but once

again, the fact that the sequence {φ

(t)} converges uniformly is sufﬁcient to allow us to

take the limiting operation inside the integral sign. Next, we wish to take the limit inside

the function f , which would give

φ(t) =

f [s, lim

n→∞

(s)]ds (28)

and hence

φ(t) =

f [s, φ(s)]ds. (29)

The statement that

lim

n→∞

f [s, φ

(s)]=f [s, lim

n→∞

(s)]

is equivalent to the statement that f is continuous in its second variable,which is known by

hypothesis. Hence Eq. (29) is valid, and the function φ satisﬁes the integral equation (3).

Thus φ is also a solution of the initial value problem (2).

4. Are there other solutions of the integral equation (3) besides y = φ(t)?

To show the uniqueness of the solution y = φ(t),wecan proceed much as in the example.

First, assume the existence of another solution y = ψ(t). It is then possible to show (see

Problem 19) that the difference φ(t) − ψ(t) satisﬁes the inequality

|φ(t) − ψ(t)|≤A

|φ(s) − ψ(s)|ds (30)

for 0 ≤ t ≤ h and a suitable positive number A. From this point the argument is identical

to that given in the example, and we conclude that there is no solution of the initial value

problem (2) other than the one generated by the method of successive approximations.

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120 Chapter 2. First Order Differential Equations

PROBLEMS In each of Problems 1 and 2, transform the given initial value problem into an equivalent

problem with the initial point at the origin.

1. dy/dt = t

+ y

, y(1) = 22.dy/dt = 1 − y

, y(−1) = 3

In each of Problems 3 through 6, let φ

(t) = 0 and deﬁne {φ

(t)} by the method of succes-

sive approximations

(a) Determine φ

(t) for an arbitrary value of n.

(b) Plot φ

(t) for n = 1, ..., 4. Observe whether the iterates appear to be converging.

n→∞

(t) = φ(t) in terms of elementary functions; that is, solve the given initial

value problem.

(d) Plot |φ(t) − φ

(t)| for n = 1, ..., 4. For each of φ

(t), ..., φ

(t), estimate the interval in

which it is a reasonably good approximation to the actual solution.

′

= 2(y + 1), y(0) = 04.y

′

=−y − 1, y(0) = 0

′

=−y/2 + t, y(0) = 06. y

′

= y + 1 − t, y(0) = 0

In each of Problems 7 and 8, let φ

(t) = 0 and use the method of successive approximations to

solve the given initial value problem.

(a) Determine φ

(t) for an arbitrary value of n.

(b) Plot φ

(t) for n = 1, ..., 4. Observe whether the iterates appear to be converging.

(t)} converges.

′

= ty + 1, y(0) = 08.y

′

= t

y − t, y(0) = 0

In each of Problems 9 and 10, let φ

(t) = 0 and use the method of successive approximations

to approximate the solution of the given initial value problem.

(a) Calculate φ

(t), ..., φ

(t).

(b) Plot φ

(t), ..., φ

(t) and observe whether the iterates appear to be converging.

′

= t

+ y

, y(0) = 0 10. y

′

= 1 − y

, y(0) = 0

In each of Problems 11 and 12, let φ

(t) = 0 and use the method of successive approximations

to approximate the solution of the given initial value problem.

(a) Calculate φ

(t), ..., φ

(t), or (if necessary) Taylor approximations to these iterates. Keep

terms up to order six.

(b) Plot the functions you found in part (a) and observe whether they appear to be converging.

11.

′

=−sin y + 1, y(0) = 0 12. y

′

= (3t

+ 4t + 2)/2(y − 1), y(0) = 0

13. Let φ

(x) = x

for 0 ≤ x ≤ 1 and show that

lim

n→∞

(x) =

0, 0 ≤ x < 1,

1, x = 1.

This example shows that a sequence of continuous functions may converge to a limit

function that is discontinuous.

14. Consider the sequence φ

(x) = 2nxe

−nx

,0≤ x ≤ 1.

(a) Show that lim

n→∞

(x) = 0 for 0 ≤ x ≤ 1; hence

lim

n→∞

(x) dx = 0.

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2.8 The Existence and Uniqueness Theorem 121

(b) Show that

2nxe

−nx

dx = 1 − e

−n

; hence

lim

n→∞

(x) dx = 1.

Thus, in this example,

lim

n→∞

(x) dx ̸=

lim

n→∞

(x) dx,

even though lim

n→∞

(x) exists and is continuous.

In Problems 15 through 18, we indicate how to prove that the sequence {φ

(t)}, deﬁned by

Eqs. (4) through (7), converges.

15. If ∂f /∂y is continuous in the rectangle D,show that there is a positive constant K such that

|f (t, y

) − f (t, y

)|≤K|y

− y

|, (i)

where (t, y

) and (t, y

) are any two points in D having the same t coordinate.Thisinequality

is known as a Lipschitz

condition.

Hint: Hold t ﬁxed and use the mean value theorem on f as a function of y only. Choose K

to be the maximum value of |∂f /∂y| in D.

16. If φ

n−1

(t) and φ

(t) are members of the sequence {φ

(t)}, use the result of Problem 15 to

show that

|f [t, φ

(t)]−f [t, φ

n−1

(t)]| ≤ K|φ

(t) − φ

n−1

(t)|.

17. (a) Show that if |t|≤h, then

|φ

(t)|≤M|t|,

where M is chosen so that |f (t, y)|≤M for (t, y) in D.

(b) Use the results of Problem 16 and part (a) of Problem 17 to show that

|φ

(t) − φ

(t)|≤

MK|t|

|φ

(t) − φ

n−1

(t)|≤

n−1

|t|

≤

n−1

18. Note that

(t) = φ

(t) +[φ

(t) − φ

(t)]+···+[φ

(t) − φ

n−1

(t)].

(a) Show that

|φ

(t)|≤|φ

(t)|+|φ

(t) − φ

(t)|+···+|φ

(t) − φ

n−1

(t)|.

(b) Use the results of Problem 17 to show that

|φ

(t)|≤

Kh +

(Kh)

+···+

(Kh)

also converges as n →∞. Conclude therefore that the sequence {φ

(t)} converges since

it is the sequence of partial sums of a convergent inﬁnite series.

The German mathematician Rudolf Lipschitz (1832–1903), professor at the University of Bonn for many

years, worked in several areas of mathematics. The inequality (i) can replace the hypothesis that ∂f /∂y is

continuous in Theorem 2.8.1; this results in a slightly stronger theorem.

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122 Chapter 2. First Order Differential Equations

19. In this problem we deal with the question of uniqueness of the solution of the integral

equation (3)

φ(t) =

f [s, φ(s)]ds.

(a) Suppose that φ and ψ are two solutions of Eq. (3). Show that, for t ≥ 0,

φ(t) − ψ(t) =

{f [s, φ(s)]−f [s, ψ(s)]}ds.

(b) Show that

|φ(t) − ψ(t)|≤

|f [s, φ(s)]−f [s, ψ(s)]|ds.

|φ(t) − ψ(t)|≤K

|φ(s) − ψ(s)|ds,

where K is an upper bound for |∂f /∂y| in D. This is the same as Eq. (30), and the rest of

the proof may be constructed as indicated in the text.

2.9 First Order Difference Equations

Although a continuous model leading to a differential equation is reasonable and

attractive for many problems, there are some cases in which a discrete model may

be more natural. For instance, the continuous model of compound interest used in

Section 2.3 is only an approximation to the actual discrete process. Similarly, some-

times population growth may be described more accurately by a discrete than by

a continuous model. This is true, for example, of species whose generations do not

overlap and that propagate at regular intervals, such as at particular times of the cal-

endar year.Then the population y

n+1

of the species in the year n + 1 is some function

of n and the population y

in the preceding year; that is,

n+1

= f (n, y

), n = 0, 1, 2, .... (1)

Equation (1) is called a ﬁrst order difference equation. It is ﬁrst order because the

value of y

n+1

depends on the value of y

but not on earlier values y

n−1

, y

n−2

, and so

forth. As for differential equations, the difference equation (1) is linear if f is a linear

function of y

; otherwise, it is nonlinear.Asolution of the difference equation (1) is

a sequence of numbers y

, y

, ...that satisfy the equation for each n. In addition

to the difference equation itself, there may also be an initial condition

= α (2)

that prescribes the value of the ﬁrst term of the solution sequence.

We now assume temporarily that the function f in Eq. (1) depends only on y

, but

not on n. In this case

n+1

= f ( y

), n = 0, 1, 2, .... (3)

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2.9 First Order Difference Equations 123

If y

is given, then successive terms of the solution can be found from Eq. (3). Thus

= f ( y

and

= f (y

) = f [f (y

)].

The quantity f [f (y

)] is called the second iterate of the difference equation and is

sometimes denoted by f

). Similarly, the third iterate y

is given by

= f (y

) = f {f [f (y

)]} = f

and so on. In general, the nth iterate y

= f ( y

n−1

) = f

This procedure is referred to as iterating the difference equation. It is often of pri-

mary interest to determine the behavior of y

as n →∞. In particular, does y

approach a limit, and if so, what is it?

Solutions for which y

has the same value for all n are called equilibrium solutions.

They arefrequently ofspecial importance,justas inthe studyof differential equations.

If equilibrium solutions exist, you can ﬁnd them by setting y

n+1

equal to y

in Eq. (3)

and solving the resulting equation

= f (y

) (4)

for y

Linear Equations. Suppose that the population of a certain species in a given region in

year n + 1, denoted by y

n+1

, is a positive multiple ρ

of the population y

in year n;

that is,

n+1

= ρ

, n = 0, 1, 2, .... (5)

Note that the reproduction rate ρ

may differ from year to year. The difference

equation (5) is linear and can easily be solved by iteration. We obtain

= ρ

and, in general,

= ρ

n−1

···ρ

, n = 1, 2, .... (6)

Thus, if the initial population y

is given, then the population of each succeeding

generation is determined by Eq. (6). Although for a population problem ρ

is intrin-

sically positive, the solution (6) is also valid if ρ

is negative for some or all values of

n. Note, however, that if ρ

is zero for some n, then y

n+1

and all succeeding values of

y are zero; in other words, the species has become extinct.

If the reproduction rate ρ

has the same value ρ for each n, then the difference

equation (5) becomes

n+1

= ρy

(7)

and its solution is

= ρ

. (8)

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124 Chapter 2. First Order Differential Equations

Equation (7) also has an equilibrium solution,namely,y

= 0 for all n,corresponding

to the initial value y

= 0. The limiting behavior of y

is easy to determine from

Eq. (8). In fact,

lim

n→∞

⎧

⎪

⎨

⎪

⎩

0, if |ρ| < 1;

,if ρ = 1;

does not exist, otherwise.

(9)

In other words, the equilibrium solution y

= 0 is asymptotically stable for |ρ| < 1

and unstable for |ρ| > 1.

Now we will modify the population model represented by Eq. (5) to include the

effect of immigration or emigration. If b

is the net increase in population in year

n due to immigration, then the population in year n + 1 is the sum of the part of

the population resulting from natural reproduction and the part due to immigration.

Thus

n+1

= ρy

+ b

, n = 0, 1, 2, ..., (10)

where we are now assuming that the reproduction rate ρ is constant. We can solve

Eq. (10) by iteration in the same manner as before. We have

= ρy

+ b

= ρ(ρy

+ b

) + b

= ρ

+ ρ b

+ b

= ρ(ρ

+ ρ b

+ b

) + b

= ρ

+ ρ

+ ρ b

+ b

and so forth. In general, we obtain

= ρ

+ ρ

n−1

+···+ρb

n−2

+ b

n−1

= ρ

n−1

j =0

n−1−j

. (11)

Note that the ﬁrst term on the right side of Eq. (11) represents the descendants of

the original population, while the other terms represent the population in year n

resulting from immigration in all preceding years.

In the special case where b

= b ̸= 0 for all n, the difference equation is

n+1

= ρy

+ b, (12)

and from Eq. (11) its solution is

= ρ

+ (1 + ρ + ρ

+···+ρ

n−1

)b. (13)

If ρ ̸= 1, we can write this solution in the more compact form

= ρ

1 − ρ

b, (14)

where again the two terms on the right side are the effects of the original population

and of immigration, respectively. Rewriting Eq. (14) as

= ρ

−

1 − ρ

(

1 − ρ

(15)

makes the long-time behavior of y

more evident. It follows from Eq. (15) that

→ b/(1 − ρ) if |ρ| < 1. If |ρ| > 1orifρ =−1 then y

has no limit unless

= b/(1 − ρ). The quantity b/(1 − ρ), for ρ ̸= 1, is an equilibrium solution of

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2.9 First Order Difference Equations 125

Eq. (12), as can readily be seen directly from that equation. Of course, Eq. (14)

is not valid for ρ = 1. To deal with that case, we must return to Eq. (13) and let ρ = 1

there. It follows that

= y

+ nb, (16)

so in this case y

becomes unbounded as n →∞.

The same model also provides a framework for solving many problems of a ﬁnan-

cial character. For such problems, y

is the account balance in the nth time period,

= 1 + r

, where r

is the interest rate for that period, and b

is the amount

deposited or withdrawn. The following example is typical.

EXAMPLE

A recent college graduate takes out a $10,000 loan to purchase a car. If the interest rate is

12%, what monthly payment is required to pay off the loan in 4 years?

The relevant difference equation is Eq. (12), where y

is the loan balance outstanding in the

nth month, ρ = 1 + r, where r is the interest rate per month and b is the effect of the monthly

payment. Note that ρ = 1.01, corresponding to a monthly interest rate of 1%. Since payments

reduce the loan balance, b must be negative; the actual payment is

The solution of the difference equation (12) with this value for ρ and the initial condition

= 10,000 is given by Eq. (15); that is,

= (1.01)

(10,000 + 100b) − 100b. (17)

The value of b needed to pay off the loan in 4 years is found by setting y

= 0 and solving

for b. This gives

b =−100

(1.01)

− 1

=−263.34. (18)

The total amount paid on the loan is 48 times |b|, or $12,640.32. Of this amount, $10,000 is

repayment of the principal and the remaining $2640.32 is interest.

Nonlinear Equations. Nonlinear difference equations are much more complicated and

have much more varied solutions than linear equations.We will restrict our attention

to a single equation, the logistic difference equation

n+1

= ρy

)

1 −

, (19)

which is analogous to the logistic differential equation

= ry

)

1 −

(20)

that was discussed in Section 2.5. Note that if the derivative dy/dt in Eq. (20) is

replaced by the difference (y

n+1

− y

)/h, then Eq. (20) reduces to Eq. (19) with

ρ = 1 + hr and k = (1 + hr)K/hr. To simplify Eq. (19) a little more, we can scale the

variable y

by introducing the new variable u

= y

/k. Then Eq. (19) becomes

n+1

= ρu

(1 − u

), (21)

where ρ is a positive parameter.

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126 Chapter 2. First Order Differential Equations

We begin our investigation of Eq. (21) by seeking the equilibrium,or constant,solu-

tions. These can be found by setting u

n+1

equal to u

in Eq. (21), which corresponds

to setting dy/dt equal to zero in Eq. (20). The resulting equation is

= ρu

− ρ u

, (22)

so it follows that the equilibrium solutions of Eq. (21) are

= 0, u

ρ − 1

. (23)

The next question is whether the equilibrium solutions are asymptotically stable or

unstable. That is, for an initial condition near one of the equilibrium solutions, does

the resulting solution sequence approach or depart from the equilibrium solution?

One way to examine this question is by approximating Eq. (21) by a linear equation

in the neighborhood of an equilibrium solution. For example, near the equilibrium

solution u

= 0, the quantity u

is small compared to u

itself, so we assume that we

can neglect the quadratic term in Eq. (21) in comparison with the linear terms. This

leaves us with the linear difference equation

n+1

= ρu

, (24)

which is presumably a good approximation to Eq. (21) for u

sufﬁciently near zero.

However, Eq. (24) is the same as Eq. (7), and we have already concluded, in Eq. (9),

that u

→ 0asn →∞if and only if |ρ| < 1, or (since ρ must be positive) for

0 <ρ<1.Thus the equilibrium solution u

= 0 is asymptotically stable for the linear

approximation (24) for this set of ρ values, so we conclude that it is also asymptoti-

cally stable for the full nonlinear equation (21). This conclusion is correct, although

our argument is not complete. What is lacking is a theorem stating that the solutions

of the nonlinear equation (21) resemble those of the linear equation (24) near the

equilibrium solution u

= 0. We will not take time to discuss this issue here; the same

question is treated for differential equations in Section 9.3.

Now consider the other equilibrium solution u

= (ρ − 1)/ρ. To study solutions in

the neighborhood of this point, we write

ρ − 1

+ v

, (25)

where we assume that v

is small. By substituting from Eq. (25) in Eq. (21) and

simplifying the resulting equation, we eventually obtain

n+1

= (2 − ρ)v

− ρv

. (26)

Since v

is small, we again neglect the quadratic term in comparison with the linear

terms and thereby obtain the linear equation

n+1

= (2 − ρ)v

. (27)

Referring to Eq. (9) once more, we ﬁnd that v

→ 0asn →∞for |2 − ρ| < 1, or in

other words for 1 <ρ<3. Therefore, we conclude that for this range of values of ρ,

the equilibrium solution u

= (ρ − 1)/ρ is asymptotically stable.

Figure 2.9.1 contains the graphs of solutions of Eq. (21) for ρ = 0 .8, ρ = 1.5,

and ρ = 2.8, respectively. Observe that the solution converges to zero for ρ = 0.8

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2.9 First Order Difference Equations 127

and to the nonzero equilibrium solution for ρ = 1.5 and ρ = 2.8. The convergence is

monotone for ρ = 0.8 and ρ = 1.5 and is oscillatory for ρ = 2.8. The graphs shown

are for particular initial conditions, but the graphs for other initial conditions are

similar.

2648

( a)

0.8

0.6

0.4

0.2

2648

( b)

0.8

0.6

0.4

0.2

2648

( c)

0.8

0.6

0.4

0.2

= = 0.6429

1.8

2.8

FIGURE 2.9.1 Solutions of u

n+1

= ρu

(1 − u

):(a) ρ = 0.8; (b) ρ = 1.5; (c) ρ = 2.8.

Another way of displaying the solution of a difference equation is shown in

Figure 2.9.2. In each part of this ﬁgure, the graphs of the parabola y = ρx(1 − x)

and of the straight line y = x are shown. The equilibrium solutions correspond to

the points of intersection of these two curves. The piecewise linear graph consist-

ing of successive vertical and horizontal line segments, sometimes called a stairstep

diagram, represents the solution sequence. The sequence starts at the point u

the x-axis. The vertical line segment drawn upward to the parabola at u

corresponds

to thecalculation of ρu

(1 − u

) = u

.This value isthen transferred from the y-axis to

the x-axis; this step is represented by the horizontal line segment from the parabola

to the line y = x. Then the process is repeated over and over again. Clearly, the

sequence converges to the origin in Figure 2.9.2a and to the nonzero equilibrium

solution in the other two cases.

To summarize our results so far: the difference equation (21) has two equilib-

rium solutions, u

= 0 and u

= (ρ − 1)/ρ; the former is asymptotically stable for

0 ≤ ρ<1, and the latter is asymptotically stable for 1 <ρ<3. When ρ = 1, the two

equilibrium solutions coincide at u = 0; this solution can be shown to be asymptoti-

cally stable. In Figure 2.9.3 the parameter ρ is plotted on the horizontal axis and u on

the vertical axis. The equilibrium solutions u = 0 and u = (ρ − 1)/ρ are shown. The

intervals in which each one is asymptotically stable are indicated by the solid portions

of the curves. There is an exchange of stability from one equilibrium solution to the

other at ρ = 1.

For ρ>3, neither of the equilibrium solutions is stable, and the solutions of

Eq. (21) exhibit increasing complexity as ρ increases. For ρ somewhat greater than 3,

the sequence u

rapidly approaches a steady oscillation of period 2; that is, u

oscil-

lates back and forth between two distinct values. For ρ = 3.2, a solution is shown in

Figure 2.9.4. For n greater than about 20, the solution alternates between the values

0.5130 and 0.7995. The graph is drawn for the particular initial condition u

= 0.3,

but it is similar for all other initial values between 0 and 1. Figure 2.9.4b also shows

the same steady oscillation as a rectangular path that is traversed repeatedly in the

clockwise direction.

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128 Chapter 2. First Order Differential Equations

( a)

0.8

0.6

0.4

0.2

0.2 0.4 0.6 0.8 1

y = x

= 0.3

y = x (1 – x)

( b)

0.2 0.4 0.6 0.8 1

y = x

= 0.85

y = x (1 – x)

0.8

0.6

0.4

0.2

0.2 0.4 0.6 0.8

( c)

y = x

(0.6429..., 0.6429...)

= 0.3

y = x (1 – x)

0.8

0.6

0.4

0.2

FIGURE 2.9.2 Iterates of u

n+1

= ρu

(1 − u

):(a) ρ = 0.8; (b) ρ = 1.5; (c) ρ = 2.8.

0.5

–0.5

123

u = 0

Asymptotically stable

Unstable

u = ( – 1)/

ρρ

FIGURE 2.9.3 Exchange of stability for u

n+1

= ρu

(1 − u

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2.9 First Order Difference Equations 129

0.8

0.6

0.4

0.2

10 20 30 40

(a)

0.7995

0.5130

0.8

0.6

0.4

0.2

0.2 0.4 0.6 0.8 1

y = x

0.5130 0.7995

0.5130

0.7995

(b)

y = x (1 – x)

FIGURE 2.9.4 A solution of u

n+1

= ρu

(1 − u

) for

ρ = 3.2; period 2. (a) u

versus n;(b) a two-cycle.

At about ρ = 3.449, each state in the oscillation of period 2 separates into two

distinct states,andthe solution becomes periodic with period 4;see Figure 2.9.5,which

shows a solution of period 4 for ρ = 3.5. As ρ increases further, periodic solutions

of period 8, 16, ... appear. The appearance of a new solution at a certain parameter

value is called a bifurcation.

The ρ-values at which the successive period doublings occur approach a limit that

is approximately 3.57. For ρ>3.57, the solutions possess some regularity but no

discernible detailed pattern for most values of ρ. For example,a solution for ρ = 3.65

is shown in Figure 2.9.6. It oscillates between approximately 0.3 and 0.9, but its ﬁne

structure is unpredictable. The term chaotic is used to describe this situation. One of

the features of chaotic solutions is extreme sensitivity to the initial conditions. This

is illustrated in Figure 2.9.7, where two solutions of Eq. (21) for ρ = 3.65 are shown.

One solution is the same as that in Figure 2.9.6 and has the initial value u

= 0.3,while

the other solution has the initial value u

= 0.305. For about 15 iterations the two

solutions remain close and are hard to distinguish from each other in the ﬁgure. After

that,although they continue to wander about in approximately the same set of values,

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130 Chapter 2. First Order Differential Equations

0.8

0.4

841216202428323640

(a)

0.5

y = x (1 – x)

y = x

0.5009

0.3828

0.8269

0.8750

(b)

FIGURE 2.9.5 A solution of u

n+1

= ρu

(1 − u

) for ρ = 3.5;

period 4. (a) u

versus n;(b) a four-cycle.

0.3

10 20 30 40 50 60

0.4

0.5

0.6

0.7

0.8

0.9

FIGURE 2.9.6 A solution of u

n+1

= ρu

(1 − u

) for ρ = 3.65; a chaotic solution.

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2.9 First Order Difference Equations 131

0.8

0.9

0.7

0.6

0.3

0.4

0.5

10 20 30 40 50 60

FIGURE 2.9.7 Two solutions of u

n+1

= ρu

(1 − u

) for ρ = 3.65; u

= 0.3 and u

= 0.305.

their graphs are quite dissimilar. It would certainly not be possible to use one of these

solutions to estimate the value of the other for values of n larger than about 15.

It is only comparativelyrecently that chaoticsolutions ofdifference anddifferential

equations have become widely known. Equation (20) was one of the ﬁrst instances of

mathematical chaos to be found and studied in detail, by Robert May

in 1974. On

the basis of his analysis of this equation as a model of the population of certain insect

species, May suggested that if the growth rate ρ is too large, then it will be impossible

to make effective long-range predictions about these insect populations. The occur-

rence of chaotic solutions in seemingly simple problems has stimulated an enormous

amount of research, but many questions remain unanswered. It is increasingly clear,

however, that chaotic solutions are much more common than was suspected at ﬁrst

and that they may be a part of the investigation of a wide range of phenomena.

PROBLEMS In each of Problems 1 through 6, solve the given difference equation in terms of the initial

value y

. Describe the behavior of the solution as n →∞.

1. y

n+1

=−0.9y

2. y

n+1

n + 1

n + 2

3. y

n+1

n + 3

n + 1

4. y

n+1

= (−1)

n+1

5. y

n+1

= 0.5y

+ 66.y

n+1

=−0.5y

+ 6

Robert M. May (1938– ) was born in Sydney, Australia, and received his education at the University of

Sydney with a doctorate in theoretical physics in 1959. His interests soon turned to population dynamics

and theoretical ecology; the work cited in the text is described in two papers listed in the References at

the end of this chapter. He has held professorships at Sydney, at Princeton, at Imperial College (London),

and (since 1988) at Oxford.

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132 Chapter 2. First Order Differential Equations

7. Find the effective annual yield of a bank account that pays interest at a rate of 7%,

compounded daily; that is, divide the difference between the ﬁnal and initial balances by

the initial balance.

8. An investor deposits $1000 in an account paying interest at a rate of 8%, compounded

monthly, and also makes additional deposits of $25 per month. Find the balance in the

account after 3 years.

9. A certain college graduate borrows $8000 to buy a car. The lender charges interest at an

annual rate of 10%. What monthly payment rate is required to pay off the loan in 3 years?

Compare your result with that of Problem 9 in Section 2.3.

10. A homebuyer wishesto take outa mortgage of$100,000 for a30-year period.What monthly

payment is required if the interest rate is (a) 9%, (b) 10%, (c) 12%?

11. A homebuyer takes out a mortgage of $100,000 with an interest rate of 9%.What monthly

payment is required to pay off the loan in 30 years? In 20 years? What is the total amount

paid during the term of the loan in each of these cases?

12. If the interest rate on a 20-year mortgage is ﬁxed at 10% and if a monthly payment of

$1000 is the maximum that the buyer can afford, what is the maximum mortgage loan that

can be made under these conditions?

13. A homebuyer wishes to ﬁnance the purchase with a $95,000 mortgage with a 20-year term.

What is the maximum interest rate the buyer can afford if the monthly payment is not to

exceed $900?

The Logistic Difference Equation. Problems 14 through 19 deal with the difference equation

(21), u

n+1

= ρu

(1 − u

14. Carry out the details in the linear stability analysis of the equilibrium solution

= (ρ − 1)/ρ. That is, derive the difference equation (26) in the text for the pertur-

bation v

15.

(a) For ρ = 3.2, plot or calculate the solution of the logistic equation (21) for several

initial conditions, say, u

= 0.2, 0.4, 0.6, and 0.8. Observe that in each case the solution

approaches a steady oscillation between the same two values. This illustrates that the

long-term behavior of the solution is independent of the initial conditions.

(b) Make similar calculations and verify that the nature of the solution for large n is

independent of the initial condition for other values of ρ, such as 2.6, 2.8, and 3.4.

16. Assume that ρ>1 in Eq. (21).

(a) Draw a qualitatively correct stairstep diagram and thereby show that if u

< 0, then

→−∞as n →∞.

(b) In a similar way, determine what happens as n →∞if u

> 1.

17.

The solutions of Eq. (21) change from convergent sequences to periodic oscillations of

period 2 as the parameter ρ passes through the value 3. To see more clearly how this

happens, carry out the following calculations.

(a) Plot or calculate the solution for ρ = 2.9, 2.95, and 2.99, respectively, using an initial

value u

of your choice in the interval (0, 1). In each case estimate how many iterations

are required for the solution to get “very close” to the limiting value. Use any convenient

interpretation of what “very close” means in the preceding sentence.

(b) Plot or calculate the solution for ρ = 3.01, 3.05, and 3.1, respectively, using the same

initial condition as in part (a). In each case estimate how many iterations are needed to

reach a steady-state oscillation. Also ﬁnd or estimate the two values in the steady-state

oscillation.

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2.9 First Order Difference Equations 133

18. By calculating or plotting the solution of Eq. (21) for different values of ρ, estimate the

value of ρ at which the solution changes from an oscillation of period 2 to one of period

4. In the same way, estimate the value of ρ at which the solution changes from period 4 to

period 8.

19.

Let ρ

be the value of ρ at which the solution of Eq. (21) changes from period 2

k−1

period 2

. Thus, as noted in the text, ρ

= 3, ρ

∼

3.449, and ρ

∼

3.544.

(a) Using these values of ρ

, ρ

, and ρ

, or those you found in Problem 18, calculate

(ρ

− ρ

)/(ρ

− ρ

(b) Letδ

= (ρ

− ρ

n−1

)/(ρ

n+1

− ρ

). It can beshown that δ

approachesa limit δ as n →∞,

where δ

∼

4.6692 is known as the Feigenbaum

number. Determine the percentage

difference between the limiting value δ and δ

, as calculated in part (a).

= δ and use this relation to estimate ρ

,the valueof ρ at whichsolutions

of period 16 appear.

(d) By plotting or calculating solutions near the value of ρ

found in part (c), try to detect

the appearance of a period 16 solution.

(e) Observe that

= ρ

+ (ρ

− ρ

) + (ρ

− ρ

) +···+(ρ

− ρ

n−1

Assuming that (ρ

− ρ

) = (ρ

− ρ

)δ

−1

, (ρ

− ρ

) = (ρ

− ρ

)δ

−2

, and so forth, express ρ

as a geometric sum. Then ﬁnd the limit of ρ

as n →∞. This is an estimate of the value of

ρ at which the onset of chaos occurs in the solution of the logistic equation (21).

PROBLEMS Miscellaneous Problems. One of the difﬁculties in solving ﬁrst order equations is that there

are several methods of solution, each of which can be used on a certain type of equation. It

may take some time to become proﬁcient in matching solution methods with equations. The

ﬁrst 32 of the following problems are presented to give you some practice in identifying the

method or methods applicable to a given equation. The remaining problems involve certain

types of equations that can be solved by specialized methods.

In each of Problems 1 through 32, solve the given differential equation. If an initial condition

is given, also ﬁnd the solution that satisﬁes it.

− 2y

1 + cos x

2 − sin y

2x + y

3 + 3y

− x

, y(0) = 04.

= 3 − 6x + y − 2xy

=−

2xy + y

+ 1

+ 2xy

6. x

+ xy = 1 − y, y(1) = 0

+ 1

y(2 + 3y)

8. x

+ 2y =

sin x

, y(2) = 1

This result for the logistic difference equation was discovered in August 1975 by Mitchell Feigenbaum

(1944– ), while he was working at the LosAlamos National Laboratory. Within a few weeks he had estab-

lished that the same limiting value also appears in a large class of period-doubling difference equations.

Feigenbaum, who has a doctorate in physics from M.I.T., is now at Rockefeller University.

August 7, 2012 21:03 c02 Sheet number 104 Page number 134 cyan black

134 Chapter 2. First Order Differential Equations

=−

2xy + 1

+ 2y

10. (x

y + xy − y) + (x

y − 2x

)

= 0

11. (x

+ y) + (x + e

)

= 0 12.

+ y =

1 + e

13.

= 1 + 2x + y

+ 2xy

14. (x + y) + (x + 2y)

= 0, y(2) = 3

15. (e

+ 1)

= y − ye

16.

−x

cos y − e

cos x

−e

−x

sin y + 2e

sin x

17.

= e

+ 3y 18.

+ 2y = e

−x

−2x

, y(0) = 3

19.

− 2y − y

2x + 3xy

20. y

′

= e

x+y

21.

+ 6xy − 4

+ 4xy + 3y

= 0 22.

− 1

+ 1

, y(−1) = 1

23. t

+ (t + 1)y = e

24. 2 sin y sin x cos x + cos y sin

= 0

25.

−

+ y

(

+ y

−

(

= 0

26. xy

′

= y + xe

y/x

27.

y + y

Hint: Let u = x

28. (2y + 3x) =−x

29.

x + y

x − y

30. (3y

+ 2xy) − (2xy + x

)

= 0 31.

=−

y + y

+ 3xy

, y(1) =−2

32. xy

′

+ y − y

= 0, y(1) = 2

33. Riccati Equations. The equation

= q

(t) + q

(t)y + q

(t)y

is known as a Riccati

equation. Suppose that some particular solution y

of this equation

is known. A more general solution containing one arbitrary constant can be obtained

through the substitution

y = y

(t) +

v(t)

Show that v(t) satisﬁes the ﬁrst order linear equation

=−(q

+ 2q

)v − q

Note that v(t) will contain a single arbitrary constant.

Riccati equations are named for Jacopo Francesco Riccati (1676–1754), a Venetian nobleman, who

declined university appointments in Italy,Austria, and Russia to pursue his mathematical studies privately

at home. Riccati studied these equations extensively; however, it was Euler (in 1760) who discovered the

result stated in this problem.

August 7, 2012 21:03 c02 Sheet number 105 Page number 135 cyan black

2.9 First Order Difference Equations 135

34. Using the method of Problem 33 and the given particular solution, solve each of the

following Riccati equations:

(a) y

′

= 1 + t

− 2ty + y

; y

(t) = t (b) y

′

=−

−

+ y

; y

(t) =

(c)

2 cos

t − sin

t + y

2 cos t

; y

(t) = sin t

35. The propagation of a single action in a large population (for example, drivers turning on

headlights at sunset) often depends partly on external circumstances (gathering darkness)

and partly on a tendency to imitate others who have already performed the action in

question. In this case the proportion y(t) of people who have performed the action can be

described

by the equation

dy/dt = (1 − y)[x(t) + by], (i)

where x(t) measures the external stimulus and b is the imitation coefﬁcient.

(a) Observe that Eq. (i) is a Riccati equation and that y

(t) = 1 is one solution. Use the

transformation suggested in Problem 33, and ﬁnd the linear equation satisﬁed by v(t).

(b) Find v(t) in the case that x(t) = at, where a is a constant. Leave your answer in the

form of an integral.

Some Special Second Order Equations. Second order equations involve the second deriva-

tive of the unknown function and have the general form y

′′

= f (t, y, y

′

). Usually such equations

cannot be solved by methods designed for ﬁrst order equations. However, there are two types

of second order equations that can be transformed into ﬁrst order equations by a suitable

change of variable. The resulting equation can sometimes be solved by the methods presented

in this chapter. Problems 36 through 51 deal with these types of equations.

Equations with the Dependent Variable Missing. For a second order differential equation

of the form y

′′

= f (t, y

′

), the substitution v = y

′

, v

′

= y

′′

leads to a ﬁrst order equation of the

form v

′

= f (t, v). If this equation can be solved for v, then y can be obtained by integrating

dy/dt = v. Note that one arbitrary constant is obtained in solving the ﬁrst order equation for

v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use

this substitution to solve the given equation.

36. t

′′

+ 2ty

′

− 1 = 0, t > 0 37. ty

′′

+ y

′

= 1, t > 0

38. y

′′

+ t(y

′

)

= 0 39. 2t

′′

+ (y

′

)

= 2ty

′

, t > 0

40. y

′′

+ y

′

= e

−t

41. t

′′

= (y

′

)

, t > 0

Equations with the Independent Variable Missing. Consider second order differential equa-

tions of the form y

′′

= f (y, y

′

), in which the independent variable t does not appear explicitly.

If we let v = y

′

,then we obtain dv/dt = f (y,v). Since the right side of this equation depends on

y and v, rather than on t and v,this equation contains too many variables. However, if we think

of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy).

Hence the original differential equation can be written as v(dv/dy) = f (y, v). Provided that

this ﬁrst order equation can be solved, we obtain v as a function of y. A relation between y

and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two

See Anatol Rapoport,“Contribution to the Mathematical Theory of Mass Behavior: I. The Propagation

of Single Acts,” Bulletin of Mathematical Biophysics 14 (1952), pp. 159–169, and John Z. Hearon, “Note

on the Theory of Mass Behavior,” Bulletin of Mathematical Biophysics 17 (1955), pp. 7–13.

August 7, 2012 21:03 c02 Sheet number 106 Page number 136 cyan black

136 Chapter 2. First Order Differential Equations

arbitrary constants in the ﬁnal result. In each of Problems 42 through 47, use this method to

solve the given differential equation.

42. yy

′′

+ (y

′

)

= 0 43. y

′′

+ y = 0

44. y

′′

+ y(y

′

)

= 0 45. 2y

′′

+ 2y(y

′

)

= 1

46. yy

′′

− (y

′

)

= 0 47. y

′′

+ (y

′

)

= 2e

−y

Hint: In Problem 47 the transformed equation is a Bernoulli equation. See Problem 27 in

Section 2.4.

In each of Problems 48 through 51, solve the given initial value problem using the methods of

Problems 36 through 47.

48. y

′

′′

= 2, y(0) = 1, y

′

(0) = 2

49. y

′′

− 3y

= 0, y(0) = 2, y

′

(0) = 4

50. (1 + t

′′

+ 2ty

′

+ 3t

−2

= 0, y(1) = 2, y

′

(1) =−1

51. y

′

′′

− t = 0, y(1) = 2, y

′

(1) = 1

REFERENCES The two books mentioned in Section 2.5 are

Bailey, N. T. J.,The Mathematical Theory of Infectious Diseases and Its Applications (2nd ed.) (New York:

Hafner Press, 1975).

Clark, Colin W., Mathematical Bioeconomics (2nd ed.) (New York: Wiley-Interscience, 1990).

A good introduction to population dynamics in general is

Frauenthal, J. C., Introduction to Population Modeling (Boston: Birkhauser, 1980).

A fuller discussion of the proof of the fundamental existence and uniqueness theorem can be found in

many more advanced books on differential equations. Two that are reasonably accessible to elementary

readers are

Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice-

Hall, 1961; New York: Dover, 1989).

Brauer,F.,and Nohel,J.,The Qualitative Theory of Ordinary Differential Equations (New York: Benjamin,

1969; New York: Dover, 1989).

A valuable compendium of methods for solving differential equations is

Zwillinger, D., Handbook of Differential Equations (3rd ed.) (San Diego:Academic Press, 1998).

For further discussion and examples of nonlinear phenomena, including bifurcations and chaos, see

Strogatz, Steven H., Nonlinear Dynamics and Chaos (Reading, MA:Addison-Wesley, 1994).

A general reference on difference equations is

Mickens, R. E., Difference Equations, Theory and Applications (2nd ed.) (New York: Van Nostrand

Reinhold, 1990).

Two papers by Robert May cited in the text are

R. M. May,“Biological Populations with Nonoverlapping Generations:Stable Points,Stable Cycles,and

Chaos,” Science 186 (1974), pp. 645–647; “Biological Populations Obeying Difference Equations: Stable

Points, Stable Cycles, and Chaos,” Journal of Theoretical Biology 51 (1975), pp. 511–524.

An elementary treatment of chaotic solutions of difference equations is

Devaney, R. L., Chaos, Fractals, and Dynamics (Reading, MA:Addison-Wesley, 1990).

August 7, 2012 21:04 c03 Sheet number 1 Page number 137 cyan black

137

CHAPTER

Second Order Linear

Equations

Linear equations of second order are of crucial importance in the study of differential

equations for two main reasons. The ﬁrst is that linear equations have a rich theo-

retical structure that underlies a number of systematic methods of solution. Further,

a substantial portion of this structure and of these methods is understandable at a

fairly elementary mathematical level. In order to present the key ideas in the sim-

plest possible context, we describe them in this chapter for second order equations.

Another reason to study second order linear equations is that they are vital to any

serious investigation of the classical areas of mathematical physics. One cannot go

very far in the development of ﬂuid mechanics, heat conduction, wave motion, or

electromagnetic phenomena without ﬁnding it necessary to solve second order lin-

ear differential equations. As an example, we discuss the oscillations of some basic

mechanical and electrical systems at the end of the chapter.

3.1 Homogeneous Equations with Constant Coefﬁcients

A second order ordinary differential equation has the form

= f

t, y,

(

, (1)

where f is some given function. Usually, we will denote the independent variable

by t since time is often the independent variable in physical problems, but some-

times we will use x instead. We will use y, or occasionally some other letter, to

designate the dependent variable. Equation (1) is said to be linear if the function f

August 7, 2012 21:04 c03 Sheet number 2 Page number 138 cyan black

138 Chapter 3. Second Order Linear Equations

has the form

t, y,

(

= g(t) − p(t)

− q(t)y, (2)

that is, if f is linear in y and dy/dt. In Eq. (2) g, p, and q are speciﬁed functions of

the independent variable t but do not depend on y. In this case we usually rewrite

Eq. (1) as

′′

+ p(t)y

′

+ q(t)y = g(t), (3)

where the primes denote differentiation with respect to t. Instead of Eq. (3),we often

see the equation

P(t)y

′′

+ Q(t)y

′

+ R(t)y = G(t). (4)

Of course, if P(t) ̸= 0, we can divide Eq. (4) by P(t) and thereby obtain Eq. (3) with

p(t) =

Q(t)

P(t)

, q(t) =

R(t)

P(t)

, g(t) =

G(t)

P(t)

. (5)

In discussing Eq. (3) and in trying to solve it, we will restrict ourselves to intervals in

which p, q, and g are continuous functions.

If Eq. (1) is not of the form (3) or (4), then it is called nonlinear. Analytical investi-

gations of nonlinear equationsare relativelydifﬁcult,so we willhave littleto sayabout

them in this book. Numerical or geometical approaches are often more appropriate,

and these are discussed in Chapters 8 and 9.

An initial value problem consists of a differential equation such as Eq. (1), (3), or

(4) together with a pair of initial conditions

y(t

) = y

, y

′

) = y

′

, (6)

where y

and y

′

are given numbers prescribing values for y and y

′

at the initial

point t

. Observe that the initial conditions for a second order equation identify not

only a particular point (t

, y

) through which the graph of the solution must pass,

but also the slope y

′

of the graph at that point. It is reasonable to expect that two

initial conditions are needed for a second order equation because, roughly speaking,

two integrations are required to ﬁnd a solution and each integration introduces an

arbitrary constant. Presumably, two initial conditions will sufﬁce to determine values

for these two constants.

A second order linear equation is said to be homogeneous if the term g(t) in

Eq. (3), or the term G(t) in Eq. (4), is zero for all t. Otherwise, the equation is

called nonhomogeneous. As a result, the term g(t),orG(t), is sometimes called the

nonhomogeneous term.We beginour discussion with homogeneous equations,which

we will write in the form

P(t)y

′′

+ Q(t)y

′

+ R(t)y = 0. (7)

Later, in Sections 3.5 and 3.6, we will show that once the homogeneous equation

has been solved, it is always possible to solve the corresponding nonhomogeneous

There is a corresponding treatment of higher order linear equations in Chapter 4. If you wish, you may

read the appropriate parts of Chapter 4 in parallel with Chapter 3.

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3.1 Homogeneous Equations with Constant Coefﬁcients 139

equation (4), or at least to express the solution in terms of an integral. Thus the

problem of solving the homogeneous equation is the more fundamental one.

In this chapter we will concentrate our attention on equations in which the

functions P, Q, and R are constants. In this case, Eq. (7) becomes

′′

+ by

′

+ cy = 0, (8)

where a, b, and c are given constants. It turns out that Eq. (8) can always be solved

easily in terms of the elementary functions of calculus. On the other hand, it is usu-

ally much more difﬁcult to solve Eq. (7) if the coefﬁcients are not constants, and

a treatment of that case is deferred until Chapter 5. Before taking up Eq. (8), let

us ﬁrst gain some experience by looking at a simple example that in many ways is

typical.

EXAMPLE

Solve the equation

′′

− y = 0, (9)

and also ﬁnd the solution that satisﬁes the initial conditions

y(0) = 2, y

′

(0) =−1. (10)

Observe that Eq. (9) is just Eq. (8) with a = 1, b = 0, and c =−1. In words, Eq. (9) says that

we seek a function with the property that the second derivative of the function is the same as the

function itself. Do any of the functions that you studied in calculus have this property? A little

thought will probably produce at least one such function, namely, y

(t) = e

, the exponential

function. A little more thought may also produce a second function, y

(t) = e

−t

. Some further

experimentation reveals that constant multiples of these two solutions are also solutions. For

example, the functions 2e

and 5e

−t

also satisfy Eq. (9), as you can verify by calculating their

second derivatives. In the same way, the functions c

(t) = c

and c

(t) = c

−t

satisfy the

differential equation (9) for all values of the constants c

and c

Next, it is vital to notice that the sum of any two solutions of Eq. (9) is also a solution.

In particular, since c

(t) and c

(t) are solutions of Eq. (9) for any values of c

and c

, so is

the function

y = c

(t) + c

(t) = c

+ c

−t

. (11)

Again, this can be veriﬁed by calculating the second derivative y

′′

from Eq. (11). We have

′

= c

− c

−t

and y

′′

= c

+ c

−t

; thus y

′′

is the same as y, and Eq. (9) is satisﬁed.

Letus summarize what we havedone so far in thisexample.Oncewe notice that the functions

(t) = e

and y

(t) = e

−t

are solutions of Eq. (9),it follows that the general linear combination

(11) of these functions is also a solution. Since the coefﬁcients c

and c

in Eq. (11) are arbitrary,

this expression represents an inﬁnite family of solutions of the differential equation (9).

It is now possible to consider how to pick out a particular member of this inﬁnite family of

solutions that also satisﬁes a given set of initial conditions (10). In other words, we seek the

solution that passes through the point (0, 2) and at that point has the slope −1. First, we set

t = 0 and y = 2 in Eq. (11); this gives the equation

+ c

= 2. (12)

Next, we differentiate Eq. (11) with the result that

′

= c

− c

−t

August 7, 2012 21:04 c03 Sheet number 4 Page number 140 cyan black

140 Chapter 3. Second Order Linear Equations

Then, setting t = 0 and y

′

=−1, we obtain

− c

=−1. (13)

By solving Eqs. (12) and (13) simultaneously for c

and c

, we ﬁnd that

, c

. (14)

Finally, inserting these values in Eq. (11), we obtain

y =

−t

, (15)

the solution of the initial value problem consisting of the differential equation (9) and the

initial conditions (10).

What conclusions can we draw from the preceding example that will help us to

deal with the more general equation (8),

′′

+ by

′

+ cy = 0,

whose coefﬁcients a, b, and c are arbitrary (real) constants? In the ﬁrst place, in the

example the solutions were exponential functions. Further, once we had identiﬁed

two solutions, we were able to use a linear combination of them to satisfy the given

initial conditions as well as the differential equation itself.

It turns out that by exploiting these two ideas, we can solve Eq. (8) for any values

of its coefﬁcients and also satisfy any given set of initial conditions for y and y

′

We start by seeking exponential solutions of the form y = e

, where r is a parameter

to be determined. Then it follows that y

′

= re

and y

′′

= r

. By substituting these

expressions for y, y

′

, and y

′′

in Eq. (8), we obtain

(ar

+ br + c)e

= 0,

or, since e

̸= 0,

+ br + c = 0. (16)

Equation (16) is called the characteristic equation for the differential equation (8).

Its signiﬁcance lies in the fact that if r is a root of the polynomial equation (16), then

y = e

is a solution of the differential equation (8). Since Eq. (16) is a quadratic

equation with real coefﬁcients, it has two roots, which may be real and different, real

but repeated, or complex conjugates. We consider the ﬁrst case here and the latter

two cases in Sections 3.3 and 3.4.

Assuming that the roots of the characteristic equation (16) are real and different,

let them be denoted by r

and r

, where r

̸= r

. Then y

(t) = e

and y

(t) = e

are

two solutions of Eq. (8). Just as in Example 1, it now follows that

y = c

(t) + c

(t) = c

+ c

(17)

is also a solution of Eq. (8).To verify that this is so,we can differentiate the expression

in Eq. (17); hence

′

= c

+ c

(18)

and

′′

= c

+ c

. (19)

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3.1 Homogeneous Equations with Constant Coefﬁcients 141

Substituting these expressions for y, y

′

, and y

′′

in Eq. (8) and rearranging terms, we

obtain

′′

+ by

′

+ cy = c

(ar

+ br

+ c)e

+ c

(ar

+ br

+ c)e

. (20)

The quantities in the two sets of parentheses on the right-hand side of Eq. (20) are

zero because r

and r

are roots of Eq. (16);therefore, y as given by Eq. (17) is indeed

a solution of Eq. (8), as we wished to verify.

Now suppose that we want to ﬁnd the particular member of the family of solutions

(17) that satisﬁes the initial conditions (6)

y(t

) = y

, y

′

) = y

′

By substituting t = t

and y = y

in Eq. (17), we obtain

+ c

= y

. (21)

Similarly, setting t = t

and y

′

= y

′

in Eq. (18) gives

+ c

= y

′

. (22)

On solving Eqs. (21) and (22) simultaneously for c

and c

, we ﬁnd that

′

− y

− r

−r

, c

− y

′

− r

−r

. (23)

Recall that r

− r

̸= 0 so that the expressions in Eq. (23) always make sense. Thus,

no matter what initial conditions are assigned—that is, regardless of the values of

, y

, and y

′

in Eqs. (6)—it is always possible to determine c

and c

so that the

initial conditions are satisﬁed. Moreover, there is only one possible choice of c

and

for each set of initial conditions. With the values of c

and c

given by Eq. (23), the

expression (17) is the solution of the initial value problem

′′

+ by

′

+ cy = 0, y(t

) = y

, y

′

) = y

′

. (24)

It is possible to show, on the basis of the fundamental theorem cited in the next

section,that all solutions of Eq. (8) are included in the expression (17),at least for the

case in which the roots of Eq. (16) are real and different. Therefore, we call Eq. (17)

the general solution of Eq. (8). The fact that any possible initial conditions can be

satisﬁed by the proper choice of the constants in Eq. (17) makes more plausible the

idea that this expression does include all solutions of Eq. (8).

Let us now look at some further examples.

EXAMPLE

Find the general solution of

′′

+ 5y

′

+ 6y = 0. (25)

We assume that y = e

, and it then followsthat r must bea root ofthe characteristicequation

+ 5r + 6 = (r + 2)(r + 3) = 0.

Thus the possible values of r are r

=−2 and r

=−3; the general solution of Eq. (25) is

y = c

−2t

+ c

−3t

. (26)

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142 Chapter 3. Second Order Linear Equations

EXAMPLE

Find the solution of the initial value problem

′′

+ 5y

′

+ 6y = 0, y(0) = 2, y

′

(0) = 3. (27)

The general solution of the differential equation was found in Example 2 and is given by

Eq. (26). To satisfy the ﬁrst initial condition, we set t = 0 and y = 2 in Eq. (26); thus c

and c

must satisfy

+ c

= 2. (28)

To use the second initial condition, we must ﬁrst differentiate Eq. (26). This gives

′

=−2c

−2t

− 3c

−3t

. Then, setting t = 0 and y

′

= 3, we obtain

−2c

− 3c

= 3. (29)

By solving Eqs. (28) and (29), we ﬁnd that c

= 9 and c

=−7. Using these values in the

expression (26), we obtain the solution

y = 9e

−2t

− 7e

−3t

(30)

of the initial value problem (27). The graph of the solution is shown in Figure 3.1.1.

0.5 1 1.5 2

y = 9e

–2t

– 7e

–3t

FIGURE 3.1.1 Solution of the initial value problem (27):

′′

+ 5y

′

+ 6y = 0, y(0) = 2, y

′

(0) = 3.

EXAMPLE

Find the solution of the initial value problem

′′

− 8y

′

+ 3y = 0, y(0) = 2, y

′

(0) =

. (31)

If y = e

, then we obtain the characteristic equation

− 8r + 3 = 0

whose roots are r = 3/2 and r = 1/2. Therefore, the general solution of the differential

equation is

y = c

3t/2

+ c

t/2

. (32)

Applying the initial conditions, we obtain the following two equations for c

and c

+ c

= 2,

The solution of these equations is c

=−

, c

, so the solution of the initial value problem

(31) is

y =−

3t/2

t/2

. (33)

Figure 3.1.2 shows the graph of the solution.

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3.1 Homogeneous Equations with Constant Coefﬁcients 143

–1

0.5 1 1.5 2

y = – e

3t/2

+ e

t/2

FIGURE 3.1.2 Solution of the initial value problem (31):

′′

− 8y

′

+ 3y = 0, y(0) = 2, y

′

(0) = 0.5.

EXAMPLE

The solution (30) of the initial value problem (27) initially increases (because its initial slope

is positive) but eventually approaches zero (because both terms involve negative exponential

functions). Therefore, the solution must have a maximum point, and the graph in Figure 3.1.1

conﬁrms this. Determine the location of this maximum point.

The coordinates of the maximum point can be estimated from the graph, but to ﬁnd

them more precisely, we seek the point where the solution has a horizontal tangent line.

By differentiating the solution (30), y = 9e

−2t

− 7e

−3t

, with respect to t, we obtain

′

=−18e

−2t

+ 21e

−3t

. (34)

Setting y

′

equal to zero and multiplying by e

,we ﬁnd that the critical value t

satisﬁes e

= 7/6;

hence

= ln(7/6)

∼

0.15415. (35)

The corresponding maximum value y

is given by

= 9e

−2t

− 7e

−3t

108

∼

2.20408. (36)

In this example the initial slope is 3, but the solution of the given differential equation

behaves in a similar way for any other positive initial slope. In Problem 26 you are asked to

determine how the coordinates of the maximum point depend on the initial slope.

Returning to the equation ay

′′

+ by

′

+ cy = 0 with arbitrary coefﬁcients, recall

that when r

̸= r

, its general solution (17) is the sum of two exponential functions.

Therefore, the solution has a relatively simple geometrical behavior: as t increases,

the magnitude of the solution either tends to zero (when both exponents are nega-

tive) or else grows rapidly (when at least one exponent is positive). These two cases

are illustrated by the solutions of Examples 3 and 4, which are shown in Figures 3.1.1

and 3.1.2, respectively. There is also a third case that occurs less often: the solution

approaches a constant when one exponent is zero and the other is negative.

In Sections 3.3 and 3.4, respectively, we return to the problem of solving the

equation ay

′′

+ by

′

+ cy = 0 when the roots of the characteristic equation either

are complex conjugates or are real and equal. In the meantime, in Section 3.2,

we provide a systematic discussion of the mathematical structure of the solutions

of all second order linear homogeneous equations.

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144 Chapter 3. Second Order Linear Equations

PROBLEMS In each of Problems 1 through 8, ﬁnd the general solution of the given differential equation.

1. y

′′

+ 2y

′

− 3y = 02.y

′′

+ 3y

′

+ 2y = 0

3. 6y

′′

− y

′

− y = 04.2y

′′

− 3y

′

+ y = 0

5. y

′′

+ 5y

′

= 06.4y

′′

− 9y = 0

7. y

′′

− 9y

′

+ 9y = 08.y

′′

− 2y

′

− 2y = 0

In each of Problems 9 through 16, ﬁnd the solution of the given initial value problem. Sketch

the graph of the solution and describe its behavior as t increases.

9. y

′′

+ y

′

− 2y = 0, y(0) = 1, y

′

(0) = 1

10. y

′′

+ 4y

′

+ 3y = 0, y(0) = 2, y

′

(0) =−1

11. 6y

′′

− 5y

′

+ y = 0, y(0) = 4, y

′

(0) = 0

12. y

′′

+ 3y

′

= 0, y(0) =−2, y

′

(0) = 3

13. y

′′

+ 5y

′

+ 3y = 0, y(0) = 1, y

′

(0) = 0

14. 2y

′′

+ y

′

− 4y = 0, y(0) = 0, y

′

(0) = 1

15. y

′′

+ 8y

′

− 9y = 0, y(1) = 1, y

′

(1) = 0

16. 4y

′′

− y = 0, y(−2) = 1, y

′

(−2) =−1

17. Find a differential equation whose general solution is y = c

+ c

−3t

18. Find a differential equation whose general solution is y = c

−t/2

+ c

−2t

19.

Find the solution of the initial value problem

′′

− y = 0, y(0) =

, y

′

(0) =−

Plot the solution for 0 ≤ t ≤ 2 and determine its minimum value.

20. Find the solution of the initial value problem

′′

− 3y

′

+ y = 0, y(0) = 2, y

′

(0) =

Then determine the maximum value of the solution and also ﬁnd the point where the

solution is zero.

21. Solve the initial value problem y

′′

− y

′

− 2y = 0, y(0) = α, y

′

(0) = 2. Then ﬁnd α so that

the solution approaches zero as t →∞.

22. Solve the initial value problem 4y

′′

− y = 0, y(0) = 2, y

′

(0) = β. Then ﬁnd β so that the

solution approaches zero as t →∞.

In each of Problems 23 and 24, determine the values of α, if any, for which all solutions tend to

zero as t →∞; also determine the values of α,if any, for which all (nonzero) solutions become

unbounded as t →∞.

23. y

′′

− (2α − 1)y

′

+ α(α − 1)y = 0

24. y

′′

+ (3 − α)y

′

− 2(α − 1)y = 0

25.

Consider the initial value problem

′′

+ 3y

′

− 2y = 0, y(0) = 1, y

′

(0) =−β,

where β>0.

(a) Solve the initial value problem.

(b) Plot the solution when β = 1. Find the coordinates (t

, y

) of the minimum point of

the solution in this case.

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3.2 Solutions of Linear Homogeneous Equations; the Wronskian 145

26. Consider the initial value problem (see Example 5)

′′

+ 5y

′

+ 6y = 0, y(0) = 2, y

′

(0) = β,

where β>0.

(a) Solve the initial value problem.

(b) Determine the coordinates t

and y

of the maximum point of the solution as func-

tions of β.

≥ 4.

(d) Determine the behavior of t

and y

as β →∞.

27. Consider the equation ay

′′

+ by

′

+ cy = d, where a, b, c, and d are constants.

(a) Find all equilibrium, or constant, solutions of this differential equation.

(b) Let y

denote an equilibrium solution, and let Y = y − y

. Thus Y is the deviation of

a solution y from an equilibrium solution. Find the differential equation satisﬁed by Y.

28. Consider the equation ay

′′

+ by

′

+ cy = 0, where a, b, and c are constants with a > 0. Find

conditions on a, b, and c such that the roots of the characteristic equation are:

(a) real, different, and negative.

(b) real with opposite signs.

3.2 Solutions of Linear Homogeneous Equations; the Wronskian

In the preceding section we showed how to solve some differential equations of the

form

′′

+ by

′

+ cy = 0,

where a, b, and c are constants. Now we build on those results to provide a clearer

picture of the structure of the solutions of all second order linear homogeneous

equations. In turn, this understanding will assist us in ﬁnding the solutions of other

problems that we will encounter later.

To discuss general properties of linear differential equations, it is helpful to intro-

duce a differential operator notation. Let p and q be continuous functions on an

open interval I—that is, for α<t <β. The cases for α =−∞,orβ =∞, or both,

are included. Then, for any function φ that is twice differentiable on I, we deﬁne the

differential operator L by the equation

L[φ]=φ

′′

+ pφ

′

+ qφ. (1)

Note that L[φ] is a function on I. The value of L[φ] at a point t is

L[φ](t) = φ

′′

(t) + p(t)φ

′

(t) + q(t)φ(t).

For example, if p(t) = t

, q(t) = 1 + t, and φ(t) = sin 3t, then

L[φ](t) = (sin 3t)

′′

+ t

(sin 3t)

′

+ (1 + t) sin 3t

=−9 sin 3t + 3t

cos 3t + (1 + t) sin 3t.

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146 Chapter 3. Second Order Linear Equations

The operator L is often written as L = D

+ pD + q, where D is the derivative

operator.

In this section we study the second order linear homogeneous equation

L[φ](t) = 0. Since it is customary to use the symbol y to denote φ(t), we will usually

write this equation in the form

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0. (2)

With Eq. (2) we associate a set of initial conditions

y(t

) = y

, y

′

) = y

′

, (3)

where t

is any point in the interval I, and y

and y

′

are given real numbers.

We would like to know whether the initial value problem (2), (3) always has a solu-

tion, and whether it may have more than one solution. We would also like to know

whether anything can be said about the form and structure of solutions that might

be helpful in ﬁnding solutions of particular problems. Answers to these questions are

contained in the theorems in this section.

The fundamental theoretical result for initial value problems for second order

linear equations is stated in Theorem 3.2.1, which is analogous to Theorem 2.4.1

for ﬁrst order linear equations. The result applies equally well to nonhomogeneous

equations, so the theorem is stated in that form.

Theorem 3.2.1

(Existence and Uniqueness Theorem)

Consider the initial value problem

′′

+ p(t)y

′

+ q(t)y = g(t), y(t

) = y

, y

′

) = y

′

, (4)

where p, q, and g are continuous on an open interval I that contains the point t

Then there is exactly one solution y = φ(t) of this problem, and the solution exists

throughout the interval I.

We emphasize that the theorem says three things:

1. The initial value problem has a solution; in other words, a solution exists.

2. The initial value problem has only one solution; that is, the solution is unique.

3. The solution φ is deﬁned throughout the interval I where the coefﬁcients are continuous

and is at least twice differentiable there.

For some problems some of these assertions are easy to prove. For instance, we

found in Example 1 of Section 3.1 that the initial value problem

′′

− y = 0, y(0) = 2, y

′

(0) =−1 (5)

has the solution

y =

−t

. (6)

The fact that we found a solution certainly establishes that a solution exists for this

initial value problem. Further,the solution (6) is twice differentiable,indeed differen-

tiable any number of times, throughout the interval (−∞, ∞) where the coefﬁcients

in the differential equation are continuous. On the other hand, it is not obvious, and

is more difﬁcult to show, that the initial value problem (5) has no solutions other

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3.2 Solutions of Linear Homogeneous Equations; the Wronskian 147

than the one given by Eq. (6). Nevertheless, Theorem 3.2.1 states that this solution

is indeed the only solution of the initial value problem (5).

For most problems of the form (4), it is not possible to write down a useful expres-

sion for the solution. This is a major difference between ﬁrst order and second order

linear equations.Therefore,all parts of the theorem must be proved by general meth-

ods that do not involve having such an expression.The proof ofTheorem 3.2.1 is fairly

difﬁcult, and we do not discuss it here.

We will, however, accept Theorem 3.2.1 as

true and make use of it whenever necessary.

EXAMPLE

Find the longest interval in which the solution of the initial value problem

− 3t)y

′′

+ ty

′

− (t + 3)y = 0, y(1) = 2, y

′

(1) = 1

is certain to exist.

If the given differential equation is written in the form of Eq. (4), then p(t) = 1/(t − 3),

q(t) =−(t + 3)/t(t − 3), and g(t) = 0. The only points of discontinuity of the coefﬁcients are

t = 0 and t = 3. Therefore, the longest open interval, containing the initial point t = 1, in

which all the coefﬁcients are continuous is 0 < t < 3. Thus, this is the longest interval in which

Theorem 3.2.1 guarantees that the solution exists.

EXAMPLE

Find the unique solution of the initial value problem

′′

+ p(t)y

′

+ q(t)y = 0, y(t

) = 0, y

′

) = 0,

where p and q are continuous in an open interval I containing t

The function y = φ(t) = 0 for all t in I certainly satisﬁes the differential equation and ini-

tial conditions. By the uniqueness part of Theorem 3.2.1, it is the only solution of the given

problem.

Let us now assume that y

and y

are two solutions of Eq. (2); in other words,

L[y

]=y

′′

+ py

′

+ qy

= 0,

and similarly for y

. Then, just as in the examples in Section 3.1, we can generate

more solutions by forming linear combinations of y

and y

. We state this result as a

theorem.

Theorem 3.2.2

(Principle of Superposition)

If y

and y

are two solutions of the differential equation (2),

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0,

then the linear combination c

+ c

is also a solution for any values of the

constants c

and c

A proof of Theorem 3.2.1 can be found, for example, in Chapter 6, Section 8 of the book by Coddington

listed in the references at the end of this chapter.

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148 Chapter 3. Second Order Linear Equations

A special case of Theorem 3.2.2 occurs if either c

or c

is zero. Then we conclude

that any constant multiple of a solution of Eq. (2) is also a solution.

To prove Theorem 3.2.2, we need only substitute

y = c

(t) + c

(t) (7)

for y in Eq. (2). By calculating the indicated derivatives and rearranging terms, we

obtain

L[c

+ c

]=[c

+ c

]

′′

+ p[c

+ c

]

′

+ q[c

+ c

]

= c

′′

+ c

′′

+ c

′

+ c

′

+ c

= c

′′

+ py

′

+ qy

]+c

′′

+ py

′

+ qy

]

= c

L[y

]+c

L[y

Since L[y

]=0 and L[y

]=0, it follows that L[c

+ c

]=0 also. Therefore,

regardless of the values of c

and c

, y as given by Eq. (7) satisﬁes the differential

equation (2), and the proof of Theorem 3.2.2 is complete.

Theorem 3.2.2 states that, beginning with only two solutions of Eq. (2), we can

construct an inﬁnite family of solutions by means of Eq. (7). The next question is

whether all solutions of Eq. (2) are included in Eq. (7) or whether there may be

other solutions of a different form. We begin to address this question by examining

whether the constants c

and c

in Eq. (7) can be chosen so as to satisfy the initial

conditions (3). These initial conditions require c

and c

to satisfy the equations

) + c

) = y

(8)

′

) + c

′

) = y

′

The determinant of coefﬁcients of the system (8) is

W =

) y

)

′

) y

′

)

= y

′

) − y

′

). (9)

If W ̸= 0, then Eqs. (8) have a unique solution (c

, c

) regardless of the values of

and y

′

. This solution is given by

′

) − y

′

)

′

) − y

′

)

, c

−y

′

) + y

′

)

′

) − y

′

)

, (10)

or, in terms of determinants,

)

′

)

) y

)

′

) y

′

)

, c

) y

′

) y

′

) y

)

′

) y

′

)

. (11)

With these values for c

and c

, the linear combination y = c

(t) + c

(t) satis-

ﬁes the initial conditions (3) as well as the differential equation (2). Note that the

denominator in the expressions for c

and c

is the nonzero determinant W.

On the other hand, if W = 0, then the denominators appearing in Eqs. (10) and

(11) are zero. In this case Eqs. (8) have no solution unless y

and y

′

have values that

also make the numerators in Eqs. (10) and (11) equal to zero.Thus,if W = 0,there are

many initial conditions that cannot be satisﬁed no matter how c

and c

are chosen.

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3.2 Solutions of Linear Homogeneous Equations; the Wronskian 149

The determinantW iscalledtheWronskian

determinant,orsimply theWronskian,

of the solutionsy

and y

. Sometimes weuse themore extended notationW(y

, y

)(t

)

to stand for the expression on the right side of Eq. (9), thereby emphasizing that the

Wronskian depends on the functions y

and y

, and that it is evaluated at the point

. The preceding argument establishes the following result.

Theorem 3.2.3

Suppose that y

and y

are two solutions of Eq. (2)

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0,

and that the initial conditions (3)

y(t

) = y

, y

′

) = y

′

are assigned. Then it is always possible to choose the constants c

, c

so that

y = c

(t) + c

(t)

satisﬁes the differential equation (2) and the initial conditions (3) if and only if the

Wronskian

W = y

′

− y

′

is not zero at t

EXAMPLE

In Example 2 of Section 3.1 we found that y

(t) = e

−2t

and y

(t) = e

−3t

are solutions of the

differential equation

′′

+ 5y

′

+ 6y = 0.

Find the Wronskian of y

and y

The Wronskian of these two functions is

W =

−2t

−3t

−2e

−2t

−3e

−3t

=−e

−5t

Since W is nonzero for all values of t, the functions y

and y

can be used to construct solutions

of the given differential equation, together with initial conditions prescribed at any value of t.

One such initial value problem was solved in Example 3 of Section 3.1.

The next theorem justiﬁes the term “general solution” that we introduced in

Section 3.1 for the linear combination c

+ c

Theorem 3.2.4

Suppose that y

and y

are two solutions of the differential equation (2),

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0.

Then the family of solutions

y = c

(t) + c

(t)

with arbitrary coefﬁcients c

and c

includes every solution of Eq. (2) if and only if

there is a point t

where the Wronskian of y

and y

is not zero.

Wronskian determinants are named for Jósef Maria Hoëné-Wronski (1776–1853), who was born in Pol-

and but spent most of his life in France. Wronski was a gifted but troubled man, and his life was marked

by frequent heated disputes with other individuals and institutions.

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150 Chapter 3. Second Order Linear Equations

Let φ be any solution of Eq. (2).To prove the theorem, we must determine whether

φ is included in the linear combinations c

+ c

. That is, we must determine

whether there are values of the constants c

and c

that make the linear combi-

nation the same as φ. Let t

be a point where the Wronskian of y

and y

is nonzero.

Then evaluate φ and φ

′

at this point and call these values y

and y

′

, respectively; thus

= φ(t

), y

′

= φ

′

Next, consider the initial value problem

′′

+ p(t)y

′

+ q(t)y = 0, y(t

) = y

, y

′

) = y

′

. (12)

The function φ is certainly a solution of this initial value problem. Further, because

we are assuming that W(y

, y

)(t

) is nonzero, it is possible (by Theorem 3.2.3)

to choose c

and c

such that y = c

(t) + c

(t) is also a solution of the initial

value problem (12). In fact, the proper values of c

and c

are given by Eqs. (10) or

(11). The uniqueness part of Theorem 3.2.1 guarantees that these two solutions of

the same initial value problem are actually the same function; thus, for the proper

choice of c

and c

φ(t) = c

(t) + c

(t), (13)

and therefore φ is included in the family of functions c

+ c

. Finally,since φ is an

arbitrary solution of Eq. (2), it follows that every solution of this equation is included

in this family.

Now suppose that there is no point t

where the Wronskian is nonzero. Thus

W(y

, y

)(t

) = 0 no matter which point t

is selected. Then (by Theorem 3.2.3) there

are values of y

and y

′

such that the system (8) has no solution for c

and c

. Select

a pair of such values and choose the solution φ(t) of Eq. (2) that satisﬁes the initial

condition (3). Observe that such a solution is guaranteed to exist by Theorem 3.2.l.

However, this solution is not included in the family y = c

+ c

. Thus this linear

combination does not include all solutions of Eq. (2) if W(y

, y

) = 0.This completes

the proof of Theorem 3.2.4.

Theorem 3.2.4states that,if and onlyif theWronskian ofy

and y

is not everywhere

zero, then the linear combination c

+ c

contains all solutions of Eq. (2). It is

therefore natural (and we have already done this in the preceding section) to call the

expression

y = c

(t) + c

(t)

with arbitrary constant coefﬁcients the general solution of Eq. (2). The solutions

and y

are said to form a fundamental set of solutions of Eq. (2) if and only if their

Wronskian is nonzero.

We can restate the result of Theorem 3.2.4 in slightly different language: to ﬁnd

the general solution, and therefore all solutions, of an equation of the form (2), we

need only ﬁnd two solutions of the given equation whose Wronskian is nonzero.

We did precisely this in several examples in Section 3.1, although there we did not

calculate the Wronskians. You should now go back and do that, thereby verifying

that all the solutions we called “general solutions” in Section 3.1 do satisfy the nec-

essary Wronskian condition. Alternatively, the following example includes all those

mentioned in Section 3.1, as well as many other problems of a similar type.

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3.2 Solutions of Linear Homogeneous Equations; the Wronskian 151

EXAMPLE

Suppose that y

(t) = e

and y

(t) = e

are two solutions of an equation of the form (2). Show

that they form a fundamental set of solutions if r

̸= r

We calculate the Wronskian of y

and y

W =

= (r

− r

) exp[(r

+ r

)t].

Since the exponential function is never zero, and since we are assuming that r

− r

̸= 0, it

follows that W is nonzero for every value of t. Consequently, y

and y

form a fundamental

set of solutions.

EXAMPLE

Show that y

(t) = t

1/2

and y

(t) = t

−1

form a fundamental set of solutions of

′′

+ 3ty

′

− y = 0, t > 0. (14)

We will show how to solve Eq. (14) later (see Problem 34 in Section 3.3). However, at

this stage we can verify by direct substitution that y

and y

are solutions of the differential

equation. Since y

′

(t) =

−1/2

and y

′′

(t) =−

−3/2

, we have

(−

−3/2

) + 3t(

−1/2

) − t

1/2

= (−

− 1)t

1/2

= 0.

Similarly, y

′

(t) =−t

−2

and y

′′

(t) = 2t

−3

, so

(2t

−3

) + 3t(−t

−2

) − t

−1

= (4 − 3 − 1)t

−1

= 0.

Next we calculate the Wronskian W of y

and y

W =

1/2

−1

−1/2

−t

−2

=−

−3/2

. (15)

Since W ̸= 0 for t > 0, we conclude that y

and y

form a fundamental set of solutions there.

In several cases we have been able to ﬁnd a fundamental set of solutions, and

therefore the general solution,of a given differential equation. However,this is often

a difﬁcult task, and the question arises as to whether a differential equation of the

form (2) always has a fundamental set of solutions. The following theorem provides

an afﬁrmative answer to this question.

Theorem 3.2.5

Consider the differential equation (2),

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0,

whose coefﬁcients p and q are continuous on some open interval I. Choose some

point t

in I. Let y

be the solution of Eq. (2) that also satisﬁes the initial conditions

y(t

) = 1, y

′

) = 0,

and let y

be the solution of Eq. (2) that satisﬁes the initial conditions

y(t

) = 0, y

′

) = 1.

Then y

and y

form a fundamental set of solutions of Eq. (2).

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152 Chapter 3. Second Order Linear Equations

First observe that the existence of the functions y

and y

is ensured by the existence

part ofTheorem 3.2.1.To show that they form a fundamental set of solutions,we need

only calculate their Wronskian at t

W(y

, y

)(t

) =

) y

)

′

) y

′

)

= 1.

Since their Wronskian is not zero at the point t

, the functions y

and y

do form a

fundamental set of solutions, thus completing the proof of Theorem 3.2.5.

Note that the potentially difﬁcult part of this proof, demonstrating the existence

of a pair of solutions, is taken care of by reference to Theorem 3.2.1. Note also that

Theorem 3.2.5 does not address the question of how to ﬁnd the solutions y

and y

by solving the speciﬁed initial value problems. Nevertheless, it may be reassuring to

know that a fundamental set of solutions always exists.

EXAMPLE

Find the fundamental set of solutions y

and y

speciﬁed by Theorem 3.2.5 for the differential

equation

′′

− y = 0, (16)

using the initial point t

= 0.

In Section 3.1 we noted that two solutions of Eq. (16) are y

(t) = e

and y

(t) = e

−t

.The

Wronskian of these solutions is W(y

, y

)(t) =−2 ̸= 0, so they form a fundamental set of

solutions. However,they are not the fundamental solutions indicated byTheorem 3.2.5 because

they do not satisfy the initial conditions mentioned in that theorem at the point t = 0.

To ﬁnd the fundamental solutions speciﬁed by the theorem, we need to ﬁnd the solutions

satisfying the proper initial conditions. Let us denote by y

(t) the solution of Eq. (16) that

satisﬁes the initial conditions

y(0) = 1, y

′

(0) = 0. (17)

The general solution of Eq. (16) is

y = c

+ c

−t

, (18)

and the initial conditions (17) are satisﬁed if c

= 1/2 and c

= 1/2. Thus

(t) =

−t

= cosh t.

Similarly, if y

(t) satisﬁes the initial conditions

y(0) = 0, y

′

(0) = 1, (19)

then

(t) =

−

−t

= sinh t.

Since the Wronskian of y

and y

W(y

, y

)(t) = cosh

t − sinh

t = 1,

August 7, 2012 21:04 c03 Sheet number 17 Page number 153 cyan black

3.2 Solutions of Linear Homogeneous Equations; the Wronskian 153

these functions also form a fundamental set of solutions,as stated by Theorem 3.2.5. Therefore,

the general solution of Eq. (16) can be written as

y = k

cosh t + k

sinh t, (20)

as well as in the form (18). We have used k

and k

for the arbitrary constants in Eq. (20)

because they are not the same as the constants c

and c

in Eq. (18). One purpose of this

example is to make it clear that a given differential equation has more than one fundamental

set of solutions; indeed, it has inﬁnitely many; see Problem 21. As a rule, you should choose

the set that is most convenient.

In the next section we will encounter equations that have complex-valued solu-

tions. The following theorem is fundamental in dealing with such equations and their

solutions.

Theorem 3.2.6

Consider again the equation (2),

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0,

where p and q are continuous real-valued functions. If y = u(t) + iv(t) is a complex-

valued solution of Eq. (2), then its real part u and its imaginary part v are also

solutions of this equation.

To prove this theorem we substitute u(t) + iv(t) for y in L[y], obtaining

L[y]=u

′′

(t) + iv

′′

(t) + p(t)[u

′

(t) + iv

′

(t)]+q(t)[u(t) + iv(t)]. (21)

Then, by separating Eq. (21) into its real and imaginary parts (and this is where we

need to know that p(t) and q(t) are real-valued), we ﬁnd that

L[y]=u

′′

(t) + p(t)u

′

(t) + q(t)u(t) + i[v

′′

(t) + p(t)v

′

(t) + q(t)v(t)]

= L[u](t) + iL[v](t).

Recall that a complex number is zero if and only if its real and imaginary parts

are both zero. We know that L[y]=0 because y is a solution of Eq. (2). Therefore,

L[u](t) = 0 and L[v](t) = 0 also; consequently, u and v are also solutions of Eq. (2),

so the theorem is established. We will see examples of the use of Theorem 3.2.6 in

Section 3.3.

Incidentally, the complex conjugate

y of a solution y is also a solution. This is a

consequence of Theorem 3.2.2 since

y = u(t) − iv(t) is a linear combination of two

solutions.

Now let us examine further the properties of the Wronskian of two solutions

of a second order linear homogeneous differential equation. The following the-

orem, perhaps surprisingly, gives a simple explicit formula for the Wronskian of

any two solutions of any such equation, even if the solutions themselves are not

known.

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154 Chapter 3. Second Order Linear Equations

Theorem 3.2.7

(Abel’s Theorem)

If y

and y

are solutions of the differential equation

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0, (22)

where p and q are continuous on an open interval I,then theWronskian W(y

, y

)(t)

is given by

W(y

, y

)(t) = c exp

−

p(t) dt

, (23)

where c is a certain constant that depends on y

and y

, but not on t. Further,

W(y

, y

)(t) either is zero for all t in I (if c = 0) or else is never zero in I (if c ̸= 0).

To prove Abel’s theorem, we start by noting that y

and y

satisfy

′′

+ p(t)y

′

+ q(t)y

= 0,

(24)

′′

+ p(t)y

′

+ q(t)y

= 0.

If we multiply the ﬁrst equation by −y

, multiply the second by y

, and add the

resulting equations, we obtain

′′

− y

′′

) + p(t)(y

′

− y

′

) = 0. (25)

Next, we let W(t) = W(y

, y

)(t) and observe that

′

= y

′′

− y

′′

. (26)

Then we can write Eq. (25) in the form

′

+ p(t)W = 0. (27)

Equation (27) can be solved immediately since it is both a ﬁrst order linear equation

(Section 2.1) and a separable equation (Section 2.2). Thus

W(t) = c exp

−

p(t) dt

, (28)

where c is a constant. The value of c depends on which pair of solutions of Eq. (22)

is involved. However, since the exponential function is never zero, W(t) is not

zero unless c = 0, in which case W(t) is zero for all t. This completes the proof of

Theorem 3.2.7.

Note that the Wronskians of any two fundamental sets of solutions of the same dif-

ferential equation can differ only by a multiplicative constant,andthat theWronskian

of any fundamentalset of solutionscan be determined,up toa multiplicative constant,

The result in Theorem 3.2.7 was derived by the Norwegian mathematician Niels Henrik Abel (1802–

1829) in 1827 and is known as Abel’s formula. Abel also showed that there is no general formula for

solving a quintic, or ﬁfth degree, polynomial equation in terms of explicit algebraic operations on the

coefﬁcients, thereby resolving a question that had been open since the sixteenth century. His greatest

contributions, however, were in analysis, particularly in the study of elliptic functions. Unfortunately, his

work was not widely noticed until after his death. The distinguished French mathematician Legendre

called it a “monument more lasting than bronze.”

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3.2 Solutions of Linear Homogeneous Equations; the Wronskian 155

without solving the differential equation. Further,since under the conditions of The-

orem 3.2.7 the Wronskian W is either always zero or never zero, you can determine

which case actually occurs by evaluating W at any single convenient value of t.

EXAMPLE

In Example 5 we veriﬁed that y

(t) = t

1/2

and y

(t) = t

−1

are solutions of the equation

′′

+ 3ty

′

− y = 0, t > 0. (29)

Verify that the Wronskian of y

and y

is given by Eq. (23).

From the example just cited we know that W(y

, y

)(t) =−(3/2)t

−3/2

. To use Eq. (23), we

must write the differential equation (29) in the standard form with the coefﬁcient of y

′′

equal

to 1. Thus we obtain

′′

′

−

y = 0,

so p(t) = 3/2t. Hence

W(y

, y

)(t) = c exp

−

= c exp

−

ln t

(

= ct

−3/2

. (30)

Equation (30) gives the Wronskian of any pair of solutions of Eq. (29). For the particular

solutions given in this example, we must choose c =−3/2.

Summar y. We can summarize the discussion in this section as follows: to ﬁnd the

general solution of the differential equation

′′

+ p(t)y

′

+ q(t)y = 0, α<t <β,

we must ﬁrst ﬁnd two functions y

and y

that satisfy the differential equation in

α<t <β. Then we must make sure that there is a point in the interval where the

Wronskian W of y

and y

is nonzero. Under these circumstances y

and y

form a

fundamental set of solutions, and the general solution is

y = c

(t) + c

(t),

where c

and c

are arbitrary constants. If initial conditions are prescribed at a point

in α<t <β, then c

and c

can be chosen so as to satisfy these conditions.

PROBLEMS In each of Problems 1 through 6, ﬁnd the Wronskian of the given pair of functions.

1. e

, e

−3t/2

2. cos t, sin t

3. e

−2t

, te

−2t

4. x, xe

5. e

sin t, e

cos t 6. cos

θ,1+ cos 2θ

In each of Problems 7 through 12, determine the longest interval in which the given initial

value problem is certain to have a unique twice-differentiable solution. Do not attempt to ﬁnd

the solution.

7. ty

′′

+ 3y = t, y(1) = 1, y

′

(1) = 2

8. (t − 1)y

′′

− 3ty

′

+ 4y = sin t, y(−2) = 2, y

′

(−2) = 1

9. t(t − 4)y

′′

+ 3ty

′

+ 4y = 2, y(3) = 0, y

′

(3) =−1

10. y

′′

+ (cos t)y

′

+ 3(ln |t|)y = 0, y(2) = 3, y

′

(2) = 1

August 7, 2012 21:04 c03 Sheet number 20 Page number 156 cyan black

156 Chapter 3. Second Order Linear Equations

11. (x − 3)y

′′

+ xy

′

+ (ln |x|)y = 0, y(1) = 0, y

′

(1) = 1

12. (x − 2)y

′′

+ y

′

+ (x − 2)(tan x)y = 0, y(3) = 1, y

′

(3) = 2

13. Verify that y

(t) = t

and y

(t) = t

−1

are two solutions of the differential equation

′′

− 2y = 0 for t > 0. Then show that y = c

+ c

−1

is also a solution of this equation

for any c

and c

14. Verify that y

(t) = 1 and y

(t) = t

1/2

are solutions of the differential equation

′′

+ (y

′

)

= 0 for t > 0.Then show that y = c

+ c

1/2

is not, in general, a solution of this

equation. Explain why this result does not contradict Theorem 3.2.2.

15. Show that if y = φ(t) is a solution of the differential equation y

′′

+ p(t)y

′

+ q(t)y = g(t),

where g(t) is not always zero, then y = cφ(t), where c is any constant other than 1, is not a

solution. Explain why this result does not contradict the remark following Theorem 3.2.2.

16. Can y = sin(t

) be a solution on an interval containing t = 0 of an equation

′′

+ p(t)y

′

+ q(t)y = 0 with continuous coefﬁcients? Explain your answer.

17. If the Wronskian W of f and g is 3e

, and if f (t) = e

, ﬁnd g(t).

18. If the Wronskian W of f and g is t

, and if f (t) = t, ﬁnd g(t).

19. If W(f , g) is the Wronskian of f and g, and if u = 2f − g, v = f + 2g, ﬁnd the Wronskian

W(u, v) of u and v in terms of W(f , g).

20. If the Wronskian of f and g is t cos t − sin t, and if u = f + 3g, v = f − g, ﬁnd the

Wronskian of u and v.

21. Assume that y

and y

are a fundamental set of solutions of y

′′

+ p(t)y

′

+ q(t)y = 0 and let

= a

+ a

and y

= b

+ b

,where a

,and b

are any constants. Show that

W(y

, y

) = (a

− a

)W(y

, y

Are y

and y

also a fundamental set of solutions? Why or why not?

In each of Problems 22 and 23, ﬁnd the fundamental set of solutions speciﬁed by Theorem

3.2.5 for the given differential equation and initial point.

22. y

′′

+ y

′

− 2y = 0, t

= 0

23. y

′′

+ 4y

′

+ 3y = 0, t

= 1

In each of Problems 24 through 27,verify that the functions y

and y

are solutions of the given

differential equation. Do they constitute a fundamental set of solutions?

24. y

′′

+ 4y = 0; y

(t) = cos 2t, y

(t) = sin 2t

25. y

′′

− 2y

′

+ y = 0; y

(t) = e

, y

(t) = te

26. x

′′

− x(x + 2)y

′

+ (x + 2)y = 0, x > 0; y

(x) = x, y

(x) = xe

27. (1 − x cot x)y

′′

− xy

′

+ y = 0, 0 < x <π; y

(x) = x, y

(x) = sin x

28. Consider the equation y

′′

− y

′

− 2y = 0.

(a) Show that y

(t) = e

−t

and y

(t) = e

form a fundamental set of solutions.

(b) Let y

(t) =−2e

, y

(t) = y

(t) + 2y

(t), and y

(t) = 2y

(t) − 2y

(t). Are y

(t), y

(t),

and y

(t) also solutions of the given differential equation?

(t), y

(t)]; [y

(t), y

(t)]; [y

(t), y

(t)]; [y

(t), y

(t)].

In each of Problems 29 through 32,ﬁnd theWronskian of two solutions of the given differential

equation without solving the equation.

29. t

′′

− t(t + 2)y

′

+ (t + 2)y = 0 30. (cos t)y

′′

+ (sin t)y

′

− ty = 0

31. x

′′

+ xy

′

+ (x

− ν

)y = 0, Bessel’s equation

32. (1 − x

′′

− 2xy

′

+ α(α + 1)y = 0, Legendre’s equation

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3.2 Solutions of Linear Homogeneous Equations; the Wronskian 157

33. Show that if p is differentiable and p(t)>0, then the Wronskian W(t) of two solutions of

[p(t)y

′

]

′

+ q(t)y = 0 is W(t) = c/p(t), where c is a constant.

34. If the differential equation ty

′′

+ 2y

′

+ te

y = 0 has y

and y

as a fundamental set of

solutions and if W(y

, y

)(1) = 2, ﬁnd the value of W(y

, y

)(5).

35. If the differential equation t

′′

− 2y

′

+ (3 + t)y = 0 has y

and y

as a fundamental set of

solutions and if W(y

, y

)(2) = 3, ﬁnd the value of W(y

, y

)(4).

36. If the Wronskian of any two solutions of y

′′

+ p(t)y

′

+ q(t)y = 0 is constant,what does this

imply about the coefﬁcients p and q?

37. If f , g, and h are differentiable functions, show that W(fg, fh) = f

W(g, h).

In Problems 38 through 40, assume that p and q are continuous and that the functions y

and

are solutions of the differential equation y

′′

+ p(t)y

′

+ q(t)y = 0 on an open interval I.

38. Prove that if y

and y

are zero at the same point in I, then they cannot be a fundamental

set of solutions on that interval.

39. Prove that if y

and y

have maxima or minima at the same point in I, then they cannot

be a fundamental set of solutions on that interval.

40. Prove that if y

and y

have a common point of inﬂection t

in I, then they cannot be a

fundamental set of solutions on I unless both p and q are zero at t

41. Exact Equations. The equation

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0

is said to be exact if it can be written in the form

[P(x)y

′

]

′

+[f (x)y]

′

= 0,

where f (x) is to be determined in terms of P(x), Q(x), and R(x). The latter equation can

be integrated once immediately, resulting in a ﬁrst order linear equation for y that can be

solved as in Section 2.1. By equating the coefﬁcients of the preceding equations and then

eliminating f (x), show that a necessary condition for exactness is

′′

(x) − Q

′

(x) + R(x) = 0.

It can be shown that this is also a sufﬁcient condition.

In each of Problems 42 through 45, use the result of Problem 41 to determine whether the

given equation is exact. If it is, then solve the equation.

42. y

′′

+ xy

′

+ y = 0 43. y

′′

+ 3x

′

+ xy = 0

44. xy

′′

− (cos x)y

′

+ (sin x)y = 0, x > 0 45. x

′′

+ xy

′

− y = 0, x > 0

46. The Adjoint Equation. If a second order linearhomogeneous equationis not exact,it can

be made exact by multiplying by an appropriate integrating factor µ(x). Thus we require

that µ(x) be such that

µ(x)P(x)y

′′

+ µ(x)Q(x)y

′

+ µ(x)R(x)y = 0

can be written in the form

[µ(x)P(x)y

′

]

′

+[f (x )y]

′

= 0.

By equating coefﬁcients in these two equations and eliminating f (x), show that the

function µ must satisfy

Pµ

′′

+ (2P

′

− Q)µ

′

+ (P

′′

− Q

′

+ R)µ = 0.

August 7, 2012 21:04 c03 Sheet number 22 Page number 158 cyan black

158 Chapter 3. Second Order Linear Equations

This equation is known as the adjoint of the original equation and is important in

the advanced theory of differential equations. In general, the problem of solving the

adjoint differential equation is as difﬁcult as that of solving the original equation, so only

occasionally is it possible to ﬁnd an integrating factor for a second order equation.

In each of Problems 47 through 49, use the result of Problem 46 to ﬁnd the adjoint of the given

differential equation.

47. x

′′

+ xy

′

+ (x

− ν

)y = 0, Bessel’s equation

48. (1 − x

′′

− 2xy

′

+ α(α + 1)y = 0, Legendre’s equation

49. y

′′

− xy = 0, Airy’s equation

50. For the second order linear equation P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0, show that the adjoint

of the adjoint equation is the original equation.

51. A second order linear equation P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0 is said to be self-adjoint if

its adjoint is the same as the original equation. Show that a necessary condition for this

equation to be self-adjoint is that P

′

(x) = Q(x). Determine whether each of the equations

in Problems 47 through 49 is self-adjoint.

3.3 Complex Roots of the Characteristic Equation

We continue our discussion of the equation

′′

+ by

′

+ cy = 0, (1)

where a, b, and c are given real numbers. In Section 3.1 we found that if we seek

solutions of the form y = e

, then r must be a root of the characteristic equation

+ br + c = 0. (2)

We showed in Section 3.1 that if the roots r

and r

are real and different, which

occurs whenever the discriminant b

− 4ac is positive, then the general solution of

Eq. (1) is

y = c

+ c

. (3)

Suppose now that b

− 4ac is negative. Then the roots of Eq. (2) are conjugate

complex numbers; we denote them by

= λ + iµ, r

= λ − iµ, (4)

where λ and µ are real. The corresponding expressions for y are

(t) = exp[(λ + iµ)t], y

(t) = exp[(λ − iµ)t]. (5)

Our ﬁrst task is to explore what is meant by these expressions, which involve

evaluating the exponential function for a complex exponent. For example, if

λ =−1, µ = 2, and t = 3, then from Eq. (5),

(3) = e

−3+6i

. (6)

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3.3 Complex Roots of the Characteristic Equation 159

What does it mean to raise the number e to a complex power? The answer is provided

by an important relation known as Euler’s formula.

Euler’s Formula. To assign a meaning to the expressions in Eqs. (5), we need to give

a deﬁnition of the complex exponential function. Of course, we want the deﬁnition

to reduce to the familiar real exponential function when the exponent is real. There

are several ways to discover how this extension of the exponential function should

be deﬁned. Here we use a method based on inﬁnite series; an alternative is outlined

in Problem 28.

Recall from calculus that the Taylor series for e

about t = 0 is

∞

n=0

, −∞ < t < ∞. (7)

If we now assume that we can substitute it for t in Eq. (7), then we have

∞

n=0

(it)

∞

n=0

(−1)

(2n)!

+ i

∞

n=1

(−1)

n−1

2n−1

(2n − 1)!

, (8)

where we have separated the sum into its real and imaginary parts, making use of the

fact that i

=−1, i

=−i, i

= 1, and so forth. The ﬁrst series in Eq. (8) is precisely

the Taylor series for cos t about t = 0, and the second is the Taylor series for sin t

about t = 0. Thus we have

= cos t + i sin t. (9)

Equation(9) is known asEuler’sformulaand is an extremelyimportant mathematical

relationship.Although ourderivation of Eq.(9) is based on theunveriﬁed assumption

that the series (7) can be used for complex as well as real values of the independent

variable, our intention is to use this derivation only to make Eq. (9) seem plausible.

We now put matters on a ﬁrm foundation by adopting Eq. (9) as the deﬁnition of e

In other words, whenever we write e

, we mean the expression on the right side of

Eq. (9).

There are some variations of Euler’s formula that are also worth noting. If we

replace t by −t in Eq. (9) and recall that cos(−t) = cos t and sin(−t) =−sin t, then

we have

−it

= cos t − i sin t. (10)

Further, if t is replaced by µt in Eq. (9), then we obtain a generalized version of

Euler’s formula, namely,

iµt

= cos µt + i sin µt. (11)

Next, we want to extend the deﬁnition of the exponential function to arbitrary

complex exponents of the form (λ + iµ)t. Since we want the usual properties

of the exponential function to hold for complex exponents, we certainly want

exp[(λ + iµ)t] to satisfy

(λ+iµ)t

= e

λt

iµt

. (12)

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160 Chapter 3. Second Order Linear Equations

Then, substituting for e

iµt

from Eq. (11), we obtain

(λ+iµ)t

= e

λt

(cos µt + i sin µt)

= e

λt

cos µt + ie

λt

sin µt. (13)

We now take Eq. (13) as the deﬁnition of exp[(λ + iµ)t].The value of the exponential

function with a complex exponent is a complex number whose real and imaginary

parts are given by the terms on the right side of Eq. (13). Observe that the real

and imaginary parts of exp[(λ + iµ)t] are expressed entirely in terms of elementary

real-valued functions. For example, the quantity in Eq. (6) has the value

−3+6i

= e

−3

cos 6 + ie

−3

sin 6

∼

0.0478041 − 0.0139113i.

With the deﬁnitions (9) and (13), it is straightforward to show that the usual laws of

exponents are valid for the complex exponential function. You can also use Eq. (13)

to verify that the differentiation formula

) = re

(14)

holds for complex values of r.

EXAMPLE

Find the general solution of the differential equation

′′

+ y

′

+ 9.25y = 0, (15)

Also ﬁnd the solution that satisﬁes the initial conditions

y(0) = 2, y

′

(0) = 8, (16)

and draw its graph.

The characteristic equation for Eq. (15) is

+ r + 9.25 = 0

so its roots are

=−

+ 3i, r

=−

− 3i.

Therefore, two solutions of Eq. (15) are

(t) = exp

;)

−

+ 3i

= e

−t/2

(cos 3t + i sin 3t) (17)

and

(t) = exp

;)

−

− 3i

= e

−t/2

(cos 3t − i sin 3t). (18)

You can verify that the Wronskian W(y

, y

)(t) =−6ie

−t

, which is not zero, so the general

solution of Eq. (15) can be expressed as a linear combination of y

(t) and y

(t) with arbitrary

coefﬁcients.

However, the initial value problem (15), (16) has only real coefﬁcients, and it is often

desirable to express the solution of such a problem in terms of real-valued functions.

To do this we can make use of Theorem 3.2.6, which states that the real and imaginary parts of

August 7, 2012 21:04 c03 Sheet number 25 Page number 161 cyan black

3.3 Complex Roots of the Characteristic Equation 161

a complex-valued solution of Eq. (15) are also solutions of Eq. (15). Thus, starting from either

(t) or y

(t), we obtain

u(t) = e

−t/2

cos 3t, v(t) = e

−t/2

sin 3t (19)

as real-valued solutions

of Eq. (15). On calculating the Wronskian of u(t) and v(t), we ﬁnd

that W(u, v)(t) = 3e

−t

, which is not zero; thus u(t) and v(t) form a fundamental set of solutions,

and the general solution of Eq. (15) can be written as

y = c

u(t) + c

v(t) = e

−t/2

cos 3t + c

sin 3t), (20)

where c

and c

are arbitrary constants.

To satisfy the initial conditions (16), we ﬁrst substitute t = 0 and y = 2 in Eq. (20) with the

result that c

= 2. Then, by differentiating Eq. (20), setting t = 0, and setting y

′

= 8, we obtain

−

+ 3c

= 8, so that c

= 3. Thus the solution of the initial value problem (15), (16) is

y = e

−t/2

(2 cos 3t + 3 sin 3t). (21)

The graph of this solution is shown in Figure 3.3.1.

From the graph we see that the solution of this problem is a decaying oscillation. The sine

and cosine factors control the oscillatory nature of the solution, and the negative exponential

factor in each term causes the magnitude of the oscillations to diminish as time increases.

–1

264810

FIGURE 3.3.1 Solution of the initial value problem (15), (16):

′′

+ y

′

+ 9.25y = 0, y(0) = 2, y

′

(0) = 8.

Complex Roots; The General Case. The functions y

(t) and y

(t), given by Eqs. (5) and

with the meaning expressed by Eq. (13), are solutions of Eq. (1) when the roots of

the characteristic equation (2) are complex numbers λ ± iµ. However, the solutions

and y

are complex-valued functions, whereas in general we would prefer to have

If you are not completely sure that u(t) and v(t) are solutions of the given differential equation, you

should substitute these functions into Eq. (15) and conﬁrm that they satisfy it.

August 7, 2012 21:04 c03 Sheet number 26 Page number 162 cyan black

162 Chapter 3. Second Order Linear Equations

real-valued solutions because the differential equation itself has real coefﬁcients. Just

as in Example 1, we can use Theorem 3.2.6 to ﬁnd a fundamental set of real-valued

solutions by choosing the real and imaginary parts of either y

(t) or y

(t). In this way

we obtain the solutions

u(t) = e

λt

cos µt, v(t) = e

λt

sin µt. (22)

By direct computation you can show that the Wronskian of u and v is

W(u, v)(t) = µe

2λt

. (23)

Thus, as long as µ ̸= 0, the Wronskian W is not zero, so u and v form a fundamental

set of solutions. (Of course, if µ = 0, then the roots are real and the discussion in this

section is not applicable.) Consequently, if the roots of the characteristic equation

are complex numbers λ ± iµ, with µ ̸= 0, then the general solution of Eq. (1) is

y = c

λt

cos µt + c

λt

sin µt, (24)

where c

and c

are arbitrary constants. Note that the solution (24) can be written

down as soon as the values of λ and µ are known. Let us now consider some further

examples.

EXAMPLE

Find the solution of the initial value problem

16y

′′

− 8y

′

+ 145y = 0, y(0) =−2, y

′

(0) = 1. (25)

The characteristic equation is 16r

− 8r + 145 = 0 and its roots are r = 1/4 ± 3i. Thus the

general solution of the differential equation is

y = c

t/4

cos 3t + c

t/4

sin 3t. (26)

To apply the ﬁrst initial condition, we set t = 0 in Eq. (26); this gives

y(0) = c

=−2.

For the second initial condition, we must differentiate Eq. (26) and then set t = 0. In this way

we ﬁnd that

′

(0) =

+ 3c

= 1,

from which c

= 1/2. Using these values of c

and c

in Eq. (26), we obtain

y =−2e

t/4

cos 3t +

t/4

sin 3t (27)

as the solution of the initial value problem (25). The graph of this solution is shown in

Figure 3.3.2.

In this case we observe that the solution is a growing oscillation. Again the trigonometric

factors in Eq. (27) determine the oscillatory part of the solution, while the exponential factor

(with a positive exponent this time) causes the magnitude of the oscillation to increase with

time.

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3.3 Complex Roots of the Characteristic Equation 163

–5

–10

2468

y = –2e

t/4

cos 3t + e

t/4

sin 3t

FIGURE 3.3.2 Solution of the initial value problem (25):

16y

′′

− 8y

′

+ 145y = 0, y(0) =−2, y

′

(0) = 1.

EXAMPLE

Find the general solution of

′′

+ 9y = 0. (28)

The characteristic equation is r

+ 9 = 0 with the roots r =±3i; thus λ = 0 and µ = 3. The

general solution is

y = c

cos 3t + c

sin 3t; (29)

–1

–2

–3

640182

FIGURE 3.3.3 Two typical solutions of Eq. (28): y

′′

+ 9y = 0.

August 7, 2012 21:04 c03 Sheet number 28 Page number 164 cyan black

164 Chapter 3. Second Order Linear Equations

note that if the real part of the roots is zero, as in this example, then there is no exponential

factor in the solution. Figure 3.3.3 shows the graph of two typical solutions of Eq. (28). In each

case the solution is a pure oscillation whose amplitude is determined by the initial conditions.

Since there is no exponential factor in the solution (29), the amplitude of each oscillation

remains constant in time.

PROBLEMS In each of Problems 1 through 6, use Euler’s formula to write the given expression in the form

a + ib.

1. exp(1 + 2i) 2. exp(2 − 3i)

3. e

iπ

4. e

2−(π/2)i

5. 2

1−i

6. π

−1+2i

In each of Problems 7 through 16, ﬁnd the general solution of the given differential equation.

7. y

′′

− 2y

′

+ 2y = 08.y

′′

− 2y

′

+ 6y = 0

9. y

′′

+ 2y

′

− 8y = 0 10. y

′′

+ 2y

′

+ 2y = 0

11. y

′′

+ 6y

′

+ 13y = 0 12. 4y

′′

+ 9y = 0

13. y

′′

+ 2y

′

+ 1.25y = 0 14. 9y

′′

+ 9y

′

− 4y = 0

15. y

′′

+ y

′

+ 1.25y = 0 16. y

′′

+ 4y

′

+ 6.25y = 0

In each of Problems 17 through 22, ﬁnd the solution of the given initial value problem. Sketch

the graph of the solution and describe its behavior for increasing t.

17. y

′′

+ 4y = 0, y(0) = 0, y

′

(0) = 1

18. y

′′

+ 4y

′

+ 5y = 0, y(0) = 1, y

′

(0) = 0

19. y

′′

− 2y

′

+ 5y = 0, y(π/2) = 0, y

′

(π/2) = 2

20. y

′′

+ y = 0, y(π/3) = 2, y

′

(π/3) =−4

21. y

′′

+ y

′

+ 1.25y = 0, y(0) = 3, y

′

(0) = 1

22. y

′′

+ 2y

′

+ 2y = 0, y(π/4) = 2, y

′

(π/4) =−2

23.

Consider the initial value problem

′′

− u

′

+ 2u = 0, u(0) = 2, u

′

(0) = 0.

(a) Find the solution u(t) of this problem.

(b) For t > 0, ﬁnd the ﬁrst time at which |u(t)|=10.

24.

Consider the initial value problem

′′

+ 2u

′

+ 7u = 0, u(0) = 2, u

′

(0) = 1.

(a) Find the solution u(t) of this problem.

(b) Find the smallest T such that |u(t)|≤0.1 for all t > T.

25.

Consider the initial value problem

′′

+ 2y

′

+ 6y = 0, y(0) = 2, y

′

(0) = α ≥ 0.

(a) Find the solution y(t) of this problem.

(b) Find α such that y = 0 when t = 1.

(d) Determine the limit of the expression found in part (c) as α →∞.

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3.3 Complex Roots of the Characteristic Equation 165

26. Consider the initial value problem

′′

+ 2ay

′

+ (a

+ 1)y = 0, y(0) = 1, y

′

(0) = 0.

(a) Find the solution y(t) of this problem.

(b) For a = 1 ﬁnd the smallest T such that |y(t)| < 0.1 for t > T.

(d) Using the results of parts (b) and (c),plot T versus a and describe the relation between

T and a.

27. Show that W(e

λt

cos µt, e

λt

sin µt) = µe

2λt

28. In this problem we outline a different derivation of Euler’s formula.

(a) Show that y

(t) = cos t and y

(t) = sin t are a fundamental set of solutions of

′′

+ y = 0; that is, show that they are solutions and that their Wronskian is not zero.

(b) Show (formally) that y = e

is also a solution of y

′′

+ y = 0. Therefore,

= c

cos t + c

sin t (i)

for some constants c

and c

. Why is this so?

= 1.

(d) Assuming that Eq. (14) is true,differentiate Eq. (i) and then set t = 0 to conclude that

= i. Use the values of c

and c

in Eq. (i) to arrive at Euler’s formula.

29. Using Euler’s formula, show that

cos t = (e

+ e

−it

)/2, sin t = (e

− e

−it

)/2i.

30. If e

is given by Eq. (13), show that e

= e

for any complex numbers r

and r

31. If e

is given by Eq. (13), show that

= re

for any complex number r.

32. Consider the differential equation

′′

+ by

′

+ cy = 0,

where b

− 4ac < 0 and the characteristic equation has complex roots λ ± iµ. Substitute

the functions

u(t) = e

λt

cos µt and v(t) = e

λt

sin µt

for y in the differential equation and thereby conﬁrm that they are solutions.

33. If the functions y

and y

are a fundamental set of solutions of y

′′

+ p(t)y

′

+ q(t)y = 0,

show that between consecutive zeros of y

there is one and only one zero of y

. Note

that this result is illustrated by the solutions y

(t) = cos t and y

(t) = sin t of the equation

′′

+ y = 0.

Hint: Suppose that t

and t

are two zeros of y

between which there are no zeros of y

Apply Rolle’s theorem to y

to reach a contradiction.

Change of Variables. Sometimes a differential equation with variable coefﬁcients,

′′

+ p(t)y

′

+ q(t)y = 0, (i)

can be put in a more suitable form for ﬁnding a solution by making a change of the independent

variable. We explore these ideas in Problems 34 through 46. In particular, in Problem 34 we

show that a class of equations known as Euler equations can be transformed into equations

with constant coefﬁcients bya simple changeof theindependent variable. Problems 35 through

August 7, 2012 21:04 c03 Sheet number 30 Page number 166 cyan black

166 Chapter 3. Second Order Linear Equations

42 are examples of this type of equation. Problem 43 determines conditions under which the

more general Eq. (i) can be transformed into a differential equation with constant coefﬁcients.

Problems 44 through 46 give speciﬁc applications of this procedure.

34. Euler Equations. An equation of the form

+ αt

+ βy = 0, t > 0, (ii)

where α and β are real constants, is called an Euler equation.

(a) Let x = ln t and calculate dy/dt and d

y/dt

in terms of dy/dx and d

y/dx

(b) Use the results of part (a) to transform Eq. (ii) into

+ (α − 1)

+ βy = 0. (iii)

Observe that Eq. (iii) has constant coefﬁcients. If y

(x) and y

(x) form a fundamental set

of solutions of Eq. (iii), then y

(ln t) and y

(ln t) form a fundamental set of solutions of

Eq. (ii).

In each of Problems 35 through 42, use the method of Problem 34 to solve the given equation

for t > 0.

35. t

′′

+ ty

′

+ y = 0 36. t

′′

+ 4ty

′

+ 2y = 0

37. t

′′

+ 3ty

′

+ 1.25y = 0 38. t

′′

− 4ty

′

− 6y = 0

39. t

′′

− 4ty

′

+ 6y = 0 40. t

′′

− ty

′

+ 5y = 0

41. t

′′

+ 3ty

′

− 3y = 0 42. t

′′

+ 7ty

′

+ 10y = 0

43. In this problem we determine conditions on p and q that enable Eq. (i) to be transformed

into an equation with constant coefﬁcients by a change of the independent variable. Let

x = u(t) be the new independent variable, with the relation between x and t to be speciﬁed

later.

(a) Show that

(

(b) Show that the differential equation (i) becomes

(

+ p(t)

(

+ q(t)y = 0. (iv)

y/dx

and of

y must be proportional. If q(t)>0, then we can choose the constant of proportionality to

be 1; hence

x = u(t) =

[q(t)]

1/2

dt. (v)

(d) With x chosen as in part (c), show that the coefﬁcient of dy/dx in Eq. (iv) is also a

constant, provided that the expression

′

(t) + 2p(t)q(t)

2[q(t)]

3/2

(vi)

is a constant. Thus Eq. (i) can be transformed into an equation with constant coefﬁcients

by a change of the independent variable, provided that the function (q

′

+ 2pq)/q

3/2

is a

constant. How must this result be modiﬁed if q(t)<0?

August 7, 2012 21:04 c03 Sheet number 31 Page number 167 cyan black

3.4 Repeated Roots; Reduction of Order 167

In each of Problems 44 through 46, try to transform the given equation into one with constant

coefﬁcients by the method of Problem 43. If this is possible, ﬁnd the general solution of the

given equation.

44. y

′′

+ ty

′

+ e

−t

y = 0, −∞ < t < ∞

45. y

′′

+ 3ty

′

+ t

y = 0, −∞ < t < ∞

46. ty

′′

+ (t

− 1)y

′

+ t

y = 0, 0 < t < ∞

3.4 Repeated Roots; Reduction of Order

In earlier sections we showed how to solve the equation

′′

+ by

′

+ cy = 0 (1)

when the roots of the characteristic equation

+ br + c = 0 (2)

either are real and different or are complex conjugates. Now we consider the third

possibility, namely, that the two roots r

and r

are equal. This case is transitional

between the other two and occurs when the discriminant b

− 4ac is zero. Then it

follows from the quadratic formula that

= r

=−b/2a. (3)

The difﬁculty is immediately apparent; both roots yield the same solution

(t) = e

−bt/2a

(4)

of the differential equation (1), and it is not obvious how to ﬁnd a second solution.

EXAMPLE

Solve the differential equation

′′

+ 4y

′

+ 4y = 0. (5)

The characteristic equation is

+ 4r + 4 = (r + 2)

= 0,

so r

= r

=−2. Therefore, one solution of Eq. (5) is y

(t) = e

−2t

. To ﬁnd the general solution

of Eq. (5), we need a second solution that is not a constant multiple of y

. This second solution

can be found in several ways (see Problems 20 through 22); here we use a method originated

by D’Alembert

in the eighteenth century. Recall that since y

(t) is a solution of Eq. (1), so is

(t) for any constant c. The basic idea is to generalize this observation by replacing c by a

Jean d’Alembert (1717–1783), a French mathematician, was a contemporary of Euler and Daniel

Bernoulli and is known primarily for his work in mechanics and differential equations. D’Alembert’s

principle in mechanics and d’Alembert’s paradox in hydrodynamics are named for him, and the wave

equation ﬁrst appeared in his paper on vibrating strings in 1747. In his later years he devoted himself

primarily to philosophy and to his duties as science editor of Diderot’s Encyclopédie.

August 7, 2012 21:04 c03 Sheet number 32 Page number 168 cyan black

168 Chapter 3. Second Order Linear Equations

function v(t) and then trying to determine v(t) so that the product v(t)y

(t) is also a solution

of Eq. (1).

To carryout this program,we substitute y = v(t)y

(t) in Eq. (5)and use the resultingequation

to ﬁnd v(t). Starting with

y = v(t)y

(t) = v(t)e

−2t

, (6)

we have

′

= v

′

(t)e

−2t

− 2v(t)e

−2t

(7)

and

′′

= v

′′

(t)e

−2t

− 4v

′

(t)e

−2t

+ 4v(t)e

−2t

. (8)

By substituting the expressions in Eqs. (6), (7), and (8) in Eq. (5) and collecting terms, we

obtain

′′

(t) − 4v

′

(t) + 4v(t) + 4v

′

(t) − 8v(t) + 4v(t)]e

−2t

= 0,

which simpliﬁes to

′′

(t) = 0 . (9)

Therefore,

′

(t) = c

and

v(t) = c

t + c

, (10)

where c

and c

are arbitrary constants. Finally, substituting for v(t) in Eq. (6), we obtain

y = c

−2t

+ c

−2t

. (11)

The second term on the right side of Eq. (11) corresponds to the original solution

(t) = exp(−2t), but the ﬁrst term arises from a second solution, namely, y

(t) = t exp(−2t).

We can verify that these two solutions form a fundamental set by calculating their Wronskian:

W(y

, y

)(t) =

−2t

−2e

−2t

(1 − 2t)e

−2t

= e

−4t

− 2te

−4t

+ 2te

−4t

= e

−4t

̸= 0.

Therefore,

(t) = e

−2t

, y

(t) = te

−2t

(12)

form a fundamental set of solutions ofEq. (5),and the general solution of that equation is given

by Eq. (11). Note that both y

(t) and y

(t) tend to zero as t →∞; consequently, all solutions

of Eq. (5) behave in this way. The graph of a typical solution is shown in Figure 3.4.1.

0.5 1 1.5 2

FIGURE 3.4.1 A typical solution of Eq. (5): y

′′

+ 4y

′

+ 4y = 0.

August 7, 2012 21:04 c03 Sheet number 33 Page number 169 cyan black

3.4 Repeated Roots; Reduction of Order 169

The procedure used in Example 1 can be extended to a general equation whose

characteristic equation has repeated roots. That is, we assume that the coefﬁcients in

Eq. (1) satisfy b

− 4ac = 0, in which case

(t) = e

−bt/2a

is a solution. To ﬁnd a second solution, we assume that

y = v(t)y

(t) = v(t)e

−bt/2a

(13)

and substitute for y in Eq. (1) to determine v(t). We have

′

= v

′

(t)e

−bt/2a

−

v(t)e

−bt/2a

(14)

and

′′

= v

′′

(t)e

−bt/2a

−

′

(t)e

−bt/2a

v(t)e

−bt/2a

. (15)

Then, by substituting in Eq. (1), we obtain

′′

(t) −

′

(t) +

v(t)

+ b

′

(t) −

v(t)

+ cv(t)

−bt/2a

= 0. (16)

Canceling the factor exp(−bt/2a), which is nonzero, and rearranging the remaining

terms, we ﬁnd that

′′

(t) + (−b + b)v

′

(t) +

−

+ c

(

v(t) = 0. (17)

The term involving v

′

(t) is obviously zero. Further, the coefﬁcient of v(t) is

c − (b

/4a), which is also zero because b

− 4ac = 0 in the problem that we are

considering. Thus, just as in Example 1, Eq. (17) reduces to

′′

(t) = 0,

v(t) = c

+ c

Hence, from Eq. (13), we have

y = c

−bt/2a

+ c

−bt/2a

. (18)

Thus y is a linear combination of the two solutions

(t) = e

−bt/2a

, y

(t) = te

−bt/2a

. (19)

The Wronskian of these two solutions is

W(y

, y

)(t) =

−bt/2a

−

−bt/2a

1 −

(

−bt/2a

= e

−bt/a

. (20)

Since W(y

, y

)(t) is never zero, the solutions y

and y

given by Eq. (19) are a

fundamental set of solutions. Further, Eq. (18) is the general solution of Eq. (1)

when the roots of the characteristic equation are equal. In other words, in this case

there is one exponential solution corresponding to the repeated root and a second

solution that is obtained by multiplying the exponential solution by t.

August 7, 2012 21:04 c03 Sheet number 34 Page number 170 cyan black

170 Chapter 3. Second Order Linear Equations

EXAMPLE

Find the solution of the initial value problem

′′

− y

′

+ 0.25y = 0, y(0) = 2, y

′

(0) =

. (21)

The characteristic equation is

− r + 0.25 = 0,

so the roots are r

= r

= 1/2. Thus the general solution of the differential equation is

y = c

t/2

+ c

t/2

. (22)

The ﬁrst initial condition requires that

y(0) = c

= 2.

To satisfy the second initial condition, we ﬁrst differentiate Eq. (22) and then set t = 0. This

gives

′

(0) =

+ c

so c

=−2/3. Thus the solution of the initial value problem is

y = 2e

t/2

−

t/2

. (23)

The graph of this solution is shown by the blue curve in Figure 3.4.2.

–1

123

y'(0) = 2: y = 2e

t/2

+ te

t/2

y'(0) = : y = 2e

t/2

– te

t/2

FIGURE 3.4.2 Solutions of y

′′

− y

′

+ 0.25y = 0, y(0) = 2, with

′

(0) = 1/3 (blue curve) and with y

′

(0) = 2 (black curve), respectively.

Let us now modify the initial value problem (21) by changing the initial slope; to be speciﬁc,

let the second initial condition be y

′

(0) = 2. The solution of this modiﬁed problem is

y = 2e

t/2

+ te

t/2

and its graph is shown by the black curve in Figure 3.4.2.The graphs shown in this ﬁgure suggest

that there is a critical initial slope, with a value between

and 2, that separates solutions that

grow positively from those that ultimately grow negatively. In Problem 16 you are asked to

determine this critical initial slope.

August 7, 2012 21:04 c03 Sheet number 35 Page number 171 cyan black

3.4 Repeated Roots; Reduction of Order 171

The geometrical behavior of solutions is similar in this case to that when the

roots are real and different. If the exponents are either positive or negative, then

the magnitude of the solution grows or decays accordingly; the linear factor t has

little inﬂuence. A decaying solution is shown in Figure 3.4.1 and growing solutions in

Figure 3.4.2. However, if the repeated root is zero, then the differential equation is

′′

= 0 and the general solution is a linear function of t.

Summar y. We can now summarize the results that we have obtained for second order

linear homogeneous equations with constant coefﬁcients

′′

+ by

′

+ cy = 0. (1)

Let r

and r

be the roots of the corresponding characteristic equation

+ br + c = 0. (2)

If r

and r

are real but not equal, then the general solution of the differential

equation (1) is

y = c

+ c

. (24)

If r

and r

are complex conjugates λ ± iµ, then the general solution is

y = c

λt

cos µt + c

λt

sin µt. (25)

If r

= r

, then the general solution is

y = c

+ c

. (26)

Reduction of Order. It is worth noting that the procedure used in this section for equa-

tions with constant coefﬁcients is more generally applicable. Suppose that we know

one solution y

(t), not everywhere zero, of

′′

+ p(t)y

′

+ q(t)y = 0. (27)

To ﬁnd a second solution, let

y = v(t)y

(t); (28)

then

′

= v

′

(t)y

(t) + v(t)y

′

(t)

and

′′

= v

′′

(t)y

(t) + 2v

′

(t)y

′

(t) + v(t)y

′′

(t).

Substituting for y, y

′

, and y

′′

in Eq. (27) and collecting terms, we ﬁnd that

′′

+ (2y

′

+ py

′

+ (y

′′

+ py

′

+ qy

)v = 0. (29)

Since y

is a solution of Eq. (27),the coefﬁcient of v in Eq. (29) is zero,so that Eq. (29)

becomes

′′

+ (2y

′

+ py

′

= 0. (30)

August 7, 2012 21:04 c03 Sheet number 36 Page number 172 cyan black

172 Chapter 3. Second Order Linear Equations

Despite its appearance, Eq. (30) is actually a ﬁrst order equation for the function v

′

and can be solved either as a ﬁrst order linear equation or as a separable equation.

Once v

′

has been found, then v is obtained by an integration. Finally, y is determined

from Eq. (28).This procedure is called the method of reduction of order,because the

crucial step is the solution of a ﬁrst order differential equation for v

′

rather than

the original second order equation for y. Although it is possible to write down a

formula for v(t), we will instead illustrate how this method works by an example.

EXAMPLE

Given that y

(t) = t

−1

is a solution of

′′

+ 3ty

′

− y = 0, t > 0, (31)

ﬁnd a fundamental set of solutions.

We set y = v(t)t

−1

; then

′

= v

′

−1

− vt

−2

, y

′′

= v

′′

−1

− 2v

′

−2

+ 2vt

−3

Substituting for y, y

′

, and y

′′

in Eq. (31) and collecting terms, we obtain

′′

−1

− 2v

′

−2

+ 2vt

−3

) + 3t(v

′

−1

− vt

−2

) − vt

−1

= 2tv

′′

+ (−4 + 3)v

′

+ (4t

−1

− 3t

−1

− t

−1

= 2tv

′′

− v

′

= 0. (32)

Note that the coefﬁcient ofv is zero,as it shouldbe;this provides a useful checkon our algebraic

calculations.

If we let w = v

′

, then Eq. (32) becomes

2tw

′

− w = 0.

Separating the variables and solving for w(t), we ﬁnd that

w(t) = v

′

(t) = ct

1/2

;

then

v(t) =

3/2

+ k.

It follows that

y = v(t)t

−1

1/2

+ kt

−1

, (33)

where c and k are arbitrary constants.The second term on the right side of Eq. (33) is a multiple

of y

(t) and can be dropped, but the ﬁrst term provides a new solution y

(t) = t

1/2

. You can

verify that the Wronskian of y

and y

W(y

, y

)(t) =

−3/2

̸= 0 for t > 0. (34)

Consequently, y

and y

form a fundamental set of solutions of Eq. (31) for t > 0.

PROBLEMS In each of Problems 1 through 10, ﬁnd the general solution of the given differential equation.

1. y

′′

− 2y

′

+ y = 02.9y

′′

+ 6y

′

+ y = 0

3. 4y

′′

− 4y

′

− 3y = 04.4y

′′

+ 12y

′

+ 9y = 0

5. y

′′

− 2y

′

+ 10y = 06.y

′′

− 6y

′

+ 9y = 0

August 7, 2012 21:04 c03 Sheet number 37 Page number 173 cyan black

3.4 Repeated Roots; Reduction of Order 173

7. 4y

′′

+ 17y

′

+ 4y = 08.16y

′′

+ 24y

′

+ 9y = 0

9. 25y

′′

− 20y

′

+ 4y = 0 10. 2y

′′

+ 2y

′

+ y = 0

In each of Problems 11 through 14, solve the given initial value problem. Sketch the graph of

the solution and describe its behavior for increasing t.

11. 9y

′′

− 12y

′

+ 4y = 0, y(0) = 2, y

′

(0) =−1

12. y

′′

− 6y

′

+ 9y = 0, y(0) = 0, y

′

(0) = 2

13. 9y

′′

+ 6y

′

+ 82y = 0, y(0) =−1, y

′

(0) = 2

14. y

′′

+ 4y

′

+ 4y = 0, y(−1) = 2, y

′

(−1) = 1

15.

Consider the initial value problem

′′

+ 12y

′

+ 9y = 0, y(0) = 1, y

′

(0) =−4.

(a) Solve the initial value problem and plot its solution for 0 ≤ t ≤ 5.

(b) Determine where the solution has the value zero.

, y

) of the minimum point.

(d) Change the second initial condition to y

′

(0) = b and ﬁnd the solution as a function

of b. Then ﬁnd the critical value of b that separates solutions that always remain positive

from those that eventually become negative.

16. Consider the following modiﬁcation of the initial value problem in Example 2:

′′

− y

′

+ 0.25y = 0, y(0) = 2, y

′

(0) = b.

Find the solution as a function of b, and then determine the critical value of b that

separates solutions that grow positively from those that eventually grow negatively.

17.

Consider the initial value problem

′′

+ 4y

′

+ y = 0, y(0) = 1, y

′

(0) = 2.

(a) Solve the initial value problem and plot the solution.

(b) Determine the coordinates (t

, y

) of the maximum point.

′

(0) = b > 0 and ﬁnd the solution as a function

of b.

(d) Find the coordinates (t

, y

) of the maximum point in terms of b. Describe the

dependence of t

and y

on b as b increases.

18. Consider the initial value problem

′′

+ 12y

′

+ 4y = 0, y(0) = a > 0, y

′

(0) =−1.

(a) Solve the initial value problem.

(b) Find the critical value of a that separates solutions that become negative from those

that are always positive.

19. Consider the equation ay

′′

+ by

′

+ cy = 0. If the roots of the corresponding characteristic

equation are real, show that a solution to the differential equation either is everywhere

zero or else can take on the value zero at most once.

Problems 20 through 22 indicate other ways of ﬁnding the second solution when the

characteristic equation has repeated roots.

20. (a) Consider the equation y

′′

+ 2ay

′

+ a

y = 0. Show that the roots of the characteristic

equation are r

= r

=−a, so that one solution of the equation is e

−at

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174 Chapter 3. Second Order Linear Equations

(b) Use Abel’s formula [Eq. (23) of Section 3.2] to show that the Wronskian of any two

solutions of the given equation is

W(t) = y

(t)y

′

(t) − y

′

(t)y

(t) = c

−2at

where c

is a constant.

(t) = e

−at

and use the result of part (b) to obtain a differential equation satisﬁed

by a second solution y

(t). By solving this equation, show that y

(t) = te

−at

21. Suppose that r

and r

are roots of ar

+ br + c = 0 and that r

̸= r

; then exp(r

and exp(r

t) are solutions of the differential equation ay

′′

+ by

′

+ cy = 0. Show that

φ(t; r

, r

) =[exp(r

t) − exp(r

t)]/(r

− r

) is also a solution of the equation for r

̸= r

Then think of r

as ﬁxed, and use L’Hôpital’s rule to evaluate the limit of φ(t; r

, r

) as

→ r

, thereby obtaining the second solution in the case of equal roots.

22. (a) If ar

+ br + c = 0 has equal roots r

, show that

L[e

]=a(e

)

′′

+ b(e

)

′

+ ce

= a(r − r

)

. (i)

Since the right side of Eq. (i) is zero when r = r

, it follows that exp(r

t) is a solution of

L[y]=ay

′′

+ by

′

+ cy = 0.

(b) Differentiate Eq. (i) with respect to r, and interchange differentiation with respect to

r and with respect to t, thus showing that

∂

∂r

L[e

]=L

∂

∂r

= L[te

]=ate

(r − r

)

+ 2ae

(r − r

). (ii)

Since the right side of Eq. (ii) is zero when r = r

, conclude that t exp(r

t) is also a solution

of L[y]=0.

In each of Problems 23 through 30, use the method of reduction of order to ﬁnd a second

solution of the given differential equation.

23. t

′′

− 4ty

′

+ 6y = 0, t > 0; y

(t) = t

24. t

′′

+ 2ty

′

− 2y = 0, t > 0; y

(t) = t

25. t

′′

+ 3ty

′

+ y = 0, t > 0; y

(t) = t

−1

26. t

′′

− t(t + 2)y

′

+ (t + 2)y = 0, t > 0; y

(t) = t

27. xy

′′

− y

′

+ 4x

y = 0, x > 0; y

(x) = sin x

28. (x − 1)y

′′

− xy

′

+ y = 0, x > 1; y

(x) = e

29. x

′′

− (x − 0 .1875)y = 0, x > 0; y

(x) = x

1/4

√

30. x

′′

+ xy

′

+ (x

− 0.25)y = 0, x > 0; y

(x) = x

−1/2

sin x

31. The differential equation

′′

+ δ(xy

′

+ y) = 0

arises in the study of the turbulent ﬂow of a uniform stream past a circular cylinder. Verify

that y

(x) = exp(−δx

/2) is one solution, and then ﬁnd the general solution in the form of

an integral.

32. The method of Problem 20 can be extended to second order equations with variable

coefﬁcients. If y

is a known nonvanishing solution of y

′′

+ p(t)y

′

+ q(t)y = 0, show that a

second solution y

satisﬁes (y

)

′

= W(y

, y

)/y

, where W(y

, y

) is the Wronskian of

and y

. Then use Abel’s formula [Eq. (23) of Section 3.2] to determine y

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3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 175

In each of Problems 33 through 36,use the method of Problem 32 to ﬁnd a second independent

solution of the given equation.

33. t

′′

+ 3ty

′

+ y = 0, t > 0; y

(t) = t

−1

34. ty

′′

− y

′

+ 4t

y = 0, t > 0; y

(t) = sin(t

)

35. (x − 1)y

′′

− xy

′

+ y = 0, x > 1; y

(x) = e

36. x

′′

+ xy

′

+ (x

− 0.25)y = 0, x > 0; y

(x) = x

−1/2

sin x

Behavior of Solutions as t →∞. Problems 37 through 39 are concerned with the behavior

of solutions as t →∞.

37. If a, b, and c are positive constants, show that all solutions of ay

′′

+ by

′

+ cy = 0 approach

zero as t →∞.

38. (a) If a > 0 and c > 0, but b = 0, show that the result of Problem 37 is no longer true, but

that all solutions are bounded as t →∞.

(b) If a > 0 and b > 0, but c = 0, show that the result of Problem 37 is no longer true, but

that all solutions approach a constant that depends on the initial conditions as t →∞.

Determine this constant for the initial conditions y(0) = y

, y

′

(0) = y

′

39. Show that y = sin t is a solution of

′′

+ (k sin

t)y

′

+ (1 − k cos t sin t)y = 0

for any value of the constant k.If 0< k < 2,show that 1 − k cos t sin t > 0 and k sin

t ≥ 0.

Thusobserve that even though the coefﬁcients ofthis variable-coefﬁcient differentialequa-

tion are nonnegative (and the coefﬁcient of y

′

is zero only at the points t = 0, π,2π, ...),

it has a solution that does not approach zero as t →∞. Compare this situation with the

result of Problem 37. Thus we observe a not unusual situation in the study of differential

equations: equations that are apparently very similar can have quite different properties.

Euler Equations. In each of Problems 40 through 45, use the substitution introduced in

Problem 34 in Section 3.3 to solve the given differential equation.

40. t

′′

− 3ty

′

+ 4y = 0, t > 0

41. t

′′

+ 2ty

′

+ 0.25y = 0, t > 0

42. 2t

′′

− 5ty

′

+ 5y = 0, t > 0

43. t

′′

+ 3ty

′

+ y = 0, t > 0

44. 4t

′′

− 8ty

′

+ 9y = 0, t > 0

45. t

′′

+ 5ty

′

+ 13y = 0, t > 0

3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients

We now return to the nonhomogeneous equation

L[y]=y

′′

+ p(t)y

′

+ q(t)y = g(t), (1)

where p, q, and g are given (continuous) functions on the open interval I.

The equation

L[y]=y

′′

+ p(t)y

′

+ q(t)y = 0, (2)

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176 Chapter 3. Second Order Linear Equations

in which g(t) = 0 and p and q are the same as in Eq. (1), is called the homogeneous

equation corresponding to Eq. (1).The following two results describe the structure of

solutions of the nonhomogeneous equation (1) and provide a basis for constructing

its general solution.

Theorem 3.5.1

If Y

and Y

are two solutions of the nonhomogeneous equation (1), then their

difference Y

− Y

is a solution of the corresponding homogeneous equation (2).

If, in addition, y

and y

are a fundamental set of solutions of Eq. (2), then

(t) − Y

(t) = c

(t) + c

(t), (3)

where c

and c

are certain constants.

To prove this result, note that Y

and Y

satisfy the equations

L[Y

](t) = g(t), L[Y

](t) = g(t). (4)

Subtracting the second of these equations from the ﬁrst, we have

L[Y

](t) − L[Y

](t) = g(t) − g(t) = 0. (5)

However,

L[Y

]−L[Y

]=L[Y

− Y

so Eq. (5) becomes

L[Y

− Y

](t) = 0. (6)

Equation (6) states that Y

− Y

is a solution of Eq. (2). Finally, since by Theorem

3.2.4 all solutions of Eq. (2) can be expressed as linear combinations of a fundamental

set of solutions, it follows that the solution Y

− Y

can be so written. Hence Eq. (3)

holds and the proof is complete.

Theorem 3.5.2

The general solution of the nonhomogeneous equation (1) can be written in the

form

y = φ(t) = c

(t) + c

(t) + Y(t), (7)

where y

and y

are a fundamental set of solutions of the corresponding homoge-

neous equation (2),c

and c

are arbitrary constants, and Y is some speciﬁc solution

of the nonhomogeneous equation (1).

The proof of Theorem 3.5.2 follows quickly from the preceding theorem. Note

that Eq. (3) holds if we identify Y

with an arbitrary solution φ of Eq. (1) and Y

with

the speciﬁc solution Y. From Eq. (3) we thereby obtain

φ(t) − Y(t) = c

(t) + c

(t), (8)

which is equivalentto Eq.(7). Sinceφ isan arbitrarysolution of Eq.(1),the expression

on the right side of Eq. (7) includes all solutions of Eq. (1); thus it is natural to call it

the general solution of Eq. (1).

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3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 177

In somewhat different words, Theorem 3.5.2 states that to solve the nonhomoge-

neous equation (1), we must do three things:

1. Find the general solution c

(t) + c

(t) of the corresponding homogeneous equation.

This solution is frequently called the complementary solution and may be denoted by y

(t).

2. Find some single solution Y(t) of the nonhomogeneous equation. Often this solution is

referred to as a particular solution.

3. Form the sum of the functions found in steps 1 and 2.

We have already discussed how to ﬁnd y

(t),at least when the homogeneous equa-

tion (2) has constant coefﬁcients. Therefore, in the remainder of this section and in

the next, we will focus on ﬁnding a particular solution Y(t) of the nonhomogeneous

equation (1). There are two methods that we wish to discuss. They are known as

the method of undetermined coefﬁcients (discussed here) and the method of varia-

tion of parameters (see Section 3.6). Each has some advantages and some possible

shortcomings.

Method of Undetermined Coefficients. The method of undetermined coefﬁcients requires

us to make an initial assumption about the form of the particular solution Y(t), but

with the coefﬁcients left unspeciﬁed. We then substitute the assumed expression into

Eq. (1) and attempt to determine the coefﬁcients so as to satisfy that equation. If we

are successful, then we have found a solution of the differential equation (1) and can

use it for the particular solution Y(t). If we cannot determine the coefﬁcients, then

this means that there is no solution of the form that we assumed. In this case we may

modify the initial assumption and try again.

The mainadvantage of themethod of undeterminedcoefﬁcients is thatit isstraight-

forward to execute once the assumption is made about the form of Y(t). Its major

limitation is that it is useful primarily for equations for which we can easily write

down the correct form of the particular solution in advance. For this reason, this

method is usually used only for problems in which the homogeneous equation has

constant coefﬁcients and the nonhomogeneous term is restricted to a relatively small

class of functions. In particular,we consider only nonhomogeneous terms that consist

of polynomials, exponential functions, sines, and cosines. Despite this limitation, the

method of undetermined coefﬁcients is useful for solving many problems that have

important applications. However, the algebraic details may become tedious, and a

computer algebra system can be very helpful in practical applications. We will illus-

trate the method of undetermined coefﬁcients by several simple examples and then

summarize some rules for using it.

EXAMPLE

Find a particular solution of

′′

− 3y

′

− 4y = 3e

. (9)

We seek a function Y such that the combination Y

′′

(t) − 3Y

′

(t) − 4Y(t) is equal to 3e

Since the exponential function reproduces itself through differentiation, the most plausible

way to achieve the desired result is to assume that Y (t) is some multiple of e

Y(t) = Ae

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178 Chapter 3. Second Order Linear Equations

where the coefﬁcient A is yet to be determined. To ﬁnd A, we calculate

′

(t) = 2Ae

, Y

′′

(t) = 4Ae

and substitute for y, y

′

, and y

′′

in Eq. (9). We obtain

(4A − 6A − 4A)e

= 3e

Hence −6Ae

must equal 3e

, so A =−1/2. Thus a particular solution is

Y(t) =−

. (10)

EXAMPLE

Find a particular solution of

′′

− 3y

′

− 4y = 2 sin t. (11)

By analogy with Example 1, let us ﬁrst assume that Y(t ) = A sin t, where A is a constant to

be determined. On substituting in Eq. (11) we obtain

−A sin t − 3A cos t − 4A sint = 2 sin t,

or, by rearranging terms,

(2 + 5A) sin t + 3A cos t = 0. (12)

We want Eq. (12) to hold for all t. Thus it must hold for two speciﬁc points, such as t = 0

and t = π/2. At these points Eq. (12) reduces to 3A = 0 and 2 + 5A = 0, respectively. These

contradictory requirements mean that there is no choice of the constant A that makes

Eq. (12) true for t = 0 and t = π/2, much less for all t. Thus we conclude that our assumption

concerning Y(t) is inadequate.

The appearance of the cosine term in Eq. (12) suggests that we modify our original

assumption to include a cosine term in Y(t); that is,

Y(t) = A sin t + B cos t ,

where A and B are

be determined. Then

′

(t) = A cos t − B sin t, Y

′′

(t) =−A sin t − B cos t.

By substituting these expressions for y, y

′

, and y

′′

in Eq. (11) and collecting terms, we obtain

(−A + 3B − 4A) sin t + (−B − 3A − 4B) cos t = 2 sin t. (13)

To satisfy Eq. (13), we must match the coefﬁcients of sin t and cos t on each side of the equation;

thus A and B must satisfy the equations

−5A + 3B = 2, −3A − 5B = 0.

By solving these equations for A and B,we obtain A =−5/17 and B = 3/17;hence a particular

solution of Eq. (11) is

Y(t) =−

sin t +

cos t.

The method illustrated in the preceding examples can also be used when the right

side of the equation is a polynomial. Thus, to ﬁnd a particular solution of

′′

− 3y

′

− 4y = 4t

− 1, (14)

we initially assume that Y(t) is a polynomial of the same degree as the nonhomoge-

neous term; that is, Y(t) = At

+ Bt + C.

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3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 179

To summarize our conclusions up to this point: if the nonhomogeneous term g(t)

in Eq. (1) is an exponential function e

αt

, then assume that Y(t) is proportional to

the same exponential function; if g(t) is sin βt or cos β t, then assume that Y(t) is a

linear combination of sin βt and cos βt;ifg(t) is a polynomial, then assume that Y(t)

is a polynomial of like degree. The same principle extends to the case where g (t) is

a product of any two, or all three, of these types of functions, as the next example

illustrates.

EXAMPLE

Find a particular solution of

′′

− 3y

′

− 4y =−8e

cos 2t. (15)

In this case we assume that Y(t) is the product of e

and a linear combination of cos 2t and

sin 2t; that is,

Y(t) = Ae

cos 2t + Be

sin 2t.

The algebra is more tedious in this example, but it follows that

′

(t) = (A + 2B)e

cos 2t + (−2A + B)e

sin 2t

and

′′

(t) = (−3A + 4B)e

cos 2t + (−4A − 3B)e

sin 2t.

By substituting these expressions in Eq. (15), we ﬁnd that A and B must satisfy

10A + 2B = 8, 2A − 10B = 0.

Hence A = 10/13 and B = 2/13; therefore, a particular solution of Eq. (15) is

Y(t) =

cos 2t +

sin 2t.

Now suppose that g(t) is the sum of two terms, g(t) = g

(t) + g

(t), and suppose

that Y

and Y

are solutions of the equations

′′

+ by

′

+ cy = g

(t) (16)

and

′′

+ by

′

+ cy = g

(t), (17)

respectively. Then Y

+ Y

is a solution of the equation

′′

+ by

′

+ cy = g(t). (18)

To prove this statement, substitute Y

(t) + Y

(t) for y in Eq. (18) and make use

of Eqs. (16) and (17). A similar conclusion holds if g(t) is the sum of any ﬁnite

number of terms. The practical signiﬁcance of this result is that for an equation

whose nonhomogeneous function g(t) can be expressed as a sum, you can consider

instead several simpler equations and then add the results together. The following

example is an illustration of this procedure.

EXAMPLE

Find a particular solution of

′′

− 3y

′

− 4y = 3e

+ 2 sin t − 8e

cos 2t. (19)

By splitting up the right side of Eq. (19), we obtain the three equations

′′

− 3y

′

− 4y = 3e

′′

− 3y

′

− 4y = 2 sin t,

August 7, 2012 21:04 c03 Sheet number 44 Page number 180 cyan black

180 Chapter 3. Second Order Linear Equations

and

′′

− 3y

′

− 4y =−8e

cos 2t.

Solutions of these three equations have been found in Examples 1, 2, and 3, respectively.

Therefore, a particular solution of Eq. (19) is their sum, namely,

Y(t) =−

cos t −

sin t +

cos 2t +

sin 2t.

The procedure illustrated in these examples enables us to solve a fairly large class

of problems in a reasonably efﬁcient manner. However, there is one difﬁculty that

sometimes occurs. The next example illustrates how it arises.

EXAMPLE

Find a particular solution of

′′

− 3y

′

− 4y = 2e

−t

. (20)

Proceeding as in Example 1, we assume that Y(t) = Ae

−t

. By substituting in Eq. (20), we

obtain

(A + 3A − 4A)e

−t

= 2e

−t

. (21)

Since the left side of Eq. (21) is zero, there is no choice of A that satisﬁes this equation.

Therefore, there is no particular solution of Eq. (20) of the assumed form. The reason for this

possibly unexpected result becomes clear if we solve the homogeneous equation

′′

− 3y

′

− 4y = 0 (22)

that corresponds to Eq. (20). A fundamental set of solutions of Eq. (22) is y

(t) = e

−t

and

(t) = e

. Thus our assumed particular solution of Eq. (20) is actually a solution of the

homogeneous equation (22); consequently, it cannot possibly be a solution of the nonho-

mogeneous equation (20). To ﬁnd a solution of Eq. (20), we must therefore consider functions

of a somewhat different form.

At this stage, we have several possible alternatives. One is simply to try to guess the proper

form of the particular solution of Eq. (20). Another is to solve this equation in some different

way and then to use the result to guide ourassumptions if this situation arises again in the future;

see Problems 29 and 35 for other solution methods. Still another possibility is to seek a simpler

equation where this difﬁculty occurs and to use its solution to suggest how we might proceed

with Eq. (20). Adopting the latter approach, we look for a ﬁrst order equation analogous to

Eq. (20). One possibility is the linear equation

′

+ y = 2e

−t

. (23)

If we try to ﬁnd a particular solution of Eq. (23) of the form Ae

−t

, we will fail because e

−t

is a solution of the homogeneous equation y

′

+ y = 0. However, from Section 2.1 we already

know how to solve Eq. (23). An integrating factor is µ(t) = e

, and by multiplying by µ(t) and

then integrating both sides, we obtain the solution

y = 2te

−t

+ ce

−t

. (24)

The second term on the right side of Eq. (24) is the general solution of the homogeneous

equation y

′

+ y = 0, but the ﬁrst term is a solution of the full nonhomogeneous equation (23).

Observe that it involves the exponential factor e

−t

multiplied by the factor t. This is the clue

that we were looking for.

We now return to Eq. (20) and assume a particular solution of the form Y(t) = Ate

−t

. Then

′

(t) = Ae

−t

− Ate

−t

, Y

′′

(t) =−2Ae

−t

+ Ate

−t

. (25)

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3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 181

Substituting these expressions for y, y

′

, and y

′′

in Eq. (20), we obtain

(−2A − 3A)e

−t

+ (A + 3A − 4A)te

−t

= 2e

−t

The coefﬁcient of te

−t

is zero, so from the terms involving e

we have −5A = 2, or A =−2/5.

Thus a particular solution of Eq. (20) is

Y(t) =−

−t

. (26)

The outcome of Example 5 suggests a modiﬁcation of the principle stated pre-

viously: if the assumed form of the particular solution duplicates a solution of the

corresponding homogeneous equation, then modify the assumed particular solution

by multiplying it by t. Occasionally, this modiﬁcation will be insufﬁcient to remove

all duplication with the solutions of the homogeneous equation, in which case it is

necessary to multiply by t a second time. For a second order equation, it will never

be necessary to carry the process further than this.

Summar y. We now summarize the steps involved in ﬁnding the solution of an initial

value problem consisting of a nonhomogeneous equation of the form

′′

+ by

′

+ cy = g(t), (27)

where the coefﬁcients a, b, and c are constants, together with a given set of initial

conditions.

1. Find the general solution of the corresponding homogeneous equation.

2. Make sure that the function g(t) in Eq. (27) belongs to the class of functions discussed

in this section; that is, be sure it involves nothing more than exponential functions, sines,

cosines, polynomials, or sums or products of such functions. If this is not the case, use the

method of variation of parameters (discussed in the next section).

3. If g(t) = g

(t) +···+g

(t)—that is, if g(t) is a sum of n terms—then form n subproblems,

each of which contains only one of the terms g

(t), ..., g

(t).Theith subproblem consists

of the equation

′′

+ by

′

+ cy = g

(t),

where i runs from 1 to n.

4. For the ith subproblem assume a particular solution Y

(t) consisting of the appropriate

exponential function, sine, cosine, polynomial, or combination thereof. If there is any

duplication in the assumed form of Y

(t) with the solutions of the homogeneous equation

(found in step 1), then multiply Y

(t) by t, or (if necessary) by t

, so as to remove the

duplication. See Table 3.5.1.

5. Find a particular solution Y

(t) for each of the subproblems. Then the sum

(t) +···+Y

(t) is a particular solution of the full nonhomogeneous equation (27).

6. Form the sum of the general solution of the homogeneous equation (step 1) and the

particular solution of the nonhomogeneous equation (step 5). This is the general solution

of the nonhomogeneous equation.

7. Use the initial conditions to determine the values of the arbitrary constants remaining in

the general solution.

For some problems this entire procedure is easy to carry out by hand, but often

the algebraic calculations are lengthy. Once you understand clearly how the method

works, a computer algebra system can be of great assistance in executing the details.

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182 Chapter 3. Second Order Linear Equations

TABLE 3.5.1 The Particular Solution of ay

′′

+ by

′

+ cy = g

(t)

(t) Y

(t)

(t) = a

+ a

n−1

+···+a

+ A

n−1

+···+A

)

(t)e

αt

+ A

n−1

+···+A

αt

(t)e

αt

sin βt

cos βt

[(A

+ A

n−1

+···+A

αt

cos βt

+ (B

+ B

n−1

+···+B

αt

sin βt]

Notes. Here s is the smallest nonnegative integer (s = 0, 1, or 2) that will ensure that no

term in Y

(t) is a solution of the corresponding homogeneous equation. Equivalently,

for the three cases,s is the number of times 0 is a root of the characteristic equation,α is

a root of the characteristic equation, and α + iβ is a root of the characteristic equation,

respectively.

The method of undetermined coefﬁcients is self-correcting in the sense that if you

assume too little for Y(t), then a contradiction is soon reached that usually points the

way to the modiﬁcation that is needed in the assumed form. On the other hand,if you

assume too many terms, then some unnecessary work is done and some coefﬁcients

turn out to be zero, but at least the correct answer is obtained.

Proo f of the Me tho d of Un det er min ed Coe ffic ien ts . In the preceding discussion we have

described the method of undetermined coefﬁcients on the basis of several exam-

ples. To prove that the procedure always works as stated, we now give a general

argument, in which we consider three cases corresponding to different forms for the

nonhomogeneous term g(t).

Case 1: g(t) = P

(t) = a

+ a

n−1

+···+a

. In this case Eq. (27) becomes

′′

+ by

′

+ cy = a

+ a

n−1

+···+a

. (28)

To obtain a particular solution, we assume that

Y(t) = A

+ A

n−1

+···+A

n−2

+ A

n−1

t + A

. (29)

Substituting in Eq. (28), we obtain

a[n(n − 1)A

n−2

+···+2A

n−2

]+b(nA

n−1

+···+A

n−1

)

+ c(A

+ A

n−1

+···+A

) = a

+···+a

. (30)

Equating the coefﬁcients of like powers of t, beginning with t

, leads to the following

sequence of equations:

= a

+ nbA

= a

+ bA

n−1

+ 2aA

n−2

= a

Provided that c ̸= 0,the solution of the ﬁrst equation is A

= a

/c, and the remaining

equations determine A

, ..., A

successively. If c = 0 but b ̸= 0, then the polynomial

on the left side of Eq. (30) is of degree n − 1, and we cannot satisfy Eq. (30). To be

August 7, 2012 21:04 c03 Sheet number 47 Page number 183 cyan black

3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 183

sure that aY

′′

(t) + bY

′

(t) is a polynomial of degree n, we must choose Y(t) to be a

polynomial of degree n + 1. Hence we assume that

Y(t) = t(A

+···+A

There is no constant term in this expression for Y(t), but there is no need to include

such a term since a constant is a solution of the homogeneous equation when c = 0.

Since b ̸= 0, we have A

= a

/b(n + 1), and the other coefﬁcients A

, ..., A

can be

determined similarly. If both c and b are zero, we assume that

Y(t) = t

+···+A

The term aY

′′

(t) gives rise to a term of degree n,and we can proceed as before.Again

the constant and linear terms in Y(t) are omitted, since in this case they are both

solutions of the homogeneous equation.

Case 2: g(t) = e

αt

(t). The problem of determining a particular solution of

′′

+ by

′

+ cy = e

αt

(t) (31)

can be reduced to the preceding case by a substitution. Let

Y(t) = e

αt

u(t);

then

′

(t) = e

αt

′

(t) + αu(t)]

and

′′

(t) = e

αt

′′

(t) + 2αu

′

(t) + α

u(t)].

Substituting for y, y

′

, and y

′′

in Eq. (31), canceling the factor e

αt

, and collecting terms,

we obtain

′′

(t) + (2aα + b)u

′

(t) + (aα

+ bα + c)u(t) = P

(t). (32)

The determinationof aparticular solution of Eq.(32) isprecisely thesame problem,

except for the names of the constants, as solving Eq. (28). Therefore, if aα

+ bα + c

is not zero, we assume that u(t) = A

+···+A

; hence a particular solution of

Eq. (31) is of the form

Y(t) = e

αt

+ A

n−1

+···+A

). (33)

On the other hand,if aα

+ bα + c is zero but 2aα + b is not,we must take u(t) to be of

the form t(A

+···+A

).The corresponding form for Y(t) is t times the expression

on the right side of Eq. (33). Note that if aα

+ bα + c is zero, then e

αt

is a solution

of the homogeneous equation. If both aα

+ bα + c and 2aα + b are zero (and this

implies that both e

αt

and te

αt

are solutions of the homogeneous equation), then the

correct form for u(t) is t

+···+A

). Hence Y(t) is t

times the expression on

the right side of Eq. (33).

Case 3: g(t) = e

αt

(t) cos βt or e

αt

(t) sin βt. These two cases are similar, so we

consider only the latter. We can reduce this problem to the preceding one by noting

that, as a consequence of Euler’s formula, sin βt = (e

iβt

− e

−iβt

)/2i. Hence g(t) is of

the form

g(t) = P

(t)

(α+iβ)t

− e

(α−iβ)t

August 7, 2012 21:04 c03 Sheet number 48 Page number 184 cyan black

184 Chapter 3. Second Order Linear Equations

and we should choose

Y(t) = e

(α+iβ)t

+···+A

) + e

(α−iβ)t

+···+B

or, equivalently,

Y(t) = e

αt

+···+A

) cos βt + e

αt

+···+B

) sin βt.

Usually, the latter form is preferred. If α ± iβ satisfy the characteristic equation cor-

responding to the homogeneous equation, we must, of course, multiply each of the

polynomials by t to increase their degrees by one.

If the nonhomogeneous function involves both cos βt and sin βt, it is usually con-

venient to treat these terms together,since each one individually may give rise to the

same form for a particular solution. For example, if g(t) = t sin t + 2 cos t, the form

for Y(t) would be

Y(t) = (A

t + A

) sin t + (B

t + B

) cos t,

provided that sin t and cos t are not solutions of the homogeneous equation.

PROBLEMS In each of Problems 1 through 14, ﬁnd the general solution of the given differential equation.

1. y

′′

− 2y

′

− 3y = 3e

2. y

′′

+ 2y

′

+ 5y = 3 sin 2t

3. y

′′

− y

′

− 2y =−2t + 4t

4. y

′′

+ y

′

− 6y = 12e

+ 12e

−2t

5. y

′′

− 2y

′

− 3y =−3te

−t

6. y

′′

+ 2y

′

= 3 + 4 sin 2t

7. y

′′

+ 9y = t

+ 68.y

′′

+ 2y

′

+ y = 2e

−t

9. 2y

′′

+ 3y

′

+ y = t

+ 3 sin t 10. y

′′

+ y = 3 sin 2t + t cos 2t

11. u

′′

+ ω

u = cos ωt, ω

̸= ω

12. u

′′

+ ω

u = cos ω

13. y

′′

+ y

′

+ 4y = 2 sinh t 14. y

′′

− y

′

− 2y = cosh 2t

Hint: sinh t = (e

− e

−t

)/2 Hint: cosh t = (e

+ e

−t

)/2

In each of Problems 15 through 20, ﬁnd the solution of the given initial value problem.

15. y

′′

+ y

′

− 2y = 2t, y(0) = 0, y

′

(0) = 1

16. y

′′

+ 4y = t

+ 3e

, y(0) = 0, y

′

(0) = 2

17. y

′′

− 2y

′

+ y = te

+ 4, y(0) = 1, y

′

(0) = 1

18. y

′′

− 2y

′

− 3y = 3te

, y(0) = 1, y

′

(0) = 0

19. y

′′

+ 4y = 3 sin 2t, y(0) = 2, y

′

(0) =−1

20. y

′′

+ 2y

′

+ 5y = 4e

−t

cos 2t, y(0) = 1, y

′

(0) = 0

In each of Problems 21 through 28:

(a) Determine a suitable form for Y(t) if the method of undetermined coefﬁcients is to be

used.

(b) Use a computer algebra system to ﬁnd a particular solution of the given equation.

21.

′′

+ 3y

′

= 2t

+ t

−3t

+ sin 3t

22.

′′

+ y = t(1 + sin t)

23.

′′

− 5y

′

+ 6y = e

cos 2t + e

(3t + 4) sin t

24.

′′

+ 2y

′

+ 2y = 3e

−t

+ 2e

−t

cos t + 4e

−t

sin t

25.

′′

− 4y

′

+ 4y = 2t

+ 4te

+ t sin 2t

26.

′′

+ 4y = t

sin 2t + (6t +7) cos 2t

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3.5 Nonhomogeneous Equations; Method of Undetermined Coefﬁcients 185

27. y

′′

+ 3y

′

+ 2y = e

+ 1) sin 2t + 3e

−t

cos t + 4e

28. y

′′

+ 2y

′

+ 5y = 3te

−t

cos 2t − 2te

−2t

cos t

29. Consider the equation

′′

− 3y

′

− 4y = 2e

−t

(i)

from Example 5. Recall that y

(t) = e

−t

and y

(t) = e

are solutions of the corresponding

homogeneous equation. Adapting the method of reduction of order (Section 3.4), seek a

solution of the nonhomogeneous equation of the form Y (t) = v(t)y

(t) = v(t)e

−t

, where

v(t) is to be determined.

(a) Substitute Y(t), Y

′

(t), and Y

′′

(t) into Eq. (i) and show that v(t) must satisfy

′′

− 5v

′

= 2.

(b) Let w(t) = v

′

(t) and show that w(t) must satisfy w

′

− 5w = 2. Solve this equation

for w(t).

Y(t) =−

−t

+ c

−t

The ﬁrst term on the right side is the desired particular solution of the nonhomogeneous

equation. Note that it is a product of t and e

−t

30. Determine the general solution of

′′

+ λ

y =

m=1

sin mπt,

where λ>0 and λ ̸= mπ for m = 1, ...,N.

31.

In many physical problems the nonhomogeneous term may be speciﬁed by different

formulas in different time periods. As an example, determine the solution y = φ(t) of

′′

+ y =

t,0≤ t ≤ π,

πe

π−t

, t >π,

satisfying the initial conditions y(0) = 0 and y

′

(0) = 1. Assume that y and y

′

are also con-

tinuous at t = π. Plot the nonhomogeneous term and the solution as functions of time.

Hint: First solve the initial value problem for t ≤ π; then solve for t >π, determining the

constants in the latter solution from the continuity conditions at t = π.

32.

Follow the instructions in Problem 31 to solve the differential equation

′′

+ 2y

′

+ 5y =

1, 0 ≤ t ≤ π/2,

0, t >π/2

with the initial conditions y(0) = 0 and y

′

(0) = 0.

Behavior of Solutions as t →∞. In Problems 33 and 34, we continue the discussion started

with Problems 37 through 39 of Section 3.4. Consider the differential equation

′′

+ by

′

+ cy = g(t), (i)

where a, b, and c are positive.

33. If Y

(t) and Y

(t) are solutions of Eq. (i), show that Y

(t) − Y

(t) → 0 as t →∞. Is this

result true if b = 0?

34. If g(t) = d, a constant, show that every solution of Eq. (i) approaches d/c as t →∞. What

happens if c = 0? What if b = 0 also?

August 7, 2012 21:04 c03 Sheet number 50 Page number 186 cyan black

186 Chapter 3. Second Order Linear Equations

35. In this problem we indicate an alternative procedure

for solving the differential equation

′′

+ by

′

+ cy = (D

+ bD + c)y = g(t), (i)

where b and c are constants, and D denotes differentiation with respect to t. Let r

and r

be the zeros of the characteristic polynomial of the corresponding homogeneous equation.

These roots may be real and different, real and equal, or conjugate complex numbers.

(a) Verify that Eq. (i) can be written in the factored form

(D − r

)(D − r

)y = g(t),

where r

+ r

=−b and r

= c.

(b) Let u = (D − r

)y. Then show that the solution of Eq (i) can be found by solving the

following two ﬁrst order equations:

(D − r

)u = g(t), (D − r

)y = u(t).

In each of Problems 36 through 39,use the method of Problem 35 to solve the given differential

equation.

36. y

′′

− 3y

′

− 4y = 3e

(see Example 1)

37. 2y

′′

+ 3y

′

+ y = t

+ 3 sin t (see Problem 9)

38. y

′′

+ 2y

′

+ y = 2e

−t

(see Problem 8)

39. y

′′

+ 2y

′

= 3 + 4 sin 2t (see Problem 6)

3.6 Variation of Parameters

In this section we describe another method of ﬁnding a particular solution of a non-

homogeneous equation.Thismethod,variation of parameters,is due to Lagrange and

complements the method of undetermined coefﬁcients rather well. The main advan-

tage of variation of parameters is that it is a general method;in principle at least,it can

be applied to any equation, and it requires no detailed assumptions about the form

of the solution. In fact,later in this section we use this method to derive a formula for

a particular solution of an arbitrary second order linear nonhomogeneous differen-

tial equation. On the other hand, the method of variation of parameters eventually

requires us to evaluate certain integrals involving the nonhomogeneous term in the

differential equation, and this may present difﬁculties. Before looking at this method

in the general case, we illustrate its use in an example.

EXAMPLE

Find a particular solution of

′′

+ 4y = 3 csc t. (1)

Observe that this problem is not a good candidate for the method of undetermined coefﬁ-

cients, as described in Section 3.5, because the nonhomogeneous term g(t) = 3 csc t involves

R. S. Luthar, “Another Approach to a Standard Differential Equation,”Two Year College Mathematics

Journal 10 (1979), pp. 200–201. Also see D. C. Sandell and F. M. Stein, “Factorization of Operators of

Second Order Linear Homogeneous Ordinary Differential Equations,” Two Year College Mathematics

Journal 8 (1977), pp. 132–141, for a more general discussion of factoring operators.

August 7, 2012 21:04 c03 Sheet number 51 Page number 187 cyan black

3.6 Variation of Parameters 187

a quotient (rather than a sum or a product) of sin t or cos t. Therefore, we need a different

approach. Observe also that the homogeneous equation corresponding to Eq. (1) is

′′

+ 4y = 0, (2)

and that the general solution of Eq. (2) is

(t) = c

cos 2t + c

sin 2t. (3)

The basic idea in the method of variation of parameters is to replace the constants c

and c

in Eq. (3) by functions u

(t) and u

(t), respectively, and then to determine these functions so

that the resulting expression

y = u

(t) cos 2t + u

(t) sin 2t (4)

is a solution of the nonhomogeneous equation (1).

To determine u

and u

, we need to substitute for y from Eq. (4) in Eq. (1). However, even

without carrying out this substitution,we can anticipate that the result will be a single equation

involving some combination of u

, u

, and their ﬁrst two derivatives. Since there is only one

equation and two unknown functions, we can expect that there are many possible choices of

and u

that will meet our needs. Alternatively, we may be able to impose a second condition

of our own choosing, thereby obtaining two equations for the two unknown functions u

and

. We will soon show (following Lagrange) that it is possible to choose this second condition

in a way that makes the computation markedly more efﬁcient.

Returning now to Eq. (4), we differentiate it and rearrange the terms, thereby obtaining

′

=−2u

(t) sin 2t + 2u

(t) cos 2t + u

′

(t) cos 2t + u

′

(t) sin 2t. (5)

Keeping in mind the possibility of choosing a second condition on u

and u

, let us require the

sum of the last two terms on the right side of Eq. (5) to be zero; that is, we require that

′

(t) cos 2t + u

′

(t) sin 2t = 0. (6)

It then follows from Eq. (5) that

′

=−2u

(t) sin 2t + 2u

(t) cos 2t. (7)

Although the ultimate effect of the condition (6) is not yet clear,at the very least it has simpliﬁed

the expression for y

′

. Further, by differentiating Eq. (7) we obtain

′′

=−4u

(t) cos 2t − 4u

(t) sin 2t − 2u

′

(t) sin 2t + 2u

′

(t) cos 2t. (8)

Then, substituting for y and y

′′

in Eq. (1) from Eqs. (4) and (8), respectively, we ﬁnd that

′′

+ 4y =−4u

(t) cos 2t − 4u

(t) sin 2t − 2u

′

(t) sin 2t + 2u

′

(t) cos 2t

+ 4u

(t) cos 2t + 4u

(t) sin 2t = 3 csc t.

Hence u

and u

must satisfy

−2u

′

(t) sin 2t + 2u

′

(t) cos 2t = 3 csc t. (9)

Summarizing our results to this point, we want to choose u

and u

so as to satisfy Eqs. (6)

and (9). These equations can be viewed as a pair of linear algebraic equations for the two

unknown quantities u

′

(t) and u

′

(t). Equations (6) and (9) can be solved in various ways. For

example, solving Eq. (6) for u

′

(t), we have

′

(t) =−u

′

(t)

cos 2t

sin 2t

. (10)

August 7, 2012 21:04 c03 Sheet number 52 Page number 188 cyan black

188 Chapter 3. Second Order Linear Equations

Then, substituting for u

′

(t) in Eq. (9) and simplifying, we obtain

′

(t) =−

3 csc t sin 2t

=−3 cos t. (11)

Further,putting this expression for u

′

(t) back in Eq. (10) and using the double-angle formulas,

we ﬁnd that

′

(t) =

3 cos t cos 2t

sin 2t

3(1 − 2 sin

2 sin t

csc t − 3 sin t. (12)

Having obtained u

′

(t) and u

′

(t), we next integrate so as to ﬁnd u

(t) and u

(t). The result is

(t) =−3 sin t + c

(13)

and

(t) =

ln |csct − cot t|+3 cos t + c

. (14)

On substituting these expressions in Eq. (4), we have

y =−3 sin t cos 2t +

ln |csct − cot t|sin 2t + 3 cos t sin 2t

+ c

cos 2t + c

sin 2t.

Finally, by using the double-angle formulas once more, we obtain

y = 3 sin t +

ln |csct − cot t|sin 2t + c

cos 2t + c

sin 2t. (15)

The terms in Eq. (15) involving the arbitrary constants c

and c

are the general solution of the

corresponding homogeneous equation, while the remaining terms are a particular solution of

the nonhomogeneous equation (1). Thus Eq. (15) is the general solution of Eq. (1).

In the preceding example the method of variation of parameters worked well in

determining a particular solution,and hence the general solution,of Eq. (1).The next

question is whether this method can be applied effectively to an arbitrary equation.

Therefore, we consider

′′

+ p(t)y

′

+ q(t)y = g(t), (16)

where p, q, and g are given continuous functions. As a starting point, we assume that

we know the general solution

(t) = c

(t) + c

(t) (17)

of the corresponding homogeneous equation

′′

+ p(t)y

′

+ q(t)y = 0. (18)

This is a major assumption. So far we have shown how to solve Eq. (18) only if it has

constant coefﬁcients. If Eq. (18) has coefﬁcients that depend on t, then usually the

methods described in Chapter 5 must be used to obtain y

(t).

The crucial idea, as illustrated in Example 1, is to replace the constants c

and c

in Eq. (17) by functions u

(t) and u

(t), respectively; thus we have

y = u

(t)y

(t) + u

(t)y

(t). (19)

Then we try to determine u

(t) and u

(t) so that the expression in Eq. (19) is a solution

of the nonhomogeneous equation (16) rather than the homogeneous equation (18).

Thus we differentiate Eq. (19), obtaining

′

= u

′

(t)y

(t) + u

(t)y

′

(t) + u

′

(t)y

(t) + u

(t)y

′

(t). (20)

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3.6 Variation of Parameters 189

As in Example 1, we now set the terms involving u

′

(t) and u

′

(t) in Eq. (20) equal to

zero; that is, we require that

′

(t)y

(t) + u

′

(t)y

(t) = 0. (21)

Then, from Eq. (20), we have

′

= u

(t)y

′

(t) + u

(t)y

′

(t). (22)

Further, by differentiating again, we obtain

′′

= u

′

(t)y

′

(t) + u

(t)y

′′

(t) + u

′

(t)y

′

(t) + u

(t)y

′′

(t). (23)

Now we substitute for y, y

′

, and y

′′

in Eq. (16) from Eqs. (19), (22), and (23),

respectively. After rearranging the terms in the resulting equation, we ﬁnd that

(t)

′′

(t) + p(t)y

′

(t) + q(t)y

(t)

+ u

(t)

′′

(t) + p(t)y

′

(t) + q(t)y

(t)

+ u

′

(t)y

′

(t) + u

′

(t)y

′

(t) = g(t). (24)

Each of the expressions in square brackets in Eq. (24) is zero because both y

and y

are solutions of the homogeneous equation (18). Therefore, Eq. (24) reduces to

′

(t)y

′

(t) + u

′

(t)y

′

(t) = g(t). (25)

Equations (21) and (25) form a system of two linear algebraic equations for the

derivatives u

′

(t) and u

′

(t) of the unknown functions. They correspond exactly to

Eqs. (6) and (9) in Example 1.

By solving the system (21), (25) we obtain

′

(t) =−

(t)g(t)

W(y

, y

)(t)

, u

′

(t) =

(t)g(t)

W(y

, y

)(t)

, (26)

where W(y

, y

) is the Wronskian of y

and y

. Note that division by W is permissible

since y

and y

are a fundamental set of solutions, and therefore their Wronskian

is nonzero. By integrating Eqs. (26), we ﬁnd the desired functions u

(t) and u

(t),

namely,

(t) =−

(t)g(t)

W(y

, y

)(t)

dt + c

, u

(t) =

(t)g(t)

W(y

, y

)(t)

dt + c

. (27)

If the integrals in Eqs. (27) can be evaluated in terms of elementary functions,then we

substitute the results in Eq. (19), thereby obtaining the general solution of Eq. (16).

More generally, the solution can always be expressed in terms of integrals, as stated

in the following theorem.

Theorem 3.6.1

If the functions p, q, and g are continuous on an open interval I,and if the functions

and y

are a fundamental set of solutions of the homogeneous equation (18)

corresponding to the nonhomogeneous equation (16)

′′

+ p(t)y

′

+ q(t)y = g(t),

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190 Chapter 3. Second Order Linear Equations

then a particular solution of Eq. (16) is

Y(t) =−y

(t)

(s)g(s)

W(y

, y

)(s)

ds + y

(t)

(s)g(s)

W(y

, y

)(s)

ds, (28)

where t

is any conveniently chosen point in I. The general solution is

y = c

(t) + c

(t) + Y(t), (29)

as prescribed by Theorem 3.5.2.

By examining the expression (28) and reviewing the process by which we derived

it,we can see that there may be two major difﬁculties in using the method of variation

of parameters. As we have mentioned earlier, one is the determination of y

(t) and

(t), a fundamental set of solutions of the homogeneous equation (18), when the

coefﬁcients in that equation are not constants. The other possible difﬁculty lies in

the evaluation of the integrals appearing in Eq. (28). This depends entirely on the

nature of the functions y

, y

, and g. In using Eq. (28), be sure that the differential

equation is exactly in the form (16); otherwise, the nonhomogeneous term g(t) will

not be correctly identiﬁed.

A major advantage of the method of variation of parameters is that Eq. (28) pro-

vides an expression for the particular solution Y(t) in terms of an arbitrary forcing

function g(t). This expression is a good starting point if you wish to investigate the

effect of variations in the forcing function, or if you wish to analyze the response of

a system to a number of different forcing functions.

PROBLEMS In each of Problems 1 through 4, use the method of variation of parameters to ﬁnd a particular

solution of the given differential equation. Then check your answer by using the method of

undetermined coefﬁcients.

1. y

′′

− 5y

′

+ 6y = 2e

2. y

′′

− y

′

− 2y = 2e

−t

3. y

′′

+ 2y

′

+ y = 3e

−t

4. 4y

′′

− 4y

′

+ y = 16e

t/2

In each of Problems 5 through 12, ﬁnd the general solution of the given differential equation.

In Problems 11 and 12, g is an arbitrary continuous function.

5. y

′′

+ y = tan t,0< t <π/26. y

′′

+ 9y = 9 sec

3t,0< t <π/6

7. y

′′

+ 4y

′

+ 4y = t

−2

−2t

, t > 08.y

′′

+ 4y = 3 csc 2t,0< t <π/2

9. 4y

′′

+ y = 2 sec(t/2), −π<t <π 10. y

′′

− 2y

′

+ y = e

/(1 + t

)

11. y

′′

− 5y

′

+ 6y = g(t) 12. y

′′

+ 4y = g(t)

In each of Problems 13 through 20, verify that the given functions y

and y

satisfy the corre-

sponding homogeneous equation;then ﬁnd a particular solution of the givennonhomogeneous

equation. In Problems 19 and 20, g is an arbitrary continuous function.

13. t

′′

− 2y = 3t

− 1, t > 0; y

(t) = t

, y

(t) = t

−1

14. t

′′

− t(t + 2)y

′

+ (t + 2)y = 2t

, t > 0; y

(t) = t, y

(t) = te

15. ty

′′

− (1 + t)y

′

+ y = t

, t > 0; y

(t) = 1 + t, y

(t) = e

16. (1 − t)y

′′

+ ty

′

− y = 2(t − 1)

−t

,0< t < 1; y

(t) = e

, y

(t) = t

17. x

′′

− 3xy

′

+ 4y = x

ln x, x > 0; y

(x) = x

, y

(x) = x

ln x

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3.6 Variation of Parameters 191

18. x

′′

+ xy

′

+ (x

− 0.25)y = 3x

3/2

sin x, x > 0;

(x) = x

−1/2

sin x, y

(x) = x

−1/2

cos x

19. (1 − x)y

′′

+ xy

′

− y = g(x),0< x < 1; y

(x) = e

, y

(x) = x

20. x

′′

+ xy

′

+ (x

− 0.25)y = g(x), x > 0; y

(x) = x

−1/2

sin x, y

(x) = x

−1/2

cos x

21. Show that the solution of the initial value problem

L[y]=y

′′

+ p(t)y

′

+ q(t)y = g(t), y(t

) = y

, y

′

) = y

′

(i)

can be written as y = u(t) + v(t), where u and v are solutions of the two initial value

problems

L[u]=0, u(t

) = y

, u

′

) = y

′

, (ii)

L[v]=g(t), v(t

) = 0, v

′

) = 0, (iii)

respectively. In other words, the nonhomogeneities in the differential equation and in the

initialconditions can be dealt with separately. Observe that u is easy to ﬁnd if afundamental

set of solutions of L[u]=0 is known.

22. By choosing the lower limit of integration in Eq. (28) in the text as the initial point t

,show

that Y(t) becomes

Y(t) =

(s)y

(t) − y

(t)y

(s)

(s)y

′

(s) − y

′

(s)y

(s)

g(s) ds.

Show that Y(t) is a solution of the initial value problem

L[y]=g(t), y(t

) = 0, y

′

) = 0.

Thus Y can be identiﬁed with v in Problem 21.

23. (a) Use the result of Problem 22 to show that the solution of the initial value problem

′′

+ y = g(t), y(t

) = 0, y

′

) = 0 (i)

y =

sin(t − s)g(s) ds. (ii)

(b) Use the result of Problem 21 to ﬁnd the solution of the initial value problem

′′

+ y = g(t), y(0) = y

, y

′

(0) = y

′

24. Use the result of Problem 22 to ﬁnd the solution of the initial value problem

L[y]=(D − a)(D − b)y = g(t), y(t

) = 0, y

′

) = 0,

where a and b are real numbers with a ̸= b.

25. Use the result of Problem 22 to ﬁnd the solution of the initial value problem

L[y]=[D

− 2λD + (λ

+ µ

)]y = g(t), y(t

) = 0, y

′

) = 0.

Note that the roots of the characteristic equation are λ ± iµ.

26. Use the result of Problem 22 to ﬁnd the solution of the initial value problem

L[y]=(D − a)

y = g(t), y(t

) = 0, y

′

) = 0,

where a is any real number.

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192 Chapter 3. Second Order Linear Equations

27. By combining the results of Problems 24 through 26, show that the solution of the initial

value problem

L[y]=(D

+ bD + c)y = g(t), y(t

) = 0, y

′

) = 0,

where b and c are constants, has the form

y = φ(t) =

K(t − s)g(s) ds. (i)

The function K depends onlyon the solutions y

and y

of the corresponding homogeneous

equation and is independent of the nonhomogeneous term. Once K is determined, all

nonhomogeneous problems involving the same differential operator L are reduced to the

evaluation of an integral. Note also that although K depends on both t and s,only the com-

bination t − s appears, so K is actually a function of a single variable.When we think of g(t)

as the input to the problem and of φ(t) as the output, it follows from Eq. (i) that the output

depends on the input over the entire interval from the initial point t

to the current value t.

Theintegral in Eq.(i) is called the convolution of K and g,andK is referred to as the kernel.

28. The method of reduction of order (Section 3.4) can also be used for the nonhomogeneous

equation

′′

+ p(t)y

′

+ q(t)y = g(t), (i)

provided one solution y

of the corresponding homogeneous equation is known. Let

y = v(t)y

(t) and show that y satisﬁes Eq. (i) if v is a solution of

(t)v

′′

+[2y

′

(t) + p(t)y

(t)]v

′

= g(t). (ii)

Equation (ii) is a ﬁrst order linear equation for v

′

. By solving this equation, integrating

the result, and then multiplying by y

(t), you can ﬁnd the general solution of Eq. (i).

In each of Problems 29 through 32, use the method outlined in Problem 28 to solve the given

differential equation.

29. t

′′

− 2ty

′

+ 2y = 4t

, t > 0; y

(t) = t

30. t

′′

+ 7ty

′

+ 5y = t, t > 0; y

(t) = t

−1

31. ty

′′

− (1 + t)y

′

+ y = t

, t > 0; y

(t) = 1 + t (see Problem 15)

32. (1 − t)y

′′

+ ty

′

− y = 2(t − 1)

−t

,0< t < 1; y

(t) = e

(see Problem 16)

3.7 Mechanical and Electrical Vibrations

One of the reasons why second order linear equations with constant coefﬁcients

are worth studying is that they serve as mathematical models of some important

physical processes. Two important areas of application are the ﬁelds of mechanical

and electrical oscillations. For example, the motion of a mass on a vibrating spring,

the torsional oscillations of a shaft with a ﬂywheel, the ﬂow of electric current in

a simple series circuit, and many other physical problems are all described by the

solution of an initial value problem of the form

′′

+ by

′

+ cy = g(t), y(0) = y

, y

′

(0) = y

′

. (1)

This illustrates a fundamental relationship between mathematics and physics:

many physical problems may have the same mathematical model.Thus,once we know

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3.7 Mechanical and Electrical Vibrations 193

how to solve the initial value problem (1), it is only necessary to make appropriate

interpretations of the constants a, b, and c, and of the functions y and g, to obtain

solutions of different physical problems.

We will study the motion of a mass on a spring in detail because understanding the

behavior of this simple system is the ﬁrst step in the investigation of more complex

vibrating systems. Further, the principles involved are common to many problems.

Consider a mass m hanging at rest on the end of a vertical spring of original length

l, as shown in Figure 3.7.1. The mass causes an elongation L of the spring in the

downward (positive) direction. In this static situation there are two forces acting at

the point where the mass is attached to the spring; see Figure 3.7.2. The gravitational

force, or weight of the mass, acts downward and has magnitude mg, where g is the

acceleration due to gravity.There is also a force F

,due to the spring,that acts upward.

If we assume that the elongation L of the spring is small,the spring force is very nearly

proportional to L; this is known as Hooke’s

law. Thus we write F

=−kL, where

the constant of proportionality k is called the spring constant, and the minus sign is

due to the fact that the spring force acts in the upward (negative) direction. Since

the mass is in equilibrium, the two forces balance each other, which means that

mg − kL = 0. (2)

For a given weight w = mg, you can measure L and then use Eq. (2) to determine k .

Note that k has the units of force/length.

l + L + u

FIGURE 3.7.1 A spring–mass system.

= –kL

w = mg

FIGURE 3.7.2 Force diagram for a spring–mass system.

Robert Hooke (1635–1703) was an English scientist with wide-ranging interests. His most important

book, Micrographia, was published in 1665 and described a variety of microscopical observations. Hooke

ﬁrst published his law of elastic behavior in 1676 as ceiiinosssttuv; in 1678 he gave the interpretation ut

tensio sic vis, which means, roughly, “as the force so is the displacement.”

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194 Chapter 3. Second Order Linear Equations

In the corresponding dynamic problem, we are interested in studying the motion

of the mass when it is acted on by an external force or is initially displaced. Let

u(t), measured positive downward, denote the displacement of the mass from its

equilibrium position at time t; see Figure 3.7.1. Then u(t) is related to the forces

acting on the mass through Newton’s law of motion

′′

(t) = f (t), (3)

where u

′′

is the acceleration of the mass and f is the net force acting on the mass.

Observe that both u and f are functions of time. In this dynamic problem there are

now four separate forces that must be considered.

1. The weight w = mg of the mass always acts downward.

2. The spring force F

is assumed to be proportional to the total elongation L + u of the

spring and always acts to restore the spring to its natural position. If L + u > 0, then

the spring is extended, and the spring force is directed upward. In this case

=−k(L + u). (4)

On the other hand, if L + u < 0, then the spring is compressed a distance |L + u|, and

the spring force, which is now directed downward, is given by F

= k|L + u|. However,

when L + u < 0, it follows that |L + u|=−(L + u),soF

is again given by Eq. (4). Thus,

regardless of the position of the mass, the force exerted by the spring is always expressed

by Eq. (4).

3. The damping or resistive force F

always acts in the direction opposite to the direction of

motion of the mass. This force may arise from several sources: resistance from the air or

other medium in which the mass moves, internal energy dissipation due to the extension

or compression of the spring, friction between the mass and the guides (if any) that con-

strain its motion to one dimension,ora mechanical device (dashpot) that imparts a resistive

force to the mass. In any case, we assume that the resistive force is proportional to the

speed |du/dt| of the mass; this is usually referred to as viscous damping. If du/dt > 0,

then u is increasing, so the mass is moving downward. Then F

is directed upward and is

given by

(t) =−γu

′

(t), (5)

where γ is a positive constant of proportionality known as the damping constant. On

the other hand, if du/dt < 0, then u is decreasing, the mass is moving upward, and F

directed downward. In this case, F

= γ|u

′

(t)|; since |u

′

(t)|=−u

′

(t), it follows that F

(t)

is again given by Eq. (5). Thus, regardless of the direction of motion of the mass, the

damping force is always expressed by Eq. (5).

The damping force may be rather complicated, and the assumption that it is modeled

adequately by Eq. (5) may be open to question. Some dashpots do behave as Eq. (5)

states, and if the other sources of dissipation are small, it may be possible to neglect

them altogether or to adjust the damping constant γ to approximate them. An important

beneﬁt of the assumption (5) is that it leads to a linear (rather than a nonlinear) differential

equation. In turn, this means that a thorough analysis of the system is straightforward, as

we will show in this section and the next.

4. An applied external force F(t) is directed downward or upward as F(t) is positive or

negative. This could be a force due to the motion of the mount to which the spring is

attached, or it could be a force applied directly to the mass. Often the external force

is periodic.

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3.7 Mechanical and Electrical Vibrations 195

Taking account of these forces, we can now rewrite Newton’s law (3) as

′′

(t) = mg + F

(t) + F

(t) + F(t)

= mg − k[L + u(t)]−γu

′

(t) + F(t). (6)

Since mg − kL = 0 by Eq. (2), it follows that the equation of motion of the mass is

′′

(t) + γu

′

(t) + ku(t) = F(t), (7)

where the constants m, γ, and k are positive. Note that Eq. (7) has the same form as

Eq. (1).

It is important to understand that Eq. (7) is only an approximate equation for the

displacement u(t). In particular, both Eqs. (4) and (5) should be viewed as approx-

imations for the spring force and the damping force, respectively. In our derivation

we have also neglected the mass of the spring in comparison with the mass of the

attached body.

The complete formulation of the vibration problem requires that we specify two

initial conditions,namely, the initial position u

and the initial velocity v

of the mass:

u(0) = u

, u

′

(0) = v

. (8)

It follows from Theorem 3.2.1 that these conditions give a mathematical problem

that has a unique solution. This is consistent with our physical intuition that if the

mass is set in motion with a given initial displacement and velocity, then its posi-

tion will be determined uniquely at all future times. The position of the mass is

given (approximately) by the solution of Eq. (7) subject to the prescribed initial

conditions (8).

EXAMPLE

A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional

6 in displacement in the positive direction and then released. The mass is in a medium that

exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumptions

discussed in this section, formulate the initial value problem that governs the motion of the

mass.

The required initial value problem consists of the differential equation (7) and initial condi-

tions (8), so our task is to determine the various constants that appear in these equations. The

ﬁrst step is to choose the units of measurement. Based on the statement of the problem, it is

natural to use the English rather than the metric system of units. The only time unit mentioned

is the second, so we will measure t in seconds. On the other hand, both the foot and the inch

appear in the statement as units of length. It is immaterial which one we use, but having made

a choice, we must be consistent. To be deﬁnite, let us measure the displacement u in feet.

Since nothing is said in the statement of the problem about an external force, we assume

that F(t) = 0. To determine m, note that

m =

4 lb

32 ft/s

lb·s

The damping coefﬁcient γ is determined from the statement that γu

′

is equal to 6 lb when u

′

is 3 ft/s. Therefore,

γ =

6 lb

3 ft/s

= 2

lb·s

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196 Chapter 3. Second Order Linear Equations

The spring constant k is found from the statement that the mass stretches the spring by 2 in,

or 1/6 ft. Thus

k =

4 lb

1/6 ft

= 24

Consequently, Eq. (7) becomes

′′

+ 2u

′

+ 24u = 0,

′′

+ 16u

′

+ 192u = 0. (9)

The initial conditions are

u(0) =

, u

′

(0) = 0. (10)

The second initial condition is implied by the word “released” in the statement of the problem,

which we interpret to mean that the mass is set in motion with no initial velocity.

Undamped Free Vibrations. If there is no external force, then F(t) = 0 in Eq. (7). Let

us also suppose that there is no damping, so that γ = 0; this is an idealized conﬁg-

uration of the system, seldom (if ever) completely attainable in practice. However,

if the actual damping is very small, then the assumption of no damping may yield

satisfactory results over short to moderate time intervals. In this case the equation

of motion (7) reduces to

′′

+ ku = 0. (11)

The characteristic equation for Eq. (11) is

+ k = 0

and its roots are r =±i

k/m. Thus the general solution of Eq. (11) is

u = A cosω

t + B sin ω

t, (12)

where

= k/m. (13)

The arbitrary constants A and B can be determined if initial conditions of the form

(8) are given.

In discussing the solution of Eq. (11), it is convenient to rewrite Eq. (12) in the

form

u = R cos(ω

t − δ), (14)

u = R cosδ cos ω

t + R sin δ sin ω

t. (15)

By comparing Eq. (15) with Eq. (12), we ﬁnd that A, B, R, and δ are related by the

equations

A = R cosδ, B = R sin δ. (16)

Thus

R =

+ B

, tan δ = B/A. (17)

In calculating δ, we must take care to choose the correct quadrant; this can be done

by checking the signs of cos δ and sin δ in Eqs. (16).

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3.7 Mechanical and Electrical Vibrations 197

The graphof Eq. (14),or the equivalent Eq.(12),fora typicalset of initialconditions

is shown in Figure 3.7.3.Thegraph isa displacedcosine wavethat describesa periodic,

or simple harmonic, motion of the mass. The period of the motion is

T =

2π

= 2π

)

1/2

. (18)

The circular frequency ω

k/m, measured in radians per unit time, is called the

natural frequency of the vibration. The maximum displacement R of the mass from

equilibrium is the amplitude of the motion. The dimensionless parameter δ is called

the phase,or phase angle,and measures the displacement of the wave from its normal

position corresponding to δ = 0.

–R

R cos

+ 2

FIGURE 3.7.3 Simple harmonic motion; u = R cos(ω

t − δ).

Note that the motion described by Eq. (14) has a constant amplitude that does not

diminish with time. This reﬂects the fact that, in the absence of damping, there is no

way for the system to dissipate the energy imparted to it by the initial displacement

and velocity. Further, for a given mass m and spring constant k, the system always

vibrates at the same frequency ω

, regardless of the initial conditions. However, the

initial conditions do help to determine the amplitude of the motion. Finally, observe

from Eq. (18) that T increases as m increases, so larger masses vibrate more slowly.

On the other hand, T decreases as k increases, which means that stiffer springs cause

the system to vibrate more rapidly.

EXAMPLE

Suppose that a mass weighing 10 lb stretches a spring 2 in. If the mass is displaced an additional

2 in and is then set in motion with an initial upward velocity of 1 ft/s, determine the position

of the mass at any later time. Also determine the period, amplitude, and phase of the motion.

The spring constant is k = 10 lb/2in= 60 lb/ft, and the mass is m = w/g = 10/32 lb·s

/ft.

Hence the equation of motion reduces to

′′

+ 192u = 0, (19)

and the general solution is

u = A cos(8

√

3t) + B sin(8

√

3t).

The solution satisfying the initial conditions u(0) = 1/6 ft and u

′

(0) =−1 ft/s is

u =

cos(8

√

3t) −

√

sin(8

√

3t). (20)

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198 Chapter 3. Second Order Linear Equations

The natural frequency is ω

√

192

∼

13.856 rad/s, so the period is T = 2π/ω

∼

0.45345 s.

The amplitude R and phase δ are found from Eqs. (17). We have

192

576

,soR

∼

0.18162 ft.

The second of Eqs. (17) yields tan δ =−

√

3/4. There are two solutions of this equation, one in

the second quadrant and one in the fourth. In the present problem, cos δ>0 and sin δ<0, so

δ is in the fourth quadrant. In fact,

δ =−arctan(

√

3/4)

∼

−0.40864 rad.

The graph of the solution (20) is shown in Figure 3.7.4.

0.2

– 0.2

1.5

T = 0.453

u = 0.182 cos(8√3 t + 0.409)

R = 0.182

FIGURE 3.7.4 An undamped free vibration: u

′′

+ 192u = 0, u(0) = 1/6, u

′

(0) =−1.

Damped Free Vibrations. If we include the effect of damping, the differential equation

governing the motion of the mass is

′′

+ γu

′

+ ku = 0. (21)

We are especially interested in examining the effect of variations in the damping

coefﬁcient γ for given values of the mass m and spring constant k. The corresponding

characteristic equation is

+ γr + k = 0,

and its roots are

, r

−γ ±

− 4km

−1 ±

1 −

4km

. (22)

Depending on the sign of γ

− 4km, the solution u has one of the following forms:

− 4km > 0, u = Ae

+ Be

; (23)

− 4km = 0, u = (A + Bt)e

−γt/2m

; (24)

− 4km < 0, u = e

−γt/2m

(A cos µt + B sin µt), µ =

(4km − γ

)

1/2

> 0. (25)

Since m, γ, and k are positive, γ

− 4km is always less than γ

. Hence,

if γ

− 4km ≥ 0, then the values of r

and r

given by Eq. (22) are negative.If

− 4km < 0, then the values of r

and r

are complex, but with negative real part.

Thus, in all cases, the solution u tends to zero as t →∞; this occurs regardless of the

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3.7 Mechanical and Electrical Vibrations 199

values of the arbitrary constants A and B—that is, regardless of the initial conditions.

This conﬁrms our intuitive expectation, namely, that damping gradually dissipates

the energy initially imparted to the system, and consequently the motion dies out

with increasing time.

The most important case is the third one, which occurs when the damping is small.

If we let A = R cosδ and B = R sin δ in Eq. (25), then we obtain

u = Re

−γt/2m

cos(µt − δ). (26)

The displacement u lies between the curves u =±Re

−γt/2m

; hence it resembles a

cosine wave whose amplitude decreases as t increases. A typical example is sketched

in Figure 3.7.5. The motion is called a damped oscillation or a damped vibration. The

amplitude factor R depends on m, γ, k, and the initial conditions.

+ 3

R cos

– t/2m

−Re

– t/2m

δ π

+ 2

δ π

FIGURE 3.7.5 Damped vibration; u = Re

−γt/2m

cos(µt − δ).

Although the motion is not periodic, the parameter µ determines the frequency

with which the mass oscillates back and forth; consequently, µ is called the quasi

frequency. By comparing µ with the frequency ω

of undamped motion, we ﬁnd that

(4km − γ

)

1/2

/2m

k/m

1 −

4km

(

1/2

∼

1 −

8km

. (27)

The last approximation is valid when γ

/4km is small; we refer to this situation as

“small damping.”Thus the effect of small damping is to reduce slightly the frequency

of the oscillation. By analogy with Eq. (18), the quantity T

= 2π/µ is called the

quasi period. It is the time between successive maxima or successive minima of

the position of the mass, or between successive passages of the mass through its

equilibrium position while going in the same direction. The relation between T

and

T is given by

1 −

4km

(

−1/2

∼

1 +

8km

(

, (28)

where again the lastapproximation is valid whenγ

/4km is small.Thussmall damping

increases the quasi period.

Equations (27) and (28) reinforce the signiﬁcance of the dimensionless ratio

/4km. It is not the magnitude of γ alone that determines whether damping is large

or small, but the magnitude of γ

compared to 4km. When γ

/4km is small, then

damping has a small effect on the quasi frequency and quasi period of the motion.

August 7, 2012 21:04 c03 Sheet number 64 Page number 200 cyan black

200 Chapter 3. Second Order Linear Equations

On the other hand, if we want to study the detailed motion of the mass for all time,

then we can never neglect the damping force, no matter how small.

As γ

/4km increases, the quasi frequency µ decreases and the quasi period T

increases. In fact, µ → 0 and T

→∞as γ → 2

√

km. As indicated by Eqs. (23), (24),

and (25), the nature of the solution changes as γ passes through the value 2

√

km.

The motion with γ = 2

√

km is said to be critically damped. For larger values of γ

the motion is said to be overdamped. In these cases, given by Eqs. (24) and (23),

respectively, the mass may pass through its equilibrium position at most once (see

Figure3.7.6) andthen creeps back toit.The massdoes notoscillate about equilibrium,

as it does for small γ. Two typical examples of critically damped motion are shown

in Figure 3.7.6, and the situation is discussed further in Problems 21 and 22.

–1

2 4 6 10

u(0) = , u'(0) =

u = + 2t e

–t/2

(

)

u(0) = , u'

u = – t e

–t/2

(

)

FIGURE 3.7.6 Critically damped motions: u

′′

+ u

′

+ 0.25u = 0; u = (A + Bt)e

−t/2

EXAMPLE

The motion of a certain spring–mass system is governed by the differential equation

′′

+ 0.125u

′

+ u = 0, (29)

where u is measured in feet and t in seconds. If u(0) = 2 and u

′

(0) = 0, determine the position

of the mass at any time. Find the quasi frequency and the quasi period, as well as the time

at which the mass ﬁrst passes through its equilibrium position. Also ﬁnd the time τ such that

|u(t)| < 0 .1 for all t >τ.

The solution of Eq. (29) is

u = e

−t/16

A cos

√

255

t + B sin

√

255

To satisfy the initial conditions, we must choose A = 2 and B = 2/

√

255; hence the solution of

the initial value problem is

u = e

−t/16

2 cos

√

255

t +

√

255

sin

√

255

√

255

−t/16

cos

√

255

t − δ

, (30)

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3.7 Mechanical and Electrical Vibrations 201

–1

–2

u" + u

u" + 0.125 u' + u = 0

u(0) = 2, u' (0) = 0

10 20 30 40 50

FIGURE 3.7.7 Vibration with small damping (solid curve)

and with no damping (dashed curve).

where tan δ = 1/

√

255, so δ

∼

0.06254. The displacement of the mass as a function of time is

shown in Figure 3.7.7. For purposes of comparison, we also show the motion if the damping

term is neglected.

The quasi frequency is µ =

√

255/16

∼

0.998, and the quasi period is T

= 2π/µ

∼

6.295 s.

These values differ only slightly from the corresponding values (1 and 2π, respectively) for

the undamped oscillation. This is evident also from the graphs in Figure 3.7.7, which rise

and fall almost together. The damping coefﬁcient is small in this example: only one-sixteenth

of the critical value, in fact. Nevertheless, the amplitude of the oscillation is reduced rather

rapidly. Figure 3.7.8 shows the graph of the solution for 40 ≤ t ≤ 60, together with the graphs

of u =±0.1. From the graph it appears that τ is about 47.5, and by a more precise calculation

we ﬁnd that τ

∼

47.5149 s.

0.1

0.05

–0.05

–0.1

–0.15

40 45 50 55 60

u = –0.1

u = 0.1

u = e

–t/16

cos t – 0.06254

√255

()

FIGURE 3.7.8 Solution of Example 3; determination of the time τ after which |u(t)| < 0.1.

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202 Chapter 3. Second Order Linear Equations

To ﬁnd the time at which the mass ﬁrst passes through its equilibrium position, we refer to

Eq. (30) and set

√

255t/16 − δ equal to π/2, the smallest positive zero of the cosine function.

Then, by solving for t, we obtain

t =

√

255

)

+ δ

∼

1.637 s.

Electric Circuits. A second example of the occurrence of second order linear differen-

tial equations with constant coefﬁcients is their use as a model of the ﬂow of electric

current in the simple series circuit shown in Figure 3.7.9. The current I, measured

in amperes (A), is a function of time t. The resistance R in ohms (4), the capaci-

tance C in farads (F), and the inductance L in henrys (H) are all positive and are

assumed to be known constants. The impressed voltage E in volts (V) is a given func-

tion of time. Another physical quantity that enters the discussion is the total charge

Q in coulombs (C) on the capacitor at time t. The relation between charge Q and

current I is

I = dQ/dt. (31)

Resistance R Capacitance C

Inductance L

Impressed voltage E(t)

FIGURE 3.7.9 A simple electric circuit.

The ﬂow of current in the circuit is governed by Kirchhoff’s

second law:In a closed

circuit the impressed voltage is equal to the sum of the voltage drops in the rest of the

circuit.

According to the elementary laws of electricity, we know that

The voltage drop across the resistor is IR.

The voltage drop across the capacitor is Q/C.

The voltage drop across the inductor is LdI/dt.

Hence, by Kirchhoff’s law,

+ RI +

Q = E(t). (32)

Gustav Kirchhoff (1824–1887) was a German physicist and professor at Breslau, Heidelberg, and Berlin.

He formulated the basic laws of electric circuits about 1845 while still a student at Albertus University in

his native Königsberg. In 1857 he discovered that an electric current in a resistanceless wire travels at the

speed of light. He is also famous for fundamental work in electromagnetic absorption and emission and

was one of the founders of spectroscopy.

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3.7 Mechanical and Electrical Vibrations 203

The units for voltage, resistance, current, charge, capacitance, inductance, and time

are all related:

1 volt = 1 ohm · 1 ampere = 1 coulomb/1 farad = 1 henry · 1 ampere/1 second.

Substituting for I from Eq. (31), we obtain the differential equation

′′

+ RQ

′

Q = E(t) (33)

for the charge Q. The initial conditions are

Q(t

) = Q

, Q

′

) = I(t

) = I

. (34)

Thus we must know the charge on the capacitor and the current in the circuit at some

initial time t

Alternatively, we can obtain a differential equation for the current I by differen-

tiating Eq. (33) with respect to t, and then substituting for dQ/dt from Eq. (31). The

result is

′′

+ RI

′

I = E

′

(t), (35)

with the initial conditions

I(t

) = I

, I

′

) = I

′

. (36)

From Eq. (32) it follows that

′

E(t

) − RI

− (1/C)Q

. (37)

Hence I

′

is also determined by the initial charge and current, which are physically

measurable quantities.

The most important conclusion from this discussion is that the ﬂow of current in the

circuit is described by an initial value problem of precisely the same form as the one

that describes the motion of a spring–mass system. This is a good example of the uni-

fying role of mathematics:once you know how to solve second order linear equations

with constant coefﬁcients, you can interpret the results in terms of mechanical vibra-

tions, electric circuits, or any other physical situation that leads to the same problem.

PROBLEMS In each of Problems 1 through 4, determine ω

, R, and δ so as to write the given expression in

the form u = R cos(ω

t − δ).

1. u = 3 cos 2t + 4 sin 2t 2. u =−cos t +

√

3 sin t

3. u = 4 cos 3t − 2 sin 3t 4. u =−2 cos πt − 3 sin πt

A mass weighing 2 lb stretches a spring 6 in. If the mass is pulled down an additional 3 in

and then released, and if there is no damping, determine the position u of the mass at any

time t. Plot u versus t. Find the frequency, period, and amplitude of the motion.

6. A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium posi-

tion with a downwardvelocity of 10cm/s,andif there isno damping,determine theposition

u of the mass at any time t. When does the mass ﬁrst return to its equilibrium position?

7. A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting

the spring a distance of 1 in, and then set in motion with a downward velocity of 2 ft/s,

and if there is no damping, ﬁnd the position u of the mass at any time t. Determine the

frequency, period, amplitude, and phase of the motion.

August 7, 2012 21:04 c03 Sheet number 68 Page number 204 cyan black

204 Chapter 3. Second Order Linear Equations

8. A series circuit has a capacitor of 0.25 × 10

−6

F and an inductor of 1 H. If the initial charge

on the capacitor is 10

−6

C and there is no initial current, ﬁnd the charge Q on the capacitor

at any time t.

A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous

damper with a damping constant of 400 dyn·s/cm. If the mass is pulled down an additional

2 cm and then released,ﬁnd its position u at any time t. Plot u versus t. Determine the quasi

frequency and the quasi period. Determine the ratio of the quasi period to the period of

the corresponding undamped motion. Also ﬁnd the time τ such that |u(t)| < 0.05 cm for

all t >τ.

10.

A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper

with a damping constant of 2 lb·s/ft. If the mass is set in motion from its equilibrium posi-

tion with a downward velocity of 3 in/s, ﬁnd its position u at any time t. Plot u versus t.

Determine when the mass ﬁrst returns to its equilibrium position. Also ﬁnd the time τ

such that |u(t)| < 0.01 in for all t >τ.

11. A spring is stretched 10 cm by a force of 3 N. A mass of 2 kg is hung from the spring and

is also attached to a viscous damper that exerts a force of 3 N when the velocity of the

mass is 5 m/s. If the mass is pulled down 5 cm below its equilibrium position and given an

initial downward velocity of 10 cm/s, determine its position u at any time t. Find the quasi

frequency µ and the ratio of µ to the natural frequency of the corresponding undamped

motion.

12. A series circuit has a capacitor of 10

−5

F, a resistor of 3 × 10

4, and an inductor of 0.2 H.

The initial charge on the capacitor is 10

−6

C and there is no initial current. Find the charge

Q on the capacitor at any time t.

13. A certain vibrating system satisﬁes the equation u

′′

+ γu

′

+ u = 0. Find the value of the

damping coefﬁcient γ for which the quasi period of the damped motion is 50% greater

than the period of the corresponding undamped motion.

14. Show that the period of motion of an undamped vibration of a mass hanging from a ver-

tical spring is 2π

L/g, where L is the elongation of the spring due to the mass, and g is

the acceleration due to gravity.

15. Show that the solution of the initial value problem

′′

+ γu

′

+ ku = 0, u(t

) = u

, u

′

) = u

′

can be expressed as the sum u = v + w, where v satisﬁes the initial conditions

v(t

) = u

, v

′

) = 0, w satisﬁes the initial conditions w(t

) = 0, w

′

) = u

′

, and both v

and w satisfy the same differential equation as u. This is another instance of superposing

solutions of simpler problems to obtain the solution of a more general problem.

16. Show that A cos ω

t + B sin ω

t can be written in the form r sin(ω

t − θ). Determine r and

θ in terms of A and B.IfR cos(ω

t − δ) = r sin(ω

t − θ), determine the relationship among

R, r, δ, and θ.

17. A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a damper with

coefﬁcient γ. Determine the value of γ for which the system is critically damped; be sure

to give the units for γ.

18. If a series circuit has a capacitor of C = 0.8 × 10

−6

F and an inductor of L = 0.2 H, ﬁnd

the resistance R so that the circuit is critically damped.

19. Assume that the system described by the equation mu

′′

+ γu

′

+ ku = 0 is either critically

damped or overdamped. Show that the mass can pass through the equilibrium position at

most once, regardless of the initial conditions.

Hint: Determine all possible values of t for which u = 0.

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3.7 Mechanical and Electrical Vibrations 205

20. Assume thatthe system describedby the equationmu

′′

+ γu

′

+ ku = 0 iscritically damped

and that the initial conditionsare u(0) = u

, u

′

(0) = v

.Ifv

= 0,show that u → 0ast →∞

but that u is never zero. If u

is positive, determine a condition on v

that will ensure that

the mass passes through its equilibrium position after it is released.

21. Logarithmic Decrement. (a) For the damped oscillation described by Eq. (26), show

that the time between successive maxima is T

= 2π/µ.

(b) Show that the ratio of the displacements at two successive maxima is given by

exp(γT

/2m). Observe that this ratio does not depend on which pair of maxima

is chosen. The natural logarithm of this ratio is called the logarithmic decrement and is

denoted by 5.

a mechanical system, this result provides a convenient and practical method for determin-

ing the damping constant of the system, which is more difﬁcult to measure directly. In par-

ticular,for the motion of a vibrating mass in a viscous ﬂuid,the damping constant depends

on the viscosity of the ﬂuid; for simple geometric shapes the form of this dependence is

known, and the preceding relation allows the experimental determination of the viscosity.

This is one of the most accurate ways of determining the viscosity of a gas at high pressure.

22. Referring to Problem 21, ﬁnd the logarithmic decrement of the system in Problem 10.

23. For the system in Problem 17, suppose that 5 = 3 and T

= 0.3 s. Referring to Problem

21, determine the value of the damping coefﬁcient γ.

24. The position of a certain spring–mass system satisﬁes the initial value problem

′′

+ ku = 0, u(0) = 2, u

′

(0) = v.

If the period andamplitude of theresulting motionare observed tobe π and 3,respectively,

determine the values of k and v.

25.

Consider the initial value problem

′′

+ γu

′

+ u = 0, u(0) = 2, u

′

(0) = 0.

We wish to explore how long a time interval is required for the solution to become “neg-

ligible” and how this interval depends on the damping coefﬁcient γ. To be more precise,

let us seek the time τ such that |u(t)| < 0.01 for all t >τ. Note that critical damping for

this problem occurs for γ = 2.

(a) Let γ = 0.25 and determine τ, or at least estimate it fairly accurately from a plot of

the solution.

(b) Repeat part (a) for several other values of γ in the interval 0 <γ<1.5. Note that τ

steadily decreases as γ increases for γ in this range.

Is the graph a smooth curve?

(d) Repeat part (b) for values of γ between 1.5 and 2. Show that τ continues to decrease

until γ reaches a certain critical value γ

, after which τ increases. Find γ

and the

corresponding minimum value of τ to two decimal places.

(e) Another way to proceed is to write the solution of the initial value problem in

the form (26). Neglect the cosine factor and consider only the exponential factor and the

amplitude R. Then ﬁnd an expression for τ as a function of γ. Compare the approximate

results obtained in this way with the values determined in parts (a), (b), and (d).

26. Consider the initial value problem

′′

+ γu

′

+ ku = 0, u(0) = u

, u

′

(0) = v

Assume that γ

< 4km.

August 7, 2012 21:04 c03 Sheet number 70 Page number 206 cyan black

206 Chapter 3. Second Order Linear Equations

(a) Solve the initial value problem.

(b) Write the solution in the form u(t) = R exp(−γt/2m) cos(µt − δ). Determine R in

terms of m, γ, k, u

, and v

other parameters.

27. A cubic block of side l and mass density ρ per unit volume is ﬂoating in a ﬂuid of mass den-

sity ρ

per unit volume, where ρ

>ρ. If the block is slightly depressed and then released,

it oscillates in the vertical direction. Assuming that the viscous damping of the ﬂuid and

air can be neglected, derive the differential equation of motion and determine the period

of the motion.

Hint: Use Archimedes’

principle: an object that is completely or partially submerged

in a ﬂuid is acted on by an upward (buoyant) force equal to the weight of the displaced

ﬂuid.

28.

The position of a certain undamped spring–mass system satisﬁes the initial value problem

′′

+ 2u = 0, u(0) = 0, u

′

(0) = 2.

(a) Find the solution of this initial value problem.

(b) Plot u versus t and u

′

versus t on the same axes.

′

versus u; that is, plot u(t) and u

′

(t) parametrically with t as the parameter.

This plot is known as a phase plot, and the uu

′

-plane is called the phase plane. Observe

that a closed curve in the phase plane corresponds to a periodic solution u(t). What is the

direction of motion on the phase plot as t increases?

29.

The position of a certain spring–mass system satisﬁes the initial value problem

′′

′

+ 2u = 0, u(0) = 0, u

′

(0) = 2.

(a) Find the solution of this initial value problem.

(b) Plot u versus t and u

′

versus t on the same axes.

′

versus u in the phase plane (see Problem 28). Identify several corresponding

points on the curves in parts (b) and (c). What is the direction of motion on the phase plot

as t increases?

30. In the absence of damping, the motion of a spring–mass system satisﬁes the initial value

problem

′′

+ ku = 0, u(0) = a, u

′

(0) = b.

(a) Show that the kinetic energy initially imparted to the mass is mb

/2 and that the

potential energy initially stored in the spring is ka

/2, so that initially the total energy in

the system is (ka

+ mb

)/2.

(b) Solve the given initial value problem.

Your result should conﬁrm the principle of conservation of energy for this system.

Archimedes (287–212 BC) was the foremost of the ancient Greek mathematicians. He lived in Syracuse

on the island of Sicily. His most notable discoveries were in geometry, but he also made important contri-

butions to hydrostatics and other branches of mechanics. His method of exhaustion is a precursor of the

integral calculus developed by Newton and Leibniz almost two millennia later. He died at the hands of a

Roman soldier during the Second Punic War.

August 7, 2012 21:04 c03 Sheet number 71 Page number 207 cyan black

3.8 Forced Vibrations 207

31. Suppose that a mass m slides without friction on a horizontal surface.The mass is attached

to a spring with spring constant k , as shown in Figure 3.7.10, and is also subject to viscous

air resistance with coefﬁcient γ. Show that the displacement u(t) of the mass from its

equilibrium position satisﬁes Eq. (21). How does the derivation of the equation of motion

in this case differ from the derivation given in the text?

u(t)

FIGURE 3.7.10 A spring–mass system.

32.

In the spring–mass system of Problem 31, suppose that the spring force is not given by

Hooke’s law but instead satisﬁes the relation

=−(ku + ϵu

where k > 0 and ϵ is small but may be of either sign.The spring is called a hardening spring

if ϵ>0 and a softening spring if ϵ<0. Why are these terms appropriate?

(a) Show that the displacement u(t) of the mass from its equilibrium position satisﬁes the

differential equation

′′

+ γu

′

+ ku + ϵu

= 0.

Suppose that the initial conditions are

u(0) = 0, u

′

(0) = 1.

In the remainder of this problem, assume that m = 1, k = 1, and γ = 0.

(b) Find u(t) when ϵ = 0 and also determine the amplitude and period of the motion.

to be periodic? Estimate the amplitude and period.

(d) Repeat part (c) for ϵ = 0.2 and ϵ = 0.3.

(e) Plot your estimated values of the amplitude A and the period T versus ϵ. Describe

the way in which A and T, respectively, depend on ϵ.

(f) Repeat parts (c), (d), and (e) for negative values of ϵ.

3.8 Forced Vibrations

We will now investigate the situation in which a periodic external force is applied

to a spring–mass system. The behavior of this simple system models that of many

oscillatory systems with an external force due, for example, to a motor attached to

the system. We will ﬁrst consider the case in which damping is present and will look

later at the idealized special case in which there is assumed to be no damping.

Forced Vibrations with Damping. The algebraic calculations can be fairly complicated in

this kind of problem, so we will begin with a relatively simple example.

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208 Chapter 3. Second Order Linear Equations

EXAMPLE

Suppose that the motion of a certain spring–mass system satisﬁes the differential equation

′′

+ u

′

+ 1.25u = 3 cos t (1)

and the initial conditions

u(0) = 2, u

′

(0) = 3. (2)

Find the solution of this initial value problem and describe the behavior of the solution for

large t.

The homogeneous equation corresponding to Eq. (1) has the characteristic equation

+ r + 1.25 = 0 with roots r =−0.5 ± i. Thus the general solution u

(t) of this homogeneous

equation is

(t) = c

−t/2

cos t + c

−t/2

sin t. (3)

A particular solution of Eq. (1) has the form U(t) = A cos t + B sin t, where A and B

are found by substituting U(t) for u in Eq. (1). We have U

′

(t) =−A sin t + B cos t and

′′

(t) =−A cos t − B sin t. Thus, from Eq. (1) we obtain

(0.25A + B) cos t + (−A + 0.25B) sin t = 3 cos t.

Consequently, A and B must satisfy the equations

0.25A + B = 3, −A + 0.25B = 0,

with the result that A = 12/17 and B = 48/17. Therefore, the particular solution is

U(t) =

cos t +

sin t, (4)

and the general solution of Eq. (1) is

u = u

(t) + U(t) = c

−t/2

cos t + c

−t/2

sin t +

cos t +

sin t. (5)

The remaining constants c

and c

are determined by the initial conditions (2). From Eq. (5)

we have

u(0) = c

= 2, u

′

(0) =−

+ c

= 3,

so c

= 22/17 and c

= 14/17. Thus we ﬁnally obtain the solution of the given initial value

problem, namely,

u =

−t/2

cos t +

−t/2

sin t +

cos t +

sin t. (6)

The graph of the solution (6) is shown by the black curve in Figure 3.8.1.

It is important to note that the solution consists of two distinct parts. The ﬁrst two terms on

the right side of Eq. (6) contain the exponential factor e

−t/2

; as a result they rapidly approach

zero. It is customary to call these terms transient. The remaining terms in Eq. (6) involve only

sines and cosines, so they represent an oscillation that continues indeﬁnitely. We refer to them

as a steady state. The solid and dashed blue curves in Figure 3.8.1 show the transient and the

steady state parts of the solution, respectively. The transient part comes from the solution

of the homogeneous equation corresponding to Eq. (1) and is needed to satisfy the initial

conditions. The steady state is the particular solution of the full nonhomogeneous equation.

After a fairly short time, the transient is vanishingly small and the full solution is essentially

indistinguishable from the steady state.

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3.8 Forced Vibrations 209

–1

–2

–3

128614

Full solution

Steady state

Transient

FIGURE 3.8.1 Solution of the initial value problem (1), (2):

′′

+ u

′

+ 1.25u = 3 cos t, u(0) = 2, u

′

(0) = 3.

The equation of motion of a general spring–mass system subject to an external

force F(t) is [Eq. (7) in Section 3.7]

′′

(t) + γu

′

(t) + ku(t) = F(t), (7)

where m,γ,and k are the mass,damping coefﬁcient,and springconstant of the spring–

mass system. Suppose now that the external force is given by F

cos ωt, where F

and

ω are positive constants representing the amplitude and frequency, respectively, of

the force. Then Eq. (7) becomes

′′

+ γu

′

+ ku = F

cos ωt. (8)

Solutions of Eq. (8) behave very much like the solution in the preceding example.

The general solution of Eq. (8) must have the form

u = c

(t) + c

(t) + A cos ω t + B sin ωt = u

(t) + U(t). (9)

The ﬁrst two terms on the right side of Eq. (9) are the general solution u

(t) of

the homogeneous equation corresponding to Eq. (8), and the latter two terms are

a particular solution U(t) of the full nonhomogeneous equation. The coefﬁcients

A and B can be found, as usual, by substituting these terms into the differen-

tial equation (8), while the arbitrary constants c

and c

are available to satisfy

initial conditions, if any are prescribed. The solutions u

(t) and u

(t) of the homo-

geneous equation depend on the roots r

and r

of the characteristic equation

+ γr + k = 0. Since m, γ, and k are all positive, it follows that r

and r

either

are real and negative or are complex conjugates with negative real part. In either

case, both u

(t) and u

(t) approach zero as t →∞. Since u

(t) dies out as t increases,

it is called the transient solution. In many applications, it is of little importance and

(depending on the value of γ) may well be undetectable after only a few seconds.

The remaining terms in Eq. (9)—namely, U(t) = A cos ωt + B sin ωt—do not die

out as t increases but persist indeﬁnitely, or as long as the external force is applied.

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210 Chapter 3. Second Order Linear Equations

They represent a steady oscillation with the same frequency as the external force

and are called the steady state solution or the forced response.The transient solution

enables us to satisfy whatever initial conditions may be imposed. With increasing

time, the energy put into the system by the initial displacement and velocity is dis-

sipated through the damping force, and the motion then becomes the response of

the system to the external force. Without damping,the effect of the initial conditions

would persist for all time.

It is convenient to express U(t) as a single trigonometric term rather than as a sum

of two terms. Recall that we did this for other similar expressions in Section 3.7. Thus

we write

U(t) = R cos(ωt − δ). (10)

The amplitude R and phase δ depend directly on A and B and indirectly on the

parameters in the differential equation (8). It is possible to show, by straightforward

but somewhat lengthy algebraic computations, that

R =

, cos δ =

m(ω

− ω

)

, sin δ =

γω

, (11)

where

5 =

(ω

− ω

)

+ γ

and ω

= k/m. (12)

Recall that ω

is the natural frequency of the unforced system in the absence of

damping.

We now investigate how the amplitude R of the steady state oscillation depends on

the frequency ω of the external force. Substituting from Eq. (12) into the expression

for R in Eq. (11) and executing some algebraic manipulations, we ﬁnd that

= 1

1 −

(

+ 2

1/2

, (13)

where 2 = γ

/mk. Observe that the quantity Rk/F

is the ratio of the amplitude R

of the forced response to F

/k, the static displacement of the spring produced by a

force F

For low frequency excitation—that is, as ω → 0—it follows from Eq. (13) that

Rk/F

→ 1orR → F

/k. At the other extreme, for very high frequency excitation,

Eq. (13) implies that R → 0asω →∞. At an intermediate value of ω the amplitude

may have amaximum.To ﬁnd this maximumpoint,wecan differentiateR with respect

to ω and set the result equal to zero. In this way we ﬁnd that the maximum amplitude

occurs when ω = ω

max

, where

max

= ω

−

= ω

1 −

2mk

(

. (14)

Note that ω

max

<ω

and that ω

max

is close to ω

when γ is small. The maximum

value of R is

max

γω

1 − (γ

/4mk)

∼

γω

1 +

8mk

(

, (15)

where the last expression is an approximation for small γ.Ifγ

/mk > 2, then ω

max

given by Eq. (14) is imaginary; in this case the maximum value of R occurs for ω = 0,

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3.8 Forced Vibrations 211

and R is a monotone decreasing function of ω. Recall that critical damping occurs

when γ

/mk = 4.

For small γ it follows from Eq. (15) that R

max

∼

/γω

. Thus, for lightly damped

systems, the amplitude R of the forced response when ω is near ω

is quite large

even for relatively small external forces, and the smaller the value of γ, the more

pronounced is this effect. This phenomenon is known as resonance, and it is often

an important design consideration. Resonance can be either good or bad, depending

on the circumstances. It must be taken very seriously in the design of structures,

such as buildings and bridges, where it can produce instabilities that might lead to

the catastrophic failure of the structure. On the other hand, resonance can be put

to good use in the design of instruments, such as seismographs, that are intended to

detect weak periodic incoming signals.

Figure 3.8.2 contains some representative graphs of Rk/F

versus ω/ω

for several

values of 2 = γ

/mk. The graph corresponding to 2 = 0.015625 is included because

this is the value of 2 that occurs in Example 2 below. Note particularly the sharp

peak in the curve corresponding to 2 = 0.015625 near ω/ω

= 1. The limiting case

as 2 → 0 is also shown. It follows from Eq. (13), or from Eqs. (11) and (12), that

R → F

/m|ω

− ω

| as γ → 0 and hence Rk/F

is asymptotic to the vertical line

ω = ω

, as shown in the ﬁgure. As the damping in the system increases, the peak

response gradually diminishes.

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Γ = 2

Γ = 0.5

Rk/F

ωω

Γ = 0.1

Γ = 0.015625

→

FIGURE 3.8.2 Forced vibration with damping: amplitude of steady

state response versus frequency of driving force; 2 = γ

/mk.

Figure3.8.2 also illustrates the usefulnessof dimensionless variables.Youcan easily

verify thateach ofthe quantitiesRk/F

,ω/ω

,and 2 is dimensionless.The importance

of this observation is that the number of signiﬁcant parameters in the problem has

been reduced to three rather than the ﬁve that appear in Eq. (8). Thus only one

family of curves, of which a few are shown in Figure 3.8.2, is needed to describe the

response-versus-frequency behavior of all systems governed by Eq. (8).

The phase angle δ also depends in aninteresting way on ω.Forω near zero,it follows

from Eqs. (11) and (12) that cos δ

∼

1 and sin δ

∼

0. Thus δ

∼

0, and the response is

nearly in phase with the excitation, meaning that they rise and fall together and,

in particular, assume their respective maxima nearly together and their respective

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212 Chapter 3. Second Order Linear Equations

minima nearly together. For ω = ω

we ﬁnd that cos δ = 0 and sin δ = 1, so δ = π/2.

In this case the response lags behind the excitation by π/2; that is, the peaks of

the response occur π/2 later than the peaks of the excitation, and similarly for the

valleys. Finally, for ω very large,we have cos δ

∼

−1 and sin δ

∼

0. Thus δ

∼

π, so that

the response is nearly out of phase with the excitation; this means that the response

is minimum when the excitation is maximum, and vice versa. Figure 3.8.3 shows

the graphs of δ versus ω/ω

for several values of 2. For small damping, the phase

transition from near δ = 0 to near δ = π occurs rather abruptly, whereas for larger

values of the damping parameter, the transition takes place more gradually.

Γ = 0.1

Γ = 0.5

δπ

Γ = 2

Γ = 0.015625

FIGURE 3.8.3 Forced vibration with damping: phase of steady

state response versus frequency of driving force; 2 = γ

/mk.

EXAMPLE

Consider the initial value problem

′′

+ 0.125u

′

+ u = 3 cos ωt, u(0) = 2, u

′

(0) = 0. (16)

Show plots of the solution for different values of the forcing frequency ω, and compare them

with corresponding plots of the forcing function.

For this system we have ω

= 1 and 2 = 1/64 = 0.015625. Its unforced motion was discussed

in Example 3 of Section 3.7, and Figure 3.7.7 shows the graph of the solution of the unforced

problem. Figures 3.8.4,3.8.5,and 3.8.6 show the solution of the forced problem (16) forω = 0.3,

ω = 1, and ω = 2, respectively. The graph of the corresponding forcing function is also shown

in each ﬁgure. In this example the static displacement, F

/k, is equal to 3.

Figure 3.8.4 shows the low frequency case,ω/ω

= 0.3. After the initial transient response is

substantially damped out, the remaining steady state response is essentially in phase with the

excitation, and the amplitude of the response is somewhat larger than the static displacement.

To be speciﬁc, R

∼

3.2939 and δ

∼

0.041185.

The resonant case,ω/ω

= 1,is shown in Figure 3.8.5. Here the amplitude of the steady state

response is eight times the static displacement, and the ﬁgure also shows the predicted phase

lag of π/2 relative to the external force.

The case of comparatively high frequency excitation is shown in Figure 3.8.6. Observe that

the amplitude of the steady forced response is approximately one-third the static displacement

andthat the phasedifference between theexcitation and theresponse is approximately π. More

precisely, we ﬁnd that R

∼

0.99655 and that δ

∼

3.0585.

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3.8 Forced Vibrations 213

–3

–2

–1

10 20 30 40 50 60 70 80

Solution Forcing function

FIGURE 3.8.4 A forced vibration with damping; solution of

Eq. (16) with ω = 0.3: u

′′

+ 0.125u

′

+ u = 3 cos 0.3t, u(0) = 2, u

′

(0) = 0.

–20

–10

SolutionForcing function

10 20 30 40 50 60

FIGURE 3.8.5 A forced vibration with damping; solution of

Eq. (16) with ω = 1: u

′′

+ 0.125u

′

+ u = 3 cos t, u(0) = 2, u

′

(0) = 0.

–3

–2

–1

SolutionForcing function

10 20 30 40 50

FIGURE 3.8.6 A forced vibration with damping; solution of

Eq. (16) with ω = 2: u

′′

+ 0.125u

′

+ u = 3 cos 2t, u(0) = 2, u

′

(0) = 0.

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214 Chapter 3. Second Order Linear Equations

Forced Vibrations Without Damping. We now assume that γ = 0 in Eq. (8), thereby

obtaining the equation of motion of an undamped forced oscillator,

′′

+ ku = F

cos ωt. (17)

The form of the general solution of Eq. (17) is different, depending on whether the

forcing frequency ω is different from or equal to the natural frequency ω

k/m

of the unforced system. First consider the case ω ̸= ω

; then the general solution of

Eq. (17) is

u = c

cos ω

t + c

sin ω

t +

m(ω

− ω

)

cos ωt. (18)

The constants c

and c

are determined by the initial conditions.The resulting motion

is, in general, the sum of two periodic motions of different frequencies (ω

and ω)

and different amplitudes as well.

It is particularly interesting to suppose that the mass is initially at rest, so that

the initial conditions are u(0) = 0 and u

′

(0) = 0. Then the energy driving the sys-

tem comes entirely from the external force, with no contribution from the initial

conditions. In this case it turns out that the constants c

and c

in Eq. (18) are

given by

=−

m(ω

− ω

)

, c

= 0, (19)

and the solution of Eq. (17) is

u =

m(ω

− ω

)

(cos ωt − cos ω

t). (20)

This is the sum of two periodic functions of different periods but the same

amplitude. Making use of the trigonometric identities for cos(A ± B) with

A = (ω

+ ω)t/2 and B = (ω

− ω)t/2, we can write Eq. (20) in the form

u =

m(ω

− ω

)

sin

(ω

− ω)t

sin

(ω

+ ω)t

. (21)

If |ω

− ω | is small, then ω

+ ω is much greater than |ω

− ω |. Consequently,

sin(ω

+ ω)t/2 is a rapidly oscillating function compared to sin(ω

− ω)t/2. Thus the

motion is a rapid oscillation with frequency (ω

+ ω)/2 but with a slowly varying

sinusoidal amplitude

m|ω

− ω

sin

(ω

− ω)t

This type of motion, possessing a periodic variation of amplitude, exhibits what is

called a beat. For example, such a phenomenon occurs in acoustics when two tuning

forks of nearly equal frequency are excited simultaneously. In this case the peri-

odic variation of amplitude is quite apparent to the unaided ear. In electronics, the

variation of the amplitude with time is called amplitude modulation.

EXAMPLE

Solve the initial value problem

′′

+ u = 0.5 cos 0.8t, u(0) = 0, u

′

(0) = 0, (22)

and plot the solution.

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3.8 Forced Vibrations 215

In this case ω

= 1, ω = 0.8, and F

= 0.5, so from Eq. (21) the solution of the given

problem is

u = 2.77778(sin 0.1t)(sin 0.9t). (23)

A graph of this solution is shown in Figure 3.8.7. The amplitude variation has a slow frequency

of 0.1 and a corresponding slow period of 20π. Note that a half-period of 10π corresponds to a

single cycle of increasing and then decreasing amplitude.The displacement of the spring–mass

system oscillates with a relatively fast frequency of 0.9, which is only slightly less than the

natural frequency ω

Now imagine that the forcing frequency ω is increased, say, to ω = 0.9. Then the slow

frequency is halved to 0.05, and the corresponding slow half-period is doubled to 20π.The

multiplier 2.7778 also increases substantially, to 5.2632. However, the fast frequency is only

marginally increased, to 0.95. Can you visualize what happens as ω takes on values closer and

closer to the natural frequency ω

= 1?

–1

–3

–2

u = 2.77778 sin 0.1t

u = –2.77778 sin 0.1t

u = 2.77778 sin 0.1t sin 0.9t

10 20 30 40 50 60

FIGURE 3.8.7 A beat; solution of Eq. (22):

′′

+ u = 0.5 cos 0.8t, u(0) = 0, u

′

(0) = 0; u = 2.77778(sin 0.1t)(sin 0.9t).

Now let us return to Eq. (17) and consider the case of resonance, where ω = ω

;

that is,the frequency of the forcing function is the same asthe naturalfrequency of the

system. Then the nonhomogeneous term F

cos ωt is a solution of the homogeneous

equation. In this case the solution of Eq. (17) is

u = c

cos ω

t + c

sin ω

t +

2mω

t sin ω

t. (24)

Consider the following example.

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216 Chapter 3. Second Order Linear Equations

EXAMPLE

Solve the initial value problem

′′

+ u = 0.5 cos t, u(0) = 0, u

′

(0) = 0, (25)

and plot the graph of the solution.

The general solution of the differential equation is

u = c

cos t + c

sin t + 0.25t sin t,

and the initial conditions require that c

= c

= 0. Thus the solution of the given initial value

problem is

u = 0.25t sin t. (26)

The graph of the solution is shown in Figure 3.8.8.

10 20 30 40

–5

–10

u = 0.25t sin t

u = 0.25t

u = – 0.25t

FIGURE 3.8.8 Resonance; solution of Eq. (25):

′′

+ u = 0.5 cos t, u(0) = 0, u

′

(0) = 0; u = 0.25t sin t.

Because of the term t sin ω

t,the solution (24) predicts that the motion will become

unbounded as t →∞regardless of the values of c

and c

, and Figure 3.8.8 bears this

out. Of course, in reality, unbounded oscillations do not occur, because the spring

cannot stretch inﬁnitely far. Moreover, as soon as u becomes large, the mathemat-

ical model on which Eq. (17) is based is no longer valid, since the assumption that

the spring force depends linearly on the displacement requires that u be small. As

we have seen, if damping is included in the model, the predicted motion remains

bounded; however, the response to the input function F

cos ωt may be quite large if

the damping is small and ω is close to ω

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3.8 Forced Vibrations 217

PROBLEMS In each of Problems 1 through 4, write the given expression as a product of two trigonometric

functions of different frequencies.

1. cos 9t − cos 7t 2. sin 7t − sin 6t

3. cos πt + cos 2πt 4. sin 3t + sin 4t

5. A mass weighing 4 lb stretches a spring 1.5 in. The mass is given a positive displacement of

2 in from its equilibrium position and released with no initial velocity.Assuming that there

is no damping and that the mass is acted on by an external force of 2 cos 3t lb, formulate

the initial value problem describing the motion of the mass.

6. A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of

10 sin(t/2) N (newtons) and moves in a medium that imparts a viscous force of 2 N

when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium

position with an initial velocity of 3 cm/s, formulate the initial value problem describing

the motion of the mass.

(a) Find the solution of Problem 5.

(b) Plot the graph of the solution.

of ω for which resonance occurs.

(a) Find the solution of the initial value problem in Problem 6.

(b) Identify the transient and steady state parts of the solution.

(d) If the given external force is replaced by a force of 2 cos ωt of frequency ω, ﬁnd the

value of ω for which the amplitude of the forced response is maximum.

9. If an undamped spring–mass system with a mass that weighs 6 lb and a spring constant

1 lb/in is suddenly set in motion at t = 0 by an external force of 4 cos 7t lb, determine the

position of the mass at any time, and draw a graph of the displacement versus t.

10. A mass that weighs 8 lb stretches a spring 6 in. The system is acted on by an external force

of 8 sin 8t lb. If themass is pulled down 3 in and then released,determine the position of the

mass at any time. Determine the ﬁrst four times at which the velocity of the mass is zero.

11. A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot

mechanism that has a damping constant of 0.25 lb·s/ft and is acted on by an external force

of 4 cos 2t lb.

(a) Determine the steady state response of this system.

(b) If the given mass is replaced by a mass m, determine the value of m for which the

amplitude of the steady state response is maximum.

12. A spring–mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the

spring, and the motion takes place in a viscous ﬂuid that offers a resistance numerically

equal to the magnitude of the instantaneous velocity. If the system is driven by an external

force of (3 cos 3t − 2 sin 3t) N, determine the steady state response. Express your answer

in the form R cos(ωt − δ).

13. In this problem we ask you to supply some of the details in the analysis of a forced damped

oscillator.

(a) Derive Eqs. (10), (11), and (12) for the steady state solution of Eq. (8).

(b) Derive the expression in Eq. (13) for Rk/F

max

and R

max

are given by Eqs. (14) and (15), respectively.

14. Find the velocity of the steady state response given by Eq. (10).Then show that the velocity

is maximum when ω = ω

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218 Chapter 3. Second Order Linear Equations

15. Find the solution of the initial value problem

′′

+ u = F(t), u(0) = 0, u

′

(0) = 0,

where

F(t) =

⎧

⎪

⎨

⎪

⎩

t,0≤ t ≤ π,

(2π − t), π<t ≤ 2π,

0, 2π<t.

Hint: Treat each time interval separately, and match the solutions in the different

intervals by requiring u and u

′

to be continuous functions of t.

16. A series circuit has a capacitor of 0.25 × 10

−6

F, a resistor of 5 × 10

4, and an inductor of

1 H. The initial charge on the capacitor is zero. If a 12-volt battery is connected to the cir-

cuit and the circuit is closed at t = 0, determine the charge on the capacitor at t = 0.001 s,

at t = 0.01 s, and at any time t. Also determine the limiting charge as t →∞.

17.

Consider a vibrating system described by the initial value problem

′′

′

+ 2u = 2 cos ωt, u(0) = 0, u

′

(0) = 2.

(a) Determine the steady state part of the solution of this problem.

(b) Find the amplitude A of the steady state solution in terms of ω.

(d) Find the maximum value of A and the frequency ω for which it occurs.

18.

Consider the forced but undamped system described by the initial value problem

′′

+ u = 3 cos ωt, u(0) = 0, u

′

(0) = 0.

(a) Find the solution u(t) for ω ̸= 1.

(b) Plot the solution u(t) versus t for ω = 0.7, ω = 0.8, and ω = 0.9. Describe how the

response u(t) changes as ω varies in this interval. What happens as ω takes on val-

ues closer and closer to 1? Note that the natural frequency of the unforced system

is ω

= 1.

19.

Consider the vibrating system described by the initial value problem

′′

+ u = 3 cos ωt, u(0) = 1, u

′

(0) = 1.

(a) Find the solution for ω ̸= 1.

(b) Plot the solution u(t) versus t for ω = 0.7, ω = 0.8, and ω = 0.9. Compare the results

with those of Problem 18; that is, describe the effect of the nonzero initial conditions.

20.

For the initial value problem in Problem 18, plot u

′

versus u for ω = 0.7, ω = 0.8, and

ω = 0.9. Such a plot is called a phase plot. Use a t interval that is long enough so that the

phase plot appears as a closed curve. Mark your curve with arrows to show the direction

in which it is traversed as t increases.

Problems 21 through 23 deal with the initial value problem

′′

+ 0.125u

′

+ 4u = F(t), u(0) = 2, u

′

(0) = 0.

In each of these problems:

(a) Plot the given forcing function F(t) versus t, and also plot the solution u(t) versus t on the

same set of axes. Use a t interval that is long enough so the initial transients are substantially

eliminated. Observe the relation between the amplitude and phase of the forcing term and

the amplitude and phase of the response. Note that ω

k/m = 2.

(b) Draw the phase plot of the solution; that is, plot u

′

versus u.

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3.8 Forced Vibrations 219

21. F(t) = 3 cos(t/4)

22.

F(t) = 3 cos 2t

23.

F(t) = 3 cos 6t

24.

A spring–mass system with a hardening spring (Problem 32 of Section 3.7) is acted on by

a periodic external force. In the absence of damping, suppose that the displacement of the

mass satisﬁes the initial value problem

′′

+ u +

= cos ωt, u(0) = 0, u

′

(0) = 0.

(a) Let ω = 1 and plot a computer-generated solution of the given problem. Does the

system exhibit a beat?

(b) Plot the solution for several values of ω between 1/2 and 2. Describe how the solution

changes as ω increases.

25.

Suppose that the system of Problem 24 is modiﬁed to include a damping term and that

the resulting initial value problem is

′′

′

+ u +

= cos ωt, u(0) = 0, u

′

(0) = 0.

(a) Plot a computer-generated solution of the given problem for several values of ω

between 1/2 and 2, and estimate the amplitude R of the steady response in each case.

(b) Using the data from part (a), plot the graph of R versus ω. For what frequency ω is

the amplitude greatest?

spring.

REFERENCES Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice-

Hall, 1961; New York: Dover, 1989).

There are many books on mechanical vibrations and electric circuits. One that deals with both is

Close,C. M., and Frederick,D. K.,Modeling and Analysis of Dynamic Systems (3rd ed.) (NewYork:Wiley,

2001).

A classic book on mechanical vibrations is

Den Hartog,J. P.,Mechanical Vibrations (4th ed.) (NewYork:McGraw-Hill,1956;New York;Dover,1985).

A more recent, intermediate-level book is

Thomson, W. T., Theory of Vibrations with Applications (5th ed.) (Englewood Cliffs, NJ: Prentice-Hall,

1997).

An elementary book on electric circuits is

Bobrow, L. S., Elementary Linear Circuit Analysis (New York: Oxford University Press, 1996).

August 7, 2012 21:04 c03 Sheet number 84 Page number 220 cyan black

August 7, 2012 21:04 C04 Sheet number 1 Page number 221 cyan black

221

CHAPTER

Higher Order Linear

Equations

The theoretical structure and methods of solution developed in the preceding chapter

for second order linear equations extend directly to linear equations of third and

higher order. In this chapter we brieﬂy review this generalization, taking particular

note of those instances where new phenomena may appear, because of the greater

variety of situations that can occur for equations of higher order.

4.1 General Theory of nth Order Linear Equations

An nth order linear differential equation is an equation of the form

(t)

+ P

(t)

n−1

+···+P

n−1

(t)

+ P

(t)y = G(t). (1)

We assume that the functions P

, ..., P

, and G are continuous real-valued functions

on some interval I: α<t <β, and that P

is nowhere zero in this interval. Then,

dividing Eq. (1) by P

(t), we obtain

L[y]=

+ p

(t)

n−1

+···+p

n−1

(t)

+ p

(t)y = g(t). (2)

The linear differential operator L of order n deﬁned by Eq. (2) is similar to the second

order operator introduced in Chapter 3. The mathematical theory associated with

Eq. (2) is completely analogous to that for the second order linear equation; for this

reason we simply state the results for the nth order problem. The proofs of most of

the results are also similar to those for the second order equation and are usually left

as exercises.

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222 Chapter 4. Higher Order Linear Equations

Since Eq. (2) involves the nth derivative of y with respect to t, it will, so to speak,

require n integrations to solve Eq. (2). Each of these integrations introduces an

arbitrary constant. Hence we expect that to obtain a unique solution it is necessary

to specify n initial conditions

y(t

) = y

, y

′

) = y

′

, ..., y

(n−1)

) = y

(n−1)

, (3)

where t

may be any point in the interval I and y

, y

′

, ..., y

(n−1)

is any set of prescribed

real constants. The following theorem, which is similar to Theorem 3.2.1, guarantees

that the initial value problem (2), (3) has a solution and that it is unique.

Theorem 4.1.1

If the functions p

, ..., p

,and g are continuous on the open interval I,then there

exists exactly one solution y = φ(t) of the differential equation (2) that also satisﬁes

the initial conditions (3), where t

is any point in I. This solution exists throughout

the interval I.

We will not give a proof of this theorem here. However,if the coefﬁcients p

, ..., p

are constants, then we can construct the solution of the initial value problem (2), (3)

much as in Chapter3;seeSections 4.2through 4.4.Even though we mayﬁnd a solution

in this case, we do not know that it is unique without the use of Theorem 4.1.1. A

proof of the theorem can be found in Ince (Section 3.32) or Coddington (Chapter 6).

The Homogeneous Equation. As in the corresponding second order problem, we ﬁrst

discuss the homogeneous equation

L[y]=y

(n)

+ p

(t)y

(n−1)

+···+p

n−1

(t)y

′

+ p

(t)y = 0. (4)

If the functions y

, y

, ..., y

are solutions of Eq. (4), then it follows by direct

computation that the linear combination

y = c

(t) + c

(t) +···+c

(t), (5)

where c

, ..., c

are arbitrary constants, is also a solution of Eq. (4). It is then natural

to ask whether every solution of Eq. (4) can be expressed as a linear combination

of y

, ..., y

. This will be true if, regardless of the initial conditions (3) that are

prescribed, it is possible to choose the constants c

, ..., c

so that the linear com-

bination (5) satisﬁes the initial conditions. That is, for any choice of the point t

in I,

and for any choice of y

, y

′

, ..., y

(n−1)

, we must be able to determine c

, ..., c

so that

the equations

) +···+c

) = y

′

) +···+c

′

) = y

′

(6)

(n−1)

) +···+c

(n−1)

) = y

(n−1)

are satisﬁed. Equations (6) can be solved uniquely for the constants c

, ..., c

provided that the determinant of coefﬁcients is not zero. On the other hand, if the

determinant of coefﬁcients is zero, then it is always possible to choose values of

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4.1 General Theory of nth Order Linear Equations 223

, y

′

, ..., y

(n−1)

so that Eqs. (6) do not have a solution. Therefore a necessary and

sufﬁcient condition for the existence of a solution of Eqs. (6) for arbitrary values of

, y

′

, ..., y

(n−1)

is that the Wronskian

W(y

, ..., y

) =

··· y

′

··· y

′

(n−1)

··· y

(n−1)

(7)

is not zero at t = t

. Since t

can be any point in the interval I, it is necessary and

sufﬁcient that W(y

, y

, ..., y

) be nonzero at every point in the interval. Just as for

the second order linear equation, it can be shown that if y

, y

, ..., y

are solutions

of Eq. (4), then W(y

, y

, ..., y

) either is zero for every t in the interval I or else is

never zero there; see Problem 20. Hence we have the following theorem.

Theorem 4.1.2

If the functions p

, p

, ..., p

are continuous on the open interval I, if the functions

, y

, ..., y

are solutions of Eq. (4), and if W(y

, y

, ..., y

)(t) ̸= 0 for at least one

point in I, then every solution of Eq. (4) can be expressed as a linear combination

of the solutions y

, y

, ..., y

A set of solutions y

, ..., y

of Eq. (4) whose Wronskian is nonzero is referred to

as a fundamental set of solutions.The existence of a fundamental set of solutions can

be demonstrated in precisely the same way as for the second order linear equation

(seeTheorem 3.2.5). Since all solutions of Eq. (4) are of the form (5), we use the term

general solution to refer to an arbitrary linear combination of any fundamental set

of solutions of Eq. (4).

Linear Dependence and Independence. We now explore the relationship between funda-

mental sets of solutions and the concept of linear independence,a central idea in the

study of linear algebra. The functions f

, f

, ..., f

are said to be linearly dependent

on an interval I if there exists a set of constants k

, k

, ..., k

, not all zero, such that

(t) + k

(t) +···+k

(t) = 0 (8)

for all t in I. The functions f

, ..., f

are said to be linearly independent on I if they

are not linearly dependent there.

EXAMPLE

Determine whether the functions f

(t) = 1, f

(t) = t, and f

(t) = t

are linearly independent or

dependent on the interval I :−∞< t < ∞.

Form the linear combination

(t) + k

(t) = k

+ k

t + k

and set it equal to zero to obtain

+ k

t + k

= 0. (9)

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224 Chapter 4. Higher Order Linear Equations

If Eq. (9) is to hold for all t in I, then it must certainly be true at any three distinct points in I.

Any three points will serve our purpose, but it is convenient to choose t = 0,t = 1, and t =−1.

Evaluating Eq. (9) at each of these points, we obtain the system of equations

= 0,

+ k

= 0, (10)

− k

+ k

= 0.

From the ﬁrst of Eqs. (10) we note that k

= 0; then from the other two equations it follows

that k

= k

= 0 as well. Therefore, there is no set of constants k

, k

, not all zero, for

which Eq. (9) holds even at the three chosen points, much less throughout I. Thus the given

functions are not linearly dependent on I, so they must be linearly independent. Indeed, they

are linearly independent on any interval.Thiscan be established just as in this example,possibly

using a different set of three points.

EXAMPLE

Determine whether the functions

(t) = 1, f

(t) = 2 + t, f

(t) = 3 − t

, and f

(t) = 4t + t

are linearly independent or dependent on any interval I.

Form the linear combination

(t) + k

(t) = k

+ k

(2 + t) + k

(3 − t

) + k

(4t + t

)

= (k

+ 2k

+ 3k

) + (k

+ 4k

)t + (−k

+ k

. (11)

For this expression to be zero throughout an interval, it is certainly sufﬁcient to require that

+ 2k

+ 3k

= 0, k

+ 4k

= 0, −k

+ k

= 0.

These three equations, with four unknowns, have many solutions. For instance, if k

= 1, then

= 1, k

=−4, and k

= 5. If we use these values for the coefﬁcients in Eq. (11), then we

have

(t) − 4f

(t) + f

(t) = 0

for each value of t. Thus the given functions are linearly dependent on every interval.

The concept of linear independence provides an alternative characterization of

fundamental sets of solutions of the homogeneous equation (4). Suppose that the

functions y

, ..., y

are solutions of Eq. (4) on an interval I,and consider the equation

(t) +···+k

(t) = 0. (12)

By differentiating Eq. (12) repeatedly, we obtain the additional n − 1 equations

′

(t) +···+k

′

(t) = 0,

. (13)

(n−1)

(t) +···+k

(n−1)

(t) = 0.

The system consisting of Eqs. (12) and (13) is a system of n linear algebraic equations

for the n unknowns k

, ..., k

. The determinant of coefﬁcients for this system is the

Wronskian W(y

, ..., y

)(t) of y

, ..., y

. This leads to the following theorem.

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4.1 General Theory of nth Order Linear Equations 225

Theorem 4.1.3

If y

(t), ..., y

(t) is a fundamental set of solutions of Eq. (4)

L[y]=y

(n)

+ p

(t)y

(n−1)

+···+p

n−1

(t)y

′

+ p

(t)y = 0

on an interval I, then y

(t), ..., y

(t) are linearly independent on I. Conversely, if

(t), ..., y

(t) are linearly independent solutions of Eq. (4) on I, then they form a

fundamental set of solutions on I.

To prove this theorem, ﬁrst suppose that y

(t), ..., y

(t) is a fundamental set of

solutions of Eq. (4) on I. Then the Wronskian W(y

, ..., y

)(t) ̸= 0 for every t in

I. Hence the system (12), (13) has only the solution k

=···=k

= 0 for every t

in I. Thus y

(t), ..., y

(t) cannot be linearly dependent on I and must therefore be

linearly independent there.

To demonstrate the converse, let y

(t), ..., y

(t) be linearly independent on I.To

show that they form a fundamental set of solutions, we need to show that their

Wronskian is never zero in I. Suppose that this is not true; then there is at least one

point t

where theWronskian is zero.At this point the system (12),(13) has a nonzero

solution; let us denote it by k

∗

, ..., k

∗

. Now form the linear combination

φ(t) = k

∗

(t) +···+k

∗

(t). (14)

Then φ(t) satisﬁes the initial value problem

L[y]=0, y(t

) = 0, y

′

) = 0, ..., y

(n−1)

) = 0. (15)

The function φ satisﬁes the differential equation because it is a linear combination of

solutions; it satisﬁes the initial conditions because these are just the equations in the

system (12), (13) evaluated at t

. However, the function y(t) = 0 for all t in I is also

a solution of this initial value problem, and by Theorem 4.1.1, the solution is unique.

Thus φ(t) = 0 for all t in I. Consequently, y

(t), ..., y

(t) are linearly dependent on

I, which is a contradiction. Hence the assumption that there is a point where the

Wronskian is zero is untenable. Therefore, the Wronskian is never zero on I, as was

to be proved.

Note that for a set of functions f

, ..., f

that are not solutions of Eq. (4), the con-

verse part ofTheorem 4.1.3 is not necessarily true. They may be linearly independent

on I even though the Wronskian is zero at some points, or even every point, but

with different sets of constants k

, ..., k

at different points. See Problem 25 for an

example.

The Nonhomogeneous Equation. Now consider the nonhomogeneous equation (2)

L[y]=y

(n)

+ p

(t)y

(n−1)

+···+p

(t)y = g(t).

If Y

and Y

are any two solutions of Eq. (2), then it follows immediately from the

linearity of the operator L that

L[Y

− Y

](t) = L[Y

](t) − L[Y

](t) = g(t) − g(t) = 0.

Hence the difference of any two solutions of the nonhomogeneous equation (2) is a

solution of the homogeneous equation (4). Since any solution of the homogeneous

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226 Chapter 4. Higher Order Linear Equations

equation can be expressed as a linear combination of a fundamental set of solutions

, ..., y

, it follows that any solution of Eq. (2) can be written as

y = c

(t) + c

(t) +···+c

(t) + Y(t), (16)

where Y is some particular solution of the nonhomogeneous equation (2). The linear

combination (16) is called the general solution of the nonhomogeneous equation (2).

Thus the primary problem is to determine a fundamental set of solutions y

, ..., y

of the homogeneous equation (4). If the coefﬁcients are constants, this is a fairly

simple problem;it is discussed in the next section. If the coefﬁcients are not constants,

it is usually necessary to use numerical methods such as those in Chapter 8 or series

methods similar to those in Chapter 5. These tend to become more cumbersome as

the order of the equation increases.

To ﬁnd a particular solution Y(t) in Eq. (16), the methods of undetermined coef-

ﬁcients and variation of parameters are again available. They are discussed and

illustrated in Sections 4.3 and 4.4, respectively.

The method of reduction of order (Section 3.4) also applies to nth order linear

equations. If y

is one solution of Eq. (4), then the substitution y = v(t)y

(t) leads to

a linear differential equation of order n − 1 for v

′

(see Problem 26 for the case when

n = 3). However, if n ≥ 3, the reduced equation is itself at least of second order, and

only rarely will it be signiﬁcantly simpler than the original equation. Thus, in practice,

reduction of order is seldom useful for equations of higher than second order.

PROBLEMS In each of Problems 1 through 6, determine intervals in which solutions are sure to exist.

1. y

(4)

+ 4y

′′′

+ 3y = t 2. ty

′′′

+ (sin t)y

′′

+ 3y = cos t

3. t(t − 1)y

(4)

+ e

′′

+ 4t

y = 04.y

′′′

+ ty

′′

+ t

′

+ t

y = ln t

5. (x − 1)y

(4)

+ (x + 1)y

′′

+ (tan x)y = 0 6. (x

− 4)y

(6)

+ x

′′′

+ 9y = 0

In each of Problems 7 through 10, determine whether the given functions are linearly depen-

dent or linearly independent. If they are linearly dependent, ﬁnd a linear relation among

them.

7. f

(t) = 2t − 3, f

(t) = t

+ 1, f

(t) = 2t

− t

8. f

(t) = 2t − 3, f

(t) = 2t

+ 1, f

(t) = 3t

+ t

9. f

(t) = 2t − 3, f

(t) = t

+ 1, f

(t) = 2t

− t, f

(t) = t

+ t + 1

10. f

(t) = 2t − 3, f

(t) = t

+ 1, f

(t) = 2t

− t, f

(t) = t

+ t + 1

In each of Problems 11 through 16, verify that the given functions are solutions of the

differential equation, and determine their Wronskian.

11. y

′′′

+ y

′

= 0; 1, cos t, sin t

12. y

(4)

+ y

′′

= 0; 1, t, cos t, sin t

13. y

′′′

+ 2y

′′

− y

′

− 2y = 0; e

, e

−t

, e

−2t

14. y

(4)

+ 2y

′′′

+ y

′′

= 0; 1, t, e

−t

, te

−t

15. xy

′′′

− y

′′

= 0; 1, x, x

16. x

′′′

+ x

′′

− 2xy

′

+ 2y = 0; x, x

,1/x

17. Show that W(5, sin

t, cos 2t) = 0 for all t. Can you establish this result without direct

evaluation of the Wronskian?

18. Verify that the differential operator deﬁned by

L[y]=y

(n)

+ p

(t)y

(n−1)

+···+p

(t)y

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4.1 General Theory of nth Order Linear Equations 227

is a linear differential operator. That is, show that

L[c

+ c

]=c

L[y

]+c

L[y

where y

and y

are n-times-differentiable functions and c

and c

are arbitrary constants.

Hence, show that if y

, y

, ..., y

are solutions of L[y]=0, then the linear combination

+···+c

is also a solution of L[y]=0.

19. Let the linear differential operator L be deﬁned by

L[y]=a

(n)

+ a

(n−1)

+···+a

where a

, a

, ..., a

are real constants.

(a) Find L[t

(b) Find L[e

(4)

− 5y

′′

+ 4y = 0. Do you think the four

solutions form a fundamental set of solutions? Why?

20. In this problem we show how to generalize Theorem 3.2.7 (Abel’s theorem) to higher

order equations. We ﬁrst outline the procedure for the third order equation

′′′

+ p

(t)y

′′

+ p

(t)y

′

+ p

(t)y = 0 .

Let y

, y

, and y

be solutions of this equation on an interval I.

(a) If W = W(y

, y

), show that

′

′′′

Hint: The derivative of a 3-by-3 determinant is the sum of three 3-by-3 determinants

obtained by differentiating the ﬁrst, second, and third rows, respectively.

(b) Substitute for y

′′′

, y

′′′

, and y

′′′

from the differential equation; multiply the ﬁrst row by

, multiply the second row by p

, and add these to the last row to obtain

′

=−p

(t)W.

W(y

, y

)(t) = c exp

−

(t) dt

It follows that W is either always zero or nowhere zero on I.

(d) Generalize this argument to the nth order equation

(n)

+ p

(t)y

(n−1)

+···+p

(t)y = 0

with solutions y

, ..., y

. That is, establish Abel’s formula

W(y

, ..., y

)(t) = c exp

−

(t) dt

for this case.

In each of Problems 21 through 24, use Abel’s formula (Problem 20) to ﬁnd the Wronskian of

a fundamental set of solutions of the given differential equation.

21. y

′′′

+ 2y

′′

− y

′

− 3y = 0 22. y

(4)

+ y = 0

23. ty

′′′

+ 2y

′′

− y

′

+ ty = 0 24. t

(4)

+ ty

′′′

+ y

′′

− 4y = 0

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228 Chapter 4. Higher Order Linear Equations

25. (a) Show that the functions f (t) = t

|t| and g(t) = t

are linearly dependent on

0 < t < 1 and on −1 < t < 0.

(b) Show that f (t) and g(t) are linearly independent on −1 < t < 1.

26. Show that if y

is a solution of

′′′

+ p

(t)y

′′

+ p

(t)y

′

+ p

(t)y = 0,

then the substitution y = y

(t)v(t) leads to the following second order equation for v

′

′′′

+ (3y

′

+ p

′′

+ (3y

′′

+ 2p

′

+ p

′

= 0.

In each of Problems 27 and 28, use the method of reduction of order (Problem 26) to solve

the given differential equation.

27. (2 − t)y

′′′

+ (2t − 3)y

′′

− ty

′

+ y = 0, t < 2; y

(t) = e

28. t

(t + 3)y

′′′

− 3t(t + 2)y

′′

+ 6(1 + t)y

′

− 6y = 0, t > 0; y

(t) = t

, y

(t) = t

4.2 Homogeneous Equations with Constant Coefﬁcients

Consider the nth order linear homogeneous differential equation

L[y]=a

(n)

+ a

(n−1)

+···+a

n−1

′

+ a

y = 0, (1)

where a

, a

, ..., a

are real constants and a

̸= 0. From our knowledge of second

order linear equations with constant coefﬁcients,it is natural to anticipate that y = e

is a solution of Eq. (1) for suitable values of r. Indeed,

L[e

]=e

+ a

n−1

+···+a

n−1

r + a

) = e

Z(r) (2)

for all r, where

Z(r) = a

+ a

n−1

+···+a

n−1

r + a

. (3)

For those values of r for which Z(r) = 0, it follows that L[e

]=0 and y = e

is a

solution of Eq. (1). The polynomial Z(r) is called the characteristic polynomial, and

the equation Z(r) = 0 is the characteristic equation of the differential equation (1).

Since a

̸= 0,weknow that Z(r) is a polynomialof degree n and thereforehas n zeros,

say, r

, r

, ..., r

, some of which may be equal. Hence we can write the characteristic

polynomial in the form

Z(r) = a

(r − r

)(r − r

) ···(r − r

). (4)

An important question in mathematics for more than 200 years was whether every polynomial equation

has at least one root. The afﬁrmative answer to this question, the fundamental theorem of algebra, was

given by Carl Friedrich Gauss (1777–1855) in his doctoral dissertation in 1799, although his proof does

not meet modern standards of rigor. Several other proofs have been discovered since, including three by

Gauss himself.Today, students often meet the fundamental theorem of algebra in a ﬁrst course on complex

variables, where it can be established as a consequence of some of the basic properties of complex analytic

functions.

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4.2 Homogeneous Equations with Constant Coefﬁcients 229

Real and Unequal Roots. If the roots of the characteristic equation are real and no

two are equal, then we have n distinct solutions e

, e

, ..., e

of Eq. (1). If these

functions are linearly independent, then the general solution of Eq. (1) is

y = c

+ c

+···+c

. (5)

One way to establish the linear independence of e

, e

, ..., e

is to evaluate their

Wronskian determinant. Another way is outlined in Problem 40.

EXAMPLE

Find the general solution of

(4)

+ y

′′′

− 7y

′′

− y

′

+ 6y = 0. (6)

Also ﬁnd the solution that satisﬁes the initial conditions

y(0) = 1, y

′

(0) = 0, y

′′

(0) =−2, y

′′′

(0) =−1 (7)

and plot its graph.

Assuming that y = e

, we must determine r by solving the polynomial equation

+ r

− 7r

− r + 6 = 0. (8)

The roots of this equation are r

= 1, r

=−1, r

= 2, and r

=−3. Therefore, the general

solution of Eq. (6) is

y = c

+ c

−t

+ c

−3t

. (9)

The initial conditions (7) require that c

, ..., c

satisfy the four equations

+ c

= 1,

− c

+ 2c

− 3c

= 0,

(10)

+ c

+ 4c

+ 9c

=−2,

− c

+ 8c

− 27c

=−1.

By solving this system of four linear algebraic equations, we ﬁnd that

, c

=−

, c

=−

Thus the solution of the initial value problem is

y =

−t

−

−3t

. (11)

The graph of the solution is shown in Figure 4.2.1.

–1

0.5 1

FIGURE 4.2.1 Solution of the initial value problem (6), (7):

(4)

+ y

′′′

− 7y

′′

− y

′

+ 6y = 0, y(0) = 1, y

′

(0) = 0, y

′′

(0) =−2, y

′′′

(0) =−1.

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230 Chapter 4. Higher Order Linear Equations

As Example 1 illustrates, the procedure for solving an nth order linear differential

equation with constant coefﬁcients depends on ﬁnding the roots of a corresponding

nth degree polynomial equation. If initial conditions are prescribed, then a system

of n linear algebraic equations must be solved to determine the proper values of

the constants c

, ..., c

. Each of these tasks becomes much more complicated as n

increases, and we have omitted the detailed calculations in Example 1. Computer

assistance can be very helpful in such problems.

For third and fourth degree polynomials there are formulas,

analogous to the

formula for quadraticequations but morecomplicated,thatgive exact expressionsfor

the roots. Root-ﬁnding algorithms are readily available on calculators and computers.

Sometimes they are included in the differential equation solver,so that the process of

factoring the characteristic polynomial is hidden and the solution of the differential

equation is produced automatically.

If you are faced with the need to factor the characteristic polynomial by hand,here

is one result that is sometimes helpful. Suppose that the polynomial

+ a

n−1

+···+a

n−1

r + a

= 0 (12)

has integer coefﬁcients. If r = p/q is a rational root, where p and q have no common

factors, then p must be a factor of a

, and q must be a factor of a

. For example, in

Eq. (8) the factors of a

are ±1 and the factors of a

are ±1, ±2, ±3, and ±6. Thus

the only possible rational roots of this equation are ±1, ±2, ±3, and ±6. By testing

these possible roots, we ﬁnd that 1, −1, 2, and −3 are actual roots. In this case there

are no other roots, since the polynomial is of fourth degree. If some of the roots are

irrational or complex, as is usually the case, then this process will not ﬁnd them, but

at least the degree of the polynomial can be reduced by dividing the polynomial by

the factors corresponding to the rational roots.

If the roots of the characteristic equation are real and different, we have seen

that the general solution (5) is simply a sum of exponential functions. For large values

of t the solution is dominated by the term corresponding to the algebraically largest

root. If this rootis positive,then solutions become exponentially unbounded,whereas

if it is negative, then solutions tend exponentially to zero. Finally, if the largest root

is zero, then solutions approach a nonzero constant as t becomes large. Of course,

for certain initial conditions, the coefﬁcient of the otherwise dominant term may be

zero; then the nature of the solution for large t is determined by the next largest root.

Complex Roots. If the characteristic equation has complex roots, they must occur in

conjugate pairs, λ ± iµ, since the coefﬁcients a

, a

, ..., a

are real numbers. Pro-

vided that none of the roots is repeated, the general solution of Eq. (1) is still of the

The method for solving the cubic equation was apparently discovered by Scipione dal Ferro (1465–

1526) about 1500, although it was ﬁrst published in 1545 by Girolamo Cardano (1501–1576) in his Ars

Magna.This book also contains a method for solving quartic equations that Cardano attributes to his pupil

Ludovico Ferrari (1522–1565). The question of whether analogous formulas exist for the roots of higher

degree equations remained open for more than two centuries, until 1826, when Niels Abel showed that

no general solution formulas can exist for polynomial equations of degree ﬁve or higher. A more general

theory was developed by Evariste Galois (1811–1832) in 1831,but unfortunately it did not become widely

known for several decades.

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4.2 Homogeneous Equations with Constant Coefﬁcients 231

form of Eq. (5). However, just as for the second order equation (Section 3.3), we can

replace the complex-valued solutions e

(λ+iµ)t

and e

(λ−iµ)t

by the real-valued solutions

λt

cos µt, e

λt

sin µt (13)

obtained as the real and imaginary parts of e

(λ+iµ)t

. Thus, even though some of

the roots of the characteristic equation are complex, it is still possible to express the

general solution of Eq. (1) as a linear combination of real-valued solutions.

EXAMPLE

Find the general solution of

(4)

− y = 0. (14)

Also ﬁnd the solution that satisﬁes the initial conditions

y(0) = 7/2, y

′

(0) =−4, y

′′

(0) = 5/2, y

′′′

(0) =−2 (15)

and draw its graph.

Substituting e

for y, we ﬁnd that the characteristic equation is

− 1 = (r

− 1)(r

+ 1) = 0.

Therefore, the roots are r = 1, −1, i, −i, and the general solution of Eq. (14) is

y = c

+ c

−t

+ c

cos t + c

sin t.

If we impose the initial conditions (15), we obtain

= 0, c

= 3, c

= 1/2, c

=−1;

thus the solution of the given initial value problem is

y = 3e

−t

cos t − sin t. (16)

The graph of this solution is shown in Figure 4.2.2.

4 6 8 10 12

–2

142

FIGURE 4.2.2 Solution of the initial value problem (14), (15):

(4)

− y = 0, y(0) = 7/2, y

′

(0) =−4, y

′′

(0) = 5/2, y

′′′

(0) =−2.

Observe that the initial conditions (15) cause the coefﬁcient c

of the exponentially growing

term in the general solution to be zero. Therefore, this term is absent in the solution (16),

which describes an exponential decay to a steady oscillation, as Figure 4.2.2 shows. However,

if the initial conditions are changed slightly, then c

is likely to be nonzero, and the nature of

the solution changes enormously. For example, if the ﬁrst three initial conditions remain the

same,but the value of y

′′′

(0) is changed from −2to−15/8, then the solution of the initial value

problem becomes

y =

−t

cos t −

sin t. (17)

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232 Chapter 4. Higher Order Linear Equations

The coefﬁcients in Eq. (17) differ only slightly from those in Eq. (16), but the exponentially

growing term, even with the relatively small coefﬁcient of 1/32, completely dominates the

solution by the time t is larger than about 4 or 5. This is clearly seen in Figure 4.2.3, which

shows the graphs of the two solutions (16) and (17).

FIGURE 4.2.3 The blue curve is the solution of the initial value problem (14), (15)

and is the same as the curve in Figure 4.2.2. The black curve is the solution of the

modiﬁed problem in which the last initial condition is changed to y

′′′

(0) =−15/8.

Repeated Roots. If the roots of the characteristic equation are not distinct—that is,

if some of the roots are repeated—then the solution (5) is clearly not the general

solution of Eq. (1). Recall that if r

is a repeated root for the second order linear

equation a

′′

+ a

′

+ a

y = 0, then two linearly independent solutions are e

and

. For an equation of order n, if a root of Z(r) = 0, say r = r

, has multiplicity s

(where s ≤ n), then

, te

, t

, ..., t

s−1

(18)

are corresponding solutions of Eq. (1). See Problem 41 for a proof of this statement,

which is valid whether the repeated root is real or complex.

Note that a complex root can be repeated only if the differential equation (1) is

of order four or higher. If a complex root λ + iµ is repeated s times, the complex

conjugate λ − iµ is also repeated s times. Corresponding to these 2s complex-valued

solutions, we can ﬁnd 2s real-valued solutions by noting that the real and imaginary

parts of e

(λ+iµ)t

, te

(λ+iµ)t

, ..., t

s−1

(λ+iµ)t

are also linearly independent solutions:

λt

cos µt, e

λt

sin µt, te

λt

cos µt, te

λt

sin µt,

..., t

s−1

λt

cos µt, t

s−1

λt

sin µt.

Hence thegeneral solution of Eq. (1) can alwaysbe expressed as a linear combination

of n real-valued solutions. Consider the following example.

EXAMPLE

Find the general solution of

(4)

+ 2y

′′

+ y = 0. (19)

The characteristic equation is

+ 2r

+ 1 = (r

+ 1)(r

+ 1) = 0.

The roots are r = i, i, −i, −i , and the general solution of Eq. (19) is

y = c

cos t + c

sin t + c

t cos t + c

t sin t.

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4.2 Homogeneous Equations with Constant Coefﬁcients 233

In determining the roots of the characteristic equation, it may be necessary to

compute the cube roots, the fourth roots, or even higher roots of a (possibly com-

plex) number. This can usually be done most conveniently by using Euler’s formula

= cos t + i sin t and the algebraic laws given in Section 3.3. This is illustrated in the

following example.

EXAMPLE

Find the general solution of

(4)

+ y = 0. (20)

The characteristic equation is

+ 1 = 0.

To solve the equation, we must compute the fourth roots of −1. Now −1, thought of as a

complex number, is −1 + 0i. It has magnitude 1 and polar angle π. Thus

−1 = cos π + i sin π = e

iπ

Moreover, the angle is determined only up to a multiple of 2π. Thus

−1 = cos(π + 2mπ) + i sin(π + 2mπ) = e

i(π+2mπ)

where m is zero or any positive or negative integer. Thus

(−1)

1/4

= e

i(π/4+mπ/2)

= cos

)

mπ

+ i sin

)

mπ

The four fourth roots of −1 are obtained by setting m = 0, 1, 2, and 3; they are

1 + i

√

−1 + i

√

−1 − i

√

1 − i

√

It is easy to verify that, for any other value of m, we obtain one of these four roots. For

example, corresponding to m = 4, we obtain (1 + i)/

√

2. The general solution of Eq. (20) is

y = e

√

cos

√

+ c

sin

√

(

+ e

−t/

√

cos

√

+ c

sin

√

(

. (21)

In conclusion, we note that the problem of ﬁnding all the roots of a polynomial

equation may not be entirely straightforward, even with computer assistance. For

instance, it may be difﬁcult to determine whether two roots are equal or merely

very close together. Recall that the form of the general solution is different in these

two cases.

If the constants a

, a

, ..., a

in Eq. (1) are complex numbers,the solution of Eq. (1)

is still of the form (4). In this case, however, the roots of the characteristic equation

are, in general, complex numbers, and it is no longer true that the complex conjugate

of a root is also a root. The corresponding solutions are complex-valued.

PROBLEMS In each of Problems 1 through 6, express the given complex number in the form

R(cos θ + i sin θ) = Re

iθ

1. 1 + i 2. −1 +

√

3. −34. −i

√

3 − i 6. −1 − i

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234 Chapter 4. Higher Order Linear Equations

In each of Problems 7 through 10, follow the procedure illustrated in Example 4 to determine

the indicated roots of the given complex number.

7. 1

1/3

8. (1 − i)

1/2

9. 1

1/4

10. [2(cos π/3 + i sinπ/3)]

1/2

In each of Problems 11 through 28, ﬁnd the general solution of the given differential equation.

11. y

′′′

− y

′′

− y

′

+ y = 0 12. y

′′′

− 3y

′′

+ 3y

′

− y = 0

13. 2y

′′′

− 4y

′′

− 2y

′

+ 4y = 0 14. y

(4)

− 4y

′′′

+ 4y

′′

= 0

15. y

(6)

+ y = 0 16. y

(4)

− 5y

′′

+ 4y = 0

17. y

(6)

− 3y

(4)

+ 3y

′′

− y = 0 18. y

(6)

− y

′′

= 0

19. y

(5)

− 3y

(4)

+ 3y

′′′

− 3y

′′

+ 2y

′

= 0 20. y

(4)

− 8y

′

= 0

21. y

(8)

+ 8y

(4)

+ 16y = 0 22. y

(4)

+ 2y

′′

+ y = 0

23. y

′′′

− 5y

′′

+ 3y

′

+ y = 0 24. y

′′′

+ 5y

′′

+ 6y

′

+ 2y = 0

25.

18y

′′′

+ 21y

′′

+ 14y

′

+ 4y = 0 26. y

(4)

− 7y

′′′

+ 6y

′′

+ 30y

′

− 36y = 0

27.

12y

(4)

+ 31y

′′′

+ 75y

′′

+ 37y

′

+ 5y = 0 28. y

(4)

+ 6y

′′′

+ 17y

′′

+ 22y

′

+ 14y = 0

In each of Problems 29 through 36, ﬁnd the solution of the given initial value problem, and

plot its graph. How does the solution behave as t →∞?

29.

′′′

+ y

′

= 0; y(0) = 0, y

′

(0) = 1, y

′′

(0) = 2

30.

(4)

+ y = 0; y(0) = 0, y

′

(0) = 0, y

′′

(0) =−1, y

′′′

(0) = 0

31.

(4)

− 4y

′′′

+ 4y

′′

= 0; y(1) =−1, y

′

(1) = 2, y

′′

(1) = 0, y

′′′

(1) = 0

32.

′′′

− y

′′

+ y

′

− y = 0; y(0) = 2, y

′

(0) =−1, y

′′

(0) =−2

33.

(4)

− y

′′′

− 9y

′′

+ 4y

′

+ 4y = 0; y(0) =−2, y

′

(0) = 0, y

′′

(0) =−2, y

′′′

(0) = 0

34.

′′′

+ y

′

+ 5y = 0; y(0) = 2, y

′

(0) = 1, y

′′

(0) =−1

35.

′′′

+ 5y

′′

+ y

′

= 0; y(0) =−2, y

′

(0) = 2, y

′′

(0) = 0

36.

(4)

+ 6y

′′′

+ 17y

′′

+ 22y

′

+ 14y = 0; y(0) = 1, y

′

(0) =−2, y

′′

(0) = 0,

′′′

(0) = 3

37. Show that the general solution of y

(4)

− y = 0 can be written as

y = c

cos t + c

sin t + c

cosh t + c

sinh t.

Determine the solution satisfying the initial conditions y(0) = 0, y

′

(0) = 0, y

′′

(0) = 1,

′′′

(0) = 1. Why is it convenient to use the solutions cosh t and sinh t rather than e

and

−t

38. Consider the equation y

(4)

− y = 0.

(a) Use Abel’s formula [Problem 20(d) of Section 4.1] to ﬁnd the Wronskian of a

fundamental set of solutions of the given equation.

(b) Determine the Wronskian of the solutions e

, e

−t

, cos t, and sin t.

39. Consider the spring–mass system, shown in Figure 4.2.4, consisting of two unit masses

suspended from springs with spring constants 3 and 2, respectively. Assume that there is

no damping in the system.

(a) Show that the displacements u

and u

of the masses from their respective equilibrium

positions satisfy the equations

′′

+ 5u

= 2u

, u

′′

+ 2u

= 2u

. (i)

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4.2 Homogeneous Equations with Constant Coefﬁcients 235

(b) Solve the ﬁrst of Eqs. (i) for u

and substitute into the second equation, thereby

obtaining the following fourth order equation for u

(4)

+ 7u

′′

+ 6u

= 0. (ii)

Find the general solution of Eq. (ii).

(0) = 1, u

′

(0) = 0, u

(0) = 2, u

′

(0) = 0. (iii)

Use the ﬁrst of Eqs. (i) and the initial conditions (iii) to obtain values for u

′′

(0) and u

′′′

(0).

Then show that the solution of Eq. (ii) that satisﬁes the four initial conditions on u

(t) = cos t. Show that the corresponding solution u

is u

(t) = 2 cos t.

(d) Now suppose that the initial conditions are

(0) =−2, u

′

(0) = 0, u

(0) = 1, u

′

(0) = 0. (iv)

Proceed as in part (c) to show that the corresponding solutions are u

(t) =−2 cos

√

6 t

and u

(t) = cos

√

6 t.

(e) Observe that the solutions obtained in parts (c) and (d) describe two distinct modes

of vibration. In the ﬁrst, the frequency of the motion is 1, and the two masses move in

phase, both moving up or down together; the second mass moves twice as far as the ﬁrst.

The second motion has frequency

√

6, and the masses move out of phase with each other,

one moving down while the other is moving up, and vice versa. In this mode the ﬁrst mass

moves twice as far as the second. For other initial conditions, not proportional to either

of Eqs. (iii) or (iv), the motion of the masses is a combination of these two modes.

= 3

= 1

= 2

FIGURE 4.2.4 A two-spring, two-mass system.

40. In this problem we outline one way to show that if r

, ..., r

are all real and different, then

, ..., e

are linearly independent on −∞ < t < ∞. To do this, we consider the linear

relation

+···+c

= 0, −∞ < t < ∞ (i)

and show that all the constants are zero.

(a) Multiply Eq. (i) by e

−r

and differentiate with respect to t, thereby obtaining

− r

−r

+···+c

− r

−r

= 0.

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236 Chapter 4. Higher Order Linear Equations

(b) Multiply the result of part (a) by e

−(r

−r

and differentiate with respect to t to obtain

− r

)(r

− r

−r

+···+c

− r

)(r

− r

−r

= 0.

− r

n−1

) ···(r

− r

−r

n−1

= 0.

Hence c

= 0, and therefore,

+···+c

n−1

= 0.

(d) Repeat the preceding argument to show that c

n−1

= 0. In a similar way it follows that

n−2

=···=c

= 0. Thus the functions e

, ..., e

are linearly independent.

41. In this problem we indicate one way to show that if r = r

is a root of multiplicity s

of the characteristic polynomial Z(r), then e

, te

, ..., t

s−1

are solutions of Eq. (1).

This problem extends to nth order equations the method for second order equations given

in Problem 22 of Section 3.4. We start from Eq. (2) in the text

L[e

]=e

Z(r) (i)

and differentiate repeatedly with respect to r, setting r = r

after each differentiation.

(a) Observe that if r

is a root of multiplicity s, then Z(r) = (r − r

)

q(r), where q(r) is

a polynomial of degree n − s and q(r

) ̸= 0. Show that Z(r

), Z

′

), ..., Z

(s−1)

) are all

zero, but Z

(s)

) ̸= 0.

(b) By differentiating Eq. (i) repeatedly with respect to r, show that

∂

∂r

L[e

]=L

∂

∂r

= L[te

∂

s−1

∂r

s−1

L[e

]=L[t

s−1

, te

, ..., t

s−1

are solutions of Eq. (1).

4.3 The Method of Undetermined Coefﬁcients

A particular solution Y of the nonhomogeneous nth order linear equation with

constant coefﬁcients

L[y]=a

(n)

+ a

(n−1)

+···+a

n−1

′

+ a

y = g(t) (1)

can be obtained by the method of undetermined coefﬁcients, provided that g(t) is

of an appropriate form. Although the method of undetermined coefﬁcients is not

as general as the method of variation of parameters described in the next section, it

is usually much easier to use when it is applicable.

Just as for the second order linear equation, when the constant coefﬁcient lin-

ear differential operator L is applied to a polynomial A

+ A

m−1

+···+A

,an

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4.3 The Method of Undetermined Coefﬁcients 237

exponential function e

αt

, a sine function sin βt, or a cosine function cos βt, the result

is a polynomial, an exponential function, or a linear combination of sine and cosine

functions, respectively. Hence,if g(t) is a sum of polynomials, exponentials, sines, and

cosines, or products of such functions, we can expect that it is possible to ﬁnd Y(t)

by choosing a suitable combination of polynomials, exponentials, and so forth, mul-

tiplied by a number of undetermined constants. The constants are then determined

by substituting the assumed expression into Eq. (1).

The main difference in using this method for higher order equations stems from

the fact that roots of the characteristic polynomial equation may have multiplicity

greater than 2. Consequently, terms proposed for the nonhomogeneous part of the

solution may need to be multiplied by higher powers of t to make them different from

terms in the solution of the corresponding homogeneous equation. The following

examples illustrate this. In these examples we have omitted numerous straightfor-

ward algebraic steps, because our main goal is to show how to arrive at the correct

form for the assumed solution.

EXAMPLE

Find the general solution of

′′′

− 3y

′′

+ 3y

′

− y = 4e

. (2)

The characteristic polynomial for the homogeneous equation corresponding to Eq. (2) is

− 3r

+ 3r − 1 = (r − 1)

so the general solution of the homogeneous equation is

(t) = c

+ c

. (3)

To ﬁnd a particular solution Y(t) of Eq. (2), we start by assuming that Y(t) = Ae

. However,

since e

, te

, and t

are all solutions of the homogeneous equation, we must multiply this

initial choice by t

. Thus our ﬁnal assumption is that Y(t) = At

, where A is an undeter-

mined coefﬁcient. To ﬁnd the correct value for A, we differentiate Y(t) three times, substitute

for y and its derivatives in Eq. (2), and collect terms in the resulting equation. In this way we

obtain

6Ae

= 4e

Thus A =

and the particular solution is

Y(t) =

. (4)

The general solution of Eq. (2) is the sum of y

(t) from Eq. (3) and Y(t) from Eq. (4):

y = c

+ c

EXAMPLE

Find a particular solution of the equation

(4)

+ 2y

′′

+ y = 3 sin t − 5 cos t. (5)

The general solution of the homogeneous equation was found in Example 3 of Section 4.2;

it is

(t) = c

cos t + c

sin t + c

t cos t + c

t sin t, (6)

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238 Chapter 4. Higher Order Linear Equations

corresponding to the roots r = i,i,−i,and −i of the characteristic equation. Our initial assump-

tion for a particular solution is Y(t) = A sin t + B cos t , but we must multiply this choice by t

make it different from all solutions of the homogeneous equation. Thus our ﬁnal assumption is

Y(t) = At

sin t + Bt

cos t.

Next, we differentiate Y(t) four times, substitute into the differential equation (4), and collect

terms, obtaining ﬁnally

−8A sin t − 8B cos t = 3 sin t − 5 cos t.

Thus A =−

, B =

, and the particular solution of Eq. (4) is

Y(t) =−

sin t +

cos t. (7)

If g(t) is a sum of several terms, it may be easier in practice to compute separately

the particular solution corresponding to each term in g(t). As for the second order

equation,the particular solution of the complete problem is the sum of the particular

solutions of the individual component problems. This is illustrated in the following

example.

EXAMPLE

Find a particular solution of

′′′

− 4y

′

= t + 3 cos t + e

−2t

. (8)

First we solve the homogeneous equation. The characteristic equation is r

− 4r = 0, and

the roots are r = 0, ±2; hence

(t) = c

+ c

−2t

We can write a particular solution of Eq. (8)as the sum of particular solutions of the differential

equations

′′′

− 4y

′

= t, y

′′′

− 4y

′

= 3 cos t, y

′′′

− 4y

′

= e

−2t

Our initial choice for a particular solution Y

(t) of the ﬁrst equation is A

t + A

,but a constant

is a solution of the homogeneous equation, so we multiply by t. Thus

(t) = t(A

t + A

For the second equation we choose

(t) = B cos t + C sin t,

and there is no need to modify this initial choice since sin t and cos t are not solutions of the

homogeneous equation. Finally, for the third equation, since e

−2t

is a solution of the homo-

geneous equation, we assume that

(t) = Ete

−2t

The constants are determined by substituting into the individual differential equations; they

are A

=−

, A

= 0, B = 0, C =−

, and E =

. Hence a particular solution of Eq. (8) is

Y(t) =−

−

sin t +

−2t

. (9)

You should keep in mind that the amount of algebra required to calculate

the coefﬁcients may be quite substantial for higher order equations, especially if the

nonhomogeneous term is even moderately complicated. A computer algebra system

can be extremely helpful in executing these algebraic calculations.

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4.3 The Method of Undetermined Coefﬁcients 239

The method of undetermined coefﬁcients can be used whenever it is possible to

guess the correct form for Y(t). However, this is usually impossible for differential

equations not having constant coefﬁcients, or for nonhomogeneous terms other than

the type described previously. For more complicatedproblems wecan use the method

of variation of parameters, which is discussed in the next section.

PROBLEMS In each of Problems 1 through 8, determine the general solution of the given differential

equation.

1. y

′′′

− y

′′

− y

′

+ y = 2e

−t

+ 32.y

(4)

− y = 3t + cos t

3. y

′′′

+ y

′′

+ y

′

+ y = e

−t

+ 4t 4. y

′′′

− y

′

= 2 sin t

5. y

(4)

− 4y

′′

= t

+ e

6. y

(4)

+ 2y

′′

+ y = 3 + cos 2t

7. y

(6)

+ y

′′′

= t 8. y

(4)

+ y

′′′

= sin 2t

In each of Problems 9 through 12, ﬁnd the solution of the given initial value problem. Then

plot a graph of the solution.

′′′

+ 4y

′

= t; y(0) = y

′

(0) = 0, y

′′

(0) = 1

10.

(4)

+ 2y

′′

+ y = 3t + 4; y(0) = y

′

(0) = 0, y

′′

(0) = y

′′′

(0) = 1

11.

′′′

− 3y

′′

+ 2y

′

= t + e

; y(0) = 1, y

′

(0) =−

, y

′′

(0) =−

12. y

(4)

+ 2y

′′′

+ y

′′

+ 8y

′

− 12y = 12 sin t − e

−t

; y(0) = 3, y

′

(0) = 0,

′′

(0) =−1, y

′′′

(0) = 2

In each of Problems 13 through 18, determine a suitable form for Y(t) if the method of

undetermined coefﬁcients is to be used. Do not evaluate the constants.

13. y

′′′

− 2y

′′

+ y

′

= t

+ 2e

14. y

′′′

− y

′

= te

−t

+ 2 cos t

15. y

(4)

− 2y

′′

+ y = e

+ sin t 16. y

(4)

+ 4y

′′

= sin 2t + te

+ 4

17. y

(4)

− y

′′′

− y

′′

+ y

′

= t

+ 4 + t sin t 18. y

(4)

+ 2y

′′′

+ 2y

′′

= 3e

+ 2te

−t

+ e

−t

sin t

19. Consider the nonhomogeneous nth order linear differential equation

(n)

+ a

(n−1)

+···+a

y = g(t), (i)

where a

, ..., a

are constants. Verify that if g(t) is of the form

αt

+···+b

then the substitution y = e

αt

u(t) reduces Eq. (i) to the form

(n)

+ k

(n−1)

+···+k

u = b

+···+b

, (ii)

where k

, ..., k

are constants. Determine k

and k

in terms of the a’s and α. Thus the

problem of determining a particular solution of the original equation is reduced to the sim-

pler problem of determining a particular solution of an equation with constant coefﬁcients

and a polynomial for the nonhomogeneous term.

Method of Annihilators. In Problems 20 through 22, we consider another way of arriving at

the proper form of Y (t) for use in the method of undetermined coefﬁcients. The procedure

is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and

products of such terms) can be viewed as solutions of certain linear homogeneous differential

equations with constant coefﬁcients. It is convenient to use the symbol D for d/dt. Then, for

example, e

−t

is a solution of (D + 1)y = 0; the differential operator D + 1 is said to annihilate,

or to be an annihilator of, e

−t

. In the same way, D

+ 4 is an annihilator of sin 2t or cos 2t,

(D − 3)

= D

− 6D + 9 is an annihilator of e

or te

, and so forth.

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240 Chapter 4. Higher Order Linear Equations

20. Show that linear differential operators with constant coefﬁcients obey the commutative

law. That is, show that

(D − a)(D − b)f = (D − b)(D − a)f

for any twice-differentiable function f and any constants a and b. The result extends at

once to any ﬁnite number of factors.

21. Consider the problem of ﬁnding the form of a particular solution Y(t) of

(D − 2)

(D + 1)Y = 3e

− te

−t

, (i)

where the left side of the equation is written in a form corresponding to the factorization

of the characteristic polynomial.

(a) Show that D − 2 and (D + 1)

, respectively, are annihilators of the terms on the right

side of Eq. (i), and that the combined operator (D − 2)(D + 1)

annihilates both terms on

the right side of Eq. (i) simultaneously.

(b) Apply the operator (D − 2)(D + 1)

to Eq. (i) and use the result of Problem 20 to

obtain

(D − 2)

(D + 1)

Y = 0. (ii)

Thus Y is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that

Y(t) = c

+ c

−t

+ c

−t

+ c

−t

, (iii)

where c

, ..., c

are constants, as yet undetermined.

, te

, t

, and e

−t

are solutions of the homogeneous equation cor-

responding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous

equation. Therefore, choose c

, c

, and c

to be zero in Eq. (iii), so that

Y(t) = c

+ c

−t

+ c

−t

. (iv)

This is the form of the particular solution Y of Eq. (i). The values of the coefﬁcients c

, c

and c

can be found by substituting from Eq. (iv) in the differential equation (i).

Summary. Suppose that

L(D)y = g(t), (v)

where L(D) is a linear differential operator with constant coefﬁcients, and g(t) is a sum or

product of exponential, polynomial, or sinusoidal terms. To ﬁnd the form of a particular

solution of Eq. (v), you can proceed as follows:

(a) Find a differential operator H(D) with constant coefﬁcients that annihilates g(t)—that is,

an operator such that H(D)g(t) = 0.

(b) Apply H(D) to Eq. (v), obtaining

H(D)L(D)y = 0, (vi)

which is a homogeneous equation of higher order.

(d) Eliminate from the solution found in step (c) the terms that also appear in the solution

of L(D)y = 0. The remaining terms constitute the correct form of a particular solution of

Eq. (v).

22. Use the method of annihilators to ﬁnd the form of a particular solution Y(t) for each of

the equations in Problems 13 through 18. Do not evaluate the coefﬁcients.

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4.4 The Method of Variation of Parameters 241

4.4 The Method of Variation of Parameters

The method of variation of parameters for determining a particular solution of the

nonhomogeneous nth order linear differential equation

L[y]=y

(n)

+ p

(t)y

(n−1)

+···+p

n−1

(t)y

′

+ p

(t)y = g(t) (1)

is a direct extension of the method for the second order differential equation (see

Section 3.6). As before, to use the method of variation of parameters, it is ﬁrst nec-

essary to solve the corresponding homogeneous differential equation. In general,

this may be difﬁcult unless the coefﬁcients are constants. However, the method of

variation of parameters is still more general than the method of undetermined coef-

ﬁcients in that it leads to an expression for the particular solution for any continuous

function g, whereas the method of undetermined coefﬁcients is restricted in practice

to a limited class of functions g.

Suppose then that we know a fundamental set of solutions y

, y

, ..., y

of the

homogeneous equation. Then the general solution of the homogeneous equation is

(t) = c

(t) + c

(t) +···+c

(t). (2)

The method of variationof parameters for determininga particular solution ofEq. (1)

rests on the possibility of determining n functions u

, u

, ..., u

such that Y(t) is of

the form

Y(t) = u

(t)y

(t) + u

(t)y

(t) +···+u

(t)y

(t). (3)

Since we have n functions to determine, we will have to specify n conditions. One

of these is clearly that Y satisfy Eq. (1). The other n − 1 conditions are chosen so

as to make the calculations as simple as possible. Since we can hardly expect a sim-

pliﬁcation in determining Y if we must solve high order differential equations for

, ..., u

, it is natural to impose conditions to suppress the terms that lead to higher

derivatives of u

, ..., u

. From Eq. (3) we obtain

′

= (u

′

+ u

′

+···+u

′

) + (u

′

+ u

′

+···+u

′

), (4)

where we have omitted the independent variable t on which each function in Eq. (4)

depends. Thus the ﬁrst condition that we impose is that

′

+ u

′

+···+u

′

= 0. (5)

It follows that the expression (4) for Y

′

reduces to

′

= u

′

+ u

′

+···+u

′

. (6)

We continue this process by calculating the successive derivatives Y

′′

, ..., Y

(n−1)

After each differentiation we set equal to zero the sum of terms involving derivatives

of u

, ..., u

. In this way we obtain n − 2 further conditions similar to Eq. (5); that is,

′

(m)

+ u

′

(m)

+···+u

′

(m)

= 0, m = 1, 2, ..., n − 2. (7)

As a result of these conditions, it follows that the expressions for Y

′′

, ..., Y

(n−1)

reduce to

(m)

= u

(m)

+ u

(m)

+···+u

(m)

, m = 2, 3, ..., n − 1, (8)

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242 Chapter 4. Higher Order Linear Equations

Finally, we need to impose the condition that Y must be a solution of Eq. (1). By

differentiating Y

(n−1)

from Eq. (8), we obtain

(n)

= (u

(n)

+···+u

(n)

) + (u

′

(n−1)

+···+u

′

(n−1)

). (9)

To satisfy the differential equation we substitute for Y and its derivatives in Eq. (1)

from Eqs. (3), (6), (8), and (9). Then we group the terms involving each of the

functions y

, ..., y

and their derivatives. It then follows that most of the terms in the

equation drop out because each of y

, ..., y

is a solution of Eq. (1) and therefore

L[y

]=0, i = 1, 2, ..., n. The remaining terms yield the relation

′

(n−1)

+ u

′

(n−1)

+···+u

′

(n−1)

= g. (10)

Equation (10), Eq. (5), and the n − 2 equations (7) provide n simultaneous linear

nonhomogeneous algebraic equations for u

′

, u

′

, ..., u

′

+ y

′

+···+y

′

= 0,

′

+ y

′

+···+y

′

= 0,

′′

′

+ y

′′

′

+···+y

′′

′

= 0, (11)

(n−1)

′

+···+y

(n−1)

′

= g.

The system (11) is a linear algebraic system for the unknown quantities u

′

, ..., u

′

By solving this system and then integrating the resulting expressions, you can obtain

the coefﬁcients u

, ..., u

. A sufﬁcient condition for the existence of a solution of the

system of equations (11) is that the determinant of coefﬁcients is nonzero for each

value of t. However,the determinant of coefﬁcients is precisely W(y

, y

, ..., y

),and

it is nowhere zero since y

, ..., y

is a fundamental set of solutions of the homoge-

neous equation. Hence it is possible to determine u

′

, ..., u

′

. Using Cramer’s

rule,

we can write the solution of the system of equations (11) in the form

′

(t) =

g(t)W

(t)

W(t)

, m = 1, 2, ..., n. (12)

Here W(t) = W(y

, y

, ..., y

)(t), and W

is the determinant obtained from W by

replacing the mth column by the column (0, 0, ...,0,1).With this notation a particular

solution of Eq. (1) is given by

Y(t) =

m=1

(t)

g(s)W

(s)

W(s)

ds, (13)

where t

is arbitrary. Although the procedure is straightforward, the algebraic com-

putations involved in determining Y(t) from Eq. (13) become more and more

Cramer’s rule is credited to the Swiss mathematician Gabriel Cramer (1704–1752), professor at the

Académie de Calvin in Geneva, who published it in a general form (but without proof) in 1750. For small

systems the result had been known earlier.

August 7, 2012 21:04 C04 Sheet number 23 Page number 243 cyan black

4.4 The Method of Variation of Parameters 243

complicated as n increases. In some cases the calculations may be simpliﬁed to some

extent by using Abel’s identity (Problem 20 of Section 4.1),

W(t) = W(y

, ..., y

)(t) = c exp

−

(t) dt

The constant c can be determined by evaluating W at some convenient point.

EXAMPLE

Given that y

(t) = e

, y

(t) = te

, and y

(t) = e

−t

are solutions of the homogeneous equation

corresponding to

′′′

− y

′′

− y

′

+ y = g(t), (14)

determine a particular solution of Eq. (14) in terms of an integral.

We use Eq. (13). First, we have

W(t) = W(e

, te

, e

−t

)(t) =

−t

(t + 1)e

−e

−t

(t + 2)e

−t

Factoring e

from each of the ﬁrst two columns and e

−t

from the third column, we obtain

W(t) = e

1 t 1

1 t + 1 −1

1 t + 21

Then, by subtracting the ﬁrst row from the second and third rows, we have

W(t) = e

1 t 1

01−2

020

Finally, evaluating the latter determinant by minors associated with the ﬁrst column, we ﬁnd

that

W(t) = 4e

Next,

(t) =

0 te

−t

0 (t + 1)e

−e

−t

1 (t + 2)e

−t

Using minors associated with the ﬁrst column, we obtain

(t) =

−t

(t + 1)e

−e

−t

=−2t − 1.

In a similar way,

(t) =

0 e

−t

0 −e

−t

1 e

−t

=−

−t

−e

−t

= 2

August 7, 2012 21:04 C04 Sheet number 24 Page number 244 cyan black

244 Chapter 4. Higher Order Linear Equations

and

(t) =

(t + 1)e

(t + 2)e

(t + 1)e

= e

Substituting these results in Eq. (13), we have

Y(t) = e

g(s)(−1 − 2s)

ds + te

g(s)(2)

ds + e

−t

g(s)e

t−s

[−1 + 2(t − s)]+e

−(t−s)

g(s) ds. (15)

Depending on the speciﬁc function g(t),it may or may not be possible to evaluate the integrals

in Eq. (15) in terms of elementary functions.

PROBLEMS In each of Problems 1 through 6, use the method of variation of parameters to determine the

general solution of the given differential equation.

1. y

′′′

+ y

′

= tan t, −π/2 < t <π/22.y

′′′

− y

′

= t

3. y

′′′

− 2y

′′

− y

′

+ 2y = e

4. y

′′′

+ y

′

= sec t, −π/2 < t <π/2

5. y

′′′

− y

′′

+ y

′

− y = e

−t

sin t 6. y

(4)

+ 2y

′′

+ y = sin t

In each of Problems 7 and 8, ﬁnd the general solution of the given differential equation. Leave

your answer in terms of one or more integrals.

7. y

′′′

− y

′′

+ y

′

− y = sec t, −π/2 < t <π/2

8. y

′′′

− y

′

= csc t,0< t <π

In each of Problems 9 through 12, ﬁnd the solution of the given initial value problem. Then

plot a graph of the solution.

′′′

+ y

′

= sec t; y(0) = 2, y

′

(0) = 1, y

′′

(0) =−2

10.

(4)

+ 2y

′′

+ y = sin t; y(0) = 2, y

′

(0) = 0, y

′′

(0) =−1, y

′′′

(0) = 1

11.

′′′

− y

′′

+ y

′

− y = sec t; y(0) = 2, y

′

(0) =−1, y

′′

(0) = 1

12.

′′′

− y

′

= csc t; y(π/2) = 2, y

′

(π/2) = 1, y

′′

(π/2) =−1

13. Given that x, x

, and 1/x are solutions of the homogeneous equation corresponding to

′′′

+ x

′′

− 2xy

′

+ 2y = 2x

, x > 0,

determine a particular solution.

14. Find a formula involving integrals for a particular solution of the differential equation

′′′

− y

′′

+ y

′

− y = g(t).

15. Find a formula involving integrals for a particular solution of the differential equation

(4)

− y = g(t).

Hint: The functions sin t, cos t, sinh t, and cosh t form a fundamental set of solutions of

the homogeneous equation.

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4.4 The Method of Variation of Parameters 245

16. Find a formula involving integrals for a particular solution of the differential equation

′′′

− 3y

′′

+ 3y

′

− y = g(t).

If g(t) = t

−2

, determine Y(t).

17. Find a formula involving integrals for a particular solution of the differential equation

′′′

− 3x

′′

+ 6xy

′

− 6y = g(x), x > 0.

Hint: Verify that x, x

, and x

are solutions of the homogeneous equation.

REFERENCES Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice-

Hall, 1961; New York: Dover, 1989).

Coddington, E. A. and Carlson, R., Linear Ordinary Differential Equations (Philadelphia, PA: Society for

Industrial and Applied Mathematics, 1997).

August 7, 2012 21:04 C04 Sheet number 26 Page number 246 cyan black

August 7, 2012 21:04 c05 Sheet number 1 Page number 247 cyan black

247

CHAPTER

Series Solutions of

Second Order

Linear Equations

Finding the general solution of a linear differential equation depends on determin-

ing a fundamentalset ofsolutions of thehomogeneous equation.So far,wehave given

a systematic procedure for constructing fundamental solutions only if the equation

has constant coefﬁcients. To deal with the much larger class of equations that have

variable coefﬁcients, it is necessary to extend our search for solutions beyond the

familiar elementary functions of calculus. The principal tool that we need is the rep-

resentation of a given function by a power series. The basic idea is similar to that in

the method of undetermined coefﬁcients:we assume that the solutions of a given dif-

ferential equation have power series expansions, and then we attempt to determine

the coefﬁcients so as to satisfy the differential equation.

5.1 Review of Power Series

In this chapter we discuss the use of power series to construct fundamental sets of

solutions of second order linear differential equations whose coefﬁcients are func-

tions of the independent variable.We begin by summarizing very brieﬂy the pertinent

results about power series that we need. Readers who are familiar with power series

may go on to Section 5.2. Those who need more details than are presented here

should consult a book on calculus.

1. A power series

∞

n=0

(x − x

)

is said to converge at a point x if

lim

m→∞

n=0

(x − x

)

August 7, 2012 21:04 c05 Sheet number 2 Page number 248 cyan black

248 Chapter 5. Series Solutions of Second Order Linear Equations

exists for that x. The series certainly converges for x = x

; it may converge for all x,orit

may converge for some values of x and not for others.

2. The series

∞

n=0

(x − x

)

is said to converge absolutely at a point x if the series

∞

n=0

(x − x

)

∞

n=0

||x − x

converges. It can be shown that if the series converges absolutely, then the series

also converges; however, the converse is not necessarily true.

3. One of the most useful tests for the absolute convergence of a power series is the ratio

test. If a

̸= 0, and if, for a ﬁxed value of x,

lim

n→∞

n+1

(x − x

)

n+1

(x − x

)

=|x − x

| lim

n→∞

n+1

=|x − x

|L,

then the power series converges absolutely at that value of x if |x − x

|L < 1 and diverges

if |x − x

|L > 1. If |x − x

|L = 1, the test is inconclusive.

EXAMPLE

For which values of x does the power series

∞

n=1

(−1)

n+1

n(x − 2)

converge?

To test for convergence, we use the ratio test. We have

lim

n→∞

(−1)

n+2

(n + 1)(x − 2)

n+1

(−1)

n+1

n(x − 2)

=|x − 2| lim

n→∞

n + 1

=|x − 2|.

According to statement 3, the series converges absolutely for |x − 2| < 1, or 1 < x < 3, and

diverges for |x − 2| > 1. The values of x corresponding to |x − 2|=1 are x = 1 and x = 3. The

series diverges for each of these values of x since the nth term of the series does not approach

zero as n →∞.

4. If the power series

∞

n=0

(x − x

)

converges at x = x

, it converges absolutely for

|x − x

| < |x

− x

|; and if it diverges at x = x

, it diverges for |x − x

| > |x

− x

5. For a typical power series,such as the one in Example 1,there is a positive number ρ,called

the radius of convergence, such that

∞

n=0

(x − x

)

converges absolutely for |x − x

| <ρ

and diverges for |x − x

| >ρ.The interval|x − x

| <ρis called the interval of convergence;

it is indicated by the hatched lines in Figure 5.1.1.The series may either converge or diverge

when |x − x

|=ρ. Many important power series converge for all values of x. In this case

it is customary to say that ρ is inﬁnite and the interval of convergence is the entire real

line. It is also possible for a power series to converge only at x

. For such a series we say

that ρ = 0 and the series has no interval of convergence. When these exceptional cases are

taken into account, every power series has a nonnegative radius of convergence ρ, and if

ρ>0, then there is a (ﬁnite or inﬁnite) interval of convergence centered at x

August 7, 2012 21:04 c05 Sheet number 3 Page number 249 cyan black

5.1 Review of Power Series 249

Series

diverges

Series

diverges

Series

converges

absolutely

Series may

converge or diverge

–

ρρ

FIGURE 5.1.1 The interval of convergence of a power series.

EXAMPLE

Determine the radius of convergence of the power series

∞

n=1

(x + 1)

We apply the ratio test:

lim

n→∞

(x + 1)

n+1

(n + 1)2

n+1

(x + 1)

|x + 1|

lim

n→∞

n + 1

|x + 1|

Thus the series converges absolutely for |x + 1| < 2, or −3 < x < 1, and diverges for

|x + 1| > 2. The radius of convergence of the power series is ρ = 2. Finally, we check the

endpoints of the interval of convergence. At x = 1 the series becomes the harmonic series

∞

n=1

which diverges. At x =−3 we have

∞

n=1

(−3 + 1)

∞

n=1

(−1)

which converges but does not converge absolutely. The series is said to converge condition-

ally at x =−3. To summarize, the given power series converges for −3 ≤ x < 1 and diverges

otherwise. It converges absolutely for −3 < x < 1 and has a radius of convergence 2.

Suppose that

∞

n=0

(x − x

)

and

∞

n=0

(x − x

)

converge to f (x) and g(x), respec-

tively, for |x − x

| <ρ, ρ>0.

6. The two series can be added or subtracted termwise, and

f (x) ± g(x) =

∞

n=0

± b

)(x − x

)

;

the resulting series converges at least for |x − x

| <ρ.

7. The two series can be formally multiplied, and

f (x)g(x) =

∞

n=0

(x − x

)

∞

n=0

(x − x

)

∞

n=0

(x − x

)

where c

= a

+ a

n−1

+···+a

. The resulting series converges at least for

|x − x

| <ρ.

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250 Chapter 5. Series Solutions of Second Order Linear Equations

Further, if g(x

) ̸= 0, the series for f (x) can be formally divided by the series for g(x), and

f (x)

g(x)

∞

n=0

(x − x

)

In most cases the coefﬁcients d

can be most easily obtained by equating coefﬁcients in

the equivalent relation

∞

n=0

(x − x

)

∞

n=0

(x − x

)

∞

n=0

(x − x

)

∞

n=0

k=0

n−k

(x − x

)

In the case of division, the radius of convergence of the resulting power series may be less

than ρ.

8. The function f is continuous and has derivatives of all orders for |x − x

| <ρ. Moreover,

′

, f

′′

, ...can be computed by differentiating the series termwise; that is,

′

(x) = a

+ 2a

(x − x

) +···+na

(x − x

)

n−1

+···

∞

n=1

(x − x

)

n−1

′′

(x) = 2a

+ 6a

(x − x

) +···+n(n − 1)a

(x − x

)

n−2

+···

∞

n=2

n(n − 1)a

(x − x

)

n−2

and so forth, and each of the series converges absolutely for |x − x

| <ρ.

9. The value of a

is given by

(n)

)

The series is called the Taylor

series for the function f about x = x

10. If

∞

n=0

(x − x

)

∞

n=0

(x − x

)

for each x in some open interval with center x

, then

= b

for n = 0, 1, 2, 3, .... In particular, if

∞

n=0

(x − x

)

= 0 for each such x, then

= a

=···=a

=···=0.

A function f that has a Taylor series expansion about x = x

f (x) =

∞

n=0

(n)

)

(x − x

)

Brook Taylor (1685–1731), English mathematician, received his education at Cambridge University. His

book Methodus incrementorum directa et inversa, published in 1715, includes a general statement of the

expansiontheorem that is named for him.Thisisabasicresultinall branches of analysis,butitsfundamental

importance was not recognized until 1772 (by Lagrange). Taylor was also the ﬁrst to use integration by

parts, was one of the founders of the calculus of ﬁnite differences, and was the ﬁrst to recognize the

existence of singular solutions of differential equations.

August 7, 2012 21:04 c05 Sheet number 5 Page number 251 cyan black

5.1 Review of Power Series 251

with a radius of convergence ρ>0, is said to be analytic at x = x

. All of the familiar

functions of calculus are analytic except perhaps at certain easily recognized points.

For example, sin x and e

are analytic everywhere, 1/x is analytic except at x = 0, and

tan x is analytic except at odd multiples of π/2. According to statements 6 and 7, if f

and g are analytic at x

, then f ± g, f · g, and f /g [provided that g(x

) ̸= 0] are also

analytic at x = x

. In many respects the natural context for the use of power series

is the complex plane. The methods and results of this chapter nearly always can be

directly extended to differential equations in which the independent and dependent

variables are complex-valued.

Shift of Index of Summation. The index of summation in an inﬁnite series is a dummy

parameter just as the integration variable in a deﬁnite integral is a dummy variable.

Thus it is immaterial which letter is used for the index of summation. For example,

∞

n=0

∞

j =0

Just as we make changes of the variable of integration in a deﬁnite integral, we ﬁnd

it convenient to make changes of summation indices in calculating series solutions of

differential equations. We illustrate by several examples how to shift the summation

index.

EXAMPLE

Write

∞

n=2

as a series whose ﬁrst term corresponds to n = 0 rather than n = 2.

Let m = n − 2; then n = m + 2, and n = 2 corresponds to m = 0. Hence

∞

n=2

∞

m=0

m+2

. (1)

By writing out the ﬁrst few terms of each of these series, you can verify that they contain

precisely the same terms. Finally, in the series on the right side of Eq. (1), we can replace the

dummy index m by n, obtaining

∞

n=2

∞

n=0

n+2

. (2)

In effect, we have shifted the index upward by 2 and have compensated by starting to count

at a level 2 lower than originally.

EXAMPLE

Write the series

∞

n=2

(n + 2)(n + 1)a

(x − x

)

n−2

(3)

as a series whose generic term involves (x − x

)

rather than (x − x

)

n−2

Again, we shift the index by 2 so that n is replaced by n + 2 and start counting 2 lower. We

obtain

∞

n=0

(n + 4)(n + 3)a

n+2

(x − x

)

. (4)

You can readily verify that the terms in the series (3) and (4) are exactly the same.

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252 Chapter 5. Series Solutions of Second Order Linear Equations

EXAMPLE

Write the expression

∞

n=0

(r + n)a

r+n−1

(5)

as a series whose generic term involves x

r+n

First, take the x

inside the summation, obtaining

∞

n=0

(r + n)a

r+n+1

. (6)

Next, shift the index down by 1 and start counting 1 higher. Thus

∞

n=0

(r + n)a

r+n+1

∞

n=1

(r + n − 1)a

n−1

r+n

. (7)

Again,you can easily verify that the two series in Eq. (7) are identical and that both are exactly

the same as the expression (5).

EXAMPLE

Assume that

∞

n=1

n−1

∞

n=0

(8)

for all x, and determine what this implies about the coefﬁcients a

We want to use statement 10 to equate corresponding coefﬁcients in the two series. In order

to do this, we must ﬁrst rewrite Eq. (8) so that the series display the same power of x in their

generic terms. For instance, in the series on the left side of Eq. (8), we can replace n by n + 1

and start counting 1 lower. Thus Eq. (8) becomes

∞

n=0

(n + 1)a

n+1

∞

n=0

. (9)

According to statement 10, we conclude that

(n + 1)a

n+1

= a

, n = 0, 1, 2, 3, ...

n+1

n + 1

, n = 0, 1, 2, 3, .... (10)

Hence, choosing successive values of n in Eq. (10), we have

= a

, a

and so forth. In general,

, n = 1, 2, 3, .... (11)

Thus the relation (8) determines all the following coefﬁcients in terms of a

. Finally, using the

coefﬁcients given by Eq. (11), we obtain

∞

n=0

= a

∞

n=0

= a

where we have followed the usual convention that 0!=1.

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5.1 Review of Power Series 253

PROBLEMS In each of Problems 1 through 8, determine the radius of convergence of the given power

series.

∞

n=0

(x − 3)

∞

n=0

∞

n=0

∞

n=0

∞

n=1

(2x + 1)

∞

n=1

(x − x

)

∞

n=1

(−1)

(x + 2)

∞

n=1

n!x

In each of Problems 9 through 16, determine the Taylor series about the point x

for the given

function. Also determine the radius of convergence of the series.

9. sin x, x

= 0 10. e

, x

= 0

11. x, x

= 1 12. x

, x

=−1

13. ln x, x

= 1 14.

1 + x

, x

= 0

15.

1 − x

, x

= 0 16.

1 − x

, x

= 2

17. Given that y =

∞

n=0

, compute y

′

and y

′′

and write out the ﬁrst four terms of each series,

as well as the coefﬁcient of x

in the general term.

18. Given that y =

∞

n=0

, compute y

′

and y

′′

and write out the ﬁrst four terms of each

series, as well as the coefﬁcient of x

in the general term. Show that if y

′′

= y, then the

coefﬁcients a

and a

are arbitrary, and determine a

and a

in terms of a

and a

. Show

that a

n+2

= a

/(n + 2)(n + 1), n = 0, 1, 2, 3, ....

In each of Problems 19 and 20, verify the given equation.

19.

∞

n=0

(x − 1)

n+1

∞

n=1

n−1

(x − 1)

20.

∞

k=0

k+1

∞

k=0

k+1

= a

∞

k=1

k+1

+ a

k−1

In each of Problems 21 through 27, rewrite the given expression as a sum whose generic term

involves x

21.

∞

n=2

n(n − 1)a

n−2

22.

∞

n=0

n+2

23. x

∞

n=1

n−1

∞

k=0

24. (1 − x

)

∞

n=2

n(n − 1)a

n−2

25.

∞

m=2

m(m − 1)a

m−2

+ x

∞

k=1

k−1

26.

∞

n=1

n−1

+ x

∞

n=0

27. x

∞

n=2

n(n − 1)a

n−2

∞

n=0

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254 Chapter 5. Series Solutions of Second Order Linear Equations

28. Determine the a

so that the equation

∞

n=1

n−1

+ 2

∞

n=0

= 0

is satisﬁed. Try to identify the function represented by the series

∞

n=0

5.2 Series Solutions Near an Ordinary Point, Part I

In Chapter 3 we described methods of solving second order linear differential equa-

tions with constant coefﬁcients. We now consider methods of solving second order

linear equations when the coefﬁcients are functions of the independent variable. In

this chapter we will denote the independent variable by x. It is sufﬁcient to consider

the homogeneous equation

P(x)

+ Q(x)

+ R(x)y = 0, (1)

since the procedure for the corresponding nonhomogeneous equation is similar.

Many problems in mathematical physics lead to equations of the form (1) having

polynomial coefﬁcients; examples include the Bessel equation

′′

+ xy

′

+ (x

− ν

)y = 0,

where ν is a constant, and the Legendre equation

(1 − x

′′

− 2xy

′

+ α(α + 1)y = 0,

where α isa constant.For thisreason,as well asto simplify thealgebraic computations,

we primarily consider the case in which the functions P, Q, and R are polynomials.

However, as we will see, the method of solution is also applicable when P, Q, and R

are general analytic functions.

For the present, then, suppose that P, Q, and R are polynomials and that there is

no factor (x − c) that is common to all three of them. If there is such a factor (x − c),

then divide it out before proceeding. Suppose also that we wish to solve Eq. (1) in

the neighborhood of a point x

. The solution of Eq. (1) in an interval containing x

is closely associated with the behavior of P in that interval.

A point x

such that P(x

) ̸= 0 is called an ordinary point. Since P is continuous, it

follows that there is an interval about x

in which P(x) is never zero. In that interval

we can divide Eq. (1) by P(x) to obtain

′′

+ p(x)y

′

+ q(x)y = 0, (2)

where p(x) = Q(x)/P(x) and q(x) = R(x)/P(x) are continuous functions. Hence,

according to the existence and uniqueness Theorem 3.2.1, there exists in that inter-

val a unique solution of Eq. (1) that also satisﬁes the initial conditions y(x

) = y

′

) = y

′

for arbitrary values of y

and y

′

. In this and the following section, we

discuss the solution of Eq. (1) in the neighborhood of an ordinary point.

On the other hand, if P (x

) = 0, then x

is called a singular point of Eq. (1). In

this case at least one of Q(x

) and R(x

) is not zero. Consequently, at least one of

August 7, 2012 21:04 c05 Sheet number 9 Page number 255 cyan black

5.2 Series Solutions Near an Ordinary Point, Part I 255

the coefﬁcients p and q in Eq. (2) becomes unbounded as x → x

, and therefore

Theorem 3.2.1 does not apply in this case. Sections 5.4 through 5.7 deal with ﬁnding

solutions of Eq. (1) in the neighborhood of a singular point.

We now take up the problem of solving Eq. (1) in the neighborhood of an ordinary

point x

. We look for solutions of the form

y = a

+ a

(x − x

) +···+a

(x − x

)

+···=

∞

n=0

(x − x

)

(3)

and assume that the series converges in the interval |x − x

| <ρfor some ρ>0.

While at ﬁrst sight it may appear unattractive to seek a solution in the form of a

power series, this is actually a convenient and useful form for a solution. Within their

intervals of convergence, power series behave very much like polynomials and are

easy to manipulate both analytically and numerically. Indeed, even if we can obtain

a solution in terms of elementary functions, such as exponential or trigonometric

functions, we are likely to need a power series or some equivalent expression if we

want to evaluate them numerically or to plot their graphs.

The most practical way to determine the coefﬁcients a

is to substitute the series

(3) and its derivatives for y,y

′

,and y

′′

in Eq. (1).The following examples illustrate this

process. The operations, such as differentiation, that are involved in the procedure

are justiﬁed so long as we stay within the interval of convergence. The differential

equations in these examples are also of considerable importance in their own right.

EXAMPLE

Find a series solution of the equation

′′

+ y = 0, −∞ < x < ∞. (4)

As we know, sin x and cos x form a fundamental set of solutions of this equation, so

series methods are not needed to solve it. However, this example illustrates the use of

power series in a relatively simple case. For Eq. (4), P(x) = 1, Q(x) = 0, and R(x) = 1; hence

every point is an ordinary point.

We look for a solution in the form of a power series about x

= 0

y = a

+ a

x + a

+···+a

+···=

∞

n=0

(5)

and assume that the series converges in some interval |x| <ρ. Differentiating Eq. (5) term by

term, we obtain

′

= a

+ 2a

x +···+na

n−1

+···=

∞

n=1

n−1

(6)

and

′′

= 2a

+···+n(n − 1)a

n−2

+···=

∞

n=2

n(n − 1)a

n−2

. (7)

Substituting the series (5) and (7) for y and y

′′

in Eq. (4) gives

∞

n=2

n(n − 1)a

n−2

∞

n=0

= 0.

August 7, 2012 21:04 c05 Sheet number 10 Page number 256 cyan black

256 Chapter 5. Series Solutions of Second Order Linear Equations

To combine the two series, we need to rewrite at least one of them so that both series display

the same generic term. Thus, in the ﬁrst sum, we shift the index of summation by replacing n

by n + 2 and starting the sum at 0 rather than 2. We obtain

∞

n=0

(n + 2)(n + 1)a

n+2

∞

n=0

= 0

∞

n=0

[

(n + 2)(n + 1)a

n+2

+ a

]

= 0.

For this equation to be satisﬁed for all x, the coefﬁcient of each power of x must be zero; hence

we conclude that

(n + 2)(n + 1)a

n+2

+ a

= 0, n = 0, 1, 2, 3, .... (8)

Equation (8) is referred to as a recurrence relation. The successive coefﬁcients can be

evaluated one by one by writing the recurrence relation ﬁrst for n = 0, then for n = 1, and

so forth. In this example Eq. (8) relates each coefﬁcient to the second one before it. Thus

the even-numbered coefﬁcients (a

, a

, ...)and the odd-numbered ones (a

, a

, ...)are

determined separately. For the even-numbered coefﬁcients we have

=−

2 · 1

=−

, a

=−

4 · 3

, a

=−

6 · 5

=−

, ....

These results suggest that in general, if n = 2k, then

= a

(−1)

(2k)!

, k = 1, 2, 3, .... (9)

We can prove Eq. (9) by mathematical induction. First, observe that it is true for k = 1. Next,

assume that it is true for an arbitrary value of k and consider the case k + 1. We have

2k+2

=−

(2k + 2)(2k + 1)

=−

(−1)

(2k + 2)(2k + 1)(2k)!

(−1)

k+1

(2k + 2)!

Hence Eq. (9) is also true for k + 1, and consequently it is true for all positive integers k.

Similarly, for the odd-numbered coefﬁcients

=−

2 · 3

=−

, a

=−

5 · 4

, a

=−

7 · 6

=−

, ...,

and in general, if n = 2 k + 1, then

= a

2k+1

(−1)

(2k + 1)!

, k = 1, 2, 3, .... (10)

Substituting these coefﬁcients into Eq. (5), we have

The result given in Eq. (10) and other similar formulas in this chapter can be proved by an induction

argument resembling the one just given for Eq. (9). We assume that the results are plausible and omit the

inductive argument hereafter.

August 7, 2012 21:04 c05 Sheet number 11 Page number 257 cyan black

5.2 Series Solutions Near an Ordinary Point, Part I 257

y = a

+ a

x −

−

+···+

(−1)

(2n)!

(−1)

(2n + 1)!

2n+1

+···

= a

1 −

+···+

(−1)

(2n)!

+···

+ a

x −

+···+

(−1)

(2n + 1)!

2n+1

+···

= a

∞

n=0

(−1)

(2n)!

+ a

∞

n=0

(−1)

(2n + 1)!

2n+1

. (11)

Now that we have formally obtained two series solutions of Eq. (4), we can test them for

convergence. Using the ratio test, we can show that each of the series in Eq. (11) converges

for all x, and this justiﬁes retroactively all the steps used in obtaining the solutions. Indeed,

we recognize that the ﬁrst series in Eq. (11) is exactly the Taylor series for cos x about x = 0

and that the second is the Taylor series for sin x about x = 0. Thus, as expected, we obtain the

solution y = a

cos x + a

sin x.

Notice that no conditions are imposed on a

and a

; hence they are arbitrary. From Eqs. (5)

and (6) we see that y and y

′

evaluated at x = 0 are a

and a

, respectively. Since the initial

conditions y(0) and y

′

(0) can be chosen arbitrarily, it follows that a

and a

should be arbitrary

until speciﬁc initial conditions are stated.

Figures 5.2.1 and 5.2.2 show how the partial sums of the series in Eq. (11) approximate

cos x and sin x. As the number of terms increases, the interval over which the approximation is

satisfactory becomes longer, and for each x in this interval the accuracy of the approximation

improves. However, you should always remember that a truncated power series provides only

a local approximation of the solution in a neighborhood of the initial point x = 0; it cannot

adequately represent the solution for large |x|.

246

–1

–2

8 10

y = cos x

n = 4 n = 8 n = 20n = 16n = 12

n = 2 n = 6 n = 10 n = 14 n = 18

FIGURE 5.2.1 Polynomial approximations to cos x. The value

of n is the degree of the approximating polynomial.

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258 Chapter 5. Series Solutions of Second Order Linear Equations

246

–1

–2

810

y = sin x

n = 5 n = 9 n = 13 n = 17 n = 21

n = 3 n = 7 n = 11 n = 15 n = 19

FIGURE 5.2.2 Polynomial approximations to sin x. The value

of n is the degree of the approximating polynomial.

In Example 1 we knew from the start that sin x and cos x form a fundamental set of

solutions of Eq. (4). However,if we had not known this and had simply solved Eq. (4)

using series methods, we would still have obtained the solution (11). In recognition

of the fact that the differential equation (4) often occurs in applications, we might

decide to give the two solutions of Eq. (11) special names, perhaps

C(x) =

∞

n=0

(−1)

(2n)!

, S(x) =

∞

n=0

(−1)

(2n + 1)!

2n+1

. (12)

Then we might ask what properties these functions have. For instance,can we be sure

that C(x) and S(x) form a fundamental set of solutions? It follows at once from the

series expansions that C(0) = 1 and S(0) = 0. By differentiating the series for C(x)

and S(x) term by term, we ﬁnd that

′

(x) = C(x), C

′

(x) =−S(x). (13)

Thus, at x = 0 we have S

′

(0) = 1 and C

′

(0) = 0. Consequently, the Wronskian of C

and S at x = 0 is

W(C, S)(0) =

= 1, (14)

so these functions do indeed form a fundamental set of solutions. By substituting −x

for x in each of Eqs. (12), we obtain C(−x) = C(x) and S(−x) =−S(x). Moreover,

by calculating with the inﬁnite series,

we can show that the functions C(x) and S(x )

have all the usual analytical and algebraic properties of the cosine and sine functions,

respectively.

Such an analysis is given in Section 24 of K. Knopp,Theory and Applications of Inﬁnite Series (NewYork:

Hafner, 1951).

August 7, 2012 21:04 c05 Sheet number 13 Page number 259 cyan black

5.2 Series Solutions Near an Ordinary Point, Part I 259

Although you probably ﬁrst saw the sine and cosine functions deﬁned in a more

elementary manner in terms of right triangles, it is interesting that these functions

can be deﬁned as solutions of a certain simple second order linear differential equa-

tion. To be precise, the function sin x can be deﬁned as the unique solution of the

initial value problem y

′′

+ y = 0, y(0) = 0, y

′

(0) = 1; similarly, cos x can be deﬁned

as the unique solution of the initial value problem y

′′

+ y = 0, y(0) = 1, y

′

(0) = 0.

Many other functions that are important in mathematical physics are also deﬁned

as solutions of certain initial value problems. For most of these functions there is no

simpler or more elementary way to approach them.

EXAMPLE

Find a series solution in powers of x of Airy’s

equation

′′

− xy = 0, −∞ < x < ∞. (15)

For this equation P(x) = 1,Q(x) = 0, and R(x) =−x;hence every point is an ordinary point.

We assume that

y =

∞

n=0

(16)

and that the series converges in some interval |x| <ρ. The series for y

′′

is given by Eq. (7); as

explained in the preceding example, we can rewrite it as

′′

∞

n=0

(n + 2)(n + 1)a

n+2

. (17)

Substituting the series (16) and (17) for y and y

′′

in Eq. (15), we obtain

∞

n=0

(n + 2)(n + 1)a

n+2

= x

∞

n=0

∞

n=0

n+1

. (18)

Next, we shift the index of summation in the series on the right side of Eq. (18) by replacing

n by n − 1 and starting the summation at 1 rather than zero. Thus we have

2 · 1a

∞

n=1

(n + 2)(n + 1)a

n+2

∞

n=1

n−1

Again, for this equation to be satisﬁed for all x in some interval, the coefﬁcients of like powers

of x must be equal; hence a

= 0, and we obtain the recurrence relation

(n + 2)(n + 1)a

n+2

= a

n−1

for n = 1, 2, 3, .... (19)

Sincea

n+2

isgiven interms of a

n−1

,the a’saredetermined insteps of three.Thusa

determines

, which in turn determines a

, ...; a

determines a

, which in turn determines a

, ...; and a

determines a

, which in turn determines a

, .... Since a

= 0, we immediately conclude that

= a

=···=0.

Sir George Biddell Airy (1801–1892), an English astronomer and mathematician, was director of the

Greenwich Observatory from 1835 to 1881. He studied the equation named for him in an 1838 paper on

optics. One reason why Airy’s equation is of interest is that for x negative the solutions are similar to

trigonometric functions, and for x positive they are similar to hyperbolic functions. Can you explain why

it is reasonable to expect such behavior?

August 7, 2012 21:04 c05 Sheet number 14 Page number 260 cyan black

260 Chapter 5. Series Solutions of Second Order Linear Equations

For the sequence a

, a

, ...we set n = 1, 4, 7, 10, ...in the recurrence relation:

2 · 3

, a

5 · 6

2 · 3 · 5 · 6

, a

8 · 9

2 · 3 · 5 · 6 · 8 · 9

, ....

These results suggest the general formula

2 · 3 · 5 · 6 ···(3n − 1)(3n)

, n ≥ 4.

For the sequence a

, a

, ..., we set n = 2, 5, 8, 11, ...in the recurrence relation:

3 · 4

, a

6 · 7

3 · 4 · 6 · 7

, a

9 · 10

3 · 4 · 6 · 7 · 9 · 10

, ....

In general, we have

3n+1

3 · 4 · 6 · 7 ···(3n)(3n + 1)

, n ≥ 4.

Thus the general solution of Airy’s equation is

y = a

1 +

2 · 3

2 · 3 · 5 · 6

+···+

2 · 3 ···(3n − 1)(3n)

+···

+ a

x +

3 · 4

3 · 4 · 6 · 7

+···+

3n+1

3 · 4 ···(3n)(3n + 1)

+···

. (20)

Having obtained these two series solutions, we can now investigate their convergence.

Because of the rapid growth of the denominators of the terms in the series (20), we might

expect these series to have a large radius of convergence. Indeed, it is easy to use the ratio test

to show that both of these series converge for all x; see Problem 20.

Assuming for the moment that the series do converge for all x, let y

and y

denote the

functions deﬁned by the expressions in the ﬁrst and second sets of brackets, respectively,

in Eq. (20). Then, by choosing ﬁrst a

= 1, a

= 0 and then a

= 0, a

= 1, it follows that y

and y

are individually solutions of Eq. (15). Notice that y

satisﬁes the initial conditions

(0) = 1, y

′

(0) = 0 and that y

satisﬁes the initial conditions y

(0) = 0, y

′

(0) = 1. Thus

W(y

, y

)(0) = 1 ̸= 0, and consequently y

and y

are a fundamental set of solutions. Hence

the general solution of Airy’s equation is

y = a

(x) + a

(x), −∞ < x < ∞.

In Figures 5.2.3 and 5.2.4, respectively, we show the graphs of the solutions y

and y

Airy’s equation, as well as graphs of several partial sums of the two series in Eq. (20). Again,

the partial sums provide local approximations to the solutions in a neighborhood of the origin.

Although the quality of the approximation improves as the number of terms increases, no

polynomial can adequately represent y

and y

for large |x|. A practical way to estimate the

interval in which a given partial sum is reasonably accurate is to compare the graphs of that

partial sum and the next one, obtained by including one more term. As soon as the graphs

begin to separate noticeably, we can be conﬁdent that the original partial sum is no longer

accurate. For example, in Figure 5.2.3 the graphs for n = 24 and n = 27 begin to separate

at about x =−9/2. Thus, beyond this point, the partial sum of degree 24 is worthless as an

approximation to the solution.

August 7, 2012 21:04 c05 Sheet number 15 Page number 261 cyan black

5.2 Series Solutions Near an Ordinary Point, Part I 261

–2

–2–4–6–8

n = 48 36 24 12

n = 45 33 21 9

39 27 15

42 30 18 6

y = y

(x)

n ≥ 6

n = 3

–10

FIGURE 5.2.3 Polynomial approximations to the solution y

(x) of Airy’s equation.

The value of n is the degree of the approximating polynomial.

–2

–2–6–8–10

n = 46

n ≥ 4

40 16 4

n = 49

31 19 7

25 13

–4

y = y

(x)

FIGURE 5.2.4 Polynomial approximations to the solution y

(x) of Airy’s equation.

The value of n is the degree of the approximating polynomial.

Observe that both y

and y

are monotone for x > 0 and oscillatory for x < 0. You can

also see from the ﬁgures that the oscillations are not uniform but, rather, decay in amplitude

and increase in frequency as the distance from the origin increases. In contrast to Example 1,

the solutions y

and y

of Airy’s equation are not elementary functions that you have already

encountered in calculus. However, because of their importance in some physical applications,

these functions have been extensively studied, and their properties are well known to applied

mathematicians and scientists.

EXAMPLE

Find a solution of Airy’s equation in powers of x − 1.

The point x = 1 is an ordinary point of Eq. (15), and thus we look for a solution of the form

y =

∞

n=0

(x − 1)

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262 Chapter 5. Series Solutions of Second Order Linear Equations

where we assume that the series converges in some interval |x − 1| <ρ. Then

′

∞

n=1

(x − 1)

n−1

∞

n=0

(n + 1)a

n+1

(x − 1)

and

′′

∞

n=2

n(n − 1)a

(x − 1)

n−2

∞

n=0

(n + 2)(n + 1)a

n+2

(x − 1)

Substituting for y and y

′′

in Eq. (12), we obtain

∞

n=0

(n + 2)(n + 1)a

n+2

(x − 1)

= x

∞

n=0

(x − 1)

. (21)

Now to equate the coefﬁcients of like powers of (x − 1), we must express x, the coefﬁcient of y

in Eq. (15), in powers of x − 1; that is, we write x = 1 + (x − 1). Note that this is precisely the

Taylor series for x about x = 1. Then Eq. (21) takes the form

∞

n=0

(n + 2)(n + 1)a

n+2

(x − 1)

=[1 + (x − 1)]

∞

n=0

(x − 1)

∞

n=0

(x − 1)

∞

n=0

(x − 1)

n+1

Shifting the index of summation in the second series on the right gives

∞

n=0

(n + 2)(n + 1)a

n+2

(x − 1)

∞

n=0

(x − 1)

∞

n=1

n−1

(x − 1)

Equating coefﬁcients of like powers of x − 1, we obtain

= a

(3 · 2)a

= a

+ a

(4 · 3)a

= a

+ a

(5 · 4)a

= a

+ a

The general recurrence relation is

(n + 2)(n + 1)a

n+2

= a

+ a

n−1

for n ≥ 1. (22)

Solving for the ﬁrst few coefﬁcients a

in terms of a

and a

, we ﬁnd that

, a

120

Hence

y = a

1 +

(x − 1)

+···

+ a

(x − 1) +

(x − 1)

120

+···

. (23)

In general, when the recurrence relation has more than two terms, as in Eq. (22), the deter-

mination of a formula for a

in terms a

and a

will be fairly complicated, if not impossible. In

this example such a formula is not readily apparent. Lacking such a formula,we cannot test the

two series in Eq. (23) for convergence by direct methods such as the ratio test. However, we

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5.2 Series Solutions Near an Ordinary Point, Part I 263

shall see in Section 5.3 that even without knowing the formula for a

, it is possible to establish

that the two series in Eq. (23) converge for all x. Further, they deﬁne functions y

and y

that

are a fundamental set of solutions of the Airy equation (15). Thus

y = a

(x) + a

(x)

is the general solution of Airy’s equation for −∞ < x < ∞.

It is worth emphasizing, as we saw in Example 3, that if we look for a solution

of Eq. (1) of the form y =

∞

n=0

(x − x

)

, then the coefﬁcients P(x), Q(x), and R(x)

in Eq. (1) must also be expressed in powers of x − x

. Alternatively, we can make

the change of variable x − x

= t, obtaining a new differential equation for y as a

function of t, and then look for solutions of this new equation of the form

∞

n=0

When we have ﬁnished the calculations, we replace t by x − x

(see Problem 19).

In Examples 2 and 3 we have found two sets of solutions of Airy’s equation. The

functions y

and y

deﬁned by the series in Eq. (20) are a fundamental set of solutions

of Eq.(15) forall x,and this is also true for the functions y

and y

deﬁned bythe series

in Eq. (23). According to the general theory of second order linear equations, each

of the ﬁrst two functions can be expressed as a linear combination of the latter two

functions, and vice versa—a result that is certainly not obvious from an examination

of the series alone.

Finally,we emphasize that it is not particularly important if,as in Example 3,we are

unable to determine the general coefﬁcient a

in terms of a

and a

. What is essential

is that we can determine as many coefﬁcients as we want. Thus we can ﬁnd as many

terms in the two series solutions as we want, even if we cannot determine the general

term. While the task of calculating several coefﬁcients in a power series solution is

not difﬁcult, it can be tedious. A symbolic manipulation package can be very helpful

here; some are able to ﬁnd a speciﬁed number of terms in a power series solution in

response to a single command. With a suitable graphics package we can also produce

plots such as those shown in the ﬁgures in this section.

PROBLEMS In each of Problems 1 through 14:

(a) Seek power series solutions of the given differential equation about the given point x

;

ﬁnd the recurrence relation.

(b) Find the ﬁrst four terms in each of two solutions y

and y

(unless the series terminates

sooner).

, y

)(x

), show that y

and y

form a fundamental set

of solutions.

(d) If possible, ﬁnd the general term in each solution.

1. y

′′

− y = 0, x

= 02.y

′′

− xy

′

− y = 0, x

= 0

3. y

′′

− xy

′

− y = 0, x

= 14.y

′′

+ k

y = 0, x

= 0, k a constant

5. (1 − x)y

′′

+ y = 0, x

= 06.(2 + x

′′

− xy

′

+ 4y = 0, x

= 0

7. y

′′

+ xy

′

+ 2y = 0, x

= 08.xy

′′

+ y

′

+ xy = 0, x

= 1

9. (1 + x

′′

− 4xy

′

+ 6y = 0, x

= 0 10. (4 − x

′′

+ 2y = 0, x

= 0

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264 Chapter 5. Series Solutions of Second Order Linear Equations

11. (3 − x

′′

− 3xy

′

− y = 0, x

= 0

12. (1 − x)y

′′

+ xy

′

− y = 0, x

= 0

13. 2y

′′

+ xy

′

+ 3y = 0, x

= 0

14. 2y

′′

+ (x + 1)y

′

+ 3y = 0, x

= 2

In each of Problems 15 through 18:

(a) Find the ﬁrst ﬁve nonzero terms in the solution of the given initial value problem.

(b) Plot the four-term and the ﬁve-term approximations to the solution on the same axes.

reasonably accurate.

15.

′′

− xy

′

− y = 0, y(0) = 2, y

′

(0) = 1; see Problem 2

16.

(2 + x

′′

− xy

′

+ 4y = 0, y(0) =−1, y

′

(0) = 3; see Problem 6

17.

′′

+ xy

′

+ 2y = 0, y(0) = 4, y

′

(0) =−1; see Problem 7

18.

(1 − x)y

′′

+ xy

′

− y = 0, y(0) =−3, y

′

(0) = 2; see Problem 12

19. (a) By making the change of variable x − 1 = t and assuming that y has a Taylor series in

powers of t, ﬁnd two series solutions of

′′

+ (x − 1)

′

+ (x

− 1)y = 0

in powers of x − 1.

(b) Show that you obtain the same result by assuming that y has a Taylor series in powers

of x − 1 and also expressing the coefﬁcient x

− 1 in powers of x − 1.

20. Show directly, using the ratio test, that the two series solutions of Airy’s equation about

x = 0 converge for all x; see Eq. (20) of the text.

21. The Hermite Equation. The equation

′′

− 2xy

′

+ λy = 0, −∞ < x < ∞,

where λ is a constant, is known as the Hermite

equation. It is an important equation in

mathematical physics.

(a) Find the ﬁrst four terms in each of two solutions about x = 0 and show that they form

a fundamental set of solutions.

(b) Observe that if λ is a nonnegative even integer, then one or the other of the series

solutions terminates and becomes a polynomial. Find the polynomial solutions for λ = 0,

2, 4, 6, 8, and 10. Note that each polynomial is determined only up to a multiplicative

constant.

(x) is deﬁned as the polynomial solution of the Hermite

equation with λ = 2 n for which the coefﬁcient of x

is 2

. Find H

(x), ..., H

(x).

22. Consider the initial value problem y

′

1 − y

, y(0) = 0.

(a) Show that y = sin x is the solution of this initial value problem.

(b) Look for a solution of the initial value problem in the form of a power series about

x = 0. Find the coefﬁcients up to the term in x

in this series.

Charles Hermite (1822–1901) was an inﬂuential French analyst and algebraist. An inspiring teacher, he

was professor at the École Polytechnique and the Sorbonne. He introduced the Hermite functions in 1864

and showed in 1873 that e is a transcendental number (that is, e is not a root of any polynomial equation

with rational coefﬁcients). His name is also associated with Hermitian matrices (see Section 7.3), some of

whose properties he discovered.

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5.3 Series Solutions Near an Ordinary Point, Part II 265

In each of Problems 23 through 28, plot several partial sums in a series solution of the given

initial value problem about x = 0,thereby obtaining graphs analogous to those in Figures 5.2.1

through 5.2.4.

23.

′′

− xy

′

− y = 0, y(0) = 1, y

′

(0) = 0; see Problem 2

24.

(2 + x

′′

− xy

′

+ 4y = 0, y(0) = 1, y

′

(0) = 0; see Problem 6

25.

′′

+ xy

′

+ 2y = 0, y(0) = 0, y

′

(0) = 1; see Problem 7

26.

(4 − x

′′

+ 2y = 0, y(0) = 0, y

′

(0) = 1; see Problem 10

27.

′′

+ x

y = 0, y(0) = 1, y

′

(0) = 0; see Problem 4

28.

(1 − x)y

′′

+ xy

′

− 2y = 0, y(0) = 0, y

′

(0) = 1

5.3 Series Solutions Near an Ordinary Point, Part II

In the preceding section we considered the problem of ﬁnding solutions of

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0, (1)

where P, Q, and R are polynomials, in the neighborhood of an ordinary point x

Assuming that Eq. (1) does have a solution y = φ(x) and that φ has a Taylor series

y = φ(x) =

∞

n=0

(x − x

)

(2)

that converges for |x − x

| <ρ,where ρ>0, we found that the a

can be determined

by directly substituting the series (2) for y in Eq. (1).

Let us now consider how we might justify the statement that if x

is an ordinary

point of Eq. (1), then there exist solutions of the form (2). We also consider the

question of the radius of convergence of such a series. In doing this, we are led to a

generalization of the deﬁnition of an ordinary point.

Suppose, then, that there is a solution of Eq. (1) of the form (2). By differentiating

Eq. (2) m times and setting x equal to x

, we obtain

m!a

= φ

(m)

Hence, to compute a

in the series (2), we must show that we can determine φ

(n)

)

for n = 0, 1,2, ...from the differential equation (1).

Suppose that y = φ(x) is a solution of Eq. (1) satisfying the initial conditions

y(x

) = y

′

) = y

′

.Then a

= y

and a

= y

′

. If we are solely interested in ﬁnding

a solution of Eq. (1) without specifying any initial conditions, then a

and a

remain

arbitrary. To determine φ

(n)

) and the corresponding a

for n = 2, 3,..., we turn to

Eq. (1). Since φ is a solution of Eq. (1), we have

P(x)φ

′′

(x) + Q(x )φ

′

(x) + R(x )φ(x) = 0.

For the interval about x

for whichP is nonzero,we can write this equation in the form

′′

(x) =−p(x)φ

′

(x) − q(x )φ(x), (3)

where p(x) = Q(x)/P(x) and q(x) = R(x)/P(x). Setting x equal to x

in Eq. (3) gives

′′

) =−p(x

)φ

′

) − q(x

)φ(x

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266 Chapter 5. Series Solutions of Second Order Linear Equations

Hence a

is given by

2!a

= φ

′′

) =−p(x

− q(x

. (4)

To determine a

, we differentiate Eq. (3) and then set x equal to x

, obtaining

3!a

= φ

′′′

) =−[pφ

′′

+ (p

′

+ q)φ

′

+ q

′

φ]

x=x

=−2!p(x

−[p

′

) + q(x

)]a

− q

′

. (5)

Substituting for a

from Eq. (4) gives a

in terms of a

and a

. Since P, Q, and R are

polynomials and P(x

) ̸= 0, all the derivatives of p and q exist at x

. Hence, we can

continue to differentiate Eq. (3) indeﬁnitely, determining after each differentiation

the successive coefﬁcients a

, a

, ...by setting x equal to x

Notice that the important property that we used in determining the a

was that

we could compute inﬁnitely many derivatives of the functions p and q. It might seem

reasonable to relax our assumption that the functions p and q are ratios of polyno-

mials and simply require that they be inﬁnitely differentiable in the neighborhood of

. Unfortunately, this condition is too weak to ensure that we can prove the conver-

gence of the resulting series expansion for y = φ(x). What is needed is to assume that

the functions p and q are analytic at x

;that is, they have Taylor series expansions that

converge to them in some interval about the point x

p(x) = p

+ p

(x − x

) +···+p

(x − x

)

+···=

∞

n=0

(x − x

)

, (6)

q(x) = q

+ q

(x − x

) +···+q

(x − x

)

+···=

∞

n=0

(x − x

)

. (7)

With this idea in mind, we can generalize the deﬁnitions of an ordinary point and a

singular point of Eq. (1) as follows:if the functions p = Q/P and q = R/P are analytic

at x

, then the point x

is said to be an ordinary point of the differential equation (1);

otherwise, it is a singular point.

Now let us turn to the question of the interval of convergence of the series solution.

One possibility is actually to compute the series solution for each problem and then

to apply one of the tests for convergence of an inﬁnite series to determine its radius

of convergence. Unfortunately, these tests require us to obtain an expression for the

general coefﬁcient a

as a function of n, and this task is often quite difﬁcult, if not

impossible; recall Example 3 in Section 5.2. However, the question can be answered

at once for a wide class of problems by the following theorem.

Theorem 5.3.1

If x

is an ordinary point of the differential equation (1)

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0,

that is, if p = Q/P and q = R/P are analytic at x

, then the general solution of

Eq. (1) is

y =

∞

n=0

(x − x

)

= a

(x) + a

(x), (8)

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5.3 Series Solutions Near an Ordinary Point, Part II 267

where a

and a

are arbitrary, and y

and y

are two power series solutions that are

analytic at x

. The solutions y

and y

form a fundamental set of solutions. Further,

the radius of convergence for each of the series solutions y

and y

is at least as

large as the minimum of the radii of convergence of the series for p and q.

To see that y

and y

are a fundamental set of solutions, note that they have the

form y

(x) = 1 + b

(x − x

)

+··· and y

(x) = (x − x

) + c

(x − x

)

+···, where

+ c

= a

. Hence y

satisﬁes the initial conditions y

) = 1, y

′

) = 0, and y

satisﬁes the initial conditions y

) = 0, y

′

) = 1. Thus W(y

, y

)(x

) = 1.

Also note that although calculating the coefﬁcients by successively differenti-

ating the differential equation is excellent in theory, it is usually not a practical

computational procedure. Rather,you should substitute the series (2) for y in the dif-

ferential equation (1) and determine the coefﬁcients so that the differential equation

is satisﬁed, as in the examples in the preceding section.

We will not prove this theorem, which in a slightly more general form was estab-

lished by Fuchs.

What is important for our purposes is that there is a series solution

of the form (2) and that the radius of convergence of the series solution cannot be

less than the smaller of the radii of convergence of the series for p and q; hence we

need only determine these.

This can be done in either of two ways. Again, one possibility is simply to compute

the power series for p and q and then to determine the radii of convergence by

using one of the convergence tests for inﬁnite series. However, there is an easier

way when P, Q, and R are polynomials. It is shown in the theory of functions of a

complex variable that the ratio of two polynomials,say, Q/P, has a convergent power

series expansion about a point x = x

if P(x

) ̸= 0. Further, if we assume that any

factors common to Q and P have been canceled, then the radius of convergence of

the power series for Q/P about the point x

is precisely the distance from x

to the

nearest zero of P. In determining this distance,we must remember that P(x) = 0 may

have complex roots, and these must also be considered.

EXAMPLE

What is the radius of convergence of the Taylor series for (1 + x

)

−1

about x = 0?

One way to proceed is to ﬁnd the Taylor series in question, namely,

1 + x

= 1 − x

+ x

− x

+···+(−1)

+···.

Then it can be veriﬁed by the ratio test that ρ = 1.Another approach is to note that the zeros of

1 + x

are x =±i. Since the distance in the complex plane from 0 to i or to −i is 1, the radius

of convergence of the power series about x = 0 is 1.

Lazarus Immanuel Fuchs (1833–1902), a German mathematician, was a student and later a professor at

the University of Berlin. He proved the result of Theorem 5.3.1 in 1866. His most important research was

on singular points of linear differential equations. He recognized the signiﬁcance of regular singular points

(Section 5.4), and equations whose only singularities, including the point at inﬁnity, are regular singular

points are known as Fuchsian equations.

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268 Chapter 5. Series Solutions of Second Order Linear Equations

EXAMPLE

What is the radius of convergence of the Taylor series for (x

− 2x + 2)

−1

about x = 0? about

x = 1?

First notice that

− 2x + 2 = 0

has solutions x = 1 ± i. The distance in the complex plane from x = 0 to either x = 1 + i or

x = 1 − i is

√

2; hence the radius of convergence of the Taylor series expansion

∞

n=0

about

x = 0 is

√

The distance in the complex plane from x = 1 to either x = 1 + i or x = 1 − i is 1; hence the

radius of convergence of the Taylor series expansion

∞

n=0

(x − 1)

about x = 1 is 1.

According to Theorem 5.3.1, the series solutions of the Airy equation in Examples

2 and 3 of the preceding section converge for all values of x and x − 1, respectively,

since in each problem P(x) = 1 and hence is never zero.

A series solution may converge for a wider range of x than indicated by Theorem

5.3.1, so the theorem actually gives only a lower bound on the radius of convergence

of the series solution. This is illustrated by the Legendre polynomial solution of the

Legendre equation given in the next example.

EXAMPLE

Determine a lower bound for the radius of convergence of series solutions about x = 0 for the

Legendre equation

(1 − x

′′

− 2xy

′

+ α(α + 1)y = 0,

where α is a constant.

Note that P(x) = 1 − x

, Q(x) =−2x, and R(x) = α(α + 1) are polynomials, and that the

zeros of P, namely, x =±1, are a distance 1 from x = 0. Hence a series solution of the form

∞

n=0

converges at least for |x| < 1,and possibly for larger values ofx. Indeed,it can be shown

that if α is a positive integer,one of the series solutions terminates after a ﬁnite number of terms

and hence converges not just for |x | < 1 but for all x. For example, if α = 1, the polynomial

solution is y = x. See Problems 22 through 29 at the end of this section for a further discussion

of the Legendre equation.

EXAMPLE

Determine a lower bound for the radius of convergence of series solutions of the differential

equation

(1 + x

′′

+ 2xy

′

+ 4x

y = 0 (9)

about the point x = 0; about the point x =−

Again P, Q, and R are polynomials, and P has zeros at x =±i. The distance in the complex

plane from 0 to ±i is 1, and from −

to ±i is

1 +

√

5/2. Hence in the ﬁrst case the

series

∞

n=0

converges at least for |x| < 1, and in the second case the series

∞

n=0

x +

converges at least for |x +

| <

√

5/2.

An interesting observation that we can make about Eq. (9) follows from Theorems 3.2.1

and 5.3.1. Suppose that initial conditions y(0) = y

and y

′

(0) = y

′

are given. Since 1 + x

̸= 0

for all x, we know from Theorem 3.2.1 that there exists a unique solution of the initial value

problem on −∞ < x < ∞. On the other hand,Theorem 5.3.1 only guarantees a series solution

of the form

∞

n=0

(with a

= y

, a

= y

′

) for −1 < x < 1. The unique solution on the interval

−∞ < x < ∞ may not have a power series about x = 0 that converges for all x.

August 7, 2012 21:04 c05 Sheet number 23 Page number 269 cyan black

5.3 Series Solutions Near an Ordinary Point, Part II 269

EXAMPLE

Can we determine a series solution about x = 0 for the differential equation

′′

+ (sin x)y

′

+ (1 + x

)y = 0,

and if so, what is the radius of convergence?

For this differential equation, p(x) = sin x and q(x) = 1 + x

. Recall from calculus that sin x

has a Taylor series expansion about x = 0 that converges for all x. Further, q also has a Taylor

series expansion about x = 0, namely, q(x) = 1 + x

, that converges for all x. Thus there is

a series solution of the form y =

∞

n=0

with a

and a

arbitrary, and the series converges

for all x.

PROBLEMS In each of Problems 1 through 4, determine φ

′′

), φ

′′′

), and φ

(4)

) for the given point x

if y = φ(x) is a solution of the given initial value problem.

1. y

′′

+ xy

′

+ y = 0; y(0) = 1, y

′

(0) = 0

2. y

′′

+ (sin x)y

′

+ (cos x)y = 0; y(0) = 0, y

′

(0) = 1

3. x

′′

+ (1 + x)y

′

+ 3(ln x)y = 0; y(1) = 2, y

′

(1) = 0

4. y

′′

+ x

′

+ (sin x)y = 0; y(0) = a

, y

′

(0) = a

In each of Problems 5 through 8, determine a lower bound for the radius of convergence of

series solutions about each given point x

for the given differential equation.

5. y

′′

+ 4y

′

+ 6xy = 0; x

= 0, x

= 4

6. (x

− 2x − 3)y

′′

+ xy

′

+ 4y = 0; x

= 4, x

=−4, x

= 0

7. (1 + x

′′

+ 4xy

′

+ y = 0; x

= 0, x

= 2

8. xy

′′

+ y = 0; x

= 1

9. Determine a lower bound for the radius of convergence of series solutions about the given

for each of the differential equations in Problems 1 through 14 of Section 5.2.

10. The Chebyshev Equation. The Chebyshev

differential equation is

(1 − x

′′

− xy

′

+ α

y = 0,

where α is a constant.

(a) Determine two solutions in powers of x for |x| < 1, and show that they form a

fundamental set of solutions.

(b) Show that if α is a nonnegative integer n, then there is a polynomial solution of

degree n. These polynomials, when properly normalized, are called the Chebyshev

polynomials. They are very useful in problems that require a polynomial approximation

to a function deﬁned on −1 ≤ x ≤ 1.

Pafnuty L. Chebyshev (1821–1894), the most inﬂuential nineteenth-century Russian mathematician,was

for 35 years professor at the University of St. Petersburg, which produced a long line of distinguished

mathematicians. His study of Chebyshev polynomials began in about 1854 as part of an investigation of

the approximation of functions by polynomials. Chebyshev is also known for his work in number theory

and probability.

August 7, 2012 21:04 c05 Sheet number 24 Page number 270 cyan black

270 Chapter 5. Series Solutions of Second Order Linear Equations

For each of the differential equations in Problems 11 through 14, ﬁnd the ﬁrst four nonzero

terms in each of two power series solutions about the origin. Show that they form a fundamental

set of solutions. What do you expect the radius of convergence to be for each solution?

11. y

′′

+ (sin x)y = 0 12. e

′′

+ xy = 0

13. (cos x)y

′′

+ xy

′

− 2y = 0 14. e

−x

′′

+ ln(1 + x)y

′

− xy = 0

15. Let x and x

be solutions of a differential equation P (x)y

′′

+ Q(x)y

′

+ R(x)y = 0. Can you

say whether the point x = 0 is an ordinary point or a singular point? Prove your answer.

First Order Equations. The series methods discussed in this section are directly applicable

to the ﬁrst order linear differential equation P(x)y

′

+ Q(x)y = 0 at a point x

, if the function

p = Q/P has a Taylor series expansion about that point. Such a point is called an ordinary

point, and further, the radius of convergence of the series y =

∞

n=0

(x − x

)

is at least as large

as the radius of convergence of the series for Q/P. In each of Problems 16 through 21,solve the

given differential equation by a series in powers of x and verify that a

is arbitrary in each case.

Problems 20 and 21 involve nonhomogeneous differential equations to which series methods

can be easily extended. Where possible, compare the series solution with the solution obtained

by using the methods of Chapter 2.

16. y

′

− y = 0 17. y

′

− xy = 0

18. y

′

= e

y, three terms only 19. (1 − x)y

′

= y

20. y

′

− y = x

21. y

′

+ xy = 1 + x

The Legendre Equation. Problems 22 through 29 deal with the Legendre

equation

(1 − x

′′

− 2xy

′

+ α(α + 1)y = 0.

As indicated in Example 3,thepoint x = 0 is an ordinary point of this equation,andthe distance

from the origin to the nearest zero of P (x) = 1 − x

is 1. Hence the radius of convergence of

series solutions about x = 0 is at least 1. Also notice that we need to consider only α>−1

because if α ≤−1, then the substitution α =−(1 + γ), where γ ≥ 0, leads to the Legendre

equation (1 − x

′′

− 2xy

′

+ γ(γ + 1)y = 0.

22. Show that two solutions of the Legendre equation for |x| < 1 are

(x) = 1 −

α(α + 1)

α(α − 2)(α + 1)(α + 3)

∞

m=3

(−1)

α ···(α − 2m + 2)(α + 1) ···(α + 2m − 1)

(2m)!

(x) = x −

(α − 1)(α + 2)

(α − 1)(α − 3)(α + 2)(α + 4)

∞

m=3

(−1)

(α − 1) ···(α − 2m + 1)(α + 2) ···(α + 2m)

(2m + 1)!

2m+1

Adrien-Marie Legendre (1752–1833) held various positions in the French Académie des Sciences from

1783 onward. His primary work was in the ﬁelds of elliptic functions and number theory. The Legendre

functions,solutions of Legendre’s equation,ﬁrst appeared in 1784 in his study of the attraction of spheroids.

August 7, 2012 21:04 c05 Sheet number 25 Page number 271 cyan black

5.3 Series Solutions Near an Ordinary Point, Part II 271

23. Show that if α is zero or a positive even integer 2n, the series solution y

reduces to a

polynomial of degree 2n containing only even powers of x. Find the polynomials corre-

sponding to α = 0, 2, and 4. Show that if α is a positive odd integer 2n + 1, the series

solution y

reduces to a polynomial of degree 2n + 1 containing only odd powers of x.

Find the polynomials corresponding to α = 1, 3, and 5.

24.

The Legendre polynomial P

(x) is deﬁned as the polynomial solution of the Legendre

equation with α = n that also satisﬁes the condition P

(1) = 1.

(a) Using the results of Problem 23, ﬁnd the Legendre polynomials P

(x), ..., P

(x).

(b) Plot the graphs of P

(x), ..., P

(x) for −1 ≤ x ≤ 1.

(x), ..., P

(x).

25. It can be shown that the general formula for P

(x) is

(x) =

⌊n/2⌋

k=0

(−1)

(2n − 2k)!

k!(n − k)!(n − 2k)!

n−2k

where ⌊n/2 ⌋ denotes the greatest integer less than or equal to n/2. By observing the form

of P

(x) for n even and n odd, show that P

(−1) = (−1)

26. TheLegendre polynomials play an important role in mathematical physics. For example,in

solving Laplace’s equation (the potential equation) in spherical coordinates,we encounter

the equation

F(ϕ)

dϕ

+ cot ϕ

dF(ϕ)

dϕ

+ n(n + 1)F(ϕ) = 0, 0 <ϕ<π,

where n is a positive integer. Show that the change of variable x = cos ϕ leads to the

Legendre equation with α = n for y = f (x) = F(arccos x).

27. Show that for n = 0, 1, 2, 3, the corresponding Legendre polynomial is given by

(x) =

− 1)

This formula, known as Rodrigues’s formula,

is true for all positive integers n.

28. Show that the Legendre equation can also be written as

[(1 − x

′

]

′

=−α(α + 1)y.

Then it follows that

[(1 − x

′

(x)]

′

=−n(n + 1)P

(x) and [(1 − x

′

(x)]

′

=−m(m + 1)P

(x).

By multiplying the ﬁrst equation by P

(x) and the second equation by P

(x), integrating

by parts, and then subtracting one equation from the other, show that

−1

(x)P

(x) dx = 0 if n ̸= m.

This property of the Legendre polynomials is known as the orthogonality property. If

m = n, it can be shown that the value of the preceding integral is 2/(2n + 1).

Benjamin Olinde Rodrigues (1795–1851) published this result as part of his doctoral thesis from the

University of Paris in 1815. He then became a banker and social reformer but retained an interest in

mathematics. Unfortunately, his later papers were not appreciated until the late twentieth century.

August 7, 2012 21:04 c05 Sheet number 26 Page number 272 cyan black

272 Chapter 5. Series Solutions of Second Order Linear Equations

29. Given a polynomial f of degree n, it is possible to express f as a linear combination of

, P

, ..., P

f (x) =

k=0

(x).

Using the result of Problem 28, show that

2k + 1

−1

f (x)P

(x) dx.

5.4 Euler Equations; Regular Singular Points

In this section we will begin to consider how to solve equations of the form

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0 (1)

in the neighborhood of a singular point x

. Recall that if the functions P, Q, and R

are polynomials having no factors common to all three of them, then the singular

points of Eq. (1) are the points for which P(x) = 0.

Euler Equations. A relatively simple differential equation that has a singular point is

the Euler equation

L[y]=x

′′

+ αxy

′

+ βy = 0, (2)

where α and β are real constants. In this case P(x) = x

,sox = 0 is the only singu-

lar point for Eq. (2); all other points are ordinary points. For convenience we ﬁrst

consider the interval x > 0; later we extend our results to the interval x < 0.

Observe that (x

)

′

= rx

r−1

and (x

)

′′

= r(r − 1)x

r−2

. Hence, if we assume that

Eq. (2) has a solution of the form

y = x

, (3)

then we obtain

L[x

]=x

)

′′

+ αx(x

)

′

+ βx

= x

r(r − 1)x

r−2

+ αx(rx

r−1

) + βx

= x

[r(r − 1) + αr + β]. (4)

If r is a root of the quadratic equation

F(r ) = r(r − 1) + αr + β = 0, (5)

then L[x

] is zero, and y = x

is a solution of Eq. (2). The roots of Eq. (5) are

, r

−(α − 1) ±

(α − 1)

− 4β

, (6)

and F(r) = (r − r

)(r − r

). As for second order linear equations with constant coef-

ﬁcients,we consider separately the cases in which the roots are real and different,real

This equation is sometimes called the Cauchy–Euler equation or the equidimensional equation. Euler

studied it in about 1740, but its solution was known to Johann Bernoulli before 1700.

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5.4 Euler Equations; Regular Singular Points 273

but equal, and complex conjugates. Indeed, the entire discussion of Euler equations

is similar to the treatment of second order linear equations with constant coefﬁcients

in Chapter 3, with e

replaced by x

Real, Distinct Roots. If F (r) = 0 has real roots r

and r

, with r

̸= r

, then y

(x) = x

and y

(x) = x

are solutions of Eq. (2). Since

W(x

, x

) = (r

− r

−1

is nonzero for r

̸= r

and x > 0, it follows that the general solution of Eq. (2) is

y = c

+ c

, x > 0. (7)

Note that if r is not a rational number, then x

is deﬁned by x

= e

r ln x

EXAMPLE

Solve

′′

+ 3xy

′

− y = 0, x > 0. (8)

Substituting y = x

in Eq. (8) gives

[2r(r − 1) + 3r − 1]=x

(2r

+ r − 1) = x

(2r − 1)(r + 1) = 0.

Hence r

and r

=−1, so the general solution of Eq. (8) is

y = c

1/2

+ c

−1

, x > 0. (9)

Equal Roots. If the roots r

and r

are equal, then we obtain only one solution

(x) = x

of the assumed form. A second solution can be obtained by the method

of reduction of order, but for the purpose of our future discussion we consider an

alternative method. Since r

= r

, it follows that F(r) = (r − r

)

. Thus in this case,

not only does F(r

) = 0 but also F

′

) = 0. This suggests differentiating Eq. (4) with

respect to r and then setting r equal to r

. By differentiating Eq. (4) with respect to

r, we obtain

∂

∂r

L[x

∂

∂r

F(r )]=

∂

∂r

(r − r

)

]

= (r − r

)

ln x + 2(r − r

. (10)

However,by interchanging differentiation with respect to x and with respect to r,we

also obtain

∂

∂r

L[x

]=L

∂

∂r

= L[x

ln x].

The right side of Eq. (10) is zero for r = r

; consequently, L[x

ln x]=0 also.

Therefore,

(x) = x

ln x, x > 0 (11)

is a second solution of Eq. (2). By evaluating the Wronskian of y

and y

, we ﬁnd that

W(x

, x

ln x) = x

−1

Hence x

and x

ln x are a fundamental set of solutions for x > 0, and the general

solution of Eq. (2) is

y = (c

+ c

ln x)x

, x > 0. (12)

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274 Chapter 5. Series Solutions of Second Order Linear Equations

EXAMPLE

Solve

′′

+ 5xy

′

+ 4y = 0, x > 0. (13)

Substituting y = x

in Eq. (13) gives

[r(r − 1) + 5r + 4]=x

+ 4r + 4) = 0.

Hence r

= r

=−2, and

y = x

−2

+ c

ln x), x > 0 (14)

is the general solution of Eq. (13).

Complex Roots. Finally, suppose that the roots r

and r

of Eq. (5) are complex con-

jugates, say, r

= λ + iµ and r

= λ − iµ, with µ ̸= 0. We must now explain what is

meant by x

when r is complex. Remembering that

= e

r ln x

(15)

when x > 0 and r is real, we can use this equation to deﬁne x

when r is complex.

Then, using Euler’s formula for e

iµ ln x

, we obtain

λ+iµ

= e

(λ+iµ) ln x

= e

λ ln x

iµ ln x

= x

iµ ln x

= x

[cos(µ ln x) + i sin(µ ln x)], x > 0. (16)

With this deﬁnition of x

for complex values of r, it can be veriﬁed that the usual

laws of algebra and the differential calculus hold, and hence x

and x

are indeed

solutions of Eq. (2). The general solution of Eq. (2) is

y = c

λ+iµ

+ c

λ−iµ

. (17)

The disadvantage of this expression is that the functions x

λ+iµ

and x

λ−iµ

are complex-

valued. Recall that we had a similar situation for the second order differential

equation with constant coefﬁcients when the roots of the characteristic equation

were complex. Just as we did then, we can use Theorem 3.2.6 to obtain real-valued

solutions of Eq. (2) by taking the real and imaginary parts of x

λ+iµ

, namely,

cos(µ ln x) and x

sin(µ ln x). (18)

A straightforward calculation shows that

W[x

cos(µ ln x), x

sin(µ ln x)]=µx

2λ−1

Hence these solutions form a fundamental set of solutions for x > 0, and the general

solution of Eq. (2) is

y = c

cos(µ ln x) + c

sin(µ ln x), x > 0. (19)

EXAMPLE

Solve

′′

+ xy

′

+ y = 0. (20)

Substituting y = x

in Eq. (20) gives

[r(r − 1) + r + 1]=x

+ 1) = 0.

Hence r =±i, and the general solution is

y = c

cos(ln x) + c

sin(ln x), x > 0. (21)

The factor x

does not appear explicitly in Eq. (21) because in this example λ = 0 and x

= 1.

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5.4 Euler Equations; Regular Singular Points 275

Now let us consider the qualitative behavior of the solutions of Eq. (2) near the

singular point x = 0. This depends entirely on the values of the exponents r

and r

First, if r is real and positive, then x

→ 0asx tends to zero through positive values.

On the other hand, if r is real and negative, then x

becomes unbounded. Finally, if

r = 0, then x

= 1. Figure 5.4.1 shows these possibilities for various values of r.Ifr is

complex,thena typicalsolution is x

cos(µ ln x).This function becomesunbounded or

approaches zero if λ is negative or positive, respectively, and also oscillates more and

more rapidly as x → 0. This behavior is shown in Figures 5.4.2 and 5.4.3 for selected

values of λ and µ.Ifλ = 0, the oscillation is of constant amplitude. Finally, if there

are repeated roots, then one solution is of the form x

ln x,which tends to zero if r > 0

and becomes unbounded if r ≤ 0. An example of each case is shown in Figure 5.4.4.

y = x

–1/2

y = x

1/2

y = x

–3/2

y = x

3/2

FIGURE 5.4.1 Solutions of an Euler equation; real roots.

y = x

–1/4

cos(5 ln x)

0.125 0.25 0.375 0.5

–2

FIGURE 5.4.2 Solution of anEuler equation;

complex roots with negative real part.

–1

0.5 1 1.5 2

y = x

1/2

cos(5 ln x)

FIGURE 5.4.3 Solution of an Eulerequation;

complex roots with positive real part.

y = x

–1

ln x

y = x ln x

–1

0.5 1 1.5 2

FIGURE 5.4.4 Typical second solutions of an Euler equation with equal roots.

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276 Chapter 5. Series Solutions of Second Order Linear Equations

The extension of the solutions of Eq. (2) into the interval x < 0 can be carried out

in a relatively straightforward manner. The difﬁculty lies in understanding what is

meant by x

when x is negative and r is not an integer; similarly, ln x has not been

deﬁned for x < 0. The solutions of the Euler equation that we have given for x > 0

can be shown to be valid for x < 0, but in general they are complex-valued. Thus in

Example 1 the solution x

1/2

is imaginary for x < 0.

It is always possible to obtain real-valued solutions of the Euler equation (2) in the

interval x < 0 by making the following change of variable. Let x =−ξ, where ξ>0,

and let y = u(ξ). Then we have

d ξ

=−

d ξ

−

d ξ

(

d ξ

. (22)

Thus, for x < 0, Eq. (2) takes the form

d ξ

+ αξ

d ξ

+ βu = 0, ξ>0. (23)

But except for names of the variables,this is exactly the same as Eq. (2);from Eqs. (7),

(12), and (19), we have

u(ξ) =

⎧

⎪

⎨

⎪

⎩

+ c

ln ξ)ξ

cos(µ ln ξ) + c

sin(µ ln ξ),

(24)

depending on whether the zeros of F(r) = r(r − 1) + αr + β are real and different,

real and equal, or complex conjugates. To obtain u in terms of x, we replace ξ by −x

in Eqs. (24).

We can combine the results for x > 0 and x < 0 by recalling that |x|=x when x > 0

and that |x|=−x when x < 0. Thus we need only replace x by |x|in Eqs. (7), (12), and

(19) to obtain real-valued solutions valid in any interval not containing the origin.

Hence the general solution of the Euler equation (2)

′′

+ αxy

′

+ βy = 0

in any interval not containing the origin is determined by the roots r

and r

of the

equation

F(r ) = r(r − 1) + αr + β = 0

as follows. If the roots are real and different, then

y = c

|x|

+ c

|x|

. (25)

If the roots are real and equal, then

y = (c

+ c

ln |x|)|x|

. (26)

If the roots are complex conjugates, then

y =|x|

[

cos(µ ln |x|) + c

sin(µ ln |x|)

]

, (27)

where r

, r

= λ ± iµ.

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5.4 Euler Equations; Regular Singular Points 277

The solutions of an Euler equation of the form

(x − x

)

′′

+ α(x − x

′

+ βy = 0 (28)

are similar. If welook for solutions of theform y = (x − x

)

,thenthe generalsolution

is given by Eq. (25), Eq. (26), or Eq. (27) with x replaced by x − x

. Alternatively,

we can reduce Eq. (28) to the form of Eq. (2) by making the change of independent

variable t = x − x

Regular Singular Points. We now return to a consideration of the general equation (1)

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0,

where x

is a singular point. This means that P(x

) = 0 and that at least one of Q and

R is not zero at x

Unfortunately, if we attempt to use the methods of the preceding two sections to

solve Eq. (1) in the neighborhood of a singular point x

, we ﬁnd that these methods

fail.This is because the solution of Eq. (1) is often not analytic at x

and consequently

cannot be represented by a Taylor series in powers of x − x

. Examples 1, 2, and 3

illustrate this fact; in each of these examples, the solution fails to have a power series

expansion about the singular point x = 0. Therefore, to have any chance of solving

Eq. (1) in the neighborhood of a singular point, we must use a more general type of

series expansion.

Since the singular points of a differential equation are usually few in number, we

might ask whether we can simply ignore them, especially since we already know how

to construct solutions about ordinary points. However, this is not feasible. The sin-

gular points determine the principal features of the solution to a much larger extent

than you might at ﬁrst suspect. In the neighborhood of a singular point the solution

often becomes large in magnitude or experiences rapid changes in magnitude. For

example,the solutions found in Examples 1, 2,and 3 are illustrations of this fact.Thus

the behavior of a physical system modeled by a differential equation frequently is

most interesting in the neighborhood of a singular point. Often geometric singular-

ities in a physical problem, such as corners or sharp edges, lead to singular points

in the corresponding differential equation. Thus, although at ﬁrst we might want to

avoid the few points where a differential equation is singular, it is precisely at these

points that it is necessary to study the solution most carefully.

As an alternative to analytical methods, we can consider the use of numerical

methods, which are discussed in Chapter 8. However, these methods are ill suited

for the study of solutions near a singular point. Thus, even if we adopt a numerical

approach,it is advantageous to combine it with the analytical methods of this chapter

in order to examine the behavior of solutions near singular points.

Without any additional information about the behavior of Q/P and R/P in the

neighborhood of the singular point, it is impossible to describe the behavior

of the solutions of Eq. (1) near x = x

. It may be that there are two distinct solu-

tions of Eq. (1) that remain bounded as x → x

(as in Example 3); or there may be

only one, with the other becoming unbounded as x → x

(as in Example 1); or they

may both become unbounded as x → x

(as in Example 2). If Eq. (1) has solutions

that become unbounded as x → x

, it is often important to determine how these

solutions behave as x → x

. For example, does y →∞in the same way as (x − x

)

−1

or |x − x

−1/2

, or in some other manner?

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278 Chapter 5. Series Solutions of Second Order Linear Equations

Our goal is to extend the method already developed for solving Eq. (1) near an

ordinary point so that it also applies to the neighborhood of a singular point x

To do this in a reasonably simple manner, it is necessary to restrict ourselves to cases

in which the singularities in the functions Q/P and R/P at x = x

are not too severe—

that is, to what we might call “weak singularities.”At this stage it is not clear exactly

what is an acceptable singularity. However, as we develop the method of solution,

you will see that the appropriate conditions (see also Section 5.6, Problem 21) to

distinguish “weak singularities” are

lim

x→x

(x − x

)

Q(x)

P(x)

is ﬁnite (29)

and

lim

x→x

(x − x

)

R(x)

P(x)

is ﬁnite. (30)

This means that the singularity in Q/P can be no worse than (x − x

)

−1

and the singu-

larity in R/P can be no worse than (x − x

)

−2

. Such a point is called a regular singular

point of Eq. (1). For equations with more general coefﬁcients than polynomials, x

is a regular singular point of Eq. (1) if it is a singular point and if both

(x − x

)

Q(x)

P(x)

and (x − x

)

R(x)

P(x)

(31)

have convergentTaylorseries about x

—that is,if the functions inEq. (31)are analytic

at x = x

. Equations (29) and (30) imply that this will be the case when P, Q, and R

are polynomials. Any singular point of Eq. (1) that is not a regular singular point is

called an irregular singular point of Eq. (1).

Observe that the conditions in Eqs. (29) and (30) are satisﬁed bythe Eulerequation

(28). Thus the singularity in an Euler equation is a regular singular point. Indeed, we

will see that all equations of the form (1) behave very much like Euler equations near

a regular singular point. That is, solutions near a regular singular point may include

powers of x with negative or nonintegral exponents, logarithms, or sines or cosines

of logarithmic arguments.

In the following sections we discuss how to solve Eq. (1) in the neighborhood of a

regular singular point. A discussion of the solutions of differential equations in the

neighborhood of irregular singular points is more complicated and may be found in

more advanced books.

EXAMPLE

Determine the singular points of the Legendre equation

(1 − x

′′

− 2xy

′

+ α(α + 1)y = 0 (32)

and determine whether they are regular or irregular.

The functions given in Eq. (31) may not be deﬁned at x

, in which case their values at x

are to be

assigned as their limits as x → x

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5.4 Euler Equations; Regular Singular Points 279

In this case P(x) = 1 − x

, so the singular points are x = 1 and x =−1. Observe that when

we divide Eq. (32) by 1 − x

,the coefﬁcients of y

′

and y are −2x/(1 − x

) and α(α + 1)/(1 − x

respectively. We consider the point x = 1 ﬁrst. Thus, from Eqs. (29) and (30), we calculate

lim

x→1

(x − 1)

−2x

1 − x

= lim

x→1

(x − 1)(−2x)

(1 − x)(1 + x)

= lim

x→1

1 + x

= 1

and

lim

x→1

(x − 1)

α(α + 1)

1 − x

= lim

x→1

(x − 1)

α(α + 1)

(1 − x)(1 + x)

= lim

x→1

(x − 1)(−α)(α + 1)

1 + x

= 0.

Since these limits are ﬁnite, the point x = 1 is a regular singular point. It can be shown in a

similar manner that x =−1 is also a regular singular point.

EXAMPLE

Determine the singular points of the differential equation

2x(x − 2)

′′

+ 3xy

′

+ (x − 2)y = 0

and classify them as regular or irregular.

Dividing the differential equation by 2x(x − 2)

, we have

′′

2(x − 2)

′

2x(x − 2)

y = 0,

so p(x) = Q(x)/P(x) = 3/2(x − 2)

and q(x) = R(x)/P(x) = 1/2x(x − 2). The singular points

are x = 0 and x = 2. Consider x = 0. We have

lim

x→0

xp(x) = lim

x→0

2(x − 2)

= 0,

and

lim

x→0

q(x) = lim

x→0

2x(x − 2)

= 0.

Since these limits are ﬁnite, x = 0 is a regular singular point. For x = 2 we have

lim

x→2

(x − 2)p(x) = lim

x→2

(x − 2)

2(x − 2)

= lim

x→2

2(x − 2)

so the limit does not exist; hence x = 2 is an irregular singular point.

EXAMPLE

Determine the singular points of

)

x −

′′

+ (cos x)y

′

+ (sin x)y = 0

and classify them as regular or irregular.

The only singular point is x = π/2. To study it, we consider the functions

)

x −

p(x) =

)

x −

Q(x)

P(x)

cos x

x − π/2

and

)

x −

q(x) =

)

x −

R(x)

P(x)

= sin x.

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280 Chapter 5. Series Solutions of Second Order Linear Equations

Starting from the Taylor series for cos x about x = π/2, we ﬁnd that

cos x

x − π/2

=−1 +

(x − π/2)

−

(x − π/2)

+···,

which converges for all x. Similarly, sin x is analytic at x = π/2. Therefore, we conclude that

π/2 is a regular singular point for this equation.

PROBLEMS In each of Problems 1 through 12, determine the general solution of the given differential

equation that is valid in any interval not including the singular point.

1. x

′′

+ 4xy

′

+ 2y = 0 2. (x + 1)

′′

+ 3(x + 1)y

′

+ 0.75y = 0

3. x

′′

− 3xy

′

+ 4y = 0 4. x

′′

+ 3xy

′

+ 5y = 0

5. x

′′

− xy

′

+ y = 0 6. (x − 1)

′′

+ 8(x − 1)y

′

+ 12y = 0

7. x

′′

+ 6xy

′

− y = 0 8. 2x

′′

− 4xy

′

+ 6y = 0

9. x

′′

− 5xy

′

+ 9y = 0 10. (x − 2)

′′

+ 5(x − 2)y

′

+ 8y = 0

11. x

′′

+ 2xy

′

+ 4y = 0 12. x

′′

− 4xy

′

+ 4y = 0

In each of Problems 13 through 16, ﬁnd the solution of the given initial value problem. Plot

the graph of the solution and describe how the solution behaves as x → 0.

13. 2x

′′

+ xy

′

− 3y = 0, y(1) = 1, y

′

(1) = 4

14.

′′

+ 8xy

′

+ 17y = 0, y(1) = 2, y

′

(1) =−3

15.

′′

− 3xy

′

+ 4y = 0, y(−1) = 2, y

′

(−1) = 3

16.

′′

+ 3xy

′

+ 5y = 0, y(1) = 1, y

′

(1) =−1

In each of Problems 17 through 34,ﬁnd all singular points of the given equation and determine

whether each one is regular or irregular.

17. xy

′′

+ (1 − x)y

′

+ xy = 0 18. x

(1 − x)

′′

+ 2xy

′

+ 4y = 0

19. x

(1 − x)y

′′

+ (x − 2)y

′

− 3xy = 0 20. x

(1 − x

′′

+ (2/x)y

′

+ 4y = 0

21. (1 − x

)

′′

+ x(1 − x)y

′

+ (1 + x)y = 0

22. x

′′

+ xy

′

+ (x

− ν

)y = 0, Bessel equation

23. (x + 3)y

′′

− 2xy

′

+ (1 − x

)y = 0

24. x(1 − x

)

′′

+ (1 − x

)

′

+ 2(1 + x)y = 0

25. (x + 2)

(x − 1)y

′′

+ 3(x − 1)y

′

− 2(x + 2)y = 0

26. x(3 − x)y

′′

+ (x + 1)y

′

− 2y = 0

27. (x

+ x − 2)y

′′

+ (x + 1)y

′

+ 2y = 0 28. xy

′′

+ e

′

+ (3 cos x)y = 0

29. y

′′

+ (ln |x|)y

′

+ 3xy = 0 30. x

′′

+ 2(e

− 1)y

′

+ (e

−x

cos x)y = 0

31. x

′′

− 3(sin x)y

′

+ (1 + x

)y = 0 32. xy

′′

+ y

′

+ (cot x)y = 0

33. (sin x)y

′′

+ xy

′

+ 4y = 0 34. (x sin x)y

′′

+ 3y

′

+ xy = 0

35. Find all values of α for which all solutions of x

′′

+ αxy

′

+ (5/2)y = 0 approach zero as

x → 0.

36. Find all values of β for which all solutions of x

′′

+ βy = 0 approach zero as x → 0.

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5.4 Euler Equations; Regular Singular Points 281

37. Find γ so that the solution of the initial value problem x

′′

− 2y = 0, y(1) = 1, y

′

(1) = γ

is bounded as x → 0.

38. Find all values of α for which all solutions of x

′′

+ αxy

′

+ (5/2)y = 0 approach zero as

x →∞.

39. Consider the Euler equation x

′′

+ αxy

′

+ βy = 0. Find conditions on α and β so that:

(a) All solutions approach zero as x → 0.

(b) All solutions are bounded as x → 0.

(d) All solutions are bounded as x →∞.

(e) All solutions are bounded both as x → 0 and as x →∞.

40. Using the method of reduction of order, show that if r

is a repeated root of

r(r − 1) + αr + β = 0,

then x

and x

ln x are solutions of x

′′

+ αxy

′

+ βy = 0 for x > 0.

In each of Problems 41 and 42, show that the point x = 0 is a regular singular point. In each

problem try to ﬁnd solutions of the form

∞

n=0

. Show that (except for constant multiples)

there is only one nonzero solution of this form in Problem 41 and that there are no nonzero

solutions of this form in Problem 42. Thus in neither case can the general solution be found in

this manner. This is typical of equations with singular points.

41. 2xy

′′

+ 3y

′

+ xy = 0

42. 2x

′′

+ 3xy

′

− (1 + x)y = 0

43. Singularities at Inﬁnity. The deﬁnitions of an ordinary point and a regular singular point

given in the preceding sections apply only if the point x

is ﬁnite. In more advanced work

in differential equations, it is often necessary to consider the point at inﬁnity. This is done

by making the change of variable ξ = 1/x and studying the resulting equation at ξ = 0.

Show that, for the differential equation

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0,

the point at inﬁnity is an ordinary point if

P(1/ξ)

2P(1/ξ)

−

Q(1/ξ)

and

R(1/ξ)

P(1/ξ)

have Taylor series expansions about ξ = 0. Show also that the point at inﬁnity is a regular

singular point if at least one of the above functions does not have aTaylor series expansion,

but both

P(1/ξ)

2P(1/ξ)

−

Q(1/ξ)

and

R(1/ξ)

P(1/ξ)

do have such expansions.

In each of Problems 44 through 49, use the results of Problem 43 to determine whether the

point at inﬁnity is an ordinary point, a regular singular point, or an irregular singular point of

the given differential equation.

44. y

′′

+ y = 0

45. x

′′

+ xy

′

− 4y = 0

46. (1 − x

′′

− 2xy

′

+ α(α + 1)y = 0, Legendre equation

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282 Chapter 5. Series Solutions of Second Order Linear Equations

47. x

′′

+ xy

′

+ (x

− ν

)y = 0, Bessel equation

48. y

′′

− 2xy

′

+ λy = 0, Hermite equation

49. y

′′

− xy = 0, Airy equation

5.5 Series Solutions Near a Regular Singular Point, Part I

We now consider the question of solving the general second order linear equation

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0 (1)

in the neighborhood of a regular singular point x = x

. For convenience we assume

that x

= 0. If x

̸= 0, we can transform the equation into one for which the regular

singular point is at the origin by letting x − x

equal t.

The assumption that x = 0 is a regular singular point of Eq. (1) means that

xQ(x)/P(x) = xp(x ) and x

R(x)/P(x) = x

q(x) have ﬁnite limits as x → 0 and are

analytic at x = 0. Thus they have convergent power series expansions of the form

xp(x) =

∞

n=0

, x

q(x) =

∞

n=0

, (2)

on some interval |x| <ρabout the origin, where ρ>0. To make the quantities xp(x)

and x

q(x) appear in Eq. (1), it is convenient to divide Eq. (1) by P(x) and then to

multiply by x

, obtaining

′′

+ x[xp(x)]y

′

+[x

q(x)]y = 0, (3)

′′

+ x(p

+ p

x +···+p

+···)y

′

+ (q

+ q

x +···+q

+···)y = 0. (4)

If all of the coefﬁcients p

and q

are zero, except possibly

= lim

x→0

xQ(x)

P(x)

and q

= lim

x→0

R(x)

P(x)

, (5)

then Eq. (4) reduces to the Euler equation

′′

+ p

′

+ q

y = 0, (6)

which was discussed in the preceding section. In general, of course,some of the coef-

ﬁcients p

and q

, n ≥ 1, are not zero. However, the essential character of solutions

of Eq. (4) in the neighborhood of the singular point is identical to that of solu-

tions of the Euler equation (6). The presence of the terms p

x +···+p

+···and

x +···+q

+···merely complicates the calculations.

We restrict our discussion primarily to the interval x > 0. The interval x < 0 can

be treated, just as for the Euler equation, by making the change of variable x =−ξ

and then solving the resulting equation for ξ>0.

The coefﬁcients in Eq. (4) can be viewed as“Euler coefﬁcients”times power series.

To see this, you can write the coefﬁcient of y

′

in Eq. (4) as

x[1 + (p

)x + (p

+···+(p

+···],

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5.5 Series Solutions Near a Regular Singular Point, Part I 283

and similarly for the coefﬁcient of y. Thus it may seem natural to seek solutions of

Eq. (4) in the form of “Euler solutions” times power series. Hence we assume that

y = x

+ a

x +···+a

+···) = x

∞

n=0

∞

n=0

r+n

, (7)

where a

̸= 0. In other words, r is the exponent of the ﬁrst nonzero term in the series,

and a

is its coefﬁcient. As part of the solution, we have to determine:

1. The values of r for which Eq. (1) has a solution of the form (7).

2. The recurrence relation for the coefﬁcients a

3. The radius of convergence of the series

∞

n=0

The general theory was constructed by Frobenius

and is fairly complicated.

Rather than trying to present this theory, we simply assume, in this and the next

two sections, that there does exist a solution of the stated form. In particular, we

assume that any power series in an expression for a solution has a nonzero radius of

convergence and concentrate on showing how to determine the coefﬁcients in such

a series. To illustrate the method of Frobenius, we ﬁrst consider an example.

EXAMPLE

Solve the differential equation

′′

− xy

′

+ (1 + x)y = 0. (8)

It is easy to show that x = 0 is a regular singular point of Eq. (8). Further, xp(x) =−1/2

and x

q(x) = (1 + x)/2. Thus p

=−1/2, q

= 1/2, q

= 1/2, and all other p’s and q’s are zero.

Then, from Eq. (6), the Euler equation corresponding to Eq. (8) is

′′

− xy

′

+ y = 0. (9)

To solve Eq. (8), we assume that there is a solution of the form (7). Then y

′

and y

′′

are

given by

′

∞

n=0

(r + n)x

r+n−1

(10)

and

′′

∞

n=0

(r + n)(r + n − 1)x

r+n−2

. (11)

By substituting the expressions for y, y

′

, and y

′′

in Eq. (8), we obtain

′′

− xy

′

+ (1 + x)y =

∞

n=0

(r + n)(r + n − 1)x

r+n

−

∞

n=0

(r + n)x

r+n

∞

n=0

r+n

∞

n=0

r+n+1

. (12)

Ferdinand Georg Frobenius (1849–1917) grew up in the suburbs of Berlin,received his doctorate in 1870

from the University of Berlin, and returned as professor in 1892. For most of the intervening years he was

professor at the Eidgenössische Polytechnikum at Zürich. He showed how to construct series solutions

about regular singular points in 1874. His most distinguished work,however,was in algebra, where he was

one of the foremost early developers of group theory.

August 7, 2012 21:04 c05 Sheet number 38 Page number 284 cyan black

284 Chapter 5. Series Solutions of Second Order Linear Equations

The last term in Eq. (12) can be written as

∞

n=1

n−1

r+n

, so by combining the terms in Eq. (12),

we obtain

′′

− xy

′

+ (1 + x)y = a

[2r(r − 1) − r + 1]x

∞

n=1

[

2(r + n)(r + n − 1) − (r + n) + 1

]

+ a

n−1

r+n

= 0. (13)

If Eq. (13) is to be satisﬁed for all x, the coefﬁcient of each power of x in Eq. (13) must be zero.

From the coefﬁcient of x

we obtain, since a

̸= 0,

2r(r − 1) − r + 1 = 2r

− 3r + 1 = (r − 1)(2r − 1) = 0. (14)

Equation (14) is called the indicial equation for Eq. (8). Note that it is exactly the polynomial

equation we would obtain for the Euler equation (9) associated with Eq. (8). The roots of the

indicial equation are

= 1, r

= 1/2. (15)

These values of r are called the exponents at the singularity for the regular singular point

x = 0. They determine the qualitative behavior of the solution (7) in the neighborhood of the

singular point.

Now we return to Eq. (13) and set the coefﬁcient of x

r+n

equal to zero.This gives the relation

[

2(r + n)(r + n − 1) − (r + n) + 1

]

+ a

n−1

= 0, n ≥ 1, (16)

=−

n−1

2(r + n)

− 3(r + n) + 1

=−

n−1

[(r + n) − 1][2(r + n) − 1]

, n ≥ 1. (17)

For each root r

and r

of the indicial equation,we usethe recurrencerelation (17)to determine

a set of coefﬁcients a

, a

, ....For r = r

= 1, Eq. (17) becomes

=−

n−1

(2n + 1)n

, n ≥ 1.

Thus

=−

3 · 1

=−

5 · 2

(3 · 5)(1 · 2)

and

=−

7 · 3

=−

(3 · 5 · 7)(1 · 2 · 3)

In general, we have

(−1)

[3 · 5 · 7 ···(2n + 1)]n!

, n ≥ 4. (18)

If we multiply both the numerator and denominator of the right side of Eq. (18) by

2 · 4 · 6 ···2n = 2

n!, we can rewrite a

(−1)

(2n + 1)!

, n ≥ 1.

August 7, 2012 21:04 c05 Sheet number 39 Page number 285 cyan black

5.5 Series Solutions Near a Regular Singular Point, Part I 285

Hence, if we omit the constant multiplier a

, one solution of Eq. (8) is

(x) = x

1 +

∞

n=1

(−1)

(2n + 1)!

, x > 0. (19)

To determine the radius of convergence of the series in Eq. (19), we use the ratio test:

lim

n→∞

n+1

= lim

n→∞

2|x|

(2n + 2)(2n + 3)

= 0

for all x. Thus the series converges for all x.

Corresponding to the second root r = r

, we proceed similarly. From Eq. (17) we have

=−

n−1

n −

=−

n−1

n(2n − 1)

, n ≥ 1.

Hence

=−

1 · 1

=−

2 · 3

(1 · 2)(1 · 3)

=−

3 · 5

=−

(1 · 2 · 3)(1 · 3 · 5)

and, in general,

(−1)

n![1 · 3 · 5 ···(2n − 1)]

, n ≥ 4. (20)

Just as in the case of the ﬁrst root r

, we multiply the numerator and denominator by

2 · 4 · 6 ···2n = 2

n!. Then we have

(−1)

(2n)!

, n ≥ 1.

Again omitting the constant multiplier a

, we obtain the second solution

(x) = x

1/2

1 +

∞

n=1

(−1)

(2n)!

, x > 0. (21)

As before, we can show that the series in Eq. (21) converges for all x. Since y

and y

behave

like x and x

1/2

, respectively, near x = 0, they form a fundamental set of solutions. Hence the

general solution of Eq. (8) is

y = c

(x) + c

(x), x > 0.

The preceding example illustrates that if x = 0 is a regular singular point, then

sometimes there are two solutions of the form (7) in the neighborhood of this point.

Similarly,if there is a regular singular point at x = x

,then there may be two solutions

of the form

y = (x − x

)

∞

n=0

(x − x

)

(22)

that are valid near x = x

. However, just as an Euler equation may not have two

solutions of the form y = x

, so a more general equation with a regular singular point

may not have two solutions of the form (7) or (22). In particular,we show in the next

August 7, 2012 21:04 c05 Sheet number 40 Page number 286 cyan black

286 Chapter 5. Series Solutions of Second Order Linear Equations

section that if the roots r

and r

of the indicial equation are equal or differ by an

integer, then the second solution normally has a more complicated structure. In all

cases, though, it is possible to ﬁnd at least one solution of the form (7) or (22); if r

and r

differ by an integer, this solution corresponds to the larger value of r. If there

is only one such solution, then the second solution involves a logarithmic term, just

as for the Euler equation when the roots of the characteristic equation are equal.The

method of reduction of order or some other procedure can be invoked to determine

the second solution in such cases. This is discussed in Sections 5.6 and 5.7.

If the roots of the indicial equation are complex,then they cannot be equal or differ

by an integer,so thereare always twosolutions ofthe form (7)or (22).Of course,these

solutions are complex-valued functions of x. However, as for the Euler equation, it

is possible to obtain real-valued solutions by taking the real and imaginary parts of

the complex solutions.

Finally, we mention a practical point. If P, Q, and R are polynomials, it is often

much better to work directly with Eq. (1) than with Eq. (3). This avoids the necessity

of expressing xQ(x)/P(x) and x

R(x)/P(x) as power series. For example, it is more

convenient to consider the equation

x(1 + x )y

′′

+ 2y

′

+ xy = 0

than to write it in the form

′′

1 + x

′

1 + x

y = 0,

which would entail expanding 2x/(1 + x) and x

/(1 + x) in power series.

PROBLEMS In each of Problems 1 through 10:

(a) Show that the given differential equation has a regular singular point at x = 0.

(b) Determine the indicial equation, the recurrence relation, and the roots of the indicial

equation.

(d) If the roots are unequal and do not differ by an integer, ﬁnd the series solution

corresponding to the smaller root also.

1. 2xy

′′

+ y

′

+ xy = 02.x

′′

+ xy

′

−

y = 0

3. xy

′′

+ y = 04.xy

′′

+ y

′

− y = 0

5. 3x

′′

+ 2xy

′

+ x

y = 06.x

′′

+ xy

′

+ (x − 2)y = 0

7. xy

′′

+ (1 − x)y

′

− y = 08.2x

′′

+ 3xy

′

+ (2x

− 1)y = 0

9. x

′′

− x(x + 3)y

′

+ (x + 3)y = 0 10. x

′′

y = 0

11. The Legendre equation of order α is

(1 − x

′′

− 2xy

′

+ α(α + 1)y = 0.

The solution of this equation near the ordinary point x = 0 was discussed in Problems 22

and 23 of Section 5.3. In Example 4 of Section 5.4, it was shown that x =±1 are regular

singular points.

(a) Determine the indicial equation and its roots for the point x = 1.

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5.5 Series Solutions Near a Regular Singular Point, Part I 287

(b) Find a series solution in powers of x − 1 for x − 1 > 0.

Hint: Write 1 + x = 2 + (x − 1) and x = 1 + (x − 1). Alternatively, make the change of

variable x − 1 = t and determine a series solution in powers of t.

12. The Chebyshev equation is

(1 − x

′′

− xy

′

+ α

y = 0,

where α is a constant; see Problem 10 of Section 5.3.

(a) Show that x = 1 and x =−1 are regular singular points, and ﬁnd the exponents at

each of these singularities.

(b) Find two solutions about x = 1.

13. The Laguerre

differential equation is

′′

+ (1 − x)y

′

+ λy = 0.

(a) Show that x = 0 is a regular singular point.

(b) Determine the indicial equation, its roots, and the recurrence relation.

to a polynomial. When properly normalized, this polynomial is known as the Laguerre

polynomial, L

(x).

14. The Bessel equation of order zero is

′′

+ xy

′

+ x

y = 0.

(a) Show that x = 0 is a regular singular point.

(b) Show that the roots of the indicial equation are r

= r

= 0.

(x) = 1 +

∞

n=1

(−1)

(n!)

(d) Show that the series for J

(x) converges for all x. The function J

is known as the

Bessel function of the ﬁrst kind of order zero.

15. Referring to Problem 14, use the method of reduction of order to show that the second

solution of the Bessel equation of order zero contains a logarithmic term.

Hint: If y

(x) = J

(x)v(x), then

(x) = J

(x)

x[J

(x)]

Find the ﬁrst term in the series expansion of 1/x[J

(x)]

16. The Bessel equation of order one is

′′

+ xy

′

+ (x

− 1)y = 0.

(a) Show that x = 0 is a regular singular point.

(b) Show that the roots of the indicial equation are r

= 1 and r

=−1.

Edmond Nicolas Laguerre (1834–1886),a French geometer and analyst, studied the polynomials named

for him about 1879. He is also known for an algorithm for calculating roots of polynomial equations.

August 7, 2012 21:04 c05 Sheet number 42 Page number 288 cyan black

288 Chapter 5. Series Solutions of Second Order Linear Equations

(x) =

∞

n=0

(−1)

(n + 1)!n!2

(d) Show that the series for J

(x) converges for all x. The function J

is known as the

Bessel function of the ﬁrst kind of order one.

(e) Show that it is impossible to determine a second solution of the form

−1

∞

n=0

, x > 0.

5.6 Series Solutions Near a Regular Singular Point, Part II

Now let us consider the general problem of determining a solution of the equation

L[y]=x

′′

+ x[xp(x)]y

′

+[x

q(x)]y = 0, (1)

where

xp(x) =

∞

n=0

, x

q(x) =

∞

n=0

, (2)

and both series converge in an interval |x| <ρfor some ρ>0. The point x = 0isa

regular singular point, and the corresponding Euler equation is

′′

+ p

′

+ q

y = 0. (3)

We seek a solution of Eq. (1) for x > 0 and assume that it has the form

y = φ(r, x ) = x

∞

n=0

∞

n=0

r+n

, (4)

where a

̸= 0, and we have written y = φ(r, x) to emphasize that φ depends on r as

well as x. It follows that

′

∞

n=0

(r + n)a

r+n−1

, y

′′

∞

n=0

(r + n)(r + n − 1)a

r+n−2

. (5)

Then, substituting from Eqs. (2), (4), and (5) in Eq. (1) gives

r(r − 1)x

+ a

(r + 1)rx

r+1

+···+a

(r + n)(r + n − 1)x

r+n

+···

+ (p

+ p

x +···+p

+···)

×[a

+ a

(r + 1)x

r+1

+···+a

(r + n)x

r+n

+···]

+ (q

+ q

x +···+q

+···)

× (a

+ a

r+1

+···+a

r+n

+···) = 0.

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5.6 Series Solutions Near a Regular Singular Point, Part II 289

Multiplying the inﬁnite series together and then collecting terms, we obtain

F(r )x

+[a

F(r + 1) + a

r + q

)]x

r+1

F(r + 2) + a

r + q

) + a

(r + 1) + q

]

r+2

+···+

F(r + n) + a

r + q

) + a

n−1

(r + 1) + q

n−1

]

+···+a

n−1

(r + n − 1) + q

]

r+n

+···=0,

or, in a more compact form,

L[φ](r, x) = a

F(r )x

∞

n=1

F(r + n)a

n−1

k=0

[(r + k)p

n−k

+ q

n−k

]

r+n

= 0, (6)

where

F(r ) = r(r − 1) + p

r + q

. (7)

For Eq. (6) to be satisﬁed for all x > 0, the coefﬁcient of each power of x must be

zero.

Since a

̸= 0, the term involving x

yields the equation F(r) = 0. This equation is

called the indicial equation; note that it is exactly the equation we would obtain in

looking for solutions y = x

of the Euler equation (3). Let us denote the roots of

the indicial equation by r

and r

with r

≥ r

if the roots are real. If the roots are

complex, the designation of the roots is immaterial. Only for these values of r can

we expect to ﬁnd solutions of Eq. (1) of the form (4). The roots r

and r

are called

the exponents at the singularity; they determine the qualitative nature of the solution

in the neighborhood of the singular point.

Setting the coefﬁcient of x

r+n

in Eq. (6) equal to zero gives the recurrence relation

F(r + n)a

n−1

k=0

[(r + k)p

n−k

+ q

n−k

]=0, n ≥ 1. (8)

Equation (8) shows that, in general, a

depends on the value of r and all the pre-

ceding coefﬁcients a

, a

, ..., a

n−1

. It also shows that we can successively compute

, a

, ..., a

, ...in terms of a

and the coefﬁcients in the series for xp(x) and x

q(x),

provided that F(r + 1), F(r + 2), ..., F(r + n), ... are not zero. The only values of r

for which F(r) = 0 are r = r

and r = r

; since r

≥ r

, it follows that r

+ n is not

equal to r

or r

for n ≥ 1. Consequently, F(r

+ n) ̸= 0 for n ≥ 1. Hence we can

always determine one solution of Eq. (1) in the form (4), namely,

(x) = x

1 +

∞

n=1

, x > 0. (9)

Here we have introduced the notation a

) to indicate that a

has been determined

from Eq. (8) with r = r

. To specify the arbitrary constant in the solution, we have

taken a

to be 1.

August 7, 2012 21:04 c05 Sheet number 44 Page number 290 cyan black

290 Chapter 5. Series Solutions of Second Order Linear Equations

If r

is not equal to r

, and r

− r

is not a positive integer, then r

+ n is not equal

to r

for any value of n ≥ 1; hence F(r

+ n) ̸= 0, and we can also obtain a second

solution

(x) = x

1 +

∞

n=1

, x > 0. (10)

Just as for the series solutions about ordinary points discussed in Section 5.3, the

series in Eqs. (9) and (10) converge at least in the interval |x| <ρwhere the series for

both xp(x) and x

q(x) converge. Within their radii of convergence, the power series

1 +

∞

n=1

and 1 +

∞

n=1

deﬁne functions that are analytic at x = 0. Thus

the singular behavior, if there is any, of the solutions y

and y

is due to the factors

and x

that multiply these two analytic functions. Next, to obtain real-valued

solutions for x < 0, we can make the substitution x =−ξ with ξ>0. As we might

expect from our discussion of the Euler equation, it turns out that we need only

replace x

in Eq. (9) and x

in Eq. (10) by |x|

and |x|

, respectively. Finally, note

that if r

and r

are complex numbers, then they are necessarily complex conjugates

and r

̸= r

+ N for any positive integer N. Thus, in this case we can always ﬁnd two

series solutions of the form (4); however, they are complex-valued functions of x.

Real-valued solutions can be obtained by taking the real and imaginary parts of

the complex-valued solutions. The exceptional cases in which r

= r

or r

− r

= N,

where N is a positive integer,require more discussion and will be considered later in

this section.

It is important to realize that r

and r

,the exponents at the singular point,are easy

to ﬁnd and that they determine the qualitative behavior of the solutions.To calculate

and r

, it is only necessary to solve the quadratic indicial equation

r(r − 1) + p

r + q

= 0, (11)

whose coefﬁcients are given by

= lim

x→0

xp(x), q

= lim

x→0

q(x). (12)

Note that these are exactly the limits that must be evaluated in order to classify the

singularity as a regular singular point; thus they have usually been determined at an

earlier stage of the investigation.

Further, if x = 0 is a regular singular point of the equation

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0, (13)

where the functions P, Q, and R are polynomials, then xp(x) = xQ(x)/P(x) and

q(x) = x

R(x)/P(x). Thus

= lim

x→0

Q(x)

P(x)

, q

= lim

x→0

R(x)

P(x)

. (14)

Finally, the radii of convergence for the series in Eqs. (9) and (10) are at least equal

to the distance from the origin to the nearest zero of P other than x = 0 itself.

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5.6 Series Solutions Near a Regular Singular Point, Part II 291

EXAMPLE

Discuss the nature of the solutions of the equation

2x(1 + x)y

′′

+ (3 + x)y

′

− xy = 0

near the singular points.

This equation is of the form (13) with P(x) = 2x(1 + x), Q(x) = 3 + x, and R(x) =−x.The

points x = 0 and x =−1 are the only singular points. The point x = 0 is a regular singular

point, since

lim

x→0

Q(x)

P(x)

= lim

x→0

3 + x

2x(1 + x)

lim

x→0

R(x)

P(x)

= lim

x→0

−x

2x(1 + x)

= 0.

Further, from Eq. (14), p

and q

= 0. Thus the indicial equation is r(r −1) +

r = 0, and

the roots are r

= 0, r

=−

. Since these roots are not equal and do not differ by an integer,

there are two solutions of the form

(x) = 1 +

∞

n=1

(0)x

and y

(x) =|x|

−1/2

1 +

∞

n=1

−

for 0 < |x| <ρ. A lower bound for the radius of convergence of each series is 1, the distance

from x = 0tox =−1, the other zero of P(x). Note that the solution y

is bounded as x → 0,

indeed is analytic there, and that the second solution y

is unbounded as x → 0.

The point x =−1 is also a regular singular point, since

lim

x→−1

(x + 1)

Q(x)

P(x)

= lim

x→−1

(x + 1)(3 + x)

2x(1 + x)

=−1,

lim

x→−1

(x + 1)

R(x)

P(x)

= lim

x→−1

(x + 1)

(−x)

2x(1 + x)

= 0.

In this case p

=−1,q

= 0,so the indicial equation is r(r − 1) − r = 0. Theroots of the indicial

equation are r

= 2 and r

= 0. Corresponding to the larger root there is a solution of the form

(x) = (x + 1)

1 +

∞

n=1

(2)(x + 1)

The series converges at least for |x + 1| < 1, and y

is an analytic function there. Since the two

roots differ by a positive integer, there may or may not be a second solution of the form

(x) = 1 +

∞

n=1

(0)(x + 1)

We cannot say more without further analysis.

Observe that no complicated calculations were required to discover the information about

the solutions presented in this example. All that was needed was to evaluate a few limits and

solve two quadratic equations.

We now consider the cases in which the roots of the indicial equation are equal or

differ by a positive integer,r

− r

= N.As we have shown earlier,there is always one

solution of the form (9) corresponding to the larger root r

of the indicial equation.

By analogy with the Euler equation, we might expect that if r

= r

, then the second

solution contains a logarithmic term. This may also be true if the roots differ by an

integer.

August 7, 2012 21:04 c05 Sheet number 46 Page number 292 cyan black

292 Chapter 5. Series Solutions of Second Order Linear Equations

Equal Roots. The method of ﬁnding the second solution is essentially the same as the

one we used in ﬁnding the second solution of the Euler equation (see Section 5.4)

when the roots of the indicial equation were equal. We consider r to be a continuous

variable and determine a

as a function of r by solving the recurrence relation (8).

For this choice of a

(r) for n ≥ 1, the terms in Eq. (6) involving x

r+1

, x

r+2

, x

r+3

, ...all

have coefﬁcients equal to zero. Therefore, Eq. (6) reduces to

L[φ](r, x) = a

F(r )x

= a

(r − r

)

, (15)

since r

is a repeated root of F (r). Setting r = r

in Eq. (15), we ﬁnd that

L[φ](r

, x) = 0; hence, as we already know, y

(x) given by Eq. (9) is one solution

of Eq. (1). But more important, it also follows from Eq. (15), just as for the Euler

equation, that

∂φ

∂r

, x) = a

∂

∂r

(r − r

)

]

r=r

= a

[(r − r

)

ln x + 2(r − r

]

r=r

= 0. (16)

Hence, a second solution of Eq. (1) is

(x) =

∂φ(r, x)

∂r

r=r

∂

∂r

∞

n=1

(r)x

-C

r=r

= (x

ln x)

∞

n=1

+ x

∞

n=1

′

= y

(x) ln x + x

∞

n=1

′

, x > 0, (17)

where a

′

) denotes da

/dr evaluated at r = r

Although Eq. (17) provides an explicit expression for a second solution y

(x),it

may turn out that it is difﬁcult to determine a

(r) as a function of r from the recur-

rence relation (8) and then to differentiate the resulting expression with respect to r.

An alternative is simply to assume that y has the form of Eq. (17).That is, assume that

y = y

(x) ln x + x

∞

n=1

, x > 0, (18)

where y

(x) has already been found. The coefﬁcients b

are calculated, as usual, by

substituting into the differential equation, collecting terms, and setting the coefﬁ-

cient of each power of x equal to zero. A third possibility is to use the method of

reduction of order to ﬁnd y

(x) once y

(x) is known.

Roots r

and r

Differing by an Integer N. For this case the derivation of the second solu-

tion is considerably more complicated and will not be given here. The form of this

solution is stated in Eq. (24) in the following theorem. The coefﬁcients c

) in

Eq. (24) are given by

) =

[(r − r

(r)]

r=r

, n = 1, 2, ..., (19)

August 7, 2012 21:04 c05 Sheet number 47 Page number 293 cyan black

5.6 Series Solutions Near a Regular Singular Point, Part II 293

where a

(r) is determined from the recurrence relation (8) with a

= 1. Further, the

coefﬁcient a in Eq. (24) is

a = lim

r→r

(r − r

(r). (20)

If a

) is ﬁnite, then a = 0 and there is no logarithmic term in y

. A full derivation

of formulas (19) and (20) may be found in Coddington (Chapter 4).

In practice, the best way to determine whether a is zero in the second solution

is simply to try to compute the a

corresponding to the root r

and to see whether it is

possible to determine a

). If so, there is no further problem. If not, we must use

the form (24) with a ̸= 0.

When r

− r

= N, there are again three ways to ﬁnd a second solution. First, we

can calculate a and c

) directly by substituting the expression (24) for y in Eq. (1).

Second, we can calculate c

) and a of Eq. (24) using the formulas (19) and (20).

If this is the planned procedure, then in calculating the solution corresponding to

r = r

, be sure to obtain the general formula for a

(r) rather than just a

).The

third alternative is to use the method of reduction of order.

The followingtheorem summarizes the resultsthat we have obtainedin this section.

Theorem 5.6.1

Consider the differential equation (1)

′′

+ x[xp(x)]y

′

+[x

q(x)]y = 0,

where x = 0 is a regular singular point. Then xp(x) and x

q(x) are analytic at x = 0

with convergent power series expansions

xp(x) =

∞

n=0

, x

q(x) =

∞

n=0

for |x| <ρ, where ρ>0 is the minimum of the radii of convergence of the power

series for xp(x) and x

q(x). Let r

and r

be the roots of the indicial equation

F(r ) = r(r − 1) + p

r + q

= 0,

with r

≥ r

if r

and r

are real.Then in either the interval −ρ<x < 0 or the interval

0 < x <ρ, there exists a solution of the form

(x) =|x|

1 +

∞

n=1

, (21)

where the a

) are given by the recurrence relation (8) with a

= 1 and r = r

If r

− r

is not zero or a positive integer, then in either the interval −ρ<x < 0

or the interval 0 < x <ρ, there exists a second solution of the form

(x) =|x|

1 +

∞

n=1

. (22)

August 7, 2012 21:04 c05 Sheet number 48 Page number 294 cyan black

294 Chapter 5. Series Solutions of Second Order Linear Equations

The a

) are also determined by the recurrence relation (8) with a

= 1 and

r = r

. The power series in Eqs. (21) and (22) converge at least for |x| <ρ.

If r

= r

, then the second solution is

(x) = y

(x) ln |x |+|x|

∞

n=1

. (23)

If r

− r

= N, a positive integer, then

(x) = ay

(x) ln |x |+|x|

1 +

∞

n=1

. (24)

The coefﬁcients a

), b

), and c

) and the constant a can be determined by

substituting the form of the series solutions for y in Eq. (1). The constant a may

turn out to be zero, in which case there is no logarithmic term in the solution (24).

Each of the series in Eqs. (23) and (24) converges at least for |x| <ρand deﬁnes a

function that is analytic in some neighborhood of x = 0.

In all three cases, the two solutions y

(x) and y

(x) form a fundamental set of

solutions of the given differential equation.

PROBLEMS In each of Problems 1 through 12:

(a) Find all the regular singular points of the given differential equation.

(b) Determine the indicial equation and the exponents at the singularity for each regular

singular point.

1. xy

′′

+ 2xy

′

+ 6e

y = 0 2. x

′′

− x(2 + x)y

′

+ (2 + x

)y = 0

3. x(x − 1)y

′′

+ 6x

′

+ 3y = 04.y

′′

+ 4xy

′

+ 6y = 0

5. x

′′

+ 3(sin x)y

′

− 2y = 06.2x(x + 2)y

′′

+ y

′

− xy = 0

7. x

′′

(x + sin x)y

′

+ y = 08.(x + 1)

′′

+ 3(x

− 1)y

′

+ 3y = 0

9. x

(1 − x)y

′′

− (1 + x)y

′

+ 2xy = 0 10. (x − 2)

(x + 2)y

′′

+ 2xy

′

+ 3(x − 2)y = 0

11. (4 − x

′′

+ 2xy

′

+ 3y = 0 12. x(x + 3)

′′

− 2(x + 3)y

′

− xy = 0

In each of Problems 13 through 17:

(a) Show that x = 0 is a regular singular point of the given differential equation.

(b) Find the exponents at the singular point x = 0.

about x = 0.

13. xy

′′

+ y

′

− y = 0

14. xy

′′

+ 2xy

′

+ 6e

y = 0; see Problem 1

15. x(x − 1)y

′′

+ 6x

′

+ 3y = 0; see Problem 3

16. xy

′′

+ y = 0

17. x

′′

+ (sin x)y

′

− (cos x)y = 0

18. (a) Show that

(ln x)y

′′

′

+ y = 0

has a regular singular point at x = 1.

August 7, 2012 21:04 c05 Sheet number 49 Page number 295 cyan black

5.6 Series Solutions Near a Regular Singular Point, Part II 295

(b) Determine the roots of the indicial equation at x = 1.

∞

n=0

(x − 1)

r+n

corresponding

to the larger root. Take x − 1 > 0.

(d) What would you expect the radius of convergence of the series to be?

19. In several problems in mathematical physics, it is necessary to study the differential

equation

x(1 − x)y

′′

+[γ − (1 + α + β)x]y

′

− αβy = 0, (i)

where α, β, and γ are constants. This equation is known as the hypergeometric equation.

(a) Show that x = 0 is a regular singular point and that the roots of the indicial equation

are 0 and 1 − γ.

(b) Show that x = 1 is a regular singular point and that the roots of the indicial equation

are 0 and γ − α − β.

one solution of Eq. (i) is

(x) = 1 +

αβ

γ · 1!

x +

α(α + 1)β(β + 1)

γ(γ + 1)2!

+···.

What would you expect the radius of convergence of this series to be?

(d) Assuming that 1 − γ is not an integer or zero, show that a second solution for

0 < x < 1 is

(x) = x

1−γ

1 +

(α − γ + 1)(β − γ + 1)

(2 − γ)1!

(α − γ + 1)(α − γ + 2)(β − γ + 1)(β − γ + 2)

(2 − γ)(3 − γ)2!

+···

(e) Show that the point at inﬁnity is a regular singular point and that the roots of the

indicial equation are α and β. See Problem 43 of Section 5.4.

20. Consider the differential equation

′′

+ αxy

′

+ βy = 0,

where α and β are real constants and α ̸= 0.

(a) Show that x = 0 is an irregular singular point.

(b) By attempting to determine a solution of the form

∞

n=0

r+n

, show that the indicial

equation for r is linear and that, consequently, there is only one formal solution of the

assumed form.

fore is an actual solution. For other values of β/α,show that the formal series solution has

a zero radius of convergence and so does not represent an actual solution in any interval.

21. Consider the differential equation

′′

′

y = 0, (i)

where α ̸= 0 and β ̸= 0 are real numbers, and s and t are positive integers that for the

moment are arbitrary.

August 7, 2012 21:04 c05 Sheet number 50 Page number 296 cyan black

296 Chapter 5. Series Solutions of Second Order Linear Equations

(a) Show that if s > 1 or t > 2, then the point x = 0 is an irregular singular point.

(b) Try to ﬁnd a solution of Eq. (i) of the form

y =

∞

n=0

r+n

, x > 0. (ii)

Show that if s = 2 and t = 2, then there is only one possible value of r for which there is a

formal solution of Eq. (i) of the form (ii).

(d) Show that the maximum values of s and t for which the indicial equation is quadratic in

r [and hence we can hope to ﬁnd two solutions of the form (ii)] are s = 1 and t = 2. These

are precisely the conditions that distinguish a “weak singularity,” or a regular singular

point, from an irregular singular point, as we deﬁned them in Section 5.4.

As a note of caution, we point out that although it is sometimes possible to obtain a formal

series solution of the form (ii) at an irregular singular point, the series may not have a

positive radius of convergence. See Problem 20 for an example.

5.7 Bessel’s Equation

In this section we illustrate the discussion in Section 5.6 by considering three special

cases of Bessel’s

equation,

′′

+ xy

′

+ (x

− ν

)y = 0, (1)

where ν is a constant. It is easy to show that x = 0 is a regular singular point of Eq. (1).

We have

= lim

x→0

Q(x)

P(x)

= lim

x→0

= 1,

= lim

x→0

R(x)

P(x)

= lim

x→0

− ν

=−ν

Thus the indicial equation is

F(r ) = r(r − 1) + p

r + q

= r(r − 1) + r − ν

= r

− ν

= 0,

with the roots r =±ν. We will consider the three cases ν = 0, ν =

, and ν = 1 for

the interval x > 0.

Bessel Equation of Order Zero. In this case ν = 0, so Eq. (1) reduces to

L[y]=x

′′

+ xy

′

+ x

y = 0, (2)

Friedrich Wilhelm Bessel (1784–1846) left school at the age of 14 to embark on a career in the import-

export business but soon became interested in astronomy and mathematics. He was appointed director

of the observatory at Königsberg in 1810 and held this position until his death. His study of planetary

perturbations led him in 1824 to make the ﬁrst systematic analysis of the solutions, known as Bessel

functions, of Eq. (1). He is also famous for making, in 1838,the ﬁrst accurate determination of the distance

from the earth to a star.

August 7, 2012 21:04 c05 Sheet number 51 Page number 297 cyan black

5.7 Bessel’s Equation 297

and the roots of the indicial equation are equal: r

= r

= 0. Substituting

y = φ(r, x ) = a

∞

n=1

r+n

(3)

in Eq. (2), we obtain

L[φ](r, x) =

∞

n=0

[(r + n)(r + n − 1) + (r + n)]x

r+n

∞

n=0

r+n+2

= a

[r(r − 1) + r]x

+ a

[(r + 1)r + (r + 1)]x

r+1

∞

n=2

[(r + n)(r + n − 1) + (r + n)]+a

n−2

r+n

= 0. (4)

As we have already noted, the roots of the indicial equation F(r) = r(r − 1) + r = 0

are r

= 0 and r

= 0. The recurrence relation is

(r) =−

n−2

(r)

(r + n)(r + n − 1) + (r + n)

=−

n−2

(r)

(r + n)

, n ≥ 2. (5)

To determine y

(x), we set r equal to 0. Then, from Eq. (4), it follows that for

the coefﬁcient of x

r+1

to be zero we must choose a

= 0. Hence, from Eq. (5),

= a

=···=0. Further,

(0) =−a

n−2

(0)/n

, n = 2, 4, 6, 8, ...,

or, letting n = 2 m, we obtain

(0) =−a

2m−2

(0)/(2m)

, m = 1, 2, 3, ... .

Thus

(0) =−

, a

(0) =

, a

(0) =−

(3 · 2)

and, in general,

(0) =

(−1)

(m!)

, m = 1, 2, 3, .... (6)

Hence

(x) = a

1 +

∞

m=1

(−1)

(m!)

, x > 0. (7)

The function in brackets is known as the Bessel function of the ﬁrst kind of order

zero and is denoted by J

(x). It follows from Theorem 5.6.1 that the series converges

for all x and that J

is analytic at x = 0. Some of the important properties of J

are

discussed in the problems. Figure 5.7.1 shows the graphs of y = J

(x) and some of

the partial sums of the series (7).

August 7, 2012 21:04 c05 Sheet number 52 Page number 298 cyan black

298 Chapter 5. Series Solutions of Second Order Linear Equations

–1

4 6 8 10

n = 4

n = 2 n = 6 n = 10 n = 14 n = 18

n = 8 n = 12 n = 16 n = 20

y = J

(x)

FIGURE 5.7.1 Polynomial approximations to J

(x), the Bessel function of the ﬁrst kind

of order zero. The value of n is the degree of the approximating polynomial.

To determine y

(x), we will calculate

′

(0). First we note from the coefﬁcient of

r+1

in Eq. (4) that (r + 1)

(r) = 0. Thus a

(r) = 0 for all r near r = 0. So not

only does a

(0) = 0 but also a

′

(0) = 0. From the recurrence relation (5) it fol-

lows that a

′

(0) = a

′

(0) =···=a

′

2n+1

(0) =···=0; hence we need only compute

′

(0), m = 1, 2, 3,....From Eq. (5) we have

(r) =−a

2m−2

(r)/(r + 2m)

, m = 1, 2, 3, ....

By solving this recurrence relation, we obtain

(r) =−

(r + 2)

, a

(r) =

(r + 2)

(r + 4)

and, in general,

(r) =

(−1)

(r + 2)

···(r + 2m)

, m ≥ 3. (8)

The computation of a

′

(r) can be carried out most conveniently by noting that if

f (x) = (x − α

)

(x − α

)

(x − α

)

···(x − α

)

and if x is not equal to α

, α

, ..., α

, then

′

(x)

f (x)

x − α

+···+

x − α

Applying this result to a

(r) from Eq. (8), we ﬁnd that

′

(r)

=−2

r + 2

r + 4

+···+

r + 2m

(

Problem 10 outlines an alternative procedure,in which we simply substitute the form (23) of Section 5.6

in Eq. (2) and then determine the b

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5.7 Bessel’s Equation 299

and setting r equal to 0, we obtain

′

(0) =−2

+···+

(0).

Substituting for a

(0) from Eq. (6), and letting

= 1 +

+···+

, (9)

we obtain, ﬁnally,

′

(0) =−H

(−1)

(m!)

, m = 1, 2, 3, ....

The second solution of the Bessel equation of order zero is found by setting a

= 1

and substituting for y

(x) and a

′

(0) = b

(0) in Eq. (23) of Section 5.6. We obtain

(x) = J

(x) ln x +

∞

m=1

(−1)

m+1

(m!)

, x > 0. (10)

Instead of y

,the second solution is usually taken to be a certain linear combination

of J

and y

. It is known as the Bessel function of the second kind of order zero

and is denoted by Y

. Following Copson (Chapter 12), we deﬁne

(x) =

(x) + (γ − ln 2)J

(x)]. (11)

Here γ is a constant known as the Euler–Máscheroni

constant; it is deﬁned by the

equation

γ = lim

n→∞

− ln n)

∼

0.5772. (12)

Substituting for y

(x) in Eq. (11), we obtain

(x) =

)

γ + ln

(x) +

∞

m=1

(−1)

m+1

(m!)

, x > 0. (13)

The general solution of the Bessel equation of order zero for x > 0 is

y = c

(x) + c

(x).

Note that J

(x) → 1asx → 0 and that Y

(x) has a logarithmic singularity at x = 0;

that is, Y

(x) behaves as (2/π) ln x when x → 0 through positive values. Thus, if we

are interested in solutions of Bessel’s equation of order zero that are ﬁnite at the

origin, which is often the case,we must discard Y

. The graphs of the functions J

and

are shown in Figure 5.7.2.

Other authors use other deﬁnitions for Y

.The present choice for Y

is also known as theWeber function,

after Heinrich Weber (1842–1913), who taught at several German universities.

The Euler-Máscheroni constant ﬁrst appeared in 1734 in a paper by Euler. Lorenzo Máscheroni

(1750–1800) was an Italian priest and professor at the University of Pavia. He correctly calculated the ﬁrst

19 decimal places of γ in 1790.

August 7, 2012 21:04 c05 Sheet number 54 Page number 300 cyan black

300 Chapter 5. Series Solutions of Second Order Linear Equations

–0.5

2 4 6 8 10 12 14

0.5

y = Y

(x)

y = J

(x)

FIGURE 5.7.2 The Bessel functions of order zero: J

and Y

It is interesting to note from Figure 5.7.2 that for x large, both J

(x) and Y

(x) are

oscillatory. Such a behavior might be anticipated from the original equation; indeed

it is true for the solutions of the Bessel equation of order ν. If we divide Eq. (1) by

, we obtain

′′

′

1 −

(

y = 0.

For x very large, it is reasonable to conjecture that the terms (1/x)y

′

and (ν

)y are

small and hence can be neglected. If this is true, then the Bessel equation of order ν

can be approximated by

′′

+ y = 0.

The solutions of this equation are sin x and cos x; thus we might anticipate that the

solutions of Bessel’s equation for large x are similar to linear combinations of sin x

and cos x. This is correct insofar as the Bessel functions are oscillatory; however, it is

only partly correct. For x large the functions J

and Y

also decay as x increases; thus

the equation y

′′

+ y = 0 does not provide an adequate approximation to the Bessel

equation for large x, and a more delicate analysis is required. In fact, it is possible to

show that

(x)

∼

πx

(

1/2

cos

)

x −

as x →∞ (14)

and that

(x)

∼

πx

(

1/2

sin

)

x −

as x →∞. (15)

These asymptotic approximations, as x →∞, are actually very good. For example,

Figure 5.7.3 shows that the asymptotic approximation (14) to J

(x) is reasonably

accurate for all x ≥ 1. Thus to approximate J

(x) over the entire range from zero to

inﬁnity, you can use two or three terms of the series (7) for x ≤ 1 and the asymptotic

approximation (14) for x ≥ 1.

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5.7 Bessel’s Equation 301

y = J

(x)

–1

Asymptotic approximation: y = (2/ x)

1/2

cos(x – /4)

ππ

FIGURE 5.7.3 Asymptotic approximation to J

(x).

Bessel Equation of Order One-Half. This case illustrates the situation in which the roots

of the indicial equation differ by a positive integer but there is no logarithmic term

in the second solution. Setting ν =

in Eq. (1) gives

L[y]=x

′′

+ xy

′

−

y = 0. (16)

When we substitute the series (3) for y = φ(r, x), we obtain

L[φ](r, x) =

∞

n=0

(r + n)(r + n − 1) + (r + n) −

r+n

∞

n=0

r+n+2

−

(r + 1)

−

r+1

∞

n=2

(r + n)

−

+ a

n−2

r+n

= 0. (17)

The roots of the indicial equation are r

, r

=−

; hence the roots differ by an

integer. The recurrence relation is

(r + n)

−

=−a

n−2

, n ≥ 2. (18)

Corresponding to the larger root r

, we ﬁnd, from the coefﬁcient of x

r+1

Eq. (17), that a

= 0. Hence, from Eq. (18), a

= a

=···=a

2n+1

=···=0. Fur-

ther, for r =

=−

n−2

n(n + 1)

, n = 2, 4, 6 ...,

or, letting n = 2 m, we obtain

=−

2m−2

2m(2m + 1)

, m = 1, 2, 3, ....

By solving this recurrence relation, we ﬁnd that

=−

, a

, ...

August 7, 2012 21:04 c05 Sheet number 56 Page number 302 cyan black

302 Chapter 5. Series Solutions of Second Order Linear Equations

and, in general,

(−1)

(2m + 1)!

, m = 1, 2, 3, ....

Hence, taking a

= 1, we obtain

(x) = x

1/2

1 +

∞

m=1

(−1)

(2m + 1)!

= x

−1/2

∞

m=0

(−1)

2m+1

(2m + 1)!

, x > 0. (19)

The second power series in Eq. (19) is precisely the Taylor series for sin x; hence one

solution of the Bessel equation of order one-half is x

−1/2

sin x. The Bessel function

of the ﬁrst kind of order one-half, J

1/2

, is deﬁned as (2/π)

1/2

. Thus

1/2

(x) =

πx

(

1/2

sin x, x > 0. (20)

Corresponding to the root r

=−

, it is possible that we may have difﬁculty in

computing a

since N = r

− r

= 1. However, from Eq. (17) for r =−

, the coefﬁ-

cients of x

and x

r+1

are both zero regardless of the choice of a

and a

. Hence a

and a

can be chosen arbitrarily. From the recurrence relation (18), we obtain a set

of even-numbered coefﬁcients corresponding to a

and a set of odd-numbered coef-

ﬁcients corresponding to a

. Thus no logarithmic term is needed to obtain a second

solution in this case. It is left as an exercise to show that, for r =−

(−1)

(2n)!

, a

2n+1

(−1)

(2n + 1)!

, n = 1, 2, ....

Hence

(x) = x

−1/2

∞

n=0

(−1)

(2n)!

+ a

∞

n=0

(−1)

2n+1

(2n + 1)!

= a

cos x

1/2

+ a

sin x

1/2

, x > 0. (21)

The constant a

simply introduces a multiple of y

(x). The second solution of the

Bessel equation of order one-half is usually taken to be the solution for which

= (2/π)

1/2

and a

= 0. It is denoted by J

−1/2

. Then

−1/2

(x) =

πx

(

1/2

cos x, x > 0. (22)

The general solution of Eq. (16) is y = c

1/2

(x) + c

−1/2

(x).

By comparing Eqs. (20) and (22) with Eqs. (14) and (15), we see that, except for a

phase shift of π/4, the functions J

−1/2

and J

1/2

resemble J

and Y

, respectively, for

large x. The graphs of J

1/2

and J

−1/2

are shown in Figure 5.7.4.

Bessel Equation of Order One. This case illustrates the situation in which the roots of

the indicial equation differ by a positive integer and the second solution involves a

logarithmic term. Setting ν = 1 in Eq. (1) gives

L[y]=x

′′

+ xy

′

+ (x

− 1)y = 0. (23)

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5.7 Bessel’s Equation 303

–0.5

4142

0.5

6 12

–1/2

(x)

1/2

(x)

8 10

FIGURE 5.7.4 The Bessel functions J

1/2

and J

−1/2

If we substitute the series (3) for y = φ(r, x) and collect terms as in the preceding

cases, we obtain

L[φ](r, x) = a

− 1)x

+ a

[(r + 1)

− 1]x

r+1

∞

n=2

(r + n)

− 1

+ a

n−2

r+n

= 0. (24)

The roots of the indicial equation are r

= 1 and r

=−1. The recurrence relation is

[(r + n)

− 1]a

(r) =−a

n−2

(r), n ≥ 2. (25)

Corresponding to the larger root r = 1, the recurrence relation becomes

=−

n−2

(n + 2)n

, n = 2, 3, 4, ....

We also ﬁnd, from the coefﬁcient of x

r+1

in Eq. (24), that a

= 0; hence, from the

recurrence relation, a

= a

=···=0. For even values of n, let n = 2m; then

=−

2m−2

(2m + 2 )(2m)

=−

2m−2

(m + 1)m

, m = 1, 2, 3, ....

By solving this recurrence relation, we obtain

(−1)

(m + 1)!m!

, m = 1, 2, 3, .... (26)

The Bessel function of the ﬁrst kind of order one, denoted by J

, is obtained by

choosing a

= 1/2. Hence

(x) =

∞

m=0

(−1)

(m + 1)!m!

. (27)

The series converges absolutely for all x, so the function J

is analytic everywhere.

In determining a second solution of Bessel’s equation of order one, we illustrate

the method of direct substitution. The calculation of the general term in Eq. (28)

August 7, 2012 21:04 c05 Sheet number 58 Page number 304 cyan black

304 Chapter 5. Series Solutions of Second Order Linear Equations

below is rather complicated, but the ﬁrst few coefﬁcients can be found fairly easily.

According to Theorem 5.6.1, we assume that

(x) = aJ

(x) ln x + x

−1

1 +

∞

n=1

, x > 0. (28)

Computing y

′

(x), y

′′

(x), substituting in Eq. (23), and making use of the fact that J

a solution of Eq. (23), we obtain

2axJ

′

(x) +

∞

n=0

[

(n − 1)(n − 2)c

+ (n − 1)c

− c

]

n−1

∞

n=0

n+1

= 0, (29)

where c

= 1. Substituting for J

(x) from Eq. (27), shifting the indices of summation

in the two series, and carrying out several steps of algebra, we arrive at

−c

+[0 · c

+ c

]x +

∞

n=2

[(n

− 1)c

n+1

+ c

n−1

=−a

x +

∞

m=1

(−1)

(2m + 1)x

2m+1

(m + 1)!m!

. (30)

From Eq. (30) we observe ﬁrst that c

= 0, and a =−c

=−1. Further, since there

are only odd powers of x on the right, the coefﬁcient of each even power of x on the

left must be zero. Thus, since c

= 0, we have c

= c

=···=0. Corresponding to

the odd powers of x, we obtain the following recurrence relation [let n = 2m + 1

in the series on the left side of Eq. (30)]:

[(2m + 1)

− 1]c

2m+2

+ c

(−1)

(2m + 1)

(m + 1)!m!

, m = 1, 2, 3, ... . (31)

When we set m = 1 in Eq. (31), we obtain

− 1)c

+ c

= (−1)3/(2

· 2!).

Notice that c

can be selected arbitrarily, and then this equation determines c

. Also

notice that in the equation for the coefﬁcient of x, c

appeared multiplied by 0, and

that equation was used to determine a. That c

is arbitrary is not surprising, since c

is the coefﬁcient of x in the expression x

−1

1 +

∞

n=1

. Consequently, c

simply

generates a multiple of J

, and y

is determined only up to an additive multiple of J

In accordance with the usual practice, we choose c

= 1/2

. Then we obtain

−1

· 2

+ 1

−1

1 +

(

+ 1

(−1)

· 2!

+ H

It is possible to show that the solution of the recurrence relation (31) is

(−1)

m+1

+ H

m−1

)

m!(m − 1)!

, m = 1, 2, ...

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5.7 Bessel’s Equation 305

with the understanding that H

= 0. Thus

(x) =−J

(x) ln x +

1 −

∞

m=1

(−1)

+ H

m−1

)

m!(m − 1)!

, x > 0. (32)

The calculation of y

(x) using the alternative procedure [see Eqs. (19) and (20)

of Section 5.6] in which we determine the c

) is slightly easier. In particular, the

latter procedure yields the general formula for c

without the necessity of solving a

recurrence relation of the form (31) (see Problem 11). In this regard, you may also

wish to compare the calculations of the second solution of Bessel’s equation of order

zero in the text and in Problem 10.

The second solution of Eq. (23), the Bessel function of the second kind of order

one, Y

, is usually taken to be a certain linear combination of J

and y

. Following

Copson (Chapter 12), Y

is deﬁned as

(x) =

[−y

(x) + (γ − ln 2)J

(x)], (33)

where γ is deﬁned in Eq. (12). The general solution of Eq. (23) for x > 0 is

y = c

(x) + c

(x).

Notice that although J

is analytic at x = 0, the second solution Y

becomes

unbounded in the same manner as 1/x as x → 0. The graphs of J

and Y

are shown

in Figure 5.7.5.

–0.5

2 4 8 10 14

0.5

612

y = J

(x)

y = Y

(x)

FIGURE 5.7.5 The Bessel functions J

and Y

PROBLEMS

In each of Problems 1through 4,show that the givendifferential equationhas a regularsingular

point at x = 0, and determine two solutions for x > 0.

1. x

′′

+ 2xy

′

+ xy = 0 2. x

′′

+ 3xy

′

+ (1 + x)y = 0

3. x

′′

+ xy

′

+ 2xy = 0 4. x

′′

+ 4xy

′

+ (2 + x)y = 0

5. Find two solutions (not multiples of each other) of the Bessel equation of order

′′

+ xy

′

−

y = 0, x > 0.

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306 Chapter 5. Series Solutions of Second Order Linear Equations

6. Show that the Bessel equation of order one-half

′′

+ xy

′

−

y = 0, x > 0

can be reduced to the equation

′′

+ v = 0

by the change of dependent variable y = x

−1/2

v(x). From this, conclude that

(x) = x

−1/2

cos x and y

(x) = x

−1/2

sin x are solutions of the Bessel equation of order

one-half.

7. Show directly that the series for J

(x), Eq. (7), converges absolutely for all x.

8. Show directly that the series for J

(x), Eq. (27), converges absolutely for all x and that

′

(x) =−J

(x).

9. Consider the Bessel equation of order ν

′′

+ xy

′

+ (x

− ν

)y = 0, x > 0,

where ν is real and positive.

(a) Show that x = 0 is a regular singular point and that the roots of the indicial equation

are ν and −ν.

(b) Corresponding to the larger root ν, show that one solution is

(x) = x

1 −

1!(1 + ν)

)

2!(1 + ν)(2 + ν)

)

∞

m=3

(−1)

m!(1 + ν) ···(m + ν)

)

(x) = x

−ν

1 −

1!(1 − ν)

)

2!(1 − ν)(2 − ν)

)

∞

m=3

(−1)

m!(1 − ν) ···(m − ν)

)

Note that y

(x) → 0 as x → 0, and that y

(x) is unbounded as x → 0.

(d) Verify by direct methods that the power series in the expressions for y

(x) and y

(x)

converge absolutely for all x. Also verify that y

is a solution, provided only that ν is not

an integer.

10. In this section we showed that one solution of Bessel’s equation of order zero

L[y]=x

′′

+ xy

′

+ x

y = 0

is J

, where J

(x) is given by Eq. (7) with a

= 1. According to Theorem 5.6.1, a second

solution has the form (x > 0)

(x) = J

(x) ln x +

∞

n=1

(a) Show that

L[y

](x) =

∞

n=2

n(n − 1)b

∞

n=1

∞

n=1

n+2

+ 2xJ

′

(x). (i)

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5.7 Bessel’s Equation 307

(b) Substituting the series representation for J

(x) in Eq. (i), show that

x + 2

∞

n=3

+ b

n−2

=−2

∞

n=1

(−1)

2nx

(n!)

. (ii)

= b

=···=0, b

= 1/2

(1!)

, and that

(2n)

+ b

2n−2

=−2(−1)

(2n)/2

(n!)

, n = 2, 3, 4, ....

Deduce that

=−

1 +

(

and b

1 +

(

The general solution of the recurrence relation is b

= (−1)

n+1

(n!)

. Substituting

for b

in the expression for y

(x), we obtain the solution given in Eq. (10).

11. Find a second solution of Bessel’s equation of order one by computing the c

) and a

of Eq. (24) of Section 5.6 according to the formulas (19) and (20) of that section. Some

guidelines along the way of this calculation are the following. First, use Eq. (24) of this

section to show that a

(−1) and a

′

(−1) are 0. Then show that c

(−1) = 0 and, from the

recurrence relation, that c

(−1) = 0 for n = 3, 5, ....Finally, use Eq. (25) to show that

(r) =−

(r + 1)(r + 3)

, a

(r) =

(r + 1)(r + 3)(r + 3)(r + 5)

and that

(r) =

(−1)

(r + 1) ···(r + 2m − 1)(r + 3) ···(r + 2m + 1)

, m ≥ 3.

Then show that

(−1) = (−1)

m+1

+ H

m−1

)/2

m!(m − 1)!, m ≥ 1.

12. By a suitable change of variables it is sometimes possible to transform another differential

equation into a Bessel equation. For example, show that a solution of

′′

2β

− ν

y = 0, x > 0

is given by y = x

1/2

f (αx

), where f (ξ) is a solution of the Bessel equation of order ν.

13. Using the result of Problem 12, show that the general solution of the Airy equation

′′

− xy = 0, x > 0

is y = x

1/2

(

3/2

) + c

(

3/2

)], where f

(ξ) and f

(ξ) are a fundamental set of

solutions of the Bessel equation of order one-third.

14. It can be shown that J

has inﬁnitely many zeros for x > 0. In particular, the ﬁrst three

zeros are approximately 2.405, 5.520, and 8.653 (see Figure 5.7.1). Let λ

, j = 1, 2,3, ...,

denote the zeros of J

; it follows that

(λ

x) =

1, x = 0,

0, x = 1.

Verify that y = J

(λ

x) satisﬁes the differential equation

′′

′

+ λ

y = 0, x > 0.

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308 Chapter 5. Series Solutions of Second Order Linear Equations

Hence show that

(λ

x)J

(λ

x) dx = 0 if λ

̸= λ

This important property of J

(λ

x), which is known as the orthogonality property, is useful

in solving boundary value problems.

Hint: Write the differential equation for J

(λ

x). Multiply it by xJ

(λ

x) and subtract it

from xJ

(λ

x) times the differential equation for J

(λ

x). Then integrate from 0 to 1.

REFERENCES Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ:

Prentice-Hall, 1961; New York: Dover, 1989).

Coddington, E. A., and Carlson, R., Linear Ordinary Differential Equations (Philadelphia, PA: Society

for Industrial and Applied Mathematics, 1997).

Copson, E. T., An Introduction to the Theory of Functions of a Complex Variable (Oxford: Oxford

University Press, 1935).

Proofs of Theorems 5.3.1 and 5.6.1 can be found in intermediate or advanced books; for example, see

Chapters 3 and 4 of Coddington, Chapters 5 and 6 of Coddington and Carlson, or Chapters 3 and 4 of

Rainville, E. D., Intermediate Differential Equations (2nd ed.) (New York: Macmillan, 1964).

Also see these texts for a discussion of the point at inﬁnity, which was mentioned in Problem 43 of

Section 5.4. The behavior of solutions near an irregular singular point is an even more advanced topic; a

brief discussion can be found in Chapter 5 of

Coddington,E.A.,and Levinson,N.,Theory of Ordinary Differential Equations (New York:McGraw-Hill,

1955; Malabar, FL: Krieger, 1984).

Fuller discussions of the Bessel equation, the Legendre equation, and many of the other named equa-

tions can be found in advanced books on differential equations, methods of applied mathematics, and

special functions. A text dealing with special functions such as the Legendre polynomials and the Bessel

functions is

Hochstadt, H., Special Functions of Mathematical Physics (New York: Holt, 1961).

An excellent compilation of formulas, graphs, and tables of Bessel functions, Legendre functions, and

other special functions of mathematical physics may be found in

Abramowitz, M., and Stegun, I. A. (eds.), Handbook of Mathematical Functions with Formulas, Graphs,

and Mathematical Tables (New York: Dover, 1965); originally published by the National Bureau of

Standards, Washington, DC, 1964.

The digital successor to Abramowitz and Stegun is

Digital Library of Mathematical Functions. Release date 2011-08-29. National Institute of Standards and

Technology from http://dlmf.nist.gov/.

August 7, 2012 21:04 c06 Sheet number 1 Page number 309 cyan black

309

CHAPTER

The Laplace

Transform

Many practical engineering problems involve mechanical or electrical systems acted

on by discontinuous or impulsive forcing terms. For such problems the methods

described in Chapter 3 are often rather awkward to use.Another method that is espe-

cially well suited to these problems, although useful much more generally, is based

on the Laplace transform. In this chapter we describe how this important method

works, emphasizing problems typical of those that arise in engineering applications.

6.1 Deﬁnition of the Laplace Transform

Improper Integrals. Since the Laplace transform involves an integral from zero to inﬁn-

ity, a knowledge of improper integrals of this type is necessary to appreciate the

subsequent development of theproperties ofthe transform.Weprovide a brief review

of such improper integrals here. If you are already familiar with improper integrals,

you may wish to skip over this review. On the other hand, if improper integrals are

new to you, then you should probably consult a calculus book, where you will ﬁnd

many more details and examples.

An improper integral over an unbounded interval is deﬁned as a limit of integrals

over ﬁnite intervals; thus

∞

f (t) dt = lim

A→∞

f (t) dt, (1)

where A is a positive real number. If the integral from a to A exists for each A > a,

and if the limit as A →∞exists, then the improper integral is said to converge to

that limiting value. Otherwise the integral is said to diverge, or to fail to exist. The

following examples illustrate both possibilities.

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310 Chapter 6. The Laplace Transform

EXAMPLE

Let f (t) = e

, t ≥ 0, where c is a real nonzero constant. Then

∞

dt = lim

A→∞

dt = lim

A→∞

= lim

A→∞

− 1).

It follows that the improper integral converges to the value −1/c if c < 0 and diverges if c > 0.

If c = 0, the integrand f (t) is the constant function with value 1. In this case

lim

A→∞

1 dt = lim

A→∞

(A − 0) =∞,

so the integral again diverges.

EXAMPLE

Let f (t) = 1/t, t ≥ 1. Then

∞

= lim

A→∞

= lim

A→∞

ln A.

Since lim

A→∞

ln A =∞, the improper integral diverges.

EXAMPLE

Let f (t) = t

−p

, t ≥ 1, where p is a real constant and p ̸= 1; the case p = 1 was considered in

Example 2. Then

∞

−p

dt = lim

A→∞

−p

dt = lim

A→∞

1 − p

1−p

− 1).

As A →∞, A

1−p

→ 0ifp > 1, but A

1−p

→∞if p < 1. Hence

∞

−p

dt converges to the

value 1/(p − 1) for p > 1 but (incorporating the result of Example 2) diverges for p ≤ 1.

These results are analogous to those for the inﬁnite series

∞

n=1

−p

Before discussing the possible existence of

∞

f (t) dt, it is helpful to deﬁne

certain terms. A function f is said to be piecewise continuous on an inter-

val α ≤ t ≤ β if the interval

can be partitioned by a ﬁnite number of points

α = t

< t

< ···< t

= β so that

1. f is continuous on each open subinterval t

i−1

< t < t

2. f approaches a ﬁnite limit as the endpoints of each subinterval are approached from within

the subinterval.

In other words, f is piecewise continuous on α ≤ t ≤ β if it is continuous there except

for a ﬁnite number of jump discontinuities. If f is piecewise continuous on α ≤ t ≤ β

for every β>α, then f is said to be piecewise continuous on t ≥ α. An example of a

piecewise continuous function is shown in Figure 6.1.1.

It is not essential that the interval be closed; the same deﬁnition applies if the interval is open at one or

both ends.

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6.1 Deﬁnition of the Laplace Transform 311

FIGURE 6.1.1 A piecewise continuous function y = f (t).

The integral of a piecewise continuous function on a ﬁnite interval is just the sum

of the integrals on the subintervals created by the partition points. For instance, for

the function f (t) shown in Figure 6.1.1, we have

f (t) dt =

f (t) dt +

f (t) dt. (2)

For the function shown in Figure 6.1.1, we have assigned values to the function at

the endpoints α and β and at the partition points t

and t

. However, as far as the

integrals in Eq. (2) are concerned, it does not matter whether f (t) is deﬁned at these

points, or what values may be assigned to f (t) at them. The values of the integrals in

Eq. (2) remain the same regardless.

Thus, if f is piecewise continuous on the interval a ≤ t ≤ A, then

f (t) dt exists.

Hence, if f is piecewise continuous for t ≥ a, then

f (t) dt exists for each A > a.

However,piecewise continuity is not enough to ensure convergence of the improper

integral

∞

f (t) dt, as the preceding examples show.

If f cannot be integrated easily in terms of elementary functions, the deﬁnition of

convergence of

∞

f (t) dt may be difﬁcult to apply. Frequently, the most convenient

way to test the convergence or divergence of an improper integral is by the following

comparison theorem, which is analogous to a similar theorem for inﬁnite series.

Theorem 6.1.1

If f is piecewise continuous for t ≥ a, if |f (t)|≤g(t) when t ≥ M for some positive

constant M, and if

∞

g(t) dt converges, then

∞

f (t) dt also converges. On the

other hand, if f (t) ≥ g(t) ≥ 0 for t ≥ M, and if

∞

g(t) dt diverges, then

∞

f (t) dt

also diverges.

The proof of this result from calculus will not be given here. It is made plausible,

however, by comparing the areas represented by

∞

g(t) dt and

∞

|f (t)|dt.The

functions most useful for comparison purposes are e

and t

−p

, which we considered

in Examples 1, 2, and 3.

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312 Chapter 6. The Laplace Transform

The Laplace Transform. Among the tools that are very useful for solving linear differ-

ential equations are integral transforms. An integral transform is a relation of the

form

F(s) =

K(s, t)f (t) dt, (3)

where K(s, t) is a given function, called the kernel of the transformation, and the

limits of integration α and β are also given. It is possible that α =−∞,orβ =∞,

or both. The relation (3) transforms the function f into another function F, which is

called the transform of f .

There are several integral transforms that are useful in applied mathematics, but

in this chapter we consider only the Laplace

transform. This transform is deﬁned

in the following way. Let f (t) be given for t ≥ 0, and suppose that f satisﬁes certain

conditions to be stated a little later. Then the Laplace transform of f , which we will

denote by

L{f (t)} or by F(s), is deﬁned by the equation

L{f (t)}=F(s) =

∞

−st

f (t) dt, (4)

whenever this improper integral converges. The Laplace transform makes use of

the kernel K(s, t) = e

−st

. Since the solutions of linear differential equations with

constant coefﬁcients are based on the exponential function, the Laplace transform is

particularly useful forsuch equations.The general ideain usingthe Laplace transform

to solve a differential equation is as follows:

1. Use the relation (4) to transform an initial value problem for an unknown function f in

the t-domain into a simpler problem (indeed,an algebraic problem) for F in the s-domain.

2. Solve this algebraic problem to ﬁnd F.

3. Recover the desired function f from its transform F. This last step is known as “inverting

the transform.”

In general, the parameter s may be complex, and the full power of the Laplace

transform becomes available only when we regard F(s) as a function of a complex

variable. However, for the problems discussed here, it is sufﬁcient to consider only

real values of s. The Laplace transform F of a function f exists if f satisﬁes certain

conditions, such as those stated in the following theorem.

Theorem 6.1.2

Suppose that

1. f is piecewise continuous on the interval 0 ≤ t ≤ A for any positive A.

2. |f (t)|≤Ke

when t ≥ M. In this inequality, K, a, and M are real constants, K and M

necessarily positive.

Then the Laplace transform L{f (t)}=F(s), deﬁned by Eq. (4), exists for s > a.

The Laplace transform is named for the eminent Frenchmathematician P.S. Laplace,who studied the rela-

tion (3) in 1782. However, the techniques described in this chapter were not developed until a century or

more later. We owe them mainly to Oliver Heaviside (1850–1925), an innovative self-taught English elec-

trical engineer,who made signiﬁcant contributions to the development and application of electromagnetic

theory. He was also one of the developers of vector calculus.

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6.1 Deﬁnition of the Laplace Transform 313

To establish this theorem, we must show that the integral in Eq. (4) converges for

s > a. Splitting the improper integral into two parts, we have

∞

−st

f (t) dt =

−st

f (t) dt +

∞

−st

f (t) dt. (5)

The ﬁrst integral on the right side of Eq. (5) exists by hypothesis (1) of the theorem;

hence the existence of F(s) depends on the convergence of the second integral. By

hypothesis (2) we have, for t ≥ M,

−st

f (t)|≤Ke

−st

= Ke

(a−s)t

and thus, by Theorem 6.1.1, F(s) exists provided that

∞

(a−s)t

dt converges. Refer-

ring to Example 1 with c replaced by a − s, we see that this latter integral converges

when a − s < 0, which establishes Theorem 6.1.2.

In this chapter (except in Section 6.5), we deal almost exclusively with functions

that satisfy the conditions ofTheorem 6.1.2. Such functions are described aspiecewise

continuous and of exponential order as t →∞. Note that there are functions that

are not of exponential order as t →∞. One such function is f (t) = e

.Ast →∞,

this function increases faster than Ke

regardless of how large the constants K and

a may be.

The Laplace transforms of some important elementary functions are given in the

following examples.

EXAMPLE

Let f (t) = 1, t ≥ 0. Then, as in Example 1,

L{1}=

∞

−st

dt =−lim

A→∞

−st

, s > 0.

EXAMPLE

Let f (t) = e

, t ≥ 0. Then, again referring to Example 1,

L{e

∞

−st

dt =

∞

−(s−a)t

s − a

, s > a.

EXAMPLE

Let

f (t) =

⎧

⎪

⎨

⎪

⎩

1, 0 ≤ t < 1,

k, t = 1,

0, t > 1,

where k is a constant. In engineering contexts f (t) often represents a unit pulse, perhaps of

force or voltage.

Note that f is a piecewise continuous function. Then

L{f (t)}=

∞

−st

f (t) dt =

−st

dt =−

−st

1 − e

−s

, s > 0.

Observe that L{f (t)} does not depend on k, the function value at the point of discontinuity.

Even if f (t) is not deﬁned at this point, the Laplace transform of f remains the same. Thus

there are many functions, differing only in their value at a single point, that have the same

Laplace transform.

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314 Chapter 6. The Laplace Transform

EXAMPLE

Let f (t) = sin at, t ≥ 0. Then

L{sin at}=F(s) =

∞

−st

sin at dt, s > 0.

Since

F(s) = lim

A→∞

−st

sin at dt,

upon integrating by parts, we obtain

F(s) = lim

A→∞

⎡

⎣

−

−st

cos at

−

−st

cos at dt

⎤

⎦

−

∞

−st

cos at dt.

A second integration by parts then yields

F(s) =

−

∞

−st

sin at dt

−

F(s).

Hence, solving for F(s), we have

F(s) =

+ a

, s > 0.

Now let us suppose that f

and f

are two functions whose Laplace transforms exist

for s > a

and s > a

, respectively. Then,for s greater than the maximum of a

and a

L{c

(t) + c

(t)}=

∞

−st

(t) + c

(t)]dt

= c

∞

−st

(t) dt + c

∞

−st

(t) dt;

hence

L{c

(t) + c

(t)}=c

L{f

(t)}+c

L{f

(t)}. (6)

Equation (6) states that the Laplace transform is a linear operator, and we make

frequent use of this property later. The sum in Eq. (6) can be readily extended to an

arbitrary number of terms.

EXAMPLE

Find the Laplace transform of f (t) = 5e

−2t

− 3 sin 4t, t ≥ 0.

Using Eq. (6), we write

L{f (t)}=5L{e

−2t

}−3L{sin 4t}.

Then, from Examples 5 and 7, we obtain

L{f (t)}=

s + 2

−

+ 16

, s > 0.

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6.1 Deﬁnition of the Laplace Transform 315

PROBLEMS In each of Problems 1 through 4, sketch the graph of the given function. In each case deter-

mine whether f is continuous, piecewise continuous, or neither on the interval 0 ≤ t ≤ 3.

1. f (t) =

⎧

⎨

⎩

,0≤ t ≤ 1

2 + t,1< t ≤ 2

6 − t,2< t ≤ 3

2. f (t) =

⎧

⎨

⎩

,0≤ t ≤ 1

(t − 1)

−1

,1< t ≤ 2

1, 2 < t ≤ 3

3. f (t) =

⎧

⎨

⎩

,0≤ t ≤ 1

1, 1 < t ≤ 2

3 − t,2< t ≤ 3

4. f (t) =

⎧

⎨

⎩

t,0≤ t ≤ 1

3 − t,1< t ≤ 2

1, 2 < t ≤ 3

5. Find the Laplace transform of each of the following functions:

(a) f (t) = t

(b) f (t) = t

, where n is a positive integer

6. Find the Laplace transform of f (t) = cos at, where a is a real constant.

Recall that cosh bt = (e

+ e

−bt

)/2 and sinh bt = (e

− e

−bt

)/2. In each of Problems 7 through

10, ﬁnd the Laplace transform of the given function; a and b are real constants.

7. f (t) = cosh bt 8. f (t) = sinh bt

9. f (t) = e

cosh bt 10. f (t) = e

sinh bt

Recall that cos bt = (e

ibt

+ e

−ibt

)/2 and that sin bt = (e

ibt

− e

−ibt

)/2i. In each of Problems 11

through 14, ﬁnd the Laplace transform of the given function; a and b are real constants.

Assume that the necessary elementary integration formulas extend to this case.

11. f (t) = sin bt 12. f (t) = cos bt

13. f (t) = e

sin bt 14. f (t) = e

cos bt

In each of Problems 15 through 20, use integration by parts to ﬁnd the Laplace transform of

the given function; n is a positive integer and a is a real constant.

15. f (t) = te

16. f (t) = t sin at

17. f (t) = t cosh at 18. f (t) = t

19. f (t) = t

sin at 20. f (t) = t

sinh at

In each of Problems 21 through 24, ﬁnd the Laplace transform of the given function.

21. f (t) =

1, 0 ≤ t <π

0, π ≤ t < ∞

22. f (t) =

t,0≤ t < 1

0, 1 ≤ t < ∞

23. f (t) =

t,0≤ t < 1

1, 1 ≤ t < ∞

24. f (t) =

⎧

⎪

⎨

⎪

⎩

t,0≤ t < 1

2 − t,1≤ t < 2

0, 2 ≤ t < ∞

In each of Problems25 through 28,determine whether thegiven integralconverges or diverges.

25.

∞

+ 1)

−1

dt 26.

∞

−t

27.

∞

−2

dt 28.

∞

−t

cos tdt

August 7, 2012 21:04 c06 Sheet number 8 Page number 316 cyan black

316 Chapter 6. The Laplace Transform

29. Suppose that f and f

′

are continuous for t ≥ 0 and of exponential order as t →∞. Use

integration by parts to show that if F (s) = L{f (t)},then lim

s→∞

F(s) = 0.The result is actually

true under less restrictive conditions, such as those of Theorem 6.1.2.

30. The Gamma Function. The gamma function is denoted by 2(p) and is deﬁned by the

integral

2(p + 1) =

∞

−x

dx. (i)

The integral converges as x →∞for all p.Forp < 0 it is also improper at x = 0,

because the integrand becomes unbounded as x → 0. However, the integral can be shown

to converge at x = 0 for p > −1.

(a) Show that, for p > 0,

2(p + 1) = p2(p).

(b) Show that 2(1) = 1.

2(n + 1) = n!.

Since 2(p) is also deﬁned when p is not an integer, this function provides an extension

of the factorial function to nonintegral values of the independent variable. Note that it is

also consistent to deﬁne 0!=1.

(d) Show that, for p > 0,

p(p + 1)(p + 2) ···(p + n − 1) = 2(p + n)/ 2(p).

hus 2(

p) can be determined for all positive values of p if 2(p) is known in a single interval

of unit length—say,0 < p ≤ 1. It is possible to show that 2

√

π. Find 2

and 2

31. Consider the Laplace transform of t

, where p > −1.

(a) Referring to Problem 30, show that

L{t

∞

−st

dt =

p+1

∞

−x

= 2( p + 1)/s

p+1

, s > 0.

(b) Let p be a positive integer n in part (a); show that

L{t

}=n!/s

n+1

, s > 0.

L{t

−1/2

√

∞

−x

dx, s > 0.

It is possible to show that

∞

−x

dx =

√

;

hence

L{t

−1/2

π/s, s > 0.

(d) Show that

L{t

1/2

√

π/(2s

3/2

), s > 0.

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6.2 Solution of Initial Value Problems 317

6.2 Solution of Initial Value Problems

In this section we show how the Laplace transform can be used to solve initial value

problems for linear differential equations with constant coefﬁcients. The usefulness

of the Laplacetransform for thispurpose rests primarily on thefact that the transform

of f

′

is related in a simple way to the transform of f . The relationship is expressed in

the following theorem.

Theorem 6.2.1

Suppose that f is continuous and f

′

is piecewise continuous on any interval

0 ≤ t ≤ A. Suppose further that there exist constants K, a, and M such that

|f (t)|≤Ke

for t ≥ M. Then L{f

′

(t)} exists for s > a, and moreover,

L{f

′

(t)}=sL{f (t)}−f (0). (1)

To prove this theorem, we consider the integral

−st

′

(t) dt,

whose limit as A →∞,if it exists,is the Laplace transform of f

′

.To calculate this limit

we ﬁrst need to write the integral in a suitable form. If f

′

has points of discontinuity

in the interval 0 ≤ t ≤ A, let them be denoted by t

, t

, ..., t

. Then we can write the

integral as

−st

′

(t) dt =

−st

′

(t) dt +

−st

′

(t) dt +···+

−st

′

(t) dt.

Integrating each term on the right by parts yields

−st

′

(t) dt = e

−st

f (t)

+ e

−st

f (t)

+···+e

−st

f (t)

+ s

−st

f (t) dt +

−st

f (t) dt +···+

−st

f (t) dt

Since f is continuous, the contributions of the integrated terms at t

, t

, ..., t

cancel.

Further, the integrals on the right side can be combined into a single integral, so that

we obtain

−st

′

(t) dt = e

−sA

f (A) − f (0) + s

−st

f (t) dt. (2)

Now we let A →∞in Eq. (2). The integral on the right side of this equa-

tion approaches

L{f (t)}. Further, for A ≥ M, we have |f (A)|≤Ke

; consequently,

−sA

f (A)|≤Ke

−(s−a)A

. Hence e

−sA

f (A) → 0asA →∞whenever s > a. Thus the

right side of Eq. (2) has the limit s

L{f (t)}−f (0) . Consequently, the left side of

Eq. (2) also has a limit, and as noted above, this limit is

L{f

′

(t)}. Therefore, for s > a,

we conclude that

L{f

′

(t)}=sL{f (t)}−f (0),

which establishes the theorem.

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318 Chapter 6. The Laplace Transform

If f

′

and f

′′

satisfy the same conditions that are imposed on f and f

′

, respectively,

in Theorem 6.2.1, then it follows that the Laplace transform of f

′′

also exists for s > a

and is given by

L{f

′′

(t)}=sL{f

′

(t)}−f

′

(0)

= s[s

L{f (t)}−f (0)]−f

′

(0)

= s

L{f (t)}−sf (0) − f

′

(0). (3)

Indeed, provided the function f and its derivatives satisfy suitable conditions, an

expression for the transform of the nth derivative f

(n)

can be derived by n successive

applications of this theorem. The result is given in the following corollary.

Corollary 6.2.2

Suppose that the functions f , f

′

, ..., f

(n−1)

are continuous and that f

(n)

is piecewise

continuous on any interval 0 ≤ t ≤ A. Suppose further that there exist constants

K, a, and M such that |f (t)|≤Ke

, |f

′

(t)|≤Ke

, ..., |f

(n−1)

(t)|≤Ke

for t ≥ M.

Then

L{f

(n)

(t)} exists for s > a and is given by

L{f

(n)

(t)}=s

L{f (t)}−s

n−1

f (0) −···−sf

(n−2)

(0) − f

(n−1)

(0). (4)

We now show how the Laplace transform can be used to solve initial value prob-

lems. It is most usefulfor problems involvingnonhomogeneous differential equations,

as we will demonstrate in later sections of this chapter. However,we begin by looking

at some homogeneous equations, which are a bit simpler.

EXAMPLE

Consider the differential equation

′′

− y

′

− 2y = 0 (5)

and the initial conditions

y(0) = 1, y

′

(0) = 0. (6)

This problem is easily solved by the methods of Section 3.1. The characteristic equation is

− r − 2 = (r − 2)(r + 1) = 0,

and consequently, the general solution of Eq. (5) is

y = c

−t

+ c

. (7)

To satisfy the initial conditions (6), we must have c

+ c

= 1 and −c

+ 2c

= 0; hence c

and c

, so the solution of the initial value problem (5) and (6) is

y = φ(t) =

−t

. (8)

Now let us solve the same problem by using the Laplace transform. To do this, we must

assume that the problem has a solution y = φ(t),which with its ﬁrst two derivatives satisﬁes the

conditions of Corollary 6.2.2. Then, taking the Laplace transform of the differential equation

(5), we obtain

L{y

′′

}−L{y

′

}−2L{y}=0, (9)

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6.2 Solution of Initial Value Problems 319

where we have used the linearity of the transform to write the transform of a sum as the sum

of the separate transforms. Upon using the corollary to express L{y

′′

} and L{y

′

} in terms of

L{y}, we ﬁnd that Eq. (9) becomes

L{y}−sy(0) − y

′

(0) −[sL{y}−y(0)]−2L{y}=0,

− s − 2)Y (s) + (1 − s)y(0) − y

′

(0) = 0, (10)

where Y(s) = L{y}. Substituting for y(0) and y

′

(0) in Eq. (10) from the initial conditions (6),

and then solving for Y(s), we obtain

Y(s) =

s − 1

− s − 2

s − 1

(s − 2)(s + 1)

. (11)

We have thus obtained an expression for the Laplace transform Y(s) of the solution y = φ(t) of

the given initial value problem. To determine the function φ, we must ﬁnd the function whose

Laplace transform is Y(s), as given by Eq. (11).

This can be done most easily by expanding the right side of Eq. (11) in partial fractions.

Thus we write

Y(s) =

s − 1

(s − 2)(s + 1)

s − 2

s + 1

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

, (12)

where the coefﬁcients a and b are to be determined. By equating numerators of the second

and fourth members of Eq. (12), we obtain

s − 1 = a(s + 1) + b(s − 2),

an equation that must hold for all s. In particular, if we set s = 2, then it follows that a =

Similarly, if we set s =−1, then we ﬁnd that b =

. By substituting these values for a and b,

respectively, we have

Y(s) =

1/3

s − 2

2/3

s + 1

. (13)

Finally, if we use the result of Example 5 of Section 6.1, it follows that

has the transform

(s − 2)

−1

; similarly,

−t

has the transform

(s + 1)

−1

. Hence, by the linearity of the Laplace

transform,

y = φ(t) =

−t

has the transform (13) and is therefore the solution of the initial value problem (5),(6).Observe

that it does satisfy the conditions of Corollary 6.2.2, as we assumed initially. Of course, this is

the same solution that we obtained earlier.

The same procedure can be applied to the general second order linear equation

with constant coefﬁcients

′′

+ by

′

+ cy = f (t). (14)

Assuming that the solution y = φ(t) satisﬁes the conditions of Corollary 6.2.2 for

n = 2, we can take the transform of Eq. (14) and thereby obtain

a[s

Y(s) − sy(0) − y

′

(0)]+b[sY(s) − y(0)]+cY(s) = F(s), (15)

where F(s) is the transform of f (t). By solving Eq. (15) for Y(s), we ﬁnd that

Y(s) =

(as + b)y(0) + ay

′

(0)

+ bs + c

F(s)

+ bs + c

. (16)

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320 Chapter 6. The Laplace Transform

The problem is then solved, provided that we can ﬁnd the function y = φ(t) whose

transform is Y(s).

Even at this early stage of our discussion we can point out some of the essential

featuresofthe transform method. In the ﬁrstplace,the transform Y(s) of the unknown

function y = φ(t) is found by solving an algebraic equation rather than a differential

equation,Eq. (10) rather than Eq. (5) in Example 1,or in general Eq. (15) rather than

Eq. (14). This is the key to the usefulness of Laplace transforms for solving linear,

constant coefﬁcient, ordinary differential equations—the problem is reduced from

a differential equation to an algebraic one. Next, the solution satisfying given initial

conditions is automatically found, so that the task of determining appropriate values

for the arbitrary constants in the general solution does not arise. Further,as indicated

in Eq. (15), nonhomogeneous equations are handled in exactly the same way as

homogeneous ones; it is not necessary to solve the corresponding homogeneous

equation ﬁrst. Finally, the method can be applied in the same way to higher order

equations, as long as we assume that the solution satisﬁes the conditions of Corollary

6.2.2 for the appropriate value of n.

Observe that the polynomial as

+ bs + c in the denominator on the right side of

Eq. (16) is precisely the characteristic polynomial associated with Eq. (14). Since the

use of a partial fraction expansion of Y(s) to determine φ(t) requires us to factor this

polynomial, the use of Laplace transforms does not avoid the necessity of ﬁnding

roots of the characteristic equation. For equations of higher than second order, this

may require a numerical approximation, particularly if the roots are irrational or

complex.

The main difﬁculty that occurs in solving initial value problems by the transform

method lies in the problem of determining the function y = φ(t) corresponding to

the transform Y(s ). This problem is known as the inversion problem for the Laplace

transform;φ(t) is called the inverse transform corresponding to Y(s),and the process

of ﬁndingφ(t) from Y(s) is known as inverting the transform.We alsouse the notation

−1

{Y(s)} to denote the inverse transform of Y(s). There is a general formula for

the inverse Laplace transform, but its use requires a familiarity with functions of a

complex variable, and we do not consider it in this book. However, it is still possible

to develop many important properties of the Laplace transform, and to solve many

interesting problems, without the use of complex variables.

In solving the initial value problem (5), (6), we did not consider the question of

whether there may be functions other than the one given by Eq. (8) that also have

the transform (13). By Theorem 3.2.1 we know that the initial value problem has no

other solutions.We also know that the unique solution (8) of the initial value problem

is continuous. Consistent with this fact, it can be shown that if f and g are continuous

functions with the same Laplace transform, then f and g must be identical. On the

other hand, if f and g are only piecewise continuous, then they may differ at one or

more points of discontinuity and yet have the same Laplace transform;see Example 6

in Section 6.1.This lack of uniqueness of the inverse Laplace transform for piecewise

continuous functions is of no practical signiﬁcance in applications.

Thus there is essentially a one-to-one correspondence between functions and their

Laplace transforms. This fact suggests the compilation of a table, such as Table 6.2.1,

giving the transforms of functions frequently encountered,and vice versa.The entries

in the second column of Table 6.2.1 are the transforms of those in the ﬁrst column.

Perhaps more important, the functions in the ﬁrst column are the inverse transforms

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6.2 Solution of Initial Value Problems 321

TABLE 6.2.1 Elementary Laplace Transforms

f (t) = L

−1

{F(s)} F (s) = L{f (t)} Notes

1. 1

, s > 0 Sec. 6.1; Ex. 4

2. e

s − a

, s > a Sec. 6.1; Ex. 5

3. t

, n = positive integer

n+1

, s > 0 Sec. 6.1; Prob. 31

4. t

, p > −1

2(p + 1)

p+1

, s > 0 Sec. 6.1; Prob. 31

5. sin at

+ a

, s > 0 Sec. 6.1; Ex. 7

6. cos at

+ a

, s > 0 Sec. 6.1; Prob. 6

7. sinh at

− a

, s > |a| Sec. 6.1; Prob. 8

8. cosh at

− a

, s > |a| Sec. 6.1; Prob. 7

9. e

sin bt

(s − a)

+ b

, s > a Sec. 6.1; Prob. 13

10. e

cos bt

s − a

(s − a)

+ b

, s > a Sec. 6.1; Prob. 14

11. t

, n = positive integer

(s − a)

n+1

, s > a Sec. 6.1; Prob. 18

12. u

(t)

−cs

, s > 0 Sec. 6.3

13. u

(t)f (t − c) e

−cs

F(s) Sec. 6.3

14. e

f (t) F(s − c) Sec. 6.3

15. f (ct)

)

, c > 0 Sec. 6.3; Prob. 25

16.

f (t − τ)g(τ) dτ F (s)G(s) Sec. 6.6

17. δ(t − c) e

−cs

Sec. 6.5

18. f

(n)

(t) s

F(s) − s

n−1

f (0) −···−f

(n−1)

(0) Sec. 6.2; Cor. 6.2.2

19. (−t)

f (t) F

(n)

(s) Sec. 6.2; Prob. 29

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322 Chapter 6. The Laplace Transform

of those in the second column. Thus, for example, if the transform of the solution of

a differential equation is known, the solution itself can often be found merely by

looking it up in the table. Some of the entries in Table 6.2.1 have been used as

examples, or appear as problems in Section 6.1, while others will be developed later

in the chapter. The third column of the table indicates where the derivation of the

given transforms may be found.AlthoughTable6.2.1 is sufﬁcientfor theexamples and

problems in this book, much larger tables are also available (see the list of references

at the end of the chapter). Transforms and inverse transforms can also be readily

obtained electronically by using a computer algebra system.

Frequently, a Laplace transform F(s) is expressible as a sum of several terms

F(s) = F

(s) + F

(s) +···+F

(s). (17)

Suppose that f

(t) = L

−1

(s)}, ..., f

(t) = L

−1

(s)}. Then the function

f (t) = f

(t) +···+f

(t)

has the Laplace transform F(s). By the uniqueness property stated previously, no

other continuous function f has the same transform. Thus

−1

{F(s)}=L

−1

(s)}+···+L

−1

(s)}; (18)

that is, the inverse Laplace transform is also a linear operator.

In many problems it is convenient to make use of this property by decomposing a

given transform into a sum of functions whose inverse transforms are already known

or can be found in the table. Partial fraction expansions are particularly useful for

this purpose, and a general result covering many cases is given in Problem 39. Other

useful properties of Laplace transforms are derived later in this chapter.

As further illustrations of the technique of solving initial value problems by means

of the Laplace transform and partial fraction expansions, consider the following

examples.

EXAMPLE

Find the solution of the differential equation

′′

+ y = sin 2t (19)

satisfying the initial conditions

y(0) = 2, y

′

(0) = 1. (20)

We assume that this initial value problem has a solution y = φ(t), which with its ﬁrst two

derivatives satisﬁes the conditions of Corollary 6.2.2. Then, taking the Laplace transform of

the differential equation, we have

Y(s) − sy(0) − y

′

(0) + Y(s) = 2/(s

+ 4),

where the transform of sin 2t has been obtained from line 5 ofTable 6.2.1. Substituting for y(0)

and y

′

(0) from the initial conditions and solving for Y(s), we obtain

Y(s) =

+ s

+ 8s + 6

+ 1)(s

+ 4)

. (21)

Using partial fractions, we can write Y(s) in the form

Y(s) =

as + b

+ 1

cs + d

+ 4

(as + b)(s

+ 4) + (cs + d)(s

+ 1)

+ 1)(s

+ 4)

. (22)

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6.2 Solution of Initial Value Problems 323

By expanding the numerator on the right side of Eq. (22) and equating it to the numerator in

Eq. (21), we ﬁnd that

+ s

+ 8s + 6 = (a + c)s

+ (b + d)s

+ (4a + c)s + (4b + d)

for all s. Then, comparing coefﬁcients of like powers of s, we have

a + c = 2, b + d = 1,

4a + c = 8, 4b + d = 6.

Consequently, a = 2, c = 0, b =

, and d =−

, from which it follows that

Y(s) =

+ 1

5/3

+ 1

−

2/3

+ 4

. (23)

From lines 5 and 6 of Table 6.2.1, the solution of the given initial value problem is

y = φ(t) = 2 cos t +

sin t −

sin 2t. (24)

EXAMPLE

Find the solution of the initial value problem

(4)

− y = 0, (25)

y(0) = 0, y

′

(0) = 1, y

′′

(0) = 0, y

′′′

(0) = 0. (26)

In this problem we need to assume that the solution y = φ(t) satisﬁes the conditions of

Corollary 6.2.2 for n = 4. The Laplace transform of the differential equation (25) is

Y(s) − s

y(0) − s

′

(0) − sy

′′

(0) − y

′′′

(0) − Y(s) = 0.

Then, using the initial conditions (26) and solving for Y(s), we have

Y(s) =

− 1

. (27)

A partial fraction expansion of Y(s) is

Y(s) =

as + b

− 1

cs + d

+ 1

and it follows that

(as + b)(s

+ 1) + (cs + d)(s

− 1) = s

(28)

for all s. By setting s = 1 and s =−1, respectively, in Eq. (28), we obtain the pair of equations

2(a + b) = 1, 2(−a + b) = 1,

and therefore a = 0 and b =

.Ifwesets = 0 in Eq. (28), then b − d = 0, so d =

. Finally,

equating the coefﬁcients of the cubic terms on each side of Eq. (28), we ﬁnd that a + c = 0, so

c = 0. Thus

Y(s) =

1/2

− 1

1/2

+ 1

, (29)

and from lines 7 and 5 of Table 6.2.1, the solution of the initial value problem (25), (26) is

y = φ(t) =

sinh t + sin t

. (30)

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324 Chapter 6. The Laplace Transform

The most important elementary applications of the Laplace transform are in the

study of mechanical vibrations and in the analysis of electric circuits; the govern-

ing equations were derived in Section 3.7. A vibrating spring–mass system has the

equation of motion

+ γ

+ ku = F(t), (31)

where m is the mass, γ the damping coefﬁcient, k the spring constant, and F(t) the

applied external force. The equation that describes an electric circuit containing an

inductance L, a resistance R, and a capacitance C (an LRC circuit) is

+ R

Q = E(t), (32)

where Q(t) is the charge on the capacitor and E(t) is the applied voltage. In terms of

the current I(t) = dQ(t)/dt, we can differentiate Eq. (32) and write

+ R

I =

(t). (33)

Suitable initial conditions on u, Q, or I must also be prescribed.

We have noted previously, in Section 3.7, that Eq. (31) for the spring–mass system

and Eqs. (32) or (33) for the electric circuit are identical mathematically, differing

only in the interpretation of the constants and variables appearing in them. There

are other physical problems that also lead to the same differential equation. Thus,

once the mathematical problem is solved, its solution can be interpreted in terms of

whichever corresponding physical problem is of immediate interest.

In the problem lists following this and other sections in this chapter are numerous

initial value problems for second order linear differential equations with constant

coefﬁcients. Many can be interpreted as models of particular physical systems, but

usually we do not point this out explicitly.

PROBLEMS In each of Problems 1 through 10, ﬁnd the inverse Laplace transform of the given function.

1. F(s) =

+ 4

2. F(s) =

(s − 1)

3. F(s) =

+ 3s − 4

4. F(s) =

− s − 6

5. F(s) =

2s + 2

+ 2s + 5

6. F(s) =

2s − 3

− 4

7. F(s) =

2s + 1

− 2s + 2

8. F(s) =

− 4s + 12

s(s

+ 4)

9. F(s) =

1 − 2s

+ 4s + 5

10. F(s) =

2s − 3

+ 2s + 10

In each of Problems 11 through 23, use the Laplace transform to solve the given initial value

problem.

11. y

′′

− y

′

− 6y = 0; y(0) = 1, y

′

(0) =−1

12. y

′′

+ 3y

′

+ 2y = 0; y(0) = 1, y

′

(0) = 0

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6.2 Solution of Initial Value Problems 325

13. y

′′

− 2y

′

+ 2y = 0; y(0) = 0, y

′

(0) = 1

14. y

′′

− 4y

′

+ 4y = 0; y(0) = 1, y

′

(0) = 1

15. y

′′

− 2y

′

+ 4y = 0; y(0) = 2, y

′

(0) = 0

16. y

′′

+ 2y

′

+ 5y = 0; y(0) = 2, y

′

(0) =−1

17. y

(4)

− 4y

′′′

+ 6y

′′

− 4y

′

+ y = 0; y(0) = 0, y

′

(0) = 1, y

′′

(0) = 0, y

′′′

(0) = 1

18. y

(4)

− y = 0; y(0) = 1, y

′

(0) = 0, y

′′

(0) = 1, y

′′′

(0) = 0

19. y

(4)

− 4y = 0; y(0) = 1, y

′

(0) = 0, y

′′

(0) =−2, y

′′′

(0) = 0

20. y

′′

+ ω

y = cos 2t, ω

̸= 4; y(0) = 1, y

′

(0) = 0

21. y

′′

− 2y

′

+ 2y = cos t; y(0) = 1, y

′

(0) = 0

22. y

′′

− 2y

′

+ 2y = e

−t

; y(0) = 0, y

′

(0) = 1

23. y

′′

+ 2y

′

+ y = 4e

−t

; y(0) = 2, y

′

(0) =−1

In each of Problems 24 through 27, ﬁnd the Laplace transform Y(s) = L{y} of the solution of

the given initial value problem. A method of determining the inverse transform is developed

in Section 6.3. You may wish to refer to Problems 21 through 24 in Section 6.1.

24. y

′′

+ 4y =

1, 0 ≤ t <π,

0, π ≤ t < ∞;

y(0) = 1, y

′

(0) = 0

25. y

′′

+ y =

t,0≤ t < 1,

0, 1 ≤ t < ∞;

y(0) = 0, y

′

(0) = 0

26. y

′′

+ 4y =

t,0≤ t < 1,

1, 1 ≤ t < ∞;

y(0) = 0, y

′

(0) = 0

27. y

′′

+ y =

⎧

⎪

⎨

⎪

⎩

t,0≤ t < 1,

2 − t,1≤ t < 2,

0, 2 ≤ t < ∞;

y(0) = 0, y

′

(0) = 0

28. The Laplace transforms of certain functions can be found conveniently from their Taylor

series expansions.

(a) Using the Taylor series for sin t

sin t =

∞

n=0

(−1)

2n+1

(2n + 1)!

and assuming that the Laplace transform of this series can be computed term by term,

verify that

L{sin t}=

+ 1

, s > 1.

(b) Let

f (t) =

(sin t)/t, t ̸= 0,

1, t = 0.

Find the Taylor series for f about t = 0. Assuming that the Laplace transform of this

function can be computed term by term, verify that

L{f (t)}=arctan(1/s), s > 1.

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326 Chapter 6. The Laplace Transform

, has the Taylor series (see

Section 5.7)

(t) =

∞

n=0

(−1)

(n!)

Assuming that the following Laplace transforms can be computed term by term, verify

that

L{J

(t)}=(s

+ 1)

−1/2

, s > 1

and

L{J

(

√

t)}=s

−1

−1/(4s)

, s > 0.

Problems 29 through 37 are concerned with differentiation of the Laplace transform.

29. Let

F(s) =

∞

−st

f (t) dt.

It is possible to show that as long as f satisﬁes the conditions of Theorem 6.1.2, it is

legitimate to differentiate under the integral sign with respect to the parameter s when

s > a.

(a) Show that F

′

(s) = L{−tf (t)}.

(b) Show that F

(n)

(s) = L{(−t)

f (t)}; hence differentiating the Laplace transform corre-

sponds to multiplying the original function by −t.

In each of Problems 30 through 35, use the result of Problem 29 to ﬁnd the Laplace transform

of the given function; a and b are real numbers and n is a positive integer.

30. f (t) = te

31. f (t) = t

sin bt

32. f (t) = t

33. f (t) = t

34. f (t) = te

sin bt 35. f (t) = te

cos bt

36. Consider Bessel’s equation of order zero

′′

+ y

′

+ ty = 0.

Recall from Section 5.7 thatt = 0 is a regular singularpoint for thisequation,and therefore

solutions may become unbounded ast → 0. However,let us try todetermine whether there

are any solutions that remain ﬁnite at t = 0 and have ﬁnite derivatives there. Assuming

that there is such a solution y = φ(t), let Y(s) = L{φ(t)}.

(a) Show that Y(s) satisﬁes

(1 + s

′

(s) + sY(s) = 0 .

(b) Show that Y(s) = c(1 + s

)

−1/2

, where c is an arbitrary constant.

)

−1/2

= s

−1

(1 + s

−2

)

−1/2

, expanding in a binomial series valid for s > 1,

and assuming that it is permissible to take the inverse transform term by term, show that

y = c

∞

n=0

(−1)

(n!)

= cJ

(t),

where J

is the Bessel function of the ﬁrst kind of order zero. Note that J

(0) = 1 and that

has ﬁnite derivatives of all orders at t = 0. It was shown in Section 5.7 that the second

solution of this equation becomes unbounded as t → 0.

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6.3 Step Functions 327

37. For each of the following initial value problems, use the results of Problem 29 to ﬁnd

the differential equation satisﬁed by Y(s) = L{φ(t)}, where y = φ(t) is the solution of the

given initial value problem.

(a) y

′′

− ty = 0; y(0) = 1, y

′

(0) = 0 (Airy’s equation)

(b) (1 − t

′′

− 2ty

′

+ α(α + 1)y = 0; y(0) = 0, y

′

(0) = 1 (Legendre’s equation)

Note that the differential equation for Y(s) is of ﬁrst order in part (a), but of second order

in part (b). This is due to the fact that t appears at most to the ﬁrst power in the equation

of part (a), whereas it appears to the second power in that of part (b). This illustrates that

the Laplace transform is not often useful in solving differential equations with variable

coefﬁcients, unless all the coefﬁcients are at most linear functions of the independent

variable.

38. Suppose that

g(t) =

f (τ) dτ.

If G(s) and F(s) are the Laplace transforms of g(t) and f (t), respectively, show that

G(s) = F(s)/s.

39. In this problem we show how a general partial fraction expansion can be used to calculate

many inverse Laplace transforms. Suppose that

F(s) = P(s)/Q(s),

where Q(s) is a polynomial of degree n with distinct zeros r

, ..., r

, and P(s) is a

polynomial of degree less than n. In this case it is possible to show that P(s)/Q(s) has

a partial fraction expansion of the form

P(s)

Q(s)

s − r

+···+

s − r

, (i)

where the coefﬁcients A

, ..., A

must be determined.

(a) Show that

= P(r

)/Q

′

), k = 1, ..., n. (ii)

Hint: One way to do this is to multiply Eq. (i) by s − r

and then to take the limit as s → r

(b) Show that

−1

{F(s)}=

k=1

P(r

)

′

)

. (iii)

6.3 Step Functions

In Section 6.2 we outlined the general procedure involved in solving initial value

problems by means of the Laplace transform. Some of the most interesting elemen-

tary applications of the transform method occur in the solution of linear differential

equations with discontinuous or impulsive forcing functions. Equations of this type

frequently arise in the analysis of the ﬂow of current in electric circuits or the vibra-

tions of mechanical systems. In this section and the following ones, we develop some

additional properties of the Laplace transform that are useful in the solution of such

problems. Unless a speciﬁc statement is made to the contrary,all functions appearing

August 7, 2012 21:04 c06 Sheet number 20 Page number 328 cyan black

328 Chapter 6. The Laplace Transform

below will be assumed to be piecewise continuous and of exponential order, so that

their Laplace transforms exist, at least for s sufﬁciently large.

To deal effectively with functions having jump discontinuities, it is very helpful

to introduce a function known as the unit step function or Heaviside function. This

function will be denoted by u

and is deﬁned by

(t) =

0, t < c,

1, t ≥ c.

(1)

Since the Laplace transform involves values of t in the interval [0, ∞), we are also

interested only in nonnegative values of c. The graph of y = u

(t) is shown in Figure

6.3.1. We have somewhat arbitrarily assigned the value one to u

at t = c. However,

for a piecewise continuous function such as u

, the value at a discontinuity point is

usually irrelevant. The step can also be negative. For instance,Figure 6.3.2 shows the

graph of y = 1 − u

(t).

FIGURE 6.3.1 Graph of y = u

(t).

FIGURE 6.3.2 Graph of y=1−u

(t).

EXAMPLE

Sketch the graph of y = h(t), where

h(t) = u

(t) − u

2π

(t), t ≥ 0.

From the deﬁnition of u

(t) in Eq. (1), we have

h(t) =

⎧

⎪

⎨

⎪

⎩

0 − 0 = 0, 0 ≤ t <π,

1 − 0 = 1, π ≤ t < 2π,

1 − 1 = 0, 2π ≤ t < ∞.

Thus the equation y = h(t) has the graph shown in Figure 6.3.3. This function can be thought

of as a rectangular pulse.

ππ

FIGURE 6.3.3 Graph of y = u

(t) − u

2π

(t).

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6.3 Step Functions 329

EXAMPLE

Consider the function

f (t) =

⎧

⎪

⎨

⎪

⎩

2, 0 ≤ t < 4,

5, 4 ≤ t < 7,

−1, 7 ≤ t < 9,

1, t ≥ 9,

(2)

whose graph is shown in Figure 6.3.4. Express f (t) in terms of u

(t).

21018246

–1

FIGURE 6.3.4 Graph of the function in Eq. (2).

We start with the function f

(t) = 2, which agrees with f (t) on [0, 4). To produce the jump

of three units at t = 4, we add 3u

(t) to f

(t), obtaining

(t) = 2 + 3u

(t),

which agrees with f (t) on [0, 7). The negative jump of six units at t = 7 corresponds to

adding −6u

(t), which gives

(t) = 2 + 3u

(t) − 6u

(t).

Finally, we must add 2u

(t) to match the jump of two units at t = 9. Thus we obtain

f (t) = 2 + 3u

(t) − 6u

(t) + 2u

(t). (3)

The Laplace transform of u

for c ≥ 0 is easily determined:

L{u

(t)}=

∞

−st

(t) dt =

∞

−st

−cs

, s > 0. (4)

For a given function f deﬁned for t ≥ 0, we will often want to consider the related

function g deﬁned by

y = g(t) =

0, t < c,

f (t − c), t ≥ c,

August 7, 2012 21:04 c06 Sheet number 22 Page number 330 cyan black

330 Chapter 6. The Laplace Transform

which represents a translation of f a distance c in the positive t direction; see Fig-

ure 6.3.5. In terms of the unit step function we can write g(t) in the convenient

form

g(t) = u

(t)f (t − c).

f (0)

(a)

f (0)

(b)

FIGURE 6.3.5 A translation of the given function. (a) y = f (t);(b) y = u

(t)f (t − c).

The unit step function is particularly important in transform use because of the fol-

lowing relation between the transform of f (t) and that of its translation u

(t)f (t − c).

Theorem 6.3.1

If F(s) = L{f (t)} exists for s > a ≥ 0, and if c is a positive constant, then

L{u

(t)f (t − c)}=e

−cs

L{f (t)}=e

−cs

F(s), s > a. (5)

Conversely, if f (t) =

−1

{F(s)}, then

(t)f (t − c) = L

−1

−cs

F(s)}. (6)

Theorem 6.3.1 simply states that the translation of f (t) a distance c in the positive

t direction corresponds to the multiplication of F(s) by e

−cs

. To prove Theorem 6.3.1,

it is sufﬁcient to compute the transform of u

(t)f (t − c):

L{u

(t)f (t − c)}=

∞

−st

(t)f (t − c) dt

∞

−st

f (t − c) dt.

Introducing a new integration variable ξ = t − c, we have

L{u

(t)f (t − c)}=

∞

−(ξ+c)s

f (ξ) dξ = e

−cs

∞

−sξ

f (ξ) dξ

= e

−cs

F(s).

Thus Eq. (5) is established; Eq. (6) follows by taking the inverse transform of both

sides of Eq. (5).

A simple example of this theorem occurs if we take f (t) = 1. Recalling that

L{1}=1/s, we immediately have from Eq. (5) that L{u

(t)}=e

−cs

/s. This result

agrees with that of Eq. (4). Examples 3 and 4 illustrate further how Theorem 6.3.1

can be used in the calculation of transforms and inverse transforms.

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6.3 Step Functions 331

EXAMPLE

If the function f is deﬁned by

f (t) =

sin t,0≤ t <π/4,

sin t + cos(t − π/4), t ≥ π/4,

ﬁnd L{f (t)}. The graph of y = f (t) is shown in Figure 6.3.6.

y = sin t + u

(t)cos(t – )

0.5

0.5 1 1.5 2 2.5 3

FIGURE 6.3.6 Graph of the function in Example 3.

Note that f (t) = sin t + g(t), where

g(t) =

0, t <π/4,

cos(t − π/4), t ≥ π/4.

Thus

g(t) = u

π/4

(t) cos(t − π/4)

and

L{f (t)}=L{sin t}+L{u

π/4

(t) cos(t − π/4)}

= L{sin t}+e

−πs/4

L{cos t}.

Introducing the transforms of sin t and cos t, we obtain

L{f (t)}=

+ 1

+ e

−πs/4

+ 1

1 + se

−πs/4

+ 1

You should compare this method with the calculation of L{f (t)} directly from the deﬁnition.

EXAMPLE

Find the inverse transform of

F(s) =

1 − e

−2s

From the linearity of the inverse transform, we have

f (t) = L

−1

{F(s)}=L

−1

− L

−1

−2s

= t − u

(t)(t − 2).

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332 Chapter 6. The Laplace Transform

The function f may also be written as

f (t) =

t,0≤ t < 2,

2, t ≥ 2.

The following theorem contains another very useful property of Laplace trans-

forms that is somewhat analogous to that given in Theorem 6.3.1.

Theorem 6.3.2

If F(s) = L{f (t)} exists for s > a ≥ 0, and if c is a constant, then

L{e

f (t)}=F(s − c), s > a + c. (7)

Conversely, if f (t) =

−1

{F(s)}, then

f (t) = L

−1

{F(s − c)}. (8)

According to Theorem 6.3.2, multiplication of f (t) by e

results in a translation of

the transform F(s) a distance c in the positive s direction, and conversely. To prove

this theorem, we evaluate

L{e

f (t)}. Thus

L{e

f (t)}=

∞

−st

f (t) dt =

∞

−(s−c)t

f (t) dt

= F(s − c),

which is Eq. (7).The restriction s > a + c follows fromthe observation that,according

to hypothesis (ii) of Theorem 6.1.2, |f (t)|≤Ke

; hence |e

f (t)|≤Ke

(a+c)t

. Equation

(8) is obtained by taking the inverse transform of Eq. (7), and the proof is complete.

The principal application of Theorem 6.3.2 is in the evaluation of certain inverse

transforms, as illustrated by Example 5.

EXAMPLE

Find the inverse transform of

G(s) =

− 4s + 5

By completing the square in the denominator, we can write

G(s) =

(s − 2)

+ 1

= F(s − 2),

where F(s) = (s

+ 1)

−1

. Since L

−1

{F(s)}=sin t, it follows from Theorem 6.3.2 that

g(t) = L

−1

{G(s)}=e

sin t.

The results of this section are often useful in solving differential equations, partic-

ularly those that have discontinuous forcing functions. The next section is devoted to

examples illustrating this fact.

August 7, 2012 21:04 c06 Sheet number 25 Page number 333 cyan black

6.3 Step Functions 333

PROBLEMS In each of Problems 1 through 6, sketch the graph of the given function on the interval t ≥ 0.

1. g(t) = u

(t) + 2u

(t) − 6u

(t) 2. g(t) = (t − 3)u

(t) − (t − 2)u

(t)

3. g(t) = f (t − π)u

(t), where f (t) = t

4. g(t) = f (t − 3)u

(t), where f (t) = sin t

5. g(t) = f (t − 1)u

(t), where f (t) = 2t

6. g(t) = (t − 1)u

(t) − 2(t − 2)u

(t) + (t − 3)u

(t)

In each of Problems 7 through 12:

(a) Sketch the graph of the given function.

(b) Express f (t) in terms of the unit step function u

(t).

7. f (t) =

⎧

⎪

⎨

⎪

⎩

0, 0 ≤ t < 3,

−2, 3 ≤ t < 5,

2, 5 ≤ t < 7,

1, t ≥ 7.

8. f (t) =

⎧

⎪

⎨

⎪

⎩

1, 0 ≤ t < 1,

−1, 1 ≤ t < 2,

1, 2 ≤ t < 3,

−1, 3 ≤ t < 4,

0, t ≥ 4.

9. f (t) =

1, 0 ≤ t < 2,

−(t−2)

, t ≥ 2.

10. f (t) =

,0≤ t < 2,

1, t ≥ 2.

11. f (t) =

⎧

⎪

⎨

⎪

⎩

t,0≤ t < 1,

t − 1, 1 ≤ t < 2,

t − 2, 2 ≤ t < 3,

0, t ≥ 3.

12. f (t) =

⎧

⎪

⎨

⎪

⎩

t,0≤ t < 2,

2, 2 ≤ t < 5,

7 − t,5≤ t < 7,

0, t ≥ 7.

In each of Problems 13 through 18, ﬁnd the Laplace transform of the given function.

13. f (t) =

0, t < 2

(t − 2)

, t ≥ 2

14. f (t) =

0, t < 1

− 2t + 2, t ≥ 1

15. f (t) =

⎧

⎪

⎨

⎪

⎩

0, t <π

t − π, π ≤ t < 2π

0, t ≥ 2π

16. f (t) = u

(t) + 2u

(t) − 6u

(t)

17. f (t) = (t − 3)u

(t) − (t − 2)u

(t) 18. f (t) = t − u

(t)(t − 1), t ≥ 0

In each of Problems 19 through 24, ﬁnd the inverse Laplace transform of the given function.

19. F(s) =

(s − 2)

20. F(s) =

−2s

+ s − 2

21. F(s) =

2(s − 1)e

−2s

− 2s + 2

22. F(s) =

−2s

− 4

23. F(s) =

(s − 2)e

−s

− 4s + 3

24. F(s) =

−s

+ e

−2s

− e

−3s

− e

−4s

25. Suppose that F (s) = L{f (t)} exists for s > a ≥ 0.

(a) Show that if c is a positive constant, then

L{f (ct)}=

)

, s > ca.

August 7, 2012 21:04 c06 Sheet number 26 Page number 334 cyan black

334 Chapter 6. The Laplace Transform

(b) Show that if k is a positive constant, then

−1

{F(ks)}=

(

−1

{F(as + b)}=

−bt/a

(

In each of Problems 26 through 29, use the results of Problem 25 to ﬁnd the inverse Laplace

transform of the given function.

26. F(s) =

n+1

27. F(s) =

2s + 1

+ 4s + 5

28. F(s) =

− 12s + 3

29. F(s) =

−4s

2s − 1

In each of Problems 30 through 33, ﬁnd the Laplace transform of the given function. In

Problem 33, assume that term-by-term integration of the inﬁnite series is permissible.

30. f (t) =

1, 0 ≤ t < 1

0, t ≥ 1

31. f (t) =

⎧

⎪

⎨

⎪

⎩

1, 0 ≤ t < 1

0, 1 ≤ t < 2

1, 2 ≤ t < 3

0, t ≥ 3

32. f (t) = 1 − u

(t) +···+u

(t) − u

2n+1

(t) = 1 +

2n+1

k=1

(−1)

(t)

33. f (t) = 1 +

∞

k=1

(−1)

(t). See Figure 6.3.7.

12345

FIGURE 6.3.7 The function f (t) in Problem 33; a square wave.

34. Let f satisfy f (t + T) = f (t) for all t ≥ 0 and for some ﬁxed positive number T; f is said

to be periodic with period T on 0 ≤ t < ∞. Show that

L{f (t)}=

−st

f (t) dt

1 − e

−sT

In each of Problems 35 through 38, use the result of Problem 34 to ﬁnd the Laplace transform

of the given function.

35. f (t) =

1, 0 ≤ t < 1,

0, 1 ≤ t < 2;

f (t + 2) = f (t).

Compare with Problem 33.

36. f (t) =

1, 0 ≤ t < 1,

−1, 1 ≤ t < 2;

f (t + 2) = f (t).

See Figure 6.3.8.

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6.3 Step Functions 335

–1

1234 5

FIGURE 6.3.8 The function f (t) in Problem 36; a square wave.

37. f (t) = t,0≤ t < 1;

f (t + 1) = f (t).

See Figure 6.3.9.

38. f (t) = sin t,0≤ t <π;

f (t + π) = f (t).

See Figure 6.3.10.

1234

FIGURE 6.3.9 The function f (t) in

Problem 37; a sawtooth wave.

ππ

FIGURE 6.3.10 The function f (t) in

Problem 38; a rectiﬁed sine wave.

39. (a) If f (t) = 1 − u

(t), ﬁnd L{f (t)}; compare with Problem 30. Sketch the graph of

y = f (t).

(b) Let g(t) =

f (ξ) dξ, where the function f is deﬁned in part (a). Sketch the graph of

y = g(t) and ﬁnd L{g(t)}.

(t)g(t − 1), where g is deﬁned in part (b). Sketch the graph of

y = h(t) and ﬁnd L{h(t)}.

40. Consider the function p deﬁned by

p(t) =

t,0≤ t < 1,

2 − t,1≤ t < 2;

p(t + 2) = p(t).

(a) Sketch the graph of y = p(t).

(b) Find L{p(t)} by noting that p is the periodic extension of the function h in

Problem 39(c) and then using the result of Problem 34.

p(t) =

f (t) dt,

where f is the function in Problem 36, and then using Theorem 6.2.1.

August 7, 2012 21:04 c06 Sheet number 28 Page number 336 cyan black

336 Chapter 6. The Laplace Transform

6.4 Differential Equations with Discontinuous Forcing Functions

In this section we turn our attention to some examples in which the nonhomogeneous

term, or forcing function, is discontinuous.

EXAMPLE

Find the solution of the differential equation

′′

+ y

′

+ 2y = g(t), (1)

where

g(t) = u

(t) − u

(t) =

1, 5 ≤ t < 20,

0, 0 ≤ t < 5 and t ≥ 20.

(2)

Assume that the initial conditions are

y(0) = 0, y

′

(0) = 0. (3)

This problem governs the charge on the capacitor in a simple electric circuit with a unit

voltage pulse for 5 ≤ t < 20.Alternatively,y may represent the response of a damped oscillator

subject to the applied force g(t).

The Laplace transform of Eq. (1) is

Y(s) − 2sy(0) − 2y

′

(0) + sY(s) − y(0) + 2Y(s) = L{u

(t)}−L{u

(t)}

= (e

−5s

− e

−20s

)/s.

Introducing the initial values (3) and solving for Y(s ), we obtain

Y(s) =

−5s

− e

−20s

s(2s

+ s + 2)

. (4)

To ﬁnd y = φ(t), it is convenient to write Y(s) as

Y(s) = (e

−5s

− e

−20s

)H(s), (5)

where

H(s) =

s(2s

+ s + 2)

. (6)

Then, if h(t) = L

−1

{H(s)}, we have

y = φ(t) = u

(t)h(t − 5) − u

(t)h(t − 20). (7)

Observe that we have used Theorem 6.3.1 to write the inverse transforms of e

−5s

H(s) and

−20s

H(s), respectively. Finally, to determine h(t), we use the partial fraction expansion of

H(s):

H(s) =

bs + c

+ s + 2

. (8)

Upon determining the coefﬁcients, we ﬁnd that a =

, b =−1, and c =−

. Thus

H(s) =

1/2

−

s +

+ s + 2

1/2

−

(

s +

1/2

−

(

⎡

⎢

⎣

s +

)

√

15/4

s +

)

√

⎤

⎥

⎦

. (9)

August 7, 2012 21:04 c06 Sheet number 29 Page number 337 cyan black

6.4 Differential Equations with Discontinuous Forcing Functions 337

Then, by referring to lines 9 and 10 of Table 6.2.1, we obtain

h(t) =

−

;

−t/4

cos(

√

15 t/4) + (

√

15/15)e

−t/4

sin(

√

15 t/4)

. (10)

In Figure 6.4.1 the graph of y = φ(t) from Eqs. (7) and (10) shows that the solution consists

of three distinct parts. For 0 < t < 5, the differential equation is

′′

+ y

′

+ 2y = 0, (11)

and the initial conditions are given by Eq. (3). Since the initial conditions impart no energy

to the system, and since there is no external forcing, the system remains at rest; that is, y = 0

for 0 < t < 5. This can be conﬁrmed by solving Eq. (11) subject to the initial conditions (3). In

particular,evaluating the solution and its derivative at t = 5,or,more precisely,as t approaches

5 from below, we have

y(5) = 0, y

′

(5) = 0. (12)

Once t > 5, the differential equation becomes

′′

+ y

′

+ 2y = 1, (13)

whose solution is the sum of a constant (the response to the constant forcing function) and

a damped oscillation (the solution of the corresponding homogeneous equation). The plot

in Figure 6.4.1 shows this behavior clearly for the interval 5 ≤ t ≤ 20. An expression for this

portion of the solution can be found by solving the differential equation (13) subject to the

initial conditions (12). Finally, for t > 20 the differential equation becomes Eq. (11) again,

and the initial conditions are obtained by evaluating the solution of Eqs. (13), (12) and its

derivative at t = 20. These values are

y(20)

∼

0.50162, y

′

(20)

∼

0.01125. (14)

The initial value problem (11), (14) contains no external forcing, so its solution is a damped

oscillation about y = 0, as can be seen in Figure 6.4.1.

0.6

0.4

0.8

0.2

–0.2

04020130

FIGURE 6.4.1 Solution of the initial value problem (1), (2), (3):

′′

+ y

′

+ 2y = u

(t) − u

(t), y(0) = 0, y

′

(0) = 0.

August 7, 2012 21:04 c06 Sheet number 30 Page number 338 cyan black

338 Chapter 6. The Laplace Transform

Although it may be helpful to visualize the solution shown in Figure 6.4.1 as composed of

solutions of three separate initial value problems in three separate intervals, it is somewhat

tedious to ﬁnd the solution by solving these separate problems. Laplace transform methods

provide a much more convenient and elegant approach to this problem and to others that

have discontinuous forcing functions.

The effect of the discontinuity in the forcing function can be seen if we examine the

solution φ(t) of Example 1 more closely. According to the existence and uniqueness

Theorem 3.2.1, the solution φ and its ﬁrst two derivatives are continuous except

possibly at the points t = 5 and t = 20, where g is discontinuous. This can also be

seen at once from Eq. (7). One can also show by direct computation from Eq. (7)

that φ and φ

′

are continuous even at t = 5 and t = 20. However, if we calculate φ

′′

we ﬁnd that

lim

t→5−

′′

(t) = 0, lim

t→5+

′′

(t) = 1/2.

Consequently, φ

′′

(t) has a jump of 1/2att = 5. In a similar way, we can show that

′′

(t) has a jump of −1/2att = 20. Thus the jump in the forcing term g(t) at these

points is balanced by a corresponding jump in the highest order term 2y

′′

on the left

side of the equation.

Consider now the general second order linear equation

′′

+ p(t)y

′

+ q(t)y = g(t), (15)

where p and q are continuous on some interval α<t <β, but g is only piecewise

continuous there. If y = ψ(t) is a solution of Eq. (15), then ψ and ψ

′

are continuous

on α<t <β,but ψ

′′

has jump discontinuities at the same points as g. Similar remarks

apply to higher order equations; the highest derivative of the solution appearing in

the differential equation has jump discontinuities at the same points as the forcing

function,but the solution itself and its lower derivatives are continuous even at those

points.

EXAMPLE

Describe the qualitative nature of the solution of the initial value problem

′′

+ 4y = g(t), (16)

y(0) = 0, y

′

(0) = 0, (17)

where

g(t) =

⎧

⎪

⎨

⎪

⎩

0, 0 ≤ t < 5,

(t − 5)/5, 5 ≤ t < 10,

1, t ≥ 10,

(18)

and then ﬁnd the solution.

In this example the forcing function has the graph shown in Figure 6.4.2 and is known as

ramp loading. It is relatively easy to identify the general form of the solution. For t < 5 the

solution is simply y = 0. On the other hand, for t > 10 the solution has the form

y = c

cos 2t + c

sin 2t + 1/4. (19)

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6.4 Differential Equations with Discontinuous Forcing Functions 339

The constant 1/4 is a particular solution of the nonhomogeneous equation, while the other

two terms are the general solution of the corresponding homogeneous equation. Thus the

solution (19) is a simple harmonic oscillation about y = 1/4. Similarly, in the intermediate

range 5 < t < 10,the solution is an oscillation about a certain linear function. In an engineering

context, for example, we might be interested in knowing the amplitude of the eventual steady

oscillation.

0.5

10 15 20

y = g(t)

FIGURE 6.4.2 Ramp loading; y = g(t) from Eq. (18) or Eq. (20).

To solve the problem, it is convenient to write

g(t) =

(t)(t − 5) − u

(t)(t − 10)

/5, (20)

as you may verify. Then we take the Laplace transform of the differential equation and use

the initial conditions, thereby obtaining

+ 4)Y(s) = (e

−5s

− e

−10s

)/5s

Y(s) = (e

−5s

− e

−10s

)H(s)/5, (21)

where

H(s) =

+ 4)

. (22)

Thus the solution of the initial value problem (16), (17), (18) is

y = φ(t) =

(t)h(t − 5) − u

(t)h(t − 10)

/5, (23)

where h(t) is the inverse transform of H(s). The partial fraction expansion of H(s) is

H(s) =

1/4

−

1/4

+ 4

, (24)

and it then follows from lines 3 and 5 of Table 6.2.1 that

h(t) =

t −

sin 2t. (25)

The graph of y = φ(t) is shown in Figure 6.4.3. Observe that it has the qualitative form that

we indicated earlier. To ﬁnd the amplitude of the eventual steady oscillation, it is sufﬁcient to

locate one of the maximum or minimum points for t > 10. Setting the derivative of the solution

(23) equal to zero, we ﬁnd that the ﬁrst maximum is located approximately at (10.642, 0.2979),

so the amplitude of the oscillation is approximately 0.0479.

August 7, 2012 21:04 c06 Sheet number 32 Page number 340 cyan black

340 Chapter 6. The Laplace Transform

0.30

0.20

0.10

5101520

FIGURE 6.4.3 Solution of the initial value problem (16), (17), (18).

Note that in this example, the forcing function g is continuous but g

′

is discontinuous at

t = 5 and t = 10. It follows that the solution φ and its ﬁrst two derivatives are continuous

everywhere, but φ

′′′

has discontinuities at t = 5 and at t = 10 that match the discontinuities in

′

at those points.

PROBLEMS In each of Problems 1 through 13:

(a) Find the solution of the given initial value problem.

(b) Draw the graphs of the solution and of the forcing function; explain how they are related.

′′

+ y = f (t); y(0) = 0, y

′

(0) = 1; f (t) =

1, 0 ≤ t < 3π

0, 3π ≤ t < ∞

′′

+ 2y

′

+ 2y = h(t); y(0) = 0, y

′

(0) = 1; h(t) =

1, π ≤ t < 2π

0, 0 ≤ t <π and t ≥ 2π

′′

+ 4y = sin t − u

2π

(t) sin(t − 2π); y(0) = 0, y

′

(0) = 0

′′

+ 4y = sin t + u

(t) sin(t − π); y(0) = 0, y

′

(0) = 0

′′

+ 3y

′

+ 2y = f (t); y(0) = 0, y

′

(0) = 0; f (t) =

1, 0 ≤ t < 10

0, t ≥ 10

′′

+ 3y

′

+ 2y = u

(t); y(0) = 0, y

′

(0) = 1

′′

+ y = u

3π

(t); y(0) = 1, y

′

(0) = 0

′′

+ y

′

y = t − u

π/2

(t)(t − π/2 ); y(0) = 0, y

′

(0) = 0

′′

+ y = g(t); y(0) = 0, y

′

(0) = 1; g(t) =

t/2, 0 ≤ t < 6

3, t ≥ 6

10.

′′

+ y

′

y = g(t); y(0) = 0, y

′

(0) = 0; g(t) =

sin t,0≤ t <π

0, t ≥ π

11.

′′

+ 4y = u

(t) − u

3π

(t); y(0) = 0, y

′

(0) = 0

August 7, 2012 21:04 c06 Sheet number 33 Page number 341 cyan black

6.4 Differential Equations with Discontinuous Forcing Functions 341

12. y

(4)

− y = u

(t) − u

(t); y(0) = 0, y

′

(0) = 0, y

′′

(0) = 0, y

′′′

(0) = 0

13.

(4)

+ 5y

′′

+ 4y = 1 − u

(t); y(0) = 0, y

′

(0) = 0, y

′′

(0) = 0, y

′′′

(0) = 0

14. Find an expression involving u

(t) for a function f that ramps up from zero at t = t

to the

value h at t = t

+ k.

15. Find an expression involving u

(t) for a function g that ramps up from zero at t = t

to the

value h at t = t

+ k and then ramps back down to zero at t = t

+ 2k.

16.

A certain spring–mass system satisﬁes the initial value problem

′′

′

+ u = kg(t), u(0) = 0, u

′

(0) = 0,

where g(t) = u

3/2

(t) − u

5/2

(t) and k > 0 is a parameter.

(a) Sketch the graph of g(t). Observe that it is a pulse of unit magnitude extending over

one time unit.

(b) Solve the initial value problem.

solution and how they depend on k.

(d) Find, to two decimal places,the smallest value of k for which the solution u(t) reaches

the value 2.

(e) Suppose k = 2. Find the time τ after which |u(t)| < 0.1 for all t >τ.

17.

Modify the problem in Example 2 of this section by replacing the given forcing function

g(t) by

f (t) =

(t)(t − 5) − u

5+k

(t)(t − 5 − k)

/k.

(a) Sketch the graph of f (t) and describe how it depends on k. For what value of k is f (t)

identical to g(t) in the example?

(b) Solve the initial value problem

′′

+ 4y = f (t), y(0) = 0, y

′

(0) = 0.

a simple harmonic oscillation about y = 1/4. Try to decide how the amplitude of this

eventual oscillation depends on k. Then conﬁrm your conclusion by plotting the solution

for a few different values of k.

18.

Consider the initial value problem

′′

′

+ 4y = f

(t), y(0) = 0, y

′

(0) = 0,

where

(t) =

1/2k,4− k ≤ t < 4 + k

0, 0 ≤ t < 4 − k and t ≥ 4 + k

and 0 < k < 4.

(a) Sketch the graph of f

(t). Observe that the area under the graph is independent of k.

If f

(t) represents a force, this means that the product of the magnitude of the force and

the time interval during which it acts does not depend on k.

(b) Write f

(t) in terms of the unit step function and then solve the given initial value

problem.

. Describe how the solution depends on k.

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342 Chapter 6. The Laplace Transform

Resonance and Beats. In Section 3.8 we observed that an undamped harmonic oscillator

(such as a spring–mass system) with a sinusoidal forcing term experiences resonance if the

frequency of the forcing term is the same as the natural frequency. If the forcing frequency

is slightly different from thenatural frequency,then the system exhibits a beat.In Problems

19 through 23 we explore the effect of some nonsinusoidal periodic forcing functions.

19.

Consider the initial value problem

′′

+ y = f (t), y(0) = 0, y

′

(0) = 0,

where

f (t) = u

(t) + 2

k=1

(−1)

kπ

(t).

(a) Draw the graph of f (t) on an interval such as 0 ≤ t ≤ 6π.

(b) Find the solution of the initial value problem.

and explain why it behaves as it does.

(d) Investigate how the solution changes as n increases. What happens as n →∞?

20.

Consider the initial value problem

′′

+ 0.1y

′

+ y = f (t), y(0) = 0, y

′

(0) = 0,

where f (t) is the same as in Problem 19.

(a) Plot the graph of the solution. Use a large enough value of n and a long enough

t-interval so that the transient part of the solution has become negligible and the steady

state is clearly shown.

(b) Estimate the amplitude and frequency of the steady state part of the solution.

oscillator.

21.

Consider the initial value problem

′′

+ y = g(t), y(0) = 0, y

′

(0) = 0,

where

g(t) = u

(t) +

k=1

(−1)

kπ

(t).

(a) Draw the graph of g(t) on an interval such as 0 ≤ t ≤ 6π. Compare the graph with

that of f (t) in Problem 19(a).

(b) Find the solution of the initial value problem.

and explain why it behaves as it does. Compare it with the solution of Problem 19.

(d) Investigate how the solution changes as n increases. What happens as n →∞?

22.

Consider the initial value problem

′′

+ 0.1y

′

+ y = g(t), y(0) = 0, y

′

(0) = 0,

where g(t) is the same as in Problem 21.

(a) Plot the graph of the solution. Use a large enough value of n and a long enough t-

interval so that the transient part of the solution has become negligible and the steady

state is clearly shown.

(b) Estimate the amplitude and frequency of the steady state part of the solution.

August 7, 2012 21:04 c06 Sheet number 35 Page number 343 cyan black

6.5 Impulse Functions 343

a sinusoidally forced oscillator.

23.

Consider the initial value problem

′′

+ y = h(t), y(0) = 0, y

′

(0) = 0,

where

f (t) = u

(t) + 2

k=1

(−1)

11k/4

(t).

Observe that this problem is identical to Problem 19 except that the frequency of the

forcing term has been increased somewhat.

(a) Find the solution of this initial value problem.

(b) Let n ≥ 33 and plot the solution for 0 ≤ t ≤ 90 or longer. Your plot should show a

clearly recognizable beat.

oscillator.

(d) For a sinusoidally forced oscillator, it was shown in Section 3.8 that the “slow fre-

quency” is given by |ω − ω

|/2, where ω

is the natural frequency of the system and ω is

the forcing frequency. Similarly, the “fast frequency” is (ω + ω

)/2. Use these expressions

to calculate the “fast period” and the “slow period” for the oscillator in this problem. How

well do the results compare with your estimates from part (c)?

6.5 Impulse Functions

In some applications it is necessary to deal with phenomena of an impulsive nature—

for example, voltages or forces of large magnitude that act over very short time

intervals. Such problems often lead to differential equations of the form

′′

+ by

′

+ cy = g(t), (1)

where g(t) is large during a short interval t

− τ<t < t

+ τ for some τ>0, and is

otherwise zero.

The integral I(τ), deﬁned by

I(τ) =

+τ

−τ

g(t) dt, (2)

or, since g(t) = 0 outside of the interval (t

− τ, t

+ τ), by

I(τ) =

∞

−∞

g(t) dt, (3)

is a measure of the strength of the forcing function. In a mechanical system, where

g(t) is a force, I(τ) is the total impulse of the force g(t) over the time interval

− τ, t

+ τ). Similarly, if y is the current in an electric circuit and g(t) is the time

derivative of the voltage, then I(τ) represents the total voltage impressed on the

circuit during the interval (t

− τ, t

+ τ).

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344 Chapter 6. The Laplace Transform

In particular, let us suppose that t

is zero and that g(t) is given by

g(t) = d

(t) =

1/(2τ), −τ<t <τ,

0, t ≤−τ or t ≥ τ,

(4)

where τ is a small positive constant (see Figure 6.5.1). According to Eq. (2) or (3),

it follows immediately that in this case, I(τ) = 1 independent of the value of τ,

as long as τ ̸= 0. Now let us idealize the forcing function d

by prescribing it to act

over shorter and shorter time intervals; that is, we require that τ → 0

, as indicated

in Figure 6.5.2. As a result of this limiting operation, we obtain

lim

τ→0

(t) = 0, t ̸= 0. (5)

Further, since I(τ) = 1 for each τ ̸= 0, it follows that

lim

τ→0

I(τ) = 1. (6)

ττ

FIGURE 6.5.1 Graph of y = d

(t).

FIGURE 6.5.2 Graphs of y = d

(t) as τ → 0

Equations (5) and (6) can be used to deﬁne an idealized unit impulse function δ,

which imparts an impulse of magnitude one at t = 0 but is zero for all values of t

other than zero. That is, the “function” δ is deﬁned to have the properties

δ(t) = 0, t ̸= 0; (7)

∞

−∞

δ(t) dt = 1. (8)

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6.5 Impulse Functions 345

There is no ordinary function of the kind studied in elementary calculus that satisﬁes

both Eqs. (7) and (8). The “function” δ, deﬁned by those equations, is an example of

what are known as generalized functions;it is usually called the Dirac

delta function.

Since δ(t) corresponds to a unit impulse at t = 0, a unit impulse at an arbitrary point

t = t

is given by δ(t − t

). From Eqs. (7) and (8), it follows that

δ(t − t

) = 0, t ̸= t

; (9)

∞

−∞

δ(t − t

) dt = 1. (10)

The delta function does not satisfy the conditions of Theorem 6.1.2, but its Laplace

transform can nevertheless be formally deﬁned. Since δ(t) is deﬁned as the limit

of d

(t) as τ → 0

, it is natural to deﬁne the Laplace transform of δ as a similar

limit of the transform of d

. In particular, we will assume that t

> 0 and will deﬁne

L{δ(t − t

)} by the equation

L{δ(t − t

)}= lim

τ→0

L{d

(t − t

)}. (11)

To evaluate the limit in Eq. (11), we ﬁrst observe that if τ<t

, which must eventually

be the case as τ → 0

, then t

− τ>0. Since d

(t − t

) is nonzero only in the interval

from t

− τ to t

+ τ, we have

L{d

(t − t

)}=

∞

−st

(t − t

) dt

+τ

−τ

−st

(t − t

) dt.

Substituting for d

(t − t

) from Eq. (4), we obtain

L{d

(t − t

)}=

2τ

+τ

−τ

−st

dt =−

2sτ

−st

t=t

+τ

t=t

−τ

2sτ

−st

sτ

− e

−sτ

)

L{d

(t − t

)}=

sinh sτ

sτ

−st

. (12)

PaulA. M. Dirac (1902–1984),Englishmathematical physicist,received his Ph.D. from Cambridge in 1926

and was professor of mathematics there until 1969. He was awarded the Nobel Prize for Physics in 1933

(with Erwin Schrödinger) for fundamental work in quantum mechanics. His most celebrated result was the

relativistic equation for the electron, published in 1928. From this equation he predicted the existence of

an“anti-electron,”or positron, which was ﬁrst observed in 1932. Following his retirement from Cambridge,

Dirac moved to the United States and held a research professorship at Florida State University.

August 7, 2012 21:04 c06 Sheet number 38 Page number 346 cyan black

346 Chapter 6. The Laplace Transform

The quotient (sinh sτ)/sτ is indeterminate as τ → 0

, but its limit can be evaluated

by L’Hôpital’s

rule. We obtain

lim

τ→0

sinh sτ

sτ

= lim

τ→0

s cosh sτ

= 1.

Then from Eq. (11) it follows that

L{δ(t − t

)}=e

−st

. (13)

Equation (13) deﬁnes

L{δ(t − t

)} for any t

> 0. We extend this result, to allow t

be zero, by letting t

→ 0

on the right side of Eq. (13); thus

L{δ(t)}= lim

→0

−st

= 1. (14)

In a similar way, it is possible to deﬁne the integral of the product of the delta

function and any continuous function f . We have

∞

−∞

δ(t − t

)f (t) dt = lim

τ→0

∞

−∞

(t − t

)f (t) dt. (15)

Using the deﬁnition (4) of d

(t) and the mean value theorem for integrals, we ﬁnd

that

∞

−∞

(t − t

)f (t) dt =

2τ

+τ

−τ

f (t) dt

2τ

· 2τ · f (t

∗

) = f (t

∗

where t

− τ<t

∗

< t

+ τ. Hence t

∗

→ t

as τ → 0

,and it follows from Eq. (15) that

∞

−∞

δ(t − t

)f (t) dt = f (t

). (16)

The following example illustrates the use of the delta function in solving an initial

value problem with an impulsive forcing function.

EXAMPLE

Find the solution of the initial value problem

′′

+ y

′

+ 2y = δ(t − 5), (17)

y(0) = 0, y

′

(0) = 0. (18)

This initial value problem arises from the study of the same electric circuit or mechanical

oscillator as in Example 1 of Section 6.4. The only difference is in the forcing term.

To solve the given problem, we take the Laplace transform of the differential equation and

use the initial conditions, obtaining

(2s

+ s + 2)Y (s) = e

−5s

Marquis Guillaume de L’Hôpital (1661–1704) was a French nobleman with deep interest in mathematics.

For a time he employed Johann Bernoulli as his private tutor in calculus. L’Hôpital published the ﬁrst

textbook on differential calculus in 1696; in it appears the differentiation rule that is named for him.

August 7, 2012 21:04 c06 Sheet number 39 Page number 347 cyan black

6.5 Impulse Functions 347

Thus

Y(s) =

−5s

+ s + 2

−5s

s +

. (19)

By Theorem 6.3.2 or from line 9 of Table 6.2.1,

−1

s +

√

−t/4

sin

√

t. (20)

Hence, by Theorem 6.3.1, we have

y = L

−1

{Y(s)}=

√

(t)e

−(t−5)/4

sin

√

(t − 5), (21)

which is the formal solution of the given problem. It is also possible to write y in the form

y =

⎧

⎪

⎨

⎪

⎩

0, t < 5,

√

−(t−5)/4

sin

√

(t − 5), t ≥ 5.

(22)

The graph of Eq. (22) is shown in Figure 6.5.3. Since the initial conditions at t = 0 are

homogeneous and there is no external excitation until t = 5,there is no response in the interval

0 < t < 5. The impulse at t = 5 produces a decaying oscillation that persists indeﬁnitely. The

response is continuous at t = 5 despite the singularity in the forcing function at that point.

However, the ﬁrst derivative of the solution has a jump discontinuity at t = 5, and the second

derivative has an inﬁnite discontinuity there. This is required by the differential equation (17),

since a singularity on one side of the equation must be balanced by a corresponding singularity

on the other side.

0.3

0.2

0.1

5 10 15 20

–0.1

FIGURE 6.5.3 Solution of the initial value problem (17), (18):

′′

+ y

′

+ 2y = δ(t − 5), y(0) = 0, y

′

(0) = 0.

In dealing with problems that involve impulsive forcing, the use of the delta

function usually simpliﬁes the mathematical calculations, often quite signiﬁcantly.

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348 Chapter 6. The Laplace Transform

However, if the actual excitation extends over a short, but nonzero, time interval,

then an error will be introduced by modeling the excitation as taking place instan-

taneously. This error may be negligible, but in a practical problem it should not be

dismissed without consideration. In Problem 16 you are asked to investigate this

issue for a simple harmonic oscillator.

PROBLEMS In each of Problems 1 through 12:

(a) Find the solution of the given initial value problem.

(b) Draw a graph of the solution.

′′

+ 2y

′

+ 2y = δ(t − π); y(0) = 1, y

′

(0) = 0

′′

+ 4y = δ(t − π) − δ(t − 2π); y(0) = 0, y

′

(0) = 0

′′

+ 3y

′

+ 2y = δ(t − 5) + u

(t); y(0) = 0, y

′

(0) = 1/2

′′

− y =−20δ(t − 3); y(0) = 1, y

′

(0) = 0

′′

+ 2y

′

+ 3y = sin t + δ(t − 3π); y(0) = 0, y

′

(0) = 0

′′

+ 4y = δ(t − 4π); y(0) = 1/2, y

′

(0) = 0

′′

+ y = δ(t − 2π) cos t; y(0) = 0, y

′

(0) = 1

′′

+ 4y = 2δ(t − π/4); y(0) = 0, y

′

(0) = 0

′′

+ y = u

π/2

(t) + 3δ(t − 3π/2) − u

2π

(t); y(0) = 0, y

′

(0) = 0

10.

′′

+ y

′

+ 4y = δ(t − π/6) sin t; y(0) = 0, y

′

(0) = 0

11.

′′

+ 2y

′

+ 2y = cos t + δ(t − π/2); y(0) = 0, y

′

(0) = 0

12.

(4)

− y = δ(t − 1); y(0) = 0, y

′

(0) = 0, y

′′

(0) = 0, y

′′′

(0) = 0

13.

Consider again the system in Example 1 of this section, in which an oscillation is excited

by a unit impulse at t = 5. Suppose that it is desired to bring the system to rest again after

exactly one cycle—that is, when the response ﬁrst returns to equilibrium moving in the

positive direction.

(a) Determine the impulse kδ(t − t

) that should be applied to the system in order to

accomplish this objective. Note that k is the magnitude of the impulse and t

is the time

of its application.

(b) Solve the resulting initial value problem, and plot its solution to conﬁrm that it

behaves in the speciﬁed manner.

14.

Consider the initial value problem

′′

+ γy

′

+ y = δ(t − 1), y(0) = 0, y

′

(0) = 0,

where γ is the damping coefﬁcient (or resistance).

(a) Let γ =

. Find the solution of the initial value problem and plot its graph.

(b) Find the time t

at which the solution attains its maximum value. Also ﬁnd the

maximum value y

of the solution.

and repeat parts (a) and (b).

(d) Determine how t

and y

vary as γ decreases. What are the values of t

and y

when

γ = 0?

15.

Consider the initial value problem

′′

+ γy

′

+ y = kδ(t − 1), y(0) = 0, y

′

(0) = 0,

August 7, 2012 21:04 c06 Sheet number 41 Page number 349 cyan black

6.5 Impulse Functions 349

where k is the magnitude of an impulse at t = 1, and γ is the damping coefﬁcient (or

resistance).

(a) Let γ =

. Find the value of k for which the response has a peak value of 2; call this

value k

(b) Repeat part (a) for γ =

varies as γ decreases. What is the value of k

when γ = 0?

16.

Consider the initial value problem

′′

+ y = f

(t), y(0) = 0, y

′

(0) = 0,

where f

(t) =[u

4−k

(t) − u

4+k

(t)]/2k with 0 < k ≤ 1.

(a) Find the solution y = φ(t, k) of the initial value problem.

(b) Calculate lim

k→0

φ(t, k) from the solution found in part (a).

k→0

(t) = δ(t − 4). Find the solution φ

(t) of the given initial value

problem with f

(t) replaced by δ(t − 4). Is it true that φ

(t) = lim

k→0

φ(t, k)?

(d) Plot φ(t,1/2), φ(t,1/4), and φ

(t) on the same axes. Describe the relation between

φ(t, k) and φ

(t).

Problems 17 through 22 deal with the effect of a sequence of impulses on an undamped

oscillator. Suppose that

′′

+ y = f (t), y(0) = 0, y

′

(0) = 0.

For each of the following choices for f (t):

(a) Try to predict the nature of the solution without solving the problem.

(b) Test your prediction by ﬁnding the solution and drawing its graph.

17.

f (t) =

k=1

δ(t − kπ) 18. f (t) =

k=1

(−1)

k+1

δ(t − kπ)

19.

f (t) =

k=1

δ(t − kπ/2) 20. f (t) =

k=1

(−1)

k+1

δ(t − kπ/2)

21.

f (t) =

k=1

δ[t − (2k − 1)π] 22. f (t) =

k=1

(−1)

k+1

δ(t − 11k/4)

23.

The position of a certain lightly damped oscillator satisﬁes the initial value problem

′′

+ 0.1y

′

+ y =

k=1

(−1)

k+1

δ(t − kπ), y(0) = 0, y

′

(0) = 0.

Observe that, except for the damping term, this problem is the same as Problem 18.

(a) Try to predict the nature of the solution without solving the problem.

(b) Test your prediction by ﬁnding the solution and drawing its graph.

24.

Proceed as in Problem 23 for the oscillator satisfying

′′

+ 0.1y

′

+ y =

k=1

δ[t − (2k − 1)π], y(0) = 0, y

′

(0) = 0.

Observe that, except for the damping term, this problem is the same as Problem 21.

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350 Chapter 6. The Laplace Transform

25. (a) By the method of variation of parameters, show that the solution of the initial value

problem

′′

+ 2y

′

+ 2y = f (t); y(0) = 0, y

′

(0) = 0

y =

−(t−τ)

f (τ) sin(t − τ) dτ.

(b) Show that if f (t) = δ(t − π), then the solution of part (a) reduces to

y = u

(t)e

−(t−π)

sin(t − π).

f (t) = δ(t − π), and conﬁrm that the solution agrees with the result of part (b).

6.6 The Convolution Integral

Sometimes it is possible to identify a Laplace transform H(s) as the product of two

other transforms F(s ) and G(s), the latter transforms corresponding to known func-

tions f and g, respectively. In this event, we might anticipate that H(s) would be

the transform of the product of f and g. However, this is not the case; in other

words, the Laplace transform cannot be commuted with ordinary multiplication.

On the other hand, if an appropriately deﬁned “generalized product” is introduced,

then the situation changes, as stated in the following theorem.

Theorem 6.6.1

If F(s) = L{f (t)} and G(s) = L{g(t)} both exist for s > a ≥ 0, then

H(s) = F(s)G(s) =

L{h(t)}, s > a, (1)

where

h(t) =

f (t − τ)g(τ) dτ =

f (τ)g(t − τ) dτ. (2)

The function h is known as the convolution of f and g; the integrals in Eq. (2) are

called convolution integrals.

The equality of the two integrals in Eq. (2) follows by making the change of vari-

able t − τ = ξ in the ﬁrst integral. Before giving the proof of this theorem, let us

make some observations about the convolution integral. According to this theorem,

the transform of the convolution of two functions, rather than the transform of their

ordinary product, is given by the product of the separate transforms. It is conven-

tional to emphasize that the convolution integral can be thought of as a “generalized

product” by writing

h(t) = (f ∗ g)(t). (3)

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6.6 The Convolution Integral 351

In particular, the notation (f ∗ g)(t) serves to indicate the ﬁrst integral appearing

in Eq. (2).

The convolution f ∗ g has many of the properties of ordinary multiplication. For

example, it is relatively simple to show that

f ∗ g = g ∗ f (commutative law) (4)

f ∗ (g

+ g

) = f ∗ g

+ f ∗ g

(distributive law) (5)

(f ∗ g) ∗ h = f ∗ (g ∗ h) (associative law) (6)

f ∗ 0 = 0 ∗ f = 0. (7)

In Eq. (7) the zeros denote not the number 0 but the function that has the value 0

for each value of t. The proofs of these properties are left to you as exercises.

However,there are other properties ofordinary multiplication that theconvolution

integral does not have. For example, it is not true in general that f ∗ 1 is equal to f .

To see this, note that

(f ∗ 1)(t) =

f (t − τ) · 1 dτ =

f (t − τ) dτ.

If, for example, f (t) = cos t, then

(f ∗ 1)(t) =

cos(t − τ) dτ =−sin(t − τ)

τ=t

τ=0

=−sin 0 + sin t

= sin t.

Clearly, (f ∗ 1)(t) ̸= f (t) in this case. Similarly, it may not be true that f ∗ f is

nonnegative. See Problem 3 for an example.

Convolution integrals arise in various applications in which the behavior of the

system at time t depends not only on its state at time t but also on its past history.

Systemsofthiskind are sometimes called hereditary systems and occurin such diverse

ﬁelds as neutron transport, viscoelasticity, and population dynamics, among others.

Turning now to the proof of Theorem 6.6.1, we note ﬁrst that if

F(s) =

∞

−sξ

f (ξ) dξ

and

G(s) =

∞

−sτ

g(τ) dτ,

then

F(s)G(s) =

∞

−sξ

f (ξ) dξ

∞

−sτ

g(τ) dτ. (8)

Since the integrand of the ﬁrst integral does not depend on the integration variable

of the second, we can write F(s)G(s) as an iterated integral

F(s)G(s) =

∞

−sτ

g(τ)

∞

−sξ

f (ξ) dξ

d τ

∞

g(τ)

∞

−s(ξ+τ)

f (ξ) dξ

d τ. (9)

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352 Chapter 6. The Laplace Transform

The latter integral can be put into a more convenient form by introducing a change

of variable. Let ξ = t − τ, for ﬁxed τ, so that dξ = dt. Further, ξ = 0 corresponds to

t = τ, and ξ =∞corresponds to t =∞; then the integral with respect to ξ in Eq. (9)

is transformed into one with respect to t:

F(s)G(s) =

∞

g(τ)

∞

−st

f (t − τ) dt

d τ. (10)

The iterated integral on the right side of Eq. (10) is carried out over the shaded

wedge-shaped region extending to inﬁnity in the tτ-plane shown in Figure 6.6.1.

Assuming that the order of integration can be reversed, we rewrite Eq. (10) so that

the integration with respect to τ is executed ﬁrst. In this way we obtain

F(s)G(s) =

∞

−st

f (t − τ)g(τ) dτ

dt (11)

F(s)G(s) =

∞

−st

h(t) dt = L{h(t)}, (12)

where h(t) is deﬁned by Eq. (2). This completes the proof of Theorem 6.6.1.

= 0

→

t =

= t

FIGURE 6.6.1 Region of integration in F (s)G(s).

EXAMPLE

Find the inverse transform of

H(s) =

+ a

)

. (13)

It is convenient to think of H(s) as the product of s

−2

and a/(s

+ a

), which, according to

lines 3 and 5 of Table 6.2.1, are the transforms of t and sin at, respectively. Hence, by Theorem

6.6.1, the inverse transform of H(s) is

h(t) =

(t − τ) sin aτ dτ =

at − sin at

. (14)

You can verify that the same result is obtained if h(t) is written in the alternative form

h(t) =

τ sin a(t − τ) dτ,

which conﬁrms Eq. (2) in this case. Of course, h(t) can also be found by expanding H(s) in

partial fractions.

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6.6 The Convolution Integral 353

EXAMPLE

Find the solution of the initial value problem

′′

+ 4y = g(t), (15)

y(0) = 3, y

′

(0) =−1. (16)

By taking the Laplace transform of the differential equation and using the initial conditions,

we obtain

Y(s) − 3s + 1 + 4Y(s) = G(s)

Y(s) =

3s − 1

+ 4

G(s)

+ 4

. (17)

Observe that the ﬁrst and second terms on the right side of Eq. (17) contain the dependence

of Y(s) on the initial conditions and forcing function, respectively. It is convenient to write

Y(s) in the form

Y(s) = 3

+ 4

−

+ 4

G(s). (18)

Then, using lines 5 and 6 of Table 6.2.1 and Theorem 6.6.1, we obtain

y = 3 cos 2t −

sin 2t +

sin 2(t − τ)g(τ) dτ. (19)

If a speciﬁc forcing function g is given, then the integral in Eq. (19) can be evaluated (by

numerical means, if necessary).

Example 2 illustrates the power of the convolution integral as a tool for writing

the solution of an initial value problem in terms of an integral. In fact, it is possible

to proceed in much the same way in more general problems. Consider the problem

consisting of the differential equation

′′

+ by

′

+ cy = g(t), (20)

where a, b, and c are real constants and g is a given function, together with the initial

conditions

y(0) = y

, y

′

(0) = y

′

. (21)

The transform approach yields some important insights concerning the structure

of the solution of any problem of this type.

The initial value problem (20),(21) is often referred to as an input–output problem.

The coefﬁcients a, b, and c describe the properties of some physical system, and g(t)

is the input to the system. The values y

and y

′

describe the initial state, and the

solution y is the output at time t.

By taking the Laplace transform of Eq. (20) and using the initial conditions (21),

we obtain

(as

+ bs + c)Y(s) − (as + b)y

− ay

′

= G(s).

If we let

7(s) =

(as + b)y

+ ay

′

+ bs + c

, 8(s) =

G(s)

+ bs + c

, (22)

then we can write

Y(s) = 7(s) + 8(s). (23)

Consequently,

y = φ(t) + ψ(t), (24)

August 7, 2012 21:04 c06 Sheet number 46 Page number 354 cyan black

354 Chapter 6. The Laplace Transform

where φ(t) = L

−1

{7(s)} and ψ(t) = L

−1

{8(s)}. Observe that φ(t) is the solution of

the initial value problem

′′

+ by

′

+ cy = 0, y(0) = y

, y

′

(0) = y

′

, (25)

obtained from Eqs. (20) and (21) by setting g(t) equal to zero. Similarly, ψ(t) is the

solution of

′′

+ by

′

+ cy = g(t), y(0) = 0, y

′

(0) = 0, (26)

in which the initial values y

and y

′

are each replaced by zero.

Once speciﬁc values of a, b, and c are given, we can ﬁnd φ(t) =

−1

{7(s)} by using

Table 6.2.1, possibly in conjunction with a translation or a partial fraction expansion.

To ﬁnd ψ(t) =

−1

{8(s)}, it is convenient to write 8(s) as

8(s) = H(s)G(s), (27)

where H(s) = (as

+ bs + c)

−1

. The function H is known as the transfer function

and depends only on the properties of the system under consideration; that is, H(s) is

determined entirely by the coefﬁcients a, b, and c. On the other hand, G (s ) depends

only on the external excitation g(t) that is applied to the system. By the convolution

theorem we can write

ψ(t) =

−1

{H(s)G(s)}=

h(t − τ)g(τ) dτ, (28)

where h(t) =

−1

{H(s)}, and g(t) is the given forcing function.

To obtain a better understanding of the signiﬁcance of h(t), we consider the case in

which G(s) = 1; consequently, g(t) = δ(t) and 8(s) = H(s). This means that y = h(t)

is the solution of the initial value problem

′′

+ by

′

+ cy = δ(t), y(0) = 0, y

′

(0) = 0, (29)

obtained from Eq. (26) by replacing g(t) by δ(t). Thus h(t) is the response of the

system to a unit impulse applied at t = 0, and it is natural to call h(t) the impulse

response of the system. Equation (28) then says that ψ(t) is the convolution of the

impulse response and the forcing function.

Referring to Example 2, we note that in that case, the transfer function is

H(s) = 1/(s

+ 4) and the impulse response is h(t) = (sin 2t)/2. Also, the ﬁrst two

terms on the right side of Eq. (19) constitute the function φ(t), the solution of the

corresponding homogeneous equation that satisﬁes the given initial conditions.

PROBLEMS 1. Establish the commutative, distributive, and associative properties of the convolution

integral.

(a) f ∗g = g ∗ f

(b) f ∗(g

+ g

) = f ∗g

+ f ∗ g

This terminology arises from the fact that H(s) is the ratio of the transforms of the output and the input

of the problem (26).

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6.6 The Convolution Integral 355

2. Find an example different from the one in the text showing that (f ∗ 1)(t) need not be

equal to f (t).

3. Show, by means of the example f (t) = sin t, that f ∗f is not necessarily nonnegative.

In each of Problems 4 through 7, ﬁnd the Laplace transform of the given function.

4. f (t) =

(t − τ)

cos 2τ dτ 5. f (t) =

−(t−τ)

sin τ dτ

6. f (t) =

(t − τ)e

dτ 7. f (t) =

sin(t − τ) cos τ dτ

In each of Problems 8 through 11, ﬁnd the inverse Laplace transform of the given function by

using the convolution theorem.

8. F(s) =

+ 1)

9. F(s) =

(s + 1)(s

+ 4)

10. F(s) =

(s + 1)

+ 4)

11. F(s) =

G(s)

+ 1

12. (a) If f (t) = t

and g(t) = t

, where m and n are positive integers, show that

f ∗g = t

m+n+1

(1 − u)

du.

(b) Use the convolution theorem to show that

(1 − u)

du =

m!n!

(m + n + 1)!

necessarily integers.

In each of Problems 13 through 20, express the solution of the given initial value problem in

terms of a convolution integral.

13. y

′′

+ ω

y = g(t); y(0) = 0, y

′

(0) = 1

14. y

′′

+ 2y

′

+ 2y = sin αt; y(0) = 0, y

′

(0) = 0

15. 4y

′′

+ 4y

′

+ 17y = g(t); y(0) = 0, y

′

(0) = 0

16. y

′′

+ y

′

y = 1 − u

(t); y(0) = 1, y

′

(0) =−1

17. y

′′

+ 4y

′

+ 4y = g(t); y(0) = 2, y

′

(0) =−3

18. y

′′

+ 3y

′

+ 2y = cos αt; y(0) = 1, y

′

(0) = 0

19. y

(4)

− y = g(t); y(0) = 0, y

′

(0) = 0, y

′′

(0) = 0, y

′′′

(0) = 0

20. y

(4)

+ 5y

′′

+ 4y = g(t); y(0) = 1, y

′

(0) = 0, y

′′

(0) = 0, y

′′′

(0) = 0

21. Consider the equation

φ(t) +

k(t − ξ)φ(ξ) dξ = f (t),

in which f and k are known functions, and φ is to be determined. Since the unknown

function φ appears under an integral sign, the given equation is called an integral equation;

inparticular,it belongs toa class of integral equations known asVolterraintegral equations.

Take the Laplace transform of the given integral equation and obtain an expression for

L{φ(t)} in terms of the transforms L{f (t)} and L{k(t)} of the given functions f and k.The

inverse transform of L{φ(t)} is the solution of the original integral equation.

August 7, 2012 21:04 c06 Sheet number 48 Page number 356 cyan black

356 Chapter 6. The Laplace Transform

22. Consider the Volterra integral equation (see Problem 21)

φ(t) +

(t − ξ)φ(ξ) dξ = sin 2t. (i)

(a) Solve the integral equation (i) by using the Laplace transform.

(b) By differentiating Eq. (i) twice, show that φ(t) satisﬁes the differential equation

′′

(t) + φ(t) =−4 sin 2t.

Show also that the initial conditions are

φ(0) = 0, φ

′

(0) = 2.

the one in part (a).

In each of Problems 23 through 25:

(a) Solve the given Volterra integral equation by using the Laplace transform.

(b) Convert the integral equation into an initial value problem, as in Problem 22(b).

one in part (a).

23. φ(t) +

(t − ξ)φ(ξ) dξ = 1 24. φ(t) −

(t − ξ)φ(ξ) dξ = 1

25. φ(t) + 2

cos(t − ξ)φ(ξ) dξ = e

−t

There are also equations, known as integro-differential equations, in which both derivatives

and integrals of the unknown function appear. In each of Problems 26 through 28:

(a) Solve the given integro-differential equation by using the Laplace transform.

(b) By differentiating the integro-differential equation a sufﬁcient number of times, convert

it into an initial value problem.

one in part (a).

26. φ

′

(t) +

(t − ξ)φ(ξ) dξ = t, φ(0) = 0

27. φ

′

(t) −

(t − ξ)

φ(ξ) dξ =−t, φ(0) = 1

28. φ

′

(t) + φ(t) =

sin(t − ξ)φ(ξ) dξ, φ(0) = 1

29. The Tautochrone. A problem of interest in the history of mathematics is that of ﬁnding

the tautochrone

—the curve down which a particle will slide freely under gravity alone,

reaching the bottom in the same time regardless of its starting point on the curve. This

problem arose in the construction of a clock pendulum whose period is independent of

the amplitude of its motion. The tautochrone was found by Christian Huygens (1629–

1695) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using

analytical arguments. Bernoulli’s solution (in 1690) was one of the ﬁrst occasions on which

The word “tautochrone” comes from the Greek words tauto, which means “same,” and chronos, which

means “time.”

August 7, 2012 21:04 c06 Sheet number 49 Page number 357 cyan black

6.6 The Convolution Integral 357

a differential equation was explicitly solved. The geometric conﬁguration is shown in

Figure 6.6.2. The starting point P(a, b) is joined to the terminal point (0, 0) by the arc C.

Arc length s is measured from the origin, and f (y) denotes the rate of change of s with

respect to y:

f (y) =

1 +

(

1/2

. (i)

Then it follows from the principle of conservation of energy that the time T(b) required

for a particle to slide from P to the origin is

T(b) =

f (y)

b − y

dy. (ii)

P(a, b)

FIGURE 6.6.2 The tautochrone.

(a) Assume that T(b) = T

, a constant, for each b. By taking the Laplace transform of

Eq. (ii) in this case, and using the convolution theorem, show that

F(s) =

√

; (iii)

then show that

f (y) =

√

. (iv)

Hint: See Problem 31 of Section 6.1.

(b) Combining Eqs. (i) and (iv), show that

2α − y

, (v)

where α = gT

/π

(θ/2) to solve Eq. (v), and show that

x = α(θ + sin θ), y = α(1 − cos θ). (vi)

Equations (vi) can be identiﬁed as parametric equations of a cycloid.Thus the tautochrone

is an arc of a cycloid.

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358 Chapter 6. The Laplace Transform

REFERENCES The books listed below contain additional information on the Laplace transform and its applications.

Churchill, R. V., Operational Mathematics (3rd ed.) (New York: McGraw-Hill, 1971).

Doetsch, G., Introduction to the Theory and Application of the Laplace Transform (trans. W. Nader) (New

York: Springer, 1974).

Kaplan,W., Operational Methods for Linear Systems (Reading, MA:Addison-Wesley, 1962).

Kuhﬁttig, P. K. F., Introduction to the Laplace Transform (New York: Plenum, 1978).

Miles, J. W., Integral Transforms in Applied Mathematics (London: Cambridge University Press, 2008).

Rainville, E. D.,The Laplace Transform: An Introduction (New York: Macmillan, 1963).

Each of the books just mentioned contains a table of transforms. Extensive tables are also available.

See, for example,

Erdelyi,A. (ed.), Tables of Integral Transforms (Vol. 1) (New York: McGraw-Hill, 1954).

Roberts, G. E., and Kaufman, H., Table of Laplace Transforms (Philadelphia: Saunders, 1966).

A further discussion of generalized functions can be found in

Lighthill, M. J., An Introduction to Fourier Analysis and Generalized Functions (London: Cambridge

University Press, 1958).

August 7, 2012 21:04 c07 Sheet number 1 Page number 359 cyan black

359

CHAPTER

Systems of First

Order Linear

Equations

There are many physical problems that involve a number of separate elements

linked together in some manner. For example, electrical networks have this char-

acter, as do many problems in mechanics and in other ﬁelds. In these and similar

cases, the corresponding mathematical problem consists of a system of two or

more differential equations, which can always be written as ﬁrst order equations.

In this chapter we focus on systems of ﬁrst order linear equations, and in par-

ticular equations having constant coefﬁcients, utilizing some of the elementary

aspects of linear algebra to unify the presentation. In many respects this chap-

ter follows the same lines as the treatment of second order linear equations in

Chapter 3.

7.1 Introduction

Systems of simultaneous ordinary differential equations arise naturally in problems

involving several dependent variables, each of which is a function of the same single

independent variable. We will denote the independent variable by t and will let

, x

, ... represent dependent variables that are functions of t. Differentiation

with respect to t will be denoted by a prime.

For example, consider the spring–mass system in Figure 7.1.1. The two masses

move on a frictionless surface under the inﬂuence of external forces F

(t) and F

(t),

and they are also constrained by the three springs whose constants are k

, k

, and k

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360 Chapter 7. Systems of First Order Linear Equations

respectively. Using arguments similar to those in Section 3.7, we ﬁnd the following

equations for the coordinates x

and x

of the two masses:

= k

− x

) − k

+ F

(t)

=−(k

+ k

+ F

(t),

(1)

=−k

− k

− x

) + F

(t)

= k

− (k

+ k

+ F

(t).

A derivation of Eqs. (1) is outlined in Problem 17.

(t) F

(t)

FIGURE 7.1.1 A two-mass, three-spring system.

Next, consider the parallel LRC circuit shown in Figure 7.1.2. Let V be the voltage

drop across the capacitor and I the current through the inductor. Then, referring

to Section 3.7 and to Problem 19 of this section, we can show that the voltage and

current are described by the system of equations

(2)

=−

−

where L is the inductance, C is the capacitance, and R is the resistance.

FIGURE 7.1.2 A parallel LRC circuit.

One reason why systems of ﬁrst order equations are particularly important is that

equations of higher order can always be transformed into such systems.This is usually

required if a numerical approach is planned, because almost all codes for generat-

ing numerical approximations to solutions of differential equations are written for

August 7, 2012 21:04 c07 Sheet number 3 Page number 361 cyan black

7.1 Introduction 361

systems of ﬁrst order equations. The following example illustrates how easy it is to

make the transformation.

EXAMPLE

The motion of a certain spring–mass system (see Example 3 of Section 3.7) is described by the

second order differential equation

′′

+ 0.125u

′

+ u = 0. (3)

Rewrite this equation as a system of ﬁrst order equations.

Let x

= u and x

= u

′

. Then it follows that x

′

= x

. Further, u

′′

= x

′

. Then, by substituting

for u, u

′

, and u

′′

in Eq. (3), we obtain

′

+ 0.125x

+ x

= 0.

Thus x

and x

satisfy the following system of two ﬁrst order differential equations:

′

= x

(4)

′

=−x

− 0.125x

The general equation of motion of a spring–mass system

′′

+ γu

′

+ ku = F(t) (5)

can be transformed into a system of ﬁrst order equations in the same manner. If we

let x

= u and x

= u

′

, and proceed as in Example 1, we quickly obtain the system

′

= x

(6)

′

=−(k/m)x

− (γ/m)x

+ F(t)/m.

To transform an arbitrary nth order equation

(n)

= F(t, y, y

′

, ..., y

(n−1)

) (7)

into a system of n ﬁrst order equations, we extend the method of Example 1 by

introducing the variables x

, x

, ..., x

deﬁned by

= y, x

= y

′

, x

= y

′′

, ..., x

= y

(n−1)

. (8)

It then follows immediately that

′

= x

′

= x

(9)

′

n−1

= x

and, from Eq. (7),

′

= F(t, x

, x

, ..., x

). (10)

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362 Chapter 7. Systems of First Order Linear Equations

Equations (9) and (10) are a special case of the more general system

′

= F

(t, x

, x

, ..., x

′

= F

(t, x

, x

, ..., x

(11)

′

= F

(t, x

, x

, ..., x

In asimilar way,the system (1) can be reduced to a system offour ﬁrst order equations

of the form (11), and the system (2) is already in this form. In fact, systems of the

form (11) include almost all cases of interest, so much of the more advanced theory

of differential equations is devoted to such systems.

A solution of the system (11) on the interval I: α<t <βis a set of n functions

= φ

(t), x

= φ

(t), ..., x

= φ

(t) (12)

that are differentiable at all points in the interval I and that satisfy the system of equa-

tions (11) at all points in this interval. In addition to the given system of differential

equations, there may also be given initial conditions of the form

) = x

, x

) = x

, ..., x

) = x

, (13)

where t

is a speciﬁed value of t in I, and x

, ..., x

are prescribed numbers. The

differential equations (11) and the initial conditions (13) together form an initial

value problem.

A solution (12) can be viewed as a set of parametric equations in an n-dimensional

space. For a given value of t, Eqs. (12) give values for the coordinates x

, ..., x

of a

point inthe space. As t changes,the coordinates ingeneral also change.The collection

of points corresponding to α<t <βforms a curve in the space. It is often helpful to

think of the curve as the trajectory, or path, of a particle moving in accordance with

the system of differential equations (11). The initial conditions (13) determine the

starting point of the moving particle.

The following conditions on F

, F

, ..., F

, which are easily checked in speciﬁc

problems,are sufﬁcient to ensure that the initial value problem (11),(13) has a unique

solution. Theorem 7.1.1 is analogous to Theorem 2.4.2, the existence and uniqueness

theorem for a single ﬁrst order equation.

Theorem 7.1.1

Let each of the functions F

, ..., F

and the partial derivatives ∂F

/∂x

, ...,

∂F

/∂x

, ..., ∂F

/∂x

, ..., ∂F

/∂x

be continuous in a region R of tx

···x

space deﬁned by α<t <β, α

< x

<β

, ..., α

< x

<β

, and let the point

, x

, ..., x

) be in R. Then there is an interval |t − t

| < h in which there exists

a unique solution x

= φ

(t), ..., x

= φ

(t) of the system of differential equations

(11) that also satisﬁes the initial conditions (13).

The proof of this theorem can be constructed by generalizing the argument in

Section 2.8, but we do not give it here. However, note that, in the hypotheses of the

theorem, nothing is said about the partial derivatives of F

, ..., F

with respect to

the independent variable t. Also, in the conclusion, the length 2h of the interval in

which the solution exists is not speciﬁed exactly, and in some cases it may be very

August 7, 2012 21:04 c07 Sheet number 5 Page number 363 cyan black

7.1 Introduction 363

short. Finally, the same result can be established on the basis of somewhat weaker but

more complicated hypotheses, so the theorem as stated is not the most general one

known, and the given conditions are sufﬁcient, but not necessary, for the conclusion

to hold.

If each of the functions F

, ..., F

in Eqs. (11) is a linear function of the dependent

variables x

, ..., x

, then the system of equations is said to be linear; otherwise, it is

nonlinear. Thus the most general system of n ﬁrst order linear equations has the form

′

= p

(t)x

+···+p

(t)x

+ g

(t),

′

= p

(t)x

+···+p

(t)x

+ g

(t),

(14)

′

= p

(t)x

+···+p

(t)x

+ g

(t).

If each of the functions g

(t), ..., g

(t) is zero for all t in the interval I, then the

system (14) is said to be homogeneous; otherwise, it is nonhomogeneous. Observe

that the systems (1) and (2) are both linear.The system (1) is nonhomogeneous unless

(t) = F

(t) = 0, while the system (2) is homogeneous. For the linear system (14),

the existence and uniqueness theorem is simpler and also has a stronger conclusion.

It is analogous to Theorems 2.4.1 and 3.2.1.

Theorem 7.1.2

If the functions p

, p

, ..., p

, g

, ..., g

are continuous on an open interval

I: α<t <β, then there exists a unique solution x

= φ

(t), ..., x

= φ

(t) of the

system (14) that also satisﬁes the initial conditions (13), where t

is any point in I,

and x

, ..., x

are anyprescribed numbers. Moreover,the solution existsthroughout

the interval I.

Note that, in contrast to the situation for a nonlinear system, the existence and

uniqueness of the solution of a linear system are guaranteed throughout the interval

in which the hypotheses are satisﬁed. Furthermore, for a linear system the initial

values x

, ..., x

at t = t

are completely arbitrary, whereas in the nonlinear case the

initial point must lie in the region R deﬁned in Theorem 7.1.1.

The rest of this chapter is devoted to systems of linear ﬁrst order equations (non-

linear systems are included in the discussion in Chapters 8 and 9). Our presentation

makes use of matrix notation and assumes that you have some familiarity with the

properties of matrices. The basic facts about matrices are summarized in Sections 7.2

and 7.3, and some more advanced material is reviewed as needed in later sections.

PROBLEMS In each of Problems 1 through 4, transform the given equation into a system of ﬁrst order

equations.

1. u

′′

+ 0.5u

′

+ 2u = 02.u

′′

+ 0.5u

′

+ 2u = 3 sin t

3. t

′′

+ tu

′

+ (t

− 0.25)u = 04. u

(4)

− u = 0

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364 Chapter 7. Systems of First Order Linear Equations

In each of Problems 5 and 6, transform the given initial value problem into an initial value

problem for two ﬁrst order equations.

5. u

′′

+ 0.25u

′

+ 4u = 2 cos 3t, u(0) = 1, u

′

(0) =−2

6. u

′′

+ p(t)u

′

+ q(t)u = g(t), u(0) = u

, u

′

(0) = u

′

7. Systems of ﬁrst order equations can sometimes be transformed into a single equation of

higher order. Consider the system

′

=−2x

+ x

, x

′

= x

− 2x

(a) Solve the ﬁrst equation for x

and substitute into the second equation,thereby obtain-

ing a second order equation for x

. Solve this equation for x

and then determine x

also.

(b) Find the solution of the given system that also satisﬁes the initial conditions x

(0) = 2,

(0) = 3.

and x

obtained in part (b).

In each of Problems 8 through 12, proceed as in Problem 7.

(a) Transform the given system into a single equation of second order.

(b) Find x

and x

that also satisfy the given initial conditions.

-plane for t ≥ 0.

8. x

′

= 3x

− 2x

, x

(0) = 3

′

= 2x

− 2x

, x

(0) =

9. x

′

= 1.25x

+ 0.75x

, x

(0) =−2

′

= 0.75x

+ 1.25x

, x

(0) = 1

10. x

′

= x

− 2x

, x

(0) =−1

′

= 3x

− 4x

, x

(0) = 2

11. x

′

= 2x

, x

(0) = 3

′

=−2x

, x

(0) = 4

12. x

′

=−0.5x

+ 2x

, x

(0) =−2

′

=−2x

− 0.5x

, x

(0) = 2

13. Transform Eqs. (2) for the parallel circuit into a single second order equation.

14. Show that if a

, a

, and a

are constants with a

and a

not both zero, and if the

functions g

and g

are differentiable, then the initial value problem

′

= a

+ a

+ g

(t), x

(0) = x

′

= a

+ a

+ g

(t), x

(0) = x

can be transformed into an initial value problem for a single second order equation. Can

the same procedure be carried out if a

, ..., a

are functions of t?

15. Consider the linear homogeneous system

′

= p

(t)x + p

(t)y,

′

= p

(t)x + p

(t)y.

Show that if x = x

(t), y = y

(t) and x = x

(t), y = y

(t) are two solutions of the given

system, then x = c

(t) + c

(t), y = c

(t) + c

(t) is also a solution for any constants

and c

. This is the principle of superposition.

August 7, 2012 21:04 c07 Sheet number 7 Page number 365 cyan black

7.1 Introduction 365

16. Let x = x

(t), y = y

(t) and x = x

(t), y = y

(t) be any two solutions of the linear

nonhomogeneous system

′

= p

(t)x + p

(t)y + g

(t),

′

= p

(t)x + p

(t)y + g

(t).

Show that x = x

(t) − x

(t), y = y

(t) − y

(t) is a solution of the corresponding homoge-

neous system.

17. Equations (1) can be derived by drawing a free-body diagram showing the forces acting

on each mass. Figure 7.1.3a shows the situation when the displacements x

and x

of the

two masses are both positive (to the right) and x

> x

.Then springs 1 and 2 are elongated

and spring 3 is compressed, giving rise to forces as shown in Figure 7.1.3b. Use Newton’s

law (F = ma) to derive Eqs. (1).

(a)

(t)

– x

)

– x

) k

(t)

(b)

FIGURE 7.1.3 (a) The displacements x

and x

are both positive.

(b) The free-body diagram for the spring–mass system.

18. Transform the system (1) into a system of ﬁrst order equations by letting y

= x

, y

= x

′

, and y

= x

′

Electric Circuits. The theory of electric circuits, such as that shown in Figure 7.1.2, consisting

of inductors, resistors, and capacitors, is based on Kirchhoff’s laws: (1) The net ﬂow of current

into each node (or junction) is zero, and (2) the net voltage drop around each closed loop

is zero. In addition to Kirchhoff’s laws, we also have the relation between the current I in

amperes through each circuit element and the voltage drop V in volts across the element:

V = RI, R = resistance in ohms;

= I, C = capacitance in farads

;

= V, L = inductance in henrys.

Kirchhoff’s laws and the current–voltage relation for each circuit element provide a system of

algebraic and differential equations from which the voltage and current throughout the circuit

can be determined. Problems 19 through 21 illustrate the procedure just described.

Actual capacitors typically have capacitances measured in microfarads. We use farad as the unit for

numerical convenience.

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366 Chapter 7. Systems of First Order Linear Equations

19. Consider the circuit shown in Figure 7.1.2. Let I

, I

, and I

be the currents through the

capacitor, resistor, and inductor, respectively. Likewise, let V

, V

, and V

be the corre-

sponding voltage drops. The arrows denote the arbitrarily chosen directions in which

currents and voltage drops will be taken to be positive.

(a) Applying Kirchhoff’s second law to the upper loop in the circuit, show that

− V

= 0. (i)

In a similar way, show that

− V

= 0. (ii)

(b) Applying Kirchhoff’s ﬁrst law to either node in the circuit, show that

+ I

= 0. (iii)

equations

′

= I

, V

= RI

, LI

′

= V

. (iv)

(d) Eliminate V

, V

, I

, and I

among Eqs. (i) through (iv) to obtain

′

=−I

−

, LI

′

= V

. (v)

Observe that if we omit the subscripts in Eqs. (v), then we have the system (2) of this

section.

20. Consider the circuit shown in Figure 7.1.4. Use the method outlined in Problem 19 to show

that the current I through the inductor and the voltage V across the capacitor satisfy the

system of differential equations

=−I − V,

= 2I − V.

R = 1 ohm

R = 2 ohms

L = 1 henry

C = farad

FIGURE 7.1.4 The circuit in Problem 20.

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7.1 Introduction 367

21. Consider the circuit shown in Figure 7.1.5. Use the method outlined in Problem 19 to show

that the current I through the inductor and the voltage V across the capacitor satisfy the

system of differential equations

=−R

I − V, C

= I −

FIGURE 7.1.5 The circuit in Problem 21.

22. Consider the two interconnected tanks shown in Figure 7.1.6. Tank 1 initially contains

30 gal of water and 25 oz of salt, and Tank 2 initially contains 20 gal of water and 15 oz

of salt. Water containing 1 oz/gal of salt ﬂows into Tank 1 at a rate of 1.5 gal/min. The

mixture ﬂows fromTank 1 to Tank 2 at a rate of 3 gal/min.Water containing 3 oz/gal of salt

also ﬂows into Tank 2 at a rate of 1 gal/min (from the outside). The mixture drains from

Tank 2 at a rate of 4 gal/min, of which some ﬂows back into Tank 1 at a rate of 1.5 gal/min,

while the remainder leaves the system.

(a) Let Q

(t) and Q

(t), respectively, be the amount of salt in each tank at time t. Write

down differential equations and initial conditions that model the ﬂow process. Observe

that the system of differential equations is nonhomogeneous.

(b) Find the values of Q

and Q

for which the system is in equilibrium—that is, does not

change with time. Let Q

and Q

be the equilibrium values. Can you predict which tank

will approach its equilibrium state more rapidly?

= Q

(t) − Q

and x

= Q

(t) − Q

. Determine an initial value problem for x

and x

. Observe that the system of equations for x

and x

is homogeneous.

1.5 gal/min

1 gal/min

3 gal/min

3 oz/gal

2 knaT1 knaT

(t) oz salt

30 gal water

(t) oz salt

20 gal water

2.5 gal/min

1 oz/gal

1.5 gal/min

FIGURE 7.1.6 Two interconnected tanks (Problem 22).

August 7, 2012 21:04 c07 Sheet number 10 Page number 368 cyan black

368 Chapter 7. Systems of First Order Linear Equations

23. Consider two interconnected tanks similar to those in Figure 7.1.6. Initially,Tank 1 contains

60 gal of water and Q

oz of salt, and Tank 2 contains 100 gal of water and Q

oz of salt.

Water containing q

oz/gal of salt ﬂows into Tank 1 at a rate of 3 gal/min. The mixture

in Tank 1 ﬂows out at a rate of 4 gal/min, of which half ﬂows into Tank 2, while the

remainder leaves the system.Water containing q

oz/gal of salt also ﬂows intoTank 2 from

the outside at the rate of 1 gal/min. The mixture in Tank 2 leaves it at a rate of 3 gal/min, of

which some ﬂows back into Tank 1 at a rate of 1 gal/min, while the rest leaves the system.

(a) Draw a diagram that depicts the ﬂow process described above. Let Q

(t) and Q

(t),

respectively,be the amount of salt in each tank at time t. Write down differential equations

and initial conditions for Q

and Q

that model the ﬂow process.

(b) Find the equilibrium values Q

and Q

in terms of the concentrations q

and q

) to obtain Q

= 60 and Q

= 50 as anequilibrium

state?

(d) Describe which equilibrium states are possible for this system for various values of q

and q

7.2 Review of Matrices

For both theoretical and computational reasons, it is advisable to bring some of the

results of matrix algebra

to bear on the initial value problem for a system of linear

differential equations. For reference purposes, this section and the next are devoted

to a brief summary of the facts that will be needed later. More details can be found

in any elementary book on linear algebra. We assume, however, that you are familiar

with determinants and how to evaluate them.

We designate matrices by boldfaced capitals A, B, C, ..., occasionally using bold-

faced Greek capitals #, $, ....A matrix A consists of a rectangular array of numbers,

or elements, arranged in m rows and n columns—that is,

A =

⎛

⎜

⎝

··· a

⎞

⎟

⎠

. (1)

We speak of A as an m × n matrix. Although later in the chapter we will often assume

that the elements of certain matrices are real numbers, in this section we assume that

The properties of matrices were ﬁrst extensively explored in 1858 in a paper by the English algebraist

Arthur Cayley (1821–1895),although the word“matrix”was introduced by his good friend James Sylvester

(1814–1897) in 1850. Cayley did some of his best mathematical work while practicing law from 1849 to

1863; he then became professor of mathematics at Cambridge, a position he held for the rest of his life.

After Cayley’s groundbreaking work,the development of matrix theory proceeded rapidly,with signiﬁcant

contributions by Charles Hermite, Georg Frobenius, and Camille Jordan, among others.

August 7, 2012 21:04 c07 Sheet number 11 Page number 369 cyan black

7.2 Review of Matrices 369

the elements of matrices may be complex numbers. The element lying in the ith row

and jth column is designated by a

, the ﬁrst subscript identifying its row and the

second its column. Sometimes the notation (a

) is used to denote the matrix whose

generic element is a

Associated with each matrix A is the matrix A

, which is known as the transpose

of A and is obtained from A by interchanging the rows and columns of A.Thus,if

A = (a

), then A

= (a

). Also, we will denote by a

the complex conjugate of a

and by

A the matrix obtained from A by replacing each element a

by its conjugate

. The matrix A is called the conjugate of A. It will also be necessary to consider

the transpose of the conjugate matrix

. This matrix is called the adjoint of A and

will be denoted by A

∗

For example, let

A =

32− i

4 + 3i −5 + 2i

Then

34+ 3i

2 − i −5 + 2i

, A =

32+ i

4 − 3i −5 − 2i

∗

34− 3i

2 + i −5 − 2i

We are particularly interested in two somewhat special kinds of matrices: square

matrices, which have the same number of rows and columns—that is, m = n; and

vectors (or column vectors), which can be thought of as n × 1 matrices, or matri-

ces having only one column. Square matrices having n rows and n columns are

said to be of order n. We denote (column) vectors by boldfaced lowercase letters:

x, y, ξ, η, ....The transpose x

of an n × 1 column vector is a 1 × n row vector—that

is, the matrix consisting of one row whose elements are the same as the elements in

the corresponding positions of x.

Prop ert ie s of Mat ri ces .

1. Equality. Two m × n matrices A and B are said to be equal if all corresponding

elements are equal—that is, if a

= b

for each i and j.

2. Zero. The symbol 0 will be used to denote the matrix (or vector) each of whose

elements is zero.

3. Addition. The sum of two m × n matrices A and B is deﬁned as the matrix

obtained by adding corresponding elements:

A + B = (a

) + (b

) = (a

+ b

). (2)

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370 Chapter 7. Systems of First Order Linear Equations

With this deﬁnition, it follows that matrix addition is commutative and associative,

so that

A + B = B +A, A + (B + C) = (A + B) + C. (3)

4. Multiplication by a Number. The product of a matrix A by a real or complex

number α is deﬁned as follows:

αA = α(a

) = (αa

); (4)

that is, each element of A is multiplied by α. The distributive laws

α(A + B) = αA + αB, (α + β)A = αA + βA (5)

are satisﬁed for this type of multiplication. In particular, the negative of A, denoted

by −A, is deﬁned by

−A = (−1)A. (6)

5. Subtraction. The difference A − B of two m × n matrices is deﬁned by

A − B = A + (−B). (7)

Thus

A − B = (a

) − (b

) = (a

− b

), (8)

which is similar to Eq. (2).

6. Multiplication. The product AB of two matrices is deﬁned whenever the num-

ber of columns in the ﬁrst factor is the same as the number of rows in the second.

If A and B are m × n and n × r matrices, respectively, then the product C = AB is an

m × r matrix. The element in the ith row and jth column of C is found by multiplying

each element of the ith row of A by the corresponding element of the jth column of

B and then adding the resulting products. In symbols,

k=1

. (9)

By direct calculation, it can be shown that matrix multiplication satisﬁes the

associative law

(AB)C = A(BC) (10)

and the distributive law

A(B + C) = AB +AC. (11)

However, in general, matrix multiplication is not commutative. For both products

AB and BA to exist and to be of the same size, it is necessary that A and B be square

matrices of the same order. Even in that case the two products are usually unequal,

so that, in general,

AB ̸= BA. (12)

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7.2 Review of Matrices 371

EXAMPLE

To illustrate the multiplication of matrices, and also the fact that matrix multiplication is not

necessarily commutative, consider the matrices

A =

⎛

⎜

⎝

1 −21

02−1

211

⎞

⎟

⎠

, B =

⎛

⎜

⎝

21−1

1 −10

2 −11

⎞

⎟

⎠

From the deﬁnition of multiplication given in Eq. (9), we have

AB =

⎛

⎜

⎝

2 − 2 + 21+ 2 − 1 −1 + 0 + 1

0 + 2 − 20− 2 + 10+ 0 − 1

4 + 1 + 22− 1 − 1 −2 + 0 + 1

⎞

⎟

⎠

⎛

⎜

⎝

220

0 −1 −1

70−1

⎞

⎟

⎠

Similarly

we ﬁnd that

BA =

⎛

⎜

⎝

0 −30

1 −42

4 −54

⎞

⎟

⎠

Clearly, AB ̸= BA.

7. Multiplication of Vectors. There are several ways of forming a product of two

vectors x and y, each with n components. One is a direct extension to n dimensions

of the familiar dot product from physics and calculus; we denote it by x

y and write

y =

i=1

. (13)

The result of Eq. (13) isa realor complex number,and it follows directly fromEq. (13)

that

y = y

x, x

(y + z) = x

y + x

z, (αx)

y = α(x

y) = x

(αy). (14)

There is another vector product that is also deﬁned for any two vectors having the

same number of components. This product, denoted by (x, y), is called the scalar or

inner product and is deﬁned by

(x, y) =

i=1

. (15)

The scalar product is also a real or complex number, and by comparing Eqs. (13) and

(15), we see that

(x, y) = x

y. (16)

Thus,if all the elements of y are real,then the two products (13) and (15) are identical.

From Eq. (15) it follows that

(x, y) =

(y, x), (x, y + z) = (x, y) + (x, z),

(17)

(αx, y) = α(x, y), (x, αy) =

α(x, y).

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372 Chapter 7. Systems of First Order Linear Equations

Note that even if the vector x has elements with nonzero imaginary parts,the scalar

product of x with itself yields a nonnegative real number

(x, x) =

i=1

. (18)

The nonnegative quantity (x, x)

1/2

, often denoted by ∥x∥, is called the length, or

magnitude,ofx.If(x, y) = 0, then the two vectors x and y are said to be orthogonal.

For example, the unit vectors i, j, k of three-dimensional vector geometry form an

orthogonal set. On the other hand, if some of the elements of x are not real, then the

product

x =

i=1

(19)

may not be a real number.

For example, let

x =

⎛

⎜

⎝

−2

1 + i

⎞

⎟

⎠

, y =

⎛

⎜

⎝

2 − i

⎞

⎟

⎠

Then

y = (i)(2 − i) + (−2)(i) + (1 + i)(3) = 4 + 3i,

(x, y) = (i)(2 + i) + (−2)(−i) + (1 + i )(3) = 2 + 7i,

x = (i)

+ (−2)

+ (1 + i)

= 3 + 2i,

(x, x) = (i)(−i ) + (−2)(−2) + (1 + i)(1 − i) = 7.

8. Identity. The multiplicative identity, or simply the identity matrix I, is given by

I =

⎛

⎜

⎝

10··· 0

01··· 0

00··· 1

⎞

⎟

⎠

. (20)

From the deﬁnition of matrix multiplication, we have

AI = IA = A (21)

for any (square) matrix A. Hence the commutative law does hold for square matrices

if one of the matrices is the identity.

9. Inverse. The square matrix A is said to be nonsingular or invertible if there

is another matrix B such that AB = I and BA = I

, where I is the identity. If there is

such

a B, it can be shown that there is only one. It is called the multiplicative inverse,

or simply the inverse, of A, and we write B = A

−1

. Then

−1

= A

−1

A = I. (22)

Matrices that do not have an inverse are called singular or noninvertible.

There are various ways to compute A

−1

from A, assuming that it exists. One way

involves the use of determinants. Associated with each element a

of a given matrix

August 7, 2012 21:04 c07 Sheet number 15 Page number 373 cyan black

7.2 Review of Matrices 373

is the minor M

, which is the determinant of the matrix obtained by deleting the

ith row and jth column of the original matrix—that is, the row and column con-

taining a

. Also associated with each element a

is the cofactor C

deﬁned by the

equation

= (−1)

i+j

. (23)

If B = A

−1

, then it can be shown that the general element b

is given by

detA

. (24)

Although Eq. (24) is not an efﬁcient way

to calculate A

−1

, it does suggest a con-

dition that A must satisfy for it to have an inverse. In fact, the condition is both

necessary and sufﬁcient: A is nonsingular if and only if det A ̸= 0. If detA = 0, then

A is singular.

Another (and usually better) way to compute A

−1

is by means of elementary row

operations. There are three such operations:

1. Interchange of two rows.

2. Multiplication of a row by a nonzero scalar.

3. Addition of any multiple of one row to another row.

The transformation of a matrix by a sequence of elementary row operations is

referred to as row reduction or Gaussian

elimination. Any nonsingular matrix A

can be transformed into the identity I by a systematic sequence of these operations.

It is possible to show that if the same sequence of operations is then performed on I,it

is transformed into A

−1

. It is most efﬁcient to perform the sequence of operations on

both matrices at the same time by forming the augmented matrix A |I. The following

example illustrates the calculation of an inverse matrix in this way.

EXAMPLE

Find the inverse of

A =

⎛

⎜

⎝

1 −1 −1

3 −12

223

⎞

⎟

⎠

For large n the number of multiplications required to evaluate A

−1

by Eq. (24) is proportional to n!. If we

use a more efﬁcient method, such as the row reduction procedure described in this section, the number

of multiplications is proportional only to n

. Even for small values of n (such as n = 4), determinants are

not an economical tool in calculating inverses, and row reduction methods are preferred.

Carl Friedrich Gauss (1777–1855) was born in Brunswick (Germany) and spent most of his life as pro-

fessor of astronomy and director of the Observatory at the University of Göttingen. Gauss made major

contributions to many areas of mathematics, including number theory, algebra, non-Euclidean and dif-

ferential geometry, and analysis, as well as to more applied ﬁelds such as geodesy, statistics, and celestial

mechanics. He is generally considered to be among the half-dozen best mathematicians of all time.

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374 Chapter 7. Systems of First Order Linear Equations

We begin by forming the augmented matrix A |I:

A |I =

⎛

⎜

⎝

1 −1 −1

100

3 −12

010

223

001

⎞

⎟

⎠

The matrix A can be transformed into I by the following sequence of operations, and at the

same time, I is transformed into A

−1

. The result of each step appears below the statement.

(a) Obtain zeros in the off-diagonal positions in the ﬁrst column by adding (−3) times the

ﬁrst row to the second row and adding (−2) times the ﬁrst row to the third row.

⎛

⎜

⎝

1 −1 −1

100

025

−310

045

−201

⎞

⎟

⎠

(b) Obtain a 1 in the diagonal position in the second column by multiplying the second row

⎛

⎜

⎝

1 −1 −1

100

−

045

−201

⎞

⎟

⎠

to the ﬁrst row and adding (−4) times the second row to the third row.

⎛

⎜

⎝

−

00−5

4 −21

⎞

⎟

⎠

(d) Obtaina1inthediagonal position in the third column by multiplying the third row by

(−

⎛

⎜

⎝

−

001

−

⎞

⎟

⎠

(e) Obtain zeros in the off-diagonal positions in the third column by adding (−

) times the

third row to the ﬁrst row and adding (−

) times the third row to the second row.

⎛

⎜

⎝

100

−

010

−

001−

−

⎞

⎟

⎠

The last of these matrices is I |A

−1

, a fact that can be veriﬁed by direct multiplication with

the original matrix A.

This example was made slightly simpler by the fact that the given matrix A had a 1

in the upper left corner (a

= 1). If this is not the case,then the ﬁrst step is to produce

a 1 there by multiplying the ﬁrst row by 1/a

, as long as a

̸= 0. If a

= 0, then the

ﬁrst row must be interchanged with some other row to bring a nonzero element

into the upper left position before proceeding. If this cannot be done, because every

element in the ﬁrst column is zero, then the matrix has no inverse and is singular.

August 7, 2012 21:04 c07 Sheet number 17 Page number 375 cyan black

7.2 Review of Matrices 375

A similar situation may occur at later stages of the process as well, and the remedy

is the same: interchange the given row with a lower row so as to bring a nonzero

element to the desired diagonal location. If this cannot be done, then the original

matrix is singular.

Matrix Functions. We sometimes need to consider vectors or matrices whose elements

are functions of a real variable t. We write

x(t) =

⎛

⎜

⎝

(t)

⎞

⎟

⎠

, A(t) =

⎛

⎜

⎝

(t) ··· a

(t)

(t) ··· a

(t)

⎞

⎟

⎠

, (25)

respectively.

The matrix A(t) is said to be continuous at t = t

or on an interval α<t <βif each

element of A is a continuous function at the given point or on the given interval.

Similarly,A(t) is said to be differentiable if each of its elements is differentiable, and

its derivative dA/dt is deﬁned by

(

; (26)

that is, each element of dA/dt is the derivative of the corresponding element of A.

In the same way, the integral of a matrix function is deﬁned as

A(t) dt =

(t) dt

(

. (27)

For example, if

A(t) =

sin tt

1 cos t

(

then

′

(t) =

cos t 1

0 −sin t

(

A(t) dt =

2 π

π 0

(

Many of the rules of elementary calculus extend easily to matrix functions; in partic-

ular,

(CA) = C

, where C is a constant matrix; (28)

(A + B) =

d B

; (29)

(AB) = A

d B

B. (30)

In Eqs. (28) and (30), care must be taken in each term to avoid interchanging the

order of multiplication. The deﬁnitions expressed by Eqs. (26) and (27) also apply as

special cases to vectors.

To conclude this section:some important operations on matrices are accomplished

by applyingthe operation separately to each element ofthe matrix. Examples include

August 7, 2012 21:04 c07 Sheet number 18 Page number 376 cyan black

376 Chapter 7. Systems of First Order Linear Equations

multiplication by a number,differentiation, and integration. However,this is not true

of many other operations. For instance, the square of a matrix is not calculated by

squaring each of its elements.

PROBLEMS

1. If A =

⎛

⎜

⎝

1 −20

32−1

−213

⎞

⎟

⎠

and B =

⎛

⎜

⎝

4 −23

−150

612

⎞

⎟

⎠

, ﬁnd

(a) 2A + B (b) A − 4B

2. If A =

1 + i −1 + 2i

3 + 2i 2 − i

and B =

i 3

2 −2i

, ﬁnd

(a) A −2B (b) 3A + B

3. If A =

⎛

⎜

⎝

−212

10−3

2 −11

⎞

⎟

⎠

and B =

⎛

⎜

⎝

3 −

1 −1

−210

⎞

⎟

⎠

, ﬁnd

(a) A

(b) B

+ B

(d) (A + B)

4. If A =

3 − 2i 1 + i

2 − i −2 + 3i

, ﬁnd

(a) A

(b) A (c) A

∗

5. If A =

⎛

⎜

⎝

32−1

2 −12

121

⎞

⎟

⎠

and B =

⎛

⎜

⎝

21−1

−233

102

⎞

⎟

⎠

, verify that 2(A + B) = 2A + 2B.

6. If A =

⎛

⎜

⎝

1 −20

32−1

−203

⎞

⎟

⎠

, B =

⎛

⎜

⎝

21−1

−233

102

⎞

⎟

⎠

, and C =

⎛

⎜

⎝

210

122

01−1

⎞

⎟

⎠

, verify that

(a) (AB)C = A(BC) (b) (A + B) + C = A + (B + C)

Prove

each of the following laws of matrix algebra:

(a) A + B = B +A (b) A + (B + C) = (A + B) + C

(e) A (BC) = (AB)C (f) A(B + C) = AB +AC

8. If x =

⎛

⎜

⎝

1 − i

⎞

⎟

⎠

and y =

⎛

⎜

⎝

−1 + i

3 − i

⎞

⎟

⎠

ﬁnd

(a) x

y (b) y

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7.2 Review of Matrices 377

9. If x =

⎛

⎜

⎝

1 − 2i

⎞

⎟

⎠

and y =

⎛

⎜

⎝

3 − i

1 + 2i

⎞

⎟

⎠

, show that

(a) x

y = y

x (b) (x,y) = (y, x)

In each of Problems 10 through 19, either compute the inverse of the given matrix, or else

show that it is singular.

10.

−23

11.

3 −1

12.

⎛

⎜

⎝

123

245

356

⎞

⎟

⎠

13.

⎛

⎜

⎝

11−1

2 −11

112

⎞

⎟

⎠

14.

⎛

⎜

⎝

121

−218

1 −2 −7

⎞

⎟

⎠

15.

⎛

⎜

⎝

210

021

002

⎞

⎟

⎠

16.

⎛

⎜

⎝

1 −1 −1

210

3 −21

⎞

⎟

⎠

17.

⎛

⎜

⎝

231

−121

4 −1 −1

⎞

⎟

⎠

18.

⎛

⎜

⎝

100−1

0 −110

−10

01−

⎞

⎟

⎠

19.

⎛

⎜

⎝

1 −120

−12−42

1013

−220−1

⎞

⎟

⎠

20. IfA is a square matrix,and if there are two matrices B and C such that AB = I and CA = I,

show that B = C. Thus, if a matrix has an inverse, it can have only one.

21. If A(t) =

⎛

⎜

⎝

−t

−e

−t

⎞

⎟

⎠

and B(t) =

⎛

⎜

⎝

−t

−e

−t

−e

−t

−e

⎞

⎟

⎠

, ﬁnd

(a) A + 3B (b) AB

A(t) dt

In each of Problems 22 through 24, verify that the given vector satisﬁes the given differential

equation.

22. x

′

3 −2

2 −2

x, x =

23. x

′

2 −1

3 −2

x +

−1

, x =

+ 2

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378 Chapter 7. Systems of First Order Linear Equations

24. x

′

⎛

⎜

⎝

111

21−1

0 −11

⎞

⎟

⎠

x, x =

⎛

⎜

⎝

−8

−4

⎞

⎟

⎠

−t

+ 2

⎛

⎜

⎝

−1

⎞

⎟

⎠

In each of Problems 25 and 26, verify that the given matrix satisﬁes the given differential

equation.

25. $

′

4 −2

$, $(t) =

−3t

−4e

−3t

26. $

′

⎛

⎜

⎝

1 −14

32−1

21−1

⎞

⎟

⎠

$, $(t) =

⎛

⎜

⎝

−2t

−4e

−e

−2t

−e

−2t

⎞

⎟

⎠

7.3 Systems of Linear Algebraic Equations; Linear Independence,

Eigenvalues, Eigenvectors

In this section we review some results from linear algebra that are important for the

solution of systems of linear differential equations. Some of these results are easily

proved and others are not;since we are interested simply in summarizing some useful

information in compact form, we give no indication of proofs in either case. All the

results in this section depend on some basic facts about the solution of systems of

linear algebraic equations.

Systems of Linear Algebraic Equations. A set ofn simultaneouslinear algebraic equations

in n variables

+ a

+···+a

= b

. (1)

+ a

+···+a

= b

can be written as

Ax = b, (2)

where the n × n matrix A and the vector b are given, and the components of x are

to be determined. If b = 0, the system is said to be homogeneous; otherwise, it is

nonhomogeneous.

If the coefﬁcient matrix A is nonsingular—that is, if detA is not zero—then there

is a unique solution of the system (2). Since A is nonsingular, A

−1

exists, and the

solution can be found by multiplying each side of Eq. (2) on the left by A

−1

; thus

x = A

−1

b. (3)

In particular, the homogeneous problem Ax = 0, corresponding to b = 0 in Eq. (2),

has only the trivial solution x = 0.

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7.3 Systems of Linear Equations; Linear Independence, Eigenvalues, Eigenvectors 379

On the other hand, if A is singular—that is, if detA is zero—then solutions of

Eq. (2) either do not exist, or do exist but are not unique. Since A is singular, A

−1

does not exist, so Eq. (3) is no longer valid. The homogeneous system

Ax = 0 (4)

has (inﬁnitely many) nonzero solutions in addition to the trivial solution. The situ-

ation for the nonhomogeneous system (2) is more complicated. This system has no

solution unless the vector b satisﬁes a certain further condition.This condition is that

(b, y) = 0, (5)

for all vectors y satisfying A

∗

y = 0, where A

∗

is the adjoint of A. If condition (5) is

met, then the system (2) has (inﬁnitely many) solutions. These solutions are of the

form

x = x

(0)

+ ξ, (6)

where x

(0)

is a particular solution of Eq. (2), and ξ is the most general solution of the

homogeneous system (4). Note the resemblance between Eq. (6) and the solution of

a nonhomogeneous linear differential equation. The proofs of some of the preceding

statements are outlined in Problems 26 through 30.

The results in the preceding paragraph are important as a means of classifying the

solutions of linear systems. However,for solving particular systems,it is generally best

to use row reduction to transform the system into a much simpler one from which

the solution(s), if there are any, can be written down easily. To do this efﬁciently, we

can form the augmented matrix

A |b =

⎛

⎜

⎝

··· a

⎞

⎟

⎠

(7)

by adjoining the vector b to the coefﬁcient matrix A as an additional column. The

vertical line replaces the equals sign and is said to partition the augmented matrix.

We now perform row operations on the augmented matrix so as to transform A into

an upper triangular matrix—that is,a matrix whose elements belowthe maindiagonal

are all zero. Once this is done, it is easy to see whether the system has solutions, and

to ﬁnd them if it does. Observe that elementary row operations on the augmented

matrix (7) correspond to legitimate operations on the equations in the system (1).

The following examples illustrate the process.

EXAMPLE

Solve the system of equations

− 2x

+ 3x

= 7,

−x

+ x

− 2x

=−5, (8)

− x

= 4.

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380 Chapter 7. Systems of First Order Linear Equations

The augmented matrix for the system (8) is

⎛

⎜

⎝

1 −23

−11−2

−5

2 −1 −1

⎞

⎟

⎠

. (9)

We now perform row operations on the matrix (9) with a view to introducing zeros in the

lower left part of the matrix. Each step is described and the result recorded below.

(a) Add the ﬁrst row to the second row, and add (−2) times the ﬁrst row to the third row.

⎛

⎜

⎝

1 −23

0 −11

03−7

−10

⎞

⎟

⎠

(b) Multiply the second row by −1.

⎛

⎜

⎝

1 −23

01−1

−2

03−7

−10

⎞

⎟

⎠

⎛

⎜

⎝

1 −23

01−1

−2

00−4

−4

⎞

⎟

⎠

(d) Divide the third row by −4.

⎛

⎜

⎝

1 −23

01−1

−2

001

⎞

⎟

⎠

The matrix obtained in this manner corresponds to the system of equations

− 2x

+ 3x

= 7,

− x

=−2, (10)

= 1,

which is equivalent to the original system (8). Note that the coefﬁcients in Eqs. (10) form

a triangular matrix. From the last of Eqs. (10) we have x

= 1, from the second equation

=−2 + x

=−1, and from the ﬁrst equation x

= 7 + 2x

− 3x

= 2. Thus we obtain

x =

⎛

⎜

⎝

−1

⎞

⎟

⎠

which is the solution of the given system (8). Incidentally, since the solution is unique, we

conclude that the coefﬁcient matrix is nonsingular.

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7.3 Systems of Linear Equations; Linear Independence, Eigenvalues, Eigenvectors 381

EXAMPLE

Discuss solutions of the system

− 2x

+ 3x

= b

−x

+ x

− 2x

= b

, (11)

− x

+ 3x

= b

for various values of b

, b

, and b

Observe that the coefﬁcientsin thesystem (11)are the same as those in thesystem (8)except

for the coefﬁcient of x

in the third equation. The augmented matrix for the system (11) is

⎛

⎜

⎝

1 −23

−11−2 b

2 −13b

⎞

⎟

⎠

. (12)

By performing steps (a), (b), and (c) as in Example 1, we transform the matrix (12) into

⎛

⎜

⎝

1 −23

01−1 −b

− b

000b

+ 3b

+ b

⎞

⎟

⎠

. (13)

The equation corresponding to the third row of the matrix (13) is

+ 3b

+ b

= 0; (14)

thus the system (11) has no solution unless the condition (14) is satisﬁed by b

, b

, and b

.Itis

possible to show that this condition is just Eq. (5) for the system (11).

Let us now assume that b

= 2,b

= 1, and b

=−5,in which case Eq. (14) is satisﬁed. Then

the ﬁrst two rows of the matrix (13) correspond to the equations

− 2x

+ 3x

= 2,

(15)

− x

=−3.

To solve the system (15), we can choose one of the unknowns arbitrarily and then solve for

the other two. If we let x

= α, where α is arbitrary, it then follows that

= α − 3,

= 2(α − 3) − 3α + 2 =−α − 4.

If we write the solution in vector notation, we have

x =

⎛

⎜

⎝

−α − 4

α − 3

⎞

⎟

⎠

= α

⎛

⎜

⎝

−1

⎞

⎟

⎠

⎛

⎜

⎝

−4

−3

⎞

⎟

⎠

. (16)

It is easy to verify that the second term on the right side of Eq. (16) is a solution of the nonho-

mogeneous system (11) and that the ﬁrst term is the most general solution of the homogeneous

system corresponding to (11).

Row reduction is also useful in solving homogeneous systems and systems in which

the number of equations is different from the number of unknowns.

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382 Chapter 7. Systems of First Order Linear Equations

Linear Dependence and Independence. A set of k vectors x

(1)

, ..., x

(k)

is said to be linearly

dependent if there exists a set of real or complex numbers c

, ..., c

, at least one of

which is nonzero, such that

(1)

+···+c

(k)

= 0. (17)

In other words, x

(1)

, ..., x

(k)

are linearly dependent if there is a linear relation among

them. On the other hand, if the only set c

, ..., c

for which Eq. (17) is satisﬁed is

= c

=···=c

= 0, then x

(1)

, ..., x

(k)

are said to be linearly independent.

Consider now a set of n vectors, each of which has n components. Let x

= x

(j )

the ith component of the vector x

(j )

,and let X = (x

).Then Eq. (17) can be written as

⎛

⎜

⎝

(1)

+···+x

(n)

(1)

+···+x

(n)

⎞

⎟

⎠

⎛

⎜

⎝

+···+x

⎞

⎟

⎠

= 0,

or, equivalently,

Xc = 0. (18)

If det X ̸= 0, then the only solution of Eq. (18) is c = 0, but if det X = 0, there are

nonzero solutions. Thus the set of vectors x

(1)

, ..., x

(n)

is linearly independent if and

only if det X ̸= 0.

EXAMPLE

Determine whether the vectors

(1)

⎛

⎜

⎝

−1

⎞

⎟

⎠

, x

(2)

⎛

⎜

⎝

⎞

⎟

⎠

, x

(3)

⎛

⎜

⎝

−4

−11

⎞

⎟

⎠

(19)

are linearly independent or linearly dependent. If they are linearly dependent, ﬁnd a linear

relation among them.

To determine whether x

(1)

, x

(2)

, and x

(3)

are linearly dependent, we seek constants c

, c

, and

such that

(1)

+ c

(2)

+ c

(3)

= 0. (20)

Equation (20) can also be written in the form

⎛

⎜

⎝

12−4

21 1

−13−11

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

(21)

and solved by means of elementary row operations starting from the augmented matrix

⎛

⎜

⎝

12−4

21 1

−13−11

⎞

⎟

⎠

. (22)

We proceed as in Examples 1 and 2.

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7.3 Systems of Linear Equations; Linear Independence, Eigenvalues, Eigenvectors 383

(a) Add (−2) times the ﬁrst row to the second row, and add the ﬁrst row to the third row.

⎛

⎜

⎝

12−4

0 −39

05−15

⎞

⎟

⎠

(b) Divide the second row by −3; then add (−5) times the second row to the third row.

⎛

⎜

⎝

12−4

01−3

000

⎞

⎟

⎠

Thus we obtain the equivalent system

+ 2c

− 4c

= 0,

(23)

− 3c

= 0.

From the second of Eqs. (23) we have c

= 3c

, and then from the ﬁrst we obtain

= 4c

− 2c

=−2c

. Thus we have solved for c

and c

in terms of c

, with the latter remain-

ing arbitrary. If we choose c

=−1 for convenience, then c

= 2 and c

=−3. In this case the

relation (20) becomes

(1)

− 3x

(2)

− x

(3)

= 0,

and the given vectors are linearly dependent.

Alternatively, we can compute det(x

), whose columns are the components of x

(1)

, x

(2)

, and

(3)

, respectively. Thus

det(x

) =

12−4

21 1

−13−11

and direct calculation shows that it is zero. Hence x

(1)

, x

(2)

, and x

(3)

are linearly dependent.

However, if the coefﬁcients c

, c

, and c

are required, we still need to solve Eq. (20) to ﬁnd

them.

Frequently, it is useful to think of the columns (or rows) of a matrix A as vectors.

These column (or row) vectors are linearly independent if and only if detA ̸= 0.

Further, if C = AB, then it can be shown that det C = (det A)(det B). Therefore, if

the columns (or rows) of both A and B are linearly independent, then the columns

(or rows) of C are also linearly independent.

Now let us extend the concepts of linear dependence and independence to a set

of vector functions x

(1)

(t), ..., x

(k)

(t) deﬁned on an interval α<t <β. The vectors

(1)

(t), ..., x

(k)

(t) are said to be linearly dependent on α<t <βif there exists a set

of constants c

, ..., c

, not all of which are zero, such that

(1)

(t) +···+c

(k)

(t) = 0 for all t in the interval.

Otherwise, x

(1)

(t), ..., x

(k)

(t) are said to be linearly independent. Note that if

(1)

(t), ..., x

(k)

(t) are linearly dependent on an interval, they are linearly dependent

at each point in the interval. However, if x

(1)

(t), ..., x

(k)

(t) are linearly independent

on an interval, they may or may not be linearly independent at each point; they may,

August 7, 2012 21:04 c07 Sheet number 26 Page number 384 cyan black

384 Chapter 7. Systems of First Order Linear Equations

in fact, be linearly dependent at each point, but with different sets of constants at

different points. See Problem 15 for an example.

Eigenvalues and Eigenvectors. The equation

Ax = y (24)

can be viewed as a linear transformation that maps (or transforms) a given vector x

into a new vector y. Vectors that are transformed into multiples of themselves are

important in many applications.

To ﬁnd such vectors, we set y = λx, where λ is a

scalar proportionality factor, and seek solutions of the equation

Ax = λx, (25)

(A − λI)x = 0. (26)

The latter equation has nonzero solutions if and only if λ is chosen so that

det(A − λI) = 0. (27)

Equation (27) is a polynomial equation of degree n in λ and is called the characteristic

equation of the matrix A. Values of λ that satisfy Eq. (27) may be either real- or

complex-valued and are called eigenvalues of A. The nonzero solutions of Eq. (25)

or (26) that are obtained by using such a value of λ are called the eigenvectors

corresponding to that eigenvalue.

If A is a 2 × 2 matrix, then Eq. (26) is

− λ a

− λ

(

(28)

and Eq. (27) becomes

− λ)(a

− λ) − a

= 0,

− (a

+ a

)λ + a

− a

= 0. (29)

The following example illustrates how eigenvalues and eigenvectors are found.

EXAMPLE

Find the eigenvalues and eigenvectors of the matrix

A =

3 −1

4 −2

. (30)

The eigenvalues λ and eigenvectors x satisfy the equation (A − λI)x = 0, or

3 − λ −1

4 −2 − λ

. (31)

For example, this problem is encountered in ﬁnding the principal axes of stress or strain in an elastic

body, and in ﬁnding the modes of free vibration in a conservative system with a ﬁnite number of degrees

of freedom.

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7.3 Systems of Linear Equations; Linear Independence, Eigenvalues, Eigenvectors 385

The eigenvalues are the roots of the equation

det(A − λI) =

3 − λ −1

4 −2 − λ

= λ

− λ − 2 = (λ − 2)(λ + 1) = 0. (32)

Thus the eigenvalues are λ

= 2 and λ

=−1.

To ﬁnd the eigenvectors, we return to Eq. (31) and replace λ by each of the eigenvalues in

turn. For λ = 2 we have

1 −1

4 −4

. (33)

Hence each row of this vector equation leads to the condition x

− x

= 0, so x

and x

are

equal but their value is not determined. If x

= c, then x

= c also, and the eigenvector x

(1)

= c

, c ̸= 0. (34)

Thus there is an inﬁnite family of eigenvectors,indexed by the arbitrary constant c,correspond-

ing to the eigenvalue λ

. We will choose a single member of this family as a representative of

the rest; in this example it seems simplest to let c = 1. Then, instead of Eq. (34), we write

(1)

(35)

and remember that any nonzero multiple of this vector is also an eigenvector. We say that x

(1)

is the eigenvector corresponding to the eigenvalue λ

= 2.

Now, setting λ =−1 in Eq. (31), we obtain

4 −1

. (36)

Again we obtain a single condition on x

and x

, namely, 4x

− x

= 0. Thus the eigenvector

corresponding to the eigenvalue λ

=−1 is

(2)

(37)

or any nonzero multiple of this vector.

As Example 4 illustrates, eigenvectors are determined only up to an arbitrary

nonzero multiplicative constant; if this constant is speciﬁed in some way, then the

eigenvectors are said to be normalized. In Example 4, we chose the constant c so

that the components of the eigenvectors would be small integers. However, any

other choice of c is equally valid, although perhaps less convenient. Sometimes it

is useful to normalize an eigenvector x by choosing the constant so that its length

∥x∥=(x, x)

1/2

= 1.

Since the characteristic equation (27) for an n × n matrix A is a polynomial equa-

tion of degree n in λ, each such matrix has n eigenvalues λ

, ..., λ

, some of which

may be repeated. If a given eigenvalue appears m times as a root of Eq. (27), then

that eigenvalue is said to have algebraic multiplicity m. Each eigenvalue has at least

one associated eigenvector,and an eigenvalue of algebraic multiplicity m may have q

August 7, 2012 21:04 c07 Sheet number 28 Page number 386 cyan black

386 Chapter 7. Systems of First Order Linear Equations

linearly independent eigenvectors. The integer q is called the geometric multiplicity

of the eigenvalue, and it is possible to show that

1 ≤ q ≤ m. (38)

Further, examples demonstrate that q may be any integer in this interval. If each

eigenvalue of A is simple (has algebraic multiplicity 1),then each eigenvalue also has

geometric multiplicity 1.

It is possible to show that if λ

and λ

are two eigenvalues of A and if λ

̸= λ

,then

their corresponding eigenvectors x

(1)

and x

(2)

are linearly independent (Problem 34).

This result extends to any set λ

, ..., λ

of distinct eigenvalues: their eigenvectors

(1)

, ..., x

(k)

are linearly independent. Thus, if each eigenvalue of an n × n matrix

is simple, then the n eigenvectors of A, one for each eigenvalue, are linearly inde-

pendent. On the other hand, if A has one or more repeated eigenvalues, then there

may be fewer than n linearly independent eigenvectors associated with A, since for a

repeated eigenvalue we may have q < m. As we will see in Section 7.8, this fact may

lead to complications later on in the solution of systems of differential equations.

EXAMPLE

Find the eigenvalues and eigenvectors of the matrix

A =

⎛

⎜

⎝

011

101

110

⎞

⎟

⎠

. (39)

The eigenvalues λ and eigenvectors x satisfy the equation (A − λI)x = 0, or

⎛

⎜

⎝

−λ 11

1 −λ 1

11−λ

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

. (40)

The eigenvalues are the roots of the equation

det(A − λI) =

−λ 11

1 −λ 1

11−λ

=−λ

+ 3λ + 2 = 0. (41)

The roots of Eq. (41) are λ

= 2, λ

=−1, and λ

=−1. Thus 2 is a simple eigenvalue, and −1

is an eigenvalue of algebraic multiplicity 2, or a double eigenvalue.

To ﬁnd the eigenvector x

(1)

corresponding to the eigenvalue λ

, we substitute λ = 2in

Eq. (40); this gives the system

⎛

⎜

⎝

−211

1 −21

11−2

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

. (42)

We can reduce this to the equivalent system

⎛

⎜

⎝

2 −1 −1

01−1

000

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

(43)

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7.3 Systems of Linear Equations; Linear Independence, Eigenvalues, Eigenvectors 387

by elementary row operations. Solving this system yields the eigenvector

(1)

⎛

⎜

⎝

⎞

⎟

⎠

. (44)

For λ =−1, Eqs. (40) reduce immediately to the single equation

+ x

= 0. (45)

Thus values for two of the quantities x

, x

can be chosen arbitrarily, and the third is

determined from Eq. (45). For example, if x

= c

and x

= c

, then x

=−c

− c

. In vector

notation we have

x =

⎛

⎜

⎝

−c

− c

⎞

⎟

⎠

= c

⎛

⎜

⎝

−1

⎞

⎟

⎠

+ c

⎛

⎜

⎝

−1

⎞

⎟

⎠

. (46)

For example, by choosing c

= 1 and c

= 0, we obtain the eigenvector

(2)

⎛

⎜

⎝

−1

⎞

⎟

⎠

. (47)

Any nonzero multiple of x

(2)

is also an eigenvector, but a second independent eigenvector can

be found by making another choice of c

and c

—for instance, c

= 0 and c

= 1. In this case

we obtain

(3)

⎛

⎜

⎝

−1

⎞

⎟

⎠

, (48)

which is linearly independent of x

(2)

. Therefore, in this example, two linearly independent

eigenvectors are associated with the double eigenvalue.

An important special class of matrices, called self-adjoint or Hermitian matrices,

are those for which A

∗

= A;that is, a

= a

. Hermitian matrices include as a subclass

real symmetric matrices—that is, matrices that have real elements and for which

= A. The eigenvalues and eigenvectors of Hermitian matrices always have the

following useful properties:

1. All eigenvalues are real.

2. There always exists a full set of n linearly independent eigenvectors, regardless of the

algebraic multiplicities of the eigenvalues.

3. If x

(1)

and x

(2)

are eigenvectors that correspond to different eigenvalues, then

(1)

, x

(2)

) = 0. Thus, if all eigenvalues are simple, then the associated eigenvectors form an

orthogonal set of vectors.

4. Corresponding to an eigenvalue of algebraic multiplicity m, it is possible to choose m

eigenvectors that are mutually orthogonal. Thus the full set of n eigenvectors can always

be chosen to be orthogonal as well as linearly independent.

The proofsof statements 1 and 3 aboveare outlined in Problems 32 and33. Example

5 involves a real symmetric matrix and illustrates properties 1, 2, and 3, but the choice

we have made for x

(2)

and x

(3)

does not illustrate property 4. However, it is always

possible to choose an x

(2)

and x

(3)

so that (x

(2)

, x

(3)

) = 0. For instance, in Example 5

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388 Chapter 7. Systems of First Order Linear Equations

we could have chosen x

(2)

as before and x

(3)

by using c

= 1 and c

=−2 in Eq. (46).

In this way we obtain

(2)

⎛

⎜

⎝

−1

⎞

⎟

⎠

, x

(3)

⎛

⎜

⎝

−2

⎞

⎟

⎠

as the eigenvectors associated with the eigenvalue λ =−1. These eigenvectors are

orthogonal to each other as well as to the eigenvector x

(1)

that corresponds to the

eigenvalue λ = 2.

PROBLEMS In each of Problems 1 through 6, either solve the given system of equations, or else show that

there is no solution.

1. x

− x

= 0

+ x

= 1

−x

+ x

+ 2x

= 2

2. x

+ 2x

− x

= 1

+ x

= 1

− x

+ 2x

= 1

3. x

+ 2x

− x

= 2

+ x

= 1

− x

+ 2x

=−1

4. x

+ 2x

− x

= 0

+ x

= 0

− x

+ 2x

= 0

5. x

− x

= 0

+ x

= 0

−x

+ x

+ 2x

= 0

6. x

+ 2x

− x

=−2

−2x

− 4x

+ 2x

= 4

+ 4x

− 2x

=−4

In each of Problems 7 through 11, determine whether the members of the given set of vectors

are linearly independent. If they are linearly dependent, ﬁnd a linear relation among them.

The vectors are written as row vectors to save space but may be considered as column vectors;

that is, the transposes of the given vectors may be used instead of the vectors themselves.

7. x

(1)

= (1, 1, 0), x

(2)

= (0, 1, 1), x

(3)

= (1, 0, 1)

8. x

(1)

= (2, 1, 0), x

(2)

= (0, 1, 0), x

(3)

= (−1, 2, 0)

9. x

(1)

= (1, 2, 2,3), x

(2)

= (−1, 0, 3,1), x

(3)

= (−2, −1, 1,0), x

(4)

= (−3, 0, −1,3)

10. x

(1)

= (1, 2, −1,0), x

(2)

= (2, 3, 1,−1), x

(3)

= (−1, 0, 2,2), x

(4)

= (3, −1, 1,3)

11. x

(1)

= (1, 2, −2), x

(2)

= (3, 1, 0), x

(3)

= (2, −1, 1), x

(4)

= (4, 3, −2)

12. Suppose that each of the vectors x

(1)

, ..., x

(m)

has n components, where n < m. Show that

(1)

, ..., x

(m)

are linearly dependent.

In each of Problems 13 and 14, determine whether the members of the given set of vectors are

linearly independent for −∞ < t < ∞. If they are linearly dependent, ﬁnd the linear relation

among them.As in Problems 7 through 11,the vectors are written as row vectors to save space.

13. x

(1)

(t) = (e

−t

,2e

−t

), x

(2)

(t) = (e

−t

, e

−t

), x

(3)

(t) = (3e

−t

,0)

14. x

(1)

(t) = (2 sin t, sin t), x

(2)

(t) = (sin t, 2 sin t)

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7.3 Systems of Linear Equations; Linear Independence, Eigenvalues, Eigenvectors 389

15. Let

(1)

(t) =

, x

(2)

(t) =

Show that x

(1)

(t) and x

(2)

(t) are linearly dependent at each point in the interval 0 ≤ t ≤ 1.

Nevertheless, show that x

(1)

(t) and x

(2)

(t) are linearly independent on 0 ≤ t ≤ 1.

In each of Problems 16 through 25, ﬁnd all eigenvalues and eigenvectors of the given matrix.

16.

5 −1

17.

3 −2

4 −1

18.

−21

1 −2

19.

1 i

−i 1

20.

√

3 −1

21.

−33/4

−51

22.

⎛

⎜

⎝

100

21−2

321

⎞

⎟

⎠

23.

⎛

⎜

⎝

322

141

−2 −4 −1

⎞

⎟

⎠

24.

⎛

⎜

⎝

11/9 −2/98/9

−2/92/910/9

8/910/95/9

⎞

⎟

⎠

25.

⎛

⎜

⎝

324

202

423

⎞

⎟

⎠

Problems 26 through 30 deal with the problem of solving Ax = b when detA = 0.

26. (a) Suppose that A is a real-valued n × n matrix. Show that (Ax, y) = (x,A

y) for any

vectors x and y.

Hint: You may ﬁnd it simpler to consider ﬁrst the case n = 2; then extend the result to an

arbitrary value of n.

(b) If A is not necessarily real, show that (Ax, y) = (x,A

∗

y) for any vectors x and y.

27. Suppose that, for a given matrix A, there is a nonzero vector x such that Ax = 0. Show

that there is also a nonzero vector y such that A

∗

y = 0.

28. Suppose that detA = 0 and thatAx = b has solutions. Show that (b, y) = 0, where y is any

solution ofA

∗

y = 0. Verify that this statement is true for the set of equations in Example 2.

Hint: Use the result of Problem 26(b).

29. Suppose that det A = 0 and that x = x

(0)

is a solution ofAx = b. Show that if ξ is a solution

of Aξ = 0 and α is any constant, then x = x

(0)

+ αξ is also a solution of Ax = b.

30. Suppose that detA = 0 and that y is a solution ofA

∗

y = 0. Show that if (b, y) = 0 for every

such y, then Ax = b has solutions. Note that this is the converse of Problem 28; the form

of the solution is given by Problem 29.

Hint: What does the relation A

∗

y = 0 say about the rows of A? Again, it may be helpful

to consider the case n = 2 ﬁrst.

August 7, 2012 21:04 c07 Sheet number 32 Page number 390 cyan black

390 Chapter 7. Systems of First Order Linear Equations

31. Prove that λ = 0 is an eigenvalue of A if and only if A is singular.

32. In this problem we show that the eigenvalues of a Hermitian matrix A are real. Let x be

an eigenvector corresponding to the eigenvalue λ.

(a) Show that (Ax, x) = (x, Ax). Hint: See Problem 26(c).

(b) Show that λ(x, x) =

λ(x, x). Hint: Recall that Ax = λx.

λ; that is, the eigenvalue λ is real.

33. Show that if λ

and λ

are eigenvalues of a Hermitian matrix A, and if λ

̸= λ

, then the

corresponding eigenvectors x

(1)

and x

(2)

are orthogonal.

Hint: Use the results of Problems 26(c) and 32 to show that (λ

− λ

)(x

(1)

, x

(2)

) = 0.

34. Show that if λ

and λ

are eigenvalues of any matrix A, and if λ

̸= λ

, then the corre-

sponding eigenvectors x

(1)

and x

(2)

are linearly independent.

Hint: Start from c

(1)

+ c

(2)

= 0; multiply by A to obtain c

(1)

+ c

(2)

= 0. Then

show that c

= c

= 0.

7.4 Basic Theory of Systems of First Order Linear Equations

The general theory of a system of n ﬁrst order linear equations

′

= p

(t)x

+···+p

(t)x

+ g

(t),

. (1)

′

= p

(t)x

+···+p

(t)x

+ g

(t)

closely parallels that of a single linear equation of nth order. The discussion in this

section therefore follows the same general lines as that in Sections 3.2 and 4.1.

To discuss the system (1) most effectively, we write it in matrix notation. That is, we

consider x

= φ

(t), ..., x

= φ

(t) to be components of a vector x = φ(t); similarly,

(t), ..., g

(t) are components of a vector g(t), and p

(t), ..., p

(t) are elements of

an n × n matrix P(t). Equation (1) then takes the form

′

= P(t)x + g(t). (2)

The use of vectors and matrices not only saves a great deal of space and facilitates

calculations but also emphasizes the similarity between systems of equations and

single (scalar) equations.

A vector x = φ(t) is said to be a solution of Eq. (2) if its components satisfy the sys-

tem of equations (1). Throughout this section we assume that P and g are continuous

on some interval α<t <β; that is, each of the scalar functions p

, ..., p

, g

, ..., g

is continuous there. According to Theorem 7.1.2, this is sufﬁcient to guarantee the

existence of solutions of Eq. (2) on the interval α<t <β.

It is convenient to consider ﬁrst the homogeneous equation

′

= P(t)x (3)

obtained from Eq. (2) by setting g(t) = 0. Once the homogeneous equation has been

solved, there are several methods that can be used to solve the nonhomogeneous

August 7, 2012 21:04 c07 Sheet number 33 Page number 391 cyan black

7.4 Basic Theory of Systems of First Order Linear Equations 391

equation (2); this is taken up in Section 7.9. We use the notation

(1)

(t) =

⎛

⎜

⎝

(t)

⎞

⎟

⎠

, ..., x

(k)

(t) =

⎛

⎜

⎝

(t)

⎞

⎟

⎠

, ... (4)

to designate speciﬁc solutions of the system (3). Note that x

(t) = x

(j )

(t) refers to

the i th component of the jth solution x

(j )

(t). The main facts about the structure of

solutions of the system (3) are stated inTheorems 7.4.1 to 7.4.5.They closely resemble

the corresponding theorems in Sections 3.2 and 4.1; some of the proofs are left to

you as exercises.

Theorem 7.4.1

If the vector functions x

(1)

and x

(2)

are solutions of the system (3), then the linear

combination c

(1)

+ c

(2)

is also a solution for any constants c

and c

This is the principle of superposition; it is proved simply by differentiating

(1)

+ c

(2)

and using the fact that x

(1)

and x

(2)

satisfy Eq. (3). As an example,

it can be veriﬁed that

(1)

(t) =

(

, x

(2)

(t) =

−t

−2e

−t

(

−2

(

−t

(5)

satisfy the equation

′

(

x. (6)

Then, according to Theorem 7.4.1,

x = c

(

+ c

−2

(

−t

= c

(1)

(t) + c

(2)

(t) (7)

also satisﬁes Eq. (6).

By repeated application of Theorem 7.4.1, we can conclude that if x

(1)

, ..., x

(k)

are

solutions of Eq. (3), then

x = c

(1)

(t) +···+c

(k)

(t) (8)

is also a solution for any constants c

, ..., c

. Thus every ﬁnite linear combination

of solutions of Eq. (3) is also a solution. The question that now arises is whether all

solutions of Eq. (3) can be found in this way. By analogy with previous cases, it is

reasonable to expect that for the system (3) of n ﬁrst order equations, it is sufﬁcient to

form linear combinations of n properly chosen solutions. Therefore, let x

(1)

, ..., x

(n)

August 7, 2012 21:04 c07 Sheet number 34 Page number 392 cyan black

392 Chapter 7. Systems of First Order Linear Equations

be n solutions of the system (3), and consider the matrix X(t) whose columns are the

vectors x

(1)

(t), ..., x

(n)

(t):

X(t) =

⎛

⎜

⎝

(t) ··· x

(t)

(t) ··· x

(t)

⎞

⎟

⎠

. (9)

Recall from Section 7.3 that the columns of X(t) are linearly independent for a given

value of t if and only if det X ̸= 0 for that value of t. This determinant is called the

Wronskian of the n solutions x

(1)

, ..., x

(n)

and is also denoted by W[x

(1)

, ..., x

(n)

];

that is,

W[x

(1)

, ..., x

(n)

](t) = det X(t). (10)

The solutions x

(1)

, ..., x

(n)

are then linearly independent at a point if and only if

W[ x

(1)

, ..., x

(n)

] is not zero there.

Theorem 7.4.2

If the vector functions x

(1)

, ..., x

(n)

are linearly independent solutions of the sys-

tem (3) for each point in the interval α<t <β, then each solution x = φ(t) of the

system (3) can be expressed as a linear combination of x

(1)

, ..., x

(n)

φ(t) = c

(1)

(t) +···+c

(n)

(t) (11)

in exactly one way.

Before proving Theorem 7.4.2, note that according to Theorem 7.4.1, all expres-

sions of the form (11) are solutions of the system (3), while by Theorem 7.4.2 all

solutions of Eq. (3) can be written in the form (11). If the constants c

, ..., c

are

thought of as arbitrary, then Eq. (11) includes all solutions of the system (3), and it is

customary to call it the general solution. Any set of solutions x

(1)

, ..., x

(n)

of Eq. (3)

that is linearly independent at each point in the interval α<t <βis said to be a

fundamental set of solutions for that interval.

To prove Theorem 7.4.2, we will show that any solution φ of Eq. (3) can be written

as φ(t) = c

(1)

(t) +···+c

(n)

(t) for suitable values of c

, ..., c

. Let t = t

be some

point in the interval α<t <βand let ξ = φ(t

). We now wish to determine whether

there is any solution of the form x = c

(1)

(t) +···+c

(n)

(t) that also satisﬁes the

same initial condition x(t

) = ξ. That is, we wish to know whether there are values of

, ..., c

such that

(1)

) +···+c

(n)

) = ξ, (12)

or, in scalar form,

) +···+c

) = ξ

. (13)

) +···+c

) = ξ

The necessary and sufﬁcient condition that Eqs. (13) possess a unique solution

, ..., c

is precisely the nonvanishing of the determinant of coefﬁcients, which is

the Wronskian W[ x

(1)

, ..., x

(n)

] evaluated at t = t

. The hypothesis that x

(1)

, ..., x

(n)

are linearly independent throughout α<t <βguarantees that W[x

(1)

, ..., x

(n)

] is

August 7, 2012 21:04 c07 Sheet number 35 Page number 393 cyan black

7.4 Basic Theory of Systems of First Order Linear Equations 393

not zero at t = t

, and therefore there is a (unique) solution of Eq. (3) of the form

x = c

(1)

(t) +···+c

(n)

(t) that also satisﬁes the initial condition (12). By the

uniqueness part of Theorem 7.1.2, this solution is identical to φ(t), and hence

φ(t) = c

(1)

(t) +···+c

(n)

(t), as was to be proved.

Theorem 7.4.3

If x

(1)

, ..., x

(n)

are solutions of Eq. (3) on the interval α<t <β, then in this interval

W[x

(1)

, ..., x

(n)

] either is identically zero or else never vanishes.

The signiﬁcance of Theorem 7.4.3 lies in the fact that it relieves us of the necessity

of examining W[x

(1)

, ..., x

(n)

] at all points in the interval of interest and enables us

to determine whether x

(1)

, ..., x

(n)

form a fundamental set of solutions merely by

evaluating their Wronskian at any convenient point in the interval.

Theorem 7.4.3 is proved by ﬁrst establishing that the Wronskian of x

(1)

, ..., x

(n)

satisﬁes the differential equation (see Problem 2)

=[p

(t) + p

(t) +···+p

(t)]W. (14)

Hence

W(t) = c exp

(t) +···+p

(t)]dt

, (15)

where c is an arbitrary constant, and the conclusion of the theorem follows imme-

diately. The expression for W(t) in Eq. (15) is known as Abel’s formula; note the

similarity of this result to Theorem 3.2.7 and especially to Eq. (23) of Section 3.2.

Alternatively, Theorem 7.4.3 can be established by showing that if n solu-

tions x

(1)

, ..., x

(n)

of Eq. (3) are linearly dependent at one point t = t

, then they

must be linearly dependent at each point in α<t <β (see Problem 8). Conse-

quently, if x

(1)

, ..., x

(n)

are linearly independent at one point, they must be linearly

independent at each point in the interval.

The next theorem states that the system (3) always has at least one fundamental

set of solutions.

Theorem 7.4.4

Let

(1)

⎛

⎜

⎝

⎞

⎟

⎠

, e

(2)

⎛

⎜

⎝

⎞

⎟

⎠

, ..., e

(n)

⎛

⎜

⎝

⎞

⎟

⎠

;

further, let x

(1)

, ..., x

(n)

be the solutions of the system (3) that satisfy the initial

conditions

(1)

) = e

(1)

, ..., x

(n)

) = e

(n)

, (16)

respectively, where t

is any point in α<t <β. Then x

(1)

, ..., x

(n)

form a funda-

mental set of solutions of the system (3).

August 7, 2012 21:04 c07 Sheet number 36 Page number 394 cyan black

394 Chapter 7. Systems of First Order Linear Equations

To prove this theorem, note that the existence and uniqueness of the solutions

(1)

, ..., x

(n)

mentioned in Theorem 7.4.4 are ensured by Theorem 7.1.2. It is not

hard to see that the Wronskian of these solutions is equal to 1 when t = t

; therefore

(1)

, ..., x

(n)

are a fundamental set of solutions.

Once one fundamental set of solutions has been found, other sets can be gener-

ated by forming (independent) linear combinations of the ﬁrst set. For theoretical

purposes, the set given by Theorem 7.4.4 is usually the simplest.

Finally, it may happen (just as for second order linear equations) that a system

whose coefﬁcients are all real may give rise to solutions that are complex-valued.

In this case, the following theorem is analogous to Theorem 3.2.6 and enables us to

obtain real-valued solutions instead.

Theorem 7.4.5

Consider the system (3)

′

= P(t)x,

where each element of P is a real-valued continuous function. If x = u(t) + iv(t) is

a complex-valued solution of Eq. (3), then its real part u(t) and its imaginary part

v(t) are also solutions of this equation.

To prove this result, we substitute u(t) + iv(t) for x in Eq. (3), thereby obtaining

′

− P(t)x = u

′

(t) − P(t)u(t) + i

[

′

(t) − P(t)v(t)

]

= 0. (17)

We have used the assumption that P(t) is real-valued to separate Eq. (17) into

its real and imaginary parts. Since a complex number is zero if and only if its

real and imaginary parts are both zero, we conclude that u

′

(t) − P(t)u(t) = 0 and

′

(t) − P(t)v(t) = 0. Therefore, u(t) and v(t) are solutions of Eq. (3).

To summarize the results of this section:

1. Any set of n linearly independent solutions of the system x

′

= P(t)x constitutes a

fundamental set of solutions.

2. Under the conditions given in this section, such fundamental sets always exist.

3. Every solution of the system x

′

= P(t)x can be represented as a linear combination of any

fundamental set of solutions.

PROBLEMS 1. Prove the generalization of Theorem 7.4.1, as expressed in the sentence that includes

Eq. (8), for an arbitrary value of the integer k.

2. In this problem we outline a proof of Theorem 7.4.3 in the case n = 2. Let x

(1)

and x

(2)

be solutions of Eq. (3) for α<t <β, and let W be the Wronskian of x

(1)

and x

(2)

(a) Show that

(1)

(2)

(1)

(2)

(1)

(2)

(1)

(2)

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7.4 Basic Theory of Systems of First Order Linear Equations 395

(b) Using Eq. (3), show that

= (p

+ p

)W.

to obtain the conclusion stated in Theorem 7.4.3.

(d) ProveTheorem 7.4.3 for an arbitrary value of n by generalizing the procedure of parts

(a), (b), and (c).

3. Show that the Wronskians of two fundamental sets of solutions of the system (3) can differ

at most by a multiplicative constant.

Hint: Use Eq. (15).

4. If x

= y and x

= y

′

, then the second order equation

′′

+ p(t)y

′

+ q(t)y = 0 (i)

corresponds to the system

′

= x

′

=−q(t)x

− p(t)x

. (ii)

Show that if x

(1)

and x

(2)

are a fundamental set of solutions of Eqs. (ii), and if y

(1)

and y

(2)

are a fundamental set of solutions of Eq. (i), then W[y

(1)

, y

(2)

]=cW[x

(1)

, x

(2)

], where c is a

nonzero constant.

Hint: y

(1)

(t) and y

(2)

(t) must be linear combinations of x

(t) and x

(t).

5. Show that the general solution of x

′

= P(t)x + g(t) is the sum of any particular solution x

(p)

of this equation and the general solution x

(c)

of the corresponding homogeneous equation.

6. Consider the vectors x

(1)

(t) =

(

and x

(2)

(t) =

(

(a) Compute the Wronskian of x

(1)

and x

(2)

(b) In what intervals are x

(1)

and x

(2)

linearly independent?

differential equations satisﬁed by x

(1)

and x

(2)

(d) Find this system of equations and verify the conclusions of part (c).

7. Consider the vectors x

(1)

(t) =

(

and x

(2)

(t) =

(

, and answer the same questions as

in Problem 6.

The following two problems indicate an alternative derivation of Theorem 7.4.2.

8. Let x

(1)

, ..., x

(m)

be solutions of x

′

= P(t)x on the interval α<t <β. Assume that P is

continuous, and let t

be an arbitrary point in the given interval. Show that x

(1)

, ..., x

(m)

are linearly dependent for α<t <βif (and only if) x

(1)

), ..., x

(m)

) are linearly depen-

dent. In other words x

(1)

, ..., x

(m)

are linearly dependent on the interval (α, β) if they are

linearly dependent at any point in it.

Hint: There are constants c

, ..., c

that satisfy c

(1)

) +···+c

(m)

) = 0. Let

z(t) = c

(1)

(t) +···+c

(m)

(t), and use the uniqueness theorem to show that z(t) = 0

for each t in α<t <β.

9. Let x

(1)

, ..., x

(n)

be linearly independent solutions of x

′

= P(t)x, where P is continuous on

α<t <β.

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396 Chapter 7. Systems of First Order Linear Equations

(a) Show that any solution x = z(t) can be written in the form

z(t) = c

(1)

(t) +···+c

(n)

(t)

for suitable constants c

, ..., c

Hint: Use the result of Problem 12 of Section 7.3, and also Problem 8 above.

(b) Show that the expression for the solution z(t) in part (a) is unique; that is, if

z(t) = k

(1)

(t) +···+k

(n)

(t), then k

= c

, ..., k

= c

Hint: Show that (k

− c

(1)

(t) +···+(k

− c

(n)

(t) = 0 for each t in α<t <β, and use

the linear independence of x

(1)

, ..., x

(n)

7.5 Homogeneous Linear Systems with Constant Coefﬁcients

We will concentrate most of our attention on systems of homogeneous linear

equations with constant coefﬁcients—that is, systems of the form

′

= Ax, (1)

where A is a constant n × n matrix. Unless stated otherwise, we will assume further

that all the elements of A are real (rather than complex) numbers.

If n = 1, then the system reduces to a single ﬁrst order equation

= ax, (2)

whose solution is x = ce

. Note that x = 0 is the only equilibrium solution if a ̸= 0.

If a < 0, then other solutions approach x = 0ast increases, and in this case we say

that x = 0 is an asymptotically stable equilibrium solution. On the other hand, if

a > 0, then x = 0 is unstable, since other solutions depart from it with increasing t.

For systems of n equations,the situation is similar but more complicated. Equilibrium

solutions are found by solving Ax = 0. We usually assume that det A ̸= 0, so x = 0

is the only equilibrium solution. An important question is whether other solutions

approach this equilibrium solution or depart from it as t increases; in other words, is

x = 0 asymptotically stable or unstable? Or are there still other possibilities?

The case n = 2 is particularly important and lends itself to visualization in the

-plane, called the phase plane. By evaluating Ax at a large number of points

and plotting the resulting vectors, we obtain a direction ﬁeld of tangent vectors to

solutions of the system of differential equations. A qualitative understanding of the

behavior of solutions can usually be gained from a direction ﬁeld. More precise

information results from including in the plot some solution curves, or trajectories.

A plot that shows a representative sample of trajectories for a given system is called

a phase portrait. A well-constructed phase portrait provides easily understood infor-

mation about all solutions of a two-dimensional system in a single graphical display.

Although creating quantitatively accurate phase portraits requires computer assis-

tance, it is usually possible to sketch qualitatively accurate phase portraits by hand,

as we demonstrate in Examples 2 and 3 below.

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7.5 Homogeneous Linear Systems with Constant Coefﬁcients 397

Our ﬁrst task, however, is to show how to ﬁnd solutions of systems such as Eq. (1).

We start with a particularly simple example.

EXAMPLE

Find the general solution of the system

′

0 −3

x. (3)

The most important feature of this system is that the coefﬁcient matrix is a diagonal matrix.

Thus, by writing the system in scalar form, we obtain

′

= 2x

, x

′

=−3x

Each of these equations involves only one of the unknown variables, so we can solve the two

equations separately. In this way we ﬁnd that

= c

, x

= c

−3t

where c

and c

are arbitrary constants. Then, by writing the solution in vector form, we have

x =

−3t

= c

+ c

−3t

= c

+ c

−3t

. (4)

Now we deﬁne the two solutions x

(1)

and x

(2)

so that

(1)

(t) =

, x

(2)

(t) =

−3t

, (5)

The Wronskian of these solutions is

W[x

(1)

, x

(2)

](t) =

0 e

−3t

= e

−t

, (6)

which is never zero. Therefore, x

(1)

and x

(2)

form a fundamental set of solutions, and the

general solution of Eq. (3) is given by Eq. (4).

In Example 1 we found two independent solutions of the given system (3) in

the form of an exponential function multiplied by a vector. This was perhaps to be

expected since we have found other linear equations with constant coefﬁcients to

have exponential solutions, and the unknown x in the system (3) is a vector. So let us

try to extend this idea to the general system (1) by seeking solutions of the form

x = ξe

, (7)

where the exponent r and the vector ξ are to be determined. Substituting from Eq. (7)

for x in the system (1) gives

rξe

= Aξe

Upon canceling the nonzero scalar factor e

, we obtain Aξ = rξ, or

(A − rI)ξ = 0, (8)

where I is the n × n identity matrix. Thus, to solve the system of differential equa-

tions (1), we must solve the system of algebraic equations (8). This latter problem is

precisely the one that determines the eigenvalues and eigenvectors of the matrix A.

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398 Chapter 7. Systems of First Order Linear Equations

Therefore, the vector x given by Eq. (7) is a solution of Eq. (1), provided that r is an

eigenvalue and ξ an associated eigenvector of the coefﬁcient matrix A.

The following two examples are typical of 2 × 2 systems with eigenvalues that are

real and different. In each example we will solve the system and construct a corre-

sponding phase portrait. We will see that solutions have very distinct geometrical

patterns depending on whether the eigenvalues have the same sign or different signs.

Later in the section we return to a further discussion of the general n × n system.

EXAMPLE

Consider the system

′

x. (9)

Plot a direction ﬁeld and determine the qualitative behavior of solutions.Then ﬁnd the general

solution and draw a phase portrait showing several trajectories.

A direction ﬁeld for this system is shown in Figure 7.5.1. By following the arrows in this

ﬁgure, you can see that a typical solution in the second quadrant eventually moves into the

ﬁrst or third quadrant, and likewise for a typical solution in the fourth quadrant. On the

other hand, no solution leaves either the ﬁrst or the third quadrant. Further, it appears that

a typical solution departs from the neighborhood of the origin and ultimately has a slope of

approximately 2.

–1

–2

–2 –1 1 2

FIGURE 7.5.1 Direction ﬁeld for the system (9).

To ﬁnd solutions explicitly, we assume that x = ξe

and substitute for x in Eq. (9). We are

led to the system of algebraic equations

1 − r 1

41− r

. (10)

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7.5 Homogeneous Linear Systems with Constant Coefﬁcients 399

Equations (10) have a nontrivial solution if and only if the determinant of coefﬁcients is zero.

Thus, allowable values of r are found from the equation

1 − r 1

41− r

= (1 − r)

− 4

= r

− 2r − 3 = (r − 3)(r + 1) = 0. (11)

Equation (11) has the roots r

= 3 and r

=−1; these are the eigenvalues of the coefﬁcient

matrix in Eq. (9). If r = 3, then the system (10) reduces to the single equation

−2ξ

+ ξ

= 0. (12)

Thus ξ

= 2ξ

, and the eigenvector corresponding to r

= 3 can be taken as

(1)

. (13)

Similarly, corresponding to r

=−1, we ﬁnd that ξ

=−2ξ

, so the eigenvector is

(2)

−2

. (14)

The corresponding solutions of the differential equation are

(1)

(t) =

, x

(2)

(t) =

−2

−t

. (15)

The Wronskian of these solutions is

W[x

(1)

, x

(2)

](t) =

−t

−2e

−t

=−4e

, (16)

which is never zero. Hence the solutions x

(1)

and x

(2)

form a fundamental set, and the general

solution of the system (9) is

x = c

(1)

(t) + c

(2)

(t)

= c

+ c

−2

−t

, (17)

where c

and c

are arbitrary constants.

To visualize the solution (17), it is helpful to consider its graph in the x

-plane for various

values of the constants c

and c

. We start with x = c

(1)

(t) or, in scalar form,

= c

, x

= 2c

By eliminating t between these two equations, we see that this solution lies on the straight line

= 2x

;see Figure 7.5.2a.This is the line through the origin in the direction of the eigenvector

(1)

. If we look on the solution as the trajectory of a moving particle, then the particle is in the

ﬁrst quadrant when c

> 0 and in the third quadrant when c

< 0. In either case the particle

departs from the origin as t increases. Next consider x = c

(2)

(t), or

= c

−t

, x

=−2c

−t

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400 Chapter 7. Systems of First Order Linear Equations

This solution lies on the line x

=−2x

, whose direction is determined by the eigenvector ξ

(2)

The solution is in the fourth quadrant when c

> 0 and in the second quadrant when c

< 0, as

shown in Figure 7.5.2a. In both cases the particle moves toward the origin as t increases. The

solution (17) is a combination of x

(1)

(t) and x

(2)

(t). For large t the term c

(1)

(t) is dominant

and the term c

(2)

(t) becomes negligible.Thus all solutions for which c

̸= 0 are asymptotic to

the line x

= 2x

as t →∞. Similarly, all solutions for which c

̸= 0 are asymptotic to the line

=−2x

as t →−∞.A phase portrait for the system including the graphs of severalsolutions

is shown in Figure 7.5.2a. The pattern of trajectories in this ﬁgure is typical of all 2 × 2 systems

′

= Ax for which the eigenvalues are real and of opposite signs. The origin is called a saddle

point in this case. Saddle points are always unstable because almost all trajectories depart

from them as t increases.

1–1–2 2

–1

–2

(2)

(t)

(1)

(t)

(a)

0.5

–2

(b)

FIGURE 7.5.2 (a) A phase portrait for the system (9); the origin is a saddle point.

(b) Typical plots of x

versus t for the system (9).

In the preceding paragraph, we have described how to draw by hand a qualitatively correct

sketch of the trajectories of a system such as Eq. (9), once the eigenvalues and eigenvectors

have been determined. However, to produce a detailed and accurate drawing, such as Figure

7.5.2a and other ﬁgures that appear later in this chapter, a computer is extremely helpful, if

not indispensable.

As an alternative to Figure 7.5.2a, you can also plot x

or x

as a function of t; some typical

plots of x

versus t are shown in Figure 7.5.2b, and those of x

versus t are similar. For certain

initial conditions it follows that c

= 0 in Eq. (17),so that x

= c

−t

and x

→ 0ast →∞. One

such graph is shownin Figure 7.5.2b,corresponding to a trajectory that approachesthe originin

Figure 7.5.2a. For most initial conditions, however, c

̸= 0 and x

is given by x

= c

+ c

−t

Then the presence of the positive exponential term causes x

to grow exponentially in mag-

nitude as t increases. Several graphs of this type are shown in Figure 7.5.2b, corresponding to

trajectories that depart from the neighborhood of the origin in Figure 7.5.2a. It is important to

understand the relation between parts (a) and (b) of Figure 7.5.2 and other similar ﬁgures that

appear later, since you may want to visualize solutions either in the x

-plane or as functions

of the independent variable t.

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7.5 Homogeneous Linear Systems with Constant Coefﬁcients 401

EXAMPLE

Consider the system

′

−3

√

2 −2

x. (18)

Draw a direction ﬁeld for this system and ﬁnd its general solution. Then plot a phase portrait

showing several typical trajectories in the phase plane.

Thedirection ﬁeld for the system (18) in Figure7.5.3 shows clearly that all solutions approach

the origin. To ﬁnd the solutions, we assume that x = ξe

; then we obtain the algebraic system

−3 − r

√

2 −2 − r

. (19)

The eigenvalues satisfy

(−3 − r)(−2 − r) − 2 = r

+ 5r + 4

= (r + 1)(r + 4) = 0, (20)

so r

=−1 and r

=−4. For r =−1, Eq. (19) becomes

−2

√

2 −1

. (21)

Hence ξ

√

2 ξ

, and the eigenvector ξ

(1)

corresponding to the eigenvalue r

=−1 can be

taken as

(1)

√

. (22)

Similarly, corresponding to the eigenvalue r

=−4 we have ξ

=−

√

2 ξ

, so the eigenvector is

(2)

−

√

. (23)

Thus a fundamental set of solutions of the system (18) is

(1)

(t) =

√

−t

, x

(2)

(t) =

−

√

−4t

, (24)

–1

–2

–2 –1 1 2

FIGURE 7.5.3 Direction ﬁeld for the system (18).

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402 Chapter 7. Systems of First Order Linear Equations

and the general solution is

x = c

(1)

(t) + c

(2)

= c

√

−t

+ c

−

√

−4t

. (25)

A phase portrait for the system (18) is constructed by drawing graphs of the solution (25)

for several values of c

and c

, as shown in Figure 7.5.4a. The solution x

(1)

(t) approaches

the origin along the line x

√

2 x

, and the solution x

(2)

(t) approaches the origin along the

line x

=−

√

2 x

. The directions of these lines are determined by the eigenvectors ξ

(1)

and

(2)

, respectively. In general, we have a combination of these two fundamental solutions. As

t →∞, the solution x

(2)

(t) is negligible compared to x

(1)

(t). Thus, unless c

= 0, the solution

(25) approaches the origin tangent to the line x

√

. The pattern of trajectories shown

in Figure 7.5.4a is typical of all 2 × 2 systems x

′

= Ax for which the eigenvalues are real,

different, and of the same sign. The origin is called a node for such a system. If the eigenvalues

were positive rather than negative, then the trajectories would be similar but traversed in

the outward direction. Nodes are asymptotically stable if the eigenvalues are negative and

unstable if the eigenvalues are positive.

–2

–1

(a)

(1)

(t)

(2)

(t)

0.5 1

–1

(b)

FIGURE 7.5.4 (a) A phase portrait for the system (18); the origin is an asynptotically

stable node. (b) Typical plots of x

versus t for the system (18).

Although Figure 7.5.4a was computer-generated, a qualitatively correct sketch of the tra-

jectories can be drawn quickly by hand on the basis of a knowledge of the eigenvalues and

eigenvectors.

Some typical plots of x

versus t are shown in Figure 7.5.4b. Observe that each of the

graphs approaches the t-axis asymptotically as t increases, corresponding to a trajectory that

approaches the origin in Figure 7.5.2a. The behavior of x

as a function of t is similar.

Examples 2 and 3illustrate thetwo maincases for2 × 2 systems having eigenvalues

that are real and different. The eigenvalues have either opposite signs (Example 2)

or the same sign (Example 3). The other possibility is that zero is an eigenvalue,

but in this case it follows that det A = 0, which violates the assumption made at the

beginning of this section. However, see Problems 7 and 8.

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7.5 Homogeneous Linear Systems with Constant Coefﬁcients 403

Returning to the general system (1), we proceed as in the examples. To ﬁnd solu-

tions of the differential equation (1), we must ﬁnd the eigenvalues and eigenvectors

of A from the associated algebraic system (8). The eigenvalues r

, ..., r

(which need

not all be different) are roots of the nth degree polynomial equation

det(A − rI) = 0. (26)

The nature of the eigenvalues and the corresponding eigenvectors determines the

nature of the general solution of the system (1). If we assume that A is a real-valued

matrix, then we must consider the following possibilities for the eigenvalues of A:

1. All eigenvalues are real and different from each other.

2. Some eigenvalues occur in complex conjugate pairs.

3. Some eigenvalues, either real or complex, are repeated.

If the n eigenvalues are all real and different, as in the three preceding examples,

then associated with each eigenvalue r

is a real eigenvector ξ

(i)

, and the n eigen-

vectors ξ

(1)

, ..., ξ

(n)

are linearly independent. The corresponding solutions of the

differential system (1) are

(1)

(t) = ξ

(1)

, ..., x

(n)

(t) = ξ

(n)

. (27)

To show that these solutions form a fundamental set, we evaluate their Wronskian:

W[x

(1)

, ..., x

(n)

](t) =

(1)

··· ξ

(n)

(1)

··· ξ

(n)

= e

+···+r

(1)

··· ξ

(n)

(1)

··· ξ

(n)

. (28)

First, we observe that the exponential function is never zero. Next, since the eigen-

vectors ξ

(1)

, ..., ξ

(n)

are linearly independent, the determinant in the last term of

Eq. (28) is nonzero. As a consequence, the Wronskian W[x

(1)

, ..., x

(n)

](t) is never

zero; hence x

(1)

, ..., x

(n)

form a fundamental set of solutions. Thus the general

solution of Eq. (1) is

x = c

(1)

+···+c

(n)

. (29)

If A is real and symmetric (a special case of Hermitian matrices), recall from Sec-

tion 7.3 that all the eigenvalues r

, ..., r

must be real. Further, even if some of the

eigenvalues are repeated, there is always a full set of n eigenvectors ξ

(1)

, ..., ξ

(n)

that

are linearly independent (in fact, orthogonal). Hence the corresponding solutions of

the differentialsystem (1)given byEq. (27) again form a fundamental set of solutions,

and the general solution is again given by Eq. (29). The following example illustrates

this case.

EXAMPLE

Find the general solution of

′

⎛

⎜

⎝

011

101

110

⎞

⎟

⎠

x. (30)

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404 Chapter 7. Systems of First Order Linear Equations

Observe that the coefﬁcient matrix is real and symmetric.The eigenvalues and eigenvectors

of this matrix were found in Example 5 of Section 7.3:

= 2, ξ

(1)

⎛

⎜

⎝

⎞

⎟

⎠

; (31)

=−1, r

=−1; ξ

(2)

⎛

⎜

⎝

−1

⎞

⎟

⎠

, ξ

(3)

⎛

⎜

⎝

−1

⎞

⎟

⎠

. (32)

Hence a fundamental set of solutions of Eq. (30) is

(1)

(t) =

⎛

⎜

⎝

⎞

⎟

⎠

, x

(2)

(t) =

⎛

⎜

⎝

−1

⎞

⎟

⎠

−t

, x

(3)

(t) =

⎛

⎜

⎝

−1

⎞

⎟

⎠

−t

, (33)

and the general solution is

x = c

⎛

⎜

⎝

⎞

⎟

⎠

+ c

⎛

⎜

⎝

−1

⎞

⎟

⎠

−t

+ c

⎛

⎜

⎝

−1

⎞

⎟

⎠

−t

. (34)

This example illustrates the fact that even though an eigenvalue (r =−1) has algebraic mul-

tiplicity 2, it may still be possible to ﬁnd two linearly independent eigenvectors ξ

(2)

and ξ

(3)

and, as a consequence, to construct the general solution (34).

The behavior of the solution (34) depends critically on the initial conditions. For large t the

ﬁrst term on the right side of Eq. (34) is the dominant one; therefore, if c

̸= 0, all components

of x become unbounded as t →∞. On the other hand, for certain initial points c

will be zero.

In this case, the solution involves only the negative exponential terms, and x → 0 as t →∞.

The initial points that cause c

to be zero are precisely those that lie in the plane determined by

the eigenvectors ξ

(2)

and ξ

(3)

corresponding to the two negative eigenvalues.Thus solutions that

start in this plane approach the origin as t →∞, while all other solutions become unbounded.

If some of the eigenvalues occur in complex conjugate pairs, then there are still n

linearly independent solutions of the form (27), provided that all the eigenvalues are

different. Of course, the solutions arising from complex eigenvalues are complex-

valued. However, as in Section 3.3, it is possible to obtain a full set of real-valued

solutions. This is discussed in Section 7.6.

More serious difﬁculties can occur if an eigenvalue is repeated. In this event the

number of corresponding linearly independent eigenvectors may be smaller than

the algebraic multiplicity of the eigenvalue. If so,the number of linearly independent

solutions of the form ξe

will be smaller than n. To construct a fundamental set

of solutions, it is then necessary to seek additional solutions of another form. The

situation issomewhat analogousto thatfor an nth order linear equation with constant

coefﬁcients; a repeated root of the characteristic equation gave rise to solutions of

the form e

, te

, t

, ....The case of repeated eigenvalues is treated in Section 7.8.

Finally, if A is complex, then complex eigenvalues need not occur in conjugate

pairs, and the eigenvectors are normally complex-valued even though the associated

eigenvalue may be real. The solutions of the differential equation (1) are still of the

August 7, 2012 21:04 c07 Sheet number 47 Page number 405 cyan black

7.5 Homogeneous Linear Systems with Constant Coefﬁcients 405

form (27),provided that there are n linearly independent eigenvectors, but in general

all the solutions are complex-valued.

PROBLEMS In each of Problems 1 through 6:

(a) Find the general solution of the given system of equations and describe the behavior of

the solution as t →∞.

(b) Draw a direction ﬁeld and plot a few trajectories of the system.

′

3 −2

2 −2

x 2.

′

1 −2

3 −4

′

2 −1

3 −2

x 4.

′

4 −2

′

−21

1 −2

x 6.

′

In each of Problems 7 and 8:

(a) Find the general solution of the given system of equations.

(b) Draw a direction ﬁeld and a few of the trajectories. In each of these problems, the coef-

ﬁcient matrix has a zero eigenvalue. As a result, the pattern of trajectories is different from

those in the examples in the text.

′

4 −3

8 −6

x 8.

′

−1 −2

In each of Problems 9 through 14, ﬁnd the general solution of the given system of equations.

9. x

′

1 i

−i 1

x 10. x

′

22+ i

−1 −1 − i

11. x

′

⎛

⎜

⎝

112

121

211

⎞

⎟

⎠

x 12. x

′

⎛

⎜

⎝

324

202

423

⎞

⎟

⎠

13. x

′

⎛

⎜

⎝

111

21−1

−8 −5 −3

⎞

⎟

⎠

x 14. x

′

⎛

⎜

⎝

1 −14

32−1

21−1

⎞

⎟

⎠

In each of Problems 15 through 18,solve the given initial value problem. Describe the behavior

of the solution as t →∞.

15. x

′

5 −1

x, x(0) =

−1

16. x

′

−21

−54

x, x(0) =

17. x

′

⎛

⎜

⎝

112

022

−113

⎞

⎟

⎠

x, x(0) =

⎛

⎜

⎝

⎞

⎟

⎠

18. x

′

⎛

⎜

⎝

00−1

200

−124

⎞

⎟

⎠

x, x(0) =

⎛

⎜

⎝

⎞

⎟

⎠

19. The system tx

′

= Ax is analogous to the second order Euler equation (Section 5.4).

Assuming that x = ξt

, where ξ is a constant vector, show that ξ and r must satisfy

(A − rI)ξ = 0 in order to obtain nontrivial solutions of the given differential equation.

August 7, 2012 21:04 c07 Sheet number 48 Page number 406 cyan black

406 Chapter 7. Systems of First Order Linear Equations

Referring to Problem 19, solve the given system of equations in each of Problems 20 through

23. Assume that t > 0.

20. tx

′

2 −1

3 −2

x 21. tx

′

5 −1

22. tx

′

4 −3

8 −6

x 23. tx

′

3 −2

2 −2

In each of Problems 24 through 27, the eigenvalues and eigenvectors of a matrix A are given.

Consider the corresponding system x

′

= Ax.

(a) Sketch a phase portrait of the system.

(b) Sketch the trajectory passing through the initial point (2, 3).

versus t and of x

versus t on the

same set of axes.

24. r

=−1, ξ

(1)

−1

; r

=−2, ξ

(2)

25. r

= 1, ξ

(1)

−1

; r

=−2, ξ

(2)

26. r

=−1, ξ

(1)

−1

; r

= 2, ξ

(2)

27. r

= 1, ξ

(1)

; r

= 2, ξ

(2)

−2

28. Consider a 2 × 2 system x

′

= Ax. If we assume that r

̸= r

, the general solution is

x = c

(1)

+ c

(2)

,provided that ξ

(1)

and ξ

(2)

are linearly independent. In this prob-

lem we establish the linear independence of ξ

(1)

and ξ

(2)

by assuming that they are linearly

dependent and then showing that this leads to a contradiction.

(a) Note that ξ

(1)

satisﬁes the matrix equation (A − r

I)ξ

(1)

= 0; similarly, note that

(A − r

I)ξ

(2)

= 0.

(b) Show that (A − r

I)ξ

(1)

= (r

− r

)ξ

(1)

and ξ

(2)

are linearly dependent. Then c

(1)

+ c

(2)

= 0 and at least

one of c

and c

(say c

) is not zero. Show that (A − r

I)(c

(1)

+ c

(2)

) = 0,and also show

that (A − r

I)(c

(1)

+ c

(2)

) = c

− r

)ξ

(1)

. Hence c

= 0, which is a contradiction.

Therefore, ξ

(1)

and ξ

(2)

are linearly independent.

(d) Modify the argument of part (c) if we assume that c

̸= 0.

(e) Carry out a similar argument for the case in which the order n is equal to 3; note that

the procedure can be extended to an arbitrary value of n.

29. Consider the equation

′′

+ by

′

+ cy = 0, (i)

where a, b, and c are constants with a ̸= 0. In Chapter 3 it was shown that the general

solution depended on the roots of the characteristic equation

+ br + c = 0. (ii)

(a) Transform Eq. (i) into a system of ﬁrst order equations by letting x

= y, x

= y

′

. Find

the system of equations x

′

= Ax satisﬁed by x =

(

August 7, 2012 21:04 c07 Sheet number 49 Page number 407 cyan black

7.5 Homogeneous Linear Systems with Constant Coefﬁcients 407

(b) Find the equation that determines the eigenvalues of the coefﬁcient matrix A in

part (a). Note that this equation is just the characteristic equation (ii) of Eq. (i).

30.

The two-tank system of Problem 22 in Section 7.1 leads to the initial value problem

′

−

x, x(0) =

−17

−21

where x

and x

are the deviations of the salt levels Q

and Q

from their respective

equilibria.

(a) Find the solution of the given initial value problem.

(b) Plot x

versus t and x

versus t on the same set of axes.

(t)|≤0 .5 and |x

(t)|≤0.5 for all t ≥ T.

31. Consider the system

′

−1 −1

−α −1

(a) Solve the system for α = 0.5. What are the eigenvalues of the coefﬁcient matrix?

Classify the equilibrium point at the origin as to type.

(b) Solve the system for α = 2.Whatare the eigenvaluesof the coefﬁcient matrix? Classify

the equilibrium point at the origin as to type.

ior. Find the eigenvalues of the coefﬁcient matrix in terms of α, and determine the value

of α between 0.5 and 2 where the transition from one type of behavior to the other occurs.

Electric Circuits. Problems 32 and 33 are concerned with the electric circuit described by the

system of differential equations in Problem 21 of Section 7.1:

⎛

⎜

⎝

−

⎞

⎟

⎠

. (i)

32. (a) Find the general solution of Eq. (i) if R

= 1 4, R

4, L = 2 H, and C =

(b) Show that I(t) → 0 and V(t) → 0ast →∞, regardless of the initial values I(0) and

V(0).

33. Consider the preceding system of differential equations (i).

(a) Find a condition on R

, R

, C, and L that must be satisﬁed if the eigenvalues of the

coefﬁcient matrix are to be real and different.

(b) If the condition found in part (a) is satisﬁed, show that both eigenvalues are negative.

Then show that I(t) → 0 and V(t) → 0 as t →∞, regardless of the initial conditions.

complex or repeated. Do you think that I(t) → 0 and V(t) → 0ast →∞in these cases

as well?

Hint: In part (c), one approach is to change the system (i) into a single second order

equation. We also discuss complex and repeated eigenvalues in Sections 7.6 and 7.8.

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408 Chapter 7. Systems of First Order Linear Equations

7.6 Complex Eigenvalues

In this section we consider again a system of n linear homogeneous equations with

constant coefﬁcients

′

= Ax, (1)

where the coefﬁcient matrixA is real-valued. If we seek solutions of the form x = ξe

then it follows, as in Section 7.5, that r must be an eigenvalue and ξ a corresponding

eigenvector of the coefﬁcient matrix A. Recall that the eigenvalues r

, ..., r

of A

are the roots of the characteristic equation

det(A − rI) = 0 (2)

and that the corresponding eigenvectors satisfy

(A − rI)ξ = 0. (3)

If A is real, then the coefﬁcients in the polynomial equation (2) for r are real, and

any complex eigenvalues must occur in conjugate pairs. For example, if r

= λ + iµ,

where λ and µ are real, is an eigenvalue of A, then so is r

= λ − iµ. To explore the

effect of complex eigenvalues, we begin with an example.

EXAMPLE

Find a fundamental set of real-valued solutions of the system

′

−

−1 −

x. (4)

Plot a phase portrait and graphs of components of typical solutions.

A direction ﬁeld for the system (4) is shown in Figure 7.6.1. This plot suggests that the

trajectories in the phase plane spiral clockwise toward the origin.

To ﬁnd a fundamental set of solutions, we assume that

x = ξe

(5)

and obtain the set of linear algebraic equations

−

− r 1

−1 −

− r

(6)

for the eigenvalues and eigenvectors of A . The characteristic equation is

−

− r 1

−1 −

− r

= r

+ r +

= 0; (7)

therefore the eigenvalues are r

=−

+ i and r

=−

− i. From Eq. (6) a straightforward

calculation shows that the corresponding eigenvectors are

(1)

, ξ

(2)

−i

. (8)

Observe that the eigenvectors ξ

(1)

and ξ

(2)

are also complex conjugates. Hence a fundamental

set of solutions of the system (4) is

(1)

(t) =

(−1/2+i)t

, x

(2)

(t) =

−i

(−1/2−i)t

. (9)

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7.6 Complex Eigenvalues 409

–2

–1

12–2 –1

FIGURE 7.6.1 A direction ﬁeld for the system (4).

To obtain a set of real-valued solutions, we can (by Theorem 7.4.5) choose the real and

imaginary parts of either x

(1)

or x

(2)

. In fact,

(1)

(t) =

−t/2

(cos t + i sin t) =

−t/2

cos t

−e

−t/2

sin t

+ i

−t/2

sin t

−t/2

cos t

. (10)

Hence a set of real-valued solutions of (Eq. 4) is

u(t) = e

−t/2

cos t

−sin t

, v(t) = e

−t/2

sin t

cos t

. (11)

To verify that u(t) and v(t) are linearly independent, we compute their Wronskian:

W(u , v)(t) =

−t/2

cos te

−t/2

sin t

−e

−t/2

sin te

−t/2

cos t

= e

−t

(cos

t + sin

t) = e

−t

TheWronskian W(u, v)(t) is never zero,so it follows that u(t) and v(t) constitute a fundamental

set of (real-valued) solutions of the system (4).

The graphs of the solutions u(t) and v(t) are shown in Figure 7.6.2a. Since

u(0) =

, v(0) =

the graphs of u(t) and v(t) pass through the points (1, 0 ) and (0,1),respectively. Other solutions

of the system (4) are linear combinations of u(t) and v (t),and graphs of a few of these solutions

are also shown in Figure 7.6.2a; this ﬁgure is a phase portrait for the system (4). Each trajectory

approaches the origin along a spiral path as t →∞, making inﬁnitely many circuits about the

origin; this is due to the fact that the solutions (11) are products of decaying exponential and

sine or cosine factors. Some typical graphs of x

versus t are shown in Figure 7.6.2b; each one

represents a decaying oscillation in time.

August 7, 2012 21:04 c07 Sheet number 52 Page number 410 cyan black

410 Chapter 7. Systems of First Order Linear Equations

u(t)

v(t)

–1

–2

21 –1–2

(a)

–1

(b)

FIGURE 7.6.2 (a) A phase portrait for the system (4); the origin is a spiral point.

(b) Plots of x

versus t for the system (4); graphs of x

versus t are similar.

Figure 7.6.2a is typical of all 2 × 2 systems x

′

= Ax whose eigenvalues are complex with

negative real part. The origin is called a spiral point and is asymptotically stable because all

trajectories approach it as t increases. For a system whose eigenvalues have a positive real

part, the trajectories are similar to those in Figure 7.6.2a, but the direction of motion is away

from the origin, and the trajectories become unbounded. In this case, the origin is unstable. If

the real part of the eigenvalues is zero, then the trajectories neither approach the origin nor

become unbounded but instead repeatedly traverse a closed curve about the origin. Examples

of this behavior can be seen in Figures 7.6.3b and 7.6.4b below. In this case the origin is called

a center and is said to be stable, but not asymptotically stable. In all three cases, the direction

of motion may be either clockwise, as in this example, or counterclockwise, depending on the

elements of the coefﬁcient matrix A.

The phase portrait in Figure 7.6.2a was drawn by a computer, but it is possible to produce

a useful sketch of the phase portrait by hand. We have noted that when the eigenval-

ues λ ± iµ are complex, then the trajectories either spiral in (λ<0), spiral out (λ > 0),or

repeatedly traverse a closed curve (λ = 0). To determine whether the direction of motion

is clockwise or counterclockwise, we only need to determine the direction of motion at a

single convenient point. For instance, in the system (4) we might choose x = (0, 1)

. Then

Ax = (1, −

)

. Thus at the point (0,1) in the phase plane the tangent vector x

′

to the tra-

jectory at that point has a positive x

-component and therefore is directed from the second

quadrant into the ﬁrst. The direction of motion is therefore clockwise for the trajectories of

this system.

Returning to the general equation (1)

′

= Ax,

we can proceed just as in the example. Suppose that there is a pair of complex

conjugate eigenvalues, r

= λ + iµ and r

= λ − iµ. Then the corresponding eigen-

vectors ξ

(1)

and ξ

(2)

are also complex conjugates. To see that this is so, recall that

August 7, 2012 21:04 c07 Sheet number 53 Page number 411 cyan black

7.6 Complex Eigenvalues 411

and ξ

(1)

satisfy

(A − r

I)ξ

(1)

= 0. (12)

On taking the complex conjugate of this equation and noting that A and I are real-

valued, we obtain

(A −

I)ξ

(1)

= 0, (13)

where

and ξ

(1)

are the complex conjugates of r

and ξ

(1)

, respectively. In other

words, r

= r

is also an eigenvalue, and ξ

(2)

= ξ

(1)

is a corresponding eigenvector.

The corresponding solutions

(1)

(t) = ξ

(1)

, x

(2)

(t) = ξ

(1)

(14)

of the differential equation (1) are then complex conjugates of each other.Therefore,

as in Example 1, we can ﬁnd two real-valued solutions of Eq. (1) corresponding to

the eigenvalues r

and r

by taking the real and imaginary parts of x

(1)

(t) or x

(2)

(t)

given by Eq. (14).

Let us write ξ

(1)

= a + ib, where a and b are real; then we have

(1)

(t) = (a + ib)e

(λ+iµ)t

= (a + ib)e

λt

(cos µt + i sin µt). (15)

Upon separating x

(1)

(t) into its real and imaginary parts, we obtain

(1)

(t) = e

λt

(a cos µt − b sin µt) + ie

λt

(a sin µt + b cos µt). (16)

If we write x

(1)

(t) = u(t) + iv(t), then the vectors

u(t) = e

λt

(a cos µt − b sin µt),

(17)

v(t) = e

λt

(a sin µt + b cos µt)

are real-valued solutions of Eq. (1). It is possible to show that u and v are linearly

independent solutions (see Problem 27).

For example, suppose that the matrix A has two complex eigenvalues r

= λ + iµ,

= λ − iµ, and that r

, ..., r

are all real and distinct. Let the corresponding eigen-

vectors be ξ

(1)

= a + ib, ξ

(2)

= a − ib, ξ

(3)

, ..., ξ

(n)

. Then the general solution of

Eq. (1) is

x = c

u(t) + c

v(t) + c

(3)

+···+c

(n)

, (18)

where u(t) and v(t) are given by Eqs. (17). We emphasize that this analysis applies

only if the coefﬁcient matrix A in Eq. (1) is real, for it is only then that complex

eigenvalues and eigenvectors must occur in conjugate pairs.

For 2 × 2 systems with real coefﬁcients, we have now completed our description

of the three main cases that can occur.

1. Eigenvalues are real and have opposite signs; x = 0 is a saddle point.

2. Eigenvalues are real and have the same sign but are unequal; x = 0 is a node.

3. Eigenvalues are complex with nonzero real part; x = 0 is a spiral point.

August 7, 2012 21:04 c07 Sheet number 54 Page number 412 cyan black

412 Chapter 7. Systems of First Order Linear Equations

Other possibilities are of less importance and occur as transitions between two

of the cases just listed. For example, a zero eigenvalue occurs during the transition

between a saddle point and a node. Purely imaginary eigenvalues occur during a

transition between asymptotically stable and unstable spiral points. Finally, real and

equal eigenvalues appear during the transition between nodes and spiral points.

EXAMPLE

The system

′

α 2

−20

x (19)

contains a parameter α. Describe how the solutions depend qualitatively on α; in particular,

ﬁnd the critical values of α at which the qualitative behavior of the trajectories in the phase

plane changes markedly.

The behavior of the trajectories is controlled by the eigenvalues of the coefﬁcient matrix.

The characteristic equation is

− αr + 4 = 0, (20)

so the eigenvalues are

r =

α ±

√

− 16

. (21)

From Eq. (21) it follows that the eigenvalues are complex conjugates for −4 <α<4 and are

real otherwise. Thus two critical values are α =−4 and α = 4, where the eigenvalues change

from real to complex, or vice versa. For α<−4 both eigenvalues are negative, so all trajecto-

ries approach the origin, which is an asymptotically stable node. For α>4 both eigenvalues

are positive, so the origin is again a node, this time unstable; all trajectories (except x = 0)

become unbounded. In the intermediate range, −4 <α<4, the eigenvalues are complex and

the trajectories are spirals. However, for −4 <α<0 the real part of the eigenvalues is

negative, the spirals are directed inward, and the origin is asymptotically stable, whereas for

0 <α<4 the real part of the eigenvalues is positive and the origin is unstable. Thus α = 0is

also a critical value where the direction of the spirals changes from inward to outward. For

this value of α, the origin is a center and the trajectories are closed curves about the origin,

corresponding to solutions that are periodic in time. The other critical values, α =±4, yield

eigenvalues that are real and equal. In this case the origin is again a node,but the phase portrait

differs somewhat from those in Section 7.5. We take up this case in Section 7.8.

AMultipleSpring–MassSystem. Consider the system of two masses and three springs

shown in Figure 7.1.1, whose equations of motion are given by Eqs. (1) in Section

7.1. If we assume that there are no external forces, then F

(t) = 0, F

(t) = 0, and the

resulting equations are

=−(k

+ k

(22)

= k

− (k

+ k

These equations can be solved as a system of two second order equations (see

Problem 29), but,as is consistent with our approach in this chapter, we will transform

them into a system of four ﬁrst order equations. Let y

= x

, y

= x

, y

= x

′

, and

= x

′

. Then

′

= y

, y

′

= y

, (23)

August 7, 2012 21:04 c07 Sheet number 55 Page number 413 cyan black

7.6 Complex Eigenvalues 413

and, from Eqs. (22),

′

=−(k

+ k

, m

′

= k

− (k

+ k

. (24)

The following example deals with a particular case of this two-mass, three-spring

system.

EXAMPLE

Suppose that m

= 2, m

= 9/4, k

= 1, k

= 3, and k

= 15/4 in Eqs. (23) and (24) so that

these equations become

′

= y

, y

′

= y

, y

′

=−2y

, y

′

− 3y

. (25)

Analyze the possible motions described by Eqs. (25), and draw graphs showing typical

behavior.

We can write the system (25) in matrix form as

′

⎛

⎜

⎝

00 1 0

00 0 1

−2

−300

⎞

⎟

⎠

y = Ay . (26)

Keep in mind that y

and y

are the positions of the two masses, relative to their equilibrium

positions,and that y

and y

are their velocities.We assume, as usual,that y = ξe

,where r must

be an eigenvalue of the matrix A and ξ a corresponding eigenvector. It is possible, though a bit

tedious, to ﬁnd the eigenvalues and eigenvectors of A by hand, but it is easy with appropriate

computer software. The characteristic polynomial of A is

+ 5r

+ 4 = (r

+ 1)(r

+ 4), (27)

so the eigenvalues are r

= i, r

=−i, r

= 2i, and r

=−2i. The corresponding eigenvectors

are

(1)

⎛

⎜

⎝

⎞

⎟

⎠

, ξ

(2)

⎛

⎜

⎝

−3i

−2i

⎞

⎟

⎠

, ξ

(3)

⎛

⎜

⎝

−4

−8i

⎞

⎟

⎠

, ξ

(4)

⎛

⎜

⎝

−4

−6i

⎞

⎟

⎠

. (28)

The complex-valued solutions ξ

(1)

and ξ

(2)

−it

are complex conjugates, so two real-valued

solutions can be found by ﬁnding the real and imaginary parts of either of them. For instance,

we have

(1)

⎛

⎜

⎝

⎞

⎟

⎠

(cos t + i sin t)

⎛

⎜

⎝

3 cos t

2 cos t

−3 sin t

−2 sin t

⎞

⎟

⎠

+ i

⎛

⎜

⎝

3 sin t

2 sin t

3 cos t

2 cos t

⎞

⎟

⎠

= u

(1)

(t) + iv

(1)

(t). (29)

August 7, 2012 21:04 c07 Sheet number 56 Page number 414 cyan black

414 Chapter 7. Systems of First Order Linear Equations

In a similar way, we obtain

(3)

2it

⎛

⎜

⎝

−4

−8i

⎞

⎟

⎠

(cos 2t + i sin2t)

⎛

⎜

⎝

3 cos 2t

−4 cos 2t

−6 sin 2t

8 sin 2t

⎞

⎟

⎠

+ i

⎛

⎜

⎝

3 sin 2t

−4 sin 2t

6 cos 2t

−8 cos 2t

⎞

⎟

⎠

= u

(2)

(t) + iv

(2)

(t). (30)

We leave it to you to verify that u

(1)

, v

(1)

, u

(2)

, and v

(2)

are linearly independent and therefore

form a fundamental set of solutions. Thus the general solution of Eq. (26) is

y = c

⎛

⎜

⎝

3 cos t

2 cos t

−3 sin t

−2 sin t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

3 sin t

2 sin t

3 cos t

2 cos t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

3 cos 2t

−4 cos 2t

−6 sin 2t

8 sin 2t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

3 sin 2t

−4 sin 2t

6 cos 2t

−8 cos 2t

⎞

⎟

⎠

, (31)

where c

, c

, and c

are arbitrary constants.

The phase space for this system is four-dimensional, and each solution, obtained by a par-

ticular set of values for c

, ..., c

in Eq. (31), corresponds to a trajectory in this space. Since

each solution, given by Eq. (31), is periodic with period 2π, each trajectory is a closed curve.

No matter where the trajectory starts at t = 0, it returns to that point at t = 2π, t = 4π, and

so forth, repeatedly traversing the same curve in each time interval of length 2π.Wedonot

attempt to show any of these four-dimensional trajectories here. Instead, in the ﬁgures below

we show projections of certain trajectories in the y

-ory

-plane, thereby showing the

motion of each mass separately.

The ﬁrst two terms on the right side of Eq. (31) describe motions with frequency 1 and period

2π. Note that y

= (2/3)y

in these terms and that y

= (2/3)y

.This means that the two masses

move back and forth together, always going in the same direction, but with the second mass

moving only two-thirds as far as the ﬁrst mass. If we focus on the solution u

(1)

(t) and plot y

versus t and y

versus t on the same axes, we obtain the cosine graphs of amplitude 3 and 2,

respectively, shown in Figure 7.6.3a. The trajectory of the ﬁrst mass in the y

-plane lies on

the circle of radius 3 shown in Figure 7.6.3b, traversed clockwise starting at the point (3, 0)

and completing a circuit in time 2π. Also shown in this ﬁgure is the trajectory of the second

mass in the y

-plane, which lies on the circle of radius 2, also traversed clockwise starting at

(2, 0) and also completing a circuit in time 2π. The origin is a center in the respective y

- and

-planes. Similar graphs (with an appropriate shift in time) are obtained from v

(1)

or from

a linear combination of u

(1)

and v

(1)

The remaining terms on the right side of Eq. (31) describe motions with frequency 2 and

period π. Observe that in this case, y

=−(4/3)y

and y

=−(4/3)y

. This means that the two

masses are always moving in opposite directions and that the second mass moves four-thirds

as far as the ﬁrst mass. If we look only at u

(2)

(t) and plot y

versus t and y

versus t on the

same axes, we obtain Figure 7.6.4a. There is a phase difference of π, and the amplitude of y

is four-thirds that of y

, conﬁrming the preceding statements about the motions of the masses.

Figure 7.6.4b shows a superposition of the trajectories for the two masses in their respective

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7.6 Complex Eigenvalues 415

phase planes. Both graphs are ellipses, the inner one corresponding to the ﬁrst mass and the

outer one to the second. The trajectory on the inner ellipse starts at (3, 0), and the trajectory

on the outer ellipse starts at (−4, 0). Both are traversed clockwise, and a circuit is completed

in time π. The origin is a center in the respective y

- and y

-planes. Once again, similar

graphs are obtained from v

(2)

or from a linear combination of u

(2)

and v

(2)

–3

–2

–1

5 10 15

(a)

)

–3

–2

–1

–3 –2 –1 1 2

)

+ y

= 4

+ y

= 9

(b)

FIGURE 7.6.3 (a) A plot of y

versus t and y

versus t for the solution

(1)

(t).(b) Superposition of projections of trajectories in the

- and y

-planes for the solution u

(1)

(t).

–4

–2

26810124

(a)

–8

–6

–4

–2

–4 –2 2 4

)

+ y

= 64

+ y

= 36

(b)

FIGURE 7.6.4 (a) A plot of y

versus t and y

versus t for the solution

(2)

(t).(b) Superposition of projections of trajectories in the

- and y

-planes for the solution u

(2)

(t).

The types of motion described in the two preceding paragraphs are called fundamental

modes of vibration for the two-mass system. Each of them results from fairly special initial

conditions. For example, to obtain the fundamental mode of frequency 1, both of the con-

stants c

and c

in Eq. (31) must be zero. This occurs only for initial conditions in which

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416 Chapter 7. Systems of First Order Linear Equations

(0) = 2y

(0) and 3y

(0) = 2y

(0). Similarly, the mode of frequency 2 is obtained only when

both of the constants c

and c

in Eq. (31) are zero—that is, when the initial conditions are

such that 3y

(0) =−4y

(0) and 3y

(0) =−4y

(0).

For more general initial conditions the solution is a combination of the two fundamental

modes. A plot of y

versus t for a typical case is shown in Figure 7.6.5a, and the projection

of the corresponding trajectory in the y

-plane is shown in Figure 7.6.5b. Observe that this

latter ﬁgure may be a bit misleading in that it shows the projection of the trajectory crossing

itself. This cannot be the case for the actual trajectory in four dimensions, because it would

violate the general uniqueness theorem: there cannot be two different solutions issuing from

the same initial point.

–4

–2

5 10 15

(a)

–6

–4

–2

42–2–4

(b)

FIGURE 7.6.5 A solution of the system (25) satisfying the

initial condition y(0) = (−1, 4,1, 1)

.(a) A plot of y

versus t.(b) The

projection of the trajectory in the y

-plane. As stated in the text, the

actual trajectory in four dimensions does not intersect itself.

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7.6 Complex Eigenvalues 417

PROBLEMS In each of Problems 1 through 6:

(a) Express the general solution of the given system of equations in terms of real-valued

functions.

(b) Also draw a direction ﬁeld, sketch a few of the trajectories, and describe the behavior of

the solutions as t →∞.

′

3 −2

4 −1

x 2.

′

−1 −4

1 −1

′

2 −5

1 −2

x 4.

′

2 −

−1

′

1 −1

5 −3

x 6.

′

−5 −1

In each of Problems 7 and 8, express the general solution of the given system of equations in

terms of real-valued functions.

7. x

′

⎛

⎜

⎝

100

21−2

321

⎞

⎟

⎠

x 8. x

′

⎛

⎜

⎝

−302

1 −10

−2 −10

⎞

⎟

⎠

In each of Problems 9 and 10, ﬁnd the solution of the given initial value problem. Describe the

behavior of the solution as t →∞.

9. x

′

1 −5

1 −3

x, x(0) =

10. x

′

−32

−1 −1

x, x(0) =

−2

In each of Problems 11 and 12:

(a) Find the eigenvalues of the given system.

(b) Choose an initial point (other than the origin) and draw the corresponding trajectory in

the x

-plane.

versus t and of x

versus t.

(d) For your trajectory in part (b), draw the corresponding graph in three-dimensional

-space.

11.

′

−2

1 −

x 12.

′

−

−1

In each of Problems 13 through 20, the coefﬁcient matrix contains a parameter α. In each of

these problems:

(a) Determine the eigenvalues in terms of α.

(b) Find the critical value or values of α where the qualitative nature of the phase portrait for

the system changes.

each critical value.

13.

′

α 1

−1 α

x 14.

′

0 −5

1 α

15.

′

2 −5

α −2

x 16.

′

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418 Chapter 7. Systems of First Order Linear Equations

17. x

′

−1 α

−1 −1

x 18.

′

3 α

−6 −4

19.

′

α 10

−1 −4

x 20.

′

4 α

8 −6

In each of Problems 21 and 22, solve the given system of equations by the method of Problem

19 of Section 7.5. Assume that t > 0.

21. tx

′

−1 −1

2 −1

x 22. tx

′

2 −5

1 −2

In each of Problems 23 and 24:

(a) Find the eigenvalues of the given system.

(b) Choose an initial point (other than the origin) and draw the corresponding trajectory in

the x

-plane. Also draw the trajectories in the x

- and x

-planes.

-space.

23.

′

⎛

⎜

⎝

−

−1 −

00−

⎞

⎟

⎠

x 24.

′

⎛

⎜

⎝

−

−1 −

⎞

⎟

⎠

25. Consider the electric circuit shown in Figure 7.6.6. Suppose that R

= R

= 4 4,

C =

F, and L = 8 H.

(a) Show that this circuit is described by the system of differential equations

−

2 −

, (i)

where I is the current through the inductor and V is the voltage drop across the capacitor.

Hint: See Problem 20 of Section 7.1.

(b) Find the general solution of Eqs. (i) in terms of real-valued functions.

(d) Determine the limiting values of I(t) and V(t) as t →∞. Do these limiting values

depend on the initial conditions?

FIGURE 7.6.6 The circuit in Problem 25.

26. Theelectric circuit shown in Figure7.6.7 is described by the system of differential equations

⎛

⎜

⎝

−

⎞

⎟

⎠

, (i)

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7.6 Complex Eigenvalues 419

where I is the current through the inductor and V is the voltage drop across the capacitor.

These differential equations were derived in Problem 19 of Section 7.1.

(a) Show that the eigenvalues of the coefﬁcient matrix are real and different if L > 4R

show that they are complex conjugates if L < 4R

(b) Suppose that R = 1 4 , C =

F, and L = 1 H. Find the general solution of the

system (i) in this case.

(d) For the circuit of part (b) determine the limiting values of I(t) and V(t) as t →∞.Do

these limiting values depend on the initial conditions?

FIGURE 7.6.7 The circuit in Problem 26.

27. In this problem we indicate how to show that u(t) and v(t), as given by Eqs. (17), are

linearly independent. Letr

= λ + iµ and r

= λ − iµ be a pair of conjugate eigenvaluesof

the coefﬁcient matrixA of Eq. (1);let ξ

(1)

= a + ib and ξ

(1)

= a − ib be the corresponding

eigenvectors. Recall that it was stated in Section 7.3 that two different eigenvalues have

linearly independent eigenvectors,so if r

̸= r

,then ξ

(1)

and ξ

(1)

are linearly independent.

(a) First we show that a and b are linearly independent. Consider the equation

a + c

b = 0. Express a and b in terms of ξ

(1)

and ξ

(1)

, and then show that

− ic

)ξ

(1)

+ (c

+ ic

)ξ

(1)

= 0.

(b) Show that c

− ic

= 0 and c

+ ic

= 0 and then that c

= 0 and c

= 0. Consequently,

a and b are linearly independent.

u(t

) + c

v(t

) = 0, where t

is an arbitrary point. Rewrite this equation in terms of a

and b, and then proceed as in part (b) to show that c

= 0 and c

= 0. Hence u(t) and v(t)

are linearly independent at the arbitrary point t

.Therefore, they are linearly independent

at every point and on every interval.

28. A mass m on a spring with constant k satisﬁes the differential equation (see Section 3.7)

′′

+ ku = 0,

where u(t) is the displacement at time t of the mass from its equilibrium position.

(a) Let x

= u, x

= u

′

, and show that the resulting system is

′

−k/m 0

(b) Find the eigenvalues of the matrix for the system in part (a).

the corresponding graphs of x

versus t and x

versus t. Sketch both graphs on one set of

axes.

(d) What is the relation between the eigenvalues of the coefﬁcient matrix and the natural

frequency of the spring–mass system?

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420 Chapter 7. Systems of First Order Linear Equations

29. Consider the two-mass,three-spring system of Example 3 in the text. Instead of converting

the problem into a system of four ﬁrst order equations, we indicate here how to proceed

directly from Eqs. (22).

(a) Show that Eqs. (22) can be written in the form

′′

−2

−3

x = Ax . (i)

(b) Assume that x = ξ e

and show that

(A − r

I)ξ = 0.

Note that r

(rather than r) is an eigenvalue of A corresponding to an eigenvector ξ.

(d) Write down expressions for x

and x

. There should be four arbitrary constants in

these expressions.

(e) By differentiating the results from part (d),write down expressions for x

′

and x

′

.Your

results from parts (d) and (e) should agree with Eq. (31) in the text.

30.

Consider the two-mass, three-spring system whose equations of motion are Eqs. (22) in

the text. Let m

= 1, m

= 4/3, k

= 1, k

= 3, and k

= 4/3.

(a) As in the text, convert the system to four ﬁrst order equations of the form y

′

= Ay.

Determine the coefﬁcient matrix A.

(b) Find the eigenvalues and eigenvectors of A.

(d) Describe the fundamental modes of vibration. For each fundamental mode draw

graphs of y

versus t and y

versus t. Also draw the corresponding trajectories in the y

and y

-planes.

(e) Consider the initial conditions y(0) = (2, 1, 0, 0)

. Evaluate the arbitrary constants in

the general solution in part (c). What is the period of the motion in this case? Plot graphs

of y

versus t and y

versus t. Also plot the corresponding trajectories in the y

- and

-planes. Be sure you understand how the trajectories are traversed for a full period.

(f) Consider other initial conditions of your own choice, and plot graphs similar to those

requested in part (e).

31.

Consider the two-mass, three-spring system whose equations of motion are Eqs. (22) in

the text. Let m

= m

= 1 and k

= k

= 1.

(a) As in the text, convert the system to four ﬁrst order equations of the form y

′

= Ay.

Determine the coefﬁcient matrix A.

(b) Find the eigenvalues and eigenvectors of A.

(d) Describe the fundamental modes of vibration. For each fundamental mode draw

graphs of y

versus t and y

versus t. Also draw the corresponding trajectories in the y

and y

-planes.

(e) Consider the initial conditions y(0 ) = (−1, 3, 0, 0)

. Evaluate the arbitrary constants

in the general solution in part (c). Plot y

versus t and y

versus t. Do you think the solution

is periodic? Also draw the trajectories in the y

- and y

-planes.

(f) Consider other initial conditions of your own choice, and plot graphs similar to those

requested in part (e).

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7.7 Fundamental Matrices 421

7.7 Fundamental Matrices

The structure of the solutions of systems of linear differential equations can be

further illuminated by introducing the idea of a fundamental matrix. Suppose that

(1)

(t), ..., x

(n)

(t) form a fundamental set of solutions for the equation

′

= P(t)x (1)

on some interval α<t <β. Then the matrix

$(t) =

⎛

⎜

⎝

(1)

(t) ··· x

(n)

(t)

(1)

(t) ··· x

(n)

(t)

⎞

⎟

⎠

, (2)

whose columns are the vectors x

(1)

(t), ..., x

(n)

(t), is said to be a fundamental matrix

for the system (1). Note that a fundamental matrix is nonsingular since its columns

are linearly independent vectors.

EXAMPLE

Find a fundamental matrix for the system

′

x. (3)

In Example 2 of Section 7.5, we found that

(1)

(t) =

, x

(2)

(t) =

−t

−2e

−t

are linearly independent solutions of Eq. (3). Thus a fundamental matrix for the system (3) is

$(t) =

−t

−2e

−t

. (4)

The solution of an initial value problem can be written very compactly in terms of

a fundamental matrix. The general solution of Eq. (1) is

x = c

(1)

(t) +···+c

(n)

(t) (5)

or, in terms of $(t),

x = $(t)c, (6)

where c is a constant vector with arbitrary components c

, ..., c

. For an initial value

problem consisting of the differential equation (1) and the initial condition

x(t

) = x

, (7)

where t

is a given point in α<t <βand x

is a given initial vector,it is only necessary

to choose the vector c in Eq. (6) so as to satisfy the initial condition (7). Hence c must

satisfy

$(t

)c = x

. (8)

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422 Chapter 7. Systems of First Order Linear Equations

Therefore, since $(t

) is nonsingular,

c = $

−1

(9)

and

x = $(t)$

−1

(10)

is the solution of the initial value problem (1), (7). We emphasize, however, that to

solve a giveninitial value problem,we would ordinarily solveEq. (8) by rowreduction

and then substitute for c in Eq. (6), rather than compute $

−1

) and use Eq. (10).

Recall that each column of the fundamental matrix $ is a solution of Eq. (1). It

follows that $ satisﬁes the matrix differential equation

′

= P(t)$. (11)

This relation is readily conﬁrmed by comparing the two sides of Eq. (11) column by

column.

Sometimes it is convenient to make use of the special fundamental matrix,denoted

by #(t),whose columns are thevectors x

(1)

(t), ..., x

(n)

(t) designated inTheorem 7.4.4.

Besides the differential equation (1), these vectors satisfy the initial conditions

(j )

) = e

(j )

, (12)

where e

(j )

is the unit vector, deﬁned in Theorem 7.4.4, with a one in the jth position

and zeros elsewhere. Thus #(t) has the property that

#(t

) =

⎛

⎜

⎝

10··· 0

01··· 0

00··· 1

⎞

⎟

⎠

= I. (13)

We will always reserve the symbol # to denote the fundamental matrix satisfying the

initial condition (13) and use $ when an arbitrary fundamental matrix is intended.

In terms of #(t), the solution of the initial value problem (1), (7) is even simpler in

appearance; since #

−1

) = I, it follows from Eq. (10) that

x = #(t)x

. (14)

Although the fundamental matrix #(t) is often more complicated than $(t),itis

especially helpful if the same system of differential equations is to be solved repeat-

edly subject to many different initial conditions. This corresponds to a given physical

system that can be started from many different initial states. If the fundamental

matrix #(t) has been determined, then the solution for each set of initial condi-

tions can be found simply by matrix multiplication, as indicated by Eq. (14). The

matrix #(t) thus represents a transformation of the initial conditions x

into the solu-

tion x(t) at an arbitrary time t. Comparing Eqs. (10) and (14) makes it clear that

#(t) = $(t)$

−1

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7.7 Fundamental Matrices 423

EXAMPLE

For the system (3)

′

in Example 1, ﬁnd the fundamental matrix # such that #(0) = I.

The columns of # are solutions of Eq. (3) that satisfy the initial conditions

(1)

(0) =

(

, x

(2)

(0) =

(

. (15)

Since the general solution of Eq. (3) is

x = c

(

+ c

−2

(

−t

we can ﬁnd the solution satisfying the ﬁrst set of these initial conditions by choosing

= c

; similarly, we obtain the solution satisfying the second set of initial conditions

by choosing c

and c

=−

. Hence

#(t) =

⎛

⎝

−t

−

−t

− e

−t

⎞

⎠

. (16)

Note that the elements of #(t) are more complicated than those of the fundamental matrix

$(t) given by Eq. (4); however, it is now easy to determine the solution corresponding to any

set of initial conditions.

The Matrix exp(At). Recall that the solution of the scalar initial value problem

′

= ax, x(0) = x

, (17)

where a is a constant, is

x = x

exp(at). (18)

Now consider the corresponding initial value problem for an n × n system, namely,

′

= Ax, x(0) = x

, (19)

where A is a constant matrix. Applying the results of this section to the problem (19),

we can write its solution as

x = #(t)x

, (20)

where #(0) = I. Comparing the problems (17) and (19), and their solutions, sug-

gests that the matrix #(t) might have an exponential character. We now explore this

possibility.

The scalar exponential function exp(at) can be represented by the power series

exp(at) = 1 +

∞

n=1

, (21)

which converges for all t. Let us now replace the scalar a by the n × n constant matrix

A and consider the corresponding series

I +

∞

n=1

= I +At +

+···+

+···. (22)

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424 Chapter 7. Systems of First Order Linear Equations

Each term in the series (22) is an n × n matrix. It is possible to show that each element

of this matrix sum converges for all t as n →∞. Thus the series (22) deﬁnes as its

sum a new matrix, which we denote by exp(At); that is,

exp(At) = I +

∞

n=1

, (23)

analogous to the expansion (21) of the scalar function exp(at).

By differentiating the series (23) term by term, we obtain

[exp(At)]=

∞

n=1

n−1

(n − 1)!

= A

I +

∞

n=1

= A exp(At). (24)

Therefore, exp(At) satisﬁes the differential equation

exp(At) = A exp(At). (25)

Further,by setting t = 0 in Eq. (23) we ﬁnd that exp(At) satisﬁes the initial condition

exp(At)

t=0

= I. (26)

The fundamental matrix # satisﬁes the same initial value problem as exp(At),namely,

′

= A#, #(0) = I. (27)

Then, by the uniqueness part of Theorem 7.1.2 (extended to matrix differential equa-

tions),we conclude that exp(At) and the fundamental matrix #(t) are the same.Thus

we can write the solution of the initial value problem (19) in the form

x = exp(At)x

, (28)

which is analogous to the solution (18) of the initial value problem (17).

Inorderto justify more conclusively the useof exp(At) for the sumof the series (22),

we should demonstrate that this matrix function does indeed have the properties we

associate with the exponential function. One way to do this is outlined in Problem 15.

Diagonalizable Matrices. The basic reason why a system of linear (algebraic or differ-

ential) equations presents some difﬁculty is that the equations are usually coupled.

In other words, some or all of the equations involve more than one—typically all—of

the unknown variables. Hence the equations in the system must be solved simultane-

ously. In contrast, if each equation involves only a single variable,then each equation

can be solved independently of all the others, which is a much easier task. This obser-

vation suggests that one way to solve a system of equations might be by transforming

it into an equivalent uncoupled system in which each equation contains only one

unknown variable. This corresponds to transforming the coefﬁcient matrix A into a

diagonal matrix.

Eigenvectors are useful in accomplishing such a transformation. Suppose that the

n × n matrix A has a full set of n linearly independent eigenvectors. Recall that this

will certainly be the case if the eigenvalues of A are all different, or if A is Hermitian.

August 7, 2012 21:04 c07 Sheet number 67 Page number 425 cyan black

7.7 Fundamental Matrices 425

Letting ξ

(1)

, ..., ξ

(n)

denote these eigenvectors and λ

, ..., λ

the corresponding

eigenvalues, form the matrix T whose columns are the eigenvectors—that is,

T =

⎛

⎜

⎝

(1)

··· ξ

(n)

(1)

··· ξ

(n)

⎞

⎟

⎠

. (29)

Since the columns of T are linearly independent vectors, detT ̸= 0; hence T is non-

singular and T

−1

exists. A straightforward calculation shows that the columns of the

matrix AT are just the vectors Aξ

(1)

, ..., Aξ

(n)

. Since Aξ

(k)

= λ

(k)

, it follows that

AT =

⎛

⎜

⎝

(1)

··· λ

(n)

(1)

··· λ

(n)

⎞

⎟

⎠

= TD, (30)

where

D =

⎛

⎜

⎝

0 ··· 0

0 λ

··· 0

00··· λ

⎞

⎟

⎠

(31)

is a diagonal matrix whose diagonal elements are the eigenvalues ofA. From Eq. (30)

it follows that

−1

AT = D. (32)

Thus, if the eigenvalues and eigenvectors of A are known, A can be transformed

into a diagonal matrix by the process shown in Eq. (32). This process is known as a

similarity transformation, and Eq. (32) is summed up in words by saying that A is

similar to the diagonal matrix D. Alternatively, we may say that A is diagonalizable.

Observe that a similarity transformation leaves the eigenvalues of A unchanged and

transforms its eigenvectors into the coordinate vectors e

(1)

, ..., e

(n)

If A is Hermitian, then the determination of T

−1

is very simple. The eigenvectors

(1)

, ..., ξ

(n)

of A are known to be mutually orthogonal, so let us choose them so

that they are also normalized by (ξ

(i)

, ξ

(i)

) = 1 for each i. Then it is easy to verify that

−1

= T

∗

; in other words, the inverse of T is the same as its adjoint (the transpose of

its complex conjugate).

Finally, we note that if A has fewer than n linearly independent eigenvectors, then

there is no matrix T such that T

−1

AT = D. In this case,A is not similar to a diagonal

matrix and is not diagonalizable.

EXAMPLE

Consider the matrix

A =

. (33)

Find the similarity transformation matrix T and show that A can be diagonalized.

In Example 2 of Section 7.5, we found that the eigenvalues and eigenvectors of A are

= 3, ξ

(1)

; r

=−1, ξ

(2)

−2

. (34)

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426 Chapter 7. Systems of First Order Linear Equations

Thus the transformation matrix T and its inverse T

−1

are

T =

2 −2

; T

−1

−

. (35)

Consequently, you can check that

−1

AT =

0 −1

= D. (36)

Now let us turn again to the system

′

= Ax, (37)

where A is a constant matrix. In Sections 7.5 and 7.6, we described how to solve

such a system by starting from the assumption that x = ξe

. Now we provide another

viewpoint, one based on diagonalizing the coefﬁcient matrix A.

Accordingto the results stated justabove,it ispossible to diagonalizeA wheneverA

has a full set of n linearly independent eigenvectors. Let ξ

(1)

, ..., ξ

(n)

be eigenvectors

of A corresponding to the eigenvalues r

, ..., r

and form the transformation matrix

T whose columns are ξ

(1)

, ..., ξ

(n)

. Then, deﬁning a new dependent variable y by the

relation

x = Ty, (38)

we have from Eq. (37) that

′

= ATy. (39)

Multiplying by T

−1

, we then obtain

′

= (T

−1

AT)y, (40)

or, using Eq. (32),

′

= Dy. (41)

Recall that D is the diagonal matrix with the eigenvalues r

, ..., r

of A along the

diagonal. A fundamental matrix for the system (41) is the diagonal matrix (see

Problem 16)

Q(t) = exp(Dt) =

⎛

⎜

⎝

0 ... 0

0 e

... 0

00··· e

⎞

⎟

⎠

. (42)

A fundamental matrix $ for the system (37) is then found from Q by the transfor-

mation (38)

$ = TQ; (43)

that is,

$(t) =

⎛

⎜

⎝

(1)

··· ξ

(n)

(1)

··· ξ

(n)

⎞

⎟

⎠

. (44)

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7.7 Fundamental Matrices 427

The columns of $(t) are the same as the solutions in Eq. (27) of Section 7.5. Thus

the diagonalization procedure does not offer any computational advantage over the

method of Section 7.5, since in either case it is necessary to calculate the eigenvalues

and eigenvectors of the coefﬁcient matrix in the system of differential equations.

EXAMPLE

Consider again the system of differential equations

′

= Ax, (45)

where A is given by Eq. (33). Using the transformation x = Ty, where T is given by Eq. (35),

you can reduce the system (45) to the diagonal system

′

0 −1

y = Dy. (46)

Obtain a fundamental matrix for the system (46), and then transform it to obtain a fundamen-

tal matrix for the original system (45).

By multiplying D repeatedly with itself, we ﬁnd that

, D

27 0

0 −1

, .... (47)

Therefore, it follows from Eq. (23) that exp(Dt) is a diagonal matrix with the entries e

and

−t

on the diagonal; that is,

0 e

−t

. (48)

Finally, we obtain the required fundamental matrix $(t) by multiplying T and exp(Dt):

$(t) =

2 −2

0 e

−t

−2e

−t

. (49)

Observe that this fundamental matrix is the same as the one found in Example 1.

PROBLEMS In each of Problems 1 through 10:

(a) Find a fundamental matrix for the given system of equations.

(b) Also ﬁnd the fundamental matrix #(t) satisfying #(0) = I.

1. x

′

3 −2

2 −2

x 2. x

′

−

3. x

′

2 −1

3 −2

x 4. x

′

4 −2

5. x

′

2 −5

1 −2

x 6. x

′

−1 −4

1 −1

7. x

′

5 −1

x 8. x

′

1 −1

5 −3

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428 Chapter 7. Systems of First Order Linear Equations

9. x

′

⎛

⎜

⎝

111

21−1

−8 −5 −3

⎞

⎟

⎠

x 10. x

′

⎛

⎜

⎝

1 −14

32−1

21−1

⎞

⎟

⎠

11. Solve the initial value problem

′

2 −1

3 −2

x, x(0) =

−1

by using the fundamental matrix #(t) found in Problem 3.

12. Solve the initial value problem

′

−1 −4

1 −1

x, x(0) =

by using the fundamental matrix #(t) found in Problem 6.

13. Show that #(t) = $(t)$

−1

), where #(t) and $(t) are as deﬁned in this section.

14. The fundamental matrix #(t) for the system (3) was found in Example 2. Show that

#(t)#(s) = #(t + s) by multiplying #(t) and #(s).

15. Let #(t) denote the fundamental matrix satisfying #

′

= A#, #(0) = I. In the text we also

denoted this matrix by exp(At). In this problem we show that # does indeed have the

principal algebraic properties associated with the exponential function.

(a) Show that #(t)#(s) = #(t + s); that is, show that exp(At) exp(As) = exp[A(t + s)].

Hint: Show that if s is ﬁxed and t is variable, then both #(t)#(s) and #(t + s) satisfy the

initial value problem Z

′

= AZ, Z(0) = #(s).

(b) Show that #(t)#(−t) = I; that is, exp(At) exp[A(−t)]=I. Then show that

#(−t) = #

−1

(t).

−1

(s).

16. Show that if A is a diagonal matrix with diagonal elements a

, a

, ..., a

, then exp(At) is

also a diagonal matrix with diagonal elements exp(a

t), exp(a

t), ..., exp(a

t).

17. Consider an oscillator satisfying the initial value problem

′′

+ ω

u = 0, u(0) = u

, u

′

(0) = v

. (i)

(a) Let x

= u, x

= u

′

, and transform Eqs. (i) into the form

′

= Ax, x(0) = x

. (ii)

(b) By using the series (23), show that

expAt = I cos ωt +A

sin ωt

. (iii)

18. The method of successive approximations (see Section 2.8) can also be applied to systems

of equations. For example, consider the initial value problem

′

= Ax, x (0) = x

, (i)

where A is a constant matrix and x

is a prescribed vector.

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7.8 Repeated Eigenvalues 429

(a) Assuming that a solution x = φ(t) exists,show that it must satisfy the integral equation

φ(t) = x

Aφ(s ) ds. (ii)

(b) Start with the initial approximation φ

(0)

(t) = x

. Substitute this expression for φ(s) in

the right side of Eq. (ii) and obtain a new approximation φ

(1)

(t). Show that

(1)

(t) = (I +At)x

. (iii)

(0)

, φ

(1)

(2)

, ..., φ

(n)

, ....Use an inductive argument to show that

(n)

(t) =

I +At +A

+···+A

(

. (iv)

(d) Let n →∞and show that the solution of the initial value problem (i) is

φ(t) = exp(At)x

. (v)

7.8 Repeated Eigenvalues

We conclude our consideration of the linear homogeneous system with constant

coefﬁcients

′

= Ax (1)

with a discussion of the case in which the matrix A has a repeated eigenvalue. Recall

that in Section 7.3 we stated that a repeated eigenvalue with algebraic multiplicity

m ≥ 2 may have a geometric multiplicity less than m. In other words, there may be

fewer than m linearly independent eigenvectors associated with this eigenvalue. The

following example illustrates this possibility.

EXAMPLE

Find the eigenvalues and eigenvectors of the matrix

A =

1 −1

. (2)

The eigenvalues r and eigenvectors ξ satisfy the equation (A − rI)ξ = 0 , or

1 − r −1

13− r

. (3)

The eigenvalues are the roots of the equation

det(A − rI) =

1 − r −1

13− r

= r

− 4r + 4 = (r − 2)

= 0. (4)

Thus the two eigenvalues are r

= 2 and r

= 2; that is, the eigenvalue 2 has algebraic multi-

plicity 2.

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430 Chapter 7. Systems of First Order Linear Equations

To determine the eigenvectors,we must return to Eq. (3) and use for r the value 2.This gives

−1 −1

. (5)

Hence we obtain the single condition ξ

+ ξ

= 0, which determines ξ

in terms of ξ

, or vice

versa. Thus the eigenvector corresponding to the eigenvalue r = 2 is

(1)

−1

, (6)

or any nonzero multiple of this vector. Observe that there is only one linearly independent

eigenvector associated with the double eigenvalue.

Returning to the system (1), suppose that r = ρ is an m-fold root of the charac-

teristic equation

det(A − rI) = 0. (7)

Then ρ is an eigenvalue of algebraic multiplicity m of the matrix A. In this event,

there are two possibilities: either there are m linearly independent eigenvectors

corresponding to the eigenvalue ρ, or else there are fewer than m such eigenvectors.

In the ﬁrst case, let ξ

(1)

, ..., ξ

(m)

be m linearly independent eigenvectors associ-

ated with the eigenvalue ρ of algebraic multiplicity m. Then there are m linearly

independent solutions x

(1)

(t) = ξ

(1)

ρt

, ..., x

(m)

(t) = ξ

(m)

ρt

of Eq. (1). Thus in this

case it makes no difference that the eigenvalue r = ρ is repeated; there is still a fun-

damental set of solutions of Eq. (1) of the form ξe

. This case always occurs if the

coefﬁcient matrix A is Hermitian (or real and symmetric).

However,if the coefﬁcient matrix is not Hermitian,then there may be fewer than m

independent eigenvectors corresponding to an eigenvalue ρ of algebraic multiplicity

m, and if so, there will be fewer than m solutions of Eq. (1) of the form ξe

ρt

associ-

ated with this eigenvalue. Therefore, to construct the general solution of Eq. (1), it is

necessary to ﬁnd other solutions of a different form. Recall that a similar situation

occurred in Section 3.4 for the linear equation ay

′′

+ by

′

+ cy = 0 when the charac-

teristic equation has a double root r. In that case we found one exponential solution

(t) = e

, but a second independent solution had the form y

(t) = te

. With that

result in mind, consider the following example.

EXAMPLE

Find a fundamental set of solutions of

′

= Ax =

1 −1

x (8)

and draw a phase portrait for this system.

A direction ﬁeld for the system (8) is shown in Figure 7.8.1. From this ﬁgure it appears that

all nonzero solutions depart from the origin.

To solve the system,observe that the coefﬁcient matrix A is the same as the matrix in Exam-

ple 1.Thus weknow that r = 2 isa double eigenvalueand thatit has onlya single corresponding

eigenvector, which we may take as ξ

= (1, −1). Thus one solution of the system (8) is

(1)

(t) =

−1

, (9)

but there is no second solution of the form x = ξe

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7.8 Repeated Eigenvalues 431

–1

–2

–2 –1 12

FIGURE 7.8.1 A direction ﬁeld for the system (8).

Based on the procedure used for second order linear equations in Section 3.4, it may be

natural to attempt to ﬁnd a second independent solution of the system (8) of the form

x = ξte

, (10)

where ξ is a constant vector to be determined. Substituting for x in Eq. (8), we obtain

2ξte

+ ξe

−Aξte

= 0. (11)

For Eq. (11) to be satisﬁed for all t, it is necessary for the coefﬁcients of te

and e

both to be

zero. From the term in e

we ﬁnd that

ξ = 0. (12)

Hence there is no nonzero solution of the system (8) of the form (10).

Since Eq. (11) contains terms in both te

and e

, it appears that in addition to ξte

, the

second solution must contain a term of the form ηe

; in other words, we need to assume that

x = ξte

+ ηe

, (13)

where ξ and η are constant vectors to be determined. Upon substituting this expression for x

in Eq. (8), we obtain

2ξte

+ (ξ + 2η)e

= A(ξte

+ ηe

). (14)

Equating coefﬁcients of te

and e

on each side of Eq. (14) gives the two conditions

(A − 2I)ξ = 0 (15)

and

(A − 2I)η = ξ (16)

for the determination of ξ and η. Equation (15) is satisﬁed if ξ is an eigenvector of A

corresponding to the eigenvalue r = 2, such as ξ

= (1, −1). Since det(A − 2I) is zero,

Eq. (16) is solvable only if the right side ξ satisﬁes a certain condition. Fortunately, ξ and

August 7, 2012 21:04 c07 Sheet number 74 Page number 432 cyan black

432 Chapter 7. Systems of First Order Linear Equations

its multiples are exactly the vectors that allow Eq. (16) to be solved. The augmented matrix for

Eq. (16) is

−1 −1

−1

The second row of this matrix is proportional to the ﬁrst, so the system is solvable. We have

−η

− η

= 1,

so if η

= k, where k is arbitrary, then η

=−k − 1. If we write

η =

−1 − k

−1

+ k

−1

, (17)

then by substituting for ξ and η in Eq. (13), we obtain

x =

−1

(

−1

(

+ k

−1

(

. (18)

The last term in Eq. (18) is merely a multiple of the ﬁrst solution x

(1)

(t) and may be ignored,

but the ﬁrst two terms constitute a new solution:

(2)

(t) =

−1

. (19)

An elementary calculation shows that W[x

(1)

, x

(2)

](t) =−e

̸= 0, and therefore x

(1)

and x

(2)

form a fundamental set of solutions of the system (8). The general solution is

x = c

(1)

(t) + c

(2)

(t)

= c

−1

+ c

−1

. (20)

The main features of a phase portrait for the solution (20) follow from the presence of the

exponential factor e

in every term.Therefore x → 0 as t →−∞and,unless both c

and c

are

zero,x becomes unbounded as t →∞.Ifc

and c

are not both zero,then along any trajectory

we have

lim

t→−∞

(t)

= lim

t→−∞

−c

− c

t − c

+ c

=−1.

Therefore, as t →−∞, every trajectory approaches the origin tangent to the line x

=−x

determined by the eigenvector; this behavior is clearly evident in Figure 7.8.2a. Further, as

t →∞, the slope of each trajectory also approaches −1. However, it is possible to show that

trajectories do not approach asymptotes as t →∞. Several trajectories of the system (8) are

shown in Figure 7.8.2a, and some typical plots of x

versus t are shown in Figure 7.8.2b.The

pattern of trajectories in this ﬁgure is typical of 2 × 2 systems x

′

= Ax with equal eigenvalues

and only one independent eigenvector. The origin is called an improper node in this case. If

the eigenvalues are negative, then the trajectories are similar but are traversed in the inward

direction. An improper node is asymptotically stable or unstable, depending on whether the

eigenvalues are negative or positive.

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7.8 Repeated Eigenvalues 433

(1)

(t)

(2)

–2 –1

–1

–2

(a)

x (t)

(b)

1 2

–1

–2

FIGURE 7.8.2 (a) Phase portrait of the system (8); the origin is an

improper node. (b) Plots of x

versus t for the system (8).

One difference between a system of two ﬁrst order equations and a single second

order equation is evident from the preceding example. For a second order linear

equation with a repeated root r

of the characteristic equation,a term ce

in the sec-

ond solution is not required since it is a multiple of the ﬁrst solution. On the other

hand, for a system of two ﬁrst order equations, the term ηe

of Eq. (13) with r

= 2

is not, in general, a multiple of the ﬁrst solution ξe

, so the term ηe

must be

retained.

Example 2 is entirely typical of the general case when there is a double eigenvalue

and a single associated eigenvector. Consider again the system (1), and suppose that

r = ρ isa doubleeigenvalue ofA,but that thereis onlyone correspondingeigenvector

ξ. Then one solution [similar to Eq. (9)] is

(1)

(t) = ξe

ρt

, (21)

where ξ satisﬁes

(A − ρ I)ξ = 0. (22)

By proceeding as in Example 2, we ﬁnd that a second solution [similar to Eq. (19)] is

(2)

(t) = ξte

ρt

+ ηe

ρt

, (23)

where ξ satisﬁes Eq. (22) and η is determined from

(A − ρ I)η = ξ. (24)

Even though det(A − ρI) = 0, it can be shown that it is always possible to solve

Eq. (24) for η. Note that if we multiply Eq. (24) by A − ρ I and use Eq. (22), then we

obtain

(A − ρ I)

η = 0.

The vector η is called a generalized eigenvector of the matrix A corresponding to the

eigenvalue ρ.

Fundamental Matrices. As explained in Section 7.7, fundamental matrices are formed

by arranging linearly independent solutions in columns. Thus, for example, a

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434 Chapter 7. Systems of First Order Linear Equations

fundamental matrix for the system (8) can be formed from the solutions x

(1)

(t) and

(2)

(t) from Eqs. (9) and (19), respectively:

$(t) =

−e

−te

− e

(

= e

1 t

−1 −1 − t

(

. (25)

The particular fundamental matrix # that satisﬁes #(0) = I can also be readily found

from the relation #(t) = $(t)$

−1

(0). For Eq. (8) we have

$(0) =

−1 −1

(

, $

−1

(0) =

−1 −1

(

, (26)

and then

#(t) = $(t)$

−1

(0) = e

1 t

−1 −1 − t

−1 −1

(

= e

1 − t −t

t 1 + t

(

. (27)

The latter matrix is also the exponential matrix exp(At).

Jordan Forms. An n × n matrixA canbe diagonalized as discussed in Section 7.7 only if

it has a full complement of n linearly independent eigenvectors. If there is a shortage

of eigenvectors (because of repeated eigenvalues),then A can always be transformed

into a nearly diagonal matrix called its Jordan

form, which has the eigenvalues of

A on the main diagonal, ones in certain positions on the diagonal above the main

diagonal, and zeros elsewhere.

Consider again the matrix A given by Eq. (2). To transform A into its Jordan form,

we construct the transformation matrix T with the single eigenvector ξ from Eq. (6)

in its ﬁrst column and the generalized eigenvector η from Eq. (17) with k = 0 in the

second column. Then T and its inverse are given by

T =

−1 −1

(

, T

−1

−1 −1

(

. (28)

As you can verify, it follows that

−1

AT =

(

= J. (29)

The matrix J in Eq. (29) is the Jordan form of A. It is typical of all Jordan forms in that

it has a 1 above the main diagonal in the column corresponding to the eigenvector

that is lacking (and is replaced in T by the generalized eigenvector).

If we start again from Eq. (1)

′

= Ax,

Marie Ennemond Camille Jordan (1838–1922) was professor at the École Polytechnique and the Collège

de France. He is known for his important contributions to analysis and to topology (the Jordan curve

theorem) and especially for his foundational work in group theory. The Jordan form of a matrix appeared

in his inﬂuential book Traité des substitutions et des équations algébriques, published in 1870.

August 7, 2012 21:04 c07 Sheet number 77 Page number 435 cyan black

7.8 Repeated Eigenvalues 435

the transformation x = Ty, where T is given by Eq. (28), produces the system

′

= Jy, (30)

where J is given by Eq. (29). In scalar form the system (30) is

′

= 2y

+ y

, y

′

= 2y

. (31)

These equations can be solved readily in reverse order—that is, by starting with the

equation for y

. In this way we obtain

= c

, y

= c

+ c

. (32)

Thus two independent solutions of the system (30) are

(1)

(t) =

(

, y

(2)

(t) =

(

, (33)

and the corresponding fundamental matrix is

$(t) =

0 e

(

. (34)

Since

$(0) = I, we can also identify the matrix in Eq. (34) as exp(Jt).The same result

can be reached by calculating powers of J and substituting them into the exponential

series (see Problems 20 through 22). To obtain a fundamental matrix for the original

system, we now form the product

$(t) = T exp(Jt) =

−e

− te

(

, (35)

which is the same as the fundamental matrix given in Eq. (25).

We will not discuss n × n systems x

′

= Ax in more detail here. For large n it is

possible that there may be eigenvalues of high algebraic multiplicity m, perhaps with

much lower geometric multiplicity q, thus giving rise to m − q generalized eigenvec-

tors. For n ≥ 4 there may also be repeated complex eigenvalues. A full discussion

of the Jordan form of a general n × n matrix requires a greater background in linear

algebra than we assume for most readers of this book. Problems 18 through 22 ask

you to explore the use of Jordan forms for systems of three equations.

The amount of arithmetic required in the analysis of a general n × n system may

be prohibitive to do by hand even if n is no greater than 3 or 4. Consequently,suitable

computer software should be used routinely in most cases. This does not overcome

all difﬁculties by any means, but it does make many problems much more tractable.

Finally, for a set of equations arising from modeling a physical system, it is likely

that some of the elements in the coefﬁcient matrix A result from measurements of

some physical quantity. The inevitable uncertainties in such measurements lead to

uncertainties in the values of the eigenvalues of A. For example, in such a case it

may not be clear whether two eigenvalues are actually equal or are merely close

together.

For example, see the books listed in the References at the end of this chapter.

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436 Chapter 7. Systems of First Order Linear Equations

PROBLEMS In each of Problems 1 through 4:

(a) Draw a direction ﬁeld and sketch a few trajectories.

(b) Describe how the solutions behave as t →∞.

′

3 −4

1 −1

x 2.

′

4 −2

8 −4

′

−

x 4.

′

−3

−

In each of Problems 5 and 6, ﬁnd the general solution of the given system of equations.

5. x

′

⎛

⎜

⎝

111

21−1

0 −11

⎞

⎟

⎠

x 6. x

′

⎛

⎜

⎝

011

101

110

⎞

⎟

⎠

In each of Problems 7 through 10:

(a) Find the solution of the given initial value problem.

(b) Draw the trajectory of the solution in the x

-plane, and also draw the graph of x

versus t.

′

1 −4

4 −7

x, x(0) =

(

′

−

x, x(0) =

−1

′

−

−1

x, x(0) =

−2

10.

′

−1 −3

x, x(0) =

In each of Problems 11 and 12:

(a) Find the solution of the given initial value problem.

(b) Draw the corresponding trajectory in x

-space, and also draw the graph of x

versus t.

11.

′

⎛

⎜

⎝

100

−410

362

⎞

⎟

⎠

x, x(0) =

⎛

⎜

⎝

−1

−30

⎞

⎟

⎠

12.

′

⎛

⎜

⎝

−

1 −

11−

⎞

⎟

⎠

x, x(0) =

⎛

⎜

⎝

−1

⎞

⎟

⎠

In each of Problems 13 and 14, solve the given system of equations by the method of Problem

19 of Section 7.5. Assume that t > 0.

13. tx

′

3 −4

1 −1

x 14. tx

′

1 −4

4 −7

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7.8 Repeated Eigenvalues 437

15. Show that all solutions of the system

′

approach zero as t →∞if and only if a + d < 0 and ad − bc > 0. Compare this result

with that of Problem 37 in Section 3.4.

16. Consider again the electric circuit in Problem 26 of Section 7.6. This circuit is described

by the system of differential equations

⎛

⎜

⎝

−

⎞

⎟

⎠

(a) Show that the eigenvalues are real and equal if L = 4R

(b) Suppose that R = 1 4, C = 1 F, and L = 4 H. Suppose also that I(0) = 1 A and

V(0) = 2 V. Find I(t) and V(t).

17. Consider again the system

′

= Ax =

1 −1

x (i)

that we discussed in Example 2.We found there that A has a double eigenvalue r

= r

= 2

with a single independent eigenvector ξ

(1)

= (1, −1)

, or any nonzero multiple thereof.

Thus one solution of the system (i) is x

(1)

(t) = ξ

(1)

and a second independent solution

has the form

(2)

(t) = ξte

+ ηe

where ξ and η satisfy

(A − 2I)ξ = 0, (A − 2I)η = ξ. (ii)

In the text we solved the ﬁrst equation for ξ and then the second equation for η. Here we

ask you to proceed in the reverse order.

(a) Show that η satisﬁes (A − 2I)

η = 0.

(b) Show that (A − 2I)

= 0.Thus the generalized eigenvector η can be chosen arbitrarily,

except that it must be independent of ξ

(1)

. Then determine ξ from the second of Eqs. (ii) and observe that

ξ = (1, −1)

= ξ

(1)

. This choice of η reproduces the solution found in Example 2.

(d) Let η = (1, 0)

and determine the corresponding eigenvector ξ.

(e) Let η = (k

, k

)

, where k

and k

are arbitrary numbers. Then determine ξ. How is

it related to the eigenvector ξ

(1)

Eigenvalues of Multiplicity 3. If the matrix A has an eigenvalue of algebraic multiplicity 3,

then there may be either one, two, or three corresponding linearly independent eigenvectors.

The general solution of the system x

′

= Ax is different, depending on the number of eigenvec-

tors associated with the triple eigenvalue. As noted in the text, there is no difﬁculty if there

are three eigenvectors, since then there are three independent solutions of the form x = ξe

The following two problems illustrate the solution procedure for a triple eigenvalue with one

or two eigenvectors, respectively.

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438 Chapter 7. Systems of First Order Linear Equations

18. Consider the system

′

= Ax =

⎛

⎜

⎝

111

21−1

−324

⎞

⎟

⎠

x. (i)

(a) Show that r = 2 is an eigenvalue of algebraic multiplicity 3 of the coefﬁcient matrix

A and that there is only one corresponding eigenvector, namely,

(1)

⎛

⎜

⎝

−1

⎞

⎟

⎠

(b) Using the information in part (a), write down one solution x

(1)

(t) of the system (i).

There is no other solution of the purely exponential form x = ξe

+ ηe

. Show that ξ and η satisfy the

equations

(A − 2I)ξ = 0, (A − 2I)η = ξ.

Since ξ has already been found in part (a), solve the second equation for η. Neglect the

multiple of ξ

(1)

that appears in η, since it leads only to a multiple of the ﬁrst solution x

(1)

Then write down a second solution x

(2)

(t) of the system (i).

(d) To ﬁnd a third solution, assume that x = ξ(t

/2)e

+ ηte

+ ζe

. Show that ξ, η, and ζ

satisfy the equations

(A − 2I)ξ = 0, (A − 2I)η = ξ, (A − 2I)ζ = η.

The ﬁrst two equations are the same as in part (c), so solve the third equation for ζ, again

neglecting the multiple of ξ

(1)

that appears. Then write down a third solution x

(3)

(t) of the

system (i).

(e) Write down a fundamental matrix $(t) for the system (i).

(f) Form a matrix T with the eigenvector ξ

(1)

in the ﬁrst column and the generalized

eigenvectors η and ζ in the second and third columns. Then ﬁnd T

−1

and form the product

J = T

−1

AT. The matrix J is the Jordan form of A.

19. Consider the system

′

= Ax =

⎛

⎜

⎝

5 −3 −2

8 −5 −4

−433

⎞

⎟

⎠

x. (i)

(a) Show that r = 1 is a triple eigenvalue of the coefﬁcient matrix A and that there are

only two linearly independent eigenvectors, which we may take as

(1)

⎛

⎜

⎝

⎞

⎟

⎠

, ξ

(2)

⎛

⎜

⎝

−3

⎞

⎟

⎠

. (ii)

Write down two linearly independent solutions x

(1)

(t) and x

(2)

(t) of Eq. (i).

(b) To ﬁnd a third solution, assume that x = ξte

+ ηe

; then show that ξ and η must satisfy

(A − I)ξ = 0, (iii)

(A − I)η = ξ. (iv)

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7.8 Repeated Eigenvalues 439

ξ to be a suitable linear combination of ξ

(1)

and ξ

(2)

so that Eq. (iv) is solvable, and then

to solve that equation for η. However, let us proceed in a different way and follow the

pattern of Problem 17. First, show that η satisﬁes

(A − I)

η = 0.

Further, show that (A − I)

= 0. Thus η can be chosen arbitrarily, except that it must be

independent of ξ

(1)

and ξ

(2)

(d) A convenient choice for η is η = (0, 0, 1)

. Find the corresponding ξ from Eq. (iv).

Verify that ξ is an eigenvector.

(e) Write down a fundamental matrix $(t) for the system (i).

(f) Form a matrix T with the eigenvector ξ

(1)

in the ﬁrst column and with the eigenvector

ξ from part (d) and the generalized eigenvector η in the other two columns. Find T

−1

and

form the product J = T

−1

AT. The matrix J is the Jordan form of A.

20. Let J =

λ 1

0 λ

, where λ is an arbitrary real number.

(a) Find J

, J

, and J

(b) Use an inductive argument to show that J

nλ

n−1

0 λ

(d) Use exp(Jt) to solve the initial value problem x

′

= Jx, x (0) = x

21. Let

J =

⎛

⎜

⎝

λ 00

0 λ 1

00λ

⎞

⎟

⎠

where λ is an arbitrary real number.

(a) Find J

, J

, and J

(b) Use an inductive argument to show that

⎛

⎜

⎝

0 λ

nλ

n−1

00 λ

⎞

⎟

⎠

(d) Observe that if you choose λ = 1, then the matrix J in this problem is the same as

the matrix J in Problem 19(f). Using the matrix T from Problem 19(f), form the product

T exp(Jt) with λ = 1. Is the resulting matrix the same as the fundamental matrix $(t) in

Problem 19(e)? If not, explain the discrepancy.

22. Let

J =

⎛

⎜

⎝

λ 10

0 λ 1

00λ

⎞

⎟

⎠

where λ is an arbitrary real number.

(a) Find J

, J

, and J

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440 Chapter 7. Systems of First Order Linear Equations

(b) Use an inductive argument to show that

⎛

⎜

⎝

nλ

n−1

[n(n − 1)/2]λ

n−2

0 λ

nλ

n−1

00 λ

⎞

⎟

⎠

(d) Note that if you choose λ = 2, then the matrix J in this problem is the same as the

matrix J in Problem 18(f). Using the matrix T from Problem 18(f), form the product

T exp(Jt) with λ = 2. The resulting matrix is the same as the fundamental matrix $(t) in

Problem 18(e).

7.9 Nonhomogeneous Linear Systems

In this section we turn to the nonhomogeneous system

′

= P(t)x + g(t), (1)

where the n × n matrix P(t) and n × 1 vector g(t) are continuous for α<t <β.By

the same argument as in Section 3.5 (see also Problem 16 in this section), the general

solution of Eq. (1) can be expressed as

x = c

(1)

(t) +···+c

(n)

(t) + v(t), (2)

where c

(1)

(t) +···+c

(n)

(t) is the general solution of the homogeneous system

′

= P(t)x, and v(t) is a particular solution of the nonhomogeneous system (1). We

will brieﬂy describe several methods for determining v(t).

Diagonalization. We begin with systems of the form

′

= Ax + g(t), (3)

where A is an n × n diagonalizable constant matrix. By diagonalizing the coefﬁcient

matrix A, as indicated in Section 7.7, we can transform Eq. (3) into a system of

equations that is readily solvable.

Let T be the matrix whose columns are the eigenvectors ξ

(1)

, ..., ξ

(n)

of A, and

deﬁne a new dependent variable y by

x = Ty. (4)

Then, substituting for x in Eq. (3), we obtain

′

= ATy + g(t).

When we multiply by T

−1

, it follows that

′

= (T

−1

AT)y +T

−1

g(t) = Dy + h(t), (5)

where h (t) = T

−1

g(t) and where D is the diagonal matrix whose diagonal entries

are the eigenvalues r

, ..., r

of A, arranged in the same order as the corresponding

eigenvectors ξ

(1)

, ..., ξ

(n)

that appear as columns of T. Equation (5) is a system of

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7.9 Nonhomogeneous Linear Systems 441

n uncoupled equations for y

(t), ..., y

(t); as a consequence, the equations can be

solved separately. In scalar form, Eq. (5) has the form

′

(t) = r

(t) + h

(t), j = 1, ..., n, (6)

where h

(t) is a certain linear combination of g

(t), ..., g

(t). Equation (6) is a ﬁrst

order linear equation and can be solved by the methods of Section 2.1. In fact, we

have

(t) = e

−r

(s) ds + c

, j = 1, ..., n, (7)

where the c

are arbitrary constants. Finally,the solution x of Eq. (3) is obtained from

Eq. (4). When multiplied by the transformation matrix T, the second term on the

right side of Eq. (7) produces the general solution of the homogeneous equation

′

= Ax, while the ﬁrst term on the right side of Eq. (7) yields a particular solution of

the nonhomogeneous system (3).

EXAMPLE

Find the general solution of the system

′

−21

1 −2

x +

−t

= Ax + g(t). (8)

Proceeding as in Section 7.5, we ﬁnd that the eigenvalues of the coefﬁcient matrix are

=−3 and r

=−1 and that the corresponding eigenvectors are

(1)

−1

, ξ

(2)

. (9)

Thus the general solution of the homogeneous system is

x = c

−1

−3t

+ c

−t

. (10)

Before writing down the matrix T of eigenvectors, we recall that eventually we must ﬁnd T

−1

The coefﬁcient matrix A is real and symmetric, so we can use the result stated just above

Example 3 in Section 7.7: T

−1

is simply the adjoint or (since T is real) the transpose of T,

provided that the eigenvectors of A are normalized so that (ξ, ξ) = 1. Hence,upon normalizing

(1)

and ξ

(2)

, we have

T =

√

−11

, T

−1

√

1 −1

. (11)

Letting x = Ty and substituting for x in Eq. (8),we obtain the following system of equations

for the new dependent variable y:

′

= Dy +T

−1

g(t) =

−30

0 −1

y +

√

−t

− 3t

−t

+ 3t

. (12)

Thus

′

+ 3y

√

−t

−

√

(13)

′

+ y

√

−t

√

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442 Chapter 7. Systems of First Order Linear Equations

Each of Eqs. (13) is a ﬁrst order linear equation and so can be solved by the methods of Section

2.1. In this way we obtain

√

−t

−

√

(

−

+ c

−3t

(14)

√

2te

−t

√

(t − 1) + c

−t

Finally, we write the solution in terms of the original variables:

x = Ty =

√

+ y

−y

+ y

⎛

⎜

⎝

√

2)e

−3t

;

√

2) +

−t

+ t −

+ te

−t

−(c

√

2)e

−3t

;

√

2) −

−t

+ 2t −

+ te

−t

⎞

⎟

⎠

= k

−1

−3t

+ k

−t

−1

−t

t −

, (15)

where k

= c

√

2 and k

= c

√

2. The ﬁrst two terms on the right side of Eq. (15) form the

general solution of the homogeneous system corresponding to Eq. (8). The remaining terms

are a particular solution of the nonhomogeneous system.

If the coefﬁcient matrix A in Eq. (3) is not diagonalizable (because of repeated

eigenvalues and a shortage of eigenvectors), it can nevertheless be reduced to

a Jordan form J by a suitable transformation matrix T involving both eigen-

vectors and generalized eigenvectors. In this case the differential equations for

, ..., y

are not totally uncoupled since some rows of J have two nonzero ele-

ments: an eigenvalue in the diagonal position anda1intheadjacent position to

the right. However, the equations for y

, ..., y

can still be solved consecutively,

starting with y

. Then the solution of the original system (3) can be found by the

relation x = Ty.

Undetermined Coefficients. A second way to ﬁnd a particular solution of the nonho-

mogeneous system (1) is the method of undetermined coefﬁcients that we ﬁrst

discussed in Section 3.5. To use this method, we assume the form of the solution

with some or all of the coefﬁcients unspeciﬁed, and then seek to determine these

coefﬁcients so as to satisfy the differential equation. As a practical matter, this

method is applicable only if the coefﬁcient matrix P is a constant matrix, and if

the components of g are polynomial, exponential, or sinusoidal functions, or sums

or products of these. In these cases the correct form of the solution can be pre-

dicted in a simple and systematic manner. The procedure for choosing the form of

the solution is substantially the same as that given in Section 3.5 for linear second

order equations. The main difference is illustrated by the case of a nonhomogeneous

term of the form ue

λt

, where λ is a simple root of the characteristic equation. In

this situation, rather than assuming a solution of the form ate

λt

, it is necessary to

use ate

λt

+ be

λt

, where a and b are determined by substituting into the differential

equation.

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7.9 Nonhomogeneous Linear Systems 443

EXAMPLE

Use the method of undetermined coefﬁcients to ﬁnd a particular solution of

′

−21

1 −2

x +

−t

= Ax + g(t). (16)

This is the same system of equations as in Example 1. To use the method of undetermined

coefﬁcients, we write g(t) in the form

g(t) =

−t

t. (17)

Observe that r =−1 is an eigenvalue of the coefﬁcient matrix, and therefore we must include

both ate

−t

and be

−t

in the assumed solution. Thus we assume that

x = v(t) = ate

−t

+ be

−t

+ ct + d, (18)

where a, b, c, and d are vectors to be determined. By substituting Eq. (18) into Eq. (16) and

collecting terms, we obtain the following algebraic equations for a, b, c, and d:

Aa =−a,

Ab = a − b −

(19)

Ac =−

Ad = c.

From the ﬁrst of Eqs. (19),we see that a is an eigenvector of A corresponding to the eigenvalue

r =−1. Thus a

= (α, α), where α is any nonzero constant. Then we ﬁnd that the second of

Eqs. (19) can be solved only if α = 1 and that in this case,

b = k

−

(20)

for any constant k. The simplest choice is k = 0, from which b

= (0, −1). Then the third and

fourth of Eqs. (19) yield c

= (1, 2) and d

= (−

, −

), respectively. Finally, from Eq. (18) we

obtain the particular solution

v(t) =

−t

−

−t

t −

. (21)

The particular solution (21) is not identical to the one contained in Eq. (15) of Example 1

because the term in e

−t

is different. However,if we choose k =

in Eq. (20),then b

= (

, −

)

and the two particular solutions agree.

Variation of Parameters. Now let us turn to more general problems in which the

coefﬁcient matrix is not constant or not diagonalizable. Let

′

= P(t)x + g(t), (22)

where P(t) and g(t) are continuous on α<t <β. Assume that a fundamental matrix

$(t) for the corresponding homogeneous system

′

= P(t)x (23)

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444 Chapter 7. Systems of First Order Linear Equations

has been found. We use the method of variation of parameters to construct a partic-

ular solution, and hence the general solution, of the nonhomogeneous system (22).

Since the general solution of the homogeneous system (23) is $(t)c, it is natural to

proceed as in Section 3.6 and to seek a solution of the nonhomogeneous system (22)

by replacing the constant vector c by a vector function u(t). Thus we assume that

x = $(t)u(t), (24)

where u(t) is a vector to be found. Upon differentiating x as given by Eq. (24) and

requiring that Eq. (22) be satisﬁed, we obtain

′

(t)u(t) + $(t)u

′

(t) = P(t)$(t)u(t) + g(t). (25)

Since $(t) is a fundamental matrix, $

′

(t) = P(t)$(t); hence Eq. (25) reduces to

$(t)u

′

(t) = g(t). (26)

Recall that $(t) is nonsingular on any interval where P is continuous. Hence $

−1

(t)

exists, and therefore

′

(t) = $

−1

(t)g(t). (27)

Thus for u(t) we can select any vector from the class of vectors that satisfy Eq. (27).

These vectors are determined only up to an arbitrary additive constant vector;

therefore, we denote u(t) by

u(t) =

−1

(t)g(t) dt + c, (28)

where the constant vector c is arbitrary. If the integrals in Eq. (28) can be evaluated,

then the general solution of the system (22) is found by substituting for u(t) from

Eq. (28) in Eq. (24). However, even if the integrals cannot be evaluated, we can still

write the general solution of Eq. (22) in the form

x = $(t)c + $(t)

−1

(s)g(s) ds, (29)

where t

is any point in the interval (α, β).The ﬁrst term on the right side of Eq. (29) is

the general solution of the corresponding homogeneous system (23), and the second

term is a particular solution of Eq. (22).

Now let us consider the initial value problem consisting of the differential equation

(22) and the initial condition

x(t

) = x

. (30)

We can ﬁnd the solution of this problem most conveniently if we choose the lower

limit of integration in Eq. (29) to be the initial point t

. Then the general solution of

the differential equation is

x = $(t)c + $(t)

−1

(s)g(s) ds. (31)

For t = t

the integral in Eq. (31) is zero, so the initial condition (30) is also satisﬁed

if we choose

c = $

−1

. (32)

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7.9 Nonhomogeneous Linear Systems 445

Therefore,

x = $(t)$

−1

+ $(t)

−1

(s)g(s) ds (33)

is the solution of the given initial value problem. Again, although it is helpful to use

−1

to write the solutions (29) and (33),it is usually better in particular cases to solve

the necessary equations by row reduction than to calculate $

−1

and substitute into

Eqs. (29) and (33).

The solution (33) takes a slightly simpler form if we use the fundamental matrix

#(t) satisfying #(t

) = I. In this case we have

x = #(t)x

+ #(t)

−1

(s)g(s) ds. (34)

Equation (34) can be simpliﬁed further if the coefﬁcient matrix P(t) is a constant

matrix (see Problem 17).

EXAMPLE

Use the method of variation of parameters to ﬁnd the general solution of the system

′

−21

1 −2

x +

−t

= Ax + g(t). (35)

This is the same system of equations as in Examples 1 and 2.

The general solution of the corresponding homogeneous system was given in Eq. (10). Thus

$(t) =

−3t

−t

−e

−3t

−t

(36)

is a fundamental matrix. Then the solution x of Eq. (35) is given by x = $(t)u(t), where u(t)

satisﬁes $ (t)u

′

(t) = g(t), or

−3t

−t

−e

−3t

−t

′

−t

. (37)

Solving Eq. (37) by row reduction, we obtain

′

= e

−

′

= 1 +

Hence

(t) =

−

+ c

(t) = t +

−

+ c

and

x = $(t)u(t)

= c

−1

−3t

+ c

−t

−1

−t

t −

, (38)

which is the same as the solution obtained in Example 1 [Eq. (15)] and is equivalent to the

solution obtained in Example 2 [Eq. (21)].

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446 Chapter 7. Systems of First Order Linear Equations

Laplace Transforms. We used the Laplace transform in Chapter 6 to solve linear equa-

tions of arbitrary order. It can also be used in very much the same way to solve

systems of equations. Since the transform is an integral, the transform of a vector is

computed component by component. Thus

L{x(t)} is the vector whose components

are the transforms of the respective components of x(t), and similarly for

L{x

′

(t)}.

We will denote

L{x(t)} by X(s). Then, by an extension of Theorem 6.2.1 to vectors,

we also have

L{x

′

(t)}=sX(s) − x(0). (39)

EXAMPLE

Use the method of Laplace transforms to solve the system

′

−21

1 −2

x +

−t

= Ax + g(t). (40)

This is the same system of equations as in Examples 1 through 3.

We take the Laplace transform of each term in Eq. (40), obtaining

sX(s) − x(0) = AX(s) + G(s), (41)

where G(s) is the transform of g(t). The transform G(s) is given by

G(s) =

2/(s + 1)

3/s

. (42)

To proceed further we need to choose the initial vector x(0). For simplicity let us choose

x(0) = 0. Then Eq. (41) becomes

(sI −A)X(s) = G(s), (43)

where, as usual, I is the identity matrix. Consequently, X(s) is given by

X(s) = (sI −A)

−1

G(s). (44)

The matrix (sI −A)

−1

is called the transfer matrix because multiplying it by the transform of

the input vector g(t) yields the transform of the output vector x(t). In this example we have

sI −A =

s + 2 −1

−1 s + 2

, (45)

and by a straightforward calculation, we obtain

(sI −A)

−1

(s + 1)(s + 3)

s + 21

1 s + 2

. (46)

Then, substituting from Eqs. (42) and (46) in Eq. (44) and carrying out the indicated multi-

plication, we ﬁnd that

X(s) =

⎛

⎜

⎝

2(s + 2)

(s + 1)

(s + 3)

(s + 1)(s + 3)

(s + 1)

(s + 3)

3(s + 2)

(s + 1)(s + 3)

⎞

⎟

⎠

. (47)

Finally, we need to obtain the solution x(t) from its transform X(s). This can be done by

expanding the expressions in Eq. (47) in partial fractions and using Table 6.2.1, or (more

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7.9 Nonhomogeneous Linear Systems 447

efﬁciently) by using appropriate computer software. In any case, after some simpliﬁcation the

result is

x(t) =

−t

−

−1

−3t

−t

t −

. (48)

Equation (48) gives the particular solution of the system (40) that satisﬁes the initial condition

x(0) = 0. As a result, it differs slightly from the particular solutions obtained in the preceding

three examples. To obtain the general solution of Eq. (40), you must add to the expression in

Eq. (48) the general solution (10) of the homogeneous system corresponding to Eq. (40).

Each of the methods for solving nonhomogeneous equations has some advantages

and disadvantages. The method of undetermined coefﬁcients requires no integra-

tion, but it is limited in scope and may entail the solution of several sets of algebraic

equations. The method of diagonalization requires ﬁnding the inverse of the trans-

formation matrix and the solution of a set of uncoupled ﬁrst order linear equations,

followed by a matrix multiplication. Its main advantage is that for Hermitian coefﬁ-

cient matrices, the inverse of the transformation matrix can be written down without

calculation—a feature that is more important for large systems. The method of

Laplace transforms involves a matrix inversion to ﬁnd the transfer matrix, followed

by a multiplication, and ﬁnally by the determination of the inverse transform of each

term in the resulting expression. It is particularly useful in problems with forcing

functions that involve discontinuous or impulsive terms. Variation of parameters is

the most general method. On the other hand, it involves the solution of a set of lin-

ear algebraic equations with variable coefﬁcients, followed by an integration and a

matrix multiplication, so it may also be the most complicated from a computational

viewpoint. For many small systems with constant coefﬁcients, such as the one in the

examples in this section,all of these methods work well,and theremay be little reason

to select one over another.

PROBLEMS In each of Problems 1 through 12 ﬁnd the general solution of the given system of equations.

1. x

′

2 −1

3 −2

x +

2. x

′

√

3 −1

x +

√

3 e

−t

3. x

′

2 −5

1 −2

x +

−cos t

sin t

4. x

′

4 −2

x +

−2t

−2e

5. x

′

4 −2

8 −4

x +

−3

−t

−2

, t > 0

6. x

′

−42

2 −1

x +

−1

+ 4

, t > 0

7. x

′

x +

−1

8. x

′

2 −1

3 −2

x +

−1

9. x

′

−

x +

10. x

′

−3

√

2 −2

x +

−1

−t

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448 Chapter 7. Systems of First Order Linear Equations

11. x

′

2 −5

1 −2

x +

cos t

,0< t <π

12. x

′

2 −5

1 −2

x +

csc t

sec t

< t <π

13. Theelectric circuit shown in Figure7.9.1 is described by the system of differential equations

−

2 −

x +

I(t), (i)

where x

is the current through the inductor, x

is the voltage drop across the capacitor,

and I(t) is the current supplied by the external source.

(a) Determine a fundamental matrix $(t) for the homogeneous system corresponding to

Eq. (i). Refer to Problem 25 of Section 7.6.

(b) If I(t) = e

−t/2

, determine the solution of the system (i) that also satisﬁes the initial

conditions x(0) = 0.

I(t)

R = 4 ohms R = 4 ohms

L = 8 henrys

C = farad

FIGURE 7.9.1 The circuit in Problem 13.

In each of Problems 14 and 15, verify that the given vector is the general solution of the

corresponding homogeneous system, and then solve the nonhomogeneous system. Assume

that t > 0.

14. tx

′

2 −1

3 −2

x +

1 − t

, x

(c)

= c

t + c

−1

15. tx

′

3 −2

2 −2

x +

−2t

− 1

, x

(c)

= c

−1

+ c

16. Let x = φ(t) be the general solution of x

′

= P(t)x + g(t), and let x = v(t) be some

particular solution of the same system. By considering the difference φ(t) − v(t), show

that φ(t) = u(t) + v(t), where u(t) is the general solution of the homogeneous system

′

= P(t)x.

17. Consider the initial value problem

′

= Ax + g(t), x(0) = x

(a) By referring to Problem 15(c) in Section 7.7, show that

x = #(t)x

#(t − s)g(s) ds.

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7.9 Nonhomogeneous Linear Systems 449

(b) Show also that

x = exp(At)x

exp[A(t − s)]g(s) ds.

Compare these results with those of Problem 27 in Section 3.6.

18. Use the Laplace transform to solve the system

′

−21

1 −2

x +

−t

= Ax + g(t) (i)

used in the examples in this section. Instead of using zero initial conditions, as in Example

4, let

x(0) =

, (ii)

where α

and α

are arbitrary. How mustα

and α

be chosen so that the solution isidentical

to Eq. (38)?

REFERENCES Further information on matrices and linear algebra is available in any introductory book on the subject.

Here is a representative sample:

Anton, H. and Rorres, C., Elementary Linear Algebra (10th ed.) (Hoboken, NJ: Wiley, 2010).

Johnson, L. W., Riess, R. D., and Arnold, J. T., Introduction to Linear Algebra (6th ed.) (Boston: Addison-

Wesley, 2008).

Kolman, B. and Hill, D. R.,Elementary Linear Algebra (8th ed.) (Upper Saddle River, NJ: Pearson, 2004).

Lay, D. C., Linear Algebra and Its Applications(4th ed.)(Boston:Addison-Wesley, 2012).

Leon, S. J., Linear Algebra with Applications (8th ed.) (Upper Saddle River, NJ: Pearson/Prentice-Hall,

2010).

Strang, G., Linear Algebra and Its Applications (4th ed.)(Belmont, CA:Thomson, Brooks/Cole, 2006).

A more extended treatment of systems of ﬁrst order linear equations may be found in many books,

including the following:

Coddington, E. A. and Carlson, R., Linear Ordinary Differential Equations (Philadelphia, PA: Society for

Industrial and Applied Mahematics, 1997).

Hirsch, M. W., Smale, S., and Devaney, R. L., Differential Equations, Dynamical Systems, and an Introduc-

tion to Chaos (2nd ed.)(San Diego, CA:Academic Press, 2004).

The following book treats elementary differential equations with a particular emphasis on systems of

ﬁrst order equations:

Brannan, J. R. and Boyce, W. E., Differential Equations: An Introduction to Modern Methods and

Applications (2nd ed.) (New York: Wiley, 2011).

August 7, 2012 21:04 c07 Sheet number 92 Page number 450 cyan black

August 7, 2012 21:05 c08 Sheet number 1 Page number 451 cyan black

451

CHAPTER

Numerical

Methods

Up tothis pointwe havediscussed methods for solving differential equations by using

analytical techniques such as integration or series expansions. Usually, the emphasis

was on ﬁnding an exact expression for the solution. Unfortunately, there are many

important problems in engineering and science, especially nonlinear ones, to which

these methods either do not apply or are very complicated to use. In this chapter

we discuss an alternative approach, the use of numerical approximation methods to

obtain an accurate approximation to the solution of an initial value problem. We

present these methods in the simplest possible context, namely, a single scalar ﬁrst

order equation. However, they can readily be extended to systems of ﬁrst order

equations, and this is outlined brieﬂy in Section 8.5. The procedures described here

can be executed easily on personal computers.

8.1 The Euler or Tangent Line Method

To discuss the development and use of numerical approximation procedures, we

will concentrate mainly on the ﬁrst order initial value problem consisting of the

differential equation

= f (t, y) (1)

and the initial condition

y(t

) = y

. (2)

We assumethat the functions f and f

are continuous on somerectangle in the ty-plane

containing the point (t

, y

). Then, by Theorem 2.4.2, there exists a unique solution

y = φ(t) of the given problem in some interval about t

. If Eq. (1) is nonlinear, then

August 7, 2012 21:05 c08 Sheet number 2 Page number 452 cyan black

452 Chapter 8. Numerical Methods

the interval of existence of the solution may be difﬁcult to determine and may have

no simple relationship to the function f . However, in all our discussions we assume

that there is a unique solution of the initial value problem (1), (2) in the interval of

interest.

In Section 2.7 we described the oldest and simplest numerical approximation

method, namely, the Euler or tangent line method. To derive this method, let us

write the differential equation (1) at the point t = t

in the form

d φ

) = f [t

, φ(t

)]. (3)

Then we approximate the derivative in Eq. (3) by the corresponding (forward)

difference quotient, obtaining

φ(t

n+1

) − φ(t

)

n+1

− t

∼

f [t

, φ(t

)]. (4)

Finally, if we replace φ(t

n+1

) and φ(t

) by their approximate values y

n+1

and y

respectively, and solve for y

n+1

, we obtain the Euler formula

n+1

= y

+ f (t

, y

)(t

n+1

− t

), n = 0, 1, 2, .... (5)

If the step size t

n+1

− t

has a uniform value h for all n and if we denote f (t

, y

) by

, then Eq. (5) simpliﬁes to

n+1

= y

+ hf

, n = 0, 1, 2, .... (6)

Euler’s method consists of repeatedly evaluating Eq. (5) or (6), using the result

of each step to execute the next step. In this way we obtain a sequence of values

, y

, ..., y

, ... that approximate the values of the solution φ(t) at the points

, t

, ..., t

, ....

A computer program for Euler’s method has a structure such as that shown below.

The speciﬁc instructions can be written in any convenient programming language.

The Euler Method

Step 1. deﬁne f (t, y)

Step 2. input initial values t0 and y0

Step 3. input step size h and number of steps n

Step 4. output t0 and y0

Step 5. for j from 1 to n do

Step 6. k1 = f (t, y)

y = y + h ∗ k1

t = t + h

Step 7. output t and y

Step 8. end

August 7, 2012 21:05 c08 Sheet number 3 Page number 453 cyan black

8.1 The Euler or Tangent Line Method 453

Some examples of Euler’s method appear in Section 2.7. As another example,

consider the initial value problem

′

= 1 − t + 4y, (7)

y(0) = 1. (8)

Equation (7) is a ﬁrst order linear equation,and you can easily verify that the solution

satisfying the initial condition (8) is

y = φ(t) =

t −

. (9)

Since the exact solution is known, we do not need numerical methods to approx-

imate the solution of the initial value problem (7), (8). On the other hand, the

availability of the exact solution makes it easy to determine the accuracy of any

numerical procedure that we use on this problem. We will use this problem through-

out the chapter to illustrate and compare different numerical methods. The solutions

of Eq. (7) diverge rather rapidly from each other, so we should expect that it will

be fairly difﬁcult to approximate the solution (9) well over any interval of moder-

ate length. Indeed, this is the reason for choosing this particular problem; it will be

relatively easy to observe the beneﬁts of using more efﬁcient methods.

EXAMPLE

Using the Euler formula (6) and step sizes h = 0.05, 0.025, 0.01, and 0.001, determine approx-

imate values of the solution y = φ(t) of the problem (7), (8) on the interval 0 ≤ t ≤ 2.

The indicated calculationswere carriedout on acomputer,andsome ofthe results areshown

in Table 8.1.1. Their accuracy is not particularly impressive. For h = 0.01 the percentage error

is 3.85% at t = 0.5, 7.49% at t = 1.0, and 14.4% at t = 2.0. The corresponding percentage

errors for h = 0.001 are 0.40%, 0.79%, and 1.58%, respectively. Observe that if h = 0.001,

then it requires 2000 steps to traverse the interval from t = 0tot = 2. Thus considerable

computation is needed to obtain even reasonably good accuracy for this problem using the

Euler method. When we discuss other numerical approximation methods later in this chapter,

we will ﬁnd that it is possible to obtain comparable or better accuracy with much larger step

sizes and many fewer computational steps.

TABLE 8.1.1 A Comparison of Results for the Numerical Approximation of the Solution

of y

′

= 1 − t + 4y, y(0) = 1 Using the Euler Method for Different Step Sizes h

th= 0.05 h = 0.025 h = 0.01 h = 0.001 Exact

0.01.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.11.5475000 1.5761188 1.5952901 1.6076289 1.6090418

0.22.3249000 2.4080117 2.4644587 2.5011159 2.5053299

0.33.4333560 3.6143837 3.7390345 3.8207130 3.8301388

0.45.0185326 5.3690304 5.6137120 5.7754845 5.7942260

0.57.2901870 7.9264062 8.3766865 8.6770692 8.7120041

1.045.588400 53.807866 60 .037126 64.382558 64.897803

1.5 282.07187 361.75945 426.40818 473.55979 479.25919

2.0

1745

.6662 2432.7878 3029.3279 3484.1608 3540.2001

August 7, 2012 21:05 c08 Sheet number 4 Page number 454 cyan black

454 Chapter 8. Numerical Methods

To begin to investigate the errors in using numerical approximations, and also to

suggest ways to construct more accurate algorithms, it is helpful to mention some

alternative ways to look at the Euler method.

One way to proceed is to write the problem as an integral equation. Let y = φ(t)

be the solution of the initial value problem (1), (2); then, by integrating from t

n+1

, we obtain

n+1

′

(t) dt =

n+1

f [t, φ(t)]dt,

φ(t

n+1

) = φ(t

) +

n+1

f [t, φ(t)]dt. (10)

The integral in Eq. (10) is represented geometrically as the area under the curve in

Figure 8.1.1 between t = t

and t = t

n+1

. If we approximate the integral by replacing

f [t, φ(t)] by its value f [t

, φ(t

)] at t = t

, then we are approximating the actual area

by the area of the shaded rectangle. In this way we obtain

φ(t

n+1

)

∼

φ(t

) + f [t

, φ(t

)](t

n+1

− t

)

= φ(t

) + hf [t

, φ(t

)]. (11)

Finally,to obtain an approximation y

n+1

for φ(t

n+1

),we make a second approximation

by replacing φ(t

) by its approximate value y

in Eq. (11).This gives the Euler formula

n+1

= y

+ hf (t

, y

). A more accurate algorithm can be obtained by approximating

the integral more accurately. This is discussed in Section 8.2.

f [t

, (t

)]

n+1

y' = f [t, (t)]

FIGURE 8.1.1 Integral derivation of the Euler method.

Another approach is to assume that the solution y = φ(t) has a Taylor series about

the point t

. Then

φ(t

+ h) = φ(t

) + φ

′

)h + φ

′′

)

+···,

φ(t

n+1

) = φ(t

) + f [t

, φ(t

)]h + φ

′′

)

+···. (12)

If the series is terminated after the ﬁrst two terms, and φ(t

n+1

) and φ(t

) are replaced

by their approximate values y

n+1

and y

, we again obtain the Euler formula (6). If

more terms in the series are retained, a more accurate formula is obtained. Further,

by using a Taylor series with a remainder, it is possible to estimate the magnitude of

the error in the formula. This is discussed later in this section.

August 7, 2012 21:05 c08 Sheet number 5 Page number 455 cyan black

8.1 The Euler or Tangent Line Method 455

The Backward Euler Formula. A variation on the Euler formula can be obtained by

using the backward difference quotient [φ(t

) − φ(t

n−1

)]/h to approximate the deri-

vative in Eq. (3) instead of the forward difference quotient used in Eq. (4). In this

way we obtain

φ(t

) − φ(t

n−1

)

∼

hf (t

, y

= y

n−1

+ hf (t

, y

Stepping the index up from n to n + 1, we obtain the backward Euler formula

n+1

= y

+ hf (t

n+1

, y

n+1

). (13)

Assuming that y

is known and y

n+1

is to be calculated,observe that Eq. (13) does not

provide an explicit formula for y

n+1

. Rather, it is an equation that implicitly deﬁnes

n+1

and must be solved to determine the value of y

n+1

. How difﬁcult this is depends

entirely on the nature of the function f .

EXAMPLE

Use the backward Euler formula (13) and step sizes h = 0.05, 0.025, 0.01, and 0.001 to ﬁnd

approximate values of the solutionof theinitial value problem(7),(8) on the interval 0 ≤ t ≤ 2.

For this problem, the backward Euler formula (13) becomes

n+1

= y

+ h(1 − t

n+1

+ 4y

n+1

We will show the ﬁrst two steps in detail so that it will be clear how the method works. At the

ﬁrst step we have

= y

+ h(1 − t

+ 4y

) = 1 + (0.05)(1 − 0.05 + 4y

Solving this equation for y

, we obtain

= 1.0475/0.8 = 1.309375.

Observe that because the differential equation is linear, the implicit equation for y

is also

linear and therefore easy to solve. Next,

= y

+ h(1 − t

+ 4y

) = 1.309375 + (0.05)(1 − 0.1 + 4y

which leads to

= 1.354375/0.8 = 1.69296875.

Continuing the computations on a computer, we obtain the results shown in Table 8.1.2.

The values given by the backward Euler method are uniformly too large for this problem,

TABLE 8.1.2 A Comparison of Results for the Numerical Approximation of the Solution of

′

= 1 − t + 4y, y(0) = 1 Using the Backward Euler Method for Different Step Sizes h

th= 0 .05 h = 0.025 h = 0.01 h = 0.001 Exact

01.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.11.6929688 1.6474375 1.6236638 1.6104634 1.6090418

0.22.7616699 2.6211306 2.5491368 2.5095731 2.5053299

0.34.4174530 4 .0920886 3.9285724 3.8396379 3.8301388

0.46.9905516 6.3209569 5.9908303 5.8131282 5.7942260

0.510.996956 9.7050002 9.0801473 8.7472667 8.7120041

1.0 103.06171 80.402761 70.452395 65.419964 64.897803

1.5 959.44236 661.00731 542.12432 485.05825 479.25919

2.0

8934.

0696 5435.7294 4172.7228 3597.4478 3540.2001

August 7, 2012 21:05 c08 Sheet number 6 Page number 456 cyan black

456 Chapter 8. Numerical Methods

whereas the values obtained from the Euler method were too small. In this problem the errors

are somewhat larger for the backward Euler method than for the Euler method, although for

small values of h the differences are insigniﬁcant. Since the backward Euler method appears

to be no more accurate than the Euler method, and is somewhat more complicated, a natural

question is why it should even be mentioned. The answer is that it is the simplest example of a

class of methods known as backward differentiation formulas that are very useful for certain

types of differential equations. We will return to this issue later in this chapter.

Errors in Numerical Approximations. The use of a numerical procedure,such as the Euler

formula, on an initial value problem raises a number of questions that must be

answered before the approximate numerical solution can be accepted as satisfac-

tory. One of these is the question of convergence. That is, as the step size h tends

to zero, do the values of the numerical approximation y

, y

, ..., y

, ...approach the

corresponding values of the actual solution? If we assume that the answer is afﬁr-

mative, there remains the important practical question of how rapidly the numerical

approximation converges to the solution. In other words, how small a step size is

needed in order to guarantee a given level of accuracy? We want to use a step size

that is small enough to ensure the required accuracy, but not too small. An unneces-

sarily small step size slows down the calculations and in some cases may even cause

a loss of accuracy.

There are three fundamental sources of error in approximating the solution of an

initial value problem numerically.

1. The formula, or algorithm, used in the calculations is an approximate one. For instance,

the Euler formula uses straight-line approximations to the actual solution.

2. Except for the ﬁrst step, the input data used in the calculations are only approximations

to the actual values of the solution at the speciﬁed points.

3. The computer used for the calculations has ﬁnite precision; in other words, at each stage

only a ﬁnite number of digits can be retained.

Let us temporarily assume that our computer can execute all computations exactly;

that is, it can retain inﬁnitely many digits (if necessary) at each step. Then the dif-

ference E

between the solution y = φ(t) of the initial value problem (1), (2) and its

numerical approximation y

at the point t = t

is given by

= φ(t

) − y

. (14)

The error E

is known as the global truncation error. It arises entirely from the

ﬁrst two error sources listed above—that is, by applying an approximate formula to

approximate data.

However, in reality we must carry out the computations using ﬁnite precision

arithmetic, which means that we can keep only a ﬁnite number of digits at each step.

This leads to a round-off error R

deﬁned by

= y

− Y

, (15)

where Y

is the value actually computed from the given numerical method.

The absolute value of the total error in computing φ(t

) is given by

|φ(t

) − Y

|=|φ(t

) − y

+ y

− Y

|. (16)

August 7, 2012 21:05 c08 Sheet number 7 Page number 457 cyan black

8.1 The Euler or Tangent Line Method 457

Making use of the triangle inequality, |a + b|≤|a|+|b|, we obtain, from Eq. (16),

|φ(t

) − Y

|≤|φ(t

) − y

|+|y

− Y

≤|E

|+|R

|. (17)

Thus the total error is bounded by the sum of the absolute values of the global

truncation and round-off errors. For the numerical procedures discussed in this book,

it is possible to obtain useful estimates of the global truncation error. The round-off

error is more difﬁcult to analyze, since it depends on the type of computer used, the

sequence in which the computations are carried out, the method of rounding off, and

so forth. A careful examination of round-off error is beyond the scope of this book,

but see,for example,the book by Henrici listed in the references. Some of the dangers

from round-off error are discussed in Problems 25 through 27 and in Section 8.6.

It is often useful to consider separately the part of the global truncation error that

is due only to the use of an approximate formula. We can do this by assuming at the

nth step that the input data are accurate—that is, that y

= φ(t

). This error is known

as the local truncation error; we will denote it by e

Local Truncation Error for the Euler Method. Let us assume that the solution y = φ(t) of

the initial value problem (1), (2) has a continuous second derivative in the interval

of interest. To ensure this, we can assume that f , f

, and f

are continuous. For if f has

these properties and if φ is a solution of the initial value problem (1), (2), then

′

(t) = f [t, φ(t)],

and, by the chain rule,

′′

(t) = f

[t, φ(t)]+f

[t, φ(t)]φ

′

(t)

= f

[t, φ(t)]+f

[t, φ(t)]f [t, φ(t)]. (18)

Since the right side of this equation is continuous, φ

′′

is also continuous.

Then, making use of a Taylor polynomial with a remainder to expand φ about t

we obtain

φ(t

+ h) = φ(t

) + φ

′

)h +

′′

, (19)

where

is some point in t

< t

+ h. Then, noting that φ(t

+ h) = φ(t

n+1

) and

′

) = f [t

, φ(t

)], we can rewrite Eq. (19) as

φ(t

n+1

) = φ(t

) + hf [t

, φ(t

)]+

′′

. (20)

Now let us use the Euler formula to calculate an approximation to φ(t

n+1

) under

the assumption that we know the correct value for y

at t

, namely y

= φ(t

).The

result is

∗

n+1

= φ(t

) + hf [t

, φ(t

)], (21)

where the asterisk is used to designate this hypothetical approximate value for

φ(t

n+1

). The difference between φ(t

n+1

) and y

∗

n+1

is the local truncation error for the

(n + 1)st step in the Euler method, which we will denote by e

n+1

.Thus,by subtracting

Eq. (21) from Eq. (20), we ﬁnd that

n+1

= φ(t

n+1

) − y

∗

n+1

′′

, (22)

since the remaining terms in Eqs. (20) and (21) cancel.

August 7, 2012 21:05 c08 Sheet number 8 Page number 458 cyan black

458 Chapter 8. Numerical Methods

Thus thelocal truncation error forthe Euler method isproportional to the squareof

the step size h,and the proportionality factor depends on the second derivative of the

solution φ. The expression given by Eq. (22) depends on n and,in general, is different

for each step. A uniform bound, valid on an interval [a, b], is given by

|≤Mh

/2, (23)

where M is the maximum of |φ

′′

(t)| on the interval [a, b]. Since Eq. (23) is based

on a consideration of the worst possible case—that is, the largest possible value of

|φ

′′

(t)|—it may well be a considerable overestimate of the actual local truncation

error in some parts of the interval [a, b].

One use of Eq. (23) is to choose a step size that will result in a local truncation error

no greater than some given tolerance level. For example, if the local truncation

error must be no greater than ϵ, then from Eq. (23) we have

/2 ≤ ϵ or h ≤

2ϵ/M. (24)

The primary difﬁculty in using Eq. (22), (23), or (24) lies in estimating |φ

′′

(t)| or M.

However, the central fact expressed by these equations is that the local truncation

error is proportional to h

. For example,if a new value of h is used that is one-half of

its original value, then the resulting error will be reduced to one-fourth of its previous

value.

More important than the local truncation error is the global truncation error E

The analysis for estimating E

is much more difﬁcult than that for e

. Nevertheless,

it can be shown that the global truncation error in using the Euler method on a ﬁnite

interval is no greater than a constant times h. Thus

|≤Kh (25)

for some constant K; see Problem 23 for more details. The Euler method is called

a ﬁrst order method because its global truncation error is proportional to the ﬁrst

power of the step size.

Because it is more accessible,we will hereafter use the local truncation error as our

principal measure of the accuracy of a numerical method and for comparing different

methods. If we have a priori information about the solution of the given initial value

problem,we can use the result (22) to obtain more precise information about how the

local truncation error varies with t. As an example, consider the illustrative problem

′

= 1 − t + 4y, y(0) = 1 (26)

on the interval 0 ≤ t ≤ 2. Let y = φ(t) be the solution of the initial value problem

(26). Then, as noted previously,

φ(t) = (4t − 3 + 19e

)/16

and therefore

′′

(t) = 19e

Equation (22) then states that

n+1

19e

, t

< t

+ h. (27)

August 7, 2012 21:05 c08 Sheet number 9 Page number 459 cyan black

8.1 The Euler or Tangent Line Method 459

The appearance of the factor 19 and the rapid growth of e

explain why the results

in Table 8.1.1 are not very accurate.

For instance, for h = 0.05 the error in the ﬁrst step is

= φ(t

) − y

19e

(0.0025)

,0<

< 0.05.

It is clear that e

is positive, and since e

< e

0.2

, we have

≤

19e

0.2

(0.0025)

∼

0.02901. (28)

Note also that e

> 1; hence e

> 19(0.0025)/2 = 0.02375. The actual error is

0.02542. It follows from Eq. (27) that the error becomes progressively worse

with increasing t; this is also clearly shown by the results in Table 8.1.1. Similar

computations for bounds for the local truncation error give

1.0617

∼

19e

3.8

(0.0025)

≤ e

≤

19e

(0.0025)

∼

1.2967 (29)

in going from 0.95 to 1.0 and

57.96

∼

19e

7.8

(0.0025)

≤ e

≤

19e

(0.0025)

∼

70.80 (30)

in going from 1.95 to 2.0.

These results indicate that for this problem, the local truncation error is about 2500

times larger near t = 2 than near t = 0. Thus, to reduce the local truncation error to

an acceptable level throughout 0 ≤ t ≤ 2, we must choose a step size h based on an

analysis near t = 2. Of course, this step size will be much smaller than necessary near

t = 0. For example, to achieve a local truncation error of 0.01 for this problem, we

need a step size of about 0.00059 near t = 2 and a step size of about 0.032 near t = 0.

The use of a uniform step size that is smaller than necessary over much of the interval

results in more calculations than necessary, more time consumed, and possibly more

danger of unacceptable round-off errors.

Another approach is to keep the local truncation error approximately constant

throughout the interval by gradually reducing the step size as t increases. In the

example problem, we would need to reduce h by a factor of about 50 in going from

t = 0tot = 2.A method that provides for variations in the step size is called adaptive.

All modern computer codes for solving differential equations have the capability of

adjusting the step size as needed. We will return to this question in the next section.

August 7, 2012 21:05 c08 Sheet number 10 Page number 460 cyan black

460 Chapter 8. Numerical Methods

PROBLEMS In each of Problems 1 through 6, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.1, 0.2, 0.3, and 0.4.

(a) Use the Euler method with h = 0.05.

(b) Use the Euler method with h = 0.025.

(d) Use the backward Euler method with h = 0.025.

′

= 3 + t − y, y(0) = 12.y

′

= 5t − 3

√

y, y(0) = 2

′

= 2y − 3t, y(0) = 14.y

′

= 2t + e

−ty

, y(0) = 1

′

+ 2ty

3 + t

, y(0) = 0.56. y

′

= (t

− y

) sin y, y(0) =−1

In each of Problems 7 through 12, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.5, 1.0, 1.5, and 2.0.

(a) Use the Euler method with h = 0.025.

(b) Use the Euler method with h = 0.0125.

(d) Use the backward Euler method with h = 0.0125.

′

= 0.5 − t + 2y, y(0) = 18.y

′

= 5t − 3

√

y, y(0) = 2

′

√

t + y, y(0) = 3 10. y

′

= 2t + e

−ty

, y(0) = 1

11.

′

= (4 − ty)/(1 + y

), y(0) =−2

12.

′

= (y

+ 2ty)/(3 + t

), y(0) = 0.5

13.

Complete the calculations leading to the entries in columns three and four of Table 8.1.1.

14.

Complete the calculations leading to the entries in columns three and four of Table 8.1.2.

15. Using three terms in the Taylor series given in Eq. (12) and taking h = 0.1, determine

approximate values of the solution of the illustrative example y

′

= 1 − t + 4y, y(0) = 1at

t = 0.1 and 0.2. Compare the results with those using the Euler method and with the exact

values.

Hint: If y

′

= f (t, y), what is y

′′

In each of Problems 16 and 17, estimate the local truncation error for the Euler method in

terms of the solution y = φ(t). Obtain a bound for e

n+1

in terms of t and φ(t) that is valid on

the interval 0 ≤ t ≤ 1. By using a formula for the solution, obtain a more accurate error bound

for e

n+1

.Forh = 0.1 compute a bound for e

and compare it with the actual error at t = 0.1.

Also compute a bound for the error e

in the fourth step.

16. y

′

= 2y − 1, y(0) = 1 17. y

′

− t + 2y, y(0) = 1

In each of Problems 18 through 21,obtain a formula for the local truncation error for the Euler

method in terms of t and the solution φ.

18. y

′

= t

+ y

, y(0) = 1 19. y

′

= 5t − 3

√

y, y(0) = 2

20. y

′

√

t + y, y(1) = 3 21. y

′

= 2t + e

−ty

, y(0) = 1

22.

Consider the initial value problem

′

= cos 5πt, y(0) = 1.

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8.1 The Euler or Tangent Line Method 461

(a) Determine the solution y = φ(t), and draw a graph of y = φ(t) for 0 ≤ t ≤ 1.

(b) Determine approximate values of φ(t) at t = 0.2, 0.4, and 0.6 using the Euler method

with h = 0.2. Draw a broken-line graph for the approximate solution,and compare it with

the graph of the exact solution.

(d) Show by computing the local truncation error that neither of these step sizes is sufﬁ-

ciently small. Determine a value of h to ensure that the local truncation error is less than

0.05 throughout the interval 0 ≤ t ≤ 1.That such a small value of h is required results from

the fact that max |φ

′′

(t)| is large.

23. In this problem we discuss the global truncation error associated with the Euler method

for the initial value problem y

′

= f (t, y), y(t

) = y

. Assuming that the functions f and f

are continuous in a closed,bounded region R of the ty-plane that includes the point (t

, y

it can be shown that there exists a constant L such that |f (t, y) − f (t,

y| < L|y −

y|, where

(t, y) and (t,

y) are any two points in R with the same t coordinate (see Problem 15 of

Section 2.8). Further, we assume that f

is continuous, so the solution φ has a continuous

second derivative.

(a) Using Eq. (20), show that

n+1

|≤|E

|+h|f [t

, φ(t

)]−f (t

, y

)|+

|φ

′′

)|≤α|E

|+βh

, (i)

where α = 1 + hL and β = max |φ

′′

(t)|/2 on t

≤ t ≤ t

(b) Assume that if E

= 0, and if |E

| satisﬁes Eq. (i), then |E

|≤βh

(α

− 1)/(α − 1)

for α ̸= 1. Use this result to show that

|≤

(1 + hL)

− 1

βh. (ii)

Equation (ii) gives a bound for |E

| in terms of h, L, n, and β. Notice that for a ﬁxed h, this

error bound increases with increasing n; that is, the error bound increases with distance

from the starting point t

≤ e

nhL

; hence

|≤

nhL

− 1

βh.

If we select an ending point T greater than t

and then choose the step size h so that n

steps are required to traverse the interval [t

, T], then nh = T − t

, and

|≤

(T−t

− 1

βh = Kh,

which is Eq. (25). Note that K depends on the length T − t

of the interval and on the

constants L and β that are determined from the function f .

24. Derive an expression analogous to Eq. (22) for the local truncation error for the backward

Euler formula.

Hint: Construct a suitable Taylor approximation to φ(t) about t = t

n+1

25.

Using a step size h = 0.05 and the Euler method,but retaining only three digits throughout

the computations, determine approximate values of the solution at t = 0.1, 0.2, 0.3, and

0.4 for each of the following initial value problems:

(a) y

′

= 1 − t + 4y, y(0) = 1

(b) y

′

= 3 + t − y, y(0) = 1

′

= 2y − 3t, y(0) = 1

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462 Chapter 8. Numerical Methods

Compare the results with those obtained in Example 1 and in Problems 1 and 3. The

small differences between some of those results rounded to three digits and the present

results are due to round-off error. The round-off error would become important if the

computation required many steps.

26. The following problem illustrates a danger that occurs because of round-off error when

nearly equal numbers are subtracted and the difference is then multiplied by a large

number. Evaluate the quantity

1000 ·

6.010 18.04

2.004 6.000

in the following ways:

(a) First round each entry in the determinant to two digits.

(b) First round each entry in the determinant to three digits.

27. The distributive law a(b − c) = ab − ac does not hold, in general, if the products are

rounded off to a smaller number of digits. To show this in a speciﬁc case, take a = 0.22,

b = 3.19, and c = 2.17. After each multiplication, round off the last digit.

8.2 Improvements on the Euler Method

Since for many problems the Euler method requires a very small step size to produce

sufﬁciently accurate results, much effort has been devoted to the development of

more efﬁcient methods. In the next three sections, we will discuss some of these

methods. Consider the initial value problem

′

= f (t, y), y(t

) = y

(1)

and let y = φ(t) denote its solution. Recall from Eq. (10) of Section 8.1 that by

integrating the given differential equation from t

to t

n+1

, we obtain

φ(t

n+1

) = φ(t

) +

n+1

f [t, φ(t)]dt. (2)

The Euler formula

n+1

= y

+ hf (t

, y

) (3)

is obtained by replacing f [t, φ(t)] in Eq. (2) by its approximate value f (t

, y

) at the

left endpoint of the interval of integration.

Improved Euler Formula. A better approximate formula can be obtained if the

integrand in Eq. (2) is approximated more accurately. One way to do this is to

replace the integrand by the average of its values at the two endpoints, namely,

{f [t

, φ(t

)]+f [t

n+1

, φ(t

n+1

)]}/2. This is equivalent to approximating the area under

the curve in Figure 8.2.1 between t = t

and t = t

n+1

by the area of the shaded

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8.2 Improvements on the Euler Method 463

trapezoid. Further, we replace φ(t

) and φ(t

n+1

) by their respective approximate

values y

and y

n+1

. In this way we obtain, from Eq. (2),

n+1

= y

f (t

, y

) + f (t

n+1

, y

n+1

)

h. (4)

Since the unknown y

n+1

appears as one of the arguments of f on the right side of

Eq. (4), this equation deﬁnes y

n+1

implicitly rather than explicitly. Depending on

the nature of the function f , it may be fairly difﬁcult to solve Eq. (4) for y

n+1

. This

difﬁculty can be overcome by replacing y

n+1

on the right side of Eq. (4) by the value

obtained using the Euler formula (3). Thus

n+1

= y

f (t

, y

) + f [t

+ h, y

+ hf (t

, y

)]

= y

+ f (t

+ h, y

+ hf

)

h, (5)

where t

n+1

has been replaced by t

+ h.

{f [t

, (t

)] +

y' = f [t, (t)]

f [t

, (t

)]

f [t

n+1

, (t

n+1

)]}

n+1

, (t

n+1

)]

n+1

FIGURE 8.2.1 Derivation of the improved Euler method.

Equation (5) gives an explicit formula for computing y

n+1

, the approximate value

of φ(t

n+1

), in terms of the data at t

. This formula is known as the improved Euler

formula or the Heun

formula. The improved Euler formula is an example of a two-

stage method; that is, we ﬁrst calculate y

+ hf

from the Euler formula and then

use this result to calculate y

n+1

from Eq. (5). The improved Euler formula (5) is an

improvement over the Euler formula (3) because the local truncation error in using

Eq. (5) is proportional to h

, while for the Euler method it is proportional to h

. This

error estimate forthe improvedEuler formula isestablished in Problem14. It canalso

be shown that for a ﬁnite interval,the global truncation error for the improved Euler

formula is bounded by a constant times h

, so this method is a second order method.

Note that this greater accuracy is achieved at the expense of more computational

work, since it is now necessary to evaluate f (t, y) twice in order to go from t

to t

n+1

The formula is named for the German mathematician Karl Heun (1859–1929), who was a professor at

Technische Hochschule Karlsruhe.

August 7, 2012 21:05 c08 Sheet number 14 Page number 464 cyan black

464 Chapter 8. Numerical Methods

If f (t, y) depends only on t and not on y, then solving the differential equation

′

= f (t, y) reduces to integrating f (t). In this case the improved Euler formula (5)

becomes

n+1

− y

[f (t

) + f (t

+ h)], (6)

which is just the trapezoid rule for numerical integration.

EXAMPLE

Use the improved Euler formula (5) to calculate approximate values of the solution of the

initial value problem

′

= 1 − t + 4y, y(0) = 1. (7)

To make clear exactly what computations are required, we show a couple of steps in detail.

For this problem f (t, y) = 1 − t + 4y; hence

= 1 − t

+ 4y

and

f (t

+ h, y

+ hf

) = 1 − (t

+ h) + 4(y

+ hf

Further, t

= 0, y

= 1, and f

= 1 − t

+ 4y

= 5. If h = 0.025, then

f (t

+ h, y

+ hf

) = 1 − 0.025 + 4[1 + (0.025)(5)]=5.475.

Then, from Eq. (5),

= 1 + (0.5)(5 + 5.475)(0.025) = 1.1309375. (8)

At the second step we must calculate

= 1 − 0.025 + 4(1.1309375) = 5.49875,

+ hf

= 1.1309375 + (0.025)(5.49875) = 1.26840625,

and

f (t

, y

+ hf

) = 1 − 0.05 + 4(1.26840625) = 6.023625.

Then, from Eq. (5),

= 1.1309375 + (0.5)(5.49875 + 6.023625)(0.025) = 1.2749671875. (9)

Further results for 0 ≤ t ≤ 2 obtained by using the improved Euler method with h = 0.025

and h = 0.01 are given inTable 8.2.1.To compare the results of the improved Euler method with

those of the Euler method, note that the improved Euler method requires two evaluations of

f at each step, while the Euler method requires only one. This is signiﬁcant because typically,

most of the computing time in each step is spent in evaluating f , so counting these evalua-

tions is a reasonable way to estimate the total computing effort. Thus, for a given step size h,

the improved Euler method requires twice as many evaluations of f as the Euler method.

Alternatively, the improved Euler method for step size h requires the same number of eval-

uations of f as the Euler method for step size h/2.

By referring toTable 8.2.1,you can see that the improved Euler method with h = 0.025 gives

much better results than the Euler method with h = 0.01. Note that to reach t = 2 with these

step sizes, the improved Euler method requires 160 evaluations of f , while the Euler method

requires 200. More noteworthy is that the improved Euler method with h = 0.025 is also

slightly more accurate than the Euler method with h = 0.001 (2000 evaluations of f ). In other

words, with something like one-twelfth of the computing effort, the improved Euler method

yields results for this problem that are comparable to, or a bit better than, those generated by

the Euler method. This illustrates that, compared to the Euler method, the improved Euler

method is clearly more efﬁcient, yielding substantially better results or requiring much less

total computing effort, or both.

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8.2 Improvements on the Euler Method 465

TABLE 8.2.1 A Comparison of Results Using the Euler and Improved Euler Methods for

the Initial Value Problem y

′

= 1 − t + 4y, y(0) = 1

Euler Improved Euler

th= 0.01 h = 0.001 h = 0.025 h = 0.01 Exact

01.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.11.5952901 1.6076289 1.6079462 1.6088585 1.6090418

0.22.4644587 2.5011159 2.5020618 2.5047827 2.5053299

0.33.7390345 3.8207130 3.8228282 3.8289146 3.8301388

0.45.6137120 5.7754845 5.7796888 5.7917911 5.7942260

0.58.3766865 8.6770692 8.6849039 8.7074637 8.7120041

1.060.037126 64.382558 64.497931 64.830722 64.897803

1.5 426.40818 473.55979 474.83402 478.51588 479.25919

2.0

3029

.3279 3484.1608 3496.6702 3532.8789 3540.2001

The percentage errors at t = 2 for the improved Euler method are 1.23% for h = 0.025 and

0.21% for h = 0.01.

A computer program for the Euler method can be readily modiﬁed to implement

the improved Euler method instead. All that is required is to replace Step 6 in the

algorithm in Section 8.1 by the following:

The Improved Euler Method

Step 6. k1 = f (t, y)

k2 = f (t + h, y + h ∗ k1)

y = y + (h/2) ∗ (k1 + k2)

t = t + h

Variation of Step Size. In Section 8.1 we mentioned the possibility of adjusting the

step size as a calculation proceeds so as to maintain the local truncation error at

a more or less constant level. The goal is to use no more steps than necessary

and, at the same time, to keep some control over the accuracy of the approxi-

mation. Here we will describe how this can be done. First we choose the error

tolerance ϵ, which is the local truncation error that we are willing to accept. Sup-

pose that after n steps we have reached the point (t

, y

). We choose a step size

h and calculate y

n+1

. Next we need to estimate the error we have made in calcu-

lating y

n+1

. Not knowing the actual solution, the best that we can do is to use a

more accurate method and repeat the calculation starting from (t

, y

). For exam-

ple, if we used the Euler method for the original calculation, we might repeat it

with the improved Euler method. Then the difference between the two calculated

values is an estimate e

est

n+1

of the error in using the original method. If the esti-

mated error is different from the error tolerance ϵ, then we adjust the step size

and repeat the calculation. The key to making this adjustment efﬁciently is know-

ing how the local truncation error e

n+1

depends on the step size h. For the Euler

method,thelocal truncation erroris proportional toh

,soto bring theestimated error

down (or up) to the tolerance level ϵ, we must multiply the original step size by the

factor

√

ϵ/e

est

n+1

August 7, 2012 21:05 c08 Sheet number 16 Page number 466 cyan black

466 Chapter 8. Numerical Methods

To illustrate this procedure, consider the example problem (7):

′

= 1 − t + 4y, y(0) = 1.

Suppose that we choose the error tolerance ϵ to be 0.05. You can verify that after one

step with h = 0.1, we obtain the values 1.5 and 1.595 from the Euler method and the

improved Euler method, respectively. Thus the estimated error in using the Euler

method is 0.095. Since this is larger than the tolerance level of 0.05, we need to adjust

the step size downward by the factor

0.05/0.095

∼

0.73. Rounding downward to

be conservative, let us choose the adjusted step size h = 0 .07. Then, from the Euler

formula, we obtain

= 1 + (0.07)f (0, 1) = 1.35

∼

φ(0.07).

Using the improved Euler method, we obtain y

= 1.39655, so the estimated error in

using the Euler formula is 0.04655, which is slightly less than the speciﬁed tolerance.

The actual error, based on a comparison with the solution itself, is somewhat greater,

namely, 0.05122.

We can follow the same procedure at each step of the calculation, thereby keeping

the local truncation error approximately constant throughout the entire numerical

process. Modern adaptive codes for solving differential equations adjust the step

size in very much this way as they proceed, although they usually use more accurate

formulas than the Euler and improved Euler formulas. Consequently, they are able

to achieve both efﬁciency and accuracy by using very small steps only where they

are really needed.

PROBLEMS In each of Problems 1 through 6, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.1, 0.2, 0.3, and 0.4. Compare the results with those obtained by the

Euler method and the backward Euler method in Section 8.1 and with the exact solution (if

available).

(a) Use the improved Euler method with h = 0 .05.

(b) Use the improved Euler method with h = 0.025.

′

= 3 + t − y, y(0) = 12.y

′

= 5t − 3

√

y, y(0) = 2

′

= 2y − 3t, y(0) = 14.y

′

= 2t + e

−ty

, y(0) = 1

′

+ 2ty

3 + t

, y(0) = 0.56. y

′

= (t

− y

) sin y, y(0) =−1

In each of Problems 7 through 12, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.5, 1.0, 1.5, and 2.0.

(a) Use the improved Euler method with h = 0 .025.

(b) Use the improved Euler method with h = 0.0125.

′

= 0.5 − t + 2y, y(0) = 18.y

′

= 5t − 3

√

y, y(0) = 2

′

√

t + y, y(0) = 3 10. y

′

= 2t + e

−ty

, y(0) = 1

11.

′

= (4 − ty)/(1 + y

), y(0) =−2

12.

′

= (y

+ 2ty)/(3 + t

), y(0) = 0.5

13.

Complete the calculations leading to the entries in columns four and ﬁve of Table 8.2.1.

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8.2 Improvements on the Euler Method 467

14. In this problem we establish that the local truncation error for the improved Euler for-

mula is proportional to h

. If we assume that the solution φ of the initial value problem

′

= f (t, y), y(t

) = y

has derivatives that are continuous through the third order ( f has

continuous second partial derivatives), then it follows that

φ(t

+ h) = φ(t

) + φ

′

)h +

′′

)

′′′

)

where t

< t

+ h. Assume that y

= φ(t

(a) Show that, for y

n+1

as given by Eq. (5),

n+1

= φ(t

n+1

) − y

n+1

′′

)h −{f [t

+ h, y

+ hf (t

, y

)]−f (t

, y

)}

h +

′′′

. (i)

(b) Making use of the facts that φ

′′

(t) = f

[t, φ(t)]+f

[t, φ(t)]φ

′

(t) and that the Taylor

approximation with a remainder for a function F(t, y) of two variables is

F(a + h, b + k) = F (a, b) + F

(a, b)h + F

(a, b)k

+ 2hkF

+ k

)

x=ξ,y=η

where ξ lies between a and a + h, and η lies between b and b + k, show that the ﬁrst term

on the right side of Eq. (i) is proportional to h

plus higher order terms. This is the desired

result.

n+1

= φ

′′′

/6 for some t

with

< t

n+1

Hint: What are f

, f

, and f

15. Consider the improved Euler method for solving the illustrative initial value problem

′

= 1 − t + 4y, y(0) = 1. Using the result of Problem 14(c) and the exact solution of the

initial value problem, determine e

n+1

and a bound for the error at any step on 0 ≤ t ≤ 2.

Compare this error with the one obtained in Eq. (27) of Section 8.1 using the Euler method.

Also obtain a bound for e

for h = 0.05, and compare it with Eq. (28) of Section 8.1.

In each of Problems 16 and 17, use the actual solution φ(t) to determine e

n+1

and a bound

for e

n+1

at any step on 0 ≤ t ≤ 1 for the improved Euler method for the given initial value

problem. Also obtain a bound for e

for h = 0.1, and compare it with the similar estimate for

the Euler method and with the actual error for the improved Euler method.

16. y

′

= 2y − 1, y(0) = 1 17. y

′

= 0.5 − t + 2y, y(0) = 1

In each of Problems 18 through 21, carry out one step of the Euler method and of the

improved Euler method, using the step size h = 0.1. Suppose that a local truncation error

no greater than 0.0025 is required. Estimate the step size that is needed for the Euler method

to satisfy this requirement at the ﬁrst step.

18. y

′

= 0.5 − t + 2y, y(0) = 1 19. y

′

= 5t − 3

√

y, y(0) = 2

20. y

′

√

t + y, y(0) = 3 21. y

′

= (y

+ 2ty)/(3 + t

), y(0) = 0.5

22. The modiﬁed Euler formula for the initial value problem y

′

= f (t, y), y(t

) = y

is given by

n+1

= y

+ hf [t

h, y

hf (t

, y

)].

Following the procedure outlined in Problem 14, show that the local truncation error in

the modiﬁed Euler formula is proportional to h

August 7, 2012 21:05 c08 Sheet number 18 Page number 468 cyan black

468 Chapter 8. Numerical Methods

In each of Problems 23 through 26,use the modiﬁed Euler formula of Problem 22 with h = 0.05

to compute approximate values of the solution of the given initial value problem at t = 0.1,

0.2, 0.3, and 0.4. Compare the results with those obtained in Problems 1 through 4.

23.

′

= 3 + t − y, y(0) = 1 24. y

′

= 5t − 3

√

y, y(0) = 2

25.

′

= 2y − 3t, y(0) = 1 26. y

′

= 2t + e

−ty

, y(0) = 1

27. Show that the modiﬁed Euler formula of Problem 22 is identical to the improved Euler

formula of Eq. (5) for y

′

= f (t, y) if f is linear in both t and y.

8.3 The Runge–Kutta Method

In preceding sections we have introduced the Euler formula, the backward Euler

formula, and the improved Euler formula as ways to approximate the solution of the

initial value problem

′

= f (t, y), y(t

) = y

(1)

numerically. The local truncation errors for these methods are proportional to h

, h

and h

, respectively. The Euler and improved Euler methods belong to what is now

called the Runge–Kutta

class of methods.

In this section we discuss the method originally developed by Runge and Kutta.

This method is now called the classic fourth order four-stage Runge–Kutta method,

but it is often referred to simply as the Runge–Kutta method, and we will follow this

practice for brevity. This method has a local truncation error that is proportional

to h

. Thus it is two orders of magnitude more accurate than the improved Euler

method and three orders of magnitude better than the Euler method. It is relatively

simple to use and is sufﬁciently accurate to handle many problems efﬁciently. This

is especially true of adaptive Runge–Kutta methods, in which provision is made to

vary the step size as needed. We return to this issue at the end of the section.

The Runge–Kutta formula involves a weighted average of values of f (t, y) at

different points in the interval t

≤ t ≤ t

n+1

. It is given by

n+1

= y

+ h

+ 2k

+ k

(

, (2)

where

= f (t

, y

= f (t

h, y

(3)

= f (t

h, y

= f (t

+ h, y

+ hk

Carl David Runge (1856–1927), a German mathematician and physicist, worked for many years in

spectroscopy.The analysis of data led him to consider problems in numerical computation,and the Runge–

Kutta method originated in his paper on the numerical solution of differential equations in 1895. The

method was extended to systems of equations in 1901 by Martin Wilhelm Kutta (1867–1944). Kutta was

a German mathematician and aerodynamicist who is also well known for his important contributions to

classical airfoil theory.

August 7, 2012 21:05 c08 Sheet number 19 Page number 469 cyan black

8.3 The Runge–Kutta Method 469

The sum (k

+ 2k

+ k

)/6 can be interpreted as an average slope. Note

that k

is the slope at the left end of the interval, k

is the slope at the midpoint

using the Euler formula to go from t

to t

+ h/2, k

is a second approximation

to the slope at the midpoint, and k

is the slope at t

+ h using the Euler formula

and the slope k

to go from t

to t

+ h.

Although in principle it is not difﬁcult to show that Eq. (2) differs from the Taylor

expansion of the solution φ by terms that are proportional to h

, the algebra is

rather lengthy.

Thus we simply state without proof that the local truncation error

in using Eq. (2) is proportional to h

and that for a ﬁnite interval the global trunca-

tion error is at most a constant times h

. The earlier description of this method as a

fourth order four-stage method reﬂects the facts that the global truncation error is

of fourth order in the step size h and that there are four intermediate stages in the

calculation (the calculation of k

, ..., k

Clearly the Runge–Kutta formula, Eqs. (2) and (3), is more complicated than any

of the formulas discussed previously. This is of relatively little signiﬁcance, however,

since it is not hard to write a computer program to implement this method. Such a

program has the same structure as the algorithm for the Euler method outlined in

Section 8.1. To be speciﬁc, the lines in Step 6 in the Euler algorithm must be replaced

by the following:

The Runge–Kutta Method

Step 6. k1 = f (t, y)

k2 = f (t + 0.5 ∗ h, y + 0.5 ∗ h ∗ k1)

k3 = f (t + 0.5 ∗ h, y + 0.5 ∗ h ∗ k2)

k4 = f (t + h, y + h ∗ k3)

y = y + (h/6) ∗ (k1 + 2 ∗ k2 + 2 ∗ k3 + k4)

t = t + h

Note that if f does not depend on y, then

= f (t

), k

= k

= f (t

+ h/2), k

= f (t

+ h), (4)

and Eq. (2) reduces to

n+1

− y

[f (t

) + 4f (t

+ h/2) + f (t

+ h)]. (5)

Equation (5) can be identiﬁed as Simpson’s

rule for the approximate evaluation of

the integral of y

′

= f (t). The fact that Simpson’s rule has an error proportional to h

is consistent with the local truncation error in the Runge–Kutta formula.

See, for example, Chapter 3 of the book by Henrici listed in the references.

Simpson’s rule is named for Thomas Simpson (1710–1761), an English mathematician and textbook

author, who published it in 1743.

August 7, 2012 21:05 c08 Sheet number 20 Page number 470 cyan black

470 Chapter 8. Numerical Methods

EXAMPLE

Use the Runge–Kutta method to calculate approximate values of the solution y = φ(t) of the

initial value problem

′

= 1 − t + 4y, y(0) = 1. (6)

Taking h = 0.2, we have

= f (0, 1) = 5; hk

= 1.0,

= f (0 + 0.1, 1 + 0.5) = 6.9; hk

= 1.38,

= f (0 + 0.1, 1 + 0.69) = 7.66; hk

= 1.532,

= f (0 + 0.2, 1 + 1.532) = 10.928.

Thus

= 1 +

0.2

[5 + 2(6.9) + 2(7.66) + 10.928]

= 1 + 1.5016 = 2.5016.

Further results using the Runge–Kutta method with h = 0.2,h = 0.1,and h = 0.05 are given

in Table 8.3.1. Note that the Runge–Kutta method yields a value at t = 2 that differs from the

exact solution by only 0.122% if the step size is h = 0.1, and by only 0.00903% if h = 0.05.

In the latter case, the error is less than one part in 10,000, and the calculated value at t = 2is

correct to four digits.

For comparison, note that both the Runge–Kutta method with h = 0.05 and the improved

Euler method with h = 0.025 require 160 evaluations of f to reach t = 2. The improved Euler

method yields a result at t = 2 that is in error by 1.23%. Although this error may be acceptable

for some purposes, it is more than 135 times the error yielded by the Runge–Kutta method

with comparable computing effort. Note also that the Runge–Kutta method with h = 0.2,

40 evaluations of f , produces a value at t = 2 with an error of 1.40%, which is only slightly

greater than the error in the improved Euler method with h = 0.025, or 160 evaluations of f .

Thus we see again that a more accurate algorithm is more efﬁcient; it produces better results

with similar effort, or similar results with less effort.

TABLE 8.3.1 A Comparison of Results for the Numerical Approximation of the Solution

of the Initial Value Problem y

′

= 1 − t + 4y, y(0) = 1

Improved

Euler Runge–Kutta Exact

th= 0.025 h = 0.2 h = 0.1 h = 0.05

01.0000000 1.0000000 1.0000000 1.0000000 1.0000000

0.11.6079462 1.6089333 1.6090338 1.6090418

0.22.5020618 2.5016000 2.5050062 2.5053060 2.5053299

0.33.8228282 3.8294145 3.8300854 3.8301388

0.45.7796888 5.7776358 5.7927853 5.7941197 5.7942260

0.58.6849039 8.7093175 8.7118060 8.7120041

1.064.497931 64.441579 64.858107 64.894875 64.897803

1.5 474.83402 478.81928 479.22674 479.25919

2.0 3496.6702 3490.5574 3535.8667

3539

.8804 3540.2001

The classic Runge–Kutta methodsuffers fromthe sameshortcoming as other meth-

ods with a ﬁxed step size for problems in which the local truncation error varies

widely over the interval of interest. That is, a step size that is small enough to achieve

August 7, 2012 21:05 c08 Sheet number 21 Page number 471 cyan black

8.3 The Runge–Kutta Method 471

satisfactory accuracy in some parts of the interval may be much smaller than neces-

sary in other parts of the interval. This has stimulated the development of adaptive

Runge–Kutta methods that provide for modifying the step size automatically as the

computation proceeds, so as to maintain the local truncation error near or below a

speciﬁed tolerance level. As explained in Section 8.2, this requires the estimation of

the local truncation error at each step. One way to do this is to repeat the compu-

tation with a ﬁfth order method—which has a local truncation error proportional

to h

—and then to use the difference between the two results as an estimate of the

error. If this is done in a straightforward manner, then the use of the ﬁfth order

method requires at least ﬁve more evaluations of f at each step, in addition to those

required originally by the fourth order method. However,if we make an appropriate

choice of the intermediate points and the weighting coefﬁcients in the expressions for

, ..., k

in a certain fourth order Runge–Kutta method, then these expressions

can be used again, together with one additional stage, in a corresponding ﬁfth order

method. This results in a substantial gain in efﬁciency. It turns out that this can be

done in more than one way.

The ﬁrst fourth and ﬁfth order Runge–Kutta pair was developed by Erwin

Fehlberg

in the late 1960s and is now called the Runge–Kutta–Fehlberg, or RKF,

method. The popularity of the RKF method was considerably enhanced by the

appearance in 1977 of its Fortran implementation RKF45 by Lawrence F. Shampine

and H.A.Watts.The RKF methodand other adaptive Runge–Kutta methods are very

powerful and efﬁcient means of approximating numerically the solutions of an enor-

mous class of initial value problems. Speciﬁc implementations of one or more of them

are widely available in commercial software packages.

PROBLEMS In each of Problems 1 through 6, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.1, 0.2, 0.3, and 0.4. Compare the results with those obtained by using

other methods and with the exact solution (if available).

(a) Use the Runge-Kutta method with h = 0.1.

(b) Use the Runge-Kutta method with h = 0.05.

′

= 3 + t − y, y(0) = 12. y

′

= 5t − 3

√

y, y(0) = 2

′

= 2y − 3t, y(0) = 14.y

′

= 2t + e

−ty

, y(0) = 1

′

+ 2ty

3 + t

, y(0) = 0.56. y

′

= (t

− y

) sin y, y(0) =−1

In each of Problems 7 through 12, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.5, 1.0, 1.5, and 2.0. Compare the results with those obtained by other

methods.

(a) Use the Runge–Kutta method with h = 0.1.

(b) Use the Runge–Kutta method with h = 0.05.

Erwin Fehlberg (1911–1990) was born in Germany, received his doctorate from the Technical University

of Berlin in 1942, emigrated to the United States after World War II, and was employed by NASA for

many years. The Runge–Kutta–Fehlberg method was ﬁrst published in a NASATechnical Report in 1969.

The details of the RKF method may be found, for example, in the books by Ascher and Petzold and by

Mattheij and Molenaar that are listed in the references.

August 7, 2012 21:05 c08 Sheet number 22 Page number 472 cyan black

472 Chapter 8. Numerical Methods

7. y

′

= 0.5 − t + 2y, y(0) = 18.y

′

= 5t − 3

√

y, y(0) = 2

′

√

t + y, y(0) = 3 10. y

′

= 2t + e

−ty

, y(0) = 1

11.

′

= (4 − ty)/(1 + y

), y(0) =−2

12.

′

= (y

+ 2ty)/(3 + t

), y(0) = 0.5

13.

Conﬁrm the results in Table 8.3.1 by executing the indicated computations.

14.

Consider the initial value problem

′

= t

+ y

, y(0) = 1.

(a) Draw a direction ﬁeld for this equation.

(b) Use the Runge–Kutta method or another method to ﬁnd approximate values of the

solution at t = 0.8, 0.9, and 0.95. Choose a small enough step size so that you believe your

results are accurate to at least four digits.

solution at t = 1. If you encounter difﬁculties in doing this, explain why you think this

happens. The direction ﬁeld in part (a) may be helpful.

15.

Consider the initial value problem

′

= 3t

/(3y

− 4), y(0) = 0.

(a) Draw a direction ﬁeld for this equation.

(b) Estimate how far the solution can be extended to theright. Let t

be the right endpoint

of the interval of existence of this solution. What happens at t

to prevent the solution

from continuing farther?

value of t

(d) If you continue the computation beyond t

, you can continue to generate values of y.

What signiﬁcance, if any, do these values have?

(e) Suppose that the initial condition is changed to y(0) = 1. Repeat parts (b) and (c) for

this problem.

8.4 Multistep Methods

In previous sections we have discussed numerical procedures for approximating the

solution of the initial value problem

′

= f (t, y), y(t

) = y

, (1)

in which data at the point t = t

are used to calculate an approximate value of the

solution φ(t

n+1

) at the next mesh point t = t

n+1

. In other words, the calculated value

of φ at any mesh point depends only on the data at the preceding mesh point. Such

methods are called one-step methods. However, once approximate values of the

solution y = φ(t) have been obtained at a few points beyond t

, it is natural to ask

whether we can make use of some of this information, rather than just the value at

the last point, to calculate the value of φ(t) at the next point. Speciﬁcally, if y

at t

, ..., y

at t

are known, how can we use this information to determine y

n+1

August 7, 2012 21:05 c08 Sheet number 23 Page number 473 cyan black

8.4 Multistep Methods 473

n+1

? Methods that use information at more than the last mesh point are referred to

as multistep methods. In this section we will describe two types of multistep methods:

Adams

methods and backward differentiation methods. Within each type, we can

achieve various levels of accuracy,depending on the number of preceding data points

that are used. For simplicity, we will assume throughout our discussion that the step

size h is constant.

Adams Methods. Recall that

φ(t

n+1

) − φ(t

) =

n+1

′

(t) dt, (2)

where φ(t) is the solution of the initial value problem (1).The basic idea of an Adams

method is to approximate φ

′

(t) by a polynomial P

(t) of degree k and to use the

polynomial to evaluate the integral on the right side of Eq. (2). The coefﬁcients in

(t) are determined by using k + 1 previously calculated data points. For example,

suppose that we wish to use a ﬁrst degree polynomial P

(t) = At + B. Then we need

only the two data points (t

, y

) and (t

n−1

, y

n−1

).ForP

to be an approximation to

′

, we require that P

) = f (t

, y

) and that P

n−1

) = f (t

n−1

, y

n−1

). Recall that we

denote f (t

, y

) by f

for an integer j. Then A and B must satisfy the equations

+ B = f

(3)

n−1

+ B = f

n−1

Solving for A and B, we obtain

A =

− f

n−1

, B =

n−1

− f

n−1

. (4)

Replacing φ

′

(t) by P

(t) and evaluating the integral in Eq. (2), we ﬁnd that

φ(t

n+1

) − φ(t

) =

n+1

− t

) + B(t

n+1

− t

Finally, we replace φ(t

n+1

) and φ(t

) by y

n+1

and y

, respectively, and carry out some

algebraic simpliﬁcation. For a constant step size h, we obtain

n+1

= y

−

n−1

. (5)

Equation (5) is the second orderAdams–Bashforth

formula. It is an explicit formula

for y

n+1

in terms of y

and y

n−1

and has a local truncation error proportional to h

We note in passing that the ﬁrst order Adams–Bashforth formula, based on the

polynomial P

(t) = f

of degree zero, is just the original Euler formula.

John Couch Adams (1819–1892),an English mathematician and astronomer, is most famous as codiscov-

erer, with Joseph Leverrier, of the planet Neptune in 1846. He was associated with Cambridge University

for most of his life, as student (1839–1843), fellow, Lowdean Professor, and director of the Observa-

tory. Adams was extremely skilled at computation; his procedure for numerical integration of differential

equations appeared in 1883 in a book with Francis Bashforth on capillary action.

Francis Bashforth (1819–1912), English mathematician and Anglican priest, was a classmate of J. C.

Adams at Cambridge. He was particularly interested in ballistics and invented the Bashforth chronograph

for measuring the velocity of artillery projectiles.

August 7, 2012 21:05 c08 Sheet number 24 Page number 474 cyan black

474 Chapter 8. Numerical Methods

More accurate Adams formulas can be obtained by following the procedure out-

lined above, but using a higher degree polynomial and correspondingly more data

points. For example, suppose that a polynomial P

(t) of degree three is used. The

coefﬁcients are determined from the four points (t

, y

), (t

n−1

, y

n−1

), (t

n−2

, y

n−2

), and

n−3

, y

n−3

). Substituting this polynomial for φ

′

(t) in Eq. (2), evaluating the integral,

and simplifying the result, we eventually obtain the fourth order Adams–Bashforth

formula

n+1

= y

+ (h/24)(55f

− 59f

n−1

+ 37f

n−2

− 9f

n−3

). (6)

The local truncation error of this fourth order formula is proportional to h

A variation on the derivation of the Adams–Bashforth formulas gives another set

of formulas called the Adams–Moulton

formulas. To see the difference, let us again

considerthesecondorder case.Again we use aﬁrst degree polynomial Q

(t) = αt + β,

but we determine the coefﬁcients by using the points (t

, y

) and (t

n+1

, y

n+1

). Thus α

and β must satisfy

αt

+ β = f

(7)

αt

n+1

+ β = f

n+1

and it follows that

α =

n+1

− f

, β =

n+1

− f

n+1

. (8)

Substituting Q

(t) for φ

′

(t) in Eq. (2) and simplifying, we obtain

n+1

= y

hf (t

n+1

, y

n+1

), (9)

which is the second order Adams–Moulton formula. We have written f (t

n+1

, y

n+1

)

in the last term to emphasize that the Adams–Moulton formula is implicit, rather

than explicit, since the unknown y

n+1

appears on both sides of the equation. The

local truncation error for the second order Adams–Moulton formula is proportional

to h

The ﬁrst order Adams–Moulton formula is just the backward Euler formula, as

you might anticipate by analogy with the ﬁrst order Adams–Bashforth formula.

More accurate higher order formulas can be obtained by using an approximating

polynomial of higher degree. The fourth orderAdams–Moulton formula,with a local

truncation error proportional to h

, is

n+1

= y

+ (h/24)(9f

n+1

+ 19f

− 5f

n−1

+ f

n−2

). (10)

Observe that this is also an implicit formula, because y

n+1

appears in f

n+1

Although both the Adams–Bashforth and Adams–Moulton formulas of the same

order have local truncation errors proportional to the same power of h, the Adams–

Moulton formulas of moderate order are in fact considerably more accurate.

For example, for the fourth order formulas (6) and (10), the proportionality constant

Forest Ray Moulton (1872–1952), an American astronomer and administrator of science, was for many

years professor of astronomy at the University of Chicago. During World War I, he was in charge of

the Ballistics Branch of the U.S. Army at Aberdeen (MD) Proving Ground. In the course of calculating

ballistics trajectories he devised substantial improvements in the Adams formula.

August 7, 2012 21:05 c08 Sheet number 25 Page number 475 cyan black

8.4 Multistep Methods 475

for the Adams–Moulton formula is less than 1/10 of the proportionality constant

for the Adams–Bashforth formula. Thus the question arises: is it better to use the

explicit (and faster) Adams–Bashforth formula or the more accurate but implicit

(and slower) Adams–Moulton formula? The answer depends on whether, by using

the more accurate formula, you can increase the step size, and thereby reduce the

number of steps enough to compensate for the additional computations required at

each step.

In fact, numerical analysts have attempted to achieve both simplicity and accuracy

by combining the two formulas in what is called a predictor–corrector method. Once

n−3

, y

n−2

, y

n−1

, and y

are known, we can compute f

n−3

, f

n−2

, f

n−1

, and f

, and then

use the Adams–Bashforth (predictor) formula (6) to obtain a ﬁrst value for y

n+1

Then we compute f

n+1

and use the Adams–Moulton (corrector) formula (10), which

is no longer implicit,to obtain an improved value of y

n+1

. We can, of course,continue

to use the corrector formula (10) if the change in y

n+1

is too large. However, if it is

necessary to use the corrector formula more than once or perhaps twice, it means

that the step size h is too large and should be reduced.

In order to use any of the multistep methods, it is necessary ﬁrst to calculate a few

by some other method. For example, the fourth order Adams–Moulton method

requires values for y

and y

, while the fourth order Adams–Bashforth method also

requires avalue for y

. Oneway to proceed is to use aone-step method of comparable

accuracy to calculate the necessary starting values. Thus, for a fourth order multistep

method,we might use the fourth order Runge–Kutta method to calculate the starting

values. This is the method used in the next example.

Another approach is to use a low order method with a very small h to calculate y

and then to increase gradually both the order and the step size until enough starting

values have been determined.

EXAMPLE

Consider again the initial value problem

′

= 1 − t + 4y, y(0) = 1. (11)

With a step size of h = 0.1, determine an approximate value of the solution y = φ(t) at t = 0.4

using the fourth order Adams–Bashforth formula, the fourth order Adams–Moulton formula,

and the predictor–corrector method.

For starting data, we use the values of y

, y

, and y

found by the Runge–Kutta method.

These are tabulated in Table 8.3.1. Next, calculating the corresponding values of f (t, y),we

obtain

= 1, f

= 5,

= 1.6089333, f

= 7.3357332,

= 2.5050062, f

= 10.820025,

= 3.8294145, f

= 16.017658.

Then, from the Adams–Bashforth formula, Eq. (6), we ﬁnd that y

= 5.7836305. The exact

value of the solution at t = 0.4, correct through eight digits, is 5.7942260, so the error is

−0.0105955.

The Adams–Moulton formula, Eq. (10), leads to the equation

= 4.9251275 + 0.15y

from which it follows that y

= 5.7942676 with an error of only 0 .0000416.

August 7, 2012 21:05 c08 Sheet number 26 Page number 476 cyan black

476 Chapter 8. Numerical Methods

Finally, using the result from the Adams–Bashforth formula as a predicted value of φ(0.4),

we can then use Eq. (10) as a corrector. Corresponding to the predicted value of y

, we ﬁnd

that f

= 23.734522. Hence, from Eq. (10), the corrected value of y

is 5.7926721. This result is

in error by −0.0015539.

Observe that the Adams–Bashforth method is the simplest and fastest of these methods,

since it involves only the evaluation of a single explicit formula. It is also the least accurate.

Using the Adams–Moulton formula as a corrector increases the amount of calculation that

is required, but the method is still explicit. In this problem the error in the corrected value

of y

is reduced by approximately a factor of 7 when compared to the error in the predicted

value. The Adams–Moulton method alone yields by far the best result, with an error that is

about 1/40 as large as the error from the predictor–corrector method. Remember, however,

that the Adams–Moulton method is implicit, which means that an equation must be solved

at each step. In the problem considered here this equation is linear, so the solution is quickly

found, but in other problems this part of the procedure may be much more time-consuming.

The Runge–Kutta method with h = 0.1 gives y

= 5.7927853 with an error of −0.0014407;

see Table 8.3.1. Thus, for this problem, the Runge–Kutta method is comparable in accuracy to

the predictor–corrector method.

Backward Differentiation Formulas. Another type of multistep methoduses a polynomial

(t) to approximate the solution φ(t) of the initial value problem (1) rather than

its derivative φ

′

(t), as in the Adams methods. We then differentiate P

(t) and set

′

n+1

) equal to f (t

n+1

, y

n+1

) to obtain an implicit formula for y

n+1

. These are called

backward differentiation formulas. These methods became widely used in the 1970s

because of the work of C. William Gear

on so-called stiff differential equations,

whose solutions are very difﬁcult to approximate by the methods discussed up to

now; see Section 8.6.

The simplest case uses a ﬁrst degree polynomial P

(t) = At + B. The coefﬁcients

are chosen to match the computed values of the solution y

and y

n+1

. Thus A and B

must satisfy

+ B = y

(12)

n+1

+ B = y

n+1

Since P

′

(t) = A, the requirement that

′

n+1

) = f (t

n+1

, y

n+1

)

is just

A = f (t

n+1

, y

n+1

). (13)

Another expression for A comes from subtracting the ﬁrst of Eqs. (12) from the

second, which gives

A = (y

n+1

− y

)/h.

C.William Gear (1935– ), born in London,England, received his undergraduate education at Cambridge

University and his doctorate in 1960 from the University of Illinois. He was a member of the faculty at the

University of Illinois for most of his career and made signiﬁcant contributions to both computer design

and numerical analysis. His inﬂuential book on numerical methods for differential equations is listed in

the references.

August 7, 2012 21:05 c08 Sheet number 27 Page number 477 cyan black

8.4 Multistep Methods 477

Substituting this value of A into Eq. (13) and rearranging terms, we obtain the ﬁrst

order backward differentiation formula

n+1

= y

+ hf (t

n+1

, y

n+1

). (14)

Note that Eq. (14) is just the backward Euler formula that we ﬁrst saw in Section 8.1.

By using higher order polynomials and correspondingly more data points, we can

obtain backward differentiation formulas of any order. The second order formula is

n+1

[

− y

n−1

+ 2hf (t

n+1

, y

n+1

)

]

, (15)

and the fourth order formula is

n+1

[

48y

− 36y

n−1

+ 16y

n−2

− 3y

n−3

+ 12hf (t

n+1

, y

n+1

)

]

. (16)

These formulas have local truncation errors proportional to h

and h

, respectively.

EXAMPLE

Use the fourth order backward differentiation formula with h = 0.1 and the data given in

Example 1 to determine an approximate value of the solution y = φ(t) at t = 0.4 for the initial

value problem (11).

Using Eq. (16) with n = 3, h = 0.1, and with y

, ..., y

given in Example 1, we obtain the

equation

= 4.6837842 + 0.192y

Thus

= 5.7967626.

Comparing the calculated value with the exact value φ(0.4) = 5.7942260, we ﬁnd that the error

is 0.0025366. This is somewhat better than the result using the Adams–Bashforth method, but

it is not as good as the result using the predictor–corrector method, and not nearly as good as

the result using the Adams–Moulton method.

A comparison between one-step and multistep methods must take several factors

into consideration. The fourth order Runge–Kutta method requires four evaluations

of f at each step, while the fourth order Adams–Bashforth method (once past the

starting values) requires only one, and the predictor–corrector method only two.

Thus, for a given step size h, the latter two methods may well be considerably faster

than Runge–Kutta. However,if Runge–Kutta is more accurate and therefore can use

fewer steps, then the difference in speed will be reduced and perhaps eliminated.The

Adams–Moulton and backward differentiation formulas also require that the difﬁ-

culty in solving the implicit equation at each step be taken into account. All multistep

methods have the possible disadvantage that errors in earlier steps can feed back into

later calculations with unfavorable consequences. On the other hand, the underly-

ing polynomial approximations in multistep methods make it easy to approximate

the solution at points between the mesh points, should this be desirable. Multistep

methods have become popular largely because it is relatively easy to estimate the

error at each step and to adjust the order or the step size to control it. For a further

discussion of such questions as these, see the books listed at the end of this chapter;

in particular, Shampine (1994) is an authoritative source.

August 7, 2012 21:05 c08 Sheet number 28 Page number 478 cyan black

478 Chapter 8. Numerical Methods

PROBLEMS In each of Problems 1 through 6, determine an approximate value of the solution at t = 0.4

and t = 0.5 using the speciﬁed method. For starting values,use the values given by the Runge–

Kutta method; see Problems 1 through 6 of Section 8.3. Compare the results of the various

methods with each other and with the actual solution (if available).

(a) Use the fourth order predictor–corrector method with h = 0.1. Use the corrector formula

once at each step.

(b) Use the fourth order Adams–Moulton method with h = 0.1.

′

= 3 + t − y, y(0) = 12.y

′

= 5t − 3

√

y, y(0) = 2

′

= 2y − 3t, y(0) = 14.y

′

= 2t + e

−ty

, y(0) = 1

′

+ 2ty

3 + t

, y(0) = 0.56. y

′

= (t

− y

) sin y, y(0) =−1

In each of Problems 7 through 12, ﬁnd approximate values of the solution of the given initial

value problem at t = 0.5, 1.0, 1.5, and 2.0, using the speciﬁed method. For starting values,

use the values given by the Runge–Kutta method; see Problems 7 through 12 in Section 8.3.

Compare the results of the various methods with each other and with the actual solution (if

available).

(a) Use the fourth orderpredictor–corrector method with h = 0.05. Use the correctorformula

once at each step.

(b) Use the fourth order Adams–Moulton method with h = 0.05.

′

= 0.5 − t + 2y, y(0) = 18.y

′

= 5t − 3

√

y, y(0) = 2

′

√

t + y, y(0) = 3 10. y

′

= 2t + e

−ty

, y(0) = 1

11.

′

= (4 − ty)/(1 + y

), y(0) =−2

12.

′

= (y

+ 2ty)/(3 + t

), y(0) = 0.5

13. Show that the ﬁrst order Adams–Bashforth method is the Euler method and that the ﬁrst

order Adams–Moulton method is the backward Euler method.

14. Show that the third order Adams–Bashforth formula is

n+1

= y

+ (h/12)(23f

− 16f

n−1

+ 5f

n−2

15. Show that the third order Adams–Moulton formula is

n+1

= y

+ (h/12)(5f

n+1

+ 8f

− f

n−1

16. Derive the second order backwarddifferentiation formula givenby Eq. (15)in this section.

8.5 Systems of First Order Equations

In the preceding sections we discussed numerical methods for approximating the

solution of an initial value problem associated with a ﬁrst order differential equation.

These methods can also be applied to a system of ﬁrst order equations. Since a

higher order equation can always be reduced to a system of ﬁrst order equations,

August 7, 2012 21:05 c08 Sheet number 29 Page number 479 cyan black

8.5 Systems of First Order Equations 479

it is sufﬁcient to deal with systems of ﬁrst order equations alone. For simplicity, we

consider a system of two ﬁrst order equations

′

= f (t, x, y), y

′

= g(t, x, y), (1)

with the initial conditions

x(t

) = x

, y(t

) = y

. (2)

The functions f and g are assumed to satisfy the conditions of Theorem 7.1.1 so that

the initial value problem (1), (2) has a unique solution in some interval of the t-axis

containing the point t

. We wish to determine approximate values x

, x

, ..., x

, ...

and y

, y

, ..., y

, ...of the solution x = φ(t), y = ψ(t) at the points t

= t

+ nh with

n = 1, 2,....

In vector notation, the initial value problem (1), (2) can be written as

′

= f(t, x), x(t

) = x

, (3)

where x is the vector with components x and y, f is the vector function with com-

ponents f and g, and x

is the vector with components x

and y

. The methods of

the previous sections can be readily generalized to handle systems of two (or more)

equations. All that is needed (formally) is to replace the scalar variable x by the vec-

tor x and the scalar function f by the vector function f in the appropriate equations.

For example, the Euler formula becomes

n+1

= x

+ hf

, (4)

or, in component form,

n+1

(

+ h

f (t

, x

, y

)

g(t

, x

, y

)

(

. (5)

The initialconditions are used todetermine f

,whichis thevector tangent to thegraph

of the solution x = φ(t) at the initial point x

in the xy-plane. We move in the direc-

tion of this tangent vector for a time step h in order to ﬁnd the next point x

. Then

we calculate a new tangent vector f

at x

, move along it for a time step h to ﬁnd x

and so forth.

In a similar way, the Runge–Kutta method can be extended to a system. For the

step from t

to t

n+1

we have

n+1

= x

+ (h/6)(k

+ 2k

+ k

), (6)

where

= f(t

, x

= f[t

+ (h/2), x

+ (h/2)k

(7)

= f[t

+ (h/2), x

+ (h/2)k

= f(t

+ h, x

+ hk

The formulas for the Adams–Moulton predictor–corrector method as it applies to

the initial value problem (1), (2) are given in Problem 9.

The vector equations (3), (4), (6), and (7) are, in fact, valid in any number of

dimensions. All that is needed is to interpret the vectors as having n components

rather than two.

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480 Chapter 8. Numerical Methods

EXAMPLE

Determine approximate values of the solution x = φ(t), y = ψ(t) of the initial value problem

′

= x − 4y, y

′

=−x + y, (8)

x(0) = 1, y(0) = 0, (9)

at the point t = 0.2. Use the Euler method with h = 0.1 and the Runge–Kutta method with

h = 0.2. Compare the results with the values of the exact solution:

φ(t) =

−t

+ e

, ψ(t) =

−t

− e

. (10)

Let us ﬁrst use the Euler method. For this problem, f

= x

− 4y

and g

=−x

+ y

; hence

= 1 − (4)(0) = 1, g

=−1 + 0 =−1.

Then, from the Euler formulas (4) and (5), we obtain

= 1 + (0.1)(1) = 1.1, y

= 0 + (0.1)(−1) =−0.1.

At the next step,

= 1.1 − (4)(−0.1) = 1.5, g

=−1.1 + (−0.1) =−1.2.

Consequently,

= 1.1 + (0.1)(1.5) = 1.25, y

=−0.1 + (0.1)(−1.2) =−0.22.

The values of the exact solution, correct to eight digits, are φ(0.2) = 1.3204248 and

ψ(0.2) =−0.25084701. Thus the values calculated from the Euler method are in error by

about 0.0704 and 0.0308, respectively, corresponding to percentage errors of about 5.3% and

12.3%.

Now let us use the Runge–Kutta method to approximate φ(0.2) and ψ(0.2). With h = 0.2

we obtain the following values from Eqs. (7):

f (1, 0)

g(1, 0)

−1

;

f (1.1, −0.1)

g(1.1, −0.1)

1.5

−1.2

;

f (1.15, −0.12)

g(1.15, −0.12)

1.63

−1.27

;

f (1.326, −0.254)

g(1.326, −0.254)

2.342

−1.580

Then, substituting these values in Eq. (6), we obtain

(

0.2

9.602

−7.52

(

1.3200667

−0.25066667

(

These values of x

and y

are in error by about 0.000358 and 0.000180, respectively, with

percentage errors much less than one-tenth of 1%.

This example again illustrates the great gains in accuracy that are obtainable by using a more

accurate approximation method,such as the Runge–Kuttamethod. In the calculations we have

just outlined, the Runge–Kutta method requires only twice as many function evaluations as

the Euler method, but the error in the Runge–Kutta method is about 200 times less than that

in the Euler method.

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8.5 Systems of First Order Equations 481

PROBLEMS In each of Problems 1 through 6, determine approximate values of the solution x = φ(t),

y = ψ(t) of the given initial value problem at t = 0.2, 0.4, 0.6, 0.8, and 1.0. Compare the results

obtained by different methods and different step sizes.

(a) Use the Euler method with h = 0.1.

(b) Use the Runge–Kutta method with h = 0.2.

′

= x + y + t, y

′

= 4x − 2y; x(0) = 1, y(0) = 0

′

= 2x + ty, y

′

= xy; x(0) = 1, y(0) = 1

′

=−tx − y − 1, y

′

= x; x(0) = 1, y(0) = 1

′

= x − y + xy, y

′

= 3x − 2y − xy; x(0) = 0, y(0) = 1

′

= x(1 − 0.5x − 0.5y), y

′

= y(−0.25 + 0.5x); x(0) = 4, y(0) = 1

′

= exp(−x + y) − cos x, y

′

= sin(x − 3y); x(0) = 1, y(0) = 2

Consider the example problem x

′

= x − 4y, y

′

=−x + y with the initial conditions

x(0) = 1 and y(0) = 0. Use the Runge–Kutta method to ﬁnd approximate values of

the solution of this problem on the interval 0 ≤ t ≤ 1. Start with h = 0.2, and then

repeat the calculation with step sizes h = 0.1, 0.05, ..., each half as long as in the preceding

case. Continue the process until the ﬁrst ﬁve digits of the solution at t = 1 are unchanged

for successive step sizes. Determine whether these digits are accurate by comparing them

with the exact solution given in Eqs. (10) in the text.

Consider the initial value problem

′′

+ t

′

+ 3x = t, x(0) = 1, x

′

(0) = 2.

Convert this problem to a system of two ﬁrst order equations, and determine approximate

values of the solution at t = 0.5 and t = 1.0 using the Runge–Kutta method with h = 0.1.

Consider the initial value problem x

′

= f (t, x, y) and y

′

= g(t, x, y) with x(t

) = x

and

y(t

) = y

. The generalization of the Adams–Moulton predictor–corrector method of

Section 8.4 is

n+1

= x

h(55f

− 59f

n−1

+ 37f

n−2

− 9f

n−3

n+1

= y

h(55g

− 59g

n−1

+ 37g

n−2

− 9g

n−3

)

and

n+1

= x

h(9f

n+1

+ 19f

− 5f

n−1

+ f

n−2

n+1

= y

h(9g

n+1

+ 19g

− 5g

n−1

+ g

n−2

Determine an approximate value of the solution at t = 0.4 for the example initial value

problem x

′

= x − 4y, y

′

=−x + y with x(0) = 1, y(0) = 0. Take h = 0.1. Correct the pre-

dicted value once. For the values of x

, ..., y

use the values of the exact solution rounded

to six digits:x

= 1.12735, x

= 1.32042, x

= 1.60021, y

=−0.111255, y

=−0.250847,

and y

=−0.429696.

August 7, 2012 21:05 c08 Sheet number 32 Page number 482 cyan black

482 Chapter 8. Numerical Methods

8.6 More on Errors; Stability

In Section 8.1 we discussed some ideas related to the errors that can occur in a

numerical approximation of the solution of the initial value problem

′

= f (t, y), y(t

) = y

. (1)

In this section we continue that discussion and also point out some other difﬁculties

that can arise. Some of the points that we wish to make are fairly difﬁcult to treat in

detail, so we will illustrate them by means of examples.

Truncation an d Round-off Errors. Recall that for the Euler method we showed that the

localtruncationerror is proportional to h

andthatfor a ﬁnite interval,theglobal trun-

cation error is at most a constant times h. In general,for a method of order p,the local

truncation error is proportional to h

p+1

and the global truncation error on a ﬁnite

interval is bounded by a constant times h

.To achieve high accuracy,we normally use

a numerical procedure for which p is fairly large, perhaps 4 or higher. As p increases,

the formula used in computing y

n+1

normally becomes more complicated, and hence

more calculations are required at each step. However, this is usually not a serious

problem unless f (t, y) is very complicated or the calculation must be repeated very

many times. If the step size h is decreased, the global truncation error is decreased by

the same factor raised to the power p. However,as we mentioned in Section 8.1,if h is

very small,a great many steps will be required to cover a ﬁxed interval,and the global

round-off error may be larger than the global truncation error.The situation is shown

schematically in Figure 8.6.1. We assume that the round-off error R

is proportional

to the number of computations performed and therefore is inversely proportional to

the step size h. On the other hand,the truncation error E

is proportional to a positive

power of h. From Eq. (17) of Section 8.1, we know that the total error is bounded by

|+|R

|; hence we wish to choose h so as to minimize this quantity. The optimum

value of h occurs when the rate of increase of the truncation error (as h increases) is

balanced by the rate of decrease of the round-off error, as indicated in Figure 8.6.1.

Error

⏐E

⏐ + ⏐R

⏐

⏐R

⏐

⏐E

⏐

opt

FIGURE 8.6.1 The dependence of truncation and round-off errors on the step size h.

EXAMPLE

Consider the example problem

′

= 1 − t + 4y, y(0) = 1. (2)

Using the Euler method with various step sizes, calculate approximate values for the solution

φ(t) at t = 0.5 and t = 1. Try to determine the optimum step size.

August 7, 2012 21:05 c08 Sheet number 33 Page number 483 cyan black

8.6 More on Errors; Stability 483

Keeping only four digits in order to shorten the calculations, we obtain the data shown in

Table 8.6.1. The ﬁrst two columns are the step size h and the number of steps N required to

traverse the interval 0 ≤ t ≤ 1. Then y

N/2

and y

are approximations to φ(0.5) = 8.712 and

φ(1) = 64 .90, respectively. These quantities appear in the third and ﬁfth columns. The fourth

and sixth columns display the differences between the calculated values and the actual value

of the solution.

TABLE 8.6.1 Approximations to the Solution of the Initial Value

Problem y

′

= 1 − t + 4y, y(0) = 1 Using the Euler Method with

Different Step Sizes

hNy

N/2

Error y

Error

0.01 100 8.390 0.322 60.12 4.78

0.005 200 8.551 0.161 62.51 2.39

0.002 500 8.633 0.079 63.75 1.15

0.001 1000 8.656 0.056 63.94 0.96

0.0008 1250 8.636 0.076 63.78 1.12

0.000625 1600 8.616 0.096 64.35 0.55

0.0005 2000 8.772 0.060 64 .00 0.90

0.0004 2500 8.507 0.205 63.40 1.50

0.00025 4000 8.231 0.481 56.77 8.13

For relatively large step sizes, the round-off error is much less than the global truncation

error. Consequently, the total error is approximately the same as the global truncation error,

which for the Euler method is bounded by a constant times h. Thus, as the step size is reduced,

the error is reduced proportionally.Theﬁrst three lines inTable 8.6.1 show this type of behavior.

For h = 0.001 the error has been further reduced, but much less than proportionally; this

indicates that round-off error is becoming important. As h is reduced still more, the error

begins to ﬂuctuate, and further improvements in accuracy become problematical. For values

of h less than 0.0005, the error is clearly increasing, which indicates that round-off error is

now the dominant part of the total error.

These results can also be expressed in terms of the number of steps N.ForN less than

about 1000, accuracy is improved by taking more steps, while for N greater than about 2000,

using more steps has an adverse effect. Thus for this problem it is best to use an N somewhere

between 1000 and 2000. For the calculations shown in Table 8.6.1, the best result at t = 0.5

occurs for N = 1000, while at t = 1.0 the best result is for N = 1600.

You should be careful not to read too much into the results shown in Example 1.

The optimum ranges for h and N depend on the differential equation, the numerical

method that is used, and the number of digits that are retained in the calculation.

Nevertheless, it is generally true that if too many steps are required in a calculation,

then eventually round-off error is likely to accumulate to the point where it seriously

degrades the accuracy of the procedure. For many problems this is not a concern:

for them, any of the fourth order methods we have discussed in Sections 8.3 and 8.4

will produce good results with a number of steps far less than the level at which

round-off error becomes important. For some problems, however, round-off error

does become vitally important. For such problems, the choice of method may be

crucial. This is also one reason why modern codes provide a means of adjusting

the step size as they go along, using a larger step size wherever possible and a very

small step size only where necessary.

August 7, 2012 21:05 c08 Sheet number 34 Page number 484 cyan black

484 Chapter 8. Numerical Methods

Vertical Asymptotes. As a second example, consider the problem of approximating the

solution y = φ(t) of

′

= t

+ y

, y(0) = 1. (3)

Since the differential equation is nonlinear,the existence and uniqueness theorem

(Theorem 2.4.2) guarantees only that there is a solution in some interval about t = 0.

Suppose that we try to compute an approximation to the solution of the initial value

problem on the interval 0 ≤ t ≤ 1 using different numerical procedures.

If we use the Euler method with h = 0.1, 0.05, and 0.01, we ﬁnd the follow-

ing approximate values at t = 1: 7.189548, 12.32093, and 90.75551, respectively. The

large differences among the computed values are convincing evidence that we should

use a more accurate numerical procedure—the Runge–Kutta method, for exam-

ple. Using the Runge–Kutta method with h = 0.1, we ﬁnd the approximate value

735.0991at t = 1,whichis quite differentfrom those obtainedusing the Euler method.

Repeating the calculations using step sizes of h = 0.05 and h = 0.01, we obtain the

information shown in Table 8.6.2.

TABLE 8.6.2 Approximations to the Solution

of the Initial Value Problem y

′

= t

+ y

y(0) = 1 Using the Runge–Kutta Method

ht= 0.90 t = 1.0

0.114.02182 735.0991

0.05 14.27117 1.75863 × 10

0.01 14.30478 2.0913 × 10

2893

0.001 14.30486

The values at t = 0.90 are reasonable, and we might well believe that the solution

has a value of about 14.305 at t = 0.90. However, it is not clear what is happen-

ing between t = 0.9 and t = 1.0. To help clarify this, let us turn to some analytical

approximations to thesolution of theinitial value problem(3). Note thaton 0 ≤ t ≤ 1,

≤ t

+ y

≤ 1 + y

. (4)

This suggests that the solution y = φ

(t) of

′

= 1 + y

, y(0) = 1 (5)

and the solution y = φ

(t) of

′

= y

, y(0) = 1 (6)

are upper and lower bounds, respectively, for the solution y = φ(t) of the original

problem,since all these solutions pass through the same initial point. Indeed,it can be

shown (for example, by the iteration method of Section 2.8) that φ

(t) ≤ φ(t) ≤ φ

(t)

August 7, 2012 21:05 c08 Sheet number 35 Page number 485 cyan black

8.6 More on Errors; Stability 485

as long as these functions exist. The important thing to note is that we can solve

Eqs. (5) and (6) for φ

and φ

by separation of variables. We ﬁnd that

(t) = tan

)

t +

, φ

(t) =

1 − t

. (7)

Thus φ

(t) →∞as t → 1, and φ

(t) →∞as t → π/4

∼

0.785. These calculations

show that the solution of the original initial value problem exists at least for

0 ≤ t <π/4 and at most for 0 ≤ t < 1. The solution of the problem (3) has a vertical

asymptote for some t in π/4 ≤ t ≤ 1 and thus does not exist on the entire interval

0 ≤ t ≤ 1.

Our numerical calculations, however, suggest that we can go beyond t = π/4, and

probably beyond t = 0.9. Assuming that the solution of the initial value problem

exists at t = 0.9 and has the value 14.305, we can obtain a more accurate estimate of

what happens for larger t by considering the initial value problems (5) and (6) with

y(0) = 1 replaced by y(0.9) = 14.305. Then we obtain

(t) = tan(t + 0.60100), φ

(t) = 1/(0.96991 − t), (8)

where ﬁve decimal places have been kept in the calculation.Therefore, φ

(t) →∞as

t → π/2 − 0.60100

∼

0.96980 and φ

(t) →∞as t → 0.96991. We conclude that the

asymptote of the solution of the initial value problem (3) lies between these two

values. This example illustrates the sort of information that can be obtained by a

judicious combination of analytical and numerical work.

Stability. The concept of stability is associated with the possibility that small errors

that are introduced in the course of a mathematical procedure may die out as the

procedure continues. Conversely, instability occurs if small errors tend to increase,

perhaps without bound. For example, in Section 2.5 we identiﬁed equilibrium solu-

tions of a differential equation as (asymptotically) stable or unstable, depending on

whether solutions that were initiallynear theequilibrium solution tendedto approach

it or to depart from it as t increased. Somewhat more generally, the solution of an

initial value problem is asymptotically stable if initially nearby solutions tend to

approach the given solution, and unstable if they tend to depart from it. Visually, in

an asymptotically stable problem the graphs of solutions will come together, while

in an unstable problem they will separate.

If we are investigating an initial value problem numerically, the best that we can

hope for is that the numerical approximation will mimic the behavior of the actual

solution. We cannot make an unstable problem into a stable one merely by approx-

imating its solution numerically. However, it may well happen that a numerical

procedure will introduce instabilities that were not part of the original problem, and

this can cause trouble in approximating the solution. Avoidance of such instabilities

may require us to place restrictions on the step size h.

To illustrate what can happen in the simplest possible context, consider the

differential equation

dy/dt = ry, (9)

where r is a constant. Suppose that in approximating the solution of this equation,

we have reached the point (t

, y

). Let us compare the exact solution of Eq. (9) that

August 7, 2012 21:05 c08 Sheet number 36 Page number 486 cyan black

486 Chapter 8. Numerical Methods

passes through this point, namely,

y = y

exp[r(t − t

)], (10)

with numerical approximations obtained from the Euler formula

n+1

= y

+ hf (t

, y

) (11)

and from the backward Euler formula

n+1

= y

+ hf (t

n+1

, y

n+1

). (12)

From the Euler formula (11) we obtain

n+1

= y

+ hry

= y

(1 + rh). (13)

Similarly, from the backward Euler formula (12) we have

n+1

= y

+ hry

n+1

1 − rh

= y

[1 + rh + (rh)

+···]. (14)

Finally, evaluating the solution (10) at t

+ h, we ﬁnd that

n+1

= y

exp(rh) = y

1 + rh +

(rh)

+···

. (15)

Comparing Eqs. (13), (14), and (15), we see that the errors in both the Euler formula

and the backward Euler formula are of order h

, as the theory predicts.

Now suppose that we change the value y

to y

+ δ. Think, if you wish, of δ as the

error that has accumulated by the time we reach t = t

. The question is whether this

error increases or decreases in going one more step to t

n+1

For the exact solution (15), the change in y

n+1

due to the change δ in y

is just

δ exp(rh). This quantity is less than δ if exp(rh)<1, or in other words if r < 0. This

conﬁrms our conclusion in Chapter 2 that Eq. (9) is asymptotically stable if r < 0,

and is unstable if r > 0.

For thebackward Euler method,the change iny

n+1

in Eq. (14) due toδ isδ/(1 − rh).

For r < 0 the quantity 1/(1 − rh) is always nonnegative and less than 1. Thus, if the

differential equation is stable, then so is the backward Euler method for an arbitrary

step size h.

On the other hand, for the Euler method, the change in y

n+1

in Eq. (13) due to

δ is δ(1 + rh).Forr < 0 we can write 1 + rh as 1 −|r|h. Then the requirement that

|1 + rh| < 1 is equivalent to

−1 < 1 −|r|h < 1, or 0 < |r|h < 2.

Consequently, h must satisfy h < 2/|r|. Thus the Euler method is not stable for this

problem unless h is sufﬁciently small.

The restriction on the step size h in using the Euler method in the preceding

example is rather mild unless |r| is quite large. Nonetheless, the example illustrates

that it may be necessary to restrict h in order to achieve stability in the numerical

method, even though the initial value problem itself is stable for all values of h.

Problems for which a much smaller step size is needed for stability than for accuracy

August 7, 2012 21:05 c08 Sheet number 37 Page number 487 cyan black

8.6 More on Errors; Stability 487

are called stiff. The backward differentiation formulas described in Section 8.4 (of

which the backward Euler formula is the lowest order example) are the most popular

formulas for dealing with stiff problems.The following example illustrates the kind of

instability that can occur when we try to approximate the solution of a stiff problem.

EXAMPLE

AStiff

Problem

Consider the initial value problem

′

=−100y + 100t + 1, y(0) = 1. (16)

Find numerical approximations to the solution for 0 ≤ t ≤ 1 using the Euler, backward Euler,

and Runge–Kutta methods. Compare the numerical results with the exact solution.

Since the differential equation is linear, it is easy to solve, and the solution of the initial

value problem (16) is

y = φ(t) = e

−100t

+ t. (17)

Some values of the solution φ(t),correct to sixdecimal places,are given in the second column of

Table 8.6.3,and a graph of the solution is shown in Figure 8.6.2.There is a thin layer (sometimes

called a boundary layer) to the right of t = 0 in which the exponential term is signiﬁcant

and the solution varies rapidly. Once past this layer, however, φ(t)

∼

t and the graph of the

solution is essentially a straight line. The width of the boundary layer is somewhat arbitrary,

but it is certainly small. At t = 0.1, for example, exp(−100t)

∼

0.000045.

TABLE 8.6.3 Numerical Approximations to the Solution of the Initial Value Problem

′

=−100y + 100t + 1, y(0) = 1

t Exact Euler Euler Runge–Kutta Runge–Kutta Backward Euler

0.025 0.0166 ... 0.0333 ... 0.025 0.1

0.01.000000 1.000000 1.000000 1.000000 1.000000 1.000000

0.05 0.056738 2.300000 −0.246296 0.470471

0.10.100045 5.162500 0.187792 10.6527 0.276796 0.190909

0.20.200000 25.8289 0.207707 111.559 0.231257 0.208264

0.40.400000 657.241 0.400059 1.24 × 10

0.400977 0.400068

0.60.600000 1.68 × 10

0.600000 1.38 × 10

0.600031 0.600001

0.80.800000 4.31 × 10

0.800000 1.54 × 10

0.800001 0.800000

1.01.000000 1.11 × 10

1.000000 1.71 × 10

1.000000 1.000000

0.8

0.6

0.4

0.2

0.8 10.60.40.2

FIGURE 8.6.2 The solution of the initial value problem y

′

=−100y + 100t + 1, y(0) = 1.

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488 Chapter 8. Numerical Methods

If we plan to approximate the solution (17) numerically, we might intuitively expect that a

small step size will be needed only in the boundary layer. To make this expectation a bit more

precise, recall from Section 8.1 that the local truncation errors for the Euler and backward

Euler methods are proportional to φ

′′

(t). For this problem φ

′′

(t) = 10

−100t

, which varies from

a value of 10

at t = 0 to nearly zero for t > 0.2. Thus a very small step size is needed for

accuracy near t = 0, but a much larger step size is adequate once t is a little larger.

On the other hand, the stability analysis in Eqs. (9) through (15) also applies to this problem.

Since r =−100 for Eq. (16), it follows that for stability we need h < 0.02 for the Euler method,

but there is no corresponding restriction for the backward Euler method.

Some results obtained from the Euler method are shown in columns three and four of

Table 8.6.3. The values for h = 0.025 are worthless because of instability, while those for

h = 0.01666 ... are reasonably accurate for t ≥ 0.2. However, comparable accuracy for this

range of t is obtained for h = 0.1 by using the backward Euler method, as shown by the results

in column seven of the table.

The situation is not improved by using, instead of the Euler method, a more accurate one,

such as Runge–Kutta. For this problem the Runge–Kutta method is unstable for h = 0.033 ...

but stable for h = 0.025, as shown by the results in columns ﬁve and six of Table 8.6.3.

The results given in the table for t = 0.05 and for t = 0.1 show that in the boundary layer,

a smaller step size is needed to obtain an accurate approximation. You are invited to explore

this matter further in Problem 3.

The following example illustrates some other difﬁculties that may be encountered

in dealing with numerical approximations for unstable differential equations.

EXAMPLE

Consider the task of ﬁnding numerical approximations to two linearly independent solutions

of the second order linear equation

′′

− 10π

y = 0 (18)

for t > 0. Note any difﬁculties that may arise.

Since we contemplate a numerical approach to this problem, we ﬁrst convert Eq. (18) into

a system of two ﬁrst order equations so that we can use the methods of Section 8.5. Thus we

let x

= y and x

= y

′

. In this way we obtain the system

′

= x

, x

′

= 10π

or, if x = (x

, x

)

′

10π

x. (19)

The eigenvalues and eigenvectors of the coefﬁcient matrix in Eq. (19) are

√

10π, ξ

(1)

√

10π

; r

=−

√

10π, ξ

(2)

−

√

10π

, (20)

so two linearly independent solutions of the system (19) are

(1)

(t) =

√

10π

√

10πt

, x

(2)

(t) =

−

√

10π

−

√

10πt

. (21)

August 7, 2012 21:05 c08 Sheet number 39 Page number 489 cyan black

8.6 More on Errors; Stability 489

The corresponding solutions of the second order equation (18) are the ﬁrst components of

(1)

(t) and x

(2)

(t): y

(t) = e

√

10πt

and y

(t) = e

−

√

10πt

, respectively.

We also wish to consider another pair of linearly independent solutions obtained by forming

linear combinations of x

(1)

(t) and x

(2)

(t):

(3)

(t) =

(1)

(t) +

(2)

(t) =

cosh

√

10πt

√

10π sinh

√

10πt

(22)

and

(4)

(t) =

(1)

(t) −

(2)

(t) =

sinh

√

10πt

√

10π cosh

√

10πt

(23)

Although the expressions for x

(3)

(t) and x

(4)

(t) are quite different, recall that for large t we

have cosh

√

10πt ∼ e

√

10πt

/2 and sinh

√

10πt ∼ e

√

10πt

/2. Thus if t is large enough and only a

ﬁxed number of digits are retained,then numerically the two vector functions x

(3)

(t) and x

(4)

(t)

look exactly the same. For instance, correct to eight digits, we have, for t = 1,

sinh

√

10π = cosh

√

10π = 10,315.894.

If we retain only eight digits, then the two solutions x

(3)

(t) and x

(4)

(t) are identical at t = 1 and

indeed for all t > 1. Even if we retain more digits, eventually the two solutions will appear

(numerically) to be identical. This phenomenon is called numerical dependence.

For the system (19) we can avoid the issue of numerical dependence by calculating instead

the solutions x

(1)

(t) and x

(2)

(t). From Eq. (21) we know that x

(1)

(t) is proportional to e

√

10πt

while x

(2)

(t) is proportional to e

−

√

10πt

, so they behave very differently as t increases. Even so,

we encounter difﬁculty in calculating x

(2)

(t) correctly on a large interval. Observe that x

(2)

(t)

is the solution of Eq. (19) subject to the initial condition

x(0) =

−

√

10π

. (24)

If we attempt to approximate the solution of the initial value problem (19), (24) numerically,

then at each step of the calculation we introduce truncation and round-off errors. Thus, at

any point t

, the data to be used in going to the next point are not precisely the values of the

components of x

(2)

). The solution of the initial value problem with these data at t

involves

not only e

−

√

10πt

but also e

√

10πt

. Because the error in the data at t

is small, the latter function

appears with a very small coefﬁcient. Nevertheless, since e

−

√

10πt

tends to zero and e

√

10πt

grows

very rapidly, the latter eventually dominates, and ultimately the calculated solution is very far

from x

(2)

(t).

To be speciﬁc, suppose that we try to approximate the solution of the initial value problem

(19),(24),the ﬁrst component of which is the solution y

(t) = e

−

√

10πt

of the second order initial

value problem

′′

− 10π

y = 0, y(0) = 1, y

′

(0) =−

√

10π. (25)

Using the Runge–Kutta method with a step size h = 0.01 and keeping eight digits in the

calculations,we obtain the results inTable 8.6.4. It is clear from these results that the numerical

approximation begins to deviate signiﬁcantly from the exact solution for t > 0.5, and soon

differs from it by many orders of magnitude. The reason is the presence, in the numerical

August 7, 2012 21:05 c08 Sheet number 40 Page number 490 cyan black

490 Chapter 8. Numerical Methods

approximation,of a small component of the exponentially growing quantity e

√

10πt

.With eight-

digit arithmetic we can expect a round-off error of the order of 10

−8

at each step. Since

√

10πt

grows by a factor of 3.7 × 10

from t = 0tot = 5, an error of order 10

−8

near t = 0

can produce an error of order 10

at t = 5 even if no further errors are introduced in the

intervening calculations. The results given in Table 8.6.4 demonstrate that this is exactly what

happens.

TABLE 8.6.4 Exact Solution of y

′′

− 10π

y = 0,

y(0) = 1, y

′

(0) =−

√

10π and Numerical

Approximation Using the Runge–Kutta Method

with h = 0.01

t Numerical Exact

0.01.01.0

0.25 8.3439 × 10

−2

8.3438 × 10

−2

0.56.9623 × 10

−3

6.9620 × 10

−3

0.75 5.8409 × 10

−4

5.8089 × 10

−4

1.08.6688 × 10

−5

4.8469 × 10

−5

1.55.4900 × 10

−3

3.3744 × 10

−7

2.07.8852 × 10

−1

2.3492 × 10

−9

2.51.1326 × 10

1.6355 × 10

−11

3.01.6268 × 10

1.1386 × 10

−13

3.52.3368 × 10

7.9272 × 10

−16

4.03.3565 × 10

5.5189 × 10

−18

4.54.8211 × 10

3.8422 × 10

−20

5.06.9249 × 10

2.6749 × 10

−22

You should bear in mind that the numerical values of the entries in the second column of

Table 8.6.4 are extremely sensitive to slight variations in how the calculations are executed.

Regardless of such details, however, the exponential growth of the approximation will be

clearly evident.

Equation(18) ishighly unstable,and thebehavior shownin this example is typical of unstable

problems. We can track a solution accurately for a while, and the interval can be extended by

using smaller step sizes or more accurate methods,but eventually the instability in the problem

itself takes over and leads to large errors.

Some Comments on Numerical Methods. In this chapter we have introduced several nu-

merical methods for approximating the solution of an initial value problem. We have

tried to emphasize some important ideas while maintaining a reasonable level of

complexity. For one thing, we have always used a uniform step size, whereas produc-

tion codes that are currently in use provide for varying the step size as the calculation

proceeds.

There are several considerations that must be taken into account in choosing step

sizes. Of course, one is accuracy; too large a step size leads to an inaccurate result.

Normally, an error tolerance is prescribed in advance, and the step size at each step

must be consistent with this requirement. As we have seen, the step size must also be

August 7, 2012 21:05 c08 Sheet number 41 Page number 491 cyan black

8.6 More on Errors; Stability 491

chosen so that the method is stable. Otherwise,small errors will grow and soon render

the results worthless. Finally,for implicit methods an equation must be solved at each

step, and the method used to solve the equation may impose additional restrictions

on the step size.

In choosing a method,one must alsobalance theconsiderations of accuracy and sta-

bility against the amount of time required to execute each step. An implicit method,

such as the Adams–Moulton method, requires more calculations for each step, but

if its accuracy and stability permit a larger step size (and consequently fewer steps),

then this may more than compensate for the additional calculations. The backward

differentiation formulas of moderate order (say, four) are highly stable and are

therefore indicated for stiff problems, for which stability is the controlling factor.

Some current production codes also permit the order of the method to be varied,

as well as the step size, as the calculation proceeds. The error is estimated at each

step, and the order and step size are chosen to satisfy the prescribed error tolerance.

In practice,Adams methods up to order twelve and backward differentiation formu-

las up to order ﬁve are in use. Higher order backward differentiation formulas are

unsuitable because of a lack of stability.

Finally, we note that the smoothness of the function f —that is, the number of

continuous derivatives that it possesses—is a factor in choosing the order of the

method to be used. High order methods lose some of their accuracy if f is not smooth

to a corresponding order.

PROBLEMS 1. To obtain some idea of the possible dangers of small errors in the initial conditions, such

as those due to round-off, consider the initial value problem

′

= t + y − 3, y(0) = 2.

(a) Show that the solution is y = φ

(t) = 2 − t.

(b) Suppose that in the initial condition a mistake is made, and 2.001 is used instead of 2.

Determine the solution y = φ

(t) in this case, and compare the difference φ

(t) − φ

(t) at

t = 1 and as t →∞.

2. Consider the initial value problem

′

= t

+ e

, y(0) = 0. (i)

Using the Runge–Kutta method with step size h, we obtain the results in Table 8.6.5.

Theseresults suggest that the solution has a vertical asymptote between t = 0.9 and t = 1.0.

TABLE 8.6.5 Approximations to the

Solution of the Initial Value Problem

′

= t

+ e

, y(0) = 0 Using the

Runge–Kutta Method

ht= 0.90 t = 1.0

0.02 3.42985 > 10

0.01 3.42982 > 10

August 7, 2012 21:05 c08 Sheet number 42 Page number 492 cyan black

492 Chapter 8. Numerical Methods

(a) Let y = φ(t) be the solution of initial value problem (i). Further, let y = φ

(t) be the

solution of

′

= 1 + e

, y(0) = 0, (ii)

and let y = φ

(t) be the solution of

′

= e

, y(0) = 0. (iii)

Show that

(t) ≤ φ(t) ≤ φ

(t) (iv)

on some interval, contained in 0 ≤ t ≤ 1, where all three solutions exist.

(b) Determine φ

(t) and φ

(t). Then show that φ(t) →∞for some t between

t = ln 2

∼

0.69315 and t = 1.

′

= e

and y

′

= 1 + e

, respectively, with the initial

condition y(0.9) = 3.4298. Use the results to show that φ(t) →∞when t

∼

0.932.

Consider again the initial value problem (16) from Example 2. Investigate how small a

step size h must be chosen to ensure that the error at t = 0.05 and at t = 0.1 is less than

0.0005.

(a) Use the Euler method.

(b) Use the backward Euler method.

Consider the initial value problem

′

=−10y + 2.5t

+ 0.5t, y(0) = 4.

(a) Find the solution y = φ(t) and draw its graph for 0 ≤ t ≤ 5.

(b) The stability analysis in the text suggests that for this problem, the Euler method is

stable only for h < 0.2. Conﬁrm that this is true by applying the Euler method to this

problem for 0 ≤ t ≤ 5 with step sizes near 0.2.

What can you conclude about the stability of this method?

(d) Apply the backward Euler method to this problem for 0 ≤ t ≤ 5 with various step

sizes. What step size is needed to ensure that the error at t = 5 is less than 0.01?

In each of Problems 5 and 6:

(a) Find a formula for the solution of the initial value problem,and note that it is independent

of λ.

(b) Use the Runge–Kutta method with h = 0.01 to compute approximate values of the

solution for 0 ≤ t ≤ 1 for various values of λ such as λ = 1, 10, 20, and 50.

imations.

′

− λy = 1 − λt, y(0) = 06. y

′

− λy = 2t − λt

, y(0) = 0

August 7, 2012 21:05 c08 Sheet number 43 Page number 493 cyan black

References 493

REFERENCES There are many books of varying degrees of sophistication that deal with numerical analysis in general

and the numerical approximation of solutions of ordinary differential equations in particular. Among

these are:

Ascher, Uri M., and Petzold, Linda R., Computer Methods for Ordinary Differential Equations and

Differential-Algebraic Equations (Philadelphia: Society for Industrial and Applied Mathematics,

1998).

Atkinson, Kendall E., Han, Weimin, and Stewart, David, Numerical Solution of Ordinary Differential

Equations (Hoboken, NJ:Wiley, 2009).

Gear, C. William,Numerical Initial Value Problems in Ordinary Differential Equations (Englewood Cliffs,

NJ: Prentice-Hall, 1971).

Henrici, Peter, Discrete Variable Methods in Ordinary Differential Equations (New York: Wiley, 1962).

Henrici,Peter,Error Propagation for Difference Methods (NewYork:Wiley,1963;Huntington,NY:Krieger,

1977).

Mattheij, Robert, and Molenaar, Jaap, Ordinary Differential Equations in Theory and Practice

(New York: Wiley, 1996; Philadelphia: Society for Industrial and Applied Mathematics, 2002).

Shampine,Lawrence F.,Numerical Solution of Ordinary Differential Equations (New York: Chapman and

Hall, 1994).

A detailed exposition of Adams predictor–corrector methods, including practical guidelines for

implementation, may be found in

Shampine, L. F., and Gordon, M. K., Computer Solution of Ordinary Differential Equations: The Initial

Value Problem (San Francisco: Freeman, 1975).

Many books on numerical analysis have chapters on differential equations. For example, at an

elementary level, see

Burden, Richard L., and Faires, J. Douglas, Numerical Analysis (9th ed.) (Boston: Brooks/Cole, Cengage

Learning, 2011).

The following two books are at a slightly higher level and include information on implementing the

algorithms in MATLAB.

Atkinson, Kendall E., and Han, Weimin, Elementary Numerical Analysis (3rd ed.) (Hoboken, NJ:

Wiley, 2004).

Shampine, L. F., Gladwell, I., and Thompson, S., Solving ODEs with MATLAB (New York: Cambridge

University Press, 2003).

August 7, 2012 21:05 c08 Sheet number 44 Page number 494 cyan black

August 7, 2012 21:05 c09 Sheet number 1 Page number 495 cyan black

495

CHAPTER

Nonlinear

Differential

Equations and

Stability

There are many differential equations, especially nonlinear ones, that are not sus-

ceptible to analytical solution in any reasonably convenient manner. Numerical

approximation methods, such as those discussed in the preceding chapter, provide

one means of dealing with these equations. Another approach, presented in this

chapter, is geometrical in character and leads to a qualitative understanding of the

behavior of solutions rather than to detailed quantitative information.

9.1 The Phase Plane: Linear Systems

Since many differential equations cannot be solved conveniently by analytical meth-

ods, it is important to consider what qualitative

information can be obtained about

their solutions without actually solving the equations. The questions that we consider

in this chapter are associated with the idea of stability of a solution, and the methods

that we employ are basically geometrical. Both the concept of stability and the use

The qualitative theory of differential equations was created by Henri Poincaré (1854–1912) in several

major papers between 1880 and 1886. Poincaré was a professor at the University of Paris and is generally

considered the leading mathematician of his time. He made fundamental discoveries in several differ-

ent areas of mathematics, including complex function theory, partial differential equations, and celestial

mechanics. In a series of papers beginning in 1894, he initiated the use of modern methods in topology. In

differential equations he was a pioneer in the use of asymptotic series, one of the most powerful tools of

contemporary applied mathematics. Among other things, he used asymptotic expansions to obtain solu-

tions about irregular singular points, thereby extending the work of Fuchs and Frobenius discussed in

Chapter 5.

August 7, 2012 21:05 c09 Sheet number 2 Page number 496 cyan black

496 Chapter 9. Nonlinear Differential Equations and Stability

of geometrical analysis were introduced in Chapter 1 and used in Section 2.5 for ﬁrst

order autonomous equations

dy/dt = f (y). (1)

In this chapter wereﬁne the ideasand extendthe discussionto autonomous systemsof

equations. We are particularly interested in nonlinear systems because they typically

cannot be solved in terms of elementary functions. Further, we consider primarily

systems of two equations because they lend themselves to geometrical analysis in a

plane, rather than in a higher-dimensional space.

However, before taking up nonlinear systems, we want to summarize some of the

results that we obtained in Chapter 7 for two-dimensional systems of ﬁrst order linear

homogeneous equations with constant coefﬁcients. Such a system has the form

d x/dt = Ax, (2)

where A is a 2 × 2 constant matrix and x is a 2 × 1 vector. In Sections 7.5 through 7.8,

we found that we could solve such systems by seeking solutions of the form x = ξe

By substituting for x in Eq. (2), we ﬁnd that

(A − rI)ξ = 0. (3)

Thus r must be an eigenvalue and ξ a corresponding eigenvector of the coefﬁcient

matrix A. The eigenvalues are the roots of the polynomial equation

det(A − rI) = 0, (4)

and the eigenvectors are determined from Eq. (3) up to an arbitrary multiplicative

constant. Although we presented graphs of solutions of equations of the form (2)

in Sections 7.5, 7.6, and 7.8, our main emphasis there was on ﬁnding a convenient

expression for the general solution. Our goal in this section is to bring together the

geometrical information for linear systems in one place. We use this information in

the rest of the chapter to investigate the much more difﬁcult nonlinear systems.

In Section 2.5 we found that points where the right side of Eq. (1) is zero are

of special importance. Such points correspond to constant solutions, or equilibrium

solutions, of Eq. (1) and are often called critical points. Similarly, for the system (2),

points where Ax = 0 correspond to equilibrium (constant) solutions, and again they

are called critical points. We will assume that A is nonsingular, or that detA ̸= 0. It

follows that x = 0 is the only critical point of the system (2).

Recall that a solution of Eq. (2) is a vector function x = φ(t) that satisﬁes the differ-

ential equation. Such a function can be viewed as a parametric representation for a

curve in the x

-plane. It is often useful to regard this curve as the path,or trajectory,

traversed by a moving particle whose velocity dx/dt is speciﬁed by the differential

equation. The x

-plane itself is called the phase plane, and a representative set of

trajectories is referred to as a phase portrait.

In analyzing the system (2), we must consider several different cases, depending

on the nature of the eigenvalues of A. We will characterize the differential equation

according to the geometric pattern formed by its trajectories. In each case we discuss

the behavior of the trajectories in general and illustrate it with an example. It is

important that you become familiar with the types of behavior that the trajectories

have for each case, because these are the basic ingredients of the qualitative theory

of differential equations.

August 7, 2012 21:05 c09 Sheet number 3 Page number 497 cyan black

9.1 The Phase Plane: Linear Systems 497

CASE 1 Real, Unequal Eigenvalues of the Same Sign. In this case, the general solution of

Eq. (2) is

x = c

(1)

+ c

(2)

, (5)

where r

and r

are either both positive or both negative. Suppose ﬁrst that

< r

< 0 and that the eigenvectors ξ

(1)

and ξ

(2)

are as shown in Figure 9.1.1a.

It follows from Eq. (5) that x → 0 as t →∞regardless of the values of c

and c

;in

other words, all solutions approach the critical point at the origin as t →∞. If the

solution starts at an initial point on the line through ξ

(1)

, then c

= 0. Consequently,

the solution remains on the line through ξ

(1)

for all t and approaches the origin as

t →∞. Similarly, if the initial point is on the line through ξ

(2)

, then the solution

approaches the origin along that line. In the general situation, it is helpful to rewrite

Eq. (5) in the form

x = e

(1)

−r

+ c

(2)

. (6)

Observe that r

− r

< 0. Therefore, as long as c

̸= 0, the term c

(1)

exp[(r

− r

)t]

is of negligible magnitude compared to that of c

(2)

for t sufﬁciently large. Thus,

as t →∞, the trajectory not only approaches the origin but also tends toward the

line through ξ

(2)

. Hence all solutions are tangent to ξ

(2)

at the critical point except

for those solutions that start exactly on the line through ξ

(1)

. Several trajectories are

sketched in Figure 9.1.1a. Typical graphs of x

versus t are shown in Figure 9.1.1b,

illustrating that all solutions exhibit exponential decay in time. The behavior of x

versus t is similar. This type of critical point is called a node or a nodal sink .

(a)

(2)

(1)

(b)

FIGURE 9.1.1 A node; r

< r

< 0. (a) The phase plane. (b) x

versus t.

Let us now look backward in time and inquire what happens as t →−∞. Still

assuming that r

< r

< 0, we note that if c

̸= 0, then the dominant term in Eq. (5)

as t →−∞is the term involving e

. Thus, except for the trajectories lying along the

line through ξ

(2)

, for large negative t the trajectories have slopes that are very nearly

that of the eigenvector ξ

(1)

. This is also indicated in Figure 9.1.1a.

If r

and r

are both positive, and 0 < r

< r

, then the trajectories have the same

pattern as in Figure 9.1.1a, but the direction of motion is away from, rather than

toward, the critical point at the origin. In this case x

and x

grow exponentially as

functions of t. Again the critical point is called a node or a nodal source.

An example of a node occurs in Example 3 of Section 7.5, and its trajectories are

shown in Figure 7.5.4.

August 7, 2012 21:05 c09 Sheet number 4 Page number 498 cyan black

498 Chapter 9. Nonlinear Differential Equations and Stability

CASE 2 Real Eigenvalues of Opposite Sign. The general solution of Eq. (2) is

x = c

(1)

+ c

(2)

, (7)

where r

> 0 and r

< 0. Suppose that the eigenvectors ξ

(1)

and ξ

(2)

are as shown

in Figure 9.1.2a. If the solution starts at an initial point on the line through ξ

(1)

then it follows that c

= 0. Consequently, the solution remains on the line through

(1)

for all t, and since r

> 0, ∥x∥→∞as t →∞. If the solution starts at an ini-

tial point on the line through ξ

(2)

, then it always remains on this line and ∥x∥→0

as t →∞because r

< 0. For solutions starting at other initial points the positive

exponential is the dominant term in Eq. (7) for large t, so eventually all these solu-

tions approach inﬁnity asymptotic to the line determined by the eigenvector ξ

(1)

corresponding to the positive eigenvalue r

. The only solutions that approach the

critical point at the origin are those that start precisely on the line determined by

(2)

. For large negative t, the dominant term in Eq. (7) is the negative exponen-

tial, so a typical solution is asymptotic to the line through the eigenvector ξ

(2)

t →−∞. The exceptions are those solutions that lie exactly on the line through the

eigenvector ξ

(1)

; these solutions approach the origin as t →−∞. The phase portrait

shown in Figure 9.1.2a is typical of the case in which the eigenvalues are real and of

opposite sign. The origin is called a saddle point in this case.

Figure 9.1.2b shows some typical graphs of x

versus t. For certain initial conditions

the positive exponential term is absent from the solution, so x

→ 0ast →∞.For

all other initial conditions the positive exponential term eventually dominates and

causes x

to be unbounded. The behavior of x

is similar.

An example of a saddle point occurs in Example 2 of Section 7.5,and its trajectories

are shown in Figure 7.5.2.

(a)

(2)

(1)

(b)

FIGURE 9.1.2 A saddle point; r

> 0, r

< 0. (a) The phase plane. (b) x

versus t.

CASE 3 Equal Eigenvalues. We now suppose that r

= r

= r. We consider the case in

which the eigenvalues are negative; if they are positive, the trajectories are similar

but the direction of motionis reversed.There are twosubcases,depending on whether

the repeated eigenvalue has two independent eigenvectors or only one.

(a) Two independent eigenvectors. The general solution of Eq. (2) is

x = c

(1)

+ c

(2)

, (8)

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9.1 The Phase Plane: Linear Systems 499

where ξ

(1)

and ξ

(2)

are the independent eigenvectors. The ratio x

is independent

of t but depends on the components of ξ

(1)

and ξ

(2)

and on the arbitrary constants

and c

. Thus every trajectory lies on a straight line through the origin, as shown

in Figure 9.1.3a. Typical graphs of x

or x

versus t are shown in Figure 9.1.3b.The

critical point is called a proper node, or sometimes a star point.

(a)

(2)

(1)

(b)

FIGURE 9.1.3 A proper node, two independent eigenvectors; r

= r

< 0.

(a) The phase plane. (b) x

versus t.

(b) One independent eigenvector. As shown in Section 7.8, the general solution

of Eq. (2) in this case is

x = c

ξe

+ c

(ξte

+ ηe

), (9)

where ξ is the eigenvector and η is the generalized eigenvector associated with the

repeated eigenvalue. For large t the dominant term in Eq. (9) is c

ξte

.Thus,as

t →∞, every trajectory approaches the origin tangent to the line through the eigen-

vector. This is true even if c

= 0, for then the solution x = c

ξe

lies on this line.

Similarly, for large negative t the term c

ξte

is again the dominant one, so as

t →−∞, the slope of each trajectory approaches the slope of the eigenvector ξ.

The orientation of the trajectories depends on the relative positions of ξ and η.

One possible situation is shown in Figure 9.1.4a.To locate the trajectories,it is helpful

to write the solution (9) in the form

x =[(c

ξ + c

η) + c

ξt]e

= ye

, (10)

where y = (c

ξ + c

η) + c

ξt. Observe that the vector y determines the direction of

x, whereas the scalar quantity e

affects only the magnitude of x. Also note that for

ﬁxed values of c

and c

, the expression for y is a vector equation of the straight line

through the point c

ξ + c

η and parallel to ξ.

To sketch the trajectory corresponding to a given pair of values of c

and c

,you can

proceed in the following way. First,draw the line given by (c

ξ + c

η) + c

ξt and note

the direction of increasing t on this line.Two such lines are shown in Figure 9.1.4a,one

for c

> 0 and the other for c

< 0. Next,note that the given trajectory passes through

the point c

ξ + c

η when t = 0. Further, as t increases, the direction of the vector x

given by Eq. (10) follows the direction of increasing t on the line,but the magnitude of

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500 Chapter 9. Nonlinear Differential Equations and Stability

x rapidly decreases and approaches zero because of the decaying exponential factor

. Finally, as t decreases toward −∞, the direction of x is determined by points on

the corresponding part of the line, and the magnitude of x approaches inﬁnity. In

this way we obtain the black trajectories in Figure 9.1.4a. A few other trajectories

are sketched as well, to help complete the diagram. Typical graphs of x

versus t are

shown in Figure 9.1.4b.

The other possible situationis shown in Figure 9.1.4c,where therelative orientation

of ξ and η is reversed. As indicated in the ﬁgure, this results in a reversal in the

orientation of the trajectories.

(a)

Increasing t

< 0

> 0

ξ + c

(b)

(c)

Increasing t

< 0

> 0

ξ + c

FIGURE 9.1.4 An improper node, one independent eigenvector; r

= r

< 0.

(a) The phase plane. (b) x

versus t.(c) The phase plane for the same ξ and a different η.

If r

= r

> 0, you can sketch the trajectories by following the same procedure. In

this event the trajectories are traversed in the outward direction, and the orientation

of the trajectories with respect to that of ξ and η is also reversed.

When a double eigenvalue has only a single independent eigenvector, the critical

point is called an improper or degenerate node. A speciﬁc example of this case is

Example 2 in Section 7.8; the trajectories are shown in Figure 7.8.2.

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9.1 The Phase Plane: Linear Systems 501

CASE 4 Complex Eigenvalues with Nonzero Real Part. Suppose that the eigenvalues are

λ ± iµ, where λ and µ are real, λ ̸= 0, and µ>0. In Section 7.6 it was shown that it is

possible to write the general solution in terms of the eigenvalues and eigenvectors.

However, we proceed in a different way.

Let us consider the system

′

λµ

−µλ

(

x, (11)

whose scalar form is

′

= λx

+ µx

, x

′

=−µx

+ λx

. (12)

It is sufﬁcient to examine the system (11) since every 2 × 2 system with eigenvalues

λ ± iµ can be converted into the form (11) by a linear transformation (an example

of how this can be done is found in Problem 22). We introduce the polar coordinates

r, θ given by

= x

+ x

, tan θ = x

By differentiating these equations, we obtain

′

= x

′

+ x

′

, (sec

θ)θ

′

= (x

′

− x

′

)/x

. (13)

Substituting from Eqs. (12) in the ﬁrst of Eqs. (13), we ﬁnd that

′

= λr, (14)

and hence

r = ce

λt

, (15)

where c is a constant. Similarly, substituting from Eqs. (12) in the second of Eqs. (13),

and using the fact that sec

θ = r

, we have

′

=−µ. (16)

Hence

θ =−µt + θ

, (17)

where θ

is the value of θ when t = 0.

Equations (15) and (17) are parametric equations in polar coordinates of the tra-

jectories of the system (11). Since µ>0, it follows from Eq. (17) that θ decreases

as t increases, so the direction of motion on a trajectory is clockwise. As t →∞,we

see from Eq. (15) that r → 0ifλ<0 and r →∞if λ>0. Thus the trajectories are

spirals, which approach or recede from the origin, depending on the sign of λ. Both

possibilities are shown in Figure 9.1.5, along with some typical graphs of x

versus t.

The critical point is called a spiral point in this case. Frequently, the terms spiral sink

and spiral source, respectively, are used to refer to spiral points whose trajectories

approach, or depart from, the critical point.

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502 Chapter 9. Nonlinear Differential Equations and Stability

(a)

(b)

(c)

(d)

FIGURE 9.1.5 A spiral point; r

= λ + iµ, r

= λ − iµ.(a) λ<0, the phase plane.

(b) λ<0, x

versus t.(c) λ>0, the phase plane. (d) λ>0, x

versus t.

More generally, the trajectories are spirals for any system with complex eigenval-

ues λ ± iµ, where λ ̸= 0. The spirals are traversed in the inward or outward direction,

depending on whether λ is negative or positive. They may be elongated and skewed

with respect to the coordinate axes, and the direction of motion may be either clock-

wise or counterclockwise. Further,it is easy to obtain a general idea of the orientation

of the trajectories directly from the differential equations. Suppose that

dx/dt

dy/dt

(

(18)

has complex eigenvalues λ ± iµ, and look at the point (0, 1) on the positive y-axis. At

this point it follows from Eqs. (18) that dx/dt = b and dy/dt = d. Depending on the

signs of b and d,we can infer the direction of motion and the approximate orientation

of thetrajectories. For instance,if both b and d are negative,then the trajectories cross

the positive y-axis soas tomove downand into the secondquadrant. Ifλ<0 also,then

the trajectories must be inward-pointing spirals resembling the one in Figure 9.1.6.

Another case was given in Example 1 of Section 7.6, whose trajectories are shown in

Figure 7.6.2.

CASE 5 Pure Imaginary Eigenvalues. In this case λ = 0, and the system (11) reduces to

′

0 µ

−µ 0

(

x (19)

with eigenvalues ±iµ. Using the same argument as in Case 4, we ﬁnd that

′

= 0, θ

′

=−µ, (20)

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9.1 The Phase Plane: Linear Systems 503

FIGURE 9.1.6 A spiral point; r = λ ± iµ with λ<0.

and consequently,

r = c, θ =−µt + θ

, (21)

where c and θ

are constants.Thus the trajectories are circles,with center at the origin,

that are traversed clockwise if µ>0 and counterclockwise if µ<0. A complete

circuit about the origin is made in a time interval of length 2π/µ, so all solutions are

periodic with period 2π/µ. The critical point is called a center.

In general, when the eigenvalues are pure imaginary, it is possible to show (see

Problem 19) that the trajectories are ellipses centered atthe origin.A typical situation

is shown in Figure 9.1.7, which also includes some typical graphs of x

versus t. See

also Example 3 in Section 7.6, especially Figures 7.6.3 and 7.6.4.

(a)

(b)

FIGURE 9.1.7 A center; r

= iµ, r

=−iµ.(a) The phase plane. (b) x

versus t.

By reﬂecting on these ﬁve cases and by examining the corresponding ﬁgures, we

can make several observations:

1. After a long time,each individual trajectoryexhibits one ofonly three typesof behavior.As

t →∞, each trajectory approaches the critical point x = 0, repeatedly traverses a closed

curve (corresponding to a periodic solution) that surrounds the critical point, or becomes

unbounded.

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504 Chapter 9. Nonlinear Differential Equations and Stability

2. Viewed as a whole, the pattern of trajectories in each case is relatively simple. To be

more speciﬁc, through each point (x

, y

) in the phase plane there is only one trajectory;

thus the trajectories do not cross each other. Do not be misled by the ﬁgures, in which

it sometimes appears that many trajectories pass through the critical point x = 0. In fact,

the only solution passing through the origin is the equilibrium solution x = 0. The other

solutions that appear to pass through the origin actually only approach this point as t →∞

or t →−∞.

3. In each case, the set of all trajectories is such that one of three situations occurs.

(a) All trajectories approach the critical point x = 0 as t →∞. This is the case if the

eigenvalues are real and negative or complex with negative real part. The origin is

either a nodal sink or a spiral sink.

(b) All trajectories remain bounded but do not approach the origin as t →∞. This is the

case if the eigenvalues are pure imaginary. The origin is a center.

t →∞.Thisisthe case ifat least one of the eigenvalues is positiveor if theeigenvalues

have positive real part.The origin is a nodal source,a spiral source,or a saddle point.

The situations described in 3(a),(b),and (c) above illustrate the concepts of asymp-

totic stability, stability, and instability, respectively, of the equilibrium solution x = 0

of the system (2). The precise deﬁnitions of these terms are given in Section 9.2,

but their basic meaning should be clear from the geometrical discussion in this sec-

tion. The information that we have obtained about the system (2) is summarized in

Table 9.1.1. Also see Problems 20 and 21.

TABLE 9.1.1 Stability Properties of Linear Systems x

′

= Ax with

det(A − rI) = 0 and detA ̸= 0

Eigenvalues Type of Critical Point Stability

> r

> 0 Node Unstable

< r

< 0 Node Asymptotically stable

< 0 < r

Saddle point Unstable

= r

> 0 Proper or improper node Unstable

= r

< 0 Proper or improper node Asymptotically stable

, r

= λ ± iµ Spiral point

λ>0 Unstable

λ<0 Asymptotically stable

= iµ, r

=−iµ Center Stable

The analysis in this section applies only to second order systems x

′

= Ax whose

solutions can be visualized as curves in the phase plane. A similar, though more

complicated, analysis can be carried out for an nth order system, with an n × n

coefﬁcient matrix A, whose solutions are represented geometrically by curves in

an n-dimensional phase space. The cases that can occur in higher order systems are

essentially combinations of those we have seen in two dimensions. For instance, in

a third order system with a three-dimensional phase space, one possibility is that

solutions in a certain plane may spiral toward the origin, while other solutions may

tend to inﬁnity along a line transverse to this plane. This will be the case if the coef-

ﬁcient matrix has two complex eigenvalues with negative real part and one positive

real eigenvalue. However, because of their complexity, we will not discuss systems of

higher than second order.

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9.1 The Phase Plane: Linear Systems 505

PROBLEMS For each of the systems in Problems 1 through 12:

(a) Find the eigenvalues and eigenvectors.

(b) Classify the critical point (0, 0) as to type,and determinewhether itis stable,asymptotically

stable, or unstable.

versus t.

(d) Use a computer to plot accurately the curves requested in part (c).

3 −2

2 −2

x 2.

5 −1

2 −1

3 −2

x 4.

1 −4

4 −7

1 −5

1 −3

x 6.

2 −5

1 −2

3 −2

4 −1

x 8.

−1 −1

0 −0.25

3 −4

1 −1

x 10.

−5 −1

11.

−10

0 −1

x 12.

2 −

−1

In each of Problems 13 through 16, determine the critical point x = x

, and then classify its

type and examine its stability by making the transformation x = x

+ u.

13.

1 −1

x −

14.

−21

1 −2

x +

−2

15.

−1 −1

2 −1

x +

−1

16.

0 −β

δ 0

x +

−γ

; α, β, γ, δ>0

17. The equation of motion of a spring–mass system with damping (see Section 3.7) is

+ c

+ ku = 0,

where m, c, and k are positive. Write this second order equation as a system of two ﬁrst

orderequations for x = u, y = du/dt.Show thatx = 0, y = 0 is a criticalpoint,andanalyze

the nature and stability of the critical point as a function of the parameters m, c, and k.A

similar analysis can be applied to the electric circuit equation (see Section 3.7)

+ R

I = 0.

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506 Chapter 9. Nonlinear Differential Equations and Stability

18. Consider the system x

′

= Ax, and suppose that A has one zero eigenvalue.

(a) Show that det(A) = 0.

(b) Show that x = 0 is a critical point and that,in addition,every point on a certain straight

line through the origin is also a critical point.

= 0 and r

̸= 0, and let ξ

(1)

and ξ

(2)

be corresponding eigenvectors. Show that

the trajectories are as indicated in Figure 9.1.8. What is the direction of motion on the

trajectories?

(2)

(1)

FIGURE 9.1.8 Nonisolated critical points; r

= 0, r

̸= 0. Every point

on the line through ξ

(1)

is a critical point.

19. In this problem we indicate how to show that the trajectories are ellipses when the

eigenvalues are pure imaginary. Consider the system

′

. (i)

(a) Show that the eigenvalues of the coefﬁcient matrix are pure imaginary if and only if

+ a

= 0, a

− a

> 0. (ii)

(b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single

equation

dy/dt

dx/dt

x + a

. (iii)

Use the ﬁrst of Eqs. (ii) to show that Eq. (iii) is exact.

+ 2a

xy − a

= k, (iv)

where k is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an

ellipse.

Hint: What is the discriminant of the quadratic form in Eq. (iv)?

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9.1 The Phase Plane: Linear Systems 507

20. Consider the linear system

dx/dt = a

x + a

y, dy/dt = a

x + a

where a

, a

, and a

are real constants. Let p = a

+ a

, q = a

− a

, and

5 = p

− 4q. Observe that p and q are the trace and determinant, respectively, of the

coefﬁcient matrix of the given system. Show that the critical point (0, 0) is a

(a) Node if q > 0 and 5 ≥ 0; (b) Saddle point if q < 0;

Hint: These conclusions can be reached by studying the eigenvalues r

and r

. It may also

be helpful to establish, and then to use, the relations r

= q and r

+ r

= p.

21. Continuing Problem 20, show that the critical point (0, 0) is

(a) Asymptotically stable if q > 0 and p < 0;

(b) Stable if q > 0 and p = 0;

The results of Problems 20 and 21 are summarized visually in Figure 9.1.9.

Asymp. stable, spiral point

Unstable, spiral point

Unstable

node

Asymp. stable

node

∆ = p

– 4q > 0

∆ = p

– 4q = 0

Stable

center

Unstable, saddle point

∆ = p

– 4q < 0

FIGURE 9.1.9 Stability diagram.

22. In this problem we illustrate how a 2 × 2 system with eigenvalues λ ± iµ can be

transformed into the system (11). Consider the system in Problem 12:

′

2 −2.5

1.8 −1

x = Ax . (i)

(a) Show that the eigenvalues of this system are r

= 0.5 + 1.5i and r

= 0.5 − 1.5i.

(b) Show that the eigenvector corresponding to r

can be chosen as

(1)

3 − 3i

+ i

−3

. (ii)

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508 Chapter 9. Nonlinear Differential Equations and Stability

(1)

. Thus

P =

3 −3

. (iii)

Let x = Py and substitute for x in Eq. (i). Show that

′

= (P

−1

AP)y. (iv)

(d) Find P

−1

and show that

−1

AP =

0.51.5

−1.50.5

. (v)

Thus Eq. (v) has the form of Eq. (11).

9.2 Autonomous Systems and Stability

In this section we begin to draw together, and to expand on, the geometrical ideas

introduced in Section 2.5 for certain ﬁrst order equations and in Section 9.1 for

systems of two ﬁrst order linear homogeneous equations with constant coefﬁcients.

These ideas concern the qualitative study of differential equations and the concept

of stability, an idea that will be deﬁned precisely later in this section.

Autonomous Systems. We are concerned with systems of two simultaneous differential

equations of the form

dx/dt = F(x, y), dy/dt = G(x, y). (1)

We assume that the functions F and G are continuous and have continuous partial

derivatives in some domain D of the xy-plane. If (x

, y

) is a point in this domain,

then by Theorem 7.1.1 there exists a unique solution x = φ(t), y = ψ(t) of the system

(1) satisfying the initial conditions

x(t

) = x

, y(t

) = y

. (2)

The solution is deﬁned in some time interval I that contains the point t

Frequently, we will write the initial value problem (1), (2) in the vector form

d x/dt = f(x), x(t

) = x

, (3)

where x = (x, y)

, f(x) = (F(x, y), G(x, y))

, and x

= (x

, y

)

. In this case the solu-

tion is expressed as x = φ(t), where φ(t) = (φ(t), ψ(t))

. As usual, we interpret a

solution x = φ(t) as a curve traced by a moving point in the xy-plane, the phase

plane.

Observe that the functions F and G in Eqs. (1) do not depend on the independent

variable t, but only on the dependent variables x and y. A system with this property

is said to be autonomous. The system

′

= Ax, (4)

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9.2 Autonomous Systems and Stability 509

where A is a constant 2 × 2 matrix, is a simple example of a two-dimensional

autonomous system. On the other hand, if one or more of the elements of the

coefﬁcient matrix A is a function of the independent variable t, then the system

is nonautonomous. The distinction between autonomous and nonautonomous sys-

tems is important, because the geometrical qualitative analysis in Section 9.1 can be

effectively extended to two-dimensional autonomous systems in general but is not

nearly as useful for nonautonomous systems.

In particular, the autonomous system (1) has an associated direction ﬁeld that is

independent of time. Consequently,there is only one trajectory passing through each

point (x

, y

) in the phase plane. In other words, all solutions that satisfy an initial

condition of the form (2) lie on the same trajectory, regardless of the time t

at which

they pass through (x

, y

). Thus, just as for the constant coefﬁcient linear system (4),

a single phase portrait simultaneously displays important qualitative information

about all solutions of the system (1). We will see this fact conﬁrmed repeatedly in

this chapter.

Autonomous systems occur frequently in applications. Physically, an autonomous

system is one whose conﬁguration,including physical parameters and external forces

oreffects,isindependentof time.Theresponse of thesystem to giveninitial conditions

is then independent of the time at which the conditions are imposed.

Stability and Instability. The concepts of stability, asymptotic stability, and instability

have already been mentioned several times in this book. It is now time to give a

precise mathematical deﬁnition of these concepts, at least for autonomous systems

of the form

′

= f(x). (5)

In the following deﬁnitions, and elsewhere, we use the notation ∥x∥ to designate the

length, or magnitude, of the vector x.

The points, if any, where f(x) = 0 are called critical points of the autonomous

system (5). At such points x

′

= 0 also, so critical points correspond to constant, or

equilibrium, solutions of the system of differential equations. A critical point x

the system (5) is said to be stable if, given any ϵ>0, there is a δ>0 such that every

solution x = φ(t) of the system (1), which at t = 0 satisﬁes

∥φ(0 ) − x

∥ <δ, (6)

exists for all positive t and satisﬁes

∥φ(t) − x

∥ <ϵ (7)

for all t ≥ 0. This is illustrated geometrically in Figures 9.2.1a and 9.2.1b. These

mathematical statements say that all solutions that start “sufﬁciently close” (that

is, within the distance δ)tox

stay “close” (within the distance ϵ)tox

. Note that in

Figure 9.2.1a the trajectory is within the circle ∥x − x

∥=δ at t = 0 and, although

it soon passes outside of this circle, it remains within the circle ∥x − x

∥=ϵ for all

t ≥ 0. However, the trajectory of the solution does not have to approach the critical

point x

as t →∞; it only has to remain within the circle of radius ϵ, as illustrated in

Figure 9.2.1b. A critical point that is not stable is said to be unstable.

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510 Chapter 9. Nonlinear Differential Equations and Stability

φψ

( (0), (0))

(a)

(b)

φψ

( (0), (0))

FIGURE 9.2.1 (a) Asymptotic stability. (b) Stability.

A critical point x

is said to be asymptotically stable if it is stable and if there exists

a δ

(δ

> 0) such that if a solution x = φ(t) satisﬁes

∥φ(0 ) − x

∥ <δ

, (8)

then

lim

t→∞

φ(t) = x

. (9)

Thus trajectories that start “sufﬁciently close” to x

not only must stay “close” but

must eventually approach x

as t →∞. This is the case for the trajectory in Figure

9.2.1a,but not for the one in Figure 9.2.1b. Note that asymptotic stability is a stronger

property than stability, since a critical point must be stable before we can even con-

sider whether it might be asymptotically stable. On the other hand,the limit condition

(9), which is an essential feature of asymptotic stability, does not by itself imply even

ordinary stability. Indeed, examples can be constructed in which all the trajectories

approach x

as t →∞, but for which x

is not a stable critical point. Geometrically,

all that is needed is a family of trajectories having members that start arbitrarily close

to x

and then depart an arbitrarily large distance before eventually approaching x

as t →∞.

In this chapterwe areconcentrating on systemsof twoequations,but the deﬁnitions

just given are independent of the size of the system. If you interpret the vectors in

Eqs. (5) through (9) as n-dimensional, then the deﬁnitions of stability, asymptotic

stability, and instability apply also to systems of n equations. The concepts expressed

in these deﬁnitions can be seen more clearly byinterpreting themin termsof a speciﬁc

physical problem.

The Oscillating Pendulum. The concepts of asymptotic stability, stability, and instability

can be easily visualized in terms of an oscillating pendulum. Consider the conﬁgura-

tion shown in Figure 9.2.2, in which a mass m is attached to one end of a rigid, but

weightless, rod of length L. The other end of the rod is supported at the origin O,

and the rod is free to rotate in the plane of the paper. The position of the pendu-

lum is described by the angle θ between the rod and the downward vertical direction,

with the counterclockwise direction taken as positive. The gravitational force mg acts

downward,while the damping force c|dθ/dt|,where c is positive, is always opposite to

the direction of motion. We assume that both θ and dθ/dt are positive. The equation

of motion can be quickly derived from the principle of angular momentum, which

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9.2 Autonomous Systems and Stability 511

states that the time rate of change of angular momentum about any point is equal

to the moment of the resultant force about that point.The angular momentum about

the origin is mL

(d θ/dt), so the governing equation is

=−cL

d θ

− mgL sin θ. (10)

The factors L and L sin θ on the right side of Eq. (10) are the moment arms of

the resistive force and of the gravitational force,respectively; the minus signs are due

to the fact that the two forces tend to make the pendulum rotate in the clockwise

(negative) direction. You should verify, as an exercise, that the same equation is

obtained for the other three possible sign combinations of θ and dθ/dt.

L sin

L(1 – cos )

d /dt

FIGURE 9.2.2 An oscillating pendulum.

By straightforward algebraic operations,we can write Eq. (10) in the standard form

d θ

sin θ = 0, (11)

+ γ

d θ

+ ω

sin θ = 0, (12)

where γ = c/mL and ω

= g/L. To convert Eq. (12) to a system of two ﬁrst order

equations, we let x = θ and y = dθ/dt; then

= y,

=−ω

sin x − γy. (13)

Since γ and ω

are constants, the system (13) is an autonomous system of the

form (1).

The critical points of Eqs. (13) are found by solving the equations

y = 0, −ω

sin x − γy = 0.

We obtain y = 0 and x =±nπ, where n is an integer. These points correspond to two

physical equilibrium positions, one with the mass directly below the point of support

(θ = 0) and the other with the mass directly above the point of support (θ = π). Our

intuition suggests that the ﬁrst is stable and the second is unstable.

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512 Chapter 9. Nonlinear Differential Equations and Stability

More precisely, if the mass is slightly displaced from the lower equilibrium posi-

tion, it will oscillate back and forth with gradually decreasing amplitude, eventually

converging to the equilibrium position as the initial potential energy is dissipated by

the damping force. This type of motion illustrates asymptotic stability and is shown

in Figure 9.2.3a.

(a)

(b)

(c)

FIGURE 9.2.3 Qualitative motion of a pendulum. (a) With air resistance.

(b) With or without air resistance. (c) Without air resistance.

On the other hand, if the mass is slightly displaced from the upper equilib-

rium position, it will rapidly fall, under the inﬂuence of gravity, and will ultimately

converge to the lower equilibrium position in this case also. This type of motion

illustrates instability. See Figure 9.2.3b. In practice, it is impossible to maintain the

pendulum in its upward equilibrium position for any extended length of time with-

out an external constraint mechanism, since the slightest perturbation will cause the

mass to fall.

Finally, consider the ideal situation in which the damping coefﬁcient c (or γ) is

zero. In this case, if the mass is displaced slightly from its lower equilibrium position,

it will oscillate indeﬁnitely with constant amplitude about the equilibrium position.

Since there is no dissipation in the system, the mass will remain near the equilib-

rium position but will not approach it asymptotically. This type of motion is stable

but not asymptotically stable, as indicated in Figure 9.2.3c. In general, this motion is

impossible to achieve experimentally, because the slightest degree of air resistance

or friction at the point of support will eventually cause the pendulum to converge to

its rest position.

Solutions of the pendulum equations are discussed in more detail in the next

section.

The Importance of Critical Points. Critical points correspond to equilibrium solutions—

that is, solutions for which x(t) and y(t) are constant. For such a solution, the system

described by x and y is not changing; it remains in its initial state forever. It might

seem reasonable to conclude that such points are not very interesting. Recall, how-

ever, that for linear homogeneous systems with constant coefﬁcients, x

′

= Ax, the

nature of the critical point at the origin determines to a large extent the behavior of

trajectories throughout the xy-plane.

For nonlinear autonomous systems this is no longer true, for at least two reasons.

First, there may be several, or many, critical points that are competing, so to speak,

for inﬂuence on the trajectories. Second, the nonlinearities in the system are also of

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9.2 Autonomous Systems and Stability 513

great importance, especially far away from the critical points. Nevertheless, critical

points of nonlinear autonomous systems can be classiﬁed just as for linear systems.

We will discuss this in detail in Section 9.3. Here we illustrate how it can be done

graphically, assuming that you have software that can construct direction ﬁelds and

perhaps plot good numerical approximations to a few trajectories.

EXAMPLE

Consider the system

dx/dt =−(x − y)(1 − x − y), dy/dt = x(2 + y). (14)

Find the critical points for this system, and draw direction ﬁelds on rectangles containing the

critical points. By inspecting the direction ﬁelds, classify each critical point as to type,and state

whether it is asymptotically stable, stable, or unstable.

The critical points are found by solving the algebraic equations

(x − y)(1 − x − y) = 0, x(2 + y) = 0. (15)

One way to satisfy the second equation is by choosing x = 0. Then the ﬁrst equation becomes

y(1 − y) = 0, so y = 0ory = 1. More solutions can be found by choosing y =−2 in the second

equation. Then the ﬁrst equation becomes (x + 2)(3 − x) = 0, so x =−2 or x = 3. Thus we

have obtained the four critical points (0, 0), (0, 1), (−2, −2),

and (

3, −2).

Figure 9.2.4 shows a direction ﬁeld containing the ﬁrst two of the critical points. Comparing

this ﬁgure with those in Section 9.1 and in Chapter 7 should make it clear that the origin is

a saddle point and that (0, 1) is a spiral point. Of course, the saddle point is unstable. The

trajectories near the spiral point appear to be approaching this point, so we conclude that it is

asymptotically stable. A direction ﬁeld for the other two critical points is shown in Figure 9.2.5.

Each of these points is a node. The arrows point toward the point (−2, −2) and away from the

point (3, −2); thus the former is asymptotically stable and the latter is unstable.

10.5–0.5–1

0.5

–0.5

–1

1.5

FIGURE 9.2.4 Direction ﬁeld for the

system (14) containing the critical

points (0, 0) and (0, 1); the former is a

saddle point and the latter is a spiral point.

43213–2–1–

–1.6

–1.8

–2

–2.2

–2.4

FIGURE 9.2.5 Direction ﬁeld for the

system (14) containing the critical points

(−2, −2) and (3, −2); both of these points

are nodes.

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514 Chapter 9. Nonlinear Differential Equations and Stability

For a two-dimensional autonomous system with at least one asymptotically stable

critical point, it is often of interest to determine where in the phase plane the tra-

jectories lie that ultimately approach a given critical point. Let P be a point in the

xy-plane with the property that a trajectory passing through P ultimately approaches

the critical point as t →∞. Then this trajectory is said to be attracted by the critical

point. Further, the set of all such points P is called the basin of attraction or the

region of asymptotic stability of the critical point. A trajectory that bounds a basin

of attraction is called a separatrix because it separates trajectories that approach

a particular critical point from other trajectories that do not do so. Determination

of basins of attraction is important in understanding the large-scale behavior of the

solutions of an autonomous system.

EXAMPLE

Consider again the system (14) from Example 1. Describe the basin of attraction for each of

the asymptotically stable critical points.

Figure 9.2.6 shows a phase portrait for this system with a direction ﬁeld in the background.

Observe that there are two trajectories that approach the saddle point at the origin as t →∞,

One of these lies in the fourth quadrant and is almost a straight line from the unstable node

at (3, −2). The other also originates at the unstable node, then passes into the ﬁrst quadrant,

loops around the spiral point, and ultimately approaches the saddle point through the second

quadrant. These two trajectories are separatrices; the region between them (but not including

the separatrices themselves) is the basin of attraction for the spiral point at (0, 1). This region

is shaded in Figure 9.2.6.

–2

42–2–4

FIGURE 9.2.6 Trajectories of the system (14). The basis of attraction

for the spiral point (0, 1) is shaded.

The basin of attraction for the asymptotically stable node at (−2, −2) consists of the rest of

the xy-plane,with only a handful of exceptions. The separatrices approach the saddle point, as

we have noted already, rather than the node.The saddle point itself and the unstable node are

equilibrium solutions and thus remain ﬁxed for all time. Finally, there is a trajectory lying on

the line y =−2 for x > 3 on which the direction of motion is always to the right;this trajectory

also does not approach the point (−2, −2).

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9.2 Autonomous Systems and Stability 515

Figures 9.2.4, 9.2.5, and 9.2.6 show that in the immediate vicinity of a critical point,

the direction ﬁeld and pattern of trajectories resemble those for a linear system with

constant coefﬁcients. This becomes even more unmistakable if you use your software

to zoom in closer and closer to a critical point. Thus we have visual evidence that a

nonlinear system behaves very much like a linear system,at least in the neighborhood

of a critical point. We will pursue this idea in the next section.

Determination of Trajectories. The trajectoriesof a two-dimensional autonomous system

dx/dt = F(x, y), dy/dt = G(x, y) (16)

can sometimes be found by solving a related ﬁrst order differential equation. From

Eqs. (16) we have

dy/dt

dx/dt

G(x, y)

F(x, y)

, (17)

which is a ﬁrst order equation in the variables x and y. Observe that such a reduction

is not usually possible if F and G depend also on t. If Eq. (17) can be solved by any

of the methods of Chapter 2, and if we write solutions (implicitly) in the form

H(x, y) = c, (18)

then Eq. (18) is an equation for the trajectories of the system (16). In other words, the

trajectories lie on the level curves of H(x, y). Keep in mind that there is no general

way of solving Eq. (17) to obtain the function H, so this approach is applicable only

in special cases.

EXAMPLE

Find the trajectories of the system

dx/dt = y, dy/dt = x. (19)

In this case, Eq. (17) becomes

. (20)

This equation is separable since it can be written as

ydy = xdx,

and its solutions are given by

H(x, y) = y

− x

= c, (21)

where c is arbitrary. Therefore, the trajectories of the system (19) are the hyperbolas shown

in Figure 9.2.7. The direction of motion on the trajectories can be inferred from the fact that

both dx/dt and dy dt are positive in the ﬁrst quadrant. The only critical point is the saddle

point at the origin.

Another way to obtain the trajectories is to solve the system (19) by the methods of

Section 7.5. We omit the details, but the result is

x = c

+ c

−t

, y = c

− c

−t

Eliminating t between these two equations again leads to Eq. (21).

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516 Chapter 9. Nonlinear Differential Equations and Stability

–2 –1 120

–1

–2

FIGURE 9.2.7 Trajectories of the system (19); the origin is a saddle point.

EXAMPLE

Find the trajectories of the system

= 4 − 2y,

= 12 − 3x

. (22)

From the equations

4 − 2y = 0, 12 − 3x

= 0

we ﬁnd that the critical points of the system (22) are the points (−2, 2) and (2, 2).To

determine the trajectories, note that for this system, Eq. (17) becomes

12 − 3x

4 − 2y

. (23)

Separating the variables in Eq. (23) and integrating, we ﬁnd that solutions satisfy

H(x, y) = 4y − y

− 12x + x

= c, (24)

where c is an arbitrary constant. A computer plotting routine is helpful in displaying the

level curves of H(x, y), some of which are shown in Figure 9.2.8. The direction of motion

on the trajectories can be determined by drawing a direction ﬁeld for the system (22), or

by evaluating dx/dt and dy/dt at one or two selected points. From Figure 9.2.8 you can

see that the critical point (2, 2) is a saddle point and the point (−2, 2) is a center. Observe

that there is a separatrix (shown in black) that leaves the saddle point (as t →−∞), loops

around the center, and returns to the saddle point (as t →+∞). Within the separatrix are

closed trajectories, or periodic solutions, that surround the center. Outside the separatrix,

trajectories become unbounded, except for the trajectory that enters the saddle point from

the right.

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9.2 Autonomous Systems and Stability 517

2–2–4 4

–2

–4

FIGURE 9.2.8 Trajectories of the system (22). The point (−2, 2) is a center,

and the point (2, 2) is a saddle point. The black curve is a separatrix.

PROBLEMS In each of Problems 1 through 4, sketch the trajectory corresponding to the solution satisfying

the speciﬁed initial conditions, and indicate the direction of motion for increasing t.

1. dx/dt =−x, dy/dt =−2y; x(0) = 4, y(0) = 2

2. dx/dt =−x, dy/dt = 2y; x(0) = 4, y(0) = 2 and x(0) = 4, y(0) = 0

3. dx/dt =−y, dy/dt = x; x(0) = 4, y(0) = 0 and x(0) = 0, y(0) = 4

4. dx/dt = ay, dy/dt =−bx, a > 0, b > 0; x(0) =

√

a, y(0) = 0

For each of the systems in Problems 5 through 16:

(a) Find all the critical points (equilibrium solutions).

(b) Use a computer to draw a direction ﬁeld and phase portrait for the system.

stable, or unstable, and classify it as to type.

(d) Describe the basin of attraction for each asymptotically stable critical point.

dx/dt = x − xy, dy/dt = y + 2xy

dx/dt = 1 + 2 y, dy/dt = 1 − 3x

7. dx/dt = 2x − x

− xy, dy/dt = 3y − 2y

− 3xy

dx/dt =−(2 + y)(x + y), dy/dt =−y(1 − x)

dx/dt = y(2 − x − y), dy/dt =−x − y − 2xy

10.

dx/dt = (2 + x)(y − x), dy/dt = y(2 + x − x

)

11.

dx/dt =−x + 2xy, dy/dt = y − x

− y

12. dx/dt = y, dy/dt = x −

−

13.

dx/dt = (2 + x)(y − x), dy/dt = (4 − x)(y + x)

14.

dx/dt = (2 − x)(y − x), dy/dt = y(2 − x − x

)

15.

dx/dt = x(2 − x − y), dy/dt =−x + 3y − 2xy

16.

dx/dt = x(2 − x − y), dy/dt = (1 − y)(2 + x)

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518 Chapter 9. Nonlinear Differential Equations and Stability

In each of Problems 17 through 24:

(a) Find an equation of the form H(x, y) = c satisﬁed by the trajectories.

(b) Plot several level curves of the function H. These are trajectories of the given system.

Indicate the direction of motion on each trajectory.

17.

dx/dt = 2y, dy/dt = 8x

18.

dx/dt = 2y, dy/dt =−8x

19.

dx/dt = y, dy/dt = 2x + y

20.

dx/dt =−x + y, dy/dt =−x − y

21.

dx/dt =−x + y + x

, dy/dt = y − 2xy

22.

dx/dt = 2x

y − 3x

− 4y, dy/dt =−2xy

+ 6xy

23.

Undamped pendulum: dx/dt = y, dy/dt =−sin x

24.

Dufﬁng’s

equations: dx/dt = y, dy/dt =−x + (x

/6)

25. Given that x = φ(t), y = ψ(t) is a solution of the autonomous system

dx/dt = F (x, y), dy/dt = G(x, y)

for α<t <β, show that x = 7(t) = φ(t − s), y = 8(t) = ψ(t − s) is a solution for

α + s < t <β+s for any real number s.

26. Prove that for the system

dx/dt = F (x, y), dy/dt = G(x, y)

there is at most one trajectory passing through a given point (x

, y

Hint: Let C

be the trajectory generated by the solution x = φ

(t), y = ψ

(t), with

) = x

, ψ

) = y

, and let C

be the trajectory generated by the solution x = φ

(t),

y = ψ

(t), with φ

) = x

, ψ

) = y

. Use the fact that the system is autonomous, and

also the existence and uniqueness theorem, to show that C

and C

are the same.

27. Prove that if a trajectory starts at a noncritical point of the system

dx/dt = F (x, y), dy/dt = G(x, y),

then it cannot reach a critical point (x

, y

) in a ﬁnite length of time.

Hint: Assume the contrary; that is, assume that the solution x = φ(t), y = ψ(t) satisﬁes

φ(a) = x

, ψ(a) = y

. Then use the fact that x = x

, y = y

is a solution of the given system

satisfying the initial condition x = x

, y = y

at t = a.

28. Assuming that the trajectory corresponding to a solution x = φ(t),y = ψ(t),−∞ < t < ∞,

of an autonomous system is closed, show that the solution is periodic.

Hint: Since the trajectory is closed, there exists at least one point (x

, y

) such that

φ(t

) = x

, ψ(t

) = y

and a number T > 0 such that φ(t

+ T) = x

, ψ(t

+ T) = y

. Show

that x = 7(t) = φ(t + T) and y = 8(t) = ψ(t + T) is a solution,and then use the existence

and uniqueness theorem to show that 7(t) = φ(t) and 8(t) = ψ(t) for all t.

Georg Dufﬁng (1861–1944), a German engineer, was a pioneer in the study of the oscillations

of nonlinear mechanical systems. His most important work was the inﬂuential monograph Erzwungene

Schwingungen bei veränderlicher Eigenfrequenz und ihre technische Bedeutung, published in 1918.

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9.3 Locally Linear Systems 519

9.3 Locally Linear Systems

In Section 9.1 we described the stability properties of the equilibrium solution x = 0

of the two-dimensional linear system

′

= Ax. (1)

The results are summarized in Table 9.1.1. Recall that we required that detA ̸= 0,

so x = 0 is the only critical point of the system (1). Now that we have deﬁned the

concepts of asymptotic stability,stability,and instabilitymore precisely,we can restate

these results in the following theorem.

Theorem 9.3.1

The critical point x = 0 of the linear system (1) is asymptotically stable if the eigen-

values r

, r

are real and negative or have negative real part; is stable, but not

asymptotically stable, if r

and r

are pure imaginary; and is unstable if r

and r

are

real and either is positive, or if they have positive real part.

Effect of Small Perturbations. It is apparent from this theorem or from Table 9.1.1 that

the eigenvalues r

, r

of the coefﬁcient matrix A determine the type of critical point

at x = 0 and its stability characteristics. In turn, the values of r

and r

depend on

the coefﬁcients in the system (1).When such a system arises in some applied ﬁeld, the

coefﬁcients usually result from the measurements of certain physical quantities. Such

measurements areoften subject to small uncertainties,so it is of interest toinvestigate

whether small changes (perturbations) in the coefﬁcients can affect the stability or

instability of a critical point and/or signiﬁcantly alter the pattern of trajectories.

Recall that the eigenvalues r

, r

are the roots of the polynomial equation

det(A − rI) = 0. (2)

It is possible to show that small perturbations in some or all the coefﬁcients are

reﬂected in small perturbations in the eigenvalues. The most sensitive situation

occurs when r

= iµ and r

=−iµ. Then the critical point is a center and the tra-

jectories are closed curves (ellipses) surrounding it. If a slight change is made in the

coefﬁcients, then the eigenvalues r

and r

will take on new values r

′

= λ

′

+ iµ

′

and

′

= λ

′

− iµ

′

, where λ

′

is small in magnitude and µ

′

∼

µ (see Figure 9.3.1). If λ

′

̸= 0,

which almost always occurs, then the trajectories of the perturbed system are spi-

rals rather than ellipses. The system is asymptotically stable if λ

′

< 0 but unstable if

′

> 0. Thus, in the case of a center, small perturbations in the coefﬁcients may well

change a stable system into an unstable one and, in any case, may be expected to

change the trajectories from ellipses to spirals (see Problem 27).

Another slightly less sensitive case occurs if the eigenvalues r

and r

are equal; in

this case the critical point is a node. Small perturbations in the coefﬁcients will nor-

mally cause the two equal roots to separate (bifurcate). If the separated roots are real,

then the critical point of the perturbed system remains a node, but if the separated

roots are complex conjugates, then the critical point becomes a spiral point. These

two possibilities are shown schematically in Figure 9.3.2. In this case the stability or

instability of the system is not affected by small perturbations in the coefﬁcients, but

the type of the critical point may be changed (see Problem 28).

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520 Chapter 9. Nonlinear Differential Equations and Stability

= i

= –i

= ' – i '

= ' + i '

= i

= –i

= ' – i '

= ' + i '

FIGURE 9.3.1 Schematic perturbation of r

= iµ, r

=−iµ.

= r

' = + i

' = – i

FIGURE 9.3.2 Schematic perturbation of r

= r

In all other cases the stability or instability of the system is not changed, nor is the

type of critical point altered, by sufﬁciently small perturbations in the coefﬁcients

of the system. For example, if r

and r

are real, negative, and unequal, then a small

change in the coefﬁcients will neither change the sign of r

and r

nor allow them to

coalesce. Thus the critical point remains an asymptotically stable node.

Linear Approximations to Nonlinear Systems. Nowletus consider a nonlinear autonomous

two-dimensional system

′

= f(x). (3)

Our main object is to investigate the behavior of trajectories of the system (3) near a

criticalpointx

.Recallthatin Example 1 in Section 9.2 we notedthat near each critical

point of that nonlinear system the pattern of trajectories resembles the trajectories

of a certain linear system. This suggests that near a critical point we may be able

to approximate the nonlinear system (3) by an appropriate linear system, whose

trajectories are easy to describe. The crucial question is whether and how we can

ﬁnd an approximating linear system whose trajectories closely match those of the

nonlinear system near the critical point

It is convenient to choose the critical point to be the origin. This involves no loss

of generality, since if x

̸= 0, it is always possible to make the substitution u = x − x

in Eq. (3). Then u will satisfy an autonomous system with a critical point at the origin.

First, let us consider what it means for a nonlinear system (3) to be “close” to a

linear system (1). Accordingly, suppose that

′

= Ax + g(x) (4)

and that x = 0 is an isolated critical point of the system (4). This means that there

is some circle about the origin within which there are no other critical points. In

addition, we assume that detA ̸= 0, so that x = 0 is also an isolated critical point

of the linear system x

′

= Ax. For the nonlinear system (4) to be close to the linear

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9.3 Locally Linear Systems 521

system x

′

= Ax, we must assume that g(x) is small. More precisely, we assume that

the components of g have continuous ﬁrst partial derivatives and satisfy the limit

condition

∥g(x)∥/∥x∥→0 as x → 0; (5)

that is, ∥g∥ is small in comparison to ∥x∥ itself near the origin. Such a system is called

a locally linear system in the neighborhood of the critical point x = 0.

It may be helpful to express the condition (5) in scalar form. If we let

= (x, y), then ∥x∥=(x

+ y

)

1/2

= r. Similarly, if g

(x) = (g

(x, y), g

(x, y)),

then ∥g(x)∥=[g

(x, y) + g

(x, y)]

1/2

. Then it follows that condition (5) is satisﬁed if

and only if

(x, y)/r → 0, g

(x, y)/r → 0 as r → 0. (6)

EXAMPLE

Determine whether the system

′

00.5

−x

− xy

−0.75xy − 0.25y

(7)

is locally linear in the neighborhood of the origin.

Observe that the system (7) is of the form (4), that (0, 0) is a critical point, and that

detA ̸= 0. It is not hard to show that the other critical points of Eqs. (7) are (0, 2), (1, 0),

and (0.5, 0.5); consequently, the origin is an isolated critical point. In checking the condition

(6), it is convenient to introduce polar coordinates by letting x = r cos θ, y = r sin θ. Then

(x, y)

−x

− xy

−r

cos

θ − r

sin θ cos θ

=−r(cos

θ + sin θ cos θ) → 0

as r → 0. In a similar way you can show that g

(x, y)/r → 0asr → 0. Hence the system (7) is

locally linear near the origin.

EXAMPLE

The motion of a pendulum is described by the system [see Eq. (13) of Section 9.2]

= y,

=−ω

sin x − γy. (8)

The critical points are (0, 0), (±π,0), (±2π,0), ..., so the origin is an isolated critical point of

this system. Show that the system is locally linear near the origin.

To compare Eqs. (8) with Eq. (4),we must rewrite the former so that the linearand nonlinear

terms are clearly identiﬁed. If we write sin x = x + (sin x − x) and substitute this expression in

the second of Eqs. (8), we obtain the equivalent system

(

′

−ω

−γ

− ω

sin x − x

. (9)

On comparing Eq. (9) and Eq. (4), we see that g

(x, y) = 0 and g

(x, y) =−ω

(sin x − x).

From the Taylor series for sin x, we know that sin x − x behaves like −x

/3!=−(r

cos

θ)/3!

when x is small. Consequently, (sin x − x)/r → 0asr → 0. Thus the conditions (6) are

satisﬁed, and the system (9) is locally linear near the origin.

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522 Chapter 9. Nonlinear Differential Equations and Stability

Let us now return to the general nonlinear system (3), which we write in the scalar

form

′

= F(x, y), y

′

= G(x, y); (10)

that is, x = (x, y)

and f(x) = (F(x, y), G(x, y))

. The system (10) is locally linear in

the neighborhood of a critical point (x

, y

) whenever the functions F and G have

continuous partial derivatives up to order two.To show this,we useTaylor expansions

about the point (x

, y

) to write F(x, y) and G(x, y) in the form

F(x, y) = F(x

, y

) + F

, y

)(x − x

) + F

, y

)(y − y

) + η

(x, y),

G(x, y) = G(x

, y

) + G

, y

)(x − x

) + G

, y

)(y − y

) + η

(x, y),

where η

(x, y)/[(x − x

)

+ (y − y

)

]

1/2

→ 0as(x, y) → (x

, y

), and similarly for

. Note that F(x

, y

) = G(x

, y

) = 0, and that dx/dt = d(x − x

)/dt and

dy/dt = d(y − y

)/dt. Then the system (10) reduces to

x − x

y − y

(

, y

) F

, y

)

, y

) G

, y

)

x − x

y − y

(

(x, y)

(

, (11)

or, in vector notation,

d u

d f

d x

)u + η(x), (12)

where u = (x − x

, y − y

)

and η = (η

, η

)

The signiﬁcance of this result is twofold. First, if the functions F and G are twice

differentiable, then the system (10) is locally linear, and it is unnecessary to resort

to the limiting process used in Examples 1 and 2. Second, the linear system that

approximates the nonlinear system (10) near (x

, y

) is given by the linear part of

Eqs. (11) or (12):

(

, y

) F

, y

)

, y

) G

, y

)

(

, (13)

where u

= x − x

and u

= y − y

. Equation (13) provides a simple and general

method for ﬁnding the linear system corresponding to a locally linear system near a

given critical point.

The matrix

J =

(

, (14)

which appears as the coefﬁcient matrix in Eq. (13), is called the Jacobian

matrix

of the functions F and G with respect to x and y. We need to assume that det(J) is

not zero at (x

, y

) so that this point is also an isolated critical point of the linear

system (13).

Carl Gustav Jacob Jacobi (1804–1851), a German analyst who was professor and lecturer at the Uni-

versities of Königsberg and Berlin, made important contributions to the theory of elliptic functions. The

determinant of J and its extension to n functions of n variables is called the Jacobian because of his notable

paper in 1841 on the properties of this determinant. The corresponding matrix is also named for Jacobi,

even though matrices were not developed until after his death.

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9.3 Locally Linear Systems 523

EXAMPLE

Use Eq. (13) to ﬁnd the linear system corresponding to the pendulum equations (8) near the

origin; near the critical point (π,0).

In this case we have, from Eq. (8),

F(x, y) = y, G(x, y) =−ω

sin x − γy; (15)

since these functions are differentiable as many times as necessary, the system (8) is locally

linear near each critical point. The derivatives of F and G are

= 0, F

= 1, G

=−ω

cos x, G

=−γ. (16)

Thus, at the origin the corresponding linear system is

−ω

−γ

, (17)

which agrees with Eq. (9).

Similarly, evaluating the partial derivatives in Eq. (16) at (π,0), we obtain

−γ

, (18)

where u = x − π,v = y. This is the linear system corresponding to Eqs. (8) near the point (π,0).

We now return to the locally linear system (4). Since the nonlinear term g(x) is

small compared to the linear term Ax when x is small, it is reasonable to hope that

the trajectories of the linear system (1) are good approximations to those of the

nonlinear system (4), at least near the origin. This turns out to be true in many (but

not all) cases, as the following theorem states.

Theorem 9.3.2

Let r

and r

be the eigenvalues of the linear system (1) corresponding to

the locally linear system (4). Then the type and stability of the critical point

(0, 0) of the linear system (1) and the locally linear system (4) are as shown in

Table 9.3.1.

TABLE 9.3.1 Stability and Instability Properties of Linear and Locally Linear Systems

Linear System Locally Linear System

, r

Type Stability Type Stability

> r

> 0 N Unstable N Unstable

< r

< 0 N Asymptotically N Asymptotically

stable stable

< 0 < r

SP Unstable SP Unstable

= r

> 0 PN or IN Unstable N or SpP Unstable

= r

< 0 PN or IN Asymptotically N or SpP Asymptotically

stable stable

, r

= λ ± iµ

λ>0 SpP Unstable SpP Unstable

λ<0 SpP Asymptotically SpP Asymptotically

stable stable

= iµ, r

=−iµ C Stable C or SpP Indeterminate

Note: N, node; IN, improper node; PN, proper node; SP, saddle point; SpP, spiral point;

C, center.

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524 Chapter 9. Nonlinear Differential Equations and Stability

At this stage, the proof of Theorem 9.3.2 is too difﬁcult to give, so we will accept

the results without proof. The statements for asymptotic stability and instability

follow as a consequence of a result discussed in Section 9.6,and a proof is sketched in

Problems 10 to 12 of that section. Essentially,Theorem 9.3.2 says that for small x (or

x − x

) the nonlinear terms are also small and do not affect the stability and type of

critical point as determined by the linear terms,except in two sensitive cases. If r

and

are pure imaginary, then the small nonlinear terms may change the stable center

into a spiral point, which may be either asymptotically stable or unstable. If r

and r

are real and equal, then the nonlinear terms may change the node into a spiral point,

but its asymptotic stability or instability remains unchanged. Recall that earlier in

this section we stated that small perturbations in the coefﬁcients of the linear system

(1), and hence in the eigenvalues r

and r

, can alter the type and stability of the crit-

ical point only in these two cases. It is reasonable to expect that the small nonlinear

term in Eq. (4) might have a similar substantial effect, at least in these two cases. This

is so, but the main signiﬁcance of Theorem 9.3.2 is that in all other cases the small

nonlinear term does not alter the type or stability of the critical point. Thus, except

in the two sensitive cases, the type and stability of the critical point of the nonlinear

system (4) can be determined from a study of the much simpler linear system (1).

Even if the critical point is of the same type as that of the linear system, the

trajectories of the locally linear system may be considerably different in appearance

from those of the corresponding linear system, except very near the critical point.

However, it can be shown that the slopes at which trajectories “enter” or “leave” the

critical point are given correctly by the linear system.

Damped Pendulum. We continue the discussion of the damped pendulum begun in

Examples 2 and 3. Near the origin the nonlinear equations (8) are approximated

by the linear system (17), whose eigenvalues are

, r

−γ ±

− 4ω

. (19)

The nature of the solutions of Eqs. (8) and (17) depends on the sign of γ

− 4ω

follows:

1. If γ

− 4ω

> 0, then the eigenvalues are real, unequal, and negative. The critical point

(0, 0) is an asymptotically stable node of the linear system (17) and of the locally linear

system (8).

2. If γ

− 4ω

= 0, then the eigenvalues are real, equal, and negative. The critical point (0, 0)

is an asymptotically stable (proper or improper) node of the linear system (17). It may be

either an asymptotically stable node or spiral point of the locally linear system (8).

3. If γ

− 4ω

< 0, then the eigenvalues are complex with negative real part. The critical

point (0, 0) is an asymptotically stable spiral point of the linear system (17) and of the

locally linear system (8).

Thus the critical point (0, 0) is a spiral point of the system (8) if the damping is small

and a node if the damping is large enough. In either case, the origin is asymptotically

stable.

Let us now consider the case γ

− 4ω

< 0, corresponding to small damping, in

more detail.The direction of motion on the spirals near (0, 0) can be obtained directly

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9.3 Locally Linear Systems 525

from Eqs. (8). Consider the point at which a spiral intersects the positive y-axis (x = 0

and y > 0). It follows from Eqs. (8) that at such a point, dx/dt > 0. Thus the point

(x, y) on the trajectory is moving to the right, so the direction of motion on the spirals

is clockwise.

The behavior of the pendulum near the critical points (±nπ,0), with n even, is

the same as its behavior near the origin. We expect this on physical grounds since

all these critical points correspond to the downward equilibrium position of the

pendulum. The conclusion can be conﬁrmed by repeating the analysis carried out

above for the origin. Figure 9.3.3 shows the clockwise spirals at a few of these critical

points.

–2

–

FIGURE 9.3.3 Asymptotically stable spiral points for the damped pendulum.

Now let us consider the critical point (π,0). Here the nonlinear equations (8) are

approximated by the linear system (18), whose eigenvalues are

, r

−γ ±

+ 4ω

. (20)

One eigenvalue (r

) is positive and the other (r

) is negative. Therefore, regardless

of the amount of damping, the critical point x = π, y = 0 is an unstable saddle point

both of the linear system (18) and of the locally linear system (8).

To examine the behavior of trajectories near the saddle point (π,0) in more detail,

we write down the general solution of Eqs. (18), namely,

(

= C

(

+ C

(

, (21)

where C

and C

are arbitrary constants. Since r

> 0 and r

< 0, it follows that the

solution that approaches zero as t →∞corresponds to C

= 0. For this solution

v/u = r

, so the slope of the entering trajectories is negative; one lies in the second

quadrant (C

< 0), and the other lies in the fourth quadrant (C

> 0).ForC

= 0

we obtain the pair of trajectories “exiting” from the saddle point. These trajectories

have slope r

> 0; one lies in the ﬁrst quadrant (C

> 0), and the other lies in the

third quadrant (C

< 0).

The situation is the same at other critical points (nπ,0) with n odd. These all

correspond to the upward equilibrium position of the pendulum, so we expect them

to be unstable. The analysis at (π,0) can be repeated to show that they are saddle

points oriented in the same way as the one at (π,0). Diagrams of the trajectories in

the neighborhood of two saddle points are shown in Figure 9.3.4.

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526 Chapter 9. Nonlinear Differential Equations and Stability

–

FIGURE 9.3.4 Unstable saddle points for the damped pendulum.

EXAMPLE

The equations of motion of a certain pendulum are

dx/dt = y, dy/dt =−9 sin x −

y, (22)

where x = θ and y = dθ/dt. Draw a phase portrait for this system, and explain how it shows

the possible motions of the pendulum.

By plotting the trajectories starting at various initial points in the phase plane, we obtain

the phase portrait shown in Figure 9.3.5. As we have seen, the critical points (equilibrium

solutions) are the points (nπ,0), where n = 0, ±1, ±2, ....Even values of n, including zero,

correspond to the downward position of the pendulum, while odd values of n correspond to

the upward position. Near each of the asymptotically stable critical points, the trajectories

are clockwise spirals that represent a decaying oscillation about the downward equilibrium

position. The wavy horizontal portions of the trajectories that occur for larger values of |y|

represent whirling motions of the pendulum. Note that a whirling motion cannot continue

indeﬁnitely, no matter how large |y| is; eventually, the angular velocity is so much reduced by

the damping term that the pendulum can no longer go over the top and, instead, begins to

oscillate about its downward position.

6486–4–

–10

–8 –2 2

–5

FIGURE 9.3.5 Phase portrait for the damped pendulum of Example 4.

The basin of attraction for the origin is the shaded region in Figure 9.3.5. It is bounded by

the trajectories that enter the two adjacent saddle points at (π,0) and (−π,0). The bound-

ing trajectories are separatrices. Each asymptotically stable critical point has its own basin of

attraction, which is bounded by the separatrices entering the two neighboring saddle points.

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9.3 Locally Linear Systems 527

All of the basins of attraction are congruent to the shaded one; the only difference is that

they are translated horizontally by appropriate distances. Note that it is mathematically pos-

sible (but physically unrealizable) to choose initial conditions exactly on a separatrix so that

the resulting motion leads to a balanced pendulum in a vertically upward position of unstable

equilibrium.

An important difference between nonlinear autonomous systems and the linear

systems discussed in Section 9.1 is illustrated by the pendulum equations. Recall that

the linear system (1) has only the single critical point x = 0 if det A ̸= 0. Thus, if

the origin is asymptotically stable, then not only do trajectories that start close to the

origin approach it, but, in fact, every trajectory approaches the origin. In this case

the critical point x = 0 is said to be globally asymptotically stable. This property of

linear systems is not, in general, true for nonlinear systems, even if the nonlinear

system has only one asymptotically stable critical point. Therefore, for nonlinear

systems, it is important to determine (or to estimate) the basin of attraction for each

asymptotically stable critical point.

PROBLEMS In each of Problems 1 through 4, verify that (0, 0) is a critical point, show that the system is

locally linear, and discuss the type and stability of the critical point (0, 0) by examining the

corresponding linear system.

1. dx/dt = x − y

, dy/dt = x − 2y + x

2. dx/dt =−x + y + 2xy, dy/dt =−4x − y + x

− y

3. dx/dt = (1 + x) sin y, dy/dt = 1 − x − cos y

4. dx/dt = x + y

, dy/dt = x + y

In each of Problems 5 through 18:

(a) Determine all critical points of the given system of equations.

(b) Find the corresponding linear system near each critical point.

the nonlinear system?

(d) Draw a phase portrait of the nonlinear system to conﬁrm your conclusions, or to extend

them in those cases where the linear system does not provide deﬁnite information about the

nonlinear system.

dx/dt = (2 + x)(y − x), dy/dt = (4 − x)(y + x)

dx/dt = x − x

− xy, dy/dt = 3y − xy − 2y

7. dx/dt = 1 − y, dy/dt = x

− y

8. dx/dt = x − x

− xy, dy/dt =

y −

−

dx/dt = (2 + y)(y − 0.5x), dy/dt = (2 − x)(y + 0.5x)

10.

dx/dt = x + x

+ y

, dy/dt = y − xy

11.

dx/dt = 2x + y + xy

, dy/dt = x − 2y − xy

12.

dx/dt = (1 + x ) sin y, dy/dt = 1 − x − cos y

13.

dx/dt = x − y

, dy/dt = y − x

14. dx/dt = 1 − xy, dy/dt = x − y

15. dx/dt =−2x − y − x(x

+ y

), dy/dt = x − y + y(x

+ y

)

16.

dx/dt = y + x (1 − x

− y

), dy/dt =−x + y(1 − x

− y

)

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528 Chapter 9. Nonlinear Differential Equations and Stability

17. dx/dt = 4 − y

, dy/dt = (1.5 + x)(y − x)

18.

dx/dt = (1 − y)(2 x − y), dy/dt = (2 + x)(x − 2y)

19. Consider the autonomous system

dx/dt = y, dy/dt = x + 2x

(a) Show that the critical point (0, 0) is a saddle point.

(b) Sketch the trajectories forthe correspondinglinear systemby integratingthe equation

for dy/dx. Show from the parametric form of the solution that the only trajectory on which

x → 0, y → 0 as t →∞is y =−x.

dy/dx. Sketch the trajectories for the nonlinear system that correspond to y =−x and

y = x for the linear system.

20. Consider the autonomous system

dx/dt = x, dy/dt =−2y + x

(a) Show that the critical point (0, 0) is a saddle point.

(b) Sketch the trajectories for the corresponding linear system, and show that the

trajectory for which x → 0, y → 0 as t →∞is given by x = 0.

tion for dy/dx. Show that the trajectory corresponding to x = 0 for the linear system is

unaltered, but that the one corresponding to y = 0isy = x

/5. Sketch several of the

trajectories for the nonlinear system.

21.

The equation of motion of an undamped pendulum is d

θ/dt

+ ω

sin θ = 0, where

= g/L. Let x = θ, y = dθ/dt to obtain the system of equations

dx/dt = y, dy/dt =−ω

sin x.

(a) Show that the critical points are (±nπ,0), n = 0, 1,2, ..., and that the system is locally

linear in the neighborhood of each critical point.

(b) Show that the critical point (0, 0) is a (stable) center of the corresponding linear

system. Using Theorem 9.3.2, what can you say about the nonlinear system? The

situation is similar at the critical points (±2nπ,0), n = 1, 2, 3, .... What is the physical

interpretation of these critical points?

ing linear system. What conclusion can you draw about the nonlinear system? The

situation is similar at the critical points [±(2n − 1)π,0], n = 1, 2, 3, ....What is the physical

interpretation of these critical points?

(d) Choose a value for ω

and plot a few trajectories of the nonlinear system in the

neighborhood of the origin. Can you now draw any further conclusion about the nature

of the critical point at (0, 0) for the nonlinear system?

(e) Using the value of ω

from part (d), draw a phase portrait for the pendulum. Compare

your plot with Figure 9.3.5 for the damped pendulum.

22. (a) By solving the equation for dy/dx, show that the equation of the trajectories of the

undamped pendulum of Problem 21 can be written as

+ ω

(1 − cos x) = c, (i)

where c is a constant of integration.

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9.3 Locally Linear Systems 529

(b) Multiply Eq. (i) by mL

. Then express the result in terms of θ to obtain

dθ

(

+ mgL(1 − cos θ) = E, (ii)

where E = mL

the second term is the potential energy due to gravity. Thus the total energy E of the

pendulum is constant along any trajectory; its value is determined by the initial conditions.

23.

The motion of a certain undamped pendulum is described by the equations

dx/dt = y, dy/dt =−4 sin x.

If the pendulum is set in motion with an angular displacement A and no initial velocity,

then the initial conditions are x(0) = A, y(0) = 0.

(a) Let A = 0.25 and plot x versus t. From the graph, estimate the amplitude R and

period T of the resulting motion of the pendulum.

(b) Repeat part (a) for A = 0.5, 1.0, 1.5, and 2.0.

position A? Draw a graph to show each of these relationships. Can you say anything about

the limiting value of the period as A → 0?

(d) Let A = 4 and plot x versus t. Explain why this graph differs from those in parts (a)

and (b). For what value of A does the transition take place?

24.

Consider again the pendulum equations (see Problem 23)

dx/dt = y, dy/dt =−4 sin x.

If the pendulum is set in motion from its downward equilibrium position with angular

velocity v, then the initial conditions are x(0) = 0, y(0) = v.

(a) Plot x versus t for v = 2 and also for v = 5. Explain the differing motions of the

pendulum that these two graphs represent.

(b) There is a critical value of v, which we denote by v

, such that one type of motion

occurs for v<v

and the other for v>v

. Estimate the value of v

25.

This problem extends Problem 24 to a damped pendulum. The equations of motion are

dx/dt = y, dy/dt =−4 sin x − γy,

where γ is the damping coefﬁcient, with the initial conditions x(0) = 0, y(0) = v.

(a) For γ = 1/4, plot x versus t for v = 2 and for v = 5. Explain these plots in terms of

the motions of the pendulum that they represent.Also explain how they are related to the

corresponding graphs in Problem 24(a).

(b) Estimate the critical value v

of the initial velocity where the transition from one type

of motion to the other occurs.

depends on γ.

26. Theorem 9.3.2 provides no information about the stability of a critical point of a locally

linear system if that point is a center of the corresponding linear system. That this must

be the case is illustrated by the systems

dx/dt = y + x (x

+ y

(i)

dy/dt =−x + y(x

+ y

)

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530 Chapter 9. Nonlinear Differential Equations and Stability

and

dx/dt = y − x (x

+ y

(ii)

dy/dt =−x − y(x

+ y

(a) Show that (0, 0) is a critical point of each system and, furthermore, is a center of the

corresponding linear system.

(b) Show that each system is locally linear.

= x

+ y

, and note that xdx/dt + ydy/dt = rdr/dt. For system (ii), show that

dr/dt < 0 and that r → 0ast →∞; hence the critical point is asymptotically stable. For

system (i), show that the solution of the initial value problem for r with r = r

at t = 0

becomes unbounded as t → 1/2r

, and hence the critical point is unstable.

27. In this problem we show how small changes in the coefﬁcients of a system of linear

equations can affect a critical point that is a center. Consider the system

′

−10

Show that the eigenvalues are ±i so that (0, 0) is a center. Now consider the system

′

ϵ 1

−1 ϵ

where |ϵ| is arbitrarily small. Show that the eigenvalues are ϵ ± i. Thus no matter how

small |ϵ|̸=0 is,the center becomes a spiral point. If ϵ<0,the spiral point is asymptotically

stable; if ϵ>0, the spiral point is unstable.

28. In this problem we show how small changes in the coefﬁcients of a system of linear equa-

tions can affect the nature of a critical point when the eigenvalues are equal. Consider the

system

′

−11

0 −1

Show that the eigenvalues are r

=−1, r

=−1 so that the critical point (0, 0) is an

asymptotically stable node. Now consider the system

′

−11

−ϵ −1

where |ϵ| is arbitrarily small. Show that if ϵ>0, then the eigenvalues are −1 ± i

√

ϵ,

so that the asymptotically stable node becomes an asymptotically stable spiral point.

If ϵ<0, then the roots are −1 ±

√

|ϵ|, and the critical point remains an asymptotically

stable node.

29.

In this problem we derive a formula for the natural period of an undamped nonlinear

pendulum [c = 0 in Eq. (10) of Section 9.2]. Suppose that the bob is pulled through a

positive angle α and then released with zero velocity.

(a) We usually think of θ and dθ/dt as functions of t. However, if we reverse the roles of

t and θ, we can regard t as a function of θ and, consequently, can also think of dθ/dt as a

function of θ. Then derive the following sequence of equations:

dθ

(

=−mgL sin θ,

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9.4 Competing Species 531

dθ

(

= mgL(cos θ − cos α),

dt =−

dθ

√

cos θ − cos α

Why was the negative square root chosen in the last equation?

(b) If T is the natural period of oscillation, derive the formula

=−

dθ

√

cos θ − cos α

(θ/2) and cos α = 1 − 2 sin

(α/2),followed by

the change of variable sin(θ/2) = k sin φ with k = sin(α/2), show that

T = 4

π/2

dφ

1 − k

sin

The integral is called the elliptic integral of the ﬁrst kind. Note that the period depends

on the ratio L/g and also on the initial displacement α through k = sin(α/2).

(d) By evaluating the integral in the expression for T, obtain values for T that you can

compare with the graphical estimates you obtained in Problem 23.

30. A generalization of the damped pendulum equation discussed in the text, or a damped

spring–mass system, is the Liénard

equation

+ c(x)

+ g(x ) = 0.

If c(x) is a constant and g(x) = kx, then this equation has the form of the linear pen-

dulum equation [replace sin θ with θ in Eq. (12) of Section 9.2]; otherwise, the damping

force c(x) dx/dt and the restoring force g(x) are nonlinear. Assume that c is continuously

differentiable, g is twice continuously differentiable, and g(0) = 0.

(a) Write the Liénard equation as a system of two ﬁrst order equations by introducing

the variable y = dx /dt.

(b) Show that (0, 0) is a critical point and that the system is locally linear in the

neighborhood of (0, 0).

′

(0)>0, then the critical point is asymptotically stable,and

that if c(0)<0 or g

′

(0)<0, then the critical point is unstable.

Hint: Use Taylor series to approximate c and g in the neighborhood of x = 0.

9.4 Competing Species

In this section and the next, we explore the application of phase plane analysis to

some problems in population dynamics. These problems involve two interacting

populations and are extensions of those discussed in Section 2.5, which dealt with

Alfred-Marie Liénard (1869–1958), a French physicist and engineer, was professor at l’École des Mines

in Paris. He worked primarily in electricity, mechanics, and applied mathematics. The results of his

investigation of this differential equation were published in 1928.

August 7, 2012 21:05 c09 Sheet number 38 Page number 532 cyan black

532 Chapter 9. Nonlinear Differential Equations and Stability

a single population. Although the equations discussed here are extremely simple

compared to the very complex relationships that exist in nature, it is still possible to

acquire some insight into ecological principles from a study of these model problems.

The same, or similar, models have also been used to study other types of competitive

situations—for instance, businesses competing in the same market.

Suppose that in some closed environment there are two similar species competing

for a limited food supply—for example, two species of ﬁsh in a pond that do not prey

on each other but do compete for the available food. Let x and y be the populations of

the two species at time t. As discussed in Section 2.5, we assume that the population

of each of the species, in the absence of the other, is governed by a logistic equation.

Thus

dx/dt = x (ϵ

− σ

x), (1a)

dy/dt = y(ϵ

− σ

y), (1b)

respectively, where ϵ

and ϵ

are the growth rates of the two populations, and ϵ

/σ

and ϵ

/σ

are their saturation levels. However, when both species are present, each

will tend to diminish the available food supply for the other. In effect, they reduce

each other’s growth rates and saturation populations. The simplest expression for

reducing the growth rate of species x due to the presence of species y is to replace the

growth rate factor ϵ

− σ

x in Eq. (1a) by ϵ

− σ

x − α

y, where α

is a measure of

the degree to whichspecies y interferes withspecies x. Similarly,in Eq. (1b)we replace

− σ

y by ϵ

− σ

y − α

x. Thus we have the system of equations

dx/dt = x (ϵ

− σ

x − α

y),

(2)

dy/dt = y(ϵ

− σ

y − α

x).

The values of the positive constants ϵ

, σ

, α

, ϵ

, σ

, and α

depend on the particular

species under consideration and in general must be determined from observations.

We are interested in solutions of Eqs. (2) for which x and y are nonnegative. In the

following two examples we discuss two typical problems in some detail. At the end

of the section we return to the general equations (2).

EXAMPLE

Discuss the qualitative behavior of solutions of the system

dx/dt = x(1 − x − y),

(3)

dy/dt = y(0.75 − y − 0.5x).

We ﬁnd the critical points by solving the system of algebraic equations

x(1 − x − y) = 0, y(0.75 − y − 0.5x) = 0. (4)

The ﬁrst equation can be satisﬁed by choosing x = 0; then the second equation requires that

y = 0ory = 0.75. Similarly, the second equation can be satisﬁed by choosing y = 0, and then

the ﬁrst equation requires that x = 0orx = 1.Thus we have found three critical points, namely,

(0, 0), (0, 0.75), and (1, 0). If neither x nor y is zero, then Eqs. (4) are also satisﬁed by solutions

of the system

1 − x − y = 0,

75 − y − 0.5x = 0, (5)

which leads to a fourth critical point (0.5, 0.5). These four critical points correspond to equi-

librium solutions of the system (3). The ﬁrst three of these points involve the absence of one

or both species; only the last corresponds to the presence of both species. Other solutions

are represented as curves or trajectories in the xy-plane that describe the evolution of the

August 7, 2012 21:05 c09 Sheet number 39 Page number 533 cyan black

9.4 Competing Species 533

populations in time. To begin to discover their qualitative behavior, we can proceed in the

following way.

First, observe that the coordinate axes are themselves trajectories. This follows directly

from Eqs. (3) since dx/dt = 0 on the y-axis (where x = 0) and, similarly, dy/dt = 0 on the

x-axis (where y = 0). Thus no other trajectories can cross the coordinate axes. For a popu-

lation problem only nonnegative values of x and y are signiﬁcant, and we conclude that any

trajectory that starts in the ﬁrst quadrant remains there for all time.

A direction ﬁeld for the system (3) in the positive quadrant is shown in Figure 9.4.1; the

black dots in this ﬁgure are the critical points or equilibrium solutions. Based on the direction

ﬁeld, it appears that the point (0.5, 0.5) attracts other solutions and is therefore asymptotically

stable, while the other three critical points are unstable. To conﬁrm these conclusions, we can

look at the linear approximations near each critical point.

0.75

0.5

0.25

1.2510.750.50.250

FIGURE 9.4.1 Critical points and direction ﬁeld for the system (3).

The system (3) is locally linear in the neighborhood of each critical point. There are two

ways to obtain the linear system near a critical point (X, Y). First, we can use the substitution

x = X + u, y = Y + v in Eqs. (3), retaining only the terms that are linear in u and v. Alter-

natively, we can evaluate the Jacobian matrix J at each critical point to obtain the coefﬁcient

matrix in the approximating linear system; see Eq. (13) in Section 9.3. When several critical

points are to be investigated, it is usually better to use the Jacobian matrix. For the system (3),

we have

F(x, y) = x(1 − x − y), G(x, y) = y(0.75 − y − 0.5x), (6)

J =

1 − 2x − y −x

−0.5y 0.75 − 2y − 0.5x

. (7)

We will now examine each critical point in turn.

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534 Chapter 9. Nonlinear Differential Equations and Stability

x = 0, y = 0. This critical point corresponds to the state in which neither species is present.

To determine what happens near the origin we can set x = y = 0 in Eq. (7), which leads to the

corresponding linear system

00.75

. (8)

The eigenvalues and eigenvectors of the system (8) are

= 1, ξ

(1)

; r

= 0.75, ξ

(2)

, (9)

so the general solution of the system is

= c

+ c

0.75t

. (10)

Thus the origin is an unstable node of both the linear system (8) and the nonlinear system

(3). In the neighborhood of the origin, all trajectories are tangent to the y-axis except for one

trajectory that lies along the x-axis. If either or both of the species are present in small numbers,

the population(s) will grow.

x = 1, y = 0. This corresponds to a state in which species x is present but species y is not.

By evaluating J from Eq. (7) at (1, 0), we ﬁnd that the corresponding linear system is

−1 −1

00.25

. (11)

Its eigenvalues and eigenvectors are

=−1, ξ

(1)

; r

= 0.25, ξ

(2)

−5

, (12)

and its general solution is

= c

−t

+ c

−5

0.25t

. (13)

Since the eigenvalues have opposite signs,the point (1, 0) is a saddle point,and so it is an unsta-

ble equilibrium point of the linear system (11) and of the nonlinear system (3). The behavior

of the trajectories near (1, 0) can be seen from Eq. (13). If c

= 0, then there is one pair of

trajectories that approaches the critical point along the x-axis. In other words, if the y pop-

ulation is initially zero, then it remains zero forever. All other trajectories depart from the

neighborhood of (1, 0);ify is initially small and positive, then the y population grows with

time. As t →−∞, one trajectory approaches the saddle point tangent to the eigenvector ξ

(2)

whose slope is −1.25.

x = 0, y = 0.75. This critical point is a state where species y is present but x is not. The

analysis is similar to that for the point (1, 0). The corresponding linear system is

0.25 0

−0.375 −0.75

. (14)

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9.4 Competing Species 535

The eigenvalues and eigenvectors are

= 0.25, ξ

(1)

−3

; r

=−0.75, ξ

(2)

, (15)

so the general solution of Eq. (14) is

= c

−3

0.25t

+ c

−0.75t

. (16)

Thus the point (0, 0.75) is also a saddle point. All trajectories leave the neighborhood of this

point except one pair that approaches along the y-axis. The trajectory that approaches the

saddle point as t →−∞is tangent to the line with slope −0.375 determined by the eigenvector

(1)

.Ifthex population is initially zero, it will remain zero, but a small positive x population

will grow.

x = 0.5, y = 0.5. This critical point corresponds to a mixed equilibrium state, or coexis-

tence, in the competition between the two species. The eigenvalues and eigenvectors of the

corresponding linear system

−0.5 −0.5

−0.25 −0.5

(17)

are

= (−2 +

√

2)/4

∼

−0.146, ξ

(1)

√

−1

;

(18)

= (−2 −

√

2)/4

∼

−0.854, ξ

(2)

√

Therefore, the general solution of Eq. (17) is

= c

√

−1

−0.146t

+ c

√

−0.854t

. (19)

Since both eigenvalues are negative, the critical point (0.5, 0.5) is an asymptotically sta-

ble node of the linear system (17) and of the nonlinear system (3). All nearby trajectories

approach the critical point as t →∞. One pair of trajectories approaches the critical point

along the line with slope

√

2/2 determined from the eigenvector ξ

(2)

. All other trajecto-

ries approach the critical point tangent to the line with slope −

√

2/2 determined from the

eigenvector ξ

(1)

A phase portrait for the system (3) is shown in Figure 9.4.2. By looking closely at the

trajectories near each critical point, you can see that they behave in the manner predicted by

the linear system near that point. In addition, note that the quadratic terms on the right side

of Eqs. (3) are all negative. Since for x and y large and positive these terms are the dominant

ones, it follows that far from the origin in the ﬁrst quadrant both x

′

and y

′

are negative; that

is, the trajectories are directed inward. Thus all trajectories that start at a point (x

, y

) with

> 0 and y

> 0 eventually approach the point (0.5, 0.5). In other words, the entire open ﬁrst

quadrant is the basin of attraction for (0.5, 0.5).

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536 Chapter 9. Nonlinear Differential Equations and Stability

0.75

0.5

0.25

1.2510.750.50.25

FIGURE 9.4.2 A phase portrait of the system (3).

EXAMPLE

Discuss the qualitative behavior of the solutions of the system

dx/dt = x(1 − x − y),

(20)

dy/dt = y(0.5 − 0.25y − 0 .75x),

when x and y are nonnegative. Observe that this system is also a special case of the system (2)

for two competing species.

Once again, there are four critical points, namely, (0, 0), (1, 0), (0, 2), and (0.5, 0.5), corre-

sponding to equilibrium solutions of the system (20). Figure 9.4.3 shows a direction ﬁeld for

the system (20),togetherwith the four critical points. Fromthe direction ﬁeld itappears that the

mixed equilibrium solution (0.5, 0.5) is a saddle point, and therefore unstable,while the points

(1, 0) and (0, 2) are asymptotically stable. Thus, for competition described by Eqs. (20), one

species will eventually overwhelm the other and drive it to extinction. The surviving species is

determined by the initial state of the system. To conﬁrm these conclusions, we can look at the

linear approximations near each critical point. For later use, we record the Jacobian matrix J

for the system (20):

J =

(x, y) F

(x, y)

(x, y) G

(x, y)

1 − 2x − y −x

−0.75y 0.5 − 0.5y − 0.75x

. (21)

x = 0, y = 0. Using the Jacobian matrix J from Eq. (21) evaluated at (0, 0), we obtain the

linear system

00.5

, (22)

which is valid near the origin. The eigenvalues and eigenvectors of the system (22) are

= 1, ξ

(1)

; r

= 0.5, ξ

(2)

, (23)

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9.4 Competing Species 537

1.5

0.5

1.2510.750.50.25

FIGURE 9.4.3 Critical points and direction ﬁeld for the system (20).

so the general solution is

= c

+ c

0.5t

. (24)

Therefore, the origin is an unstable node of the linear system (22) and also of the nonlinear

system (20). All trajectories leave the neighborhood of the origin tangent to the y-axis except

for one trajectory that lies along the x-axis.

x = 1, y = 0. The corresponding linear system is

−1 −1

0 −0.25

. (25)

Its eigenvalues and eigenvectors are

=−1, ξ

(1)

; r

=−0.25, ξ

(2)

−3

, (26)

and its general solution is

= c

−t

+ c

−3

−0.25t

. (27)

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538 Chapter 9. Nonlinear Differential Equations and Stability

The point (1, 0) is an asymptotically stable node of the linear system (25) and of the nonlinear

system (20). If the initial values of x and y are sufﬁciently close to (1, 0), then the interaction

process will lead ultimately to that state—that is, to the survival of species x and the extinction

of species y. There is one pair of trajectories that approaches the critical point along the x-axis.

All other trajectories approach (1, 0) tangent to the line with slope −3/4 that is determined

by the eigenvector ξ

(2)

x = 0, y = 2. The analysis in this case is similar to that for the point (1, 0). The appropriate

linear system is

−10

−1.5 −0.5

. (28)

The eigenvalues and eigenvectors of this system are

=−1, ξ

(1)

; r

=−0.5, ξ

(2)

, (29)

and its general solution is

= c

−t

+ c

−0.5t

. (30)

Thus the critical point (0, 2) is an asymptotically stable node of both the linear system (28) and

the nonlinear system (20). All nearby trajectories approach the critical point tangent to the

y-axis except for one trajectory that approaches along the line with slope 3.

x = 0.5, y = 0.5. The corresponding linear system is

−0.5 −0.5

−0.375 −0.125

. (31)

The eigenvalues and eigenvectors are

−5 +

√

∼

0.1594, ξ

(1)

(−3 −

√

57)/8

∼

−1.3187

(32)

−5 −

√

∼

−0.7844, ξ

(2)

(−3 +

√

57)/8

∼

0.5687

so the general solution is

= c

−1.3187

0.1594t

+ c

0.5687

−0.7844t

. (33)

Since the eigenvalues are of opposite signs, the critical point (0.5, 0.5) is a saddle point and

therefore is unstable, as we had surmised earlier. All trajectories depart from the neighbor-

hood of the critical point except for one pair that approaches the saddle point as t →∞.

As they approach the critical point, the entering trajectories are tangent to the line with slope

(

√

57 − 3)/8

∼

0.5687 determined from the eigenvector ξ

(2)

. There is also a pair of trajectories

that approach the saddle point as t →−∞.These trajectories are tangent to the line with slope

−1.3187 corresponding to ξ

(1)

A phase portrait for the system (20) is shown in Figure 9.4.4. Near each of the critical

points, the trajectories of the nonlinear system behave as predicted by the corresponding

linear approximation. Of particular interest is the pair of trajectories that enter the saddle

point. These trajectories form a separatrix that divides the ﬁrst quadrant into two basins of

August 7, 2012 21:05 c09 Sheet number 45 Page number 539 cyan black

9.4 Competing Species 539

attraction. Trajectories starting above the separatrix ultimately approach the node at (0, 2),

while trajectories starting below the separatrix approach the node at (1, 0). If the initial state

lies precisely on the separatrix,then the solution (x, y) will approach the saddle point as t →∞.

However, the slightest perturbation of the point (x, y) as it follows this trajectory will dislodge

the point from the separatrix and cause it to approach one of the nodes instead. Thus, in

practice, one species will survive the competition and the other will not.

1.5

0.5

Separatrix

1.2510.750.50.25

FIGURE 9.4.4 A phase portrait of the system (20). The black curve is the separatrix.

Examples 1 and 2 show that in some cases the competition between two species

leads to an equilibrium state of coexistence, while in other cases the competition

results in the eventual extinction of one of the species. To understand more clearly

how and why this happens, and to learn how to predict which situation will occur,it is

useful to look again at the general system (2). There are four cases to be considered,

depending on the relative orientation of the lines

− σ

x − α

y = 0 and ϵ

− σ

y − α

x = 0, (34)

as shown in Figure 9.4.5. These lines are called the x- and y-nullclines, respectively,

because x

′

is zero on the ﬁrst and y

′

is zero on the second. In each part of Figure 9.4.5,

the x-nullcline is the solid line and the y-nullcline is the dashed line.

Let (X , Y) denote any critical point in any one of the four cases. As in Examples

1 and 2, the system (2) is locally linear in the neighborhood of this point because the

right side of each differential equation is a quadratic polynomial. To study the system

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540 Chapter 9. Nonlinear Differential Equations and Stability

(a)

(b)

(c)

(d)

FIGURE 9.4.5 The various cases for the competing-species system (2). The x-nullcline is

the solid line, and the y-nullcline is the dashed line.

(2) in the neighborhood of this critical point, we can look at the corresponding linear

system obtained from Eq. (13) of Section 9.3:

(

− 2σ

X − α

Y −α

−α

Y ϵ

− 2σ

Y − α

(

. (35)

We now use Eq. (35) to determine the conditions under which the model described

by Eqs. (2) permits the coexistence of the two species x and y. Of the four possible

cases shown in Figure 9.4.5, coexistence is possible only in cases (c) and (d). In these

cases, the nonzero values of X and Y are obtained by solving the algebraic equations

(34); the result is

X =

− ϵ

− α

, Y =

− ϵ

− α

. (36)

Further, since ϵ

− σ

X − α

Y = 0 and ϵ

− σ

Y − α

X = 0, Eq. (35) immediately

reduces to

(

−σ

X −α

−α

Y −σ

(

. (37)

The eigenvalues of the system (37) are found from the equation

+ (σ

X + σ

Y)r + (σ

− α

)XY = 0. (38)

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9.4 Competing Species 541

Thus

1,2

−(σ

X + σ

Y) ±

(σ

X + σ

− 4(σ

− α

)XY

. (39)

If σ

− α

< 0, then the radicand of Eq. (39) is positive and greater than

(σ

X + σ

. Thus the eigenvalues are real and opposite in sign. Consequently,

the critical point (X , Y) is an (unstable) saddle point, and coexistence is not pos-

sible. This is the case in Example 2, where σ

= 1, α

= 1, σ

= 0.25, α

= 0.75, and

− α

=−0.5.

On the other hand, if σ

− α

> 0, then the radicand of Eq. (39) is less than

(σ

X + σ

. Thus the eigenvalues are real, negative, and unequal, or complex with

negative real part. A straightforward analysis of the radicand of Eq. (39) shows that

the eigenvalues cannot be complex (see Problem 7). Thus the critical point is an

asymptotically stable node, and sustained coexistence is possible. This is illustrated

by Example 1, where σ

= 1, α

= 1, σ

= 1, α

= 0.5, and σ

− α

= 0.5.

Let us relate this result to Figures 9.4.5c and 9.4.5d. In Figure 9.4.5c we have

or ϵ

>ϵ

and

or ϵ

>ϵ

. (40)

These inequalities, coupled with the condition that X and Y given by Eqs. (36) be

positive, yield the inequality σ

<α

. Hence in this case the critical point is a

saddle point. On the other hand, in Figure 9.4.5d we have

or ϵ

<ϵ

and

or ϵ

<ϵ

. (41)

Now the condition that X and Y be positive yields σ

>α

. Hence the critical

point is asymptotically stable. For this case we can also show that the other

critical points (0, 0), (ϵ

/σ

,0), and (0, ϵ

/σ

) are unstable.Thus for any positive initial

values of x and y, the two populations approach the equilibrium state of coexistence

given by Eqs. (36).

Equations (2) provide the biological interpretation of the result that whether coex-

istence occurs depends on whether σ

− α

is positive or negative. The σ’s are

a measure of the inhibitory effect that the growth of each population has on itself,

whereas the α’s are a measure of the inhibiting effect that the growth of each popu-

lation has on the other species. Thus, when σ

>α

, interaction (competition) is

“weak” and the species can coexist; when σ

<α

, interaction (competition)

is “strong” and the species cannot coexist—one must die out.

PROBLEMS Each of Problems 1 through 6 can be interpreted as describing the interaction of two species

with populations x and y. In each of these problems, carry out the following steps.

(a) Draw a direction ﬁeld and describe how solutions seem to behave.

(b) Find the critical points.

eigenvectors of the linear system; classify each critical point as to type,and determine whether

it is asymptotically stable, stable, or unstable.

(d) Sketch the trajectories in the neighborhood of each critical point.

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542 Chapter 9. Nonlinear Differential Equations and Stability

(e) Compute and plot enough trajectories of the given system to show clearly the behavior of

the solutions.

(f) Determine the limiting behavior of x and y as t →∞, and interpret the results in terms

of the populations of the two species.

dx/dt = x(1.5 − x − 0.5y)

dy/dt = y(2 − y − 0.75x)

2. dx/dt = x(1.5 − x − 0.5y)

dy/dt = y(2 − 0.5y − 1.5x)

dx/dt = x(1.5 − 0.5x − y)

dy/dt = y(2 − y − 1.125x )

4. dx/dt = x(1.5 − 0.5x − y)

dy/dt = y(0.75 − y − 0.125x)

dx/dt = x(1 − x − y)

dy/dt = y(1.5 − y − x)

6. dx/dt = x(1 − x + 0.5y)

dy/dt = y(2.5 − 1.5y + 0.25x)

7. Consider the eigenvalues given by Eq. (39) in the text. Show that

(σ

X + σ

− 4(σ

− α

)XY = (σ

X − σ

+ 4α

XY.

Hence conclude that the eigenvalues can never be complex.

8. Two species of ﬁsh that compete with each other for food, but do not prey on each other,

are bluegill and redear. Suppose that a pond is stocked with bluegill and redear, and let

x and y be the populations of bluegill and redear, respectively, at time t. Suppose further

that the competition is modeled by the equations

dx/dt = x(ϵ

− σ

x − α

y),

dy/dt = y(ϵ

− σ

y − α

x).

(a) If ϵ

/α

>ϵ

/σ

and ϵ

/σ

>ϵ

/α

, show that the only equilibrium populations in the

pond are no ﬁsh, no redear, or no bluegill. What will happen for large t?

(b) If ϵ

/σ

>ϵ

/α

and ϵ

/α

>ϵ

/σ

, show that the only equilibrium populations in the

pond are no ﬁsh, no redear, or no bluegill. What will happen for large t?

9. Consider the competition between bluegill and redear mentioned in Problem 8. Suppose

that ϵ

/α

>ϵ

/σ

and ϵ

/α

>ϵ

/σ

, so,as shown in the text, there is a stable equilibrium

point at which both species can coexist. It is convenient to rewrite the equationsof Problem

8 in terms of the carrying capacity of the pond for bluegill (B = ϵ

/σ

) in the absence of

redear and its carrying capacity for redear (R = ϵ

/σ

) in the absence of bluegill.

(a) Show that the equations of Problem 8 take the form

= ϵ

1 −

x −

(

= ϵ

1 −

y −

(

where γ

= α

/σ

and γ

= α

/σ

. Determine the coexistence equilibrium point (X, Y) in

terms of B, R, γ

, and γ

(b) Now suppose that an angler ﬁshes only for bluegill with the effect that B is reduced.

What effect does this have on the equilibrium populations? Is it possible, by ﬁshing, to

reduce the population of bluegill to such a level that they will die out?

10. Consider the system (2) in the text, and assume that σ

− α

= 0.

(a) Find all the critical points of the system. Observe that the result depends on whether

− α

is zero.

(b) If σ

− α

> 0, classify each critical point and determine whether it is asymptoti-

cally stable, stable, or unstable. Note that Problem 5 is of this type. Then do the same if

− α

< 0.

− α

= 0.

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9.4 Competing Species 543

11. Consider the system (3) in Example 1 of the text. Recall that this system has an asymptot-

ically stable critical point at (0.5, 0.5), corresponding to the stable coexistence of the two

population species. Now suppose that immigration or emigration occurs at the constant

rates of δa and δb for the species x and y, respectively. In this case Eqs. (3) are replaced by

dx/dt = x(1 − x − y) + δa,

(i)

dy/dt = y(0.75 − y − 0.5x) + δb.

The question is what effect this has on the location of the stable equilibrium point.

(a) To ﬁnd the new critical point, we must solve the equations

x(1 − x − y) + δa = 0,

(ii)

y(0.75 − y − 0.5x) + δb = 0.

One way to proceed is to assume that x and y are given by power series in the parameter

δ; thus

x = x

+ x

δ +···, y = y

+ y

δ +···. (iii)

Substitute Eqs. (iii) into Eqs. (ii) and collect terms according to powers of δ.

(b) From the constant terms (the terms not involving δ), show that x

= 0.5 and y

= 0.5,

thus conﬁrming that in the absence of immigration or emigration, the critical point is

(0.5, 0.5).

= 4a − 4b, y

=−2a + 4b. (iv)

(d) Suppose that a > 0 and b > 0 so that immigration occurs for both species. Show that

the resulting equilibrium solution may represent an increase in both populations, or an

increase in one but a decrease in the other. Explain intuitively why this is a reasonable

result.

12.

The system

′

=−y, y

′

=−γy − x(x − 0.15)(x − 2)

results from an approximation to the Hodgkin–Huxley

equations, which model the

transmission of neural impulses along an axon.

(a) Find the critical points, and classify them by investigating the approximate linear

system near each one.

(b) Draw phase portraits for γ = 0.8 and for γ = 1.5.

which this trajectory ultimately approaches the origin as t →∞. Draw a phase portrait

for this value of γ.

Bifurcation Points. Consider the system

′

= F(x, y, α), y

′

= G(x, y, α), (i)

where α is a parameter. The equations

F(x, y, α) = 0, G(x, y, α) = 0 (ii)

Sir Alan L. Hodgkin (1914–1998) and Sir Andrew F. Huxley (1917– ), British physiologists and bio-

physicists, studied the excitation and transmission of neural impulses at Cambridge University and the

Marine Biological Association Laboratory in Plymouth. This work was both theoretical (resulting in a

system of nonlinear differential equations) and experimental (involving measurements on the giant axon

of the Atlantic squid). They were awarded the Nobel Prize in Physiology or Medicine in 1963.

August 7, 2012 21:05 c09 Sheet number 50 Page number 544 cyan black

544 Chapter 9. Nonlinear Differential Equations and Stability

determine the x- and y-nullclines, respectively; any point where an x-nullcline and a

y-nullcline intersect is a critical point. As α varies and the conﬁguration of the nullclines

changes, it may well happen that, at a certain value of α, two critical points coalesce into one.

For further variation in α, the critical point may once again separate into two critical points,

or it may disappear altogether. Or the process may occur in reverse: For a certain value of α,

two formerly nonintersecting nullclines may come together, creating a critical point, which,

for further changes in α, may split into two. A value of α at which such phenomena occur is

a bifurcation point. It is also common for a critical point to experience a change in its type

and stability properties at a bifurcation point. Thus both the number and the kind of critical

points may change abruptly as α passes through a bifurcation point. Since a phase portrait of a

system is very dependent on the location and nature of the critical points, an understanding of

bifurcations is essential to an understanding of the global behavior of the system’s solutions.

In each of Problems 13 through 16:

(a) Sketch the nullclines and describe how the critical points move as α increases.

(b) Find the critical points.

linear system. Draw a phase portrait in a rectangle containing the critical points.

(d) Find the bifurcation point α

at which the critical points coincide. Locate this critical point,

and ﬁnd the eigenvalues of the approximate linear system. Draw a phase portrait.

(e) For α>α

there are no critical points. Choose such a value of α and draw a phase portrait.

13.

′

=−4x + y + x

, y

′

α − y

14.

′

α − y, y

′

=−4x + y + x

15. x

′

=−4x + y + x

, y

′

=−α − x + y

16.

′

=−α − x + y, y

′

=−4x + y + x

Problems 17 through 19 deal with competitive systems much like those in Examples 1 and 2,

except that some coefﬁcients depend on a parameter α. In each of these problems,assume that

x, y, and α are always nonnegative. In each of Problems 17 through 19:

(a) Sketch the nullclines in the ﬁrst quadrant, as in Figure 9.4.5. For different ranges of α your

sketch may resemble different parts of Figure 9.4.5.

(b) Find the critical points.

(d) Find the Jacobian matrix J, and evaluate it for each of the critical points.

(e) Determine the type and stability property of each critical point. Pay particular attention

to what happens as α passes through a bifurcation point.

(f) Draw phase portraits for the system for selected values of α to conﬁrm your conclusions.

17. dx/dt = x(1 − x − y), dy/dt = y(α − y − 0.5x)

18. dx/dt = x(1 − x − y), dy/dt = y(0.75 − αy − 0.5x )

19. dx/dt = x(1 − x − y), dy/dt = y[α − y − (2α − 1)x]

9.5 Predator–Prey Equations

In the preceding section we discussed a model of two species that interact by

competing for a common food supply or other natural resource. In this section we

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9.5 Predator–Prey Equations 545

investigate the situation inwhich one species (thepredator) preys on theother species

(the prey), while the prey lives on a different source of food. For example, consider

foxes and rabbits in a closed forest. The foxes prey on the rabbits, the rabbits live

on the vegetation in the forest. Other examples are bass in a lake as predators and

redear as prey, and ladybugs as predators and aphids as prey. We emphasize again

that a model involving only two species cannot fully describe the complex relation-

ships among species that actually occur in nature. Nevertheless, the study of simple

models is the ﬁrst step toward an understanding of more complicated phenomena.

We will denote by x and y the populations of the prey and predator, respectively,

at time t. In constructing a model of the interaction of the two species, we make the

following assumptions:

1. In the absence of the predator, the prey grows at a rate proportional to the current

population; thus dx/dt = ax, a > 0, when y = 0.

2. In the absence of the prey, the predator dies out; thus dy/dt =−cy, c > 0, when x = 0.

3. The number of encounters between predator and prey is proportional to the product

of their populations. Each such encounter tends to promote the growth of the predator

and to inhibit the growth of the prey. Thus the growth rate of the predator is increased by

a term of the form γxy, while the growth rate of the prey is decreased by a term −αxy,

where γ and α are positive constants.

As a consequence of these assumptions, we are led to the equations

dx/dt = ax − αxy = x(a − αy),

(1)

dy/dt =−cy + γxy = y(−c + γx).

The constants a, c, α, and γ are all positive; a and c are the growth rate of the prey

and the death rate of the predator,respectively,and α and γ are measures of the effect

of the interaction between the two species. Equations (1) are known as the Lotka–

Volterra equations.Theywere developed inpapers byLotka

in 1925 and byVolterra

in 1926.Although these are rather simple equations, they do characterize a wide class

of problems. Ways of making them more realistic are discussed at the end of this

section and in the problems. Our goal here is to determine the qualitative behavior

of the solutions (trajectories) of the system (1) for arbitrary positive initial values of x

and y. We ﬁrst do this for a speciﬁc example and then return to the general equations

(1) at the end of the section.

Alfred J. Lotka (1880–1949),anAmericanbiophysicist,was born in what is now Ukraine and was educated

mainly in Europe. He is remembered chieﬂy for his formulation of the Lotka–Volterra equations. He was

also the author, in 1924, of the ﬁrst book on mathematical biology; it is now available as Elements of

Mathematical Biology (New York: Dover, 1956).

Vito Volterra (1860–1940), a distinguished Italian mathematician, held professorships at Pisa, Turin,

and Rome. He is particularly famous for his work in integral equations and functional analysis. Indeed,

one of the major classes of integral equations is named for him; see Problem 21 of Section 6.6. His theory

of interacting species was motivated by data collected by a friend, D’Ancona, concerning ﬁsh catches

in the Adriatic Sea. A translation of his 1926 paper can be found in an appendix to R. N. Chapman,

Animal Ecology with Special Reference to Insects (New York: McGraw-Hill, 1931).

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546 Chapter 9. Nonlinear Differential Equations and Stability

EXAMPLE

Discuss the solutions of the system

dx/dt = x(1 − 0.5y) = x − 0.5xy = F(x, y),

(2)

dy/dt = y(−0.75 + 0.25x) =−0.75y + 0.25xy = G(x, y)

for x and y positive.

The critical points of this system are the solutions of the algebraic equations

x(1 − 0.5y) = 0, y(−0.75 + 0.25x) = 0, (3)

namely, the points (0, 0) and (3, 2).

Figure 9.5.1 shows the critical points and a direction ﬁeld

for

the system (2). From this ﬁgure it appears that trajectories in the ﬁrst quadrant encircle the

critical point (3, 2). Whether the trajectories are actually closed curves, or whether they slowly

spiral in or out, cannot be deﬁnitely determined from the direction ﬁeld. The origin appears to

be a saddle point. Just as for the competition equations in Section 9.4, the coordinate axes are

trajectories of Eqs. (1) or (2). Consequently, no other trajectory can cross a coordinate axis,

which means that every solution starting in the ﬁrst quadrant remains there for all time.

7654321

FIGURE 9.5.1 Critical points and direction ﬁeld for the predator–prey system (2).

Next we examine the local behavior of solutions near each critical point. Near the origin

we can neglect the nonlinear terms in Eqs. (2) to obtain the corresponding linear system

0 −0.75

. (4)

The eigenvalues and eigenvectors of Eq. (4) are

= 1, ξ

(1)

; r

=−0.75, ξ

(2)

, (5)

so its general solution is

= c

+ c

−0.75t

. (6)

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9.5 Predator–Prey Equations 547

Thus the origin is a saddle point both of the linear system (4) and of the nonlinear system (2)

and therefore is unstable. One pair of trajectories enters the origin along the y-axis; all other

trajectories depart from the neighborhood of the origin.

To examine the critical point (3, 2), we can use the Jacobian matrix

J =

(x, y) F

(x, y)

(x, y) G

(x, y)

1 − 0.5y −0.5x

0.25y −0.75 + 0.25x

. (7)

Evaluating J at the point (3, 2), we obtain the linear system

0 −1.5

0.50

, (8)

where u = x − 3 and v = y − 2. The eigenvalues and eigenvectors of this system are

√

3 i

, ξ

(1)

−i/

√

(

; r

=−

√

3 i

, ξ

(2)

√

. (9)

Since the eigenvalues are imaginary, the critical point (3, 2) is a center of the linear system

(8) and is therefore a stable critical point for that system. Recall from Section 9.3 that this

is one of the cases in which the behavior of the linear system may or may not carry over to

the nonlinear system, so the nature of the point (3, 2) for the nonlinear system (2) cannot be

determined from this information.

The simplest way to ﬁnd the trajectories of the linear system (8) is to divide the second of

Eqs. (8) by the ﬁrst so as to obtain the differential equation

dv/dt

du/dt

0.5u

−1.5v

=−

udu+ 3v dv = 0. (10)

Consequently,

+ 3v

= k, (11)

where k is an arbitrary nonnegative constant of integration. Thus the trajectories of the linear

system (8) are ellipses centered at the critical point and elongated somewhat in the horizontal

direction.

Now let us return to the nonlinear system (2). Dividing the second of Eqs. (2) by the ﬁrst,

we obtain

y(−0.75 + 0.25x)

x(1 − 0.5y)

. (12)

Equation (12) is a separable equation and can be put in the form

1 − 0.5y

dy =

−0.75 + 0.25x

dx,

from which it follows that

0.75 ln x + ln y − 0.5y − 0.25x = c, (13)

where c is a constant of integration. Although by using only elementary functions we cannot

solve Eq. (13) explicitly for either variable in terms of the other, it is possible to show that the

graph of the equation for a ﬁxed value of c is a closed curve surrounding the critical point

(3, 2). Thus the critical point is also a center of the nonlinear system (2), and the predator and

prey populations exhibit a cyclic variation.

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548 Chapter 9. Nonlinear Differential Equations and Stability

Figure 9.5.2 shows a phase portrait of the system (2). For some initial conditions, the tra-

jectory represents small variations in x and y about the critical point and is almost elliptical

in shape, as the linear analysis suggests. For other initial conditions, the oscillations in x and y

are more pronounced, and the shape of the trajectory is signiﬁcantly different from an ellipse.

Observe that the trajectories are traversed in the counterclockwise direction.The dependence

of x and y on t for a typical set of initial conditions is shown in Figure 9.5.3. Note that x and

y are periodic functions of t, as they must be since the trajectories are closed curves. Fur-

ther, the oscillation of the predator population lags behind that of the prey. Starting from a

state in which both predator and prey populations are relatively small, the prey ﬁrst increase

because there is little predation. Then the predators, with abundant food, increase in pop-

ulation also. This causes heavier predation, and the prey tend to decrease. Finally, with a

diminished food supply, the predator population also decreases, and the system returns to the

original state.

2467135

FIGURE 9.5.2 A phase portrait of the system (2).

, y

5 10 15 20 25

Prey

x(t)

Predator

y(t)

FIGURE 9.5.3 Variations of the prey and predator populations with time for the system (2).

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9.5 Predator–Prey Equations 549

The general system (1) can be analyzed in exactly the same way as in the example.

The critical points of the system (1) are the solutions of

x(a − αy) = 0, y(−c + γx) = 0,

that is, the points (0, 0) and (c/γ, a/α). We ﬁrst examine the solutions of the

corresponding linear system near each critical point.

In the neighborhood of the origin, the corresponding linear system is

(

a 0

0 −c

(

. (14)

The eigenvalues and eigenvectors are

= a, ξ

(1)

(

; r

=−c , ξ

(2)

(

, (15)

so the general solution is

(

= c

(

+ c

(

−ct

. (16)

Thus the origin is a saddle point and hence unstable. Entrance to the saddle point is

along the y-axis; all other trajectories depart from the neighborhood of the critical

point.

Next consider the critical point (c/γ, a/α). The Jacobian matrix is

J =

a − αy −αx

γy −c + γx

Evaluating J at (c/γ, a/α), we obtain the approximate linear system

(

0 −αc/γ

γa/α 0

(

, (17)

where u = x − (c/γ) and v = y − (a/α). The eigenvalues of the system (17) are

r =±i

√

ac, so the critical point is a (stable) center of the linear system. To ﬁnd

the trajectories of the system (17), we can divide the second equation by the ﬁrst to

obtain

d v

d v/dt

du/dt

=−

(γa/α)u

(αc/γ)v

, (18)

au du + α

cv dv = 0. (19)

Consequently,

+ α

= k, (20)

where k is a nonnegative constant of integration. Thus the trajectories of the linear

system (17) are ellipses, just as in the example.

Returning brieﬂy to the nonlinear system (1),observe that it can be reduced to the

single equation

dy/dt

dx/dt

y(−c + γx)

x(a − αy)

. (21)

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550 Chapter 9. Nonlinear Differential Equations and Stability

Equation (21) is separable and has the solution

a ln y − αy + c ln x − γx = C, (22)

where C is a constant of integration. Again, it is possible to show that for ﬁxed C, the

graph of Eq. (22) is a closed curve surrounding the critical point (c/γ, a/α). Thus this

critical point is also a center for the general nonlinear system (1).

The cyclic variation of the predator and prey populations can be analyzed in more

detail when the deviations from the point (c/γ, a/α) are small and the linear system

(17) can be used. The solution of the system (17) can be written in the form

u =

K cos(

√

ac t + φ), v =

K sin(

√

ac t + φ), (23)

where the constants K and φ are determined by the initial conditions. Thus

x =

K cos(

√

ac t + φ),

(24)

y =

K sin(

√

ac t + φ).

These equations are good approximations for the nearly elliptical trajectories close

to the critical point (c/γ, a/α). We can use them to draw several conclusions about

the cyclic variation of the predator and prey on such trajectories.

1. The sizes of the predator and prey populations vary sinusoidally with period 2π/

√

ac. This

period of oscillation is independent of the initial conditions.

2. The predator and prey populations are out of phase by one-quarter of a cycle. The prey

leads and the predator lags, as explained in the example.

3. The amplitudes of the oscillations are Kc/γ for the prey and a

√

cK/α

√

a for the predator

and hence depend on the initial conditions as well as on the parameters of the problem.

4. The average populations of predator and prey over one complete cycle are c/γ and a/α,

respectively. These are the same as the equilibrium populations; see Problem 10.

Cyclic variations of predator and prey as predicted by Eqs. (1) have been observed

in nature. One striking example is described by Odum (pp. 191–192); based on the

records of the Hudson’s Bay Company of Canada, the abundance of lynx and snow-

shoe hare, as indicated by the number of pelts turned in over the period 1845–1935,

shows a distinct periodic variation with period of 9 to 10 years. The peaks of abun-

dance are followed by very rapid declines, and the peaks of abundance of the lynx

and hare are out of phase, with that of the hare preceding that of the lynx by a year

or more.

Since the critical point (c/γ, a/α) is a center, we expect that small perturbations

of the Lotka–Volterra equations may well lead to solutions that are not periodic.

To put it another way, unless the Lotka–Volterra equations exactly describe a given

predator–prey relationship, the actual ﬂuctuations of the populations may differ

substantially from those predicted by the Lotka-Volterra equations, due to small

inaccuracies in the model equations. This has led to many attempts

to replace the

See the book by Brauer and Castillo-Chávez listed in the references for an extensive discussion of

alternative models for predator–prey relationships.

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9.5 Predator–Prey Equations 551

Lotka–Volterra equations by other systems that are less susceptible to the effects of

small perturbations. Problem 13 introduces one such alternative model.

Another criticism of the Lotka–Volterra equations is that in the absence of

the predator, the prey will grow without bound. This can be corrected by allow-

ing for the natural inhibiting effect that an increasing population has on the growth

rate of that population. For example, the ﬁrst of Eqs. (1) can be modiﬁed so that

when y = 0, it reduces to a logistic equation for x. The effects of this modiﬁcation are

explored in Problems 11 and 12. Problems 14 through 16 deal with harvesting in a

predator–prey relationship. The results may seem rather counterintuitive.

Finally, we repeat a warning stated earlier: relationships among species in the

natural world are often complex and subtle. You should not expect too much of a

simple system of two differential equations in describing such relationships. Even if

you are convinced that the general form of the equations is sound, the determination

of numerical values for the coefﬁcients may present serious difﬁculties.

PROBLEMS Each of Problems 1 through 5 can be interpreted as describing the interaction of two species

with population densities x and y. In each of these problems, carry out the following steps.

(a) Draw a direction ﬁeld and describe how solutions seem to behave.

(b) Find the critical points.

eigenvectors of the linear system; classify each critical point as to type,and determine whether

it is asymptotically stable, stable, or unstable.

(d) Sketch the trajectories in the neighborhood of each critical point.

(e) Draw a phase portrait for the system.

(f) Determine the limiting behavior of x and y as t →∞, and interpret the results in terms

of the populations of the two species.

dx/dt = x(1.5 − 0.5y)

dy/dt = y(−0.5 + x)

2. dx/dt = x(1 − 0.5y)

dy/dt = y(−0.25 + 0.5x)

dx/dt = x(1 − 0.5x − 0.5y)

dy/dt = y(−0.25 + 0.5x)

4. dx/dt = x(1.125 − x − 0.5y)

dy/dt = y(−1 + x)

dx/dt = x(−1 + 2.5x − 0.3y − x

)

dy/dt = y(−1.5 + x)

6. In this problem we examine the phase difference between the cyclic variations of the

predator and prey populations as given by Eqs. (24) of this section. Suppose we assume

that K > 0 and that t is measured from the time that the prey population x is a maximum;

then φ = 0.

(a) Show that the predator population y is a maximum at t = π/2

√

ac = T/4, where T is

the period of the oscillation.

(b) When is the prey population increasing most rapidly? decreasing most rapidly? a

minimum?

(d) Draw a typical elliptic trajectory enclosing the point (c/γ, a/α), and mark on it the

points found in parts (a), (b), and (c).

(a) Findthe ratio of theamplitudes of the oscillations ofthe prey and predator populations

about the critical point (c/γ, a/α), using the approximation (24), which is valid for small

oscillations. Observe that the ratio is independent of the initial conditions.

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552 Chapter 9. Nonlinear Differential Equations and Stability

(b) Evaluate the ratio found in part (a) for the system (2).

Figure 9.5.3. Does the result agree with that obtained from the linear approximation?

(d) Determine the prey–predator amplitude ratio for other solutions of the system (2)

—that is, for solutions satisfying other initial conditions. Is the ratio independent of the

initial conditions?

(a) Find the period of the oscillations of the prey and predator populations, using the

approximation (24), which is valid for small oscillations. Note that the period is inde-

pendent of the amplitude of the oscillations.

(b) For the solution of the nonlinear system (2) shown in Figure 9.5.3,estimate the period

as well as possible. Is the result the same as for the linear approximation?

conditions—and determine their periods. Is the period the same for all initial conditions?

Consider the system

dx/dt = ax[1 − (y/2)], dy/dt = by[−1 + (x/3)],

where a and b are positive constants. Observe that this system is the same as in the example

in the text if a = 1 and b = 0.75. Suppose the initial conditions are x(0) = 5 and y(0) = 2.

(a) Let a = 1 and b = 1. Plot the trajectory in the phase plane, and determine (or esti-

mate) the period of the oscillation.

(b) Repeat part (a) for a = 3 and a = 1/3, with b = 1.

(d) Describe how the period and the shape of the trajectory depend on a and b.

10.

The average sizes of the prey and predator populations are deﬁned as

x =

A+T

x(t) dt, y =

A+T

y(t) dt,

respectively, where T is the period of a full cycle, and A is any nonnegative constant.

(a) Using the approximation (24), which is valid near the critical point, show that

x = c/γ

and

y = a/α.

(b) For the solution of the nonlinear system (2) shown in Figure 9.5.3, estimate

x and y as

well as you can. Try to determine whether

x and y are given by c/γ and a/α, respectively,

in this case.

Hint: Consider how you might estimate the value of an integral even though you do not

have a formula for the integrand.

conditions—and determine

x and y for these solutions. Are the values of x and y the same

for all solutions?

In Problems 11 and 12, we consider the effect of modifying the equation for the prey x by

including a term −σx

so that this equation reduces to a logistic equation in the absence of

the predator y. Problem 11 deals with a speciﬁc system of this kind, and Problem 12 takes

up this modiﬁcation to the general Lotka–Volterra system. The systems in Problems 3 and

4 are other examples of this type.

11. Consider the system

′

= x(1 − σx − 0.5y), y

′

= y(−0.75 + 0.25x),

where σ>0. Observe that this system is a modiﬁcation of the system (2) in Example 1.

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9.5 Predator–Prey Equations 553

(a) Find all of the critical points. How does their location change as σ increases from zero?

Observe that there is a critical point in the interior of the ﬁrst quadrant only if σ<1/3.

(b) Determine the type and stability property of each critical point. Find the value

< 1/3 where the nature of the critical point in the interior of the ﬁrst quadrant changes.

Describe the change that takes place in this critical point as σ passes through σ

; for a

value of σ between σ

and 1/3.

(d) Describe the effect on the two populations as σ increases from zero to 1/3.

12. Consider the system

dx/dt = x(a − σx − αy), dy/dt = y(−c + γx),

where a, σ, α, c, and γ are positive constants.

(a) Find all critical points of the given system. How does their location change as σ

increases from zero? Assume that a/σ > c/γ, that is, σ<aγ/c. Why is this assumption

necessary?

(b) Determine the nature and stability characteristics of each critical point.

interior of the ﬁrst quadrant changes from a spiral point to a node.

(d) Describe the effect on the two populations as σ increases from zero to aγ/c.

13. In the Lotka–Volterra equations, the interaction between the two species is modeled by

terms proportional to the product xy of the respective populations. If the preypopulation is

much larger than the predator population,this may overstate the interaction;for example,

a predator may hunt only when it is hungry and ignore the prey at other times. In this

problem we consider an alternative model proposed by Rosenzweig and MacArthur.

(a) Consider the system

′

= x

1 − 0.2x −

x + 6

(

, y

′

= y

−0.25 +

x + 6

(

Find all of the critical points of this system.

(b) Determine the type and stability characteristics of each critical point.

Harvesting in a Predator–Prey Relationship. In a predator–prey situation,it may happen

that one or perhaps both species are valuable sources of food. Or, the prey species may

be regarded as a pest, leading to efforts to reduce its numbers. In a constant-effort model

of harvesting, we introduce a term −E

x in the prey equation and a term −E

y in the

predator equation, where E

and E

are measures of the effort invested in harvesting

the respective species. A constant-yield model of harvesting is obtained by including the

term −H

in the prey equation and the term −H

in the predator equation. The constants

, E

, H

, and H

are always nonnegative. Problems 14 and 15 deal with constant-effort

harvesting, and Problem 16 deals with constant-yield harvesting.

14. Applying a constant-effort model of harvesting to the Lotka–Volterra equations (1), we

obtain the system

′

= x(a − αy − E

), y

′

= y(−c + γx − E

When there is no harvesting, the equilibrium solution is (c/γ, a/α).

See the book by Brauer and Castillo-Chávez for further details.

August 7, 2012 21:05 c09 Sheet number 60 Page number 554 cyan black

554 Chapter 9. Nonlinear Differential Equations and Stability

(a) Before doing any mathematical analysis, think about the situation intuitively. How do

you think the populations will change if the prey alone is harvested? if the predator alone

is harvested? if both are harvested?

(b) How does the equilibrium solution change if the prey is harvested, but not the preda-

tor (E

> 0, E

= 0)?

prey (E

= 0, E

> 0)?

(d) How does the equilibrium solution change if both are harvested (E

> 0, E

> 0)?

15. If we modify the Lotka–Volterra equations by including a self-limiting term −σx

in the

prey equation, and then assume constant-effort harvesting, we obtain the equations

′

= x(a − σx − αy − E

), y

′

= y(−c + γx − E

In the absence of harvesting, the equilibrium solution of interest is x = c/γ,

y = (a/α) − (σc)/(αγ).

(a) How does the equilibrium solution change if the prey is harvested, but not the preda-

tor (E

> 0, E

= 0)?

(b) How does the equilibrium solution change if the predator is harvested, but not the

prey (E

= 0, E

> 0)?

> 0, E

> 0)?

16. In this problem we apply a constant-yield modelof harvesting to the situation in Example 1.

Consider the system

′

= x(1 − 0.5y) − H

, y

′

= y(−0.75 + 0.25x) − H

where H

and H

are nonnegative constants. Recall that if H

= H

= 0, then (3, 2) is an

equilibrium solution for this system.

(a) Before doing any mathematical analysis, think about the situation intuitively. How do

you think the populations will change if the prey alone is harvested? if the predator alone

is harvested? if both are harvested?

(b) How does the equilibrium solution change if the prey is harvested, but not the preda-

tor (H

> 0, H

= 0)?

prey (H

= 0, H

> 0)?

(d) How does the equilibrium solution change if both are harvested (H

> 0, H

> 0)?

9.6 Liapunov’s Second Method

In Section 9.3 we showed how the stability of a critical point of a locally linear

system can usually be determined from a study of the corresponding linear system.

However, no conclusion can be drawn when the critical point is a center of the

corresponding linear system. Examples of this situation are the undamped pendulum,

Eqs. (1) and (2) below, and the predator–prey problem discussed in Section 9.5. Also,

for an asymptotically stable critical point, it may be important to investigate the

basin of attraction—that is, the domain such that all solutions starting within that

domain approach the critical point. The theory of locally linear systems provides no

information about this question.

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9.6 Liapunov’s Second Method 555

In this section we discuss another approach,known as Liapunov’s

second method

or direct method.The method is referred to as a direct method because no knowledge

of the solution of the system of differential equations is required. Rather,conclusions

about the stability or instability of a critical point are obtained by constructing a suit-

able auxiliary function. The technique is a very powerful one that provides a more

global type of information—for example, an estimate of the extent of the basin of

attraction of a critical point. Liapunov’s second method can also be used to study

systems of equations that are not locally linear; however, we will not discuss such

problems.

The Pendulum Equations. Basically, Liapunov’s second method is a generalization of

two physical principles for conservative systems, namely, (i) a rest position is stable

if the potential energy is a local minimum, otherwise it is unstable, and (ii) the total

energy is a constant during any motion. To illustrate these concepts, again consider

the undamped pendulum (a conservative mechanical system), which is governed

by the equation

sin θ = 0. (1)

The corresponding system of ﬁrst order equations is

= y,

=−

sin x, (2)

where x = θ and y = dθ/dt. If we omit an arbitrary constant, the potential energy U

is the work done in lifting the pendulum above its lowest position, namely,

U(x, y) = mgL(1 − cos x); (3)

see Figure 9.2.2. The critical points of the system (2) are x =±nπ, y = 0, for

n = 0, 1,2, 3, ..., corresponding to θ =±nπ, dθ/dt = 0. Physically, we expect the

points x = 0,y = 0;x =±2π,y = 0; ...,corresponding to θ = 0, ±2π, ..., to be stable,

since for them the pendulum bob is vertical with the weight down. Further,we expect

the points x =±π, y = 0; x =±3π, y = 0; ..., corresponding to θ =±π, ±3π, ...,to

unstable, since for them the pendulum bob is vertical with the weight up. This

agrees with statement (i), for at the former points U is a minimum equal to zero, and

at the latter points U is a maximum equal to 2mgL.

Next consider the total energy V, which is the sum of the potential energy U and

the kinetic energy

(d θ/dt)

. In terms of x and y,

V(x, y) = mgL(1 − cos x) +

. (4)

Alexander M. Liapunov (1857–1918), a student of Chebyshev at St. Petersburg, taught at the Uni-

versity of Kharkov from 1885 to 1901, when he became an academician in applied mathematics at the

St. Petersburg Academy of Sciences. In 1917 he moved to Odessa because of his wife’s frail health. His

research in stability encompassed both theoretical analysis and applications to various physical problems.

His second method formed part of his most inﬂuential work, General Problem of Stability of Motion,

published in 1892.

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556 Chapter 9. Nonlinear Differential Equations and Stability

On a trajectory corresponding to a solution x = φ(t), y = ψ(t) of Eqs. (2), V can be

considered a function of t. The derivative of V[φ(t), ψ(t)] with respect to t is called

the rate of change of V following the trajectory. By the chain rule,

dV[φ(t), ψ(t)]

= V

[φ(t), ψ(t)]

d φ(t)

+ V

[φ(t), ψ(t)]

d ψ(t)

= (mgL sin x)

+ mL

, (5)

where it is understood that x = φ(t), y = ψ(t). Finally,substituting in Eq. (5) for dx/dt

and dy/dt from Eqs. (2), we ﬁnd that dV/dt = 0. Hence V is a constant along any

trajectory of the system (2), which is statement (ii).

It is important to note that at any point (x, y), the rate of change of V along the

trajectory through that point was computed without actually solving the system (2).

It is precisely this fact that enables us to use Liapunov’s second method for systems

whose solution we do not know, which is the main reason for its importance.

At the stable critical points, x =±2nπ, y = 0, n = 0, 1, 2, ..., the energy V is zero.

If the initial state (x

, y

) of the pendulum is sufﬁciently near a stable critical point,

then the energy V(x

, y

) is small, and the motion (trajectory) associated with this

energy stays close to the critical point. It can be shown that if V(x

, y

) is sufﬁciently

small, then the trajectory is closed and contains the critical point. For example, sup-

pose that (x

, y

) is near (0, 0) and that V(x

, y

) is very small. The equation of the

trajectory with energy V(x

, y

) is

V(x, y) = mgL(1 − cos x) +

= V(x

, y

For x small, we have 1 − cos x = 1 − (1 − x

/2!+···)

∼

/2. Thus the equation of

the trajectory is approximately

mgLx

= V(x

, y

2V(x

, y

)/mgL

2V(x

, y

)/mL

= 1.

This is an ellipse enclosing the critical point (0, 0);the smaller V(x

, y

) is, the smaller

are the major and minor axes of the ellipse. Physically, the closed trajectory corre-

sponds to a solution that is periodic in time—the motion is a small oscillation about

the equilibrium point.

If damping is present, however, it is natural to expect that the amplitude of the

motion decays in time and that the stable critical point (center) becomes an asymp-

totically stable critical point (spiral point). See the phase portrait for the damped

pendulum in Figure 9.3.5. This can almost be argued from a consideration of dV/dt.

For the damped pendulum, the total energy is still given by Eq. (4), but now, from

Eqs. (13) of Section 9.2,dx /dt = y and dy/dt =−(g/L) sin x − (c/mL)y. Substituting

for dx/dt and dy/dt in Eq. (5) gives dV/dt =−cLy

≤ 0. Thus the energy is nonin-

creasing along any trajectory, and except for the line y = 0, the motion is such that

the energy decreases. Hence each trajectory must approach a point of minimum

energy—a stable equilibrium point. If dV/dt < 0 instead of dV/dt ≤ 0, it is reason-

able to expect that this would be true for all trajectories that start sufﬁciently close

to the origin.

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9.6 Liapunov’s Second Method 557

General Systems. To pursue these ideas further, consider the autonomous system

dx/dt = F(x, y), dy/dt = G(x, y), (6)

and suppose that the point x = 0, y = 0 is an asymptotically stable critical point.

Then there exists some domain D containing (0, 0) such that every trajectory that

starts in D must approach the origin as t →∞. Suppose that there exists an“energy”

function V such that V ≥ 0 for (x, y) in D with V = 0 only at the origin. Since each

trajectory in D approaches the origin as t →∞, then following any particular trajec-

tory, V decreases to zero as t approaches inﬁnity. The type of result we want to prove

is essentially the converse: if, on every trajectory, V decreases to zero as t increases,

then the trajectories must approach the origin as t →∞, and hence the origin is

asymptotically stable. First, however, it is necessary to make several deﬁnitions.

Let V be deﬁned on some domain D containing the origin. Then V is said to

be positive deﬁnite on D if V(0, 0) = 0 and V(x, y)>0

for all other points in D

Similarly, V is said to be negative deﬁnite on D if V(0, 0) = 0 and V(x, y)<0 for all

other points in D. If the inequalities > and < are replaced by ≥ and ≤, then V is

said to be positive semideﬁnite and negative semideﬁnite,respectively.We emphasize

that when we speak of a positive deﬁnite (negative deﬁnite, …) function on a domain

D containing the origin,the functionmust bezero atthe origin in additionto satisfying

the proper inequality at all other points in D.

EXAMPLE

The function

V(x, y) = sin(x

+ y

)

is positive deﬁnite on x

+ y

<π/2 since V(0, 0) = 0 and V(x, y)>0 for 0 < x

+ y

<π/2.

However, the function

V(x, y) = (x + y)

is only positive semideﬁnite since V(x, y) = 0 on the line y =−x.

We also want to consider the function

V(x, y) = V

(x, y)F(x, y) + V

(x, y)G(x, y), (7)

where F and G are the same functions as in Eqs. (6). We choose this notation

because

V(x, y) can be identiﬁed as the rate of change of V along the trajectory

of the system (6) that passes through the point (x , y). That is, if x = φ(t), y = ψ(t) is

a solution of the system (6), then

dV[φ(t), ψ(t)]

= V

[φ(t), ψ(t)]

d φ(t)

+ V

[φ(t), ψ(t)]

d ψ(t)

= V

(x, y)F(x, y) + V

(x, y)G(x, y)

V(x, y). (8)

The function

V is sometimes referred to as the derivative of V with respect to the

system (6).

We now state two Liapunov theorems, the ﬁrst dealing with stability, the second

with instability.

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558 Chapter 9. Nonlinear Differential Equations and Stability

Theorem 9.6.1

Suppose that the autonomous system (6) has an isolated critical point at the ori-

gin. If there exists a function V that is continuous and has continuous ﬁrst partial

derivatives, that is positive deﬁnite, and for which the function

V given by Eq. (7)

is negative deﬁnite on some domain D in the xy-plane containing (0, 0), then the

origin is an asymptotically stable critical point. If

V is negative semideﬁnite, then

the origin is a stable critical point.

Theorem 9.6.2

Let the origin be an isolated critical point of the autonomous system (6). Let V be a

function that is continuous and has continuous ﬁrst partial derivatives. Suppose that

V(0, 0) = 0 and that in every neighborhood of the origin there is at least one point

at which V is positive (negative). If there exists a domain D containing the origin

such that the function

V given by Eq. (7) is positive deﬁnite (negative deﬁnite) on

D, then the origin is an unstable critical point.

The function V is called a Liapunov function. Before sketching geometrical argu-

ments forTheorems 9.6.1 and 9.6.2,we note that the difﬁculty in using these theorems

is that they tell us nothing about how to construct a Liapunov function, assuming that

one exists. In cases where the autonomous system (6) represents a physical prob-

lem, it is natural to consider ﬁrst the actual total energy function of the system as

a possible Liapunov function. However, Theorems 9.6.1 and 9.6.2 are applicable in

cases where the concept of physical energy is not pertinent. In such cases a judicious

trial-and-error approach may be necessary.

Now consider the second part of Theorem 9.6.1—that is, the case

V ≤ 0. Let

c ≥ 0 be a constant, and consider the curve in the xy-plane given by V(x, y) = c.

For c = 0 the curve reduces to the single point x = 0, y = 0. We assume that if

0 < c

< c

, then the curve V(x, y) = c

contains the origin and lies within the curve

V(x, y) = c

, as illustrated in Figure 9.6.1a. We show that a trajectory starting inside

a closed curve V(x , y) = c cannot cross to the outside. Thus, given a circle of radius

ϵ about the origin, by taking c sufﬁciently small, we can ensure that every trajectory

starting inside the closed curve V(x, y) = c stays within the circle of radius ϵ; indeed,

it stays within the closed curve V(x, y) = c itself. Thus the origin is a stable critical

point.

To show this, recall from calculus that the vector

∇V(x, y) = V

(x, y)i + V

(x, y)j, (9)

known as the gradient of V, is normal to the curve V(x, y) = c and points in the direc-

tion of increasing V. In the present case, V increases outward from the origin, so ∇V

points away from the origin, as indicated in Figure 9.6.1b. Next, consider a trajectory

x = φ(t), y = ψ(t) of the system (6), and recall that the vector T(t) = φ

′

(t)i + ψ

′

(t)j

is tangent to the trajectory at each point; see Figure 9.6.1b. Let x

= φ(t

), y

= ψ(t

)

be a point of intersection of the trajectory and a closed curve V(x, y) = c. At this

point φ

′

) = F(x

, y

), ψ

′

) = G(x

, y

), so from Eq. (7) we obtain

V(x

, y

) = V

, y

)φ

′

) + V

, y

)ψ

′

)

=[V

, y

)i + V

, y

)j] · [φ

′

)i + ψ

′

)j]

= ∇V(x

, y

) · T(t

). (10)

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9.6 Liapunov’s Second Method 559

(a)

c = 0

0 < c

< c

(b)

x = (t)

y = (t)

V(x, y) = c

, y

)

, y

) i + V

, y

) j = V(x

, y

)

d (t

)

d (t

)

i + j = T(t

)

FIGURE 9.6.1 (a) The curves V(x, y) = c

and V(x, y) = c

with 0 < c

< c

(b) Geometrical interpretation of Liapunov’s second method.

Thus

V(x

, y

) is the scalar product of the vector ∇V(x

, y

) and the vector

T(t

). Since

V(x

, y

) ≤ 0, it follows that the cosine of the angle between ∇V(x

, y

)

and T(t

) is also less than or equal to zero; hence the angle itself is in the range

[π/2, 3π/2]. Thus the direction of motion on the trajectory is inward with respect to

V(x

, y

) = c or, at worst, tangent to this curve. Trajectories starting inside a closed

curve V(x

, y

) = c (no matter how small c is) cannot escape, so the origin is a sta-

ble point. If

V(x

, y

)<0, then the trajectories passing through points on the curve

are actually pointed inward. As a consequence, it can be shown that trajectories

starting sufﬁciently close to the origin must approach the origin; hence the origin is

asymptotically stable.

A geometrical argument for Theorem 9.6.2 follows in a somewhat similar manner.

Brieﬂy,suppose that

V is positive deﬁnite,and suppose that given any circle about the

origin, there is an interior point (x

, y

) at which V(x

, y

)>0. Consider a trajectory

that starts at (x

, y

). It follows from Eq. (8) that along this trajectory,V must increase

since

V(x

, y

)>0; furthermore, since V(x

, y

)>0, the trajectory cannot approach

the origin because V(0, 0) = 0. This shows that the origin cannot be asymptotically

stable. By further exploiting the fact that

V(x, y)>0, it is possible to show that the

origin is an unstable point; however, we will not pursue this argument.

EXAMPLE

Use Theorem 9.6.1 to show that (0, 0) is a stable critical point for the undamped pendulum

equations (2). Also use Theorem 9.6.2 to show that (π ,0) is an unstable critical point.

Let V be the total energy given by Eq. (4):

V(x, y) = mgL(1 − cos x) +

. (4)

If we take D to be the domain −π/2 < x <π/2, −∞ < y < ∞, then V is positive there

except at the origin, where it is zero. Thus V is positive deﬁnite on D. Further, as we have

already seen,

V = (mgL sin x)(y) + (mL

y)(−g sin x)/L = 0

for all x and y. Thus

V is negative semideﬁnite on D. Consequently, by the last statement

in Theorem 9.6.1, the origin is a stable critical point for the undamped pendulum. Observe

that this conclusion cannot be obtained from Theorem 9.3.2 because (0, 0) is a center for the

corresponding linear system.

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560 Chapter 9. Nonlinear Differential Equations and Stability

Now consider the critical point (π,0). The Liapunov function given by Eq. (4) is no longer

suitable because Theorem 9.6.2 calls for a function V for which

V is either positive or negative

deﬁnite. To analyze the point (π,0), it is convenient to move this point to the origin by the

change of variables x = π + u, y = v. Then the differential equations (2) become

du/dt = v, dv/dt =

sin u, (11)

and the critical point is (0, 0) in the uv-plane. Consider the function

V(u, v) = v sinu (12)

and let D be the domain −π/4 < u <π/4, −∞ <v<∞. Then

V = (v cos u)(v) + (sin u)[(g/L) sin u]=v

cos u + (g/L) sin

u (13)

is positive deﬁnite in D. The only remaining question is whether there are points in every

neighborhood of the origin where V itself is positive. From Eq. (12) we see that V(u, v) > 0

in the ﬁrst quadrant (where both sin u and v are positive) and in the third quadrant (where

both are negative). Thus the conditions of Theorem 9.6.2 are satisﬁed, and the point (0, 0) in

the uv-plane, corresponding to the point (π,0) in the xy-plane, is unstable.

The damped pendulum equations are discussed in Problem 7.

From a practical point of view, we are often interested in the basin of attraction.

The following theorem provides some information on this subject.

Theorem 9.6.3

Let the origin be an isolated critical point of the autonomous system (6). Let the

function V be continuous and have continuous ﬁrst partial derivatives. If there is a

bounded domain D

containing the origin where V(x, y)<K for some positive K,

V is positive deﬁnite, and

V is negative deﬁnite, then every solution of Eqs. (6) that

starts at a point in D

approaches the origin as t approaches inﬁnity.

In other words, Theorem 9.6.3 says that if x = φ(t), y = ψ(t) is the solution of

Eqs. (6) for initial data lying in D

, then (x, y) approaches the critical point (0, 0) as

t →∞. Thus D

gives a region of asymptotic stability; of course, it may not be the

entire basin of attraction. This theorem is proved by showing that (i) there are no

periodic solutions of the system (6) in D

, and (ii) there are no other critical points

in D

. It then follows that trajectories starting in D

cannot escape and therefore

must tend to the origin as t tends to inﬁnity.

Theorems 9.6.1 and 9.6.2 give sufﬁcient conditions for stability and instability,

respectively, but these conditions are not necessary. Also, our failure to ﬁnd a suit-

able Liapunov function does not mean that there is no such function. Unfortunately,

there are no general methods for the construction of Liapunov functions; how-

ever, there has been extensive work on the construction of Liapunov functions for

special classes of equations. An elementary algebraic result that is often useful in

constructing positive deﬁnite or negative deﬁnite functions is stated without proof

in the following theorem.

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9.6 Liapunov’s Second Method 561

Theorem 9.6.4

The function

V(x, y) = ax

+ bxy + cy

(14)

is positive deﬁnite if, and only if,

a > 0 and 4 ac − b

> 0 (15)

and is negative deﬁnite if, and only if,

a < 0 and 4 ac − b

> 0. (16)

The use of Theorem 9.6.4 is illustrated in the following example.

EXAMPLE

Show that the critical point (0, 0) of the autonomous system

dx/dt =−x − xy

, dy/dt =−y − x

y (17)

is asymptotically stable.

We try to construct a Liapunov function of the form (14). Then V

(x, y) = 2 ax + by,

(x, y) = bx + 2cy, so

V(x, y) = (2 ax + by)(−x − xy

) + (bx + 2cy)(−y − x

=−

2a(x

+ x

) + b(2xy + xy

+ x

y) + 2c(y

+ x

)

If we choose b = 0, and a and c to be any positive numbers, then

V is negative deﬁnite and V

is positive deﬁnite by Theorem 9.6.4. Thus, by Theorem 9.6.1, the origin is an asymptotically

stable critical point.

EXAMPLE

Consider the system

dx/dt = x(1 − x − y),

(18)

dy/dt = y(0.75 − y − 0.5x).

In Example 1 of Section 9.4, we found that this system models a certain pair of competing

species and that the critical point (0.5, 0.5) is asymptotically stable. Conﬁrm this conclusion by

ﬁnding a suitable Liapunov function.

It is helpful to transform the point (0.5, 0.5) to the origin. To this end, let

x = 0.5 + u, y = 0.5 + v. (19)

Then, substituting for x and y in Eqs. (18), we obtain the new system

du/dt =−0.5u − 0.5v − u

− uv,

(20)

dv/dt =−0.25u − 0.5v − 0.5uv − v

To keep the calculations relatively simple,consider the function V(u, v) = u

+ v

as a possible

Liapunov function. This function is clearly positive deﬁnite, so we need only to determine

August 7, 2012 21:05 c09 Sheet number 68 Page number 562 cyan black

562 Chapter 9. Nonlinear Differential Equations and Stability

whether there is a region containing the origin in the uv-plane where the derivative

V with

respect to the system (20) is negative deﬁnite. We compute

V(u, v) and ﬁnd that

V(u, v) = V

+ V

= 2u(−0.5u − 0.5v − u

− uv) + 2v(−0 .25u − 0.5v − 0.5uv − v

V(u, v) =−

+ 1.5uv + v

) + (2u

+ 2u

v + uv

+ 2v

)

, (21)

where we have collected together the quadratic and cubic terms. We want to show that the

expression in square brackets in Eq. (21) is positive deﬁnite, at least for u and v sufﬁciently

small. Observe that the quadratic terms can be written as

+ 1.5uv + v

= 0.25(u

+ v

) + 0.75(u + v)

, (22)

so these terms are positive deﬁnite. On the other hand, the cubic terms in Eq. (21) may be of

either sign. Thus we must show that, in some neighborhood of u = 0, v = 0, the cubic terms

are smaller in magnitude than the quadratic terms; that is,

+ 2u

v + uv

+ 2v

< 0.25(u

+ v

) + 0.75(u + v)

. (23)

To estimate the left side of Eq. (23), we introduce polar coordinates u = r cos θ, v = r sin θ.

Then

+ 2u

v + uv

+ 2v

= r

2 cos

θ + 2 cos

θ sin θ + cos θ sin

θ + 2 sin

≤ r

2|cos

θ|+2 cos

θ|sin θ|+|cos θ|sin

θ + 2|sin

θ|

≤ 7r

since |sin θ|, |cos θ|≤1. To satisfy Eq. (23), it is now certainly sufﬁcient to satisfy the more

stringent requirement

< 0.25(u

+ v

) = 0.25r

which yields r < 1/28. Thus, at least in this disk, the hypotheses of Theorem 9.6.1 are satisﬁed,

so the origin is an asymptotically stable critical point of the system (20). The same is then true

of the critical point (0.5, 0.5) of the original system (18).

If we refer to Theorem 9.6.3, the preceding argument also shows that the disk with center

(0.5, 0.5) and radius 1/28 is a region of asymptotic stability for the system (18). This is a

severe underestimate of the full basin of attraction, as the discussion in Section 9.4 shows.

To obtain a better estimate of the actual basin of attraction from Theorem 9.6.3, we would

have to estimate the terms in Eq. (23) more accurately, or use a better (and presumably more

complicated) Liapunov function, or both.

PROBLEMS In each of Problems 1 through 4, construct a suitable Liapunov function of the form ax

+ cy

where a and c are to be determined. Then show that the critical point at the origin is of the

indicated type.

1. dx/dt =−x

+ xy

, dy/dt =−2x

y − y

; asymptotically stable

2. dx/dt =−

+ 2xy

, dy/dt =−y

; asymptotically stable

3. dx/dt =−x

+ 2y

, dy/dt =−2xy

; stable (at least)

4. dx/dt = x

− y

, dy/dt = 2xy

+ 4x

y + 2y

; unstable

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9.6 Liapunov’s Second Method 563

5. Consider the system of equations

dx/dt = y − xf (x, y), dy/dt =−x − yf (x, y),

where f is continuous and has continuous ﬁrst partial derivatives. Show that if f (x, y)>0

in some neighborhood of the origin, then the origin is an asymptotically stable critical

point, and that if f (x, y)<0 in some neighborhood of the origin, then the origin is an

unstable critical point.

Hint: Construct a Liapunov function of the form c(x

+ y

6. A generalization of the undamped pendulum equation is

u/dt

+ g(u) = 0, (i)

where g(0) = 0, g (u)>0 for 0 < u < k, and g(u)<0 for −k < u < 0; that is, ug(u)>0

for u ̸= 0, −k < u < k. Notice that g(u) = sin u has this property on (−π/2, π/2).

(a) Letting x = u, y = du/dt, write Eq. (i) as a system of two equations, and show that

x = 0, y = 0 is a critical point.

(b) Show that

V(x, y) =

g(s) ds, −k < x < k (ii)

is positive deﬁnite, and use this result to show that the critical point (0, 0) is stable.

Note that the Liapunov function V given by Eq. (ii) corresponds to the energy function

V(x, y) =

+ (1 − cos x) for the case g(u) = sin u.

7. By introducing suitable dimensionless variables, we can write the system of nonlinear

equations for the damped pendulum [Eqs. (8) of Section 9.3] as

dx/dt = y, dy/dt =−y − sin x.

(a) Show that the origin is a critical point.

(b) Show that althoughV(x, y) = x

+ y

is positive deﬁnite,

V(x, y) takes on both positive

and negative values in any domain containing the origin, so V is not a Liapunov function.

Hint: x − sin x > 0 for x > 0, and x − sin x < 0 for x < 0. Consider these cases with y

positive but y so small that y

can be ignored compared to y.

+ (1 − cos x) mentioned in Problem 6(b),

show that the origin is a stable critical point. Since there is damping in the system, we

can expect that the origin is asymptotically stable. However, it is not possible to draw this

conclusion using this Liapunov function.

(d) To show asymptotic stability, it is necessary to construct a better Liapunov function

than the one used in part (c). Show that V(x, y) =

(x + y)

+ x

is such a Liapunov

function, and conclude that the origin is an asymptotically stable critical point.

Hint: From Taylor’s formula with a remainder, it follows that sin x = x − αx

/3!, where α

depends on x but 0 <α<1 for −π/2 < x <π/2. Then,letting x = r cos θ, y = r sin θ, show

that

V(r cos θ, r sinθ) =−r

[1 + h(r, θ)], where |h(r, θ)| < 1 if r is sufﬁciently small.

8. The Liénard equation (Problem 30 of Section 9.3) is

+ c(u)

+ g(u) = 0,

where g satisﬁes the conditions of Problem 6 and c (u) ≥ 0. Show that the point u = 0,

du/dt = 0 is a stable critical point.

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564 Chapter 9. Nonlinear Differential Equations and Stability

9. (a) A special case of the Liénard equation of Problem 8 is

+ g(u) = 0,

where g satisﬁes the conditions of Problem 6. Letting x = u, y = du/dt, show that the

origin is a critical point of the resulting system.This equation can be interpreted as describ-

ing the motion of a spring–mass system with damping proportional to the velocity and a

nonlinear restoring force. Using the Liapunov function of Problem 6, show that the origin

is a stable critical point, but note that even with damping, we cannot conclude asymptotic

stability using this Liapunov function.

(b) Asymptotic stability of the critical point (0, 0) can be shown by constructing a better

Liapunov function, as was done in part (d) of Problem 7. However, the analysis for a

general function g is somewhat sophisticated, and we mention only that an appropriate

form for V is

V(x, y) =

+ Ayg(x ) +

g(s) ds,

where A is a positive constant to be chosen so that V is positive deﬁnite and

V is negative

deﬁnite. For the pendulum problem g(x) = sin x,use V as given by the preceding equation

with A =

to show that the origin is asymptotically stable.

Hint: Use sin x = x − αx

/3! and cos x = 1 − βx

/2!, where α and β depend on x, and

0 <α<1 and 0 <β<1 for −π/2 < x <π/2; let x = r cos θ, y = r sin θ, and show

that

V(r cos θ, r sinθ) =−

[1 +

sin 2θ + h(r, θ)], where |h(r, θ)| <

if r is sufﬁciently

small. To show that V is positive deﬁnite, use cos x = 1 − x

/2 + γx

/4!, where γ depends

on x, and 0 <γ<1 for −π/2 < x <π/2.

In Problems 10 and 11, we will prove part of Theorem 9.3.2: If the critical point (0, 0) of the

locally linear system

dx/dt = a

x + a

y + F

(x, y), dy/dt = a

x + a

y + G

(x, y) (i)

is an asymptotically stable critical point of the corresponding linear system

dx/dt = a

x + a

y, dy/dt = a

x + a

y, (ii)

then it is an asymptotically stable critical point of the locally linear system (i). Problem 12

deals with the corresponding result for instability.

10. Consider the linear system (ii).

(a) Since (0, 0) is an asymptotically stable critical point, show that a

+ a

< 0 and

− a

> 0. (See Problem 21 of Section 9.1.)

(b) Construct a Liapunov function V(x, y) = Ax

+ Bxy + Cy

such that V is positive

deﬁnite and

V is negative deﬁnite. One way to ensure that

V is negative deﬁnite is to

choose A, B, and C so that

V(x, y) =−x

− y

. Show that this leads to the result

A =−

+ a

+ (a

− a

)

, B =

+ a

C =−

+ a

+ (a

− a

)

where 5 = (a

+ a

)(a

− a

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9.7 Periodic Solutions and Limit Cycles 565

are required) that

4AC − B

+ a

)(a

− a

) + 2(a

− a

)

> 0.

Thus, by Theorem 9.6.4, V is positive deﬁnite.

11. In this problem we show that the Liapunov function constructed in the preceding problem

is also a Liapunov function for the locally linear system (i). We must show that there is

some region containing the origin for which

V is negative deﬁnite.

(a) Show that

V(x, y) =−(x

+ y

) + (2Ax + By)F

(x, y) + (Bx + 2Cy)G

(x, y).

(b) Recall that F

(x, y)/r → 0 and G

(x, y)/r → 0asr = (x

+ y

)

1/2

→ 0. This means

that, given any ϵ>0, there exists a circle r = R about the origin such that for 0 < r < R,

(x, y)| <ϵr and |G

(x, y)| <ϵr. Letting M be the maximum of |2A|, |B|, and |2C|, show

by introducing polar coordinates that R can be chosen so that

V(x, y)<0 for r < R.

Hint: Choose ϵ sufﬁciently small in terms of M.

12. In this problem we prove a part of Theorem 9.3.2 related to instability.

(a) Show that if a

+ a

> 0 and a

− a

> 0, then the critical point (0, 0) of the

linear system (ii) is unstable.

(b) The same result holds for the locally linear system (i). As in Problems 10 and 11,

construct a positive deﬁnite function V such that

V(x, y) = x

+ y

and hence is positive

deﬁnite, and then invoke Theorem 9.6.2.

9.7 Periodic Solutions and Limit Cycles

In this section we discuss further the possible existence of periodic solutions of two-

dimensional autonomous systems

′

= f(x). (1)

Such solutions satisfy the relation

x(t + T) = x(t) (2)

for all t and for some positive constant T called the period. The corresponding

trajectories are closed curves in the phase plane. Periodic solutions often play an

important role in physical problems because they represent phenomena that occur

repeatedly. In many situations a periodic solution represents a “ﬁnal state” that

is approached by all “neighboring” solutions as the transients due to the initial

conditions die out.

Aspecial case of a periodic solutionis a constant solutionx = x

,which corresponds

to a critical point of the autonomous system. Such a solution is clearly periodic

with any period. In this section, when we speak of a periodic solution, we mean

a nonconstant periodic solution. In this case the period T is usually chosen as the

smallest positive number for which Eq. (2) is valid.

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566 Chapter 9. Nonlinear Differential Equations and Stability

Recall that the solutions of the linear autonomous system

′

= Ax (3)

are periodic if and only if the eigenvalues of A are pure imaginary. In this case the

critical point at the origin is a center, as discussed in Section 9.1. We emphasize

that if the eigenvalues of A are pure imaginary, then every solution of the linear

system (3) is periodic, while if the eigenvalues are not pure imaginary, then there

are no (nonconstant) periodic solutions. The predator–prey equations discussed in

Section 9.5, although nonlinear, behave similarly: all solutions in the ﬁrst quadrant

are periodic. The following example illustrates a different way in which periodic

solutions of nonlinear autonomous systems can occur.

EXAMPLE

Discuss the solutions of the system

′

x + y − x(x

+ y

)

−x + y − y(x

+ y

)

. (4)

It is not difﬁcult to show that (0, 0 ) is the only critical point of the system (4) and also that

the system is locally linear in the neighborhood of the origin. The corresponding linear system

′

−11

(5)

has eigenvalues 1 ± i. Therefore, the origin is an unstable spiral point for both the linear

system (5) and the nonlinear system (4). Thus any solution that starts near the origin in the

phase plane will spiral away from the origin. Since there are no other critical points, we might

think that all solutions of Eqs. (4) correspond to trajectories that spiral out to inﬁnity. However,

we now show that this is incorrect,because far away from theorigin the trajectories aredirected

inward.

It is convenient to introduce polar coordinates r and θ, where

x = r cos θ, y = r sin θ, (6)

and r ≥ 0. If we multiply the ﬁrst of Eqs. (4) by x, multiply the second by y, and add, we then

obtain

+ y

= (x

+ y

) − (x

+ y

)

. (7)

Since r

= x

+ y

and r(dr/dt) = x(dx/dt) + y(dy/dt), it follows from Eq. (7) that

= r

(1 − r

). (8)

This equation is similar to the equations discussed in Section 2.5. The critical points (for r ≥ 0)

are the origin and the point r = 1, which corresponds to the unit circle in the phase plane.

From Eq. (8) it follows that dr/dt > 0ifr < 1 and dr/dt < 0ifr > 1. Thus, inside the unit

circle the trajectories are directed outward, while outside the unit circle they are directed

inward. Apparently, the circle r = 1 is a limiting trajectory for this system.

To determine an equation for θ, we multiply the ﬁrst of Eqs. (4) by y, multiply the second

by x, and subtract, obtaining

− x

= x

+ y

. (9)

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9.7 Periodic Solutions and Limit Cycles 567

Upon calculating dx/dt and dy/dt from Eqs. (6) and simplifying, we ﬁnd that the left side of

Eq. (9) is −r

(dθ/dt), so Eq. (9) reduces to

dθ

=−1. (10)

The system of equations (8), (10) for r and θ is equivalent to the original system (4). One

solution of the system (8), (10) is

r = 1, θ =−t + t

, (11)

where t

is an arbitrary constant. As t increases, a point satisfying Eqs. (11) moves clockwise

around the unit circle.Thus the autonomous system (4) has a periodic solution. Other solutions

can be obtained by solving Eq. (8) by separation of variables; if r ̸= 0 and r ̸= 1, then

r(1 − r

)

= dt. (12)

Equation (12) can be solved by using partial fractions to rewrite the left side and then inte-

grating. By performing these calculations, we ﬁnd that the solution of Eqs. (10) and (12) is

r =

1 + c

−2t

, θ =−t + t

, (13)

where c

and t

are arbitrary constants. The solution (13) also contains the solution (11), which

is obtained by setting c

= 0 in the ﬁrst of Eqs. (13).

The solution satisfying the initial conditions r = ρ, θ = α at t = 0 is given by

r =

1 +[(1/ρ

) − 1]e

−2t

, θ =−(t − α). (14)

If ρ<1, then r → 1 from the inside as t →∞;ifρ>1,then r → 1 from the outside as t →∞.

Thus in all cases, the trajectories spiral toward the circle r = 1ast →∞. Several trajectories

are shown in Figure 9.7.1.

–1

FIGURE 9.7.1 Trajectories of the system (4); the circle r = 1 is a limit cycle.

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568 Chapter 9. Nonlinear Differential Equations and Stability

In this example, the circle r = 1 not only corresponds to periodic solutions of

the system (4), but it also attracts other nonclosed trajectories that spiral toward it

as t →∞. In general, a closed trajectory in the phase plane such that other non-

closed trajectories spiral toward it, from either the inside or the outside, as t →∞,

is called a limit cycle. Thus the circle r = 1 is a limit cycle for the system (4). If

all trajectories that start near a closed trajectory (both inside and outside) spiral

toward the closed trajectory as t →∞, then the limit cycle is asymptotically stable.

Since the limiting trajectory is itself a periodic orbit rather than an equilibrium point,

this type of stability is often called orbital stability. If the trajectories on one side

spiral toward the closed trajectory,whilethose on the other sidespiral away as t →∞,

then the limit cycle is said to be semistable. If the trajectories on both sides of the

closed trajectory spiral away as t →∞, then the closed trajectory is unstable.Itis

also possible to have closed trajectories that other trajectories neither approach nor

depart from—for example, the periodic solutions of the predator–prey equations in

Section 9.5. In this case the closed trajectory is stable.

In Example 1 the existence of an asymptotically stable limit cycle was established

by solving the equations explicitly. Unfortunately, this is usually not possible, so it

is worthwhile to know general theorems concerning the existence or nonexistence

of limit cycles of nonlinear autonomous systems. In discussing these theorems, it is

convenient to rewrite the system (1) in the scalar form

dx/dt = F(x, y), dy/dt = G(x, y). (15)

Theorem 9.7.1

Let the functions F and G have continuous ﬁrst partial derivatives in a domain D

of the xy-plane. A closed trajectory of the system (15) must necessarily enclose at

least one critical (equilibrium) point. If it encloses only one critical point,the critical

point cannot be a saddle point.

Although we omit the proof of this theorem, it is easy to show examples of it.

One is given by Example 1 and Figure 9.7.1, in which the closed trajectory encloses

the critical point (0, 0), a spiral point. Another example is the system of predator–

prey equations in Section 9.5; see Figure 9.5.2. Each closed trajectory surrounds the

critical point (3, 2); in this case the critical point is a center.

Theorem 9.7.1 is also useful in a negative sense. If a given region contains no

critical points, then there can be no closed trajectory lying entirely in the region. The

same conclusion is true if the region contains only one critical point, and this point

is a saddle point. For instance, in Example 2 of Section 9.4, which deals with two

competing species, the only critical point in the interior of the ﬁrst quadrant is the

saddle point (0.5, 0.5). Therefore, this system has no closed trajectory lying in the ﬁrst

quadrant.

A second result about the nonexistence of closed trajectories is given in the

following theorem.

Theorem 9.7.2

Let the functions F and G have continuous ﬁrst partial derivatives in a simply

connected domain D of the xy-plane. If F

+ G

has the same sign throughout D,

then there is no closed trajectory of the system (15) lying entirely in D.

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9.7 Periodic Solutions and Limit Cycles 569

A simply connected two-dimensional domain is one with no holes. Theorem 9.7.2

is a straightforward consequence of Green’s theorem in the plane; see Problem 13.

Note that if F

+ G

changes sign in the domain, then no conclusion can be drawn;

there may or may not be closed trajectories in D.

To illustrate Theorem 9.7.2, consider the system (4). A routine calculation

shows that

(x, y) + G

(x, y) = 2 − 4(x

+ y

) = 2(1 − 2r

), (16)

where, as usual, r

= x

+ y

. Hence F

+ G

is positive for 0 ≤ r < 1/

√

2, so there

is no closed trajectory in this circular disk. Of course, we showed in Example 1 that

there is no closed trajectory in the larger region r < 1. This illustrates that the infor-

mation given by Theorem 9.7.2 may not be the best possible result. Again referring

to Eq. (16), note that F

+ G

< 0 for r > 1/

√

2. However, the theorem is not appli-

cable in this case because this annular region is not simply connected. Indeed, as

shown in Example 1, it does contain a limit cycle.

The following theorem gives conditions that guarantee the existence of a closed

trajectory.

Theorem 9.7.3

(Poincaré–Bendixson

Theorem)

Let the functions F and G have continuous ﬁrst partial derivatives in a domain D

of the xy-plane. Let D

be a bounded subdomain in D, and let R be the region that

consists of D

plus its boundary (all points of R are in D). Suppose that R contains

no critical point of the system (15). If there exists a constant t

such that x = φ(t),

y = ψ(t) is a solution of the system (15) that exists and stays in R for all t ≥ t

then either x = φ(t), y = ψ(t) is a periodic solution (closed trajectory), or x = φ(t),

y = ψ(t) spirals toward a closed trajectory as t →∞. In either case,the system (15)

has a periodic solution in R.

Note that if R does contain a closed trajectory, then necessarily, by Theorem 9.7.1,

this trajectory must enclose a critical point. However, this critical point cannot be in

R. Thus R cannot be simply connected; it must have a hole.

As an application of the Poincaré–Bendixson theorem, consider again the system

(4). Since the origin is a critical point, it must be excluded. For instance, we can

consider the region R deﬁned by 0.5 ≤ r ≤ 2. Next, we must show that there is a

solution whose trajectory stays in R for all t greater than or equal to some t

. This

follows immediately from Eq. (8). For r = 0.5, dr/dt > 0, so r increases, while for

r = 2, dr/dt < 0, so r decreases. Thus any trajectory that crosses the boundary of R

is entering R. Consequently, any solution of Eqs. (4) that starts in R at t = t

cannot

leave but must stay in R for t > t

. Of course, other numbers could be used instead

of 0.5 and 2; all that is important is that r = 1 is included.

One should not infer from this discussion of the preceding theorems that it is

easy to determine whether a given nonlinear autonomous system has periodic solu-

tions; often it is not a simple matter at all. Theorems 9.7.1 and 9.7.2 are frequently

inconclusive, and for Theorem 9.7.3 it is often difﬁcult to determine a region R and

a solution that always remains within it.

Ivar Otto Bendixson (1861–1935) a Swedish mathematician, received his doctorate from Uppsala

University and was professor and for many years rector of Stockholm University. This theorem improved

on an earlier result of Poincaré and was published in a paper by Bendixson in Acta Mathematica in 1901.

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570 Chapter 9. Nonlinear Differential Equations and Stability

We close this section with another example of a nonlinear system that has a limit

cycle.

EXAMPLE

The van der Pol

equation

′′

− µ(1 − u

′

+ u = 0, (17)

where µ is a nonnegative constant, describes the current u in a triode oscillator. Discuss the

solutions of this equation.

If µ = 0, Eq. (17) reduces to u

′′

+ u = 0, whose solutions are sine or cosine waves of period

2π.Forµ>0, the second term on the left side of Eq. (17) must also be considered. This is the

resistance term, proportional to u

′

, with a coefﬁcient −µ(1 − u

) that depends on u. For large

u, this term is positive and acts as usual to reduce the amplitude of the response. However,for

small u, the resistance term is negative and so causes the response to grow. This suggests that

perhaps there is a solution of intermediate size that other solutions approach as t increases.

To analyze Eq. (17) more carefully, we write it as a system of two equations by introducing

the variables x = u, y = u

′

. Then it follows that

′

= y, y

′

=−x + µ(1 − x

)y. (18)

The only critical point of the system (18) is the origin. Near the origin the corresponding linear

system is

′

−1 µ

, (19)

whose eigenvalues are (µ ±

− 4)/2. Thus the origin is an unstable spiral point for

0 <µ<2 and an unstable node for µ ≥ 2. In all cases, a solution that starts near the origin

grows as t increases.

With regard to periodic solutions, Theorems 9.7.1 and 9.7.2 provide only partial informa-

tion. From Theorem 9.7.1 we conclude that if there are closed trajectories, they must enclose

the origin. Next we calculate F

(x, y) + G

(x, y), with the result that

(x, y) + G

(x, y) = µ(1 − x

). (20)

Then it follows from Theorem 9.7.2 that closed trajectories, if there are any, are not contained

in the strip |x| < 1 where F

+ G

> 0.

The application of the Poincaré–Bendixson theorem to this problem is not nearly as simple

as for Example 1. If we introduce polar coordinates, we ﬁnd that the equation for the radial

variable r is

′

= µ(1 − r

cos

θ)r sin

θ. (21)

Again, consider an annular region R given by r

≤ r ≤ r

, where r

is small and r

is large.

When r = r

, the linear term on the right side of Eq. (21) dominates, and r

′

> 0 except on

the x-axis, where sin θ = 0 and consequently r

′

= 0 also. Thus trajectories are entering R at

every point on the circle r = r

, except possibly for those on the x-axis, where the trajecto-

ries are tangent to the circle. When r = r

, the cubic term on the right side of Eq. (21) is

the dominant one. Thus r

′

< 0, except for points on the x-axis where r

′

= 0 and for points

near the y-axis where r

cos

θ<1 and the linear term makes r

′

> 0. Thus, no matter how

large a circle is chosen, there will be points on it (namely, the points on or near the y-axis)

where trajectories are leaving R.Therefore,the Poincaré–Bendixson theorem is not applicable

unless we consider more complicated regions.

Balthasar van der Pol (1889–1959) was a Dutch physicist and electrical engineer who worked at the

Philips Research Laboratory in Eindhoven. He was a pioneer in the experimental study of nonlinear

phenomena and investigated the equation that bears his name in a paper published in 1926.

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9.7 Periodic Solutions and Limit Cycles 571

It is possible to show, by a more intricate analysis, that the van der Pol equation does have a

unique limit cycle. However, we will not follow this line of argument further. We turn instead

to a different approach in which we plot numerically computed approximations to solutions.

Experimental observations indicate that the van der Pol equation has an asymptotically stable

periodic solution whose period and amplitude depend on the parameter µ. By looking at

graphs of trajectories in the phase plane and of u versus t, we can gain some understanding of

this periodic behavior.

Figure 9.7.2 shows two trajectories of the van der Pol equation in the phase plane for

µ = 0.2. The trajectory starting near the origin spirals outward in the clockwise direction;

this is consistent with the behavior of the linear approximation near the origin. The other

trajectory passes through (−3, 2) and spirals inward, again in the clockwise direction.

Both trajectories approach a closed curve that corresponds to a stable periodic solution. In

Figure 9.7.3 we show the plots of u versus t for the solutions corresponding to the trajectories

in Figure 9.7.2. The solution that is initially smaller gradually increases in amplitude, while

the larger solution gradually decays. Both solutions approach a stable periodic motion that

corresponds to the limit cycle. Figure 9.7.3 also shows that there is a phase difference between

the two solutions as they approach the limit cycle. The plots of u versus t are nearly sinusoidal

in shape, consistent with the nearly circular limit cycle in this case.

–2

–1

–2

(–3, 2)

FIGURE 9.7.2 Trajectories of the van der Pol equation (17) for µ = 0.2.

–1

–2

10 20 30 40 50

FIGURE 9.7.3 Plots of u versus t for the trajectories in Figure 9.7.2.

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572 Chapter 9. Nonlinear Differential Equations and Stability

Figures 9.7.4 and 9.7.5 show similar plots for the case µ = 1. Trajectories again move clock-

wise in the phase plane, but the limit cycle is considerably different from a circle. The plots

of u versus t tend more rapidly to the limiting oscillation, and again show a phase difference.

The oscillations are somewhat less symmetric in this case, rising somewhat more steeply than

they fall.

1–1

–1

–2

2–2

FIGURE 9.7.4 Trajectories of the van der Pol equation (17) for µ = 1.

–1

–2

10 20 30 40 50

FIGURE 9.7.5 Plots of u versus t for the trajectories in Figure 9.7.4.

Figure 9.7.6 shows the phase plane for µ = 5. The motion remains clockwise, and the limit

cycle is even more elongated, especially in the y direction. Figure 9.7.7 is a plot of u versus t.

Although the solution starts far from the limit cycle, the limiting oscillation is virtually reached

in a fraction of a period. Starting from one of its extreme values on the x-axis in the phase

plane, the solution moves toward the other extreme position slowly at ﬁrst, but once a certain

point on the trajectory is reached, the remainder of the transition is completed very swiftly.

The process is then repeated in the opposite direction. The waveform of the limit cycle, as

shown in Figure 9.7.7, is quite different from a sine wave.

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9.7 Periodic Solutions and Limit Cycles 573

–5

–10

–2

FIGURE 9.7.6 Trajectories of the van der Pol equation (17) for µ = 5.

–1

–2

10 20 30 40 50

FIGURE 9.7.7 Plot of u versus t for the outward spiralling trajectory in Figure 9.7.6.

These graphs clearly show that in the absence of external excitation, the van der Pol

oscillator has a certain characteristic mode of vibration for each value of µ. The graphs of

u versus t show that the amplitude of this oscillation changes very little with µ, but the period

increases as µ increases. At the same time, the waveform changes from one that is very nearly

sinusoidal to one that is much less smooth.

The presence of a single periodic motion that attracts all (nearby) solutions—that is, an

asymptotically stable limit cycle—is one of the characteristic phenomena associated with

nonlinear differential equations.

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574 Chapter 9. Nonlinear Differential Equations and Stability

PROBLEMS In each of Problems 1 through 6, an autonomous system is expressed in polar coordinates.

Determine all periodic solutions, all limit cycles, and the stability characteristics of all peri-

odic solutions.

1. dr/dt = r

(1 − r

), dθ/dt = 12.dr/dt = r(1 − r)

, dθ/dt =−1

3. dr/dt = r(r − 1)(r − 3), dθ/dt = 14.dr/dt = r(1 − r)(r − 2), dθ/dt =−1

5. dr/dt = sin πr, dθ/dt = 1

6. dr/dt = r|r − 2|(r − 3), dθ/dt =−1

7. If x = r cos θ, y = r sin θ, show that y(dx/dt) − x(dy/dt) =−r

(dθ/dt).

8. (a) Show that the system

dx/dt =−y + xf (r)/r, dy/dt = x + yf (r)/r

has periodic solutions corresponding to the zeros of f (r). What is the direction of motion

on the closed trajectories in the phase plane?

(b) Let f (r) = r(r − 2)

− 4r + 3). Determine all periodic solutions, and determine

their stability characteristics.

9. Determine the periodic solutions, if any, of the system

= y +

+ y

− 2),

=−x +

+ y

− 2).

10. Using Theorem 9.7.2, show that the linear autonomous system

dx/dt = a

x + a

y, dy/dt = a

x + a

does not have a periodic solution (other than x = 0, y = 0) if a

+ a

̸= 0.

In each of Problems 11 and 12, show that the given system has no periodic solutions other than

constant solutions.

11. dx/dt = x + y + x

− y

, dy/dt =−x + 2y + x

y + y

12. dx/dt =−2x − 3y − xy

, dy/dt = y + x

− x

13. Prove Theorem 9.7.2 by completing the following argument. According to Green’s the-

orem in the plane, if C is a sufﬁciently smooth simple closed curve, and if F and G are

continuous and have continuous ﬁrst partial derivatives, then

F(x, y) dy − G(x, y) dx

(x, y) + G

(x, y)

dA,

where C is traversed counterclockwise and R is the region enclosed by C. Assume that

x = φ(t), y = ψ(t) is a solution of the system (15) that is periodic with period T. Let C be

the closed curve given by x = φ(t), y = ψ(t) for 0 ≤ t ≤ T. Show that for this curve, the

line integral is zero. Then show that the conclusion of Theorem 9.7.2 must follow.

14.

(a) By examining the graphs of u versus t in Figures 9.7.3, 9.7.5, and 9.7.7, estimate the

period T of the van der Pol oscillator in these cases.

(b) Calculate and plot the graphs of solutions of the van der Pol equation for other values

of the parameter µ. Estimate the period T in these cases also.

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9.7 Periodic Solutions and Limit Cycles 575

15. The equation

′′

− µ(1 −

′2

′

+ u = 0

is often called the Rayleigh

equation.

(a) Write the Rayleigh equation as a system of two ﬁrst order equations.

(b) Show that the origin is the only critical point of this system. Determine its type and

whether it is asymptotically stable, stable, or unstable.

the system on an interval such as 0 ≤ t ≤ 20 or longer. Plot u versus t, and also plot the

trajectory in the phase plane. Observe that the trajectory approaches a closed curve (limit

cycle). Estimate the amplitude A and the period T of the limit cycle.

(d) Repeat part (c) for other values of µ, such as µ = 0.2, 0.5, 2, and 5. In each case

estimate the amplitude A and the period T.

(e) Describe how the limit cycle changes as µ increases. For example, make a table of

values and/or plot A and T as functions of µ.

16. Consider the system of equations

′

= µx + y − x(x

+ y

), y

′

=−x + µy − y(x

+ y

), (i)

where µ is a parameter. Observe that this system is the same as the one in Example 1,

except for the introduction of µ.

(a) Show that the origin is the only critical point.

(b) Find the linear system that approximates Eqs. (i) near the origin, and ﬁnd its eigen-

values. Determine the type and stability of the critical point at the origin. How does this

classiﬁcation depend on µ?

(d) Show that when µ>0, there is a periodic solution r =

√

µ. By solving the system

found in part (c), or by plotting numerically computed approximate solutions, conclude

that this periodic solution attracts all other nonzero solutions.

Note: As the parameter µ increases through the value zero, the previously asymptot-

ically stable critical point at the origin loses its stability, and simultaneously a new

asymptotically stable solution (the limit cycle) emerges. Thus the point µ = 0isa

bifurcation point; this type of bifurcation is called a Hopf

bifurcation.

17.

Consider the van der Pol system

′

= y, y

′

=−x + µ(1 − x

)y,

where now we allow the parameter µ to be any real number.

(a) Show that the origin is the only critical point. Determine its type, its stability property,

and how these depend on µ.

John William Strutt (1842–1919), the third Lord Rayleigh, made notable contributions in several areas

of mathematical physics. His theory of scattering (1871) provided the ﬁrst correct explanation of why

the sky is blue, and his two volume treatise The Theory of Sound, published in 1877 and 1878,is one of the

classics of applied mathematics. Apart from ﬁve years as Cavendish Professor of Physics at Cambridge,

he worked primarily in his private laboratory at home. He was awarded the Nobel Prize for Physics in

1904 for the discovery of argon.

Eberhard Hopf (1902–1983) was born in Austria and educated at the University of Berlin but spent

much of his life in the United States, mainly at Indiana University. He was one of the founders of ergodic

theory. Hopf bifurcations are named for him in honor of his rigorous treatment of them in a 1942 paper.

August 7, 2012 21:05 c09 Sheet number 82 Page number 576 cyan black

576 Chapter 9. Nonlinear Differential Equations and Stability

(b) Let µ =−1; draw a phase portrait, and conclude that there is a periodic solution that

surrounds the origin. Observe that this periodic solution is unstable. Compare your plot

with Figure 9.7.4.

of the periodic solution changes with µ.

(d) Consider small positive or negative values of µ. By drawing phase portraits,determine

how the periodic solution changes as µ → 0. Compare the behavior of the van der Pol

system as µ increases through zero with the behavior of the system in Problem 16.

Problems 18 and 19 extend the consideration of the Rosenzweig–MacArthur predator–

prey model introduced in Problem 13 of Section 9.5.

18.

Consider the system

′

= x

2.4 − 0.2x −

x + 6

(

, y

′

= y

−0.25 +

x + 6

(

Observe that this system differs from that in Problem 13 of Section 9.5 only in the growth

rate for the prey.

(a) Find all of the critical points.

(b) Determine the type and stability of each critical point.

stable limit cycle. Thus this model predicts a stable long-term oscillation of the prey and

predator populations.

19.

Consider the system

′

= x

a − 0.2x −

x + 6

(

, y

′

= y

−0.25 +

x + 6

(

where a is a positive parameter. Observe that this system includes the one in Problem 18

above and also the one in Problem 13 in Section 9.5.

(a) Find all of the critical points.

(b) Consider the critical point in the interior of the ﬁrst quadrant. Find the eigenvalues of

the approximate linear system. Determine the value a

where this critical point changes

from asymptotically stable to unstable.

. Observe that a limit

cycle has appeared. How does the limit cycle change as a increases further?

20.

There are certain chemical reactions in which the constituent concentrations oscillate

periodically over time. The system

′

= 1 − (b + 1)x + x

y/4, y

′

= bx − x

y/4

is a special case of a model, known as the Brusselator, of this kind of reaction. Assume

that b is a positive parameter, and consider solutions in the ﬁrst quadrant of the xy-plane.

(a) Show that the only critical point is (1, 4b).

(b) Find the eigenvalues of the approximate linear system at the critical point.

on b?

(d) As b increases through a certain value b

, the critical point changes from asymptoti-

cally stable to unstable. What is that value b

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9.8 Chaos and Strange Attractors: The Lorenz Equations 577

(e) Plot trajectories in the phase plane for values of b slightly less than and slightly greater

than b

. Observe the limit cycle when b > b

; the Brusselator has a Hopf bifurcation point

at b

(f) Plot trajectories for several values of b > b

and observe how the limit cycle deforms

as b increases.

21.

The system

′

= 3(x + y −

− k), y

′

=−

(x + 0.8y − 0.7)

is a special case of the Fitzhugh–Nagumo

equations, which model the transmission of

neural impulses along an axon. The parameter k is the external stimulus.

(a) Show that the system has one critical point regardless of the value of k.

(b) Find the critical point for k = 0, and show that it is an asymptotically stable spiral

point. Repeat the analysis for k = 0.5, and show that the critical point is now an unstable

spiral point. Draw a phase portrait for the system in each case.

where the critical point changes from asymptotically stable to

unstable. Find the critical point and draw a phase portrait for the system for k = k

(d) For k > k

the system exhibits an asymptotically stable limit cycle; the system has a

Hopf bifurcation point at k

. Draw a phase portrait for k = 0.4, 0.5, and 0.6; observe that

the limit cycle is not small when k is near k

. Also plot x versus t and estimate the period

T in each case.

(e) As k increases further, there is a value k

at which the critical point again becomes

asymptotically stable and the limit cycle vanishes. Find k

9.8 Chaos and Strange Attractors: The Lorenz Equations

In principle,the methods described in this chapter for second order autonomous sys-

tems can be applied to higher order systems as well. In practice, several difﬁculties

arise when we try to do this. One problem is that there is simply a greater num-

ber of possible cases that can occur, and the number increases with the number of

equations in the system (and the dimension of the phase space). Another problem is

the difﬁculty of graphing trajectories accurately in a phase space of more than two

dimensions; even in three dimensions it may not be easy to construct a clear and

understandable plot of the trajectories, and it becomes more difﬁcult as the number

of variables increases. Finally—and this has been widely recognized only since the

1970s—there are different and very complex phenomena that can occur, and fre-

quently do occur, in systems of third and higher order that are not present in second

order systems. Our goal in this section is to provide a brief introduction to some of

these phenomena by discussing one particular three-dimensional autonomous sys-

tem that has been intensively studied. In some respects, the presentation here is

similar to the treatment of the logistic difference equation in Section 2.9.

Richard Fitzhugh (1922–2007) of the United States Public Health Service and Jin-Ichi Nagumo (1926–

1999) of the University of Tokyo independently proposed a simpliﬁcation of the Hodgkin–Huxley model

of neural transmission around 1961.

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578 Chapter 9. Nonlinear Differential Equations and Stability

An important problem in meteorology, and in other applications of ﬂuid dynam-

ics, concerns the motion of a layer of ﬂuid, such as the earth’s atmosphere, that is

warmer at the bottom than at the top; see Figure 9.8.1. If the vertical temperature

difference 5T is small, then there is a linear variation of temperature with altitude

but no signiﬁcant motion of the ﬂuid layer. However, if 5T is large enough, then the

warmer air rises, displacing the cooler air above it, and a steady convective motion

results. If the temperature difference increases further, then eventually the steady

convective ﬂow breaks up and a more complex and turbulent motion ensues.

Cooler

Warmer

Temperature

difference ∆

FIGURE 9.8.1 A layer of ﬂuid heated from below.

While investigating this phenomenon, Edward N. Lorenz

was led (by a pro-

cess too involved to describe here) to the nonlinear autonomous three-dimensional

system

dx/dt = σ(−x + y),

dy/dt = rx − y − xz, (1)

dz/dt =−bz + xy.

Equations (1) are now commonly referred to as the Lorenz equations.

Observe that

the second and third equations involve quadratic nonlinearities. However,except for

being a system of three equations, superﬁcially the Lorenz equations appear no more

complicated than the competing species and predator–prey equations discussed in

Sections 9.4 and 9.5. The variable x in Eqs. (1) is related to the intensity of the

ﬂuid motion, while the variables y and z are related to the temperature variations

in the horizontal and vertical directions. The Lorenz equations also involve three

parameters σ, r, and b, all of which are real and positive. The parameters σ and b

depend on the material and geometrical properties of the ﬂuid layer. For the earth’s

atmosphere, reasonable values of these parameters are σ = 10 and b = 8/3; they

will be assigned these values in much of what follows in this section. The parameter r,

on the other hand,is proportional to the temperature difference 5T,and our purpose

is to investigate how the nature of the solutions of Eqs. (1) changes with r.

Edward N. Lorenz (1917–2008), an American meteorologist, received his Ph.D. from the Massachusetts

Institute of Technology in 1948 and was associated with that institution throughout his scientiﬁc career.

His ﬁrst studies of the system (1) appeared in a famous 1963 paper (cited in the references) dealing with

the stability of ﬂuid ﬂows in the atmosphere.

A very thorough treatment of the Lorenz equations appears in the book by Sparrow listed in the

references at the end of this chapter.

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9.8 Chaos and Strange Attractors: The Lorenz Equations 579

Before proceeding further, we note that for an autonomous system of three ﬁrst

order equations

dx/dt = F(x, y, z), dy/dt = G(x, y, z), dz/dt = H(x, y, z), (2)

the Jacobian matrix J is deﬁned by

J =

⎛

⎜

⎝

⎞

⎟

⎠

. (3)

Thus, for the Lorenz equations (1), the Jacobian matrix is

J =

⎛

⎜

⎝

−σσ0

r − z −1 −x

yx−b

⎞

⎟

⎠

. (4)

The ﬁrst step in analyzing the Lorenz equations is to locate the critical points by

solving the algebraic system

σx − σy = 0,

rx − y − xz = 0, (5)

−bz + xy = 0.

From the ﬁrst equation we have y = x. Then, eliminating y from the second and

third equations, we obtain

x(r − 1 − z) = 0, (6)

−bz + x

= 0. (7)

One way to satisfy Eq. (6) is to choose x = 0. Then it follows that y = 0 and,

from Eq. (7), z = 0. Alternatively, we can satisfy Eq. (6) by choosing z = r − 1. Then

Eq. (7) requires that x =±

√

b(r − 1) and then y =±

√

b(r − 1) also. Observe that

these expressions for x and y are real only when r ≥ 1. Thus (0, 0, 0), which we will

denote by P

, is a critical point for all values of r, and it is the only critical point

for r < 1. However, when r > 1, there are also two other critical points, namely,

(

√

b(r − 1),

√

b(r − 1), r − 1) and (−

√

b(r − 1), −

√

b(r − 1), r − 1). We will denote

the latter two points by P

and P

, respectively. Note that all three critical points

coincide when r = 1. As r increases through the value 1, the critical point P

at the

origin bifurcates, and the critical points P

and P

come into existence.

Next we will determine the local behavior of solutions in the neighborhood of

each critical point. Although much of the following analysis can be carried out for

arbitrary values of σ and b, we will simplify our work by using the values σ = 10 and

b = 8/3. Near the origin (the critical point P

), the approximating linear system is

⎛

⎜

⎝

⎞

⎟

⎠

′

⎛

⎜

⎝

−10 10 0

r −10

00−8/3

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

. (8)

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580 Chapter 9. Nonlinear Differential Equations and Stability

The eigenvalues

are determined from the equation

−10 − λ 10 0

r −1 − λ 0

00−8/3 − λ

=−(8/3 + λ)[λ

+ 11λ − 10(r − 1)]=0. (9)

Therefore,

=−

, λ

−11 −

√

81 + 40r

, λ

−11 +

√

81 + 40r

. (10)

Note that all three eigenvalues are negative for r < 1; for example, when r = 1/2,

the eigenvalues are λ

=−8/3, λ

=−10.52494, λ

=−0.47506. Hence the origin is

asymptotically stable for this range of r both for the linear approximation (8) and

for the original system (1). However, λ

changes sign when r = 1 and is positive for

r > 1. The value r = 1 corresponds to the initiation of convective ﬂow in the physical

problem described earlier. The origin is unstable for r > 1; all solutions starting near

the origin tend to grow, except for those lying precisely in the plane determined

by the eigenvectors associated with λ

and λ

[or, for the nonlinear system (1), in a

certain surface tangent to this plane at the origin].

The second critical point is P

(

8(r − 1)/3,

8(r − 1)/3, r − 1) for r > 1. To con-

sider the neighborhood of this critical point, let u, v, and w be the perturbations from

the critical point in the x-, y-, and z-directions, respectively.The approximating linear

system is

⎛

⎜

⎝

⎞

⎟

⎠

′

⎛

⎜

⎝

−10 10 0

1 −1 −

8(r − 1)/3

8(r − 1)/3 −8/3

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

. (11)

The eigenvalues of the coefﬁcient matrix of Eq. (11) are determined from the

equation

3λ

+ 41λ

+ 8(r + 10)λ + 160(r − 1) = 0, (12)

which is obtained by straightforward algebraic steps that are omitted here. The

solutions of Eq. (12) depend on r in the following way:

For 1 < r < r

∼

1.3456 there are three negative real eigenvalues.

For r

< r < r

∼

24.737 there areone negative realeigenvalue and two complex

eigenvalues with negative real part.

For r

< r there are one negative real eigenvalue and two complex eigenvalues

with positive real part.

The same results are obtained for the critical point P

. Thus there are several

different situations.

For 0 < r < 1 the only critical point is P

and it is asymptotically stable. All

solutions approach this point (the origin) as t →∞.

Since r appears as a parameter in the Lorenz equations, we will use λ to denote the eigenvalues.

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9.8 Chaos and Strange Attractors: The Lorenz Equations 581

For 1 < r < r

the critical points P

and P

are asymptotically stable and P

unstable. All nearby solutions approach one or the other of the points P

and P

exponentially.

For r

< r < r

the critical points P

and P

are asymptotically stable and P

unstable. All nearby solutions approach one or the other of the points P

and P

;

most of them spiral inward to the critical point.

For r

< r all three critical points are unstable. Most solutions near P

or P

spiral

away from the critical point.

However, this is by no means the end of the story. Let us consider solutions for r

somewhat greater than r

. In this case P

has one positive eigenvalue, and each of

and P

has a pair of complex eigenvalues with positive real part. A trajectory can

approach any one of the critical points only on certain highly restricted paths. The

slightest deviation from these paths causes the trajectory to depart from the critical

point. Since none of thecritical pointsis stable,one might expect thatmost trajectories

would approach inﬁnity for large t. However,it can be shown that all solutions remain

bounded ast →∞;see Problem 5. In fact,it canbe shown that all solutions ultimately

approach a certain limiting set of points that has zero volume. Indeed, this is true not

only for r > r

but for all positive values of r.

A plot of computed values of x versus t for a typical solution with r > r

is shown

in Figure 9.8.2. Note that the solution oscillates back and forth between positive and

negative values in a rather erratic manner. Indeed, the graph of x versus t resem-

bles a random vibration, although the Lorenz equations are entirely deterministic

and the solution is completely determined by the initial conditions. Nevertheless,

the solution also exhibits a certain regularity in that the frequency and amplitude

of the oscillations are essentially constant in time.

–8

–16

10 20

FIGURE 9.8.2 A plot of x versus t for the Lorenz equations (1) with r = 28;

the initial point is (5, 5, 5).

The solutions of the Lorenz equations are also extremely sensitive to perturbations

in the initial conditions. Figure 9.8.3 shows the graphs of computed values of x versus

t for the two solutions whose initial points are (5, 5, 5) and (5.01, 5, 5). The dashed

graph is the same as the one in Figure 9.8.2, while the solid graph starts at a nearby

point. The two solutions remain close until t is near 10, after which they are quite

different and, indeed, seem to have no relation to each other. It was this property

that particularly attracted the attention of Lorenz in his original study of these

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582 Chapter 9. Nonlinear Differential Equations and Stability

–8

10 20

–16

FIGURE 9.8.3 Plots of x versus t for two initially nearby solutions of Lorenz equations with

r = 28; the initial point is (5, 5, 5) for the dashed curve and is (5.01, 5, 5) for the solid curve.

equations, and it caused him to conclude that accurate detailed long-range weather

predictions are probably not possible.

The attracting set in this case, although of zero volume, has a rather complicated

structure and is called a strange attractor. The term chaotic has come into general

use to describe solutions such as those shown in Figures 9.8.2 and 9.8.3.

To determine how and when the strange attractor is created, it is illuminating to

investigate solutions for smaller values of r.Forr = 21,solutions starting at three dif-

ferent initial points are shown in Figure 9.8.4. For the initial point (3, 8, 0) the solution

begins to converge to the point P

almost at once; see Figure 9.8.4a. For the second

initial point (5, 5, 5) there is a fairly short interval of transient behavior, after which

the solution converges to P

; see Figure 9.8.4b. However, as shown in Figure 9.8.4c,

for the third initial point (5, 5, 10) there is a much longer interval of transient chaotic

behavior before the solution eventually converges to P

.Asr increases, the duration

of the chaotic transient behavior also increases. When r = r

∼

24.06, the chaotic

transients appear to continue indeﬁnitely,and the strange attractor comes into being.

We can also show the trajectories of the Lorenz equations in the three-

dimensional phase space, or at least projections of them in various planes. Figures

9.8.5 and 9.8.6 show projections in the xy- and xz-planes, respectively, of the tra-

jectory starting at (5, 5, 5). Observe that the graphs in these ﬁgures appear to cross

over themselves repeatedly, but this cannot be true for the actual trajectories in

three-dimensional space because of the general uniqueness theorem. The apparent

crossings are due wholly to the two-dimensional character of the ﬁgures.

The sensitivity of solutions to perturbations of the initial data also has impli-

cations for numerical computations, such as those reported here. Different step

sizes, different numerical algorithms, and even the execution of the same algorithm

on different machines will introduce small differences in the computed solution,

which eventually lead to large deviations. For example, the exact sequence of pos-

itive and negative loops in the calculated solution depends strongly on the precise

numerical algorithm and its implementation, as well as on the initial conditions.

However, the general appearance of the solution and the structure of the attracting

set are independent of all these factors.

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9.8 Chaos and Strange Attractors: The Lorenz Equations 583

–8

–16

12 24

(a)

12 24

(b)

–8

–16

6321

60 72

(c)

–8

–16

FIGURE 9.8.4 Plots of x versus t for three solutions of Lorenz equations with r = 21.

(a) Initial point is (3, 8, 0). (b) Initial point is (5, 5, 5). (c) Initial point is (5, 5, 10).

–20

–10

–20

FIGURE 9.8.5 Projections of a trajectory of the Lorenz equations

(with r = 28) in the xy-plane.

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584 Chapter 9. Nonlinear Differential Equations and Stability

–10 10

FIGURE 9.8.6 Projections of a trajectory of the Lorenz equations

(with r = 28) in the xz-plane.

Solutions of the Lorenz equations for other parameter ranges exhibit other

interesting types of behavior. For example, for certain values of r greater than

, intermittent bursts of chaotic behavior separate long intervals of apparently

steady periodic oscillation. For other ranges of r,solutions show the period-doubling

property that we saw in Section 2.9 for the logistic difference equation. Some of these

features are taken up in the problems.

Since about 1975 the Lorenz equations and other higher-dimensional autonomous

systems have been studied intensively, and this is one of the most active areas of cur-

rent mathematical research. Chaotic behavior of solutions appears to be much more

common than was suspected at ﬁrst, and many questions remain unanswered. Some

of these are mathematical in nature, while others relate to the physical applications

or interpretations of solutions.

PROBLEMS Problems 1 through 3 ask you to ﬁll in some of the details of the analysis of the Lorenz equations

in this section.

1. (a) Show that the eigenvalues of the linear system (8), valid near the origin, are given by

Eq. (10).

(b) Determine the corresponding eigenvectors.

(a) Show that the linear approximation valid near the critical point P

is given by Eq. (11).

(b) Show that the eigenvalues of the system (11) satisfy Eq. (12).

(a) By solving Eq. (12) numerically, show that the real part of the complex roots changes

sign when r

∼

24.737.

(b) Show that a cubic polynomial x

+ Ax

+ Bx + C has one real zero and two pure

imaginary zeros only if AB = C.

roots changes sign when r = 470/19.

4. Use the Liapunov function V(x, y, z) = x

+ σy

+ σz

to show that the origin is a globally

asymptotically stable critical point for the Lorenz equations (1) if r < 1.

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9.8 Chaos and Strange Attractors: The Lorenz Equations 585

5. Consider the ellipsoid

V(x, y, z) = rx

+ σy

+ σ(z − 2r)

= c > 0.

(a) Calculate dV/dt along trajectories of the Lorenz equations (1).

(b) Determine a sufﬁcient condition on c so that every trajectory crossing V(x, y, z) = c

is directed inward.

Problems 6 through 10 suggest some further investigations of the Lorenz equations.

For r = 28, plot x versus t for the cases shown in Figures 9.8.2 and 9.8.3. Do your graphs

agree with those shown in the ﬁgures? Recall the discussion of numerical computation in

the text.

For r = 28, plot the projections in the xy- and xz-planes, respectively, of the trajectory

starting at the point (5, 5, 5). Do the graphs agree with those in Figures 9.8.5 and 9.8.6?

(a) For r = 21,plot x versus t for the solutions starting at the initial points (3, 8, 0),(5, 5,5),

and (5, 5,10). Use a t-interval of at least 0 ≤ t ≤ 30. Compare your graphs with those in

Figure 9.8.4.

(b) Repeat the calculation in part (a) for r = 22,r = 23, and r = 24. Increase the t interval

as necessary so that you can determine when each solution begins to converge to one of

the critical points. Record the approximate duration of the chaotic transient in each case.

Describe how this quantity depends on the value of r.

Try to estimate the value of r for which the duration of the chaotic transient approaches

inﬁnity.

For certain r intervals, or windows, the Lorenz equations exhibit a period-doubling prop-

erty similar to that of the logistic difference equation discussed in Section 2.9. Careful

calculations may reveal this phenomenon.

(a) One period-doubling window contains the value r = 100. Let r = 100 and plot the

trajectory starting at (5, 5, 5) or some other initial point of your choice. Does the solution

appear to be periodic? What is the period?

(b) Repeat the calculation in part (a) for slightly smaller values of r. When r

∼

99.98, you

may be able to observe that the period of the solution doubles. Try to observe this result

by performing calculations with nearby values of r.

doubling occurs at about r = 99.629. Try to observe this by plotting trajectories for nearby

values of r.

10.

Now consider values of r slightly larger than those in Problem 9.

(a) Plot trajectories of the Lorenz equations for values of r between 100 and 100.78. You

should observe a steady periodic solution for this range of r values.

(b) Plot trajectories for values of r between 100.78 and 100.8. Determine as best you can

how and when the periodic trajectory breaks up.

The Rössler

System. The system

′

=−y − z, y

′

= x + ay, z

′

= b + z(x − c), (i)

Otto E. Rössler (1940– ), a German medical doctor and biochemist, was a student and later became

a faculty member at the University of Tübingen. The equations named for him ﬁrst appeared in a paper

he published in 1976.

August 7, 2012 21:05 c09 Sheet number 92 Page number 586 cyan black

586 Chapter 9. Nonlinear Differential Equations and Stability

where a, b, and c are positive parameters, is known as the Rössler system.

It is a relatively

simple system, consisting of two linear equations and a third equation with a single quadratic

nonlinearity. In Problems 11 through 15, we ask you to carry out some numerical investigations

of this system, with the goal of exploring its period-doubling property. To simplify matters, set

a = 0.25, set b = 0.5, and let c > 0 remain arbitrary.

11.

(a) Show that there are no critical points when c <

√

0.5, there is one critical point for

c =

√

0.5, and there are two critical points when c >

√

0.5.

(b) Find the critical point(s) and determine the eigenvalues of the associated Jacobian

matrix when c =

√

0.5 and when c = 1.

starting at the origin. Does it behave the way that you expected?

(d) Choose one or two other initial points, and plot the corresponding trajectories. Do

these plots agree with your expectations?

12.

(a) Let c = 1.3. Find the critical points and the corresponding eigenvalues. What

conclusions, if any, can you draw from this information?

(b) Plot the trajectory starting at the origin. What is the limiting behavior of this trajec-

tory? To see the limiting behavior clearly, you may wish to choose a t-interval for your

plot so that the initial transients are eliminated.

the limiting behavior(s) the same as in part (b)?

(d) Observe that there is a limit cycle whose basin of attraction is fairly large (although

not all of xyz-space). Draw a plot of x, y,orz versus t,and estimate the period T

of motion

around the limit cycle.

13.

The limit cycle found in Problem 12 comes into existence as a result of a Hopf bifurcation

at a value c

of c between 1 and 1.3. Determine, or at least estimate more precisely, the

value of c

. There are several ways in which you might do this.

(a) Draw plots of trajectories for different values of c.

(b) Calculate eigenvalues at critical points for different values of c .

14.

(a) Let c = 3. Find the critical points and the corresponding eigenvalues.

(b) Plot the trajectory starting at the point (1, 0, −2). Observe that the limit cycle now

consists of two loops before it closes; it is often called a 2-cycle.

of motion on the 2-cycle is very

nearly double the period T

of the simple limit cycle in Problem 12. There has been a

period-doubling bifurcation of cycles for a certain value of c between 1.3 and 3.

15.

(a) Let c = 3.8. Find the critical points and the corresponding eigenvalues.

(b) Plot the trajectory starting at the point (1, 0, −2). Observe that the limit cycle is now a

4-cycle. Find the period T

of motion. Another period-doubling bifurcation has occurred

for c between 3 and 3.8.

to eight times the period of the simple limit cycle in Problem 12.

Note: As c increases further, there is an accelerating cascade of period-doubling bifurca-

tions. The bifurcation values of c converge to a limit, which marks the onset of chaos.

See the book by Strogatz for a more extensive discussion and further references.

August 7, 2012 21:05 c09 Sheet number 93 Page number 587 cyan black

References 587

REFERENCES There are many books that expand on the material treated in this chapter. They include

Drazin, P. G., Nonlinear Systems (Cambridge: Cambridge University Press, 1992).

Glendinning, P., Stability, Instability, and Chaos (Cambridge: Cambridge University Press, 1994).

Grimshaw, R., Nonlinear Ordinary Differential Equations (Oxford:Blackwell Scientiﬁc Publications, 1990;

New York: CRC Press, 1991).

Hirsch, M. W., Smale, S., and Devaney, R. L., Differential Equations, Dynamical Systems, and an Introduc-

tion to Chaos (2nd edition) (San Diego, CA:Academic Press, 2004).

Hubbard, J. H., and West, B. H., Differential Equations: A Dynamical Systems Approach, Higher Dimen-

sional Systems (New York/Berlin: Springer-Verlag, 1995)

Two books that are especially notable from the point of view of applications are

Danby, J. M. A., Computer Applications to Differential Equations (Englewood Cliffs, NJ: Prentice-Hall,

1985).

Strogatz, S. H., Nonlinear Dynamics and Chaos (Reading, MA: Addison-Wesley, 1994).

A good reference on Liapunov’s second method is

LaSalle,J.,and Lefschetz,S.,Stability by Liapunov’s Direct Method with Applications (NewYork:Academic

Press, 1961).

Among the large number of more comprehensive books on differential equations are

Arnol’d, V. I., Ordinary Differential Equations (New York/Berlin: Springer-Verlag, 1992). Translation of

the third Russian edition by Roger Cooke.

Brauer, F., and Nohel, J., Qualitative Theory of Ordinary Differential Equations (New York: Benjamin,

1969; New York: Dover, 1989).

Guckenheimer, J. C., and Holmes, P., Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vec-

tor Fields (New York/Berlin: Springer-Verlag, 1983).

A classic reference on ecology is

Odum,E. P.,and Barrett,G.W.,Fundamentals of Ecology (5th ed.) (Belmont,CA:Thompson Brooks/Cole,

2005).

Three books dealing with ecology and population dynamics on a more mathematical level are

Brauer, Fred and Castillo-Chávez, Carlos, Mathematical Models in Population Biology and Epidemiology

(New York/Berlin: Springer-Verlag, 2001).

May, R. M., Stability and Complexity in Model Ecosystems (Princeton, NJ: Princeton University Press,

1973).

Pielou, E. C., Mathematical Ecology (New York: Wiley, 1977).

The original paper on the Lorenz equations is

Lorenz, E. N., “Deterministic Nonperiodic Flow,” Journal of the Atmospheric Sciences 20 (1963),

pp. 130–141.

A very detailed treatment of the Lorenz equations is

Sparrow, C., The Lorenz Equations: Bifurcations, Chaos, and Strange Attractors (New York/Berlin:

Springer-Verlag, 1982).

An excellent nontechnical exposition of the early development of chaos may be found in

Gleick, James, Chaos: Making a New Science (New York: Viking Penguin, 1987).

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August 7, 2012 21:05 c10 Sheet number 1 Page number 589 cyan black

589

CHAPTER

Partial

Differential

Equations and

Fourier Series

In many important physical problems there are two or more independent variables,

so the corresponding mathematical models involve partial, rather than ordinary,

differential equations. This chapter treats one important method for solving par-

tial differential equations, a method known as separation of variables. Its essential

feature is the replacement of the partial differential equation by a set of ordinary

differential equations, which must be solved subject to given initial or boundary con-

ditions. The ﬁrst section of this chapter deals with some basic properties of boundary

value problems for ordinary differential equations. The desired solution of the par-

tial differential equation is expressed as a sum, usually an inﬁnite series, formed from

solutions of the ordinary differential equations. In many cases we ultimately need

to deal with a series of sines and/or cosines, so part of the chapter is devoted to a

discussion of such series,which are known as Fourier series.With the necessary math-

ematical background in place, we then illustrate the use of separation of variables in

a variety of problems arising from heat conduction, wave propagation, and potential

theory.

10.1 Two-Point Boundary Value Problems

Up to this point in the book we have dealt with initial value problems, consisting

of a differential equation together with suitable initial conditions at a given point.

A typical example, which was discussed at length in Chapter 3, is the differential

equation

′′

+ p(t)y

′

+ q(t)y = g(t), (1)

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590 Chapter 10. Partial Differential Equations and Fourier Series

with the initial conditions

y(t

) = y

, y

′

) = y

′

. (2)

Physical applications often lead to another type of problem,one in which the value

of the dependent variable y or its derivative is speciﬁed at two different points. Such

conditions are called boundary conditions to distinguish them from initial conditions

that specify the value of y and y

′

at the same point. A differential equation and

suitable boundary conditions form a two-point boundary value problem. A typical

example is the differential equation

′′

+ p(x)y

′

+ q(x)y = g(x) (3)

with the boundary conditions

y(α) = y

, y(β) = y

. (4)

The natural occurrence of boundary value problems usually involves a space coordi-

nate as the independent variable, so we have used x rather than t in Eqs. (3) and (4).

To solve the boundary value problem (3),(4),we need to ﬁnd a function y = φ(x) that

satisﬁes the differential equation (3) in the interval α<x <βand that takes on the

speciﬁed values y

and y

at the endpoints of the interval. Usually, we ﬁrst seek

the general solution of thedifferential equation and then usethe boundary conditions

to determine the values of the arbitrary constants.

Boundary value problems can also be posed for nonlinear differential equations,

butwe will restrict ourselves toa consideration of linearequations only. Itis important

to determine whether a linear boundary value problem is homogeneous or nonho-

mogeneous. If the function g has the value zero for each x,and if the boundary values

and y

are also zero, then the problem (3), (4) is called homogeneous. Otherwise,

the problem is nonhomogeneous.

Although the initial value problem (1), (2) and the boundary value problem (3),

(4) may superﬁcially appear to be quite similar, they actually differ in some very

important ways. Under mild conditions on the coefﬁcients, initial value problems are

certain to have a unique solution. On the other hand,boundary value problems under

similar conditions may have a unique solution, but they may also have no solution

or, in some cases, inﬁnitely many solutions. In this respect, linear boundary value

problems resemble systems of linear algebraic equations.

Let us recall some facts (see Section 7.3) about the system

Ax = b, (5)

where A is a given n × n matrix, b is a given n × 1 vector, and x is an n × 1 vector to

be determined. If A is nonsingular, then the system (5) has a unique solution for any

b. However, if A is singular, then the system (5) has no solution unless b satisﬁes a

certain additional condition, in which case the system has inﬁnitely many solutions.

Now consider the corresponding homogeneous system

Ax = 0, (6)

obtained from the system (5) when b = 0. The homogeneous system (6) always has

the solution x = 0,which is often referred to as the trivial solution. If A is nonsingular,

then this is the only solution, but if A is singular, then there are inﬁnitely many

nonzero, or nontrivial, solutions. Note that it is impossible for the homogeneous

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10.1 Two-Point Boundary Value Problems 591

system to have no solution. These results can also be stated in the following way: the

nonhomogeneous system (5) has a unique solution if and only if the homogeneous

system (6) has only the solution x = 0, and the nonhomogeneous system (5) has

either no solution or inﬁnitely many solutions if and only if the homogeneous system

(6) has nonzero solutions.

We now turn to some examples of linear boundary value problems that illustrate

very similar behavior. A more general discussion of linear boundary value problems

appears in Chapter 11.

EXAMPLE

Solve the boundary value problem

′′

+ 2y = 0, y(0) = 1, y(π) = 0. (7)

The general solution of the differential equation (7) is

y = c

cos

√

2 x + c

sin

√

2 x. (8)

The ﬁrst boundary condition requires that c

= 1. The second boundary condition implies

that c

cos

√

2 π + c

sin

√

2 π = 0, so c

=−cot

√

2 π

∼

−0.2762. Thus the solution of the

boundary value problem (7) is

y = cos

√

2 x − cot

√

2 π sin

√

2 x. (9)

This example illustrates the case of a nonhomogeneous boundary value problem with a unique

solution.

EXAMPLE

Solve the boundary value problem

′′

+ y = 0, y(0) = 1, y(π) = a, (10)

where a is a given number.

The general solution of this differential equation is

y = c

cos x + c

sin x, (11)

and from the ﬁrst boundary condition we ﬁnd that c

= 1. The second boundary condition

now requires that −c

= a. These two conditions on c

are incompatible if a ̸=−1, so the

problem has no solution in that case. However, if a =−1, then both boundary conditions are

satisﬁed provided that c

= 1, regardless of the value of c

. In this case there are inﬁnitely

many solutions of the form

y = cos x + c

sin x, (12)

where c

remains arbitrary. This example illustrates that a nonhomogeneous boundary value

problem may have no solution—andalso that under special circumstances it may have inﬁnitely

many solutions.

The nonhomogeneous boundary value problem (3), (4) has a corresponding

homogeneous problem consisting of the differential equation

′′

+ p(x)y

′

+ q(x)y = 0 (13)

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592 Chapter 10. Partial Differential Equations and Fourier Series

and the boundary conditions

y(α) = 0, y(β) = 0. (14)

Observe that this problem has the solution y = 0 for all x, regardless of the coefﬁ-

cients p(x) and q(x). This (trivial) solution is rarely of interest. What we usually want

to know is whether the problem has other,nonzero solutions. Consider the following

two examples.

EXAMPLE

Solve the boundary value problem

′′

+ 2y = 0, y(0) = 0, y(π) = 0. (15)

The general solution of the differential equation is again given by Eq. (8),

y = c

cos

√

2 x + c

sin

√

2 x.

The ﬁrst boundary condition requires that c

= 0, and the second boundary condition leads to

sin

√

2 π = 0. Since sin

√

2 π ̸= 0, it follows that c

= 0 also. Consequently, y = 0 for all x is

the only solution of the problem (15). This example illustrates that a homogeneous boundary

value problem may have only the trivial solution y = 0.

EXAMPLE

Solve the boundary value problem

′′

+ y = 0, y(0) = 0, y(π) = 0. (16)

The general solution is given by Eq. (11),

y = c

cos x + c

sin x,

and the ﬁrst boundary condition requires that c

= 0. Since sin π = 0, the second boundary

condition is also satisﬁed when c

= 0, regardless of the value of c

. Thus the solution of

the problem (16) is y = c

sin x, where c

remains arbitrary. This example illustrates that a

homogeneous boundary value problem may have inﬁnitely many solutions.

Examples 1 through 4 illustrate (but of course do not prove) that there is the same

relationship between homogeneous and nonhomogeneous linear boundary value

problems as there is between homogeneous and nonhomogeneous linear algebraic

systems. The nonhomogeneous boundary value problem (Example 1) has a unique

solution, and the corresponding homogeneous problem (Example 3) has only the

trivial solution. Further, the nonhomogeneous problem (Example 2) has either no

solution or inﬁnitely many, and the corresponding homogeneous problem (Example

4) has nontrivial solutions.

Eigenvalue Problems. Recall the matrix equation

Ax = λx (17)

that we discussed in Section 7.3. Equation (17) has the solution x = 0 for every value

of λ, but for certain values of λ, called eigenvalues, there are also nonzero solutions,

called eigenvectors. The situation is similar for boundary value problems.

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10.1 Two-Point Boundary Value Problems 593

Consider the problem consisting of the differential equation

′′

+ λy = 0, (18)

together with the boundary conditions

y(0) = 0, y(π) = 0. (19)

Observe that the problem (18), (19) is the same as the problems in Examples 3

and 4 if λ = 2 and λ = 1, respectively. Recalling the results of these examples, we

note that for λ = 2, Eqs. (18), (19) have only the trivial solution y = 0, while for

λ = 1, the problem (18), (19) has other, nontrivial solutions. By extension of the

terminology associated with Eq. (17), the values of λ for which nontrivial solutions

of (18), (19) occur are called eigenvalues,and the nontrivial solutions themselves are

called eigenfunctions. Restating the results of Examples 3 and 4, we have found that

λ = 1 is an eigenvalue of the problem (18), (19) and that λ = 2 is not. Further, any

nonzero multiple of sin x is an eigenfunction corresponding to the eigenvalue λ = 1.

Let us now turn to the problem of ﬁnding other eigenvalues and eigenfunctions

of the problem (18), (19). We need to consider separately the cases λ>0, λ = 0, and

λ<0, because the form of the solution of Eq. (18) is different in each of these cases.

λ>0. To avoid the frequent appearance of radical signs, it is convenient in this case

to let λ = µ

and to rewrite Eq. (18) as

′′

+ µ

y = 0. (20)

The characteristic polynomial equation for Eq. (20) is r

+ µ

= 0 with roots

r =±iµ, so the general solution is

y = c

cos µx + c

sin µx. (21)

Note that µ is nonzero (since λ>0) and there is no loss of generality if we also

assume that µ is positive.The ﬁrst boundary condition requires that c

= 0, and then

the second boundary condition reduces to

sin µπ = 0. (22)

We are seeking nontrivial solutions, so we must require that c

̸= 0. Consequently,

sin µπ must be zero, and our task is to choose µ so that this will occur. We know that

the sine function has the value zero at every integer multiple of π , so we can choose

µ to be any (positive) integer. The corresponding values of λ are the squares of the

positive integers, so we have determined that

= 1, λ

= 4, λ

= 9, ..., λ

= n

, ... (23)

are eigenvalues of the problem (18), (19). The eigenfunctions are given by Eq. (21)

with c

= 0, so they are just multiples of the functions sin nx for n = 1, 2, 3, ....

Observe that the constant c

in Eq. (21) is never determined, so eigenfunctions are

determined only up to an arbitrary multiplicative constant [just as are the eigenvec-

tors of the matrix problem (17)]. We will usually choose the multiplicative constant

to be 1 and write the eigenfunctions as

(x) = sin x, y

(x) = sin 2x, ..., y

(x) = sin nx, ..., (24)

remembering that multiples of these functions are also eigenfunctions.

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594 Chapter 10. Partial Differential Equations and Fourier Series

λ<0. In this case we let λ =−µ

, so that Eq. (18) becomes

′′

− µ

y = 0. (25)

The characteristic equation for Eq. (25) is r

− µ

= 0 with roots r =±µ, so its

general solution can be written as

y = c

cosh µx + c

sinh µx. (26)

We have chosen the hyperbolic functions cosh µx and sinh µx, rather than the

exponential functions exp(µx) and exp(−µx), as a fundamental set of solutions

for convenience in applying the boundary conditions. The ﬁrst boundary condition

requires that c

= 0; then the second boundary condition gives c

sinh µπ = 0. Since

µ ̸= 0, it follows that sinh µπ ̸= 0, and therefore we must have c

= 0. Consequently,

y = 0 and there are no nontrivial solutions for λ<0. In other words, the problem

(18), (19) has no negative eigenvalues.

λ = 0. Now Eq. (18) becomes

′′

= 0, (27)

and its general solution is

y = c

x + c

. (28)

The boundary conditions (19) can be satisﬁed only by choosing c

= 0 and c

= 0,

so there is only the trivial solution y = 0 in this case as well. That is, λ = 0 is not an

eigenvalue.

To summarize our results: we have shown that the problem (18), (19) has an

inﬁnite sequence of positive eigenvalues λ

= n

for n = 1, 2, 3, ...and that the cor-

responding eigenfunctions are proportional to sin nx. Further,there are no other real

eigenvalues.There remains the possibility that theremight be some complexeigenval-

ues; recall that a matrix with real elements may very well have complex eigenvalues.

In Problem 23 we outline an argument showing that the particular problem (18),(19)

cannot have complex eigenvalues. Later,in Section 11.2,we discuss an important class

of boundary value problems that includes (18), (19). One of the useful properties of

this class of problems is that all their eigenvalues are real.

In later sections of this chapter, we will often encounter the problem

′′

+ λy = 0, y(0) = 0, y(L) = 0, (29)

which differs from the problem (18),(19) only in that the second boundary condition

is imposed at an arbitrary point x = L rather than at x = π. The solution process

for λ>0 is exactly the same as before, up to the step where the second boundary

condition is applied. For the problem (29) this condition requires that

sin µL = 0 (30)

rather than Eq. (22), as in the former case. Consequently, µL must be an integer

multiple of π,soµ = nπ/L, where n is a positive integer. Hence the eigenvalues and

eigenfunctions of the problem (29) are given by

= n

, y

(x) = sin(nπx/L), n = 1, 2, 3,.... (31)

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10.1 Two-Point Boundary Value Problems 595

As usual, the eigenfunctions y

(x) are determined only up to an arbitrary multiplica-

tive constant. In the same way as for the problem (18), (19), you can show that the

problem (29) has no eigenvalues or eigenfunctions other than those in Eq. (31).

The problems following this section explore to some extent the effect of differ-

ent boundary conditions on the eigenvalues and eigenfunctions. A more systematic

discussion of two-point boundary and eigenvalue problems appears in Chapter 11.

PROBLEMS

In each of Problems 1 through 13, either solve the given boundary value problem or else show

that it has no solution.

1. y

′′

+ y = 0, y(0) = 0, y

′

(π) = 12.y

′′

+ 2y = 0, y

′

(0) = 1, y

′

(π) = 0

3. y

′′

+ y = 0, y(0) = 0, y(L) = 04.y

′′

+ y = 0, y

′

(0) = 1, y(L) = 0

5. y

′′

+ y = x, y(0) = 0, y(π) = 06.y

′′

+ 2y = x, y(0) = 0, y(π) = 0

7. y

′′

+ 4y = cos x, y(0) = 0, y(π) = 08.y

′′

+ 4y = sin x, y(0) = 0, y(π) = 0

9. y

′′

+ 4y = cos x, y

′

(0) = 0, y

′

(π) = 0 10. y

′′

+ 3y = cos x, y

′

(0) = 0, y

′

(π) = 0

11. x

′′

− 2xy

′

+ 2y = 0, y(1) =−1, y(2) = 1

12. x

′′

+ 3xy

′

+ y = x

, y(1) = 0, y(e) = 0

13. x

′′

+ 5xy

′

+ (4 + π

)y = ln x, y(1) = 0, y(e) = 0

In each of Problems 14 through 20, ﬁnd the eigenvalues and eigenfunctions of the given

boundary value problem. Assume that all eigenvalues are real.

14. y

′′

+ λy = 0, y(0) = 0, y

′

(π) = 0 15. y

′′

+ λy = 0, y

′

(0) = 0, y(π) = 0

16. y

′′

+ λy = 0, y

′

(0) = 0, y

′

(π) = 0 17. y

′′

+ λy = 0, y

′

(0) = 0, y(L) = 0

18. y

′′

+ λy = 0, y

′

(0) = 0, y

′

(L) = 0 19. y

′′

− λy = 0, y(0) = 0, y

′

(L) = 0

20. x

′′

− xy

′

+ λy = 0, y(1) = 0, y(L) = 0, L > 1

21. The axially symmetric laminar ﬂow of a viscous incompressible ﬂuid through a long

straight tube of circular cross section under a constant axial pressure gradient is known as

Poiseuille

ﬂow. The axial velocity w is a function of the radial variable r only and satisﬁes

the boundary value problem

′′

′

=−

, w(R) = 0, w(r) bounded for 0 < r < R,

where R is the radius of the tube, G is the pressure gradient, and µ is the coefﬁcient of

viscosity of the ﬂuid.

(a) Find the velocity proﬁle w(r).

(b) By integrating w(r) over a cross section, show that the total ﬂow rate Q is given by

Q = π R

G/8µ.

Since Q, R, and G can be measured, this result provides a practical way to determine the

viscosity µ.

reduction in Q? This result has implications for blood ﬂow through arteries constricted by

plaque.

Jean Louis Marie Poiseuille (1797–1869) was a French physician who was also trained in mathematics

and physics. He was particularly interested in the ﬂow of blood and published his ﬁrst paper on the subject

in 1840.

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596 Chapter 10. Partial Differential Equations and Fourier Series

22. Consider a horizontal metal beam of length L subject to a vertical load f (x) per unit length.

The resulting vertical displacement in the beam y(x) satisﬁes the differential equation

= f (x),

where E is Young’s modulus and I is the moment of inertia of the cross section about

an axis through the centroid perpendicular to the xy-plane. Suppose that f (x)/EI is a

constant k. For each of the boundary conditions given below, solve for the displacement

y(x), and plot y versus x in the case that L = 1 and k =−1.

(a) Simply supported at both ends: y(0) = y

′′

(0) = y(L) = y

′′

(L) = 0

(b) Clamped at both ends: y(0) = y

′

(0) = y(L) = y

′

(L) = 0

′

(0) = y

′′

(L) = y

′′′

(L) = 0

23. In this problem we outline a proof that the eigenvalues of the boundary value problem

(18), (19) are real.

(a) Write the solution of Eq. (18) as y = k

exp(iµx) + k

exp(−iµx), where λ = µ

, and

impose the boundary conditions (19). Show that nontrivial solutions exist if and only if

exp(iµπ) − exp(−iµπ) = 0. (i)

(b) Let µ = ν + iσ and use Euler’s relation exp(iνπ) = cos(νπ) + i sin(νπ) to determine

the real and imaginary parts of Eq. (i).

σ = 0. Consequently, µ is real and so is λ.

10.2 Fourier Series

Later inthis chapteryou willﬁnd that you can solve many important problems involv-

ing partial differential equations, provided that you can express a given function as

an inﬁnite sum of sines and/or cosines. In this and the following two sections we

explain in detail how this can be done. These trigonometric series are called Fourier

series

; they are somewhat analogous to Taylor series in that both types of series pro-

vide a means of expressing quite complicated functions in terms of certain familiar

elementary functions.

Jean Baptiste Joseph Fourier (1768–1830) was twice imprisoned during the French Revolution, later

served as scientiﬁc advisor in Napoleon’s army in Egypt, and was prefect of the department of Isère

(Grenoble) from 1801 to 1815. He made the ﬁrst systematic use, although not a completely rigorous

investigation, of trigonometric series in 1807 and 1811 in his papers on heat conduction. The papers were

not published due to objections from the referees, principally Lagrange. Although it turned out that

Fourier’s claim of generality was somewhat too strong, his results inspired a ﬂood of important research

that has continued to the present day. See Grattan-Guinness or Carslaw (Historical Introduction) for a

detailed history of Fourier series.

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10.2 Fourier Series 597

We begin with a series of the form

∞

m=1

)

cos

mπx

+ b

sin

mπx

. (1)

On the set of points where the series (1) converges, it deﬁnes a function f , whose

value at each point is the sum of the series for that value of x. In this case the series

(1) is said to be the Fourier series for f . Our immediate goals are to determine what

functions can be represented as a sum of a Fourier series and to ﬁnd some means

of computing the coefﬁcients in the series corresponding to a given function. The

ﬁrst term in the series (1) is written as a

/2 rather than as a

to simplify a formula

for the coefﬁcients that we derive below. Besides their association with the method

of separation of variables and partial differential equations, Fourier series are also

useful in various other ways,such as in the analysis of mechanical or electrical systems

acted on by periodic external forces.

Periodicity of the Sine and Cosine Functions. To discuss Fourier series, it is necessary

to develop certain properties of the trigonometric functions sin(mπx/L) and

cos(mπx/L), where m is a positive integer. The ﬁrst property is their periodic charac-

ter. A function f is said to be periodic with period T > 0 if the domain of f contains

x + T whenever it contains x, and if

f (x + T) = f (x) (2)

for every value of x. An example of a periodic function is shown in Figure 10.2.1. It

follows immediately from the deﬁnition that if T is a period of f , then 2T is also a

period, and so indeed is any integral multiple of T. The smallest value of T for which

Eq. (2) holds is called the fundamental period of f . A constant function is a periodic

function with an arbitrary period but no fundamental period.

FIGURE 10.2.1 A periodic function of period T.

If f and g are any two periodic functions with common period T, then any linear

combination c

f + c

g is also periodic with period T. To prove this statement, let

F(x) = c

f (x) + c

g(x); then, for any x,

F(x + T) = c

f (x + T) + c

g(x + T) = c

f (x) + c

g(x) = F(x). (3)

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598 Chapter 10. Partial Differential Equations and Fourier Series

Moreover, it can be shown that the sum of any ﬁnite number, or even the sum of a

convergent inﬁnite series, of functions of period T is also periodic with period T.In

a similar way, you can show that the product fg is periodic with period T.

In particular, the functions sin(mπx/L) and cos(mπx/L), m = 1, 2, 3, ..., are peri-

odic with fundamental period T = 2L/m. To see this, recall that sin x and cos x have

fundamental period 2π and that sin αx and cos αx have fundamental period 2π/α.If

we choose α = mπ/L, then the period T of sin(mπx/L) and cos(mπx/L) is given by

T = 2πL/mπ = 2L/m.

Note also that, since every positive integral multiple of a period is also a period,

each of the functions sin(mπx/L) and cos(mπx/

L) has the common period 2L.

Orthogonality of the Sine and Cosine Functions. To describe a second essential property of

the functions sin(mπx/L) and cos(mπx /L), we generalize the concept of orthogonal-

ity of vectors (see Section 7.2). The standard inner product (u, v) of two real-valued

functions u and v on the interval α ≤ x ≤ β is deﬁned by

(u, v) =

u(x)v(x) dx. (4)

The functions u and v are said to be orthogonal on α ≤ x ≤ β if their inner product

is zero—that is, if

u(x)v(x) dx = 0. (5)

A set of functions is said to be mutually orthogonal if each distinct pair of functions

in the set is orthogonal.

The functions sin(mπx/L) and cos(mπx/L), m = 1, 2, ...form a mutually orthog-

onal set of functions on the interval −L ≤ x ≤ L. In fact, they satisfy the following

orthogonality relations:

−L

cos

mπx

cos

nπx

dx =

0, m ̸= n,

L, m = n;

(6)

−L

cos

mπx

sin

nπx

dx = 0, all m, n; (7)

−L

sin

mπx

sin

nπx

dx =

0, m ̸= n,

L, m = n.

(8)

These results can be obtained by direct integration. For example, to derive Eq. (8),

note that

−L

sin

mπx

sin

nπx

dx =

−L

cos

(m − n)πx

− cos

(m + n)πx

sin[(m − n)πx/L]

m − n

−

sin[(m + n)πx/L]

m + n

−L

= 0

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10.2 Fourier Series 599

as long as m + n and m − n are not zero. Since m and n are positive, m + n ̸= 0. On

the other hand, if m − n = 0, then m = n, and the integral must be evaluated in a

different way. In this case

−L

sin

mπx

sin

nπx

dx =

−L

)

sin

mπx

−L

1 − cos

2mπx

x −

sin(2mπx/L)

2mπ/L

−L

= L.

This establishes Eq. (8); Eqs. (6) and (7) can be veriﬁed by similar computations.

The Euler–Fourier Formulas. Now let us suppose that a series of the form (1) converges,

and let us call its sum f (x):

f (x) =

∞

m=1

)

cos

mπx

+ b

sin

mπx

. (9)

The coefﬁcientsa

and b

can be related tof (x) as a consequence ofthe orthogonality

conditions (6), (7), and (8). First multiply Eq. (9) by cos(nπx/L), where n is a ﬁxed

positive integer (n > 0), and integrate with respect to x from −L to L. Assuming that

the integration can be legitimately carried out term by term,

we obtain

−L

f (x) cos

nπx

dx =

−L

cos

nπx

dx +

∞

m=1

−L

cos

mπx

cos

nπx

∞

m=1

−L

sin

mπx

cos

nπx

dx. (10)

Keeping in mind that n is ﬁxed whereas m ranges over the positive integers, it follows

from the orthogonality relations (6) and (7) that the only nonzero term on the right

side of Eq. (10) is the one for which m = n in the ﬁrst summation. Hence

−L

f (x) cos

nπx

dx = La

, n = 1, 2, .... (11)

To determine a

, we can integrate Eq. (9) from −L to L, obtaining

−L

f (x) dx =

−L

dx +

∞

m=1

−L

cos

mπx

dx +

∞

m=1

−L

sin

mπx

= La

, (12)

This is a nontrivial assumption, because not all convergent series with variable terms can be so integrated.

For the special case of Fourier series, however, term-by-term integration can always be justiﬁed.

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600 Chapter 10. Partial Differential Equations and Fourier Series

since each integral involving a trigonometric function is zero. Thus

−L

f (x) cos

nπx

dx, n = 0, 1, 2, .... (13)

By writing the constant term in Eq. (9) as a

/2, it is possible to compute all the a

from Eq. (13). Otherwise, a separate formula would have to be used for a

A similar expression for b

may be obtained by multiplying Eq. (9) by sin(nπx/L),

integrating termwise from −L to L, and using the orthogonality relations (7) and (8);

thus

−L

f (x) sin

nπx

dx, n = 1, 2, 3, .... (14)

Equations (13) and (14) are known as the Euler–Fourier formulas for the coefﬁcients

in a Fourier series. Hence, if the series (9) converges to f (x), and if the series can be

integrated term by term, then the coefﬁcients must be given by Eqs. (13) and (14).

Note that Eqs. (13) and (14) are explicit formulas for a

and b

in terms of f , and

that the determination of any particular coefﬁcient is independent of all the other

coefﬁcients. Of course, the difﬁculty in evaluating the integrals in Eqs. (13) and (14)

depends very much on the particular function f involved.

Note also that the formulas (13) and (14) depend only on the values of f (x) in the

interval −L ≤ x ≤ L. Since each of the terms in the Fourier series (9) is periodic with

period 2L, the series converges for all x whenever it converges in −L ≤ x ≤ L, and

its sum is also a periodic function with period 2L. Hence f (x) is determined for all x

by its values in the interval −L ≤ x ≤ L.

It is possible to show (see Problem 27) that if g is periodic with period T,then every

integral of g over an interval of length T has the same value. If we apply this result to

the Euler–Fourier formulas (13) and (14), it follows that the interval of integration,

−L ≤ x ≤ L, can be replaced, if it is more convenient to do so, by any other interval

of length 2L.

EXAMPLE

Assume that there is a Fourier series converging to the function f deﬁned by

f (x) =

−x, −2 ≤ x < 0,

x,0≤ x < 2;

(15)

f (x + 4) = f (x ).

Determine the coefﬁcients in this Fourier series.

This function represents a triangular wave (see Figure 10.2.2) and is periodic with period 4.

Thus in this case L = 2, and the Fourier series has the form

f (x) =

∞

m=1

)

cos

mπx

+ b

sin

mπx

, (16)

where the coefﬁcients are computed from Eqs. (13) and (14) with L = 2. Substituting for f (x)

in Eq. (13) with m = 0, we have

−2

(−x) dx +

xdx = 1 + 1 = 2. (17)

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10.2 Fourier Series 601

–6 –4 –2 2

FIGURE 10.2.2 The triangular wave in Example 1.

For m > 0, Eq. (13) yields

−2

(−x) cos

mπx

dx +

x cos

mπx

dx.

These integrals can be evaluated through integration by parts, with the result that

−

mπ

x sin

mπx

−

mπ

(

cos

mπx

−2

mπ

x sin

mπx

mπ

(

cos

mπx

−

mπ

(

mπ

(

cos mπ +

mπ

(

cos mπ −

mπ

(

(mπ)

(cos mπ − 1), m = 1, 2, ...

−8/(mπ)

, m odd,

0, m even.

(18)

Finally, from Eq. (14) it follows in a similar way that

= 0, m = 1, 2, .... (19)

By substituting the coefﬁcients from Eqs. (17), (18), and (19) in the series (16), we obtain the

Fourier series for f :

f (x) = 1 −

cos

πx

cos

3πx

cos

5πx

+···

(

= 1 −

∞

m=1,3,5,...

cos(mπx/2)

= 1 −

∞

n=1

cos(2n − 1)πx/2

(2n − 1)

. (20)

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602 Chapter 10. Partial Differential Equations and Fourier Series

EXAMPLE

Let

f (x) =

⎧

⎪

⎨

⎪

⎩

0, −3 < x < −1,

1, −1 < x < 1,

0, 1 < x < 3

(21)

and suppose that f (x + 6) = f (x); see Figure 10.2.3. Find the coefﬁcients in the Fourier

series for f .

–7 –5 –3 –1 1

357

FIGURE 10.2.3 Graph of f (x) in Example 2.

Note that f (x) is not assigned a value at the points of discontinuity, such as x =−1 and

x = 1. This has no effect on the values of the Fourier coefﬁcients, because they result from the

evaluation of integrals,and the value of an integral is not affected by the value of the integrand

at a single point or at a ﬁnite number of points. Thus the coefﬁcients are the same regardless

of what value, if any, f (x) is assigned at a point of discontinuity.

Since f has period 6, it follows that L = 3 in this problem. Consequently, the Fourier series

for f has the form

f (x) =

∞

n=1

)

cos

nπx

+ b

sin

nπx

, (22)

where the coefﬁcients a

and b

are given by Eqs. (13) and (14) with L = 3. We have

−3

f (x) dx =

−1

1 dx =

. (23)

Similarly,

−1

cos

nπx

dx =

nπ

sin

nπx

−1

nπ

sin

nπ

, n = 1, 2, ..., (24)

and

−1

sin

nπx

dx =−

nπ

cos

nπx

−1

= 0, n = 1, 2, .... (25)

Thus the Fourier series for f is

f (x) =

∞

n=1

nπ

sin

nπ

cos

nπx

√

cos(πx/3) +

cos(2πx/3)

−

cos(4π x/3)

−

cos(5πx/3)

+···

. (26)

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10.2 Fourier Series 603

EXAMPLE

Consider again the function in Example 1 and its Fourier series (20). Investigate the speed

with which the series converges. In particular, determine how many terms are needed so that

the error is no greater than 0.01 for all x.

The mth partial sum in this series

(x) = 1 −

n=1

cos(2n − 1)πx/2

(2n − 1)

(27)

can be used to approximate the function f . The coefﬁcients diminish as (2n − 1)

−2

,so the series

converges fairly rapidly. This is borne out by Figure 10.2.4, where the partial sums for m = 1

and m = 2 are plotted. To investigate the convergence in more detail,we can consider the error

(x) = f (x) − s

(x). Figure 10.2.5 shows a plot of |e

(x)| versus x for 0 ≤ x ≤ 2. Observe that

(x)| is greatest at the points x = 0 and x = 2, where the graph of f (x) has corners. It is more

difﬁcult for the series to approximate the function near these points, resulting in a larger error

there for a given m. Similar graphs are obtained for other values of m.

m = 2

m = 1

–4 –2

FIGURE 10.2.4 Partial sums in the Fourier series, Eq. (20), for the triangular wave.

0.025

0.020

0.015

0.010

0.005

(x)

0.030

0.035

1.51 20.5

FIGURE 10.2.5 Plot of |e

(x)| versus x for the triangular wave.

Once you realize that the maximum error always occurs at x = 0orx = 2, you can obtain

a uniform error bound for each m simply by evaluating |e

(x)| at one of these points. For

example,for m = 6 we have e

(2) = 0.03370,so |e

(x)| < 0.034 for 0 ≤ x ≤ 2 and consequently

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604 Chapter 10. Partial Differential Equations and Fourier Series

for all x. Table 10.2.1 shows corresponding data for other values of m; these data are plotted

in Figure 10.2.6. From this information you can begin to estimate the number of terms that

are needed in the series in order to achieve a given level of accuracy in the approximation.

For example, to guarantee that |e

(x)|≤0.01, we need to choose m = 21, for which the error

is e

(2) = 0.00965.

TABLE 10.2.1 Values of the

Error e

(2) for the Triangular

Wave

(2)

2 0.09937

4 0.05040

6 0.03370

10 0.02025

15 0.01350

20 0.01013

25 0.00810

0.08

0.06

0.04

0.02

(2)

0.10

1510 20 255

FIGURE 10.2.6 Plot of e

(2) versus m for the triangular wave.

In this book,Fourier series appear mainly as a means of solving certain problems in

partial differential equations. However, such series have much wider application

in science and engineering and, in general, are valuable tools in the investigation of

periodic phenomena. A basic problem is to resolve an incoming signal into its har-

monic components, which amounts to constructing its Fourier series representation.

In some frequency ranges, the separate terms correspond to different colors or to

different audible tones. The magnitude of the coefﬁcient determines the amplitude

of each component. This process is referred to as spectral analysis.

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10.2 Fourier Series 605

PROBLEMS In each of Problems 1 through 8, determine whether the given function is periodic. If so, ﬁnd

its fundamental period.

1. sin 5x 2. cos 2πx 3. sinh 2x 4. sin πx/L 5. tan πx 6. x

7. f (x) =

0, 2n − 1 ≤ x < 2n,

1, 2n ≤ x < 2n + 1;

n = 0, ±1, ±2, ...

8. f (x) =

(−1)

,2n − 1 ≤ x < 2n,

1, 2n ≤ x < 2n + 1;

n = 0, ±1, ±2, ...

9. If f (x) =−x for −L < x < L,and iff (x + 2L) = f (x),ﬁnd a formula for f (x) in the interval

L < x < 2L; in the interval −3L < x < −2L.

10. If f (x) =

x + 1, −1 < x < 0,

x,0< x < 1,

and if f (x + 2) = f (x),

ﬁnd

a formula for f (x) in the

interval 1 < x < 2; in the interval 8 < x < 9.

11. If f (x) = L − x for 0 < x < 2L, and if f (x + 2L) = f (x ), ﬁnd a formula for f (x) in the

interval −L < x < 0.

12. Verify Eqs. (6) and (7) in this section by direct integration.

In each of Problems 13 through 18:

(a) Sketch the graph of the given function for three periods.

(b) Find the Fourier series for the given function.

13. f (x) =−x, −L ≤ x < L; f (x + 2L) = f (x)

14. f (x) =

1, −L ≤ x < 0,

0 ≤ x < L

;

f (x + 2L) = f (x)

15. f (x) =

x, −π ≤ x < 0,

0, 0 ≤ x <π;

f (x + 2π) = f (x)

16. f (x) =

x + 1, −1 ≤ x < 0,

1 − x,0≤ x < 1;

f (x + 2) = f (x)

17. f (x) =

x + L, −L ≤ x ≤ 0,

L,0< x < L;

f (x + 2L) = f (x)

18. f (x) =

⎧

⎪

⎨

⎪

⎩

0, −2 ≤ x ≤−1,

x, −1 < x < 1,

1 ≤ x < 2;

f (

x + 4) = f (x)

In each of Problems 19 through 24:

(a) Sketch the graph of the given function for three periods.

(b) Find the Fourier series for the given function.

(x) versus x for m = 5, 10, and 20.

(d) Describe how the Fourier series seems to be converging.

19.

f (x) =

−1, −2 ≤ x < 0,

1, 0 ≤ x < 2;

f (x + 4) = f (x )

20.

f (x) = x, −1 ≤ x < 1; f (x + 2) = f (x)

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606 Chapter 10. Partial Differential Equations and Fourier Series

21. f (x) = x

/2, −2 ≤ x ≤ 2; f (x + 4) = f (x)

22.

f (x) =

x + 2, −2 ≤ x < 0,

2 − 2x,0≤ x < 2;

f (x + 4) = f (x )

23.

f (x) =

−

x, −2 ≤ x < 0,

2x −

,0≤ x < 2;

f (x + 4) = f (x )

24.

f (x) =

0, −3 ≤ x ≤ 0,

(3 − x),0< x < 3;

f (x + 6) = f (x)

25.

Consider the function f deﬁned in Problem 21, and let e

(x) = f (x) − s

(x).

(a) Plot |e

(x)| versus x for 0 ≤ x ≤ 2 for several values of m.

(b) Find the smallest value of m for which |e

(x)|≤0.01 for all x.

26.

Consider the function f deﬁned in Problem 24, and let e

(x) = f (x) − s

(x).

(a) Plot |e

(x)| versus x for 0 ≤ x ≤ 3 for several values of m.

(b) Find the smallest value of m for which |e

(x)|≤0.1 for all x.

27. Suppose that g is an integrable periodic function with period T.

(a) If 0 ≤ a ≤ T, show that

g(x) dx =

a+T

g(x) dx.

Hint: Show ﬁrst that

g(x) dx =

a+T

g(x) dx. In the second integral, consider the change

of variable s = x − T.

(b) Show that for any value of a, not necessarily in 0 ≤ a ≤ T,

g(x) dx =

a+T

g(x) dx.

a+T

g(x) dx =

b+T

g(x) dx.

28. If f is differentiable and is periodic with period T, show that f

′

is also periodic with

period T. Determine whether

F(x) =

f (t) dt

is always periodic.

29. In this problem we indicate certain similarities between three-dimensional geometric

vectors and Fourier series.

(a) Let v

, v

, and v

be a set of mutually orthogonal vectors in three dimensions, and let

u be any three-dimensional vector. Show that

u = a

+ a

, (i)

where

u · v

· v

, i = 1, 2, 3. (ii)

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10.3 The Fourier Convergence Theorem 607

Show that a

can be interpreted as the projection of u in the direction of v

divided by the

length of v

(b) Deﬁne the inner product (u, v) by

(u, v) =

−L

u(x)v(x) dx. (iii)

Also let

(x) = cos(nπx/L), n = 0, 1, 2, ...;

(iv)

(x) = sin(nπx/L), n = 1, 2, ....

Show that Eq. (10) can be written in the form

(f , φ

) =

(φ

, φ

) +

∞

m=1

(φ

, φ

) +

∞

m=1

(ψ

, φ

). (v)

), together with the orthogonal-

ity relations, to show that

(f , φ

)

(φ

, φ

)

, n = 0, 1, 2, ...; b

(f , ψ

)

(ψ

, ψ

)

, n = 1, 2, .... (vi)

Note the resemblance between Eqs. (vi) and Eq. (ii). The functions φ

and ψ

play a role

for functions similar to that of the orthogonal vectors v

, v

, and v

in three-dimensional

space. The coefﬁcients a

and b

can be interpreted as projections of the function f onto

the base functions φ

and ψ

Observe also that any vector in three dimensions can be expressed as a linear combi-

nation of three mutually orthogonal vectors. In a somewhat similar way, any sufﬁciently

smooth function deﬁned on −L ≤ x ≤ L can be expressed as a linear combination of the

mutually orthogonal functions cos(nπx/L) and sin(nπx/L), that is, as a Fourier series.

10.3 The Fourier Convergence Theorem

In the preceding section we showed that if the Fourier series

∞

m=1

)

cos

mπx

+ b

sin

mπx

(1)

converges and thereby deﬁnes a function f ,then f is periodic with period 2L, and the

coefﬁcients a

and b

are related to f (x) by the Euler–Fourier formulas:

−L

f (x) cos

mπx

dx, m = 0, 1, 2, ...; (2)

−L

f (x) sin

mπx

dx, m = 1, 2, .... (3)

In this section we suppose that a function f is given. If this function is periodic with

period 2L and integrable on the interval [−L, L], then a set of coefﬁcients a

and

can be computed from Eqs. (2) and (3), and a series of the form (1) can be

formally constructed. The question is whether this series converges for each value

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608 Chapter 10. Partial Differential Equations and Fourier Series

of x and, if so, whether its sum is f (x). Examples have been discovered showing that

the Fourier series corresponding to a function f may not converge to f (x) and may

even diverge. Functions whose Fourier series do not converge to the value of the

function at isolated points are easily constructed, and examples will be presented

later in this section. Functions whose Fourier series diverge at one or more points

are more pathological, and we will not consider them in this book.

To guarantee convergence of a Fourier series to the function from which its

coefﬁcients were computed, it is essential to place additional conditions on the func-

tion. From a practical point of view,such conditions should be broad enough to cover

all situations of interest, yet simple enough to be easily checked for particular func-

tions. Through the years, several sets of conditions have been devised to serve this

purpose.

Before stating a convergence theorem for Fourier series, we deﬁne a term

that appears in the theorem. A function f is said to be piecewise continuous on

an interval a ≤ x ≤ b if the interval can be partitioned by a ﬁnite number of points

a = x

< x

< ···< x

= b so that

1. f is continuous on each open subinterval x

i−1

< x < x

2. f approaches a ﬁnite limit as the endpoints of each subinterval are approached from within

the subinterval.

The graph of a piecewise continuous function is shown in Figure 10.3.1.

FIGURE 10.3.1 A piecewise continuous function.

The notation f (c+) is used to denote the limit of f (x) as x → c from the right;

similarly, f (c−) denotes the limit of f (x) as x approaches c from the left.

Note that it is not essential that the function even be deﬁned at the partition points

. For example, in the following theorem we assume that f

′

is piecewise continuous;

but certainly f

′

does not exist at those points where f itself is discontinuous. It is also

not essential that the interval be closed; it may also be open, or open at one end and

closed at the other.

Theorem 10.3.1

Suppose that f and f

′

are piecewise continuouson the interval −L ≤ x < L. Further,

suppose that f is deﬁned outside the interval −L ≤ x < L so that it is periodic with

period 2L. Then f has a Fourier series

f (x) =

∞

m=1

)

cos

mπx

+ b

sin

mπx

, (4)

whose coefﬁcients are given by Eqs. (2) and (3). The Fourier series converges to

f (x) at all points where f is continuous, and it converges to [f (x+) + f (x−)]/2 at all

points where f is discontinuous.

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10.3 The Fourier Convergence Theorem 609

Note that [f (x+) + f (x−)]/2 is the mean value of the right- and left-hand limits

at the point x. At any point where f is continuous, f (x+) = f (x −) = f (x). Thus it is

correct to say that the Fourier series converges to [f (x+) + f (x−)]/2 at all points.

Whenever we say that a Fourier series converges to a function f , we always mean

that it converges in this sense.

We emphasize that the conditions given in this theorem are only sufﬁcient for the

convergence of a Fourier series; they are by no means necessary. Nor are they

the most general sufﬁcient conditions that have been discovered. In spite of this,

the proof of the theorem is fairly difﬁcult, and we do not discuss it here.

Under more

restrictive conditions a much simpler convergence proof is possible; see Problem 18.

To obtain a better understanding of the content of the theorem, it is helpful to con-

sider some classes of functions that fail to satisfy the assumed conditions. Functions

that are not included in the theorem are primarily those with inﬁnite discontinuities

in the interval [−L, L],such as 1/x

as x → 0,or ln |x − L|as x → L. Functions having

an inﬁnite number of jump discontinuities in this interval are also excluded;however,

such functions are rarely encountered.

It is noteworthy that a Fourier series may converge to a sum that is not differ-

entiable, or even continuous, in spite of the fact that each term in the series (4) is

continuous, and even differentiable inﬁnitely many times. The example below is an

illustration of this, as is Example 2 in Section 10.2.

EXAMPLE

Let

f (x) =

0, −L < x < 0,

L,0< x < L,

(5)

and let f be deﬁned outside this interval so that f (x + 2L) = f (x ) for all x. We will temporarily

leave open the deﬁnition of f at the points x = 0, ±L. Find the Fourier series for this function

and determine where it converges.

x3L2LL

–L–2L–3L

FIGURE 10.3.2 The square wave in Example 1.

The equation y = f (x) has the graph shown in Figure 10.3.2, extended to inﬁnity in both

directions. It can be thought of as representing a square wave. The interval [−L, L] can be

partitioned into the two open subintervals (−L,0) and (0, L).In(0, L), f (x) = L and f

′

(x) = 0.

Clearly, both f and f

′

are continuous and furthermore have limits as x → 0 from the right

and as x → L from the left. The situation in (−L,0) is similar. Consequently, both f and f

′

are piecewise continuous on [−L, L),sof satisﬁes the conditions of Theorem 10.3.1. If the

Proofs of the convergence of a Fourier series can be found in most books on advanced calculus. See, for

example, Kaplan (Chapter 7) or Buck (Chapter 6).

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610 Chapter 10. Partial Differential Equations and Fourier Series

coefﬁcients a

and b

are computed from Eqs. (2) and (3), the convergence of the resulting

Fourier series to f (x) is ensured at all points where f is continuous. Note that the values of a

and b

are the same regardless of the deﬁnition of f at its points of discontinuity. This is true

because the value of an integral is unaffected by changing the value of the integrand at a ﬁnite

number of points. From Eq. (2),

−L

f (x) dx =

dx = L

and

−L

f (x) cos

mπx

dx =

cos

mπx

mπ

(sin mπ − 0) = 0, m ̸= 0.

Similarly, from Eq. (3),

−L

f (x) sin

mπx

dx =

sin

mπx

mπ

(1 − cos mπ) =

0, m even;

2L/mπ, m odd.

Hence

f (x) =

sin

πx

sin

3πx

sin

5πx

+···

(

∞

m=1,3,5,...

sin(mπx/L)

∞

n=1

sin(2n − 1)πx/L

2n − 1

. (6)

At the points x = 0, ±nL, where the function f in the example is not continuous, all terms in

the series after the ﬁrst vanish and the sum is L/2. This is the mean value of the limits from the

right and left, as it should be. Thus we might as well deﬁne f at these points to have the value

L/2. If we choose to deﬁne it otherwise, the series still gives the value L/2 at these points,

since all of the preceding calculations remain valid. The series simply does not converge to

the function at those points unless f is deﬁned to have the value L/2. This illustrates the

possibility that the Fourier series corresponding to a function may not converge to it at points

of discontinuity unless the function is suitably deﬁned at such points.

The manner in which the partial sums

(x) =

sin

πx

+···+

2n − 1

sin

(2n − 1)πx

(

, n = 1, 2, ...

of the Fourier series (6) converge to f (x) is indicated in Figure 10.3.3,where L has been chosen

to be 1 and the graph of s

(x) is plotted.The ﬁgure suggests that at points where f is continuous,

the partial sums do approach f (x) as n increases. However, in the neighborhood of points of

discontinuity, such as x = 0 and x = L, the partial sums do not converge smoothly to the mean

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10.3 The Fourier Convergence Theorem 611

value. Instead,they tend to overshoot the mark at each end of the jump, as though they cannot

quite accommodate themselves to the sharp turn required at this point.This behavior is typical

of Fourier series at points of discontinuity and is known as the Gibbs

phenomenon.

Additional insight is attained by considering the error e

(x) = f (x) − s

(x). Figure 10.3.4

shows a plot of |e

(x)| versus x for n = 8 and for L = 1. The least upper bound of |e

(x)| is 0.5

and is approached as x → 0 and as x → 1. As n increases, the error decreases in the interior

of the interval [where f (x) is continuous], but the least upper bound does not diminish with

increasing n. Thus we cannot uniformly reduce the error throughout the interval by increasing

the number of terms.

Figures 10.3.3 and 10.3.4 also show that the series in this example converges more slowly

than the one in Example 1 in Section 10.2. This is due to the fact that the coefﬁcients in the

series (6) are proportional only to 1/(2n − 1).

2 2–11–

n = 8

FIGURE 10.3.3 The partial sum s

(x) in the Fourier series, Eq. (6), for the square wave.

10.4 0.6 8.02.0

0.4

0.3

0.2

0.1

(x)

0.5

FIGURE 10.3.4 A plot of the error |e

(x)| versus x for the square wave.

The Gibbs phenomenon is named after Josiah Willard Gibbs (1839–1903), who is better known for his

work on vector analysis and statistical mechanics. Gibbs was professor of mathematical physics at Yale

and one of the ﬁrst American scientists to achieve an international reputation. The Gibbs phenomenon is

discussed in more detail by Carslaw (Chapter 9).

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612 Chapter 10. Partial Differential Equations and Fourier Series

PROBLEMS In each of Problems 1 through 6, assume that the given function is periodically extended

outside the original interval.

(a) Find the Fourier series for the extended function.

(b) Sketch the graph of the function to which the series converges for three periods.

1. f (x) =

−1, −1 ≤ x < 0,

1, 0 ≤ x < 1

2. f (x) =

0, −π ≤ x < 0,

x,0≤ x <π

3. f (x) =

L + x, −L ≤ x < 0,

L − x,0≤ x < L

4. f (x) = 1 − x

, −1 ≤ x < 1

5. f (x) =

⎧

⎪

⎨

⎪

⎩

0, −π ≤ x < −π/2,

1, −π/2 ≤ x <π/2,

0, π/2 ≤ x <π

6. f (x) =

0, −1 ≤ x < 0,

,0≤ x < 1

In each of Problems 7 through 12, assume that the given function is periodically extended

outside the original interval.

(a) Find the Fourier series for the given function.

(b) Let e

(x) = f (x) − s

(x). Find the least upper bound or the maximum value (if it exists)

of |e

(x)| for n = 10, 20, and 40.

(x)|≤0.01 for all x.

f (x) =

x, −π ≤ x < 0,

0, 0 ≤ x <π;

f (x + 2π) = f (x) (see Section 10.2, Problem 15)

f (x) =

x + 1, −1 ≤ x < 0,

1 − x,0≤ x < 1;

f (x + 2) = f (x) (see Section 10.2, Problem 16)

f (x) = x, −1 ≤ x < 1; f (x + 2) = f (x) (see Section 10.2, Problem 20)

10.

f (x) =

x + 2, −2 ≤ x < 0,

2 − 2x,0≤ x < 2;

f (x + 4) = f (x ) (see Section 10.2, Problem 22)

11.

f (x) =

0, −1 ≤ x < 0,

,0≤ x < 1;

f (x + 2) = f (x) (see Problem 6)

12.

f (x) = x − x

, −1 ≤ x < 1; f (x + 2) = f (x)

Periodic Forcing Terms. In this chapter we are concerned mainly with the use of Fourier series

to solve boundary value problems for certain partial differential equations. However, Fourier

series are also useful in many other situations where periodic phenomena occur. Problems 13

through 16 indicate how they can be employed to solve initial value problems with periodic

forcing terms.

13. Find the solution of the initial value problem

′′

+ ω

y = sin nt, y(0) = 0, y

′

(0) = 0,

where n is a positive integer and ω

̸= n

. What happens if ω

= n

14. Find the formal solution of the initial value problem

′′

+ ω

y =

∞

n=1

sin nt, y(0) = 0, y

′

(0) = 0,

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10.3 The Fourier Convergence Theorem 613

where ω>0 is not equal to a positive integer. How is the solution altered if ω = m, where

m is a positive integer?

15. Find the formal solution of the initial value problem

′′

+ ω

y = f (t), y(0) = 0, y

′

(0) = 0,

where f is periodic with period 2π and

f (t) =

⎧

⎪

⎨

⎪

⎩

1, 0 < t <π;

0, t = 0, π,2π;

−1, π<t < 2π.

See Problem 1.

16. Find the formal solution of the initial value problem

′′

+ ω

y = f (t), y(0) = 1, y

′

(0) = 0,

where f is periodic with period 2 and

f (t) =

1 − t,0≤ t < 1;

−1 + t,1≤ t < 2.

See Problem 8.

17. Assuming that

f (x) =

∞

n=1

)

cos

nπx

+ b

sin

nπx

, (i)

show formally that

−L

[f (x)]

dx =

∞

n=1

+ b

This relation between a function f and its Fourier coefﬁcients is known as Parseval’s

equation. This relation is very important in the theory of Fourier series; see Problem 9 in

Section 11.6.

Hint: Multiply Eq. (i) by f (x), integrate from −L to L,and use the Euler–Fourier formulas.

18. This problem indicates a proof of convergence of a Fourier series under conditions more

restrictive than those in Theorem 10.3.1.

(a) If f and f

′

are piecewise continuous on −L ≤ x < L, and if f is periodic with period

2L, show that na

and nb

are bounded as n →∞.

Hint: Use integration by parts.

(b) If f is continuous on −L ≤ x ≤ L and periodic with period 2L, and if f

′

and f

′′

are

piecewise continuous on −L ≤ x < L, show that n

and n

are bounded as n →∞.If

f is continuous on the closed interval, then it is continuous for all x.Why is this important?

Hint: Again, use integration by parts.

Marc-Antoine Parseval (1755–1836) was a relatively obscure French mathematician for whom an impor-

tant result has been named. He presented a forerunner of this result in 1799, though not in the context of

Fourier series.

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614 Chapter 10. Partial Differential Equations and Fourier Series

∞

n=1

| and

∞

n=1

| converge.

(d) From the result in part (c), show that the Fourier series (4) converges absolutely

for

all x.

Acceleration of Convergence. In the next problem, we show how it is sometimes possible to

improve the speed of convergence of a Fourier series.

19. Suppose that we wish to calculate values of the function g, where

g(x) =

∞

n=1

(2n − 1)

1 + (2n − 1)

sin(2n − 1)πx. (i)

It is possible to show that this series converges,albeit rather slowly. However,observe that

for large n the terms in the series (i) are approximately equal to [sin(2n − 1)πx]/(2n − 1)

and that the latter terms are similar to those in the example in the text, Eq. (6).

(a) Show that

∞

n=1

[sin(2n − 1)πx]/(2n − 1) = (π/2)

f (x) −

, (ii)

where f is the square wave in the example with L = 1.

(b) Subtract Eq. (ii) from Eq. (i) and show that

g(x) =

f (x) −

−

∞

n=1

sin(2n − 1)πx

(2n − 1)[1 + (2n − 1)

]

. (iii)

The series (iii) converges much faster than the series (i) and thus provides a better way to

calculate values of g(x).

10.4 Even and Odd Functions

Before looking at further examples of Fourier series, it is useful to distinguish two

classes of functions for which the Euler–Fourier formulas can be simpliﬁed. These

are even and odd functions, which are characterized geometrically by the property of

symmetry with respect to the y-axis and to the origin, respectively (see Figure 10.4.1).

Analytically, f is an even function if its domain contains the point −x whenever it

contains the point x, and if

f (−x) = f (x) (1)

for each x in the domain of f . Similarly, f is an odd function if its domain contains

−x whenever it contains x, and if

f (−x) =−f (x) (2)

It also converges uniformly; for an explanation of what this means, see a book on advanced calculus or

analysis.

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10.4 Even and Odd Functions 615

for each x in the domain of f . Examples of even functions are 1, x

, cos nx, |x|, and

. The functions x, x

, sin nx, and x

2n+1

are examples of odd functions. Note that

according to Eq. (2), f (0) must be zero if f is an odd function whose domain contains

the origin. Most functions are neither even nor odd; an example is e

. Only one

function, f identically zero, is both even and odd.

(a)

(b)

FIGURE 10.4.1 (a) An even function. (b) An odd function.

Elementary properties of even and odd functions include the following:

1. The sum (difference) and product (quotient) of two even functions are even.

2. The sum (difference) of two odd functions is odd; the product (quotient) of two odd

functions is even.

3. The sum (difference) of an odd function and an even function is neither even nor odd; the

product (quotient) of two such functions is odd.

The proofs of all these assertions are simple and follow directly from the deﬁnitions.

For example, if both f

and f

are odd, and if g(x) = f

(x) + f

(x), then

g(−x) = f

(−x) + f

(−x) =−f

(x) − f

(x)

=−[f

(x) + f

(x)]=−g(x), (3)

so f

+ f

is an odd function also. Similarly, if h(x) = f

(x)f

(x), then

h(−x) = f

(−x)f

(−x) =[−f

(x)][−f

(x)]=f

(x)f

(x) = h(x), (4)

so that f

is even.

Also of importance are the following two integral properties of even and odd

functions:

4. If f is an even function, then

−L

f (x) dx = 2

f (x) dx. (5)

These statements may need to be modiﬁed if either function vanishes identically.

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616 Chapter 10. Partial Differential Equations and Fourier Series

5. If f is an odd function, then

−L

f (x) dx = 0. (6)

These properties are intuitively clear from the interpretation of an integral in

terms of area under a curve, and they also follow immediately from the deﬁnitions.

For example, if f is even, then

−L

f (x) dx =

−L

f (x) dx +

f (x) dx.

Letting x =−s in the ﬁrst term on the right side and using Eq. (1), we obtain

−L

f (x) dx =−

f (s) ds +

f (x) dx = 2

f (x) dx.

The proof of the corresponding property for odd functions is similar.

Even and odd functions are particularly important in applications of Fourier series

since their Fourier series have special forms, which occur frequently in physical

problems.

Cosine Series. Suppose that f and f

′

are piecewise continuous on −L ≤ x < L and

that f is an even periodic function with period 2L. Then it follows from properties 1

and 3 that f (x) cos(nπx/L) is even and f (x) sin(nπx/L) is odd. As a consequence of

Eqs. (5) and (6), the Fourier coefﬁcients of f are then given by

f (x) cos

nπx

dx, n = 0, 1, 2, ...;

(7)

= 0, n = 1, 2, ....

Thus f has the Fourier series

f (x) =

∞

n=1

cos

nπx

In other words, the Fourier series of any even function consists only of the even

trigonometric functions cos(nπx/L) and the constant term; it is natural to call such

a series a Fourier cosine series. From a computational point of view, observe that

only the coefﬁcients a

, for n = 0, 1, 2, ..., need to be calculated from the integral

formula (7). Each of the b

, for n = 1, 2, ..., is automatically zero for any even

function and so does not need to be calculated by integration.

Sine Series. Suppose that f and f

′

are piecewise continuous on −L ≤ x < L and that

f is an odd periodic function of period 2L. Then it follows from properties 2 and 3

that f (x) cos(nπx/L) is odd and f (x) sin(nπx /L) is even. Thus, from Eqs. (5) and (6),

the Fourier coefﬁcients of f are

= 0, n = 0, 1, 2, ...,

(8)

f (x) sin

nπx

dx, n = 1, 2, ...,

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10.4 Even and Odd Functions 617

and the Fourier series for f is

f (x) =

∞

n=1

sin

nπx

Thus the Fourier series for any odd function consists only of the odd trigonometric

functions sin(nπx/L); such a series is called a Fourier sine series. Again observe that

only half of the coefﬁcients need to be calculated by integration, since each a

, for

n = 0, 1,2, ..., is zero for any odd function.

EXAMPLE

Let f (x) = x, −L < x < L, and let f (−L) = f (L) = 0. Let f be deﬁned elsewhere so that it is

periodic of period 2L (see Figure 10.4.2). The function deﬁned in this manner is known as a

sawtooth wave. Find the Fourier series for this function.

–3L –2L

–L

L 2L 3L

FIGURE 10.4.2 The sawtooth wave in Example 1.

Since f is an odd function, its Fourier coefﬁcients are, according to Eq. (8),

= 0, n = 0, 1, 2, ...;

x sin

nπx

nπ

(

sin

nπx

−

nπx

cos

nπx

nπ

(−1)

n+1

, n = 1, 2, ....

Hence the Fourier series for f , the sawtooth wave, is

f (x) =

∞

n=1

(−1)

n+1

sin

nπx

. (9)

Observe that the periodic function f is discontinuous at the points ±L, ±3L, ..., as shown in

Figure 10.4.2. At these points the series (9) converges to the mean value of the left and right

limits, namely, zero. The partial sum of the series (9) for n = 9 is shown in Figure 10.4.3. The

Gibbs phenomenon (mentioned in Section 10.3) again occurs near the points of discontinuity.

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618 Chapter 10. Partial Differential Equations and Fourier Series

–2L

–L

n = 9

FIGURE 10.4.3 A partial sum in the Fourier series, Eq. (9), for the sawtooth wave.

Note that in this example f (−L) = f (L) = 0, as well as f (0) = 0. This is required

if the function f is to be both odd and periodic with period 2L. When we speak

of constructing a sine series for a function deﬁned on 0 ≤ x ≤ L, it is understood

that, if necessary, we must ﬁrst redeﬁne the function to be zero at x = 0 and

x = L.

It is worthwhile to observe that the triangular wave function (Example 1 of Section

10.2) and the sawtooth wave function just considered are identical on the interval

0 ≤ x < L. Therefore, their Fourier series converge to the same function, f (x) = x,

on this interval. Thus, if it is required to represent the function f (x) = x on 0 ≤ x < L

by a Fourier series, it is possible to do this by either a cosine series or a sine series.In

the former case, f is extended as an even function into the interval −L < x < 0 and

elsewhere periodically (the triangular wave). In the latter case, f is extended into

−L < x < 0asanodd function and elsewhere periodically (the sawtooth wave). If

f is extended in any other way, the resulting Fourier series will still converge to x in

0 ≤ x < L but will involve both sine and cosine terms.

In solving problems in differential equations, it is often useful to expand in a

Fourier series of period 2L a function f originally

deﬁned only on the interval [0, L].

As indicated previously for the function f (x) = x, several alternatives are available.

Explicitly, we can

1. Deﬁne a function g of period 2L so that

g(x) =

f (x),0≤ x ≤ L,

f (−x), −L < x < 0.

(10)

The function g is thus the even periodic extension of f . Its Fourier series, which is a cosine

series, represents f on [0, L].

2. Deﬁne a function h of period 2L so that

h(x) =

⎧

⎪

⎨

⎪

⎩

f (x),0< x < L,

0, x = 0, L,

−f (−x), −L < x < 0.

(11)

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10.4 Even and Odd Functions 619

The function h is thus the odd periodic extension of f . Its Fourier series, which is a sine

series, also represents f on (0, L).

3. Deﬁne a function k of period 2L so that

k(x) = f (x),0≤ x ≤ L, (12)

and let k(x) be deﬁned for (−L,0) in any way consistent with the conditions of Theorem

10.3.1. Sometimes it is convenient to deﬁne k(x) to be zero for −L < x < 0. The Fourier

series for k, which involves both sine and cosine terms, also represents f on [0, L], regard-

less of the manner in which k(x) is deﬁned in (−L,0). Thus there are inﬁnitely many such

series, all of which converge to f (x) in the original interval.

Usually,the form of the expansion to be used will be dictated (or at least suggested)

by the purpose for which it is needed. However, if there is a choice as to the kind

of Fourier series to be used, the selection can sometimes be based on the rapidity

of convergence. For example, the cosine series for the triangular wave [Eq. (20) of

Section 10.2] converges more rapidly than the sine series for the sawtooth wave

[Eq. (9) in this section], although both converge to the same function for 0 ≤ x < L.

This is because the triangular wave is a smoother function than the sawtooth wave

and is therefore easier to approximate. In general, the more continuous derivatives

possessed by a function over the entire interval −∞ < x < ∞, the faster its Fourier

series will converge. See Problem 18 of Section 10.3.

EXAMPLE

Suppose that

f (x) =

1 − x,0< x ≤ 1,

0, 1 < x ≤ 2.

(13)

As indicated previously, we can represent f either by a cosine series or by a sine series. Sketch

the graph of the sum of each of these series for −6 ≤ x ≤ 6.

In this example L = 2, so the cosine series for f converges to the even periodic extension

of f of period 4, whose graph is sketched in Figure 10.4.4.

–6

–4

–2

–1

FIGURE 10.4.4 Even periodic extension of f (x) given by Eq. (13).

Similarly, the sine series for f converges to the odd periodic extension of f of period 4. The

graph of this function is shown in Figure 10.4.5.

–6

–4

–2

–1

FIGURE 10.4.5 Odd periodic extension of f (x ) given by Eq. (13).

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620 Chapter 10. Partial Differential Equations and Fourier Series

PROBLEMS

In each of Problems 1 through 6,determine whether the given function is even, odd,or neither.

1. x

− 2x 2. x

− 2x + 1

3. tan 2x 4. sec x

5. |x|

6. e

−x

In each of Problems 7 through 12, a function f is given on an interval of length L. In each case

sketch the graphs of the even and odd extensions of f of period 2L.

7. f (x) =

x,0≤ x < 2,

1, 2 ≤ x < 3

8. f (x) =

0, 0 ≤ x < 1,

x − 1, 1 ≤ x < 2

9. f (x) = 2 − x ,0< x < 2 10. f (x) = x − 3, 0 < x < 4

11. f (x) =

0, 0 ≤ x < 1,

1, 1 ≤ x < 2

12. f (x) = 4 − x

,0< x < 1

13. Prove that any function can be expressed as the sum of two other functions, one of which

is even and the other odd.That is,for any function f , whose domain contains −x whenever

it contains x, show that there are an even function g and an odd function h such that

f (x) = g(x) + h(x).

Hint: What can you say about f (x) + f (−x)?

14. Find the coefﬁcients in the cosine and sine series described in Example 2.

In each of Problems 15 through 22:

(a) Find the required Fourier series for the given function.

(b) Sketch the graph of the function to which the series converges over three periods.

15. f (x) =

1, 0 < x < 1,

0, 1 < x < 2;

cosine series, period 4

Compare with Example 1 and Problem 5 of Section 10.3.

16. f (x) =

x,0≤ x < 1,

1, 1 ≤ x < 2;

sine series, period 4

17. f (x) = 1, 0 ≤ x ≤ π; cosine series, period 2π

18. f (x) = 1,

0 < x <π;

sine series, period 2π

19. f (x) =

⎧

⎪

⎨

⎪

⎩

0, 0 < x <π,

1, π<x < 2π,

2, 2π<x < 3π;

sine series, period 6π

20. f (x) = x,0≤ x < 1; series of period 1

21. f (x) = L − x,0≤ x ≤ L; cosine series, period 2L

Compare with Example 1 of Section 10.2.

22. f (x) = L − x,0< x < L; sine series, period 2L

In each of Problems 23 through 26:

(a) Find the required Fourier series for the given function.

(b) Sketch the graph of the function to which the series converges for three periods.

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10.4 Even and Odd Functions 621

23. f (x) =

x,0< x <π,

0, π<x < 2π ;

cosine series, period 4π

24.

f (x) =−x, −π<x < 0; sine series, period 2π

25.

f (x) = 2 − x

,0< x < 2; sine series, period 4

26.

f (x) = x

− 2x,0< x < 4; cosine series, period 8

In each of Problems 27 through 30, a function is given on an interval 0 < x < L.

(a) Sketch the graphs of the even extension g(x) and the odd extension h(x) of the given

function of period 2L over three periods.

(b) Find the Fourier cosine and sine series for the given function.

(d) For each series, investigate the dependence on n of the maximum error on [0, L].

27.

f (x) = 3 − x,0< x < 3

28.

f (x) =

x,0< x < 1,

0, 1 < x < 2

29.

f (x) = (4x

− 4x − 3)/4, 0 < x < 2

30.

f (x) = x

− 5x

+ 5x + 1, 0 < x < 3

31. Prove that if f is an odd function, then

−L

f (x) dx = 0.

32. Prove properties 2 and 3 of even and odd functions, as stated in the text.

33. Prove that the derivative of an even function is odd and that the derivative of an odd

function is even.

34. Let F(x) =

f (t) dt. Show that if f is even, then F is odd, and that if f is odd, then F is

even.

35. From the Fourier series for the square wave in Example 1 of Section 10.3, show that

= 1 −

−

+···=

∞

n=0

(−1)

2n + 1

This relation between π and the odd positive integers was discovered by Leibniz in 1674.

36. From the Fourier series for the triangular wave (Example 1 of Section 10.2), show that

= 1 +

+···=

∞

n=0

(2n + 1)

37. Assume that f has a Fourier sine series

f (x) =

∞

n=1

sin(nπx/L),0≤ x ≤ L.

(a) Show formally that

[f (x)]

dx =

∞

n=1

Compare this result (Parseval’s equation) with that of Problem 17 in Section 10.3. What

is the corresponding result if f has a cosine series?

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622 Chapter 10. Partial Differential Equations and Fourier Series

(b) Apply the result of part (a) to the series for the sawtooth wave given in Eq. (9), and

thereby show that

= 1 +

+···=

∞

n=1

This relation was discovered by Euler in about 1735.

More Specialized Fourier Series. Let f be a function originally deﬁned on 0 ≤ x ≤ L and

satisfying there the continuity conditions of Theorem 10.3.1. In this section we have shown

that it is possible to represent f by either a sine series or a cosine series by constructing odd or

even periodic extensions of f , respectively. Problems 38 through 40 concern some other, more

specialized Fourier series that converge to the given function f on (0, L).

38. Let f be extended into (L,2L] in an arbitrary manner subject to the continuity conditions

ofTheorem 10.3.1.Then extend the resulting function into (−2L,0) as an odd function and

elsewhere as a periodic function of period 4L (see Figure 10.4.6). Show that this function

has a Fourier sine series in terms of the functions sin(nπx/2L), n = 1, 2, 3, ...; that is,

f (x) =

∞

n=1

sin(nπx/2L),

where

f (x) sin(nπx/2L) dx.

This series converges to the original function on (0, L).

– L–2L

FIGURE 10.4.6 Graph of the function in Problem 38.

39. Let f ﬁrst be extended into (L,2L) so that it is symmetric about x = L. Then f satisﬁes

f (2L − x) = f (x) for 0 ≤ x < L. Let the resulting function be extended into (−2L,0) as an

odd function and elsewhere as a periodic function of period 4L (see Figure 10.4.7). Show

that this function has a Fourier series in terms of the functions sin(πx/2L), sin(3πx/2L),

sin(5πx/2L), ...; that is,

f (x) =

∞

n=1

sin

(2n − 1)πx

where

f (x) sin

(2n − 1)πx

dx.

This series converges to the original function on (0, L].

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10.5 Separation of Variables; Heat Conduction in a Rod 623

2LL– L

–2L

FIGURE 10.4.7 Graph of the function in Problem 39.

40. (a) How should f , originally deﬁned on [0, L], be extended so as to obtain a Fourier

series involving only the functions cos(πx/2L), cos(3πx/2L), cos(5πx/2L), ...? Refer to

Problems 38 and 39.

(b) If f (x) = x for 0 ≤ x ≤ L, sketch the function to which the Fourier series converges

for −4L ≤ x ≤ 4L.

10.5 Separation of Variables; Heat Conduction in a Rod

The basic partial differential equations of heat conduction, wave propagation, and

potential theory that we discuss in this chapter are associated with three distinct

types of physical phenomena: diffusive processes, oscillatory processes, and time-

independent or steady processes. Consequently, they are of fundamental importance

in many branches of physics. They are also of considerable signiﬁcance from a

mathematical point of view. The partial differential equations whose theory is best

developed and whose applications are most signiﬁcant and varied are the linear

equations of second order. All such equations can be classiﬁed into one of three cat-

egories:the heat conduction equation,the wave equation,and the potential equation,

respectively,are prototypes of these categories.Thus a study of these three equations

yields much information about more general second order linear partial differential

equations.

During the last two centuries, several methods have been developed for solving

partial differential equations. The method of separation of variables is the oldest

systematic method, having been used by D’Alembert, Daniel Bernoulli, and Euler

about 1750 in their investigations of waves and vibrations. It has been considerably

reﬁned and generalized in the meantime,and it remainsa methodof great importance

and frequent use today. To show how the method of separation of variables works,

we considerﬁrst abasic problemof heat conduction in a solid body.The mathematical

study of heat conduction originated

about 1800, and it continues to command the

attention of modern scientists. For example, analysis of the dissipation and transfer

The ﬁrst important investigation of heat conduction was carried out by Joseph Fourier. He presented basic

papers on the subject to the Academy of Sciences of Paris in 1807 and 1811. Although these papers were

controversial and were not published at the time, Fourier continued to develop his ideas and eventually

wrote one of the classics of applied mathematics,Théorie analytique de la chaleur, published in 1822.

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624 Chapter 10. Partial Differential Equations and Fourier Series

of heat away from its sources in high-speed machinery is frequently an important

technological problem.

Let us now consider a heat conduction problem for a straight bar of uniform cross

section and homogeneous material. Let the x-axis be chosen to lie along the axis

of the bar, and let x = 0 and x = L denote the ends of the bar (see Figure 10.5.1).

Suppose further that the sides of the bar are perfectly insulated so that no heat passes

through them. We also assume that the cross-sectional dimensions are so small that

the temperature u can be considered constant on any given cross section. Then u is

a function only of the axial coordinate x and the time t.

u(x, t)

x = 0 x = L

FIGURE 10.5.1 A heat-conducting solid bar.

The variation of temperature in the bar is governed by a partial differential

equation whose derivation appears in Appendix A at the end of this chapter. The

equation is called the heat conduction equation and has the form

= u

,0< x < L, t > 0, (1)

where α

is a constant known as the thermal diffusivity. The parameter α

depends

only on the material from which the bar is made and is deﬁned by

= κ/ρs, (2)

where κ is the thermal conductivity, ρ is the density, and s is the speciﬁc heat of the

material in the bar. The units of α

are (length)

/time. Typical values of α

are given

in Table 10.5.1.

TABLE 10.5.1 Values of the Thermal

Diffusivity for Some Common

Materials

Material α

(cm

/s)

Silver 1.71

Copper 1.14

Aluminum 0.86

Cast iron 0.12

Granite 0.011

Brick 0.0038

Water 0.00144

In addition, we assume that the initial temperature distribution in the bar is given;

thus

u(x,0) = f (x),0≤ x ≤ L, (3)

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10.5 Separation of Variables; Heat Conduction in a Rod 625

where f is a given function. Finally,we assume that the ends of thebar are held at ﬁxed

temperatures: the temperature T

at x = 0 and the temperature T

at x = L. How-

ever, it turns out that we need only consider the case where T

= T

= 0. We show

in Section 10.6 how to reduce the more general problem to this special case. Thus, in

this section we will assume that u is always zero when x = 0 or x = L:

u(0, t) = 0, u(L, t) = 0, t > 0. (4)

The fundamental problem of heat conduction is to ﬁnd u(x, t) that satisﬁes the differ-

ential equation (1) for 0 < x < L and for t > 0, the initial condition (3) when t = 0,

and the boundary conditions (4) at x = 0 and x = L.

The problem described by Eqs. (1), (3), and (4) is an initial value problem in the

time variable t; an initial condition is given, and the differential equation governs

what happens later. However, with respect to the space variable x, the problem is

a boundary value problem; boundary conditions are imposed at each end of the

bar, and the differential equation describes the evolution of the temperature in the

interval between them. Alternatively, we can consider the problem as a boundary

value problem in the xt-plane (see Figure 10.5.2). The solution u(x, t) of Eq. (1) is

sought in the semi-inﬁnite strip 0 < x < L, t > 0, subject to the requirement that

u(x, t) must assume a prescribed value at each point on the boundary of this strip.

x = L

u(0, t) = 0

u(x,0) = f (x)

u(L, t) = 0

= u

FIGURE 10.5.2 Boundary value problem for the heat conduction equation.

The heat conduction problem (1), (3), (4) is linear since u appears only to the

ﬁrst power throughout. The differential equation and boundary conditions are also

homogeneous.This suggests thatwe might approachthe problem byseeking solutions

of the differential equation and boundary conditions, and then superposing them to

satisfy the initial condition. The remainder of this section describes how this plan can

be implemented.

Method of Separation of Variables. One solution of the differential equation (1) that sat-

isﬁes the boundary conditions (4) is the function u(x, t) = 0, but this solution does

not satisfy the initial condition (3) except in the trivial case in which f (x) is also zero.

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626 Chapter 10. Partial Differential Equations and Fourier Series

Thus our goal is to ﬁnd other, nonzero solutions of the differential equation and

boundary conditions.To ﬁnd the needed solutions,we start by making a basic assump-

tion about the form of the solutions that has far-reaching, and perhaps unforeseen,

consequences. The assumption is that u(x, t) is a product of two functions, one

depending only on x and the other depending only on t; thus

u(x, t) = X(x)T(t). (5)

We will simplify the following derivation by often omitting the independent variables

x and t; our notation is intended to indicate that the functions X and T depend on

the variables x and t, respectively.

Substituting from Eq. (5) for u in the differential equation (1) yields

′′

T = XT

′

, (6)

where primes refer to ordinary differentiation with respect to the independent

variable, whether x or t. Equation (6) is equivalent to

′′

′

, (7)

in which the variables are separated; that is, the left side depends only on x and the

right side only on t.

It is now crucial to realize that for Eq. (7) to be valid for 0 < x < L, t > 0, it is

necessary that both sides of Eq. (7) must be equal to the same constant. Otherwise,

if one independent variable (say, x) were kept ﬁxed and the other were allowed to

vary, one side (the left in this case) of Eq. (7) would remain unchanged while the

other varied, thus violating the equality. If we call this separation constant −λ, then

Eq. (7) becomes

′′

′

=−λ. (8)

Hence we obtain the following two ordinary differential equations for X(x) and T(t):

′′

+ λX = 0, (9)

′

+ α

λT = 0. (10)

We denote the separation constant by −λ (rather than λ) because in this problem

it turns out that it must be negative, and it is convenient to exhibit the minus sign

explicitly.

The assumption (5) has led to the replacement of the partial differential equation

(1) by the two ordinary differential equations (9) and (10). Each of these equations is

linear and homogeneous, with constant coefﬁcients, and so can be readily solved for

any value of λ.The product of two solutions of Eq. (9) and (10),respectively,provides

a solution of the partial differential equation (1). However, we are interested only in

those solutions of Eq. (1) that also satisfy the boundary conditions (4). As we now

show, this severely restricts the possible values of λ.

Substituting for u(x, t) from Eq. (5) in the boundary condition at x = 0, we obtain

u(0, t) = X(0)T(t) = 0. (11)

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10.5 Separation of Variables; Heat Conduction in a Rod 627

If Eq. (11) is satisﬁed by choosing T(t) to be zero for all t, then u(x , t) is zero for

all x and t, and we have already rejected this possibility. Therefore, Eq. (11) must be

satisﬁed by requiring that

X(0) = 0. (12)

Similarly, the boundary condition at x = L requires that

X(L) = 0. (13)

We now want to consider Eq. (9) subject to the boundary conditions (12) and (13).

This is an eigenvalue problem and, in fact, is the same problem that we discussed in

detail at the end of Section 10.1; see especially the paragraph following Eq. (29) in

that section. The only difference is that the dependent variable there was called y

rather than X. If we refer to the results obtained earlier [Eq. (31) of Section 10.1],

the only nontrivial solutions of Eqs. (9), (12), and (13) are the eigenfunctions

(x) = sin(nπx/L), n = 1, 2, 3,... (14)

associated with the eigenvalues

= n

, n = 1, 2, 3, .... (15)

Turning now to Eq. (10) for T(t) and substituting n

for λ, we have

′

+ (n

)T = 0. (16)

Thus T(t) is proportional to exp(−n

t/L

). Hence, multiplying solutions of

Eqs. (9) and (10) together, and neglecting arbitrary constants of proportionality, we

conclude that the functions

(x, t) = e

−n

t/L

sin(nπx/L), n = 1, 2, 3,... (17)

satisfy the partial differential equation (1) and the boundary conditions (4) for each

positive integer value of n. The functions u

are sometimes called fundamental

solutions of the heat conduction problem (1), (3), and (4).

It remains only to satisfy the initial condition (3):

u(x,0) = f (x),0≤ x ≤ L. (18)

Recall that we have often solved an initial value problem by using the principle of

superposition; that is, we form a linear combination of a set of fundamental solutions

of the differential equation and then choose the coefﬁcients to satisfy the initial

conditions.The analogous step in the present problem is to form a linear combination

of the functions (17) and then to choose the coefﬁcients to satisfy Eq. (18). The main

difference from earlier problems is that there are inﬁnitely many functions (17), so a

general linear combination of them is an inﬁnite series. Thus we assume that

u(x, t) =

∞

n=1

(x, t) =

∞

n=1

−n

t/L

sin

nπx

, (19)

where the coefﬁcients c

are as yet undetermined. The individual terms in the series

(19) satisfy the differential equation (1) and boundary conditions (4).We will assume

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628 Chapter 10. Partial Differential Equations and Fourier Series

that the inﬁnite series of Eq. (19) converges and also satisﬁes Eqs. (1) and (4). To

satisfy the initial condition (3), we must have

u(x,0) =

∞

n=1

sin

nπx

= f (x). (20)

In other words,weneed to choosethe coefﬁcients c

so that theseries of sinefunctions

in Eq. (20) converges to the initial temperature distribution f (x) for 0 ≤ x ≤ L. The

series in Eq. (20) is just the Fourier sine series for f ; according to Eq. (8) of Section

10.4, its coefﬁcients are given by

f (x) sin

nπx

dx. (21)

Hence the solution of the heat conduction problem of Eqs. (1), (3), and (4) is given

by the series in Eq. (19) with the coefﬁcients computed from Eq. (21).

EXAMPLE

Find the temperature u(x, t) at any time in a metal rod 50 cm long,insulated on the sides,which

initially has a uniform temperature of 20

◦

C throughout and whose ends are maintained at 0

◦

for all t > 0.

The temperature in the rod satisﬁes the heat conduction problem (1), (3), (4) with L = 50

and f (x) = 20 for 0 < x < 50. Thus, from Eq. (19), the solution is

u(x, t) =

∞

n=1

−n

t/2500

sin

nπx

, (22)

where, from Eq. (21),

sin

nπx

dx =

nπ

(1 − cos nπ) =

80/nπ, n odd;

0, n even.

(23)

Finally, by substituting for c

in Eq. (22), we obtain

u(x, t) =

∞

n=1,3,5,...

−n

t/2500

sin

nπx

. (24)

The expression (24) for the temperature is moderately complicated, but the negative expo-

nential factor in each term of the series causes the series to converge quite rapidly, except for

small values of t or α

. Therefore, accurate results can usually be obtained by using only the

ﬁrst few terms of the series.

In order to display quantitative results, let us measure t in seconds; then α

has the units

of cm

/s. If we choose α

= 1 for convenience, this corresponds to a rod of a material whose

thermal properties are somewhere between copper and aluminum. The behavior of the solu-

tion can be seen from the graphs in Figures 10.5.3 through 10.5.5. In Figure 10.5.3 we show

the temperature distribution in the bar at several different times. Observe that the tempera-

ture diminishes steadily as heat in the bar is lost through the end points. The way in which the

temperature decays at a given point in the bar is indicated in Figure 10.5.4,where temperature

is plotted against time for a few selected points in the bar. Finally, Figure 10.5.5 is a three-

dimensional plot of u versus both x and t. Observe that we obtain the graphs in Figures 10.5.3

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10.5 Separation of Variables; Heat Conduction in a Rod 629

5020 30 0401

t = 0

t = 20

t = 50

t = 150

t = 300

FIGURE 10.5.3 Temperature distributions at several times

for the heat conduction problem of Example 1.

500200 300 004001

x = 5

x = 15

x = 25

FIGURE 10.5.4 Dependence of temperature on time at several locations

for the heat conduction problem of Example 1.

and 10.5.4 by intersecting the surface in Figure 10.5.5 by planes on which either t or x is con-

stant. The slight waviness in Figure 10.5.5 at t = 0 results from using only a ﬁnite number of

terms in the series for u(x, t) and from the slow convergence of the series for t = 0.

A problem with possible practical implications is to determine the time τ at which the

entire bar has cooled to a speciﬁed temperature. For example, when is the temperature in

the entire bar no greater than 1

◦

C? Because of the symmetry of the initial temperature distri-

bution and the boundary conditions, the warmest point in the bar is always the center. Thus τ

is found by solving u(25, t) = 1 for t. Using one term in the series expansion (24), we obtain

τ =

2500

ln(80/π)

∼

820 s.

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630 Chapter 10. Partial Differential Equations and Fourier Series

5010

100

150

200

250

FIGURE 10.5.5 Plot of temperature u versus x and t for

the heat conduction problem of Example 1.

PROBLEMS In each of Problems 1 through 6, determine whether the method of separation of variables

can be used to replace the given partial differential equation by a pair of ordinary differential

equations. If so, ﬁnd the equations.

1. xu

+ u

= 02.tu

+ xu

= 0

3. u

+ u

= 04.[p(x)u

]

− r(x)u

= 0

5. u

+ (x + y)u

= 06.u

+ u

+ xu = 0

7. Find the solution of the heat conduction problem

100u

= u

,0< x < 1, t > 0;

u(0, t) = 0, u(1, t) = 0, t > 0;

u(x,0) = sin 2πx − sin 5πx,0≤ x ≤ 1.

8. Find the solution of the heat conduction problem

= 4u

,0< x < 2, t > 0;

u(0, t) = 0, u(2, t) = 0, t > 0;

u(x,0) = 2 sin(πx/2) − sin πx + 4 sin 2πx,0≤ x ≤ 2.

Consider the conduction of heat in a rod 40 cm in length whose ends are maintained at 0

◦

for all t > 0. In each of Problems 9 through 12, ﬁnd an expression for the temperature u(x, t)

if the initial temperature distribution in the rod is the given function. Suppose that α

= 1.

9. u(x,0) = 50, 0 < x < 40

10. u(x,0) =

x,0≤ x < 20,

40 − x,20≤ x ≤ 40

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10.5 Separation of Variables; Heat Conduction in a Rod 631

11. u(x,0) =

⎧

⎪

⎨

⎪

⎩

0, 0 ≤ x < 10,

50, 10 ≤ x ≤ 30,

0, 30 < x ≤ 40

12. u(x,0) = x,0< x < 40

13.

Consider again the rod in Problem 9. For t = 5 and x = 20, determine how many terms are

needed to ﬁnd the solution correct to three decimal places. A reasonable way to do this is

to ﬁnd n so that including one more term does not change the ﬁrst three decimal places of

u(20, 5). Repeat for t = 20 and t = 80. Form a conclusion about the speed of convergence

of the series for u(x, t).

14.

For the rod in Problem 9:

(a) Plot u versus x for t = 5, 10, 20, 40, 100, and 200. Put all of the graphs on the same

set of axes and thereby obtain a picture of the way in which the temperature distribution

changes with time.

(b) Plot u versus t for x = 5, 10, 15, and 20.

(d) How long does it take for the entire rod to cool off to a temperature of no more than

◦

15.

Follow the instructions in Problem 14 for the rod in Problem 10.

16.

Follow the instructions in Problem 14 for the rod in Problem 11.

17.

For the rod in Problem 12:

(a) Plot u versus x for t = 5, 10, 20, 40, 100, and 200.

(b) For each value of t used in part (a), estimate the value of x for which the temperature

is greatest. Plot these values versus t to see how the location of the warmest point in the

rod changes with time.

(d) Draw a three-dimensional plot of u versus x and t.

(e) How long does it take the entire rod to cool to a temperature of no more than 1

◦

18.

Let a metallic rod 20 cm long be heated to a uniform temperature of 100

◦

C. Suppose that

at t = 0, the ends of the bar are plunged into an ice bath at 0

◦

C and thereafter maintained

at this temperature, but that no heat is allowed to escape through the lateral surface. Find

an expression for the temperature at any point in the bar at any later time. Determine the

temperature at the center of the bar at time t = 30 s if the bar is made (a) of silver, (b) of

aluminum, (c) of cast iron.

19.

For the rod of Problem 18, ﬁnd the time that will elapse before the center of the bar cools

to a temperature of 5

◦

C if the bar is made (a) of silver, (b) of aluminum, (c) of cast iron.

20. In solving differential equations, the computations can almost always be simpliﬁed by the

use of dimensionless variables.

(a) Show that if the dimensionless variable ξ = x/L is introduced, the heat conduction

equation becomes

∂

∂ξ

∂u

∂t

,0<ξ<1, t > 0.

(b) Since L

/α

has the units of time, it is convenient to use this quantity to deﬁne a di-

mensionless time variable τ = (α

)t. Then show that the heat conduction equation

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632 Chapter 10. Partial Differential Equations and Fourier Series

reduces to

∂

∂ξ

∂u

∂τ

,0<ξ<1, τ>0.

21. Consider the equation

− bu

+ cu = 0, (i)

where a, b, and c are constants.

(a) Let u(x, t) = e

δt

w(x, t), where δ is constant, and ﬁnd the corresponding partial

differential equation for w.

(b) If b ̸= 0, show that δ can be chosen so that the partial differential equation found in

part (a) has no term in w. Thus, by a change of dependent variable, it is possible to reduce

Eq. (i) to the heat conduction equation.

22. The heat conduction equation in two space dimensions is

+ u

) = u

Assuming that u(x, y, t) = X(x)Y(y)T(t), ﬁnd ordinary differential equations that are

satisﬁed by X (x), Y(y), and T(t).

23. The heat conduction equation in two space dimensions may be expressed in terms of polar

coordinates as

+ (1/r)u

+ (1/r

θθ

]=u

Assuming that u(r, θ, t) = R(r);(θ)T(t), ﬁnd ordinary differential equations that are

satisﬁed by R(r), ;(θ), and T(t).

10.6 Other Heat Conduction Problems

In Section10.5 we considered the problem consisting ofthe heat conduction equation

= u

,0< x < L, t > 0, (1)

the boundary conditions

u(0, t) = 0, u(L, t) = 0, t > 0, (2)

and the initial condition

u(x,0) = f (x),0≤ x ≤ L. (3)

We found the solution to be

u(x, t) =

∞

n=1

−n

t/L

sin

nπx

, (4)

where the coefﬁcients c

are the same as in the series

f (x) =

∞

n=1

sin

nπx

. (5)

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10.6 Other Heat Conduction Problems 633

The series in Eq. (5) is just the Fourier sine series for f ; according to Section 10.4, its

coefﬁcients are given by

f (x) sin

nπx

dx. (6)

Hence the solution of the heat conduction problem, Eqs. (1) to (3), is given by the

series in Eq. (4) with the coefﬁcients computed from Eq. (6).

We emphasize that, at this stage, the solution (4) must be regarded as a formal

solution;that is,we obtained it without rigorous justiﬁcation of the limiting processes

involved. Such a justiﬁcation is beyond the scope of this book. However, once the

series (4) has been obtained, it is possible to show that in 0 < x < L, t > 0, it con-

verges to a continuous function; that the derivatives u

and u

can be computed

by differentiating the series (4) term by term; and that the heat conduction equa-

tion (1) is indeed satisﬁed. The argument relies heavily on the fact that each term

of the series (4) contains a negative exponential factor, and this results in relatively

rapid convergence of the series. A further argument establishes that the function u

given by Eq. (4) also satisﬁes the boundary and initial conditions; this completes the

justiﬁcation of the formal solution.

It is interesting to note that although f satisﬁes the conditions of the Fourier con-

vergence theorem (Theorem 10.3.1), it may have points of discontinuity. In this case

the initial temperature distribution u(x,0) = f (x) is discontinuous at one or more

points. Nevertheless, the solution u(x, t) is continuous for arbitrarily small values of

t > 0. This illustrates the fact that heat conduction is a diffusive process that instantly

smooths out any discontinuities that may be present in the initial temperature dis-

tribution. Finally, since f is bounded, it follows from Eq. (6) that the coefﬁcients c

are also bounded. Consequently, the presence of the negative exponential factor in

each term of the series (4) guarantees that

lim

t→∞

u(x, t) = 0 (7)

for all x regardless of the initial condition. This is in accord with the result expected

from physical intuition.

We now consider two other problems of one-dimensional heat conduction that can

be handled by the method developed in Section 10.5.

Nonhomogeneous Boundary Conditions. Suppose now that one end of the bar is held at a

constant temperature T

and the other is maintained at a constant temperature T

Then the boundary conditions are

u(0, t) = T

, u(L, t) = T

, t > 0. (8)

The differential equation (1) and the initial condition (3) remain unchanged.

This problem is only slightly more difﬁcult, because of the nonhomogeneous

boundary conditions, than the one in Section 10.5. We can solve it by reducing it

to a problem having homogeneous boundary conditions, which can then be solved

as in Section 10.5. The technique for doing this is suggested by the following physical

argument.

After a long time—that is, as t →∞—we anticipate that a steady temperature

distribution v(x) will be reached, which is independent of the time t and the initial

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634 Chapter 10. Partial Differential Equations and Fourier Series

conditions. Since v(x) does not depend on t, its time derivative v

is zero; the heat

conduction equation (1) for v then becomes

′′

(x) = 0, 0 < x < L. (9)

Hence the steady state temperature distribution is a linear function of x. Further,

v(x) must satisfy the boundary conditions

v(0) = T

, v(L) = T

, (10)

which are valid even as t →∞. The solution of Eq. (9) satisfying Eqs. (10) is

v(x) = (T

− T

)

+ T

. (11)

Returning to the original problem,Eqs. (1),(3),and (8),we will try to express u(x, t)

as the sum of the steady state temperature distribution v(x) and another (transient)

temperature distribution w(x, t); thus we write

u(x, t) = v(x) + w(x, t). (12)

Since v(x) is given by Eq. (11), the problem will be solved provided that we can

determine w(x, t). The boundary value problem for w(x, t) is found by substituting

the expression in Eq. (12) for u(x, t) in Eqs. (1), (3), and (8).

From Eq. (1) we have

(v + w)

= (v + w)

;

it follows that

= w

, (13)

since v

= 0 and v

= 0. Similarly, from Eqs. (12), (8), and (10),

w(0, t) = u(0, t) − v(0) = T

− T

= 0,

(14)

w(L, t) = u(L, t) − v(L) = T

− T

= 0.

Finally, from Eqs. (12) and (3),

w(x,0) = u(x,0) − v(x) = f (x) − v(x), (15)

where v(x) is given by Eq. (11). Thus the transient part of the solution to the original

problem is found by solving the problem consisting of Eqs. (13), (14), and (15). This

latter problem is precisely the one solved in Section 10.5, provided that f (x) − v(x)

is now regarded as the initial temperature distribution. Hence

u(x, t) = (T

− T

)

+ T

∞

n=1

−n

t/L

sin

nπx

, (16)

where

;

f (x) − (T

− T

)

− T

sin

nπx

dx. (17)

This is another case in which a more difﬁcult problem is solved by reducing it to a

simpler problem that has already been solved. The technique of reducing a problem

with nonhomogeneous boundary conditions to one with homogeneous boundary

conditions by subtracting the steady state solution has wide application.

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10.6 Other Heat Conduction Problems 635

EXAMPLE

Consider the heat conduction problem

= u

,0< x < 30, t > 0, (18)

u(0, t) = 20, u(30, t) = 50, t > 0, (19)

u(x,0) = 60 − 2x,0< x < 30. (20)

Find the steady state temperature distribution and the boundary value problem that

determines the transient distribution.

The steady state temperature satisﬁes v

′′

(x) = 0 and the boundary conditions v(0) = 20 and

v(30) = 50. Thus v(x) = 20 + x. The transient distribution w(x, t) satisﬁes the heat conduction

equation

= w

, (21)

the homogeneous boundary conditions

w(0, t) = 0, w(30, t) = 0, (22)

and the modiﬁed initial condition

w(x,0) = 60 − 2x − (20 + x ) = 40 − 3x. (23)

Note that this problem is of the form (1), (2), (3) with f (x) = 40 − 3x , α

= 1, and L = 30. Thus

the solution is given by Eqs. (4) and (6).

Figure 10.6.1 shows a plot of the initial temperature distribution 60 − 2x, the ﬁnal temper-

ature distribution 20 + x, and the temperature at three intermediate times found by solving

Eqs. (21) through (23). Note that the intermediate temperature satisﬁes the boundary condi-

tions (19) for any t > 0. As t increases, the effect of the boundary conditions gradually moves

from the ends of the bar toward its center.

10 510252350

t = 0

t = 30

t = 10

t = 2

t →

∞

FIGURE 10.6.1 Temperature distributions at several times

for the heat conduction problem of Example 1.

Bar with Insulated Ends. A slightly different problem occurs if the ends of the bar are

insulated so that there is no passage of heat through them. According to Eq. (2) in

Appendix A, the rate of ﬂow of heat across a cross section is proportional to the rate

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636 Chapter 10. Partial Differential Equations and Fourier Series

of change of temperature in the x direction. Thus, in the case of no heat ﬂow, the

boundary conditions are

(0, t) = 0, u

(L, t) = 0, t > 0. (24)

The problem posed by Eqs. (1), (3), and (24) can also be solved by the method of

separation of variables. If we let

u(x, t) = X(x)T(t), (25)

and substitute for u in Eq. (1), then it follows, as in Section 10.5, that

′′

′

=−λ, (26)

where λ is a constant. Thus we obtain again the two ordinary differential equations

′′

+ λX = 0, (27)

′

+ α

λT = 0. (28)

For any value of λ, a product of solutions of Eqs. (27) and (28) is a solution of the

partial differential equation (1). However, we are interested only in those solutions

that also satisfy the boundary conditions (24).

If we substitute for u(x, t) from Eq. (25) in the boundary condition at x = 0, we

obtain X

′

(0)T(t) = 0. We cannot permit T(t) to be zero for all t, since then u(x, t)

would also be zero for all t. Hence we must have

′

(0) = 0. (29)

Proceeding in the same way with the boundary condition at x = L, we ﬁnd that

′

(L) = 0. (30)

Thus we wish to solve Eq. (27) subject to the boundary conditions (29) and (30). It

is possible to show that nontrivial solutions of this problem can exist only if λ is real.

One way to show this is indicated in Problem 18; alternatively, we can appeal to a

more general theory to be discussed in Section 11.2. We will assume that λ is real and

will consider in turn the three cases λ<0, λ = 0, and λ>0.

If λ<0, we let λ =−µ

, where µ is real and positive, so that Eq. (27) becomes

′′

− µ

X = 0. It is convenient to write its general solution as

X(x) = k

sinh µx + k

cosh µx (31)

so that the boundary conditions are easier to apply. In this case the boundary con-

ditions can be satisﬁed only by choosing k

= k

= 0. Since this is unacceptable, it

follows that λ cannot be negative; in other words, the problem (27), (29), (30) has no

negative eigenvalues.

If λ = 0, then Eq. (27) is X

′′

= 0, and therefore,

X(x) = k

x + k

. (32)

The boundary conditions (29) and (30) require that k

= 0 but do not determine k

Thus λ = 0 is an eigenvalue,corresponding to the eigenfunction X (x) = 1. For λ = 0,

it follows from Eq. (28) that T(t) is also a constant, which can be combined with k

Hence, for λ = 0, we obtain the constant solution u(x, t) = k

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10.6 Other Heat Conduction Problems 637

Finally, if λ>0, let λ = µ

, where µ is real and positive. Then Eq. (27) becomes

′′

+ µ

X = 0, and consequently,

X(x) = k

sin µx + k

cos µx. (33)

The boundary condition (29) requires that k

= 0, and the boundary condition (30)

requires that µ = nπ/L for n = 1, 2, 3, ... but leaves k

arbitrary. Thus the problem

(27), (29),(30) has an inﬁnite sequence of positive eigenvalues λ = n

with the

correspondingeigenfunctions X(x) = cos(nπx/L). Forthese valuesof λ,thesolutions

T(t) of Eq. (28) are proportional to exp(−n

t/L

Combining all these results, we have the following fundamental solutions for the

problem (1), (3), and (24):

(x, t) = 1,

(34)

(x, t) = e

−n

t/L

cos

nπx

, n = 1, 2, ...,

where arbitrary constants of proportionality have been dropped. Each of these func-

tions satisﬁes the differential equation (1) and the boundary conditions (24). Because

both the differential equation and the boundary conditions are linear and homoge-

neous, any ﬁnite linear combination of the fundamental solutions satisﬁes them. We

will assume that this is true for convergent inﬁnite linear combinations of fundamen-

tal solutions as well. Thus, to satisfy the initial condition (3), we assume that u(x , t)

has the form

u(x, t) =

(x, t) +

∞

n=1

(x, t)

∞

n=1

−n

t/L

cos

nπx

. (35)

The coefﬁcients c

are determined by the requirement that

u(x,0) =

∞

n=1

cos

nπx

= f (x). (36)

Thus the unknown coefﬁcients in Eq. (35) must be the coefﬁcients in the Fourier

cosine series of period 2L for f . Hence

f (x) cos

nπx

dx, n = 0, 1, 2, .... (37)

With this choice of the coefﬁcients c

, c

, ..., the series (35) provides the solution

to the heat conduction problem for a rod with insulated ends, Eqs. (1), (3), and (24).

It is worth observing that the solution (35) can also be thought of as the sum

of a steady state temperature distribution (given by the constant c

/2), which is

independent of time t, and a transient distribution (given by the rest of the inﬁnite

series) that approaches zero in the limit as t approaches inﬁnity. That the steady state

is a constant is consistent with the expectation that the process of heat conduction

will gradually smooth out the initial temperature distribution in the bar as long as no

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638 Chapter 10. Partial Differential Equations and Fourier Series

heat is allowed to enter from, or to escape to, the outside. The physical interpretation

of the term

f (x) dx (38)

is that it is the mean value of the original temperature distribution.

EXAMPLE

Find the temperature u(x, t) in a metal rod of length 25 cm that is insulated on the ends as well

as on the sides and whose initial temperature distribution is u(x,0) = x for 0 < x < 25.

The temperature in the rod satisﬁes the heat conduction problem (1), (3), (24) with L = 25.

Thus, from Eq. (35), the solution is

u(x, t) =

∞

n=1

−n

t/625

cos

nπx

, (39)

where the coefﬁcients are determined from Eq. (37). We have

xdx = 25 (40)

and, for n ≥ 1,

x cos

nπx

= 50(cos nπ −1)/(nπ)

−100/(nπ)

, n odd;

0, n even.

(41)

Thus

u(x, t) =

−

100

∞

n=1,3,5,...

−n

t/625

cos(nπx/25) (42)

is the solution of the given problem.

For α

= 1, Figure 10.6.2 shows plots of the temperature distribution in the bar at several

times. Again, the convergence of the series is rapid so that only a relatively few terms are

needed to generate the graphs.

2510 15 025

t = 0

t = 10

t = 40

t = 100

t →

∞

FIGURE 10.6.2 Temperature distributions at several times

for the heat conduction problem of Example 2.

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10.6 Other Heat Conduction Problems 639

More General Problems. The method of separation of variables can also be used to

solve heat conduction problems with boundary conditions other than those given

by Eqs. (8) and Eqs. (24). For example, the left end of the bar might be held at a ﬁxed

temperature T while the other end is insulated. In this case the boundary conditions

are

u(0, t) = T, u

(L, t) = 0, t > 0. (43)

The ﬁrst step in solving this problem is to reduce the given boundary conditions to

homogeneous ones by subtracting the steady state solution. The resulting problem

is solved by essentially the same procedure as in the problems previously consid-

ered. However, the extension of the initial function f outside of the interval [0, L] is

somewhat different from that in any case considered so far (see Problem 15).

A more general type of boundary condition occurs when the rate of heat ﬂow

through the end of the bar is proportional to the temperature. It is shown in

Appendix A that the boundary conditions in this case are of the form

(0, t) − h

u(0, t) = 0, u

(L, t) + h

u(L, t) = 0, t > 0, (44)

where h

and h

are nonnegative constants. If we apply the method of separation of

variables to the problem consisting of Eqs. (1), (3), and (44), we ﬁnd that X(x) must

be a solution of

′′

+ λX = 0, X

′

(0) − h

X(0) = 0, X

′

(L) + h

X(L) = 0, (45)

where λ is the separation constant. Once again it is possible to show that nontrivial

solutions can exist only for certain nonnegative real values of λ, the eigenvalues, but

these values are not given by a simple formula (see Problem 20). It is also possible

to show that the corresponding solutions of Eqs. (45), the eigenfunctions, satisfy an

orthogonality relation and that we can satisfy the initial condition (3) by superposing

solutions of Eqs. (45). However, the resulting series is not included in the discussion

of this chapter. There is a more general theory that covers such problems, and it is

outlined in Chapter 11.

PROBLEMS In each of Problems 1 through 8, ﬁnd the steady state solution of the heat conduction equation

= u

that satisﬁes the given set of boundary conditions.

1. u(0, t) = 10, u(50, t) = 40 2. u(0, t) = 30, u(40, t) =−20

3. u

(0, t) = 0, u(L, t) = 04. u

(0, t) = 0, u(L, t) = T

5. u(0, t) = 0, u

(L, t) = 06.u(0, t) = T, u

(L, t) = 0

7. u

(0, t) − u(0, t) = 0, u(L, t) = T 8. u(0, t) = T, u

(L, t) + u(L, t) = 0

Let an aluminum rod of length 20 cm be initially at the uniform temperature of 25

◦

Suppose that at time t = 0, the end x = 0 is cooled to 0

◦

C while the end x = 20 is heated

to 60

◦

C, and both are thereafter maintained at those temperatures.

(a) Find the temperature distribution in the rod at any time t.

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640 Chapter 10. Partial Differential Equations and Fourier Series

(b) Plot the initial temperature distribution, the ﬁnal (steady state) temperature distribu-

tion, and the temperature distributions at two representative intermediate times on the

same set of axes.

(d) Determine how much time must elapse before the temperature at x = 5 cm comes

(and remains) within 1% of its steady state value.

10.

(a) Let the ends of a copper rod 100 cm long be maintained at 0

◦

C. Suppose that the

center of the bar is heated to 100

◦

C by an external heat source and that this situation is

maintained until a steady state results. Find this steady state temperature distribution.

(b) At a time t = 0 [after the steady state of part (a) has been reached],let the heat source

be removed. At the same instant let the end x = 0 be placed in thermal contact with a

reservoir at 20

◦

C, while the other end remains at 0

◦

C. Find the temperature as a function

of position and time.

(d) What limiting value does the temperature at the center of the rod approach after a

long time? How much time must elapse before the center of the rod cools to within 1

◦

of its limiting value?

11.

Consider a rod of length 30 for which α

= 1. Suppose the initial temperature distribution

is given by u(x,0) = x(60 − x)/30 and that the boundary conditions are u(0, t) = 30 and

u(30, t) = 0.

(a) Find the temperature in the rod as a function of position and time.

(b) Plot u versus x for several values of t. Also plot u versus t for several values of x.

while, and ﬁnally decreases to approach its steady state value. Explain physically why this

behavior occurs at this point.

12.

Consider a uniform rod of length L with an initial temperature given by

u(x,0) = sin(πx/L),0≤ x ≤ L. Assume that both ends of the bar are insulated.

(a) Find the temperature u(x , t).

(b) What is the steady state temperature as t →∞?

= 1 and L = 40. Plot u versus x for several values of t. Also plot u versus t for

several values of x.

(d) Describe brieﬂy how the temperature in the rod changes as time progresses.

13.

Consider a bar of length 40 cm whose initial temperature is given by u(x,0) = x(60 − x)/30.

Suppose that α

= 1/4 cm

/s and that both ends of the bar are insulated.

(a) Find the temperature u(x , t).

(b) Plot u versus x for several values of t. Also plot u versus t for several values of x.

(d) Determine how much time must elapsebefore the temperature at x = 40 comes within

◦

C of its steady state value.

14.

Consider a bar 30 cm long that is made of a material for which α

= 1 and whose ends are

insulated. Suppose that the initial temperature is zero except for the interval 5 < x < 10,

where the initial temperature is 25

◦

(a) Find the temperature u(x , t).

(b) Plot u versus x for several values of t. Also plot u versus t for several values of x.

rically located with respect to the initial temperature pulse, yet their temperature plots

are signiﬁcantly different. Explain physically why this is so.

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10.6 Other Heat Conduction Problems 641

15. Consider a uniform bar of length L having an initial temperature distribution given by

f (x),0≤ x ≤ L. Assume that the temperature at the end x = 0 is held at 0

◦

C, while the

end x = L is insulated so that no heat passes through it.

(a) Show that the fundamental solutions of the partial differential equation and boundary

conditions are

(x, t) = e

−(2n−1)

t/4L

sin[(2n − 1)πx/2L], n = 1, 2, 3, ....

(b) Find a formal series expansion for the temperature u(x, t)

u(x, t) =

∞

n=1

(x, t)

that also satisﬁes the initial condition u(x,0) = f (x).

Hint: Even though the fundamental solutions involve only the odd sines, it is still possi-

ble to represent f by a Fourier series involving only these functions. See Problem 39 of

Section 10.4.

16.

In the bar of Problem 15, suppose that L = 30, α

= 1, and the initial temperature distri-

bution is f (x) = 30 − x for 0 < x < 30.

(a) Find the temperature u(x , t).

(b) Plot u versus x for several values of t. Also plot u versus t for several values of x.

of the warmest point in the bar change as t increases? Draw

a graph of x

versus t.

(d) Plot the maximum temperature in the bar versus t.

17.

Suppose that the conditions are as in Problems 15 and 16 except that the boundary

condition at x = 0 is u(0, t) = 40.

(a) Find the temperature u(x , t).

(b) Plot u versus x for several values of t. Also plot u versus t for several values of x.

how the change in the boundary condition at x = 0 causes the observed differences in the

behavior of the temperature in the bar.

18. Consider the problem

′′

+ λX = 0, X

′

(0) = 0, X

′

(L) = 0. (i)

Let λ = µ

, where µ = ν + iσ with ν and σ real. Show that if σ ̸= 0, then the only solution

of Eqs. (i) is the trivial solution X(x) = 0.

Hint: Use an argument similar to that in Problem 23 of Section 10.1.

19. The right end of a bar of length a with thermal conductivity κ

and cross-sectional area A

is joined to the left end of a bar of thermal conductivity κ

and cross-sectional area A

The composite bar has a total length L. Suppose that the end x = 0 is held at temperature

zero, while the end x = L is held at temperature T. Find the steady state temperature in

the composite bar, assuming that the temperature and rate of heat ﬂow are continuous at

x = a.

Hint: See Eq. (2) of Appendix A.

20. Consider the problem

= u

,0< x < L, t > 0;

u(0, t) = 0, u

(L, t) + γu(L, t) = 0, t > 0; (i)

u(x,0) = f (x),0≤ x ≤ L.

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642 Chapter 10. Partial Differential Equations and Fourier Series

(a) Let u(x, t) = X(x)T(t), and show that

′′

+ λX = 0, X(0) = 0, X

′

(L) + γX(L) = 0, (ii)

and

′

+ λα

T = 0,

where λ is the separation constant.

(b) Assume that λ is real, and show that problem (ii) has no nontrivial solutions if λ ≤ 0.

with µ>0. Show that problem (ii) has nontrivial solutions only if

µ is a solution of the equation

µ cos µL + γ sin µL = 0. (iii)

(d) Rewrite Eq. (iii) as tan µL =−µ/γ. Then, by drawing the graphs of y = tan µL and

y =−µL/γL for µ>0 on the same set of axes, show that Eq. (iii) is satisﬁed by inﬁnitely

many positive values of µ; denote these by µ

, µ

, ..., µ

, ..., ordered in increasing size.

(e) Determine the set of fundamental solutions u

(x, t) corresponding to the values µ

found in part (d).

An External Heat Source. Consider the heat conduction problem in a bar that is in thermal

contact with an external heat source or sink. Then the modiﬁed heat conduction equation is

= α

+ s(x), (i)

where the term s(x) describes the effect of the external agency;s(x) is positive for a source and

negative for a sink. Suppose that the boundary conditions are

u(0, t) = T

, u(L, t) = T

(ii)

and the initial condition is

u(x,0) = f (x). (iii)

Problems 21 through 23 deal with this kind of problem.

21. Write u(x, t) = v(x) + w(x, t), where v and w are the steady state and transient parts of the

solution,respectively. State theboundary value problems that v(x) and w(x, t),respectively,

satisfy. Observe that the problem for w is the fundamental heat conduction problem

discussed in Section 10.5, with a modiﬁed initial temperature distribution.

22.

(a) Suppose that α

= 1 and s(x ) = k, a constant, in Eq. (i). Find v(x).

(b) Assume that T

= 0, T

= 0, L = 20, k = 1/5, and that f (x) = 0 for 0 < x < L. Deter-

mine w(x, t). Then plot u(x, t) versus x for several values of t; on the same axes, also plot

the steady state part of the solution v(x).

23.

(a) Let α

= 1 and s(x ) = kx/L, where k is a constant, in Eq. (i). Find v(x).

(b) Assume that T

= 10, T

= 30, L = 20, k = 1/2, and that f (x) = 0 for 0 < x < L.

Determine w(x, t). Then plot u(x, t) versus x for several values of t; on the same axes,

also plot the steady state part of the solution v(x).

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10.7 The Wave Equation: Vibrations of an Elastic String 643

10.7 The Wave Equation: Vibrations of an Elastic String

A second partial differential equation that occurs frequently in applied mathematics

is the wave

equation. Some form of this equation, or a generalization of it, almost

inevitably arises in any mathematical analysis of phenomena involving the propaga-

tion of waves in a continuous medium. For example, the studies of acoustic waves,

water waves,electromagnetic waves,and seismic waves are all based on this equation.

Perhaps the easiest situation to visualize occurs in the investigation of mechanical

vibrations. Suppose that an elastic string of length L is tightly stretched between two

supports at the same horizontal level, so that the x-axis lies along the string (see

Figure 10.7.1).

u(x, t )

x = 0 x = L

FIGURE 10.7.1 A vibrating string.

The elastic string may be thought of as a guitar string, a guy wire, or possibly an

electric power line. Suppose that the string is set in motion (by plucking,for example)

so that it vibrates in a vertical plane, and let u(x, t) denote the vertical displacement

experienced by the string at the point x at time t. If damping effects, such as air

resistance, are neglected, and if the amplitude of the motion is not too large, then

u(x, t) satisﬁes the partial differential equation

= u

(1)

in the domain 0 < x < L, t > 0. Equation (1) is known as the one-dimensional wave

equation and is derived in Appendix B at the end of the chapter. The constant

coefﬁcient a

appearing in Eq. (1) is given by

= T/ρ, (2)

where T is the tension (force) in the string, and ρ is the mass per unit length of the

string material. It follows that a has the units of length/time—that is, of velocity. In

Problem 14 it is shown that a is the velocity of propagation of waves along the string.

The solution of the wave equation was one of the major mathematical problems of the mid-eighteenth

century. The wave equation was ﬁrst derived and studied by D’Alembert in 1746. It also attracted the

attention of Euler (1748), Daniel Bernoulli (1753), and Lagrange (1759). Solutions were obtained in

several different forms, and the merits of, and relations among, these solutions were argued, sometimes

heatedly, in a series of papers extending over more than 25 years. The major points at issue concerned

the nature of a function and the kinds of functions that can be represented by trigonometric series. These

questions were not resolved until the nineteenth century.

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644 Chapter 10. Partial Differential Equations and Fourier Series

To describe the motion of the string completely, it is necessary also to specify

suitable initial and boundary conditions for the displacement u(x, t). The ends are

assumed to remain ﬁxed, and therefore the boundary conditions are

u(0, t) = 0, u(L, t) = 0, t ≥ 0. (3)

Since the differential equation (1) is of second order with respect to t, it is plausible

to prescribe two initial conditions. These are the initial position of the string

u(x,0) = f (x),0≤ x ≤ L (4)

and its initial velocity

(x,0) = g(x ),0≤ x ≤ L, (5)

where f and g are given functions. In order for Eqs. (3), (4), and (5) to be consistent,

it is also necessary to require that

f (0) = f (L) = 0, g(0) = g(L) = 0. (6)

The mathematical problem then is to determine the solution of the wave equation

(1) that also satisﬁes the boundary conditions (3) and the initial conditions (4) and

(5). Like the heat conduction problem of Sections 10.5 and 10.6, this problem is an

initial value problem in the time variable t and a boundary value problem in the space

variable x. Alternatively, it can be considered as a boundary value problem in the

semi-inﬁnite strip 0 < x < L, t > 0 of the xt-plane (see Figure 10.7.2). One condition

is imposed at each point on the semi-inﬁnite sides, and two are imposed at each point

on the ﬁnite base.

x = L

u(0, t) = 0

u(L, t) = 0

u(x, 0) = f (x)

(x, 0) = g(x)

= u

FIGURE 10.7.2 Boundary value problem for the wave equation.

It is important to realize that Eq. (1) governs a large number of other wave prob-

lems besides the transverse vibrations of an elastic string. For example, it is only

necessary to interpret the function u and the constant a appropriately to have prob-

lems dealing with water waves in an ocean, acoustic or electromagnetic waves in

the atmosphere, or elastic waves in a solid body. If more than one space dimension

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10.7 The Wave Equation: Vibrations of an Elastic String 645

is signiﬁcant, then Eq. (1) must be slightly generalized. The two-dimensional wave

equation is

+ u

) = u

. (7)

This equation would arise, for example, if we considered the motion of a thin elastic

sheet, such as a drumhead. Similarly, in three dimensions the wave equation is

+ u

) = u

. (8)

In connection with the latter two equations, the boundary and initial conditions must

also be suitably generalized.

We now solve three boundary value problems involving the wave equation in one

dimension.

Elastic String with Nonzero Initial Displacement. First suppose that the string is disturbed

from its equilibrium position and then released at time t = 0 with zero veloc-

ity to vibrate freely. Then the vertical displacement u(x, t) must satisfy the wave

equation (1)

= u

,0< x < L, t > 0;

the boundary conditions (3)

u(0, t) = 0, u(L, t) = 0, t ≥ 0;

and the initial conditions

u(x,0) = f (x), u

(x,0) = 0, 0 ≤ x ≤ L, (9)

where f is a given function describing the conﬁguration of the string at t = 0.

The method of separation of variables can be used to obtain the solutionof Eqs. (1),

(3), and (9). Assuming that

u(x, t) = X(x)T(t) (10)

and substituting for u in Eq. (1), we obtain

′′

=−λ, (11)

where λ is a separation constant. Thus we ﬁnd that X(x) and T(t) satisfy the ordinary

differential equations

′′

+ λX = 0, (12)

′′

+ a

λT = 0. (13)

Further, by substituting from Eq. (10) for u(x, t) in the boundary conditions (3), we

ﬁnd that X(x) must satisfy the boundary conditions

X(0) = 0, X(L) = 0. (14)

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646 Chapter 10. Partial Differential Equations and Fourier Series

Finally, by substituting from Eq. (10) into the second of the initial conditions (9), we

also ﬁnd that T(t) must satisfy the initial condition

′

(0) = 0. (15)

Our next task is to determine X (x), T(t), and λ by solving Eq. (12) subject to the

boundary conditions (14) and solving Eq. (13) subject to the initial condition (15).

The problem of solving the differential equation (12) subject to the boundary

conditions (14) is precisely the same problem that arose in Section 10.5 in connection

with the heat conduction equation. Thus we can use the results obtained there and

at the end of Section 10.1: the problem (12), (14) has nontrivial solutions if and only

if λ is an eigenvalue

λ = n

, n = 1, 2, ..., (16)

and X(x) is proportional to the corresponding eigenfunction sin(nπ x/L).

Using the values of λ given by Eq. (16) in Eq. (13), we obtain

′′

T = 0. (17)

Therefore,

T(t) = k

cos

nπat

+ k

sin

nπat

, (18)

where k

and k

are arbitrary constants.The initialcondition (15)requires that k

= 0,

so T(t) must be proportional to cos(nπat/L).

Thus the functions

(x, t) = sin

nπx

cos

nπat

, n = 1, 2, ... (19)

satisfy the partial differential equation (1), the boundary conditions (3), and the sec-

ond initial condition (9). These functions are the fundamental solutions for the given

problem.

To satisfy the remaining (nonhomogeneous) initial condition (9), we will consider

a superposition of the fundamental solutions (19) with properly chosen coefﬁcients.

Thus we assume that u(x, t) has the form

u(x, t) =

∞

n=1

(x, t) =

∞

n=1

sin

nπx

cos

nπat

, (20)

where the constants c

remain to be chosen. The initial condition u(x,0) = f (x)

requires that

u(x,0) =

∞

n=1

sin

nπx

= f (x). (21)

Consequently, the coefﬁcients c

must be the coefﬁcients in the Fourier sine series of

period 2L for f ; hence

f (x) sin

nπx

dx, n = 1, 2, .... (22)

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10.7 The Wave Equation: Vibrations of an Elastic String 647

Thus the formal solution of the problem of Eqs. (1), (3), and (9) is given by Eq. (20)

with the coefﬁcients calculated from Eq. (22).

For a ﬁxed value of n, the expression sin(nπx /L) cos(nπat/L) in Eq. (19) is peri-

odic in time t with the period 2L/na; it therefore represents a vibratory motion of the

string having this period, or having the frequency nπa/L. The quantities nπa/L for

n = 1, 2,...are the natural frequencies of the string—that is, the frequencies at which

the string will freely vibrate. The factor sin(nπ x/L) represents the displacement pat-

ternoccurring in the stringwhen it isexecuting vibrations ofthe given frequency.Each

displacement pattern is called a natural mode of vibration and is periodic in the space

variable x; the spatial period 2L/n is called the wavelength of the mode of frequency

nπa/L. Thus the eigenvalues n

of the problem (12), (14) are proportional to

the squares of the natural frequencies, and the eigenfunctions sin(nπ x/L) give the

natural modes. The ﬁrst three natural modes are sketched in Figure 10.7.3. The total

motion of the string,given by the function u(x, t) of Eq. (20), is thus a combination of

thenatural modes of vibrationand is alsoa periodic functionof time withperiod 2L/a.

–1

(a)

–1

(b)

–1

(c)

FIGURE 10.7.3 First three fundamental modes of vibration of an elastic string.

(a) Frequency = πa/L, wavelength = 2L;

(b) frequency = 2πa/L, wavelength = L;

EXAMPLE

Consider a vibrating string of length L = 30 that satisﬁes the wave equation

= u

,0< x < 30, t > 0. (23)

Assume that the ends of the string are ﬁxed and that the string is set in motion with no initial

velocity from the initial position

u(x,0) = f (x) =

x/10, 0 ≤ x ≤ 10,

(30 − x)/20, 10 < x ≤ 30.

(24)

Find the displacement u(x , t) of the string, and describe its motion through one period.

The solution is given by Eq. (20) with a = 2 and L = 30; that is,

u(x, t) =

∞

n=1

sin

nπx

cos

2nπt

, (25)

where c

is calculated from Eq. (22). Substituting from Eq. (24) into Eq. (22), we obtain

sin

nπx

dx +

30 − x

sin

nπx

dx. (26)

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648 Chapter 10. Partial Differential Equations and Fourier Series

By evaluating the integrals in Eq. (26), we ﬁnd that

sin

nπ

, n = 1, 2, .... (27)

The solution (25), (27) gives the displacement of the string at any point x at any time t.The

motion is periodic intime with period30,so it is sufﬁcientto analyzethe solution for0 ≤ t ≤ 30.

The best way to visualize the solution is by a computer animation showing the dynamic

behavior of the vibrating string. Here we indicate the motion of the string in Figures 10.7.4,

10.7.5, and 10.7.6. Plots of u versus x for t = 0, 4, 7.5, 11, and 15 are shown in Figure 10.7.4.

Observe that the maximum initial displacement is positive and occurs at x = 10, while at

t = 15, a half-period later, the maximum displacement is negative and occurs at x = 20. The

string then retraces its motion and returns to its original conﬁguration at t = 30. Figure 10.7.5

shows the behavior of the points x = 5, 10, 15, and 25 by plots of u versus t for these ﬁxed

values of x. The plots conﬁrm that the motion is indeed periodic with period 30 and illustrate

that each point remains motionless for a substantial part of each period. Figure 10.7.6 shows a

three-dimensional plot of u versus both x and t, from which the overall nature of the solution

is apparent. Of course, the curves in Figures 10.7.4 and 10.7.5 lie on the surface shown in

Figure 10.7.6.

x2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

0.8

0.6

0.4

0.2

–0.2

–0.4

–0.6

–0.8

–1

t = 0

t = 4

t = 7.5

t = 11

t = 15

FIGURE 10.7.4 Plots of u versus x for ﬁxed values of t for the string in Example 1.

0.5

10 20 30 40 50 60

–0.5

–1

x =

FIGURE 10.7.5 Plots of u versus t for x = 5, 10, 15, and 25 for the string in Example 1.

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10.7 The Wave Equation: Vibrations of an Elastic String 649

–1

FIGURE 10.7.6 Plot of u versus x and t for the string in Example 1.

Justification of the Solution. As in the heat conduction problem considered earlier,

Eq. (20) with the coefﬁcients c

given by Eq. (22) is only a formal solution of

Eqs. (1), (3), and (9). To determine whether Eq. (20) actually represents the solution

of the given problem requires some further investigation. As in the heat conduction

problem,it is tempting to try to show this directly by substituting Eq. (20) for u(x, t) in

Eqs. (1), (3), and (9). However, upon formally computing u

, for example,we obtain

(x, t) =−

∞

n=1

)

nπ

sin

nπx

cos

nπat

;

due to the presence of the n

factor in the numerator, this series may not converge.

This would not necessarily mean that the series (20) for u(x, t) is incorrect, but only

that the series (20) cannot be used to calculate u

and u

.A basic difference between

solutions of the wave equation and those of the heat conduction equation is that

the latter contain negative exponential terms that approach zero very rapidly with

increasing n, which ensures the convergence of the series solution and its derivatives.

In contrast, series solutions of the wave equation contain only oscillatory terms that

do not decay with increasing n.

However,there is an alternative way to establish the validity of Eq. (20) indirectly.

At the same time, we will gain additional information about the structure of the

solution. First we will show that Eq. (20) is equivalent to

u(x, t) =

h(x − at) + h(x + at)

, (28)

where h is the function obtained by extending the initial data f into (−L,0) as an

odd function, and to other values of x as a periodic function of period 2L. That is,

h(x) =

f (x),0≤ x ≤ L,

−f (−x), −L < x < 0;

(29)

h(x + 2 L) = h(x ).

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650 Chapter 10. Partial Differential Equations and Fourier Series

To establish Eq. (28), note that h has the Fourier series

h(x) =

∞

n=1

sin

nπx

, (30)

where c

is given by Eq. (22). Then, using the trigonometric identities for the sine of

a sum or difference, we obtain

h(x − at) =

∞

n=1

sin

nπx

cos

nπat

− cos

nπx

sin

nπat

(

h(x + at) =

∞

n=1

sin

nπx

cos

nπat

+ cos

nπx

sin

nπat

(

and Eq. (28) follows immediately upon adding the last two equations. From Eq. (28)

we see that u(x, t) is continuous for 0 < x < L, t > 0, provided that h is continuous

on the interval (−∞, ∞). This requires f to be continuous on the original interval

[0, L]. Similarly, u is twice continuously differentiable with respect to either variable

in 0 < x < L, t > 0, provided that h is twice continuously differentiable on (−∞, ∞).

This requires f

′

and f

′′

to be continuous on [0, L]. Furthermore, since h

′′

is the odd

extension of f

′′

, we must also have f

′′

(0) = f

′′

(L) = 0. However, since h

′

is the even

extension of f

′

, no further conditions are required on f

′

. Provided that these con-

ditions are met, u

and u

can be computed from Eq. (28), and it is an elementary

exercise to show that these derivatives satisfy the wave equation. Some of the details

of the argument just indicated are given in Problems 19 and 20.

If some of the continuity requirements stated in the last paragraph are not met,

then u is not differentiable at some points in the semi-inﬁnite strip 0 < x < L, t > 0,

and thus is a solution of the wave equation only in a somewhat restricted sense.

An important physical consequence of this observation is that if there are any dis-

continuities present in the initial data f , then they will be preserved in the solution

u(x, t) for all time. In contrast, in heat conduction problems,initial discontinuities are

instantly smoothed out (Section 10.6). Suppose that the initial displacement f has

a jump discontinuity at x = x

,0≤ x

≤ L. Since h is a periodic extension of f , the

same discontinuity is present in h(ξ) at ξ = x

+ 2nL and at ξ =−x

+ 2nL, where

n is any integer. Thus h(x − at) is discontinuous when x − at = x

+ 2nL or when

x − at =−x

+ 2nL. For a ﬁxed x in [0, L], the discontinuity that was originally at

will reappear in h(x − at) at t = (x ± x

− 2nL)/a. Similarly, h(x + at) is discon-

tinuous at the point x at t = (−x ± x

+ 2mL)/a, where m is any integer. If we refer

to Eq. (28), it then follows that the solution u(x, t) is also discontinuous at the given

point x at these values of t. Since the physical problem is posed for t > 0, only those

values of m and n that yield positive values of t are of interest.

Elastic String with Nonzero Initial Velocity. Let us modify the problem just considered by

supposing that the string is set in motion from its equilibrium position with a given

velocity. Then the vertical displacement u(x, t) must satisfy the wave equation (1)

= u

,0< x < L, t > 0;

the boundary conditions (3)

u(0, t) = 0, u(L, t) = 0, t ≥ 0;

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10.7 The Wave Equation: Vibrations of an Elastic String 651

and the initial conditions

u(x,0) = 0, u

(x,0) = g(x ),0≤ x ≤ L, (31)

where g(x) is the initial velocity at the point x of the string.

The solution of this new problem can be obtained by following the procedure

described above for the problem (1), (3), and (9). Upon separating variables, we ﬁnd

that theproblem forX(x) isexactly the same as before.Thus,once again,λ = n

and X(x) is proportional to sin(nπx/L). The differential equation for T(t) is again

Eq. (17),

′′

T = 0, (17)

but the associated initial condition is now

T(0) = 0, (32)

correspondingtothe ﬁrst of the initial conditions(31).The general solution of Eq. (17)

is given by Eq. (18),

T(t) = k

cos

nπat

+ k

sin

nπat

, (18)

but now the initial condition (32) requires that k

= 0. Therefore, T(t) is now pro-

portional to sin(nπat/L), and the fundamental solutions for the problem (1),(3), and

(31) are

(x, t) = sin

nπx

sin

nπat

, n = 1, 2, 3, .... (33)

Each of the functions u

(x, t) satisﬁes the wave equation (1),the boundary conditions

(3),and theﬁrst of theinitial conditions (31).The mainconsequence of using theinitial

conditions (31) rather than (9) is that the time-dependent factor in u

(x, t) involves

a sine rather than a cosine.

To satisfythe remaining (nonhomogeneous)initial condition,we assume that u(x, t)

can be expressed as a linear combination of the fundamental solutions (33); that is,

u(x, t) =

∞

n=1

(x, t) =

∞

n=1

sin

nπx

sin

nπat

. (34)

To determine the values of the coefﬁcients k

, we differentiate Eq. (34) with respect

to t, set t = 0, and use the second initial condition (31); this gives the equation

(x,0) =

∞

n=1

nπa

sin

nπx

= g(x). (35)

Hence the quantities (nπa/L)k

are the coefﬁcients in the Fourier sine series of

period 2L for g. Therefore,

nπa

g(x) sin

nπx

dx, n = 1, 2, .... (36)

Thus Eq. (34), with the coefﬁcients given by Eq. (36), constitutes a formal solution

to the problem of Eqs. (1), (3), and (31). The validity of this formal solution can

be established by arguments similar to those previously outlined for the solution of

Eqs. (1), (3), and (9).

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652 Chapter 10. Partial Differential Equations and Fourier Series

General Problem for the Elastic String. Finally, we turn to the problem consisting of the

wave equation (1), the boundary conditions (3), and the general initial conditions

(4), (5):

u(x,0) = f (x), u

(x,0) = g(x ),0< x < L, (37)

where f (x) and g(x) are the given initial position and velocity, respectively, of the

string. Although this problem can be solved by separating variables, as in the cases

discussed previously,it is important to note that it can also be solved simply by adding

together the two solutions that we obtained above.To show that this is true, let v(x, t)

be the solution of the problem (1), (3), and (9), and let w(x, t) be the solution of the

problem (1), (3), and (31). Thus v(x, t) is given by Eqs. (20) and (22), and w(x, t) is

given by Eqs. (34) and (36). Now let u(x, t) = v(x, t) + w(x, t); what problem does

u(x, t) satisfy? First, observe that

− u

= (a

− v

) + (a

− w

) = 0 + 0 = 0, (38)

so u(x, t) satisﬁes the wave equation (1). Next, we have

u(0, t) = v(0, t) + w(0, t) = 0 + 0 = 0, u(L, t) = v(L, t) + w(L, t) = 0 + 0 = 0,

(39)

so u(x, t) also satisﬁes the boundary conditions (3). Finally, we have

u(x,0) = v(x,0) + w(x,0) = f (x) + 0 = f (x) (40)

and

(x,0) = v

(x,0) + w

(x,0) = 0 + g (x) = g(x). (41)

Thus u(x, t) satisﬁes the general initial conditions (37).

We can restate the result we have just obtained in the following way. To solve the

wave equation withthe general initial conditions (37),you cansolve instead the some-

what simpler problems with the initial conditions (9) and (31), respectively, and then

add together the two solutions. This is another use of the principle of superposition.

PROBLEMS Consider an elastic string of length L whose ends are held ﬁxed. The string is set in motion

with no initial velocity from an initial position u(x,0) = f (x). In each of Problems 1 through 4,

carry out the following steps. Let L = 10 and a = 1 in parts (b) through (d).

(a) Find the displacement u(x , t) for the given initial position f (x).

(b) Plot u(x, t) versus x for 0 ≤ x ≤ 10 and for several values of t between t = 0 and t = 20.

(d) Construct an animation of the solution in time for at least one period.

(e) Describe the motion of the string in a few sentences.

f (x) =

2x/L,0≤ x ≤ L/2,

2(L − x)/L, L/2 < x ≤ L

f (x) =

⎧

⎪

⎨

⎪

⎩

4x/L,0≤ x ≤ L/4,

1, L/4 < x < 3L/4,

4(L − x)/L,3L/4 ≤ x ≤ L

f (x) = 8x(L − x)

4. f (x) =

1, L/2 − 1 < x < L/2 + 1 (L > 2),

0, otherwise

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10.7 The Wave Equation: Vibrations of an Elastic String 653

Consider an elastic string of length L whose ends are held ﬁxed. The string is set in motion

from its equilibrium position with an initial velocity u

(x,0) = g(x). In each of Problems 5

through 8, carry out the following steps. Let L = 10 and a = 1 in parts (b) through (d).

(a) Find the displacement u(x , t) for the given g(x).

(b) Plot u(x, t) versus x for 0 ≤ x ≤ 10 and for several values of t between t = 0 and t = 20.

(d) Construct an animation of the solution in time for at least one period.

(e) Describe the motion of the string in a few sentences.

g(x) =

2x/L,0≤ x ≤ L/2,

2(L − x)/L, L/2 < x ≤ L

g(x) =

⎧

⎪

⎨

⎪

⎩

4x/L,0≤ x ≤ L/4,

1, L/4 < x < 3L/4,

4(L − x)/L,3L/4 ≤ x ≤ L

g(x) = 8x(L − x)

8. g(x) =

1, L/2 − 1 < x < L/2 + 1 (L > 2),

0, otherwise

9. If an elastic string is free at one end, the boundary condition to be satisﬁed there is that

= 0. Find the displacement u(x , t) in an elastic string of length L, ﬁxed at x = 0 and

free at x = L, set in motion with no initial velocity from the initial position u(x,0) = f (x),

where f is a given function.

Hint: Show that the fundamental solutions for this problem, satisfying all conditions

except the nonhomogeneous initial condition, are

(x, t) = sin λ

x cos λ

at,

where λ

= (2n − 1)π/(2L), n = 1, 2, .... Compare this problem with Problem 15 of Sec-

tion 10.6; pay particular attention to the extension of the initial data out of the original

interval [0, L].

10.

Consider an elastic string of length L. The end x = 0 is held ﬁxed, while the end x = L

is free; thus the boundary conditions are u(0, t) = 0 and u

(L, t) = 0. The string is set in

motion with no initial velocity from the initial position u(x,0) = f (x), where

f (x) =

1, L/2 − 1 < x < L/2 + 1 (L > 2),

0, otherwise.

(a) Find the displacement u(x , t).

(b) With L = 10 and a = 1, plot u versus x for 0 ≤ x ≤ 10 and for several values of t. Pay

particular attention to values of t between 3 and 7. Observe how the initial disturbance is

reﬂected at each end of the string.

(d) Construct an animation of the solution in time for at least one period.

(e) Describe the motion of the string in a few sentences.

11.

Suppose that the string in Problem 10 is started instead from the initial position

f (x) = 8x(L − x)

. Follow the instructions in Problem 10 for this new problem.

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654 Chapter 10. Partial Differential Equations and Fourier Series

12. Dimensionless variables can be introduced into the wave equation a

= u

in the

following manner:

(a) Let s = x/L and show that the wave equation becomes

= L

(b) Show that L/a has the dimensions of time and thus can be used as the unit on the

time scale. Thus, let τ = at/L and show that the wave equation then reduces to

= u

ττ

Problems 13 and 14 indicate the form of the general solution of the wave equation and the

physical signiﬁcance of the constant a.

13. (a) Show that the wave equation

= u

can be reduced to the form u

ξη

= 0 by the change of variables ξ = x − at, η = x + at.

(b) Show that u(x, t) can be written as

u(x, t) = φ(x − at) + ψ(x + at),

where φ and ψ are arbitrary functions.

14. (a) Plot the value of φ(x − at) for t = 0, 1/a,2/a, and t

/a if φ(s) = sin s. Note that for any

t ̸= 0, the graph of y = φ(x − at) is the same as that of y = φ(x) when t = 0,but displaced a

distance at in the positive x direction.Thus a represents the velocity at which a disturbance

moves along the string.

(b) What is the interpretation of φ(x + at)?

15. A steel wire 5 ft in length is stretched by a tensile force of 50 lb. The wire has a weight per

unit length of 0.026 lb/ft.

(a) Find the velocity of propagation of transverse waves in the wire.

(b) Find the natural frequencies of vibration.

the natural modes also changed?

16. Consider the wave equation

= u

in an inﬁnite one-dimensional medium subject to the initial conditions

u(x,0) = f (x), u

(x,0) = 0, −∞ < x < ∞.

(a) Using the form of the solution obtained in Problem 13, show that φ and ψ must satisfy

φ(x) + ψ(x) = f (x),

−φ

′

(x) + ψ

′

(x) = 0.

(b) Solve the equations of part (a) for φ and ψ, and thereby show that

u(x, t) =

f (x − at) + f (x + at)

This form of the solution was obtained by D’Alembert in 1746.

Hint: Note that the equation ψ

′

(x) = φ

′

(x) is solved by choosing ψ(x) = φ(x) + c.

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10.7 The Wave Equation: Vibrations of an Elastic String 655

f (x) =

2, −1 < x < 1,

0, otherwise.

Show that

f (x − at) =

2, −1 + at < x < 1 + at,

0, otherwise.

Also determine f (x + at).

(d) Sketch the solution found in part (b) at t = 0, t = 1/2a, t = 1/a, and t = 2/a, obtain-

ing the results shown in Figure 10.7.7. Observe that an initial displacement produces two

waves moving in opposite directions away from the original location; each wave consists

of one-half of the initial displacement.

1–1

–2–3

–1

–2

–1 1

–1

t = 0

t =

–

t =

FIGURE 10.7.7 Propagation of an initial disturbance in an inﬁnite one-dimensional medium.

17. Consider the wave equation

= u

in an inﬁnite one-dimensional medium subject to the initial conditions

u(x,0) = 0, u

(x,0) = g(x), −∞ < x < ∞.

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656 Chapter 10. Partial Differential Equations and Fourier Series

(a) Using the form of the solution obtained in Problem 13, show that

φ(x) + ψ(x) = 0,

−aφ

′

(x) + aψ

′

(x) = g(x).

(b) Use the ﬁrst equation of part (a) to show that ψ

′

(x) =−φ

′

(x). Then use the second

equation to show that −2aφ

′

(x) = g(x) and therefore that

φ(x) =−

g(ξ) dξ + φ(x

where x

is arbitrary. Finally, determine ψ(x).

u(x, t) =

x+at

x−at

g(ξ) dξ.

18. By combining the results of Problems 16 and 17, show that the solution of the problem

= u

u(x,0) = f (x), u

(x,0) = g(x), −∞ < x < ∞

is given by

u(x, t) =

f (x − at) + f (x + at)

x+at

x−at

g(ξ) dξ.

Problems 19 and 20 indicate how the formal solution (20), (22) of Eqs. (1), (3), and (9) can be

shown to constitute the actual solution of that problem.

19. By using the trigonometric identity sin A cos B =

sin(A + B) + sin(A − B)

, show that

the solution (20) of the problem of Eqs. (1), (3), and (9) can be written in the form (28).

20. Let h(ξ) represent the initial displacement in [0, L], extended into (−L,0) as an odd func-

tion and extended elsewhere as a periodic function of period 2L. Assuming that h, h

′

, and

′′

are continuous,show by direct differentiation that u(x, t) as given in Eq. (28) satisﬁes the

wave equation (1) and also the initial conditions (9). Note also that since Eq. (20) clearly

satisﬁes the boundary conditions (3), the same is true of Eq. (28). Comparing Eq. (28)

with the solution of the corresponding problem for the inﬁnite string (Problem 16), we see

that they have the same form, provided that the initial data for the ﬁnite string, deﬁned

originally only on the interval 0 ≤ x ≤ L,are extended in the given manner over the entire

x-axis. If this is done, the solution for the inﬁnite string is also applicable to the ﬁnite one.

21. The motion of a circular elastic membrane, such as a drumhead, is governed by the two-

dimensional wave equation in polar coordinates

+ (1/r)u

+ (1/r

θθ

= a

−2

Assuming that u(r, θ, t) = R(r);(θ)T(t), ﬁnd ordinary differential equations satisﬁed by

R(r), ;(θ), and T(t).

22. The total energy E(t) of the vibrating string is given as a function of time by

E(t) =

ρu

(x, t) +

(x, t)

dx; (i)

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10.7 The Wave Equation: Vibrations of an Elastic String 657

the ﬁrst term is the kinetic energy due to the motion of the string, and the second term is

the potential energy created by the displacement of the string away from its equilibrium

position.

For the displacement u(x, t) given by Eq. (20)—that is, for the solution of the string

problem with zero initial velocity—show that

E(t) =

∞

n=1

. (ii)

Note that the right side of Eq. (ii) does not depend on t. Thus the total energy E is a

constant and therefore is conserved during the motion of the string.

Hint: Use Parseval’s equation (Problem 37 of Section 10.4 and Problem 17 of Section 10.3),

and recall that a

= T/ρ.

23. Dispersive Waves. Consider the modiﬁed wave equation

−2

+ γ

u = u

,0< x < L, t > 0 (i)

with the boundary conditions

u(0, t) = 0, u(L, t) = 0, t > 0 (ii)

and the initial conditions

u(x,0) = f (x), u

(x,0) = 0, 0 < x < L. (iii)

(a) Show that the solution can be written as

u(x, t) =

∞

n=1

cos

+ γ

sin

nπx

where

f (x) sin

nπx

dx.

(b) By using trigonometric identities, rewrite the solution as

u(x, t) =

∞

n=1

;

sin

nπ

(x + a

t) + sin

nπ

(x − a

Determine a

, the speed of wave propagation.

, found in part (b), depends on n. This means that components of

different wave lengths (or frequencies) are propagated at different speeds, resulting in a

distortion of the original wave form over time.This phenomenon is called dispersion. Find

the condition under which a

is independent of n—that is, there is no dispersion.

24.

Consider the situation in Problem 23 with a

= 1, L = 10, and

f (x) =

⎧

⎪

⎨

⎪

⎩

x − 4, 4 ≤ x ≤ 5,

6 − x,5≤ x ≤ 6,

0, otherwise.

(a) Determine the coefﬁcients c

in the solution of Problem 23(a).

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658 Chapter 10. Partial Differential Equations and Fourier Series

(b) Plot

n=1

sin

nπx

for 0 ≤ x ≤ 10,

choosing N large enough so that the plot accurately displays the graph of f (x). Use this

value of N for the remaining plots called for in this problem.

(d) Let γ = 1/8. Plot u(x, t) versus x for t = 20, 40, and 60.

(e) Let γ = 1/4. Plot u(x, t) versus x for t = 20, 40, and 60.

10.8 Laplace’s Equation

One of the most important of all partial differential equations occurring in applied

mathematics is that associated with the name of Laplace

: in two dimensions

+ u

= 0, (1)

and in three dimensions

+ u

= 0. (2)

For example, in a two-dimensional heat conduction problem, the temperature

u(x, y, t) must satisfy the differential equation

+ u

) = u

, (3)

where α

is the thermal diffusivity. If a steadystate exists,u isa functionof x and y only,

and the time derivative vanishes; in this case Eq. (3) reduces to Eq. (1). Similarly, for

the steady state heat conduction problem in three dimensions, the temperature must

satisfy the three-dimensional form of Laplace’s equation. Equations (1) and (2) also

occur in other branches of mathematical physics. In the consideration of electrostatic

ﬁelds, the electric potential function in a dielectric medium containing no electric

charges must satisfy either Eq. (1) or Eq. (2), depending on the number of space

dimensions involved. Similarly, the potential function of a particle in free space acted

on only by gravitational forces satisﬁes the same equations. Consequently, Laplace’s

equation is often referred to as the potential equation. Another example arises in

the study of the steady (time-independent), two-dimensional, inviscid, irrotational

motion of an incompressible ﬂuid. This study centers on two functions, known as the

velocity potential function and the stream function, both of which satisfy Eq. (1).

In elasticity, the displacements that occur when a perfectly elastic bar is twisted are

described in terms of the so-called warping function, which also satisﬁes Eq. (1).

Laplace’s equation is named for Pierre-Simon de Laplace, who, beginning in 1782, studied its solutions

extensively while investigating the gravitational attraction of arbitrary bodies in space. However, the

equation ﬁrst appeared in 1752 in a paper by Euler on hydrodynamics.

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10.8 Laplace’s Equation 659

Since there is no time dependence in any of the problems just mentioned, there

are no initial conditions to be satisﬁed by the solutions of Eq. (1) or (2). They must,

however,satisfy certain boundary conditions on the bounding curve or surface of the

region in which the differential equation is to be solved. Since Laplace’s equation

is of second order, it might be plausible to expect that two boundary conditions

would be required to determine the solution completely. This, however, is not the

case. Recall that in the heat conduction problem for the ﬁnite bar (Sections 10.5

and 10.6), it was necessary to prescribe one condition at each end of the bar—that

is, one condition at each point of the boundary. If we generalize this observation

to multidimensional problems, it is then natural to prescribe one condition on the

function u at each point on the boundary of the region in which a solution of Eq. (1)

or (2) is sought. The most common boundary condition occurs when the value of u

is speciﬁed at each boundary point; in terms of the heat conduction problem, this

corresponds to prescribing the temperature on the boundary. In some problems the

value of the derivative,or rate of change,of u in the direction normal to the boundary

is speciﬁed instead; the condition on the boundary of a thermally insulated body,

for example, is of this type. It is entirely possible for more complicated boundary

conditions to occur; for example, u might be prescribed on part of the boundary and

its normal derivative speciﬁed on the remainder. The problem of ﬁnding a solution

of Laplace’s equation that takes on given boundary values is known as a Dirichlet

problem,in honor of P. G. L. Dirichlet.

On the other hand,if the values of thenormal

derivative are prescribed on the boundary, the problem is said to be a Neumann

problem, in honor of C. G. Neumann.

The Dirichlet and Neumann problems are

also known as the ﬁrst and second boundary value problems of potential theory,

respectively.

Physically, it is plausible to expect that the types of boundary conditions just

mentioned will be sufﬁcient to determine the solution completely. Indeed, it is pos-

sible to establish the existence and uniqueness of the solution of Laplace’s equation

under the boundary conditions mentioned, provided that the shape of the bound-

ary and the functions appearing in the boundary conditions satisfy certain very mild

requirements. However, the proofs of these theorems, and even their accurate state-

ment, are beyond the scope of the present book. Our only concern will be solving

some typical problems by means of separation of variables and Fourier series.

While the problems chosen as examples have interesting physical interpretations

(in terms of electrostatic potentials or steady state temperature distributions, for

instance), our purpose here is primarily to point out some of the features that may

occur during their mathematical solution. It is also worth noting again that more

complicated problems can sometimes be solved by expressing the solution as the

sum of solutions of several simpler problems (see Problems 3 and 4).

Peter Gustav Lejeune Dirichlet (1805–1859) was born in Germany of Belgian ancestry. He was a pro-

fessor at Berlin and, after the death of Gauss, at Göttingen. In 1829 he provided the ﬁrst set of conditions

sufﬁcient to guarantee the convergence of a Fourier series. The deﬁnition of function that is usually used

today in elementary calculus is essentially the one given by Dirichlet in 1837. Although he is best known

for his work in analysis and differential equations, Dirichlet was also one of the leading number theorists

of the nineteenth century.

Carl Gottfried Neumann (1832–1925),a German mathematician and professor at Leipzig for more than

forty years, made contributions to differential equations, integral equations, and complex variables.

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660 Chapter 10. Partial Differential Equations and Fourier Series

Dirichlet Problem for a Rectangle. Consider the mathematical problem of ﬁnding the

function u satisfying Laplace’s equation (1)

+ u

= 0,

in the rectangle 0 < x < a,0< y < b, and also satisfying the boundary conditions

u(x,0) = 0, u(x, b) = 0, 0 < x < a,

(4)

u(0, y) = 0, u(a, y) = f (y),0≤ y ≤ b,

where f is a given function on 0 ≤ y ≤ b (see Figure 10.8.1).

u(0, y ) = 0

u(x, b ) = 0

u(a, y ) = f (y)

u(x, 0) = 0

(a, b)

+ u

= 0

FIGURE 10.8.1 Dirichlet problem for a rectangle.

To solve this problem,we wish to construct a fundamental set of solutions satisfying

the partial differential equation and the homogeneous boundary conditions; then we

will superpose these solutions so as to satisfy the remaining boundary condition. Let

us assume that

u(x, y) = X(x)Y(y) (5)

and substitute for u in Eq. (1). This yields

′′

=−

′′

= λ,

where λ is the separation constant. Thus we obtain the two ordinary differential

equations

′′

− λX = 0, (6)

′′

+ λY = 0. (7)

If we now substitute for u from Eq. (5) in each of the homogeneous boundary

conditions, we ﬁnd that

X(0) = 0 (8)

and

Y(0) = 0, Y(b) = 0. (9)

We will ﬁrst determine the solution of the differential equation (7) subject to

the boundary conditions (9). However, this problem is essentially identical to one

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10.8 Laplace’s Equation 661

encountered previously in Sections 10.1, 10.5, and 10.7. We conclude that there are

nontrivial solutions if and only if λ is an eigenvalue, namely,

λ = (nπ/b)

, n = 1, 2, ..., (10)

and Y(y) is proportional to the corresponding eigenfunction sin(nπy/b). Next, we

substitute from Eq. (10) for λ in Eq. (6), obtaining

′′

− (nπ/b)

X = 0 .

To solve this equation subject to the boundary condition (8), it is convenient to write

the general solution as

X(x) = k

cosh(nπx/b) + k

sinh(nπx/b). (11)

The boundary condition (8) then requires that k

= 0. Therefore, X (x) must be

proportional to sinh(nπx/b). Thus we obtain the fundamental solutions

(x, y) = sinh

nπx

sin

nπy

, n = 1, 2, .... (12)

These functions satisfy the differential equation (1) and all the homogeneous

boundary conditions for each value of n.

To satisfythe remaining nonhomogeneousboundary conditionat x = a,we assume,

as usual, that we can represent the solution u(x, y) in the form

u(x, y) =

∞

n=1

(x, y) =

∞

n=1

sinh

nπx

sin

nπy

. (13)

The coefﬁcients c

are determined by the boundary condition

u(a, y) =

∞

n=1

sinh

nπa

sin

nπy

= f (y). (14)

Therefore, the quantities c

sinh(nπa/b) must be the coefﬁcients in the Fourier sine

series of period 2b for f and are given by

sinh

nπa

f (y) sin

nπy

dy. (15)

Thus the solution of the partial differential equation (1) satisfying the boundary

conditions (4) is given by Eq. (13) with the coefﬁcients c

computed from Eq. (15).

From Eqs. (13) and (15), we can see that the nth term of the solution contains the

factor sinh(nπx/b)/ sinh(nπa/b). To estimate this quantity for large n, we can use the

approximation sinh ξ

∼

and thereby obtain

sinh(nπx/b)

sinh(nπa/b)

∼

exp(nπx/b)

exp(nπa/b)

= exp[−nπ(a − x)/b].

Thus this factor has the character of a negative exponential; consequently, the series

(13) converges quite rapidly unless a − x is very small.

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662 Chapter 10. Partial Differential Equations and Fourier Series

EXAMPLE

To illustrate these results, let a = 3, b = 2, and

f (y) =

y,0≤ y ≤ 1,

2 − y,1≤ y ≤ 2.

(16)

By evaluating c

from Eq. (15), we ﬁnd that

8 sin(nπ/2)

sinh(3nπ/2)

. (17)

Then u(x, y) is given by Eq. (13). Keeping 20 terms in the series, we can plot u versus x and y,

as shown in Figure 10.8.2. Alternatively, we can construct a contour plot showing level curves

of u(x, y); Figure 10.8.3 is such a plot, with an increment of 0.1 between adjacent curves.

0.2

0.4

0.6

0.8

1.0

FIGURE 10.8.2 Plot of u versus x and y for Example 1.

u = 0

u = 0.9

u = 0.7

u = 0.5

u = 0.4

u = 0.2

u = 0.6

u = 0.8

u = 0.3

u = 0.1

321

FIGURE 10.8.3 Level curves of u(x, y) for Example 1.

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10.8 Laplace’s Equation 663

Dirichlet Problem for a Circle. Consider the problem of solving Laplace’s equation in a

circular region r < a subject to the boundary condition

u(a, θ) = f (θ), (18)

where f is a given function on 0 ≤ θ<2π (see Figure 10.8.4). In polar coordinates

(r, θ), Laplace’s equation has the form

θθ

= 0. (19)

To complete the statement of the problem,we note that for u(r, θ) to be single-valued,

it is necessary that u be periodic in θ with period 2π. Moreover, we state explicitly

that u(r, θ) must be bounded for r ≤ a, since this will become important later.

u(a, ) = f ( )

+ u

+ u = 0

θθ

FIGURE 10.8.4 Dirichlet problem for a circle.

To apply the method of separation of variables to this problem, we assume that

u(r, θ) = R(r);(θ) (20)

and substitute for u in the differential equation (19). This yields

′′

; +

′

; +

′′

= 0,

′′

+ r

′

=−

;

′′

;

= λ, (21)

where λ is the separation constant. Thus we obtain the two ordinary differential

equations

′′

+ rR

′

− λR = 0, (22)

;

′′

+ λ; = 0. (23)

In this problem there are no homogeneous boundary conditions; recall, however,

that solutions must be bounded and also periodic in θ with period 2π. It is possible

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664 Chapter 10. Partial Differential Equations and Fourier Series

to show (Problem 9) that the periodicity condition requires λ to be real. We will

consider in turn the cases in which λ is negative, zero, and positive.

If λ<0, let λ =−µ

, where µ>0. Then Eq. (23) becomes ;

′′

− µ

; = 0, and

consequently,

;(θ) = c

µθ

+ c

−µθ

. (24)

Thus ;(θ) can be periodic only if c

= c

= 0, and we conclude that λ cannot be

negative.

If λ = 0, then Eq. (23) becomes ;

′′

= 0, and thus

;(θ) = c

+ c

θ. (25)

For ;(θ) to be periodic we must have c

= 0, so that ;(θ) is a constant. Further, for

λ = 0, Eq. (22) becomes

′′

+ rR

′

= 0. (26)

This equation is of the Euler type and has the solution

R(r) = k

+ k

ln r. (27)

The logarithmic term cannot be accepted if u(r, θ) is to remain bounded as r → 0;

hence k

= 0. Thus, corresponding to λ = 0, we conclude that u(r, θ) must be a

constant—that is, proportional to the solution

(r, θ) = 1. (28)

Finally, if λ>0, we let λ = µ

, where µ>0. Then Eqs. (22) and (23) become

′′

+ rR

′

− µ

R = 0 (29)

and

;

′′

+ µ

; = 0, (30)

respectively. Equation (29) is an Euler equation and has the solution

R(r) = k

+ k

−µ

, (31)

while Eq. (30) has the solution

;(θ) = c

sin µθ + c

cos µθ. (32)

In order for ; to be periodic with period 2π, it is necessary for µ to be a positive

integer n. With µ = n it follows that the solution r

−µ

in Eq. (31) must be discarded

since it becomes unbounded as r → 0. Consequently, k

= 0 and the appropriate

solutions of Eq. (19) are

(r, θ) = r

cos nθ, v

(r, θ) = r

sin nθ, n = 1, 2, .... (33)

These functions, together with u

(r, θ) = 1, form a set of fundamental solutions for

the present problem.

In the usual way, we now assume that u can be expressed as a linear combination

of the fundamental solutions; that is,

u(r, θ) =

∞

n=1

cos nθ + k

sin nθ). (34)

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10.8 Laplace’s Equation 665

The boundary condition (18) then requires that

u(a, θ) =

∞

n=1

cos nθ + k

sin nθ) = f (θ) (35)

for 0 ≤ θ<2π. The function f may be extended outside this interval so that it is

periodic with period 2π and therefore has a Fourier series of the form (35). Since

the extended function has period 2π, we may compute its Fourier coefﬁcients by

integrating over any period of the function. In particular, it is convenient to use the

original interval (0, 2π); then

2π

f (θ) cos nθ dθ, n = 0, 1, 2, ...; (36)

2π

f (θ) sin nθ dθ, n = 1, 2, .... (37)

With this choice of the coefﬁcients, Eq. (34) represents the solution of the boundary

value problem of Eqs. (18) and (19). Note that in this problem we needed both sine

and cosine terms in the solution. This is because the boundary data were given on

0 ≤ θ<2π and have period 2π. As a consequence, the full Fourier series is required,

rather than sine or cosine terms alone.

PROBLEMS 1. (a) Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,0< y < b,

that satisﬁes the boundary conditions

u(0, y) = 0, u(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x, b) = g(x),0≤ x ≤ a.

(b) Find the solution if

g(x) =

x,0≤ x ≤ a/2,

a − x, a/2 ≤ x ≤ a.

several

values of x.

(d) Plot u versus both x and y in three dimensions. Also draw a contour plot showing

several level curves of u(x, y) in the xy-plane.

2. Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,0< y < b, that

satisﬁes the boundary conditions

u(0, y) = 0, u(a,y) = 0, 0 < y < b,

u(x,0) = h(x), u(x, b) = 0, 0 ≤ x ≤ a.

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666 Chapter 10. Partial Differential Equations and Fourier Series

3. (a) Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,0< y < b,

that satisﬁes the boundary conditions

u(0, y) = 0, u(a, y) = f (y),0< y < b,

u(x,0) = h(x), u(x, b) = 0, 0 ≤ x ≤ a.

Hint: Consider the possibility of adding the solutions of two problems, one with homo-

geneous boundary conditions except for u(a, y) = f (y), and the other with homogeneous

boundary conditions except for u(x,0) = h(x).

(b) Find the solution if h(x) = (x/a)

and f (y) = 1 − (y/b).

both x and y, and a contour plot.

4. Show how to ﬁnd the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,

0 < y < b, that satisﬁes the boundary conditions

u(0, y) = k(y), u(a, y) = f (y),0< y < b,

u(x,0) = h(x), u(x, b) = g(x),0≤ x ≤ a.

Hint: See Problem

5. Find the solution u(r, θ) of Laplace’s equation

+ (1/r)u

+ (1/r

θθ

= 0

outside the circle r = a, that satisﬁes the boundary condition

u(a, θ) = f (θ),0≤ θ<2π,

on the circle. Assume that u(r, θ) is single-valued and bounded for r > a.

(a) Find the solution u(r, θ) of Laplace’s equation in the semicircular region r < a,

0 <θ<π, that satisﬁes the boundary conditions

u(r,0) = 0, u(r, π) = 0, 0 ≤ r < a,

u(a, θ) = f (θ),0≤ θ ≤ π.

Assume that u is single-valued and bounded in the given region.

(b) Find the solution if f (θ) = θ(π − θ).

and θ, and a contour plot.

7. Find the solution u(r, θ) of Laplace’s equation in the circular sector 0 < r < a,

0 <θ<α, that satisﬁes the boundary conditions

u(r,0) = 0, u(r, α) = 0,

0 ≤ r < a

u(a, θ) = f (θ),0≤ θ ≤ α.

Assume that u is single-valued and bounded in the sector and that 0 <α<2π.

(a) Find the solution u(x, y) of Laplace’s equation in the semi-inﬁnite strip 0 < x < a,

y > 0, that satisﬁes the boundary conditions

u(0, y) = 0, u(a, y) = 0, y > 0,

u(x,0) = f (x),0≤ x ≤ a

and the additional condition that u(x, y) → 0 as y →∞.

(b) Find the solution if f (x) = x(a − x).

for which u(x, y) ≤ 0.1 for all y ≥ y

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10.8 Laplace’s Equation 667

9. Show that Eq. (23) has periodic solutions only if λ is real.

Hint: Let λ =−µ

, where µ = ν + iσ with ν and σ real.

10. Consider the problem of ﬁnding a solution u(x, y) of Laplace’s equation in the rectangle

0 < x < a,0< y < b, that satisﬁes the boundary conditions

(0, y) = 0, u

(a, y) = f (y),0< y < b,

(x,0) = 0, u

(x, b) = 0, 0 ≤ x ≤ a.

This is an example of a Neumann problem.

(a) Show that Laplace’s equation and the homogeneous boundary conditions determine

the fundamental set of solutions

(x, y) = c

cosh(nπx/b) cos(nπy/b), n = 1, 2, 3, ....

(b) By superposing the fundamental solutions of part (a), formally determine a function

u satisfying the nonhomogeneous boundary condition u

(a, y) = f (y). Note that when

(a, y) is calculated, the constant term in u(x, y) is eliminated, and there is no condition

from which to determine c

. Furthermore, it must be possible to express f by means of a

Fourier cosine series of period 2b, which does not have a constant term. This means that

f (y) dy = 0

is a necessary condition for the given problem to be solvable. Finally, note that c

remains

arbitrary, and hence the solution is determined only up to this additive constant. This is a

property of all Neumann problems.

11. Find a solution u(r, θ) of Laplace’s equation inside the circle r = a that satisﬁes the

boundary condition on the circle

(a, θ) = g(θ),0≤ θ<2π.

Note that this is a Neumann problem and that its solution is determined only up to an

arbitrary additive constant. State a necessary condition on g(θ) for this problem to be

solvable by the method of separation of variables (see Problem 10).

12.

(a) Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,

0 < y < b, that satisﬁes the boundary conditions

u(0, y) = 0, u(a, y) = 0, 0 < y < b,

(x,0) = 0, u(x, b) = g(x ),0≤ x ≤ a.

Note that this is neither a Dirichlet nor a Neumann problem, but a mixed problem in

which u is prescribed on part of the boundary and its normal derivative on the rest.

(b) Find the solution if

g(x) =

x,0≤ x ≤ a/2,

a − x, a/2 ≤ x ≤ a.

of Problem 1.

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668 Chapter 10. Partial Differential Equations and Fourier Series

13. (a) Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,

0 < y < b, that satisﬁes the boundary conditions

u(0, y) = 0, u(a, y) = f (y),0< y < b,

u(x,0) = 0, u

(x, b) = 0, 0 ≤ x ≤ a.

Hint: Eventually, it will be necessary to expand f (y) in a series that makes use of the

functions sin(πy/2b), sin(3πy/2b), sin(5πy/2b), ...(see Problem 39 of Section 10.4).

(b) Find the solution if f (y) = y(2b − y).

14.

(a) Find the solution u(x, y) of Laplace’s equation in the rectangle 0 < x < a,

0 < y < b, that satisﬁes the boundary conditions

(0, y) = 0, u

(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x, b) = g(x),0≤ x ≤ a.

(b) Find the solution if g(x) = 1 + x

(x − a)

15. By writing Laplace’s equation in cylindrical coordinates r, θ, and z and then assuming that

the solution is axially symmetric (no dependence on θ), we obtain the equation

+ (1/r)u

+ u

= 0.

Assuming that u(r, z) = R(r)Z(z), show that R and Z satisfy the equations

′′

+ R

′

+ λ

rR = 0, Z

′′

− λ

Z = 0.

The equation for R is Bessel’s equation of order zero with independent variable λr.

16.

Flow in an Aquifer. Consider the ﬂow of water in a porous medium, such as sand, in an

aquifer. The ﬂow is driven by the hydraulic head, a measure of the potential energy of the

water above the aquifer. Let R : 0 < x < a,0< z < b be a vertical section of an aquifer. In

a uniform, homogeneous medium, the hydraulic head u(x, z) satisﬁes Laplace’s equation

+ u

= 0 in R. (i)

If water cannot ﬂow through the sides and bottom of R, then the boundary conditions

there are

(0, z) = 0, u

(a, z) = 0, 0 ≤ z ≤ b (ii)

(x,0) = 0, 0 ≤ x ≤ a. (iii)

Finally, suppose that the boundary condition at z = b is

u(x, b) = b + αx,0≤ x ≤ a, (iv)

where α is the slope of the water table.

(a) Solve the given boundary value problem for u(x, z).

(b) Let a = 1000, b = 500, and α = 0.1. Draw a contour plot of the solution in R; that is,

plot some level curves of u(x, z).

direction of decreasing u. Plot some of the ﬂow paths.

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Appendix A 669

APPENDIX

Derivation of the Heat Conduction Equation. In this section we derive the differ-

ential equation that, to a ﬁrst approximation at least, governs the conduction of heat

in solids. It is important to understand that the mathematical analysis of a physical

situation or process such as this ultimately rests on a foundation of empirical knowl-

edge of the phenomenon involved. The mathematician must have a place to start, so

to speak, and this place is furnished by experience. Consider a uniform rod insulated

on the lateral surfaces so that heat can ﬂow only in the axial direction. It has been

demonstrated many times that if two parallel cross sections of the same area A and

different temperatures T

and T

, respectively, are separated by a small distance d,

an amount of heat per unit time will pass from the warmer section to the cooler one.

Moreover, this amount of heat is proportional to the area A and to the temperature

difference |T

− T

| and is inversely proportional to the separation distance d. Thus

Amount of heat per unit time = κA|T

− T

|/d , (1)

where the positive proportionality factor κ is called the thermal conductivity and

depends primarily on the material

of the rod. The relation (1) is often called

Fourier’s law of heat conduction. We repeat that Eq. (1) is an empirical, not a theo-

retical, result and that it can be, and often has been, veriﬁed by careful experiment.

It is the basis of the mathematical theory of heat conduction.

Now consider a straight rod of uniform cross section and homogeneous material,

oriented so that the x-axis lies along the axis of the rod (see Figure 10.A.1). Let x = 0

and x = L designate the ends of the bar.

x = x

+ ∆x

H = – Au

H = Au

FIGURE 10.A.1 Conduction of heat in an element of a rod.

We will assume that the sides of the bar are perfectly insulated so that there is no

passage of heat through them. We will also assume that the temperature u depends

only on the axial position x and the time t, and not on the lateral coordinates y and z.

In other words,we assume that the temperature remains constant on any cross section

of the bar. This assumption is usually satisfactory when the lateral dimensions of the

rod are small compared to its length.

The differential equation governing the temperature in the bar is an expression of

a fundamental physical balance; the rate at which heat ﬂows into any portion of the

bar is equal to the rate at which heat is absorbed in that portion of the bar.The terms

in the equation are called the ﬂux (ﬂow) term and the absorption term, respectively.

Actually,κ also depends on the temperature,but if the temperature range is not too great,it is satisfactory

to assume that κ is independent of temperature.

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670 Chapter 10. Partial Differential Equations and Fourier Series

We will ﬁrst calculate the ﬂux term. Consider an element of the bar lying between

the cross sections x = x

and x = x

+ 5x, where x

is arbitrary and 5x is small. The

instantaneous rate of heat transfer H(x

, t) from left to right across the cross section

x = x

is given by

H(x

, t) =−lim

d→0

κA

u(x

+ d/2, t) − u(x

− d/2, t)

=−κAu

, t). (2)

The minus sign appears in this equation because there will be a positive ﬂow of heat

from left to right only if the temperature is greater to the left of x = x

than to the

right; in this case u

, t) is negative. In a similar manner, the rate at which heat

passes from left to right through the cross section x = x

+ 5x is given by

H(x

+ 5x, t) =−κAu

+ 5x, t). (3)

The net rate at which heat ﬂows into the segment of the bar between x = x

and

x = x

+ 5x is thus given by

Q = H(x

, t) − H(x

+ 5x, t) = κA[u

+ 5x, t) − u

, t)], (4)

and the amount of heat entering this bar element in time 5t is

Q5t = κA[u

+ 5x, t) − u

, t)]5t. (5)

Let us now calculate the absorption term. The average change in temperature 5u,

in the time interval 5t, is proportional to the amount of heat Q5t introduced and is

inversely proportional to the mass 5m of the element. Thus

5u =

Q5t

sρA5x

, (6)

where the constant of proportionality s is known as the speciﬁc heat of the material of

the bar, and ρ is its density.

The average temperature change 5u in the bar element

under consideration is the actual temperature change at some intermediate point

x = x

+ θ5x, where 0 <θ<1. Thus Eq. (6) can be written as

u(x

+ θ5x, t + 5t) − u(x

+ θ5x, t) =

Q5t

sρA5x

(7)

or as

Q5t =[u(x

+ θ5x, t + 5t) − u(x

+ θ5x, t)]sρA5x. (8)

To balance the ﬂux and absorption terms, we equate the two expressions for Q5t:

κA[u

+ 5x, t) − u

, t)]5t

= sρA[u(x

+ θ5x, t + 5t) − u(x

+ θ5x, t)]5x. (9)

Upon dividing Eq. (9) by 5x 5t and then letting 5x → 0 and 5t → 0, we obtain the

heat conduction or diffusion equation

= u

. (10)

The dependence of the density and speciﬁc heat on temperature is relatively small and will be neglected.

Thus both ρ and s will be considered as constants.

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Appendix A 671

The quantity α

deﬁned by

= κ/ρs (11)

is called the thermal diffusivity and is a parameter that depends only on the mate-

rial of the bar. The units of α

are (length)

/time. Typical values of α

are given in

Table 10.5.1.

Several relatively simple conditions may be imposed at the ends of the bar. For

example, the temperature at an end may be maintained at some constant value T.

This might be accomplished by placing the end of the bar in thermal contact with

some reservoir of sufﬁcient size so that any heat that ﬂows between the bar and

the reservoir does not appreciably alter the temperature of the reservoir. At an end

where this is done, the boundary condition is

u = T. (12)

Another simple boundary condition occurs if the end is insulated so that no heat

passes through it. Recalling the expression (2) for the amount of heat crossing any

cross section of the bar, we conclude that the condition of insulation is that this

quantity vanish. Thus

= 0 (13)

is the boundary condition at an insulated end.

A more general typeof boundarycondition occursif therate ofﬂow of heatthrough

an end of the bar is proportional to the temperature there. Let us consider the end

x = 0, where the rate of ﬂow of heat from left to right is given by −κAu

(0, t); see

Eq. (2). Hence the rate of heat ﬂow out of the bar (from right to left) at x = 0is

κAu

(0, t). If this quantity is proportional to the temperature u(0, t), then we obtain

the boundary condition

(0, t) − h

u(0, t) = 0, t > 0, (14)

where h

is a nonnegative constant of proportionality. Note that h

= 0 corresponds

to an insulated end and that h

→∞corresponds to an end held at zero temperature.

If heat ﬂow is taking place at the right end of the bar (x = L), then in a similar way

we obtain the boundary condition

(L, t) + h

u(L, t) = 0, t > 0, (15)

where again h

is a nonnegative constant of proportionality.

Finally, to determine completely the ﬂow of heat in the bar, it is necessary to state

the temperature distribution at one ﬁxed instant, usually taken as the initial time

t = 0. This initial condition is of the form

u(x,0) = f (x),0≤ x ≤ L. (16)

The problem then is to determine the solution of the differential equation (10)

subject to one of the boundary conditions (12) to (15) at each end and to the initial

condition (16) at t = 0.

Several generalizations of the heat equation (10) also occur in practice. First, the

bar material may be nonuniform, and the cross section may not be constant along

the length of the bar. In this case, the parameters κ, ρ, s, and A may depend on the

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672 Chapter 10. Partial Differential Equations and Fourier Series

axial variable x. Going back to Eq. (2), we see that the rate of heat transfer from left

to right across the cross section at x = x

is now given by

H(x

, t) =−κ(x

)A(x

, t) (17)

with a similar expression for H(x

+ 5x, t). If we introduce these quantities into

Eq. (4) and eventually into Eq. (9), and proceed as before, we obtain the partial

differential equation

(κAu

)

= sρAu

. (18)

We will usually write Eq. (18) in the form

r(x)u

=[p(x)u

]

, (19)

where p(x) = κ(x)A(x) and r(x) = s(x)ρ(x)A(x). Note that both of these quantities

are intrinsically positive.

A second generalization occurs if there are other ways in which heat enters or

leaves the bar. Suppose that there is a source that adds heat to the bar at a rate

G(x, t, u) per unit time per unit length, where G(x, t, u)>0. In this case we must add

the term G(x, t, u)5x 5t to the left side of Eq. (9), and this leads to the differential

equation

r(x)u

=[p(x)u

]

+ G(x, t, u). (20)

If G(x, t, u)<0, then we speak of a sink that removes heat from the bar at the rate

G(x, t, u) per unit time per unit length. To make the problem tractable, we must

restrict the form of the function G. In particular, we assume that G is linear in u and

that the coefﬁcient of u does not depend on t. Thus we write

G(x, t, u) = F(x, t) − q(x)u. (21)

The minus sign in Eq. (21) has been introduced so that certain equations that appear

later will have their customary forms. Substituting from Eq. (21) into Eq. (20), we

obtain

r(x)u

=[p(x)u

]

− q(x)u + F(x, t). (22)

This equation is sometimes called the generalized heat conduction equation. Bound-

ary value problems for Eq. (22) will be discussed to some extent in Chapter 11.

Finally, if instead of a one-dimensional bar, we consider a body with more than

one signiﬁcant space dimension, then the temperature is a function of two or three

space coordinates rather than of x alone. Considerations similar to those leading to

Eq. (10) can be employed to derive the heat conduction equation in two dimensions

+ u

) = u

, (23)

or in three dimensions

+ u

) = u

. (24)

The boundary conditions corresponding to Eqs. (12) and (13) for multidimensional

problems correspond to a prescribed temperature distribution on the boundary, or to

an insulated boundary. Similarly, the initial temperature distribution will in general

be a function of x and y for Eq. (23) and a function of x, y, and z for Eq. (24).

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Appendix B 673

APPENDIX

Derivation of the Wave Equation. In this appendix we derive the wave equation

in one space dimension as it applies to the transverse vibrations of an elastic string,

or cable; the elastic string may be thought of as a guitar string, a guy wire, or possi-

bly an electric power line. The same equation, however, with the variables properly

interpreted, occurs in many other wave problems having only one signiﬁcant space

variable.

Consider a perfectly ﬂexible elastic string stretched tightly between supports ﬁxed

at the same horizontal level (see Figure 10.B.1a). Let the x-axis lie along the string

with the endpoints located at x = 0 and x = L. If the string is set in motion at some

initial time t = 0 (by plucking, for example) and is thereafter left undisturbed, it will

vibrate freely in a vertical plane,provided that damping effects, such as air resistance,

are neglected. To determine the differential equation governing this motion, we will

consider the forces acting on a small element of the string of length 5x lying between

the points x and x + 5x (see Figure 10.B.1b).We assume that the motion of the string

is small and that, as a consequence, each point on the string moves solely in a vertical

line. We denote by u(x, t) the vertical displacement of the point x at the time t. Let

the tension in the string, which always acts in the tangential direction, be denoted by

T(x, t), and let ρ denote the mass per unit length of the string.

(a)

x = 0 x = L



x x + ∆x

(b)

T (x + ∆x, t)

θθ

+ ∆

T(x, t)

(c)

V = T sin

H = T cos

FIGURE 10.B.1 (a) An elastic string under tension.

(b) An element of the displaced string.

Newton’s law, as it applies to the element 5x of the string, states that the net

external force, due to the tension at the ends of the element, must be equal to the

product of the mass of the element and the acceleration of its mass center. Since

there is no horizontal acceleration, the horizontal components must satisfy

T(x + 5x, t) cos(θ + 5θ) − T(x, t) cos θ = 0. (1)

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674 Chapter 10. Partial Differential Equations and Fourier Series

If we denote the horizontal component of the tension (see Figure 10.B.1c)byH,then

Eq. (1) states that H is independent of x.

On the other hand, the vertical components satisfy

T(x + 5x, t) sin(θ + 5θ) − T(x, t) sin θ = ρ5xu

(x, t), (2)

where

x is the coordinate of the center of mass of the element of the string under

consideration. Clearly,

x lies in the interval x < x < x + 5x. The weight of the

string, which acts vertically downward, is assumed to be negligible and has been

neglected in Eq. (2).

If the vertical component of T is denoted by V, then Eq. (2) can be written as

V(x + 5x, t) − V(x, t)

= ρu

(x, t).

Passing to the limit as 5x → 0 gives

(x, t) = ρu

(x, t). (3)

To express Eq. (3) entirely in terms of u, we note that

V(x, t) = H(t) tan θ = H(t)u

(x, t).

Hence Eq. (3) becomes

(Hu

)

= ρu

or, since H is independent of x,

= ρu

. (4)

For small motions of the string, it is permissible to replace H = T cos θ by T. Then

Eq. (4) takes its customary form

= u

, (5)

where

= T/ρ. (6)

We will assume further that a

is a constant, although this is not required in our

derivation, even for small motions. Equation (5) is called the wave equation for one

space dimension. Since T has the dimension of force, and ρ that of mass/length, it

follows that the constant a has the dimension of velocity. It is possible to identify

a as the velocity with which a small disturbance (wave) moves along the string.

According to Eq. (6), the wave velocity a varies directly with the tension in the string,

but inversely with the density of the string material. These facts are in agreement

with experience.

As in the case of the heat conduction equation,there are various generalizations of

the wave equation (5). One important equation is known as the telegraph equation

and has the form

+ cu

+ ku = a

+ F(x, t), (7)

where c and k are nonnegative constants. The terms cu

, ku, and F(x, t) arise from a

viscous damping force, an elastic restoring force, and an external force, respectively.

Note the similarity of Eq. (7), except for the term a

, to the equation for the

spring–mass system derived in Section 3.7; the additional term a

arises from a

consideration of internal elastic forces.

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Appendix B 675

The telegraph equation also governs the ﬂow of voltage, or current, in a trans-

mission line (hence its name); in this case the coefﬁcients are related to electrical

parameters in the line.

For a vibrating system with more than one signiﬁcant space coordinate, it may be

necessary to consider the wave equation in two dimensions

+ u

) = u

, (8)

or in three dimensions

+ u

) = u

. (9)

REFERENCES The following books contain additional information on Fourier series:

Buck, R. C., and Buck, E. F., Advanced Calculus (3rd ed.) (New York: McGraw-Hill, 1978).

Carslaw,H. S.,Introduction to the Theory of Fourier’s Series and Integrals (3rded.) (Cambridge:Cambridge

University Press, 1930; New York: Dover, 1952).

Courant,R.,and John,F.,Introduction to Calculus andAnalysis (2 vols.) (Berlin/NewYork:Springer-Verlag,

1999).

Kaplan,W., Advanced Calculus (5th ed.) (Boston: Addison-Wesley, 2003).

A brief biography of Fourier and an annotated copy of his 1807 paper are contained in

Grattan-Guinness, I., Joseph Fourier 1768–1830:A Survey of his Life and Work (Cambridge, MA: MIT

Press, 1972).

Useful references on partial differential equations and the method of separation of variables include

the following:

Churchill, R. V., and Brown, J. W., Fourier Series and Boundary Value Problems (8th ed.) (New York:

McGraw-Hill, 2012).

Haberman, R., Elementary Applied Partial Differential Equations: With Fourier Series and Boundary

Value Problems (4th ed.) (Upper Saddle River, NJ: Pearson Prentice-Hall, 2004).

Pinsky, M. A., Partial Differential Equations and Boundary Value Problems with Applications (3rd ed.)

(Boston:WCB/McGraw-Hill, 1998; Providence, RI: American Mathematical Society, 2011).

Powers, D. L., Boundary Value Problems and Partial Differential Equations (6th ed.) (Boston: Academic

Press, 2010).

Strauss, W. A., Partial Differential Equations, an Introduction (2nd ed.) (New York:Wiley, 2009).

Weinberger, H. F.,A First Course in Partial Differential Equations with Complex Variables and Transform

Methods (New York: Blaisdell, 1965; New York: Dover, 1995).

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August 7, 2012 21:05 c11 Sheet number 1 Page number 677 cyan black

677

CHAPTER

Boundary Value

Problems and

Sturm–Liouville

Theory

As a result of separating variables in a partial differential equation in Chapter 10, we

repeatedly encountered the differential equation

′′

+ λX = 0, 0 < x < L

with the boundary conditions

X(0) = 0, X(L) = 0.

This boundary value problem is the prototype of a large class of problems that are

important in applied mathematics. These problems are known as Sturm–Liouville

boundary value problems. In this chapter we discuss the major properties of Sturm–

Liouville problems and their solutions; in the process we are able to generalize

somewhat the method of separation of variables for partial differential equations.

11.1 The Occurrence of Two-Point Boundary Value Problems

In Chapter 10 we described the method of separation of variables as a means of

solving certain problems involvingpartial differentialequations.The heat conduction

problem consisting of the partial differential equation

= u

,0< x < L, t > 0 (1)

subject to the boundary conditions

u(0, t) = 0, u(L, t) = 0, t > 0 (2)

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678 Chapter 11. Boundary Value Problems

and the initial condition

u(x,0) = f (x),0≤ x ≤ L (3)

is typical of the problems considered there. A crucial part of the process of solv-

ing such problems is to ﬁnd the eigenvalues and eigenfunctions of the differential

equation

′′

+ λX = 0, 0 < x < L (4)

with the boundary conditions

X(0) = 0, X(L) = 0 (5)

or perhaps

′

(0) = 0, X

′

(L) = 0. (6)

The sine or cosine functions that result from solving Eq. (4) subject to the boundary

conditions (5) or (6) are used to expand the initial temperature distribution f (x) in a

Fourier series.

In this chapter we extend and generalize the results of Chapter 10. Our main goal

is to show how the method of separation of variables can be used to solve problems

somewhat more general than that of Eqs. (1), (2), and (3). We are interested in three

types of generalizations.

First,we wish to consider more general partial differential equations—for example,

the equation

r(x)u

=[p(x)u

]

− q(x)u + F(x, t). (7)

This equation can arise,as indicated inAppendixA of Chapter 10,in the study of heat

conduction in a bar of variable material properties in the presence of heat sources.

If p(x) and r(x) are constants, and if the source terms q(x)u and F(x, t) are zero, then

Eq. (7) reduces to Eq. (1). The partial differential equation (7) also occurs in the

investigation of other phenomena of a diffusive character.

A second generalization is to allow more general boundary conditions. In

particular, we wish to consider boundary conditions of the form

(0, t) − h

u(0, t) = 0, u

(L, t) + h

u(L, t) = 0. (8)

Such conditions occur when the rate of heat ﬂow through an end of the bar is pro-

portional to the temperature there. Usually, h

and h

are nonnegative constants,but

in some cases they may be negative or depend on t. The boundary conditions (2)

are obtained in the limit as h

→∞and h

→∞. The other important limiting case,

= h

= 0, gives the boundary conditions for insulated ends.

The ﬁnal generalization that we discuss in this chapter concerns the geometry of

the region in which the problem is posed.The results of Chapter 10 are adequate only

for a rather restricted class of problems, chieﬂy those in which the region of interest

is rectangular or, in a few cases, circular. Later in this chapter we consider certain

problems posed in a few other geometrical regions.

Let us consider the equation

r(x)u

=[p(x)u

]

− q(x)u (9)

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11.1 Two-Point Boundary Value Problems 679

obtained by setting the term F(x, t) in Eq. (7) equal to zero.To separate the variables,

we assume that

u(x, t) = X(x)T(t) (10)

and substitute for u in Eq. (9). We obtain

r(x)XT

′

=[p(x)X

′

]

′

T − q(x)XT, (11)

or, upon dividing by r(x)XT,

′

[p(x)X

′

]

′

r(x)X

−

q(x)

r(x)

=−λ. (12)

We have denoted the separation constant by −λ in anticipation of the fact that usually

it will turn out to be real and negative. From Eq. (12) we obtain the following two

ordinary differential equations for X and T:

[p(x)X

′

]

′

− q(x)X +λr(x)X = 0, (13)

′

+ λT = 0. (14)

If we substitute from Eq. (10) for u in Eqs. (8) and assume that h

and h

are constants,

then we obtain the boundary conditions

′

(0) − h

X(0) = 0, X

′

(L) + h

X(L) = 0. (15)

To proceed further we need to solve Eq. (13) subject to the boundary conditions

(15). Although this is a more general linear homogeneous two-point boundary value

problem than the problem consisting of the differential equation (4) and the bound-

ary conditions (5) or (6), the solutions behave in very much the same way. For every

value of λ, the problem (13),(15) has the trivial solution X (x) = 0. For certain values

of λ, called eigenvalues, there are also other, nontrivial solutions called eigenfunc-

tions.These eigenfunctions form the basis for series solutions of a variety of problems

in partial differential equations, such as the generalized heat conduction equation (9)

subject to the boundary conditions (8) and the initial condition (3).

In this chapter wediscuss some of theproperties of solutions oftwo-point boundary

value problems for second order linear equations. Sometimes we consider the general

linear homogeneous equation

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0, (16)

investigated in Chapter 3. However,for most purposes it is better to discuss equations

in which the ﬁrst and second derivative terms are related as in Eq. (13). It is always

possible to transform the general equation (16) so that the derivative terms appear

as in Eq. (13) (see Problem 11).

Boundary value problems with higher order differential equations can also occur;

in them the number of boundary conditions must equal the order of the differen-

tial equation. As a rule, the order of the differential equation is even, and half the

boundary conditions are given at each end of the interval. It is also possible for a

single boundary condition to involve values of the solution and/or its derivatives at

both boundary points; for example,

y(0) − y(L) = 0. (17)

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680 Chapter 11. Boundary Value Problems

The following example involves one boundary condition of the form (15) and is

therefore more complicated than the problems in Section 10.1.

EXAMPLE

Find the eigenvalues and the corresponding eigenfunctions of the boundary value problem

′′

+ λy = 0, (18)

y(0) = 0, y

′

(1) + y(1) = 0. (19)

One place where this problem occurs is in the heat conduction problem in a bar of unit length.

The boundary condition at x = 0 corresponds to a zero temperature there. The boundary

condition at x = 1 corresponds to a rate of heat ﬂow that is proportional to the temperature

there, and units are chosen so that the constant of proportionality is 1 (see Appendix A of

Chapter 10).

The solution of the differential equation may have one of several forms, depending on λ,so

it is necessary to consider several cases. First, if λ = 0, the general solution of the differential

equation is

y = c

x + c

. (20)

The two boundary conditions require that

= 0, 2c

+ c

= 0, (21)

respectively. The only solution of Eqs. (21) is c

= c

= 0, so the boundary value problem has

no nontrivial solution in this case. Hence λ = 0 is not an eigenvalue.

If λ>0, then the general solution of the differential equation (18) is

y = c

sin

√

λ x + c

cos

√

λ x, (22)

where

√

λ>0. The boundary condition at x = 0 requires that c

= 0; from the boundary

condition at x = 1 we then obtain the equation

(sin

√

λ +

√

λ cos

√

λ) = 0.

For a nontrivial solution y we must have c

̸= 0, and thus λ must satisfy

sin

√

λ +

√

λ cos

√

λ = 0. (23)

Note that if λ is such that cos

√

λ = 0, then sin

√

λ ̸= 0, and Eq. (23) is not satisﬁed. Hence we

may assume that cos

√

λ ̸= 0; dividing Eq. (23) by cos

√

λ, we obtain

√

λ =−tan

√

λ. (24)

The solutions ofEq. (24) can be determinednumerically.Theycan also befound approximately

by sketching the graphs of f (

√

λ) =

√

λ and g(

√

λ) =−tan

√

λ for

√

λ>0 on the same set of

axes and identifying the points of intersection of the two curves (see Figure 11.1.1). The point

√

λ = 0 is speciﬁcally excluded from this argument because the solution (22) is valid only

for

√

λ ̸= 0. Despite the fact that the curves intersect there, λ = 0 is not an eigenvalue, as we

have already shown.The ﬁrstthree positive solutions ofEq. (24) are

√

∼

2.029,

√

∼

4.913,

and

√

∼

7.979. As can be seen from Figure 11.1.1, the other roots are given with reasonable

accuracy by

√

∼

(2n − 1)π/2 for n = 4, 5, ..., the precision of this estimate improving as n

increases. Hence the eigenvalues are

∼

4.116, λ

∼

24.14,

(25)

∼

63.66, λ

∼

(2n − 1)

/4 for n = 4, 5, ....

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11.1 Two-Point Boundary Value Problems 681

Finally, since c

= 0, the eigenfunction corresponding to the eigenvalue λ

(x, λ

) = k

sin

x; n = 1, 2, ..., (26)

where the constant k

remains arbitrary.

3 /2

5 /2

7 /2

5 /2

–2/

–

3 /2

–

√

u = g(√ )

= – tan √

u = f (√ )

= √

FIGURE 11.1.1 Graphical solution of

√

λ =−tan

√

λ.

Next consider λ<0. In this case let λ =−µ so that µ>0. Then Eq. (14) becomes

′′

− µy = 0, (27)

and its general solution is

y = c

sinh

√

µ x + c

cosh

√

µ x, (28)

where

√

µ>0. Proceeding as in the previous case, we ﬁnd that µ must satisfy the equation

√

µ =−tanh

√

µ. (29)

From Figure 11.1.2 it is clear that the graphs of f (

√

µ) =

√

µ and g(

√

µ) =−tanh

√

µ intersect

only at the origin. Hence there are no positive values of

√

µ that satisfy Eq. (29), and hence

the boundary value problem (18), (19) has no negative eigenvalues.

Finally, it is necessary to consider the possibility that λ may be complex. It is possible to

show by direct calculation that the problem (18), (19) has no complex eigenvalues. However,

in Section 11.2 we consider in more detail a large class of problems that includes this example.

One of the things we show there is that every problem in this class has only real eigenvalues.

Therefore, we omit the discussion of the nonexistence of complex eigenvalues here. Thus all

the eigenvalues and eigenfunctions of the problem (18), (19) are given by Eqs. (25) and (26).

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682 Chapter 11. Boundary Value Problems

u = f (√ )

µµ

= √

√

u = g(√ )

= –tanh√

–1

FIGURE 11.1.2 Graphical solution of

√

µ =−tanh

√

µ.

PROBLEMS In each of Problems 1 through 6, state whether the given boundary value problem is homo-

geneous or nonhomogeneous.

1. y

′′

+ 4y = 0, y(−1) = 0, y(1) = 0

2. [(1 + x

′

]

′

+ 4y = 0, y(0) = 0, y(1) = 1

3. y

′′

+ 4y = sin x, y(0) = 0, y(1) = 0

4. −y

′′

+ x

y = λy, y

′

(0) − y(0) = 0, y

′

(1) + y(1) = 0

5. −[(1 + x

′

]

′

= λy + 1, y(−1) = 0, y(1) = 0

6. −y

′′

= λ(1 + x

)y, y(0) = 0, y

′

(1) + 3y(1) = 0

In each of Problems 7 through 10, assume that all eigenvalues are real.

(a) Determine the form of the eigenfunctions and the determinantal equation satisﬁed by the

nonzero eigenvalues.

(b) Determine whether λ = 0 is an eigenvalue.

and λ

,the nonzero eigenvalues of smallest absolute value.

(d) Estimate λ

for large values of n.

7. y

′′

+ λy = 0, y(0) = 0, y(π) + y

′

(π) = 0

8. y

′′

+ λy = 0, y

′

(0) = 0, y(1) + y

′

(1) = 0

9. y

′′

+ λy = 0, y(0) − y

′

(0) = 0, y(1) + y

′

(1) = 0

10. y

′′

− λy = 0, y(0) + y

′

(0) = 0, y(1) = 0

11. Consider the general linear homogeneous second order equation

P(x)y

′′

+ Q(x)y

′

+ R(x)y = 0. (i)

We seek an integrating factor µ(x) such that, upon multiplying Eq. (i) by µ(x), we can

write the resulting equation in the form

[µ(x)P(x)y

′

]

′

+ µ(x)R(x)y = 0 . (ii)

(a) By equating coefﬁcients of y

′

in Eqs. (i) and (ii), show that µ must be a solution of

Pµ

′

= (Q − P

′

)µ. (iii)

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11.1 Two-Point Boundary Value Problems 683

(b) Solve Eq. (iii) and thereby show that

µ(x) =

P(x)

exp

Q(s)

P(s)

ds. (iv)

Compare this result with that of Problem 41 in Section 3.2.

In each of Problems 12 through 15, use the method of Problem 11 to transform the given

equation into the form [p(x)y

′

]

′

+ q(x)y = 0.

12. y

′′

− 2xy

′

+ λy = 0 (Hermite equation)

13. x

′′

+ xy

′

+ (x

− ν

)y = 0 (Bessel equation)

14. xy

′′

+ (1 − x)y

′

+ λy = 0 (Laguerre equation)

15. (1 − x

′′

− xy

′

+ α

y = 0 (Chebyshev equation)

16. The equation

+ cu

+ ku = a

+ F(x, t), (i)

where a

> 0, c ≥ 0, and k ≥ 0 are constants, is known as the telegraph equation. It arises

in the study of an elastic string under tension (seeAppendix B of Chapter 10). Equation (i)

also occurs in other applications.Assumingthat F (x, t) = 0,let u(x, t) = X(x)T(t),separate

the variables in Eq. (i), and derive ordinary differential equations for X and T.

17. Consider the boundary value problem

′′

− 2y

′

+ (1 + λ)y = 0, y(0) = 0, y(1) = 0.

(a) Introduce a new dependent variable u by the relation y = s(x)u. Determine s(x) so

that the differential equation for u has no u

′

term.

(b) Solve the boundary value problem for u and thereby determine the eigenvalues and

eigenfunctions of the original problem. Assume that all eigenvalues are real.

18. Consider the boundary value problem

′′

+ 4y

′

+ (4 + 9λ)y = 0, y(0) = 0, y

′

(L) = 0.

(a) Determine, at least approximately, the real eigenvalues and the corresponding

eigenfunctions by proceeding as in Problem 17(a, b).

(b) Also solve the given problem directly (without introducing a new variable).

Hint: In part (a) be sure to pay attention to the boundary conditions as well as the

differential equation.

The differential equations in Problems 19 and 20 differ from those in previous problems in

that the parameter λ multiplies the y

′

term as well as the y term. In each of these problems,

determine the real eigenvalues and the corresponding eigenfunctions.

19. y

′′

+ y

′

+ λ(y

′

+ y) = 0, y

′

(0) = 0, y(1) = 0

20. x

′′

− λ(xy

′

− y) = 0, y(1) = 0, y(2) − y

′

(2) = 0

21. Consider the problem

′′

+ λy = 0, 2y(0) + y

′

(0) = 0, y(1) = 0 .

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684 Chapter 11. Boundary Value Problems

(a) Find the determinantal equation satisﬁed by the positive eigenvalues.

(b) Show that there is an inﬁnite sequence of such eigenvalues.

and λ

. Then show that λ

∼

[(2n + 1)π/2]

for large n.

(d) Find the determinantal equation satisﬁed by the negative eigenvalues.

(e) Show that there is exactly one negative eigenvalue and ﬁnd its value.

22. Consider the problem

′′

+ λy = 0, αy(0) + y

′

(0) = 0, y(1) = 0,

where α is a given constant.

(a) Show that for all values of α there is an inﬁnite sequence of positive eigenvalues.

(b) If α<1,showthat all(real) eigenvalues are positive.Show thatthe smallest eigenvalue

approaches zero as α approaches 1 from below.

(d) If α>1, show that there is exactly one negative eigenvalue and that this eigenvalue

decreases as α increases.

23. Consider the problem

′′

+ λy = 0, y(0) = 0, y

′

(L) = 0.

Show that if φ

and φ

are eigenfunctions corresponding to the eigenvalues λ

and λ

respectively, with λ

̸= λ

, then

(x)φ

(x) dx = 0.

Hint: Note that

′′

+ λ

= 0, φ

′′

+ λ

= 0.

Multiply the ﬁrst of these equations by φ

, the second by φ

, and integrate from 0 to L,

using integration by parts. Finally, subtract one equation from the other.

24. In this problem we consider a higher order eigenvalue problem. The analysis of transverse

vibrations of a uniform elastic bar is based on the differential equation

(4)

− λy = 0,

where y is the transverse displacement and λ = mω

/EI; m is the mass per unit length of

the rod, E is Young’s modulus, I is the moment of inertia of the cross section about an

axis through the centroid perpendicular to the plane of vibration, and ω is the frequency

of vibration. Thus, for a bar whose material and geometrical properties are given, the

eigenvalues determine the natural frequencies of vibration. Boundary conditions at each

end are usually one of the following types:

y = y

′

= 0, clamped end,

y = y

′′

= 0, simply supported or hinged end,

′′

= y

′′′

= 0, free end.

For each of the following three cases, ﬁnd the form of the eigenfunctions and the equation

satisﬁed by the eigenvalues of this fourth order boundary value problem. Determine

and λ

, the two eigenvalues of smallest magnitude. Assume that the eigenvalues are

real and positive.

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11.2 Sturm–Liouville Boundary Value Problems 685

(a) y(0) = y

′′

(0) = 0, y(L) = y

′′

(L) = 0

(b) y(0) = y

′′

(0) = 0, y(L) = y

′

(L) = 0

′

(0) = 0, y

′′

(L) = y

′′′

(L) = 0 (cantilevered bar)

25. This problem illustrates that the eigenvalueparameter sometimes appears in the boundary

conditions as well as in the differential equation. Consider the longitudinal vibrations of a

uniform straight elastic bar of length L. It can be shown that the axial displacement u(x, t)

satisﬁes the partial differential equation

(E/ρ)u

= u

;0< x < L, t > 0, (i)

where E is Young’s modulus and ρ is the mass per unit volume. If the end x = 0 is ﬁxed,

then the boundary condition there is

u(0, t) = 0, t > 0. (ii)

Suppose that the end x = L is rigidly attached to a mass m but is otherwise unrestrained.

We can obtain the boundary condition here by writing Newton’s law for the mass. From

the theory of elasticity, it can be shown that the force exerted by the bar on the mass is

given by −EAu

(L, t). Hence the boundary condition is

EAu

(L, t) + mu

(L, t) = 0, t > 0. (iii)

(a) Assume that u(x, t) = X(x)T(t), and show that X(x) and T (t) satisfy the differential

equations

′′

+ λX = 0, (iv)

′′

+ λ(E/ρ)T = 0. (v)

(b) Show that the boundary conditions are

X(0) = 0, X

′

(L) − γλLX(L) = 0, (vi)

where γ = m/ρ AL is a dimensionless parameter that gives the ratio of the end mass to

the mass of the bar.

Hint:Use the differential equation for T(t) in simplifying the boundary condition at x = L.

eigenvalues of Eqs. (iv) and (vi).

(d) Find the ﬁrst two eigenvalues λ

and λ

if γ = 0.5.

11.2 Sturm–Liouville Boundary Value Problems

We now consider two-point boundary value problems of the type obtained in Section

11.1 by separating the variables in a heat conduction problem for a bar of variable

material properties and with a source term proportional to the temperature. This

kind of problem also occurs in many other applications.

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686 Chapter 11. Boundary Value Problems

These boundary value problems are commonly associated with the names of Sturm

and Liouville.

They consist of a differential equation of the form

[p(x)y

′

]

′

− q(x)y + λr(x)y = 0 (1)

on the interval 0 < x < 1, together with the boundary conditions

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0 (2)

at the endpoints. It is often convenient to introduce the linear homogeneous differ-

ential operator L deﬁned by

L[y]=−[p(x)y

′

]

′

+ q(x)y. (3)

Then the differential equation (1) can be written as

L[y]=λr(x)y. (4)

We assume that the functions p,p

′

,q,and r are continuous on the interval 0 ≤ x ≤ 1

and, further, that p(x )>0 and r(x)>0 at all points in 0 ≤ x ≤ 1. In this case the

boundary value problem is said to be regular. These assumptions are necessary to

render the theory as simple as possible while retaining considerable generality. It

turns out that these conditions are satisﬁed in many signiﬁcant problems in math-

ematical physics. For example, the equation y

′′

+ λy = 0, which arose repeatedly in

the preceding chapter, is of the form (1) with p(x) = 1, q(x) = 0, and r(x) = 1.

The boundary conditions (2) are said to be separated;that is,each involves only one

of the boundary points. These are the most general separated boundary conditions

that are possible for a second order differential equation. Note that for the ﬁrst

boundary condition to impose any restriction on y, it is necessary for at least one of

the numbers α

and α

to be nonzero. Similarly, at least one of the numbers β

and

must be nonzero.

Before proceeding to establishsome ofthe propertiesof theSturm–Liouville prob-

lem (1), (2), it is necessary to derive an identity, known as Lagrange’s identity, that is

basic to the study of linear boundary value problems. Let u and v be functions having

continuous second derivatives on the interval 0 ≤ x ≤ 1. Then

L[u]v dx =

[−(pu

′

)

′

v + quv]dx.

Charles-François Sturm (1803–1855) and Joseph Liouville (1809–1882), in a series of papers in 1836 and

1837, set forth many properties of the class of boundary value problems associated with their names,

including the results stated in Theorems 11.2.1 to 11.2.4. Sturm was born in Geneva, Switzerland, but

spent almost his entire adult life in Paris. He is also famous for a theorem on the number of real zeros

of a polynomial and, in addition, did extensive work in physics and mechanics. Liouville was a French

mathematicianwho did notable research in analysis,algebra,andnumbertheory. One of hismost important

results was the proof (in 1844) of the existence of transcendental numbers. He was also the founder, and

for 39 years the editor, of the inﬂuential Journal de mathématiques pures et appliquées.

For brevity, we sometimes use the notation

fdxrather than

f (x) dx in this chapter.

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11.2 Sturm–Liouville Boundary Value Problems 687

Integrating the ﬁrst term on the right side twice by parts, we obtain

L[u]v dx =−p(x)u

′

(x)v(x)

+ p(x)u(x)v

′

(x)

[−u(pv

′

)

′

+ uqv]dx

=−p(x)

[

′

(x)v(x) − u(x)v

′

(x)

]

uL[v]dx.

Hence, upon transposing the integral on the right side, we have

L[u]v − uL[v]

dx =−p(x)

′

(x)v(x) − u(x)v

′

(x)

, (5)

which is Lagrange’s identity.

Now let us suppose that the functions u and v in Eq. (5) also satisfy the boundary

conditions (2). Then, if we assume that α

̸= 0 and β

̸= 0, the right side of Eq. (5)

becomes

−p(x)

′

(x)v(x) − u(x)v

′

(x)

=−p(1)

′

(1)v(1) − u(1)v

′

(1)

+ p(0)

′

(0)v(0) − u(0 )v

′

(0)

=−p(1)

−

u(1)v(1) +

u(1)v(1)

+ p(0)

−

u(0)v(0) +

u(0)v(0)

= 0.

The same result holds if either α

or β

is zero; the proof in this case is even simpler

and is left for you. Thus, if the differential operator L is deﬁned by Eq. (3), and if

the functions u and v satisfy the boundary conditions (2), then Lagrange’s identity

reduces to

L[u]v − uL[v]

dx = 0. (6)

Let us now write Eq. (6) in a slightly different way. In Eq. (4) of Section 10.2 we

introduced the inner product (u, v) of two real-valued functions u and v on a given

interval; using the interval 0 ≤ x ≤ 1, we have

(u, v) =

u(x)v(x) dx. (7)

In this notation, Eq. (6) becomes

(L[u], v) − (u, L[v]) = 0. (8)

In proving Theorem 11.2.1 below, it is necessary to deal with complex-valued func-

tions. By analogy with the deﬁnition in Section 7.2 for vectors, we deﬁne the inner

product of two complex-valued functions on 0 ≤ x ≤ 1 as

(u, v) =

u(x)v(x) dx, (9)

where

v is the complex conjugate of v. Clearly, Eq. (9) coincides with Eq. (7) if u(x)

and v(x) are real. It is important to know that Eq. (8) remains valid under the stated

conditions if u and v are complex-valued functions and if the inner product is deﬁned

as in Eq. (9). To see this, you can start with the quantity

L[u]v dx and retrace the

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688 Chapter 11. Boundary Value Problems

steps leading to Eq. (6), making use of the fact that p(x), q(x), α

, α

, β

, and β

are

all real quantities (see Problem 22).

We now consider some of the implications of Eq. (8) for the Sturm–Liouville

boundary value problem (1), (2). We assume without proof

that this problem actu-

ally has eigenvalues and eigenfunctions. In Theorems 11.2.1 to 11.2.4 below, we

state several of their important, but relatively elementary, properties. Each of these

properties is illustrated by the basic Sturm–Liouville problem

′′

+ λy = 0, y(0) = 0, y(1) = 0, (10)

whose eigenvalues areλ

,withcorresponding eigenfunctions φ

(x) = sin nπx.

Theorem 11.2.1

All the eigenvalues of the Sturm–Liouville problem (1), (2) are real.

To prove this theorem, let us suppose that λ is a (possibly complex) eigenvalue

of the problem (1), (2) and that φ is a corresponding eigenfunction, also possibly

complex-valued. Let us write λ = µ + iν and φ(x) = U(x) + iV(x), where µ, ν, U(x),

and V(x) are real. Then, if we let u = φ and also v = φ in Eq. (8), we have

(L[φ], φ) = (φ, L[φ]). (11)

However, we know that L[φ]=λrφ, so Eq. (11) becomes

(λrφ, φ) = (φ, λrφ). (12)

Writing out Eq. (12) in full, using the deﬁnition (9) of the inner product, we obtain

λr(x)φ(x)φ(x) dx =

φ(x)λr(x)φ(x) dx. (13)

Since r(x) is real,

r(x) = r(x), and Eq. (13) reduces to

(λ −

λ)

r(x)φ(x)φ(x) dx = 0.

However, φ(x)

φ(x) = U

(x) + V

(x), so we have

(λ −

λ)

r(x)[U

(x) + V

(x)]dx = 0 . (14)

The integrand in Eq. (14) is nonnegative and not identically zero. Since the inte-

grand is also continuous, it follows that the integral is positive. Therefore, the factor

λ −

λ = 2iν must be zero. Hence ν = 0 and λ is real, so the theorem is proved.

An important consequence of Theorem 11.2.1 is that in ﬁnding eigenvalues and

eigenfunctions of a Sturm–Liouville boundary value problem, we need look only for

real eigenvalues. Recall that this is what we did in Chapter 10. It is also possible to

show that the eigenfunctions of the boundary value problem (1), (2) are real. A proof

is sketched in Problem 23.

For a proof of this statement, see the books by Sagan (Chapter 5) or by Birkhoff and Rota (Chapter 10)

listed in the references at the end of this chapter.

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11.2 Sturm–Liouville Boundary Value Problems 689

Theorem 11.2.2

If φ

and φ

are two eigenfunctions of the Sturm–Liouville problem (1), (2)

corresponding to eigenvalues λ

and λ

, respectively, and if λ

̸= λ

, then

r(x)φ

(x)φ

(x) dx = 0 . (15)

This theorem expresses the property of orthogonality of the eigenfunctions with

respect to the weight function r. To prove the theorem,we note that φ

and φ

satisfy

the differential equations

L[φ

]=λ

rφ

(16)

and

L[φ

]=λ

rφ

, (17)

respectively. If we let u = φ

, let v = φ

, and substitute for L[u] and L[v] in Eq. (8),

we obtain

(λ

rφ

, φ

) − (φ

, λ

rφ

) = 0,

or, using Eq. (9),

r(x)φ

(x)φ

(x) dx − λ

(x)r(x)φ

(x) dx = 0 .

Because λ

, r(x), and φ

(x) are real, this equation becomes

(λ

− λ

)

r(x)φ

(x)φ

(x) dx = 0 . (18)

Since by hypothesis λ

̸= λ

, it follows that φ

and φ

must satisfy Eq. (15), and the

theorem is proved.

Theorem 11.2.3

The eigenvalues of the Sturm–Liouville problem (1), (2) are all simple; that is, to

each eigenvalue there corresponds only one linearly independent eigenfunction.

Further, the eigenvalues form an inﬁnite sequence and can be ordered according

to increasing magnitude so that

<λ

< ···<λ

< ···.

Moreover, λ

→∞as n →∞.

The proof of this theorem is somewhat more advanced than those of the two

previous theorems and will be omitted. However, a proof that the eigenvalues are

simple is outlined in Problem 20.

Again we note that all the properties stated in Theorems 11.2.1 to 11.2.3 are

exempliﬁed by the eigenvalues λ

= n

and eigenfunctions φ

(x) = sin nπx of the

example problem (10). Clearly, the eigenvalues are real. The eigenfunctions satisfy

the orthogonality relation

(x)φ

(x) dx =

sin mπx sin nπxdx = 0, m ̸= n, (19)

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690 Chapter 11. Boundary Value Problems

which was established in Section 10.2 by direct integration. Further, the eigenval-

ues can be ordered so that λ

<λ

< ···, and λ

→∞as n →∞. Finally, to each

eigenvalue there corresponds a single linearly independent eigenfunction.

We will now assume that the eigenvalues of the Sturm–Liouville problem (1),

(2) are ordered as indicated in Theorem 11.2.3. Associated with the eigenvalue λ

is a corresponding eigenfunction φ

, determined up to a multiplicative constant.

It is often convenient to choose the arbitrary constant multiplying each eigenfunction

so as to satisfy the condition

r(x)φ

(x) dx = 1, n = 1, 2, .... (20)

Equation (20) is called a normalization condition, and eigenfunctions satisfying this

condition are said to be normalized. Indeed, in this case, the eigenfunctions are said

to form an orthonormal set (with respect to the weight function r) since they already

satisfy the orthogonality relation (15). It is sometimes useful to combine Eqs. (15)

and (20) into a single equation. To this end we introduce the symbol δ

, which is

known as the Kronecker

delta and deﬁned by

0, if m ̸= n,

1, if m = n.

(21)

Making use of the Kronecker delta, we can write Eqs. (15) and (20) as

r(x)φ

(x)φ

(x) dx = δ

. (22)

EXAMPLE

Determine the normalized eigenfunctions of the problem (10):

′′

+ λy = 0, y(0) = 0, y(1) = 0.

The eigenvalues of this problem are λ

= π

, λ

= 4π

, ..., λ

= n

, ..., and the corre-

sponding eigenfunctions are k

sin πx, k

sin 2πx, ..., k

sin nπx, ..., respectively. In this case

the weight function is r(x) = 1. To satisfy Eq. (20), we must choose k

so that

sin nπx)

dx = 1 (23)

for each value of n. Since

sin

nπxdx = k

−

cos 2nπx

dx =

Eq. (23) is satisﬁed if k

is chosen to be

√

2 for each value of n. Hence the normalized

eigenfunctions of the given boundary value problem are

(x) =

√

2 sin nπx, n = 1, 2, 3, .... (24)

The German mathematician Leopold Kronecker (1823–1891), a student of Dirichlet,was associated with

the University of Berlin for most of his life, although (since he was independently wealthy) he held a

faculty position only from 1883 onward. He worked in number theory, elliptic functions, algebra, and their

interconnections.

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11.2 Sturm–Liouville Boundary Value Problems 691

EXAMPLE

Determine the normalized eigenfunctions of the problem

′′

+ λy = 0, y(0) = 0, y

′

(1) + y(1) = 0. (25)

In Example 1 of Section 11.1, we found that the eigenvalues λ

satisfy the equation

sin

cos

= 0 (26)

and that the corresponding eigenfunctions are

(x) = k

sin

x, (27)

where k

is arbitrary. We can determine k

from the normalization condition (20). Since

r(x) = 1 in this problem, we have

(x) dx = k

sin

xdx

= k

)

−

cos 2

dx = k

−

sin 2

√

(

= k

√

− sin 2

√

= k

√

− sin

√

cos

√

= k

1 + cos

√

where in the last step we have used Eq. (26). Hence, to normalize the eigenfunctions φ

,we

must choose

1 + cos

√

(

1/2

. (28)

The normalized eigenfunctions of the given problem are

(x) =

√

2 sin

√

(1 + cos

√

)

1/2

; n = 1, 2, .... (29)

We now turn to the question of expressing a given function f as a series of eigen-

functions of the Sturm–Liouville problem (1), (2). We have already seen examples

of such expansions in Sections 10.2 to 10.4. For example, it was shown there that if

f is continuous and has a piecewise continuous derivative on 0 ≤ x ≤ 1, and satisﬁes

the boundary conditions f (0) = f (1) = 0, then f can be expanded in a Fourier sine

series of the form

f (x) =

∞

n=1

sin nπx. (30)

The functions sin nπx, n = 1, 2, ..., are precisely the eigenfunctions of the boundary

value problem (10). The coefﬁcients b

are given by

= 2

f (x) sin nπxdx (31)

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692 Chapter 11. Boundary Value Problems

and the series (30) converges for each x in 0 ≤ x ≤ 1. In a similar way, f can be

expanded in a Fourier cosine series using the eigenfunctions cos nπx, n = 0, 1, 2, ...,

of the boundary value problem y

′′

+ λy = 0, y

′

(0) = 0, y

′

(1) = 0.

Nowsuppose that a given functionf ,satisfying suitable conditions,can be expanded

in an inﬁnite series of eigenfunctions of the more general Sturm–Liouville problem

(1), (2). If this can be done, then we have

f (x) =

∞

n=1

(x), (32)

where the functions φ

(x) satisfy Eqs. (1), (2) and also the orthonormality condition

(22).To compute the coefﬁcients in the series (32),we multiply Eq. (32) by r(x )φ

(x),

where m is a ﬁxed positive integer, and integrate from x = 0tox = 1. Assuming that

the series can be integrated term by term, we obtain

r(x)f (x)φ

(x) dx =

∞

n=1

r(x)φ

(x)φ

(x) dx =

∞

n=1

. (33)

By the deﬁnition of δ

the last sum in Eq. (33) reduces to the single term c

. Hence,

using the deﬁnition of the inner product, we have

r(x)f (x)φ

(x) dx = (f , rφ

), m = 1, 2, .... (34)

The coefﬁcients in the series (32) have thus been formally determined. Equation (34)

has the same structure as the Euler–Fourier formulas for the coefﬁcients in a Fourier

series, and the eigenfunction series (32) also has convergence properties similar to

those of Fourier series. The following theorem is analogous to Theorem 10.3.1.

Theorem 11.2.4

Let φ

, φ

, ..., φ

, ... be the normalized eigenfunctions of the Sturm–Liouville

problem (1), (2):

[p(x)y

′

]

′

− q(x)y + λr(x)y = 0,

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0.

Let f and f

′

be piecewise continuous on 0 ≤ x ≤ 1. Then the series (32) whose

coefﬁcients c

are given by Eq. (34) converges to [f (x +) + f (x−)]/2 at each point

in the open interval 0 < x < 1.

If f satisﬁes further conditions, then a stronger conclusion can be established.

Suppose that, in addition to the hypotheses of Theorem 11.2.4, the function f is

continuous on 0 ≤ x ≤ 1. If α

= 0 in the ﬁrst of Eqs. (2) [so that φ

(0) = 0], then

assume that f (0) = 0. Similarly, if β

= 0 in the second of Eqs. (2), assume that

f (1) = 0. Otherwise, no boundary conditions need be prescribed for f . Then the

series (32) converges to f (x) at each point in the closed interval 0 ≤ x ≤ 1.

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11.2 Sturm–Liouville Boundary Value Problems 693

EXAMPLE

Expand the function

f (x) = x,0≤ x ≤ 1 (35)

in terms of the normalized eigenfunctions φ

(x) of the problem (25).

In Example 2 we found the normalized eigenfunctions to be

(x) = k

sin

x, (36)

where k

is given by Eq. (28) and λ

satisﬁes Eq. (26). To ﬁnd the expansion for f in terms of

the eigenfunctions φ

, we write

f (x) =

∞

n=1

(x), (37)

where the coefﬁcients are given by Eq. (34). Thus

f (x)φ

(x) dx = k

x sin

xdx.

Integrating by parts, we obtain

= k

sin

√

−

cos

√

(

= k

2 sin

√

where we have used Eq. (26) in the last step. Upon substituting for k

from Eq. (28),we obtain

√

2 sin

√

(1 + cos

√

)

1/2

. (38)

Thus

f (x) = 4

∞

n=1

sin

√

sin

√

(1 + cos

√

)

. (39)

Observe that although the right side of Eq. (39) is a series of sines, it is not included in the

discussion of Fourier sine series in Section 10.4.

Self-Adjoint Problems. Sturm–Liouville boundary value problems are of great impor-

tance in their own right, but they can also be viewed as belonging to a much more

extensive class of problems that have many of the same properties. For example,there

are many similarities between Sturm–Liouville problems and the algebraic system

Ax = λx, (40)

where the n × n matrix A is real symmetric or Hermitian. Comparing the results

mentioned in Section 7.3 with those of this section, we note that in both cases the

eigenvalues are real, and the eigenfunctions or eigenvectors form an orthogonal set.

Further, the eigenfunctions or eigenvectors can be used as the basis for expressing

an essentially arbitrary function or vector,respectively, as a sum. The most important

difference is that a matrix has only a ﬁnite number of eigenvalues and eigenvectors,

while a Sturm–Liouville problem has inﬁnitely many. It is interesting and of fun-

damental importance in mathematics that these seemingly different problems—the

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694 Chapter 11. Boundary Value Problems

matrix problem (40) and the Sturm–Liouville problem (1), (2)—which arise in dif-

ferent ways, are actually parts of a single underlying theory. This theory is usually

referred to as linear operator theory and is part of the subject of functional analysis.

We now point out some ways in which Sturm–Liouville problems can be gen-

eralized, while still preserving the main results of Theorems 11.2.1 to 11.2.4—the

existence of a sequence of real eigenvalues tending to inﬁnity, the orthogonality of

the eigenfunctions, and the possibility of expressing an arbitrary function as a series

of eigenfunctions. These generalizations depend on the continued validity of the

crucial relation (8).

Let us consider the boundary value problem consisting of the differential equation

L[y]=λr(x)y,0< x < 1, (41)

where

L[y]=P

(x)

+···+P

(x)

+ P

(x)y, (42)

and n linear homogeneous boundary conditions at the endpoints. If Eq. (8) is valid for

every pair of sufﬁciently differentiable functions that satisfy the boundary conditions,

then thegiven problemis saidto be self-adjoint. It isimportant toobserve thatEq. (8)

involves restrictions on both the differential equation and the boundary conditions.

The differentialoperator L must be such thatthe same operatorappears in bothterms

of Eq (8). This requires L to be of even order. Further, a second order operator must

have the form (3), a fourth order operator must have the form

L[y]=[p(x)y

′′

]

′′

−[q(x)y

′

]

′

+ s(x)y, (43)

and higher order operators must have an analogous structure. In addition,the bound-

ary conditions must be such as to eliminate the boundary terms that arise during the

integration by parts used in deriving Eq. (8). For example,in a second order problem,

this is true for the separated boundary conditions (2) and also in certain other cases,

one of which is given in Example 4 below.

Let us suppose that we have a self-adjoint boundary value problem for Eq. (41),

where L[y] is given now by Eq. (43). We assume that p, q, r, and s are continuous on

0 ≤ x ≤ 1 and that the derivatives of p and q indicated in Eq. (43) are also continuous.

If in addition p(x)>0 and r(x)>0 for 0 ≤ x ≤ 1, then there is an inﬁnite sequence

of real eigenvalues tending to +∞, the eigenfunctions are orthogonal with respect

to the weight function r, and an arbitrary function can be expressed as a series of

eigenfunctions. However, the eigenvalues may not be simple in these more general

problems.

We turn now to the relation between Sturm–Liouville problems and Fourier series.

We have noted previously that Fourier sine and cosine series can be obtained by using

the eigenfunctions of certain Sturm–Liouville problems involving the differential

equation y

′′

+ λy = 0.This raises the question of whether we can obtain a full Fourier

series, including both sine and cosine terms, by choosing a suitable set of boundary

conditions. The answer is provided by the following example, which also serves to

illustrate the occurrence of nonseparated boundary conditions.

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11.2 Sturm–Liouville Boundary Value Problems 695

EXAMPLE

Find the eigenvalues and eigenfunctions of the boundary value problem

′′

+ λy = 0, (44)

y(−L) − y(L) = 0, y

′

(−L) − y

′

(L) = 0. (45)

This is not a Sturm–Liouville problem because the boundary conditions are not separated.

The boundary conditions (45) are called periodic boundary conditions since they require that

y and y

′

assume the same values at x = L as at x =−L. Nevertheless, it is straightforward to

show that the problem (44), (45) is self-adjoint. A simple calculation establishes that λ

= 0is

an eigenvalue and that the corresponding eigenfunction is φ

(x) = 1. Further, there are addi-

tional eigenvalues λ

= (π/L)

, λ

= (2π/L)

, ..., λ

= (nπ/L)

, ....To each of these nonzero

eigenvalues there correspond two linearly independent eigenfunctions; for example, corre-

sponding to λ

are the two eigenfunctions φ

(x) = cos(nπx/L) and ψ

(x) = sin(nπx/L). This

illustrates that the eigenvalues may not be simple when the boundary conditions are not sepa-

rated. Further,if we seek to expand a given function f of period 2L in a series of eigenfunctions

of the problem (44), (45), we obtain the series

f (x) =

∞

n=1

)

cos

nπx

+ b

sin

nπx

which is just the Fourier series for f .

We will not give further consideration to problems that have nonseparated bound-

ary conditions, nor will we deal with problems of higher than second order, except in

a few problems. There is, however, one other kind of generalization that we do wish

to discuss. That is the case in which the coefﬁcients p, q, and r in Eq. (1) do not quite

satisfy the rather strict continuity and positivity requirements laid down at the begin-

ning of this section. Such problems are called singular Sturm–Liouville problems and

are the subject of Section 11.4.

PROBLEMS In each of Problems 1 through 5, determine the normalized eigenfunctions of the given

problem.

1. y

′′

+ λy = 0, y(0) = 0, y

′

(1) = 0

2. y

′′

+ λy = 0, y

′

(0) = 0, y(1) = 0

3. y

′′

+ λy = 0, y

′

(0) = 0, y

′

(1) = 0

4. y

′′

+ λy = 0, y

′

(0) = 0, y

′

(1) + y(1) = 0; see Section 11.1, Problem 8.

5. y

′′

− 2y

′

+ (1 + λ)y = 0, y(0) = 0, y(1) = 0; see Section 11.1, Problem 17.

In each of Problems 6 through 9, ﬁnd the coefﬁcients in the eigenfunction expansion

∞

n=1

(x) of the given function, using the normalized eigenfunctions of Problem 1.

6. f (x) = 1, 0 ≤ x ≤ 17.f (x) = x,0≤ x ≤ 1

8. f (x) =

1, 0 ≤ x <

≤ x ≤ 1

9. f (x) =

2x,0≤ x <

≤ x ≤ 1

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696 Chapter 11. Boundary Value Problems

In each of Problems 10 through 13, ﬁnd the coefﬁcients in the eigenfunction expansion

∞

n=1

(x) of the given function, using the normalized eigenfunctions of Problem 4.

10. f (x) = 1, 0 ≤ x ≤ 1 11. f (x) = x,0≤ x ≤ 1

12. f (x) = 1 − x,0≤ x ≤ 1 13. f (x) =

1, 0 ≤ x <

≤ x ≤ 1

In each of Problems 14 through 18, determine whether the given boundary value problem is

self-adjoint.

14. y

′′

+ y

′

+ 2y = 0, y(0) = 0, y(1) = 0

15. (1 + x

′′

+ 2xy

′

+ y = 0, y

′

(0) = 0, y(1) + 2y

′

(1) = 0

16. y

′′

+ y = λy, y(0) − y

′

(1) = 0, y

′

(0) − y(1) = 0

17. (1 + x

′′

+ 2xy

′

+ y = λ(1 + x

)y, y(0) − y

′

(1) = 0, y

′

(0) + 2y(1) = 0

18. y

′′

+ λy = 0, y(0) = 0, y(π) + y

′

(π) = 0

19. Show that if the functions u and v satisfy Eqs. (2), and either α

= 0 or β

= 0, or both,

then

p(x)

′

(x)v(x) − u(x)v

′

(x)

= 0.

20. In this problem we outline a proof of the ﬁrst part of Theorem 11.2.3: that the eigenvalues

of the Sturm–Liouville problem (1), (2) are simple. The proof is by contradiction.

(a) Suppose that a given eigenvalue λ is not simple. Then there exist two corresponding

eigenfunctions φ

and φ

that are linearly independent—that is, not multiples of each

other.

(b) Compute the Wronskian W(φ

, φ

)(x), and use the boundary conditions (2) to show

that W(φ

, φ

)(0) = 0.

must be simple, as asserted in Theorem 11.2.3.

21. Consider the Sturm–Liouville problem

−[p(x)y

′

]

′

+ q(x)y = λr(x)y,

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0,

where p, q, and r satisfy the conditions stated in the text.

(a) Show that if λ is an eigenvalue and φ a corresponding eigenfunction, then

rφ

dx =

(pφ

′2

+ qφ

) dx +

p(1)φ

(1) −

p(0)φ

(0),

provided that α

̸= 0 and β

̸= 0. How must this result be modiﬁed if α

= 0 or β

= 0?

(b) Show that if q(x) ≥ 0 and if β

/β

and −α

/α

are nonnegative, then the eigenvalue λ

is nonnegative.

= β

= 0 and q(x) = 0 for each x in 0 ≤ x ≤ 1.

22. Derive Eq. (8) using the inner product (9) and assuming that u and v are complex-valued

functions.

Hint: Consider the quantity

L[u]v dx, split u and v into real and imaginary parts, and

proceed as in the text.

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11.2 Sturm–Liouville Boundary Value Problems 697

23. In this problem we outline a proof that the eigenfunctions of the Sturm–Liouville problem

(1), (2) are real.

(a) Let λ be an eigenvalue and let φ(x) = U(x) + iV(x) be a corresponding eigenfunction.

Show that U and V are also eigenfunctions corresponding to λ.

(b) Using Theorem 11.2.3, or the result of Problem 20, show that U and V are linearly

dependent.

complex.

24. Consider the problem

′′

= λ(xy

′

− y), y(1) = 0, y(2) = 0.

Note that λ appears as a coefﬁcient of y

′

as well as of y itself. It is possible to extend

the deﬁnition of self-adjointness to this type of problem and to show that this particular

problem is not self-adjoint. Show that the problem has eigenvalues but that none of them

is real.This illustrates that in general, nonself-adjoint problems may have eigenvalues that

are not real.

Buckling of an Elastic Column. An investigation of the buckling of a uniform elastic column

of length L by an axial load P (Figure 11.2.1a) leads to the differential equation

(4)

+ λy

′′

= 0, 0 < x < L. (i)

x = 0

x = L

(a)

(b)

FIGURE 11.2.1 (a) A column under compression. (b) Shape of the buckled column.

The parameter λ is equal to P/EI, where E is Young’s modulus and I is the moment of

inertia of the cross section about an axis through the centroid perpendicular to the xy-plane.

The boundary conditions at x = 0 and x = L depend on how the ends of the column are

supported. Typical boundary conditions are

y = y

′

= 0, clamped end,

y = y

′′

= 0, simply supported (hinged) end.

The bar shown in Figure 11.2.1a is simply supported at x = 0 and clamped at x = L.Itis

desired to determine the eigenvalues and eigenfunctions of Eq. (i) subject to suitable boundary

conditions. In particular,the smallest eigenvalue λ

gives the load at which the column buckles,

or can assume a curved equilibrium position, as shown in Figure 11.2.1b. The corresponding

eigenfunction describes the conﬁguration of the buckled column. Note that the differential

equation (i) does not fall within the theory discussed in this section. It is possible to show,

however,that in each of the cases given here all the eigenvalues are real and positive. Problems

25 and 26 deal with column-buckling problems.

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698 Chapter 11. Boundary Value Problems

25. For each of the following boundary conditions, ﬁnd the smallest eigenvalue (the buckling

load) of y

(4)

+ λy

′′

= 0, and also ﬁnd the corresponding eigenfunction (the shape of the

buckled column).

(a) y(0) = y

′′

(0) = 0, y(L) = y

′′

(L) = 0

(b) y(0) = y

′′

(0) = 0, y(L) = y

′

(L) = 0

′

(0) = 0, y(L) = y

′

(L) = 0

26. In some buckling problems the eigenvalue parameter appears in the boundary conditions

as well as in the differential equation. One such case occurs when one end of the column

is clamped and the other end is free. In this case the differential equation y

(4)

+ λy

′′

= 0

must be solved subject to the boundary conditions

y(0) = 0, y

′

(0) = 0, y

′′

(L) = 0, y

′′′

(L) + λy

′

(L) = 0.

Find the smallest eigenvalue and the corresponding eigenfunction.

27.

Solutes in an aquifer are transported by two separate mechanisms. The process by which

a solute is transported by the bulk motion of the ﬂowing groundwater is called advection.

In addition, the solute is spread by small-scale ﬂuctuations in the groundwater velocity

along the tortuous ﬂow paths within individual pores, a process called mechanical disper-

sion. The one-dimensional form of the advection–dispersion equation for a nonreactive

dissolved solute in a saturated, homogeneous, isotropic porous medium under steady,

uniform ﬂow is

+ vc

= Dc

,0< x < L, t > 0, (i)

where c(x, t ) is the concentration of the solute,v is the average linear groundwater velocity,

D is the coefﬁcient of hydrodynamic dispersion,and L is the length of the aquifer. Suppose

that the boundary conditions are

c(0, t) = 0, c

(L, t) = 0, t > 0 (ii)

and that the initial condition is

c(x,0) = f (x),0< x < L, (iii)

where f (x) is the given initial concentration of the solute.

(a) Assume that c(x, t) = X(x)T(t), use the method of separation of variables, and ﬁnd

the equations satisﬁed by X(x) and T(t), respectively. Show that the problem for X(x) can

be written in the Sturm–Liouville form

[p(x)X

′

]

′

+ λr(x)X = 0, 0 < x < L, (iv)

X(0) = 0, X

′

(L) = 0, (v)

where p(x) = r(x ) = exp(−vx/D). Hence the eigenvalues are real, and the eigenfunctions

are orthogonal with respect to the weight function r(x).

(b) Let µ

= λ − (v

/4D

). Show that the eigenfunctions are

(x) = e

vx/2D

sin µ

x, (vi)

where µ

satisﬁes the equation

tan µL =−2Dµ/v. (vii)

∼

(2n − 1)π/2L for large n.

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11.3 Nonhomogeneous Boundary Value Problems 699

(d) Show that

r(x)X

(x) dx =

4Dµ

sin

(e) Find a formal solution of the problem (i), (ii), (iii) in terms of a series of the

eigenfunctions X

(x).

(f) Let v = 1, D = 0. 5, L = 10, and f (x) = δ(x − 3), where δ is the Dirac delta

function.

Using the solution found in part (e), plot c(x, t) versus x for several values of t, such as

t = 0.5, 1,3,6,and 10.Also plot c(x, t) versus t for several values of x. Note that the number

of terms that are needed to obtain an accurate plot depends strongly on the values of t

and x.

(g) Describe in a few words how the solution evolves as time advances.

28.

A nonreactive tracer at concentration c

is continuously introduced into a steady ﬂow at

the upstream end of a column of length L packed with a homogeneous granular medium.

Assuming that the tracer concentration in the column is initially zero, the boundary value

problem that models this process is

+ vc

= Dc

,0< x < L, t > 0,

c(0, t) = c

, t > 0,

(L, t) = 0, t > 0,

c(x,0) = 0, 0 < x < L,

where c(x , t), v, and D are as in Problem 27.

(a) Assuming that c(x, t) = c

+ u(x, t), ﬁnd the boundary value problem satisﬁed by

u(x, t).

(b) Proceeding as in Problem 27, ﬁnd u(x, t) in terms of an eigenfunction expansion.

= 1, and L = 10. Plot c(x, t) versus x for several values of t, and

also plot c(x, t) versus t for several values of x.

(d) Describe in a few words how the solution evolves with time. For example, about how

long does it take for the steady state solution to be essentially attained?

11.3 Nonhomogeneous Boundary Value Problems

In this section we discuss how to solve nonhomogeneous boundary value problems

for both ordinary and partial differential equations. Most of our attention is directed

toward problems in which the differential equation alone is nonhomogeneous, while

the boundary conditions are homogeneous. We assume that the solution can be

expanded in a series of eigenfunctions of a related homogeneous problem, and then

we determine the coefﬁcients in this series so that the nonhomogeneous problem

is satisﬁed. We ﬁrst describe this method as it applies to boundary value problems

for second order linear ordinary differential equations. Later we illustrate its use for

partial differential equations by solving a heat conduction problem in a bar with

variable material properties and in the presence of source terms.

See Section 6.5, especially Eq. (16) of that section.

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700 Chapter 11. Boundary Value Problems

Nonhomogeneous Sturm–Liouville Problems. Consider the boundary value problem con-

sisting of the nonhomogeneous differential equation

L[y]=−[p(x)y

′

]

′

+ q(x)y = µr(x)y + f (x), (1)

where µ is a given constant and f is a given function on 0 ≤ x ≤ 1, and the boundary

conditions

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0. (2)

As in Section 11.2, we assume that p, p

′

, q, and r are continuous on 0 ≤ x ≤ 1 and

that p(x)>0 and r(x)>0 there. We will solve the problem (1), (2) by making use

of the eigenfunctions of the corresponding homogeneous problem consisting of the

differential equation

L[y]=λr(x)y (3)

and the boundary conditions (2). Let λ

<λ

< ···<λ

< ··· be the eigenvalues

of this problem, and let φ

, φ

, ..., φ

, ... be the corresponding normalized eigen-

functions.

We now assume that the solution y = φ(x) of the nonhomogeneous problem (1),

(2) can be expressed as a series of the form

φ(x) =

∞

n=1

(x). (4)

From Eq. (34) of Section 11.2, we know that

r(x)φ(x)φ

(x) dx, n = 1, 2, .... (5)

However, since we do not know φ(x), we cannot use Eq. (5) to calculate b

. Instead,

we will try to determine b

so that the problem (1), (2) is satisﬁed and then use

Eq. (4) to ﬁnd φ(x). Note ﬁrst that φ as given by Eq. (4) always satisﬁes the boundary

conditions (2) since each φ

does.

Now consider the differential equation that φ must satisfy. This is just Eq. (1) with

y replaced by φ:

L[φ](x ) = µr(x)φ(x) + f (x ). (6)

We substitutethe series(4) into thedifferential equation(6) andattempt to determine

so that the differential equation is satisﬁed. The term on the left side of Eq. (6)

becomes

L[φ](x ) = L

∞

n=1

(x) =

∞

n=1

L[φ

](x) =

∞

n=1

r(x)φ

(x), (7)

where we have assumed that we can interchange the operations of summation and

differentiation.

Note that the function r appears in Eq. (7) and also in the term µr(x)φ(x) in Eq. (6).

This suggests that we rewrite the nonhomogeneous term in Eq. (6) as r(x)[f (x)/r(x)]

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11.3 Nonhomogeneous Boundary Value Problems 701

so that r(x) also appears as a multiplier in this term. If the function f /r satisﬁes the

conditions of Theorem 11.2.4, then

f (x)

r(x)

∞

n=1

(x), (8)

where, using Eq. (5) with φ replaced by f /r,

r(x)

f (x)

r(x)

(x) dx =

f (x)φ

(x) dx, n = 1, 2, .... (9)

Upon substituting for φ(x), L[φ](x), and f (x) in Eq. (6) from Eqs. (4), (7), and (8),

respectively, we ﬁnd that

∞

n=1

r(x)φ

(x) = µr(x)

∞

n=1

(x) + r(x)

∞

n=1

(x).

After collecting terms and canceling the common nonzero factor r(x), we have

∞

n=1

(λ

− µ)b

− c

(x) = 0. (10)

If Eq. (10) is to hold for each x in the interval 0 ≤ x ≤ 1, then the coefﬁcient of φ

(x)

must be zero for each n; see Problem 14 for a proof of this fact. Hence

(λ

− µ)b

− c

= 0, n = 1, 2, .... (11)

We must now distinguish two main cases, one of which also has two subcases.

First suppose that µ ̸= λ

for n = 1, 2, 3, ...; that is, µ is not equal to any eigen-

value of the corresponding homogeneous problem. Then

− µ

, n = 1, 2, 3, ..., (12)

and

y = φ(x) =

∞

n=1

− µ

(x). (13)

Equation (13), with c

given by Eq. (9), is a formal solution of the nonhomogeneous

boundary value problem (1), (2). Our argument does not prove that the series (13)

converges. However, any solution of the boundary value problem (1), (2) clearly

satisﬁes the conditions of Theorem 11.2.4; indeed, it satisﬁes the more stringent con-

ditions given in the paragraph following that theorem.Thus it is reasonable to expect

that the series (13) does converge at each point, and this fact can be established,

provided, for example, that f is continuous.

Now suppose that µ is equal to one of the eigenvalues of the corresponding homo-

geneous problem, say, µ = λ

; then the situation is quite different. In this event, for

n = m, Eq. (11) has the form 0 · b

− c

= 0. Again we must consider two cases.

If µ = λ

and c

̸= 0, then there is no value of b

that satisﬁes Eq. (11), and

therefore the nonhomogeneous problem (1), (2) has no solution.

If µ = λ

and c

= 0, then Eq. (11) is satisﬁed regardless of the value of b

;in

other words, b

remains arbitrary. In this case the boundary value problem (1), (2)

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702 Chapter 11. Boundary Value Problems

does have a solution, but it is not unique, since it contains an arbitrary multiple of

the eigenfunction φ

Since c

is given by Eq. (9), the condition c

= 0 means that

f (x)φ

(x) dx = 0 . (14)

Thus,if µ = λ

, the nonhomogeneous boundary value problem (1),(2) can be solved

only if f is orthogonal to the eigenfunction corresponding to the eigenvalue λ

The results we have formally obtained are summarized in the following theorem.

Theorem 11.3.1

The nonhomogeneous boundary value problem (1), (2) has a unique solution for

each continuous f whenever µ is different from all the eigenvalues of the corre-

sponding homogeneous problem; the solution is given by Eq. (13), and the series

convergesfor each x in0 ≤ x ≤ 1. Ifµ is equalto an eigenvalueλ

ofthe correspond-

ing homogeneousproblem,then thenonhomogeneous boundary value problem has

no solution unless condition (14) holds; that is, unless f is orthogonal to φ

. In that

case, the solution is not unique and contains an arbitrary multiple of φ

(x).

The main part of Theorem 11.3.1 is sometimes stated in the following way:

Theorem 11.3.2

For a given value of µ, either the nonhomogeneous problem (1), (2) has a unique

solution for each continuous f (if µ is not equal to any eigenvalue λ

of the corre-

sponding homogeneous problem), or else the homogeneous problem (3), (2) has a

nontrivial solution (the eigenfunction corresponding to λ

This latterform ofthe theoremis known asthe Fredholm

alternative theorem.This

is one of the basic theorems of mathematical analysis and occurs in many different

contexts. You may be familiar with it in connection with sets of linear algebraic

equations where the vanishing or nonvanishing of the determinant of coefﬁcients

replaces the statements about µ and λ

. See the discussion in Section 7.3.

EXAMPLE

Solve the boundary value problem

′′

+ 2y =−x, (15)

y(0) = 0, y(1) + y

′

(1) = 0. (16)

The Swedish mathematician Erik Ivar Fredholm (1866–1927), professor at the University of Stockholm,

established the modern theory of integral equations in a fundamental paper in 1903. Fredholm’s work

emphasized the similarities between integral equations and systems of linear algebraic equations. There

are also many interrelations between differential and integral equations; for example, see Section 2.8 and

Problem 22 of Section 6.6.

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11.3 Nonhomogeneous Boundary Value Problems 703

This particular problem can be solved directly in an elementary way and has the solution

y =

sin

√

2 x

sin

√

2 +

√

2 cos

√

−

. (17)

The method of solution described below illustrates the use of eigenfunction expansions, a

method that can be employed in many problems not accessible by elementary procedures.

We begin by rewriting Eq. (15) as

−y

′′

= 2y + x (18)

so that it will have the same form as Eq. (1). We seek the solution of the given problem as a

series of normalized eigenfunctions φ

of the corresponding homogeneous problem

′′

+ λy = 0, y(0) = 0, y(1) + y

′

(1) = 0. (19)

These eigenfunctions were found in Example 2 of Section 11.2 and are

(x) = k

sin

x, (20)

where

1 + cos

√

(

1/2

(21)

and λ

satisﬁes

sin

cos

= 0. (22)

Recall that in Example 1 of Section 11.1, we found that

∼

4.116, λ

∼

24.14,

∼

63.66, λ

∼

(2n − 1)

/4 for n = 4, 5, ... .

We assume that y is given by Eq. (4)

y =

∞

n=1

(x),

and it follows that the coefﬁcients b

are found from Eq. (12)

− 2

where the c

are the expansion coefﬁcients of the nonhomogeneous term f (x) = x in Eq. (18)

in terms of the eigenfunctions φ

. These coefﬁcients were found in Example 3 of Section 11.2

and are

√

2 sin

√

(1 + cos

√

)

1/2

. (23)

Putting everything together, we ﬁnally obtain the solution

y = 4

∞

n=1

sin

√

(λ

− 2)(1 + cos

√

)

sin

x. (24)

Although Eqs. (17) and (24) are quite different in appearance,they areactually twodifferent

expressions for the same function. This follows from the uniqueness part of Theorem 11.3.1

or 11.3.2, since λ = 2 is not an eigenvalue of the homogeneous problem (19). Alternatively,

you can show the equivalence of Eqs. (17) and (24) by expanding the right side of Eq. (17)

August 7, 2012 21:05 c11 Sheet number 28 Page number 704 cyan black

704 Chapter 11. Boundary Value Problems

in terms of the eigenfunctions φ

(x). For this problem it is fairly obvious that Eq. (17) is a

more convenient expression for the solution than Eq. (24). However,we emphasize again that

in other problems we may not be able to obtain the solution except by series (or numerical

approximation) methods.

Nonhomogeneous Heat Conduction Problems. To show how eigenfunction expansions can

be used to solve nonhomogeneous problems for partial differential equations, let us

consider the generalized heat conduction equation

r(x)u

=[p(x)u

]

− q(x)u + F(x, t) (25)

with the boundary conditions

(0, t) − h

u(0, t) = 0, u

(1, t) + h

u(1, t) = 0 (26)

and the initial condition

u(x,0) = f (x). (27)

This problem was previously discussed in Appendix A of Chapter 10 and in Section

11.1. In the latter section we let u(x, t) = X(x)T(t) in the homogeneous equation

obtained by setting F(x, t) = 0 and showed that X(x) must be a solution of the

boundary value problem

−[p(x)X

′

]

′

+ q(x)X = λr(x)X, (28)

′

(0) − h

X(0) = 0, X

′

(1) + h

X(1) = 0. (29)

If we assume that p, q, and r satisfy the proper continuity requirements and that

p(x) and r(x) are always positive, the problem (28), (29) is a Sturm–Liouville

problem as discussed in Section 11.2. Thus we obtain a sequence of eigenvalues

<λ

< ···<λ

< ··· and the corresponding normalized eigenfunctions

(x), φ

(x), ..., φ

(x), ....

We will solve the given nonhomogeneous boundary value problem (25) to (27) by

assuming that u(x, t) can be expressed as a series of eigenfunctions

u(x, t) =

∞

n=1

(t)φ

(x) (30)

and then showing how to determine the coefﬁcients b

(t). The procedure is basi-

cally the same as that used in the nonhomogeneous Sturm–Liouville problem (1),

(2) that we considered earlier, although it is more complicated in certain respects.

For instance, the coefﬁcients b

must now depend on t, because otherwise u would

be a function of x only. Note that the boundary conditions (26) are automatically

satisﬁed by an expression of the form (30) because each φ

(x) satisﬁes the boundary

conditions (29).

Next we substitute from Eq. (30) for u in Eq. (25). From the ﬁrst two terms on the

right side of Eq. (25), we formally obtain

[p(x)u

]

− q(x)u =

∂

∂x

p(x)

∞

n=1

(t)φ

′

(x)

− q(x)

∞

n=1

(t)φ

(x)

∞

n=1

(t)

p(x)φ

′

(x)

′

− q(x)φ

(x)

. (31)

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11.3 Nonhomogeneous Boundary Value Problems 705

Since

p(x)φ

′

(x)

′

− q(x)φ

(x) =−λ

r(x)φ

(x), we obtain ﬁnally

p(x)u

− q(x)u =−r(x)

∞

n=1

(t)λ

(x). (32)

Now consider the term on the left side of Eq. (25). We have

r(x)u

= r(x)

∂

∂t

∞

n=1

(t)φ

(x) = r(x)

∞

n=1

′

(t)φ

(x). (33)

We must also express the nonhomogeneous term in Eq. (25) as a series of eigen-

functions. Once again, it is convenient to look at the ratio F(x, t)/r(x) and to write

F(x, t)

r(x)

∞

n=1

(t)φ

(x), (34)

where the coefﬁcients are given by

(t) =

r(x)

F(x, t)

r(x)

(x) dx =

F(x, t)φ

(x) dx, n = 1, 2, .... (35)

Since F(x, t) is given, we can consider the functions γ

(t) to be known.

Gathering all these results together, we substitute from Eqs. (32), (33), and (34) in

Eq. (25) and ﬁnd that

r(x)

∞

n=1

′

(t)φ

(x) =−r(x)

∞

n=1

(t)λ

(x) + r(x)

∞

n=1

(t)φ

(x). (36)

To simplify Eq. (36), we cancel the common nonzero factor r(x) from all terms and

write everything in one summation:

∞

n=1

′

(t) + λ

(t) − γ

(t)

(x) = 0. (37)

Once again, if Eq. (37) is to hold for all x in 0 < x < 1, it is necessary for the quantity

in square brackets to be zero for each n (again see Problem 14). Hence b

(t) is a

solution of the ﬁrst order linear ordinary differential equation

′

(t) + λ

(t) = γ

(t), n = 1, 2, ..., (38)

where γ

(t) is given by Eq. (35). To determine b

(t) completely, we must have an

initial condition

(0) = B

, n = 1, 2, ... (39)

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706 Chapter 11. Boundary Value Problems

for Eq. (38). This we obtain from the initial condition (27). Setting t = 0 in Eq. (30)

and using Eq. (27), we have

u(x,0) =

∞

n=1

(0)φ

(x) =

∞

n=1

(x) = f (x). (40)

Thus the initial values B

are the coefﬁcients in the eigenfunction expansion for f (x).

Therefore,

r(x)f (x)φ

(x) dx, n = 1, 2, .... (41)

Note that everything on the right side of Eq. (41) is known, so we can consider B

known.

The initial value problem (38), (39) is solved by the methods of Section 2.1. The

integrating factor is µ(t) = exp(λ

t), and it follows that

(t) = B

−λ

(t−s)

(s) ds, n = 1, 2, .... (42)

The details of this calculation are left to you. Note that the ﬁrst term on the right side

of Eq. (42) depends on the function f through the coefﬁcients B

, while the second

depends on the nonhomogeneous term F through the coefﬁcients γ

(s).

Thus an explicit solution of the boundary value problem (25) to (27) is given by

Eq. (30)

u(x, t) =

∞

n=1

(t)φ

(x),

where the coefﬁcients b

(t) are determined from Eq. (42). The quantities B

and

(s) in Eq. (42) are found in turn from Eqs. (41) and (35), respectively.

Summarizing, to use this method to solve a boundary value problem such as that

given by Eqs. (25) to (27), we must

1. Find the eigenvalues λ

and the normalized eigenfunctions φ

of the homogeneous

problem (28), (29).

2. Calculate the coefﬁcients B

and γ

(t) from Eqs. (41) and (35), respectively.

3. Evaluate the integral in Eq. (42) to determine b

(t).

4. Sum the inﬁnite series (30).

Since any or all of these steps may be difﬁcult, the entire process can be quite

formidable. One redeeming feature is that often the series (30) converges rapidly, in

which case onlya very fewterms may be needed toobtain an adequate approximation

to the solution.

EXAMPLE

Find the solution of the heat conduction problem

= u

+ xe

−t

, (43)

u(0, t) = 0, u

(1, t) + u(1, t) = 0, (44)

u(x,0) = 0. (45)

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11.3 Nonhomogeneous Boundary Value Problems 707

Again we use the normalized eigenfunctions φ

of the problem (19) and assume that u is

given by Eq. (30)

u(x, t) =

∞

n=1

(t)φ

(x).

The coefﬁcients b

are determined from the differential equation

′

+ λ

= γ

(t), (46)

where λ

is the nth eigenvalue of problem (19) and γ

(t) is the nth expansion coefﬁcient of the

nonhomogeneous term xe

−t

in terms of the eigenfunctions φ

. Thus we have

(t) =

−t

(x) dx = e

−t

xφ

(x) dx = c

−t

, (47)

where c

xφ

(x) dx is given by Eq. (23). The initial condition for Eq. (46) is

(0) = 0 (48)

since the initial temperature distribution (45) is zero everywhere. The solution of the initial

value problem (46), (48) is

(t) = e

−λ

−s

ds = c

−λ

(λ

−1)t

− 1

−t

− e

−λ

). (49)

Thus the solution of the heat conduction problem (43) to (45) is given by

u(x, t) = 4

∞

n=1

(sin

√

)(e

−t

− e

−λ

) sin

√

(λ

− 1)(1 + cos

√

)

. (50)

The solution given by Eq. (50) is exact but complicated. To judge whether a satisfactory

approximation to the solution can be obtained by using only a few terms in this series,we must

estimate its speed of convergence. First we split the right side of Eq. (50) into two parts:

u(x, t) = 4e

−t

∞

n=1

sin

√

sin

√

(λ

− 1)(1 + cos

√

)

− 4

∞

n=1

−λ

sin

√

sin

√

(λ

− 1)(1 + cos

√

)

. (51)

Recall from Example 1 in Section 11.1 that the eigenvalues λ

are very nearly proportional to

. In the ﬁrst series on the right side of Eq. (51), the trigonometric factors are all bounded

as n →∞; thus this series converges similarly to the series

∞

n=1

−2

∞

n=1

−4

. Hence at most

two or three terms are required for us to obtain an excellent approximation to this part of

the solution. The second series contains the additional factor e

−λ

, so its convergence is even

more rapid for t > 0; all terms after the ﬁrst are almost surely negligible.

Further Discussion. Eigenfunction expansions can be used to solve a much greater

variety of problems than the preceding discussion and examples may suggest. For

example, time-independent nonhomogeneous boundary conditions can be handled

much as in Section 10.6. To reduce the problem to one with homogeneous boundary

conditions, subtract from u a function v that is chosen to satisfy the given bound-

ary conditions. Then the difference w = u − v satisﬁes a problem with homogeneous

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708 Chapter 11. Boundary Value Problems

boundary conditions, but with a modiﬁed forcing term and initial condition. This

problem can be solved by the procedure described in this section.

One potential difﬁculty in using eigenfunction expansions is that the normalized

eigenfunctions of the corresponding homogeneous problem must be found. For a

differential equation with variable coefﬁcients this may be difﬁcult,if not impossible.

In such a case it is sometimes possible to use other functions, such as eigenfunctions

of a simpler problem, that satisfy the same boundary conditions. For instance, if the

boundary conditions are

u(0, t) = 0, u(1, t) = 0, (52)

then it may be convenient to replace the functions φ

(x) in Eq. (30) by sin nπx. These

functions at least satisfy the correct boundary conditions, although in general they

are not solutions of the corresponding homogeneous differential equation. Next we

expand the nonhomogeneous term F(x, t) in a series of the form (34), again with

(x) replaced by sin nπx, and then substitute for both u and F in Eq. (25). Upon

collecting the coefﬁcients of sin nπx for each n, we have an inﬁnite set of linear ﬁrst

order differential equations from which to determine b

(t), b

(t), ....The essential

difference between this case and the one considered earlier is that now the equations

for the functions b

(t) are coupled. Thus they cannot be solved one by one, as before,

but must be dealt with simultaneously. In practice, the inﬁnite system is replaced by

an approximating ﬁnite system, from which approximations to a ﬁnite number of

coefﬁcients are calculated.

Boundary value problems for equations of higher than second order can also often

be solved by eigenfunction expansions. In some cases the procedure parallels almost

exactly that for second order problems. However,a variety of complications can also

arise.

Finally, we emphasize that the discussion in this section has been purely formal.

Separate and sometimes elaborate arguments must be used to establish convergence

of eigenfunction expansions or to justify some of the steps used,such as term-by-term

differentiation of eigenfunction series.

There are also other, altogether different methods for solving nonhomogeneous

boundary value problems. One of these leads to a solution expressed as a deﬁnite

integral rather than as an inﬁnite series. This approach involves certain functions

known as Green’s functions and, for ordinary differential equations, is the subject of

Problems 28 through 36.

PROBLEMS In each of Problems 1 through 5, solve the given problem by means of an eigenfunction

expansion.

1. y

′′

+ 2y =−x, y(0) = 0, y(1) = 0

2. y

′′

+ 2y =−x, y(0) = 0, y

′

(1) = 0; see Section 11.2, Problem 7.

3. y

′′

+ 2y =−x, y

′

(0) = 0, y

′

(1) = 0; see Section 11.2, Problem 3.

4. y

′′

+ 2y =−x, y

′

(0) = 0, y

′

(1) + y(1) = 0; see Section 11.2, Problem 11.

5. y

′′

+ 2y =−1 +|1 − 2x|, y(0) = 0, y(1) = 0

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11.3 Nonhomogeneous Boundary Value Problems 709

In each of Problems 6 through 9, determine a formal eigenfunction series expansion for the

solution of the given problem. Assume that f satisﬁes the conditions of Theorem 11.3.1. State

the values of µ for which the solution exists.

6. y

′′

+ µy =−f (x), y(0) = 0, y

′

(1) = 0

7. y

′′

+ µy =−f (x), y

′

(0) = 0, y(1) = 0

8. y

′′

+ µy =−f (x), y

′

(0) = 0, y

′

(1) = 0

9. y

′′

+ µy =−f (x), y

′

(0) = 0, y

′

(1) + y(1) = 0

In each of Problems 10 through 13, determine whether there is any value of the constant a for

which the problem has a solution. Find the solution for each such value.

10. y

′′

+ π

y = a + x, y(0) = 0, y(1) = 0

11. y

′′

+ 4π

y = a + x, y(0) = 0, y(1) = 0

12. y

′′

+ π

y = a, y

′

(0) = 0, y

′

(1) = 0

13. y

′′

+ π

y = a − cos πx, y(0) = 0, y(1) = 0

14. Let φ

, φ

, ..., φ

, ...be the normalized eigenfunctions of the differential equation (3) sub-

ject to the boundary conditions (2). If

∞

n=1

(x) converges to f (x), where f (x) = 0 for

each x in 0 ≤ x ≤ 1, show that c

= 0 for each n.

Hint: Multiply by r(x)φ

(x), integrate, and use the orthogonality property of the

eigenfunctions.

15. Let L be a second order linear differential operator. Show that the solution y = φ(x) of

the problem

L[y]=f (x),

y(0) + α

′

(0) = a, β

y(1) + β

′

(1) = b

can be written as y = u + v, where u = φ

(x) and v = φ

(x) are solutions of the problems

L[u]=0,

u(0) + α

′

(0) = a, β

u(1) + β

′

(1) = b

and

L[v]=f (x),

v(0) + α

′

(0) = 0, β

v(1) + β

′

(1) = 0,

respectively.

16. Show that the problem

′′

+ π

y = π

x, y(0) = 1, y(1) = 0

has the solution

y = c

sin πx + cos πx + x.

Also show that this solution cannot be obtained by splitting the problem as suggested in

Problem 15, since neither of the two subsidiary problems can be solved in this case.

17. Consider the problem

′′

+ p(x)y

′

+ q(x)y = 0, y(0) = a, y(1) = b.

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710 Chapter 11. Boundary Value Problems

Let y = u + v, where v is any twice-differentiable function satisfying the boundary con-

ditions (but not necessarily the differential equation). Show that u is a solution of the

problem

′′

+ p(x)u

′

+ q(x)u = g(x), u(0) = 0, u(1) = 0,

where g(x ) =−[v

′′

+ p(x)v

′

+ q(x)v] and is known once v is chosen. Thus nonhomo-

geneities can be transferred from the boundary conditions to the differential equation.

Find a function v for this problem.

18. Using the method of Problem 17, transform the problem

′′

+ 2y = 2 − 4x, y(0) = 1, y(1) + y

′

(1) =−2

into a new problem in which the boundary conditions are homogeneous. Solve the latter

problem by reference to Example 1 of the text.

In each of Problems 19 through 22, use eigenfunction expansions to ﬁnd the solution of the

given boundary value problem.

19. u

= u

− x, u(0, t) = 0, u

(1, t) = 0, u(x,0) = sin(πx/2);

see Problem 2.

20. u

= u

+ e

−t

, u

(0, t) = 0, u

(1, t) + u(1, t) = 0, u(x,0) = 1 − x;

see Section 11.2, Problems 10 and 12.

21. u

= u

+ 1 −|1 − 2x|, u(0, t) = 0, u(1, t) = 0, u(x,0) = 0;

see Problem 5.

22. u

= u

+ e

−t

(1 − x), u(0, t) = 0, u

(1, t) = 0, u(x,0) = 0;

see Section 11.2, Problems 6 and 7.

23. Consider the boundary value problem

r(x)u

=[p(x)u

]

− q(x)u + F (x),

u(0, t) = T

, u(1, t) = T

, u(x,0) = f (x).

(a) Let v(x) be a solution of the problem

[p(x)v

′

]

′

− q(x)v =−F(x), v(0) = T

, v(1) = T

If w(x, t) = u(x, t) − v(x), ﬁnd the boundary value problem satisﬁed by w. Note that this

problem can be solved by the method of this section.

(b) Generalize the procedure of part (a) to the case where u satisﬁes the boundary

conditions

(0, t) − h

u(0, t) = T

, u

(1, t) + h

u(1, t) = T

In each of Problems 24 and 25, use the method indicated in Problem 23 to solve the given

boundary value problem.

24. u

= u

− 2,

u(0, t) = 1, u(1, t) = 0,

u(x,0) = x

− 2x + 2

25. u

= u

− π

cos πx,

(0, t) = 0, u(1, t) = 1,

u(x,0) = cos(3πx/2) − cos πx

26. The method of eigenfunction expansions is often useful for nonhomogeneous problems

related to the wave equation or its generalizations. Consider the problem

r(x)u

=[p(x)u

]

− q(x)u + F (x, t), (i)

(0, t) − h

u(0, t) = 0, u

(1, t) + h

u(1, t) = 0, (ii)

u(x,0) = f (x), u

(x,0) = g(x). (iii)

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11.3 Nonhomogeneous Boundary Value Problems 711

This problem can arise in connection with generalizations of the telegraph equation (Prob-

lem 16 in Section 11.1) or the longitudinal vibrations of an elastic bar (Problem 25 in

Section 11.1).

(a) Let u(x, t) = X(x)T (t) in the homogeneous equation corresponding to Eq. (i), and

show that X(x) satisﬁes Eqs. (28) and (29) of the text. Let λ

and φ

(x) denote the

eigenvalues and normalized eigenfunctions of this problem.

(b) Assume that u(x, t) =

∞

n=1

(t)φ

(x), and show that b

(t) must satisfy the initial value

problem

′′

(t) + λ

(t) = γ

(t), b

(0) = α

, b

′

(0) = β

where α

, β

, and γ

(t) are the expansion coefﬁcients for f (x), g(x), and F (x, t)/r(x) in

terms of the eigenfunctions φ

(x), ..., φ

(x), ....

27. In this problem we explore a little further the analogy between Sturm–Liouville bound-

ary value problems and Hermitian matrices. Let A be an n × n Hermitian matrix with

eigenvalues λ

, ..., λ

and corresponding orthogonal eigenvectors ξ

(1)

, ..., ξ

(n)

Consider the nonhomogeneous system of equations

Ax − µx = b, (i)

where µ is a given real number and b is a given vector. We will point out a way of solving

Eq. (i) that is analogous to the method presented in the text for solving Eqs. (1) and (2).

(a) Show that b =

i=1

(i)

, where b

= (b, ξ

(i)

(b) Assume that x =

i=1

(i)

and show that for Eq. (i) to be satisﬁed, it is necessary that

= b

/(λ

− µ). Thus

x =

i=1

(b, ξ

(i)

)

− µ

(i)

, (ii)

provided that µ is not one of the eigenvalues of A, µ ̸= λ

for i = 1, ..., n. Compare this

result with Eq. (13).

Green’s

Functions. Consider the nonhomogeneous system of algebraic equations

Ax − µx = b, (i)

where A is an n × n Hermitian matrix,µ is a given real number,and b is a given vector. Instead

of using an eigenvector expansion as in Problem 27, we can solve Eq. (i) by computing the

inverse matrix (A −µI)

−1

, which exists if µ is not an eigenvalue of A. Then

x = (A − µI)

−1

b. (ii)

Green’s functions are named after George Green (1793–1841) of England. He was almost entirely self-

taught in mathematics and made signiﬁcant contributions to electricity and magnetism, ﬂuid mechanics,

and partial differential equations. His most important work was an essay on electricity and magnetism

that was published privately in 1828. In this paper Green was the ﬁrst to recognize the importance of

potential functions. He introduced the functions now known as Green’s functions as a means of solving

boundary value problems and developed the integral transformation theorems,of which Green’s theorem

in the plane is a particular case. However, these results did not become widely known until Green’s essay

was republished in the 1850s through the efforts of William Thomson (Lord Kelvin).

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712 Chapter 11. Boundary Value Problems

Problems 28 through 36 indicate a way of solving nonhomogeneous boundary value problems

that is analogous to using the inverse matrix for a system of linear algebraic equations. The

Green’s function plays a part similar to the inverse of the matrix of coefﬁcients. This method

leadsto solutions expressed as deﬁnite integrals ratherthan as inﬁniteseries.Except in Problem

35, we will assume that µ = 0 for simplicity.

28. (a) Show by the method of variation of parameters that the general solution of the

differential equation

−y

′′

= f (x)

can be written in the form

y = φ(x) = c

+ c

x −

(x − s)f (s) ds,

where c

and c

are arbitrary constants.

(b) Let y = φ(x) also be required to satisfy the boundary conditions y(0) = 0,

y(1) = 0. Show that in this case

= 0, c

(1 − s)f (s) ds.

φ(x) =

s(1 − x)f (s) ds +

x(1 − s)f (s) ds.

(d) Deﬁning

G(x, s) =

s(1 − x),0≤ s ≤ x,

x(1 − s), x ≤ s ≤ 1,

show that the solution can be written as

φ(x) =

G(x, s)f (s) ds .

The function G(x, s) appearing under the integral sign is a Green’s function.The usefulness

of a Green’s function solution rests on the fact that the Green’s function is indepen-

dent of the nonhomogeneous term in the differential equation. Thus, once the Green’s

function is determined, the solution of the boundary value problem for any nonhomoge-

neous term f (x) is obtained by a single integration. Note further that no determination

of arbitrary constants is required, since φ(x) as given by the Green’s function integral

formula automatically satisﬁes the boundary conditions.

29. By a procedure similar to that in Problem 28, show that the solution of the boundary value

problem

−(y

′′

+ y) = f (x), y(0) = 0, y(1) = 0

y = φ(x) =

G(x, s)f (s) ds ,

where

G(x, s) =

⎧

⎪

⎨

⎪

⎩

sin s sin(1 − x)

sin 1

,0≤ s ≤ x,

sin x sin(1 − s)

sin 1

, x ≤ s ≤ 1.

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11.3 Nonhomogeneous Boundary Value Problems 713

30. It is possible to show that the Sturm–Liouville problem

L[y]=−[p(x)y

′

]

′

+ q(x)y = f (x), (i)

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0 (ii)

has a Green’s function solution

y = φ(x) =

G(x, s)f (s) ds , (iii)

provided that λ = 0 is not an eigenvalue of L[y]=λy subject to the boundary conditions

(ii). Further, G(x, s) is given by

G(x, s) =

−y

(s)y

(x)/p(x)W(y

, y

)(x),0≤ s ≤ x,

−y

(x)y

(s)/p(x)W(y

, y

)(x), x ≤ s ≤ 1,

(iv)

where y

is a solution of L[y]=0 satisfying the boundary condition at x = 0,y

is a solution

of L[y]=0 satisfying the boundary condition at x = 1, and W(y

, y

) is the Wronskian of

and y

(a) Verify that the Green’s function obtained in Problem 28 is given by formula (iv).

(b) Verify that the Green’s function obtained in Problem 29 is given by formula (iv).

, y

)(x) is a constant by showing that its derivative is zero.

(d) Using Eq. (iv) and the result of part (c), show that G(x , s) = G(s, x).

(e) Verify that y = φ(x) from Eq.(iii) with G(x, s) givenby Eq. (iv)satisﬁes the differential

equation (i) and the boundary conditions (ii).

In each of Problems 31 through 34, solve the given boundary value problem by determining

the appropriate Green’s function and expressing the solution as a deﬁnite integral. Use Eqs. (i)

to (iv) of Problem 30.

31. −y

′′

= f (x), y

′

(0) = 0, y(1) = 0

32. −y

′′

= f (x), y(0) = 0, y(1) + y

′

(1) = 0

33. −(y

′′

+ y) = f (x), y

′

(0) = 0, y(1) = 0

34. −y

′′

= f (x), y(0) = 0, y

′

(1) = 0

35. Consider the boundary value problem

L[y]=−[p(x)y

′

]

′

+ q(x)y = µr(x)y + f (x), (i)

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0. (ii)

According to the text, the solution y = φ(x) is given by Eq. (13), where c

is deﬁned by

Eq. (9),provided that µ is not an eigenvalue of the corresponding homogeneous problem.

In this case it can also be shown that the solution is given by a Green’s function integral

of the form

y = φ(x) =

G(x, s, µ)f (s) ds. (iii)

Note that in this problem the Green’s function also depends on the parameter µ.

(a) Show that if these two expressions for φ(x) are to be equivalent, then

G(x, s, µ) =

∞

i=1

(x)φ

(s)

− µ

, (iv)

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714 Chapter 11. Boundary Value Problems

where λ

and φ

are the eigenvalues and eigenfunctions, respectively, of Eqs. (3), (2) of the

text. Again we see from Eq. (iv) that µ cannot be equal to any eigenvalue λ

(b) Derive Eq. (iv) directly by assuming that G(x, s, µ) has the eigenfunction expansion

G(x, s, µ) =

∞

i=1

(x, µ)φ

(s). (v)

Determine a

(x, µ) by multiplying Eq. (v) by r(s)φ

(s) and integrating with respect to s

from s = 0 to s = 1.

Hint: Show ﬁrst that λ

and φ

satisfy the equation

(x) = (λ

− µ)

G(x, s, µ)r(s)φ

(s) ds. (vi)

36. Consider the boundary value problem

−d

y/ds

= δ(s − x), y(0) = 0, y(1) = 0,

where s is the independent variable, s = x is a deﬁnite point in the interval 0 < s < 1, and

δ is the Dirac delta function (see Section 6.5). Show that the solution of this problem is

the Green’s function G(x, s) obtained in Problem 28.

In solving the given problem, note that δ(s − x) = 0 in the intervals 0 ≤ s < x and

x < s ≤ 1. Note further that −dy/ds experiences a jump of magnitude 1 as s passes through

the value x.

This problem illustrates a general property, namely, that the Green’s function G (x, s)

can be identiﬁed as the response at the point s to a unit impulse at the point x.A more gen-

eral nonhomogeneous term f on 0 ≤ x ≤ 1 can be regarded as a continuous distribution of

impulses with magnitude f (x) at the point x. The solution of a nonhomogeneous boundary

value problem in terms of a Green’s function integral can then be interpreted as the result

of superposing the responses to the set of impulses represented by the nonhomogeneous

term f (x).

11.4 Singular Sturm–Liouville Problems

In the preceding sections of this chapter, we considered Sturm–Liouville boundary

value problems: the differential equation

L[y]=−[p(x)y

′

]

′

+ q(x)y = λr(x)y,0< x < 1, (1)

together with boundary conditions of the form

y(0) + α

′

(0) = 0, (2)

y(1) + β

′

(1) = 0. (3)

Until now, we have always assumed that the problem is regular. That is, we have

assumed that p is differentiable, that p

′

, q, and r are continuous, and that p(x)>0

and r(x)>0 at all points in the closed interval. However, there are also equations of

physical interest in which some of these conditions are not satisﬁed.

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11.4 Singular Sturm–Liouville Problems 715

For example, suppose that we wish to study Bessel’s equation of order ν on the

interval 0 < x < 1. This equation is sometimes written in the form

−(xy

′

)

′

y = λxy (4)

so that p(x) = x, q(x) = ν

/x, and r(x) = x. Thus p(0) = 0, r(0) = 0, and q(x) is

unbounded and hence discontinuous as x → 0. However, the conditions imposed

on regular Sturm–Liouville problems are met elsewhere in the interval.

Similarly, for Legendre’s equation we have

−[(1 − x

′

]

′

= λy, −1 < x < 1, (5)

where λ = α(α + 1), p(x ) = 1 − x

, q(x) = 0, and r(x) = 1. Here the required condi-

tions on p, q, and r are satisﬁed in the interval 0 ≤ x ≤ 1 except at x = 1, where p is

zero.

We use the term “singular Sturm–Liouville problem” to refer to a certain class of

boundary value problems for the differential equation (1) in which the functions p, q,

and r satisfy the conditions stated earlier on the open interval 0 < x < 1, but at least

one of these functions fails to satisfy them at one or both of the boundary points.

We also prescribe suitable separated boundary conditions of a kind to be described

in more detail later in this section. Singular problems also occur if the interval is

unbounded, for example, 0 ≤ x < ∞. We do not consider this latter kind of singular

problem in this book.

As an example of a singular problem on a ﬁnite interval, consider the equation

′′

+ y

′

+ λxy = 0, (6)

−(xy

′

)

′

= λxy, (7)

on the interval 0 < x < 1, and suppose that λ>0. This equation arises in the study of

free vibrations of a circular elastic membrane and is discussed further in Section 11.5.

If we introduce the new independent variable t deﬁned by t =

√

λ x, then

√

= λ

Hence Eq. (6) becomes

√

+ λ

√

y = 0,

or, if we cancel the common factor

√

λ in each term,

+ ty = 0. (8)

Equation (8) is Bessel’s equation of order zero (see Section 5.7).The general solution

of Eq. (8) for t > 0 is

y = c

(t) + c

(t);

The substitution t =

√

λ x reduces Eq. (4) to the standard form t

′′

+ ty

′

+ (t

− ν

)y = 0.

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716 Chapter 11. Boundary Value Problems

hence the general solution of Eq. (7) for x > 0 is

y = c

(

√

λ x) + c

(

√

λ x), (9)

where J

and Y

denote the Bessel functions of the ﬁrst and second kinds of order

zero. From Eqs. (7) and (13) of Section 5.7, we have

(

√

λ x) = 1 +

∞

m=1

(−1)

(m!)

, x > 0, (10)

(

√

λ x) =

γ + ln

√

λ x

(

√

λ x) +

∞

m=1

(−1)

m+1

(m!)

, x > 0,

(11)

where H

= 1 + (1/2) +···+(1/m) and γ = lim

m→∞

− ln m). The graphs of

y = J

(x) and y = Y

(x) are given in Figure 5.7.2.

Suppose that we seek a solution of Eq. (7) that also satisﬁes the boundary con-

ditions

y(0) = 0, (12)

y(1) = 0, (13)

which are typical of those we have met in other problems in this chapter. Since

(0) = 1 and Y

(x) →−∞as x → 0, the condition y(0) = 0 can be satisﬁed only by

choosing c

= c

= 0 in Eq. (9). Thus the boundary value problem (7), (12), (13) has

only the trivial solution.

One interpretation of this result is that the boundary condition (12) at x = 0is

too restrictive for the differential equation (7). This illustrates the general situation,

namely, that at a singular boundary point it is necessary to consider a modiﬁed type

of boundary condition. In the present problem, suppose that we require only that

the solution (9) and its derivative remain bounded. In other words, we take as the

boundary condition at x = 0 the requirement

y, y

′

bounded as x → 0. (14)

This condition can be satisﬁed by choosing c

= 0 in Eq. (9), so as to eliminate the

unbounded solution Y

. The second boundary condition, y(1) = 0, then yields

(

√

λ) = 0. (15)

It is possible to show

that Eq. (15) has an inﬁnite set of discrete positive roots,

which yield the eigenvalues 0 <λ

<λ

< ···<λ

< ···of the given problem. The

corresponding eigenfunctions are

(x) = J

)

, (16)

The function J

is well tabulated;the roots of Eq. (15) can be found in tables such as those in Abramowitz

and Stegun. You can also use a computer algebra system to compute them quickly. The ﬁrst three roots

of Eq. (15) are

√

λ = 2.405, 5.520, and 8.654, respectively, to four signiﬁcant ﬁgures;

√

∼

(n − 1/4)π for

large n.

August 7, 2012 21:05 c11 Sheet number 41 Page number 717 cyan black

11.4 Singular Sturm–Liouville Problems 717

determined only up to a multiplicative constant. The boundary value problem (7),

(13), (14) is an example of a singular Sturm–Liouville problem. This example illus-

trates that if the boundary conditions are relaxed in an appropriate way, then a

singular Sturm–Liouville problem may have an inﬁnite sequence of eigenvalues and

eigenfunctions, just as a regular Sturm–Liouville problem does.

Because of their importance in applications,it is worthwhile to investigate singular

boundary value problems a little further. There are two main questions that are of

concern:

1. Precisely what type of boundary conditions can be allowed in a singular Sturm–Liouville

problem?

2. To what extent do the eigenvalues and eigenfunctions of a singular problem share the

properties of eigenvalues and eigenfunctions of regular Sturm–Liouville problems? In

particular, are the eigenvalues real, are the eigenfunctions orthogonal, and can a given

function be expanded as a series of eigenfunctions?

Both of these questions can be answered by a study of the identity

{L[u]v − uL[v]}dx = 0, (17)

which played an essential part in the development of the theory of regular Sturm–

Liouville problems.We therefore investigate the conditions under which this relation

holds for singular problems, where the integral in Eq. (17) may now have to be

examinedasanimproper integral.Tobedeﬁnite,we consider the differential equation

(1) and assume that x = 0 is a singular boundary point but that x = 1 is not. The

boundary condition (3) is imposed at the nonsingular boundary point x = 1, but we

leave unspeciﬁed, for the moment, the boundary condition at x = 0. Indeed, our

principal objective is to determine what kinds of boundary conditions are allowable

at a singular boundary point if Eq. (17) is to hold.

Since the boundary value problem under investigation is singular at x = 0, we

choose ϵ>0 and consider the integral

L[u]v dx, instead of

L[u]v dx,asin

Section 11.2. Afterwards we let ϵ approach zero. Assuming that u and v have at least

two continuous derivatives on ϵ ≤ x ≤ 1, and integrating twice by parts, we ﬁnd that

L[u]v − uL[v]

dx =−p(x)

′

(x)v(x) − u(x)v

′

(x)

. (18)

The boundary term at x = 1 is again eliminated if both u and v satisfy the boundary

condition (3), and thus

L[u]v − uL[v]

dx = p(ϵ)

′

(ϵ)v(ϵ) − u(ϵ)v

′

(ϵ)

. (19)

Taking the limit as ϵ → 0 yields

L[u]v − uL[v]

dx = lim

ϵ→0

p(ϵ)

′

(ϵ)v(ϵ) − u(ϵ)v

′

(ϵ)

. (20)

Hence Eq. (17) holds if and only if, in addition to the assumptions stated previously,

lim

ϵ→0

p(ϵ)

′

(ϵ)v(ϵ) − u(ϵ)v

′

(ϵ)

= 0 (21)

August 7, 2012 21:05 c11 Sheet number 42 Page number 718 cyan black

718 Chapter 11. Boundary Value Problems

for every pair of functions u and v in the class under consideration. Equation (21) is

therefore the criterion that determines what boundary conditions are allowable at

x = 0 if that point is a singular boundary point. A similar condition applies at x = 1

if that boundary point is singular, namely,

lim

ϵ→0

p(1 − ϵ)

′

(1 − ϵ)v(1 − ϵ) − u(1 − ϵ)v

′

(1 − ϵ)

= 0. (22)

In summary, as in Section 11.2, a singular boundary value problem for Eq. (1) is

said to be self-adjoint if Eq. (17) is valid, possibly as an improper integral, for each

pair of functions u and v with the following properties: they are twice continuously

differentiable on the open interval 0 < x < 1, they satisfy a boundary condition of

the form (2) at each regular boundary point, and they satisfy a boundary condition

sufﬁcient to ensure Eq. (21) if x = 0 is a singular boundary point, or Eq. (22) if

x = 1 is a singular boundary point. If at least one boundary point is singular, then

the differential equation (1), together with two boundary conditions of the type just

described, are said to form a singular Sturm–Liouville problem .

For example, for Eq. (7) we have p(x) = x. If both u and v satisfy the boundary

condition (14) at x = 0, it is clear that Eq. (21) will hold. Hence the singular bound-

ary value problem,consisting of the differential equation (7),the boundary condition

(14) at x = 0, and any boundary condition of the form (3) at x = 1, is self-adjoint.

The most striking difference between regular and singular Sturm–Liouville prob-

lems is that in a singular problem the eigenvalues may not be discrete. That is, the

problem may have nontrivial solutions for every value of λ, or for every value of λ in

some interval. In such a case the problem is said to have a continuous spectrum.It

may happen that a singular problem has a mixture of discrete eigenvalues and also

a continuous spectrum. Finally, it is possible that only a discrete set of eigenvalues

exists, just as in the regular case discussed in Section 11.2. For example, this is true

of the problem consisting of Eqs. (7), (13), and (14). In general, it may be difﬁcult to

determine which case actually occurs in a given problem.

A systematic discussion of singular Sturm–Liouville problems is quite sophisti-

cated

indeed, requiring a substantial extension of the methods presented in this

book. We restrict ourselves to some examples related to physical applications; in

each of these examples it is known that there is an inﬁnite set of discrete eigenvalues.

If a singular Sturm–Liouville problem does have only a discrete set of eigenvalues

and eigenfunctions,then Eq. (17) can be used,just as in Section 11.2,to prove that the

eigenvalues of such a problem are real and that the eigenfunctions are orthogonal

with respect to the weight function r. The expansion of a given function in terms of

a series of eigenfunctions then follows as in Section 11.2.

Such expansions are useful, as in the regular case, for solving nonhomoge-

neous boundary value problems. The procedure is very similar to that described in

Section 11.3. Some examples for ordinary differential equations are indicated

in Problems 1 to 4, and some problems for partial differential equations appear in

Section 11.5.

See, for example, Chapter 5 of the book by Yosida listed in the references at the end of this chapter.

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11.4 Singular Sturm–Liouville Problems 719

For instance,the eigenfunctions φ

(x) = J

(

√

x) of the singular Sturm–Liouville

problem

−(xy

′

)

′

= λxy,0< x < 1,

y, y

′

bounded as x → 0, y(1) = 0

satisfy the orthogonality relation

xφ

(x)φ

(x) dx = 0, m ̸= n (23)

with respect to the weight function r(x) = x. Then, if f is a given function, we assume

that

f (x) =

∞

n=1

(

x). (24)

Multiplying Eq. (24) by xJ

(

√

x) and integrating term by term from x = 0tox = 1,

we obtain

xf (x)J

(

x) dx =

∞

n=1

(

x) J

(

x) dx. (25)

Because of the orthogonality condition (23), the right side of Eq. (25) collapses to a

single term; hence

xf (x)J

(

x) dx

(

x) dx

, (26)

which determines the coefﬁcients in the series (24).

The convergence of the series (24) is established by an extension ofTheorem 11.2.4

to cover this case. This theorem can also be shown to hold for other sets of Bessel

functions,which are solutions of appropriate boundary value problems;for Legendre

polynomials; and for solutions of a number of other singular Sturm–Liouville pro-

blems of considerable interest.

It must be emphasized that the singular problems mentioned here are not nec-

essarily typical. In general, singular boundary value problems are characterized by

continuous spectra, rather than by discrete sets of eigenvalues. The corresponding

sets of eigenfunctions are therefore not denumerable, and series expansions of the

type described in Theorem 11.2.4 do not exist. They are replaced by appropriate

integral representations.

PROBLEMS 1. Find a formal solution of the nonhomogeneous boundary value problem

−(xy

′

)

′

= µxy + f (x),

y, y

′

bounded as x → 0, y(1) = 0,

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720 Chapter 11. Boundary Value Problems

where f is a given continuous function on 0 ≤ x ≤ 1, and µ is not an eigenvalue of the

corresponding homogeneous problem.

Hint: Use a series expansion similar to those in Section 11.3.

2. Consider the boundary value problem

−(xy

′

)

′

= λxy,

y, y

′

bounded as x → 0, y

′

(1) = 0.

(a) Show that λ

= 0 is an eigenvalue of this problem corresponding to the eigenfunction

(x) = 1. If λ>0, show formally that the eigenfunctions are given by φ

(x) = J

(

√

x),

where

√

is the nth positive root (in increasing order) of the equation J

′

(

√

λ) = 0. It is

possible to show that there is an inﬁnite sequence of such roots.

(b) Show that if m, n = 0, 1, 2, ..., then

xφ

(x)φ

(x) dx = 0, m ̸= n.

−(xy

′

)

′

= µxy + f (x),

y, y

′

bounded as x → 0, y

′

(1) = 0,

where f is a given continuous function on 0 ≤ x ≤ 1, and µ is not an eigenvalue of the

corresponding homogeneous problem.

3. Consider the problem

−(xy

′

)

′

+ (k

/x)y = λxy,

y, y

′

bounded as x → 0, y(1) = 0,

where k is a positive integer.

(a) Using the substitution t =

√

λ x, show that the given differential equation reduces to

Bessel’s equation of order k (see Problem 9 of Section 5.7). One solution is J

(t); a second

linearly independent solution, denoted by Y

(t), is unbounded as t → 0.

(b) Show formally that the eigenvalues λ

, λ

, ... of the given problem are the

squares of the positive zeros of J

(

√

λ) and that the corresponding eigenfunctions are

(x) = J

(

√

x). It is possible to show that there is an inﬁnite sequence of such zeros.

(x) satisfy the orthogonality relation

xφ

(x)φ

(x) dx = 0, m ̸= n.

(d) Determine the coefﬁcients in the formal series expansion

f (x) =

∞

n=1

(x).

(e) Find a formal solution of the nonhomogeneous problem

−(xy

′

)

′

+ (k

/x)y = µxy + f (x),

y, y

′

bounded as x → 0, y(1) = 0,

where f is a given continuous function on 0 ≤ x ≤ 1, and µ is not an eigenvalue of the

corresponding homogeneous problem.

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11.5 A Bessel Series Expansion 721

4. Consider Legendre’s equation (see Problems 22 through 24 of Section 5.3)

−[(1 − x

′

]

′

= λy

subject to the boundary conditions

y(0) = 0, y, y

′

bounded as x → 1.

The eigenvalues for this problem are λ

= 2, λ

= 4 · 3, ..., λ

= 2n(2n − 1), ... and the

eigenfunctions are the corresponding odd Legendre polynomials

(x) = P

(x) = x, φ

(x) = P

(x) = (5x

− 3x)/2, ..., φ

(x) = P

2n−1

(x), ...

(a) Show that

(x)φ

(x) dx = 0, m ̸= n.

(b) Find a formal solution of the nonhomogeneous problem

−[(1 − x

′

]

′

= µy + f (x),

y(0) = 0, y, y

′

bounded as x → 1,

where f is a given continuous function on 0 ≤ x ≤ 1, and µ is not an eigenvalue of the

corresponding homogeneous problem.

5. The equation

(1 − x

′′

− xy

′

+ λy = 0 (i)

is Chebyshev’s equation; see Problem 10 of Section 5.3.

(a) Show that Eq. (i) can be written in the form

−[(1 − x

)

1/2

′

]

′

= λ(1 − x

)

−1/2

y, −1 < x < 1. (ii)

(b) Consider the boundary conditions

y, y

′

bounded as x →−1, y, y

′

bounded as x → 1. (iii)

Show that the boundary value problem (ii), (iii) is self-adjoint.

= 0, λ

= 1, λ

= 4, ..., λ

= n

, .... The corresponding eigenfunctions are the

Chebyshev polynomials T

(x): T

(x) = 1, T

(x) = x, T

(x) = 1 − 2x

, ....Show that

−1

(x)T

(x)

(1 − x

)

1/2

dx = 0, m ̸= n. (iv)

Note that this is a convergent improper integral.

11.5 Further Remarks on the Method of Separation

of Variables: A Bessel Series Expansion

In this chapter we are interested in extending the method of separation of vari-

ables developed in Chapter 10 to a larger class of problems—to problems involving

more general differential equations, more general boundary conditions, or different

geometrical regions. We indicated in Section 11.3 how to deal with a class of more

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722 Chapter 11. Boundary Value Problems

general differential equations or boundary conditions. Here we concentrate on prob-

lems posed in various geometrical regions,with emphasis on those leading to singular

Sturm–Liouville problems when the variables are separated.

Because of its relative simplicity,as well as the considerable physical signiﬁcance of

many problems to which it is applicable,the method of separation of variables merits

its important place in the theory and application of partial differential equations.

However, this method does have certain limitations that should not be forgotten. In

the ﬁrst place, the problem must be linear so that the principle of superposition can

be invoked to construct additional solutions by forming linear combinations of the

fundamental solutions of an appropriate homogeneous problem.

As a practical matter, we must also be able to solve the ordinary differential equa-

tions, obtained after separating the variables, in a reasonably convenient manner.

In some problems to which the method of separation of variables can be applied in

principle, it is of very limited practical value due to a lack of information about the

solutions of the ordinary differential equations that appear.

Furthermore, the geometry of the region involved in the problem is subject to

rather severe restrictions. On the one hand, a coordinate system must be employed

in which the variables can be separated,and the partial differential equation replaced

by a set of ordinary differential equations. For Laplace’s equation there are about a

dozen such coordinate systems; only rectangular, circular cylindrical, and spherical

coordinates are likely to be familiar to most readers of this book. On the other

hand, the boundary of the region of interest must consist of coordinate curves or

surfaces—that is, curves or surfaces on which one variable remains constant. Thus, at

an elementary level, we are limited to regions bounded by straight lines or circular

arcs in two dimensions, or by planes, circular cylinders, circular cones, or spheres in

three dimensions.

In three-dimensional problems, the separation of variables in Laplace’s operator

+ u

leads to the equation X

′′

+ λX = 0 in rectangular coordinates, to

Bessel’s equation in cylindrical coordinates, and to Legendre’s equation in spherical

coordinates. It is this fact that is largely responsible for the intensive study that has

been made of these equations and the functions deﬁned by them. It is also noteworthy

that two of the three most important situations lead to singular, rather than regular,

Sturm–Liouville problems. Thus singular problems are by no means exceptional and

may be of even greater interest than regular ones. The remainder of this section is

devoted to an example involving an expansion of a given function as a series of Bessel

functions.

The Vibrations of a Circular Elastic Membrane. In Section 10.7 [Eq. (7)] we noted that the

transverse vibrations of athin elastic membraneare governedby thetwo-dimensional

wave equation

+ u

) = u

. (1)

To study the motion of a circular membrane, it is convenient to write Eq. (1) in polar

coordinates:

θθ

(

= u

. (2)

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11.5 A Bessel Series Expansion 723

We will assume that the membrane has unit radius, that it is ﬁxed securely around

its circumference, and that initially it occupies a displaced position independent of

the angular variable θ, from which it is released at time t = 0. Because of the circular

symmetry of the initial and boundary conditions, it is natural to assume also that u is

independent of θ; hence u is a function of r and t only. In this event the differential

equation (2) becomes

(

= u

,0< r < 1, t > 0. (3)

The boundary condition at r = 1 is

u(1, t) = 0, t ≥ 0, (4)

and the initial conditions are

u(r,0) = f (r),0≤ r ≤ 1, (5)

(r,0) = 0, 0 ≤ r ≤ 1, (6)

where f (r) describes the initial conﬁguration of the membrane. For consistency we

also require that f (1) = 0. Finally,we state explicitly the requirement that u(r, t) must

be bounded for 0 ≤ r ≤ 1.

Assuming that u(r, t) = R(r)T(t), and substituting for u(r, t) in Eq. (3), we obtain

′′

+ (1/r)R

′

′′

=−λ

. (7)

We have anticipated that the separation constant mustbe negativeby writingit as −λ

with λ>0.

Then Eq. (7) yields the following two ordinary differential equations:

′′

+ rR

′

+ λ

R = 0, (8)

′′

+ λ

T = 0. (9)

Thus, from Eq. (9),

T(t) = k

sin λat + k

cos λat. (10)

Introducing the new independent variable ξ = λr into Eq. (8), we obtain

d ξ

+ ξ

d ξ

+ ξ

R = 0, (11)

which is Bessel’s equation of order zero. Thus

R = c

(ξ) + c

(ξ), (12)

where J

and Y

are Bessel functions of the ﬁrst and second kinds, respectively, of

order zero (see Section 11.4). In terms of r, we have

R = c

(λr) + c

(λr). (13)

By denoting the separation constant by −λ

, rather than simply by −λ, we avoid the appearance of

numerous radical signs in the following discussion.

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724 Chapter 11. Boundary Value Problems

The boundedness condition on u(r, t) requires that R remain bounded as r → 0. Since

(λr) →−∞as r → 0, we must choose c

= 0. The boundary condition (4) then

requires that

(λ) = 0. (14)

Consequently, the allowable values of the separation constant are obtained from the

roots of the transcendental equation (14). Recall from Section 11.4 that J

(λ) has

an inﬁnite set of discrete positive zeros, which we denote by λ

, λ

, ..., λ

, ...,

ordered in increasing magnitude. Further, the functions J

(λ

r) are the eigenfunc-

tions of a singular Sturm–Liouville problem and can be used as the basis of a series

expansion for the given function f . The fundamental solutions of this problem,

which satisfy the partial differential equation (3), the boundary condition (4), and

the boundedness condition, are

(r, t) = J

(λ

r) sin λ

at, n = 1, 2, ..., (15)

(r, t) = J

(λ

r) cos λ

at, n = 1, 2, .... (16)

Next we assume that u(r, t) can be expressed as an inﬁnite linear combination of the

fundamental solutions (15), (16):

u(r, t) =

∞

n=1

(r, t) + c

(r, t)

∞

n=1

(λ

r) sin λ

at + c

(λ

r) cos λ

at]. (17)

The initial conditions require that

u(r,0) =

∞

n=1

(λ

r) = f (r), (18)

(r,0) =

∞

n=1

(λ

r) = 0. (19)

From Eq. (26) of Section 11.4, we obtain

= 0, c

rf (r)J

(λ

r) dr

r[J

(λ

r)]

; n = 1, 2, .... (20)

Thus the solution of the partial differential equation (3) satisfying the boundary

condition (4) and the initial conditions (5) and (6) is given by

u(r, t) =

∞

n=1

(λ

r) cos λ

at (21)

with the coefﬁcients c

deﬁned by Eq. (20).

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11.5 A Bessel Series Expansion 725

PROBLEMS 1. Consider Laplace’s equation u

+ u

= 0 in the parallelogram whose vertices are (0, 0),

(2, 0), (3, 2), and (1, 2). Suppose that the boundary condition u(x,2) = f (x) is imposed on

the side y = 2 for 1 ≤ x ≤ 3, and that on the other three sides u = 0 (see Figure 11.5.1).

(a) Show that there are no nontrivial solutions of the partial differential equation of the

form u(x, y) = X (x)Y(y) that also satisfy the homogeneous boundary conditions.

(b) Let ξ = x −

y, η = y. Show that the given parallelogram in the xy-plane transforms

into the square 0 ≤ ξ ≤ 2, 0 ≤ η ≤ 2 in the ξη-plane. Show that the differential equation

transforms into

ξξ

− u

ξη

+ u

ηη

= 0.

How are the boundary conditions transformed?

u(ξ, η) = U(ξ)V(η).

Thus in the xy-plane, the shape of the boundary precludes a solution by the method of the

separation of variables, while in the ξη-plane, the region is acceptable but the variables in

the differential equation can no longer be separated.

(a)

(1, 2)

(3, 2)

B (2, 0)

A (0, 0)

(b)

(0, 2)

(2, 2)

B' (2, 0)

A' (0, 0)

FIGURE 11.5.1 The region in Problem 1.

2. Find the displacement u(r, t) in a vibrating circular elastic membrane of radius 1 that

satisﬁes the boundary condition

u(1, t) = 0, t ≥ 0,

and the initial conditions

u(r,0) = 0, u

(r,0) = g(r),0≤ r ≤ 1,

where g(1) = 0.

Hint: The differential equation to be satisﬁed is Eq. (3) of this section.

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726 Chapter 11. Boundary Value Problems

3. Find the displacement u(r, t) in a vibrating circular elastic membrane of radius 1 that

satisﬁes the boundary condition

u(1, t) = 0, t ≥ 0,

and the initial conditions

u(r,0) = f (r), u

(r,0) = g(r),0≤ r ≤ 1,

where f (1) = g(1) = 0.

4. The wave equation in polar coordinates is

+ (1/r)u

+ (1/r

θθ

= a

−2

Show that if u(r, θ, t) = R(r);(θ)T(t), then R, ;, and T satisfy the ordinary differential

equations

′′

+ rR

′

+ (λ

− n

)R = 0,

;

′′

+ n

; = 0,

′′

+ λ

T = 0.

5. In the circular cylindrical coordinates r, θ, z deﬁned by

x = r cos θ, y = r sin θ, z = z,

Laplace’s equation is

+ (1/r)u

+ (1/r

θθ

+ u

= 0.

(a) Show that if u(r, θ, z) = R(r);(θ)Z(z),thenR,;,and Z satisfy the ordinary differential

equations

′′

+ rR

′

+ (λ

− n

)R = 0,

;

′′

+ n

; = 0,

′′

− λ

Z = 0.

(b) Show that if u(r, θ, z) is independent of θ, then the ﬁrst equation in part (a) becomes

′′

+ rR

′

+ λ

R = 0,

the second is omitted altogether, and the third is unchanged.

6. Find the steady state temperature in a semi-inﬁnite rod 0 < z < ∞,0≤ r < 1, if the tem-

perature is independent of θ and approaches zero as z →∞.Assume that the temperature

u(r, z) satisﬁes the boundary conditions

u(1, z) = 0, z > 0,

u(r,0) = f (r),0≤ r ≤ 1.

Hint: Refer to Problem 5.

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11.5 A Bessel Series Expansion 727

7. The equation

+ v

+ k

v = 0

is a generalization of Laplace’s equation and is sometimes called the Helmholtz

equation.

(a) In polar coordinates the Helmholtz equation is

+ (1/r)v

+ (1/r

θθ

+ k

v = 0.

If v(r, θ) = R(r);(θ), show that R and ; satisfy the ordinary differential equations

′′

+ rR

′

+ (k

− λ

)R = 0, ;

′′

+ λ

; = 0.

(b) Consider the Helmholtz equation in the disk r < c. Find the solution that remains

bounded at all points in the disk, that is periodic in θ with period 2π, and that satisﬁes the

boundary condition v(c, θ) = f (θ), where f is a given function on 0 ≤ θ<2π.

Hint: The equation for R is a Bessel equation. See Problem 3 of Section 11.4.

8. Consider the ﬂow of heat in a cylinder 0 ≤ r < 1, 0 ≤ θ<2π, −∞ < z < ∞ of radius 1

and of inﬁnite length. Let the surface of the cylinder be held at temperature zero, and let

the initial temperature distribution be a function of the radial variable r only. Then the

temperature u is a function of r and t only and satisﬁes the heat conduction equation

+ (1/r)u

]=u

,0< r < 1, t > 0,

and the following initial and boundary conditions:

u(r,0) = f (r),0≤ r ≤ 1,

u(1, t) = 0, t > 0.

Show that

u(r, t) =

∞

n=1

(λ

r)e

−α

where J

(λ

) = 0. Find a formula for c

9. In the spherical coordinates ρ,θ, φ (ρ>0, 0 ≤ θ<2π,0 ≤ φ ≤ π) deﬁned by the equations

x = ρ cos θ sin φ, y = ρ sinθ sin φ, z = ρ cos φ,

Laplace’s equation is

ρρ

+ 2ρu

+ (csc

φ)u

θθ

+ u

φφ

+ (cot φ)u

= 0.

(a) Show that if u(ρ, θ, φ) = P(ρ);(θ)7(φ), then P, ;, and 7 satisfy ordinary differential

equations of the form

′′

+ 2ρP

′

− µ

P = 0,

;

′′

+ λ

; = 0,

(sin

φ)7

′′

+ (sin φ cos φ)7

′

+ (µ

sin

φ − λ

)7 = 0.

The German scientist Hermann von Helmholtz (1821–1894) was trained in medicine and physiology;

early in his career he made important contributions to physiological optics and acoustics, including the

invention of the ophthalmoscope in 1851. Later his interests turned to physics, especially ﬂuid mechan-

ics and electrodynamics. During his lifetime, he held chairs in physiology or physics at several German

universities.

August 7, 2012 21:05 c11 Sheet number 52 Page number 728 cyan black

728 Chapter 11. Boundary Value Problems

The ﬁrst of these equations is of the Euler type, while the third is related to Legendre’s

equation.

(b) Show that if u(ρ, θ, φ) is independent of θ, then the ﬁrst equation in part (a) is

unchanged, the second is omitted, and the third becomes

(sin

φ)7

′′

+ (sin φ cos φ)7

′

+ (µ

sin

φ)7 = 0.

7 in part (b) becomes

(1 − s

)

− 2s

+ µ

7 = 0, −1 ≤ s ≤ 1.

Note that this is Legendre’s equation.

10. Find the steady state temperature u(ρ, φ) in a sphere of unit radius if the temperature is

independent of θ and satisﬁes the boundary condition

u(1, φ) = f (φ),0≤ φ ≤ π.

Hint: Refer to Problem 9 and to Problems 22 through 29 of Section 5.3. Use the fact that

the only solutions of Legendre’s equation that are ﬁnite at both ±1 are the Legendre

polynomials.

11.6 Series of Orthogonal Functions: Mean Convergence

In Section 11.2 we stated that under certain restrictions, a given function f can be

expanded in a series of eigenfunctions of a Sturm–Liouville boundary value problem,

the series converging to [f (x+) + f (x−)]/2 at each point in the open interval. Under

somewhat more restrictive conditions, the series converges to f (x) at each point in

the closed interval. This type of convergence is referred to as pointwise convergence.

In this section we describe a different kind of convergence that is especially useful

for series of orthogonal functions, such as eigenfunctions.

Suppose that we are given the set of functions φ

, φ

, ..., φ

, that are continuous

on the interval 0 ≤ x ≤ 1 and satisfy the orthonormality condition

r(x)φ

(x)φ

(x) dx =

0, i ̸= j,

1, i = j,

(1)

where r is a nonnegative weight function. Suppose also that we wish to approximate

a given function f , deﬁned on 0 ≤ x ≤ 1, by a linear combination of φ

, ..., φ

. That

is, if

(x) =

i=1

(x), (2)

we wish to choose the coefﬁcients a

, ..., a

so that the function S

will best approx-

imate f on 0 ≤ x ≤ 1. The ﬁrst problem that we must face in doing this is to state

precisely what we mean by “best approximate f on 0 ≤ x ≤ 1.” There are several

reasonable meanings that can be attached to this phrase.

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11.6 Mean Convergence 729

1. We can choose n points x

, ..., x

in the interval 0 ≤ x ≤ 1 and require that S

(x) have the

same value as f (x) at each of these points. The coefﬁcients a

, ..., a

are found by solving

the set of linear algebraic equations

i=1

) = f (x

), j = 1, ..., n. (3)

This procedure is known as the method of collocation. It has the advantage that it is very

easy to write down Eqs. (3); we need only to evaluate the functions involved at the points

, ..., x

. If these points are well chosen, and if n is fairly large, then presumably S

(x)

not only will be equal to f (x) at the chosen points but will be reasonably close to it at

other points as well. However,collocation has several deﬁciencies. One is that if one more

base function φ

n+1

is added, then one more point x

n+1

is required, and all the coefﬁcients

must be recomputed. Thus it is inconvenient to improve the accuracy of a collocation

approximation by including additional terms. Further, the coefﬁcients a

depend on the

location of the points x

, ..., x

, and it is not obvious how best to select these points.

2. Alternatively, we can consider the difference |f (x) − S

(x)| and try to make it as small as

possible.Thetrouble hereis that |f (x) − S

(x)|is a functionof x as well asof thecoefﬁcients

, ..., a

, and it is not clear how to calculate a

. The choice of a

that makes |f (x) − S

(x)|

small at one point may make it large at another. One way to proceed is to consider instead

the least upper bound

of |f (x) − S

(x)| for x in 0 ≤ x ≤ 1, and then to choose a

, ..., a

so as to make this quantity as small as possible. That is, if

, ..., a

) = lub

0≤x≤1

|f (x) − S

(x)|, (4)

then choose a

, ..., a

so as to minimize E

. This approach is intuitively appealing and is

often used in theoretical calculations. In practice, however, it is usually very hard, if not

impossible, to write down an explicit formula for E

, ..., a

). Further, this procedure

shares one of the disadvantages of collocation: upon adding an additional term to S

(x),

we must recompute all the preceding coefﬁcients. Thus it is not often useful in practical

problems.

3. Another way to proceed is to consider

, ..., a

) =

r(x)|f (x) − S

(x)|dx. (5)

If r(x) = 1, then I

is the area between the graphs of y = f (x) and y = S

(x) (see

Figure 11.6.1). We can then determine the coefﬁcients a

so as to minimize I

. To avoid

the complications resulting from calculations with absolute values, it is more convenient

to consider instead

, ..., a

) =

r(x)

;

f (x) − S

(x)

dx (6)

as our measure of the quality of approximation of the linear combination S

(x) to f (x).

Although R

is clearly similar in some ways to I

, it lacks the simple geometric interpre-

tation of the latter. Nevertheless, it is much easier mathematically to deal with R

than

with I

. The quantity R

is called the mean square error of the approximation S

to f .If

, ..., a

are chosen so as to minimize R

, then S

is said to approximate f in the mean

square sense.

The least upper bound (lub) is an upper bound that is smaller than any other upper bound. The lub of

a bounded function always exists and is equal to the function’s maximum if it has one.

August 7, 2012 21:05 c11 Sheet number 54 Page number 730 cyan black

730 Chapter 11. Boundary Value Problems

y = S

(x)

y = f (x)

FIGURE 11.6.1 Approximation of f (x) by S

(x).

To choose a

, ..., a

so as to minimize R

, we must satisfy the necessary conditions

∂R

/∂a

= 0, i = 1, ..., n. (7)

Writing out Eq. (7) and noting that ∂S

(x; a

, ..., a

)/∂a

is equal to φ

(x), we obtain

−

∂R

∂a

= 2

r(x)[f (x) − S

(x)]φ

(x) dx = 0 . (8)

Substituting for S

(x) from Eq. (2) and making use of the orthogonality relation (1),

we ﬁnd that

r(x)f (x)φ

(x) dx, i = 1, ..., n. (9)

The coefﬁcients deﬁned by Eq. (9) are called the Fourier coefﬁcients of f with respect

to the orthonormal set φ

, φ

, ..., φ

and the weight function r. Since the conditions

(7) are only necessary and not sufﬁcient for R

to be a minimum,a separate argument

is required to show that R

is actually minimized if the a

are chosen by Eq. (9). This

argument is outlined in Problem 5.

Notethat the coefﬁcients (9) are thesame as those inthe eigenfunction series whose

convergence, under certain conditions, was stated in Theorem 11.2.4. Thus S

(x) is

the nth partial sum in this series and constitutes the best mean square approximation

to f (x) that is possible with the functions φ

, ..., φ

. We will assume hereafter that

the coefﬁcients a

in S

(x) are given by Eq. (9).

Equation (9) is noteworthy in two other important respects. In the ﬁrst place, it

gives a formula foreach a

separately,ratherthan aset oflinear algebraicequations for

, ..., a

as inthe method of collocation,for example.This is due to the orthogonality

of the base functions φ

, ..., φ

. Further, the formula for a

is independent of n, the

number of terms in S

(x). The practical signiﬁcance of this is as follows. Suppose

that, to obtain a better approximation to f , we desire to use an approximation with

more terms—say, k terms, where k > n. It is then unnecessary to recompute the ﬁrst

n coefﬁcients in S

(x).All that is required is to compute,from Eq. (9), the coefﬁcients

n+1

, ..., a

arising from the additional base functions φ

n+1

, ..., φ

. Of course, if f , r,

and the φ

are complicated functions, it may be necessary to evaluate the integrals

numerically.

Now let us suppose that there is an inﬁnite sequence of functions φ

, ..., φ

, ...,

that are continuous and orthonormal on the interval 0 ≤ x ≤ 1. Suppose further that

August 7, 2012 21:05 c11 Sheet number 55 Page number 731 cyan black

11.6 Mean Convergence 731

as n increases without bound, the mean square error R

approaches zero. In this

event the inﬁnite series

∞

i=1

(x)

is said to converge in the mean square sense (or, more simply, in the mean)tof (x).

Mean convergence is an essentially different type of convergence from the pointwise

convergence considered up to now. A series may converge in the mean without

converging at each point. This is plausible geometrically because the area between

two curves, which behaves in the same way as the mean square error, may be zero

even though the functions are not the same at every point. They may differ on any

ﬁnite set of points, for example, without affecting the mean square error. It is less

obvious,but also true,that even if an inﬁniteseries converges at every point,it may not

converge in the mean. Indeed, the mean square error may even become unbounded.

An example of this phenomenon is given in Problem 4.

Now suppose that we wish to know what class of functions, deﬁned on 0 ≤ x ≤ 1,

can be represented as an inﬁnite series of the orthonormal set φ

, i = 1, 2, ....The

answer depends on what kind of convergence we require. We say that the set

, ..., φ

, ... is complete with respect to mean square convergence for a set of

functions

F if, for each function f in F, the series

f (x) =

∞

i=1

(x), (10)

with coefﬁcients given by Eq. (9), converges in the mean. There is a similar deﬁnition

for completeness with respect to pointwise convergence.

Theorems having to do with the convergence of series such as that in Eq. (10) can

now be restated in terms of the idea of completeness. For example,Theorem 11.2.4

can be restated as follows:The eigenfunctions of the Sturm–Liouville problem

−[p(x)y

′

]

′

+ q(x)y = λr(x)y,0< x < 1, (11)

y(0) + α

′

(0) = 0, β

y(1) + β

′

(1) = 0 (12)

are complete with respect to ordinary pointwise convergence for the set of functions

that are continuous on 0 ≤ x ≤ 1 and that have piecewise continuous derivatives

there.

If pointwise convergence is replaced by mean convergence,Theorem 11.2.4 can be

considerably generalized. Before we state such a companion theorem to Theorem

11.2.4, we ﬁrst deﬁne what is meant by a square integrable function. A function f is

said to be square integrable on the interval 0 ≤ x ≤ 1 if both f and f

are integrable

on that interval. The following theorem is similar to Theorem 11.2.4 except that it

involves mean convergence.

For the Riemann integral used in elementary calculus, the hypotheses that f and f

are integrable

are independent; that is, there are functions such that f is integrable but f

is not, and conversely (see

Problem 6).A generalized integral, known as the Lebesgue integral,has the property (among others) that

if f

is integrable, then f is also necessarily integrable. The term square integrable came into common use

in connection with this type of integration.

August 7, 2012 21:05 c11 Sheet number 56 Page number 732 cyan black

732 Chapter 11. Boundary Value Problems

Theorem 11.6.1

The eigenfunctions φ

of the Sturm–Liouville problem (11), (12) are complete with

respect to mean convergence for the set of functions that are square integrable on

0 ≤ x ≤ 1. In other words, given any square integrable function f , the series (10),

whose coefﬁcients are given by Eq. (9), converges to f (x) in the mean square sense.

It is signiﬁcant that the class of functions speciﬁed in Theorem 11.6.1 is very

large indeed. The class of square integrable functions contains some functions with

many discontinuities, including some kinds of inﬁnite discontinuities, as well as some

functions that are not differentiable at any point.All these functions have mean con-

vergent expansions in the eigenfunctions of the boundary value problem (11), (12).

However, in many cases these series do not converge pointwise, at least not at every

point. Thus mean convergence is more naturally associated with series of orthogonal

functions, such as eigenfunctions, than ordinary pointwise convergence.

The theory of Fourier series discussed in Chapter 10 is just a special case of the

general theory of Sturm–Liouville problems. For instance, the functions

(x) =

√

2 sin nπx (13)

are the normalized eigenfunctions of the Sturm–Liouville problem

′′

+ λy = 0, y(0) = 0, y(1) = 0. (14)

Thus, if f is a given square integrable function on 0 ≤ x ≤ 1, then according to

Theorem 11.6.1, the series

f (x) =

∞

m=1

(x) =

√

∞

m=1

sin mπx, (15)

where

f (x)φ

(x) dx =

√

f (x) sin mπxdx, (16)

converges in the mean. The series (15) is precisely the Fourier sine series discussed

in Section 10.4. If f satisﬁes the further conditions stated inTheorem 11.2.4, then this

series converges pointwise, as well as in the mean. Similarly, a Fourier cosine series

is associated with the Sturm–Liouville problem

′′

+ λy = 0, y

′

(0) = 0, y

′

(1) = 0. (17)

EXAMPLE

Let f (x) = 1 for 0 < x < 1. Expand f (x) using the eigenfunctions (13), and discuss the point-

wise and mean square convergence of the resulting series.

The series has the form (15), and its coefﬁcients b

are given by Eq. (16). Thus

√

sin mπxdx=

√

mπ

(1 − cos mπ), (18)

August 7, 2012 21:05 c11 Sheet number 57 Page number 733 cyan black

11.6 Mean Convergence 733

and the nth partial sum of the series is

(x) = 2

m=1

1 − cos mπ

mπ

sin mπx. (19)

The mean square error is then

[f (x) − S

(x)]

dx. (20)

By calculating R

for several values of n and plotting the results, we obtain Figure 11.6.2.

This ﬁgure indicates that R

steadily decreases as n increases. Of course,Theorem 11.6.1 asserts

that R

→ 0asn →∞. Pointwise, we know that S

(x) → f (x) = 1asn →∞for 0 < x < 1;

further, S

(x) has the value zero for x = 0orx = 1 for every n. Although the series converges

pointwise for each value of x, the least upper bound of the error does not diminish as n grows

larger. For each n there are points close to x = 0 and x = 1 where the error is arbitrarily close

to 1. The situation is similar to Example 1 in Section 10.3; see Figures 10.3.3 and 10.3.4.

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18

0.20

0246801214161

FIGURE 11.6.2 Dependence of the mean square error R

on n in Example 1.

Theorem 11.6.1 can be extended to cover self-adjoint boundary value problems

having periodic boundary conditions, such as the problem

′′

+ λy = 0, (21)

y(−L) − y(L) = 0, y

′

(−L) − y

′

(L) = 0 (22)

considered in Example 4 of Section 11.2.The eigenfunctions of the problem (21),(22)

are φ

(x) = cos(nπx/L) for n = 0, 1, 2, ... and ψ

(x) = sin(nπx/L) for n = 1, 2, ....

August 7, 2012 21:05 c11 Sheet number 58 Page number 734 cyan black

734 Chapter 11. Boundary Value Problems

If f is a given square integrable function on −L ≤ x ≤ L, then its expansion in terms

of the eigenfunctions φ

and ψ

is of the form

f (x) =

∞

n=1

)

cos

nπx

+ b

sin

nπx

, (23)

where

−L

f (x) cos

nπx

dx, n = 0, 1, 2, ..., (24)

−L

f (x) sin

nπx

dx, n = 1, 2, .... (25)

This expansion is exactly the Fourier series for f discussed in Sections 10.2 and

10.3. According to the generalization of Theorem 11.6.1, the series (23) converges

in the mean for any square integrable function f , even though f may not satisfy the

conditions of Theorem 10.3.1, which ensure pointwise convergence.

PROBLEMS 1. Extend the results of Example 1 by ﬁnding the smallest value of n for which R

< 0.02,

where R

is given by Eq. (20).

Let f (x) = x for 0 < x < 1, and let φ

(x) =

√

2 sin mπx.

(a) Find the coefﬁcients b

in the expansion of f (x) in terms of φ

(x), φ

(x), ....

(b) Calculate the mean square error R

for several values of n and plot the results.

< 0.01.

Follow the instructions for Problem 2 using f (x) = x(1 − x) for 0 < x < 1.

4. In this problem we show that pointwise convergence of a sequence S

(x) does not imply

mean convergence, and conversely.

(a) Let S

(x) = n

√

−nx

,0≤ x ≤ 1. Show that S

(x) → 0asn →∞for each x in

0 ≤ x ≤ 1. Show also that

[0 − S

(x)]

dx =

(1 − e

−n

)

and hence R

→∞as n →∞. Thus pointwise convergence does not imply mean con-

vergence.

(b) Let S

(x) = x

for 0 ≤ x ≤ 1, and let f (x) = 0 for 0 ≤ x ≤ 1. Show that

[f (x) − S

(x)]

dx =

2n + 1

and hence S

(x) converges to f (x) in the mean. Also show that S

(x) does not converge to

f (x) pointwise throughout 0 ≤ x ≤ 1. Thus mean convergence does not imply pointwise

convergence.

August 7, 2012 21:05 c11 Sheet number 59 Page number 735 cyan black

11.6 Mean Convergence 735

5. Suppose that the functions φ

, ..., φ

satisfy the orthonormality relation (1) and that a

given function f is to be approximated by S

(x) = c

(x) +···+c

(x), where the coef-

ﬁcients c

are not necessarily those of Eq. (9). Show that the mean square error R

given

by Eq. (6) may be written in the form

r(x)f

(x) dx −

i=1

− a

)

where the a

are the Fourier coefﬁcients given by Eq. (9). Show that R

is minimized if

= a

for each i.

6. In this problem we show by examples that the (Riemann) integrability of f and of f

are

independent.

(a) Let f (x) =

−1/2

,0< x ≤ 1,

0, x = 0.

Show that

f (x) dx exists as an improper integral, but

(x) dx does not.

(b) Let f (x) =

1, x rational,

−1, x irrational.

Show that

(x) dx exists, but

f (x) dx does not.

7. Suppose that it is desired to construct a set of polynomials f

(x), f

(x), ..., f

(x), ...,

where f

(x) is of degree k, that are orthonormal on the interval 0 ≤ x ≤ 1. That is, the set

of polynomials must satisfy

, f

) =

(x)f

(x) dx = δ

(a) Find f

(x) by choosing the polynomial of degree zero such that (f

, f

) = 1.

(b) Find f

(x) by determining the polynomial of degree one such that (f

, f

) = 0 and

, f

) = 1.

(x).

(d) The normalization condition (f

, f

) = 1 is somewhat awkward to apply. Let g

(x),

(x), ..., g

(x), ...be the sequence of polynomials that are orthogonal on 0 ≤ x ≤ 1 and

that are normalized by the condition g

(1) = 1. Find g

(x), g

(x), and g

(x), and compare

them with f

(x), f

(x), and f

(x).

8. Suppose that it is desired to construct a set of polynomials P

(x), P

(x), ..., P

(x), ...,

where P

(x) is of degree k, that are orthogonal on the interval −1 ≤ x ≤ 1; see Problem

7. Suppose further that P

(x) is normalized by the condition P

(1) = 1. Find P

(x), P

(x),

(x), and P

(x). Note that these are the ﬁrst four Legendre polynomials (see Problem 24

of Section 5.3).

9. This problem develops some further results associated with mean convergence. Let

, ..., a

), S

(x), and a

be deﬁned by Eqs. (6), (2), and (9), respectively.

(a) Show that

r(x)f

(x) dx −

i=1

Hint: Substitute for S

(x) in Eq. (6) and integrate, using the orthogonality relation (1).

August 7, 2012 21:05 c11 Sheet number 60 Page number 736 cyan black

736 Chapter 11. Boundary Value Problems

(b) Show that

i=1

≤

r(x)f

(x) dx. This result is known as Bessel’s inequality.

∞

i=1

converges.

(d) Show that lim

n→∞

r(x)f

(x) dx −

∞

i=1

(e) Show that

∞

i=1

(x) converges to f (x) in the mean if and only if

r(x)f

(x) dx =

∞

i=1

This result is known as Parseval’s equation.

In Problems 10 through 12, let φ

, φ

, ..., φ

, ... be the normalized eigenfunctions of the

Sturm–Liouville problem (11), (12).

10. Show that if a

is the nth Fourier coefﬁcient of a square integrable function f , then

lim

n→∞

= 0.

Hint: Use Bessel’s inequality, Problem 9(b).

11. Show that the series

(x) + φ

(x) +···+φ

(x) +···

cannot be the eigenfunction series for any square integrable function.

Hint: See Problem 10.

12. Show that the series

(x) +

(x)

√

+···+

(x)

√

+···

is not the eigenfunction series for any square integrable function.

Hint: Use Bessel’s inequality, Problem 9(b).

13. Show that Parseval’s equation in Problem 9(e) is obtained formally by squaring the series

(10) corresponding to f , multiplying by the weight function r, and integrating term by

term.

REFERENCES The following books were mentioned in the text in connection with certain theorems about Sturm–

Liouville problems:

Birkhoff, G., and Rota, G.-C., Ordinary Differential Equations (4th ed.) (New York: Wiley, 1989).

Sagan, H., Boundary and Eigenvalue Problems in Mathematical Physics (New York: Wiley, 1961; New

York: Dover, 1989).

Weinberger, H. F.,A First Course in Partial Differential Equations with Complex Variables and Transform

Methods (New York: Blaisdell, 1965; New York: Dover, 1995).

Yosida, K., Lectures on Differential and Integral Equations (New York: Interscience Publishers, 1960; New

York: Dover, 1991).

August 7, 2012 21:05 c11 Sheet number 61 Page number 737 cyan black

References 737

The following book is a convenient source of numerical and graphical data about Bessel and Legendre

functions:

Abramowitz, M., and Stegun, I. A. (eds.), Handbook of Mathematical Functions with Formulas, Graphs,

and Mathematical Tables (New York: Dover, 1965); originally published by the National Bureau of

Standards, Washington, DC, 1964.

The digital successor to Abramowitz and Stegun is

Digital Library of Mathematical Functions. Release date 2011-08-29. National Institute of Standards and

Technology from http://dlmf.nist.gov/.

The following books also contain much information about Sturm–Liouville problems:

Cole, R. H.,Theory of Ordinary Differential Equations (New York: Appleton-Century-Crofts, 1968).

Hochstadt,H.,Differential Equations:A ModernApproach (NewYork:Holt,Rinehart,andWinston,1964;

New York: Dover, 1975).

Miller,R.K.,andMichel,A. N.,OrdinaryDifferential Equations (NewYork:Academic Press,1982;Mineola,

NY: Dover, 2007).

Tricomi, F. G., Differential Equations (translated by Elizabeth A. McHarg) (New York: Hafner, 1961;

Mineola, NY: Dover, 2012).

August 7, 2012 21:05 c11 Sheet number 62 Page number 738 cyan black

August 2, 2012 13:11 bansw Sheet number 1 Page number 739 cyan black

739

Answers to

Problems

CHAPTER 1 Section 1.1, page 7

1. y → 3/2 as t →∞ 2. y diverges from 3/2 as t →∞

3. y diverges from −3/2 as t →∞ 4. y →−1/2 as t →∞

5. y diverges from −1/2 as t →∞ 6. y diverges from −2 as t →∞

7. y

′

= 3 − y 8. y

′

= 2 − 3y

9. y

′

= y − 2 10. y

′

= 3y − 1

11. y = 0 and y = 4 are equilibrium solutions;y → 4 if initial value is positive;y diverges from

0 if initial value is negative.

12. y = 0 and y = 5 are equilibrium solutions; y diverges from 5 if initial value is greater than

5; y → 0 if initial value is less than 5.

13. y = 0 is equilibrium solution; y → 0 if initial value is negative; y diverges from 0 if initial

value is positive.

14. y = 0 and y = 2 are equilibrium solutions; y diverges from 0 if initial value is negative;

y → 2 if initial value is between 0 and 2; y diverges from 2 if initial value is greater than 2.

15. (j) 16. (c) 17. (g) 18. (b) 19. (h) 20. (e)

21. (a) dq/dt = 300(10

−2

− q10

−6

); q in g, t in h (b) q → 10

g; no

22. dV/dt =−kV

2/3

for some k > 0.

23. du/dt =−0.05(u − 70); u in

◦

F, t in min

24. (a) dq/dt = 500 − 0.4q; q in mg, t in h (b) q → 1250 mg

25. (a) mv

′

= mg − kv

(b) v →

mg/k (c) k = 2/49

26. y is asymptotic to t − 3 as t →∞

27. y → 0 as t →∞

28. y →∞,0,or−∞ depending on the initial value of y

29. y →∞or −∞ depending on the initial value of y

30. y →∞or −∞ or y oscillates depending on the initial value of y

31. y →−∞or is asymptotic to

√

2t − 1 depending on the initial value of y

32. y → 0 and then fails to exist after some t

≥ 0

33. y →∞or −∞ depending on the initial value of y

August 2, 2012 13:11 bansw Sheet number 2 Page number 740 cyan black

740 Answers to Problems

Section 1.2, page 15

1. (a) y = 5 + (y

− 5)e

−t

(b) y = (5/2) +[y

− (5/2)]e

−2t

− 5)e

−2t

Equilibrium solution is y = 5 in (a) and (c),y = 5/2 in(b); solution approaches equilibrium

faster in (b) and (c) than in (a).

2. (a) y = 5 + (y

− 5)e

(b) y = (5/2) +[y

− (5/2)]e

− 5)e

Equilibrium solution is y = 5 in (a) and (c), y = 5/2 in (b); solution diverges from

equilibrium faster in (b) and (c) than in (a).

3. (a) y = ce

−at

+ (b/a)

(iii) Equilibrium remains the same and is approached more rapidly.

4. (a) y

= b/a (b) Y

′

= aY

5. (a) y

(t) = ce

(b) y = ce

+ (b/a)

6. y = ce

−at

+ (b/a)

7. (a) T = 2ln18

∼

5.78 months (b) T = 2ln[900/(900 − p

)] months

= 900(1 − e

−6

)

∼

897.8

8. (a) r = (ln 2)/30 day

−1

(b) r = (ln 2)/N day

−1

9. (a) T = 5ln50

∼

19.56 s (b) 718.34 m

10. (a) dv/dt = 9.8, v(0) = 0 (b) T =

300/4.9

∼

7.82 s (c) v

∼

76.68 m/s

11. (b) v = 49 tanh(t/5) m/s (e) x = 245 ln cosh(t/5) m (f) T

∼

9.48 s

12. (a) r

∼

0.02828 day

−1

(b) Q(t) = 100e

−0.02828t

∼

24.5d

14. 1620 ln(4/3)/ ln 2

∼

672.4 yr

15. (a) u = T + (u

− T)e

−kt

(b) kτ = ln 2

16. 6.69 h

17. (a) Q(t) = CV(1 − e

−t/RC

) (b) Q(t) → CV = Q

)/RC]

18. (a) Q

′

= 3(1 − 10

−4

Q), Q(0) = 0

(b) Q(t) = 10

(1 − e

−3t/10

), t in h; after 1 year Q

∼

9277.77 g

′

=−3Q/10

, Q(0) = 9277.77

(d) Q(t) = 9277.77e

−3t/10

, t in h; after 1 year Q

∼

670.07 g (e) T

∼

2.60 yr

19. (a) q

′

=−q/300, q(0) = 5000 g (b) q(t) = 5000e

−t/300

(d) T = 300 ln(25/6)

∼

428.13 min

∼

7.136 h (e) r = 250 ln(25/6)

∼

356.78 gal/min

Section 1.3, page 24

1. Second order, linear 2. Second order, nonlinear

3. Fourth order, linear 4. First order, nonlinear

5. Second order, nonlinear 6. Third order, linear

15. r =−2 16. r =±1

17. r = 2, −3 18. r = 0, 1, 2

19. r =−1, −2 20. r = 1, 4

21. Second order, linear 22. Second order, nonlinear

23. Fourth order, linear 24. Second order, nonlinear

CHAPTER 2 Section 2.1, page 39

1. (c) y = ce

−3t

+ (t/3) − (1/9) + e

−2t

; y is asymptotic to t/3 − 1/9 as t →∞

2. (c) y = ce

+ t

/3; y →∞as t →∞

3. (c) y = ce

−t

+ 1 + t

−t

/2; y → 1 as t →∞

4. (c) y = (c/t) + (3 cos 2t)/4 t + (3 sin 2t)/2; y is asymptotic to (3 sin 2t)/2 as t →∞

5. (c) y = ce

− 3e

; y →∞or −∞ as t →∞

6. (c) y = (c − t cos t + sin t)/t

; y → 0 as t →∞

7. (c) y = t

−t

+ ce

−t

; y → 0 as t →∞

August 2, 2012 13:11 bansw Sheet number 3 Page number 741 cyan black

Answers to Pr oblems 741

8. (c) y = (arctan t + c)/(1 + t

)

; y → 0 as t →∞

9. (c) y = ce

−t/2

+ 3t − 6; y is asymptotic to 3t − 6 as t →∞

10. (c) y =−te

−t

+ ct; y →∞,0,or−∞ as t →∞

11. (c) y = ce

−t

+ sin 2t − 2 cos 2t; y is asymptotic to sin 2t − 2 cos 2t as t →∞

12. (c) y = ce

−t/2

+ 3t

− 12t + 24; y is asymptotic to 3t

− 12t + 24 as t →∞

13. y = 3e

+ 2(t − 1)e

14. y = (t

− 1)e

−2t

15. y = (3t

− 4t

+ 6t

+ 1)/12t

16. y = (sin t)/t

17. y = (t + 2)e

18. y = t

−2

[(π

/4) − 1 − t cos t + sin t]

19. y =−(1 + t)e

−t

, t ̸= 0

20. y = (t − 1 + 2e

−t

)/t, t ̸= 0

21. (b) y =−

cos t +

sin t + (a +

t/2

; a

=−

22. (b) y =−3e

t/3

+ (a + 3)e

t/2

; a

=−3 (c) y →−∞for a = a

23. (b) y =[2 + a(3π + 4)e

2t/3

− 2e

−πt/2

]/(3π + 4 ); a

=−2/(3π + 4 )

24. (b) y = te

−t

+ (ea − 1)e

−t

/t; a

= 1/e (c) y → 0 as t → 0 for a = a

25. (b) y =−(cos t)/t

+ π

a/4t

; a

= 4/π

as t → 0 for a = a

26. (b) y = (e

− e + a sin 1)/ sin t; a

= (e − 1)/ sin 1 (c) y → 1 for a = a

27. (t, y) = (1.364312, 0.820082)

28. y

=−1.642876

29. (b) y = 12 +

cos 2t +

sin 2t −

788

−t/4

; y oscillates about 12 as t →∞

30. y

=−5/2

31. y

=−16/3; y →−∞as t →∞for y

=−16/3

39. See Problem 2.

40. See Problem 4.

41. See Problem 6.

42. See Problem 12.

Section 2.2, page 48

1. 3y

− 2x

= c; y ̸= 0

2. 3y

− 2ln|1 + x

|=c; x ̸=−1, y ̸= 0

3. y

−1

+ cos x = c if y ̸= 0; also y = 0; everywhere

4. 3y + y

− x

+ x = c; y ̸=−3/2

5. 2 tan 2y − 2x − sin 2x = c if cos 2y ̸= 0; also y =±(2n + 1)π/4 for any integer n;

everywhere

6. y = sin[ln |x|+c] if x ̸= 0 and |y| < 1; also y =±1

7. y

− x

+ 2(e

− e

−x

) = c; y + e

̸= 0

8. 3y + y

− x

= c; everywhere

9. (a) y = 1/(x

− x − 6) (c) −2 < x < 3

10. (a) y =−

√

2x − 2x

+ 4 (c) −1 < x < 2

11. (a) y =[2(1 − x)e

− 1]

1/2

12. (a) r = 2/(1 − 2lnθ) (c) 0 <θ<

√

13. (a) y =−[2ln(1 + x

) + 4]

1/2

14. (a) y =

;

3 − 2

√

1 + x

−1/2

√

15. (a) y =−

√

− 15 (c) x >

√

16. (a) y =−

+ 1)/2 (c) −∞ < x < ∞

17. (a) y = 5/2 −

− e

+ 13/4 (c) −1.4445 < x < 4.6297 approximately

August 2, 2012 13:11 bansw Sheet number 4 Page number 742 cyan black

742 Answers to Problems

18. (a) y =−

√

65 − 8e

− 8e

−x

19. (a) y =[π − arcsin (3 cos

x)]/3 (c) |x − π/2| < 0 .6155

20. (a) y =

;

(arcsin x)

+ 1

1/3

21. y

− 3y

− x − x

+ 2 = 0, |x| < 1

22. y

− 4y − x

=−1, |x

− 1| < 16/3

√

3 or −1.28 < x < 1.60

23. y =−1/(x

/2 + 2x − 1); x =−2

24. y =−3/2 +

2x − e

+ 13/4; x = ln 2

25. y =−3/2 +

sin 2x + 1/4; x = π/4 26. y = tan(x

+ 2x); x =−1

27. (a) y → 4 if y

> 0; y = 0 if y

= 0; y →−∞if y

< 0 (b) T = 3.29527

28. (a) y → 4 as t →∞ (b) T = 2.84367 (c) 3.6622 < y

< 4.4042

29. x =

y +

ad − bc

ln |ay + b|+k; a ̸= 0, ay + b ̸= 0

30. (e) |y + 2x|

|y − 2x|=c

31. (b) arctan (y/x) − ln |x|=c

32. (b) x

+ y

− cx

= 0 33. (b) |y − x|=c|y + 3x|

; also y = x

34. (b) |y + x||y + 4x|

= c

35. (b) 2x/(x + y) + ln |x + y|=c; also y =−x

36. (b) x/(x + y) + ln |x|=c; also y =−x

37. (b) |x|

− 5y

|=c

38. (b) c|x|

=|y

− x

Section 2.3, page 60

1. t = 100 ln 100 min

∼

460.5 min 2. Q(t) = 120γ[1 − exp(−t/60)]; 120γ

3. Q = 50e

−0.2

(1 − e

−0.2

) lb

∼

7.42 lb

4. Q(t) = 200 + t −[100(200)

/(200 + t)

] lb, t < 300; c = 121/125 lb/gal;

lim

t→∞

c = 1 lb/gal

5. (a) Q(t) =

63,150

2501

−t/50

+ 25 −

625

2501

cos t +

5002

sin t

√

2501/5002

∼

0.24995

6. (c) 130.41 s

7. (a) (ln 2)/r yr (b) 9.90 yr (c) 8.66%

8. (a) k(e

− 1)/r (b) k

∼

$3930 (c) 9.77%

9. k = $3086.64/yr; $1259.92

10. (a) $209,641.74 (20 years); $250,410.33 (30 years)

(b) $150,358.26 (20 years); $289,589.67 (30 years)

11. (a) $1788.77 (20 years); $1497.54 (30 years)

(b) $179,303.83 (20 years); $289,115.13 (30 years)

12. (a) t

∼

146.54 months (b) $246,758.02

13. (a) 0.00012097 yr

−1

(b) Q

exp(−0.00012097t), t in yr (c) 13,305 yr

14. (a) τ

∼

2.9632; no (b) τ = 10 ln 2

∼

6.9315 (c) τ = 6.3805

15. (b) y

∼

0.83

16. t = ln

/ ln

min

∼

6.07 min

17. (a) u(t) = 2000/(1 + 0.048t)

1/3

∼

750.77 s

18. (a) u(t) = ce

−kt

+ T

+ kT

(k cos ωt + ω sinωt)/(k

+ ω

)

(b) R

∼

9.11

◦

F; τ

∼

3.51 h

√

+ ω

; τ = (1/ω) arctan (ω/k)

19. (a) c = k + (P/r) +[c

− k − (P/r)]e

−rt/V

; lim

t→∞

c = k + (P/r)

(b) T = (V ln 2)/r; T = (V ln 10)/r

20. (a) 50.408 m (b) 5.248 s

21. (a) 45.783 m (b) 5.129 s

August 2, 2012 13:11 bansw Sheet number 5 Page number 743 cyan black

Answers to Pr oblems 743

22. (a) 48.562 m (b) 5.194 s

23. (a) 176.7 ft/s (b) 1074.5 ft (c) 15 ft/s (d) 256.6 s

24. (a) d v/dx =−µv (b) µ = (66/25) ln 10 mi

−1

∼

6.0788 mi

−1

∼

35.533 s

25. (a) x

=−

1 +

(

; t

1 +

(

26. (a) v =−(mg/k) +[v

+ (mg/k)]exp(−kt/m) (b) v = v

− gt; yes

27. (a) v

= 2a

g(ρ − ρ

′

)/9µ (b) e = 4πa

g(ρ − ρ

′

)/3E

28. (a) 11.58 m/s (b) 13.45 m (c) k ≥ 0 .2394 kg/s

29. (a) v = R

2g/(R + x) (b) 50.6 h

30. (b) x = ut cos A, y =−gt

/2 + ut sin A + h

(d) −16L

/(u

cos

A) + L tan A + 3 ≥ H

(e) 0.63 rad ≤ A ≤ 0.96 rad (f) u = 106.89 ft/s, A = 0.7954 rad

31. (a) v = (u cos A)e

−rt

, w =−g/r + (u sin A + g/r)e

−rt

(b) x = u cos A(1 − e

−rt

)/r, y =−gt/r + (u sin A + g/r)(1 − e

−rt

)/r + h

(d) u = 145.3 ft/s, A = 0.644 rad

32. (d) k = 2.193

Section 2.4, page 76

1. 0 < t < 3 2. 0 < t < 4

3. π/2 < t < 3π/2 4. −∞ < t < −2

5. −2 < t < 2 6. 1 < t <π

7. 2t + 5y > 0 or 2t + 5y < 0 8. t

+ y

< 1

9. 1 − t

+ y

> 0 or 1 − t

+ y

< 0, t ̸= 0, y ̸= 0

10. Everywhere

11. y ̸= 0, y ̸= 3

12. t ̸= nπ for n = 0, ±1, ±2, ...; y ̸=−1 13. y =±

− 4t

if y

̸= 0; |t| < |y

|/2

14. y =[(1/y

) − t

]

−1

if y

̸= 0; y = 0 if y

= 0;

interval is |t| < 1/

√

if y

> 0; −∞ < t < ∞ if y

≤ 0

15. y = y

2ty

+ 1 if y

̸= 0; y = 0 if y

= 0;

interval is −1/2y

< t < ∞ if y

̸= 0; −∞ < t < ∞ if y

= 0

16. y =±

ln(1 + t

) + y

; −[1 − exp(−3y

/2)]

1/3

< t < ∞

17. y → 3 if y

> 0; y = 0 if y

= 0; y →−∞if y

< 0

18. y →−∞if y

< 0; y → 0 if y

≥ 0

19. y → 0 if y

≤ 9; y →∞if y

> 9

20. y →−∞if y

< y

≈−0.019; otherwise y is asymptotic to

√

t − 1

21. (a) No (b) Yes; set t

= 1/2 in Eq. (19) in text. (c) |y|≤(4/3)

3/2

∼

1.5396

22. (a) y

(t) is a solution for t ≥ 2; y

(t) is a solution for all t

(b) f

is not continuous at (2, −1)

26. (a) y = c

µ(t)

µ(s)g(s) ds

28. y =±[5t/(2 + 5ct

)]

1/2

29. y = r/(k + cre

−rt

)

30. y =±[ϵ/(σ + cϵe

−2ϵt

)]

1/2

31. y =±

µ(t)

µ(s) ds + c

1/2

, where µ(t) = exp(22 sin t + 2Tt)

32. y =

(1 − e

−2t

) for 0 ≤ t ≤ 1; y =

− 1)e

−2t

for t > 1

33. y = e

−2t

for 0 ≤ t ≤ 1; y = e

−(t+1)

for t > 1

August 2, 2012 13:11 bansw Sheet number 6 Page number 744 cyan black

744 Answers to Problems

Section 2.5, page 88

1. y = 0 is unstable

2. y =−a/b is asymptotically stable, y = 0 is unstable

3. y = 1 is asymptotically stable, y = 0 and y = 2 are unstable

4. y = 0 is unstable

5. y = 0 is asymptotically stable

6. y = 0 is asymptotically stable

7. (c) y =[y

+ (1 − y

)kt]/[1 + (1 − y

)kt]

8. y = 1 is semistable

9. y =−1 is asymptotically stable, y = 0 is semistable, y = 1 is unstable

10. y =−1 and y = 1 are asymptotically stable, y = 0 is unstable

11. y = 0 is asymptotically stable, y = b

is unstable

12. y = 2 is asymptotically stable, y = 0 is semistable, y =−2 is unstable

13. y = 0 and y = 1 are semistable

15. (a) τ = (1/r) ln 4; 55.452 yr (b) T = (1/r) ln[β(1 − α)/(1 − β)α]; 175.78 yr

16. (a) y = 0 is unstable, y = K is asymptotically stable

(b) Concave up for 0 < y ≤ K/e, concave down for K/e ≤ y < K

17. (a) y = K exp{[ln(y

/K)]e

−rt

} (b) y(2)

∼

0.7153K

∼

57.6 × 10

∼

2.215 yr

18. (b) (h/a)

k/απ; yes (c) k/α ≤ πa

19. (b) k

/2g(αa)

20. (c) Y = Ey

= KE[1 − (E/r)] (d) Y

= Kr/4 for E = r/2

21. (a) y

1,2

= K[1 ∓

1 − (4h/rK)]/2

22. (a) y = 0 is unstable, y = 1 is asymptotically stable

(b) y = y

/[y

+ (1 − y

−αt

]

23. (a) y = y

−βt

(b) x = x

exp[−αy

(1 − e

−βt

)/β] (c) x

exp(−αy

/β)

24. (b) z = 1/[ν + (1 − ν)e

βt

] (c) 0.0927

25. (a,b) a = 0: y = 0 is semistable.

a > 0: y =

√

a is asymptotically stable and y =−

√

a is unstable.

26. (a) a ≤ 0: y = 0 is asymptotically stable.

a > 0: y = 0 is unstable; y =

√

a and y =−

√

a are asymptotically stable.

27. (a) a < 0: y = 0 is asymptotically stable and y = a is unstable.

a = 0: y = 0 is semistable.

a > 0: y = 0 is unstable and y = a is asymptotically stable.

28. (a) lim

t→∞

x(t) = min(p, q); x(t) =

pq[e

α(q−p)t

− 1]

α(q−p)t

− p

(b) lim

t→∞

x(t) = p; x(t) =

αt

pαt + 1

Section 2.6, page 101

1. x

+ 3x + y

− 2y = c 2. Not exact

3. x

− x

y + 2x + 2y

+ 3y = c 4. x

+ 2xy = c

5. ax

+ 2bxy + cy

= k 6. Not exact

7. e

sin y + 2y cos x = c; also y = 0 8. Not exact

9. e

cos 2x + x

− 3y = c 10. y ln x + 3x

− 2y = c

11. Not exact 12. x

+ y

= c

13. y =

;

x +

√

28 − 3x

/2, |x| <

28/3

14. y =[x − (24x

+ x

− 8x − 16)

1/2

]/4, x > 0.9846

August 2, 2012 13:11 bansw Sheet number 7 Page number 745 cyan black

Answers to Pr oblems 745

15. b = 3; x

+ 2x

y = c 16. b = 1; e

2xy

+ x

= c

19. x

+ 2ln|y|−y

−2

= c; also y = 0 20. e

sin y + 2y cos x = c

21. xy

− (y

− 2y + 2)e

= c 22. x

sin y = c

24. µ(t) = exp

R(t) dt, where t = xy 25. µ(x) = e

; (3x

y + y

= c

26. µ(x) = e

−x

; y = ce

+ 1 + e

27. µ(y) = y; xy + y cos y − sin y = c

28. µ(y) = e

/y; xe

− ln |y|=c; also y = 0

29. µ(y) = sin y; e

sin y + y

= c

30. µ(y) = y

; x

+ 3xy + y

= c

31. µ(x, y) = xy; x

y + 3x

+ y

= c

Section 2.7, page 110

1. (a) 1.2, 1.39, 1.571, 1.7439 (b) 1.1975, 1.38549, 1.56491, 1.73658

2. (a) 1.1, 1.22, 1.364, 1.5368 (b) 1.105, 1.23205, 1.38578, 1.57179

3. (a) 1.25, 1.54, 1.878, 2.2736 (b) 1.26, 1.5641, 1.92156, 2.34359

4. (a) 0.3, 0.538501, 0.724821, 0.866458 (b) 0.284813, 0.513339, 0.693451, 0.831571

5. Converge for y ≥ 0; undeﬁned for y < 0 6. Converge for y ≥ 0; diverge for y < 0

7. Converge

8. Converge for |y(0)| < 2.37 (approximately); diverge otherwise

9. Diverge

10. Diverge

11. (a) 2.30800, 2.49006, 2.60023, 2.66773, 2.70939, 2.73521

(b) 2.30167, 2.48263, 2.59352, 2.66227, 2.70519, 2.73209

(d) 2.29686, 2.47691, 2.58830, 2.65798, 2.70185, 2.72959

12. (a) 1.70308, 3.06605, 2.44030, 1.77204, 1.37348, 1.11925

(b) 1.79548, 3.06051, 2.43292, 1.77807, 1.37795, 1.12191

(d) 1.87734, 3.05607, 2.42672, 1.78224, 1.38150, 1.12411

13. (a) −1.48849, −0.412339, 1.04687, 1.43176, 1.54438, 1.51971

(b) −1.46909, −0.287883, 1.05351,

1.42003, 1.53000, 1.50549

(d) −1.45212, −0.173376, 1.05941, 1.41197, 1.51949, 1.49490

14. (a) 0.950517, 0.687550, 0.369188, 0.145990, 0.0421429, 0.00872877

(b) 0.938298, 0.672145, 0.362640, 0.147659, 0.0454100, 0.0104931

(d) 0.928649, 0.660463, 0.357783, 0.148848, 0.0478492, 0.0118978

15. (a) −0.166134, −0.410872, −0.804660, 4.15867

(b) −0.174652, −0.434238, −0.889140, −3.09810

16. A reasonable estimate for y at t = 0.8 is between 5.5 and 6. No reliable estimate is possible

at t = 1 from the speciﬁed data.

17. A reasonable estimate for y at t = 2.5 is between 18 and 19. No reliable estimate is possible

at t = 3 from the speciﬁed data.

18. (b) 2.37 <α

< 2.38 19. (b) 0.67 <α

< 0.68

August 2, 2012 13:11 bansw Sheet number 8 Page number 746 cyan black

746 Answers to Problems

Section 2.8, page 120

1. dw/ds = (s + 1)

+ (w + 2)

, w(0) = 0 2. dw/ds = 1 − (w + 3)

, w(0) = 0

3. (a) φ

(t) =

k=1

n→∞

(t) = e

− 1

4. (a) φ

(t) =

k=1

(−1)

n→∞

(t) = e

−t

− 1

5. (a) φ

(t) =

k=1

(−1)

k+1

/(k + 1)!2

k−1

n→∞

(t) = 4e

−t/2

+ 2t − 4

6. (a) φ

(t) = t −

n+1

(n + 1)!

n→∞

(t) = t

7. (a) φ

(t) =

k=1

2k−1

1 · 3 · 5 ···(2k − 1)

8. (a) φ

(t) =−

k=1

3k−1

2 · 5 · 8 ···(3k − 1)

9. (a) φ

(t) =

; φ

(t) =

7 · 9

; φ

(t) =

7 · 9

3 · 7 · 9 · 11

(7 · 9)

· 15

10. (a) φ

(t) = t; φ

(t) = t −

; φ

(t) = t −

4 · 7

−

16 · 10

64 · 13

11. (a) φ

(t) = t, φ

(t) = t −

−

+ O(t

(t) = t −

−

14t

+ O(t

(t) = t −

−

31t

+ O(t

)

12. (a) φ

(t) =−t − t

−

(t) =−t −

−

+ O(t

(t) =−t −

−

+ O(t

(t) =−t −

−

+ O(t

)

Section 2.9, page 131

1. y

= (−1)

(0.9)

; y

→ 0 as n →∞

2. y

= y

/(n + 1); y

→ 0 as n →∞

3. y

= y

(n + 2)(n + 1)/2; y

→∞as n →∞

4. y

,ifn = 4k or n = 4k − 1;

−y

,ifn = 4k − 2 or n = 4k −3;

has no limit as n →∞

5. y

= (0.5)

− 12) + 12; y

→ 12 as n →∞

6. y

= (−1)

(0.5)

− 4) + 4; y

→ 4 as n →∞

7. 7.25%

8. $2283.63

9. $258.14

10. (a) $804.62 (b) $877.57 (c) $1028.61

11. 30 yrs: $804.62/mo; $289,663.20 total 20 yrs: $899.73/mo; $215,935.20 total

12. $103,624.62

13. 9.73%

16. (b) u

→−∞as n →∞

19. (a) 4.7263 (b) 1.223% (c) 3.5643 (e) 3.5699

August 2, 2012 13:11 bansw Sheet number 9 Page number 747 cyan black

Answers to Pr oblems 747

Miscellaneous Problems, page 133

1. y = (c/x

) + (x

/5) 2. 2y + cos y − x − sin x = c

3. x

+ xy − 3y − y

= 0 4. y =−3 + ce

x−x

5. x

y + xy

+ x = c 6. y = x

−1

(1 − e

1−x

)

7. x

+ x − y

− y

= c 8. y = (4 + cos 2 − cos x)/x

9. x

y + x + y

= c 10. x + ln |x|+x

−1

+ y − 2ln|y|=c; also y = 0

11. (x

/3) + xy + e

= c 12. y = ce

−x

+ e

−x

ln(1 + e

)

13. y = tan(x + x

+ c) 14. x

+ 2xy + 2y

= 34

15. y = c/cosh

(x/2) 16. e

−x

cos y + e

sin x = c

17. y = ce

− e

18. y = e

−2x

−s

ds + 3e

−2x

19. 2xy + xy

− x

= c 20. e

+ e

−y

= c

21. 2xy

+ 3x

y − 4x + y

= c 22. y

+ 3y − x

+ 3x = 2

23. y =

+ c

−t

24. sin y sin

x = c

25. (x

/y) + arctan(y/x) = c 26. e

−y/x

+ ln |x|=c

27. (x

+ y

+ 1)e

−y

= c 28. x

+ x

y = c

29. arctan(y/x) − ln

+ y

= c 30. (y

) + (y/x

) = c

31. x

+ xy

=−4 32.

=−x

ds +

34. (a) y = t + (c − t)

−1

(b) y = t

−1

+ 2t(c − t

)

−1

sin t)

−1

35. (a) v

′

−[x(t) + b]v = b

(b) v =

;

µ(t) dt + c

/µ(t), µ(t) = exp[−(at

/2) − bt]

36. y = c

−1

+ c

+ ln t

37. y = c

ln t + c

+ t

38. y = (1/k) ln |(k − t)/(k + t)|+c

if c

= k

> 0; y = (2/k) arctan(t/k) + c

=−k

< 0; y =−2t

−1

+ c

if c

= 0; also y = c

39. y =±

(t − 2c

)

√

t + c

+ c

; also y = c Hint: µ(v) = v

−3

is an integrating factor.

40. y = c

−t

+ c

− te

−t

41. c

y = c

t − ln |1 + c

t|+c

if c

̸= 0; y =

+ c

if c

= 0; also y = c

42. y

= c

t + c

43. y = c

sin(t + c

) = k

sin t + k

cos t

44.

− 2c

y + c

= 2t; also y = c 45. t + c

=±

(y − 2c

)(y + c

)

1/2

46. y ln |y|−y + c

y + t = c

; also y = c 47. e

= (t + c

)

+ c

48. y =

(t + 1)

3/2

−

49. y = 2(1 − t)

−2

50. y = 3lnt −

ln(t

+ 1) − 5 arctan t + 2 +

ln 2 +

51. y =

CHAPTER 3 Section 3.1, page 144

1. y = c

+ c

−3t

2. y = c

−t

+ c

−2t

3. y = c

t/2

+ c

−t/3

4. y = c

t/2

+ c

5. y = c

+ c

−5t

6. y = c

3t/2

+ c

−3t/2

7. y = c

exp[(9 + 3

√

5)t/2]+c

exp[(9 − 3

√

5)t/2]

8. y = c

exp[(1 +

√

3)t]+c

exp[(1 −

√

3)t]

9. y = e

; y →∞as t →∞

10. y =

−t

−

−3t

; y → 0 as t →∞

11. y = 12e

t/3

− 8e

t/2

; y →−∞as t →∞

12. y =−1 − e

−3t

; y →−1 as t →∞

13. y =

(13 + 5

√

13) exp[(−5 +

√

13)t/2]+

(13 − 5

√

13) exp[(−5 −

√

13)t/2]; y → 0 as

t →∞

August 2, 2012 13:11 bansw Sheet number 10 Page number 748 cyan black

748 Answers to Problems

14. y = (2/

√

33) exp[(−1 +

√

33)t/4]−(2/

√

33) exp[(−1 −

√

33)t/4]; y →∞as t →∞

15. y =

−9(t−1)

t−1

; y →∞as t →∞

16. y =−

(t+2)/2

−(t+2)/2

; y →−∞as t →∞

17. y

′′

+ y

′

− 6y = 0

18. 2y

′′

+ 5y

′

+ 2y = 0

19. y =

+ e

−t

; minimum is y = 1 at t = ln 2

20. y =−e

+ 3e

t/2

; maximum is y =

at t = ln(9/4 ), y = 0 at t = ln 9

21. α =−2

22. β =−1

23. y → 0 for α<0; y becomes unbounded for α>1

24. y → 0 for α<1; there is no α for which all nonzero solutions become unbounded

25. (a) y =

(1 + 2β)e

−2t

(4 − 2β)e

t/2

(b) y

∼

0.71548 when t =

ln 6

∼

0.71670

26. (a) y = (6 + β)e

−2t

− (4 + β)e

−3t

(b) t

= ln[(12 + 3β)/(12 + 2β)], y

(6 + β)

/(4 + β)

√

∼

16.3923 (d) t

→ ln(3/2), y

→∞

27. (a) y = d/c (b) aY

′′

+ bY

′

+ cY = 0

28. (a) b > 0 and 0 < c < b

/4a (b) c < 0 (c) b < 0 and 0 < c < b

/4a

Section 3.2, page 155

1. −

t/2

2. 1

3. e

−4t

4. x

5. −e

6. 0

7. 0 < t < ∞ 8. −∞ < t < 1

9. 0 < t < 4 10. 0 < t < ∞

11. 0 < x < 3 12. 2 < x < 3π/2

14. The equation is nonlinear. 15. The equation is nonhomogeneous.

16. No 17. 3te

+ ce

18. te

+ ct 19. 5W(f , g)

20. −4(t cos t − sin t)

21. y

and y

are a fundamental set of solutions if and only if a

− a

̸= 0.

22. y

(t) =

−2t

, y

(t) =−

−2t

23. y

(t) =−

−3(t−1)

−(t−1)

, y

(t) =−

−3(t−1)

−(t−1)

24. Yes

25. Yes

26. Yes 27. Yes

28. (b) Yes (c) [y

(t), y

(t)] and [y

(t), y

(t)] are fundamental sets of solutions;

(t), y

(t)] and [y

(t), y

(t)] are not

29. ct

30. c cos t

31. c/x 32. c/(1 − x

)

34. 2/25 35. 3

√

∼

4.946

36. p(t) = 0 for all t

40. If t

is an inﬂection point, and y = φ(t) is a solution, then from the differential equation

p(t

)φ

′

) + q(t

)φ(t

) = 0.

42. Yes, y = c

−x

dt + c

−x

43. No

44. Yes, y =

µ(x)

µ(t)

dt + c

, µ(x) = exp

−

cos x

(

45. Yes, y = c

−1

+ c

47. x

′′

+ 3xµ

′

+ (1 + x

− ν

)µ = 0

48. (1 − x

)µ

′′

− 2xµ

′

+ α(α + 1)µ = 0 49. µ

′′

− xµ = 0

51. The Legendre and Airy equations are self-adjoint.

August 2, 2012 13:11 bansw Sheet number 11 Page number 749 cyan black

Answers to Pr oblems 749

Section 3.3, page 164

1. e cos 2 + ie sin 2

∼

−1.1312 + 2.4717i 2. e

cos 3 − ie

sin 3

∼

−7.3151 − 1.0427i

3. −1

4. e

cos(π/2) − ie

sin(π/2) =−e

∼

−7.3891i

5. 2 cos(ln 2) − 2i sin(ln 2)

∼

1.5385 − 1.2779i

6. π

−1

cos(2lnπ) + iπ

−1

sin(2lnπ)

∼

−0.20957 + 0.23959i

7. y = c

cos t + c

sin t

8. y = c

cos

√

5 t + c

sin

√

5 t

9. y = c

+ c

−4t

10. y = c

−t

cos t + c

−t

sin t

11. y = c

−3t

cos 2t + c

−3t

sin 2t 12. y = c

cos(3t/2) + c

sin(3t/2)

13. y = c

−t

cos(t/2) + c

−t

sin(t/2) 14. y = c

t/3

+ c

−4t/3

15. y = c

−t/2

cos t + c

−t/2

sin t 16. y = c

−2t

cos(3t/2) + c

−2t

sin(3t/2)

17. y =

sin 2t; steady oscillation

18. y = e

−2t

cos t + 2e

−2t

sin t; decaying oscillation

19. y =−e

t−π/2

sin 2t; growing oscillation

20. y = (1 + 2

√

3) cos t − (2 −

√

3) sin t; steady oscillation

21. y = 3e

−t/2

cos t +

−t/2

sin t; decaying oscillation

22. y =

√

2 e

−(t−π/4)

cos t +

√

2 e

−(t−π/4)

sin t; decaying oscillation

23. (a) u = 2e

t/6

cos(

√

23 t/6) − (2/

√

23)e

t/6

sin(

√

23 t/6) (b) t = 10.7598

24. (a) u = 2e

−t/5

cos(

√

34 t/5) + (7/

√

34) e

−t/5

sin(

√

34 t/5) (b) T = 14.5115

25. (a) y = 2e

−t

cos

√

5 t +[(α + 2)/

√

5]e

−t

sin

√

5 t (b) α = 1.50878

√

5/(2 + α)]}/

√

5 (d) π/

√

26. (a) y = e

−at

cos t + ae

−at

sin t (b) T = 1.8763

, T = 7.4284; α =

, T = 4.3003; α = 2, T = 1.5116

35. y = c

cos(ln t) + c

sin(ln t)

36. y = c

−1

+ c

−2

37. y = c

−1

cos(

ln t) + c

−1

sin(

ln t) 38. y = c

+ c

−1

39. y = c

+ c

40. y = c

t cos(2lnt) + c

t sin(2lnt)

41. y = c

t + c

−3

42. y = c

−3

cos(ln t) + c

−3

sin(ln t)

44. Yes, y = c

cos x + c

sin x, x =

−t

45. No

46. Yes, y = c

−t

cos(

√

3 t

/4) + c

−t

sin(

√

3 t

/4)

Section 3.4, page 172

1. y = c

+ c

2. y = c

−t/3

+ c

−t/3

3. y = c

−t/2

+ c

3t/2

4. y = c

−3t/2

+ c

−3t/2

5. y = c

cos 3t + c

sin 3t 6. y = c

+ c

7. y = c

−t/4

+ c

−4t

8. y = c

−3t/4

+ c

−3t/4

9. y = c

2t/5

+ c

2t/5

10. y = e

−t/2

cos(t/2) + c

−t/2

sin(t/2)

11. y = 2e

2t/3

−

2t/3

, y →−∞as t →∞

12. y = 2te

, y →∞as t →∞

13. y =−e

−t/3

cos 3t +

−t/3

sin 3t, y → 0 as t →∞

14. y = 7e

−2(t+1)

+ 5te

−2(t+1)

, y → 0 as t →∞

15. (a) y = e

−3t/2

−

−3t/2

(b) t =

= 16/15, y

=−

−8/5

∼

−0.33649

(d) y = e

−3t/2

+ (b +

)te

−3t/2

; b =−

16. y = 2e

t/2

+ (b − 1)te

t/2

; b = 1

17. (a) y = e

−t/2

(b) t

, y

= 5e

−4/5

∼

2.24664

−t/2

+ (b +

)te

−t/2

(d) t

= 4b/(1 + 2b) → 2 as b →∞; y

= (1 + 2b) exp[−2b/(1 + 2b)]→∞

as b →∞

August 2, 2012 13:11 bansw Sheet number 12 Page number 750 cyan black

750 Answers to Problems

18. (a) y = ae

−2t/3

+ (

a − 1)te

−2t/3

(b) a =

23. y

(t) = t

24. y

(t) = t

−2

25. y

(t) = t

−1

ln t 26. y

(t) = te

27. y

(x) = cos x

28. y

(x) = x

29. y

(x) = x

1/4

−2

√

30. y

(x) = x

−1/2

cos x

31. y = c

−δx

δs

ds + c

−δx

32. y

(t) = y

(t)

−2

(s) exp

−

p(r) dr

33. y

(t) = t

−1

ln t 34. y

(t) = cos t

35. y

(x) = x 36. y

(x) = x

−1/2

cos x

38. (b) y

+ (a/b)y

′

40. y = c

+ c

ln t

41. y = c

−1/2

+ c

−1/2

ln t 42. y = c

t + c

5/2

43. y = c

−1

+ c

−1

ln t 44. y = c

3/2

+ c

3/2

ln t

45. y = c

−2

cos(3lnt) + c

−2

sin(3lnt)

Section 3.5, page 184

1. y = c

+ c

−t

− e

2. y = c

−t

cos 2t + c

−t

sin 2t +

sin 2t −

cos 2t

3. y = c

−t

+ c

−

+ 3t − 2t

4. y = c

+ c

−3t

+ 2e

− 3e

−2t

5. y = c

+ c

−t

6. y = c

+ c

−2t

t −

sin 2t −

cos 2t

7. y = c

cos 3t + c

sin 3t +

162

(9t

− 6t + 1)e

8. y = c

−t

+ c

−t

+ t

−t

9. y = c

−t

+ c

−t/2

+ t

− 6t + 14 −

sin t −

cos t

10. y = c

cos t + c

sin t −

t cos 2t −

sin 2t

11. u = c

cos ω

t + c

sin ω

t + (ω

− ω

)

−1

cos ωt

12. u = c

cos ω

t + c

sin ω

t + (1/2ω

)t sin ω

13. y = c

−t/2

cos(

√

15 t/2) + c

−t/2

sin(

√

15 t/2) +

−

−t

14. y = c

−t

+ c

−2t

15. y = e

−

−2t

− t −

16. y =

sin 2t −

cos 2t +

−

17. y = 4te

− 3e

+ 4

18. y = e

−t

−

− te

19. y = 2 cos 2t −

sin 2t −

t cos 2t

20. y = e

−t

cos 2t +

−t

sin 2t + te

−t

sin 2t

21. (a) Y(t) = t(A

+ A

t + A

) + t(B

+ B

t + B

−3t

+ D sin 3t + E cos 3t

(b) A

= 2/15, A

=−2/9, A

= 8/27, A

=−8/27, A

= 16/81, B

=−1/9,

=−1/9, B

=−2/27, D =−1/18, E =−1/18

22. (a) Y(t) = A

t + A

+ t(B

t + B

) sin t + t(D

t + D

) cos t

(b) A

= 1, A

= 0, B

= 1/4, D

=−1/4, D

= 0

23. (a) Y(t) = e

(A cos 2t + B sin 2t) + (D

t + D

sin t + (E

t + E

cos t

(b) A =−1/20, B =−3/20, D

=−3/2, D

=−5, E

= 3/2, E

= 1/2

24. (a) Y(t) = Ae

−t

+ t(B

+ B

t + B

−t

cos t + t(D

+ D

t + D

−t

sin t

(b) A = 3, B

=−2/3, B

= 0, B

= 1, D

= 0, D

= 1, D

= 1

25. (a) Y(t) = A

+ A

t + A

+ t

t + B

+ (D

t + D

) sin 2t + (E

t + E

) cos 2t

(b) A

= 1/2, A

= 1, A

= 3/4, B

= 2/3, B

= 0, D

=−1/16,

= 1/8, E

= 1/16

26. (a) Y(t) = t(A

+ A

t + A

) sin 2t + t(B

+ B

t + B

) cos 2t

(b) A

= 0, A

= 13/16, A

= 7/4, B

=−1/12, B

= 0, B

= 13/32

27. (a) Y(t) = (A

+ A

t + A

sin 2t + (B

+ B

t + B

cos 2t +

−t

(D cos t + E sin t) + Fe

(b) A

= 1/52, A

= 10/169, A

=−1233/35,152, B

=−5/52, B

= 73/676,

=−4105/35,152, D =−3/2, E = 3/2, F = 2/3

August 2, 2012 13:11 bansw Sheet number 13 Page number 751 cyan black

Answers to Pr oblems 751

28. (a) Y(t) = t(A

t + A

−t

cos 2t + t(B

t + B

−t

sin 2t + (D

t + D

−2t

cos t +

t + E

−2t

sin t

(b) A

= 0, A

= 3/16, B

= 3/8, B

= 0, D

=−2/5, D

=−7/25, E

= 1/5,

= 1/25

29. (b) w =−

+ c

30. y = c

cos λt + c

sin λt +

m=1

/(λ

− m

)]sin mπt

31. y =

t,0≤ t ≤ π

−(1 + π/2) sin t − (π/2) cos t + (π/2)e

π−t

, t >π

32. y =

−

−t

sin 2t −

−t

cos 2t,0≤ t ≤ π/2

−

(1 + e

π/2

−t

cos 2t −

(1 + e

π/2

−t

sin 2t, t >π/2

33. No

36. y = c

+ c

−t

−

Section 3.6, page 190

1. Y(t) = e

2. Y(t) =−

−t

3. Y(t) =

−t

4. Y(t) = 2t

t/2

5. y = c

cos t + c

sin t − (cos t) ln(tan t + sec t)

6. y = c

cos 3t + c

sin 3t + (sin 3t) ln(tan 3t + sec 3t) − 1

7. y = c

−2t

+ c

−2t

− e

−2t

ln t

8. y = c

cos 2t + c

sin 2t +

(sin 2t) ln sin 2t −

t cos 2t

9. y = c

cos(t/2) + c

sin(t/2) + t sin(t/2) + 2[ln cos(t/2)]cos(t/2)

10. y = c

+ c

−

ln(1 + t

) + te

arctan t

11. y = c

+ c

3(t−s)

− e

2(t−s)

]g(s) ds

12. y = c

cos 2t + c

sin 2t +

[sin 2(t − s)]g (s ) ds

13. Y(t) =

+ t

ln t

14. Y(t) =−2t

15. Y(t) =

(t − 1)e

16. Y(t) =−

(2t − 1)e

−t

17. Y(x) =

(ln x)

18. Y(x) =−

1/2

cos x

19. Y(x) =

− te

(1 − t)

g(t) dt 20. Y(x) = x

−1/2

−3/2

sin(x − t)g(t) dt

23. (b) y = y

cos t + y

′

sin t +

sin(t − s)g(s) ds

24. y = (b − a)

−1

b(t−s)

− e

a(t−s)

]g(s) ds

25. y = µ

−1

λ(t−s)

sin µ(t − s)g(s) ds

26. y =

(t − s)e

a(t−s)

g(s) ds 29. y = c

t + c

+ 4t

ln t

30. y = c

−1

+ c

−5

t 31. y = c

(1 + t) + c

(t − 1)e

32. y = c

+ c

t −

(2t − 1)e

−t

Section 3.7, page 203

1. u = 5 cos(2t − δ), δ = arctan(4/3)

∼

0.9273

2. u = 2 cos(t − 2π/3)

3. u = 2

√

5 cos(3t − δ), δ =−arctan(1/2)

∼

−0.4636

4. u =

√

13 cos(πt − δ), δ = π + arctan(3/2)

∼

4.1244

5. u =

cos 8t ft, t in s; ω = 8 rad/s, T = π/4s, R = 1/4 ft

6. u =

sin 14t cm, t in s; t = π/14 s

August 2, 2012 13:11 bansw Sheet number 14 Page number 752 cyan black

752 Answers to Problems

7. u = (1/4

√

2) sin(8

√

2 t) −

cos(8

√

2 t) ft, t in s; ω = 8

√

2 rad/s,

T = π/4

√

2s, R =

11/288

∼

0.1954 ft, δ = π − arctan(3/

√

∼

2.0113

8. Q = 10

−6

cos 2000t C, t in s

9. u = e

−10t

[2 cos(4

√

6 t) + (5/

√

6) sin(4

√

6 t)] cm, t in s;

µ = 4

√

6 rad/s, T

= π/2

√

6s, T

/T = 7/2

√

∼

1.4289, τ

∼

0.4045 s

10. u = (1/8

√

31)e

−2t

sin(2

√

31 t) ft, t in s; t = π/2

√

31 s , τ

∼

1.5927 s

11. u

∼

0.057198e

−0.15t

cos(3.87008 t − 0.50709) m, t in s; µ = 3.87008 rad/s,

µ/ω

= 3.87008/

√

∼

0.99925

12. Q = 10

−6

(2e

−500t

− e

−1000t

) C; t in s

13. γ =

20/9

∼

1.4907

16. r =

√

+ B

, r cos θ = B, r sin θ =−A; R = r; δ = θ + (4n + 1)π/2,

n = 0, 1, 2,...

17. γ = 8 lb·s/ft 18. R = 10

20. v

< −γu

/2m 22. 2π/

√

23. γ = 5 lb·s/ft 24. k = 6, v =±2

√

25. (a) τ

∼

41.715 (d) γ

∼

1.73, min τ

∼

4.87 (e) τ = (2/γ) ln(400/

4 − γ

)

26. (a) u(t) = e

−γt/2m

;

4km − γ

cos µt + (2mv

+ γu

) sin µt

4km − γ

(b) R

= 4m(ku

+ γu

+ mv

)/(4km − γ

)

27. ρlu

′′

+ ρ

gu = 0, T = 2π

ρl/ρ

28. (a) u =

√

2 sin

√

2 t (c) clockwise

29. (a) u = (16/

√

127)e

−t/8

sin(

√

127 t/8) (c) clockwise

30. (b) u = a cos(

k/mt) + b

m/k sin(

k/mt)

32. (b) u = sin t, A = 1, T = 2π (c) A = 0.98, T = 6.07

(d) ϵ = 0.2, A = 0.96, T = 5.90; ϵ = 0.3, A = 0.94, T = 5.74

(f) ϵ =−0.1, A = 1.03, T = 6.55; ϵ =−0.2, A = 1.06, T = 6.90; ϵ =−0.3,

A = 1.11, T = 7.41

Section 3.8,

page 217

1. −2 sin 8t sin t 2. 2 sin(t/2) cos(13t/2)

3. 2 cos(3πt/2) cos(πt/2) 4. 2 sin(7t/2) cos(t/2)

5. u

′′

+ 256u = 16 cos 3t, u(0) =

, u

′

(0) = 0, u in ft, t in s

6. u

′′

+ 10u

′

+ 98u = 2 sin(t/2), u(0) = 0, u

′

(0) = 0.03, u in m, t in s

7. (a) u =

151

1482

cos 16t +

247

cos 3t (c) ω = 16 rad/s

8. (a) u =

153,281

[160e

−5t

cos(

√

73 t) +

383,443

7300

−5t

sin(

√

73 t) − 160 cos(t/2) +

3128 sin(t/2)] (b) The ﬁrst two terms are the transient. (d) ω = 4

√

3 rad/s

9. u =

(cos 7t − cos 8t) =

128

sin(t/2) sin(15t/2) ft, t in s

10. u = (cos 8t + sin 8t − 8t cos 8t)/4 ft, t in sec; 1/8, π/8, π/4, 3π/8s

11. (a)

901

(30 cos 2t + sin 2t) ft, t in s (b) m = 4 slugs

12. u = (

√

2/6) cos(3t − 3π/4) m, t in s

15. u =

⎧

⎨

⎩

(t − sin t),0≤ t ≤ π

[(2π − t) − 3 sin t], π<t ≤ 2π

−4F

sin t,2π<t < ∞

16. Q(t) = 10

−6

−4000t

− 4e

−1000t

+ 3) C, t in s, Q(0.001)

∼

1.5468 × 10

−6

;

Q(0.01)

∼

2.9998 × 10

−6

; Q(t) → 3 × 10

−6

as t →∞

17. (a) u =[32(2 − ω

) cos ωt + 8ω sin ωt]/(64 − 63ω

+ 16ω

) (b) A = 8/

√

64 − 63ω

+ 16ω

(d) ω = 3

√

14/8

∼

1.4031, A = 64/

√

127

∼

5.6791

18. (a) u = 3(cos t − cos ωt)/(ω

− 1)

19. (a) u =[(ω

+ 2) cos t − 3 cos ωt]/(ω

− 1) + sin t

August 2, 2012 13:11 bansw Sheet number 15 Page number 753 cyan black

Answers to Pr oblems 753

CHAPTER 4 Section 4.1, page 226

1. −∞ < t < ∞ 2. t > 0 or t < 0

3. t > 1, or 0 < t < 1, or t < 0 4. t > 0

5. ..., −3π/2 < x < −π/2, −π/2 < x < 1, 1 < x <π/2, π/2 < x < 3π/2, ...

6. −∞ < x < −2, −2 < x < 2, 2 < x < ∞

7. Linearly independent

8. Linearly dependent; f

(t) + 3f

(t) − 2f

(t) = 0

9. Linearly dependent; 2f

(t) + 13f

(t) − 3f

(t) − 7f

(t) = 0

10. Linearly independent

11. 1

12. 1 13. −6e

−2t

14. e

−2t

15. 6x

16. 6/x 17. sin

t =

(5) −

cos 2t

19. (a) a

[n(n − 1)(n − 2) ···1]+a

[n(n − 1) ···2]t +···+a

(b) (a

+ a

n−1

+···+a

, e

−t

, e

−2t

; yes, W(e

, e

−t

, e

−2t

) ̸= 0, −∞ < t < ∞

21. W(t) = ce

−2t

22. W(t) = c

23. W(t) = c/t

24. W(t) = c/t

27. y = c

+ c

t + c

28. y = c

+ c

(t + 1)

Section 4.2, page 233

√

2 e

i[(π/4)+2mπ]

2. 2e

i[(2π/3)+2mπ]

3. 3e

i(π+2mπ)

4. e

i[(3π/2)+2mπ]

5. 2e

i[(11π/6)+2mπ]

√

i[(5π/4)+2mπ]

7. 1,

(−1 + i

√

3),

(−1 − i

√

3) 8. 2

1/4

−πi/8

1/4

7πi/8

9. 1, i, −1, −i 10.

)

√

3 + i

√

2, −

)

√

3 + i

√

11. y = c

+ c

−t

12. y = c

+ c

13. y = c

+ c

−t

14. y = c

+ c

t + c

+ c

15. y = c

cos t + c

sin t + e

√

3t/2

cos

t + c

sin

t) + e

−

√

3t/2

cos

t + c

sin

16. y = c

+ c

−t

+ c

−2t

17. y = c

+ c

−t

+ c

−t

+ c

−t

18. y = c

+ c

t + c

+ c

−t

+ c

cos t + c

sin t

19. y = c

+ c

cos t + c

sin t

20. y = c

+ c

+ e

−t

cos

√

3 t + c

sin

√

3 t)

21. y = e

[(c

+ c

t) cos t + (c

+ c

t) sin t]+e

−t

[(c

+ c

t) cos t + (c

+ c

t) sin t]

22. y = (c

+ c

t) cos t + (c

+ c

t) sin t

23. y = c

+ c

(2+

√

5)t

+ c

(2−

√

5)t

24. y = c

−t

+ c

(−2+

√

2)t

+ c

(−2−

√

2)t

25. y = c

−t/2

+ c

−t/3

cos(t/

√

3) + c

−t/3

sin(t/

√

26. y = c

+ c

−2t

+ c

(3+

√

3)t

+ c

(3−

√

3)t

27. y = c

−t/3

+ c

−t/4

+ c

−t

cos 2t + c

−t

sin 2t

28. y = c

−t

cos t + c

−t

sin t + c

−2t

cos(

√

3 t) + c

−2t

sin(

√

3 t)

29. y = 2 − 2 cost + sin t

30. y =

−t/

√

sin(t/

√

2) −

√

sin(t/

√

31. y = 2t − 3 32. y = 2 cos t − sin t

33. y =−

−

−2t

−

−t/2

34. y =

−t

t/2

cos t +

t/2

sin t

35. y = 8 − 18e

−t/3

+ 8e

−t/2

36. y =

−t

cos t −

−t

sin t −

−2t

cos(

√

3 t) +

√

−2t

sin(

√

3 t)

August 2, 2012 13:11 bansw Sheet number 16 Page number 754 cyan black

754 Answers to Problems

37. y =

(cosh t − cos t) +

(sinh t − sin t)

38. (a) W(t) = c, a constant (b) W(t) =−8 (c) W(t) = 4

39. (b) u

= c

cos t + c

sin t + c

cos

√

6 t + c

sin

√

6 t

Section 4.3, page 239

1. y = c

+ c

−t

+ 3

2. y = c

+ c

−t

+ c

cos t + c

sin t − 3t −

t sin t

3. y = c

−t

+ c

cos t + c

sin t +

−t

+ 4(t − 1)

4. y = c

+ c

−t

+ cos t

5. y = c

+ c

t + c

−2t

+ c

−

6. y = c

cos t + c

sin t + c

t cos t + c

t sin t + 3 +

cos 2t

7. y = c

+ c

t + c

+ c

−t

+ e

t/2

cos(

√

3 t/2) + c

sin(

√

3 t/2)]+

8. y = c

+ c

t + c

+ c

−t

sin 2t +

cos 2t

9. y =

(1 − cos 2t) +

10. y = (t − 4) cost − (

t + 4) sin t + 3t + 4

11. y = 1 +

+ 3t) − te

12. y =−

cos t −

sin t +

−t

520

−3t

cos 2t −

130

sin 2t

13. Y(t) = t(A

+ A

t + A

) + Bt

14. Y(t) = t(A

t + A

−t

+ B cos t + C sin t

15. Y(t) = At

+ B cos t + C sin t

16. Y(t) = At

+ (B

t + B

+ t(C cos 2t + D sin 2t)

17. Y(t) = t(A

+ A

t + A

) + (B

t + B

) cos t + (C

t + C

) sin t

18. Y(t) = Ae

+ (B

t + B

−t

+ te

−t

(C cos t + D sin t)

19. k

= a

, k

= a

+ a

n−1

+···+a

n−1

α + a

Section 4.4, page 244

1. y = c

+ c

cos t + c

sin t − ln cos t − (sin t) ln(sec t + tan t)

2. y = c

+ c

−t

−

3. y = c

+ c

−t

+ c

4. y = c

+ c

cos t + c

sin t + ln(sec t + tan t) − t cos t + (sin t) ln cos t

5. y = c

+ c

cos t + c

sin t −

−t

cos t

6. y = c

cos t + c

sin t + c

t cos t + c

t sin t −

sin t

7. y = c

+ c

cos t + c

sin t −

(cos t) ln cos t +

(sin t) ln cos t −

t cos t

−

t sin t +

−s

/ cos s

(

8. y = c

+ c

−t

− ln sin t + ln(cos t + 1) +

−s

/ sin s

(

−t

/ sin s

(

9. c

= 0, c

= 2, c

= 1 in answer to Problem 4

10. c

= 2, c

, c

=−

, c

in answer to Problem 6

11. c

, c

=−

, t

= 0 in answer to Problem 7

12. c

= 3, c

= 0, c

=−e

π/2

, t

= π/2 in answer to Problem 8

13. Y(x) = x

/15

14. Y(t) =

t−s

− sin(t − s) − cos(t − s)]g(s) ds

15. Y(t) =

[sinh(t − s) − sin(t − s)]g(s) ds

August 2, 2012 13:11 bansw Sheet number 17 Page number 755 cyan black

Answers to Pr oblems 755

16. Y(t) =

(t−s)

(t − s)

g(s) ds; Y(t) =−te

ln |t|

17. Y(x) =

[(x/t

) − 2(x

) + (x

)]g(t) dt

CHAPTER 5 Section 5.1, page 253

1. ρ = 1 2. ρ = 2

3. ρ =∞ 4. ρ =

5. ρ =

6. ρ = 1

7. ρ = 3 8. ρ = e

∞

n=0

(−1)

2n+1

(2n + 1)!

, ρ =∞ 10.

∞

n=0

, ρ =∞

11. 1 + (x − 1), ρ =∞ 12. 1 − 2(x + 1) + (x + 1)

, ρ =∞

13.

∞

n=1

(−1)

n+1

(x − 1)

, ρ = 1 14.

∞

n=0

(−1)

, ρ = 1

15.

∞

n=0

, ρ = 1 16.

∞

n=0

(−1)

n+1

(x − 2)

, ρ = 1

17. y

′

= 1 + 2

x + 3

+ 4

+···+(n + 1)

+···

′′

= 2

+ 3

· 2x + 4

· 3x

+ 5

· 4x

+···+(n + 2)

(n + 1)x

+···

18. y

′

= a

+ 2a

x + 3a

+ 4a

+···+(n + 1)a

n+1

+···

∞

n=1

n−1

∞

n=0

(n + 1)a

n+1

′′

= 2a

+ 6a

x + 12a

+ 20a

+···+(n + 2)(n + 1)a

n+2

+···

∞

n=2

n(n − 1)a

n−2

∞

n=0

(n + 2)(n + 1)a

n+2

21.

∞

n=0

(n + 2)(n + 1)a

n+2

22.

∞

n=2

n−2

23.

∞

n=0

(n + 1)a

24.

∞

n=0

[(n + 2)(n + 1)a

n+2

− n(n − 1)a

25.

∞

n=0

[(n + 2)(n + 1)a

n+2

+ na

26. a

∞

n=1

[(n + 1)a

n+1

+ a

n−1

27.

∞

n=0

[(n + 1)na

n+1

+ a

28. a

= (−2)

/n!, n = 1, 2, ...; a

−2x

Section 5.2, page 263

1. (a) a

n+2

= a

/(n + 2)(n + 1)

(b, d) y

(x) = 1 +

+···=

∞

n=0

(2n)!

= cosh x

(x) = x +

+···=

∞

n=0

2n+1

(2n + 1)!

= sinh x

August 2, 2012 13:11 bansw Sheet number 18 Page number 756 cyan black

756 Answers to Problems

2. (a) a

n+2

= a

/(n + 2)

(b, d) y

(x) = 1 +

2 · 4

2 · 4 · 6

+···=

∞

n=0

(x) = x +

3 · 5

3 · 5 · 7

+···=

∞

n=0

n!x

2n+1

(2n + 1)!

3. (a) (n + 2)a

n+2

− a

n+1

− a

= 0

(b) y

(x) = 1 +

(x − 1)

+···

(x) = (x − 1) +

(x − 1)

+···

4. (a) a

n+4

=−k

/(n + 4)(n + 3); a

= a

= 0

(b, d) y

(x) = 1 −

3 · 4

3 · 4 · 7 · 8

−

3 · 4 · 7 · 8 · 11 · 12

+···

= 1 +

∞

m=0

(−1)

m+1

)

m+1

3 · 4 · 7 · 8 ···(4m + 3)(4m + 4)

(x) = x −

4 · 5

4 · 5 · 8 · 9

−

4 · 5 · 8 · 9 · 12 · 13

+···

= x

1 +

∞

m=0

(−1)

m+1

)

m+1

4 · 5 · 8 · 9 ···(4m + 4)(4m + 5)

Hint: Let n = 4m in the recurrence relation, m = 1, 2, 3, ... .

5. (a) (n + 2)(n + 1)a

n+2

− n(n + 1)a

n+1

+ a

= 0, n ≥ 1; a

=−

(b) y

(x) = 1 −

−

+···, y

(x) = x −

−

+···

6. (a) a

n+2

=−(n

− 2n + 4)a

/[2(n + 1)(n + 2)], n ≥ 2; a

=−a

, a

=−

(b) y

(x) = 1 − x

−

+···,

(x) = x −

160

−

1920

+···

7. (a) a

n+2

=−a

/(n + 1), n = 0, 1, 2, ...

(b, d) y

(x) = 1 −

1 · 3

−

1 · 3 · 5

+···=1 +

∞

n=1

(−1)

1 · 3 · 5 ···(2n − 1)

(x) = x −

2 · 4

−

2 · 4 · 6

+···=x +

∞

n=1

(−1)

2n+1

2 · 4 · 6 ···(2n)

8. (a) a

n+2

=−[(n + 1)

n+1

+ a

n−1

]/(n + 1)(n + 2), n = 1, 2, ...

=−(a

+ a

)/2

(b) y

(x) = 1 −

(x − 1)

−

(x − 1)

+···

(x) = (x − 1) −

(x − 1)

−

(x − 1)

+···

9. (a) (n + 2)(n + 1)a

n+2

+ (n − 2)(n − 3)a

= 0; n = 0, 1, 2, ...

(b) y

(x) = 1 − 3x

, y

(x) = x − x

10. (a) 4(n + 2)a

n+2

− (n − 2)a

= 0; n = 0, 1, 2, ...

(b, d) y

(x) = 1 −

, y

(x) = x −

−

240

−

2240

−···−

2n+1

(2n − 1)(2n + 1)

−···

11. (a) 3(n + 2)a

n+2

− (n + 1)a

= 0; n = 0, 1, 2, ...

(b, d) y

(x) = 1 +

432

+···+

3 · 5 ···(2n − 1)

· 2 · 4 ···(2n)

+···

(x) = x +

135

945

+···+

2 · 4 ···(2 n)

· 3 · 5 ···(2n + 1)

2n+1

+···

12. (a) (n + 2)(n + 1)a

n+2

− (n + 1)na

n+1

+ (n − 1)a

= 0; n = 0, 1, 2, ...

(b, d) y

(x) = 1 +

+···+

+···, y

(x) = x

August 2, 2012 13:11 bansw Sheet number 19 Page number 757 cyan black

Answers to Pr oblems 757

13. (a) 2(n + 2)(n + 1)a

n+2

+ (n + 3)a

= 0; n = 0, 1, 2, ...

(b, d) y

(x) = 1 −

−

384

+···+(−1)

3 · 5 ···(2n + 1)

(2n)!

+···

(x) = x −

−

210

+···+(−1)

4 · 6 ···(2n + 2)

(2n + 1)!

2n+1

+···

14. (a) 2(n + 2)(n + 1)a

n+2

+ 3(n + 1)a

n+1

+ (n + 3)a

= 0; n = 0, 1, 2, ...

(b) y

(x) = 1 −

(x − 2)

+···

(x) = (x − 2) −

(x − 2)

+···

15. (a) y = 2 + x + x

+··· (c) about |x| < 0.7

16. (a) y =−1 + 3x + x

−

+··· (c) about |x| < 0.7

17. (a) y = 4 − x − 4x

+··· (c) about |x| < 0.5

18. (a) y =−3 + 2x −

−

+··· (c) about |x| < 0.9

19. (a) y

(x) = 1 −

(x − 1)

−

(x − 1)

+···

(x) = (x − 1) −

(x − 1)

−

(x − 1)

+···

21. (a) y

(x) = 1 −

λ(λ − 4)

−

λ(λ − 4)(λ − 8)

+···

(x) = x −

λ − 2

(λ − 2)(λ − 6)

−

(λ − 2)(λ − 6)(λ − 10)

+···

(b) 1, x,1− 2x

, x −

,1− 4x

, x −

− 2, 8x

− 12x, 16x

− 48x

+ 12, 32x

− 160x

+ 120x

22. (b) y = x − x

/6 +···

Section 5.3, page 269

1. φ

′′

(0) =−1, φ

′′′

(0) = 0, φ

(4)

(0) = 3

2. φ

′′

(0) = 0, φ

′′′

(0) =−2, φ

(4)

(0) = 0

3. φ

′′

(1) = 0, φ

′′′

(1) =−6, φ

(4)

(1) = 42

4. φ

′′

(0) = 0, φ

′′′

(0) =−a

, φ

(4)

(0) =−4a

5. ρ =∞, ρ =∞

6. ρ = 1, ρ = 3, ρ = 1

7. ρ = 1, ρ =

√

8. ρ = 1

9. (a) ρ =∞ (b) ρ =∞ (c) ρ =∞ (d) ρ =∞ (e) ρ = 1

(f) ρ =

√

2 (g) ρ =∞ (h) ρ = 1 (i) ρ = 1 (j) ρ = 2

(k) ρ =

√

3 (l) ρ = 1 (m) ρ =∞ (n) ρ =∞

10. (a) y

(x) = 1 −

−

− α

)α

−

− α

)(2

− α

)α

−···

−

[(2m − 2)

− α

]···(2

− α

)α

(2m)!

−···

(x) = x +

1 − α

− α

)(1 − α

)

+···

[(2m − 1)

− α

]···(1 − α

)

(2m + 1)!

2m+1

+···

(b) y

(x) or y

(x) terminates with x

as α = n is even or odd

; n = 3, y = x −

11. y

(x) = 1 −

120

180

+···, y

(x) = x −

180

504

+···,

ρ =∞

12. y

(x) = 1 −

−

+···, y

(x) = x −

−

+···,

ρ =∞

August 2, 2012 13:11 bansw Sheet number 20 Page number 758 cyan black

758 Answers to Problems

13. y

(x) = 1 + x

120

+···, y

(x) = x +

560

+···,

ρ = π/2

14. y

(x) = 1 +

−

120

+···, y

(x) = x −

120

+···,

ρ = 1

15. Cannot specify arbitrary initial conditions at x = 0; hence x = 0 is a singular point.

16. y = 1 + x +

+···+

+···=e

17. y = 1 +

2 · 4

2 · 4 · 6

+···+

· n!

+···

18. y = 1 + x +

+···

19. y = 1 + x + x

+···+x

+···=

1 − x

20. y = a

1 + x +

+···+

+···

(

+ 2

+···+

+···

(

= a

+ 2

− 1 − x −

(

= ce

− 2 − 2x − x

21. y = a

1 −

−

+···+

(−1)

+···

(

x +

−

2 · 4

3 · 5

+···

(

= a

−x

x +

−

2 · 4

3 · 5

+···

(

23. 1, 1 − 3x

,1− 10x

; x, x −

, x −

24. (a) 1, x, (3x

− 1)/2, (5x

− 3x)/2, (35x

− 30x

+ 3)/8, (63x

− 70x

+ 15x)/8

,0; P

, ±0.57735; P

,0,±0.77460; P

, ±0.33998, ±0.86114;

,0,±0.53847, ±0.90618

Section 5.4, page 280

1. y = c

−1

+ c

−2

2. y = c

|x + 1|

−1/2

+ c

|x + 1|

−3/2

3. y = c

+ c

ln |x| 4. y = c

−1

cos(2ln|x|) + c

−1

sin(2ln|x|)

5. y = c

x + c

x ln |x| 6. y = c

(x − 1)

−3

+ c

(x − 1)

−4

7. y = c

|x|

(−5+

√

29)/2

+ c

|x|

(−5−

√

29)/2

8. y = c

|x|

3/2

cos(

√

3ln|x|) + c

|x|

3/2

sin(

√

3ln|x|)

9. y = c

+ c

ln |x|

10. y = c

(x − 2)

−2

cos(2ln|x − 2|) + c

(x − 2)

−2

sin(2ln|x − 2|)

11. y = c

|x|

−1/2

cos(

√

15 ln |x|) + c

|x|

−1/2

sin(

√

15 ln |x|)

12. y = c

x + c

13. y = 2x

3/2

− x

−1

14. y = 2x

−1/2

cos(2lnx) − x

−1/2

sin(2lnx) 15. y = 2x

− 7x

ln |x|

16. y = x

−1

cos(2lnx) 17. x = 0, regular

18. x = 0, regular; x = 1, irregular 19. x = 0, irregular; x = 1, regular

20. x = 0, irregular; x =±1, regular 21. x = 1, regular; x =−1, irregular

22. x = 0, regular 23. x =−3, regular

24. x = 0, −1, regular; x = 1, irregular 25. x = 1, regular; x =−2, irregular

26. x = 0, 3, regular 27. x = 1, −2, regular

28. x = 0, regular 29. x = 0, irregular

30. x = 0, regular 31. x = 0, regular

32. x = 0, ±nπ, regular 33. x = 0, ±nπ,

regular

34. x = 0,

irregular; x =±nπ, regular 35. α<1

36. β>0 37. γ = 2

August 2, 2012 13:11 bansw Sheet number 21 Page number 759 cyan black

Answers to Pr oblems 759

38. α>1

39. (a) α<1 and β>0 (b) α<1 and β ≥ 0, or α = 1 and β>0

(e) α = 1 and β>0

41. y = a

1 −

2 · 5

2 · 4 · 5 · 9

−···

(

44. Irregular singular point

45. Regular singular point 46. Regular singular point

47. Irregular singular point 48. Irregular singular point

49. Irregular singular point

Section 5.5, page 286

1. (b) r(2r − 1) = 0; a

=−

n−2

(n + r)[2(n + r) − 1]

; r

, r

= 0

(x) = x

1/2

1 −

2 · 5

2 · 4 · 5 · 9

−

2 · 4 · 6 · 5 · 9 · 13

+···

(−1)

n!5 · 9 · 13 ···(4n + 1)

+···

(d) y

(x) = 1 −

2 · 3

2 · 4 · 3 · 7

−

2 · 4 · 6 · 3 · 7 · 11

+···

(−1)

n!3 · 7 · 11 ···(4n − 1)

+···

2. (b) r

−

= 0; a

=−

n−2

(n + r)

−

; r

, r

=−

(x) = x

1/3

1 −

1!(1 +

)

2!(1 +

)(2 +

)

+···

(−1)

m!(1 +

)(2 +

) ···(m +

)

+···

(d) y

(x) = x

−1/3

1 −

1!(1 −

)

2!(1 −

)(2 −

)

+···

(−1)

m!(1 −

)(2 −

) ···(m −

)

+···

Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....

3. (b) r(r − 1) = 0; a

=−

n−1

(n + r)(n + r − 1)

; r

= 1, r

= 0

(x) = x

1 −

1!2!

2!3!

+···+

(−1)

n!(n + 1)!

+···

4. (b) r

= 0; a

n−1

(n + r)

; r

= r

= 0

(x) = 1 +

(1!)

(2!)

+···+

(n!)

+···

5. (b) r(3r − 1) = 0; a

=−

n−2

(n + r)[3(n + r) − 1]

; r

, r

= 0

(x) = x

1/3

1 −

1!7

(

2!7 · 13

(

+···

(−1)

m!7 · 13 ···(6m + 1)

(

+···

August 2, 2012 13:11 bansw Sheet number 22 Page number 760 cyan black

760 Answers to Problems

(d) y

(x) = 1 −

1!5

(

2!5 · 11

(

+···+

(−1)

m!5 · 11 ···(6m − 1)

(

+···

Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....

6. (b) r

− 2 = 0; a

=−

n−1

(n + r)

− 2

; r

√

2, r

=−

√

(x) = x

√

1 −

1(1 + 2

√

2!(1 + 2

√

2)(2 + 2

√

+···

(−1)

n!(1 + 2

√

2)(2 + 2

√

2) ···(n + 2

√

+···

(d) y

(x) = x

−

√

1 −

1(1 − 2

√

2!(1 − 2

√

2)(2 − 2

√

+···

(−1)

n!(1 − 2

√

2)(2 − 2

√

2) ···(n − 2

√

+···

7. (b) r

= 0; (n + r)a

= a

n−1

; r

= r

= 0

(x) = 1 + x +

+···+

+···=e

8. (b) 2r

+ r − 1 = 0; (2n + 2r − 1)(n + r + 1)a

+ 2a

n−2

= 0;

, r

=−1

(x) = x

1/2

1 −

2!7 · 11

−···+

(−1)

m!7 · 11 ···(4m + 3)

+···

(

(d) y

(x) = x

−1

1 − x

2!5

−···+

(−1)

m!5 · 9 ···(4m − 3)

+···

(

9. (b) r

− 4r + 3 = 0; (n + r − 3)(n + r − 1)a

− (n + r − 2)a

n−1

= 0; r

= 3, r

= 1

(x) = x

1 +

x +

+···+

n!(n + 2)

+···

(

10. (b) r

− r +

= 0; (n + r −

)

+ a

n−2

= 0; r

= r

= 1/2

(x) = x

1/2

1 −

−···+

(−1)

(m!)

+···

(

11. (a) r

= 0; r

= 0, r

= 0

(b) y

(x) = 1 +

α(α + 1)

2 · 1

(x − 1) −

α(α + 1)[1 · 2 − α(α + 1)]

(2 · 1

)(2 · 2

)

(x − 1)

+···

+(−1)

n+1

α(α + 1)[1 · 2 − α(α + 1)]···[n(n − 1) − α(α + 1)]

(n!)

(x − 1)

+···

12. (a) r

, r

= 0 at both x =±1

(b) y

(x) =|x − 1|

1/2

1 +

∞

n=1

(−1)

(1 + 2α) ···(2n − 1 + 2α)(1 − 2α) ···(2n − 1 − 2α)

(2n + 1)!

(x − 1)

(x) = 1

∞

n=1

(−1)

α(1 + α) ···(n − 1 + α)(−α)(1 − α) ···(n − 1 − α)

n!1 · 3 · 5 ···(2n − 1)

(x − 1)

13. (b) r

= 0; r

= 0, r

= 0; a

(n − 1 − λ)a

n−1

August 2, 2012 13:11 bansw Sheet number 23 Page number 761 cyan black

Answers to Pr oblems 761

(x) = 1 +

−λ

(1!)

x +

(−λ)(1 − λ)

(2!)

+···+

(−λ)(1 − λ) ···(n − 1 − λ)

(n!)

+···

For λ = n, the coefﬁcients of all terms past x

are zero.

16. (e) [(n − 1)

− 1]b

=−b

n−2

, and it is impossible to determine b

Section 5.6, page 294

1. (a) x = 0; (b) r(r − 1) = 0; r

= 1, r

= 0

2. (a) x = 0; (b) r

− 3r + 2 = 0; r

= 2, r

= 1

3. (a) x = 0; (b) r(r − 1) = 0; r

= 1, r

= 0

(a) x = 1; (b) r(r + 5) = 0; r

= 0, r

=−5

4. None

5. (a) x = 0; (b) r

+ 2r − 2 = 0; r

=−1 +

√

∼

0.732, r

=−1 −

√

∼

−2.73

6. (a) x = 0; (b) r(r −

) = 0; r

, r

= 0

(a) x =−2; (b) r(r −

) = 0; r

, r

= 0

7. (a) x = 0; (b) r

+ 1 = 0; r

= i, r

=−i

8. (a) x =−1; (b) r

− 7r + 3 = 0; r

= (7 +

√

37)/2

∼

6.54, r

= (7 −

√

37)/2

∼

0.459

9. (a) x = 1; (b) r

+ r = 0; r

= 0, r

=−1

10. (a) x =−2; (b) r

− (5/4)r = 0; r

= 5/4, r

= 0

11. (a) x = 2; (b) r

− 2r = 0; r

= 2, r

= 0

(a) x =−2; (b) r

− 2r = 0; r

= 2, r

= 0

12. (a) x = 0; (b) r

− (5/3)r = 0; r

= 5/3, r

= 0

(a) x =−3; (b) r

− (r/3) − 1 = 0; r

= (1 +

√

37)/6

∼

1.18,

= (1 −

√

37)/6

∼

−0.847

13. (b) r

= 0, r

= 0

(x) = 1 + x +

+···, y

(x) = y

(x) ln x − 2x −

−

108

+···

14. (b) r

= 1, r

= 0

(x) = x − 4x

−

+···, y

(x) =−6y

(x) ln x + 1 − 33x

449

+···

15. (b) r

= 1, r

= 0

(x) = x +

+···, y

(x) = 3y

(x) ln x + 1 −

−

+···

16. (b) r

= 1, r

= 0

(x) = x −

−

144

+···

(x) =−y

(x) ln x + 1 −

−

1728

+···

17. (b) r

= 1, r

=−1

(x) = x −

720

+···, y

(x) =−

(x) ln x + x

−1

−

+···

18. (b) r

, r

= 0

(x) = (x − 1)

1/2

[1 −

(x − 1) +

480

(x − 1)

+···], (d) ρ = 1

19. (c) Hint: (n − 1)(n − 2) + (1 + α + β)(n − 1) + αβ = (n − 1 + α)(n − 1 + β)

(d) Hint: (n − γ)(n − 1 − γ) + (1 + α + β)(n − γ) + αβ = (n − γ + α)(n − γ + β)

Section 5.7, page 305

1. y

(x) =

∞

n=0

(−1)

n!(n + 1)!

, y

(x) =−y

(x) ln x +

1 −

∞

n=1

+ H

n−1

n!(n − 1)!

(−1)

2. y

(x) =

∞

n=0

(−1)

(n!)

, y

(x) = y

(x) ln x −

∞

n=1

(−1)

(n!)

3. y

(x) =

∞

n=0

(−1)

(n!)

, y

(x) = y

(x) ln x − 2

∞

n=1

(−1)

(n!)

August 2, 2012 13:11 bansw Sheet number 24 Page number 762 cyan black

762 Answers to Problems

4. y

(x) =

∞

n=0

(−1)

n!(n + 1)!

, y

(x) =−y

(x) ln x +

1 −

∞

n=1

+ H

n−1

n!(n − 1)!

(−1)

5. y

(x) = x

3/2

1 +

∞

m=1

(−1)

m!(1 +

)(2 +

) ···(m +

)

(x) = x

−3/2

1 +

∞

m=1

(−1)

m!(1 −

)(2 −

) ···(m −

)

Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....

For r =−

, a

= 0 and a

is arbitrary.

CHAPTER 6 Section 6.1, page 315

1. Piecewise continuous 2. Neither

3. Continuous 4. Piecewise continuous

5. (a) F(s) = 1/s

, s > 0 (b) F (s) = 2/s

, s > 0 (c) F (s) = n!/s

n+1

, s > 0

6. F(s) = s/(s

+ a

), s > 0

7. F(s) =

− b

, s > |b|

8. F(s) =

− b

, s > |b| 9. F(s) =

s − a

(s − a)

− b

, s − a > |b|

10. F(s) =

(s − a)

− b

, s − a > |b| 11. F(s) =

+ b

, s > 0

12. F(s) =

+ b

, s > 0 13. F(s) =

(s − a)

+ b

, s > a

14. F(s) =

s − a

(s − a)

+ b

, s > a 15. F(s) =

(s − a)

, s > a

16. F(s) =

2as

+ a

)

, s > 0 17. F(s ) =

+ a

(s − a)

(s + a)

, s > |a|

18. F(s) =

(s − a)

n+1

, s > a 19. F(s) =

2a(3s

− a

)

+ a

)

, s > 0

20. F(s) =

2a(3s

+ a

)

− a

)

, s > |a| 21. F (s) =

1 − e

−πs

22. F(s) =

1 − (s + 1)e

−s

23. F(s) =

1 − e

−s

24. F(s) =

1 − 2e

−s

+ e

−2s

25. Converges

26. Converges 27. Diverges

28. Converges

30. (d) 2(3/2) =

√

π/2; 2(11/2) = 945

√

π/32

Section 6.2, page 324

1. f (t) =

sin 2t 2. f (t ) = 2t

3. f (t) =

−

−4t

4. f (t) =

−2t

5. f (t) = 2e

−t

cos 2t 6. f (t) = 2 cosh 2t −

sinh 2t

7. f (t) = 2e

cos t + 3e

sin t 8. f (t ) = 3 − 2 sin 2t + 5 cos 2t

9. f (t) =−2e

−2t

cos t + 5e

−2t

sin t 10. f (t) = 2e

−t

cos 3t −

−t

sin 3t

11. y =

+ 4e

−2t

) 12. y = 2e

−t

− e

−2t

13. y = e

sin t 14. y = e

− te

15. y = 2e

cos

√

3 t − (2/

√

3)e

sin

√

3 t 16. y = 2e

−t

cos 2t +

−t

sin 2t

17. y = te

− t

18. y = cosh t

19. y = cos

√

2 t 20. y = (ω

− 4)

−1

[(ω

− 5) cos ωt + cos 2t]

August 2, 2012 13:11 bansw Sheet number 25 Page number 763 cyan black

Answers to Pr oblems 763

21. y =

(cos t − 2 sin t + 4e

cos t − 2e

sin t) 22. y =

−t

− e

cos t + 7e

sin t)

23. y = 2e

−t

+ te

−t

+ 2t

−t

24. Y(s) =

+ 4

1 − e

−πs

s(s

+ 4)

25. Y(s) =

+ 1)

−

−s

(s + 1)

+ 1)

26. Y(s) =

1 − e

−s

+ 4)

27. Y(s) =

1 − 2e

−s

+ e

−2s

+ 1)

30. F(s) = 1/(s − a)

31. F(s) = 2b(3s

− b

)/(s

+ b

)

32. F(s) = n!/s

n+1

33. F(s) = n!/(s − a)

n+1

34. F(s) = 2b(s − a)/[(s − a)

+ b

]

35. F(s) =[(s − a)

− b

]/[(s − a)

+ b

]

37. (a) Y

′

+ s

Y = s (b) s

′′

+ 2sY

′

−[s

+ α(α + 1)]Y =−1

Section 6.3, page 333

7. (b) f (t) =−2u

(t) + 4u

(t) − u

(t)

8. (b) f (t) = 1 − 2u

(t) + 2u

(t) − 2u

(t) + u

(t)

9. (b) f (t) = 1 + u

(t)[e

−(t−2)

− 1]

10. (b) f (t) = t

+ u

(t)(1 − t

)

11. (b) f (t) = t − u

(t) − u

(t)(t − 2)

12. (b) f (t) = t + u

(t)(2 − t) + u

(t)(5 − t) − u

(t)(7 − t)

13. F(s) = 2e

−s

14. F(s) = e

−s

+ 2)/s

15. F(s) =

−πs

−

−2πs

(1 + πs ) 16. F(s) =

−s

+ 2e

−3s

− 6e

−4s

)

17. F(s) = s

−2

[(1 − s)e

−2s

− (1 + s)e

−3s

] 18. F(s) = (1 − e

−s

)/s

19. f (t) = t

20. f (t) =

(t)[e

t−2

− e

−2(t−2)

]

21. f (t) = 2u

(t)e

t−2

cos(t − 2) 22. f (t) = u

(t) sinh 2(t − 2)

23. f (t) = u

(t)e

2(t−1)

cosh(t − 1) 24. f (t) = u

(t) + u

(t) − u

(t)

26. f (t) = 2(2t)

27. f (t) =

−t/2

cos t

28. f (t) =

t/3

2t/3

− 1) 29. f (t) =

t/2

(t/2)

30. F(s) = s

−1

(1 − e

−s

), s > 0 31. F(s) = s

−1

(1 − e

−s

+ e

−2s

− e

−3s

), s > 0

32. F(s) =

[1 − e

−s

+···+e

−2ns

− e

−(2n+1)s

1 − e

−(2n+2)s

s(1 + e

−s

)

, s > 0

33. F(s) =

∞

n=0

(−1)

−ns

1/s

1 + e

−s

, s > 0

35. L{f (t)}=

1/s

1 + e

−s

, s > 0 36. L{f (t)}=

1 − e

−s

s(1 + e

−s

)

, s > 0

37. L{f (t)}=

1 − (1 + s)e

−s

(1 − e

−s

)

, s > 0 38. L{f (t)}=

1 + e

−πs

(1 + s

)(1 − e

−πs

)

, s > 0

39. (a) L{f (t)}=s

−1

(1 − e

−s

), s > 0 (b) L{g(t)}=s

−2

(1 − e

−s

), s > 0

−2

(1 − e

−s

)

, s > 0

40. (b) L{p(t)}=

1 − e

−s

(1 + e

−s

)

, s > 0

Section 6.4, page 340

1. (a) y = 1 − cos t + sin t − u

3π

(t)(1 + cos t)

2. (a) y = e

−t

sin t +

(t)[1 + e

−(t−π)

cos t + e

−(t−π)

sin t]

−

2π

(t)[1 − e

−(t−2π)

cos t − e

−(t−2π)

sin t]

3. (a) y =

[1 − u

2π

(t)](2 sin t − sin 2t)

4. (a) y =

(2 sin t − sin 2t) −

(t)(2 sin t + sin 2t)

5. (a) y =

−2t

− e

−t

− u

(t)[

−2(t−10)

− e

−(t−10)

]

August 2, 2012 13:11 bansw Sheet number 26 Page number 764 cyan black

764 Answers to Problems

6. (a) y = e

−t

− e

−2t

+ u

(t)[

− e

−(t−2)

−2(t−2)

]

7. (a) y = cos t + u

3π

(t)[1 − cos(t − 3π)]

8. (a) y = h(t) − u

π/2

(t)h(t − π/2 ), h(t) =

(−4 + 5t + 4e

−t/2

cos t − 3e

−t/2

sin t)

9. (a) y =

sin t +

t −

(t)[t − 6 − sin(t − 6)]

10. (a) y = h(t) + u

(t)h(t − π), h(t) =

[−4 cos t + sin t + 4e

−t/2

cos t + e

−t/2

sin t]

11. (a) y = u

(t)[

−

cos(2t − 2π)]−u

3π

(t)[

−

cos(2t − 6π)]

12. (a) y = u

(t)h(t − 1) − u

(t)h(t − 2), h(t) =−1 + (cos t + cosh t)/2

13. (a) y = h(t) − u

(t)h(t − π), h(t) = (3 − 4 cos t + cos 2t)/12

14. f (t) =[u

(t)(t − t

) − u

(t)(t − t

− k)](h/k)

15. g(t) =[u

(t)(t − t

) − 2u

(t)(t − t

− k) + u

+2k

(t)(t − t

− 2k)](h/k)

16. (b) u(t) = 4 ku

3/2

(t)h(t −

) − 4ku

5/2

(t)h(t −

h(t) =

− (

√

7/84) e

−t/8

sin(3

√

7 t/8) −

−t/8

cos(3

√

7 t/8)

(d) k = 2.51 (e) τ = 25.6773

17. (a) k = 5 (b) y =[u

(t)h(t − 5) − u

5+k

(t)h(t − 5 − k)]/k, h(t) =

t −

sin 2t

18. (b) f

(t) =[u

4−k

(t) − u

4+k

(t)]/2k;

y =[u

4−k

(t)h(t − 4 + k) − u

4+k

(t)h(t − 4 − k)]/2k,

h(t) =

−

−t/6

cos(

√

143 t/6) − (

√

143/572) e

−t/6

sin(

√

143 t/6)

19. (b) y = 1 − cos t + 2

k=1

(−1)

kπ

(t)[1 − cos(t − kπ)]

21. (b) y = 1 − cos t +

k=1

(−1)

kπ

(t)[1 − cos(t − kπ)]

23. (a) y = 1 − cos t + 2

k=1

(−1)

11k/4

(t)[1 − cos(t − 11k/4)]

Section 6.5, page 348

1. (a) y = e

−t

cos t + e

−t

sin t + u

(t)e

−(t−π)

sin(t − π)

2. (a) y =

(t) sin 2(t − π) −

2π

(t) sin 2(t − 2π)

3. (a) y =−

−2t

−t

+ u

(t)[−e

−2(t−5)

+ e

−(t−5)

]+u

(t)[

−2(t−10)

− e

−(t−10)

]

4. (a) y = cosh(t) − 20u

(t) sinh(t − 3)

5. (a) y =

sin t −

cos t +

−t

cos

√

2 t + (1/

√

2) u

3π

(t)e

−(t−3π)

sin

√

2(t − 3π)

6. (a) y =

cos 2t +

4π

(t) sin 2(t − 4π)

7. (a) y = sin t + u

2π

(t) sin(t − 2π)

8. (a) y = u

π/4

(t) sin 2(t − π/4)

9. (a) y = u

π/2

(t)[1 − cos(t − π/2 )]+3u

3π/2

(t) sin(t − 3π/2) − u

2π

(t)[1 − cos(t − 2π)]

10. (a) y = (1/

√

31) u

π/6

(t) exp[−

(t − π/6)]sin(

√

31/4)(t − π/6)

11. (a) y =

cos t +

sin t −

−t

cos t −

−t

sin t + u

π/2

(t)e

−(t−π/2)

sin(t − π/2)

12. (a) y = u

(t)[sinh(t − 1) − sin(t − 1)]/2

13. (a) −e

−T/4

δ(t − 5 − T), T = 8π/

√

14. (a) y = (4/

√

15) u

(t)e

−(t−1)/4

sin(

√

15/4)(t − 1) (b) t

∼

2.3613, y

∼

0.71153

√

7/21) u

(t)e

−(t−1)/8

sin(3

√

7/8)(t − 1); t

∼

2.4569, y

∼

0.83351

(d) t

= 1 + π/2

∼

2.5708, y

= 1

15. (a) k

∼

2.8108 (b) k

∼

2.3995 (c) k

= 2

16. (a) φ(t, k) =[u

4−k

(t)h(t − 4 + k) − u

4+k

(t)h(t − 4 − k)]/2k, h(t) = 1 − cos t

(b) φ

(t) = u

(t) sin(t − 4) (c) Yes

17. (b) y =

k=1

kπ

(t) sin(t − kπ)

18. (b) y =

k=1

(−1)

k+1

kπ

(t) sin(t − kπ)

19. (b) y =

k=1

kπ/2

(t) sin(t − kπ/2) 20. (b) y =

k=1

(−1)

k+1

kπ/2

(t) sin(t − kπ/2)

August 2, 2012 13:11 bansw Sheet number 27 Page number 765 cyan black

Answers to Pr oblems 765

21. (b) y =

k=1

(2k−1)π

(t) sin[t − (2k − 1)π] 22. (b) y =

k=1

(−1)

k+1

11k/4

(t) sin(t − 11k/4)

23. (b) y =

√

399

k=1

(−1)

k+1

kπ

(t)e

−(t−kπ)/20

sin[

√

399(t − kπ)/20]

24. (b) y =

√

399

k=1

(2k−1)π

(t)e

−[t−(2k−1)π]/20

sin{

√

399[t − (2k − 1)π]/20}

Section 6.6, page 354

3. sin t ∗ sin t =

(sin t − t cos t) is negative when t = 2π, for example.

4. F(s) = 2/s

+ 4)

5. F(s) = 1/(s + 1)(s

+ 1)

6. F(s) = 1/s

(s − 1) 7. F (s) = s/(s

+ 1)

8. f (t) =

(t − τ)

sin τ dτ 9. f (t) =

−(t−τ)

cos 2τ dτ

10. f (t) =

(t − τ)e

−(t−τ)

sin 2τ dτ 11. f (t) =

sin(t − τ)g(τ) dτ

12. (c)

(1 − u)

du =

2(m + 1)2(n + 1)

2(m + n + 2)

13. y =

sin ωt +

sin ω(t − τ)g(τ) dτ 14. y =

−(t−τ)

sin(t − τ) sin ατ dτ

15. y =

−(t−τ)/2

sin 2(t − τ)g(τ) dτ

16. y = e

−t/2

cos t −

−t/2

sin t +

−(t−τ)/2

sin(t − τ)[1 − u

(τ)]dτ

17. y = 2e

−2t

+ te

−2t

(t − τ)e

−2(t−τ)

g(τ) dτ

18. y = 2e

−t

− e

−2t

−(t−τ)

− e

−2(t−τ)

]cos ατ dτ

19. y =

[sinh(t − τ) − sin(t − τ)]g(τ) dτ

20. y =

cos t −

cos 2t +

[2 sin(t − τ) − sin 2(t − τ)]g(τ) dτ

21. 7(s) =

F(s)

1 + K(s)

22. (a) φ(t) =

(4 sin 2t − 2 sin t )

23. (a) φ(t) = cos t (b) φ

′′

(t) + φ(t) = 0, φ(0) = 1, φ

′

(0) = 0

24. (a) φ(t) = cosh(t) (b) φ

′′

(t) − φ(t) = 0, φ(0) = 1, φ

′

(0) = 0

25. (a) φ(t) = (1 − 2t + t

−t

(b) φ

′′

(t) + 2φ

′

(t) + φ(t) = 2e

−t

, φ(0) = 1, φ

′

(0) =−3

26. (a) φ(t) =

−t

−

t/2

cos(

√

3t/2) +

√

t/2

sin(

√

3t/2)

(b) φ

′′′

(t) + φ(t) = 0, φ(0) = 0, φ

′

(0) = 0, φ

′′

(0) = 1

27. (a) φ(t) = cos t

(b) φ

(4)

(t) − φ(t) = 0, φ(0) = 1, φ

′

(0) = 0, φ

′′

(0) =−1, φ

′′′

(0) = 0

28. (a) φ(t) = 1 −

√

−t/2

sin(

√

3t/2)

(b) φ

′′′

(t) + φ

′′

(t) + φ

′

(t) = 0, φ(0) = 1, φ

′

(0) =−1, φ

′′

(0) = 1

CHAPTER 7 Section 7.1, page 363

1. x

′

= x

, x

′

=−2x

− 0.5x

2. x

′

= x

, x

′

=−2x

− 0.5x

+ 3 sin t

3. x

′

= x

, x

′

=−(1 − 0.25t

−2

− t

−1

4. x

′

= x

, x

′

= x

, x

′

= x

, x

′

= x

5. x

′

= x

, x

′

=−4x

− 0.25x

+ 2 cos 3t, x

(0) = 1, x

(0) =−2

August 2, 2012 13:11 bansw Sheet number 28 Page number 766 cyan black

766 Answers to Problems

6. x

′

= x

, x

′

=−q(t)x

− p(t)x

+ g(t); x

(0) = u

, x

(0) = u

′

7. (a) x

= c

−t

+ c

−3t

, x

= c

−t

− c

−3t

(b) c

= 5/2, c

=−1/2 in solution in (a)

= x

8. (a) x

′′

− x

′

− 2x

= 0 (b) x

−

−t

, x

−

−t

= 2x

in the ﬁrst quadrant.

9. (a) 2x

′′

− 5x

′

+ 2x

= 0 (b) x

=−

t/2

−

, x

t/2

−

= x

in the third quadrant.

10. (a) x

′′

+ 3x

′

+ 2x

= 0 (b) x

=−7e

−t

+ 6e

−2t

, x

=−7e

−t

+ 9e

−2t

= x

11. (a) x

′′

+ 4x

= 0 (b) x

= 3 cos 2t + 4 sin 2t, x

=−3 sin 2t + 4 cos2t

12. (a) x

′′

+ x

′

+ 4.25x

= 0

(b) x

=−2e

−t/2

cos 2t + 2e

−t/2

sin 2t, x

= 2e

−t/2

cos 2t + 2e

−t/2

sin 2t

13. LRCI

′′

+ LI

′

+ RI = 0

18. y

′

= y

, y

′

= y

, m

′

=−(k

+ k

+ F

(t),

′

= k

− (k

+ k

+ F

(t)

22. (a) Q

′

−

, Q

(0) = 25, Q

′

= 3 +

−

, Q

(0) = 15

(b) Q

= 42, Q

= 36

′

=−

, x

(0) =−17, x

′

−

, x

(0) =−21

23. (a) Q

′

= 3q

−

100

, Q

(0) = Q

′

= q

−

100

, Q

(0) = Q

(b) Q

= 6(9q

+ q

), Q

= 20(3q

+ 2q

) (c) No (d)

≤ Q

≤

Section 7.2, page 376

1. (a)

⎛

⎝

6 −63

59−2

238

⎞

⎠

(b)

⎛

⎝

−15 6 −12

7 −18 −1

−26 −3 −5

⎞

⎠

(c)

⎛

⎝

6 −12 3

437

9120

⎞

⎠

(d)

⎛

⎝

−8 −911

14 12 −5

5 −85

⎞

⎠

2. (a)

1 − i −7 + 2i

−1 + 2i 2 + 3i

(

(b)

3 + 4i 6i

11 + 6i 6 − 5i

(

(c)

−3 + 5i 7 + 5i

2 + i 7 + 2i

(

(d)

8 + 7i 4 − 4i

6 − 4i − 4

(

3. (a)

⎛

⎝

−21

10−

2 −31

⎞

⎠

(b)

⎛

⎝

13−2

2 −11

3 −10

⎞

⎠

(c), (d)

⎛

⎝

−14 0

3 −10

5 −41

⎞

⎠

4. (a)

3 − 2i 2 − i

1 + i −2 + 3i

(

(b)

3 + 2i 1 − i

2 + i −2 − 3i

(

(c)

3 + 2i 2 + i

1 − i −2 − 3i

(

⎛

⎝

6 −

0410

446

⎞

⎠

August 2, 2012 13:11 bansw Sheet number 29 Page number 767 cyan black

Answers to Pr oblems 767

6. (a)

⎛

⎝

7 −11 −3

11 20 17

−43−12

⎞

⎠

(b)

⎛

⎝

50−1

274

−114

⎞

⎠

(c)

⎛

⎝

6 −8 −11

915 6

−5 −15

⎞

⎠

8. (a) 4i (b) 12 − 8i (c) 2 + 2i (d) 16

10.

−

11.

−

12.

⎛

⎝

1 −32

−33−1

2 −10

⎞

⎠

13.

⎛

⎜

⎝

−

⎞

⎟

⎠

14. Singular 15.

⎛

⎜

⎝

−

⎞

⎟

⎠

16.

⎛

⎜

⎝

−

⎞

⎟

⎠

17. Singular

18.

⎛

⎜

⎝

1101

1011

1111

0101

⎞

⎟

⎠

19.

⎛

⎜

⎝

−

0 −

−2 −

−

⎞

⎟

⎠

21. (a)

⎛

⎝

−t

10e

−e

−t

0 −e

⎞

⎠

(b)

⎛

⎝

− 2 + 3e

1 + 4e

−2t

− e

+ 2e

− e

− 1 − 3e

2 + 2e

−2t

+ e

−2e

− 3 + 6e

−1 + 6e

−2t

− 2e

−3e

+ 3e

− 2e

⎞

⎠

(c)

⎛

⎝

−2e

−t

−e

−t

−2e

−e

−3e

−t

⎞

⎠

(d) (e − 1)

⎛

⎜

⎝

12e

−1

(e + 1)

2 e

−1

−

(e + 1)

−13e

−1

e + 1

⎞

⎟

⎠

Section 7.3, page 388

1. x

=−

, x

=−

2. No solution

3. x

=−c, x

= c + 1, x

= c, where c is arbitrary

4. x

= c, x

=−c, x

=−c, where c is arbitrary

5. x

= 0, x

= 0 6. x

= c

, x

= c

, x

= c

+ 2c

+ 2

7. Linearly independent 8. x

(1)

− 5x

(2)

+ 2x

(3)

= 0

9. 2x

(1)

− 3x

(2)

+ 4x

(3)

− x

(4)

= 0 10. Linearly independent

11. x

(1)

+ x

(2)

− x

(4)

= 0 13. 3x

(1)

(t) − 6x

(2)

(t) + x

(3)

(t) = 0

14. Linearly independent 16. λ

= 2, x

(1)

(

; λ

= 4, x

(2)

(

August 2, 2012 13:11 bansw Sheet number 30 Page number 768 cyan black

768 Answers to Problems

17. λ

= 1 + 2i, x

(1)

1 − i

(

; λ

= 1 − 2i, x

(2)

1 + i

(

18. λ

=−3, x

(1)

−1

(

; λ

=−1, x

(2)

(

19. λ

= 0, x

(1)

(

; λ

= 2, x

(2)

−i

(

20. λ

= 2, x

(1)

√

(

; λ

=−2, x

(2)

−

√

(

21. λ

=−1/2, x

(1)

(

; λ

=−3/2, x

(2)

(

22. λ

= 1, x

(1)

⎛

⎝

−3

⎞

⎠

; λ

= 1 + 2i, x

(2)

⎛

⎝

−i

⎞

⎠

; λ

= 1 − 2i, x

(3)

⎛

⎝

⎞

⎠

23. λ

= 1, x

(1)

⎛

⎝

−1

⎞

⎠

; λ

= 2, x

(2)

⎛

⎝

−2

⎞

⎠

; λ

= 3, x

(3)

⎛

⎝

−1

⎞

⎠

24. λ

= 1, x

(1)

⎛

⎝

−2

−1

⎞

⎠

; λ

= 2, x

(2)

⎛

⎝

⎞

⎠

; λ

=−1, x

(3)

⎛

⎝

−2

⎞

⎠

25. λ

=−1, x

(1)

⎛

⎝

−4

⎞

⎠

; λ

=−1, x

(2)

⎛

⎝

−1

⎞

⎠

; λ

= 8, x

(3)

⎛

⎝

⎞

⎠

Section 7.4, page 394

2. (c) W(t) = c exp

(t) + p

(t)]dt

6. (a) W(t) = t

(b) x

(1)

and x

(2)

are linearly independent at each point except t = 0; they are linearly

independent on every interval.

(d) x

′

−2t

−2

−1

(

7. (a) W(t) = t(t − 2)e

(b) x

(1)

and x

(2)

are linearly independent at each point except t = 0 and t = 2; they are

linearly independent on every interval.

(d) x

′

⎛

⎝

2 − 2t

− 2t

− 2

− 2t

⎞

⎠

Section 7.5, page 405

1. (a) x = c

(

−t

+ c

(

2. (a) x = c

(

−t

+ c

(

−2t

3. (a) x = c

(

+ c

(

−t

4. (a) x = c

−4

(

−3t

+ c

(

August 2, 2012 13:11 bansw Sheet number 31 Page number 769 cyan black

Answers to Pr oblems 769

5. (a) x = c

−1

(

−3t

+ c

(

−t

6. (a) x = c

−1

(

t/2

+ c

(

7. (a) x = c

(

+ c

(

−2t

8. (a) x = c

−2

(

+ c

−3

(

9. x = c

(

+ c

−i

(

10. x = c

2 + i

−1

(

+ c

−1

(

−it

11. x = c

⎛

⎝

⎞

⎠

+ c

⎛

⎝

−2

⎞

⎠

+ c

⎛

⎝

−1

⎞

⎠

−t

12. x = c

⎛

⎝

−4

⎞

⎠

−t

+ c

⎛

⎝

−1

⎞

⎠

−t

+ c

⎛

⎝

⎞

⎠

13. x = c

⎛

⎝

−5

−7

⎞

⎠

−2t

+ c

⎛

⎝

−4

−2

⎞

⎠

−t

+ c

⎛

⎝

−1

⎞

⎠

14. x = c

⎛

⎝

−4

−1

⎞

⎠

+ c

⎛

⎝

−1

⎞

⎠

−2t

+ c

⎛

⎝

⎞

⎠

15. x =−

(

16. x =

(

−t

(

17. x =

⎛

⎝

−2

⎞

⎠

+ 2

⎛

⎝

⎞

⎠

18. x = 6

⎛

⎝

−1

⎞

⎠

+ 3

⎛

⎝

−2

⎞

⎠

−t

−

⎛

⎝

−8

⎞

⎠

20. x = c

(

t + c

(

−1

21. x = c

(

+ c

(

22. x = c

(

+ c

(

−2

23. x = c

(

−1

+ c

(

29. (a) x

′

= x

, x

′

=−(c/a)x

− (b/a)x

30. (a) x =−

(

−t/20

−2

(

−t/4

∼

74.39

31. (a) x = c

−

√

(

(−2+

√

2)t/2

+ c

√

(

(−2−

√

2)t/2

; r

1,2

= (−2 ±

√

2)/2; node

(b) x = c

−1

√

(

(−1+

√

2)t

+ c

√

(

(−1−

√

2)t

; r

1,2

=−1 ±

√

2; saddle point

1,2

=−1 ±

√

α; α = 1

32. (a)

(

= c

(

−2t

+ c

(

−t

33. (a)

−

(

−

> 0

Section 7.6, page 417

1. (a) x = c

cos 2t

cos 2t + sin 2t

(

+ c

sin 2t

−cos 2t + sin 2t

(

2. (a) x = c

−t

2 cos 2t

sin 2t

(

+ c

−t

−2 sin 2t

cos 2t

(

August 2, 2012 13:11 bansw Sheet number 32 Page number 770 cyan black

770 Answers to Problems

3. (a) x = c

5 cos t

2 cos t + sin t

(

+ c

5 sin t

−cos t + 2 sin t

(

4. (a) x = c

t/2

5 cos

3(cos

t + sin

(

+ c

t/2

5 sin

3(−cos

t + sin

(

5. (a) x = c

−t

cos t

2 cos t + sin t

(

+ c

−t

sin t

−cos t + 2 sin t

(

6. (a) x = c

−2 cos 3t

cos 3t + 3 sin 3t

(

+ c

−2 sin 3t

sin 3t − 3 cos 3t

(

7. x = c

⎛

⎝

−3

⎞

⎠

+ c

⎛

⎝

cos 2t

sin 2t

⎞

⎠

+ c

⎛

⎝

sin 2t

−cos 2t

⎞

⎠

8. x = c

⎛

⎝

−2

⎞

⎠

−2t

+ c

−t

⎛

⎝

−

√

2 sin

√

2 t

cos

√

2 t

−cos

√

2 t −

√

2 sin

√

2 t

⎞

⎠

+ c

−t

⎛

⎝

√

2 cos

√

2 t

sin

√

2 t

√

2 cos

√

2 t − sin

√

2 t

⎞

⎠

9. x = e

−t

cos t − 3 sin t

cos t − sin t

(

10. x = e

−2t

cos t − 5 sin t

−2 cos t − 3 sin t

(

11. (a) r =−

± i 12. (a) r =

± i

13. (a) r = α ± i (b) α = 0

14. (a) r = (α ±

√

− 20)/2 (b) α =−

√

20, 0,

√

15. (a) r =±

√

4 − 5α (b) α = 4/5

16. (a) r =

√

3α (b) α = 0, 25/12

17. (a) r =−1 ±

√

−α (b) α =−1, 0

18. (a) r =−

√

49 − 24α (b) α = 2, 49/24

19. (a) r =

α − 2 ±

√

+ 8α − 24 (b) α =−4 − 2

√

10, −4 + 2

√

10, 5/2

20. (a) r =−1 ±

√

25 + 8α (b) α =−25/8, −3

21. x = c

−1

cos(

√

2lnt)

√

2 sin(

√

2lnt)

(

+ c

−1

sin(

√

2lnt)

−

√

2 cos(

√

2lnt)

(

22. x = c

5 cos(ln t)

2 cos(ln t) + sin(ln t)

(

+ c

5 sin(ln t)

−cos(ln t) + 2 sin(ln t)

(

23. (a) r =−

± i, −

24. (a) r =−

± i,

25. (b)

(

= c

−t/2

cos(t/2)

4 sin(t/2)

(

+ c

−t/2

sin(t/2)

−4 cos(t/2)

(

= 2, c

=−

in answer to part (b).

(d) lim

t→∞

I(t) = lim

t→∞

V(t) = 0; no

26. (b)

(

= c

−t

cos t

−cos t − sin t

+ c

−t

sin t

−sin t + cos t

(

= 2 and c

= 3 in answer to part (b).

(d) lim

t→∞

I(t) = lim

t→∞

V(t) = 0; no

28. (b) r =±i

k/m (d) |r| is the natural frequency.

29. (c) r

=−1, ξ

(1)

(

; r

=−4, ξ

(2)

−4

(

(d) x

= 3c

cos t + 3c

sin t + 3c

cos 2t + 3c

sin 2t,

= 2c

cos t + 2c

sin t − 4c

cos 2t − 4c

sin 2t

August 2, 2012 13:11 bansw Sheet number 33 Page number 771 cyan black

Answers to Pr oblems 771

(e) x

′

=−3c

sin t + 3c

cos t − 6c

sin 2t + 6c

cos 2t,

′

=−2c

sin t + 2c

cos t + 8c

sin 2t − 8c

cos 2t

30. (a) A =

⎛

⎜

⎝

0010

0001

−4300

9/4 −13/400

⎞

⎟

⎠

(b) r

= i, ξ

(1)

⎛

⎜

⎝

⎞

⎟

⎠

; r

=−i, ξ

(2)

⎛

⎜

⎝

−i

⎞

⎟

⎠

;

i, ξ

(3)

⎛

⎜

⎝

−3

10i

−

⎞

⎟

⎠

; r

=−

i, ξ

(4)

⎛

⎜

⎝

−3

−10i

⎞

⎟

⎠

⎛

⎜

⎝

cos t

−sin t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

sin t

cos t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

4 cos

−3 cos

−10 sin

sin

⎞

⎟

⎠

+ c

⎛

⎜

⎝

4 sin

−3 sin

10 cos

−

cos

⎞

⎟

⎠

(e) c

, c

= 0, c

, c

= 0. period = 4π.

31. (a) A =

⎛

⎜

⎝

0010

0001

−2100

1 −200

⎞

⎟

⎠

(b) r

= i, ξ

(1)

⎛

⎜

⎝

⎞

⎟

⎠

; r

=−i, ξ

(2)

⎛

⎜

⎝

−i

⎞

⎟

⎠

;

√

3i, ξ

(3)

⎛

⎜

⎝

−1

√

−

√

⎞

⎟

⎠

; r

=−

√

3i, ξ

(4)

⎛

⎜

⎝

−1

−

√

⎞

⎟

⎠

⎛

⎜

⎝

cos t

−sin t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

sin t

cos t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

cos

√

3 t

−cos

√

3 t

−

√

3 sin

√

3 t

√

3 sin

√

3 t

⎞

⎟

⎠

+ c

⎛

⎜

⎝

sin

√

3 t

−sin

√

3 t

√

3 cos

√

3 t

−

√

3 cos

√

3 t

⎞

⎟

⎠

(e) c

= 1, c

= 0, c

=−2, c

= 0.

Section 7.7, page 427

1. (b) 7(t) =

−

−t

−

−t

−

2. (b) 7(t) =

−t/2

−t

−t/2

− e

−t

−t/2

−

−t

−t/2

−t

3. (b) 7(t) =

−

−t

−

−t

−

−t

−

−t

August 2, 2012 13:11 bansw Sheet number 34 Page number 772 cyan black

772 Answers to Problems

4. (b) 7(t) =

−3t

−

−3t

−

−3t

5. (b) 7(t) =

cos t + 2 sin t −5 sin t

sin t cos t − 2 sin t

(

6. (b) 7(t) =

−t

cos 2t −2e

−t

sin 2t

−t

sin 2te

−t

cos 2t

(

7. (b) 7(t) =

−

8. (b) 7(t) =

−t

cos t + 2e

−t

sin t −e

−t

sin t

−t

sin te

−t

cos t − 2e

−t

sin t

9. (b) 7(t) =

⎛

⎜

⎝

−2e

−2t

+ 3e

−t

−e

−2t

+ e

−t

−e

−2t

+ e

−t

−2t

− 4e

−t

−2t

−

−t

−2t

−

−t

−2t

− 2e

−t

−

−2t

−

−t

−

−2t

−

−t

−

⎞

⎟

⎠

10. (b) 7(t) =

⎛

⎜

⎝

−2t

−

−2t

− e

−2t

−

−2t

+ e

−

−2t

−2e

+ e

−2t

+ e

−

−2t

−

−2t

−

+ e

−2t

⎞

⎟

⎠

11. x =

(

−

(

−t

12. x =

(

−t

cos 2t +

−2

3/2

(

−t

sin 2t

17. (c) x =

(

cos ωt +

−ω

(

sin ωt

Section 7.8, page 436

1. (c) x = c

(

+ c

(

2. (c) x = c

(

+ c

(

t −

3. (c) x = c

(

−t

+ c

(

−t

(

−t

4. (c) x = c

(

−t/2

+ c

(

−t/2

(

−t/2

5. x = c

⎛

⎝

−3

⎞

⎠

−t

+ c

⎛

⎝

−1

⎞

⎠

+ c

⎡

⎣

⎛

⎝

−1

⎞

⎠

⎛

⎝

⎞

⎠

⎤

⎦

6. x = c

⎛

⎝

⎞

⎠

+ c

⎛

⎝

−1

⎞

⎠

−t

+ c

⎛

⎝

−1

⎞

⎠

−t

7. (a) x =

3 + 4t

2 + 4t

(

−3t

8. (a) x =

−1

(

−t

− 6

(

−t

August 2, 2012 13:11 bansw Sheet number 35 Page number 773 cyan black

Answers to Pr oblems 773

9. (a) x =

−2

(

t/2

−1

(

t/2

10. (a) x = 2

(

+ 14

−1

(

11. (a) x =

⎛

⎝

−1

−33

⎞

⎠

+ 4

⎛

⎝

−6

⎞

⎠

+ 3

⎛

⎝

⎞

⎠

12. (a) x =

⎛

⎝

⎞

⎠

−t/2

⎛

⎝

−7

⎞

⎠

−7t/2

13. x = c

(

t + c

(

t ln t +

(

14. x = c

(

−3

+ c

(

−3

ln t −

(

−3

16. (b)

(

=−

−2

(

−t/2

−2

(

−t/2

(

−t/2

17. (d) ξ =−ξ

(1)

(e) ξ =−(k

+ k

)ξ

(1)

, k

+ k

̸= 0

18. (b) x

(1)

(t) =

⎛

⎝

−1

⎞

⎠

(2)

(t) =

⎛

⎝

−1

⎞

⎠

⎛

⎝

⎞

⎠

(d) x

(3)

(t) =

⎛

⎝

−1

⎞

⎠

/2)e

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

(e) 8(t) = e

⎛

⎜

⎝

01 t + 2

1 t + 1 (t

/2) + t

−1 −t −(t

/2) + 3

⎞

⎟

⎠

(f) T =

⎛

⎜

⎝

012

110

−103

⎞

⎟

⎠

, T

−1

⎛

⎜

⎝

−332

3 −2 −2

−111

⎞

⎟

⎠

J =

⎛

⎜

⎝

210

021

002

⎞

⎟

⎠

19. (a) x

(1)

(t) =

⎛

⎝

⎞

⎠

, x

(2)

(t) =

⎛

⎝

−3

⎞

⎠

(d) x

(3)

(t) =

⎛

⎝

−2

−4

⎞

⎠

⎛

⎝

⎞

⎠

August 2, 2012 13:11 bansw Sheet number 36 Page number 774 cyan black

774 Answers to Problems

(e) 8(t) = e

⎛

⎝

10 −2t

02 −4t

2 −32t + 1

⎞

⎠

or e

⎛

⎝

1 −2 −2t

0 −4 −4t

222t + 1

⎞

⎠

(f) T =

⎛

⎜

⎝

1 −20

0 −40

221

⎞

⎟

⎠

, T

−1

⎛

⎜

⎝

1 −1/20

0 −1/40

−23/21

⎞

⎟

⎠

J =

⎛

⎜

⎝

100

011

001

⎞

⎟

⎠

20. (a) J

2λ

0 λ

, J

3λ

0 λ

, J

4λ

0 λ

λt

1 t

(d) x = exp(Jt)x

21. (c) exp(Jt) = e

λt

⎛

⎜

⎝

100

01 t

001

⎞

⎟

⎠

22. (c) exp(Jt) = e

λt

⎛

⎜

⎝

1 tt

01 t

00 1

⎞

⎟

⎠

Section 7.9, page 447

1. x = c

(

+ c

(

−t

(

−

(

t −

(

2. x = c

√

(

+ c

−

√

(

−2t

−

2/3

√

(

−1

√

(

−t

3. x = c

5 cos t

2 cos t + sin t

(

+ c

5 sin t

−cos t + 2 sin t

(

t cos t −

(

t sin t

−

(

cos t

4. x = c

−4

(

−3t

+ c

(

−

(

−2t

(

5. x = c

(

+ c

(

t −

− 2

(

ln t +

(

−1

−

(

−2

6. x = c

(

+ c

−2

(

−5t

(

ln t +

(

t +

−2

(

7. x = c

(

+ c

−2

(

−t

−8

(

8. x = c

(

+ c

(

−t

(

+ 2

(

9. x = c

(

−t/2

+ c

−1

(

−2t

t −

August 2, 2012 13:11 bansw Sheet number 37 Page number 775 cyan black

Answers to Pr oblems 775

10. x = c

√

(

−t

+ c

−

√

(

−4t

−

√

2 − 1

2 −

√

(

−t

2 +

√

−1 −

√

(

−t

11. x = c

5 cos t

2 cos t + sin t

(

+ c

5 sin t

−cos t + 2 sin t

(

1/2

(

t cos t −

5/2

(

t sin t

−

5/2

(

cos t

12. x =[

ln(sin t) − ln(−cos t) −

t + c

]

5 cos t

2 cos t + sin t

(

ln(sin t) −

t + c

]

5 sin t

−cos t + 2 sin t

(

13. (a) 8(t) =

−t/2

cos

−t/2

sin

−t/2

sin

t −4e

−t/2

cos

(b) x = e

−t/2

sin

4 − 4 cos

14. x = c

(

t + c

(

−1

−

(

t −

(

t ln t −

(

15. x = c

(

+ c

(

−1

(

t +

−2

(

−

(

18. [3(α

− α

) − 4]/6 = c

, (α

+ α

+ 3)/2 = c

CHAPTER 8 Section 8.1, page 460

1. (a) 1.1975, 1.38549, 1.56491, 1.73658

(b) 1.19631, 1.38335, 1.56200, 1.73308

(d) 1.19405, 1.37925, 1.55644, 1.72638

2. (a) 1.59980, 1.29288, 1.07242, 0.930175

(b) 1.61124, 1.31361, 1.10012, 0.962552

(d) 1.63301, 1.35295, 1.15267, 1.02407

3. (a) 1.2025, 1.41603, 1.64289, 1.88590

(b) 1.20388, 1.41936, 1.64896, 1.89572

(d) 1.20693, 1.42683, 1.66265, 1.91802

4. (a) 1.10244, 1.21426, 1.33484, 1.46399

(b) 1.10365, 1.21656, 1.33817, 1.46832

(d) 1.10603, 1.22110, 1.34473, 1.47688

5. (a) 0.509239, 0.522187, 0.539023, 0.559936

(b) 0.509701, 0.523155, 0.540550, 0.562089

(d) 0.510645, 0.525138, 0.543690, 0.566529

6. (a) −0.920498, −0.857538, −0.808030, −0.770038

(b) −0.922575, −0.860923, −0.812300, −0.774965

(d) −0

.926341, −0.867163, −0.820279, −0.784275

7. (a) 2.90330,

7.53999,

19.4292, 50.5614

(b) 2.93506, 7.70957, 20.1081, 52.9779

(d) 3.00306, 8.07933, 21.6163, 58.4462

August 2, 2012 13:11 bansw Sheet number 38 Page number 776 cyan black

776 Answers to Problems

8. (a) 0.891830, 1.25225, 2.37818, 4.07257

(b) 0.908902, 1.26872, 2.39336, 4.08799

(d) 0.942261, 1.30153, 2.42389, 4.11908

9. (a) 3.95713, 5.09853, 6.41548, 7.90174

(b) 3.95965, 5.10371, 6.42343, 7.91255

(d) 3.96473, 5.11411, 6.43937, 7.93424

10. (a) 1.60729, 2.46830, 3.72167, 5.45963

(b) 1.60996, 2.47460, 3.73356, 5.47774

(d) 1.61528, 2.48723, 3.75742, 5.51404

11. (a) −1.45865, −0.217545, 1.05715, 1.41487

(b) −1.45322, −0.180813, 1.05903, 1.41244

(d) −1.44190, −0.105737, 1.06290, 1.40789

12. (a) 0.587987, 0.791589, 1.14743, 1.70973

(b) 0.589440, 0.795758, 1.15693, 1.72955

(d) 0.592396, 0.804319, 1.17664, 1.77111

15. 1.595, 2.4636

16. e

n+1

=[2φ(t

) − 1]h

, |e

n+1

|≤

[

1 + 2 max

0≤t≤1

|φ(t)|

]

n+1

= e

, |e

|≤0.012, |e

|≤0.022

17. e

n+1

=[2φ(t

) − t

, |e

n+1

|≤

[

1 + 2 max

0≤t≤1

|φ(t)|

]

n+1

= 2e

, |e

|≤0.024, |e

|≤0.045

18. e

n+1

=[t

+ t

φ(t

) + φ

)]h

19. e

n+1

=[19 − 15t

−1/2

)]h

20. e

n+1

={1 +[t

+ φ(t

)]

−1/2

21. e

n+1

={2 −[φ(t

) + 2t

]exp[−t

φ(t

)]−t

exp[−2t

φ(t

)]}h

22. (a) φ(t) = 1 + (1/5π) sin5πt (b) 1.2, 1.0, 1.2 (c) 1.1, 1.1, 1.0, 1.0

(d) h < 1/

√

50π

∼

0.08

24. e

n+1

=−

′′

25. (a) 1.55, 2.34, 3.46, 5.07 (b) 1.20, 1.39, 1.57, 1.74 (c) 1.20, 1.42, 1.65, 1.90

26. (a) 0 (b) 60 (c) −92.16

27. 0.224 ̸= 0.225

Section 8.2, page 466

1. (a) 1.19512, 1.38120, 1.55909, 1.72956

(b) 1.19515, 1.38125, 1.55916, 1.72965

2. (a) 1.62283, 1.33460, 1.12820, 0.995445

(b) 1.62243, 1.33386, 1.12718, 0.994215

3. (a) 1.20526, 1.42273, 1.65511, 1.90570

(b) 1.20533, 1.42290, 1.65542, 1.90621

4. (a) 1.10483, 1.21882, 1.34146, 1.47263

(b) 1.10484, 1.21884, 1.34147, 1.47262

August 2, 2012 13:11 bansw Sheet number 39 Page number 777 cyan black

Answers to Pr oblems 777

5. (a) 0.510164, 0.524126, 0.542083, 0.564251

(b) 0.510168, 0.524135, 0.542100, 0.564277

6. (a) −0.924650, −0.864338, −0.816642, −0.780008

(b) −0.924550, −0.864177, −0.816442, −0.779781

7. (a) 2.96719, 7.88313, 20.8114, 55.5106

(b) 2.96800, 7.88755, 20.8294, 55.5758

8. (a) 0.926139, 1.28558, 2.40898, 4.10386

(b) 0.925815, 1.28525, 2.40869, 4.10359

9. (a) 3.96217, 5.10887, 6.43134, 7.92332

(b) 3.96218, 5.10889, 6.43138, 7.92337

10. (a) 1.61263, 2.48097, 3.74556, 5.49595

(b) 1.61263, 2.48092, 3.74550, 5.49589

11. (a) −1.44768, −0.144478, 1.06004, 1.40960

(b) −1.44765, −0.143690, 1.06072, 1.40999

12. (a) 0.590897, 0.799950, 1.16653, 1.74969

(b) 0.590906,

0.799988,

1.16663, 1.74992

15. e

n+1

= (38h

/3) exp(4t

), |e

n+1

|≤37, 758.8h

on 0 ≤ t ≤ 2, |e

|≤0.00193389

16. e

n+1

= (2h

/3) exp(2t

), |e

n+1

|≤4.92604h

on 0 ≤ t ≤ 1, |e

|≤0.000814269

17. e

n+1

= (4h

/3) exp(2t

), |e

n+1

|≤9.85207h

on 0 ≤ t ≤ 1, |e

|≤0.00162854

18. h

∼

0.036

19. h

∼

0.023

20. h

∼

0.081 21. h

∼

0.117

23. 1.19512, 1.38120, 1.55909, 1.72956 24. 1.62268, 1.33435, 1.12789, 0.995130

25. 1.20526, 1.42273, 1.65511, 1.90570 26. 1.10485, 1.21886, 1.34149, 1.47264

Section 8.3, page 471

1. (a) 1.19516, 1.38127, 1.55918, 1.72968

(b) 1.19516, 1.38127, 1.55918, 1.72968

2. (a) 1.62231, 1.33362, 1.12686, 0.993839

(b) 1.62230, 1.33362, 1.12685, 0.993826

3. (a) 1.20535, 1.42295, 1.65553, 1.90638

(b) 1.20535, 1.42296, 1.65553, 1.90638

4. (a) 1.10484, 1.21884, 1.34147, 1.47262

(b) 1.10484, 1.21884, 1.34147, 1.47262

5. (a) 0.510170, 0.524138, 0.542105, 0.564286

(b) 0.520169, 0.524138, 0.542105, 0.564286

6. (a) −0.924517, −0.864125, −0.816377, −0.779706

(b) −0.924517, −0.864125, −0.816377, −0.779706

7. (a) 2.96825,

7.88889,

20.8349, 55.5957

(b) 2.96828, 7.88904, 20.8355, 55.5980

8. (a) 0.925725, 1.28516, 2.40860, 4.10350

(b) 0.925711, 1.28515, 2.40860, 4.10350

9. (a) 3.96219, 5.10890, 6.43139, 7.92338

(b) 3.96219, 5.10890, 6.43139, 7.92338

10. (a) 1.61262, 2.48091, 3.74548, 5.49587

(b) 1.61262, 2.48091, 3.74548, 5.49587

11. (a) −1.44764, −0.143543, 1.06089, 1.41008

(b) −1.44764, −0.143427, 1.06095, 1.41011

12. (a) 0.590909, 0.800000, 1.166667, 1.75000

(b) 0.590909, 0.800000, 1.166667, 1.75000

August 2, 2012 13:11 bansw Sheet number 40 Page number 778 cyan black

778 Answers to Problems

Section 8.4, page 478

1. (a) 1.7296801, 1.8934697 2. (a) 0.993852, 0.925764

(b) 1.7296802, 1.8934698 (b) 0.993846, 0.925746

3. (a) 1.906382, 2.179567 4. (a) 1.4726173, 1.6126215

(b) 1.906391, 2.179582 (b) 1.4726189, 1.6126231

5. (a) 0.56428577, 0.59090918 6. (a) −0.779693, −0 .753135

(b) 0.56428581, 0.59090923 (b) −0.779692, −0.753137

7. (a) 2.96828, 7.88907, 20.8356, 55.5984

(b) 2.96829, 7.88909, 20.8357, 55.5986

8. (a) 0.9257133, 1.285148, 2.408595, 4.103495

(b) 0.9257124, 1.285148, 2.408595, 4.103495

9. (a) 3.962186, 5.108903, 6.431390, 7.923385

(b) 3.962186, 5.108903, 6.431390, 7.923385

10. (a) 1.612622, 2.480909, 3.745479, 5.495872

(b) 1.612622, 2.480909, 3.745479, 5.495873

11. (a) −1.447639, −0.1436281, 1.060946, 1.410122

(b) −1.447638, −0.1436762

1.060913, 1.410103

12. (a) 0.5909091, 0.8000000, 1.166667, 1.750000

(b) 0.5909091, 0.8000000, 1.166667, 1.750000

Section 8.5, page 481

1. (a) 1.26, 0.76; 1.7714, 1.4824; 2.58991, 2.3703; 3.82374, 3.60413;

5.64246, 5.38885

(b) 1.32493, 0.758933; 1.93679, 1.57919; 2.93414, 2.66099; 4.48318, 4.22639;

6.84236, 6.56452

6.8444, 6.56684

2. (a) 1.451, 1.232; 2.16133, 1.65988; 3.29292, 2.55559; 5.16361, 4.7916;

8.54951, 12.0464

(b) 1.51844, 1.28089; 2.37684, 1.87711; 3.85039, 3.44859; 6.6956, 9.50309;

15.0987, 64.074

15.6384, 70.3792

3. (a) 0.582, 1.18; 0.117969, 1.27344; −0.336912, 1.27382; −0.730007, 1.18572;

−1.02134, 1.02371

(b) 0.568451, 1.15775; 0.109776, 1.22556; −0.32208, 1.20347;

−0.681296, 1.10162; −0.937852, 0.937852

−0.681291, 1.10161; −0.937841, 0.93784

4. (a) −0.198, 0.618; −0.378796, 0.28329; −0.51932, −0.0321025;

−0.594324, −0.326801; −0.588278, −0.57545

August 2, 2012 13:11 bansw Sheet number 41 Page number 779 cyan black

Answers to Pr oblems 779

(b) −0.196904, 0.630936; −0.372643, 0.298888; −0.501302, −0.0111429;

−0.561270, −0.288943; −0.547053, −0.508303

−0.561292, −0.28907; −0.547031, −0.508427

5. (a) 2.96225, 1.34538; 2.34119, 1.67121; 1.90236, 1.97158; 1.56602, 2.23895;

1.29768, 2.46732

(b) 3.06339, 1.34858; 2.44497, 1.68638; 1.9911, 2.00036; 1.63818, 2.27981;

1.3555, 2.5175

1.35514, 2.51827

6. (a) 1.42386, 2.18957; 1.82234, 2.36791; 2.21728, 2.53329; 2.61118, 2.68763;

2.9955, 2.83354

(b) 1.41513, 2.18699; 1.81208, 2.36233; 2.20635, 2.5258; 2.59826, 2.6794;

2.97806, 2.82487

2.97806, 2.82488

7. For h = 0.05 and 0.025: x = 10.227, y =−4.9294; these results agree with the exact

solution to ﬁve digits.

8. 1.543, 0.0707503; 1.14743, −1.3885 9. 1.99521, −0.662442

Section 8.6, page 491

1. (b) φ

(t) − φ

(t) = 0 .001e

→∞as t →∞

2. (b) φ

(t) = ln[e

/(2 − e

)]; φ

(t) = ln[1/(1 − t)]

3. (a, b) h = 0.00025 is sufﬁcient. (c) h = 0.005 is sufﬁcient.

4. (a) y = 4e

−10t

+ (t

/4) (c) Runge–Kutta is stable for h = 0.25 but unstable for h = 0.3.

(d) h = 5/13

∼

0.384615 is small enough.

5. (a) y = t 6. (a) y = t

CHAPTER 9 Section 9.1, page 505

1. (a) r

=−1, ξ

(1)

= (1, 2)

; r

= 2, ξ

(2)

= (2, 1)

(b) saddle point, unstable

2. (a) r

= 2, ξ

(1)

= (1, 3)

; r

= 4, ξ

(2)

= (1, 1)

(b) node, unstable

3. (a) r

=−1, ξ

(1)

= (1, 3)

; r

= 1, ξ

(2)

= (1, 1)

(b) saddle point, unstable

4. (a) r

= r

=−3, ξ

(1)

= (1, 1)

(b) improper node, asymptotically stable

5. (a) r

, r

=−1 ± i; ξ

(1)

, ξ

(2)

= (2 ± i,1)

(b) spiral point, asymptotically stable

6. (a) r

, r

=±i; ξ

(1)

, ξ

(2)

= (2 ± i,1)

(b) center, stable

7. (a) r

, r

= 1 ± 2i; ξ

(1)

, ξ

(2)

= (1, 1 ∓ i)

(b) spiral point, unstable

8. (a) r

=−1, ξ

(1)

= (1, 0)

; r

=−1/4, ξ

(2)

= (4, −3)

(b) node, asymptotically stable

9. (a) r

= r

= 1, ξ

(1)

= (2, 1)

(b) improper node, unstable

10. (a) r

, r

=±3i; ξ

(1)

, ξ

(2)

= (2, −1 ± 3i)

(b) center, stable

11. (a) r

= r

=−1; ξ

(1)

= (1, 0)

, ξ

(2)

= (0, 1)

(b) proper node, asymptotically stable

12. (a) r

, r

= (1 ± 3i)/2; ξ

(1)

, ξ

(2)

= (5, 3 ∓ 3i)

(b) spiral point, unstable

13. x

= 1, y

= 1; r

√

2, r

=−

√

2; saddle point, unstable

14. x

=−1, y

= 0; r

=−1, r

=−3; node, asymptotically stable

15. x

=−2, y

= 1; r

, r

=−1 ±

√

2 i; spiral point, asymptotically stable

16. x

= γ/δ, y

= α/β; r

, r

=±

√

βδ i; center, stable

17. c

> 4km, node, asymptotically stable; c

= 4km, improper node, asymptotically stable;

< 4km, spiral point, asymptotically stable

August 2, 2012 13:11 bansw Sheet number 42 Page number 780 cyan black

780 Answers to Problems

Section 9.2, page 517

1. x = 4e

−t

, y = 2e

−2t

, y = x

2. x = 4e

−t

, y = 2e

, y = 32x

−2

; x = 4e

−t

, y = 0

3. x = 4 cos t, y = 4 sin t, x

+ y

= 16; x =−4 sin t, y = 4 cost, x

+ y

= 16

4. x =

√

a cos

√

ab t, y =−

√

b sin

√

ab t; (x

/a) + (y

/b) = 1

5. (a, c) (−

,1), saddle point, unstable; (0, 0), (proper) node, unstable

6. (a, c) (−

√

3/3, −

), saddle point, unstable; (

√

3/3, −

), center, stable

7. (a, c) (0, 0), node, unstable; (2, 0), node, asymptotically stable;

(0,

), saddle point, unstable; (−1, 3), node, asymptotically stable

8. (a, c) (0, 0), node, asymptotically stable; (1, −1), saddle point, unstable;

(1, −2), spiral point, asymptotically stable

9. (a, c) (0, 0), spiral point, asymptotically stable;

(1 −

√

2, 1 +

√

2), saddle point, unstable; (1 +

√

2, 1 −

√

2), saddle point, unstable

10. (a, c) (0, 0), saddle point, unstable; (2, 2), spiral point, asymptotically stable;

(−1, −1), spiral point, asymptotically stable; (−2, 0), saddle point, unstable

11. (a, c) (0, 0), saddle point, unstable; (0, 1), saddle point, unstable;

(

), center, stable; (−

), center, stable

12. (a, c) (0, 0), saddle point, unstable; (

√

6, 0), spiral point, asymptotically stable;

(−

√

6, 0), spiral point, asymptotically stable

13. (a, c) (0, 0), saddle point, unstable; (−2, 2), node, unstable;

(4, 4), spiral point, asymptotically stable

14. (a, c) (0, 0), saddle point, unstable; (2, 0), saddle point, unstable;

(1, 1), spiral point, asymptotically stable; (−2, −2 ), spiral point, asymptotically stable

15. (a, c) (0, 0), node, unstable; (1, 1), saddle point, unstable;

(3, −1), spiral point, asymptotically stable

16. (a, c) (0, 1), saddle point, unstable; (1, 1), node, asymptotically stable;

(−2, 4), spiral point, unstable

17. (a) 4x

− y

= c 18. (a) 4x

+ y

= c

19. (a) (y − 2x)

(x + y) = c 20. (a) arctan(y/x) − ln

+ y

= c

21. (a) 2x

y − 2xy + y

= c 22. (a) x

− 3x

y − 2y

= c

23. (a) (y

/2) − cos x = c 24. (a) x

+ y

− (x

/12) = c

Section 9.3, page 527

1. linear and nonlinear: saddle point, unstable

2. linear and nonlinear: spiral point, asymptotically stable

3. linear: center, stable; nonlinear: spiral point or center, indeterminate

4. linear: improper node, unstable; nonlinear: node or spiral point, unstable

5. (a, b, c) (0, 0); u

′

=−2u + 2v, v

′

= 4u + 4v; r = 1 ±

√

17;

saddle point, unstable

(−2, 2); u

′

= 4u, v

′

= 6u + 6v; r = 4, 6; node, unstable

(4, 4); u

′

=−6u + 6v, v

′

=−8u; r =−3 ±

√

39 i; spiral point, asymptotically

stable

6. (a, b, c) (0, 0); u

′

= u, v

′

= 3v; r = 1, 3; node, unstable

(1, 0); u

′

=−u − v, v

′

= 2v; r =−1, 2; saddle point, unstable

(0,

); u

′

=−

u, v

′

=−

u − 3v; r =−

, −3; node, asymptotically stable

(−1, 2); u

′

= u + v, v

′

=−2u − 4v; r = (−3 ±

√

17)/2; saddle point, unstable

August 2, 2012 13:11 bansw Sheet number 43 Page number 781 cyan black

Answers to Pr oblems 781

7. (a, b, c) (1, 1); u

′

=−v, v

′

= 2u − 2v; r =−1 ± i; spiral point, asymptotically

stable

(−1, 1); u

′

=−v, v

′

=−2u − 2v; r =−1 ±

√

3; saddle point, unstable

8. (a, b, c) (0, 0); u

′

= u, v

′

v; r = 1,

; node, unstable

(0, 2); u

′

=−u, v

′

=−

u −

v; r =−1, −

; node, asymptotically stable

(1, 0); u

′

=−u − v, v

′

=−

v; r =−1, −

; node, asymptotically stable

(

); u

′

=−

u −

v, v

′

=−

u −

v; r = (−5 ±

√

57)/16;

saddle point, unstable

9. (a, b, c) (0, 0); u

′

=−u + 2v, v

′

= u + 2v; r = (1 ±

√

17/2; saddle point, unstable

(2, 1); u

′

=−

u + 3v, v

′

=−2u; r = (−3 ±

√

87i)/4; spiral point, asymptotically

stable

(2, −2); u

′

=−3v, v

′

= u; r =±

√

3 i; center or spiral point, indeterminate

(4, −2); u

′

=−4v, v

′

=−u − 2v; r =−1 ±

√

5; saddle point, unstable

10. (a, b, c) (0, 0); u

′

= u, v

′

= v; r = 1, 1; node or spiral point, unstable

(−1, 0); u

′

=−u, v

′

= 2v; r =−1, 2; saddle point, unstable

11. (a, b, c) (0, 0); u

′

= 2u + v, v

′

= u − 2v; r =±

√

5; saddle point, unstable

(−1.1935, −1.4797); u

′

=−1.2399u − 6.8393v, v

′

= 2.4797u − 0.80655v;

r =−1.0232 ± 4.1125i; spiral point, asymptotically stable

12. (a, b, c) (0, ±2nπ), n = 0, 1,2, ...; u

′

= v, v

′

=−u;

r =±i; center or spiral point, indeterminate

[2, ±(2n − 1)π], n = 1, 2, 3, ...; u

′

=−3v, v

′

=−u; r =±

√

saddle point, unstable

13. (a, b, c) (0, 0); u

′

= u, v

′

= v; r = 1, 1; node or spiral point, unstable

(1, 1); u

′

= u − 2v, v

′

=−2u + v; r = 3, −1; saddle point, unstable

14. (a, b, c) (1, 1); u

′

=−u − v, v

′

= u − 3v; r =−2, −2;

node or spiral point, asymptotically stable

(−1, −1); u

′

= u + v, v

′

= u − 3v; r =−1 ±

√

5; saddle point, unstable

15. (a, b, c) (0, 0); u

′

=−2u − v, v

′

= u − v; r = (−3 ±

√

3 i)/2;

spiral point, asymptotically stable

(−0.33076, 1.0924) and (0.33076, −1.0924); u

′

=−3.5216u − 0.27735v,

′

= 0.27735u + 2.6895v; r =−3.5092, 2.6771; saddle point, unstable

16. (a, b, c) (0, 0); u

′

= u + v, v

′

=−u + v; r = 1 ± i; spiral point, unstable

17. (a,b,c) (2, 2); u

′

=−4v, v

′

=−

u +

v; r = (7 ±

√

273)/4; saddlepoint, unstable

(−2, −2); u

′

= 4v, v

′

u −

v; r = (−1 ±

√

33)/4; saddle point, unstable

(−

,2); u

′

=−4v, v

′

u; r =±

√

14 i; center or spiral point, indeterminate

(−

, −2); u

′

= 4v, v

′

=−

u; r =±

√

2 i; center or spiral point, indeterminate

18. (a, b, c) (0, 0); u

′

= 2u − v, v

′

= 2u − 4v; r =−1 ±

√

7; saddle point, unstable

(2, 1); u

′

=−3v, v

′

= 4u − 8v; r =−2, −6; node, asymptotically stable

(−2, 1); u

′

= 5v, v

′

=−4u; r =±2

√

5 i; center or spiral point, indeterminate

(−2, −4); u

′

= 10u − 5v, v

′

= 6u; r = 5 ±

√

5 i; spiral point, unstable

21. (b, c) Refer to Table 9.3.1.

23. (a) R = A, T

∼

3.17 (b) R = A, T

∼

3.20, 3.35, 3.63, 4.17

24. (b) v

∼

4.00 25. (b) v

∼

4.51

30. (a) dx/dt = y, dy/dt =−g(x) − c(x)y

(b) The linear system is dx/dt = y, dy/dt =−g

′

(0)x − c(0)y.

+ c(0)r + g

′

(0) = 0.

August 2, 2012 13:11 bansw Sheet number 44 Page number 782 cyan black

782 Answers to Problems

Section 9.4, page 541

1. (b, c) (0, 0); u

′

u, v

′

= 2v; r =

, 2; node, unstable

(0, 2); u

′

u, v

′

=−

u − 2v; r =

, −2; saddle point, unstable

(

,0); u

′

=−

u −

v, v

′

v; r =−

; saddle point, unstable

(

); u

′

=−

u −

v, v

′

=−

u −

v; r = (−22 ±

√

204)/20;

node, asymptotically stable

2. (b, c) (0, 0); u

′

u, v

′

= 2v; r =

, 2; node, unstable

(0, 4); u

′

=−

u, v

′

=−6u − 2v; r =−

, −2; node, asymptotically stable

(

,0); u

′

=−

u −

v, v

′

=−

v; r =−

, −

; node, asymptotically stable

(1, 1); u

′

=−u −

v, v

′

=−

u −

v; r = (−3 ±

√

13)/4;

saddle point, unstable

3. (b, c) (0, 0); u

′

u, v

′

= 2v; r =

, 2; node, unstable

(0, 2); u

′

=−

u, v

′

=−

u − 2v; r =−

, −2; node, asymptotically stable

(3, 0); u

′

=−

u − 3v, v

′

=−

v; r =−

, −

; node, asymptotically stable

(

); u

′

=−

u −

v, v

′

=−

u −

v; r =−1.80475, 0.30475;

saddle point, unstable

4. (b, c) (0, 0); u

′

u, v

′

v; r =

; node, unstable

(0,

); u

′

u, v

′

=−

v; r =±

; saddle point, unstable

(3, 0); u

′

=−

u − 3v, v

′

v; r =−

; saddle point, unstable

(2,

); u

′

=−u − 2v, v

′

=−

u −

v; r =−1.18301, −0.31699;

node, asymptotically stable

5. (b, c) (0, 0); u

′

= u, v

′

v; r = 1,

; node, unstable

(0,

); u

′

=−

u, v

′

=−

u −

v; r =−

, −

; node, asymptotically stable

(1, 0); u

′

=−u − v, v

′

v; r =−1,

; saddle point, unstable

6. (b, c) (0, 0); u

′

= u, v

′

v; r = 1,

; node, unstable

(0,

); u

′

u, v

′

u −

v; r =

, −

; saddle point, unstable

(1, 0); u

′

=−u +

v, v

′

v; r =−1,

; saddle point, unstable

(2, 2); u

′

=−2u + v, v

′

u − 3v; r = (−5 ±

√

3)/2;

node, asymptotically stable

8. (a) Critical points are x = 0, y = 0; x = ϵ

/σ

, y = 0; x = 0, y = ϵ

/σ

x → 0, y → ϵ

/σ

as t →∞; the redear survive.

(b) Same as part (a) except x → ϵ

/σ

, y → 0 as t →∞; the bluegill survive.

9. (a) X = (B − γ

R)/(1 − γ

), Y = (R − γ

B)/(1 − γ

)

(b) X is reduced, Y is increased; yes, if B becomes less than γ

R, then x → 0 and y → R

as t →∞.

10. (a) σ

− α

̸= 0: (0, 0), (0, ϵ

/σ

), (ϵ

/σ

,0)

− α

= 0: (0, 0), and all points on the line σ

x + α

y = ϵ

(b) σ

− α

> 0: (0, 0) is unstable node; (ϵ

/σ

,0) is saddle point;

(0, ϵ

/σ

) is asymptotically stable node.

− α

< 0: (0, 0) is unstable node; (0, ϵ

/σ

) is saddle point;

(ϵ

/σ

,0) is asymptotically stable node.

x + α

y = ϵ

are stable,nonisolated critical

points.

12. (a) (0, 0), saddle point; (0.15, 0), spiral point if γ

< 1.11,node if γ

≥ 1.11; (2, 0), saddle

point

∼

1.20

August 2, 2012 13:11 bansw Sheet number 45 Page number 783 cyan black

Answers to Pr oblems 783

13. (b) (2 −

4 −

α,

α), (2 +

4 −

α,

α)

(d) α

= 8/3; critical point is (2, 4); λ = 0, −1

14. (b) (2 −

4 −

α,

α), (2 +

4 −

α,

α)

(d) α

= 8/3; critical point is (2, 4); λ = 0, 1

15. (b) ([3 −

√

9 − 4α]/2, [3 + 2α −

√

9 − 4α]/2),

([3 +

√

9 − 4α]/2, [3 + 2α +

√

9 − 4α]/2)

(d) α

= 9/4; critical point is (3/2, 15/4); λ = 0, 0

16. (b) ([3 −

√

9 − 4α]/2, [3 + 2α −

√

9 − 4α]/2),

([3 +

√

9 − 4α]/2, [3 + 2α +

√

9 − 4α]/2)

a saddle point

(d) α

= 9/4; critical point is (3/2, 15/4); λ = 0, 0

17. (b) P

(0, 0), P

(1, 0), P

(0, α), P

(2 − 2α, −1 + 2α). P

is in the ﬁrst quadrant for

0.5 ≤ α ≤ 1.

coincides with P

. α = 0.5;P

coincides with P

. α = 1;P

coincides with

(d) J =

1 − 2x − y −x

−0.5y α − 2y − 0.5x

(e) P

is an unstable node for α>0. P

is an asymptotically stable node for 0 <α<0.5

and a saddle point for α>0.5. P

is a saddle point for 0 <α<1 and an asymptotically

stable node for α>1. P

is an asymptotically stable node for 0.5 <α<1.

18. (b) P

(0, 0), P

(1, 0), P

(0, 0.75/α), P

[(4α − 3)/(4α − 2),1/(4α − 2)]. P

is in the

ﬁrst quadrant for α ≥ 0.75.

coincides with P

(d) J =

1 − 2x − y −x

−0.5y 0.75 − 2αy − 0.5x

(e) P

is an unstable node. P

is a saddle point. P

is an asymptotically stable node for

0 <α<0.75 and a saddle point for α>0.75. P

is an asymptotically stable node for

α>0.75.

19. (b) P

(0, 0), P

(1, 0), P

(0, α), P

(0.5, 0.5). In addition, for α = 1 every point on the

line x + y = 1 is a critical point.

coincides with P

. Also α = 1.

(d) J =

1 − 2x − y −x

−(2α − 1)y α − 2y − (2α − 1)x

(e) P

is an unstable node for α>0. P

and P

are saddle points for 0 <α<1 and asymp-

totically stable nodes for α>1. P

is an asymptotically stable spiral point for

0 <α<0.5, an asymptotically stable node for 0.5 ≤ α<1, and a saddle point for α>1.

Section 9.5, page 551

1. (b, c) (0, 0); u

′

u, v

′

=−

v; r =

, −

; saddle point, unstable

(

,3); u

′

=−

v, v

′

= 3u; r =±

√

3 i/2; center or spiral point, indeterminate

2. (b, c) (0, 0); u

′

= u, v

′

=−

v; r = 1, −

; saddle point, unstable

(

,2); u

′

=−

v, v

′

= u; r =±

i; center or spiral point, indeterminate

August 2, 2012 13:11 bansw Sheet number 46 Page number 784 cyan black

784 Answers to Problems

3. (b, c) (0, 0); u

′

= u, v

′

=−

v; r = 1, −

; saddle point, unstable

(2, 0); u

′

=−u − v, v

′

v; r =−1,

; saddle point, unstable

(

); u

′

=−

u −

v, v

′

u; r = (−1 ±

√

11 i)/8; spiral point,

asymptotically stable

4. (b, c) (0, 0); u

′

u, v

′

=−v; r =

, −1; saddle point, unstable

(

,0); u

′

=−

u −

v, v

′

v; r =−

; saddle point, unstable

(1,

); u

′

=−u −

v, v

′

u; r = (−1 ±

√

0.5)/2; node, asymptotically stable

5. (b, c) (0, 0); u

′

=−u, v

′

=−

v; r =−1, −

; node, asymptotically stable

(

,0); u

′

u −

v, v

′

=−v; r =−1,

; saddle point, unstable

(2, 0); u

′

=−3u −

v, v

′

v; r =−3,

; saddle point, unstable

(

); u

′

=−

u −

v, v

′

u; r = (−3 ±

√

39 i)/8; spiral point,

asymptotically stable

6. (b, c) t = 0, T,2T, ...: H is a max., dP/dt is a max.

t = T/4, 5T/4, ...: dH/dt is a min., P is a max.

t = T/2, 3T/2, ...: H is a min., dP/dt is a min.

t = 3T/4, 7T/4, ...: dH/dt is a max., P is a min.

7. (a)

√

c α/

√

a γ (b)

√

(d) The ratio of prey amplitude to predator amplitude increases very slowly as the initial

point moves away from the equilibrium point.

8. (a) 4π/

√

∼

7.2552

9. (a) T

∼

6.5 (b) T

∼

3.7, T

∼

11.5 (c) T

∼

3.8, T

∼

11.1

11. (a) P

(0, 0), P

(1/σ,0), P

(3, 2 − 6σ); P

moves to the left and P

moves down; they

coincide at (3, 0) when σ = 1/3.

(b) P

is a saddle point. P

is a saddle point for σ<1/3 and an asymptotically stable node

for σ>1/3. P

is an asymptotically stable spiral point forσ<σ

= (

7/3 − 1)/2

∼

0.2638,

an asymptotically stable node for σ

<σ<1/3, and a saddle point for σ>1/3.

12. (a) P

(0, 0), P

(a/σ,0), P

[c/γ, (a/α) − (cσ/αγ)]; P

moves to the left and P

moves

down; they coincide at (c/γ,0) when σ = aγ/c.

(b) P

is a saddle point. P

is a saddle point for σ<aγ/c and an asymptotically stable

node for σ>aγ/c. P

is an asymptotically stable spiral point for sufﬁciently small values

of σ and becomes an asymptotically stable node at a certain value σ

< aγ/c. P

is a saddle

point for σ>aγ/c.

13. (a, b) P

(0, 0) is a saddle point; P

(5, 0) is a saddle point; P

(2, 2.4) is an asymptotically

stable spiral point.

14. (b) same prey, fewer predators (c) more prey, same predators

(d) more prey, fewer predators

15. (b) same prey, fewer predators (c) more prey, fewer predators

(d) more prey, even fewer predators

16. (b) same prey, fewer predators (c) more prey, same predators

(d) more prey, fewer predators

Section 9.7, page 574

1. r = 1, θ = t + t

, stable limit cycle 2. r = 1, θ =−t + t

, semistable limit cycle

3. r = 1, θ = t + t

, stable limit cycle; r = 3, θ = t + t

, unstable periodic solution

4. r = 1, θ =−t + t

, unstable periodic solution; r = 2, θ =−t + t

, stable limit cycle

5. r = 2n − 1, θ = t +t

, n = 1, 2, 3,..., stable limit cycle;

r = 2n, θ = t + t

, n = 1, 2, 3, ..., unstable periodic solution

August 2, 2012 13:11 bansw Sheet number 47 Page number 785 cyan black

Answers to Pr oblems 785

6. r = 2, θ =−t + t

, semistable limit cycle;

r = 3, θ =−t + t

, unstable periodic solution

8. (a) Counterclockwise

(b) r = 1, θ = t + t

, stable limit cycle; r = 2, θ = t + t

, semistable limit cycle;

r = 3, θ = t + t

, unstable periodic solution

9. r =

√

2, θ =−t + t

, unstable periodic solution

14. (a) µ = 0.2, T

∼

6.29; µ = 1, T

∼

6.66; µ = 5, T

∼

11.60

15. (a) x

′

= y, y

′

=−x + µy − µy

(b) 0 <µ<2, unstable spiral point; µ ≥ 2, unstable node

∼

2.16, T

∼

6.65

(d) µ = 0.2, A

∼

1.99, T

∼

6.31; µ = 0.5, A

∼

2.03, T

∼

6.39;

µ = 2, A

∼

2.60, T

∼

7.65; µ = 5, A

∼

4.36, T

∼

11.60

16. (b) x

′

= µx + y, y

′

=−x + µy; λ = µ ± i ; the origin is an asymptotically stable

spiral point for µ<0 and an unstable spiral point for µ>0.

′

= r(µ − r

), θ

′

=−1

17. (a) The origin is an asymptotically stable node for µ<−2, an asymptotically stable spiral

point for −2 <µ<0, an unstable spiral point for 0 <µ<2, and an unstable node for

µ>2.

18. (a, b) (0, 0) is a saddle point; (12, 0) is a saddle point; (2, 8) is an unstable spiral point.

19. (a) (0, 0), (5a,0), (2, 4a − 1.6)

(b) r =−0.25 + 0.125a ± 0.025

√

220 − 400a + 25a

; a

= 2

20. (b) λ =

;

−(5/4 − b) ±

(5/4 − b)

− 1

point; 5/4 < b < 9/4: unstable spiral point; 9/4 < b; unstable node

(d) b

= 5/4

21. (b) k = 0, (1.1994, −0.62426); k = 0.5, (0.80485, −0.13106)

∼

0.3465, (0.95450, −0.31813)

(d) k = 0.4, T

∼

11.23; k = 0.5, T

∼

10.37; k = 0.6, T

∼

9.93

(e) k

∼

1.4035

Section 9.8, page 584

1. (b) λ = λ

, ξ

(1)

= (0, 0, 1)

; λ = λ

, ξ

(2)

= (20, 9 −

√

81 + 40r,0)

;

λ = λ

, ξ

(3)

= (20, 9 +

√

81 + 40r,0)

∼

−2.6667, ξ

(1)

= (0, 0, 1)

; λ

∼

−22.8277, ξ

(2)

∼

(20, −25.6554, 0)

;

∼

11.8277, ξ

(3)

∼

(20, 43.6554, 0)

2. (c) λ

∼

−13.8546; λ

, λ

∼

0.0939556 ± 10.1945i

5. (a) dV/dt =−2σ[rx

+ y

+ b(z − r)

− br

]

11. (b) c =

√

0.5 : P

(

√

2/4, −

√

2); λ = 0, −0.05178 ± 1.5242i

c = 1 : P

= (0.8536, −3.4142, 3.4142); λ = 0.1612, −0 .02882 ± 2.0943i

(0.1464, −0.5858, 0.5858); λ =−0.5303, −0.03665 ± 1.1542i

12. (a) P

(1.1954, −4.7817, 4.7817); λ = 0.1893, −0.02191 ± 2.4007i

(0.1046, −0.4183, 0.4183); λ =−0.9614, 0.007964 ± 1.0652i

(d) T

∼

5.9

13. (a, b, c) c

∼

1.243

14. (a) P

(2.9577, −11.8310, 11.8310); λ = 0.2273, −0.009796 ± 3.5812i

(0.04226, −0.1690, 0.1690); λ =−2.9053, 0.09877 ± 0.9969i

∼

11.8

15. (a) P

(3.7668, −15.0673, 15.0673); λ = 0.2324, −0.007814 ± 4.0078i

(0.03318, −0.1327, 0.1327); λ =−3.7335, 0.1083 ± 0.9941i

(b) T

∼

23.6

August 2, 2012 13:11 bansw Sheet number 48 Page number 786 cyan black

786 Answers to Problems

CHAPTER 10 Section 10.1, page 595

1. y =−sin x 2. y = (cot

√

2π cos

√

2x + sin

√

2x)/

√

3. y = 0 for all L; y = c

sin x if sin L = 0

4. y =−tan L cos x + sin x if cos L ̸= 0; no solution if cos L = 0

5. No solution

6. y = (−π sin

√

2x + x sin

√

2π)/2 sin

√

2π

7. No solution 8. y = c

sin 2x +

sin x

9. y = c

cos 2x +

cos x 10. y =

cos x

11. y =−

x +

12. y =−

−1

(1 − e

−1

ln x +

13. No solution

14. λ

=[(2n − 1)/2]

, y

(x) = sin[(2n − 1)x/2]; n = 1, 2, 3, ...

15. λ

=[(2n − 1)/2]

, y

(x) = cos[(2n − 1)x/2]; n = 1, 2, 3, ...

16. λ

= 0, y

(x) = 1; λ

= n

, y

(x) = cos nx ; n = 1, 2, 3, ...

17. λ

=[(2n − 1)π/2L]

, y

(x) = cos[(2n − 1)πx/2L]; n = 1, 2, 3, ...

18. λ

= 0, y

(x) = 1; λ

= (nπ/L)

, y

(x) = cos(nπx/L); n = 1, 2, 3, ...

19. λ

=−[(2n − 1)π/2L]

, y

(x) = sin[(2n − 1)πx/2L]; n = 1, 2, 3, ...

20. λ

= 1 + (nπ/ ln L)

, y

(x) = x sin(nπ ln x/ ln L); n = 1, 2, 3, ...

21. (a) w(r) = G(R

− r

)/4µ (c) Q is reduced to 0.3164 of its original value.

22. (a) y = k(x

− 2Lx

+ L

x)/24 (b) y = k(x

− 2Lx

+ L

)/24

− 4Lx

+ 6L

)/24

Section 10.2, page 605

1. T = 2π/5 2. T = 1

3. Not periodic 4. T = 2L

5. T = 1 6. Not periodic

7. T = 2 8. T = 4

9. f (x) = 2L − x in L < x < 2L; f (x) =−2L − x in −3L < x < −2L

10. f (x) = x − 1 in 1 < x < 2; f (x) = x − 8 in 8 < x < 9

11. f (x) =−L − x in −L < x < 0

13. (b) f (x) =

∞

n=1

(−1)

sin

nπx

14. (b) f (x) =

−

∞

n=1

sin[(2n − 1)πx/L]

2n − 1

15. (b) f (x) =−

∞

n=1

2 cos(2n − 1)x

π(2n − 1)

(−1)

n+1

sin nx

16. (b) f (x) =

∞

n=1

cos(2n − 1)πx

(2n − 1)

17. (b) f (x) =

∞

n=1

2L cos[(2n − 1)πx/L]

(2n − 1)

(−1)

n+1

L sin(nπx/L)

nπ

18. (b) f (x) =

∞

n=1

−

nπ

cos

nπ

(

sin

nπ

sin

nπx

19. (b) f (x) =

∞

n=1

sin[(2n − 1)πx/2]

2n − 1

20. (b) f (x) =

∞

n=1

(−1)

n+1

sin nπx

21. (b) f (x) =

∞

n=1

(−1)

cos

nπx

August 2, 2012 13:11 bansw Sheet number 49 Page number 787 cyan black

Answers to Pr oblems 787

22. (b) f (x) =

∞

n=1

cos[(2n − 1)πx/2]

(2n − 1)

∞

n=1

(−1)

sin

nπx

23. (b) f (x) =

∞

n=1

(−1)

− 5

cos

nπx

∞

n=1

4[1 − (−1)

]

−

(−1)

nπ

sin

nπx

24. (b) f (x) =

∞

n=1

162[(−1)

− 1]

−

27(−1)

cos

nπx

−

∞

n=1

108(−1)

+ 54

sin

nπx

25. (b) m = 81

26. (b) m = 27

28.

f (t) dt may not be periodic; for example, let f (t) = 1 + cos t.

Section 10.3, page 612

1. (a) f (x) =

∞

n=1

sin(2n − 1)πx

2n − 1

2. (a) f (x) =

−

∞

n=1

(2n − 1)

cos(2n − 1)x +

(−1)

sin nx

3. (a) f (x) =

∞

n=1

cos[(2n − 1)πx/L]

(2n − 1)

4. (a) f (x) =

∞

n=1

(−1)

n+1

cos nπx

5. (a) f (x) =

∞

n=1

(−1)

n−1

2n − 1

cos(2n − 1)x

6. (a) f (x) =

∞

n=1

cos nπx + b

sin nπx);

, a

2(−1)

, b

−1/nπ, n even

1/nπ − 4 /n

, n odd

7. (a) f (x) =−

∞

n=1

1 − cos nπ

πn

cos nx −

(−1)

sin nx

(b) n = 10; max|e|=1.6025 at x =±π

n = 20; max|e|=1.5867 at x =±π

n = 40; max|e|=1.5788 at x =±π

8. (a) f (x) =

∞

n=1

1 − cos nπ

cos nπx

(b) n = 10; max|e|=0.02020 at x = 0, ±1

n = 20; max|e|=0 .01012 at x = 0, ±1

n = 40; max|e|=0.005065 at x = 0, ±1

9. (a) f (x) =

∞

n=1

(−1)

n+1

sin nπx

(b) n = 10, 20, 40; max|e |=1 at x ± 1

10. (a) f (x) =

∞

n=1

6(1 − cos nπ)

cos

nπx

2 cos nπ

nπ

sin

nπx

August 2, 2012 13:11 bansw Sheet number 50 Page number 788 cyan black

788 Answers to Problems

(b) n = 10; lub|e|=1.0606 as x → 2

n = 20; lub|e|=1.0304 as x → 2

n = 40; lub|e|=1.0152 as x → 2

11. (a) f (x) =

∞

n=1

2 cos nπ

cos nπx −

2 − 2 cos nπ + n

cos nπ

sin nπx

(b) n = 10; lub|e|=0.5193 as x → 1

n = 20; lub|e|=0.5099 as x → 1

n = 40; lub|e|=0.5050 as x → 1

12. (a) f (x) =−

∞

n=1

(−1)

sin nπx

(b) n = 10; max|e|=0.001345 at x =±0.9735

n = 20; max|e|=0.0003534 at x =±0.9864

n = 40; max|e|=0.00009058 at x =±0.9931

13. y = (ω sin nt − n sin ωt)/ω(ω

− n

), ω

̸= n

y = (sin nt − nt cos nt)/2n

, ω

= n

14. y =

∞

n=1

(ω sin nt − n sin ωt)/ω(ω

− n

), ω ̸= 1, 2, 3, ...

y =

∞

n=1

n̸=m

(m sin nt − n sin mt)/m(m

− n

) + b

(sin mt − mt cos mt)/2m

, ω = m

15. y =

∞

n=1

− (2n − 1)

2n − 1

sin(2n − 1)t −

sin ωt

16. y = cos ωt +

2ω

(1 − cos ωt) +

∞

n=1

cos(2n − 1)πt − cos ωt

(2n − 1)

[ω

− (2n − 1)

]

Section 10.4, page 620

1. Odd 2. Neither

3. Odd 4. Even

5. Even 6. Neither

14. f (x) =

∞

n=1

1 − cos(nπ/2)

cos

nπx

f (x) =

∞

n=1

(nπ/2) − sin(nπ/2)

sin

nπx

15. (a) f (x) =

∞

n=1

(−1)

n−1

2n − 1

cos

(2n − 1)πx

16. (a) f (x) =

∞

n=1

nπ

−cos nπ +

nπ

sin

nπ

(

sin

nπx

17. (a) f (x) = 1

18. (a) f (x) =

∞

n=1

sin(2n − 1)x

2n − 1

19. (a) f (x) =

∞

n=1

nπ

cos

nπ

+ cos

2nπ

− 2 cos nπ

(

sin

August 2, 2012 13:11 bansw Sheet number 51 Page number 789 cyan black

Answers to Pr oblems 789

20. (a) f (x) =

−

∞

n=1

sin 2nπx

21. (a) f (x) =

∞

n=1

cos[(2n − 1)πx/L]

(2n − 1)

22. (a) f (x) =

∞

n=1

sin(nπx/L)

23. (a) f (x) =

∞

n=1

2π

sin

nπ

)

cos

nπ

− 1

cos

24. (a) f (x) = 2

∞

n=1

(−1)

sin nx

25. (a) f (x) =

∞

n=1

(1 + cos nπ)

16(1 − cos nπ)

sin

nπx

26. (a) f (x) =

∞

n=1

1 + 3 cos nπ

cos

nπx

27. (b) g(x) =

∞

n=1

1 − cos nπ

cos

nπx

h(x) =

∞

n=1

sin

nπx

28. (b) g(x) =

∞

n=1

4 cos(nπ/2) + 2nπ sin(nπ/2) − 4

cos

nπx

h(x) =

∞

n=1

4 sin(nπ/2) − 2nπ cos(nπ/2)

sin

nπx

29. (b) g(x) =−

∞

n=1

12 cos nπ + 4

cos

nπx

h(x) =−

∞

n=1

(3 + 5 cos nπ) + 32(1 − cos nπ)

sin

nπx

30. (b) g(x) =

∞

n=1

(2 cos nπ − 5) + 324(1 − cos nπ)

cos

nπx

h(x) =

∞

n=1

4 cos nπ + 2

nπ

144 cos nπ + 180

sin

nπx

40. (a) Extend f (x) antisymmetrically into (L,2L]; that is, f (2L − x) =−f (x) for 0 ≤ x < L.

Then extend this function as an even function into (−2L,0).

Section 10.5, page 630

1. xX

′′

− λX = 0, T

′

+ λT = 0 2. X

′′

− λxX = 0, T

′

+ λtT = 0

3. X

′′

− λ(X

′

+ X ) = 0, T

′

+ λT = 0 4. [p(x)X

′

]

′

+ λr(x)X = 0, T

′′

+ λT = 0

5. Not separable 6. X

′′

+ (x + λ)X = 0, Y

′′

− λY = 0

7. u(x, t) = e

−400π

sin 2πx − e

−2500π

sin 5πx

8. u(x, t) = 2e

−π

t/16

sin(πx/2) − e

−π

t/4

sin πx + 4e

−π

sin 2πx

August 2, 2012 13:11 bansw Sheet number 52 Page number 790 cyan black

790 Answers to Problems

9. u(x, t) =

100

∞

n=1

1 − cos nπ

−n

t/1600

sin

nπx

10. u(x, t) =

160

∞

n=1

sin(nπ/2)

−n

t/1600

sin

nπx

11. u(x, t) =

100

∞

n=1

cos(nπ/4) − cos(3nπ/4)

−n

t/1600

sin

nπx

12. u(x, t) =

∞

n=1

(−1)

n+1

−n

t/1600

sin

nπx

13. t = 5, n = 16; t = 20, n = 8; t = 80, n = 4

14. (d) t = 673.35

15. (d) t = 451.60

16. (d) t = 617.17

17. (b) t = 5, x = 33.20; t = 10, x = 31.13; t = 20, x = 28.62; t = 40, x = 25.73;

t = 100, x = 21.95; t = 200, x = 20.31

(e) t = 524.81

18. u(x, t) =

200

∞

n=1

1 − cos nπ

−n

t/400

sin

nπx

(a) 35.91

◦

C (b) 67.23

◦

C (c) 99.96

◦

19. (a) 76.73 s (b) 152.56 s (c) 1093.36 s

21. (a) aw

− bw

+ (c − bδ)w = 0 (b) δ = c/b if b ̸= 0

22. X

′′

+ µ

X = 0, Y

′′

+ (λ

− µ

)Y = 0, T

′

+ α

T = 0

23. r

′′

+ rR

′

+ (λ

− µ

)R = 0, ;

′′

+ µ

; = 0, T

′

+ α

T = 0

Section 10.6, page 639

1. u = 10 +

x 2. u = 30 −

3. u = 0 4. u = T

5. u = 0 6. u = T

7. u = T(1 + x)/(1 + L) 8. u = T(1 + L − x)/(1 + L)

9. (a) u(x, t) = 3x +

∞

n=1

70 cos nπ + 50

nπ

−0.86n

t/400

sin

nπx

(d) 160.29 s

10. (a) f (x) = 2x,0≤ x ≤ 50; f (x) = 200 − 2x,50< x ≤ 100

(b) u(x, t) = 20 −

∞

n=1

−1.14n

t/(100)

sin

nπx

100

, c

800

sin

nπ

−

nπ

(d) u(50, t) → 10 as t →∞; 3754 s

11. (a) u(x, t) = 30 − x +

∞

n=1

−n

t/900

sin

nπx

[2(1 − cos nπ) − n

(1 + cos nπ)]

12. (a) u(x, t) =

∞

n=1

−n

t/L

cos

nπx

0, n odd;

−4/(n

− 1)π, n even

(b) lim

t→∞

u(x, t) = 2/π

August 2, 2012 13:11 bansw Sheet number 53 Page number 791 cyan black

Answers to Pr oblems 791

13. (a) u(x, t) =

200

∞

n=1

−n

t/6400

cos

nπx

=−

160

(3 + cos nπ)

14. (a) u(x, t) =

∞

n=1

−n

t/900

cos

nπx

nπ

)

sin

nπ

− sin

nπ

15. (b) u(x, t) =

∞

n=1

−(2n−1)

t/4L

sin

(2n − 1)πx

f (x) sin

(2n − 1)πx

16. (a) u(x, t) =

∞

n=1

−(2n−1)

t/3600

sin

(2n − 1)πx

120

(2n − 1)

[2 cos nπ + (2n − 1)π]

increases from x = 0 and reaches x = 30 when t = 104.4.

17. (a) u(x, t) = 40 +

∞

n=1

−(2n−1)

t/3600

sin

(2n − 1)πx

(2n − 1)

[6 cos nπ − (2n − 1)π]

19. u(x) =

⎧

⎪

⎨

⎪

⎩

ξ + (L/a) − 1

,0≤ x ≤ a,

1 −

L − x

ξ + (L/a) − 1

, a ≤ x ≤ L,

where ξ = κ

/κ

20. (e) u

(x, t) = e

−µ

sin µ

21. α

′′

+ s(x) = 0; v(0) = T

, v(L) = T

= α

; w(0, t) = 0, w(L, t) = 0, w(x,0) = f (x) − v(x)

22. (a) v(x) = T

+ (T

− T

)(x/L) + kLx/2 − kx

(b) w(x, t) =

∞

n=1

−n

t/400

sin

nπx

, c

160 (cos nπ − 1)

23. (a) v(x) = T

+ (T

− T

)x/L + kLx/6 − kx

/6L

(b) w(x, t) =

∞

n=1

−n

t/400

sin

nπx

(3 cos(nπ) − 1) + 60 cos nπ

Section 10.7, page 652

1. (a) u(x, t) =

∞

n=1

sin

nπ

sin

nπx

cos

nπat

2. (a) u(x, t) =

∞

n=1

sin

nπ

+ sin

3nπ

(

sin

nπx

cos

nπat

August 2, 2012 13:11 bansw Sheet number 54 Page number 792 cyan black

792 Answers to Problems

3. (a) u(x, t) =

∞

n=1

2 + cos nπ

sin

nπx

cos

nπat

4. (a) u(x, t) =

∞

n=1

sin(nπ/2) sin(nπ/L)

sin

nπx

cos

nπat

5. (a) u(x, t) =

aπ

∞

n=1

sin

nπ

sin

nπx

sin

nπat

6. (a) u(x, t) =

aπ

∞

n=1

sin(nπ/4) + sin(3nπ/4)

sin

nπx

sin

nπat

7. (a) u(x, t) =

32L

aπ

∞

n=1

cos nπ + 2

sin

nπx

sin

nπat

8. (a) u(x, t) =

aπ

∞

n=1

sin(nπ/2) sin(nπ/L)

sin

nπx

sin

nπat

9. u(x, t) =

∞

n=1

sin

(2n − 1)πx

cos

(2n − 1)πat

f (x) sin

(2n − 1)πx

10. (a) u(x, t) =

∞

n=1

2n − 1

sin

(2n − 1)π

sin

(2n − 1)π

sin

(2n − 1)πx

cos

(2n − 1)πat

11. (a) u(x, t) =

512

∞

n=1

(2n − 1)π + 3 cos nπ

(2n − 1)

sin

(2n − 1)πx

cos

(2n − 1)πat

14. (b) φ(x + at) represents a wave moving in the negative x direction with speed a > 0.

15. Using g =32.2 ft/s

: (a) 249 ft/s (b) 49.8nπ rad/s

21. r

′′

+ rR

′

+ (λ

− µ

)R = 0, ;

′′

+ µ

; = 0, T

′′

+ λ

T = 0

23. (b) a

= a

1 + (γ

)(c)γ= 0

24. (a) c

2 sin

nπ

− sin

2nπ

− sin

3nπ

(

Section 10.8, page 665

1. (a) u(x, y) =

∞

n=1

sin

nπx

sinh

nπy

, c

2/a

sinh(nπb/a)

g(x) sin

nπx

(b) u(x, y) =

∞

n=1

sin(nπ/2)

sinh(nπb/a)

sin

nπx

sinh

nπy

2. u(x, y) =

∞

n=1

sin

nπx

sinh

nπ(b − y)

, c

2/a

sinh(nπb/a)

h(x) sin

nπx

3. (a) u(x, y) =

∞

n=1

(1)

sinh

nπx

sin

nπy

∞

n=1

(2)

sin

nπx

sinh

nπ(b − y)

(1)

2/b

sinh(nπa/b)

f (y) sin

nπy

dy, c

(2)

2/a

sinh(nπb/a)

h(x) sin

nπx

(b) c

(1)

nπ sinh(nπa/b)

, c

(2)

=−

− 2) cos nπ + 2

sinh(nπb/a)

5. u(r, θ) =

∞

n=1

−n

cos nθ + k

sin nθ);

2π

f (θ) cos nθ dθ, k

2π

f (θ) sin nθ dθ

August 2, 2012 13:11 bansw Sheet number 55 Page number 793 cyan black

Answers to Pr oblems 793

6. (a) u(r, θ) =

∞

n=1

sin nθ, c

πa

f (θ) sin nθ dθ

(b) c

πa

1 − cos nπ

7. u(r, θ) =

∞

n=1

nπ/α

sin

nπθ

, c

= (2/α)a

−nπ/α

f (θ) sin

nπθ

dθ

8. (a) u(x, y) =

∞

n=1

−nπy/a

sin

nπx

, c

f (x) sin

nπx

(b) c

(1 − cos nπ) (c) y

∼

6.6315

10. (b) u(x, y) = c

∞

n=1

cosh

nπx

cos

nπy

, c

2/nπ

sinh(nπa/b)

f (y) cos

nπy

11. u(r, θ) = c

∞

n=1

cos nθ + k

sin nθ),

nπa

n−1

2π

g(θ) cos nθ dθ, k

nπa

n−1

2π

g(θ) sin nθ dθ;

necessary condition is

2π

g(θ) dθ = 0.

12. (a) u(x, y) =

∞

n=1

sin

nπx

cosh

nπy

, c

2/a

cosh(nπb/a)

g(x) sin

nπx

(b) c

4a sin(nπ/2)

cosh(nπb/a)

13. (a) u(x, y) =

∞

n=1

sinh

(2n − 1)πx

sin

(2n − 1)πy

2/b

sinh[(2n − 1)πa/2b]

f (y) sin

(2n − 1)πy

(b) c

32b

(2n − 1)

sinh[(2n − 1)πa/2b]

14. (a) u(x, y) =

∞

n=1

cos

nπx

sinh

nπy

g(x) dx, c

2/a

sinh(nπb/a)

g(x) cos

nπx

(b) c

1 +

(

, c

=−

24a

(1 + cos nπ)

sinh(nπb/a)

16. (a) u(x, z) = b +

αa

−

4αa

∞

n=1

cos[(2n − 1)πx/a]cosh[(2n − 1)πz/a]

(2n − 1)

cosh[(2n − 1)πb/a]

CHAPTER 11 Section 11.1, page 682

1. Homogeneous 2. Nonhomogeneous

3. Nonhomogeneous 4. Homogeneous

5. Nonhomogeneous 6. Homogeneous

7. (a) φ

(x) = sin

√

x, where

√

satiﬁes

√

λ =−tan

√

λπ; (b) No

∼

0.6204, λ

∼

2.7943

(d) λ

∼

(2n − 1)

/4 for large n

August 2, 2012 13:11 bansw Sheet number 56 Page number 794 cyan black

794 Answers to Problems

8. (a) φ

(x) = cos

√

x, where

√

satisﬁes

√

λ = cot

√

λ; (b) No

∼

0.7402, λ

∼

11.7349

(d) λ

∼

(n − 1)

for large n

9. (a) φ

(x) = sin

√

x +

√

cos

√

x, where

√

satisﬁes

(λ − 1) sin

√

λ − 2

√

λ cos

√

λ = 0; (b) No

∼

1.7071, λ

∼

13.4924

(d) λ

∼

(n − 1)

for large n

10. (a) For n = 1, 2,3, ..., φ

(x) = sin µ

x − µ

cos µ

x and λ

=−µ

, where µ

satisﬁes

µ = tan µ.

(b) Yes; λ

= 0, φ

(x) = 1 − x

∼

−20.1907, λ

∼

−59.6795

(d) λ

∼

−(2n + 1)

/4 for large n

12. µ(x) = e

−x

13. µ(x) = 1/x

14. µ(x) = e

−x

15. µ(x) = (1 − x

)

−1/2

16. X

′′

+ λX = 0, T

′′

+ cT

′

+ (k + λa

)T = 0

17. (a) s(x) = e

(b) λ

= n

, φ

(x) = e

sin nπx; n = 1, 2, 3,...

18. Positive eigenvalues λ = λ

, where

√

satisﬁes

√

λ =

tan 3

√

λL; corresponding eigen-

functions are φ

(x) = e

−2x

sin 3

√

x.IfL =

, λ

= 0 is eigenvalue, φ

(x) = xe

−2x

eigenfunction;if L ̸=

, λ = 0 is noteigenvalue.IfL ≤

,there are no negative eigenvalues;

if L >

, there is one negative eigenvalue λ =−µ

, where µ is a root of µ =

tanh 3µL;

corresponding eigenfunction is φ

−1

(x) = e

−2x

sinh 3µx.

19. No real eigenvalues.

20. Only eigenvalue is λ = 0; eigenfunction is φ(x) = x − 1.

21. (a) 2 sin

√

λ −

√

λ cos

√

λ = 0 (c) λ

∼

18.2738, λ

∼

57.7075

(d) 2 sinh

√

µ −

√

µ cosh

√

µ = 0, µ =−λ (e) λ

−1

∼

−3.6673

24. (a) λ

= µ

, where µ

is a root of sin µL sinh µL = 0, hence λ

= (nπ/L)

;

∼

97.409/L

, λ

∼

1558.5/L

, φ

(x) = sin(nπx/L)

(b) λ

= µ

, where µ

is a root of sin µL cosh µL − cos µL sinh µL = 0;

∼

237.72/L

, λ

∼

2496.5/L

, φ

sin µ

x sinh µ

L − sin µ

L sinh µ

sinh µ

= µ

, where µ

is a root of 1 + cosh µL cos µL = 0; λ

∼

12.362/L

∼

485.52/L

(x) =

[(sin µ

x − sinh µ

x)(cos µ

L + cosh µ

L) + (sin µ

L + sinh µ

L)(cosh µ

x − cos µ

x)]

cos µ

L + cosh µ

25. (c) φ

(x) = sin

√

x, where λ

satisﬁes cos

√

L − γ

√

L sin

√

L = 0

(d) λ

∼

1.1597/L

, λ

∼

13.276/L

Section 11.2, page 695

1. φ

(x) =

√

2 sin(n −

)πx; n = 1, 2, ...

2. φ

(x) =

√

2 cos(n −

)πx; n = 1, 2, ...

3. φ

(x) = 1, φ

(x) =

√

2 cos nπx; n = 1, 2, ...

4. φ

(x) =

√

2 cos

√

(1 + sin

√

)

1/2

, where λ

satisﬁes cos

√

−

√

sin

√

= 0

5. φ

(x) =

√

2 e

sin nπx; n = 1, 2, ...

6. a

√

(2n − 1)π

; n = 1, 2, ...

7. a

√

2(−1)

n−1

(2n − 1)

; n = 1, 2, ...

8. a

√

(2n − 1)π

{1 − cos[(2n − 1)π/4]}; n = 1, 2, ...

August 2, 2012 13:11 bansw Sheet number 57 Page number 795 cyan black

Answers to Pr oblems 795

9. a

√

2 sin

n −

(n −

)

; n = 1, 2, ...

In Problems 10 through 13, α

= (1 + sin

√

)

1/2

and cos

√

−

√

sin

√

= 0.

10. a

√

2 sin

√

; n = 1, 2, ...

11. a

√

2(2 cos

√

− 1)

; n = 1, 2, ...

12. a

√

2(1 − cos

√

)

; n = 1, 2, ... 13. a

√

2 sin(

√

/2)

√

; n = 1, 2, ...

14. Not self-adjoint 15. Self-adjoint

16. Not self-adjoint 17. Self-adjoint

18. Self-adjoint

21. (a) If a

= 0 or b

= 0, then the corresponding boundary term is missing.

25. (a) λ

= π

; φ

(x) = sin(πx/L)

(b) λ

∼

(4.4934)

; φ

(x) = sin

√

x −

√

x cos

√

= (2π)

; φ

(x) = 1 − cos(2πx/L)

26. λ

= π

/4L

; φ

(x) = 1 − cos(πx/2L)

27. (a) X

′′

− (v/D)X

′

+ λX = 0, X (0) = 0, X

′

(L) = 0; T

′

+ λDT = 0

(e) c(x, t) =

∞

n=1

−λ

vx/2D

sin µ

x, where λ

= µ

+ (v

/4D

);

4Dµ

−vx/2D

f (x) sinµ

xdx

(2LDµ

+ v sin

28. (a) u

+ vu

= Du

, u(0, t) = 0, u

(L, t) = 0, u(x,0) =−c

(b) u(x, t) =

∞

n=1

−λ

vx/2D

sin µ

x, where λ

= µ

+ (v

/4D

);

(2Dµ

−vL/2D

cos(µ

L) + ve

−vL/2D

sin(µ

L) − 2Dµ

)

+ 4D

)(2LDµ

+ v sin

(µ

L))

Section 11.3, page 708

1. y = 2

∞

n=1

(−1)

n+1

sin nπx

− 2)nπ

2. y = 2

∞

n=1

(−1)

n+1

sin(n −

)πx

[(n −

)

− 2](n −

)

3. y =−

− 4

∞

n=1

cos(2n − 1)πx

[(2n − 1)

− 2](2n − 1)

4. y = 2

∞

n=1

(2 cos

√

− 1) cos

√

(λ

− 2)(1 + sin

√

)

5. y = 8

∞

n=1

sin(nπ/2) sin nπx

− 2)n

6–9. For each problem the solution is

y =

∞

n=1

− µ

(x), c

f (x)φ

(x) dx, µ ̸= λ

August 2, 2012 13:11 bansw Sheet number 58 Page number 796 cyan black

796 Answers to Problems

where φ

(x) is given in Problems 1–4, respectively, in Section 11.2, and λ

is the corresponding

eigenvalue. In Problem 8 summation starts at n = 0.

10. a =−

, y =

2π

cos πx +

x −

(

+ c sin πx

11. No solution

12. a is arbitrary, y = c cos πx + a/π

13. a = 0, y = c sin πx − (x/2π) sin π x

17. v(x) = a + (b − a)x

18. v(x) = 1 −

19. u(x, t) =

√

−

√

(

−π

t/4

sin

πx

−

√

∞

n=2

(2n − 1)

[1 − e

−(n−1/2)

]sin(n −

)πx,

√

2(−1)

n+1

(2n − 1)

, n = 1, 2, ...

20. u(x, t) =

√

∞

n=1

− 1

−t

− e

−λ

) + α

−λ

cos

√

(1 + sin

√

)

1/2

√

2 sin

√

(1 + sin

√

)

1/2

, α

√

2(1 − cos

√

)

(1 + sin

√

)

1/2

and λ

satisﬁes cos

√

−

√

sin

√

= 0.

21. u(x, t) = 8

∞

n=1

sin(nπ/2)

(1 − e

−n

) sin nπx

22. u(x, t) =

√

∞

n=1

−t

− e

−(n−1/2)

) sin(n −

)πx

(n −

)

− 1

√

2(2n − 1)π + 4

√

2(−1)

(2n − 1)

23. (a) r(x)w

=[p(x)w

]

− q(x)w, w(0, t) = 0, w(1, t) = 0, w(x,0) = f (x) − v(x)

24. u(x, t) = x

− 2x + 1 +

∞

n=1

−(2n−1)

sin(2n − 1)πx

2n − 1

25. u(x, t) =−cos πx + e

−9π

t/4

cos(3πx/2)

31–34. In all cases solution is y =

G(x, s)f (s) ds , where G(x, s) is given below.

31. G(x, s ) =

1 − x,0≤ s ≤ x

1 − s, x ≤ s ≤ 1

32. G(x, s ) =

s(2 − x)/2, 0 ≤ s ≤ x

x(2 − s)/2, x ≤ s ≤ 1

33. G(x, s ) =

cos s sin(1 − x)/ cos 1, 0 ≤ s ≤ x

sin(1 − s) cos x/ cos

1, x ≤ s ≤ 1

34. G(

x, s) =

s,0≤ s ≤ x

x, x ≤ s ≤ 1

Section 11.4, page 719

1. y =

∞

n=1

− µ

(

√

x), c

f (x)J

(

√

x) dx

(

√

x) dx,

√

satisﬁes J

(

√

λ) = 0.

August 2, 2012 13:11 bansw Sheet number 59 Page number 797 cyan black

Answers to Pr oblems 797

2. (c) y =−

∞

n=1

− µ

(

√

x);

= 2

f (x) dx; c

f (x)J

(

√

x) dx

(

√

x) dx, n = 1, 2, ...;

√

satisﬁes J

′

(

√

λ) = 0.

3. (d) a

(

√

x)f (x) dx

(

√

x) dx

(e) y =

∞

n=1

− µ

(

√

x), c

f (x)J

(

√

x) dx

(

√

x) dx

4. (b) y =

∞

n=1

− µ

2n−1

(x), c

f (x)P

2n−1

(x) dx

2n−1

(x) dx

Section 11.5, page 725

1. (b) u(ξ,2) = f (ξ + 1), u(ξ,0) = 0, 0 ≤ ξ ≤ 2

u(0, η) = u(2, η) = 0, 0 ≤ η ≤ 2

2. u(r, t) =

∞

n=1

(λ

r) sin λ

at, k

(λ

r)g(r) dr

(λ

r) dr

3. Superpose the solution of Problem 2 and the solution [Eq. (21)] of the example in the text.

6. u(r, z) =

∞

n=1

−λ

(λ

r), c

(λ

r)f (r) dr

(λ

r) dr,

and λ

satisﬁes J

(λ) = 0.

7. (b) v(r, θ) =

(kr) +

∞

m=1

(kr)(b

sin mθ + c

cos mθ),

πJ

(kc)

2π

f (θ) sin mθ dθ; m = 1, 2, ...

πJ

(kc)

2π

f (θ) cos mθ dθ; m = 0, 1, 2,...

8. c

rf (r)J

(λ

r) dr

(λ

r) dr

10. u(ρ, s) =

∞

n=0

(s), where c

−1

f (arccos s)P

(s) ds

−1

(s) ds;

is the nth Legendre polynomial and s = cos φ.

Section 11.6, page 734

1. n = 21 2. (a) b

= (−1)

m+1

√

2/mπ (c) n = 20

3. (a) b

= 2

√

2(1 − cos mπ)/m

7. (a) f

(x) = 1 (b) f

(x) =

√

3(1 − 2x) (c) f

(x) =

√

5(−1 + 6x − 6x

)

(d) g

(x) = 1, g

(x) = 2x − 1, g

(x) = 6x

− 6x + 1

8. P

(x) = 1, P

(x) = x, P

(x) = (3x

− 1)/2, P

(x) = (5x

− 3x)/2

August 2, 2012 13:11 bansw Sheet number 60 Page number 798 cyan black

August 8, 2012 13:57 bindex Sheet number 1 Page number 799 cyan black

799

INDEX

Abel, Niels Henrik, 154, 230

Abel’s formula, 154, 174, 227,

243

for systems of equations, 393

Acceleration of convergence,

614

Adams, John Couch, 473

Adams–Bashforth formula

fourth order, 474

second order, 473

Adams–Moulton formula

fourth order, 474

second order, 474

Adaptive numerical method,

459, 465, 471

Adjoint

differential equation, 157

matrix, 369

Advection, 698

Airy, George Biddell, 259

Airy equation, 158, 259–263,

268, 282, 307, 327

Amplitude modulation, 214

Amplitude

of simple harmonic motion,

197

Analytic function, 251, 266

Angular momentum

principle of, 26, 510

Annihilators,

method of, 239–240

Aquifer

ﬂow in, 668, 698

Archimedes, 206

Asymptotic stability, see

Stability

Augmented matrix, 373, 379

Autonomous

equation, 78

system, 508

Backward differentiation

formulas, 476–477

Backward Euler formula,

455–456

Bashforth, Francis, 473

Basin of attraction, 514, 526,

539, 560–562

Beat, 214, 342–343

Bendixson, Ivar Otto, 569

Bernoulli, Daniel, 27, 92, 623,

643

Bernoulli, Jakob, 27, 68, 77, 356

Bernoulli, Johann, 27, 68, 272,

346

Bernoulli equation, 77

Bessel, Friedrich Wilhelm, 296

Bessel equation of order

k, 720

ν, 156, 158, 254, 280, 282, 296,

306, 683, 715

one, 287, 302–305, 307

one-half, 301–302, 306

zero, 287, 296–300, 306, 326,

668, 715, 723

Bessel functions, 27

(x), 287, 297, 300, 306, 326,

716, 719, 720, 723, 727

asymptotic approximation

to, 300

Laplace transform of, 326

zeros of, 307, 716, 724

(x), 288, 303, 306

−1/2

(x), 302

1/2

(x), 302

(x), 299, 300, 716, 723

asymptotic approximation

to, 300

(x), 305

orthogonality of, 308, 719, 720

Bessel inequality, 736

Bessel series expansion, 719, 724

Bifurcation diagram, 93

Bifurcation point, 93–94, 129,

519, 543–544, 575–577, 579

Hopf, 575

pitchfork, 94

saddle-node, 93

transcritical, 94

Boltzmann, Ludwig, 63

Boundary conditions, 590

for elastic string, 644, 653

for heat conduction equation,

625, 633, 636, 671

August 8, 2012 13:57 bindex Sheet number 2 Page number 800 cyan black

800 Index

for Laplace’s equation, 659

nonhomogeneous, 633–635,

707

periodic, 695, 733

separated, 686

Boundary layer, 487

Boundary value problems

heat conduction equation,

623–642, 669–672,

704–707

homogeneous, 590, 685–699

Laplace’s equation, 658–668

nonhomogeneous, 590,

699–714

self-adjoint, 693–694, 718

singular, 714–721

Sturm–Liouville, 685–693

two-point, 589–596, 679

wave equation, 643–658,

673–675, 710, 722–724