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Page 1: iiinis.jinr.ru/sl/vol1/uh/_ready/mathematics... · Preface This short book is intended for a one-semester course for students in the sciences and engineering after they have taken

Partial Di�erential Equationsand Fourier Analysis | A

Short Introduction

Ka Kit TungProfessor of Applied Mathematics

University of Washington

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Preface

This short book is intended for a one-semester course for students in thesciences and engineering after they have taken one year of calculus and oneterm of ordinary differential equations. For universities on quarter systems,sections labelled “optional” can be omitted without loss of continuity. Acourse based on this book can be offered to sophomores and juniors.

Examples used in this book are drawn from traditional application areassuch as physics and engineering, as well as from biology, music, financeand geophysics. I have tried, whenever appropriate, to emphasize physicalmotivation and have generally avoided theorems and proofs. I also havetried to teach solution techniques, which will be useful in a student’s othercourses, instead of the theory of partial differential equations.

I believe that the subjects of partial differential equations and Fourieranalysis should be taught as early as feasible in an undergraduate’s curricu-lum. Towards this end the book is written to present the subject matteras simply as possible. Ample worked examples are given at the end ofthe chapters as a further learning aid. Exercises are provided for the pur-pose of reinforcing standard techniques learned in class. Tricky problems,whose purpose is mainly to test the student’s mental dexterity, are generallyavoided.

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Contents

1 Introduction 11.1 Review of Ordinary Differential Equations . . . . . . . . . . . 1

1.1.1 First and Second Order Equations . . . . . . . . . . . 11.2 Nonhomogeneous Ordinary Differential Equations . . . . . . . 7

1.2.1 First-Order Equations: . . . . . . . . . . . . . . . . . . 71.2.2 Second-Order Equations: . . . . . . . . . . . . . . . . 10

1.3 Summary of ODE solutions . . . . . . . . . . . . . . . . . . . 111.4 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Exercise I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Solutions to Exercise I . . . . . . . . . . . . . . . . . . . . . . 141.7 Exercise II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8 Solutions to Exercise II . . . . . . . . . . . . . . . . . . . . . 18

2 Physical Origins of Some PDEs 192.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Conservation Laws: . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2.1 Diffusion of a tracer: . . . . . . . . . . . . . . . . . . . 202.2.2 Advection of a tracer: . . . . . . . . . . . . . . . . . . 222.2.3 Nonlinear advection: . . . . . . . . . . . . . . . . . . . 232.2.4 Heat conduction in a rod: . . . . . . . . . . . . . . . . 242.2.5 Ubiquity of the Diffusion Equation . . . . . . . . . . . 25

2.3 Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3.1 A drunken sailor . . . . . . . . . . . . . . . . . . . . . 262.3.2 Price of stocks as a random walk . . . . . . . . . . . . 27

2.4 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . 272.5 Multiple Dimensions . . . . . . . . . . . . . . . . . . . . . . . 292.6 Types of second-order PDEs . . . . . . . . . . . . . . . . . . . 302.7 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . 302.8 Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . 32

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iv CONTENTS

2.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Method of Similarity Variables (Optional) 373.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Similarity Variable . . . . . . . . . . . . . . . . . . . . . . . . 373.3 The “drunken sailor” problem solved . . . . . . . . . . . . . . 413.4 The fundamental solution . . . . . . . . . . . . . . . . . . . . 433.5 Examples from Fluid Mechanics . . . . . . . . . . . . . . . . . 44

3.5.1 The Rayleigh problem . . . . . . . . . . . . . . . . . . 443.5.2 Diffusion of vorticity . . . . . . . . . . . . . . . . . . . 46

3.6 The Age of the Earth . . . . . . . . . . . . . . . . . . . . . . 473.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4 Simple Plane-Wave Solutions 534.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Linear homogeneous equations . . . . . . . . . . . . . . . . . 53

4.2.1 Three different types of behaviors . . . . . . . . . . . . 544.2.2 Rossby waves in the atmosphere . . . . . . . . . . . . 564.2.3 Laplace’s equation in a circular disk . . . . . . . . . . 57

4.3 Further developments (optional) . . . . . . . . . . . . . . . . 624.3.1 The wave equation . . . . . . . . . . . . . . . . . . . . 624.3.2 The diffusion equation . . . . . . . . . . . . . . . . . . 64

4.4 Forced oscillation . . . . . . . . . . . . . . . . . . . . . . . . . 654.4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 654.4.2 Subsoil temperature . . . . . . . . . . . . . . . . . . . 664.4.3 Why the earth does not have a corona . . . . . . . . . 67

4.5 A comment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.6 Exercise I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.7 Solutions to Exercise I . . . . . . . . . . . . . . . . . . . . . . 714.8 Exercise II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.9 Solutions to Exercise II . . . . . . . . . . . . . . . . . . . . . 74

5 d’Alembert’s Solution 775.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2 d’Alembert’s approach . . . . . . . . . . . . . . . . . . . . . . 775.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.4 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

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CONTENTS v

5.5 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6 Separation of Variables 896.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2 An example of heat conduction in a rod: . . . . . . . . . . . . 896.3 Separation of variables: . . . . . . . . . . . . . . . . . . . . . 906.4 Physical interpretation of the solution: . . . . . . . . . . . . . 966.5 A vibrating string problem: . . . . . . . . . . . . . . . . . . . 976.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

7 Fourier Sine Series 1037.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.2 Finding the Fourier coefficients . . . . . . . . . . . . . . . . . 1037.3 An Example: . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.4 Some comments: . . . . . . . . . . . . . . . . . . . . . . . . . 1097.5 A mathematical curiosity . . . . . . . . . . . . . . . . . . . . 1107.6 Representing the cosine by sines . . . . . . . . . . . . . . . . 1107.7 Application to the Heat Conduction Problem . . . . . . . . . 1117.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1137.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8 Fourier Cosine Series 1218.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1218.2 Finding the Fourier coefficients . . . . . . . . . . . . . . . . . 1218.3 Application to PDE with Neumann Boundary Conditions . . 123

9 Eigenfunction Expansion 1279.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1279.2 Eigenfunctions and boundary conditions . . . . . . . . . . . . 1289.3 Orthognality Condition for Xn . . . . . . . . . . . . . . . . . 132

10 Nonhomogeneous Partial Differential Equations 13510.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13510.2 Eigenfunction expansion . . . . . . . . . . . . . . . . . . . . . 13510.3 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

11 Collapsing Bridges 14111.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14111.2 Marching soldiers on a bridge, a simple model . . . . . . . . . 14111.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

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vi CONTENTS

11.4 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14511.5 A different forcing function . . . . . . . . . . . . . . . . . . . 14511.6 Tacoma Narrows Bridge . . . . . . . . . . . . . . . . . . . . . 14711.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14811.8 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

12 Fourier Series 15112.1 Introdution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15112.2 Periodic Eigenfunctions . . . . . . . . . . . . . . . . . . . . . 15112.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 15312.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

12.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15612.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15812.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

12.5 Complex Fourier series . . . . . . . . . . . . . . . . . . . . . . 162

13 Fourier Series, Fourier Transform and Laplace Transform 16513.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16513.2 Dirichlet Theorem . . . . . . . . . . . . . . . . . . . . . . . . 16513.3 Fourier integrals . . . . . . . . . . . . . . . . . . . . . . . . . 16613.4 Fourier transform and inverse transform . . . . . . . . . . . . 16713.5 Laplace transform and inverse transform . . . . . . . . . . . . 168

14 Fourier Transform and Its Application to Partial DifferentialEquations 17114.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17114.2 Fourier transform of some simple functions . . . . . . . . . . 17114.3 Application to PDEs . . . . . . . . . . . . . . . . . . . . . . . 17414.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

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ii

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Contents

1 Introduction 11.1 Review of Ordinary Differential Equations . . . . . . . . . . . 1

1.1.1 First and Second Order Equations . . . . . . . . . . . 11.2 Nonhomogeneous Ordinary Differential Equations . . . . . . . 7

1.2.1 First-Order Equations: . . . . . . . . . . . . . . . . . . 71.2.2 Second-Order Equations: . . . . . . . . . . . . . . . . 10

1.3 Summary of ODE solutions . . . . . . . . . . . . . . . . . . . 111.4 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Exercise I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Solutions to Exercise I . . . . . . . . . . . . . . . . . . . . . . 141.7 Exercise II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8 Solutions to Exercise II . . . . . . . . . . . . . . . . . . . . . 18

2 Physical Origins of Some PDEs 192.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Conservation Laws: . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2.1 Diffusion of a tracer: . . . . . . . . . . . . . . . . . . . 202.2.2 Advection of a tracer: . . . . . . . . . . . . . . . . . . 222.2.3 Nonlinear advection: . . . . . . . . . . . . . . . . . . . 232.2.4 Heat conduction in a rod: . . . . . . . . . . . . . . . . 242.2.5 Ubiquity of the Diffusion Equation . . . . . . . . . . . 25

2.3 Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3.1 A drunken sailor . . . . . . . . . . . . . . . . . . . . . 262.3.2 Price of stocks as a random walk . . . . . . . . . . . . 27

2.4 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . 272.5 Multiple Dimensions . . . . . . . . . . . . . . . . . . . . . . . 292.6 Types of second-order PDEs . . . . . . . . . . . . . . . . . . . 302.7 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . 302.8 Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . 32

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iv CONTENTS

2.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Method of Similarity Variables (Optional) 373.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Similarity Variable . . . . . . . . . . . . . . . . . . . . . . . . 373.3 The “drunken sailor” problem solved . . . . . . . . . . . . . . 413.4 The fundamental solution . . . . . . . . . . . . . . . . . . . . 433.5 Examples from Fluid Mechanics . . . . . . . . . . . . . . . . . 44

3.5.1 The Rayleigh problem . . . . . . . . . . . . . . . . . . 443.5.2 Diffusion of vorticity . . . . . . . . . . . . . . . . . . . 46

3.6 The Age of the Earth . . . . . . . . . . . . . . . . . . . . . . 473.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4 Simple Plane-Wave Solutions 534.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Linear homogeneous equations . . . . . . . . . . . . . . . . . 53

4.2.1 Three different types of behaviors . . . . . . . . . . . . 544.2.2 Rossby waves in the atmosphere . . . . . . . . . . . . 564.2.3 Laplace’s equation in a circular disk . . . . . . . . . . 57

4.3 Further developments (optional) . . . . . . . . . . . . . . . . 624.3.1 The wave equation . . . . . . . . . . . . . . . . . . . . 624.3.2 The diffusion equation . . . . . . . . . . . . . . . . . . 64

4.4 Forced oscillation . . . . . . . . . . . . . . . . . . . . . . . . . 654.4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 654.4.2 Subsoil temperature . . . . . . . . . . . . . . . . . . . 664.4.3 Why the earth does not have a corona . . . . . . . . . 67

4.5 A comment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.6 Exercise I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.7 Solutions to Exercise I . . . . . . . . . . . . . . . . . . . . . . 714.8 Exercise II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.9 Solutions to Exercise II . . . . . . . . . . . . . . . . . . . . . 74

5 d’Alembert’s Solution 775.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2 d’Alembert’s approach . . . . . . . . . . . . . . . . . . . . . . 775.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.4 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

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CONTENTS v

5.5 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6 Separation of Variables 896.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2 An example of heat conduction in a rod: . . . . . . . . . . . . 896.3 Separation of variables: . . . . . . . . . . . . . . . . . . . . . 906.4 Physical interpretation of the solution: . . . . . . . . . . . . . 966.5 A vibrating string problem: . . . . . . . . . . . . . . . . . . . 976.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

7 Fourier Sine Series 1037.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.2 Finding the Fourier coefficients . . . . . . . . . . . . . . . . . 1037.3 An Example: . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.4 Some comments: . . . . . . . . . . . . . . . . . . . . . . . . . 1097.5 A mathematical curiosity . . . . . . . . . . . . . . . . . . . . 1107.6 Representing the cosine by sines . . . . . . . . . . . . . . . . 1107.7 Application to the Heat Conduction Problem . . . . . . . . . 1117.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1137.9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8 Fourier Cosine Series 1218.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1218.2 Finding the Fourier coefficients . . . . . . . . . . . . . . . . . 1218.3 Application to PDE with Neumann Boundary Conditions . . 123

9 Eigenfunction Expansion 1279.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1279.2 Eigenfunctions and boundary conditions . . . . . . . . . . . . 1289.3 Orthognality Condition for Xn . . . . . . . . . . . . . . . . . 132

10 Nonhomogeneous Partial Differential Equations 13510.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13510.2 Eigenfunction expansion . . . . . . . . . . . . . . . . . . . . . 13510.3 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

11 Collapsing Bridges 14111.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14111.2 Marching soldiers on a bridge, a simple model . . . . . . . . . 14111.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

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vi CONTENTS

11.4 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14511.5 A different forcing function . . . . . . . . . . . . . . . . . . . 14511.6 Tacoma Narrows Bridge . . . . . . . . . . . . . . . . . . . . . 14711.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14811.8 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

12 Fourier Series 15112.1 Introdution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15112.2 Periodic Eigenfunctions . . . . . . . . . . . . . . . . . . . . . 15112.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 15312.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

12.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15612.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15812.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

12.5 Complex Fourier series . . . . . . . . . . . . . . . . . . . . . . 162

13 Fourier Series, Fourier Transform and Laplace Transform 16513.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16513.2 Dirichlet Theorem . . . . . . . . . . . . . . . . . . . . . . . . 16513.3 Fourier integrals . . . . . . . . . . . . . . . . . . . . . . . . . 16613.4 Fourier transform and inverse transform . . . . . . . . . . . . 16713.5 Laplace transform and inverse transform . . . . . . . . . . . . 168

14 Fourier Transform and Its Application to Partial DifferentialEquations 17114.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17114.2 Fourier transform of some simple functions . . . . . . . . . . 17114.3 Application to PDEs . . . . . . . . . . . . . . . . . . . . . . . 17414.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

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Chapter 5

d’Alembert’s Solution

5.1 Introduction

Consider the wave equation in an infinite domain:

PDE: utt = c2uxx, −∞ < x <∞, t > 0 (5.1)BCs: u(x, t) → 0, x→ ±∞, t > 0 (5.2)ICs: u(x, 0) = f(x), −∞ < x <∞ (5.3)

ut(x, 0) = g(x), −∞ < x <∞. (5.4)

The problem can be solved using a Fourier transform, as we will show later.However, the first person to solve this problem was not Fourier (1768-1830),but d’Alembert (1746). d’Alembert had a simpler method, which works onlyon the wave equation. It also works whether or not the domain is infinite,but it can get complicated quickly when finite boundaries are included.

5.2 d’Alembert’s approach

d’Alembert recognized that any function of either x+ct or x−ct will satisfythe wave equation. We can simply verify that any trial solution

u(x, t) = L(x+ ct) +R(x− ct) (5.5)

satisfies the PDE (5.1) for any functions L and R (which have second deriva-tives, of course).

To show this, let us write

ξ ≡ x+ ct

η ≡ x− ct,

77

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78 CHAPTER 5. D’ALEMBERT’S SOLUTION

andu(x, t) = L(ξ) +R(η).

We use a prime to denote ordinary differentiation with respect to the argu-ment of the function, e.g.

L′(ξ) ≡ d

dξL(ξ), R′(η) ≡ d

dηR(η).

The partial derivatives are calculated in the following way:

∂tL(ξ) = L′(ξ)

∂ξ

∂t= cL′(ξ)

∂xL(ξ) = L′(ξ)

∂ξ

∂x= L′(ξ)

∂2

∂t2L(ξ) =

∂t(cL′(ξ)) =

d

∂ξ(cL′(ξ))

∂ξ

∂t= c2L′′(ξ)

=∂

∂x(L′(ξ)) =

d

dξ(L′(ξ))

∂ξ

∂x= L′′(ξ).

Similarly,

∂tR(η) = R′(η)

∂η

∂t= −cR′(η)

∂xR(η) = R′(η)

∂η

∂x= R′(η)

∂2

∂t2R(η) =

∂η

∂t

d

dη(−cR′(η)) = c2R′′(η)

∂2

∂x2R(η) =

∂η

∂x

d

dη(R′(η)) = R′′(η).

It is seen that both L(ξ) and R(η) satisfy the wave equations, i.e.(∂2

∂t2− c2

∂x2

)L(ξ) = 0,

and (∂2

∂t2− c2

∂2

∂x2

)R(η) = 0.

Therefore the combination, (5.5), also satisfies the wave equation.The functional forms of L and R should be determined from initial con-

ditions.

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5.3. EXAMPLE 79

Since

u(x, t) = L(x+ ct) +R(x− ct), ut(x, t) = cL′(x+ ct)− cR′(x− ct),

it follows from the initial conditions that,

u(x, 0) = L(x) +R(x) = f(x), −∞ < x <∞ (5.6)ut(x, 0) = cL′(x)− cR′(x) = g(x), −∞ < x <∞. (5.7)

The second equation, (5.7), can be integrated with respect to x to yield

L(x)−R(x) =1c

∫ x

0g(x)dx+K, (5.8)

where K is an arbitrary constant. Adding (5.8) to (5.6) yields

L(x) =12f(x) +

12c

∫ x

0g(x)dx+

K

2, (5.9)

and subtracting (5.8) from (5.6) yields

R(x) =12f(x)− 1

2c

∫ x

0g(x)dx− K

2. (5.10)

We change x to ξ in (5.9) and x to η in (5.10), and then add:

u(x, t) = L(ξ) +R(η).

Thus

u(x, t) =12f(x+ ct) +

12f(x− ct) +

12c

∫ x+ct

x−ctg(x)dx .

(5.11)

This is the full solution to the wave equation in an infinite spatial domain.

5.3 Example

Solve:

PDE: utt = c2uxx, −∞ < x <∞, t > 0

ICs: u(x, 0) = f(x) =

{1, −1 < x < 10, elsewhere

ut(x, 0) = 0, −∞ < x <∞.

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80 CHAPTER 5. D’ALEMBERT’S SOLUTION

Since g ≡ 0, d’Alembert’s solution is

u(x, t) =12[f(x+ ct) + f(x− ct)].

We simply divide the initial shape in two halves. Let one half move to theleft with speed c and the other half to the right with speed c. In the intervalwhere the two halves overlap we add their amplitudes.

−6 −4 −2 0 2 4 60

0.5

1

1.5

2

2.5

3

x

u(x,t)

ct=0

−6 −4 −2 0 2 4 60

0.5

1

1.5

2

2.5

3

x

u(x,t)

ct=0.5

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5.4. REFLECTION 81

−6 −4 −2 0 2 4 60

0.5

1

1.5

2

2.5

3

x

u(x,t)

ct=1

−6 −4 −2 0 2 4 60

0.5

1

1.5

2

2.5

3

x

u(x,t)

ct=2

Figure 5.1. Wave propagation in an infinite domain.

5.4 Reflection

Consider the following problem in a semi-infinite domain:

PDE: utt = c2uxx, 0 < x <∞, t > 0. (5.12)BCs: u(0, t) = 0, u(x, t) → 0 as x→∞. (5.13)ICs: u(x, 0) = f(x), 0 < x <∞. (5.14)

ut(x, 0) = g(x), 0 < x <∞. (5.15)

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82 CHAPTER 5. D’ALEMBERT’S SOLUTION

We can proceed as before and easily show that

u(x, t) = L(x+ ct) +R(x− ct)

satisfies the PDE (5.12). To satisfy the ICs, we have, as before (cf. (5.6)and (5.7)):

u(x, 0) = L(x) +R(x) = f(x), 0 < x <∞, (5.16)ut(x, 0) = cL′(x)− cR′(x) = g(x), 0 < x <∞. (5.17)

An important difference between these ((5.16), (5.17)) and the earlier ((5.6),(5.7)) is that (5.16) and (5.17) define L(x) and R(x) only for x > 0. We donot know L(x) and R(x) for x < 0; hence we do not know R(x − ct) whenct > x.

For ct < x, we can proceed as before. Integrate (5.17) and add to orsubtract from (5.16) to obtain (5.9) and (5.10).

So far we have not used the boundary conditions. These provide themissing information

u(0, t) = 0 = L(ct) +R(−ct), t > 0. (5.18)

Equation (5.18) relates R with a negative argument to L with a positiveargument:

R(η) = −L(−η), η < 0 (5.19)

The solution can be summarized as follows. For ξ = x+ ct, η = x− ct,

u(x, t) = L(ξ) +R(η), (5.20)

where

L(x) =12f(x) +

12c

∫ x

0g(x)dx, x > 0, (5.21)

and

R(x) =12f(x)− 1

2c

∫ x

0g(x)dx, x > 0. (5.22)

In addition,

R(x) = −L(−x), x < 0. (5.23)

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5.5. EXAMPLE 83

If g(x) ≡ 0, the solution is simplified to:

u(x, t) =

{12 [f(x+ ct) + f(x− ct)] for x > ct12 [f(x+ ct)− f(ct− x)] for x < ct.

(5.24)

A simpler way of writing (5.24) is to define the initial condition for negativevalues of x by

f(x) = −f(−x) for −∞ < x < 0 (5.25)

and the solution in (5.24) can be rewritten as

u(x, t) =12[f(x+ ct) + f(x− ct)] (5.26)

for all x > 0 and ct > 0.For x > ct, (5.26) agrees with (5.24). For x < ct, f(x−ct) has a negative

argument, and is now defined according to (5.25). With the change in signdictated by (5.25), (5.26) is then the same as the second line in (5.24). Since(5.26) is the same as the solution in the infinite domain problem, we canpretend that there is no boundary at x = 0, provided we define f(x) fornegative x according to (5.26), find the solution in the infinite domain andignore the final result for x < 0.

5.5 Example

PDE: utt = c2uxx, 0 < x <∞, t > 0BCs: u(0, t) = 0, u(x, t) → 0, as x→∞

ICs: u(x, 0) = f(x) =

{2 if −1 < x− 4 < 10 elsewhere

ut(x, 0) = 0, 0 < x <∞.

d’Alembert’s solution is

u(x, t) =12[f(x+ ct) + f(x− ct)].

This solution is valid for all x > 0, ct > 0, provided that the initial conditionis redefined so that

f(x) = −f(−x) for negative x.

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84 CHAPTER 5. D’ALEMBERT’S SOLUTION

The initial condition is depicted in Figure 5.2(a). We now pretend this isan infinite domain problem and let the initial shape be split into two halvespropagating to the left and right with speed c.

−10 −8 −6 −4 −2 0 2 4 6 8 10−2

−1.5

−1

−0.5

0

0.5

1

1.5

2u(x,0)=f(x), all x

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−1

−0.5

0

0.5

1

1.5

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5.5. EXAMPLE 85

−10 −8 −6 −4 −2 0 2 4 6 8 10−2

−1.5

−1

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0

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−10 −8 −6 −4 −2 0 2 4 6 8 10−2

−1.5

−1

−0.5

0

0.5

1

1.5

2ct=4

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86 CHAPTER 5. D’ALEMBERT’S SOLUTION

−10 −8 −6 −4 −2 0 2 4 6 8 10−2

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x

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At this time part of the image initially at x < 0 moves to the real domainat x > 0, and cancels with the left going wave initially from x > 0.

−10 −8 −6 −4 −2 0 2 4 6 8 10−2

−1.5

−1

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0

0.5

1

1.5

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5.5. EXAMPLE 87

−10 −5 0 5 10−2

−1.5

−1

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2ct=6. Ignore x < 0.

Figure 5.2. Wave propagation and reflection in a semi-infinite domain.

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88 CHAPTER 5. D’ALEMBERT’S SOLUTION

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Chapter 8

Fourier Cosine Series

8.1 Introduction

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bn cosnπx

L, 0 < x < L. (8.1)

Here the summation starts from n = 0, because cos 0 = 1 is not zero. Wewill delay the motivation for wanting to write f(x) in this form until later.Here we discuss only how to find the Fourier cosine series coefficients bnassuming that f(x) can be represented in the form of (8.1).

8.2 Finding the Fourier coefficients

We multiply both sides of (8.1) by cos mπxL , where m is any integer, m = 0,

1, 2, 3, . . . , and integrate from 0 to L:∫ L

0f(x) cos

mπx

Ldx =

∞∑n=0

bn

∫ L

0cos

mπx

Lcos

nπx

Ldx. (8.2)

There is an orthogonality condition for the cosines which can be written as

Imn ≡∫ L

0cos

mπx

Lcos

nπx

Ldx =

0 if m 6= nL2 if m = n 6= 0L if m = n = 0.

(8.3)

121

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122 CHAPTER 8. FOURIER COSINE SERIES

To show this, we note the trigonometric identity

cos a cos b =12

cos(a− b) +12

cos(a+ b),

so∫ L

0cos

mπx

Lcos

nπx

Ldx =

12

∫ L

0cos

(m− n)πxL

dx+12

∫ L

0cos

(m+ n)πxL

dx

=sin (m−n)πx

L

2(m− n)π/L

∣∣∣∣∣L

0

+sin (m+n)πx

L

2(m+ n)π/2

∣∣∣∣∣L

0

= 0 if m 6= n.

When n = m 6= 0

cosmπx

Lcos

nπx

L=

12

(1 + cos

2mπxL

).

The integral from 0 to L of the first term, 12 , is L/2, while the integral of

the second term, −12 cos 2mπx

L , is zero. When m = n = 0,

cosmπx

Lcos

nπx

L= 1,

so its integral from 0 to L is L. Thus we have derived the identity in (8.3).Substituting (8.3) into (8.2) then yields∫ L

0f(x) cos

mπx

Ldx = bmImn =

{b0L if m = 0bm

L2 if m 6= 0.

Thus we have the Fourier cosine series representation for f(x) in the form

f(x) =∞∑

n=0

bn cosnπx

L, 0 < x < L ,

where,

b0 =1L

∫ L

0f(x)dx

bn =2L

∫ L

0f(x) cos

nπx

Ldx, n = 1, 2, 3, 4, . . . . (8.4)

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8.3. APPLICATION TO PDE WITH NEUMANN BOUNDARY CONDITIONS123

8.3 Application to PDE with Neumann Boundary

Conditions

Consider heat conduction in a rod of length L whose initial temperature isgiven as

IC: u(x, 0) = f(x), 0 < x < L . (8.5)

Find the evolution of u(x, t) for t > 0 if the ends of the rod are insulated,i.e.

BCs: ux(0, t) = 0, ux(L, t) = 0, t > 0 . (8.6)

Assume heat conduction is governed by the heat equation:

PDE: ut = α2uxx, 0 < x < L . (8.7)

The usual method of separation of variables will lead us to the solution inthe form:

u(x, t) =∑n

Tn(t)Xn(x), (8.8)

where the “eigenfunction”, Xn(x), satisfies

d2

dx2Xn(x) + λ2

nXn(x) = 0. (8.9)

The only difference between this case and the previous one in Chapter 3,section 2, is the boundary conditions. Here the Neumann condition. (8.6)implies

d

dxXn(0) = 0,

d

dxXn(L) = 0. (8.10)

Nontrivial solutions to (8.9) and (8.10) are

Xn(x) = cos λnx (8.11)

providedλn =

L, n = 0, 1, 2, 3, . . . .

The Tn(t) satisfies, as in section 3.2:

d

dtTn(t) + α2λ2

nTn(t) = 0. (8.12)

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124 CHAPTER 8. FOURIER COSINE SERIES

So

Tn(t) = Tn(0)e−α2λ2nt. (8.13)

The general solution, satisfying the PDE and the BCs, is

u(x, t) =∞∑

n=0

Tn(0)e−α2λ2nt cos

nπx

L. (8.14)

To satisfy the IC, we require, at t = 0,

u(x, 0) = f(x) =∞∑

n=0

Tn(0) cosnπx

L, 0 < x < L. (8.15)

(8.15) implies that the constants, Tn(0)’s, are the Fourier cosine coefficientsfor f(x). Thus

Tn(0) = bn =2L

∫ L

0f(x) cos

nπx

Ldx, n = 1, 2, 3, . . .

T0(0) = b0 =1L

∫ L

0f(x)dx.

The problem is now completely solved, assuming f(x) is given.

An Example:

SolvePDE: ut = uxx, 0 < x < 1, t > 0BCs: ux(0, t) = 0, ux(1, t) = 0, t > 0

IC: u(x, 0) = x, 0 < x < 1.

Since the boundary conditions are homogeneous Neumann, try a cosineseries expansion of the solution

u(x, t) =∞∑

n=0

Tn(t)Xn(x), (8.16)

where Xn(x) = cosnπx. Substituting the assumed form (7.16) into the PDEyields:

d

dtTn(t) = −(nπ)2Tn(t), n = 0, 1, 2, 3, . . . . (8.17)

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8.3. APPLICATION TO PDE WITH NEUMANN BOUNDARY CONDITIONS125

The solution of (8.17) is

Tn(t) = Tn(0)e−(nπ)2t.

Therefore,

u(x, t) =∞∑

n=0

Tn(0)e−(nπ)2t cosnπx, 0 < x < 1.

To satisfy the IC, we require

x =∞∑

n=0

Tn(0) cos nπx, 0 < x < 1.

So the Tn(0)’s are the Fourier cosine coefficients of the function x, and thus

Tn(0) =∫ 1

0xdx =

12

Tn(0) = 2∫ 1

0x cosnπxdx =

{0 if n = even− 4

(nπ)2if n = odd.

Finally,

u(x, t) =12− 4π2

∞∑n=1

n odd

1n2e−(nπ)2t cosnπx.

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126 CHAPTER 8. FOURIER COSINE SERIES

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164 CHAPTER 12. FOURIER SERIES

12.6 Example, Laplace’s equation in a circular disk

Consider again the solution of Laplace’s equation in a region bounded by acircle of radius a:

PDE:∇2u = 0, 0 ≤ r < a

BC:u(r, θ) = f(θ) for r = a

and is periodic in θ with period 2π. In polar coordinates, the Laplaceoperator is (see Figure 4.1)

∇2 =∂2

∂r2+

1r

∂r+

1r2

∂2

∂θ2,

where θ is the angle and r is the radius.Since f(θ) is periodic with period 2π, we can expand it in a periodic

Fourier series:

f(θ) =∞∑

n=−∞cne

−inθ,

with cn given by (from 12.26)

cn =12π

∫ π

−πf(θ)einθdθ.

Since the solution u(r, θ) should also be periodic with period 2π, it too canbe expanded in a periodic Fourier series:

u(r, θ) =∞∑

n=−∞Rn(r)e−inθ.

Substituting this assumed form for u into the PDE yields:

R′′n(r) +1rR′n(r)− n2

r2Rn(r) = 0.

As explained in Section 4.2.3, this ODE belongs to the “equi-dimensionaltype” and the solution is of the form rb. Substituting this assumed forminto the ODE we find that b = ±n. Thus the solution is

Rn(r) = αnrn + βnr

−n

In order that Rn(r) be finite at r = 0, we set βn = 0 for n > 0 and setαn = 0 for n < 0. Also to satisfy the BC at r = a, we want Rn(a) = cn.Thus

Rn(r) = cn(r/a)|n|, n = 0,±1,±2,±3, . . .

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Chapter 9

Eigenfunction Expansion

9.1 Introduction

The Fourier sine and cosine series discussed in the previous two chapters arespecial cases of a more general method of “eigenfunction expansions”. Inthis method, the eigenfunctions Xn(x) obtained from separation of variablesare used to expand any function of x as a linear combination of the Xn(x)’s:

f(x) =∑n

anXn(x), 0 < x < L,

where the coefficients an of the expansion, are constants. For a function of xand t, the coefficients of expansion are in general function of t. Thus, u(x, t)can be expanded as

u(x, t) =∑n

an(t)Xn(x), 0 < x < L,

which we re-write as

u(x, t) =∑

n

Tn(t)Xn(x), 0 < x < L.

[We write Tn(t) instead of an(t) simply to be consistent with our previousnotation]

What eigenfunctions Xn(x) to use depends very much on the type ofboundary conditions and the type of PDE we are given. Below we shallreview different cases using the diffusion equation as an example.

127

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128 CHAPTER 9. EIGENFUNCTION EXPANSION

9.2 Eigenfunctions and boundary conditions

To solve the diffusion equation

PDE: ut = α2uxx, 0 < x < L, t > 0IC: u(x, 0) = f(x), 0 < x < L. (9.1)

Case A.

Step (1) If the boundary conditions are of the homogeneous Dirichlet type,i.e.:

u(0, t) = 0, u(L, t) = 0, t > 0 (9.2)

use a sine series expansion of the form:

u(x, t) =∞∑

n=1

Tn(t) sinnπx

L, 0 < x < L. (9.3)

Which automatically satisfies the boundary conditions.

Step (2) Substitute this assumed form into the PDE (9.1). Note the partialderivatives

ut =∞∑

n=1

T ′n(t) sinnπx

L

α2uxx = α2∞∑

n=1

Tn(t)(−(nπL

)2)

sinnπx

L.

Therefore (9.1) becomes

∞∑n=1

T ′n(t) sinnπx

L= −

∞∑n=1

α2(nπL

)2Tn(t) sin

nπx

L, 0 < x < L.

The above equation is satisfied if

T ′n(t) = −α2(nπL

)2Tn(t), for n = 1, 2, 3, . . . .

(9.4)

The solution of (9.4) is

Tn(t) = Tn(0)e−α2(nπL )2

t,

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9.2. EIGENFUNCTIONS AND BOUNDARY CONDITIONS 129

and so (9.3) implies

u(x, t) =∞∑

n=1

Tn(0)e−α2(nπL )2

t sinnπx

L.

Step (3) To satisfy the IC, set

f(x) = u(x, 0) =∞∑

n=1

Tn(0) sinnπx

L, 0 < x < L.

Thus the Tn(0) are the Fourier sine series coefficients of the givenfunction f(x), i.e.

Tn(0) =2L

∫ L

0f(x) sin

nπx

Ldx. (9.5)

Step (4) Find Tn(0) by evaluating the integral in (9.5). This then com-pletes the solution.

Case B.

Step (1) If the boundary conditions are of the homogeneous Neumanntype, i.e.

∂u

∂x= 0 at x = 0,

∂u

∂x= 0 at x = L, (9.6)

use a cosine series expansion instead:

u(x, t) =∞∑

n=0

Tn(t) cosnπx

L, 0 < x < L. (9.7)

This assumed form automatically satisfies the boundary condi-tions.

Step (2) To satisfy the PDE, substitute (9.7) into (9.1)

∞∑n=0

T ′n(t) cosnπx

L= −

∞∑n=0

α2(nπL

)2Tn(t) cos

nπx

L, 0 < x < L.

The above equation is satisfied if

T ′n(t) = −α2(nπL

)2Tn(t), for n = 0, 1, 2, 3, . . . .

(9.8)

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130 CHAPTER 9. EIGENFUNCTION EXPANSION

The solutionTn(t) = Tn(0)e−α2(nπ

L )2t.

Profides the full solution (9.7) is

u(x, t) =∞∑

n=0

Tn(0)e−α2(nπL )2

t cosnπx

L, 0 < x < L.

(9.9)

Step (3) To satisfy IC, we must have

Tn(0) =2L

∫ L

0f(x) cos

nπx

Ldx, n = 1, 2, 3, . . .

Tn(0) =1L

∫ L

0f(x)dx. (9.10)

Case C.

A more systematic way to proceed is to use the method of separation ofvariables. Here we shall emonstrate this method on a more general boundarycondition (Robin condition):

α1∂

∂xu(0, t) + β1u(0, t) = 0

α2∂

∂xu(L, t) + β2u(L, t) = 0. (9.11)

For α1 = 0, α2 = 0, (9.11) becomes the Dirichlet condition (9.2). Forβ1 = 0, β2 = 0, (9.11) becomes the Neumann condition (9.6). So whatfollows applies to both cases A and B.

Step (1) Assume a solution of the separated form

u(x, t) = T (t)X(x).

Substituting into the PDE (9.1) yields

T ′(t)X(x) = α2X ′′(x)T.

Divide by TX:T ′(x)T (t)

= α2X′′(x)

X(x).

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9.2. EIGENFUNCTIONS AND BOUNDARY CONDITIONS 131

Step (2) LHS being a function of t only and the RHS being a function ofx only, the only way that the LHS can be equal to the RHS is foreach to be equal to a constant. Let that constant be denoted by−α2λ2, so

T ′(t)T (t)

=α2X ′′(x)

X(x)= −α2λ2,

and

T ′(t) = −α2λ2T (t) (9.12)

X ′′(x) = −λ2X(x). (9.13)

We now solve (9.13) subject to the boundary condition

α1X′(0) + β1X(0) = 0 (9.14)

α2X′(L) + β2X(L) = 0. (9.15)

The general solution to (9.13) is

X(x) = A sinλx+Bcosλx.

Boundary condition (9.14) implies that

α1λA+ β1B = 0. (9.16)

Boundary condition (9.15) implies

α2λ(A cos λL−B sinλL) + β2(A sin λL+B cos λL) = 0.(9.17)

(9.16) yields B = −α1λA/β1, so (9.17) becomes:

A{[α2λ cos λL+ β2 sinλL] +α1λ

β1[α2λ sinλL− β2 cosλL]} = 0.

Step (3) For a nontrivial solution, we must have

{[α2λ cos λL+ β2 sinλL] +α1λ

β1[α2λ sinλL− β2 cos λL]} = 0.

(9.18)

(9.18) is called the eigenvalue equation. Its solutions are expectedto be discrete

λ = λn, n = 0, 1, 2, 3, . . . . (9.19)

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132 CHAPTER 9. EIGENFUNCTION EXPANSION

These are called eigenvalues. For each eigenvalue there is a cor-responding eigenfunction

X(x) = Xn(x) = An sinλnx+Bn cos λnx.(9.20)

The explicit forms of λn and Xn are obtainable for specific valuesof α1, α2, β1 and β2.

Step (4) The most general solution is obtained by superimposing all sep-arated solutions Tn(t)Xn(x):

u(x, t) =∞∑

n=0

Tn(t)Xn(x), (9.21)

where Tn(t) is determined from (9.12) to be

Tn(t) = Tn(0)e−α2λ2nt.

(9.21) already satisfies the boundary conditions and the PDE.

Step (5) To satisfy the initial condition, we require

f(x) = u(x, 0) =∞∑

n=1

Tn(0)Xn(x),

so

Tn(0) =1∫ L

0 X2ndx

∫ L

0f(x)Xn(x)dx. (9.22)

This then completes the formal solution of this problem. (Thederivation of the orthogonality relationship for the Xn’s used in(9.22) will be derived in the next section. It does not need to bere-derived each time you solve a problem.)

9.3 Orthognality Condition for Xn

In (9.22) we have used the orthogonality relationship for the eigenfunctionsXn, subject to the general BCs (9.14) and (9.15). A derivation is given here.From (9.13), Xn(x) is an eigenfunction corresponding to the eigenvalue λn,satisfying (9.13), (9.14) and (9.15):

d2

dx2Xn + λ2

nXn(x) = 0, 0 < x < L (9.23)

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9.3. ORTHOGNALITY CONDITION FOR XN 133

α1X′n(0) + β1Xn(0) = 0 (9.24)

α2X′n(L) + β2Xn(L) = 0. (9.25)

Let Xm(x) be the eigenfunction corresponding to the eigenvalue λm,where m is not necessarily the same as n. So Xm(x) satisfies

d2

dx2Xm + λ2

mXm(x) = 0, 0 < x < L (9.26)

α1X′m(0) + β1Xm(0) = 0 (9.27)

α2X′m(L) + β2Xm(L) = 0. (9.28)

Multiply (9.23) by Xm and (9.26) by Xn and subtract. We get

Xmd2

dx2Xn −Xn

d2

dx2Xm = (λ2

m − λ2n)XmXn.

Integrate both sides from 0 to L:∫ L

0(Xm

d

dx2Xn −Xn

d2

dx2Xm)dx = (λ2

m − λ2n)∫ L

0XmXndx.

(9.29)

The left-hand side of (9.29) can be evaluated through an integration byparts, to yield:

Xmd

dxXn

∣∣∣L0−Xn

d

dxXm

∣∣∣L0,

which is zero when the boundary conditions (9.24), (9.25), (9.27) and (9.28)are applied. We therefore have, from (9.29):

(λ2m − λ2

n)∫ L

0XmXndx = 0, (9.30)

so ∫ L

0XmXndx = 0 if m 6= n. (9.31)

(9.31) is the orthogonality relationship we were seeking.

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134 CHAPTER 9. EIGENFUNCTION EXPANSION

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X(é(\ §�¨�©�ª�«N¬�­¯®(®(°±§�²m³S³ ©�ª�´�µw¶�·¹¸m­�µ;º]·m¬+´

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

sin(2

*pi*x

)

plot of the resonating mode

687D9B;�<�>%?�?BAD�BAX�+>�I�LBK8NQJ�7DK=9^S^L�Y�>`A

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Chapter 12

Fourier Series

12.1 Introdution

We discussed the Fourier sine series in Chapter 7 and the Fourier cosineseries in Chapter 8. Now, we shall combine the two to form a periodic Fourierseries (or simply called the Fourier series). Before we do so however, we firsttry to motivate the need for such a series by looking for the eigenfunctionssatisfying periodic boundary conditions.

12.2 Periodic Eigenfunctions

Consider heat conduction in a circular ring. Let us denote the circumferenceof the ring by 2L. Denote any point on the ring by x = 0. Then the pointsx = −L and x = L are actually the same point. The problem is to be solvedin the domain −L < x < L, subject to the boundary condition that thesolution should be the same at x = −L and x = L.

PDE: ut = α2uxx, −L < x < L, t > 0 (12.1)BCs: u(−L, t) = u(L, t) (12.2)

ux(−L, t) = ux(L, t) (12.3)IC: u(x, 0) = f(x), −L < x < L. (12.4)

As we will show, that these boundary conditions are sufficient to define aperiodic function. That is, we can look for a solution for all x, −∞ < x <∞,with the condition that it repeats itself with period 2L, i.e.

u(x, t) = u(x+ 2L, t) . (12.5)

151

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152 CHAPTER 12. FOURIER SERIES

We will get the same result as (12.2) and (12.3).We shall again use the method of separation of variables, and we first

tryu(x, t) = T (t)X(x).

Substituting into the PDE yields

T ′(t)α2T (t)

=X ′′(x)X(x)

= −λ2,

where −λ2 is the separation constant. The eigenfunction X(x) = Xn(x) isdetermined from

X ′′(x) + λ2X(x) = 0 (12.6)X(−L) = X(L), X ′(−L) = X ′(L). (12.7)

The boundary conditions for X comes from (12.2) and (12.3). The solutionto (12.6) is

X(x) = a sinλx+ b cos λx, (12.8)

so the first boundary condition in (12.7) implies

−a sinλL+ b cos λL = a sinλL+ b cos λL,

i.e.

2a sin λL = 0. (12.9)

The second boundary condition

λa cos λL+ λb sinλL = λa cos λL− λb sinλL,

is

2bλ sin λL = 0. (12.10)

Both (12.9) and (12.10) can be satisfied if sin λL = 0, or by taking

λ = nπ/L ≡ λn, n = 0, 1, 2, 3, . . . . (12.11)

So the eigenfunction corresponding to λn is

X(x) = an sinλnx+ bn cos λnx ≡ Xn(x), n = 1, 2, 3, . . . .(12.12)

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12.3. FOURIER SERIES 153

Superposition over all n then yields the general solution of the PDE in theform

u(x, t) =∑n

Tn(t)Xn(x).

Thus

u(x, t) =∞∑

n=0

[an sinnπx

L+ bn cos

nπx

L]e−(nπα/L)2t .

(12.13)

To satisfy the IC, we need

f(x) =∞∑

n=0

[an sinnπx

L+ bn cos

nπx

L], −L < x < L .

(12.14)

The right-hand side of (12.14) is the Fourier series expansion of f(x) in thedomain −L < x < L. It is periodic with period 2L in −∞ < x < L.

12.3 Fourier Series

We now return to the mathematical problem of representing an arbitrary,piecewise continuous, function f(x) in a Fourier series of period 2L,

f(x) =∞∑

n=0

[an sinnπx

L+ bn cos

nπx

L], −L < x < L.

(12.15)

If f(x) is itself periodic with period 2L, then the representation is good forall x, −∞ < x <∞. If f(x) is not periodic outside the interval −L < x < L,or if f(x) is not defined beyond this interval, the representation is good onlyin the restricted interval.

There are two ways to find the coefficients an and bn of the Fourier series.The standard way is to use the orthogonality conditions between sines andcosines. They are, for integers m and n:∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

{0 if m 6= n

L if m = n(12.16)

∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

0 if m 6= n

L if m = n 6= 02L if m = n = 0 (12.17)∫ L

−Lsin

mπx

Lcos

nπx

Ldx = 0 for all m and n (12.18)

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154 CHAPTER 12. FOURIER SERIES

Of the three orthogonality relations (12.16), (12.17) and (12.18), only thelast one is really new. We derived the first two previously when the integra-tion was over half the domain, from 0 to L. Since sines are odd functions ofx and cosines are even function of x, the integrands in (12.16) and (12.17)are even in x. Thus,∫ L

−Lsin

mπx

Lsin

nπx

Ldx = 2

∫ L

0sin

mπx

Lsin

nπx

Ldx∫ L

−Lcos

mπx

Lcos

nπx

Ldx = 2

∫ L

0cos

mπx

Lcos

nπx

Ldx.

(12.16) then follows from our previous results, namely (6.7), and (12.17)follows from (8.3). The last identity, (12.18) follows from the fact that theintegrand in (12.18) is odd in x and so the integral over positive and negativevalues of x yields zero.

Using these orthogonality relations, we can now obtain the coefficientsan and bn in the following way. Multiply both sides of (12.15) by sin mπx

Land integrate with respect to x from −L to L to get∫ L

−Lf(x) sin

mπx

Ldx =

∞∑n=0

[an

∫ L

−Lsin

mπx

Lsin

nπx

Ldx

+ bn

∫ L

−Lsin

mπx

Lcos

nπx

Ldx]

= amL.

so

am =1L

∫ L

−Lf(x) sin

mπx

Ldx, m = 0, 1, 2, 3, . . . .

Multiplying (12.15) by cos mπxL and integrating with respect to x from −L

to L yields, similarly,

bm =1L

∫ L

−Lf(x) cos

nπx

Ldx, m 6= 0

b0 =1

2L

∫ L

−Lf(x)dx.

Summary: The Fourier series representation of a piecewise continuous func-tion f(x) in the interval −L < x < L is given by

f(x) =∞∑

n=0

[an sinnπx

L+ bn cos

nπx

L], −L < x < L ,

(12.19)

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12.3. FOURIER SERIES 155

where

an =1L

∫ L

−Lf(x) sin

nπx

Ldx

b0 =1

2L

∫ L

−Lf(x)dx

bn =1L

∫ L

−Lf(x) cos

nπx

Ldx, n = 1, 2, 3, . . . .

Notes: (i) If f(x) is an odd function of x, i.e.

f(−x) = −f(x), −L < x < L,

then all the cosine coefficients bn are zero, and the Fourier series (12.19)becomes a sine series:

f(x) =∞∑

n=1

an sinnπx

L, −L < x < L,

where

an =1L

∫ L

−Lf(x) sin

nπx

Ldx =

2L

∫ L

0f(x) sin

nπx

Ldx

(ii) If f(x) is an even function of x, i.e.

f(−x) = f(x),

then all the sine coefficients an are zero. The Fourier series (12.19) becomesa cosine series:

f(x) =∞∑

n=0

bn cosnπx

L, −L < x < L,

where

bn =1L

∫ L

−Lf(x) cos

nπx

Ldx =

2L

∫ 2

0f(x) cos

nπx

Ldx

b0 =1

2L

∫ L

−Lf(x)dx =

1L

∫ L

0f(x)dx.

(iii) The above discussion suggests a second way for obtaining the coefficientsof the Fourier series (12.19). Since any function f(x) can be written as

f(x) = fsym(x) + fanti(x), (12.20)

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156 CHAPTER 12. FOURIER SERIES

where fsym(x) ≡ 12(f(x) + f(−x)) is symmetric about x = 0, and

fanti(x) ≡ 12(f(x)− f(−x))

is antisymmetric about x = 0.Now, the symmetric function can be represented by a cosine series in

−L < x < L:

fsym(x) =∞∑

n=0

bn cosnπx

L, (12.21)

where, from (8.4),

bn =2L

∫ L

0fsym(x) cos

nπx

Ldx =

1L

∫ L

−Lfsym(x) cos

nπx

Ldx =

1L

∫ L

−Lf(x) cos

nπx

Ldx

and

b0 =1L

∫ L

0fsym(x)dx =

12L

∫ L

−Lfsym(x)dx =

12L

∫ L

−Lf(x)dx.

Similarly, the antisymmetric function can be represented by a sine seriesin −L < x < L:

fanti(x) =∞∑

n=1

an sinnπx

L, (12.22)

where from (7.5),

an =2L

∫ L

0fanti(x) sin

nπx

Ldx =

1L

∫ L

−Lfanti(x) sin

nπx

Ldx

=1L

∫ L

−Lf(x) sin

nπx

Ldx.

Combining (12.21) and (12.22) into (12.20) then yields (12.19).

12.4 Examples

12.4.1

(a) Represent f(x) = 1, as a Fourier sine series in 0 < x < L. We let

fs(x) ≡∞∑

n=1

an sinnπx

L,

an =2L

∫ L

0f(x) sin

nπx

Ldx =

2nπ

[1− cosnπ].

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12.4. EXAMPLES 157

The Fourier sine series representation, obtained by combining sineswith coefficients determined above is an antisymmetric function andso looks like:

−4 −3 −2 −1 0 1 2 3 4−1.5

−1

−0.5

0

0.5

1

1.5

x/L

f(x)=

1

Fs

Figure 12.1 Fourier series representation of f(x) = 1, as a function of x/L; 50 terms used

in the sum.

Thus, fs(x) looks like f(x) only in the interval 0 < x < L. It isantisymmetric about x = 0 and periodic with period 2L.

(b) Represent f(x) = 1 as a Fourier cosine series in 0 < x < L. We let

fc(x) ≡∞∑

n=0

bn cosnπx

L,

bn =2L

∫ L

0f(x) cos

nπx

Ldx = 0, n 6= 0

b0 =1L

∫ L

01dx = 1.

fc(x) = b0 is actually a one-term cosine series. With b0 = 1 it is aperfect representation of f(x) in 0 < x < L.

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158 CHAPTER 12. FOURIER SERIES

(c) Represent f(x) = 1 as a Fourier series in −L < x < L. We let

fsc(x) ≡∞∑

n=0

[an sinnπx

L+ bn cos

nπx

L],

an =1L

∫ L

−Lf(x) sin

nπx

Ldx = 0

bn =1L

∫ L

−Lf(x) cos

nπx

Ldx = 0

b0 =12L

∫ L

−Lf(x)dx = 1.

In this case fsc(x) = fc(x) = f(x) in −L < x < L.

−4 −3 −2 −1 0 1 2 3 40

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

x

f(x)=1

Fc

Figure 12.2 Fourier series representation of f(x) = 1.

12.4.2

(a) Represent f(x) = 1 + x as a Fourier sine series in 0 < x < 1.

The sine series representation, fs(x), is plotted in Figure 12.3, for allx.

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12.4. EXAMPLES 159

−4 −3 −2 −1 0 1 2 3 4−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

x

f(x)=

1+x

Fs

Figure 12.3 Fourier sine series representation of f(x) = 1 + x; 50 terms used in the sum.

(b) Represent f(x) = 1 + x as a Fourier cosine series in 0 < x < 1. Thecosine series representation, fc(x), is plotted in Figure 12.4 for all x.

−4 −3 −2 −1 0 1 2 3 41

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

x

f(x)=1

+x

Fc

Figure 12.4 The Fourier cosine series representation of f(x) = 1 + x; using 50 terms in

the sum.

(c) Represent f(x) = 1 + x as a Fourier series in −1 < x < 1. fsc(x)is plotted in Figure 12.5 for all x. Notice that fsc(x) is different romfc(x) or fs(x) beyond the interval 0 < x < 1.

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160 CHAPTER 12. FOURIER SERIES

−4 −3 −2 −1 0 1 2 3 4−0.5

0

0.5

1

1.5

2

2.5

x

f(x)=

1+x

Fsc

Figure 12.5 Fourier series representation of f(x) = 1 + x; 50 terms used in the sum.

12.4.3

(a) Represent f(x) = ex as a Fourier sine series in 0 < x < 1. fs(x) isplotted for all x in Figure 12.6

−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x

f(x)=e

xp(x)

Fs

Figure 12.6 Fourier sine series representation of f(x) = ex; 50 terms used in the sum.

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12.4. EXAMPLES 161

(b) Represent f(x) = ex as a Fourier cosine series in 0 < x < 1. fc(x) isplotted for all x in Figure 12.7

−4 −3 −2 −1 0 1 2 3 41

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

x

f(x)=e

xp(x)

Fc

Figure 12.7 Fourier cosine series representation of f(x) = ex; 50 terms used in the sum.

(c) Represent f(x) = ex as a Fourier series in −1 < x < 1. fsc(x) isplotted for all x in Figure 12.8.

−4 −3 −2 −1 0 1 2 3 40

0.5

1

1.5

2

2.5

3

x

f(x)=e

xp(x)

Fsc

Figure 12.8 Fourier series representation of f(x) = ex; 50 terms used in the sum.

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162 CHAPTER 12. FOURIER SERIES

12.5 Complex Fourier series

In this section we will discuss other forms of Fourier series (12.19).In (12.19), the Fourier coefficients are

an =1L

∫ L

−Lf(x) sin

nπx

Ldx

bn =1L

∫ L

−Lf(x) cos

nπx

Ldx, n = 1, 2, 3, . . .

b0 =1

2L

∫ L

−Lf(x)dx.

If we change n to −n in the above definition for an and bn, we will find

a−n = −an, b−n = bn.

Therefore we can rewrite the sum in (12.19) as

f(x) =12

∞∑n=−∞

[an sinnπx

L+ bn cos

nπx

L], −L < x < L,

(12.23)

where now, for n = 0, ±1, ±2, ±3, . . .

an =1L

∫ L

−Lf(x) sin

nπx

Ldx,

bn =1L

∫ L

−Lf(x) cos

nπx

Ldx,

[Note that b0 defined this way is twice as big as the definition obtained in(12.19). This is an advantage, since b0 now has the same form as the rest ofthe bn’s.]

The form (12.23) is equivalent to (12.19) but is sometimes preferredbecause the coefficients are easier to remember.

(12.23) can be further written more compactly using the complex nota-tion:

f(x) =∞∑

n=−∞cne

−inπx/L, −L < x < L, (12.24)

where

cn =1

2L

∫ L

−Lf(x)einπx/Ldx.

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12.5. COMPLEX FOURIER SERIES 163

To show this, we use Euler’s identity

eiθ = cos θ + i sin θ.

Changing θ to −θ gives

e−iθ = cos θ − i sin θ.

Adding yields

cos θ =12(eiθ + e−iθ).

Subtracting yields

sin θ =12i

(eiθ − e−iθ).

We now also rewrite sin nπxL and cos nπx

L in terms of e±inπx/L. (12.23) be-comes

f(x) =12

∞∑n=−∞

[bn12(einπx/L + e−inπx/L)

+ an12i

(einπx/L − e−inπx/L)]

=12

∞∑n=−∞

12(bn − ian)einπx/L +

12

∞∑n=−∞

12(bn + ian)e−inπx/L.

In the first sum we change n to −n, which is permitted since n is a dummyvariable. We then see that the first sum is exactly the same as the secondsum. Thus,

f(x) =∞∑

n=−∞cne

−inπx/L, −L < x < L , (12.25)

where

cn =12(bn + ian) =

12

{1L

∫ L

−Lf(x)(cos

nπx

L+ i sin

nπx

L)dx}.

Thus,

cn =1

2L

∫ L

−Lf(x)einπx/Ldx, n = 0, ±1, ±2, . . . .

(12.26)

The complex form (12.25) appears to be the most convenient to use.

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164 CHAPTER 12. FOURIER SERIES

12.6 Example, Laplace’s equation in a circular disk

Consider again the solution of Laplace’s equation in a region bounded by acircle of radius a:

PDE:∇2u = 0, 0 ≤ r < a

BC:u(r, θ) = f(θ) for r = a

and is periodic in θ with period 2π. In polar coordinates, the Laplaceoperator is (see Figure 4.1)

∇2 =∂2

∂r2+

1r

∂r+

1r2

∂2

∂θ2,

where θ is the angle and r is the radius.Since f(θ) is periodic with period 2π, we can expand it in a periodic

Fourier series:

f(θ) =∞∑

n=−∞cne

−inθ,

with cn given by (from 12.26)

cn =12π

∫ π

−πf(θ)einθdθ.

Since the solution u(r, θ) should also be periodic with period 2π, it too canbe expanded in a periodic Fourier series:

u(r, θ) =∞∑

n=−∞Rn(r)e−inθ.

Substituting this assumed form for u into the PDE yields:

R′′n(r) +1rR′n(r)− n2

r2Rn(r) = 0.

As explained in Section 4.2.3, this ODE belongs to the “equi-dimensionaltype” and the solution is of the form rb. Substituting this assumed forminto the ODE we find that b = ±n. Thus the solution is

Rn(r) = αnrn + βnr

−n

In order that Rn(r) be finite at r = 0, we set βn = 0 for n > 0 and setαn = 0 for n < 0. Also to satisfy the BC at r = a, we want Rn(a) = cn.Thus

Rn(r) = cn(r/a)|n|, n = 0,±1,±2,±3, . . .

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12.6. EXAMPLE, LAPLACE’S EQUATION IN A CIRCULAR DISK 165

Finally the solution to the PDE, satisfying all BCs, is:

u(r, θ) =∞∑

n=−∞cn(r/a)|n|e−inθ

wherecn =

12π

∫ π

−πf(θ)einθdθ.

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166 CHAPTER 12. FOURIER SERIES

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Chapter 13

Fourier Series, FourierTransform and LaplaceTransform

13.1 Introduction

In the previous chapter, we discussed the Fourier series

f(x) =∞∑

n=−∞cne

−inπx/L, −L < x < L, (13.1)

for f(x) in the interval −L < x < L where

cn =1

2L

∫ L

−Lf(x)einπx/Ldx. (13.2)

Previously, when sine and cosine series were discussed, we alluded to Dirich-let’s Theorem, which tells us conditions under which Fourier series is a sat-isfactory representation of the original function f(x). The full Dirichlet’sTheorem is stated below.

13.2 Dirichlet Theorem

If f(x) is a bounded and piecewise continuous function in −L < x < L, itsFourier series representation converges to f(x) at each point x in the intervalwhere f(x) is continuous, and to the average of the left- and right-hand limits

167

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168CHAPTER 13. FOURIER SERIES, FOURIER TRANSFORM AND LAPLACE TRANSFOR

of f(x) at those points where f(x) is discontinuous. If f(x) is periodic withperiod 2L, the above statement applies throughout −∞ < x <∞.

This theorem is easy to understand. The Fourier series, consisting ofsines and cosines of period 2L, has period 2L. If f(x) itself is also has thesame period, then the Fourier series can be a good representation of f(x)over the whole real axis −∞ < x <∞. If on the other hand, f(x) is eithernot periodic, or not defined beyond the interval −L < x < L, the Fourierseries gives a good representation of f(x) only in the stated interval. Beyond−L < x < L, the Fourier series is periodic and so simply repeats itself, butf(x) may or may not be so. The above statements apply where f(x) iscontinuous. When f(x) takes a jump at a point, say x0, the value of f(x) atx0 is not defined. The value which the Fourier seris of f(x) converges to isthe average of the value immediately to the left of x0 and to the right of x0,i.e. to limε→0

12(f(x0− ε)+ 1

2f(x0 + ε)). We have already demonstrated thiswith the Fourier sine series. The same behavior applies to the full Fourierseries.

13.3 Fourier integrals

Unless f(x) is periodic, the Fourier series representation of f(x) is an appro-priate representation of f(x) only over the interval −L < x < L. Question:Can we take L → ∞, so as to obtain a good representation of f(x) overthe whole interval −∞ < x <∞? The positive answer leads to the Fourierintegral, and hence the Fourier transform, provided f(x) is “integrable” overthe whole domain.

We first rewrite (13.1) as a Riemann sum by letting

∆ω = π/L and ωn = nπ/L.

Thus, (13.1) becomes

f(x) =12π

∞∑n=−∞

∆ωFn · e−iωnx, (13.3)

where

Fn ≡ (2Lcn) =∫ L

−Lf(x′)eiωnx′

dx′. (13.4)

[We have changed the dummy variable in (13.4) from x to x′, to avoidconfusion later.]

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13.4. FOURIER TRANSFORM AND INVERSE TRANSFORM 169

In the limit L→∞, ωn becomes ω, which takes on continuous values in−∞ < ω <∞. So Fn → F (ω), where

F (ω) =∫ ∞

−∞f(x′)eiωx′

dx′, (13.5)

and (13.3) becomes the integral:

f(x) =12π

∫ ∞

−∞F (ω)e−iωxdω. (13.6)

Substituting (13.5) into (13.6) leads to the Fourier integral formula:

f(x) =12π

∫ ∞

−∞

[∫ ∞

−∞f(x′)eiωx′

dx′]e−iωxdω . (13.7)

The validity of the formula (13.7) is subject to the integrability of the func-tion f(x) in (13.5). Furthermore, the left-hand side of (13.7) must be modi-fied at points of discontinuity of f(x) to be the average of the left- and right-hand limits at the discontinuity, because the Fourier series, upon which the(13.7) is based, has this property.

The formula shows that the operation of integrating f(x)eiωx over all xis “reversible”, by multiplying it by e−iωx and integrating over all ω. (13.7)allows us to define Fourier transforms and inverse transforms.

13.4 Fourier transform and inverse transform

Let the Fourier transform of f(x) be denoted by F [f(x)]. We should use(13.5) for a definition of such an operation:

F (ω) = F [f(x)] ≡∫ ∞

−∞f(x)eiωxdx . (13.8)

Let F−1 be the inverse Fourier transform, which recovers the original func-tion f(x) from F (ω). (13.6) tells us that the inverse transform is givenby

f(x) = F−1[F (ω)] ≡ 12π

∫ ∞

−∞F (ω)e−iωxdω . (13.9)

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170CHAPTER 13. FOURIER SERIES, FOURIER TRANSFORM AND LAPLACE TRANSFOR

Equations (13.8) and (13.9) form the transform pair. Not too many integralscan be “reversed”. Take for example∫ ∞

−∞f(x)dx.

Once integrated, the information about f(x) is lost and cannot be recovered.Therefore the relationships such as (13.8) and (13.9) are rather special andhave wide application in solving PDEs.

Note that our definition of the Fourier transform and inverse transformis not unique. One could, as in some textbooks, put the factor 1

2π in (13.8)instead of in (13.9), or split it as 1√

2πin (13.8) and 1√

2πin (13.9). The

only thing that matters is that in the Fourier integral formula there is thefactor 1

2π when the two integrals are both carried out. Also, in (13.7), wecan change ω to −ω without changing the form of (13.7). So, the Fouriertransform in (13.8) can alternatively be defined with a negative sign in frontof ω in the exponent, and the inverse transform in (13.9) with a positivesign in front of ω in the exponent. It does not matter to the final result aslong as the transform and inverse transform have opposite signs in front ofω in their exponents.

13.5 Laplace transform and inverse transform

The Laplace transform is often used to transform a function of time, f(t)for t > 0. It is defined as

L[f(t)] =∫ ∞

0f(t)e−stdt ≡ f̃(s) . (13.10)

Mathematically, it does not matter whether we denote our independent vari-able by the symbol t or x; nor does it matter whether we call t time andx space or vice versa. What does matter for Laplace transforms is the in-tegration ranges only over a semi-infinite interval, 0 < t < ∞. We do notconsider what happens before t = 0. In fact, as we will see, we need to takef(t) = 0 for t < 0.

Functions which are zero for t < 0 are called one-sided functions. Forone-sided f(t), we see that the Laplace transform (13.10) is the same as theFourier transform (13.8) if we replace x by t and ω by is. That is, from

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13.5. LAPLACE TRANSFORM AND INVERSE TRANSFORM 171

(13.8)

F (is) =∫ ∞

−∞f(t)e−stdt =

∫ ∞

0f(t)e−stdt

≡ f̃(s). (13.11)

Since the Laplace transform is essentially the same as the Fourier transform,we can use the Fourier inverse transform (13.9) to recover f(t) from itsLaplace transform f̃(s). Let

f(t) = L−1[f̃(s)].

Then the operation L−1, giving the inverse Laplace transform, must bedefined by

f(t) =1

2πi

∫ i∞

−i∞f̃(s)estds. (13.12)

This is because

f(t) = F−1[F (ω)] =12π

∫ ∞

−∞F (ω)e−iωtdω

=1

2πi

∫ i∞

−i∞f̃(s)estds

through the change from ω to is. Since the inverse Laplace transform in-volves an integration in the complex plane, it is usually not discussed inelementary mathematics courses which deal with real integrals. Tables ofLaplace transforms and inverse transforms are used instead. Our purposehere is simply to point out the connection between Fourier and Laplacetransform, and the origin of both in Fourier series.

Note: The inverse Laplace transform formula in (13.12) was obtainedfrom the Fourier integral formula, which applies only to integrable functions.For “nonintegrable”, but one-sided f(t), (13.12) should be modified, withthe limits of integration changed to α − i∞ and α + i∞, where α is somepositive real constant bounding the exponential growth of f allowed.

To show this, suppose f(t) is not integrable because it grows as t→∞.Suppose that for some positive α the product

g(t) ≡ f(t)e−αt → 0 as t→∞

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172CHAPTER 13. FOURIER SERIES, FOURIER TRANSFORM AND LAPLACE TRANSFOR

and is thus integrable. We proceed to find the Laplace transform of g(t):

g̃(s) ≡ L[g(t)] =∫ ∞

0f(t)e−αte−stdτ

=∫ ∞

0f(t)e−(α+s)tdt.

Thus iff̃(s) = L[f(t)] =

∫ ∞

0f(t)e−stdt,

theng̃(s) = f̃(α+ s).

The inverse of g̃(s) is

g(t) = L−1[g̃(s)] =1

2πi

∫ i∞

−i∞g̃(s)estds

=1

2πi

∫ i∞

−i∞f̃(α+ s)estds

=1

2πi

∫ α+i∞

α−i∞f̃(s′)e−αtes

′tds′

where we have made the substitution s′ = α+ s. Since g(t) ≡ f(t)e−αt, weobtain, on cancelling out the e−αt factor:

f(t) =1

2πi

∫ α+i∞

α−i∞f̃(s′)es

′tds′.

This is a modified, and more general, formula for Laplace transform of aone-sided function f(t), whether or not it decays as t → ∞, as long asf(t)e−αt is integrable.

f̃(s) = L[f(t)] =∫ ∞

0f(t)e−stdt, Re s ≥ α

f(t) = L−1[f̃(s)] =1

2πi

∫ α+i∞

α−i∞f̃(s)estds.

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