MAT 417: Introduction to Partial Di erential Equations

MAT 417: Introduction to Partial Differential Equations

James V. Lambers

January 5, 2021

Contents

1 Introduction 5

1.1 Introduction to Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 What are PDEs? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.2 Why are PDEs Useful? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.3 How Do You Solve a PDE? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.4 Kinds of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Diffusion-Type Problems 7

2.1 Diffusion-Type Problems (Parabolic Equations) . . . . . . . . . . . . . . . . . . . . . 7

2.1.1 A Simple Heat-Flow Experiment . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.2 The Mathematical Model of the Heat-Flow Experiment . . . . . . . . . . . . 7

2.1.3 More Diffusion-Type Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Boundary Conditions for Diffusion-Type Problems . . . . . . . . . . . . . . . . . . . 8

2.2.1 Type 1 BC (Temperature specified on the boundary) . . . . . . . . . . . . . . 8

2.2.2 Type 2 BC (Temperature of the surrounding medium specified) . . . . . . . . 8

2.2.3 Type 3 BC (Flux specified) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Derivation of the Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.2 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.5 Homogenizing Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5.1 Transforming Time-Independent BCs . . . . . . . . . . . . . . . . . . . . . . 14

2.5.2 Transforming Time-Dependent BCs . . . . . . . . . . . . . . . . . . . . . . . 15

2.5.3 The General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.6 Solving More Complicated Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.6.1 Heat-Flow Problem with Derivative BC . . . . . . . . . . . . . . . . . . . . . 17

2.6.2 Sturm-Liouville Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.7 Solving Nonhomogeneous PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.8 Integral Transforms (Sine and Cosine Transforms) . . . . . . . . . . . . . . . . . . . 24

2.8.1 Solution of an Infinite-Diffusion Problem via the Sine Transform . . . . . . . 24

2.8.2 Solution Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.9 The Fourier Series and Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.9.1 Discrete Frequency Spectrum of a Periodic Function . . . . . . . . . . . . . . 28

2.9.2 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3

4 CONTENTS

2.10 The Fourier Transform and its Application to PDEs . . . . . . . . . . . . . . . . . . 302.10.1 Properties of the Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 302.10.2 Solution of an Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . 31

3 Hyperbolic-Type Problems 333.1 The One-Dimensional Wave Equation (Hyperbolic Equations) . . . . . . . . . . . . . 33

3.1.1 Vibrating-String Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.2 Intuitive Interpretation of the Wave Equation . . . . . . . . . . . . . . . . . . 333.1.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2 The D’Alembert Solution of the Wave Equation . . . . . . . . . . . . . . . . . . . . . 343.3 More on the D’Alembert Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.3.1 The Space-Time Interpretation of D’Alembert’s Solution . . . . . . . . . . . . 363.3.2 Solution of the Semi-Infinite String via the D’Alembert Solution . . . . . . . 37

3.4 Boundary Conditions for the Wave Equation . . . . . . . . . . . . . . . . . . . . . . 383.5 The Finite Vibrating String . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.6 Classification of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.7 The Wave Equation in Two and Three Dimensions . . . . . . . . . . . . . . . . . . . 44

3.7.1 Waves in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.7.2 Two-Dimensional Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.8 The Finite Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.8.1 Examples of the Sine Transform . . . . . . . . . . . . . . . . . . . . . . . . . 483.8.2 Properties of the Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.8.3 Solving Problems via Finite Transforms . . . . . . . . . . . . . . . . . . . . . 49

3.9 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.9.1 Superposition Used to Break an IBVP Into Two Simpler Problems . . . . . . 503.9.2 Separation of Variables and Integral Transforms as Superpositions . . . . . . 50

3.10 Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.11 Nonlinear First-Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4 Elliptic-Type Problems 554.1 The Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.1.1 Interpretation of the Laplacian in Two Dimensions . . . . . . . . . . . . . . . 554.1.2 Intuitive Meanings of Basic Laws of Physics . . . . . . . . . . . . . . . . . . . 564.1.3 Changing Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.2 Addendum: Correction of Polar Coordinate Conversion . . . . . . . . . . . . . . . . 584.3 The Interior Dirichlet Problem for a Circle . . . . . . . . . . . . . . . . . . . . . . . . 58

4.3.1 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.3.2 The Poisson Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Chapter 1

Introduction

1.1 Introduction to Partial Differential Equations

1.1.1 What are PDEs?

PDEs are equations involving partial derivatives of an unknown function (the solution) of severalvariables. Recall that partial derivatives can be denoted using subscripts.

A Few Well-Known PDEs

Heat equation ut = ∆u, Laplace’s equation ∆u = 0, wave equation utt = ∆u (∆u = uxx+uyy+uzzis the Laplacian in 3-D, and ∆u = urr + 1/rur + 1/r2uθθ in polar coordinates)

Note on the Examples

The solution is the dependent variable; it depends on the independent variables.

1.1.2 Why are PDEs Useful?

Natural laws (and other laws) can be expressed as PDEs. Derivatives represent quantities such asvelocity, acceleration, force, friction, flux, current, etc.

1.1.3 How Do You Solve a PDE?

• Separation of Variables: reduces a PDE in n variables to nl ODEs

• Integral Transforms: reduces PDE in n variables to one in n− 1 variables

• Eigenfunction expansions: the solution is expressed as a linear combination of solutions to aneigenvalue problem, which is obtained from the original PDE

• Numerical methods: construct approximate solutions using various approaches, includingfinite differences or linear combinations of basis functions. Basic ideas covered in MAT 461

5

6 CHAPTER 1. INTRODUCTION

1.1.4 Kinds of PDEs

• Order: from highest-order derivative

• Number of Variables: number of independent variables on which the solution depends

• Linearity: if u1, u2 are solutions, so is u1 + αu2 where α is a constant

• Homogeneity: PDE is homogeneous if zero is a solution

• Kinds of Coefficients: constant or variable

• Types of Linear Equations: General second-order linear equation

Auxx +Buxy + Cuyy +Dux + Euy + Fu = G

is parabolic, hyperbolic or parabolic if B2 − 4AC is 0, > 0, or < 0

Examples: classify well-known PDEs; heat is parabolic, wave is hyperbolic, Laplace’s is elliptic

Chapter 2

Diffusion-Type Problems

2.1 Diffusion-Type Problems (Parabolic Equations)

2.1.1 A Simple Heat-Flow Experiment

Consider a rod of length L whose lateral sides are insulated, so heat can flow in and out only atthe ends. This rod is then warmed or cooled, as necessary, to a fixed temperature T0. Then, twotemperature elements are attached to the ends of the rod, to keep the left and right ends of the rodat fixed temperatures T1 and T2. Finally, the temperature profile of the entire rod is monitoredover time.

2.1.2 The Mathematical Model of the Heat-Flow Experiment

The mathematical description of the physical problem requires a PDE, boundary conditions, andinitial conditions.

The Heat Equation

The PDE is

ut = α2uxx, 0 < x < L, 0 < t <∞

Using a centered difference, we get

ut ≈2α2

∆x2

[u(x+ ∆x, t) + u(x−∆x, t)

2− u(x, t)

]which indicates the direction of heat flow into x based on whether u(x, t) compares to the averageof the neighboring temperatures

Boundary Conditions

Boundary conditions (BCs) are needed to uniquely determine the solution. In our experiment, theboundary conditions are

u(0, t) = T1, u(L, t) = T2, 0 < t <∞

7

8 CHAPTER 2. DIFFUSION-TYPE PROBLEMS

Initial Conditions

PDE, BCs and IC (initial condition) together make up an IBVP (initial-boundary-value problem),which has a unique solution. In our experiment, the initial condition is

u(x, 0) = T0, 0 < x < L

2.1.3 More Diffusion-Type Equations

Lateral Heat Loss Proportional to the Temperature Difference

ut = α2uxx − β(u− u0), β > 0

Heat loss if u > u0, heat gain if u < u0

Internal Heat Source

ut = α2uxx + f(x, t)

Equation is inhomogeneous; f(x, t) denotes heat source

Diffusion-convection Equation

ut = α2uxx − νux

Example: particles smoke rising from a smoke stack are convected upward into the hot air anddiffuse within the air currents

2.2 Boundary Conditions for Diffusion-Type Problems

2.2.1 Type 1 BC (Temperature specified on the boundary)

If the temperatures on the left and right boundaries are g1(t) and g2(t) respectively, then

u(0, t) = g1(t), u(L, t) = g2(t)

2.2.2 Type 2 BC (Temperature of the surrounding medium specified)

Newton’s law of cooling states that the outward flux at x = 0 is h[u(0, t)− g1(t)], and the outwardflux at x = L is h[u(L, t) − g2(t)], where g1(t) and g2(t) are the medium temperatures at theendpoints, and h is a heat-exchange coefficient. Fourier’s law of cooling states that this outwardflux is proportional to the inward normal derivative. Therefore

kux(0, t) = h[u(0, t)− g1(t)], −kux(L, t) = h[u(L, t)− g2(t)], 0 < t <∞,

where k is the thermal conductivity of the medium, which is small for poorly conducting materials.

Heat always flows from high temperatures to low temperatures, so if ux < 0, then heat is flowingfrom left to right, and the flux is positive. This is consistent with the specification of flux from leftto right as −kux everywhere in the rod, not just at the right boundary x = L.

2.3. DERIVATION OF THE HEAT EQUATION 9

2.2.3 Type 3 BC (Flux specified)

One-dimensional rod with insulated ends:

ux(0, t) = 0, ux(L, t) = 0, 0 < t <∞

In higher dimensions, the normal derivative is used to specify the flux on the boundary.

Typical BCs for One-Dimensional Heat Flow

What are the boundary conditions for a problem in which one end of the rod is insulated, whilethe other is immersed in water that has a constant temperature?

2.3 Derivation of the Heat Equation

We will now derive the heat equation with an external source,

ut = α2uxx + F (x, t), 0 < x < L, t > 0,

where u is the temperature in a rod of length L, α2 is a diffusion coefficient, and F (x, t) representsan external heat source. We begin with the following assumptions:

• The rod is made of a homogeneous material.

• The rod is laterally insulated, so that heat flows only in the x-direction.

• The rod is sufficiently thin so that the temperature within any particular cross-section isconstant.

These last two assumptions are used to allow us to treat the problem as one-dimensional. Aswe will see, the first assumption is not absolutely necessary, but it does simplify certain solutiontechniques.

From the principle of conservation of energy, it follows that the heat within a segment of therod [x, x+ ∆x] satisfies the following:

Net change inside [x, x+ ∆x] = Net inward flux across boundaries +

Total heat generated inside [x, x+ ∆x]

The total amount of heat, in calories, in any segment [a, b] is given by∫ b

acρAu(s, t) ds,

where c is the thermal capacity of the rod (also known as the specific heat), ρ is the density ofthe rod, and A is the cross-sectional area of the rod. In view of our assumptions, c, ρ and A areconstants. Also, recall that the flux from left to right at x = a is given by −kux(a, t), where k isthe thermal conductivity of the rod.


Putting all of these facts together, we can translate the conservation relation into the equation

cρA

∫ x+∆x

xut(s, t) ds = kA[ux(x+ ∆x, t)− ux(x, t) +A

∫ x+∆x

xf(s, t) ds,

where f(x, t) is the amount of heat generated by the external source per unit of length per unit oftime. Note that we must use inward flux, which is why the flux term at x = L must be negated.

Applying the Fundamental Theorem of Calculus “in reverse”,

f(b)− f(a) =

∫ b

af ′(s) ds,

we obtain, after dividing both sides by A,

cρ

∫ x+∆x

xut(s, t) ds =

∫ x+∆x

xkuxx(s, t) + f(s, t) ds.

Rearranging yields ∫ x+∆x

xut(s, t)− α2uxx(s, t) + F (s, t) ds = 0,

where

α2 =k

cρ, F (x, t) =

1

cρf(x, t),

are the diffusivity of the rod and the heat source density, respectively.Since this equation holds on an arbitrary segment of the rod, it follows that the integrand must

vanish everywhere in the rod, which yields the equation

ut = α2uxx + F (x, t).

It is worth noting that the diffusivity α2 = kcρ is proportional to the conductivity, but inversely

proportional to the thermal capacity and the density. Physically, this makes sense because themore an object tends to store heat, and the denser it is, the more difficult it should be for heatenergy to diffuse through the object, whereas the better the ability of the material to conduct heat,the easier it should be for heat energy to move through the object and diffuse.

2.4 Separation of Variables

2.4.1 Overview

We will now present the simplest technique for solving PDEs, known as separation of variables. Thebasic idea behind this technique is to separate a PDE with n independent variables into n ODEs,and construct the solution of the PDE using solutions of the ODEs. Separation of variables can onlybe applied to IBVPs in which both the PDE and BCs are linear and homogeneous. Furthermore,the PDE and BCs should have constant coefficients.

In the case of a PDE containing two independent variables x and t, we assume a solution of theform

u(x, t) =

∞∑n=1

AnUn(x, t),

2.4. SEPARATION OF VARIABLES 11

where the constants {An}∞n=1 are determined by the initial condition and the functions {Un(x, t)}∞n=1

are solutions of the PDE that also satisfy the BCs. Because the PDE and BCs are linear, any linearcombination of these solutions is also a solution. We also assume that each solution Un(x, t) hasthe form

Un(x, t) = Xn(x)Tn(t),

where the functions Xn(x) and Tn(t) are obtained by solving ODEs that are derived from the PDE.

2.4.2 Separation of Variables

We will now apply separation of variables to solve the IBVP

ut = α2uxx, 0 < x < 1, t > 0,

u(x, 0) = φ(x), 0 < x < 1,

u(0, t) = 0, u(1, t) = 0, t > 0.

Step 1: Finding Elementary Solutions of the PDE

We substitute Un(x, t) = Xn(x)Tn(t) into the PDE and obtain

Xn(x)T ′n(t) = α2U ′′n(x)Tn(t).

Rearranging yieldsT ′n(t)

α2Tn(t)=X ′′n(x)

Xn(x).

The left side is a function of t, and the right side is a function of x. Because these expressions mustbe equal for all x and t in the domain of the PDE, and x and t are independent variables, bothsides must equal a constant k. It follows that Xn(x) and Tn(t) satisfy the ODEs

X ′′n − kXn = 0,

T ′n − kα2Tn = 0

where k is a constant. Furthermore, Xn(x) must satisfy the BCs

Xn(0) = 0, Xn(1) = 0.

We now consider three scenarios:

1. k = 0. We then have X ′′n = 0, which has the general solution

Xn(x) = Anx+Bn

for constants An and Bn. Applying the BCs yields the equations

Bn = 0, An +Bn = 0,

which has the trivial solutions An = Bn = 0. However, this means that Xn(x) = 0, so weobtain only the trivial solution of the ODE.


2. k > 0. We then have the general solution

Xn(x) = Ane√kx +Bne

−√kx

for constants An and Bn. Applying the BCs yields the equations

An +Bn = 0, Ane√k +Bne

−√k = 0,

which has the trivial solutions An = Bn = 0. Once again, we obtain only the trivial solutionXn(x) = 0.

3. k < 0. In this case, we have the general solution

Xn(x) = An sin√−kx+Bn cos

√−kx.

For convenience, we let k = −λ2, so we instead write

Xn(x) = An sinλx+Bn cosλx.

We will see that the third scenario yields a nontrivial solution to the spatial ODE and BCs. Finally,the general solution to the temporal ODE is

Tn(t) = Cne−λ2α2t,

where Cn is a constant. Since Xn(x) already has undetermined constants, we set Cn = 1 forconvenience. We conclude that the PDE has elementary solutions

Un(x, t) = e−λ2α2t[An sinλx+Bn cosλx].

Step 2: Finding Solutions of the PDE and BCs

We now ensure that Un(x, t) satisfies the BCs. Substituting x = 0 and x = 1 yields the equations

e−λ2α2tBn = 0, e−λ

2α2t[An sinλ+Bn cosλ] = 0.

From the first equation, we obtain Bn = 0. Then, from the second equation, we obtain the condition

λ = nπ

where n is a positive integer. We conclude that solutions of the PDE that satisfy the BCs have theform

Un(x, t) = Ane−(nπ)2α2t sinnπx, n = 1, 2, . . . .

It is not necessary to consider negative integers, because sin(−nπx) = − sinnπx. That is, solutionscorresponding to negative values of n are not linearly independent of those corresponding to positivevalues of n.

2.4. SEPARATION OF VARIABLES 13

Step 3: Finding Solutions of the PDE, BCs and IC

Now, our overall solution to the IBVP has the form

u(x, t) =∞∑n=1

Ane−(nπα)2t sinnπx.

Setting t = 0 and applying the initial condition yields

φ(x) =∞∑n=1

An sinnπx.

We will now use this equation to obtain the values of the constants {An}∞n=1.Suppose that m and n are positive integers. If m 6= n, then∫ 1

0sinmπx sinnπx dx =

1

2

∫ 1

0cos(m− n)πx− cos(m+ n)πx dx

=1

2

[sin(m− n)πx

(m− n)π− sin(m+ n)πx

(m+ n)π

]∣∣∣∣10

=1

2

[sin(m− n)π

(m− n)π− sin(m+ n)π

(m+ n)π

]= 0,

whereas if m = n, we have∫ 1

0sinmπx sinnπx dx =

∫ 1

0sin2 nπx dx

=1

2

∫ 1

01− cos 2nπx dx

=1

2

[x− sin 2nπx

2nπ

]∣∣∣∣10

=1

2

[1− sin 2nπ

2nπ

]=

1

2.

That is, the functions {sinnπx}∞n=1 are orthgonal on the interval [0, 1].Now, if we multiply both sides of the equation describing φ(x) by sinmπx and then integrate

from 0 to 1, we obtain∫ 1

0φ(x) sinmπxdx =

∫ 1

0

∞∑n=1

An sinnπx sinmπxdx

=∞∑n=1

An

∫ 1

0sinnπx sinmπxdx

=1

2Am.


We conclude that the solution of our IBVP is

u(x, t) =∞∑n=1

Ane−(nπx)2t sinnπx,

where

An = 2

∫ 1

0φ(x) sinnπx dx.

The series

φ(x) =

∞∑n=1

An sinnπx

is called a Fourier sine series. If φ is continuous on the interval [0, 1], and is 1-periodic, thenthis series converges to φ(x) for every x ∈ [0, 1]. Because of the orthogonality of the functions{sinnπx}∞n=1, if φ(x) is a linear combination of these functions, then the coefficients {An}∞n=1 aresimply the coefficients of this linear combination.

As t increases, the factors e−(nπα)2t rapidly decay to zero, especially for larger values of n. Itfollows that the solution can be well-approximated by the first term of its Fourier sine series. Thatis,

u(x, t) ≈ A1e−(πα)2t sinπx,

and this approximation improves as t increases.

2.5 Homogenizing Boundary Conditions

In order to use separation of variables to solve an IBVP, it is essential that the boundary conditions(BCs) be homogeneous. If they are not, then it is possible to transform the IBVP into an equivalentproblem in which the BCs are homogeneous. We illustrate this process with some examples.

2.5.1 Transforming Time-Independent BCs

Consider the IBVP

ut = α2uxx, 0 < x < L, t > 0,

u(0, t) = k1, u(L, t) = k2, t > 0,

u(x, 0) = φ(x), 0 < x < L.

We express the solution u(x, t) in the form

u(x, t) = U(x, t) +B(x),

where B(x) satisfies the boundary conditions, and therefore U(x, t) satisfies boundary conditionsof the same form as those for u(x, t), except that they are homogeneous. The function B dependsonly on x, not t, because the boundary values k1 and k2 do not depend on t either.

We also require that U(x, t) satisfies the same PDE as u(x, t). Because the PDE is linear, itfollows that B(x) must be a solution of the PDE as well. However, because B is independent of t,Bt = 0, and therefore we must also have Bxx = 0. which means B(x) must be a linear function of x.

2.5. HOMOGENIZING BOUNDARY CONDITIONS 15

In order to meet this requirement, and the requirement that B(x) satisfy the boundary conditions,we can use Lagrange interpolation to obtain

B(x) = k1x− L0− L

+ k2x− 0

L− 0= k1 +

x

L(k2 − k1).

We conclude that U(x, t) must be a solution of the IBVP

Ut = α2Uxx, 0 < x < L, t > 0,

U(0, t) = 0, U(L, t) = 0, t > 0,

U(x, 0) = φ(x)−B(x), 0 < x < L.

This IBVP can be solved using separation of variables. Then, the solution U(x, t) can be added toB(x) to obtain the solution u(x, t) of the original IBVP.

2.5.2 Transforming Time-Dependent BCs

We now consider an IBVP in wihch the boundary conditions involve functions that vary over time:

ut = α2uxx, 0 < x < L, t > 0,

u(0, t) = g1(t), ux(L, t) + hu(L, t) = g2(t), t > 0,

u(x, 0) = φ(x), 0 < x < L.

Proceeding as before, we express the solution u(x, t) in the form

u(x, t) = U(x, t) + S(x, t),

where S(x, t) satisfies the boundary conditions, and therefore U(x, t) satisfies boundary conditionsof the same form as those for u(x, t), except that they are homogeneous. Unlike the previousexample, the function S depends on both x and t, because the boundary values g1(t) and g2(t)depends on t.

We would like U(x, t) to satisfy a PDE of the same form as the one satisfied by u(x, t), but itwill not be exactly the same PDE in this case. As in the previous example, we can construct alinear function of x that will satisfy the boundary conditions, but it will also depend on t. UsingLagrange interpolation as before, we prescribe that S(x, t) has the form

S(x, t) = g1(t)x− L0− L

+ k2(t)x− 0

L− 0= g1(t) +

x

L[k2(t)− g1(t)],

where the function k2(t) is to be determined. Substituting this form for S(x, t) into the BC atx = L yields

Sx(L, t) + hS(L, t) =1

L[k2(t)− g1(t)] + hg1(t) + h[k2(t)− g1(t)] = g2(t),

which yields

k2(t) = g1(t) +L

1 + Lh[g2(t)− hg1(t)],


and thereforeS(x, t) = g1(t) +

x

1 + Lh[g2(t)− hg1(t)] .

Because S(x, t) is a linear function of x, we have Sxx = 0. By substituting the above form ofu(x, t) into the original IBVP, we obtain the new IBVP

Ut = α2Uxx − St, 0 < x < L, t > 0,

U(0, t) = 0, Ux(L, t) + hU(L, t) = 0, t > 0,

U(x, 0) = φ(x)− S(x, 0), 0 < x < L.

The resulting PDE is no longer homogeneous, so we cannot apply separation of variables directly.Later, we will learn how to handle this kind of inhomogeneity.

2.5.3 The General Case

To homogenize the general boundary conditions

α1u(0, t) + β1ux(0, t) = g1(t),

α2u(L, t) + β2ux(L, t) = g2(t),

we need a function S(x, t) that is linear in x, and satisfies these boundary conditions. UsingLagrange interpolation, we obtain

S(x, t) = k1(t)L− xL

+ k2(t)x

L,

where k1(t) and k2(t) are to be determined. This form of S(x, t) ensures that

S(0, t) = k1(t), S(L, t) = k2(t).

We also have

Sx(x, t) =1

L[k2(t)− k1(t)].

Substituting S(x, t) into the boundary conditions for u(x, t) yields

α1k1(t) +β1

L[k2(t)− k1(t)] = g1(t),

α2k2(t) +β2

L[k2(t)− k1(t)] = g2(t).

Rearranging, we obtain (α1 −

β1

L

)k1(t) +

β1

Lk2(t) = g1(t),

−β2

Lk1(t) +

(α2 +

β2

L

)k2(t) = g2(t).

In matrix-vector form, this system of linear equations can be written as

1

L

[α1L− β1 β1

−β2 α2L+ β2

] [k1(t)k2(t)

]=

[g1(t)g2(t)

].

2.6. SOLVING MORE COMPLICATED PROBLEMS 17

Using the formula for the inverse of a 2× 2 matrix,[a bc d

]−1

=1

ad− bc

[d −b−c a

],

we obtain the solutions[k1(t)k2(t)

]= L

[α1L− β1 β1

−β2 α2L+ β2

]−1 [g1(t)g2(t)

]=

L

(α1L− β1)(α2L+ β2) + β1β2

[α2L+ β2 −β1

β2 α1L− β1

] [g1(t)g2(t)

]=

L

α1α2L2 + L(α1β2 − α2β1)

[(α2L+ β2)g1(t)− β1g2(t)β2g1(t) + (α1L− β1)g2(t)

].

We can see from these solution formulas that not all boundary conditions can be homogenizedin this way, because the solutions cannot be found if the determinant of the coefficient matrix,α1α2L

2 + L(α1β2 − α2β1), is zero.

2.6 Solving More Complicated Problems

We have seen a simple example of solving the heat equation using separation of variables, but theboundary conditions can sometimes complicate the solution process. We illustrate this with anexample.

2.6.1 Heat-Flow Problem with Derivative BC

Consider the problem of modeling the temperature in a rod of length 1, in which the temperatureat the left boundary x = 0 is fixed at 0, but the right boundary, at x = 1, is immersed in a solutionof water that is maintained at temperature 0. This leads to the boundary conditions

u(0, t) = 0, ux(1, t) = −hu(1, t),

where, as before, h is the heat exchange coefficient, and we assume for simplicity that the thermalconductivity of the rod is k = 1. If the initial temperature of the rod is equal to the position alongthe rod, we obtain the IBVP

ut = α2uxx, 0 < x < 1, t > 0,

u(0, t) = 0, ux(1, t) + hu(1, t) = 0, t > 0,

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

The PDE and boundary conditions are both constant-coefficient and homogeneous, so we can applyseparation of variables. This leads to the ODEs

T ′ − µα2T = 0, X ′′ − µX = 0,

where µ is a constant, and X(x) must also satisfy the boundary conditions

X(0) = 0, X ′(1) + hX(1) = 0.


We consider the cases where the separation constant µ is positive, zero, or negative. If µ > 0,then we have

T (t) = eµα2t,

which grows exponentially as t increases. This is not physically reasonable, so we dismiss thispossibility. On the other hand, if µ = 0, then X ′′ = 0, which yields

X(x) = Ax+B.

Applying the boundary conditions yields the system of equations

B = 0, A+ h(A+B) = 0,

which means we must have A = B = 0. This leaves only the trivial solution, so µ = 0 is not feasibleeither.

We therefore must have µ < 0. Setting µ = −λ2, we obtain

T (t) = Ae−(λα)2t, X(x) = B sin(λx) + C cos(λx),

where A, B and C are constants to be determined. Applying the boundary conditions to X yieldsthe system of equations

C = 0, Bλ cosλ+ hB sinλ = 0.

B must be nonzero, or we would again have the trivial solution. Therefore, λ must satisfy

tanλ = −λh.

This equation has infinitely many solutions λ1, λ2, . . .; however, they can only be computed numer-ically.

We therefore have the set of solutions of the PDE

un(x, t) = Xn(x)Tn(t) = e−(λnα)2t sin(λnx),

that also satisfy the BCs. The constants {λn}∞n=1 are the positive solutions of the equation tanλn =−λn/h. We only consider the positive solutions of this equation because the negative solutions donot lead to linearly independent solutions of the PDE and BCs, in view of the identity tan(−θ) =− tan θ.

It remains to satisfy the initial conditions. Our solution is of the form

u(x, t) =

∞∑n=1

ane(−λnα)2t sin(λnx),

where the constants {an}∞n=1 are chosen so as to satisfy the initial condition u(x, 0) = x. Substi-tuting t = 0 yields the equation

x =∞∑n=1

an sin(λnx).

2.6. SOLVING MORE COMPLICATED PROBLEMS 19

We multiply both sides of this equation by sin(λmx) and integrate from 0 to 1, which yields

∫ 1

0x sin(λmx) dx =

∞∑n=1

an

∫ 1

0sin(λnx) sin(λmx) dx.

The integral on the left side can be evaluated using integration by parts. As for the integral onthe right side, we first consider the case m 6= n. Using trigonometric identities, we obtain

∫ 1

0sin(λnx) sin(λmx) dx =

1

2

∫ 1

0cos[(λn − λm)x]− cos[(λm + λn)x] dx

=1

2

[sin[(λn − λm)x]

λn − λm− sin[(λm + λn)x]

λm + λn

]∣∣∣∣10

=1

2

[sin(λn − λm)

λn − λm− sin(λm + λn)

λm + λn

]=

1

2(λ2n − λ2

m)[(λn + λm) sin(λn − λm)− (λn − λm) sin(λn + λn)]

= − h

2(λ2n − λ2

m)[(tanλn + tanλm) sin(λn − λm)−

(tanλn − tanλm) sin(λn + λn)]

= − h

2(λ2n − λ2

m)

[(sinλncosλn

+sinλmcosλm

)[sinλn cosλm − cosλn sinλm]−(

sinλncosλn

− sinλmcosλm

)[sinλn cosλm + cosλn sinλm]

]= − h

2(λ2n − λ2

m)

[sin2 λn cosλm

cosλn+ sinλm sinλn − sinλm sinλn−

sin2 λm cosλncosλm

− sin2 λn cosλmcosλn

+ sinλm sinλn − sinλm sinλn+

sin2 λm cosλncosλm

]= 0.


That is, the functions {sin(λnx)}∞n=1 are orthogonal. For m = n, we have∫ 1

0sin(λnx) sin(λmx) dx =

∫ 1

0sin2(λmx) dx

=1

2

∫ 1

01− cos(2λmx) dx

=1

2

[x− sin(2λmx)

2λm

]∣∣∣∣10

=1

2

[1− sin(2λm)

2λm

]=

1

2

[1− sinλm cosλm

λm

]=

λm − sinλm cosλm2λm

.

It follows that∫ 1

0x sin(λmx) dx =

∞∑n=1

an

∫ 1

0sin(λnx) sin(λmx) dx = am

λm − sinλm cosλm2λm

.

We conclude that the solution of our IBVP is

u(x, t) =

∞∑n=1

ane−(λnα)2t sin(λnx),

where

an =2λn

λn − sinλn cosλn

∫ 1

0x sin(λnx) dx.

This solution can be approximated by computing the first few values λ1, λ2, . . . numerically andthen computing the first few terms of the above series, as terms for larger values of n will rapidlydecay to zero as t increases.

2.6.2 Sturm-Liouville Problems

The ODE that was solved in the preceding example,

X ′′ + λ2X = 0, X(0) = 0, X ′(1) + hX(1) = 0,

is a special case of the general Sturm-Liouville problem

−(p(x)y′)′ + q(x)y = λw(x)y, 0 < x < 1,

α1y(0) + β1y′(0) = 0, α2y(1) + β2y

′(1) = 0.

The values of λ for which the Sturm-Liouville problem has a solution, that satisfies the boundaryconditions, are called eigenvalues, and the accompanying solutions are called eigenfunctions. The

2.7. SOLVING NONHOMOGENEOUS PDES 21

eigenvalues λ1, λ2, . . . form a countably infinite set, as we have seen in examples. Furthermore, theeigenfunctions y1(x), y2(x), . . . are orthogonal, in the sense that∫ 1

0yn(x)ym(x)w(x) dx = 0, m 6= n.

The function w(x) is called a weight function, and the Sturm-Liouville problem, as described above,is self-adjoint, which ensures that the eigenvalues are real. If p(x) > 0 and q(x) ≥ 0, then they arealso positive; we then say that the differential operator

L[y] = −(p(x)y′)′ + q(x)y

is positivei definite. The fact that the Sturm-Liouville problem is self-adjoint and has real eigen-values and orthogonal eigenfunctions corresponds closely to the fact, from linear algebra, that asymmetric matrix has real eigenvalues and orthogonal eigenvectors.

2.7 Solving Nonhomogeneous PDEs

Separation of variables can only be applied directly to homogeneous PDE. However, it can begeneralized to nonhomogeneous PDE with homogeneous boundary conditions by solving nonhomo-geneous ODE in time.

We consider a general diffusive, second-order, self-adjoint linear IBVP of the form

ut = (p(x)ux)x − q(x)u+ f(x, t), 0 < x < L,

α1u(0, t) + β1ux(0, t) = 0, β1u(L, t) + β2ux(L, t) = 0, t > 0,

u(x, 0) = φ(x), 0 < x < L.

We assume that the corresponding homogeneous PDE,

vt = (p(x)v′)′ − q(x)v, 0 < x < L,

has a solution of the form

v(x, t) = X(x)S(t).

Substituting into the PDE yields

S′(t)X(x) = S(t)(p(x)X ′(x))′ − q(x)S(t)X(x).

Rearranging yieldsS′(t)

S(t)=

(p(x)X ′(x))′ − q(x)X(x)

X(x)= −λ,

where λ is a constant. We then obtain the ODE

S′(t) + λS(t) = 0,

−(p(x)X ′(x))′ + q(x)X(x) = λX(x).


The ODE for X(x) is a Sturm-Liouville problem, which has eigenvalues {λn}∞n=1 and corre-sponding eigenfunctions {Xn(x)}∞n=1 that satisfy the boundary conditions. Furthermore, theseeigenfunctions are orthogonal, in the sense that∫ L

0Xn(x)Xm(x) dx = 0, m 6= n.

They also form a basis for the function space consisting of functions that satisfy the boundaryconditions, meaning that any such function can be expressed as a linear combination of theseeigenfunctions.

Returning to the original nonhomogeneous PDE, we expand the solution u(x, t) and the externalsource term f(x, t) in the basis of eigenfunctions. That is,

u(x, t) =∞∑n=1

un(t)Xn(x), f(x, t) =

∞∑n=1

fn(t)Xn(x),

where

un(t) =

∫ L0 u(x, t)Xn(x)∫ L

0 Xn(x)2 dx, fn(t) =

∫ L0 f(x, t)Xn(x) dx∫ L

0 Xn(x)2 dx.

Then, to solve the PDE, we multiply both sides by Xm(x) and integrate from 0 to L. This yields∫ L

0ut(x, t)Xm(x) dx =

∫ L

0(p(x)ux(x, t))x − q(x)u(x, t)Xm(x) dx+

∫ L

0f(x, t)Xm(x) dx.

Applying the above expansions, and the fact that Xm(x) is an eigenfunction of the spatial ODE,yields

u′m(t) = −λmum(t) + fm(t),

and from the initial condition, we obtain

um(0) = am =

∫ L0 φ(x)Xm(x) dx∫ L

0 Xm(x)2 dx.

This nonhomogeneous ODE can be solved using an integrating factor µ(t) = eλmt. We thenhave

um(t) = e−λmtam +

∫ t

0e−λm(t−s)fm(s) ds.

In summary, the solution of the IBVP is,

u(x, t) =

∞∑n=1

un(t)Xn(x),

where Xn(x) is the solution of

−(p(x)X ′n(x))′ + q(x)Xn(x) = λnXn(x)

2.7. SOLVING NONHOMOGENEOUS PDES 23

that also satisfies the boundary conditions, and the coefficients un(t) of the eigenfunction expansionof u(x, t) are given by

un(t) = e−λntan +

∫ t

0e−λn(t−s)fn(s) ds,

where

an =

∫ L0 φ(x)Xn(x) dx∫ L

0 Xn(x)2 dx, fn(t) =

∫ L0 f(x, t)Xn(x) dx∫ L

0 Xn(x)2 dx.

This approach to finding u(x, t) is known as the method of eigenfunction expansions.

Example We will use the method of eigenfunction expansions to solve the IBVP

ut = α2uxx + sin(3πx), 0 < x < 1,

u(0, t) = 0, u(1, t) = 0, t > 0,

u(x, 0) = sin(πx), 0 < x < 1.

Applying separation of variables to the corresponding homogeneous PDE,

ut = α2uxx,

yields the ODET ′ + α2λT = 0,

X ′′ + λX = 0,

with boundary conditionsX(0) = 0, X(1) = 0.

The spatial ODE has the solutions, or eigenfunctions,

Xn(x) = sin(nπx), n = 1, 2, . . . ,

with corresponding eigenvalues λn = (nπ)2, n = 1, 2, . . . .Expanding the source term, initial data, and solution in the basis of eigenfunctions yields

u(x, t) =

∞∑n=1

un(t) sin(nπx),

sin(πx) =∞∑n=1

an sin(nπx),

sin(3πx) =

∞∑n=1

fn sin(nπx),

where, because of the orthogonality of the eigenfunctions, a1 = 1, f3 = 1, and an = 0 for all n 6= 1and fn = 0 for all n 6= 3. From the formula

un(t) = e−(nπα)2tan +

∫ t

0e−(nπα)2(t−s)fn ds,


we obtain

u1(t) = e−(πα)2ta1 = e−(πα)2t,

u3(t) = e−(3πα)2ta3 + f3

∫ t

0e−(3πα)2(t−s) ds

=1

(3πα)2e−(3πα)2te(3πα)2s

∣∣∣∣t0

=1

(3πα)2[1− e−(3πα)2t],

and un(t) = 0 for all n 6= 1, 3. We conclude that the solution of the IBVP is

u(x, t) = e−(πα)2t sinπx+1

(3πα)2[1− e−(3πα)2t] sin 3πx.

Letting t→∞, we obtain the steady-state solution 1(3πα)2

sin 3πx. 2

2.8 Integral Transforms (Sine and Cosine Transforms)

An integral transformation, or integral transform, maps a function f(t) to a function F (s) using aformula of the form

F (s) =

∫ b

aK(s, t)f(t) dt

for some function K(s, t) that is known as a kernel. For differential equations, integral transformsare used to convert a differential equation into an algebraic equation, which is much easier to solve.This is possible if the integral transform can be used to easily express the transform of f ′(t) interms of the transform of f(t).

Integral transforms are generally applied to IBVP as follows: a transform is applied to one ofthe independent variables of the PDE, which eliminates all partial derivatives with respect to thatvariable. This yields a new PDE in one less variable, which can then be solved directly, or can betransformed using a second integral transformation. Once all transforms have been applied and theresulting equation is solved, this solution must be converted to the solution of the original IBVPby inverting all of the previously applied transforms. That is, for each integral transform that hasbeen applied, we apply the corresponding inverse transform to our solution. The result of theseinverse transforms is the solution of the IBVP.

There are many integral transforms to choose from. Generally, the choice of transform fora particular IBVP is determined by the boundary conditions. The initial condition is generallytransformed to initial conditions of ODEs that arise from transforming the PDE. Inverse transformsare normally performed by reversing known rules for transforming certain functions, or are carriedout numerically.

Table 2.1 lists certain frequently used integral transforms and corresponding inverse transforms.

2.8.1 Solution of an Infinite-Diffusion Problem via the Sine Transform

We illustrate the use of integral transforms on the IBVP

ut = α2uxx, x > 0, t > 0,

2.8. INTEGRAL TRANSFORMS (SINE AND COSINE TRANSFORMS) 25

Name Formula Domain, BCs

Fourier F [f ] = F (ω) =1√2π

∫ ∞−∞

f(x)e−iωx dx (−∞,∞)

Inverse Fourier F−1[F ] = f(x) =1√2π

∫ ∞−∞

F (ω)eiωx dω

Fourier sine Fs[f ] = F (ω) =2

π

∫ ∞0

f(x) sin(ωx) dx (0,∞), Dirichlet

Inverse Fourier sine F−1s [F ] = f(x) =

∫ ∞0

F (ω) sin(ωx) dω

Fourier cosine Fc[f ] = F (ω) =2

π

∫ ∞0

f(x) cos(ωx) dx (0,∞), Neumann

Inverse Fourier cosine F−1c [F ] = f(x) =

∫ ∞0

F (ω) cos(ωx) dω

Finite sine Fs[f ] = Sn =2

L

∫ L

0f(x) sin(nπx/L) dx (0, L), Dirichlet

Inverse finite sine F−1s [Sn] = f(x) =

∞∑n=1

Sn sin(nπx/L)

Finite cosine Fc[f ] = Cn =2

L

∫ L

0f(x) cos(nπx/L) dx (0, L), Neumann

Inverse finite cosine F−1c [Cn] = f(x) =

C0

2+

∞∑n=1

Cn cos(nπx/L)

Table 2.1: Integral transforms and their inverses

u(0, t) = A, t > 0,

u(x, 0) = 0, x > 0.

Because the problem is defined on a semi-infinite domain (0,∞) in space, and has a Dirichletboundary condition u(x, 0) = A, the ideal transform for this problem is the Fourier sine transform.

We apply this transform to both sides of the PDE. Using the definition from Table 2.1, anddenoting the Fourier sine transform of u(x, t) by U(ω, t), we obtain

Fs[ut] =2

π

∫ ∞0

ut(x, t) sin(ωx) dx =∂

∂t

[2

π

∫ ∞0

u(x, t) sin(ωx) dx

]=

∂

∂tU(ω, t),


Fs[α2uxx] =

2

π

∫ ∞0

α2uxx(x, t) sin(ωx) dx

= α2 2

π

∫ ∞0

uxx(x, t) sin(ωx) dx

= α2Fs[uxx]

= α2 2

π

[sin(ωx)ux(x, t)|∞0 − ω

∫ ∞0

ux(x, t) cos(ωx) dx

]= −α2 2ω

π

[cos(ωx)u(x, t)|∞0 + ω

∫ ∞0

u(x, t) sin(ωx) dx

]= α2

[−ω2Fs[u] +

2ω

πu(0, t)

]= α2

[−ω2U(ω, t) +

2ω

πA

].

We thus obtain the ODE∂

∂tU(ω, t) = −α2ω2U(ω, t) +

2α2ωA

π,

with initial condition obtained from that of the IBVP,

U(ω, 0) = Fs[u(x, 0)] = Fs[0] = 0.

This ODE has the solution

U(ω, t) =

∫ t

0e−α

2ω2(t−s) 2α2ωA

πds =

2A

πω[1− e−α2ω2t].

Thus U(ω, t) is the Fourier sine transform of the solution u(x, t) of the original IBVP. To obtainu(x, t), we must apply the inverse Fourier sine transform. This yields

u(x, t) =

∫ ∞0

U(ω, t) sin(ωx) dω

=2A

πω

∫ ∞0

[1− e−α2ω2t] sin(ωx) dω

= A

∫ ∞0

2

π

[1− e−(α2t)ω2

ω

]sin(ωx) dω.

Using the known transform

F−1s

{2

π

[1− e−aω2

ω

]}= erfc(x/2

√a),

where erfc(x) is the complementary error function

erfc(x) =2√π

∫ ∞x

e−t2dt,

we obtainu(x, t) = Aerfc(x/2

√α2t) = Aerfc(x/2α

√t).

2.8. INTEGRAL TRANSFORMS (SINE AND COSINE TRANSFORMS) 27

If the boundary condition had been a Neumann boundary condition of the form ux(0, t) = g(t),then the Fourier cosine transform would have been used instead. The following rules are essentialfor obtaining the ODE that is satisfied by the Fourier sine or cosine transform U(ω, t) of the solutionu(x, t).

Fs[f′′] = −ω2F [f ] +

2ω

πf(0)

Fc[f′′] = −ω2F [f ]− 2

πf ′(0)

Note that both rules account for whatever boundary condition is imposed at x = 0. Using theserules, we obtain an initial value problem for U(ω, t), for which the initial condition is obtainedby taking the appropriate transform (the same one as for the PDE) of both sides of the initialcondition of the IBVP. Once this initial value problem is solved, all that remains is to perform theinverse integral transform.

2.8.2 Solution Formulas

In general, the solution of the IBVP with Dirichlet boundary condition at x = 0,

ut = α2uxx, x > 0, t > 0,

u(0, t) = g(t), t > 0,

u(x, 0) = φ(x), x > 0,

is given by

u(x, t) =

∫ ∞0

U(ω, t) sin(ωx) dω

where

U(ω, t) = e−α2ω2tΦ(ω) +

2α2ω

π

∫ t

0e−α

2ω2(t−s)g(s) ds

and Φ(ω) = Fs[φ]. On the other hand, if the boundary condition for this problem is changed tothe Neumann boundary condition

ux(0, t) = h(t), t > 0,

then the solution is

u(x, t) =

∫ ∞0

U(ω, t) cos(ωx) dω

where

U(ω, t) = e−α2ω2tΦ(ω)− 2α2

π

∫ t

0e−α

2ω2(t−s)h(s) ds.


2.9 The Fourier Series and Transform

Let f(x) be a function that is periodic, with period 2L. Then, it can be represented as a Fourierseries

f(x) =a0

2+

∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)],

where

an =1

L

∫ L

−Lf(x) cos

(nπxL

)dx, n = 0, 1, 2, . . . ,

bn =1

L

∫ L

−Lf(x) sin

(nπxL

)dx, n = 1, 2, . . .

represent the discrete Fourier transform of f(x). These formulas result from the orthogonality ofthe basis functions cos(nπx/L) and sin(nπx/L) on the interval (−L,L).

The Fourier series for a function f(x) converges to the value of f(x) for each x ∈ (−L,L) iff(x) is continuous on (−L,L). Furthermore, if f(−L) = f(L), then the series converges to f(x) atx = −L,L as well; otherwise, it converges to the average of f(−L) and f(L). This same behaviorapplies at any point in (−L,L) at which f(x) has a jump discontinuity; this result is known asDirichlet’s Theorem. More precisely, if

f(a+) = limx→a+

f(x), f(a−) = limx→a−

f(x),

and f(a+) 6= f(a−), then

limx→a

{a0

2+

∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]}=f(a−) + f(a+)

2.

When f(x) is discontinuous at a point, the truncated Fourier series exhibits oscillations near thediscontinuity. This behavior is called Gibbs’ phenomenon.

Fourier series are very useful for solving an IBVP on a bounded interval, especially if the PDEhas constant coefficients. We have seen that the basis functions cos(nπx/L) and sin(nπx/L) areeigenfunctions of the constant-coefficient Sturm-Liouville problem

X ′′ + λX = 0

obtained using separation of variables. The coefficients {an}∞n=0 and {bn}∞n=1 are determined bythe initial condition and the boundary conditions.

2.9.1 Discrete Frequency Spectrum of a Periodic Function

The discrete frequency spectrum of a periodic function f(x) with a Fourier series as above is givenby

cn =√a2n + b2n, n = 0, 1, 2, . . . ,

where we define b0 = 0. For each n, cn is the amplitude of the wave of frequency n in the Fourierseries of f(x). Taken as a set, the spectrum of f(x) describes its frequency content.

2.9. THE FOURIER SERIES AND TRANSFORM 29

2.9.2 The Fourier Transform

As we let L→∞, the Fourier series becomes the Fourier integral representation

f(x) =

∫ ∞0

a(ξ) cos(ξx) dξ +

∫ ∞0

b(ξ) sin(ξx) dξ,

where the coefficients a(ξ) and b(ξ) represent the Fourier cosine transform

a(ξ) =1

π

∫ ∞−∞

f(x) cos(ξx) dx

and the Fourier sine transform

b(ξ) =1

π

∫ ∞−∞

f(x) sin(ξx) dx.

In the case where f(x) is an even function (f(−x) = f(x)) or an odd function (f(−x) = −f(x)),these formulas are consistent with the ones given for the Fourier cosine and sine transform, respec-tively, of a function defined on (0,∞).

By analogy with the discrete frequency spectrum, we define the (continuous) frequency spectrumof f(x) on (−∞,∞) as

C(ξ) =√

[a(ξ)]2 + [b(ξ)]2.

Using Euler’s equationeiθ = cos θ + i sin θ

and related formulas

cos θ =eiθ + e−iθ

2, sin θ =

eiθ − e−iθ

2i,

we obtain

f(x) =

∫ ∞0

a(ξ) cos(ξx) dξ +

∫ ∞0

b(ξ) sin(ξx) dξ

=

∫ ∞0

1

π

∫ ∞−∞

f(y) cos(ξy) dy cos(ξx) dξ +

∫ ∞0

1

π

∫ ∞−∞

f(y) sin(ξy) dy sin(ξx) dξ

=

∫ ∞0

1

π

∫ ∞−∞

f(y)eiξy + e−iξy

2dyeiξx + e−iξx

2dξ +∫ ∞

0

1

π

∫ ∞−∞

f(y)eiξy − e−iξy

2idyeiξx − e−iξx

2idξ

=1

4π

∫ ∞0

∫ ∞−∞

f(y)[(eiξy + e−iξy)(eiξx + e−iξx)− (eiξy − e−iξy)(eiξx − e−iξx)

]dy dξ

=1

2π

∫ ∞0

∫ ∞−∞

f(y)[eiξ(x−y) + eiξ(y−x)

]dy dξ

=1

2π

∫ ∞−∞

∫ ∞−∞

f(y)eiξ(x−y) dy dξ

=1√2π

∫ ∞−∞

[1√2π

∫ ∞−∞

f(y)e−iξy dy

]eiξx dξ

=1√2π

∫ ∞−∞

F (ξ)eiξx dξ,


where

F (ξ) =1√2π

∫ ∞−∞

f(y)e−iξx dx

is the Fourier transform of f(x). The reconstruction of f(x) from F (ξ),

f(x) =1√2π

∫ ∞−∞

F (ξ)eiξx dξ,

is the inverse Fourier transform. The Fourier transform is useful for solving PDE defined on theentire real number line (−∞,∞).

2.10 The Fourier Transform and its Application to PDEs

2.10.1 Properties of the Fourier Transform

1. The inverse Fourier transform, when applied to the Fourier transform F (ξ) of a function f(x),yields f(x). That is,

f(x) =1√2π

∫ ∞−∞

F (ξ)eiξx dξ =1√2π

∫ ∞−∞

[1√2π

∫ ∞−∞

f(y)e−iξy dy

]eiξx dξ,

or, more concisely,

F−1[F [f ]] = f.

2. The Fourier transform is a linear transformation. That is, if f and g are functions and α andβ are constants, then

F [αf + βg] = αF [f ] + βF [f ].

3. There is a simple relationship between the Fourier transform of a function and that of itsderivative:

F [f ′] =1√2π

∫ ∞−∞

f ′(x)e−iξx dx

=1√2πf(x)e−iξx

∣∣∣∣∞−∞− (iξ)

1√2π

∫ ∞−∞

f(x)e−iξx dx

= iξF [f ].

That is, differentiation with respect to x is transformed to multiplication by iξ. It followsthat

F [uxx] = (iξ)2F [u] = −xi2F [u].

On the other hand, differentiation with respect to t is simply passed through to the Fouriertransform. That is, if F [u] = U(ξ, t), then

F [ut] =∂

∂tU(ξ, t).

2.10. THE FOURIER TRANSFORM AND ITS APPLICATION TO PDES 31

4. The Fourier transform of a product is not the product of the Fourier transforms. We have

F−1[F [f ]F [g]] =1√2π

∫ ∞−∞

F (ξ)G(ξ)eiξx dξ

=1√2π

∫ ∞−∞

F (ξ)

[1√2π

∫ ∞−∞

g(y)e−iξy dy

]eiξx dξ

=1√2π

∫ ∞−∞

g(y)

[1√2π

∫ ∞−∞

F (ξ)eiξ(x−y) dξ

]dy

=1√2π

∫ ∞−∞

f(x− y)g(y) dy

= (f ∗ g)(x).

That is, the Fourier transform of the convolution of f and g, denoted by f ∗ g and defined by

(f ∗ g)(x) =1√2π

∫ ∞−∞

f(x− y)g(y) dy,

is the product of the Fourier transforms of f and g:

F [f ∗ g] = F [f ]F [g].

Example Let f(x) = x and g(x) = e−x2. Then

(f ∗ g)(x) =1√2π

∫ ∞−∞

(x− y)e−y2dy =

x√2π

∫ ∞−∞

e−y2dy =

x√2π

√π =

x√2.

2

2.10.2 Solution of an Initial Value Problem

We will use the Fourier transform to solve the IVP, or Cauchy problem

ut = α2uxx, −∞ < x <∞, t > 0,

u(x, 0) = φ(x), −∞ < x <∞.

Taking the Fourier transform of both sides of the PDE yields the ODE

Ut(ξ, t) = −α2ξ2U(ξ, t),

with initial condition

U(ξ, 0) = Φ(ξ),

where Φ(ξ) is the Fourier transform of φ(x). The solution of this initial value problem is

U(ξ, t) = Φ(ξ)e−α2ξ2t.


It follows that the solution of the original Cauchy problem is

u(x, t) = F−1[U ]

= F−1[Φ(ξ)e−α2ξ2t]

= F−1[Φ(ξ)] ∗F−1[e−α2ξ2t]

= φ(x) ∗[

1

α√

2te−x

2/4α2t

]=

1

2α√πt

∫ ∞−∞

φ(y)e−(x−y)2/4α2t dy.

That is, the solution u(x, t) is the convolution of the initial temperature φ(x) with the impulse-response function, or Green’s function,

G(x, y) = e−(x−y)2/4α2t.

The temperature u(x, t) is a superposition of the response at x to the impulse φ(y), which is givenby φ(y)G(x, y), for each y along the real number line.

Chapter 3

Hyperbolic-Type Problems

3.1 The One-Dimensional Wave Equation (Hyperbolic Equations)

3.1.1 Vibrating-String Problem

Newton’s second law applied to an arbitrary segment [x, x+ ∆x] of a vibrating string yields

∆xρutt = T [ux(x+ ∆x, t)− ux(x, t)] + ∆xF (x, t)−∆xβut(x, t)−∆xγu(x, t),

where u(x, t) is the displacement of the string from equilibrium, ρ is the density, T is the tension,F (x, t) is the external force, −βut is the frictional force against the string, and −γu is the restoringforce. Dividing by ∆x and letting ∆x→ 0 yields the telephone or telegraph equation

utt = α2uxx − βut − γu+ F (x, t),

where α2 = T/ρ and β, γ and F (x, t) have been relabeled after dividing by ρ. This equationdescribes the transverse vibration of the string. It should be noted that because of the secondderivative with respect to time, the wave equation has two initial conditions, imposed on u(x, 0)and ut(x, 0), which are the initial position and initial velocity, respectively.

The net force on the segment [x, x+ ∆x] due to the tension is

T sin θ2 − T sin θ1 ≈ T [ux(x+ ∆x, t)− ux(x, t)],

where θ1 and θ2 are the angles that the string makes with the x-axis at x and x+ ∆x, respectively.The approximations by ux follow from right-triangle trigonometry. If the rod has a variable densityρ(x), then the term of the wave equation arising from the net force due to tension is (α2(x)ux)x,rather than α2uxx.

3.1.2 Intuitive Interpretation of the Wave Equation

The wave equation states that the acceleration of the string is proportional to the tension in thestring, which is given by its concavity.

33

34 CHAPTER 3. HYPERBOLIC-TYPE PROBLEMS

3.1.3 Applications

Other applications of the one-dimensional wave equation are:

• Modeling the longitudinal and torsional vibration of a rod, or of sound waves. In this case,the coefficient α2 is called Young’s modulus, which is a measure of the elasticity of the rod.

• Modeling electric current along a wire. This model actually yields the transmission-lineequations, which are then manipulated to obtain two wave equations, one for the voltageand one for the current. The coefficients α2 is inversely proportional to the capacitance andself-inductance per unit length.

3.2 The D’Alembert Solution of the Wave Equation

The solution of the Cauchy problem for the wave equation in one space dimension,

utt = c2uxx, −∞ < x <∞, t > 0,

u(x, 0) = f(x), ut(x, 0) = g(x), −∞ < x <∞,

is known as d’Alembert’s solution. We will derive this solution in two different ways.For the first approach, we introduce the change of variables

ξ = x+ ct, η = x− ct.

Then we have

ux = uξξx + uηηx

= uξ + uη,

uxx = (ux)ξ + (ux)η

= uξξ + 2uξη + uηη,

ut = uξξt + uηηt

= cuξ − cuηutt = c(ut)ξ − c(ut)η

= c2uξξ − 2c2uξη + c2uηη.

Substituting these expressions for uxx and utt into the wave equation yields the very simple PDE

uξη = 0.

By integrating with respect to ξ, and then with respect to η, we obtain the general solution

u(ξ, η) = Φ(η) + ψ(ξ)

where the functions Φ(η) and ψ(ξ) are chosen so as to satisfy the initial conditions.Substituting this expression into the initial conditions yields the equations

Φ(x) + ψ(x) = f(x),

−cΦ′(x) + cψ′(x) = g(x).

3.2. THE D’ALEMBERT SOLUTION OF THE WAVE EQUATION 35

This system of equations has the solutions

Φ(x) =1

2f(x)− 1

2c

∫ x

x0

g(s) ds, ψ(x) =1

2f(x) +

1

2c

∫ x

x0

g(s) ds

for some x0. It follows that

u(x, t) = Φ(x− ct) + ψ(x+ ct) =f(x+ ct) + f(x− ct)

2+

1

2c

∫ x+ct

x−ctg(s) ds.

For the second approach, we consider the IBVP

utt = c2uxx, 0 < x < L, t > 0,

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L,

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

We assume a solution of the formu(x, t) = X(x)T (t).

Separation of variables yields the ODEs

X ′′ + λ2X = 0, X(0) = 0, X(L) = 0,

T ′′ + c2λ2T = 0.

The boundary value problem for X has the solutions

Xn(x) = sin(nπxL

), n = 1, 2, . . . ,

and therefore the ODE for T has the general solutions

Tn(t) = An cos

(nπct

L

)+Bn sin

(nπct

L

), n = 1, 2, . . . ,

where An and Bn are constants.It follows that the solution of our IBVP has the form

u(x, t) =

∞∑n=1

An cos

(nπct

L

)sin(nπxL

)+Bn sin

(nπct

L

)sin(nπxL

)=

1

2

∞∑n=1

An

[sin

(nπ(x+ ct)

L

)+ sin

(nπ(x− ct)

L

)]−

Bn

[cos

(nπ(x+ ct)

L

)− cos

(nπ(x− ct)

L

)].

Substituting this solution into the initial conditions yields

f(x) =∞∑n=1

An sin(nπxL

),

g(x) = c

∞∑n=1

Bnnπ

Lsin(nπxL

).


Applying the Fundamental Theorem of Calculus “in reverse” to the cosine terms in the solutionyields

u(x, t) =f(x+ ct) + f(x− ct)

2+

1

2c

∫ x+ct

x−ctg(s) ds.

3.3 More on the D’Alembert Solution

Earlier, we learned that the solution of the initial value problem

utt = c2uxx, −∞ < x <∞, t > 0,

u(x, 0) = f(x), ut(x, 0) = g(x), −∞ < x <∞

is given by D’Alembert’s solution

u(x, t) =1

2[f(x− ct) + f(x+ ct)] +

1

2c

∫ x+ct

x−ctg(s) ds.

We now examine how this solution can be interpreted.

3.3.1 The Space-Time Interpretation of D’Alembert’s Solution

First, we consider the case of a zero initial velocity, which has initial conditions

u(x, 0) = f(x), ut(x, 0) = 0, −∞ < x <∞.

Then, the solution is

u(x, t) =1

2[f(x− ct) + f(x+ ct)].

It follows that at any point (x0, t0), the solution is equal to the average of the initial displacementu(x, 0) = f(x) at the two points obtained by backtracking along the lines

x− ct = x0 − ct0, x+ ct = x0 + ct0

back to the x-axis.For example, suppose the initial displacement is given by

f(x) =

{1 −1 < x < 10 |x| ≥ 1

.

Then, u(x, t) = 1/2 between the lines x+ ct = ±1, and between the lines x− ct = ±1. Where theseregions overlap, these values of u(x, t) are added, and the solution is equal to 1. Outside of theseregions, u(x, t) = 0.

Next, we consider the case of a zero initial displacement,

u(x, 0) = 0, ut(x, 0) = g(x), −∞ < x <∞.

Then, the solution is

u(x, t) =1

2c

∫ x+ct

x−ctg(s) ds.

3.3. MORE ON THE D’ALEMBERT SOLUTION 37

That is, the solution at (x0, t0) is obtained by integrating the initial velocity ut(x, 0) = g(x) alongthe x-axis from x0 − ct0 to x0 + ct0.

Therefore, if the initial velocity is given by

g(x) =

{1 −1 < x < 10 |x| ≥ 1

,

then u(x, t) = (1 + x + ct)/(2c) between the lines x + ct = ±1, and u(x, t) = (1 − x + ct)/(2c)between the lines x − ct = ±1. Where these regions overlap, the solution is equal to t. Betweenthese two regions, the solution is equal to 1/c; everywhere else, it is equal to 0.

3.3.2 Solution of the Semi-Infinite String via the D’Alembert Solution

We now consider a vibrating semi-infinite string with a fixed end, modeled by the IBVP

utt = c2uxx, 0 < x <∞, t > 0,

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x <∞,u(0, t) = 0, t > 0.

As with the infinite string, using the change of variables

ξ = x+ ct, η = x− ct,

we obtain the much simpler PDEuξη = 0,

which has the general solution

u(x, t) = φ(η) + ψ(ξ) = φ(x− ct) + ψ(x+ ct).

As before, we substitute this form of the solution into the initial conditions, and obtain

φ(x− ct) =1

2f(x− ct)− 1

2c

∫ x−ct

x0

g(s) ds, ψ(x+ ct) =1

2f(x+ ct) +

1

2c

∫ x+ct

x0

g(s) ds.

However, we can only evaluate f(x) and g(x) wherever x > 0, which presents a problem whenx− ct < 0. To get around this, we apply the boundary condition to the form of u(x, t) to obtain

u(0, t) = φ(−ct) + ψ(ct) = 0,

orφ(−ct) = −ψ(ct).

This yields

φ(x− ct) = −ψ(ct− x) = −1

2f(ct− x)− 1

2c

∫ ct−x

x0

g(s) ds,

and therefore

u(x, t) = ψ(x+ ct)− ψ(ct− x) =1

2[f(x+ ct)− f(ct− x)] +

1

2c

∫ x+ct

ct−xg(s) ds, 0 < x < ct.

When x ≥ ct, we simply use D’Alembert’s solution as before,

u(x, t) =1

2[f(x+ ct) + f(x− ct)] +

1

2c

∫ x+ct

x−ctg(s) ds, x ≥ ct.

This solution exhibits reflection at the boundary x = 0.


3.4 Boundary Conditions for the Wave Equation

We now consider a finite vibrating string, modeled using the PDE

utt = c2uxx, 0 < x < L, t > 0

and initial conditionsu(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L.

Typically, we impose boundary conditions of one of the following three forms:

1. Controlled end points: When the ends of the string are specified, we use Dirichlet boundaryconditions of the form

u(0, t) = g1(t), u(L, t) = g2(t), t > 0.

2. Force specified on the boundaries: The vertical forces on the string at the endpoints are givenby Tux(0, t) and −Tux(L, t), where T is the tension in the string. By specifying these forces,we obtain Neumann boundary conditions. For example, if the ends of the string are allowedto slide vertically on frictionless sleeves, the boundary conditions become

ux(0, t) = 0, ux(L, t) = 0, t > 0.

3. Elastic attachment: Suppose that the ends of the string are attached to springs. Then, thevertical forces on the string are proportional to the displacements, with constant of propor-tionality given by the spring constant h. This yields the mixed, or Robin boundary conditions

Tux(0, t) = hu(0, t), −Tux(L, t) = hu(L, t), t > 0.

If, in addition, the displacement of the spring attachments at the left and right ends is specifiedby functions θ1(t) and θ2(t), respectively, then the boundary conditions become

Tux(0, t) = h[u(0, t)− θ1(t)], −Tux(L, t) = h[u(L, t)− θ2(t)], t > 0.

Combinations of different boundary conditions are possible. For example, when modeling thelongitudinal vibration in a spring with the end at x = 0 fastened and the end at x = L free, theboundary conditions are

u(0, t) = 0, ux(L, t) = 0, t > 0.

3.5 The Finite Vibrating String

We now consider the wave equation on a finite interval,

utt = c2uxx, 0 < x < L, t > 0,

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L,

with homogeneous Dirichlet boundary conditions

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

3.5. THE FINITE VIBRATING STRING 39

We can solve this IBVP using separation of variables, as we did for the heat equation. We assumea solution of the form

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

and substitute this form into the PDE to obtain

XT ′′ = c2X ′′T.

Dividing both sides by c2XT yieldsT ′′

c2T=X ′′

X= k,

where k must be a constant because the first expression is a function of t and the second is afunction of x. We then obtain the ODEs

X ′′ − kX = 0, X(0) = X(L) = 0,

T ′′ − kc2T = 0.

As with the heat equation, we cannot have k = 0, because then X(x) would be a linear function,which would then equal zero everywhere due to the boundary conditions. On the other hand, ifk > 0, then T (t) would include an exponential term that would grow without bound over time,which is not physically reasonable for modeling the displacement of a vibrating string. We concludethat k < 0, and therefore let k = −λ2. We then have the ODEs

X ′′ + λ2X = 0, X(0) = X(L) = 0,

T ′′ + c2λ2T = 0.

These ODEs have the solutions

X(x) = A cos(λx) +B sin(λx), T (t) = C cos(cλt) +D sin(cλt).

Applying the boundary conditions to X(x) yields the equations

A = 0, B sin(λL) = 0.

It follows that λL = nπ, where n is a positive integer, which yields the solutions

Xn(x) = sin(nπxL

), n = 1, 2, . . . ,

which are eigenfunctions of the problem

−X ′′ = λ2nX, n = 1, 2, . . . ,

with corresponding eigenvalues λ2n = (nπ/L)2. We then have

Tn(t) = Cn cos

(cnπt

L

)+Dn sin

(cnπt

L

), n = 1, 2, . . . .


Combining these solutions to our ODEs, we obtain the general solution to the PDE that alsosatisfies the boundary conditions,

u(x, t) =

∞∑n=1

sin(nπxL

)[Cn cos

(cnπt

L

)+Dn sin

(cnπt

L

)],

where the constants Cn, Dn, n = 1, 2, . . . , are chosen in order to satisfy the initial conditions.Substituting this form of u(x, t) into the initial conditions yields the equations

f(x) =

∞∑n=1

Cn sin(nπxL

), g(x) =

∞∑n=1

cnπDn

Lsin(nπxL

).

That is, Cn and cnπDn/L are the coefficients of the Fourier sine series of f(x) and g(x), respec-tively. It follows that

Cn =2

L

∫ L

0f(x) sin

(nπxL

)dx, Dn =

2

cnπ

∫ L

0g(x) sin

(nπxL

)dx.

This completes the solution of the IBVP.

If the IBVP has homogeneous Neumann boundary conditions

ux(0, t) = 0, ux(L, t) = 0, t > 0,

then the eigenfunctions are

Xn(x) = cos(nπxL

), n = 0, 1, 2, . . .

with corresponding eigenvalues λ2n = (nπ/L)2 for n = 0, 1, 2, . . . . We also have

Tn(t) = Cn cos

(cnπt

L

)+Dn sin

(cnπt

L

), n = 1, 2, . . . , T0(t) = C0.

The solution then has the form

u(x, t) =C0

2+

∞∑n=1

cos(nπxL

)[Cn cos

(cnπt

L

)+Dn sin

(cnπt

L

)].

Proceeding as before, we obtain

Cn =2

L

∫ L

0f(x) cos

(nπxL

)dx, n = 0, 1, 2, . . . ,

Dn =2

cnπ

∫ L

0g(x) cos

(nπxL

)dx, n = 1, 2, . . . .

Non-homogeneous boundary conditions can be handled by transforming the problem into one withhomogeneous boundary conditions, as with the heat equation.

3.6. CLASSIFICATION OF PDES 41

3.6 Classification of PDEs

Previously, we learned how to classify a general second-order linear PDE of the form

Auxx +Buxy + Cuyy +Dux + Euy + Fu = G

as hyperbolic if B2 − 4AC > 0, parabolic if B2 − 4AC = 0, and elliptic if B2 − 4AC < 0. We willnow use this classification system to transform such PDEs into simpler ones, to facilitate analysisand solution.

This transformation process consists of choosing new independent variables ξ, η that lead tosimplified coefficients. From the Chain Rule, we obtain

ux = uξξx + uηηx,

uy = uξξy + uηηy,

uxx = uξξξ2x + 2uξηξxηx + uηηη

2x + uξξxx + uηηxx,

uxy = uξξξxξy + uξη[ξxηy + ξyηx] + uηηηxηy + uξξxy + uηηxy,

uyy = uξξξ2y + 2uξηξyηy + uηηη

2y + uξξyy + uηηyy.

We then obtain a new PDE of the form

Auξξ +Buξη + Cuηη +Duξ + Euη + Fu = G,

where

A = Aξ2x +Bξxξy + Cξ2

y ,

B = 2Aξxηx +B[ξxηy + ξyηx] + 2Cξyηy,

C = Aη2x +Bηxηy + Cη2

y ,

D = Aξxx +Bξxy + Cξyy +Dξx + Eξy,

E = Aηxx +Bηxy + Cηyy +Dηx + Eηy,

F = F,

G = G.

We will now solve for coordinates ξ and η so that A = C = 0, in order to reduce the PDE to acanonical form in which the only second-order coefficient is B. From the above equations for thecoefficients of the transformed PDE, we obtain the equations

A[ξx/ξy]2 +B[ξx/ξy] + C = 0,

A[ηx/ηy]2 +B[ηx/ηy] + C = 0.

These are quadratic equations in [ξx/ξy] and [ηx/ηy], each of which has two solutions. Since thecoefficients of both equations are the same, we must choose different solutions for each unknown,in order to avoid having both variables ξ and η being the same. We therefore obtain

[ξx/ξy] =−B +

√B2 − 4AC

2A, [ηx/ηy] =

−B −√B2 − 4AC

2A.


These solutions yield first-order ODEs that can now be solved to obtain the coordinates ξ and η interms of x and y. From the equations ξ = constant and η = constant, we obtain the equations

dξ = ξx dx+ ξy dy = 0,

dη = ηx dx+ ηy dy = 0,

which yields the ODEsdy

dx= −ξx

ξy,

dy

dx= −ηx

ηy.

These ODE can be solved using standard techniques, such as those applied to separable equations,linear equations or exact equations. Integrating factors can be used if needed.

The new coefficient B can be rewritten as

B = ξyηy

[2A

ξxξy

ηxηy

+B

(ξxξy

+ηxηy

)+ 2C

].

Using the fact that the sum of the roots of the general quadratic equation

Ax2 +Bx+ C = 0

is equal to −B/A, and that the product of the roots is equal to C/A, we obtain

B = ξyηy

[2A

C

A+B

(−BA

)+ 2C

]= ξyηy

[4C − B2

A

]= −ξyηy

A[B2 − 4AC].

The PDE can now be written in the canonical form

Buξη +Duξ + Euη + Fu = G.

The canonical form is useful because much theory related to second-order linear PDE, as well asnumerical methods for their solution, assume that a PDE is already in canonical form.

It is worth noting the relationship between the characteristic variables ξ, η and the classificationof the PDE as hyperbolic, parabolic, or elliptic. If the PDE is hyperbolic, then these characteristicvariables can be obtained, as there are two distinct real roots of the quadratic equation usedto compute their slopes. If the PDE is parabolic, there is a double root, so there is only onecharacteristic variable; therefore, the only characteristic surface is a plane η = constant or ξ =constant. If the PDE is elliptic, there are no real roots, so elliptic PDE do not have characteristiccurves.

Example Consider the hyperbolic PDE

y2uxx − x2uyy = 0,

in which A = y2, B = 0, and C = −x2. To transform this into canonical form, we introduce thenew variables ξ, η and obtain

ξx/ξy =−B +

√B2 − 4AC

2A=x

y,

ηx/ηy =−B −

√B2 − 4AC

2A= −x

y.

3.6. CLASSIFICATION OF PDES 43

We then solve for η and ξ by solving the ODEs

dy

dx= −ξx

ξy= −x

y,

dy

dx= −ηx

ηy=x

y

that arise by implicit differentiation of the equations ξ = constant and η = constant, respectively.These ODEs are separable. Rewriting them as

y dy = −x dx, y dy = x dx

and integrating both sides yields

ξ = y2 + x2, η = y2 − x2.

It follows that the coefficients of the transformed PDE are

A = 0,

B = 2Aξxηx +B[ξxηy + ξyηx] + 2Cξyηy

= 2y2(2x)(−2x) + 0[(2x)(2y) + (2y)(−2x)] + 2(−x2)(2y)(2y)

= −16x2y2,

C = 0,

D = Aξxx +Bξxy + Cξyy +Dξx + Eξy

= y2(2) + 0(0)− x2(2) + 0(2x) + 0(2y)

= 2(y2 − x2),

E = Aηxx +Bηxy + Cηyy +Dηx + Eηy

= y2(−2) + 0(0)− x2(2) + 0(−2x) + 0(2y)

= −2(y2 + x2),

F = F = 0,

G = G = 0.

We conclude that the transformed PDE, in the original variables, is

−16x2y2uξη + 2(y2 − x2)uξ − 2(y2 + x2)uη = 0.

Rewriting the coefficients in terms of the new variables yields

2(η2 − ξ2)uξη + ηuξ − ξuη = 0.

2

An alternative canonical form can be obtained by introducing the new variables

α = ξ + η,

β = ξ − η.


From the Chain Rule, we obtain

uξ = uααξ + uββξ = uα + uβ,

uη = uααη + uββη = uα − uβ,uξη = uαααη + uαββη + uβααη + uβββη

= uαα − uαβ + uβα − uββ= uαα − uββ .

This leads to the canonical form

Buαα −Buββ + (D + E)uα + (D − E)uβ + Fu = G.

Example The alternative canonical form of the PDE from the previous example is

2(η2 − ξ2)(uαα − uββ)− (ξ − η)uα + (ξ + η)uβ = 0,

or, in the new variables,

−2αβ(uαα − uββ)− βuα + αuβ = 0.

2

3.7 The Wave Equation in Two and Three Dimensions

3.7.1 Waves in Three Dimensions

Consider the IVP

utt = c2∇2u, (x, y, z) ∈ R3, t > 0,

u(x, y, z, 0) = 0, ut(x, y, z, 0) = ψ(x, y, z),

where ∇2u is the Laplacian

∇2u = ∇ · ∇u = uxx + uyy + uzz.

Using the 3-D Fourier transform,

u(ω1, ω2, ω3, t) =1

(2π)3/2

∫R3

e−i(ω1x+ω2y+ω3z)u(x, y, z, t) dV,

we obtain the ODE

utt = −c2(ω21 + ω2

2 + ω23)u,

with initial conditions

u(ω1, ω2, ω3, 0) = 0, ut(ω1, ω2, ω3, 0) = ψ(ω1, ω2, ω3),

where ψ is the Fourier transform of ψ. This IVP has the solution

u(ω1, ω2, ω3, t) =1

c‖~ω‖ψ(ω1, ω2, ω3) sin(c‖~ω‖t),

3.7. THE WAVE EQUATION IN TWO AND THREE DIMENSIONS 45

where ‖~ω‖ =√ω2

1 + ω22 + ω2

3. To compute the inverse Fourier transform, we first compute

F−1

[sin(c‖~ω‖t)c‖~ω‖

]=

1

(2π)3/2

∫ 2π

0

∫ π

0

∫ ∞0

eirρ cosφ sin(cρt)

cρρ2 sinφdρ dφ dθ

=1

c√

2π

∫ ∞0

ρ sin(cρt)

∫ π

0eirρ cosφ sinφdφ dρ

=1

c√

2π

∫ ∞0

ρ sin(cρt)

∫ 1

−1eirρu du dρ, u = cosφ

=1

c√

2π

∫ ∞0

sin(cρt)1

ir[eirρ − e−irρ] dρ

=1

cr

√2

π

∫ ∞0

sin(cρt) sin(rρ) dρ

=1

cr√

2π

∫ ∞0

cos[ρ(r − ct)]− cos[ρ(r + ct)] dρ

=π

cr√

2πδ(r − ct)

where r =√x2 + y2 + z2 and δ(x) is the Dirac delta function. We have used spherical coordinates,

with the “north pole” φ = 0 pointing in the direction of the position vector (x, y, z) = (0, 0, r). Wethen apply the convolution theorem to obtain

u(x, y, z, t) =1

(2π)3/2

∫R3

ψ(x′, y′, z′)

[π

cr√

2πδ(r′ − ct)

]dx′ dy′ dz′

=1

4πc2t

∫ 2π

0

∫ π

0ψ(x+ ct cos θ sinφ, y + ct sin θ sinφ, z + ct cosφ)(ct)2 sinφdφ dθ

= tψ,

where r′ =√

(x− x′)2 + (y − y′)2 + (z − z′)2 and

ψ =1

4πc2t2

∫ 2π

0

∫ π

0ψ(x+ ct cos θ sinφ, y + ct sin θ sinφ, z + ct cosφ)(ct)2 sinφdφ dθ

is the average of ψ over the sphere of center (x, y, z) and radius ct.

Now, we consider the IVP

utt = c2∇2u, (x, y, z) ∈ R3, t > 0,

u(x, y, z, 0) = φ(x, y, z), ut(x, y, z, 0) = 0.

Let v be the solution of

vtt = c2∇2v, (x, y, z) ∈ R3, t > 0,

v(x, y, z, 0) = 0, vt(x, y, z, 0) = φ(x, y, z).

Then

v(x, y, z, t) = tφ,


where φ is the average of φ over the sphere centered at (x, y, z) with radius ct. We then have

(vt)tt = (vtt)t = (c2∇2v)t = c2∇2(vt),

and

vt(x, y, z, 0) = φ(x, y, z), (vt)t(x, y, z, 0) = 2φt∣∣t=0

= 0,

which means that u = vt, as it satisfies the PDE and initial conditions of the IVP for u.

We conclude that the solution of the IVP

utt = c2∇2u, (x, y, z) ∈ R3, t > 0,

u(x, y, z, 0) = φ(x, y, z), ut(x, y, z, 0) = ψ,

is

u(x, y, z, t) =∂

∂t[tφ] + tψ.

This formula for the solution is known as Kirchhoff’s formula, as well as Poisson’s formula.

Because the solution depends on integrals over a sphere of radius ct, it follows that if the initialdata are zero except within a small sphere, then the solution is zero at any point (x0, y0, z0) outsidethis sphere until ct is large enough so that the sphere centered at (x0, y0, z0) with radius ct overlapsthe sphere within which the initial data is nonzero. That is, the solution exhibits a sharp leadingedge. Then, once ct is so large that the sphere centered at (x0, y0, z0) with radius ct contains thesphere within which the initial data is nonzero, the solution at (x0, y0, z0) is zero again, and willremain zero. That is, the solution also exhibits a sharp trailing edge. This is known as Huygen’sprinciple. It holds in dimensions 3, 5, 7, and so on, but not in another dimension. We have alreadyseen that it does not hold in one dimension, as D’Alembert’s solution integrates ut(x, 0) = g(x)from x− ct to x+ ct.

3.7.2 Two-Dimensional Wave Equation

The solution of the wave equation in two dimensions can be obtained by solving the three-dimensional wave equation in the case where the initial data depends only on x and y, but not z.In this case, the three-dimensional solution consists of cylindrical waves.

We first consider the IVP

utt = c2∇2u, (x, y) ∈ R2, t > 0,

u(x, y, 0) = 0, ut(x, y) = ψ(x, y).

Extending this problem to three dimensions, we obtain the solution u = tψ, where

ψ =1

4πc2t2

∫ 2π

0

∫ π

0ψ(x+ ct cos θ sinφ, y + ct sin θ sinφ, z + ct cosφ)(ct)2 sinφdφ dθ.

To obtain the solution of the original two-dimensional problem, we need to convert this integralover a sphere of radius ct to an integral over a disc of radius ct. This approach to obtaining thesolution is called the method of descent.

3.8. THE FINITE FOURIER TRANSFORMS 47

We use a substitution to change the spherical coordinate variable φ to the polar coordinatevariable r. From the relation between them,

φ = cos−1( zct

)= cos−1

(√(ct)2 − r2

ct

)

we obtain

dφ =1√

1− (ct)2−r2(ct)2

1

ct

1

2

1√(ct)2 − r2

(−2r) =1√

(ct)2 − r2dr.

From

sinφ =r

ct

our solution becomes

tψ =t

4πc2t2

∫ 2π

0

∫ π

0ψ(x+ ct cos θ sinφ, y + ct sin θ sinφ)(ct)2 sinφdφ dθ

=2

4πc2t

∫ 2π

0

∫ ct

0

ψ(x+ r cos θ, y + r sin θ)√(ct)2 − r2

(ct)2 r

ctdr dθ

=1

2πc

∫ 2π

0

∫ ct

0


r dr dθ.

It follows from Kirchhoff’s formula that the solution of the IVP

utt = c2∇2u, (x, y) ∈ R2, t > 0,

u(x, y, 0) = φ(x, y), ut(x, y) = ψ(x, y).

is

u(x, t) =1

2πc

∫ 2π

0

∫ ct

0


r dr dθ +

∂

∂t

[1

2πc

∫ 2π

0

∫ ct

0

φ(x+ r cos θ, y + r sin θ)√(ct)2 − r2

r dr dθ

].

Because these integrals are over the interior of the disc, as opposed to the boundary of a spherein the three-dimensional case, it can be concluded that Huygen’s principle does not hold in twodimensions.

3.8 The Finite Fourier Transforms

When solving a PDE on a finite interval 0 < x < L, whether it be the heat equation or waveequation, it can be very helpful to use a finite Fourier transform. In particular, we have the finitesine transform

Sn = S[f ] =2

L

∫ L

0f(x) sin(nπx/L) dx, n = 1, 2, . . . ,


with its inverse sine transform

S−1[Sn] = f(x) =∞∑n=1

Sn sin(nπx/L).

This transform should be used with Dirichlet boundary conditions, that specify the value of u atx = 0 and x = L.

When Neumann boundary conditions are used, that specify the value of ux at x = 0 and x = L,it is best to use the finite cosine transform

Cn = C[f ] =2

L

∫ L

0f(x) cos(nπx/L) dx, n = 0, 1, 2, . . . ,

with its inverse sine transform

C−1[Cn] = f(x) =C0

2+

∞∑n=1

Cn cos(nπx/L).

Both of these transforms can be used to reduce a PDE to an ODE.

3.8.1 Examples of the Sine Transform

Consider the function f(x) = 1 on (0, 1). If we apply the finite sine transform to this function, weobtain

Sn = 2

∫ 1

0sin(nπx) dx

= − 2

nπcos(nπx)

∣∣∣∣10

=

{ 4

nπn odd

0 n even.

Applying the inverse sine transform yields

1 =4

π

∞∑n=1

1

2n− 1sin[(2n− 1)πx].

3.8.2 Properties of the Transforms

To apply these transforms to PDEs, we must know how to transform appropriate derivatives. Wehave the following rules:

S[ut] =dS[u]

dt, S[utt] =

d2S[u]

dt2,

C[ut] =dC[u]

dt, C[utt] =

d2C[u]

dt2,

S[uxx] = −[nπ/L]2S[u] +2nπ

L2[u(0, t) + (−1)n+1u(L, t)],

C[uxx] = −[nπ/L]2C[u]− 2

L[ux(0, t) + (−1)n+1ux(L, t)].

3.8. THE FINITE FOURIER TRANSFORMS 49

The last two rules can be obtained by applying integration by parts twice.

3.8.3 Solving Problems via Finite Transforms

We illustrate the use of finite Fourier transforms by solving the IBVP

utt = uxx + sin(πx), 0 < x < 1, t > 0,

u(0, t) = 0, u(1, t) = 0, t > 0,

u(x, 0) = 1, ut(x, 0) = 0, 0 < x < 1.

Because this problem has Dirichlet boundary conditions, we use the finite sine transform. Fromthe preceding example, the transform of the initial conditions are

Sn(0) =

{ 4

nπn odd

0 n even, S′n(0) = 0.

Using the definition and aforementioned properties, we obtain the transform of the PDE,

S′′1 (t) = −π2S1(t) + 1,

S′′n(t) = −(nπ)2Sn(t), n = 2, 3, . . . .

The ODE for S1(t) is nonhomogeneous, and can be solved using either the method of undeterminedcoefficients or variation of parameters. The general solution is

S1(t) = A cos(πt) +B sin(pit) + C,

where A,B and C are constants. Substituting this form of the solution into the ODE and initialconditions yields

S1(t) =

(4

π− 1

π2

)cos(πt) +

1

π2.

The ODEs for Sn(t), n > 1, are homogeneous and can easily be solved to obtain

Sn(t) =

{ 4

nπcos(nπt) n = 3, 5, 7, . . . ,

0 n = 2, 4, 6, . . ..

Applying the inverse sine transform, we conclude that the solution is

u(x, t) =

[(4

π− 1

π2

)cos(πt) +

1

π2

]sin(πx) +

4

π

∞∑n=1

1

2n+ 1cos[(2n+ 1)πt] sin[(2n+ 1)πx].


3.9 Superposition

3.9.1 Superposition Used to Break an IBVP Into Two Simpler Problems

Based on the principle of superposition, the solution of the IBVP

ut = uxx + sin(πx), 0 < x < 1, t > 0,

u(0, t) = 0, u(1, t) = 0, t > 0,

u(x, 0) = sin(2πx), 0 < x < 1,

is given by the solution of the simpler IBVP

ut = uxx + sin(πx), 0 < x < 1, t > 0,

u(0, t) = 0, u(1, t) = 0, t > 0,

u(x, 0) = 0, 0 < x < 1,

added to solution of the second simpler IBVP

ut = uxx, 0 < x < 1, t > 0,

u(0, t) = 0, u(1, t) = 0, t > 0,

u(x, 0) = sin(2πx), 0 < x < 1.

The solution of the original IBVP is

u(x, t) =1

π2(1− e−π2t) sin(πx) + e−(2π)2t sin(2πx).

The first term is the solution of the first simpler IBVP, and the second term is the solution of thesecond simpler IBVP.

3.9.2 Separation of Variables and Integral Transforms as Superpositions

Consider the IBVP

ut = uxx + f(x, t), 0 < x < 1, t > 0,

u(0, t) = 0, u(1, t) = 0, t > 0,

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

In view of the boundary conditions, we represent f(x, t) in terms of its finite Fourier sine series,

f(x, t) =

∞∑n=1

Fn(t) sin(nπx)

where

Fn(t) = 2

∫ 1

0f(x, t) sin(nπx) dx, n = 1, 2, . . . .

3.10. METHOD OF CHARACTERISTICS 51

Using the principle of superposition, this IBVP can be decomposed into the IBVPs

(un)t = (un)xx + Fn(t) sin(nπx), 0 < x < 1, t > 0,

un(0, t) = 0, un(1, t) = 0, t > 0,

un(x, 0) = 0, 0 < x < 1

We assume that each of these solutions has the form

un(x, t) = Un(t) sin(nπx).

Substitution of this form into the PDE, and dividing through by sin(nπx), yields the ODE

U ′n(t) = −(nπ)2Un(t) + Fn(t), t > 0,

with initial conditionUn(0) = 0.

3.10 Method of Characteristics

Consider the general first-order linear initial value problem

a(x, t)ux + b(x, t)ut + cu = 0, −∞ < x <∞, t > 0,

u(x, 0) = f(x), −∞ < x <∞.

To solve this problem, we will change independent variables (x, t) to new variables (τ, s) in orderto obtain a new PDE that is easier to solve. To define these new variables, we solve the IVPs

dx

ds= a(x, t), x(0) = τ,

dt

ds= b(x, t), t(0) = 0.

Then, by the Chain Rule, the PDE reduces to the ODE

a(x, t)ux + b(x, t)ut + cu = uxxs + utts + cu =du

ds+ c(x(τ, s), t(τ, s))u = 0,

with initial conditionu(τ, 0) = f(τ).

Once we solve this problem, we change back to the original variables (x, t) to obtain the solution.This approach is known as the method of characteristics. The new variables (τ, s) define char-

acteristic curves {x(s), t(s)}, each of which has the initial point (τ, 0). The initial data propagatesalong these curves. The s-coordinate indicates points along a characteristic curve, whereas theτ -coordinate indicates the initial point (t = 0) for a given curve. For each τ , the initial value f(τ)is evolved along the curve starting at the point (x, t) = (τ, 0) according to the ODE obtained byexpressing the PDE in characteristic variables.


Example We will solve the IVP

ux + ut + 2u = 0, −∞ < x <∞, t > 0,

u(x, 0) = sinx, −∞ < x <∞.The characteristic coordinates satsify the equations

dx

ds= 1, x(0) = τ,

dt

ds= 1, t(0) = 0.

These equations have the solutionsx = s+ τ, t = s.

The PDE thus reduces to the ODE

du

ds+ 2u = 0, u(τ, 0) = sin τ,

which has the solutionu(τ, s) = e−2s sin τ.

Changing back to the original variables yields

u(x, t) = e−2t sin(x− t).

2

Example We will solve the IVP

xux + ut + tu = 0, −∞ < x <∞, t > 0,

u(x, 0) = F (x).

The characteristic coordinates satsify the equations

dx

ds= x, x(0) = τ,

dt

ds= 1, t(0) = 0.

These equations have the solutionsx = τes, t = s.

The PDE thus reduces to the ODE

du

ds+ su = 0, u(τ, 0) = F (τ),

which has the solutionu(τ, s) = e−s

2/2F (τ).

Changing back to the original variables yields

u(x, t) = e−t2/2F (xe−t).

2

3.11. NONLINEAR FIRST-ORDER EQUATIONS 53

3.11 Nonlinear First-Order Equations

We consider the problem of modeling traffic flow. Let u(x, t) represent the density of cars at thepoint x of a highway at time t. Then u satisfies the following conservation equation: on any segment[a, b] of the highway,

d

dt

∫ b

au(x, t) dx = f(a, t)− f(b, t),

where f(x, t) is the flux, which is the cars per minute passing the point x. Both sides of theequation represent the change in the number of cars within [a, b]. Using the Fundamental Theoremof Calculus, we obtain ∫ b

aut(x, t) dx = −

∫ b

afx(x, t) dt

which is rearranged to ∫ b

aut(x, t) + fx(x, t) dx = 0.

Because the segment [a, b] is arbitrary, we conclude that u is a solution of the PDE

ut + fx = 0.

We now attempt to solve the conservation equation

ut + fx = 0, −∞ < x <∞, t > 0,

u(x, 0) = φ(x), −∞ < x <∞.

Using the Chain Rule, we rewrite the PDE as

ut +df

duux = 0, −∞ < x <∞, t > 0,

or

ut + g(u)ux = 0, −∞ < x <∞, t > 0,

where g(u) = df/du.

The characteristic coordinates satisfy the equations

dx

ds= g(u(τ, 0)), x(0) = τ,

dt

ds= 1, t(0) = 0

and the IVP reduces todu

ds= 0, u(τ, 0) = φ(τ).

Since the solution is independent of s, the characteristic curves are defined by

x = τ + g(φ(τ))s, t = s.


The reduced IVP for u has the solution

u(τ, s) = φ(τ),

and therefore the original IVP has the implicitly defined solution

u(x, t) = φ(x− g(u)t).

In general, this equation can not be solved to obtain an explicit formula for u(x, t). However,it is still possible to understand the behavior of the solution by obtaining the equations of thecharacteristic curves and using the fact that the initial data is propagating along the curves.

Specifically, consider the characteristic curve that begins at (x0, 0). This curve has the equation

t =x− x0

g(φ(x0))

when g(φ(x0)) 6= 0; otherwise, it is simply the vertical line x = x0. Along this curve, the solutionis equal to φ(x0). However, for a nonlinear PDE, unlike the linear case, the characteristic curvesmay intersect, which causes a discontinuity in the solution. This is referred to as a shock. Whenthis occurs, the discontinuity continues to propagate with speed

dx

dt=f(uR)− f(uL)

uR − uL,

where uL and uR are the values of the solution on the left and right sides of the discontinuity,respectively.

Chapter 4

Elliptic-Type Problems

4.1 The Laplacian

The Laplacian of u, defined in 3-D as

∆u = ∇2u = uxx + uyy + uzz,

is one of the most important operators in mathematical physics, as it is featured in several essentialPDEs.

4.1.1 Interpretation of the Laplacian in Two Dimensions

The Laplacian has a very useful intuitive meaning, which can easily be derived in the 1-D case. In1-D, the Laplacian is simply the second derivative. From a Taylor expansion,

u(x) = u(x0) + u′(x0)(x− x0) +1

2∇2u′′(x0)(x− x0)2 + · · · ,

we integrate over the interval [x0 − h, x0 + h] to obtain∫ x0+h

x0−hu(x) dx = 2hu(x0) + u′(x0)

∫ x0+h

x0−h(x− x0) dx+

1

2∇2u(x0)

∫ x0+h

x0−h(x− x0)2 dx+ · · ·

= 2hu(x0) + h2u′(x0)

∫ 1

−1s ds+

h3

2∇2u(x0)

∫ 1

−1s2 ds+ · · · , s = (x− x0)/h,

= 2hu(x0) + 0 + h3∇2u(x0)

∫ 1

0s2 ds+ · · ·

= 2hu(x0) +h3

3∇2u(x0) + · · ·

It follows that

Avgu = u(x0) +h2

6∇2u(x0) + · · · ,

where the average is taken over the interval [x0 − h, x0 + h], and we therefore conclude that

• When ∇2u > 0, u is smaller than its average of neighboring values,

55

56 CHAPTER 4. ELLIPTIC-TYPE PROBLEMS

• When ∇2u < 0, u is larger than its average of neighboring values, and

• When ∇2u = 0, u is equal to its average of neighboring values.

A similar result can be obtained in higher dimensions by integrating over, say, the interior of acircle or a sphere.

4.1.2 Intuitive Meanings of Basic Laws of Physics

The Laplacian is featured in the following prominent PDEs:

• Heat equation: ut = α2uxx states that the rate of change of the temperature u is proportionalto the Laplacian. That is, the temperature is increasing over time when it is less than theaverage of the temperature at neighboring points.

• Wave equation: utt = c2uxx states that the acceleration of the displacement of a vibrat-ing string (in 1-D) or a drumhead (in 2-D) is proportional to the Laplacian. That is, thedisplacement is accelerating upward when it is less than the displacement at neighboringpoints.

• Laplace’s equation: ∇2u = 0 states that the average of u is always equal to its average atneighboring points. This equation can be interpreted as describing the steady-state solutionof the heat equation or wave equation.

• Poisson’s equation: ∇2u = −f describes electrostatic potential when f represents the chargedensity, or the steady-state temperature when f represents a heat source (when f is positive)or heat sink (when f is negative).

• Helmholtz equation: ∇2u + λu = 0 arises from separation of variables applied to the heatequation or wave equation, but as a standalone equation.

4.1.3 Changing Coordinates

Often, it is necesssary to express the Laplacian in a different coordinate system. We illustrate theprocess of changing coordinates for the case of polar coordinates,

x = r cos θ, y = r sin θ,

or, equivalently,

r =√x2 + y2, tan θ =

y

x.

Applying implicit differentiation to these formulas, we obtain

rx = cos θ, θx = −sin θ

r,

ry = sin θ, θy =cos θ

r,

rxx =sin2 θ

r, θxx =

2 cos θ sin θ

r2,

4.1. THE LAPLACIAN 57

ryy =cos2 θ

r, θyy = −2 cos θ sin θ

r2,

which yields

ux = urrx + uθθx

= cos θur −sin θ

ruθ,

uy = urry + uθθy

= sin θur +cos θ

ruθ,

uxx = urxrx + urrxx + uθxθx + uθθxx

= urrr2x + 2urθrxθx + urrxx + uθθθ

2x + uθθxx

= cos2 θurr −2 cos θ sin θ

rurθ +

sin2 θ

rur +

sin2 θ

r2uθθ +

2 cos θ sin θ

r2uθ

uyy = uryry + urryy + uθyθy + uθθyy

= urrr2y + 2urθryθy + urryy + uθθθ

2y + uθθyy

= sin2 θurr +2 cos θ sin θ

rurθ +

cos2 θ

rur +

cos2 θ

r2uθθ −

2 cos θ sin θ

r2uθ.

Combining these results, we obtain

∇2u = urr +1

rur +

1

r2uθθ.

Similarly, for cylindrical coordinates

x = r cos θ, y = r sin θ, z = z,

we have

∇2u = urr +1

rur +

1

r2uθθ + uzz,

and for spherical coordinates

x = ρ cos θ sinφ, y = ρ sin θ sinφ, z = ρ cosφ,

we have

∇2u = uρρ +2

ruρ +

1

r2uφφ +

cot θ

r2uφ +

1

ρ2 sin2 φuθθ.


4.2 Addendum: Correction of Polar Coordinate Conversion

From class:

uxx =

(cos θur −

sin θ

ruθ

)x

=

(cos θur −

sin θ

ruθ

)r

rx +

(cos θur −

sin θ

ruθ

)θ

θx

=

[cos θurr +

sin θ

r2uθ −

sin θ

rurθ

]cos θ +[

− sin θur + cos θurθ −cos θ

ruθ −

sin θ

ruθθ

](−sin θ

r

)=

[cos2 θurr +

sin θ cos θ

r2uθ −

sin θ cos θ

rurθ

]+[

sin2 θ

rur −

sin θ cos θ

rurθ +

sin θ cos θ

r2uθ +

sin2 θ

r2uθθ

]= cos2 θurr +

2 sin θ cos θ

r2uθ −

2 sin θ cos θ

rurθ +

sin2 θ

rur +

sin2 θ

r2uθθ

4.3 The Interior Dirichlet Problem for a Circle

We consider the boundary value problem (BVP) on a circle of radius R,

∇2u = 0, x2 + y2 < R2,

which a Dirichlet boundary condition

u = g, x2 + y2 = R2.

It is more convenient to work in polar coodinates,

urr +1

rur +

1

r2uθθ = 0, 0 < r < R, 0 ≤ θ < 2π,

u(R, θ) = g(θ), 0 ≤ θ < 2π.

4.3.1 Separation of Variables

We first solve this problem using separation of variables. We assume the solution has the form

u(r, θ) = R(r)Θ(θ).

Substituting this form of the solution into the PDE yields

R′′Θ +1

rR′Θ +

1

r2Θ′′ = 0,

4.3. THE INTERIOR DIRICHLET PROBLEM FOR A CIRCLE 59

which can be rearranged to obtain

−r2R′′ + rR′

R=

Θ′′

Θ= k,

where k is the separation constant. We then obtain the ODEs

Θ′′ − kΘ = 0,

r2R′′ + rR′ + kR = 0.

We consider the three possibilities for k:

1. k = 0: Then Θ = A+Bθ, where A and B are constants. However, Θ must be a 2π-periodicfunction, which is not possible unless B = 0.

2. k > 0: Then Θ = Aekθ + Be−kθ, where A and B are constants. This function cannot be2π-periodic unless A = B = 0, so this scenario is not considered.

3. k < 0: Then we let k = −λ2, in which case

Θ = A cos(λθ) +B sin(λθ)

is the general solution. In order to be 2π-periodic, we must require that λ be a positiveinteger.

We conclude that the ODE for Θ has the solutions

Θn(θ) = An cos(nθ) +Bn sin(nθ), n = 0, 1, 2, . . . ,

where An and Bn are constants.We then consider the ODE for R,

r2R′′ + rR′ − n2R = 0,

which is an Euler equation. Using the substitution z = lnx, we can transform this equation into asecond-order linear ODE with constant coefficients, which yields the general solution

R(r) = Arn +Br−n,

where A and B are constants. In order for the solution to be well-defined at the center of the circle,we set B = 0. It follows that a solution of Laplace’s equation on the circle is

un(r, θ) =

(r

R

)n[An cos(nθ) +Bn sin(nθ)], n = 0, 1, 2, . . . ,

where we have added the constant factor 1/Rn for convenience. Thus we have the general solutionof our BVP,

u(r, θ) =A0

2+

∞∑n=1

(r

R

)n[An cos(nθ) +Bn sin(nθ)],


where again we have added the constant factor of 1/2 in the first term for convenience.

We must choose the constants An and Bn to satisfy the boundary condition. Substituting r = Ryields the equation

g(θ) =A0

2+∞∑n=1

An cos(nθ) +Bn sin(nθ).

It follows that these constants are the coefficients of the Fourier series of g(θ),

An =1

π

∫ 2π

0g(α) cos(nα) dα, n = 0, 1, 2, . . . ,

Bn =1

π

∫ 2π

0g(α) sin(nα) dα, n = 1, 2, . . . .

This completes the solution of the BVP.

4.3.2 The Poisson Integral Formula

Using the formulas for the constants An and Bn in the above formula for the solution, we canobtain an alternative formula. Using R in place of R (not to be confused with the function R(r)from separation of variables), we have

u(r, θ) =1

π

∫ 2π

0

[1

2+∞∑n=1

( rR

)n[cos(nθ) cos(nα) + sin(nθ) sin(nα)]

]g(α) dα

=1

π

∫ 2π

0

[1

2+

∞∑n=1

( rR

)ncos[n(θ − α)]

]g(α) dα

=1

2π

∫ 2π

0

[1 +

∞∑n=1

( rR

)n (ein(θ−α) + e−in(θ−α)

)]g(α) dα

=1

2π

∫ 2π

0

[1 +

∞∑n=1

( rRei(θ−α)

)n+∞∑n=1

( rRe−i(θ−α)

)n]g(α) dα

=1

2π

∫ 2π

0

[1 +

rRe

i(θ−α)

1−(rRe

i(θ−α)) +

rRe−i(θ−α)

1−(rRe−i(θ−α)

)] g(α) dα

=1

2π

∫ 2π

0

[1 +

rei(θ−α)

R− rei(θ−α)+

re−i(θ−α)

R− re−i(θ−α)

]g(α) dα

=1

2π

∫ 2π

0

[1 +

rei(θ−α)(R− re−i(θ−α)) + re−i(θ−α)(R− rei(θ−α))

R2 − 2rR cos(θ − α) + r2

]g(α) dα

=1

2π

∫ 2π

0

[1 +

2rR cos(θ − α)− 2r2

R2 − 2rR cos(θ − α) + r2

]g(α) dα

=1

2π

∫ 2π

0

[R2 − r2

R2 − 2rR cos(θ − α) + r2

]g(α) dα.

4.3. THE INTERIOR DIRICHLET PROBLEM FOR A CIRCLE 61

This formula is called the Poisson integral formula. It is interesting to note that from this formula,it can easily be seen that

u(0, 0) =1

2π

∫ 2π

0g(α) dα.

That is, the value of the solution at the center of the circle is equal to the average value of g onthe boundary. Furthermore, by the Law of Cosines, the denominator in the integrand is the lengthof the side opposite the angle |θ − α| of the triangle with vertices (0, 0), (r, θ), and (R,α).

For a BVP defined on a non-circular domain, this formula can be applied by first conformallymapping the domain to a circle, using the formula on the transformed problem, and then trans-forming back to the original domain.

MAT 417: Introduction to Partial Di erential Equations

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