partial di erential equations - iit kanpurhome.iitk.ac.in/~tmk/courses/mth203/mso203b.pdf · 7.2...

Partial Differential Equations

T. [email protected]

November 13, 2015

Contents

Notations vii

1 Introduction 1

1.1 Multi-Index Notations . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Classification of PDE . . . . . . . . . . . . . . . . . . . . . . . 4

2 Introduction Continued... 7

2.1 Solution of PDE . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Well-posedness of PDE . . . . . . . . . . . . . . . . . . . . . . 9

3 First Order PDE 11

3.1 Linear Transport Equation . . . . . . . . . . . . . . . . . . . . 11

3.2 Method of Characteristics . . . . . . . . . . . . . . . . . . . . 14

4 Method of Characteristics: Continued... 17

5 Classification of Second Order PDE 23

5.1 Semilinear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 Classification of SOPDE: Continued 29

6.1 Quasilinear . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.2 Why Characteristic Curves? . . . . . . . . . . . . . . . . . . . 29

6.3 Cauchy Boundary Condition . . . . . . . . . . . . . . . . . . . 31

7 Classification of SOPDE: Continued 33

7.1 Invariance of Discriminant . . . . . . . . . . . . . . . . . . . . 33

7.2 Standard or Canonical Forms . . . . . . . . . . . . . . . . . . 34

7.3 Reduction to Standard Form . . . . . . . . . . . . . . . . . . . 35

iii

CONTENTS iv

8 The Laplacian 41

8.1 Properties of Laplacian . . . . . . . . . . . . . . . . . . . . . . 41

8.2 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . 43

8.3 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . 45

9 Properties of Harmonic Functions 47

9.0.1 Existence and Uniqueness of Solution . . . . . . . . . . 48

10 Sturm-Liouville Problems 51

10.1 Eigen Value Problems . . . . . . . . . . . . . . . . . . . . . . 51

10.2 Sturm-Liouville Problems . . . . . . . . . . . . . . . . . . . . 52

11 Spectral Results 55

12 Singular Sturm-Liouville Problem 59

12.0.1 EVP of Legendre Operator . . . . . . . . . . . . . . . . 59

12.0.2 EVP of Bessel’s Operator . . . . . . . . . . . . . . . . 61

13 Orthogonality of Eigen Functions 63

13.1 Eigen Function Expansion . . . . . . . . . . . . . . . . . . . . 67

14 Fourier Series 69

14.1 Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . . 69

14.2 Fourier Coefficients and Fourier Series . . . . . . . . . . . . . 70

15 Fourier Series: Continued... 77

15.1 Piecewise Smooth Functions . . . . . . . . . . . . . . . . . . . 77

15.2 Complex Fourier Coefficients . . . . . . . . . . . . . . . . . . . 78

15.3 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

15.3.1 Odd and Even functions . . . . . . . . . . . . . . . . . 82

15.4 Fourier Sine-Cosine Series . . . . . . . . . . . . . . . . . . . . 82

15.5 Fourier Transform and Integral . . . . . . . . . . . . . . . . . 84

16 Standing Waves: Separation of Variable 87

16.1 Elliptic Equations . . . . . . . . . . . . . . . . . . . . . . . . . 91

17 Parabolic: Heat Equation 99

17.1 Inhomogeneous Equation . . . . . . . . . . . . . . . . . . . . . 102

CONTENTS v

18 Travelling Waves 10518.1 Domain of Dependence and Influence . . . . . . . . . . . . . . 108

Appendices 109

A Divergence Theorem 111

B Normal Vector of a Surface 113

C Duhamel’s Principle 115

Bibliography 117

Index 119

CONTENTS vi

Notations

Symbols

N denotes the set of natural numbers

Ω denotes an open subset of Rn, not necessarily bounded

∂Ω denotes the boundary of Ω

R denotes the set of real numbers

Rn denotes the n-dimensional Euclidean space

∆∑n

i=1∂2

∂x2i

Dα ∂α1∂x1α1

. . . ∂αn

∂xnαnand α = (α1, . . . , αn). In particular, for α = (1, 1, . . . , 1),

D = ∇ = D(1,1,,...,1) =(

∂∂x1, ∂∂x2, . . . , ∂

∂xn

)Function Spaces

C(X) is the class of all continuous functions on X

Ck(X) is the class of Ck functions which admit a continuous extension tothe boundary of X

Ck(X) is the class of all k-times (k ≥ 1) continuously differentiable functionson X

C∞(X) is the class of all infinitely differentiable functions on X

Cj,k(X × Y ) is the class of all j-times (j ≥ 0) continuously differentiablefunctions on X and k-times (k ≥ 0) continuously differentiable func-tions on Y

vii

NOTATIONS viii

C∞c (X) is the class of all infinitely differentiable functions on X with com-pact support

General Conventions

∇x, ∆x or D2x When a PDE involves both the space variable x and time vari-

able t, the quantities like ∇, ∆, D2, etc. are always taken with respectto the space variable x only. This is a standard convention. Some-times the suffix, like ∇x or ∆x, is used to indicate that the operationis taken w.r.t x.

BVP Boundary value problem

IVP Initial value problem

w.r.t with respect to

Lecture 1

Introduction

A partial differential equation (PDE) is an equation involving an unknownfunction u of two or more variables and some or all of its partial derivatives.The partial differential equation is a tool to analyse the models of nature. Theprocess of understanding natural system can be divided in to three stages:

(i) Modelling the problem or deriving the mathematical equation (formu-lating a PDE) describing the natural system. The derivation process isa result of physical laws such as Newton’s law, momentum, conservationlaws, balancing forces etc.

(ii) Solving the equation (PDE). What constitutes as a solution to a PDE?

(iii) Studying the properties of a solution. Most often the solution of a PDEmay not have a nice formula or representation. How much informationabout the solution can one extract without any representation of asolution? In this text, one encounters similar situation while studyingharmonic functions.

1.1 Multi-Index Notations

Let Ω be an open subset of R. Recall that the derivative of a functionu : Ω→ R, at x ∈ Ω, is defined as

u′(x) := limh→0

u(x+ h)− u(x)

h

1

LECTURE 1. INTRODUCTION 2

provided the limit exists. Now, let Ω be an open subset of Rn. The directionalderivative of u : Ω → R, at x ∈ Ω and in the direction of a given vectorξ ∈ Rn, is defined as

∂u

∂ξ(x) := lim

h→0

u(x+ hξ)− u(x)

h

provided the limit exists. Let the n-tuple ei := (0, 0, . . . , 1, 0, . . . , 0), where 1is in the i-th place, denote the standard basis vectors of Rn. The i-th partialderivative of u at x is the directional derivative of u, at x ∈ Ω and along thedirection ei, and is denoted as

uxi(x) =∂u

∂xi(x) = lim

h→0

u(x+ hei)− u(x)

h.

A multi-index α = (α1, . . . , αn) is a n-tuple where αi, for each 1 ≤ i ≤n, is a non-negative integer. Let |α| := α1 + . . . + αn. If α and β aretwo multi-indices, then α ≤ β means αi ≤ βi, for all 1 ≤ i ≤ n, andα ± β = (α1 ± β1, . . . , αn ± βn). Also, α! = α1! . . . αn! and, for any x ∈ Rn,xα = xα1

1 . . . xαnn . The multi-index notation, introduced by L. Schwartz, isquite handy in representing multi-variable equations in a concise form. Forinstance, a k-degree polynomial in n-variables can be written as∑

|α|≤k

aαxα.

The partial differential operator of order α is denoted as

Dα =∂α1

∂x1α1. . .

∂αn

∂xnαn=

∂|α|

∂x1α1 . . . ∂xnαn

.

One adopts the convention that, among the similar components of α, theorder in which differentiation is performed is irrelevant. This is not a re-strictive convention because the independence of order of differentiation isvalid for smooth1 functions. For instance, if α = (1, 1, 2) then one adopts theconvention that

∂4

∂x1∂x2∂x32

=∂4

∂x2∂x1∂x32.

1smooth, usually, refers to as much differentiability as required.


For each k ∈ N, Dku(x) := Dαu(x) | |α| = k. The k = 1 case is

D1u(x) =(D(1,0,...,0)u(x), D(0,1,0,...,0)u(x), . . . , D(0,0,...,0,1)u(x)

)=

(∂u(x)

∂x1

,∂u(x)

∂x2

, . . . ,∂u(x)

∂xn

).

The operator D1 is called the gradient operator and is denoted as D or ∇.Thus, ∇u(x) = (ux1(x), ux2(x), . . . , uxn(x)). The directional derivative alonga vector ξ ∈ Rn satisfies the identity

∂u

∂ξ(x) = ∇u(x) · ξ.

The normal derivative is the directional derivative along the normal directionν(x), at x, with respect to the surface in which x lies. The divergence of avector function u = (u1, . . . , un), denoted as div(u), is defined as div(u) :=∇ · u. The k = 2 case is

D2u(x) =

∂2u(x)

∂x21. . . ∂2u(x)

∂x1∂xn∂2u(x)∂x2∂x1

. . . ∂2u(x)∂x2∂xn

.... . .

...∂2u(x)∂xn∂x1

. . . ∂2u(x)∂x2n

n×n

.

The matrix D2u is called the Hessian matrix. Observe that the Hessianmatrix is symmetric due to the independence hypothesis of the order inwhich partial derivatives are taken. The Laplace operator, denoted as ∆, isdefined as the trace of the Hessian operator, i.e., ∆ :=

∑ni=1

∂2

∂x2i. Note that

∆ = ∇ · ∇. Further, for a k-times differentiable function u, the nk-tensorDku(x) := Dαu(x) | |α| = k may be viewed as a map from Rn to Rnk .Thus, the magnitude of Dku(x) is

|Dku(x)| :=

∑|α|=k

|Dαu(x)|2 1

2

.

In particular, |∇u(x)| = (∑n

i=1 u2xi

(x))12 or |∇u(x)|2 = ∇u(x) · ∇u(x) and

|D2u(x)| = (∑n

i,j=1 u2xixj

(x))12 .


Example 1.1. Let u(x, y) : R2 → R be defined as u(x, y) = ax2 + by2. Then

∇u = (ux, uy) = (2ax, 2by)

and

D2u =

(uxx uyxuxy uyy

)=

(2a 00 2b

).

Observe that ∇u : R2 → R2 and D2u : R2 → R4 = R22 .

1.2 Classification of PDE

Definition 1.2.1. Let Ω be an open subset of Rn. A k-th order partialdifferential equation of an unknown function u : Ω→ R is of the form

F(Dku(x), Dk−1u(x), . . . Du(x), u(x), x

)= 0, (1.2.1)

for each x ∈ Ω, where F : Rnk × Rnk−1 × . . . × Rn × R × Ω → R is a givenmap such that F depends, at least, on one k-th partial derivative u and isindependent of (k + j)-th partial derivatives of u for all j ∈ N.

In short, the order of a PDE is the highest partial derivative order thatoccurs in the PDE. A first order PDE with two unknown variable (x, y) isrepresented as F (ux, uy, u, x, y) = 0 with F depending, at least, on one of uxand uy. Similarly, a first order PDE with three variable unknown functionu(x, y, z) is written as F (ux, uy, uz, u, x, y, z) = 0 with F depending, at least,on one of ux, uy and uz. Note the abuse in the usage of x. In the n-variablecase, x ∈ Rn is a vector. In the two and three variable case x is the firstcomponent of the vector. The usage should be clear from the context.

A PDE is a mathematical description of a physical process and solvinga PDE for an unknown u helps in predicting the behaviour of the physicalprocess. The level of difficulty in solving a PDE may depend on its order kand linearity of F . We begin by classifying PDEs in a scale of linearity.

Definition 1.2.2. (i) A k-th order PDE is linear if F in (1.2.1) has theform

G(u(x)) = f(x)

where G(u(x)) :=∑|α|≤k aα(x)Dαu(x) for given functions f and aα’s.

It is called linear because G is linear in u for all derivatives , i.e.,G(λu1 + µu2) = λG(u1) + µG(u2) for λ, µ ∈ R. In addition, if f ≡ 0then the PDE is linear and homogeneous.


(ii) A k-th order PDE is semilinear if F is linear only in the highest (k-th)order, i.e., F has the form∑

|α|=k

aα(x)Dαu(x) + f(Dk−1u(x), . . . , Du(x), u(x), x) = 0.

(iii) A k-th order PDE is quasilinear if F has the form∑|α|=k

aα(Dk−1u(x), . . . , u(x), x)Dαu+ f(Dk−1u(x), . . . , u(x), x) = 0,

i.e., the coefficient of its highest (k-th) order derivative depends on uand its derivative only upto the previous (k − 1)-th orders.

(iv) A k-th order PDE is fully nonlinear if it depends nonlinearly on thehighest (k-th) order derivatives.

Observe that, for a semilinear PDE, f is never linear in u and its deriva-tives, otherwise it reduces to being linear. For a quasilinear PDE, aα (with|α| = k), cannot be independent of u or its derivatives, otherwise it reducesto being semilinear or linear.

Example 1.2. (i) xuy − yux = u is linear.

(ii) xux + yuy = x2 + y2 is linear.

(iii) utt − c2uxx = f(x, t) is linear.

(iv) y2uxx + xuyy = 0 is linear.

(v) ux + uy − u2 = 0 is semilinear.

(vi) ut + uux + uxxx = 0 is semilinear.

(vii) u2tt + uxxxx = 0 is semilinear.

(viii) ux + uuy − u2 = 0 is quasilinear.

(ix) uux + uy = 2 is quasilinear.

(x) uxuy − u = 0 is nonlinear.

Lecture 2

Introduction Continued...

2.1 Solution of PDE

Definition 2.1.1. We say u : Ω → R is a (classical) solution to the k-thorder PDE (1.2.1),

• if Dαu exists for all |α| explicitly present in (1.2.1);

• and u satisfies the equation (1.2.1).

Example 2.1. Consider the first order equation ux(x, y) = 0 in R2. Freezingthe y-variable, the PDE can be viewed as an ODE in x-variable. On integrat-ing both sides w.r.t x, u(x, y) = f(y) for any arbitrary function f : R → R.Therefore, for every choice of f : R → R, there is a solution u of the PDE.Note that the solution u is not necessarily in C1(R2) to solve the first orderPDE as is the case in solving an ODE. In fact, choosing a discontinuousfunction f , we obtain a solution which is discontinuous in the y-direction.Similarly, the solution of uy(x, y) = 0 is u(x, y) = f(x) for any choice off : R → R. We say u is a classical solution if u satisfies the equation andu ∈ C1,0(R× R).

Example 2.2. Consider the first order equation ut(x, t) = u(x, t) in R×(0,∞)such that u(x, t) 6= 0, for all (x, t). Freezing the x-variable, the PDE can beviewed as an ODE in t-variable. Integrating both sides w.r.t t we obtainu(x, t) = f(x)et, for some arbitrary function f : R→ R.

Example 2.3. Consider the second order PDE uxy(x, y) = 0 in R2. In contrastto the previous two examples, the PDE involves derivatives in both variables.

7

LECTURE 2. INTRODUCTION CONTINUED... 8

On integrating both sides w.r.t x we obtain uy(x, y) = F (y), for any arbitraryintegrable function F : R→ R. Now, integrating both sides w.r.t y, u(x, y) =f(y)+g(x) for an arbitrary g : R→ R and a f ∈ C1(R)1. But the u obtainedabove is not a solution to uyx(x, y) = 0 if g is not differentiable. Since weassume mixed derivatives to be same we need to assume f, g ∈ C1(R) for thesolution to exist.

Example 2.4. Consider the first order equation ux(x, y) = uy(x, y) in R2.On first glance, the PDE does not seem simple to solve. But, by changeof coordinates, the PDE can be rewritten in a simpler form. Choose thecoordinates w = x + y and z = x − y and, by chain rule, ux = uw + uzand uy = uw − uz. In the new coordinates, the PDE becomes uz(w, z) = 0which is in the form considered in Example 2.1. Therefore, its solution isu(w, z) = f(w) for any arbitrary f : R → R and, hence, u(x, y) = f(x + y).But now an arbitrary f cannot be a solution. We impose that f ∈ C1(R).

The family of solutions, obtained in the above examples, may not bethe only family that solves the given PDE. Following example illustrates asituation where three different family of solutions exist (more may exist too)for the same PDE.

Example 2.5. Consider the second order PDE ut(x, t) = uxx(x, t).

(i) Note that u(x, t) = c is a solution of the PDE, for any constant c ∈ R.This is a family of solutions indexed by c ∈ R.

(ii) The function u : R2 → R defined as u(x, t) = x2

2+t+c, for any constant

c ∈ R, is also a family of solutions of the PDE. Because ut = 1, ux = xand uxx = 1. This family is not covered in the first case.

(iii) The function u(x, t) = ec(x+ct) is also a family of solutions to the PDE,for each c ∈ R. Because ut = c2u, ux = cu and uxx = c2u. This familyis not covered in the previous two cases.

Recall that the family of solutions of an ODE is indexed by constants. Incontrast to ODE, observe that the family of solutions of a PDE is indexedby either functions or constants.

1In fact, it is enough to assume f is differentiable a.e. which is beyond the scope ofthis text


2.2 Well-posedness of PDE

It has been illustrated via examples that a PDE has a family of solutions.The choice of one solution from the family of solutions is made by imposingboundary conditions (boundary value problem) or initial conditions (initialvalue problem). If too many initial/boundary conditions are specified, thenthe PDE may have no solution. If too few initial/boundary conditions arespecified, then the PDE may have many solutions. Even with right amountof initial/boundary conditions, but at wrong places, the solution may failto be stable, i.e., may not depend continuously on the initial or boundarydata. It is, usually, desirable to solve a well-posed problem, in the sense ofHadamard. A PDE, along with the boundary condition or initial condition,is said to be well-posedness if the PDE

(a) has a solution (existence);

(b) the solution is unique (uniqueness);

(c) and the solution depends continuously on the data given (stability).

Any PDE not meeting the above criteria is said to be ill-posed. If the PDE(with boundary/initial conditions) is viewed as a map then the well-posednessof the PDE is expressed in terms of the surjectivity, injectivity and continuityof the “inverse” map. The existence and uniqueness condition depends on thenotion of solution in consideration. There are three notions of solution, viz.,classical solutions, weak solutions and strong solutions. This textbook, forthe most part, is in the classical situation. Further, the stability conditionmeans that a small “change” in the data reflects a small “change” in thesolution. The change is measured using a metric or “distance” in the functionspace of data and solution, respectively. Though in this text we study onlywell-posed problems there are ill-posed problems which are also of interest.

The following example illustrates the idea of continuous dependence ofsolution on data in the uniform metric on the space of continuous functions.

Example 2.6. The initial value problem (IVP)utt(x, t) = uxx(x, t) in R× (0,∞)u(x, 0) = ut(x, 0) = 0


has the trivial solution u(x, t) = 0. Consider the IVP with a small change indata,

utt(x, t) = uxx(x, t) in R× (0,∞)u(x, 0) = 0ut(x, 0) = ε sin

(xε

)which has the unique2 solution uε(x, t) = ε2 sin(x/ε) sin(t/ε). The change insolution of the IVP is measured using the uniform metric as

sup(x,t)

|uε(x, t)− u(x, t)| = ε2 sup(x,t)

|sin(x/ε) sin(t/ε)| = ε2.

Thus, a small change in data induces a small enough change in solution underthe uniform metric3.

Example 2.7 (Ill-posed). The IVPutt(x, t) = −uxx(x, t) in R× (0,∞)u(x, 0) = ut(x, 0) = 0

has the trivial solution u(x, t) = 0. Consider the IVP with a small change indata,

utt(x, t) = −uxx(x, t) in R× (0,∞)u(x, 0) = 0ut(x, 0) = ε sin

(xε

)which has the unique solution uε(x, t) = ε2 sin(x/ε) sinh(t/ε). The solutionof the IVP is not stable because the data change is small, i.e.,

supx|uεt(x, 0)− ut(x, 0)| = ε sup

x|sin(x/ε)| = ε

and the solution change is not at all small, i.e.,

limt→∞

supx|uε(x, t)− u(x, t)| = lim

t→∞ε2 |sinh(t/ε)| = +∞.

In fact, the solution will not converge in any reasonable metric.

2This claim will be proved in later chapters.3The space R× (0,∞) is not compact and the metric is not complete. The example is

only to explain the notion of stability at an elementarty level.

Lecture 3

First Order PDE

The aim of this chapter is to find the general solution and to solve the Cauchyproblem associated with the first order PDE of the form

F (∇u(x), u(x), x) = 0 x ∈ Rn.

3.1 Linear Transport Equation

The transport of a substance in a fluid flowing (one dimenisonal flow) withconstant speed b, with neither source or sink of substance, is given by

ut(x, t) + bux(x, t) + du(x, t) = cuxx(x, t) (x, t) ∈ R× (0,∞)

where c is the diffusive coefficient of the substance and d is the rate of decayof the substance. The function u(x, t) denotes the concentration/density ofthe substance at (x, t). Note that the case of no diffusion (c = 0) is a linearfirst order equation which will be studied in this section.

Example 3.1 (One Dimension with no decay). The one space dimension trans-port equation is

ut(x, t) + bux(x, t) = 0; (x, t) ∈ R× (0,∞)

with b ∈ R. The c = d = 0 case describes the transport of an insoluble1

substance O in a fluid flowing with constant speed b. To solve this we considertwo observers, one a fixed observer A and another observer B, moving with

1assuming no diffusion and no decay of substance.

11

LECTURE 3. FIRST ORDER PDE 12

Figure 3.1: Transport of initial data g

speed b and in the same direction as the substance O. For B, the substanceO would appear stationary while for A, the fixed observer, the substance Owould appear to travel with speed b. What is the equation of the transportof the “stationary” substance O from the viewpoint of the moving observerB? The answer to this question lies in identifying the coordinate systemfor B relative to A. Fix a point x at time t = 0. After time t, the pointx remains as x for the fixed observer A, while for the moving observer B,the point x is now x − bt. Therefore, the coordinate system for B is (w, z)where w = x− bt and z = t. Let v(w, z) describe the motion of O from B’sperspective. Since B sees O as stationary, the PDE describing the motionof O is vz(w, z) = 0. Therefore, v(w, z) = g(w), for some arbitrary functiong (sufficiently differentiable), is the solution from B’s perspective. To solvethe problem from A’s perspective, note that

ut = vwwt + vzzt = −bvw + vz and

ux = vwwx + vzzx = vw.

Therefore, ut + bux = −bvw + vz + bvw = vz and, hence, u(x, t) = v(w, z) =g(w) = g(x − bt) (cf. Fig 3.1). The choice of g is based on our restrictionto be in a classical solution set-up. Note that, for any choice of g, we have


Figure 3.2: Path of substance, over time, placed at x0 with b > 0

g(x) = u(x, 0). The line x − bt = x0, for some constant x0, in the xt-planetracks the flow of the substance placed at x0 at time t = 0 (cf. Fig 3.2). Also,observe that 0 = ut + bux = (ux, ut) · (b, 1) is precisely that the directionalderivative along the vector (b, 1) is zero. This means that u is constant ifwe move along the direction (b, 1). Thus, the value of u(x, t) on the linex− bt = x0 is constant.

Example 3.2 (Transport equation in first quadrant). The transport equationis

ut(x, t) + bux(x, t) = 0; (x, t) ∈ (0,∞)× (0,∞)

with b ∈ R. As before, we obtain u(x, t) = g(x − bt) where g(x) = u(x, 0).This problem is uniquely solvable in the given region only for b < 0. Forb > 0, g defined on x-axis is not adequate to solve the problem. The problemis not well-posed! The given data is enough only to solve for u in the region(x, t) ∈ (0,∞) × (0,∞) | x > bt when b > 0. To compute u in (x, t) ∈(0,∞)× (0,∞) | x > bt we need to provide data on the t-axis (0, t).

Example 3.3 (Transport equation in semi-infinite strip). The transport equa-tion is

ut(x, t) + bux(x, t) = 0; (x, t) ∈ (0, L)× (0,∞)


with b ∈ R. As before, we obtain u(x, t) = g(x− bt) where g(x) = u(x, 0). Ifb > 0 then the problem is well-posed when the data given on x and t axes.If b < 0 then the problem is well-posed when the data is given on x-axis and(L, t)-axis.

3.2 Method of Characteristics

The method of characteristics gives the equation, a system of ODE, of thecharacteristic curves. The method of characteristics reduces a first orderPDE to a system of ODE. For illustration, consider the two variable firstorder quasi-linear equation:

A(x, y, u)ux +B(x, y, u)uy = C(x, y, u). (3.2.1)

Solving for u(x, y) in the above equation is equivalent to finding the surfaceS ≡ (x, y, u(x, y)) generated by u in R3. If u is a solution of (3.2.1), ateach (x, y) in the domain of u, then

A(x, y, u)ux +B(x, y, u)uy = C(x, y, u)

A(x, y, u)ux +B(x, y, u)uy − C(x, y, u) = 0

(A(x, y, u), B(x, y, u), C(x, y, u)) · (ux, uy,−1) = 0

(A(x, y, u), B(x, y, u), C(x, y, u)) · (∇u(x, y),−1) = 0.

But (∇u(x, y),−1) is normal to S at the point (x, y) (cf. Appendix B).Hence, the coefficients (A(x, y, u), B(x, y, u), C(x, y, u)) are perpendicular tothe normal and, therefore, (A(x, y, u), B(x, y, u), C(x, y, u)) lie on the tangentplane to S at (x, y, u(x, y)).

Definition 3.2.1. A smooth curve in Rn is said to be an integral or charac-teristic curve w.r.t a given vector field, if the vector field is tangential to thecurve at each of its point.

Definition 3.2.2. A smooth surface in Rn is said to be an integral surfacew.r.t a given vector field, if the vector field is tangential to the surface at eachof its point.

In the spirit of above definition and arguments, finding a solution u to(3.2.1) is equivalent to determining an integral surface S corresponding to the


coefficient vector field V (x, y) = (A(x, y, u), B(x, y, u), C(x, y, u)) of (3.2.1).Let s denote the parametrization of the characteristic curves w.r.t V . Forconvenience, let z(s) := u(x(s), y(s)). Then the characteristic curves can befound by solving the system of ODEs

dx

ds= A(x(s), y(s), z(s)),

dy

ds= B(x(s), y(s), z(s)),

dz

ds= C(x(s), y(s), z(s)).

(3.2.2)The three ODE’s obtained are called characteristic equations. The union ofthese characteristic (integral) curves give us the integral surface S. The unionis in the sense that every point in the integral surface belongs to exactly onecharacteristic.

Example 3.4 (Linear Transport Equation). The linear transport equation isalready solved earlier using elementary method. Let us solve the same usingthe method of characteristics. Consider the linear transport equation in twovariable,

ut + bux = 0, x ∈ R and t ∈ (0,∞),

where the constant b ∈ R is given. Thus, the given vector field V (x, t) =(b, 1, 0). The characteristic equations are

dx

ds= b,

dt

ds= 1, and

dz

ds= 0.

Solving the 3 ODE’s, we get

x(s) = bs+ c1, t(s) = s+ c2, and z(s) = c3.

Note that solving the system of ODEs requires some initial condition. Wehave already observe that the solution of the transport equation dependedon the value of u at time t = 0, i.e., the value of u on the curve (x, 0) inthe xt-plane. Thus, the problem of finding a function u solving a first orderPDE such that u is known on a curve Γ in the xy-plane is called the Cauchyproblem.

Example 3.5. Let g be given (smooth enough) function g : R→ R. Considerthe linear transport equation

ut + bux = 0 x ∈ R and t ∈ (0,∞)u(x, 0) = g(x) x ∈ R. (3.2.3)


We parametrize the curve Γ with r-variable, i.e., Γ = γ1(r), γ2(r) =(r, 0). The characteristic equations are:

dx(r, s)

ds= b,

dt(r, s)

ds= 1, and

dz(r, s)

ds= 0

with initial conditions,

x(r, 0) = r, t(r, 0) = 0, and z(r, 0) = g(r).

Solving the ODE’s, we get

x(r, s) = bs+ c1(r), t(r, s) = s+ c2(r)

and z(r, s) = c3(r) with initial conditions

x(r, 0) = c1(r) = r

t(r, 0) = c2(r) = 0, and z(r, 0) = c3(r) = g(r).

Therefore,

x(r, s) = bs+ r, t(r, s) = s, and z(r, s) = g(r).

The idea is to solve for r and s in terms of x, t. Let us set u(x, t) =z(r(x, t), s(x, t)). In this case we can solve for r and s in terms of x andt, to get

r(x, t) = x− bt and s(x, t) = t.

Therefore, u(x, t) = z(r, s) = g(r) = g(x− bt).The above example shows how the information on data curve Γ is reduced

as initial condition for the characteristic ODEs.

Lecture 4

Method of Characteristics:Continued...

Let us study the Example 3.5 with a different data curve Γ.

Example 4.1. Consider the linear transport equationut + bux = 0 x ∈ R and t ∈ (0,∞)u(bt, t) = g(t) t ∈ (0,∞).

(4.0.1)

We parametrize the data curve Γ with r-variable, i.e., Γ = γ1(r), γ2(r) =(br, r). The characteristic equations are:

dx(r, s)

ds= b,

dt(r, s)

ds= 1, and

dz(r, s)

ds= 0


x(r, 0) = br, t(r, 0) = r, and z(r, 0) = g(r).


x(r, s) = bs+ c1(r), t(r, s) = s+ c2(r)

and z(r, s) = c3(r) with initial conditions

x(r, 0) = c1(r) = br

t(r, 0) = c2(r) = r, and z(r, 0) = c3(r) = g(r).

17

LECTURE 4. METHOD OF CHARACTERISTICS: CONTINUED... 18

Therefore,

x(r, s) = b(s+ r), t(r, s) = s+ r, and z(r, s) = g(r).

Note that in this case we cannot solve for r and s in terms of x, t.

The above examples suggest that solving a Cauchy problem dependson the curve on which u is prescribed. One way is to prescribe values ofu on a data curve Γ parametrized as (γ1(r), γ2(r)), for r ∈ I ⊆ R, suchthat all the characteristic curves (parametrised with s variable) intersect the(γ1(r), γ2(r), u(γ1, γ2)) at s = 0. Let u(γ1(r), γ2(r)) = g(r). Thus, solving theCauchy problem is now reduced to solving for x(r, s), y(r, s), z(r, s) in (3.2.2)with the initial conditions x(r, 0) = γ1(r), y(r, 0) = γ2(r) and z(r, 0) = g(r),r ∈ I. The system of ODE can be solved uniquely for x(r, s), y(r, s) andz(r, s), in a neighbourhood of s = 0 and all r ∈ I. At this juncture, one mayask the following questions:

(i) Can the three solutions be used to define a function z = u(x, y)?

(ii) If yes to (i), then is the solution z = u(x, y) unique for the Cauchyproblem? The answer is yes because two integral surface intersectingin Γ must contain the same charaterictics (beyond the scope of thiscourse).

Let us answer (i) in a neighbourhood of (r, 0). Set x(r, 0) = γ1(r) = x0,y(r, 0) = γ2(r) = y0 and z(r, 0) = g(r) = z0. Note that the answer to(i) is in affirmation if we can solve for r and s in terms of x and y, i.e.,r = R(x, y) and s = S(x, y) such that R(x0, y0) = 0 and S(x0, y0) = 0. Thenz = z(R(x, y), S(x, y)) = u(x, y). The inverse function theorem tells us thatr and s can be solved in terms of x and y in a neighbourhood of (x0, y0) if

J(r, 0) =

∣∣∣∣ xr(r, 0) yr(r, 0)xs(r, 0) ys(r, 0)

∣∣∣∣ =

∣∣∣∣ γ′1(r) γ′2(r)A(x0, y0, z0) B(x0, y0, z0)

∣∣∣∣ 6= 0.

The quantity J(r, 0) 6= 0 means that the vectors (A(x0, y0, z0), B(x0, y0, z0))and (γ′1(r), γ′2(r)) are not parallel.

What happens in the case of J(r, 0) = 0, i.e., when the associated vectorsare parallel? The condition given on Γ is u(γ1(r), γ2(r)) = g(r). If u is a C1

solution to the Cauchy problem then on differentiation, w.r.t r, the Cauchy


condition yields g′(r) = ux(γ1(r), γ2(r))γ′1(r) + uy(γ1(r), γ2(r))γ′2(r). Since uis solution, at (x0, y0, z0), of the algebraic system(

A(x0, y0, z0) B(x0, y0, z0)γ′1(r) γ′2(r)

)(ux(x0, y0)uy(x0, y0)

)=

(C(x0, y0, z0)

g′(r)

)a necessary condition is that

rank

(A(x0, y0, z0) B(x0, y0, z0) C(x0, y0, z0)

γ′1(r) γ′2(r) g′(r)

)= 1.

If the above rank condition is satisfied then the data curve Γ is said to be acharacteristic at (x0, y0, z0). Thus, we have the following possibilities:

(a) J(r, 0) 6= 0 for all r ∈ I. Note that, when J(r, 0) 6= 0, then the rankcondition is not satisfied1 and, hence, the data curve Γ does not haveany characteristic points (Γ is not parallel at all points). Then, in aneighborhood of Γ, there exists a unique solution u = u(x, y) of theCauchy problem given by the system of ODEs.

(b) J(r0, 0) = 0, for some r0 ∈ I, and Γ is characteristic at the point P0 =(γ1(r0), γ2(r0), g(r0)). Then a C1 solution may exist in a neighborhoodof P0.

(c) J(r0, 0) = 0 for some r0 ∈ I and Γ is not characteristic at P0. There areno C1 solutions in a neighborhood of P0. There may exist less regularsolutions.

(d) If Γ is a characteristic then there exists infinitely many C1 solutions ina neighborhood of Γ.

Example 4.2. Consider the Burgers’ equation given asut + uux = 0 x ∈ R and t ∈ (0,∞)u(x, 0) = g(x) x ∈ R.

The parametrization of the curve Γ with r-variable, i.e., Γ = γ1(r), γ2(r) =(r, 0). The characteristic equations are:

dx(r, s)

ds= z,

dt(r, s)

ds= 1, and

dz(r, s)

ds= 0

1because rank is 2 in this case



x(r, 0) = r, t(r, 0) = 0, and z(r, 0) = g(r).

Solving the ODE corresponding to z, we get z(r, s) = c3(r) with initialconditions z(r, 0) = c3(r) = g(r). Thus, z(r, s) = g(r). Using this in theODE of x, we get

dx(r, s)

ds= g(r).


x(r, s) = g(r)s+ c1(r), t(r, s) = s+ c2(r)

with initial conditions

x(r, 0) = c1(r) = r and t(r, 0) = c2(r) = 0.

Therefore,

x(r, s) = g(r)s+ r, t(r, s) = s and z(r, s) = g(r).

Let us compute J(r, s) = 1 + sg′(r)− 0 · 1 = 1 + sg′(r) and J(r, 0) = 1 6= 0,the data curve Γ does not have characteristic points. Hence, one can solvefor r and s, in terms of x, t and z. Thus, we get s = t and r = x − zt.Therefore, u(x, t) = g(x− tu) is the solution in the implicit form.

Example 4.3. Consider the Burgers’ equation given asut + uux = 1 x ∈ R and t ∈ (0,∞)

u(t2

4, t)

= t2

t > 0.

Note that the data curve is the parabola x = t2/4. The parametrizationof the curve Γ with r-variable, i.e., Γ = γ1(r), γ2(r) = (r2, 2r). Thecharacteristic equations are:

dx(r, s)

ds= z,

dt(r, s)

ds= 1, and

dz(r, s)

ds= 1


x(r, 0) = r2, t(r, 0) = 2r, and z(r, 0) = r.


Solving the ODE corresponding to z, we get z(r, s) = s + c3(r) with initialconditions z(r, 0) = c3(r) = r. Thus, z(r, s) = s + r. Using this in the ODEof x, we get

dx(r, s)

ds= s+ r.


x(r, s) =s2

2+ rs+ c1(r), t(r, s) = s+ c2(r)


x(r, 0) = c1(r) = r2 and t(r, 0) = c2(r) = 2r.

Therefore,

x(r, s) =s2

2+ rs+ r2, t(r, s) = s+ 2r and z(r, s) = s+ r.

Let us compute J(r, s) = 2r + s − 2(s + r) = −s and J(r, 0) = 0 for all r.The rank of the matrix, at (r, 0),(

r 1 12r 2 1

)= 2 6= 1

and, hence, Γ does not have any characteristic points. We know in this casewe cannot have a C1 solution but might have less regular solutions. Let us

solve for r and s, in terms of x, t and z. Thus, we get s = ∓2√x− t2

4and

r = t2±√x− t2

4. Therefore, u(x, t) = t

2±√x− t2

4are two solutions in the

region x > t2/4 and are not differentiable on Γ.

Example 4.4. Consider the Burgers’ equation given asut + uux = 1 x ∈ R and t ∈ (0,∞)

u(t2

2, t)

= t t > 0.

Note that the data curve is the parabola x = t2/2. The parametrizationof the curve Γ with r-variable, i.e., Γ = γ1(r), γ2(r) = (r2/2, r). Thecharacteristic equations are:

dx(r, s)

ds= z,

dt(r, s)

ds= 1, and

dz(r, s)

ds= 1



x(r, 0) =r2

2, t(r, 0) = r, and z(r, 0) = r.

Solving the ODE corresponding to z, we get z(r, s) = s + c3(r) with initialconditions z(r, 0) = c3(r) = r. Thus, z(r, s) = s + r. Using this in the ODEof x, we get

dx(r, s)

ds= s+ r.


x(r, s) =s2

2+ rs+ c1(r), t(r, s) = s+ c2(r)


x(r, 0) = c1(r) =r2

2and t(r, 0) = c2(r) = r.

Therefore,

x(r, s) =s2 + r2

2+ rs, t(r, s) = s+ r and z(r, s) = s+ r.

Let us compute J(r, s) = r + s− (s+ r) = 0 for all r and s. Further, Γ is acharacteristic (at all points) because the rank of the matrix, at (r, 0),(

r 1 1r 1 1

)= 1

In this case there may exist infinitely many C1 solutions. For instance,u(x, t) = t is a solution satisfying Cauchy data. Also, u(x, t) = ±

√2x.

Remark 4.0.1. If the coefficients a and b are independent of u, then thecharacteristic curves are lying in the xy-plane. If the coefficients a and b areconstants (independent of both x and u) then the characteristic curves arestraight lines. In the linear case, the characteristics curves will not intersect.Because, if the curves intersect then, at the point of intersection, they havethe same tangent, which is not possible!

Lecture 5

Classification of Second OrderPDE

A general second order PDE is of the form F (D2u(x), Du(x), u(x), x) = 0,for each x ∈ Ω ⊂ Rn and u : Ω → R is the unknown. A Cauchy problem is:given the knowledge of u on a smooth hypersurface Γ ⊂ Ω, can one find thesolution u of the PDE? The knowledge of u on Γ is said to be the Cauchydata.

What should be the minimum required Cauchy data for the Cauchy prob-lem to be solved? Viewing the Cauchy problem as an initial value problemcorresponding to ODE, there is a unique solution to the second order ODE

y′′(x) + P (x)y′(x) +Q(x)y(x) = 0 x ∈ Iy(x0) = y0

y′(x0) = y′0.

where P and Q are continuous on I (assume I closed interval of R) and forany point x0 ∈ I. This motivates us to define the Cauchy problem for secondorder PDE as:

F (D2u(x), Du(x), u(x), x) = 0 x ∈ Ωu(x) = g(x) x ∈ Γ

Du(x) · ν(x) = h(x) x ∈ Γ(5.0.1)

where ν is the outward unit normal vector on the hypersurface Γ and g, hare known functions on Γ.

23

LECTURE 5. CLASSIFICATION OF SECOND ORDER PDE 24

5.1 Semilinear

Consider the Cauchy problem for the second order semilinear PDE in twovariables (x, y) ∈ Ω ⊂ R2,

A(x, y)uxx + 2B(x, y)uxy + C(x, y)uyy = D (x, y) ∈ Ωu(x, y) = g(x, y) (x, y) ∈ Γux(x, y) = h1(x, y) (x, y) ∈ Γuy(x, y) = h2(x, y) (x, y) ∈ Γ.

(5.1.1)

where D(x, y, u, ux, uy) may be non-linear and Γ is a smooth1 curve in Ω.Also, one of the coefficients A,B or C is identically non-zero (else the PDEis not of second order). Let r 7→ (γ1(r), γ2(r)) be a parametrisation of thecurve Γ. Then we have the compatibility condition that

g′(r) = h1γ′1(r) + h2γ

′2(r).

By computing the second derivatives of u on Γ and considering uxx, uyyand uxy as unknowns, we have the linear system of three equations in threeunknowns on Γ,

Auxx +2Buxy +Cuyy = Dγ′1(r)uxx +γ′2(r)uxy = h′1(r)

γ′1(r)uxy +γ′2(r)uyy = h′2(r).

This system of equation is solvable if the determinant of the coefficients arenon-zero, i.e., ∣∣∣∣∣∣

A 2B Cγ′1 γ′2 00 γ′1 γ′2

∣∣∣∣∣∣ 6= 0.

Definition 5.1.1. We say a curve Γ ⊂ Ω ⊂ R2 is characteristic (w.r.t(5.1.1)) if

A(γ′2)2 − 2Bγ′1γ′2 + C(γ′1)2 = 0.

where (γ1(r), γ2(r)) is a parametrisation of Γ.

Note that the geometry hidden in the above definition is very similarto that we encountered in first order equation. Since ν = (−γ′2, γ′1) is the

1twice differentiable


normal to Γ at each point, the above definition says that the curve is non-characteristic if

2∑i,j=1

Aijνiνj = A(γ′2)2 − 2Bγ′1γ′2 + C(γ′1)2 6= 0

where A11 = A, A12 = A21 = B and A22 = C. If y = y(x) is a representationof the curve Γ (locally, if necessary), we have γ1(r) = r and γ2(r) = y(r).Then the characteristic equation reduces as

A

(dy

dx

)2

− 2Bdy

dx+ C = 0.

Therefore, the characteristic curves of (5.1.1) are given by the graphs whoseequation is

dy

dx=B ±

√B2 − ACA

.

Thus, we have three situations arising depending on the sign of the dis-criminant, B2 − AC. This classifies the given second order PDE based onthe sign of its discriminant d = B2 − AC.

Definition 5.1.2. We say a second order PDE is of

(a) hyperbolic type if d > 0,

(b) parabolic type if d = 0 and

(c) elliptic type if d < 0.

The hyperbolic PDE have two families of characteristics, parabolic PDEhas one family of characteristic and elliptic PDE have no characteristic. Wecaution here that these names are no indication of the shape of the graph ofthe solution of the PDE.

Note that the classification depends on the determinant of the coefficientmatrix (

A BB C

)For every (x, y) ∈ Ω, the matrix is symmetric and hence diagonalisable.If λ1, λ2 are the diagonal entries, then d = −λ1λ2. Thus, a equation ishyperbolic at a point (x, y) if the eigen values have opposite sign. It is ellipicif the eigenvalues have same sign and is parabolic if, at least, one of theeigenvalue is zero.


Example 5.1 (Wave Equation). For a given non-zero c ∈ R, uyy−c2uxx = 0 ishyperbolic. Since A = −c2, B = 0 and C = 1, we have d = B2−AC = c2 > 0.The eigen values of the coefficient matrix are 1,−c2 which have opposite sign.

Example 5.2 (Heat Equation). For a given c ∈ R, uy − cuxx = 0 is parabolic.Since A = −c, B = 0 and C = 0, thus d = B2 − AC = 0. The eigen valuesof the coefficient matrix are 0,−c has a zero eigenvalue.

Example 5.3 (Laplace equation). uxx+uyy = 0 is elliptic. Since A = 1, B = 0and C = 1, thus d = B2 −AC = −1 < 0. The eigen values of the coefficientmatrix are 1, 1 which have same sign.

Example 5.4 (Velocity Potential Equation). In the equation (1 −M2)uxx +uyy = 0, A = (1−M2), B = 0 and C = 1. Then d = B2−AC = −(1−M2).The eigen values of the coefficient matrix are 1 −M2, 1. Thus, for M > 1(opposite sign), the equation is hyperbolic (supersonic flow), for M = 1 (zeroeigenvalue) it is parabolic (sonic flow) and for M < 1 (same sign) it is elliptic(subsonic flow).

Note that the classification of PDE is dependent on its coefficients. Thus,for constant coefficients the type of PDE remains unchanged throughoutthe region Ω. However, for variable coefficients, the PDE may change itsclassification from region to region.

Example 5.5. An example is the Tricomi equation , uxx + xuyy = 0. Thediscriminant of the Tricomi equation is d = −x. The eigenvalues are 1, x.Thus, tricomi equation is hyperbolic when x < 0, elliptic when x > 0 anddegenerately parabolic when x = 0, i.e., y-axis. Such equations are calledmixed type.

Example 5.6. Let us compute the family of characteristic curves of secondorder PDE, whenever they exist. For instance, recall that elliptic equationwill not have any real characteristic curves.

(i) For a given non-zero c ∈ R, uyy−c2uxx = 0. We have already noted thatthe equation is hyperbolic and, hence, should admit two characteristiccurves. Recall that the characteristic curves are given by the equation

dy

dx=B ±

√B2 − ACA

=±√c2

−c2=∓1

c.

Thus, cy ± x = a constant is the equation for the two characteristiccurves. Note that the characteristic curves y = ∓x/c+y0 are boundaryof two cones in R2 with vertex at (0, y0).


(ii) For any given c ∈ R, consider uy−cuxx = 0. We have already noted thatthe equation is parabolic and, hence, should admit one characteristiccurve. The characteristic curve is given by the equation

dy

dx=B ±

√B2 − ACA

= 0.

Thus, y = a constant is the equation of the characteristic curve. i.e.,any horizontal line in R2 is a charateristic curve.

(iii) We have already noted that the equation uxx + uyy = 0 is elliptic and,hence, will have no real characteristics.

(iv) The equation uxx + xuyy = 0 is of mixed type. In the region x > 0, thecharacteristic curves are y ∓ 2x3/2/3 = a constant.

Lecture 6

Classification of SOPDE:Continued

6.1 Quasilinear

The notion of classification of second order semilinear PDE could be gen-eralised to quasilinear PDE A(x, u(x), Du(x)), non-linear PDE and systemof ODE. However, in these cases the classification may also depend on thesolution u. The solutions to characteristic equation for a quasilinear equationdepends on the solution considered.

Example 6.1. Consider the quasilinear PDE uxx−uuyy = 0. The discriminantis d = u. The eigenvalues are 1,−u(x). It is hyperbolic for u > 01, ellipticwhen u < 0 and parabolic when u = 0.Example 6.2. Consider the quasilinear PDE

(c2 − u2x)uxx − 2uxuyuxy + (c2 − u2

y)uyy = 0

where c > 0. Then d = B2 − AC = c2(u2x + u2

y − c2) = c2(|∇u|2 − c2). It ishyperbolic if |∇u| > c, parabolic if |∇u| = c and elliptic if |∇u| < c.

6.2 Why Characteristic Curves?

Recall that a second order ODE

y′′(x) + P (x)y′(x) +Q(x)y(x) = 0, x ∈ I1The notation u > 0 means x ∈ Ω | u(x) > 0

29

LECTURE 6. CLASSIFICATION OF SOPDE: CONTINUED 30

can have other types of boundary conditions, in addition to the initial (orCauchy) condition

y(x0) = y0 and y′(x0) = y′0,

such as, if I = (a, b) then

(a) Dirichlet condition

y(a) = y0 and y(b) = y1.

(b) Neumann condition

y′(a) = y0 and y′(b) = y1.

(c) Periodic condition

y(a) = y(b) and y′(a) = y′(b).

(d) Robin condition

αy(a) + βy′(a) = y0 and γy(b) + δy′(b) = y1.

In contrast to Initial (Cauchy boundary) condition, boundary value problemsmay be ill-posed. These boundary conditions also can be generalised tosecond order PDE. The classification described tells us the right amountof initial/boundary condition to be imposed for a second order PDE to bewell-posed.

For hyperbolic, which has two real characteristics, requires as many ini-tial condition as the number of characteristics emanating from initial timeand as many boundary conditions as the number of characteristics that passinto the spatial boundary. Thus, hyperbolic equations will take the Cauchyboundary conditions on an open surface. For parabolic, which has exactlyone real characteristic, we need one boundary condition at each point of thespatial boundary and one initial condition at initial time. Thus, parabolicequation will take either Dirichlet or Neumann on an open surface. For ellip-tic, which has no real characteristic curves, we need one boundary conditionat each point of the spatial boundary. Thus, elliptic equations will take eitherDirichlet or Neumann in a closed surface enclosing the domain of interest.


6.3 Cauchy Boundary Condition

Recall that for a second order Cauchy problem we need to know both uand its normal derivative on a data curve Γ contained in Ω. However, theCauchy problem for Laplacian (more generally for elliptic equations) is notwell-posed. In fact, the Cauchy problem for Laplace equation on a boundeddomain Ω is over-determined.

Example 6.3 (Hadamard). Consider the Cauchy problem for Laplace equa-tion

uxx + uyy = 0

u(0, y) = cos kyk2

ux(0, y) = 0,

where k > 0 is an integer. It is easy to verify that there is a unique solution

uk(x, y) =cosh(kx) cos(ky)

k2

of the Cauchy problem. Note that for any x0 > 0,

|uk(x0, nπ/k)| = cosh(kx0)

k2.

Since, as k →∞, nπ/k → 0 and |uk(x0, nπ/k)| → ∞ the Cauchy problem isnot stable, and hence not well-posed.

Exercise 1. Show that the Cauchy problem for Laplace equationuxx + uyy = 0u(x, 0) = 0uy(x, 0) = k−1 sin kx,

where k > 0, is not well-posed. (Hint: Compute explicit solution using sep-aration of variable. Note that, as k → ∞, the Cauchy data tends to zerouniformly, but the solution does not converge to zero for any y 6= 0. There-fore, a small change from zero Cauchy data (with corresponding solutionbeing zero) may induce bigger change in the solution.)

This issue of ill-posedness of the Cauchy problem is very special to secondorder elliptic equations. In general, any hyperbolic equation Cauchy problemis well-posed, as long as the hyperbolicity is valid in the full neighbourhoodof the data curve.


Example 6.4. Consider the Cauchy problem for the second order hyperbolicequation

y2uxx − yuyy + 12uy = 0 y > 0

u(x, 0) = f(x)uy(x, 0) = g(x).

The general solution to this problem can be computed as

u(x, y) = F

(x+

2

3y3/2

)+G

(x− 2

3y3/2

).

On y = 0 u(x, 0) = F (x) +G(x) = f(x). Further,

uy(x, y) = y1/2F ′(x+

2

3y3/2

)− y1/2G′

(x− 2

3y3/2

)and uy(x, 0) = 0. Thus, the Cauchy problem has no solution unless g(x) = 0.If g ≡ 0 then the solution is

u(x, y) = F

(x+

2

3y3/2

)− F

(x− 2

3y3/2

)+ f

(x− 2

3y3/2

)for arbitrary F ∈ C2. Therefore, when g ≡ 0 the solution is not unique.The Cauchy problem is not well-posed because the equation is hyperbolic(B2 − AC = y3) not in the full neighbourhood of the data curve y = 0.

Lecture 7

Classification of SOPDE:Continued

7.1 Invariance of Discriminant

The classification of second order semilinear PDE is based on the discriminantB2−AC. In this section, we note that the classification is independent of thechoice of coordinate system (to represent a PDE). Consider the two-variablesemilinear PDE

A(x, y)uxx+2B(x, y)uxy+C(x, y)uyy = D(x, y, u, ux, uy) (x, y) ∈ Ω (7.1.1)

where the variables (x, y, u, ux, uy) may appear non-linearly in D and Ω ⊂ R2.Also, one of the coefficients A,B or C is identically non-zero (else the PDE isnot of second order). We shall observe how (7.1.1) changes under coordinatetransformation.

Definition 7.1.1. For any PDE of the form (7.1.1) we define its discrimi-nant as B2 − AC.

Let T : R2 → R2 be the coordinate transformation with componentsT = (w, z), where w, z : R2 → R. We assume that w(x, y), z(x, y) are suchthat w, z are both continuous and twice differentiable w.r.t (x, y), and theJacobian J of T is non-zero,

J =

∣∣∣∣ wx wyzx zy

∣∣∣∣ 6= 0.

33


We compute the derivatives of u in the new variable,

ux = uwwx + uzzx,

uy = uwwy + uzzy,

uxx = uwww2x + 2uwzwxzx + uzzz

2x + uwwxx + uzzxx

uyy = uwww2y + 2uwzwyzy + uzzz

2y + uwwyy + uzzyy

uxy = uwwwxwy + uwz(wxzy + wyzx) + uzzzxzy + uwwxy + uzzxy

Substituting above equations in (7.1.1), we get

a(w, z)uww + 2b(w, z)uwz + c(w, z)uzz = d(w, z, u, uw, uz).

where D transforms in to d and

a(w, z) = Aw2x + 2Bwxwy + Cw2

y (7.1.2)

b(w, z) = Awxzx +B(wxzy + wyzx) + Cwyzy (7.1.3)

c(w, z) = Az2x + 2Bzxzy + Cz2

y . (7.1.4)

Note that the coefficients in the new coordinate system satisfy

b2 − ac = (B2 − AC)J2.

Since J 6= 0, we have J2 > 0. Thus, both b2 − ac and B2 − AC have thesame sign. Thus, the sign of the discriminant is invariant under coordinatetransformation. All the above arguments can be carried over to quasilinearand non-linear PDE.

7.2 Standard or Canonical Forms

The advantage of above classification helps us in reducing a given PDE intosimple forms. Given a PDE, one can compute the sign of the discriminantand depending on its clasification we can choose a coordinate transformation(w, z) such that

(i) For hyperbolic, a = c = 0 or b = 0 and a = −c.

(ii) For parabolic, c = b = 0 or a = b = 0. We conveniently choosec = b = 0 situation so that a 6= 0 (so that division by zero is avoided inthe equation for characteristic curves).


(iii) For elliptic, b = 0 and a = c.

If the given second order PDE (7.1.1) is such that A = C = 0, then(7.1.1) is of hyperbolic type and a division by 2B (since B 6= 0) gives

uxy = D(x, y, u, ux, uy)

where D = D/2B. The above form is the first standard form of second orderhyperbolic equation. If we introduce the linear change of variable X = x+ yand Y = x − y in the first standard form, we get the second standard formof hyperbolic PDE

uXX − uY Y = D(X, Y, u, uX , uY ).

If the given second order PDE (7.1.1) is such that A = B = 0, then(7.1.1) is of parabolic type and a division by C (since C 6= 0) gives

uyy = D(x, y, u, ux, uy)

where D = D/C. The above form is the standard form of second orderparabolic equation.

If the given second order PDE (7.1.1) is such that A = C and B = 0,then (7.1.1) is of elliptic type and a division by A (since A 6= 0) gives

uxx + uyy = D(x, y, u, ux, uy)

where D = D/A. The above form is the standard form of second orderelliptic equation.

Note that the standard forms of the PDE is an expression with no mixedderivatives.

7.3 Reduction to Standard Form

Consider the second order semilinear PDE (7.1.1) not in standard form. Welook for transformation w = w(x, y) and z = z(x, y), with non-vanishingJacobian, such that the reduced form is the standard form.

If B2 − AC > 0, we have two characteristics. We are looking for thecoordinate system w and z such that a = c = 0. This implies from equation(7.1.2) and (7.1.4) that we need to find w and z such that

wxwy

=−B ±

√B2 − ACA

=zxzy.


Therefore, we need to find w and z such that along the slopes of the charac-teristic curves,

dy

dx=B ±

√B2 − ACA

=−wxwy

.

This means that, using the parametrisation of the characteristic curves,wxγ

′1(r) + wyγ

′2(r) = 0 and w′(r) = 0. Similarly for z. Thus, w and z

are chosen such that they are constant on the characteristic curves.The characteristic curves are found by solving

dy

dx=B ±

√B2 − ACA

and the coordinates are then chosen such that along the characteristic curvew(x, y) = a constant and z(x, y) = a constant. Note that wxzy − wyzx =wyzy

(2A

√B2 − AC

)6= 0.

Example 7.1. For a non-zero constant c ∈ R, let us reduce to canonicalform the PDE uxx − c2uyy = 0. Note that A = 1, B = 0, C = −c2 andB2 − AC = c2 and the equation is hyperbolic. The characteristic curves aregiven by the equation

dy

dx=B ±

√B2 − ACA

= ±c.

Solving we get y ∓ cx = a constant. Thus, w = y + cx and z = y − cx. Nowwriting


2x + uwwxx + uzzxx

= c2(uww − 2uwz + uzz)


2y + uwwyy + uzzyy

= uww + 2uwz + uzz

−c2uyy = −c2(uww + 2uwz + uzz)

Substituting into the given PDE, we get

0 = 4c2uwz

= uwz.

Example 7.2. Let us reduce the PDE uxx − x2yuyy = 0 given in the region(x, y) | x ∈ R, x 6= 0, y > 0 to its canonical form. Note that A = 1, B = 0,


C = −x2y and B2 − AC = x2y. In the given region x2y > 0, hence theequation is hyperbolic. The characteristic curves are given by the equation

dy

dx=B ±

√B2 − ACA

= ±x√y.

Solving we get x2/2 ∓ 2√y = a constant. Thus, w = x2/2 + 2

√y and

z = x2/2− 2√y. Now writing

ux = uwwx + uzzx = x(uw + uz)

uy = uwwy + uzzy =1√y

(uw − uz)


2x + uwwxx + uzzxx

= x2(uww + 2uwz + uzz) + uw + uz


2y + uwwyy + uzzyy

=1

y(uww − 2uwz + uzz)−

1

2y√y

(uw − uz)

−x2yuyy = −x2(uww − 2uwz + uzz) +x2

2√y

(uw − uz)


0 = 4x2uwz +2√y + x2

2√y

uw +2√y − x2

2√y

uz

= 8x2√yuwz + (x2 + 2√y)uw + (2

√y − x2)uz.

Note that w + z = x2 and w − z = 4√y. Now, substituting x, y in terms of

w, z, we get

0 = 2(w2 − z2)uwz +

(w + z +

w − z2

)uw +

(w − z

2− w − z

)uz

= uwz +

(3w + z

4(w2 − z2)

)uw −

(w + 3z

4(w2 − z2)

)uz.

In the parabolic case, B2 − AC = 0, we have a single characteristic. Weare looking for a coordinate system such that b = c = 0.


Example 7.3. Let us reduce the PDE e2xuxx + 2ex+yuxy + e2yuyy = 0 to itscanonical form. Note that A = e2x, B = ex+y, C = e2y and B2 − AC = 0.The PDE is parabolic. The characteristic curves are given by the equation

dy

dx=B

A=ey

ex.

Solving, we get e−y − e−x = a constant. Thus, w = e−y − e−x. Now, wechoose z such that wxzy − wyzx 6= 0. For instance, z = x is one such choice.Then

ux = e−xuw + uz

uy = −e−yuwuxx = e−2xuww + 2e−xuwz + uzz − e−xuwuyy = e−2yuww + e−yuw

uxy = −e−y(e−xuww − uwz)


exe−yuzz = (e−y − e−x)uw

Replacing x, y in terms of w, z gives

uzz =w

1 + wezuw.

In the elliptic case, B2 − AC < 0, we have no real characteristics. Thus,we choose w, z to be the real and imaginary part of the solution of thecharacteristic equation.

Example 7.4. Let us reduce the PDE x2uxx + y2uyy = 0 given in the region(x, y) ∈ R2 | x > 0, y > 0 to its canonical form. Note that A = x2,B = 0, C = y2 and B2 −AC = −x2y2 < 0. The PDE is elliptic. Solving thecharacteristic equation

dy

dx= ±iy

xwe get ln x± i ln y = c. Let w = lnx and z = ln y. Then

ux = uw/x

uy = uz/y

uxx = −uw/x2 + uww/x2

uyy = −uz/y2 + uzz/y2


Substituting into the PDE, we get

uww + uzz = uw + uz.

Example 7.5. Let us reduce the PDE uxx+2uxy+5uyy = xux to its canonicalform. Note that A = 1, B = 1, C = 5 and B2 − AC = −4 < 0. The PDE iselliptic. The characteristic equation is

dy

dx= 1± 2i.

Solving we get x− y ± i2x = c. Let w = x− y and z = 2x. Then

ux = uw + 2uz

uy = −uwuxx = uww + 4uwz + 4uzz

uyy = uww

uxy = −(uww + 2uwz)

Substituting into the PDE, we get

uww + uzz = x(uw + 2uz)/4.

Replacing x, y in terms of w, z gives

uww + uzz =z

8(uw + 2uz).

Lecture 8

The Laplacian

We introduced in Lecture 1 the Laplacian to be the trace of the Hessainmatrix, ∆ :=

∑ni=1

∂2

∂x2i. The Laplace operator usually appears in physical

models associated with dissipative effects (except wave equation). The im-portance of Laplace operator can be realised by its appearance in variousphysical models. For instance, in the

(a) heat equation ∂∂t−∆,

(b) the wave equation ∂2

∂t2−∆,

(c) and the Schrodinger’s equation i ∂∂t

+∆.

8.1 Properties of Laplacian

In cartesian coordiantes, the n-dimensional Laplacian is ∆ :=∑n

i=1∂2

∂x2i. Note

that in one dimension, i.e. n = 1, ∆ = d2

dx2. In two dimension polar coordi-

nates, the Laplacian is given as

∆ :=1

r

∂

∂r

(r∂

∂r

)+

1

r2

∂2

∂θ2,

where r is the magnitude component (0 ≤ r < ∞) and θ is the directioncomponent (0 ≤ θ < 2π). The direction component is also called the azimuthangle or polar angle. This is easily seen by using the relation x = r cos θ and

41

LECTURE 8. THE LAPLACIAN 42

y = r sin θ. Then ∂x∂r

= cos θ, ∂y∂r

= sin θ and ∂u∂r

= cos θ ∂u∂x

+ sin θ ∂u∂y

. Also,

∂2u

∂r2= cos2 θ

∂2u

∂x2+ sin2 θ

∂2u

∂y2+ 2 cos θ sin θ

∂2u

∂x∂y.

Similarly, ∂x∂θ

= −r sin θ, ∂y∂θ

= r cos θ, ∂u∂θ

= r cos θ ∂u∂y− r sin θ ∂u

∂xand

1

r2

∂2u

∂θ2= sin2 θ

∂2u

∂x2+ cos2 θ

∂2u

∂y2− 2 cos θ sin θ

∂2u

∂x∂y− 1

r

∂u

∂r.

Therefore, ∂2u∂r2

+ 1r2∂2u∂θ2

= ∂2u∂x2

+ ∂2u∂y2− 1

r∂u∂r

and, hence,

∆u =∂2u

∂r2+

1

r2

∂2u

∂θ2+

1

r

∂u

∂r.

Further, in three dimension cylindrical coordinates, the Laplacian is given as

∆ :=1

r

∂

∂r

(r∂

∂r

)+

1

r2

∂2

∂θ2+

∂2

∂z2

where r ∈ [0,∞), θ ∈ [0, 2π) and z ∈ R. In three dimension sphericalcoordinates, the Laplacian is given as

∆ :=1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sinφ

∂

∂φ

(sinφ

∂

∂φ

)+

1

r2 sin2 φ

∂2

∂θ2

where r ∈ [0,∞), φ ∈ [0, π] (zenith angle or inclination) and θ ∈ [0, 2π)(azimuth angle).

Theorem 8.1.1. Let n ≥ 2 and u be a radial function, i.e., u(x) = v(r)where x ∈ Rn and r = |x|, then

∆u(x) =d2v(r)

dr2+

(n− 1)

r

dv(r)

dr.

Proof. Note that

∂r

∂xi=∂|x|∂xi

=∂(√x2

1 + . . .+ x2n)

∂xi=

1

2(x2

1 + . . .+ x2n)−1/2(2xi) =

xir.


Thus,

∆u(x) =n∑i=1

∂

∂xi

(∂u(x)

∂xi

)=

n∑i=1

∂

∂xi

(dv(r)

dr

xir

)=

n∑i=1

xi∂

∂xi

(1

r

dv(r)

dr

)+n

r

dv(r)

dr

=n∑i=1

x2i

r

d

dr

(dv(r)

dr

1

r

)+n

r

dv(r)

dr

=n∑i=1

x2i

r

1

r

d2v(r)

dr2− 1

r2

dv(r)

dr

+n

r

dv(r)

dr

=r2

r

1

r

d2v(r)

dr2− 1

r2

dv(r)

dr

+n

r

dv(r)

dr

=d2v(r)

dr2− 1

r

dv(r)

dr+n

r

dv(r)

dr

=d2v(r)

dr2+

(n− 1)

r

dv(r)

dr.

Hence the result proved.

More generally, the Laplacian in Rn may be written in polar coordinatesas

∆ :=∂2

∂r2+n− 1

r

∂

∂r+

1

r2∆Sn−1

where ∆Sn−1 is a second order differential operator in angular variables only.The angular part of Laplacian is called the Laplace-Beltrami operator actingon Sn−1 (unit sphere of Rn) with Riemannian metric induced by the standardEuclidean metric in Rn.

8.2 Boundary Conditions

Let Ω be a bounded open subset of Rn with boundary denoted as ∂Ω. Tomake the over-determined Cauchy problem of an elliptic equation well-posed,it is reasonable to specify one of the following conditions on the boundary∂Ω:

(i) (Dirichlet condition) u = g;


(ii) (Neumann condition) ∇u·ν = g, where ν(x) is the unit outward normalof x ∈ ∂Ω;

(iii) (Robin condition) ∇u · ν + cu = g for any c > 0.

(iv) (Mixed condition) u = g on Γ1 and∇u·ν = h on Γ2, where Γ1∪Γ2 = ∂Ωand Γ1 ∩ Γ2 = ∅.

The elliptic equation with Neumann boundary condition naturally im-poses a compatibility condition. By Gauss divergence theorem (cf. Corol-lary A.0.5), if u is a solution of the Neumann problem then u satisfies, forevery connected component ω of Ω,∫

ω

∆u =

∫∂ω

∇u · ν (Using GDT)

−∫ω

f =

∫∂ω

g.

The second equality is called the compatibility condition. Thus, for an inho-mogeneous Laplace equation with Neumann boundary condition, the givendata f, g must necessarily satisfy the compatibility condition. Otherwise, theNeumann problem does not make any sense.

The aim of this chapter is to solve, for any open bounded subset Ω ⊂ Rn,−∆u(x) = f(x) in Ωone of the above inhomogeneous boudary condition on ∂Ω.

By the linearity of Laplacian, u = v + w where v is a solution of∆v(x) = 0 in Ωone of the above inhomogeneous boudary condition on ∂Ω,

and w is a solution of−∆w(x) = f(x) in Ωone of the above homogeneous boudary condition on ∂Ω.

Therefore, we shall solve for u by solving for v and w separately.


8.3 Harmonic Functions

The one dimensional Laplace equation is an ODE ( d2

dx2) and is solvable with

solutions u(x) = ax+b for some constants a and b. But in higher dimensionssolving Laplace equation is not so simple. For instance, a two dimensionalLaplace equation

uxx + uyy = 0

has the trivial solution, u(x, y) = ax + by + c, all one degree polynomials oftwo variables. In addition, xy, x2 − y2, x3 − 3xy2, 3x2y − y3, ex sin y andex cos y are all solutions to the two variable Laplace equation. In Rn, it istrivial to check that all polynomials up to degree one, i.e.∑

|α|≤1

aαxα

is a solution to ∆u = 0 in Rn. But we also have functions of higher degreeand functions not expressible in terms of elementary functions as solutionsto Laplace equation. For instance, note that u(x) =

∏ni=1 xi is a solution to

∆u = 0 in Rn.

Definition 8.3.1. Let Ω be an open subset of Rn. A function u ∈ C2(Ω) issaid to be harmonic on Ω if ∆u(x) = 0 in Ω.

Gauss was the first to deduce some important properties of harmonicfunctions and thus laid the foundation for Potential theory and HarmonicAnalysis. Due to the linearity of ∆, sum of any finite number of har-monic functions is harmonic and a scalar multiple of a harmonic function isharmonic.

In two dimension, one associates with a harmonic function u(x, y), aconjugate harmonic function, v(x, y) defined as the solution of a first ordersystem of PDE called the Cauchy-Riemann equations,

ux(x, y) = vy(x, y) and uy(x, y) = −vx(x, y).

Harmonic functions and holomorphic functions (differentiable complex func-tions) are related in the sense that, for any pair (u, v), harmonic and its conju-gate, gives a holomorphic function f(z) = u(x, y)+ iv(x, y) where z = x+ iy.Conversely, for any holomorphic function f , its real part and imaginary partare conjugate harmonic functions. This observation gives us more examples


of harmonic functions, for instance, since all complex polynomials f(z) = zm

are holomorphic we have (using the polar coordinates) u(r, θ) = rm cosmθand v(r, θ) = rm sinmθ are harmonic functions in R2 for all m ∈ N. Simi-larly, since f(z) = log z = ln r+ iθ is holomorphic in certain region, we haveu(r, θ) = ln r and v(r, θ) = θ are harmonic in R2 \ (0, 0) and R2 \ θ = 0,respectively.

Lecture 9

Properties of HarmonicFunctions

In this section we shall study properties of harmonic functions. We shall as-sume the divergence theorems from multivariable calculus (cf. Appendix A).Also, note that if u is a harmonic function on Ω then, by Gauss divergencetheorem (cf. Theorem A.0.4), ∫

∂Ω

∂u

∂νdσ = 0.

Theorem 9.0.1 (Maximum Principle). Let Ω be an open, bounded subset ofRn. Let u ∈ C(Ω) be harmonic in Ω. Then

maxy∈Ω

u(y) = maxy∈∂Ω

u(y).

Proof. Since ∂Ω ⊂ Ω, we have max∂Ω u ≤ maxΩ u. It only remains to provethe other equality. For the given harmonic function u and for a fixed ε > 0,we set vε(x) = u(x) + ε|x|2, for each x ∈ Ω. For each x ∈ Ω, ∆vε =∆u+ 2nε > 0. Recall that1 if a function v attains local maximum at a pointx ∈ Ω, then in each direction its second order partial derivative vxixi(x) ≤ 0,for all i = 1, 2, . . . , n. Therefore ∆v(x) ≤ 0. Thus, we argue that vε does notattain (even a local) maximum in Ω. But vε has to have a maximum in Ω,

1v ∈ C2(a, b) has a local maximum at x ∈ (a, b) then v′(x) = 0 and v′′(x) ≤ 0

47

LECTURE 9. PROPERTIES OF HARMONIC FUNCTIONS 48

hence it should be attained at some point x? ∈ ∂Ω, on the boundary. For allx ∈ Ω,

u(x) ≤ vε(x) ≤ vε(x?) = u(x?) + ε|x?|2 ≤ max

x∈∂Ωu(x) + εmax

x∈∂Ω|x|2.

The above inequality is true for all ε > 0. Thus, u(x) ≤ maxx∈∂Ω u(x), for allx ∈ Ω. Therefore, maxΩ u ≤ maxx∈∂Ω u(x). and hence we have equality.

9.0.1 Existence and Uniqueness of Solution

A consequence of the maximum principle is the uniqueness of the harmonicfunctions.

Theorem 9.0.2 (Uniqueness of Harmonic Functions). Let Ω be an open,bounded subset of Rn. Let u1, u2 ∈ C2(Ω)∩C(Ω) be harmonic in Ω such thatu1 = u2 on ∂Ω, then u1 = u2 in Ω.

Proof. Note that u1 − u2 is a harmonic function and hence, by maximumprinciple, should attain its maximum on ∂Ω. But u1 − u2 = 0 on ∂Ω. Thusu1 − u2 ≤ 0 in Ω. Now, repeat the argument for u2 − u1, we get u2 − u1 ≤ 0in Ω. Thus, we get u1 − u2 = 0 in Ω.

In fact harmonic functions satisfy a much stronger maximum principlewhose proof is beyond the scope of this course.

Theorem 9.0.3 (Strong Maximum Principle). Let Ω be an open, connected(domain) subset of Rn. Let u be harmonic in Ω and M := maxy∈Ω u(y).Then

u(x) < M ∀x ∈ Ω

or u ≡M is constant in Ω.

By the strong maximum principle (cf. Theorem 9.0.3), if Ω is connectedand g ≥ 0 and g(x) > 0 for some x ∈ ∂Ω then u(x) > 0 for all x ∈ Ω.

Theorem 9.0.4. Let Ω be an open bounded connected subset of Rn andg ∈ C(∂Ω). Then the Dirichlet problem (9.0.1) has atmost one solutionu ∈ C2(Ω) ∩ C(Ω). Moreover, if u1 and u2 are solution to the Dirichletproblem corresponding to g1 and g2 in C(∂Ω), respectively, then

(a) (Comparison) g1 ≥ g2 on ∂Ω and g1(x0) > g2(x0) for some x ∈ ∂Ωimplies that u1 > u2 in Ω.


(b) (Stability) |u1(x)− u2(x)| ≤ maxy∈∂Ω |g1(y)− g2(y)| for all x ∈ Ω.

Proof. The fact that there is atmost one solution to the Dirichlet problemfollows from the Theorem 9.0.2. Let w = u1 − u2. Then w is harmonic.

(a) Note that w = g1 − g2 ≥ 0 on ∂Ω. Since g1(x0) > g2(x0) for somex0 ∈ ∂Ω, then w(x) > 0 for all x ∈ ∂Ω. This proves the comparisonresult.

(b) Again, by maximum principle, we have

±w(x) ≤ maxy∈∂Ω|g1(y)− g2(y)|∀x ∈ Ω.

This proves the stability result.

We remark that the uniqueness result is not true for unbounded domains.

Example 9.1. Let u ∈ C2(Ω) ∩ C(Ω) be a solution of the Dirichlet problem∆u(x) = 0 x ∈ Ωu(x) = g(x) x ∈ ∂Ω.

(9.0.1)

Let Ω = x ∈ Rn | |x| > 1 and g ≡ 0. Obviously, u = 0 is a solution. Butwe also have a non-trivial solution

u(x) =

ln |x| n = 2

|x|2−n − 1 n ≥ 3.

Example 9.2. Consider the problem (9.0.1) with g ≡ 0 and Ω = x ∈ Rn |xn > 0. Obviously, u = 0 is a solution. But we also have a non-trivialsolution u(x) = xn.

We have shown above that if a solution exists for (9.0.1) then it is unique(cf. Theorem 9.0.2). So it only remains to show the existence of solutionof (9.0.1), for any given domain Ω. In the modern theory, there are threedifferent methods to address the question of existence, viz., Perron’s Method,Layer Potential (Integral Equations) and L2 methods which are beyond thescope of this course.


Example 9.3 (Non-existence of Solutions). In 1912, Lebesgue gave an exampleof a domain on which the classical Dirichlet problem is not solvable. Thedomain is

Ω := (x, y, z) ∈ R3 | r2 + z2 < 1; r > e−1/2z for z > 0.

Note that Ω is the unit ball in R3 with a sharp inward cusp, called Lebesguespine, at the origin (0, 0, 0).

Example 9.4. There are domains with inward cusps for which the classicalproblem is solvable. For instance, consider

Ω := (x, y, z) ∈ R3 | r2 + z2 < 1; r > z2k for z > 0,

for any positive integer k. The proof of this fact involves the theory ofcapacities, beyond the scope of this course.

Remark 9.0.5 (Neumann Boundary Condition). The Neumann problem isstated as follows: Given f : Ω → R and g : ∂Ω → R, find u : Ω → R suchthat

−∆u = f in Ω∂u∂ν

= g on ∂Ω(9.0.2)

where ∂u∂ν

:= ∇u · ν and ν = (ν1, . . . , νn) is the outward pointing unit normalvector field of ∂Ω. Thus, the boundary imposed is called the Neumannboundary condition. The solution of a Neumann problem is not necessarilyunique. If u is any solution of (9.0.2), then u+ c for any constant c is also asolution of (9.0.2). More generally, for any v such that v is constant on theconnected components of Ω, u+ v is a solution of (9.0.2).

Lecture 10

Sturm-Liouville Problems

10.1 Eigen Value Problems

Definition 10.1.1. Let L denote a linear differential operator and I ⊂ R.Then we say Ly(x) = λy(x) on I is an eigenvalue problem (EVP) corre-sponding to L when both λ and y : I → R are unknown.

Example 10.1. For instance, if L = −d2dx2

then its corresponding eigenvalueproblem is −y′′ = λy.

If λ ∈ R is fixed then one can obtain a general solution. But, in an EVP1

we need to find all λ ∈ R for which the given ODE is solvable. Note thaty ≡ 0 is a trivial solution, for all λ ∈ R.

Definition 10.1.2. A λ ∈ R, for which the EVP corresponding to L admitsa non-trivial solution yλ is called an eigenvalue of the operator L and yλ issaid to be an eigen function corresponding to λ. The set of all eigenvalues ofL is called the spectrum of L.

Exercise 2. Note that for a linear operator L, if yλ is an eigen functioncorresponding to λ, then αyλ is also an eigen function corresponding to λ,for all α ∈ R.

As a consequence of the above exercise note that, for a linear operatorL, the set of all eigen functions corresponding to λ forms a vector spaceWλ, called the eigen space corresponding to λ. Let V denote the set of allsolutions of the EVP corresponding to a linear operator L. Necessarily, 0 ∈ Vand V ⊂ C2(I). Note that V = ∪λWλ where λ’s are eigenvalues of L.

1compare an EVP with the notion of diagonalisation of matrices from Linear Algebra

51

LECTURE 10. STURM-LIOUVILLE PROBLEMS 52

Exercise 3. Show that any second order ODE of the form

y′′ + P (x)y′ +Q(x)y(x) = R(x)

can be written in the form

d

dx

(p(x)

dy

dx

)+ q(x)y(x) = r(x).

(Find p, q and r in terms of P , Q and R).

Proof. Let us multiply the original equation by a function p(x), which willbe chosen appropriately later. Thus, we get

p(x)y′′ + p(x)P (x)y′ + p(x)Q(x)y(x) = p(x)R(x).

We shall choose µ such that

p′ = p(x)P (x).

Hence, p(x) = e∫P (x) dx. Thus, by setting q(x) = p(x)Q(x) and r(x) =

p(x)R(x), we have the other form.

10.2 Sturm-Liouville Problems

Given a finite interval (a, b) ⊂ R, the Sturm-Liouville (S-L) problem is givenas

ddx

(p(x) dy

dx

)+ q(x)y + λr(x)y = 0 x ∈ (a, b)c1y(a) + c2y

′(a) = 0. c21 + c2

2 > 0d1y(b) + d2y

′(b) = 0 d21 + d2

2 > 0.(10.2.1)

The function y(x) and λ are unknown quantities. The pair of boundaryconditions given above is called separated. The boundary conditions corre-sponds to the end-point a and b, respectively. Note that both c1 and c2

cannot be zero simultaneously and, similar condition on d1 and d2.

Definition 10.2.1. The Sturm-Liouville problem with separated boundaryconditions is said to be regular if:

(a) p, p′, q, r : [a, b]→ R are continuous functions

(b) p(x) > 0 and r(x) > 0 for x ∈ [a, b].


We say the S-L problem is singular if either the interval (a, b) is un-bounded or one (or both) of the regularity condition given above fails.

We say the S-L problem is periodic if p(a) = p(b) and the separatedboundary conditions are replaced with the periodic boundary condition y(a) =y(b) and y′(a) = y′(b).

Example 10.2. Examples of regular S-L problem:

(a) −y′′(x) = λy(x) x ∈ (0, a)

y(0) = y(a) = 0.

We have chosen c1 = d1 = 1 and c2 = d2 = 0. Also, q ≡ 0 and p ≡ r ≡ 1.

(b) −y′′(x) = λy(x) x ∈ (0, a)

y′(0) = y′(a) = 0.

We have chosen c1 = d1 = 0 and c2 = d2 = 1. Also, q ≡ 0 and p ≡ r ≡ 1.

(c) −y′′(x) = λy(x) x ∈ (0, a)y′(0) = 0

cy(a) + y′(a) = 0,

where c > 0 is a constant.

(d) − (x2y′(x))′

= λy(x) x ∈ (1, a)y(1) = 0y(a) = 0,

where p(x) = x2, q ≡ 0 and r ≡ 1.

Remark 10.2.2. In a singular Sturm-Liouville problem, the boundary con-dition at an (or both) end(s) is dropped if p vanishes in (or both) the cor-responding end(s). This is because when p vanishes, the equation at thatpoint is no longer second order. Note that dropping a boundary conditioncorresponding to a end-point is equivalent to taking both constants zero (forinstance, c1 = c2 = 0, in case of left end-point).

Example 10.3. Examples of singular S-L problem:


(a) For each n = 0, 1, 2, . . ., consider the Bessel’s equation− (xy′(x))′ =

(−n2

x+ λx

)y(x) x ∈ (0, a)

y(a) = 0,

where p(x) = r(x) = x, q(x) = −n2/x. This equation is not regularbecause p(0) = r(0) = 0 and q is not continuous in the closed interval[0, a], since q(x) → −∞ as x → 0. Note that there is no boundarycondition corresponding to 0.

(b) The Legendre equation

−[(1− x2)y′(x)

]′= λy(x) x ∈ (−1, 1)

with no boundary condition. Here p(x) = 1− x2, q ≡ 0 and r ≡ 1. Thisequation is not regular because p(−1) = p(1) = 0. Note that there is noboundary conditions because p vanishes at both the end-points.

Example 10.4. Examples of periodic S-L problem:−y′′(x) = λy(x) x ∈ (−π, π)y(−π) = y(π)y′(−π) = y′(π).

Lecture 11

Spectral Results

We shall now state without proof the spectral theorem for regular S-L prob-lem. Our aim, in this course, is to check the validity of the theorem throughsome examples.

Theorem 11.0.1. For a regular S-L problem, there exists an increasing se-quence of eigenvalues 0 < λ1 < λ2 < λ3 < . . . < λk < . . . with λk → ∞, ask →∞.

Exercise 4. Let Wk = Wλk be the eigen space corresponding λk. Show thatfor regular S-L problem Wk is one dimensional, i.e., corresponding to eachλk, there cannot be two or more linearly independent eigen vectors.

Example 11.1. Consider the boundary value problem,y′′ + λy = 0 x ∈ (0, a)

y(0) = y(a) = 0.

This is a second order ODE with constant coeffcients. Its characteristicequation is m2 + λ = 0. Solving for m, we get m = ±

√−λ. Note that

the λ can be either zero, positive or negative. If λ = 0, then y′′ = 0 andthe general solution is y(x) = αx + β, for some constants α and β. Sincey(0) = y(a) = 0 and a 6= 0, we get α = β = 0. Thus, we have no non-trivialsolution corresponding to λ = 0.

If λ < 0, then ω = −λ > 0. Hence y(x) = αe√ωx + βe−

√ωx. Using the

boundary condition y(0) = y(a) = 0, we get α = β = 0 and hence we haveno non-trivial solution corresponding to negative λ’s.

If λ > 0, then m = ±i√λ and y(x) = α cos(

√λx) + β sin(

√λx). Using

the boundary condition y(0) = 0, we get α = 0 and y(x) = β sin(√λx).

55

LECTURE 11. SPECTRAL RESULTS 56

Using y(a) = 0 (and β = 0 yields trivial solution), we assume sin(√λa) = 0.

Thus, λ = (kπ/a)2 for each non-zero k ∈ N (since λ > 0). Hence, for eachk ∈ N, there is a solution (yk, λk) with

yk(x) = sin

(kπx

a

),

and λk = (kπ/a)2. Notice the following properties of the eigenvalues λk andeigen functions yk

(i) We have discrete set of λ’s such that 0 < λ1 < λ2 < λ3 < . . . andλk →∞, as k →∞.

(ii) The eigen functions yλ corresponding to λ form a subspace of dimensionone.

In particular, in the above example, when a = π the eigenvalues, for eachk ∈ N, are (yk, λk) where yk(x) = sin(kx) and λk = k2.

Theorem 11.0.2. For a periodic S-L problem, there exists an increasingsequence of eigenvalues 0 < λ1 < λ2 < λ3 < . . . < λk < . . . with λk → ∞,as k → ∞. Moreover, W1 = Wλ1, the eigen space corresponding to the firsteigen value is one dimensional.

Example 11.2. Consider the boundary value problem,y′′ + λy = 0 in (−π, π)y(−π) = y(π)y′(−π) = y′(π).

The characteristic equation is m2 +λ = 0. Solving for m, we get m = ±√−λ.

Note that the λ can be either zero, positive or negative.If λ = 0, then y′′ = 0 and the general solution is y(x) = αx+ β, for some

constants α and β. Since y(−π) = y(π), we get α = 0. Thus, for λ = 0, y ≡a constant is the only non-trivial solution.

If λ < 0, then ω = −λ > 0. Hence y(x) = αe√ωx + βe−

√ωx. Using

the boundary condition y(−π) = y(π), we get α = β and using the otherboundary condition, we get α = β = 0. Hence we have no non-trivial solutioncorresponding to negative λ’s.


If λ > 0, then m = ±i√λ and y(x) = α cos(

√λx) + β sin(

√λx). Using

the boundary condition, we get

α cos(−√λπ) + β sin(−

√λπ) = α cos(

√λπ) + β sin(

√λπ)

and

−α sin(−√λπ) + β cos(−

√λπ) = −α sin(

√λπ) + β cos(

√λπ).

Thus, β sin(√λπ) = α sin(

√λπ) = 0.

For a non-trivial solution, we must have sin(√λπ) = 0. Thus, λ = k2 for

each non-zero k ∈ N (since λ > 0).Hence, for each k ∈ N ∪ 0, there is a solution (yk, λk) with

yk(x) = αk cos kx+ βk sin kx,

and λk = k2.

Lecture 12

Singular Sturm-LiouvilleProblem

Singular S-L, in general, have continuous spectrum. However, the exampleswe presented viz. Bessel’s equation and Legendre equation have a discretespectrum, similar to the regular S-L problem.

12.0.1 EVP of Legendre Operator

Consider the Legendre equation

d

dx

((1− x2)

dy

dx

)+ λy = 0 for x ∈ [−1, 1].

Note that, equivalently, we have the form

(1− x2)y′′ − 2xy′ + λy = 0 for x ∈ [−1, 1].

The function p(x) = 1− x2 vanishes at the endpoints x = ±1.

Definition 12.0.1. A point x0 is a singular point of

y′′(x) + P (x)y′(x) +Q(x)y(x) = R(x)

if either P or Q (or both) are not analytic at x0. A singular point x0 is saidto be regular if (x− x0)P (x) and (x− x0)2Q(x) are analytic at x0.

59

LECTURE 12. SINGULAR STURM-LIOUVILLE PROBLEM 60

The end points x = ±1 are regular singular point. The coefficients P (x) =−2x1−x2 and R(x) = λ

1−x2 are analytic at x = 0, with radius of convergence 1.

We look for power series form of solutions y(x) =∑∞

k=0 akxk. Differentiating

(twice) the series term by term, substituting in the Legendre equation and

equating like powers of x, we get a2 = −λa02

, a3 = (2−λ)a16

and for k ≥ 2,

ak+2 =(k(k + 1)− λ)ak(k + 2)(k + 1)

.

Thus, the constants a0 and a1 can be fixed arbitrarily and the remainingconstants are defined as per the above relation. For instance, if a1 = 0, weget the non-trivial solution of the Legendre equation as

y1 = a0 +∞∑k=1

a2kx2k

and if a0 = 0, we get the non-trivial solution as

y2 = a1x+∞∑k=1

a2k+1x2k+1,

provided the series converge. Note from the recurrence relation that if acoefficient is zero at some stage, then every alternate coefficient, subsequently,is zero. Thus, there are two possibilities of convergence here:

(i) the series terminates after finite stage to become a polynomial

(ii) the series does not terminate, but converges.

Suppose the series does not terminate, say for instance, in y1. Thena2k 6= 0, for all k. Consider the ratio

limk→∞

∣∣∣∣a2(k+1)x2(k+1)

a2kx2k

∣∣∣∣ = limk→∞

∣∣∣∣ 2k(2k + 1)x2

(2k + 2)(2k + 1)

∣∣∣∣ = limk→∞

∣∣∣∣ 2kx2

(2k + 2)

∣∣∣∣ = x2.

The term involving λ tends to zero. Therefore, by ratio test, y1 converges inx2 < 1 and diverges in x2 > 1. Also, it can be shown that when x2 = 1 theseries diverges (beyond the scope of this course).

Since, Legendre equation is a singular S-L problem, we try to find solutiony such that y and its derivative y′ are continuous in the closed interval [−1, 1].


Thus, the only such possible solutions will be terminating series becomingpolynomials.

Note that, for k ≥ 2,

ak+2 =(k(k + 1)− λ)ak(k + 2)(k + 1)

.

Hence, for any n ≥ 2, if λ = n(n+1), then an+2 = 0 and hence every alternateterm is zero. Also, if λ = 1(1 + 1) = 2, then a3 = 0. If λ = 0(0 + 1) = 0, thena2 = 0. Thus, for each n ∈ N ∪ 0, we have λn = n(n + 1) and one of thesolution y1 or y2 is a polynomial. Thus, for each n ∈ N ∪ 0, we have theeigen value λn = n(n+ 1) and the Legendre polynmial Pn of degree n whichis a solution to the Legendre equation.

12.0.2 EVP of Bessel’s Operator

Consider the EVP, for each fixed n = 0, 1, 2, . . .,− (xy′(x))′ =

(−n2

x+ λx

)y(x) x ∈ (0, a)

y(a) = 0.

As before, since this is a singular S-L problem we shall look for solutions ysuch that y and its derivative y′ are continuous in the closed interval [0, a].We shall assume that the eigenvalues are all real1. Thus, λ may be zero,positive or negative.

When λ = 0, the given ODE reduces to the Cauchy-Euler form

− (xy′(x))′+n2

xy(x) = 0

or equivalently,

x2y′′(x) + xy′(x)− n2y(x) = 0.

The above second order ODE with variable coefficients can be converted toan ODE with constant coefficients by the substitution x = es (or s = lnx).Then, by chain rule,

y′ =dy

dx=dy

ds

ds

dx= e−s

dy

ds

1needs proof


and

y′′ = e−sdy′

ds= e−s

d

ds

(e−s

dy

ds

)= e−2s

(d2y

ds2− dy

ds

).

Therefore,y′′(s)− n2y(s) = 0,

where y is now a function of the new variable s. For n = 0, the generalsolution is y(s) = αs+β, for some arbitrary constants. Thus, y(x) = α lnx+β. The requirement that both y and y′ are continuous on [0, a] forces α = 0.Thus, y(x) = β. But y(a) = 0 and hence β = 0, yielding the trivial solution.

Now, let n > 0 be positive integers. Then the general solution is y(s) =αens + βe−ns. Consequently, y(x) = αxn + βx−n. Since y and y′ has to becontinuous on [0, a], β = 0. Thus, y(x) = αxn. Now, using the boundarycondition y(a) = 0, we get α = 0 yielding the trivial solution. Therefore,λ = 0 is not an eigenvalue for all n = 0, 1, 2, . . ..

When λ > 0, the given ODE reduces to

x2y′′(x) + xy′(x) + (λx2 − n2)y(x) = 0.

Using the change of variable s2 = λx2, we get y′(x) =√λy′(s) and y′′(x) =

λy′′(s). Then the given ODE is transformed into the Bessel’s equation

s2y′′(s) + sy′(s) + (s2 − n2)y(s) = 0.

Using the power series form of solution, we know that the general solutionof the Bessel’s equation is

y(s) = αJn(s) + βYn(s),

where Jn and Yn are the Bessel functions of first and second kind, respectively.Therefore, y(x) = αJn(

√λx) + βYn(

√λx). The continuity assumptions of

y and y′ force that β = 0, because Yn(√λx) is discontinuous at x = 0.

Thus, y(x) = αJn(√λx). Using the boundary condition y(a) = 0, we get

Jn(√λa) = 0.

Theorem 12.0.2. For each non-negative integer n, Jn has infinitely manypositive zeroes.

For each n ∈ N ∪ 0, let znm be the m-th zero of Jn, m ∈ N. Hence√λa = znm and so λnm = z2

nm/a2 and the corresponding eigen functions are

ynm(x) = Jn(znmx/a).For λ < 0, there are no eigen values. Observing this fact is beyond the

scope of this course, hence we assume this fact.

Lecture 13

Orthogonality of EigenFunctions

Observe that for a regular S-L problem the differential operator can be writ-ten as

L =−1

r(x)

d

dx

(p(x)

d

dx

)− q(x)

r(x).

Let V denote the set of all solutions of (10.2.1). Necessarily, 0 ∈ V andV ⊂ C2(a, b). We define the inner product1 〈·, ·〉 : V × V → R on V as,

〈f, g〉 :=

∫ b

a

r(x)f(x)g(x) dx.

Definition 13.0.1. We say two functions f and g are perpendicular ororthogonal with weight r if 〈f, g〉 = 0. We say f is of unit length if its norm‖f‖ =

√〈f, f〉 = 1.

Theorem 13.0.2. With respect to the inner product defined above in V , theeigen functions corresponding to distinct eigenvalues of the S-L problem areorthogonal.

Proof. Let yi and yj are eigen functions corresponding to distinct eigenvaluesλi and λj. We need to show that 〈yi, yj〉 = 0. Recall that L is the S-L operator

1a generalisation of the usual scalar product of vectors

63

LECTURE 13. ORTHOGONALITY OF EIGEN FUNCTIONS 64

and hence Lyk = λkyk, for k = i, j. Consider

λi〈yi, yj〉 = 〈Lyi, yj〉 =

∫ b

a

rLyiyj dx

= −∫ b

a

d

dx

(p(x)

dyidx

)yj(x)−

∫ b

a

q(x)yiyj dx

=

∫ b

a

p(x)dyidx

dyj(x)

dxdx− [p(b)y′i(b)yj(b)− p(a)y′i(a)yj(b)]

−∫ b

a

q(x)yiyj dx

= −∫ b

a

yi(x)d

dx

(p(x)

dyjdx

)+[p(b)y′j(b)yi(b)− p(a)y′j(a)yi(b)

]− [p(b)y′i(b)yj(b)− p(a)y′i(a)yj(b)]−

∫ b

a

q(x)yiyj dx

= 〈yi, Lyj〉+ p(b)[y′j(b)yi(b)− y′i(b)yj(b)

]−p(a)

[y′j(a)yi(a)− y′i(a)yj(a)

]= λj〈yi, yj〉+ p(b)

[y′j(b)yi(b)− y′i(b)yj(b)

]−p(a)


].

Thus,

(λi−λj)〈yi, yj〉 = p(b)[y′j(b)yi(b)− y′i(b)yj(b)

]−p(a)


].

For regular S-L problem, the boundary condition corresponding to the end-point b is the system of equations

d1yi(b) + d2y′i(b) = 0

d1yj(b) + d2y′j(b) = 0

such that d21 + d2

2 = 0. Therefore, the determinant of the coefficient matrixyi(b)y

′j(b)− yj(b)y′i(b) = 0. Similar, argument is also valid for the boundary

condition corresponding to a. Thus, (λi − λj)〈yi, yj〉 = 0. But λi − λj 6= 0,hence 〈yi, yj〉 = 0.

For periodic S-L problem, p(a) = p(b), yk(a) = yk(b) and y′k(a) = y′k(b),for k = i, j. Then the RHS vanishes and 〈yi, yj〉 = 0.

For singular S-L problems such that either p(a) = 0 or p(b) = 0 or bothhappens, then again RHS vanishes. This is because if p(a) = 0, we drop theboundary condition corresponding to the end-point a.


Let us examine the orthogonality of the eigenvectors computed in theexamples earlier.

Example 13.1. We computed in Example 11.1 the eigenvalues and eigenvec-tors of the regular S-L problem,

y′′ + λy = 0 x ∈ (0, a)y(0) = y(a) = 0

to be (yk, λk) where

yk(x) = sin

(kπx

a

)and λk = (kπ/a)2, for each k ∈ N. For m,n ∈ N such that m 6= n, we needto check that ym and yn are orthogonal. Since r ≡ 1, we consider

〈ym(x), yn(x)〉 =

∫ a

0

sin(mπx

a

)sin(nπx

a

)dx

Exercise 5. Show that, for any n ≥ 0 and m positive integer,

(i) ∫ π

−πcosnt cosmtdt =

π, for m = n

0, for m 6= n.

(ii) ∫ π

−πsinnt sinmtdt =

π, for m = n

0, for m 6= n.

(iii) ∫ π

−πsinnt cosmtdt = 0.

Consequently, show that cos kt√π

and sin kt√π

are of unit length.

Proof. (i) Consider the trigonometric identities

cos((n+m)t) = cosnt cosmt− sinnt sinmt (13.0.1)

andcos((n−m)t) = cosnt cosmt+ sinnt sinmt. (13.0.2)


Adding (13.0.1) and (13.0.2), we get

1

2cos((n+m)t) + cos((n−m)t) = cosnt cosmt.

Integrating both sides from −π to π, we get∫ π


1

2

∫ π

−π(cos((n+m)t) + cos((n−m)t)) dt.

But ∫ π

−πcos kt dt =

1

ksin kt|π−π = 0, for k 6= 0.

Thus, ∫ π


π, for m = n

0, for m 6= n.

Further,‖ cos kt‖ = 〈cos kt, cos kt〉1/2 =

√π.

Therefore, cos kt√π

is of unit length.

(ii) Subtract (13.0.1) from (13.0.2) and use similar arguments as above.

(iii) Arguments are same using the identities (13.0.1) and (13.0.2) corre-sponding to sin.

Exercise 6. Show that∫ π

−πeimte−int dt =

2π, for m = n

0, for m 6= n.

Example 13.2. We computed in Example 11.2 the eigenvalues and eigenvec-tors of the periodic S-L problem,

y′′ + λy = 0 in (−π, π)y(−π) = y(π)y′(−π) = y′(π)

to be, for each k ∈ N ∪ 0, (yk, λk) where

yk(x) = αk cos kx+ βk sin kx,

and λk = k2. Again r ≡ 1 and the orthogonality follows from the exerciseabove.


Example 13.3. The orthogonality of Legendre polynomial and Bessel functionmust have been discussed in your course on ODE. Recall that the Legendrepolynomials has the property∫ 1

−1

Pm(x)Pn(x) dx =

0, if m 6= n

22n+1

, if m = n

and the Bessel functions have the property∫ 1

0

xJn(znix)Jn(znjx) dx =

0, if m 6= n12[Jn+1(zni)]

2, if m = n

where zni is the i-th positive zero of the Bessel function (of order n) Jn.

13.1 Eigen Function Expansion

Observe that an eigenvector yk, for any k, can be normalised (unit norm) inits inner-product by dividing yk by its norm ‖yk‖. Thus, yk/‖yk‖, for anyk, is a unit vector. For instance, in view of Exercise 5, cos kt√

πand sin kt√

πare

functions of unit length.

Definition 13.1.1. Any given function f : (a, b)→ R is said to be have theeigen function expansion corresponding to the S-L problem (10.2.1), if

f(x) ≈∞∑k=0

akyk,

for some constants ak and yk are the normalised eigenvectors correspondingto (10.2.1).

We are using the “≈” symbol to highlight the fact that the issue of con-vergence of the series is ignored.

If the eigenvectors (or eigen functions) yk involves only sin or cos terms,as in regular S-L problem (cf. Example 11.1), then the series is called FourierSine or Fourier Cosine series.

If the eigen functions yk involve both sin and cos, as in periodic S-Lproblem (cf. Example 11.2), then the series is called Fourier series. In thecase of the eigen functions being Legendre polynomial or Bessel function, wecall it Fourier-Legendre or Fourier-Bessel series, respectively.

Lecture 14

Fourier Series

At the end of previous chapter, we introduced the Fourier series of a functionf . A natural question that arises at this moment is: what classes of functionsadmit a Fourier series expansion? We attempt to answer this question in thischapter.

14.1 Periodic Functions

We isolate the properties of the trigonometric functions, viz., sin, cos, tanetc.

Definition 14.1.1. A function f : R → R is said to be periodic of periodT , if T > 0 is the smallest number such that

f(t+ T ) = f(t) ∀t ∈ R.

Such functions are also called T -periodic functions.

Example 14.1. The trigonometric functions sin t and cos t are 2π-periodicfunctions, while sin 2t and cos 2t are π-periodic functions.

Given a L-periodic real-valued function g on R, one can always constructa T -periodic function as: f(t) = g(Lt/T ). For instance, f(t) = sin

(2πtT

)is a

T -periodic function.

sin

(2π(t+ T )

T

)= sin

(2πt

T+ 2π

)= sin

(2πt

T

).

In fact, for any positive integer k, sin(

2πktT

)and cos

(2πktT

)are T -periodic

functions.

69

LECTURE 14. FOURIER SERIES 70

Exercise 7. If f : R→ R is a T -periodic function, then show that

(i) f(t− T ) = f(t), for all t ∈ R.

(ii) f(t+ kT ) = f(t), for all k ∈ Z.

(iii) g(t) = f(αt+ β) is (T/α)-periodic, where α > 0 and β ∈ R.

Exercise 8. Show that for a T -periodic integrable function f : R→ R,∫ α+T

α

f(t) dt =

∫ T

0

f(t) dt ∀α ∈ R.

14.2 Fourier Coefficients and Fourier Series

Without loss of generality, to simplify our computation, let us assume thatf is a 2π-periodic1 function on R. Suppose that f : (−π, π) → R, extendedto all of R as a 2π-periodic function, is such that the infinite series

a0 +∞∑k=1

(ak cos kt+ bk sin kt)

converges uniformly2 to f . Then,

f(t) = a0 +∞∑k=1

(ak cos kt+ bk sin kt). (14.2.1)

and integrating both sides of (14.2.1), from −π to π, we get∫ π

−πf(t) dt =

∫ π

−π

(a0 +

∞∑k=1


)dt

= a0(2π) +

∫ π

−π

(∞∑k=1


)dt

Since the series converges uniformly to f , the interchange of integral andseries is possible. Therefore,∫ π

−πf(t) dt = a0(2π) +

∞∑k=1

(∫ π

−π(ak cos kt+ bk sin kt) dt

)1similar idea will work for any T -periodic function2note the uniform convergence hypothesis


From Exercise 5, we know that∫ π

−πsin kt dt =

∫ π

−πcos kt dt = 0, ∀k ∈ N.

Hence,

a0 =1

2π

∫ π

−πf(t) dt.

To find the coefficients ak, for each fixed k ∈ N, we multiply both sidesof (14.2.1) by cos kt and integrate from −π to π. Consequently,∫ π

−πf(t) cos kt dt = a0

∫ π

−πcos kt dt

+∞∑j=1

∫ π

−π(aj cos jt cos kt+ bj sin jt cos kt) dt

=

∫ π

−πak cos kt cos kt dt = πak.

Similar argument, after multiplying by sin kt, gives the formula for bk. Thus,we have derived , for all k ∈ N,

ak =1

π

∫ π

−πf(t) cos kt dt

bk =1

π

∫ π

−πf(t) sin kt dt

a0 =1

2π

∫ π

−πf(t) dt.

These are the formulae for Fourier coefficients of a 2π-periodic functions f ,in terms of f . Similarly, if f is a T -periodic function extended to R, then itsFourier series is

f(t) = a0 +∞∑k=1

[ak cos

(2πkt

T

)+ bk sin

(2πkt

T

)],

where

ak =2

T

∫ T

0

f(t) cos

(2πkt

T

)dt (14.2.2a)


bk =2

T

∫ T

0

f(t) sin

(2πkt

T

)dt (14.2.2b)

a0 =1

T

∫ T

0

f(t) dt. (14.2.2c)

The above discussion motivates us to give the following definition.

Definition 14.2.1. If f : R→ R is any T -periodic integrable function thenwe define the Fourier coefficients of f , a0, ak and bk, for all k ∈ N, by (14.2.2)and the Fourier series of f is given by

f(x) ≈ a0 +∞∑k=1

[ak cos

(2πkt

T

)+ bk sin

(2πkt

T

)]. (14.2.3)

Note the use of “≈” symbol in (14.2.3). This is because we have thefollowing issues once we have the definition of Fourier series of f , viz.,

(a) Will the Fourier series of f always converge?

(b) If it converges, will it converge to f?

(c) If so, is the convergence point-wise or uniform3.

Answering these question, in all generality, is beyond the scope of thiscourse. However, we shall state some results, in the next section, that willget us in to working mode. We end this section with some simple exampleson computing Fourier coefficients of functions.

Example 14.2. Consider the constant function f ≡ c on (−π, π). Then

a0 =1

2π

∫ π

−πc dt = c.

For each k ∈ N,

ak =1

π

∫ π

−πc cos kt dt = 0

and

bk =1

π

∫ π

−πc sin kt dt = 0.

3because our derivation of formulae for Fourier coefficients assumed uniform conver-gence of the series


Example 14.3. Consider the trigonometric function f(t) = sin t on (−π, π).Then

a0 =1

2π

∫ π

−πsinx dt = 0.

For each k ∈ N,

ak =1

π

∫ π

−πsin t cos kt dt = 0

and

bk =1

π

∫ π

−πsin t sin kt dt =

0 k 6= 1

1 k = 1.

Similarly, for f(t) = cos t on (−π, π), all Fourier coefficients are zero, excepta1 = 1.

Example 14.4. Consider the function f(t) = t on (−π, π). Then

a0 =1

2π

∫ π

−πt dt = 0.

For each k ∈ N,

ak =1

π

∫ π

−πt cos kt dt =

1

kπ

[−∫ π

−πsin kt dt+ (π sin kπ − (−π) sin k(−π))

]and hence ak = 0, for all k.

bk =1

π

∫ π

−πt sin kt dt =

1

kπ

[∫ π

−πcos kt dt− (π cos kπ − (−π) cos k(−π))

]=

1

kπ

[0−

(π(−1)k + π(−1)k

)]=

(−1)k+12

k

Therefore, t as a 2π-periodic function defined in (−π, π) has the Fourier seriesexpansion

t ≈ 2∞∑k=1

(−1)k+1

ksin kt

Example 14.5. Let us consider the same function f(t) = t, as in previousexample, but defined on (0, π). Viewing this as π-periodic function, we com-pute

a0 =1

π

∫ π

0

t dt =π

2.


For each k ∈ N,

ak =2

π

∫ π

0

t cos 2kt dt =2

2kπ

[−∫ π

0

sin 2kt dt+ (π sin 2kπ − 0)

]=

1

kπ

[1

2k(cos 2kπ − cos(0))

]= 0

and

bk =2

π

∫ π

0

t sin 2kt dt =2

2kπ

[∫ π

0

cos 2kt dt− (π cos 2kπ − 0)

]=

1

kπ

[1

2k(sin 2kπ − sin(0))− π

]=−1

k.

Therefore, t as a π-periodic function defined on (0, π) has the Fourier seriesexpansion

t ≈ π

2−∞∑k=1

1

ksin 2kt.

Note that difference in Fourier expansion of the same function when theperiodicity changes.

Exercise 9. Find the Fourier coefficients and Fourier series of the function

f(t) =

0 if t ∈ (−π, 0]

t if t ∈ (0, π).

Theorem 14.2.2 (Riemann-Lebesgue Lemma). Let f be a continuous func-tion in [−π, π]. Show that the Fourier coefficients of f converges to zero,i.e.,

limk→∞

ak = limk→∞

bk = 0.

Proof. Observe that |ak| and |bk| are bounded sequences, since

|ak|, |bk| ≤∫ π

−π|f(t)| dt < +∞.


We need to show that these bounded sequences, in fact, converges to zero.Now set x = t− π/k and hence

bk =

∫ π

−πf(t) sin kt dt =

∫ π−π/k

−π−π/kf(x+ π/k) sin(kx+ π) dx

= −∫ π−π/k

−π−π/kf(x+ π/k) sin kx dx.

Therefore, after reassigning x as t,

2bk =

∫ π

−πf(t) sin kt dt−

∫ π−π/k

−π−π/kf(t+ π/k) sin kt dt

= −∫ −π−π−π/k

f(t+ π/k) sin kt dt+

∫ π−π/k

−π(f(t)− f(t+ π/k)) sin kt dt

+

∫ π

π−π/kf(t) sin kt dt

= I1 + I2 + I3.

Thus, |2bk| ≤ |I1|+ |I2|+ |I3|. Consider

|I3| =

∣∣∣∣∫ π

π−π/kf(t) sin kt dt

∣∣∣∣≤

∫ π

π−π/k|f(t)| dt

≤(

maxt∈[−π,π]

|f(t)|)π

k=Mπ

k.

Similar estimate is also true for I1. Let us consider,

|I2| =

∣∣∣∣∣∫ π−π/k

−π(f(t)− f(t+ π/k)) sin kt dt

∣∣∣∣∣≤

(max

t∈[−π,π−π/k]|f(t)− f(t+ π/k)|

)(2π − π

k

)By the uniform continuity of f on [−π, π], the maximum will tend to zero ask →∞. Hence |bk| → 0. Exactly, similar arguments hold for ak.

Lecture 15

Fourier Series: Continued...

15.1 Piecewise Smooth Functions

Definition 15.1.1. A function f : [a, b]→ R is said to be piecewise contin-uously differentiable if it has a continuous derivative f ′ in (a, b), except atfinitely many points in the interval [a, b] and at each these finite points, theright-hand and left-hand limit for both f and f ′ exist.

Example 15.1. Consider f : [−1, 1] → R defined as f(t) = |t| is continuous.It is not differentiable at 0, but it is piecewise continuously differentiable.

Example 15.2. Consider the function f : [−1, 1]→ R defined as

f(t) =

−1, for − 1 < t < 0,

1, for 0 < t < 1,

0, for t = 0, 1,−1.

It is not continuous, but is piecewise continuous. It is also piecewise contin-uously differentiable.

Exercise 10 (Riemann-Lebesgue Lemma). Let f be a piecewise continuousfunction in [−π, π] such that∫ π

−π|f(t)| dt < +∞.

Show that the Fourier coefficients of f converges to zero, i.e.,

limk→∞

ak = limk→∞

bk = 0.

77

LECTURE 15. FOURIER SERIES: CONTINUED... 78

Theorem 15.1.2. If f is a T -periodic piecewise continuously differentiablefunction, then the Fourier series of f converges to f(t), for every t at whichf is smooth. Further, at a non-smooth point t0, the Fourier series of f willconverge to the average of the right and left limits of f at t0.

Corollary 15.1.3. If f : R→ R is a continuously differentiable (derivativef ′ exists and is continuous) T -periodic function, then the Fourier series of fconverges to f(t), for every t ∈ R.

Example 15.3. For a given constant c 6= 0, consider the piecewise function

f(t) =

0 if t ∈ (−π, 0)

c if t ∈ (0, π).

Then,

a0 =1

2π

∫ π

0

c dt =c

2.

For each k ∈ N,

ak =1

π

∫ π

0

c cos kt dt = 0

and

bk =1

π

∫ π

0

c sin kt dt =c

π

[1

k(− cos kπ + cos(0))

]=c(1 + (−1)k+1)

kπ.

Therefore,

f(t) ≈ c

2+∞∑k=1

c(1 + (−1)k+1)

kπsin kt.

The point t0 = 0 is a non-smooth point of the function f . Note that theright limit of f at t0 = 0 is c and the left limit of f at t0 = 0 is 0. Note thatthe Fourier series of f ay t0 = 0 converges to c/2, the average of c and 0.

15.2 Complex Fourier Coefficients

The Fourier series of a 2π-periodic function f : R → R as given in (14.2.1),can be recast in complex number notation using the formulae

cos t =eıt + e−ıt

2, sin t =

eıt − e−ıt

2i.


Note that we can rewrite the Fourier series expansion of f as

f(t) =a0

2+∞∑k=1


with a factor 2 in denominator of a0 and make the formulae of the Fouriercoefficient having uniform factor. Thus,

f(t) =a0

2+∞∑k=1


=a0

2+∞∑k=1

[ak2

(eıkt + e−ıkt

)− ıbk

2

(eıkt − e−ıkt

)]f(t) =

a0

2+∞∑k=1

([ak − ıbk

2

]eıkt +

[ak + ıbk

2

]e−ıkt

)= c0 +

∞∑k=1

(cke

ıkt + c−ke−ıkt)

=∞∑

k=−∞

ckeıkt.

Exercise 11. Given a 2π-periodic function f such that

f(t) =∞∑

k=−∞

ckeıkt, (15.2.1)

where the convergence is uniform. Use the integral formulae from Exercise 6to show that, for all k ∈ Z,

ck =1

2π

∫ π

−πf(t)e−ıkt dt.

Proof. Fix a k. To find the coefficient ck, multiply both sides of (15.2.1) bye−ıkt and integrate from −π to π.

Using the real Fourier coefficients one can write down the complex Fouriercoefficients using the relations

c0 =a0

2, ck =

ak − ıbk2

and c−k =ak + ıbk

2


and if one can compute directly the complex Fourier series of a periodicfunction f , then one can write down the real Fourier coefficients using theformula,

a0 = 2c0, ak = ck + c−k and bk = ı(ck − c−k).

Exercise 12. Find the complex Fourier coefficients (directly) of the functionf(t) = t for t ∈ (−π, π] extended to R periodically with period 2π. Use thecomplex Fourier coefficients to find the real Fourier coefficients of f .

Proof. We use the

ck =1

2π

∫ π

−πte−ıkt dt.

for all k = 0,±1,±2, . . .. For k = 0, we get c0 = 0. For k 6= 0, we applyintegration by parts to get

ck =1

2π

[(ıt

ke−ıkt

)π−π−∫ π

−π

ı

ke−ıkt dt

]=

1

2π

[ ıπke−ıkπ +

ıπ

keıkπ]

=ı

kcos kπ =

ı

k(−1)k.

Hence the real Fourier coefficients are

a0 = 0, ak = (ck + c−k) = 0 and bk = ı(ck − c−k) = 2(−1)k+1

k.

15.3 Orthogonality

Let V be the class of all 2π-periodic real valued continuous function on R.

Exercise 13. Show that V is a vector space over R.

We introduce an inner product on V . For any two elements f, g ∈ V , wedefine:

〈f, g〉 :=

∫ π

−πf(t)g(t) dt.

The inner product generalises to V the properties of scalar product on Rn.


Exercise 14. Show that the inner product defined on V as

〈f, g〉 =

∫ π

−πf(t)g(t) dt

satisfies the properties of a scalar product in Rn, viz., for all f, g ∈ V ,

(a) 〈f, g〉 = 〈g, f〉.

(b) 〈f + g, h〉 = 〈f, h〉+ 〈g, h〉.

(c) 〈αf, g〉 = α〈f, g〉 ∀α ∈ R.

(d) 〈f, f〉 ≥ 0 and 〈f, f〉 = 0 implies that f ≡ 0.

Definition 15.3.1. We say two functions f and g are perpendicular ororthogonal if 〈f, g〉 = 0. We say f is of unit length if its norm ‖f‖ =√〈f, f〉 = 1.

Consider, for k ∈ N, the following elements in V

e0(t) =1√2π, ek(t) =

cos kt√π

and fk(t) =sin kt√π.

Example 15.4. e0, ek and fk are all of unit length. 〈e0, ek〉 = 0 and 〈e0, fk〉 =0. Also, 〈em, en〉 = 0 and 〈fm, fn〉 = 0, for m 6= n. Further, 〈em, fn〉 = 0for all m,n. Check and compare these properties with the standard basisvectors of Rn!

In this new formulation, we can rewrite the formulae for the Fouriercoefficients as:

a0 =1√2π〈f, e0〉, ak =

1√π〈f, ek〉 and bk =

1√π〈f, fk〉.

and the Fourier series of f has the form,

f(t) = 〈f, e0〉1√2π

+1√π

∞∑k=1

(〈f, ek〉 cos kt+ 〈f, fk〉 sin kt) .


15.3.1 Odd and Even functions

Definition 15.3.2. We say a function f : R → R is odd if f(−t) = −f(t)and even if f(−t) = f(t).

Example 15.5. All constant functions are even functions. For all k ∈ N, sin ktare odd functions and cos kt are even functions.

Exercise 15. Any odd function is always orthogonal to an even function.

The Fourier series of an odd or even functions will contain only sine orcosine parts, respectively. The reason being that, if f is odd

〈f, 1〉 = 0 and 〈f, cos kt〉 = 0

and hence a0 = 0 and ak = 0, for all k. If f is even

〈f, sin kt〉 = 0

and bk = 0, for all k.

15.4 Fourier Sine-Cosine Series

Let f : (0, T )→ R be a piecewise smooth function. To compute the FourierSine series of f , we extend f , as an odd function fo, to (−T, T )

fo(t) =

f(t), for t ∈ (0, T )

−f(−t) , for t ∈ (−T, 0).

Note that fo is a 2T -periodic function and is an odd function. Since fo isodd, the cosine coefficients ak and the constant term a0 vanishes in Fourierseries expansion of fo. The restriction of the Fourier series of fo to f in theinterval (0, T ) gives the Fourier sine series of f . We derive the formulae forFourier sine coefficient of f .

f(t) =∞∑k=1

bk sin

(πkt

T

)where (15.4.1)


bk =1

T

⟨fo, sin

(πkt

T

)⟩=

1

T

∫ T

−Tfo(t) sin

(πkt

T

)dt

=1

T

[∫ 0

−T−f(−t) sin

(πkt

T

)dt+

∫ T

0

f(t) sin

(πkt

T

)dt

]=

1

T

[∫ 0

T

−f(t) sin

(πkt

T

)dt+

∫ T

0

f(t) sin

(πkt

T

)dt

]=

2

T

∫ T

0

f(t) sin

(πkt

T

)dt.

Example 15.6. Let us consider the function f(t) = t on (0, π). To compute theFourier sine series of f , we extend f to (−π, π) as an odd function fo(t) = ton (−π, π). For each k ∈ N,

bk =2

π

∫ π

0

t sin kt dt =2

kπ

[∫ π

0

cos kt dt− (π cos kπ − 0)

]=

2

kπ

[1

k(sin kπ − sin(0)) + π(−1)k+1

]=

(−1)k+12

k.

Therefore, the Fourier sine series expansion of f(t) = t on (0, π) is

t ≈ 2∞∑k=1

(−1)k+1

ksin kt

Compare the result with Example 14.4.

For computing the Fourier cosine series of f , we extend f as an evenfunction to (−T, T ),

fe(t) =

f(t), for t ∈ (0, T )

f(−t) , for t ∈ (−T, 0).

The function fe is a 2T -periodic function extended to all of R. The Fourierseries of fe has no sine coefficients, bk = 0 for all k. The restriction of theFourier series of fe to f in the interval (0, T ) gives the Fourier cosine seriesof f . We derive the formulae for Fourier cosine coefficient of f .

f(t) = a0 +∞∑k=1

ak cos

(πkt

T

)(15.4.2)


where

ak =2

T

∫ T

0

f(t) cos

(πkt

T

)dt

and

a0 =1

T

∫ T

0

f(t) dt.

Example 15.7. Let us consider the function f(t) = t on (0, π). To computethe Fourier cosine series of f , we extend f to (−π, π) as an even functionfe(t) = |t| on (−π, π). Then,

a0 =1

π

∫ π

0

t dt =π

2.

For each k ∈ N,

ak =2

π

∫ π

0

t cos kt dt =2

kπ

[−∫ π

0

sin kt dt+ (π sin kπ − 0)

]=

2

kπ

[1

k(cos kπ − cos(0))

]=

2[(−1)k − 1]

k2π.

Therefore, the Fourier cosine series expansion of f(t) = t on (0, π) is

t ≈ π

2+ 2

∞∑k=1

(−1)k − 1

k2πcos kt.

Compare the result with the Fourier series of the function f(t) = |t| on(−π, π).

15.5 Fourier Transform and Integral

Recall that we had computed the Fourier series expansion of periodic func-tions. The periodicity was assumed due to the periodicity of sin and cosfunctions. The question we shall address in this section is: Can we generalisethe notion of Fourier series of f , to non-periodic functions?

The answer is a “yes”! Note that the periodicity of f is captured by theinteger k appearing in the arguments of sin and cos. To generalise the notionof Fourier series to non-periodic functions, we shall replace k, a positiveinteger, with a real number ξ. Note that when we replace k with ξ, thesequences ak, bk become functions of ξ, a(ξ) and b(ξ) and the series form isreplaced by an integral form over R.


Definition 15.5.1. If f : R → R is a piecewise continuous function whichvanishes outside a finite interval, then its Fourier integral is defined as

f(t) =

∫ ∞0

(a(ξ) cos ξt+ b(ξ) sin ξt) dξ,

where

a(ξ) =1

π

∫ ∞−∞

f(t) cos ξt dt

b(ξ) =1

π

∫ ∞−∞

f(t) sin ξt dt.

Lecture 16

Standing Waves: Separation ofVariable

The method of separation of variables was introduced by d’Alembert (1747)and Euler (1748) for the wave equation. This technique was also employedby Laplace (1782) and Legendre (1782) while studying the Laplace equationand also by Fourier while studying the heat equation.

Recall the set-up of the vibrating string given by the equation utt = uxx,we have normalised the constant c. Initially at time t, let us say the stringhas the shape of the graph of v, i.e., u(x, 0) = v(x). The snapshot of thevibrating string at each time are called the “standing waves”. The shape ofthe string at time t0 can be thought of as some factor (depending on time)of v. This observation motivates the idea of “separation of variable”, i.e.,u(x, t) = v(x)w(t), where w(t) is the factor depending on time, which scalesv at time t to fit with the shape of u(x, t).

The fact that endpoints are fixed is given by the boundary condition

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

We are also given the initial position u(x, 0) = g(x) (at time t = 0) and initialvelocity of the string at time t = 0, ut(x, 0) = h(x). Given g, h : [0, L] → Rsuch that g(0) = g(L) = 0 and h(0) = h(L), we need to solve the initial value

87

LECTURE 16. STANDING WAVES: SEPARATION OF VARIABLE 88

problemutt(x, t)− c2uxx(x, t) = 0 in (0, L)× (0,∞)

u(x, 0) = g(x) in [0, L]ut(x, 0) = h(x) in [0, L]u(0, t) = φ(t) in (0,∞)u(L, t) = ψ(t) in (0,∞),

(16.0.1)

where φ, ψ, g, h satisfies the compatibility condition

g(0) = φ(0), g′′(0) = φ′′(0), h(0) = φ′(0)

andg(L) = ψ(0), g′′(L) = ψ′′(0), h(L) = ψ′(0).

Let φ = ψ ≡ 0. Let us seek for solutions u(x, t) whose variables can beseparated. Let u(x, t) = v(x)w(t). Differentiating and substituting in thewave equation, we get

v(x)w′′(t) = c2v′′(x)w(t)

Hencew′′(t)

c2w(t)=v′′(x)

v(x).

Since RHS is a function of x and LHS is a function t, they must equal aconstant, say λ. Thus,

v′′(x)

v(x)=

w′′(t)

c2w(t)= λ.

Using the boundary condition u(0, t) = u(L, t) = 0, we get

v(0)w(t) = v(L)w(t) = 0.

If w ≡ 0, then u ≡ 0 and this cannot be a solution to (16.0.1). Hence, w 6≡ 0and v(0) = v(L) = 0. Thus, we need to solve the eigen value problem for thesecond order differential operator.

v′′(x) = λv(x), x ∈ (0, L)v(0) = v(L) = 0,

Note that the λ can be either zero, positive or negative. If λ = 0, thenv′′ = 0 and the general solution is v(x) = αx + β, for some constants α and


β. Since v(0) = 0, we get β = 0, and v(L) = 0 and L 6= 0 implies that α = 0.Thus, v ≡ 0 and hence u ≡ 0. But, this cannot be a solution to (16.0.1).

If λ > 0, then v(x) = αe√λx + βe−

√λx. Equivalently,

v(x) = c1 cosh(√λx) + c2 sinh(

√λx)

such that α = (c1 + c2)/2 and β = (c1− c2)/2. Using the boundary conditionv(0) = 0, we get c1 = 0 and hence

v(x) = c2 sinh(√λx).

Now using v(L) = 0, we have c2 sinh√λL = 0. Thus, c2 = 0 and v(x) = 0.

We have seen this cannot be a solution.Finally, if λ < 0, then set ω =

√−λ. We need to solve the simple

harmonic oscillator problemv′′(x) + ω2v(x) = 0 x ∈ (0, L)

v(0) = v(L) = 0.

The general solution is

v(x) = α cos(ωx) + β sin(ωx).

Using v(0) = 0, we get α = 0 and hence v(x) = β sin(ωx). Now usingv(L) = 0, we have β sinωL = 0. Thus, either β = 0 or sinωL = 0. Butβ = 0 does not yield a solution. Hence ωL = kπ or ω = kπ/L, for all non-zero k ∈ Z. Since ω > 0, we can consider only k ∈ N. Hence, for each k ∈ N,there is a solution (vk, λk) for the eigen value problem with

vk(x) = βk sin

(kπx

L

),

for some constant bk and λk = −(kπ/L)2. It now remains to solve w for eachof these λk. For each k ∈ N, we solve for wk in the ODE

w′′k(t) + (ckπ/L)2wk(t) = 0.


wk(t) = ak cos

(ckπt

L

)+ bk sin

(ckπt

L

).


For each k ∈ N, we have

uk(x, t) =

[ak cos

(ckπt

L

)+ bk sin

(ckπt

L

)]sin

(kπx

L

)for some constants ak and bk. The situation corresponding to k = 1 is calledthe fundamental mode and the frequency of the fundamental mode is

c√−λ1

2π=

1

2π

cπ

L=

c

2L=

√T/ρ

2L.

The frequency of higher modes are integer multiples of the fundamental fre-quency. Note that the frequency of the vibration is related to eigenvalues ofthe second order differential operator.

The general solution of (16.0.1), by principle of superposition, is

u(x, t) =∞∑k=1

[ak cos

(ckπt

L

)+ bk sin

(ckπt

L

)]sin

(kπx

L

).

Note that the solution is expressed as series, which raises the question ofconvergence of the series. Another concern is whether all solutions of (16.0.1)have this form. We ignore these two concerns at this moment.

Since we know the initial position of the string as the graph of g, we get

g(x) = u(x, 0) =∞∑k=1

ak sin

(kπx

L

).

This expression is again troubling and rises the question: Can any arbitraryfunction g be expressed as an infinite sum of trigonometric functions? An-swering this question led to the study of “Fourier series”. Let us also, asusual, ignore this concern for time being. Then, can we find the the con-stants ak with knowledge of g. By multiplying sin

(lπxL

)both sides of the

expression of g and integrating from 0 to L, we get∫ L

0

g(x) sin

(lπx

L

)dx =

∫ L

0

[∞∑k=1

ak sin

(kπx

L

)]sin

(lπx

L

)dx

=∞∑k=1

ak

∫ L

0

sin

(kπx

L

)sin

(lπx

L

)dx


Therefore, the constants ak are given as

ak =2

L

∫ L

0

g(x) sin

(kπx

L

).

Finally, by differentiating u w.r.t t, we get

ut(x, t) =∞∑k=1

ckπ

L

[bk cos

ckπt

L− ak sin

ckπt

L

]sin

(kπx

L

).

Employing similar arguments and using ut(x, 0) = h(x), we get

h(x) = ut(x, 0) =∞∑k=1

bkkcπ

Lsin

(kπx

L

)and hence

bk =2

kcπ

∫ L

0

h(x) sin

(kπx

L

).

16.1 Elliptic Equations

Theorem 16.1.1 (Laplacian in 2D Rectangle). Let Ω = (x, y) ∈ R2 | 0 <x < a and 0 < y < b be a rectangle in R2. Let g : ∂Ω → R which vanisheson three sides of the rectangle, i.e., g(0, y) = g(x, 0) = g(a, y) = 0 andg(x, b) = h(x) where h is a continuous function h(0) = h(a) = 0. Then thereis a unique solution to (9.0.1) on this rectangle with given boundary value g.

Proof. We begin by looking for solution u(x, y) whose variables are separated,i.e., u(x, y) = v(x)w(y). Substituting this form of u in the Laplace equation,we get

v′′(x)w(y) + v(x)w′′(y) = 0.

Hencev′′(x)

v(x)= −w

′′(y)

w(y).

Since LHS is function of x and RHS is function y, they must equal a constant,say λ. Thus,

v′′(x)

v(x)= −w

′′(y)

w(y)= λ.


Using the boundary condition on u, u(0, y) = g(0, y) = g(a, y) = u(a, y) =0, we get v(0)w(y) = v(a)w(y) = 0. If w ≡ 0, then u ≡ 0 which is not asolution to (9.0.1). Hence, w 6≡ 0 and v(0) = v(a) = 0. Thus, we need tosolve,

v′′(x) = λv(x), x ∈ (0, a)v(0) = v(a) = 0,

the eigen value problem for the second order differential operator. Note thatthe λ can be either zero, positive or negative.

If λ = 0, then v′′ = 0 and the general solution is v(x) = αx+ β, for someconstants α and β. Since v(0) = 0, we get β = 0, and v(a) = 0 and a 6= 0implies that α = 0. Thus, v ≡ 0 and hence u ≡ 0. But, this can not be asolution to (9.0.1).

If λ > 0, then v(x) = αe√λx + βe−

√λx. Equivalently,

v(x) = c1 cosh(√λx) + c2 sinh(

√λx)

such that α = (c1 + c2)/2 and β = (c1− c2)/2. Using the boundary conditionv(0) = 0, we get c1 = 0 and hence

v(x) = c2 sinh(√λx).

Now using v(a) = 0, we have c2 sinh√λa = 0. Thus, c2 = 0 and v(x) = 0.

We have seen this cannot be a solution.If λ < 0, then set ω =

√−λ. We need to solve

v′′(x) + ω2v(x) = 0 x ∈ (0, a)v(0) = v(a) = 0.

(16.1.1)


v(x) = α cos(ωx) + β sin(ωx).

Using the boundary condition v(0) = 0, we get α = 0 and hence v(x) =β sin(ωx). Now using v(a) = 0, we have β sinωa = 0. Thus, either β = 0or sinωa = 0. But β = 0 does not yield a solution. Hence ωa = kπ orω = kπ/a, for all non-zero k ∈ Z. Hence, for each k ∈ N, there is a solution(vk, λk) for (16.1.1), with

vk(x) = βk sin

(kπx

a

),


for some constant βk and λk = −(kπ/a)2. We now solve w corresponding toeach λk. For each k ∈ N, we solve for wk in the ODE

w′′k(y) =(kπa

)2wk(y), y ∈ (0, b)

w(0) = 0.

Thus, wk(y) = ck sinh(kπy/a). Therefore, for each k ∈ N,

uk = δk sin

(kπx

a

)sinh

(kπy

a

)is a solution to (9.0.1). The general solution is of the form (principle ofsuperposition) (convergence?)

u(x, y) =∞∑k=1

δk sin

(kπx

a

)sinh

(kπy

a

).

The constant δk are obtained by using the boundary condition u(x, b) = h(x)which yields

h(x) = u(x, b) =∞∑k=1

δk sinh

(kπb

a

)sin

(kπx

a

).

Since h(0) = h(a) = 0, the function h admits a Fourier Sine series. Thusδk sinh

(kπba

)is the k-th Fourier sine coefficient of h, i.e.,

δk =

(sinh

(kπb

a

))−12

a

∫ a

0

h(x) sin

(kπx

a

).

Theorem 16.1.2 (2D Disk). Let Ω = (x, y) ∈ R2 | x2 + y2 < R2 be thedisk of radius R in R2. Let g : ∂Ω→ R is a continuous function. Then thereis a unique solution to (9.0.1) on the unit disk with given boundary value g.

Proof. Given the nature of the domain, we shall use the Laplace operator inpolar coordinates,

∆ :=1

r

∂

∂r

(r∂

∂r

)+

1

r2

∂2

∂θ2


where r is the magnitude component and θ is the direction component. Then∂Ω is the circle of radius one. Then, solving for u(x, y) in the Dirichletproblem is to equivalent to finding U(r, θ) : Ω→ R such that

1r∂∂r

(r ∂U∂r

)+ 1

r2∂2U∂θ2

= 0 in ΩU(r, θ + 2π) = U(r, θ) in Ω

U(R, θ) = G(θ) on ∂Ω

(16.1.2)

where U(r, θ) = u(r cos θ, r sin θ), G : [0, 2π) → R is G(θ) = g(cos θ, sin θ).Note that both U and G are 2π periodic w.r.t θ. We will look for solutionU(r, θ) whose variables can be separated, i.e., U(r, θ) = v(r)w(θ) with bothv and w non-zero. Substituting it in the polar form of Laplacian, we get

w

r

d

dr

(rdv

dr

)+v

r2

d2w

dθ2= 0

and hence−rv

d

dr

(rdv

dr

)=

1

w

(d2w

dθ2

).

Since LHS is a function of r and RHS is a function of θ, they must equal aconstant, say λ. We need to solve the eigen value problem,

w′′(θ)− λw(θ) = 0 θ ∈ Rw(θ + 2π) = w(θ) ∀θ.

Note that the λ can be either zero, positive or negative. If λ = 0, thenw′′ = 0 and the general solution is w(θ) = αθ + β, for some constants α andβ. Using the periodicity of w,

αθ + β = w(θ) = w(θ + 2π) = αθ + 2απ + β

implies that α = 0. Thus, the pair λ = 0 and w(θ) = β is a solution. Ifλ > 0, then

w(θ) = αe√λθ + βe−

√λθ.

If either of α and β is non-zero, then w(θ)→ ±∞ as θ →∞, which contra-dicts the periodicity of w. Thus, α = β = 0 and w ≡ 0, which cannot be asolution. If λ < 0, then set ω =

√−λ and the equation becomes

w′′(θ) + ω2w(θ) = 0 θ ∈ Rw(θ + 2π) = w(θ) ∀θ


Its general solution is

w(θ) = α cos(ωθ) + β sin(ωθ).

Using the periodicity of w, we get ω = k where k is an integer. For eachk ∈ N, we have the solution (wk, λk) where

λk = −k2 and wk(θ) = αk cos(kθ) + βk sin(kθ).

For the λk’s, we solve for vk, for each k = 0, 1, 2, . . .,

rd

dr

(rdvkdr

)= k2vk.

For k = 0, we get v0(r) = α ln r + β. But ln r blows up as r → 0, but anysolution U and, hence v, on the closed unit disk (compact subset) has to bebounded. Thus, we must have the α = 0. Hence v0 ≡ β. For k ∈ N, we needto solve for vk in

rd

dr

(rdvkdr

)= k2vk.

Use the change of variable r = es. Then es dsdr

= 1 and ddr

= ddsdsdr

= 1es

dds

.Hence r d

dr= d

ds. vk(e

s) = αeks + βe−ks. vk(r) = αrk + βr−k. Since r−k blowsup as r → 0, we must have β = 0. Thus, vk = αrk. Therefore, for eachk = 0, 1, 2, . . .,

Uk(r, θ) = akrk cos(kθ) + bkr

k sin(kθ).


U(r, θ) =a0

2+∞∑k=1

(akr

k cos(kθ) + bkrk sin(kθ)

).

To find the constants, we must use U(R, θ) = G(θ). If G ∈ C1[0, 2π], then Gadmits Fourier series expansion. Therefore,

G(θ) =a0

2+∞∑k=1

[Rkak cos(kθ) +Rkbk sin(kθ)

]where

ak =1

Rkπ

∫ π

−πG(θ) cos(kθ) dθ,


bk =1

Rkπ

∫ π

−πG(θ) sin(kθ) dθ.

Using this in the formula for U and the uniform convergence of Fourier series,we get

U(r, θ) =1

π

∫ π

−πG(η)

[1

2+∞∑k=1

( rR

)k(cos kη cos kθ + sin kη sin kθ)

]dη

=1

π

∫ π

−πG(η)

[1

2+∞∑k=1

( rR

)kcos k(η − θ)

]dη.

Using the relation

∞∑k=1

( rR

)kcos k(η − θ) = Re

[∞∑k=1

( rRei(η−θ)

)k]= Re

[rRei(η−θ)

1− rRei(η−θ)

]

=R2 − rR cos(η − θ)

R2 + r2 − 2rR cos(η − θ)− 1

=rR cos(η − θ)− r2

R2 + r2 − 2rR cos(η − θ)

in U(r, θ) we get

U(r, θ) =R2 − r2

2π

∫ π

−π

G(η)

R2 + r2 − 2rR cos(η − θ)dη.

Note that the formula derived above for U(r, θ) can be rewritten in Carte-sian coordinates and will have the form

u(x) =R2 − |x|2

2πR

∫SR(0)

g(y)

|x− y|2dy.

This can be easily seen, by setting y = R(x10 cos η+x2

0 sin η), we get dy = Rdηand |x− y|2 = R2 + r2 − 2rR cos(η − θ). This is called the Poisson formula.More generally, the unique solution to the Dirichlet problem on a ball ofradius R centred at x0 in Rn is given by Poisson formula

u(x) =R2 − |x− x0|2

ωnR

∫SR(x0)

g(y)

|x− y|ndy.

We will derive this general form later (cf. (??)).


Theorem 16.1.3 (3D Sphere). Let Ω = (x, y, z) ∈ R3 | x2 + y2 + z2 < 1be the unit sphere in R3. Let g : ∂Ω → R is a continuous function. Thenthere is a unique solution to (9.0.1) on the unit sphere with given boundaryvalue g.

Proof. Given the nature of domain, the Laplace operator in spherical coor-dinates,

∆ :=1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sinφ

∂

∂φ

(sinφ

∂

∂φ

)+

1

r2 sin2 φ

∂2

∂θ2.

where r is the magnitude component, φ is the inclination (zenith or elevation)in the vertical plane and θ is the azimuth angle (in the direction in horizontalplane). Solving for u in (9.0.1) is equivalent to finding U(r, φ, θ) : Ω → Rsuch that

1r2

∂∂r

(r2 ∂U

∂r

)+ 1

r2 sinφ∂∂φ

(sinφ∂U

∂φ

)+ 1r2 sin2 φ

∂2U∂θ2

= 0 in Ω

U(1, φ, θ) = G(φ, θ) on ∂Ω

(16.1.3)

where U(r, φ, θ) and G(φ, θ) are appropriate spherical coordinate functioncorresponding to u and g. We will look for solution U(r, φ, θ) whose variablescan be separated, i.e., U(r, φ, θ) = v(r)w(φ)z(θ) with v, w and z non-zero.Substituting it in the spherical form of Laplacian, we get

wz

r2

d

dr

(r2dv

dr

)+

vz

r2 sinφ

d

dφ

(sinφ

dw

dφ

)+

vw

r2 sin2 φ

d2z

dθ2= 0

and hence

1

v

d

dr

(r2dv

dr

)=−1

w sinφ

d

dφ

(sinφ

dw

dφ

)− 1

z sin2 φ

d2z

dθ2.

Since LHS is a function of r and RHS is a function of (φ, θ), they must equala constant, say λ. If Azimuthal symmetry is present then z(θ) is constantand hence dz

dθ= 0. We need to solve for w,

sinφw′′(φ) + cosφw′(φ) + λ sinφw(φ) = 0, φ ∈ (0, π)

Set x = cosφ. Then dxdφ

= − sinφ.

w′(φ) = − sinφdw

dxand w′′(φ) = sin2 φ

d2w

dx2− cosφ

dw

dx


In the new variable x, we get the Legendre equation

(1− x2)w′′(x)− 2xw′(x) + λw(x) = 0 x ∈ [−1, 1].

We have already seen that this is a singular problem (while studying S-Lproblems). For each k ∈ N ∪ 0, we have the solution (wk, λk) where

λk = k(k + 1) and wk(φ) = Pk(cosφ).

For the λk’s, we solve for vk, for each k = 0, 1, 2, . . .,

d

dr

(r2dvkdr

)= k(k + 1)vk.

For k = 0, we get v0(r) = −α/r+β. But 1/r blows up as r → 0 and U mustbe bounded in the closed sphere. Thus, we must have the α = 0. Hencev0 ≡ β. For k ∈ N, we need to solve for vk in

d

dr

(r2dvkdr

)= k(k + 1)vk.

Use the change of variable r = es. Then es dsdr

= 1 and ddr

= ddsdsdr

= 1es

dds

.Hence r d

dr= d

ds. Solving for m in the quadratic equation m2 +m = k(k+ 1).

m1 = k and m2 = −k − 1. vk(es) = αeks + βe(−k−1)s. vk(r) = αrk + βr−k−1.

Since r−k−1 blows up as r → 0, we must have β = 0. Thus, vk = αrk.Therefore, for each k = 0, 1, 2, . . .,

Uk(r, φ, θ) = akrkPk(cosφ).


U(r, φ, θ) =∞∑k=0

akrkPk(cosφ).

Since we have azimuthal symmetry, G(φ, θ) = G(φ). To find the constants,we use U(1, φ, θ) = G(φ), hence

G(φ) =∞∑k=0

akPk(cosφ).

Using the orthogonality of Pk, we have

ak =2k + 1

2

∫ π

0

G(φ)Pk(cosφ) sinφ dφ.

Lecture 17

Parabolic: Heat Equation

Theorem 17.0.1 (Heat Flow on a Bar). Let Ω = (0, L) be a homogeneousrod of length L insulated along sides and its ends are kept at zero temperature.The temperature zero at the end points of the rod is given by the Dirichletboundary condition u(0, t) = u(L, t) = 0. The initial temperature of the rod,at time t = 0, is given by u(x, 0) = g(x), where g : [0, L] → R be such thatg(0) = g(L) = 0. Then there is a solution u of

ut(x, t)− c2uxx(x, t) = 0 in (0, L)× (0,∞)u(0, t) = u(L, t) = 0 in (0,∞)

u(x, 0) = g(x) on [0, L]

where c is a constant.

Proof. We begin with the ansatz that u(x, t) = v(x)w(t) (variable separated).Substituting u in separated form in the equation, we get

v(x)w′(t) = c2v′′(x)w(t)

and, hence,w′(t)

c2w(t)=v′′(x)

v(x).

Since LHS, a function of t, and RHS, a function x, are equal they must beequal to some constant, say λ. Thus,

w′(t)

c2w(t)=v′′(x)

v(x)= λ.

99

LECTURE 17. PARABOLIC: HEAT EQUATION 100

Therefore, we need to solve two ODE to obtain v and w,

w′(t) = λc2w(t) and v′′(x) = λv(x).

We first solve the eigenvalue problem involving v. For each k ∈ N, thereis a pair (λk, vk) which solves the eigenvalue problem involving v, whereλk = −(kπ)2/L2 and vk(x) = sin

(kπxL

). For each k ∈ N, we solve for wk to

getlnwk(t) = λkc

2t+ lnα,

where α is integration constant. Thus, wk(t) = αe−(kcπ/L)2t. Hence,

uk(x, t) = vk(x)wk(t) = βk sin

(kπx

L

)e−(kcπ/L)2t,

for some constants βk, is a solution to the heat equation. By superpositionprinciple, the general solution is

u(x, t) =∞∑k=1

uk(x, t) =∞∑k=1

βk sin

(kπx

L

)e−(kcπ/L)2t.

We now use the initial temperature of the rod, given as g : [0, L] → R tocompute the constants. Since u(x, 0) = g(x),

g(x) = u(x, 0) =∞∑k=1

βk sin

(kπx

L

).

Further, g(0) = g(L) = 0. Thus, g admits a Fourier Sine expansion andhence its coefficients βk are given as

βk =2

L

∫ L

0

g(x) sin

(kπx

L

).

Theorem 17.0.2 (Circular Wire). Let Ω be a circle (circular wire) of radiusone insulated along its sides. Let the initial temperature of the wire, at timet = 0, be given by a 2π-periodic function g : R→ R. Then there is a solutionu(r, θ) of

ut(θ, t)− c2uθθ(θ, t) = 0 in R× (0,∞)u(θ + 2π, t) = u(θ, t) in R× (0,∞)

u(θ, 0) = g(θ) on R× t = 0

where c is a constant.


Proof. Note that u(θ, t) is 2π-periodic in θ-variable, i.e., u(θ+2π, t) = u(θ, t)for all θ ∈ R and t ≥ 0. We begin with ansatz u(θ, t) = v(θ)w(t) withvariables separated. Substituting for u in the equation, we get

w′(t)

c2w(t)=v′′(θ)

v(θ)= λ.

For each k ∈ N∪0, the pair (λk, vk) is a solution to the eigenvalue problemwhere λk = −k2 and

vk(θ) = ak cos(kθ) + bk sin(kθ).

For each k ∈ N ∪ 0, we get wk(t) = αe−(kc)2t. For k = 0

u0(θ, t) = a0/2 (To maintain consistency with Fourier series)

and for each k ∈ N, we have

uk(θ, t) = [ak cos(kθ) + bk sin(kθ)] e−k2c2t.

Therefore, the general solution is

u(θ, t) =a0

2+∞∑k=1

[ak cos(kθ) + bk sin(kθ)] e−k2c2t.

We now use the initial temperature on the circle to find the constants. Sinceu(θ, 0) = g(θ),

g(θ) = u(θ, 0) =a0

2+∞∑k=1

[ak cos(kθ) + bk sin(kθ)] .

Further, g is 2π-periodic and, hence, admits a Fourier series expansion. Thus,

ak =1

π

∫ π

−πg(θ) cos(kθ) dθ

and

bk =1

π

∫ π

−πg(θ) sin(kθ) dθ.

Note that as t → ∞ the temperature of the wire approaches a constanta0/2.


17.1 Inhomogeneous Equation

In this section we solve the inhomogeneous heat equation, using Duhamel’sprinciple. The Duhamel’s principle states that one can obtain a solution ofthe inhomogeneous IVP for heat from its homogeneous IVP. The motivationfor Duhamel’s principle is given in Appendix C.

For a given f , let u(x, t) be the solution of the inhomogeneous heat equa-tion,

ut(x, t)− c2∆u(x, t) = f(x, t) in Ω× (0, T )

u(x, t) = 0 in ∂Ω× (0, T )u(x, 0) = 0 in Ω.

(17.1.1)

As a first step, for each s ∈ (0,∞), consider w(x, t; s) as the solution of thehomogeneous problem (auxiliary)

wst (x, t)− c2∆ws(x, t) = 0 in Ω× (s, T )

ws(x, t) = 0 in ∂Ω× (s, T )ws(x, s) = f(x, s) on Ω× s.

Since t ∈ (s, T ), introducing a change of variable r = t−s, we have ws(x, t) =w(x, t− s) which solves

wt(x, r)− c2∆w(x, r) = 0 in Ω× (0, T − s)

w(x, r) = 0 in ∂Ω× (0, T − s)w(x, 0) = f(x, s) on Ω.

Theorem 17.1.1 (Duhamel’s Principle). The function u(x, t) defined as

u(x, t) :=

∫ t

0

ws(x, t) ds =

∫ t

0

w(x, t− s) ds

solves (17.1.1).


Proof. Suppose w is C2,1(Rn × (0, T )), we get

ut(x, t) =∂

∂t

∫ t

0

w(x, t− s) ds

=

∫ t

0

wt(x, t− s) ds+ w(x, t− t)d(t)

dt

− w(x, t− 0)d(0)

dt

=

∫ t

0

wt(x, t− s) ds+ w(x, 0)

=

∫ t

0

wt(x, t− s) ds+ f(x, t).

Similarly,

∆u(x, t) =

∫ t

0

∆w(x, t− s) ds.

Thus,

ut − c2∆u = f(x, t) +

∫ t

0

(wt(x, t− s)− c2∆w(x, t− s)

)ds

= f(x, t).

Lecture 18

Travelling Waves

Consider the wave equation utt = c2uxx on R× (0,∞), describing the vibra-tion of an infinite string. We have already seen in Chapter 5 that the equationis hyperbolic and has the two characteristics x ± ct= a constant. Introducethe new coordinates w = x + ct, z = x− ct and set u(w, z) = u(x, t). Thus,we have the following relations, using chain rule:

ux = uwwx + uzzx = uw + uz

ut = uwwt + uzzt = c(uw − uz)uxx = uww + 2uzw + uzz

utt = c2(uww − 2uzw + uzz)

In the new coordinates, the wave equation satisfies uwz = 0. Integrating1

this twice, we have u(w, z) = F (w) + G(z), for some arbitrary functions Fand G. Thus, u(x, t) = F (x + ct) + G(x − ct) is a general solution of thewave equation.

Consider the case where G is chosen to be zero function. Then u(x, t) =F (x+ct) solves the wave equation. At t = 0, the solution is simply the graphof F and at t = t0 the solution is the graph of F with origin translated to theleft by ct0. Similarly, choosing F = 0 and G = F , we have u(x, t) = F (x−ct)also solves wave equation and at time t is the translation to the right of thegraph of F by ct. This motivates the name “travelling waves” and “waveequation”. The graph of F is shifted to right or left with a speed of c.

1We are assuming the function is integrable, which may be false

105

LECTURE 18. TRAVELLING WAVES 106

Now that we have derived the general form of the solution of wave equa-tion, we return to understand the physical system of a vibrating infinitestring. The initial shape (position at initial time t = 0) of the string is givenas u(x, 0) = g(x), where the graph of g on R2 describes the shape of thestring. Since we need one more data to identify the arbitrary functions, wealso prescribe the initial velocity of the string, ut(x, 0) = h(x).

Another interesting property that follows from the general solution isthat for any four points A,B,C and D that form a rectangle bounded bycharacteristic curves in R × R+ then u(A) + u(C) = u(B) + u(D) becauseu(A) = F (α) +G(β), u(C) = F (γ) +G(δ), u(B) = F (α) +G(δ) and u(D) =F (γ) +G(β).

Theorem 18.0.1. Given g ∈ C2(R) and h ∈ C1(R), there is a unique C2

solution u of the Cauchy initial value problem (IVP) of the wave equation,utt(x, t)− c2uxx(x, t) = 0 in R× (0,∞)

u(x, 0) = g(x) in Rut(x, 0) = h(x) in R,

(18.0.1)

which is given by the d’Alembert’s formula

u(x, t) =1

2(g(x+ ct) + g(x− ct)) +

1

2c

∫ x+ct

x−cth(y) dy. (18.0.2)

Proof. The general solution is u(x, t) = F (x + ct) + G(x − ct) with F,G ∈C2(R). Using the initial position we get

F (x) +G(x) = g(x).

Thus, g should be C2(R). Now, ut(x, t) = c (F ′(w)−G′(z)) and puttingt = 0, we get

F ′(x)−G′(x) =1

ch(x).

Thus, h should be C1(R). Now solving for F ′ and G′, we get 2F ′(x) =g′(x) + h(x)/c. Similarly, 2G′(x) = g′(x) − h(x)/c. Integrating2 both theseequations, we get

F (x) =1

2

(g(x) +

1

c

∫ x

0

h(y) dy

)+ c1

2assuming they are integrable and the integral of their derivatives is itself


and

G(x) =1

2

(g(x)− 1

c

∫ x

0

h(y) dy

)+ c2.

Since F (x) +G(x) = g(x), we get c1 + c2 = 0. Therefore, the solution to thewave equation is given by (18.0.2).

Aliter. Let us derive the d’Alembert’s formula in an alternate way. Notethat the wave equation can be factored as(

∂

∂t+ c

∂

∂x

)(∂

∂t− c ∂

∂x

)u = utt − c2uxx = 0.

We set v(x, t) =(∂∂t− c ∂

∂x

)u(x, t) and hence

vt(x, t) + cvx(x, t) = 0 in R× (0,∞).

Notice that the above first order PDE obtained is in the form of homogeneouslinear transport equation (cf. (??)), which we have already solved. Hence,for some smooth function φ,

v(x, t) = φ(x− ct)

and φ(x) := v(x, 0). Using v in the original equation, we get the inhomoge-neous transport equation,

ut(x, t)− cux(x, t) = φ(x− ct).

Recall the formula for inhomogenoeus transport equation (cf. (??))

u(x, t) = g(x− at) +

∫ t

0

φ(x− a(t− s), s) ds.

Since u(x, 0) = g(x) and a = −c, in our case the solution reduces to,

u(x, t) = g(x+ ct) +

∫ t

0

φ(x+ c(t− s)− cs) ds

= g(x+ ct) +

∫ t

0

φ(x+ ct− 2cs) ds

= g(x+ ct) +−1

2c

∫ x−ct

x+ct

φ(y) dy

= g(x+ ct) +1

2c

∫ x+ct

x−ctφ(y) dy.


But φ(x) = v(x, 0) = ut(x, 0) − cux(x, 0) = h(x) − cg′(x) and substitutingthis in the formula for u, we get

u(x, t) = g(x+ ct) +1

2c

∫ x+ct

x−ct(h(y)− cg′(y)) dy

= g(x+ ct) +1

2(g(x− ct)− g(x+ ct))

+1

2c

∫ x+ct

x−cth(y) dy

=1

2(g(x− ct) + g(x+ ct)) +

1

2c

∫ x+ct

x−cth(y) dy

For c = 1, the d’Alembert’s formula takes the form

u(x, t) =1

2(g(x− t) + g(x+ t)) +

1

2

∫ x+t

x−th(y) dy.

A useful observation from the d’Alembert’s formula is that the regularity ofu is same as the regularity of its initial value g.

18.1 Domain of Dependence and Influence

Note that the solution u(x, t) depends only on the interval [x − ct, x + ct]because g takes values only on the end-points of this interval and h takesvalues between this interval. The interval [x− ct, x+ ct] is called the domainof dependence. Thus, the region of R × (0,∞) on which the value of u(x, t)depends forms a triangle with base [x− ct, x + ct] and vertex at (x, t). Thedomain of dependence of (x, t) is marked in x-axis by the characteristic curvespassing through (x, t).

Given a point p on the x-axis what values of u on (x, t) will depend onthe value of g(p) and h(p). This region turns out to be a cone with vertexat p and is called the domain of influence. The domain of influence is theregion bounded by the characteristic curves passing through p.

If the initial data g and h are supported in the interval Bx0(R) then thesolution u at (x, t) is supported in the region Bx0(R+ ct). Consequently, if gand h have compact support then the solution u has compact support in Rfor all time t > 0. This phenomenon is called the finite speed of propagation.

Appendices

109

Appendix A

Divergence Theorem

Definition A.0.1. For an open set Ω ⊂ Rn, its boundary ∂Ω is said to be Ck

(k ≥ 1) if, for every point x ∈ ∂Ω, there is a r > 0 and a Ck diffeomorphism1

φ : Br(x)→ B1(0) such that

1. φ(∂Ω ∩Br(x)) ⊂ B1(0) ∩ x ∈ Rn | xn = 0 and

2. φ(Ω ∩Br(x)) ⊂ B1(0) ∩ x ∈ Rn | xn > 0

The boundary ∂Ω is said to be C∞ if ∂Ω is Ck, for all k ∈ N, and ∂Ω isanalytic if φ is analytic.

Equivalently, ∂Ω is Ck if, for every point x ∈ ∂Ω, there exists a neigh-bourhood Ux of x and a Ck function φ : Rn−1 → R such that

Ω ∩Bx = x ∈ Bx | xn > φ(x1, x2, . . . , xn−1).

Theorem A.0.2. Let Ω be an open bounded subset of Rn with C1 boundary.If v ∈ C1(Ω) then ∫

Ω

∂v

∂xidx =

∫∂Ω

vνi dσ

where ν = (ν1, . . . , νn) is the unit vector pointing outward and dσ is thesurface measure of ∂Ω.

The hypothesis that Ω is bounded can be relaxed provided |v| and∣∣∣ ∂v∂xi ∣∣∣

decays as |x| → ∞. Much weaker hypotheses on ∂Ω and v are considered ingeometric measure theory.

1φ−1 exists and both φ and φ−1 are k-times continuously differentiable

111

APPENDIX A. DIVERGENCE THEOREM 112

Theorem A.0.3 (Integration by parts). Let Ω be an open bounded subset ofRn with C1 boundary. If u, v ∈ C1(Ω) then∫

Ω

u∂v

∂xidx+

∫Ω

v∂u

∂xidx =

∫∂Ω

uvνi dσ.

Hint. Set v := uv in the theorem above.

Theorem A.0.4 (Gauss). Let Ω be an open bounded subset of Rn with C1

boundary. If V = (v1, . . . , vn) on Ω is a vector field such that vi ∈ C1(Ω), forall 1 ≤ i ≤ n, then ∫

Ω

∇ · V dx =

∫∂Ω

V · ν dσ. (A.0.1)

The divergence of a vector field is the measure of the magnitude (outgoingnature) of all source (of the vector field) and absorption in the region. Thedivergence theorem was discovered by C. F. Gauss in 18132 which relates theoutward flow (flux) of a vector field through a closed surface to the behaviourof the vector field inside the surface (sum of all its “source” and “sink”). Thedivergence theorem is the mathematical formulation of the conservation law.

Theorem A.0.5 (Green’s Identities). Let Ω be an open bounded subset ofRn with C1 boundary. If u, v ∈ C2(Ω) then

(i) ∫Ω

(v∆u+∇v · ∇u) dx =

∫∂Ω

v∂u

∂νdσ,

where ∂u∂ν

:= ∇u · ν;

(ii) ∫Ω

(v∆u− u∆v) dx =

∫∂Ω

(v∂u

∂ν− u∂v

∂ν

)dσ.

Hint. Apply divergence theorem to V = v∇u to get the first formula. To getsecond formula apply divergence theorem for both V = v∇u and V = u∇vand subtract one from the other.

2J. L. Lagrange might have discovered this, before Gauss, in 1762

Appendix B

Normal Vector of a Surface

Let S(x, y, z) = 0 be the equation of a surface G in R3. Fix p0 = (x0, y0, z0) ∈G. What is the normal vector at p0? Fix an arbitrary curve C lying on Gand passing through p0. Let r(t) = (x(t), y(t), z(t)) be the parametric formof C with r(t0) = p0. Since C lies on G, S(r(t)) = S(x(t), y(t), z(t)) = 0, forall t. Differentiating w.r.t t (using chain rule),

∂S

∂x

dx(t)

dt+∂S

∂y

dy(t)

dt+∂S

∂z

dz(t)

dt= 0

(Sx, Sy, Sz) · (x′(t), y′(t), z′(t)) = 0

∇S(r(t)) · r′(t) = 0.

In particular, ∇S(p0) · r′(t0) = 0. Since r′(t0) is the slope of the tangent,at t0, to the curve C, the vector∇S(p0) is perpendicular to the tangent vectorat p0. Since the argument is valid for any curve in G that passes throughp0, ∇S(p0) is normal vector to the tangent plane at p0. If, in particular,the equation of the surface is given as S(x, y, z) = u(x, y) − z, for someu : R2 → R, then

∇S(p0) = (Sx(p0), Sy(p0), Sz(p0))

= (ux(x0, y0), uy(x0, y0),−1) = (∇u(x0, y0),−1).

113

APPENDIX B. NORMAL VECTOR OF A SURFACE 114

Appendix C

Duhamel’s Principle

Consider the first order inhomogeneous ODEx′(t) + ax(t) = f(t) in (0,∞)

x(0) = x0.(C.0.1)

Multiplying the integration factor eat both sides, we get

[eatx(t)]′ = eatf(t)

and

x(t) = e−at∫ t

0

easf(s) ds+ ce−at.

Using the initial condition x(0) = x0, we get

x(t) = x0e−at +

∫ t

0

ea(s−t)f(s) ds.

Notice that x0e−at is a solution of the homogeneous ODE. Thus, the solution

x(t) can be given as

x(t) = S(t)x0 +

∫ t

0

S(t− s)f(s) ds

where S(t) is a solution operator of the linear equation, given as S(t) = e−at.Consider the second order inhomogeneous ODE

x′′(t) + a2x(t) = f(t) in (0,∞)x(0) = x0

x′(0) = x1.(C.0.2)

115

APPENDIX C. DUHAMEL’S PRINCIPLE 116

We introduce a new function y such that

x′(t) = ay(t).

Then

y′(t) =f(t)

a− ax(t)

and the second order ODE can be rewritten as a system of first order ODE

X ′(t) + AX(t) = F (t)

where X = (x, y), F = (0, f/a) and

A =

(0 −aa 0

)with the initial condition X0 := X(0) = (x0, x1/a). We introduce the matrix

exponential eAt =∑∞

n=1(At)n

n!. Then, multiplying the integration factor eAt

both sides, we get[eAtX(t)]′ = eAtF (t)

and

X(t) = X0e−At +

∫ t

0

eA(s−t)F (s) ds.

Notice that X0e−At is a solution of the homogeneous ODE. Thus, the solution

X(t) can be given as

X(t) = S(t)X0 +

∫ t

0

S(t− s)F (s) ds

where S(t) is a solution operator of the linear equation, given as S(t) = e−At.

Bibliography

117

BIBLIOGRAPHY 118

Index

characteristic curve, 14

directional derivative, 2divergence, 3

elliptic PDE, 25equation

velocity potential, 26heat, 26Laplace, 26tricomi, 26wave, 26

Gauss divergence result, 112gradient, 3Green’s identities, 112

Hadamard, 9wellposed, 9

harmonic function, 45Hessian matrix, 3hyperbolic PDE, 25

integral curve, 14integral surface, 14

Laplace operator, 41Laplace-Beltrami operator, 43Laplacian, 3

maximum principlestrong, 48weak, 47

method of characteristics, 14

Neumann boundary condition, 50Neumann Problem, 50

parabolic PDE, 25

tensor, 3

119

partial di erential equations - iit kanpurhome.iitk.ac.in/~tmk/courses/mth203/mso203b.pdf · 7.2...

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