pde . quing

8/16/2019 PDE . QUING

http://slidepdf.com/reader/full/pde-quing 1/203

An Introduction to

Linear Partial Differential Equations

Qing Han

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


To Yansu, Raymond and Tommy

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


Contents

Preface vii

Chapter 1. Introduction 11.1. Notations 1

1.2. Well-Posed Problems 3

1.3. Overview 4

Chapter 2. First-Order Differential Equations 72.1. Non-Characteristic Hypersurfaces 72.2. The Method of Characteristics 112.3. A Priori Estimates 23

Exercises 34

Chapter 3. An Overview of Second-Order PDEs 37

3.1. Classifications of Second Order Equations 373.2. Energy Estimates 433.3. Separation of Variables 49Exercises 56

Chapter 4. Laplace Equations 59

4.1. Dirichlet Problems 594.2. Fundamental Solutions 654.3. Mean-Value Properties 714.4. The Maximum Principle 76

4.5. Schauder Estimates 864.6. Weak Solutions 91Exercises 97

Chapter 5. Heat Equations 1015.1. Initial-Value Problems 1015.2. Regularity of Solutions 112

5.3. The Maximum Principle 1175.4. Harnack Inequalities 123Exercises 129

Chapter 6. Wave Equations 1336.1. One-Dimensional Wave Equations 1336.2. Higher Dimensional Wave Equations 142

6.3. Energy Estimates 153Exercises 160

v

8/16/2019 PDE . QUING


vi CONTENTS

Chapter 7. First-Order Differential Systems 165

7.1. Reductions to First-Order Differential Systems 165

7.2. Cauchy-Kowalevski Theorem 1717.3. Hyperbolic Differential Systems 180Exercises 188

Bibliography 191

Index 193

8/16/2019 PDE . QUING


Preface

Is it really necessary to classify partial differential equations (PDEs) and toemploy different methods to discuss different types of equations? Why is it im-

portant to derive a priori estimates of solutions before even proving the existenceof solutions? These are only a few questions any students who just start studyingPDEs might ask. Students may find answers to these questions only at the end of a

one-semester course in basic PDEs, sometimes after they have already lost interestin the subject. In this book, we attempt to address these issues at the beginning .There are several notable features in this book.

First, the importance of a priori estimates is addressed at the beginning andemphasized throughout this book. This is well illustrated by the first chapter on

first-order PDEs. Although first-order linear PDEs can b e solved by the method of characteristics, we provide a detailed analysis of a priori estimates of solutions insup-norms and in integral norms. To emphasize the importance of these estimates,we demonstrate how to prove the existence of weak solutions with the help of basic

results from functional analysis. The setting here is easy, since Sobolev spaces arenot required. Meanwhile, all important ideas are in full display. In this book, wedo attempt to derive explicit expressions for solutions whenever possible. However,

these explicit expressions of solutions of special equations usually serve mostly tosuggest the correct form of estimates for solutions of general equations.

The second feature is the illustration of the necessity of classifying second-order PDEs at the beginning. In the second chapter, immediately after classifyingsecond order PDEs into elliptic, parabolic and hyperbolic type, we discuss vari-

ous boundary-value problems and initial/boundary-value problems for the Laplaceequation, the heat equation and the wave equation. We discuss energy methods forproving uniqueness and find solutions in the plane by the method of separation of variables. The explicit expressions of solutions demonstrate different properties of

solutions of different types of PDEs. Such differences clearly indicate that there isunlikely to be a unified approach to study PDEs.

Third, we focus on simple models of PDEs and study these equations in detail.

We have chapters devoted to the Laplace equation, the heat equation and the waveequation, and use several methods to study each equation. For example, for theLaplace equation, we use three different methods to study harmonic functions, thefundamental solution, the mean-value property and the maximum principle. Foreach method, we indicate its advantages and its shortcomings. General equationsare not forgotten. We also discuss maximum principles for general elliptic and

parabolic equations and energy estimates for general hyperbolic equations.

vii

8/16/2019 PDE . QUING


viii PREFACE

The book is designed for a one-semester course at the graduate level. As a

result, many important topics have to be omitted. One topic notably missing from

this book is Sobolev spaces. It is hard, if not impossible, to introduce Sobolevspaces and discuss applications in a one-semester course on basic PDEs. How-ever, we seize every opportunity we can to address the importance of the missing

topic. For example, in Chapter 3 we solve initial/boundary-value problems for the1-dimensional heat equation and for the 1-dimensional wave equation by using sep-aration of variables. An important role is played by the eigenvalue problem for

the operator d2

dx2 over an interval. After successfully solving 1-dimensional prob-

lems, we point out that, in order to solve higher dimensional versions, we need tosolve first the eigenvalue problem for the Laplace operator in a bounded smoothdomain. For that, Sobolev spaces play an essential role. Without filling in details,we introduce Sobolev spaces and discuss how to find weak solutions of the Dirichlet

problem of the Poisson equation in Section 4.6.The choice of topics in this book, as in any books, is influenced by personaltastes of the author. Some topics in this book may not be viewed as basic byothers. For example, differential Harnack inequality for the heat equation usually

is not found in PDE textbooks at a comparable level. It is included here simplybecause of its importance. Attempts have been made to give a balanced coverageof different classes of partial differential equations.

Acknowledgments. This book is based on the one-semester course the author

taught at University of Notre Dame in the fall of 2007. The author would like tothank Ms. Hairong Liu for a wonderful job of typing the manuscript.

Qing Han

8/16/2019 PDE . QUING


CHAPTER 1

Introduction

This chapter serves as an introduction of entire book.

1.1. Notations

Most of the time, we denote by x points in Rn and write x = (x1, · · · , xn) in

terms of its coordinates. For any x

∈Rn, we denote by

|x

| the standard Euclidean

norm, unless otherwise stated. Namely, for any x = (x1, · · · , xn), we have

|x| = n

i=1

x2i

12 .

Sometimes, we need to distinguish one particular direction as the time directionand write points in Rn+1 by (x, t) for x ∈ Rn and t ∈ R. In this case, we call

x = (x1, · · · , xn) ∈ Rn the space variable and t ∈ R the time variable. In R2, wealso denote points by (x, y).

Let Ω be a domain in Rn, an open and connected subset in Rn. We denote byC (Ω) the collection of all continuous functions in Ω, by C m(Ω) the collection of all

functions with continuous derivatives up to order m, for any integer m ≥ 1, and byC ∞(Ω) the collection of all functions with continuous derivatives of arbitrary order.

For any u ∈ C m(Ω), we denote by ∇mu the collection of all partial derivatives of uof order m. In particular,

∇u = (ux1 , · · · , uxn).

This is the gradient of u. For second derivatives, we usually write in the matrixform as follows

∇2u =

ux1x1 ux1x2 · · · ux1xn

ux2x1 ux2x2 · · · ux2xn

......

. . . ...

uxnx1 uxnx2 · · · uxnxn

.

This is a symmetric matrix, called the Hessian matrix of u. For derivatives of orderhigher than two, we need to use multi-indices. A multi-index α

∈ Zn

+ is given by

α = (α1, · · · , αn) for nonnegative integers α1, · · · , αn. We write

|α| =

ni=1

αi.

For any vector ξ = (ξ 1, · · · , ξ n) ∈ Rn, we denote

ξ α = ξ α1

1 · · · ξ αnn .

The partial derivative ∂ αu is defined by

∂ αu = ∂ α1x1

· · · ∂ αnxn

u,

1

8/16/2019 PDE . QUING


2 1. INTRODUCTION

and its order is |α|. For any positive integer m, we define

|∇m

u| = |α|=m |∂

α

u|2 1

2

.

In particular,

|∇u| = n

i=1

u2xi

12 ,

and

|∇2u| = n

i,j=1

u2xixj

12 .

A hypersurface in Rn is a surface of n−1 dimension. Locally, a C m-hypersurfacecan be expressed by ϕ = 0 for a C m-function ϕ with ∇ϕ = 0. Alternatively, bya rotation, we may take ϕ(x) = xn

−ψ(x1,

· · · , xn−1) for a C m-function ψ of n

−1

variables. A domain Ω ⊂ Rn is C m if its boundary ∂ Ω is an C m-hypersurface.A partial differential equation (henceforth abbreviated as PDE) in a domain

Ω ⊂ Rn is a relation of independent variables x ∈ Ω, an unknown function udefined in Ω and a finite number of its partial derivatives. Solving a PDE is to find

this unknown function. The order of a PDE is the order of the highest derivativein the relation. Hence for a positive integer m, the general form of an m-th orderPDE in a domain Ω ⊂ Rn is given by

F

x,u, ∇u(x), ∇2u(x), · · · , ∇mu(x)

= 0 for any x ∈ Ω.

Here F is a function which is continuous in all its arguments and u is a C m-function in Ω. A C m-solution u satisfying the above equation in the pointwise

sense in Ω is often called a classical solution . Sometimes, we need to relax regularity

requirements for solutions when classical solutions are not known to exist. Insteadof going to details, we only mention that it is an important method to establishthe existence of weak solutions first, functions with less regularity than C m and

satisfying the equation in some weak sense, and then to prove that these weaksolutions actually possess required regularity to be classical solutions.

A PDE is linear if it is linear in the unknown functions and their derivatives,with coefficients depending on independent variables x. A general m-th order linear

PDE in Ω is given by |α|≤m

aα(x)∂ αu = f (x) for x ∈ Ω.

Here aα is called the coefficient of ∂ αu. A PDE of order m is quasilinear if it is linear

in derivatives of solutions of order m, with coefficients depending on independentvariables x and derivatives of solutions of order < m. In general, an m-th orderquasilinear PDE in Ω is given by

|α|=m

aα(x,u, · · · , ∇m−1u)∂ αu = f (x,u, · · · , ∇m−1u) for x ∈ Ω.

Several PDEs involving one or more unknown functions and their derivatives

form a differential system. We define linear and quasilinear partial differentialsystems accordingly.

8/16/2019 PDE . QUING


1.2. WELL-POSED PROBLEMS 3

In this book, we will focus on first-order and second-order linear PDEs and

first-order linear differential systems. In a few occasions, we diverge to nonlinear

PDEs.

1.2. Well-Posed Problems

What is the meaning of solving partial differential equations? Ideally, we obtain

explicit solutions in terms of elementary functions. In practice this is only possiblefor very simple PDEs or very simple solutions of more general PDEs. In general, itis impossible to find explicit expressions of all solutions of all PDEs. In the absenceof explicit solutions, we need to seek methods to prove existence of solutions of

PDEs and discuss properties of these solutions.A given PDE may not have solutions at all or may have many solutions. When it

has many solutions, we intend to find side conditions to pick up the most reasonablesolutions. Hadamard introduced the notion of wel l-posed problems . A PDE and side

conditions are called well-posed if (i) there admits a solution;(ii) this solution is unique;(iii) solutions depend continuously in some suitable sense on side conditions,

i.e., solutions change a little if side conditions change a little.Now the basic question can be formulated as follows: Given an equation, find

side conditions to have the well-posedness. We usually refer to (i), (ii) and (iii) asthe existence, uniqueness and the continuous dependence respectively.

In practice, both the uniqueness and the continuous dependence are proved by a priori estimates . Namely, we assume solutions already exist and then derive certainnorms of solutions in terms of known functions in the equation and side conditions.It is important to note that establishing a priori estimates is in fact the first step

in proving the existence of solutions. A closely related issue here is regularity of solutions such as continuity and differentiability. Solutions of a particular PDEcan only be obtained if the right kind of regularity, or the right kind of norms, areemployed. Two classes of norms are used often, sup-norms and L2-norms.

Let Ω be a domain in Rn. For any bounded function u in Ω, we define thesup-norm of u in Ω by

|u|L∞(Ω) = supΩ

|u|.Let m be a nonnegative integer. For any function u in Ω with bounded derivativesup to order m, we define the C m-norm of u in Ω by

|u|C m(Ω) =

|α|≤m

|∂ αu|L∞(Ω).

If Ω is a bounded C m-domain in Rn, then C m(Ω), the collection of functions which

are C m in Ω, is a Banach space equipped with the C m-norm.Next, for any Lebesgue measurable function u in Ω, we define the L2-norm of

u in Ω by

uL2(Ω) =

Ω

u2 12 ,

where integration is in the Lebesgue sense. The L2-space in Ω is the collectionof all Lebesgue measurable functions in Ω with finite L2-norms and is denoted by

8/16/2019 PDE . QUING


4 1. INTRODUCTION

L2(Ω). We learned from real analysis that L2(Ω) is a Banach space equipped with

the L2-norm.

Other norms will also be used. We will introduce them when needed.In deriving a priori estimates, we follow a common practice and use the variable

constant convention. The same letter C is used to denote constants which may

change from line to line, as long as it is clear from context on what quantities theconstants depend upon. In most cases, we are not interested in the value of theconstant, but only in its existence.

1.3. Overview

There are 7 chapters in this book.

The main topic in Chapter 2 is first-order PDEs. In Section 2.1, we introducethe basic notion of characteristic hypersurfaces. In Section 2.2, we use the method

of integral curves to study initial value problems with initial values prescribed onnon-characteristic hypersurfaces. In Section 2.3, we derive estimates of solutions in

L∞-norms and in L2-norms without using explicit expressions of solutions.Chapter 3 should be considered as an introduction to the theory of second-order

linear PDEs. In Section 3.1, we introduce the Laplace equation, the heat equationand the wave equation. We also introduce their general forms, elliptic equations,

parabolic equations and hyperbolic equations, which will be studied in detail insubsequent chapters. We use energy methods to discuss the uniqueness of certainboundary value problems in Section 3.2 and use separation of variables to solvethese problems in the plane in Section 3.3.

In Chapter 4, we discuss the Laplace equation and the Poisson equation. TheLaplace equation is probably the most important PDE with the widest range of

applications. In Section 4.1, we solve Dirichlet problems for the Laplace equa-tion in balls and derive Poisson integral formula. In the next three sections, we

study harmonic functions, (i.e., solutions of the Laplace equation), by three dif-ferent methods: fundamental solutions, mean-value properties and the maximumprinciple. These three sections are relatively independent of each other. In Section4.5, we study the Poisson equation ∆u = f and derive Schauder estimates for its

solutions. Then in Section 4.6, we briefly discuss weak solutions of the Poissonequation.

In Chapter 5, we study the heat equation and discuss properties of its solutions.This equation describes the temperature of a body conducting heat, when the

density is constant, In Section 5.1, we discuss the fundamental solution of the heatequation and solve some initial-value problems. Then in Section 5.2, we use the

fundamental solution to discuss regularity of arbitrary solutions. In Section 5.3,we discuss the maximum principle for the heat equation and its applications. In

particular, we use the maximum principle to derive interior gradient estimates. InSection 5.4, we discuss Harnack inequalities.

In Chapter 6, we study the n-dimensional wave equation, which representsvibrations of strings or propagation of sound waves in tubes for n = 1, waves onthe surface of shallow water for n = 2, and acoustic or light waves for n = 3. In

Section 6.1, we discuss initial-value problems and various initial/boundary-valueproblems for the one-dimensional wave equation. In Section 6.2, we study initial

8/16/2019 PDE . QUING


1.3. OVERVIEW 5

value problems for the wave equation in higher dimensional spaces. Then in Section

6.3, we derive energy estimates for solutions of initial-value problems. Chapter 6 is

relatively independent of Chapter 4 and Chapter 5 and can be taught after Chapter3.

In Chapter 7, we discuss partial differential systems of first-order and focus

on non-characteristic initial-value problems. In Section 7.1, we introduce non-characteristic hypersurfaces for partial differential equations and systems of ar-bitrary order. We also demonstrate that partial differential systems of arbitraryorder can always be changed to those of first-order. In Section 7.2, we discuss the

Cauchy-Kowalevski Theorem, which asserts the existence of analytic solutions of non-characteristic initial-value problems for analytic differential systems and initialvalues on analytic non-characteristic hypersurfaces. In Section 7.3, we discuss hy-perbolic differential systems and derive energy estimates for symmetric hyperbolic

differential systems.

Each chapter, except this one, ends with exercises. Level of difficulty variesconsiderably. Some exercises, at the most difficult level, may require efforts fordays.

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


CHAPTER 2

First-Order Differential Equations

The main topic in this chapter is first-order PDEs. In Section 2.1, we intro-duce the basic notion of non-characteristic hypersurfaces. In Section 2.2, we use

the method of characteristics to study initial value problems with initial valuesprescribed on non-characteristic hypersurfaces. In Section 2.3, we derive estimatesof solutions in L∞-norms and in L2-norms without using explicit expressions of

solutions.

2.1. Non-Characteristic Hypersurfaces

Let Ω be a domain in Rn. A first-order PDE in Ω is given by

(2.1) F

x,u, ∇u(x)

= 0 for any x ∈ Ω,

where F = F (x,u,p) is continuous in (x,u,p) ∈ Ω ×R×Rn and u is a C 1-functionin Ω.

We start our study with the simplest case. Given a point in Ω, say the origin,

find a reasonable condition so that (2.1) admits a solution in a neighborhood of theorigin. Let Σ be a hypersurface in Rn, a surface of dimension n − 1, passing the

origin. We will prescribe conditions on Σ to find a solution.To illustrate, we consider first-order linear PDEs. Let Ω be a domain in Rn

containing the origin and L be a first-order linear differential operator given by

Lu =n

i=1

ai(x)uxi + b(x)u,

where ai are b are continuous functions in Ω, for any i = 1, · · · , n. Here ai and bare called the coefficients for uxi

and u respectively.For a given function f in Ω, we consider the equation

(2.2) Lu = f (x) in Ω.

The function f is called the non-homogeneous term . If f

≡ 0, (2.2) is called a

homogeneous equation.Let Σ be the hyperplane xn = 0. For x ∈ Rn, we write x = (x, xn) for

x = (x1, · · · , xn−1) ∈ Rn−1. We intend to prescribe u on Σ to solve (2.2) in a

neighborhood of the origin. We first check whether we can find all derivatives of uat the origin. For a given function u0 in a neighborhood of the origin in Rn−1, weset

(2.3) u(x, 0) = u0(x) for any small x ∈ Rn−1.

Then we can find all x-derivatives of u at the origin. In particular, we can

7

8/16/2019 PDE . QUING


8 2. FIRST-ORDER DIFFERENTIAL EQUATIONS

Figure 2.1. Initial values.

determine all first order derivatives of u at the origin except uxn. To find this, we

need to use the equation. If we assume

an(0) = 0,

then we can find uxn (0) from (2.2). In fact, we can compute all derivatives of u of any order at the origin by using u0 and differentiating (2.2).

We usually call Σ an initial hypersurface and u0 an initial value or a Cauchyvalue. The problem of solving (2.2) together with (2.3) is called an initial-valueproblem or a Cauchy problem.

More generally, consider a hypersurface Σ given by ϕ = 0 for a C 1-functionϕ in a neighborhood of the origin with ∇ϕ = 0, with the origin on the hypersurfaceΣ, i.e., ϕ(0) = 0. We note that ∇ϕ is simply a normal vector of the hypersurface Σ.Without loss of generality, we assume ϕxn

(0) = 0. Then by the implicit function

theorem, we solve ϕ = 0 around x = 0 for xn = ψ(x1, · · · , xn−1). We consider achange of variables

x

→y = x1,

· · · , xn−1, ϕ(x).

This is a well defined transform in a neighborhood of the origin with a nonsingular

Jacobian. In fact, the Jacobian J is given by

J = ∂ (y1, · · · , yn)

∂ (x1, · · · , xn) =

0

Id...

0ϕx1 · · · ϕxn−1 ϕxn

,

and hence det J (0) = ϕxn(0) = 0. Now we write the operator L in new variables.

Note

uxi =

n

k=1

yk,xiuyk

.

Therefore,

Lu =

nk=1

ni=1

ai(x(y))yk,xi

uyk

+ b(x(y))u.

In new coordinates, the initial hypersurface Σ is given by yn = 0 and the coeffi-cient of uyn

is given byn

i=1

ai(x)ϕxi.

8/16/2019 PDE . QUING


2.1. NON-CHARACTERISTIC HYPERSURFACES 9

We recall that ∇ϕ = (ϕx1 , · · · , ϕxn) is simply a normal vector of Σ = ϕ = 0.

When Σ = xn = 0, or ϕ(x) = xn, then ∇ϕ = (0, · · · , 0, 1) and

ni=1

ai(x)ϕxi = an(x).

This reduces to the special case we just discussed.

Definition 2.1. For a linear operator L as in (2.2) defined in a neighborhoodof x0 ∈ Rn, a C 1-hypersurface Σ is non-characteristic at x0 ∈ Σ if

(2.4)

ni=1

ai(x0)ξ i = 0,

where ξ = (ξ 1, · · · , ξ n) is a normal vector of Σ at x0. Otherwise, it is calledcharacteristic at x0.

Figure 2.2. Non-characteristic hypersurfaces.

A hypersurface is non-characteristic if it is non-characteristic at every point.Strictly speaking, a hypersurface is characteristic if it is not non-characteristic, i.e.,if it is characteristic at some point. In this book, we will abuse this notion. Whenwe say a hypersurface is characteristic, we mean it is characteristic everywhere .

This should cause few confusions. When n = 2, hypersurfaces are simply curves. Itis more appropriate to call them characteristic curves and non-characteristic curves.

The non-characteristics condition has a simple geometric interpretation. If we

view a = (a1, · · · , an) as a vector in Rn, then (2.4) holds if and only if a(x0) isnot a tangent vector to Σ at x0. We note that condition (2.4) is maintained underC 1-changes of local coordinates. This condition assures that we can compute allderivatives of solutions at x0.

The concept of the non-characteristics can also be generalized to linear partialdifferential systems. Consider in a neighborhood of x0 ∈ Rn

ni=1

Ai(x)uxi + B(x)u = f (x),

where Ai and B are N × N matrices and u and f are N -column vectors. A

hypersurface Σ is non-characteristic at x0 ∈ Σ if

det n

i=1

ξ iAi(x0) = 0,

8/16/2019 PDE . QUING



where ξ = (ξ 1, · · · , ξ n) is a normal vector of Σ at the origin.

Next, we turn to general nonlinear partial differential equations

(2.5) F (x,u, ∇u) = 0.

We ask the same question as for linear equations. Given a hypersurface Σ and aninitial value on it, can we compute all derivatives of solutions? First, we considerthe case of hyperplane xn = 0.

Example 2.2. Considern

i=1

u2xi

= 1,

and

u(x, 0) = u0(x).

It is obvious that u = xi is a solution for u0(x) = xi, i = 1, · · · , n − 1. However, if |∇xu0(x)|2 > 1, there are no solutions for such an initial value.

Hence, in the case of nonlinear PDEs, the concept of non-characteristics alsodepends on initial values.

Suppose an initial value is given on Σ = xn = 0 by

u(x, 0) = u0(x).

We ask whether (2.5) admits a solution in a neighborhood of the origin with thegiven initial value. To answer this question, we first assume that there is a C 1-

function v in a neighborhood of the origin having the given initial value and satis-

fying F = 0 at the origin, i.e.,F

0, v(0), ∇v(0)

= 0.

As in the discussion of linear PDEs, we ask whether we can find uxn at the origin.

By the implicit function theorem, this is possible if

F uxn

0, v(0), ∇v(0)

= 0.

Now we reconsider the equation in Example 2.2. We set

F (x,u,p) = | p|2 − 1 for any p ∈ Rn.

Suppose the initial value u0 satisfies

|∇xu0(0)| < 1.Let v = u0 +cxn for a constant to be determined. Then |∇v(0)|2 = |∇xu0(0)|2 +c2.By choosing

c = ±

1 − |∇xu0(0)|2 = 0,

v satisfies the equation at x = 0. For such two choices of v , we have

F uxn(0, v(0), ∇v(0)) = 2vxn

(0) = 2c = 0.

Hence, the notion of non-characteristics depends on the choice of v .

8/16/2019 PDE . QUING


2.2. THE METHOD OF CHARACTERISTICS 11

Definition 2.3. Let F = 0 be a first-order nonlinear PDE as in (2.5) in

a neighborhood of x0 ∈ Rn and Σ be a hypersurface. With an initial value u0

prescribed on Σ, suppose there exists a function v such that v = u0 on Σ andF (x0, v(x0), ∇v(x0)) = 0. Then the hypersurface Σ is non-characteristic at x0 ∈ Σwith respect to the given initial value u0 and with respect to v if

ni=1

F uxi

x0, v(x0), ∇v(x0)

ξ i = 0,

where ξ = (ξ 1, · · · , ξ n) is a normal vector of Σ at x0.

2.2. The Method of Characteristics

In this section, we use the method of characteristics to solve first-order PDEs.

We demonstrate that solutions of any first-order PDEs with initial values prescribedon non-characteristic hypersurfaces can be obtained by solving systems of OrdinaryDifferential Equations (ODEs). This is not true for higher-order PDEs or for sys-

tems of first-order PDEs.Let Ω ⊂ Rn be a domain. The general form of first-order PDEs in Ω is given

by

F (x,u, ∇u) = 0 for x ∈ Ω,

where F is continuous in Ω ×R×Rn. Let Σ be a hypersurface in Rn and an initialvalue be prescribed on Σ by

u = u0 on Σ.

In the following, we assume Ω is a domain containing the origin and Σ is non-

characteristic at the origin.As shown in the previous section, for nonlinear differential equations, the non-

characteristics condition depends on initial values. For example, for a quasi-linearPDE of the form

ni=1

ai(x, u)uxi = f (x, u),

the non-characteristics condition at 0 is given by

ni=1

ai

0, u0(0)

ξ i = 0,

where ξ = (ξ 1

,· · ·

, ξ n

) is a normal vector of Σ at 0.We first consider a simple case where the hypersurface Σ is given by the hyper-

plane xn = 0. Obviously, a normal vector of xn = 0 is given by (0, · · · , 0, 1).Our goal is to solve the following initial value problem

F (x,u, ∇u) = 0,

u(x, 0) = u0(x),

for x ∈ Rn close to the origin. Here, we write x = (x, xn) for x ∈ Rn−1. Solvingfor u locally around the origin means to find values of u(x) for x close to the origin.

8/16/2019 PDE . QUING



To motivate, we consider initial-value problems of linear homogeneous equa-

tions

ni=1

ai(x)uxi = 0,

u(x, 0) = u0(x),

(2.6)

where ai is C 1 in a neighborhood of 0 ∈ Rn, i = 1, · · · , n, and u0 is at least C 1 ina neighborhood of 0 ∈ Rn−1. By introducing a = (a1, · · · , an), we simply write the

equation in (2.6) as

a(x) · ∇u = 0.

Here a(x) is regarded as a vector field in Rn. Then a(x)·∇ is a directional derivativealong a(x) at x.

In the following, we assume the hyperplane xn = 0 is non-characteristic at

the origin, i.e., an(0) = 0.

Our strategy is as follows. For any x ∈ Rn close to the origin, we construct a

special curve along which u is constant. If such a curve starts from x and intersectsRn−1 × 0 at (y, 0) for a small y ∈ Rn−1, then u(x) = u0(y).

To find such a curve x(t), we consider the restriction of u on it, which givesa one-variable function u(x(t)). Now we calculate the t-derivative of this function

and haved

dt

u(x(t))

=

ni=1

uxi

dxi

dt .

In order to have a constant value of u along this curve, we require

d

dtu(x(t)) = 0.

A simple comparison with the equation in (2.6) yields the following sufficient con-dition

dxi

dt = ai(x) for i = 1, · · · , n.

This naturally leads to the following definition.

Figure 2.3. Integral curves.

Definition 2.4. Let x = x(t) be a C 1-curve in Rn. It is an integral curve of (2.6) if

(2.7) dx(t)

dt = a

x(t)

.

8/16/2019 PDE . QUING



The calculation preceding Definition 2.4 shows that solutions u of (2.6) are

constant along integral curves of the coefficient vector. This yields the following

method of solving (2.6). For any x ∈ Rn near the origin, we find an integral curveof the coefficient vector through x by solving

dx

dt = a(x),

x(0) = x.(2.8)

If it intersects the hyperplane xn = 0 at some (y, 0), then we let u(x) = u0(y).It is easy to verify that, for any x ∈ Rn close to the origin, there is a unique in-

tegral curve, i.e., a solution of (2.8), starting from x and intersecting the hyperplanexn = 0 near the origin at a unique point.

Since (2.8) is an autonomous system (i.e., the independent variable t does notappear explicitly), we consider the following system instead of (2.8)

dxdt

= a(x),

x(0) = (y, 0).(2.9)

Starting with any y ∈ Rn−1 close to the origin, we expect the integral curve x(y, t)

to reach any x ∈ Rn close to the origin for small t. This is confirmed by thefollowing result.

Lemma 2.5. Let k ≥ 2 be an integer and a be a C k−1-vector field in a neigh-borhood of the origin with an(0) = 0. Then for any small y ∈ Rn−1 and any small t, the solution x = x(y, t) of (2.9) defines a C k-diffeomorphism in a neighborhood of the origin in Rn.

Proof. This follows from the implicit function theorem easily. By standard

results in ordinary differential equations, (2.9) admits a C k-solution x = x(y, t) forsmall (y, t) ∈ Rn−1 ×R. We treat it as a map (y, t) → x and calculate its JacobianJ at (y, t) = (0, 0). By x(y, 0) = (y, 0), we have

J (0) = ∂x

∂ (y, t)|(y,t)=(0,0) =

a1(0)

Id...

an−1(0)0 · · · 0 an(0)

.

Hence det J (0) = an(0) = 0.

Hence, for any small x, we can solve

x(y, t) = xuniquely for small y and t. Then u(x) = u0(y) yields a solution of (2.6). Note

that t is not present in the expression of solutions. Hence the value of the solutionu(x) depends only the initial value u0 at (y, 0) and, meanwhile, the initial valueu0 at (y, 0) influences the solution u along the integral curve starting from (y, 0).Therefore, we say the domain of dependence of the solution u(x) on the initial valueis represented by the single point (y, 0) and the domain of influence of the initial

value at a particular point (y, 0) on solutions consists of the integral curve startingfrom (y, 0).

8/16/2019 PDE . QUING



Figure 2.4. Solutions by integral curves.

For n = 2, integral curves are exactly characteristic curves. This can be seeneasily by (2.7) and Definition 2.1. Hence the ODE (2.7) is often referred to asthe characteristic ODE . This term is adopted for arbitrary dimensions. We have

demonstrated how to solve homogeneous first-order linear PDEs by using charac-teristics ODEs. Such a method is called the method of characteristics . Later on,we will develop a similar method to solve general first-order PDEs.

We need to emphasize that solutions constructed by the method of characteris-

tics are only local. In other words, they exist only in a neighborhood of the origin.A natural question here is whether there exists a global solution for globally de-fined a and u0. There are several reasons that local solutions cannot be extendedglobally. First, u(x) cannot be evaluated at x ∈ Rn if x is not on integral curves

Figure 2.5. Undetermined u(x).

from the initial hypersurface, or equivalently, integral curves from x does not in-tersect the initial hypersurface. Second, u(x) cannot be evaluated at x ∈ Rn if theintegral curve starting from x intersects the initial hypersurface more than once. Inthis case, we should have a compatibility condition for initial values and we cannot

prescribe initial values arbitrarily.

Example 2.6. Consider in R2 = (x, y)uy + ux =0,

u(x, 0) =u0(x).

8/16/2019 PDE . QUING



Figure 2.6. Undetermined u(x).

Note that the vector a is given by a = (1, 1) and hence

y = 0

is non-characteristic.

Characteristic ODEs are given byx = 1,

y = 1,and

x|t=0 = x0,

y|t=0 = 0.

Hence

x =t + x0,

y =t.

By eliminating t, we have

x − y = x0.

Along this curve, u is constant. Hence

u(x, y) = u0(x − y).

We can interpret the fact that u is constant along the straight line x −y = x0 in thefollowing way. If we interpret y as time, the graph of the solution represents a wavepropagating to the right with velocity 1 without changing shape. The solution uexists globally in R2.

Next, we discuss initial-value problems of quasi-linear PDEs. Let Ω ⊂ Rn be adomain and consider

ni=1

ai(x, u)uxi = f (x, u) in Ω,

u(x, 0) = u0(x) for x

∈Ω,

(2.10)

where ai and f are C 1-functions in Ω×R and u0 is C 1 in Ω = x ∈ Rn−1; (x, 0) ∈Ω. Assume the hyperplane xn = 0 is non-characteristic at the origin, i.e.,

an

0, u0(0)

= 0.

Suppose (2.10) admits a solution u. We first examine integral curves

dx

dt =a

x, u

,

x|t=0 =(y, 0),

8/16/2019 PDE . QUING



where y ∈ Rn−1. Contrary to the case of homogenous linear equations we studied

earlier, we are unable to solve this ODE since u, the unknown function we intend

to find, is present. However, viewing u as a function of t along these curves, we cancalculate how u changes. A similar calculation as before yields

du

dt =f (x, u),

u|t=0 =u0(y).

Hence we have an ordinary differential system for x and u. This leads to the

following method.Consider the following ordinary differential system

dx

dt =a(x, u),

du

dt =f (x, u),

with initial values

x|t=0 =(y, 0),

u|t=0 =u0(y),

where y ∈ Rn−1. In formulating this system, we treat x and u as functions of t only.This system consists of n + 1 equations for n + 1 functions and is the characteristic ODE of the quasilinear first-order PDE (2.10). By solving the characteristic ODE,we have a solution given by

x = x(y, t), u = ϕ(y, t).

As in the proof of Lemma 2.5, we can prove that the map (y, t) → x is a diffeomor-phism. Hence, for any x ∈ Rn close to the origin, there exists a unique y ∈ Rn−1

and t ∈ R close to the origin such that

x = x(y, t).

Then the solution u at x is given by

u(x) = ϕ(y, t).

Similarly, we discuss initial-value problems when initial hypersurfaces are nothyperplanes. Let Σ b e a hypersurface containing the origin and u0 be a functionon Σ. Instead of (2.10), we consider

ni=1

ai(x, u)uxi = f (x, u),

u = u0 on Σ.

We assume that Σ is non-characteristic at 0 with respect to u0, i.e.,

ni=1

ai

0, u0(0)

ξ i = 0,

8/16/2019 PDE . QUING



where ξ = (ξ 1, · · · , ξ n) is a normal vector of Σ at 0. We consider the characteristic

ODE in the following form

dx

dt =a(x, u),

du

dt =f (x, u),

with initial values

x|t=0 =x0,

u|t=0 =u0(x0),

for x0 ∈ Σ. Suppose we have a solution

x = x(x0, t), u = ϕ(x0, t).

As before, the map (x0, t) → x is a diffeomorphism. Hence for any given x closeto the origin, there exists a unique (x0, t) ∈ Σ ×R in a neighborhood of the origin

such that

x(x0, t) = x.

Then define

u(x) = ϕ(x0, t).

This is the desired solution.

Figure 2.7. Solutions by the method of characteristics.

Example 2.7. Consider in R2 = (x, y)

uy + uux =0,u(x, 0) = − x.

Note that the vector a is given by a = (u, 1) and hence y = 0 is non-characteristic.The characteristic ODE is given by

x = u,

y = 1,

u = 0,

and

x|t=0 = x0,

y|t=0 = 0,

u|t=0 = −x0.

8/16/2019 PDE . QUING



Hence

x =−

tx0 + x0,

y =t,

u = − x0.

By eliminating t and x0, we have

t = y, x0 = x

1 − t =

x

1 − y.

Therefore, the solution u is given by

u(x, y) = x

y − 1.

We note that this solution is not defined at y = 1. In fact, the integral curve γ x0passing through the point (x0, 0) is given by the straight line

x = x0 − x0y,

and the solution on this line is given by u = −x0. Each γ x0 passes the point (0, 1).

Figure 2.8. Integral curves in Example 2.7.

In general, smooth solutions of nonlinear PDEs are not expected to exist glob-ally, as shown in Example 2.7. The equation discussed in Example 2.7 is calledBurgers’ equation.

We consider a more general equation

uy + g(u)ux = 0,

with an initial value is given on y = 0 by

u(x, 0) = u0(x).

The characteristic ODE is given byx = g(u),

y = 1,

u = 0,

and

x|t=0 = x0,

y|t=0 = 0,

u|t=0 = u0(x0).

First, u = u0(x0), which is constant, along the integral curve starting from (x0, 0).Then on this integral curve,

x = g

u0(x0)

, y = 1.

8/16/2019 PDE . QUING



So the integral curve is a straight line with the slope 1

g(u0(x0)). Consider x0 < x1.

If the slope of the straight line from x1 is greater than that from x0, these twostraight lines will intersect. For example, for g(u) = u, the slope is 1u0(x) . Therefore

we have a non-global solution if u0 is strictly decreasing. See Example 2.7.


Example 2.8. Consider

uy + uux = 1,

u = 0 on y = x2.

By setting ϕ(x, y) = x2 − y, we have ∇ϕ(x, y) = (2x, −1) and hence for a = (u, 1)

a · ∇ϕ|ϕ=0 = (u, 1) · (2x, −1)|ϕ=0 = −1 = 0.

Therefore, y = x2 is non-characteristic. The characteristic ODE is given by

x = u,

y = 1,

u = 1,

and x|t=0 = x0,

y|t=0 = x20,

u|t=0 = 0.

Hence, we have

x =t2

2 + x0,

y =t + x20,

u =t.

By eliminating t and x0, we obtain an implicit expression for u

y = u + (x

− u2

2

)2.

So far in our discussion, initial values are prescribed on non-characteristic hy-persurfaces. It becomes complicated when initial values are prescribed on charac-

teristic hypersurfaces.Let Ω ⊂ Rn be a domain and consider

ni=1

ai(x, u)uxi = f (x, u) in Ω,

u(x, 0) = u0(x) for x ∈ Ω,

8/16/2019 PDE . QUING



where ai and f are C 1 in Ω ×R and u0 is C 1 in Ω = x ∈ Rn−1; (x, 0) ∈ Ω. For

y0 ∈ Ω, we assume the hyperplane Σ = xn = 0 is characteristic at y0, i.e.,

any0, 0, u0(y0) = 0.

Then we should have the following compatibility condition

(2.11)

n−1i=1

ai

y0, 0, u0(y0)

u0xi

(y0) = f

y0, 0, u0(y0)

.

Even if y0 is the only point where xn = 0 is characteristic, solutions may not existin a neighborhood of (y0, 0) for initial values satisfying the compatibility condition(2.11) at y0.

Now we discuss general nonlinear first-order partial differential equations. Let

Ω ⊂ Rn be a domain containing the origin and F = F (x,u,p) be a C 1-function in(x,u,p)

∈Ω

×R

×Rn. Consider

(2.12) F (x,u, ∇u) = 0 for x ∈ Ω,

and prescribe an initial value on xn = 0(2.13) u(x, 0) = u0(x) for x with (x, 0) ∈ Ω.

Here we write x = (x, xn). Assume there is a scalar a0 such that

F

0, u0(0), ∇xu0(0), a0

= 0.

We seek a solution u in a neighborhood of the origin with the given initial value

and uxn(0) = a0. The non-characteristics condition with respect to u0 and a0 is

given by

(2.14) F pn0, u0(0), ∇xu0(0), a0 = 0.

Note that by this condition, using the differential equation with the help of theimplicit function theorem, we can solve for a(x) for small x ∈ Rn−1, with a(0) =a0, such that

F

x, 0, u0(x), ∇xu0(x), a(x)

= 0.

We also require

uxn(x, 0) = a(x) for x small.

We start with a formal consideration. Suppose we have a solution u. Set

(2.15) pi = uxi for i = 1, · · · , n.

Then(2.16) F (x1, · · · , xn, u , p1, · · · , pn) = 0.

Differentiating (2.16) with respect to xi, we have

(2.17)n

j=1

F pj pi,xj

= −F xi − F uuxi

for i = 1, · · · , n,

where we used pj,xi = pi,xj

. We view (2.17) as a first-order quasilinear equationfor pi, for each fixed i = 1, · · · , n. An important feature here is that the coefficient

8/16/2019 PDE . QUING



for pi,xj is F pj

, independent of i. Characteristic ODEs associated with (2.17) are

given by

xj =F pj , j = 1, · · · , n,

˙ pi = − F u pi − F xi, i = 1, · · · , n.

We also have

u =

nj=1

uxj xj =

nj=1

pj F pj.

Now we collect ordinary differential equations for xj , u and pi.Consider the following ordinary differential system

xj =F pj(x,u,p), j = 1, · · · , n,

˙ pi = − F u(x,u,p) pi − F xi(x,u,p), i = 1, · · · , n

u =

n

j=1

pj F pj (x,u,p),

(2.18)

with initial values at t = 0

x|t=0 =(y, 0),

u|t=0 =u0(y),

pi|t=0 =u0xi(y), i = 1, · · · , n − 1,

pn|t=0 =a(y),

(2.19)

where y ∈ Rn−1. This is the characteristic ODE for general nonlinear first-orderpartial differential equations. It is an ordinary differential system of 2n+1 equationsfor 2n + 1 functions x, u and p. Here we view x, u and p as functions of t. Compare

this with a similar ordinary differential system of n +1 equations for n +1 functions

x and u for quasilinear equations. Solving (2.18) with (2.19), we have

x = x(y, t), u = ϕ(y, t), p = p(y, t).

We will prove that the map (y, t) → x is a diffeomorphism. Hence for any given xnear the origin, there exists a unique y ∈ Rn−1 and t ∈ R such that

x = x(y, t).

Then we define u by

u(x) = ϕ(y, t).

Theorem 2.9. The function u defined above is a solution of (2.12)-(2.13).

We should note that this solution u depends on the choice of the scalar a0 and

the function a(x

).Proof. The proof consists of several steps.Step 1. The map (y, t) → x is a diffeomorphism. This is proved as in the proof

of Lemma 2.5. In fact, the Jacobian of the map (y, t) → x at (0, 0) is given by

J (0) = ∂x

∂ (y, t)|y=0,t=0 =

∗

Id...

∗0 · · · 0 xn(0, 0)

,

8/16/2019 PDE . QUING



where

xn(0, 0) = F pn(0, u0(0), u0x1(0),

· · · , u0xn−1 , a0)

= 0.

Hence det J (0) = 0 by the non-characteristics condition (2.14). By the implicitfunction theorem, for any x ∈ Rn close to the origin, we can solve x = x(y, t)uniquely for y and t. Then define

u(x) = ϕ(y, t).

We will prove that this is the desired solution and

pi(y, t) = uxi(x(y, t)).

Step 2. We claim

F

x(y, t), ϕ(y, t), p(y, t) ≡ 0.

Denote by f (t) the function at the left hand side. Then f (0) = 0 since

f (0) = F y, 0, u0(y),∇

xu0(y), a(y)) = 0.

Next, we have by (2.18)

df (t)

dt =

d

dtF

x(y, t), ϕ(y, t), p(y, t)

=n

i=1

F xi

dxi

dt + F u

du

dt +

nj=1

F pj

dpj

dt

=n

i=1

F xiF pi

+ F u

nj=1

pjF pj +

nj=1

F pj(−F u pj − F xj

) = 0.

Hence f (t) ≡ 0.Step 3. We claim

pi(y, t) = uxi (x(y, t)).Let

wi(t) = uxi(x(y, t)) − pi(y, t).

We will prove

wi(t) = 0 for any t and i = 1, · · · , n.

First, consider t = 0. For i = 1, · · · , n − 1, wi(0) = 0 by initial values (2.19).For i = n, we will show uxn

(y, 0) = a(y). To see this, we first note by (2.18)

(2.20) 0 = u −n

j=1

pjF pj =

nj=1

uxj

xj − pj F pj) =

nj=1

F pj(uxj

− pj ).

Since wi(0) = 0 for i = 1, · · · , n−1, we have F pnwn|t=0 = 0. This implies wn(0) = 0

since F pn |t=0 = 0 by the non-characteristics condition (2.14).To prove wi(t) = 0 for any t, we show wi is a linear combination of wl, l =

1, · · · , n, i.e.,

wi =n

l=1

ailwl.

To prove this, we differentiate (2.20) with respect to xi and getn

j=1

F pj · (uxixj

− pj,xi) +

nj=1

(F pj)xi

wj = 0.

8/16/2019 PDE . QUING


2.3. A PRIORI ESTIMATES 23

Then

wi =

n

j=1

uxixj F pj + F u pi + F xi

=n

j=1

F pj pj,xi

−n

j=1

(F pj)xi

wj + F u pi + F xi.

By Step 2,

F

x, u(x), p1(x), · · · , pn(x)

= 0.

Differentiating with respect to xi, we have

F xi + F uuxi

+n

j=1

F pj pj,xi

= 0.

Hence

wi = − F xi − F uuxi

−n

j=1

(F pj)xi

wj + F u pi + F xi

= − F uwi −n

j=1

(F pj)xi

wj,

or

wi = −n

j=1

F uδ ij + (F pj

)xi

wj .

This ends the proof of Step 3.

Step 2 and Step 3 imply that u is the desired solution.

To end this section, we briefly compare methods we used to solve first-orderlinear or quasilinear PDEs and general first-order nonlinear PDEs. In solving afirst-order quasilinear PDE, we formulate an ordinary differential system of n + 1equations for n + 1 functions x and u. For a general first-order nonlinear PDE,

the corresponding ordinary differential system consists of 2n +1 equations for 2n +1 functions x, u and ∇u. Here, we need to take into account the gradient of uby adding n more equations for ∇u. In other words, we may regard our first-order nonlinear PDE as a relation for (u, p) with a constraint p = ∇u. We should

emphasize this is a unique feature for single first-order PDEs. For PDEs of higherorder or first-order partial differential systems, nonlinear equations are dramaticallydifferent from linear equations. In the rest of the book, we concentrate only on linearequations.

2.3. A Priori Estimates

A priori estimates play a fundamental role in partial differential equations.Usually, they are the starting point for the existence and regularity of solutions. Toderive a priori estimates, we first assume solutions already exist and then controlcertain norms of solutions by those of known functions in equations, such as non-homogenous terms and coefficients. Two frequently used norms are L∞-norms and

L2-norms. The importance of L2-norm estimates lies in the Hilbert space structureof the L2-space. Once L2-estimates of solutions and their derivatives are derived,

8/16/2019 PDE . QUING



standard results in Hilbert spaces such as the Riesz representation theorem can be

utilized. In many cases, this is how the existence of solutions is established.

In this section, we will use first-order linear PDEs to demonstrate how to derivea priori estimates in L∞-norms and L2-norms. We first examine briefly first-orderlinear ordinary differential equations. Let β be a constant and f = f (t) be a

function. Considerdu

dt − βu = f (t).

A simple calculation shows

u(t) = eβt u(0) +

t

0

eβ(t−s)f (s)ds.

For any T > 0, we have

|u(t)| ≤ eβt

|u(0)| + T sup

[0,T ]

|f |

for any t ∈ (0, T ).

Here, we control the sup-norm of u in [0, T ] by the initial value u(0) and the sup-norm of the nonhomogeneous term f in [0, T ].

Now we turn to PDEs. For convenience, we work in Rn ×R+ and denote pointsby (x, t), with x ∈ Rn and t ∈ R+. Let ai, b and f be functions in Rn × R+. We

consider

ut +

ni=1

ai(x, t)uxi + b(x, t)u =f (x, t) in Rn ×R+,

u(x, 0) =u0(x) in Rn,

(2.21)

where a = (a1, · · · , an) satisfies

(2.22)

|a

| ≤ 1

κ

in Rn

×R+,

for a positive constant κ. Obviously, the initial hypersurface t = 0 is non-characteristic. We may write the equation in (2.21) as

ut + a(x, t) · ∇xu + b(x, t)u = f (x, t).

We note that a(x, t)·∇x+∂ t is a directional derivative along the direction (a(x, t), 1).

With (2.22), it is easy to see that the vector (a(x, t), 1) (starting from the origin)is in fact in the cone given by

(y, s); κ|y| ≤ s ⊂ Rn ×R.

This is a cone opening upward and with vertex at the origin.For any point P = (X, T ) ∈ Rn × R+, consider the cone C κ(P ) (opening

downward) with vertex at P defined by

C κ(P ) = (x, t); 0 < t < T, κ|x − X | < T − t.

We denote by ∂ sC κ(P ) and ∂ −C κ(P ) the side and bottom of the boundary respec-tively, i.e.,

∂ sC κ(P ) = (x, t); 0 < t ≤ T, κ|x − X | = T − t,

∂ −C κ(P ) = (x, 0); κ|x − X | ≤ T .

We note that ∂ −C κ(P ) is simply a closed disc centered at X with radius T /κ inRn × 0. For any (x, t) ∈ ∂ sC κ(P ), let a(x, t) be a vector in Rn satisfying (2.22).

8/16/2019 PDE . QUING



Figure 2.10. The cone with the vertex at the origin.

Then the vector (a(x, t), 1), if positioned at (x, t), points outward from the inside

of the cone C κ(P ). Hence for a function defined only in C κ(P ), it makes sense to

calculate ut + a · ∇xu at (x, t). This is in particular true when (x, t) is the vertexP .

Figure 2.11. The cone C κ(P ) and positions of vectors.

Now we calculate the unit outward normal vector of ∂ sC κ(P ) \ P . Set

ϕ(x, t) = κ|x − X | − (T − t).

Obviously, ∂ sC κ(P )\P is a part of ϕ = 0. Then for any (x, t) ∈ ∂ sC κ(P )\P

∇ϕ = (∇xϕ, ϕt) =

κ x − X

|x − X | , 1

.

Therefore, the unit outward normal vector γ of ∂ sC κ(P ) \ P at (x, t) is given by

γ = 1

√ κ2

+ 1 κ x − X

|x − X |, 1.

For n = 1, the cone C κ(P ) is a triangle bounded by straight lines ±κ(x − X ) =

T − t and t = 0. The side of the cone consists of two line segments

the left segment: − κ(x − X ) = T − t, 0 < t ≤ T, with a normal vector (−κ, 1),

the right segment: κ(x − X ) = T − t, 0 < t ≤ T, with a normal vector (κ, 1).

It is easy to see that the integral curve associated with (2.21) starting from P and going to the initial surface Rn ×0 stays in C κ(P ). In fact, this is true for anypoint (x, t) ∈ C κ(P ). This suggests that solutions in C κ(P ) should depend only

8/16/2019 PDE . QUING



on f in C κ(P ) and the initial value u0 on ∂ −C κ(P ). The following result, proved

using a maximum principle type argument, confirms this.

Figure 2.12. The domain of dependence.

Theorem 2.10. Let ai, b and f be continuous functions in Rn ×R+ satisfying (2.22) and u be a C 1-solution of (2.21) in Rn × R+. Then for any P = (X, T ) ∈Rn × R+,

supC κ(P )

|e−βt u| ≤ sup∂ −C κ(P )

|u0| + T supC κ(P )

|e−βt f |,where β is a constant such that

b ≥ −β in C κ(P ).

If b ≥ 0, we take β = 0 and have

sup

C κ(P ) |u

| ≤ sup

∂ −C κ(P ) |u0

|+ T sup

C κ(P ) |f

|.

Proof. Take any positive number β > β and let

M = sup∂ −C κ(P )

|u0|, F = supC κ(P )

|e−βtf |.

We will prove

|e−βtu(x, t)| ≤ M + tF for any (x, t) ∈ C κ(P ).

For the upper bound, we consider

w(x, t) = e−βtu(x, t) − M − tF.

A simple calculation shows

wt +

ni=1

ai(x, t)wxi + b(x, t) + β w = −b(x, t) + β

(M + tF ) + e−βt

f − F.

The expression in the right hand side is nonpositive since b + β > 0. Hence

wt + a · ∇xw +

b + β )w ≤ 0 in C κ(P ).

Let w attain its maximum in C κ(P ) at (x0, t0) ∈ C κ(P ). We prove w(x0, t0) ≤ 0.First, it is obvious if (x0, t0) ∈ ∂ −C κ(P ), since w(x0, t0) = u0(x0) − M ≤ 0 by the

definition of M . If (x0, t0) ∈ C κ(P ), then

(wt + a · ∇xw)|(x0,t0) = 0.

8/16/2019 PDE . QUING



If (x0, t0) ∈ ∂ sC κ(P ), by the position of the vector (a(x0, t0), 1) relative to the cone

C κ(P ), we have

(wt + a · ∇xw)|(x0,t0) ≥ 0.Hence in both cases, we obtain

(b + β )w|(x0,t0) ≤ 0.

Since b + β > 0, this implies w(x0, t0) ≤ 0. (We need the positivity of b + β here!)Hence w(x0, t0) ≤ 0 in all three cases. Therefore, w ≤ 0 in C κ(P ), or,

u(x, t) ≤ eβt(M + tF ) for any (x, t) ∈ C κ(P ).

We simply let β → β to get the desired upper bound. For the lower bound, weconsider

v(x, t) = e−βtu(x, t) + M + tF.

The argument is similar and hence omitted.

For n = 1, (2.21) has the form

ut + a(x, t)ux + b(x, t)u = f (x, t).

In this case, it is straightforward to see that

(wt + awx)|(x0,t0) ≥ 0,

if w assumes its maximum at (x0, t0) ∈ ∂ sC κ(P ). We first note that ∂ t + 1

κ∂ x and

∂ t − 1

κ∂ x are directional derivatives along straight lines t − t0 = κ(x − x0) and

t − t0 = −κ(x − x0) respectively. Since w assumes its maximum at (x0, t0), we have

(wt + 1

κ wx)|(x0,t0) ≥ 0, (wt − 1

κ wx)|(x0,t0) ≥ 0.

In fact, one of them is zero if (x0, t0) ∈ ∂ sC κ(P ) \ P . Then we obtain

wt(x0, t0) ≥ 1

k|wx|(x0, t0) ≥ |awx|(x0, t0).

Theorem 2.10 implies the uniqueness of solutions of (2.21). Let u1 and u2 betwo solutions of (2.21). Then u1 − u2 satisfies (2.21) with f = 0 in C κ(P ) and

u0 = 0 on ∂ −C κ(P ). Hence u1 − u2 = 0 in Cκ(P ).

Theorem 2.10 also shows that the value u(P ) depends only on f in C κ(P ) andu0 on ∂ −C κ(P ). Hence C κ(P ) contains the domain of dependence of u(P ) on f and∂ −C κ(P ) contains the domain of dependence of u(P ) on u0. In fact, the domain of dependence of u(P ) on f is the integral curve through P in C κ(P ) and the domain

of dependence of u(P ) on u0 is the intercept of this integral curve with the initialhyperplane t = 0. We consider this from another point of view. For simplicity,we assume f is identically zero and the initial value u0 at t = 0 is zero outside abounded domain D0 ⊂ Rn. Then for any t > 0, u(·, t) = 0 outside

Dt = (x, t); κ|x − x0| < t for some x0 ∈ D0.

In other words, u0 only influences u in ∪t>0Dt. This is the so-called finite-speed

propagation.Now we use the same idea to estimate derivatives.

8/16/2019 PDE . QUING



Figure 2.13. The domain of influence.

Theorem 2.11. Let ai, b and f be C 1-functions in Rn

×R+ satisfying (2.22)

and u be a C 2-solution of (2.21) in Rn ×R+. Then for any P = (X, T ) ∈ Rn ×R+,

|u|C 1(C κ(P )) ≤ C |u0|C 1(∂ −C κ(P )) + |f |C 1(C κ(P ))

,

where C is a positive constant depending only on T and the C 1-norms of ai and bin C κ(P ).

Proof. In the following, we prove

supC κ(P )

|∇xu| ≤ C

sup∂ −C κ(P )

|∇xu0| + supC κ(P )

|∇xf | + supC κ(P )

|u|,

where C is a positive constant depending only on T and the C 1-norms of ai and

b in C κ(P ). Then, (2.21) implies a similar estimate for ut in C κ(P ). Hence, we

obtain the desired estimate by Theorem 2.10. The proof for for n = 1 is easier andwe will do that first.

Proof for n = 1. In this case, the equation in (2.21) has the form

ut + a(x, t)ux + b(x, t)u = f (x, t).

By differentiating with respect to x, we have

(ux)t + a(ux)x + (b + ax)ux = f x − bxu.

We view this as an equation for ux, which has the structure as the equation for u.Then by Theorem 2.10, we obtain

supC

κ(P )

|ux| ≤ eβ1T

sup∂ −

C κ

(P )|u0x| + T sup

C κ

(P )|f x − bxu|,

where β 1 is a nonnegative constant satisfying

b + ax ≥ −β 1 in C κ(P ).

Proof for n > 1. Now we consider the general case. Differentiating the equationin (2.21) with respect to xk, we obtain

(uxk)t +

ni=1

ai(uxk)xi

+ buxk +

ni=1

ai,xkuxi

= f xk − bxk

u.

8/16/2019 PDE . QUING



We note that this is not an equation for uxk as other derivatives of u are present.

To proceed, we set

v = 12

nk=1

u2xk

.

Then

vt =

nk=1

uxkutxk

, vxi =

nk=1

uxkuxixk

.

By multiplying the equation above by uxk and summing over k = 1, · · · , n, we

obtain

vt +n

i=1

aivxi + 2bv +

ni,k=1

ai,xkuxi

uxk =

nk=1

(f xk − bxk

u)uxk.

This is not an equation for v as products of two first order derivatives of u are

present in the fourth term in the left hand side. However, any such product can becontrolled by v . In fact, the Cauchy inequality yields

|uxiuxk

| ≤ 1

2

u2

xi + u2

xk

.

Hence, a summation impliesn

i,k=1

|uxiuxk

| ≤ nn

i=1

u2xi

= 2nv.

Then we have

vt +n

i=1

aivxi + 2bv − 2n

ni=1

|∇xai|L∞(C κ(P ))v ≤n

k=1

(f xk − bxk

u)uxk.

We apply the Cauchy inequality in a similar way to the expression in the right handside. Therefore, we obtain

vt +n

i=1

aivxi + (2b − 2n

ni=1

|∇xai|L∞(C κ(P )) − 1)v ≤ 1

2

nk=1

(f xk − bxk

u)2.

This is a first order partial differential inequality rather than an equation for v. Wecannot apply Theorem 2.10 directly. However, the method in the proof still applies.

We only need an estimate on the upper bound of v since v is nonnegative. Hence,

supC κ(P )

v ≤ eβ1T

sup∂ −C κ(P )

v + T

2 supC κ(P )

nk=1

(f xk − bxk

u)2

,

where β 1 is a nonnegative constant such that

2b − 2nn

i=1

|∇xai|L∞(C κ(P )) − 1 ≥ −β 1.

With the definition of v , we get the desired estimate for ∇xu easily.

Next, we derive an estimate of the L2-norm of u in terms of the L2-norms of f and u0. We will do this in another class of domains rather than cones. For fixedT, t > 0, consider

Dκ,T,t = (x, t); κ|x| < t − t, 0 < t < T .

8/16/2019 PDE . QUING



In other words,

Dκ,T,t = C κ(0, t)

∩ 0 < t < T

.

If T ≥ t, then Dκ,T,t = C κ(0, t).

Figure 2.14. An integral domain.

In the following, we are only interested in the case T < t. We denote by∂ −Dκ,T,t, ∂ sDκ,T,t and ∂ +Dκ,T,t the bottom, the side and the top of the boundary,i.e.,

∂ −Dκ,T,t =(x, 0); κ|x| < t,

∂ sDκ,T,t =(x, t); κ|x| = t − t, 0 < t < T ,

∂ +Dκ,T,t =(x, T ); κ|x| < t − T .

We recall the Green’s formula. Let Ω be a piecewise C 1-domain in Rn and

γ = (γ 1, · · · , γ n) be the unit exterior normal vector to ∂ Ω. Then for any u ∈C 1(Ω) ∩ C (Ω),

Ω

uxi =

∂ Ω

uγ i for i = 1, · · · , n.

Theorem 2.12. Let ai be C 1-functions satisfying (2.22), b and f be continuous functions and u be a C 1-solution of (2.21) in Rn × R+. Then for any 0 < T < t,

Dκ,T,t

e−αtu2 ≤

∂ −Dκ,T,t

u20 +

Dκ,T,t

e−αtf 2,

where α is a positive constant depending only on the C 1-norms of ai and the sup-norm of b− in Dκ,T,t.

Proof. For a nonnegative α, we multiply the equation in (2.21) by 2e−αtu.

By

2e−αtuut =(e−αtu2)t + αe−αtu2,

2aie−αtuuxi =(e−αtaiu2)xi

− e−αtai,xiu2,

we have

(e−αtu2)t +n

i=1

(e−αtaiu2)xi + e−αt(α + 2b −

ni=1

ai,xi)u2 = 2e−αtuf.

8/16/2019 PDE . QUING



We simply write D = Dκ,T,t. An integration in D yields

∂ +D

e−αtu2 + ∂ sD

e−αt(γ t +

ni=1

aiγ i)u2 + D

e−αt(α + 2b −n

i=1ai,xi )u2

=

∂ −D

u20 +

D

2e−αtuf,

where (γ 1, · · · , γ n, γ t) is the unit exterior normal vector of ∂ sD given by

(γ 1, · · · , γ n, γ t) = 1√ 1 + κ2

κ

x

|x| , 1

.

First, we note

γ t +n

i=1

aiγ i ≥ 0 on ∂ sD.

Next, we choose α such that

α + 2b −n

i=1

ai,xi ≥ 2 in D.

Then

2

D

e−αtu2 ≤

∂ −D

u20 +

D

2e−αtuf.

Here we simply dropped integrals over ∂ +D and ∂ sD since they are nonnegative.

The Cauchy inequality implies D

2e−αtuf ≤

D

e−αtu2 +

D

e−αtf 2.

We then have the desired result.

The domain Dκ,T,t is important for us to discuss properties of solutions inRn × (0, T ) for a fixed T . We note that t enters the estimate in Theorem 2.12only through the domain Dκ,T,t. Hence, we may let t → ∞ to get the followingconsequence.

Corollary 2.13. Let ai be C 1-functions satisfying (2.22), b be continuous functions and u be a C 1-solution of (2.21) in Rn × R+. For any T > 0, if f ∈L2(Rn × (0, T )) and u0 ∈ L2(Rn), then

Rn×(0,T )

e−αtu2 ≤ Rn

u20 +

Rn×(0,T )

e−αtf 2,

where α is a positive constant depending only on the C 1-norms of ai and the sup-norm of b− in Rn × (0, T ).

We point out that there are no decay assumptions on u.A natural question for anyone who just starts studying PDEs is why we need

a priori estimates. Now we use (2.21) to demonstrate that with the estimate inCorollary 2.13 we can actually prove the existence of (weak) solutions of initial

value problems (2.21). We use the homogeneous initial value, i.e., u0 = 0, for anillustration.

8/16/2019 PDE . QUING



We first introduce the notion of weak solutions. In the following, we fix a T > 0

and consider functions in

DT = Rn × (0, T ).Denote by C ∞0 (DT ) the collection of smooth functions in DT with compact supports

in DT and by C ∞0 (DT ) the collection of smooth functions in DT with compactsupports in x-directions. In other words, functions in C ∞0 (DT ) vanish for large x

and for t close to 0 and T and functions in C ∞0 (DT ) vanish only for large x.Set

(2.23) Lu = ut +n

i=1

ai(x, t)uxi + b(x, t)u in DT .

For any u, v ∈ C ∞0 , we integrate vLu in DT . To this end, we write

vLu = (uv)t +

n

i=1

(aiuv)

xi −uv

t +

n

i=1

(aiv)

xi −bv.

By a simple integration in DT , we have DT

vLu =

Rn×t=T

uv − Rn×t=0

uv −

DT

u

vt +n

i=1

(aiv)xi − bv

.

We note that there are no derivatives of u in the right-hand side.

Next, we define the adjoint differential operator L∗ of L by

L∗v = −vt −n

i=1

(aiv)xi + bv = −vt −

ni=1

aivxi + (b −

ni=1

ai,xi)v.

Then we have

DT

vLu = DT

uL∗v + Rn×t=T

uv − Rn×t=0

uv for any u, v ∈ C ∞0 (DT ).

Definition 2.14. An L2-function u is a weak solution of Lu = f in DT if DT

uL∗v =

DT

f v for any v ∈ C ∞0 (DT ).

Now we are ready to prove the existence of weak solutions of (2.21) with ho-mogeneous initial values. The proof requires the Hahn-Banach Theorem and theRiesz Representation Theorem in functional analysis. In the following, we denote

by (·, ·) the L2-inner product in DT .We first note that, with L in (2.23), we can rewrite the estimate in Corollary

2.13 as

(2.24) uL2(Rn×(0,T )) ≤ C u(·, 0)L2(Rn) + LuL2(Rn×(0,T )),

where C is a positive constant depending only on T , the C 1-norms of ai and thesup-norm of b− in Rn × (0, T ). This holds for any u ∈ C 1(Rn × [0, T )) with

Lu ∈ L2(Rn × (0, T )) and u(·, 0) ∈ L2(Rn).

Theorem 2.15. Let ai be C 1-functions satisfying (2.22) and b a continuous function in DT . Then for any f ∈ L2(DT ), there exists a u ∈ L2(DT ) such that

(u, L∗v) = (f, v) for any v ∈ C ∞0 (DT ) with v = 0 on t = T .

8/16/2019 PDE . QUING



Moreover,

u

L2(DT )

≤C

f

L2(DT ),

where C is a positive constant depending only on T , the C 1-norms of ai and the sup-norm of b− in Rn × (0, T ).

The function u in Theorem 2.15 is called a weak solution of (2.21) with u0 = 0.

Proof. We first note that L∗ has a similar structure as L except differentsigns for the principal part, the part involving derivatives. Applying (2.24) to L∗,we have

vL2(DT ) ≤ C L∗vL2(DT ) for any v ∈ C ∞0 (DT ) with v = 0 on t = T ,

where C is a positive constant depending only on T , the C 1-norms of ai and the

sup-norm of b. Here we view t = T as the initial curve for the domain DT and v

has a homogeneous initial value. We denote by C ∞(DT ) the collection of functions

v ∈ C ∞0 (DT ) with v = 0 on t = T .Consider the linear functional F : L∗ C ∞(DT ) → R given by

F (L∗v) = (f, v) for any v ∈ C ∞(DT ).

Then

|F (L∗v)| ≤ f L2(DT )vL2(DT ) ≤ C f L2(DT )L∗vL2(DT ).

Hence F is a well-defined bounded linear functional on the subspace L∗ C ∞(DT )

of L2(DT ). Thus we apply the Hahn-Banach Theorem to obtain a bounded linearextension of F defined on L2(DT ) such that

F ≤ C f L2(DT ).

Here,

F

is the norm of the linear functional F . By the Riesz RepresentationTheorem, there exists a u ∈ L2(DT ) such that

F (z) = (u, z) for any z ∈ L2(DT ),

and

uL2(DT ) ≤ C f L2(DT ).

Now we restrict z back to L∗ C ∞(DT ) to obtain the desired result.

Theorem 2.15 asserts the existence of weak solutions of (2.21) with u0 = 0. Now

we illustrate that u is indeed a classical solution if u is C 1 in DT and continuousup to t = 0. Under these extra assumptions on u, we integrate by parts (u, L∗v) toget

DT

vLu + Rn×t=0

uv = DT

f v for any v ∈ C ∞0 (DT ) with v = 0 on t = T .

There are no boundary integrals on vertical sides and on the upper side since vvanishes there. In particular,

DT

vLu =

DT

f v for any v ∈ C ∞0 (DT ).

Since C ∞0 (DT ) is dense in L2(DT ), we conclude

Lu = f in DT .

8/16/2019 PDE . QUING



Therefore,

Rn×t=0 uv = 0 for any v ∈ C

∞

0 (DT ) with v = 0 on t = T ,or

Rn

u(0)ϕ = 0 for any ϕ ∈ C ∞0 (Rn).

Again by the density of C ∞0 (Rn) in L2(Rn), we conclude

u(·, 0) = 0 on Rn.

Therefore, a crucial step in passing weak solutions to classical solutions is to improve

the regularity of weak solutions.Now we summarize the process to establish solutions by using energy estimates.Step 1. Prove a priori estimates;Step 2. Prove the existence of weak solutions by methods in functional analysis;

Step 3. Improve regularities of weak solutions.In fact, the last step is also closely related to a priori estimates of derivatives of solutions.

Next, we make a crucial observation. The requirement that u possess contin-uous derivatives can be weakened. It suffices to assume that u has derivatives in

the L2-sense so that the integration by parts can be performed. Then we concludeLu = f almost everywhere. This clearly suggests the need to introduce a new spaceof functions, functions with only L2-derivatives. This is the so-called Sobolev space,which plays a fundamental role in PDEs and will be studied in details in the future.

Exercises

(1) Find solutions of the following initial-value problems in R2:

(a) 2uy − ux + xu = 0 with u(x, 0) = 2xex2/2.

(b) uy + (1 + x2)ux − u = 0 with u(x, 0) = arctan x.

(2) Solve the following initial-value problems:

(a) uy + ux = u2 with u(x, 0) = h(x).

(b) uz + xux + yuy = u with u(x,y, 0) = h(x, y).

(3) Let B1 be the unit disc in R2 and a and b be continuous functions in B1

with a(x, y)x + b(x, y)y > 0 on ∂ B1. Assume u is a C 1-solution of

a(x, y)ux + b(x, y)uy = −u in B1.

Prove that u vanishes identically.

(4) Find an equation in R2

of the form a(x, y)ux + b(x, y)uy = 0 with smoothcoefficients such that(a) any horizontal line y = c is non-characteristic;(b) there does not exist a solution in entireR2 for any nonconstant initial

value prescribed on y = 0.

(5) Let α be a number and h = h(x) be a continuous function in R. Consider

yux + xuy =αu,

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

8/16/2019 PDE . QUING


EXERCISES 35

(a) Find all points on y = 0 where y = 0 is characteristic. What is

the compatibility condition of h at these points?

(b) Away from points in (a), find the solution of the initial-value problem.What is the domain of this solution in general?

(c) For cases h(x) = x, α = 1 and h(x) = x, α = 3, check whether this

solution can be extended over points in (a).(d) For each point in (a), find all characteristic curves passing it. What

is the relation of these curves and the domain in (b)?

(6) Let α ∈ R be a real number and h = h(x) be continuous in R and C 1 inR \ 0. Consider

xux + yuy =αu,

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

(a) Check that the straight line

y = 0

is characteristic at each point.(b) Find all h satisfying the compatibility condition on y = 0. (Con-

sider three cases, α > 0, α = 0 and α < 0.)(c) For α > 0, find two solutions with the given initial value on y = 0.

(This is easy to do simply by inspecting the equation, especially forα = 2.)

(7) In the plane, solve uy = 4u2x near the origin with u(x, 0) = x2 on y = 0.

(8) In the plane, find two solutions of the initial-value problem

xux + yuy + 1

2(u2

x + u2y) = u,

u(x, 0) = 1

2

(1

−x2).

(9) In the plane, find two solutions of the initial-value problem

1

4u2

x + uuy = u,

u(x, 1

2x2) = −1

2x2.

(10) Let ai, b and f be continuous functions satisfying (2.22) and u be a C 1-solution of (2.21) in Rn × R+. Prove for any P = (X, T ) ∈ Rn ×R+

supC κ(P )

|e−αtu| ≤ sup∂ −C κ(P )

|u0| + 1

α + inf C κ(P ) b supC κ(P )

|e−αtf |,

where α is a constant such thatα + inf

C κ(P )b > 0.

(11) Consider the following first-order differential system for (u, v) in R× R+

ut − a(x, t)vx + b11(x, t)u + b12(x, t)v = f 1(x, t),

vt − a(x, t)ux + b21(x, t)u + b22(x, t)v = f 2(x, t),

with

u(x, 0) = u0(x), v(x, 0) = v0(x),

8/16/2019 PDE . QUING



where a satisfies

|a(x, t)

| ≤

1

κ

.

Derive an L2-estimate of (u, v) in appropriate cones.

8/16/2019 PDE . QUING


CHAPTER 3

An Overview of Second-Order PDEs

This chapter should be considered as an introduction to second-order linearPDEs. In Section 3.1, we introduce the Laplace equation, the heat equation and

the wave equation. We also introduce their general forms, elliptic equations, par-abolic equations and hyperbolic equations. We use energy methods to discussthe uniqueness of boundary-value problems and initial/boundary-value problems

in Section 3.2 and use separation of variables to solve these problems in the planein Section 3.3.

3.1. Classifications of Second Order Equations

The main focus in this section is second-order linear PDEs. We proceed simi-larly as in Section 2.1.

Let Ω be a domain in Rn containing the origin and L be a second-order lineardifferential operator given by

Lu =n

i,j=1

aij (x)uxixj +

ni=1

bi(x)uxi + c(x)u in Ω,

where aij , bi, c are continuous in Ω. Here aij , bi, c are called coefficients of uxixj ,uxi

, u respectively. We usually assume that (aij ) is a symmetric matrix in Ω. Forthe operator L, we define its principal symbol by

p(x; ξ ) =n

i,j=1

aij (x)ξ iξ j for any x ∈ Ω, ξ ∈ Rn.

For a given function f in Ω, we consider the equation

(3.1) Lu = f (x) in Ω.

The function f is called the non-homogeneous term .Let Σ be the hyperplane xn = 0. We now prescribe values of u and its

derivatives on Σ so that we can at least find all derivatives of u at the origin.Let u0, u1 be functions defined in a neighborhood of the origin in Rn−1. Now weprescribe

(3.2) u(x, 0) = u0(x), uxn(x, 0) = u1(x) for any small x ∈ Rn−1.

Then we can find all x-derivatives of u and all x-derivatives of uxn at the origin.

In particular, we can determine all derivatives of u of order up to 2 at the origin

except uxnxn. To find this, we need to use the equation. If we assume

ann(0) = 0,

37

8/16/2019 PDE . QUING


38 3. AN OVERVIEW OF SECOND-ORDER PDES

then we can find uxnxn(0) from (3.1). In this case, we can compute all derivatives

of u of any order at the origin by using values u0, u1 and differentiating (3.1).

We usually call Σ the initial hypersurface and u0, u1 initial values or Cauchy values . The problem of solving (3.1) together with (3.2) is called the initial-value problem or the Cauchy problem .

More generally, consider the hypersurface Σ given by ϕ = 0 for a C 2-functionϕ in a neighborhood of the origin with ∇ϕ = 0, with the origin on the hypersurfaceΣ, i.e., ϕ(0) = 0. We note that ∇ϕ is simply a normal vector of the hypersurface Σ.Without loss of generality, we assume ϕxn


theorem, we solve ϕ = 0 around x = 0 for xn = ψ(x1, · · · , xn−1). Consider a changeof variables

x → y = (x1, · · · , xn−1, ϕ(x)).


Jacobian. Now we write the operator L in new variables y . Note

uxi =

nk=1

yk,xiuyk

,

and

uxixj =

nk,l=1

yk,xiyl,xj

uykyl +

nk=1

yk,xixjuyk

.

Hence,

Lu =n

k,l=1

ni,j=1

aij yk,xiyl,xj

)uykyl +

nk=1

bk +

ni,j=1

aij yk,xixj

uyk

+ cu.

The initial hypersurface Σ is given by yn = 0 in new coordinates. With yn = ϕ,

the coefficient of uynyn is given byn

i,j=1

aij (x)ϕxiϕxj

.

This is the principal symbol p(x, ξ ) evaluated at ξ = ∇ϕ(x).

Definition 3.1. For a linear operator L as in (3.1) defined in a neighborhoodof x0 ∈ Rn, a C 1-hypersurface Σ passing x0 is non-characteristic at x0 if

(3.3)n

i,j=1

aij (x0)ξ iξ j = 0,

where ξ is a normal vector of Σ at x0. Otherwise, it is characteristic at x0. A

hypersurface is non-characteristic if it is non-characteristic at every point.When the hypersurface Σ is given by ϕ = 0 with ∇ϕ = 0, its normal vector

is given by ∇ϕ = (ϕx1 , · · · , ϕxn). Hence we may take ξ = ∇ϕ(x0) in (3.3). We

note that the condition (3.3) is maintained under C 2-changes of local coordinates.By this condition, we can find successively values of all derivatives of u at x0, as faras they exist. Then, we could write formal power series at x0 for solutions of initialvalue problems. It would be actual representations of solutions u in a neighborhood

of x0, if u were known to be analytic. This process can be carried out for analyticinitial values and analytic coefficients and nonhomogeneous terms, and the result

8/16/2019 PDE . QUING


3.1. CLASSIFICATIONS OF SECOND ORDER EQUATIONS 3 9

is referred to as the Cauchy-Kowalevski theorem. We will discuss it later in this

book.

Now we introduce a special class of linear differential operators.

Definition 3.2. The linear differential operator L in (3.1) is elliptic at x0 if n

i,j=1

aij (x0)ξ iξ j = 0 for any ξ ∈ Rn \ 0.

Hence, linear differential equations are elliptic if there are no characteristic

hypersurfaces. In other words, every hypersurface is non-characteristic. We alreadyassumed that (aij ) is an n×n symmetric matrix. Then L is elliptic at x0 if (aij (x0))is a definite matrix, positive definite or negative definite.

We now turn our attention to second-order linear differential equations in R2,

where complete classifications are available. Let Ω be a domain in R2 and consider

(3.4) Lu =2

i,j=1

aij uxixj +

2i=1

bi(x)uxi + c(x)u = f (x) in Ω.

Here we assume (aij ) is a 2 × 2 symmetric matrix.

Definition 3.3. Let L be a differential operator defined in Ω ⊂ R2 as in (3.4).(1) L is elliptic at x ∈ Ω if det(aij (x)) > 0;(2) L is hyperbolic at x ∈ Ω if det(aij (x)) < 0;

(3) L is degenerate at x ∈ Ω if det(aij (x)) = 0.

It is obvious that the ellipticity defined here coincides with that in Definition3.2 for n = 2.

For the operator L in (3.4), the symmetric matrix (aij ) always has two eigen-

values. Then

L is elliptic if the two eigenvalues have the same sign;

L is hyperbolic if the two eigenvalues have different signs;

L is degenerate if one of the eigenvalues vanishes.

The number of characteristic curves are determined by the type of operators.For the operator L in (3.4),

there are two characteristic curves if L is hyperbolic;

there is one characteristic curve if L is degenerate;

there are no characteristic curves if L is elliptic.

Now we study several important linear differential operators in R2

.In R2, the Laplace operator ∆ is give by

∆u = ux1x1 + ux2x2 .

Obviously, the Laplace operator is elliptic. In polar coordinates

x1 = r cos θ, x2 = r sin θ,

the Laplace operator ∆ can be expressed by

∆u = urr + 1

rur +

1

r2uθθ .

8/16/2019 PDE . QUING



The equation

∆u = 0

is called the Laplace equation and its solutions are called harmonic functions . Bywriting x = x1 and y = x2, we can associate with a harmonic function u(x, y) aconjugate harmonic function v(x, y) such that u and v satisfy the first order systemof Cauchy-Riemann equations

ux = vy , uy = −vx.

Any such a pair gives an analytic function

f (z) = u(x, y) + iv(x, y)

of the complex argument z = x + iy. Physically, (u, −v) is the velocity field of anirrotational, incompressible flow. Conversely, for any analytic function f , functions

u = Ref and v = Imf are harmonic. In this way, we can find many nontrivial

harmonic functions in the plane. For example, for any positive integer k, Re(x +iy)k and Im(x + iy)k are homogeneous harmonic polynomials of degree k. Next,with ez = ex+iy = ex cos y + iex sin y, we know ex cos y and ex sin y are harmonic

functions.Although there are no characteristic curves for the Laplace operator, initial

value problems are not well-posed.

Example 3.4. Consider the Laplace equation in R2

uxx + uyy = 0.

Set

uk(x, y) = 1

k sin(kx)eky .

Then u is harmonic. Moreover,

uk,x(x, y) = cos(kx)eky , uk,y (x, y) = sin(kx)eky ,

and hence

u2k,x(x, y) + u2

k,y(x, y) = e2ky .

Therefore,

u2k,x(x, 0) + u2

k,y (x, 0) = 1 for any x ∈ R and any k ,

and

u2k,x(x, y) + u2

k,y (x, y) → ∞ as k → ∞, for any x ∈ R and y > 0.

There is no continuous dependence on initial values.

In R2, the wave operator is give by

u = ux2x2 − ux1x1 .

Obviously, the wave operator is hyperbolic. It is actually called the one-dimensional

wave operator. This is because the wave equation u = 0 in R2 represents vibra-tions of strings or propagation of sound waves in tubes. Because of its physicalinterpretation, we write u as a function of two independent variables x and t. Thevariable x is commonly identified with position and t with time. Then the wave

equation in R2 has the form

utt − uxx = 0.

8/16/2019 PDE . QUING


3.1. CLASSIFICATIONS OF SECOND ORDER EQUATIONS 4 1

Two families of straight lines t = ±x+constant are characteristic everywhere.

In R2, an important class of degenerate operators is given by

Lu = ux2 − ux1x1 .

This is called the heat operator. The heat equation ux2 − ux1x1 = 0 is satisfied bythe temperature distribution in a heat-conducting insulated wire. As for the waveequation, we write u as a function of two independent variables x and t. Then the

heat equation in R2 has the form

ut − uxx = 0.

Note that t = 0 is characteristic everywhere. If we prescribe u(x, 0) = u0(x)in an interval of t = 0, then using the equation we can compute all derivativesthere. However, u0 does not determine uniquely a solution even in a neighborhood

of this interval. We will see later on that we need initial values on the entire initialline

t = 0

to compute local solutions. Hence, the problem with Cauchy values

prescribed on characteristic hypersurfaces is not well-posed.

Definition 3.5. Let L be a differential operator defined in Ω ⊂ R2 as in (3.4).Then L is of mixed-type if L is elliptic in a subdomain in Ω and hyperbolic in

another subdomain.

Example 3.6. Consider the Tricomi equation

ux2x2 + x2ux1x1 = f in R2.

It is elliptic if x2 > 0, hyperbolic if x2 < 0 and degenerate if x2 = 0.

Characteristic curves arise also naturally in connection with the propagation

of singularities. We only consider a simple case.

Theorem 3.7. Let Ω be a domain in R2 and Γ be a curve in Ω. Assume

aij , bi, c , f are continuous functions in Ω and u ∈ C 1(Ω) ∩ C 2(Ω \ Γ) satisfies

2i,j=1

aij uxixj +

2i=1

bi(x)uxi + c(x)u = f (x) in Ω \ Γ.

If ∇2u has a jump across Γ, then Γ is a characteristic curve.

Proof. Without loss of generality, we assume Γ divides Ω into two parts Ω+

and Ω−. For any function w in Ω and any x0 ∈ Γ, we set

w−(x0) = limx→x0, x∈Ω−

w(x), w+(x0) = limx→x0, x∈Ω+

w(x),

and [w] = w+ − w− on Γ.

Then

[u] = [ux1 ] = [ux2 ] = 0 on Γ.

Let (ξ 1, ξ 2) be a normal vector along Γ. Then ξ 2∂ x1−ξ 1∂ x2 is a directional derivativealong Γ. Hence on Γ

(ξ 2∂ x1 − ξ 1∂ x2)[ux1 ] = ξ 2[ux1x1 ] − ξ 1[ux1x2 ] = 0,

(ξ 2∂ x1 − ξ 1∂ x2)[ux2 ] = ξ 2[ux1x2 ] − ξ 1[ux2x2 ] = 0.

8/16/2019 PDE . QUING



Rigorously, we need a limiting process to justify these two identities. By the con-

tinuity of aij , bi, c and f in Ω, we have

a11[ux1x2 ] + 2a12[ux1x2 ] + a22[ux2x2 ] = 0 on Γ.

The nontrivial vector ([ux1x1 ], [ux1x2 ], [ux2x2 ]) satisfies a 3 × 3 homogeneous linearsystem on Γ. Hence the coefficient matrix is singular. We then have on Γ

det

ξ 2 −ξ 1 00 ξ 2 −ξ 1

a11 2a12 a22

= 0,

or

a11ξ 21 + 2a12ξ 1ξ 2 + a22ξ 22 = 0.

This yields the desired result.

The Laplace operator, the wave operator and the heat operator can be gener-

alized to higher dimensions.Example 3.8. The n-dimensional Laplace operator in Rn is defined by

∆u =n

i=1

uxixi,

and the Laplace equation is given by ∆u = 0. Solutions are called harmonic func-tions. In this case, (aij ) is the identity matrix. Obviously, ∆ is elliptic. Note that

∆ is invariant under rotations. If x = Ay for an orthogonal matrix A, thenn

i=1

uxixi =

ni=1

uyiyi.

For nonzero function f , we call equation ∆u = f the Poisson equation .

Example 3.9. We denote points in Rn+1 by (x1, · · · , xn, t). The heat operatorin Rn+1 is given by

Lu = ut − ∆xu.

It is often called the n-dimensional heat operator. Its principal symbol is given by

p(x, t; ξ, τ ) = −|ξ |2 for any ξ ∈ Rn, τ ∈ R.

A hypersurface ϕ(x1, · · · , xn, t) = 0 is non-characteristic if

−|∇xϕ|2 = 0.

Likewise, a hypersurface ϕ(x, t) = 0 is characteristic everywhere if ∇xϕ = 0 and

ϕt = 0. For example, any horizontal hyperplane t = t0, for a fixed t0 ∈ R, ischaracteristic everywhere.

Example 3.10. We denote points in Rn+1 by (x1, · · · , xn, t). The wave oper-

ator in Rn+1 is given byu = utt − ∆xu.

It is often called the n-dimensional wave operator. Its principal symbol is given by

p(x, t; ξ, τ ) = τ 2 − |ξ |2 for any ξ ∈ Rn, τ ∈ R.

A hypersurface ϕ(x1, · · · , xn, t) = 0 is non-characteristic if

ϕ2t − |∇xϕ|2 = 0.

8/16/2019 PDE . QUING


3.2. ENERGY ESTIMATES 43

For any (x0, t0) ∈ Rn × R, the surface

|x

−x0

|2 = (t

−t0)2

is characteristic everywhere except at (x0, t0). We note that this surface, smoothexcept at (x0, t0), is the boundary of two cones.

Figure 3.1. The characteristic surface.

The heat equation and the wave equation can be generalized to parabolic equa-tions and hyperbolic equations in arbitrary dimensions. Again, we denote pointsin Rn+1 by (x1, · · · , xn, t). Let aij , bi, c and f be functions defined in a domainΩ ⊂ Rn+1, i, j = 1, · · · , n. We assume (aij (x, t)) is an n × n positive definite

matrix in Ω. A parabolic equation in Ω has the form

ut −n

i=1

aij (x, t)uxixj +

ni=1

bi(x, t)uxi + c(x, t)u = f (x, t) in Ω,

and a hyperbolic equation has the form

utt −n

i=1aij (x, t)uxixj +

n

i=1bi(x, t)uxi + c(x, t)u = f (x, t) in Ω.

3.2. Energy Estimates

In this section, we discuss the uniqueness of solutions of boundary-value prob-lems for the Laplace equation and initial/boundary-value problems for the heatequation and the wave equation. Our main tool is the energy estimates. Specifi-cally, we derive estimates of L2-norms of solutions in terms of those of boundary

values and/or initial values.We first recall Green’s formulas. Let Ω be a C 1-domain in Rn and γ =

(γ 1, · · · , γ n) be the unit exterior normal vector to ∂ Ω. Then for any u ∈ C 1(Ω) ∩C (Ω),

Ω

uxi = ∂ Ω

uγ i for i = 1, · · · , n.

This is referred to as the divergence theorem. Now for any w ∈ C 2(Ω) ∩ C 1(Ω) and

v ∈ C 1(Ω) ∩ C (Ω), substitute u above by vwxi to get

Ω

(vwxixi + vxi

wxi) =

∂ Ω

vwxiγ i.

By summing up for i = 1, · · · , n, we get the Green’s formula of the following form Ω

v∆w + ∇v · ∇w

=

∂ Ω

v∂w

∂n,

8/16/2019 PDE . QUING



where ∂w

∂n is the normal derivative of w on ∂ Ω. If v ≡ 1, we get the divergence

theorem Ω

∆w = ∂ Ω

∂w∂n

.

For any v , w ∈ C 2(Ω) ∩ C 1(Ω), we have another form of the Green’s formula Ω

v∆w − w∆v

=

∂ Ω

(v∂w

∂n − w

∂v

∂n).

Figure 3.2. A smooth domain in Rn.

Now we consider the Laplace equation. Assume Ω ⊂ Rn is a bounded C 1-domain. The Dirichlet problem for the Laplace equation has the following form

∆u =0 in Ω,

u =ϕ on ∂ Ω,

for a continuous function ϕ on ∂ Ω. We now prove that a C 2-solution, if exists, is

unique. In fact, the difference w of any two C 2-solutions satisfies

∆w =0 in Ω,

w =0 on ∂ Ω.

We multiply the homogeneous Laplace equation by w and integrate in Ω. By theGreen’s formula, we have

0 =

Ω

w∆w = −

Ω

|∇w|2 +

∂ Ω

w∂w

∂n.

With the homogeneous boundary value, we have Ω

|∇w|2 = 0,

or ∇

w = 0 in Ω. Hence w is constant and this constant is zero since w is zeroon the boundary. Obviously, the above argument applies to Dirichlet problems forPoisson equations ∆u = f . In general, we have the following result.

Lemma 3.11. Assume Ω ⊂ Rn is a bounded C 1-domain, f is a continuous function in Ω and ϕ is a continuous function on ∂ Ω. Then

∆u =f in Ω,

u =ϕ on ∂ Ω,

admits at most one solution in C 2(Ω) ∩ C 1(Ω).

8/16/2019 PDE . QUING



By the maximum principle, the solution is in fact unique in C 2(Ω) ∩ C (Ω). We

will discuss this in Chapter 4.

For the Neumann problem, we prescribe normal derivatives on the boundary.It has the following form

∆u =0 in Ω,

∂u

∂n =ψ on ∂ Ω,

for a continuous function ψ on ∂ Ω. Similarly, we can prove that solutions are uniqueup to additive constants. We note that if there exists a solution of the Neumannproblem, then ψ necessarily satisfies

∂ Ω

ψ = 0.

This can be seen easily by the Green’s formula.Next, we derive an estimate of solutions of Dirichlet boundary value problems

of the Poisson equation. We need the following result, which is referred to as thePoincare Lemma.

Lemma 3.12. Let Ω be a bounded C 1-domain in Rn and u be a C 1-function in Ω with u = 0 on ∂ Ω. Then

uL2(Ω) ≤ diam (Ω)∇uL2(Ω).

Here diam(Ω) denotes the diameter of Ω and is defined by

diam(Ω) = supx,y∈Ω

|x − y|.

Proof. For any x0 ∈Rn−1, let l

x0be the straight line through x

0 and normal

to Rn−1 × 0. Consider those x0 such that lx0 ∩ Ω = ∅. Let I be an open interval

on lx0 given by

I = (x0, xn); a < xn < bsuch that I ⊂ Ω and (x0, a), (x0, b) ∈ ∂ Ω. Since u(x0, a) = 0, then

Figure 3.3. An integration along lx0 .

u(x0, xn) =

xn

a

uxn(x0, s)ds for any x ∈ (a, b).

8/16/2019 PDE . QUING



The Holder inequality yields

u2

(x

0, xn) ≤ (xn − a) xn

a u2

xn (x

0, s)ds for any x ∈ (a, b).

By a simple integration along I , we have b

a

u2(x0, xn)dxn ≤ (b − a)2

b

a

u2xn

(x0, xn)dxn.

By adding all possible intervals like I on lx0 , we then obtain lx0

∩Ω

u2(x0, xn)dxn ≤ C 2x0

lx0

∩Ω

u2xn

(x0, xn)dxn,

where C x0 is the length of lx0 in Ω. Now a simple integration over x0 yields thedesired result.

Now consider ∆u =f in Ω,

u =0 on ∂ Ω.(3.5)

We note that u has a homogeneous Dirichlet boundary value on ∂ Ω.

Theorem 3.13. Let Ω ⊂ Rn be a bounded C 1-domain and u ∈ C 2(Ω) ∩ C 1(Ω)be a solution of (3.5). Then

uL2(Ω) + ∇uL2(Ω) ≤ C f L2(Ω),

where C is a positive constant depending only on Ω.

Proof. Multiplying the equation in (3.5) by u and integrating over Ω, we

obtain Ω

|∇u|2 = − Ω

uf.

The Holder inequality yields

(

Ω

|∇u|2)2 ≤

Ω

u2 ·

Ω

f 2.

By Lemma 3.12, we get Ω

|∇u|2 ≤ diam(Ω)

2

Ω

f 2.

With Lemma 3.12 again, we have the desired estimate.

Now we study initial/boundary-value problems for the heat equation. SupposeΩ is a bounded C 1- domain in Rn, f is continuous in Ω

×[0,

∞) and u0 is continuous

in Ω. Considerut − ∆u =f in Ω × (0, ∞),

u(·, 0) =u0 in Ω,

u =0 on ∂ Ω × (0, ∞).

(3.6)

The geometric boundary of Ω × (0, ∞) consists of two parts, Ω × 0 and ∂ Ω ×(0, ∞). We treat Ω × 0 and ∂ Ω × (0, ∞) as the initial hypersurface and as

the boundary respectively. We refer to values prescribed on these two parts asinitial values and boundary values respectively. Problems of this type are usually

8/16/2019 PDE . QUING



called initial/boundary value problems or mixed problems . We note that u has a

homogeneous Dirichlet boundary value on ∂ Ω × (0, ∞). We derive an estimate of

the L2-norms of solutions. In the following, we write u(t) = u(·, t) as a function inΩ for any t ≥ 0.

Theorem 3.14. Let u ∈ C 2(Ω × (0, ∞)) ∩ C (Ω × [0, ∞)) be a solution of (3.6).Then

u(t)L2(Ω) ≤ u0L2(Ω) +

t

0

f (s)L2(Ω)ds for any t > 0.

Theorem 3.14 yields the uniqueness of solutions of (3.6). In fact, if f ≡ 0 and

u0 ≡ 0, then u ≡ 0. We also have the continuous dependence on initial values.

Proof. Multiply the equation in (3.6) by u and integrate over Ω for each fixedt > 0. Note

uut − u∆u = 12

(u2)t −n

i=1

(uuxi)xi

+ |∇u|2.

With u(t) = 0 on ∂ Ω, we have

1

2

d

dt

Ω

u2(t) +

Ω

|∇u(t)|2 =

Ω

f (t)u(t).

Then for any t > 0 Ω

|u(t)|2 + 2

t

0

Ω

|∇u|2 =

Ω

u20 + 2

t

0

Ω

f u.

Set

E (t) =

u(t)

L2(Ω).

Then E (t)

2+ 2

t

0

Ω

|∇u|2 =

E (0)2

+ 2

t

0

Ω

f u.

Differentiating with respect to t, we have

2E (t)E (t) ≤2E (t)E (t) + 2

Ω

|∇u(t)|2 = 2

Ω

f (t)u(t)

≤2f (t)L2(Ω) · u(t)L2(Ω).

Hence

E (t) ≤ f (t)L2(Ω).

Integrating from 0 to t, we obtain the desired estimate.

Now we study initial/boundary-value problems for the wave equation. Againsuppose Ω is a bounded C 1-domain in Rn, f is continuous in Ω × (0, ∞), u0 is C 1

in Ω and u1 is continuous in Ω. Consider

utt − ∆u = f in Ω × (0, ∞),

u(·, 0) = u0, ut(·, ) = u1 in Ω,

u = 0 on ∂ Ω × (0, ∞).

(3.7)

8/16/2019 PDE . QUING



Theorem 3.15. Let u ∈ C 2(Ω×(0, ∞))∩C 1(Ω× [0, ∞)) be a solution of (3.7).

Then ut(t)2L2(Ω)+∇xu(t)2

L2(Ω) 12 ≤ u12

L2(Ω)+∇xu02L2(Ω)

12 + t

0

f (s)L2(Ω)ds,

and

u(t)L2(Ω) ≤ u0L2(Ω) + tu12

L2(Ω) + ∇xu02L2(Ω)

12 +

t

0

(t − s)f (s)L2(Ω)ds.

Hence we also have the uniqueness and continuous dependence on initial values.

Proof. Multiply the equation by ut and integrate over Ω for each fixed t > 0.Note

ututt − ut∆u = 1

2(u2

t + |∇u|2)t −n

i=1

(utuxi)xi

.

Then 1

2

d

dt

Ω

|ut(t)|2 + |∇xu(t)|2− ∂ Ω

ut(t)∂u

∂n(t) =

Ω

f (t)ut(t).

Note ut = 0 on ∂ Ω × (0, ∞) since u = 0 on ∂ Ω × (0, ∞). Then

1

2

d

dt

Ω

|ut(t)|2 + |∇xu(t)|2

=

Ω

f (t)ut(t).

Define the energy by

E (t) =

Ω

|ut(t)|2 + |∇xu(t)|2

.

If f ≡ 0, thend

dtE (t) = 0.

HenceE (t) = E (0) =

1

2

Ω

|u1|2 + |∇xu0|2

.

This is the conservation of energy. In general,

E (t) = E (0) + 2

t

0

Ω

f ut.

To get an estimate on the energy E (t), set

J (t) =

E (t) 12 .

Then

J (t)

2

=

J (0)

2

+ 2

t

0 Ωf ut.

Differentiating with respect to t, we get

2J (t)J (t) = 2

Ω

f (t)ut(t) ≤ 2f (t)L2(Ω)ut(t)L2(Ω) ≤ 2J (t)f (t)L2(Ω).

Hence

J (t) ≤ f (t)L2(Ω).

Integrating from 0 to t, we obtain

J (t) ≤ J (0) +

t

0

f (s)L2(Ω)ds.

8/16/2019 PDE . QUING


3.3. SEPARATION OF VARIABLES 49

This is the desired estimate on the energy. Next, to estimate the L2-norm of u, we

set

F (t) = u(t)L2(Ω),

i.e., F (t)

2=

Ω

|u(t)|2.

A simple differentiation yields

2F (t)F (t) = 2

Ω

u(t)ut(t) ≤ 2u(t)L2(Ω)ut(t)L2(Ω).

Hence

F (t) ≤ utL2(Ω) ≤ J (0) +

t

0

f (s)L2(Ω)ds.

Integrating from 0 to t, we get

u(t)L2(Ω) ≤ u0L2(Ω) + tJ (0) +

t

0

t

0

f (s)L2(Ω)dsdt.

This implies the desired estimate on u itself easily.

There are other forms of estimates on energies. By squaring the first estimate

in Theorem 3.15, we obtain Ω

|ut(t)|2 + |∇xu(t)|2 ≤ 2

Ω

|u1|2 + |∇xu0|2

+ 2t

t

0

Ω

|f |2.

Integrating from 0 to t, we get

t

0

Ω|ut|2 + |∇xu|2 ≤ 2t

Ω|u1|2 + |∇xu0|2 + t2 t

0

Ω|f |2.

To end this section, we briefly review methods used in deriving estimates inTheorems 3.13-3.15. In the proof of Theorems 3.13-3.14, we multiply the Laplace

equation and the heat equation by u and integrate over Ω. While in the proof of Theorem 3.15, we multiply the wave equation by ut and integrate over Ω. Thesestrategies also work for general elliptic equations, parabolic equations and hyper-bolic equations.

3.3. Separation of Variables

In this section, we use the separation of variables to solve initial/boundary-

value problems of the heat equation and the wave equation in R×R+.We first discuss initial/boundary-value problems for the 1-dimensional heat

equation of the following form

ut − uxx = 0 for any x ∈ (0, π) and any t > 0,

u(x, 0) = u0(x) for any x ∈ (0, π),

u(0, t) = u(π, t) = 0 for any t > 0.

(3.8)

Physically, u represents the temperature in an insulated rod with ends kept at 0temperature.

8/16/2019 PDE . QUING



We now use separation of variables to find solutions of

ut

−uxx = 0 for any x

∈(0, π) and any t > 0,

u(0, t) = u(π, t) = 0 for any t > 0.

Set

u(x, t) = a(t)w(x).

Then

a(t)w(x) − a(t)w(x) = 0,

ora(t)

a(t) − w(x)

w(x) = 0.

The first term is a function of t and the second term is a function of x. Since theirdifference is zero, they should be equal to a constant. We denote this common

constant by −λ. Then

a(t) + λa(t) = 0 for any t > 0,

and

w(x) + λw(x) = 0 for any x ∈ (0, π),

w(0) = w(π) = 0.

The latter is the homogeneous eigenvalue problem of − d2

dx2 in (0, π). It is well

known that λk = k2 and wk(x) = sin kx are a sequence of solutions and wk forms

an orthogonal basis for Fourier series in L2(0, π). For any v ∈ L2(0, π), we have

v(x) =

∞k=1

ak sin kx,

where

ak = 2

π

π

0

v(x)sin kxdx.

The convergence for the series of v is in the L2-sense. By the completeness of

sin kx in L2(0, π), we have

vL2(0,π) =π

2

∞k=1

a2k

12 .

Back to a(t), we have for each λ = k2

a(t) = ae−k2t,

where a is a constant. Now we set for each integer k ≥ 1

uk(x, t) = e−k2t sin kx for any (x, t) ∈ (0, π) × (0, ∞).

Then uk satisfies the heat equation and the boundary value in (3.8). In order toget a solution satisfying the equation, the boundary value and the initial value

in (3.8), we consider an infinite linear combination of uk and choose coefficientsappropriately.

8/16/2019 PDE . QUING



Now we are ready to solve the initial/boundary-value problem (3.8). We treat

t as a parameter and expand u(x, t) as a function of x ∈ (0, π) in its Fourier series

with respect to sin kx. Formally, we write

u(x, t) =

∞k=1

ak(t)sin kx.

Then

ut − uxx =∞

k=1

ak(t) + k2ak(t)

sin kx.

With

ak(t) + k2ak(t) = 0 for any k = 1, 2, · · · ,

we have

ak(t) = ak(0)e−k2t for any k = 1, 2,

· · · ,

and hence

u(x, t) =

∞k=1

ak(0)e−k2t sin kx.

Let

u0(x) =

∞k=1

u0k sin kx,

where

u0k = 2

π

π

0

u0(x)sin kxdx.

Then ak(0) = u0k and the solution u is formally given by

(3.9) u(x, t) =∞

k=1

u0ke−k2t sin kx.

Next we prove that u in (3.9) indeed solves (3.8).

Theorem 3.16. If u0 ∈ L2(0, π), then u given by (3.9) is smooth in [0, π] ×(0, ∞) and satisfies

ut − uxx = 0 for any x ∈ (0, π) and any t > 0,

u(0, t) = u(π, t) = 0 for any t > 0,

and

limt→0+

u(

·, t)

−u0

L2(0,π) = 0.

Proof. By u0 ∈ L2(0, π), we have

u02L2(0,π) =

π

2

∞k=1

|u0k|2 < ∞.

For any nonnegative integers i and j , any x ∈ [0, π] and t ∈ (0, ∞), we have

|∂ ix∂ jt u(x, t)| ≤∞

k=1

|u0k|ki+2j

ek2t .

8/16/2019 PDE . QUING



Hence the Cauchy inequality implies

|∂ ix∂ jt u(x, t)| ≤

∞k=1

|u0k|2 1

2 ∞k=1

k2i+4j

e2k2t 12 ≤ C i,j,tu0L2(0,π),

where C i,j,t is a positive constant depending only on i, j and t. This shows the

absolute convergence of series for ∂ ix∂ jt u(x, t), for any x ∈ [0, π], t > 0 and anynonnegative integers i and j. Hence u is smooth in [0, π] × (0, ∞). The proof forthe convergence as t → 0+ is left as an exercise.

By examining the proof, we have the following estimate. For any integer m ≥ 0and any t0 > 0

uC m([0,π]×(t0,∞)) ≤ C u0L2(0,π),

where C is a positive constant depending only on m and t0. This estimate controlsthe C m-norm of u in (0, π)

×(t0,

∞) for any t0 > 0 in terms of the L2-norm of u0

on (0, π). It is referred to as an interior estimate (with respect to t). We note thatu becomes smooth immediately after t = 0 even if the initial value u0 is only L2.

Naturally, we ask whether u in Theorem 3.16 is continuous up to t = 0, orgenerally whether u is smooth up to t = 0. In order to have continuity of u up tot = 0, first we should require u0 ∈ C [0, π]. Next, u0 should satisfy a compatibilitycondition. By comparing the initial value with the homogeneous boundary value

at the corner, we have

u0(0) = 0, u0(π) = 0.

In general, in order to have smoothness of u up to t = 0, we assume u0 ∈ C ∞[0, π].Then from the heat equation, we have uxx(0, 0) = 0 and uxx(π, 0) = 0. Hence

u0 (0) = 0, u0 (π) = 0.

Continuing this process, we have a necessary condition

(3.10) u(2)0 (0) = 0, u

(2)0 (π) = 0 for any = 0, 1, · · · .

Now, we prove this is also a sufficient condition.

Theorem 3.17. If u0 ∈ C ∞[0, π] and (3.10) holds, then u given by (3.9) is smooth in [0, π] × [0, ∞) with u(·, 0) = u0.

Proof. In order to prove the smoothness up to t = 0, we need to bound

∂ ix∂ jt u(x, t) independent of t for t small, for any nonnegative integers i and j. To thisend, we first improve estimates of u0k. With (3.10), we have by simple integrationsby parts

u0k = 2

π

π

0

u0(x)sin kxdx = 2

π

π

0

u0(x)cos kx

k dx = − 2

π

π

0

u0 (x)sin kx

k2 dx.

We note that values at end points are not present since u0(0) = u0(π) = 0 i nthe first integration by parts and sin kx = 0 at x = 0 and x = π in the second

integration by parts. Hence for any m ≥ 1, we continue this process with the helpof (3.10) for = 0, · · · , [(m − 1)/2] and obtain

u0k = (−1)m−12

2

π

π

0

u(m)0 (x)

cos kx

km dx if m is odd,

8/16/2019 PDE . QUING



and

u0k = (

−1)

m2

2

π π

0

u(m)0 (x)

sin kx

km

dx if m is even.

By the Cauchy inequality, we get for any m ≥ 0

|u0k| ≤

2

πu(m)

0 L2(0,π)1

km for any k = 1, 2, · · · ,

where we used π

0

sin2 kxdx =

π

0

cos2 kxdx = π

2.

Now for any x ∈ [0, π] and t > 0

|∂ ix∂ jt (x, t)| ≤∞

j=1

|u0k|ki+2j ≤

2

πu(m)

0 L2(0,π)

∞k=1

ki+2j

km < ∞,

which is independent of t, if we take m = i + 2 j + 2. This implies that ∂

i

x∂

j

t (x, t) iscontinuous in [0, π] × [0, ∞).

If we are only interested in the continuity of u up to t = 0, we have the following

result.

Corollary 3.18. If u0 ∈ C 1[0, π] with u0 ∈ L2(0, π) and u0(0) = u0(π) = 0,

then u given by (3.9) is smooth in [0, π] × (0, ∞) and continuous in [0, π] × [0, ∞)and satisfies (3.8).

Proof. It follows from Theorem 3.16 that u is smooth in [0, π] × (0, ∞) andsatisfies the heat equation and the homogeneous boundary condition in (3.8). Thecontinuity of u up to t = 0 follows from the proof of Theorem 3.17 with i = j = 0

and m = 2.

The regularity assumption on u0 in Corollary 3.18 does not seem to be optimal.It is natural to ask whether it suffices to assume u0 ∈ C [0, π]. To answer thisquestion, we should analyze whether the series in (3.9) converges uniformly in tsince we need to take limit t → 0. We will not pursue this issue here.

Now we provide another expression of u in (3.9). With explicit expressions of u0k in terms of u0, we can write

(3.11) u(x, t) =

π

0

G(x, y; t)u0(y)dy,

where

G(x, y; t) = 2

π

∞

k=1

e−k2t sin kx sin ky for any x, y

∈[0, π] and t > 0.

The function G is called the Green’s function associated with the initial-boundary

value problem (3.8). For each fixed t, the series for G is uniformly convergent forx, y ∈ [0, π]. In fact, this uniform convergence justifies the exchange of summationand integration in obtaining (3.11). The Green’s function G satisfies the followingproperties:

(G1) Symmetry: G(x, y; t) = G(y, x; t).

(G2) Smoothness: G(x, y; t) is smooth in x, y ∈ [0, π] and t > 0.(G3) Solution of the heat equation: Gt − Gxx = 0.

8/16/2019 PDE . QUING



(G4) Homogeneous boundary values: G(0, y; t) = G(π, y; t) = 0.

These properties follow easily from the explicit expression for G. They imply

that u in (3.11) is a smooth function for t > 0 and satisfies the heat equationwith homogeneous boundary values. We can also prove with help of the explicitexpression of G that u in (3.11) is continuous up to t = 0 and satisfies u(·, 0) = u0

under appropriate assumptions on u0. We point out that G can also be expressedin terms of the fundamental solution of the heat equation. See Chapter 5 fordiscussions of the fundamental solution.

Next, we discuss initial/boundary-value problems for the 1-dimensional waveequation of the following form

utt − uxx = 0 for any x ∈ (0, π) and any t > 0,

u(x, 0) = u0(x), ut(x, 0) = u1(x) for any x ∈ (0, π),

u(0, t) = u(π, t) = 0 for any t > 0.

(3.12)

We proceed as for the heat equation. Set

u(x, t) =

∞k=1

T k(t)sin kx.

Then

utt − uxx =

∞k=1

T k (t) + k2T k(t)

sin kx.

With

T k (t) + k2T k(t) = 0 for any k = 1, 2, · · · ,

we have

T k(t) = ak sin kt + bk cos kt for any k = 1, 2, · · · .Therefore,

(3.13) u(x, t) =

∞k=1

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

If we write

u0(x) =∞

k=1

u0k sin kx, u1(x) =∞

k=1

u1k sin kx,

we can determine ak, bk from u0 and u1 by

ak = u1k

k

= 2

kπ π

0

u1(x)sin kxdx,

bk = u0k = 2

π

π

0

u0(x)sin kxdx.

Unlike the case in the heat equation, in order to get regularity now, we need to

impose similar regularity conditions on initial values. Preceding as for the heatequation, we note that if u is a C 2-solution, then

u0(0) = 0, u1(0) = 0, u0 (0) = 0,

u0(π) = 0, u1(π) = 0, u0 (π) = 0.(3.14)

8/16/2019 PDE . QUING



Theorem 3.19. If u0 ∈ C 2[0, π], u0 ∈ L2(0, π), u1 ∈ C 1[0, π], u1 ∈ L2(0, π)

and u0, u1 satisfy (3.14), then u given by (3.13) is C 2 in [0, π] × [0, ∞) and is a

solution of (3.12).

Proof. We first improve estimates for ak and bk. By (3.14) and integration

by parts, we have

ak = 2

kπ

π

0

u1(x)sin kxdx = − 2

πk3

π

0

u1 (x)sin kxdx,

bk = 2

π

π

0

u0(x)sin kxdx = − 2

πk3

π

0

u0 (x)cos kxdx.

In other words, k3ak is the Fourier coefficients of −u1 (x) with respect to sin kxand k3bk is the Fourier coefficients of −u0 (x) with respect to 1, cos kx. Hence

∞

k=1

(k6a2

k + k6b2

k) <

∞.

Note for any integers i, j with 0 ≤ i + j ≤ 2 and any x ∈ [0, π] and t > 0

|∂ ix∂ jt u(x, t)| ≤∞

k=1

ki+j (|ak| + |bk|),

and hence

|∂ ix∂ jt u(x, t)| ≤ ∞k=1

k2(i+j+1)(|ak| + |bk|)2 12 · ∞

k=1

1

k2

12 < ∞.

Therefore, u is C 2 in [0, π] × [0, ∞). Then u satisfies the wave equation and theinitial condition by a simple differentiation term by term.

By examining the proof, we have

uC 2([0,π]×(0,∞)) ≤ C 3

i=0

u(i)0 L2(0,π) +

2i=0

u(i)1 L2(0,π)

,

where C is a positive constant independent of u.

In fact, in order to get a C 2-solution of (3.12), it suffices to assume u0 ∈ C 2[0, π]and u1 ∈ C 1[0, π]. We will prove this for a more general initial/boundary-valueproblem of the wave equation in Section 6.1. See Theorem 6.3.

Now, we compare the regularity results in Theorems 3.16, 3.17 and 3.19. For

initial/boundary-value problems of the heat equation in Theorem 3.16, solutionsbecome smooth immediately after t = 0, even for L2-initial values. This is the

so-called interior smoothness (with respect to time). We also proved in Theorem3.17 that solutions are smooth up to t = 0 if initial values are also smooth with a

compatibility condition. This property is called the global smoothness . However,solutions of initial/boundary-value problems of the wave equation exhibit a differ-ent property. We proved in Theorem 3.19 that solutions have a similar degree of regularity as initial values. In fact, this is a general result. Solutions of the waveequation do not have better regularity than initial values, and in some cases they

are less regular than initial values. We will see in Chapter 6 how solutions of thewave equation depend on initial values.

8/16/2019 PDE . QUING



To end this section, we point out that methods employed in this section to

solve initial/boundary-value problems for the one-dimensional heat equation and

wave equation can actually be generalized to higher dimensions. We use the heatequation as an illustration. Let Ω be a bounded smooth domain in Rn. We consider

ut − ∆u = 0 in Ω × (0, ∞),

u(·, 0) = u0 in Ω,

u = 0 on ∂ Ω × (0, ∞),

(3.15)

for a continuous function u0 in Ω. To solve (3.15) by separation of variables, weneed solutions of the eigenvalue problem of −∆ in Ω with homogeneous boundaryvalues, i.e.,

∆ϕ + λϕ = 0 in Ω,

ϕ = 0 on ∂ Ω.

(3.16)

Of course, this is much harder to solve than its one-dimensional counterpart. Nev-ertheless, a similar result still holds. In fact, solutions of (3.16) are given by asequence (λk, ϕk) where λk is a sequence of nondecreasing positive numbers such

that λk → ∞ as k → ∞ and ϕk is a sequence of smooth functions in Ω which formsa basis in L2(Ω). Then we can use a similar method to find a solution of (3.15) of the form

u(x, t) =

∞k=1

ake−λktϕk(x) for any (x, t) ∈ Ω × (0, ∞).

We should remark that solving (3.16) is a complicated process. We need to work in

Sobolev spaces, spaces of functions with L2

-integrable derivatives. The process issimilar as that indicated at the end of Chapter 2. First, we prove the existence of weak solutions of (3.16) and then improve regularity of solutions. A brief discussioncan be found in Section 4.6.

Exercises

(1) Classify the following second-order PDEs:

(a)n

i=1

uxixi +

1≤i<j≤n

uxixj = 0.

(b) 1≤i<j≤n

uxixj = 0.

(2) Discuss the uniqueness of the following problems by energy methods:

(a)

∆u − u3 = f in Ω,

u = ϕ on ∂ Ω.

(b)

∆u − u

Ω

u2(y)dy = f in Ω,

u = ϕ on ∂ Ω.

8/16/2019 PDE . QUING


EXERCISES 57

(3) Let Ω be a bounded C 1-domain in Rn and u be a C 2-function in Ω× [0, T ]satisfying

ut − ∆u =f in Ω × (0, ∞),

u(·, 0) =u0 in Ω,

u =0 on ∂ Ω × (0, ∞).

Prove

sup0≤t≤T

Ω

|∇u(·, t)|2 +

T

0

Ω

u2t ≤ C

Ω

|∇u0|2 +

T

0

Ω

f 2

,

where C is a positive constant independent of u, u0 and f .

(4) Prove the convergence part in Theorem 3.16.

(5) For u0 ∈ L2(0, π), let u be given in (3.9). Then for any nonnegative

integers i and jsup[0,π]

|∂ ix∂ jt u(·, t)| → 0 as t → ∞.

(6) Let G be defined in (3.11). Prove

|G(x, y; t)| ≤ 1√ πt

for any x, y ∈ [0, π] and t > 0.

(7) Solve the following problem by separation of variables

ut − uxx = 0 for any 0 < x < π,t > 0,

u(x, 0) = u0(x) for any 0 < x < π,

ux(0, t) = ux(π, t) = 0 for any t > 0.

(8) For any u0 ∈ L2(0, π) and f ∈ L2((0, π) × (0, ∞)), find a formal explicitexpression of a solution of the following problem

ut − uxx = f for any x ∈ (0, π) and any t > 0,

u(x, 0) = u0(x) for any x ∈ (0, π),

u(0, t) = u(π, t) = 0 for any t > 0.

(9) For any u0, u1 ∈ L2(0, π) and f ∈ L2((0, π)×(0, ∞)), find a formal explicitexpression of a solution of the following problem

utt − uxx = f for any x ∈ (0, π) and any t > 0,

u(x, 0) = u0(x), ut(x, 0) = u1(x) for any x ∈ (0, π),u(0, t) = u(π, t) = 0 for any t > 0.

(10) Let Ω be a bounded C 1-domain in Rn and u be C 2 in x and C 1 in t in

Ω × [0, T ] for a constant T > 0. Suppose u satisfies

ut − ∆u = 0 in Ω × (0, T ),

u(·, T ) = 0 in Ω,

u = 0 on ∂ Ω × (0, T ).

8/16/2019 PDE . QUING



Then u = 0 in Ω × (0, T ).

Hint: The function J (t) = log Ω u2(·, t) is a decreasing convex function.

8/16/2019 PDE . QUING


CHAPTER 4

Laplace Equations

The Laplace equation ∆u = 0 is probably the most important PDE with thewidest range of applications. In Section 4.1, we solve Dirichlet problems for the

Laplace equation in balls and derive Poisson integral formula. In the next threesections, we study harmonic functions, solutions of the Laplace equation, by threedifferent methods: fundamental solutions, mean-value properties and the maximum

principle. These three sections are relatively independent of each other. In Section4.5, we study the Poisson equation ∆u = f and derive Schauder estimates for itssolutions. Then in Section 4.6, we briefly discuss weak solutions of the Poissonequation.

4.1. Dirichlet Problems

The Laplace operator ∆ is defined on a C 2-function u in a domain in Rn by

∆u =

ni=1

uxixi.

The equation ∆u = 0 is called the Laplace equation and its C 2-solutions are called

harmonic functions.One of important properties of the Laplace equation is its spherical symmetry.

The equation is preserved by rotations about some point a ∈ Rn. Hence, it is

plausible that there exist special solutions that are invariant under rotations abouta. We begin this section by seeking a harmonic function u in Rn which dependsonly on r = |x − a| for some fixed a ∈ Rn. By setting v(r) = u(x), we have

∆u = v + n − 1

r v = 0,

and hence

v(r) =

c1 + c2 log r for n = 2,

c3 + c4r2−n for n

≥3,

where ci are constants for i = 1, 2, 3, 4. We are interested in solutions with asingularity such that

∂Br

∂v

∂rdσ = 1 for any r > 0.

Hence we set for any fixed a ∈ Rn

Γ(a, x) =

12π log |a − x| for n = 2,

1ωn(2−n) |a − x|2−n for n ≥ 3,

59

8/16/2019 PDE . QUING


60 4. LAPLACE EQUATIONS

where ωn is the surface area of the unit sphere in Rn. In summary, for any fixed

a ∈ Rn, Γ(a, ·) is harmonic in Rn \ a, i.e.,

∆xΓ(a, x) = 0 for any x = a,

and has a singularity at x = a. Moreover, ∂Br(a)

∂ Γ

∂nx(a, x) dS x = 1 for any r > 0.

The function Γ is called the fundamental solution of the Laplace operator.Now we prove the Green’s identity, which plays an important role in discussions

of harmonic functions. In the next result, we replace a, x by x, y and write Γ(x, y)instead of Γ(a, x).

Theorem 4.1. Suppose Ω is a bounded C 1-domain in Rn and that u ∈ C 1(Ω)∩C 2(Ω). Then for any x

∈Ω

u(x) = Ω

Γ(x, y)∆u(y)dy −

∂ Ω

Γ(x, y)

∂u

∂ny(y) − u(y)

∂ Γ

∂ny(x, y)

dS y.

Remark 4.2. (i) For any x ∈ Ω, Γ(x, ·) is integrable in Ω although it has a

singularity.(ii) For x /∈ Ω, the expression in the right hand side is zero. This follows from

the Green’s formula in Ω and ∆yΓ(x, y) = 0 if x /∈ Ω.(iii) By letting u = 1, we have

∂ Ω

∂ Γ

∂ny(x, y)dS y = 1 for any x ∈ Ω.

Proof. By applying the Green’s formula to u and Γ(x, ·) in Ω\Br(x) for smallr > 0, we get

Ω\Br (x)

(Γ∆u − u∆Γ)dy =

∂ Ω

Γ

∂u

∂n − u

∂ Γ

∂n

dS y −

∂Br (x)

Γ

∂u

∂n − u

∂ Γ

∂n

dS y .

Noting ∆Γ = 0 in Ω \ Br(x), we have Ω

Γ∆udy =

∂ Ω

Γ

∂u

∂n − u

∂ Γ

∂n

dS y − lim

r→0

∂Br (x)

Γ

∂u

∂n − u

∂ Γ

∂n

dS y .

For n ≥ 3, we get by the definition of Γ

∂Br(x)

Γ

∂u

∂n dS y = 1

(2 − n)ωn r2−n

∂Br(x)

∂u

∂n dS y≤ r

n − 2 sup∂Br(x)

|∇u| → 0 as r → 0,

and ∂Br(x)

u∂ Γ

∂n dS y =

1

ωnrn−1

∂Br(x)

udS y → u(x) as r → 0.

For n = 2, we get the same conclusion similarly.

8/16/2019 PDE . QUING


4.1. DIRICHLET PROBLEMS 61

Now we use Theorem 4.1 to discuss the Dirichlet boundary-value problem

∆u =f in Ω,

u =ϕ on ∂ Ω,(4.1)

for some f ∈ C (Ω) and ϕ ∈ C (∂ Ω). Lemma 3.11 asserts the uniqueness of a solutionin C 2(Ω)∩C 1(Ω). Another method to obtain this result is by the maximum principle

to be discussed later in this chapter. If u solves (4.1), then by Theorem 4.1 u can

be expressed in terms of f and ϕ, with one unknown term ∂u

∂n on ∂ Ω. We intend to

eliminate this term by adjusting Γ. We need to emphasize that we cannot prescribe∂u

∂n on ∂ Ω together with u on ∂ Ω.

For any fixed x ∈ Ω, we consider a function Φ(x, ·) ∈ C 2(Ω) ∩ C 1(Ω) with∆yΦ(x, y) = 0 in Ω. Then by the Green’s formula,

0 = Ω

Φ(x, y)∆u(y)dy − ∂ Ω

Φ(x, y) ∂u∂ny

(y) − u(y) ∂ Φ∂ny

(x, y)dS y.

By adding this to the Green’s identity in Theorem 4.1, we obtain for any x ∈ Ω

u(x) =

Ω

γ (x, y)∆u(y)dy −

∂ Ω

γ (x, y)

∂u

∂ny(y) − u(y)

∂γ

∂ny(x, y)

dS y,

where

γ (x, y) = Γ(x, y) + Φ(x, y).

Now we choose Φ appropriately so that γ (x, ·) = 0 on ∂ Ω. This leads to theimportant concept of Green’s functions.

For each fixed x ∈ Ω, we consider Φ(x, ·) ∈ C 1

(Ω) ∩ C 2

(Ω) such that

∆yΦ(x, y) = 0 for y ∈ Ω

Φ(x, y) = −Γ(x, y) for y ∈ ∂ Ω.

By Corollary 4.7, Φ(x, ·) is smooth in Ω for each fixed x if it exists. Denote byG(x, y) the resulting γ (x, y), which is called the Green’s function. If such a Gexists, then the solution u of the Dirichlet problem (4.1) can be expressed by

(4.2) u(x) =

Ω

G(x, y)f (y)dy +

∂ Ω

ϕ(y) ∂G

∂ny(x, y)dS y .

Note that the Green’s function G(x, y) is defined as a function of y ∈ Ω for each

fixed x ∈ Ω. Now we discuss properties of G as a function of x and y. We shouldpoint out that the Green’s function is unique. This follows from Lemma 3.11 orCorollary 4.16 to be established, since a difference of any two Green’s functions isharmonic with a zero boundary value.

Lemma 4.3. The Green’s function G(x, y) is symmetric in Ω × Ω, i.e., G(x, y)= G(y, x) for x = y ∈ Ω.

Proof. For any x1, x2 ∈ Ω with x1 = x2, take r > 0 small such that Br(x1) ∩Br(x2) = φ. Set Gi(y) = G(xi, y) and Γi(y) = Γ(xi, y). We apply the Green’s

8/16/2019 PDE . QUING



formula in Ω \ Br(x1) ∪ Br(x2) and get

Ω\Br(x1)∪Br (x2) G1∆G2 − G2∆G1 = ∂ Ω G1

∂G2

∂n − G2

∂G1

∂n −

∂Br(x1)

G1

∂G2

∂n − G2

∂G1

∂n

−

∂Br (x2)

G1

∂G2

∂n − G2

∂G1

∂n

.

Since Gi is harmonic for y = xi, i = 1, 2, and vanishes on ∂ Ω, we have ∂Br(x1)

G1

∂G2

∂n − G2

∂G1

∂n

+

∂Br (x2)

G1

∂G2

∂n − G2

∂G1

∂n

= 0.

Now we replace G1 in the first integral by Γ1 and replace G2 in the second integralby Γ2. Since G1 − Γ1 is harmonic in Ω and G2 is harmonic in Ω \ Br(x2), we have

∂Br(x1) (G1

−Γ1)

∂G2

∂n −G2

∂ (G1 − Γ1)

∂n →0 as r

→0.

Similarly, ∂Br(x2)

G1

∂ (G2 − Γ2)

∂n − (G2 − Γ2)

∂G1

∂n

→ 0 as r → 0.

Therefore, we obtain ∂Br(x1)

Γ1

∂G2

∂n − G2

∂ Γ1

∂n

+

∂Br (x2)

G1

∂ Γ2

∂n − Γ2

∂G1

∂n

→ 0,

as r → 0. On the other hand, we have

∂Br (x1)

Γ1∂G2

∂n →0, ∂Br (x2)

Γ2∂G1

∂n →0 as r

→0,

and ∂Br(x1)

G2∂ Γ1

∂n → G2(x1),

∂Br (x2)

G1∂ Γ2

∂n → G1(x2) as r → 0.

The two limits above can be proved similarly as the last limit in the proof of Theorem 4.1. This implies G2(x1) − G1(x2) = 0, or G(x2, x1) = G(x1, x2).

Finding Green’s functions involves solving a Dirichlet problem for the Laplaceequation. Meanwhile, Green’s functions are introduced to solve Dirichlet problems.We need to break this cycle. In fact, we can construct Green’s functions explicitlyfor special domains. The next result yields an expression for Green’s functions in

balls.

Theorem 4.4. The Green’s function in the ball BR ⊂ Rn is given by

G(x, y) = 1

2π

log |x − y| − log

R

|x|x − |x|R

y

for n = 2,

and

G(x, y) = 1

(2 − n)ωn

|x − y|2−n −

R

|x|x − |x|R

y

2−n for n ≥ 3.

8/16/2019 PDE . QUING


4.1. DIRICHLET PROBLEMS 63

For x = 0, we have

G(0, y) =

1

(2 − n)ωn |y|2−n

− R2−n for n ≥ 3

and

G(0, y) = 1

2π

log |y| − log |R| for n = 2.

Proof. We need to adjust the fundamental solution by adding a harmonicfunction so that the sum vanishes on the boundary. Fix an x = 0 with |x| < R, and

consider X ∈ Rn \ BR given by X = R2

|x|2x. In other words, X and x are reflexive of

each other with respect to the sphere ∂BR. Note that the map x → X is conformal,i.e., preserves angles. For |y| = R, we have by the similarity of triangles

|x|R

= R

|X | =

|y − x||y − X |

,

which implies

(4.3) |y − x| = |x|

R|y − X | =

|x|R

y − R

|x|x

for any y ∈ ∂BR.

Therefore, in order to have a zero boundary value, we take for n ≥ 3

Figure 4.1. The reflection about the sphere.

G(x, y) = 1

(2 − n)ωn

1

|x − y|n−2 −

R

|x|n−2

1

|y − X |n−2

.

We point out that the second term is smooth for y ∈ BR, since X /∈ BR. The casen = 2 is similar.

Next, we calculate normal derivatives of the Green’s function on spheres.

Corollary 4.5. Suppose G is the Green’s function in BR as in Theorem 4.4.Then

∂G

∂ny(x, y) =

R2 − |x|2

ωnR|x − y|n for any x ∈ BR and y ∈ ∂BR.

Proof. We only consider the case n ≥ 3. With X = R2x/|x|2, we have

G(x, y) = 1

(2 − n)ωn

|y − x|2−n − R

|x|n−2|y − X |2−n

,

8/16/2019 PDE . QUING



for any x ∈ BR and y ∈ ∂BR. Hence we get for such x and y

Gyi (x, y) = 1

ωn yi

−xi

|y − x|n − R

|x|n−2

· yi

−X i

|y − X |n = yi

ωnR2

R2

− |x|2

|x − y|n ,

by (4.3) in the proof of Theorem 4.4. With ni = yi

R for |y| = R, we obtain

∂G

∂ny(x, y) =

ni=1

niGyi(x, y) =

1

ωnR · R2 − |x|2

|x − y|n .


Denote by K (x, y) the function in Corollary 4.5, i.e.,

K (x, y) = R2 − |x|2

ωnR|x − y|n for any x ∈ BR, y ∈ ∂BR.

It is called the Poisson kernel and has the following properties:(K1) K (x, y) is smooth for any x ∈ BR and y ∈ ∂BR.(K2) K (x, y) > 0 for any x ∈ BR and y ∈ ∂BR.(K3) For any fixed x0 ∈ ∂BR, limx→x0,|x|<R K (x, y) = 0 uniformly in y for

|y − x0| > δ > 0.

(K4) ∆xK (x, y) = 0 for any x ∈ BR and y ∈ ∂BR.(K5)

∂BR

K (x, y)dS y = 1 for any x ∈ BR.

Here (K1), (K2) and (K3) follow easily from the explicit expression for K

and (K4) follows easily from the definition K (x, y) = ∂

∂nyG(x, y) and the fact that

G(x, y) is harmonic in x. An easy derivation of (K5) is based on (4.2). By takinga C 2( BR) harmonic function u in (4.2), we conclude

u(x) =

∂BR

K (x, y)u(y)dS y for any |x| < R.

This is called the Poisson integral formula. Then we have (K5) easily by takingu ≡ 1.

Now we are ready to solve the Laplace equation in balls with prescribed Dirich-let boundary values.

Theorem 4.6. For any ϕ ∈ C (∂BR), the function u defined by

(4.4) u(x) =

∂BR

R2 − |x|2

ωnR|x − y|nϕ(y)dS y for any x ∈ BR

is smooth in BR and continuous up to ∂BR and satisfies ∆u = 0 in BR,

u = ϕ on ∂BR.

Proof. By (K1) and (K4), we conclude easily that u defined by (4.4) is smoothand harmonic in BR. We only need to prove the continuity of u up to the boundary∂BR. Let x0 ∈ ∂BR and x ∈ BR. By (K5), we have

u(x) − ϕ(x0) =

∂BR

K (x, y)

ϕ(y) − ϕ(x0)

dS y = I 1 + I 2,

8/16/2019 PDE . QUING


4.2. FUNDAMENTAL SOLUTIONS 65

where

I 1 = |y−x0|<δ,y∈∂BR · · · , I 2 = |y−x0|>δ,y∈∂BR · · ·

,

for a positive constant δ to be determined. For any given ε > 0, we choose δ =δ (ε) > 0 so small that

|ϕ(y) − ϕ(x0)| < ε for any y ∈ ∂BR with |y − x0| < δ,

by the continuity of ϕ. Then |I 1| ≤ ε by (K2) and (K5). Let M = sup∂BR|ϕ|. By

(K3), we find a δ such that

K (x, y) ≤ ε

2MωnRn−1 for any |x − x0| < δ , |y − x0| > δ,

where δ depends on ε and δ = δ (ε), and hence only on ε. Then |I 2| < ε. Hence

|u(x) − ϕ(x0)| < 2ε for any x ∈ BR with |x − x0| < δ .

This shows the continuity of u at x0 ∈ ∂BR.

For n = 2, with x = (r cos θ, r sin θ) and y = (R cos η, R sin η) in (4.4), we have

u(r cos θ, r sin θ) = 1

2π

2π

0

R2 − r2

R2 − 2Rr cos(θ − η) + r2ϕ(R cos η, R sin η)dη.

Now we discuss some properties of the solution (4.4). First, u(x) in (4.4) issmooth for |x| < R, even if the boundary value ϕ is simply continuous. In fact,this is a general fact. Any harmonic function is smooth. We will prove this result

in the next section.In (4.4), by letting x = 0, we have

u(0) = 1

ωnRn−1 ∂BR

u(y)dS y.

This is the so-called mean-value property. The value of harmonic functions at centerof spheres is equal to their average on spheres.

Moreover, by (K2) and (K5), u in (4.4) satisfies

inf ∂BR

ϕ ≤ u ≤ sup∂BR

ϕ in BR.

In other words, harmonic functions in balls are bounded from above by their maxi-mum on boundary and bounded from below by their minimum on boundary. Sucha result is referred to as the maximum principle . Again, this is a general fact and

will be proved for any harmonic functions in any bounded domains.The mean-value property and the maximum principle are main topics in Section

4.3 and Section 4.4.

4.2. Fundamental Solutions

In this section, we use fundamental solutions of the Laplace equation to discussregularity of harmonic functions.

First, as an application of Theorem 4.1, we prove that harmonic functions aresmooth.

Lemma 4.7. Let Ω be a domain in Rn and u ∈ C 2(Ω) be a harmonic function in Ω. Then u is smooth in Ω.

8/16/2019 PDE . QUING



We note that, in its definition, a harmonic function is required only to be C 2.

Proof. For any x0 ∈ Ω, we take a bounded C

1

-domain Ω

in Ω such thatx ∈ Ω and Ω is a compact subset of Ω. Obviously, u is C 1 in Ω and ∆u = 0 inΩ. Then by Theorem 4.1, we have

u(x) = −

∂ Ω

Γ(x, y)

∂u

∂ny(y) − u(y)

∂ Γ

∂ny(x, y)

dS y for any x ∈ Ω.

There is no singularity in the integrand since x ∈ Ω and y ∈ ∂ Ω. This implieseasily that u is smooth in Ω.

In the following, we will prove that harmonic functions are analytic. As the first

step, we estimate derivatives of harmonic functions. For convenience, we considerharmonic functions in balls.

The following result is referred to as interior gradient estimates. It asserts

that derivatives of a harmonic function at any point is controlled by its maximalvalue in a ball centered around the point.

Lemma 4.8. Suppose u ∈ C

BR(x0)

is harmonic in BR(x0) ⊂ Rn. Then

|∇u(x0)| ≤ C

R maxBR(x0)

|u|,

where C is a positive constant depending only on n.

Proof. Without loss of generality, we assume x0 = 0.We first consider R = 1 and employ the local version of the Green’s identity.

Choose a cut-off function ϕ ∈ C ∞0 (B3/4) such that ϕ = 1 in B1/2 and 0 ≤ ϕ ≤ 1.

For y = x, Γ(x, y) is harmonic in y and hence

∆y

ϕ(y)Γ(x, y)

= ∆yϕ(y)Γ(x, y) + 2∇yϕ(y) · ∇yΓ(x, y).

This is zero for |y| < 1/2 and 3/4 < |y| < 1 since ϕ is constant there. Apply theGreen’s formula to u and ϕΓ(x, ·) in B1 \ Bρ(x) for x ∈ B1/4 and ρ small enough.We proceed as in the proof of Theorem 4.1 to obtain

u(x) = − 12 <|y|< 3

4

u(y)

∆yϕ(y)Γ(x, y) + 2∇yϕ(y) · ∇yΓ(x, y)

dy,

for any x ∈ B1/4. Then

∇u(x) = −

12 <|y|< 34

u(y)

∆yϕ(y)∇xΓ(x, y) + 2∇yϕ(y) · ∇x∇yΓ(x, y)

dy,

for any x ∈ B1/4. There is no singularity in the integrand for |x| < 1/4 and1/2 < |y| < 3/4. Hence, we obtain

supB 1

4

|Du| ≤ C supB1

|u|,

where C is a constant depending only on n.The general case follows from a simple dilation. Define

u(x) = u(Rx) for any x ∈ B1.

8/16/2019 PDE . QUING



Then u is a harmonic function in B1. By applying the result we just proved to u,

we obtain

supB 1

4R

|∇u| ≤ C R supBR

|u|.

This implies the desired result.

Next, we estimate derivatives of harmonic functions of arbitrary order.


BR(x0)

is harmonic in BR(x0) ⊂ Rn. Then for

any multi-index α with |α| = m

|∂ αu(x0)| ≤ C mem−1m!

Rm sup

BR(x0)

|u|,


Proof. We prove by an induction on m ≥ 1. It holds for m = 1 by Lemma4.17. We assume it holds for m and consider m +1. Let v be an arbitrary derivativeof u of order m. Obviously, it is harmonic in BR(x0). For 0 < θ < 1, we applyLemma 4.17 to v in B(1−θ)R(x0) and get

|∇v(x0)| ≤ C

(1 − θ)R max

B(1−θ)R(x0)|v|.

For any x ∈ B(1−θ)R(x0), we have BθR(x) ⊂ BR(x0). By the induction assumption,we obtain

|v(x)| ≤ C mem−1m!

(θR)m max

BθR(x)|u| for any x ∈ B(1−θ)R(x0),

and hencemax

B(1−θ)R(x0)|v| ≤ C mem−1m!

(θR)m max

BR(x0)|u|.

Therefore,

|∇v(x0)| ≤ C m+1em−1m!

(1 − θ)θmRm+1 maxBR(x0)

|u|.

By taking θ = m

m + 1, we have

1

(1 − θ)θm =

1 +

1

m

m

(m + 1) < e(m + 1).

Hence the desired result is established for any derivatives of order m + 1.

Harmonic functions are not only smooth but also analytic in their domains.Real analytic functions will be studied in Section 7.2. Now we simply introduce thenotion. A function u : Rn → R is called analytic near 0 if its Taylor series about 0is convergent to u in a neighborhood of 0 ∈ Rn, i.e.,

u(x) =

α

1

α!∂ αu(0)xα for any |x| < r.

Theorem 4.10. Harmonic functions are analytic.

8/16/2019 PDE . QUING



Proof. Suppose u is a harmonic function in Ω ⊂ Rn. For any fixed x ∈ Ω,

we prove that u is equal to its Taylor series in a neighborhood of x. To do this, we

take B2R(x) ⊂ Ω and h ∈ Rn with |h| ≤ R. For any integer m ≥ 1, we have by theTaylor expansion

u(x + h) = u(x) +

m−1i=1

1

i!

(h1∂ x1 + · · · + hn∂ xn

)i u

(x) + Rm(h),

where

Rm(h) = 1

m! [(h1∂ x1 + · · · + hn∂ xn

)m u] (x1 + θh1, . . . , xn + θhn),

for some θ ∈ (0, 1). Note that x + h ∈ BR(x) for |h| < R. Hence by Lemma 4.9,we obtain

|Rm(h)

| ≤ 1

m! |h|m

·nm

· C mem−1m!

Rm

maxB2R(x) |

u| ≤

Cne|h|R

m

maxB2R(x) |

u|.

Then for any h with C ne|h| < R/2, Rm(h) → 0 as m → ∞.

The proof of Theorem 4.10 shows that the radius of convergence is uniform forharmonic functions, depending only on n. In other words, any harmonic functionsdefined in a fixed ball are equal to their Taylor expansions at least in balls with afixed radius and such a radius depends only on the dimension n and is indepen-

dent of harmonic functions. It is easy to see that each homogenous polynomial of fixed degree in Taylor expansions of harmonic functions is still harmonic, which iscalled a homogenous harmonic polynomial. Homogeneous harmonic polynomialsconstitute an important subject in the study of harmonic functions. Because of

the homogeneity, homogeneous harmonic polynomials in Rn can be identified with

corresponding spherical harmonics, their restrictions on the unit sphere Sn−1 in Rn.To end this section, we classify homogeneous harmonic polynomials in R2 and

R3 and study their zero sets. For any integer m ≥ 1, we denote by Hm(Rn) the

collection of all homogeneous harmonic polynomials of degree m in Rn. Obviously,Hm(Rn) is a linear space.

We start with n = 2 and denote points in R2 by (x, y). A general homogeneouspolynomial P m of degree m in R2 contains m + 1 coefficients. Then ∆P m is a

homogeneous polynomial of degree m−2 and therefore contains m−1 terms. Theseterms have to vanish if P m is a harmonic function. This gives m−1 relations amongthe m + 1 constants if P m is a harmonic function; so that these constants can beexpressed linearly in terms of (m+1) −(m−1) or 2 of them. Therefore, Hm(R2) is alinear space of dimension 2. We note that Re(x + iy)m and Im(x + iy)m are linearly

independent homogeneous harmonic polynomials and hence form a basis in Hm(R2).In polar coordinates (r, θ), these two homogeneous harmonic polynomials are givenby rm cos mθ and rm sin mθ. Hence, any homogeneous harmonic polynomial P m of degree m can be expressed by

P m = c1rm sin mθ + c2rm cos mθ = arm cos(mθ + θ0),

for some constants c1, c2, a and θ0 ∈ [0, 2π). We note that the nodal set, or the zero

set, P −1m (0) of P m consists of m straight lines passing the origin and forming equal

angles by any two consecutive lines.

8/16/2019 PDE . QUING



Now we turn to n = 3. A general homogeneous polynomial P m of degree m inR3 contains 1

2 (m +1)(m + 2) coefficients. Then ∆P m is a homogeneous polynomial

of degree m − 2 and therefore contains 12 m(m − 1) terms. These terms have tovanish if P m is a harmonic function. This gives 1

2m(m − 1) relations among the

12

(m +1)(m + 2) constants if P m is a harmonic function; so that these constants can

be expressed linearly in terms of 12(m +1)(m + 2) − m(m − 1) or 2m + 1 of them.

Therefore, Hm(R3) is a linear space of dimension 2m + 1. A basis of this linearspace is given in terms of Legendre functions. We now discuss how to construct

such a basis.Let (x,y,z) and (r,θ,φ) be rectangular coordinates and corresponding polar

coordinates in R3, i.e.,

x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ.

Let P m = rmQm(θ, ϕ) be a homogeneous harmonic polynomial of degree m in R3.

Thenm(m + 1)Qm +

1

sin θ

∂

∂θ

sin θ

∂Qm

∂θ

+

1

sin2 θ

∂ 2Qm

∂ϕ2 = 0.

By writing µ = cos θ, we have

m(m + 1)Qm + ∂

∂µ

(1 − µ2)

∂Qm

∂µ

+

1

1 − µ2

∂ 2Qm

∂ϕ2 = 0.

By setting Qm(θ, ϕ) = f (θ)g(ϕ), we obtain

m(m + 1) + 1

f

∂

∂µ

(1 − µ2)

∂f

∂µ

+

1

1 − µ2

1

g

∂ 2g

∂ϕ2 = 0.

The first two terms in this equation are independent of g, and therefore so is the

last. Hence, the value of 1

g

∂ 2g

∂ϕ2

must be a constant. Since g is 2π-periodic in ϕ, we

take this constant to be −k2 for an integer k . Thus

∂ 2g

∂ϕ2 = −k2g,

and henceg(ϕ) = A cos kϕ + B sin kϕ,

where A and B are constants. Then f satisfies

∂

∂µ

(1 − µ2)

∂f

∂µ

+

m(m + 1) − k2

1 − µ2

f = 0,

which is known as Legendre’s associated equation . If f m,k(µ) is a solution, then

(4.5) rm

A cos kϕ + B sin kϕf m,k(cos θ)

is a harmonic function wherever it is defined in R3. We are interested in only those

f m,k such that (4.5) gives a homogeneous polynomial of degree m in R3. A lengthycalculation then yields for k = 0, 1, · · · , m

(4.6) f m,k(µ) = (1 − µ2)k2

dm+k

dµm+k(1 − µ2)m.

We omit details here. For k = 0, we have

f m,0(µ) = dm

dµm(1 − µ2)m.

8/16/2019 PDE . QUING



This is the Legendre function . For each fixed positive integer m, the collection in

(4.5) with f m,k given by (4.6) for 0 ≤ k ≤ m consists of 2m+ 1 linearly independent

homogeneous harmonic polynomials of degree m and hence forms a basis of Hm(R3).(We note that there is only one function for k = 0 in (4.5). )

Now, we demonstrate that these homogenous harmonic polynomials in (4.5)

exhibit different properties for different k by plotting their nodal sets. Because of the homogeneity, it is convenient to consider restrictions of nodal sets to the unitsphere S2, which is called the nodal curves .

If k = 0, the corresponding harmonic polynomial is a constant multiple of the

Legendre function f m,0(cos θ). A simple calculus argument shows that f m,0(µ) hasm distinct zeros between −1 and 1, arranged symmetrically about µ = 0. Hencef m,0(cos θ) has m distinct zeros between 0 and π, arranged symmetrically aboutθ = π/2. Accordingly on S2, the function f m,0(cos θ) vanishes on m latitude circles.

These circles are symmetrically situated with respect to the equator θ = π/2, and,

if m is an odd number, the equator itself is one of these circles. Similarly, level sets

Figure 4.2. Nodal curves of zonal harmonics.

of this function consist of latitude circles. Because of this division of the sphere intozones by sets of latitude circles, the function f m,0(cos θ) and its constant multiples

are called zonal harmonics .If 0 < k < m, the spherical harmonic is of the form

A cos kϕ + B sin kϕ

sink θ dm+k

dµm+k(µ2 − 1)m|µ=cos θ.

The first factor vanishes when A cos kϕ + B sin kϕ = 0, i.e., when tan kϕ = −A/B,and on S2 this corresponds to k great circles through the pole θ = 0, the angle

between the planes of any two consecutive great circles being π/k. The secondfactor vanishes at the points θ = 0 and θ = π, and the third on m − k latitudecircles, arranged like the corresponding circles in the case of zonal harmonics. Sincethe two sets of circles intersect orthogonally, these harmonics are called tesseral harmonics .

Finally, if k = m, the spherical harmonic is of the formA cos mϕ + B sin mϕ

sinm θ,

which vanishes when θ = 0 or π, or when tan mϕ = −A/B. This corresponds onS2 to the points θ = 0 and θ = π, and to m great circles through these points, the

angle between the planes of any two consecutive great circles being π /m. As S2 isthus divided up into 2m sectors, these functions called sectorial harmonics . We

8/16/2019 PDE . QUING


4.3. MEAN-VALUE PROPERTIES 71

Figure 4.3. Nodal curves of tesseral harmonics.

Figure 4.4. Nodal curves of sectorial harmonics.

note that sectorial harmonics are simply linear combinations of Re(x + iy)m and

Im(x + iy)m.In summary, zonal harmonics, tesseral harmonics and sectorial harmonics ex-

hibit different nodal curves. Generally, we do not expect a simple pattern for nodalsets of arbitrary spherical harmonics.

Finally, we point out that it is beyond the reach of this book to classify homo-geneous harmonic polynomials in arbitrary dimensions.

4.3. Mean-Value Properties

As a simple consequence of the Poisson integral formula, harmonic functions areequal to their mean values over arbitrary spheres. This is the so-called mean-value

property. In this section, we will use the mean-value property to discuss harmonicfunctions. The fundamental solution is not used throughout this section.

We first prove by a simple calculation that harmonic functions satisfy the mean-value property.

Theorem 4.11. Let Ω be a domain in Rn and u ∈ C 2(Ω) be harmonic in Ω.Then

u(x) = 1

ωnrn−1

∂Br(x)

u(y)dS y for any Br(x) ⊂ Ω,

where ωn is the area of the unit sphere ∂B1 in Rn.

8/16/2019 PDE . QUING



Proof. Take any ball Br(x) ⊂ Ω. For any ρ ∈ (0, r), we apply the Green’s

formula in Bρ(x) and get Bρ(x)

u =

∂Bρ(x)

∂u∂n

dS = ρn−1 ∂B1

∂u∂ρ

(x + ρw)dS w

= ρn−1 ∂

∂ρ

∂B1

u(x + ρw)dS w.

(4.7)

Hence for the harmonic function u, we have for any ρ ∈ (0, r)

∂

∂ρ

∂B1

u(x + ρw) dS w = 0.

Integrating from 0 to r , we obtain ∂B1

u(x + rw) dS w =

∂B1

u(x) dS w = u(x)ωn,

oru(x) =

1

ωn

∂B1

u(x + rw) dS w.

A simple change of variables yields the desired result.

Theorem 4.11 asserts that harmonic functions equal to their mean values onspheres. There are two versions of mean-value properties, mean values on spheresand mean values on balls.

Definition 4.12. We assume that Ω is a domain in Rn. For u ∈ C (Ω), (i) usatisfies the first mean-value property if

u(x) = 1

ωnrn−1 ∂Br(x)

u(y)dS y for any Br(x) ⊂ Ω;

(ii) u satisfies the second mean-value property if

u(x) = n

ωnrn

Br(x)

u(y)dy for any Br(x) ⊂ Ω,

where ωn denotes the surface area of the unit sphere in Rn.

We note that ωnrr−1 is the surface area of the sphere ∂Br(x) and ωnrn/n isthe volume of the ball Br(x).

These two definitions are equivalent. In fact, if we write (i) as

u(x)rn−1 = 1

ωn ∂Br(x)

u(y)dS y ,

we integrate with respect to r to get (ii). If we write (ii) as

u(x)rn = n

ωn

Br(x)

u(y)dy,

we differentiate to get (i).

By a change of variables, we also write mean-value properties in the followingequivalent forms:

8/16/2019 PDE . QUING



(i) u satisfies the first mean-value property if

u(x) =

1

ωn ∂B1

u(x + ry)dS y for any Br(x) ⊂ Ω;

(ii) u satisfies the second-mean value property if

u(x) = n

ωn

B1

u(x + ry) dy for any Br(x) ⊂ Ω.

For a function u satisfying mean-value properties, u is required only to becontinuous. However, a harmonic function is required to be C 2. We now provethese two are equivalent. We already proved that harmonic functions satisfy the

mean-value property in Theorem 4.11. We now prove that any function with themean-value property is harmonic.

Theorem 4.13. If u ∈

C (Ω) has the mean-value property in Ω, then u is

smooth and harmonic in Ω.

Proof. For the smoothness, we prove that u is equal to the convolution of

itself with some smooth function. To this end, we choose a function ϕ ∈ C ∞0 (B1)with

B1

ϕ = 1 and ϕ(x) = ψ(|x|), i.e.,

ωn

1

0

rn−1ψ(r) dr = 1.

The existence of such a function can be verified easily. We define ϕε(x) = 1

εnϕx

ε

for any ε > 0. Obviously, suppϕε ⊂ Bε. We claim

u(x) = Ω u(y)ϕε(y−

x)dy for any x∈

Ω with d(x, ∂ Ω) > ε.

Then it follows that u is smooth. Moreover, by (4.7) in the proof of Theorem 4.11and the mean-value property, we have for any Br(x) ⊂ Ω

Br(x)

∆u = rn−1 ∂

∂r

∂B1

u(x + rw)dS w = rn−1 ∂

∂r

ωnu(x)

= 0.

This implies ∆u = 0 in Ω.Now we prove the claim. For any x ∈ Ω and ε < dist (x, ∂ Ω), we have

Ω

u(y)ϕε(y − x)dy =

Bε

u(x + y)ϕε(y)dy = 1

εn

Bε

u(x + y)ϕy

ε

dy

= B1

u(x + εy)ϕ(y)dy

=

1

0

rn−1dr

∂B1

u(x + εrw)ϕ(rw)dS w

=

1

0

ψ(r)rn−1dr

∂B1

u(x + εrw)dS w

= u(x)ωn

1

0

ψ(r)rn−1dr = u(x),

where in the last equality we used the mean-value property.

8/16/2019 PDE . QUING



By combining Theorems 4.11 and 4.13, we conclude the following result.

Corollary 4.14. Harmonic functions are smooth and satisfy the mean-value property.

Now we prove the maximum principle for harmonic functions.

Theorem 4.15. Let u ∈ C 2(Ω) ∩ C (Ω) be harmonic in Ω. Then u assumes its

maximum and minimum only on ∂ Ω unless u is constant.

Proof. We prove only for the maximum. Set

D =

x ∈ Ω; u(x) = M ≡ max

Ωu

⊂ Ω.

It is obvious that D is relatively closed; namely, for any sequence xm ⊂ D, if xm → x ∈ Ω, then x ∈ D. This follows easily from the continuity of u. Next we

show that D is open. For any x0 ∈ D, take ¯Br(x0) ⊂ Ω for some r > 0. By themean-value property, we have

M = u(x0) = n

ωnrn

Br(x0)

u(y)dy ≤ M n

ωnrn

Br (x0)

dy = M.

This implies u = M in Br(x0). Hence D is both relatively closed and open in Ω.Therefore either D = φ or D = Ω. A similar argument can be used to prove for theminimum.

A consequence is the uniqueness of solutions of the Dirichlet problem in abounded domain.

Corollary 4.16. For any f ∈ C (Ω) and ϕ ∈ C (∂ Ω), there exists at most one

solution u ∈ C 2

(Ω) ∩ C (Ω) of the following problem ∆u = f in Ω,

u = ϕ on ∂ Ω.

Proof. Let w be the difference of any two solutions. Then ∆w = 0 in Ω andw = 0 on ∂ Ω. Theorem 4.15 implies w = 0 in Ω.

Compare Corollary 4.16 with Lemma 3.11, where the uniqueness was proved

by energy estimates for solutions u ∈ C 2(Ω) ∩ C 1(Ω).In general, the uniqueness does not hold for unbounded domains. Consider the

Dirichlet problem in an unbounded domain Ω

∆u = 0 in Ω

u = 0 on ∂ Ω.

First, we consider the case Ω = x ∈ Rn; |x| > 1. We have a nontrivial solution ugiven by

u(x) =

log |x| for n = 2;

|x|2−n − 1 for n ≥ 3.

Note that u → ∞ as |x| → ∞ for n = 2 and u is bounded for n ≥ 3. Next,

we consider the upper half space Ω = x ∈ Rn; xn > 0. Then u(x) = xn is anontrivial solution, which is unbounded.

8/16/2019 PDE . QUING



Next, we employ the mean-value property to prove interior gradient estimates.

Lemma 4.17. Suppose u ∈ C BR(x0) is harmonic in BR(x0). Then

|∇u(x0)| ≤ n

R maxBR(x0)

|u|.

We note that Lemma 4.17 improves Lemma 4.8.

Proof. For simplicity, we assume u ∈ C 1( BR(x0)). Since u is smooth, then∆(uxi

) = 0, i.e., uxi is also harmonic in BR(x0). Hence uxi

satisfies the mean-value

property. By the divergence theorem, we have

uxi(x0) =

n

ωnRn

BR(x0)

uxi(y)dy =

n

ωnRn

∂BR(x0)

u(y) ν idS y ,

and hence

|uxi(x0)| ≤ n

ωnRn max∂BR(x0)

|u| · ωnRn−1 ≤ nR

maxBR(x0)

|u|.


When u is nonnegative, we can improve Lemma 4.17.


BR(x0)

is a nonnegative harmonic function in BR(x0). Then

|∇u(x0)| ≤ n

Ru(x0).

This result is often referred to as the differential Harnack inequality. It hasmany important consequences.

Proof. As in the proof of Lemma 4.17, by the divergence theorem and thenonnegativeness of u, we have

|uxi(x0)| ≤ n

ωnRn

∂BR(x0)

u(y) dS y = n

Ru(x0),

where in the last equality we used the mean-value property.

As an application, we prove the following result which is referred to as theLiouville Theorem.

Corollary 4.19. A harmonic function in Rn bounded from above or below is constant.

Proof. Suppose u is a harmonic function in Rn with u

≥ 0. Then for any

x ∈ Rn, we apply Lemma 4.18 to u in BR(x) and then let R → ∞. We conclude∇u(x) = 0 for any x ∈ Rn and hence u is constant.

As another application, we prove the Harnack inequality, which asserts thatnonnegative harmonic functions have comparable values in compact subsets.

Lemma 4.20. Suppose u is a nonnegative harmonic function in BR(x0). Then

u(x) ≤ Cu(y) for any x, y ∈ B R2

(x0),


8/16/2019 PDE . QUING



Proof. By considering u + ε for any ε > 0, we assume u > 0 in BR(x0). For

any x ∈ BR/2(x0), we apply Lemma 4.18 to u in BR/2(x) and get

|∇u(x)| ≤ 2nR

u(x),

or

|∇ log u(x)| ≤ 2n

R .

For any x, y ∈ BR/2(x0),

log u(x)

u(y) =

1

0

d

dt log u(ty + (1 − t)x)dt.

With ty +(1−t)x ∈ BR/2(x0) for any t ∈ [0, 1] and |x−y| ≤ R, a simple integrationyields

log

u(x)

u(y) ≤ |x − y| 1

0 |∇ log u(ty + (1 − t)x)|dt ≤ 2n

R |x − y| ≤ 2n.

Therefore

u(x) ≤ e2nu(y).


In fact, Lemma 4.20 can be proved directly by the mean-value property.

Another proof of Lemma 4.20. Consider any B4r(x) ⊂ BR(x0). We firstclaim

u(x) ≤ 2nu(y) for any x, y ∈ Br(x).

To see this, we note that Br(x) ⊂ B2r(y) ⊂ B4r(x) for any x, y ∈ Br(x). Then the

mean-value properties imply

u(x) = n

ωnrn

Br(x)

u ≤ n

ωnrn

B2r (y)

u ≤ 2nu(y).

With r = R/8, we choose finitely many x1, · · · , xN ∈ BR/2(x0) such that Br(xi)covers BR/2(x0). Obviously, B4r(xi) ⊂ BR(x0) and N is a constant depending only

on n. Then we have the desired result by choosing C = 2nN .

4.4. The Maximum Principle

One of the important tools in studying harmonic functions is the maximumprinciple. In this section, we discuss the maximum principle for a class of ellipticdifferential equations slightly more general than the Laplace equation and derive apriori estimates for boundary-value problems, interior gradient estimates and the

Harnack inequality.Throughout this section, we assume Ω is a bounded domain in Rn and c is a

continuous function in Ω. Sometimes, we assume c ≤ 0 in Ω. For u ∈ C 2(Ω)∩C (Ω)and f ∈ C (Ω), we consider

(4.8) ∆u + c(x)u = f (x) in Ω.

Obviously, u is harmonic if c = f = 0. A C 2-function u is called a subsolution

(or supersolution) of (4.8) if ∆u + cu ≥ f (or ∆u + cu ≤ f ). If c = 0 andf = 0, subsolutions (or supersolutions) are called subharmonic (or superharmonic)

8/16/2019 PDE . QUING


4.4. THE MAXIMUM PRINCIPLE 77

functions. In other words, a C 2-function u is subharmonic (or superharmonic) if

∆u ≥ 0 (or ∆u ≤ 0).

Now we start to prove the maximum principle without using mean-value prop-erties. We prove a simple result first.

Lemma 4.21. Suppose u ∈ C 2(Ω) ∩ C (Ω) satisfies ∆u + cu > 0 in Ω with

c(x) ≤ 0 in Ω. If u has a nonnegative maximum in Ω, then u cannot attain this maximum in Ω.

Proof. Suppose u attains its nonnegative maximum of Ω at x0 ∈ Ω. Then

∇u(x0) = 0 and the Hessian matrix∇2u(x0)

is nonpositive definite. By taking a

trace, we have

∆u(x0) = tr(∇2u(x0)) ≤ 0,

and hence

(∆u + cu)(x0)

≤0,

which is a contradiction.

Remark 4.22. If c(x) ≡ 0, the requirement for the nonnegativeness on u canbe removed. This remark holds for many results in the rest of this section.

Remark 4.23. Lemma 4.21 in fact holds for more general elliptic differentialequations. For u ∈ C 2(Ω) ∩ C (Ω) and f ∈ C (Ω), consider

Lu ≡n

i,j=1

aij (x)uxixj +

ni=1

bi(x)uxi + c(x)u = f (x) in Ω.

We assume that aij , bi and c are continuous and hence bounded in Ω and that Lis uniformly elliptic in Ω in the following sense

ni,j=1

aij (x)ξ iξ j ≥ λ|ξ |2 for any x ∈ Ω and any ξ ∈ Rn,

for some positive constant λ. The uniform ellipticity means a uniform positive lowerbound of all eigenvalues of (aij ) in Ω. Many results in this section hold for Lu = f .

We now prove the maximum principle for subsolutions.

Theorem 4.24. Suppose u ∈ C 2(Ω) ∩ C (Ω) satisfies ∆u + cu ≥ 0 in Ω with c(x) ≤ 0 in Ω. Then u attains on ∂ Ω its nonnegative maximum in Ω, i.e.,

supΩ

u ≤ sup∂ Ω

u+.

A continuous function in Ω always attains its maximum in Ω. Theorem 4.24

asserts that any subsolution continuous up to the boundary attains its nonnegativemaximum on the boundary ∂ Ω, possibly also in Ω. Theorem 4.24 is often called

the weak maximum principle. A stronger version asserts that subsolutions attaintheir maximum only on the boundary. We will prove the strong maximum principlelater.

Proof. Without loss of generality, we assume Ω is contained in a ball of radiusR and centered at the origin, i.e., Ω ⊂ BR. For any ε > 0, consider

uε(x) = u(x) − ε(R2 − |x|2).

8/16/2019 PDE . QUING



By c ≤ 0 and |x| < R in Ω, we have

∆uε + cuε = ∆u + cu + 2nε

−εc(R2

− |x

|2)

≥2nε > 0 in Ω.

By Lemma 4.21, uε attains its nonnegative maximum only on ∂ Ω, i.e.,

supΩ

uε ≤ sup∂ Ω

u+ε .

Then we obtain

supΩ

u ≤ supΩ

uε + εR2 ≤ sup∂ Ω

u+ε + εR2 ≤ sup

∂ Ωu+ + εR2.

We finish the proof by letting ε → 0.

The following result is often referred to as the comparison principle.

Corollary 4.25. Suppose c ≤ 0 in Ω and u, v ∈ C 2(Ω) ∩ C (Ω) satisfy

∆u + cu

≥∆v + cv in Ω

u ≤v on ∂ Ω.

Then u ≤ v in Ω.

Proof. The difference w = u − v satisfies ∆w + cw ≥ 0 in Ω and w ≤ 0 on

∂ Ω. Theorem 4.24 implies w ≤ 0 in Ω.

The comparison principle provides a reason that functions u satisfying ∆u +cu ≥ f are called subsolutions. They are less than solutions v of ∆v + cv = f with

the same boundary values.A consequence of the comparison principle is the uniqueness of solutions of

Dirichlet problems.

Corollary 4.26. Let Ω be a bounded domain in Rn and u

∈C 2(Ω)

∩C (Ω) be

a solution of

∆u + cu = f in Ω

u = ϕ on ∂ Ω,

for some f ∈ C (Ω) and ϕ ∈ C (∂ Ω). If c(x) ≤ 0 in Ω, then u is the unique solution.

The boundedness of domains Ω is essential, since it guarantees the existenceof maximum and minimum of u in Ω. The uniqueness does not hold if domains

are unbounded. Examples are given in Section 4.3. Equally important is thenonpositiveness of the coefficient c. For example, u = sin x is a nontrivial solutionof the problem

u + u = 0 in (0, π)

u(0) = u(π) = 0.

The weak maximum principle asserts subsolutions of elliptic differential equa-

tions with nonpositive zero order coefficients attain on boundary their nonnegativemaximum. In fact, they can attain their nonnegative maximum only on boundary,unless they are constant. This is the strong maximum principle. To prove this,we need the following Hopf lemma. Let u be a subsolution in Ω which attains its

maximum on ∂ Ω, say at x0 ∈ ∂ Ω. Then ∂u

∂n(x0) ≥ 0. The Hopf lemma asserts that

it is in fact positive.

8/16/2019 PDE . QUING



Theorem 4.27. Let B be an open ball in Rn with x0 ∈ ∂B . Suppose u ∈C 2(B) ∩ C 1(B ∪ x0) satisfies ∆u + cu ≥ 0 in B with c(x) ≤ 0 in B. Assume in

addition that u(x) < u(x0) for any x ∈ B and u(x0) ≥ 0.

Then ∂u

∂n(x0) > 0.

Proof. We assume that B is centered at the origin with radius r. We assume

further that u ∈ C ( B) and u(x) < u(x0) for any x ∈ B\x0, since we can constructa ball B∗ ⊂ B tangent to B at x0.

Set D = B ∩ Br/2(x0) and for some α to be determined

h(x) = e−α|x|2 − e−αr2 .

A direct calculation yields

∆h + ch =e−α|x|2

4α2|x|2 − 2nα + c− ce−αr2

≥e−α|x|2

4α2|x|2 − 2nα + c

,

where we used c ≤ 0 in B . By |x| ≥ r/2 in D and choosing α sufficiently large, weconclude

∆h + ch > 0 in D.

Consider v(x) = u(x) + εh(x). Then,

∆v + cv = ∆u + cu + ε(∆h + ch) > 0 in D,

for any ε > 0. By Lemma 4.21, v cannot attain its nonnegative maximum in D .Next, we prove, for some small ε > 0, v attains at x0 its nonnegative maximum.

Consider v on the boundary ∂ D.(i) For x ∈ ∂D∩B, since u(x) < u(x0), so u(x) < u(x0)−δ for some δ > 0. Take

ε small such that εh < δ on ∂D ∩ B. Hence, for such an ε, we have v(x) < u(x0)for any x ∈ ∂D ∩ B.

(ii) On ∂D ∩ ∂B , h(x) = 0 and u(x) < u(x0) for x = x0. Hence v(x) < u(x0)on ∂D ∩ ∂B \ x0 and v(x0) = u(x0).

Therefore, we conclude

∂v

∂n(x0) ≥ 0,

or∂u

∂n(x0) ≥ −ε

∂h

∂n(x0) = 2εαre−αr2 > 0.


Now, we are ready to prove the strong maximum principle.

Theorem 4.28. Let u ∈ C 2(Ω) ∩ C (Ω) satisfy ∆u + cu ≥ 0 with c(x) ≤ 0 in

Ω. Then the nonnegative maximum of u in Ω can be attained only on ∂ Ω unless uis a constant.

8/16/2019 PDE . QUING



Proof. Let M be the nonnegative maximum of u in Ω. Set D = x ∈Ω; u(x) = M . It is relatively closed in Ω. We prove D = Ω by contradiction.

If D is a proper subset of Ω, we can find an open ball B ⊂ Ω \ D with a point onits boundary belonging to D. In fact, we may choose a point p ∈ Ω \ D such thatd( p, D) < d( p, ∂ Ω) first, construct a small ball in Ω \ D of center p and then extend

this ball. It hits D before hitting ∂ Ω. Suppose x0 ∈ ∂ B ∩ D. Obviously, we have∆u + cu ≥ 0 in B and

u(x) < u(x0) for any x ∈ B and u(x0) = M ≥ 0.

Theorem 4.27 implies ∂u

∂n(x0) > 0, where n is the outward normal vector at x0 to

the ball B . While x0 is an interior maximal point of Ω, we have ∇u(x0) = 0. This

leads to a contradiction.

The following result improves Corollary 4.25.

Corollary 4.29. Suppose u ∈ C 2(Ω) ∩ C (Ω) satisfies ∆u + cu ≥ 0 in Ω with c(x) ≤ 0 in Ω. If u ≤ 0 on ∂ Ω, then either u < 0 in Ω or u ≡ 0 in Ω.

In order to discuss boundary-value problems with general boundary conditions,we need the following result, which is a corollary of Theorem 4.27 and Theorem4.28.

Corollary 4.30. Suppose Ω is a bounded C 1-domain in Rn and that u ∈C 2(Ω) ∩ C 1(Ω) satisfies ∆u + cu ≥ 0 in Ω with c(x) ≤ 0. Assume u attains its nonnegative maximum at x0 ∈ Ω. Then x0 ∈ ∂ Ω and

∂u

∂n(x0) > 0,

unless u is a constant in Ω.

Now we discuss the uniqueness of solutions of a class of boundary-value prob-

lems with general boundary conditions.

Corollary 4.31. Suppose Ω is a bounded C 1-domain in Rn, c is a continuous

function in Ω and α is a continuous function on ∂ Ω. Let u ∈ C 2(Ω) ∩ C 1(Ω) be a solution of the boundary-value problem

∆u + cu =f in Ω

∂u

∂n + αu =ϕ on ∂ Ω,

for some f ∈ C (Ω) and ϕ ∈ C (∂ Ω). Assume in addition that c ≤ 0 in Ω and α ≥ 0

on ∂ Ω. Then u is the unique solution if c ≡ 0 or α ≡ 0. If c ≡ 0 and α ≡ 0, u is unique up to additive constants.

Proof. Suppose u is a solution of the homogeneous problem

∆u + cu = 0 in Ω

∂u

∂n + αu = 0 on ∂ Ω.

Case 1. c ≡ 0 or α ≡ 0. Suppose that u has a positive maximum at x0 ∈ Ω.If u is a positive constant, there leads to a contradiction to the condition c ≡ 0 in

8/16/2019 PDE . QUING



Ω or α ≡ 0 on ∂ Ω. Otherwise x0 ∈ ∂ Ω and ∂u∂n (x0) > 0 by Corollary 4.30, which

contradicts the boundary value. Therefore u ≡ 0.

Case 2 . c ≡ 0 and α ≡ 0. Suppose u is a nonconstant solution. Then itsmaximum in Ω is assumed only on ∂ Ω by Theorem 4.28, say at x0 ∈ ∂ Ω. AgainCorollary 4.30 implies ∂u

∂n (x0) > 0. This contradiction shows that u is a constant.

Proving the uniqueness of solutions of boundary-value problems is an impor-

tant application of maximum principles. Equally important or more important isto derive a priori estimates. In derivations of a priori estimates, essential stepsconsist of constructions of auxiliary functions. Next, we derive a priori estimatesfor solutions of Dirichlet problems.

Theorem 4.32. Suppose u ∈ C 2(Ω) ∩ C (Ω) satisfies

∆u + cu = f in Ω,

u = ϕ on ∂ Ω,

for some f ∈ C (Ω) and ϕ ∈ C (∂ Ω). If c(x) ≤ 0, then

maxΩ

|u| ≤ max∂ Ω

|ϕ| + C maxΩ

|f |,where C is a positive constant depending only on n and diam(Ω).

If Ω = BR(x0), then

maxBR(x0)

|u| ≤ max∂BR(x0)

|ϕ| + R2

2n maxBR(x0)

|f |.

This follows from the proof below easily.

Proof. We construct an auxiliary function w in Ω such that(i) (∆ + c)(w ± u) = (∆ + c)w ± f ≤ 0, or (∆ + c)w ≤ ∓f in Ω;(ii) w ± u = w ± ϕ ≥ 0, or w ≥ ∓ϕ on ∂ Ω.

Set

F = maxΩ

|f |, Φ = max∂ Ω

|ϕ|.We need

∆w + cw ≤ −F in Ω,

w ≥ Φ on ∂ Ω.

Without loss of generality, we assume Ω is contained in a ball of radius R andcentered at the origin, i.e., Ω ⊂ BR. Set

w = Φ + F 2n (R2 − |x|2).

Then by c ≤ 0 and |x| ≤ R in Ω, we have

∆w + cw = −F + cΦ + cF

2n(R2 − |x|2) ≤ −F.

We also have w ≤ Φ on ∂ Ω. Hence w satisfies (i) and (ii). By Corollary 4.25, thecomparison principle, we conclude −w ≤ u ≤ w in Ω, and in particular,

|u(x)| ≤ Φ + 1

2n(R2 − |x|2)F for any x ∈ Ω.

8/16/2019 PDE . QUING




Now we consider a class of general boundary-value problems.Theorem 4.33. Suppose u ∈ C 2(Ω) ∩ C 1(Ω) satisfies

∆u + cu = f in Ω,

∂u

∂n + αu = ϕ on ∂ Ω,

for some f ∈ C (Ω) and ϕ ∈ C (∂ Ω). If c(x) ≤ 0 in Ω and α(x) ≥ α0 on ∂ Ω for a

positive constant α0, then

maxΩ

|u| ≤ C

max∂ Ω

|ϕ| + maxΩ

|f |,

where C is a positive constant depending only on α0 and diam(Ω).

If c ≡ 0 in Ω and α ≡ 0 on ∂ Ω, the homogeneous problem in Theorem 4.33(with f ≡ 0 in Ω and ϕ ≡ 0 on ∂ Ω) admits a nontrivial solution u ≡ 1 in Ω. Hence

there does not hold a sup-norm estimate in this case.

Proof. By assuming Ω ⊂ BR for some R > 0, we prove

supΩ

|u| ≤ 1

α0max

∂ Ω|ϕ| +

1

2n

1 + R2

α0+ R2

max

Ω|f |.

Set

Φ = max∂ Ω

|ϕ|, F = maxΩ

|f |,and

v(x) =

1

α0 Φ +

1

2n1 + R2

α0 + R

2

− |x|2F.

Then

∆v + cv = −F + cv ≤ −F in Ω,

and

∂v

∂n + αv =

α

α0Φ +

F

2n

− 2x · n + α(1 + R2

α0+ R2 − |x|2)

≥Φ +

F

2n(−2x · n + 1 + R2) ≥ Φ on ∂ Ω,

where n is the unit exterior normal vector of ∂ Ω. With

w = v ± u,

we have

∆w + cw ≤ 0 in Ω,

∂w

∂n + αw ≥ 0 on ∂ Ω.

It is easy to prove w ≥ 0 in Ω, or

|u| ≤ v in Ω.


8/16/2019 PDE . QUING



In the following, we derive gradient estimates, estimates of derivatives. The

basic method to derive gradient estimates, the so-called the Bernstein method,

involves a differentiation of the equation with respect to xk, k = 1, . . . , n, followedby a multiplication by uxk

and summation over k . The maximum principle is thenapplied to the resulting equation in the function v = |∇u|2, possibly with some

modifications. There are two kinds of gradient estimates, global gradient estimatesand interior gradient estimates. We use the Laplace equation to demonstrate ideasof proving interior gradient estimates. Compare with Lemma 4.17.

Theorem 4.34. Suppose u ∈ C ( B1) is harmonic in B1. Then

supB 1

2

|∇u| ≤ C sup∂B1

|u|,


Proof. A direct calculation shows

(|∇u|2) = 2n

i,j=1

u2xixj

+ 2n

i=1

uxi(u)xi

= 2n

i,j=1

u2xixj

,

where we used u = 0 in B1. Hence |∇u|2 is a subharmonic function. To getinterior estimates, we need to introduce a cut-off function. For any nonnegativefunction ϕ ∈ C 10 (B1), we have

(ϕ|∇u|2) = (ϕ)|∇u|2 + 4

ni,j=1

ϕxiuxj

uxixj + 2ϕ

ni,j=1

u2xixj

.

By the Cauchy inequality, we have

4|ϕxi uxj uxixj | ≤ 2ϕu2xixj +

2

ϕ ϕ2xi u

2xj .

Then

(ϕ|∇u|2) ≥ ∆ϕ − 2|∇ϕ|2

ϕ

|∇u|2.

To get a bounded |∇ϕ|2/ϕ in B1, we take ϕ = η2 for some η ∈ C 10 (B1). Moreover,we require η ≡ 1 in B1/2. Then

(η2|∇u|2) ≥ 2ηη − 6|∇η|2

|∇u|2 ≥ −C |∇u|2,

where C is a positive constant depending only on η and n. Note

(u2) = 2|∇u|2 + 2uu = 2|∇u|2,

since u is harmonic. By taking a constant α large enough, we get

(η2|∇u|2 + αu2) ≥ (2α − C )|∇u|2 ≥ 0.

We apply Theorem 4.24, the maximum principle, to get

supB1

(η2|∇u|2 + αu2) ≤ sup∂B1

(η2|∇u|2 + αu2).

This implies the desired result easily since η = 0 on ∂ B1 and η = 1 in B1/2.

Next we derive the differential Harnack inequality for harmonic functions.Compare this with Lemma 4.18.

8/16/2019 PDE . QUING



Theorem 4.35. Suppose u is a positive harmonic function in B1. Then

supB 1

2 |∇log u

| ≤C,


Proof. Set v = log u. A direct calculation shows

v = −|∇v|2.

Next, we prove an interior gradient estimate for v . By setting w = |∇v|2, we get

w + 2n

i=1

vxiwxi

= 2n

i,j=1

v2xixj

.

As before, we need to introduce a cut-off function. First note

(4.9)n

i,j=1

v2xixj

≥ ni=1

v2xixi

≥ 1n

(v)2 = |∇v|4

n =

w2

n .

Take a nonnegative function ϕ ∈ C 10 (B1). As in the proof of Theorem 4.34, we have

(ϕw) + 2n

i=1

vxi(ϕw)xi

=2ϕn

i,j=1

v2xixj

+ 4n

i,j=1

ϕxivxj

vxixj + 2w

ni=1

ϕxivxi

+ (ϕ)w

≥ϕn

i,j=1

v2xixj

− 2|∇ϕ||∇v|3 +

ϕ − 4|∇ϕ|2

ϕ |∇v|2.

Here we keep one v2xixj

instead of dropping it entirely in the proof of Theorem 4.34.

To get a bounded |∇ϕ|2/ϕ in B1, we take ϕ = η4 for some η ∈ C 10 (B1). We obtainby (4.9)

(η4w) + 2n

i=1

vxi(η4w)xi

≥ 1

nη4|∇v|4 − 8η3|∇η||∇v|3 + 4η2(ηη − 13|∇η|2)|∇v|2.

Note1

2nt4 − 8|∇η|t3 + 4(ηη − 13|∇η|2)t2 ≥ −C for any t ∈ R,

where C is a positive constant depending only on n and η . Hence with t = η|∇v|,we get

(η4w) + 2n

i=1

vxi(η4w)xi

≥ 1

2nη4w2 − C.

where C is a positive constant depending only on n and η .Suppose η4w attains its maximum at x0 ∈ B1. Then ∇(η4w) = 0 a n d

(η4w) ≤ 0 at x0. Hence

η4w2(x0) ≤ C.

8/16/2019 PDE . QUING



If w(x0) ≥ 1, then η4w(x0) ≤ C . Otherwise η4w(x0) ≤ η4(x0). In both cases we

conclude

η4w ≤ C ∗ in B1,where C ∗ is a positive constant depending only on n and η. With the help of thedefinition of w and η = 1 in B1/2, this implies the desired result.

The following result is called the Harnack inequality. Compare it with Lemma

4.20.

Corollary 4.36. Suppose u is a nonnegative harmonic function in B1. Then

u(x1) ≤ Cu(x2) for any x1, x2 ∈ B 12

,


The proof is identical to the first proof of Lemma 4.20 and is omitted.To end this section, we discuss isolated singularities of harmonic functions.

We note that the fundamental solution of the Laplace operator has an isolatedsingularity and is harmonic elsewhere. The next result asserts that an isolated

singularity of harmonic functions, if “better” than that of the fundamental solution,can be removed.

Theorem 4.37. Suppose u is harmonic in BR \ 0 ⊂ Rn and satisfies

u(x) =

o(log |x|), n = 2

o(|x|2−n), n ≥ 3as |x| → 0.

Then u can be defined at 0 so that it is C 2 and harmonic in BR.

Proof. Assume u is continuous in 0 < |x| ≤ R. Let v solve

∆v = 0 in BR,v = u on ∂ BR.

The existence of v is guaranteed by Theorem 4.6. By the maximum principle,

|v| ≤ M in BR,

where M = max∂BR|u|. We prove u = v in BR \ 0. Set w = v − u in BR \ 0

and M r = max∂Br|w| for any r < R. We only consider the case n ≥ 3. First, we

have

|w(x)| ≤ M r · rn−2

|x|n−2 for any x ∈ ∂Br.

Note that w and 1

|x

|n−2

are harmonic in BR \ Br with w = 0 on ∂BR. Hence the

maximum principle implies

|w(x)| ≤ M r · rn−2

|x|n−2 for any x ∈ BR \ Br,

whereM r = max

∂Br

|v − u| ≤ max∂Br

|v| + max∂Br

|u| ≤ M + max∂Br

|u|.Then for each fixed x = 0,

|w(x)| ≤ rn−2

|x|n−2M +

1

|x|n−2rn−2 max

∂Br

|u| → 0 as r → 0.

8/16/2019 PDE . QUING



This implies w = 0 in BR \ 0.

4.5. Schauder Estimates

In this section, we discuss regularity of solutions of the Poisson equation. Let Ω

be a domain in Rn and f be a continuous function in Ω. Then the Poisson equationhas the form

(4.10) ∆u = f.

The Laplace operator ∆ acts on C 2-functions and ∆u is continuous for any C 2-function u. Conversely, we ask whether u is C 2 if f is continuous. The followingexample provides a negative answer.

Example 4.38. For any R < 1, consider in BR ⊂ R2

(4.11) ∆u = x2

2 −x2

12|x|2 4

(− log |x|)1/2 + 1

2(− log |x|)3/2.

By letting f (x) be the function in the right hand side of (4.11), we note that f is

continuous in BR if we set it equal to zero at the origin. The function

u(x) = (x21 − x2

2)(− log |x|)1/2 ∈ C ( BR) ∩ C ∞( BR \ 0)

satisfies (4.11) in BR \ 0. Note that u is not in C 2(BR) since limx→0 ux1x1 = ∞.In fact, (4.11) has no C 2-solutions. Assume on the contrary that there exists aC 2-solution v in BR. Then the function w = u − v is harmonic and bounded inBR \ 0. By Theorem 4.37, w may be redefined at the origin so that w = 0 in

BR and therefore belongs to C 2(BR). In particular, the (finite) limit limx→0 ux1x1

exists, which is a contradiction.

It turns out that u fails to be C 2 because there is no control on the module of continuity of f . If there is a better assumption than the continuity of f , we can

control the module of continuity of ∇2u. Here, controlling the module of continuitymeans imposing conditions on how fast functions approach constants. The simplestway to do this is known by Holder norms.

In the following, we discuss the Schauder estimates for Poisson equations.

Schauder estimates are among the most important results in the theory of ellipticpartial differential equations. They give Holder regularity estimates for solutionswith Holder continuous data and form the basis of general existence theorems.There are two classes of Schauder estimates, interior estimates and boundary es-

timates. In this section, we only discuss interior Schauder estimates for Poissonequations. In its simplest form, the interior Schauder estimate asserts that theC α-norm of any second-order derivatives of u in a ball is estimated by the sum of the C α-norm of ∆u and the sup-norm of u in a larger ball. It is an optimal result

in the sense that u is as regular as the data allow . Even though ∆u is just oneparticular combination of second-order derivatives of u, the Holder continuity of ∆u implies the same regularity for all second-order derivatives.

We first define pointwise Holder continuity.

Definition 4.39. Let k be a positive integer and α ∈ (0, 1) be a constant.Suppose u is a function defined in B1.

8/16/2019 PDE . QUING


4.5. SCHAUDER ESTIMATES 87

(1) A function u is C α at 0 if

|u(x)

−u(0)

| ≤c

|x

|α for any x

∈B1,

for a positive constant c. The Holder semi-norm of u at 0 is defined by

[u]C α (0) ≡ sup|x|≤1

|u(x) − u(0)||x|α

.

(2) A function u is C k,α at 0 if there exists a polynomial p(x) of degree k suchthat

|u(x) − p(x)| ≤ c|x|k+α for any x ∈ B1,

for a positive constant c.

We note that the definition of Holder continuity here gives a quantitative speedof how fast functions approach constant or a polynomial. We are only interested

in asymptotic behavior of u as x →

0. The ball B1 can be replaced by any Br

for small r. This affects only the definition of Holder semi-norms. The asymptoticbehavior of u as x → 0 remains the same.

We also note that the polynomial p in (2) is unique if it exists. We point out

that u in Definition 4.39(2) is not necessarily C k in a neighborhood of 0. If u is,then p is evidently the Taylor expansion of order k of u at 0. In particular, fork = 2,

p(x) =u(0) + ∇u(0) · x + 1

2xT ∇2u(0)x

=u(0) +n

i=1

uxi(0)xi +

1

2

ni,j=1

uxixj(0)xixj .

We first state a global estimate on solutions in terms of nonhomogeneous termsand boundary values.

Lemma 4.40. Suppose u ∈ C 2(B1) ∩ C ( B1) satisfies

∆u = f in B1,

for some f ∈ C (B1). Then

supB1

|u| ≤ sup∂B1

|u| + 1

2n sup

B1

|f |.

This is a special case of Theorem 4.32.Now we begin to estimate second derivatives of solutions.

Lemma 4.41. Suppose u

∈C 2(B1) satisfies

∆u = f in B1,

for some f ∈ C (B1). Then for any α ∈ (0, 1), there exist constants c0 > 0, µ ∈ (0, 1)and ε0 > 0, depending only on n and α, such that, if |u| ≤ 1 and |f | ≤ ε0 in B1,then there exists a second order harmonic polynomial

p(x) = 1

2xT Ax + B · x + C,

satisfying

|u(x) − p(x)| ≤ µ2+α for any |x| ≤ µ,

8/16/2019 PDE . QUING



and

|A

|+

|B

|+

|C

| ≤c0.

Here, A is an n × n matrix, B is a vector and C is a constant.

Proof. Without loss of generality, we assume u ∈ C ( B1). Suppose v is theharmonic function satisfying

∆v = 0 in B1,

v = u on ∂ B1.

The existence of v is guaranteed by Theorem 4.6. Clearly |v| ≤ 1 in B1, by themaximum principle since |u| ≤ 1 on ∂ B1. Then Lemma 4.9 implies

3

k=0 |∇kv

|L∞(B 1

2

)

≤c(n),

where c(n) is a positive constant depending only on n, which is often called a

universal constant. Hence the second order Taylor polynomial of v at 0

p(x) = 1

2xT Ax + B · x + C

has universal bounded coefficients and is harmonic. Now the function u −v satisfies

∆(u − v) = f in B1,

u − v = 0 on ∂ B1.

By Lemma 4.40, we have

|u(x) − v(x)| ≤ 12n

supB1

|f | for any x ∈ B1.

By the mean value theorem, we have for any x ∈ B1/2

|u(x) − p(x)| ≤ |v(x) − p(x)| + 1

2n sup

B1

|f | ≤ c|x|3 supB 1

2

|∇3v| + 1

2n sup

B1

|f |

≤ c(n)|x|3 + 1

2n sup

B1

|f |.

Now take µ small enough such that

c(n)µ1−α ≤ 1

2,

and then take ε0 such that

ε0 ≤ µ2+α.

We then have

|u(x) − p(x)| ≤ 1

2µ2+α +

1

2ε0 ≤ µ2+α for any x ∈ Bµ.

This is the desired estimate.

Now we prove the pointwise Schauder estimate.

8/16/2019 PDE . QUING


4.5. SCHAUDER ESTIMATES 89

Theorem 4.42. Suppose u ∈ C 2(B1) satisfies

∆u = f in B1,

where f is H¨ older continuous at 0. Then, u is C 2,α at 0. Moreover, there exists a

second order polynomial

p(x) = 1

2xT Ax + Bx + C

satisfying

|u(x) − p(x)| ≤ c0|x|2+α|u|L∞ + |f (0)| + [f ]C α (0)

for any x ∈ B1,

and

|A| + |B| + |C | ≤ c0

|u|L∞ + |f (0)| + [f ]C α (0)

,

where c0 is a positive constant depending only on n and α.

Obviously, for u

∈C 2(B1), we have

p(x) = u(0) + ∇u(0) · x + 1

2xT ∇2u(0)x.

Proof. Without loss of generality, we assume f (0) = 0. In general, we set

v = u− 12n f (0)|x|2. Then ∆v = f −f (0) in B1. Furthermore, we assume |u| ≤ 1 and

[f ]C α (0) ≤ ε0 for small ε0 > 0. The general case can be recovered by consideringu

|u|L∞ + 1ε0

[f ]C α (0).

First we claim that there exist harmonic polynomials for any k = 1, 2, · · · ,

P k(x) = 1

2xT Akx + Bkx + C k,

satisfying

|u(x) − P k(x)| ≤ µ(2+α)k for any |x| ≤ µk,

and

|Ak − Ak+1| ≤ cµαk,

|Bk − Bk+1| ≤ cµ(α+1)k,

|C k − C k+1| ≤ cµ(α+2)k,

where µ ∈ (0, 1) and c are positive constants depending only on n and α.Note that the case k = 1 corresponds to Lemma 4.41. Assume it holds for some

k ≥ 1. Set

w(y) = (u − P k)(µky)

µ(2+α)k for any y ∈ B1.

Then

∆w(y) = f (µky)

µαk for any y ∈ B1,

with

supy∈B1

|f (µky)|µαk

≤ [f ]C 0(0) ≤ ε0.

By Lemma 4.41, there is a harmonic polynomial p0 with bounded coefficients suchthat

|w(y) − p0(y)| ≤ µ2+α for |y| ≤ µ.

8/16/2019 PDE . QUING



Now we scale back to get

|u(x) − P k(x) − µ(2+α)k

p0 x

µk | ≤ µ(k+1)(2+α)

for |x| ≤ µk+1

.

Clearly we proved the (k + 1)-th step by letting

P k+1(x) = P k(x) + µ(2+α)k p0

x

µk

.

It is easy to see that Ak, Bk and C k converge and the limiting polynomial

p(x) = 1

2xT A∞x + B∞x + C ∞

satisfies

|P k(x) − p(x)| ≤ c

|x|2µαk + |x|µ(α+1)k + µ(α+2)k

≤ cµ(2+α)k,

for any |x| ≤ µk

. Hence|u(x) − p(x)| ≤ |u(x) − P k(x)| + |P k(x) − p(x)| ≤ cµ(α+2)k for any |x| ≤ µk.

For any x ∈ B1, there exists a nonnegative integer k such that µk+1 ≤ |x| ≤ µk.This implies

|u(x) − p(x)| ≤ c|x|2+α for any x ∈ B1,

which yields the desired result.

Next, we put Theorem 4.42 in the classical form of Schauder estimates. We

first introduce Hoder continuous functions in domains.

Definition 4.43. Let k be a nonnegative integer, α ∈ (0, 1) be a constant andΩ be a domain in Rn.

(1) A function u is C α in Ω, denoted by u ∈ C α(Ω), if

|u(x) − u(y)| ≤ c|x − y|α for any x, y ∈ Ω,

for some constant c. The Holder semi-norm of u in Ω is defined by

[u]C α(Ω) ≡ supx,y∈Ω

|u(x) − u(y)||x − y|α

.

(2) A function u is C k,α in Ω, denoted by u ∈ C k,α(Ω), if u ∈ C k(Ω) and

∇ku ∈ C α(Ω).(3) The C k,α-norm in Ω is defined by

|u|C k,α(Ω) = |u|C k(Ω) + [∇ku]C α(Ω).

In Definition 4.43, we may replace Ω by Ω and define C k,α(Ω) similarly. It iseasy to check that C k,α(Ω) is a Banach space equipped with the C k,α-norm.

For u ∈ C α(Ω), it is easy to see that u is C α at x for any x ∈ Ω as in Definition4.39. In fact, the converse is also true. If u is C α at x for any x ∈ Ω as in Definition4.39, then u ∈ C α(Ω). In other words, the (global) Holder continuity is equivalentto the everywhere pointwise Holder continuity. We will leave verification as anexercise.

The following result is often referred to as the interior Schauder regularity orthe interior Schauder estimate.

8/16/2019 PDE . QUING


4.6. WEAK SOLUTIONS 91

Theorem 4.44. Let α ∈ (0, 1) be a constant and u ∈ C 2(B1) satisfy

∆u = f in B1.

If f ∈ C α(B1), then u ∈ C 2,α(B1). Moreover,

|u|C 2,α(B 12

) ≤ c|u|L∞(B1) + |f |C α(B1)

,

where c is a positive constant depending only on n and α.

Proof. We only sketch the proof. For any x ∈ B1/2, we apply Theorem 4.42in B(1−|x|)(x) to obtain a quadratic polynomial px such that

the absolute values of coefficients of px are bounded by c(|u|L∞(B1) + |f |C α(B1))

and

|u(y) − px(y − x)| ≤ c(|u|L∞(B1) + |f |C α(B1))|y − x|2+α for any y ∈ B 1−|x|2

(x),

where c is a positive constant depending only on n and α. Then we have the desiredresult easily by the equivalence of the (global) Holder continuity and the everywhere

pointwise Holder continuity.

To end this section, we mention briefly another approach for Schauder esti-

mates. Let Ω be a domain in Rn and f be a continuous function in Ω. We considerthe Poisson equation (4.10) in Ω.

If u ∈ C ∞0 (Ω) is a solution of (4.10), then f evidently has a compact supportin Ω and hence by Theorem 4.1

u(x) =

Ω

Γ(x, y)f (y)dy,

where Γ(x, y) is the fundamental solution of ∆ in Rn.

Now let Ω be a bounded domain in Rn and f be a bounded function in Ω. Wedefine

(4.12) uf (x) =

Ω

Γ(x, y)f (y)dy.

This is called the potential of f in Ω. If uf is C 2 and satisfies (4.10) in Ω, i.e.,∆uf = f in Ω, then any solution of (4.10) differs from uf by an addition of a

harmonic function. Since harmonic functions are analytic, regularity of arbitrarysolutions of (4.10) is determined by those of uf defined in (4.12). The classicalapproach for the Schauder theory is to prove uf ∈ C 2,α(Ω) and ∆uf = f in Ω if f ∈ C α(Ω) for some α ∈ (0, 1). We will not provide proofs here. Example 4.38

shows that uf is not necessarily C 2 if f is only continuous.

4.6. Weak Solutions

In this section, we discuss briefly how to extend the notion of classical solutionsof the Poisson equation to less regularized solutions, solutions with derivatives onlyin integral sense, the so-called weak solutions. The same process can be applied togeneral linear elliptic equations of second order, even nonlinear elliptic equations, of divergence form. This section serves only as an introduction to this important ad-

vanced topic in the theory of elliptic differential equations. A complete presentationwill constitute a book much thicker than this one.

8/16/2019 PDE . QUING



To introduce weak solutions, we make use of divergence structure or variation

structure of the Laplace operator. Namely, we write the Laplace operator as

∆u = div(∇u).

Then the Laplace equation can be considered as an Euler equation of the Dirichletenergy.

For any domain Ω ⊂ Rn, we define the Dirichlet energy of u in Ω by

E (u) = 1

2

Ω

|∇u|2.

Suppose u is a C 1-minimizer of E (u) among all C 1-functions in Ω with the same

boundary value. In particular, a small perturbation of u in any subregion of Ωincreases the energy. Hence, for any C 1-function ϕ with a compact support in Ω,

1

2 Ω

|∇(u + εϕ)|2 ≥ 1

2 Ω

|∇u|2 for any ε.

In other words,

F (ε) ≡ 1

2

Ω

|∇(u + εϕ)|2

has a minimum at ε = 0. This implies F (0) = 0, or Ω

∇u · ∇ϕ = 0 for any ϕ ∈ C 10 (Ω).

In general, for any f ∈ L2(Ω), we consider

(4.13) J (u) = 1

2 Ω |∇u

|2 + Ω uf.

If u is a C 1-minimizer, we obtain similarly

−

Ω

∇u · ∇ϕ =

Ω

f ϕ for any ϕ ∈ C 10 (Ω).

This suggests the following definition.

Definition 4.45. Let f ∈ L2(Ω). Then u ∈ C 1(Ω) is a weak solution of ∆u = f in Ω if

−

Ω

∇u · ∇ϕ =

Ω


In Definition 4.45, ϕ is called a test function. If u is a weak C 2

-solution, weobtain by a simple integration by parts Ω

ϕ∆u =

Ω


This implies easily that

∆u = f a.e. in Ω.

If f is continuous in Ω, then ∆u = f in Ω. In other words, a weak C 2-solution of the Poisson equation is a classical solution.

8/16/2019 PDE . QUING



Next, we illustrate how to find a weak solution of the Poisson equation with a

homogeneous Dirichlet boundary value, i.e.,

∆u =f in Ω,

u =0 on ∂ Ω.(4.14)

We intend to minimize J (u) given in (4.13) in the space

C = u ∈ C 1(Ω) ∩ C (Ω); u = 0 on ∂ Ω.

We first demonstrate that J has a lower bound in C. First, by the Cauchy inequality,we have

Ω

|uf | ≤ Ω

u2 12

Ω

f 2 12 .

By Lemma 3.12, the Poincare inequality, we have for any u ∈ C

Ω u2

≤ C Ω |∇u|2

,

where C is a positive constant depending only on Ω. Then Ω

|uf | ≤√

C

Ω

|∇u|2 12

Ω

f 2 12 ≤ 1

4

Ω

|∇u|2 + C

Ω

f 2.

Hence for any u ∈ C

(4.15) J (u) ≥ 1

4

Ω

|∇u|2 − C

Ω

f 2,

and in particular

J (u) ≥ −C

Ω

f 2.

Therefore, J has a lower bound in C. We set

J 0 = inf J (u); u ∈ C.

If J 0 is realized by some u ∈ C, then u is a minimizer of J and hence a weak solutionof ∆u = f .

To discuss whether J realizes J 0 in C, we consider a minimizing sequence uk ⊂C with J (uk) → J 0 as k → ∞ and ask whether there exists a function u ∈ C suchthat J (uk) → J (u). If such a u exists, then J 0 is realized by u by the convergence of J (uk). Here we do not require that uk converge to u in the C 1-norm. However, inorder to construct u, uk has to converge in some sense. So what is a reasonable

norm to provide such a convergence? To answer this question, we examine theexpression of J (u) in (4.13). First, we have by (4.15)

Ω

|∇uk|2 ≤ 4J (uk) + 4C

Ω

f 2.

Since J (uk) is bounded, there is a uniform bound on the L2-norms of gradients of uk. Hence a convergence of uk, if exists, should be related to the L2-norm. Forthis, we need to introduce Sobolev norms.

For any u ∈ C 10 (Ω), we define the H 10 -norm of u by

uH 10=

Ω

|∇u|2 12 .

8/16/2019 PDE . QUING



Here in H 10 , the super-index 1 indicates the order of differentiation and the sub-index

0 refers to functions with compact support in Ω. We then complete the space C 10 (Ω)

with the H 10 -norm. The resulting space is called the Sobolev space H 10 (Ω) and itselements are called H 10 -functions in Ω. By the Poincare inequality, we can proveH 10 (Ω) ⊂ L2(Ω). In general, functions in H 10 (Ω) may not have classical derivatives.

However, it has weak derivatives in an integral sense. For any u ∈ H 10 (Ω) and anyi = 1, · · · , n, there exists a function vi ∈ L2(Ω) such that

Ω

uϕxi = −

Ω

viϕ for any ϕ ∈ C 10 (Ω).

Here vi is called a weak xi-derivative of u and is often denoted by uxi, the same

notion of classical derivatives. Weak derivatives are unique by the density of contin-

uous functions in the L2-space. Hence, uH 10can be computed for any u ∈ H 10 (Ω)

by using weak derivatives of u. By a simple integration by parts, classical deriva-

tives of C 1

-functions are weak derivatives. We also point out that H 10 (Ω) is in fact

a Hilbert space with the inner product

(u, v) =

Ω

∇u · ∇v for any u, v ∈ H 10 (Ω),

where ∇u and ∇v are weak gradients of u and v.

The following compactness result, called Rellich’s Theorem, plays an importantrole in the theory of Sobolev spaces. Any sequence in H 10 (Ω) with bounded H 10 (Ω)-norms has a subsequence convergent in the L2(Ω)-norm to some function in H 10 (Ω).We emphasize that the limit function is in H 10 (Ω) although the convergence is only

in L2. In general, we do not have a convergence in the H 10 -norm.It is easy to see that the functional J (u) in (4.13) can be extended to functions

u ∈ H 1

0 (Ω), with classical gradients replaced by weak gradients. Instead of minimiz-ing J in C 10 (Ω), we minimize J in H 10 (Ω). One advantage of H 10 (Ω) over C 10 (Ω) is

that H 10 (Ω) is complete under the H 10 (Ω)-norm, which is more naturally associatedwith the functional J (u) in (4.13). Another important aspect of the Sobolev spaceH 10 (Ω) in studies of elliptic differential equations lies in the following simple fact.The concept of weak solutions in Definition 4.45 can be extended to H 10 -functions.We also point out that the Poincare inequaltiy holds for H 10 (Ω)-functions by a sim-

ple approximation of H 10 (Ω)-functions by C 10 (Ω)-functions. Hence, (4.15) also holdsfor H 10 (Ω)-functions.

We return to our minimizing problem. First, it is easy to see

J 0 = inf u∈H 10(Ω)

J (u).

Let uk be a minimizing sequence of J (u) in H 10 (Ω). Then (4.15) shows thatukH 10

is uniformly bounded. By Rellich’s theorem, there exists a subsequence

uk and a u ∈ H 10 (Ω) such that

uk → u in the L2-norm as k → ∞.

Next, with the help of the Hilbert space structure of H 10 (Ω) it is not difficult toprove

J (u) ≤ lim inf k→∞

J (uk).

8/16/2019 PDE . QUING



This implies J (u) = J 0. In fact, we can also prove uk → u in the H 10 (Ω)-norm

as k → ∞. Now we try to interpret in what sense u solves (4.14). First, u is a

minimizer of J in H 10 (Ω), i.e.,

J (u + ϕ) ≥ J (u) for any ϕ ∈ C 10 (Ω).

Then by the same analysis preceding Definition 4.45, we have

−

Ω

∇u · ∇ϕ =

Ω


Here ∇u is the weak gradient of u. Hence, u is a weak solution of ∆u = f .Concerning the boundary value, we point out that u is not defined in the pointwise

sense. We cannot conclude u = 0 at each point on ∂ Ω. Here the boundary conditionu = 0 on ∂ Ω is interpreted precisely by the fact u ∈ H 10 (Ω), i.e., u is the limit of a sequence of C 10 (Ω)-functions in the H 10 (Ω)-norm. One consequence is that u|∂ Ω

is a well-defined zero function in L2

(∂ Ω). This finishes the existence part of thevariational approach. We have found a weak solution of (4.14), the Poisson equationwith a homogeneous Dirichlet boundary value.

Now we ask whether u possesses a better regularity. The answer is yes. Withf ∈ L2(Ω), the solution u has weak derivatives of second order in the following

sense. For any 1 ≤ i, j ≤ n, there exists a function vij ∈ L2(Ω) such that Ω

uϕxixj =

Ω

vij ϕ for any ϕ ∈ C 20 (Ω).

We also denote vij by uxixj. The solution u is a so-called H 2-function in Ω. With

weak derivatives uxixj defined as L2-functions, we also conclude

∆u = f a.e. in Ω.

In fact, if f is an H k-function for any k ≥ 1, a function with weak derivatives of order up to k as L2-functions, then u is an H k+2-function. In particular, if f is

smooth, then u is smooth. We omit details of discussions.With similar methods, we can solve the eigenvalue problem of the Laplace

operator in a smooth bounded domain Ω in Rn. Consider

−∆u =λu in Ω,

u =0 on ∂ Ω.

First, set

λ1 = inf

Ω

|∇u|2

Ω u2 ; u ∈ H 10 (Ω) \ 0

.

With a minimizing sequence, we can prove that λ1 is achieved in H 10 (Ω), say by

u1 ∈ H 10 (Ω). We further assume

Ωu2

1 = 1. Then for any ϕ ∈ C 10 (Ω),

F (ε) =

Ω

|∇u1 + ε∇ϕ|2 Ω

(u1 + εϕ)2

has a minimum at ε = 0. Hence F (0) = 0, or Ω

∇u1 · ∇ϕ = λ1

Ω

u1ϕ for any ϕ ∈ C 10 (Ω).

8/16/2019 PDE . QUING



It is easy to check that λ1 > 0. Here λ1 is the first eigenvalue of the Laplace

operator and u1 is a corresponding eigenfunction in the weak sense. In fact, we can

prove u1 is a smooth function in Ω and satisfies

−∆u1 =λ1u1 in Ω,

u1 =0 on ∂ Ω

in the classical sense. The definition of λ1 yields Ω

|∇u|2 Ω

u2 ≥ λ1 for any u ∈ H 10 (Ω),

or Ω

u2 ≤ 1

λ1

Ω

|∇u|2 for any u ∈ H 10 (Ω).

Therefore, 1/λ1 is the best constant in the Poincare inequality. See Lemma 3.12.

To find the next eigenvalue, we set

λ2 = inf

Ω|∇u|2 Ω

u2 ; u ∈ H 10 (Ω) \ 0,

Ω

uu1 = 0

.

Obviously, λ2 ≥ λ1. With a similar minimizing process, we can prove that λ2

is achieved in H 10 (Ω), say by u2 ∈ H 10 (Ω) with

Ωu1u2 = 0. We further assume

Ωu2

2 = 1. We can check similarly Ω

∇u2 · ∇ϕ = λ2

Ω

u2ϕ for any ϕ ∈ C 10 (Ω).

Next, we set

λ3 = inf Ω|∇u|2

Ω u2

; u

∈H 10 (Ω)

\ 0

, Ω uui = 0, i = 1, 2 .

By continuing this process, we obtain an increasing sequence λi and a sequence

ui ⊂ H 10 (Ω) with

Ωuiuj = δ ij such that

Ω

∇ui · ∇ϕ = λi

Ω

uiϕ for any ϕ ∈ C 10 (Ω).

We can prove that ui in fact is a smooth function in Ω for any i = 1, 2, · · · , andsatisfies

−∆ui =λiui in Ω,

ui =0 on ∂ Ω

in the classical sense. Moreover,

ui

forms a complete sequence in L2(Ω). In other

words, for any v ∈ L2(Ω)

v =

∞i=1

(v, ui)L2(Ω)ui,

where (·, ·)L2(Ω) is the L2-inner product in Ω and the convergence of this series is

in the L2-norm in Ω.With such a sequence of eigenvalues and eigenfunctions in a bounded smooth

domain, we can solve initial/boundary-value problems of the heat equation and thewave equation in arbitrary dimension, as described at the end of Section 3.3.

8/16/2019 PDE . QUING


EXERCISES 97

Exercises

(1) Suppose u(x) is harmonic in some domain in Rn. Then

v(x) = |x|2−nu x

|x|2

is also harmonic in a suitable domain.

(2) For n = 2, find the Green’s function for the Laplace operator on the first

quadrant.

(3) Find the Green’s function for the Laplace operator in the upper half spacexn > 0 and then derive a formal integral representation for a solutionof the Dirichlet problem

∆u = 0 in xn > 0,

u = ϕ on

xn = 0

.

(4) Let λ be a positive constant. Find the fundamental solution for ∆3u+λu =0 in R3.

(5) (a) Prove by the Poisson formula the following Harnack inequality: Sup-pose u is harmonic in BR(x0) and u ≥ 0. Then

R

R + r

n−2R − r

R + ru(x0) ≤ u(x) ≤

R

R − r

n−2R + r

R − ru(x0),

where r = |x − x0| < R.

(b) Prove by (a) the Liouville Theorem: If u is a harmonic function inRn and bounded above or below, then u ≡ const.

(6) Let u be a harmonic function in Rn with Rn |u| p < ∞ for some p ∈ (1, ∞).Then u ≡ 0.

(7) Let u be a harmonic function in Rn and u(x) = O(|x|m) as |x| → ∞ for

a positive integer m. Then u is a polynomial of degree m.

(8) Suppose u ∈ C ( B+1 ) is harmonic in B+

1 = x ∈ B1; xn > 0 with u = 0on xn = 0 ∩ B1. Then the odd extension of u in B1 is harmonic in B1.

(9) Suppose u is harmonic in the open upper half space xn > 0 and contin-uous in its closure. If u = 0 on xn = 0 and u(x) = O(|x|) as |x| → ∞,then u(x) = cxn for some constant c.

(10) Let Ω be a domain in Rn and u ∈ C 2(Ω) ∩ C (Ω) satisfies ∆u + λu = 0 in

Ω for a constant λ > 0. Then u is analytic in Ω.(11) Suppose Ω is a domain in Rn and um is a sequence of uniformly bounded

harmonic functions in Ω. Prove that um has a subsequence convergentto a function u uniformly on any compact subsets of Ω and that u is

harmonic in Ω.

(12) Prove that the square of a nonnegative subharmonic function is subhar-monic. Give an example to show that the condition nonnegative cannotbe omitted.

8/16/2019 PDE . QUING



(13) Suppose u is harmonic in the open upper half space xn > 0 and contin-

uous and bounded in its closure. Then

supxn≥0

u = supxn=0

u.

(14) Let u be a C 2-solution of

∆u = 0 in Rn \ BR,

u = 0 on ∂ BR.

Prove that u ≡ 0 if

lim|x|→∞

u(x)

ln |x| = 0 for n = 2,

lim|x|→∞

u(x) = 0 for n ≥ 3.

(15) Suppose u ∈ C 2(Ω) ∩ C 1(Ω) satisfies

−∆u + u3 = 0 in Ω,

∂u

∂n + αu = ϕ on ∂ Ω,

for continuous functions ϕ and α on ∂ Ω with α(x) ≥ α0 > 0. Prove

maxΩ

|u| ≤ 1

α0max

∂ Ω|ϕ|.

(16) Let Ω be a smooth bounded domain in Rn with ∂ Ω = Γ1 ∪ Γ2, c be acontinuous function in Ω with c ≤ 0 and α be a continuous function on

∂ Ω with α ≥ 0. Discuss the uniqueness of the following problem

∆u + cu = f in Ω,∂u

∂n + αu = ϕ on Γ1,

u = ψ on Γ2.

(17) Let k and λ > 0 be constants and B+1 be the upper half disc in R2.

Suppose u ∈ C 2(B+1 ) ∩ C ( B+

1 ) is a solution of the following problem

∆u + k

yuy − λu = f in B1,

u = 0 on ∂ B+1 .

Then

maxB+1

|u| ≤ 1

λ maxB+1

|f |.

(18) Let u be a nonzero harmonic function in B1 ⊂ Rn and set

N (r) =r

Br|∇u|2

∂Bru2

for any r ∈ (0, 1).

(a) Prove N (r) is a nondecreasing function in r ∈ (0, 1) and identify

limr→0+

N (r).

8/16/2019 PDE . QUING


EXERCISES 99

(b) Prove for any 0 < r < R < 1

1

Rn−1 ∂BR

u2 ≤ R

r 2N (R)

1

rn−1 ∂Br

u2.

Remark: The quantity N (r) is called the frequency . The estimate in (b)is referred to as the doubling condition for R = 2r.

(19) Let k be a nonnegative integer, α ∈ (0, 1) be a constant and u be a C k-function in [0, 1]. Suppose for any x ∈ [0, 1]

|u(y) − T k(y; x)| ≤ C |y − x|k+α for any y ∈ [0, 1],

where T k(·; x) is the k-th Taylor expansion of u at x. Then u ∈ C k,α[0, 1]and [∇ku]C α[0,1] ≤ C .

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


CHAPTER 5

Heat Equations

The n-dimensional heat equation is given by ut − ∆u = 0 for functions u =u(x, t), with x ∈ Rn and t ∈ R. Here, x is called the space variable and t the time

variable. The heat equation models the temperature of a body conducting heat,when the density is constant. In this chapter, we discuss properties of solutionsof the heat equation. In Section 5.1, we discuss the fundamental solution of the

heat equation and solve initial-value problems. Then in Section 5.2, we use thefundamental solution to discuss regularity of arbitrary solutions. In Section 5.3,we discuss the maximum principle for the heat equation and its applications. Inparticular, we use the maximum principle to derive interior gradient estimates. InSection 5.4, we discuss Harnack inequalities.

5.1. Initial-Value Problems

In this section, we discuss initial-value problems of the heat equation and derivean explicit expression for their solutions.

We first examine the equation itself. The n-dimensional heat equation is givenby

(5.1) ut − ∆u = 0,for u = u(x, t) with x ∈ Rn and t ∈ R. We note that (5.1) is not preserved bythe change t → −t. This indicates that the heat equation describes an irreversibleprocess and distinguishes between past and future. This fact will be well illustratedby Harnack inequalities which we will derive in Section 5.4. Next, (5.1) is preserved

under linear transforms x = λx and t = λ2t for any nonzero constant λ, whichleave the quotient |x|2/t invariant. Because of this, the expression |x|2/t appearsfrequently in connection with the heat equation (5.1). In fact, the fundamentalsolution we will derive has such an expression.

If u is a solution of (5.1) in a domain in Rn × R, then for any (x0, t0) in this

domain and any r > 0

u(x, t) = u(x0 + rx,t0 + r2t)

is a solution of (5.1) in an appropriate domain in Rn ×R.Let u be a polynomial in Rn ×R. It is of p-homogeneous degree d if

u(λx,λ2t) = λdu(x, t) for any (x, t) ∈ Rn ×R and λ > 0.

Now we let P be a homogeneous polynomial of degree d in Rn and proceed to finda p-homogeneous polynomial of degree d such that

ut − ∆u =0 in Rn × R,

u(·, 0) =P in Rn.

101

8/16/2019 PDE . QUING


102 5. HEAT EQUATIONS

To do this, we simply expand u as a power function of t with coefficients as functions

of x, i.e.,

u(x, t) =∞

k=0

ak(x)tk.

Then a simple calculation shows that

a0 = P, ak = 1

k∆ak−1 for any k ≥ 1.

Since P is a polynomial of degree d, then ∆[d/2]+1P = 0, where [d/2] is the integralpart of d/2, i.e., [d/2] = d/2 if d is an even integer and [d/2] = (d − 1)/2 if d is anodd integer. Hence

u(x, t) =

[ d2 ]

k=0

∆kP (x)tk.

For n = 1, let ud be a p-homogeneous polynomial of degree d in R × R satisfyingthe heat equation and ud(x, 0) = xd. The first five such polynomials are given by

u1(x, t) =x,

u2(x, t) =x2 + 2t,

u3(x, t) =x3 + 6xt,

u4(x, t) =x4 + 12x2t + 12t2,

u5(x, t) =x5 + 20x3t + 60xt2.

In the following, we discuss initial-value problem of the heat equation and derive

an explicit expression for its solution.We first introduce briefly Fourier transforms. Fourier transforms are an impor-tant subject and has a close connection with many fields in mathematics, especiallywith partial differential equations.

By allowing functions to be complex valued, we define the Schwartz class S asthe collection of all functions u ∈ C ∞(Rn;C) such that

(1 + |x|2)m2 ∂ αx u(x) is bounded in Rn for any α ∈ Zn

+ and m ∈ Z+.

In other words, the Schwartz class consists of smooth functions whose arbitrary

derivatives decay faster than any polynomials. For example, u(x) = e−|x|2

is in theSchwartz class.

For any u ∈ S , define a new function u in Rn by

u(ξ ) = 1

(2π)n2

Rn

e−ix·ξu(x)dx for any ξ ∈ Rn.

The operator u → u is called the Fourier transform operator and u is called theFourier transform of u.

We now state some properties of Fourier transforms.

(F1) Linearity: The Fourier transform operator is linear, i.e., for any u1, u2 ∈ S and c1, c2 ∈ C

(cu1 + c2u2)∧ = c1u1 + c2u2.

8/16/2019 PDE . QUING


5.1. INITIAL-VALUE PROBLEMS 103

(F2) Differentiation: For any u ∈ S and any j = 1, · · · , n, we have by integra-

tion by parts ∂ j u(ξ ) = 1(2π)

n2 Rn

e−ix·ξ∂ xj u(x)dx = iξ j u(ξ ),

and hence for any multi-index α ∈ Zn+∂ αx u(ξ ) = (iξ )αu(ξ ).

For any polynomial P (ξ ) on Rn given by

P (ξ ) =|α|≤m

aαξ α,

we define a differential operator P (∂ ) by

P (∂ )u =

|α|≤α

aα∂ αu.

Then P (∂ )u(ξ ) = P (iξ )u(ξ ).

(F3) Multiplication by polynomials: It is easy to see that u ∈ C ∞(Rn) for anyu ∈ S . Then for any j = 1, · · · , n

∂ ξju(ξ ) = − i

(2π)n2

Rn

e−ix·ξxj u(x)dx = −ixj u(ξ ),

and hence for any multi-index β ∈ Zn+

∂ βξ u(ξ ) = (−i)|β|xβu(ξ ).

As a consequence, we have u ∈ S for u ∈ S , i.e., |ξ α∂ βξ u(ξ )| is bounded in Rn

for any α, β ∈Zn

+. To see this, we note by (F2) and (F3)ξ α∂ βξ u(ξ ) =(−i)|β|ξ αxβ u(ξ ) = (−i)|α|+|β|(iξ )αxβ u(ξ ) = (−i)|α|+|β| ∂ αx (xβu)(ξ )

= 1

(2π)n2

(−i)|α|+|β| Rn

e−ix·ξ∂ αx

xβu(x)

dx.

Hence

supξ∈Rn

|ξ α∂ βξ u(ξ )| ≤ 1

(2π)n2

Rn

|∂ αx

xβ u(x)|dx < ∞,

since each term in the integrand decays faster than any polynomials.(F4) Translation and dilation: For any u ∈ S , a ∈ Rn and k ∈ R,

u(· − a)(ξ ) = e−iξ·au(ξ ),

and u(k·)(ξ ) =

1

|k|n u(

ξ

k).

(F5) Convolution: For any u, v ∈ S , it is easy to check u ∗ v ∈ S , where u ∗ vis the convolution of u and v defined by

(u ∗ v)(x) =

Rn

u(x − y)v(y)dy.

We then have

u ∗ v(ξ ) = (2π)n2 u(ξ )v(ξ ).

8/16/2019 PDE . QUING



We leave verification of Properties (F1)-(F5) as an exercise.

The following example will be useful in discussions of the heat equation. We

first note ∞−∞

e−x2dx =√

π.

Example 5.1. Consider u(x) = e−|x|2

in Rn. Then u ∈ S and a direct calcu-lation yields

u(ξ ) = 1

2n2

e−14 |ξ|

2

.

Therefore by (F4), for f (x) = e−A|x|2 with A > 0,

f (ξ ) = 1

(2A)n2

e−|ξ|24A .

The Fourier transform is an important subject in mathematics. Now we at-tempt to use it to solve certain classes of partial differential equations. Let P bea polynomial in Rn of degree m. Consider the following linear partial differentialequation of degree m

P (∂ )u = f in Rn.

By applying Fourier transforms and (F2), we have

P (iξ )u(ξ ) = f (ξ ),

and then

u(ξ ) =f (ξ )

P (iξ ).

Hence a solution u is a function whose Fourier transform is f (ξ )/P (iξ ). Thisnaturally leads to inverse Fourier transforms.

Theorem 5.2. Suppose u ∈ S . Then

u(x) = 1

(2π)n2

Rn

eix·ξu(ξ )dξ.

The right hand side is simply the Fourier transform of u evaluated at −x.

Hence, u(x) = ˆu(−x).

Proof. First, by the definition of the Fourier transform, we have

R.H.S = 1

(2π)n Rn Rn

eix·ξe−iy·ξu(y)dy dξ.

We cannot simply change the order of integrations since there is no absolute conver-gence. Instead, we introduce a limiting process to have the absolute convergence.

Then

R.H.S. = 1

(2π)n limλ→∞

Rn

Rn

ei(x−y)·ξe−|ξ|2

λ u(y)dydξ

= 1

(2π)n limλ→∞

Rn

Rn

ei(x−y)·ξ−|ξ|2

λ u(y)dξdy,

8/16/2019 PDE . QUING



where we changed the order of integrations in the last step. A simple calculation

of the exponential factor yields

R.H.S. = 1(2π)n

limλ→∞

Rn

u(y)e−λ4 |x−y|2

Rn

e−n

j=1 i(xj−yj)√ λ

2 + ξj√

λ2dξdy.

For any a ∈ R, we first note R

e−

i a√

λ2 + η√

λ

2dη =

R

e−η2

λ dη =√

λ

R

e−η2dη =√

λπ.

This integral is independent of a. Hence Rn

e−

nj=1

i(xj−yj)

√ λ

2 + ξj√

λ

2dξ =

nj=1

R

e−

i(xj−yj)

√ λ

2 + ξj√

λ

2dξ j = (λπ)

n2 .

Then

R.H.S. = 1

(4π)n

2

limλ→∞ Rn

λn2 e−

λ|x−y|24 u(y)dy.

Let

K λ(z) = 1

(4π)n2

λn2 e−

|z|2λ4 .

Then K λ > 0 in Rn and Rn K λ = 1. It is straightforward to prove

R.H.S. = limλ→∞

Rn

K λ(x − y)u(y)dy = u(x).

We will provide detail in the proof of Theorem 5.5.

Next, we set

(u, v)L2(Rn) = Rn

uv for any u, v ∈ L2(Rn).

Here u and v may be complex-valued. The next result is referred to as the Parseval’sidentity.

Theorem 5.3. For any u, v ∈ S ,(u, v)L2(Rn) = (u, v)L2(Rn).

In particular, for any u ∈ S uL2(Rn) = uL2(Rn).

Proof. We note

(u, v)L2(Rn) =

Rn

u(ξ )v(ξ )dξ =

Rn Rn

v(ξ )e−ix·ξu(x)dxdξ

= Rn

Rn

v(ξ )eix·ξu(x)dξdx = Rn

u(x)v(x)dx = (u, v)L2(Rn),

where we applied Theorem 5.2 to v. This yields the desired result.

We now extend Fourier transforms to a larger class of functions than S . Note

that the Fourier transform is a linear operator from S to S . Since S is dense inL2(Rn), we can complete the space S under · L2(Rn) to get L2(Rn). Hence, the

Fourier transform can be extended to a bounded linear operator in L2(Rn). Sinceit is an isometry in S with respect to the L2-norm, it is also an isometry in L2(Rn).

8/16/2019 PDE . QUING



We note that the Fourier transform also extends to functions in L1(Rn). For

any u ∈ S ,u(ξ ) = 1

(2π)n2 Rn

e−ix·ξu(x)dx.

Hence

|u(ξ )| ≤ 1

(2π)n2

Rn

|u(x)|dx,

or

supRn

|u| ≤ 1

(2π)n2uL1(Rn).

Note that S is dense in L1(Rn). By completing S under the L1(Rn), we concludethat the Fourier transform operator is a bounded linear operator from L1(Rn) to

the set of bounded continuous functions in Rn.

Now we are ready to derive an expression for solutions of general initial-value

problems of the heat equation. We considerut − ∆u = 0 in Rn × (0, ∞),

u(·, 0) = u0 in Rn.(5.2)

Here we require u ∈ C 2(Rn × (0, ∞)) ∩ C (Rn × [0, ∞)). Although called an initial-

value problem, (5.2) is not the type of initial-value problems we discussed in Section3.1. The heat equation is of the second order, while only one condition is prescribedon the initial hypersurface t = 0, which is characteristic.

We first use Fourier transforms to derive a formal solution. In the following,

we employ the Fourier transform of u with respect to x ∈ Rn and write

u(ξ, t) = 1

(2π)n2 Rn

e−ix·ξu(x, t)dx.

Then by (F2)

ut + |ξ |2u = 0 in Rn × (0, ∞),

u(·, 0) = u0 in Rn.

This is an initial-value problem for an ODE with ξ as a parameter and its solutionis given by

u(ξ, t) = u0(ξ )e−|ξ|2t.

Now we treat t as a parameter. Let K (x, t) satisfy

K (ξ, t) = 1

(2π)n2

e−|ξ|2t for t > 0.

Then u(ξ, t) = (2π)n2 K (ξ, t)u0(ξ ).

Therefore Theorem 5.2 and (F5) imply

u(x, t) =

Rn

K (x − y, t)u0(y)dy.

This is called the Poisson integral formula. By Theorem 5.2 and Example 5.1, wehave

K (x, t) = 1

(2π)n

Rn

eix·ξe−|ξ|2tdξ,

8/16/2019 PDE . QUING



or

(5.3) K (x, t) = 1

(4πt)n2

e−|x|24t for any x

∈Rn, t > 0,

which is called the fundamental solution of the heat equation.Now we verify directly that the Poisson integral formula defines a solution under

suitable conditions on u0. The proof follows the same line as that of the Poissonintegral formula for the Laplace equation in Theorem 4.6.

We first note that the fundamental solution K satisfies the following properties:(K1) K (x, t) is smooth for any x ∈ Rn and t > 0.(K2) K (x, t) > 0 for any x ∈ Rn and t > 0.(K3) (∂ t − ∆x)K (x, t) = 0 for any x ∈ Rn and t > 0.

(K4) Rn K (·, t) = 1 for any t > 0.

(K5) For any δ > 0,

limt→0+

|x|>δ

K (x, t)dx = 0.

Figure 5.1. Graphs of fundamental solutions for t2 > t1 > 0.

Here (K1) and (K2) follow from the explicit expression for K in (5.3). We may

also get (K3) from (5.3) by a straightforward calculation. However, an easy way toget (K3) is to use the integral expression of K preceding (5.3). For (K4) and (K5),we simply note

|x|>δ

K (x, t)dx = 1

πn2

|η|> δ

2√

t

e−|η|2

dη.

This implies (K4) for δ = 0 and (K5) for δ > 0.

Now we are ready to solve initial-value problems of the heat equation.

Theorem 5.4. For any bounded u0 ∈ C (Rn), the function u defined by

(5.4) u(x, t) =

Rn

K (x − y, t)u0(y)dy

is smooth in Rn × (0, ∞) and continuous up to t = 0 and satisfies

ut − ∆u = 0 in Rn × (0, ∞),

u(·, 0) = u0 in Rn.

8/16/2019 PDE . QUING



Proof. Step 1. We first prove that u(x, t) is smooth for any x ∈ Rn and t > 0.

It suffices to check Rn

|x − y|me− |x−

y|2

4t |u0(y)|dy < ∞,

for any m ≥ 0. This follows from the exponential decay of the integrand. Then u is

a solution of the heat equation by a simple differentiation. We point out for futurereferences that we only use the boundedness of u0.

Step 2. For any fixed x0 ∈ Rn, we claim

lim(x,t)→(x0,0)

u(x, t) = u0(x0).

For simplicity, we assume x0 = 0. By (K4), we have

u(x, t) − u0(0) =

Rn

K (x − y, t)

u0(y) − u0(0)

dy.

Then|u(x, t) − u0(0)| ≤ 1

(4πt)n2

Rn

e−|x−y|2

4t |u0(y) − u0(0)|dy.

For any ε > 0, by the continuity of u0, there exists a positive constant δ = δ (ε)such that

|u0(y) − u0(0)| < ε for any |y| ≤ δ.

Next, we assume |u0| ≤ M by the boundedness of u0. Then by (K2) and (K4)

|u(x, t) − u0(0)|

≤

Bδ

K (x − y, t)|u0(y) − u0(0)|dy +

Rn\Bδ

K (x − y, t)|u0(y) − u0(0)|dy

≤ε + 2M Rn\Bδ

K (x − y, t)dy.

We note |x − y| ≥ δ/2 for any y ∈ Rn \ Bδ and x ∈ Bδ/2. Hence

|u(x, t) − u0(0)| ≤ ε + 2M

Rn\B δ

2

K (z, t)dz for any x ∈ B δ2

.

By (K5), there exists a t0 > 0 depending only on ε, M and δ such that

|u(x, t) − u0(0)| ≤ 2ε for any x ∈ B δ2

, t ∈ (0, t0).


Now we discuss a result more general than Theorem 5.4. The boundedness

assumption on u0 in Theorem 5.4 can be relaxed. To seek a reasonable assumptionon initial values, we examine the expression of the fundamental solution K . We notethat K in (5.3) has an exponential decay in space variables with a large decay ratefor small time. This suggests that we can allow an exponential growth for initial

values. In the convolution (5.4), a fixed exponential growth from initial values canbe offset by the fast exponential decay in the fundamental solution at least for ashort time period. To see this clearly, we consider an example. For any α > 0, set

G(x, t) = 1

(1 − 4αt)n2

e α1−4αt

|x|2 for any x ∈ Rn, t < 1

4α.

8/16/2019 PDE . QUING



It is straightforward to check

Gt − ∆G = 0 for any x ∈Rn

, t <

1

4α .Note

G(x, 0) = eα|x|2 for any x ∈ Rn.

Hence, G has an exponential growth initially for t = 0, and in fact for any t < 1/4α.

The growth rate becomes arbitrarily large as t approaches 1/4α and G does notexist beyond t = 1/4α.

Figure 5.2. Graphs of G for different time.

Now we formulate a general result. If u0 is continuous and has an exponentialgrowth, then (5.4) still defines a solution of the initial-value problem in a short

period of time.

Theorem 5.5. Suppose u0 ∈ C (Rn) satisfy

|u0(x)| ≤ Me

A|x|2

for any x ∈Rn

, for some constants M, A ≥ 0. Then u defined in (5.4) is a C ∞(Rn × (0, T )) ×C (Rn × [0, T ))-solution of

ut − ∆u = 0 in Rn × (0, T ],

u(·, 0) = u0 in Rn,

for any T < 14A . Moreover,

|u(x, t)| ≤ M 1eA1|x|2

for any x ∈ Rn, 0 < t ≤ T,

where A1 ≥ 0 and M 1 > 0 are constants depending only on A, M and T .

We first prove the following lemma.

Lemma 5.6

. Suppose f, g ∈ C (

Rn

) satisfy |f (x)| ≤ MeA|x|2 , |g(x)| ≤ M eA|x|2 ,

for some constants M, M > 0 and some A, A. If A + A < 0, then

h(x) =

Rn

g(x − y)f (y)dy

is continuous in Rn and

|h(x)| ≤ MeA|x|2 ,

where M and A are constants depending only on M, M , A and A.

8/16/2019 PDE . QUING



Proof. With A + A < 0, we have

|h(x)| ≤ M ¯M Rn e

A|y|2+ A|x−y|2

dy

= M MeA A

A+ A|x|2

Rn

e(A+ A)|y− A

A+ Ax|2dy

= M M −π

A + A

n2 e

A AA+ A

|x|2 .

This yields the desired estimate.

Now we prove Theorem 5.5. The proof follows the same line as that of Theorem5.4.

Proof of Theorem 5.5. Step 1. The case A = 0 is covered by Theorem 5.4.We only consider A > 0 and take T such that − 1

4T + A < 0 or T < 14A . Then

Lemma 5.6 implies u is continuous in Rn

× (0, T ]. To show that u has continuousderivatives of arbitrary order in Rn × (0, T ], we only need to verify

Rn

|x − y|me−|x−y|2

4t |u0(y)|dy < ∞,

for any m ≥ 0. This can be proved similarly as the proof of Lemma 5.6.

Step 2. For any fixed x0 ∈ Rn, we claim

lim(x,t)→(x0,0)

u(x, t) = u0(x0).

For simplicity, we assume x0 = 0. By (K4), we have

u(x, t) − u0(0) = 1

(4πt)n2 Rn

e−|x−y|2

4t

u0(y) − u0(0)

dy.

Set v(y) = u0(y) − u0(0). Then

|u(x, t) − u0(0)| ≤ 1

(4πt)n2

Rn

e−|x−y|2

4t |v(y)|dy.

For any ε > 0, by the continuity of v , there exists a δ = δ (ε) such that

|v(y)| < ε for any |y| ≤ δ.

Note |v(y)| ≤ MeA|y|2 for some constants M and A. Then there exists a constantB > 0 such that

|v(y)| ≤ εeB|y|2 for any |y| ≥ δ.

In fact, we can choose B such that εe(B−A)δ2 ≥ M. Then by eB|y|2 ≥ 1, we have

|v(y)| ≤ εeB|y|

2

for any y ∈ Rn.

Therefore,

|u(x, t) − u0(0)| ≤ ε

(4πt)n2

Rn

e−|x−y|2

4t eB|y|2dy.

A direct calculation then yields

|u(x, t) − u0(0)| ≤ ε

(1 − 4Bt)n2

e B1−4Bt

|x|2 for any t < 1

4B.

We have the desired result by letting (x, t) → (0, 0).

8/16/2019 PDE . QUING



Now we discuss some properties of the solution (5.4) of the initial-value problem

(5.2). First from (5.4), u(x, t) for t > 0 depends on values of u0 at all points.

Equivalently, values of u0 near a point x0 ∈ Rn affect the value of u(x, t) at all xas long as t > 0. Thus values here travel at infinite speed.

Next, the function u(x, t) in (5.4) becomes smooth for t > 0, even if the initial

value u0 is simply bounded. This is well illustrated in Step 1 in the proof of Theorem5.4. We did not use any regularity assumption on u0 there. Compare with Theorem3.16. Later on, we will prove a general result that any solutions of the heat equationin a domain in Rn × (0, ∞) are smooth away from the boundary. Moreover, for

each fixed time, any solutions of the heat equation are analytic in space variablesaway from the boundary.

We need to point out that (5.4) represents only one of infinitely many solutions.Solutions are not unique without further conditions on u, like the boundedness

assumption or the exponential growth assumption. In fact, there exists a nontrivial

solution u ∈ C ∞(Rn × R) of ut − ∆u = 0, with u ≡ 0 for t ≤ 0. In the following,we construct such a solution of the 1-dimensional heat equation due to Tychonov.

Proposition 5.7. There exists a nonzero smooth function u ∈ C ∞(R× [0, ∞))satisfying

ut − uxx =0 in R× [0, ∞),

u(·, 0) =0 in R.

Proof. We construct a smooth function in R × R such that ut − uxx = 0 inR×R and u ≡ 0 for t < 0. We treat x = 0 as the initial curve and attempt to finda smooth solution of the following initial-value problem

ut − uxx = 0 in R× R,

u(0, t) = a(t), ux(0, t) = 0 for any t ∈ R,

for an appropriate function a in R. We write u as a power series in x as

u(x, t) =

∞k=1

ak(t)xk.

By a simple substitution in ut = uxx and a comparison of coefficients of power of

x, we have

a0 = a, a1 = 0, ak = (k + 1)(k + 2)ak+2 for any k ≥ 0.

Therefore, we have a formal solution

u(x, t) =

∞

k=0

1

(2k)!

a(k)(t)x2k.

We choose a(t) appropriately so that u(x, t) defined above is a smooth function and

is identically zero for t < 0. To this end, we define

a(t) =

e−

1

t2 for t > 0,

0 for t ≤ 0.

Then it is straightforward to verify that the series defining u is absolutely convergent

in R×R. This implies that u is continuous. Similarly, we can prove that u is smooth.We skip details and leave the rest of the proof as an exercise.

8/16/2019 PDE . QUING



To end this section, we discuss briefly terminal-value problems. For a fixed

constant T > 0, we consider

ut − uxx = 0 in R× (0, T ),

u(·, T ) = ϕ in R.

Here the initial value ϕ is prescribed at the terminal time T . We are interested in

the well-posedness of this problem. First, we ask whether there exists a function ain R such that the solution u of the initial-value problem

ut − uxx = 0 in R× (0, T ),

u(·, 0) = a in R

satisfies

u(·, T ) = ϕ in R.

Obviously, a necessary condition is given by ϕ ∈

C ∞(R). In fact, ϕ has to be

analytic, as will be shown by Corollary 5.11.Now we consider an example. For any positive integer m, we set

ϕm(x) = um(x, T ) = ε sin(mx).

Then, a solution is given by

um(x, t) = εem2(T −t) sin(mx),

with

am(x) = vm(x, 0) = εem2T sin(mx).

We note

maxR

|ϕm| = ε,

andmaxR

|am| = εem2T → ∞ as m → ∞.

There is no continuous dependence on initial values (prescribed at the terminaltime T ).

5.2. Regularity of Solutions

In this section, we discuss regularity of solutions of the heat equation with the

help of the fundamental solution.We often discuss the heat equation ut − ∆u = 0 in cylinders of the form

Ω × (t0, t1], where Ω is a domain in Rn and t0 < t1 are two scalars. The parabolic boundary ∂ p(Ω

×(t0, t1]) is defined by

∂ pΩ × (t0, t1] = Ω × t0 ∪ ∂ Ω × [t0, t1).

In other words, parabolic boundary consists of the bottom and the side of the

boundary. For any (x0, t0) ∈ Rn × R and any r > 0, we define

Qr(x0, t0) = Br(x0) × (t0 − r2, t0].

We note that Qr(x0, t0) for the heat equation play the same role as balls for theLaplace equation. If u is a solution of the heat equation ut −∆u = 0 in Qr(0), then

ur(x, t) = u(rx,r2t)

8/16/2019 PDE . QUING


5.2. REGULARITY OF SOLUTIONS 113

Figure 5.3. The region QR(x0, t0).

is a solution of the heat equation in Q1(0). For any (x, t) and (y, s), the parabolic

distance is defined by

dP

(x, t), (y, s)

=|x − y|2 + |t − s| 1

2 .

In the following, we denote by C 2,1 the collection of functions which are C 2 inx and C 1 in t. We first have the following regularity result for solutions of the heatequation.

Theorem 5.8. Let u be a C 2,1-solution of ut − ∆u = 0 in QR(x0, t0) for some

R > 0. Then u is smooth in QR(x0, t0).

Proof. We claim for any (x, t) ∈ QR(x0, t0)

u(x, t) = BR(x0)

K x

−y, t

−(t0

−R2)u(y, t0

−R2)dy

+

t

t0−R2

∂BR(x0)

K (x − y, t − s)

∂u

∂ny(y, s) − u(y, s)

∂K

∂ny(x − y, t − s)

dS yds.

This implies the smoothness of u in QR(x0, t0) easily since the boundary integrals

in the right-hand side are only on the parabolic boundary of BR(x0) × (t0 − R2, t]and there is no singularity for integrands for (x, t) ∈ QR(x0, t0).

We denote by (y, s) points in QR(x0, t0). Set

K = K (x − y, t − s) = 1

4π(t − s)n2

e−|x−y|24(t−s) for s < t.

Then

K s + ∆yK = 0,

and hence

0 =K (us − ∆yu) = (uK )s +

ni=1

(uK yi − Kuyi

)yi − u(K s + ∆yK )

=(uK )s +n

i=1

(uK yi − Kuyi

)yi.

8/16/2019 PDE . QUING



For any fixed (x, t) ∈ QR(x0, t0) and any ε > 0, we integrate with respect to (y, s)

in BR(x0) × (t0 − R2, t − ε). Then BR(x0)

K (x − y, ε)u(y, t − ε)dy = BR(x0)

K x − y, t − (t0 − R2)u(y, t0 − R2)dy

+

t−ε

t0−R2

∂BR(x0)

K (x − y, t − s)

∂u

∂ny(y, s) − u(y, s)

∂K

∂ny(x − y, t − s)

dS yds.

Now it suffices to prove

limε→0+

BR(x0)

K (x − y, ε)u(y, t − ε) = u(x, t).

The proof is almost identical to Step 2 in the proof of Theorem 5.4. The integral

over a finite domain here introduces few changes. We omit details.

Now we prove the interior gradient estimates by a similar method.

Theorem 5.9. Let u be a C 2,1-solution of ut − ∆u = 0 in QR(x0, t0) for some R > 0. Then

|∇xu(x0, t0)| ≤ c

R sup

QR(x0,t0)|u|,

where c is a positive constant depending only on n.

Figure 5.4. Domains for interior gradient estimates.

Proof. We first modify the proof of Theorem 5.8 to express u using the fun-damental solution and cutoff functions. We denote by (y, s) points in QR(x0, t0).For any smooth function v is BR(x0) × (t0 − R2, t0], we have

0 = v(us − ∆yu) = (uv)s +n

i=1

(uvyi − vuyi

)yi − u(vs + ∆yv).

We first choose a cutoff function ϕ ∈ C ∞(QR(x0, t0)) such that

suppϕ ⊂ Q 34 R(x0, t0), ϕ ≡ 1 in Q 1

2 R(x0, t0).

Let K be the fundamental solution of the heat equation as in (5.3) and set for anyfixed (x, t) ∈ QR/4(x0, t0)

v = ϕK = ϕ(y, s)K (x − y, t − s) for s < t.

For any ε > 0, we integrate with respect to (y, s) in BR(x0) × (t0 − R2, t − ε).We first note that there is no boundary integral over the parabolic boundary of

8/16/2019 PDE . QUING


5.2. REGULARITY OF SOLUTIONS 115

BR(x0) × (t0 − R2, t − ε) since the support of ϕ has an empty intersection with this

part of the boundary. Hence BR(x0)

(ϕu)(y, t − ε)K (x − y, ε)dy =

BR(x0)×(t0−R2,t−ε)

u(∂ s + ∆y)(ϕK )dyds.

Then similarly as in the proof of Theorem 5.4, with ε → 0 we have

ϕ(x, t)u(x, t) =

BR(x0)×(t0−R2,t)

u(∂ s + ∆y)(ϕK )dyds.

With

K = 1

4π(t − s)n2

e−|x−y|24(t−s) for s < t,

we note

K s + ∆yK = 0.

Hence for any (x, t) ∈ QR/4(x0, t0)

u(x, t) =

BR(x0)×(t0−R2,t)

u(ϕsK + ∆yϕK + 2∇yϕ · ∇yK )dyds.

We note that each term in the integrand involves a derivative of ϕ, which is zero inQR/2(x0, t0) since ϕ ≡ 1 there. Then the integral domain D is given by

D = B 34 R(x0) × (t0 − (

3

4R)2, t] \ B 1

2 R(x0) × (t0 − (1

2R)2, t].

Hence the distance between any (y, s) ∈ D and any (x, t) ∈ QR/4(x0, t0) has apositive lower bound. Therefore, the integrand has no singularity in the integraldomain. (This gives an alternate proof of the smoothness of u in QR/4(x0, t0).)

Next, we have for any (x, t) ∈ QR/4(x0, t0)

∇xu(x, t) =

D

u

(ϕs + ∆yϕ)∇xK + 2∇yϕ · ∇x∇yK

dyds.

For the cutoff function ϕ in QR(x0, t0), we require further

|∇yϕ| ≤ c

R, |ϕs| + |∇2

yϕ| ≤ c

R2,

where c is a positive constant depending only on n. With the explicit expression of K , we have

|∇xK | ≤ c |x − y|(t − s)

n2 +1

e−|x−y|24(t−s) ,

and

|∇2xK | ≤ c

|x − y|2 + (t − s)

(t − s)n2 +2

e−|x−y|24(t−s) .

Obviously, for any (x, t) ∈ QR/4(x0, t0) and any (y, s) ∈ D,

|x − y| ≤ R, 0 < t − s ≤ R2.

Therefore, for any (x, t) ∈ QR/4(x0, t0),

|∇xu(x, t)| ≤ c

D

1

R(t − s)n2 +1

e−|x−y|24(t−s) +

R

(t − s)n2 +2

e−|x−y|24(t−s)

|u(y, s)|dyds.

8/16/2019 PDE . QUING



Now we claim

D 1

(t − s) n2 +i e

− |x−y|24(t

−s)

dyds ≤ c

R2i−2 for i = 1, 2.

Then we obtain easily

|∇xu(x, t)| ≤ c

R sup

QR(x0,t0)|u| for any (x, t) ∈ Q R

4(x0, t0).

Next, we prove the claim. We decompose D into two parts

D1 =B R2

(x0) × t0 − (

3

4R)2, t0 − (

1

2R)2

,

D2 =

B 34R(x0) \ B 1

2 R(x0)×

t0 − (3

4R)2, t

.

For any (x, t)

∈QR/4(x0, t0) and (y, s)

∈D1, we have

t − s ≥ 1

4R2.

Hence for any (x, t) ∈ QR/4(x0, t0) D1

1

(t − s)n2 +i

e−|x−y|24(t−s) dyds ≤

D1

2n+2i

Rn+2idyds ≤ c

R2i−2,

where we used the fact that the volume of D1 has an order of Rn+2. Now weconsider D2. For any (x, t) ∈ QR/4(x0, t0) and (y, s) ∈ D2, we have

|y − x| ≥ 1

4R,

and hence 1

(t − s)n2 +i

e−|x−y|24(t−s) ≤ 1

(t − s)n2 +i

e− R2

43(t−s) .

Then for any (x, t) ∈ QR/4(x0, t0) and (y, s) ∈ D2,

I ≡

D2

1

(t − s)n2 +i

e−|x−y|24(t−s) dyds ≤ cRn

t

t0−( 34 R)2

1

(t − s)n2 +i

e− R2

43(t−s) ds.

With t − s = R2τ , we obtain

I ≤ c

R2i−2

( 34 )2

0

1

τ n2 +i

e− 143τ dτ ≤ c

R2i−2.

This finishes the proof of the claim.

Next, we estimate derivatives of arbitrary order.

Corollary 5.10. Let u be a C 2,1-solution of ut − ∆u = 0 in QR(x0, t0) for some R > 0. Then

|∂ kt ∇mx u(x0, t0)| ≤ cm+2k

Rm+2knkem+2k−1(m + 2k)! sup

QR(x0,t0)|u|,

where c is a positive constant depending only on n.

8/16/2019 PDE . QUING



Proof. For x-derivatives, we proceed as in the proof of Theorem 4.9 and

obtain for any multi-index α with |α| = m

|∂ αx u(x0, t0)| ≤ cmem−1m!

Rm sup

QR(x0,t0)|u|.

For t-derivatives, we have ut = ∆u and hence ∂ kt u = ∆ku for any nonnegativeinteger k. We note that there are nk terms of x-derivatives of u of order 2k in ∆ku.

Hence

|∂ kt ∇mx u(x0, t0)| ≤ nk max

|β|=m+2k|∂ βx u(x0, t0)|.

This implies the desired result easily.

Now we prove the analyticity of solutions of the heat equation on any timeslice.

Corollary 5.11. Let u be a C 2,1

-solution of ut − ∆u = 0 in QR(x0, t0) for some R > 0. Then u(·, t) is analytic in BR(x0) for any t ∈ (t0 − R2, t0]. Moreover, for each nonnegative integer k, ∂ kt u(·, t) is analytic in BR(x0) for any t ∈ (t0 −R2, t0].

The proof is identical to that of Theorem 4.10 and is omitted. As shown in

Proposition 5.7, solutions of ut − ∆u = 0 are not analytic in t in general.

5.3. The Maximum Principle

In this section, we discuss the maximum principle for the heat equation and its

applications. Suppose Ω ⊂ Rn is a bounded smooth domain. For any T > 0, set

QT = Ω

×(0, T ] =

(x, t); x

∈Ω, 0 < t

≤T

,

and

∂ pQT = (Ω × t = 0) ∪ (∂ Ω × [0, T ]).

As in the previous section, ∂ pQT is called the parabolic boundary of QT . We denoteby C 2,1(QT ) the collection of functions which are C 2 in x and C 1 in t in QT .Without confusion, we simply write Q instead QT .

We first prove the maximum principle, which asserts that any subsolutions of

the heat equation attains their maximum on the parabolic boundary.

Theorem 5.12. For u ∈ C 2,1(Q) ∩ C ( Q), if ut − ∆u ≤ 0, then the maximum of u in Q is attained on ∂ pQ, i.e.,

maxQ

u = max∂ pQ

u.

Proof. 1) We first consider a special case. If ut − ∆u < 0, we prove bycontradiction that the maximum of u in Q cannot be attained in Q. Suppose there

exists a point P 0 ∈ Q such that

u(P 0) = maxQ

u.

Then uxi(P 0) = 0 and uxixj

(P 0) is a nonpositive definite matrix. Moreover,

ut(P 0) = 0 for t ∈ (0, T ) and ut(P 0) ≥ 0 for t = T . Hence ut −∆u ≥ 0 at P 0, whichleads to a contradiction.

8/16/2019 PDE . QUING



2) For any ε > 0, we consider an auxiliary function

v(x, t) = u(x, t)−

εt.

Then

vt − ∆v = ut − ∆u − < 0.

By 1), v cannot attain its maximum inside Q. Hence

maxQ

v = max∂ pQ

v.

Then

maxQ

u(x, t) = maxQ

(v(x, t) + εt)) ≤ maxQ

v(x, t) + εT

=max∂ pQ

v(x, t) + εT ≤ max∂ pQ

u(x, t) + εT.

Letting ε → 0, we get the desired result.

The following result is referred to as the comparison principle.

Corollary 5.13. Suppose u, v ∈ C 2,1(Q) ∩ C ( Q) satisfy ut − ∆u ≤ vt − ∆vin Q and u ≤ v on ∂ pQ. Then u ≤ v in Q.

Next, we consider a slightly more general operator given by

Lu ≡ ut − ∆u + cu = f in Q.

We prove the following weak maximum principle for L.

Theorem 5.14. Suppose u ∈ C 2,1(Q) ∩ C ( Q) satisfies Lu ≤ 0 with c(x, t) ≥ 0.Then the nonnegative maximum of u must be attained on ∂ pQ, i.e.,

maxQ

u ≤ max∂ pQ

0, u.

The proof is a simple modification of that of Theorem 5.12 and is omitted.

Now, we consider a more general case.

Theorem 5.15. Suppose c(x, t) ≥ −c0 in Q for a nonnegative constant c0 and u ∈ C 2,1(Q) ∩ C ( Q) satisfies Lu ≤ 0. If u ≤ 0 on ∂ pQ, then u ≤ 0 in Q.

Continuous functions in Q always have global minimum. Therefore, c ≥ −c0

always holds in Q for some nonnegative constant c0 if c is continuous in Q. Such a

condition is introduced to emphasize the role of the minimum of c.

Proof. Let v(x, t) = e−c0tu(x, t). Then

vt − ∆v + (c(x, t) + c0)v = e−c0tLu ≤ 0.

Since c(x, t) + c0 ≥ 0, by Theorem 5.14, we get

maxQ

v(x, t) ≤ max∂ pQ

0, v(x, t) ≤ max∂ pQ

0, e−c0tu(x, t) ≤ 0.

Hence u(x, t) ≤ 0 in Q.

The following result is referred to as the comparison principle, which generalizesCorollary 5.13.

8/16/2019 PDE . QUING



Corollary 5.16. Suppose c(x, t) ≥ −c0 in Q for a nonnegative constant c0

and u, v ∈ C 2,1(Q) ∩ C ( Q) satisfy Lu ≤ Lv in Q and u ≤ v on ∂ pQ. Then u ≤ v

in Q.

There is also a strong maximum principle for parabolic differential equations.We will not discuss it here. In the following, we discuss applications of maximumprinciples.

We first derive an estimate in sup-norms of solutions of initial/boundary-valueproblems with Dirichlet boundary values. Compare this with the estimates in

integral norms in Theorem 3.14.

Theorem 5.17. Suppose u ∈ C 2,1(Q) ∩ C ( Q) is a solution of

Lu ≡ ut − ∆u + cu =f in QT ,

u(·, 0) =u0 in Ω,

u =ϕ on ∂ Ω × (0, T ),

for some f ∈ C ( Q), u0 ∈ C (Ω) and ϕ ∈ C (∂ Ω × [0, T ]). If c(x, t) ≥ −c0 in Q for a nonnegative constant c0, then

maxQ

|u| ≤ ec0T

maxsupΩ

|u0|, sup∂ Ω×(0,T )

|ϕ| + T supQ

|f |.

Proof. Set

B = maxsupΩ

|u0|, sup∂ Ω×(0,T )

|ϕ|, F = supQ

|f |,

and

v(x, t) = ec0t(F t + B).

ThenLv = (c0 + c)ec0t(F t + B) + ec0tF ≥ F ≥ ±Lu in Q,

and

v ≥ B ≥ ±u on ∂ pQ.

Here we used c + c0 ≥ 0 and ec0t ≥ 1 in Q. By Corollary 5.16, we obtain

v ≥ ±u in Q,

or

|u(x, t)| ≤ ec0t(F t + B) for any (x, t) ∈ QT .

This implies the desired estimate.

In the following, we study a priori estimates for initial-value problems.

Theorem 5.18. Suppose u is a bounded C 2-solution of

Lu ≡ ut − ∆u + cu =f in Rn × (0, T ],

u(·, 0) =u0 in Rn,

for some bounded f ∈ C (Rn ×(0, T ]) and u0 ∈ C (Rn). If c ≥ −c0 for a nonnegative constant c0, then

supRn×(0,T )

|u| ≤ ec0T ( supRn×(0,T )

|u0| + T supRn×(0,T )

|f |).

8/16/2019 PDE . QUING



Proof. Set

H = Rn

×(0, T ] =

(x, t); x

∈Rn, t

∈(0, T ]

,

and

F = supH

|f |, B = supRn

|u0|.We assume supH |u| ≤ M for a positive constant M since u is bounded. For any

R > 0, set

QR = BR × (0, T ].

Consider an auxiliary function

w(x, t) = ec0t(F t + B) + vR(x, t) ± u(x, t) in QR,

where vR is a function to be chosen. Then

Lw = (c + c0)ec0t(F t + B) + ec0tF

±Lu + LvR

≥LvR in QR.

Here we used c + c0 ≥ 0 and ec0t ≥ 1. Moreover,

w(·, 0) = B + vR(·, 0) ± u0 in BR,

and

w ≥ vR ± u on ∂ BR × (0, T ].

We require

LvR ≥0 in QR,

vR(·, 0) ≥0 in BR,

vR ± u ≥0 on ∂ BR × (0, T ].

To construct such a vR, we consider

vR(x, t) = M

R2ec0t(2nt + |x|2).

Obviously, vR ≥ 0 for t = 0 and vR ≥ M on |x| = R. Next,

LvR = M

R2ec0t(c + c0)(2nt + |x|2) ≥ 0 in QR.

With such a vR, we have

Lw ≥ 0 in QR,

w ≥ 0 on ∂ pQR.

Hence Corollary 5.16 yields w ≥ 0 in QR. Therefore for any fixed (x, t), by taking

R large such that (x, t)

∈QR, we have

|u(x, t)| ≤ ec0t(F t + B) + M

R2ec0t(2nt + |x|2).

By letting R → +∞, we have

|u(x, t)| ≤ ec0t(F t + B) for any (x, t) ∈ H.

This yields the desired estimate.

Next, we prove the uniqueness of solutions of the heat equation under an as-sumption of exponential growth.

8/16/2019 PDE . QUING



Theorem 5.19. Let u ∈ C 2,1(Rn × (0, T )) ∩ C (Rn × [0, T ]) satisfy

ut

−∆u =0 in Rn

×(0, T ),

u(·, 0) =0 in Rn.

If

|u(x, t)| ≤ MeA|x|2 for any (x, t) ∈ Rn × (0, T ),

for some positive constants M and A, then u ≡ 0 in Rn × (0, T ).

Proof. For any constant α > A, we prove

u(x, t) = 0 for any (x, t) ∈ Rn × (0, 1

4α).

Our strategy is as follows. We construct an appropriate auxiliary function vR inBR × (0, 1/4α) so that

|u(x, t)| ≤ vR(x, t) for any (x, t) ∈ BR × 0,

1

4α,

and for any fixed (x, t) ∈ BR × (0, 1/4α)

vR(x, t) → 0 as R → ∞.

The function vR is constructed with the help of the fundamental solution. Now we

start our proof.For any constant R > 0, consider

vR(x, t) = M e(A−α)R2

(1 − 4αt)n2

eα|x|21−4αt for (x, t) ∈ BR ×

0, 1

4α

.

We note that vR is simply the example we discussed preceding Theorem 5.5. Then

∂ tvR − ∆vR = 0 in BR × 0,

1

4α.Obviously,

vR(·, 0) ≥ 0 = ±u(·, 0) in BR.

Next,

vR(x, t) ≥Me(A−α)R2

eαR2

= M eAR2 ≥ ±u(x, t)

for any (x, t) ∈ ∂BR × 0,

1

4α

.

By Corollary 5.13, the comparison principle, we have

±u(x, t) ≤ vR(x, t) for any (x, t) ∈ BR × 0,

1

4α

,

or

|u(x, t)| ≤ vR(x, t) for any (x, t) ∈ BR × 0, 14α

.

For any fixed (x, t) ∈ Rn × (0, 1/4α), we choose R > |x|. We note vR(x, t) → 0 asR → ∞, by α > A. This implies easily u(x, t) = 0.

As the final application of the maximum principle, we provide another proof of interior gradient estimates by the maximum principle. We do this only for solutions

of the heat equation. Recall for any r

Qr = Br × (−r2, 0].

8/16/2019 PDE . QUING



Theorem 5.20. Suppose u ∈ C 2,1(Q1) ∩ C ( Q1) satisfies ut − ∆u = 0 in Q1.

Then

supQ 1

2

|∇xu| ≤ C sup∂ pQ1

|u|,


Proof. Set v = |∇xu|2. Then

vt − ∆v = −2

ni,j=1

u2xixj

+ 2

ni=1

uxi(ut − u)xi

= −2

ni,j=1

u2xixj

.

To get interior estimates, we intorduce a cut-off function. For any smooth function

ϕ in C ∞(Q1) with suppϕ ⊂ Q3/4, we have

(ϕv)t −

∆(ϕv) = (ϕt −

ϕ)|∇x

u|2

−4

n

i,j=1

ϕxi

uxj

uxixj −

2ϕn

i,j=1

u2

xixj

.

By taking ϕ = η2 for some η ∈ C ∞(Q1) with η ≡ 1 in Q1/2 and suppη ⊂ Q3/4, wehave

(η2|∇xu|2)t − (η2|∇xu|2)

=(2ηηt − 2ηη − 2|∇xη|2)|∇xu|2 − 8ηn

i,j=1

ηxiuxj

uxixj − 2η2

ni,j=1

u2xixj

.

By the Cauchy inequality, we obtain

8|ηηxiuxj

uxixj| ≤ 4η2

xiu2

xj + 2η2u2

xixj,

and hence

(η2|∇xu|2)t − (η2|∇xu|2) ≤

2ηηt − 2ηη + 6|∇xη|2

|∇xu|2 ≤ C |∇xu|2,

where C is a positive constant depending only on η and n. Note

(u2)t − (u2) = −2|∇xu|2 + 2u(ut − u) = −2|∇xu|2.

By taking a constant α large enough, we get

(∂ t − ∆)(η2|∇xu|2 + αu2) ≤ (C − 2α)|∇xu|2 ≤ 0.

We apply Corollary 5.13 to get

sup

Q1

(η2

|∇xu

|2 + αu2)

≤ sup

∂ pQ1

(η2

|∇xu

|2 + αu2).

This implies the desired result since η = 0 on ∂ pQ1 and η = 1 in Q1/2.

To end our discussions, we compare maximum principles for elliptic and para-bolic equations of the following forms

Leu = −∆u + c(x)u in Ω,

and

L pu = ut − ∆u + c(x, t)u in QT ≡ Ω × (0, T ).

8/16/2019 PDE . QUING


5.4. HARNACK INEQUALITIES 123

If c ≥ 0, then

Leu

≤0 implies u attains its nonnegative maximum on ∂ Ω,

L pu ≤ 0 implies u attains its nonnegative maximum on ∂ pQ.

If c ≡ 0, the nonnegativity condition can be removed. For c ≥ 0, comparisonprinciples can be stated as follows:

Leu ≤ Lew in Ω, u ≤ v on ∂ Ω ⇒ u ≤ v in Ω,

L pu ≤ L pw in QT , u ≤ v on ∂ pQT ⇒ u ≤ v in QT .

We should note that comparison principles for parabolic equations hold for c ≥ −c0,for a nonnegative constant c0.

In practice, we need to construct auxiliary functions for comparisons. Usually,

we take |x|2 or e±α|x|2 for elliptic equations and K t + |x|2 for parabolic equations.Sometimes, auxiliary functions are constructed with the help of the fundamental

solutions for the Laplace equation and the heat equation.

5.4. Harnack Inequalities

For positive harmonic functions, the Harnack inequality asserts that values incompact subsets are comparable. In this section, we study the Harnack inequalityfor positive solutions of the heat equation. In order to investigate a proper form of the Harnack inequality for solutions of the heat equation, we begin our discussion

with the fundamental solution of the heat equation.Consider for any ξ ∈ Rn

u(x, t) = 1

(4πt)n2

e−|x−ξ|2

4t for any x ∈ Rn, t > 0.

Then u satisfies the heat equation ut − ∆u = 0 in Rn

× (0, ∞). Hence, for any(x1, t1) and (x2, t2) ∈ Rn × (0, ∞)

u(x1, t1)

u(x2, t2) =

t2

t1

n2

e|x2−ξ|2

4t2−|x1−ξ|2

4t1 .

Note( p + q )2

a + b ≤ p2

a +

q 2

b for any a,b > 0,

where equality holds if and only if bp = aq . This implies for any t2 > t1 > 0

|x2 − ξ |2

t2≤ |x2 − x1|2

t2 − t1+

|x1 − ξ |2

t1,

where equality holds if and only if

ξ = t2x1 − t1x2

t2 − t1.

Therefore,

u(x1, t1) ≤

t2

t1

n2

e|x2−x1|24(t2−t1) u(x2, t2) for any t2 > t1 > 0,

where equality holds if ξ is chosen as above. This simple calculation suggests that

the Harnack inequality for the heat equation has an “evolution” feature: the valueof a positive solution at a certain time is controlled from above by the value at

8/16/2019 PDE . QUING



Figure 5.5. Values of the fundamental solution at different time.

a later time. In general, we cannot estimate the value of a positive solution ata certain time from above by values at a previous time. Hence if we attempt toestablish the following estimate

u(x1, t1) ≤ Cu(x2, t2),

the constant C should depend on t2/t1, |x2−x1| and most importantly (t2−t1)−1

(>0). Usually, we choose following regions for points (x1, t1) ∈ D1 and (x2, t2) ∈ D2.

Figure 5.6. Domains for Harnack inequalities.

Suppose u is a positive function and set v = log u. In order to derive an estimatefor the quotient

u(x1, t1)

u(x2, t2),

it suffices to get an estimate for the difference

v(x1, t1) − v(x2, t2).

To this end, we need an estimate of vt and |∇v|. For a hint of proper forms, weagain turn our attention to the fundamental solution of the heat equation.

Consider

u(x, t) = 1

(4πt)n2 e

− |x|2

4t for any x ∈ Rn

, t > 0.

Then

v(x, t) = log u(x, t) = −n

2 log(4πt) − |x|2

4t ,

and hence

vt = − n

2t + |∇v|2.

We have the following result for arbitrary positive solutions of the heat equa-tion.

8/16/2019 PDE . QUING



Theorem 5.21. Suppose u ∈ C 2,1(Rn × (0, T ]) satisfies

ut = ∆u, u > 0 in Rn

×(0, T ].

Then v = log u satisfies

(5.5) vt + n

2t ≥ |∇v|2 in Rn × (0, T ].

The inequality (5.5) is referred to as the differential Harnack inequality . We

first prove a corollary.

Corollary 5.22. Suppose u ∈ C 2,1(Rn × (0, T ]) satisfies

ut = ∆u, u > 0 in Rn × (0, T ].

Then for any (x1, t1), (x2, t2) ∈ Rn × (0, T ] with t2 > t1 > 0

(5.6)

u(x1, t1)

u(x2, t2) ≤ t2

t1n2

exp|x2

−x1

|2

4(t2 − t1) .

Proof. Along any path

t, x(t)

with x(ti) = xi, i = 1, 2, we have by Theorem

5.21d

dtv(x(t), t) = vt + ∇v · dx

dt ≥ |∇v|2 + ∇v · dx

dt − n

2t.

By completing square, we obtain

d

dtv(x(t), t) ≥ −1

4

dx

dt

2

− n

2t,

and hence

v(x2, t2) ≥ v(x1, t1) − n

2 log

t2

t1− 1

4 t2

t1 dx

dt 2

dt.

To seek an optimal path which makes the last integral minimal, we require

d2x

dt2 = 0

along the path. Hence we set

x = at + b for some a, b ∈ Rn.

By xi = ati + b, i = 1, 2, we take

a = x2 − x1

t2 − t1and b =

t2x1 − t1x2

t2 − t1.

Then,

t2

t1dx

dt 2

dt = |x2

−x1

|2

t2 − t1 .

Therefore, we obtain

v(x2, t2) ≥ v(x1, t1) − n

2 log

t2

t1− 1

4

|x2 − x1|2

t2 − t1,

oru(x2, t2)

u(x1, t1) ≥

t1

t2

n2

exp−|x2 − x1|2

4(t2 − t1).

This is the desired estimate.

8/16/2019 PDE . QUING



Now we start to prove the differential Harnack inequality. The basic idea is to

apply the maximum principle to an appropriate combination of derivatives of v. In

our case, we consider |∇v|2 − vt and intend to derive an upper bound. First, weneed to derive a parabolic equation satisfied by |∇v|2 −vt. A careful analysis showsthere are uncontrollable terms in this equation. Therefore, we should introduce a

parameter α ∈ (0, 1) and consider α|∇v|2−vt instead. After we apply the maximumprinciple, we let α → 1−. Since the proof below is quite involved, we divide it intoseveral steps.

Proof of Theorem 5.21. Step 1. We first derive some equations involvingderivatives of v = log u. A simple calculation shows

vt = ∆v + |∇v|2.

Note that the differential Harnack inequality asserts

∆v +

n

2t ≥ 0,

which implies ∆v is almost concave.Consider w = ∆v. Then we have

wt = ∆vt = ∆(∆v + |∇v|2) = ∆w + ∆|∇v|2.

Since

∆|∇v|2 = 2|∇2v|2 + 2∇v · ∇(∆v) = 2|∇2v|2 + 2∇v · ∇w,

then

(5.7) wt − ∆w − 2∇v · ∇w = 2|∇2v|2.

Note that the uncontrolled expression ∇v appears as a coefficient of the equationin (5.7). It is convenient to derive an equation for it. Set w =

|∇v

|2. Then,

wt = 2∇v · ∇vt = 2∇v · ∇(∆v + |∇v|2)

= 2∇v · ∇(∆v) + 2∇v · ∇w

= ∆|∇v|2 − 2|∇2v|2 + 2∇v · ∇w

= ∆w + 2∇v · ∇w − 2|∇2v|2.

Therefore, w satisfies

(5.8) wt − ∆w − 2∇v · ∇w = −2|∇2v|2.

Note by the Cauchy inequality

|∇2v|2 =n

i,j=1

v2xixj

≥n

i=1

v2xixi

≥ 1

n(

n

i=1

vxixi)2 =

1

n(∆v)2.

So (5.7) implies

wt − ∆w − 2∇v · ∇w ≥ 2

nw2.

Step 2. Set for a constant α ∈ (0, 1]

F = α|∇v|2 − vt.

Then

F = α|∇v|2 − ∆v − |∇v|2 = −∆v − (1 − α)|∇v|2 = −w − (1 − α)w,

8/16/2019 PDE . QUING



and hence by (5.7) and (5.8)

F t−

∆F −

2∇

v· ∇

F =−

2α|∇

2v|2.

Next, we estimate |∇2v|2 by F . Note

|∇2v|2 ≥ 1

n(∆v)2 =

1

n

|∇v|2 − vt

2=

1

n((1 − α)|∇v|2 + F )2

= 1

n

F 2 + (1 − α)2|∇v|4 + 2(1 − α)|∇v|2F

≥ 1

n

F 2 + 2(1 − α)|∇v|2F

.

We obtain

(5.9) F t − ∆F − 2∇v · ∇F ≤ −2α

n

F 2 + 2(1 − α)|∇v|2F

.

We should point out that the term −|∇v|2

at the right-hand side plays an importantrole later on.

Step 3. Now we introduce a cut-off function. Set

G = tFϕ,

where ϕ ≥ 0 and ϕ ∈ C ∞0 (Rn). We derive a differential inequality for G. Note

Gt = F ϕ + tF tϕ,

∇G = t∇F ϕ + tF ∇ϕ,

∆G = t∆F ϕ + 2t∇F · ∇ϕ + tF ∆ϕ.

Then,

tF tϕ = Gt − Gt

,

t∇F ϕ = ∇G − ∇ϕ

ϕ G,

t∆F ϕ = ∆G − 2

ϕ∇ϕ ·

∇G − ∇ϕ

ϕ G

− ∆ϕ

ϕ G

= ∆G − 2∇ϕ

ϕ · ∇G + 2

|∇ϕ|2

ϕ2 G − ∆ϕ

ϕ G.

Multiplying (5.9) by t2ϕ2 and substituting above inequalities, we obtain

tϕ(Gt − ∆G) + 2t(∇ϕ − ϕ∇v) · ∇G

≤ G ϕ − 2α

n G + t2 |∇ϕ|2

ϕ − ∆ϕ − 4α(1

−α)

n ϕ|∇v|2 − 2∇v · ∇ϕ .

To eliminate |∇v| from the right-hand side, we complete square for the last two

terms. Here we need α < 1! Otherwise, we cannot control the expression −2∇v ·∇ϕin the right-hand side. Then we have

tϕ(Gt − ∆G) + 2t(∇ϕ − ϕ∇v) · ∇G

≤ G

ϕ − 2α

n G + t

2|∇ϕ|2

ϕ − ∆ϕ − n

4α(1 − α)

|∇ϕ|2

ϕ

.

8/16/2019 PDE . QUING



We point out that there are no unknown expressions in the right-hand side except

G. By choosing ϕ = η2 for some η ≥ 0 and η ∈ C ∞0 (Rn), we get

tη2(Gt − ∆G) + 2t(2η∇η − η2∇v) · ∇G

≤ G

η2 − 2α

n G + t

6|∇η|2 − 2η∆η − n

α(1 − α)|∇η|2

.

Now we fix a cut-off function η0 ∈ C ∞0 (B1), with 0 < η0 ≤ 1 in B1 and η0 = 1

in B1/2. For an arbitrary R ≥ 1, we consider η(x) = η0(x/R). Then we obtain inBR × (0, T )

tη2(Gt − ∆G) + 2t(2η∇η − η2∇v) · ∇G ≤ G

1 − 2α

n G +

C αt

R2

,

where C α is a positive constant depending only on α and η0.Step 4. We claim

(5.10) 1 − 2αn

G + C αtR2

≥ 0 in BR × (0, T ).

Note that G vanishes on the parabolic boundary of BR × (0, T ) since G = tF ϕ.

Suppose (5.10) does not hold. Then 1 − 2α

n G +

C αt

R2 has a negative minimum at

(x0, t0) ∈ BR × (0, T ]. Moreover, we have

G(x0, t0) > 0,

and

Gt ≥ 0, ∇G = 0, ∆G ≤ 0 at (x0, t0).

Then at (x0, t0), we get

0

≤ tη 2(Gt

−∆G) + 2t(2η

∇η

−η2

∇v)

· ∇G

≤ G

1 − 2α

n G +

C αt

R2

< 0.

This is a contradiction. Hence (5.10) holds in BR × (0, T ).Therefore, we obtain

(5.11) 1 − 2α

n tη2(α|∇v|2 − vt) +

C αt

R2 ≥ 0 in BR × (0, T ).

For any fixed (x, t) ∈ Rn × (0, T ), choose R > |x|. Recall η(x) = η0(x/R) andη0 = 1 in B1/2. Letting R → +∞, we obtain

1 − 2α

n t(α|∇v|2 − vt) ≥ 0.

We then let α→

1−

and get the desired estimate.

By a simple modification of the proof above, we obtain the following differentialHarnack inequality for positive solutions in a finite region.

Theorem 5.23. Suppose u satisfies

ut − ∆u = 0, u > 0 in B1 × (0, 1).

Then for any α ∈ (0, 1), v = log u satisfies

vt − α|∇v|2 + n

2αt + C ≥ 0 in B 1

2× (0, 1),

8/16/2019 PDE . QUING


EXERCISES 129

where C is a positive constant depending only on n and α.

Proof. We simply take R = 1 in (5.11).

Now we state the Harnack inequality in a finite region.

Corollary 5.24. Suppose u satisfies

ut − ∆u = 0, u > 0 in B1 × (0, 1).

Then for any (x1, t1), (x2, t2) ∈ B1/2 × (0, 1) with t2 > t1,

u(x1, t1) ≤ Cu(x2, t2),

where C is a positive constant depending only on t2/t1 and (t2 − t1)−1.

The proof is left as an exercise.

Exercises

(1) Prove the following statements by straightforward calculations.

(a) K (x, t) = t−n2 e−

|x|24t satisfies the heat equation for t > 0.

(b) For any α > 0, G(x, t) = (1−4αt)−n2 e

α|x|21−4αt satisfies the heat equation

for t < 1/4α.

(2) For u0 ∈ C 0(Rn) and u defined in (5.4), prove limt→∞ u(x, t) = 0 uni-

formly in x.

(3) For u0 ∈ C 0(Rn) and u defined in (5.4), find an appropriate condition onu0 so that u is C k up to t = 0 and give a proof.

(4) Let u0 be a bounded and continuous function in [0 , ∞) with u0(0) = 0.Find an integral representation for the solution of the following problem

ut − uxx =0 for x > 0, t > 0,

u(x, 0) =u0(x) for x > 0,

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

(5) Prove that u constructed in the proof of Proposition 5.7 is smooth inR×R.

(6) Suppose the initial temperature u0 of an infinite rod is given by

u0 = 1 x ∈ [0, 1],

0 otherwise.At what time does a point with coordinate x attain the highest tempera-

ture? In particular, consider points x = 1/2 and x = 3/2.

(7) Suppose u ∈ C 2(QT ) ∩ C ( QT ) satisfies

ut − ∆u + c(x, t)u = − u2 in QT = Ω × (0, T ),

u|t=0 =u0,

u|∂ Ω×(0,T ) =0,

8/16/2019 PDE . QUING



where c(x, t) ∈ C ( QT ) with c ≥ −c0 and u0 ∈ C (Ω) with u0 ≥ 0. Prove

0

≤u

≤ec0t sup

Ω |u0

| in Ω.

(8) Suppose u ∈ C 2(QT ) ∩ C ( QT ) satisfies

ut − ∆u =e−u − f (x) in QT ,

u|t=0 =u0 in Ω,

u|∂ Ω×(0,T ) =ϕ.

Prove

−M ≤ u ≤ eMT + M in Ω,

whereM = T sup

Ω|f | + supsup

Ω|u0|, sup

∂ Ω×(0,T )|ϕ|.

(9) Let Q = (0, l) × (0, ∞) and u ∈ C 3(Q) ∩ C 1( Q) satisfy

ut − uxx =0 in Q,

u|t=0 =ϕ in (0, l)

u|x=0 = 0, u|x=l =0 for any t > 0,

for a function ϕ ∈ C 1[0, l] with ϕ(0) = ϕ(l). Prove

supQ

|ux| ≤ sup[0,l]

|ϕ|.

(10) Let h(t) be a continuous increasing function with h(0) ≥ 0 and

QT = (x, t); 0 ≤ x ≤ h(t), 0 ≤ t ≤ T .

Let g be an increasing nonnegative function in [0, T ] and ϕ be a decreasingC 1-function in [0, h(0)] with where ϕ(0) = g(0) and ϕ(h(0)) = 0. Supposeu ∈ C 2,1(Q) is a solution of

ut − uxx =0 in Q,

u(x, 0) =ϕ(x) for x ∈ [0, h(0)],

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

u(h(t), t) =0,

Prove ∂ xu ≤ 0 in QT .

Figure 5.7. A domain.

8/16/2019 PDE . QUING


EXERCISES 131

(11) Let Q = (0, l) × (0, ∞) and uh = uh(x, t) be a solution of

ut

−uxx =0 in Q,

u(x, 0) =0 in (0, l),

u|x=l =0,∂ xu + h(u0 − u)

|x=0 =0,

for positive constants u0 and h. Prove(a) 0 ≤ u ≤ u0 in (0, l) × (0, ∞);(b) uh is monotone increasing for h.

(12) Let u1, · · · , um be C 2,1-functions in Q = Ω × (0, T ] ⊂ Rn ×R+ satisfying

∂ tui = ∆ui in Q, for i = 1, · · · , m.

Suppose f is a convex function in Rm. Then

supQ

f (u1, · · · , um) ≤ sup∂ pQ

f (u1, · · · , um).

(13) Let u ∈ C 2,1(Rn × (0, T ]) satisfy

ut − ∆u = 0 in Rn × (0, T ].

Suppose u and ∇u are bounded in Rn × (0, T ]. Prove

supRn×(0,T ]

|∇u| ≤ 1√ 2t

supRn

|u(·, 0)|.

Hint: With |u| ≤ M at t = 0, consider

w = u2 + 2t|∇u|2 − M 2.

(14) Prove Corollary 5.24.

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


CHAPTER 6

Wave Equations

In this chapter, we study the wave equation in Rn × R+. The wave equationrepresents vibrations of strings or propagation of sound waves in tubes for n = 1,

waves on the surface of shallow water for n = 2, and acoustic or light waves forn = 3. In Section 6.1, we discuss initial-value problems and initial/boundary-valueproblems for the one-dimensional wave equation. In Section 6.2, we study initial-

value problems for the wave equation in higher dimensional spaces. Then in Section6.3, we derive energy estimates for solutions of initial-value problems.

6.1. One-Dimensional Wave Equations

In this section, we discuss initial-value problems and various initial/boundary-

value problems for the one-dimensional wave equation. We first study initial-valueproblems.

For f ∈ C (R × R+), ϕ ∈ C 2(R) and ψ ∈ C 1(R), we seek a solution u ∈C 2(R×R+) of the following problem

utt − uxx = f in R×R+,

u(

·, 0) = ϕ, ut(

·, 0) = ψ on R.

(6.1)

We will use different methods to derive expressions of solutions in various special

cases.As discussed in Section 3.1, characteristic curves for the one-dimensional wave

equation are given by straight lines t = ±x+c. In particular, for any (x, t) ∈ R×R+,there are two characteristic curves through (x, t) given by

t − x = t − x and t + x = t + x.

We first consider the case f ≡ 0. By setting

ξ = x − t, η = x + t,

we have

uξη = 0,and hence

u(ξ, η) = g(ξ ) + h(η),

for some functions g and h in R. Therefore,

(6.2) u(x, t) = g(x − t) + h(x + t).

With initial values, we have

ϕ(x) = g(x) + h(x), ψ(x) = −g(x) + h(x).

133

8/16/2019 PDE . QUING


134 6. WAVE EQUATIONS

Then

g(x) =

1

2 ϕ(x) − 1

2 x

0 ψ(s)ds + c

h(x) = 1

2ϕ(x) +

1

2

x

0

ψ(s)ds − c,

for a constant c. Hence,

(6.3) u(x, t) = 1

2

ϕ(x − t) + ϕ(x + t)

+

1

2

x+t

x−t

ψ(s)ds.

This is called the d’Alembert’s formula . Such a formula clearly shows that regularityof u(·, t) for any t > 0 is the same as that of the initial value u(·, 0). There is noimprovement of regularity.

We see from (6.3) that u(x, t) is determined uniquely by initial values in the

interval [x− t, x + t] of the x-axis whose end points are cut out by the characteristiccurves through the point (x, t). This interval represents the domain of dependence for the solution at the point (x, t). Conversely, the initial values at a point (x0, 0)of the x-axis influence u(x, t) at points (x, t) in the wedge-shaped region bounded

by characteristic curves through (x0, 0), i.e., for x0 − t < x < x0 + t.

Figure 6.1. The domain of influence and the domain of dependence.

Now we derive an important formula for the solution of the wave equation. Letu be a C 2-solution of

utt − uxx = 0 in R× R+,

and A,B, C, D be four vertices of a parallelogram bounded by four characteristiccurves in R× R+ as follows. (This parallelogram is in fact a rectangle.) Then

(6.4) u(A) + u(D) = u(B) + u(C ).

In other words, the sums of the values of u in opposite vertices are equal. Thisfollows easily from (6.2).

Next, we use the method of characteristics to solve (6.1) if f ≡ 0 and ϕ ≡ 0.

By writing

utt − uxx = (∂ t + ∂ x)(∂ t − ∂ x)u,

we decompose (6.1) to two initial-value problems for first order PDEs as follows

ut − ux = v in R× R+,

u(·, 0) = 0 on R,(6.5)

8/16/2019 PDE . QUING


6.1. ONE-DIMENSIONAL WAVE EQUATIONS 135

Figure 6.2. A rectangle.

and

vt + vx = 0 in R× R+,

v(·, 0) = ψ on R.

(6.6)

We note that integral curves are exactly characteristic curves for first order linear

PDEs in the plane. First, we discuss (6.6). Its associated integral curves are givenby x = t + c. Along these curves, the equation in (6.6) becomes

dv(x(t), t)

dt = 0.

Hence

v(x(t), t) = v(x(0), 0) = ψ(x(0)).

For any (x, t), we have x(t) = t + (x − t) and hence

v(x, t) = v(x(t), t) = ψ(x(0)) = ψ(x − t).

Therefore, (6.6) has a solutionv(x, t) = ψ(x − t).

For (6.5), its integral curves are given by x = −t + c and hence x(t) = −t + x + tfor any (x, t). Along these curves, the equation in (6.5) becomes

du(x(t), t)

dt = v(x(t), t) = ψ(x(t) − t).

Then

u

x(t), t

=

t

0

ψ

x(τ ) − τ

dτ,

and hence

u(x,¯t) =

t

0 ψx(τ ) − τ dτ = t

0 ψ(x + ¯t − 2τ )dτ =

1

2 x+t

x−t ψ(s)ds.

Therefore,

u(x, t) = 1

2

x+t

x−t

ψ(s)ds.

This is simply a special case of the D’Alemberts formula.Now we derive an expression of solutions in the general case. For any (x, t) ∈

R×R+, consider the triangle

C 1(x, t) = (y, τ ); |y − x| < t − τ , τ > 0.

8/16/2019 PDE . QUING



This is exactly the cone we introduced in Section 2.3 for n = 1. The boundary of

C 1(x, t) consists of three parts

L+ = (y, τ ); τ = −y + x + t, 0 < τ < t,

L− = (y, τ ); τ = y − x + t, 0 < τ < t,

and

L0 = (y, 0); x − t < y < x + t.

We note that L+ and L− are parts of the characteristic curves through (x, t). Letγ = (γ 1, γ 2) be the unit exterior normal vector of ∂C 1(x, t). Then

γ =

(1, 1)/√

2 on L+,

(1, −1)/√

2 on L−,

(0, −1) on L0.

Hence by the Green’s formula, we have


C 1(x,t)

f =

C 1(x,t)

(utt − uxx) =

∂C 1(x,t)

(utγ 2 − uxγ 1)

= − x+t

x−t

ut(s, 0)ds +

L+

1√ 2

(ut − ux) +

L−

1√ 2

(ut + ux),

where the orientation of integrals on L+ and L− is counterclockwise. Note that

(∂ t − ∂ x)/√

2 is a directional derivative along L+ with unit length and with thedirection matching the orientation of the integral on L+. Hence

L+

1√ 2

(ut − ux) = u(x, t) − u(x + t, 0).

On the other hand, (∂ t + ∂ x)/√ 2 is a directional derivative along L− with unitlength and with the opposite direction matching the orientation of the integral onL−. Hence

L−

1√ 2

(ut + ux) = −u(x − t, 0) − u(x, t)

.

Therefore, a simple substitution yields

(6.7) u(x, t) = 1

2

ϕ(x+t)+ϕ(x−t)

+

1

2

x+t

x−t

ψ(s)ds+1

2

t

0

dτ

x+(t−τ )

x−(t−τ )

f (y, τ )dy.

8/16/2019 PDE . QUING



Theorem 6.1. Suppose ϕ ∈ C 2(R), ψ ∈ C 1(R) and f ∈ C 1(R× [0, ∞)). Then

u in (6.7) is a C 2-solution of (6.1).

The proof is a straightforward calculation and is omitted. Obviously, C 2-solutions of (6.1) are unique.

The formula (6.7) illustrates that the value u(x, t) is determined by f in thetriangle C 1(x, t) and by ϕ and ψ on the interval [x − t, x + t] × 0. This has the

same feature as solutions of initial value problems of first-order linear differentialequations discussed in Section 2.3.

Without using the explicit expression of solutions in (6.7), we can also deriveenergy estimates, the estimates of the L2-norms of solutions of (6.1) and theirderivatives. For any constants 0 < T < t, we use the following domain for energy

estimates

(x, t); |x| < t − t, 0 < t < T .

We postpone derivation until the final section of this chapter.

Figure 6.4. A domain of integration.

In the rest of this section, we study initial/boundary-value problems. For sim-

plicity, we only discuss homogeneous wave equations.First, we study half-space problems. Assume ϕ ∈ C 2[0, ∞), ψ ∈ C 1[0, ∞) and

α ∈ C 2[0, ∞), and consider

utt − uxx = 0 in [0, ∞) × (0, ∞),

u(·, 0) = ϕ, ut(·, 0) = ψ on [0, ∞),

u(0, t) = α(t) for t > 0.

(6.8)

We construct a C 2-solution under appropriate compatibility conditions. If (6.8)

admits a solution which is C 2 in [0, ∞) × [0, ∞), a simple calculation shows

(6.9) ϕ(0) = α(0), ψ(0) = α(0), ϕ(0) = α(0).

We first consider the case α ≡ 0 and use the method of extension to solve (6.8).

The compatibility condition (6.9) has the following form

ϕ(0) = 0, ψ(0) = 0, ϕ(0) = 0.

Now we assume this holds and proceed to construct a C 2-solution of (6.8). Weextend ϕ and ψ as odd functions in R. The extended ϕ and ψ are C 2 and C 1 in R

8/16/2019 PDE . QUING



respectively. Let u be the unique C 2-solution of

utt

−uxx = 0 in R

×R+,

u(·, 0) = ϕ, ut(·, 0) = ψ in R.

In fact, u is given by the D’Alemberts formula (6.3). We now prove that u(x, t) is

a solution of (6.8) when we restrict x ∈ [0, ∞). We only need to prove

u(0, t) = 0 for any t > 0.

In fact, for v(x, t) = −u(−x, t), a simple calculation yields

vtt − vxx = 0 in R×R+,

v(·, 0) = ϕ, vt(·, 0) = ψ in R.

By the uniqueness, u(x, t) = v(x, t) = −u(−x, t) and hence u(0, t) = 0.Now we consider the general case of (6.8) and use the method of reflection to

construct a solution in [0, ∞)× [0, ∞). To do this, we divide [0, ∞)× [0, ∞) into twoparts by the straight line t = x and construct u in each region. First, we consider

Ω1 = (x, t); x > t > 0.

By (6.3), the solution u1 in Ω1 is given by

u1(x, t) = 1

2

ϕ(x + t) + ϕ(x − t)

+

1

2

x+t

x−t

ψ(s)ds for any (x, t) ∈ Ω1.

Set

γ (x) = u1(x, x) = 1

2

ϕ(2x) + ϕ(0)

+

1

2

2x

0

ψ(s)ds.

Next, we set

Ω2 = (x, t); t > x > 0,and consider

utt − uxx = 0 in Ω2

u(0, t) = α(t), u(x, x) = γ (x).

We denote its solution by u2. For any (x, t) ∈ Ω2, consider a rectangle formed by(x, t), (0, t − x), ( t−x

2 , t−x

2 ) and ( t+x

2 , t+x

2 ). By (6.4), we have

u2(x, t) = α(−x + t) − γ (−x + t

2 ) + γ (

x + t

2 )

= α(−x + t) + 1

2

ϕ(x + t) − ϕ(−x + t)

+

1

2

x+t

−x+t

ψ(s)ds.

Set u = u1 in Ω1 and u = u2 in Ω2. Now we check u, ut, ux, utt, uxx, utx arecontinuous along t = x. By a direct calculation, we have

u1(x, t)|t=x − u2(x, t)|t=x = γ (0) − α(0) = ϕ(0) − α(0),

∂ xu1(x, t)|t=x − ∂ xu2(x, t)|t=x = −ψ(0) + α(0),

∂ 2xu1(x, t)|t=x − ∂ 2xu2(x, t)|t=x = −ϕ(0) + α(0).

Then (6.9) implies

u1 = u2, ∂ xu1 = ∂ xu2, ∂ 2xu1 = ∂ 2xu2 on t = x.

8/16/2019 PDE . QUING



Figure 6.5. The method of reflection.

It is easy to get ∂ tu1 = ∂ tu2 on t = x by u1 = u2 and ∂ xu1 = ∂ xu2 on t = x.Similarly, we get ∂ xtu1 = ∂ xtu2 and ∂ ttu1 = ∂ ttu2 on

t = x

. Therefore, u is C 2

across t = x. Hence, we obtain the following result.

Theorem 6.2. Suppose ϕ ∈ C 2[0, ∞), ψ ∈ C 1[0, ∞), α ∈ C 2[0, ∞) and sat-isfy the compatibility condition (6.9). Then there exists a C 2-solution of (6.8) in

[0, ∞) × [0, ∞).

As for solutions of initial-value problems, we can also derive a priori energyestimates for solutions of (6.8). For any constants 0 < T < t, we use the following

domain for energy estimates

(x, t); 0 < x < t − t, 0 < t < T .

Figure 6.6. A domain of integration.

Now we consider the initial/boundary-value problem discussed in Section 3.3.

For a positive l > 0, assume ϕ

∈C 2[0, l], ψ

∈C 1[0, l] and α, β

∈C 2[0,

∞). Consider

utt − uxx = 0 in [0, l] × (0, ∞),

u(·, 0) = ϕ, ut(·, 0) = ψ on [0, l],

u(0, t) = α(t), u(l, t) = β (t) for t > 0.

(6.10)

The compatibility condition is given by

ϕ(0) = α(0), ψ(0) = α(0), ϕ(0) = α(0),

ϕ(l) = β (0), ψ(l) = β (0), ϕ(l) = β (0).(6.11)

8/16/2019 PDE . QUING



For α = β ≡ 0, we can use the method of extension to construct solutions. We

first extend ϕ(x) = −ϕ(−x) for any x ∈ [0, l] and then extend ϕ(x + 2l) = ϕ(x)

for any x. So ϕ is odd and 2l-periodic in R. We extend ψ similarly. The extendedfunctions ϕ and ψ are C 2 and C 1 on R respectively. Let u be the unique solutionof

utt − uxx = 0 in R×R+,

u(·, 0) = ϕ, ut(·, 0) = ψ in R.

We now prove that u(x, t) is a solution of (6.10) when we restrict x ∈ [0, l], i.e.,

u(0, t) = 0, u(l, t) = 0 for any t > 0.

The proof is similar to that of Theorem 6.2. We use v(x, t) = −u(−x, t) to proveu(0, t) = 0 and w(x, t) = −u(2l − x, t) to prove u(l, t) = 0.

Figure 6.7. Extensions of initial values.

Next we use the method of reflection to construct a solution of (6.10) in the

general case. We divide [0, l] × [0, l] into four regions by t = x and t = −x + l andconstruct u in each region. The process is similar and hence is omitted. Therefore,

Figure 6.8. A method of reflection.

we obtain the following result.

Theorem 6.3. Suppose ϕ ∈ C 2[0, ∞), ψ ∈ C 1[0, ∞), α , β ∈ C 2[0, ∞) and sat-

isfy the compatibility condition (6.11). Then there exists a C 2-solution of (6.10) in [0, ∞) × [0, ∞).

8/16/2019 PDE . QUING



Theorem 6.3 includes Theorem 3.19 in Chapter 2 as a special case.

Now we summarize various problems discussed in this section. We emphasize

that characteristic curves play an important role in studies of the 1-dimensionalwave equation.

First, presentation of problems depends on characteristic curves. Let Ω be apiecewise smooth domain in R2 whose boundary is not characteristic. We intend

to prescribe appropriate boundary values to ensure the well-posedness for the waveequation. To do this, we take an arbitrary point on the boundary and examinecharacteristic curves through this point. We then count how many characteristiccurves enter the domain Ω in the positive t-direction. In this section, we discussedcases where Ω is given by the upper half space R×R+, the first quadrant R+ ×R+

and I ×R+ for a finite integral I . We note that the number of boundary values isthe same as the number of characteristic curves entering the domain in the positivet-direction. In summary, we have

u|t=0 = ϕ, ut|t=0 = ψ for initial-value problems;

u|t=0 = ϕ, ut|t=0 = ψ, u|x=0 = α for half-space problems;

u|t=0 = ϕ, ut|t=0 = ψ, u|x=0 = α, u|x=l = β for initial/boundary-value problems.

Figure 6.9. Characteristic directions.

Second, characteristic curves determine the domain of influence and the do-

main of dependence. In fact, as illustrated by (6.7), initial values propagate alongcharacteristic curves.

Figure 6.10. The domain of influence and the domain of dependence.

Last, characteristic curves also determine domains for energy estimates. Wewill explore this in detail in the final section of this chapter.

8/16/2019 PDE . QUING



6.2. Higher Dimensional Wave Equations

In this section, we discuss initial-value problems for the wave equation in higherdimensions. Let ϕ and ψ be C 2 and C 1 functions in Rn respectively and f be acontinuous function in Rn ×R+. Consider

utt − ∆u = f in Rn × R+,

u(·, 0) = ϕ, ut(·, 0) = ψ in Rn.(6.12)

We intend to derive an expression for C 2-solution u in Rn × [0, ∞). To do this, wedecompose (6.12) to three problems

utt − ∆u = 0 in Rn × R+,

u(·, 0) = ϕ, ut(·, 0) = 0 in Rn,(6.13)

utt − ∆u = 0 in Rn ×R+,

u(·, 0) = 0, ut(·, 0) = ψ in Rn,(6.14)

and

utt − ∆u = f in Rn × R+,

u(·, 0) = 0, ut(·, 0) = 0 in Rn.(6.15)

Obviously, a sum of solutions of (6.13)-(6.15) yields a solution of (6.12).The next result is referred to as the Duhamel’s Principle.

Theorem 6.4. Suppose u2 = M ψ(x, t) is a solution of (6.14). Then

u1(x, t) = ∂ tM ϕ(x, t),

u3(x, t) = t

0

M f τ (x, t

−τ )dτ where f τ = f (

·, τ )

are solutions of (6.13) and (6.15) respectively.

The proof is based on straightforward calculations.

Proof. First, M ϕ(x, t) satisfies

∂ ttM ϕ − ∆M ϕ = 0 in Rn ×R+,

M ϕ|t=0 = 0, ∂ tM ϕ|t=0 = ϕ in Rn.

Then

∂ ttu1 − ∆u1 = (∂ tt − ∆)∂ tM ϕ = ∂ t(∂ ttM ϕ − ∆M ϕ) = 0 in Rn × R+,

u1

|t=0 = ∂ tM ϕ(x, t)

|t=0 = ϕ in Rn,

∂ tu1|t=0 = (∂ ttM ϕ)|t=0 = (∆M ϕ)|t=0 = 0 in Rn.

Next, w(x, t) = M f τ (x, t − τ ) satisfies

wtt − ∆w = 0 in Rn ×R+,

w|t=τ = 0, ∂ tw|t=τ = f (·, τ ) in Rn.

Then

∂ tu3 = M f τ (x, t − τ )|τ =t +

t

0

∂ tM f τ (x, t − τ )dτ =

t

0

∂ tM f τ (x, t − τ )dτ,

8/16/2019 PDE . QUING


6.2. HIGHER DIMENSIONAL WAVE EQUATIONS 143

and

∂ ttu3 = ∂ tM f τ (x, t − τ )|τ =t + t

0 ∂ ttM f τ (x, t − τ )dτ

= f (x, t) +

t

0

∆M f τ (x, t − τ )dτ = f (x, t) + ∆

t

0

M f τ (x, t − τ )dτ

= f (x, t) + ∆u3.

Hence ∂ ttu3 − ∆u3 = f in Rn ×R+ and u3|t=0 = 0, ∂ tu3|t=0 = 0 in Rn.

In the following, we concentrate on (6.14). Suppose u is a C 2-solution of (6.14)in Rn × [0, ∞), i.e.,

utt − ∆u = 0 in Rn ×R+,

u(·, 0) = 0, ut(·, 0) = ψ in Rn.

We solve this initial-value problem by the method of spherical average due to Pois-son. Set for any t > 0 and r > 0

I (r, t) = 1

ωnrn−1

|y−x|=r

u(y, t)dS y,

where ωn is the surface area of the unit sphere in Rn. Obviously, I (r, t) is theaverage of u(·, t) over the sphere ∂Br(x) and also depends on x. First, I (r, t) is

defined for r > 0. By changing it to the form

I (r, t) = 1

ωn

|ω|=1

u(x + rω,t)dS ω,

we note that I (r, t) is defined for all r ∈ (−∞, ∞) and is even for r, since replacingr by

−r does not change the value of the integral. It is clear that u can be recovered

from I , since

I (0, r) = limr→0

I (r, t) = 1

ωn

|ω|=1

u(x, t)dS ω = u(x, t).

Next, we change the equation of u into an equation of I (r, t). We claim thatI (r, t) satisfies the following Euler-Poisson-Darboux equation

I tt = I rr + n − 1

r I r for r > 0, t > 0,

with initial values

I (r, 0) = 0, I t(r, 0) = Ψ(r) for r > 0,

where

Ψ(r) = 1ωn

|ω|=1

ψ(x + rω)dS ω for r > 0.

To verify the equation satisfied by I , we first have by differentiating under theintegral sign

I r(r, t) = 1

ωn

|ω|=1

∂u

∂n(x + rω,t)dS ω =

1

ωnrn−1

|y−x|=r

∂u

∂n(y, t)dS y

= 1

ωnrn−1

|y−x|≤r

∆u(y, t)dy.

8/16/2019 PDE . QUING



Then

rn−1I r = 1

ωn |y−x|≤r

∆u(y, t)dy = 1

ωn |y−x|≤r

utt(y, t)dy.

Hence

(rn−1I r)r = 1

ωn

|y−x|=r

utt(y, t)dS y = 1

ωn∂ tt

|y−x|=r

u(y, t)dS y = rn−1I tt.

For initial values, we simply note for any r > 0

I (r, 0) = 1

ωn

|ω|=1

u(x + rω, 0)dS ω = 0,

I t(r, 0) = 1

ωn

|ω|=1

ψ(x + rω)dS ω.

In general, it is a tedious process to solve initial-value problems for the Euler-

Poisson-Darboux equation for general dimension. However, this process is relativelyeasy for n = 3. If n = 3, we have

I tt = I rr + 2

rI r,

or

(rI )tt = (rI )rr for r > 0, t > 0,

with

(rI )(r, 0) = 0, (rI )t(r, 0) = rΨ(r) for r > 0.

This is a half-space problem for rI (r, t) studied in the previous section for the one-dimensional wave equation. We note (rΨ)|r=0 = 0 and (rI )|r=0 = 0. Hence thecompatibility condition in Theorem 6.2 is satisfied. In order to find a solution rI ,

we extend rΨ(r) to r ∈ (−∞, ∞) as an odd function. (In fact, Ψ(r) as the averageof ψ over ∂ Br(x) can be extended as an even function for r ∈ (−∞, ∞) as before.)Then

(rI )(r, t) = 1

2

r+t

r−t

sΨ(s)ds = 1

2

r+t

t−r

sΨ(s)ds,

where we changed r − t to t − r for the lower integral limit by the oddness of the

integrand. Hence,

I (r, t) = 1

2r

t+r

t−r

sΨ(s)ds.

Note that the area of the unit sphere in R3 is 4π. Then we have

limr→0

I (r, t) = tΨ(t) = t

4π |ω|=1

ψ(x + tω)dS ω = 1

4πt |y−x|=t

ψ(y)dS y.

Therefore, we obtain formally an expression of a solution of (6.14) in R3 ×R+ given

by

(6.16) u(x, t) = 1

4πt

|y−x|=t

ψ(y)dS y.

We note that u(x, t) depends on the initial value in the sphere ∂ Bt(x).

Any C 2-solution u of the initial-value problem (6.14) in R3 × R+ is given bythe formula (6.16), hence is unique. Now we prove it yields the classical solution.

8/16/2019 PDE . QUING



Theorem 6.5. For any ψ ∈ C 2(R3), u in (6.16) is a C 2-solution of (6.14) in R3 ×R+.

Proof. By setting

Ψ(x, t) = 1

4π

|ω|=1

ψ(x + tω)dS ω,

we have by (6.16)

u(x, t) = tΨ(x, t).

Then

ut = Ψ + tΨt, utt = 2Ψt + tΨtt.

We calculate derivatives of Ψ as follows

Ψt = 1

4π

|ω|=1

∂ψ

∂n(x + tω)dS ω =

1

4πt2

|y−x|=t

∂ψ

∂n(y)dS y

= 1

4πt2

|y−x|≤t

∆ψ(y)dy = 1

4πt2

t

0

dρ

|y−x|=ρ

∆ψ(y)dS y ,

Ψtt(t) = − 1

2πt3

|y−x|≤t

∆ψ(y)dy + 1

4πt2

|y−x|=t

∆ψ(y)dS y

= − 2

tΨt(t) +

1

4πt2

|y−x|=t

∆yψ(y)dS y.

By y = x + ωt, we have

utt = t

4πt2

|y−x|=t

∆yψ(y)dS y = t

4π

|ω|=1

∆xψ(x + tω)dS ω = ∆u,

where we used ∆xψ(x + tω) = ∆yψ(y). It is easy to see u(·, 0) = 0 and ut(·, 0) = ψin R3.

Now we consider the initial-value problem (6.12) of the wave equation in R3 as

follows

utt − ∆u = f in R3 ×R+,

u(·, 0) = ϕ, ut(·, 0) = ψ in R3.

By Theorem 6.4, we have

u(x, t) =∂ t

1

4πt

|y−x|=t

ϕ(y)dS y

+

1

4πt

|y−x|=t

ψ(y)dS y

+ 1

4π

|y−x|≤t

f (y, t − |y − x|)|y − x| dy.

(6.17)

8/16/2019 PDE . QUING



In fact, we only need to verify the final term in (6.17). We note

1

4π t

0 (t − τ ) |ω|=1 f (x + ω(t − τ ), τ )dS ωdτ (with t − τ = τ

)

= 1

4π

t

0

τ

|ω|=1

f (x + τ ω , t − τ )dS ωdτ

= 1

4π

t

0

|y−x|=τ

f (y, t − τ )

τ dS ydτ

= 1

4π

|y−x|≤t

f (y, t − |y − x|)|y − x| dy.

Hence we obtain the following result.

Theorem 6.6. For any ϕ ∈ C 3(R3), ψ ∈ C 2(R3) and f ∈ C 2(R3 × R+), the

function u given by (6.17) is a C

2

-solution of (6.12) in R3

×R

+.We can also express the final term in (6.17) as

1

4π

t

0

1

t − τ

|y−x|=t−τ

f (y, τ )dS ydτ.

Hence the value of the solution u at (x, t) depends only on the values of f in points(y, τ ) with

|y − x| = t − τ, 0 < τ < t.

This is exactly the backward characteristic cone with vertex (x, t).We now use the method of descent to solve initial-value problems of the wave

equation in R2 × R+. As for three dimensional case, we first consider (6.14). Let

ψ be C 2 in R2 and u(x, t) be C 2 in R2

×[0,

∞) and satisfy (6.14), i.e.,

utt − ∆u = 0 in R2 × R+,

u(·, 0) = 0, ut(·, 0) = ψ in R2.

Any solutions in R2 can be viewed as solutions of the same problem in R3 whichare independent of the third space variable. Then for any (x, t) ∈ R2 × R+ (withx3 = 0), we have by (6.16)

u(x, t) = 1

4πt

|y−x|=t

ψ(y)dS y,

where y = (y1, y2, y3) = (y, y3). The integral here is a surface integral in R3. Nowwe evaluate it as a 2-dimensional integral by eliminating y3. The sphere |y − x| = tin R3 has two pieces given by

y3 = ± t2 − |y − x|2,

and its surface area element is

dS y =

1 + (∂ y1y3)2 + (∂ y2y3)2 12 dy1dy2 =

t t2 − |y − x|2

dy1dy2.


(6.18) u(x, t) = 1

2π

|y−x|≤t

ψ(y) t2 − |y − x|2

dy.

8/16/2019 PDE . QUING



We note that u(x, t) depends on initial values in the solid disc Bt(x).

Theorem 6.7. For any ψ ∈

C 2(R2), u in (6.18) is a C 2-solution of (6.14) in R2 ×R+.

Next, we discuss briefly how to obtain explicit expressions for solutions of thewave equation in arbitrary dimension. For odd dimension, we seek an appropriate

combination of I (r, t) and its derivatives to satisfy the 1-dimensional wave equationand then proceed as for n = 3. For even dimension, we again use the method of descent.

Let n ≥ 3 be an odd integer. Then the spherical average I (r, t) satisfies

(6.19) I tt = I rr + n − 1

r I r for r > 0, t > 0.

First, we write (6.19) as

I tt = 1

r rI rr + (n − 1)I r.

By

(rI )rr = rI rr + 2I r,

we obtain

I tt = 1

r

(rI )rr + (n − 3)I r

,

or

(6.20) (rI )tt = (rI )rr + (n − 3)I r.

If n = 3, then rI satisfies the 1-dimensional wave equation. This is how we solved

the wave equation in 3-dimension. By differentiating (6.19) with respect to r, wehave

I rtt = I rrr + n − 1r

I rr − n − 1r2

I r = 1r2r2I rrr + (n − 1)rI rr − (n − 1)I r.

By

(r2I r)rr = r2I rrr + 4rI rr + 2I r,

we obtain

I rtt = 1

r2

(r2I )rr + (n − 5)rI rr − (n + 1)I r

,

or

(6.21) (r2I r)tt = (r2I r)rr + (n − 5)rI rr − (n + 1)I r.

The second term in the right-hand side of (6.21) has a coefficient n − 5, which is2 less than the coefficient of the second term in the right-hand side of (6.20). Also

the third term involving I r in the right-hand side of (6.21) has a similar expressionas the second term in the right-hand side of (6.20). Therefore an appropriatecombination of (6.20) and (6.21) eliminates terms involving I r. In particular, forn = 5, we have

(r2I r + 3rI )tt = (r2I r + 3rI )rr .

In other words, r2I r + 3rI satisfies the 1-dimensional wave equation. We cancontinue this process to obtain appropriate forms for all odd dimensions. Next, we

note

r2I r + 3rI = 1

r(r3I )r.

8/16/2019 PDE . QUING



It turns out that the correct combination of I and its derivatives for arbitrary odd

dimension n is given by

J (r, t) =1

r

∂

∂r

n−32

rn−2I (r, t)

.

We leave derivation as an exercise.

We need to point out that the method of spherical average is by no means theonly way to derive explicit expressions of solutions of the wave equation. Othermethods are available. Refer to an exercise for an alternative approach to solvingthree-dimensional wave equation.

Now we compare several formulas we obtained in the previous section and inthis section. Let u be a C 2-solution of the initial-value problem

utt − ∆u = 0 in Rn × R+,

u(·, 0) = ϕ, ut(·, 0) = ψ in Rn.(6.22)

We write un for dimension n. Then

u1(x, t) = 1

2

ϕ(x + t) + ϕ(x − t)

+

1

2

x+t

x−t

ψ(y)dy,

u2(x, t) = ∂ t

1

2π

|y−x|≤t

ϕ(y) t2 − |y − x|2

dy

+

1

2π

|y−x|≤t

ψ(y) t2 − |y − x|2

dy,

u3(x, t) = ∂ t

1

4πt

|y−x|=t

ϕ(y)dS y

+

1

4πt

|y−x|=t

ψ(y)dS y.

These formulas display many important properties of solutions u.

According to these expressions, the value u at (x, t) depends on the values of ϕ and ψ on the interval [x − t, x + t] for n = 1, on the solid disc Bt(x) of centerx and radius t for n = 2, and on the sphere ∂Bt(x) of center x and radius t forn = 3. These regions are the domain of dependence of solutions on initial values.Conversely, initial values ϕ and ψ at a point x0 on the initial hypersurface t = 0

Figure 6.11. The domain of dependence.

influence u in points (x, t) in the solid cone |x −x0| ≤ t for n = 2 and only the cone|x − x0| = t for n = 3 at a later time t.

8/16/2019 PDE . QUING



Figure 6.12. The domain of influence.

The central issue here is that the solution at a given point is determined byinitial values in a proper subset of the initial hypersurface. An important conse-

quence is that the process of solving initial-value problems for the wave equationcan be localized in space. Specifically, changing initial values outside domain of dependence of a point does not change values of solutions there. This is a uniqueproperty of the wave equation which distinguishes it from the heat equation.

Before exploring the difference between n = 2 and n = 3, we first note that

it takes time (literally) for initial values to make influences. Suppose the initialvalues ϕ, ψ have their support contained in a ball Br(x0). Then at a later time t,the support of u(·, t) is contained in the union of all balls Bt(x) for x ∈ Br(x0). Itis easy to see that such a union is in fact the ball with center x0 and radius r + t.

The support of u spreads at a finite speed. To put it in another perspective, we fixan x /∈ Br(x0). Then u(x, t) = 0 for t < |x − x0| − r. This is the so called finite speed propagation .

For n = 2, if the support of ϕ and ψ takes the entire disc Br(x0), then thesupport of u(·, t) will take the entire disc Br+t(x0) in general. The influence frominitial values never disappears in a finite time at any particular point, like thesurface waves arising from a stone dropped into water.

For n = 3, the behavior of solutions is different. Again, we assume the support

of ϕ and ψ are contained in a ball Br(x0). Then at a later time t, the support of u(·, t) is in fact contained in the union of all spheres ∂ Bt(x) for x ∈ Br(x0). Such aunion is a ball Bt+r(x0) for t ≤ r, as in the two-dimensional case, and an annularregion with center x0 and outer and inner radius t + r and t − r respectively for

t > r. Such an annular region has a thickness 2r and spreads at a finite speed. In

other words, u(x, t) is not zero only if

t−

r <|x

−x0

|< t + r,

or

|x − x0| − r < t < |x − x0| + r.

Therefore, for a fixed x ∈ R3, u(x, t) = 0 for t < |x − x0| − r (corresponding to thefinite speed propagation) and for t > |x − x0| + r. So, influence from initial valueslasts only for an interval of length 2r in time. This phenomenon is called Huygens’ principle for the wave equation.

In fact, Huygens’ principle holds for the wave equation in every odd spacedimension n except n = 1 and does not hold in even space dimension.

8/16/2019 PDE . QUING



Figure 6.13. A comparison between n = 2 and n = 3.

Next, we discuss whether solutions decay as t → ∞. In this aspect, there is asharp difference between 1-dimension and higher dimensions. By the d’Alembert’s

formula, it is obvious that solutions to the 1-dimensional homogeneous wave equa-tion do not decay as t → ∞. However, solutions in higher dimensions have a

different behavior.

Theorem 6.8. Let ψ be a smooth function with compact support in Rn and ube a solution of

utt − ∆u = 0 in Rn × R+,

u(·, 0) = 0, ut(·, 0) = ψ on Rn.

Then for any t > 1

|u(·, t)|L∞(Rn) ≤ C √ t

ψL1(Rn) + ∇ψL1(Rn)

for n = 2,

and

|u(·, t)|L∞(Rn) ≤ C

t ∇ψL1(Rn) for n = 3,

where C is a positive constant independent of u and ψ.

We note that decay rates vary according to dimensions. In fact, solutions to the

n-dimensional wave equation have a decay rate of t−n−12 . These decay estimates

play an important role in study of global solutions of nonlinear wave equations.

Proof. We first consider n = 3. By (6.16), the solution u is given by

u(x, t) = t

4π

|ω|=1

ψ(x + tω)dS ω.

Since ψ has compact support, we have

ψ(x + tω) = − ∞

t

∂

∂sψ(x + sω)ds.

Then

u(x, t) = − t

4π

∞t

|ω|=1

∂

∂sψ(x + sω)dS ωds.

For s ≥ t, we write t ≤ s2/t and hence

|u(x, t)| ≤ 1

4πt

∞t

s2

|ω|=1

|∇ψ(x + sω)|dS ωds ≤ 1

4πt∇ψL1(R3).

8/16/2019 PDE . QUING



This implies the desired result for n = 3. In fact, it holds for any t > 0.

Now we consider n = 2. By (6.18) and a change of variables, we have

u(x, t) = 1

2π

t

0

r√ t2 − r2

|ω|=1

ψ(x + rω)dS ωdr.

For some ε > 0 to be determined, we write u as

u(x, t) = 1

2π

t−ε

0

+

t

t−ε

=

1

2π

I 1 + I 2

.

For I 1, we have

|I 1| = t−ε

0

r√ t2 − r2

|ω|=1

ψ(x + rω)dS ωdr

≤ 1 t2 − (t − ε)2

t−ε

0 r |ω|=1 |ψ(x + rω)|dS ωdr

≤ 1√ 2εt − ε2

ψL1(R2).

Next, for r ∈ (t − ε, t), as in the proof for n = 3, we have |ω|=1

ψ(x + rω)dS ω = − ∞

r

|ω|=1

∂

∂sψ(x + sω)dS ωds,

and hencer

|ω|=1

ψ(x + rω)dS ω≤ ∞

r

s

|ω|=1

|∇ψ(x + sω)|dS ωds ≤ ∇ψL1(R2).

Then

|I 2| = t

t−ε

r√ t2 − r2

|ω|=1

ψ(x + rω)dS ωdr

≤ t

t−ε

1√ t2 − r2

dr · supr∈[t−ε,t]

r

|ω|=1

ψ(x + rω)dS ω

≤ t

t−ε

1√ t2 − r2

dr · ∇ψL1(R2).

A simple calculation yields

t

t−ε

1

√ t2

− r2

dr = t

t−ε

1

(t + r)(t − r)

dr

≤ 1

√ t t

t−ε

1

√ t − r

dr = 2

√ ε

√ t.


|u(x, t)| ≤ 1

2π

1√ 2εt − ε2

ψL1(R2) + 2

√ ε√ t

∇ψL1(R2)

.

For any t > 1, we take ε = 1/2 and obtain the desired result.

Decay estimates in Theorem 6.8 are optimal for large t. We have the followingresult for arbitrary t.

8/16/2019 PDE . QUING



Theorem 6.9. Let ψ be a smooth function with compact support in Rn and ube a solution of

utt − ∆u = 0 in Rn × R+,

u(·, 0) = 0, ut(·, 0) = ψ on Rn.

Then for any t > 0

|u(·, t)|L∞(Rn) ≤ C ∇ψL1(Rn) for n = 2,

and

|u(·, t)|L∞(Rn) ≤ C ∇2ψL1(Rn) for n = 3,

where C is a positive constant independent of u and ψ.

The proof is left as an exercise.

Now we compare regularity of solutions for n = 1 and n = 3. For n = 1, the

regularity of u is clearly the same as u(·, 0) and one order better than ut(·, 0). Inother words, u ∈ C m and ut ∈ C m−1 initially at t = 0 guarantee u ∈ C m at a later

time. However, such a result does not hold for n = 3. By writing spherical averagesas integrals over the unit sphere |ω| = 1 and differentiating under integral sign, wehave

u3(x, t) = 1

4πt2

|y−x|=t

ϕ(y) +

3i=1

ϕyi(y)(yi − xi) + tψ(y)

dS y.

This formula indicates that u can be less regular than initial values. There is apossible loss of one order of differentiability. Namely, u ∈ C m and ut ∈ C m−1

initially at t = 0 only guarantee u ∈ C m−1 at a later time.Next, we construct a solution of the 3-dimensional homogeneous wave equation

with a loss of differentiation described above. Let u = u(r, t) be a sphericallysymmetric solution of the wave equation in R3 × R+. Then

utt − urr − 2

rur = 0,

or

(ru)tt − (ru)rr = 0.

In other words, ru(r, t) is a solution of the 1-dimensional wave equation. Hence

u(r, t) = 1

r (f (t + r) + g(t − r)) for any r = 0,

for some functions f and g on R. In the following, we consider f = g. For anypositive integer m, we define

f (r) = 0 for r ≤ 1,(r − 1)2m+1 for 1 < r < 2.

We extend f smoothly for r > 1. We note that f is C 2m in R and not C 2m+1 atr = 1. Then for t = 0,

u(r, 0) = 1

r (f (r) + f (−r)) , ut(r, 0) =

1

r (f (r) + f (−r)) .

Hence u(·, 0) ∈ C 2m(R), ut(·, 0) ∈ C 2m−1(R), and u(r, 0) is not C 2m+1 and ut(r, 0)is not C 2m at r = 1. We now claim that u(r, t) is not C 2m at (r, t) = (0, 1). The

8/16/2019 PDE . QUING



intuitive idea is that the singularity of initial values at r = 1 propagates along a

characteristic hypersurface and focuses at t = 1. We note that (r, t) = (0, 1) is the

vertex of the characteristic hypersurface (x, t); t < |x| which intersects t = 0 onr = 1. First, u(r, 0) = ut(r, 0) = 0 for r ≤ 1. Then u(x, t) = 0 for 0 < t ≤ |x|. Inparticular,

limt→1−

∂ 2mt u(0, t) = 0.

If u were C 2m at (r, t) = (0, 1), then ∆mu = ∂ 2mt u = 0 at (0, 1). Next for r > 0

sufficiently small, we have 1 + r > 1 and 1 − r < 1, and hence

u(r, 1) = 1

rf (1 + r) = r2m = |x|2m.

Therefore for r > 0 sufficiently small,

∆mx u(r, 1) = ∂ rr +

2

r

∂ rm

r2m = (2m + 1)!.

We conclude that u is not C 2m at (r, t) = (0, 1).This example demonstrates that solutions of the higher dimensional wave equa-

tion do not have good point-wise behavior. However, the differentiability is pre-served in the L2-sense. Before we move on to energy estimates in the next section,

we show by a simple case what is involved.Suppose u is a C 2-solution of (6.22) for general n. We assume that ϕ and ψ have

compact support. By finite speed propagation, u(·, t) also has compact support forany t > 0. We multiply the equation in (6.22) by ut and integrate in BR × (0, t).

Here we choose R sufficiently large such that the support of u(·, s) is contained inBR for any s ∈ (0, t). Note

ututt − ut∆u = 12 (u2t + |∇xu|2)t −n

i=1(utuxi )xi .

Then a simple integration in BR × (0, t) yields

1

2

Rn×t

(u2t + |∇xu|2) =

1

2

Rn×0

(u2t + |∇xu|2).

This is the conservation of energy: the L2-norm of derivatives at each time slice isconstant, independent of time.

6.3. Energy Estimates

In this section, we derive energy estimates of solutions of initial-value problems

for a slightly more general hyperbolic equations instead of just wave equations. Fixa positive number T . Let a, c and f be continuous functions in Rn × [0, T ] andϕ and ψ be continuous functions in Rn. We consider the following initial-valueproblem

utt − a∆u + cu = f in Rn × [0, T ],

u(·, 0) = ϕ, ut(·, 0) = ψ in Rn.(6.23)

We assume a is a positive function satisfying

(6.24) λ ≤ a(x, t) ≤ Λ for any (x, t) ∈ Rn × [0, T ] and ξ ∈ Rn,

8/16/2019 PDE . QUING



for some positive constants λ and Λ. If the principal part of the equation is given

by the wave operator utt − ∆u, then a = 1 and we can choose λ = Λ = 1 in (6.24).

With κ = 1/√ Λ, we introduce for fixed t > T

Dκ,T,t = (x, t); κ|x| < t − t, 0 < t < T .

We denote by ∂ −Dκ,T,t, ∂ sDκ,T,t and ∂ +Dκ,T,t the bottom, the side and the top of the boundary. See discussions in Section 2.3 for details.

Figure 6.14. An integral domain.

Theorem 6.10. Assume a is C 1, c and f are continuous in Rn × [0, T ], ϕ is

C 1 in Rn and ψ is continuous in Rn. Let (6.24) be assumed and u be a C 2-solution of (6.23). Then for any η ≥ η0

∂ +Dκ,T,t

e−ηt(u2 + u2t + a|∇u|2) + (η − η0)

Dκ,T,t

e−ηt(u2 + u2t + a|∇u|2)

≤

∂ −Dκ,T,t

(ϕ2 + ψ2 + a|∇ϕ|2) +

Dκ,T,t

e−ηtf 2,

where η0 is a positive constant depending only on λ, the C 1-norm of a and the L∞-norm of c in Dκ,T,t.

Proof. In the following, we simply set D = Dκ,T,t. We multiply the equation

in (6.23) by 2e−ηtut and integrate in D , for a scalar η to be determined. First, wenote

2e

−ηt

ututt = e

−ηt

(u

2

t )t + ηe

−ηt

u

2

t ,and

−2e−ηtaut∆u =n

i=1

− 2(e−ηtautuxi)xi

+ 2e−ηtauxiutxi

+ 2e−ηtaxiutuxi

=

ni=1

− 2(e−ηtautuxi)xi

+ (e−ηtau2xi

)t + 2e−ηtaxiutuxi

+ ηe−ηtau2xi

− e−ηtatu2xi

,

8/16/2019 PDE . QUING



where we used 2uxiutxi

= (u2xi

)t. Therefore, we obtain

(e−ηtu2t + e−ηta|∇u|2)t − 2

ni=1

(e−ηtutuxi )xi + ηe−ηt(u2t + a|∇u|2)

+n

i=1

2e−ηtaxiutuxi

− e−ηtat|∇u|2 + 2e−ηtcuut = 2e−ηtutf.

To control the final term in the left hand side involving uut, we note

(e−ηtu2)t + ηe−ηtu2 − 2e−ηtuut = 0.

Then a simple addition yields

e−ηt(u2 + u2

t + a|∇u|2)

t−

ni=1

2(e−ηtautuxi)xi

+ ηe−ηt(u2 + u2t + a|∇u|2)

= −2e−ηt ni=1

axiutuxi

+ e−ηtat|∇u|2 − 2e−ηt(c − 1)uut + 2e−ηtutf.

The first three terms in the right hand side is quadratic in ut, uxi and u. Then by

(6.24) and the Cauchy inequality applied to each term in the right hand side, wehave

e−ηt(u2 + u2t + a|∇u|2)

t− 2

ni=1

(e−ηtautuxi)xi

+ (η − η0)e−ηt(u2 + u2t + a|∇u|2) ≤ e−ηtf 2,

where η0 is a positive constant depending only on λ, the C 1-norm of a and theL∞-norm of c. By integrating over D , we obtain

∂ +D

e−ηt(u2 + u2t + a|∇u|2) + (η − η0)

D

e−ηt(u2 + u2t + a|∇u|2)

+

∂ sD

e−ηt

(u2 + u2t + a|∇u|2)γ t − 2

ni=1

autuxiγ i

≤

∂ −D

(u2 + u2t + a|∇u|2) +

D

e−ηtf 2,

where γ = (γ 1, · · · , γ n, γ t) is the unit exterior normal vector along ∂ sD. We onlyneed to prove that the integrand for ∂ sD is nonnegative. We claim

(u2

t

+ a|∇

u|2)γ t

−2

n

i=1

autuxiγ i

≥0 on ∂ sD.

To prove this, we first note by the Cauchy inequality

|n

i=1

uxiγ i| ≤ n

i=1

u2xi

)12 · n

i=1

γ 2i 12 = |∇u|

1 − γ 2t .

By the explicit expression of γ (see Section 2.3), we have

γ t = 1√ 1 + κ2

,

8/16/2019 PDE . QUING



and hence with (6.24) and κ = 1/√

Λ

(u2t + a|∇u|

2

)γ t − 2

ni=1

autuxi γ i ≥ 1

√ 1 + κ2 u2t + a|∇u|

2

− 2κa|ut| · |∇u|≥ 1√

1 + κ2

u2

t + a|∇u|2 − 2√

a|ut| · |∇u|,

which is nonnegative. Therefore, the boundary integral on ∂ sD is nonnegative.

A consequence of Theorem 6.10 is the uniqueness of solutions of (6.23). Wecan also discuss domain of dependence and domain of influence as in the previoussection.

Next, we consider (6.23) in a general domain

D = (x, t); h−(x) < t < h+(x), x ∈ Ω,

where Ω is a bounded domain in Rn and h− and h+ are two piecewise C 1-functions

in Ω with h− = h+ on ∂ Ω. We now perform a similar integration in D as in the

Figure 6.15. A general domain.

above proof to get ∂ +D

e−ηt

(u2 + u2t + a|∇u|2)γ +t − 2

ni=1

autuxiγ +i

+ (η − η0)

D

e−ηt(u2 + u2t + a|∇u|2)

≤

∂ −D

e−ηt

(u2 + u2t + a|∇u|2)γ −t − 2

ni=1

autuxiγ −i

+

D

e−ηtf 2,

where γ ± = (γ ±1, · · · , γ ±n, γ ±t) are unit normal vector pointing to positive t-

direction along ∂ ±D. We are interested in whether the integrand for ∂ +D is non-negative. As in the proof above, we have by the Cauchy inequality

|n

i=1

uxiγ +i| ≤ n

i=1

u2xi

)12 · n

i=1

γ 2+i

12 = |∇u|

1 − γ 2+t.

Then it is easy to see

(u2t + a|∇u|2)γ +t − 2

ni=1

autuxiγ +i

≥(u2t + a|∇u|2)γ +t − 2

a(1 − γ 2+t) · √

a|∇u| ≥ 0 on ∂ +D

8/16/2019 PDE . QUING



if

γ +t

≥ a(1

−γ 2+t),

or

(6.25) γ +t ≥

a

1 + a on ∂ +D.

With (6.25), we obtain ∂ +D

e−ηtu2γ +t + (η − η0)

D

e−ηt(u2 + u2t + a|∇u|2)

≤

∂ −D

e−ηt

(u2 + u2t + a|∇u|2)γ −t − 2

ni=1

autuxiγ −i

+

D

e−ηtf 2.

In particular, if u = ut = 0 on ∂ −D and f = 0 in D, then u = 0 in D .

Now we introduce the notion of space-like and time-like surfaces.

Definition 6.11. Let Σ be a C 1-hypersurface in Rn ×R+ and γ = (γ x, γ t) be

a unit normal vector with γ t ≥ 0. Then Σ is space-like at (x, t) if

γ t(x, t) >

a(x, t)

1 + a(x, t);

Σ is time-like at (x, t) if

γ t(x, t) <

a(x, t)

1 + a(x, t).

If the hypersurface Σ is given by t = t(x), it is easy to check that Σ is space-likeat (x, t(x)) if

|∇t(x)| < 1 a(x, t(x)) .

For the wave equation

(6.26) utt − ∆u = f,

we have a = 1. Then the hypersurface Σ is space-like at (x, t) if γ t(x, t) > 1/√

2. If

Σ is given by t = t(x), then Σ is space-like at (x, t(x)) if

|∇t(x)| < 1.

In the following, we use the wave equation to discuss the importance of space-likesurfaces.

Let Σ be a space-like surface. Then for any (x0, t0) ∈ Σ, the domain of influenceof (x0, t0) is given by the cone

(x, t); t

−t0 >

|x

−x0

| and hence is always above

Σ. It follows that prescribing initial values on space-like surfaces yields a well-posed problem. In fact, integral domains for energy estimates can be constructed

accordingly. However, we cannot prescribe initial values on time-like hypersurfacesin general. This can be explained as follows. Let Σ be a time-like hypersurface.For some point p on Σ, the domain of influence of p includes other points on Σ.Values of solutions at those points, say q , depend on those at p. Hence, initialvalues cannot be prescribed arbitrarily at points like q .

Of course, the above illustration is valid only for n > 1. The one-dimensionalwave equation utt − uxx = 0 has a symmetric form with respect to x and t, and

8/16/2019 PDE . QUING



Figure 6.16. Space-like and time-like surfaces.

Figure 6.17. An integral domain for space-like initial hypersurfaces.

we can always turn around the influence domain if we are allowed to exchange therole of space and time. We emphasize that this works only for initial-value prob-

Figure 6.18. An illustration for n = 1.

lems. For initial/boundary-value problems for the one-dimensional wave equation

we discussed in Section 6.1, we do not have freedom to exchange the role of spaceand time. The space and time variables are already fixed. The initial curve is notnecessarily given by t = 0. It can be any space-like curve, i.e., any curve t = t(x)with |t(x)| < 1. The two vertical lines are time-like. We point out that we did notprescribe initial values there. (Recall that initial values consist of two conditions

for a second-order PDE.) Instead, we prescribed Dirichlet boundary values u = 0there.

8/16/2019 PDE . QUING



Figure 6.19. Initial/boundary-value problems.

To end our discussion of space-like and time-like surfaces, we consider an ex-

ample of initial-value problems with initial values prescribed on a time-like hyper-surface. Consider

utt = uxx + uyy for x > 0, y , t ∈ R,

u = 1m2 sin my, ∂u∂n = 1m sin my on x = 0.

Here we treat x = 0 as the initial hypersurface. We note that x = 0 is time-like.A solution is given by

um(x, y) = 1

m2emx sin my.

Note

um → 0 on x = 0, ∂um

∂n → 0 on x = 0 as m → ∞.

Meanwhile,

supR2

|um(x, ·)| = 1

m2emx → ∞ as m → ∞ for any x > 0.

Therefore, there is no continuous dependence on initial values.Now we return to Theorem 6.10. As in Theorem 2.12 and Corollary 2.13, we

note that t enters the estimate only through the domain Dκ,T,t. Hence, we let

t → ∞ and obtain the following result.

Theorem 6.12. Assume a is C 1, c and f are continuous in Rn × [0, T ], ϕ is

C 1 in Rn and ψ is continuous in Rn. Let (6.24) be assumed and u be a C 2-solution of (6.23). If f ∈ L2(Rn × (0, T )) and ϕ, ∇xϕ, ψ ∈ L2(Rn), then for any η ≥ η0

Rn×T

e−ηt(u2 + u2t + a|∇u|2) + (η − η0)

Rn×(0,T )

e−ηt(u2 + u2t + a|∇u|2)

≤ Rn×0

(ϕ2 + ψ2 + a|∇

ϕ|2) + Rn×(0,T )

e−ηtf 2,

where η0 is a positive constant depending only on λ, the C 1-norm of a and the L∞-norm of c in Rn × [0, T ].

With Theorem 6.12, we can prove the existence of weak solutions of (6.23) bya similar process used in the proof of Theorem 2.15. However, there is a significantdifference. The weak solutions in Theorem 2.15 are in L2 because estimates of

the L2-norms of solutions are established in Corollary 2.13. In the present situa-tion, Theorem 6.12 establishes an estimate of the L2-norms of solutions and their

8/16/2019 PDE . QUING



derivatives . This naturally leads to a new norm defined by

uH 1(Rn×(0,T )) = Rn×(0,T )

(u2 + u2t + |∇xu|2) 1

2

.

The superscript 1 in H 1 indicates the order of derivatives. With such a norm, we

can define the Sobolev space H 1(Rn ×(0, T )) as the completion of smooth functionsof finite H 1-norms with respect to the H 1-norm. Obviously, H 1(Rn×(0, T )) definedin this way is complete. In fact, it is a Hilbert space, since the H 1-norm is naturallyinduced by an H 1-inner product given by

(u, v)H 1(Rn×(0,T )) =

Rn×(0,T )

(uv + utvt + ∇xu · ∇xv).

Then proceeding as in the proof of Theorem 2.15, we can prove (6.23) admits a

weak H 1-solution in Rn × (0, T ) if ϕ = ψ = 0. We will not provide details here.In fact, we did not define the notion of weak H 1-solutions. The purpose of theshort discussion here, similar as that at the end of Section 3.3, is to demonstratethe importance of Sobolev spaces in PDEs. We will study Sobolev spaces in detailin the future.

Exercises

(1) Let l be a positive number and consider

utt − uxx = 0 in (0, l) × (0, ∞),

u(·, 0) = ϕ, ut(·, 0) = ψ in [0, l],

u(0, t) = 0, ux(l, t) = 0 for t > 0.

Find a compatibility condition and prove the existence of a C 2-solution

under such a condition.

(2) Let ϕ1 and ϕ2 be functions defined in x < 0 and x > 0 respectively.Consider the characteristic initial-value problem

utt − uxx = 0 for t > |x|,u(x, −x) = ϕ1(x) for x < 0,

u(x, x) = ϕ2(x) for x > 0.

Solve and find the domain of dependence for any point ( x, t) with t >

|x

|.

(3) Let ϕ1 and ϕ2 be functions defined in x > 0. Consider the Goursatproblem

utt − uxx = 0 for 0 < t < x,

u(x, 0) = ϕ1(x), u(x, x) = ϕ2(x) for x > 0.

Solve and find the domain of dependence for any point (x, t) with 0 < t <x.

8/16/2019 PDE . QUING


EXERCISES 161

(4) Let α be a constant and ϕ and ψ be C 2-functions on R+ which vanish

near x = 0. Consider

utt − uxx = 0 for x > 0, t > 0,

u(x, 0) = ϕ(x), ut(x, 0) = ψ(x) for x > 0,

ut(0, t) = αux(0, t) for t > 0.

Find a solution for α = −1 and prove that there exist no solutions ingeneral for α = −1.

(5) Let λ be a positive constant. Use the method of descent to solve the

following initial-value problems for n = 2

utt = ∆u + λ2u in R2 ×R+,

u(·, 0) = 0, ut(·, 0) = ψ in R2,

andutt = ∆u − λ2u in R2 ×R+,

u(·, 0) = 0, ut(·, 0) = ψ in R2.

Hint: You need to use complex functions temporarily to solve the secondproblem.

(6) Let u be a solution of the following initial-value problem

utt − ∆u = 0 in Rn ×R+,

u(·, 0) = 0, ut(·, 0) = ψ on Rn.

(a) Let n ≥ 3 be odd and I (r, t) be the spherical average of u at (x, t).Set

J (r, t) =1

r

∂

∂r

n−32 rn−2I (r, t)

.

Prove J (r, t) satisfies the 1-dimensional wave equation, i.e., J tt −J rr = 0.

(b) Let n ≥ 3 be odd. Derive a formula for u(x, t).(c) Let n ≥ 2 be even. Derive a formula for u(x, t) by the method of

descent.

(7) Let ψ be a smooth function with compact support in Rn and u be asolution of

utt − ∆u = 0 in Rn ×R+,

u(·, 0) = 0, ut(·, 0) = ψ on Rn

.Prove for any t > 1

|u(·, t)|L∞(Rn) ≤ Ct−n−12

n−22

i=0

∇iψL1(Rn) for even n ≥ 2,

and

|u(·, t)|L∞(Rn) ≤ Ct−n−12 ∇ n−1

2 ψL1(Rn) for odd n ≥ 3,


8/16/2019 PDE . QUING



(8) Let ψ be a smooth function with compact support in Rn and u be a

solution of

utt − ∆u = 0 in Rn ×R+,

u(·, 0) = 0, ut(·, 0) = ψ on Rn.

Prove for any n ≥ 2 and for any t > 0

|u(·, t)|L∞(Rn) ≤ C ∇n−1ψL1(Rn),


(9) Let u be a solution of the following initial-value problem

utt − ∆u = 0 in R3 × R+,

u(·, 0) = ϕ, ut(·, 0) = ψ on R3.

(a) For any fixed (x0, t0) ∈ R3 ×R+, set for any x ∈ Bt0(x0) \ x0

v(x) =

∇xu(x, t)

|x − x0| + x − x0

|x − x0|3u(x, t) +

x − x0

|x − x0|2ut(x, t)

t=t0−|x−x0|

.

Prove div v = 0.(b) Derive an expression of u(x0, t0) in terms of ϕ and ψ by integrating

divv in Bt0(x0) \ Bε(x0) and then letting ε → 0.Remark: This exercise gives an alternative approach to solving the 3-

dimensional wave equation.

(10) Let a be a positive constant and u be a C 2-solution of the following char-acteristic initial-value problem

utt − ∆u = 0 in (x, t) ∈ R3 × R+; t > |x| > a,

u(x, |x|) = 0 for any |x| > a.

(a) For any fixed (x0, t0) ∈ R3 × R+ with t0 > |x0| > a, integrate divv

(introduced in the previous problem) in a region bounded by |x −x0| + |x| = t0, |x| = a and |x − x0| = ε. By letting ε → 0, expressu(x0, t0) by an integral over ∂ Ba.

(b) For any ω ∈ S2 and τ > 0, prove the limit

limr→∞

ru(rω,r + τ )

exists and the convergence is uniform for ω ∈ S2 and τ ∈ (0, τ 0] for

any fixed τ 0 > 0.Remark: The limit in (b) is called the radiation field .

(11) Set QT = (x, t); 0 < x < l, 0 < t < T . Consider the equation

Lu ≡ 2utt + 3utx + uxx = 0.

(a) Give a correct presentation of the boundary-value problem in QT ;(b) Find an explicit expression of a solution with prescribed boundary

values;(c) Derive an estimate of the integral of u2

x + u2t in QT .

8/16/2019 PDE . QUING


EXERCISES 163

Hint: For (b), divide QT into three regions separated by characteristic

curves from (0, 0). For (c), integrate an appropriate linear combination of

utLu and uxLu to make integrands on [0, l] × t and l × [0, t] positivedefinite.

(12) For some constant a > 0, consider the following characteristic initial-value

problem for the wave equation

utt − ∆u =f (x, t) in a < |x| < t + a,

u =ϕ(x, t) on |x| > a, t = |x| − a,

u =ψ(x, t) on |x| = a, t > 0,

where f is a C 1-function in a < |x| < t + a, ϕ is a C 1-function on

r0 < |x| = t − a and ψ is a C 1-function on |x| = a and t > 0. Derive anenergy estimate in an appropriate domain in a < |x| < t + a.

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


CHAPTER 7

First-Order Differential Systems

In this chapter, we discuss partial differential systems of first-order and focuson non-characteristic initial-value problems. In Section 7.1, we introduce non-

characteristic hypersurfaces for partial differential equations and systems of ar-bitrary order. We also demonstrate that partial differential systems of arbitraryorder can always be changed to those of first-order. In Section 7.2, we discuss the

Cauchy-Kowalevski Theorem, which asserts the existence of analytic solutions of non-characteristic initial-value problems for analytic differential systems and initialvalues on analytic non-characteristic hypersurfaces. In Section 7.3, we discuss hy-perbolic differential systems and derive energy estimates for symmetric hyperbolicdifferential systems.

7.1. Reductions to First-Order Differential Systems

The main focus in this section is linear partial differential systems. We startwith linear PDEs of arbitrary order and proceed similarly as in Section 2.1 and

Section 3.1.Let Ω be a domain in Rn containing the origin and L be an m-th order linear

differential operator given by

Lu =|α|≤m

aα(x)∂ αu in Ω,

where aα is continuous in Ω, for any α ∈ Zn+. Here aα is called the coefficient for

∂ αu.

Definition 7.1. The principal part L0 and the principal symbol p of L are

defined by

L0u =|α|=m

aα(x)∂ αu in Ω,

and

p(x; ξ ) = |α|=m

aα(x)ξ α for any x ∈ Ω and ξ ∈ Rn.

The principal symbol p plays an important role in prescribing appropriate con-ditions associated with L. We usually write first-order and second-order linear

differential operators in following forms respectively

Lu =n

i=1

ai(x)uxi + b(x)u,

165

8/16/2019 PDE . QUING


166 7. FIRST-ORDER DIFFERENTIAL SYSTEMS

and

Lu =n

i,j=1

aij (x)uxixj +

n

i=1

bi(x)uxi + c(x)u.

Their principal symbols are given by

p(x; ξ ) =n

i=1

ai(x)ξ i,

and

p(x; ξ ) =n

i,j=1

aij (x)ξ iξ j .

For second-order differential operators, we usually assume that (aij ) is a symmetric

matrix in Ω.For a given function f in Ω, we consider the equation

(7.1) Lu = f (x) in Ω.

The function f is called the non-homogeneous term .Let Σ be the hyperplane xn = 0. We now prescribe values of u and its

derivatives on Σ so that we can at least find all derivatives of u at the origin.Let u0, u1, · · · , um−1 be functions defined in a neighborhood of the origin in Rn−1.First, we prescribe

u(x, 0) = u0(x) for any small x ∈ Rn−1.

Then we can find all x-derivatives of u at the origin. Similarly, for any j =

1, · · · , m − 1, we prescribe

∂ jxnu(x, 0) = uj (x) for any small x ∈ Rn−1.

Then we can find all x-derivatives of ∂ jxn u at the origin for j = 1, · · · , m−1. Hencewith the help of u0(x), u1(x), · · · , um−1(x) on Σ, we can determine all derivativesof u of order up to m at the origin except ∂ mxn

u. To find this, we need to use theequation. If we assume

a(0,··· ,0,m)(0) = 0,

then we can find ∂ mxnu(0) from (7.1). In this case, we can compute all derivatives

of u of any order at the origin by using values u0, u1, · · · , um−1 and differentiating(7.1).

Now we summarize these conditions on Σ = xn = 0 by writing

(7.2) ∂ jxnu(x, 0) = uj(x) for any small x ∈ Rn−1 and j = 0, 1, · · · , m − 1.

We usually call Σ the initial hypersurface and u0,

· · · , um−1 initial values or Cauchy

values. The problem of solving (7.1) together with (7.2) is called the initial valueproblem or the Cauchy problem.

More generally, consider the hypersurface Σ given by ϕ = 0 for a C m-functionϕ in a neighborhood of the origin with ∇ϕ = 0, with the origin on the hypersurfaceΣ, i.e., ϕ(0) = 0. We note that ∇ϕ is simply a normal vector of the hypersurface Σ.Without loss of generality, we assume ϕxn


theorem, we solve ϕ = 0 around x = 0 for xn = ψ(x1, · · · , xn−1). Consider a changeof variables

x → y = (x1, · · · , xn−1, ϕ(x)).

8/16/2019 PDE . QUING


7.1. REDUCTIONS TO FIRST-ORDER DIFFERENTIAL SYSTEMS 167


Jacobian. Now we write the operator L in new variables y as

Lu = |α|≤m

aα(y)∂ αy u.

The initial hypersurface Σ is given by yn = 0 in new coordinates. We are inter-ested in the coefficient a(0,··· ,0,m) of ∂ myn

u. Note

uxi =

nk=1

yk,xiuyk

= ϕxiuyn

+ terms not involving uyn,

and hence

∂ αx u = ∂ α1x1 · · · ∂ αn

xnu = ϕα1

x1 · · · ϕαnxn

∂ mynu + terms not involving ∂ myn

u.

Therefore,

Lu = |α|≤m

aα(x)∂ αx u = |α|=m

aαx(y)ϕα1x1 · · · ϕαn

xn∂ myn

u

+ terms not involving ∂ mynu.

The coefficient of ∂ mynu is given by

|α|=m

aα

x(y)

ϕα1

x1 · · · ϕαn

xn.

Back to x-variables, we note that this is simply

p

x; ∇ϕ(x)

,

where p is the principal symbol of the operator L.

Definition 7.2. For a linear operator L as in (7.1) defined in a neighborhoodof x0 ∈ Rn, a C 1-hypersurface Σ passing x0 is non-characteristic at x0 if

(7.3) p(x0; ξ ) =|α|=m

aα(x0)ξ α = 0,

where ξ is a normal vector of Σ at x0. Otherwise, it is characteristic at x0. A

hypersurface is non-characteristic if it is non-characteristic at every point.

When the hypersurface Σ is given by ϕ = 0 with ∇ϕ = 0, its normal vector

is given by ∇ϕ = (ϕx1 , · · · , ϕxn). Hence we may take ξ = ∇ϕ(x0) in (7.3). We

note that the condition (7.3) is maintained under C m-changes of local coordinates.By this condition, we can find successively values of all derivatives of u at x0, as

far as they exist. Then, we could write formal power series at x0 for solutionsof initial-value problems. It would be actual representations of solutions u in aneighborhood of x0, if u were known to be analytic. This process can be carriedout for analytic initial values and analytic coefficients and nonhomogeneous termsand the result is referred to as the Cauchy-Kowalevski theorem. We will discuss it

in Section 7.2.Now we introduce a special class of linear differential operators.

Definition 7.3. The linear differential operator L in (7.1) is elliptic at x0 if p(x0; ξ ) = 0 for any ξ ∈ Rn \ 0, where p is the principal symbol of L.

8/16/2019 PDE . QUING



Hence, linear differential equations are elliptic if there are no characteristic

hypersurfaces. In other words, every hypersurface is non-characteristic.

For example, consider a first-order linear differential operator of the form

Lu =

ni=1

ai(x)uxi + b(x)u in Ω ⊂ Rn.

Its principal symbol is given by

p(x; ξ ) =n

i=1

ai(x)ξ i for any x ∈ Ω and any ξ ∈ Rn \ 0.

Hence first-order linear differential equations with real coefficients are never elliptic.

Complex coefficients may yield elliptic equations. For example, we take a1 = 1/2and a2 = 1/2i in R2. Then ∂ z = (∂ x + i∂ y)/2 is elliptic.

For second-order linear differential operators, we write

Lu =n

i,j=1

aij (x)uxixj +

ni=1

bi(x)uxi + c(x)u in Ω ⊂ Rn.

Its principal symbol is given by

p(x; ξ ) =

ni,j=1

aij (x)ξ iξ j for any x ∈ Ω and any ξ ∈ Rn \ 0.

Then L is elliptic at x ∈ Ω if

n

i,j=1

aij (x)ξ iξ j = 0 for any ξ ∈ Rn \ 0.

If (aij (x)) is a real-valued n × n symmetric matrix, L is elliptic at x if (aij (x)) is a

definite matrix at x, positive definite or negative definite.

The concept of non-characteristics can also be generalized to linear partialdifferential systems. Let Ω ⊂ Rn be a domain. Consider a linear partial differential

systems of N equations for N unknowns as follows

(7.4) Lu ≡|α|≤m

Aα(x)∂ αu = f in Ω,

where Aα are N × N matrices and u and f are N -column vectors.

We define principle parts, principle symbols and non-characteristic hypersur-faces similarly as for single differential equations.

Definition 7.4. The principal part L0 and the principal symbol p of L aredefined by

L0u =|α|=m

Aα(x)∂ αu in Ω,

and

p(x; ξ ) = det |α|=m

Aα(x)ξ α

for any x ∈ Ω and ξ ∈ Rn.

8/16/2019 PDE . QUING


7.1. REDUCTIONS TO FIRST-ORDER DIFFERENTIAL SYSTEMS 169

Definition 7.5. For a linear operator L as in (7.4) defined in a neighborhood

of x0 ∈ Rn, a C 1-hypersurface Σ passing x0 is non-characteristic at x0 if

p(x0; ξ ) = det |α|=m

Aα(x0)ξ α = 0,

where ξ is a normal vector of Σ at x0. Otherwise, it is characteristic at x0. Ahypersurface is non-characteristic if it is non-characteristic at every point.

Let Σ be a non-characteristic hypersurface. We prescribe initial values on Σ asfollows. Let ν be a normal vector on Σ. We prescribe

(7.5) ∂ j u

∂ν j = uj on Σ for j = 0, 1, · · · , m − 1,

where u0, u1, · · · , um−1 are N -column vectors on Σ.We now demonstrate that we can always reduce the order of differential systems

to 1 by increasing the number of equations and the number of components of solution vectors.

Proposition 7.6. Let Σ be a non-characteristic hypersurface at x0 ∈ Σ. Then the initial-value problem (7.4)-(7.5) in a neighborhood of x0 is equivalent to an initial-value problem of a first-order differential system with initial values prescribed on Σ.

Proof. We assume x0 is the origin. In the following, we write x = (x, xn)and α = (α, αn).

Step 1. Straightening initial hypersurfaces. Assume Σ is given by ϕ = 0 fora C m-function ϕ in a neighborhood of the origin with ϕxn

= 0. Then we introduce

a change of coordinates x = (x, xn) →

(x, ϕ(x)). In the new coordinates, stilldenoted by x, the hypersurface Σ is given by xn = 0 and the initial condition(7.5) is given by

∂ jxnu(x, 0) = uj (x) for j = 0, 1, · · · , m − 1.

Step 2. Reductions to canonical forms and zero initial values. In the newcoordinates, xn = 0 is non-characteristic at 0. Then, the coefficient matrix

A(0,··· ,0,m)(x) is a nonsingular matrix at x = 0 and hence also in a neighborhood of the origin. Multiplying the partial differential system by the inverse of this matrix,we assume that A(0,··· ,0,m) is the identity matrix in a neighborhood of the origin.Next, we may assume

uj (x) = 0 for j = 0, 1,

· · · , m

−1.

To this end, we introduce a function v such that

u = v +m−1j=0

1

j!uj (x)xj

n.

Then the differential system for v is the same as that for u with f replaced by

f (x) −m−1j=0

|α|≤m

Aα(x)∂ α 1

j!uj (x)xj

n

.

8/16/2019 PDE . QUING



With Step 1 and Step 2, we assume (7.4)-(7.5) have the following form

∂ mxn u +

m−1αn=0

|α|≤m−αn

Aα∂ αu = f,

with

∂ jxnu(x, 0) = 0 for j = 0, 1, · · · , m − 1.

Step 3. Lowering the order. We now replace the above system by an equivalentone of order m − 1. Introduce new functions

U 0 = u, U i = uxi, i = 1, · · · , n,

and

U = (U 0, U 1, · · · , U n)T .

Then

U 0,xn = U n, U i,xn = U n,xi , i = 1, · · · , n − 1.

Hence

∂ m−1xn

U 0 − ∂ m−2xn

U n = 0,

∂ m−1xn

U i − ∂ xi∂ m−2

xn U n = 0, i = 1, · · · , n − 1.

To get an (m − 1)-th order differential equation for U n, we write the equation foru as

∂ mxnu +

m−1αn=1

|α|≤m−αn

Aα∂ αu +

|α|≤m

A(α,0)∂ (α,0)u = f.

We substitute U n = uxn in the first two terms in the right-hand side to get

∂ m−1xn

U n +m−2

αn=0

|α|≤m−αn−1

Aα∂ αU n +

|α|≤m

A(α,0)∂ (α,0)u = f.

In the last summation in the left hand side, any m-th order term of u can bechanged to an (m − 1)-th order term for U i for some i = 1, · · · , n − 1, since noderivatives with respect to xn are involved. Now we can write a differential system

for U in the following form

∂ m−1xn

U +m−2

αn=0

|α|≤m−αn−1

A(1)α ∂ αU = F (1).

The initial value for U is given by

∂ jxn U (x, 0) = 0 for j = 0, 1, · · · , m − 2.

Now we prove that the initial-value problem for the new (m−1)-th order differentialsystem is equivalent to that for the original m-th order differential system. In

particular, if U is a solution of the new (m − 1)-th order differential system, thenU 0 is a solution of the original m-th order differential system. To this end, we proveU i = U 0,xi

, for i = 1, · · · , n. First, by the first equation in the system for U andinitial conditions for U , we have

∂ m−2xn

(U n − U 0,xn) = 0,

8/16/2019 PDE . QUING


7.2. CAUCHY-KOWALEVSKI THEOREM 171

and on t = 0

∂ jxn(U n

−U 0,xn

) = 0 for j = 0,

· · · , m

−3.

This implies easily U n = U 0,xn . Next, for i = 1, · · · , n − 1,

∂ m−1xn

U i − ∂ xi∂ m−2

xn U n = ∂ m−1

xn U i − ∂ xi

∂ m−1xn

U 0 = ∂ m−1xn

(U i − ∂ xiU 0).

By the second equation in the system for U and initial conditions, we have

∂ m−1xn

(U i − ∂ xiU 0) = 0,

and on t = 0

∂ jxn(U i − ∂ xi

U 0) = 0 for j = 0, · · · , m − 2.

Hence, U i = U 0,xi, for i = 1, · · · , n−1. Substituting U i = U 0,xi

in the final equationin the system for U , we conclude that U 0 is a solution for the original m-th orderdifferential system.

Now, we can repeat the procedure to reduce m to 1.

We point that straightening initial hypersurfaces and reducing initial values to

zero are frequently used techniques in discussions of initial-value problems.

7.2. Cauchy-Kowalevski Theorem

For any given first-order linear partial differential system in a neighborhoodof x0 ∈ Rn and an initial value u0 prescribed on a hypersurface Σ passing x0, wefirst intend to find a solution u formally. To this end, we need to determine all

derivatives of u at x0, in terms of the initial value u0 and all known functions in theequation. Obviously, all tangential derivatives (with respect to Σ) of u are given byderivatives of u0. In order to find derivatives of u involving the normal direction, weneed help of the equation. It has been established that, if Σ is non-characteristic at

x0, the initial-value problem leads to evaluations of all derivatives of u at x0. Thisis clearly a necessary first step to the determination of a solution of the initial-valueproblem. If the coefficient matrices and initial values are analytic, a Taylor seriessolution could be developed for u. The Cauchy-Kowalevski Theorem asserts the

convergence of this Taylor series in a neighborhood of x0.To motivate our discussion, we study an example of first-order partial differen-

tial systems which may not have solutions in any neighborhood of the origin, unlessinitial values prescribed on analytic non-characteristic hypersurfaces are analytic.

Example 7.7. Let g = g(x) be a real valued function in R. Consider thefollowing system in R2

+ = (x, y); y > 0ux

−vy =0,

uy + vx =0,(7.6)

with prescribed initial values given by

u = g(x), v = 0 on y = 0.

Note that (7.6) is simply the Cauchy-Riemann equation in C = R2. It can bewritten in the matrix form as follows

1 00 1

uv

y

+

0 1−1 0

uv

x

=

00

.

8/16/2019 PDE . QUING



Note that y = 0 is non-characteristic. In fact, there are no characteristic curves.

To see this, we take any ξ = (ξ 1, ξ 2) ∈ R2 and then have

ξ 11 0

0 1

+ ξ 2

0 1−1 0

= ξ 1 ξ 2

−ξ 2 ξ 1

.

Its determinant is ξ 21 + ξ 22 , which is not zero for any ξ = 0. Therefore there areno characteristic curves. We now write (7.6) in complex form. Suppose we have a

solution (u, v) for (7.6) with the given initial values and let w = u + iv. Then

wx + iwy = 0 for y > 0 with w|y=0 = g(x).

Therefore, w is (complex) analytic in the upper half plane and its imaginary part iszero on the x-axis. By the Schwartz reflection principle, w can be extended acrossy = 0 to an analytic function in C = R2. This implies in particular that g is (real)

analytic since w|y=0 = g. We conclude that (7.6) does not admit any solutions withthe given initial value on y = 0 unless g is real analytic.

Example 7.7 naturally leads to discussions of analytic solutions. We introducedreal analyticity in Section 4.3. We now discuss this subject in detail.

To introduce (real) analytic functions, we need to study convergence of multipleinfinite series of the form

α

cα,

where cα are real numbers defined for all multi-indices α ∈ Zn+. Throughout this

section, the term convergence always refers to absolute convergence . Hence, aseries

α cα is convergent if and only if

α |cα| < ∞. Here, the summation is over

all multi-indices α ∈ Zn+.

Definition 7.8. A function u : Rn → R is called analytic near 0 if there existsan r > 0 and constants uα such that

u(x) =

α

uαxα for any |x| < r.

If u is analytic near 0, then u is smooth near 0. Moreover, the constants uα

are given by

uα = 1

α!∂ αu(0) for any α ∈ Zn

+.

Thus u equals its Taylor series about 0, i.e.,

u(x) =

α

1

α!∂ αu(0)xα for any |x| < r.

Now we give an important analytic function.

Example 7.9. For r > 0, set

u(x) = r

r − (x1 + · · · + xn) for any |x| <

r√ n

.

8/16/2019 PDE . QUING



Then

u(x) = 1 − x1 +

· · ·+ xn

r −1

=

∞

k=0

x1 +

· · ·+ xn

r k

=

∞k=0

1

rk

|α|=k

kα

xα =

α

|α|!r|α|α!

xα.

This power series is absolutely convergent for |x| < r/√

n sinceα

|α|!r|α|α!

|xα| =

∞k=0

|x1| + · · · + |xn|r

k< ∞,

for |x1| + · · · + |xn| ≤ |x|√ n < r. We also note

∂ αu(0) = |α|!r|α|

for any α ∈ Zn+.

We point out that all derivatives of u at 0 are positive.

An effective method to prove analyticity of functions is to control their deriva-tives by derivatives of functions known to be analytic. For this, we first introduce

the following terminology.

Definition 7.10. Let u and v be two C ∞-functions defined in a neighborhoodof the origin in Rn. Then v majorizes u, denoted by v u or u v, if

∂ αv(0) ≥ |∂ αu(0)| for any α ∈ Zn+.

We also call v a majorant of u.

We have the following simple method to verify analyticity.

Lemma 7.11. If v u and v is analytic near 0, then u is analytic near 0.Proof. We prove that the Taylor series of u about 0 converges for |x| < r if

the Taylor series of v about 0 converges for |x| < r. We simply noteα

1

α!|∂ αu(0)xα| ≤

α

1

α!∂ αv(0)|xα| < ∞ for any |x| < r.

We hence have the desired convergence for u.

Next, we prove that every analytic function has a majorant.

Lemma 7.12. If the Taylor series of u is convergent for |x| < r and 0 < s√

n <r, then u has an analytic majorant for |x| ≤ s/

√ n.

Proof. By setting y = s(1,· · ·

, 1), we have |

y|

= s√

n < r and henceα

1

α!∂ αu(0)yα

is a convergent series. Then there exists a constant C such that

| 1

α!∂ αu(0)yα| ≤ C for any α ∈ Zn

+,

and in particular,

| 1

α!∂ αu(0)| ≤ C

yα1

1 · · · yαnn

≤ C |α|!s|α|α!

.

8/16/2019 PDE . QUING



Now set

v(x)

≡ Cs

s − (x1 + · · · + xn)

= C α

|α|!s|α|

α!

.

Then v majorizes u for |x| ≤ s/√

n.

So far, our discussions are limited to scalar-valued functions. All results can begeneralized to vector-valued functions easily. For example, a vector-valued function

u = (u1, · · · , uN ) is analytic if each of its component is analytic.We have following results for compositions of functions.

Lemma 7.13. Let u, v be C ∞-functions of a neighborhood of 0 ∈ Rn into Rm

and f, g be C ∞-functions of a neighborhood of 0 ∈ Rm into RN , with u(0) = 0,

f (0) = 0, u v and f g. Then f u g v.

Lemma 7.14. Let u be an analytic function of a neighborhood of 0 ∈ Rn into

R

m

and f be an analytic function of a neighborhood of u(0) ∈ Rm

into RN

. Then f u is an analytic function in a neighborhood of 0 ∈ Rn.

We leave proof as an exercise.Now we are ready to discuss real analytic solutions of non-characteristic initial-

value problems for real analytic equations and initial values. We study first-orderquasilinear partial differential systems of N equations for N unknowns with initial

values prescribed on non-characteristic hyperplanes.For convenience, we study initial-value problems in Rn+1 = (x, t) with initial

values prescribed on t = 0. Consider

(7.7) ut =

n

j=1

Aj (x,t,u)uxj + F (x,t,u),

with

(7.8) u = u0 on t = 0,

where A1, · · · , An are N × N matrices in (x, t) and F are N -column vectors. Weassume all functions are analytic in their arguments.

The next result is referred to as the Cauchy-Kowalevski theorem.

Theorem 7.15. Let A1, · · · , An be analytic N × N matrices and F be an an-alytic N -column vector near (0, 0, u0(0)) ∈ Rn+1+N and u0 be an analytic function near 0 ∈ Rn. Then (7.7)-(7.8) admits an analytic solution u near 0 ∈ Rn+1.

Proof. Without loss of generality, we assume u0(x) = 0. To this end, we

introduce an analytic function v by v (x, t) = u(x, t)

−u0(x). Then the differential

system for v is similar as that for u. Next, we add t as an additional componentof u by introducing uN +1 such that uN +1,t = 1 and uN +1|t=0 = 0. This increasesthe number of equations and the number of components of solution vectors in (7.7)by 1 and meanwhile deletes t from A1, · · · , An and F . We still denote by N the

number of equations and the number of components of solution vectors.In the following, we study

(7.9) ut =n

j=1

Aj (x, u)uxj + F (x, u),

8/16/2019 PDE . QUING



with

(7.10) u = 0 on

t = 0

,

where A1, · · · , An are analytic N × N matrices and F is an analytic N -columnvector in a neighborhood of the origin in Rn+N . We seek an analytic solution uin a neighborhood of the origin in Rn+1. To this end, we will compute ∂ αu(0) interms of derivatives of A1, · · · , An and F at (0, 0) ∈ Rn+N and then prove that the

Taylor series of u at 0 converges in a neighborhood of 0 ∈ Rn+1. We note that tdoes not appear explicitly in the right hand side of (7.9).

Since u = 0 on t = 0, we have

∂ αx u(0) = 0 for any α ∈ Zn+.

Fix i = 1, · · · , n and differentiate (7.9) with respect to xi to get

uxi

t =n

j=1

(Aj uxi

xj

+ Aj,xi

uxj

+ Aj,uuxi

uxj

) + F uuxi

+ F xi

.

In view of (7.10), we have

uxit(0) = F xi(0, 0).

More generally, we obtain by induction

∂ αx ∂ tu(0) = ∂ αx F (0, 0) for any α ∈ Zn+.

Next, for any α ∈ Zn+, we have

∂ αx ∂ 2t u =∂ αx (ut)t = ∂ αx n

j=1

Aj uxj + F

t

=∂ αx

n

j=1

(Aj

uxj t

+ Aj,u

utu

xj

) + F u

ut.

Here we used the fact that Aj and F are independent of t. Thus,

∂ αx ∂ 2t u(0) = ∂ αx n

j=1

(Aj uxj t + Aj,uutuxj) + F uut

x=0,t=0,u=0

.

The expression on the right hand side can be worked out to be a polynomialwith nonnegative coefficients involving various derivatives of A1, · · · , An and F and derivatives ∂ βx ∂ ltu with |β | + l ≤ |α| + 2 and l ≤ 1.

More generally, for any α ∈ Zn+ and k ≥ 0, we have

(7.11) ∂ αx ∂ kt u(0) = pα,k(∂ ηx ∂ γ u A1, · · · , ∂ ηx ∂ γ

u An, ∂ ηx ∂ γ u F, ∂ βx ∂ ltu)

x=0,t=0,u=0

,

where pα,k is a polynomial with nonnegative coefficients and indices η, γ , β rangeover η, β ∈ Zn

+ and γ ∈ ZN + with |η|+ |γ | ≤ |α|+k−1, |β |+ l ≤ |α|+k and l ≤ k−1.

We need to point out that pα,k(∂ ηx ∂ γ u A1, · · · ) is considered as a polynomial of

components of ∂ ηx ∂ γ u A1, · · · . We denote by pα,k(|∂ ηx ∂ γ

u A1|, · · · ) the value of pα,k

when all components of ∂ ηx ∂ γ u A1, · · · are replaced by their absolute values. Since

pα,k has nonnegative coefficients, we conclude pα,k(∂ ηx ∂ γ u A1, · · · , ∂ ηx ∂ γ

u An, ∂ ηx ∂ γ u F, ∂ βx ∂ ltu)

x=0,t=0,u=0

≤ pα,k(|∂ ηx ∂ γ

u A1|, · · · , |∂ ηx ∂ γ u An|, |∂ ηx ∂ γ

u F |, |∂ βx ∂ ltu|)x=0,t=0,u=0

.(7.12)

8/16/2019 PDE . QUING



We now consider a new differential system

vt =

nj=1

Bj (x, v)vxi + G(x, v)

v =0 on t = 0,

(7.13)

where B1, · · · , BN are analytic N × N matrices and G is an analytic N -columnvector in a neighborhood of the origin in Rn+N . We will choose B1, · · · , Bn and Gsuch that

(7.14) Bj Aj for j = 1, · · · , n and G F.

Hence, for any (η, γ ) ∈ Zn+N + ,

∂ ηx ∂ γ u Bj (0) ≥ |∂ ηx ∂ γ

u Aj (0)| for j = 1, · · · , n and ∂ ηx ∂ γ u G(0) ≥ |∂ ηx ∂ γ

u F (0)|.The above inequalities should be understood as for each components.

Let v be a solution of (7.13). We now claim

|∂ αx ∂ kt u(0)| ≤ ∂ αx ∂ kt v(0) for any (α, k) ∈ Zn+1+ .

The proof is by induction on the order of t-derivatives. The general step followssince

|∂ αx ∂ kt u(0)| =| pα,k(∂ ηx ∂ γ uA1, · · · , ∂ ηx ∂ γ

uAn, ∂ ηx ∂ γ u F, ∂ βx ∂ ltu)

x=0,t=0,u=0

|≤ pα,k(|∂ ηx ∂ γ

uA1|, · · · , |∂ ηx ∂ γ uAn|, |∂ ηx ∂ γ

u F |, |∂ βx ∂ ltu|)x=0,t=0,u=0

≤ pα,k(∂ ηx ∂ γ uB1, · · · , ∂ ηx ∂ γ

u Bn, ∂ ηx ∂ γ u G, ∂ βx ∂ ltv)

x=0,t=0,v=0

=∂ αx ∂ kt v(0),

where we used (7.11), (7.12) and the fact that pα,k has positive coefficients. Thus

(7.15) v u.

It remains to prove the Taylor series of v at 0 converges in a neighborhood of 0 ∈ Rn+1.

To this end, we consider

B1 = · · · = Bn = Cr

r − (x1 + · · · + xn + u1 + · · · + uN )

1 · · · 1...

. . . ...

1 · · · 1

,

and

G = Cr

r − (x1 + · · · + xn + u1 + · · · + uN )

1...1

.

for positive constants C and r with |x| + |u| < r/√

n + N . As demonstrated in theproof of Lemma 7.12, we may choose C sufficiently large and r sufficiently smallsuch that (7.14) holds.

Set

v = w

1...1

,

8/16/2019 PDE . QUING



for some scalar-valued function w in a neighborhood of 0 ∈ Rn+1. Then (7.13) is

reduced to

wt = Cr

r − (x1 + · · · + xn + N w)(N

ni=1

wxi + 1),

w =0 on t = 0.

This is a (single) first-order quasilinear partial differential equation and has a solu-

tion of the form

w(x1, · · · , xn, t) = w(x1 + · · · + xn, t).

Then w = w(z, t) satisfies

wt = Cr

r − z − N w(nN wz + 1),

w =0 on

t = 0

.

By using the method of characteristics as in Section 2.2, we have an explicit solution

w(z, t) = 1

(n + 1)N

r − z − [(r − z)2 − 2Cr (n + 1)N t]

12

.

and hence

w(x, t) = 1

(n + 1)N

r −

ni=1

xi − [(r −n

i=1

xi)2 − 2Cr (n + 1)Nt]12

.

This expression is analytic for |(x, t)| < s, for sufficiently small s > 0. Hence, thecorresponding solution v of (7.13) is analytic for |(x, t)| < s. By Lemma 7.11 and

(7.15), u is analytic for |(x, t)| < s. Since the Taylor series of the analytic functionsut and

nj=1 Aj (x, u)uxj

+ F (x, u) have the same coefficients at the origin, they

agree throughout the region |(x, t)| < s.

At the beginning of the proof, we introduced an extra component for the solu-tion vector to get rid of t in coefficient matrices of the differential system. Had wechosen to preserve t, we would have to solve the initial-value problem

wt = Cr

r − z − t − Nw(nN wz + 1),

w =0 on t = 0.

It is difficult, if not impossible, to find an explicit expression of the solution w.The solution given in Theorem 7.15 is the only analytic solution since all deriva-

tives of the solution are computed at the origin and they uniquely determine the

analytic solution. A natural question is whether non-analytic solution is also unique.For this, we discuss initial value problems of linear differential systems as follows

A0(x, t)ut +n

j=1

Aj (x, t)uxj + B(x, t)u =F (x, t),

u(x, 0) = 0,

(7.16)

where A0, A1, · · · , An, B are analytic N ×N matrices and F is an analytic N -column

vector in a neighborhood of the origin in Rn+1.The next result is referred to as the local Holmgren uniqueness theorem.

8/16/2019 PDE . QUING



Theorem 7.16. Let A0, A1, · · · , An, B be analytic N ×N matrices and F be an

analytic N -column vector near the origin in Rn+1 and u0 be an analytic function

near 0 ∈ Rn. If t = 0 is non-characteristic at the origin, then any C 1-solution of (7.16) is analytic in a sufficiently small neighborhood of the origin in Rn+1.

In the proof, we need a notion of adjoint operators. Let L be a differentialoperator defined by

Lu = A0(x, t)ut +n

i=1

Ai(x, t)uxi + B(x, t)u.

For any vector-valued functions u and v , we write

vT Lu = (vT A0u)t +n

i=1

(vT Aiu)xi −

AT 0 vt +

ni=1

(AT i v)xi

− BT vT

u.

We define the adjoint operator L∗ of L by

L∗v = − (AT 0 v)t −

ni=1

(AT i v)xi

+ BT v

= − AT 0 vt −

ni=1

AT i vxi

+ (BT − AT 0,t −

ni=1

AT i,xi

)v.

Then

vT Lu = (vT A0u)t +n

i=1

(vT Aiu)xi +

L∗v

T u.

Proof. We introduce an analytic change of coordinates so that the initial

hypersurface Σ = t = 0 becomes a paraboloid

t =

ni=1

x2i .

We will prove that any C 1-solution u of Lu = 0 with a zero initial value on Σ is infact zero. For any ε > 0, we set

Ωε = (x, t);n

i=1

x2i < t < ε.

We will prove u = 0 in Ωε for a sufficiently small ε. In the following, we denote by∂ +Ωε and ∂ −Ωε the upper and lower boundary of Ωε respectively, i.e.,

∂ +Ωε =(x, t);n

i=1

x2i < t = ε,

∂ −Ωε =(x, t);n

i=1

x2i = t < ε.

We note that det(A0(0)) = 0 since Σ is non-characteristic at the origin. Hence

A0 is nonsingular in a neighborhood of the origin. By multiplying the equation in(7.16) by A−1

0 , we assume A0 = I .

8/16/2019 PDE . QUING



For any vector-valued function v in a neighborhood of the origin containing Ωε,

we have

0 = Ωε

vT Lu = Ωε

uT L∗v + ∂ +Ωε

uvT .

There is no boundary integral on ∂ −Ωε since u = 0 there. Let P k = P k(x) bean arbitrary polynomial in Rn , k = 1, · · · , n, and form P = (P 1, · · · , P N ). Weconsider the initial value problem

L∗v =0 in Br,

v =P on Br ∩ t = ε,

where Br is the ball with center at the origin and radius r in Rn+1. The principalpart of L∗ is the same as that of L, except a different sign and a transpose. We fix

r so that t = ε ∩ Br is non-characteristic for L∗, for each small ε. By Theorem

7.15, an analytic solution v exists in Br for any ε small. We need to point outthat the domain of convergence of v is independent of P , whose components arepolynomials. We choose ε small such that Ωε ⊂ Br. Then we have

∂ +Ωε

uP T = 0.

By Weierstrass approximation theorem, any continuous function in a compact do-main can be approximated in the L∞-norm by a sequence of polynomials. There-fore, u = 0 on ∂ +Ωε for any small ε and hence in Ωε.

Theorem 7.15 guarantees the existence of solutions of initial-value problems inthe analytic setting. As the next example shows, we do not expect any estimatesof solutions in terms of initial values.

Example 7.17. In R2, consider the first-order homogeneous linear differentialsystem (7.6)

ux − vy =0,

uy + vx =0.

Note that all coefficients are constant. As shown in Example 7.7, there are no

characteristic curves. Consider for any integer k ≥ 1

uk(x, y) = sin(kx)eky , vk(x, y) = cos(kx)eky .

Then uk and vk satisfy (7.6) and on y = 0uk(x, 0) = sin(kx), vk(x, 0) = cos(kx).

Obviously,

u2(x, 0) + v2(x, 0) = 1,

and

u2k(x, y) + v2

k(x, y) = e2ky → ∞ as k → ∞ for any y > 0.

Therefore, there is no continuous dependence on initial values. This illustrates thatinitial-value problems are not well posed for (7.6).

8/16/2019 PDE . QUING



7.3. Hyperbolic Differential Systems

In this section, we study the non-characteristic initial-value problem for partialdifferential systems without analyticity and introduce an important class of first-order partial differential systems – hyperbolic differential systems. The basic idea

underlying the hyperbolicity of a differential system is that the initial-value problemshould be well-posed. Specifically, initial values on non-characteristic hypersurfaceΣ are sufficient to determine unique solutions that depend continuously on valuesspecified at points on Σ.

We consider a first-order linear differential system in Rn × R = (x, t) in thefollowing form

(7.17) A0(x, t)ut +

nk=1

Ak(x, t)uxk + B(x, t)u = f (x, t),

where u and f are column vectors with N elements and A0, A1, · · · , An and B areN × N matrices. We prescribe an initial value on the hyperplane t = 0 by

(7.18) u(x, 0) = u0(x).

In the following, we always assume that A0(x, t) is nonsigular for any (x, t), i.e.,

det(A0(x, t)) = 0.

Hence, the hypersurface t = 0 is non-characteristic.For (7.17), principal symbols defined in Definition 7.4 have the following form

(7.19) p(x, t; ξ, λ) = det n

k=1

ξ kAk(x, t) + λA0(x, t)

,

for any (x, t) ∈ Rn+1 and (ξ, λ) ∈ Rn+1. For any ξ ∈ Rn, p(x, t; ξ, λ) is a polynomialof λ of degree N . If A0 = I , p(x, t; ξ, ·) is simply the characteristic polynomial of

the matrix

−n

k=1

ξ kAk(x, t).

Now, we start to investigate solvability of initial value problems associated with(7.17). As discussed in Section 7.2, we can find all derivatives of u at t = 0, in termsof A0, A1, · · · , An, B , f and u0. This determines a formal solution for small t. If thecoefficient matrices and initial values are analytic, a Taylor series solution could bedeveloped for u. The Cauchy-Kowaleski Theorem asserts the convergence of this

Taylor series at least in a neighborhood of any point (x, 0).

In this section, we discuss the non-analytic case. Extra assumptions are neededfor the well-posedness of initial-value problems.As a motivation, we study a constant coefficient system

(7.20) A0ut +n

k=1

Akuxk = 0.

For some constant λ ∈ C and constant vectors ξ ∈ Rn and v ∈ CN , we seek asolution u of the form

u(x, t) = ei(ξ·x+λt)v.

8/16/2019 PDE . QUING


7.3. HYPERBOLIC DIFFERENTIAL SYSTEMS 181

An easy calculation shows that u is a solution of (7.20) if and only if

(7.21) nk=1

ξ kAk + λA0v = 0.

Suppose (7.21) holds for some λ ∈ C \ R, ξ ∈ Rn \ 0 and v ∈ CN , with Imλ < 0.We consider a family of solutions

ur(x, t) = 1√

rReeir(ξ·x+λt)v =

1√ r

e−tImλReeir(ξ·x+tReλ)v,

for r > 0. Then,

ur(x, 0) → 0 as r → ∞,

and

ur(x, t)

→ ∞ as r

→ ∞ for any t > 0.

This simple example illustrates that the initial-value problem for (7.20) is not well-posed if Imλ = 0.

In summary, a necessary condition for initial-value problems (7.17) to be well-

posed is that the principal symbol p(x, t; ξ, ·) admits only real roots for any ξ ∈Rn \ 0.

We are led naturally to the following definition of hyperbolicity.

Definition 7.18. The differential system (7.17) is hyperbolic at (x, t) if for anyunit vector ξ ∈ Rn the principal symbol p(x, t; ξ, ·) has N real zeros λ1, · · · , λN and

there exist N linearly independent v1, · · · , vN in RN satisfying

n

k=1

ξ kAk(x, t) + λiA0(x, t)vi = 0, i = 1,· · ·

, N.

The differential system (7.17) is strictly hyperbolic at (x, t) if p(x, t; ξ, ·) has N distinct real zeros λ1, · · · , λN .

The notion of hyperbolicity is invariant under the change of coordinates of the

form

x = x(y, s), t = s.

An important class of hyperbolic differential system is symmetric hyperbolic

differential systems.

Definition 7.19. The differential system (7.17) is symmetric hyperbolic at

(x, t) if A0(x, t), A1(x, t), · · · , An(x, t) are symmetric and A0(x, t) is positive defi-nite.

Obviously, symmetric hyperbolic differential systems are indeed hyperbolic.The canonical form of (7.17) is given when A0 = I . In general, we can obtain

canonical forms easily. Since A0 is assumed to be nonsingular, we may multiply(7.17) by A−1

0 to get an equivalent system

ut +

nk=1

Ak(x, t)uxk + B(x, t)u = f (x, t).

8/16/2019 PDE . QUING



Its principal symbol is simply given by

p(x, t; ξ, λ) = det nk=1

ξ k Ak(x, t) + λI .

If the differential system (7.17) is symmetric hyperbolic, we wish to preserve thesymmetry. Since A0 is positive definite, we may write

A0 = M tM,

where M (x, t) is an N ×N nonsingular matrix. Introduce a new dependent variable

vector v by

v = M u.

Then v satisfies

vt +

n

k=1

˜Ak(x, t)vxk +

˜B(x, t)v =

˜f (x, t),

where Ak = M −tAkM −1 is still symmetric.

Now we study solvability of the initial-value problem (7.17)-(7.18). If coeffi-cient matrices and nonhomogeneous terms in (7.17) and initial values in (7.18) areanalytic, then analytic solutions exist locally in any neighborhood of (x, 0). This is

the Cauchy-Kawaleski Theorem. No hyperbolicity condition is required. However,we do not have any estimates for solutions in general, as Example 7.17 suggests.

If N = 1, the system (7.17) is reduced to a single equation for a scalar-valuedfunction u. The hyperbolicity condition is redundant. The initial-value problem(7.17)-(7.18) can be solved by integrating along integral curves. An energy estimate

was derived in Section 2.3.

If N > 1, the symmetry plays an essential role in solving the initial-value prob-lem (7.17)-(7.18). Symmetric hyperbolic differential systems in general dimensionsbehave like single differential equations of a similar form. An energy estimate can

be derived similarly and hence the problem (7.17)-(7.18) can be solved in weaksense. Now, we derive such an estimate.

We consider the initial-value problem for a first-order linear differential systemin Rn × R = (x, t) in the following form

A0(x, t)ut +n

k=1

Ak(x, t)uxk + B(x, t)u =f (x, t),

u(x, 0) =u0(x),

(7.22)

where u and f are column vectors with N elements and A1, · · · , An and B areN × N matrices. In the following, we always assume

A0, A1, · · · , An are symmetric in Rn × (0, T ),

and

A0 is positive definite.

We take positive constants a and κ such that

(7.23) A0 ≥ aI in Rn × R+,

8/16/2019 PDE . QUING



and

(7.24) A0 + κ

n

i=1

ξ iAi ≥ 0 in Rn

×R+,

for any ξ ∈ Rn with |ξ | ≤ 1.We recall the domain Dκ,T,t defined in Section 2.3

Dκ,T,t = (x, t); κ|x| < t − t, 0 < t < T ,

for fixed T , t > 0. Its boundary consists of several pieces given by

∂ −Dκ,T,t =(x, 0); κ|x| < t,

∂ sDκ,T,t =(x, t); κ|x| = t − t, 0 < t < T ,

∂ +Dκ,T,t =(x, T ); κ|x| < t − T .

Theorem 7.20. Let A0, A1,· · ·

, An be C 1 symmetric N ×

N matrices satisfying (7.23)-(7.24), B a continuous N × N matrix, f a continuous function and u be a

C 1-solution of (7.22) in Rn × R+. Then for any 0 < T < t, Dκ,T,t

e−αt|u|2 ≤ C

∂ −Dκ,T,t

|u0|2 +

Dκ,T,t

e−αt|f |2

,

where α and C are positive constants depending only on a in (7.23), the C 1-norms of Ai and the sup-norm of B in Dκ,T,t.

The estimate in Theorem 7.20 implies the uniqueness of solutions and the

property of finite speed propagation.Before proving Theorem 7.20, we point out the role of symmetry. The proof

proceeds similarly as that for scalar equations Section 2.3. We take an inner productof the equation (7.22) with 2u and hence need to analyze terms such as 2uT A0ut

and similar terms for x-derivatives. If A0 is symmetric, then

2uT A0ut = uT t A0u + uT A0ut = (uT A0u)t − uT A0,tu.

It is not necessary to assume symmetry for B since no derivatives of u are involvedwith B . We simply note

2uT Bu = uT (B + BT )u,

and B T + B is always symmetric.

Proof. For a nonnegative α, we multiply the equation in (7.22) by 2e−αtuT .By

2e−αt

uT

A0ut =(e−αt

uT

A0u)t + αe−αt

|u|2

− e−αt

uT

A0,tu,2e−αtuT Aiuxi

=(e−αtuT Aiu)xi − e−αtuT Ai,xi

u,

we have

(e−αtuT A0u)t +

ni=1

(e−αtuT Aiu)xi

+e−αtuT (αA0 −n

i=1

Ai,xi + B + BT )u = 2e−αtuT f.

8/16/2019 PDE . QUING



We write D = Dκ,T,t. An integration in D yields

∂ +D

e−αtuT A0u + ∂ sD

e−αtuT (γ t +

ni=1

Aiγ i)u

+

D

e−αtuT (αA0 −n

i=1

Ai,xi + B + BT )u

=

∂ −D

uT A0u +

D

2e−αtuT f,

where (γ 1, · · · , γ n, γ t) is the unit exterior normal vector of ∂ sD given by

(γ 1, · · · , γ n, γ t) = 1√ 1 + κ2

κ

x

|x| , 1

.

First, we note by (7.24)

γ t +n

i=1

aiγ i = 1√ 1 + κ2

(A0 + κn

i=1

xi

|x|Ai) ≥ 0 on ∂ sD.

Next, we choose α such that

αA0 −n

i=1

Ai,xi + B + BT ≥ 2aI in D.

Then

2a

D

e−αt|u|2 ≤ sup |A0|

∂ −D

|u0|2 +

D

2e−αtuT f.

Here we simply dropped integrals over ∂ +D and ∂ sD since they are nonnegative.The Cauchy inequality implies

D

2e−αtuT f ≤ a D

e−αt|u|2 + 1

a D

e−αt|f |2.


By letting t → ∞, we have the following result.

Theorem 7.21. Let A0, A1, · · · , An be C 1 symmetric N ×N matrices satisfying (7.23)-(7.24), B a continuous N × N matrix, f a continuous functions and u be a C 1-solution of (7.22) in Rn × R+. For any T > 0, if f ∈ L2(Rn × (0, T )) and u0 ∈ L2(Rn), then

Rn×(0,T )

e−αt|u|2 ≤ C

Rn

|u0|2 +

Rn×(0,T )

e−αt|f |2

,

where α and C are positive constants depending only on a, the C 1

-norms of Ai and the sup-norm of B in Rn × (0, T ).

Proceeding as in Section 2.3, we can discuss weak solutions of (7.22). For afixed T > 0, we consider functions in

DT = Rn × (0, T ).

Denote by C ∞0 (DT ) the collection of smooth functions in DT with compact supports

in DT and by C ∞0 (DT ) the collection of smooth functions in DT with compactsupports in x-directions.

8/16/2019 PDE . QUING



Set

(7.25) Lu = A0(x, t)ut +

n

i=1

Ai(x, t)uxi + B(x, t)u in DT .

We introduce the adjoint operator L∗ of L by

L∗v = − (A0v)t −n

i=1

(Aiv)xi + BT v

= − A0vt −n

i=1

Aivxi + (BT − A0,t −

ni=1

Ai,xi)v.

For any u, v ∈ C ∞0 , we have

vT Lu = (vT A0u)t +

n

i=1

(vT Aiu)xi −

L∗v

T

u.

By a simple integration in DT , we have DT

vT Lu =

DT

uT L∗v +

Rn×t=T

vT A0u − Rn×t=0

vT A0u

for any u, v ∈ C ∞0 (DT ).

We note that there are no derivatives of u in the right-hand side.

Definition 7.22. An L2-function u is a weak solution of Lu = f in DT if DT

uT L∗v =

DT

f T v for any v ∈ C ∞0 (DT ).

Now we can prove the existence of weak solutions of (7.22) with homogeneous

initial values by the Hahn-Banach Theorem and the Riesz Representation Theoremas in Section 2.3.

Theorem 7.23. Let A0, A1, · · · , An be C 1 symmetric N ×N matrices satisfying (7.23)-(7.24) and B a continuous N × N matrix. Then for any f ∈ L2(DT ), there exists a u ∈ L2(DT ) such that

DT

uT L∗v =

DT

f T v for any v ∈ C ∞0 (DT ) with v = 0 on t = T .

Moreover,

uL2(DT ) ≤ C f L2(DT ),

where C is a positive constant depending only on a, the C 1-norms of Ai and the sup-norm of B in Rn

×(0, T ).

The symmetry plays an essential role in Theorem 7.20, Theorem 7.21 andTheorem 7.23. The remaining question is whether we can change an arbitraryhyperbolic differential system into a symmetric hyperbolic differential system. The

discussion of this question in general case is beyond the scope of this book. Nowwe study a special case.

When the space dimension is one, we can change strictly hyperbolic differentialsystems to symmetric hyperbolic differential systems easily. For n = 1, we consider

(7.26) ut + A(x, t)ux + B(x, t)u = f (x, t).

8/16/2019 PDE . QUING



The principal symbol is given by

p(x, t; ξ, λ) = det ξA(x, t) + λI .

The hyperbolicity condition means that all eigenvalues of A are real and eigenvec-tors span RN . If the system (7.26) is a constant coefficient homogeneous system of the form

ut + Aux = 0,

then solution can be written explicitly. Suppose A has N real eigenvalues λ1, · · · , λN

and corresponding eigenvectors v1, · · · , vN , which span RN . Then solution u isgiven by

u(x, t) =

N k=1

ϕk(x − λkt)vk,

where functions ϕ1, · · · , ϕN are determined by the initial condition

u(x, 0) =N

k=1

ϕk(x)vk.

Back to the general system (7.26), we introduce a new solution vector v by

u = M v,

where M is an N ×N nonsingular matrix. Then a straightforward calculation yields

vt + M −1AMvx + (M −1BM + M −1AM x + M −1M t)v = M −1f.

This is a symmetric hyperbolic differential system if M −1AM is symmetric. Now

we take M whose columns consist of eigenvectors of A. Then M −1

AM is a diago-nal matrix with eigenvalues of A as diagonal entries, which is obviously symmetric.Therefore, v satisfies a symmetric hyperbolic differential system. If M is C 1, anenergy estimate can be derived. In fact, we can obtain a solution v by integratingalong characteristic curves. However, the matrix M consisting of eigenvectors of

A may not be C 1 when eigenvalues of high multiplicity are present. One condi-tion to ensure regularity of eigenvectors is that all eigenvalues are simple, i.e., thedifferential system (7.26) is strictly hyperbolic.

To end this section, we demonstrate that second-order hyperbolic differential

equations can always be changed to symmetric hyperbolic differential systems. Weconsider

(7.27) wtt −n

i,j=1aij (x, t)wxixj −

ni=1

bi(x, t)wxi − c(x, t)w = f,

where (aij ) is positive definite in Rn × [0, T ]. Here we assume aij , bi and c are

smooth, with

(7.28) λI ≤ (aij ) ≤ ΛI in Rn × [0, T ],

for some positive constants λ ≤ Λ. For convenience, we write

A = (aij ),

8/16/2019 PDE . QUING



and

ai =(ai1,· · ·

, ain) for any i = 1,· · ·

, n,

b =(b1, · · · , bn).

Set

(7.29) U ≡

u−1

u0

u1

...un

=

wwt

wx1

...wxn

.

This is an (n + 2)-vector. We write the equation (7.27) as

u−1,t = u0,

u0,t =n

i,j=1

aij uj,xi +

ni=1

biui + cu−1 + f,

nj=1

aij uj,t =

nj=1

aij u0,xj, i = 1, · · · , n.

Then we put it in a matrix form

(7.30) A0U t +n

i=1

AiU xi + BU = F,

where

A0 = I 2×2

A ,

Ai = −0 0 0

0 0 ai

0 aT i

0n×n

, i = 1, · · · , n,

B = −0 1 0

c 0 b

0 0 0n×n

,

and

F T =

0 f 0 · · · 0

.

Obviously, (7.30) is symmetric hyperbolic since A0, A1,

· · · , An are symmetric and

A0 is positive definite.We now consider the initial-value problem

wtt −n

i,j=1

aij wxixj −

ni=1

biwxi − cw =f in Rn × (0, T ),

w(·, 0) = w0, wt(·, 0) =w1 on Rn.

(7.31)

Then U defined in (7.29) satisfies (7.30) and initial conditions

(7.32) U = (w0, w1, w0,x1 , · · · , w0,xn)T on Rn.

8/16/2019 PDE . QUING



Now we verify that the initial-value problem (7.31) is equivalent to the initial-

value problem (7.30) and (7.32). We only need to prove that a solution U =

(u−1, u0, u1, · · · , un)T of (7.30) and (7.32) yields a solution w = u−1 of (7.31). First,by the first equation in the system (7.30) and the corresponding initial condition,we have u0 = u−1,t. Hence we have u−1,t = w1 on t = 0. Next, by the last nequations in (7.30), we have uj,t = u0,xj

= u−1,xj t, or (uj − u−1,xj)t = 0. Then we

get uj = u−1,xj since they match on t = 0. Hence w = u−1 is a solution of (7.31).

We then have the following result as a consequence of Theorem 7.21.

Theorem 7.24. Let aij be C 1 and bi, c be continuous in Rn×(0, T ), (aij ) satisfy (7.28) and w be a C 2-solution of (7.31) in Rn × (0, T ). If f ∈ L2(Rn × (0, T )) and

w0, ∇w0, w1 ∈ L2(Rn), then

Rn×(0,T )

e−αt(w2 + |∇w|2 + w2t ) ≤ C

Rn

(|w0|2 + |∇w0|2 + w21)

+ Rn×(0,T )

e−αt|f |2,

where α and C are positive constants depending only on λ, Λ, the C 1-norms of aij

and the sup-norm of bi, c in Rn × (0, T ).

Theorem 6.12 in Section 6.3 is a special case of Theorem 7.24.

Exercises

(1) Classify the following 4-th order equation in R3

2∂ 4xu + 2∂ 2x∂ 2y u + ∂ 4y u − 2∂ 2x∂ 2z u + ∂ 4z u = f.

(2) Prove Lemma 7.13 and Lemma 7.14.

(3) Consider the initial-value problem

utt − uxx − u = 0 in R× R+,

u(x, 0) = x, ut(x, 0) = −x.

Find a solution in a power series expansion about the origin and identifythis solution.

(4) Consider the initial-value problem for u : R× (0, T ) → RN as follows

ut + A(x, t)ux = f (x,t,u) in R× (0, T ),

withu(·, 0) = 0 on R,

where A is an N ×N diagonal C 1-matrix on R×(0, T ) and f : R×(0, T )×RN → RN is a C 2-map. Under appropriate conditions on f , prove theabove initial-value problem admits a C 1-solution by using the contractionmapping principle.Hint: It may be helpful to write in a system of equations instead of in a

matrix form.

8/16/2019 PDE . QUING


EXERCISES 189

(5) Consider in D = (x, t); x > 0, t > 0 ⊂ R2

ut + aux + b11u + b12v =f

vx + b12u + b22v =g,

with the condition

u(x, 0) = ϕ(x) for x > 0 and v(0, t) = ψ(t) for t > 0.

Assume a is C 1 and bij are continuous in D.(a) Assume a ≤ 0 in D. Derive an energy estimate for (u, v) in an

appropriate domain in D .(b) Assume a ≤ 0 in D. For sufficiently small T , derive an estimate for

sup[0,T ] |u(0, ·)| in terms of sup-norms of f , g , ϕ and ψ .

(c) Discuss whether similar estimates can be derived if a is positive some-where along (0, t); t > 0.

(6) Consider in a neighborhood of the origin in R2 = (x, t)ut + aux + b11u + b12v =f

vx + b12u + b22v =g,

with the condition

u(x, 0) = ϕ(x) and v(0, t) = ψ(t).

Assume a, bij are analytic in a neighborhood of 0 ∈ R2 and ϕ, ψ areanalytic in a neighborhood of 0 ∈ R.

(a) Prove all derivatives of u and v at 0 can be expressed in terms of derivatives of a, bij , ϕ and ψ at 0.

(b) Prove there exists an analytic solution (u, v) in a neighborhood of

0 ∈ R2

.

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


Bibliography

[1] Arnolds, V. I., Lectures on Partial Differential Equations, Universitext, Springer, 2004.

[2] Caffarelli, L., Cabre, X., Fully Nonlinear Elliptic Equations, AMS Colloquium Publications,

Vol. 43, American Math. Society, Providence, RI, 1993.

[3] Evans, L., Partial Differential Equations, Graduate Studies in Mathematics, Vol. 19, Amer-

ican Math. Society, Providence, RI, 1998.

[4] Friedman, A., Partial Differential Equations of Parabolic Type, Prentice-Hall, Englewood

Cliffs, 1964.[5] Gilbarg, D., Trudinger, N., Elliptic Partial Differential Equations of Second Order , (2nd ed.),

Grundlehren der Mathematischen Wisenschaften, Vol. 224, Spinger-Verlag, Berlin-New York,

1983.

[6] Han, Q., Lin, F.-H., Elliptic Partial Differential Equations, Courant Institute Lecture Notes,

Volume 1, American Math. Society, Providence, RI, 2000.

[7] John, F., Partial Differential Equations, (4th ed.), Applied Math. Sciences, Vol. 1, Springer-

Verlag, New York, 1991.

[8] MacRobert, T. M., Spherical Harmonics, An Elementary Treatise on Harmonic Functions

with Applications, Pergamon Press, Oxford-New York-Toronto, 1967.

[9] Taylor, M., Partial Differential Equations I: Basic Theory , Applied Math. Sciences, Vol. 115,

Springer-Verlag, New York, 1996.

191

8/16/2019 PDE . QUING


8/16/2019 PDE . QUING


Index

a priori estimates, 3

adjoint differential operators, 32, 178

analytic functions, 67, 172

auxiliary functions, 81, 120

Bernstein method, 83

Burgers’ equation, 18

canonical forms

first-order systems, 181

Cauchy problems, 8, 38, 166

Cauchy values, 8, 38, 166, 169

Cauchy-Kowalevski theorem, 174

characteristic curves, 9

characteristic hypersurfaces, 9, 38

non-characteristic hypersurfaces, 167, 169

characteristic ODEs, 14, 16, 21

compact supports, 32

comparison principles, 78, 80, 118

compatibility conditions, 20, 52, 54, 137,

139

conservation of energies, 48, 153

convergence of series, 67, 172

absolute convergence, 172

convolutions, 103

d’Alembert’s formula, 134

decay estimates, 150

degenerate differential equations, 39

diameters, 45

differential Harnack inequalities, 75

Dirichlet energy, 92

Dirichlet problems, 44, 61, 74Green’s function, 61

divergence theorem, 43

domain of influence, 13

domains, 1

domains of dependence, 13, 27, 134, 148

domains of influence, 27, 134, 148

doubling condition, 99

Duhamel’s principle, 142

eigenvalue problems, 50, 56

elliptic differential equations, 39, 168

energy estimates

first-order PDEs, 31

second-order hyperbolic equations, 159symmetric hyperbolic differential

systems, 184

wave equations, 47

Euclidean norms, 1

Euler-Poisson-Darboux equation, 143

finite speed propagation, 149

finite-speed propagation, 27

first-order linear differential systems, 180

canonical forms, 181

initial-value problems, 180

first-order linear PDEs, 7

initial-value problems, 24

Fourier series, 50Fourier transforms, 102

inverse Fourier transforms, 104

frequency, 99

fundamental solutions, 107

Laplace equations, 60

gradient estimates

interior gradient estimates, 66, 75, 83,

114, 121

gradients, 1

Green’s formula, 30, 43

Green’s function, 53, 61

Green’s function in balls, 62

Green’s identity, 60

Holder continuity, 90

pointwise Holder continuity, 86

half-space problems, 137

harmonic functions, 40, 59

conjugate harmonic functions, 40

converegence of Taylor series, 67

differential Harnack inequalities, 75, 83

doubling condition, 99

frequency, 99

Harnack inequalities, 75, 85

193

8/16/2019 PDE . QUING


194 INDEX

interior gradient estimates, 66, 75, 83

Liouville Theorem, 75

mean-value properties, 72nodal sets, 68

removable singularity, 85

superharmonic functions, 77

Harnack inequalities, 75, 85, 129

differential Harnack inequalities, 75, 83,

125, 128

heat equation


heat equations

1 dimension, 41

analyticity of solutions, 117

differential Harnack inequalities, 125, 128

Harnack inequalities, 129

initial/boundary-value problems, 46, 49

interior gradient estimates, 114, 121

n dimensions, 42

Poisson integral formula, 106

Hessian matrices, 1

Holmgren uniqueness theorem, 177

Hopf lemma, 78

Huygens’ principle, 149

hyperbolic differential equations, 39, 43

hyperbolic differential systems, 181

strictly hyperbolic systems, 181

symmetric hyperbolic systems, 181

hypersurfaces, 2

initial hypersurfaces, 8, 38, 166

initial values, 8, 38, 166, 169initial-value problems, 166

first-order PDEs, 8, 11

first-order systems, 180

second-order PDEs, 38

wave equations, 133, 142

initial/boundary-value problems

heat equations, 46, 49

wave equations, 47, 54, 139

integral curves, 13

Laplace equations, 39, 42


Green’s indentity, 60

Poisson integral formula, 64Poisson kernel, 64

Legendre equations, 69

Legendre functions, 70

linear differential systems

m-th order, 168

first-order, 180

linear PDEs, 2

first-order, 7

m-th order, 165

second-order, 37

Liouville Theorem, 75

loss of differentiations, 152

majorants, 173

maximum principles, 74

strong maximum principles, 79

weak maximum principles, 77, 117, 118

mean-value properties, 72

method of characteristics, 14

method of descent, 146

method of extensions, 137, 140

method of reflections, 138, 140

method of spherical averages, 143

mixed problems, 47

multi-indices, 1

Neumann problems, 45

nodal curves, 70nodal sets, 68

non-characteristic curves, 9

non-characteristic hypersurfaces, 9, 11, 38

non-homogeneous terms, 7, 37, 166

parabolic boundaries, 112, 117

parabolic differential equations, 43

parabolic distance, 113

Parseval’s identity, 105

partial differential equations (PDEs), 2

linear PDEs, 2

quasilinear PDEs, 2

PDEs of mixed type, 41

Poincare Lemma, 45

Poisson equations, 42, 86

weak solutions, 92

Poisson integral formula, 64, 106

Poisson kernel, 64

principal parts, 165, 168

principal symbols, 37, 165, 168, 180

propagation of singularities, 41

quasilinear PDEs, 2

radiation field, 162

Rellich’s theorem, 94

removable singularity, 85

Schauder estimates, 90pointwise Schauder estimates, 88

Schwartz class, 102

second-order linear PDEs, 37

in the plane, 39

separation of variables, 49

space variables, 1

space-like surfaces, 157

spherical harmonics, 68

sectorial harmonics, 70

tesseral harmonics, 70

8/16/2019 PDE . QUING


INDEX 195

zonal harmonics, 70

strictly hyperbolic systems, 181

subsolutions, 77subharmonic functions, 77

supersolutions, 77

superharmonic functions, 77

symmetric hyperbolic systems, 181

Taylor series, 67, 172

terminal value problems, 112

time variables, 1

time-time surfaces, 157

Tricomi equation, 41

uniform ellipticity, 77

wave equations

1 dimension, 40, 133

2 dimensions, 146

3 dimensions 144

pde . quing

Documents