Partial Differential Equations (1)

Partial Differential Equations

of Applied Mathematics

Lecture Notes, Math 713

Fall, 2003

−3 −2 −1 0 1 2 3−4

−2

0

2

4

6

8

10

12

14

n=12

n=4

D.H. Sattinger

Department of Mathematics

Yale Univerity

i

Preface

We have comprehended some auspicious characters.

Sheriff Dogberry, in Much Ado about Nothing

These lecture notes have arisen over a period of years teachingcourses in partial differential equations at the University of Minnesotaand Yale University. They are, by and large, a plain vanilla approachto the subject, emphasizing emphasize basic elements of partial differ-ential equations and applied analysis and their applications to the basicequations of mathematical physics and applied mathematics. They areintended to be selective rather than encyclopedic, and to illustrate theinterrelationship between mathematics and physics. The main prereq-uisites for the course are a solid foundation in undergraduate analysis,vector analysis, some background in physics, and an interest in theapplication of mathematics to natural phenomena. They are aimed atstudents in mathematics, physics, and engineering.

D.H. Sattinger

New Haven

September, 2003

ii

Contents

1 The Heat Equation 11.1 The flow of heat . . . . . . . . . . . . . . . . . . . . . . 11.2 The fundamental solution . . . . . . . . . . . . . . . . 31.3 Fourier series & separation of variables. . . . . . . . . 121.4 Brownian Motion . . . . . . . . . . . . . . . . . . . . . 191.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 Laplace’s Equation 292.1 Conservative vector fields and potentials . . . . . . . . 292.2 Polar and Spherical coordinates . . . . . . . . . . . . . 372.3 Analytic function theory. . . . . . . . . . . . . . . . . . 402.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3 The Wave Equation 473.1 The wave equation in 1,2,3 dimensions . . . . . . . . . 473.2 Characteristic curves in the plane. . . . . . . . . . . . . 543.3 Characteristic surfaces in higher dimensions. . . . . . . 603.4 The wave equation in Rn. . . . . . . . . . . . . . . . . 623.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4 Equations of Fluid and Gas Dynamics 734.1 Conservation Laws . . . . . . . . . . . . . . . . . . . . 734.2 Bernoulli’s Theorem . . . . . . . . . . . . . . . . . . . 824.3 D’Alembert’s Paradox . . . . . . . . . . . . . . . . . . 874.4 Hyperbolic Conservation Laws . . . . . . . . . . . . . . 884.5 Shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 94

iii

iv CONTENTS

5 The Maximum Principle 975.1 Elliptic and parabolic inequalities . . . . . . . . . . . . 975.2 Monotone methods . . . . . . . . . . . . . . . . . . . . 1035.3 Crash Course in Elliptic Equations . . . . . . . . . . . 110

5.3.1 Regularity theory for elliptic equations . . . . . 1105.3.2 The Fredholm Alternative . . . . . . . . . . . . 1125.3.3 The Krein-Rutman Theorem . . . . . . . . . . . 113

5.4 The method of moving planes . . . . . . . . . . . . . . 1145.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 117

List of Figures

1.1 Fundamental solution of the heat equation . . . . . . . 61.2 Discontinuous data . . . . . . . . . . . . . . . . . . . . 111.3 Gibbs phenomenon . . . . . . . . . . . . . . . . . . . . 141.4 Dirichlet function . . . . . . . . . . . . . . . . . . . . . 17

3.1 Domain of Dependence . . . . . . . . . . . . . . . . . . 483.2 Forward and Backward Light Cone . . . . . . . . . . . 493.3 Backward light cone in R3 . . . . . . . . . . . . . . . . 523.4 Jump conditions across a characteristic. . . . . . . . . . 563.5 The Goursat problem . . . . . . . . . . . . . . . . . . . 573.6 Conditions along a non-characteristic curve . . . . . . . 583.7 Backward ray cone in three dimensions . . . . . . . . . 703.8 Shock discontinuity of a nonlinear hyperbolic equation 71

5.1 Boundary point lemma . . . . . . . . . . . . . . . . . . 99

v

vi LIST OF FIGURES

Chapter 1

The Heat Equation

1.1 The flow of heat

The heat equation describes the flow of heat in a homogeneous isotropicmedium. In three space dimensions it is

∂u

∂t= K∆u, ∆u = uxx + uyy + uzz. (1.1)

The operator ∆ is called the Laplacian and plays a fundamental rolein mathematical physics and in partial differential equations. The con-stant K is determined by physical properties of the medium, and iscalled the diffusion coefficient.

Throughout these notes we denote the time variable by t, and thespatial variable by x. When x ∈ R3, we write x = (x1, x2, x3), and weabbreviate the partial derivatives of u by

ut =∂u

∂t, uj =

∂u

∂xj

,

etc.The heat equation models not only the flow of heat energy through a

medium, but also models diffusion processes, such as Brownian motion,the statistical motion of very small particles suspended in a fluid. Wefirst derive the heat equation from the point of view of continuummechanics, then discuss methods of solution and some of its properties.Later we discuss the relationship of the heat equation to Brownianmotion and diffusion processes in statistical physics.

1

2 CHAPTER 1. THE HEAT EQUATION

Let u(x, t) denote the temperature in a medium at the point x attime t, and let Q denote the density of heat energy. The heat energy is afunction of the temperature: Q = Q(x, u) per unit volume. We shouldexpect that Qu > 0, and the simplest assumption that we can makeis that Q is linearly proportional to u with a fixed, positive, spatiallyindependent constant c: Q = cu. The constant c > 0 is called thespecific heat. In reality the specific heat may itself depend on x and u,but this will lead to more complicated equations; so for now we limitthe discussion to the simplest case.

If C denotes a domain (an open set) in R3, then the total heatenergy in C is given by the volume integral

∫∫∫

C

cu dx,

and the rate of change of heat energy inside C is

∂

∂t

∫∫∫

C

cu dx.

If the temperature is not uniform the heat will flow from hotter tocolder regions. This is expressed mathematically by saying that theheat flux across an oriented surface is opposite to the gradient of thetemperature on that surface. Denote the heat flux by a vector field ~F ,and think of heat as a kind of invisible fluid and ~F as the flux of thatfluid per unit area in the direction of ~F . Thus, the amount of heat perunit time flowing across a surface S in space is given by the integral

∫∫

S

~F · d~S

Here, d~S is the oriented element of surface area d~S = n dS, where n(x)is the normal unit vector to the surface S and dS is the infinitesimalarea element.

Since heat energy moves from warmer to cooler regions, the fluxis in the opposite direction of the temperature gradient and can bewritten

F = −k(u)∇u, ∇u = (u1, u2, u3)

1.2. THE FUNDAMENTAL SOLUTION 3

where k(u) > 0. Let us assume our material is a simple one, with k(u)equal to a positive constant, k.

By conservation of heat energy, the rate of increase in heat energyinside C is equal to the flux of energy flowing in across the boundaryof C, (denoted by ∂C). Thus

∂

∂t

∫∫∫

C

cu dx = −∫∫

∂C

F · d~S =

∫∫

∂C

k∇u · d~S.

By the divergence theorem,∫∫

S

k∇u · d~S = k

∫∫∫

C

∆u dx,

where ∆u = div∇u. Combining these two results we find∫∫∫

C

cut − k∆u dx = 0 (1.2)

for any smoothly bounded domain C.It is important to emphasize that this conservation law holds for

any domain C with a smooth boundary contained in the domain ofdefinition of u. For simplicity, we assume that the integrand in (1.2) iscontinuous. If it is not identically zero, then it is, say, positive at somepoint x0. By continuity it is positive in some small ball B containing x0.Applying the result (1.2) to the domain B we obtain a contradiction,for the integrand is everywhere positive, yet the integral vanishes. Thusthe integrand cannot be positive anywhere; and by a similar argument,it cannot be negative anywhere either, so it must be identically zero.Hence we obtain the heat equation (1.1) with the diffusion constantK = k/c.

1.2 The fundamental solution

We now turn to the problem of constructing a solution of the initialvalue problem for the heat equation in the case of one space variablewith K = 1:

ut = uxx, u(x, 0) = f(x). (1.3)


Recall that the solution to a linear system of ordinary differential equa-tions x = Ax, x(0) = x0 is given by x(t) = etAx0, where the exponentialof a matrix is defined by a power series expansion. The correspondingformal solution to the heat equation would be u(t) = et∆u0 where u0

and u(t) belong to some class of functions defined on Rn. The questionwe want to resolve is how to interpret the formal expression et∆ where∆ denotes the Laplacian. The representation of this operator leads tothe construction of what is known as the fundamental solution of theheat equation.

We shall show that the solution of the heat equation in R1 can bewritten in the form

u(x, t) =

∫ ∞

−∞G(x− y, t)f(y) dy.

Such an integral is called a convolution. If we denote Gt(x) = G(x, t),then we may formally write

et∆f = Gt ∗ f.

It is immediate that the function G(x, t) must be a solution of theheat equation for t > 0 and have the property that

limt→0+

∫ ∞

−∞G(x− y, t)f(y) dy = f(x), (1.4)

at least for some large class of functions f . Moreover, since the solutionof the initial value problem with f(x) = 1 is u(x, t) = 1, we should alsoexpect that ∫ ∞

−∞G(x− y, t) dy = 1

for all x, t > 0. The function G(x, t) is the fundamental solution ofthe heat equation in one dimension.

One common technique for finding special solutions of partial dif-ferential equations is to reduce the equation to an ordinary differentialequation can be solved explicitly. We do that in the present case byusing a device that works in many cases, linear as well as nonlinear.Suppose in the heat equation we rescale the variables by letting x → λxand t → λ2t. More explicitly, we consider the transformations of u de-fined by

Tλu(x, t) = u(λx, λ2t) (1.5)


The transformations Tλ, λ > 0 form a group of transformations, calledthe dilation group. It is a simple exercise to show that Tλ maps solutionsof the heat equation to solutions.

Let us see how the solution of the initial value problem transformsunder such a rescaling. The solution of (1.3) with initial data Tλf isTλu. Hence

∫ ∞

−∞G(λx− y, λ2t)f(y) dy =

∫ ∞

−∞G(λ(x− z), λ2t)f(λz)λ dz

=

∫ ∞

−∞λG(x− z, t)f(λz) dz,

andλG(λx, λ2t) = G(x, t). (1.6)

Definition A function u(x, t), x ∈ Rn, t > 0 is homogeneous ofdegree α with respect to the action of the dilation group if Tλu = λαufor all λ > 0: u(λx, λ2t) = λαu(x, t)

The argument above shows that the fundamental solution is homo-geneous of degree -1. The following result may be proved by the samegeneral argument.

Theorem 1.2.1 The fundamental solution of the heat equation in nspace dimensions is homogeneous of degree −n.

Now (1.6) holds for all λ > 0; in particular, for any fixed value of twe may choose λ = t−1/2. We then see that G has the property that

G(x, t) =1√tϕ(ξ),

whereξ =

x√t, ϕ(ξ) = u(ξ, 1).

The variable ξ is called a similarity variable.We also leave it as an exercise to show that ϕ satisfies the ordinary

differential equation

ϕ′′ +1

2(ξϕ)′ = 0, (1.7)


−5 −4 −3 −2 −1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

t=.1

t=1

Figure 1.1: The fundamental solution of the heat equation evaluated att = .1 and t = 1.

and that a solution of this equation is

ϕ = Ae−14ξ2

, (1.8)

where A is a constant of integration. The corresponding solution of theheat equation is

A√te−x2/4t. (1.9)

It is a straightforward matter to check that this expression satisfies theheat equation.

The integral of this function over the real line is constant in t, andA may be chosen so that the integral is precisely 1. We then obtain

G(x, t) =e−x2/4t

√4πt

(1.10)


as the fundamental solution of the heat equation on the line.

Theorem 1.2.2 Let f(x) be bounded and continuous on the real line.The solution of the initial value problem (1.3) is given by

u(x, t) =1√4πt

∫ ∞

−∞e−(x−y)2/4tf(y) dy. (1.11)

Proof: We leave it as an exercise to show that (1.11) is a solutionof the heat equation, and show here that

limt↓0

1√4πt

∫ ∞

−∞e−(x−y)2/4tf(y) dy = f(x)

at every point of continuity x of f . Since the fundamental solution isnormalized so that its integral over the real line is 1, we have

∆(x, t) =

∫ ∞

−∞

e−(x−y)2/4t

√4πt

f(y) dy − f(x)

=

∫ ∞

−∞

e−(x−y)2/4t

√4πt

(f(y)− f(x)) dy.

We must show that ∆(x, t) → 0 as t ↓ 0. For fixed x and any ε > 0,we choose δ > 0 so that

|f(y)− f(x)| < ε for all |x− y| < δ.

Then

∆(x, t) =

∫

|x−y|<δ

e−(x−y)2/4t

√4πt

(f(y)− f(x)) dy

+

∫

|x−y|≥δ

e−(x−y)2/4t

√4πt

(f(y)− f(x)) dy

=∆1(x, t) + ∆2(x, t).

Now

|∆1(x, t)| ≤∫

|x−y|<δ

e−(x−y)2/4t

√4πt

|f(y)− f(x)| dy < ε.


Moreover, since f is bounded, say |f(y)| ≤ M for all y,

∆2(x, t) ≤2M

∫

|x−y|≥δ

e−(x−y)2/4t

√4πt

dy

=4M

∫ ∞

δ

e−y2/4t

√4πt

dy =4M√4π

∫ ∞

δ/4√

t

e−y2

dy

→0 as t ↓ 0.

The above argument shows that lim supt→0+ ∆(x, t) ≤ ε for anyε > 0. Since ε > 0 is arbitrary, ∆(x, t) must tend to zero as t ↓ 0. ¥.

The fundamental solution of the heat equation in Rn is

u(x, t) =1

(4πt)n/2e−r2/4t, r2 = x2

1 + . . . x2n. (1.12)

The proof is left as an exercise.We now turn to a discussion of some of the properties of the solution

of the heat equation. In an appropriate sense, these properties are char-acteristic of a more general class partial differential equations knownas second order parabolic equations. The reason for the terminologywill become apparent later.

Since the fundamental solution of the heat equation is strictly posi-tive and has integral 1 for all t > 0, we immediately have the followingresult, known as the Strong Maximum Principle:

Theorem 1.2.3 Let a ≤ f(x) ≤ b for all real x, and suppose that f isnot identically constant. Then the solution (1.11) of the heat equationsatisfies the strict inequalities a < u(x, t) < b for all t > 0.

Proof: Since the integral of the fundamental solution is identically 1,

u(x, t)− a =1√4πt

∫ ∞

−∞e−(x−y)2/4t(f(y)− a) dy.

The integrand is non-negative and strictly positive on any interval onwhich f > a; hence it can never be zero for any x or any t > 0. Henceu(x, t) > a for all t > 0. Similarly u(x, t) < b for all t > 0. ¥


The strong maximum principle for the heat equation is a precisemathematical formulation of the physical observation that heat flowsfrom warmer to cooler regions. In the absence of external heat sources,the temperature cannot rise above or below the original ambient values.In this sense the heat equation accurately reflects the physics of heat.In another sense, however, it is an unphysical model; for, by the sameargument we used above, we can prove the following:

Proposition 1.2.4 Under the evolution of the heat equation, distur-bances propagate infinitely fast.

The support of a function f , denoted by supp (f), is the set ofpoints x for which f(x) 6= 0. Suppose f(x) ≥ 0 everywhere on R andsupp(f) ⊂ K, where K is a compact subset of the real line. Let usatisfy (1.3). Then by the same argument we used to prove the strongmaximum principle,we can show that u is strictly positive for all x andall t > 0. In other words, heat propagates infinitely fast.

This is clearly a nonphysical phenomenon. Thus, the heat equationis not really a fundamental equation of physics; rather, it is an approx-imate model, albeit a very good one. We will return to this later in§1.4.

The strong maximum principle holds for more general parabolicequations in n spatial dimensions. For example, consider the partialdifferential operator

L[u] =n∑

j,k=1

ajk(x, t)∂2u

∂xj∂xk

+n∑

j=1

bj(x, t)∂u

∂xj

− ∂u

∂t+ c(x, t)u.

L is said to be uniformly parabolic in a domain Ω ⊂ Rn+1 if the coeffi-cients ajk satisfy the inequality

n∑

j,k=1

ajk(x, t)ξjξk ≥ µ

n∑j=1

ξ2j

for all (x, t) ∈ Ω for some µ > 0.Let ΩT be the domain

ΩT = (x, t) : x ∈ Ω, 0 ≤ t ≤ Twhere Ω is a bounded connected domain in Rn. Then


Theorem 1.2.5 Let c(x, t) ≤ 0 in ΩT and suppose that u satisfies theuniformly parabolic partial differential inequality

L[u] ≥ 0 (x, t) ∈ ΩT .

Suppose that u ≤ M in ΩT and u(x0, T ) = M at an interior pointx0 ∈ Ω. Then u ≡ M for all x ∈ Ω and all 0 ≤ t ≤ t0.

The maximum principle also holds for nonlinear parabolic equationsas well, under appropriate conditions. For example, the coefficientsajk, bj and c in the definition of the operator L can all depend on thesolution u and its derivatives uxj

and uxjxk. A very readable account

of maximum principles for partial differential equations is given in thebook by Protter and Weinberger [15].

The heat equation does a lot of smoothing. Even if the initial dataf is only bounded and measurable, the solution of the heat equation uis real analytic in x and t for t > 0. By that we mean that the solutioncan be expanded in a convergent Taylor series about any point x0, t0where t0 > 0. We will prove a somewhat weaker result here:

Theorem 1.2.6 Let f be a bounded measurable function on the realline. Then the solution (1.11) is infinitely differentiable in both x andt for all t > 0.

Proof: The fundamental solution is infinitely differentiable in bothvariables, and it, together with all its derivatives in x and t, decayexponentially fast as x → ±∞ for any t > 0. Therefore we mayinterchange the process of differentiation and integration in the integral(1.11). The same argument holds for the solution in Rn. ¥

In Figure 1.2 we see the graph of the solution of the initial valueproblem for the heat equation with discontinuous initial data:

H(x) =

0 −∞ < x < 0;

1 0 < x < ∞.(1.13)

The function H is known as the Heaviside step function.Analyticity of the solution of the heat equation actually follows from

the analyticity of the fundamental solution.We consider now the solution of the inhomogeneous initial value

problemut = uxx + h, u(x, 0) = f


−5 −4 −3 −2 −1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t=.1

t=1

Figure 1.2: Solution of the heat equation on the line with initial u(x, 0) =H(x) at times t = .1, 1. The solution is continuous and lies strictly between0 and 1 for all positive time.

where f is defined on the real line, and h = h(x, t) is defined on t > 0.There is a standard method of solving inhomogeneous linear equations,known as Duhamel’s principle. It goes like this.

Consider an abstract equation of the form ut = Au+h, u(0) = f . Itis useful to regard the heat equation as an infinite dimensional systemof ordinary differential equations, with A = d2/dx2. If A were a matrixon a finite dimensional vector space, the solution would be given by

u(t) = etAf +

∫ t

0

e(t−s)Af(s) ds. (1.14)

In the present case, etAf is precisely the integral operator given by(1.11). For the heat equation, (1.14) should be interpreted as

u(x, t) =1√4πt

∫ ∞

−∞e−(x−y)2/4tf(y) dy

+

∫ t

0

∫ ∞

−∞

1√4π(t− s)

e−(x−y)2/4(t−s)h(y, s) dy ds

Having derived this expression formally, one can check directly that itgives the required solution.


1.3 Fourier series & separation of variables.

We noted above that partial differential equations are infinite dimen-sional in character, and that, in particular, the heat equation has aninfinite number of solutions. We apply that observation now to de-rive the solution of the initial value problem for the heat equationon a finite interval. This problem was originally posed and solved byJ. J. Fourier, Theorie Analytique de la Chaleur, and led to the de-velopment of Fourier series and integrals, one of the most significanttools of pure and applied mathematics.

Suppose we want to solve the initial value problem for the heatequation on a finite interval. For convenience, we take scale the spatialvariable so that the interval is [0, π]. Thus we have:

ut = uxx, 0 ≤ x ≤ π; u(x, 0) = f(x) (1.15)

together with the boundary conditions

u(0, t) = u(π, t) = 0.

Many other situations are possible; but we shall treat only this sim-ple case. For a more extensive discussion of the method of separationof variables, see Pinsky [14], Sommerfeld [23], Weinberger [25].As in the problem on the infinite interval, we begin by constructingspecial solutions of the homogeneous problem. We look for solutionsof ut = uxx of the form

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

This is the so-called process of separation of variables.Substituting this expression into the heat equation we obtain the

relation T ′X = X ′′T , which we write as

T ′

T=

X ′′

X.

The only way such a relationship can hold for arbitrary x and t is forboth ratios to be constant. For reasons that will quickly become clear,we take this constant to be −γ2. Then we have

T ′ = −γ2T, X ′′ + γ2X = 0.

1.3. FOURIER SERIES & SEPARATION OF VARIABLES. 13

The general solutions of this pair of equations are

T (t) = T0e−γ2t, X(x) = A sin γx + B cos γx.

In particular, T is a bounded, in fact, exponentially decreasing,function of time. The constant γ is not arbitrary, but is restrictedby the requirement that the solutions satisfy the boundary conditionX(0) = X(π) = 0. The boundary condition X(0) = 0 implies thatB = 0, while the boundary condition X(π) = 0 implies that sin γπ = 0,hence that γ = n, where n is an integer. Since sin−γπ = − sin γπ,there is no loss in generality if we restrict n to be positive.

Hence we have constructed an infinite number of independent solu-tions of the heat equation, namely

un(x, t) = e−n2t sin nx.

Since the equation is linear and homogeneous, we may form more gen-eral solutions of the heat equation by taking a linear superposition ofsuch solutions:

u(x, t) =∞∑

n=1

ane−n2t sin nx.

Since the exponential terms converge very rapidly to zero for t > 0,there is no problem proving this series converges and satisfies the heatequation under very minimal restrictions on the coefficients an. It alsosatisfies the boundary conditions.

The only remaining question is whether the an can be chosen so asto satisfy the initial conditions, that is, whether the initial data f canbe represented as an infinite sine series

f(x) =∞∑

n=1

an sin nx.

This question has to do with the completeness of the family of functionssin nx on the interval [0, π]. Since a sine series automatically repre-sents an odd, 2π periodic function, we consider the odd, 2π-periodicextension of f . The coefficients an are given by (cf. exercises)

an =1

π

∫ π

−π

f(x) sin nx dx. (1.16)


Since the method of Fourier series has been extensively discussedelsewhere, we will not go into a great amount of detail on the method.However, let us prove here that the Fourier sine series converges to f ,given some conditions on the regularity of f .

-1

-0.5

0

0.5

1

-3 -2 -1 0 1 2 3x

Figure 1.3: Partial sums S1, S2, and S5 of the Fourier series for the stepfunction (1.18). The overshooting is known as Gibbs phenomenon.


The partial sums of the Fourier sine series of f are

Sn,f (x) =1

π

n∑

k=1

sin kx

∫ π

−π

f(t) sin kt dt

=1

π

∫ π

−π

f(t)n∑

k=1

sin kx sin kt dt. (1.17)

The convergence of the Fourier series to the function is an issueof some complexity. For differentiable functions, the Fourier seriesconverge uniformly to the function itself. But when the function isdiscontinuous, e.g.

f(x) =

−1 −π ≤ x ≤ 0;

1 0 ≤ x ≤ π,(1.18)

the partial sums of the Fourier series ”overshoot” the function as shownin Figure (1.3). The overshooting of the Fourier partial sums contrastswith the behavior of the solution of the heat equation (Figure (1.2)),which strictly interpolates the discontinuity, positivity of the funda-mental solution of the heat equation.

Now

sin kx sin kt =cos k(x− t)− cos k(x + t)

2.

We leave it as an exercise to show that

Dn(θ) :=1

2+

n∑

k=1

cos kθ =1

2

sin(n + 12)θ

sin 12θ

. (1.19)

The function Dn(θ) is called Dirichlet’s kernel.The partial sum Sn can therefore be written

Sn,f (x) =1

2π

∫ π

−π

f(t)[Dn(t− x)−Dn(x + t)] dt.

By changing variables in the two terms and using the fact that f isodd and 2π-periodic, we can rewrite Sn as

Sn,f (x) =1

2π

∫ π

−π

Dn(t)[f(x + t) + f(x− t)] dt.


Since the integral of Dn(t) over [−π, π] is π, we have

Sn,f (x)− f(x) =1

π

∫ π

−π

Dn(t)fx(t) dt,

fx(t) =f(x + t) + f(x− t)− 2f(x)

2. (1.20)

Now note that

Dn(t) =1

2

(cos nt + sin nt cot

t

2

);

hence

Sn,f (x)− f(x) =1

2π

∫ π

−π

fx(t) cos nt + fx(t) cott

2sin nt dt.

The first integral on the right is the Fourier cosine coefficient of thefunction fx, while the second term is the Fourier sine coefficient of

fx(t) cott

2.

Using this observation, we prove:

Theorem 1.3.1 Suppose that for fixed x ∈ (−π, π) fx(t)/t is integrableon (−π, π), i.e.

∫ π

−π

∣∣∣∣f(x + t) + f(x− t)− 2f(t)

t

∣∣∣∣ dt < +∞.

Then limn→∞

Sn(x) = f(x).

This criterion, due to Dini, holds if f is differentiable at x, but thecriterion is far less restrictive.

Proof: Since Sn − f is expressed as the Fourier coefficients of twointegrable functions, the result follows immediately from the followingresult, known as the Riemann-Lebesgue lemma for Fourier coefficients:

Lemma 1.3.2 The Fourier coefficients of a function f ∈ L1(−π, π)tend to zero with n.


−8 −6 −4 −2 0 2 4 6 8−4

−2

0

2

4

6

8

10

12

14Dirichlet function for n=4 and n=12

n=12

n=4

Figure 1.4: The Dirichlet kernel evaluated at n = 4, 12. The oscillation inthe Dirichlet kernel gives rise to the Gibbs phenomenon.

The Riemann-Lebesgue lemma in turn follows from Bessel’s in-equality for orthonormal sequences, which we now prove. For functionsf, g ∈ L2(−π, π) define

(f, g) =1

2π

∫ π

−π

f(x)g(x) dx,

The vector space L2(−π, π) is a complex inner product space, with ( , )as a dot or inner product. It is an infinite dimensional complex analogof the real Euclidean space R3. The analog of the ordinary Euclideandistance is given by

||f ||2 = (f, f).

The functions einx form an orthonormal sequence on L2(−π, π)with respect to this inner product. That is,

(einx, eimx) =1

2π

∫ π

−π

einxe−imx dx = δnm


where δn m is the Kronecker delta.Let an = (f, einx). If ||f || < +∞, then a simple computation shows

that

0 ≤∣∣∣∣∣

∣∣∣∣∣f −∑

n

aneinx

∣∣∣∣∣

∣∣∣∣∣

2

=||f ||2 − 2∑

n

|an|2 +∑

n

|an|2 = ||f ||2 −∑

n

|an|2;

hence ∑n

|an|2 ≤ ||f ||2.

This is known as Bessel’s inequality. An immediate consequence of thisinequality is that the sum of the squares of the absolute values of theFourier coefficients of an L2 function is convergent, hence

limn→∞

an = 0.

We have given the proof here for the complex exponentials einx, butthe proof for the sine functions is the same.

To extend the result to Fourier coefficients of L1 functions, that is,to functions f satisfying the weaker condition

∫ π

−π

|f | dx < +∞,

we use a fact from Lebesgue integration theory that, given any functionf ∈ L1 and given ε > 0, f can be decomposed into the sum of abounded function and a function whose L1 norm is less than ε. Thatis, f = f1 + f2, where,

||f1||∞ = supx|f1| < +∞ and

∫|f2| dx < ε.

The Fourier coefficients are correspondingly decomposed into an =an,1 + an,2, corresponding to the Fourier coefficients of f1 and f2 re-spectively.

Since f1 is bounded, it belongs to L2, and its Fourier coefficients,an,1, tend to zero by the Bessel inequality. On the other hand,

|an,2| = 1

2π

∣∣∣∫ π

−π

f2(x)einx dx∣∣∣ ≤ 1

2π

∫ π

−π

|f2(x)| dx < ε.

1.4. BROWNIAN MOTION 19

Hence

lim supn→∞

|an| ≤ ε

for any ε > 0, and the limit must in fact be zero. ¥Dini’s test gives a simple criterion for the pointwise convergence

of the partial sums of a Fourier series. For a more comprehensivediscussion of the convergence of Fourier series, see A. Zygmund [28].

The history of Jean-Joseph Fourier (1768-1830) is a fascinating one.‘Citizen’ Fourier served as Secretary of the Commission of Arts andSciences in Napolean Bonaparte’s expedition to Egypt in 1798. His tourof duty in Egypt affected his health, and when he returned to Grenoble,in the French Alps, “he covered himself with an excessive amount ofclothing even in the heat of summer, and . . . his preoccupation withheat extended to the subject of heat propagation in solid bodies, heatloss by radiation, and heat conservation.”1

Fourier’s investigations into heat culminated with the publicationof the Theorie Analytique de la Chaleur 2:

There is no doubt that today this book stands as one of the most daring,innovative, and influential works of the nineteenth century on mathematicalphysics. . . . He worked with discontinuous functions when others dealtwith continuous ones . . . and talked about the convergence of a series offunctions before there was a definition of convergence. At the end of his1811 prize essay, he even integrated ’functions’ that have value ∞ at onepoint and are zero elsewhere. . . . It was the success of Fourier’s work inapplications that made necessary a redefinition of the concept of function,the introduction of a definition of convergence, . . . the ideas of uniformcontinuity and uniform convergence. It . . . was in the background ofideas leading to measure theory, and contained the germ of the theory ofdistributions.

1.4 Brownian Motion

The phenomenon of Brownian motion is named after the biologistRobert Brown who discovered it in 1827. Tiny particles suspendedin a fluid make small irregular jumps, due to constant bombardmentby the molecules of the fluid. The phenomenon thus provides evidence

1Gonzalez-Velasco [10].2op. cit.


for the molecular theory of matter. Albert Einstein showed in his pa-per in 1905 that the probability density function for Brownian motionsatisfies the heat equation.

Brownian motion is obtained formally as a continuum limit of arandom walk on a lattice. Consider a fair coin which comes downheads or tails with probability 1/2; and consider the process of flippingthis coin successively. The tosses are independent, in the sense thatthe outcome of the nth toss is independent of the outcomes of anyof the preceding tosses. The process is represented as a sequence ofindependent random variables Xn such that

Prob (Xn = ±1) =1

2.

The distribution function of a real valued random variable X isgiven by

F (x) = Prob (X < x).

The expectation µ = EX and variance σ2 = V ar(X) of X are thendefined by the Riemann-Stieltjes integrals

µ = E(X) =

∫ ∞

−∞xdF (x), σ2 =

∫ ∞

−∞(x− µ)2dF (x).

By the definition of distribution function, F is non-decreasing in x. IfF is differentiable, with F ′ = p, then dF (x) = p(x)dx.

The joint distribution function of two random variables X and Y isdefined by

F (x, y) = Prob (X < x, Y < y).

The random variables X and Y are independent if

E(XY ) = E(X)E(Y ).

The random variables Xn corresponding to the nth coin toss areindependent and each has mean zero and variance 1. The coin tossesgenerate a random walk on the integers in which a particle is displacedto the right or left by one, according as the coin comes down heador tails. The position of the particle after n tosses is also a randomvariable,

Sn =n∑

j=1

Xj. (1.21)


Using the independence of the Xn, it is easily seen that Sn has meanzero and variance n. In fact,

E(Sn) = E(n∑

j=1

Xj) =n∑

j=1

E(Xj) = 0,

E(S2n) = E(

n∑

j,k=1

XjXk) =n∑

j,k=1

E(XjXk) =n∑

j=1

E(X2j ) = n.

A formal passage to the continuum limit is obtained by measuringtime steps in intervals of length ∆t, and spatial steps of length ∆x. Letp(x, t) be the probability that the particle is at x at time t, assumingit began at the origin at t = 0. The particle can be at x at time t + ∆tonly if it were at x − ∆x and moved to the right, or at x + ∆x andmoved to the left, at time t. Therefore

p(x, t + ∆t) =1

2(p(x−∆x, t) + p(x + ∆x, t)) . (1.22)

We rewrite this equation as

p(x, t + ∆t)− p(x, t)

∆t

=∆x2

2∆t

(p(x−∆x, t) + p(x + ∆x, t)− 2p(x, t)

∆x2

). (1.23)

Now put

σ =∆x2

∆t,

and let ∆x and ∆t tend to zero with σ fixed. If we assume that inthe limit p(x, t) is a differentiable function, and take the limit of thesedifference quotients, we obtain

∂p

∂t=

σ

2

∂2p

∂x2,

i.e. the heat equation with σ/2 as the diffusion coefficient.The constant σ is a free parameter in the mathematical theory which

must be determined from physical principles. It was first determinedby A. Einstein in his fundamental paper of 1905 [8] to be

σ =RT

N

1

6πkP,


where N is Avogadro’s number, T is the temperature in degrees Kelvin,R is the universal constant in the gas equation, k is the coefficient ofthe viscosity of the fluid, and P is the radius of the particle (assumedspherical). The determination of a number of fundamental constantsin the molecular theory of matter ha s been made from this formulafor the diffusion constant and experimental observations.

The argument above is, of course, only formal, since one must passfrom a probability function defined only at discrete points to a proba-blity density function which is defined and differentiable on all x andt > 0. It also does not tell us which solution of the heat equation weshould take for p. As it turns out, p is precisely the fundamental solu-tion of the heat equation. In fact, the fundamental solution of the heatequation is the probability density function is fundamentally related tothe Central Limit Theorem of probability theory.

A random variable X is said to be Gaussian or normal if

Prob (X < b) =

∫ b

−∞ϕ(y) dy, ϕ(y) =

1√2π

e−y2/2 dy. (1.24)

The mean and variance of X are given by

E[X] =

∫ ∞

−∞yϕ(y) dy = 0, E[X2] =

∫ ∞

−∞y2ϕ(y) dy = 1.

Theorem 1.4.1 [Central Limit Theorem]. Let Xn be a sequenceof independent, identically distributed random variables with finite ex-pectation µ and variance σ2: µ = E[Xn] and σ2 = E[(Xn − µ)2]. LetSn =

∑nj=1 Xn. Then

limn→∞

Prob

(Sn − nµ

σn1/2< b

)=

1√2π

∫ b

−∞e−y2/2 dy.

In the case of the random walk on Z, µ = 0, σ2 = 1, so the sequenceof random variables

Sn√n

tends to a normal random variable. We rescale the random walk bymaking the particle take jumps of size ∆x at time intervals of length


∆t. Then the position of the particle at time t = n∆t is given byXt = ∆xSn. Therefore,

Xt√t

=∆x√n∆t

Sn =√

σSn√n

, σ =(∆x)2

∆t.

Letting n →∞, and ∆x, ∆t → 0, with σ fixed, we see that

Xt√σt

is a Gaussian random variable. It is then a simple exercise to showthat

Prob (Xt < x) =1√

2πσt

∫ x

−∞e−y2/2σtdy.

Hence the probability density function for Brownian motion satisfiesthe heat equation with diffusion coefficient σ/2.

We may think of a solution of the heat equation as a probabilitydensity function p(x, t):

p(x, t) =1√

2πσt

∫ ∞

−∞e−(x−y)2/2σtp(y, 0) dy

where p(x, t) is the probability density of a particle moving randomly bya diffusion process, given that the initial probability density functionof the particle is p(x, 0). Since p is a probability density, we shouldexpect that p(x, t) ≥ 0 for all t > 0 and that

∫ ∞

−∞p(x, t) dx = 1

for all positive t. As we have already seen, these two properties areimmediate consequences of the heat equation.

As a special case we may consider the situation where the particleis originally at the origin with ifs, ands, or buts, i.e. with probability1. Its distribution function is then

FX(x) = H(x) =

0 x < 0;

1 x ≥ 0(1.25)


where H is the Heaviside step function. The density function δ of sucha random variable must then have the property that

∫ ε

−ε

δ(x) dx = 1

for any ε > 0. This property describes the Dirac delta function. TheDirac function is a generalized function; it is an example of a distribu-tion. Such distributions play an important role in the modern theoryof linear partial differential equations and will be discussed later.

The delta function is obtained as the distribution limit of the fun-damental solution of the heat equation as t ↓ 0:

limt↓0

e−x2/2σt

√2πσt

= δ(x).

Thus, the Dirac delta function is the “probability density” function of aparticle which sits at the origin with probability 1, and the fundamentalsolution of the heat equation is the probability density function of thisparticle for t > 0 as it moves about randomly by diffusion.

We close with a few remarks about the Riemann-Stieltjes integral.For a more complete discussion, see [9], §25. Given a monotone non-decreasing right continuous function F , the integral

∫ b

a

f(x)dF (x) (1.26)

is defined as follows: Let Pn = a = x0 < x1 < · · · < xn = b be anypartition of the interval of [a, b]; and let

∆Fj = F (xj)− F (xj−1).

Given f be defined on [a, b] and a partition Pn define

Mj = supxj−1≤ξ≤xj

f(ξ), mj = infxj−1≤ξ≤xj

f(ξ).

We then define the upper and lower sums

U(Pn, f, F ) =n∑

j=1

Mj∆Fj, L(Pn, f, F ) =n∑

j=1

mj∆Fj,


IfsupPn

La,b,f = infPn

Ua,b,f

then f is said to be integrable with respect to F and the correspondingRiemann-Stieltjes integral (1.26) is defined to be this common value.

We leave it as an exercise to show that for any continous functionf on the real line, ∫ ∞

−∞f(x)dH(x) = f(0); (1.27)

hence, formally, the derivative of the Heaviside step function is theDirac delta function.


1.5 Exercises

1. Prove theorem 1.2.1. What is the probabilistic interpretation ofthe formula (1.12) for the fundamental solution of the heat equa-tion in Rn?

2. Find the explicit solution of the heat equation on the real linefor initial data u(x, 0) = H(x), where H(x) is the Heaviside stepfunction (1.13).

3. Let both one-sided limits f(x ± 0) = limy→x±

f(y) exist at a given

point x ∈ R, and let f be bounded and measureable on R. Then

limt↓0

1√4πt

∫ ∞

−∞e−(x−y)2/4tf(y) dy =

f(x + 0) + f(x− 0)

2.

4. Show that

u(x, t) =x

t3/2e−x2/4t

satisfies the heat equation in t > 0. What is limt↓0

u(x, t)?

5. The total variation of a function f on the real line is given by thenorm

||f ||v = sup∑

j

|f(xj+1)− f(xj)|,

where the supremum is taken over all finite sequences x1 < x2 <· · · < xn. Show that the solution of the heat equation given by(1.3) satisfies

||u(·, t)||v ≤ ||f ||v.Moreover, show that if f ∈ L1(R) then ||u(·, t)||v tends to zero ast →∞, and give an estimate of the rate of decay.

6. Prove (1.19).

7. Compute the Fourier coefficients for the function (1.18).

8. Solve the boundary-initial value problem (1.15) when the bound-ary conditions on 0 and π are given by u = 0 and ux + hu = 0

1.5. EXERCISES 27

respectively, where h is a positive constant. Show the solutioncan be represented as a Fourier series

u(x, t) =∞∑

n=0

ane−γ2nt sin γnx

where γ1 < γ2 < . . . satisfy a transcendental equation. Showthat the corresponding functions sin γnx are orthogonal on theinterval [0, π], and evaluate

∫ π

0

sin2 γnx dx.

9. State and prove an analog of the maximum principle for the differ-ence equation (1.23). What is the analog of fundamental solutionfor this difference equation?

10. Given any non-decreasing function F (x), define functional ΛF (f)by

ΛF (f) = −∫ ∞

−∞f ′(x)F (x)dx,

where f is continuously differentiable for x ∈ R and f → 0 asx → ±∞. Compute ΛH , for the Heaviside step function H.

11. Compute ∫ 1

0

x dF (x),

where F is the Cantor function.


Chapter 2

Laplace’s Equation

2.1 Conservative vector fields and potentials

Let Ω be a bounded domain with a smooth boundary in Rn, and con-sider the boundary-initial value problem for the heat equation

ut = ∆u, x ∈ Ω u(x, 0) = u0, u∣∣∣∂Ω

= f.

As t → ∞ the temperature will settle down (something we need toprove mathematically) to an equilibrium solution of the heat equation:

∆u = 0, u∣∣∣∂Ω

= f.

This is called the Dirichlet problem for Laplace’s equation. Thus,Laplace’s equation may be viewed as the equilibrium solution of theheat equation.

Laplace’s equation also arises in potential theory, as follows. Avector field F is said to be conservative if the line integral

∫ a

x

F · dy

is independent of the path of integration. This line integral representsthe work done in moving from x to a fixed reference point a (for exam-ple, to infinity). In this case we define the potential energy of the fieldto be the work done and we denote its value by Φ. It follows that

∇Φ = −F.

29

30 CHAPTER 2. LAPLACE’S EQUATION

The scaler function Φ is called the potential of the vector field F.The best-known physical example of a conservative vector field is

the inverse square law governing both the (classical, or Newtonian)gravitational field and the electric field. In both theories, the force inR3 between two point masses or charges is inversely proportional tothe square of the distance between them. Thus, the force field due toa point source at the origin is given by

F(x) = −Kx

|x|3 , (2.1)

where K is a physical constant. We have chosen the minus sign hereto represent a force of attraction. In the electrostatic case (known asCoulomb’s law) like charges repel one another, and the minus sign isdropped. As the reader may easily verify, F is a conservative vectorfield for which the potential in R3 is

Φ =K

r, r = |x|.

Gauss’ law of electrostatics states that for any (smoothly bounded)domain Ω,

∫∫

∂Ω

F · dS = sum of the charges enclosed in Ω. (2.2)

For the case of a single point charge, the integral over the sphere Bρ ofradius ρ centered at the charge itself is easily computed. The outwardunit normal is ν = x/|x|, so in spherical coordinates

∫∫

Bρ

x

|x|3 · dS =

∫∫

Bρ

x

ρ3· xρ

dS =

∫ π

0

∫ 2π

0

1

ρ2ρ2 sin θdθdϕ = 4π.

As a simple computation shows, the divergence of the vector field(2.1) is zero in the region excluding the origin; hence the integral (2.2)over any smoothly bounded region containing the point source is inde-pendent of the region. This result follows from the Gauss divergencetheorem, ∫∫∫

A

div Fdx =

∫∫

∂A

F · dS,

2.1. CONSERVATIVE VECTOR FIELDS AND POTENTIALS 31

for any smoothly bounded region A in which F is continuously differ-entiable. Simply let A be the region Ω \ Bρ, where ρ is chosen so thatBρ ⊂ Ω.

The two equations

F = −∇Φ, div F = 0

lead to Laplace’s equation for Φ:

∆Φ = 0.

Solutions of Laplace’s equation are called harmonic functions and pos-sess a number of very strong mathematical properties, which we shalldiscuss in this chapter.

The function1

4πr(2.3)

is the fundamental solution of Laplace’s equation in R3. An easy cal-culation (e.g. in spherical coordinates) shows that ∆Φ = 0 everywhereexcept at the origin. Its gradient is the vector field F, the force fielddue to a point charge.

The potential due to a sum of point charges at the points x1,x2, . . .is

n∑j=1

1

4π|x− xj| ;

and, by analogy, the potential due to a continuous distribution ofsources with density ρ is

Φ(x) =

∫∫∫

Ω

ρ(y)

4π|x− y|dy.

We shall prove that Φ satisfies Poisson’s equation, ∆Φ = −ρ.

Theorem 2.1.1 Let Ω be a smoothly bounded finite domain in R3, andlet u be a C2 function in Ω. Then we have the fundamental identity

u(x) +

∫∫∫

Ω

∆u

4πrdy +

∫∫

∂Ω

(u

∂

∂ν

1

4πr− 1

4πr

∂u

∂ν

)dSy = 0, (2.4)

where r = |x − y|, ν is the outward unit normal and dSy the elementof surface area on the boundary of Ω.


Proof: We begin by stating Green’s identity of multi-variable calculus:

∫∫∫

Ω

u∆v − v∆u dx =

∫∫

∂Ω

(u

∂v

∂ν− v

∂u

∂ν

)dS, (2.5)

where Ω is any smoothly bounded domain in R3, ν denotes the outwardunit normal to the surface ∂Ω, and u and v are continuously differen-tiable in the interior of Ω. This identity is a direct consequence of theGauss divergence theorem.

Now let v = 1/4π|x−y|, Bδ be the sphere of radius δ centered at x,and Ωδ = Ω \Bδ. Both u and v being smooth inside Ωδ, we can applyGreen’s identity. Since ∆yv = 0, y 6= x (∆y denotes the Laplacian withrespect to the y variables), we have

−∫∫∫

Ωδ

∆u

4π|x− y|dy =

∫∫

∂Ωδ

(u

∂v

∂ν− v

∂u

∂ν

)dSy.

As δ → 0, the integral on the left tends to − ∫Ω

∆u/4π|x − y| dy,since 1/r is locally integrable in R3 and ∆u is continuous. In the surfaceintegral over ∂Bδ, choose spherical coordinates centered at x. Then

∂v

∂ν

∣∣∣∂Bδ

= −∂v

∂r

∣∣∣r=δ

=1

4πδ2.

(Recall that ν is the outward unit normal from the interior of Ωδ, hencethe minus sign.) Hence

∫∫∫

∂Bδ

u∂v

∂νdSy =

1

4πδ2

∫∫

∂Bδ

u dSy.

On ∂Bδ, dSy = δ2 sin θ dθ dϕ. Since u is continuous in Ω, the surfaceintegral over ∂Bδ tends to u(x) as δ → 0.

On the other hand, the integral of v∂u/∂ν over Bδ tends to zerowith δ, since on that surface ∂u/∂ν is bounded while v = O(δ−1).Letting δ → 0 we obtain (2.4). ¥

The above calculation goes through if 1/4πr is replaced by a moregeneral kernel

G(x, y) =1

4π|x− y| + h(x, y),


where h is harmonic in y for each x. If h can be constructed so thatG(x, y) = 0 for each x ∈ Ω, and y ∈ ∂Ω, then the representation (2.4)would simplify to

u(x) = −∫∫∫

Ω

G(x, y)∆u dy −∫∫

∂Ω

u∂G

∂νy

dSy; (2.6)

In that case the solution of the boundary value problem

∆u = −ρ, x ∈ Ω; u∣∣∣∂Ω

= g,

would be given by

u(x) =

∫∫∫

Ω

G(x, y)ρ(y) dy −∫∫

∂Ω

g(y)∂G

∂νy

dSy

In particular, if u is harmonic in the interior of Ω and takes values gon the boundary, then

u(x) = −∫∫

∂Ω

g(y)∂G

∂νy

dSy.

The function G is called the Green’s function for the Dirichlet problem(on the domain Ω). The Green’s function G(x, y) is harmonic in y andvanishes for y ∈ ∂Ω.

Theorem 2.1.2 The Green’s function for a smoothly bounded domainΩ is symmetric in x and y: G(x, y) = G(y, x). It follows that theGreen’s function is also harmonic in x and vanishes for x ∈ ∂Ω.

Proof: Letting x 6= x′ ∈ Ω, we apply Green’s identity to the functionsu(y) = G(x, y) and v(y) = G(x′, y). Let Bδ and B′

δ be spheres ofradius δ about the points x and x′, with δ small enough that bothspheres lie in Ω and do not intersect. Since u and v are both harmonicin Ωδ = Ω \ (Bδ ∪B′

δ), we get

∫∫

∂Ωδ

(G(x, y)

∂G(x′, y)

∂νy

−G(x′, y)∂G(x, y)

∂νy

)dSy = 0.


Since G(x, y) and G(x′, y) both vanish on ∂Ω, the integral above re-duces to an integral over ∂Bδ ∪ ∂B′

δ. Letting δ → 0 and proceeding asin the proof of Theorem 2.1.1, we obtain G(x, x′)−G(x′, x) = 0. ¥

The proof of the existence of a Green’s function for a general domainis somewhat technical, and we won’t go into that here. However, indomains with a high degree of symmetry the Green’s function can oftenbe constructed explicitly. For example, the Green’s function for the halfspace R3

+ = x : x3 > 0 can be constructed by the method of images.Given a source at y we place an image at y′ = (y1, y2,−y3). Then forx3 > 0 the function

1

|x− y′| =1√

(x1 − y1)2 + (x2 − y2)2 + (x3 + y3)2

is a harmonic function of y in y3 > 0. The Green’s function for R3+ is

then

G(x, y) =1

4π

(1

|x− y| −1

|x− y′|)

. (2.7)

This function has a fundamental singularity at x = y and vanishes onthe plane x3 = 0. We obtain a representation of the solution of theDirichlet problem by calculating the outward normal derivative of Gon x3 = 0. By (2.6), if u is harmonic in x3 > 0 and u = f on x3 = 0,then

u(x) =x3

2π

∫∫

R2

f(y1, y2) dy1 dy2((x1 − y1)2 + (x2 − y2)2 + x2

3

)3/2(2.8)

The solution is thus given by a convolution on x3 = 0 of the boundarydata f with the kernel

K(x, z) =z

2π

1

(|x|2 + z2)3/2, x ∈ R2, z > 0.

A reflection method also works to obtain the Green’s function forballs. If B is a ball of radius R centered at the origin, and y ∈ B, thenthe reflection of y in ∂B is the point

y′ =R2

|y|2y.


We leave it as an exercise to show that the Green’s function for thesphere of radius R is

G(x, y) =1

4π|x− y| −R

|y|1

4π|x− y′| . (2.9)

If u is harmonic inside a sphere of radius R and continuous onto thesphere, then in the interior u is given by the Poisson integral

u(x) =R2 − r2

4πR

∫∫

SR

u(y)

|x− y|3dSy, r2 = |x|2. (2.10)

Three fundamental properties of harmonic functions follow imme-diately from this formula:

Theorem 2.1.3 Let u be harmonic in a domain Ω. Then at everyx ∈ Ω, u is equal to its average value on any sphere Bρ(x) centered atx and contained in Ω:

u(x) =1

4πρ2

∫∫

|y−x|=ρ

u(y)ρ2dω, dω = sin θ dθ dϕ,

This is the mean value theorem for harmonic functions.The strong maximum principle for harmonic functions follows im-

mediately from the mean value theorem.

Theorem 2.1.4 If u is harmonic in a domain Ω and u attains aninterior maximum or minimum in Ω, then u is identically constant.

Proof: Suppose m ≤ u ≤ M in Ω and u = M at an interior pointx ∈ Ω, then choose a small sphere centered at x and contained in Ω. Ifu were strictly less than M anywhere on that sphere, the average valueof u over that sphere would be strictly less than M (u is continuous);and the mean value property would be violated. Hence u must beidentically equal to M everywhere on that sphere, and so u ≡ M onthe largest sphere containing x and lying in Ω.

Now propagate this extreme value of u throughout Ω, as follows.Let Mu = x : x ∈ Ω, u(x) = M. Clearly Mu is closed, since u iscontinuous; but Mu is also open by the preceding argument. Therefore,Mu is either empty or the entire set Ω. ¥

Finally, we obtain Harnack’s inequality from (2.10).


Theorem 2.1.5 Let u be a non-negative harmonic function in a ballBR of radius R. Then for x in BR

R

R + |x|R− |x|R + |x| u0 ≤ u(x) ≤ R

R− |x|R + |x|R− |x| u0,

where u0 is the value of u at the center of the sphere.

Proof: Choose coordinates so that the sphere is centered at theorigin. For any |x| = r < R and all y, |y| = R we have

1

(R + r)3≤ 1

|x− y|3 ≤1

(R− r)3.

The Poisson kernel therefore satisfies the inequality

1

4πR

R− |x|R + |x|

1

R + |x| ≤R2 − |x|2

4πR

1

|x− y|3 ≤1

4πR

R + |x|R− |x|

1

R− |x| .

Multiplying this inequality through by u(y) > 0, integrating over thesphere |y| = R, and applying the mean value theorem, we obtain theresult. ¥

All three of these theorems are valid in any dimension n ≥ 2. Themaximum principle is valid for general second order uniformly ellipticscalar partial differential equations with continuous coefficients. TheHarnack inequality can also be extended to more general second orderelliptic equations, but not in the generality of the case of constantcoefficients. (see Moser [13] and Serrin [20]).

A number of properties of harmonic functions follow from Harnack’sinequality. These are left as exercises.

Let us prove the converse of the mean value theorem:

Theorem 2.1.6 Let u be a continuous function which satisfies themean value property on every sphere S contained in a domain Ω. Thenu is harmonic in Ω.

Proof: Let S be any sphere contained in Ω. Using the Poisson integral,we construct a harmonic function v with v = u on ∂S. Both u and vsatisfy the mean value property in the interior of S, hence w = u − vsatisfies the mean value property. Therefore w satisfies the maximumprinciple inside S, and w vanishes on the boundary. Therefore w ≡ 0inside S, and u is harmonic inside S. This argument clearly appliesanywhere in Ω, so u is harmonic in Ω. ¥

2.2. POLAR AND SPHERICAL COORDINATES 37

2.2 Polar and Spherical coordinates

The Laplacian in polar coordinates is

1

r

[∂

∂r

(r∂u

∂r

)+

1

r

∂2u

∂θ2

]= 0. (2.11)

We look for a solution of this equation by the method of separation ofvariables; thus we look for a solution of the form u(r, θ) = X(r)Θ(θ).Substituting this into Laplace’s equation and repeating the same pro-cedure we used for the heat equation, we obtain the pair of equations

r

X(rX ′)′ = −Θ′′

Θ= γ2;

hence

r2X ′′ + rX ′ − γ2X = 0, Θ′′ + γ2Θ = 0.

Since Θ should be 2π periodic in θ, we must take γ to be an integern. We then obtain the solutions Xn = r±n and Θn = cos nθ, sin nθ,and the two sets of solutions

r±n(an cos nθ + bn sin nθ), n ≥ 0.

The set of solutions with negative powers of r is regular at infinity andis used when solving the Dirichlet problem in the exterior of a disk.

As in the case of the heat equation, we look for a general solutionof the boundary value problem ∆u = 0, 0 ≤ r ≤ R; u(R, θ) = f(θ),by the method of superposition, that is,

u(r, θ) =a0

2+

∞∑n=1

(an cos nθ + bn sin nθ)rn.

The trigonometric functions satisfy the orthogonality conditions

∫ 2π

0

sin nθ cos mθ dθ = 0,

∫ 2π

0

sin nθ sin mθ dθ =

∫ 2π

0

cos nθ cos mθ dθ = δnmπ.


The boundary condition u(R, θ) = f(θ) leads to a determination of theFourier coefficients

an =1

πRn

∫ 2π

0

f(θ) cos nθ dθ, bn =1

πRn

∫ 2π

0

f(θ) sin nθ dθ.

Given these values of the Fourier coefficients, we may write u as anintegral

u(r, θ) =1

π

∫ 2π

0

f(θ′)

[1

2+

∞∑n=1

( r

R

)n

cos n(θ − θ′)

]dθ′.

This series can be summed to give

u(r, θ) =R2 − r2

2π

∫ 2π

0

f(θ′)r2 + R2 − 2rR cos(θ − θ′)

dθ′. (2.12)

Spherical coordinates in R3 are given by

x1 = r sin θ cos ϕ, x2 = r sin θ sin ϕ x3 = r cos θ

and the Laplacian in these coordinates is given by

∆u =1

r2 sin θ

[∂

∂r

(r2 sin θ

∂u

∂r

)+

∂

∂θ

(sin θ

∂u

∂θ

)+

(1

sin θ

∂2u

∂ϕ2

)].

(2.13)Consider the expansion of the Coulomb potential 1/|x − y| when

the coordinate system is chosen so that y lies on the x3 axis, i.e. y =(0, 0, R). Then

1

|x− y| =1√

r2 + R2 − 2rR cos θ=

1

R

∞∑n=0

( r

R

)n

Pn(cos θ) r < R;

1

r

∞∑n=0

(R

r

)n

P−n(cos θ) r > R.

(2.14)The functions P±n are polynomials of degree n in cos θ. The poly-

nomials Pn, n ≥ 0 are called the Legendre polynomials. They satisfya second order ordinary differential equation which may be obtainedby substituting (2.14) into Laplace’s equation. We shall see that thepolynomials P−n are multiples of Pn−1.

2.2. POLAR AND SPHERICAL COORDINATES 39

The function |x−y|−1 is a harmonic function of x for fixed y, hencethe sums in (2.14) are also harmonic functions of x. Substituting theexpression rnPn(cos θ) into Laplace’s equation in spherical coordinates,we obtain the ordinary differential equation

1

sin θ

d

dθ

(sin θ

dPn

dθ

)+ n(n + 1)Pn = 0. (2.15)

Applying the same argument to r−nPn(cos θ) instead, we find that P−n

satisfies the same equation with n replaced by n− 1.Under the change of variable z = cos θ this differential equation is

transformed into

d

dz

((1− z2)

dPn

dz

)+ n(n + 1)Pn = 0. (2.16)

This equation, called Legendre’s equation, is a special case of the hy-pergeometric equation, cf. Whittaker and Watson [27].

The spherical Laplacian is the Laplacian defined on functions re-stricted to the sphere, that is, functions which are independent of r. Itis therefore the differential operator

∆u =1

r2 sin θ

[∂

∂θ

(sin θ

∂u

∂θ

)+

∂

∂ϕ

(1

sin θ

∂u

∂ϕ

)].

The eigenfunctions of the Laplacian on the sphere are non-trivial solu-tions of the equation

1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+ λY +

1

sin2 θ

∂2Y

∂ϕ2= 0.

The Legendre polynomials Pn(cos θ) constructed above are one setof solutions of this equation, with eigenvalue λn = n(n + 1). Theseeigenfunctions are axisymmetric, that is, they are invariant under rota-tions about the z−axis. We look for more general eigenfunctions of theLaplacian of the form Y (θ, ϕ) = Pm

n (cos θ)eimϕ where m is an integer.With the substitution z = cos θ that we used before, we obtain thedifferential equation

d

dz

((1− z2)

dPmn

dz

)+

(λn,m − m2

1− z2

)Pm

n = 0. (2.17)


The eigenfunctions Pmn (cos θ)eimϕ are known as the spherical har-

monics, and the polynomials Pmn are called the associated Legendre

polynomials. It turns out that the associated eigenvalues λm,n are inde-pendent of m, and are given by n(n+1). In other words, the eigenvaluesn(n + 1) are degenerate; they are of multiplicity 2n + 1, correspondingto the 2n + 1 independent functions eimϕ, m = −n, . . . , n.

Just as the functions einθ form a complete orthogonal sequenceof functions for L2(−π, π), the spherical harmonics form a completeorthogonal basis for the Hilbert space L2(S2), where S2 is the unitsphere in R3.

2.3 Analytic function theory.

A function f = f(z), where z ∈ C is a complex variable, is said to beanalytic in a domain Ω ∈ C if it is differentiable, that is, if

f ′(z0) = limz→z0

f(z)− f(z0)

z − z0

exists for z0 ∈ Ω. By definition, this limit must be the same along anypath leading to z0; in particular, we must get the same values for thederivative as z → z0 along the paths <z = <z0 and =z = =z0. Writingz = x + iy and f(z) = u(x, y) + iv(x, y) this constraint leads to theCauchy Riemann equations

ux = vy, vx = −uy. (2.18)

This is an elliptic system of partial differential equations. One mayshow readily that both u and v are harmonic.

We develop some of the basic properties of analytic functions fromthe viewpoint of elliptic partial differential equations. We begin byintroducing the variable z = x − iy. We treat z, z as independentcoordinates in the real plane. By the chain rule,

∂

∂z=

1

2

(∂

∂x+ i

∂

∂y

),

and the Cauchy Riemann equations can be written in the compact form

∂f

∂z= 0, f(z, z) = u(z, z) + iv(z, z).

2.3. ANALYTIC FUNCTION THEORY. 41

The first order partial differential operator ∂/∂z is known as the ∂operator. We will show that its fundamental solution is

1

2πiz

A general function in the plane can be written as a function ofthe two variables z and z. Suppose that ∂f = µ. We may write theGreen-Stokes theorem in complex coordinates:

∫

∂Ω

h dt =

∫∫

Ω

d(h dt) =

∫∫

Ω

∂h

∂tdt ∧ dt.

If t = x + iy, t = x− iy, then

dt ∧ dt = (dx− idy) ∧ (dx + idy) = 2idx ∧ dy.

Now

df(t, t) dt

2πi(t− z)=

∂

∂t

f(t, t)

2πi(t− z)dt ∧ dt.

Let Ωε = Ω \ Dε where Dε is the disk of radius ε centered at z ∈ Ω.Since 1/(t− z) is an analytic function of t for t 6= z,

∂

∂t

f(t, t)

2πi(t− z)=

µ(t, t)

2πi(t− z), t ∈ Ωε.

We orient the boundary of Ωε so that the interior of Ωε lies to theleft of the contour of integration. Then

1

2πi

∫

∂Ω

f(t, t)

(t− z)dt− 1

2πi

∫

∂Dε

f(t, t)

t− zdt =

1

2πi

∫∫

Ωε

µ

(t− z)dt ∧ dt,

where ∂Dε is oriented in the counterclockwise direction. Letting t =z + εeiθ, dt = iεeiθdθ, we have

1

2πi

∫

∂Dε

f(t, t)

(t− z)dt =

1

2πi

∫ 2π

0

f(z + εeiθ, z + εe−iθ)

εeiθiεeiθdθ → f(z, z),

as ε → 0.


We leave it to the reader to show that

limε→0

∫∫

Ωε

µ

t− zdt ∧ dt =

∫∫

Ω

µ

t− zdt ∧ dt.

We thus obtain

f(z, z) =1

2πi

∫

∂Ω

f(t, t)

t− zdt− 1

2πi

∫∫

Ω

∂f

t− zdt ∧ dt.

In particular, if f is analytic inside Ω then we have the Cauchy integralformula

f(z) =1

2πi

∫

∂Ω

f(t)

t− zdt.

This is the analog of the Poisson integral for harmonic functions. Itsays that a function analytic in a domain Ω is the convolution of itsboundary values with the fundamental solution of the ∂ operator onthe bound ary of Ω.

Potential theory is used as a simple model of irrotational, incom-pressible inviscid fluid flow. If ~v denotes the velocity of a fluid, thesetwo conditions are expressed mathematically by the equations

curl~v =0, irrotational

div~v =0, incompressible.

If the domain is simply connected, then curl~v = 0 implies that ~v is thegradient of a scalar function ϕ. Then

div~v = div∇ϕ = 0,

hence ϕ is harmonic. ϕ is called the velocity potential.Since the fluid is inviscid, the only boundary condition to be sat-

isfied is that the normal velocity of the fluid vanishes on any solidobstacle. Thus, ϕ satisfies the boundary value problem

∆ϕ = 0, x ∈ Ω;∂ϕ

∂ν

∣∣∣∂Ω

= 0.

Such a problem is called a Neumann problem.

2.3. ANALYTIC FUNCTION THEORY. 43

In two space dimensions, if ϕ is harmonic, we may construct asecond harmonic function ψ from the Cauchy Riemann equations: ψx =ϕy, ψy = −ϕx. These equations form a set of overdetermined equationsfor ψ; for we cannot impose the first partial derivatives of a functionψ arbitrarily: we must have ψxy = ψyx. This leads to the integrabilitycondition ϕxx + +ϕyy = 0, which is satisfied automatically since ϕis harmonic. The function ψ is single valued if the domain is simplyconnected, since

∮dψ =

∮ψx dx + ψy dy =

∮ϕy dx− ϕx dy = −

∫∫∆ϕdxdy = 0.

The function ψ, the conjugate harmonic function, is called thestream function. Its gradient is orthogonal to the velocity field; andits level curves are the streamlines of the flow, that is, the trajectoriesof the fluid particles.

The function f = ϕ + iψ is an analytic function of z = x + iy,since the Cauchy Riemann equations are satisfied. If we write the fluidvelocity as a complex quantity, v = v1 + iv2, then v = f ′(z).

Complex variable methods can be used to construct solutions of theDirichlet or Neumann problem in a variety of domains in the plane bythe method of conformal mapping. We leave it as an exercise to showthat the stream function must be constant on the boundary an obstaclein the stream. Hence a flow problem can be expressed as a Dirichletproblem for the stream function.

As an example, consider the problem of finding the flow of an in-compressible, irrotational fluid in the exterior of the unit disk |z| ≤ 1.

We begin by noting that the potential ϕ for the uniform flow ~v =(U, 0) is given by ϕ = Ux. The stream function is then given byψ = Uy and the complex velocity potential is f(z) = Uz. We leave itas an exercise to show that the conformal transformation

z 7→ w =1

2

(z +

1

z

)(2.19)

maps the exterior of the unit disk in the z-plane one-to-one and ontothe w-plane. It follows that the potential

f(z) = U

(z +

1

z

)(2.20)


is analytic in the exterior of the unit disk in the z plane and is asymp-totic to Uz at infinity. The stream function, ψ = =f(z), is easily seento be constant on the unit circle:

ψ(1, θ) = =U(eiθ + e−iθ) = = 2U cos θ = 0.

The force on an obstacle B due to the flow around it is given bythe line integral

~f = −∮

∂B

pν ds,

where p is the hydrodynamic pressure, and ν is the outward normal toB. The pressure is not obtained directly in this simple theory. Laterwe shall prove Bernouilli’s law, namely that

1

2v2 + p

is constant along streamlines, where v is the speed of the fluid, in thiscase |f ′(z)|2. This makes it possible to calculate the lift and drag onthe unit circle due to the flow (2.20).

2.4. EXERCISES 45

2.4 Exercises

1. Show that (2.9) is the Green’s function for the sphere in R3 ofradius R: Show it is harmonic in y and vanishes on |y| = R. Findthe Poisson integral (2.10) for the sphere of radius R in R3. Thatis, calculate the outward normal derivative of (2.9) on the sphere|y| = R. (Use the diagram below.)

x

ρ

y y′

rr′

θ

r2 = ρ2 + |y|2 − 2ρ|y| cos θ

2. Prove Weierstrass’ convergence theorem: If un is a sequence ofharmonic functions continuous on the closure of a domain Ω, andif un converges uniformly on ∂Ω, then un converges uniformly toa harmonic function u in the interior of Ω.

3. Prove: A harmonic function which is non-negative in Rn is con-stant.

4. If un is a monotone sequence of harmonic functions on a domainΩ, and if un(P ) converges for some fixed P ∈ Ω, then unconverges uniformly on every compact subset of Ω; and the limitfunction is itself harmonic.

5. If un is a uniformly bounded sequence of harmonic functions onΩ then the partial derivatives ∂u/∂xj are uniformly bounded oncompact subdomains of Ω. Hence un contains a subsequencethat converges uniformly on compact subdomains of Ω.

6. Prove that the Legendre polynomials Pn(z) are orthogonal forn 6= n′ on (−1, 1). Show that this orthogonality condition, plus


the fact that Pn(z) is a polynomial of degree n, determines theLegendre polynomials up to a multiplicative constant.

7. Prove that the stream function for an incompressible, irrotationalflow is constant on a fixed boundary. Using Bernouilli’s theorem,calculate the lift and drag on the unit circle due to the flow (2.20).

8. It is possible, in a multiply connected domain, for a velocity fieldto be irrotational but not have a single-valued velocity potential.Show that the function

f(z) = −U

(z +

1

z

)+

Γ

2πilog z, Γ real, (2.21)

is multiple-valued and analytic in the exterior of the unit disk,but that it defines a single-valued velocity field. Find the streamfunction and show that f(z) defines a flow in the exterior of theunit disk; that is, that the stream function is constant on theunit circle. Calculate the circulation for this flow. What is theasymptotic velocity at infinity? What is the net force on the unitdisk exerted by this flow?

9. A function u is said to be subharmonic in Ω if ∆u ≥ 0. Showthat a subharmonic function cannot attain an interior maximum.Show that if u is harmonic, then |∇u|2 is subharmonic.

Chapter 3

The Wave Equation

3.1 The wave equation in 1,2,3 dimensions

The initial value problem for the wave equation in n space dimensionsis given by

¤ u = 0, u(x, 0) = u0(x) ut(x, 0) = u1(x). (3.1)

where ¤ denotes the D’Alembertian ¤u = utt − c2∆ u.The solution of (3.1) in R1 is due to D’Alembert. It is apparent

that the one dimensional wave equation is satisfied by any function ofthe form f(x + ct) + g(x− ct). Any function of the variable x− ct willappear as a wave-form travelling to the right with speed c; while anyfunction of x+ct will appear as a wave travelling to the left with speedc. Thus the solution of the wave equation appears as a superpositionof waves travelling to the left and right with speed c.

The initial conditions lead to the equations

f + g = u0, f ′ − g′ =1

cu1.

We solve this pair of equations to obtain

f ′ =1

2

(u′0 +

1

cu1

), g′ =

1

2

(u′0 −

1

cu1

).

As solutions to this pair of equations we take

f(x) =1

2u0(x) +

1

c

x∫

0

u1(y) dy, g(x) =1

2u0(x)− 1

c

x∫

0

u1(y) dy.

47

48 CHAPTER 3. THE WAVE EQUATION

We then obtain D’Alembert’s solution of the wave equation in one spacedimension:

u(x, t) =u0(x + ct) + u0(x− ct)

2+

1

2c

x+ct∫

x−ct

u1(y) dy. (3.2)

We see that the solution at (x, t) depends only on the values of u0 atx± ct and on the values of ut(x, 0) on the interval (x− ct, x + ct). Theregion in the triangle in Figure 3.1 is called the domain of dependenceof the point (x, t).

(x, t)

x + c tx− c t

x

t

Figure 3.1: Domain of Dependence

The fundamental solution for the wave equation in Rn is definedto be the solution Wn of the wave equation which satisfies the initialvalue problem

¤Wn = 0, Wn(x, 0) = 0,∂Wn

∂t(x, 0) = δ(x).

Equivalently, the function v = Wn ∗ f satisfies

¤v = 0, v(x, 0) = 0, vt(x, 0) = f(x). (3.3)

From (3.2) we see that

W1(x, t) =1

2c

x+ct∫

x−ct

δ(y) dy =H(x + ct)−H(x− ct)

2c,

3.1. THE WAVE EQUATION IN 1,2,3 DIMENSIONS 49

where H is the Heaviside step function

H(s) =

1 s > 012

s = 0

0 s < 0

We leave it as an exercise to show that for t > 0, W1 can be written inthe compact form

W1(x, t) =H(c2t2 − x2)

2c, t > 0. (3.4)

x

t

Figure 3.2: Forward and Backward Light Cone

The shaded area in Figure (3.2) is called the domain of influenceof the pulse. It extends forward and backward in time since the waveequation is invariant under time reversal. Unlike the heat equation,disturbances propagate at a finite speed.

Since the wave equation has constant coefficients, we should expectthat ∂W1/∂t is also a solution of the wave equation. We have already


observed that the distributional derivative of the Heaviside function isthe Dirac delta function. We leave it as an exercise to verify that

∂W1

∂t(x, 0) =

δ(x + ct) + δ(x− ct)

2, (3.5)

and that the solution of (3.1) in R1 can be written as u = W1 ∗ u1 +W1,t ∗ u0.

A representation of the explicit solution in R3 was obtained by Kir-choff, using the method of spherical means. We verify Kirchoff’ssolution here; and later, in §3.4, we present a method for finding itdeductively. The spherical mean of a function f in R3 is given by

Mf (x, t) =1

4πc2t2

∫∫

|x−y|=ct

f(y) dSy.

For any f ∈ C2(R3) consider the function

u(x, t) = 4πtMf =1

c2t

∫∫

|y−x|=ct

f(y) dSy.

We write y = x+ctω and dSy = c2t2dω, where ω is a vector on the unitsphere and dω denotes the element of surface area on the unit sphere.Then

u(x, t) = t

∫∫

|ω|=1

f(x + ctω) dω, (3.6)

and

∂u

∂t=

∫∫

|ω|=1

f(x + ctω) dω + ct

∫∫

|ω|=1

3∑j=1

∂f

∂yj

ωj dω =1

t(u + cW ) ,

where

W = t2∫∫

|ω|=1

3∑j=1

∂f

∂yj

ωj dω =1

c2

∫∫

|y−x|=t

∂f

∂νdSy.

It then follows that∂2u

∂t2=

c

t

∂W

∂t.


By the divergence theorem,

W =1

c2

∫∫∫

|x−y|<ct

∆yf(y)dy =1

c2

ct∫

0

∫∫

|ω|=1

∆xf(x + ρω) dω ρ2 dρ

=∆x1

c2

ct∫

0

∫∫

|ω|=1

f(x + ω ρ) dω ρ2 dρ;

hence∂W

∂t= ct2∆x

∫∫

|ω|=1

f(x + ctω) dω,

and

∂2u

∂t2= ∆x

c2t

∫

|ω|=1

f(x + ctω) dω

= c2∆u.

By (3.6), u(x, 0) = 0, while

ut(x, 0) =

∫∫

|ω|=1

f(x)dω = 4πf(x).

The function v(x, t) = (tMg)t also satisfies the wave equation, (pro-vided g ∈ C3(R3)) since the wave equation has constant coefficients. Itis clear that v(x, 0) = 0; moreover,

vt(x, t) =∂2

∂t2tMg = c2∆tMg = tc2∆Mg,

hence vt = 0 at t = 0. We thus see that

u(x, t) = tMu1 + (tMu0)t (3.7)

is a solution of (3.1) in R3.


t

|x− y| = t

(x, t)

t

R3

Figure 3.3: Backward light cone in R3

We leave it to the reader to verify that for t > 0, tMf = W3 ∗ f ,where

W3(r, t) =δ(r − ct)

4πrc; (3.8)

hence W3 is the fundamental solution of the wave equation in R3.The fundamental property of the wave equation in three space

dimensions known as Huyghen’s principle follows immediately fromKirchhoff’s representation. The spherical mean Mf (x, t) depends onlyon the values of f on the sphere of radius ct centered at x; while(tMg)t depends only on the values the normal derivative of g on thissame sphere. The domain of dependence of the point (x, t) is thereforethe backward light cone |x − y| = ct. This means that an observer atthe origin observes a signal at time t only when the source of the signalis precisely at a distance ct away.

The situation in R2 is quite different. The solution of the waveequation in two space dimensions was originally obtained by Poisson;but it can be obtained very quickly from the three dimensional solutionby the so-called method of descent, due to Hadamard, as follows.

A solution of the wave equation in R2 is regarded as a special case


of the solution in three dimensions for which the initial data is inde-pendent of one of the spatial variables, say x3 = z. The integration iscarried out in cylindrical coordinates:

y1 = ρ cos θ, y2 = ρ sin θ, y3 = z =√

c2t2 − ρ2.

The element of surface area is

dS =√

1 + (∇z)2 ρ dρ dθ =√

1 + z2ρ ρ dρ dθ.

Since zρ = −ρ/z, we have

dS =

√1 +

ρ2

z2ρ dρ dθ =

1

z

√z2 + ρ2ρ dρ dθ =

ct

zρ dρ dθ.

The upper and lower hemispheres make equal contributions, so tMf

reduces to the integral

21

4πc2t

∫ ct

0

∫ 2π

0

f(x1 + ρ cos θ, x2 + ρ sin θ)√c2t2 − ρ2

ct ρ dρ dθ.

This integral is a convolution W2 ∗ f , where

W2 =

1

2πc√

c2t2 − r2, r2 < c2t2;

0 r2 > c2t2.

It can be simplified to

W2 ∗ f =1

2πc

∫ ct

0

F (x; r)r√

c2t2 − r2dr,

where

F (x; r) =

∫ 2π

o

f(x1 + r cos θ, x2 + r sin θ)dθ.

Finally, we may write W2 in the compact form

W2 =1

2πc√

σ+

, σ = c2t2 − r2 (3.9)

where σ+ denotes the positive part of σ, defined to be σ when σ > 0and 0 when σ < 0.


Let us verify that the convolution integral satisfies the initial con-ditions (

W2 ∗ f)

t(x, 0) = f(x).

Under the change of variables r = ρct the integral above becomes

t

2π

∫ 1

0

ρ√1− ρ2

F (x; ρct)dρ

=tF (x; 0)

2π

∫ 1

0

ρ√1− ρ2

dρ + O(t2) = tf(x) + O(t2),

The derivative of this integral at t = 0 is therefore f(x).Note that the domain of dependence of the point (x, t) in 2 space

dimensions is the entire region |y−x| ≤ ct! Thus, in three dimensions,a flash of light, or a thunderclap immediately passes by the observer,while in two dimensions, the disturbance persists forever. For example,consider the result of dropping a stone into a calm (infinite, for the sakeof this discussion) pool of water. The ripples persist forever, althoughthey do decay.

3.2 Characteristic curves in the plane.

Consider the simple equation

ut + ux = 0, (3.10)

whose general solution is u(x, t) = f(x−t). Any solution of this partialdifferential equation is constant along the lines x− t =constant. Theselines are called the characteristics of the equation. We see that u cannotbe specified arbitrarily along a characteristic.

Moreover, we can formally extend the notion of solution of (3.10) toinclude any function of x − t, even a discontinuous one. For example,we might consider the “solution” u(x, t) = H(t − x), where H is theHeaviside step function. This solution has a jump discontinuity acrossthe characteristic x = t.

Even though the function H(t − x) is discontinuous, it satisfies(3.10) in the following weak sense. For any given domain Ω in the

3.2. CHARACTERISTIC CURVES IN THE PLANE. 55

plane, multiply (3.10) by a smooth function ϕ with compact supportin Ω, (for example, ϕ ∈ C1

0(Ω)), and integrate by parts. We get

∫∫

Ω

u(ϕx + ϕt) dx dt = 0, ∀ϕ ∈ C10(Ω). (3.11)

Conversely, if (3.11) holds on some domain Ω and u ∈ C1(Ω) then,integrating by parts, we find that

∫∫

Ω

(ut + ux)ϕdx dt = 0 ∀ϕ ∈ C10(Ω).

It follows that ut + ux must vanish everywhere in Ω, i.e. that (3.10)holds in Ω.

Thus we see that the characteristics for (3.10) have the followingtwo properties:

• A solution cannot be specified arbitrarily along a characteristic;

• A solution may have jump discontinuities across a characteristic.

The notion of a jump discontinuity requires the concept of a weaksolution of the partial differential equation.

Theorem 3.2.1 Let f(ξ) be a piecewise differentiable function with ajump discontinuity [f ] at ξ = 0, i.e.

[f ] = limξ→0+

f(ξ)− limξ→0−

f(ξ).

Then u(x, t) = f(x− t) is a weak solution of (3.10); that is, it satisfies(3.11).

Proof: Let Ω be any domain in t > 0. It is clear that u satisfies(3.10) in the strong sense in either of the domains x < t or x > t;hence (3.11) holds whenever Ω lies on one side of the line Γ := x = t.Therefore, let Ω be any domain which intersects the line x = t, and letΩ1 = Ω∩x < t and Ω2 = Ω∩x > t. We apply Green’s theorem toeach of the domains Ω1 and Ω2 separately. (See Figure (3.2).) In each


Ω1

Ω2

Γ

Figure 3.4: Jump conditions across a characteristic.

subdomain, u satisfies (3.10) in the strong sense. Since ϕ vanishes on∂Ω we get∫∫

Ω

u(ϕx + ϕt) dxdt =

∫∫

Ω1∪Ω2

(uϕ)x + (uϕ)t dxdt =

∫

Γ

ϕ[f ](dt− dx).

Since dx/dt = 1 on Γ, the one-form dt− dx = 0 on Γ. ¥A non-characteristic curve is any curve Γ which is never tangential

to any characteristic; in this case, the slope of Γ is never equal to 1.If u satisfies (3.10) and is specified along a non-characteristic curve Γ,then u is determined uniquely everywhere in the plane.

More generally, we may consider the equation

ut + c(x, t)ux = 0, (3.12)

in which we assume that c is a C1 function. In this case the character-istics are given as solutions of the ordinary differential equation

dx

dt= c(x, t), x(s, 0) = s.

Thus x(s, t) is parametrized by the time variable t and the point atwhich the curve intersects the x axis. Along any such curve, u isconstant, for

d

dtu(x(s, t), t) =

∂u

∂t+

∂u

∂x

dx

dt= ut + cux = 0.


Now consider the wave equation in one dimension, (3.1). We haveseen that its general solution is u(x, t) = f(x− t)+g(x+ t). 6mm 6mm

u(x, x) = u0(x)

u(x,−x) = u1(x)

Figure 3.5: The Goursat problem

We want to solve the wave equation in the region t2 > x2 with dataspecified along the characteristics x = ±t:

u(x, x) = u0(x), u(x,−x) = u1(x),

where u0 and u1 are given. We assume that u0 and u1 are C2 and thatu1(0) = u0(0) = a0, so that the boundary data is continuous at theorigin. Thus, we obtain

u(x, x) = f(0) + g(2x) =u0(x),

u(x,−x) = g(0) + f(2x) =u1(x).

Setting x = 0 in these two equations, we see that we must have

f(0) + g(0) = u0(0) = u1(0) = u(0, 0).

We then find

g(x) = u0

(x

2

)− f(0), f(x) = u1

(x

2

)− g(0),

hence

u(x, t) = u1

(x− t

2

)+ u0

(x + t

2

)− u(0, 0).


Γ := (x(s), y(s))

τ = (x′, y′)

ν = (−y′, x′)

Figure 3.6: Conditions along a non-characteristic curve

This is called the characteristic, or Goursat problem.We cannot, however, specify both u and ut on a characteristic, for

this would lead to an overdetermination of one of the functions f or g.For example, if we try to specify both u and ut on the line x = t, weare led to

u(x, x) = f(0) + g(2x), ut(x, x) = g′(2x)− f ′(0);

and now both g(2x) and g′(2x) are determined. This is in generalimpossible.

With this example in mind, let us consider a general second orderequation in two variables:

A(x, y)uxx + 2B(x, y)uxy + C(x, y)uyy = 0, (x, y) ∈ Ω.

Let Γ be a C2 curve lying in Ω, and suppose we try to specify bothu and its normal derivative uν on Γ. Specifying u on Γ determinesits tangential derivative uτ along the curve. If Γ is parametrized byx(s), y(s), where s is the arc length along the curve, then

uτ (s) =d

dsu(x(s), y(s)) = uxx

′ + uyy′,

uν(s) =∇u · ν = −uxy′ + uyx

′,

where x′ = dx/ds, etc, are both known functions on Γ. Since s is thearc length, the matrix (

x′ y′

−y′ x′

)

has determinant 1 and is invertible; hence both ux and uy are uniquelydetermined along Γ by the values of u and uν along the curve.


We can therefore compute the tangential derivatives of ux and uy

along Γ. These, along with the partial differential equation itself, givea set of equations for the second derivatives of u:

x′uxx + y′uxy =dux

dτ

x′uxy + y′uyy =duy

dτ

Auxx + 2Buxy + Cuyy = 0.

These equations for uxx, uxy and uyy are solvable provided the deter-minant of the 3× 3 matrix of coefficients does not vanish. In that casethe curve Γ is called non-characteristic, and u and a non-tangentialderivative can be prescribed independently along the curve.

The characteristics are determined by the vanishing of the determi-nant of the 3× 3 system above, that is, by the condition

Ay′2 − 2Bx′y′ + Cx′2 = 0.

This equation has two real solutions iff B2 − AC > 0. Using

dy

dx=

y′

x′

we can write the equation as

A

(dy

dx

)2

− 2Bdy

dx+ C = 0,

dy

dx= B ±

√B2 − AC. (3.13)

We thus obtain a pair of ordinary differential equations for the char-acteristics. For example, in the case of the standard wave equation,B = 0 and AC = −1; so the characteristics are given as solutions ofthe ordinary differential equations

dy

dx= ±1.

It is impossible to prescribe, independently, both the function anda non-tangential derivative along the curve, since these are in generalincompatible with the differential equation.

Second order equations in two variables with two real character-istics are called hyperbolic equations. An equation is called elliptic ifit has no real characteristics, and parabolic if it has precisely one realcharacteristic. Laplace’s is elliptic, and the heat equation is parabolic.


3.3 Characteristic surfaces in higher dimensions.

The determination of characteristic surfaces in higher dimensions issomewhat more complicated, but proceeds from the same notion. Con-sider a general second order partial differential equation

L[u] =n∑

j,k=1

gjkujk +n∑

j=1

bjuj + cu = 0 (3.14)

where uj = ∂u/∂xj, etc. We assume throughout the discussion that thecoefficients gjk . . . are smooth functions of x in some domain Ω ∈ Rn.

A smooth hypersurface C of codimension 1 lying in Ω is said tobe non-characteristic for L if for any solution u defined in some neigh-borhood of C, both u and its normal derivative uν can be specifiedindependently on C. Otherwise C is said to be a characteristic surfacefor the operator L.

We assume C is given as the level set ϕ(x1, . . . , xn) = 0 of somedifferentiable function ϕ, with ∇ϕ 6= 0 in a neighborhood of C. Givenu on C we can compute its n − 1 independent tangential derivatives,denoted by τ1u, . . . τn−1u, on C. These, together with uν , uniquelydetermine all n first order derivatives u1, . . . , un on C.

Let us first consider the case in which ϕ = xn and ask when xn = 0is a characteristic surface. Write the equation as

L[u] = gnnunn +n∑

j=1

n−1∑

k=1

gjkujk +n∑

j=1

bjuj + cu.

Since u and un are given on xn = 0, the first derivatives uj and the sec-ond order derivatives ujk, j = 1, . . . , n−1, k = 1, . . . , n are determined.Hence all terms in L[u] are determined except gnnunn.

If gnn = 0 on xn = 0, then (3.14) puts an additional constrainton u and un; i.e. the system is overdetermined, and u, un cannot bespecified independently. On the other hand, if gnn 6= 0 on xn = 0, thenu, un can be specified independently, and unn can be determined from(3.14).

Thus, the hypersurface xn = 0 is characteristic if and only if gnn ≡ 0there.

We now turn to the general case. Let the surface C be given byϕ = 0 and make a smooth, invertible coordinate transformation

yn = ϕ(x1, . . . xn); yj = yj(x1, . . . , xn), j = 1, . . . , n− 1.

3.3. CHARACTERISTIC SURFACES IN HIGHER DIMENSIONS. 61

The partial differential equation transforms as follows. Define w byw(y) = u(x). Then

uj = wl∂yl

∂xj

ujk = wlm∂yl

∂xj

∂ym

∂xk

+ wl∂2yl

∂xj∂xk

,

andn∑

j,k=1

gjkujk =n∑

l,m=1

glmwlm + . . .

where . . . denotes terms depending on w1, . . . wn, and

glm =n∑

j,k=1

gjk∂yl

∂xj

∂ym

∂xk

.

Since the coordinate transformation x 7→ y = y(x) is invertible,the wj may be computed as functions of the uj. Thus, the conditionthat ϕ = 0 be a characteristic surface is transformed into the equationgnn = 0, i.e.

Theorem 3.3.1 The necessary and sufficient condition that the sur-face ϕ = 0 be a characteristic surface for the second order partial dif-ferential equation (3.14) is

n∑

j,k=1

gjk∂ϕ

∂xj

∂ϕ

∂xk

= 0. (3.15)

If ϕ = 0 is non-characteristic, then the second normal derivative uνν

can be determined from the data u, uν on C and the equation L[u] = 0in a neighorhood of C.

In the case of Laplace’s equation, (3.15) becomes

(∇ϕ)2 = 0,

which has no real non-trivial solutions. Laplace’s equation is an exam-ple of an elliptic partial differential equation. It has no real character-istic surfaces.

However, (3.15) for the wave equation, with variables x, y, z, t, is

ϕ2t − c2(∇ϕ)2 = 0,


and this equation has the real non-trivial solution

σ = c2t2 − r2, r2 = x2 + y2 + z2. (3.16)

The level surfaces of σ are cones in four-dimensional space-time, calledthe light cones. These are the characteristic surfaces for the waveequation.

3.4 The wave equation in Rn.

We saw in the first two chapters that the heat and Laplace equationshave a smoothing property, in that solutions in the interior of the do-main are highly differentiable even though the boundary data may notbe differentiable. The situation is quite different for the wave equation,in that the solutions of the wave equation lose regularity. Even if theinitial data for the wave equation is C2, the solution may not be differ-entiable for t > 0. This loss of regularity increases with dimension, andone is forced to consider solutions in the class of distributions, evenfor smooth initial data. The solution of the wave equation in higherspace dimensions leads to a class of divergent integrals whose correctinterpretation, by Hadamard [11], was one of the factors leading tothe development of distribution theory1.

We have seen that the fundamental solution of Laplace’s equationin n dimensions is a scalar multiple of r−n. The quantity r is invariantunder the group of rotations about the origin, which also leaves theLaplacian invariant. The wave equation is invariant under the Lorentztransformations, the group of all linear transformations of the variablesx, y, z, t which leave the quadratic form (3.16) invariant. We thereforelook for a solution in the form W = Θ(σ).

We have

¤ W =∂

∂t

(2c2tΘ′)− c2

n∑j=1

∂

∂xj

(− 2xjΘ′) = 4c2

(σΘ′′ +

n + 1

2Θ′

).

We seek the fundamental solution of the wave equation as a solutionof the differential equation

σΘ′′ +n + 1

2Θ′ = 0 (3.17)

1The results were originally presented in series of lectures at Yale University in 1921,under the auspices of the Silliman Foundation.

3.4. THE WAVE EQUATION IN RN . 63

By direct integration, we obtain the solutions

Θn = Anσ1−n

2 .

When n = 2, this is the correct choice; however, when n = 3 we havealready seen that the fundamental solution is a delta function, so wemust be prepared to consider distribution solutions of (3.17). This isthe pattern for even and odd n. We shall see that there is a simplerecursion relation for obtaining the fundamental solutions in higherdimensions.

A function of x ∈ Rn is said to be homogeneous of degree α iff(λx) = λαf(x). Differentiating this identity with respect to λ andsetting λ = 1, we obtain Euler’s equation

n∑j=1

xj∂f

∂xj

= αf.

Therefore (3.17) is satisfied by functions Θ for which Θ′ is homogeneousof degree −(n+1)/2, and hence Θn is homogeneous of degree (1−n)/2.By differentiating (3.17) with respect to σ we obtain the equation forΘn+2. Thus, we may consider a recursion argument of the form Θn+2 =CnDΘn for some constant Cn, where D = d/dσ.

The fundamental solution of the wave equation in R1, given by(3.4), is homogeneous of degree 0, as we should expect. The followingtheorem can be proved by the same argument used in the proof oftheorem 1.2.1.

Theorem 3.4.1 The fundamental solution of the wave equation in nspace dimensions is homogeneous of degree 1− n in x1, . . . , xn, t.

According to the proposed recursion argument, we should expectW3(σ) to be a multiple of W ′

1(σ). By (3.28) in the exercises below, wehave, for t > 0, (δ is even)

δ(σ) = δ(r2 − c2t2) =δ(r − ct)

2r; (3.18)

hence, by (3.8), W3(σ) = π−1W ′1(σ). The fundamental solution for

the wave equation in all odd dimensions n is obtained by subsequentdifferentiation, and in fact, we shall see that

W2m+1(σ) =

(1

π

d

dσ

)m

W1 =1

2πmcδ[m−1](σ), m = 0, 1, . . . . (3.19)


We shall leave it as exercises to show that

W2m+1 ∗ f =1

(2π)m2c2t

[(d

dr

1

r

)m−1

r2mF (x; r)

]

r=ct

(3.20)

F (x; r) =

∫∫

|ω|=1

f(x + ωr)dω; (3.21)

and that∂

∂t(W2m+1 ∗ f)

∣∣∣t=0

= f. (3.22)

The solution of the wave equation in n = 2m+1 space dimensions thusinvolves derivatives of the initial data at the backward light cone up toorder m− 1.

The solution of the wave equation in even space dimensions raisessome new issues. We may formally raise the dimension by one bytaking a derivative of order 1/2. The fractional derivative of orderα, 0 < α < 1, is defined by2

Dαf =1

Γ(1− α)

d

dx

∫ x

0

(x− y)−αf(y) dy.

Observe that D1/2 reduces the homogeneity of a function by 1/2. Let usassume that the fractional derivative can be extended to distributions,and that Θn is a distribution homogeneous of degree (1− n)/2. ThenD1/2Θn is homogeneous of degree

1− n

2− 1 =

1− (n + 1)

2.

Thus D1/2Θn is a candidate, up to a scalar multiple, for Θn+1. Infact, we may obtain the fundamental solution of the wave equation in

2The justification for this nomenclature is obtained by taking the Laplace transformof this expression. Recall that if

L(f)(s) =

∫ ∞

0

e−stf(t)dt

is the Laplace transform of f then L(Df) = sL(f), provided that f(0) = 0. Using theconvolution theorem for the Laplace transform, one finds that L(Dαf)(s) = sαL(f)(s).


R2 by computing D1/2H(σ). First note that D1/2H(σ) has support onσ > 0 since H(σ) does. Therefore

1

Γ(1/2)

d

dσ

∫ σ

0

H(y)

2cdy =

1

2√

πc

d

dσ

∫ σ

0

dy√σ − y

=1

2√

πc

1√σ+

,

where σ+ is the positive part of σ. Hence W2(σ) = π−1/2W ′1(σ). In

view of (3.19) we might expect that

W2m =

(1

π

d

dσ

)m−1

W2 =

(1

π

d

dσ

)m−1/2

W1.

This is in fact true; and moreover, the results in even and odddimensions can be combined into a single formula.

Theorem 3.4.2 The fundamental solution for the wave equation in ndimensions is given by

Wn =

(1

π

d

dσ

)n−12

W1. (3.23)

We have already shown that the distribution in (3.23) is a solutionof the wave equation in Rn; so all that is needed is to verify that thisdistribution gives the correct solution to (3.1). This task has been leftas an exercise.

For n = 4 we obtain W4 = −σ−3/2+ /2π, which leads to the divergent

integral ∫ ct

0

∫∫∫

|ω|=1

f(x + ωr)

(c2t2 − r2)3/2r3dr dω. (3.24)

A good deal of effort was expended in the latter part of the 19th centuryin trying to regularize such singular integrals. The ultimate resolutionof the issue was obtained by Hadamard. He pointed out that in somecases such integrals can be regularized by various devices, but in the endhe comments that they would not be of interest to us, as – paradoxicalas it may seem – our proposed method will consist in not avoiding them.

Hadamard thus recognized that the integrals obtained are not tobe regularized but must instead be reinterpreted. Classical mathemat-ics had broken down for the wave equation in higher dimensions, andthe divergent integrals signified that something new was afoot; in fact,


Hadamard’s bold solution of the problem was a major impetus in thedevelopment of the modern theory of distributions. The wave equationdoes not have classical solutions unless the initial data is sufficientlyregular. Rather, it has the property that its solutions lose regularity,the loss of derivatives increasing with dimension. The solutions ob-tained are distributions rather than classical functions, for which thedivergent integrals are symbols.

Hadamard introduced what he called the ”finite part” of an integralsuch as (3.24). To illustrate the idea with a simple example, considerthe divergent integral ∫ ∞

0

y−3/2f(y) dy

for a smooth function f with compact support on the positive real line.We may try to interpret this integral by replacing the lower limit byε > 0 and letting ε → 0+. We have

∫ ∞

ε

y−3/2f(y) dy =

∫ ∞

ε

y−3/2(f(y)− f(0)) dy + f(0)

∫ ∞

ε

y−3/2 dy

=

∫ ∞

ε

y−3/2(f(y)− f(0)) dy + 2f(0)√

ε.

If |f(y) − f(0)| ≤ K|y|α, with α > 1/2, then the first term has awell-defined limit as ε → 0+. The second term, however, is unboundedas ε → 0 whenever f(0) 6= 0, and Hadamard simply threw this termaway, defining the integral to be given by its finite part3:

p.f.

∫ ∞

0

y−3/2f(y) dy =

∫ ∞

0

y−3/2(f(y)− f(0)) dy.

Hadamard constructed an elaborate calculus of such divergent inte-grals, extending the theory to multiple dimensions, and applying it tothe wave equation.

In the modern theory of distributions, Hadamard’s device is re-placed by writing x−3/2 as the derivative of −2x−1/2, and integrating

3Hadamard’s lectures were written in English, quite good English, as a matter of fact;they were later translated into French by Mlle J. Hadamard.


the improper integral by parts to obtain

p.f.

∫ ∞

0

y−3/2f(y) dy =

∫ ∞

0

y−3/2(f(y)− f(0)) dy

=− 2(y−1/2(f(y)− f(0))

)∣∣∣∞

0+ 2

∫ ∞

0

1√yf ′(y) dy

=2

∫ ∞

0

1√yf ′(y) dy.

Let us return to our discussion of the wave equation in even spacedimensions. The solution of the wave equation in 2m dimensions isformally represented by the divergent integral

∫ ct

0

[(1

π

d

dσ

)m−1

W2

]F (x; r)r2m−1dr.

As we saw above, this integral is interpreted in the sense of distributionsby formally integrating by parts and ignoring infinite terms.

For constant t,d

dσ= − 1

2r

d

dr, (3.25)

hence the integral above is interpreted as

1

(2π)mc

∫ ct

0

[(−1

r

d

dr

)m−11

σ+

]F (x; r)r2m−1dr

=1

(2π)mc

∫ ct

0

1

σ+

[(1

r

d

dr

)m−1

F (x; r)r2m−1

]dr. (3.26)

For initial data of class Cm−1 the expression (3.26) is an ordinary im-proper integral.

We leave it as an exercise to verify that (3.26) satisfies the appro-priate initial conditions. The formula for the surface area of the unitsphere in Rn, denoted by ωn, will be needed; it is given by

ωn = A(Sn−1) =2πn/2

Γ(n/2). (3.27)


3.5 Exercises

1. Let x0 = ct, x1 = x, x2 = y, x3 = z; we write xµ = (x0, x1, x2, x3),and x is called a 4-vector. Find a matrix g such that σ = gµνxµxν .Define an inner product of 4-vectors by x · y = gµνxµyν . DenoteLorentz transformations by x′ = Λx, where x and x′ are 4-vectors.The Lorentz group is the group of all 4× 4 matrices Λ which pre-serve the inner product, i.e. x ·y = x′ ·y′. Let the D’Alembertiansin the two coordinate systems be denoted by ¤ and ¤′. Provethat ¤ = ¤′, hence the wave equation is invariant under Lorentztransformations.

2. Prove the following statements:

(a) the delta function on the line is homogeneous of degree -1;

(b) the delta function is homogeneous of degree −n in Rn;

(c) the derivative of a homogeneous function of degree α is ho-mogeneous of degree α− 1;

(d) ∫ ∞

0

f(x)δ(x)dx =1

2f(0).

Find the delta function in spherical coordinates in Rn. Hint:Represent the delta function as the limit

δ(x) = limε↓0

e−x2/4ε

√4πε

.

3. Let f(x) be a smooth function such that f(x0) = 0, f ′(x0) 6= 0.Show that in the vicinity of x0,

δ(f(x)) = δ(x− x0)/|f ′(x0)|. (3.28)

Use this result to establish (3.5) and (3.8)

4. Derive (3.20) using (3.25). Verify that the solution v given byintegrals (3.20) and (3.26) satisfies the initial conditions v(x, 0) =0, vt(x, 0) = f(x).

3.5. EXERCISES 69

Solution: For n = 2m + 1, first prove that

(d

dr

1

r

)m−1

r2m =(2m− 1) · · · 3r2 =2mΓ(m + 1/2)√

πr2

=(2π)mΓ(m + 1/2)

πm+1/2r2 =

2(2π)m

ωn

r2.

Then

W2m+1 ∗ f =1

(2π)m2c2t

(d

dr

1

r

)m−1

r2mF (x; r)∣∣∣r=ct

=t

ωn

F (x; ct) + O(t2),

and the result follows. For n = 2m we have

(d

dr

1

r

)m−1

r2m−1 = 2m−1(m− 1)!r = 2m−1Γ(m)r,

hence by (3.27)

1

(2π)mc2m−1Γ(m)

∫ ct

0

r√c2t2 − r2

F (x; r)dr =t

ω2m

F (x; ct)+O(t2).

5. Given the boundary-initial value problem

utt −∆u + h(u) = 0, x ∈ Ω,

u(x, 0) = f(x), ut(x, 0) = g(x), u(·, t)∣∣∣x∈∂Ω

= 0,

prove that the energy

E =

∫∫∫

Ω

1

2(u2

t +∇u2) + H(u) dx, H(u) =

u∫

0

h(s) ds

is conserved.


6. Let u satisfy the wave equation in R3 with c = 1, and let

E(0) =1

2

∫∫∫

|x−y|=≤R

u2t + (∇u)2 dx

∣∣∣t=0

;

E(t) =1

2

∫∫∫

|x−y|+t=≤R

u2t + (∇u)2 dx,

Prove that E(t) ≤ E0. Use this to prove the following uniquenesstheorem for the wave equation: If u is a C2 solution of the waveequation in R3 which vanishes in the ball |x| ≤ R at time t = 0then u vanishes identically in the solid cone |x|+ t ≤ R.

|x| = R

|x|+ t = R

Figure 3.7: Backward ray cone in three dimensions

7. Assuming C∞ initial data, solve the equation

ut + uux = 0

3.5. EXERCISES 71

Ω

Γ

u+u−

Figure 3.8: Shock discontinuity of a nonlinear hyperbolic equation

by the method of characteristics. Show that, unless the initialdata is monotone, discontinuities in the solution must develop.Write the equation in weak form. Suppose that a weak solutionexists which is discontinuous along a smooth curve Γ in the x-tplane given by x = x(t), as in Figure (7). Suppose the limits ofu on either side of Γ are denoted by u+ and u−. Show that thespeed of the ‘shock’ Γ is given by

x =u+ + u−

2.


Chapter 4

Equations of Fluid and GasDynamics

4.1 Conservation Laws

We derive in this section the equations of fluid mechanics and a few ofthe associated conservation laws. We focus largely on the theory of in-viscid, irrotational flows, which forms the core of the classical theory offluid mechanics. The modern theory of fluid mechanics deals with theeffects of viscosity and the resulting boundary layers and turbulencein the vicinity of the boundaries of the flow domain. For a more de-tailed account of the subject, the reader may consult a number of textsand monographs, e.g. Batchelor [5], and Landau and Lifshitz,[12]; the treatise by Serrin [21] offers an historical and mathematicalperspective of the subject.

The equations of gas dynamics are the expression of conservationof mass, momentum, and energy in differential form.

Let ρ and u (u is a vector field) denote the mass density and velocityof a compressible fluid. We assume these functions to be C1 and wetake Ω to be a domain with smooth boundary in R3. The particlesfollow trajectories given by the ordinary differential equations

dxi

dt= ui(x, t).

The particles may be labelled by their positions at some reference time,e.g. t = 0, by ξ and their positions at time t by x = x(t, ξ). The

73

74 CHAPTER 4. EQUATIONS OF FLUID AND GAS DYNAMICS

variables ξ are called the material coordinates and the variables x arecalled the spatial coordinates. The trajectories of the particles arecalled the streamlines. The acceleration of the particles is given by thesecond derivative:

d2xi

dt2=

dui

dt=

∂ui

∂t+

∂ui

∂xj

dxj

dt=

∂ui

∂t+ uj ∂ui

∂xj.

If F (x, t) is a scalar valued function, its total or material derivative,moving with the fluid flow, is

dF

dt=

∂F

∂t+

3∑j=1

∂F

∂xjxj =

∂F

∂t+

3∑j=1

uj ∂F

∂xj.

The equationdF

dt= 0

means that F is constant along streamlines. The conservation laws offluid mechanics take precisely this form.

The conservation laws are expressed in terms of rates of change ofthe mass, momentum, and energy contained in a given region of theflow domain. For example, the total mass contained in Ω is

∫∫∫

Ω

ρ(x, t) dv,

where dv denotes the volume element dx1dx2dx3. If Ω moves with theflow there is no mass flux across the boundary, and conservation ofmass requires that

d

dt

∫∫∫

Ω

ρ dv = 0.

Theorem 4.1.1 When Ω = Ω(t) moves with the flow given by xi =ui(x, t) we have

d

dt

∫∫∫

Ω

ρ dv =

∫∫∫

Ω

(dρ

dt+ ρ div u

)dv. (4.1)

4.1. CONSERVATION LAWS 75

Proof: The particle trajectories are given by xi(t) = xi(t, ξ), wherexi(t, 0) = ξi. Write the integral in terms of the material variables ξ.Then we must calculate

d

dt

∫∫∫

Ω0

ρJ dξ, J =∂(x1, x2, x3)

∂(ξ1, ξ2, ξ3)

where Ω0 is a fixed domain parameterized by the ξ variables. We saythat the integral has been “pulled back” to the base manifold Ω0. Thetheorem follows from the fact that

J = Jdivu. (4.2)

and then transforming back to the coordinates x.Equation (4.2) is proved as follows.

J =∂(x1, x2, x3)

∂(ξ1, ξ2, ξ3)+

∂(x1, x2, x3)

∂(ξ1, ξ2, ξ3)+

∂(x1, x2, x3)

∂(ξ1, ξ2, ξ3)

=∂(u1, x2, x3)

∂(ξ1, ξ2, ξ3)+

∂(x1, u2, x3)

∂(ξ1, ξ2, ξ3)+

∂(x1, x2, u3)

∂(ξ1, ξ2, ξ3).

By the chain rule,

∂(u1, x2, x3)

∂(ξ1, ξ2, ξ3)=

3∑j=1

∂u1

∂xj

∂(xj, x2, x3)

∂(ξ1, ξ2, ξ3)=

∂u1

∂x1J,

etc. The other two terms are computed in the same way. This com-pletes the proof of (4.2) and hence the theorem. ¥

The equation governing the conservation of mass follows immedi-ately. Since Ω is an arbitrary domain, we must have

dρ

dt+ ρ div u =

∂ρ

∂t+ div(ρ u) = 0. (4.3)

Theorem 4.1.1 and its proof (due to Euler [21]) are a template fora much more general result, known as the transport theorem, whichis fundamental to the derivation of conservation laws. The transporttheorem provides a means to calculate the rate of change of integralsover curves or surfaces moving with the flow. Since differential forms


are the natural objects for integration theory in higher dimensions,we need to calculate the action of the total derivative on differentialforms.1

A p−form is integrated over a p−dimensional manifold. For exam-ple, the line integral of the 1-form fidxi over a path γ gives the workdone by a force field with components fi. The integral of the 3-formρ dx1 ∧ dx2 ∧ dx3 over a region Ω gives the total mass in Ω, and theintegral of

∑i<j

(∂uj

∂xi− ∂ui

∂xj

)dxi ∧ dxj

over a surface S gives the vorticity flux through S.These expressions, called exterior differential forms, have a number

of advantages over their vectorial cousins. The operations curl, gradi-ent, and divergence are not invariant under coordinate transformations,and moreover they are fundamentally tied to the metric tensor of themanifold, whereas the corresponding operation for differential forms,known as the exterior derivative, is defined in arbitrary dimensions, andtakes the same form in any coordinate system. Moreover, the calculusof differential forms keeps track of the orientation of the manifolds ofintegration.

In the language of tensors, p-forms are covariant antisymmetric ten-sors of order p on a manifold M ; they are denoted by Λ(M). The basicrules of calculation of differential forms are extremely simple, and wesummarize them here. We restrict our discussion to three operations:the wedge product, the exterior derivative, and the Lie derivative.

The wedge product is multilinear over scalar functions, anti-symmetric,and associative. Thus (dx1 ∧ dx2) ∧ dx3 = dx1 ∧ (dx2 ∧ dx3); while

dx1 ∧ dx2 = −dx2 ∧ dx1, dx1 ∧ fdx2 = fdx1 ∧ dx2,

etc. In particular, dx1 ∧ dx1 = 0. The wedgeThe exterior derivative d maps Λp to Λp+1. Its action on 0-forms

(functions), is given by

df =∂f

∂xidxi.

1The total derivative of a differential form is known in geometry as the Lie derivativewith respect to the vector field u. For a more complete account, see Spivak [24] orSattinger & Weaver [19].


Its action is uniquely extended to differential forms of any order by thethree rules

• d2 = 0;

• d is a linear map from Λp to Λp+1;

• d(ω ∧ ν) = dω ∧ ν + (−1)pω ∧ dν for ω ∈ Λp.

The theorems of Green, Gauss, and Stokes are subsumed under ageneral theorem, known as Stokes’ theorem

∫∫

Ω

dω =

∫

∂Ω

ω. (4.4)

Here, ω is a p form with differentiable coefficients, and Ω is a p + 1dimensional manifold in Rn with smooth boundary ∂Ω. It is possibleto relax these regularity conditions somewhat, but we shall not needthat for the present discussion.

We have already defined the total derivative relative to a vectorfield ~u on 0-forms, i.e. functions. The total derivative is extended toall differential forms by two simple rules:

• The Leibnitz rule:

d

dtω ∧ ν =

dω

dt∧ ν + ω ∧ dν

dt.

• The total derivative and the exterior derivative commute:

d

dtdω = d

(dω

dt

)

Note that the total derivative of a p−form is again a p−form.For example, the action of the total derivative on a one form fidxi

is

d

dtfidxi =

dfi

dtdxi + fi

d

dtdxi =

dfi

dtdxi + fidxi

=dfi

dtdxi + fidui =

dfi

dtdxi + fi

∂ui

∂xjdxj

=

(dfi

dt+ fj

∂uj

∂xi

)dxi


The total derivative of the mass density 3-form is

d

dtρ dv =

dρ

dtdv +ρ(dx1∧dx2∧dx3 +dx1∧dx2∧dx3 +dx1∧dx2∧dx3).

Now

dx1 ∧ dx2 ∧ dx3 =du1 ∧ dx2 ∧ dx3

=

(∂u1

∂x1dx1 +

∂u1

∂x2dx2 +

∂u3

∂x3dx3

)∧ dx2 ∧ dx3

=∂u1

∂x1dv

etc., sod

dt(ρ dv) =

(dρ

dt+ ρ div u

)dv.

The transport theorem states:

Theorem 4.1.2 Let ω be a p−form and Ω(t) a smooth p−dimensionalmanifold moving with the flow generated by the vector field u. Then

d

dt

∫∫

Ω

ω =

∫∫

Ω

dω

dt

The transport theorem is proved in the same way that Theorem4.1.1 was proved: the integral is “pulled back” to the base manifoldΩ0, where Ω(t) = x(t, Ω0), and the differentiation carried out there.

Let us return to the derivation of the conservation laws. The con-servation of linear momentum is derived by a similar argument. Themomentum density in Ω is given by the vector field ρu, hence the rateof change of the ith momentum component in the region Ω is2

d

dt

∫∫∫

Ω

ρui dv =

∫∫∫

Ω

d

dt

(ρ ui dv

).

Since the domain moves with the fluid, there is no transport of mo-mentum across the boundary. By Newton’s law of motion, the rate of

2This formula presupposes that momenta at different points in space can simply beadded together, a fundamental precept of Euclidean geometry; this would not be the casein relativistic fluid mechanics, for example, if one were formulating the equations of fluidflow in the interior of a very dense object, such as a star


change of momentum is equal to the total force on the body. If we as-sume there are no internal stresses in the fluid (no viscosity), then thetotal force acting on the fluid particles is the sum of the hydrodynamicpressure acting on the boundary plus the integral of any external force(such as gravity) acting throughout the interior.

The total force on Ω in the ith direction due to the hydrodynamicpressure is

−∫∫

∂Ω

pνi dS = −∫∫∫

Ω

∂p

∂xidv;

and the ith component of the total force on the fluid interior to Ω dueto an external force f is ∫∫∫

Ω

ρfi dv.

(The factor of ρ is required since f denotes the force per unit mass.)Therefore, the conservation of the ith component of the momentum

for an inviscid fluid is∫∫∫

Ω

d

dt(ρ ui dv) =

∫∫∫

Ω

(− ∂p

∂xi+ ρfi

)dv.

Since Ω is an arbitrary domain (with smooth boundary), we may con-clude that

d

dtρ ui dv = ui

d

dtρ dv + ρ

dui

dtdv =

(− ∂p

∂xi+ ρfi

)dv.

We have already shown that the total derivative of ρ dv vanishes, dueto conservation of mass, hence this equation simplifies to3

dui

dt+

1

ρ

∂p

∂xi=

∂ui

∂t+ uj ∂ui

∂xj+

1

ρ

∂p

∂xi= fi.

In terms of the total derivative, the equations of conservation ofmass and momentum are therefore

dρ

dt+ ρ div u = 0,

dui

dt+

1

ρ

∂p

∂xi= fi. (4.5)

3We use the summation convention here: repeated indices denote summation.


These equations are due to Euler in 17554

The equations above are not closed, since we have only four equa-tions in the five variables ρ, u, p. In order to close them we need eithera fifth equation or a relationship between ρ and p. If the gas is locallyin thermodynamic equilibrium, the pressure, temperature, and densityare related by an equation of state p = f(ρ, T ). If the temperature canbe assumed to be constant throughout the flow field, then the pressureand density are related by an equation p = h(ρ). In this case the flowis said to be barentropic. For example, in the case of an ideal gas withconstant specific heats, in which the entropy is constant (isentropicflow), the relationship between pressure and density is given by

p = Nργ

for constants N and γ. For air, γ = 7/5 = 1.4A second case of great importance in applications is that of in-

compressible flow. Water, for example, is essentially incompressible;but there are many applications, in geophysics for example, in whichair flow can also be considered incompressible [5]. For incompressibleflows, div~u = 0 and the first of Euler’s equations implies the density isconstant. Euler’s equations for an incompressible, inviscid fluid are

dui

dt+

1

ρ

∂p

∂xi= 0,

∂ui

∂xi= 0. (4.6)

.A one form f = fidxi is said to be exact if it is an exact differential

of a function F , i.e. fidxi = dF . In that case fi = ∂F/∂xi, and F

is said to be the potential. In vector analysis, a vector field ~f is said

to be a conservative vector field if ~f = −∇F . In the following weuse one-forms rather than vector fields; then if the external force fieldis conservative, we have f = −dF (the minus sign is in keeping withconvention).

Similarly, a potential flow is one for which the velocity field is agradient; i.e. ~u = ∇ϕ. We shall work with the corresponding one-forms and say that the flow is a potential flow if the one-form uidxi isexact, i.e. if uidxi = dϕ.5 A necessary and sufficient condition that u

4L. Euler Opera Omnia II 12. cf. Serrin op. cit.5We write the one-form as uidxi where ui are the components of the covariant tensor

obtained from the contravariant tensor ui by lowering the indices; in Cartesian coordinates,however, ui = ui.


be a potential flow is that the path integral

Γ =

∫

C

ui dxi,

called the circulation, vanish for any closed contour C. In that case wecan construct the potential via the integral

ϕ(x) =

x∫

x0

uidxi.

Theorem 4.1.3 The circulation around any closed curve C movingwith the flow is conserved for barentropic flow with a conservative forcefield. Consequently, if the flow is initially a potential flow, then itremains a potential flow.

Proof: By the transport theorem the rate of change of Γ along the flowis

Γ =d

dt

∫

C

ui dxi =

∫

C

dui

dtdxi + ui

d

dtdxi

=

∫

C

(−1

ρ

∂p

∂xi+ fi

)dxi + ui dui

=

∫

C

−dp

ρ− dF +

1

2d(u2) =

∫

C

−dp

ρ,

where fi = −∂F/∂xi. The last integral also vanishes, since we maywrite

dp

ρ= h′(ρ)dρ = dh, h′(ρ) =

p′(ρ)

ρ.

Therefore the circulation around any closed contour is invariant underthe flow, and Γ is conserved. ¥

By Stokes’ theorem, Γ =∫∫Ω

ω where ω is the two form

ω = du = dui ∧ dxi = frac12∑i<j

(∂uj

∂xi− ∂ui

∂xj

)dxi ∧ dxj.


The 2-form ω is called the vorticity, though in classical fluid mechanics,the vorticity is taken to be the vector field ω = ∇×u. The flow is saidto be irrotational if the vorticity vanishes. By the calculation used inthe proof of Theorem (4.1.3) one can show that

dω

dt= 0

when the external force field is conservative. Hence if the flow is initiallyirrotational it remains so.

If the circulation around any closed path vanishes, then the vorticityof the flow vanishes; but in a multiply connected domain, the vorticitymay vanish everywhere yet the circulation is non-zero. We said abovethat a one form is exact if it is an exact differential. Thus the oneform u = uidxi is exact if u = dϕ; in that case the circulation alwaysvanishes, and the flow is a potential flow.

On the other hand, a one form is closed if its exterior derivativevanishes. Thus u is closed if ω = du = 0. In this case the flow isirrotational. In the language of differential forms, a flow is irrotationalif uidxi is closed; and the flow is potential if uidxi is exact.

If the flow is irrotational then it can be written locally as the gra-dient of a potential; but unless the domain is simply connected, thepotential may not be single valued. The best known example is theplanar flow given by ϕ = θ, where θ = arctan y/x is the angular coor-dinate. The function θ is a local potential for the flow

u = dθ =x dy − y dx

x2 + y2= −sin θ

rdx +

cos θ

rdy.

This flow is irrotational, since du = d2θ = 0; but the circulation aroundany closed curve containing the origin is

Γ =

∫

C

dθ = 2π.

4.2 Bernoulli’s Theorem

Daniel Bernoulli derived, in 1738, an energy conservation theorem inwhich the fluid velocity is treated as kinetic energy, while the hydro-dynamic pressure is viewed as a potential energy. It sets forth the

4.2. BERNOULLI’S THEOREM 83

fundamental property of fluid flow in which variations in fluid velocitygenerate pressure differentials across a surface resulting in forces onairfoils, sails, etc.

Theorem 4.2.1 Consider a steady incompressible flow in the presenceof a conservative external force field with potential −F . Let u2 =∑

i ui,2. Then

1

2u2 +

p

ρ+ F

is constant along streamlines.

Proof: Write the momentum equation in Lagrangian form

xi +∂

∂xi

(p

ρ+ F

)= 0.

Multiplying these equations by xi and summing, we obtain

d

dt

(3∑

i=1

xi,2

2+

p

ρ+ F

)= 0.

The result follows by replacing xi by ui. ¥A similar result holds for the flow of a compressible gas, namely

1

2u2 +

∫dp

ρ+ F

is constant along streamlines. We leave this as an exercise.Bernoulli’s theorem holds for any steady flow, not just potential

flows. In particular, it holds even in the presence of vorticity. A sec-ond conservation theorem, quite similar to Bernoulli’s result, holds forunsteady potential flows.

Theorem 4.2.2 Let ~u(x, t) be an incompressible potential flow in thepresence of a conservative force field with potential −F . Let the velocitypotential be denoted by ϕ(x, t). Then, up to an additive factor C(t) inthe velocity potential, the equation

∂ϕ

∂t+

1

2(∇ϕ)2 +

p

ρ+ F = const. (4.7)


holds throughout the domain of flow.Proof: Multiplying the momentum equation in (4.5) by dxi and

summing, we obtain(

dui

dt+

1

ρ

∂p

∂xi+

∂F

∂xi

)dxi = 0.

Now

dui

dtdxi =

d

dtuidxi − ui

dxi

dt=

d

dtdϕ− uidui

=d

(dϕ

dt− 1

2u2

)= d

(∂ϕ

∂t+

1

2(∇ϕ)2

).

Therefore we obtain

d

(∂ϕ

∂t+

1

2(∇ϕ)2 +

p

ρ+ F

)= 0.

It follows that the quantity in parentheses on the left is spatially inde-pendent, hence

∂ϕ

∂t+

1

2(∇ϕ)2 +

p

ρ+ F = C(t).

Replacing the velocity potential ϕ by

ϕ +

∫ t

0

C ′(s)ds

we obtain (4.7). ¥Equation (4.7) plays a fundamental role in the dynamics of free

surface problems, for example, in the analysis of wave motion on thesurface of a body of water.

Using Bernoulli’s theorem, let us compute the force on an objectby a two dimensional incompressible flow whose asymptotic behavioras r →∞ is given by

u ∼ U i +Γ

2πr(− cos θ, sin θ).

For large r the velocity decays like 1/r to the uniform flow (U, 0).

4.2. BERNOULLI’S THEOREM 85

For incompressible flow the density is constant, and, ignoring changesin the external potential F , Bernoulli’s equation takes the form

1

2u2 +

p

ρ=

1

2u2∞ +

p∞ρ

.

Solving for p we obtain

p = p∞ +ρ

2(U2 − u2). (4.8)

The force exerted on an obstacle with finite boundary σ by thesteady flow of fluid around it is

fi = −∫∫

σ

pνi dS,

where ν is the outward normal. There are no forces due to the mo-mentum, since the normal component of the velocity vanishes on σ.Let Σ be any larger surface containing σ in the interior. By combiningthe two equations in (4.5) we have, for steady flow in the absence ofexternal forces,

∂p

∂xi+

∂(ρujui)

∂xj= 0, i = 1, 2, 3.

Integrating this expression over the region between σ and Σ, and usingthe divergence theorem, we find

fi = −∫∫

σ

pνi dS = −∫∫

Σ

pνi + (ρui)uν dS. (4.9)

Note that uν = u · ν vanishes on σ but not on Σ, since the latter is notthe boundary of an object.

Take Σ to be the circle of radius R centered at the origin, with R solarge that σ is contained within. Then ν = (cos θ, sin θ), uν = U cos θ,and the second integral in (4.9) becomes

fi = −∫ 2π

0

(pνi + ρuiU cos θ)Rdθ.


These integrals are calculated as R →∞, using the asymptotic behav-ior of the flow velocity given above. These considerations lead to

f1 = 0, f2 = −ρΓU

These results were obtained independently by Kutta in 1910 andJoukowski in 1906. In this model, the actual force on the objectis unaffected by its shape, since only the asymptotic behavior of theflows at infinity determines the force; but the effects of viscosity havebeen ignored.

Due to viscosity, the fluid must adhere to the surface of an object,so that the flow field vanishes on the boundary. When the viscosity issmall, as it is for air and water, its effects are very small away fromthe boundary, and are confined to a thin layer near the boundary,called the boundary layer. Classical fluid mechanics deals largely withinviscid irrotational flow, for which the powerful tool of potential theoryis applicable. The modern theory of fluid mechanics deals with thecomplicated effects of viscosity and the resulting turbulence within theboundary layer. An extensive treatment of the relation of viscosityand boundary layer theory to potential flow is given in the text byBatchelor.

The calculation above shows the fundamental importance of circula-tion to the theory of forces applied to an airfoil or sail. In the idealizedcase there is lift but no drag; but a mechanism is needed to create thecirculation, and that mechanism is the viscosity of the fluid and theresulting boundary layer around the wing. The optimum design of anairfoil seeks to determine a shape which will generate just the correctamount of circulation in the flow. The correct circulation is that forwhich the stagnation point sits at the trailing edge of the airfoil; thisis known as the Joukowski hypothesis of airfoil design.

The flows with finite circulation are given by (2.21) in the exteriorof the unit disk. The flow in the exterior of an airfoil is obtained fromthe conformal mapping of the exterior of the unit disk to the exteriorof the airfoil. In Joukowski’s theory of the airfoil, a conformal mappingis sought which maps the rearward stagnation point onto the trailingcusp of the airfoil.

4.3. D’ALEMBERT’S PARADOX 87

4.3 D’Alembert’s Paradox

It was proposed by D’Alembert in 1768 that the force on an objectin a three dimensional flow should be zero. This result, known asD’Alembert’s paradox,6 can be proved mathematically as a conse-quence of the fundamental difference in the asymptotic behavior ofharmonic functions in exterior domains in two and three dimensions.In two dimensions, the irrotational flow (−y, x)r−2 decays like r−1 atinfinity. It is the gradient of the multiple-valued harmonic functionθ(x, y) = tan−1 y/x. We have seen above that it is precisely this termthat generates the force on an object in the flow.

There is no such term, however, for flows in exterior domains in R3.The motion of an incompressible, irrotational fluid in the exterior ofa domain σ in R3 is obtained as the gradient of a function ϕ which isharmonic in the exterior of σ and satisfies ϕν = 0 on σ. In order thatthe flow be asymptotic to a uniform stream with velocity U i (here idenotes the unit vector in the x1 direction) at infinity, we again requireϕ ∼ Ux1 as r → ∞. In three dimensions, however, this asymptoticlimit is approached like r−2.

Lemma 4.3.1 Let ϕ be harmonic in the exterior of σ, ϕν = 0 on σ,and suppose that ϕ ∼ Ux1 as r →∞. Then

ϕ = Ux1 + O(r−2), u = ∇ϕ = U i + O(r−3).

Proof: Choose coordinates so that the origin is contained within σ,and let Σ be the sphere of radius R centered at the origin, with R largeenough that σ lies within Σ. In the exterior of any sphere containingσ the harmonic function ϕ has an expansion in inverse powers of r, sothat

ϕ = Ux1 +C

r+ O(r−2).

We shall show that C = 0. Since ϕ is harmonic in the region between

6J. L. D’Alembert, Opuscules Mathematiques 5 (1768). See also the discussion in[21]


σ and Σ, we have

0 =

∫∫

σ

∂ϕ

∂νdS =

∫∫

Σ

∂ϕ

∂rdS =

∫∫

Σ

(Uν1 − CR−2 + O(R−3))R2dω

=− 4πC + O(R−1),

since the integral of νi over any closed surface vanishes by the diver-gence theorem. It follows that C = 0. The decay of the velocity fieldthen follows by differentiation. ¥

It follows that p = p∞ + O(r−3) for flows in R3. From the integralover Σ in (4.9), we obtain

fi = −∫∫

Σ

((p∞+O(R−3))νi +(ρUδi1 +O(R−3)(Uν1 +O(R−3)

)R2dω.

Again noting that the integral of νi over any closed surface vanishes,we see that the above integral behaves like R−1. Letting R → ∞, wesee that the force must vanish.

The implication of D’Alembert’s paradox is that forces on an airfoilcan only be generated by two dimensional flows; this means that con-formal mapping techniques are relevant to the discussion of forces onobjects in fluid flows, and that three dimensional effects are, to someextent, relative – that three dimensional flows with a small aspect ratiocan be approximated by two dimensional flows.

4.4 Hyperbolic Conservation Laws

A partial differential equation is said to be in divergence form if it canbe written in the general form At +Bx = 0. Euler’s equations (4.5) forone dimensional isentropic flow, may be rewritten in divergence formas follows.

ρt + (ρu)x = 0 (ρu)t + (ρu2 + p)x = 0. (4.10)

These equations are supplemented by an equation of state p = p(ρ).Such a system of equations is called a system of conservation laws.

Equations (4.10) also form a hyperbolic system, that is, they havetwo real characteristics. We discussed characteristics of second order

4.4. HYPERBOLIC CONSERVATION LAWS 89

scalar equations in Chapter 3. The notion of characteristic for a systemis much the same. Consider a general first order system of equations

AUt + BUx + CU + D = 0, (4.11)

where U is a column vector of length n, and A, . . . D are n×n matricesdepending smoothly on x and t. For example, the system (4.10) canbe written in the form

AUt+BUx = 0, A =

(1 0u ρ

), B =

(u ρ

c2 + u2 2ρu

), U =

(ρu

),

(4.12)where c2 = p′(ρ). The quantity c(ρ) is the speed of sound.

Given a curve Γ in the x− t plane, we ask “When are the values ofU on Γ compatible with (4.11)?”. Or, to put the question another way,when are the values of U on Γ and the equation (4.11) overdetermined?If this is the case, then we say that Γ is a characteristic. To answerthis, let Γ be parameterized by smooth functions x(s), t(s). If U isspecified on Γ, then its tangential derivative Uτ along Γ is given by

Uτ =dU

ds= Uxx

′ + Utt′.

Moreover, the differential equation also gives a relationship between Ux

and Ut, so that we have a system of 2n equations in the 2n unknownsUx and Ut

AUt + BUx = f1(s), Uxx′ + Utt

′ = f2,

where f1 and f2 are entirely determined by the values of U on Γ and(4.11).

By definition, Γ is a characteristic when these equations are notuniquely solvable. This is equivalent to the statement that

(x′I t′IB A

) (U1

U2

)= 0

has a non-trivial solution. This reduces to the system (x′A−t′B)U2 = 0.Thus Γ is a characteristic if det(B − xA) = 0.

The system (4.11) is said to be hyperbolic if det(B − λA) has nreal roots λ1, . . . , λn. The roots are the characteristic speeds and the


associated eigenvectors are the characteristic directions. In the simplelinear case ut +vx = 0, vt +ux = 0 the characteristics are x± t = const.

The same concepts apply in the nonlinear case when A and B arefunctions of the dependent variables, as in the case of (4.10). In thenonlinear case, however, the characteristic speeds and directions de-pend on the solution. We leave it as an exercise to show that thecharacteristic speeds of the hyperbolic system (4.12) are x = u± c.

For the first order hyperbolic system ut + vx = 0, vt + ux = 0 thefunctions u ± v are constant along the characteristics. For example,differentiating u + v along a curve dx/dt = 1, we find

d

dt(u + v) = (u + v)t + (u + v)x

dx

dt= 0.

Similar invariants, called Riemann invariants, exist for nonlinearhyperbolic systems as well. For example, in (4.10) we look for functionsof ρ and u which are invariant along the characteristic curves dx/dt =u ± c(ρ). Let us denote by r(ρ, u) a function which is constant alongthe curve dx/dt = u + c. By the chain rule

dρ

dt= ρt +ρx

dx

dt= ρt +ρx(u+ c),

du

dt= ut +ux

dx

dt= ut +ux(u+ c),

so

dr

dt=rρ

dρ

dt+ ru

du

dt= rρ(ρt + (u + c)ρx) + ru(ut + (u + c)ux)

=rρ((u + c)ρx − (ρu)x) + ru((u + c)ux − uux − c2

ρρx)

=

(rρ − c

ρru

)(cρx − ρux).

Thus, r = r(ρ, u) is constant along the characteristics x = u + c if

ru =ρ

crρ.

It is easily seen that we may take

r(ρ, u) =1

2

(u +

∫ ρ c(ρ′)ρ′

dρ′)

.

4.5. SHOCKS 91

Similarly,

s(ρ, u) =1

2

(u−

∫ ρ ρ′

c(ρ′)dρ′

)

is invariant along the characteristics curves x = u− c. When p = Nργ,the integrations may be carried out explicitly, and we obtain

r(ρ, u) =1

2

(u +

c

γ − 1

), s(ρ, u) =

1

2

(u− c

γ − 1

).

These are the Riemann invariants for an ideal gas.

4.5 Shocks

These arguments are valid whenever the solutions are in C1; but theylead directly to the conclusion that discontinuities in the solutions mayform after a finite time, no matter how smooth the initial data. Forexample, suppose the initial data are such that s(x, 0) ≡ const. onsome interval I on the x− axis. Then s is constant on the domainof influence of I in the region t > 0, so long as the solutions remainsmooth. Since the other Riemann invariant, r, is constant on the curvesx = u + c, both u and ρ are constant along these characteristics, andthey are therefore straight lines. If now u + c is initially decreasing inx on some interval, these straight lines must intersect at some point inthe region t > 0. At this point of intersection the values of ρ and umust necessarily be different; hence at such a point the solutions musthave a discontinuity. Such discontinuities are called shocks, and playa fundamental role in the analysis of the equations of gas dynamics.We give here only a brief introduction to this very extensive subject.For further details, the reader should see the texts by Courant andFriedrichs [6], Landau and Lifshitz [12], Serrin [21], Smoller[22], Whitham [26].

Let us turn to a discussion of the hyperbolic equations (4.10) in theclass of solutions which are not C1. First consider the case where thesolutions are continuous in a domain Ω, with a discontinuity in theirderivatives along a differentiable curve γ, given by x = x(t). Let f(x, t)be any function defined in a neighborhood of γ, and denote the limitingvalues of f and its derivatives on γ from the right (x > x(t)) and left(x < x(t)) by f±, f±x , and f±t respectively. If [f ] = f+ − f−, denotes


the jump of f across the curve, etc, then the continuity of f across γis equivalent to the statement that [f ] = 0; while jumps in fx and ft

imply that [fx], [ft] are non zero.The derivative of f along the curve is df(x(t), t) = ft + fxx. Since

[f ] = 0 on the curve,

d

dt[f ] =

[df(x(t), t)

dt

]= [ft + fxx] = [ft] + x[fx] = 0.

Applying this jump relation to the variables ρ and u, we have

[ρt] + x[ρx] = 0, [ut] + x[ux] = 0.

On the other hand, from (4.5) we have

[ρt] + [(ρu)x] = 0, [ut] + u[ux] +c2(ρ)

ρ[ρx] = 0.

Using the jump relations we may eliminate the jumps in the timederivatives from the equations of motion, thus obtaining

(x− u)[ρx] = ρ[ux], (u− x)[ux] + c2 [ux]

(x− u)= 0.

If [ux] 6= 0 these equations imply that the speed of the curve γ, thatis, the speed with which the discontinuity propagates, is x = u± c. Aswe have seen, these are the characteristic speeds of the Euler equations.Thus, a jump in the derivative of either u or ρ implies a jump in theother, as well as a jump in the pressure gradient px. For this reasonthe discontinuity is interpreted as a sound wave.7

Now let us turn to an analysis of the equations (4.10) in a neigh-borhood of a discontinuity in the flow variables. Recall that the Eulerequations are expressions of the basic conservation laws in differentialform. We derive the conservation laws across a shock discontinuity.

For discontinuous solutions the partial differential equations mustbe reformulated in weak form∫∫

Ω

ρϕ1,t + ρuϕ1,x dxdt = 0,

∫∫

Ω

ρuϕ2,t + (ρu2 + p)ϕ2,x dxdt = 0.

7This derivation of the speed of sound is due to Hugoniot in 1885-1888; see thediscussion in [21], p. 212.

4.5. SHOCKS 93

These equations are to hold for all functions ϕ1, ϕ2 ∈ C10(Ω), where

Ω is an open region in the x − t plane. As usual, any weak solutionwhich is regular in any subdomain is necessarily a strong solution inthat subdomain.

The following is used to derive the jump conditions across a shock.

Lemma 4.5.1 Suppose that F and G are piecewise C1 functions in adomain Ω in the x-t plane, with jump discontinuities across a smoothcurve Γ given by x = x(t). Let Ω± denote the components of Ω onthe left and right of the curve Γ, oriented in the direction of increasingtime. Let F and G satisfy

∫∫

Ω

Fϕt + Gϕx dxdt = 0, ∀ ϕ ∈ C10(Ω).

Then[G] = [F ]x

where [F ] and [G] denote the jumps of F and G across Γ.

Proof: By the usual argument, Ft + Gx = 0 in each of the subdomainsΩ±. By Green’s theorem,

∮

∂Ω±Gϕdt− Fϕ dx =

∫∫

Ω±

(Gϕ)x + (Fϕ)t dxdt.

The line integrals are oriented so that Ω+ and Ω− lie on the left of thepath. Since ϕ vanishes on ∂Ω the only contribution to the line integralsis along the curve Γ. Adding the two equations above we get

∫

Γ

ϕ([F ]dx− [G]dt) =

∫∫

Ω+∪Ω−

(Gϕ)x + (Fϕ)t dxdt

=

∫∫

Ω

(Ft + Gx)ϕ + Fϕt + Gϕx dxdt.

The first term in the double integral vanishes since Ft + Gx = 0 in Ω+

and Ω−; and the second integral vanishes by hypothesis. Since ϕ is anarbitrary smooth function, [G]dt− [F ]dx must vanish along Γ, and thelemma is proved. ¥


When we apply this argument to the equations (4.10), we obtain

[ρu] = [ρ]x, [ρu2 + p] = [ρu]x.

Letting U = u− x be the flow velocity relative to the shock, the aboveequations can be rewritten in the form

[ρU ] = 0, [ρuU + p] = 0. (4.13)

These two jump relations express conservation of mass and momentumacross a shock, just as the Euler equations express these conservationlaws in regions where the flow is regular.

Caveat: The divergence form of the conservation laws is in generalnot unique (cf. exercise 8); hence one must be careful in choosing theweak form of the equations. The various weak forms of the equationsare not equivalent, and will lead to different conservation laws acrossthe shocks. Hence it is essential to choose the physically correct diver-gence form for the equations. The advantage of the weak formulationof the equations is that it contains all the information about the solu-tions, including the shock conditions, and remains valid even when theshock structure is extremely complicated. An alternative derivation ofthe shock conditions can be given directly by extending the transporttheorem to cases in which a shock occurs in the interior of the domainΩ.

Given the state before the shock and the speed of the shock, thatis, given ρ+, p+, U+ it is not possible to determine ρ−, p−, U− fromthese two equations alone, since they constitute two equations in threeunknowns. For, since we cannot assume that the entropy is constantacross the shock, we no longer know the relationship between ρ and pbehind the shock. Thus the resolution of the problem when there areshocks requires the introduction of thermodynamic considerations [21].

4.6 Exercises

1. A stagnation point is a point at which the velocity vanishes. Findthe stagnation points of the flow given by (2.21) for all values ofΓ/4πU . Hint: The velocity in polar coordinates is given by

vr =∂ϕ

∂r, vθ =

1

r

∂ϕ

∂r.

4.6. EXERCISES 95

2. The velocity in an irrotational flow cannot attain a maximumvalue in the interior of the domain of the flow; the pressure cannotattain an interior minimum.

3. The linearized Euler equations in three dimensions for small dis-turbances about the rest state u = 0, p = p0, ρ = ρ0 are obtainedby neglecting quadratic terms. By eliminating u show that thepressure satisfies the wave equation

ptt = c2∆p, c2 = p′(ρ0).

4. Prove the following extension of the transport theorem: Let fbe a scalar quantity, v a vector field, and Ω(t) a domain in R3

moving with the flow generated by v. Suppose f is piecewisedifferentiable in Ω with jump [f ] across a smooth surface Γ =Γ(t, s), (s=parameter) contained in Ω(t). Then

d

dt

∫∫∫

Ω

f dx =

∫∫∫

Ω

df

dtdx +

∫∫

Γ

[fU ]dS

where U = (v−Γt) · νΓ is the relative normal velocity of the fluidacross Γ.

Use this form of the transport equation to derive the conservationlaws across a shock.

5. Find the characteristic speeds for Euler’s equations

ρt + (ρu)x = 0, ρut + ρuux + c2ρx = 0.

Show that for strong solutions they are equivalent to the equationsin divergence form (4.10). Show the second equation may also bewritten in the divergence form

ut +

(u2

2+

∫c2

ρ

)

x

= 0.

What is the conservation law across the shock corresponding tothis divergence form of the momentum equation?


Chapter 5

The Maximum Principle

In this chapter we prove the strong maximum principle for second orderelliptic operators and state, but do not prove, the corresponding resultfor general parabolic operators. Maximum principles provide a unique,powerful tool, for scalar elliptic and parabolic operators of second or-der, and we shall illustrate some of the many applications later in thechapter.

5.1 Elliptic and parabolic inequalities

Throughout this section, we let L denote the following second orderdifferential operator:

Lu =n∑

j,k=1

ajk(x)ujk +n∑

j=1

bj(x)uj,

where

uj =∂u

∂xj

, ujk =∂2u

∂xj∂xk

.

The operator L is said to be uniformly elliptic in a domain Ω if thereis a positive constant µ such that

n∑

j,k=1

ajkξjξk ≥ µ

n∑j=1

ξ2j , ∀x ∈ Ω.

We assume throughout that Ω is a connected open set in Rn.We begin by proving the weak maximum principle.

97

98 CHAPTER 5. THE MAXIMUM PRINCIPLE

Theorem 5.1.1 Assume L is uniformly elliptic and that its coeffi-cients ajk, bj are continuous and uniformly bounded in Ω. If u ∈ C2(Ω)and

Lu > 0, x ∈ Ω,

then u cannot have an interior maximum in Ω.

Proof: At an interior maximum, all the first derivatives of u vanish,and the matrix ujk of second derivatives is non-positive. Thus, at aninterior maximum,

Lu = Tr au, a = ||ajk||, u = ||ujk||.Since a is positive definite and u is non-positive, the trace of theirproduct cannot be positive, hence Lu ≤ 0 at an interior maximum. ¥

We use this result to prove the following, which is known as thestrong maximum principle of E. Hopf. It generalizes the result forLaplace’s equation to uniformly elliptic, second order equations. Westate it here as a one sided inequality; this gives a more general result.If Lu ≥ 0 in Ω, then u cannot attain an interior maximum in Ω; and ifLu = 0 in Ω, then u cannot attain an interior extremum.

Theorem 5.1.2 Assume L is uniformly elliptic in a domain Ω andthat its coefficients ajk, bj are continuous and uniformly bounded in Ω.If u ∈ C2(Ω) and

Lu ≥ 0, x ∈ Ω,

then u cannot have an interior maximum in Ω unless u is identicallyconstant.

Proof: Let M be the supremum of u over Ω, and let

ΩM = x ∈ Ω : u(x) < M.Since u is continuous, ΩM is open. We are going to show that ΩM isalso closed in the relative topology of Ω. That is, if B is an open ballcontained in ΩM and u(p) = M , where p ∈ ∂B, then p ∈ ∂Ω. It thenfollows that either ΩM is empty, in which case u ≡ M , or ΩM = Ω.

Let S = ∂B and let p ∈ S, u(p) = M . We shall show that

∂u

∂ν(p) > 0, (5.1)

5.1. ELLIPTIC AND PARABOLIC INEQUALITIES 99

N

S

1

S0

S

q

p

Figure 5.1: Proof of the boundary point lemma.

where ν is the outward normal to S at p. It then follows that p ∈ ∂Ω,since all first order derivatives of u must vanish at an interior maximum.

Let N be the normal line to S through p, and choose a point q ∈ Bon N such that |p− q| = r0, where r0 is less than the radius of S. Wetake q as the origin and consider the function

h(r) = e−αr2 − e−αr20 .

It is clear that h = 0 on the sphere S0 of radius r0 centered at q, andh > 0 in the interior of S0. Moreover,

Lh = e−αr2

(4α2

n∑

j,k=1

ajkxjxk − 2αn∑

j=1

(ajj + bjxj)

);

so Lh > 0 in a neighborhood of q for sufficiently large α.Let S1 be a sphere centered at p, and let C be the intersection of

the interiors of S0 and S1, indicated by the shaded region in Figure(5.1). Put v = u + εh; then

Lv = Lu + εLh ≥ εLh > 0, x ∈ C.


By the weak maximum principle, the maximum of v must occur on theboundary of C. Since h = 0 on S0, v = u on S0, and v = M at p. Thesecond component of ∂C is a circular arc lying strictly in the interiorof S. On this arc, u is bounded away from M , so for sufficiently smallε, v = u + εh < M .

Hence the maximum of v in C is M , and this value is attained atp, and p only. At that point we must have

∂v

∂ν(p) ≥ 0,

where ν is the outward normal to S at p. Since ∂h/∂ν(p) < 0, (5.1)follows. ¥

The key point in this proof is the inequality (5.1); this result iscalled the boundary point lemma. We have shown the following:

Theorem 5.1.3 [Boundary Point Lemma] Suppose the conditions ofTheorem (5.1.2) hold on a domain Ω. Suppose that u is not identicallyconstant on Ω and attains its maximum at p ∈ ∂Ω. Suppose that thereis a sphere S whose interior lies in Ω and which is tangent to ∂Ω at p.Then

∂u

∂ν(p) > 0, (5.2)

where ν is the outward normal to Ω at p.

There are several extensions of the maximum principle worth not-ing.

Theorem 5.1.4 Let the previous assumptions on L hold, c ≤ 0 in Ω,and

Lu + cu ≥ 0, x ∈ Ω.

Then u cannot attain a non-negative interior maximum in Ω unless uis identically constant in Ω. Moreover, if u < 0 in Ω and u(p) = 0 forsome point p ∈ Ω at which the sphere condition holds, then (5.2) holds.

Proof: The proof of this theorem proceeds exactly as the proof ofthe strong maximum principle. In this case the parameter α mustbe chosen so that (L + c)h > 0. This is certainly possible when c isbounded, since 0 ≤ h ≤ 1.

5.1. ELLIPTIC AND PARABOLIC INEQUALITIES 101

Corollary 5.1.5 Let the previous assumptions on L hold, and let c ≤ 0in Ω. If

Lu + cu = 0, x ∈ Ω,

then u cannot attain an interior maximum or minimum in Ω unless uis identically constant on Ω.

We leave the proof of this result to the reader.

Theorem 5.1.6 Let the previous assumptions on L hold, and let c(x)be a bounded function (not necessarily continuous) on Ω. Suppose thatu ∈ C2(Ω) and

(L + c)u ≥ 0, u ≤ 0 x ∈ Ω.

Then either u ≡ 0 on Ω or u < 0 in Ω.Moreover, if u < 0 in Ω then (5.2) holds at any p ∈ ∂Ω at which

u(p) = 0 and the sphere condition is satisfied.

Note that there are no restrictions on the sign of c.Proof: Write c = c+(x)+ c−(x), where c+ ≥ 0 and c− ≤ 0 in Ω; and

write the inequality as

(L + c−)u ≥ −c+u ≥ 0, x ∈ Ω.

Let N = u < 0. By the argument used in the proof of Theorem(5.1.2), N is open by continuity of u, and closed in the relative topologyof Ω as a consequence of the boundary point lemma. Therefore N iseither empty or all of Ω. ¥

The maximum principle can be extended to scalar second orderparabolic operators by much the same argument. We continue to makethe same assumptions about the elliptic operator L defined above, ex-cept that now we allow the coefficients ajk and bj to depend on x andt. Let Ωτ = Ω× [0, τ ] for any τ > 0.

Theorem 5.1.7 Let u(x, t) satisfy the parabolic differential inequality

Lu− ∂u

∂t≥ 0, (x, t) ∈ ΩT .

Suppose that u attains an interior maximum M at a point (p, t0), wherep ∈ Ω and 0 < t0 ≤ T . Then u is identically equal to M in the cylinderΩt0.


Suppose that u attains its maximum at a point (p, t0) for some p ∈∂Ω, and that a sphere Sp tangent to ∂Ω at p can be constructed in Ω.Then

∂u

∂ν(p) > 0,

where ν is the outward normal to Ω at p.

Remark: The maximum of u must occur either on the sides of thecylinder, i.e. on ∂Ω × [0, T ] or in Ω at time t = 0. Physically, if udenotes the temperature and it is below freezing outside Ω and belowfreezing initially, then it is never going to get above freezing inside Ω.

The maximum principle for parabolic operators can be extended asfollows.

Theorem 5.1.8 Let h be continuous in ΩT and bounded above, andsuppose that

Lu + hu− ut ≥ 0, (x, t) ∈ ΩT , Bu ≤ 0.

Then either u ≡ 0 or u < 0 in ΩT .

Proof Put u = veλt; then

Lv + (h− λ)v − vt ≥ 0, Bv ≤ 0.

We may choose λ so that h−λ < 0 in ΩT , since h is bounded above. Ifu ≥ 0 in the interior of ΩT then v ≥ 0 somewhere also, and v must havea non-negative maximum somewhere in ΩT . At such a non-negativemaximum of v we have vt ≥ 0 and Lv ≤ 0, but

Lv − vt ≥ (λ− h)v ≥ 0.

By continuity, this inequality must hold in an open set; hence theconclusion of the theorem follows from the strong maximum principle,Theorem (5.1.7). ¥

For a full discussion of maximum principles and their applicationsto partial differential equations, see the monograph by Protter &Weinberger [15].

5.2. MONOTONE METHODS 103

5.2 Monotone methods

Maximum principles have myriad applications in partial differentialequations. Of course, they can be used to prove uniqueness theorems,but I will begin with a simple application to semi-linear equations [17].Consider the simple nonlinear boundary value problem

Lu + f(x, u) =0, x ∈ Ω (5.3)

Bu =g, Bu = u∣∣∣∂Ω

(5.4)

where L is a uniformly elliptic second order operator with Holder con-tinuous coefficients as defined in the previous section. Remark: We

require Holder continuity of the coefficients so that the solution of thelinear equation Lu = f is smooth. We assume that ∂Ω is smooth.

An upper solution for (5.3) (5.4) is a C2 function v which satisfies theinequalities

Lv + f(x, v) ≤ 0, Bv ≥ g.

A lower solution v is defined by reversing the inequalities. The followingtheorem was obtained by H. Amman [3] and D.H. Sattinger [17]; cf.also [4] and [18].

Theorem 5.2.1 Let v ≤ v be lower and upper solutions for the bound-ary value problem (5.3), (5.4). Then there exists at least one solutionu of the nonlinear boundary value problem such that v ≤ u ≤ v.

Proof: The solution is obtained by a monotone iteration scheme asfollows. Let A be a positive real number, chosen so that

fA(x, u) = f(x, u) + Au

is an increasing function of u on the interval [min v, max v]. Let u0 = vand solve the linear boundary value problem

(L− A)u1 + fA(x, u0) = 0, x ∈ Ω; Bu1 = g.

Note that

(L− A)(u1 − u0) =− (Lu0 + f(x, u0)) ≥ 0, x ∈ Ω

B(u1 − u0) =g −Bu0 ≤ 0.


Therefore, by the maximum principle, (extended to the operator L−A),the function u1−u0 cannot have a positive interior maximum. Since itis non-positive on the boundary, u1 < u0 = v everywhere in the interiorof Ω by the strong maximum principle. It follows that

Lu1 + f(x, u1) = (L− A)u1 + fA(x, u1) = fA(x, u1)− fA(x, u0) ≤ 0,

since u1 < u0 and fA is increasing. Therefore u1 is also an uppersolution.

Continuing in this way we obtain a decreasing sequence of uppersolutions

v = u0 > u1 > u2 > . . . .

Similarly, starting with u0 = v we obtain an increasing sequence oflower solutions:

v = u0 < u1 < u2 < . . . .

We leave it as an exercise to show that uj < uk for any j, k. Thereforeeach of the decreasing sequence of upper solutions is bounded belowby each member of the increasing sequence of lower solutions. By thestandard regularity theory for elliptic partial differential equations, onecan show that each of these sequences converges, and that we obtainsolutions (which may be distinct)

v ≤ u ≤ u ≤ v. ¥

We can also use the maximum principle to investigate stability ofsolutions of the initial value problem for the associated parabolic equa-tion:

Lu + f(x, u)− ut = 0, Bu = g, u(x, 0) = u0(x). (5.5)

Lemma 5.2.2 Let v, v be C2 functions which satisfy the inequalities

Lv + f(x, v)− vt ≤0

Lv + f(x, v)− vt ≥0

in the domain ΩT , and

v ≤ v on BT ,

where BT = (Ω× t = 0 ∪ (∂Ω× [0, T ]). Then

v(x, t) < v(x, t) (x, t) ∈ ΩT .


Proof: Let w = v − v; then

Lw + [f(x, v)− f(x, v)]− wt ≥ 0, (x, t) ∈ ΩT ,

w ≤ 0 on BT .

By the mean value theorem,

f(x, v)− f(x, v) =

∫ v

v

fu(x, s)ds

=

∫ 1

0

fu(x, v + zw)w dz = F (x,w)w.

Therefore, w = w(x, t) satisfies the parabolic differential inequality

Lw + Fw − wt ≥ 0, Bw ≥ 0.

We do not know F = F (x,w) explicitly, but we may regard itas a smooth coefficient in the above inequality. Since w is a smoothbounded function, it follows that F is bounded, and the lemma followsby Theorem 5.1.8. ¥

Corollary 5.2.3 Let v ≤ u0 ≤ v, where v and v are lower and uppersolutions of (5.3), (5.4), and let u satisfy the initial value problem (5.5).Then

v ≤ u(x, t) ≤ v, ∀ t > 0.

Proof: Since v and v are time independent, they satisfy the parabolicinequalities of Lemma (5.2.2), and the Corollary follows.

Theorem 5.2.4 Let v0 < v0 be lower and upper solutions of (5.3),(5.4), and let v(x, t) satisfy the initial value problem

Lv + f(x, v)− vt = 0, v(x, 0) = v0(x), Bv(·, t) = g(·).Then vt(x, t) < 0 for t > 0. If v(x, t) satisfies the initial value problemand v(x, 0) = v0(x), then vt > 0 for t > 0. Moreover, v(x, t) < v(x, t)for all t > 0.

The solution v(x, t) increases monotonically to a stationary solu-tion U(x) of (5.3), (5.4); while v(x, t) decreases monotonically to astationary solution U(x); and U(x) ≤ U(x).


Proof: Let w = vt. Then w satisfies the parabolic equation

Lw + fu(x, v)w − wt = 0, Bw = 0.

Moreover,w(x, 0) = vt(x, 0) = Lv + f(x, v) ≤ 0.

Therefore w(x, t) = vt(x, t) < 0 by Theorem (5.1.8).The same argument applies to the solution of the initial value prob-

lem when the initial data is a lower solution. The inequality v(x, t) <v(x, t) is a consequence of Theorem (5.1.8). Let w = v(x, t) − v(x, t).By the mean value theorem, w satisfies a parabolic equation

Lw + F (x, t, w)w − wt = 0, Bw = 0.

Since w(x, 0) > 0, it follows that w(x, t) > 0 for all t > 0.Since the solutions of the two initial value problems are monotone

in t and bounded, their limits exist as t →∞. Suppose that

U(x) = limt→∞

v(x, t).

Let ϕ = ϕ(x) be any smooth C20(Ω) test function. Then

(Lv, ϕ) + (f(x, v), ϕ)− (vt, ϕ) =0,

(v, L∗ϕ) + (f(x, v), ϕ)− (vt, ϕ) =0,

where L∗ is the adjoint of the elliptic operator L.We have

limT→∞

1

T

∫ T

0

v(x, t)dt = U(x);

hence

limT→∞

1

T

∫ T

0

(vt, ϕ)dt = limT→∞

(v(x, t)ϕ)− (v, ϕ)

T= 0

limT→∞

1

T

∫ T

0

(v, L∗ϕ)dt = (U , L∗ϕ)

limT→∞

1

T

∫ T

0

(f(x, v), ϕ)dt = (f(x, U), ϕ).


Hence U is a weak solution of the nonlinear boundary value problem(5.3) (5.4):

(U , L∗ϕ) + (f(x, U), ϕ) = 0, BU = g.

We have seen that weak solutions of the Dirichlet problem for theLaplacian (L = ∆) are strong solutions, and the same is true of themore general uniformly elliptic boundary value problem when the co-efficients of L are Holder continuous. ¥

A point of tangency p ∈ Ω for two functions u, v ∈ C2(Ω) is a pointat which the tangent planes to the graphs of the two functions coincide.That is, u(p) = v(p) and ∇u(p) = λ∇v(p). When p ∈ ∂Ω we define∇u(p) to be the limit of ∇u(p′) as p′ → p from the interior of Ω.

Theorem 5.2.5 Let v ≤ v be lower and upper solutions to (5.3), (5.4).If there exists a point of tangency p ∈ Ω, then v ≡ v throughout Ω. Inparticular, either v ≡ u or v < v.

Proof: Let w = v − v; then w ≥ 0 on Ω and

(L− A)w + fA(x, v)− fA(x, v) ≤ 0.

For sufficiently large A > 0, fA(x, u) is increasing in u; and so

(L− A)w ≤ 0.

Since w ≥ 0 on ∂Ω, it follows by Theorem (5.1.4) (applied to −w) thateither w > 0 or w ≡ 0 in the interior. In particular, there can be nopoint of tangency in the interior.

Now suppose there is a point p ∈ ∂Ω at which w(p) = 0. Bythe maximum principle at the boundary, Theorem (5.1.3), wν(p) < 0,hence p cannot be a point of tangency. ¥

Corollary 5.2.6 Let u1 ≤ u2 be two solutions of (5.3) (5.4). Theneither u1 < u2 in Ω or u1 ≡ u2.

We now discuss a number of examples that show the application ofthese techniques. It can be shown by a variational argument that thenonlinear elliptic boundary value problem

∆u = u2, x ∈ Ω; Bu = 0 (5.6)


has a nontrivial solution; we denote it by w. We assume throughoutthis discussion that Ω is bounded and that ∂Ω is a smooth surface in Rn,with the property that at every point p ∈ ∂Ω there is a sphere tangentto ∂Ω that is contained in Ω. By the strong maximum principle, w < 0in Ω. Let us show that w is an unstable equilibrium for the associatedparabolic equation ut = ∆u− u2.

Consider the function λw. We have

∆(λw)− (λw)2 = λ(∆w − λw2) = λ(1− λ)w2.

Hence λw is a lower solution for λ < 0 or λ > 1 and an upper solutionif 0 < λ < 1. Let uλ(x, t) satisfy the initial value problem

∂uλ

∂t= ∆uλ − u2

λ, Buλ = 0, uλ(x, 0) = λw(x).

By Theorem (5.2.4), uλ(x, t) is decreasing in time if λ > 1 or λ < 0and increasing for 0 < λ < 1.

When 0 < λ < 1, uλ(x, t) increases monotonically to a stable equi-librium. Using Theorem (5.2.5), we prove in the next paragraph thatthere can be no other equilibrium solution w2 > w. Similarly, whenλ < 0 the solution uλ(x, t) decreases monotonically to zero as t → ∞.Therefore, all solutions of the initial value problem for which the initialdata lies above w(x) tend to zero asymptotically as t → ∞. We leaveit as an exercise to show that for λ > 1 the solution uλ(x, t) blows upin finite time.

Now suppose that there are two solutions, 0 > w2 ≥ w1 of (5.6).By Theorem (5.2.5) the two solutions cannot have a point of tangency;hence w2 > w1 in Ω. Let

λ∗ = infλ : λw2 > w1.Then λ∗ > 1, hence λ∗w2 is an upper solution; while λ∗w2 and w1 musthave a point of tangency somewhere in Ω. Thus Theorem (5.2.5) isviolated, and there cannot be two solutions of (5.6), one lying entirelyon one side of the other.

Now consider the nonlinear problem

(∆ + µ)u− u3 = 0, Bu = 0, (5.7)

where µ is a parameter. We begin by showing that for µ < µ1 the trivialsolution is a stable equilibrium for the nonlinear parabolic equation

(∆ + µ)u− u3 = ut, Bu = 0.


As a comparison function for this problem we use the principaleigenfunction of the Laplacian:

∆ψ1 + µ1ψ1 = 0, Bψ1 = 0.

We have

(∆ + µ)λψ1 − (λψ1)3 = λψ1(µ− µ1 − λ2ψ2

1), Bψ1 = 0. (5.8)

By the Krein-Rutman theorem, ψ1 > 0 in Ω. For µ < µ1 it followsthat λψ1 is a lower solution for λ < 0 and an upper solution for λ > 0.Hence the origin (i.e. u = 0) is stable. These inequalities are reversedwhen µ > µ1, hence the origin becomes unstable as µ crosses µ1.

For µ < µ1 the only solution of (5.7) is the trivial solution. We usethe fact that the eigenvalues of the Laplacian decrase monotonicallyas the domain increases. Let Ω ⊂ Ω′ and let ψ′1, µ′1 be the principaleigenfunction and eigenvalue of the Laplacian on Ω′. Then ψ′1 > 0 onΩ, and µ′1 < µ1. If u is any non-trivial solution of (5.7), put u = wψ′1and substitute it into the partial differential equation to get

ψ′1∆w + 2∇ψ′1 · ∇w + w(∆ + µ)ψ′1 − w3ψ31 = 0.

Since ψ′1 > 0 on Ω we can divide through by this function to get

∆w +2

ψ′1∇ψ′1 · ∇w + w(µ− µ′1 − w2ψ′21 ) = 0.

Since µ < µ1 we choose Ω′ so that µ < µ′1 < µ1; then the coefficient ofw in the above equation is negative, and by the maximum principle, wand hence u vanishes identically.

We now show that a pair of solutions ±w, w > 0, bifurcates fromthe trivial solution as µ crosses µ1. From (5.8) we see that λψ1 is alower solution when µ > µ1 and 0 < λ < δ for some δ > 0. Applyingthe same argument to ψ′1, we get

(∆ + µ)λψ′1 − (λψ′1)3 = λψ′1(µ− µ1 − (λψ′1)

2).

Since ψ′1 > 0 on ∂Ω and since ∂Ω is compact, we can find a δ′ > 0such that ψ′1 ≥ δ′ > 0 on ∂Ω. Therefore, for λ sufficiently large, λψ′1 isan upper solution. The existence of a positive solution w of (5.7) thenfollows.


5.3 Crash Course in Elliptic Equations

In this section we summarize some of the principle facts about secondorder elliptic and parabolic boundary value problems which will beneeded.

5.3.1 Regularity theory for elliptic equations

Given a function u defined in a domain Ω ⊂ Rn the Holder norm ||u||αis defined by

||u||α = supΩ|u|+ Hα(u),

where

Hα(u) = supx,y∈Ω

|u(x)− u(y)||x− y|α , 0 < α ≤ 1.

A function u is said to be Holder continuous with exponent α inΩ if Hα(u) < +∞. The class of Holder continuous functions on Ω isdenoted by Cα(Ω). The class of functions which are Holder continuousalong with their derivatives up to order k is denoted by Ck,α(Ω). Theassociated norm is defined by

||u||k,α =∑

β≤k

||Dβu||α.

The spaces Ck,α(Ω) form a Banach algebra under the norms || · ||k,α.That is, they form a Banach space, and in addition,

||uv||k,α ≤ ||u||k,α||v||k,α.

Moreover, if α′ ≤ α then Ck,α is compactly embedded in Ck,α′ ; that is,bounded subset in Ck,α are compact in Ck,α′ .

We have previously introduced the Sobolev spaces W k,p. A gen-eral reference is Adams, [1]. The following result is the basic Sobolevembedding lemma:

Theorem 5.3.1 If Ω is a bounded domain with smooth boundary inRn, and n < p < ∞, then

||u||1−n/p ≤ C(n, p, Ω)||u||1,p.

5.3. CRASH COURSE IN ELLIPTIC EQUATIONS 111

In the following, we take L to be the uniformly elliptic second orderoperator defined in §(5.1), and assume that the coefficients of L belongto Cα(Ω) and that ∂Ω is of class C2,α. We assume g is defined on∂Ω and has a C2,α extension into the interior of Ω. We denote thatextension also by g.

Theorem 5.3.2 The linear elliptic boundary value problem

Lu = h, Bu = g

has a unique smooth solution in class C2,α; moreover u satisfies the apriori estimate

||u||2,α ≤ C(||f ||α + ||g||2,α).

The constant C depends only on Ω, n, and α.The following Lp estimates are also valid:

||u||2,p ≤ C(||f ||p + ||g||2,p).

See Agmon, Douglis, and Nirenberg [2], especially theorems 7.3 and15.2. The same estimates hold for operators L + c(x), where c ∈ Cα

and L + c has a trivial kernel, i.e. (L + c)u = 0 implies u = 0.In (5.3), (5.4) we may reduce the problem to the case g = 0, and

we do that in the following discussion. Then L is a bounded linearoperator from C2,α

0 (Ω) to Cα0 (Ω), or from W 2,p to Lp. According to

the theorem, its inverse, which we denote by G, is a continuous lineartransformation from Cα to C2,α or from Lp to W 2,p. Though we shallnot prove it here, G is an integral operator whose kernel is the Green’sfunction for L.

To show convergence of the sequence of upper solutions, write thesequence in the form

un+1 = −GAfA(x, un), GA = (L− A)−1.

Since un ↓ U(x) and un is bounded, the sequence fA(x, un) convergesin Lp(Ω) for any p. Since GA is a continuous mapping from Lp toW 2,p

0 (Ω), the solutions un converge in W 2,p. For p > n the uppersolutions are in C1−n/p, by the Sobolev embedding theorem. Since GA

maps Cα boundedly into C2,α, the upper solutions are in C2,α for any0 < α < 1. Since the upper solutions converge pointwise and areuniformly bounded in C2,α their limit U(x) ∈ C2,α.


Let u ∈ C2,α satisfy the nonlinear boundary value problem andwrite

u = −Gf(x, u).

If f is a C∞ function of (x, u), and if the coefficients of L are C∞, thenby induction we may conclude that the solution u is C∞. Likewise, if fis analytic in both variables, and the coefficients of L are analytic in theinterior, then it follows that any bounded solution u is a priori analytic.This induction argument is commonly called a bootstrap argument.

A weak solution of the nonlinear boundary value problem is a func-tion u ∈ L1

loc(Ω) such that

(u, L∗ϕ) + (f(x, u), ϕ) = 0 ∀ ϕ ∈ W 2,p(Ω). (5.9)

Theorem 5.3.3 Let u ∈ L∞(Ω) satisfy (5.9), where f is Holder con-tinuous in (x, u). Then u ∈ C2,α(Ω), and u satisfies the nonlinearboundary value problem (5.3) (5.4).

Proof: Put w = −Gf(x, u). Then w ∈ W 2,p0 (Ω) for any p > 1, and

(w,L∗ϕ) = −(Gf, L∗ϕ) = −(f,G∗L∗ϕ) = −(f, ϕ) ∀ϕ ∈ C2.

Hence (u−w,L∗ϕ) = 0 for all ϕ ∈ C2. Now take ϕ = G∗(u−w). Thenϕ ∈ W 2,p and we get

(u− w, L∗G∗(u− w) = |u− w|22 = 0.

Hence u = w a.e. and so u may be modified on a set of measure zeroso that u ∈ W 2,p

0 (Ω). By the Sobolev embedding theorem u is Holdercontinuous, and we may then repeat the bootstrap argument above. ¥

5.3.2 The Fredholm Alternative

The Fredholm alternative for elliptic operators is fundamental to non-linear analysis, especially in perturbation or bifurcation theory. Westate it here. Let f(x), c(x) ∈ Cα(Ω) and consider the elliptic bound-ary value problem

(L + c)u = f, Bu = 0. (5.10)

If c ≤ 0 in Ω then by the maximum principle the homogeneous equation

(L + c)u = 0, Bu = 0 (5.11)

5.3. CRASH COURSE IN ELLIPTIC EQUATIONS 113

has only the trivial solution. If c assumes positive values, the linearhomogeneous equation may have one or more non-trivial solutions ϕ+1, . . . ϕn. If so, then the adjoint equation

L∗u + cu = 0 Bu = 0 (5.12)

also has n solutions ϕa1st, . . . ϕ

∗n, and by the regularity theory, these lie

in C2,α.If the homogeneous equation reffellhom has only the trivial solution,

then (5.10) has a unique solution u ∈ C2,α. If the homogeneous equa-tion has n independent solutions then the adjoint equation also has nindependent solutions ϕ∗j ∈ C2,α. In that case (5.10) has a solution uif and only if ∫

Ω

f(x)ϕ∗j dx = 0, j = 1, . . . n.

One way to prove the Fredholm alternative is the following. Convert(5.10) to an integral equation by applying the Green’s operator for L:

u + Gcu = Gf.

By the a priori regularity results for elliptic operators, G is a boundedoperator from Cα to C2,α. It follows that Gf ∈ C2,α, and the integralequation above holds in C2,α.

A compact (or completely continuous) operator G on a Banachspace E is one that maps bounded sets into compact sets: that is, ifuj is a bounded sequence in E , then Guj contains a convergentsubsequence. By the elliptic regularity theory, T = Gc is a compactoperator, since it maps Cα continuously into C2,α and since C2,α is com-pactly embedded in Cα. The Fredholm alternative holds for functionalequations of the form

(I + T )u = f

on a Banach space E , where T is a compact operator. See Rieszand Nagy [16] for the theory when E is a Hilbert space, Dunford andSchwartz, volume I [7] when E is a Banach space.

5.3.3 The Krein-Rutman Theorem

Consider the eigenvalue problem

Lu + c(x)u + λu = 0, Bu = 0.


For now, we us assume that c(x) ≤ 0; hence 0 is not an eigenvalue ofthis equation, and L + c is invertible. Denoting −(L + c)−1 by G, wemay rewrite the eigenvalue problem as an integral equation

u = λGu,

where G is a compact integral operator on the Banach space C2,α.A cone in a Banach space E is a closed subset K with the properties

that i) u, v ∈ K imply that αu + βv ∈ K for all α, β ≥ 0;

ii) u, v ∈ K and u 6= 0 imply that u + v 6= 0. The set of non-

negative functions on Ω forms a cone, which we denote by K, in C2,α,or in W k,p(Ω) for k ≥ 0 and p ≥ 1. The interior of K consists of strictlypositive functions.

An operator G is said to be strongly positive relative to K is foreach u ∈ ∂K there is an integer n = n(u) such that Knu belongs tothe interior of K. We leave it as an exercise to show, using the strongmaximum principle, that −(L + c)−1 is strongly positive with n = 1for all u.

Theorem 5.3.4 [Krein-Rutman] Let K be a cone in a Banach spaceE, and let G be strongly positive with respect to K. Then G has one andonly one eigenfunction ϕ in the interior of K, and the correspondingeigenvalue is real and simple.

Corollary 5.3.5 Let L be a uniformly elliptic operator on a domain Ω,with Holder continuous coefficients. Then the eigenvalue with least realpart is real and simple, and the corresponding eigenfunction is strictlypositive on Ω.

5.4 The method of moving planes

The method of moving planes was invented by Alexandrov to prove thatglobal solutions of the nonlinear elliptic equation of constant mean cur-vature were spheres. The method was in turn extended and developedby Serrin, and later by Gidas, Ni, and Nirenberg to prove rotationalsymmetry of solutions of solutions of the semi-linear equation

∆u + f(u) = 0

5.4. THE METHOD OF MOVING PLANES 115

on Rn.Here is an example of such a result; the proof presented here is due

to J. Serrin and Henzhei Zou.

Theorem 5.4.1 Let u ∈ C2(Rn) be a positive solution of

∆u + f(u) = 0, x ∈ Rn; u(x) →∞ as |x| → ∞. (5.13)

Suppose that f is locally Lipshitz continuous on (0,∞) and is non-increasing on (0, δ) for some δ > 0.

Then u is radially symmetric about some point p ∈ Rn; and, more-over,

∂u

∂r< 0, r > 0.

Proof: For γ real and fixed, define

Σγ = x1 < γ, Γ = x1 = γ ;

xγ = (2γ − x1, x2, . . . xn).

We define uγ(x) = u(xγ), the solution reflected in the plane Γ. Sincethe Laplacian is invariant under reflections, uγ is also a solution of(5.13). Let w = wγ = u− uγ. Then

∆w + cγ(x)w = 0, cγ(x) =f(u)− f(uγ)

u− uγ.

Since f is Lipshitz, c is bounded on bounded sets in Rn. Moreover,cγ ≤ 0 when 0 < u, uγ < δ, since f is non-increasing on this interval.

The proof proceeds in several steps.

Step 1. There exists γ0 such that for all γ ≥ γ0, w(x) ≥ 0 for allx ∈ Σγ.

Since u → 0 as |x| → ∞, there exists a γ0 such that 0 < u(x) < δ forx1 > γ0. Suppose w < 0 somewhere in Σγ. Since w → 0 as |x| → ∞and w = 0 on Γγ, w must have a negative minimum at some pointy ∈ Σγ. Then

u(y)− uγ(y) = w(y) < 0;

hence 0 < u(y) < uγ(y) < δ, since yγ1 > γ0. Therefore cγ(y) ≤ 0, and in

fact, cγ(x) ≤ 0 for all x in a neighborhood of y. By Corollary (5.1.5),w is identically equal to a negative constant in a neighborhood of y, in


contradiction to the fact that w = 0 on Γγ. Step 2. Suppose w ≥ 0 in

Σγ for some γ. Then either w ≡ 0 in Σγ or

w > 0 in Σγ; and∂w

∂x1

< 0 on Γγ.

This result is a consequence of Theorem (5.1.6). Step 3. There

exists a maximal closed subinterval [µ,∞) such that w ≥ 0 in Σγ forall γ ∈ [µ,∞).

It suffices to show there exists a γ such that wγ(x) < 0 for somex ∈ Σγ. Let z ∈ Rn be fixed and choose γ < 0 so large that γ < z1 and0 < u(x) < u(z) for all x ∈ Σγ. Then zγ ∈ Σγ, and

w(zγ) = u(zγ)− u(z) < 0.

Step 4. Let [µ,∞) be the maximal closed subinterval; then wµ ≡ 0 in

Σµ; i.e., u is symmetric about the plane Γµ.If not, then by Step 2

w > 0 on Σµ;∂w

∂x1

< 0 on Γµ. (5.14)

On the other hand, since [µ,∞) is maximal, there is a sequence γkand points xk ∈ Σγk

such that

wk(γk) < 0, wk = u(x)− u(xγk).

Without loss of generality we may choose xk to be the minimum of wk

in Σγk.

There are three possibilities: i) |xk| → ∞; ii) xk → y ∈ Σµ; iii)xk → y ∈ Γµ. In the first case we find a contradiction as in Step1. In the second case, by continuity, it follows that wk(y) = 0, whichcontradicts our assumption that w > 0 in Σµ. Finally, in case iii), thereis another sequence zk such that zk → y ∈ Γµ and

∂wk

∂x1

(zk) ≥ 0.

Again, by continuity,∂w

∂x1

(y) ≥ 0,

5.5. EXERCISES 117

which contradicts the second inequality in (5.14).Thus u is symmetric about Γµ. Step 5. We have now shown that in

every direction there is a hyperplane Σµ about which u is symmetric.Choose n orthogonal hyperplanes with a common point of intersectionp ∈ Rn. Clearly ∇u(p) = 0; and moreover, u attains a maximumat p. Furthermore, any other hyperplane of symmetry must also passthrough p, since for a given direction, there is a unique maximal µfor which u is symmetric about Γµ. Thus u is symmetric about everyhyperplane passing through p; and therefore u is radially symmetricabout p. ¥

5.5 Exercises

1. Prove a uniqueness theorem for the linear boundary value problem

(L + h)u = 0, Bu = 0

under the assumption that the principal eigenvalue λ1 > 0. 2. Prove

that the solution to the initial value problem

ut = ∆u− u2, Bu = 0, u(x, 0) < w(x)

blows up in finite time, where w is the non-trivial solution to the timeindependent problem. 3. Use the maximum principle to compare the

principal eigenvalues of the Laplacian on two domains Ω ⊂ Ω′. 4. Prove

the principle of linearized stability for solutions of (5.3) (5.4). That is, astationary solution w is a stable equilibrium for the associated parabolicequation if and only if the principal eigenvalue of the linearized operator

(L + fu(x,w) + λ)ψ = 0, , Bψ = 0

is positive.

5. The nonlinear problem

∆T + λe−E/RT = 0, T∣∣∣∂Ω

= T0

arises in combustion theory. Prove that for some values of λ the prob-lem has three solutions. 6. The equation


F (x, u, Du, D2u) = 0,

where Du denotes the first order derivatives of u and D2u the secondorder derivatives, is said to be elliptic with respect to a function u ∈C2(Ω) if

n∑

j,k=1

∂F

∂ujk

ξjξk > 0, x ∈ Ω.

Formulate and prove an extension of Theorem (5.2.5) to such a fullynonlinear elliptic equation.

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