INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS, SS18

Francesca Da Lio(1)

June 10, 2018

(1)Department of Mathematics, ETH Zurich, Ramistrasse 101, 8092 Zurich, Switzerland.

Abstract

These notes are based on the course Introduction to Partial DifferentialEquations that the author held during the Spring Semester 2017 for bach-

elor and master students in mathematics and physics at ETH. They arenot supposed to replace the several books in literature on Partial Differen-tial Equations. They simply aim at being a guide for what has been done

during the course and a help for the preparation to the exam.

Contents

1 Generalities on PDEs 21.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Multi-indices and derivatives . . . . . . . . . . . . . 31.2 What is a PDE? . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Examples of PDEs . . . . . . . . . . . . . . . . . . 61.2.2 Classification of second order semilinear PDEs . . . 8

1.2.3 Well-posed problems . . . . . . . . . . . . . . . . . 9

2 First-order PDEs 122.1 The Cauchy Problem . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Linear equations with constant coefficients . . . . . 132.2 The Method of the Characteristics . . . . . . . . . . . . . 15

2.2.1 A brief description . . . . . . . . . . . . . . . . . . 152.2.2 Applications to quasilinear first order PDEs . . . . 16

2.2.3 Existence and Uniqueness for the Cauchy Problem 192.3 Derivation of characteristic ODEs in the general case . . . 22

2.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . 242.4 Local Existence: the General case . . . . . . . . . . . . . . 30

2.4.1 Applications . . . . . . . . . . . . . . . . . . . . . . 34

2.5 Introduction to Hamilton Jacobi Equation . . . . . . . . . 402.5.1 Legendre transform, Hopf-Lax formula . . . . . . . 41

3 Laplace Equation 493.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . 49

3.2 Fundamental solution . . . . . . . . . . . . . . . . . . . . . 53

i

3.2.1 The Poisson Equation in IRn . . . . . . . . . . . . . 54

3.3 Sub- and super-harmonic functions . . . . . . . . . . . . . 563.4 Mean-value formulas . . . . . . . . . . . . . . . . . . . . . 60

3.4.1 Properties of harmonic functions . . . . . . . . . . 64

3.5 Green’s Functions . . . . . . . . . . . . . . . . . . . . . . 753.5.1 Derivation of the Green’s Function . . . . . . . . . 77

3.5.2 Green’s function for a ball . . . . . . . . . . . . . . 793.5.3 Green’s function for a half space . . . . . . . . . . . 83

3.6 Perron’s Method . . . . . . . . . . . . . . . . . . . . . . . 843.7 Poisson Equations: Classical solutions . . . . . . . . . . . . 913.8 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . 97

4 Heat equation 1004.1 Motivations . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4.2 Derivation of the Fundamental solution . . . . . . . . . . . 1014.3 Homogenous Cauchy Problem . . . . . . . . . . . . . . . . 104

4.3.1 Tychonov’s counterexample . . . . . . . . . . . . . 1074.4 Nonhomogeneous Cauchy Problem . . . . . . . . . . . . . 109

4.5 Maximum Principle . . . . . . . . . . . . . . . . . . . . . . 1124.5.1 Uniqueness of bounded solutions . . . . . . . . . . 114

4.6 Mean-value formula . . . . . . . . . . . . . . . . . . . . . . 1154.6.1 Strong Maximum Principle for the Heat Equation . 1184.6.2 Regularity . . . . . . . . . . . . . . . . . . . . . . . 120

4.7 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . 121

A Convolution product 124

A.1 Properties of the Convolution . . . . . . . . . . . . . . . . 124

1

Chapter 1

Generalities on PDEs

This Chapter surveys the principal theoretical issues concerning the solvingof partial differential equations. In Section 1.1 we first list several notations

we will use throughout these notes. Then we introduce the concept ofpartial differential equation. In Section 1.2 we discuss briefly well-posed

problems for partial differential equations.

1.1 Notation

IR denotes the real numbers, C the complex numbers. We will work in

IRn and n will always denote the dimension. Points in IRn will generallybe denoted by x, y, ξ, η, the coordinates of x are (x1, . . . , xn). Occasionally

x1, x2, . . . will denote a sequence of points in IRn rather than coordinates,but this will always be clear from the context.

If Ω is a subset of IRn, Ω will denote its closure and ∂Ω its boundary.If x, y ∈ IRn we set

x · y =

n∑

i=1

xiyi,

so the Euclidean norm of x is given by

|x| = (x · x)1/2.

We use the following notation for sphere and (open) balls in IRn with radius

2

r > 0 and center at x:

S(x, r) = ∂B(x, r) = y : |y − x| = r,B(x, r) = y : |y − x| < r.

The volume of B(x, r) and the area of S(x, r) are given by

|B(x, r)| = ωn

nrn and |S(x, r)| = ωnr

n−1

where

ωn = |S(x, 1)| = nπn/2

Γ(12n+ 1)

Γ(s) =∫∞0 ts−1e−tdt is the Euler gamma function. We also denote

−∫

B(x,r)

fdy = average of f over the ball B(x, r)

and

−∫

∂B(x,r)

fdσ(y) = average of f over the sphere ∂B(x, r).

1.1.1 Multi-indices and derivatives

i) A multiindex is a vector of the form α = (α1, . . . , αn) where each

component αi is a nonnegative integer. The order of α is the number|α| = α1 + . . .+ αn .

For x ∈ IRn we set

xα = xα1

1 xα2

2 · · ·xαnn .

ii) Given a multiindex α and given u : Ω ⊂ IRn → IR define

Dαu(x) :=∂ |α|u(x)

∂α1x1 · · · ∂αnxn.

iii) If k is nonnegative integer,

Dku(x) := Dαu(x) : |α| = k .

3

Special cases

If k = 0, D0u = u.

If k = 1, we regard the elements of Du as being arranged in a vector:

Du = ∇u = (ux1, · · · , uxn

) = gradient vector .

If k = 2 we regard the elements of D2u as being arranged in a matrix:

D2u =

ux1x1

· · · ux1xn

... . . . ...

uxnx1 · · · uxnxn

= Hessian matrix

v)

∆u =

n∑

i=1

uxixi= tr(D2u) = Laplacian of u.

vi) We write the divergence of a vector field F = (F 1, . . . F n) as div(F ) =∑ni=1F

ixi. We observe that ∆u = div (Du).

vii) We sometimes employ a subscript attached to the symbols D,D2

etc. to denote the variables being differentiated . For example if

u = u(x, y), x ∈ IRn, y ∈ IRm then Dxu = (ux1, · · · , uxn

) and Dyu =(uy1, · · · , uym) .

viii) Some function spaces. Let Ω be a domain of IRn, that is an openand connected subset in IRn.

C(Ω) := u : Ω → IR, u is continuousCk(Ω) := u : Ω → IR, u is k-times continuous differentiableC∞(Ω) := u : Ω → IR, u infinitely continuous differentiableCc(Ω), C

kc (Ω), C

∞c (Ω) denote functions in C(Ω), Ck(Ω), C∞(Ω) with

compact support (the support of a function f will be denoted by

spt(f)).

4

1.2 What is a PDE?

Partial differential equations are central objects in the mathematical mod-eling of natural and social sciences (sound propagation, heat diffusion,

thermodynamics, electromagnetism, elasticity, fluid dynamics, quantummechanics, population growth, finance...etc). They were for a long time

restricted only to the study of natural phenomena or questions pertainingto geometry, before becoming over the course of time, and especially in the

last century, a field in itself.A partial differential equation (PDE) is an equation involving an un-

known function of two or more variables and certain of its partial deriva-

tives.We fix an integer k ≥ 1 and let Ω ⊆ IRn denote an open set .

Definition 1.2.1 An expression of the form

F (x, u(x), Du(x), . . . , Dku(x)) = 0, in Ω (1.1)

is called a kth-order partial differential equation, where

F : Ω× IR× IRn × · · · × IRmk → IR

is given and u : Ω → IR is the unknown, (mk is the number ofmulti-indices α of order k).

We solve the PDE if we find all u verifying (1.1) possibly only among

those functions satisfying certain auxiliary boundary conditions on somepart Γ of ∂Ω . By finding the solutions we mean, ideally, obtaining sim-ple, explicit solutions, or, failing that, deducing the existence and other

properties of solutions.

5

Definition 1.2.2 (Classification of PDE) (i) The PDE (1.1)is called linear if it has the form

∑

|α|≤k

aα(x)Dαu(x) = f(x)

for given functions aα. This linear PDE is homogeneous iff ≡ 0 .

(ii) The PDE (1.1) is called semilinear if it has the form∑

|α|=k

aα(x)Dαu(x) + a0(x, u(x), . . . , D

k−1u(x)) = 0

(iii) The PDE (1.1) is called quasilinear if it has the form

∑

|α|=k

aα(x, u(x), . . . , Dk−1u(x))Dαu(x)+a0(x, u(x), . . . , D

k−1u(x)) = 0

(iv) The PDE (1.1) is called fully nonlinear if it depends non-linearly upon the highest order derivatives.

1.2.1 Examples of PDEs

There is no general theory known concerning the solvability of all partial

differential equations. Such a theory is extremely unlikely to exist, giventhe rich variety of physical, geometry, and probabilistic phenomena whichcan be modeled by PDE.

Following is a list of some specific partial differential equations of interestin current research. We will later discuss the origin and interpretation of

these PDE.Throughout x ∈ Ω where Ω ⊆ IRn is an open set, and t ≥ 0. Also

Du = Dxu denotes the gradient of u with respect to the spatial variable xand the variable t always denotes time.

6

1. Linear Transport Equation:

ut + b(x, t) ·Du(x, t) = 0, (x, t) ∈ Ω ⊂ IRn × (0,+∞) ,

it is a linear first order equation.

2. Laplace equation:

∆u(x) =n∑

i=1

uxixi= 0, x ∈ Ω ⊂ IRn

it is a linear second order equation.

3. Heat or diffusion equation:

ut(x, t)− k∆u(x, t) = 0 (x, t) ∈ Ω ⊂ IRn × (0,+∞) .

it is a linear second order equation.

4. Wave equation:

utt(x, t)− c2∆u(x, t) = 0 (x, t) ∈ Ω ⊂ IRn × (0,+∞) .

it is a linear second order equation.

5. Poisson equation:

∆u(x) = f(x), x ∈ Ω ⊂ IRn

it is a linear second order equation.

6. Nonlinear Poisson equation:

∆u(x) = f(u), x ∈ Ω ⊂ IRn

it is a semilinear second order equation.

7. Minimal surface equation

div

(Du

(1 + |Du|2)1/2)

= 0

it is a quasilinear second order equation.

7

8. Schrodinger’s equation

iut +∆u = 0.

it is a linear second order equation.

9. Monge-Ampere equation

Det(D2u) = f(x) x ∈ Ω ⊂ IRn.

it is a fully nonlinear second order equation.

10. Hamilton-Jacobi equation

H(x, u,Du) = 0 x ∈ Ω ⊂ IRn

orut +H(x, u,Du) = 0 (x, t) ∈ Ω ⊂ IRn × (0,+∞) ,

they are fully nonlinear first order equations.

11. Scalar conservation law

ut + div(F (u)) = 0 (x, t) ∈ Ω ⊂ IRn × (0,+∞) ,

the vector field F (u) depends nonlinearly upon u.

1.2.2 Classification of second order semilinear PDEs

The second order semilinear PDEs can be classified in elliptic, parabolicand hyperbolic. To describe such a classification it is convenient to con-sider the time as a space coordinate, by setting xn+1 = t. Set N = n + 1,

the unknown function is written as u = u(x1, . . . , xN) and the equation(1.1) can be written in the form

N∑

i,j=1

Aij(x)uxixj+ A0 = 0 (1.2)

where Aij are the coefficients of a N × N symmetric matrix and A0 is afunction that may depend on u and its first derivatives. The quadratic term∑n

i,j=1Aij(x)uxixjis called principal part of the differential equation.

8

The equation (1.2) is called elliptic if the signature(1) of the matrix

Aij is (N, 0) or (0, N) (namely the matrix is positive definite or negativedefinite). The (1.2) is called parabolic if the signature of the matrix Aij is(N − 1, 0) or (0, N − 1) and it is called hyperbolic if the signature of the

matrix Aij is (N − 1, 1) or (1, N − 1). We notice that the Laplace equationis elliptic, the diffusion equation is parabolic and the wave equation is

hyperbolic.

1.2.3 Well-posed problems

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

obtain explicit solutions in terms of elementary functions. In practice thisis only possible for very simple PDEs, and in general it is impossible to find

explicit expressions of all solutions of all PDEs. In the absence of explicitsolutions, we need to seek methods to prove existence of solutions of PDEs

and discuss properties of these solutions.Hadamard introduced the notion of well-posed problems. A given prob-

lem for a partial differential equation is well-posed if

(i) there is a solution;

(ii) this solution is unique;

(iii) the solution depends continuously in some suitable sense on the datagiven in the problem, i.e., the solution changes by a small amount if

the data change by a small amount.

Clearly it would be desirable to “solve” a PDE in such a way that (i)-(iii)hold. Notice that we have not specified what we mean by a “solution”.

Should we ask, for example that a “solution” u must be real analytic orat least infinitely differentiable? This might be desirable, but perhaps weare asking too much. It would be wiser to require a solution of a PDE of

order k to be at least k times continuously differentiable. Let us informallycall a solution with such smoothness a classical solution. So by solving a

(1)Recall: the signature of Aij is the pair (p, q) where p is the number of positive eigenvalues and q isthe number of negative eigenvalues

9

partial differential equation in the classical sense we mean if possible to

write down a formula for a classical solution satisfying (i)− (iii) above.Unfortunately most PDEs cannot be solved in a classical sense and we

are forced to abandon the search for smooth classical solutions. We must

instead investigate a wider class of candidates for solutions. If from thebeginning we demand that our solutions be very regular, then we are usu-

ally going to have a really hard time finding them, as our proofs must thennecessarily include possibly intricate demonstrations that the functions we

are building are in fact smooth enough. A more reasonable strategy is toconsider as separate the existence and the regularity problems. The ideais to define for a given PDE a reasonable notion of a weak solution, with

the expectation it may be easier to establish its existence, uniqueness andcontinuous dependence on the data. For various partial differential equa-

tions this is the best that can be done. For other equations we can hopethat our weak solution may turn out after all to be smooth to qualify as a

classical solution. This leads to the question of regularity of weak solutionswhich usually rests upon many intricate calculus estimates.

Example 1.2.1 (Counterexample of Hadamard) Consider the prob-

lem of finding a harmonic function in E taking either Dirichlet data orNeumann data on a portion Σ1 of ∂E and Dirichlet data on the remaining

part Σ2 = ∂E \Σ1. Such a problem is ill-posed. Even if a solution exits, ingeneral it is not stable in any reasonable topology, as shown by the following

example due to Hadamard. The boundary value problem

uxx + uyy = 0 in (−π/2, π/2)× (0,+∞)

u(±π/2, y) = 0 y > 0u(x, 0) = 0 x ∈ (−π/2, π/2)

uy(x, 0) = e−√n cos(nx) x ∈ (−π/2, π/2)

admits the family of solutions

un(x, y) = e−√n 1

ncos(nx) sinh(ny), n odd integer.

One verifis that

‖un,y(x, 0)‖L∞(−π/2,π/2) → 0 as n → +∞

10

and for y > 0

‖un(·, y)‖L∞(−π/2,π/2) → +∞, as n→ +∞.

Example 1.2.2 (A further example of the well-posedness issue)

We considerux = c0u+ c1(x, y). (1.3)

Since the equation (1.3) contains no derivative with respect to y we canregard this variable as a parameter. We can write the general solution of(1.3) in the form

u(x, y) = ec0x[∫ x

0 e−c0ξc1(ξ, y)dξ + T (y)

](1.4)

where the function T (y) is determined by the initial condition u(x, 0).Suppose c1 ≡ 0. Then the solution (1.3) now becomes u(x, y) = ec0xT (y).i) Suppose that u(x, 0) = 2x, then T (y) must satisfy T (0) = 2xe−c0x,

which is of course impossible. Therefore in this case there is no solution.ii) Suppose that u(x, 0) = 2ec0x, then T (y) should satisfy T (0) = 2. Thus

every function satisfying T (0) = 2 will provide a solution for the equationwith this initial condition. Therefore (1.3) with c1 = 0 has infinitely many

solutions with the initial condition u(x, 0) = 2ec0x.

11

Chapter 2

First-order PDEs

In this chapter we search for explicit formulas for general first-order non-linear partial differential equations of the form

F (x, u,Du) = 0 in Ω (2.1)

subject to the boundary condition

u = g on Γ ⊂ ∂Ω (2.2)

where Ω ⊂ IRn is open. Here F : Ω × IR × Rn → IR, g : Γ → IR are givenand they are supposed smooth functions, u : Ω → IR is the unknown.

We introduce the basic notion of noncharacteristic hypersurfaces for

initial-value problems. We solve initial-value problems by the method ofcharacteristics if the initial values are prescribed on noncharacteristic hy-

persurfaces.We also introduce the Hopf-Lax formula for Hamilton-Jacobi equations

and study its regularity properties.

12

Notation. Let us write F = F (x, z, p), for p ∈ IRn, z ∈ IR,x ∈ Ω . Thus p is the name of the variable for which we substitute

the gradient Du(x) and z is the variable for which we substituteu(x). We set

DpF = (Fp1, . . . Fpn)DzF = Fz

DxF = (Fx1, . . . Fxn

) .

(2.3)

2.1 The Cauchy Problem

The problem (2.1)-(2.2) is called the Cauchy Problem associated to theequation (2.1). A particular case is a problem of the form

ut +G(x, t, u,Dxu) = 0 in Ω = U × (0,+∞)

u(x, 0) = g(x) for x ∈ U(2.4)

where U ⊂ IRn is an open set and G : U × (0,+∞)× IR × IRn → IR andg : U → IR are given functions. In this case the independent variablesare (x, t), where x = (x1, . . . , xn) is the spatial variable, and t is the time

variable, Γ = U × 0. The equation in (2.4) is associated to an operatorF = ∂t +G where G does not depend on ∂t. In this context the boundary

condition (2.2) is called initial condition.

2.1.1 Linear equations with constant coefficients

Homogeneous case

We consider the so-called transport equation with constant coefficients:

ut + b ·Du = 0 in IRn × (0,+∞), (2.5)

where b = (b1, . . . , bn) ∈ IRn is a fixed vector and u : IRn × (0,+∞) → IRis the unknown u = u(x, t). Here x = (x1, . . . , xn) ∈ IRn denotes a typical

point in space and t ≥ 0 denotes the time.

13

The equation (2.5) describes for instance the transport of a solid pollut-

ing substance along a channel; here u is the concentration of the substanceand b is the stream speed.

Which functions u solve (2.5)?

Let us suppose for the moment we are given some smooth solution uand try to compute it. We observe that given (x, t) for every s ∈ IR wehave

0 = ut(x+ sb, t+ s) + b ·Du(x+ sb, t+ s) =d

dsu(x+ sb, t+ s).(2.6)

Therefore the function z(s) = u(x + sb, t + s) is a constant function of sand consequently for each point (x, t), u is constant on the line through

(x, t) with the direction (b, 1) ∈ IRn+1. Hence if we know the value of u atany point on each such line, we know its value everywhere.

Let us consider the initial-value problemut + b ·Du(x, t) = 0 in IRn × [0,+∞)

u(x, 0) = g(x) for x ∈ IRn.(2.7)

Here b ∈ IRn and g : IRn → IR are known and the problem is to compute

u. Given (x, t), the line through (x, t) with direction (b, 1) is representedparametrically by (x + sb, t + s) (s ∈ IR). This line hits the hypersurface

Γ := IRn × 0 when s = −t at the point (x − tb, 0). Since u is constanton the line and u(x− tb, 0) = g(x− tb), we deduce

u(x, t) = g(x− tb) (x ∈ IRn, t ≥ 0). (2.8)

So if (2.7) has a sufficiently regular solution u, it must be given by (2.8).And conversely it is easy to check directly that if g ∈ C1 then u defined by

(2.8) is indeed a solution of (2.7).The formula (2.8) says that the initial data is propagated along the

curves x = x0 + bt and for this reason the equation (2.5) is known as thetransport equation.

14

Nonhomogeneous problem

Next let us look at the associated nonhomogeneous problemut + b ·Du(x, t) = f(x, t) in IRn × [0,+∞)

u(x, 0) = g(x) for x ∈ IRn.(2.9)

As before fix (x, t) ∈ Rn+1 and set z(s) = u(x+ sb, t+ s) for s ∈ IR. Then

z(s) = b ·Du(x+ sb, t+ s) + ut(x+ sb, t+ s) = f(x+ sb, t+ s).

Consequently

u(x, t)− g(x− bt) = z(0)− z(−t) =∫ 0

−t

z(s)ds

=

∫ 0

−t

f(x+ sb, t+ s)ds

=

∫ t

0

f(x+ (s− t)b, s)ds;

and so

u(x, t) = g(x− tb) +∫ t

0 f(x+ (s− t)b, s)ds (2.10)

solves the initial problem (2.9).

Remark 2.1.1 Observe that we have derived our solutions (2.8)and (2.10) by in effect converting the partial differential equations

into ordinary differential equations. This is a special case of themethod of characteristics that we develop later.

2.2 The Method of the Characteristics

2.2.1 A brief description

The method of characteristics was developed in the middle of the nine-teenth century by Hamilton. It consists in converting the Cauchy problem

15

(2.1)-(2.2) into a Cauchy problem for an appropriate system of ODEs (the

so-called characteristic equations). The solutions of the characteris-tic equations parametrize the so-called characteristic curves, which passthrough the following (n− 1)- dimensional manifold

S0 = (x, g(x)), x ∈ Γ (2.11)

which is called initial manifold.Let us denote by Cp the characteristic curve which passes through p ∈ S0.

We suppose that for some p ∈ S0 the curve Cp is not tangent to S0 . Then

by continuity the same holds for all the curves Cp for p in a neighbourhoodN of p. The construction of the characteristic equations is made in such a

way that the set of the characteristics curves Cp with p in a neighbourhoodN of p constitute a n− 1 dimensional surface corresponding to the graph

of a solution of the PDE in consideration. Such a solution is constructedin a a neighborhood N of p.

The existence of the characteristic curves and the fact that S0 is a man-ifold require some regularity assumptions on F and g. The constructionof the characteristic equation depend on the structure of the partial differ-

ential equation in consideration.We start by examining this method in thecase of quasilinear first order PDEs and we study later the general case of

fully nonlinear equations.

2.2.2 Applications to quasilinear first order PDEs

For simplicity we start with the case of quasilinear PDEs of the form

c(x, u) ·Du + b(x, u) = 0 (2.12)

where x = (x1, x2) ∈ Ω ⊆ IR2, b ∈ C1(Ω× IR) , c ∈ C1(Ω× IR).

We are interested in studying classical solutions of (2.12), namely func-tions u : D → IR of class C1 that verify c(x, u(x)) ·Du(x) + b(x, u(x)) = 0

for every x ∈ D ⊆ Ω . The graph of the solution u is given by

S = (x, u(x)) : x ∈ D

16

with D = domu. S is called integral surface for the equation (2.12).

Solving the equation (2.12) means to find all the integral surfaces. Werecall that if S is a 2-dimensional surface of IR3 then the normal unitvector to S at the generic point q = (x, u(x)) is given by

NS(q) =1√

1 + |Du|2(−Du(x), 1) .

If we define the vector field F : Ω× IR→ IR3 by setting

F(x, z) = (c(x, z),−b(x, z)) ∀x ∈ Ω, ∀z ∈ IR

the equation (2.12) can be written in the form

NS(x, u(x)) · F(x, u(x)) = 0, ∀x ∈ D . (2.13)

Thus it holds:

Proposition 2.2.1 Let S be a graph of a certain function. Then

S is an integral surface ⇔ NS(q) ⊥ F(q) ⇔ F(q) ∈ TqS

for all q ∈ S, where TqS is the tangent space to S at q .

We call characteristic direction at a point q ∈ Ω × IR the directionidentified by F. The relation (2.13) says that the integral surface is tangent

at every point to the characteristic direction in that point. This remarksuggests to build an integral surface by following the curves that solve the

systemΦ = F(Φ) . (2.14)

This system of three ODEs is the so-called system of characteristic equa-

tions . By setting Φ = (x1, x2, z) we can write the system in the followingform

x1(s) = c1(x(s), z(s))

x2(s) = c2(x(s), z(s))z(s) = −b(x(s), z(s))

(2.15)

17

We call characteristic curves the solutions to (2.15). Moreover we call

projected characteristic or simply characteristics the projection ofthe characteristic curves on the domain Ω . By the hypotheses on c andb the system (2.15) has locally a unique solution. Thus for any point

q ∈ Ω× IR there exists only one characteristic passing through it. It holdsthe following result.

Theorem 2.2.1 An integral surface is the union of characteristic

curves, namely S = ∪Φ∗ : Φ∗characteristic curve for p ∈ S .

Proof of Theorem 2.2.1. Let S be an integral surface namely S isthe graph of u : D → IR solution of (2.12). Fix a point q = (x, u(x)) ∈ Sand consider the solution (x(·), z(·)) of (2.15) such that x(0) = x and

z(0) = u(x). This solution will be defined in a certain open interval Icontaining the origin and such that x(s) ∈ Ω for all s ∈ I . It is enough

to show that the parametrized curve is completely contained in S. To thispurpose we consider the function

ϕ(s) = z(s)− u(x(s)) ∀s ∈ I.

We have

ϕ(s) = z(s)−Du(x(s)) · x(s)= −b(x(s), z(s))−Du(x(s)) · c(x(s), z(s))= −b(x(s), ϕ(s) + u(x(s)))−Du(x(s)) · c(x(s), ϕ(s) + u(x(s))) .

Thus ϕ is a solution of the Cauchy problemϕ = f(s, ϕ)ϕ(0) = 0 ,

where f is defined in a neighborhood of (0, 0) in IR2 by:

f(s, y) = −b(x(s), y + u(x(s)))−Du(x(s)) · c(x(s), y + u(x(s))) .

The function f is continuous in I×IR and it is locally Lipschitz continuouswith respect to y. This guarantees the uniqueness of the solution. It follows

18

that ϕ(s) ≡ 0 for every s ∈ I (the function identically zero is a solution

of the Cauchy problem). Thus (x(s), z(s)) ∈ S for every s ∈ I and thisconcludes the proof. 2

Philosophically speaking, one might say that the characteristic curves

take with them an initial piece of information of the initial integral surface,and propagate it with them.

Corollary 2.2.1 If two integral surfaces intersect at a point, thenthey intersect along the characteristic curve that passes through

that point.

2.2.3 Existence and Uniqueness for the Cauchy Problem

We consider now the Cauchy problemc(x, u) ·Du+ b(x, u) = 0 in Ω

u = g on Γ(2.16)

where Γ ⊂ ∂Ω is a regular curve contained in ∂Ω and g : Γ → IR is a givenfunction, which is of class C1 is a neighborhood of Γ.

We are interested in finding local solutions to (2.16) in a neighborhoodof x ∈ Γ, namely classical solutions of the equation (2.12) defined in D∩Ω

where D is an open neighborhood of x and such that u(x) = g(x), forx ∈ D ∩ Γ . We show the following result:

Theorem 2.2.2 Let x ∈ Γ be such that

c(x, g(x)) /∈ TxΓ , (2.17)

then the problem (2.16) has a unique local solution in a neighbor-hood of x .

The hypothesis (2.17) is called transversality condition and it saysthat the projected characteristic through x is not tangent to Γ. In

19

this case we also say that the curve Γ is noncharacteristic for the Cauchy

Problem (2.16) at the point x.

Proof of Theorem 2.2.2. Let ϕ : I ⊆ IR → IR2, r 7→ (ϕ1(r), ϕ2(r))be a regular (ϕ′(r) 6= 0) C1 parametrization of Γ, such that ϕ(r) = x . We

define h(r) = g(ϕ(r)) for all r ∈ I. The boundary condition can be writtenas u(ϕ(r)) = h(r) for all r ∈ I. For every r ∈ I the Cauchy problem

x1(s) = c1(x(s), z(s))x2(s) = c2(x(s), z(s))

z(s) = −b(x(s), z(s))x1(0) = ϕ1(r)x2(0) = ϕ2(r)

z(0) = h(r)

(2.18)

admits a unique solution defined in a neighborhood of 0 and it may dependon r. If r is in a compact W ⊂ I containing r then the domain of the

solution of (2.18) can be taken independent of r.(1) Let us denote by(X(r, ·), Z(r, ·)) the solution of (2.18) which is defined in an interval Jcontaining 0. From the theory of ODEs it follows that X : W × J → IR2,

and Z : W × J → IR are C1(W × J)(2) and satisfy

Xs(r, s) = c(X(r, s), Z(r, s))

Zs(r, s) = −b(X(r, s), Z(r, s))X(r, 0) = ϕ(r)

Z(r, 0) = h(r).

(2.19)

Claim 1: There is an open neighborhood W ⊆ W of r and an open

interval J ⊆ J containing 0, such that the function (r, s) 7→ X(r, s) is

invertible from W × J into X(W × J) =: D . Its inverse function is C1

(1)In the qualitative theory of ODEs the following result holds: Given an open interval I ⊂ IR and anopen set D ⊂ IRn and f : I ×D → IRn continuous and locally Lipschtiz continuous with respect to thesecond variable, then for every K ⊂ I ×D compact, there is δ > 0 such that for every (x0, t0) ∈ K theCauchy problem u(t) = f(t, u), u(t0) = x0 admits a unique solution in (t0 − δ, t0 + δ).

(2)Differentiable dependence on the data of the solution of the Cauchy Problem: Given anopen interval I ⊂ IR and an open set D ⊂ IRn and f : I ×D → IRn continuous. Suppose that for every(x0, t0) ∈ I×D there exists a unique solution φ(·, t0, x0) of the Cauchy Problem u(t) = f(t, u), u(t0) = x0

which is defined in (t0 − δ, t0 + δ) with δ > 0 independent of (t0, x0). If f is of class C1 in I ×D then themap (t, t0, x0) 7→ φ(·, t0, x0) is of class C

1 in (t0 − δ, t0 + δ)× I ×D.

20

Proof of the Claim 1. We verify that X satisfies the hypotheses of

the Inverse Function Theorem(3) in (r, 0) ∈ W × J . We know that X ∈ C1

in an open neighborhood of (r, 0). Moreover

DX(r, 0) = (Dϕ(r), c(X(r, 0), Z(r, 0)) = (Dϕ(r), c(x, g(x)) .

We observe that the matrix DX(r, 0) is invertible by the transversalitycondition. The claim is thus proved.

Hence there are functions S,R : D → IR of class C1(D) such that

x = X(R(x), S(x)), for all x ∈ D .

We define u : D → IR by setting

u(x) = Z(R(x), S(x)), for all x ∈ D .

.

Claim 2: u is a local solution of (2.12).Proof of the Claim 2.

We observe that u ∈ C1(D), since it is the composition of functions ofclass C1. If x ∈ D ∩ Γ then S(x) = 0 and if we set r = R(x) we have

ϕ(r) = X(r, 0) = x and u(x) = g(ϕ(r)) = g(x) . We can represent thegraph of u through the maps R and S:

S = (x, u(x) : x ∈ D = (X(r, s), Z(r, s)) : (r, s) ∈ W × J .

We observe that the tangent space TpS at the point p = (X(r, s), Z(r, s))

is generated by the vectors ((Xr(r, s), Zr(r, s))) and ((Xs(r, s), Zs(r, s))).Since

F (p) = (c(X(r, s), Z(r, s)),−b(X(r, s), Z(r, s))

we get that F (p) ∈ TpS . Therefore u is a solution of the equation (2.12) .

Claim 3: Uniqueness of the maximal solution.

(3)Inverse Function Theorem: Let A ⊂ IRn be an open set and x0 ∈ A and f : A → IRn be of classC1 such that Df(x0) is invertible (Jacf (x0) 6= 0). Then there exits an open neighborhood A of x0 such

that f : A → f(A) is invertible and the inverse is of class C1.

21

Proof of the Claim 3. We suppose that the solution we have built has

a maximal domain D (namely it cannot be extended in a larger domain).We take another solution u with dom(u)=D We set

S0 = (x, u(x)) : x ∈ Γ ∩ D, S0 = (x, u(x)) : x ∈ Γ ∩ D .

By the maximality of u we have S0 ⊆ S0. Then by the Corollary 2.2.1 wehave S ⊆ S. Thus the maximal solution is unique.

We can conclude the proof of Theorem 2.2.2 2

Remark 2.2.1 We observe that Theorem 2.2.2 does not providean explicit formula for the solution of the Cauchy Problem but aconstructive way to get the solution that can be adopted in several

specific case.

2.3 Derivation of characteristic ODEs in the general

case

We return to the general equation

F (x, u,Du) = 0 in Ω (2.20)

subject to the boundary condition

u = g on Γ ⊂ ∂Ω (2.21)

where Ω ⊂ IRn is open.

Also in this case we are going to convert the PDE into an appropriatesystem of ODEs. Suppose u solves (2.20)-(2.21) and fix any point x ∈ Ω.

We would like to calculate u(x) by finding some curves lying within Ω,connecting x with a point x0 ∈ Γ and along which we can compute u atthe one end x0. We hope then to be able to calculate u all along the curve,

and so particularly in x.

22

Finding the characteristic ODEs. Let us suppose the characteristic

curve is described parametrically the function x(s) = (x1(s), . . . xn(s)), theparameter s lying in some subinterval of IR . Assuming u is a C2 solutionof (2.20), we define also z(s) := u(x(s)) and in addition we set p(s) :=

Du(x(s)) . So z(s) gives the values of u along the curve and p(s) recordsthe values of the gradient Du . We must choose the function x(·) in such a

way that we can compute z(·) and p(·) .For this we differentiate p(s) and we get

pi(s) =n∑

j=1

uxixj(x(s))xj(s) . (2.22)

Now we differentiate the PDE (2.20) with respect of xi:

n∑

j=1

Fpj(x, u,Du)uxixj+ Fz(x, u,Du)uxi

+ Fxi(x, u,Du) = 0 .(2.23)

We are able to employ this identity to get rid of the second derivativesterms in (2.22) if we set for j = 1, · · · , n

xj(s) = Fpj(x(s), z(s), p(s)) . (2.24)

We evaluate (2.23) at x = x(s) and we obtain the identity

n∑

j=1

Fpj(x(s), z(s), p(s))uxixj+ Fz(x(s), z(s), p(s))p

i(s) (2.25)

+ Fxi(x(s), z(s), p(s)) = 0 .

By combining (2.24), (2.25) and (2.22) we get

pi(s) = −Fz(x(s), z(s), p(s))pi(s)− Fxi

(x(s), z(s), p(s)) . (2.26)

Finally

z(s) =

n∑

j=1

pj(s)Fpj(x(s), z(s), p(s)) . (2.27)

23

By summarizing we get the following system of 2n+ 1 ODEs:

(a)p(s) = −DzF (x(s), z(s), p(s))p(s)−DxF (x(s), z(s), p(s))(b)z(s) = DpF (x(s), z(s), p(s)) · p(s)(c)x(s) = DpF (x(s), z(s), p(s)).

(2.28)

This system of 2n+1 first-order ODE comprises the characteristic equationsof the nonlinear PDE (2.20). The functions p(·), z(·), x(·) are called thecharacteristics. We refer to x(·) as the projected characteristic: it is the

projection of the full characteristics (p(·), z(·), x(·)) onto the region Ω ⊆IRn .

We have shown:

Theorem 2.3.1 (Structure of characteristic ODEs) Let

u ∈ C2(Ω) solve the nonlinear, first-order partial differentialequation (2.1). Assume x(·) solve the ODE (2.28)(c), where

p(·) = Du(x(·)), z(·) = u(x(·)) . Then p(·) solves the ODE(2.28)(a) and z(·) solves the ODE (2.28)(b), for those s for

which x(·) ∈ Ω .

Remark 2.3.1 The key step in the derivation is the setting x =DpF so that the terms involving second derivatives drop out. We

obtain closure of the system, and in particular we are not forcedto introduce ODEs for higher derivatives of u .

2.3.1 Examples

We consider some special cases for which the structure of the characteristicequations is simple. We show how we can sometimes actually solve the

characteristic ODEs and thereby explicitily compute solutions of certainfirst-order PDEs, subject to appropriate boundary conditions.

24

a. F linear

F (x, u,Du) = b(x)u(x) + c(x) ·Du = 0.

Then F (x, z, p) = b(x)z + c(x) · p and so

DpF = c(x) .

The equation (2.28)(c) becomes

x(s) = c(x(s)) ,

an ODE involving only the function x(·). Furthermore equation (2.28)(b)becomes

z(s) = c(x(s)) · p(s) = −b(x(s))z(s) .In summary

x(s) = c(x(s))

z(s) = −b(x(s))z(s) (2.29)

comprise equations for the linear, first-order PDE (the equation for p(·) isnot needed).

a. F quasilinear

This case has been already considered in the section 2.2.2:

F (x, u,Du) = b(x, u(x)) + c(x, u(x)) ·Du = 0.

Then F (x, z, p) = b(x, z) + c(x, z) · p and so

DpF = c(x, z) .

The equation (2.28)(c) becomes

x(s) = c(x(s), z(s)) .

Furthermore equation (2.28)(b) becomes

z(s) = −b(x(s), z(s)) .

25

Consequently x(s) = c(x(s), z(s))

z(s) = −b(x(s), z(s)) (2.30)

are the characteristic equations for the quasilinear first-order PDE (once

again the equation for p() is not needed).

Example 2.3.1 We consider the simpler case of a boundary-value problem

for a semilinear PDE:ux1

+ ux2= u2 in x2 > 0

u = g in x2 = 0 (2.31)

In this case c = (1, 1) and b = −z2. Then (2.30) becomesx1(s) = 1, x2(s) = 1

z(s) = z2(2.32)

Consequently x1(s) = x0 + s, x2(s) = s

z(s) = z0

1−sz0= g(x0)

1−sg(x0)

(2.33)

where x0 ∈ IR, s ≥ 0, provided the dominator is not zero. Fix a point(x1, x2) ∈ x2 > 0 so that (x1, x2) = (x1(s), x2(s)) = (x0 + s, s), that is

x0 = x1 − x2 and s = x2. Then

u(x1, x2) = u(x1(s), x2(s)) = z(s) =g(x0)

1− sg(x0)

=g(x1 − x2)

1− x2g(x1 − x2).

This solution makes sense only if 1− x2g(x1− x2) 6= 0. In the case g(x1) =

cos(x1) then the solution to (2.31) is given by u(x1, x2) = cos(x1−x2)1−x2 cos(x1−x2)

.In this case the first blow-up occurs at x2 = 1 at the points x1 such that

cos(x1 − 1) = 1, namely x1 = 1 + 2kπ, k ∈ Z, (see Fig. 2.1).

Example 2.3.2 We consider the problem

1)

xux + yuy = x+ yey in Ωu(x, y) = ex on Γ .

26

Figure 2.1: Graph of u(x1, x2) =cos(x1−x2)

1−x2 cos(x1−x2).

where Ω = y 6= x+ 1 and Γ = y = x+ 1 .Solution.i) Parametrization of the boundary condition. The graph of the line y =

x+ 1 can be parametrized by ϕ(r) = (r, r+ 1). We set h(r) = u(r, r+ 1) .ii) Characteristic equations.

x(s) = x(s)y(s) = y(s)

z(s) = x(s) + y(s)ey(s)(2.34)

The initial conditions for the system (2.34) are

x(0) = ry(0) = r + 1

z(0) = er(2.35)

27

The general solution is

x(r, s) = res

y(r, s) = (r + 1)es

z(r, s) = res + exp((r + 1)es) + (1− e)er − r.

(2.36)

Now we want to invert (r, s) 7→ (x(r, s), y(r, s)).(iii) Invertibility of the function (r, s) 7→ (x(r, s), y(r, s)).We have to solve

the system x = res

y = (r + 1)es(2.37)

The solution is r = r(x, y) := x

y−x

s = s(x, y) := log(y − x)(2.38)

in the half-plane y > x .(iv) Writing of the solution in function of x and y. The solution is

u(x, y) = z(r(x, y), s(x, y))

= x+ ey + (1− e) exp (x

y − x)− x

y − x.

We observe that even if the coefficients of the PDE are C∞(IR2) as theinitial condition, the solution may not be defined in all IR2 .

Example 2.3.3 (Failure of the transversality condition) The follow-

ing example demonstrates the case where the transversality condition failsalong some interval.

ux + uy = 1 in IR2 \ y = xu(x, x) = x .

(2.39)

This problem has infinitely many solutions.

Proof. You first observe that the transversality condition is violatedidentically. However the characteristic direction is (1, 1, 1), and so it is the

direction of the initial curve. Hence, the initial curve is itself a character-istic curve. Set now the problem

ux + uy = 1u(x, 0) = g(x)

(2.40)

28

for an arbitrary g satisfying g(0) = 0 . The solution of (3.58) is easily found

to be u(x, y) = y + g(x− y).Notice that the Cauchy problem

ux + uy = 1u(x, x) = 1

(2.41)

on the other hand, is not solvable. To see this observe that the transver-sality condition fails again, but now the initial curve is not a characteristic

curve.

c) F fully nonlinearIn the general case, we must integrate the full characteristic equations,

if possible.

Example 2.3.4 Consider the fully nonlinear problemux1

ux2= u in x1 > 0

u = x22 in x1 = 0. (2.42)

Here F (x, z, p) = p1p2−z, and hence the characteristic ODEs (2.28) become

p1 = p1, p2 = p2

z = 2p1p2

x1 = p2, x2 = p1.

(2.43)

We integrate these equations to find

x1(s) = p02(es − 1), x2(s) = x02 + p01(e

s − 1)

z(s) = z0 + p01p02(e

2s − 1)p1(s) = p01e

s, p2(s) = p02es,

(2.44)

where 0 6= x02 ∈ IR, s ∈ IR and z0 = (x02)2.

We must determine p0 = (p01, p02). Since u = x22 on x1 = 0, p02 =

ux2(0, x02) = 2x02. Furthermore the PDE ux1

ux2= u itself implies p01p

02 =

z0 = (x02)2 and so p01 =

x02

2 . Consequently the formulas above become

29

x1(s) = x02(es − 1), x2(s) = x0

2

2 (es + 1)

z(s) = (x02)2e2s

p1(s) = x02

2es, p2(s) = 2x02e

s.

(2.45)

Fix a point (x1, x2) ∈ Ω. Select s and x0 so that

(x1, x2) = (x1(s), x2(s)) = (2x02(es − 1),

x022(es + 1))

This inequality implies x02 =4x2−x1

4 and es = x1+4x2

4x2−x1and so

u(x1, x2) = u(x1(s), x2(s)) = z(s) = (x02)2e2s

=(4x2 + x1)

2

16.

We observe that since x02 6= 0 then it should be x1 6= 4x2. Therefore u is

a solution of the equation in (2.42) in x1 > 1 \ x1 6= 4x. The triple(x0, z0, p0) = ((0, 0), 0, (0, 0)) does not verifies the transversality condition.

2.4 Local Existence: the General case

We want to solve the problem (2.20)-(2.21), at least in a small region nearan appropriate portion Γ ⊂ Ω. It is convenient first to change variables, so

as to flatten out part of the boundary ∂Ω: we first fix any point x0 ∈ ∂Ω.We can find Φ,Ψ: IRn → IRn such that Ψ = Φ−1 and Φ straightens out ∂Ω

near x0. Given u : Ω → IR we set Ω = Φ(Ω), v(y) = u(Ψ(y)). If u ∈ C1(Ω)is a solution of (2.20)-(2.21) then v solves

G(y, v,Dv) = 0 in Ω

v(y) = g(Ψ(y)) on Φ(Γ).

where

G(y, v,Dv) = F (Ψ(y), v(y), Dv(y)D(Φ(Ψ(y)))).

The point is that if we change variables to straighten out the boundary near

x0 the boundary problem (2.20)-(2.21) converts into a problem having thesame form.

30

Therefore we may assume without loss of generality that Γ is flat near

x0, lying in the plane xn = 0. We will denote x = (x, x′) where x =(x1, . . . , xn−1). We are going to construct the solution at least near x0 byusing the characteristic ODEs and for this we must discover appropriate

initial conditions

x(0) = x0, z(0) = z0, p(0) = p0. (2.46)

If the curve x(s) passes through x0, we should insists that

z0 = g(x0). (2.47)

Since u(x1, . . . , xn−1, 0) = g(x1, . . . , xn−1) we may differentiating and find

uxi(x0) = gxi

(x0), ∀i = 1, . . . , n− 1.

As we also want the PDE (2.20) to hold, we should require that p0 satisfies

p0i = gxi

(x0), ∀i = 1, . . . , n− 1F (x0, z0, p0) = 0 .

(2.48)

The conditions (2.47),(2.48) are called compatibility conditions.A triple (x0, z0, p0) ∈ IR2n+1 satisfying (2.47) and (2.48) is called ad-

missible.Now let (x0, z0, p0) be an admissible triple. Since we need to solve the

characteristic ODEs for nearby points as well, we ask if we can somehowperturb (x0, z0, p0) keeping the compatibility conditions. In other words,given a point y = (y1, . . . , yn−1, 0) ∈ Γ with y close to x0 we solve the

characteristics ODE

(a)p(s) = −DzF (x(s), z(s), p(s))p(s)−DxF (x(s), z(s), p(s))

(b)z(s) = DpF (x(s), z(s), p(s)) · p(s)(c)x(s) = DpF (x(s), z(s), p(s)),

(2.49)

with the initial conditions

p(0) = q(y), z(0) = g(y), x(0) = y. (2.50)

31

Our task is to find a function q(·) such that

q(x0) = p0 (2.51)

and (y, g(y), q(y)) is admissible, namely it satisfies

qi(y) = gxi(y) ∀i = 1, . . . , n− 1

F (y, g(y), q(y)) = 0,(2.52)

for all y ∈ Γ close to x0. With this regard the following result holds.

Lemma 2.4.1 (Noncharacteristic boundary conditions)

There exists a unique solution q(·) of (2.51)-(2.52) for all y ∈ Γclose to x0 provided

Fpn(x0, z0, p0) 6= 0. (2.53)

We say that the admissible triple (x0, z0, p0) is noncharacteristic if(2.53) holds.

Proof of Lemma 2.4.1. We apply the Implicit Function Theorem to

the mapping G : IRn × IRn → IRn with

Gi(y, p) = pi − gxi(y), i = 1, . . . , n− 1Gn(y, p) = F (y, g(y)), p).

We observe that G(x0, p0) = 0 and

DpG(x0, p0) =

1 0 0... . . . ...0 1 0

Fp1(x0, z0, p0) . . . . . . Fpn(x

0, z0, p0)

.

Therefore det(DpG(x0, p0) = Fpn(x

0, z0, p0) 6= 0. The Implicit Function The-orem thus ensures we can uniquely solve the identity G(y, p) = 0 for

p = q(y) provided y is close to x0. 2

32

Remark 2.4.1 If Γ is not flat near x0, the condition that Γ benon characteristic reads

DpF (x0, z0, p0) · ν(x0) 6= 0, (2.54)

ν(x0) denoting the outward unit normal to ∂Ω at x0.

The condition (2.54) is the so-called transversality condition that we

have already introduced in (2.17) .Let (x0, z0, p0) be an admissible noncharacteristic triple, then the fol-

lowing result holds:

Lemma 2.4.2 (Local Invertibility) Assume we have the non-

characteristic condition Fpn(x0, z0, p0) 6= 0. Then there exist an

open interval I ⊂ IR containing 0, a neighborhood W of x0 in

Γ ⊂ IRn−1 and a neighborhood V of x0 in IRn such that for eachx ∈ V there exist unique s ∈ I , y ∈ W such that

x = x(s, y).

Tha map x 7→ (s(x), s(x) is C2(V ).

Proof of Lemma 2.4.2. We have x(0, x0) = x0. Moreover x(0, y) =(y, 0), y ∈ Γ, and so

∂xj

∂yi(0, x0) =

δij (j = 1, . . . , n− 1)

0 (j = n).

From the characteristic ODEs (2.49) it follows that ∂xj

∂s(0, x0) = Fpj(x

0, z0, p0).

Therefore det(Dx(0, x0)) = Fpn(x0, z0, p0) 6= 0 and the result follows from

the Inverse Function Theorem. 2

In view of Lemma 2.4.2 for each x ∈ V we can locally solve the equation

x = x(s, y)

for y = y(x) and s = s(x).

33

Finally let us defineu(x) := z(s(x), y(x))

p(x) := p(s(x), y(x))(2.55)

for x ∈ V , y = y(x) and s = s(x).

Theorem 2.4.1 (Local Existence Theorem ) The functionu defined in (2.55) is a C2 solution of the equation

F (x, u(x), Du(x)) = 0, x ∈ V

with the boundary condition

u(x) = g(x), x ∈ Γ ∩ V.

For the proof of Theorem 2.4.1 we refer the reader to Evans’s book.

Next we show an example where the projected characteristics emanating

from distinct points in Γ may intersect outside the domain of the solution.Such an occurance usually signals the failure of our local solution to exist

within all of Ω .

2.4.1 Applications

We turn now to various special cases, to see how the local existence theorysimplifies.

a) F linear.We consider the case

F (x, u,Du) = c(x) ·Du(x) + b(x)u(x) = 0. (2.56)

The noncharacteristic assumption (2.54) at a point x0 ∈ Γ becomes

c(x0) · ν(x0) 6= 0. (2.57)

Furthermore if we specify the boundary condition

u = g, on Γ, (2.58)

34

we can solve (2.52)for q(y) if y ∈ Γ is near x0 and then apply Theorem 2.4.1

to construct a unique solution of (2.56) and (2.58) in some neighborhood Vof x0. Observe that the projected characteristics emanating from distinctpoints on Γ cannot cross, owing to the uniqueness of the solution of the

initial value problem for the ODE x(s) = c(x(s)).b) F quasilinear.

We consider the case

F (x, u,Du) = c(x, u) ·Du(x) + b(x, u(x)) = 0. (2.59)

The noncharacteristic assumption (2.54) at a point x0 ∈ Γ becomes

c(x0, z0) · ν(x0) 6= 0, (2.60)

where z0 = g(x0). If we specify as previously the boundary condition

u = g, on Γ, (2.61)

we can solve (2.52) for q(y) if y ∈ Γ is near x0 and then apply Theorem 2.4.1to construct a unique solution of (2.56) and (2.61) in some neighborhood

V of x0.In contrast with the linear case it is possible that the projected character-

istics emanating from distinct points in Γ may intersect outside V ; such an

occurance usually signals the failure of our local solution to exists withinall of Ω.

Example 2.4.1 (Characteristics for conservation law) We consider the

scalar conservation law:

ut + div(F(u)) = ut + F′(u) ·Du = 0 , (2.62)

in Ω = IRn × (0,+∞) subject to the initial condition

u = g on Γ = IRn × t = 0 (2.63)

Here F : IR → IRn, F = (F 1, . . . , F n) and we set t = xn+1. Writing q =(p, pn+1) and y = (x, t) we have

G(y, z, q) = pn+1 + F′(z) · p .

35

The transversality condition is clearly satisfied at each point y0 = (x0, 0) ∈Γ. The characteristics equations are:

(a)xi(s) = F i(z(s)), i = 1, . . . , n

(b)xn+1(s) = 1(c)z(s) = 0 .

(2.64)

Hence xn+1(s) = s in agreement with our having written xn+1 = t above.In other words, we can identify the parameter s with the time t. We have

z(s) = z0 = g(x0) Thus

x(s) = F′(g(x0))s+ x0 .

Thus the projected characteristics y(s) = (x(s), s) = (F′(g(x0))s + x0, s)are straight lines, along which u is constant.

Suppose now we apply the reasoning to a different initial point w0 ∈ Γ,where g(x0) 6= g(w0). The projected characteristics may possibly intersect

at some time t > 0. Since u ≡ g(x0) on the projected characteristicsthrough x0 and u ≡ g(w0) on the projected characteristics through w0,an apparent contradiction arises. The resolution is that the initial-value

problem (2.62)-(2.63) does not have a smooth solution existing all timest > 0 . There is the need to extend the classical solution to a kind of “weak”

or “generalized” solution.In this case since t = s we have

u(t, x(t)) = z(t) = g(x(t)− tF ′(z0))

= g(x(t)− tF ′(u(t, x(t))).

Henceu = g(x− tF ′(u)).

This gives a solution provided

1 + tg′(x− tF ′(u)) · F ′′(u) 6= 0.

36

We consider the particular case of the Burger’s equation:

ut + uux = 0 in IR× (0,+∞),u(x, 0) = g(x) on IR.

(2.65)

The characteristic ODEs and the corresponding initial values are given by

x(s) = u

t(s) = 1u(s) = 0

x(0) = x0, t(0) = 0, u(0) = g(x0).

(2.66)

Here, x, t and u are all treated as functions of s. By solving for t and ufirst and then for x, we obtain

x(s) = g(x0)s+ x0, t(s) = s, u(s) = g(x0).

By eliminating s from the expression of x and t we have

x(t) = g(x0)t+ x0. (2.67)

By the implicit function theorem, we can solve for x0 in terms of (x, t)in a neighborhood of the origin in IR2. If we denote such a function by

x0 = x0(x, t) we have a solution u = g(x0(x, t)), for any (x, t) sufficientlysmall. We may also write the solution implicitily by u = g(x − ut). Let

γx0be the characteristic curve given by (2.67). It is a straight line in IR2

with a slope 1g(x0)

along which u is the constant g(x0). For x0 < x1 with

g(x0) > g(x1), two characteristic curves γx0and γx1

intersect at (X, T ) with

T = − x0 − x1g(x0)− g(x1)

.

Hence u cannot be extended as a smooth solution up to (X, T ) even as acontinuous function. Such a positive T always exists unless g is nondecreas-

ing. In a special case where g is strictly decreasing, any two characteristiccurves intersect.

1. Let g(x) = −x. In this case γx0is given by x = x0 − x0t, and the

solution on this line is given by u = −x0. We note that each γx0contains

37

the point (x, t) = (0, 1) and hence any two characteristics intersect at (0, 1).

Then, u cannot be extended up to (0, 1) as a smooth solution. In fact wehave x0 =

x1−t and therefore u is given by

u(x, t) =x

t− 1, ∀(x, t) ∈ IR× (0, 1).

Clearly u is not defined at t = 1.

2. Let g(x) = 11+x2 . In this case the projected characteristics are

x(t) = x+1

1 + x2t.

We have

u(x(s), t(s)) = u(x+1

1 + x2s, s) =

1

1 + x2. (2.68)

We observe that x 7→ x + t1+x2 is injective if and only if t < T = 8√

27.

Actually

d

dx

(x+

t

1 + x2

)> 0 ↔ t < min

x>0

(1 + x2)2

2x=

8√27.

For t > 8√27

the derivative changes sign. This means that the projected

characteristics intersect in some point and therefore the problem (2.65) has

not anymore C1 solutions.We can see that lim inft→T− u(x, t) = −∞. Indeed from (2.68) we get

d

dxu(x+

1

1 + x2t, t) = ux(x+

1

1 + x2t, t)∂x

(x+

1

1 + x2t

)

= − 2x

(1 + x2)2.

Moreover

∂x

(x +

1

1 + x2t

)|(x=

√33 ,t= 8√

27)

=

(1− 2xt

(1 + x2)2

)|(x=

√33 ,t= 8√

27)= 0.

Then if we set x =√33 we obtain

limt→T−

ux(x+1

1 + x2t, t)) = lim

t→T−− 2x

(1 + x2)2(1− 2xt

(1 + x2)2)−1 = −∞.

38

c) F fully nonlinear. We consider the case of Hamilton Jacobi Equa-

tions of the formut +H(x,Du) = 0. (2.69)

The characteristic equations for the Hamilton-Jacobi equation are

(a)p(s) = −DxH(x(s), p(s))(b)z(s) = DpH(x(s), p(s)) · p(s)−H(x(s), p(s))

(c)x(s) = DpH(x(s), p(s)),

(2.70)

The first and the third of these equalitiesp(s) = −DxH(x(s), p(s))

x(s) = DpH(x(s), p(s)),(2.71)

are called Hamiltonian’s equations. Observe that the equation for z(·) is

trivial once x(·) and p(·) have been found by solving Hamilton’s equations.The initial-value problem for the Hamilton-Jacobi equation does not in

general have a smooth solution u lasting for all times t > 0.

Exercise 2.4.1 Find the solutions to the following problems.

1)

ut + 2tux = 0

u(x, 0) = e−x2

.

Sol. u(x, t) = e−(x−t2)2.

2)

ut + 3ux = 0

u(x, 0) = sin(x)

Sol. u(x, t) = sin(x− 3t).

3)

ut − x2tux = 0u(x, 0) = x+ 1 .

Sol. u(x, t) = (2−t2)x+22−xt2 .

4)

x2ux1

− x1ux2= u in Ω

u(x1, 0) = g(x1) on Γ .

39

where Ω = x1 > 0, x2 > 0 and Γ = x1 > 0, x2 = 0 .

5)

ux1

+ ux2= u2 in Ω

u(x1, 0) = g(x1) on Γ .

where Ω = x2 > 0 and Γ = x2 = 0 .

6)

x1ux1 + ux2 = 2x1u in Ωu(x1, 0) = g(x1) on Γ .

where Ω = x1 ∈ IR, x2 > 0 and Γ = x2 = 0 .7) Let α ∈ IR and h = (x) be a continuous function in IR. Consider

x2ux1

+ x1ux2= αu in x2 > 0

u(x1, 0) = g(x1) on x2 = 0 .

a) Find all points on x2 = 0 where x2 = 0 is characteristic. What is

the compatibility condition on h at these points?

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

c) For the case h(x) = x and α = 1, check whether this solution can beextended over the points in a).

2.5 Introduction to Hamilton Jacobi Equation

In this section we study the Hamilton-Jacobi equation:ut +H(Du) = 0 in IRn × (0,∞)

u = g in IRn × t = 0 (2.72)

Here u : IRn × [0,∞) → IR is the unknown, u = u(x, t), and Du = Dxu =(ux1

, . . . , uxn) .

The Hamiltonian H : IRn → IR and the initial function g : IRn → R aregiven.

Our goal is to find an appropriate weak or generalized solution, existingfor all times t > 0, even after the method of characteristic failed. Observe

40

that the characteristic equations associated to (2.72) can be reduced to the

system

x(s) = DpH(p(s))z(s) = DpH(p(s))−H(p(s))

p(s) = −DxH(p(s)) = 0 .

(2.73)

As for conservation laws, the problem (2.72) has not a smooth solution ulasting for all times t > 0 .

Example 2.5.1

|u′(t)|2 = 1 in (−1, 1)u(−1) = u(1) = 0

(2.74)

It is clear that by a simple application of Rolle Theorem this problem has

not C1 solution .

Hereafter we make the following assumptions on H.

(A1) H smooth

(A2) H convex

(A3) H is coercive, namely it satisfies

lim|p|→+∞

H(p)

|p| = +∞ .

2.5.1 Legendre transform, Hopf-Lax formula

a. Legendre transform. Let L : IRn → IR satisfy (A2) and (A3).

Definition 2.5.1 The Legendre transform of L is

L∗(p) = supq∈IRn

p · q − L(q) . (2.75)

We observe that the “sup” in (2.75) is really a “max”, that is there

exists some q∗ for which

L∗(p) = p · q∗ − L(q∗)

41

and the mapping q 7→ p · q−L(q) has a maximum at q = q∗. But then p =DL(q∗), provided L is differentiable at q∗. Hence the equation p = DL(q)is solvable for q in terms of p, q∗ = q(p). Therefore

L∗(p) = p · q(p)− L(q(p)).

Theorem 2.5.1 Assume L satisfies (A2) and (A3). Then(i) the mapping p 7→ L∗(p) is convex and

lim|p|→+∞

L∗(p)

|p| = +∞ .

(ii) We have L = L∗∗

b. Hopf-Lax formula We now consider the functional

J [w(·)] :=∫ t

0

L(w(s))ds (2.76)

defined for functions w ∈ A := w(·) ∈ C1([0, t]; IRn)|w(0) = y, w(t) = x .A basic problem in the calculus of variation is to find a curve x(·) ∈ A

which minimizes the functional J .One can show that if x(·) minimizes (2.76) and if p(s) = DqL(x(s)) (the

so-called generalized momentum corresponding to the position x(·)) thenx(·) and p(·) satisfies the ODEs

x(s) = DpH(p(s))p(s) = −DxH(p(s)

(2.77)

where H = L∗ . Moreover the mapping s 7→ H(p(s)) is constant.

In view of the link between a calculus variation problem and the charac-teristics equations of the associated Hamilton-Jacobi equation we conjec-

ture that there is also a connection between the Hamilton-Jacobi PDE anda calculus variation problem. To this purpose we consider the followingfunctional (which takes into account the initial condition for our PDE):

∫ t

0

L(w(s))ds+ g(w(0)). (2.78)

42

where L = H∗ . We set

u(x, t) = inf∫ t

0

L(w(s))ds+ g(y) : w(0) = y, w(t) = x, (2.79)

the infimum taken over all C1 functions w(·) with w(t) = x.The function u is called the value function associated to the minimiza-

tion problem. We propose to investigate the sense in which u defined in(2.79) solves (2.72). We henceforth suppose also g : IRn → IR is Lipschitzcontinuous.

We note that formula (2.79) can be simplified:

Theorem 2.5.2 (Hopf-Lax formula) If x ∈ IRn and t > 0,

then the solution u = u(x, t) of the minimization problem (2.79)is

u(x, t) = infy∈IRn

tL(x− y

t) + g(y) . (2.80)

Definition 2.5.2 We call the expression on the right end side of

(2.80) the Hopf-Lax formula.

Proof of Theorem 2.5.2.1. Fix any y ∈ IRn and define w(s) := y + s

t(x − y) (0 ≤ s ≤ t). The

definition of u implies

u(x, t) ≤∫ t

0

L(w(s))ds+ g(y) = tL(x− y

t) + g(y),

and so

u(x, t) ≤ infy∈IRn

tL(x− y

t) + g(y) .

2. On the other hand, if w(·) is any C1 function satisfying w(t) = x, wehave

L(1

t

∫ t

0

w(s)ds) ≤ 1

t

∫ t

0

L(w(s))ds

43

by Jensen Inequality. Thus if we write y = w(0) we find

tL(x− y

t) + g(y) ≤

∫ t

0

L(w(s))ds+ g(y);

and consequently

infy∈IRn

tL(x− y

t) + g(y) ≤ u(x, t) .

3. We have so far shown

u(x, t) = infy∈IRn

tL(x− y

t) + g(y),

and we leave it as an exercise to prove the infimum above is really a mini-

mum. 2

We are going to show that the formula provides a reasonable weak solutionof the initial-value problem (2.72).

First we record some preliminary observations.

Lemma 2.5.1 For each x ∈ IRn and 0 ≤ s ≤ t we have

u(x, t) = miny∈IRn

(t− s)L(x− y

t− s) + u(y, s) .

In other words, to compute u(·, t), we can calculate u at time s and thenuse u(·, s) as the initial condition on the remaining time interval [s, t]. It is

nothing that the so-called Dynamic Programming Principle.

Lemma 2.5.2 The function u is Lipschitz continuous in IRn ×[0,+∞) .

Now Rademacher’s Theorem asserts that a Lipschitz function is dif-ferentiable almost everywhere. Consequently in view of Lemma 2.5.2 the

function u defined by the Hopf-Lax formula is differentiable for a.e. (x, t) ∈IRn × (0,+∞).

It is easy to see that, in general, u fails to be everywhere differentiable,even if L and g are differentiable.

44

Example 2.5.2 Let us consider the problem (2.72) with n = 1, L(q) =

q2/2 and

g(x) =

−x2 if |x| < 1

1− 2|x| if |x| ≥ 1.

In this case

u(x, t) =

− x2

1−2tif t < 1/2 and |x| < 1− 2t

1− 2(|x|+ t) if |x| ≥ 1− 2t ≥ 0 or if t ≥ 1/2

Therefore u(x, t) is not differentiable in (0, t) with t ≥ 1/2.

Theorem 2.5.3 The function u defined by the Hopf-Lax formula

(2.80) is differentiable a.e. in IRn×(0,+∞) and solves the initial-value problem

ut(x, t) +H(Du(x, t)) = 0 in IRn × (0,+∞)

u(x, 0) = g(x) on IRn × t = 0Proof of Theorem 2.5.3.1. We show that u = g on IRn × t = 0.By choosing y = x in the representation formula (2.80) we discover

u(x, t) ≤ tL(0) + g(x) .

Furthermore

u(x, t) = miny∈IRn

tL(x− y

t) + g(y)

≥ g(x) + miny∈IRn

tL(x− y

t)− Lip(g)|x− y|

= g(x) + minz∈IRn

tL(z)− tLip(g)|z|= g(x)−max

z∈IRn−tL(z) + tLip(g)|z|

= g(x)− t max|w|≤Lip(g)

maxz∈IRn

w · z − L(z)

= g(x)− t max|w|≤Lip(g)

H(w) .

45

Thus

|u(x, t)− g(x)| ≤ Ct ,

where C := max(|L(0)|,maxB(0,Lip(g)) |H|) .2. Let (x, t) ∈ IRn × (0,+∞) be a differentiability point of u. We show

thatut(x, t) +H(Du(x, t)) = 0 .

2i) Fix q ∈ IRn, h > 0. Owing to Lemma 2.5.1

u(x+ hq, t+ h) = miny∈IRn

hL(x+ hq − y

h) + u(y, t)

≤ hL(q) + u(x, t) .

Henceu(x+ hq, t+ h)− u(x, t)

h≤ L(q) .

Let h→ 0+, to compute

q ·Du(x, t) + ut(x, t) ≤ L(q) .

This inequality is valid for all q ∈ IRn, and so

ut(x, t) +H(Du(x, t)) = ut(x, t) + maxq∈IRn

q ·Du(x, t)− L(q) ≤ 0 . (2.81)

The first equality holds since H = L∗ .2ii) Now we choose z such that u(x, t) = tL(x−z

t) + g(z). Fix h > 0 and

set s = t− h, y = stx+ (1− s

t )z. Thenx−zt = y−z

s and thus

u(x, t)− u(y, s) ≥ tL(x− z

t) + g(z)− [sL(

y − z

s) + g(z)]

= (t− s)L(x− z

t) .

Let h→ 0+ to compute

x− z

t·Du(x, t) + ut(x, t) ≥ L(

x− z

t) .

46

Consequently

ut(x, t) +H(Du(x, t)) = ut(x, t) + maxq∈IRn

q ·Du(x, t)− L(q)

≥ ut(x, t) +x− z

t·Du(x, t)− L(

x− z

t)

≥ 0

This complete the proof. 2

Remark 2.5.1 The problem of solving (2.72) almost everywhere is notenough to characterize the value function. Indeed such a problem can have

more than one solution in the class of Lipschitz continuous functions asthe next example shows.

Example 2.5.3 The problemut +

12u

2x = 0 in IR× (0,∞)

u = 0 in IR× t = 0 (2.82)

admits the solution u ≡ 0. However for any a > 0, the function ua defined

as

ua(x, t) =

0 if |x| ≥ at

a|x| − a2t if |x| < at

is a Lipschitz function satisfying the equation a.e. and ua(x, 0) = 0.

Exercise 2.5.1 By using the Hopf-Lax formula solve the following initial

value problems1.

ut +12|Du|2 = 0 in IRn × (0,∞)

u = |x| in IRn × t = 0 (2.83)

Solution. Set H(p) = 12|p|2. One can easy show that H∗(p) = 1

2 |p|2.

47

The Hopf-Lax formula gives

u(x, t) = infy∈IRn

|x− y|22t

+ |y| (2.84)

Observe that the function fx(y) := |x−y|22t + |y| satisfies lim|y|→+∞ fx(y) =

+∞. Therefore fx admits an absolute minimum.We observe that y = 0 is the only point where f is not differentiable and

fx(0) =|x|22t.

If y 6= 0 then Dfx(y) = −2 (x−y)2t + y

|y| = 0 if and only if |y| = |x| − t,

with |x| − t > 0. Moreover x−yt = y

|y| and fx(y) = |x| − t2.

Therefore if |x| − t > 0 u(x, t) = infy∈IRnfx(y) = |x| − t2otherwise

u(x, t) = infy∈IRnfx(y) = fx(0) =|x|22t. 2

2. ut +

12|Du|2 = 0 in IRn × (0,∞)

u = −|x| in IRn × t = 0 (2.85)

Sol. u(x, t) = −|x| − t/2 .

48

Chapter 3

Laplace Equation

The Laplace operator is probably the most important differential operatorand has a wide range of applications. We study the mean-value prop-

erty of harmonic functions. We demonstrate that the mean-value prop-erty presents an equivalent description of harmonic functions. Due to this

equivalence, the mean-value property provided another tool to study har-monic functions. We discuss the fundamental solution of the Laplace equa-

tion. We introduce the important notion of Green’s functions which aredesigned to solve Dirichlet boundary-value problems. We discuss the reg-ularity of harmonic function using fundamental solution. We finally solve

the Dirichlet problem for the Laplace equation in a large bounded domainsby Perron’s method.

3.1 Basic Properties

Among the most important of all partial differential equation is Laplace’sequation.

∆u = 0 in Ω, (3.1)

where Ω ⊆ IRn is an open set, u : Ω → IR is the unknown. Recall that

∆u =∑n

i=1 uxixi.

49

Definition 3.1.1 A function u ∈ C2(Ω) satisfying (3.1) is calleda harmonic function .

It is useful to have a physical model in mind when thinking of ∆ andperhaps the simplest among several comes from the theory of electrostatics.

According to Maxwell’s equations, an electrostatic field E in space (a vectorfield representing the electrostatic force on a unit positive charge) is related

to the charge density in space, f , by the equation divE = f and alsosatisfies curlE = 0 (in dimension n, curlE is the matrix (∂iE

j−∂jEi)). The

second condition says that at least locally E is the gradient of a function−u, called the electrostatic potential. We therefore have ∆u = −divE = f.

1) A function u(x) = x ·Ax, where A is a n× n matrice, is harmonic if

and only if trace(A) =∑n

i=1Aii = 0.

2) Harmonic functions and holomorphic functions(1)

In 2-D we identify IR2 with C by associating to every pair (x, y) ∈ IR2

the complex number z = x+ iy. The following result holds

Proposition 3.1.1 i) If f : Ω → C is holomorphic in Ω thenu = ℜ(f) and v = ℑ(f) are harmonic in Ω.

(ii) If Ω is a simply connected open subset of IR2 and u : Ω → IRis harmonic, then there exists v : Ω → IR such that u + iv is

holomorphic. v is called the harmonic conjugate of u.

Proof of Proposition 3.1.1. i) Since f is holomorphic, it is infinitely

differentiableand hence so are u and v. In particular, u and v have continuous

second partial derivatives. Moreover u and v satisfy the Cauchy-Riemann

equations ux = −vy and uy = vx in Ω. Hence

uxx + uyy = (ux)x + (uy)y = (vy)x + (−vx)y = vyx − vxy = 0

(1)A function f : Ω ⊂ C → C is holomorphic in Ω if f is complex differentiable at every point z0 ∈ Ω

50

Note that in the last step we used the fact that v has continuous sec-

ond partial derivatives. The proof that v satisfies the Laplace equation iscompletely analogous.

ii) Consider the differential form ω(x, y) = −uy(x, y)dx + ux(x, y)dy.

We observe that ω ∈ C1(Ω) and it is closed. Since Ω is simply connected,ω admits a primitive, namely there exists a function v such that vx = −uyand vy = −ux. Given z0 = (x0, y0) ∈ Ω the function v is defined by

v(z) = v(x, y) =

∫

γ

ωdℓ, (3.2)

where γ is any curve connecting z0 and z. Since Ω is simply connectedv is well-defined, namely it does not depends on γ. By construction the

function f = u+ iv is holomorphic. We can conclude the proof. 2

Since for all k ∈ IN the function f(z) = zk is holomorphic in C, there ex-ist two homogeneous polynomials of degree k which are harmonic functions

in C.

Invariance of the Laplace Equation

We say that the Laplace equation is invariant with respect a group G of C2

diffeomorphisms in IRn if for every g ∈ G and for every harmonic function

u in a open set Ω u g is harmonic in g−1Ω.

1. Invariance by translations. If u satisfies ∆u = 0 in Ω, then v(x) =

u(τa(x)) = u(x+ a) satisfies ∆v = 0 in Ω− a.

2. Invariance by dilations. If u satisfies ∆u = 0 in Ω, then v(x) =

u(δα(x)) = u(αx) satisfies ∆v = 0 in α−1Ω.

3. Invariance by orthogonal transformations. Let A be an orthogo-

nal matrix (AAT = ATA = Id, AT is the transpose of A). If u satisfies∆u = 0 in Ω, then v(x) = u(Ax) solves ∆v = 0 in ATΩ. In particularthe Laplace equation is invariant by rotations.

51

4. Invariance by inversions. In dimension n 6= 2 the Laplace equation

is not invariant with respect to the inversions. If u satisfies ∆u = 0in Ω (where Ω does not contain the origin), then v(x) = u( x

|x|2 ) isharmonic only in dimension 2. In dimension n > 2 we have to multiply

v by a suitable factor. Precisely if u satisfies ∆u = 0 in Ω thenv(x) = |x|2−nu( x

|x|2 ), x ∈ Ω−1 := x|x|2 , x ∈ Ω, x 6= 0 satisfies

∆v(x) = |x|−(n+2)∆u(x

|x|2).

In particular we deduce that since the constant u(x) = 1 is harmonicthen the function v(x) = |x|2−n is harmonic in IRn \ 0.

5. Conformal invariance in dimension 2. Two domains Ω1 andΩ2 are called conformally equivalent if there exists a diffeomorphism

f : Ω1 → Ω2 such that

|fx| = |fy| and fx · fy = 0 in Ω1. (3.3)

A function f : Ω ⊂ IR2 → IR2 of class C1 satisfying the conditions(3.3) is called conformal. A function f ∈ C1(Ω, IR2) is conformal if

and only if either f or its conjugate f is holomorphic.

Let Ω1 and Ω2 be conformally equivalent and let f : Ω1 → Ω2 be aconformal diffeomorphism. Let u : Ω1 → IR and v : Ω2 → IR such that

u = v f . Then u is harmonic in Ω1 if and only if v is harmonic in Ω2.Thanks to the Riemann Mapping Theorem (2) and the invariance of

the Laplace equation with respect to conformal maps it is equivalentto study the Laplace equation in a general simply connected domainof IR2 and in a the unit disc.

(2)In complex analysis, the Riemann mapping theorem states that if Ω is a non-empty simply connectedopen subset of the complex number plane C which is not all of C, then there exists a biholomorphicmapping f (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from Ω onto theopen unit disk D = z ∈ C : |z| < 1,

52

3.2 Fundamental solution

One good strategy for investigating any partial differential equation is firstto identify some explicit solutions. Moreover in looking for explicit so-

lutions it is often wise to restrict attention to classes of functions withcertain symmetry properties. Since Laplace’s equation is invariant under

rotations, it consequently seems advisable to search for radial solutions,that is functions of r = |x|. We attempt to find a solution u of the Laplace

equation having the form u(x) = f(r).

f ′′(r) +n− 1

rf ′(r) = 0 . (3.4)

Multiplying this equation (3.4) by rn−1 we get

(rn−1f ′(r))′ = 0 .

Consequently for r > 0 we have

f(r) = a log r + b n=2

f(r) = arn−2 + b n > 2 .

We note that f(r) has a singularity at r = 0 as long as it is not a

constant.

Definition 3.2.1 The function

Φ(x) =

− 12π log(|x|) if n = 21

n(n−2)ωn|x|2−n if n > 2

defined for 0 6= x ∈ IRn, is the fundamental solution of Laplace’sequation.

We recall that ωn is the surface area of the unit sphere in IRn.

53

The reason for the particular choice of the constants above will be clear

later. We will sometimes write Φ(x) = Φ(|x|) to emphasize that the fun-damental solution is radial.

We first observe that

|∂xiΦ(x)| ≤ C

|x|n−1and |∂2xixj

Φ(x)| ≤ C

|x|n ,

for some C > 0.

3.2.1 The Poisson Equation in IRn

By construction for any y ∈ IRn x 7→ Φ(x− y) is harmonic as a function of

x, x 6= y. Let us take now f : IRn → IR and note that x 7→ Φ(x− y)f(y),x 6= y is harmonic for each y ∈ IRn, and thus the sum of finitely many suchexpressions built for different point y.

This reasoning might suggest that the convolution u(x) = (Φ ∗ f)(x) =∫IRn Φ(x− y)f(y)dy will solve the Laplace equation.

Nevertheless we cannot just compute

∆u =

∫

IRn

∆xΦ(x− y)f(y)dy (3.5)

since D2Φ(x − y) is not summable near the singularity x = y. We mustproceed more carefully in calculating ∆u.

Theorem 3.2.1 (Solution of the Poisson’s equation) Supposef ∈ C2

c (IRn) (namely C2 with compact support) and let u = φ∗ f .

Then

(i) u ∈ C2(IRn)(ii) −∆u = f in IRn.

Proof.

1. We have

u(x) =

∫

IRn

Φ(x− y)f(y)dy =

∫

IRn

Φ(y)f(x− y)dy,

54

We observe that Φ ∈ L1loc(IR

n) and by applying the properties of convolu-

tion (see Theorem A.1.1) we get that u ∈ C2(IRn) and

∂u

∂xi(x) =

∫

IRn

Φ(y)∂f

∂xi(x− y)dy, (3.6)

∂2u

∂xi∂xj(x) =

∫

IRn

Φ(y)∂2f

∂xi∂xj(x− y)dy. (3.7)

2. Since Φ blows up at 0 we need to isolate the singularity inside a smallball. So fix ε > 0. Then

∆u(x) =

∫

B(0,ε)

Φ(y)∆xf(x− y)dy

︸ ︷︷ ︸=:Iε

+

∫

IRn\B(0,ε)

Φ(y)∆xf(x− y)dy

︸ ︷︷ ︸=:Jε

. (3.8)

Now

|Iε| ≤ C‖D2f‖L∞

∫

B(0,ε)

|Φ(y)|dy ≤Cε2 log(ε) (n = 2)

Cε2 (n ≥ 3)(3.9)

An integration by parts(3) yields

Jε =

∫

IRn\B(0,ε)

Φ(y)∆xf(x− y)dy

= −∫

IRn\B(0,ε)

DΦ(y)Dyf(x− y)dy

+

∫

∂B(0,ε)

Φ(y)∂f

∂ν(x− y)dσ(y)

=: Kε + Lε, (3.10)

where ν denote the inward unit normal along ∂B(0, ε). We can check that

|Lε| ≤ C‖Df‖L∞

∫

∂B(0,ε)

|Φ(y)|dσ(y) ≤Cε log(ε) (n = 2)

Cε (n ≥ 3)(3.11)

(3)See Green’s Formula

55

By integrating once again by part in Kε we get

Kε =

∫

IRn\B(0,ε)

∆Φ(y)f(x− y)dy −∫

∂B(0,ε)

∂Φ

∂νf(x− y)dσ(y)

= −∫

∂B(0,ε)

∂Φ

∂νf(x− y)dσ(y),

= − 1

ωnεn−1

∫

∂B(x,ε)

f(y)dσ(y) (3.12)

= −∫

∂B(x,ε)

f(y)dσ(y) → −f(x), as ε→ 0.

By combining (3.8)-(3.12) and letting ε → 0, we find −∆u(x) = f(x) and

we can conclude. 2

Remark 3.2.1 i) We sometimes write

−∆Φ = δ0, in IRn

δ0 denoting the Dirac measure on IRn giving the unit mass to

the point 0. We recall that a function u ∈ L1loc(IR

n) satisfiesthe equation −∆u = δ0 in a distributional sense if for every test

function φ ∈ C∞c (IRn) we have

∫

IRn

u(y)∆φ(y)dx = −φ(0).

ii) Theorem 3.2.1 is in fact valid under some less smoothnessassumptions on f , (see e.g. [3]).

3.3 Sub- and super-harmonic functions

Definition 3.3.1 u ∈ C2(Ω) is said sub- (resp. super-) harmonicif ∆u(x) ≥ 0 (resp. (≤ 0) for every x ∈ Ω .

56

Example 3.3.1 a) u(x) = |x|2 is sub-harmonic and ∆u(x) = 2n

b) If u ∈ C2(Ω) is sub-harmonic and f ∈ C2(IR, IR) is such that f ′ ≥ 0and f ′′ ≥ 0, then f u is sub-harmonic in Ω .

c) Let u ∈ C2(Ω) be harmonic. Then the function v(x) = u2(x) is

sub-harmonic.

Next we show some properties of sub/super harmonic functions.

Theorem 3.3.1 (Weak Maximum Principle) Let Ω ⊆ IRn be abounded open set and u ∈ C2(Ω)∩C(Ω) be a sub-harmonic func-

tion. Then

maxΩ u = max∂Ω u .

Proof. Step 1. We first suppose that ∆u(x) > 0 in Ω, and let x0 ∈ Ωbe such that u(x0) = maxΩ u(x). We clearly have x0 ∈ ∂Ω, otherwise

∆u(x0) ≤ 0 if x0 ∈ Ω . In this case we have maxΩ u = max∂Ω u .Step 2. Suppose now that ∆u(x) ≥ 0 in Ω. We set v(x) = u(x)+ ε|x|2.

We have ∆v = ∆u+ 2εn > 0 in Ω . From Step 1 it follows that maxΩ v =max∂Ω v . Thus

maxΩ

u ≤ max∂Ω

u+ εδ2 ,

where δ = max|x|, x ∈ Ω . By letting ε → 0 we get maxΩ u ≤ max∂Ω u,and we can conclude . 2

Corollary 3.3.1 (Weak Minimum Principle) Let Ω ⊆ IRn be a

bounded open set and u ∈ C2(Ω) ∩ C(Ω) be a super-harmonicfunction. Then

minΩ u = min∂Ω u .

Proof. It is enough to observe that−u is sub-harmonic and maxΩ(−u) =−minΩ u . 2

57

Corollary 3.3.2 Let Ω ⊆ IRn be a bounded open set and u ∈C2(Ω) ∩ C(Ω) be a harmonic function. Then

minΩu = min

∂Ωu and max

Ωu = max

∂Ωu .

Corollary 3.3.3 Let Ω ⊆ IRn be a bounded open set and u ∈C2(Ω) ∩ C(Ω) be a harmonic function. Then

maxΩ |u| = max∂Ω |u| .

Corollary 3.3.4 (Uniqueness of the Dirichlet Problem) Let Ω ⊆IRn be a bounded open set and let f ∈ C2(Ω), ϕ ∈ C(Ω). Then

the following problem

−∆u = f in Ω (Poisson Equation)

u = ϕ in ∂Ω(3.13)

admits at most one solution.

Proof. Let u1, u2 be two solutions of (3.13). We have ∆(u1 − u2) = 0

and thus maxΩ |u1 − u2| = max∂Ω |u1 − u2| = 0 . 2.

Exercise 3.3.1 Let Ω ⊆ IRn be a bounded open set, v : Ω → IR harmonic.If u : Ω → IR is sub-harmonic and u = v on ∂Ω then u ≤ v on Ω (that is

why the name sub-harmonic).

58

Remark 3.3.1 The maximum principle is in general false if Ωis unbounded. The function

u(x) =

log(|x|) if n = 2

1− |x|2−n if n > 2

is harmonic in Ω = x ∈ IRn : |x| > 1, it is null on ∂Ω, but

u(x) > 0 inside Ω.

The maximum principle can be used to get estimates of the solution

(which not a priori known) of a Dirichlet Problem for the Poisson equationon a certain domain, by comparing it with analogous problems where we

explicitely know the solution.

Example 3.3.2 Let QR = (−R,R)n, R > 0 and suppose there is a solu-tion of u ∈ C2(QR) ∩ C(QR) of

−∆u = 1 in QR

u = 0 in ∂QR(3.14)

Then R2

2n ≤ u(0) ≤ R2

2 .

Solution. We have B(0, R) ⊂ QR ⊆ B(0, R√n) . Let us consider the

problems −∆u1 = 1 in B(0, R)u1 = 0 in ∂B(0, R)

(3.15)

−∆u2 = 1 in B(0, R

√n)

u2 = 0 in ∂B(0, R√n)

(3.16)

We have u1(x) = R2−|x|22n , u2(x) = nR2−|x|2

2n , (see Exercise 3.3.2, n.5 ). On

B(0, R) we consider the function v = u − u1. We have ∆v = 0 and v = uon ∂B(0, R). Since u is super-harmonic on QR and u = 0 on ∂QR we haveu ≥ 0 on QR. In particular v ≥ 0 on ∂B(0, R). Thus u ≥ u1 in B(0, R),

being v harmonic. In particular u(0) ≥ u1(0) =R2

2n . In the same way one

can show that u(x) ≤ u2(x) on QR and thus u(0) ≤ u2(0) =nR2

2n . 2

59

Exercise 3.3.2 1. Let ε > 0. Show that on ∂B(0, ε) we have

∂Φ

∂ν= − 1

|∂B(0, ε)| .

Thus∫∂B(0,ε)

∂Φ∂νdS = −1 .

2. If f : Ω ⊆ C → C is holomorphic, then ℜ(f) and ℑ(f) are harmonic.3. Suppose that u ∈ C2(Ω) and x ∈ Ω. Show that

∆u(x) = limr→02nr2

[1ωn

∫∂B(0,1) u(x+ ry)dσ(y)− u(x)

].

Hint: Consider the second-order Taylor polynomial of u about x.

4. Determine a solution of the problem

−∆u(x) = |x| |x| < 1

u(x) = 0 |x| = 1

Sol. u(x) = 1−|x|33(n+1)

5. Let R > 0. Determine the radial solution of the problem

−∆u(x) = 1 |x| < R

u(x) = 0 |x| = R

Sol. u(x) = R2−|x|22n

3.4 Mean-value formulas

Consider an open set Ω ⊆ IRn and suppose that u is a harmonic function

in Ω . We next derive the important mean value formulas, which declarethat u(x) equals both the average of u over the sphere ∂B(x, r) and the

average of u over the entire ball B(x, r), provided B(x, r) ⊂ Ω.

(We can in general take r < dist(x,∂Ω)4 in order to be sure that even

B(x, r) ⊂ Ω.)

60

Definition 3.4.1 Let Ω ⊆ IRn and u is continuous function inΩ. Then

(i) u satisfies the mean-value property over spheres if for anyB(x, r) ⊂ Ω

u(x) = −∫

∂B(x,r)

u(y)dσ =1

ωnrn−1

∫

∂B(x,r)

u(y)dσ;

(ii) u satisfies the mean-value property over balls if for anyB(x, r) ⊂ Ω

u(x) = −∫

B(x,r)

u(y)dy =n

ωnrn

∫

B(x,r)

u(y)dy,

where ωnrn−1 is the surface area of the sphere ∂B(x, r) in

IRn and α(n)rn := ωnrn

n is the volume of the ball B(x, r) inIRn.

These two versions of the mean-value property are equivalent.In fact we can write (i) as

u(x)rn−1 =1

ωn

∫

∂B(x,r)

u(y)dσ;

we can integrate with respect to r and get (ii). If we write (ii) as

u(x)rn =n

ωn

∫

B(x,r)

u(y)dσ;

we can differentiate with respect to r and get (i).

61

Theorem 3.4.1 (Mean-value formulas for the Laplace’s equa-tion) If u ∈ C2(Ω) is harmonic, then

u(x) = −∫

∂B(x,r)

udσ(y) = −∫

B(x,r)

udy (3.17)

for each ball B(x, r) ⊂ Ω .

Proof. 1. Set

φ(r) := −∫

∂B(x,r)

u(y)dσ(y) = −∫

∂B(0,1)

u(x+ rz)dσ(z) .

Then

φ′(r) = −∫

∂B(0,1)

Du(x+ rz) · zdσ(z)

= −∫

∂B(x,r)

Du(y) · y − x

rdσ(y)

= −∫

∂B(x,r)

∂u

∂ν(y)dσ(y)

by Green’s formulas

=r

n−∫

B(x,r)

∆u(y)dy = 0 .

Hence φ is constant, and so

φ(r) = limt→0

φ(t) = limt→0

−∫

∂B(x,t)

u(y)dσ(y) = u(x) .

2. Observe that by employing polar coordinates we get∫

B(x,r)

udy =

∫ r

0

(∫

∂B(x,s)

udσ

)ds

= u(x)

∫ r

0

ωnsn−1ds =

ωn

nrnu(x) . 2

where ωn is the Lebesgue measure of ∂B(0, 1) . 2

62

Theorem 3.4.2 (Converse to mean-value property) If u ∈ C2(Ω)satisfies

u(x) = −∫

∂B(x,r)

udS

for each ball B(x, r) ⊂ Ω , then it is harmonic.

Proof. If ∆u 6= 0, there exists some ball B(x, r) ⊂ Ω such that ∆u > 0within B(x, r). But then for φ as above we have

0 = φ′(r) =r

n

∫

B(x,r)

− ∆u(y)dy > 0,

a contradiction. 2

As a consequence of the mean value property of the harmonic functions

we get the closure of harmonic functions with respect to uniform conver-gence.

Theorem 3.4.3 Let Ω ⊆ IRn, ukk be a sequence of harmonicfunctions on Ω which uniformly converges to a function u. Thenu is harmonic on Ω.

Proof. Let x0 ∈ Ω and r > 0 such that B(x0, r) ⊂ Ω . It holds

uk(x0) = −∫

B(x0,r)

uk(y)dy, for every k ∈ N. (3.18)

We take the limit for k → +∞ in both sides of (3.18) and since uk convergesuniformly in B(x0, r) we can pass to the limit under integral sign. We find

u(x0) = −∫

B(x0,r)

u(y)dy.

By arbitrariness of x0 we deduce that u is harmonic. 2

63

3.4.1 Properties of harmonic functions

We now present a sequence of interesting deductions about harmonic func-tions, all based on the mean-value formulas.

a. Strong Maximum Principle

Theorem 3.4.4 (Strong Maximum Principle) Let Ω ⊆ IRn bea connected bounded open set and u ∈ C2(Ω) ∩ C(Ω) be a sub-

harmonic (resp. super-harmonic) function. If there exists a pointx0 ∈ Ω such that

u(x0) = maxΩ

u (resp. u(x0) = minΩ u)

then u is constant within Ω .

Proof. Let u be a sub-harmonic function and suppose there is a pointx0 ∈ Ω with u(x0) = M := maxΩ u. Then for 0 < r < dist(x0, ∂Ω), the

mean value property asserts

M = u(x0) ≤ −∫

B(x0,r)

udy ≤M

As equality holds only if u ≡ M within B(x0, r), we see u(y) = M for all

y ∈ B(x0, r). Indeed if there is y ∈ B(x0, r) such that u(y) < M thenu(z) < M for all z ∈ B(y, s) ⊆ B(x0, r). Then

M = u(x0) ≤ −∫

B(x0,r)

udy =1

ωnrn

[∫

B(x0,r)\B(y,s)

udz +

∫

B(y,s)

udz

]< M.

and we get a contradiction.

Hence the set x ∈ Ω : u(x) = M is open and relatively closed in Ωand thus equals Ω, being Ω connected. 2.

64

Remark 3.4.1 The strong maximum principle asserts in partic-ular that if Ω is connected and u ∈ C2(Ω) ∩ C(Ω) satisfies

∆u = 0 in Ωu = g in ∂Ω

where g ≥ 0, then is positive everywhere in Ω if g is positive

somewhere on ∂Ω.

b. Regularity Now we prove that if u ∈ C2 is harmonic, then nec-essarily u ∈ C∞. Thus harmonic functions are automatically infinitely

differentiable. This sort of assertion is called a regularity theorem.

Theorem 3.4.5 (Koebe) If u ∈ C(Ω) satisfies the mean-valueproperty (3.4.1) for each B(x, r) ⊂ Ω, then

u ∈ C∞(Ω) .

Proof. Let η be a standard mollifier:

η(x) :=

C exp

(1

|x|2−1

)if |x| < 1

0 if |x| ≥ 1(3.19)

c the constant C > 0 selected so that∫IRn ηdx = 1 .

The function η is radial. Set ηε =1εnη(x

ε) and uε = ηε ∗ u in Ωε = x ∈

Ω : dist(x, ∂Ω) > ε. It is know that uε ∈ C∞(Ωε) .

65

We show that u = uε on Ωε. Indeed if x ∈ Ωε then

uε(x) =

∫

Ω

ηε(x− y)u(y)dy

=1

εn

∫

B(x,ε)

η

( |x− y|ε

)u(y)dy

=1

εn

∫ ε

0

η(rε

)(∫

∂B(x,r)

udS

)dr

=1

εnu(x)

∫ ε

0

η(rε

)nα(n)rn−1dr

= u(x)

∫

B(0,ε)

ηεdy = u(x) .

Thus u ≡ uε on Ωε and so u ∈ C∞(Ωε) for each ε > 0 . 2

Theorem 3.4.6 (Liouville)Let u ∈ C2(IRn) be a bounded harmonic in IRn. Then u is con-stant.

Proof. In Theorem 3.4.5 we show that u ∈ C∞(IRn). We fix i ∈1, . . . n, uxi

is harmonic and thus for every x0 and r > 0 we have

uxi(x0) = −

∫

B(x0,r)

uxi(ξ)dξ .

By the Divergence Theorem we have∫B(x0,r)

uxi(ξ)dξ =∫∂B(x0,r)

uνidS andthus

|uxi(x0)| ≤ 1

rnα(n)

∣∣∣∣∫

∂B(x0,r)

uνidS

∣∣∣∣

≤ 1

rnα(n)‖u‖L∞|∂B(x0, r)|

=1

rnα(n)‖u‖L∞ωnr

n−1 =n‖u‖L∞

r,

where α(n) = ωn

n = |B(0, 1)|.

66

Since limr→+∞n‖u‖L∞

r = 0, we get that for all x0 and for all i ∈ 1, . . . nwe have uxi

(x0) = 0. Hence u is constant in IRn . 2.

Theorem 3.4.7 (Representation formula) Let f ∈ C2c (IR

n) andn ≥ 3. Then any bounded solution of

∆u = f, in IRn

has the form

u(x) =

∫

IRn

Φ(x− y)f(y)dy + C,

for some constant C ∈ IR.

Proof. We observe that u(x) =∫IRn Φ(x − y)f(y)dy is bounded, since

Φ(x) → 0 as |x| → +∞. Let u ∈ C2(IRn) be another bounded solution,then Liouville Theorem implies that v = u − u is constant and we can

conclude. 2

Theorem 3.4.8 Let Ω ⊆ IRn be an open set and u harmonic inΩ . Then

|Du(x0)| ≤2n+1n

α(n)rn+1||u||L1(B(x0,r)) , (3.20)

for each ball B(x0, r) ⊂ Ω .

Proof. Fix i ∈ 1, . . . n. The function uxi is harmonic on Ω, thus

uxi(x0) =

2n

α(n)rn

∫

B(x0,r/2)

uxi(y)dy .

Thus by the same computations in the proof of Theorem 3.4.6 we get

|u(x)| ≤ 2n

r‖u‖L∞(∂B(x0,r/2)) . (3.21)

67

If x ∈ ∂B(x0, r/2), then B(x, r/2) ⊂ B(x0, r) ⊂ Ω and so

|uxi(x0)| ≤

1

α(n)

(2

r

)n

‖u‖L1(B(x0,r)) (3.22)

by the mean value formulas for harmonic functions. By combining (3.21)

and (3.22) we get (3.20) and we can conclude. 2

Theorem 3.4.9 (Estimates on derivatives of order k ≥ 0)

If u is harmonic in Ω then one can show by induction the followingestimate

|Dαu(x0)| ≤Ck

rn+k||u||L1(B(x0,r)) , (3.23)

for each ball B(x0, r) ⊂ Ω and each multiindex α of order |α| = k.

Here

C0 =1

α(n), Ck =

(2n+1nk

)k

α(n)(k ≥ 1) . (3.24)

Proof. We establish (3.23) by induction on k. The case k = 0 isimmediate being an immediate consequence of the mean value formula.

The case k = 1 has been proved in Theorem 3.4.8.Assume k ≥ 2 and that (3.23) and (3.24) are valid for all balls in Ω and

each multiindex of order less or equal to k − 1. Fix B(x0, r) ⊂ Ωand let|α| = k. Then Dαu = (Dβu)xi

for some i ∈ 1, . . . , n, |β| = k − 1. By

calculations similar to those in Theorem 3.4.8, we establish that

|Dαu(x0)| ≤nk

r‖Dβu‖L∞(∂B(x0,r/k).

If x ∈ ∂B(x0, r/k) then B(x, k−1k r) ⊂ B(x0, r) ⊂ Ω, thus (3.23) and (3.24)

for k − 1 imply

|Dβu(x)| ≤ (2n+1n(k − 1))k−1

α(n)(k−1k r)n+k−1

‖u‖L1(∂B(x0,r).

Combining the previous estimates yields the bound

|Dαu(x0)| ≤(2n+1nk)k

α(n)(r)n+k‖u‖L1(∂B(x0,r).

68

We can conclude. 2

Next we show that the unique bounded harmonic functions are theconstants.

c) Analyticity.

Theorem 3.4.10 Assume that u is harmonic in Ω . Then u is

analytic.

Proof. Let x0 ∈ Ω and r := 14dist(x0, ∂Ω) andM = 1

α(n)rn‖u‖L1(B(x0,2r)) <

+∞.Since B(x, r) ⊂ B(x0, 2r) ⊂ Ω for each x ∈ B(x0, r) we have

‖Dαu‖L∞(B(x0,r)) ≤ M

(2n+1n

r

)|α||α||α|.

From Stirling formula it follows that

|α||α| ≤ ce|α|α!.

Moreover the Multinomial Theorem implies

nk =∑

|α|=k

|α|!α!

whence |α|! ≤ n|α|α!. It follows that

‖Dαu‖L∞(B(x0,r)) ≤M

(2n+1n2e

r

)|α|α! (3.25)

The Taylor series for u at x0 is

∑

α

Dαu(x0)

α!(x− x0)

α.

Claim. This power series converges provided

|x− x0| <r

2n+2n3e.

69

Proof of the Claim.

The rest of order N of the series is given by

RN(x) =∑

|α|=N

Dαu(x0 + t(x− x0))

α!(x− x0)

α

for some t ∈ [0, 1] depending on x. We have

|RN(x)| ≤ CM∑

|α|=N

(2n+1n2e

r

)N ( r

2n+2n3e

)N

≤ CMnN1

(2n)N→ 0 as N → +∞. 2

Theorem 3.4.11 (Harnack Inequality) Let Ω ⊆ IRn be open andV ⊂ Ω be a connected open set such that V ⊂ Ω is compact. Then

there is C > 0 (depending on V ) such that

supV

u ≤ C infVu,

for every nonnegative harmonic function in Ω.

Theorem 3.4.11 says that the values of a nonnegative harmonic function

within V are all comparable. If u is small (large) in some point in V thenit is small (large) everywhere.

For the proof of Theorem 3.4.11 we refer to Theorem 11 Chap-

ter 2 in [2]. We just prove the result mentioned in [2] on thecoverings of a compact connected set which is consequence of the

following two exercises.

Exercise 3.4.1 Let V ⊆ IRn be a connected set and letU1, . . . , Um be nonempty open sets such that V = ∪m

i=1Ui. Thenthere exist i1 = 1, i2, . . . , im = m in 1, . . . , m such that

Uik−1∩ Uik 6= ∅ for k ∈ 2, . . .m− 1.

70

Solution of Exercise 3.4.1. We write i ∼ j if there exits i1 =

i, . . . , ik = j such that Ui∩Ui2 6= ∅, . . . , Uik∩Uj 6= ∅ and set I = j : j ∼ 1.If I 6= 1, . . . , m the sets A = ∪i∈IUi and B = ∪i/∈IUi are nonempty dis-joint open sets of V (if j /∈ I and i ∈ I then Uj ∩ Ui = ∅, otherwise j ∼ i,

i ∼ 1 implies j ∼ 1). This contradicts the fact that V is connected. 2

Exercise 3.4.2 Let U ⊆ IRn be a connected open set, K ⊂ U be acompact set. Then there exists a connected open set V containingK and such that V ⊂ U is compact.

Solution of Exercise 3.4.2. Let us cover K with a finite numberof open balls B1, . . . , Bm whose closure is contained in U. Since U is an

open connected set we can connect the balls through continuous curves.Let γ : [0, 1] → U be one of these curves which connect B1 to B2 andlet B(x1, r), . . . , B(xℓ, r) be such that γ([0, 1]) ⊆ ∪ℓ

i=1B(xi, r) ⊆ U and

B1 ∩ B(x1, r) 6= ∅, B(x1, r) ∩ B(x2, r) 6= ∅, . . . B2 ∩ B(xℓ, r) 6= ∅. This ispossible by Exercise 3.4.1.

It follows that B1 ∪B(x1, r)∪ . . . B(xℓ, r)∪B2 is a connected open set.We connect in the same way B2 and B3. We finally take V the union of

all the elements of these coverings. 2

Exercise 3.4.3 1. Show that if Ω is connected then the Strong Maximum

Principle implies the Weak Maximum Principle .2. Let u ∈ C2(Ω) be harmonic in Ω (connected) and let x0 ∈ Ω such

that |u(x0)| = maxΩ |u|. Show that u is constant.3. Reflection Principle. Let Ω ⊂ IRn be a bounded open subset which

is symmetric with respect to IRn−1 × 0, ((x,−xn) ∈ Ω if (x, xn) ∈ Ω).Let Ω+ denote the set x ∈ Ω, xn > 0. Assume that u ∈ C2(Ω+)∩C(Ω+)is harmonic in Ω+, with u = 0 on Ω+ ∩ xn = 0. Set

v(x) :=

u(x1, . . . , xn−1, xn) if xn > 0−u(x1, . . . , xn−1,−xn) if xn < 0

for x ∈ Ω.

71

i) Let x0 ∈ Ω ∩ (IRn−1 × 0), r > 0 such that B(x0, r) ⊂ Ω and let w

be the solution of ∆w = 0 in B(x0, r)

w = v(x) in ∂B(x0, r)(3.26)

Show that w(x,−xn) = −w(x, xn) in B(x0, r).ii) Deduce that v is harmonic.

Solution of Exercise 3.i) We set w(x, xn) = −w(x,−xn). We observe that wxixi

(x, xn) =−wxixi

(x, xn), for i = 1, . . . , n− 1 and wxnxn(x, xn) = −wxnxn

(x, xn), there-

fore ∆w = 0 in B(x0, r). It is clear that w = v on ∂B(x0, r) since v isodd. It follows that w = w by uniqueness of the solution of the Dirchlet

problem (3.26), (the existence is given by the Poisson formula).ii) v is continuous in Ω and it is harmonic in Ω+ and Ω−. It remains to

check what happens in a neighborhood of the points x0 ∈ Ω∩(IRn−1×0).We take x0 ∈ Ω and r > 0 such that B(x0, r) ⊂ Ω. Let w be the harmonicextension of v Since w is odd we have w(x, 0) = 0 on B(x0, r)∩(IRn−1×0.We B+ = B(x0, r) ∩ Ω+ and B− = B(x0, r) ∩ Ω−. We have

∆w = 0 in B−

w = v(x) in ∂B− (3.27)

and ∆w = 0 in B+

w = v(x) in ∂B+ (3.28)

By uniqueness w = v in B+ and w = v in B−. Therefore w = v in B(x0, r),hence v is harmonic.

An alternative proof. Clearly v is harmonic in Ω+ and in Ω−. More-over v ∈ C(Ω. Let x0 ∈ Ω ∩ (IRn−1 × 0), r > 0 such that B(x0, r) ⊂ Ω.

72

We have∫

B(x0,r)

v(y)dy =

∫

B+(x0,r)

v(y)dy +

∫

B−(x0,r)

v(y)dy

=

∫

B+(x0,r)

v(y)dy −∫

B−(x0,r)

v(y1, . . . , yn−1,−yn)dy

=

∫

B+(x0,r)

v(y)dy −∫

B+(x0,r)

v(y1, . . . , yn−1, yn)dy = 0.

Hence

v(x0) = 0 = −∫

B(x0,r)

v(y)dy

for every x0 ∈ Ω ∩ (IRn−1 × 0), r > 0 such that B(x0, r) ⊂ Ω. Thereforev is harmonic in Ω. 2

4. Prove that there exists a constant C > 0, depending only on n, suchthat

maxB(0,1)

|u| ≤ C( max∂B(0,1)

|u|+ maxB(0,1)

|f |) ,

where u is a smooth solution of

−∆u = f, in B(0, 1)u = g, on ∂B(0, 1)

Hint. Consider the functions v±(x) = u±C(maxB(0,1) |f(x)|)|x|2 , withC > 0 a suitable constant .

5. Generalized Liouville Theorem Suppose that u is harmonic on

IRn and |u(x)| = O(|x|m), as |x| → +∞, for some m ≥ 0. Prove that u isa polynomial of degree at most p.

Solution of Exercise 5.Since u is harmonic, it is analytic. Therefore

u(x) =∑

α

Dαu(x0)

α!(x− x0)

α, ∀x ∈ IRn

where for every r > 0 and |α| = k the following estimate holds:

|Dαu(x0)| ≤(2n+1nk)k

α(n)(r)n+k‖u‖L∞(∂B(x0,r).

73

By assumption u(x) ≤ K|r|m if x ∈ ∂B(x0, r) and r is large enough.

Therefore if |α| = k > m we get

|Dαu(x0)| ≤(2n+1nk)k

α(n)(r)n+krn+m → 0 as r → +∞.

It follows that u is a polynomial of degree at most p.6. Harnack Convergence Theorem Let ukk be an increasing

sequence of harmonic functions in an open connected set Ω ⊆ IRn. Supposethat there is y ∈ Ω such that limk→+∞ uk(y) exists finite. Then uk converges

pointwise to a harmonic function on Ω and the convergence is uniform onthe compact subsets of Ω.

Solution of Exercise 6.

Let K ⊂ Ω be compact and V an open connected set such that

K ∪ y ⊆ V ⊆ V ⊆ Ω

(which exists by Exercise 3.4.2). Let ε > 0 be fixed and N ∈ IN such that0 ≤ um(y)− uℓ(y) ≤ ε if m ≥ ℓ ≥ N . By Harnack Theorem 3.4.11 appliedto um − uℓ there exists C > 0 such that

supV

um − uℓ ≤ C infVum − uℓ ≤ Cε.

Therefore ukk uniformly converges on K. In particular limk uk(x) ex-its finite for every x ∈ Ω. We set u(x) = limk uk(x), x ∈ Ω. Since the

convergence is uniform on compact sets u is harmonic in Ω. 2

7. Removable Singularity Theorem: Suppose u ∈ C2(B(0, R) \0) ∩ C(B(0, R) \ 0) is harmonic in B(0, R) \ 0 and satisfies

u(x) =

o(log(|x|)), n = 2o(|x|2−n), n ≥ 3

as |x| → 0.

Then u can be defined at 0 so that it is harmonic in B(0, R).Solution of Exercise 7. Let u be the solution of

−∆u = 0, in B(0, R)u = u in ∂B(0, R).

74

Then u is bounded in B(0, R) (by Maximum Principle sup|x|≤R |u(x)| ≤max|x|=R |u|) and it is harmonic in B(0, R).

Claim : u = u in B(0, R) \ 0Proof of the Claim. Let w = u− u. Then w is harmonic in B(0, R) \

0. Moreover one can easily check that

limx→0

w(x)

R2−n − |x|2−n= 0, if n ≥ 3, lim

x→0

w(x)

log(R)− log(|x|) = 0 if n = 2.

Therefore for every ε > 0 there exists δ > 0 such that for all 0 < |x| ≤ δ

|w(x)| ≤ ε(R2−n− |x|2−n) if n ≥ 3, |w(x)| ≤ ε(log(R)− log(|x|)) if n = 2.

(3.29)Now we observe that ε(R2−n−|x|2−n) and ε(log(R)−log(|x|)) are harmonic

functions in B(0, R) \ B(0, δ) (resp. when n ≥ 3 and n = 2) and they arezero when |x| = R. The Weak Maximum Principle applied in B(0, R) \B(0, δ) implies that

|w(x)| ≤ ε(R2−n− |x|2−n) if n ≥ 3, |w(x)| ≤ ε(log(R)− log(|x|)) if n = 2,(3.30)

for all δ ≤ |x| ≤ R. By combining (3.30) and (3.31) we get

|w(x)| ≤ ε(R2−n− |x|2−n) if n ≥ 3, |w(x)| ≤ ε(log(R)− log(|x|)) if n = 2,(3.31)

for all 0 < |x| ≤ R. By letting ε→ 0 we get w(x) ≡ 0, namely u(x) ≡ u(x)in B(0, R) \ 0. If we ridefine u(0) := u(0) we get the conclusion.2

3.5 Green’s Functions

Let Ω ⊂ IRn be an bounded open set of class C1. We propose to obtain a

general representation formula for the solution of the Poisson’s equation

−∆u = f, in Ωu = g in ∂Ω.

75

Theorem 3.5.1 (Green’s Identity) Let Ω ⊂ IRn be an boundedopen set of class C1, u ∈ C2(Ω, IR), x ∈ Ω. Then

u(x)+

∫

Ω

Φ(y−x)∆u(y)dy =∫

∂Ω

(Φ(y−x)∂u∂ν

(y)−∂Φ∂ν

(y−x)u(y))dσ(y),(3.32)

where ∂Φ∂ν(y − x) = DΦ(y − x) · ν(y) .

Proof. We first observe that the function y 7→ Φ(y − x) has a singu-

larity in x, but since |Φ(y − x)| = cn|x−y|n−2 if (n > 2) and |Φ(y − x)| =

c2| log(|x− y|)| if (n = 2), the function y 7→ Φ(y − x)∆u(y) is integrablein Ω .

We set v(y) = Φ(y − x) for y ∈ Ω. Given ε > 0 such that B(x, ε) ⊂ Ω,

the function v is definite and harmonic on Ωε = Ω \ B(x, ε).We apply the Green formula:∫

Ωε

u(y)∆Φ(y − x)− Φ(y − x)∆u(y)dy =

∫

∂Ωε

u(y)∂Φ

∂ν(y − x)dσ(y)(3.33)

−∫

∂Ωε

Φ(y − x)(y)∂u

∂ν(y)dσ(y) .

We have ∆Φ(y − x) = 0 in Ωε and ∂Ωε = ∂Ω ∪ ∂B(x, ε) .We have

−∫

Ωε

Φ(y − x)∆u(y)dy =

∫

∂Ω

u(y)∂Φ

∂ν(y − x)dσ(y)−

∫

∂Ω

Φ(y − x)(y)∂u

∂ν(y)dσ(y)

−∫

∂B(x,ε)

u(y)∂Φ

∂ν(y − x)dσ(y) +

∫

∂B(x,ε)

Φ(y − x)(y)∂u

∂ν(y)dσ(y) . (3.34)

The following estimate holds

|∫

∂B(x,ε)

Φ(y − x)(y)∂u

∂ν(y)dσ(y)| ≤

k| log(ε)|ε, n = 2kε, n > 2 .

(3.35)

Thus |∫∂B(x,ε)Φ(y − x)(y)∂u∂ν(y)dσ(y)| → 0 as ε → 0 . On the other hand

we have∫

∂B(x,ε)

u(y)∂Φ

∂ν(y − x)dσ(y) = − 1

|∂B(x, ε)|

∫

∂B(x,ε)

u(y)dσ(y) .

76

This last estimates justifies the particular choice of the coefficients in Φ.

Thus∫∂B(x,ε) u(y)

∂Φ∂ν (y − x)dσ(y) → −u(x) as ε→ 0 .

The Dominated Convergence Theorem permits to conclude that

limε→0

∫

Ωε

Φ(y − x)∆u(y)dy =

∫

Ω

Φ(y − x)∆u(y)dy .

Thus by passing into the limit as ε→ 0 in (3.34) we get (3.32) . 2

3.5.1 Derivation of the Green’s Function

We observe that once we know ∆u in Ω, ∂u∂ν and u on ∂Ω, then the function

u is determined by the formula (3.32).

If v is a harmonic function in Ω then the Green formula gives∫

Ω

v(y)∆u(y)dy =

∫

∂Ω

v(y)∂u

∂νdσ −

∫

∂Ω

u(y)∂v

∂νdσ . (3.36)

By substracting (3.32) and (3.36) we get

u(x) = −∫

Ω

(Φ(y − x)− v(y))∆u(y)dy +

∫

∂Ω

(Φ(y − x)− v(y))∂u

∂ν(y)dσ

+

∫

∂Ω

u(∂v

∂ν− ∂Φ

∂ν)dσ .

Thus if v(y) = Φ(y − x) on ∂Ω , we get

u(x) = −∫

Ω

(Φ(y − x)− v(y))∆u(y)dy

−∫

∂Ω

∂(Φ(y − x)− v(y))

∂νu(y)dσ .

Definition 3.5.1 The Green function for the domain Ω is the

function G(x, y) = Φ(y− x)−Φx(y) where Φx(·) ∈ C2(Ω)∩C(Ω)is the function (if it exists) which solves the problem

∆Φx = 0, in ΩΦx(y) = Φ(y − x) on ∂Ω

77

From the previous computations we get

u(x) = −∫

Ω

G(x, y)∆u(y)dy −∫

∂Ω

∂G(x, y)

∂νu(y)dS , (3.37)

where∂G

∂ν(x, y) = DyG(x, y) · ν(y)

is the outer normal derivatives of G with respect to the variable y. Observethat the term ∂u

∂ν does not appear in the equation (3.37): we introducedthe corrector precisely to achieve this.

The function G(·, ·) depends (if it exists) only on Ω .Suppose now u ∈ C2(Ω) solves the boundary-value problem

−∆u = f in Ω

u = g in ∂Ω(3.38)

for given continuous functions f, g. Plugging into (3.37) we obtain

Theorem 3.5.2 (Representation formula using Green’s func-tion). If u ∈ C2(Ω) solves the problem (3.38) then

u(x) = −∫

∂Ω

g(y)∂G

∂ν(x, y)dσ(y) +

∫

Ω

f(y)G(x, y)dy, (3.39)

x ∈ Ω .

Here we have a formula for the solution of the boundary-value problem(3.38), provided we can construct Green’s function G for the given domain

Ω . This is in general a difficult matter, and can be done only when it hasa simple geometry.

Proposition 3.5.1 Let Ω be a bounded open set of IRn for whichthere exists the Green function G for the Laplacian. Then

i) G is nonnegative.

ii) G is a symmetric function .

78

We leave the proof of i) as an exercise and we refer to Theorem

13, Chapter 2 in [2] for the proof of iii).We observe that if Ω is a bounded connected open subset of IRn then

the Green function (if it exists) is strictly positive.

3.5.2 Green’s function for a ball

We consider the unit ball B(0, 1).

Definition 3.5.2 If x ∈ IRn \ 0, the point

x =x

|x|2

is called the point dual to x with respect to ∂B(0, 1). The mappingx 7→ x is inversion through the unit sphere ∂B(0, 1) . If x = 0,

we take x = ∞ .

We now employ inversion through the sphere to compute the Green’s

function for the unit ball Ω = B(0, 1). We fix x ∈ B(0, 1). We must find acorrector function Φx = Φx(y) solving

∆Φx = 0, in ΩΦx(y) = Φ(y − x) on ∂Ω

Then Green’s function will be

G(x, y) = Φ(y − x)− Φx(y) .

Assume for the moment that n ≥ 3. Now the mapping y 7→ Φ(y − x) isharmonic for y 6= x. Thus y 7→ |x|2−nΦ(y − x) is harmonic for y 6= x, and

soΦx(y) := Φ(|x|(y − x)) (3.40)

79

is harmonic in B(0, 1). Furthermore if y ∈ ∂B(0, 1) and x 6= 0, then

|x|2|y − x|2 = |x|2(|y|2 − 2y · x

|x|2 +1

|x|2)

= |x|2 − 2y · x+ 1 = |y − x|2 .

Thus (|x||y − x|)−(n−2) = |x− y|−(n−2) . Consequently

Φx(y) = Φ(y − x) y ∈ ∂B(0, 1). (3.41)

Definition 3.5.3 The Green’s function for the unit ball is

G(x, y) =

Φ(y − x)− Φ(|x|(y − x)) x 6= 0Φ(|y|)− Φ(1) x = 0

(3.42)

x, y ∈ B(0, 1), x 6= y .

The same formula is valid for n = 2 as well.

Assume now u solves the boundary-value problem

∆u = 0 in B(0, 1)u = g in ∂B(0, 1)

(3.43)

Then using (3.39) we see

u(x) = −∫

∂B(0,1)

g(y)∂G

∂ν(x, y)dσ(y). (3.44)

According to formula (3.42)

∂G

∂yi(x, y) =

∂Φ

∂yi(y − x)− ∂

∂yiΦ(|x|(y − x)) .

But∂Φ

∂yi(y − x) =

1

nα(n)

xi − yi|x− y|n ,

80

furthermore

∂

∂yiΦ(|x|(y − x)) = − 1

nα(n)

yi|x|2 − xi|x− y|n .

if y ∈ ∂B(0, 1) . Accordingly

∂G

∂ν(x, y) =

n∑

i=1

yi∂G

∂yi(x, y) = − 1

nα(n)

1− |x|2|x− y|n .

Hence formula (3.44) yields the representation formula

u(x) =1− |x|2nα(n)

∫

∂B(0,1)

g(y)

|x− y|ndσ(y) .

Suppose now u solves the boundary-value problem

∆u = 0 in B(0, r)

u = g on ∂B(0, r)(3.45)

for r > 0. Then u(x) = u(rx) solves the problem (3.43), with g(x) = g(rx)

replacing g . We change variables to obtain Poisson’s formula

u(x) =r2 − |x|2nα(n)r

∫

∂B(0,r)

g(y)

|x− y|ndσ(y) (3.46)

x ∈ B(0, r) .

The function

K(x, y) =r2 − |x|2nα(n)r

1

|x− y|n

x ∈ B(0, r), y ∈ ∂B(0, r) is called Poisson’s kernel for the ball B(0, r) .We have established (3.46) under the assumption that a smooth solution

of (3.45) exists. We next show that this formula in fact gives a solution:

81

Theorem 3.5.3 (Poisson Formula for the Ball) Assumeg ∈ C(∂B(0, r)) and define u by (3.46). Then(i) u ∈ C∞(B(0, r)),

(ii) ∆u = 0 in B(0, r),(iii) lim

x→x0

x∈B(0,r)

u(x) = g(x0) for every x0 ∈ ∂B(0, r) .

Proof. That u defined by (3.46) is harmonic in B(0, r) is evident fromthe fact that G and hence ∂G

∂ν are harmonic in x. To establish the continuityof u on ∂B(0, r) we use the Poisson formula (3.46) for the special case u = 1

to obtain the identity∫

∂B(0,r)

K(x, y)dσ(y) = 1

for all x ∈ B(0, r) .

Now let x0 ∈ ∂B(0, r) and ε > 0. Choose δ > 0 so that |g(x)−g(x0)| ≤ ε,if |x− x0| ≤ δ and let |g| ≤ M on ∂B(0, r).

Then if |x− x0| ≤ δ2 we have

|u(x)− g(x0)| =

∣∣∣∣∫

∂B(0,r)

K(x, y)(g(y)− g(x0))dσ(y)

∣∣∣∣

≤∫

|y−x0|≤δ

K(x, y)|g(y)− g(x0)|dσ(y)

+

∫

|y−x0|>δ,y∈∂B(0,r)

K(x, y)|g(y)− g(x0)|dσ(y)

≤ ε+2M(r2 − |x|2)rn−2

(δ/2)n.

If |x−x0| is sufficiently small it is clear that |u(x)− g(x0)| < 2ε and hencelimx→x0

x∈B(0,r)

u(x) = g(x0) for every x0 ∈ ∂B(0, r) . 2

82

3.5.3 Green’s function for a half space

Let us consider the half space

IRn+ = x = (x1, . . . , xn) ∈ IRn : xn > 0.

Definition 3.5.4 If x ∈ IRn+, its reflection in the plane ∂IRn

+ isthe point

x = (x1, . . . , xn−1,−xn).

In this case a solution of

∆Φx = 0, in IRn+

Φx(y) = Φ(y − x) on ∂IRn+

is given byΦx(y) = Φ(y − x), x, y ∈ IRn

+.

The idea is that the corrector Φx(y) is built from Φ by reflecting the sin-

gularity from x ∈ IRn+ to x /∈ IRn

+.

Definition 3.5.5 The Green’s function for the half-space is

G(x, y) = Φ(y − x)− Φ(y − x), x, y ∈ IRn+, x 6= y.

Suppose now u solves the boundary-value problem

∆u = 0, in IRn+

u = g on ∂IRn+

(3.47)

Then

u(x) =2xnnα(n)

∫

∂IRn+

g(y)

|x− y|n dσ(y). (3.48)

The function

83

K(x, y) =2xnnα(n)

1

|x− y|n

is the Poisson’s Kernel for IRn+ and (3.48) is the Poisson’s formula.

Theorem 3.5.4 (Poisson Formula for the Half-Space)Assume g ∈ C(Rn−1) ∩ L∞(IRn−1) and define u by (3.48). Then

(i) u ∈ C∞(Rn+) ∩ L∞(Rn

+),(ii) ∆u = 0 in Rn

+,

(iii) limx→x0x∈Rn

+

u(x) = g(x0) for every x0 ∈ ∂Rn+ .

For the Proof we refer to Theorem 14, Chapter 2 in [2].

3.6 Perron’s Method

In this section we solve the Dirichlet problem for the Laplace equation in

bounded domains by Perron’s method. The key tools are the maximumprinciple and the Poisson integral formula. The latter provides the solv-

ability of the Dirichlet problem for the Laplace equation in balls.Let Ω ⊆ IRn be a bounded open set.

Definition 3.6.1 We say that u ∈ C(Ω) is sub-harmonic (resp.super-harmonic) in Ω if for any ball B ⊂ Ω such that B ⊆ Ω andany harmonic h ∈ C2(B) ∩ C(B) such that u ≤ h (resp. u ≥ h)

on ∂B one has u ≤ h (resp. u ≥ h) in B.

Proposition 3.6.1 If u ∈ C2(Ω) then Definition 3.3.1 and Def-

inition 3.6.1 are equivalent.

84

Proof. 1) Let u ∈ C2(Ω) be such that ∆u ≥ 0 in Ω, h be harmonic

in B and u ≤ h in ∂B. Then ∆(u− h) ≥ 0. By the maximum principle wehave

maxB

(u− h) = max∂B

(u− h) ≤ 0.

Therefore u ≤ h in B.2) Conversely if ∆u(x0) < 0 at some x0 ∈ Ω, then ∆u(x) < 0 in

B := B(x0, r) ⊂ Ω. Let w solve

∆w = 0 in Bw = u in ∂B.

(3.49)

The existence of w in B is implied by the Poisson integral formula. We haveu ≤ w in B by assumption.

Next we note that

0 = ∆w > ∆u in B

w = u in ∂B(3.50)

Then by the maximum principle we have w ≤ u. Hence u = w in B, whichcontradicts the fact that ∆w > ∆u. Therefore ∆u ≥ 0. 2

The Strong Maximum Principle (resp. Strong Minimum Principle) holds

for continuous harmonic sub-harmonic (resp. super-harmonic) functions.

Proposition 3.6.2 (Strong Maximum/Minimum Principle)Let Ω ⊆ IRn be a connected open set and u ∈ C(Ω) be sub-

harmonic (resp. super-harmonic). Then for all x ∈ Ω wehave

u(x) < maxΩ

u or u(x) ≡ maxΩ

u

(resp. u(x) > minΩu or u ≡ min

Ωu).

Proof. Let u be sub-harmonic and x0 ∈ Ω be such that M = u(x0) =maxΩ u(x). If u is not constant, then the set x ∈ Ω : u(x) = M is not

85

open. We can suppose that there exists r > 0 such that B(x0, r) ⊂ Ω and

u 6≡ M on ∂B(x0, r). Let w solve

∆w = 0 in B(x0, r)w = u in ∂B(x0, r).

(3.51)

Then u ≤ w on B(x0, r). In particular

M = u(x0) ≤ w(x0) ≤ max∂B(x0,r)

w ≤M.

By the Strong Maximum Principle for harmonic functions we have w ≡u(x0) = M on B(x0, r). Hence u ≡ M on ∂B(x0, r) which is a contradic-

tion. The proof for super-harmonic functions follows by observing that ifu is super-harmonic then v = −u is sub-harmonic. 2

Corollary 3.6.1 Let Ω ⊆ IRn be a connected open set and u, v ∈C(Ω) be resp. sub-harmonic and super-harmonic in Ω with u ≤ von ∂Ω. Then either u < v in Ω or u ≡ v in Ω.

Proof. Exercise.

Next, we describe the Perron’s method.

Let Ω ⊂ IRn be a bounded domain and ϕ ∈ C(∂Ω). We will find afunction u ∈ C∞(Ω) ∩ C(Ω) such that

∆u = 0 in Ω

u = ϕ in ∂Ω.(3.52)

Suppose there exists a solution u = uϕ. Then for any v ∈ C(Ω) which is

sub-harmonic in Ω with v ≤ ϕ on ∂Ω, we obtain v ≤ uϕ in Ω.Hence for any x ∈ Ω

uϕ(x) = supv(x) : v ∈ C(Ω), is subharmonic in Ω and v ≤ ϕ on ∂Ω.(3.53)

We note that the equality holds since uϕ is an element of the set in the

right-hand side. In Perron’s method we will show that uϕ defined in (3.53)is indeed a solution of (3.52) under appropriate assumptions on the domain.

86

Proposition 3.6.3 Let u be sub-harmonic in Ω, B be a ball suchthat B ⊂ Ω and u be a solution of

∆u = 0 in B

u = u in ∂B.(3.54)

The harmonic lifting U of u defined by

U(x) =

u(x) if x ∈ B

u(x) if x ∈ Ω \ B

is sub-harmonic in Ω and u ≤ U in Ω.

Proof. Let B′ be a ball such that B′ ⊂ Ω and h be harmonic in B′,U ≤ h on ∂B′. We observe that U = u on B′ \B. Since u = U on ∂B onehas u ≤ U in B. Therefore u ≤ h on ∂B′ and u ≤ h in B′. In particular

u = U ≤ h on B′ \ B.Now we consider the set B ∩ B′. On ∂B′ ∩ B we have U ≤ h and on

∂B ∩ B′ we have U = u = u ≤ h. Both U and h are harmonic in B ∩ B′

and U ≤ h on ∂(B ∩ B′). By the maximum principle we have U ≤ h in

B ∩B′. Hence U ≤ h in B′. Therefore U is sub-harmonic in Ω. 2

Proposition 3.6.3 asserts that we obtain greater sub-harmonic functionsif we preserve the values of sub-harmonic functions outside the balls and

extend them inside the balls by the Poisson integral formula.

Proposition 3.6.4 Let u1, . . . , uN be sub-harmonic in Ω. Then

u(x) = maxu1(x), . . . , uN(x) is sub-harmonic in Ω.

Proof. Exercise.

Let ϕ ∈ L∞(∂Ω). We set

Sϕ = u ∈ C(Ω), u sub-harmonic, u ≤ ϕ on ∂Ω

87

We observe that Sϕ 6= ∅ since v(x) ≡ inf∂Ω ϕ is sub-harmonic and v ≤ ϕ

in ∂Ω.We prove next the Perron’s Theorem.

Theorem 3.6.1 [Perron’s Theorem] For every x ∈ Ω we havesupv(x) : v ∈ Sϕ < +∞ and the function u(x) =supv(x) : v ∈ Sϕ is harmonic in Ω.

Proof. If v ∈ Sϕ then v ≤ sup∂Ω ϕ in ∂Ω. Hence v ≤ sup∂Ω ϕ in Ω

and u(x) ≤ sup∂Ω ϕ for every x ∈ Ω as well. Now we fix y ∈ Ω, let (vn)nbe a sequence in Sϕ such that vn(y) → u(y). We may suppose withoutloss of generality that vn is increasing. Let B = B(y, r) ⊂ Ω and Vn be

the harmonic lifting of vn on B. Then vn ≤ Vn in B and Vn ≤ ϕ on∂Ω. Therefore Vn ∈ Sϕ, Vn ≤ u in B and vn(y) ≤ Vn(y) ≤ u(y). It

follows that limn→+∞ Vn(y) = u(y). We observe that (Vn)n is monotone:Vn = vn ≤ vn+1 = Vn+1 on ∂B, hence Vn ≤ Vn+1 on B.

By Harnack convergence theorem (see Exercise 3.4.1 8) ), the sequenceunformly converges to a harmonic function V in B. It is clear that V ≤ u in

B and V (y) = u(y). If V (ζ) < u(ζ) for some ζ ∈ B then there exists u ∈ Sϕ

such that V (ζ) < u(ζ). We set wk = maxVk, u and denote by theWk theharmonic lifting of wk. Wk uniformly converges to a harmonic function W

in B. We have u ≤ W and V ≤ W ≤ u in B. Thus V (y) = W (y) = u(y).The function V −W is harmonic in B, V −W ≤ 0 and V (y)−W (y) = 0, by

the Strong Maximum Principle V =W in B in particular V (ζ) =W (ζ) ≥u(ζ) which is a contradiction. Therefore V (ζ) = u(ζ) for every ζ ∈ B and

u is harmonic in Ω. 2

The fact that u = ϕ on ∂Ω is related to geometric properties of ∂Ω.

88

Definition 3.6.2 (Barrier) Let ξ ∈ ∂Ω. A function w ∈ C(Ω)is called barrier function at ξ relative to Ω if

i) w is super-harmonic in Ω,

ii) w > 0 in Ω \ ξ and w(ξ) = 0.

We say that a point ξ ∈ ∂Ω is regular for the laplacian if there

is a barrier function in ξ relative to Ω.

Proposition 3.6.5 Let Ω ⊆ IRn be open. We suppose that theDirichlet problem (3.52) has a solution for every ϕ ∈ C(∂Ω).

Then for every ξ ∈ ∂Ω there exists a barrier in ξ relatively to Ω.

Proof. Let ξ ∈ ∂Ω and set ψ(x) = |x − ξ| for x ∈ ∂Ω. Let U be a

harmonic function in Ω such that U = ψ on ∂Ω.Claim 1: Ψ(x) = |x− ξ| is sub-harmonic in Ω.

Proof of the Claim 1. ∆Ψ(x) = n−1|x−ξ| > 0 in Ω.

We observe that Ψ− U = 0 on ∂Ω. By maximum principle Ψ− U ≤ 0in Ω. Since U(x) ≥ Ψ(x) > 0 in Ω \ ξ we deduce that U is a barrier

function. 2

Conversely it holds:

Theorem 3.6.2 Let ϕ ∈ L∞(∂Ω) and u be the harmonic functiondefined by the Perron’s method. If ξ ∈ ∂Ω is a regular point for

the laplacian and ϕ is continuous in ξ then

limx→ξx∈Ω

u(x) = ϕ(ξ).

Proof. Let w be a barrier function in ξ. Let ε > 0 be an arbitraryconstant. By the continuity of ϕ in ξ there is δ > 0 such that

|ϕ(x)− ϕ(ξ)| ≤ ε,

89

for any x ∈ ∂Ω ∩ B(ξ, δ). We set α = min|x−ξ|≥δ w(x) and M = sup∂Ω |ϕ|.Observe that the function ψ(x) = ϕ(ξ) − ε − βw(x) is sub-harmonic andψ(x) ≤ ϕ(x) for every x ∈ ∂Ω as soon as βw(x) ≥ ϕ(ξ) − ε − ϕ(x) (thisholds for every β > 0, if |x − ξ| ≤ δ and for βw(x) > 2M and ε < M if

|x− ξ| > δ).Hence if β ≥ 2M

αwe have ψ ∈ Sϕ.

In a similar way one can prove that if β ≥ 2Mα the function ζ(x) =

ϕ(ξ) + ε + βw(x) is super-harmonic with ζ(x) ≥ ϕ(x) in ∂Ω. Since u is

harmonic we have

ψ(x) ≤ u(x) ≤ ζ(x), ∀x ∈ Ω

and therefore|u(x)− ϕ(ξ)| ≤ ε+ βw(x) ∀x ∈ Ω.

Since w(x) → 0 as x → ξ we get u(x) → ϕ(ξ) as x → ξ and we canconclude. 2

Corollary 3.6.2 The Dirichlet Problem (3.52) has a solution for

every ϕ ∈ C(∂Ω) if and only if every point ξ ∈ ∂Ω is regular forthe laplacian.

Proof. Exercise.

90

Remark 3.6.1 Barrier functions can be constructed for a largeclass of domains Ω. Take for example the case where Ω satisfies

the exterior sphere condition at ξ ∈ ∂Ω in the sense thatthere exists a ball B(y0, r) such that

Ω ∩B(y0, r) = ∅, Ω ∩B(y0, r) = ξ.

To construct a barrier function at ξ we set

wξ(y) = Φ(ξ − y0)− Φ(y − y0), for any y ∈ Ω,

where Φ is the fundamental solution.It is easy to show that wξ is a barrier function. We note that theexterior sphere condition always hold for C2 domains.

In particular if every point of ∂Ω is regular then there exists theGreen function for Ω.

In summary Perron’s method yields a solution of the Dirichlet problemfor the Laplacian equation. This method depends essentially on the max-

imum principle and the solvability of the Dirichlet problem in balls. Animportant feature here is that the interior existence problem is separated

from the boundary behavior of solutions, which is determined by the localgeometry of domains.

3.7 Poisson Equations: Classical solutions

In this section we discuss briefly the Poisson equation

−∆u = f, in Ω, (3.55)

where Ω ⊆ IRn and f ∈ C(Ω).If u is a smooth solution of (3.55) in Ω, then obviously f is smooth.

Conversely, we ask whether u is smooth if f is smooth. We note that ∆u isjust a linear combination of second derivatives of u. To proceed we define

91

wf(x) =∫ΩΦ(x− y)f(y)dy,

where Φ is the fundamental solution of the Laplace operator. The function

wf is called the Newtonian potential of f in Ω. It is well-defined if Ω isbounded and f ∈ L∞(Ω). We recall that

|DΦ(x− y)| ∼ 1

|x− y|n−1and |D2Φ(x− y)| ∼ 1

|x− y|n ,

as y → x.By differentiating under integral sign formally we have

∂xiwf(x) =

∫

Ω

∂xiΦ(x− y)f(y)dy

for any x ∈ IRn and i = 1, . . . , n. The right hand side is a well-defined

integral and defines a continuous function of x, (Exercise).We need extra conditions to infer that wf ∈ C2 and −∆wf = f.

Lemma 3.7.1 Let Ω ⊆ IRn be a bounded open set, f ∈ L∞(Ω).

Assume that f ∈ Ck−1(Ω) for some integer k ≥ 2. Then wf ∈Ck(Ω) and −∆wf = f in Ω. Moreover if f is smooth in Ω then

wf is smooth in Ω.

Proof. For simplicity we write w = wf .Step 1. Suppose f has compact support in Ω.

By the change of variables z = y − x we have

w(x) =

∫

IRn

Φ(z)f(z + x)dz.

Since f is at least C1 by a simple differentiation under integral sign andan integration by parts we obtain

wxi(x) =

∫

IRn

Φ(z)fxi(z + x)dz =

∫

IRn

Φ(z)fzi(z + x)dz

= −∫

IRn

Φzi(z)f(z + x)dz.

92

For f ∈ Ck−1(Ω), k ≥ 2 we can differentiate under the integral sign to get

∂αwxi(x) =

∫

IRn

Φzi(z)∂αf(z + x)dz,

for any multiindex α with |α| ≤ k − 1. Hence w ∈ Ck in Ω. Moreover if f

is smooth in Ω then w is smooth in Ω.Next we calculate ∆w if f is at least C1. For any x we have

∆w = −∫

IRn

n∑

i=1

Φzi(z)fzi(z + x)dz

= − limε→0

∫

IRn\B(0,ε)

n∑

i=1

Φzi(z)fzi(z + x)dz

= − limε→0

∫

∂B(0,ε)

∂Φ

∂ν(z)f(z + x)dσ(z)

= − limε→0

1

ωnεn−1

∫

∂B(0,ε)

f(z + x)dσ(z) = −f(x),

where ν is the unit exterior normal to the boundary ∂B(0, ε) of the set

IRn \ B(0, ε).Step 2. For every x0 ∈ Ω we prove that w ∈ Ck and −∆w = f in a

neighborhood of x0. To this end, we take r < dist(x0, ∂Ω) and a functionη ∈ C∞

c (IRn) with supp(η) ⊂ B(x0, r), η ≡ 1 in B(x0, r/2). Then we write

w(x) =

∫

Ω

Φ(z)(1− η(z))f(z + x)dz +

∫

Ω

Φ(z)η(z)f(z + x)dz

= w1(z) + w2(z).

We observe that w1(z) is smooth inB(x0, r/4) and ∆w1(z) = 0 inB(x0, r/4).For the second integral, η(z)f(z) is a Ck−1 function with compact support

in Ω. From Step 1 it follows that w2 ∈ Ck(Ω) and −∆w2 = ηf .Therefore w ∈ Ck(B(x0, r/4)) and

−∆w(x) = η(x)f(x) = f(x)

for any x ∈ B(x0, r/4). Moreover if f is smooth in Ω so are w2 and w issmooth in Ω. We can conclude. 2

93

We now prove a regularity for general solutions of (3.2.1).

Theorem 3.7.1 Let Ω ⊆ IRn be an open set, f ∈ C(Ω). Suppose

u ∈ C2(Ω) satisfies (3.55) in Ω. If f ∈ Ck−1(Ω) for some integerk ≥ 2. Then u ∈ Ck(Ω). Moreover if f is smooth in Ω then u is

smooth in Ω.

Proof. Take an arbitrary bounded subset Ω′ of Ω and let w = wf,Ω′ bethe Newtonian potential of f in Ω′. By Lemma 3.7.1 w ∈ Ck and −∆w = f

in Ω′. Now we set v = u−w. Since u ∈ C2, so is v and ∆v = 0. It followsthat v is smooth in Ω′. Therefore u = v + w ∈ Ck(Ω′). It is obvious thatf is smooth in Ω then w and hence u are smooth in Ω. We can conclude.

2

Theorem 3.7.1 is an optimal result concerning the smoothness.

Even though ∆u is just one particular combination of secondderivatives of u, the smoothness of ∆u implies the smoothness ofall second derivatives.

Next, we solve the Dirichlet problem for the Poisson equation.

Theorem 3.7.2 Let Ω ⊆ IRn be a bounded open set satisfying theexterior sphere condition at every boundary point, f ∈ C1(Ω) ∩L∞(Ω) and ϕ ∈ C(∂Ω). Then there exists a solution u ∈ C2(Ω)∩C(Ω) of the Dirichlet Problem

−∆u = f in Ωu = ϕ in ∂Ω.

(3.56)

Moreover if f is smooth in Ω then u is smooth in Ω.

Proof. Let w be the Newtonian potential of f in Ω. By Lemma 3.7.1for k = 2 w ∈ C2(Ω)∩C(Ω) and ∆w = f in Ω. Now consider the Dirichletproblem

94

∆v = 0 in Ωv = ϕ− w in ∂Ω.

(3.57)

Corollary 3.6.2 implies the existence of a solution v ∈ C∞(Ω)) ∩ C(Ω) of

(3.57). Then u = v + w is the desired solution of (3.56). If f is smooth inΩ then u is smooth in Ω by Theorem 3.7.1 and we conclude. 2

It is natural to ask whether the equation (3.55) admits any C2-solutions

if f is continuous. The answer turns out to be negative.

Example 3.7.1 Let f : B(0, 1) ⊂ IR2 → IR be defined by f(0) =

0 and

f(x) =x22 − x21|x|2

[2

(− log |x|)1/2 +1

4(− log |x|)3/2]

for x 6= 0. f is continuous in B(0, 1). Define the function

u(x) =

(x21 − x22)(− log |x|)1/2 x ∈ B(0, 1) \ 0

0 x = 0

One can verify that

−∆u = f in x ∈ B(0, 1) \ 0 (3.58)

and

limx→0

ux1x1= ∞.

Hence u is not in C2(B1). Actually the equation (3.58) has not

C2 solutions.

Better adapted to the Poisson equation are Holder spaces. The study of

the elliptic differential equations in Holder spaces is known as the Schaudertheory. In its simplest form, it asserts that all second derivatives of u are

Holder continuous if ∆u is, (see e.g [3]).

95

Exercise 3.7.1 Let Ω ⊂ IRn be an open set of class C2, then itsatisfies the exterior sphere condition in every ξ ∈ ∂Ω.

Solution.Step 1. We show that if there is c > 0 such that U = Ωc = (x, t) ∈

Rn−1 ×R : t > c2|x|2, then there exists r > 0 such that B = B((0, r), r) ⊆

U.

We observe that (x, t) ∈ B if and only if t ∈ (r −√r2 − |x|2, r +√

r2 − |x|2) and |x| < r. The inequality t > c2 |x|2 is satisfied for (x, t) ∈ B

if r −√r2 − |x|2 > c

2|x|2 and this holds if we take r < 1

c.

Step 2. In the general case let ξ ∈ ∂Ω be of class C2. We choosean orthonormal coordinates system so that the point lies at 0 and the

tangent hyperplane to ∂Ω at 0 is the hyperplane t = 0. By definition ofpoint of class C2 there exists an ǫ-neighborhood V of the origin on which

U is defined by t > f(x) where f is a C2 function, f(0) = f ′(0) = 0.The Hessian matrix of f at 0 is symmetric and hence diagonalizable, with

real eigenvalues λ1, . . . , λn. (These are the principal curvatures of ∂Ω atthe origin). Choose any c > maxλ1, . . . , λn. Then there is δ > 0 suchthat p(x) = c

2|x|2 ≥ f(x) for |x| < δ. Therefore by applying Step 1 if

r < min1c , δ,

ǫ2 it holds:

B((0, r), r) ⊆ (x, t) : |x| < δ, 0 < p(x) < t < ǫ ⊆ U ∩ V.

Moreover B((0, r), r)∩ ∂Ω = 0.Note: If Ω is of class Cα with 0 < α < 2 then the exterior sphere

condition may not be satisfied. If we take for instance the function f(x) =|x|3/2 it is clear that a ball B((0, r), r) cannot be contained in the epigraph

t > |x|3/2, (see Fig. 1 and Fig.2 ). 2

96

Fig.1: Graph of f(x1) = |x1|3/2 Fig.2: Graph of f(x1, x2) = (x21 + x22)3/4

3.8 Energy Methods

In this section we illustrate some “energy methods”, which is to say tech-

niques involving the L2 norms of various expressions.a. Uniqueness.Consider first the boundary-value problem

−∆u = f in Ωu = g in ∂Ω

(3.59)

We set forth a simple alternative proof of the uniqueness. Assume Ω ⊂ IRn

bounded and ∂Ω is C1.

Theorem 3.8.1 (Uniqueness) There exists at most one solution

u ∈ C2(Ω) of (3.59).

Proof. Assume u is another solution and set w = u− u. Then ∆w = 0in Ω and so integration by parts shows

0 = −∫

Ω

w∆wdx =

∫

Ω

|Dw|2dx .

Thus Dw = 0 in Ω and since w = 0 on ∂Ω, we deduce w = 0 in Ω . 2

b. Dirichlet’s principle

Next we show that a solution of (3.59) can be characterized as theminimizer of an appropriate functional. For this, we define the energy

functional

I[w] :=

∫

Ω

1

2|Dw|2 − wfdx,

97

w belonging to the admissible set

A := w ∈ C2(Ω)| w = g on ∂Ω .

Theorem 3.8.2 (Dirichlet Principle)Assume u ∈ C2(Ω) solves (3.59). Then

I[u] = minw∈A

I[w] . (3.60)

Conversely, if u ∈ A satisfies (3.60), then u solves the problem(3.59) .

In other words if u ∈ A, the PDE −∆u = f is equivalent to the state-ment that u minimizes the energy I[·].

Proof. 1. Choose w ∈ A. Then (3.59) implies

0 =

∫

Ω

(−∆u− f)(u− w)dx .

An integration by parts yields

0 =

∫

Ω

Du ·D(u− w)− f(u− w)dx,

and there is no boundary term since u− w = 0 on ∂Ω . Hence∫

Ω

|Du|2 − ufdx =

∫

Ω

Du ·Dw − wfdx∫

Ω

|Du|2dx− ufdx ≤∫

Ω

1

2|Dw|2 +

∫

Ω

1

2|Du|2 − wfdx.

Rearranging we conclude that

I[u] ≤ I[w] (w ∈ A) . (3.61)

Since u ∈ A then (3.60) follows from (3.61).

98

2. Now conversely, suppose (3.60) holds. Fix any v ∈ C2c (Ω) and write

i(τ) := I[u+ τv], (τ ∈ IR) .

Since u + τv ∈ A for each τ , the scalar function i(·) has a minimum at

zero, and thusi′(0) = 0,

provided this derivative exists. But

i(τ) =

∫

Ω

1

2|Du + τDv|2 − (u+ τv)fdx

=

∫

Ω

1

2|Du|2 + τDu ·Dv + 1

2|τDv|2 − (u+ τv)fdx .

Consequently

0 = i′(0) =

∫

Ω

Du ·Dv − vf =

∫

Ω

(−∆u− f)vdx.

This identity is valid for each function v ∈ C2c (Ω) and so −∆u = f in Ω . 2

99

Chapter 4

Heat equation

4.1 Motivations

Let us begin with some motivation. The heat equation is the simplestphysical model for the spread of mass or temperature by diffusion. Think

for example of the transport of smoke in still air. We think of smokeparticles as being much larger than the air molecules, yet sufficiently small

that they feel the random kicks of collisions with many atoms. The averagedisplacement of any particle is zero, yet there are fluctuations about this

mean that increase with time. The macroscopic manifestation of thesefluctuations is diffusion. This microscopic picture of diffusion underlies aclassical theory of diffu- sion derived by Euler (though the name usually

attach to the heat equation is that of Fourier).There are two steps in the modeling process. Let u(x, t) denote the

density of smoke at a position in space. If the smoke only gets transported(and not created or destroyed) then we may use conservation of mass to

writeut + div(F) = 0

where F denotes the flux of particles at any point in space. Since there aretwo unknowns, we need another equation. This is a constitutive relationusually called Ficks law. The flux is related to the density (or temperature)

u through F = −kDu. The experimental basis for Ficks law is that heatflows in the direction of steepest descent.

100

The nonhomogeneous heat equation or diffusion equation is

ut − k∆u = f in Ω× (0,+∞) , (4.1)

where u = u(x, t) is the unknown, k > 0, Ω ⊂ IRn and f ∈ C(Ω× (0,+∞))

is a given function . If f ≡ 0 then the equation (4.1) is called homogeneous.A solution of (4.1) is a function u : Ω×(0,+∞) → IR which is of class C2

with respect the variable x ∈ Ω and of class C1 with respect to t ∈ (0,+∞)which satisfies the equation (4.1) pointwise.

Since the heat equation can be considered as the evolution equation

associated to the Poisson equation, a good strategy to study the heatequation is to reproduce the arguments and the results obtained for the

Poisson’s equation.It will not be restrictive to study the equation (4.1) with k = 1. Indeed

u(x, t) is a solution of (4.1) if and only if v(x, t) = u(x, tk) is a solution of

ut −∆u =1

kf(x,

t

k) in Ω× (0,+∞) , (4.2)

4.2 Derivation of the Fundamental solution

We consider the Heat Equation

ut −∆u = 0 in Ω× (0,+∞) , (4.3)

We observe that the equation (4.3) is invariant by dilation with respectto x and t but with different scaling. Namely for every λ > 0, uλ(x, t) =

u(λx, λ2x) is still a solution of (4.3). We observe also that∫

IRn

uλ(x, t)dx =

∫

IRn

u(λx, λ2x)dx

= λ−n

∫

IRn

u(y, λ2t)dy .

If λ = t−1/2 then

1

tn/2

∫

IRn

u(x√t, 1)dx =

∫

IRn

u(y, 1)dy .

101

We look for a radial symmetric solution with respect to x which satisfies

the normalizing condition∫

IRn

u(x, t)dx = 1, ∀t > 0 . (4.4)

Thus it should be

1

tn/2

∫

IRn

u(x√t, 1)dx =

∫

IRn

u(x, t)dx, ∀t > 0 .

The condition (4.4) and the radial symmetry suggest that we look for asolution of the form

u(x, t) =1

tn/2v

( |x|√t

).

We construct the equation for v. We compute

ut = −n2

v

tn/2+1− 1

2

|x|v′tn/2+3/2

uxixi=

1

tn/2+1

(v′

|x| −x2iv

′

|x|3 +x2iv

′′

|x|2√t

).

Thus u verifies the equation 4.3 if and only if v satisfies

v′′ +

(n− 1

s+s

2

)v′ +

n

2v = 0 in (0,+∞) (4.5)

the normalizing condition with respect the function v is given by∫

IRn

v(|x|)dx = 1 .

We can write the equation (4.5) as follows

(2sn−1v′ + snv)′ = 0

102

thus

2sn−1v′ + snv = C

for some constant C ∈ IR . The easiest choice is to take C = 0.

Actually this is a necessary choice in order that the normalizing condition is satisfied(To check as exercise). It remains to solve the equation

v′ = −s2v

whose general solution is

v(s) = Ae−s2/4 .

The constant A is determined by imposing the normalizing condition

1 = A

∫

IRn

e−|x|2/4dx = A(2√π)n/2,

and thus A = 1(4π)n/2

.To conclude we find

u(x, t) =1

(4πt)n/2e−

|x|24t , ∀(x, t) ∈ IRn × (0,+∞) .

We define the fundamental solution of the heat equation the function

Φ(x, t) =

1

(4πt)n/2e−

|x|24t for x ∈ IRn, t > 0

0 for x ∈ IRn, t < 0 .

We observe that Φ has a singularity in (0, 0).We list the main proprieties of the fundamental solution:

1. Φ ∈ C∞ on its domain and it satisfies the heat equation (4.3)

2. Φ satisfies the normalization condition∫IRn Φ(x, t)dx = 1 for every

t > 0;

3. Φ is radially symmetric with respect to x;

103

4. Φ > 0 in IRn × (0,+∞) ;

5. Φ(·, t) → 0 uniformly in IRn as t→ +∞ ;

6. for every ε > 0 we have Φ(·, t) → 0 uniformly in IRn \ B(0, ε) ast→ 0+ ;

4.3 Homogenous Cauchy Problem

We study the Cauchy problemut −∆u = 0 in IRn × (0,+∞)u(x, 0) = g(x) in IRn (4.6)

where the function g : IRn → IR is given.

Theorem 4.3.1 Given a function g ∈ C(IRn) ∩ L∞(IRn) and lety be defined by

u(x, t) =

∫

IRn

Φ(x− y, t)g(y)dy =1

(4πt)n/2

∫

IRn

e−|x−y|2

4t g(y)dy

(4.7)

for x ∈ IRn, t > 0 . Theni) u ∈ C∞(IRn × (0,+∞));

(ii) u verifies the equation (4.3);(iii) lim(x,t)→(x0,0+) u(x, t) = g(x0) for every x0 ∈ IRn ;

(iv) ‖u(·, t)‖∞ ≤ ‖g‖∞ for every t > 0 .

Proof. 1. The function 1tn/2

e−|x|24t is infinitely differentiable and the

partial derivatives are in L1(IRn × [δ,+∞)) , for every δ > 0 .(1)

(1)One may apply the following result on the convolution of two functions: Let f ∈ Ck(IRn) andg : IRn → IR be measurable such that f ⋆ g exists in a neighborhood of x0. Suppose that there exists aneighborhood U of x0 and for every |α| ≤ k some functions γα ∈ L1(IRn) satisfying |Dαf(x− y)g(y)| ≤γα(y), ∀x ∈ U and for a.e. y ∈ IRn. Then Dα(f ⋆ g)(x0) = (Dαf) ⋆ g(x0).

104

This implies that u ∈ C∞(IRn × (0,+∞)) and

ut −∆u =

∫

IRn

[Φt(x− y, t)−∆Φ(x− y, t)]g(y)dy = 0

.

2. Fix x0 ∈ IRn, ε > 0. Choose δ > 0 such that |g(y) − g(x0)| < ε2 if

|y − x0| < δ.

Consider |x− x0| < δ2.

|u(x, t)− g(x0)| = |∫

IRn

Φ(x− y, t)(g(y)− g(x0))dy|

≤∫

B(x0,δ)

|Φ(x− y, t)(g(y)− g(x0))|dy +∫

Bc(x0,δ)

|Φ(x− y, t)(g(y)− g(x0))|dy

≤ ε

2+ 2||g||∞

∫

|y−x0|≥δ

(1

4πt)n/2e−

|x−y|24t dy

≤ ε

2+ 2||g||∞

∫

|y−x0|≥δ

(1

4πt)n/2e−

|x0−y|216t dy

by setting z = y−x0

4√t

≤ ε

2+ 2||g||∞

(4

π

)n/2 ∫

|z|≥ δ4√t

e−|z|2dz .

Since∫IRn e

−|z|2dz < +∞, there exists t0 > 0 such that for all t < t0

2||g||∞(4

π

)n/2 ∫

|z|≥ δ4√t

e−|z|2dz ≤ ε

2.

Finally if |(x, t)− (x0, 0)| ≤ min(δ/2, t0) , then

|u(x, t)− g(x0)| ≤ ε .

3. For every (x, t) ∈ IRn × (0,+∞) we have

|u(x, t)| ≤∫

IRn

|Φ(x− y, t)g(y)|dy ≤ ||g||∞∫

IRn

|Φ(x− y, t)|dy= ||g||∞

105

since Φ is positive and by the normalization condition. 2

Remark 4.3.1 1. The properties (i), (ii), (iv) of Theorem 4.3.1still hold under the assumption that g ∈ L∞ .2. If g ∈ C(IRn)∩L∞(IRn), g ≥ 0 and g 6≡0 then u(x, t) given by

(4.7) is in fact positive for all x ∈ IRn and t > 0. We interpret thisfact by saying that the heat equation forces infinite propagation

speed for disturbances .3. Even if the initial data g is discontinuous, the solution of the

Cauchy problem given by Theorem 4.3.1 is C∞ for all t > 0 Wesay that the heat equation has a regularizing effect .

4. If g ∈ L1(IRn) then u(·, t) → 0 uniformly in IRn as t→ +∞ ..Indeed

|u(x, t)| ≤ 1

(4πt)n/2

∫

IRn

e−|x−y|2

4t |g(y)|dy

≤ 1

(4πt)n/2

∫

IRn

|g(y)|dy ,

and therefore

supx∈IRn

|u(x, t)| ≤ 1

(4πt)n/2

∫

IRn

|g(y)|dy .

We observe that by Fubini Theorem and the fact that1

(4πt)n/2

∫IRn e

− |x−y|24t dx = 1 we have

∫

IRn

u(x, t)dx =

∫

IRn

g(x)dx, ∀t > 0 .

The conservation of the integral of u(x, t) with respect to x means

that the total amount of heat is conserved.

106

Exercise 4.3.1 For any x ≥ 0 define the error function

erf(x) =2√t

∫ x

0

e−s2ds .

Consider the function

g(x) =

1 |x| ≤ 1/20 otherwise

Show that the solution of (4.6) is given by

u(x, t) = −1

2erf

(x− 1/2

2√t

)+

1

2erf

(x+ 1/2

2√t

)

for all (x, t) ∈ IRn × (0 +∞). Which is the limit of u(x, t) as t→ +∞?

4.3.1 Tychonov’s counterexample

In general the solution of the Cauchy Problemut −∆u = 0 in IRn × (0,+∞)u(x, 0) = f(x) in IRn (4.8)

with f ∈ C(IRn) ∩ L∞(IRn) is not unique in the class of function C(IRn ×[0,+∞)) ∩ C2(IRn × (0,+∞)).

In dimension1, let us consider the problemut − uxx = 0 in IR× (0,+∞)

u(x, 0) = 0 in IR(4.9)

The function u ≡ 0 is clearly a solution. We are going to construct anontrivial solution of (4.9).

Let ϕ : C → C be the function

ϕ(z) =

e−

1z2 z 6= 0

0 z = 0

107

The function ϕ is holomorphic in C\0. Moreover the function t→ ϕ(t),

t ∈ IR is of class C∞(IR) and ϕ(n)(0) = 0 for all n ∈ IN . Let us considerthe series of functions

u(x, t) =∑∞

n=0 ϕ(n)(t) x2n

(2n)!, t ≥ 0, x ∈ IRn.

Claim 1. The sum defining u and the series of the derivatives of any orderconverge uniformly on any set of the form [−R,R]× [T,+∞).

Claim 2. u is a continuous function up to the boundary in the half-spacet ≥ 0.

We observe that from the second claim it follows that u attains theinitial datum 0 at the time t = 0. By the first claim we can interchange

sum and partial derivatives. Then we can compute:

uxx(x, t) =

∞∑

n=1

ϕ(n)(t)x2n−2

(2n− 2)!=

∞∑

n=0

ϕ(n+1)(t)x2n

(2n)!

=∂

∂t

∞∑

n=0

ϕ(n)(t)x2n

(2n)!= ut(x, t).

Proof of Claim 1 For fixed t > 0, by the Cauchy formula for holomor-phic functions we obtain

ϕ(n)(t) =n!

2πi

∫

|z−t|= t2

ϕ(z)

(z − t)n+1dz.

On the circle |z − t| = t2 we have |ϕ(z)| ≤ e−ℜ(1/z2) ≤ e−4/t2. Thus

|ϕ(n)(t)| ≤ n!

2π

∫ 2π

0

e−4/t2

(t/2)n+1

t

2dθ = n!2n

e−4/t2

tn.

The following inequality holds for every n ∈ IN :

n!2n

(2n)!≤ 1

n!

108

Hence we get

|u(x, t)| ≤∞∑

n=0

|ϕ(n)(t)| |x|2n

(2n)!≤

∞∑

n=0

n!2ne−4/t2

tn|x|2n(2n)!

≤ e−4/t2∞∑

n=0

1

n!

( |x|2t

)n

= e−4/t2+|x|2/t,

where the last term converges uniformly for t ≥ T > 0 and |x| ≤ R < +∞.By Weierstrass’ criterion, the sum defining u converges uniformly on

the same set. In particular we get

limt→0

|u(x, t)| ≤ limt→0

e−4/t2+|x|2/t = 0,

with uniform convergence for |x| ≤ R. This proves claim 2.The study of convergence of the series of derivatives is analogous and is

left as an exercises to the reader. 2

4.4 Nonhomogeneous Cauchy Problem

We consider now the problemut −∆u = f(x, t) in IRn × (0,+∞)u(x, 0) = 0 in IRn (4.10)

where the function f : (IRn × (0,+∞)) → IR is given.We observe that for every y ∈ IRn and 0 < s < t the function (x, t) 7→

Φ(x− y, t− s) is a solution of the heat equation.For fixed s, the function

u(x, t; s) =

∫

IRn

Φ(x− y, t− s)f(y, s)dy

solves ut(·; s)−∆(·, s) = 0 in IRn × (s,+∞)u(·, s; s) = f(·, s) in IRn × t = s (4.11)

109

which is just an initial-value problem with the starting time t = 0 replaced

by t = s and g replaced by f(·, s) . Duhamel’s principle asserts that we canbuild that we can build a solution of (4.10) by integrating u(x, t; s) withrespect to s . The idea is to consider

u(x, t) =

∫ t

0

u(x, t; s)ds (x ∈ IRn, t ≥ 0) .

Rewriting we have

u(x, t) =

∫ t

0

∫

IRn

Φ(x− y, t− s)f(y, s)dy

=

∫ t

0

1

(4π(t− s))n/2

∫

IRn

e−|x−y|24(t−s) f(y, s)dyds, (4.12)

for x ∈ IRn, t > 0 .

Theorem 4.4.1 (Solution of nonhomogeneous problem) Assume

that f ∈ C2,1(IRn × (0,+∞)) with compact support and define uby (4.12). Then

(i) u ∈ C2,1(IRn × (0,+∞)) ;(ii) ut(x, t)−∆u(x, t) = f(x, t) in IRn × (0,+∞) ;

(ii) lim(x,t)→(x0,0+) u(x, t) = 0 for each point x0 ∈ IRn .

Proof. Since Φ has a singularity we cannot directly differentiate underthe integral sign. We first make a change of variable:

u(x, t) =

∫ t

0

1

(4πs)n/2

∫

IRn

e−|y|24s f(x− y, t− s)dyds.

As f ∈ C2,1(IRn × (0,+∞)) with compact support and Φ is smooth nears = t > 0 we get

ut(x, t) =

∫ t

0

∫

IRn

Φ(y, s)ft(x− y, t− s)dyds+

∫

IRn

Φ(y, t)f(x− y, 0)dy,

110

and

uxixj(x, t) =

∫ t

0

∫

IRn

Φ(y, s)fxixj(x− y, t− s)dyds.

Therefore ut, uxixj, u, uxi

∈ C(IRn × (0,+∞)). (To check!).

ut(x, t)−∆u(x, t) =

∫ t

0

∫

IRn

Φ(y, s)(ft −∆f)(x− y, t− s)dyds+

∫

IRn

Φ(y, t)f(x−

=

∫ t

0

∫

IRn

Φ(y, s)(−fs −∆f)(x− y, t− s)dyds+

∫

IRn

Φ(y, t)f(x

=

∫ ε

0

∫

IRn

Φ(y, s)(−fs −∆f)(x− y, t− s)dyds︸ ︷︷ ︸

=:Jε

+

∫ t

ε

∫

IRn

Φ(y, s)(−fs −∆f)(x− y, t− s)dyds︸ ︷︷ ︸

=:Iε

+

∫

IRn

Φ(y, t)f(x− y, 0)dy︸ ︷︷ ︸

=:K

.

We estimate the three last terms:

|Jε| ≤ (‖ft‖L∞ + ‖D2f‖L∞)

∫ ε

0

∫

IRn

Φ(y, s)dyds ≤ Cε.

By estimating Iε we perform an integration by parts:

Iε =

∫ t

ε

∫

IRn

[Φs −∆Φ)]︸ ︷︷ ︸=0! :s≥ε>0

(y, s)(x− y, t− s)dyds+

∫

IRn

Φ(y, ε)f(x− y, t− ε)dy

−∫

IRn

Φ(y, t)f(x− y, 0)dy︸ ︷︷ ︸

=K

.

Hence

ut(x, t)−∆u(x, t) = Jε +

∫

IRn

Φ(y, ε)f(x− y, t− ε)dy.

111

By letting ε→ 0 we get

ut(x, t)−∆u(x, t) = f(x, t)

We finally note that

|u(x, t)| ≤ t‖f‖L∞

Therefore limt→0 u(x, t) = 0 and the convergence is uniform in x ∈ IRn. 2

Remark 4.4.1 We can combine Theorems 4.3.1 and 4.4.1 to dis-cover that

u(x, t) =

∫ t

0

∫

IRn

Φ(x−y, t−s)f(y, s)dyds+∫

IRn

Φ(x−y, t)g(y)dy

is, under the hypotheses on g and f as above, a solution ofut −∆u = f(x, t) in IRn × (0,+∞)u(x, 0) = g(x) in IRn (4.13)

4.5 Maximum Principle

Given a domain Ω ⊂ IRn, for every T > 0 we introduce the following

subsets of IRn × [0,+∞)

ΩT = Ω× (0, T ] = (x, t)| x ∈ Ω, 0 < t ≤ TΓT = (Ω× 0) ∪ (∂Ω× [0, T ]) .

Thus ΩT is a cylinder in IRn+1, which is called parabolic cylinder . The set

ΓT is called parabolic boundary.We interpret ΩT as being the parabolic interior of Ω× [0, T ]: note care-

fully that ΩT includes the top Ω × t = T. The parabolic boundary ΓT

comprises the bottom and vertical sides of Ω× [0, T ] but not the top.

112

Theorem 4.5.1 (Weak Maximum Principle) Let Ω ⊂ IRn be abounded open set and let u ∈ C2,1(ΩT )∩C(ΩT ∪ΓT ) be a solutionof the heat equation in ΩT . Then

maxΩT∪ΓT

u = maxΓT

u .

Proof. For every ε > 0 we define

uε(x, t) = u(x, t) + ε|x|2, ∀(x, t) ∈ ΩT ∪ ΓT .

We haveuεt −∆uε = ut −∆u− 2nε < 0, in ΩT .

This implies that the maximum of uε cannot be achieved in ΩT . Indeedif (x, t) ∈ ΩT would be a maximum point for uε then uεt(x, t) ≥ 0 and

D2uε(x, t) ≤ 0. In particular ∆uε(x, t) ≤ 0 and thus uεt(x, t)−∆uε(x, t) ≥ 0which is a contradiction. Thus

maxΩT∪ΓT

uε = maxΓT

uε .

It follows that there exists (xε, tε) ∈ ΓT such that

uε(x, t) ≤ uε(xε, tε), ∀(x, t) ∈ ΩT ∪ ΓT .

ThenmaxΩT∪ΓT

u ≤ u(xε, tε) + ε|xε|2 .

Since Γt is compact, there exists a sequence εk → 0 and a point (x, t)

such that (xεk, tεk) → (x, t). Hence

maxΩT∪ΓT

u ≤ u(x, t) ≤ maxΓT

u .

Obviously maxΓTu ≤ maxΩT∪ΓT

u and this conclude the proof. 2

113

4.5.1 Uniqueness of bounded solutions

Theorem 4.5.2 Let g ∈ C(IRn) ∩ L∞(IRn). Then the Cauchyproblem

ut −∆u = 0 in IRn × (0,+∞)

u(x, 0) = g(x) in IRn (4.14)

has a unique bounded solution.

Proof.Existence: it has already been proved in Theorem 4.3.1.

Uniqueness:Step 1. We first show that if u is a bounded solution of (4.14) then

supIRn×(0,+∞)

u(x, t) = supIRn

g(x) .

We set M = supIRn×(0,+∞) u(x, t) and M0 = supIRn g(x) and we define

uε(x, t) = u(x, t) + ε(−2nt− |x|2), ∀(x, t) ∈ IRn × [0,+∞) .

We have uεt −∆uε = 0 and

uε(x, 0) ≤ g(x)− ε|x|2 ≤M0, ∀x ∈ IRn .

Moreover for R > 0 large enough we have

sup(x,t)∈∂B(0,R)×[0,R]

uε(x, t) ≤ sup(x,t)∈∂B(0,R)×[0,R]

u(x, t)− εR2

≤ M − εR2 < M0 .

Then by the maximum principle

sup(x,t)∈B(0,R)×(0,R]

uε(x, t) ≤ M0

for all R > 0 large enough and thus

uε(x, t) ≤M0, ∀(x, t)IRn × [0,+∞) .

114

By letting ε→ 0 we get the desired estimate.

Step 2. Suppose now that u and u be two bounded solution of (4.14).Then v = u− u is a bounded solution of the Cauchy problem

ut −∆u = 0 in IRn × (0,+∞)

u(x, 0) = 0 in IRn (4.15)

By what it has been proved in Step 1 we get u ≤ u in IRn × (0,+∞) . Byarguing in the same way for the function w = u− u we find that u ≤ u in

IRn × (0,+∞) . Therefore u and u coincide. 2

Remark 4.5.1 The uniqueness for the Cauchy problem holds inthe class of bounded functions. Actually one can prove the unique-

ness in the class of unbounded functions satisfying the growth con-dition |u(x, t)| ≤ Aea|x|

2

, with A, a > 0. This growth condition isnecessary. For instance the function (Tychonov’s Example)

u(x, t) =+∞∑

k=0

x2k

(2k)!

dk

dtke−1/t2, ∀(x, t) ∈ IR× (0,+∞)

is solution of the heat equation and u(x, t) → 0 as t→ 0+ . Thissolution diverges as |x| → +∞ much more rapidly as ea|x|

2

for

every a > 0. Since v(x, t) = u(x + a, t) is a solution for everya ∈ IR we have that the Cauchy problem

ut −∆u = 0 in IR× (0,+∞)

u(x, 0) = 0 in IR(4.16)

has infinite solutions.

4.6 Mean-value formula

We want to derive a kind of analogous to the mean-value property forharmonic functions. Let us observe that for fixed x the spheres ∂B(x, r)

115

are the level sets of fundamental solution Φ(x− y) for Laplace’s equation.

This suggests that for fixed (x, t) the level sets of fundamental solutionΦ(x− y, t− s) for the heat equation may be relevant.

Definition 4.6.1 For fixed x ∈ IRn, t ∈ IR, r > 0, we define

E(x, t, r) := (y, s) ∈ IRn+1 |s ≤ t,Φ(x− y, t− s) ≥ 1

rn

This is a region in space-time, the boundary of which is a level set ofΦ(x− y, t− s). Note that the point (x, t) is at the center of the top.

E(x, t; r) is sometimes called a heat ball.

Theorem 4.6.1 ( A mean-value property for the heat equation)Let Ω ⊂ IRn be open and let u ∈ C2,1(ΩT ) solve the heat equation.Then

u(x, t) =1

4rn

∫ ∫

E(x,t;r)

u(y, s)|x− y|2(t− s)2

dyds , (4.17)

for each E(x, t; r) ⊆ ΩT .

Formula 4.17 is a sort of analogue for the heat equation of the mean-value formulas for Laplace’s equation. Observe that the right hand side

involves only u(y, s) for times s ≤ t . This is reasonable, as the value u(x, t)should not depend upon future times.

Proof. Up to changing coordinates we may suppose that (x, t) = (0, 0)

. We write E(r) = E(0, 0; r) We set

φ(r) :=1

rn

∫ ∫

E(r)

u(y, s)|y|2s2dyds

=

∫ ∫

E(1)

u(ry, r2s)|y|2s2dyds .

116

We compute

φ′(r) =

∫ ∫

E(1)

n∑

i=1

uyiyi|y|2s2

+ 2rus|y|2sdyds

=1

rn+1

∫ ∫

E(r)

n∑

i=1

uyiyi|y|2s2

+ 2rus|y|2sdyds

=: A+ B

Let us introduce the function

ψ := −n2log(−4πs) +

|y|24s

+ n log r, (4.18)

and observe ψ = 0 on ∂E(r), since Φ(y,−s) = r−n on ∂E(r).We use (4.18)

to write

B =1

rn+1

∫ ∫

E(r)

4us

n∑

i=1

yiψyidyds

= − 1

rn+1

∫ ∫

E(r)

4nusψ + 4

n∑

i=1

usyiyiψdyds;

there is no boundary term since ψ = 0 on ∂E(r). Integrating by parts withrespect to s, we discover that

B =1

rn+1

∫ ∫

E(r)

−4nusψ + 4n∑

i=1

uyiyiψsdyds

=1

rn+1

∫ ∫

E(r)

−4nusψ + 4n∑

i=1

uyiyi(−n

2s− |y|2

4s2)dyds

=1

rn+1

∫ ∫

E(r)

−4nusψ − 2n

s

n∑

i=1

uyiyidyds− A

Consequently, since u solves the heat equation,

φ′(r) =1

rn+1

∫ ∫

E(r)

−4n∆uψ − 2n

s

n∑

i=1

uyiyidyds

=1

rn+1

∫ ∫

E(r)

4nuyiψyi −2n

s

n∑

i=1

uyiyidyds = 0

117

Thus φ is constant and therefore

φ(r) = limt→0

φ(t) = u(0, 0) limt→0

1

tn

∫ ∫

E(t)

|y|2s2dyds = 4u(0, 0) .

Above we use the fact

1

tn

∫ ∫

E(t)

|y|2s2dyds = 4 ZEXERCISE

2

4.6.1 Strong Maximum Principle for the Heat Equation

Theorem 4.6.2 Let Ω ⊂ IRn be a bounded connected open set.

Assume that u ∈ C2,1(ΩT ) ∩ C(ΩT ) is a solution of the Heatequation such that

u(x0, t0) = maxΩT

u .

Then u is constant in Ωt0 .

Similar assertions are valid with “min” replacing “max”.

Remark 4.6.1 If u attains its maximum (or minimum) at aninterior point, then u is constant at earlier times. This accords

with our strong intuitive interpretation of the variable t as denot-ing time: the solution will be constant on the interval [0, t0] pro-

vided the initial and the boundary conditions are constant. Thesolution may change at times t > t0, provided the boundary con-

ditions alter after t0. The solution will however not respond tochanges in boundary conditions until these changes happen.

Proof.

118

Figure 4.1: Parabolic Balls of Radius 1 in IR3

1. Suppose there exists a point (x0, t0) ∈ ΩT with u(x0, t0) = M =:maxΩT

u. Then for sufficiently small r > 0, E(x0, t0; r) ⊂ ΩT . We employthe mean-value property to deduce

M = u(x0, t0) =1

4rn

∫ ∫

E(x0,t0;r)

u(y, s)|x0 − y|2(t0 − s)2

dyds ≤M .

Equality holds only if u is identically equal to M within E(x0, t0; r). Con-

sequently u(y, s) =M for all (y, s) ∈ E(x0, t0; r) . .Draw any line segment L in Ω connecting (x0, t0) with some other points

119

(y0, s0) ∈ ΩT with s < t . Consider

r0 := mins ≥ s : u(x, t) =M, ∀(x, t) ∈ L, s ≤ t ≤ t0.

Since u is continuous, the minimum is attained. Then u(z0, r0) = M for

some (z0, r0) ∈ L∩ΩT and so u ≡M on E(z0, r0; r) for all sufficiently smallr > 0. Since E(z0, r0; r) contains L ∩ r0 − σ ≤ t ≤ r0 for some small

σ > 0, we have a contradiction. Thus r0 = s0 and u ≡M on L .2. Now fix a point (x, t) ∈ Ωt0. There exist point x0, x1, . . . , xm = x

such that the line segments in IRm connecting xi−1 to xi lie in Ω for every

i . Select times t0 > t1 > . . . tm = t. Then the line segments in Rn+1

connecting (xi−1, ti−1) to (xi, ti) lie in ΩT . According to Step 1, u ≡M on

each segment and so u(x, t) =M .3. By continuity of the solution u(x, t0) =M for all x ∈ Ω. 2

4.6.2 Regularity

Let Ω ⊂ IRn be a bounded open set and T > 0.. We next show thatsolutions of the homogeneous heat equation in ΩT are smooth.

Theorem 4.6.3 Suppose u ∈ C2,1(ΩT ) ∩ C(ΩT ) is a solution of

the heat equation in ΩT . Then u ∈ C∞(ΩT ).

For the proof we refer the reader to Theorem 8 Section 2.3.3in [2].

Given r > 0 and (x, t) ∈ IRn+1, we define the closed parabolic cylinderas follows:

Cr(x, t) := (y, s) : |y − x|2 ≤ r, t− r2 ≤ s ≤ t

120

Theorem 4.6.4 For each pair of nonnegative integer numbersℓ, k there is a positive constant Cℓ,k such that

maxCr/2(x,t)

|DℓxD

kt u(y, s)| ≤ Cℓ,k‖u‖L1(Cr(x,t))

for all cylinder Cr/2(x, t) ⊂ Cr(x, t) ⊂ ΩT and for all solutions uof the heat equation in ΩT .

For the proof we refer the reader to Theorem 9 Section 2.3.3

in [2].

4.7 Energy Methods

a. Uniqueness.

Let us consider the initial/boundary-value problemut −∆u = f, in ΩT

u = g in ΓT(4.19)

We assume that Ω ⊂ IRn is bounded, open and of class C1. The terminal

time is given.

Theorem 4.7.1 (Uniqueness) There exists at most one solution u ∈ C2,1(ΩT )of (4.19).

Proof. 1. If u is another solution, w = u− u solveswt −∆w = 0, in ΩT

w = 0 in ΓT(4.20)

2. Set

e(t) :=

∫

Ω

w2(x, t)dx 0 ≤ t ≤ T .

121

Then

e′(t) = 2

∫

Ω

wwtdx

= 2

∫

Ω

w∆wdx

= −2

∫

Ω

|Dw|2dx ≤ 0 .

Thuse(t) ≤ e(0) = 0, 0 ≤ t ≤ T .

Consequently w = u− u = 0 . 2

b. Backwards uniqueness A rather more subtle question concerns

uniqueness backward in time for the heat equation. For this suppose thatu and u are both smooth solutions of the heat equation in ΩT , with thesame boundary conditions on ∂Ω :

wt −∆w = 0, in ΩT

w = g in ∂Ω× [0, T ](4.21)

for some function g . Note carefully that we are not supposing u = u at

time t = 0 .

Theorem 4.7.2 (Backwards uniqueness) Let g ∈ C(∂Ω×[0, T ]) . Let u1, u2 ∈C1,2(ΩT ) be two solutions

wt −∆w = 0, in ΩT

w = g in ∂Ω× [0, T ](4.22)

If u1(x, T ) = u2(x, T ) for every x ∈ Ω then u1 = u2 in ΩT .

Proof. We set w = u1 − u2. We have w ∈ C1,2(ΩT ) is a solution

wt −∆w = 0, in ΩT

w = 0 in ∂Ω× [0, T ]

w(x, T ) = 0 in Ω

(4.23)

122

We set

e(t) =

∫

Ω

w2(x, t)dx ∀t ∈ [0, T ] .

We have e(t) ≥ 0 for every t ∈ [0, T ], e(T ) = 0 and as we have already seen

e′(t) = −2

∫

Ω

|∇w|2dx .

We compute

e′′(t) = −2

∫

Ω

∂t(|∇w|2)dx = −4

∫

Ω

∇w · ∇wtdx

= 4

∫

Ω

∆w · wtdx− 4

∫

∂Ω

∂w

∂ν· wtdσ(x)

= 4

∫

Ω

|∆w|2dx .

Moreover

(e′(t))2 = 4

(∫

Ω

w∆wdx

)2

≤ e(t)e′′(t) .

Suppose now that there exits t1 ∈ [0, T ] such that e(t1) > 0 . Since e(T ) = 0it must be t1 < T and there exits t2 ∈ (t1, T ] such that e(t2) = 0 and

e(t) > 0 for all t ∈ [t1, t2). For t ∈ [t1, t2) we set f(t) = log(e(t)) and weobserve that f ′′(t) ≥ 0. Thus f is convex in [t1, t2). But f(t) → −∞ as

t → t−2 which is contradiction. Thus e(t) = 0 for every t ∈ [0, T ], namelyu1 = u2 . 2

123

Appendix A

Convolution product

Let f, g be two measurable functions in IRn. The convolution between fand g is the function f ⋆ g(x) :=

∫IRn f(x − y)g(y)dy which is defined for

all x for which y 7→ f(x − y)g(y) is summable. For instance f ⋆ g is welldefined if f ∈ Cc(IR

n) and g ∈ L1loc(IR

n).

A.1 Properties of the Convolution

We recall the following result

Proposition A.1.1 Let f ∈ L1(IRn), g ∈ Lp(IRn), 1 ≤ p ≤ +∞.

Then for a.e. x ∈ IRn the function y 7→ f(x−y)g(y) is summable,f ⋆ g ∈ Lp(IRn and

‖f ⋆ g‖Lp(IRn) ≤ ‖f‖L1(IRn)‖g‖Lp(IRn). (A.1)

For the proof of Proposition A.1.1 we refer the reader e.g to [1].

One of the most important properties of the convolution is the regular-izing effect. More precisely it holds the following result.

124

Theorem A.1.1 Let f ∈ Ck(IRn) and g : IRn → IR be measur-able. Suppose that f ⋆ g exists in a neighborhood U of x0. If for

x0 ∈ IRn there exist a neighborhood U of x0 and for all |α| ≤ ksome functions ϕα ∈ L1(IRn) such that |Dαf(x− y)g(y)| ≤ ϕα(y)

for a.e x ∈ U , then Dα(f ⋆ g)(x0) = (Dαf ⋆ g)(x0).

Proof. The proof is an application of the derivation property underintegral sign. 2

Next we see some conditions under which the hypotheses of Theorem

A.1.1 are verified.

Proposition A.1.2 The hypotheses of Theorem A.1.1 are veri-

fied in every x0 ∈ IRn if it holds:

i) f ∈ Ck(IRn), Dαf ∈ L∞(IRn) for every |α| ≤ k and g ∈L1(IRn).

ii) f ∈ Ck(IRn), |Dαf(x)| ≤ Cα

(1+|x|2)mα , mα >n2 and g ∈ L∞(IRn).

iii) f ∈ Ckc (IR

n) and g ∈ L1loc(IR

n).

Proof of Proposition A.1.2.

i) We observe that |Dαf(x − y)g(y)| ≤ ‖Dαf‖L∞|g(y)| ∈ L1(IRn) forevery x ∈ IRn.

ii) We fix x0 ∈ IRn. For all x, y ∈ IRn it holds |y−x0| ≤ |x−y|+ |x−x0|.Hence if |x−x0| ≤ 1

2 and |y−x0| ≥ 1 we get |y−x0| ≤ 2|y−x|. In particular

for |x− x0| ≤ 12 we get

|Dαf(x− y)g(y)| ≤ ϕα(y) =

Cα‖g‖L∞(1+|y−x0|2)mα if |y − x0| ≥ 1

Cα‖g‖L∞ if |y − x0| ≤ 1

125

Since mα >n2 , ϕα(y) ∈ L1(IRn).

iii) We observe that Dαf(x − y) = 0 if x − y /∈ spt(f), namely y /∈x−spt(f). Let U be a compact neighborhood of x0 and setK = U−spt(f).Then Dαf(x− y) = 0 if x ∈ U and y /∈ K. Hence

|Dαf(x− y)g(y)| ≤ ϕα(y) = ‖Dαf |L∞|g(y)|11K(y) ∈ L1(IRn).

We can conclude. 2

126

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Folland, Gerald B. Real analysis. Modern techniques and their appli-cations. Second edition. Pure and Applied Mathematics (New York).

[2] Evans, Lawrence C. Partial differential equations. GraduateStudies in Mathematics, 19. American Mathematical Society, Provi-dence, RI, 1998.

[3] Gilbarg, David; Trudinger, Neil S.Elliptic partial differential

equations of second order. Reprint of the 1998 edition.

[4] Han, Q. A Basic Course in Partial Differential Equations,Graduate Studies in Mathematics Volume: 120; 2011.

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