Lecture Notes for Partial Di erential EquationsPreface This note is about elementary partial di...

Lecture Notes for Partial Dierential Equations

December 21, 2009

Preface

This note is about elementary partial dierential equation. We mainly introduce three

important PDEs :Wave equation, Heat equation, and Laplace's equation. There are im-

portant properties for these three PDEs, like fudamental solution, mean value property,

maximum principle,...etc.. The materials for this notes is normally taken by ungradu-

ate student, who have completed a course in advanced calculus and a course in linear

algebra.

In this note, we use text book:Partial Dierential Equations, by W.A. Strauss.

We mention that Wave equation (chapter 2), Diusion equation (chapter 3), Laplace's

equation (chapter 4), and Fourier series (Chapter 5). We also refer to some books:

[1] Partial Dierential Equations, by L.C. Evans

[2] Elliptic Partial Dierential Equations of Second Order, by Gilbarg and Trudinger

[3] Partial Dierential Equations, by F.John

[4] Partial Dierential Equations, by Jost

[5] Fuctional Analysis, by P.Lax

[6] Partial Dierential Equations:methods and applications, by Robert McOwen.

In this note, we don't talk about modeling for these three equations, if you have

interest, you can refer to books from the above list.

1

Chapter 1

Introduction

1.1 Partial Dierential Equations

In general, we simply classify PDEs:

1. Elliptic: 4u = 0 (Laplace's equation)

2. Parabolic: ut −4u = 0 (Diusion equation)

3. Hyperbolic: utt −4u = 0 (Wave equation)

We concern about some problems, for example, uniqueness and existence of solution,

or stability of solution, even regularity of solution.

Denition 1. (well-posed problem)

Well-Posed problems consists of a PDE in a domain together with initial contion or

boundary contion that enjoys the following properties:

1. Existence of solution

2. Uniqueness of solution

3. Stability of solution: if the data are changed a little, the corresponding solution

changes only a little.

We also concern about regularity of solution, that is, smoothness of solution. We give

examples of ODE to show regularity of solution.

2

CHAPTER 1. INTRODUCTION 3

Example.

1. Let u ∈ C2(R) solves the equation u′′ − u = 0 together with initial contion

u(0) = 0, u′(0) = 1. Then we get the solution u(x) = 12ex− 1

2e−x. It's clearly that

the solution u ∈ C∞(R).

2. Let u ∈ C(R) solves the equation u′ = 1x2 with initial condition u(1) = 0. Then

we get the solution u(x) = − 1x

+ 1. The solution u is not smooth at x = 0.

1.2 First-order Linear Equations

For example, ∂u∂x

= 0, where u = u(x, y), is a rst order linear equation. And in this

example we can get u(x, y) = f(y), for some function f is of one variable.

We will introduce two methods to solve the rst order linear equation.

1.2.1 Constant coecient equation

Consider the problem

aux + buy = 0 (1.1)

where a, b are constants and a, b 6= 0.

Use geometric method and coordinate method to solve (1.1).

Method I: Geometric method

Let the vector v = (a, b). We rewrite (1.1) as (ux, uy) · (a, b) = ∇u · v = 0. Since

∇u · v represents the directional derivative of u in direction v, u(x, y) ≡ constant in

the direction v. So there are lines bx− ay = c on the xy plane, where c is a constant.

We call such lines characteristic lines. From this, we know that u(x, y) only depends

on bx − ay. Therefore, the solution of (1.1) is u(x, y) = f(bx − ay), for some function

f .

Method II: Coordinate method


Use change of variables: x′ = ax+ by, y′ = bx− ay. We compute ux and uy:

ux = ∂u∂x

= ∂u∂x′· ∂x′∂x

+ ∂u∂y′· ∂y′∂x

= aux′ + buy′

uy = ∂u∂y

= ∂u∂x′· ∂x′∂y

+ ∂u∂y′· ∂y′∂y

= bux′ − auy′

So aux + buy = (a2 + b2)ux′ = 0. We can get ux′ = 0 since a2 + b2 6= 0.

Hence, the solution of (1.1) is u = f(y′) = f(bx− ay), for some function f .

Example.

Solve the problem 4ux − 3uy = 0 with u(0, y) = y3. From the above, we know that

the solution is u(x, y) = f(−3x−4y). Use the condition u(0, y) = y3 to get f(y) = −y3

64.

Hence, u(x, y) = 164

(3x+ 4y)3.

1.2.2 The variable coecient equation

Consider the problem

ux + yuy = 0 (1.2)

We rewrite (1.2) as (1, y) · (ux, uy) = 0. This means that the direction derivative of

u in the direction (1, y) equals to 0, that is, u ≡ constant along the direction (1, y). Its

tangent vector has slope y on the xy plane. We get curves y = Cex on the xy plane,

for some constant C. We call such curves characteristic curves.

It's clearly that u(x, y) = u(x,Cex) satises (1.2) and u(x,Cex) = u(0, C) is inde-

pendent of x. So u(x, y) = u(x, e−xy) = f(e−xy) is general solution for some function

f .

Example.

Solve ux + yuy = 0 with u(0, y) = y3. From the above, we know that general

solution is u(x, y) = f(e−xy), then use condition u(0, y) = y3, we get the solution

u(x, y) = (e−xy)3.

Example.

Solve ux + 2xy2uy = 0. The same argument as above we can get characteristic

curves y = (c − x2)−1on the xy plane. So general solution is u(x, y) = f(x2 + 1y) for

some function f .


Exercise.

1. Solve the equation aux + buy = c, where a, b, c constants and a, b 6= 0.

2. Solve the equation aux + buy = f(x, y), where a, b constants , a, b 6= 0, and f is

given function.

3. Solve the equation aux + buy + cu = 0, where a, b, c constants and a, b, c 6= 0.

[Hint] Use the coordinate method

1.3 Type of Second-order Equations

We consider the PDE

a11uxx + 2a12uxy + a22uyy + a1ux + a2uy + a0u = 0 (1.3)

where a11, a12, a22, a1, a2, a0 constants.

How to determine type of the equation (1.3) ? We have the following theorem:

Theorem 2. By a linear transformation of independent variables , the equations can

be reduced to one of three forms, as follows.

1. Elliptic case: if a212 < a11 · a22, it's reducible to uxx + uyy + ... = 0.

2. Hyperbolic case: if a212 > a11 · a22, it's reducible to uxx − uyy + ... = 0.

3. Parabolic case: if a212 = a11 · a22, it's reducible to uxx + ... = 0 unless a11 = a12 =

a22 = 0.

We give an idea for the theorem, from the quadratic forms Ax2 + 2Bxy + Cy2 +

Dx + Ey + F = 0, where A,B,C,D,E, F constants, classify three types of geometric

graph:

1. Elliptic: B2 − AC < 0


2. Hyperbolic: B2 − AC = 0

3. Parabolic: B2 − AC > 0.

Proof. (proof of theorem 2)

For simplicity, suppose that a11 = 1, a1 = a2 = a0 = 0. Equation (1.3) becomes

(∂x + a12∂y)2u+ (a22 − a2

12)∂2yu = 0 (1.4)

If a212 < a11 · a22 = a22, dene b = a22 − a2

12 > 0, and let x = ξ, y = a12ξ + bη.

Then (1.4) becomes uξξ + uηη = 0, it's elliptic case. The procedure is similar in the

other case.

Example.

1. uxx − 5uxy = 0 is hyperbolic equation.

2. 4uxx − 12uxy + 9uyy + uy = 0 is parabolic equation.

3. 4uxx + 6uxy + 9uyy = 0 is elliptic equation.

1.4 Initial and Boundary Condition

Denition 3. (initial condition)

Physical state at a particular time t0.

Denition 4. (boundary condition)

Suppose Ω is a domain.

1. Dirichlet condition: u = 0 on ∂Ω.

2. Neumann condition: ∂u∂n

= 0 on ∂Ω, where n is unit outer normal vector of Ω.

3. Robin condition: ∂u∂n

+ a(x, y, t)u = 0 on ∂Ω, where n is unit outer normal vector

of Ω and a(x, y, t) is given function.


Example. (Heat equation)

Let ut − uxx = 0 in Ω = [0, L], t ≥ 0.

Initial condition: u(x, 0) = f(x)

Dirichlet boundary condition: u(0, t) = u(L, t) = g(t), ∀t ≥ 0

Neumann boundary condition: ∂u∂x

(0, t) = ∂u∂x

(L, t) = g(t), ∀t ≥ 0

1.5 Green's Formula

First, we introduce the integration by Parts in Rn:

Û

uxiv dx =

ˆ∂U

uvνidS −Û

uvxidx

where ν = (ν1, ν2, ..., νn) be unit outer normal vector of ∂U .

Theorem 5. (Gauss-Green Theorem)

Let u ∈ C1(U). ThenÚuxidx =

´∂UuνidS

Proof. Use integration by parts, we get the result

Theorem 6. (Green's formula)

1.Ú4u dx =

´∂U

∂u∂νdS =

´∂UDu · ν dS

2.ÚDu ·Dv dx =

´∂Uu∂v∂νdS −

Úu · 4v dx

Proof. Use integration by parts or Gauss-Green theorem, we get the results.

Chapter 2

Wave Equation

2.1 The Wave Equation on 1-D With No Source Term

First, we consider the transport equation ut + cux = 0, where c 6= 0 is a constant. From

1.2.1, we have the form of solution u(t, x) = g(x− ct) for some function g.

Consider the wave equation

utt − c2uxx = 0 (2.1)

for (x, t) ∈ R× [0,∞), and c 6= 0 is a constant.

Since utt− c2uxx = ( ∂∂t− c ∂

∂x)( ∂∂t

+ c ∂∂x

)u = vt− cvx = 0, where v = ut + cux, we get

the solution v(x, t) = f(x+ct). Then we solve the nonhomogeneous transport equation

ut + cux = v(x, t) = f(x+ ct).

Change of variables : ξ = x+ ct, η = x− ct, the equation becomes uξ = 12cf(ξ). So

u = 12c

´f(ξ) + g(η) := F (ξ) + g(η) = F (x + ct) + g(x − ct), where F ′ = 1

2cf . Hence

the general solution of (2.1) is

u = F (x+ ct) + g(x− ct) (2.2)

Another method to nd general solution of (2.1), using characteristic coordinates

method, let ξ = x + ct, η = x − ct, (2.1) becomes utt − c2uxx = −4c2uξη = 0. Then

uξη = 0 provided c 6= 0. Hence the general solution is the form of (2.2).

8

CHAPTER 2. WAVE EQUATION 9

Next, consider the initial value problemutt − c2uxx = 0

u(x, 0) = φ(x), ut(x, 0) = ψ(x)(2.3)

From (2.2), u(x, t) = F (x+ ct) + g(x− ct), and then using initial condition to getF ′ + g′ = φ′

cF ′ − cg′ = ψ⇒

F ′ = 12(φ′ + ψ

c)

g′ = 12(φ′ − ψ

c)

(2.4)

Integrating (2.4), we have

f(s) =1

2φ(s) +

1

2c

ˆ s

0

ψ +K1

g(s) =1

2φ(s)− 1

2c

ˆ s

0

ψ +K2

K1, K2 are constants.

Then K1 +K2 = 0 since f + g = φ. So the solution of (2.3)

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

2[φ(x+ ct) + φ(x− ct)] +

1

2c

ˆ x+ct

x−ctψ(s)ds (2.5)

The formula (2.5) is called d'Alembert formula.

In addition, we assume φ ∈ C2 and ψ ∈ C1, see from the formula (2.5) that the

solution u has continuous second partial derivatives in x and t.

Example. Solve the initial value problemutt − c2uxx = 0, (x, t) ∈ R× [0,∞)

u(x, 0) = sinx, ut(x, 0) = 0

Use the d'Alembert formula (2.5), we get the solution u(x, t) = sinx · cos(ct).

Example. Solve the initial value problem


utt − c2uxx = 0, (x, t) ∈ R× [0,∞)

u(x, 0) = 0, ut(x, 0) = cosx

Use the d'Alembert formula (2.5), we get the solution u(x, t) = 1csin(ct) · cosx.

Denition 7. (domain of inuence)

Initial condition at the point (x0, 0) can aect solution for t ≥ 0 only in the shaded

sector, which is called domain of inuence of (x0, 0), see gure 2-1.

(Figure 2-1)

If x a point (x, t) for t ≥ 0, how is the number u(x, t) synthesized from the initial

data φ, ψ? From formula (2.5) we see that the solution u(x, t) depends on the values

of φ, ψ at x ± ct and in the interval [x − ct, x + ct], respectively. We say that interval

(x− ct, x+ ct) of dependence of the point (x, t) on t = 0. In gure 2-2, the shaded area

is domain of dependence or the past history of the point (x, t).

(Figure 2-2)


2.2 The Wave Equation on 1-D With a Source Term

Consider the problemutt − c2uxx = f(x, t), (x, t) ∈ R× [0,∞)

u(x, 0) = φ(x), ut(x, 0) = ψ(x)(2.6)

Theorem 8. The unique solution of (2.6) is

u(x, t) =1

2[φ(x+ ct) + φ(x− ct)] +

1

2c

ˆ x+ct

x−ctψ(s)ds+

1

2c

ˆ4f(x, t)dxdt (2.7)

where 4 is the domain of dependence of the point (x, t).

Remark.

1. Eect of force f on u(x, t) is obtained by simply integrating f over the past history

of (x, t) back to the initial time t = 0.

2. Solution of nonhomogeneous equation can be decomposed homogeneous solution

and particular solution of nonhomogeneous equation.

3. Method of proving the theorem:

(a) Method of characteristic coordinate

(b) Using Green's theorem

(c) Operator method

Proof. (use method of characteristic coordinate)

Let ξ = x + ct, η = x − ct. Then utt − c2uxx = −4c2uξη = f( ξ+η2, ξ−η

2c). We have

u = − 14c2

´ ξ ´ ηfdηdξ.

Fix a point (x0, t0), then ξ0 = x0+ct0, η0 = x0−ct0 and u(x0, t0) = 14c2

´ ξ0η0

´ ξη0f( ξ+η

2, ξ−η

2c)dηdξ.

Change of variables:x = ξ+η2, t = ξ−η

2c.

Then the Jocobian J =| ∂(ξ,η)∂(x,t)

|=|1 c

1 −c|= 2c.

So u(x0, t0) = 2c4c2

´4 f(x, t)dxdt = 1

2c

´4 f(x, t)dxdt = 1

2c

´ t00

´ x0+c(t0−t)x0−c(t0−t) f(x, t)dxdt.


Now, we concern about well-posed problem for (2.6).

1. Existence: from theorem 8, we know that the existence of solution.

2. Uniqueness:

Let u, u are solutions of (2.6)

Dene w = u− u. Then w satises the equation

wtt − c2wxx = 0

w(x, 0) = wt(x, 0) = 0.

Use d'Alembert formula (2.5), we have w = 0, that is, u ≡ u.

3. Stability:

Dene the uniform norms: ‖ w ‖= maxx∈R| w(x) | and ‖ w ‖T= max

x∈R,t∈[0,T ]| w(x, t) |, T is

xed.

Suppose that ui(x, t) is the solution with data (φi, ψi, fi) for i = 1, 2. Let u = u1 − u2,

from (2.7), we have

| u(x, t) |=| 1

2[(φ1−φ2)(x− ct) + (φ1−φ2)(x+ ct)] +

1

2c

ˆ x+ct

x−ctψ1−ψ2 +

1

2c

ˆ4f1− f2 |

≤ max | φ | + 12cmax | ψ | ·2ct+ 1

2cmax | f | ·ct2

where φ = φ1 − φ2, ψ = ψ1 − ψ2, f = f1 − f2 and 4 is the domain of dependence of

the point (x, t).

Hence, ‖ u ‖T≤‖ φ ‖ +T ‖ ψ ‖ +T 2

2‖ f ‖T . If ‖ φ ‖< δ, ‖ ψ ‖< δ and ‖ f ‖T< δ for δ

small enough, ‖ u ‖T≤ δ1+T+T 2 < ε provided that δ = ε(1 + T + T 2). Since ε

arbitrarily small, this proves the stability of (2.6).

In addition, we assume φ ∈ C2, ψ ∈ C1, f is continuous with respect to the variables

x and t, then u has continuous second partial derivative in x and t.


2.3 Reections of Waves

Consider the problemVtt − c2Vxx = 0 x ∈ R+, t ∈ [0,∞)

V (x, 0) = φ(x), Vt(x, 0) = ψ(x) x ∈ R+

V (0, t) = 0 t ∈ [0,∞)

(2.8)

where R+ = [0,∞) half-line.

In this subsection, we want to nd solution of (2.8). In 2.1, we have solution on the

real line so we should use reection method to nd solution on the half-line.

Denition 9. (Odd and Even extension)

φodd(x) =

φ(x) x > 0

−φ(−x) x < 0

0 x = 0

φeven(x) =

φ(x) x ≥ 0

φ(−x) x < 0

Let u(x, t) be a solution of the equationutt − c2uxx = 0 (x, t) ∈ R× [0,∞)

u(x, 0) = φodd(x) x ∈ R

ut(x, 0) = ψodd(x) x ∈ R

Then u(x, t) is an odd function of x and u(0, t) = 0 for t ∈ [0,∞).

Dene V (x, t) = u(x, t) for x ∈ R+. From d'Alembert formula (2.5), we have

V (x, t) = u(x, t) =1

2[φodd(x+ ct) + φodd(x− ct)] +

1

2c

ˆ x+ct

x−ctψodd(s)ds

• If x > c | t |, V (x, t) = 12[φ(x+ ct) + φ(x− ct)] + 1

2c

´ x+ct

x−ct ψ(s)ds.


• If 0 < x < c | t |, V (x, t) = 12[φ(x+ ct)− φ(ct− x)] + 1

2c

´ ct+xct−x ψ(s)ds.

In gure 2-3, the shade area is the domain of dependence of (x, t) for the equation (2.8).

(Figure 2-3)

2.4 The Wave Equation on n-D (n=2,3)1

First, we introduce the method of spherical mean.

Denition 10. Let h(x) = h(x1, x2, ..., xn) is a continuous funcion in Rn. Its average

on a sphere with center x and radius r is dened by 1nωnrn−1

´∂Br(x)

h(y)dS(y), denoted

by Mh(x, r), where ωn = 2πn2

nΓ(n2

)denotes the volume of the unit ball in Rn.

Theorem 11. Let h ∈ C2(Rn). Then Mh(x, r) is a solution of the Darboux's equation

( ∂2

∂r2+ n−1

r)Mh(x, r) = 4xMh(x, r).

Proof.

Since Mh(x, r) = 1nωnrn−1

´∂Br(x)

h(y)dS(y) = 1nωn

´∂B1(0)

h(x+ rξ)dS(ξ),

∂

∂rMh(x, r) =

∂

∂r(

1

nωn

ˆ∂B1(0)

h(x+ rξ)dS(ξ)) =1

nωn

ˆ∂B1(0)

n∑i=1

hxi(x+ rξ) · ξidS(ξ)

1([1], [3])


Use Green's formula, we get

∂

∂rMh(x, r) =

1

nωn

ˆB1(0)

4xh(x+ rξ)dξ =r1−n

nωn

ˆBr(x)

4xh(y)dy

On the other hand,

r1−n

nωn

´Br(x)

4xh(y)dy = r1−n

nωn4x(´Br(x)

h(y)dy)

=r1−n

nωn4x

[ˆ r

0

dρ

ˆ∂Bρ(x)

h(y)dS(y)

]= r1−n4x

[ˆ r

0

ρn−1Mh(x, ρ)dρ

]Hence, rn−1

[∂∂rMh(x, r)

]= 4x

[´ r0ρn−1Mh(x, ρ)dρ

].

Then dierentiate the both side with respect to r, we have

∂

∂r

(rn−1

[∂

∂rMh(x, r)

])=

∂

∂r

(4x

[ˆ r

0

ρn−1Mh(x, ρ)dρ

])= rn−14xMh(x, r)

So Mh(x, r) solves the Darboux's equation ( ∂2

∂r2+ n−1

r)Mh(x, r) = 4xMh(x, r).

Denition 12. Let h is of class C2 in the half space x ∈ Rn, t ≥ 0. DeneMh(x, r; t) =1

nωnrn−1

´∂Br(x)

h(y, t)dS(y)

Theorem 13. Let h is of class C2 in the half space x ∈ Rn, t ≥ 0 and h(x, t) is a solution

of htt−c24h = 0. Then Mh(x, r; t) solves∂2

∂t2Mh(x, r; t) = c2

(∂2

∂r2+ n−1

r∂∂r

)Mh(x, r; t).

Remark. This theorem asserts that Mh(x, r; t) is a solution of the wave equation utt −c24u = 0.

Proof.

From theorem 11, we have ( ∂2

∂r2+ n−1

r)Mh(x, r; t) = 4xMh(x, r; t).

On the other hand,

4xMh(x, r; t) = 1nωn

´∂B1(0)

4xh(x+rξ, t)dS(ξ) = 1c2

1nωn

´∂B1(0)

∂2

∂t2(h(x+ rξ, t)) dS(ξ)


=1

c2

1

nωn

∂2

∂t2

[ˆ∂B1(0)

h(x+ rξ, t)dS(ξ)

]=

1

c2

∂2

∂t2Mh(x, r; t)

Hence, ∂2

∂t2Mh(x, r; t) = c2

(∂2

∂r2+ n−1

r∂∂r

)Mh(x, r; t).

Now, we consider the problemutt −4u = 0 (x, t) ∈ Rn × (0,∞)

u(x, 0) = g(x), ut(x, 0) = h(x) x ∈ Rn(2.9)

Let

U(x, r; t) =1

nωnrn−1

ˆ∂Br(x)

u(y, t)dS(y)

G(x, r) =1

nωnrn−1

ˆ∂Br(x)

g(y)dS(y)

H(x, r) =1

nωnrn−1

ˆ∂Br(x)

h(y)dS(y)

Then (2.9) becomesUtt − Urr − n−1rUr = 0 (r, t) ∈ Rn

+ × (0,∞)

U(r, 0) = G(r), Ut(r, 0) = H(r) r ∈ Rn+

(2.10)

where Rn+ = x = (x1, x2, ..., xn) : xn ≥ 0.

Consider the wave equation on R3:

Let U = rU , G = rG, H = rH. Then U solves2Utt − Urr = 0 (r, t) ∈ Rn

+ × (0,∞)

U(r, 0) = G(r), Ut(r, 0) = H(x) r ∈ Rn+

U(0, t) = 0 t ∈ (0,∞)

(2.11)

2use (2.10) to check the fact


In section 2.3, we know that the solution of (2.11) is

U(x, r; t) =1

2

(G(r + t) + G(t− r)

)+

1

2

ˆ t+r

t−rH(y)dy.

Therefore,

u(x, t) =1

nωnrn−1

∂

∂t

[t

ˆ∂Bt(x)

g(y)dS(y)

]+ t

ˆ∂Bt(x)

h(y)dS(y)

=1

nωnrn−1

ˆ∂Bt(x)

[th(y) + g(y) +Dg(y)(y − x)] dS(y)

(2.12)

is a solution of the wave equation on R3. Note that the formula (2.12) is called

Kirchho's formula.

Consider the wave equation on R2:

Use dummy variables:

Let u(x1, x2, x3, t) = u(x1, x2, t) solves the equationutt −4u = 0 (x, t) ∈ R3 × (0,∞)

u = g, ut = h (x, t) ∈ R3 × 0

where g(x1, x2, x3) = g(x1, x2) and h(x1, x2, x3) = h(x1, x2). From Kirchho's for-

mula (2.12), we have

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

nωnrn−1

ˆ∂Bt(x)

[th(y) + g(y) +Dg(y)(y − x)

]dS(y)

,

where x = (x1, x2, x3), y = (y1, y2, y3).

We directly compute 1nωnrn−1

´∂Bt(x)

g(y)dS(y), we get

1

nωnrn−1

ˆ∂Bt(x)

g(y)dS(y) =1

4πt2

ˆ∂Bt(x)

gdS =2

4πt2

ˆBt(x)

[g(y)

(1 + |Dr(y)|2

) 12

]dy


where r(y) =(t2 − |y − x|2

) 12 .

Hence,

1

nωnrn−1

ˆ∂Bt(x)

g(y)dS(y) =t

2πt2

ˆBt(x)

g(y)(t2 − |y − x|2

) 12

dy.Similarly,

1

nωnrn−1

ˆ∂Bt(x)

h(y)dS(y) =t

2πt2

ˆBt(x)

h(y)(t2 − |y − x|2

) 12

dy.So we get a solution of the wave equation on R2 is given by

u(x, t) =1

2πt2

ˆBt(x)

tg(y) + t2h(y) + tDg(y)(y − x)(t2 − |y − x|2

) 12

dy. (2.13)

The formula (2.13) is called Poisson's formula.

Exercise.

1. For n = 2, let u has continuous second partial derivative in x. Use polor coordinate

to show that

4u =

(∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂θ2

)u.

2. Check (2.10) and (2.11).

3. Show that

1

4πt2

ˆ∂Bt(x)

gdS =2

4πt2

ˆBt(x)

[g(y)

(1 + |Dr(y)|2

) 12

]dy

where r(y) =(t2 − |y − x|2

) 12 , g(x1, x2, x3) = g(x1, x2), h(x1, x2, x3) = h(x1, x2),

x = (x1, x2, x3), y = (y1, y2, y3), x = (x1, x2), and y = (y1, y2).


4. Consider the problemutt −4u = 0 (x, t) ∈ Rn × (0,∞)

u = g, ut = h (x, t) ∈ Rn × t = 0(2.14)

Let g, h ∈ C∞0 (Rn).

(a) For n = 3, the solution of (2.14) is given by Kirchho's formula. Show that

there exists a constant C independent of x and t such that |u(x, t)| ≤ Ctfor

all x ∈ R3, t > 0.

(b) Is a similar result true for n = 2?

2.5 Energy Method

Let u(x, t) be a solution of utt −4u = 0 for x ∈ Rn, t > 0. Dene the energy of u is

E(t) =1

2

ˆRn

[u2t + |Du|2

]dx.

Theorem 14. Show that conservation of energy for u as above, that is, E is a constant

independent of t.

Proof.

dE

dt=

ˆRn

(ut · utt +

n∑i=1

uxiuxit

)dx =

ˆRn

(ut · (utt −4u) +

n∑i=1

(utuxi)xi

)dx

=´

Rn (Du ·Dut + ut4u) dx=´

Rn Du ·Dutdx−´

Rn Du ·Dutdx+ (boundary term)

= 0 if u(x, t)→ 0 for |x| → ∞. (i.e. u need compact supported initial values)

Exercise.

1. Let u be a solution of (2.9) with compact supported initial values. Use energy

method to show that the solution u is unique.

Chapter 3

Diusion Equation

3.1 The Diusion Equation on 1-D With No Source

Term

Consider the problem ut = kuxx (x, t) ∈ R× (0,∞)

u(x, 0) = φ(x), (3.1)

where k > 0 is a constant.

Our method is to solve (3.1) for a particular φ(x) and then build the general solution

from this particular one. We will divide into four steps to nd a particular solution for

the equation (3.1).

Theorem 15. Let u(x, t) be a solution of (3.1). Then

1. the translate u(x− y, t) is also a solution for any xed y.

2. any derivative (ut, ux, uxx, ...) are solutios.

3. linear combination of solution is also solution.

4. integral solution is also solution, that is,´

R u(x− y, t)g(y)dy is a solution for any

g as long as this improper integral converges.

5. dilated u (√ax, at) is solution for a > 0.

20

CHAPTER 3. DIFFUSION EQUATION 21

Now, we look for the particular solution , denoted by Q(x, t), which satisesQ(x, 0) = 1 x > 0

Q(x, 0) = 0 x < 0. (3.2)

The reason for this choice is that this initial conditions does not change under

dilation.

Step 1: look for Q(x, t) of the special form

Q(x, t) = g(p), p =x√4kt

(3.3)

where g is a function of only one variable.

Explain why expect Q to have the special form:

Because (5) of theorem 15 says that (3.1) does'nt see dialation x→√ax, t→ at. If

Q depents on x and t solely through the combination x√t, for the dilation takes x√

tinto

√ax√at

= x√t, we can assume p = x√

4kt.

Step 2: look for Q(x, t) which satises equation (3.1), (3.2), and has the form

(3.3)

Use (3.3) to compute Qt, Qxx, we have Qt − kQxx = 1t

[−1

2pg′(p)− 1

4g′′(p)

]= 0. So

g′′(p) + 2pg′(p) = 0 is an O.D.E problem. Then we get

g(p) = c1

ê−p

2

dp+ c2,

where c1, c2 are constants.

Step 3: nd a completely explicit formula for Q

From step 2, we have

Q(x, t) = c1

ˆ x√4kt

0

e−p2

dp+ c2.


The formula is valid only for t > 0. Now, we use (3.2) to nd c1, c2:

• If x > 0, 1 = limt0

Q(x, t) = c1

´∞0e−p

2dp+ c2 = c1

√π

2+ c2.

• If x < 0, 0 = limt0

Q(x, t) = −c1

´ 0

−∞ e−p2dp+ c2 = −c1

√π

2+ c2.

Then we get c1 = 1√π, c2 = 1

2. So we have the explicity formula for Q is given by

Q(x, t) =1

2+

1√π

ˆ x√4kt

0

e−p2

dp (3.4)

for t > 0.

Step 4: nd solution for (3.1)

Dene S = ∂Q∂x

= 1√4πkt

e−x2

4kt is also solution of (3.1) from (2) of theorem 15.

Given any function φ(x), dene

u(x, t) =

ˆRS(x− y, t)φ(y)dy, t > 0

is also solution of (3.1) since (4) of theorem15. Hence, the solution of (3.1) is

u(x, t) =1√

4πkt

ˆRe−

(x−y)24kt φ(y)d, (3.5)

for t > 0.

Denition 16.

S(x, t) =1√

4πkte−

x2

4kt

is called source function, Green's function, fundamental solution, gaussian, diusion

kernel, heat kernel.

Proposition 17. The source function S(x, t) is dened for all x ∈ R, t > 0. Then


1. S(x, t) > 0;

2. S(x, t) is even function in x;

3.´

R S(x, t)dx = 1.

We introduce the error function.

Denition 18. The error function is dened by

erf(x) =2√π

ˆ x

0

e−p2

dp.

So the particular initial data φ(x) is expressible in terms of the error function, for

example, (3.4) is rewrited as

Q(x, t) =1

2+

1

2erf

(x√4kt

).

Some important properties for the error function are following as:

Proposition 19.

1. erf(0) = 0

2. limx→+∞

erf(x) = 1.

Exercise.

1. Prove theorem 15.

2. Prove proposition 17 and 19.

3. Show that´∞−∞ e

−x2dx =

√π.


4. Solve the problemut − kuxx + bt2u = 0 (x, t) ∈ R× (0,∞)

u(x, 0) = φ(x),

where k, b > 0 are constants.

[Hint] The solution of ODE wt+bt2w = 0 is c ·e− bt

3

3 . Change of variable: u(x, t) =

e−bt3

3 v(x, t).

5. Solve the problem ut − kuxx + bu = 0 (x, t) ∈ R× (0,∞)

u(x, 0) = φ(x),


[Hint] Change of variable: u(x, t) = e−btv(x, t).

6. Solve the reaction-diusion-advention problemut − kuxx + bux = 0 (x, t) ∈ R× (0,∞)

u(x, 0) = φ(x),


[Hint] Change of variable: y = x− bt.

3.2 The Diusion Equation on 1-D with Source Term

Consider the problemut − kuxx = f(x, t) (x, t) ∈ R× (0,∞)

u(x, 0) = φ(x)(3.6)


Idea to nd solution of (3.6):

Consider the ODE problem

du

dt+ Au(t) = f(t), u(0) = φ, (3.7)

where A is constant. Then u(t) = e−Atφ+´ t

0e(s−t)Af(s)ds.

Suppose that φ = (φ1, φ2, ..., φn), u(t) = (u1(t), u2(t), ..., un(t)), A is n × n matrix

and f(t) = (f1(t), f2(t), .., fn(t)). Then (3.7) is system of n linear ODEs.

Case 1: f(t) = 0

the solution of (3.7) is given as u(t) = S(t)φ, where S(t) = e−tA is a matrix.

Case2: f(t) 6= 0

Integrating factor for (3.7) is S(−t) = etA so (3.7) becomes

d

dt

(etAu(t)

)= f(t)etA.

Then integrating from 0 to t, we get

S(−t)u(t)− φ =

ˆ t

0

S(−s)f(s)ds.

Hence, u(t) = S(t)φ︸︷︷︸homogeneous solution

+

ˆ t

0

S(t− s)f(s)ds︸︷︷︸non−homogeneous solution

.

Return to the original problem (3.6):

We have solved the homogeneous problem in section 3.1, the homogeneous solution

is´

R S(x − y, t)φ(y)dy, where S(x, t) is the source function. From the idea of ODE

problem, we guess that the solution of (3.6) is

u(x, t) =

ˆRS(x− y, t)φ(y)dy +

ˆ t

0

ˆRS(x− y, t− s)f(y, s)dyds. (3.8)


Theorem 20. Let u is given by (3.8) which satises the equation (3.6), where S(x, t) =1√

4πkte−

x2

4kt .

Proof. Assume that φ = 0. Then (3.8) becomes

u(x, t) =

ˆ t

0

ˆRS(x− y, t− s)f(y, s)dyds.

Now, compute ∂u∂t:

∂u

∂t=

∂

∂t

(ˆ t

0

ˆRS(x− y, t− s)f(y, s)dyds

)

=

ˆ t

0

ˆR

∂S

∂t(x− y, t− s)f(y, s) + lim

s→t

ˆRS(x− y, t− s)f(y, s)

Since S(x− y, t− s) satises the heat equation and singularity of S(x− y, t− s) att− s = 0,

∂u

∂t= k

∂2

∂x2

(ˆ t

0


)+ lim

ε→0

ˆRS(x− y, ε)f(y, t+ ε)

= k∂2u

∂x2+ f(x, t).

Since S(x, t) → δ(x) as t → 0+, the last equality is valid from the fact´

R δ(x −y)f(y)dy = f(x), where δ is Dirac function. This proves (3.8) satises the equation

ut − kuxx = f(x, t). Next, we check the initial condition.

Letting t→ 0+:

limt→0+

(ˆRS(x− y, t)φ(y)dy +

ˆ t

0


)= φ(x).

So initial condition is satised.


Remark. (3.8) can be expressed by

u(x, t) =

ˆR

1√4πkt

e−|x−y|2

4kt φ(y)dy +

ˆ t

0

ˆR

1√4πk(t− s)

e−|x−y|24k(t−s)f(y, s)dyds.

Denition 21. The convolution of f with g is dened by

f ? g(x) =

ˆRf(x− y)g(y)dy =

ˆRf(y)g(x− y)dy.

In fact, the solution formula for the heat equation is of a convolution of φ with S at

xed t.

u(x, t) =

ˆRS(x− y, t)φ(y)dy =

ˆRS(z, t)φ(x− z)dz = S ? φ(x), (3.9)

where S(x, t) = 1√4πkt

e−x2

4kt .

Change of variables: p = x√kt. Then (3.12) becomes

u(x, t) =1√4π

ˆRe−

p2

4 φ(x− p√kt)dp. (3.10)

Theorem 22. Let φ(x) be bounded and continuous for x ∈ R. Then (3.10)∈ C∞ for

x ∈ R, t > 0, which satises the equation ut − kuxx = 0 and limt0

u(x, t) = φ(x).

Proof.

First, check the integral is convergent:

|u(x, t)| = 1√4π

∣∣∣∣ˆRe−

p2

4 φ(x− p√kt)dp

∣∣∣∣ ≤ 1√4πmax |φ|

ˆRe−

p2

4 dp = max |φ| <∞.

So the integral converges uniformly and absolutely.

Second, we show that ∂u∂x

exists and ∂u∂x

=´

R∂S∂x

(x−y, t)φ(y)dy provided this integral

converges absolutely.


∂u

∂x=

ˆR

∂S

∂x(x− y, t)φ(y)dy = − 1√

4πkt

ˆR

x− y2kt

e−(x−y)2

4kt φ(y)dy

=c√t

ˆRpe−

p2

4 φ(x− p√kt)dp ≤ c√

tmax |φ|

ˆRpe−

p2

4 dp <∞,

where c is a constant and p = x−y√kt.

So the integral converges uniformly and absolutely. Hence, ∂u∂x

exists and ∂u∂x

=´R∂S∂x

(x− y, t)φ(y)dy.

So all derivatives of all orders (ut, utx, uxx, utt, ....) work the same way because each

dierention brings down a power of p so that we end up with convergent integrals like´pne−

p2

4 dp. Hence, u(x, t) is dierentiable to all orders.

Finally, prove (3.10) satises the equation ut − kuxx = 0 and limt0

u(x, t) = φ(x).

Since S(x, t) satises the heat equation for x ∈ R, t > 0, so does u(x, t). It sucies

to show that the initial condition.

Consider

u(x, t)− φ(x) =

ˆRS(x− y, t) (φ(y)− φ(x)) dy

=1√4π

ˆRe−

p2

4

[φ(x− p

√kt)− φ(x)

]dp (3.11)

Fixed x, we must show that (3.11)→ 0 as t→ 0.

Idea:

1. If p√kt is small and continuity of φ, (3.11) is small.

2. If p√kt is not small, that is, p is so large, then exponential factor is so small.

Let ε > 0, δ > 0 be so small that max|y−x|≤δ

|φ(y)− φ(x)| < ε2. So we have

I :=

∣∣∣∣∣ 1√4π

ˆ|p|< δ√

kt

e−p2

4

[φ(x− p

√kt)− φ(x)

]dp

∣∣∣∣∣ ≤ 1√4π

max|y−x|≤δ

|φ(y)− φ(x)|ˆ

Re−

p2

4 dp <ε

2


and

II :=

∣∣∣∣∣ 1√4π

ˆ|p|> δ√

kt

e−p2

4

[φ(x− p

√kt)− φ(x)

]dp

∣∣∣∣∣ ≤ 2√4πmax |φ|

ˆ|p|> δ√

kt

e−p2

4 dp <ε

2

by choosing t suciently small since the integral´

R e− p

2

4 dp converges and δ is exed.

(That is,´|p|>N e

− p2

4 dp is as small as we wish if N = δ√kt

larege enough.)

Therefore, |(3.11)| ≤ I + II < ε2

+ ε2

= ε provided t small enough.

Exercise.

1. Calculate´

R e− p

2

4 dp and´

R e− bp

2

a dp for a, b are constants.

2. Explain why´

R pne−

p2

4 dp converges absolutely for n ≥ 0.

3.3 Diusion on the Half-line

Consider the problem vt − kvxx = 0 (x, t) ∈ R+ × (0,∞)

v(x, 0) = φ(x) x ∈ R+

v(0, t) = 0

(3.12)

Where R+ = [0,∞). Use the reection method which is the same as section 2.1.

Let u(x, t) be a solution ofut − kuxx = 0 (x, t) ∈ R× (0,∞)

u(x, 0) = φodd(x).


Then from section 3.1, we have u(x, t) =´

R S(x − y, t)φodd(y)dy, where S(x, t) is

a source function for the heat equation. Now, dene v(x, t) = u(x, t) for x > 0 is a

solution of (3.12). We need to check this function v(x, t) is desired solution of (3.12).

First of all, u(x, t) must be an odd function of x so u(0, t) = 0, this implies that the

boundary condition v(0, t) = 0 is satised. Second, v solves PDE as well as the initial

condition for x > 0 since it's equal to u for x > 0 and u satises the same PDE for

all x and the same initial condition for x > 0. Finally, we can write down the explicit

solution formula for (3.12) :

v(x, t) =1√

4πkt

ˆ ∞0

[e−

(x−y)24kt − e−

(x+y)2

4kt

]φ(y)dy. (3.13)

Exercise.

1. Show that the solution u(x, t) for the heat equation is an odd function of x, that

is ,u(−x, t) = −u(x, t) for all t > 0.

2. Derive the explicity solution formula (3.13) for (3.12).

3.4 The Maximum Principle for The Diusion Equa-

tion on The Finite Interval

Theorem 23. (Weak maximum principle)

Let u(x, t) satises the equation ut − kuxx = 0 for 0 ≤ x ≤ l, 0 ≤ t ≤ T . Then the

maximum of u(x, t) is assumed either initially or on the lateral sides (x = 0 or x = l).

We can use the weak maximum principle to prove the uniqueness of the solution for

the Dirichlet problem for the heat equation:ut − kuxx = f(x, t) 0 < x < l, 0 < t < T

u(x, 0) = φ(x)

u(0, t) = g(t), u(l, t) = h(t)

(3.14)


Theorem 24. (Uniqueness of the solution for the heat equation (3.14))

Let u(x, t) be a solution of (3.14). Then the solution u(x, t) is unique.

Proof.

Let u1(x, t) and u2(x, t) are solutions of (3.14). Dene w(x, t) = u1(x, t)− u2(x, t),

then w satises wt − kwxx = 0 0 < x < l, 0 < t < T

w(x, 0) = 0

w(0, t) = 0, w(l, t) = 0

. (3.15)

Let the rectangle D = (x, t) : 0 ≤ x ≤ l, 0 ≤ t ≤ T. By weak maximum principle

(theorem 23), w(x, t) has its maximum for D on its bottom or sides. Since the boundary

and initial condition, w(x, t) ≤ 0. We take w = −w into (3.15), we have w(x, t) ≥ 0.

So w(x, t) ≡ 0. This proves the uniqueness of the solution for (3.14).

Now, we prove the weak maximum principle (theorem 23).

Proof. (proof of theorem 23)

Let M = maxu(x, t) on the three sides t = 0, x = 0, and x = l. We must show that

u(x, t) ≤ M through the rectangle D = (x, t) : 0 ≤ x ≤ l, 0 ≤ t ≤ T. Let ε > 0 and

v(x, t) = u(x, t) + εx2.

WANT: v(x, t) ≤M + εl2 throughout D.

From the denition of v, we have v(x, t) ≤M + εl2 on t = 0, x = 0, and x = l. And

v satises

vt − kvxx = ut − k(u+ εx2

)xx

= ut − kuxx − 2εk = −2εk < 0. (3.16)

Assume that v(x, t) attains its maximum at an interior point (x0, y0). Then vt(x0, y0) =

0 and vxx(x0, y0) ≤ 0, this contradicts (3.16) at the point (x0, y0). So v(x, t) does not

attain its maximum at interior of D.

Assume that v(x, t) attains its maximum at a point (x0, y0) on the top edge t0 = T, 0 < x < l.Then vx(x0, t0) = 0, vxx(x0, t0) ≤ 0. And vt(x0, t0) ≥ 0 since v(x0, t0) ≥ v(x0, t0− δ) for


some δ > 0. So vt(x0, t0) − kvxx(x0, t0) ≥ 0 contradicts (3.16). Hence, v(x, t) does not

attain its maximum on the top edge t0 = T, 0 < x < l.

We conclude that v(x, t) attains its maximum on the bottom or sides. So we have

v(x, t) ≤ M + εl2. From the denition of v, we get u(x, t) ≤ M + ε(l2 − x2) for ε > 0.

Then u(x, t) ≤M since ε > 0 is arbitrary.

The other method to prove theorem 24 is the energy method :

Let u1(x, t) and u2(x, t) are solutions of (3.14). Dene w(x, t) = u1(x, t)− u2(x, t),

then w satises wt − kwxx = 0 0 < x < l, 0 < t < T

w(x, 0) = 0

w(0, t) = 0, w(l, t) = 0

.

Multiply the equation by w and using boundary condition, we get

∂

∂t

(ˆ l

0

1

2w2dx

)= −k

ˆ l

0

w2xdx ≤ 0.

This implies that the function´ l

0w2dx is nonincreasing with respect to t. Then´ l

0w2(x, t)dx ≤

´ l0w2(x, 0)dx = 0 since the initial condition. Therefore,

´ l0w2(x, t)dx =

0⇐⇒ w ≡ 0, this proves theorem 24. We dene E(t) =´ l

0w2(x, t)dx is the energy for

the heat equation.

Theorem 25. (Strong maximum principle)

Let u(x, t) satises the equation ut − kuxx = 0 for 0 ≤ x ≤ l, 0 ≤ t ≤ T . The maxi-

mum of u can't be assumed anywhere inside the rectangle D = (x, t) : 0 ≤ x ≤ l, 0 ≤ t ≤ Tbut only on the bottom or the lateral sides unless u is a constant.


Exercise.

1. Prove the strong maximum principle (theorem 25).

2. Let u(x, t) satises the equation (3.14) with g(t) = h(t) = 0. Prove the stability of

the solution for the equation (3.14) by using (1) energy method and (2) maximum

principle.

3.5 The Heat Equation on n-D (n≥ 2)1

3.5.1 Fundamental Solution for the Heat Equation

We rst scale the variable. Let u(x, t) = v(|x|2t

)= 1

tαv(xtβ

)= 1

tαv(y), where y = x

tβ.

Then v(y) = u(y, 1) and

ut −4u = αt−(α+1)v(y) + βt−(α+1)y ·Dv(y) + t−(α+2β)4v(y) = 0. (3.17)

Take β = 12into (3.17), then (3.17) becomes

ut −4u = αv(y) +1

2yDv(y) +4v(y) = 0.

Next, letting v(y) = w (|y|) = w(r), then

αw(r) +1

2rw′ + w′′ +

n− 1

rw′ = 0

=⇒(αw +

1

2rw′)

+

(w′′ +

n− 1

rw′)

= 0

=⇒ 1

2r1−2α

(r2αw

)′+ r1−n (rn−1w′

)′= 0.

Take α = n2, we get 1

2rnw + rn−1w′ = c, where c is a constant. For simplicity, we

choose a = 0, we can solve w(r) = Ke−r2

4 , where K is a constant. So we have

u(x, t) =K

tn/2e−|x|24t .

1([1], [3], [4])


Denition 26. Fundamental solution for the heat equation ut − 4u = 0, (x, t) ∈Rn × (0,∞) is dende by

S(x, t) =

1

(4πt)n/2exp

(− |x|

2

4t

)(x, t) ∈ Rn × (0,∞)

0 t ≤ 0. (3.18)

Proposition 27. Let S(x, t) be a fundamental solution for the heat equation dened by

(3.18). Then´

Rn S(x, t)dx = 1.

Proof. Left as an exercise.

3.5.2 The Dirichlet Problem for the Heat Equation

Consider the initial-valued problemut −4u = 0 (x, t) ∈ Rn × (0,∞)

u(x, 0) = g(x) x ∈ Rn. (3.19)

From (3.9), we have the solution of the equation (3.19) is

u(x, t) =

ˆRnS(x− y, t)g(y)dy =

1

(4πt)n/2

ˆRne−|x−y|2

4t g(y)dy.

Consider the nonhomogeneous problemut −4u = f(x, t) (x, t) ∈ Rn × (0,∞)

u(x, 0) = 0 x ∈ Rn. (3.20)

Fixed s, dene

u = u(x, t; s) =

ˆRnS(x− y, t− s)f(y, s)dy. (3.21)


Putting (3.21) into (3.20), we get (3.21) solvesut(x, t; s)−4u(x, t; s) = 0 (x, t) ∈ Rn × (s,∞)

u(x, t; s) = f(x, t; s) (x, t) ∈ Rn × t = s. (3.22)

We reduce the problem (3.20) to the initial problem (3.22).

Duhamel's principle asserts that the form u(x, t) =´ t

0u(x, t; s)ds is the solution of

(3.20). Hence, the solution of (3.20) is given by

u(x, t) =

ˆ t

0

ˆRnS(x− y, t− s)f(y, s)dyds

=

ˆ t

0

ˆRn

1

[4π(t− s)]n/2e−|x−y|24(t−s) f(y, s)dyds. (3.23)

Finally, we combine the problem (3.19) and (3.20).ut −4u = f(x, t) (x, t) ∈ Rn × (0,∞)

u(x, 0) = g(x) x ∈ Rn(3.24)

Hence, the solution of (3.24) is

u(x, t) =

ˆRnS(x− y, t)g(y)dy +

ˆ t

0

ˆRnS(x− y, t− s)f(y, s)dyds.

3.5.3 Mean-value Property

Denition 28. (Parabolic cylinder and parabolic boundary)

Let U ⊂ Rn be an open and bounded domain. For T > 0, dene UT = U × (0, T )

is a parabolic cylinder. And parabolic boundary ΓT is dened by UT \ UT .

Denition 29. (Heat ball)

Let S(x, t) be fundamental solution of the heat equation. Dene the heat ball as

E(x, t; r) =

(y, s) ∈ Rn−1 : s ≤ t, S(x− y, t− s) ≥ 1

rn


Theorem 30. (Mean-value property)

Let u ∈ C21(UT ) satises ut −4u = 0 for x ∈ Rn, t > 0. Then

u(x, t) =1

4rn

Ê(x,t;r)

u(y, s)|x− y|2

(t− s)2dyds

for all (x, t) ∈ UT , where E(x, t; r) be the heat ball.

Note that u ∈ C21(UT ) means that u has continuous second partial derivative with

respect to x and has continuous rst partial derivative with respect to t in parabolic

cylinder UT .

Proof. See [1].

3.5.4 Maximum Principle


Let u ∈ C21(UT ) satisfy ut −4u ≤ 0 in UT . Then max

UT

u = maxΓT

u.

Proof.

First, let ut − 4u < 0 in UT and dene Uε = (x, t) : x ∈ U, 0 < t < T − ε for

0 < ε < T .

Since u ∈ C0(Uε) and Uε is compact, ∃(x0, t0) ∈ Uε such that u(x0, t0) = maxUε

u.

Then ut(x0, t0) = 0 and 4u(x0, t0) ≤ 0. Hence, ut(x0, t0) −4u(x0, t0) ≥ 0 contradicts

to ut −4u < 0 .

Thus, (x0, t0) ∈ Γε and maxUε

u = maxΓε

u ≤ maxΓT

u.

Since every point of UT with t < T belongs to some Uε and u is continuous in UT ,

maxUT

u = maxΓT

u.

Second, let ut − 4u ≤ 0 in UT . Dene v(x, t) = u(x, t) − kt for some k > 0 is

constant. Then v ≤ u and v satises vt −4v < 0.


From the above argument, we have

maxuUT

= maxUT

(v + kt) ≤ maxΓT

(v + kt) ≤ maxΓT

u+ kT.

Letting k → 0, we have maxUT

u = maxΓT

u.


Suppose that U be connected subset of Rn. Let u ∈ C21(UT ) satisfy ut −4u ≤ 0 in

UT . If there exists (x0, t0) ∈ UT such that u(x0, t0) = maxuUT

, u is a constant in Ut0.

We can use maximum principle to prove the uniqueness of the solution for the heat

equation.

Corollary 33. (Uniqueness of solution for the heat equation)

Let u ∈ C21(UT ) ∩ C(UT ) be a solution of ut − 4u = 0 in UT together with the

boundary condition u(x, t) = f(x, t) on ΓT , where U is bounded domain. Then u is the

unique solution for the Dirichlet's problem of the heat equation.

Proof. Let u1, u2 are solutions of the Dirichlet's problem of the heat equation. Dene

w = u1 − u2. Then w satiseswt −4w = 0 (x, t) ∈ UTw = 0 (x, t) ∈ ΓT

.

Use maximum principle (theorem 31), we have w ≤ 0 in UT . On the other hand,

take w = −w into theorem31, we have w ≥ 0 in UT . We conclude that w ≡ 0 in UT .

This proves the uniqueness of the solution for the heat equation.

Now, we still have the maximum principle for the whole space, not only for the

parabolic cylinder, but we need some conditions so that the maximum principle holds

on the whole space.


Theorem 34. (Maximum principle for Cauchy problem)

Let u ∈ C21(Rn × (0, T ]) ∩ C(Rn × [0, T ]) be a solution ofut −4u ≤ 0 (x, t) ∈ Rn × (0, T )

u(x, 0) = f(x) x ∈ Rn

and satises the growth estimate u(x, t) ≤ Aeα|x|2

for x ∈ Rn, 0 ≤ t ≤ T , where

A,α > 0 constants. Then u(x, t) ≤ supz∈Rn

f(z) for x ∈ Rn, 0 ≤ t ≤ T .

Remark.

1. The result can be written as supRn×[0,T ]

u = sup fRn

if ut−4u = 0 for (x, t) ∈ Rn×(0, T )

with u(x, 0) = f(x).

2. The theorem implies that the solution of the initial-value problemut −4u = 0 (x, t) ∈ Rn × (0, T )

u(x, 0) = f(x) x ∈ Rn(3.25)

is unique provided the solution satises the growth estimate |u(x, t)| ≤ Aeα|x|2

for

x ∈ Rn, 0 ≤ t ≤ T , where A,α > 0 constants.

Proof. See [1].

3.5.5 Energy Method

Dene the energy function e(t) =Úu2(x, t)dx for 0 ≤ t ≤ T , where u satises the heat

equation ut −4u = 0.

Proposition 35. The energy function e(t) of the heat equation is nonincreasing, or

energy dissipative.

Proof.

d

dte(t) =

d

dt

Û

u2(x, t)dx = 2

Û

u · ∂u∂tdx = 2

Û

u · 4udx = −2

Û

|∇u|2 ≤ 0


since u satises the heat equation ut −4u = 0 and using integration by parts. We

assume the boundary condition u |∂U= 0 so the boundary term is equal to 0.

Exercise.

1. Prove Proposition 27.

2. Use (3.21) to verify (3.22).

3. Prove the strong maximum principle (theorem 32).

4. Use theorem 34 to prove the solution of the equation (3.25) is unique.

5. Use energy method to prove the solution of the equation ut − 4u = 0 in UT

together with the boundary condition u(x, t) = f(x, t) on ΓT is unique, where U

is bounded domain.

6. (Harnack's inequality)

Suppose that u ∈ C21(UT ) be a solution of ut −4u = 0 in UT and u ≥ 0 in UT .

Let V ⊂⊂ U is connected. Then for each 0 < t1 < t2 ≤ T , there exists a constant

C such that supVu(x, t1) ≤ Cinf

Vu(x, t2), where constant C depends only on V , t1

and t2.

Prove the Harnack's inequality.

Chapter 4

Laplace Equation

4.1 Introduction to Laplace's Equation

Denition 36. Let Ω be a domain in Rn and u ∈ C2(Ω). u is called harmonic if

4u =∑n

i=1 uxixi = 0.

Example. (Analytic function of complex variable)

Let z = x + iy and f(z) = u(z) + iv(z) = u(x + iy) + iv(x + iy). Suppose f(z) is

analytic function, then f(z) = u(z) + iv(z) =∑∞

n=0 anzn, an ∈ C for all n = 0, 1, ....

We dierentiate f(z) with respect to x and y, we get

∂u

∂x=∂v

∂y,∂u

∂y= −∂v

∂x.

It's called Cauchy-Riemann equations. In fact, we have uxx = vyx = vxy = −uyy.Hence, uxx + uyy = 0. Similarily, vxx + vyy = 0.

Thus, if functionf(z) = u(z)+ iv(z) is analytic function, real part function u(z) and

imaginary part function v(z) are harmonic.

Theorem 37. Laplace equation is invariant under all rigid motions.

Now, we solve the Laplace equation for n = 2, 3. First, we consider the problem for

n = 2.

40

CHAPTER 4. LAPLACE EQUATION 41

Use polor coodinate, we have 4u = urr + 1rur + 1

r2uθθ. Since theorem 37 says

that the harmonic function is rotationally invariant, u depends only on r. Hence,

4u = urr + 1rur = 0 can be regarded as ODE problem. Using the method of ODE,

we get a solution u = c1 ln r + c2, where c1, c2 constants. So the solution of Laplace

equation for n = 2 is

u(x) = c1 ln |x|+ c2.

Next, we consider the problem for n = 3.

Use spherical coordinate:

r =√x2 + y2 + z2 =

√s2 + z2

s =√x2 + y2

x = s cosφ = r sin θ cosφ

y = s sinφ = r sin θ sinφ

z = r cos θ

.

Change of variables: (x, y, z) → (s, φ, z) → (r, φ, θ) and by n = 2 calculation, we

have uzz + uss = urr + 1rur + 1

r2uθθ (1)

uxx + uyy = uss + 1sus + 1

s2uφφ (2)

.

(1)+(2): 4u = uxx + uyy + uzz = urr + 1rur + 1

r2uθθ + 1

sus + 1

s2uφφ.

Since s2 = r2 sin2 θ, us = ur∂r∂s

+ uθ∂θ∂s

+ uφ∂φ∂s

= ursr

+ uθcos θr. Hence,

4u = uxx + uyy + uzz = urr +2

rur +

1

r2

[uθθ + cot θuθ +

1

sin2 θuφφ

].

By theorem 37, we know that the solution depends only on r, so4u = urr+2rur = 0.

Then r2urr + 2rur = (r2ur)r = 0. So the solution of Laplace equation for n = 3 is

u(x) = − c1

|x|+ c2,


where c1, c2 constants.

Denition 38. Let Ω be a domain in Rn and u ∈ C2(Ω).

1. u is called subharmonic in Ω if 4u ≥ 0;

2. u is called superharmonic in Ω if 4u ≤ 0.

Exercise.

1. In example, derive the Cauchy-Riemann equation.

2. Prove theorem 37 for n = 2, 3.

3. Let u(x, y) = x2 + y2. Show that the function u is subharmonic.

4. Let u(x, y) = 1√x2+y2

and v(x, y, z) = 1√x2+y2+z2

. Show that v is harmonic but u

is not harmonic.

5. Let x = (x1, x2, ..., xn) be a vector in Rn. Dene the function u(x) = 1|x| =

1√x21+x2

2+...+x2n

. Show that the function u is harmonic for n ≥ 3.

4.2 Green's Function

4.2.1 Green's identity

Theorem 39. (Divergence theorem)

Let Ω be a bounded domain and ∂Ω is C1. For any w ∈ C1(Ω), we have

ˆΩ

Dwdx =

ˆ∂Ω

w · νdS (4.1)

where ν is an unit outer normal vector of Ω.

Remark. If u ∈ C2(Ω), we take w = Du into (4.1), we have

ˆΩ

4udx =

ˆ∂Ω

Du · νdS =

ˆ∂Ω

∂u

∂νdS.


Theorem 40. (Green's identity)1

Let Ω be a bounded domain and ∂Ω is C1. For any u, v ∈ C2(Ω), then

1. Green's rst identity:´

Ωv4udx+

´ΩDu ·Dvdx =

´∂Ωv ∂u∂νdS.

2. Green's second identity:´

Ω(v4u− u4v) dx =

´∂Ω

(v ∂u∂ν− u∂v

∂ν

)dS.

Proof.

Take w = vDu into (4.1), then

ˆΩ

D (vDu) dx =

ˆ∂Ω

(vDu) · νdS =

ˆ∂Ω

v∂u

∂νdS.

On the other hand,´

ΩD (vDu) dx =

´ΩDu ·Dvdx+

´Ωv4udx. Hence, we get the

Green's rst identity

ˆΩ

v4udx+

ˆΩ

Du ·Dvdx =

ˆ∂Ω

v∂u

∂νdS (4.2)

Next, we interchange u and v in Green's rst identity, we have

ˆΩ

u4vdx+

ˆΩ

Du ·Dvdx =

ˆ∂Ω

u∂v

∂νdS. (4.3)

Then (4.2)-(4.3)=´

Ω(v4u− u4v) dx =

´∂Ω


∂ν

)dS, it's Green's second

identity.

Theorem 41. (Green's representation formula for n = 3)

Let Ω be a bounded domain and ∂Ω is C1. Suppose 4u = 0 in Ω, then for any point

x0 in Ω, we have

u(x0) =1

4π

ˆ∂Ω

[−u(x)

∂

∂ν

(1

|x− x0|

)+

1

|x− x0|∂u

∂ν

]dS. (4.4)

1For simplicity, we denote Green's rst identity by (G1) and Green's second identity by (G2).


Remark.

1. In formula (4.4), the function 14π|x−x0| is the fundamental solution of Laplace's

equation for n = 3. We will mention the fundamental solution later, but from

exercise 4.1.5, we have known that the function 14π|x−x0| is harmonic.

2. The corresponding formula for n = 2 is given by

u(x0) =1

2π

ˆ∂Ω

[u(x)

∂

∂ν(ln |x− x0|)− ln |x− x0|

∂u

∂ν

]dS

whenever 4u = 0 in Ω.

Proof.

Let v(x) = −14π|x−x0| . Then 4v = 0 and it's clearly that v has singularity at x = x0.

For ε > 0 small, take Ω \Bε(x0) and v into (G2), then

ˆΩ\Bε(x0)

(v4u− u4v) dx =

ˆ∂(Ω\Bε(x0))

(v∂u

∂ν− u∂v

∂ν

)dS.

Since 4u = 0 = 4v,´∂(Ω\Bε(x0))


∂ν

)dS = 0. And

ˆ∂(Ω\Bε(x0))

(v∂u

∂ν− u∂v

∂ν

)dS =

ˆ∂Ω

(v∂u

∂ν− u∂v

∂ν

)dS+

ˆ∂Bε(x0)

(v∂u

∂ν− u∂v

∂ν

)dS︸︷︷︸

I

= 0.

(4.5)

Estimate (I):

∣∣∣∣ˆ∂Bε(x0)

v∂u

∂νdS

∣∣∣∣ =

∣∣∣∣ˆ∂Bε(x0)

−1

4π |x− x0|∂u

∂νdS

∣∣∣∣ =1

4πε

∣∣∣∣ˆ∂Bε(x0)

∂u

∂νdS

∣∣∣∣ =1

4πε

∣∣∣∣ˆ∂Bε(x0)

Du · νdS∣∣∣∣

≤ 1

4πεsup

∂Bε(x0)

|Du|ˆ∂Bε(x0)

dS = ε sup∂Bε(x0)

|Du| → 0

as ε→ 0.


On the other hand,

ˆ∂Bε(x0)

u∂v

∂νdS =

ˆ∂Bε(x0)

u

(− 1

4πε2

)dS = − 1

4πε2

ˆ∂Bε(x0)

u(x)dS.

As ε→ 0, using mean-value property, we have´∂Bε(x0)

u∂v∂νdS → −u(x0).

Hence, letting ε→ 0 into (4.5), we get

u(x0) =1

4π

ˆ∂Ω

[−u(x)

∂

∂ν

(1

|x− x0|

)+

1

|x− x0|∂u

∂ν

]dS.

4.2.2 Green's Functions

Denition 42. The Green's function G(x) for the operator −4 and the domain Ω at

the point x0 ∈ Ω is a function dened for x ∈ Ω such that

1. G(x) ∈ C2(Ω) and 4G = 0 in Ω except x = x0.

2. G(x) = 0 on ∂Ω.

3. G(x) + Γ(x) is nite at x0 and has continuous second derivative everywhere and

is harmonic at x0, where Γ(x) is fundamental solution for the Laplace equation.

Remark. Denition 42 (3) implies that the Green function G can be decomposed as

fundamental solution and a harmonic function, that is, G = Γ + h for any harmonic

function h.

Theorem 43. For any domain Ω, Green's function G(x, y) is symmetric. That is,

G(x, y) = G(y, x) for x 6= y.

Proof. ([1])

Fix x, y∈ Ω with x 6= y.

Let v(x) := G(x, z) and w(z) := G(y, z). Then 4v = 0 = 4w in Ω and v = w = 0

on ∂Ω.


Dene Ωε = Ω \ [Bε(x) ∪Bε(y)] for ε > 0 so small. Use (G2) in Ωε, we have

ˆΩε

(v4w − w4v)dx =

ˆ∂Ω

(v∂w

∂ν− w∂v

∂ν

)dS + Aε +Bε,

where Aε =´∂Bε(x)

(v ∂w∂ν− w ∂v

∂ν

)dS, Bε =

´∂Bε(y)

(v ∂w∂ν− w ∂v

∂ν

)dS.

Since 4v = 0 = 4w in Ω and v = w = 0 on ∂Ω, Aε = −Bε.

Consider∣∣∣´∂Bε(x)

v ∂w∂νdS∣∣∣ ≤ sup

∂Bε(x)

|v|´∂Bε(x)

∣∣∂w∂ν

∣∣ dS ≤ Cε2 sup∂Bε(x)

|v|, where C is a

constant since w is independent of x.

Letting ε→ 0,∣∣∣´∂Bε(x)

v ∂w∂νdS∣∣∣→ 0. Similarily,

∣∣∣´∂Bε(y)w ∂v∂νdS∣∣∣→ 0 as ε→ 0.

Let v(z) = Γ(z − x) − φx(z), where Γ(z − x) is the fundamental solution of the

Laplace equation and φx(z) is a smooth function in Ω. We take v(z) = Γ(z−x)−φx(z)

into´∂Bε(x)

w ∂v∂νdS and

´∂Bε(y)

v ∂w∂νdS. Then

limε→0

ˆ∂Bε(x)

w∂v

∂νdS = w(x). (4.6)

Similarily, limε→0

´∂Bε(y)

v ∂w∂νdS = v(y).

Finally, we passage the limit ε→ 0 on Aε and Bε, we have w(x) = v(y). This proves

G(x, y) = G(y, x).

We can use Green's function to represent the solution for the Dirichlet's problem of

the Laplace equation.

Theorem 44. For any domain Ω, let u ∈ C2(Ω) ∩ C1(Ω) satises4u = 0 x ∈ Ω

u = 0 x ∈ ∂Ω


and G(x, x0) is Green's function. Then for any point x0 ∈ Ω, we have

u(x0) =

ˆ∂Ω

u(x)∂G

∂νdS.

Proof.

Let G = Γ+h, where Γ is the fundamental solution and h is any harmonic function.

Take v := Γ = G− h into (4.4), then

u(x0) =

ˆ∂Ω

u(x)∂(G− h)

∂ν−(G−h)

∂u

∂νdS =

ˆ∂Ω

[−G∂u

∂ν+ u

∂G

∂ν

]+

[−u∂h

∂ν+ h

∂u

∂ν

]dS

=

ˆ∂Ω

[−G∂u

∂ν+ u

∂G

∂ν

]dS +

ˆΩ

(u4h− h4u) dx =

ˆ∂Ω

u∂G

∂νdS

since 4h = 0 = 4u,G = 0 on ∂Ω and using (G2).

Corollary 45. For any domain Ω, let u ∈ C2(Ω) ∩ C1(Ω) satises4u = f x ∈ Ω

u = h x ∈ ∂Ω

and G(x, x0) is Green's function. Then for any point x0 ∈ Ω, we have

u(x0) =

ˆ∂Ω

h(x)∂G

∂νdS +

ˆΩ

f(x)G(x, x0)dx.

4.2.3 Green's Function for Half-space

In R3, let D = (x, y, z) : z > 0 be a half -space. For all X = (x, y, z) ∈ D, dene its

reected point as X∗ = (x, y,−z).

Denition 46. (Green's function for half-space on R3)


The Green's function for D is

G(X,X0) = − 1

4π |X −X0|+

1

4π |X −X∗0 |. (4.7)

We verify (4.7) has properties of Green's function:

1. It's clearly that G is nite and dierentiable except X = X0 and 4G = 0 in D.

2. Let X ∈ ∂D, that is, z = 0 and |X −X0| = |X −X∗0 |, then G = 0.

3. 1

4π|X−X∗0 |has no singularity in D since X∗0 /∈ D. So G has the proper singularity

at X0.

Now, we solve the Dirichlet problem of the Laplace equation in half-space.

Example. Solve the Dirichlet problem of the Laplace equation in half-space. Consider4u = 0 X = (x, y, z) ∈ D

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

where D = (x, y, z) : z > 0.From theorem 44, for any X0 ∈ D, we have u(X0) =

´∂Du∂G∂νdS, where G is dened

as (4.7). We directly compute ∂G∂ν, we can get ∂G

∂ν= 1

2πz0

|X−X0|3. Hence, the solution is

u(X0) = z02π

´∂D

h|X−X0|3

dS.

4.2.4 Green's Function for Sphere

In R3, for any a > 0, letD = |X| < a and ∂D = |X| = a, whereX = (x, y, z) ∈ R3.

Fix X0 ∈ D, the reectd point X∗0 dened by two properties:

1. It's collinear with the origin O and the point X0;

2. Its distance from the origin is determined by |X0| · |X∗0 | = a2.

So X∗0 = X0

|X0| |X∗0 | = a2

|X0|2X0.


Denition 47. (Green's function for sphere on R3)

For X ∈ D, dene the Green's function for sphere at X0 6= 0 is

G(X,X0) = − 1

4πρ+

a

|X0|· 1

4πρ∗, (4.8)

where ρ = |X −X0| and ρ∗ = |X −X∗0 |.

Remark. If X0 = 0, the Green's function for the sphere is dened as

G(X, 0) = − 1

4π |X|+

1

4πa.

We verify (4.8) has properties of Green's function:

1. G has no singularity except X = X0 and the function 1ρ, 1ρ∗

are harmonic in D

except X = X0 .

2. G = 0 on ∂D. (need to check this)

Now, we solve the Dirichlet problem of the Laplace equation in sphere.

Example. Solve the Dirichlet problem of the Laplace equation in sphere. Consider4u = 0 X = (x, y, z) ∈ D

u(X) = h(X) X ∈ ∂D,

where D = |X| < a and ∂D = |X| = a for any a > 0.

Consider X0 6= 0. From theorem 44, for any X0 ∈ D, we have u(X0) =´∂Du∂G∂νdS,

where G is dened as (4.8). We need to compute ∂G∂ν

on ∂D:

Since ρ2 = |X −X0|2, we have ∇ρ = X−X0

ρ. Similarily, ∇ρ∗ =

X−X∗0ρ∗

. Then

∂G

∂ν= DG · ν =

(X −X0

4πρ3− a

|X0|· X −X

∗0

4π(ρ∗)3

)· Xa


=1

4πρ3

[X −X0 −

(|X0|a

)2

X +X0

]· Xa

=a2 − |X0|2

4πaρ3.

Hence, u(X0) =´∂Du∂G∂νdS = a2−|X0|2

4πa

´∂D

h(X)

|X−X0|3dS is called Poisson integral for-

mula for n = 3.

Example. Consider the Dirichlet problem of the Laplace equation for the circle on R2.

Let Green's function G for the circle on R2 is

G(X,X0) =1

2πln |X −X0|−

1

2πln

(|X0|a|X −X∗0 |

)=

1

2πln ρ− 1

2π

(|X0|aρ∗), (4.9)

where ρ = |X −X0| and ρ∗ = |X −X∗0 |.

Consider the problem4u = 0 (x, y) ∈ D = x2 + y2 < a2

u(x, y) = h(x, y) (x, y) ∈ ∂D = x2 + y2 = a2, (4.10)

where a > 0 is a real number.

The similar computation as previous example, the solution of (4.10) is

u(X0) =

ˆ∂D

u∂G

∂νdS =

a2 − |X0|2

2πa

ˆ∂D

h(X)

|X −X0|2dS,

for all X0 ∈ D. And this formula is called Poisson integral formula for n = 2.

Exercise.

1. Show that (4.6) for n = 2, 3.

2. Prove corollary 45.


3. Check the Green's function (4.8) satises G = 0 on ∂D.

4. Check (4.9) has properties of Green's function.

5. Show that the solution of the equation (4.10) is

u(X0) =

ˆ∂D

u∂G

∂νdS =

a2 − |X0|2

2πa

ˆ∂D

h(X)

|X −X0|2dS,

for any X0 ∈ D.

6. Find Green's function for the half-plane D = (x, y) : y > 0.

7. Consider the problem4u = 0 (x, y) ∈ D = (x, y) : y > 0

u(x, y) = h(x, y) (x, y) ∈ ∂D.

Use #6 to solve the problem.

8. Let D = (x, y, z) : z > 0, X = (x, y, z) and X0 = (x0, y0, z0). The solution of

the half-space problem4u = 0 X = (x, y, z) ∈ D

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

is u(X0) = z02π

´∂D

h|X−X0|3

dS for all X0 ∈ D. Suppose h(x, y) is continuous func-

tion that vanishes outside some circle. Prove u(X)→ 0 as |X| → ∞.

4.3 Maximum Principle

First, we introduce the mean-value property. In this subsection, we only consider R2

and R3. We will mention the mean-value property and maximum principle in higher

dimension in next subsection.


Theorem 48. (Mean-value property on R2)

Let u ∈ C2(B) ∩ C1(B) and 4u = 0 in B, where B := Br(x). Then

u(x) =1

2πr

ˆ∂Br(x)

u(y)dS. (4.11)

Remark.

1. The mean-value property on R3 is

u(x) =1

4πr2

ˆ∂Br(x)

u(y)dS. (4.12)

2. In fact, (4.11) is also written as u(x) = 1πr2

´Br(x)

u(y)dy. Similarily, (4.12) can be

written as u(x) = 143πr3

´Br(x)

u(y)dy.

Proof. Choose coordinates with the origin O at the center of the circle, that is, Br(x)

is replaced by Br(O). Use Poisson integral formula, we have

u(O) =r2

2πr

ˆ∂Br(O)

u(x)

r2dS =

1

2πr

ˆ∂Br(O)

u(x)dS.


Let Ω be a bounded and open subset of Rn and u ∈ C2(Ω) ∩ C1(Ω) which satises

4u = 0 in Ω. Then for all x ∈ Ω, we have inf∂Ωu ≤ u(x) ≤ sup

∂Ωu.

Proof.

We only prove the case of R2. The proof of the case of R3 is similar.

Assume at a maximum point inside Ω, that is, ∃x0 ∈ Ω such that u(x0) = supΩu. At

this point, we have uxx(x0) < 0 and uyy(x0) < 0. So 4u(x0) < 0.(→←)

Let v(x) = u(x)+ε |x|2 for any ε > 0. Then 4v = 4u+4ε = 4ε > 0 in Ω and u < v

in Ω. From the above argument, v has no interior maximum in Ω. So the maximum of

v is attained at x0 ∈ ∂Ω. For all x ∈ Ω, we have

u(x) < v(x) ≤ v(x0) = u(x0) + ε |x0|2 ≤ supu∂Ω

+ εa2,


where a > 0 is the greatest distance from ∂Ω to the origin. So u(x) ≤ supu∂Ω

for all

x ∈ Ω since ε > 0 is arbitrary.

We take u by −u into the above argument, we get inf u∂Ω≤ u(x) for all x ∈ Ω. The

proof is done.


Let Ω be a connected and open subset of Rn and u ∈ C2(Ω) ∩ C1(Ω) which satises

4u = 0 in Ω. If there exists x0 ∈ Ω such that u(x0) = supΩu, u ≡ constant.

Proof.

We only prove the case of R2. The proof of the case of R3 is similar.

Let u(x0) = supΩu = M . Then u(x) ≤M , ∀x ∈ Ω. Let Br(x0) ⊂⊂ Ω, by mean-value

property (4.11),

M = u(x0) =1

2πr

ˆ∂Br(x0)

u(x)dS ≤M.

So u(x) ≡M ∀x ∈ ∂Br(x0). This is true for any such circle. So we repeat the argument

with a dierent center and we can ll the whole domain up with these circles. Hence,

u(x) ≡M ∀x ∈ Ω.

We use maximum principle to prove the uniqueness of the Laplace equation.

Corollary 51. (Uniqueness of the Laplace equation)

Let Ω be a bounded and open subset of Rn and u ∈ C2(Ω) ∩ C1(Ω) which satises4u = f x ∈ Ω

u = g x ∈ ∂Ω.

Then the solution u is unique.

Proof. Let u1, u2 are solutions and w = u1 − u2, then w satises4w = 0 x ∈ Ω

w = 0 x ∈ ∂Ω.


We use weak maximum principle (theorem 49), we get

0 = inf∂Ωw ≤ w(x) ≤ sup

∂Ωw = 0.

Hence, w ≡ 0 in Ω. i.e. u1 ≡ u2 in Ω.

Exercise.

1. Show that the mean-value property on R3 for the Laplace equation is (4.12).

2. Use strong maximum principle to prove weak maximum principle.

3. If remove the assumption Ω is bounded of the weak maximum principle, does

the weak maximum principle still hold?

4. Recall Green's function G which satises4G = 0 (x, y) ∈ Ω

G = 0 (x, y) ∈ ∂Ω,

where Ω be a bounded subset of Rn. Prove that Green's function is unique.

4.4 Laplace's Equation on n-D (n≥ 2)2

4.4.1 Fundamental Solution for the Laplace Equation

First, we consider n = 2:

From section 4.1, we have the solution of the Laplace equation for n = 2 is

u(x) = c1 ln |x|+ c2,


2([1], [2], [3], [6])


Next, we consider n ≥ 3:

Let r = |x| =√x2

1 + x22 + ...+ x2

n. From theorem 37, we know that the solution for

the Laplace equation is radial symmetric, so we can assume u(x) = u(|x|) = u(r).

We directly compute ∂u∂xi

and ∂2u∂x2i:

∂u

∂xi= u′(r)

∂r

∂xi= u′(r)

xir

and∂2u

∂x2i

=∂

∂xi

(u′(r)

xir

)= u′′(r)

x2i

r2+ u′(r)

(1

r− x2

i

r3

).

Hence, 4u =∑n

i=1∂2u∂x2i

= u′′(r) + n−1ru′(r) = 0. We solve the ODE problem:

u′′(r) +n− 1

ru′(r) = 0

=⇒(rn−1u′(r)

)′= 0

=⇒ u(r) = c3r2−n + c4,


Note that these constants c1, c2, c3, c4 is uniquely determined. (see [6])

Denition 52. (Fundamental solution for the Laplace equation)

Γ(x) =

− 12π

ln |x| , n = 2

1n(n−2)ωn|x|n−2 , n ≥ 3

. (4.13)

4.4.2 Green's Representation

In this subsection, we just generalize the statement for the Green's representation of

the solution for the Laplace equation on Rn. It's similar as section 4.2 so we don't give

proofs for the following theorems.


Theorem 53. (Green's representation formula)

Let Ω be a bounded domain in Rn and u ∈ C2(Ω) ∩ C(Ω). Then ∀y ∈ Ω, we have

u(y) =

ˆΩ

Γ4udx+

ˆ∂Ω

(u∂Γ

∂ν− Γ

∂u

∂ν

)dS, (4.14)

where Γ is the fundamental solution of the Laplace equation dened as (4.13) and ν is

an unit outer normal vector of Ω.

Remark.

1. If u has compact support in Rn, (4.14) becomes u(y) =´

ΩΓ4udx.

2. If u is harmonic, (4.14) becomes u(y) =´∂Ω

(u∂Γ∂ν− Γ∂u

∂ν

)dS.

Theorem 54. (General Green's representation formula )

Let Ω be a bounded domain in Rn and u ∈ C2(Ω)∩C(Ω). Suppose h ∈ C2(Ω)∩C1(Ω)

with 4h = 0 in Ω and let G = Γ + h. Then ∀y ∈ Ω, we have

u(y) =

ˆΩ

G4udx+

ˆ∂Ω

(u∂G

∂ν−G∂u

∂ν

)dS, (4.15)

where Γ is the fundamental solution of the Laplace equation dened as (4.13) and ν is

an unit outer normal vector of Ω.

Remark. If G is Green's function, (4.15) becomes u(y) =´

ΩG4udx +

´∂Ωu∂G∂νdS; in

addition, u is harmonic in Ω, then (4.15) becomes u(y) =´∂Ωu∂G∂νdS.

4.4.3 Mean-value Property and Maximum Principle

Theorem 55. (Mean-value property)

Let u ∈ C2(Ω) ∩ C(Ω) is harmonic. Then for each ball Br(x) ⊂ Ω, we have

u(x) =1

nωnrn−1

ˆ∂Br(x)

u dS =1

ωnrn

ˆBr(x)

u dy.


Proof.

Let v(r) = 1nωnrn−1

´∂Br(x)

u(y)dSy. Change of variables: y = x + rξ, ξ is an unit

vector. Then

v(r) =1

nωnrn−1

ˆ∂Br(x)

u(y)dSy =1

nωn

ˆ∂B1(0)

u(x+ rξ)dSξ.

We dierentiate v with respect to r:

v′(r) =1

nωn

ˆ∂B1(0)

Du(x+ rξ) · ξ dSξ

=1

nωn

ˆB1(0)

4u(x+ rξ)dξ

=1

nωnrn−1

ˆBr(x)

4u(y)dy = 0

since using Green's formula and 4u = 0. So v(r) ≡ constant.

Since v(0) = limr→0

1nωnrn−1

´∂Br(x)

u(y)dSy = u(x), v(r) ≡ v(0) = u(x). i.e.

u(x) =1

nωnrn−1

ˆ∂Br(x)

u dS.

On the other hand, for any ρ > 0, we have nωnρn−1u(x) =

´∂Bρ(x)

u dS. Then

integrating by ρ from 0 to r,

u(x) =1

ωnrn

ˆBr(x)

u dy.

Remark. In fact, the converse statement of theorem 55 is still true. (Left as exercise.)

Theorem 56. Let u ∈ C2(Ω) ∩ C(Ω). Then for each ball Br(x) ⊂ Ω,

1. If 4u ≥ 0 in Ω, u(x) ≤ 1nωnrn−1

´∂Br(x)

u dS(

1ωnrn

´Br(x)

u dy).

2. If 4u ≤ 0 in Ω, u(x) ≥ 1nωnrn−1

´∂Br(x)

u dS(

1ωnrn

´Br(x)

u dy).



Let u ∈ C2(Ω) ∩ C(Ω) and Ω be a bounded domain in Rn. Then

1. If 4u = 0, inf∂Ωu ≤ u(x) ≤ sup

∂Ωu for all x ∈ Ω.

2. If 4u ≥ 0, sup∂Ωu = sup

Ωu.

3. If 4u ≤ 0, inf∂Ωu = inf

Ωu.


Let u ∈ C2(Ω) ∩ C(Ω) with 4u ≥ 0 in Ω and Ω be a connected domain in Rn. If

there exists y ∈ Ω such that supuΩ

= u(y), u ≡ constant in Ω.

Proof.

Let M = supuΩ

= u(y). Dene ΩM = x ∈ Ω : u(x) = M. It's clearly ΩM 6= φ and

ΩM is closed since u is continuous. Now, we prove that ΩM is relatively open to Ω.

Let z ∈ ΩM and BR(z) ⊂⊂ Ω.

WANT: BR(z) ⊂ ΩM i.e. u(y) = M for all y ∈ BR(z).

Let v = u−M . Then 4v = 4u = 0 in Ω. By mean-value property, we have

0 = v(z) = u(z)−M =1

ωnRn

ˆBR(x)

(u(y)−M) dy ≤ 0.

Hence, u ≡M in BR(z). i.e. ΩM is relatively open to Ω.

We conclude that ΩM = Ω, that is, u ≡M in Ω.

We can use maximum principle to prove the uniqueness of Dirichlet problem for

Laplace's equation and comparison priciple.

Corollary 59. (Uniqueness of Dirichlet problem for Laplace's equation)

Let u, v ∈ C2(Ω) ∩ C(Ω) and Ω be a bounded domain in Rn. If 4u = 4v = 0 in Ω

and u = v on ∂Ω, u = v in Ω.


Corollary 60. (Comparison principle)

Let u, v ∈ C2(Ω) ∩ C(Ω) and Ω be a bounded domain in Rn. If 4u = 0, 4v ≥ 0 in

Ω and u = v on ∂Ω, u ≥ v in Ω.

Now, we use another method to prove strong maximum principle. So we introduce

useful boundary point lemma or called Hopf's lemma.

Denition 61. (Interior sphere condition)

The domain Ω is said to satisfy an interior sphere condition at x0 ∈ ∂Ω if there

exists a ball B ⊂ Ω with x0 ∈ ∂B.

Lemma 62. (Hopf's lemma)

Let 4u ≥ 0 in Ω. Let x0 ∈ ∂Ω such that

1. u is continuous at x0;

2. u(x0) > u(x) ∀x ∈ Ω;

3. Ω satisfy an interior sphere condition at x0.

Then outer normal derivative of u at x0, if it exists, satises∂u∂ν

(x0) > 0.

Proof.

Since Ω satisfy an interior sphere condition at x0, ∃B = Bε(y) ⊂ Ω with x0 ∈ ∂B.For 0 < |x− y| < ε, dene

v(x) = e−α|x−y|2

− e−αε2 ≥ 0,

where α > 0 is determined later. Then

∂v

∂xi= −2α (xi − yi) e−α|x−y|

2

and∂2v

∂x2i

= 4α2 (xi − yi)2 e−α|x−y|2

− 2αe−α|x−y|2

.

So 4v =(4nα2 |x− y|2 − 2nα

)e−α|x−y|

2 ≥ 0 for α large enough.


Let A = Bε(y)∩B ε2(x0). Since u−u(x0) < 0 on ∂A, ∃η > 0 such that u−u(x0)+ηv ≤

0 on ∂A. And we have4 (u− u(x0) + ηv) = 4u+η4v ≥ 0 in A. From weak maximum

principle,

u− u(x0) + ηv ≤ supA

(u− u(x0) + ηv) = sup∂A

(u− u(x0) + ηv) ≤ 0.

So u− u(x0) + ηv ≤ 0 in A. From this, we have ∂(u+ηv)∂ν

≥ 0 at x = x0. Finally, directly

calculate the normal derivative of u at x0, we get

∂u

∂ν(x0) ≥ −η ∂v

∂ν(x0) > 0.

4.4.4 Harnack's Inequality and Liouville's Theorem

Theorem 63. (Harnack's inequality)

Let u ≥ 0 with 4u = 0 in Ω. Then for any bounded subdomain Ω′ ⊂⊂ Ω, there

exists C depend only on n, Ω, and Ω′ such that supΩ′u ≤ Cinf

Ω′u.

Proof. See [2].

Corollary 64. (Harnack's inequality in ball )

Let u ≥ 0 with 4u = 0 in BR(0). Then

Rn−2(R− |x|)(R + |x|)n−1 u(0) ≤ u(x) ≤ Rn−2(R + |x|)

(R− |x|)n−1 u(0).

Theorem 65. (Liouville's theorem)

Let u : Rn → R be a bounded and harmonic function. Then u is constant.

We want to prove the Liouville's theorem so we need some estimates for derivatives

of u.


Lemma 66.

1. Let 4u = 0 in Ω and B = Br(y) ⊂⊂ Ω. Then |Du(y)| ≤ nrsup∂B|u|.

2. Let 4u = 0 in Ω and Ω′ be any compact subset of Ω. Then for any multi-index

α, we have

supΩ′|Dαu(y)| ≤

(n |α|d

)|α|sup

Ω|u| ,

where d = dist (Ω′, ∂Ω), α = (α1, α2, ..., αn), Dα = ∂α1∂α2 ...∂αn , and |α| = α1 +

α2 + ...+ αn for all αi ∈ N.

Proof.

1. Since 4u = div(Du) = 0, 4 (Du) = div (∇ · (∇u)) = 0, that is, Du is also

harmonic in Ω. By the mean-value property, we have

Du(y) =1

ωnrn

ˆBr(y)

Dudx =1

ωnrn

ˆ∂Br(y)

u · ν dS

since Green's formula.Then

|Du(y)| ≤ 1

ωnrn

ˆ∂Br(y)

|u| dS ≤ n

rsup∂B|u| .

2. Use induction method to prove. Left as exercise.

Now, we prove the Liouville's theorem (theorem 65):

Proof. Take B = Br(y) ⊂⊂ Rn. From Lemma 66, we have |Du(y)| ≤ nrsup∂B|u| → 0 as

r →∞. So u is constant.

Exercise.

1. Prove theorem 53, theorem 54, theorem 56, and theorem 57.

2. Prove corollary 59, corollary 60, and corollary 64.

3. Prove lemma 66.


4. Prove the converse statement of mean-value property for Laplace's equation is

still true.

5. Use Hopf's lemma (lemma 62) to prove the strong maximum principle (theorem

58).

6. Let Γ is the fundamental solution for Laplace's equation. Prove

(a) |DiΓ(x− y)| ≤ 1nωn|x− y|1−n;

(b) |DijΓ(x− y)| ≤ 1ωn|x− y|−n.

Note Di = ∂∂xi

and Dij = ∂2

∂xi∂xj.

7. Consider the Poisson's equation4u = −f in Rn. Show that´

Rn Γ(x−y)f(y) dy :=

Γ ? f(x) is the solution of the Poisson's equation, where Γ is the fundamental

solution for Laplace's equation and Γ ? f is the convolution of Γ with f .

8. Let Rn+ = x = (x1, x2, ..., xn) : xn > 0 be the half-space and x = (x1, x2, ...,−xn)

is the reection point of x = (x1, x2, ..., xn). Dene the Green's function for the

half-space

G(x, y) = Γ(y − x)− Γ(y − x),

where Γ is the fundamental solution for Laplace's equation. Consider the problem4u = −f x ∈ Rn+

u = g x ∈ ∂Rn+

.

Show that for all x ∈ Rn+, the solution of the problem is u(x) = 2xn

nωn

´∂Rn+

g(y)|x−y|ndy.

9. Let x = (x1, x2, ..., xn) ∈ Rn \ 0 and x∗ = r2

|x|2x is the point dual to x with

respect to ∂Br(0). Dene the Green's function for the sphere B = Br(0)

G(x, y) = Γ(y − x)− Γ (r |x| (y − x∗)) ,


where Γ is the fundamental solution for Laplace's equation. Consider the problem4u = −f x ∈ B

u = g x ∈ ∂B.

Show that for all x ∈ B, the solution of the problem is u(x) =´∂BK(x, y)g(y) dy,

where K(x, y) = r2−|x|2nωnr

· 1|x−y|n , x ∈ B, y ∈ ∂B. Note this function K is the

Poisson's kernel and the solution formula is called the Poisson's integral formula.

4.5 Energy Method

Let u be a solution of 4u = 0 in Ω ⊂ Rn. Dene the energy function

E(u) =

ˆΩ

|Du|2 dx.

We use the energy method to prove the uniqueness of the solution for the problem4u = f x ∈ Ω

u = g x ∈ ∂Ω. (4.16)

Let u1, u2 are solutions for (4.16) and dene w = u1 − u2, then w satises4w = 0 x ∈ Ω

w = 0 x ∈ ∂Ω.

Use Green's rst identity, we have´

Ω|Dw|2 dx = 0 since 4w = 0 in Ω and w = 0 on

∂Ω. So w ≡ constant. But w vanishes somewhere. Hence, w ≡ 0. That is, u1 = u2.

Theorem 67. (Dirichlet's principle)

Let u be the unique harmoninc function in Ω and u = g on ∂Ω. Let w be any

function in Ω which satises w = g on ∂Ω. Then E(w) ≥ E(u). That is, the lowest

energy occurs for the harmonic function. We also denote by E(u) = minw∈A

E(w), where

A =w ∈ C2(Ω) : w = g on ∂Ω

.

Remark. The converse statement of Dirichlet's principle still holds.


Proof.

Let v = u− w. Then

E(w) =1

2

ˆΩ

|D(u− v)|2 dx = E(u) + E(v)−ˆ

Ω

Du ·Dv dx = E(u) + E(v)

since´

ΩDu ·Dv dx = 0 by using Green's rst identity and the fact v = 0, 4u = 0. It's

obvious that E(w) ≥ E(u) since E(v) ≥ 0.

On the other hand, let u be the function that u = g on ∂Ω and E(u) = minw∈A

E(w),

where A =w ∈ C2(Ω) : w = g on ∂Ω

. Let v be any function that vanishes on ∂Ω.

Then u + εv = g on ∂Ω, where ε is a constant. So if the energy is smallest for the

function u, we have

E(u) ≤ E(u+ εv) = E(u)− εˆ

Ω

4u · v dx+ ε2E(v).

The minimum occurs for ε = 0. So we get´

Ω4u · v dx = 0. This is valid for all v in Ω.

Let Ω′ ⊂⊂ Ω and choose the function v(x) = 1 in Ω′; v(x) = 0 in Ω \ Ω′. Note this

function v is not smooth so an approximation argument is required. So,´

Ω′4u = 0=⇒

4u = 0 in Ω. By the uniqueness, it's the only function satisfying u = g on ∂Ω that can

minimize the energy.

Exercise.

1. Let u be a solution of

4u = 0 x ∈ Ω

∂u∂ν

= 0 x ∈ ∂Ω, where ν is an unit outer normal vector

of ∂Ω. Use the energy method to prove the uniqueness of the solution.

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