Freiling G., Yurko V. Lectures on Differential Equations of Mathematical Physics: A First Course

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162 G.

Freiling and V. Yurko

hence

g(y) = g

(0)e

. (3.2.5)

virtue

of (3.2.4),

g(0) =

∞

−∞

−λ

dλ =

√

∞

−∞

−ξ

dξ =

Thus,

conclude that

u(x,t) =

√

πt

∞

−∞

ϕ(ξ)e

−

(x−ξ)

dξ. (3.2.6)

ormula

(3.2.6) is called the Poisson formula, and the integral in (3.2.6) is called the Poisson

integral.

Theorem 3.2.2. Let the function ϕ(x) be continuous and bounded. Then the solution

of the Cauchy problem (3.2.1) −(3.2.2) exists, is unique and is given by (3.2.6) .

Proof. It is sufﬁcient to prove that the function u(x,t), deﬁned by (3.2.6), is a solution

of problem (3.2.1)-(3.2.2).

1) Since |ϕ(ξ)| ≤ M, we have

|u(x,t)| ≤

√

πt

∞

−∞

−

(x−ξ)

dξ =

√

∞

−∞

−α

dα = M,

where

α =

ξ −x

√

Therefore,

the

Poisson integral converges absolutely and uniformly in

Q, and

the

function

u(x,t) is continuous and bounded in

The

replacement α =

ξ −x

√

(3.2.6)

yields

u(x,t) =

√

∞

−∞

ϕ(x + 2α

√

t)e

−α

dα. (3.2.7)

or t = 0

we get

u(x,0) = ϕ(x) ·

√

∞

−∞

−α

dα = ϕ(x),

i.e.

the

function u(x,t) satisﬁes the initial condition (3.2.2).

3) Fix δ > 0 and consider the domain G

= {(x,t) : −∞ < x < ∞, t ≥ δ}. In G

the

function u(x,t), deﬁned by (3.2.6), is inﬁnitely differentiable, i.e. it has in G

continuous

partial derivatives of all orders, and these derivatives can be obtained by differentiation

under the sign of integration. In particular,

(x,t) =

√

∞

−∞

ϕ(ξ)e

−

(x−ξ)

−

√

(x −ξ)

√

dξ,

(x,t)

√

πt

∞

−∞

ϕ(ξ)e

−

(x−ξ)

−

x −ξ

dξ,

Parabolic

Partial Differential Equations 163

(x,t)

√

∞

−∞

ϕ(ξ)e

−

(x−ξ)

−

√

(x −ξ)

√

dξ.

Consequently

the function u(x,t) satisﬁes equation (3.2.1) in G

By virtue of the arbitrariness of δ > 0 we conclude that u(x,t) satisﬁes (3.2.1) in Q.

Theorem 3.2.2 is proved. 2

Deﬁnition 3.2.2. The solution of the Cauchy problem (3.2.1)-(3.2.2) is called stable if

for each ε > 0 there exists δ > 0 such that if |ϕ(x) −

ϕ(x)| ≤ δ for all x, then |u(x,t) −

˜u(x,t)| ≤ ε in

follo

ws from (3.2.7) that

|u(x,t) − ˜u(x,t)| ≤

√

∞

−∞

|ϕ(x + 2α

√

t) −

ϕ(x + 2α

√

t)|e

−α

dα

≤ sup

|ϕ(x) −

ϕ(x)|,

and

consequently

, the solution of the Cauchy problem (3.2.1)-(3.2.2) is stable (one can take

δ = ε ). Thus, the Cauchy problem (3.2.1)-(3.2.2) is well-posed.

Chapter

Elliptic

Partial Differential

Equations

Elliptic equations usually describe stationary ﬁelds, for example, gravitational, electro-

statical and temperature ﬁelds. The most important equations of elliptic type are the Laplace

equation ∆u = 0 and the Poisson equation ∆u = f (x), where x = (x

,. .. ,x

) are spatial

variables, u(x) is an unknown function,

∆u :=

∑

k=1

∂

∂x

the

Laplace operator (or Laplacian), and f (x) is a given function. In this chapter we study

boundary value problems for elliptic partial differential equations and present methods for

their solutions.

4.1. Harmonic Functions and Their Properties

1. Basic notions

Let D ⊂ R

be a bounded domain with the boundary Σ.

Deﬁnition 4.1.1. A surface Σ is called piecewise-smooth ( Σ ∈PC

), if it consists of a

ﬁnite number of pieces with continuous tangent planes on each of them, and if each straight

line intersects Σ at no more than a ﬁnite number of points and/or segments.

Examples of surfaces of the class PC

are a ball, a parallelepiped, an ellipsoid, etc..

Everywhere below we assume that Σ ∈ PC

. As before, in the sequel the notation u(x) ∈

(D) means that the function u(x) has in D continuous partial derivatives up to the order

Deﬁnition 4.1.2. Let u ∈C

(D), x ∈Σ, n

be the outer normal to Σ at the point x. If

uniformly on Σ there exists a ﬁnite limit

lim

y→x

y=x−αn

,α>0

∂u(y)

∂n

∂u(x)

∂n

166 G.

Freiling and V. Yurko

shall

say that the function u(x) has on Σ the normal derivative

∂u(x)

∂n

(notation: u(x) ∈

−

(

D) ).

Remark

4.1.1. 1)

Clearly, if u(x) ∈C

(

D), then u(x) ∈C

−

(D) .

If u

(x) ∈C

−

(

D), then u(x) ∈C(D) (after

deﬁning u on ∂D by

continuity if necessary).

Deﬁnition 4.1.3. A function u(x) is called harmonic in the domain D, if u(x) ∈

(D) and ∆u = 0 in D.

Examples. 1) Let n = 1. Then ∆u = u

(x), and consequently, the harmonic functions

are the linear ones: u(x) = ax + b.

2) Let n = 2, i.e. x = (x

), and let z = x

+ ix

. If the function f (z) = u(x

) +

iv(x

) is analytic in D, then by virtue of the Cauchy-Riemann conditions

∂u

∂x

∂v

∂x

∂u

∂x

= −

∂v

∂x

the

functions u and v are

harmonic in D (and they are called conjugate harmonic func-

tions). Conversely, if u(x

) is harmonic in D, then there exists a conjugate harmonic

function v(x

) such that the function f = u + iv is analytic in D, and

v(x

) =

)

−

∂u

∂x

∂

∂x

(the

integral does not depend on the way of the integration since under the integral we have

the total differential of v ).

Fix x

= (x

) and denote

r = kx −x

k =

−x

)

−x

)

is easy to check that the function u(x) = ln

harmonic

everywhere except at the point

. This function is called the fundamental solution of the Laplace equation on the plane.

3) Let n = 3, i.e. x = (x

). Fix x

= (x

) and denote

r = kx −x

k =

∑

k=1

−x

)

Then

the

function u(x) =

harmonic

everywhere except at the point x

(see Lemma

2.5.1). This function is called the fundamental solution of the Laplace equation in the space

Remark 4.1.2. The importance of the fundamental solution connects with the isotropy

of the space when the physical picture depends only on the distance from the source of

energy but not on the direction. For example, the function u(x) =

represents

the

potential

of the gravitational (electro-statical) ﬁeld created by a point unit mass (point unit charge).

Similar sense has the fundamental solution of the Laplace equation on the plane: this is

the potential of the ﬁeld produced by a charge of constant linear density q = 1, distributed

uniformly along the line x

= x

, x

= x

Elliptic P

artial Differential Equations 167

2. Pr

operties of harmonic functions

Let for deﬁniteness n = 3.

Theorem 4.1.1. Let u(x),v(x) ∈C

(D) ∩C

−

(

D). Then

(u∆v −v∆u) d

x =

∂v

∂n

−v

∂u

∂n

, (4.1.1)

where n is the outer normal to the surface Σ . Formula (4.1.1) is called Green’s formula.

Proof. Let D

be the domain that is obtained from D by removing the balls K

(x) ,

x ∈ Σ of radius α around the point x. Let Σ

be the boundary of D

. Applying the

Gauß-Ostrogradskii formula for the domain D

, we get

(u∆v −v∆u) dx =

∑

k=1

∂

∂x

(uv

−vu

)

∑

k=1

(uv

−vu

)cos

(n,x

)

∑

k=1

cos(n,x

) −v

∑

k=1

cos(n,x

)

∂v

∂n

−v

∂u

∂n

As α → ∞ this yields (4.1.1). 2

Theorem 4.1.2. Let u(x) be harmonic in D and u(x) ∈C

−

(

D). Then

∂u

∂n

s = 0

. (4.1.2)

Indeed, (4.1.2) follows from (4.1.1) for v ≡ 1.

Theorem 4.1.3. Let u(x) ∈ C

(D) and assume that for any closed surface S ∈ PC

from D one has

∂u

∂n

s = 0

Then u(x) is a harmonic function in D.

Proof. It follows from (4.1.1) for v ≡ 1, Σ = S that

∆udx = 0,

where G := int S. By virtue of the arbitrariness of G this yields ∆u = 0 in D. 2

Theorem 4.1.4. Let u(x) be harmonic in D and u(x) ∈C

−

(

D). Then

∑

k=1

∂u

∂x

x =

∂

∂n

. (4.1.3)

168 G.

Freiling and V. Yurko

Relation (4.1.3)

is called the Dirichlet identity.

Proof. We apply (4.1.1) for the functions v ≡1 and u

. Since

∂(u

)

∂n

= 2u

∂u

∂n

∂(u

)

∂x

= 2u

∂u

∂x

∂

)

∂x

= 2

∂u

∂x

+ 2u

∂

∂x

arri

ve at (4.1.3). 2

Theorem 4.1.5. Let u(x) be harmonic in D and u(x) ∈C

−

(

D). Then

u(x)

4π

∂u(ξ)

∂n

−u(ξ)

∂

(

)

∂n

, x ∈D, (4.1.4)

where ξ is the variable of integration, r = kx −ξk, n

is the outer normal to Σ at the

point ξ, and ∂

means that the differentiation is performed with respect to ξ ( for a ﬁxed

x) . Formula (4.1.4) is called the basic formula for harmonic functions.

Proof. Fix x ∈ D and consider the ball K

(x) of radius δ > 0 around the point x. In

the domain D \K

(x) we apply Green’s formula (4.1.1) for v = 1/r. Since ∆u = ∆v = 0,

we have

Σ∪S

(x)

u(ξ)

∂

(

)

∂n

−

∂u(ξ)

∂n

s = 0

where S

(x) = ∂K

(x) is the sphere. On S

(x) :

∂

(

)

∂n

= −

∂(

)

∂r

, r = δ,

and

consequently

(x)

u(ξ)

∂

(

)

∂n

−

∂u(ξ)

∂n

(

u(ξ)

−

∂u(ξ)

∂n

(x)

u(ξ)d

s −

(x)

∂u(ξ)

∂n

By Theorem 4.1.2 the second integral is equal to zero, and we obtain

(x)

u(ξ)d

s =

(x)

s +

(x)

(u(ξ) −u(x)

= 4πu(x) + J

Elliptic P

artial Differential Equations 169

Since the

function u(x) is continuous, we have that for each ε > 0 there exists δ > 0 such

that |u(x) −u(ξ)|≤ ε for all ξ ∈

(x). Therefore,

≤

(x)

|u(ξ) −u(x)|d

s ≤

4πδ

= 4πε.

Thus, I

→ 4πu(x) as δ →0, and

arrive at (4.1.4). 2

Theorem 4.1.6. (Mean-Value Theorem). Let K

(x) be a ball of radius R around the

point x, and let S

(x) = ∂K

(x) be the sphere. Suppose that the function u(ξ) is harmonic

in K

(x) and continuous in

(x). Then

u(x)

4πR

(x)

u(ξ)d

. (4.1.5)

Proof. We apply the basic formula for harmonic functions (4.1.4) for the ball K

(x)

for

R < R. On S

∂

(

)

∂n

∂(

)

∂r

= −

, r =

and

consequently

u(x) =

4π

(x)

∂u(ξ)

∂n

u(ξ)

4π

(x)

∂u(ξ)

∂n

s +

4π

(x)

u(ξ)d

4π

(x)

u(ξ)d

since the ﬁrst integral here is equal to zero by Theorem 4.1.2. As

R → R we arrive at

(4.1.5). 2

Theorem 4.1.7 (Maximum Principle). Let u(x) ∈C(

D) and

let

u be harmonic in D.

Then u(x) attains its maximum and minimum on Σ. Moreover, if u(x) 6≡ const, then

min

ξ∈Σ

u(ξ) < u(x) < max

ξ∈Σ

u(ξ)

for all x ∈D, i.e. maximum and minimum cannot be attained in D.

Proof. Let u(x) 6≡ const, and let the maximum be attained at some point x ∈ D,

i.e. u(x) = max

ξ∈

u(ξ). T

e a ball K

(x) such that

(x) ⊂ D and

such

that there exists

˜x ∈ S

(x), for which u( ˜x) < u(x). Such choice is possible because u(ξ) 6≡ const. . Since

u(x) > u( ˜x), there exists d > 0 such that u(x) −u(˜x) > d > 0. By virtue of the continuity

of the function u(ξ), the inequality also holds in some neighbourhood of ˜x. Denote

(x) = {ξ ∈S

(x) : u(x) −u(ξ) ≥ d > 0},

(x) = S

(x) \S

(x).

170 G.

Freiling and V. Yurko

By the

mean-value theorem,

u(x) =

4πR

(x)

u(ξ)d

hence

4πR

(x)

(u(x) −u(ξ))d

s = 0

On the other hand,

(x)

(u(x) −u(ξ))ds ≥ dC

> 0,

(x)

(u(x) −u(ξ))ds ≥ 0.

This contradiction proves the theorem. 2

Remark 4.1.3. Let n = 2, i.e. x = (x

) ∈R

. In this case Theorems 4.1.1-4.1.4 and

4.1.7 remain valid (as for any n ), and Theorems 4.1.5-4.1.6 require small modiﬁcations,

namely:

Theorem 4.1.5’. Let u(x) be harmonic in D ⊂ R

and u(x) ∈C

−

(

D). Then

u(x)

2π

∂u(ξ)

∂n

−u(ξ)

∂

(ln

)

∂n

, x ∈D, (4.1.6)

where r = kx −ξk, n

is the outer normal to Σ at the point ξ.

Theorem 4.1.6’. Let K

(x) be the disc of radius R around the point x, and let S

(x) =

∂K

(x) be the circle. Suppose that u(ξ) is harmonic in K

(x) and continuous in

(x).

Then

u(x)

2πR

(x)

u(ξ)d

. (4.1.7)

4.2. Dirichlet and Neumann Problems

Let D ⊂ R

be a bounded domain with the boundary Σ ∈ PC

Dirichlet problem

Let a continuous function ϕ(x) be given on Σ. Find a a function u(x) which is harmonic

in D and continuous in

D and

has

on Σ an assigned value ϕ(x) :

∆u = 0 (x ∈ D),

|Σ

= ϕ(x).

)

(4.2.1)

Theorem 4.2.1. If a solution of the Dirichlet problem (4.2.1) exists, then it is unique.

Elliptic P

artial Differential Equations 171

Proof

. Let u

(x) and u

(x) be solutions of problem (4.2.1). Denote u(x) = u

(x) −

(x). Then u(x) ∈C(

D) , ∆u = 0 in D, and u

|Σ

= 0. By

the

maximum principle, u(x) ≡0

D. 2

Deﬁnition

4.2.1. The

solution of the Dirichlet problem is called stable if for each ε > 0

there exists δ > 0 such that if |ϕ(x) −

ϕ(x)| ≤ δ for all x ∈ Σ, then |u(x) − ˜u(x)| ≤ ε for

all x ∈

D. Here u and ˜u are

solutions

of the Dirichlet problems with the data ϕ and

ϕ,

respectively.

It follows from the maximum principle that the solution of the Dirichlet problem (4.2.1)

is stable (one can take δ = ε ). The question of the existence of the solution of the Dirichlet

problem is much more difﬁcult. This question will be studied later by various methods in

Sections 4.3-4.6.

Neumann problem

Let a continuous function ψ(x) be given on Σ. Find a function u(x) which is harmonic in

D and u(x) ∈C

−

(

D) ,

∂u

∂n

|Σ

= ψ :

∆u = 0 (x ∈ D),

∂u

∂n

= ψ(x).







(4.2.2)

Theor

4.2.2. If a solution of problem (4.2.2) exists, then it is unique up to an

additive constant.

Proof. Obviously, if the function u(x) is a solution of problem (4.2.2), then the

function u(x) +C is also a solution of (4.2.2). Furthermore, if u

(x) and u

(x) are solu-

tions of (4.2.2), then the function u(x) = u

(x) −u

(x) has the properties: ∆u = 0 in D ,

u(x) ∈C

−

(

D) and

∂u

∂n

|Σ

= 0. Applying

the

Dirichlet identity (see Theorem 4.1.4) we obtain

∑

k=1

∂u

∂x

x = 0

and consequently,

∂u

∂x

≡ 0, k = 1, 2,3,

i.e. u(x) ≡ cons

t. .

Theorem 4.2.2 is proved. 2

Theorem 4.2.3 (Necessary condition for the solvability of the Neumann problem).

If problem (4.2.2) has a solution, then

ψ(ξ)ds = 0.