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

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172 G.

Freiling and V. Yurko

Theorem 4.2.3

is an obvious corollary of Theorem 4.1.2, since, if u(x) is a solution of

problem (4.2.2), then

ψ(ξ)ds =

∂u

∂n

s = 0

The Dirichlet and Neumann problems can be considered also in unbounded domains.

In this case we need the additional condition u(∞) = 0 (for n = 2 the additional condition

has the form u(x) = O(1), |x| → ∞ ). For example, let D ⊂ R

be a bounded set, and

:= R

D. The

Dirichlet

and Neumann problems in the domain D

are called exterior

problems.

Exterior Dirichlet problem

∆u = 0 (x ∈ D

|Σ

= ϕ(x), u(∞) = 0.

)

(4.2.3)

A function u(x) is called a solution of problem (4.2.3) if u(x) is harmonic in D

, u(x) ∈

) , u

|Σ

= ϕ(x) and u(∞)

= 0

, i.e.

lim

R→∞

max

kxk=R

|u(x)| = 0.

Exterior Neumann problem

∆u = 0 (x ∈ D

∂u

∂n

= ψ(x), u(∞)

= 0







(4.2.4)

A function u(x) is called a solution of problem (4.2.4) if u(x) is harmonic in D

, u(x) ∈

−

(

) ,

∂u

∂n

|Σ

= ψ(x) and u(∞)

= 0

One can consider the Dirichlet and Neumann problems also in other unbounded regions

(a sector, a strip, a half-strip, a half-plane, etc.).

Let us formulate the uniqueness theorem for the exterior Dirichlet problem (4.2.3).

Theorem 4.2.4. If a solution of problem (4.2.3) exists, then it is unique.

Proof. Let u

(x) and u

(x) be solutions of (4.2.3). Denote u(x) = u

(x) −u

(x).

Then ∆u = 0 in D

, u(x) ∈C(

) , u

|Σ

= 0 and u(∞)

= 0

. Let K

(0) be a ball of radius

R around the origin such that D ⊂K

(0). Applying the maximum principle for the region

(0) \D, we obtain

:= min

ξ∈S

(0)∪Σ

u(ξ) ≤ u(x) ≤ max

ξ∈S

(0)∪Σ

u(ξ) := A

for all x ∈ K

(0) \D, where S

(0) = ∂K

(0) . Since u(∞) = 0 and u

|Σ

= 0, we have

→ 0 and A

→ 0 as R →∞. Therefore, u(x) ≡ 0, and Theorem 4.2.4 is proved. 2

Elliptic P

artial Differential Equations 173

In order

to prove the uniqueness theorem for the exterior Neumann problem we need

several assertions which also are of independent interest.

Theorem 4.2.5 (Theorem on a removable singularity). Let 0 ∈ D, and let the func-

tion u(x) be harmonic in D \{0} and lim

kxk→0

kxku(x) = 0 (i.e. u(x) = o(

kxk

), x → 0 ).

Then

the

point x = 0 is a removable singularity for u(x), i.e. the function u(x) can be

deﬁned at the point 0 such that u(x) becomes harmonic in D.

Remark 4.2.1. 1) The condition 0 ∈ D is not important and is taken for deﬁniteness;

a singularity can be removed similarly at any point.

2) The function

kxk

harmonic

everywhere except at the point 0 (see Lemma 2.5.1) and

it has at 0 an unremovable singularity. This function does not satisfy the last condition of

the theorem.

Proof. Let K

(0) be a ball around the origin such that

(0) ⊂ D. W

take the

function v(x) such that v(x) is harmonic in K

(0), continuous in

(0) and v

, where S

= ∂K

(0) is

sphere. In other words, the function v(x) is the solution

of the Dirichlet problem for the ball. Below, in Section 4.3 we will prove independently

the existence of the solution of the Dirichlet problem for a ball. Therefore, a function v(x)

with the above mentioned properties exists and is unique. Denote w(x) = v(x)−u(x). Then

w(x) is harmonic in K

(0) \{0} and w

= 0. Moreover, since

lim

kxk→0

kxku(x) = 0

and v(x) is continuous, it follows that

lim

kxk→0

kxkw(x) = 0.

Fix α > 0 and consider the function

(x) = α

kxk

−

Clearly

, Q

(x) is

harmonic in K

(0) \{0} and Q

α|S

= 0. We have

kxk(Q

±w(x)) = α −

αkxk

xkw(x)

There exists r

> 0 such that

αkxk

xkw(x)

≤ α

for all kxk≤r

. Consider the ring G

= {x : r

≤x ≤R}. The functions w(x) and Q

(x)

are harmonic in G

and on the boundary |w(x)| ≤ Q

(x). By the maximum principle this

inequality holds also in G

, i.e.

|w(x)| ≤ α

kxk

−

, x ∈G

174 G.

Freiling and V. Yurko

Fix x ∈ G

. As α →0 we get w(x)

= 0. By virtue of the arbitrariness of x we conclude

that w(x) = 0 for all x 6= 0, i.e. u(x) = v(x) for all x 6= 0. Deﬁning u(0) := v(0), we

arrive at the assertion of Theorem 4.2.5. 2

The Kelvin transform

Consider the function u(x) and make the substitution

ξ =

kxk

. (4.2.5)

The

verse transform is symmetric:

x =

kξk

Denote

v(ξ)

kξk

= kxku(x). (4.2.6)

The

function v

(ξ) is called the Kelvin transform for u(x). This transform is symmetric

since

u(x) =

kxk

Consider

bounded domain D ∈R

with the boundary Σ , D

= R

D, and

suppose

that

0 ∈D. Let D

∗

be the image of the domain D

with respect to the replacement (4.2.5).

Then D

∗

= R

∗

and Σ

∗

= ∂D

∗

are

the

images of D and Σ, respectively. For example,

if D = K

(0), then D

∗

= K

1/R

(0).

Theorem 4.2.6 (Kelvin). Let 0 ∈ D. If the function u(x) is harmonic in D

, then

its Kelvin transform v(ξ) is harmonic in D

∗

\{0}. If the function u(x) is harmonic in

D \{0}, then v(ξ) is harmonic in D

∗

Proof. Denote

∆

u(x) =

∑

k=1

∂

∂x

, ∆

v(ξ)

∑

k=1

∂

∂ξ

Using

(4.2.5)-(4.2.6)

we calculate by differentiation

∆

u(x) =

kxk

∆

v(ξ). (4.2.7)

All

assertions

of Theorem 4.2.6 follow from (4.2.7). 2

Theorem 4.2.7. Let the function u(x) be harmonic outside the ball K

(0) and u(∞) =

0 ( i.e. lim

R→∞

max

kxk=R

|u(x)| = 0) . Then

u(x) = O

kxk

∂u

∂x

= O

kxk

, x →∞.

Elliptic P

artial Differential Equations 175

Proof

. Let

v(ξ) =

kξk

the

Kelvin transform for u(x). By Theorem 4.2.6, the function v(ξ) is harmonic in

the domain K

1/R

(0) \{0}. Since u(∞) = 0, we have kξkv(ξ) → 0 as ξ → 0. Then, by

Theorem 4.2.5, the function v(ξ) is harmonic in K

1/R

(0). In particular, this yields

v(ξ) = O(1),

∂v

∂ξ

= O(1) for ξ → 0. (4.2.8)

Since

u(x)

kxk

v(ξ),

ve, by virtue of (4.2.8), that

u(x) = O

kxk

, x →∞.

Furthermore,

∂u

∂x

∂

∂x

kxk

v(ξ)

= −

kxk

v(ξ)

kxk

∑

k=1

∂v

∂ξ

∂x

Since

∂ξ

∂x

kxk

+ x

∂

∂x

kxk

∂

∂x

kxk

= −

kxk

where δ

the Kronecker delta, one gets

∂u

∂x

= −

kxk

v(ξ)

kxk

∂v

∂ξ

−

kxk

∑

k=1

∂v

∂ξ

vie

w of (4.2.8) this yields

∂u

∂x

= O

kxk

, x →∞.

Theorem

4.2.7

is proved. 2

Let us now prove the uniqueness theorem for the exterior Neumann problem (4.2.4).

Theorem 4.2.8. If the solution of problem (4.2.4) exists, then it is unique.

Proof. Let u

(x) and u

(x) be solutions of problem (4.2.4). Denote u(x) = u

(x) −

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

, u(∞) = 0 and

∂u

∂n

|Σ

= 0. Let K

(0) be

176 G.

Freiling and V. Yurko

the ball

of radius R around the origin such that D ⊂K

(0). We apply the Dirichlet identity

to the domain K

(0) \D :

(0)\D

∑

k=1

∂u

∂x

x =

∂

∂n

s, (4

.2.9)

where S

= ∂K

(0) . By Theorem 4.2.7, on S

we have

|u(x)| ≤

, |

∂u

∂n

≤

and

consequently

∂u

∂n

≤

→ 0 as R → ∞.

Then

from

(4.2.9) as R → ∞ we obtain

∑

k=1

∂u

∂x

x = 0

Therefore,

∂u

∂x

≡ 0, k = 1, 2,3,

i.e. u(x) ≡ cons

t. .

Since u(∞) = 0, we conclude that u(x) ≡ 0, and Theorem 4.2.8 is

proved. 2

Reduction of the exterior Dirichlet problem to the interior one

Let 0 ∈D. Let u(x) be the solution of problem (4.2.3) and let v(ξ) be the Kelvin transform

for u(x). Then, by Theorems 4.2.5-4.2.6, the function v(ξ) is harmonic in D

∗

. Moreover,

v(ξ) ∈C(

∗

) and v

|Σ

∗

= ϕ

∗

, where ϕ

∗

the

Kelvin transform for ϕ. Thus, v(ξ) is the

solution of the Dirichlet problem for the bounded domain D

∗

4.3. The Green’s Function Method

1. Let D ⊂ R

be a bounded domain with the boundary Σ ∈ PC

. We consider the

Dirichlet problem

∆u = 0 (x ∈ D),

|Σ

= ϕ(x),

)

(4.3.1)

where ϕ(x) ∈ C(Σ). Let u(x) be the solution of problem (4.3.1), and let v(x) be a har-

monic function in D. Moreover, suppose that u(x), v(x) ∈C

−

(

D). Then,

virtue of The-

orem 4.1.1 and 4.1.5, we obtain

u(x) =

4π

∂u(ξ)

∂n

−u(ξ)

∂

(

)

∂n

s, x ∈D, r = k

x −ξk,

Elliptic P

artial Differential Equations 177

0 =

v(ξ)

∂u(ξ)

∂n

−u(ξ)

∂v(ξ)

∂n

Therefore,

u(x) =

4π

µµ

v +

∂u

∂n

−u

∂

∂n

v +

¶¶

. (4.3.2)

We require that (v +

)

|Σ

= 0. Then

for ﬁnding the function v = v(ξ,x) we obtain the

following Dirichlet problem with respect to ξ for a ﬁxed x ∈ D :

∆

v = 0 (ξ ∈D),

|Σ

= −











(4.3.3)

Consider

the

function

G(x,ξ) = v +

where v is

the

solution of problem (4.3.3). The function G(x,ξ) is called the Green’s

function. Formula (4.3.2) takes the form

u(x) = −

4π

ϕ(ξ)

∂

G(x,ξ)

∂n

s, x ∈D. (4

.3.4)

Thus, if the solutions of the problems (4.3.1) and (4.3.3) exist (and have on Σ normal

derivatives), then formula (4.3.4) is valid.

Remark 4.3.1. In order to construct the solution of the Dirichlet problem (4.3.1) by

(4.3.4) we ﬁrst need to solve the Dirichlet problem (4.3.3). For some domains the Dirichlet

problem (4.3.3) can be solved explicitly. Below, using the Green’s function method we will

construct the solution of the Dirichlet problem for a ball.

By similar arguments one can solve also the Neumann problem

∆u = 0 (x ∈ D),

∂u

∂n

= ψ(x).

The

solution

has the form

u(x) =

4π

ψ(ξ)

G(x,ξ) d

, x ∈D,

where

G(x,ξ) = ˜v +

and

the

function ˜v(ξ,x) is the solution of the following Neumann problem with respect to

ξ :

∆

˜v = 0,

∂

˜v

∂n

= −

∂

(1/r)

∂n

178 G.

Freiling and V. Yurko

Below

we will need the following auxiliary assertion.

Lemma 4.3.1. Let l be a vector and r = kξ −xk. Then

∂

∂l

= cos(l, ¯r),

wher

e ¯r =

(ξ

−x

,ξ

−x

,ξ

−x

). Symmetrically,

∂

∂l

= −cos(l, ¯r).

Indeed,

∂

∂l

∑

k=1

∂r

∂ξ

cos(l, ξ

)

∑

k=1

−x

cos(l,

)

∑

k=1

cos(¯r,ξ

)cos(l,ξ

)

, ¯r

)

= kl

k·k¯r

kcos(l, ¯r) = cos(l, ¯r),

where

= (cos(l,ξ

),cos(l,ξ

)),

¯r

= (cos(¯r,ξ

),cos(¯r, ξ

))

are unit vectors for l and ¯r, respectively, and (l

, ¯r

) is the scalar product of the vectors

and ¯r

2. Solution of the Dirichlet problem for a ball

Let K

be a ball of radius R around the point x

, and let S

= ∂K

be the sphere. We

consider the Dirichlet problem

∆u = 0 (x ∈ K

= ϕ(x), ϕ(x) ∈C(S

)

(4.3.5)

Let x ∈K

. On the ray x

x choose the point x

such that

−xk·kx

−x

k = R

Let ξ ∈S

. Denote

r = kξ −xk, ρ = kx

−xk, r

= kξ −x

k, ρ

= kx

−x

(see ﬁg. 4.3.1).

Elliptic P

artial Differential Equations 179

Figure

4.3.1.

e note that the triangles ∆

ξx

and ∆

are similar since they have a common angle

and

Hence

∈

(

)

our

case the Dirichlet problem (4.3.3) has the form

∆

v = 0 (ξ ∈K

= −











(4.3.7)

Let

show that the solution of problem (4.3.7) has the form

v(ξ,x) =

, α −cons

The

function v(ξ,x) is harmonic with respect to ξ everywhere except at the point x

, and

in particular, it is harmonic in

. The

boundary

condition yields

= −

, ξ ∈S

Using

(4.3.6)

we calculate

α = −

Thus,

v(ξ,x)

= −

ρr

and

consequently

, the Green’s function has the form

G(x,ξ) =

−

ρr

. (4.3.8)

180 G.

Freiling and V. Yurko

Let us

show that

∂

G(x,ξ)

∂n

−R

. (4.3.9)

Indeed,

virtue of Lemma 4.3.1,

∂

∂n

= cos(n

, ¯r),

∂

∂n

= cos(n

, ¯r

). (4.3.10)

Using

the

cosine theorem we calculate

cos(n

, ¯r) =

+ r

−ρ

2Rr

, cos(n

, ¯r

)

+ r

−ρ

2Rr

. (4.3.11)

Dif

erentiating (4.3.8) and using (4.3.6), (4.3.10) and (4.3.11) we get

∂

G(x,ξ)

∂n

∂

(

)

∂n

−

∂

(

)

∂n

= −

cos( ¯n

, ¯r)

cos( ¯n

, ¯r

)

−R

−r

2Rr

+ r

−ρ

2ρr

−R

2Rr

−

2Rr

−ρ

2ρr

−R

2Rr

−(ρ

ρ)

2(ρr

)

−R

2Rr

−R

2Rr

−R

i.e.

(4.3.9)

is valid. Substituting now (4.3.9) into (4.3.4) we arrive at the formula

u(x) =

4π

ϕ(ξ)

−ρ

, x ∈K

, (4.3.12)

which is called the Poisson formula. Thus, we have proved that if the solution of problem

(4.3.5) exists, then it is given by (4.3.12).

Theorem 4.3.1. For each continuous function ϕ on S

the solution of the Dirichlet

problem (4.3.5) for the ball exists, is unique and is represented by formula (4.3.12) .

Proof. It is sufﬁcient to prove that the function u(x), deﬁned by (4.3.12), is a solution

of problem (4.3.5). First we will prove that u(x) is harmonic in K

. Let

G ⊂ K

bounded

closed domain. Then for all x ∈

G , ξ ∈ S

ve r = kx −ξk≥ d > 0, where

d is the distance from

G to S

. Therefore, u(x) ∈C

∞

(G), and

all

derivatives of u(x) can

be obtained by differentiation under the integral sign. In particular,

∆u =

4π

ϕ(ξ)∆

−ρ

Using

(4.3.10)-(4.3.11) we calculate

Clearly,

∆

−ρ

= ∆

−ρ

+ r

−∆

= ∆

−ρ

+ r

Elliptic P

artial Differential Equations 181

Using (4.3.10)-(4.3.11)

we calculate

∆

−ρ

= 2∆

cos(n

, ¯r)

= −2∆

∂

(

)

∂n

= −2

∂

∂n

∆

¶¶

= 0,

and

consequently

, ∆u(x) = 0 in

G. By

virtue

of the arbitrariness of

G we

conclude

that

u(x) ∈C

∞

) and u(x) is harmonic in K

Fix x

∗

∈ Σ. Let us show that

lim

x→x

∗

,x∈K

u(x) = ϕ(x

∗

). (4.3.13)

For this purpose we consider the auxiliary Dirichlet problem

∆w = 0 (x ∈ K

= 1.

)

(4.3.14)

Obviously, problem (4.3.14) has the unique solution w(x) ≡ 1. Hence, in view of (4.3.12),

1 =

4π

−ρ

s, (

4.3.15)

and consequently,

u(x) −ϕ(x

∗

) =

4πR

ϕ(ξ) −ϕ(x

∗

−ρ

, x ∈K

. (4.3.16)

Fix ε > 0. Since the function ϕ is continuous, there exists δ > 0 such that for kξ−x

∗

k≤δ

one has |ϕ(ξ) −ϕ(x

∗

)| ≤ ε/2. Let x ∈ K

, kx −x

∗

k ≤ δ/2. Denote

= {ξ ∈ S

: kξ −x

∗

k ≤ δ}, S

= S

Clearly, r ≥ δ/2 on S

. Let C = max

|ϕ(ξ)|. Then it follows from (4.3.15)-(4.3.16) that

for x ∈K

, kx −x

∗

k ≤ δ/2,

|u(x) −ϕ(x

∗

)| ≤

4πR

−ρ

s +

4πR

−ρ

(δ/2)

≤

−ρ

Let

w x →x

∗

. Then ρ

→ R

, i.e. there exists δ

such that for kx −x

∗

k ≤ δ

one has

−ρ

) ≤

Thus,

have proved that for each ε > 0 there exists δ

> 0 such that if kx −x

∗

k ≤ δ

then |u(x) −ϕ(x

∗

)| ≤ ε, i.e. (4.3.15) is valid. Therefore, u(x) ∈ C(

) and u

= ϕ.

Theorem

4.3.1

is proved. 2