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

192 G.

Freiling and V. Yurko

As ε →0 we deduce

that there exists lim

x→x

0

u(x), and

lim

x→x

0

u(x) = ϕ(x

0

).

Theorem 4.4.13 is proved. 2

Theorem 4.4.14. Let x

0

∈ Σ, and let there exist a closed ball

K

R

(x

1

) with

its

center

outside

D suc

h

that

K

R

(x

1

) ∩D = {x

0

} ( in

this

case we shall say that the point x

0

can

be ”reached” by a ball ) . Then for the point x

0

there exist a barrier.

Proof. We consider the function

ω(x) =

1

R

−

1

r

,

where r = kx −x

1

k. Clearly

, ω

(x) ∈ C(

D) and ω(x) is

harmonic

in D, hence ω(x)

is superharmonic. Furthermore, ω(x

0

) = 0 and ω(x) > 0 for x ∈

D, x 6= x

0

. Thus,

the

function ω

(x) is a barrier for the point x

0

. 2

Conclusion. Assume that each point of Σ can be ”reached” by a ball. Then for each

ϕ(x) ∈C(Σ) the solution of the Dirichlet problem (4.4.1) exists, is unique and it is given

by (4.4.5).

Thus, we have established the solvability of the Dirichlet problem (4.4.1) for a wide

class of domains (for example, for all convex domains). We note that one can obtain weaker

sufﬁcient conditions of regularity of a point. For example, if the point x

0

∈ Σ can be

”reached” by a non-degenerate cyclic cone, then the point x

0

is regular [4, Chapter 3].

4.5. The Dirichlet Problem for the Poisson Equation

Let D ⊂ R

3

be a bounded domain with the boundary Σ ∈ PC

1

. We consider the following

problem

∆u = f (x) (x ∈ D), (4.5.1)

u

|Σ

= ϕ(x). (4.5.2)

Suppose that we know a particular solution v(x) of equation (4.5.1). Then the solution of

problem (4.5.1)-(4.5.2) has the form u(x) = v(x) + w(x), where w(x) is the solution of the

Dirichlet problem

∆w = 0 (x ∈ D),

w

|Σ

= ϕ

1

,

)

(4.5.3)

where

ϕ

1

:= ϕ −v

|Σ

.

The Dirichlet problem (4.5.3) for the Laplace equation was studied in Sections 4.3-4.4.

Thus, in order to solve the Dirichlet problem (4.5.1)-(4.5.2) for the Poisson equation it is

sufﬁcient to ﬁnd a particular solution of equation (4.5.1). We consider the function

u(x) = −

1

4π

D

f (ξ)

r

dξ, r = kx −ξk. (4.5.4)

Elliptic P

artial Differential Equations 193

Let us

show that ∆u = f , i.e (4.5.4) gives us the desired particular solution of (4.5.1).

The integral in (4.5.4) is called the volume potential, since it describes the potential of

gravitational ﬁeld created by the body D with the density f (ξ). Denote as usual by K

δ

(x)

the ball of radius δ around the point x , S

δ

(x) = ∂K

δ

(x) is the sphere. We have

K

δ

(x)

dξ

r

=

δ

0

d

r

S

r

(x)

d

s =

δ

0

4

πr dr = 2πδ

2

,

K

δ

(x)

dξ

r

2

=

δ

0

d

r

2

S

r

(x)

d

s = 4

πδ.











(4.5.5)

Theorem 4.5.1. Let f (x) ∈C

1

(

D). Then

the

function u(x), deﬁned by (4.5.4) , is a

particular solution of the Poisson equation (4.5.1) .

Proof. 1) Since |f (ξ)|≤ M in

D, it

follo

ws from (4.5.4)-(4.5.5) that for any x ∈

D,

|u(x)|

≤

M

4π

K

δ

(x)

dξ

r

+

M

4πδ

D\K

δ

(x)

dξ ≤

Mδ

2

+

M

V

4πδ

,

where V is

the

volume of D. Therefore, the integral in (4.5.4) converges absolutely and

uniformly in

D, and

the

function u(x) exists and is bounded in

D.

Let

us

show that u(x) ∈C(

D). Let y ∈K

δ/2

(x). By

virtue

of (4.5.4),

|u(y) −u(x)| ≤

M

4π

K

δ/2

(x)

¯

1

r

1

−

1

r

¯

dξ +

M

4π

D\K

δ/2

(x)

¯

1

r

1

−

1

r

¯

dξ

=: I

δ

+ J

δ

,

where r = kx −ξk, r

1

= ky −ξk. Using

(4.5.5)

we calculate

I

δ

≤

M

4π

K

δ/2

(x)

dξ

r

1

+

M

4π

K

δ/2

(x)

dξ

r

≤

M

4π

K

δ

(y)

dξ

r

1

+

M

4π

K

δ/2

(x)

dξ

r

≤

3Mδ

2

4

.

T

ak

e ε > 0. Let δ =

p

2ε/(3M). Then I

δ

≤ ε/2. In

the

domain D \K

δ

(x) the function

1/r is continuous, and consequently, there exists δ

1

(δ

1

< δ/2) such that

¯

1

r

1

−

1

r

¯

≤

2πε

M

V

for

all y ∈K

δ

1

(x); hence J

δ

≤ ε/2. Thus,

∀ε > 0 ∃δ

1

> 0 y ∈ K

δ

1

(x) → |u(y) −u(x)| ≤ ε.

By virtue of the arbitrariness of x ∈

D we

conclude

that u(x) ∈C(

D).

2)

Denote

v

k

(x) :

=

∂u

∂x

k

, k =

1,3.

194 G.

Freiling and V. Yurko

Diff

erentiating (4.5.4) formally, we obtain

v

k

(x) = −

1

4π

D

f (ξ)

∂

∂x

k

µ

1

r

¶

dξ, x ∈

D, r = kx −ξk. (4.5.6)

Since

∂

∂x

k

µ

1

r

¶

=

ξ

k

−x

k

r

3

,

|ξ

k

−x

k

|

r

≤ 1, |f (ξ)|

≤ M,

it

follows from (4.5.6) and (4.5.5) that

|v

k

(x)| ≤

M

4π

K

δ

(x)

dξ

r

2

+

M

4πδ

2

D\K

δ

(x)

dξ ≤ Mδ +

M

V

4πδ

2

.

Therefore,

the

integral in (4.5.6) converges absolutely and uniformly in

D, and

the

func-

tions v

k

(x) =

∂u

∂x

k

e

xist

and are ﬁnite in

D.

Let

us

show that v

k

(x) ∈C(

D) , k = 1,3. Let y ∈K

δ/2

(x). By

virtue

of (4.5.6),

|v

k

(y) −v

k

(x)| ≤

M

4π

K

δ/2

(x)

¯

ξ

k

−y

k

r

3

1

−

ξ

k

−x

k

r

3

¯

dξ

+

M

4π

D\K

δ/2

(x)

¯

ξ

k

−y

k

r

3

1

−

ξ

k

−x

k

r

3

¯

dξ

= I

1

δ

+ J

1

δ

,

where r = kx −ξk, r

1

= ky −ξk. Using

(4.5.5)

we calculate

I

1

δ

≤

M

4π

K

δ/2

(x)

dξ

r

2

1

+

M

4π

K

δ/2

(x)

dξ

r

2

≤

M

4π

K

δ

(y)

dξ

r

2

1

+

M

4π

K

δ/2

(x)

dξ

r

2

≤

3Mδ

2

.

T

ak

e ε > 0. Let δ = ε/(3M). Then I

1

δ

≤ ε/2. In the domain D \K

δ/2

(x) the function

∂

∂x

k

(

1

r

) is

continuous, and consequently, there exists δ

1

(δ

1

< δ/2) such that J

1

δ

≤ ε/2 for

y ∈ K

δ

1

(x). Thus,

∀ε > 0 ∃δ

1

> 0 y ∈ K

δ

1

(x) → |v

k

(y) −v

k

(x)| ≤ ε.

Since x ∈

D is

arbitrary

, we conclude that v

k

(x) ∈C(

D), i.e. u(x) ∈C

1

(D).

3)

Denote

w

k

(x) :=

∂

2

u

∂x

2

k

, k =

1,3.

Dif

f

erentiating (4.5.4) formally, we obtain

w

k

(x) = −

1

4π

D

f (ξ)

∂

2

∂x

2

k

µ

1

r

¶

dξ, x ∈

D. (4.5.7)

Elliptic P

artial Differential Equations 195

Denote

f

k

(ξ) =

∂ f (

ξ)

∂ξ

k

.

Since

∂

∂ξ

k

µ

1

r

¶

= −

∂

∂x

k

µ

1

r

¶

,

then

using

the Gauß-Ostrogradskii formula we calculate

Q

δ,k

:= −

1

4π

K

δ

(x)

f (ξ)

∂

2

∂x

2

k

µ

1

r

¶

dξ =

1

4π

K

δ

(x)

f (ξ)

∂

∂ξ

k

µ

∂

∂x

k

µ

1

r

¶¶

dξ

=

1

4π

K

δ

(x)

∂

∂ξ

k

µ

f (ξ)

∂

∂x

k

µ

1

r

¶¶

dξ −

1

4π

K

δ

(x)

f

k

(ξ)

∂

∂x

k

µ

1

r

¶

dξ

=

1

4π

S

δ

(x)

f (ξ)

∂

∂x

k

µ

1

r

¶

cos(n,ξ

k

)d

s −

1

4π

K

δ

(x)

f

k

(ξ)

∂

∂x

k

µ

1

r

¶

dξ.

Since

∂

∂x

k

µ

1

r

¶

=

ξ

k

−x

k

r

3

,

ξ

k

−x

k

r

= cos(n, ξ

k

),

we

ha

ve

Q

δ,k

:=

1

4πδ

2

S

δ

(x)

f (ξ) cos

2

(n,ξ

k

)d

s

−

1

4π

K

δ

(x)

f

k

(ξ)

ξ

k

−x

k

r

3

dξ. (4.5.8)

There

e

xists a constant M > 0 such that |f (ξ)|≤ M, |f

k

(ξ)|≤M. Then, in view of (4.5.5),

we have

|Q

δ,k

| ≤

M

4πδ

2

S

δ

(x)

d

s +

M

4π

K

δ

(x)

dξ

r

2

≤ (1 + δ)M,

and

consequently

,

|w

k

(x)| ≤ (1 + δ)M +

M

4π

D\K

δ

(x)

¯

∂

2

∂x

2

k

µ

1

r

¶

¯

dξ ≤C

δ

.

Thus,

the

integral in (4.5.7) converges absolutely and uniformly in

D, and

the

functions

w

k

(x) :=

∂

2

u

∂x

2

k

e

xist

and are bounded in

D. Using

(4.5.7)

and (4.5.8) we calculate

∆u(x) :=

3

∑

k=1

∂

2

u

∂x

2

k

=

1

4πδ

2

S

δ

(x)

f (ξ) d

s −

1

4π

3

∑

k=1

K

δ

(x)

f

k

(ξ)

ξ

k

−x

k

r

3

dξ

= f (x)

+ Ω

δ,1

+ Ω

δ

,2

,

where

Ω

δ,1

:=

1

4πδ

2

S

δ

(x)

( f (ξ) − f (x))d

s

,

196 G.

Freiling and V. Yurko

Ω

δ,2

:= −

1

4π

3

∑

k=1

K

δ

(x)

f

k

(ξ)

ξ

k

−x

k

r

3

dξ.

By

virtue

of (4.5.5),

|Ω

δ,2

| ≤

3M

4π

K

δ

(x)

dξ

r

2

≤ 3Mδ.

T

ak

e ε > 0. Since f (ξ) ∈C(

D), there

e

xists δ > 0 such that |f (ξ) − f (x)| ≤ ε for ξ ∈

S

δ

(x). Then |Ω

δ,1

| ≤ ε. Thus, |∆u(x) − f (x)| ≤ 3Mδ + ε, and consequently, ∆u = f .

Theorem 4.5.1 is proved. 2

4.6. The Method of Integral Equations

1. Let for deﬁniteness n = 3, and let D ⊂ R

3

be a bounded domain with the boundary

Σ ∈ PC

1

, D

1

= R

3

\

D. W

e

consider the functions

Q(x) =

Σ

q(ξ)

r

d

s, F(x)

=

Σ

f (ξ)

∂

ξ

(

1

r

)

∂n

ξ

d

s

, (4.6.1)

where q(ξ), f (ξ) ∈C(Σ), r = kξ−xk and n

ξ

is the outer normal to Σ at the point ξ. The

function Q(x) is called the single-layer potential with the density q(ξ). The function F(x)

is called the double-layer potential with the density f (ξ).

Physical sense: Q(x) is the potential of the ﬁeld created by charges distributed on Σ

with the density q(ξ), and F(x) is the potential of the ﬁeld created by the dipole distribu-

tion on Σ with the density f (ξ).

Idea of the method: The basic integral formula for harmonic functions (see Section 4.1)

contains terms of the form (4.6.1). We will seek solutions of the Dirichlet and Neumann

problems in the form (4.6.1). As a result we will obtain some integral equations for ﬁnding

q and f . Beforehand we will study properties of the functions deﬁned by (4.6.1).

2. Auxiliary assertions

Everywhere in Section 4.6 we will assume that Σ ∈ C

2

, i.e. Σ has a continuous and

bounded curvature. This means that for each ﬁxed x ∈ Σ one can choose a local coordinate

system (α

1

,α

2

,α

3

) such that the origin coincides with the point x, the plane (α

1

,α

2

) co-

incides with the tangent plane to Σ at the point x, and the axis α

3

coincides with the outer

normal ¯n

x

to Σ at the point x (see ﬁg. 4.6.1). Moreover, there exists δ

0

> 0 (independent

of x ) such that the part of the surface Σ

δ

0

(x) := Σ ∩K

δ

0

(x) is represented by a single-

valued function α

3

= Φ(α

1

,α

2

) , Φ ∈C

2

and

¯

∂

2

Φ

∂α

k

∂α

j

¯

≤ Φ

0

for ρ :=

q

α

2

1

+ α

2

≤ δ

0

,

where Φ

0

does

not

depend on x (it is the maximal curvature). In the sequel, we assume

that δ

0

< 1/(8Φ

0

). Since Φ(0,0) = Φ

α

1

(0,0) = Φ

α

2

(0,0) = 0, we have by virtue of the

Taylor formula:

|Φ(α

1

,α

2

)| ≤ Φ

0

ρ

2

, |Φ

α

k

(α

1

,α

2

)| ≤ 2Φ

0

ρ for ρ :=

q

α

2

1

+ α

2

≤ δ

0

. (4.6.2)

Elliptic P

artial Differential Equations 197

6

-

6

α

3

n

x

(α

1

,α

2

)

r

1

ρ

yΣ

α

3

= Φ(α

1

,α

2

)

r

ξ

n

ξ

Á

Figure

4.6.1.

Lemma

4.6.1. Let x ∈ Σ, ξ ∈Σ

δ

0

, r = kx −ξk. Then

|sin(n

x

,n

ξ

)| ≤ 4Φ

0

r, |cos(n

x

,n

ξ

)| ≥ 1/2. (4.6.3)

Proof. Let ¯r = (α

1

,α

2

,Φ(α

1

,α

2

)) be the local coordinates of the point ξ. Then, using

(4.6.2), we calculate

|sin(n

x

,n

ξ

)| ≤ |tg(n

x

,n

ξ

)| =

¯

∂Φ(α

1

,α

2

)

∂¯r

¯

|Φ

α

1

(α

1

,α

2

)cos(α

1

, ¯r)

+ Φ

α

2

(α

1

,α

2

)cos

(α

2

, ¯r)|

≤ |Φ

α

1

(α

1

,α

2

)|+ |Φ

α

2

(α

1

,α

2

)| ≤ 4Φ

0

ρ.

Since

r =

q

α

2

1

+ α

2

+

(Φ

(α

1

,α

2

))

2

≥

q

α

2

1

+ α

2

:= ρ,

we

obtain

|sin

(n

x

,n

ξ

)| ≤ 4Φ

0

r.

From 4Φ

0

r ≤ 4Φ

0

δ

0

≤ 1/2, we infer |cos(n

x

,n

ξ

)| ≥ 1/2. 2

Lemma 4.6.2. Let x ∈ Σ, ξ ∈Σ

δ

0

, r = kx −ξk. Then

¯

∂

x

r

∂n

x

¯

≤ Φ

0

r,

¯

∂

ξ

r

∂n

ξ

¯

≤ Φ

0

r, (4.6.4)

¯

∂

x

(

1

r

)

∂n

x

¯

≤

Φ

0

r

,

¯

∂

ξ

(

1

r

)

∂n

ξ

¯

≤

Φ

0

r

. (4.6.5)

Pr

oof

. By Lemma 4.3.1,

∂

ξ

r

∂n

ξ

= cos(n

ξ

, ¯r),

∂

x

r

∂n

x

= −cos(n

x

, ¯r),

where

¯r =

(ξ

1

−x

1

,ξ

2

−x

2

,ξ

3

−x

3

)

198 G.

Freiling and V. Yurko

. Let (α

1

,α

2

,Φ

(α

1

,α

2

)) be the local coordinates of the point ξ. Then

|cos(n

x

, ¯r)| =

|Φ(α

1

,α

2

)|

r

(see

ﬁg.

4.6.1). Using (4.6.2) and the inequality ρ ≤r, we obtain

¯

∂

x

r

∂n

x

¯

= |cos(n

x

, ¯r)| =

|Φ(α

1

,α

2

)|

r

≤

Φ

0

ρ

2

r

≤ Φ

0

r.

By

symmetry

, the second formula in (4.6.4) follows from the ﬁrst one. Formulae (4.6.5) are

evident corollaries of (4.6.4). 2

Lemma 4.6.3. Let x ∈ Σ, δ ≤ δ

0

. Then

Σ

δ

(x)

ds

r

≤ 4πδ, r := kx −ξk, Σ

δ

(x) := Σ ∩K

δ

(x), ξ ∈Σ

δ

(x). (4.6.6)

Mor

eo

ver, if y ∈K

δ/2

(x), ξ ∈Σ

δ/2

(x), r

1

:= ky −ξk, then

Σ

δ/2

(x)

ds

r

1

≤ 4πδ.

Pr

oof

. Let (α

1

,α

2

,Φ(α

1

,α

2

)) be the local coordinates of the point ξ and ρ :=

q

α

2

1

+ α

2

. Denote

by σ

δ

= {

(α

1

,α

2

) : ρ ≤ δ} the disc of radius δ. Using (4.6.3) and the

inequality ρ ≤ r, we obtain

Σ

δ

(x)

ds

r

≤

Σ

δ

(x)

d

s

ρ

=

σ

δ

dα

1

dα

2

ρcos(n

x

,n

ξ

)

≤ 2

σ

δ

dα

1

dα

2

ρ

= 2

δ

0

dρ

ρ

2πρ = 4πδ.

Furthermore,

let (β

1

,β

2

,β

3

) be

the local coordinates of the point y. Then

r

1

=

q

(α

1

−β

1

)

2

+

(α

2

−β

2

)

2

+

(Φ(α

1

,α

2

) −β

3

)

2

≥

q

(α

1

−β

1

)

2

+

(α

2

−β

2

)

2

:

=

˜

ρ,

and consequently,

Σ

δ/2

(x)

ds

r

1

≤

Σ

δ/2

(x)

d

s

˜

ρ

≤ 2

σ

δ/2

dα

1

dα

2

˜

ρ

≤ 2

σ

δ

(β)

dα

1

dα

2

˜

ρ

≤ 2

σ

δ

dα

1

dα

2

ρ

= 4πδ,

where σ

δ

(β)

= {

(α

1

,α

2

) :

˜

ρ ≤ δ)} is the disc of radius δ around the point (β

1

,β

2

).

Lemma 4.6.3 is proved. 2

Elliptic P

artial Differential Equations 199

Lemma 4.6.4. Let

x ∈ Σ, δ ≤ δ

0

, ξ ∈Σ

δ

(x). Then

Σ

δ

(x)

¯

∂

ξ

(

1

r

)

∂n

ξ

¯

d

s ≤ 4

πΦ

0

δ,

Σ

δ

(x)

¯

∂

x

(

1

r

)

∂n

x

¯

d

s ≤ 4

πΦ

0

δ, r := kx −ξk. (4.6.7)

Formulae (4.6.7) are obvious consequences of the formulae (4.6.5) and (4.6.6).

Lemma 4.6.5. Let x ∈ Σ, δ ≤δ

0

, y ∈K

δ

(x), y = x + αn

x

, α ∈R, r

1

= ky −ξk. Then

Σ

δ

(x)

¯

∂

ξ

(

1

r

1

)

∂n

ξ

+

∂

y

(

1

r

1

)

∂n

x

¯

d

s ≤ 64

πΦ

0

δ. (4.6.8)

Proof. Let ξ ∈ Σ

δ

(x). Then, by Lemma 4.3.1,

∂

ξ

(r

1

)

∂n

ξ

+

∂y(r

1

)

∂n

x

= cos(n

ξ

, ¯r

1

) −cos(n

x

, ¯r

1

),

where

¯r

1

=

(ξ

1

−y

1

,ξ

2

−y

2

,ξ

3

−y

3

).

T

ogether with Lemma 4.6.1 this yields

¯

∂

ξ

(r

1

)

∂n

ξ

+

∂y(r

1

)

∂n

x

¯

≤ 2|sin(n

x

,n

ξ

)|

≤ 8

Φ

0

r, where r = kx −ξk. (4.6.9)

Let (α

1

,α

2

,Φ(α

1

,α

2

)) be the local coordinates of the point ξ. Since y = x + αn

x

, we

have ρ ≤r

1

(see ﬁg. 4.6.1). Using (4.6.2) we calculate

r =

q

α

2

1

+ α

2

+

(Φ

(α

1

,α

2

))

2

≤

q

ρ

2

+ Φ

2

0

ρ

4

≤ ρ

q

1 + Φ

2

0

δ

2

0

≤ 2ρ ≤ 2r

1

.

T

ogether

with (4.6.9) this yields

¯

∂

ξ

(r

1

)

∂n

ξ

+

∂y(r

1

)

∂n

x

¯

≤ 16Φ

0

r

1

.

Therefore

Σ

δ

(x)

¯

∂

ξ

(

1

r

1

)

∂n

ξ

+

∂

y

(

1

r

1

)

∂n

x

¯

d

s ≤ 16

Φ

0

Σ

δ

(x)

ds

r

1

≤ 16Φ

0

Σ

δ

(x)

d

s

ρ

= 16Φ

0

σ

δ

dα

1

dα

2

ρcos(n

x

,n

ξ

)

≤ 32Φ

0

σ

δ

dα

1

dα

2

ρ

= 64πΦ

0

δ,

i.e.

(4.6.8)

is valid. Lemma 4.6.5 is proved. 2

Lemma 4.6.6. Let x ∈ Σ, δ ≤δ

0

, y ∈K

δ

(x), y /∈ Σ. Then

Σ

δ

(x)

¯

∂

ξ

(

1

r

1

)

∂n

ξ

¯

d

s ≤C

∗

, r

1

:

= ky −ξk, ξ ∈Σ

δ

(x), (4.6.10)

200 G.

Freiling and V. Yurko

where

C

∗

depends on Σ and does not depend on x,y,δ.

Proof. Let Ω be a simply connected part of Σ

δ

, on which the function

∂

ξ

(

1

r

1

)

∂n

ξ

preserv

es

the sign. Let B be the cone with vertex at the point y and with the basis Ω. Fix

ε > 0 and consider the domain B

ε

= {ξ ∈ B : r

1

≥ ε} (see ﬁg. 4.6.2).

6

-

x

y

|

{z }

ε

B

ε

n

ξ

Ω

n

ξ

?

9

µ

Figure

4.6.2.

Clearly

, ∂B

ε

= Ω∪Ω

0

∪S

0

ε

(y), where Ω

0

is the lateral surface of B

ε

, S

0

ε

(y) = S

ε

(y)∩B

is the part of the sphere r

1

= ε, lying in B. In B

ε

the function 1/r

1

is harmonic with

respect to ξ. Hence, by Theorem 4.1.2,

µ

Ω

+

Ω

0

+

S

0

ε

(y)

¶

∂

ξ

(

1

r

1

)

∂n

ξ

d

s = 0

.

Since n

ξ

⊥ ¯r

1

on Ω

0

, we have

Ω

0

∂

ξ

(

1

r

1

)

∂n

ξ

d

s = −

Ω

0

cos

(n

ξ

, ¯r

1

)

r

2

1

d

s = 0

.

On S

0

ε

(y) we have

∂

ξ

(

1

r

1

)

∂n

ξ

= −

∂(

1

r

1

)

∂r

1

=

1

r

2

1

, r

1

= ε,

and

consequently

,

S

0

ε

(y)

∂

ξ

(

1

r

1

)

∂n

ξ

d

s =

1

ε

2

S

0

ε

(y)

d

s = ω

y

(Ω)

is

the solid angle under which one can see Ω from the point y. Thus,

Ω

¯

∂

ξ

(

1

r

1

)

∂n

ξ

¯

d

s = |

ω

y

(Ω)|.

Let now Σ

δ

(x) = Σ

+

δ

(x) ∪Σ

−

δ

(x), and

±

∂

ξ

(

1

r

1

)

∂n

ξ

≥ 0

Elliptic P

artial Differential Equations 201

on Σ

±

δ

(x). Then it

can be shown that

Σ

δ

(x)

¯

∂

ξ

(

1

r

1

)

∂n

ξ

¯

d

s ≤

Σ

+

δ

(x)

¯

∂

ξ

(

1

r

1

)

∂n

ξ

¯

d

s +

Σ

−

δ

(x)

¯

∂

ξ

(

1

r

1

)

∂n

ξ

¯

d

s ≤C

∗

.

Lemma

4.6.6 is proved. 2

Lemma 4.6.7. The following relations are valid

Σ

∂

ξ

(

1

r

)

∂n

ξ

d

s =







−4

π, x ∈ D,

−2π, x ∈ Σ,

0, x ∈D

1

,

r := kx −ξk, D

1

:= R

3

\

D. (4.6.11)

Pr

oof

. 1) Let x ∈ D

1

. Then the function 1/r is harmonic in

D (with

respect

to ξ ),

and consequently, by Theorem 4.1.2,

Σ

∂

ξ

(

1

r

)

∂n

ξ

d

s = 0

.

2) Let x ∈ D. In the domain D \K

δ

(x) the function 1/r is harmonic (with respect to

ξ ), and consequently, by Theorem 4.1.2,

Σ

∂

ξ

(

1

r

)

∂n

ξ

d

s +

S

δ

(x)

∂

ξ

(

1

r

)

∂n

ξ

d

s = 0

.

On S

δ

(x) :

∂

ξ

(

1

r

)

∂n

ξ

= −

∂(

1

r

)

∂r

=

1

r

2

, r = δ,

hence

S

δ

(x)

∂

ξ

(

1

r

)

∂n

ξ

d

s =

S

δ

(x)

d

s

r

2

d

s =

1

δ

2

S

δ

(x)

d

s = 4

π,

and (4.6.11) is proved.

3) Let x ∈ Σ. In the domain D \K

δ

(x) the function 1/r is harmonic with respect to ξ,

and consequently, by Theorem 4.1.2,

Σ\Σ

δ

(x)

∂

ξ

(

1

r

)

∂n

ξ

d

s +

S

0

δ

(

x)

∂

ξ

(

1

r

)

∂n

ξ

d

s = 0

,

where S

0

δ

(x) = S

δ

(x) ∩

D. Since

S

0

δ

(x)

∂

ξ

(

1

r

)

∂n

ξ

d

s =

S

0

δ

(x)

d

s

r

2

=

1

δ

2

S

0

δ

(x)

d

s → 2

π for δ → 0,

we have, in view of (4.6.7),

Σ

δ

(x)

∂

ξ

(

1

r

)

∂n

ξ

d

s → 0

for δ → 0,

and therefore we arrive at (4.6.11). Lemma 4.6.7 is proved. 2

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

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