Helms L.L. Potential Theory

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328 8 Newtonian Potential

(2 − n)σ

w(x



)









u(|(x



) − (y



)|)f



) dy













u(|(x



, −x

) − (y



, −y

)|)f



) dy













u(|(x



, −x

) − (y



)|)f



, −z

) d(−z

)













u(|(x



, −x

) − (y



)|)f



) dz





=(2− n)σ

w(x



, −x

The change from d(−z

)todz

in going from the fourth line to the ﬁfth

is justiﬁed since Γ



is replaced by Γ



. It follows that w is symmetric with

respect to R

. By the argument used above to show that D

H(h



)(x



, 0) =

0,D

w =0onΣ.

The sets Ω and Σ of the following lemma need not satisfy the standing

assumption stated at the beginning of this section.

Lemma 8.5.4 (Boundary Point Lemma [33]) If Ω is a bounded open

subset of R

,Σ is a relatively open subset of ∂Ω, u ∈ C

(Ω ∪ Σ) ∩ C

(Ω)

with Δu ≥ 0 on Ω,x

∈ Σ, u(x

) >u(x) for all x ∈ Ω,andthereisaball

B ⊂ Ω internally tangent to Σ at x

,thenD

u(x

) > 0.

Proof: Let B = B

y,ρ

,x ∈ B

y,ρ

,r = |x − y|,v(x)=e

−kr

− e

−kρ

,where

k is a constant to be chosen, and 0 <δ<ρ.Thenv =0on∂B

y,ρ

,Δv =

(4k

− 2kn)e

−kr

,andk can be chosen so that Δv ≥ 0ontheannulus

ρ,δ

= B

y,ρ

∼ B

y,δ

.Sinceu(x)−u(x

) < 0on∂B

y,δ

and v = e

−kδ

−e

−kρ

∂B

y,δ

, there is a constant >0 such that u(x)−u(x

)+v ≤ 0on∂B

y,δ

.This

inequality also holds on ∂B

y,ρ

where v =0.SinceΔ(u−u(x

)+v) ≥ Δv ≥ 0

on A

ρ,δ

and u − u(x

)+v ≤ 0on∂A

ρ,δ

,u− u(x

)+v ≤ 0onA

ρ,δ

by the

maximum principle. Thus,

u(x) − u(x

)

|x − x

≤−

v(x) − v(x

)

|x − x

,x∈ A

ρ,δ

As x → x

in A

ρ,δ

normal to Σ, the limit of the left term is the inner normal

derivative of u at x

. Thus, the outer normal derivative of u at x

u(x

) ≥−D

v(x

)=−D

r=ρ

=2kρe

−kρ

> 0.

Lemma 8.5.5 If f ∈ H

0+α

(Ω),α∈ (0, 1],andg ∈ C

(∂Ω ∼ Σ), then there

is a unique u ∈ C

(Ω

−

)∩C

(Ω∪Σ)∩C

(Ω) such that Δu = f on Ω, D

u =0

on Σ,andu = g on ∂Ω ∼ Σ.

8.5 The Reﬂection Principle 329

Proof: By Lemma 8.5.3, there is a w ∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ) ∩ C

(Ω)such

that Δw = f on Ω and D

w =0onΣ. Extending f to be 0 on R

∼ Ω,the

function w constructed by way of Lemma 8.5.2 can be regarded as an element

of C

∪ R

) by Theorem 8.2.5. In particular, w|

∂Ω

∈ C

(∂Ω ∼ Σ). By

Theorem 8.5.1, there is a v ∈ C

(Ω ∪ Σ) ∩ C

(Ω) such that Δv =0on

Ω,D

v =0onΣ,and

lim

y→x,y∈Ω

v(y)=g(x) − w|

∂Ω

(x)

for all x ∈ ∂Ω ∼ Σ.Sincev ∈ C

(Ω ∪Σ), lim

y→x,y∈Ω

v(y)existsinR for all

x ∈ Σ. As noted in Section 0.1, the function v has a continuous extension on

−

. Letting u = w + v, Δu = Δw + Δv = f on Ω, D

u = D

w + D

v =0

on Σ,and

lim

y→x,y∈Ω

u(y) = lim

y→x,y∈Ω

w(y) + lim

y→x,y∈Ω

v(y)

= w|

∂Ω

(x)+g(x) − w|

∂Ω

(x)=g(x)

for all x ∈ ∂Ω ∼ Σ.Sincew is continuous on Ω

−

and v has a continuous

extension on Ω

−

,u∈ C

(Ω

−

). To prove uniqueness, assume that both w

and

satisfy the conclusions of the lemma. The diﬀerence w = w

−w

satisﬁes

Δw =0onΩ, D

w =0onΣ,andw =0on∂Ω ∼ Σ.Sincew ∈ C

(Ω

−

), it

attains it maximum value at some point x

∈ Ω

−

.Ifx

is an interior point of

Ω,thenw is constant by the maximum principle for harmonic functions, and

therefore w =0onΩ

−

because w =0on∂Ω ∼ Σ.Ifw does not attain its

maximum value at an interior point of Ω,thenx

∈ ∂Ω and w(x) <w(x

)

for all x ∈ Ω. By the preceding lemma, x

∈ Σ because D

w(x)=0for

x ∈ Σ.Inthiscase,x

∈ ∂Ω ∼ Σ and w(x) ≤ w(x

) = 0 for all x ∈ Ω

−

Since −w has the same properties as w, w ≥ 0onΩ

−

and so w = w

−w

on Ω

−

A standard method for passing from the homogeneous boundary condition

on Σ to the nonhomogeneous condition is to incorporate the boundary func-

tion into the function f. In addition to the standing assumptions regarding

the spherical chip, another condition will be imposed in the following lem-

mas. Suppose Ω = B

= B ∩R

where B is ball with center at (0, −ρ

)and

of radius ρ where 0 <ρ

<ρ. If the condition ρ

>ρ/

√

2 is imposed and

x =(x



) ∈ Ω, then the ball in R

with center x



of radius x

will lie in

Σ. This can be seen by considering the intersection of Ω with hyperplanes

containing the x

-axis. If g : Σ → R is a boundary function, recall that

|g|

1+α,Σ

=inf{|u|

1+α,Ω

; u ∈ H

1+α

(Ω),u|

= g}.

Lemma 8.5.6 If Ω = B

where B is a ball with center at (0, −ρ

) and of

radius ρ, ρ

>ρ/

√

2,α ∈ (0, 1],andg ∈ H

1+α

(Σ), then there is a function

ψ ∈ H

2+α

(Ω) satisfying ψ =0on Σ, D

ψ = g on Σ,|ψ|

2+α,Ω

≤ C|g|

1+α,Σ

and Δψ ∈ H

0+α

(Ω).

330 8 Newtonian Potential

Proof: According to Theorem 7.2.6, it can be assumed that g ∈ H

1+α

(Ω).

Consider any η ∈ C

)withsupportinB

0,1

⊂ R

and



η(z) dz =1.

Let

ψ(x



)=x



g(x



− x

z,0)η(z) dz. (8.30)

Under the hypothesis that ρ

>ρ/

√

2, it can be seen from geometrical

arguments that the point x



− x

z ∈ Σ for all z ∈ B

0,1

⊂ R

.For

x =(x



) ∈ Ω,ψ(x



, 0) = 0,D

ψ(x



, 0) = g(x



, 0). Therefore, [ψ|

0,Ω

≤

[g]

0,Σ

≤ C|g|

1+α,Ω

.Fori = n, integrating by parts with respect to z

and

making use of the fact that η(z)=0forz outside the ball with center at x



of radius x

ψ(x



)=x



g(x



− x

z,0)η(z) dz

= −



g(x



− x

z,0)η(z) dz



g(x



− x

z,0)D

η(z) dz. (8.31)

For i, j = n,

ψ(x





g(x



− x

z,0)D

η(z) dz. (8.32)

Similarly,

ψ(x



)=x



n−1



j=1

g(x



− x

z,0)(−z

)η(z) dz



g(x



− x

z,0)η(z) dz

= x

n−1



j=1



g(x



− x

z,0)



−



(−z

η(z)) dz



g(x



− x

z,0)η(z) dz

= −

n−1



j=1



g(x



− x

z,0)D

η(z))dz



g(x



− x

z,0)η(z) dz

= −



g(x



− x

z,0)



∇η(z) · z +(n − 2)η(z)



dz. (8.33)

8.5 The Reﬂection Principle 331

By Equations (8.31) and (8.33), [ψ]

1+0,Ω

≤ C[g]

0,Σ

≤ C|g|

1+α,Ω

.Fori = n,

ψ(x



)=−



g(x



−x

z,0)



∇η(z)·z +(n−2)η(z)



dz. (8.34)

Similarly,

ψ(x







∇g(x



−x

z,0)·z



∇η(z)·z+(n−2)η(z)



dz. (8.35)

As before, [ψ]

2+0,Ω

≤ C[g]

1+0,Σ

≤ C|g|

1+α,Ω

= C|g|

1+α,Σ

.Since|(x



−x

z)−



− y

z)|≤|x − y|(1 + |z|), for i = n

ψ(x



) − D

ψ(y



|x − y|

≤



g(x



− x

z,0) − D

g(y



− y

z), 0|

|(x



− x

z) − (y



− y

z)|

(1 + |z|)

η(z)|dz

≤ C[g]

1+α,Σ

≤ C|g|

1+α,Ω

= C|g|

1+α,Σ

. (8.36)

Taking the supremum over x, x



∈ Ω,[ψ]

2+α,Ω

≤ C|g|

1+α,Σ

. Therefore,

|ψ|

2+α,Ω

≤ C|g|

1+α,Σ

Theorem 8.5.7 If Ω satisﬁes the conditions of the preceding lemma and

f ∈ H

0+α

(Ω),α ∈ (0, 1],g ∈ H

1+α

(Σ) and h ∈ C

(∂Ω ∼ Σ), then there is a

unique w ∈ C

(Ω

−

) ∩C

(Ω ∪Σ) ∩C

(Ω) satisfying Δw = f on Ω,D

w = g

on Σ,andw = h on ∂Ω ∼ Σ.

Proof: Consider the function ψ of the preceding lemma for which D

ψ = g

on Σ and Δψ ∈ H

0+α

(Ω) and the problem

(i) Δv = f −Δψ on Ω,

(ii) D

v =0onΣ,and

(iii) v = h − ψ on ∂Ω ∼ Σ.

Since ψ ∈ H

2+α

(Ω), it is uniformly continuous on Ω and has a continuous

extension to ∂Ω.Sincef − Δψ ∈ H

0+α

(Ω)andh − ψ ∈ C

(∂Ω ∼ Σ), by

Lemma 8.5.5 there is a v ∈ C

(Ω

−

) ∩C

(Ω ∪Σ) ∩C

(Ω) satisfying (i)-(iii).

Letting w = v + ψ, Δw = Δv + Δψ = f on Ω, D

w = D

v + D

ψ = g on

Σ,andw = v + ψ = h − ψ + ψ = h on ∂Ω ∼ Σ.Sinceψ ∈ H

2+α

(Ω) implies

that ψ ∈ C

(Ω

−

) ∩C

(Ω ∪Σ) ∩C

(Ω), w ∈ C

(Ω

−

) ∩C

(Ω ∪Σ) ∩C

(Ω).

Uniqueness of w follows as in the proof of Lemma 8.5.5.

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Chapter 9

Elliptic Operators

9.1 Introduction

Up to this point, all of the concepts pertain to the Dirichlet and Neumann

problems for the Laplacian. In this and later chapters, the Laplacian will

be replaced by an elliptic operator as described below and mixed boundary

conditions will be allowed that specify the value of a function on part of

the boundary and the value of its normal derivative on the remainder of the

boundary as in Theorem 8.5.7.

Let Ω be a bounded open subset of R

with a smooth boundary ∂Ω.A

precise smoothness requirement will be speciﬁed later. For i, j =1,...,n,

let a

,c be real-valued functions on Ω with a

= a

. In addition, for

i =1,...,n,letβ

,γ be real-valued functions on ∂Ω.Anelliptic operator

is an operator acting on real-valued functions u on Ω of the form

Lu(x)=



i,j=1

(x)

∂

∂x

(x)+



i=1

(x)

∂u

∂x

(x)+c(x)u(x),x∈ Ω,

with suitable conditions imposed on the coeﬃcients. In addition, a boundary

condition can be speciﬁed by an operator M deﬁned by the equation

Mu(x)=



i=1

(x)

∂u

∂x

(x)+γ(x)u(x)=0,x∈ ∂Ω.

In terms of the operators L and M, the Dirichlet problem is a special case of

the problem

Lu = f on Ω,u = g on ∂Ω,

and the Neumann problem is a special case of the problem

Lu = f on Ω,Mu = g on ∂Ω.

The latter problem is known as the oblique derivative problem.

L.L. Helms, Potential Theory, Universitext, 333

 Springer-Verlag London Limited 2009

334 9 Elliptic Operators

Each of the results of this chapter are stated under minimal conditions on

the coeﬃcients of L; all of these conditions are satisﬁed if

sup

1≤i,j≤n



a



0,Ω

+|a

; d|

(0)

α,Ω

+b



0,Ω

+|b

; d|

(1)

α,Ω

+c

0,Ω

+|c; d|

(2)

α,Ω



< +∞.

9.2 Linear Spaces

Results pertaining to the Dirichlet problem for the Laplacian operator can

be carried over to a general elliptic operator by the method of continuity.

This method requires the use of some theorems about mappings between

normed linear spaces.

AmapT from a Banach space B into a normed linear space X is bounded

if there is a constant C such that Tx

≤ Cx

. The smallest C for which

this is true is denoted by T ;thatis,T  =sup

x∈B,x=0

Tx

/x

AmapT : B → B is a contraction map if there is a constant λ<1such

that Tx − Ty

≤ λx − y

for all x, y ∈ B.ThemapT : B → X is a

linear map if T(ax + by)=aTx + bTy for all a, b ∈ R and all x, y ∈ B.

AmapT : B → B is a bounded linear operator or simply a bounded

operator on T if bounded and linear as deﬁned above.

Theorem 9.2.1 If T is a contraction map on the Banach space B,then

there is a unique x ∈ B such that Tx = x.

Proof: Let x

be an arbitrary point of B and let λ<1beaconstantsuch

that Tx≤λx for all x ∈ B. It is easy to see that T

x≤λ

x for all

j ≥ 1,x∈ B.Foreachj ≥ 1, let x

= T

.Ifj ≥ i,then

x

− x

≤



k=i+1

x

− x

k−1

 =



k=i+1

T

k−1

− T

k−1



≤



k=i+1

k−1

x

− x

≤

1 − λ

x

− x

→0asi →∞.

This shows that the sequence {x

} is Cauchy in B and therefore converges

to some x ∈ B.SinceT is a continuous map, Tx = lim

j→∞

lim

j→∞

j+1

= x.IfTy = y also, then x − y = Tx − Ty≤λx − y

with λ<1 and therefore x = y.

Theorem 9.2.2 (Method of Continuity) Let B be a Banach space, let X

be a normed linear space, and let T

, T

be bounded linear operators from B

into X.Foreacht ∈ [0, 1],letT

=(1−t)T

+ tT

. If there is a constant C,

not depending upon on t, such that

x

≤ CT

x

, for all t ∈ [0, 1],x∈ B, (9.1)

then T

maps B onto X if and only if T

maps B onto X.

9.3 Constant Coeﬃcients 335

Proof: Suppose T

maps B onto X for some s ∈ [0, 1]. By Inequality (9.1),

is one-to-one and therefore has an inverse T

−1

: X → B.Forﬁxedt ∈ [0, 1]

and y ∈ X,thereisanx ∈ B satisfying the equation T

x = y if and only if

x = y +(T

− T

)x = y +(t − s)T

− (t − s)T

This equation holds if and only if

x = T

−1

y +(t − s)T

−1

− T

)x;

that is, if and only if the map T : B → B deﬁned by

Tx = T

−1

y +(t − s)T

−1

− T

has a ﬁxed point. Since T

−1

y

≤ Cy

for all y ∈ X,

Tx

− Tx



= (t − s)T

−1

− T

)(x

− x

)

≤ C|t − s|T

− T

x

− x



≤ C|t − s|(T

 + T

) x

− x



so that T is a contraction operator on B if

|t − s| <δ=

C(T

 + T

)

By the preceding theorem, T has a ﬁxed point whenever |t − s| <δ;that

is, if |t − s| <δ,thenT

maps B onto X whenever T

maps B onto X.

By subdividing the interval [0, 1] into subintervals of length less than δ,the

mapping T

is onto for all t ∈ [0, 1] provided it is onto for some s ∈ [0, 1], in

particular if s =0ors =1.

9.3 Constant Coeﬃcients

It will be assumed in this section that α ∈ (0, 1) and that the coeﬃcients

,i,j =1,...,n, of the elliptic operator deﬁned in Section 9.1 are all con-

stants, that a

= a

,i,j =1,...,n, and that the b

,i =1,...n,andc are

all 0. In this case, the operator will be denoted by

u(x)=



i,j=1

u(x).

336 9 Elliptic Operators

It also will be assumed that there are positive constants m and M such

that

m|x|

≤



i,j=1

≤ M|x|

. (9.2)

This assumption implies that the matrix A =[a

] is positive deﬁnite.

The following facts from elementary matrix theory will be needed (c.f.,

Hohn [31]). There is an n ×n orthogonal matrix O =[o

] such that OAO

diag[λ

,...,λ

]whereλ

...,λ

are the eigenvalues of A and the superscript

t signiﬁes the transpose matrix. Regarding elements of R

as column vectors

for purposes of this discussion, if y = Ox,then

Ax =



i,j=1



i=1

It also is known from matrix theory that O preserves norms; that is, |Ox| =

|x|,x∈ R

. It follows that m =min(λ

,...,λ

)andM =max(λ

,...,λ

Now let

D =[λ

−

]=diag[1/



,...,1/



]

and deﬁne T : R

→ R

by the equation y = Tx= DOx.Since

Ax =



i=1

(Tx)

m|x|

≤



i=1

(Tx)

≤ M |x|

Thus,

|x|

≤



i=1

(Tx)

≤

|x|

and there is a constant c = c(m, M) such that

|x|≤|Tx|≤c|x|. (9.3)

If Ω is an open subset of R

,letΩ

∗

be the image of Ω under T . Recalling

that d(y)=dist(y, ∼ Ω), the d notation will be augmented by adding the

subscript Ω in order to deal with the two sets Ω and Ω

∗

simultaneously. It

is easily seen using the above inequality that

(x) ≤ d

∗

(y) ≤ cd

(x). (9.4)

Consider now any function u on Ω. The transformation T deﬁnes a function

u∗ on the image Ω

∗

of Ω by the equation u

∗

(y)=u(x)wherey = Tx.If

u ∈ C

(Ω)oru ∈ H

(Ω), then u

∗

∈ C

(Ω

∗

)oru

∗

∈ H

(Ω

∗

), respectively. If

u ∈ C

(Ω)andf ∈ H

(Ω), the equation L

u = f on Ω is transformed into

the equation

9.3 Constant Coeﬃcients 337

Δu

∗

(y)=f

∗

(y),y∈ Ω

∗

with u

∗

∈ C

(Ω

∗

)andf

∗

∈ H

(Ω

∗

). This result provides a means of solving

the equation L

u = f on the bounded open set Ω by solving the equation

Δu

∗

= f

∗

on Ω

∗

The norms |u; d

2+α,Ω

and |f ; d

(b)

2+α,Ω

are related to |u

∗

; d

∗

2+α,Ω

∗

and |f

∗

; d

∗

(b)

2+α,Ω

∗

, respectively, as in

|u; d

2+α,Ω

≤|u

∗

; d

∗

2+α,Ω

∗

≤ c|u; d

2+α,Ω

, (9.5)

|f; d

(b)

α,Ω

≤|f

∗

; d

∗

(b)

α,Ω

∗

≤ c|f; d

(b)

α,Ω

(9.6)

where c = C(n, m, M ). These inequalities follow from Inequalities (9.3) and

(9.4). Just one of the inequalities will be proven as an illustration; namely,

|u; d

2+α,Ω

≤ c|u

∗

; d

∗

2+α,Ω

∗

. Clearly, u

0,Ω

= u

∗



0.Ω

∗

. Suppose y =

Tx.Theny



j=1

(1/

√

so that ∂y

/∂x

=(1/

√

.Foreachj,

∂u

∂x



i=1

∂u

∗

∂y

∂x



i=1

∂u

∗

∂y

√

and therefore

(x)D

u(x)|≤c



i=1

∗

(y)D

∗

(y)|

√

≤ c[u

∗

; d

∗

]

1+0,Ω

∗





i=1

√



. (9.7)

With j ﬁxed, let z

=1ifo

≥ 0andz

= −1ifo

< 0. Then |z| =

√

n and

(Tz)



i=1

√



i=1

√

Thus,



i=1

√

| =(Tz)

≤|Tz|≤c|z| = c

√

n (9.8)

and it follows that

[u; d

]

1+0,Ω

≤ C(n, m, M )[u

∗

; d

∗

]

1+0,Ω

∗

Similarly, [u; d

]

2+0,Ω

≤ C(n, m, M )[u

∗

; d

∗

]

2+0,Ω

∗

and

[u; d

]

2+α,Ω

≤ C(n, m, M )[u

∗

; d

∗

]

2+α,Ω

∗

Therefore,

|u; d

2+α,Ω



j=0

[u; d

]

j+0,Ω

+[u; d

]

2+α,Ω

≤ C(n, m, M )|u

∗

; d

∗

2+α,Ω

∗