Helms L.L. Potential Theory

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318 8 Newtonian Potential

v(x)=



u(x, y))(f(y) − f(x)) dy

−f(x)



∂Ω

u(x, y)γ

(y) dσ(y),x∈ Ω,

by the same theorem. Taking i = j and summing over i,

Δv(x)=



(

(x)

u(|x − y|))(f(y) − f (x)) dy

−f(x)



i=1



∂Ω

u(|x − y|)γ

(y) dσ(y).

Since Δ

(x)

u(|x − y|)=0fory = x,

Δv(x)=−f(x)



i=1



∂Ω

u(|x − y|)γ

(y) dσ(y).

Now let Ω

= B

x,ρ

,whereρ is chosen so that Ω

−

⊂ B

x,ρ

. By Equation (8.16),

Δv(x)=−f(x)



i=1



∂B

x,ρ

−(n − 2)

− y

)

|x − y|

(y) dσ(y)

f(x)(n − 2)



i=1



∂B

x,ρ

− y

)γ

(y) dσ(y).

By Equation (8.8),

Δv(x)=

f(x)(n − 2)



i=1



x,ρ

− y

) dy

= −

f(x)n(n − 2)



|θ|=1

n−1

dθ dr

=(2− n)σ

f(x).

The function ˆv =

(2−n)σ

v therefore satisﬁes Poisson’s equation Δˆv = f

on Ω and belongs to C

(Ω

−

) by Theorem 8.2.5. Consider the continuous

boundary function h = g − ˆv|

∂Ω

and the corresponding harmonic function

.Ifw =ˆv + H

,thenw ∈ C

(Ω),Δw = Δˆv + ΔH

= f on Ω,and

lim

y→x

w(y) = lim

y→x

(ˆv(y)+H

(y)) = ˆv|

∂Ω

(x)+g(x) − ˆv|

∂Ω

(x)=g(x)for

all regular boundary points x ∈ ∂Ω. Suppose now that w

and w

satisfy the

conclusions of the theorem. Then w

− w

∈ C

(Ω),Δ(w

− w

)=0onΩ,

and lim

y→x

−w

)(y) = 0 for all regular boundary points x ∈ ∂Ω.Bythe

uniqueness of the solution to the Dirichlet problem for continuous boundary

functions, w

= w

on Ω.

8.4 H¨older Continuity of Second Derivatives 319

8.4 H¨older Continuity of Second Derivatives

It was shown in Theorem 8.3.1 that Poisson’s equation Δw = f has a solu-

tion w ∈ C

(Ω) whenever f is bounded and locally H¨older continuous with

exponent α ∈ (0, 1] on the bounded open set Ω. The purpose of the next few

resultsistoshowthatw ∈ H

2+α,loc

(Ω)whenα ∈ (0, 1).

Theorem 8.4.1 If Γ is a compact subset of the bounded open set Ω ⊂

,β =(β

,...,β

) is a multi-index, and u is a harmonic function on Ω,

then

D

u

0,Γ

≤



n|β|



|β|

u

0,Ω

where d = dist(Γ, ∂Ω).

Proof: Consider any y ∈ Γ ,letρ = d/|β|, and consider the concentric balls

,...,B

|β|

where B

= B

y,jρ

,j =1,...,|β|. Since all derivatives of u are

harmonic on Ω,

u(y)=



u(z) dz, i =1,...,n,

by the mean value property. By Equation (8.8),

u(y)=



∂B

u(z)γ

(z) dσ(z)

so that

u(y)|≤





sup

z∈B

−

|u(z)|.

Replacing u by D

u, j =1,...,n,

u(y)|≤





sup

z∈B

−

u(z)|.

Applying the same argument to each z ∈ B

−

and using the fact that B

z,ρ

⊂

u(z)|≤





sup

x∈B

−

z,ρ

|u(x)|≤





sup

x∈B

−

|u(x)|.

Thus,

u(y)|≤





sup

z∈B

−

|u(z)|.

320 8 Newtonian Potential

Continuing in this manner,

u(y)|≤





|β|

sup

z∈B

−

|β|

|u(z)|≤





|β|

u

0,Ω

Taking the supremum over y ∈ Γ and using the fact that ρ = d/|β|, D

u

0,Γ

≤ (n|β|/d)

|β|

u

0,Ω

Theorem 8.4.2 If B

= B

,ρ

and B

= B

,2ρ

are concentric balls with

closures in the bounded open set Ω ⊂ R

,f ∈ H

α,loc

(Ω),α ∈ (0, 1), u ∈

(Ω),andΔu = f on Ω,then

|u; ρ|

2+α,B

≤ C(n, α)



u

0,B

+ ρ

|f; ρ|

α,B



Proof: Note that u

0,B

< +∞ since u ∈ C

(Ω)andthatf can be replaced

by fχ

where χ

is the indicator function of B

.Extendf to be 0 outside

Ω. By Theorem 8.2.7, w =(1/(2 −n)σ

)Uf ∈ C

). Since Δw = f on B

by Theorem 8.3.1, Δ(u −w)=0onB

,v = u − w is harmonic on B

,andv

is the Dirichlet solution corresponding to the boundary function (u −w)|

∂B

A bound for w

0,B

can be found as follows, the n = 2 case being deferred

to the end of the proof. If x ∈ B

,thenB

⊂ B

x,4ρ

and

|w(x)| =



(2 − n)σ



|x − y|

n−2

f(y) dy



≤

(n − 2)σ

f

0,B



x,4ρ

|x − y|

n−2

≤ C(n)ρ

f

0,B

and

w

0,B

≤ C(n)ρ

f

0,B

. (8.21)

Therefore

v

0,B

≤v

0,B

≤u

0,B

+w

0,B

≤ C(n)(u

0,B

+ρ

f

0,B

). (8.22)

By Theorem 8.4.1, for any multi-index β with |β|≥1

D

v

0,B

≤



n|β|



|β|

v

0,B

≤ C(n)



n|β|



|β|

(u

0,B

+ ρ

f

0,B

Hence, if |β| = j = 1 or 2, then

[v]

j+0,B

≤ C(n)





(u

0,B

+ ρ

f

0,B

); (8.23)

8.4 H¨older Continuity of Second Derivatives 321

if |β| =2andx, y ∈ B

,thenforsomepointz on the line segment joining x

to y

v(x) − D

v(y)|

|x − y|

= |∇

(x)

v(z)||x − y|

1−α

≤ C(n)





v

0,B

|x − y|

1−α

≤ C(n)





(2ρ)

1−α

(u

0,B

+ ρ

f

0,B

Thus,

2+α

[v]

2+α,B

≤ C(n)(u

0,B

+ ρ

f

0,B

). (8.24)

Combining Inequalities (8.22), (8.23), and (8.24),

|v, ρ|

2+α,B



j=0

[v]

j+0,B

+ ρ

2+α

[v]

2+α,B

≤ C(n)(u

0,B

+ ρ

f

0,B

) (8.25)

≤ C(n)(u

0,B

+ ρ

|f; ρ|

α,B

By Theorem 8.2.7,

w(x)=



u(|x − y|)f(y) dy, x ∈ B

−

and by Inequality (8.18),

w(x)|≤C(n)f

0,B



|x − y|

n−1

dy x ∈ B

−

Since B

⊂ B

x,3ρ

for x ∈ B

w(x)|≤C(n)ρf 

0,B

Therefore,

[w]

1+0,B

≤ C(n)ρf

0,B

. (8.26)

By Corollary 8.2.11,

[w]

2+0,B

+ ρ

[w]

2+α,B

≤ C(n, α)(f 

0,B

+ ρ

[f]

α,B

322 8 Newtonian Potential

Using Inequalities (8.21) and (8.26),

|w; ρ|

2+α,B



j=0

[w]

j+0,B

+ ρ

2+α

[w]

2+α,B

≤ C(n, α)(ρ

f

0,B

+ ρ

f

0,B

+ ρ



f

0,B

+ ρ

[f]

α,B

)



≤ C(n, α)ρ

|f; ρ|

α,B

Combining this inequality with Inequality (8.25),

|u; ρ|

2+α,B

≤ C(n, α)(u

0,B

+ ρ

|f; ρ|

α,B

Returning to the n = 2 case, the functions u(x, y)andf(x, y) can be regarded

as functions of three variables without change to the hypotheses, the sup-

norms D

u, f 

0,B

, or the seminorms [u; ρ]

2+α

, [f ]

Corollary 8.4.3 If f ∈ H

α,loc

(Ω),α ∈ (0, 1),and{u

} is a uniformly

bounded sequence of functions in C

(Ω) satisfying Poisson’s equation Δu

f on Ω, then there is a subsequence that converges uniformly on compact

subsets of Ω toafunctionu ∈ C

(Ω) for which Δu = f on Ω.

Proof: For each x ∈ Ω,letB

x,ρ

x,2ρ

be balls with closures in Ω.Then

x,ρ

; x ∈ Ω} is an open covering of Ω and there is a countable subcov-

ering {B

,ρ

; i ≥ 1}.Notethatu

∈ C

(Ω) implies that u

restricted to

−

,ρ

belongs to C

−

,ρ

). By hypothesis, there is a constant C

such that

u



0,Ω

≤ C

for j ≥ 1. Consider the sequence {u

} on B

,ρ

.Usingthefact

that f ∈ H

−

,2ρ

) implies

|f; ρ

α,B

,2ρ

≤ (1 + ρ

)|f|

α,B

−

,2ρ

and the preceding theorem,

; ρ

2+α,B

,ρ

≤ C(n, α)(u



0,B

,2ρ

+ ρ

|f; ρ

α,B

,2ρ

)

≤ C(n, α)(C

+ ρ

|f; ρ

α,B

,2ρ

)

= M



< +∞,j≥ 1.

Recalling that D and D

are generic symbols for ﬁrst- and second-order

partials, respectively, for all j ≥ 1,

0,B

,ρ

+ ρ

[Du

]

0,B

,ρ

+ ρ

]

0,B

,ρ

+ ρ

2+α

]

α,B

,kρ

≤ M



Therefore, |u

2+α,B

,ρ

≤ M



/ min (1,ρ

,ρ

2+α

), and it follows from the

subsequence selection principle, Theorem 7.2.4, that there is a continuous u

on B

,ρ

and a subsequence {u

(1)

} of the {u

} sequence such that the se-

quences {u

(1)

}, {Du

(1)

},and{D

(1)

} converge uniformly on B

−

,ρ

to u, Du,

8.4 H¨older Continuity of Second Derivatives 323

and D

u, respectively. Since Δu

(1)

= f on Ω, Δu = f on B

−

,ρ

.Consider

now the sequence {u

(1)

} on B

,ρ

. The above argument applies to this case

also and there is a subsequence {u

(2)

} of the {u

(1)

} sequence such that the

sequences {u

(2)

}, {Du

(2)

},and{D

(2)

} converge uniformly on B

−

,ρ

to con-

tinuous functions v, Dv,andD

v, respectively, on B

−

,ρ

with Δv = f on

−

,ρ

. The functions u and v must agree on B

−

,ρ

∩ B

−

,ρ

and v can be

relabeled as u so that u is deﬁned on B

−

,ρ

∪ B

−

,ρ

. By an induction argu-

ment, for each i ≥ 2, there is a subsequence {u

(i)

} of the {u

(i−1)

} sequence

such that the sequences {u

(i)

}, {Du

(i)

},and{D

(i)

} converge uniformly



k=1

−

,ρ

to continuous functions u, Du,andD

u with Δu = f on



k=1

−

,ρ

. Finally, the diagonal sequence {u

(j)

} is a subsequence of every

sequence discussed above and there is a function u ∈ C

(Ω) such that the

sequences {u

(j)

}, {Du

(j)

},and{D

(j)

} converge uniformly on each compact

set Γ ⊂ Ω to the continuous functions u, Du,andD

u with u ∈ C

(Ω)and

Δu = f on Ω.

The requirement that f be bounded in Theorem 8.3.1 can be relaxed in

case Ω is a ball. The proof of the following theorem necessitates looking

at the eﬀect of truncation of a function satisfying a H¨older condition. If

a, b, c are any real numbers, it is easily seen that |min (a, b) − min (a, c)|≤

|b − c|.Iff ∈ H

(Ω), 0 <α<1andk is any real number, then

|min (k, f(x)) − min (k, f(y))|≤|f(x) − f (y)| from which it follows that

the function min (f, k) ∈ H

(Ω).Thesameresultistrueofmax(f, k). In

particular, the functions f

in the following proof belong to H

α,loc

(B).

Theorem 8.4.4 If B is a ball in R

,b ∈ (−1, 0),α ∈ (0, 1),andf ∈

α,loc

(B) satisﬁes [f; d]

(2+b)

0,B

< +∞, there is then a unique function u ∈

−

) ∩ C

(B) satisfying Δu = f on B and u =0on ∂B. Moreover,

[u; d]

(b)

0,B

≤ C[f; d]

(2+b)

0,B

(8.27)

where C = C(b, ρ).

Proof: Let B = B

,ρ

,r = |x − x

|,K =[f ; d]

(2+b)

0,B

,andv =(ρ

− r

)

−b

for

r ≤ ρ. Using the polar form of the Laplacian as in Section 1.3 and using the

fact that −n +2(1+b) < 0,

Δv =2b(ρ

− r

)

−b−2



n(ρ

− r

)+2(1+b)r



≤ 4b(1 + b)ρ

(ρ

− r

)

−b−2

≤ 4b(1 + b)ρ

−b

(ρ − r)

−b−2

Since d(x)

2+b

|f(x)|≤K by hypothesis and d(x)=ρ − r,

|f(x)|≤Kd(x)

−2−b

= K(ρ − r)

−b−2

≤ C

KΔv

324 8 Newtonian Potential

where C

=1/4(b(1 + b)ρ

−b

). Granted the ﬁrst assertion of the theorem, the

second assertion will be proved now. Assume that Δu = f on B and u =0

on ∂B.Then

Δ(−C

Kv ± u)(x) ≤−|f(x)|±f(x) ≤ 0onB

with −C

Kv ± u =0on∂B. By Corollary 2.3.6, −C

Kv ± u ≥ 0onB so

that

|u(x)|≤−C

Kv = −C

K(ρ

− r

)

−b

≤−C

K(2ρ)

−b

d(x)

−b

Letting C = −C

(2ρ)

−b

, d(x)

|u(x)|≤CK = C[f ; d]

(2+b)

0,B

, completing the

proof of the second assertion. Now let {B

} be an increasing sequence of

concentric balls with closures in B exhausting B.Foreachj ≥ 1, let m

sup

x∈B

|f(x)| and let

(x)=

⎧

⎨

⎩

−m

if f (x) ≤−m

f(x)if|f(x)| <m

if f (x) ≥ m

By Theorem 8.3.1, for each j ≥ 1thereisau

∈ C

−

) ∩ C

(B)

satisfying Δu

= f

on B and u

=0on∂B. By Inequality (8.27),

(x)|≤d(x)

(x)|≤C[f

; d]

(2+b)

0,B

≤ C[f; d]

(2+b)

0,B

on B,provingthat

the sequence {u

} is uniformly bounded on B. It follows from the above

corollary that there is a function u ∈ C

(B) such that Δu = f on B.Since

d(x)

|u(x)|≤C[f; d]

(2+b)

0,B

on B,lim

x→y

u(x) = 0 for all y ∈ ∂B.

Theorem 8.4.5 If Ω is a bounded open subset of R

, u ∈ C

(Ω),f ∈

(2)

(Ω; d), α ∈ (0, 1),andΔu = f on Ω,then

|u; d|

2+α,Ω

≤ C



u

0,Ω

+ |f ; d|

(2)

α,Ω



where C = C(n, α).

Proof: It can be assumed that u

0,Ω

< +∞, for otherwise the inequality

is trivial. For the time being, ﬁx x ∈ Ω and let δ = d(x)/3,B

= B

x,δ

x,2δ

.Fori, j =1,...n,

d(x)|D

u(x)| + d

(x)|D

u(x)|≤(3δ)[u]

1+0,B

+(3δ)

[u]

2+0,B

Since the sum on the right is included in the sum deﬁning |u; δ|

2+α,B

except

for a constant, by Theorem 8.4.2

d(x)|D

u(x)| + d

(x)|D

u(x)|≤C(n, α)(u

0,B

+ δ

|f; δ|

α,B

8.4 H¨older Continuity of Second Derivatives 325

Since d(y) ≥ d(x)/3=δ for y ∈ B

|f; δ|

α,B

≤ sup

y∈B

(y)|f(y)|+sup





∈B

2+α





)

|f(y



) − f (y





− y



≤ C|f ; d|

(2)

α,Ω

. (8.28)

Therefore,

|u; d|

2+0,Ω

≤ C(n, α)(u

0,Ω

+ |f ; d|

(2)

α,Ω

). (8.29)

Now consider any x, y ∈ Ω with d(x) ≤ d(y) and deﬁne δ, B

as above.

Since d(x, y)=d(x)=3δ,

2+α

(x, y)

u(x) − D

u(y)|

|x − y|

≤ (3δ)

2+α

[u]

2+α,B

2(3δ)

[u]

2+0,Ω

Since δ

2+α

[u]

2+α,B

≤ [u; δ]

2+α,B

≤|u; δ|

2+α,B

, by Theorem 8.4.2 and In-

equality (8.28)

2+α

[u]

2+α,B

≤ C(n, α)(u

0,B

+ δ

|f; δ|

α,B

)

≤ C(n, α)(u

0,Ω

+ |f ; d|

(2)

α,Ω

By Inequality (8.29)

2+α

(x, y)

u(x) − D

u(y)|

|x − y|

≤ C(n, α)(u

0,Ω

+ |f ; d|

(2)

α,Ω

This shows that

[u; d]

2+α,Ω

≤ C(n, α)(u

0,Ω

+ |f ; d|

(2)

α,Ω

Hence, |u; d|

2+α,Ω

= |u; d|

2+0,Ω

+[u; d]

2+α,Ω

≤ C(n, α)(u

0,Ω

+ |f; d|

(2)

α,Ω

Corollary 8.4.6 If Ω is a bounded open subset of R

,u∈ C

(Ω)∩C

(Ω),f∈

(2)

(Ω; d),α ∈ (0, 1),andΔu=f on Ω,thenu∈H

2+α,loc

(Ω).

Proof: Let Γ be a compact subset of Ω and let d

= d(Γ,∂Ω) > 0. By the

preceding theorem,

min (1,d

2+α

)|u|

2+α,Γ

≤|u; d|

2+α,Ω

≤ C(n, α)(u

0,Ω

+|f; d|

(2)

α,Ω

) < +∞.

Thus, u ∈ H

2+α,loc

(Ω).

326 8 Newtonian Potential

8.5 The Reﬂection Principle

The classic reﬂection principle, known as Schwarz’s Reﬂection Principle,

can be described as follows. Let Ω be an open subset of R

that is symmetric

with respect to R

with Ω

= Ω ∩ R

= ∅,letΩ

= Ω ∩ R

,andlet

−

= Ω ∩ R

−

.Ifh ∈ C

(Ω

∪ Ω

)isharmoniconΩ

with h =0onΩ

then the function

h(x



⎧

⎨

⎩

h(x



)if(x



) ∈ Ω

0if(x



) ∈ Ω

−h(x



, −x

)if(x



) ∈ Ω

−

is harmonic on Ω by Corollary 1.7.10, since it is continuous there and satisﬁes

the local averaging principle.

Throughout this section, unless noted otherwise, Ω will denote a spherical

chip; that is, Ω = B

,δ

∩ R

for some x

∈ R

−

∪ R

and δ>0. It will be

assumed that B

,δ

∩R

= ∅.Σwill denote the interior of the ﬂat portion of

the boundary of Ω. Recall that y

=(y



, −y

)ify =(y



) ∈ R

and that

= {y

; y ∈ Ω}. Each point of Σ is then an interior point of the convex

region Ω



= Ω ∪ Ω

∪ Σ. Note that each point of the boundary of Ω



is a

regular boundary point for the Dirichlet problem, since the Zaremba cone

condition is satisﬁed at each such point.

Theorem 8.5.1 If h ∈ C

(∂Ω ∼ Σ), then there is a function u ∈ C

(Ω

−

) ∩

(Ω ∪Σ) satisfying Δu =0on Ω, D

u =0on Σ,andlim

y→x,y∈Ω

u(y)=

h(x) for each x ∈ ∂Ω ∼ Σ.

Proof: Deﬁne h



on ∂Ω



by putting





h(x



)on∂Ω ∼ Σ

h(x



, −x

)on(∂Ω ∼ Σ)

Note that h



)=h



, −x

)on∂Ω



. Now consider u in the class U



of lower bounded superharmonic functions with lim inf

y→x

u(y) ≥ h



(x),x∈

∂Ω



.Thenu(x



, −x

) is a lower bounded superharmonic function on Ω



.Since

lim inf



)→(x



)

u(y



, −y

) ≥ h



, −x

)=h



u(x



, −x

)isinU



. Letting u



)=min[u(x



),u(x



, −x

)], u



is a

lower bounded superharmonic function on Ω



, and it is easily seen that

lim inf



)→(x



)



) ≥ h



Note that u



)=u



, −x

). Thus u ∈ U



⇒ u



∈ U



and u



≤ u.

Therefore,



=inf{u



: u ∈ U



} and H





)=H





, −x

) because

this is true of each u



. Similarly, H





)=H





, −x

). In particular, if

8.5 The Reﬂection Principle 327

h is continuous on ∂Ω ∼ Σ,thenH



= H



, and we put H



= H



= H



Now consider a point (x



, 0) ∈ Σ, which is an interior point of Ω



.SinceH



is harmonic on a neighborhood of (x



, 0) and





) − H





, −x

)





) − H





, 0)





, −x

) − H





, 0)

−x





, 0) = 0. Letting u be the restriction of H



to Ω, it follows that

u ∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ), since H



is harmonic on Ω and on a neighbor-

hood of each point of Σ.Lastly,ifx ∈ ∂Ω ∼ Σ, then lim

y→x,y∈Ω

u(y)=

lim

y→x,y∈Ω



(y)=h



(x)=h(x).

The following lemma results from combining Theorem 8.2.7 and Theorem

8.3.1. The set Ω need not satisfy the standing assumption stated at the

beginning of this section.

Lemma 8.5.2 Let f be bounded and locally H¨older continuous with exponent

α ∈ (0, 1] on the bounded open set Ω and let

w(x)=

(2 − n)σ



u(|x − y|)f(y) dy, x ∈ R

Then w ∈ C

(Ω

−

) ∩ C

(Ω ∪Σ) ∩ C

(Ω) with Δw = f on Ω.

The solution w of the equation Δw = f is not unique, for if w is replaced

by w + h,whereh is harmonic on R

, then another solution is obtained.

Lemma 8.5.3 If f ∈ H

0+α

(Ω),α∈ (0, 1], there is a function w ∈ C

(Ω

−

)∩

(Ω ∪ Σ) ∩ C

(Ω) such that

Δw = f on Ω and D

w =0on Σ.

Proof: Since f ∈ H

0+α

(Ω) implies that f is uniformly continuous on Ω, f

has a continuous extension to Ω

−

, also denoted by f.Let





f(x



)ifx

≥ 0

f(x



, −x

)ifx

≤ 0.

It is easily seen that f



∈ H

0+α

(Ω



). Applying the preceding lemma to f



there is a bounded function w ∈ C

(Ω

−

)∩C

(Ω



) such that Δw = f



on Ω



For each y



∈ R

,letΓ



= {y

;(y



) ∈ Ω



}. Using the fact that f



, −y

), the fact that u(|(x



) − (y



)|)=u(|(x



, −x

) − (y



, −y

)|),

and Fubini’s Theorem,