Helms L.L. Potential Theory

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18 1 Laplace’s Equation

Substituting for D

(x, z) in Equation (1.8),

u(x)=



∂B

u(z)

−|y −x|

|z − x|

dσ(z).

Suppose now that u is only harmonic on B and continuous on B

−

.Fixx ∈

y,ρ

and let {ρ

} be a sequence of positive numbers increasing to ρ with

x ∈ B

y,ρ

for all k ≥ 1. Since u is harmonic on a neighborhood of each B

y,ρ

u(x)=



∂B

y,ρ

u(z)

−|y − x|

|z −x|

dσ(z).

Letting z = ρ

θ,

u(x)=



|θ|=1

u(ρ

θ)

−|y −x|

|ρ

θ −x|

n−1

dσ(θ).

By uniform boundedness of the integrands and continuity,

u(x)=



|θ|=1

u(ρθ)

−|y −x|

|ρθ − x|

n−1

dσ(θ)



∂B

u(z)

−|y −x|

|z −x|

dσ(z).

The deﬁnition of harmonicity of a function u on an open set Ω required

that u ∈ C

(Ω). Much more is true as a consequence.

Lemma 1.5.5 Let U and C be open and compact subsets of R

, respectively,

and let u = u(x, y) be a real-valued function on U ×C with the property that

∂u/∂x

is continuous on U × C for each j =1,...,n.Then

∂

∂x



u(x, y) dμ(y)=



∂u

∂x

(x, y) dμ(y) x ∈ U, j =1,...,n,

for any ﬁnite measure μ on the Borel subsets of C.

Proof: Let ν beaunitvector.Itwillbeshownnowthat



u(x, y) dμ(y)=



u(x, y) dμ(y).

Consider a ﬁxed x ∈ U. The left side of this equation is

lim

λ→0+



u(x + λν, y) − u(x, y)

dμ(y).

Let B be a ball with center x such that B

−

⊂ U, and consider only those λ

for which x + λν ∈ B. By the mean value theorem of the calculus, for each

such λ and each y ∈ C,thereisapointξ

λ,y

on the line segment joining x to

x + λν such that

1.5 Poisson Integral Formula 19



u(x + λν) − u(x, y)

dμ(y)=



(∇u(ξ

λ,y

) · λν) dμ(y)



u(ξ

λ,y

,y) dμ(y).

Since D

u is continuous on B

−

×C, the integrand on the right is uniformly

bounded for such λ and y ∈ C.Sinceμ(C) < +∞, the Lebesgue dominated

convergence theorem can be applied as λ → 0+, sequentially, to obtain



u(x, y) dμ(y)=



u(x, y) dμ(y).

Theorem 1.5.6 If u is harmonic on the open set Ω, then all partial deriva-

tives of u are harmonic on Ω.

Proof: If y ∈ Ω,B = B

y,ρ

⊂ B

−

y,ρ

⊂ Ω,andx ∈ B,then

u(x)=



∂B

−|y −x|

|z − x|

u(z) dσ(z).

By the preceding lemma,

∂u

∂x

(x)=



∂B

∂

∂x



−|y −x|

|z − x|



u(z) dσ(z).

By calculating the partial derivative under the integral sign and using the

Lebesgue dominated convergence theorem, it can be shown that ∂u/∂x

continuous on B. In the same way, it can be shown that partial derivatives

of all orders are continuous on B.SinceB can be any ball with B

−

⊂ Ω,

partial derivatives of all orders of u are continuous on Ω.Since



i=1

∂

∂x

=0onΩ,



i=1

∂

∂x

∂

∂x

=0onΩ

for each j =1,...,n. Using continuity of the third partial derivatives, the

order of diﬀerentiation can be interchanged to obtain



i=1

∂

∂x

∂u

∂x

=0,j=1,...,n.

20 1 Laplace’s Equation

Since ∂u/∂x

has continuous second partials on Ω, ∂u/∂x

is harmonic on

Ω,aswellasallhigher-order partial derivatives.

Theorem 1.5.7 (Picard) If u is harmonic on R

and either bounded above

or bounded below, then u is a constant function.

Proof: Since −u is harmonic if u is harmonic, it can be assumed that u is

bounded below. Since the sum of a harmonic function and a constant function

is harmonic, it can be assumed that u ≥ 0. Let x and y be distinct points,

and consider balls B

x,δ

and B

y,

with B

y,

⊃ B

x,δ

. By Theorem 1.4.4,

u(x)=



x,δ

u(z) dz ≤



y,

u(z) dz = ν



u(y)

and u(x) ≤ (/δ)

u(y). Now let , δ → +∞ in such a way that /δ → 1to

obtain u(x) ≤ u(y). But since x and y are arbitrary points, u(y) ≤ u(x)and

u is a constant function.

According to the Poisson integral formula, if a harmonic function is zero

on the boundary of a ball, then it is zero on the ball. This fact suggests the

following principle, known as the maximum principle.

Deﬁnition 1.5.8

A function u deﬁned on an open connected set Ω obeys

the maximum principle if sup

x∈Ω

u(x) is not attained on Ω unless u is

constant on Ω;theminimum principle if inf

x∈Ω

u(x) is not attained on

Ω unless u is constant on Ω.

Theorem 1.5.9 If u is continuous on the open connected set Ω and for each

x ∈ Ω there is a δ

> 0 such that

u(x)=

n−1



∂B

x,δ

u(y) dσ(y)

whenever δ<δ

,thenu obeys the maximum and minimum principles.

Proof: Suppose u attains its maximum value at a point of Ω and let M =

{x ∈ Ω; u(x)=sup

y∈Ω

u(y)} = ∅.Sinceu is continuous, M is relatively

closed in Ω. It will be shown now that M is open in Ω.Forx ∈ M,

u(x)=

n−1



∂B

x,δ

u(y) dσ(y)

whenever δ<δ

.Consideranyy ∈ B

x,δ

and let δ

= |y − x| <δ

.Since

u(x)=

n−1



∂B

x,δ

u(x) dσ(y),

This version of the maximum principle will be referred to later as the strong maxi-

mum principle.

1.6 Gauss’ Averaging Principle 21



∂B

x,δ

(u(x) − u(y)) dσ(y)=0.

Since x ∈ M, u(x) − u ≥ 0on∂B

x,δ

and therefore u(x) − u = 0 a.e. on

∂B

x,δ

.Bycontinuity,u = u(x)on∂B

x,δ

and, in particular, u(y)=u(x).

This shows that y ∈ M ,thatB

x,δ

⊂ M, and that M is both relatively closed

andopeninΩ. By the connectedness of Ω,eitherM = ∅ or M = Ω.Since

the ﬁrst case has been excluded, M = Ω so that u is constant on Ω.

Corollary 1.5.10 If u is harmonic on the open connected set Ω,thenu

satisﬁes both the maximum and minimum principles.

1.6 Gauss’ Averaging Principle

Among the many contributions of Gauss to potential theory, one of the best

known is the assertion that the gravitational potential at a point in space

due to a homogeneous spherical body is the same as if the entire mass were

concentrated at the center of the body. A classical version of this property

will be stated ﬁrst.

Lemma 1.6.1 If B

x,δ

⊂ R

,then

(i) for n =2and y ∈ R

2πδ



∂B

x,δ

log

|y − z|

dσ(z)=



log

|y−x|

if |y −x| >δ

log

if |y −x|≤δ

(ii) for n ≥ 3 and y ∈ R

n−1



∂B

x,δ

|y −z|

n−2

dσ(z)=



|y−x|

n−2

if |y − x| >δ

n−2

if |y − x|≤δ.

Proof: Only (i) will be proved, (ii) being the easier of the two. Fix x ∈ R

and

δ>0, and let u

(z)=−log |y −z|,z ∈ R

. Three cases will be considered

according as y ∈ B

x,δ

,y ∈ ∂B

x,δ

,ory ∈ B

−

x,δ

. Suppose ﬁrst that y ∈ B

−

x,δ

Then u

is harmonic on a neighborhood of B

−

x,δ

and the result follows from

the mean value property. Consider now the case y ∈ ∂B

x,δ

.Foreachn ≥ 1, let

= x+(1+

)(y−x). Then |y

−x| =(1+

)|y−x| =(1+

)δ, |y

−z|≤3δ

for all z ∈ ∂B

x,δ

,and

log

− z|

≥ log

3δ

,z∈ ∂B

x,δ

22 1 Laplace’s Equation

It follows that the sequence of functions on the left is lower bounded on ∂B

x,δ

By the preceding case and Fatou’s Lemma,

2πδ



∂B

x,δ

log

|y − z|

dσ(z) ≤ lim inf

n→∞

2πδ



∂B

x,δ

log

− z|

dσ(z)

= lim inf

n→∞



log

− x|



=log

|y −x|

On the other hand, since |y

− z|≥|y −z| for z ∈ ∂B

x,δ

log

|y −x|

= lim

n→∞

log

− x|

= lim

n→∞

2πδ



∂B

x,δ

log

− z|

dσ(z)

≤

2πδ



∂B

x,δ

log

|y −z|

dσ(z),

and the assertion is true for y ∈ ∂B

x,δ

. Lastly, suppose y ∈ B

x,δ

. Letting

x,δ

denote the left side of the equation in (i) and applying Lemma 1.5.5 to

it twice, it is easily seen that u

x,δ

is harmonic on B

x,δ

. Letting γ denote the

angle between the line segment joining x to y and x to z,

x,δ

(y)=

2πδ



2π

log

|y −x|

+ δ

− 2δ|y − x|cos γ

dγ

and it is easily seen that u

x,δ

is a function of r = |y − x| only and, as in

Section 1.3, must be of the form α log r + β.Since

x,δ

(x)=

2πδ



∂B

x,δ

log

|x − z|

dσ(z)=log

α =0andβ =log(1/δ); that is, u

x,δ

(y)=log(1/δ),y ∈ B

x,δ

The preceding lemma can be interpreted as a statement about the average

value of log (1/|y − z|)or1/|y − z|

n−2

for a unit mass concentrated on the

point y. Arbitrary mass distributions will be considered now.

Some preliminary calculations will be carried out for the n =2andn ≥ 3

cases separately. If μ is a measure on the Borel subsets of R

with compact

support S

,let

(y)=



log

|y −z|

dμ(z)=



log

|y −z|

dμ(z),y∈ R

1.6 Gauss’ Averaging Principle 23

For a ﬁxed y ∈ R

, log (1/|y − z|) is lower bounded on S

and the above

integral is deﬁned as a real number or +∞. Consider a ﬁxed ball B

x,δ

.Since

log (1/|y −z|) is lower bounded on ∂B

x,δ

× S

, Tonelli’s theorem (c.f. [18])

implies that

2πδ



∂B

x,δ

(y) dσ(y)=

2πδ



∂B

x,δ



log

|y − z|

dμ(z)dσ(y)





2πδ



∂B

x,δ

log

|y −z|

dσ(y)



dμ(z).

If μ is a measure on the Borel subsets of R

,n ≥ 3, with compact support

,let

(y)=



|y −z|

n−2

dμ(z)=



|y −z|

n−2

dμ(z).

The same argument implies that

n−1



∂B

x,δ

(y) dσ(y)=





n−1



∂B

x,δ

|y − z|

n−2

dσ(y)



dμ(z).

Only the n = 2 case of the following theorem will be proved, the n ≥ 3

case being essentially the same.

Theorem 1.6.2 (Gauss’ Averaging Principle [23]) Let μ be a measure

on the Borel subsets of R

with compact support S

and let B=B

x,δ

.If

∩ B

−

=∅, then the average of U

over ∂B is U

(x);ifS

⊂ B

−

,the

average of U

over ∂B depends only upon the total mass of μ and is equal

to μ(S

)log(1/δ) in the n =2case and is equal to μ(S

)/δ

n−2

in the n ≥ 3

case.

Proof: (n = 2) From the above discussion,

2πδ



∂B

x,δ

(y) dσ(y)=





2πδ



∂B

x,δ

log

|y −z|

dσ(y)



dμ(z).

If S

⊂ B

−

, it follows from Lemma 1.6.1 that

2πδ



∂B

x,δ

(y) dσ(y)=



log

dμ(z)

= μ(S

)log(1/δ);

if S

∩ B

−

= ∅,thenlog(1/|y − z|) is a harmonic function of y and

2πδ



∂B

x,δ

(y) dσ(y)=



log

|x − z|

dμ(z)=U

(x)

by the mean value property.

24 1 Laplace’s Equation

1.7 The Dirichlet Problem for a Ball

If Ω is a nonempty open subset of R

with compact closure and f is a

real-valued function on ∂Ω,theDirichlet problem is that of ﬁnding a

harmonic function u on Ω such that lim

y→x,y∈Ω

u(y)=f (x) for all x ∈ ∂Ω.

As was noted in Section 0.1, this limiting behavior of u at the boundary

of Ω implies that f is continuous on ∂Ω. Generally speaking, the Dirichlet

problem does not have a solution even when Ω is a ball.

Theorem 1.7.1 The solution of the Dirichlet problem for a nonempty, open

connected set Ω with compact closure and a continuous boundary function f

is unique if it exists.

Proof: Let u

and u

be two solutions. Suppose there is a point z ∈ Ω such

that u

(z) >u

(z). Then lim

y→x,y∈Ω

(y) − u

(y)] = 0 for all x ∈ ∂Ω.Let

w =



− u

on Ω

0on∂Ω.

Thus, w is harmonic on Ω, continuous on Ω

−

, and zero on ∂Ω.Sincew(z) > 0

and w is continuous on Ω

−

,wmust attain a positive supremum at some point

of Ω. By the maximum principle, Corollary 1.5.10, w must be constant on

Ω.Sincew =0on∂Ω, w =0onΩ

−

, a contradiction. Therefore, u

≤ u

Ω. Interchanging u

and u

≤ u

, and the two are equal.

According to the Poisson integral formula, the value of a harmonic function

u at an interior point of a ball B is determined by values of u on ∂B, assuming

that u has a continuous extension to ∂B. It is natural to ask if a function f on

∂B determines a function u harmonic on B that has a continuous extension

to B

−

agreeing with f on ∂B. More generally, it might be asked if a measure

on the boundary of B determines a harmonic function on B.

Theorem 1.7.2 (Herglotz [29]) If μ isasignedmeasureofboundedvari-

ation on the Borel subsets of ∂B

y,ρ

,then

u(x)=



∂B

y,ρ

−|y −x|

|z −x|

dμ(z),x∈ B

y,ρ

is harmonic on B

y,ρ

Proof: Using Lemma 1.5.5, it can be shown that u has continuous second

partials and that

Δu(x)=



∂B

y,ρ

(x)

−|y −x|

|z − x|

dμ(z).

A tedious, but straightforward, diﬀerentiation shows that the integrand is

zero for x ∈ B

y,ρ

1.7 The Dirichlet Problem for a Ball 25

Corollary 1.7.3 If f is a Borel measurable function on ∂B

y,ρ

and integrable

relative to surface area, then

u(x)=



∂B

y,ρ

−|y −x|

|z − x|

f(z) dσ(z)

is harmonic on B

y,ρ

Since the Poisson integral formula will be referred to repeatedly, let

PI(μ : B)(x)=



∂B

−|y − x|

|z − x|

dμ(z),

where B = B

y,ρ

and μ is a signed measure of bounded variation on ∂B.

If B = B

y,ρ

, the dependence on the parameters y and ρ will be exhibited

by using the notation PI(μ : y, ρ). If μ is absolutely continuous relative to

surface area on ∂B, then for each Borel set M ∈ ∂B

μ(M)=



f(z) dσ(z)

for some integrable function f on ∂B.Inthiscase,letPI(μ : B)=PI(f : B).

This notation will also be used if the domain of f contains ∂B.Notethat

PI(1 : B) = 1, that PI(μ : B) is linear in μ,thatPI(μ : B) ≥ 0ifμ is a

measure, and that PI(f : B) ≥ 0iff is nonnegative.

According to Corollary 1.7.3, an integrable boundary function f deter-

mines a harmonic function u on a ball B.Inwhatwayisu related to f?For

example, is it true that lim

y→x,y∈B

u(y)=f(x)forx ∈ ∂B? The following

three lemmas answer this question for a ball B = B

y,ρ

Lemma 1.7.4 If f is Borel measurable on ∂B, integrable relative to surface

area on ∂B, u = PI(f : B) on B, and there is a constant k such that

f ≤ k a.e.(σ) on a neighborhood of x

∈ ∂B,thenlim sup

x→x

,x∈B

u(x) ≤ k.

Proof: Itcanbeassumedthatk ≥ 0 for if not, replace f by f − k.Choose

>0 such that f (z) ≤ k a.e.(σ)forz ∈ B

,

∩ ∂B. Denoting the indicator

function of B

,

by g

,

,forx ∈ B

u(x)=PI(g

,

f : B)+PI((1 − g

,

)f : B).

Since f(z) ≤ k a.e.(σ)forz ∈ B

,

∩ ∂B,

PI(g

,

f : B)(x) ≤ PI(k : B)(x)=kPI(1 : B)(x)=k.

Suppose x ∈ B

,/2

and z ∈ ∂B.Then|z − x| >/2when|z − x

| >for

otherwise, |z − x

|≤|z − x| + |x − x

|≤.Thus,

26 1 Laplace’s Equation

|PI((1 − g

,

)f : B)(x)|≤



∼B

,

∩∂B

−|y −x|

(/2)

|f(z)|dσ(z)

≤

−|y −x|

ρ(/2)



∂B

|f(z)|dσ(z).

Since |y − x|→ρ as x → x

, PI((1 − g

,

)f : B)(x) → 0asx → x

Therefore,

lim sup

x→x

u(x) ≤ lim sup

x→x

PI(g

,

f : B)(x)

+ lim sup

x→x

PI((1 − g

,

)f : B)(x) ≤ k.

Lemma 1.7.5 If f is Borel measurable on ∂B, integrable relative to surface

area on ∂B,andu = PI(f : B) on B, then for x

∈ ∂B

lim sup

x→x

,x∈B

u(x) ≤ lim sup

x→x

,x∈∂B

f(x).

Proof: It can be assumed that the right side is ﬁnite for otherwise there is

nothingtoprove.Ifk is any number greater than the right member of the

last inequality, then f (x) <kfor all x ∈ ∂B in a neighborhood of x

.Bythe

preceding lemma,

lim sup

x→x

,x∈B

u(x) ≤ k;

but since k is any number greater than lim sup

x→x

,x∈∂B

f(x), the lemma is

proved.

Lemma 1.7.6 If f is Borel measurable on ∂B, integrable relative to surface

area on ∂B, continuous at x

∈ ∂B,andu = PI(f : B) on B,then

lim

x→x

,x∈B

u(x)=f(x

Proof: By the preceding lemma, lim sup

x→x

,x∈B

u(x) ≤ f(x

). Since

PI(−f:B)=−PI(f : B),

lim sup

x→x

,x∈B

−u(x) = lim sup

x→x

,x∈B

PI(−f : B) ≤−f(x

or lim inf

x→x

,x∈B

u(x) ≥ f (x

) and the result follows.

Theorem 1.7.7 (Schwarz [56]) The Dirichlet problem is uniquely solvable

for a ball B and a continuous boundary function f. The solution is given by

PI(f : B).

Proof: Uniqueness was proven in Theorem 1.7.1.

1.7 The Dirichlet Problem for a Ball 27

Remark 1.7.8 Put another way, if f ∈ C

(∂B), then u = PI(f : B)is

harmonic on B, has a continuous extension to B

−

, and agrees with f on ∂B.

According to Theorem 1.4.3, if u is harmonic on a neighborhood of a ball,

then u has the mean value property. This property characterizes harmonic

functions.

Theorem 1.7.9 Afunctionu on the open set Ω ⊂ R

is harmonic if and

only if u ∈ C

(Ω) and

u(x)=

n−1



∂B

x,δ

u(z) dσ(z)

for every ball B

x,δ

⊂ B

−

x,δ

⊂ Ω.

Proof: The necessity follows from the continuity of harmonic functions and

Theorem 1.4.3. Consider a function u ∈ C

(Ω) that satisﬁes the above

equation for every B

x,δ

⊂ B

−

x,δ

⊂ Ω. Fix such a ball. By Lemma 1.7.6,

there is a function v ∈ C

(Ω

−

)thatagreeswithu on ∂B

x,δ

and is harmonic

on B

x,δ

. The diﬀerence u − v is then zero on ∂B

x,δ

, and since it satisﬁes the

hypothesis of Theorem 1.5.9, u −v satisﬁes both the maximum and minimum

principles on B

x,δ

. It follows that u = v on B

x,δ

and u is harmonic on B

x,δ

Since B

x,δ

is an arbitrary ball with B

−

x,δ

⊂ Ω, u is harmonic on Ω.

It is not necessary that the equation of the preceding theorem hold for

every ball B

x,δ

⊂ B

−

x,δ

⊂ Ω.

Corollary 1.7.10 Afunctionu on the open set Ω ⊂ R

is harmonic if and

only if u ∈ C

(Ω) and for each x ∈ Ω,thereisaδ

> 0 such that

u(x)=

n−1



∂B

x,δ

u(z) dσ(z) (1.10)

for all 0 <δ<δ

,inwhichcase

u(x)=

n−1



∂B

x,δ

u(z) dσ(z)=



x,δ

u(z) dz

whenever B

−

x,δ

⊂ Ω.

Proof: The necessity follows from the preceding theorem. Suppose u ∈ C

(Ω)

and for each x ∈ Ω,thereisaδ

> 0 such that Equation (1.10) holds for

all 0 <δ<δ

.IfB = B

x,ρ

⊂ B

−

x,ρ

⊂ Ω,thenu|

∂B

is continuous and the

function

v(y)=



PI(u : x, ρ)(y),y∈ B,

u(y),y∈ ∂B,

is continuous on B

−

by Lemma 1.7.6. By deﬁnition, u − v =0on∂B.Since

v is harmonic on B,foreachy ∈ B