Helms L.L. Potential Theory

Подождите немного. Документ загружается.

28 1 Laplace’s Equation

u(y) − v(y)=

n−1



∂B

y,δ

(u(z) − v(z)) dσ(z)

for all suﬃciently small δ>0. By Theorem 1.5.9, u −v satisﬁes the minimum

and maximum principles on B.Ifu − v attains its maximum at some point

of B, then it must be constant and therefore u − v =0onB;ifu − v does

not attain its maximum at a point of B,thenu −v ≤ 0onB.Ineithercase,

u ≤ v on B.Sinceu − v also satisﬁes the minimum principle on B, u = v on

B and u is harmonic on B.SinceB is arbitrary, u is harmonic on Ω.Lastly,

since ν

n = σ



x,δ

u(z) dz =



|θ|=1

u(x + ρθ)ρ

n−1

dθ dρ







|θ|=1

u(x + ρθ) dθ



n−1

dρ



u(x)ρ

n−1

dρ

= u(x).

It is possible to relax the requirement of continuity in the preceding

theorem at the expense of replacing surface averages by solid ball aver-

ages. In the proof of the following theorem, the symmetric diﬀerence of

two sets A and B is denoted by AB and is deﬁned by the equation

AB =(A ∼ B) ∪ (B ∼ A).

Lemma 1.7.11 If u is locally integrable on the open connected set Ω and

u(x)=



x,δ

u(y) dy

whenever B

x,δ

⊂ Ω,thenu is continuous and obeys the maximum and min-

imum principles on Ω.

Proof: Note ﬁrst that u is real-valued. It will be shown now that the hy-

potheses imply that u is continuous on Ω.Toseethis,consideranyx ∈ Ω

and any δ>0 such that B

−

x,2δ

⊂ Ω. For any y ∈ B

x,δ

y,δ

⊂ B

−

x,2δ

.Since

x,δ

B

y,δ

⊂ B

−

x,2δ

and u is integrable on B

−

x,2δ

|u(x) − u(y)|≤



x,δ

B

y,δ

|u(z)|dz → 0

as y → x by the absolute continuity of the Lebesgue integral, which proves

that u is continuous at x. Suppose there is a point x

∈ Ω such that u(x

inf

u. Letting Σ = {y; u(y)=inf

u},Σ is a relatively closed subset of Ω

by continuity of u. For any y ∈ Σ and δ>0withB

y,δ

⊂ Ω,

1.7 The Dirichlet Problem for a Ball 29



y,δ

(u(z) − u(y)) dz =0.

Since y ∈ Σ,u − u(y) ≥ 0onB

y,δ

and it follows that u − u(y)=0a.e. in

y,δ

;thus,u = u(y)onB

y,δ

by continuity of u. This shows that B

y,δ

⊂ Σ

and that Σ is an open subset of Ω. By the connectedness of Ω, Σ = ∅ or

Σ = Ω.IfΣ = Ω,thenu is constant on Ω;otherwise,u does not attain its

minimum at an interior point of Ω.Since−u satisﬁes the same hypotheses,

u satisﬁes the maximum principle on Ω.

Theorem 1.7.12 The function u is harmonic on the open set Ω if and only

if u is locally integrable on Ω and

u(x)=



x,δ

u(y) dy

whenever B

x,δ

⊂ Ω.

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

Theorem 1.7.9. As to the suﬃciency, let u be locally integrable on Ω and

satisfy the above equation. Since it suﬃces to prove that u is harmonic on

each component of Ω, it will be assumed that Ω is connected. Let B = B

y,ρ

be any ball with B

−

⊂ Ω. It was shown in the preceding proof that u is

continuous on Ω. Consider the harmonic function h = PI(u|

∂B

: y, ρ). By

Lemma 1.7.6, lim

z→x,z∈B

(u(z)−h(z)) = 0 for all x ∈ ∂B.Sinceu−h satisﬁes

both the minimum and maximum principles on B by Lemma 1.7.11, u = h

on B and u is harmonic on B.SinceB is an arbitrary ball with B

−

⊂ Ω, u

is harmonic on Ω.

The requirement in the preceding theorem that u satisfy a global solid ball

averaging condition cannot be relaxed. This can be seen by examining the

function u on R

deﬁned by

u(x, y)=

⎧

⎨

⎩

−1ifx<0

0ifx =0

1ifx>0.

This function is locally integrable and satisﬁes a local solid ball averaging

principle but is not harmonic.

If u

,...,u

are harmonic functions on the open set Ω and α

,...,α

are

real numbers, then it is clear from the original deﬁnition of harmonic function

that u =



i=1

is harmonic on Ω. The following theorem extends this

result to integrals.

Theorem 1.7.13 Let U and V be open subsets of R

,letμ be a measure on

U,andletH be a nonnegative function on U ×V .If(i)foreachy ∈ V,H(·,y)

is continuous on U , (ii) for each x ∈ U, H(x, ·) is harmonic on V , and (iii)

h(y)=



H(x, y) dμ(x) < +∞ for each y ∈ V ,thenh is harmonic on V .

30 1 Laplace’s Equation

Proof: By hypothesis, H(x, y) is continuous in each variable separately. This

implies that H(x, y) is jointly measurable on U ×V .Toseethis,foreachk ≥ 1

let

,...,j

)={(x

,...,x

);

≤ x

,i=1,...,n}

where each j

∈ Z, the set of integers. Deﬁne a map ψ

: U → V by letting

(x)beanyﬁxedpointofV ∩I

,...,j

) whenever x ∈ V ∩I

,...,j

Note that lim

k→∞

(x)=x for all x ∈ U and that lim

k→∞

H(ψ

(x),y)=

H(x, y)foreachx ∈ U and y ∈ V . Since each function H(ψ

(x),y)isjointly

measurable on U × V , H(x, y) is also. Therefore, H(x, y) is a nonnegative

jointly measurable function to which Tonelli’s theorem can be applied. Sup-

pose B

−

y,δ

⊂ V .Then

A(h, y, δ)=A





H(x, ·) dμ(x):y, δ





H(x, y) dμ(x)

= h(y) < +∞.

This shows that h is locally integrable and satisﬁes the hypotheses of Theo-

rem 1.7.12. Thus, h is harmonic on V .

It would appear from Theorem 1.7.2 that the class of functions harmonic

on a ball is much more extensive than the class obtained by solving the

Dirichlet problem for a Borel measurable boundary function. For example,

consider a point z

∈ ∂B

0,ρ

and a unit measure μ concentrated on z

.Then

for 0 <λ<1

u(λz

)=PI(μ : B)(λz



∂B

0,ρ

−|λz

|z −λz

dμ(z)

n−1

1 − λ

(1 − λ)

→ +∞

as λ → 1−;thatis,PI(μ : B)(x) → +∞ as x approaches z

along a radial

line. On the other hand, if the measure μ is concentrated on some point z

other than z

,then

u(λz

)=PI(μ : B)(λz

(1 − λ

)

− λz

→ 0

as λ → 1−. The function u is not determined by the boundary func-

tion f that is +∞ at z

and is 0 at all other points of ∂B

0,ρ

since

f =0a.e.(σ)andPI(f : B)=0onB.

1.7 The Dirichlet Problem for a Ball 31

Lemma 1.7.14 (Herglotz [29]) If u is harmonic on B = B

y,ρ

and

n−1



∂B

y,δ

|u|dσ ≤ k<+∞ for all δ<ρ,

then there is a signed measure μ of bounded variation on ∂B such that

u = PI(μ : B) on B.

Proof: If δ<ρand x ∈ B

y,δ

,then

u(x)=



∂B

y,δ

−|x − y|

|z − x|

u(z) dσ(z).

If M is a Borel subset of B

−

y,ρ

and δ<ρ, deﬁne

(M)=



M∩∂B

y,δ

u(z) dσ(z).

Then μ

is concentrated on ∂B

y,δ

and

μ

 =



∂B

y,δ

|u(z)|dσ(z) ≤ kσ

n−1

≤ kσ

n−1

for all δ<ρ.Let{δ

} be a sequence of positive numbers such that δ

↑ ρ.By

Theorem 0.2.5, there is a subsequence of the sequence {μ

} that converges

toasignedmeasureμ in the w

∗

-topology with μ≤kσ

n−1

.Itcanbe

assumed that the sequence {μ

} converges to μ in the w

∗

-topology. Since the

signed measure μ

is concentrated on ∂B

y,δ

and δ

↑ ρ, μ is concentrated

on ∂B

y,ρ

.Consideranyx ∈ B. By dropping a ﬁnite number of terms, if

necessary, it can be assumed that |y − x| <δ

<ρfor all j ≥ 1. Then

u(x)=



∂B

y,δ

−|y − x|

|z − x|

dμ

(z).

Since the signed measures μ

are concentrated on the spherical shell {z; δ

≤

|z − x|≤ρ} and the sequence of integrands in the last equation converges

uniformly to (ρ

−|y −x|

)/|z −x|

on this shell, by Corollary 0.2.6,

u(x)=



∂B

y,ρ

−|y − x|

|z − x|

dμ(z).

Theorem 1.7.15 (Herglotz) The harmonic function u on B = B

y,ρ

is a

diﬀerence of two nonnegative harmonic functions if and only if there is a

signed measure μ of bounded variation on ∂B such that u = PI(μ : B).

Proof: Suppose u = u

− u

,whereu

and u

are nonnegative harmonic

functions on B.If0<δ<ρ,then

32 1 Laplace’s Equation

n−1



∂B

y,δ

|u|dσ ≤

n−1



∂B

y,δ

|dσ +

n−1



∂B

y,δ

|dσ

= u

(y)+u

(y) < +∞

and the necessity follows from Lemma 1.7.14. If u = PI(μ : B), where μ is

a signed measure of bounded variation on ∂B,thenμ = μ

−μ

−

,whereμ

and μ

−

are ﬁnite measures on ∂B,andsou = PI(μ

: B) −PI(μ

−

: B). The

suﬃciency follows from Theorem 1.7.2.

1.8 Kelvin Transformation

Consider a ball B

y,ρ

⊂ R

. The transformation x → x

∗

deﬁned by

∗

= y +

|y −x|

(x − y),x= y, (1.11)

was used in the derivation of the Poisson integral formula and is called the

Kelvin transformation or inversion with respect to ∂B

y,ρ

.Thepointx

∗

lying on the radial line joining y to x, is called the inverse of x relative to

∂B

y,ρ

. Aside from the derivation of the integral formula, the Kelvin transfor-

mation is useful for solving other problems.

The following symmetry property of the Kelvin transformation will be

used later in the chapter. If x, y ∈ B

−

z,ρ

,then

|x − z||x

∗

− y| = |y − z||x − y

∗

|. (1.12)

This equation will be derived assuming that z =0.Letγ denote the angle

between the line segments joining 0 to x and 0 to y.Sincex

∗

=(ρ

/|x|

and y

∗

=(ρ

/|y|

)y,

∗

− y|

|x|

+ |y|

− 2

|x|

|y|cos γ

|y|

|x|



|y|

+ |x|

− 2

|y|

|x|cos γ



|y|

|x|

∗

− x|

Thus, |x||x

∗

− y| = |y||y

∗

− x|.

As with any other transformation, the eﬀect of the transformation on

various geometric regions can be examined. Note ﬁrst that the transformation

maps planes or spheres into planes or spheres, but not necessarily respectively.

The calculations will be much simpler if it is assumed that y =0,inwhichcase

1.9 Poisson Integral for Half-space 33

∗

|x|

x and x =

∗

Starting with the general equation of a plane or sphere in R



i=1



i=1

+ c =0,

and making the substitution x

=(ρ

/|x

∗

, the equation

aρ

+ ρ



i=1

∗

+ c



i=1

∗2

is obtained, which is again the general equation of a plane or sphere. It is

easily seen that a plane outside the sphere ∂B

0,ρ

will map into a sphere inside

the sphere ∂B

0,ρ

which passes through the origin, and conversely, a sphere

inside the sphere ∂B

0,ρ

, which passes through the origin, will map onto a

plane outside the sphere ∂B

0,ρ

The eﬀect of the Kelvin transformation on functions will also be examined.

Let Ω be an open subset of R

∼{y} and let Ω

∗

be the image of Ω under

the map x → x

∗

.Iff

∗

is a function on Ω

∗

, the equation

f(x)=

n−2

∗

(y +

(x − y),

where r = |x − y| and r

n−2

=1ifn = 2, deﬁnes a function on Ω.The

mapping f

∗

→ f deﬁned in this way will be called the Kelvin operator.

Theorem 1.8.1 The Kelvin operator preserves positivity and harmonicity.

Proof: That positivity is preserved is obvious from the deﬁnition. The proof

that f is harmonic on Ω whenever f

∗

is harmonic on Ω

∗

is accomplished by

computing the Laplacian Δ

(x)

n−2

∗

(y +

(x − y));

the computation is straightforward but tedious.

1.9 Poisson Integral for Half-space

It was shown in the preceding section that a nonnegative harmonic function

u on a ball can be represented as the Poisson integral of a measure on the

boundary of the ball. A similar result holds for half-spaces. Throughout this

section Ω will be the half-space {(x

,...,x

); x

> 0} .

34 1 Laplace’s Equation

Theorem 1.9.1 If u is a nonnegative harmonic function on the open half-

space Ω = {(x

,...,x

); x

> 0} , then there is a nonnegative constant c and

a Borel measure μ on ∂Ω such that

u(x)=cx



∂Ω

|z − x|

dμ(z) for all x ∈ Ω.

Proof: Consider an inversion relative to ∂B

y,1

where y =(0,...,0, −1). The

image of Ω under this map is the open ball Ω

∗

= {z

∗

; |z

∗

−x

∗

| < 1/2} where

∗

=(0,...,0, −

). Let u

∗

be the image of u under the inversion. Then u

∗

a nonnegative, harmonic function on the ball Ω

∗

. By Theorem 1.7.15, there

is a Borel measure μ

∗

on ∂Ω

∗

such that

∗



∂Ω

∗

−|x

∗

− x

∗

− x

∗

dμ

∗

) for all x

∗

∈ Ω

∗

Let c =(2/σ

)μ

∗

({y}) ≥ 0andletμ

∗

= μ

∗

∂Ω

∗

∼{y}

.Then

∗

)=c

−|x

∗

− x

∗

|y −x

∗



∂Ω

∗

−|x

∗

− x

∗

− x

∗

dμ

∗

Letting θ denote the angle between the line segment joining x

∗

to y and the

-axis,

−|x

∗

− x

∗

= −|x

∗

− y|

+ |x

∗

− y|cos θ

by the law of cosines. Since |x

∗

− y||x − y| = 1 by deﬁnition of x

∗

−|x

∗

− x

∗

|x − y|

(−1+|x − y|cos θ).

Since |x − y|cos θ = x

+1,wherex

is the nth component of x,

−|x

∗

− x

∗

|x − y|

Letting φ be the angle between the line segment joining z

∗

to y,wherez ∈ ∂Ω,

and the line segment joining x

∗

to y,

∗

− x

∗

= |z

∗

− y|

+ |x

∗

− y|

− 2|z

∗

− y||x

∗

− y|cos φ

|z − y|

|x − y|

−

2cosφ

|z −y||x − y|

|z − y|

|x − y|

(|x − y|

+ |z − y|

− 2|z − y||x − y|cos φ)

|z − x|

|z − y|

|x − y|

1.9 Poisson Integral for Half-space 35

Therefore,

∗

)=cx

|x − y|

n−2

|x − y|

n−2



∂Ω

∗

|z − y|

|z − x|

dμ

∗

)

|x − y|

n−2

∗



y +

(x − y)

|x − y|



= cx



∂Ω

∗

|z − y|

|z − x|

dμ

∗

Note that the left side of this equation is just u(x). Denoting the map z

∗

→ z

by T , the measure μ

∗

on ∂Ω

∗

induces a measure μ

on ∂Ω by μ

= μ

∗

−1

Deﬁning μ(E)=



|z −y|

dμ

(z) for any Borel set E, the last equation can

be written

u(x)=cx



∂Ω

|z − x|

dμ(z). (1.13)

Note: There is no reason to believe that themeasureinthelastequationis

ﬁnite.

The right side of Equation (1.13) will be denoted by PI(c, μ, Ω)(x)andis

called the Poisson integral of the pair (c, μ) for the half-space Ω.Thenumber

c can be any real number and μ any signed measure. As before, if μ is the

indeﬁnite integral of a measurable function f on ∂Ω relative to Lebesgue

measure, let PI(c, f, Ω)=PI(c, μ, Ω); that is,

PI(c, f, Ω)(x)=cx



∂Ω

f(z)

|z −x|

dz, (1.14)

where x =(x

,...,x

)withx

> 0.

A partial converse of the preceding theorem for the half-space Ω will be

taken up now. It will be shown ﬁrst that



∂Ω

|z − x|

dz < +∞ (1.15)

whenever n ≥ 2andx ∈ Ω.Toseethis,let

x be the projection of x =

,...,x

)onto∂Ω and let z ∈ ∂Ω.Then|z − x|

= x

+ |z − x|

,and



∂Ω

|z − x|

dz ≤



∂Ω∩{|z−x|<1}

dz +



∂Ω∩{|z−x|≥1}

|z −x|

dz.

Since the ﬁrst integral on the right is ﬁnite, only the second need be consid-

ered. Transforming to spherical coordinates in the (n −1)-dimensional space

∂Ω relative to the pole



∂Ω∩{|z−x|≥1}

|z − x|

dz =



∞



|θ|=1

n−2

dθdr = σ

n−1

< +∞.

36 1 Laplace’s Equation

It will be shown now that PI(0, 1,Ω)=1onΩ. By Equation (1.15),

PI(0, 1,Ω)(x)=



∂Ω

|z − x|

is ﬁnite for each x ∈ Ω. Equation (1.15) can also be used to show that

PI(0, 1,Ω)isharmoniconΩ. The argument is straightforward, but tedious,

and involves justiﬁcation of diﬀerentiation under the integral. Note that the

above integral, as a function of x, is unaﬀected by adding to x a vector in ∂Ω.

Such an operation is a translation of x in a direction normal to the x

-axis.

In other words,



∂Ω

|z − x|

−n

dz is a function of x

only. Let

g(x

)=PI(0, 1,Ω)(x)=



∂Ω

|z − x|

dz, x

> 0.

Since g is harmonic on Ω,d

g/x

=0andg(x

)=ax

+ b. It will be shown

now that a =0.Since

g(x



∂Ω

(|z|

+ x

)

n/2



∂Ω

(|z/x

+1)

n/2







∂Ω

(|z|

+1)

n/2

dz,

g(x

) is a constant function and a = 0. By transforming to spherical co-

ordinates relative to the origin in ∂Ω, using the formula for the σ

Equation (0.1), and using standard integration techniques,

b = g(1) =



∂Ω

(|z|

+1)

n/2



+∞



|θ|=1

n−2

+1)

n/2

dσ(θ)dr

2σ

n−1



+∞

n−2

+1)

n/2

=1.

Therefore, g(x

) = 1 for all x

> 0; that is,

PI(0, 1,Ω)=



∂Ω

|z − x|

dz =1 forallx ∈ Ω. (1.16)

The relationship of the harmonic function determined by a boundary func-

tion to the boundary function itself will be considered now.

1.9 Poisson Integral for Half-space 37

Theorem 1.9.2 If f is a bounded measurable function on ∂Ω and c is any

real number, then u = PI(c, f, Ω) is harmonic on Ω.Iff is continuous at

∈ ∂Ω,thenlim

x→z

u(x)=f(z

); moreover, lim

→+∞

u(x)/x

= c.

Proof: The latter statement will be proved ﬁrst. It follows from

Equation (1.16) that

lim

→+∞



∂Ω

|z −x|

dz =0.

Returning to Equation (1.14) and using the fact that f is bounded,

lim

→+∞

u(x)

= c + lim

→+∞



∂Ω

f(z)

|z − x|

dz = c.

Suppose now that f is continuous at z

∈ ∂Ω and that |f (z)|≤M on ∂Ω.

To show that lim

x→z

u(x)=f(z

), it suﬃces to prove as in Lemma 1.7.4

that f (z) ≤ k in a neighborhood of z

implies that lim sup

x→z

u(x) ≤ k for

all z,wherek>0. Suppose f(z) ≤ k for all z ∈ ∂Ω ∩ B

,

. Since the term

in Equation (1.14) has the limit zero as x → z

, it suﬃces to consider

just the integral term. The integral will be written as a sum of two integrals

by splitting the boundary ∂Ω into the two regions ∂Ω ∩{|z − z

| <} and

∂Ω ∩{|z −z

|≥}.Sincef(z) ≤ k on ∂Ω ∩{|z − z

| <},



∂Ω∩{|z−z

|<}

f(z)

|z −x|

dz ≤ kPI(0, 1,Ω)=k.

Letting

x denote the projection of x onto ∂Ω,



∂Ω∩{|z−z

|≥}

f(z)

|z − x|

dz ≤

2Mx



∂Ω∩{|z−x|

≥}

|z − x|

dz.

Note that for z ∈ ∂Ω ∼ B

,

and x ∈ B

,/2

, |z − x|≥/2. Using spherical

coordinates relative to

x,forx ∈ B

,/2



∂Ω∩{|z−z

|≥}

|z − x|

dz ≤



∂Ω∩{|z−x|≥/2}

|z − x|

≤



|θ|=1



+∞

/2

n−2

dr dθ

2σ

n−1



Therefore,

lim sup

x→z



∂Ω∩{|z−z

|≥}

f(z)

|z −x|

dz ≤ lim sup

x→z

4Mx

n−1



=0.