6

6

3 Laplace’s Equation

We now turn to studying Laplace’s equation

Δu=0

and its inhomogeneous version, Poisson’s equation,

¡Δu=f:

We say a function u satisfying Laplace’s equation is a harmonic function.

3.1 The Fundamental Solution

nConsider Laplace’s equation inR ,

nΔu=0 x2R :

Clearly, there are a lot of functions u which satisfy this equation. In particular, any

constantfunctionisharmonic. Inaddition,anyfunctionoftheform u(x)=a x +:::+a x1 1 n n

for constants a is also a solution. Of course, we can list a number of others. Here, however,i

we are interested in ﬁnding a particular solution of Laplace’s equation which will allow us

to solve Poisson’s equation.

Given the symmetric nature of Laplace’s equation, we look for a radial solution. That

nis, we look for a harmonic function u onR such that u(x) = v(jxj). In addition, to being

a natural choice due to the symmetry of Laplace’s equation, radial solutions are natural to

lookforbecausetheyreduceaPDEtoanODE,whichisgenerallyeasiertosolve. Therefore,

we look for a radial solution.

If u(x)=v(jxj), then

xi 0u = v(jxj) jxj=0;xi jxj

which implies

2 21 x xi i0 0 00u = v(jxj)¡ v(jxj)+ v (jxj) jxj=0:x xi i 3 2jxj jxj jxj

Therefore,

n¡1 0 00Δu= v(jxj)+v (jxj):

jxj

Letting r =jxj, we see that u(x) = v(jxj) is a radial solution of Laplace’s equation implies

v satisﬁes

n¡1 0 00v(r)+v (r)=0:

r

Therefore,

1¡n00 0v = v

r

00v 1¡n

=) =

0v r

0=) lnv =(1¡n)lnr+C

C0=) v(r)= ;

n¡1r

1R

R

R

6

R

6

6

which implies ‰

c lnr+c n=21 2

v(r)= c1 +c n‚3:n¡2 2(2¡n)r

From these calculations, we see that for any constants c ;c , the function1 2

‰

c lnjxj+c n=21 2

u(x)· (3.1)c1 +c n‚3:n¡2 2(2¡n)jxj

n nfor x 2 R , jxj = 0 is a solution of Laplace’s equation in R ¡f0g. We notice that the

function u deﬁned in (3.1) satisﬁes Δu(x) = 0 for x = 0, but at x = 0, Δu(0) is undeﬁned.

We claim that we can choose constants c and c appropriately so that1 2

¡Δ u=–x 0

in the sense of distributions. Recall that – is the distribution which is deﬁned as follows.0

For all `2D,

(– ;`)=`(0):0

Below,wewillprovethisclaim. Fornow,though,assumewecanprovethis. Thatis,assume

we can ﬁnd constants c ;c such that u deﬁned in (3.1) satisﬁes1 2

¡Δ u=– : (3.2)x 0

Let Φ denote the solution of (3.2). Then, deﬁne

Z

v(x)= Φ(x¡y)f(y)dy:

n

Formally, we compute the Laplacian of v as follows,

Z

¡Δ v =¡ Δ Φ(x¡y)f(y)dyx x

n

Z

=¡ Δ Φ(x¡y)f(y)dyy

n

Z

= – f(y)dy =f(x):x

n

Thatis,v isasolutionofPoisson’sequation! Ofcourse,thissetofequalitiesaboveisentirely

formal. We have not proven anything yet. However, we have motivated a solution formula

for Poisson’s equation from a solution to (3.2). We now return to using the radial solution

(3.1) to ﬁnd a solution of (3.2).

Deﬁne the function Φ as follows. Forjxj=0, let

‰

1¡ lnjxj n=2

2…Φ(x)= (3.3)1 1 n‚3;n¡2n(n¡2)ﬁ(n)jxj

nwhere ﬁ(n) is the volume of the unit ball inR . We see that Φ satisﬁes Laplace’s equation

nonR ¡f0g. As we will show in the following claim, Φ satisﬁes¡Δ Φ=– . For this reason,x 0

we call Φ the fundamental solution of Laplace’s equation.

2R

R

R

R

R

Claim 1. For Φ deﬁned in (3.3), Φ satisﬁes

¡Δ Φ=–x 0

in the sense of distributions. That is, for all g2D,

Z

¡ Φ(x)Δ g(x)dx=g(0):x

n

Proof. Let F be the distribution associated with the fundamental solution Φ. That is, letΦ

F :D!R be deﬁned such thatΦ

Z

(F ;g)= Φ(x)g(x)dxΦ

n

for all g2D. Recall that the derivative of a distribution F is deﬁned as the distribution G

such that

0(G;g)=¡(F;g)

for all g2D. Therefore, the distributional Laplacian of Φ is deﬁned as the distribution FΔΦ

such that

(F ;g)=(F ;Δg)ΔΦ Φ

for all g2D. We will show that

(F ;Δg)=¡(– ;g)=¡g(0);Φ 0

and, therefore,

(F ;g)=¡g(0);ΔΦ

which means¡Δ Φ=– in the sense of distributions.x 0

By deﬁnition, Z

(F ;Δg)= Φ(x)Δg(x)dx:Φ

n

Now, we would like to apply the divergence theorem, but Φ has a singularity at x = 0. We

get around this, by breaking up the integral into two pieces: one piece consisting of the ball

of radius – about the origin, B(0;–) and the other piece consisting of the complement of this

nball inR . Therefore, we have

Z

(F ;Δg)= Φ(x)Δg(x)dxΦ

n

Z Z

= Φ(x)Δg(x)dx+ Φ(x)Δg(x)dx

nB(0;–) ¡B(0;–)

=I +J:

3R

R

R

R

R

R

We look ﬁrst at term I. For n=2, term I is bounded as follows,

ﬂ ﬂ ﬂ ﬂZ Z

ﬂ ﬂ ﬂ ﬂ1ﬂ ﬂ ﬂ ﬂ1¡ lnjxjΔg(x)dx •CjΔgj lnjxjdxLﬂ ﬂ ﬂ ﬂ2…B(0;–) B(0;–)

ﬂ ﬂZ Z2… –ﬂ ﬂ

ﬂ ﬂ•C lnjrjrdrd ﬂ ﬂ

0 0

ﬂ ﬂZ –ﬂ ﬂ

ﬂ ﬂ•C lnjrjrdrﬂ ﬂ

0

2•Clnj–j– :

For n‚3, term I is bounded as follows,

ﬂ ﬂZ Z

ﬂ ﬂ1 1 1

ﬂ ﬂ 1Δg(x)dx •CjΔgj dxLﬂ n¡2 ﬂ n¡2n(n¡2)ﬁ(n)jxj jxjB(0;–) B(0;–)

µ ¶Z Z– 1

•C dS(y) dr

n¡2jyj0 @B(0;r)

Z µZ ¶– 1

= dS(y) dr

n¡2r0 @B(0;r)

Z – 1 n¡1= nﬁ(n)r dr

n¡2r0

Z – nﬁ(n) 2=nﬁ(n) rdr = – :

20

+Therefore, as –!0 ,jIj!0.

Next, we look at term J. Applying the divergence theorem, we have

Z Z Z

@Φ

Φ(x)Δ g(x)dx= Δ Φ(x)g(x)dx¡ g(x)dS(x)x x

@”n n n¡B(0;–) ¡B(0;–) @( ¡B(0;–))

Z

@g

+ Φ(x) dS(x)

@”n@( ¡B(0;–))

Z Z

@Φ @g

=¡ g(x)dS(x)+ Φ(x) dS(x)

n @” n @”@( ¡B(0;–)) @( ¡B(0;–))

·J1+J2:

nusing the fact that Δ Φ(x)=0 for x2R ¡B(0;–).x

We ﬁrst look at term J1. Now, by assumption, g 2 D, and, therefore, g vanishes at

1. Consequently, we only need to calculate the integral over @B(0;†) where the normal

nderivative ” is the outer normal toR ¡B(0;–). By a straightforward calculation, we see

that

x

r Φ(x)=¡ :x nnﬁ(n)jxj

nThe outer unit normal toR ¡B(0;–) on B(0;–) is given by

x

” =¡ :

jxj

4Therefore, the normal derivative of Φ on B(0;–) is given by

µ ¶ µ ¶

@Φ x x 1

= ¡ ¢ ¡ = :

n n¡1@” nﬁ(n)jxj jxj nﬁ(n)jxj

Therefore, J1 can be written as

Z Z Z

1 1

¡ g(x)dS(x)=¡ g(x)dS(x)=¡¡ g(x)dS(x):

n¡1 n¡1nﬁ(n)jxj nﬁ(n)–@B(0;–) @B(0;–) @B(0;–)

Now if g is a continuous function, then

Z

¡¡ g(x)dS(x)!¡g(0) as –!0:

Lastly,welookattermJ2. Nowusingthefactthatgvanishesasjxj!+1,weonlyneed

tointegrateover@B(0;–). Usingthefactthatg2D, and, therefore, inﬁnitelydiﬀerentiable,

we have

ﬂ ﬂ ﬂ ﬂZ Z

ﬂ ﬂ ﬂ ﬂ@g @gﬂ ﬂ ﬂ ﬂΦ(x) dS(x) • jΦ(x)jdS(x)ﬂ ﬂ ﬂ ﬂ@” @” 1@B(0;–) @B(0;–)L (@B(0;–))

Z

•C jΦ(x)jdS(x):

@B(0;–)

Now ﬁrst, for n=2,

Z Z

jΦ(x)jdS(x)=C jlnjxjjdS(x)

@B(0;–) @B(0;–)

Z

•Cjlnj–jj dS(x)

@B(0;–)

=Cjlnj–jj(2…–)•C–jlnj–jj:

Next, for n‚3,

Z Z

1

jΦ(x)jdS(x)=C dS(x)

n¡2jxj@B(0;–) @B(0;–)

Z

C

• dS(x)

n¡2– @B(0;–)

C n¡1= nﬁ(n)– •C–:

n¡2–

Therefore, we conclude that term J2 is bounded in absolute value by

C–jln–j n=2

C– n‚3:

+Therefore,jJ2j!0 as –!0 .

5R

R

R

R

R

R

Combining these estimates, we see that

Z

Φ(x)Δ g(x)dx= lim I +J1+J2=¡g(0):x

+n –!0

Therefore, our claim is proved.

Solving Poisson’s Equation. We now return to solving Poisson’s equation

n¡Δu=f x2R :

From our discussion before the above claim, we expect the function

Z

v(x)· Φ(x¡y)f(y)dy

n

to give us a solution of Poisson’s equation. We now prove that this is in fact true. First, we

make a remark.

Remark. If we hope that the function v deﬁned above solves Poisson’s equation, we must

ﬁrst verify that this integral actually converges. If we assume f has compact support on

nsome bounded set K inR , then we see that

Z Z

Φ(x¡y)f(y)dy•jfj 1 jΦ(x¡y)jdy:L

n K

If we additionally assume that f is bounded, then jfj 1 • C. It is left as an exercise toL

verify that Z

jΦ(x¡y)jdy <+1

K

on any compact set K.

2 nTheorem 2. Assume f 2C (R ) and has compact support. Let

Z

u(x)· Φ(x¡y)f(y)dy

n

where Φ is the fundamental solution of Laplace’s equation (3.3). Then

2 n1. u2C (R )

n2. ¡Δu=f inR .

Ref: Evans, p. 23.

Proof. 1. By a change of variables, we write

Z Z

u(x)= Φ(x¡y)f(y)dy = Φ(y)f(x¡y)dy:

n n

6R

R

R

R

R

Let

e =(:::;0;1;0;:::)i

n thbe the unit vector inR with a 1 in the i slot. Then

Z • ‚

u(x+he )¡u(x) f(x+he ¡y)¡f(x¡y)i i

= Φ(y) dy:

h n h

2Now f 2C implies

f(x+he ¡y)¡f(x¡y) @fi

! (x¡y) as h!0

h @xi

nuniformly onR . Therefore,

Z

@u @f

(x)= Φ(y) (x¡y)dy:

@x n @xi i

Similarly, Z

2 2@ u @ f

(x)= Φ(y) (x¡y)dy:

@xx n @xxi j i j

This function is continuous because the right-hand side is continuous.

2. By the above calculations and Claim 1, we see that

Z

Δ u(x)= Φ(y)Δ f(x¡y)dyx x

n

Z

= Φ(y)Δ f(x¡y)dyy

n

=¡f(x):

3.2 Properties of Harmonic Functions

3.2.1 Mean Value Property

In this section, we prove a mean value property which all harmonic functions satisfy. First,

we give some deﬁnitions. Let

nB(x;r)= ball of radius r about x inR

n@B(x;r)= boundary of ball of radius r about x inR

nﬁ(n)= volume of unit ball inR

nnﬁ(n)= surface area of unit ball inR :

For a function u deﬁned on B(x;r), the average of u on B(x;r) is given by

Z Z

1

¡ u(y)dy = u(y)dy:

nﬁ(n)rB(x;r) B(x;r)

7For a function u deﬁned on @B(x;r), the average of u on @B(x;r) is given by

Z Z

1

¡ u(y)dS(y)= u(y)dS(y):

n¡1nﬁ(n)r@B(x;r) @B(x;r)

n 2Theorem 3. (Mean-Value Formulas) Let Ω‰R . If u2C (Ω) is harmonic, then

Z Z

u(x)=¡ u(y)dS(y)=¡ u(y)dy

@B(x;r) B(x;r)

for every ball B(x;r)‰Ω.

2Proof. Assume u2C (Ω) is harmonic. For r >0, deﬁne

Z

`(r)=¡ u(y)dS(y):

@B(x;r)

For r = 0, deﬁne `(r) = u(x). Notice that if u is a smooth function, then lim +`(r) =r!0

0u(x), and, therefore, ` is a continuous function. Therefore, if we can show that `(r) = 0,

then we can conclude that ` is a constant function, and, therefore,

Z

u(x)=¡ u(y)dS(y):

@B(x;r)

0We prove `(r)=0 as follows. First, making a change of variables, we have

Z

`(r)=¡ u(y)dS(y)

@B(x;r)

Z

=¡ u(x+rz)dS(z):

@B(0;1)

Therefore,

Z

0`(r)=¡ ru(x+rz)¢zdS(z)

@B(0;1)

Z

y¡x

=¡ ru(y)¢ dS(y)

r@B(x;r)

Z

@u

=¡ (y)dS(y)

@”@B(x;r)

Z

1 @u

= (y)dS(y)

n¡1nﬁ(n)r @”@B(x;r)

Z

1

= r¢(ru)dy (by the Divergence Theorem)

n¡1nﬁ(n)r B(x;r)

Z

1

= Δu(y)dy =0;

n¡1nﬁ(n)r B(x;r)

8using the fact that u is harmonic. Therefore, we have proven the ﬁrst part of the theorem.

It remains to prove that Z

u(x)=¡ u(y)dy:

B(x;r)

We do so as follows, using the ﬁrst result,

Z Z µZ ¶r

u(y)dy = u(y)dS(y) ds

B(x;r) 0 @B(x;s)

Z µ Z ¶

r

n¡1= nﬁ(n)s ¡ u(y)dS(y) ds

0 @B(x;s)

Z r

n¡1= nﬁ(n)s u(x)ds

0 Z r

n¡1=nﬁ(n)u(x) s ds

0

s=rn=ﬁ(n)u(x)s j

s=0

n=ﬁ(n)u(x)r :

Therefore, Z

nu(y)dy =ﬁ(n)r u(x);

B(x;r)

which implies Z Z

1

u(x)= u(y)dy =¡ u(y)dy;

nﬁ(n)r B(x;r) B(x;r)

as claimed.

3.2.2 Converse to Mean Value Property

In this section, we prove that if a smooth function u satisﬁes the mean value property

described above, then u must be harmonic.

2Theorem 4. If u2C (Ω) satisﬁes

Z

u(x)=¡ u(y)dS(y)

@B(x;r)

for all B(x;r)‰Ω, then u is harmonic.

Proof. Let Z

`(r)=¡ u(y)dS(y):

@B(x;r)

If Z

u(x)=¡ u(y)dS(y)

@B(x;r)

90for all B(x;r)‰Ω, then `(r)=0. As described in the previous theorem,

Z

r

0`(r)= ¡ Δu(y)dy:

n B(x;r)

Suppose u is not harmonic. Then there exists some ball B(x;r)‰ Ω such that Δu > 0 or

Δu<0. Without loss of generality, we assume there is some ball B(x;r) such that Δu>0.

Therefore, Z

r0`(r)= ¡ Δu(y)dy >0;

n B(x;r)

0which contradicts the fact that `(r)=0. Therefore, u must be harmonic.

3.2.3 Maximum Principle

nIn this section, we prove that if u is a harmonic function on a bounded domain Ω in R ,

then u attains its maximum value on the boundary of Ω.

n 2Theorem5. Suppose Ω‰R is open and bounded. Suppose u2C (Ω)\C(Ω) is harmonic.

Then

1. (Maximum principle)

maxu(x)=maxu(x):

@ΩΩ

2. (Strong maximum principle) If Ω is connected and there exists a point x 2 Ω such0

that

u(x )=maxu(x);0

Ω

then u is constant within Ω.

Proof. We prove the second assertion. The ﬁrst follows from the second. Suppose there

exists a point x in Ω such that0

u(x )=M =maxu(x):0

Ω

Then for 0<r < dist(x ;@Ω), the mean value property says0

Z

M =u(x )=¡ u(y)dy•M:0

B(x ;r)0

But, therefore, Z

¡ u(y)dy =M;

B(x ;r)0

and M = max u(x). Therefore, u(y)· M for y 2 B(x ;r). To prove u· M throughout0Ω

Ω, you continue with this argument, ﬁlling Ω with balls.

10