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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
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