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# Chapter0: A Preview

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• cours - matière potentielle : linear algebra
Chapter 0 Preview 1 Chapter 0: A Preview Pythagorean Triples As an introduction to the sorts of questions that we will be studying, let us con- sider right triangles whose sides all have integer lengths. The most familiar example is the (3,4,5) right triangle, but there are many others as well, such as the (5,12,13) right triangle. Thus we are looking for triples (a, b, c) of positive integers such that a2 + b2 = c2 .
• primitive triples by arbitrary numbers
• rational points on the unit circle
• products of earlier primes
• pythagorean triples
• integer solutions
• rational points
• numbers
• circle
• equation
• list

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##### Arbitrariness

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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
(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(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.
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
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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