Analytic Methods of PDEs with Prof. Weinstein

Leonardo Abbrescia

June 3, 2014

Overview of the Course

We will begin with first order nonlinear PDEs, Laplace’s equation, the Heat equation, the Wave equa-tion, and the Schrodinger equation. In particular, we will talk about Method of characteristics, Potentialtheory (Fundamental solutions, Green’s functions, etc), Maximum principles, Weak and strong solutions,a priori estimates of solutions, Weave phenomena, Energy methods, Variational methods (Euler-Lagrangeequation, minimization, convexity).

Deriving the Heat Equation from Brownian Motion

First Class

Lets review some elementary discrete probability. Let X be a random variable that takes on values a1, . . . , anwith probabilities P (X = a1), . . . P (X = an). We will define the expected value

〈X〉 = E[x] :=

n∑j=1

aiP (X = aj).

ObviouslyE[X + Y ] = E[x] + E[Y ].

We say a random variable X and Y are independent if

P (X = a,X = b) = P (X = a)P (Y = b).

This then implies that E[X ·Y ] = E[X]E[Y ] if X and Y are independent. We define the variance of a randomvariable to be

V (X) := E[(X − E[X])2

]= 〈(X − 〈X〉)2〉.

We can also get the following identities

V (X + Y ) = E([X + Y − E(X + Y )]2

)= [(X − 〈X〉) + (〈Y − 〈Y 〉〉)]2

= (X−〉X〉)2+ (Y − 〈Y 〉)2

+ 2(X − 〈X〉)(Y − 〈Y 〉)= V (X) + V (Y ) + 2 cov(X,Y ).

Now if X and Y are independent we finally get V (X + Y ) = V (X) + V (Y ). This covers what we needfrom discrete probability. Now consider a random walk on Zδ where δ > 0. Let a particle be stationedat the origin and we say that it jumps left or right a distance δ. For i = 1, 2, . . . we define the randomvariable xi = δ if at ith step the particle jumps right, or xi = −δ if it jumps left. We give it the probabilities

1

P (xi = δ) = p and P (xi = −δ) = q = 1− p. After n steps, the location will be Xn =∑xn. We look at the

mean position :

E[Xn] =

n∑j=1

E[xi] (1)

whereE[xi] = δP (xi = δ)− δP (xi = −δ) = δ(p− q).

Then we can plug this into (1) to getE[Xn] = nδ(p− q). (2)

Now we look at the variance:

V (Xn) =

n∑i=1

V (xi) (3)

where an easy exercise can tell you that

V (xi) = E(x2i )− E(xi)

2

= δ2p+ (−δ)2q − (δ(p− q))2

= δ2(1− (p− q)2

)= 4pqδ2

We can plug this into (3) to get

V (Xn) = 4npqδ2 (4)

Now lets move on to Brownian motion and diffusion. When Brown looked into a microscope he saw a particlegoing through a jittery irregular motion. Lets call the mean displacement per unit time C. Once you havethese measurements you can calculate the variance D > 0. The hypothesis is that this motion is due torandom collisions with much smaller particles. Assume that there are r collisions per unit time. We’d liketo use our random walk to model this. Then rδ(p− q) ≈ C and 4rδ2pq ≈ D > 0. We’d like to study this inthe limit as δ → 0 and r →∞. Now assume p− q 6= 0 (this motion is not symmetric). Then from the firstrelation

δr =C

p− qand

D = 4δpq(δr)

= 4pqδC

p− q.

We intuitively see that in order for this to be true we’d need p− q = O(δ) ≈ bδ. We also have p+ q = 1 =⇒2p ≈ 1 + bδ, 2q = 1− bδ =⇒ p = 1

2 (1 + bδ) and q = 12 (1− bδ). Hence we get

p− q = bδ

(p− q)δr = (bδ)δr = bδ2r = C

4pqδ2r = D

41

4(1− b2δ2)δ2r = D

δ2r = D

bD = C

b =C

D

2

Where C and D were measured in the lab. Then we have p = 12 (1 + C

D δ) and q = 12 (1 − C

D δ). Now thequestion is to find the probability that a particle at x = 0 at time t = 0 is at position x at time t = nτ whereτ = 1

r . Hence we get

V (x, t) = P (Xn = x, t = nτ)

V (x, t+ τ) = P (Xn+1 = x, t+ τ = nτ + τ)

= P (Xn = x+ δ at time t = nτ , variable jumps left) + P (Xn = x− δ at time t = nτ , variable jumps right)

= P (Xn = x+ δ)P (jump left) + P (Xn − δ)P (jump right)

= pv(x− δ, t) + qv(x+ δ, t).

Then this gives us a difference equations. We want to look at the formal limit as δ → 0, τ → 0(r → ∞).Then we get

v(x, t+ τ) = v(x, t) + vt(x, t)τ +O(τ2)

v(x+ δ, t) = v(x, t) + vx(x, t)δ +1

2vxx(x, t)δ2 + P (δ3)

v(x− δ, t) = v(x, t)− vx(x, t)δ +1

2vxx(x, t)(−δ)2 + P (δ3)

We plug this into our difference equation to get

v(x, t) + vt(x, t)τ +O(τ2) = p(v(x, t)− vx(x, t)δ +1

2vxx(x, t)(−δ)2 + P (δ3)) + q(v(x, t)+

vx(x, t)δ +1

2vxx(x, t)δ2 + P (δ3)) + vt(x, t) +O(δ)

=δ(q − p)

τvx(x, t) +

δ2(p+ q)

2τvxx(x, t) +O(δ3/τ)

= −CD

δ2

τvx(x, t) +

δ2

2τvxx(x, t) +O(δ3/τ)

Taking δ → 0, τ → 0 gives us

vt(x, t) = −Cvx(x, t) +1

2vxx(x, t). (5)

This is the diffusion equation with diffusion coefficient D and drift coefficient C (here we used δ2/τ → D).When C = 0 and p = q = 1

2 we get out usual heat equation

vt =D

2vxx (6)

e and when D = 0, p 6= q we get the usual transport equation

vt + Cvx = 0. (7)

As a quick remark, this object is supposed to be a probability soˆ ∞−∞

v(x, t)dx = 1.

What we do is make initial values v(x, 0) with initial probability distribution. In fact we can show that if´Rn v0(x) dx = 1 (which is definitely true), then

´Rn v(x, t)dx = 1 by taking ∂/∂t.

We began our discussion of our random walk in a discrete space. In the case of PDEs, we will considerΩ to be a bounded open subset of Rn. Let ∂Ω denote the boundary of Ω and let Γ ⊂ ∂Ω. Then obviouslywe have ∂Ω = Γ ∪ (∂Ω − Γ). Then a usual boundary problem we have is ∆u(x) = 0 for x ∈ Ω, u = 1 on Γand u = 0 on ∂\Γ. Then the answer to this boundary value problem would be to examine brownian motionin Ω and consider the probability when it exits Γ.

3

1 Laplace’s Equation and Boundary Integral Methods

1.1 First Class

Lets first review some basic calculus.

Theorem 1.1 (Divergence Theorem). Let Ω ⊂ Rn be a domain (bounded open set) and ∂Ω is C∞. LetF : Ω→ Rn be a smooth vector field i.e., (x1, . . . , xn) 7→ (F1(x1, . . . , xn), · · · , Fn(x1, . . . , xn)). Denote n(x)to be the outward unit normal vector field to ∂Ω. Then

ˆΩ

∇ · Fdx =

ˆ∂Ω

F · ndS.

Let u, v ∈ C2. Then of course we know that ∆u =∑j ∂

2u/∂x2j . Then we obviously we have u∆v =

∇· (u∇v)−∇u ·∇v. Of course this is symmetric with respect to u and v. We can then subtract u∆v− v∆uand find

u∆v − v∆u = ∇ · (u∇v − v∇u) .

By the divergence theorem we can then find

ˆΩ

(u∆v − v∆u)dx =

ˆ∂Ω

(u∇v − v∇u) · ndS

=

ˆ∂Ω

(u∂v

∂n− v ∂u

∂n

)dS.

As an exercise we can show that ˆΩ

fxjdx =

ˆ∂Ω

fnjdSx.

In fact this is immediate if we think about it a little bit. Now lets go over some small notation. We willwrite ωn as the surface area of Sn−1 in Rn. Since surface area scales to the codimension of the sphere wehave that the surface area of a sphere of radius R is going to be ωnR

n−1. To find the volume of this ball weintegrate

V =

ˆ r

0

Area(|x| = s)ds

=

ˆ r

0

ωnsn−1ds

=ωnnrn.

The reader should familiarize themselves with the usual Lp and L∞ norms, convolutions of functions f ∗ g.Also recall Young’s inequality that if for f ∈ L1(Rn) and g ∈ Lp(Rn). Then

‖f ∗ g‖p ≤ ‖f‖1‖g‖p.

The proof is left as an exercise. Now lets look at some estimates of integrals.

1. Let f > 0, g > 0 and bound gmin ≤ g ≤ gmax. Then

gmin

ˆΩ

f ≤ˆ

Ω

f · g ≤ gmax

ˆΩ

f.

Letting g ≡ 1 we get

fmin Vol(Ω) ≤ˆ

Ω

f ≤ fmax Vol(Ω).

4

2. Recall the Cauchy-Schwarz inequality:∣∣∣∣ˆΩ

f · g∣∣∣∣ ≤ (ˆ

Ω

|f |2) 1

2(ˆ

Ω

|g|2) 1

2

= ‖f‖2‖g‖2.

Letting g ≡ 1 we get the bound ∣∣∣∣ˆΩ

fdx

∣∣∣∣ ≤ (Vol(Ω))12 ‖f‖2.

Now we introduce some notation. Let Ω be a region and f be a continuous function over it. Then its averageover Ω is defined as

fΩ :=1

|Ω|

ˆfdx =

fdx.

Now we can finally go to the topic of question: the Laplacian. The goal for today: Let f ∈ C20 (Rn)

and we want to solve ∆u = f . i.e. u = ∆−1f . Lets first start by finding a “fundamental solution” to thelaplacian. We seek rotationally symmetric solutions of ∆u = 0 i.e. u(x) = v(|x|). Let r = |x|. Then we caneasily see that

∆u = v′′(r) +n− 1

rv′(r) = 0. (8)

This is just a simple exercise in the chain rule. Well, this is an ODE. Writing w = v′ we have

w′ =1− nn

w.

We can solve this ODE to get log |w| = log r1−n + c i.e. v′ = w = br1−n. Solving for v(r) one can check thatthis gives us the solution

v(r) =

br2−n + c n ≥ 3b log r + c n = 2.

Note that if u is a solution to ∆u = 0 then a simple exercise will show that u(x−x0) is also a solution. Nowlet f ∈ C2

0 (Rn) and we want to construct a solution to ∆u = f . Now we define the Newtonian potential tobe

Φ(x) =

1(2−n)ωn

|x|2−n n ≥ 31

2π log |x| n = 2.

Note that Φ ∈ C∞(Rn − 0). Also by construction it is trivial to see that ∆Φ = 0 on x ∈ Rn − 0.

Theorem 1.2. Let u(x) := (Φ ∗ f)(x) for f ∈ C20 (Rn). Then u ∈ C2(Rn) and ∆u = f .

We will call this the Fundamental Theorem of Potential Theory. Some physicists say that “∆Φ(x− y) =δ(x− y).” Note that the left hand side doesn’t make sense when x = y. We will prove this next time.

1.2 Second Class

We will be interested in boundary valued problems of the following sort. Let Ω ⊂⊂ Rn be a bounded opensubset of Rn. Let ∂Ω be smooth i.e. it is locally the graph of a C∞ function. We will not work in situationswhere the boundaries have corners. We are interested in solving

∆u = f(x) x ∈ Ωu = g(x) x ∈ ∂Ω.

Another boundary value problem will be the Neumann problem, where ∂u∂n (x) = g for x ∈ ∂Ω. Here we have

∂u∂n := ∇u · n. One way to see this problems is to say what sense we take the following limit for x ∈ Ω

limx→x0

∇u(x) · n(x0).

5

The first step in solving boundary valued problems is seeing how to solve the equation ∆u(x) = f(x) forx ∈ Rn. Let f ∈ C2

0 (Ω) mean f has compact support and its partial derivatives up to order 2 are continuous.The way we will solve this problem is via the Newtonian potential:

Φ(x) :=

1(2−n)ωn

|x|2−n n ≥ 31

2π log |x| n = 2.

where ωn is defined to be the surface area of Sn−1.

Theorem 1.3 (Fundamental Theorem of Potential Theory). Let u(x) = (Φ ∗ f)(x) where

Φ ∗ f(x) :=

ˆRn

Φ(x− y)f(y)dy.

Then we have that u ∈ C2(Rn) and ∆u = f .

Proof. We will only consider the case where n ≥ 3. Notice that Φ ∈ C∞(Rn\0) and ∆Φ = 0 by construction.First lets note that u(x) is well defined: since f has compact support then |f(y)| = 0 if |y| ≥ R for some Rand so

|u(x)| =

∣∣∣∣∣ˆ|y|≤R

Φ(x− y)f(y)dy

∣∣∣∣∣≤

∣∣∣∣∣ˆ|y|≤R

1

|x− y|n−2dy

∣∣∣∣∣ max|z|≤R

|f(z)|. (9)

Then we can see that u is going to be well-defined if and only if the integral in (9) is bounded. Lets see whythis is true: for ρ small enough, y||x− y| ≤ ρ ⊂ y||y| ≤ R. Then

ˆ|y|≤R

1

|x− y|n−2dy ≤

ˆ|x−y|ρ

1

|x− y|n−2dy +

ˆ|y−x|>ρ

1

|x− y|n−2dy (10)

≤ˆSn−1

dΩ

ˆ ρ

0

1

|z|n−2|z|n−1d|z|+ C

≤ C

where we have used polar coordinates to bound the first integral by Cρ2 and the fact that the second integralin the RHS of (10) is bounded. This finishes the proof of u being well defined. The fact that u ∈ C2(Rn) isdone by difference quotients. We know that the partial derivative ∂iu is defined if

limh→0

u(x+ hei)− u(x)

h

is well defined. Note also that u =´

Φ(z)f(x− z)dz by change of variables, and so

u(x+ hei)− u(x)

h=

ˆRn

Φ(z)f(x+ hei)− f(x)

hdz.

It is left as an exercise to show that we can pass through the limit, and to show (very similarly) that thesecond derivatives also exist. Now lets show that ∆u(x) = f(x). From the exercise, we will see that

∂2u(x)

∂xi∂xj=

ˆRn

Φ(x)∂2f

∂xi∂xjdz

and view the Laplacian as

∆u(x) =

n∑i=1

∂2u(x)

∂x2i

=

ˆRn

Φ(z)∆xf(x− z)dz ?= f(x).

6

Let ε > 0 be a fixed constant and we write

∆u(x) =

ˆ|z|<ε

Φ(x)∆xf(x− z)dz +

ˆ|z|≥e

Φ(z)∆xf(x− z)dz

= Iε + Jε

We will bound I1. Recall that f has compact support and we showed∣∣∣∣∣ˆ|z|<ε

Φ(z)f(x− z)dz

∣∣∣∣∣ ≤ Cε2and so Iε will be bounded by Cε2 max |∆f | which will go to 0 when we take the limit. Now lets look atJε and note that there is no longer a singularity. Lets recall the divergence theorem (Gauss’ theorem) andrecall the following identity

u∆v = v∆u+∇ · (u∇v − v∇u)

and notice that this is the situation we are in! So we will write Jε as follows (where we use the fact that∆x = ∆z and that f vanishes outside some ball of radius Rx, and divergence theorem):

Jε =

ˆε≤|z|≤Rx

f(x− z)∆zΦ(z) +∇ · (Φ(z)∇zf(x− z)− f(x− z)∇zΦ(z)) dz

=

ˆε≤|z|≤Rx

∇ · (Φ(z)∇zf(x− z)− f(x− z)∇zΦ(z)) dz

=

ˆ|z|=ε

(Φ(z)∇zf(x− z)− f(x− z)∇zΦ(z)) · nzdS +

ˆ|z|=Rx

(Φ(z)∇zf(x− z)− f(x− z)∇zΦ(z)) · nzdS

=

ˆ|z|=ε

(Φ(z)∇zf(x− z)− f(x− z)∇zΦ(z)) · nzdS.

And so we see that

∆u(x) = Iε +

ˆ|z|=ε

Φ(z)∇xf(z) · nzdS −ˆ|z|=ε

f(x− z)∇zΦ(z) · nzdS

= Iε + J1ε + J2

ε .

We then bound J1ε as follows

|J1ε | ≤

ˆ|z|=ε

|Φ(z)||∇zf(x− z) · nz|dS

≤ˆ|z|=ε

1

(2− n)ωn

1

|z|n−2|∇zf(x− z)|dS

≤ Cn max|y|≤R

|∇f |ˆ|z|=ε

1

|z|n−2dS

= Cn‖∇f‖∞ε2−nˆ|z|=ε

1dS = Cn‖∇f‖∞ε2−nωnεn−1

= Cn‖∇f‖∞ε

7

Now we have bounded Iε and J1ε . Finally lets examine J2

ε and notice that

J2ε = −

ˆ|z|=ε

f(x− z)∇zΦ(z) · nzdS

=

ˆ|z|=ε

f(x− z)∇zΦ(z) · z|z|dS

=

ˆ|z|=ε

f(x− z)∇z(

1

ωn(2− n)|z|2−n

)z

|z|dS

=

ˆ|z|=ε

f(x− z)(

1

ωn|z|1−n z

|z|

)z

|z|dS

=1

ωnεn−1

ˆ|z|=ε

f(x− z)dS

=

|z|=ε

fdS

→ f as ε→ 0.

The last line is left as an exercise. Putting this with Iε and J1ε , we have finally that ∆u(x) = f(x).

Now lets talk about Harmonic functions. A function u is harmonic if ∆u(x) = 0 for x ∈ Ω for u ∈ C2(Ω).Let Br(x) := y||y − x| < r.Theorem 1.4 (Mean Value Property). Let u ∈ C2(Ω). Then for any Br(x) ⊂ Ω then

u(x) =

∂Br(x)

u(y)dSy =

Br(x)

u(y)dy.

Proof. Fix x and consider B(x, r). Then consider

φ(r) =

∂Br(x)

u(y)dSy

We will show that φ′(r) = 0 for all r sufficiently small. Then we will show that φ(r) = limr→0 φ(r) = u(x).Lets change variables by letting y = x+ rω and so dSy = rn−1dSω.

φ(r) =1

ωnrn−1

ˆ|y−x|=r

u(y)dSy =1

ωnrn−1

ˆ|ω|=1

u(x+ rω)rn−1dSω

=1

ωn

ˆ|ω|=1

u(x+ rω)dSω

φ′(r) =1

ωn

ˆ|ω|=1

∇xu(x− rω) · ωdSω =1

ωnrn−1

ˆ|x−y|=r

∇yu(y) · nydSy

=1

ωnrn−1

ˆ|x−y|<r

∇y∇yu(y)dSy =1

n

Br(x)

∆yudy

= 0.

So now we are done with the proof for the sphere. Lets show it for the ball:

u(x) =1

ωnrn−1

ˆ|x−y|=r

u(y)dSy

ωnrn−1

nu(x) =

ˆ r

0

ˆ|x−y|=s

dsdSy

=

Br(x)

u(y)dy

8

Theorem 1.5 (Strong Maximum Principle for Harmonic Functions). Let Ω ⊂ Rn open let u be harmonicin Ω. Then

max∂Ω

u(x) = maxx∈∂Ω

u(x).

If Ω is also connected and u(x0) = maxΩ u(x) for x0 ∈ Ω, then u ≡ .C.

Proof. Assume u attains its max at x0 ∈ Ω. Let u(x0) = M = maxΩ u and define A = x ∈ Ω|u(x) = Mand let A2 = x ∈ Ω|u(x) < M. Then of course we can write Ω = A1 t A2. Obviously A2 is open becauseu is continuous. Now note that A1 is non-empty. Then if we can show that A1 s open, then we can concludethat A2 = ∅. They key for this is the MVP. Let 0 < r < dist(x0, ∂Ω). Then by the mean value property, wehave that

M = u(x0) =

Br(x0)

u(y)dy

i.e.

0 = M − Br(x0)

u(y)dy

=

Br(x0)

Mdy − Br(x0)

u(y)dy

=

Br(x0)

(M − u(y))dy.

Now note that M − u(y) ≥ 0 for all y because M is the maximum. Then u(y) = M throughout Br(x0). Wewant to show that u ≡ M throughout A. Suppose not and ∃y0 ∈ Br(x0) such that u(y0) < M . Then bycontinuity M − u(y) > 0 by continuity =⇒M − u(y) > 0 if |y − y0| < ρ for ρ small enough. Then we have

0 =

Br(x0)

(M − u(y))dy

=

|y−y0|<ρ

(M − u(y))dy +

|y−y0|≥ρ

(M − u(y))dy

≥ |y−y0|<ρ

(M − u(y))dy

> 0,

and we have reached a contradiction. Then A1 ≡ Ω.

1.3 Third Class

Let s review very quickly. We looked at ∆u = f and defined the Newtonian potential on Rn defined as

Φ(x) =

1(2−n)ωn

1|x|n−2 n ≥ 3

12π log |x| n = 2.

We showed that for f ∈ C20 (Rn) and u defined as u(x) = Φ∗f(x) then ∆u = f . We then looked at harmonic

functions: let Ω ⊂⊂ Rn, u ∈ C2 and ∆u = 0. Then we found the mean value property for harmonic functions:Let Br(x0) ⊂ Ω. Then

u(x0) =

Br(x0)

udy =

∂Br(x0)

udSy.

We then used the mean value theorem to show the strong maximum principle: if u is harmonic and attainsits maximum at x0 ∈ Ω, then u ≡ C in Ω.

9

Now lets look at more consequences of the MVP. The following is called Instantaneous Influence. Let Ωbe open and connected, u ∈ C2(Ω),∆u = 0 for x ∈ Ω and u = g on x ∈ ∂Ω. Then suppose we assume g ≥ 0on ∂Ω and on a set of non-measure zero on the boundary ∂Ω, g is strictly positive (g > 0).

Proof. The maximum and minimum principle imply that u ≥ 0 on Ω. If u(x0) = 0 at x0 ∈ Ω then u ≡ 0 bythe strong minimum principle.

Another consequence is the following uniqueness theorem. Let g ∈ C(∂Ω) and f ∈ C(Ω) with ∆u = f inΩ and u = g on ∂Ω. Then there is at most one solution.

Proof. Let u1 and u2 be two solutions and consider w = u1 − u2. Then ∆w = 0 in Ω and w = 0 on ∂Ω.Then by the maximum principle we have that w ≡ 0 =⇒ u1 = u2.

Now lets look at some size estimates. Let u be harmonic in Ω. Then for B(x0, r) ⊂ Ω we have

|u(x0)| ≤ C0

rn‖u‖L1(B(x0,r))∣∣∣∣ ∂u∂xi (x0)

∣∣∣∣ ≤ C1

rn+1‖u‖L1(B(x0,r)).

Proof. By the mean value property we have

|u(x0)| =

∣∣∣∣∣ n

ωnrn

ˆB(x0,r)

udy

∣∣∣∣∣ ≤ n

ωnrn‖u‖L1 .

For the second estimate, note that derivatives of harmonic functions are also harmonic. Consider B(x0,r2 )

and by the mean value property and divergence theorem∣∣∣∣ ∂u∂xj∣∣∣∣ =

n

ωn(r2

)n∣∣∣∣∣ˆB(x0,

r2 )

∂u

∂yj(y)dy

∣∣∣∣∣≤ n

ωn(r2

)n∣∣∣∣∣ˆ∂B(x0,

r2 )

u(y)njdS

∣∣∣∣∣≤ n

ωn(r2

)n ˆ∂B(x0,

r2 )

njdS‖u‖L∞(B(x0,r2 ))

≤ 2n

r‖u‖L∞(B(x0,

r2 ))

≤ C1

rn+1‖u‖L1(B(x0,r)).

Where in the last line we have used our first bound.

Now lets prove Liouville’s theorem. Assume ∆u = 0 in Rn and u is bounded. Then u ≡ C on Rn.

Proof. We will use the gradient bound. Let j ∈ 1, . . . , n, x0 ∈ Rn. Then∣∣∣∣ ∂u∂xj (x0)

∣∣∣∣ ≤ C1

rn+1

ˆB(x0,r)

|u(y)|dy ≤ C1M

rn+1

ωnrn

n→ 0.

This shows that u is constant because the derivative is zero.

10

Now lets look are some generalizations for harmonic functions. We now look at strictly elliptic operatorsL defined by

Lu =

n∑i,j=1

aij(x)∂2u

∂xi∂xj+

n∑k=1

bk(x)∂u

∂xk

and A(x) = aij(x) is positive definite and symmetric for each x ∈ Ω (all the eigenvalues are strictly positive).Moreover ∃CΩ

+ and CΩ− sic that for all x ∈ Ω and ξ ∈ Rn we have

C0|ξ|2 ≤ ξTA(x)ξ ≤ C+|ξ|2.

One special case of this can be aij = δij which would make L = ∆. We then have the following theorem forsuch uniformly elliptic operators:

Theorem 1.6 (Strong Maximum Principle). If Lu ≥ 0 in Ω and u attains its maximum on the interior ofΩ, then u ≡ C. Additionally if Lu ≤ 0 in Ω and u attains its minimum on the interior of Ω, then u ≡ C.

Now lets start the topic of Green’s functions. Let f ∈ C20 (Rn) and define u := Φ ∗ f . Then we showed

that ∆u = f . Now let u ∈ C20 (Rn) and note that by the same calculations we have

u(x) =

ˆRn

Φ(x− y)∆u(y)dy. (11)

We will try to find a generalization of (11) to u ∈ C2(Ω) ∩ C1(Ω) with the following theorem.

Theorem 1.7 (Decomposition of u into single layer, double layer, and volume potentials). Any u ∈ C2(Ω)∩C(Ω) can be represented in the following as

u(x) =

ˆ∂Ω

u(y)∂Φ

∂ny(x− y)dSy −

ˆ∂Ω

∂u

∂ny(y)Φ(x− y)dSy +

ˆΩ

Φ(x− y)∆u(y)dy (12)

where the first integral is the double layer potential with surface density u(y), the second is called the singlelayer potential with surface density ∂u

∂ny(y), and the last one is the volume potential with density ∆u.

Proof. Let u ∈ C2(Ω) and fix ε > 0 so thatˆ

Ω

∆u(y)Φ(x− y)dy =

ˆBε(x)

∆u(y)Φ(x− y)dy +

ˆΩ−Bε(x)

∆u(y)Φ(x− y)dy

= I1 + I2.

Then we can bound I1 by

|I1| ≤ C‖∆u‖L∞(Ω)

ˆBε(x)

1

|x− y|n−2dy = cO(ε).

Now lets look at I2:

|I2| =

∣∣∣∣∣ˆ

Ω−Bε(x)

∆uΦ(x− y)dy

∣∣∣∣∣=

∣∣∣∣∣ˆ

Ω−Bε(x)

∇y · (∇yu(y)Φ(x− y)− u(y)∇yΦ(x− y) + ∆yΦ(x− y)u(y)) dy

∣∣∣∣∣=

∣∣∣∣∣ˆ

Ω−Bε(x)

∇y · (∇yu(y)Φ(x− y)− u(y)∇yΦ(x− y)) dy

∣∣∣∣∣=

∣∣∣∣∣ˆ∂Ω

(Φ(x− y)

∂u

∂ny(y)− u(y)

∂Φ(x− y)

∂ny

)dSy +

ˆ∂Bε(x)

(Φ(x− y)

∂u

∂ny(y)− u(y)

∂Φ(x− y)

∂ny

)dSy

∣∣∣∣∣11

And so we have thatˆΩ

Φ(x− y)∆udy −O(ε) +

ˆ∂Ω

(−Φ(x− y)

∂u

∂ny(y) + u(y)

∂Φ(x− y)

∂ny

)dSy =

ˆ∂Bε(x)

(Φ(x− y)

∂u

∂ny(y)− u(y)

∂Φ(x− y)

∂ny

)dSy

and so by the Fundamental Theorem of Potential Theory we have that the first term on the RHS is of orderε and the second term converges uniformly to u(x) as ε→ 0 and we have exactly (12).

Now lets recall our goals. We want to solve the Dirichlet problem given f at least continuous on Ω andg on ∂Ω. We wish to find u ∈ C2(Ω) such that ∆u = f in Ω and u = g on ∂Ω. By the uniqueness theorem,the solution (if it exists) of the Dirichlet boundary valued problem is completely determined by Ω, f , and gso we see that formula (12) is useless because it requires too much information (i.e. ∂u

∂ny(x)).

Now imagine that we can replace Φ with the same type of singularity such that we can eliminate thesingle layer potential term. Let φ ∈ C2(Ω) such that ∆yφ(y) = 0. Then I claim that in (12), we may replaceΦ(x− y) by Φ(x− y)− φ(y). Let y ∈ Ω and we see that

0 = −u∆yφ = ∇ · (φ∇u− u∇φ) + ∆u(y)(−φ(y))

=

ˆ∂Ω

(φ∇u− u∇φ) dSy +

ˆΩ

∆u(y)(−φ(y))dy. (13)

Then we can add (12) and (13) to get

u(x) =

ˆ∂Ω

(u(y)

∂

∂ny(Φ(x− y)− φ(y))− ∂u

∂ny(y) (Φ(x− y)− φ(y))

)dSy +

ˆΩ

∆u(y)(Φ(x− y)− φ(y))dy.

(14)

Next lets exploit this φ(y) degree of freedom and assume we can find a solution to the particular familyof Dirichlet boundary value problems parametrized by x ∈ Ω. In other words lets call φx(y) such that∆φx(y) = 0 ∀y ∈ Ω (which we can do by finding a Newtonian potential outside of Ω) such that

limy→y0∈Ω

φx(y) = Φ(x− y0).

Then we will see that the term we want to vanish in (14) will in fact vanish and we will get

u(x) =

ˆ∂Ω

(u(y)

∂

∂ny(Φ(x− y)− φ(y))

)dSy +

ˆΩ

∆u(y)(Φ(x− y)− φ(y))dy. (15)

And we will call GD(x, y; Ω) = Φ(x− y)− φx(y) the Dirichlet Green’s function. Some things on our agendaare finding GD(x, y; Ω) for domains with special symmetry (i.e. B(0, a)) and is there existence for Ω a generaldomain? The answer is yes, quite generally in fact, by using compact operators applied to integral equations.

Lets construct GD(x, y) for B(0, a) = y||y| < a. We seek a harmonic function φx(y) such that for allx ∈ B(0, a) and

limy→y0,|y0|=a

= Φ(x− y0) =1

ωn(2− n)

1

|x− y0|n−2.

Notice the key geometric property of B(0, a). Let x ∈ B(0, a) and consider the line that passes through theorigin and x. Then given any x with |x| < a, we will see that ∃x∗ with |x∗| > a such that for all y0 ∈ ∂B(0, a)such that

|x∗ − y0||x− y0|

= C(a, x) =a

|x|where the constant C(a, x) is independent of y. In fact

x∗ =a2

|x|2x.

12

This geometric property implies construction of GD(x, y;B(0, a)) as we shall now see:

1

|x− y0|= C(a, x)

1

|x∗ − y0|1

ωn(2− n)|x− y0|n−2=C(a, x)n−2

ωn(2− n)

1

|x∗ − y0|n−2

Φ(x− y0) = C(a, x)n−2Φ(x∗ − y0).

This implies that Φ(x − y0) − C(a, x)n−2Φ(x∗ − y0) = 0 and so we see that if we define GD(x, y) = Φ(x −y) − C(a, x)n−2Φ(x∗ − y) we have our desired properties. Yay! So in the end we will have on this ball aprecise solution given by

u(x) =

ˆ∂B(0,a)

g∂GD∂ny

dSy +

ˆB(0,a)

f(y)GD(x, y)dy.

For next time we will se the converse: if u is in this form, does it solve the Dirichlet boundary condition.

1.4 Fourth Class

Lets recap a little bit. We have the Newtonian potential

Φ(x) =

1(2−n)ωn

1|x|n−2 n ≥ 3

12π log |x| n = 2.

We will have Ω a bounded open subset of Rn with ∂Ω smooth. Let φ, u ∈ C2(Ω) with ∆φ = 0. Then wehave the potential representation as

u(x) =

ˆ∂Ω

(u(y)

∂

∂ny(Φ(x− y)− φ(y))− ∂u

∂ny(y) (Φ(x− y)− φ(y))

)dSy +

ˆΩ

∆u(y)(Φ(x− y)− φ(y))dy.

We would like to have a representation that solves the Dirichlet boundary value problem. We see that theproblem was the term with ∂u

∂nyand so we have to find a φ that satisfies ∆yφ

x(y) = 0 and φx(y) = Φ(x− y)

on the boundary. Then we had the following theorem.

Theorem 1.8. Let u ∈ C2(Ω) and denote the solution of ∆u = f in Ω and u = g on ∂Ω. Then

u(x) =

ˆ∂Ω

g(y)∂

∂nyGD(x, y)dSy +

ˆΩ

f(y)GD(x, y)dy

whereGD(x, y) = Φ(x, y)− φx(y; Ω)

is the Dirichlet Green’s function for the laplacian on Ω.

Lets go over some properties of the Green’s function. One is that G(x, y) = G(y, x) for x 6= y. In thespecial case that Ω = B(0, a) with |x| < a. Then from the geometric property such that for an x∗ on theline connecting x and the origin, then for any y ∈ ∂B(0, a) then

|x∗ − y||x− y|

= C(a, x) =a

|x|

and one can actually calculate to find

x∗ =a2

|x|2x.

13

Then we have

1

|x− y0|= C(a, x)

1

|x∗ − y0|1

ωn(2− n)|x− y0|n−2=C(a, x)n−2

ωn(2− n)

1

|x∗ − y0|n−2

Φ(x− y0) = C(a, x)n−2Φ(x∗ − y0).

This implies that Φ(x−y0)−C(a, x)n−2Φ(x∗−y0) = 0 and so we see that if we define GD(x, y) = Φ(x−y)−C(a, x)n−2Φ(x∗ − y) we have our desired properties. Now we want to calculate ∂Gd

∂ny(x, y) on the boundary.

We will first calculate the gradient and then dot it with the normal.

∇y1

|x− y|n−2= (2− n)|x− y|1−n∇y|x− y|

= (2− n)|x− y|1−n x− y|x− y|

(−1)

1

ωn(2− n)∇ 1

|x− y|n−2· n(y) =

1

ωn|x− y|1−n y − x

|y − x|· ya

Then we can similarly define

1

ωn(2− n)∇ 1

|x∗ − y|n−2· n(y) =

1

ωn|x∗ − y|1−n y − x

∗

|y − x∗|· ya(

a

|x|

)n−21

ωn(2− n)∇ 1

|x∗ − y|n−2· n(y) =

1

ωn|x∗ − y|1−n y − x

∗

|y − x∗|· ya

(a

|x|

)n−1

and we can take the difference to get(∇yΦ(x− y)−

(a

|x|

)n−2

∇Φ(x∗ − y)

)· ya

=1

aωn

(1

|x− y|n(|y|2 − x · y

)−(a

|x|

)n−21

|x∗ − y|n(|y|2 − x∗ · y

))

=1

aωn

1

|x− y|n

(|y|2 − x · y)− ( a

|x|

)n |x|2a2

|y|2 − a2

|x|2x · y(a|x|

)n

=1

aωn

1

|x− y|n(a2 − x · y − (|x|2 − x · y)

)=

a2 − |x|2

aωn|x− y|n

= H(x, y)

and we call H(x, y) the Poisson Kernel for B(0, a). Then for the ball we have the following representation

u(x) =

ˆ∂B(0,a)

g(y)H(x, y)dSy +

ˆB(0,a)

f(y)GD(x, y)dy.

As a corollary we can let u ≡ 1 then ∆u = 0 and u|∂B(0,a) = 1 and so we can see

1 =

ˆ∂B(0,a)

H(x, y)dSy.

Notice that this is completely obvious in terms of brownian motion because the particle will hit the boundaryat some point.

14

Now lets see something else and consider the following problem. Let ∆u = 0 in |x| < a and u = g on|x| = a where g ∈ C0(|y| = a). Define

v(x) :=

g(x) |x| = a´|y|=aH(x, y)g(y)dSy |x| < a

Then we will show that v ∈ C2(|x| < a) with ∆v = 0 in |x| < a and v ∈ C0(|x| ≤ a). In particular if |x0| = athen

limx→x0

ˆ|y|=a

H(x, y)g(y)dSy = g(x0).

This actually has a strong connection with the mean value property. If x = 0 then we can see that

u(x) =1

aωn

ˆ|y|=a

a2 − |x|2

|x− y|ng(y)dSy

u(0) =1

aωn

ˆ|y|=a

a2

|y|ng(y)dSy

=1

ωnan−1

ˆ|y|=a

g(y)dSy

=

|y|=a

udSy.

Proof. Lets first write down some properties that H satisfies. One is that if |y| ≤ a and |x| < a and y 6= x.Then H(x, y) ∈ C∞. Another is that ∆xH(x, y) = 0 for |x| < a and |y| = a. Then we also know that theintegral of H over the boundary is 1. Additionally H(x, y) > 0 for |x| < a and |y| = a. The last one is thatζ ∈ ∂B(0, a) and consider B(ζ, δ) for a fixed δ and let y ∈ ∂B(0, a) ∩B(ζ, δ). Then for x ∈ B(0, a) we have

limx→ζ

H(x, y) = 0

uniformly on y||y− ζ| ≥ δ, |y| = a. The first four properties are trivial, so lets see why the last one is true.Assume x→ ζ such that |x− ζ| ≤ δ

2 . Then note that

|x− y| = |x− ζ − y + ζ|≥ |ζ − y|+ |x− δ|

≥ δ − δ

2=δ

2

and we have a lower bound on |x− y|. We then have an upper bound on the Poisson Kernel

H(x, y) ≤ 1

aωn(a2 − |x|2)

(δ

2

)nand our claim follows. I also note that v ∈ C2(|x| < a) with ∆v = 0 in |x| < a and v ∈ C0(|x| ≤ a) is trivial(just take difference quotients and check that we can pass through the limit). So all we need to show is that

limx→x0

ˆ|y|=a

H(x, y)g(y)dSy = g(x0).

Fix ε > 0. It suffices to show that

limx→x0

∣∣∣∣∣ˆ|y|=a

H(x, y)g(y)dSy − g(x0)

∣∣∣∣∣ ≤ ε.15

Then we can see thatˆ|y|=a

H(x, y)g(y)dSy − g(x0) =

ˆ|y|=a

H(x, y)g(y)dSy − g(x0)

ˆ|y|=a

H(x, y)dSy

=

ˆ|y|=a

H(x, y)(g(y)− g(x0))dSy

By continuity of g, there exists a δ(ε) such that |g(y)− g(x0)| < ε for |y−x0| < δ(ε) for |y| = a and |x0| = a.Then we can split up our integral intoˆ|y|=a

H(x, y)(g(y)− g(x0))dSy =

ˆ|y−x0|<δ

H(x, y)(g(y)− g(x0))dSy +

ˆ|y−x0|≥δ(ε)

H(x, y)(g(y)− g(x0))dSy

= I1(δ) + I2(δ)

Then by the positivity and continuity we can have

|I1(δ)| ≤ˆ|y−x0|<δ

H(x, y)|g(y)− g(x0)|dSy

≤ max|y−x0|<δ

|g(y)− g(x0)|ˆ|y−x0|<δ

H(x, y)dSy

≤ εˆ|y|=a

H(x, y)dSy

= ε.

Now we bound I2(δ)

|I2(δ)| ≤ˆ|y−x0|≥δ(ε)

H(x, y)dSy2 max|z|=a

|g(z)|

=1

aωn

ˆ|y−x0|≥δ(ε)

a2 − |x|2

|x− y|ndSy2 max

|z|=a|g(z)|

→ 0

And so we have the desired bound.

Note that this is still for the ball. We want to extend this to general surfaces by means of integralequations and boundary integral methods. Lets look at some motivation We know that if ∆u = 0 in Ω andu = g on ∂Ω then we have the following representation

u(x) =

ˆ∂Ω

(u(y)

∂

∂ny(Φ(x− y)− φ(y))− ∂u

∂ny(y) (Φ(x− y)− φ(y))

)dSy.

Then we did some tricky thing with Φ 7→ Φ− φx, but this required too much knowledge of the domain. Sowe see a µ(y; g) defined on the boundary such that

u(x) =

ˆ∂Ω

∂Φ(x− y)

∂nyµ(y; g)dSy.

Then of course u is still going to be harmonic so the trick is to find that the boundary conditions still hold.Here is the plan.

We want to solve ∆u = 0 for xΩ and u = g on ∂Ω. Then let

u(x) =

ˆ∂Ω

∂Φ(x− y)

∂nyµ(y; g)dSy (16)

16

and we will show that for x0 ∈ ∂Ω and x ∈ Ω then

limx→x0

u(x) = −1

2µ(x) +

ˆ∂Ω

K(x, y)µ(y)dSy

where

K(x, y) =∂

∂nyΦ(x− y)

∣∣x,y∈∂Ω,x6=y.

Thus for (16) to solve the Dirchlet BVP we must have that

g(x) = −1

2µ(x) +

ˆ∂Ω

K(x, y)µ(y)dSy

for all x ∈ ∂Ω. This is a singular integral equation for a function µ(y) defined on ∂Ω. In fact we can writethis as

(−1

2I + Tk)µ = g

and we claim that (− 12I + Tk)µ = g is invertible for an appropriate space of functions. This will solve our

problem. One benefit from this is to show existence of solutions to the Dirichlet problem. It is also helpfulto find numerical solutions because these operators can be approximated by matrices.

Proposition 1.9. Let Ω be a bounded region with smooth boundary that is a subset of R3. and let µ ∈C0(∂Ω). For x ∈ R3 define

u(x) =

ˆ∂Ω

− ∂Φ

∂ny(x− y)µ(y)dSy =

ˆ∂Ω

1

4π

∂

∂ny

(1

|x− y|

)µ(y)dSy.

Then u(x) is well-defined for all x ∈ R3 and u ∈ C∞(R3 − ∂Ω) and for x ∈ R3 − ∂Ω we have ∆xu(x) = 0.

Note. This proposition is clear for x 6= ∂Ω because we have u harmonic. However, if x ∈ ∂Ω then our integralis singular. We will use the fact that ∂Ω is smooth so that we can chose coordinates x = (x1, x2, x3) suchthat locally x0 = (0, 0, 0) and x3 = φ(x1, x2). In fact one can shoe that

Theorem 1.10 (Gauss’ Theorem).

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)dSy =

−1 x ∈ Ω− 1

2 x ∈ ∂Ω0 x /∈ Ω.

1.5 Fifth Class

Before we review, note that the following material will not be in Evans, but will be in Gunther and Lee andFolland chapter 3.

Now lets review a little bit. We were dealing with two boundary valued problems. The first being theDirichlet problem i.e. Ω a bounded open subset of Rn with smooth boundary and ∆u = 0 in Ω and u = g on∂Ω where g ∈ C0(∂Ω). Sometimes this is called the interior Dirichlet problem. The other boundary valueproblem was the exterior Neuman problem i.e.

∆v = 0 x ∈ R3 − Ω∂v∂ν∗

= h x ∈ ∂Ω

|v(x)| → 0 |x| → 0

where ν∗ is the outward unit normal on ∂Ω to R3 − Ω.

Theorem 1.11. The interior Dirichlet problem and exterior Neuman problem have, for any g, h ∈ C0(∂Ω)have unique solutions.

17

Lets try to think what the motivations of boundary integral methods are (specifically to the IDP). Wewill seek a solution of the form

u(x) =1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)µ(y)dSy

where we call this integral a double layer potential with surface density µ. Note that u ∈ C∞(R3 − ∂Ω) and∆u = 0 in this domain. Now the question is: can we determine µ such that

limx→x0

u(x) = g(x0)

for x0 ∈ ∂Ω. A potential theoretic calculation that we will se in a moment is that the limit above is equal to

−1

2µ(x0) +

1

4π

ˆ∂Ω

∂

∂ny

(1

|x0 − y|

)µ(y)dSy.

Taking this as a black box, we can conclude that u(x) solves the IDP if µ ∈ C0(∂Ω) satisfies, for x ∈ ∂Ω

−1

2µ(x) +

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)µ(y)dSy = g(x).

This was for the IDP. For the ENP, we analogously introduce

v(x) =1

4π

ˆ∂Ω

1

|x− y|ρ(y)dSy

which is called the single layer potential with surface density ρ. Again it is clear that v ∈ C∞(R3−∂Ω) with∆v = 0, but the difference between the IDP is that v is continuous in all of R3. It is easy to check that forρ ∈ C0(∂Ω) then |v(x)| = O( 1

|x| ) as |x| → ∞, and

∂v

∂ν∗(x0) = lim

x→x0

ν∗ · ∇xv(x)

= limx→x0

1

4π

ˆ∂Ω

∇x1

|x− y|· ν∗(x0)ρ(y)dSy

= −1

2ρ(x0)− 1

4π

ˆ∂Ω

∂

∂νx0

(1

|x− y|

)ρ(y)dSy

We multiply by −1 and we need to choose ρ such that

1

2ρ(x) +

1

4π

ˆ∂Ω

∂

∂ν∗

(1

|x− y|

)ρ(y)dSy = −h(x). (17)

It turns out that if ρ ∈ C0(∂Ω) satisfies (17), then the single layer potential with surface density ρ satisfiesthe ENP.

Solvability of IDP and ENP reduces to solving the boundary integral equations

g(x) = −1

2µ(x) +

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)µ(y)dSy (18)

−h(x) = −1

2ρ(x)− 1

4π

ˆ∂Ω

∂

∂νx

(1

|x− y|

)ρ(y)dSy. (19)

Notice that the kernels in (18) and (19) are fundamentally different. We will write these equations by(− 1

2I + TKIDP )µ = g and (− 12I + TKENP )ρ = −h. Given K(x, y), a “kernel function,” we define for f(x) in

some sense

TK [f ](x) =

ˆ∂Ω

K(x, y)f(y)dSy

18

and these are the kernels that we wrote by TKIDP and TKENP .We will be working with the hilbert space L2(∂Ω). We have the following proposition. If µ ∈ L2(∂Ω) is

a solution of (− 12I + TKIDP )µ = g where g is continuous on ∂Ω, then µ ∈ C0(∂Ω). In fact we will see that

if g1, g2 ∈ L2(∂Ω) then

〈g1, TKIDP g2〉 = 〈T ∗KIDP g1, g2〉= 〈TKENP g1, g2〉

where 〈 , 〉 is the inner produce in L2. So we will find that T ∗KIDP = TKENP and by Fredholm theory, we’ll

see that solvability of − 12µ+ TKIDP µ = g for all g ⇐⇒ 1

2ρ+ TKENP ρ = 0 having only ρ = 0 as a solution.Now lets stop the quick overview and actually do some math.

Proposition 1.12 (G-L. pg 332). Let Ω be a region of R3 and define

u(x) =

ˆ∂Ω

1

4π

∂

∂ny

(1

|x− y|

)µ(y)dSy

where µ ∈ C0(Ω). Then we have

1. u(x) is well-defined for all x ∈ R3

2. u(x) ∈ C∞(R3 − ∂Ω)

3. ∆u = 0 in R3 − ∂Ω.

Proof. Note that we already have (2) and (3), and need to check (1) for x ∈ ∂Ω. We will do this byletting x = (x1, x2, x3) ∈ ∂Ω and analyzing the rotated domain Ω′ such that x is the new origin and in aneighborhood of 0 ∈ R3, then ∂Ω is given by a function x3 = φ(x1, x2). In fact for a small enough δ we canmore explicitly write this by representing ∂Ω ∩Bδ(0) = (∂Ω)δ as the graph of a function φ(x1, x2).

We write ∂Ω = (∂Ω)δ ∪ (∂Ω− (∂Ω)δ) and so we can write

u(x) =1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)µ(y)dSy

=1

4π

ˆ(∂Ω)δ

∂

∂ny

(1

|x− y|

)µ(y)dSy +

1

4π

ˆ∂Ω−(∂Ω)δ

∂

∂ny

(1

|x− y|

)µ(y)dSy.

Since we want to show that |u| is well defined we write this as

|u(x)| ≤Mˆ

(∂Ω)δ

∣∣∣∣ ∂∂ny(

1

|x− y|

)∣∣∣∣ dSy +M

ˆ∂Ω−(∂Ω)δ

∣∣∣∣ ∂∂ny(

1

|x− y|

)∣∣∣∣ dSy (20)

and will show that this is bounded. We take derivatives to see∣∣∣∣νy ·∆y1

|x− y|

∣∣∣∣ =

∣∣∣∣ 1

|x− y|2x− y|x− y|

νy

∣∣∣∣ ≤ 1

|x− y|2≤ 1

δ2

(where the calculations above are only valid in ∂Ω− (∂Ω)δ) to show that the second term in (20) is bounded.Now we have to use local analysis near x = 0 (this is abuse of notation, it is actually x′ = 0 from ourcoordinate change) to analyze the first term in (20).

The first term is examined as follows.ˆ

(∂Ω)δ

∣∣∣∣ ∂∂ny(

1

|0− y|

)∣∣∣∣ dSy =

ˆ(∂Ω)δ

1

|y|2

∣∣∣∣ y|y|νy∣∣∣∣ dSy. (21)

19

Notice that on the surface y = (y1, y2, φ(y1, y2)). Also the equation to the surface is φ(y1, y2)− y3 = 0 andtaking the gradient gives us ∇y(φ(y1, y2)− y3 = 0) = (φy1 , φy2 ,−1). Then we have that the integral (21) isgoing to be equal to

=

¨y21+y22≤δ2

1

|y21 + y2

2 + φ(y2, y2)2

∣∣∣∣∣∣ −(y1, y2, φ(y1, y2))√|y2

1 + y22 + φ(y2, y2)2

· (φy1(y1, y2), φy2(y1, y2),−1)√φ2y1 + φ2

y2 + 1

∣∣∣∣∣∣√φ2y1 + φ2

y2 + 1dy1dy2

=

¨y21+y22≤δ2

1

(y21 + y2

2 + φ2)32

(y1φy1 + y2φy2 − φ)dy1dy2.

We now Taylor expand φy1(y1, y2) = φy1(0, 0) +φy1y1(η1, η2)y1 +φy1y2(η1, η2)y2 such that η21 + η2

2 ≤ δ2. TheTaylor expansion for φy2 is going to be similar. Hence we get the bounds |y1φy1 | ≤ |φy1y1y1|2 + |φy1y2y1y2| ≤C(|y1|2 + |y1||y2|). Again, there are similar bounds for y2φy2 . Hence we finally get the bounds

ˆ(∂Ω)δ

≤¨y21+y22≤δ2

y21 + |y1||y2|

(y21 + y2

2 + φ2)32

+y2

2 + |y1||y2|(y2

1 + y22 + φ2)

32

+|φ(y1, y2)|

(y21 + y2

2 + φ2)32

dy1dy2.

Now lets Taylor expand φ to get φ(y1, y2) = φy1y1y21 + 2φy1y2(β1, β2)y1y2 + φy2y2(β1, β2)y2

2 . And so we canplug this in to get that the integral finally becomes bounded by

C2π

ˆ δ

0

1

δrdr = Cδ <∞

and so we finally get that our initial statement is well defined.

blah blah blah µ ≡ 1 then

Theorem 1.13 (Gauss’ Theorem).

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)dSy =

−1 x ∈ Ω− 1

2 x ∈ ∂Ω0 x /∈ Ω.

Proof. The simples case is where x /∈ Ω. Then the integral will yield

u =

ˆ∂Ω

∇y(

1

|x− y|

)· nydSy

ˆΩ

∇y∇y1

|x− y|dSy

= 0

so we are done for x 6∈ Ω. Now lets consider x ∈ Ω. Recall that from our analysis from the earlier class

u(x) =

ˆ∂Ω

(u(y)

∂

∂ny(Φ(x− y)− φ(y))− ∂u

∂ny(y) (Φ(x− y)− φ(y))

)dSy +

ˆΩ

∆u(y)(Φ(x− y)− φ(y))dy

applied with u ≡ 1 and we will trivially get our result. Now let x ∈ ∂Ω and, using our same old picture frombefore, consider Ω−Bδ(0) and the integral

0 =

ˆ∂Ω∩Bδ(0)c

∂

∂ny

(1

4π

1

|x− y|

)dSy

=

ˆ∂Ω−(∂Ω)δ

∂

∂ny

(1

4π

1

|0− y|

)dSy +

ˆSδ(0)∩Ω

∂

∂ny

(1

4π

1

|0− y|

)dSy.

20

Since this is true for any δ > 0 then we are able to say

0 = limδ→0

ˆ∂Ω−(∂Ω)δ

∂

∂ny

(1

4π

1

|y|

)dSy +

ˆδ→0

ˆSδ(0)∩Ω

∂

∂ny

(1

4π

1

|y|

)dSy.

We examine the second integral to be

1

4π

∂

∂ny

1

|y|=

1

4π|y|2=

1

4πδ2.

Since we are integrating this

So what is left? We still need to look at the limit as Ω 3 x→ x0 ∈ ∂Ω for u(x).

1.6 Sixth Class

Our agenda for now is to complete the discussion of the boundary integral approach to the Dirichlet problemon Ω. Then we will begin a short introduction to Variational Methods.

Theorem 1.14. The IDP has a solution which can be represented as a double layered potential with densityµg if µg satisfies the boundary integral equation

−1

2µ(x) +

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)µ(y)dSy = g(x)

for all x ∈ ∂Ω.

With the the double layer potential with surface density µg will have an explicit solution

u(x) =1

4π

ˆ∂Ω

1

∂ny

(1

|x− y|

)µg(y)dSy.

We similarly can write the ENP (∆v = 0 for x 6∈ Ω and ∂v∂n∗

= h for x ∈ ∂Ω with |v(x)| → 0 as |x| → ∞).This has a solution represented as a single layer potential with surface density ρh if ρh satisfies

−1

2ρ(x) +

1

4π

ˆ∂

∂n∗

(1

|x− y|

)ρ(y)dSy = −h(x).

The representation for x will give us

v(x) =1

4π

ˆ∂Ω

1

|x− y|ρ(y)dSy.

Note that these expressions not only satisfy ∆u = ∆v = 0, but also satisfy the boundary conditions.We defined the following kernels

KIDP (x, y) =1

4π

∂

∂ny

(1

|x− y|

)KENP (x, y) =

1

4π

∂

∂n∗

(1

|x− y|

)and note that KENP (x, y) = KIDP (x, y). We now define

TK [f ](x) =

ˆ∂Ω

K(x, y)f(y)dSy

21

and our problem reduces to solving the boundary integral equations(−1

2I + TKIDP

)µ = g(

−1

2I + TKENP

)ρ = −h

Lets recall very briefly how we found these equations. We used the theorem by Gauss

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)dSy =

−1 x ∈ Ω− 1

2 x ∈ ∂Ω0 x /∈ Ω.

This suggested that we do the following limit of Ω 3 x→ x0 ∈ ∂Ω

u(x) =1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)(µ(y)− µ(x0))dSy + µ(x0)

1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)dSy

=1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)(µ(y)− µ(x0))dSy − µ(x0)

= ζ(x)− µ(x0)

limx→x0

u(x) = ζ(x0)− µ(x0)

= DLP (x0)− µ(x0)1

4π

ˆ∂

∂ny

(1

|x0 − y|

)dSy − µ(x0)

= DLP (x0)− 1

2µ(x0)

Lets get back to solving our integral equations.

Proposition 1.15. Let K = KIDP or KENP . Let g ∈ C0(∂Ω) and assume µ ∈ L2(∂Ω) solves

(−1

2I + TK)µ = g.

Then µ ∈ C0(∂Ω).

Note that we need µ to be continuous because we explicitly used its continuity in constructing ourequations.

Proof. We have

TK [f ](x) =

ˆ∂Ω

K(x, z)f(z)dSz.

Let ε > 0. and consider, for x ∈ ∂Ω, define Bε(x), and let χε(z) be a cutoff function with χε ≡ 1 for|x− z| < ε

2 and zero for |x− z| ≥ ε. Then now we can write

TK [f ](x) =

ˆ∂Ω

χε(z)K(x, z)f(z)dSz +

ˆ∂Ω

(1− χε(z))K(x, z)f(z)dSz

= TχεK [f ](x)− T(1−χε)K [f ](x).

Now I claim that f ∈ L2(∂Ω). Then T(1−χε)[f ] ∈ C0(∂Ω), and this will come from the fact that our kernelis non-singular. My next claim is assuming we have

|L(x, z)| ≤ C

|x− z|

22

for x, z ∈ ∂Ω, then f being bounded implies that TL[f ] ∈ C0(∂Ω). We will apply this with L = χε(z)K(x, y).Moreover, ∃C ′ independent of f such that

‖TL[f ]‖L∞(∂Ω) ≤ C ′‖f‖L∞(∂Ω).

My last claim is that if f is bounded on ∂Ω, then

‖TχεK [f ]‖L∞(∂Ω) ≤ Cε‖f‖L∞(∂Ω).

Now lets prove some things. Recall that we have

TK [f ](x) = TχεK [f ](x)− T(1−χε)K [f ](x).

We manipulate our assumption by

−1

2µ+ TKµ = g

−1

2µ+ TχεKµ+ T(1−χε)Kµ = g

(I − 2TχεK)µ = −2g + 2T(1−χε)Kµ

where this is all done by my first claim. Lets recall some linear algebra. If T : X → X is a map of banachspaces, then (I − T )−1 exists if

‖T‖X < 1.

Also for T : X → X linear, then T is continuous if ∃ > 0 such that for all x ∈ X,

‖Tx‖ ≤ C‖x‖.

The smallest C for which this holds is called ‖T‖. The rest of this proof will not be written.

Lets review some functional analysis. An inner product space X is a vector space with a map x, y 7→〈x, y〉X ∈ C with the folioing properties

(i) 〈x, x〉 ≥ 0 and it is equal to zero if and only if x = 0

(ii) 〈x, y〉 = 〈y, x〉

(iii) 〈αx+ βy, z〉 = α∠x, z〉+ β〈y, z〉 for x, y, z ∈ X,α, β ∈ C.

We define its norm ‖x‖ =√〈x, x〉. We say X is a Hilbert space if it is a normed, complete space whose norm

is related to an inner product on x by ‖x‖X =√〈x, x〉X . We say that a bounded linear operator T : X → X

if‖Tx‖X ≤ CT ‖x‖X

and we define

‖T‖ = supx 6=0

‖Tx‖‖x‖

.

If T : X → Y then we say it is bounded if

‖Tx‖Y ≤ CT ‖x‖X .

Example 1.16. Take TK where K is a weakly singular kernel i.e.

|K(x, y)| ≤ C

|x− y|.

23

Then I claim that‖TK [f ]L2(∂Ω) ≤ CK‖f‖L2(∂Ω)

and so we have that ‖TK‖ < ∞ and so TK is a bounded operator on L2(∂Ω). Lets prove this. Recall thatwe define

TK [f ](x) =

ˆA

K(x, y)f(y)dy

and assume that K(x, y) is such that

CK ≡ supx∈A

ˆA

|K(x, y)|dy + supy∈A

ˆA

|K(x, y)|dx <∞.

Then TK : L2(A)→ L2(A) andTK [f ]‖L2(A) ≤ CK‖f‖L2(A).

Proof.

|Tk[f ](x)| ≤ˆA

|K(x, y)||f(y)|dy

=

ˆA

|K(x, y)| 12 |K(x, y)| 12 |f(y)|dy

≤(ˆ

A

|K(x, y)|dy) 1

2 (|K(x, y)||f(y)|2dy

) 12

|Tk[f ](x)|2 ≤ˆA

|K(x, y)dy

ˆA

|K(x, y)|dyˆ|K(x, y)||f(y)|2dy

ˆA

|Tk[f ](x)|2dx ≤ CKˆ ˆ

|K(x, y)||f(y)|2dydx

= CK

ˆ ˆ|K(x, y)|dx|f(y)|2dy

= C2K

ˆ|f(y)|2dy

Also, note that TK in fact satisfies our assumption, and so we are done.

Now lets talk about compact operators. Let H1, H2 be hilbert spaces. We say a bounded linear operatorT : H1 → H2 is compact if the following holds

1. Given a bounded sequence xj in H1, then Txj has a convergent subsequence in H2.

Theorem 1.17. Let Tj be a sequence of compact operators with ‖Tj − T‖ → 0 as j → ∞. Then T iscompact.

Theorem 1.18. Let T : H → H be a compact operator on the Hilbert space H. Then ∃Tm with Tm :H → H with dim ImTm <∞ such that ‖Tm − T‖ → 0 as m→∞.

What about integral operators? K(x, y) : A×A→ C is called a Hilbert - Schmidt Kernel if

¨A×A

|K(x, y)|2dxdy <∞.

We then say that K ∈ L2(A×A). Then the theorem we will use is

24

Theorem 1.19. If K is Hilbert-Schmidt then

Tk[f ](x) =

ˆA

K(x, y)f(y)dy

is compact on L2(A).

This is all turning out great! the only problem is that if K(x, y) is a weakly singular kernel i.e.

|K(x, y)| ≤ C

|x− y|

and

Tk[f ](x) =

ˆ∂Ω

K(x, y)f(y)dSy,

then ¨A×A

|K(x, y)|2dxdy =∞.

The problem is at x = y. Let δ > 0 and define

Kδ(x, y) =

K(x, y) |x− y| ≥ δ0 |x− y| < δ

and it is clear that ¨∂Ω×∂Ω

|Kδ(x, y)|2dxdy ∼ 1

δ2<∞.

This means that for each δ > 0, TKδ is compact! We now look at the difference

TK−Kδ [f ](x) = Tk[f ]− TKδ [f ]

=

ˆ∂Ω

(K(x, y)−Kδ(x, y))f(y)dy

=

ˆ|x−y|<δ

K(x, y)f(y)dy.

Taking the norms, we find (by Young’s inequality)

‖TK−Kδ [f ]‖2L2(∂Ω) ≤ C2K−Kδ‖f‖

2L2(∂Ω)

where

CK − CK−δ = supx∈∂Ω

ˆ|x−y|<δ

|K(x, y)|dy + supy∈∂Ω

ˆ|x−y|<δ

|K(x, y)|dx

≤ supx∈∂Ω

ˆ|x−y|<δ

C

|x− y|dy + sup

y∈∂Ω

ˆ|x−y|<δ

C

|x− y|dx

=

ˆ δ

0

1

|z||z|d|z|+

ˆ δ

0

1

|z||z|d|z|

= 2δ

and everything converges!

25

1.7 Seventh Class

We are looking at the boundary integral approach to BVP. Recall

IDP =

∆u = 0 x ∈ Ωu = g x ∈ ∂Ω

ENP =

∆v = 0 x ∈ Ωu = h x ∈ ∂Ω|v(x)| → 0 |x| → ∞, x 6∈ Ω

where g, h ∈ C0(∂Ω). We seek solutions of the form for x ∈ Ω

u(x) = DLP [µ](x)

=1

4π

ˆ∂Ω

∂

∂ny

(1

|x− y|

)µ(y)dSy, µ ∈ C0(∂Ω)

v(x) = SLP [ρ](x)

=1

4π

ˆ∂Ω

(ρ

|x− y|

)dSy, ρ ∈ C0(∂Ω)

We want to find ρ, µ as continuous solutions of the equations (for x ∈ Ω)

g(x) =

(−1

2I + TKIDP

)µ

−h(x) =

(−1

2I + TKENP

)ρ

where KIDP and KENP are the kernels that we’ve been using. The reason we want continuous solutions isbecause we’d have for x 6= y, x, y ∈ ∂Ω

limx→x0∈∂Ω

DLP [µ](x) = −1

2µ+ TKIDP [µ] = g.

We will reduce the solutions we seek to solutions in L2 because of Proposition 1.15.Note that L2(∂Ω) = H is a Hilbert space. Also for K = KIDP ,KENP we have

|K(x, y)| ≤ C

|x− y|

as one of our claims (for x 6= y, x, y ∈ ∂Ω). We also showed that TK is a compact linear operator betweenL2(∂Ω) and itself. Now consider the following computations

〈f, Tkg〉 =

ˆ∂Ω

f(x)Tk[g](x)dx

=

ˆ∂Ω

f(x)

ˆ∂Ω

K(x, y)g(y)dydx

=

ˆ [ˆK(x, y)f(x)dx

]g(y)dy

=

ˆ(T ∗Kf)(y)g(y)dy

where

T ∗K [f ](x) =

ˆK(y, x)f(y)dy

and so we see that 〈f, Tkg〉 = 〈T ∗k f, g〉. Since both of our K’s are real we finally have (TKIDP )∗ = TkENPbecause recalling our definitions

KIDP (x, y) =1

4π

∂

∂ny

(1

|x− y|

), KENP (x, y) =

1

4π

∂

∂nx

(1

|x− y|

).

26

Theorem 1.20 (Fredholm Alternative). Let T be a compact operator of H, a Hilbert space. Then (λI−T )x =y has a solution if and only if

y ⊥ ξ ∈ H|(λI − T ∗)ξ = 0i.e. 〈y, ξ〉H = 0 for all ξ ∈ ker(λI − T )∗.

This means that to solve the IDP for any g ∈ C0(∂Ω), it suffices to solve (− 12I + TKIDP )µ = g for

µ ∈ L2(∂Ω). Note that we already showed that TKIDP is compact. Then by the Fredholm alternative, thesolvability of (− 1

2I+TKIDP )µ = g is equivalent to show that (− 12I+T ∗KIDP )ρ = 0 has only the zero solution.

Suppose ρ ∈ L2(∂Ω) solves (− 12I+TKENP )ρ = 0. We already know that this implies that ρ is continuous,

and that ρ = 0 is a solution. We want to show that it is the only one so we now use the fact that it solvesSLP [ρ] = v(x) where

v(x) =1

4π

ˆ∂Ω

ρ(y)

|x− y|dSy.

Recall that we have ∆v = 0 for x 6∈ Ω, ∂v∂nx = 0 for x ∈ ∂Ω, and |v(x)| → 0 as |x| → ∞. We will first show

v ≡ 0, and then show that ρ ≡ 0. Consider R3 − Ω. Then I claim something without proof: ∃C > 0, R1 > 0(dependent on ∂Ω, ρ) such that

|v(x)| ≤ C

|x|,

∣∣∣∣ ∂v∂|x|∣∣∣∣ ≤ C

|x|2.

Take R > R1 and |x| < R, multiply our PDE by v and integrate to get

0 =

ˆx∈R3−Ω

v(x)∆v(x)dx

=

ˆx∈R3−Ω

∇ · (v∇v)dx−ˆx∈R3−Ω

|∇v(x)|2dx

=

ˆ|x|=R1

v(x)∂v

∂|x|(x)dSx +

ˆ∂Ω

v∂v

∂nxdSx −

ˆR3−Ω

|∇v(x)|2

=

ˆ|x|=R1

v(x)∂v

∂|x|(x)dSx −

ˆR3−Ω

|∇v(x)|2

ˆR3−Ω

|∇v(x)|2dx =

ˆ|x|=R

v(x)∂v(x)

∂|x|dSx

≤ˆ|x|=R

|v(x)|∣∣∣∣∂v(x)

∂|x|

∣∣∣∣ dSx≤ˆ|x|=R

C

R

C

R2dSx

≤ C3

R3

ˆ|x|=R

dSx

=C

R→ 0

as R → ∞. This means that the gradient equals zero, and so v ≡ C = 0 for x 6∈ Ω. Now we want to showthat ρ = 0. Now, computing the following limit x→ x0 ∈ ∂Ω gives us

∂v

∂nx

∣∣∣∣∂Ω

= 0 = −1

2ρ(x) + TKENP [ρ](x). (22)

Now recall that SLP [ρ](x) = v(x) is continuous on all of R3. This means that

limx→x0∈∂Ω

v(x) = 0

27

as x comes from inside the boundary as well!. Now recall that ∆v = 0 for x ∈ Ω and now we have v = 0 for

x ∈ ∂Ω as a new boundary condition. The only solution of this is v ≡ 0 and so by computing ∂∂nx

v(x)∣∣∣∂Ω

from the inside gives us

1

2ρ(x) + TKENP [ρ](x) = 0 (23)

and subtracting (22) from (23) gives ρ ≡ 0.

2 Variational Methods

We’ll give an introduction to the variational approach to solving PDEs. Recall our Dirichlet boundaryproblem

−∆u = f x ∈ Ωu = g x ∈ ∂Ω

We will formulate the boundary value problem as the solution to an optimization problem. Define Ag =w ∈ C2(Ω)|w = g, x ∈ ∂Ω. Define

I[w] =

ˆΩ

1

2|∇w(x)|2 − w(x)f.

We have the following theorem.

Theorem 2.1. (1) If u ∈ C2(Ω) and u = g on ∂Ω solves −∆u = f in Ω, then

I[u] = minw∈Ag

I[w].

(2) Let u ∈ Ag such that I[u] = minw∈Ag I[w]. Then −∆u = f in Ω and u = g on ∂Ω.

Note that it is not obvious that the minimum exists.

Proof. First we prove (1). We want to show that I[u] ≤ I[w] for all w ∈ Ag. We know that −∆u − f = 0.Let w ∈ Ag and observe that u− w = 0 on ∂Ω. Then we have

0 = (−∆u− f)(u− w)

=

ˆΩ

(−∆u− f)(u− w)

=

ˆΩ

−∆u(u− w) +

ˆΩ

fu+

ˆΩ

fu+

ˆΩ

fw

=

ˆΩ

∇u · ∇(u− w)−ˆ

Ω

fu+

ˆΩ

fw

=

ˆΩ

|∇u|2 −ˆ

Ω

∇u · ∇w −ˆ

Ω

fu+

ˆfw

ˆΩ

|∇u|2 −ˆfu =

ˆΩ

∇u · ∇w −ˆfw

≤ 1

2

ˆΩ

|∇u|2 + |∇w|2 −ˆfw

ˆΩ

1

2|∇u|2 − fu ≤

ˆΩ

1

2|∇w|2 − fw

and we are done with (1).

28

Now lets prove (2). Suppose u minimizes I[w] among all w ∈ Ag. Let v be any function which is C20 (Ω).

Then u+ τv(x) ∈ Ag. Therefore, I[u+ τv] is minimized at τ = 0 i.e. ddz I[u+ τz]

∣∣τ=0

= 0. We calculate

I[u+ τv] =1

2

ˆΩ

|∇u+ τ∇v|2 −ˆ

Ω

(u+ τv)f

=1

2

ˆΩ

|∇u|2 −ˆ

Ω

uf + τ

ˆΩ

∇u · ∇v − vf +τ2

2

ˆΩ

|∇v|2

d

dzI[u+ τz] =

ˆΩ

∇u · ∇v − vf + τ

ˆΩ

|∇v|2

d

dzI[u+ τz]

∣∣∣∣z=0

=

ˆΩ

∇u · ∇v − vf = 0

0 =

ˆΩ

(−∆u− f)v

and we finally have −∆u = f .

Now note that we still have not shown that this functional has a minimum, but we will (hopefully) getto it later in the course.

Eighth Class

Today we will talk about the method of characteristics. We will consider first order nonlinear differentialequations. Let x = (x1, . . . , xn) ∈ Rn, z ∈ R, p = (p1, . . . , pn) ∈ Rn and we will consider

F (Du(x), u(x), x) = 0.

The method of characteristics is a method that reduces solutions of this PDE to a solution of a family ofODEs.

Example 2.2. One linear equation∂u

∂t+ c

∂u

∂x= 0.

Example 2.3. One quasilinear equation is

∂u

∂t+ u

∂u

∂x= 0.

Example 2.4. The Eikonal equation which comes up in geometrical optics is

|∇u(x1, x2, x3)|2 = n2(x1, x2, x3).

This is a fully non-linear equation.

Lets examine example 2.2 with initial value problem u(x, 0) = g(x). We know that the solution to thisis g(x − ct) = u(x, t). An alternative way to get the solution is to view the equation as a statement abouta directional derivative of a function u(x, t). We introduce “characteristic curves.” Let α parametrize thex-axis and a point on the x-axis will be (α, 0). Then we introduce the characteristic curves

∂

∂sX(s, α) = c,

∂

∂sT (s, α) = 1

29

with the initial conditions X(0, α) = α and T (0, α) = 0. Given these characteristic curves, consideru(X(s, α), T (s, α)) and lets look at the rate of change

d

dsu(X(s, α), T (s, α)) = uX(X(s, α), T (s, α))

∂X

∂s(s, α) + uT (X(s, α), T (s, α))

= uX(X,T )c+ uT (X,T )

= 0

if u solves the PDE. We conclude that this is a constant in s and so

u(X(s, α), T (s, α)) = u(X(0, α), T (0, α))

= u(α, 0)

= g(α).

We want u(x, t), so we have to invert the mapping (s, α) 7→ (X,T ). By solving the characteristic equationswe have that X(s, α) = cs+α and T (s, α) = s. We find the inverse mapping as s(X,T ) = T and α(X,T ) =X − cT and we finally get the answer

u(X,T ) = g(α(X,T )) = g(X − cT )

which matches up with our known solution. Again lets go back to ut + cux = 0 with initial conditionu(x, 0) = g(x). We think of solutions as the surface z = u(x, t).

Now lets look at solutions of the equation F (x!, x2, u, ∂x1u, ∂x2

u) = 0. First consider

F (x, u,Du) = b(x, u) ·Du(x) + c(x, u)

=

n∑j=1

bj(x, u)∂u

∂xj(x) + c(x, u)

= 0

with Cauchy data u∣∣Γ

= g where Γ is a curve in R2. Parametrize Γ by α ∈ [0, 1] with Γ 3 (x1, x2) =(h1(α), h2(α)). We rewrite the Cauchy data as

u(h1(α), h2(α)) = g(h1(α), h2(α)) = g(α).

The characteristic equations will be

∂X1

∂s(s, α) = b1(X(s, α), z(s, α))

∂X2

∂s(s, α) = b2(X(s, α), z(s, α))

∂z

∂s(s, α) = −c(X(s, α), Z(s, α))

with initial conditions X1(0, α) = h1(α), X2(0, α) = h2(α) and z(0, α) = g(h(α)). Suppose u(x1, x2) solvesb ·Du+ c = 0. Then

d

dsu(X1(s, α), X2(s, α)) = uX1

∂X1

∂s+ uX2

∂X2

∂s= uX1

b1 + uX2b2

= −c(X1(s, α), X2(s, α))

=∂z

∂s(s, α),

30

so we do have the correct characteristic equations. Now we want to invert the map (s, α) 7→ (X1, X2) i.e. wewant s(X1, X2), α(X1, X2). and this implies that z(s(X1, X2), α(X1, X2)) = u(X1, X2) solves the equation.Let check that this is true, so suppose that we have the inverses.

b(x, u(x)) ·Du(x) = b(X(s, α), z(s, α)) ·Du(X(s, α))

=∂X

∂s(s, α) ·Du(X(s, α))

=

2∑j=1

∂Xj

∂s(s, α)

∂u

∂Xj(X(s, α))

=

2∑j=1

∂Xj

∂s(s, α)

[∂z

∂s

∂s

∂xj+∂z

∂α

∂α

∂xj

]

=

2∑j=1

∂z

∂s

∂xj∂s

∂s

∂xj+∂z

∂α

∂xj∂s

∂s

∂xj

= −c(s(X1, X2), α(X1, X2)).

So now the question is to figure out when we have an inverse. The answer comes from the Inverse FunctionTheorem, and so we have to see if the Jacobian of our mappings is invertible. Recall our map is G : (s, α) 7→(X1(s, α), X2(s, α)). Then the Jacobian will be

DG(s, α) =

(∂x1

∂s (s, α) ∂x1

∂α (s, α)∂x2

∂s (s, α) ∂x2

∂α (s, α)

).

This will need to be invertible at s = 0 i.e. along Γ.

DG(0, α) =

(∂x1

∂s (0, α) ∂x1

∂α (0, α)∂x2

∂s (0, α) ∂x2

∂α (0, α)

)=

(b1(h(α), g(α)) h′1(α)b2(h(α), g(α)) h′2(α)

).

The fact that this determinant is nonzero means that the vector field b is transverse to Γ! This is not onlya property of Γ, but also on the property of g. We say the Cauchy problem for the PDE

b(x, u) ·Du(x) + c(x, u) = 0

is non-characteristic provided that the vector field b(h(α), g(h(α))) is transverse along Γ. We also call Γnon-characteristic. Lets do an example.

Example 2.5. Consider xux + (x + y)uy = u + 1 with u(x, 0) = x2. Lets write down the characteristicODEs:

∂x

∂s= x,

∂y

∂s= x+ y,

∂z

∂s= z + 1

with the initial conditions x(0, α) = α, y(0, α) = 0, z(0, α) = α2 where y(0, α) = 0 because the initial manifoldis y = 0. The solutions for these are x(s, α) = esα, y = αses, and z = es(α2 + 1)− 1. The solution requiresus to invert these. Note that y

x = s i.e. s(x, y) = ys . Now define

u(x, y) = z(s(x, y), α(x, y)) = eyx (x2e−2 yx + 1)− 1.

and the claim is that this solves the equation.

31

Now lets do everything in complete generality. Let C1 3 F (p, z, x) : Rn × R × Rn → R and we want tosolve F (Du, u, x) = 0. We prescribe Cauchy data u|Γ = g where Γ is an (n − 1)-dimensional hypersurface.Consider the mapping [0, 1]n−1 3 α 7→ h(α) ∈ Γ. Also, the rank Dαh = n − 1. We are looking foru(x(s, α)) = z(s, α) where Du(x(s, α)) = p(s, α) = (p1(s, α), . . . pn(s, α)). Lets attempt to calculate thecharacteristic equations. First note that

∂

∂spi(s) =

∂

∂suxi(x(s)) =

n∑j=1

uxixj (x(s))∂xj∂s

(s).

Note now that this involve second derivative terms, which we haven’t dealt with before. If we differentiateour PDE with respect to xi we get

n∑j=1

Fpj (Du, u, x)uxjxi + Fz(Du, u, x)uxi + Fxi(Du, u, x) = 0.

Now choose xj(s) to satisfy∂xj∂s

(s) = Fpj (Du, u, x). (24)

Then using this we can write

∂pi(s)

∂s(s) = −Fxi(Du, u, x)− Fz(Du, u, x)pi. (25)

Finally we can differentiate z to get

∂z

∂s(s, α) =

n∑j=1

uxj (x(s, α))∂xj∂s

=

n∑j=1

Fpjpj

= p(s) · Fp(p(s), x(s), z(s)) (26)

Then (24), (25), and (26) will be our characteristic equations. Lets go back to our first example.

Example 2.6. Recall the Eikonal equation Du ·Du− n2(x) = 0. We can write this as

F (p, x) = p · p− n2(x).

This means that our characteristic equations are

Fz = 0

∂x

∂s= 2p

∂xi

∂s= 2pi

∂pi

∂s=

∂

∂xi(n2(x)).

32

Ninth Class

We were looking C∞ 3 F (p, z, x) : Rn × R× Rn → R and a solution to the equation F (Du(x), u(x), x) = 0with Cauchy data uΓ = g where Γ is an (n− 1)-dimensional hyper surface. We parametrize the curves by αand time by s. We saw that the characteristic ODEs were

d

dsp(s) = −DxF (p(s), z(s), x(s))−DzF (p(s), z(s), x(s))p(s)

d

dsz(s) = DpF (p(s), z(s), x(s))p(s)

d

dsx(s) = Fp(p(s), z(s), x(s)).

So how do we invoke the Cauchy data? we parametrize Γ by (α1, . . . , αn−1) 7→ (h1(α), . . . , hn(α)) ∈ Γ. Thenu(h(α)) = g(h(α)). Lets do some examples.

Example 2.7. We have x1∂x2u − x2∂x1u = u with Cauchy data u(x1, 0) = g(x1). The first step is to findthe characteristic equations. The first two are as follows:

dx1

ds(s) = −x2(s),

dx2

ds(s) = x1(s)

with initial conditions x1(0, α) = α, x2(0, α) = 0. The last one will be

dz

ds(s) = z(s, α)

with initial condition z(0, α) = g(α). From here we see that ∂2x1

∂s2 (s) = −x1(s) with x1(0, α) = α and∂x1

∂s (0, α) = 0. This implies that x1(s, α) = α cos(s) and similarly x2(s, α) = α sin(s). Finally, z(s, α) =esg(α). Now what is left is to invert the map (s, α) 7→ (x1, x2). Note that x2

1(s, α) + x22(s, α) = α2 =⇒

α(x1, x2) =√x2

1 + x22 where we obviously take the positive square root. To get s, note that x2

x1(s, α) = tan(s)

and so s(x1, x2) = tan−1(x2

x1). This gives finally that

u(x1, x2) = z(s(x1, x2), α(x1, x2)) = etan−1

(x2x1

)g(√x2

1 + x22).

Example 2.8 (Hamilton-Jacobi Equations). These are equations for a functions u(x, t) in the form

∂u

∂t(x, t) +H(Du(x), x) = 0

where H is called the Hamiltonian. We want to view this as an equation of the form F (Du, u, x) = 0,but we have time here so we must introduce the space time gradient. Clearly, this equation is of the formG(Du, ut, u, x, t) = 0. We write this in the form G(q, u, y) = 0 where q = (Du, ut) = (p, pn+1) and y = (x, t).The equation now becomes

G(q, z, y) = pn+1 +H(p, x) = 0.

Now see thatDqG(q, z, y) = (Hp(p, x), 1)

andDyG(q, z, y) = (Hx(p, x), 0).

So the characteristic equations are as follows:

dq

ds= −DyG−DzGq = −DyG = −(Hx(p(s), x(s)), 0) (27)

dz

ds= (Hp(p, x), 1) · (p, pn+1) = p(s)Hp(p(s), x(s)) + pn+1(s) (28)

dy

ds= (Hp(p(s), x(s)), 1). (29)

33

Now we pick out coordinates by coordinate and get

d

dsx(s) = Hp(p(s), x(s)),

d

dsxn+1(s) = 1

d

dsp(s) = −Hx(p(s), x(s)),

d

dspn+1(s) = 0

d

dsz(s) = p(s)Hp(p(s), x(s)) + pn+1(0)

where the last part has a pn+1(0) because pn+1 is constant from its characteristic equation.

Example 2.9 (Eikonal Equation). This relates to geometric optics. Here we are interested in a solutionu(x) where the PDE is

F (p, x) = p · p− n2(x) = H(p, x) = 0.

We can write this as |Du(x)|2 − n2(x) = 0 where n2(x) is a strictly positive function that is prescribed. Itis usually called the refractive index. The characteristic equations are

dx

ds(s) = Hp(p, x) = zp

dp

ds(s) = −Hx(p, x) = ∇[n2(x)]

dz

ds(s) = p · ∂H

∂p= p · 2p = 2|p|2 = 2n2(x).

Now lets show that it suffices to show transversality to prove that we have a local solution about Γ. Notethat we have for free the relations x(0, α) = h(α) and z(0, α) = g(h(α)). We’d like to know p(0, α). Fromthe compatibility condition u(h(α)) = g(h(α)) ands differentiating with respect to α gives

∂

∂αiu(h1, . . . , hn) =

∂

∂αig(h(α))

n∑j=1

uxj (h(α))∂hj

∂αi(α) =

∂

∂αig(h(α)).

Then, on Γ, the first n− 1 compatibility conditions are

n∑j=1

∂hj

∂αi(α)pj(0, α)− ∂

∂αig(h(α)) = 0

where i runs from 1 to n− 1. The last one comes from the PDE itself on Γ:

F (p1(0, α), . . . , pn(0, α), h1(g(α)), . . . , hn(g(α))) = 0.

These now all give n-nonlinear equations in n unknowns p1(0, α), . . . , pn(0, α), which are what we want. Wenow assume solvability at some α = α0 i.e. there exists compatible initial conditions p1(0, α0), . . . , pn(0, α0),g(h(α0)), h1(α0), . . . , hn(α0) for the characteristic equations. We seek a condition ensuring that in a neigh-borhood around α0 (α near α0, i.e. points x ∈ Γ near x0 = h(α0)). There are still compatible initialconditions

G(p1, . . . , pn, α1, . . . , αn) = 0.

Then the implicit function theorem gives us conditions for when this is solvable in a neighborhood for onechoice of parameters G(p0, α0) = 0 i.e. DpG(p0, x0) is invertible as an n× n matrix.

34

Now we take the Jacobian of the n compatibility conditions i.e. apply the implicit function theorem onthe n× n jacobian of G(p, α):

DG(p0, α0) =

∂h1

∂α1(α0) · · · ∂hn

∂α1(α0)

.... . .

...∂h1

∂αn−1(α0) · · · ∂hn

∂αn−1(α0)

Fp1(p0, α0) · · · Fpn(p0, α0)

.

Then the invertibility condition is that Fp(p0, α0) is not tangent to Γ at h(α0).Now lets start with Characteristics and an introduction to mathematical theory of shock waves. Consider

the density function ρ(x, t) and the PDE

∂tρ(x, t) + ∂x(Q(ρ)) = 0

where Q(ρ) = ρ(x, t)v(x, t). This is not a closed system because we need to know v(x, t)! However, weassume it to be closed and have

∂tρ(x, t) + ∂x(ρ(x, t)v(ρ(x, t))) = 0.

One case is where v(ρ) = 12ρ. Then we are able to write our PDE as ∂tρ+ ∂x( 1

2ρ2) = ∂tρ+ ρ∂xρ = 0. Then

our final initial value problem will be ∂tρ+ ρ∂xρ = 0ρ(x, 0) = F (x).

We will solve this by method of characteristics. Note that our Γ = t = 0. Our characteristic equations are

dt

ds(s, α) = 1, t(0, α) = 0

dx

ds(s, α) = z(s, α), x(0, α) = α

dz

ds(s, α) = 0, z(0, α) = F (α).

We can solve these as t(s, α) = s, z(s, α) = z(0, α) = F (α), x(s, α) = F (α)s + α. Then u(x, t) =z(s(x, t), α(x, t)) will solve our PDE. we need to invert the mapping (s, α) 7→ (t(s, α), x(s, α)) = (s, F (α)s+α).In order for this to be invertible we need

∂(t, x)

∂(s, α)= det

(∂t∂s

∂t∂α

∂x∂s

∂x∂α

)= det

(1 0

F (α) F ′(α)s+ 1

)= 1 + F ′(α)s

6= 0.

Now note that F (α) is such that |F ′(x)| ≤ C, then for all s sufficiently small, 1+F ′(α) > 0 i.e. the mapping(s, α) 7→ (t, x) is invertible i.e. we have solved our PDE. The solution is given by

ρ(x, t) = z(s(x, t), α(x, t))

= F (α(x, t))

ρ = F (x− ρt),

which is an implicit equation for ρ(x, t).

35

Tenth Class

Lets give an introduction to the mathematical theory of non-linear hyperbolic waves and shock waves.Suppose we have a tube of gas and a function ρ(x, t) that measures the density in time. Recall the law ofconservation of mass:

ˆ x2

x1

∂tρ(x, t) =d

dt

ˆ x2

x1

ρ(x, t)dx

= −[Q(ρ(x2, t), t)−Q(ρ(x1, t), t)]

= −ˆ x2

x1

∂xQ(ρ(x, t))dx.

This gives us the equation∂tρ(x, t) + ∂xQ(ρ(x, t), t) = 0. (30)

One special case we will study is Q(ρ) = ρv = ρv(ρ) = 12ρ

2. This would give us

∂tρ+ ∂x

(1

2ρ2

)= 0

∂tρ+ ρ∂xρ = 0. (31)

(31) is a first order quasilinear PDE and is called the Inviscid Burger’s equation.Now we have an exact problem that we want to solve: ∂tρ+ ρ∂xρ = 0 with ∂(x, 0) = F (x). We will solve

this by method of characteristics. The characteristic equations are as follows

dt(s, α)

ds= 1,

dx(s, α)

ds= z(s, α),

dz(s, α)

ds= 0

with initial conditions t(0, α) = 0, x(0, α) = α, and z(0, α) = F (α). The solution to these equations give ust(s, α) = s, x(s, α) = F (α)s+ α and z(s, α) = F (α). Note that solutions to these will be straight lines withslope 1/F (α0) by taking the ratio of dx/ds and dt/ds. This means that in order to solve our ODE at anarbitrary point (x0, t0) we simply need to find the characteristic F (α(x0, t0)) and trace it back because z isconstant along the curves.

From this, we will break down our problem into one case F ′(x) ≥ 0 (called Rare faction) and F ′(x) ≤ 0(called the Shock formation). Lets look at Case 1. We will look at an initial condition F that startssomewhere at 0, and increases monotonically into an F∞. Note that wherever F = 0, we see that thecharacteristic lines will be vertical lines, and they sort of collapse towards lines with constant slope 1

F∞(note

that we can always trace our way back by the inverse function theorem and finding the jacobian of ∂(t,x)∂(s,α) ).

Lets do a special case:

F (x) =

0 x ≤ 0x 0 ≤ x ≤ 11 x ≥ 1.

It’s trivial to see how the characteristics look like in (−∞, 0]∪ [1,∞) so lets examine the inside part. Recallthat in here, x(s, α) = αs + α = α(1 + s) and t(s, α) = s. We can solve to find that α(x, t) = x

1+t . Thenfrom F (α) = α we can easily solve

ρ(x, t) = 0 x ≤ 0ρ(x, t) = x

1+t 0 < x < 1 + t

ρ(x, t) = 1 x > 1 + t,

and we do have Rare Faction but note that we still need to say what it means for the PDE so be solved atx = 0 and x = 1 + t.

36

Lets look at another case of Rare Faction. Take

F (x) =

0 x ≤ 01 x > 0.

As before, the characteristics are vertical on x ≤ 0 and they turn into 45 angled lines after 0. This yieldssolutions as

ρ(x, t) =

0 x < 0?? 0 < x < t1 x > t,

so what happens between 0 < x < t? In this interval, we introduce the notion of scale invariance andself-singularity. We claim that this initial value problem is scale-invariant i.e. ρ(βx) = F (βx) = F (x) for allβ > 0. Also observe that if ρ(x, t) solves ∂tρ(x, t) + ρ(x, t)∂xρ(x, t) = 0, then ρβ(x, t) = ρ(βx, βt) also solvesthe equation. Now suppose ρ(x, t) solves the equation with data F (x), then ρβ(x, t) solves the equationwith data ρ(βx, 0) = F (βx) = F (x). In other words, ρβ(x, t) must be independent of β! This means thatρ(x, t) = G(xt ) because plugging in (βx, βt) makes the β’s cancel. Then

∂tρ(x, t) = ∂tG(xt

)= G′

(xt

)∂t

(xt

)= −G′(ξ)ξ

t

where ξ = x/t. Then similarly we get ∂tρ + ρ∂xρ = G′(ξ)t [−ξ + G(ξ)] = 0. Then either G′(ξ) = 0 (G =

constant) or G(ξ) = ξ = xt . This implies that in the general solution, we have

ρ(x, t) =

0 x < 0xt 0 < x < t1 x > t.

Now lets discuss the notion of a weak or distributional solution of a PDE. Lets start with ∂tρ+∂xQ(ρ) = 0with initial condition ρ(x, 0) = ρ0(x). Assume that ρ(x, t) ∈ C1(R× [0,∞)). Let φ(x, t) ∈ C∞0 (R× [0,∞)).We multiply the equation by φ(x, t) and integrate to get

0 = ∂t(φρ) + ∂x(φQ)− ∂tφρ− ∂xφQ(ρ)

=

ˆ ∞−∞

dx

ˆ ∞0

dt[∂t(φρ) + ∂x(φQ)− ∂tφρ− ∂xφQ(ρ)]

= −ˆ ∞−∞

dxφ(x, 0)ρ0(x)−ˆ ∞

0

dt

ˆ ∞−∞

dx[∂tφρ+ ∂xφQ(ρ)]

And so we say that our function solves our PDE if for every φ ∈ C∞0 (R× [0,∞)),ˆ ∞−∞

φ(x, 0)ρ0(x)dx = −ˆ ∞

0

ˆ ∞−∞

[∂tφρ+ ∂xφQ(ρ)].

Lets consider the PDE ρt + ρρx = 0 with initial condition

ρ(x, 0) = F (x) =

1 x < 01− x 0 < x < 10 x ≥ 1.

Then for x < 0, the characteristic curves are just lines with slope 1 and for x ≥ 1, we have vertical lines.From this, we see that these lines intersect! Recall that t(s, α) = s, x(s, α) = F (α)s+α = (1−α)s+α. Thismeans that the slope will be dt

dx = 11−α and all the lines between (0, 1) cluster up at (1, 1). Note that before

t = 1, everything is fine, so lets find the solutions there:

ρ(x, t) =

1 x < t < 11−x1−t t < x < 1.

0 x > 1 > t,

37

where we have solved for α and plugged it into ρ(x, 0) = F (α). Notice that as limt→1− gives us a discontinuityi.e. a shock. Now lets try to find a solution for t > 1. We want to extend a curve ξ(t) from t = 1 so thatto the left of it, the solution is identically 1, and to the right of it, the solution is identically zero (we areessentially transporting the shock through time). Suppose we are at t = 1 + δ and pick two points x1 andx2. To determine x = ξ(t), lets go back to the physics:

−[Q(ρ(x2, t))−Q(ρ(x1, t))] =d

dt

ˆ x2

x1

ρ(x, t)dx

=d

dt

[ˆ ξ(t)

x1

ρ(x, t)dx+

ˆ x2

ξ(t)

ρ(x, t)dx

]

= ρ(ξ(t)−, t)ξ(t) +

ˆ ξ(t)

x1

ρtdx− ρ(ξ(t)+, t)ξ(t) +

ˆ x2

ξ(t)

ρtdx

= [ρ(ξ−(t), t)− ρ(ξ(t)+, t)]ξ(t) +

ˆ x2

x1

ρtdx.

Taking the limit x1 ↑ ξ(t), x2 ↓ ξ(t) gives us

ξ(t)[(ρ(ξ(t)−, t)− ρ(ξ(t)+, t))] = Q(ρ(ξ(t)−, t))−Q(ρ(ξ(t)+, t))

ξ(t) =[Q](ξ(t))

[ρ](ξ(t))

=Q ρ(ξ+, t)−Q ρ(ξ−, t)

ρ(ξ+, t)− ρ(ξ−, t).

Now notice that we have the initial condition ξ(1) = 1 and that Q = 12ρ

2. This implies that

ξ(t) =12ρ

2R − 1

2ρ2L

ρR − ρL=

1

2(ρR + ρL) =

1

2(0 + 1) =

1

2.

This gives us the complete solution as

ρ(x, t) =

1 x ≤ t 0 ≤ t ≤ 11−x1−t t ≤ x < 1 0 ≤ t ≤ 1

0 1 ≤ x 0 ≤ t ≤ 11 x < ξ(t) 1 < t <∞0 x > ξ(t) 1 < t <∞.

38

Eleventh and Twelfth Class

Recall we were looking at nonlinear 1st order PDEs of the form ∂tρ + ∂x( 12ρ

2) = 0. Particularly, assumingρ was smooth, we have the IVP

∂tρ+ ρ∂xρ = 0ρ(x, 0) = F (x).

We saw the case where F ′ ≤ 0

ρ(x, 0) = F (x) =

1 x < 01− x 0 ≤ x < 10 x ≥ 1

and found the solution to be

ρ(x, t) =

1 x ≤ t 0 ≤ t ≤ 11−x1−t t ≤ x < 1 0 ≤ t ≤ 1

0 1 ≤ x 0 ≤ t ≤ 11 x < ξ(t) 1 < t <∞0 x > ξ(t) 1 < t <∞.

The problem is that ρ is not a smooth function, so we must introduce the notion of weak solutions. We saythat ρ is a weak solution of ∂tρ+Q(ρ(x, t)) = 0 if ∀ϕ ∈ C∞0 (R× [0,∞)),

ˆ ∞0

dt

ˆ ∞−∞

∂tϕ(x, t)ρ(x, t) + ∂xϕ(x, t)Q(ρ(x, t))dx+

ˆ ∞−∞

ϕ(x, 0)F (x)dx = 0.

The problem with weak solutions is that the class of weak solutions is too big: we might not get uniqueness.Note that smoothness guarantees uniqueness: suppose we have u1, u2 are solutions to ut + uux = 0, withu(x, 0) = F (x) ∈ C∞. Then I claim that u1 = u2. Consider δ(x, t) = u1 − u2. Then δ is smooth and itsolves the equation

0 = δt + u1u1x − u2u2x

= δt + δu1x + u2δx.

The initial value problem is δ(x, 0) = 0 and the solution to this IVP is δ ≡ 0 (left as an exercise to thereader, done by method of characteristics).

Now lets examine that weak solutions are not necessarily unique. Consider the equation ρt + (ρ2

2 )x = 0with the initial condition

ρ(x, 0) = F (x) =

0 x < 01 x ≥ 0.

We already saw one solution to be

ρ1(x, t) =

0 x < 0xt 0 < x < t1 x > t.

We claim that this is a weak solution to our PDE, and also

ρ2(x, t) =

0 x < 1

2 t1 x > 1

2 t

is a weak solution. The resolution of non-uniquness is to go back to the physics. We use Lax EntropyCondition, which says that the physical solution respects causality, and “characteristics enter the shock fromthe past.” Another resolution is to include important physical effects which we’ve neglected in the model

39

∂tρ+ ρ∂xρ = 0. We now include viscosity and call its coefficient ν (can think of it as friction). Then we getthe Burger’s equation:

∂tρ+ ρ∂xρ = ν∂2xρ.

Now the solution ρν depends on the parameter ν 1. The physical solution of ρt + ρρx = 0, ρ(x, 0) = F (x)is going to be the one which arrises as limν→0 ρ

ν .The problem is that Burger’s equation can be mapped to the linear heat equation by a non-linear change

of variables (otherwise, we’d be in trouble because this is a non-linear second order PDE)! Therefore, we’llbe able to write down an explicit solution!

Let ρ = ψx. Then

ψxt + ψxψxx = νψxxx

(ψt)x +1

2(ψ2x)x = ν(ψxx)x

ψt +1

2ψ2x = νψxx.

Now let ψ = −2ν logϕ, and calculate by an exercise, that

∂tϕ = ν∂2xϕ,

also known as our friend the heat equation! The initial condition will be

ϕ(x, 0) = e−12ν ψ(x, 0) = e−

12ν

´ x0F (y)dy

because ψ is the logarithm of ϕ, and ψ is the derivative of ρ. Lets calculate our solution now:

ϕ(x, t) =1√

4πνt

ˆ ∞−∞

exp

(− 1

2ν

ˆ ξ

0

F (y)dy − (x− ξ)2

4νt

)dξ

=1√

4πνt

ˆ ∞−∞

e−12νG(x,t,ξ)dξ

where

G(x, t, ξ) =(x− ξ)2

2t+

ˆ ξ

0

F (y)dy.

Taking derivatives and ratios now gives gives us

ϕx =1√

4πνt

ˆ ∞−∞−x− ξ

2νte−

12νG(x,t,ξ)dξ

−2νϕxϕ

=

´∞−∞

x−ξt e−

12νG(x,t,ξ)dξ´∞

−∞ e−12νG(x,t,ξ)dξ

= ρν(x, t).

Now note that the solution to our problem has been changed to the limit ν → 0 of this initial value problemto the heat equation. Lets figure out how to do this.

Lemma 2.10 (Laplace Asymptotes). Define

I(ε) =

´−∞∞l(y)e−k(y)/εdy´∞−∞ e−k(y)/εdy

where k(y), l(y) are continuous, k(y) ≤ cy2, |l(y)| ≤ c|y| as y →∞. Then.

limε→0

I(ε) = l(y0)

where y0 is defined by k(y0) = min k(y).

40

In our case, we will have to find the minimum of G. Taking derivatives gives us

∂G

∂ξ= 0

∂G

∂ξ= −x− ξ

t+ F (ξ)

ξ = x+ tF (y).

Then

limν→0

ρν =x− ξ(x, t)

t

= F (ξ(x, t)).

This is exactly the physical solution! Lets say something about the proof of Lemma 2.10. Suppose k(y0) =min k(y), and that it is unique. Pick δ > 0 small and write

ˆ ∞−∞

l(y)e−k(y)/εdy =

ˆ|y−y0|<δ

l(y)e−k(y)/εdy +

ˆ|y−y0|≥δ

l(y)e−k(y)/εdy

=

ˆ|y−y0|<δ

(l(y)− l(y0) + l(y0))e−k(y)/εdy + C

= l(y0)

ˆ|y−y0|<δ

e−k(y)/εdy +

ˆ|y−y0|<δ

(l(y)− l(y0))e−k(y)/εdy + C.

Expanding the Taylor expansion of k, we get the desired limit after taking the ratio and doing the sameprocess without l(y) in the integral.

Now lets study the heat equation in some rigor. We know it to be ut = ∆u with u(x, 0) = g(x). Thehomework helped us work out the solution as

u(x, t) = kt ∗ g(x, t) =

ˆRn

1

(4πt)n/2e−|x−y|2

4t g(y)dy.

We have the theorem from class that if g is bounded, then u ∈ C∞(Rn × (0,∞)). Moreover, if g is boundedand continuous, then g(x0) = u(x0, 0). We leave it as an exercise to show that if g has a jump discontinuityat y1, then u(y1, 0) = 1

2 (g(y+1 ) + g(y−1 )). As a typical maximum and minimum principle, we clearly see that

inf g ≤ u(x, t) ≤ sup g. This is due to g being bounded.Now lets look at Duhamel’s principle. The idea is that the solution to the non homogeneous equation

∂u

∂t= ∆u+ f(x, t)

can be reduced to solving a one-parameter family of homogeneous equations. Lets look at an elementaryexample. Consider x = ax+ f(t), where a is a constant. Then we can solve this via an integrating factor toget

x(t) = eatx0 +

ˆ t

0

ea(t−s)f(s)ds.

The first piece is the homogeneous solution, and the second is the particular solution. Note that

xparticular(t) =

ˆ t

0

u(t, s)ds

where u(t, s) satisfies ddtu(t, s) = au(t, s) and u(s, s) = f(s). This sort of stuff also holds n × n matrices:

x = Ax+ f(t) where f(t) is a vector-valued function. Then the solution will be

x(t) = eAtx0 +

ˆ t

0

eA(t−s)f(s)ds.

41

We will now view the heat equation as an ODE dudt = Au(t) + f(t) with u(0) = u0, which is an initial

value problem in time for a function u(t) in a Banach space. First lets analyze the Homogeneous problemdudt = Au(t) = ∆u(t) with u(0) = g. Then the solution will be, as expected,

u(t) = e∆tg =1

(4πt)n/2

ˆRne−|x−y|2

4t dy.

Then, by analogy of what we’ve seen, the solution of the inhomogeneous equation dudt = ∆u+ f(x, t) can be

seen as

u(t) = e∆tg +

ˆ t

0

e∆(t−s)f(s)ds

=1

(4πt)n/2

ˆRne−|x−y|2

4t dy +

ˆ t

0

1

(4π(t− s))n/2

ˆRne−|x−y|24(t−s) dy.

Now lets look at some energy estimates. Multiplying the heat equation (with initial condition u(x, 0) =g(x)→ 0 as x→ ±∞) by u and integrating, we get

uut = u∆u

(u2/2)t = u∆u

∂

∂t

ˆRn

u2

2dx = −

ˆRn|Du|2dx ≤ 0.

This tells us that the energy is decreasing with time.Now consider the equation ut = D · (D(x)Du). With D ≡ 1, we know the solution to be u(x, t) =

Kt ∗ g(x, t). What about D general where 0 < θ− ≤ D(x) ≤ θ+? In fact you can take

u(x, t) =

ˆRnKD(x, y, t)g(y)dy

where

c−e− |x−y|

2

σ−t

tn/2≤ |KD(x, y, t)| ≤ C+

e− |x−y|

2

σ+t

tn/2

where σ+, σ−, C+, C− depend on θ−, θ+.Now lets discuss the wave equation utt = c2uxx. We can factor this PDE into (∂t − c∂x)(∂t + c∂x)u = 0.

We introduce characteristic coordinates ξ = x+ ct and η = x− ct. Then the chain rule dictates that

∂

∂t= c

∂

∂ξ+ c

∂

∂η

∂

∂x=

∂

∂ξ− c ∂

∂η.

Plugging this back into the wave equation yields

∂

∂ξ

∂

∂ηu = 0.

Integrating once yields ∂∂ηu = f(η). Once more yields u(ξ, η) = F (η) +G(ξ). This gives us the solution as

u(x, t) = F (x+ ct) +G(x− ct).

Now lets look at the solution of the initial value problem u(x, 0) = f(x) and ut(x, 0) = g(x). We alreadyknow the solution must be u(x, t) = F (x + ct) + G(x − ct). Then u(x, 0) = F (x) + G(x) = f(x) and

42

ut(x, 0) = cF ′(x)− cG′(x) = g(x). Differentiating the first equation gives us F ′(x) +G′(x) = f ′(x). Solvingfor F ′ and G′ yields

F ′(x) =1

2f ′(x) +

1

2cg(x)

G′(x) =1

2f ′(x)− 1

2cg(x)

F (x) =1

2f(x) +

1

2c

ˆ x

0

g(s)ds

G(x) =1

2− 1

2c

ˆ x

0

g(s)ds.

This gives us Dalambert’s formula formula

u(x, t) =1

2(f(x+ ct) + f(x− ct)) +

1

2c

ˆ x+ct

x−ctg(s)ds.

Lets talk a little bit about the three dimensional wave equation ut = ∆u, u(x, 0) = f(x), ∂tu(x, 0) = g(x).One way to do this is to take the Fourier transform of the equation. We will take another route, by lookingat it as a version of the one dimensional wave equation. Suppose h : Rn → R is continuous. Define

Mh(x, r) =1

ωnrn−1

ˆ|y−x|=r

h(y)dSy

=1

ωn

ˆ|ξ|=1

h(x+ rξ)dSξ.

We wish to extend Mh(x, r) to all r ∈ R as an even function by using the previous equation. We concludetoday with the Euler-Poisson-Darboux identity:

∆xMh(x, r) =∂2

∂r2Mh(x, r) +

n− 1

r

∂

∂rMh(x, r).

This will be used to convert 3-d wave equation for u(x, t) in (x, t) into the 1-d wave equation for Mu(x,t)(x, r)in (r, t).

Thirteenth Class

We will discuss waves in Rn. Recall the d’Alembert’s solution of the initial value problem for 1c2 ∂

2t u = ∂2

xuas

u(x, t) =1

2(f(x+ ct) + f(x− ct)) +

1

2c

ˆ x+ct

x−ctg(s)ds

where u(x, 0) = f(x) and ut(x, 0) = g(x).We will later specialize to n = 2, 3. The strategy will be to solve 3d wave equation using the method of

spherical means. Next we’ll solve the 2d wave equation by specializing the above to initial data of the formf = f(x1, x2) and g = g(x1, x2).

Let h : Rn → R and define

Mh(x, r) =1

ωnrn−1

ˆ|x−y|=r

h(y)dSy.

Now define y = x+ rξ with |ξ| = 1. Then

Mh(x, r) =1

ωn

ˆ|ξ|=1

h(x+ rξ)dSξ.

43

The advantage for this change of variables is that now we can differentiate with respect to r easily inthe second form, and that now we can extend Mh(x, r) for values r < 0 by Mh(x, r) = mh(x,−r) (evenextension). Now consider

1

c2p2tu(x, t) = ∆xu(x, t).

We will study Mu(·,t)(x, r) by studying a PDE for it. Lets calculate

∂

∂rMh(x, r) =

1

ωn

ˆ|ξ|=1

Dxh(x+ rξ) · ξdSξ

=1

ωn

ˆ|ξ|<1

DξDxh(x+ rξ)dξ

=r

ωn

ˆ|ξ|<1

DxDxh(x+ rξ)dξ

=r

ωn∆x

ˆ|ξ|<1

h(x+ rξ)dξ

=r1−n

ωn∆x

ˆ|x−y|<1

h(y)dy

= r1−n∆x

ˆ r

0

(ˆ|y−x|=ρ

h(y)dSy1

ρn−1ωn

)ρn−1

= r1−n∆x

ˆ r

0

ρn−1Mh(x, ρ)dρ

rn−1 ∂

∂rMh(x, r) = ∆x

ˆ r

0

ρn−1Mh(x, ρ)dρ

∂

∂rrn−1 ∂

∂rMh(x, r) = ∆xMh(x, r)rn−1

∆rMh(x, r) = ∆xMh(x, r).

where we have used that for f(r), r =√x2

1 + · · ·x2n, then

∆xf(r) =1

rn−1

∂

∂rrn−1 ∂

∂rf(r).

Also note that Mh(x, r) is even w.r.t. r, ∂∂rMh(x, 0) = 0. Now suppose that u(x, t) solves the wave equation.

Consider Mu(·,t)(x, r). Then our calculations yield

∆xMu(·,t) =∂2

∂r2Mu(·,t)(x, r) +

n− 1

r

∂

∂rMu(·,t)(x, r)

∆x1

ωn

ˆ|ξ|=1

u(x+ rξ, t)dSξ =∂2

∂r2Mu(·,t)(x, r) +

n− 1

r

∂

∂rMu(·,t)(x, r)

1

ωn

ˆ|ξ|=1

∆xu(x+ rξ, t)dSξ =∂2

∂r2Mu(·,t)(x, r) +

n− 1

r

∂

∂rMu(·,t)(x, r)

1

ωn

ˆ|ξ|=1

1

c2∂2

∂t2u(x+ rξ, t)dSξ =

∂2

∂r2Mu(·,t)(x, r) +

n− 1

r

∂

∂rMu(·,t)(x, r)

1

c2∂2

∂t2Mu(·,t)(x, r) =

∂2

∂r2Mu(·,t)(x, r) +

n− 1

r

∂

∂rMu(·,t)(x, r).

Now the upshot is that for n = 3, then this can be transformed into the 1-dimensional wave equation, whichwe’ll solve using d’Alembert’s solution. A simple fact shows that

∂2

∂r2u+

2

r

∂

∂ru =

1

r

∂2

∂r2(ru).

44

Plugging this back in our equation yields

1

c2∂2tMu(·,t)(x, r) =

1

r

∂2

∂r2(rMu(·,t)(x, r))

1

c2∂2t (rMu(·,t)(x, t)) =

∂2

∂r2(rMu(·,t)(x, r)).

This is the 1-d wave equation for the function F (r, t) = rMu(·,t)(x, r)! We need to find initial conditions,

which are F (r, 0) = rMf (x, r) and ∂tF (r, 0) = rMg(x, r). Then d’Alembert’s formula for 1c2 ∂

2t F (r, t) =

∂2r

F (r, t) will yield

rMu(·,t)(x, r) =1

2((r + ct)Mf (x, r + ct) + (r − ct)Mf (x, r − ct)) +

1

2c

ˆ r+ct

r−ctsMg(x, s)ds.

We want u(x, t), but we have Mu(·,t)(x, r). We need to take the limit as r → 0 because this is the averageof u, and using evenness yields

rMu(·,t)(x, r) =1

2((r + ct)Mf (x, r + ct) + (r − ct)Mf (x, r − ct)) +

1

2c

ˆ r+ct

r−ctsMg(x, s)ds

Mu(·,t)(x, r) =1

2r((r + ct)Mf (x, r + ct)− (ct− r)Mf (x, ct− r)) +

1

c

1

2r

ˆ ct+r

ct−rsMg(x, s)ds.

Define J(r) = (ct+ r)Mf (x, ct+ r). Then

J(r)− J(−r)2t

= J ′(0) +O(r)

as r → 0 and so

J ′(r)

∣∣∣∣r=0

=∂

∂r[(ct+ r)Mf (x, ct+ r)]r=0

=

[(ct+ r)

∂

∂rMf (x, ct+ r) +Mf (x, ct+ r)

]r=0

= Mf (x, ct) +

[ct+ r

c

∂

∂rMf (x, ct+ r)

]r=0

= Mf (x, ct) + t∂

∂tMf (x, ct)

=∂

∂t(tMf (x, ct)).

The other limit of the other term is

1

c

1

2r

ˆ ct+r

ct−rsMg(x, s)ds→

1

cctMg(x, ct).

Hence the solution for the 3 − d wave equation with initial conditions u(x, 0) = f(x), ut(x, 0) = g(x) forn = 3 is

u(x, t) =∂

∂t(tMf (x, ct)) + tMg(x, ct)

=∂

∂t

[t

4π

ˆ|ξ|=1

f(x+ ctξ)dSξ

]+

t

4π

ˆ|ξ|=1

g(x+ ct)dSξ

=∂

∂t

[1

4πc2t

ˆ|x−y|=ct

f(y)dSy

]+

1

4πc2t

ˆ|x−y|=ct

g(y)dSy. (32)

Lets make some remarks that follow from these formulas.

45

Remark. There is a Degregation of point wise regularity of solutions, i.e. u(x, t) is not as “differentiable” asf, g. One way to see this is to assume f ∈ C2. Then since we are taking a derivative in the solution (33),u ∈ C1. More explicitly,

u(x, t) = tMg(x, ct) +∂

∂t(tMf (x, ct))

=t

4π

ˆ|ξ|=1

g(x+ ctξ)dSξ +1

4π

ˆ|ξ|=1

f(x+ ctξ)dSξ +t

4π

ˆ|ξ|=1

cξ ·Dxf(x+ ct)dSξ

=1

4πc2t2

ˆ|x−y|=ct

(tg(y) + f(y) +Dyf(y)(y − x)) dSy.

Remark. In an “energy sense,” the solution is just as regular as the data. By this we mean that the waveequation has a conservation law

1

c2∂2

∂t2u = ∆u

∂

∂tu

1

c2∂2

∂t2u =

∂

∂tu∆u

∂

∂tu

1

c2∂2

∂t2= D

(∂

∂tuDu

)−D ∂

∂tu ·Du

∂

∂tu

1

c2∂2

∂t2= D

(∂

∂tuDu

)− ∂

∂tDu ·Du

∂

∂tu

1

c2∂2

∂t2= D

(∂

∂tuDu

)− ∂

∂t

(Du ·Du)

2

1

c2∂

∂t

(∂u∂t

)22

= D

(∂

∂tuDu

)− ∂

∂t

(Du ·Du)

2

which leads to

∂

∂t2

[1c2

(∂u∂t

)2+Du ·Du

]2

+D

(−∂u∂tDu

)= 0.

This gives the conservation law

∂

∂te(∂tu,Du)︸ ︷︷ ︸

energy density

+ D · J(∂tu,Du)︸ ︷︷ ︸energy flux density

= 0.

Integrating yields the constant

E(t) = E(0) =1

2

ˆR3

1

c2g(x)2 + |Df(x)|2.

Remark. Lets think a little bit about the domain of independence. The representation of u(x, t) as an integralover |x− y| = ct shows that u(x, t) depends on the values of f(x),∆f(x), and g(x) on S(x, ct).

Remark. Now lets talk about the range of influence. Given a point (y, 0) ∈ R3, the surface of the forwardlight cone is what will be influenced. Now suppose f, g ∈ C∞0 (Ω) for Ω ⊂⊂ R3. Then the domain of influenceis the union of the light cones along the boundary. However, note that as time goes on, there will be a sharparrival and sharp departure of signals (it helps to draw the picture).

Remark. Lets talk about the time-decay of solutions and diffactive spreading of waves. Assume f, g vanishoutside B(0, ρ) = x : |x| < ρ. Then I claim that

|u(x, t)| ≤ C(‖g‖L∞ + ‖f‖L∞ + ‖Df‖L∞)

t.

46

Note that ˆA

h =

ˆA

hχh6=0.

This trivially implies ∣∣∣∣ˆA

h

∣∣∣∣ ≤ ‖h‖L∞Vwhere V is the volume of the support. Now, each term of the solution of the 3-d wave equation has the form

ˆ|x−y|=ct

hdSy

where h is obtained from the data and compactly supported in B0(ρ). Then∣∣∣∣∣ˆ|x−y|=ct

hdSy

∣∣∣∣∣ ≤ ‖h‖L∞ Area(|x− y| = ct ∩B0(ρ)) ≤ 4πρ2‖h‖L∞ .

Now recall that

u(x, t) =1

4πc2t2

ˆ|x−y|=ct

(tg(y) + f(y) +Dyf(y)(y − x)) dSy

and each of these terms are bounded by what I just computed. Then

|u(x, t) ≤ 1

4πc2t2t‖g‖L∞4πρ2 +

1

t2‖f‖L∞4πρ2 +

1

t2‖Df‖L∞4πρ2.

Now that we have finished talking about the 3-D wave equation, lets see how to get back the 2-D waveequation. Our goal is to solve utt = c2∆u with u(x, 0) = f(x) = f(x1, x2) and ut(x, 0) = g(x) = g(x1, x2).We will view this as a special case of the initial value problem in 3-D with initial conditions u(x1, x2, x3, 0) =f(x1, x2) and ut(x1, x2, x3, 0) = g(x1, x2). Note that ux3

(x1, x2, x3, 0) = 0 and (ut)x3(x1, x2, x3, 0) = 0

because f and g are independent of x3. Also note that if u(x1, x2, x3, t) solves the wave equation with ourprescribed initial conditions, then ux3

solves the wave equation with data fx3, gx3

. This means that ux3≡ 0

by uniqueness =⇒ u(x1, x2, x3, t) = h(x1, x2, t). This simply means we can set x3 = 0 in our solution for3-D.

u(x1, x2, x3, t) =∂

∂t

[1

4πc2t

ˆ|x−y|=ct

f(y)dSy

]+

1

4πc2t

ˆ|x−y|=ct

g(y)dSy

u(x1, x2, 0, t) = · · ·+ 1

4πc2t

ˆ(y1−x1)2+(y2−x2)2+y23=c2t2

g(y1, y2)dSy

= · · ·+ 1

4πc2t

ˆ(x1−y1)2+(x2−y2)2≤c2t2

g(y1, y2)2ctdy1dy2√

c2t2 − (x1 − y1)2 − (x2 − y2)2

u(x1, x2, t) =∂

∂t

1

2πc

¨r<ct

f(y1, y2)√c2t2 − r2

dy1dy2 +1

2πc

¨r<ct

g(y1, y2)√c2t2 − r2

dy1dy2

where r =√

(x1 − y1)2 + (x2 − y2)2.

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Fourteenth Class

For the last class, we will outline an approach to weak solutions of Dirichlet Problems. First, recall theDirichlet problem ∆u = f in Ω and u = g on ∂Ω for Ω a bounded domain with smooth boundary. Weapproached this problem through potential theory

u(x) ∼ˆ

Ω

GD(x, y)f(y)dy +

ˆ∂Ω

∂GD∂ny

(x, y)g(y)dSy

where GD(x, y) is the Dirichlet Green’s function for the Laplacian with Dirichlet boundary conditions

GD(x, y) = Φ(x− y)− φx(y)

where ∆yφx(y) = 0 in Ω and φ∗(y) = Φ(x − y) on ∂Ω. This led to our discussion of boundary integral

equations (potential theory).Now lets take another approach. Consider a hilbert space H (complete normed linear space with inner

product ‖f‖H =√

(f, f)H). Some examples are Sobolev Spaces Hk(Rn). We define these as follows.Consider C∞0 (Rn) 3 u and consider the norm

‖u‖2H1 =

ˆRn|u|2 + |Du|2, ‖u‖2Hk(Rn) =

∑|α|≤k

ˆRn|∂αx u|2dx

where we are using multi index notation. Now define Hk(Rn) = completion of C∞0 (Rn) with respect to thisnorm i.e. f ∈ Hk(Rn) =⇒ ∃fn ⊂ C∞0 (Rn) with ‖f − fn‖Hk(Rn) → 0.

Let u(ξ) denote the fourier transform of u(x). Recall the Plancherel identity ‖u‖L2 = ‖u‖L2 . Then it istrivial to see

∂αx u(ξ) = (2πiξ)αu(ξ).

Thus this implies

‖u‖2Hk ≈ˆ

(1 + |ξ|2)k|u(ξ)|2dξ.

In fact, you could take this as the definition of Sobolev spaces on Rn.Sobolev spaces are a way to characterize point wise behavior of functions (continuity, differentiability)

in terms of global integral quantities. Lets look at a simple Sobolev inequality. Let f ∈ H1(R). Then f iscontinuous and moreover ∃C such that ∀f ∈ H1,

|f(x)|2 ≤ C‖f‖2H1 .

Let f ∈ C∞0 (R). Then note that we can write

f2(x) =

ˆ x

−∞

d

dyf2(y)dy

=

ˆ x

−∞2f(y)f ′(y)dy

≤ Cˆ x

−∞f2(y) + (f ′(y))2dy

≤ C‖f‖2H1 .

Definition 2.11. A linear functional on a Hilbert Space H is a mapping φ : H → R such that φ(λu1+µu2) =λφ(u1) + µφ(u2). A linear functional φ is said to be bounded on H if ∃Mφ > 0 such that for all u ∈ H,

|φ(u)| ≤Mφ‖u‖H .

Theorem 2.12 (Riesz Representation Theorem). Let φ be a bounded linear functional on a Hilbert spaceH. Then there exists an element Vφ ∈ H such that for all u ∈ H, φ(u) = (u, vφ).

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