Lectures Pdecourse

7/28/2019 Lectures Pdecourse

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INTRODUCTION TO EQUATIONS FROM

MATHEMATICAL PHYSICS

LI MA

Date: Starting from the Summer of 1991 to 2007.1991 Mathematics Subject Classification. 35-XX.Key words and phrases. wave equation, heat equation, harmonic functions, character-

istic line method, Fourier method, maximum principle, fundamental solution.

1


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This is a lecture note for the courses of Equations from Mathemati-cal Physics or Methods of Mathematical Physics given by me to senior

students in Tsinghua University, Beijing


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EQUATIONS FROM MATHEMATICAL PHYSICS 3

To An, for patience and understanding


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Contents:

Wave equations; Heat equations; Laplace equation; Method of Char-acteristic lines; Duhamel principle; Method of separation of variables;Laplace transform; Fourier transform; Fundamental solutions; Meanvalue property; Louville theorem; Monotonicity formulas; Maximumprinciple; Variational principle.


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Preface

As I always believe that the course about equations from mathemat-ical physics is of fundamental importance to every student in college.However, it will be a hard time for us to firstly learn it. Can you imagewhat is up when you enter a new world where you know just a little?

I want to point out some key points in our course at this very be-ginning. To us, solving a problem means that we reduce the probleminto some easy problems which can be solved by us. For solving ourproblems with partial differential equations (in short, PDE), we reducethem into ordinary differential equations (in short, ODE), which canbe solved by us.

We shall derive some typical partial differential equations from the

conservation laws of mass or energy or momentum. Such equations arecalled equations from mathematical physics. The basic principle usedhere is that the change (rate) of the physical quantity in a body inspace is created from the change on the boundary and by the source inthe interior of the body. Hence, we need to use the divergence theoremfrom the differential calculus.

The equations from mathematical physics are very important objectsin both mathematics and applied mathematics. However, many of themare nonlinear equations. Even for linear partial differential equations,we can not expect to have a unifying theory to handle them. Therefore,we have to classify the equations and treat them case by case. Thecrucial observation is that many equations enjoy symmetry propertieslike translation invariant or rotation invariant, which will lead to manyidentities for solutions.

The object in our course is to try to explain some elementary partof solving simple linear partial differential equations of first order andsecond order. The principle for studying these equations is reduce theminto linear ordinary differential equations, which can be solved by toolsfrom the differential calculus and linear algebra.

The most beautiful method for solving linear partial differential equa-tions or systems of first order is the characteristic line method.

The most powerful methods for solving linear partial differentialequations or systems of second order are the methods of separationfor variables (also called Fourier method ) and Fourier transformationmethod, which can be used to translated the problems into ODE again.Theory of Fourier transformations is one of most beautiful mathemat-ical theories so that there are many books about Fourier transforma-tions. The other method is to construct the fundamental function to


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the equation with the homogeneous boundary condition and initial con-dition. In the latter method, the divergence theorem plays an essential

role.There is another important tool in the study of heat equation and

elliptic equations. It is the maximum principle trick with its brothercalled the comparison lemma. This trick is also very useful in the studyof nonlinear parabolic and elliptic equations or systems. People usuallyuse the maximum principle to derive gradient estimates or Harnackinequalities of non-negative solutions, which lead to Bernstein typetheorem.

To understand the behavior of the solutions to PDE, we use diver-gence theorem, Fourier method, or the fundamental solutions to set upenergy bounds for solutions. The nice forms of energy bounds for so-lutions are monotonicity formulae or Pohozaev type identities , whichcan be used to get Liouville type theorem or regularity theorem.

Energy bounds or gradient estimates are important to understandthe compactness of solution spaces.

The aim of this course is to let our students/readers know the basicingredients in the study of simple linear PDEs of first order and secondorder. Sections with ahead should be skipped by first reading. Justas in the study of other courses in mathematics, one can only geta good understanding of our subject by doing more exercisesand by reading more related books.

I hope you understand my words here. What did I feel in my collegelife? I did feel so lucky that I had attended many courses in math-ematics when I was in Nankai University. This experience makes mecan understand many deep lectures from some famous guys.

In conclusion, for our beginners, I hope you remember that we try tounderstand PDE by ODE! In other words, before we do PDE, letus do its partner ODE. Do you remember an old Chinese proverb?Here it is, I suddenly find that I know the world after I hear yourwords.

Written by

Li Ma

in 5-1-2007


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Notations and conventions:The functions used in our course are bounded smooth function unless

we point out explicitly. If the domain is unbounded, we assume that thefunctions is decay to zero at infinity. In other word, we assume the niceconditions for functions such that we can perform the differentiationinto integrands or the change the order of integration of variables.

We denote by Rn the Euclidean space of dimension n. Let x =(x1, ...xn) the standard coordinates in Rn.

We denote by C2(D) the space of twice differentiable smooth func-tions inside the domain DRn.

We denote by C0 (D) the space of infinitely differentiable smoothfunctions with support inside the domain DRn. Here the support of afunction f is the closure of the set

{x

D; f(x)

= 0

}.

f := (x1f,...,xnf).

jf = xjf.

rf(x) :=j

(xj pj)r

xjf(x)

is the radial derivative of the function f, where r = |x p| is the radialcoordinate with respect to the center p = (pj) Rn.

We denote by(k)x f = (

k1x1

k2x2 ...knxnf); k = k1 + ... + kn

the k-th derivatives of the function f with respect to the variablesx = (x1,...xn).

PDE=Partial Differential Equation.ODE=Ordinary Differential Equation.F-transform=Fourier transform.


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1. lecture one

1.1. Divergence theorem.Let (xi) be the standard coordinates in the Euclidean space Rn. Werecall the classical divergence theorem:

Theorem 1. Let D be a bounded domain with finite piece-wise C1

boundary D inRn. LetX : D Rn be a smooth vector field on theclosure D of the domain D. Let be the unit outer normal to D.Then we have the following formula:

D

divXdx =

D

X d.

Here divX = i Xixi for X = (Xi) , = (i) and X = i Xii.If Xi = uxi

for some smooth function, i.e., X is the gradient u :=(x1u,...,xnu) of the function u, then

divX =i

2u

(xi)2= u

which is the Laplacian of the function u.

A special case we often use is that X has only one non-zero compo-nent, say j th component,

X = (0,...0, u, 0,...0)for some smooth function u C1(D) C(D). Then, divX = ju and

D

judx =

D

ujd.

The classical divergence theorem is a basic theorem used in ourcourse.

We ask the readers to remember this basic formula. Note that whenn = 1, the divergence theorem is the famous Newtonian-Leibnizformula:

f(b) f(a) = b

af

(x)dx.

In dimension two, the divergence theorem is Greens theorem in dif-ferential calculus. Let D be a bounded smooth domain in the planeR2. Recall that the Green formula is the following: for any smoothfunction P, Q on the closure of the domain D, we have

D

(Qx Py)dx =D

P dx + Qdy.


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It is clear that the form of divergence theorem in dimension twois different from that of the Green formula. We want to prove that

they are equivalent each other. Assume that the boundary D is givenby the smooth curve p(s) = (x(s), y(s)), where s is the counter-clockarc-length parameter. Write by x = dx/ds, etc. Then T = (x, y) isthe unit tangent vector along p(s). Then N = (y, x) is the interiorunit normal to the curve p(s) such that {T, N} forms a right-handorthonormal basis system. The outer unit normal to the boundary Dis = N = (y, x). Choose X1 = Q and X2 = P. Then we have

divX = Qx Py,and along the boundary D,

P dx + Qdy = (P x + Qy)ds = X

ds

Hence, Greens formula implies thatD

divXdx =

D

X ds,which is the divergence theorem in dimension two.


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1.2. One dimensional wave equation.We consider a vibrating string of length l with its two ends fixed at

x = 0 and x = l. Let u = u(x, t) be the position function of the stringat (x, t) in the direction of y-axis. Assume that = constant is themass density of the string. Then the kinetic (or temporal) energy ofthe string is

T(u) =1

2

l0

u2tdx.

The potential (or spatial) energy is

U(u) =c

2

l0

u2xdx,

where c is a physical constant of the string.Assume that the string is acted by the exterior force with force den-sity f0(x, t) at (x, t).

Then the energy obtained by the exterior force is

U(f) = l

0

f0udx.

Then the Hamiltonian action of the string is

H(u) =

t0

d(T(u) U(u) U(f0))

= 12t

0

dl0

u2tdx

c2

t0

d

l0

u2xdx +

t0

d

l0

f0udx.

The Hamilton principle implies that the motion of the vibrating stringobeys

H(u) = 0,

that is, for every = (x, t) C20 ((0, l) (0, t1)),

< H(u), >:=d

dH(u + )

|=0 = 0.

We compute that

< H(u), > =

t0

d

l0

(utt

cuxx + f0)dx

=

t0

d

l0

(utt + cuxx + f0)dx


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Using the variational lemma (see Appendix), we then get

utt

cuxx = f0/,

which is called the one dimensional wave equation.By physical testings, we can find the initial position and velocity of

the string. So we know that

u(x, 0) = (x),

andut(x, 0) = (x);

and the boundary condition

u(0, t) = 0 = u(l, t).

The higher dimensional wave equation in its simple form isutt(x, t) = a

2u(x, t) + f(x, t),

which is a typical hyperbolic partial differential equation of secondorder.

We can also add other type of initial data and boundary data for thewave equations.


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1.3. Conservation of mass.Let = (x, t) be the mass density of a moving fluid in the region

W. Let u be the velocity field of the fluid. Let be the outer unitnormal to the boundary of the domain W.

Take a small part dx in W. Then the mass in dx is dx. Hence, thetotal mass in W at time t is the integration of dx over W, which is

W

dx.

Then, its change rate at time t is

d

dt

W

dx.

Assume that there is no fluid generating source. Then the changerate is only across fluid rate from the boundary. Take a small part din the boundary W. Then the amount of u d flow into W. Sothe total amount flowing into W across the boundary is the integrationover W:

W

u d.So we have

d

dt

W

dx = W

u d.Using the divergence theorem, we have

W

( t

+ div(u))dx = 0.

Since we can take W be any sub-domain, we have

t + div(u) = 0.

This is the continuity equation in Fluid mechanics.If u = (a(x1), 0, 0), then the continuity equation is

t + x(a(x)(x, t)) = 0,

where x = x1.


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1.4. What are important for our equations?What I want to show here is that once we have a pde from scientific

background, we must try to solve it and get the explicit expression forthe solutions in terms of initial data and boundary data. We also needto know the uniqueness and stability of our solutions to our equationsor problems.

Here we may introduce the concepts of Hilbert space and normedspace (see any textbook about Functional Analysis). A norm | | on avector space E is a function from E to non-negative numbers R+ suchthat

(1)

|f| = 0if and only if f = 0;

(2) |af| = |a||f| for any real number a;(3)

|f + g| |f| + |g|.A vector space E equipped with a norm | | is called a normed space.

If a sequence (fj) E satisfy |fj fk| 0 as j, k , then we call(fj) a Cauchy sequence in (E, | |). If every Cauchy sequence has alimit in E, then we call (E, | |) a Banach space. If further, the norm| | comes from an inner product < , >, that is, |f|2 =< f, f >, wecall (E, < , >) a Hilbert space; for example,

< f, g >= D

f(x)g(x)dx, f,g L2(D),

By definition, we say that u(x, t) is the stable solution (in the norm| |) to the problem

A[u](x, t) := (tu Lu)(x, t) = 0,where t [0, T], and

L := P(x, x,...,(k)x ) =

kj=0

aj(x)(j)x

is a partial differential operator of the variables (x), with initial datau(x, 0) if for any > 0, there exists > 0 such that for all solutionsw(x, t) to A[w] = 0 with initial data w(0),

|w(0) u(0)| ,we have

supt

|w(t) u(t)| < .


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There are also many other definitions for stabilities, which depend onwhat we wanted in reality.

To get uniqueness and stability of solutions to evolution problems,we need to find some nice norm for solutions and derive a Gronwalltype inequality. For stationary problems, we have to use maximumprinciple trick or monotonicity formulae for solutions.

In our course, we need to know some typical methods to solve linearpdes of first order and second order. Historically, the solution expres-sions were guessed by scientific workers from physical intuition. So itis important for us know the scientific background of the equations.


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2. lecture two

2.1. Heat equation and Laplace equation.Heat equation and Laplace equation are very important objects inPDE theory.

We use the energy conservation law to derive the heat equations.Let u be the heat density of the body , and let qbe the heat flowing

velocity. By Fouriers law, we have

q = ku,where k > 0 is a coefficient of physical quantity.

Let f0(x, t) be the density of heat source. Then as in the case ofcontinuity equation, we have, for a physical constant c > 0 of the

region D ,t

D

cudx =

D

q ()d +

D

f0dx.

Using the divergence theorem we haveD

q ()d = D

div(q)dx.

Hence, since D can be any domain in the body, we have

ctu = divq+ f0.Using the Fourier law, we have

ctu = div(ku) + f0,which is called the heat equation with source f0. Assume that c = 1 =k = and f0 = 0. Then we have the heat equation

ut = u, (x, t) (0, ),which is a typical parabolic partial differential equation of second order.

Physically, we can measure the heat density on the boundary ,and we may assume that u(x, t) = 0 for (x, t) (0, ). Theinitial heat density at t = 0 can also be detected. So we assume that

u(x, 0) = (x), x .The heat equation coupling with initial and boundary conditions iscalled a determined problem of heat equation.

Of course, we may have other kind of boundary conditions like

ku

(x, t) = u(x, t), (x, t) (0, ).


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When we have stationary heat in , that ut = 0, we have the follow-ing equation:

u + f = 0,where f = f0, or in other form

u = fwhich is called the Poisson equation. If f = 0, it is called the Laplaceequation in , which is a typical elliptic partial differential equation ofsecond order. In dimension one, the Laplace equation becomes

u

= 0,

So u = ax + b is a linear function of the single variable x. However, ifthe dimension n

2, it is not easy to solve the Laplace equation on

domains coupling with its boundary condition.In the following subsection, we try to solve some simple pde of first

order.


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2.2. Solve pde of first order.The principle for solving a first order PDE is to reduce it into ODE

along characteristic curves. So one should know how to solve an ODE.We shall review some ODE theory when it is necessary.

We study a first order partial differential equation. Our question isto solve the following equation (FDE):

ut + aux = f(x, t).

Assume that we have the initial condition u = u0(x) at t = 0.We introduce the C-curve C(t) (called the characteristic curve) by

solving dx/dt = a. Then we have

C := C(t) : x = at + A.

When t = 0, we have x(0) = A. On the curve C, we have

du/dt = ut + aux = f(x, t) = f(at + A, t).

Using integration along the curve C(t), we have

u(x(t), t) = u(A, 0) +

t0

f(a + A, )d.

Given the point (x, t) on the curve C(t). Then we have

x(t) = x

and we can find that initial point on the curve is x(0) = A = x

at. So

by the initial value condition, we have u(A, 0) = u0(A) = u0(x at).Inserting A = x at into the expression above, we find (FF):

u(x, t) = u0(x at) +t

0

f(x + a( t), )d.

In case when f = 0, we have

u(x, t) = u0(x at).The method above is called characteristic curve method in the litera-ture. One may study examples in our textbook.

We point out the following comparison lemma: Let u be a smoothsolution to (FDE) with the initial data u(0) = u0. Let w be a smoothfunction satisfying

wt + awx f(x, t)with the same initial data as u. Then we have w u on R [0, T].

This lemma is based on a very simple observation from the formula(FF) that if u0 = 0 and f 0, then u 0 for all t. In fact, let

g(x, t) = wt + awx f(x, t).


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Then we have g(x, t) 0 andwt + awx = f(x, t) + g(x, t).

Let v = w u. Then we havevt + avx = g(x, t) 0

with v = 0 at t = 0. Using the formula (FF) for v, we know that v 0.Hence w u.

Usually, the comparison lemma is a very important property forPDEs.


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2.3. A nonlinear PDE example.We have just used The characteristic curve method to study the

linear pde of first order. We show here that this method can also beused to study the nonlinear pde of first order. By definition, a pde(PDE)

P(u, u, D2u, ...) = 0is called a linear equation if, for any given two solutions u and v of(PDE), au + v is also a solution of (PDE) for every constant a R.Otherwise, it is called a nonlinear PDE.

Lets study the following nonlinear PDE of first order:

ut + (1 + u)ux =1

2

with the initial value condition u(x, 0) = x for x 0.We first find the characteristic curve x(t) by solving

dx

dt= 1 + u(x, t).

Note that along the characteristic curve, we have

du

dt=

1

2

and

u(x(t), t) =

t

2 + u(x(0), 0) =

t

2 + x(0)On one hand, we set x(0) = A. Then we have

u(x(t), t) =t

2+

A

Solving

A from this relation, we get

A = u(x(t), t) t2

.

On the other hand, we find the curve satisfying

x(t) = A + t + t0

u(x(), )d = A + t + t0

(

2 + A)d.Then we have

x(t) = A + t +t2

4+

At = t + (

t

2+

A)2 = t + u(x(t), t)2.

We now solve u along the curve x(t) and get

u(x(t), t) =

x(t) t.


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3. lecture three

3.1. Duhamel principle.Consider (ODE)u

+p(t)u = 0,

with initial condition u = f at t = . Then its solution to (ODE) is(S):

u(t, ) = fexp(t

p(s)ds).

We now try to solve the equation (F):

u

+p(t)u = f(t)

with initial condition u = 0 at t = 0. Let f = f() and solve (ODE)

at t = as above to get the solution (S). Then the function defined byt0

u(t, )d

is the solution to (F). This kind of method is called the Duhamel prin-ciple.


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3.2. First order ODE and a comparison lemma.For the given ODE

u = a(t)u + b(t)we can write it as

(eR t

a()u) = eR t

a()b(t)

So it has a general solution

u(t) = u(0)eR t

a() + eR t

a()

tdse

R sa()b(s).

If u(0) = 0, we have (F):

u(t) = e

R ta()

t

dseR s

a()

b(s).

One often use the argument above to prove the famous Gronwallinequality: Let > 0 and let u be a continuous function on [0,T]satisfying

u(t) a + t

0

u()d.

Then we have u(t) aet on [0,T]. Hint: Let

f(t) = a +

t0

u()d.

Then we have f = u. By assumption, u f, we getf f .

Let (t) and f(s) be two smooth functions. We consider the followingnonlinear ODE

u = (t)f(u)

with initial data u = a at t = 0.Assume that

w (t)f(w)

with initial data w = a at t = 0. Then we have w(t) u(t) for all t.Letg(t) = w (t)f(w).

Then g(t) 0. Setv = w u.

Then we have

v == (t)(f(w) f(u)) + g(t) = (t)B(t)v(t) + g(t)


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with v = 0 at t = 0 and

B(t) = f(zw + (1 z)v)dz.Let a(t) = (t)B(t) and b(t) = g(t) in (F). Then we have v 0. Thatis w u for all t.

We remark that one can use Sturm-Liouville theory and variationalprinciple to set up comparison lemma for second order ODE.


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3.3. One dimensional wave equation.We shall follow our textbook [1] for the explanation. We consider

the one-dimensional wave equation (OW):

utt uxx = 0, x R, t > 0,with initial data

u|t=0 = u0and

ut|t=0 = u1.Lets first assume that u0 = 0. Note that

2t 2x = (t x)(t + x).Then we reduce the problem into two pde of first order. Let

v = tu + xu.

andtv xv = 0.

Solving u we get

u(x, t) =

t0

v(x + ( t), )d.

Using initial condition we have v(x, 0) = u1(x). We now solve v:

v(x, t) = v0(x + t) = u1(x + t).

So,

u(x, t) =

t0

u1(x + ( t) + )d.Changing the variable

s = x + 2 t,we can get

u(x, t) =1

2[

x+txt

u1(s)ds] := Mu1 .

If u0 is non-trivial, but u1 = 0, then we can verify that

u(x, t) =d

dtMu0(x, t)

is the solution toutt uxx = 0

with initial datau|t=0 = u0

andut|t=0 = 0.


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This is the Duhamel principle for the wave equation.Hence, using the Duhamel principle, we can get the DAlembert for-

mula (EW):

u(x, t) =1

2[(u0(x + t) + u0(x t) +

x+txt

u1(s)ds].

for the solution toutt uxx = 0

with the initial datau|t=0 = u0

andut

|t=0 = u1.

From this formula we know that the solution to the problem (OW) isunique with the formula (EW).

One can show that, for fs(x) = f(x, s),

u(x, t) =

t0

Mfs(x, t s)ds

is the solution toutt uxx = f(x, t)

with the initial datau|t=0 = 0

andut|t=0 = 0.

We leave this as exercise for readers.Remark One: We remark that ifu0 = 0 and u1 0, then u 0 for

all t. For each t, |u(x, t)| M0 +tM1 is bounded provided |u0(x)| M0and |u1(x)| M1. Similar bound is also true for ux and ut at each tprovided |u0x(x)| C0 and |u1(x)| C1.

We now assume that

u0 = 0, u1 = 0

for all |x| > R for some R > 0. Then from the expression (EW), wehaveu = 0, ut = 0

for |x| > R + t. This is the finite propagation property of the waveequation.

Remark Two: Define the energy for u by

E(t) =1

2

R

(u2t + u2x)dx.


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Then we haved

dtE(t) = R(ututt + uxuxt)dx.

Using the integration by part to the last term we get

d

dtE(t) =

R

ut(utt uxx)dx = 0.

So the energy E(t) is a constant, which is

1

2

R

(u21 + u20x)dx.

Remark Three: We now study the one dimensional wave equationon half line (HWE):

utt uxx = 0, x R+, t > 0,with initial data

u|t=0 = u0and

ut|t=0 = u1.Note that u0(t + x) is the special solution to (HWE). So we can

use it to transform the problem into the case when u0 = 0. Thenwe can use the odd extension method on the x-variable to extend thefunction u1 and the equation on the whole line. Then we can use the

DAlembert formula to solve (HWE). One may also use this method totreat one dimensional wave equation on an interval, however, we have amore beautiful methodFourier method/Separation variable methodto study this case. See the next section.

Remark Four: We point out that we can use the DAlembert for-mula to treat some special case of higher dimensional wave equation.Here is the example: Consider the following problem for 2-dimensionalwave equation

utt uxx uyy = 0with the initial data

u

|t=0 = (x)

andut|t=0 = (y).

Then we split the problem int to two one-dimensional problems: Oneis

utt uxx = 0with the initial data

u|t=0 = (x)


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andut

|t=0 = 0.

The solution for this one dimensional problem is

u(x, t) =1

2[((x + t) + (x t)]

The other isutt uyy = 0

with the initial datau|t=0 = 0

andut|t=0 = (y).

Its solution is

u(x, t) =12

[y+tyt

(s)ds].

Hence, the solution to our original 2-dimensional problem is

u(x, y; t) =1

2[((x + t) + (x t) +

y+tyt

(s)ds].


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4. lecture four

4.1. The method of separation of variables 1.The separation variable method for boundary value problems is look-ing for a solution as a sum of special solutions of the form X(x)T(t).

Let u(x, t) = X(x)T(t) be a solution to the wave equation

utt = a2uxx

on [0, l] [0, ) with the homogeneous boundary conditionu(0, t) = 0 = u(1, t).

The boundary condition gives us (BC):

X(0) = 0 = X(1)

Using the wave equation we have

T

X(x) = a2T X

,

orT

a2T=

X

X.

The left is a function of variable t and the right is a function of variablex. To make them be the same, they should be the same constant. Say. Then we have one equation for T:

T

+ a2T = 0

and the other equation for X:

X

+ X = 0

with the boundary condition (BC). The latter is the simple case ofthe Sturm-Liouville problem. The key step in the Separation VariableMethod is to solve this problem at first.

Case 1: When < 0, the general solution for X is

X(x) = c1ex + c2e

x

Using (BC) we have

c1 + c2 = 0and

c1e + c2e

= 0,

which implies that c1 = c2 = 0.Case 2: When = 0, the general solution for X is

X(x) = c1 + c2x.

Again (BC) implies that c1 = c2 = 0.


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Case 3: When > 0, the general solution for X is

X(x) = c1 cos(

x) + c2 sin(

x).

The boundary condition (BC) gives us that c1 = 0 and

c2 sin(

) = 0,

and then = k22, k = 1, 2,...

HenceX(x) = Xk(x) = ck sin(kx)

for k = 1, 2,....Returning to the equation for T(t) with = k22, we get

Tk(t) = ak cos(kat) + bk sin(kat).

SoUk(x, t) = [ak cos(kat) + bk sin(kat)] sin(kx)

is a typical solution to the one-dimensional wave equation with bound-ary condition.

We remark here that a easy way to see 0 is using the integrationrelation:

|X|2 =

|X|2.

We now look for a solution of the form

u(x, t) =+k=1

[ak cos(kat) + bk sin(kat)] sin(kx)

to the wave equationutt = a

2uxx

on [0, 1] [0, ) with the initial conditionu = (x), ut = (x), t = 0

and the boundary condition

u(0, t) = 0 = u(1, t).

The reason for us to use such an expansion is the Fourier series theory.For general boundary conditions, the expansion can be used , but wehave to invoke the Sturm-Liouville theory (see Appendix).

We need to use the initial condition to determine the constant akand bk. So we have

(x) =+k=1

ak sin(kx)


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30 LI MA

and

(x) =

+

k=1

kabk sin(kx).

Using the Fourier series expansion for (x) and we have

ak = 2

()sin(k)d

and

bk =2

ka

()sin(k)d.

In this way, we get at least formally a solution to our problem. This isa basic method to solve linear pde.

If the boundary condition is not homogeneous, saying

u(0, t) = (t), ; u(1, t) = (t),

we have to make a transformation

u(x, t) = v(x, t) + (t) + x((t) (t))and reduce the problem for v with the homogeneous boundary condi-tion.


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4.2. The resonance problem. We now briefly discuss the resonanceproblem of the one dimensional wave equation. Consider

utt a2uxx = A sin(t), (x, t) [0, L] (0, )with homogeneous initial and boundary conditions

u(0, t) = u(L, t) = 0

andu(x, 0) = 0 = ut(x, 0),

where = ajL

for j = 1, 2,....Then we have

u(x, t) =

jLajaj

sin(jx

L)

t0

sin()sin(aj

L(t ))d

=j

ajj(2 2j )

( sin jt j sin t)sin(jxL

)

where

j =aj

Land

aj =2

L

L0

A(x) sin(jx

L)dx.

Because we have

2

2

jwhich may cause the coefficient

ajj(2 2j )

.

We say that we may have the resonance phenomenon caused by .


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4.3. Wave equation in two dimensional disk.We denote BR(0)

R2 the ball in the plane R2. Lets consider the

2-dimensional wave equation

utt = a2(uxx + uyy)

on BR(0) [0, ) with the homogeneous boundary conditionu(x, t) = 0, x BR(0)

and the initial conditionsu|t=0 = u0

andut|t=0 = u1.

Introduce the polar coordinates (r, ) byx = r cos , y = r sin .

Thenu = urr + r

1ur + r2u.

The reason that we use polar coordinates here is because of the bound-ary condition in our problem being on the circle. Let u = T(t)U(x, y)be the solution to our wave problem. Then

T

+ a2T = 0

and

U + U = 0, U(R, t) = 0.It is easy to see that = k2 > 0. Setting U = h()B(r) in the latterequation, we have

h

()

h()= r

2B

+ rB

+ k2r2B

B= 2.

Then we get two problems:

h

+ 2h = 0, h( + 2) = h(),

andB

+ r1B

+ (k2

2r2)B = 0, B(R) = 0.

The last equation is called the Bessel equation. As before, we find that

= n = n

andhn() = An cos(n) + Dn sin(n).

Inserting = n into the Bessel equation, we have

B

+ r1B

+ (k2 n2r2)B = 0, B(R) = 0.


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The general solutions of this equation are called the Bessel functionsof order n. Just like trigonometric polynomials, Bessel functions form

an orthonormal basis, and then we can use this basis to solve the waveproblem above. One may look for reference book for more about theBessel functions. Some people would like to define the Bessel functionsof order n as the coefficients of the Fourier expansion:

exp(ir sin ) =

n=Bn(r)e

in,

where i =1 and ein = cos(n) + i sin(n). Hence,

Bn(r) =1

2

20

exp(ir sin in)d.We remark that, if the domain is not a ball but a rectangle, we just

letU(x, y) = X(x)Y(y).


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4.4. Wave equation in R3.The other way to understand the higher dimensional wave equation

for us is to try the Duhamel principle to the wave equation in Rn

utt u = f; f or (x, t) Rn R+with the initial condition

u = (x), ut = 0, at t = 0

by assuming that we know the solution u(x, t) to the homogeneousequation

utt u = 0; f or (x, t) Rn R+with the initial condition

u = 0, ut = (x), at t = 0.

We then use the descent method to study

utt = u

in RnR in one dimension higher space Rn+1R with initial conditionsu = 0, ut = (x), at t = 0.

In dimension three, we first study the wave equation problem (3W):

utt

u = 0; R3

{t > 0

}with the initial datau = 0, ut = (x), at t = 0.

We find that

u(p, t) =1

4a2t

Sat(p)

(x)dx,

where Sat(p) = {x R3; r = |x p| = at}.In fact, let

I(x, r; g) =1

4 Br(x)ud

for g = g(y) C2(R3) andJ := J(x, r; t) = rI(x,r,u)

for u = u(y, t) be the solution to the problem (3W). Changing variables,we find that

I(x, r; g) =1

4

{||=1}

u(x + r)d


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and

BR(0) g(x + z)dz=

R0

r2dr

{||=1}

u(x + r)d

= =

R0

4r2I(x, r; g)dr.

We now compute

x

R0

4r2I(x, r; g)dr

= xBR(0)

g(x + z)dz

=

BR(0)

xg(x + z)dz

=

BR(0)

zg(x + z)dz

=

BR(0)

Rg(x + z)dz

= B1(0) Rg(x + R)R2d= = 4R2RI(x, R; g).

So, we have

x

R0

r2I(x, r; g)dr = R2RI(x, R; g).

Note that

Rx

R0

r2I(x, r; g)dr

= xRR

0

r2I(x, r; g)dr

= R2xI(x, R; g)

and

R[R2RI(x, R; g)]

= 2RRI(x, R; g) + R22RI(x, R; g).


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36 LI MA

Hence,

R2xI(x, R; g) = 2RRI(x, R; g) + R22RI(x, R; g).

Dividing both sides by R, we have

xRI(x, R; g)

= = 2RI(x, R; g) + R2RI(x, R; g)

= 2R(RI(x, R; g)).

Now,

2Jt2

= R4

B1(0)

utt(x + R,t)d

=a2R

4

B1(0)

xu(x + R,t)d

=a2

4xR

B1(0)

u(x + R,t)d

= a2x(RI(x, R; u))

= a2x(RI(x, R; u))

= a22J

R2 .

This is a one dimensional wave equation. We now determine its bound-ary condition and initial condition. Clearly, we have the boundarycondition

J(x, R; t)|R=0 = 0.

At t = 0,

J(x, R; t)|t=0 = RI(x, R; 0) = 0,

and

Jt(x, R; t)|t=0 = RI(x, R; ).

Then we have

J(x, R; t) =1

2a

at+RatR

rI(x, r; )dr.


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Hence,

u(x, t) = limR0J(x, R; t)

R

= limR0

1

2aR

at+RatR

rI(x, r; )dr

=1

a[RI(x, R; )]|R=at

= tI(x,at; )

=t

4

B1(0)

((x + at)d.

That is,

u(x, t) = 14a2t

Sat(x)

(y)dy.

Then we use the Duhamel principle in the study of the wave equation

utt u = f; R3 {t > 0}with the initial data

u = (x), ut = (x), at t = 0,

and we have the famous Kirchhoff formula for the solution:

u(p, t) =1

4a2

Bat(p)

r1f(x, t a1r)dx

+1

4a2t

Sat(p)

(x)dx

+ t[1

4a2t

Sat(p)

(x)dx]

whereBat(p) = {x R3; r = |x p| at}.

From this formula, we can see the Huygens principle that the wave atx0 at time t = 0 influences the solution only on the boundary of a coneoriginating from x0. We just remark that using the descent method,

one can get a solution formula for two dimensional wave equation.One may see other textbooks for the other understanding of the

Kirchhoff formula.


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38 LI MA

5. lecture five

5.1. The method of separation of variables 2.We begin with a typical example for heat equation with fixed bound-ary conditions.

We now describe the main idea. Let u(x, t) = X(x)T(t) be a solutionto the heat equation

ut = a2uxx

on [0, l] [0, ).Then formally, we have

T

a2T=

X

X.

The left is a function of variable t and the right is a function of variablex. To make them be the same, they should be the same constant. Say. Then we have

T

+ a2T = 0

and (EX):

X

+ X = 0.

As in the wave equation case, we impose the homogeneous boundarycondition

X(0, t) = 0 = X(1, t)

to solve X and . Multiplying both sides of (EX) by X and integrating

over [0, 1], we have

X2 =

|X|2 > 0.So > 0. Then we have

X(x) = Xk(x) = ck sin(kx)

for k = 1, 2,... and = k = k

22.

Then we solve for T and get Tk(t) = T(0)exp(a2kt) by imposinginitial conditions for T. The lesson we have learned is that the solutionto heat equation has exponential decay to 0 in t-variable as t

.

Using the superposition law, we can solve the heat equation by look-ing at solutions of the form

k

Akexp(a2k22t) sin(kx)

with the initial data

(x) =k

Ak sin(kx)


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Assume that we are given the heat equation

ut = a2uxx

with non-homogenous boundary condition, we need to use transforma-tions to make the boundary data into a homogenous condition, saying

u(0, t) = 0 = ux(L, t).

Then we have to meet the following non-homogenous heat equation

ut = a2uxx + f(x, t).

To study this problem, we need to solve for {Xj} and = j to theSturm-Liouville problem

X

+ X = 0,

with the boundary condition

X(0) = 0 = Xx(L).

Once this done, we do the expansion for f

f(x, t) =j

fj(t)Xj(x)

and look for solution u(x, t) such that

u(x, t) =j

uj(t)Xj(x).

Then we reduce the non-homogenous problem into the ODEu

j = a2juj + fj(t)with suitable initial condition. Hence, we have the expression

uj(t) = ejt(uj(0) +

t0

ejfj()d

and

u(x, t) =j

ejt(uj(0) +t

0

ejfj()d)Xj(x).

Here uj(0) are determined by

(x) = j

uj(0)Xj(x).

We omit the detail here.


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5.2. Energy bound.Let L be a partial differential operator on the domain D

Rn such

that for some 0 > j R, we haveLj(x) = jj(x),

with a fixed homogeneous boundary condition. Define the functionspace E such that the functions {j} form an orthonormal basis forthe space E with its norm given by

u =

a2j

for u =

ajj(x) E.Given a smooth function f C1,2(D [0, )). Assume that the

expansion u = uj(t)j(x) is a solution to the partial differentialequation:ut = Lu + f, t > 0,

with the initial data u = 0 at t = 0. Hence, uj(0) = 0 for all j.Assume that the known function f has the expansion

f =

fj(t)j(x)

with its norm on D [0, ) defined by

f =j

0

|fj(t)|2dt < .

Then we haveu

j(t) = juj(t) + fj(t).

Hence, as before, we have the expression

uj(t) = ejt

t0

ejfj()d.

Note that

(u

j(t) juj(t))2 = u

j(t)2 + 2juj(t)

2 2juj(t)uj(t) = fj(t)2.Then we have

T0

(u

j(t)2 + 2juj(t)

2)dt juj(T)2 = T0

fj(t)2.

Note that we have assumed that j < 0. Hence, we have0

(u

j(t)2 + 2juj(t)

2)dt

0

fj(t)2,

which implies thattu2 + 2u f2,


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where 2 = infj{2j}. This is the energy bound for the solution as wewanted.


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5.3. Fundamental solution to Heat equation.We study the heat equation (HE) on Rn

[0, +

):

ut = u,

with the initial data u = g(x) at t = 0.Assume that u is a smooth solution of (HE). Note that for > 0,

the function

u(x, t) = u(x,2t)

is also a solution of (HE).So we try to look for solutions of the form

u(x, t) = tf(|x|2/t).

Set r = |x|2/t. Then we find thatut = t

1f + t(|x|2

t2)f

= t1f rt1f,

u

=

|x|u

= 2|x|t1f

u

=2

|x|2 u

= 2t1f

+ 4

|x

|2t2f

= 2t1f + 4rt1f

u = u

+n 1|x| u

= 2t1f

+ 4rt1f

+ 2(n 1)t1f

= 4rt1f

+ 2nt1f

.

So,

r

f + f

+ 4f

+2n

rf

= 0.

Let =

n/2. Then we have

4(rn/2f

)

+ (rn/2f)

= 0.

Then for some constant A,

4rn/2f

+ rn/2f = A.

Take A = 0. Then we have

4f

= f.


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Assume that f C = const. as r +. Then we have f 0 as

r

+

. So

f = Ber/4.

Choose B = 1/(4)n/2. Define, for t > 0,

K(x, t) =1

(4t)n/2e

|x|2

4t .

This is called the Gaussian kernel, which is also called the Poissonkernel. It is easy to verify that

Rn K(x, t)dx = 1.Theorem 2. Given g(x), we let

u(x, t) =

Rn

K(x y, t)g(y)dy.

Then u(x,t) satisfies the heat equation (HE) and as t 0, u(x, t) g(x).

Roughly speaking, the theorem says that this u(x, t) is the solution

to our problem at the beginning of this subsection.

Proof. For t > 0, we can take derivatives into the integrand term, andthen u satisfies (HE) since K does.

Without loss of generality, we now assume n = 1. Assume that thereis a positive constant M > 0 such that

|g(x)| M, x R.

We use two step argument.Step one: Given an arbitrary small > 0. For any fixed point x

R,

we have some 0 > 0 such that, for |x y| < 0,|g(y) g(x)| <

2.

Step two: We take 1 > 0 small enough such that for 0 < t < 1,{|xy|0}

K(x y, t)dy < 4M

.


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44 LI MA

Then we have

|u(x, t) g(x)| = |K(x y, t)(g(y) g(x))dy| |

{|xy|0}

K(x y, t)(g(y) g(x))dy|

+ |{|xy| 0.

Proof. Fix t > 0. Choose R > 0 sufficient large. Then by our assump-tion on g(x), we have

supBR(x)|g(x)| 0, |x| .


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Note that

|u(x, t)| Rn K(x y, t)g(y)dy supBR(x)|g(x)|

BR(x)

K(x y, t)dy

+ supRnBR(x)|g(x)|RnBR(x)

K(x y, t)dy

supBR(x)|g(x)| + C(t)supRn |g(x)|R

e2/4tn1d

0 + o(R)as

|x

| . Here C(t) is a constant depending only on t and o(R) is

as small as we like for R large. Hence, we have showed that

|u(x, t)| 0,as |x| .

Using the polar coordinates (, ), we can also show that

|ju(x, t)| 0,as |x| . We omit the detail here.


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5.4. Comments on Fundamental solutions.The Gaussian kernel K(x

y, t) satisfies the heat equation for t > 0.

For t = 0, we understand it in the generalized sense that

K(x y, 0) = x(y).Here we define x(y) as the generalized function by

< x(y), g(y) >:= x(y)(g(y)) = g(x).

That is, we consider x(y) as a function defined on the function spaceC0 (R

n). Then, x is a linear functional on C0 (R). This x() is called

Diracs delta function or Dirac measure. One can also consider thedelta function 0(x) as the limit of a sequence of nice functions. Forexample, assume that n = 1 and let

qj(x) =1

2j, x [1/j, 1/j] and qj(x) = 0 |x| > 1/j.

Then R

qj(x)dx = 1,

and for x = 0, qj(x) 0 and at x = 0, qj(0) ; or for C0 (R),

< qj, > =

R

qj(x)(x)dx

=1

2j 1/j

1/j(x)dx

(0)

= < 0, > .

as j . If so, we define limqj = 0. We ask the reader use the firstconcept of limit to understand delta function. I would like to pointout that the latter limit is more rigorous definition. We may find othersequences which have 0 as its limit. For example, let

(x) = me 1

1|x|2 , for |x| < 1, (x) = 0, f or |x| 1where m > 0 is a constant such that

R (x)dx = 1,and let

t(x) =1

t(x/t),

where t > 0. Then R

t(x)dx = 1

and limt0t = 0.


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Generally speaking, for any given continuous function u(x) in Rn,we can also consider u as a linear functional on the space C0 (R

n) with

< u, >=Rn

u(x)(x)dx.

With this understanding, we say K(xy, t) is the fundamental solutionthe heat equation on RnR+. We emphasize that we call K(xy, t) isa fundamental solution to the heat equation on the whole space becauseK(xy, t) satisfies the heat equation for t > 0, and we consider K(xy, t) as a generalized function which is the delta function for t = 0.

We now consider the heat equation on the half line [0, ) with theboundary condition

u(0, t) = 0,

and the initial condition

u = (x), t = 0.

For this case, the fundamental solution is

(x,y,t) = K(x y, t) K(x + y, t).So in some sense, we use the odd extension to keep the boundarycondition. So the function defined by

u(x, t) =

R+

(x,y,t)g(y)dy

satisfies the heat equation on the half line [0, ) with boundary con-dition u(0, t) = 0 and the initial value conditionu = (x), t = 0.

If we consider the heat heat equation on half line [0, ) with theboundary data

ux(0, t) = 0.

Then the fundamental solution to his problem is

(x,y,t) = K(x y, t) + K(x + y, t).Here we have used the even extension to keep the boundary condition.


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6. lecture six

6.1. Fundamental solution to Laplace equation.Consider the Laplace equation (L) on Rn:

u = 0.

A solution to the Laplace equation in a domain is called a harmonicfunction.

Let r 0 such thatr2 =

i

(xi)2.

Then we haverdr =

ixidxi.

Thenr

xi=

xi

r.

Let u = v(r) be a solution of (L). Note that

uxi = v xi

r,

and

uxixi = v

(1

r (x

i)2

r3) + v

(xi)2

r2.

Then we have

u = v + n 1r

v .

Solve

v

+n 1

rv

= 0.

We find thatv

=c

rn1for some constant c and

v = a log r + b, f or n = 2,

and

v = ar2n + b, f or n 3,for some constants a and b.

Choose b = 0. Define

(x) =12

log r, f or n = 2,

and

(x) =1

n(n 2)|Sn1|r2n, f or n 3.


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Here |Sn1| is the volume of the sphere Sn1.Let (x, y) = (x

y) for x, y

Rn. This function (x, y) is called

the fundamental solution (or Green function) to the problem (L). Infact, we can consider

y(x y) = x(y).If so, we have

u(x) = < x(y), u(y) >

= < (x y), u(y) >=

Rn

(x y)u(y)dy

= Rn (x y)u(y)dy=

Rn

(x y)f(y)dy.To understand this, we need to use the second Green formula. Here werecall the second Green formula as follows. Given two smooth functionsdefined on a bounded domain D with piece-wise smooth boundary. Wedenote the outer unit normal to the boundary D. Then we have

D

(vu uv)dx =D

(vu uv )d.

Theorem 4. Givenf C00 (Rn), i.e, f is continuous inRn andf(x) =0 in the outside of some large ball. Let

u(x) =

Rn

(x y)f(y)dy.Then the function u satisfies the Poisson equation

u = f, in Rn.Proof. For simplicity, we assume n = 2. Take R > 1 large such thatf = 0 outside BR. Fix x R2. Using the second Green formula onthe region BR B(x), where 0, we get that

BR

(x, y)(u(y))

= lim0

BRB(x)

(x, y)(u(y))

= lim0

{|yx|=}

((x, y)ru(y)

r(x, y)f(y))d,


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50 LI MA

where r = |x y and ru(y) = ur (y).Note that

{|yx|=}|(x, y)ru(y)|d

12

{|yx|=}

ln1

r|ru(y)|

sup|u||ln| 0,

and

{|y

x

|=

}

r(x, y)u(y)d

=1

2

{|yx|=}

rln(r)u(y)d

=1

2

{|yx|=}

1

rf(y)d

=1

2

{|yx|=}

u(y)d

u(x)as 0. This shows that the function u satisfies

R2

(x, y)u(y)dy = u(x).

We ask our readers to verify the n 3 case. Lets study the Poisson equation on bounded domain. Let D

Rn be a bounded domain with piece-wise smooth boundary. We nowconsider the problem (PD):

u = f, in Dwith the boundary condition u(x) = (x) for x D. Then usingthe second Green formula and the fact v = being the fundamentalsolution on the whole Rn, we get (U1):

u(x) =D

(x y)f(y)dy

+

D

((x y)u(y) (y)(x y) )dy.This expression still contains a unknown term for u(y) on theboundary D. To overcome this drawback, we introduce, for eachfixed point x D, the harmonic function g(x, ) with its boundary


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data g(x, y) = (x y) for y D. Using the expression above forv = g(x, y), we have (U2):

0 =D

g(x, y)f(y)dy

+

D

(g(x, y)u(y) (y)g(x, y) )dy.Set

G(x, y) = (x y) + g(x, y).Note that G(x, y) has zero boundary value for y, i.e,

G(x, y) = 0, y D.Adding (U1) and (U2), we obtain that

u(x) =D

G(x, y)f(y)dy

D

(y)G(x, y) dy.

We call G(x, y) the Green function (or fundamental solution )for theproblem (PD).

Lets consider the boundary value problem for Laplace equation.Assume that u C2(D) C(D) satisfies

u = 0, i n D

Rn,

with the Dirichlet boundary condition

u(x) = (x) C(D), x D.As before, for x D, we have

u(x) =

D

(x y)(u(y))dy

+

D

((x y)u(y) (y)(x y) )dy

= D (y)(x y) dy.Let

P(x, y) = (x y) for x D and y D. We call P(x, y) the Poisson kernel to theboundary value problem in D. Then it can be verified that the functiondefined by

u(x) =

D

P(x, y)(y)dy,


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52 LI MA

which is called the Poisson formula, is the solution to the boundaryvalue problem in D.

We now point out a nice observation about the construction of har-monic functions. Given two harmonic functions u1 and u2 on the samedomain D Rn. We want to find a harmonic function U(x, y) definedon the one dimensional higher cylindrical domain D[L1, L2] such that

U(x, L1) = u1(x)

andU(x, L2) = u2(x).

This function can be achieved by setting

U(x, y) =L2 y

L2 L1u1(x) +

y L1L2 L1

u2(x).


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6.2. Mean value theorem for harmonic functions.Given a harmonic function u in a domain D. We take a ball BR(p)

in the domain D.Note that x = p + r, where r = |x p| and = xp

r. Using the

divergence theorem, we get

0 =

Bt(p)

u =

Bt(p)

u

rd.

Then we have the the right side of the identity above is

tn1B1(p)

u

r(p + t)d = tn1

t

B1(p)

u(p + t)d.

Here, we want to point out that the notationu

r(p + t)

means the radial derivative of function u taking value at x = p + t. Itis not a composite derivative for independent variables (r, . Hence,

0 = tn1

t(t1n

Bt(p)

ud).

This implies that

t1n Bt(p) ud

is a constant for all t R. Then1

|Bt(p)|Bt(p)

ud =1

|BR(p)|BR(p)

ud.

Using the mean value theorem and sending t 0, we have

u(p) =1

|BR(p)|BR(p)

ud.

Then

u(p)|BR(p)| = BR(p)

ud.

If we take integration in R, we have

u(p) =1

|BR(p)|BR(p)

udx.

Hence, we have


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Theorem 5. Given a harmonic function u in a domain D. We takea ball BR(p) in the domain D. Then we have

u(p) =1

|BR(p)|BR(p)

ud.

and

u(p) =1

|BR(p)|BR(p)

udx.

Mean value theorem is a basic property for harmonic functions. Asan application we prove the following result.

Theorem 6. (Liouville Theorem). Ifu is a bounded harmonic func-tion in the whole space Rn. Then u is a constant.

Proof. Method One. Let M > 0 such that |u(x)| M for all x Rn.Note that

u(x) =1

|BR(0)|BR(0)

u(x + y)dy.

Hence,

xiu(x) =1

|BR(0)|BR(0)

iu(x + y)dy

=1

|BR(0)|BR(0)

iu(x + y)dy,

where i = xi/R, and then,

|xiu(x)| M|BR(0)|

|BR(0)| 0as R . This implies that u is a constant.

Method Two. Fix any two point x0 and x1. Then for any R > 0, wehave

u(x0) u(x1) = 1|BR(0)|(BR(x0)

udx BR(x1)

udx)

=1

|BR(0)|(BR(x0)BR(x1) udx BR(x1)BR(x0) udx).

Hence,

|u(x0) u(x1) 1|BR(0)|(BR(x0)BR(x1)

|u|dx +BR(x1)BR(x0)

|u|dx)

2M|BR(x1) BR(x0)||BR(0)| 0


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as R (since |BR(x1) BR(x0)| Rn1 and |BR(0)| Rn). Thatis to say, u is a constant.


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6.3. Monotonicity formula for Laplace equation.This subsection is only written for readers not just for fun, but also

for its fundamental role in the regularity theory of nonlinear elliptic andparabolic equations. It is not required for the examination for studentseven though it is really important for modern PDE. Recently I andmy students have found a monotonicity formula for solutions to a verylarge class of elliptic and parabolic equations. We can not exhibit ithere, however, we would like to show the classical monotonicity formulafor harmonic functions.

Assume that u is a harmonic function in a ball BR := BR(p) withcenter at the point p and of radius R. It is easy to see that the integral

BR |u|2dx

is a non-decreasing function of R. To our surprise, so is

R2nBR

|u|2dx.

Why? let us prove it. Recall that

u = 0

in the ball BR. Let X(x) = (Xj(x)) = x p be a radial type vectorfield in BR. Recall that on BR, the outer unit normal is

= (x p)/R.We denote by

ui =u

xi= iu, uij =

2u

xixj.

We now have

0 = BR

uu X = BR

u u Xd +BR

u (u X)dx.

Then we re-write it as

BR u u Xd = BR u (u X)dx.Note that

BR

u u Xd = RBR

|ru|2.

andBR

u(uX)dx =BR

iui(juXj)dx =BR

(|u|2 +uijuiXj).


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Using integration by part to the last term, we have

BR uijuiXj =1

2 j|u|2Xj = R2 BR |u|2 1

2 n|u|2.Then we have

BR

u (u X)dx = 2 n2

BR

|u|2 + R2

BR

|u|2,

and

(n 2)BR

|u|2 = RBR

(|u|2 2|ru|2)d.We now compute

R[R2n

BR |u|2] = R2n BR |u|

2 + (2

n)R1n BR |u|

2

= 2R2nBR

|ru|2

0.This implies that the function

R2nBR

|u|2

is a monotone non-decreasing function of R. More precisely, we haveproved the following result.

Theorem 7. Assume thatu is a harmonic function in a ballBR. Then

R2nBR

|u|2 r2nBr

|u|2 = 2BRBr

|x p|2|ru|2

for all r < R. In particular, the function

r2nBr

|u|2

is a monotone non-decreasing function of r.

Monotonicity formulas play a very important role in the regular-ity theory of weak solutions to nonlinear parabolic and elliptic partial

differential equations/ systems.


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7. lecture seven

From the previous sections, we know that solving a PDE is not easymatter. To handle PDE on unbounded domains, in particular, PDE onwhole space, people try to use integration type of transformations toreduce ODE into algebraic equations and to reduce linear PDE into toODE. The famous Laplace and Fourier transforms are for this purpose.

In this section, we always assume that our functions have a gooddecay at infinity such that the integrals converge.

7.1. Laplace transforms.We define, for a function y = y(t) C0[0, ), the Laplace transform

by

L(y)(s) =

0 y(t)est

dt.

Remark:It is clear that if there is some constant M > 0 such that

|y(x)| eMx, for x > 0then the Laplace transform for y is well defined.

From the definition, we have

L(y1 + ay2) = L(y1) + aL(y2)

for every real number a R.Note that

L(y)(s) =

0

y(t)estdt = y(0) + sL(y)(s).By induction, we get

L(y(k))(s) = skL(y)(s) k

j=1

y(j1)(0)skj.

We can use this property to solve the following ODE:

y

+ ay + by = g(t),

with initial conditionsy(0) = 0 = y(0),

where a, b are constants and g(t) C0[0, ). In fact, let Y(s) =L(y)(s). Then we have

s2Y + asY + bY = L(g).

Hence, we have

Y =L(g)

s2 + as + b.


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Denote by L1 the inverse of the Laplace transform. We remark that,using complex analysis and inverse Fourier transform (see next section),

it can be proved that

L1(Y)(t) =1

2i

a+iai

Y(s)estds

where a is the real part of s = a + ib, which is a complex variable inthe plane. Then we have

y(t) = L1[L(g)

s2 + as + b].

In general, we have a list of Laplace transforms of special functions touse. So we shall invoke to special book on Laplace transforms.

Example. We now compute one example to close this part. Giveny(t) = tn.

Then we have

L(y)(s) =

0

tnestdt.

We do the computation step by step.

L(y)(s) = 1s

0

tnestd(st)

=1

s[n

0

tn1estdt]

= ...

=n!

sn+1.


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7.2. Fourier transforms.We first introduce the Schwartz space S. We say f is a Schwartz

function on R if f is smooth and for any non-negative integers m, n,

|x|m|f(n)(x)| 0as x . Then the Schwartz space is the space of Schwartz functionon R. Note that C0 (R) is a subspace ofS. The important example isthat the function

gA(x) = eAx2

is in the space of Schwartz space, where A > 0 is a constant. We denoteg(x) = g1(x).

Define the Fourier transform of a Schwartz function. Let f be a

smooth function in S. Define

f() =12

f(x)eixdx,

which is called the Fourier transform (in short, F-transform) of f in

R. Set F(f) = f.Remark: From the definition formula, we see that the Fourier trans-

form makes sense for any function f L1(Rn), that is,Rn

|f(x)|dx < .

From the definition of the F-transform, we can see that the Schwartzspace is the invariant set of F-transform, that is,

F : S S.Example 1: Lets compute the Fourier transform of the function

gA(x) above. By definition,

gA() = 12

eAx

2

eixdx.

Note that

Ax2

ix = A(x +i

2A )

2

2

4A2 ,and

eAx2

dx =

A.

Then we have

gA() = e2

4A22

eAx

2

dx =12A

e2

4A2 .


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From this we can see that, for A3 = 14

, gA is the eigenfunction of theF-transform.

For t > 0, let A = 14a2t and let g(x, t) = 1a

2te

x2

4a2t . Then

F[g(x, t)]() = ea22t.

We shall use this expression.Example 2: Let

qj(x) =1

2j, f or |x| 1/j, and qj(x) = 0, f or |x| > 1/j.

Then

qj() = 12

2j1/j1/j

eixdx = 12

sinjj

.

Hence,

qj() 0, f or = 0as j , and

qj(0) = 12

.

Given h, 0 = R. Let(hf)(x) = f(x h), (f)(x) = f(x).

Then by definition, we have

(hf)() = ei(h)f()and

(f)() = 1f(/).Hence, for

(g)(x) = e2x2,

we have

g() = 1g(/) = 12

e2

42 .

Given two functions f and g. Define their convolution by

f g(x) =

f(x y)g(y)dy.


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62 LI MA

Then we have

F(f g(x)) =

1

2 f g(x)eixdx= =

12

eixdx

f(x y)g(y)dy

=12

g(y)eiydy

f(x y)ei(xy)dx

=

2f g.The most important property is that the F-transform of the derivativefx is a multiplication:

fx() = if().

This property gives us a chance to define the inverse operator 1x ofx by the definition

1x f() = (i)1f().

Clearly, we have

1x xf = f.

Remark: One may see more interesting properties about Fouriertransforms in the subject called Harmonic/Fourier Analysis, whichis one of the beautiful theory of modern mathematics.

Lets going back. By an elementary computation, we have the fol-

lowing symmetry formula:u(x)g(x)dx =

u(y)g(y)dy.

Define

f() =12

f(x)eixdx,

which is called the inverse Fourier transform of f. In short we writeit by F(f).

By an elementary computation and the symmetry property, we have

Proposition 8.f = f

and

f = f.

Proof. Lets verify the latter equality. In fact, note that

gA(x) 1,


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as A 0 and

gA() =1

2Ae

2

4A2

20,

as A 0. Then, by symmetry property,

f(0) =12

f()d

=12

limA0

gA()f()d

=12

limA0

gA(y)f(y)dy= f(0).

In general, we let f1(y) = f(y + x). Then

f(x) = f1(0) =12

f1()d = 12

f eixd.

Using this we can prove

Theorem 9. (Plancherel Theorem): Given u S, we have

|u|2dx = |u|2dProof. In fact, using the inverse Fourier transform, we have that

u(x) =12

u(y)eixydy =u(x).

Let g = u in the above equation. So

u = g.

Using the symmetry formula. Then we have

|u|2 = uu = ug = ug = |u|2.

One can define n-dimensional Fourier transform of f in Rn by

f() =1

(

2)n

f(x)eixdx.


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64 LI MA

Here x is the inner product in Rn. The inverse Fourier transform off in Rn is defined by

f() =1

(

2)n

f(x)eixdx.

Similar properties in dimension n are also true for n-dimensionalFourier transforms. In particular, Plancherel Theorem in Rn is true.One important fact lying behind the Fourier transform is that for thelinear operator

A[u] = u

(x),

we have the spectrum property:

A[eix] = ||2eix.


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7.3. Heisenberg uncertainty principle.For any smooth function u (maybe complex-valued) vanishing at in-

finity with |u|2 = 1,

we have

(

x2|u(x)|2dx)(

2|u|2d) 1

4,

with its equality true only at

u(x) = Aex2

,

where > 0 and A4 = 2/.The inequality above is the so called Heisenberg uncertainty princi-

ple.Proof. Integrating by part, we have

1 =

|u|2 = 2

xu u 2

|x||u||u|.

Using the Cauchy-Schwartz inequality we have (H):

1 4(

x2u2)(

|u|2),

with equality atu

= xu.

The latter condition implies thatu = Aex

2/2.

Using

u2 = 1 we have = 2 and the relation between A and .Note that

|u|2 =

2|u|2.Inserting this into (H), we then get what wanted.

Notice that, as a byproduct, we have

1

2

u2 (

x2u2)1/2(

|u |2)1/2,

for every u C10 (R).There are a lot of applications of Fourier transforms in the various

branches in sciences and technologies. I hope our readers can read morebooks on Fourier transforms and Fourier analysis. Please read as moreas you can.


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66 LI MA

8. lecture eight

8.1. Application of Fourier transforms.First, let us use the F-transform to solve the wave equation

utt = u, on Rn R+

Take F-transform for the variable x on both sides. Then we have

utt(, t) = ||2u(, t).Solving this ODE, we get

u = u(, 0) cos(||t) + ut(, 0) sin(||t)|| .Using the inverse F-transform we find the solution expression:

u(x, t) = F(u(, 0) cos(||t) + ut(, 0) sin(||t)|| ).Set

u(x, 0) = f(x), ut(x, 0) = g(x).

The solution can be written as

u(x, t) =1

(

2)n

(f() cos(||t) + g() sin(||t)|| )e

ixd.

Using the relations

cos(||t) = 12

(ei||t + ei||t)

andsin(||t)

|| =1

2i||(ei||t ei||t),

we have

u(x, t) =1

(

2)n

(f+()e

i(x+||t) + f()ei(x||t))d,

where

f+() =1

2 (f() +1

i|| g())and

f() =1

2(f() 1

i|| g()).Using the similar approach, we can use properties of F-transform to

solve the heat equation

ut = u, on Rn R+


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Take F-transform for the variable x on both sides. Then we have

ut(, t) =|

|

2u(, t).

This is a first order ode, and we can solve it to get

u = u(, 0)exp(||2t).Using the inverse F-transform we have

u(x, t) = F(u(, 0)exp(||2t)).Let

gt(x) := g(x, t) =1

(2t)n/2e|x|

2/4t.

Thengt() = exp(

||

2t).

Given the initial data u(x, 0) = f(x) in Rn. Let

u(x, t) =1

(

2)nf gt(x).

Let

Ht(x) := H(x, t) =1

(4t)n/2e|x|

2/4t.

Thenu(x, t) = f Ht(x).

Hence, u(, t) = fexp(||2t), and ut(, t) = ||2u(, t). This showsthat u = f Ht is the solution to the heat equation with initial datau(x, 0) = f(x).


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8.2. Laplace equation in half space.Our problem is to solve the Laplace equation

u = uxx + uyy = 0

on the half space R2+ = {(x, y); y > 0} with boundary conditionu(x, 0) = f(x).

To use the F-transform method, we think of uy(x, 0) = 0. Then wehave

||2u(, y) + uyy(, y) = 0.Solve this and we find

u(, y) = u(, 0)exp(||y) = f()exp(||y).

Define Py(x) = F(exp(| |y))(x).

Then by an elementary computation, we have

Py(x) = F(exp(| |y))(x)

=12

e||y+ixd

=12

0

ey+ixd

+1

2 0

ey+ixd

=12

(1

y + ix+

1

y ix ).This implies that

Py(x) =12

2y

x2 + y2.

Hence

u =12f Py,

and

u(x, y) =

1

2 f Py(x)is the solution wanted for the problem above.


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8.3. Derivative bound. We ask the following question. Given asmooth function u

S(Rn). Assume the bound M0 of

|u

|and the

bound M2 of |u| on Rn. Can we give a bound of|u| on Rn in termof M0 and M2? We answer this question below.

Letu + u = f , in Rn.

Using the Fourier transform, we get

(||2 + 1)u = f .Then we have

u =1

||2 + 1 f .Define

B2(x) = 1(

2)nF( 1||2 + 1 )(x).

Then we haveu(x) = B2 f(x)

and|u(x)| = |B2 f(x)| | B2| |f|(x).

Using the knowledge from calculus we can show thatRn

|B2|(y)dy := B < .Notice that

|f(y)| max|u| + max|u|,where max means taking maximum value over the domain Rn. Hence,we have

|u(x)| B(max|u| + max|u|).Using the function u(rx) (with r > 0 to be determined) to replace u

in the inequality above, we have

r|u(rx)| B(max|u| + r2max|u|).Since r > 0 can be any positive number, we get

|u(rx)

|2

4B2max

|u

|max

|u

|.

Therefore,

max|u| 2B

max|u|max|u|.


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9. lecture nine

9.1. Maximum principle (MP) for Laplace operator.The Maximum Principle (MP) is a beautiful property for solutionsto Laplace equations and heat equations. One can use the Maximumprinciple to get uniform bound for a solution and its derivatives.

Consider a smooth solution u to the following elliptic partial differ-ential inequality in a bounded domain D in the plane R2 (IPDE):

uxx + uyy > 0.

Note that there is no maximum point of u in the interior of thedomain D. For otherwise, assume (x0, y0) is and then we have

u(x0, y0) = 0, DTDTu(x0, y0)

0,

for any direction T R2. The latter condition implies that the matrixD2u(x0, y0) is non-positive, that is,

D2u(x0, y0) 0.Hence,

(uxx + uyy)(x0, y0) 0,which contradicts to (IPDE). Therefore, we have

supDu = supDu.

Assume that we only haveuxx + uyy 0

in the domain D. Then we choose

w = u + ex

for small positive constant and large positive constant . Note that

ex > 0.

Then we have

w > 0.

Hence, we have

supDw = supDw.

Sending , we get thatsupDu = supDu.

Clearly, the above argument is true for general domains in Rn. Thisis the maximum principle for elliptic pde of second order.


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Theorem 10. Maximum Principle. Assumeu is a smooth functionsatisfying

u 0, in D.Then we have

supDu = supDu.

Remark: (1). It is clear from our argument that the function u C2(D) C(D) satisfying

uxxuyy < 0, in D,

can not reaches its maximum in the interior of the domain D. In fact,if it has the maximum point at (x0, y0) in the interior of D, then wehave

uxx 0, uyy 0at (x0, y0). Hence, we have

uxxuyy(x0, y0) 0,a contradiction! This implies that

supDu = supDu.

The same argument applies to the differential inequality:

det(D2u) < 0, in D.

(2). The Maximum principle is true for many nonlinear elliptic andparabolic partial differential equations of second order. It is a basictool in the study of these kinds of pdes.


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9.2. MP for heat equations 1.Assume that D is a bounded smooth domain in Rn. As before, we

letQ = {(x, t); x D, 0 < t T},

and letpQ = {(x, t) Q; x D, or t = 0}

be the parabolic boundary of the domain Q.We now consider the heat equation (HE):

Lu := ut a2u = f(x, t)on Q. We assume that u C2,1(Q) C(Q) is a solution to (HE). HereC2,1(Q) is the space of continuous functions which have continuous

x-derivatives up to second order and continuous t-derivative.Theorem 11. Assume that u C2,1(Q) C(Q) satisfies Lu 0 onQ. Then we have

maxQu(x, t) = maxpQu(x, t).

Proof. We first assume that Lu(x, t) < 0 in Q. Assume that there isan interior point (x0, t0) Q such that

u(x0, t0) = maxQu(x, t).

Then we have at the point (x0, t0), u = 0, D2xu 0, and ut 0. Thisimplies that u(x0, t0)

0 and

Lu(x0, t0) 0,a contradiction. Hence, u can not have an interior maximum point andthen

maxQu(x, t) = maxpQu(x, t).

For general case when

Lu(x, t) 0, (x, t) Q,we let, for > 0,

v(x, t) = u(x, t) t.Then

Lv(x, t) = Lu(x, t) < 0, (x, t) Q.Then we can use the result above to v(x, t) to get

maxQv(x, t) = maxpQv(x, t).

Note thatmaxQu(x, t) maxQv(x, t)

andmaxpQv(x, t) maxpQu(x, t) + T.


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Sending 0, we getmax

Qu(x, t)

max

pQu(x, t).

However, since pQ Q, we havemaxQu(x, t) maxpQu(x, t).

Hence,maxQu(x, t) = maxpQu(x, t).

We are done.

Lets consider an easy application of the maximum principle.

Theorem 12. Given f C(Q) with F := maxQ|f|. Assume thatu

C2,1(Q)

C(Q) satisfies

Lu = f(x, t), (x, t) Qwith u(x, t) = (x, t) on pQ. Then we have

maxQu(x, t) = F T + maxpQ|(x, t)|.Proof. Denote by

B = maxpQ|(x, t)|.Consider the function w(x, t) defined by

w(x, t) = F t + B u(x, t).Then

Lw = F f 0, o n Qand on pQ,

w 0.By the Maximum principle above, we have

w(x, t) 0, (x, t) Q,which implies the conclusion wanted.


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74 LI MA

9.3. MP for heat equations 2.We show in this subsection that the maximum principle cab be

proved to be true for the heat equation by another method.

Theorem 13. Given a bounded continuous functiong(x) on the smoothbounded domain D Rn. Let u = u(x, t) be a non-trivial smoothsolution in C(D) to the heat equation (HE):

ut = u, on D [0, T)with initial and boundary conditions

u(x, 0) = g(x), x D; u(x, t) = 0, D [0, T)where D is a bounded piece-wise smooth domain in Rn. Then,

maxxD |u(x, t)|is a monotone decreasing function of t.

Proof. Consider a new function defined by

Jk(t) =

D

|u|2kdx.

Recall that, for each t,

M(t) := supxD|u(x, t)| = limk+

Jk(t)1/2k.

First, we note that

inflimJk(t)1/2k suplimJk(t)1/2k M(t).

So we only need to show that

M(t) inflimJk(t)1/2k.In fact, For fixed t, since u is a bounded continuous function in D, wehave a sub-domain Q D such that on Q ,

M(t) |u(x, t)|Hence

(M(t)

)

|Q

|1/2k

Jk(t)

1/2k

M(t)

|D

|1/2k

Sending k +, we have(M(t) ) inflimJk(t)1/2k suplimJk(t)1/2k M(t).

Since > 0 can be any small constant, we get

limk+Jk(t) M(t).We remark that the last inequality is also true for any bounded con-tinuous function defined on a unbounded domain D.


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Note thatdJk

dt = 2k u2k1ut = 2k u2k1u.By using integration by part, we have

dJkdt

= 2k(2k 1)

u2k2|u|2 < 0.Hence Jk(t) is a monotone decreasing function of t. Therefore,

maxxD

|u(x, t)|is a monotone decreasing function of t too.

Remark 1. If we assume u 0 in Q satisfiesut u, in Q.

The same conclusion as in the theorem above is also true.Remark 2.Given a constant A > 0. If we assume u 0 in Q satisfies

ut a2u Au, in Q.Then, arguing as before, we have

Jk(t) Jk(0)e2kAt,which implies that

M(t)

M(0)e2at.

Remark 3. Note that for any bounded continuous function f definedon a unbounded domain D with

D

|f|pdx < for some p > 0, we stillhave

M := supD|f(x)| = limk

(

D

|f|kdx)1/k.According to the remark made in the proof of the theorem above, weonly need to prove that

suplimk(D

|f|kdx)1/k M.However, it is almost trivial since

suplimk(D

|f|kdx)1/k suplimk(Mkp D

|f|pdx)1/k = M.With an easy modification, we can extend the above theorem into

unbounded domain D. We omit the detail.


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10. lecture ten

10.1. Variational principle.Variational principle is a important tool in sciences. One of thebeautiful result in the calculus of variations is the Noether principle,which can not be presented here. We can only touch the the conceptof variational structure. We discuss the variational method for Laplaceequation in a region of dimension two.

Given a bounded domain D Rn. We introduce the Dirichlet energy(so called the variational structure)

I(u) =

D

|u|2dx,

for any smooth function u : D R with fixed boundary valueu(x) = (x), x D.

We denote A by such a class of functions. We look for a function w Asuch that

I(w) = infuAI(u).

Assume such a function w exists. Then for every C10 (D), we havew + t A

for every t R. So we have, for all t R,

I(w) I(w + t)and then

d

dtI(w + t)|t=0 = 0.

Note that

I(w + t) =

[|w|2 + 2tw + t2||2].

Hence,d

dtI(w + t)|t=0 = 2

w = 0.

Assume that w C2. Then using integration by part, we have(w) = 0.

Since can be any testing function in C10 (D), we get the Euler-Lagrangeequation for the functional I(), by the variational lemma (see ourtextbook [1]), that

w = 0, in D,


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with the boundary condition w = on D. This is the famous Dirichletproblem in the history. The recent study is about the obstacle problem

for harmonic functions.Remark: We mention that the existence of the infimum w can be

proved by using the Hilbert space theory, which is a beautiful part ofthe course Functional Analysis. It will be really helpful for readersto know some knowledge about Sobolev spaces. Here are some simplefacts about it. Let I = [a, b]. We consider L2(I) as the completion ofthe space C0 (I) with respect to the norm

|f| = (I

|f(x)|2dx)1/2.

In the other word, the element f (which will be called a L2-function)

in L2(I) is defined by a Cauchy sequence {fk} C0 (I), i.e.,|fk fj| 0

as k, j . We say that the element f L2(I) has derivativeg L2(I) if there is a Cauchy sequence {fk} C0 (I) such that

fk f, and (fk)x gin L2. If this is true, we often call g the strong derivative of f anddenote by g = fx. Then is easy to see that the integration by partformula

If xdx = Igdx,for all C0 (I), is true. People often use this formula to define theweak derivative off. It can be showed that the strong derivative is thesame weak derivative. Furthermore, one can show that both definitionsare equivalent each other. Hence, we simply call g the derivative of f.We denote by H10 (I) the Sobolev space which consists of L

2 functionswith derivatives. For f, h H10 (I), we define

(f, h) =

I

fxhxdx.

Then one can show that (

,

) is an inner product on H10 (I), and H

10 (I)

is a Hilbert space.


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10.2. Some comments on Nonlinear evolution problem.In some sense, we try to give a few comments about recent research

directions in the study of non-linear PDE. We assume in this sectionthat readers know the Banach fixed point theorem. We recall the FixedPoint Theorem: Let E be a Banach space and let B E be a closedconvex subset. Let f : B B be contraction mapping, i.e., there is apositive constant (0, 1) such that

|f(x) f(y)| |x y|.Then there is a fixed point x B of f, i.e., f(x) = x. Here | | isthe norm of E.

One often use Fixed Point Theorem to get the local existence ofsolutions to non-linear evolution problem. Lets see a little more. We

study the evolution problem of type

ut = u + F(u)

orutt = u + F(u)

in Rn with nice initial data. Using the Duhamel principle, we canreduce the problem into the form:

u = S(t)(u0) +

t0

S(t )F(u)d := f(u).

Here S(t)(u0) is the solution to the problem with F = 0. Let

B = BR(S(t)(u0)).

We have to show that for small R and T, the mapping f satisfies

f : B B,i.e., B is an invariant set of f and f is a contraction, that is, thereexists a positive constant (0, 1) such that |f(x) f(y)| |x y|for all x, y B. Generally speaking, to make f : B B, we need totake E to be defined by the energy method such that E is a Banachalgebra. Then we can use the interpolation inequality method to showthat f is in fact a contraction mapping.

The interesting feature in non-linear problems is that there is a blowup phenomenon. We show some simple examples below.

Example One: Consider the ODE

ut = u2,

with the initial data u(0) = a > 0. It is easy to see that the solution is

u(t) =a

1 ta ,


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which goes to infinity as t 1a

+. This is the blow up property of the

ODE.

For a non-linear evolution PDE, we can study the blow up propertyby introducing some monotone or concave function A(t) defined by theintegration like

p(t) = minDu(x, t)

or

p(t) :=

(x)u(x, t)k

or

p(t) :=

(x)|xu(x, t)|k,

where k is an integer, (x) > 0 is a solution of some elliptic equation.We can show that p(t) will satisfies some differential inequality, whichmakes p(t) go to negative as t +. This is absurd since p(t) > 0for all t. This gives us the blow up property.

Example Two: We consider the following nonlinear parabolic equa-tion

ut = u + u2, (x, t) [0, 2] [0, T)

with the non-trivial initial data

u(x, 0) = (x) > 0

and the boundary data u(x, t) = u(x + 2, t) for x [0, 2]. Letp(t) = minx[0,2]u(x, t).

Then we havep

(t) p(t)2.This gives us that

p(t) a1 ta

for a = minx[0,2](x) > 0 and u has to blow up to + in finite timebefore a1.

Example Three: Given a constant 14

. We consider the followingnonlinear parabolic equation

ut = u + u2 + u, (x, t) [0, 2] [0, T)

with the non-trivial initial data

u(x, 0) = (x) > 0, x [0, 2]and the boundary data u(0, t) = 0 = u(2, t) for x [0, 2]. By theMaximum principle, we have

u(x, t) > 0, x [0, 2] {t > 0}.


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Choose C = 14

such that

C20

sin( x2 ) = 1.

Let

p(t) = C

20

sin(x

2)u(x, t)dx.

Then, we have

p

= C

20

sin(x

2)ut

= C

20

sin(x

2)u + C

20

sin(x

2)u2 + p(t)

= C2

0

sin(x

2)u2 + ( 1

4)p(t)

p(t)2.This gives us the blow up as before in finite time.

Very recently, we have used similar idea to prove the long standingconjecture on non-linear Schrodinger equation on compact Riemannianmanifolds.


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10.3. Nonlinear PDE of first order.This is a fun model for non-linear pde of first order.

Let us first consider a nice example:

u2x + u2y = 1,

which is called the eikonal equation. Introduce the curve (x(t), y(t))by

dx

dt= ux,

dy

dt= uy.

Then by the equation, we have

(dx

dt)2 + (

dy

dt)2 = 1.

This implies that t is the arc-length parameter for the curve. Doingdifferentiation to the eikonal equation, we have

uxuxx + uyuyx = 0.

Hence along the curve,

d2x

dt2= uxx

dx

dt+ uxy

dy

dt= 0.

Similarly, we haved2y

dt2= 0.

This says that the curve is a straight line, saying,

x(t) = at + c, y(t) = bt + d, a2 + b2 = 1.

Then we have the simple equation:

ux = a, uy = b.

Sou = ax + by + C,

with a2 + b2 = 1.We now give some material for ambitious readers (who may see [2] for

more). We consider the first order PDE of the following form (FPDE):

i

Ai(x, u)uxi = B(x, u).

We look at the graphz = u(x).

Then the PDE says that the normal vector (u, 1) of the graph isorthogonal to the vector field

(A1(x, z),...,An(x, z), B(x, z)).


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This suggests us that the characteristic curve (x(t), z(t)) should satisfy

dxi

dt = Ai(x, z),

dz

dt = B(x, z).

If we can solve this system starting from a given surface (x(s), z(s)),s , of dimension n 1 to get a solution x = x(s, t), z = z(s, t).If we solve the equation x(s, t) = x for (s = x(x), t = t(x)), thenu = z(s(x), t(x)) is the solution to (FPDE).

For more, please look at the great book of D.Hilbert and R. Courant[2].


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11. Appendix A

One may consider this section as a last one to our course. We havethree parts. One is for the variational lemma. Another is about themethod of separation of variables. The other is about some open prob-lems in PDEs. Our readers may do research for these problems in thefuture.

11.1. Variational lemma.

Lemma 14. Let f C0(D) for a bounded domain in Rn. Assumethat

D

f(x)(x)dx = 0

for every C0 (D). Then f = 0.Proof. We argue by contradiction. Assume that there is a point x0 Dsuch that f(x0) = 0. Then we may assume that f(x0) > 0 and there isa small ball B(x0) such that f(x) = 0 for every x B(x0). Take

(x) = (|x x0|)where

(r) = e 1

1r2 , 0 r < and (r) = 0 for all r . Note that (|x x0|) C0 (D), and

f(x)(|x x0|) > 0, i n B(x0).Hence,

D

f(x)(x)dx =

C0 (D)

f(x)(|x x0|) > 0.

This gives a contradiction to our assumption of our lemma. Therefore,the lemma is true.

Example. Assume that u = u(t) C20 [0, L]. Define

F(u) = L

0

(1

2 |u

|2

1

3

u3

g(t)u(t))dt,

where g(t) is a given continuous function on [0, L]. Then, ifF(u) = 0,u satisfies

u = u(t)2 + g(t).In fact, by definition,

0 =< F(u), >=

L0

(u u2 g)dt


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for all C0 (0, L). Integrating by part, we get

0 = L0

(u u2 g)dt.By variational lemma, we know that

u + u2 + g = 0,

which is equivalent to our equation above.

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