INTRODUCTION TO EQUATIONS FROM MATHEMATICAL...

INTRODUCTION TO EQUATIONS FROMMATHEMATICAL 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

2 LI MA

This is a lecture note for the courses of Equations from Mathemati-cal Physics or Methods of Mathematical Physics given by me to seniorstudents in Tsinghua University, Beijing

EQUATIONS FROM MATHEMATICAL PHYSICS 3

To An, for patience and understanding

4 LI MA

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.


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 theconservation 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

6 LI MA

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

There is another important tool in the study of heat equation andelliptic 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 PDE’s 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


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 C∞0 (D) the space of infinitely differentiable smooth

functions with support inside the domain DRn. Here the support of afunction f is the closure of the set x ∈ D; f(x) 6= 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 = (∂k1

x1∂k2

x2...∂kn

xnf); 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.

8 LI MA

1. lecture one

1.1. Divergence theorem.Let (xi) be the standard coordinates in the Euclidean space Rn. We

recall the classical divergence theorem:

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

boundary ∂D in Rn. Let X : D → Rn be a smooth vector field on theclosure D of the domain D. Let ν be the unit outer normal to ∂D.Then we have the following formula:∫

D

divXdx =

∫∂D

X · νdσ.

Here divX =∑

i∂Xi

∂xi for X = (X i) , ν = (νi) and X · ν =∑

iXiνi.

If X i = ∂u∂xi 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

uνjdσ.

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

a

f′(x)dx.

In dimension two, the divergence theorem is Green’s 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

Pdx+Qdy.


It is clear that the form of divergence theorem in dimension twois different from that of the Green formula. We want to prove thatthey 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 ν = −N = (y′,−x′). Choose X1 = Q and X2 = −P . Then we have

divX = Qx − Py,

and along the boundary ∂D,

Pdx+Qdy = (Px′ +Qy′)ds = X · νdsHence, Green’s formula implies that∫

D

divXdx =

∫∂D

X · νds,

which is the divergence theorem in dimension two.

10 LI MA

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 ρ = constant is themass density of the string. Then the kinetic (or temporal) energy ofthe string is

T (u) =1

2

∫ l

0

ρu2tdx.

The potential (or spatial) energy is

U(u) =c

2

∫ l

0

ρu2xdx,

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) =

∫ t

0

dτ(T (u)− U(u)− U(f0))

=1

2

∫ t

0

dτ

∫ l

0

ρu2tdx

− c

2

∫ t

0

dτ

∫ l

0

ρu2xdx+

∫ t

0

dτ

∫ l

0

f0udx.

The Hamilton principle implies that the motion of the vibrating stringobeys

δH(u) = 0,

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

< δH(u), φ >:=d

dεH(u+ εξ)|ε=0 = 0.

We compute that

< δH(u), φ > =

∫ t

0

dτ

∫ l

0

(ρutφt

− cρuxφx + f0φ)dx

=

∫ t

0

dτ

∫ l

0

(−ρutt + cρuxx + f0)φdx


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

utt − cuxx = f0/ρ,

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) = φ(x),

andut(x, 0) = ψ(x);

and the boundary condition

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

The higher dimensional wave equation in its simple form is

utt(x, t) = a2∆u(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.

12 LI MA

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

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

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

W

ρdx.

Then, its change rate at time t is

d

dt

∫W

ρdx.

Assume that there is no fluid generating source. Then the changerate is only across fluid rate from the boundary. Take a small part dσin the boundary ∂W . Then the amount of −ρu · νdσ flow into W . Sothe total amount flowing into W across the boundary is the integrationover ∂W :

−∫

∂W

ρu · νdσ.

So we haved

dt

∫W

ρdx = −∫

∂W

ρu · νdσ.

Using the divergence theorem, we have∫W

(∂

∂tρ+ div(ρu))dx = 0.

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

∂

∂tρ+ div(ρu) = 0.

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

ρt + ∂x(a(x)ρ(x, t)) = 0,

where x = x1.


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 | = 0

if 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 ) =

k∑j=0

aj(x)∂(j)x

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

|w(0)− u(0)| ≤ δ,

we have

supt|w(t)− u(t)| < ε.

14 LI MA

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 linearpde’s 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.


2. lecture two

2.1. Heat equation and Laplace equation.Heat equation and Laplace equation are very important objects in

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

velocity. By Fourier’s law, we have

q = −k∇u,

where k > 0 is a coefficient of physical quantity.Let f0(x, t) be the density of heat source. Then as in the case of

continuity equation, we have, for a physical constant c > 0 of theregion D ⊂ Ω,

∂t

∫D

cρudx =

∫∂D

q · (−ν)dσ +

∫ ∫D

ρf0dx.

Using the divergence theorem we have∫∂D

q · (−ν)dσ = −∫

D

div(q)dx.

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

cρ∂tu = −divq + ρf0.

Using the Fourier law, we have

cρ∂tu = div(k∇u) + ρf0,

which is called the heat equation with source f0. Assume that c = 1 =k = ρ and f0 = 0. Then we have the heat equation

ut = ∆u, (x, t) ∈ Ω× (0,∞),

which is a typical parabolic partial differential equation of second order.Physically, we can measure the heat density on the boundary ∂Ω,

and we may assume that u(x, t) = 0 for (x, t) ∈ ∂Ω × (0,∞). Theinitial heat density at t = 0 can also be detected. So we assume that

u(x, 0) = φ(x), x ∈ Ω.

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

k∂u

∂ν(x, t) = αu(x, t), (x, t) ∈ ∂Ω× (0,∞).

16 LI MA

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

∆u+ f = 0,

where f = ρf0, or in other form

−∆u = f

which is called the Poisson equation. If f = 0, it is called the Laplaceequation in Ω, 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 ondomains coupling with its boundary condition.

In the following subsection, we try to solve some simple pde of firstorder.


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) +

∫ t

0

f(aτ + A, τ)dτ.

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. Soby 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(τ − t), τ)dτ.

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).

18 LI MA

Then we have g(x, t) ≥ 0 and

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

Let v = w − u. Then we have

vt + 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 forPDE’s.


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, ...) = 0

is 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.

Let’s 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)− t

2.

On the other hand, we find the curve satisfying

x(t) = A+ t+

∫ t

0

u(x(τ), τ)dτ = A+ t+

∫ t

0

(τ

2+√A)dτ.

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.

20 LI MA

Given any point (x, t) on the curve x(t), we have x(t) = x. Hence, thesolution is

u(x, t) =√x− t.


3. lecture three

3.1. Duhamel principle.Consider (ODE)

u′+ p(t)u = 0,

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

u(t, τ) = fτexp(−∫ t

τ

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 fτ = f(τ) and solve (ODE)at t = τ as above to get the solution (S). Then the function defined by∫ t

0

u(t, τ)dτ

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

22 LI MA

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

u′ = a(t)u+ b(t)

we can write it as

(e−R t a(τ)u)′ = e−

R t a(τ)b(t)

So it has a general solution

u(t) = u(0)eR t a(τ) + e

R t a(τ)

∫ t

dse−R s a(τ)b(s).

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

u(t) = eR t a(τ)

∫ t

dse−R s a(τ)b(s).

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

u(t) ≤ a+ β

∫ t

0

u(τ)dτ.

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

f(t) = a+ β

∫ t

0

u(τ)dτ.

Then we have f ′ = βu. By assumption, u ≤ f , we get

f ′ ≤ βf.

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

u′ = β(t)f(u)

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

w′ ≥ β(t)f(w)

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

g(t) = w′ − β(t)f(w).

Then g(t) ≥ 0. Set

v = w − u.

Then we have

v′ == β(t)(f(w)− f(u)) + g(t) = β(t)B(t)v(t) + g(t)


with v = 0 at t = 0 and

B(t) =

∫f ′(zw + (1− z)v)dz.

Let a(t) = β(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.

24 LI MA

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 datau|t=0 = u0

andut|t=0 = u1.

Let’s first assume that u0 = 0. Note that

∂2t − ∂2

x = (∂t − ∂x)(∂t + ∂x).

Then we reduce the problem into two pde of first order. Let

v = ∂tu+ ∂xu.

and∂tv − ∂xv = 0.

Solving u we get

u(x, t) =

∫ t

0

v(x+ (τ − t), τ)dτ.

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) =

∫ t

0

u1(x+ (τ − t) + τ)dτ.

Changing the variables = x+ 2τ − t,

we can get

u(x, t) =1

2[

∫ x+t

x−t

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


andut|t=0 = 0.


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

mula (EW):

u(x, t) =1

2[(u0(x+ t) + u0(x− t) +

∫ x+t

x−t

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) =

∫ t

0

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 if u0 = 0 and u1 ≥ 0, then u ≥ 0 for

all t. For each t, |u(x, t)| ≤M0+tM1 is bounded provided |u0(x)| ≤M0

and |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), wehave

u = 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 + u2

x)dx.

26 LI MA

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 + u2

0x)dx.

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

utt − uxx = 0, x ∈ R+, t > 0,


andut|t=0 = u1.

Note that u0(t + x) is the special solution to (HWE). So we canuse 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 theD’Alembert formula to solve (HWE). One may also use this method totreat one dimensional wave equation on an interval, however, we have amore beautiful method—Fourier method/Separation variable methodto study this case. See the next section.

Remark Four: We point out that we can use the D’Alembert 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 = 0

with the initial datau|t=0 = φ(x)

andut|t=0 = ψ(y).

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

utt − uxx = 0

with the initial datau|t=0 = φ(x)


andut|t=0 = 0.

The solution for this one dimensional problem is

u(x, t) =1

2[(φ(x+ t) + φ(x− t)]

The other isutt − uyy = 0

with the initial datau|t=0 = 0

andut|t=0 = ψ(y).

Its solution is

u(x, t) =1

2[

∫ y+t

y−t

ψ(s)ds].

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

u(x, y; t) =1

2[(φ(x+ t) + φ(x− t) +

∫ y+t

y−t

ψ(s)ds].

28 LI MA

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 condition

u(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) = a2TX

′′,

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−λ. Then we have one equation for T :

T′′

+ a2λT = 0

and the other equation for X:

X′′

+ λX = 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 λ < 0, the general solution for X is

X(x) = c1e√−λx + c2e

−√−λx

Using (BC) we havec1 + c2 = 0

andc1e

√−λ + c2e

−√−λ = 0,

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

X(x) = c1 + c2x.

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


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

X(x) = c1 cos(√λx) + c2 sin(

√λx).

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

c2 sin(√λ) = 0,

and thenλ = k2π2, k = 1, 2, ...

HenceX(x) = Xk(x) = ck sin(kπx)

for k = 1, 2, ....Returning to the equation for T (t) with λ = k2π2, we get

Tk(t) = ak cos(kπat) + bk sin(kπat).

SoUk(x, t) = [ak cos(kπat) + bk sin(kπat)] sin(kπx)

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

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

λ

∫|X|2 =

∫|X ′|2.

We now look for a solution of the form

u(x, t) =+∞∑k=1

[ak cos(kπat) + bk sin(kπat)] sin(kπx)

to the wave equationutt = a2uxx

on [0, 1]× [0,∞) with the initial condition

u = φ(x), ut = ψ(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 ak

and bk. So we have

φ(x) =+∞∑k=1

ak sin(kπx)

30 LI MA

and

ψ(x) =+∞∑k=1

kπabk sin(kπx).

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

ak = 2

∫φ(ξ)sin(kπξ)dξ

and

bk =2

kπa

∫ψ(ξ)sin(kπξ)dξ.

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) = α(t), ; u(1, t) = β(t),

we have to make a transformation

u(x, t) = v(x, t) + α(t) + x(β(t)− α(t))

and reduce the problem for v with the homogeneous boundary condi-tion.


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

utt − a2uxx = A sin(ωt), (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 ω 6= ajπL

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

u(x, t) =∑

j

Laj

ajπsin(

jπx

L)

∫ t

0

sin(ωτ) sin(ajπ

L(t− τ))dτ

=∑

j

aj

ωj(ω2 − ω2j )

(ω sinωjt− ωj sinωt) sin(jπx

L)

where

ωj =ajπ

Land

aj =2

L

∫ L

0

A(x) sin(jπx

L)dx.

Because we haveω2 − ω2

j

which may cause the coefficientaj

ωj(ω2 − ω2j )→∞.

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

32 LI MA

4.3. Wave equation in two dimensional disk.We denote BR(0) ⊂ R2 the ball in the plane R2. Let’s consider the

2-dimensional wave equation

utt = a2(uxx + uyy)

on BR(0)× [0,∞) with the homogeneous boundary condition

u(x, t) = 0, x ∈ ∂BR(0)

and the initial conditionsu|t=0 = u0

andut|t=0 = u1.

Introduce the polar coordinates (r, θ) by

x = r cos θ, y = r sin θ.

Then∆u = urr + r−1ur + r−2uθθ.

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′′

+ a2λT = 0

and∆U + λU = 0, U(R, t) = 0.

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

h′′(θ)

h(θ)= −r

2B′′

+ rB′+ k2r2B

B= −β2.

Then we get two problems:

h′′

+ β2h = 0, h(θ + 2π) = h(θ),

andB

′′+ r−1B

′+ (k2 − β2r−2)B = 0, B(R) = 0.

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

β = βn = n

andhn(θ) = An cos(nθ) +Dn sin(nθ).

Inserting β = n into the Bessel equation, we have

B′′

+ r−1B′+ (k2 − n2r−2)B = 0, B(R) = 0.


The general solutions of this equation are called the Bessel functionsof order n. Just like trigonometric polynomials, Bessel functions forman 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 θ) =∞∑

n=−∞

Bn(r)einθ,

where i =√−1 and einθ = cos(nθ) + i sin(nθ). Hence,

Bn(r) =1

2π

∫ 2π

0

exp(ir sin θ − inθ)dθ.

We remark that, if the domain is not a ball but a rectangle, we justlet

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

34 LI MA

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 ; for (x, t) ∈ Rn ×R+

with the initial condition

u = φ(x), ut = 0, at t = 0

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

utt −∆u = 0; for (x, t) ∈ Rn ×R+

with the initial condition

u = 0, ut = ψ(x), at t = 0.

We then use the descent method to study

utt = ∆u

in Rn×R in one dimension higher space Rn+1×R with initial conditions

u = 0, ut = ψ(x), at t = 0.

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

utt −∆u = 0; R3 × t > 0

with the initial data

u = 0, ut = ψ(x), at t = 0.

We find that

u(p, t) =1

4πa2t

∫Sat(p)

ψ(x)dσx,

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

I(x, r; g) =1

4π

∫∂Br(x)

udσ

for g = g(y) ∈ C2(R3) and

J := 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

4π

∫|θ|=1

u(x+ rθ)dθ


and ∫BR(0)

g(x+ z)dz

=

∫ R

0

r2dr

∫|θ|=1

u(x+ rθ)dθ

= =

∫ R

0

4πr2I(x, r; g)dr.

We now compute

∆x

∫ R

0

4πr2I(x, r; g)dr

= ∆x

∫BR(0)

g(x+ z)dz

=

∫BR(0)

∆xg(x+ z)dz

=

∫BR(0)

∆zg(x+ z)dz

=

∫∂BR(0)

∂Rg(x+ z)dσz

=

∫∂B1(0)

∂Rg(x+Rθ)R2dθ

= = 4πR2∂RI(x,R; g).

So, we have

∆x

∫ R

0

r2I(x, r; g)dr = R2∂RI(x,R; g).

Note that

∂R∆x

∫ R

0

r2I(x, r; g)dr

= ∆x∂R

∫ R

0

r2I(x, r; g)dr

= R2∆xI(x,R; g)

and

∂R[R2∂RI(x,R; g)]

= 2R∂RI(x,R; g) +R2∂2RI(x,R; g).

36 LI MA

Hence,

R2∆xI(x,R; g) = 2R∂RI(x,R; g) +R2∂2RI(x,R; g).

Dividing both sides by R, we have

∆xRI(x,R; g)

= = 2∂RI(x,R; g) +R∂2RI(x,R; g)

= ∂2R(RI(x,R; g)).

Now,

∂2J

∂t2=

R

4π

∫∂B1(0)

utt(x+Rθ, t)dθ

=a2R

4π

∫∂B1(0)

∆xu(x+Rθ, t)dθ

=a2

4π∆xR

∫∂B1(0)

u(x+Rθ, t)dθ

= a2∆x(RI(x,R;u))

= a2∆x(RI(x,R;u))

= a2 ∂2J

∂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;ψ).

Then we have

J(x,R; t) =1

2a

∫ at+R

at−R

rI(x, r;ψ)dr.


Hence,

u(x, t) = limR→0

J(x,R; t)

R

= limR→0

1

2aR

∫ at+R

at−R

rI(x, r;ψ)dr

=1

a[RI(x,R;ψ)]|R=at

= tI(x, at;ψ)

=t

4π

∫∂B1(0)

ψ((x+ atθ)dθ.

That is,

u(x, t) =1

4πa2t

∫Sat(x)

ψ(y)dσy.

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

utt −∆u = f ; R3 × t > 0with the initial data

u = φ(x), ut = ψ(x), at t = 0,

and we have the famous Kirchhoff formula for the solution:

u(p, t) =1

4πa2

∫Bat(p)

r−1f(x, t− a−1r)dx

+1

4πa2t

∫Sat(p)

ψ(x)dσx

+ ∂t[1

4πa2t

∫Sat(p)

φ(x)dσx]

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 theKirchhoff formula.

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 solution

to the heat equationut = 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−λ. Then we have

T′+ a2λT = 0

and (EX):

X′′

+ λX = 0.

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

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

to solve X and λ. Multiplying both sides of (EX) by X and integratingover [0, 1], we have

λ

∫X2 =

∫|X ′|2 > 0.

So λ > 0. Then we have

X(x) = Xk(x) = ck sin(kπx)

for k = 1, 2, ... andλ = λk = k2π2.

Then we solve for T and get Tk(t) = T (0)exp(−a2λkt) 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(−a2k2π2t) sin(kπx)


φ(x) =∑

k

Ak sin(kπx)


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 λ = λj to theSturm-Liouville problem

X′′

+ λX = 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 ODE

u′

j = −a2λjuj + fj(t)

with suitable initial condition. Hence, we have the expression

uj(t) = e−λjt(uj(0) +

∫ t

0

eλjτfj(τ)dτ

and

u(x, t) =∑

j

e−λjt(uj(0) +

∫ t

0

eλjτfj(τ)dτ)Xj(x).

Here uj(0) are determined by

φ(x) =∑

j

uj(0)Xj(x).

We omit the detail here.

40 LI MA

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

that for some 0 > λj ∈ R, we have

Lφj(x) = λjφj(x),

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

‖u‖ =√∑

a2j

for u =∑ajφj(x) ∈ E.

Given a smooth function f ∈ C1,2(D × [0,∞)). Assume that theexpansion u =

∑uj(t)φj(x) is a solution to the partial differential

equation: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)φj(x)

with its norm on D × [0,∞) defined by

‖f‖ =∑

j

∫ ∞

0

|fj(t)|2dt <∞.

Then we haveu′

j(t) = λjuj(t) + fj(t).

Hence, as before, we have the expression

uj(t) = eλjt

∫ t

0

e−λjτfj(τ)dτ.

Note that

(u′

j(t)− λjuj(t))2 = u

′

j(t)2 + λ2

juj(t)2 − 2λjuj(t)u

′

j(t) = fj(t)2.

Then we have∫ T

0

(u′

j(t)2 + λ2

juj(t)2)dt− λjuj(T )2 =

∫ T

0

fj(t)2.

Note that we have assumed that λj < 0. Hence, we have∫ ∞

0

(u′

j(t)2 + λ2

juj(t)2)dt ≤

∫ ∞

0

fj(t)2,

which implies that‖∂tu‖2 + λ2‖u‖ ≤ ‖f‖2,


where λ2 = infjλ2j. This is the energy bound for the solution as we

wanted.

42 LI MA

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 λ > 0,

the function

uλ(x, t) = u(λx, λ2t)

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

u(x, t) = tβf(|x|2/t).

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

ut = βtβ−1f + tβ(−|x|2

t2)f

′= βtβ−1f − rtβ−1f

′,

u′

=∂

∂|x|u

= 2|x|tβ−1f′

u′′

=∂2

∂|x|2u

= 2tβ−1f′+ 4|x|2tβ−2f

′′

= 2tβ−1f′+ 4rtβ−1f

′′

∆u = u′′

+n− 1

|x|u′

= 2tβ−1f′+ 4rtβ−1f

′′+ 2(n− 1)tβ−1f

′

= 4rtβ−1f′′

+ 2ntβ−1f′.

So,

−βrf + f

′+ 4f

′′+

2n

rf′= 0.

Let β = −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.


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

r → +∞. So

f = Be−r/4.

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

K(x, t) =1

(4πt)n/2e−

|x|24t .

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 solutionto 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 ε > 0. For any fixed point x ∈ R,

we have some δ0 > 0 such that, for |x− y| < δ0,

|g(y)− g(x)| < ε

2.

Step two: We take δ1 > 0 small enough such that for 0 < t < δ1,∫|x−y|≥δ0

K(x− y, t)dy <ε

4M.

44 LI MA

Then we have

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

≤ |∫|x−y|≥δ0

K(x− y, t)(g(y)− g(x))dy|

+ |∫|x−y|<δ0

K(x− y, t)(g(y)− g(x))dy|

≤ 2M

∫|x−y|≥δ0

K(x− y, t)dy

+ ε

∫|x−y|<δ0

K(x− y, t)dy

< 2Mε

4M+ε

2= ε.

Here we have used the fact that K ≥ 0 and∫K(x− y, t)dy = 1

for any t > 0. This says that u(x, t) → g(x) as t→ 0.

We now study the behavior of the solution defined above as |x| → ∞.

Theorem 3. Assume g ∈ C0(Rn) satisfies

g(x) → 0, as |x| → ∞.

Then for the solution u(x, t) defined above, we have

u(x, t) → 0, ∇xu(x, t) → 0, as |x| → ∞.

for each t > 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| → ∞.


Note that

|u(x, t)| ≤∫

Rn

K(x− y, t)g(y)dy

≤ supBR(x)|g(x)|∫

BR(x)

K(x− y, t)dy

+ supRn−BR(x)|g(x)|∫

Rn−BR(x)

K(x− y, t)dy

≤ supBR(x)|g(x)|+ C(t)supRn|g(x)|∫ ∞

R

e−ρ2/4tρn−1dρ

→ 0 + o(R)

as |x| → ∞. Here C(t) is a constant depending only on t and o(R) isas small as we like for R large. Hence, we have showed that

|u(x, t)| → 0,

as |x| → ∞.Using the polar coordinates (ρ, θ), we can also show that

|∇ju(x, t)| → 0,

as |x| → ∞. We omit the detail here.

46 LI MA

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) = δx(y).

Here we define δx(y) as the generalized function by

< δx(y), g(y) >:= δx(y)(g(y)) = g(x).

That is, we consider δx(y) as a function defined on the function spaceC∞

0 (Rn). Then, δx is a linear functional on C∞0 (R). This δx(·) is called

Dirac’s delta function or Dirac measure. One can also consider thedelta function δ0(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 6= 0, qj(x) → 0 and at x = 0, qj(0) →∞; or for φ ∈ C∞0 (R),

< qj, φ > =

∫R

qj(x)φ(x)dx

=1

2j

∫ 1/j

−1/j

φ(x)dx→ φ(0)

= < δ0, φ > .

as j →∞. If so, we define limqj = δ0. 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 δ0 as its limit. For example, let

ρ(x) = me− 1

1−|x|2 , for |x| < 1, ρ(x) = 0, for |x| ≥ 1

where m > 0 is a constant such that∫R

ρ(x)dx = 1,

and let

ρt(x) =1

tρ(x/t),

where t > 0. Then ∫R

ρt(x)dx = 1

and limt→0ρt = δ0.


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

0 (Rn) with

< u, φ >=

∫Rn

u(x)φ(x)dx.

With this understanding, we say K(x−y, t) is the fundamental solutionthe heat equation on Rn×R+. We emphasize that we call K(x−y, t) isa fundamental solution to the heat equation on the whole space becauseK(x−y, t) satisfies the heat equation for t > 0, and we consider K(x−y, 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 = φ(x), t = 0.

For this case, the fundamental solution is

Γ(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+

Γ(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 condition

u = φ(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

Γ(x, y, t) = K(x− y, t) +K(x+ y, t).

Here we have used the even extension to keep the boundary condition.

48 LI MA

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 that

r2 =∑

i

(xi)2.

Then we haverdr =

∑i

xidxi.

Then∂r

∂xi=xi

r.

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

uxi = v′ xi

r,

and

uxixi = v′(1

r− (xi)2

r3) + v

′′ (xi)2

r2.

Then we have

∆u = v′′

+n− 1

rv′.

Solve

v′′

+n− 1

rv′= 0.

We find thatv′=

c

rn−1

for some constant c and

v = a log r + b, for n = 2,

andv = ar2−n + b, for n ≥ 3,

for some constants a and b.Choose b = 0. Define

Γ(x) =−1

2πlog r, for n = 2,

and

Γ(x) =1

n(n− 2)|Sn−1|r2−n, for n ≥ 3.


Here |Sn−1| is the volume of the sphere Sn−1.Let Γ(x, y) = Γ(x− y) for x, y ∈ Rn. This function Γ(x, y) is called

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

−∆yΓ(x− y) = δx(y).

If so, we have

u(x) = < δx(y), u(y) >

= − < ∆Γ(x− y), u(y) >

= −∫

Rn

∆Γ(x− y)u(y)dy

= −∫

Rn

Γ(x− y)∆u(y)dy

=

∫Rn

Γ(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 ν the outer unit normal to the boundary ∂D. Then we have∫

D

(v∆u− u∆v)dx =

∫∂D

(v∇u · ν − u∇v · ν)dσ.

Theorem 4. Given f ∈ C00(Rn), i.e, f is continuous in Rn and f(x) =

0 in the outside of some large ball. Let

u(x) =

∫Rn

Γ(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 −Bε(x), where ε→ 0, we get that∫

BR

Γ(x, y)(−∆u(y))

= limε→0

∫BR−Bε(x)

Γ(x, y)(−∆u(y))

= limε→0

∫|y−x|=ε

(Γ(x, y)∇ru(y)

− ∇rΓ(x, y)f(y))dσ,

50 LI MA

where r = |x− y and ∇ru(y) = ∂u∂r

(y).Note that ∫

|y−x|=ε|Γ(x, y)∇ru(y)|dσ

≤ 1

2π

∫|y−x|=ε

ln1

r|∇ru(y)|

≤ sup|∇u|ε|lnε|→ 0,

and

−∫|y−x|=ε

∇rΓ(x, y)u(y)dσ

=1

2π

∫|y−x|=ε

∇rln(r)u(y)dσ

=1

2π

∫|y−x|=ε

1

rf(y)dσ

=1

2πε

∫|y−x|=ε

u(y)dσ

→ u(x)

as ε→ 0. This shows that the function u satisfies

−∫

R2

Γ(x, y)∆u(y)dy = u(x).

We ask our readers to verify the n ≥ 3 case.

Let’s 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 D

with the boundary condition u(x) = φ(x) for x ∈ ∂D. Then usingthe second Green formula and the fact v = Γ being the fundamentalsolution on the whole Rn, we get (U1):

u(x) =

∫D

Γ(x− y)f(y)dy

+

∫∂D

(Γ(x− y)∇u(y) · ν − φ(y)∇Γ(x− y) · ν)dσy.

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


data g(x, y) = −Γ(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) · ν − φ(y)∇g(x, y) · ν)dσy.

SetG(x, y) = Γ(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

φ(y)∇G(x, y) · νdσy.

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

Let’s consider the boundary value problem for Laplace equation.Assume that u ∈ C2(D) ∩ C(D) satisfies

−∆u = 0, in D ⊂ Rn,

with the Dirichlet boundary condition

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

u(x) =

∫D

Γ(x− y)(−∆u(y))dy

+

∫∂D

(Γ(x− y)∇u(y) · ν − φ(y)∇Γ(x− y) · ν)dσy

=

∫∂D

φ(y)∇Γ(x− y) · νdσy.

LetP (x, y) = ∇Γ(x− y) · ν

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)φ(y)dσy,

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 − L1

u1(x) +y − L1

L2 − L1

u2(x).


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 + rθ, where r = |x − p| and θ = x−p

r. Using the

divergence theorem, we get

0 =

∫Bt(p)

∆u =

∫∂Bt(p)

∂u

∂rdσ.

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

tn−1

∫∂B1(p)

∂u

∂r(p+ tθ)dθ = tn−1 ∂

∂t

∫∂B1(p)

u(p+ tθ)dθ.

Here, we want to point out that the notation

∂u

∂r(p+ tθ)

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

0 = tn−1 ∂

∂t(t1−n

∫∂Bt(p)

udθ).

This implies that

t1−n

∫∂Bt(p)

udθ

is a constant for all t ≤ R. Then

1

|∂Bt(p)|

∫∂Bt(p)

udθ =1

|∂BR(p)|

∫∂BR(p)

udθ.

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

u(p) =1

|∂BR(p)|

∫∂BR(p)

udθ.

Then

u(p)|∂BR(p)| =∫

∂BR(p)

udθ.

If we take integration in R, we have

u(p) =1

|BR(p)|

∫BR(p)

udx.

Hence, we have

54 LI MA

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)

udθ.

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). If u 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)

νiu(x+ y)dσy,

where νi = xi/R, and then,

|∂xiu(x)| ≤ M |∂BR(0)||BR(0)|

→ 0

as R→∞. This implies that u is a constant.Method Two. Fix any two point x0 and x1. Then for any R > 0, we

have

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


as R→∞ (since |BR(x1)−BR(x0)| ≈ Rn−1 and |BR(0)| ≈ Rn). Thatis to say, u is a constant.

56 LI MA

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

R2−n

∫BR

|∇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

ν = (x− p)/R.

We denote by

ui =∂u

∂xi= ∇iu, uij =

∂2u

∂xixj.

We now have

0 = −∫

BR

∆u∇u ·X = −∫

∂BR

∇u ·ν∇u ·Xdσ+

∫BR

∇u ·∇(∇u ·X)dx.

Then we re-write it as∫∂BR

∇u · ν∇u ·Xdσ =

∫BR

∇u · ∇(∇u ·X)dx.

Note that ∫∂BR

∇u · ν∇u ·Xdσ = R

∫∂BR

|∂ru|2.

and∫BR

∇u·∇(∇u·X)dx =

∫BR

∇iu∇i(∇juXj)dx =

∫BR

(|∇u|2+uijuiXj).


Using integration by part to the last term, we have∫BR

uijuiXj =1

2

∫∇j|∇u|2Xj =

R

2

∫∂BR

|∇u|2 − 1

2

∫n|∇u|2.

Then we have∫BR

∇u · ∇(∇u ·X)dx =2− n

2

∫BR

|∇u|2 +R

2

∫∂BR

|∇u|2,

and

(n− 2)

∫BR

|∇u|2 = R

∫∂BR

(|∇u|2 − 2|∂ru|2)dσ.

We now compute

∂R[R2−n

∫BR

|∇u|2] = R2−n

∫∂BR

|∇u|2 + (2− n)R1−n

∫BR

|∇u|2

= 2R2−n

∫∂BR

|∂ru|2

≥ 0.

This implies that the function

R2−n

∫BR

|∇u|2

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

Theorem 7. Assume that u is a harmonic function in a ball BR. Then

R2−n

∫BR

|∇u|2 − r2−n

∫Br

|∇u|2 = 2

∫BR−Br

|x− p|2|∂ru|2

for all r < R. In particular, the function

r2−n

∫Br

|∇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 partialdifferential equations/ systems.

58 LI MA

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)e−stdt.

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

|y(x)| ≤ eMx, for x > 0

then 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)e−stdt = −y(0) + sL(y)(s).

By induction, we get

L(y(k))(s) = skL(y)(s)−k∑

j=1

y(j−1)(0)sk−j.

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.


Denote by L−1 the inverse of the Laplace transform. We remark that,using complex analysis and inverse Fourier transform (see next section),it can be proved that

L−1(Y )(t) =1

2πi

∫ a+i∞

a−i∞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) = L−1[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. Given

y(t) = tn.

Then we have

L(y)(s) =

∫ ∞

0

tne−stdt.

We do the computation step by step.

L(y)(s) = −1

s

∫ ∞

0

tne−std(−st)

=1

s[n

∫ ∞

0

tn−1e−stdt]

= ...

=n!

sn+1.

60 LI MA

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)| → 0

as x→∞. Then the Schwartz space is the space of Schwartz functionon R. Note that C∞

0 (R) is a subspace of S. The important example isthat the function

gA(x) = e−Ax2

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 asmooth function in S. Define

f(ξ) =1√2π

∫f(x)e−ixξdx,

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: Let’s compute the Fourier transform of the functiongA(x) above. By definition,

gA(ξ) =1√2π

∫e−Ax2

e−ixξdx.

Note that

−Ax2 − ixξ = −A(x+i

2Aξ)2 − ξ2

4A2,

and ∫e−Ax2

dx =

√π

A.

Then we have

gA(ξ) =e−

ξ2

4A2

√2π

∫e−Ax2

dx =1√2A

e−ξ2

4A2 .


From this we can see that, for A3 = 14, gA is the eigenfunction of the

F-transform.

For t > 0, let A = 14a2t

and let g(x, t) = 1a√

2te−

x2

4a2t . Then

F [g(x, t)](ξ) = e−a2ξ2t.

We shall use this expression.Example 2: Let

qj(x) =1

2j, for |x| ≤ 1/j, and qj(x) = 0, for |x| > 1/j.

Then

qj(ξ) =1

2√

2πj

∫ 1/j

−1/j

e−ixξdx =1√2π

sin jξ

jξ.

Hence,

qj(ξ) → 0, for ξ 6= 0

as j →∞, and

qj(0) =1√2π.

Given h, 0 6= λ ∈ R. Let

(τhf)(x) = f(x− h), (δλf)(x) = f(λx).

Then by definition, we have

(τhf)(ξ) = e−i(hξ)f(ξ)

and

(δλf)(ξ) = λ−1f(ξ/λ).

Hence, for

(δλg)(x) = e−λ2x2

,

we have

δλg(ξ) = λ−1g(ξ/λ) =1√2λe−

ξ2

4λ2 .

Given two functions f and g. Define their convolution by

f ? g(x) =

∫f(x− y)g(y)dy.

62 LI MA

Then we have

F (f ? g(x)) =1√2π

∫f ? g(x)e−ixξdx

= =1√2π

∫e−ixξdx

∫f(x− y)g(y)dy

=1√2π

∫g(y)e−iyξdy

∫f(x− y)e−i(x−y)ξdx

=√

2πf · g.

The most important property is that the F-transform of the derivativefx is a multiplication:

fx(ξ) = iξf(ξ).

This property gives us a chance to define the inverse operator ∂−1x of

∂x by the definitionˆ∂−1

x f(ξ) = (iξ)−1f(ξ).

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.

Let’s 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(ξ) =1√2π

∫f(x)eixξdx,

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.ˆf = f

andˇf = f.

Proof. Let’s verify the latter equality. In fact, note that

gA(x) → 1,


as A→ 0 and

gA(ξ) =1√2A

e−ξ2

4A2 →√

2πδ0,

as A→ 0. Then, by symmetry property,

ˇf(0) =

1√2π

∫f(ξ)dξ

=1√2πlimA→0

∫gA(ξ)f(ξ)dξ

=1√2πlimA→0

∫gA(y)f(y)dy

= f(0).

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

f(x) = f1(0) =1√2π

∫f1(ξ)dξ =

1√2π

∫f eixξdξ.

Using this we can prove

Theorem 9. (Plancherel Theorem): Given u ∈ S, we have∫|u|2dx =

∫|u|2dξ

Proof. In fact, using the inverse Fourier transform, we have that

u(x) =1√2π

∫u(y)e−ixydy = 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(ξ) =1

(√

2π)n

∫f(x)e−ix·ξdx.

64 LI MA

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

f(ξ) =1

(√

2π)n

∫f(x)eix·ξdx.

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[eixξ] = −|ξ|2eixξ.


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)(

∫ξ2|u|2dξ) ≥ 1

4,

with its equality true only at

u(x) = Ae−βx2

,

where β > 0 and A4 = 2β/π.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′= λxu.

The latter condition implies that

u = Aeλx2/2.

Using∫u2 = 1 we have λ = −2β and the relation between A and β.

Note that ∫|u′|2 =

∫ξ2|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 variousbranches in sciences and technologies. I hope our readers can read morebooks on Fourier transforms and Fourier analysis. Please read as moreas you can.

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

u = u(ξ, 0) cos(|ξ|t) + ut(ξ, 0)sin(|ξ|t)|ξ|

.

Using the inverse F-transform we find the solution expression:

u(x, t) = F ?(u(ξ, 0) cos(|ξ|t) + ut(ξ, 0)sin(|ξ|t)|ξ|

).

Set

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

The solution can be written as

u(x, t) =1

(√

2π)n

∫(f(ξ) cos(|ξ|t) + g(ξ)

sin(|ξ|t)|ξ|

)eiξxdξ.

Using the relations

cos(|ξ|t) =1

2(ei|ξ|t + e−i|ξ|t)

andsin(|ξ|t)|ξ|

=1

2i|ξ|(ei|ξ|t − e−i|ξ|t),

we have

u(x, t) =1

(√

2π)n

∫(f+(ξ)ei(ξx+|ξ|t) + f−(ξ)ei(ξx−|ξ|t))dξ,

where

f+(ξ) =1

2(f(ξ) +

1

i|ξ|g(ξ))

and

f−(ξ) =1

2(f(ξ)− 1

i|ξ|g(ξ)).

Using the similar approach, we can use properties of F-transform tosolve the heat equation

ut = ∆u, on Rn ×R+


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

ut(ξ, t) = −|ξ|2u(ξ, t).This is a first order ode, and we can solve it to get

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

u(x, t) = F ?(u(ξ, 0)exp(−|ξ|2t)).Let

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

(2t)n/2e−|x|

2/4t.

Thengt(ξ) = exp(−|ξ|2t).

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

u(x, t) =1

(√

2π)nf ? gt(x).

Let

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

(4πt)n/2e−|x|

2/4t.

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

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

68 LI MA

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 condition

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

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

−|ξ|2u(ξ, y) + uyy(ξ, y) = 0.

Solve this and we find

u(ξ, y) = u(ξ, 0)exp(−|ξ|y) = f(ξ)exp(−|ξ|y).Define

Py(x) = F ?(exp(−| · |y))(x).Then by an elementary computation, we have

Py(x) = F ?(exp(−| · |y))(x)

=1√2π

∫e−|ξ|y+ixξdξ

=1√2π

∫ ∞

0

e−yξ+ixξdξ

+1√2π

∫ 0

−∞eyξ+ixξdξ

=1√2π

(1

y + ix+

1

y − ix).

This implies that

Py(x) =1√2π

2y

x2 + y2.

Hence

u =1√2πf ? Py,

and

u(x, y) =1√2πf ? Py(x)

is the solution wanted for the problem above.


8.3. Derivative bound. We ask the following question. Given asmooth function u ∈ S(Rn). Assume the bound M0 of |u| and thebound M2 of |∆u| on Rn. Can we give a bound of |∇u| on Rn in termof M0 and M2? We answer this question below.

Let−∆u+ u = f, in Rn.

Using the Fourier transform, we get

(|ξ|2 + 1)u = f .

Then we have

u =1

|ξ|2 + 1f .

Define

B2(x) =1

(√

2π)nF ?(

1

|ξ|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 that∫Rn

|∇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|.

70 LI MA

9. lecture nine

9.1. Maximum principle (MP) for Laplace operator.The Maximum Principle (MP) is a beautiful property for solutions

to 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 = sup∂Du.

Assume that we only have

uxx + uyy ≥ 0

in the domain D. Then we choose

w = u+ εeβx

for small positive constant ε and large positive constant β. Note that

∆eβx > 0.

Then we have

∆w > 0.

Hence, we have

supDw = sup∂Dw.

Sending ε→, we get that

supDu = sup∂Du.

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


Theorem 10. Maximum Principle. Assume u is a smooth functionsatisfying

∆u ≥ 0, in D.

Then we havesupDu = sup∂Du.

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 ≤ 0

at (x0, y0). Hence, we have

uxxuyy(x0, y0) ≥ 0,

a contradiction! This implies that

supDu = sup∂Du.

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 pde’s.

72 LI MA

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 let∂pQ = (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 − a2∆u = 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 continuousx-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) = max∂pQu(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. This

implies 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) = max∂pQu(x, t).

For general case when

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

v(x, t) = u(x, t)− εt.

ThenLv(x, t) = Lu(x, t)− ε < 0, (x, t) ∈ Q.

Then we can use the result above to v(x, t) to get

maxQv(x, t) = max∂pQv(x, t).

Note thatmaxQu(x, t) ≤ maxQv(x, t)

andmax∂pQv(x, t) ≤ max∂pQu(x, t) + εT.


Sending ε→ 0, we get

maxQu(x, t) ≤ max∂pQu(x, t).

However, since ∂pQ ⊂ Q, we have

maxQu(x, t) ≥ max∂pQu(x, t).

Hence,maxQu(x, t) = max∂pQu(x, t).

We are done.

Let’s 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) = φ(x, t) on ∂pQ. Then we have

maxQu(x, t) = FT +max∂pQ|φ(x, t)|.

Proof. Denote byB = max∂pQ|φ(x, t)|.

Consider the function w(x, t) defined by

w(x, t) = Ft+B ± u(x, t).

ThenLw = F ± f ≥ 0, on Q

and on ∂pQ,w ≥ 0.

By the Maximum principle above, we have

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

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 function g(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,

maxx∈D

|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) := supx∈D|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)− δ ≤ |u(x, t)|Hence

(M(t)− δ)|Q|1/2k ≤ Jk(t)1/2k ≤M(t)|D|1/2k

Sending k → +∞, we have

(M(t)− δ) ≤ inflimJk(t)1/2k ≤ suplimJk(t)

1/2k ≤M(t).

Since δ > 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.


Note thatdJk

dt= 2k

∫u2k−1ut = 2k

∫u2k−1∆u.

By using integration by part, we have

dJk

dt= −2k(2k − 1)

∫u2k−2|∇u|2 < 0.

Hence Jk(t) is a monotone decreasing function of t. Therefore,

maxx∈D

|u(x, t)|

is a monotone decreasing function of t too.

Remark 1. If we assume u ≥ 0 in Q satisfies

ut ≤ ∆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 − a2∆u ≤ Au, in Q.

Then, arguing as before, we have

Jk(t) ≤ Jk(0)e2kAt,

which implies thatM(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 still

have

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→∞(Mk−p

∫D

|f |pdx)1/k = M.

With an easy modification, we can extend the above theorem intounbounded domain D. We omit the detail.

76 LI MA

10. lecture ten

10.1. Variational principle.Variational principle is a important tool in sciences. One of the

beautiful 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 domainD ⊂ 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 value

u(x) = φ(x), x ∈ ∂D.We denote A by such a class of functions. We look for a function w ∈ Asuch that

I(w) = infu∈AI(u).

Assume such a function w exists. Then for every ξ ∈ C10(D), we have

w + tξ ∈ A

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

I(w) ≤ I(w + tξ)

and thend

dtI(w + tξ)|t=0 = 0.

Note that

I(w + tξ) =

∫[|∇w|2 + 2t∇w · ∇ξ + t2|∇ξ|2].

Hence,d

dtI(w + tξ)|t=0 = 2

∫∇w · ∇ξ = 0.

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

Since ξ can be any testing function in C10(D), we get the Euler-Lagrange

equation for the functional I(·), by the variational lemma (see ourtextbook [1]), that

∆w = 0, in D,


with the boundary condition w = φ on ∂D. This is the famous Dirichletproblem in the history. The recent study is about the obstacle problemfor harmonic functions.

Remark: We mention that the existence of the infimum w can beproved 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 C∞

0 (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 ⊂ C∞

0 (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 ⊂ C∞

0 (I) such that

fk → f, and (fk)x → g

in 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 ∫

I

fφxdx = −∫

I

gφdx,

for all φ ∈ C∞0 (I), is true. People often use this formula to define the

weak derivative of f . 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 H1

0 (I) the Sobolev space which consists of L2 functionswith derivatives. For f, h ∈ H1

0 (I), we define

(f, h) =

∫I

fxhxdx.

Then one can show that (·, ·) is an inner product on H10 (I), and H1

0 (I)is a Hilbert space.

78 LI MA

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 κ ∈ (0, 1) such that

|f(x)− f(y)| ≤ κ|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. Let’s see a little more. Westudy 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) +

∫ t

0

S(t− τ)F (u)dτ := 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 κ ∈ (0, 1) such that |f(x)− f(y)| ≤ κ|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,


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 property

by introducing some monotone or concave function A(t) defined by theintegration like

p(t) = minDu(x, t)

or

p(t) :=

∫ρ(x)u(x, t)k

or

p(t) :=

∫ρ(x)|∇xu(x, t)|k,

where k is an integer, ρ(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, 2π]× [0, T )

with the non-trivial initial data

u(x, 0) = φ(x) > 0

and the boundary data u(x, t) = u(x+ 2π, t) for x ∈ [0, 2π]. Let

p(t) = minx∈[0,2π]u(x, t).

Then we havep′(t) ≥ p(t)2.

This gives us that

p(t) ≥ a

1− tafor a = minx∈[0,2π]φ(x) > 0 and u has to blow up to +∞ in finite timebefore a−1.

Example Three: Given a constant β ≥ 14. We consider the following

nonlinear parabolic equation

ut = ∆u+ u2 + βu, (x, t) ∈ [0, 2π]× [0, T )

with the non-trivial initial data

u(x, 0) = φ(x) > 0, x ∈ [0, 2π]

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

u(x, t) > 0, x ∈ [0, 2π]× t > 0.

80 LI MA

Choose C = 14

such that

C

∫ 2π

0

sin(x

2) = 1.

Let

p(t) = C

∫ 2π

0

sin(x

2)u(x, t)dx.

Then, we have

p′

= C

∫ 2π

0

sin(x

2)ut

= C

∫ 2π

0

sin(x

2)∆u+ C

∫ 2π

0

sin(x

2)u2 + βp(t)

= C

∫ 2π

0

sin(x

2)u2 + (β − 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 standing

conjecture on non-linear Schrodinger equation on compact Riemannianmanifolds.


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 + u2

y = 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)).

82 LI MA

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 ∈ Γ, 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].


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 PDE’s. 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)ξ(x)dx = 0

for every ξ ∈ C∞0 (D). Then f = 0.

Proof. We argue by contradiction. Assume that there is a point x0 ∈ Dsuch that f(x0) 6= 0. Then we may assume that f(x0) > 0 and there isa small ball Bε(x0) such that f(x) 6= 0 for every x ∈ Bε(x0). Take

ξ(x) = ρ(|x− x0|)

where

ρ(r) = e− 1

1−r2 , 0 ≤ r < ε

and ρ(r) = 0 for all r ≥ ε. Note that ρ(|x− x0|) ∈ C∞0 (D), and

f(x)ρ(|x− x0|) > 0, in Bε(x0).

Hence, ∫D

f(x)ξ(x)dx =

∫C∞

0 (D)

f(x)ρ(|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

3u3 − g(t)u(t))dt,

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

−u′′ = u(t)2 + g(t).

In fact, by definition,

0 =< δF (u), φ >=

∫ L

0

(u′φ′ − u2φ− g)dt

84 LI MA

for all φ ∈ C∞0 (0, L). Integrating by part, we get

0 =

∫ L

0

(−u′′ − u2 − g)φdt.

By variational lemma, we know that

u′′ + u2 + g = 0,

which is equivalent to our equation above.


11.2. Sturm-Liouville theory. In the old time before Fourier, peopleoften asked if any continuous function, which is defined on the interval[0, L] with the fixed boundary condition X(0) = a,X(L) = b, can beexpressed as the expansion of trigonometric functions. The answer isyes since we know the beautiful theory of Fourier series.

In the method of separation of variables, we meet the problem if wecan use the eigenfunctions of the following eigenvalue problem (p(x)X

′)′ + [λ+ q(x)]X = 0, 0 < x < L,

−a1X′(0) + b1X(0) = 0),

a2X′(L) + b2X(L) = 0,

where the constants satisfy ai ≥ 0, bi ≥ 0 and ai + bi 6= 0, and thefunctions p(x) > 0 and q(x) are continuous on [0, L].

Luckily, the answer is yes too. This is the Sturm-Liouville theory.Roughly speaking, the Sturm-Liouville theory says that there are anincreasing eigenvalues λn → ∞ and eigenfunctions Xn (correspondingto λn) with ∫ L

0

XnXmdx = 0,

∫ L

0

|Xn|2dx = 1

for λn 6= λn, such that every smooth function f with the boundarycondition

−a1f′(0) + b1f(0) = 0),

a2f′(L) + b2f(L) = 0,

can be expressed as

f(x) =∑

n

CnXn(x).

86 LI MA

11.3. Three open problems.1. De Giorgi conjecture:In 1978, De Giorgi proposed the following conjecture:If u is a solution of the scalar Ginzburg-Landau equation:

∆u+ u(1− u2) = 0, on Rn

such that |u| ≤ 1 and ∂∂xnu > 0 on Rn, and

limxn→±∞

u(x′, xn) = ±1,

for any x′ ∈ Rn−1, then all level set of u are hyper-planes, at least forn ≤ 8.

The case when n ≤ 8 was solved recently by Savin, who is a PhDstudent of L. Caffarelli. For more reference, please see our paper:

Y.H.Du and Li Ma; Some remarks related to the De Giorgi conjec-ture, Proc. AMS, 131(2002)2415-2422.

2 Serrin conjectureIn the past years, I and my student D.Z.Chen studied the following

conjecture of Serrin:There is no bounded positive solution (u, v) to the following elliptic

system on Rn: −∆u = vq,−∆v = up,

where the indices p > 0, q > 0 satisfy

1

p+ 1+

1

q + 1> 1− 2

n.

We have some progress in this problem. Please see our paperLi Ma and D.Z.Chen, A Liouville type Theorem for an integral Sys-

tem, CPAA, 5(2006)855-859.3. A question from Yanyan Li and L.NirenbergIt is not hard to show by using the Taylor expansion that, if a non-

negative function satisfying

|u′′(x)| ≤M

on [−R,R], then we have

|u′(0)| ≤√

2u(0)M

for M ≥ 2u(0)R2 , and

|u′(0)| ≤ u(0)

R+R

2M

for M < 2u(0)R2 .


Yanyan Li and L.Nirenberg (2004) asked what is the best constant Cin the following result in the ball BR ⊂ Rn:

Assume that u is a non-negative C2 function in closure of the ballBR satisfying:

maxBR

|∆u| = M.

Then there is a constant C depending only on n such that

|∇u(x)| ≤ C√u(0)M

if R ≥√

u(0)M

≥ 2|x|,

|∇u(x)| ≤ C(u(0)

R+RM)

if ≥ 2|x| ≥ R <√

u(0)M

.

This is an open question.

88 LI MA

12. Appendix B

This is a section for part of homework exercises in my courses in thepast years.

12.1. Homework 1.Exercise 1. Assume that the vibrating string of length L moves with

two end points fixed at x = 0 and x = A. Assume its initial positionforms a triangle with the x-axis such that the middle point of thestring is the highest point and its height is b > 0. Find the determinedproblem for this string.

Answer: Let u be the height position of the vibrating string at(x, t). Then u satisfies

utt = a2uxx, for (x, t) ∈ (0, A)× (0,∞)


u(x, 0) =2b

Ax, x ∈ [0,

A

2] and u(x, 0) =

2b

A(A− x), x ∈ [

A

2, A]

and the boundary data u(0, t) = 0 and u(A, t) = 0. The relationbetween L, b, and A is

L2

4=A2

4+ b2.

Exercise 2. Assume 0 6= x ∈ Rn, n ≥ 2. Verify that

∆|x|β = (nβ + β(β − 2))|x|β−2.

Answer: Note that ∂xir = xi

r, for r = |x|.

Exercise 3. Assume 0 < u = u(x) ∈ C2. Denote by

|∇u|2 =∑

i

|∂iu|2.

(1). Assume that f : R → R is in C2. Prove that

∆f(u) = f ′(u)∆u+ f′′(u)|∇u|2.

(2). In particular, for β ∈ R, prove that

∆uβ = βuβ−1∆u+ β(β − 1)uβ−2|∇u|2.Exercise 4. Define the Gamma function Γ(s) for s > 0 by

Γ(s) =

∫ ∞

0

e−rrs−1dr.

Show that the definition is well-defined, and Γ(s+1) = sΓ(s) for s > 0.


12.2. Homework 2.Exercise 1. Solve the equation

ut + 3ux = 0, (x, t) ∈ R× (0,∞)

with initial data u = x2 at t = 0.Answer:

u(x, t) = (x− 3t)2.

Exercise 2. Solve the equation

ut + ux = u, (x, t) ∈ R× (0,∞)

with initial data u = cos x at t = 0.Answer:

u(x, t) = et cos(x− t).

Exercise 3. Solve the equation

2ut = ux − xu, (x, t) ∈ R× (0,∞)

with initial data u = 2xex2/2 at t = 0.Answer: Note that along the characteristic curve

x = − t2

+ A,

we havedu

dt= −xu = (− t

2+ A)u.

Then,u(x, t) = (2x+ t)ex2/2.

Exercise 4. Solve the problem

y2uyy − x2uxx = 0


u(x, 1) = φ(x), uy(x, 1) = ψ(x).

Answer:

u(x, y) =1

2φ(xy) +

y

2φ(x

y)

+

√xy

4

∫ xy

xy

s−32φ(s)ds

−√xy

2

∫ xy

xy

s−32ψ(s)ds.

90 LI MA

12.3. Homework 3.Exercise 1. Assume that Y is a solution to

−y′′ + y = 0, x > 0

with the initial conditions y(0) = 0 and y′(0) = 1. Using the Duhamelprinciple to find the expression for solution to the the following ODE

−y′′ + y = g(x), x > 0

with the initial conditions y(0) = A and y′(0) = B.Answer:

y(x) = AY ′(x) +BY (x)−∫ x

0

g(τ)Y (x− τ)dτ.

Exercise 2. Given constants a1, ..., ak. Assume Y (x) is a solution tothe the higher order ODE of the form

y(k) + a1y(k−1) + ...aky = 0


y(0) = 0, y′(0) = 0, ... y(k−1)(0) = 1.

Find the solution expression for the higher order ODE of the form

y(k) + a1y(k−1) + ...aky = g(x)


y(0) = b0, y′(0) = b1, ... y

(k−1)(0) = bk−1.

Answer: Define, for 0 ≤ j ≤ k − 1,

uj(x) = Y (j)(x) + a1Y(j−1)(x) + ...+ ajY (x).

Then the solution expression is

y(x) =∑

j

bjuj(x) +

∫ x

0

g(τ)Y (x− τ)dτ.


12.4. Homework 4.Exercise 1. Find a solution to the following Sturm-Liouville problem:

X′′

+ λX = 0, x ∈ (0, L),

and X(0) = X ′(L) = 0.Answer:

λk =(k + 1

2)2π2

L2

and

X = 0, or Xk(x) = B sin((k + 1

2)π

Lx)

where k = 0, 1, 2.....Exercise 2. utt = a2uxx, 0 < x < L, t > 0,

u(0, t) = 0 = ux(L, t),u(x, 0) = sin2 πx

L, ut(x, 0) = x(L− x).

Answer:

u(x, t) =∑n≥1

(An cos(anπ

Lt) +Bn sin(

anπ

Lt)) sin(

nπ

Lx),

where

An = − 8

n(n2 − 4)πfor n = 2k + 1, and An = 0 for n = 2k, and

Bn =8L3

an4π4

for n = 2k, and Bn = 0 for n = 2k + 1,

92 LI MA

12.5. Homework 5.Exercise 1. Given a domain Ω ⊂ R2. Consider the following prob-

lem:

−∆u = f(x, y), (x, y) ∈ Ω

with the boundary condition u = φ on ∂Ω. Find the Green functionfor the problem when

(1). Ω is the half upper plane.(2). Ω = x > 0, y > 0.(3). Ω = −∞ < x < ∞, 0 < y < L, where L > 0 is a fixed

constant.Answer:(1). Define

G1((x, y), (ζ, η)) = Γ((x, y), (ζ, η))− Γ((x, y), (ζ,−η)).

(2). Define

G2((x, y), (ζ, η)) = G1((x, y), (ζ, η))−G1((x, y), (−ζ, η)).

(3). Define

G3((x, y), (ζ, η)) =∞∑

k=−∞

G1((x, y), (ζ, 2kL+ η)).

Exercise 2. Assume D = (x, y); y > 0 is the upper half plane.Compute the Poisson kernel in such D.

Answer: We just remark that on the boundary ∂D, the unit ournormal is ν = (0,−1).? Exercise 3. Let B = BR(0) and assume φ ∈ C(∂B). Define

u(x) =R2 − |x|2

nωnR

∫∂B

φ(y)

|x− y|ndσy,

where ωn = |B1(0) is the volume of the unit ball. Prove that

∆u(x) = 0, x ∈ B.

? Exercise 4. Let 0 ∈ D be a bounded domain in Rn. Given u ∈C2(D) with

∆u(x) = f(x), ; x ∈ D.Define D? = I(D), where I is the inversion transform

x ∈ D → I(x) = x/|x|2 ∈ D?

on the unit sphere |x| = 1.(1). Prove that I2 = identity, that is, I(I(x)) = x for x ∈ D. This

implies that D = I(D?).


(2). Letw(x) = |x|2−nu(x/|x|2), x ∈ D?

which is called the Kelvin transform of the function u. Then

∆w(x) = |x|2−nf(x/|x|2), x ∈ D?.

(3). Using the conclusion of (2), find the Green function on B1(0).Answer: (3). Define, for x, y ∈ B1(0),

G(x, y) = Γ(x, y)− |y|2−nΓ(x, I(y)).

Then, G(x, y) is the Green function on B1(0).

94 LI MA

12.6. Homework 6.Exercise 1. Given a constant B > 0. Compute the Fourier transform

of the function f(x) = 1 for |x| ≤ B and f(x) = 0 for |x| > B.Answer:

F (f)(ξ) =

√2

B

sin(Bξ)

ξ.

Exercise 2. Given two constants a > 0, c and two bounded smoothfunctions f(x, t) and φ(x). Using the Fourier transform to solve thefollowing problem

ut = a2uxx + cu+ f(x, t), −∞ < x <∞, t > 0,u(x, 0) = φ(x).

Answer:

u(x, t) = ect[

∫K(x− ξ, t)φ(ξ)dξ

+

∫ t

0

e−cτdτ

∫K(x− ξ, t− τ)f(ξ, τ)dξ].

We can obtain this result by two methods.Method One: Let w = e−ctu and f1(x, t) = e−ctf(x, t). Then we have

wt = a2wxx + f1

with

w(x, t) = φ(x).

Method Two: Note that for u(ξ, t),

ut = (−a2|ξ|2 + c)u+ f(ξ, t),

with

u(ξ, 0) = φ(ξ).

? Exercise 3. Define, for u ∈ S,

H[u](x) = (− d2

dx2+ x2)u(x),

A[u](x) = (d

dx+ x)u(x),

C[u](x) = (− d

dx+ x)u(x).


Show that

CA[u] = (H − 1)[u],

AC[u] = (H + 1)[u],

H(e−x2/2) = e−x2/2

H(Ck(e−x2/2)) = (2k + 1)Ck(e−x2/2), k = 1, 2, ....

We remark that Ck(e−x2/2) are eigenfunctions of the operator H andthe polynomials defined by

hk(x) = ex2/2Ck(e−x2/2)

are called the Hermite functions. H is called the Hermite operator,which plays an important role in modern methods of mathematicalphysics and its higher dimensional analogue

L = −∆ + |x|2

plays an important role in Bose-Einstein condensate.? Exercise 4. Define, for u ∈ S, a new function

U(t) =∞∑

n=−∞

u(t+ 2πn),

which is called the periodization of u. Clearly, U(t) is a periodicfunction such that U(t + 2π) = U(t) on R. Denote by u the Fouriertransform of u. Using Fourier series theory, you show that

U(t) = (√

2π)−1

∞∑n=−∞

u(n)eint,

which is called the Poisson summation formula. Then setting t = 0,we get

∞∑−∞

u(n) = (√

2π)−1

∞∑−∞

u(n).

? Exercise 5. Recall that the theta function ϑ(t), t > 0, is definedby

ϑ(t) =∞∑

n=−∞

e−πn2t.

Using the Poisson summation, you show that

t−1/2ϑ(1/t) = ϑ(t).

? Exercise 6. Assume that

B(ξ) =1

1 + |ξ|2.

96 LI MA

Find B(x).Answer: First we write

1

1 + |ξ|2=

∫ ∞

0

e−s(1+|ξ|2)ds.

Then we have

B(x) =1

(√

2π)n/2

∫ ∞

0

e−sds

∫Rn

eix·ξ−s|ξ|2dξ.

Recall that ∫R

eiay−sy2

dy = (π

s)1/2e−

a2

4s .

Then we have ∫Rn

eix·ξ−s|ξ|2dξ = (π

s)n/2e−

|x|24s .

Hence, we have

B(x) = 2−n/2

∫ ∞

0

s−n/2e−s− |x|24s ds,

which is called the second Bessel potential.? Exercise 7. Given β > 0. Assume that

Bβ(ξ) =1

(1 + |ξ|2)β/2.

Find Bβ(x).Hint: One can find the expression of Bβ in E.M. Stein’s book,

Princeton Univ. press,1970, or W. P.Ziemer’s book, GTM120, Springer,1989.


12.7. Homework 7.Exercise 1. Let D be a bounded domain in Rn. Consider u ∈

C2(D) ∩ C(D) such that−∆u = f(x), x ∈ Du(x) = φ(x), x ∈ ∂D.

Then

maxD|u(x)| ≤ max∂D|φ(x)|+ d

2nsup|f(x)|,

where d = diam(D) := supx,y∈∂D|x− y|.Proof. We denote

F = sup|f(x)|, M = max∂D|φ(x)|Take p ∈ D. Let

ξ(x) =F

2n(d2 − |x− p|2) +M.

Note thatξ(x)± u(x) ≥ 0

on ∂D and−∆(ξ(x)± u(x)) ≥ 0

Applying the Maximum principle to the function

w(x) = ξ(x)± u(x),

we get that w(x) ≥ 0 on D. This implies that

maxD|u(x)| ≤ max∂D|φ(x)|+ d

2nsup|f(x)|.

98 LI MA

12.8. Homework 8.Exercise 1. Let Q = (x, t); 0 < x < L, 0 < t ≤ T. Let a > 0 be a

constant. Assume that u ∈ C2,1(Q) satisfies ut = a2uxx + f(x, t), 0 < x < L, 0 < t ≤ T,u(0, t) = 0 = ux(L, t),u(x, 0) = φ(x)

Show that maxQ|ut(x, t)| can be bounded by the suitable norm of fand φ.

Answer: There is a uniform constant C = C(T, a) such that

maxQ|ut| ≤ C[|f |C1(Q) + |φ′′|C[0,L]].

Exercise 2. Given φ ∈ C(D) and a(x, t) ∈ C(Q) withA = maxQ|a(x, t)|.Assume that u ∈ C2,1(Q) ∩ C(Q) satisfies

Lu := ut − a2∆u = −u2 + a(x, t)u, (x, t) ∈ Qand the initial data

u(x, 0) = φ(x) ≥ 0,

and the boundary data u(x, t) = 0 for x ∈ ∂D. Prove that on Q,

0 ≤ u(x, t) ≤MsupD|φ(x)|for some constant M > 0 depending only on n, T, and A.

Proof. Assume that u < 0 somewhere, saying,

u(x1, t1) = −(εt1 + ε2)2

for some arbitrary small ε and t1 > 0. Consider

w = u+ εt+ ε2.

At (x1, t1), we havew(x1, t1) = 0.

Note that∆w = ∆u

and at t = 0, w = φ + ε2 > 0. Hence, there is a small 0 < t0 ≤ t1such that w(x, t) ≥ 0 for (x, t)D× (0, t0] and for some minimum pointx0 ∈ D for w such that

w(x0, t0) = 0, ∆u(x0, t0) = ∆w(x0, t0) ≥ 0,

which implies thatu(x0, t0) = −(εt0 + ε2).

Hence, we haveLw(x0, t0) ≤ 0,


which gives us thatLu(x0, t0) ≤ −ε,

Then, using the equation for u at (x0, t0),

−ε ≥ (εt0 + ε2)2 − a(x0, t0)(εt0 + ε2).

However, this is impossible for small ε and t1. Hence, we have

u ≥ 0, on Q.

Once we have u ≥ 0, we can use the maximum principle 2 (seeSection 9.3) to conclude that

u(x, t) ≤MsupDφ(x).

Remark. One may try another method below. Assume that m :=u(x2, t2) = maxQu > 2supDφ. Then at (x2, t2), we have

Lu ≥ 0.

Using the equation, we get

−m2 + a(x2, t2)m ≥ 0.

Hence, we havem2 ≤ Am

and m ≤ A. So this method can not be used to give the result wanted.

100 LI MA

12.9. Homework 9.Exercise 1. Let F and G be two smooth functions in R. Assume

that u(x, t) ∈ C1(R× (0, T )) is a bounded function such that

limx→±∞u(x, t) = 0

uniformly in t and

∂tF (u) + ∂xG(u) = 0.

Here ∂xG(u) = G′(u)ux. Show that

d

dt

∫R

F (u)dx = 0.

We call the quantity∫RF (u)dx the conservation law.

Exercise 2. Let G be a vector-valued smooth function in R. Assumethat u(x, t) ∈ C1(Rn × (0, T )) is a bounded function such that

lim|x|→∞u(x, t) = 0

uniformly in t and

∂tu+ divxG(u) = 0.

Show thatd

dt

∫Rn

ukdx = 0

for each k = 1, 2, 3....? Exercise 3. Using the conclusions of Exercise 1 and Exercise 2 to

find at least 5 conservation laws of the KdV equation:

ut + uux + uxxx = 0.

Hint: Define

En(u) =

∫R

([u(n)x ]2 + cnu[u

(n−1)n ]2 − qn(u, ..., u(n−2)

x )dx.

for suitable constants cn and polynomials qn. Here u(n)x denotes the

x-derivative of order n for the function u.? Exercise 4. Let u = u(x, t) be a smooth solution to the equation

ut − uxx + cu2x = 0.

Find a function φ such that the function w = φ(u) satisfies the heatequation

wt − wxx = 0.

Answer: w = e−cu, which is called the Cole-Hopf transformation.? Exercise 5. Consider the Burgers’ equation

ut − uxx + uux = 0, x ∈ R


with initial data u = f(x). Let

w(x, t) =

∫ x

−∞u(y, t)dy.

Show that w satisfies

wt − wxx +1

2w2

x = 0

and use the conclusion of Exercise 4 to find an explicit expression ofu.

Answer:

u(x, t) =

∫R(x− y)e

−|x−y|24t

− f0(y)2 dy

t∫Re−|x−y|2

4t− f0(y)

2 dy,

where f0(x) =∫ x

−∞ f(y)dy.

102 LI MA

12.10. Homework 10.Exercise 1. As an exercise for pleasure, we invite readers to find the

Euler-Lagrange equation for the area functional

A(u) =

∫D

√1 + |∇u|2dx

among the class A above.Answer: From this variational structure, we know the famous

minimal surface equation:

−div( ∇u√1 + |∇u|2

) = 0.

There is a rich literature about this equation.Exercise 2. Try to use the method of separation variables to the

Minimal surface equation in dimension two in the following way. As-sume u = u(x, y) = f(x) + g(y) satisfies

∂x(ux√

1 + u2x + u2

y

) + ∂y(uy√

1 + u2x + u2

y

) = 0.

Then we havef′′

1 + f ′(x)2+

g′′

1 + g′(y)2= 0.

Prove that for some constant A > 0,

u(xu) =1

Alog|cos(Ay)

cos(Ax)|.

This solution is called the Sherk’s surface.Exercise 3. Find solution u(x, y) of the form u = f(r), r =√x2 + y2, to the equation

∂x(ux√

1 + u2x + u2

y

) + ∂y(uy√

1 + u2x + u2

y

) = 0.

Answer:

f(y) =1

Acosh(Ar +B).

Exercise 4. Find the Euler-Lagrange equation for the integral

I(u) =

∫D

L(x, u, ux1 , ..., uxn)dx.

Answer:

∂uL−n∑

j=1

∂xj(∂uxjL) = 0.


? Exercise 5. Find the Euler-Lagrange equation for the integral

J(u) =

∫D

|∇u|2dx

on the space C20(D) with the constraint∫

D

|u|2dx = 1.

Answer:−∆u = λu.

104 LI MA

Summary

This is a brief lecture note for the courses on ”Introduction to Equa-tions from Mathematical Physics” or ”Methods for Mathematical Physics”.Our motivation to write this note is to show the mainstream methodsused in solving some typical partial differential equations like waveequation, heat equations, and the Laplace equation.

For students who would like to take the final examination of thiscourse, please look my homepage at

http://faculty.math.tsinghua.edu.cn/~lma

for the highlight of the course.

Thank you very much for your patience in attending

my courses.

References

[1] Jiang Lishan, Yazhe Chen, Xiheng Liu, Fahuai Yi, Lectures on equations from Math-ematical Physics,in Chinese, Higher education press, Beijing,1999.

[2] D.Hilbert, R.Courant, Methods of Mathematical Physics, Vol. 2. Interscience Publi-caters, 1962.


Index

Bessel function, 28blow up, 60Characteristic curve method, 13continuity equation, 9D’alembert, 21Dirac’s delta function, 39divergence, 5Duhamel principle, 17Fourier, 24fundamental solution, 35Gaussian kernel, 36Green, 5, 42Gronwall, 18harmonic function, 41heat equation,11Heisenberg, 51Hermite operator, 90Huygens, 30Inversion transform, 87Kelvin transform, 88Kirchhoff, 29Laplacian, 5, 11Liouville, 67maximum principle, 51mean value theorem, 44monotonicity formula, 47partial differential operator, 10Plancherel, 59Poisson, 42Schwartz, 48string,7Sturm, 67support, 4variational lemma, 65wave equation, 7

Department of mathematical sciences, Tsinghua university, Beijing 100084,China

E-mail address: [email protected]

INTRODUCTION TO EQUATIONS FROM MATHEMATICAL...

Documents

Transcript of INTRODUCTION TO EQUATIONS FROM MATHEMATICAL...