INTRODUCTION TO EQUATIONS FROM
MATHEMATICAL PHYSICS
LI MA
Date: Starting from the Summer of 1991 to 2007.
1991 Mathematics Subject Classification. 35-XX.
Key words and phrases. wave equation, heat equation, harmonic functions, characteristic line method, Fourier method, maximum principle, fundamental solution.
1
2
LI MA
This is a lecture note for the courses of Equations from Mathematical Physics or Methods of Mathematical Physics given by me to senior
students in Tsinghua University, Beijing
EQUATIONS FROM MATHEMATICAL PHYSICS
To An, for patience and understanding
3
4
LI MA
Contents:
Wave equations; Heat equations; Laplace equation; Method of Characteristic lines; Duhamel principle; Method of separation of variables;
Laplace transform; Fourier transform; Fundamental solutions; Mean
value property; Louville theorem; Monotonicity formulas; Maximum
principle; Variational principle.
EQUATIONS FROM MATHEMATICAL PHYSICS
5
Preface
As I always believe that the course about equations from mathematical 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 image
what 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 beginning. To us, solving a problem means that we reduce the problem
into some easy problems which can be solved by us. For solving our
problems with partial differential equations (in short, PDE), we reduce
them into ordinary differential equations (in short, ODE), which can
be solved by us.
We shall derive some typical partial differential equations from the
conservation laws of mass or energy or momentum. Such equations are
called equations from mathematical physics. The basic principle used
here is that the change (rate) of the physical quantity in a body in
space is created from the change on the boundary and by the source in
the interior of the body. Hence, we need to use the divergence theorem
from the differential calculus.
The equations from mathematical physics are very important objects
in both mathematics and applied mathematics. However, many of them
are 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. The
crucial observation is that many equations enjoy symmetry properties
like translation invariant or rotation invariant, which will lead to many
identities for solutions.
The object in our course is to try to explain some elementary part
of solving simple linear partial differential equations of first order and
second order. The principle for studying these equations is reduce them
into linear ordinary differential equations, which can be solved by tools
from the differential calculus and linear algebra.
The most beautiful method for solving linear partial differential equations or systems of first order is the characteristic line method.
The most powerful methods for solving linear partial differential
equations or systems of second order are the methods of separation
for variables (also called Fourier method ) and Fourier transformation
method, which can be used to translated the problems into ODE again.
Theory of Fourier transformations is one of most beautiful mathematical theories so that there are many books about Fourier transformations. The other method is to construct the fundamental function to
6
LI MA
the equation with the homogeneous boundary condition and initial condition. In the latter method, the divergence theorem plays an essential
role.
There is another important tool in the study of heat equation and
elliptic equations. It is the maximum principle trick with its brother
called the comparison lemma. This trick is also very useful in the study
of nonlinear parabolic and elliptic equations or systems. People usually
use the maximum principle to derive gradient estimates or Harnack
inequalities of non-negative solutions, which lead to Bernstein type
theorem.
To understand the behavior of the solutions to PDE, we use divergence theorem, Fourier method, or the fundamental solutions to set up
energy bounds for solutions. The nice forms of energy bounds for solutions are monotonicity formulae or Pohozaev type identities , which
can be used to get Liouville type theorem or regularity theorem.
Energy bounds or gradient estimates are important to understand
the compactness of solution spaces.
The aim of this course is to let our students/readers know the basic
ingredients in the study of simple linear PDE’s of first order and second
order. Sections with ? ahead should be skipped by first reading. Just
as in the study of other courses in mathematics, one can only get
a good understanding of our subject by doing more exercises
and by reading more related books.
I hope you understand my words here. What did I feel in my college
life? I did feel so lucky that I had attended many courses in mathematics when I was in Nankai University. This experience makes me
can understand many deep lectures from some famous guys.
In conclusion, for our beginners, I hope you remember that we try to
understand PDE by ODE! In other words, before we do ”PDE”, let
us 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 your
words”.
Written by
Li Ma
in
5-1-2007
EQUATIONS FROM MATHEMATICAL PHYSICS
7
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 the
functions is decay to zero at infinity. In other word, we assume the nice
conditions for functions such that we can perform the differentiation
into 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 C 2 (D) the space of twice differentiable smooth functions inside the domain DRn .
We denote by C0∞ (D) the space of infinitely differentiable smooth
functions with support inside the domain DRn . Here the support of a
function f is the closure of the set {x ∈ D; f (x) 6= 0}.
∇f := (∂x1 f, ..., ∂xn f ).
∇j f = ∂xj f.
∇r f (x) :=
X (xj − pj )
∂xj f (x)
r
j
is the radial derivative of the function f , where r = |x − p| is the radial
coordinate with respect to the center p = (pj ) ∈ Rn .
We denote by
∂x(k) f = (∂xk11 ∂xk22 ...∂xknn f ); k = k1 + ... + kn
the k-th derivatives of the function f with respect to the variables
x = (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 C 1
boundary ∂D in Rn . Let X : D̄ → Rn be a smooth vector field on the
closure D̄ of the domain D. Let ν be the unit outer normal to ∂D.
Then we have the following formula:
Z
Z
divXdx =
X · νdσ.
D
∂D
∂X i
i ∂xi
i
P
Here divX =
for X = (X ) , ν = (νi ) and X · ν = i X i νi .
∂u
If X i = ∂x
i for some smooth function, i.e., X is the gradient ∇u :=
(∂x1 u, ..., ∂xn u) of the function u, then
X ∂2u
= ∆u
divX =
(∂xi )2
i
P
which is the Laplacian of the function u.
A special case we often use is that X has only one non-zero component, say j − th component,
X = (0, ...0, u, 0, ...0)
for some smooth function u ∈ C 1 (D) ∩ C(D̄). Then, divX = ∂j u and
Z
Z
uνj dσ.
∂j udx =
D
∂D
The classical divergence theorem is a basic theorem used in our
course.
We ask the readers to remember this basic formula. Note that when
n = 1, the divergence theorem is the famous Newtonian-Leibniz
formula:
Z
b
f (b) − f (a) =
0
f (x)dx.
a
In dimension two, the divergence theorem is Green’s theorem in differential calculus. Let D be a bounded smooth domain in the plane
R2 . Recall that the Green formula is the following: for any smooth
function P, Q on the closure of the domain D, we have
Z
Z
(Qx − Py )dx =
P dx + Qdy.
D
∂D
EQUATIONS FROM MATHEMATICAL PHYSICS
9
It is clear that the form of divergence theorem in dimension two
is different from that of the Green formula. We want to prove that
they are equivalent each other. Assume that the boundary ∂D is given
by the smooth curve p(s) = (x(s), y(s)), where s is the counter-clock
arc-length parameter. Write by x0 = dx/ds, etc. Then T = (x0 , y 0 ) is
the unit tangent vector along p(s). Then N = (−y 0 , x0 ) is the interior
unit normal to the curve p(s) such that {T, N } forms a right-hand
orthonormal basis system. The outer unit normal to the boundary ∂D
is ν = −N = (y 0 , −x0 ). Choose X 1 = Q and X 2 = −P . Then we have
divX = Qx − Py ,
and along the boundary ∂D,
P dx + Qdy = (P x0 + Qy 0 )ds = X · νds
Hence, Green’s formula implies that
Z
Z
divXdx =
D
X · νds,
∂D
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 string
at (x, t) in the direction of y-axis. Assume that ρ = constant is the
mass density of the string. Then the kinetic (or temporal) energy of
the string is
Z
1 l 2
T (u) =
ρut dx.
2 0
The potential (or spatial) energy is
Z
c l 2
ρux dx,
U (u) =
2 0
where c is a physical constant of the string.
Assume that the string is acted by the exterior force with force density f0 (x, t) at (x, t).
Then the energy obtained by the exterior force is
Z l
U (f ) = −
f0 udx.
0
Then the Hamiltonian action of the string is
Z t
dτ (T (u) − U (u) − U (f0 ))
H(u) =
0
Z
Z l
1 t
=
dτ
ρu2t dx
2 0
0
Z t Z l
Z t Z l
c
2
f0 udx.
ρux dx +
dτ
−
dτ
2 0
0
0
0
The Hamilton principle implies that the motion of the vibrating string
obeys
δH(u) = 0,
that is, for every ξ = ξ(x, t) ∈ C02 ((0, l) × (0, t1 )),
< δH(u), φ >:=
d
H(u + ξ)|=0 = 0.
d
We compute that
Z
< δH(u), φ > =
t
Z
dτ
0
l
(ρut φt
0
− cρux φx + f0 φ)dx
Z t Z l
=
dτ
(−ρutt + cρuxx + f0 )φdx
0
0
EQUATIONS FROM MATHEMATICAL PHYSICS
11
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),
and
ut (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 second
order.
We can also add other type of initial data and boundary data for the
wave 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 unit
normal to the boundary of the domain W .
Take a small part dx in W . Then the mass in dx is ρdx. Hence, the
total mass in W at time t is the integration of ρdx over W , which is
Z
ρdx.
W
Then, its change rate at time t is
Z
d
ρdx.
dt W
Assume that there is no fluid generating source. Then the change
rate is only across fluid rate from the boundary. Take a small part dσ
in the boundary ∂W . Then the amount of −ρu · νdσ flow into W . So
the total amount flowing into W across the boundary is the integration
over ∂W :
Z
−
ρu · νdσ.
∂W
So we have
Z
d
ρu · νdσ.
ρdx = −
dt W
∂W
Using the divergence theorem, we have
Z
∂
( ρ + div(ρu))dx = 0.
W ∂t
Since we can take W be any sub-domain, we have
∂
ρ + div(ρu) = 0.
∂t
This is the continuity equation in Fluid mechanics.
If u = (a(x1 ), 0, 0), then the continuity equation is
Z
ρt + ∂x (a(x)ρ(x, t)) = 0,
where x = x1 .
EQUATIONS FROM MATHEMATICAL PHYSICS
13
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 for
the solutions in terms of initial data and boundary data. We also need
to know the uniqueness and stability of our solutions to our equations
or problems.
Here we may introduce the concepts of Hilbert space and normed
space (see any textbook about Functional Analysis). A norm | · | on a
vector space E is a function from E to non-negative numbers R+ such
that
(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 a
limit in E, then we call (E, | · |) a Banach space. If further, the norm
| · | comes from an inner product < ·, · >, that is, |f |2 =< f, f >, we
call (E, < ·, · >) a Hilbert space; for example,
Z
< f, g >=
f (x)g(x)dx, f, g ∈ L2 (D),
D
By definition, we say that u(x, t) is the stable solution (in the norm
| · |) to the problem
A[u](x, t) := (∂t u − Lu)(x, t) = 0,
where t ∈ [0, T ], and
L :=
P (x, ∂x , ..., ∂x(k) )
=
k
X
aj (x)∂x(j)
j=0
is a partial differential operator of the variables (x), with initial data
u(x, 0) if for any > 0, there exists δ > 0 such that for all solutions
w(x, t) to A[w] = 0 with initial data w(0),
|w(0) − u(0)| ≤ δ,
we have
sup |w(t) − u(t)| < .
t
14
LI MA
There are also many other definitions for stabilities, which depend on
what 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 Gronwall
type inequality. For stationary problems, we have to use maximum
principle trick or monotonicity formulae for solutions.
In our course, we need to know some typical methods to solve linear
pde’s of first order and second order. Historically, the solution expressions were guessed by scientific workers from physical intuition. So it
is important for us know the scientific background of the equations.
EQUATIONS FROM MATHEMATICAL PHYSICS
15
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 q̄ be the heat flowing
velocity. By Fourier’s law, we have
q̄ = −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 the
region D ⊂ Ω,
Z
Z
Z Z
∂t
cρudx =
q̄ · (−ν)dσ +
ρf0 dx.
D
∂D
D
Using the divergence theorem we have
Z
Z
q̄ · (−ν)dσ = −
div(q̄)dx.
∂D
D
Hence, since D can be any domain in the body, we have
cρ∂t u = −div q̄ + ρf0 .
Using the Fourier law, we have
cρ∂t u = 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, ∞). The
initial 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 is
called a determined problem of heat equation.
Of course, we may have other kind of boundary conditions like
k
∂u
(x, t) = αu(x, t), (x, t) ∈ ∂Ω × (0, ∞).
∂ν
16
LI MA
When we have stationary heat in Ω, that ut = 0, we have the following 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 Laplace
equation in Ω, which is a typical elliptic partial differential equation of
second order. In dimension one, the Laplace equation becomes
00
u = 0,
So u = ax + b is a linear function of the single variable x. However, if
the dimension n ≥ 2, it is not easy to solve the Laplace equation on
domains coupling with its boundary condition.
In the following subsection, we try to solve some simple pde of first
order.
EQUATIONS FROM MATHEMATICAL PHYSICS
17
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 is
to 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
Z t
u(x(t), t) = u(A, 0) +
f (aτ + A, τ )dτ.
0
Given the point (x, t) on the curve C(t). Then we have
x(t) = x
and we can find that initial point on the curve is x(0) = A = x − at. So
by the initial value condition, we have u(A, 0) = u0 (A) = u0 (x − at).
Inserting A = x − at into the expression above, we find (FF):
Z t
u(x, t) = u0 (x − at) +
f (x + a(τ − t), τ )dτ.
0
In case when f = 0, we have
u(x, t) = u0 (x − at).
The method above is called characteristic curve method in the literature. One may study examples in our textbook.
We point out the following comparison lemma: Let u be a smooth
solution to (FDE) with the initial data u(0) = u0 . Let w be a smooth
function 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 for
PDE’s.
EQUATIONS FROM MATHEMATICAL PHYSICS
19
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 be
used to study the nonlinear pde of first order. By definition, a pde
(PDE)
P (u, ∇u, D2 u, ...) = 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:
1
ut + (1 + u)ux =
2
√
with the initial value condition u(x, 0) = x for x ≥ 0.
We first find the characteristic curve x(t) by solving
dx
= 1 + u(x, t).
dt
Note that along the characteristic curve, we have
1
du
=
dt
2
and
t
t p
+ u(x(0), 0) = + x(0)
2
2
On one hand, we set x(0) = A. Then we have
t √
u(x(t), t) = + A
2
√
Solving A from this relation, we get
√
t
A = u(x(t), t) − .
2
On the other hand, we find the curve satisfying
Z t
Z t
τ √
u(x(τ ), τ )dτ = A + t +
( + A)dτ.
x(t) = A + t +
0
0 2
u(x(t), t) =
Then we have
t2 √
t √
+ At = t + ( + A)2 = t + u(x(t), t)2 .
4
2
We now solve u along the curve x(t) and get
p
u(x(t), t) = x(t) − t.
x(t) = A + t +
20
LI MA
Given any point (x, t) on the curve x(t), we have x(t) = x. Hence, the
solution is
√
u(x, t) = x − t.
EQUATIONS FROM MATHEMATICAL PHYSICS
21
3. lecture three
3.1. Duhamel principle.
Consider (ODE)
0
u + p(t)u = 0,
with initial condition u = fτ at t = τ . Then its solution to (ODE) is
(S):
Z
t
u(t, τ ) = fτ exp(−
p(s)ds).
τ
We now try to solve the equation (F):
0
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
Z t
u(t, τ )dτ
0
is the solution to (F). This kind of method is called the Duhamel principle.
22
LI MA
3.2. First order ODE and a comparison lemma.
For the given ODE
u0 = a(t)u + b(t)
we can write it as
(e−
Rt
a(τ )
u)0 = e−
Rt
a(τ )
b(t)
So it has a general solution
Rt
u(t) = u(0)e
a(τ )
Rt
+e
a(τ )
Z
t
dse−
Rs
a(τ )
b(s).
If u(0) = 0, we have (F):
Rt
u(t) = e
a(τ )
Z
t
dse−
Rs
a(τ )
b(s).
One often use the argument above to prove the famous Gronwall
inequality: Let β > 0 and let u be a continuous function on [0,T]
satisfying
Z t
u(τ )dτ.
u(t) ≤ a + β
0
βt
Then we have u(t) ≤ ae
on [0,T]. Hint: Let
Z t
f (t) = a + β
u(τ )dτ.
0
0
Then we have f = βu. By assumption, u ≤ f , we get
f 0 ≤ βf.
Let β(t) and f (s) be two smooth functions. We consider the following
nonlinear ODE
u0 = β(t)f (u)
with initial data u = a at t = 0.
Assume that
w0 ≥ 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) = w0 − β(t)f (w).
Then g(t) ≥ 0. Set
v = w − u.
Then we have
v 0 == β(t)(f (w) − f (u)) + g(t) = β(t)B(t)v(t) + g(t)
EQUATIONS FROM MATHEMATICAL PHYSICS
23
with v = 0 at t = 0 and
Z
B(t) =
f 0 (zw + (1 − z)v)dz.
Let a(t) = β(t)B(t) and b(t) = g(t) in (F). Then we have v ≥ 0. That
is w ≥ u for all t.
We remark that one can use Sturm-Liouville theory and variational
principle 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 data
u|t=0 = u0
and
ut |t=0 = u1 .
Let’s first assume that u0 = 0. Note that
∂t2 − ∂x2 = (∂t − ∂x )(∂t + ∂x ).
Then we reduce the problem into two pde of first order. Let
v = ∂t u + ∂x u.
and
∂t v − ∂x v = 0.
Solving u we get
t
Z
v(x + (τ − t), τ )dτ.
u(x, t) =
0
Using initial condition we have v(x, 0) = u1 (x). We now solve v:
v(x, t) = v0 (x + t) = u1 (x + t).
So,
Z
t
u1 (x + (τ − t) + τ )dτ.
u(x, t) =
0
Changing the variable
s = x + 2τ − t,
we can get
Z
1 x+t
u(x, t) = [
u1 (s)ds] := Mu1 .
2 x−t
If u0 is non-trivial, but u1 = 0, then we can verify that
d
u(x, t) = Mu0 (x, t)
dt
is the solution to
utt − uxx = 0
with initial data
u|t=0 = u0
and
ut |t=0 = 0.
EQUATIONS FROM MATHEMATICAL PHYSICS
25
This is the Duhamel principle for the wave equation.
Hence, using the Duhamel principle, we can get the D’Alembert formula (EW):
Z x+t
1
u(x, t) = [(u0 (x + t) + u0 (x − t) +
u1 (s)ds].
2
x−t
for the solution to
utt − uxx = 0
with the initial data
u|t=0 = u0
and
ut |t=0 = u1 .
From this formula we know that the solution to the problem (OW) is
unique with the formula (EW).
One can show that, for fs (x) = f (x, s),
Z t
Mfs (x, t − s)ds
u(x, t) =
0
is the solution to
utt − uxx = f (x, t)
with the initial data
u|t=0 = 0
and
ut |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 t
provided |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), we
have
u = 0, ut = 0
for |x| > R + t. This is the finite propagation property of the wave
equation.
Remark Two: Define the energy for u by
Z
1
E(t) =
(u2 + u2x )dx.
2 R t
26
LI MA
Then we have
Z
d
E(t) = (ut utt + ux uxt )dx.
dt
R
Using the integration by part to the last term we get
Z
d
ut (utt − uxx )dx = 0.
E(t) =
dt
R
So the energy E(t) is a constant, which is
Z
1
(u2 + u20x )dx.
2 R 1
Remark Three: We now study the one dimensional wave equation
on half line (HWE):
utt − uxx = 0, x ∈ R+ , t > 0,
with initial data
u|t=0 = u0
and
ut |t=0 = u1 .
Note that u0 (t + x) is the special solution to (HWE). So we can
use it to transform the problem into the case when u0 = 0. Then
we can use the odd extension method on the x-variable to extend the
function u1 and the equation on the whole line. Then we can use the
D’Alembert formula to solve (HWE). One may also use this method to
treat one dimensional wave equation on an interval, however, we have a
more beautiful method—Fourier method/Separation variable method
to study this case. See the next section.
Remark Four: We point out that we can use the D’Alembert formula to treat some special case of higher dimensional wave equation.
Here is the example: Consider the following problem for 2-dimensional
wave equation
utt − uxx − uyy = 0
with the initial data
u|t=0 = φ(x)
and
ut |t=0 = ψ(y).
Then we split the problem int to two one-dimensional problems: One
is
utt − uxx = 0
with the initial data
u|t=0 = φ(x)
EQUATIONS FROM MATHEMATICAL PHYSICS
and
ut |t=0 = 0.
The solution for this one dimensional problem is
1
u(x, t) = [(φ(x + t) + φ(x − t)]
2
The other is
utt − uyy = 0
with the initial data
u|t=0 = 0
and
ut |t=0 = ψ(y).
Its solution is
Z
1 y+t
ψ(s)ds].
u(x, t) = [
2 y−t
Hence, the solution to our original 2-dimensional problem is
Z y+t
1
u(x, y; t) = [(φ(x + t) + φ(x − t) +
ψ(s)ds].
2
y−t
27
28
LI MA
4. lecture four
4.1. The method of separation of variables 1.
The separation variable method for boundary value problems is looking 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 = a2 uxx
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
00
00
T X(x) = a2 T X ,
or
00
00
T
X
=
.
a2 T
X
The left is a function of variable t and the right is a function of variable
x. To make them be the same, they should be the same constant. Say
−λ. Then we have one equation for T :
00
T + a2 λT = 0
and the other equation for X:
00
X + λX = 0
with the boundary condition (BC). The latter is the simple case of
the Sturm-Liouville problem. The key step in the Separation Variable
Method is to solve this problem at first.
Case 1: When λ < 0, the general solution for X is
√
X(x) = c1 e
−λx
+ c2 e−
√
−λx
Using (BC) we have
c1 + c2 = 0
and
√
√
c1 e −λ + c2 e− −λ = 0,
which implies that c1 = c2 = 0.
Case 2: When λ = 0, the general solution for X is
X(x) = c1 + c2 x.
Again (BC) implies that c1 = c2 = 0.
EQUATIONS FROM MATHEMATICAL PHYSICS
29
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 then
λ = k 2 π 2 , k = 1, 2, ...
Hence
X(x) = Xk (x) = ck sin(kπx)
for k = 1, 2, ....
Returning to the equation for T (t) with λ = k 2 π 2 , we get
Tk (t) = ak cos(kπat) + bk sin(kπat).
So
Uk (x, t) = [ak cos(kπat) + bk sin(kπat)] sin(kπx)
is a typical solution to the one-dimensional wave equation with boundary condition.
We remark here that a easy way to see λ ≥ 0 is using the integration
relation:
Z
Z
0
2
λ |X| = |X |2 .
We now look for a solution of the form
+∞
X
[ak cos(kπat) + bk sin(kπat)] sin(kπx)
u(x, t) =
k=1
to the wave equation
utt = a2 uxx
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 we
have 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
+∞
X
φ(x) =
ak sin(kπx)
k=1
30
LI MA
and
ψ(x) =
+∞
X
kπabk sin(kπx).
k=1
Using the Fourier series expansion for φ(x) and ψ we have
Z
ak = 2 φ(ξ)sin(kπξ)dξ
and
Z
2
bk =
ψ(ξ)sin(kπξ)dξ.
kπa
In this way, we get at least formally a solution to our problem. This is
a 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 condition.
EQUATIONS FROM MATHEMATICAL PHYSICS
31
4.2. The resonance problem. We now briefly discuss the resonance
problem of the one dimensional wave equation. Consider
utt − a2 uxx = A sin(ωt), (x, t) ∈ [0, L] × (0, ∞)
with homogeneous initial and boundary conditions
u(0, t) = u(L, t) = 0
and
u(x, 0) = 0 = ut (x, 0),
where ω 6=
for j = 1, 2, ....
Then we have
Z t
X Laj
jπx
ajπ
u(x, t) =
sin(
)
sin(ωτ ) sin(
(t − τ ))dτ
ajπ
L
L
0
j
X
aj
jπx
=
(ω sin ωj t − ωj sin ωt) sin(
)
2
2
ωj (ω − ωj )
L
j
ajπ
L
where
ωj =
and
2
aj =
L
Z
ajπ
L
L
A(x) sin(
0
jπx
)dx.
L
Because we have
ω 2 − ωj2
which may cause the coefficient
aj
→ ∞.
2
ωj (ω − ωj2 )
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 conditions
u|t=0 = u0
and
ut |t=0 = u1 .
Introduce the polar coordinates (r, θ) by
x = r cos θ, y = r sin θ.
Then
∆u = urr + r−1 ur + r−2 uθθ .
The reason that we use polar coordinates here is because of the boundary condition in our problem being on the circle. Let u = T (t)U (x, y)
be the solution to our wave problem. Then
00
T + a2 λT = 0
and
∆U + λU = 0, U (R, t) = 0.
It is easy to see that λ = k 2 > 0. Setting U = h(θ)B(r) in the latter
equation, we have
00
00
0
h (θ)
r2 B + rB + k 2 r2 B
=−
= −β 2 .
h(θ)
B
Then we get two problems:
00
h + β 2 h = 0, h(θ + 2π) = h(θ),
and
00
0
B + r−1 B + (k 2 − β 2 r−2 )B = 0, B(R) = 0.
The last equation is called the Bessel equation. As before, we find that
β = βn = n
and
hn (θ) = An cos(nθ) + Dn sin(nθ).
Inserting β = n into the Bessel equation, we have
00
0
B + r−1 B + (k 2 − n2 r−2 )B = 0, B(R) = 0.
EQUATIONS FROM MATHEMATICAL PHYSICS
33
The general solutions of this equation are called the Bessel functions
of order n. Just like trigonometric polynomials, Bessel functions form
an orthonormal basis, and then we can use this basis to solve the wave
problem above. One may look for reference book for more about the
Bessel functions. Some people would like to define the Bessel functions
of order n as the coefficients of the Fourier expansion:
∞
X
exp(ir sin θ) =
Bn (r)einθ ,
where i =
√
n=−∞
inθ
−1 and e
= cos(nθ) + i sin(nθ). Hence,
Z 2π
1
exp(ir sin θ − inθ)dθ.
Bn (r) =
2π 0
We remark that, if the domain is not a ball but a rectangle, we just
let
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 ; f or (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 homogeneous
equation
utt − ∆u = 0; f or (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
1
u(p, t) =
4πa2 t
Z
ψ(x)dσx ,
Sat (p)
where Sat (p) = {x ∈ R3 ; r = |x − p| = at}.
In fact, let
Z
1
I(x, r; g) =
udσ
4π ∂Br (x)
for g = g(y) ∈ C 2 (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
Z
1
I(x, r; g) =
u(x + rθ)dθ
4π {|θ|=1}
EQUATIONS FROM MATHEMATICAL PHYSICS
and
Z
g(x + z)dz
=
BR (0)
Z R
2
Z
r dr
u(x + rθ)dθ
{|θ|=1}
0
R
Z
4πr2 I(x, r; g)dr.
= =
0
We now compute
Z
R
∆x
4πr2 I(x, r; g)dr
Z0
= ∆x
g(x + z)dz
BR (0)
Z
=
∆x g(x + z)dz
BR (0)
Z
=
∆z g(x + z)dz
BR (0)
Z
=
∂R g(x + z)dσz
∂BR (0)
Z
=
∂R g(x + Rθ)R2 dθ
∂B1 (0)
= = 4πR2 ∂R I(x, R; g).
So, we have
Z
∆x
R
r2 I(x, r; g)dr = R2 ∂R I(x, R; g).
0
Note that
Z
R
r2 I(x, r; g)dr
∂R ∆x
0
Z
= ∆x ∂R
R
r2 I(x, r; g)dr
0
2
= R ∆x I(x, R; g)
and
∂R [R2 ∂R I(x, R; g)]
= 2R∂R I(x, R; g) + R2 ∂R2 I(x, R; g).
35
36
LI MA
Hence,
R2 ∆x I(x, R; g) = 2R∂R I(x, R; g) + R2 ∂R2 I(x, R; g).
Dividing both sides by R, we have
∆x RI(x, R; g)
= = 2∂R I(x, R; g) + R∂R2 I(x, R; g)
= ∂R2 (RI(x, R; g)).
Now,
∂2J
∂t2
R
=
4π
Z
a2 R
=
4π
utt (x + Rθ, t)dθ
∂B1 (0)
Z
∆x u(x + Rθ, t)dθ
∂B1 (0)
a2
∆x R
=
4π
Z
u(x + Rθ, t)dθ
∂B1 (0)
= a2 ∆x (RI(x, R; u))
= a2 ∆x (RI(x, R; u))
∂2J
= a2 2 .
∂R
This is a one dimensional wave equation. We now determine its boundary condition and initial condition. Clearly, we have the boundary
condition
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
1
J(x, R; t) =
2a
Z
at+R
rI(x, r; ψ)dr.
at−R
EQUATIONS FROM MATHEMATICAL PHYSICS
37
Hence,
J(x, R; t)
R→0
R
Z at+R
1
lim
rI(x, r; ψ)dr
R→0 2aR at−R
1
[RI(x, R; ψ)]|R=at
a
tI(x, at; ψ)
Z
t
ψ((x + atθ)dθ.
4π ∂B1 (0)
u(x, t) = lim
=
=
=
=
That is,
Z
1
ψ(y)dσy .
u(x, t) =
4πa2 t Sat (x)
Then we use the Duhamel principle in the study of the wave equation
utt − ∆u = f ; R3 × {t > 0}
with the initial data
u = φ(x), ut = ψ(x), at t = 0,
and we have the famous Kirchhoff formula for the solution:
Z
1
r−1 f (x, t − a−1 r)dx
u(p, t) =
4πa2 Bat (p)
Z
1
+
ψ(x)dσx
4πa2 t Sat (p)
Z
1
φ(x)dσx ]
+ ∂t [
4πa2 t Sat (p)
where
Bat (p) = {x ∈ R3 ; r = |x − p| ≤ at}.
From this formula, we can see the Huygens principle that the wave at
x0 at time t = 0 influences the solution only on the boundary of a cone
originating from x0 . We just remark that using the descent method,
one can get a solution formula for two dimensional wave equation.
One may see other textbooks for the other understanding of the
Kirchhoff formula.
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 boundary conditions.
We now describe the main idea. Let u(x, t) = X(x)T (t) be a solution
to the heat equation
ut = a2 uxx
on [0, l] × [0, ∞).
Then formally, we have
0
00
X
T
=
.
2
aT
X
The left is a function of variable t and the right is a function of variable
x. To make them be the same, they should be the same constant. Say
−λ. Then we have
0
T + a2 λT = 0
and (EX):
00
X + λX = 0.
As in the wave equation case, we impose the homogeneous boundary
condition
X(0, t) = 0 = X(1, t)
to solve X and λ. Multiplying both sides of (EX) by X and integrating
over [0, 1], we have
Z
Z
λ
X2 =
0
|X |2 > 0.
So λ > 0. Then we have
X(x) = Xk (x) = ck sin(kπx)
for k = 1, 2, ... and
λ = λk = k 2 π 2 .
Then we solve for T and get Tk (t) = T (0)exp(−a2 λk t) by imposing
initial conditions for T . The lesson we have learned is that the solution
to heat equation has exponential decay to 0 in t-variable as t → ∞.
Using the superposition law, we can solve the heat equation by looking at solutions of the form
X
Ak exp(−a2 k 2 π 2 t) sin(kπx)
k
with the initial data
φ(x) =
X
k
Ak sin(kπx)
EQUATIONS FROM MATHEMATICAL PHYSICS
39
Assume that we are given the heat equation
ut = a2 uxx
with non-homogenous boundary condition, we need to use transformations 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 = a2 uxx + f (x, t).
To study this problem, we need to solve for {Xj } and λ = λj to the
Sturm-Liouville problem
00
X + λX = 0,
with the boundary condition
X(0) = 0 = Xx (L).
Once this done, we do the expansion for f
X
f (x, t) =
fj (t)Xj (x)
j
and look for solution u(x, t) such that
X
u(x, t) =
uj (t)Xj (x).
j
Then we reduce the non-homogenous problem into the ODE
0
uj = −a2 λj uj + fj (t)
with suitable initial condition. Hence, we have the expression
Z t
−λj t
eλj τ fj (τ )dτ
uj (t) = e
(uj (0) +
0
and
u(x, t) =
X
−λj t
e
Z
(uj (0) +
t
eλj τ fj (τ )dτ )Xj (x).
0
j
Here uj (0) are determined by
φ(x) =
X
j
We omit the detail here.
uj (0)Xj (x).
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) = λj φj (x),
with a fixed homogeneous boundary condition. Define the function
space E such that the functions {φj } form an orthonormal basis for
the space E with its norm given by
qX
kuk =
a2j
P
for u = aj φj (x) ∈ E.
1,2
Given a smooth
P function f ∈ C (D × [0, ∞)). Assume that the
expansion 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
X
f=
fj (t)φj (x)
with its norm on D × [0, ∞) defined by
XZ ∞
kf k =
|fj (t)|2 dt < ∞.
j
0
Then we have
0
uj (t) = λj uj (t) + fj (t).
Hence, as before, we have the expression
Z t
λj t
uj (t) = e
e−λj τ fj (τ )dτ.
0
Note that
0
0
0
(uj (t) − λj uj (t))2 = uj (t)2 + λ2j uj (t)2 − 2λj uj (t)uj (t) = fj (t)2 .
Then we have
Z T
0
2
(uj (t) +
λ2j uj (t)2 )dt
2
Z
− λj uj (T ) =
0
T
fj (t)2 .
0
Note that we have assumed that λj < 0. Hence, we have
Z ∞
Z ∞
0
2
2
2
(uj (t) + λj uj (t) )dt ≤
fj (t)2 ,
0
0
which implies that
k∂t uk2 + λ2 kuk ≤ kf k2 ,
EQUATIONS FROM MATHEMATICAL PHYSICS
41
where λ2 = infj {λ2j }. 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, λ2 t)
is also a solution of (HE).
So we try to look for solutions of the form
u(x, t) = tβ f (|x|2 /t).
Set r = |x|2 /t. Then we find that
ut = βtβ−1 f + tβ (−
u
u
0
00
=
|x|2 0
0
)f = βtβ−1 f − rtβ−1 f ,
2
t
∂
u
∂|x|
= 2|x|tβ−1 f
∂2
=
u
∂|x|2
0
0
= 2tβ−1 f + 4|x|2 tβ−2 f
0
00
0
00
= 2tβ−1 f + 4rtβ−1 f
n−1 0
00
∆u = u +
u
|x|
00
= 2tβ−1 f + 4rtβ−1 f + 2(n − 1)tβ−1 f
00
0
= 4rtβ−1 f + 2ntβ−1 f .
So,
β
2n 0
0
00
− f + f + 4f + f = 0.
r
r
Let β = −n/2. Then we have
0 0
0
4(rn/2 f ) + (rn/2 f ) = 0.
Then for some constant A,
0
4rn/2 f + rn/2 f = A.
Take A = 0. Then we have
0
4f = −f.
0
EQUATIONS FROM MATHEMATICAL PHYSICS
43
0
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) =
|x|2
1
− 4t
.
e
(4πt)n/2
This is called the Gaussian kernel, which is also called the Poisson
kernel. It is easy to verify that
Z
K(x, t)dx = 1.
Rn
Theorem 2. Given g(x), we let
Z
K(x − y, t)g(y)dy.
u(x, t) =
Rn
Then u(x,t) satisfies the heat equation (HE) and as t → 0, u(x, t) →
g(x).
Roughly speaking, the theorem says that this u(x, t) is the solution
to our problem at the beginning of this subsection.
Proof. For t > 0, we can take derivatives into the integrand term, and
then u satisfies (HE) since K does.
Without loss of generality, we now assume n = 1. Assume that there
is a positive constant M > 0 such that
|g(x)| ≤ M, x ∈ R.
We use two step argument.
Step one: Given an arbitrary small > 0. For any fixed point x ∈ R,
we have some δ0 > 0 such that, for |x − y| < δ0 ,
|g(y) − g(x)| < .
2
Step two: We take δ1 > 0 small enough such that for 0 < t < δ1 ,
Z
K(x − y, t)dy <
.
4M
{|x−y|≥δ0 }
44
LI MA
Then we have
Z
|u(x, t) − g(x)| = |
K(x − y, t)(g(y) − g(x))dy|
Z
≤ |
K(x − y, t)(g(y) − g(x))dy|
{|x−y|≥δ0 }
Z
K(x − y, t)(g(y) − g(x))dy|
+ |
{|x−y|<δ0 }
Z
≤ 2M
Z
+ K(x − y, t)dy
{|x−y|≥δ0 }
K(x − y, t)dy
{|x−y|<δ0 }
< 2M
+ = .
4M
2
Here we have used the fact that K ≥ 0 and
Z
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 ∈ C 0 (Rn ) satisfies
g(x) → 0, as |x| → ∞.
Then for the solution u(x, t) defined above, we have
u(x, t) → 0, ∇x u(x, t) → 0, as |x| → ∞.
for each t > 0.
Proof. Fix t > 0. Choose R > 0 sufficient large. Then by our assumption on g(x), we have
supBR (x) |g(x)| → 0, |x| → ∞.
EQUATIONS FROM MATHEMATICAL PHYSICS
45
Note that
Z
|u(x, t)| ≤
K(x − y, t)g(y)dy
Z
≤ supBR (x) |g(x)|
K(x − y, t)dy
BR (x)
Z
+ supRn −BR (x) |g(x)|
K(x − y, t)dy
Rn −BR (x)
Z ∞
2
e−ρ /4t ρn−1 dρ
≤ supBR (x) |g(x)| + C(t)supRn |g(x)|
Rn
R
→ 0 + o(R)
as |x| → ∞. Here C(t) is a constant depending only on t and o(R) is
as small as we like for R large. Hence, we have showed that
|u(x, t)| → 0,
as |x| → ∞.
Using the polar coordinates (ρ, θ), we can also show that
|∇j u(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 space
C0∞ (Rn ). Then, δx is a linear functional on C0∞ (R). This δx (·) is called
Dirac’s delta function or Dirac measure. One can also consider the
delta function δ0 (x) as the limit of a sequence of nice functions. For
example, assume that n = 1 and let
1
qj (x) = , x ∈ [−1/j, 1/j] and qj (x) = 0 |x| > 1/j.
2j
Then
Z
qj (x)dx = 1,
R
and for x 6= 0, qj (x) → 0 and at x = 0, qj (0) → ∞; or for φ ∈ C0∞ (R),
Z
< qj , φ > =
qj (x)φ(x)dx
R
Z
1 1/j
=
φ(x)dx → φ(0)
2j −1/j
= < δ0 , φ > .
as j → ∞. If so, we define limqj = δ0 . We ask the reader use the first
concept of limit to understand delta function. I would like to point
out that the latter limit is more rigorous definition. We may find other
sequences which have δ0 as its limit. For example, let
ρ(x) = me
−
1
1−|x|2
, f or |x| < 1, ρ(x) = 0, f or |x| ≥ 1
where m > 0 is a constant such that
Z
ρ(x)dx = 1,
R
and let
where t > 0. Then
1
ρt (x) = ρ(x/t),
t
Z
ρt (x)dx = 1
R
and limt→0 ρt = δ0 .
EQUATIONS FROM MATHEMATICAL PHYSICS
47
Generally speaking, for any given continuous function u(x) in Rn ,
we can also consider u as a linear functional on the space C0∞ (Rn ) with
Z
< u, φ >=
u(x)φ(x)dx.
Rn
With this understanding, we say K(x−y, t) is the fundamental solution
the heat equation on Rn ×R+ . We emphasize that we call K(x−y, t) is
a fundamental solution to the heat equation on the whole space because
K(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 the
boundary 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 boundary
condition. So the function defined by
Z
u(x, t) =
Γ(x, y, t)g(y)dy
R+
satisfies the heat equation on the half line [0, ∞) with boundary condition 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 the
boundary 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 harmonic
function.
Let r ≥ 0 such that
X
r2 =
(xi )2 .
i
Then we have
rdr =
X
xi dxi .
i
Then
∂r
xi
=
.
∂xi
r
Let u = v(r) be a solution of (L). Note that
xi
u xi = v ,
r
0
and
i 2
(xi )2
0 1
00 (x )
u xi xi = v ( − 3 ) + v
.
r
r
r2
Then we have
00
∆u = v +
Solve
00
v +
We find that
n−1 0
v.
r
n−1 0
v = 0.
r
0
v =
c
rn−1
for some constant c and
v = a log r + b,
f or
n = 2,
and
v = ar2−n + b,
f or n ≥ 3,
for some constants a and b.
Choose b = 0. Define
−1
Γ(x) =
log r,
f or n = 2,
2π
and
1
Γ(x) =
r2−n ,
f or n ≥ 3.
n(n − 2)|S n−1 |
EQUATIONS FROM MATHEMATICAL PHYSICS
49
Here |S n−1 | is the volume of the sphere S n−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). In
fact, we can consider
−∆y Γ(x − y) = δx (y).
If so, we have
u(x) = < δx (y), u(y) >
= − < ∆Γ(x − y), u(y) >
Z
= −
∆Γ(x − y)u(y)dy
Rn
Z
= −
Γ(x − y)∆u(y)dy
Rn
Z
=
Γ(x − y)f (y)dy.
Rn
To understand this, we need to use the second Green formula. Here we
recall the second Green formula as follows. Given two smooth functions
defined on a bounded domain D with piece-wise smooth boundary. We
denote ν the outer unit normal to the boundary ∂D. Then we have
Z
Z
(v∇u · ν − u∇v · ν)dσ.
(v∆u − u∆v)dx =
∂D
D
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
Z
Γ(x − y)f (y)dy.
u(x) =
Rn
Then the function u satisfies the Poisson equation
−∆u = f, in Rn .
Proof. For simplicity, we assume n = 2. Take R > 1 large such that
f = 0 outside BR . Fix x ∈ R2 . Using the second Green formula on
the region BR − B (x), where → 0, we get that
Z
Γ(x, y)(−∆u(y))
BR
Z
= lim
Γ(x, y)(−∆u(y))
→0 B −B (x)
R
Z
= lim
(Γ(x, y)∇r u(y)
→0
{|y−x|=}
− ∇r Γ(x, y)f (y))dσ,
50
LI MA
where r = |x − y and ∇r u(y) =
Note that
Z
∂u
(y).
∂r
|Γ(x, y)∇r u(y)|dσ
{|y−x|=}
≤
Z
1
2π
1
ln |∇r u(y)|
r
{|y−x|=}
≤ sup|∇u||ln|
→ 0,
and
Z
∇r Γ(x, y)u(y)dσ
−
{|y−x|=}
=
=
=
Z
1
∇r ln(r)u(y)dσ
2π {|y−x|=}
Z
1
1
f (y)dσ
2π {|y−x|=} r
Z
1
u(y)dσ
2π {|y−x|=}
→ u(x)
as → 0. This shows that
Z the function u satisfies
−
Γ(x, y)∆u(y)dy = u(x).
R2
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 now
consider the problem (PD):
−∆u = f, in D
with the boundary condition u(x) = φ(x) for x ∈ ∂D. Then using
the second Green formula and the fact v = Γ being the fundamental
solution on the whole Rn , we get (U1):
Z
u(x) =
Γ(x − y)f (y)dy
ZD
+
(Γ(x − y)∇u(y) · ν − φ(y)∇Γ(x − y) · ν)dσy .
∂D
This expression still contains a unknown term for ∇u(y) · ν on the
boundary ∂D. To overcome this drawback, we introduce, for each
fixed point x ∈ D, the harmonic function g(x, ·) with its boundary
EQUATIONS FROM MATHEMATICAL PHYSICS
51
data g(x, y) = −Γ(x − y) for y ∈ ∂D. Using the expression above for
v = g(x, y), we have (U2):
Z
0 =
g(x, y)f (y)dy
D
Z
+
(g(x, y)∇u(y) · ν − φ(y)∇g(x, y) · ν)dσy .
∂D
Set
G(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
Z
u(x) =
G(x, y)f (y)dy
D
Z
−
φ(y)∇G(x, y) · νdσy .
∂D
We call G(x, y) the Green function (or fundamental solution )for the
problem (PD).
Let’s consider the boundary value problem for Laplace equation.
Assume that u ∈ C 2 (D) ∩ C(D̄) 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
Z
Γ(x − y)(−∆u(y))dy
u(x) =
D
Z
+
(Γ(x − y)∇u(y) · ν − φ(y)∇Γ(x − y) · ν)dσy
∂D
Z
=
φ(y)∇Γ(x − y) · νdσy .
∂D
Let
P (x, y) = ∇Γ(x − y) · ν
for x ∈ D and y ∈ ∂D. We call P (x, y) the Poisson kernel to the
boundary value problem in D. Then it can be verified that the function
defined by
Z
u(x) =
P (x, y)φ(y)dσy ,
∂D
52
LI MA
which is called the Poisson formula, is the solution to the boundary
value problem in D.
We now point out a nice observation about the construction of harmonic functions. Given two harmonic functions u1 and u2 on the same
domain D ⊂ Rn . We want to find a harmonic function U (x, y) defined
on the one dimensional higher cylindrical domain D×[L1 , L2 ] such that
U (x, L1 ) = u1 (x)
and
U (x, L2 ) = u2 (x).
This function can be achieved by setting
L2 − y
y − L1
U (x, y) =
u1 (x) +
u2 (x).
L 2 − L1
L 2 − L1
EQUATIONS FROM MATHEMATICAL PHYSICS
53
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
. Using the
r
divergence theorem, we get
Z
Z
∂u
dσ.
0=
∆u =
Bt (p)
∂Bt (p) ∂r
Then we have the the right side of the identity above is
Z
Z
∂u
n−1 ∂
n−1
(p + tθ)dθ = t
t
u(p + tθ)dθ.
∂t ∂B1 (p)
∂B1 (p) ∂r
Here, we want to point out that the notation
∂u
(p + tθ)
∂r
means the radial derivative of function u taking value at x = p + tθ. It
is not a composite derivative for independent variables (r, θ. Hence,
Z
1−n
n−1 ∂
(t
udθ).
0=t
∂t
∂Bt (p)
This implies that
1−n
Z
t
udθ
∂Bt (p)
is a constant for all t ≤ R. Then
Z
Z
1
1
udθ =
udθ.
|∂Bt (p)| ∂Bt (p)
|∂BR (p)| ∂BR (p)
Using the mean value theorem and sending t → 0, we have
Z
1
udθ.
u(p) =
|∂BR (p)| ∂BR (p)
Then
Z
u(p)|∂BR (p)| =
udθ.
∂BR (p)
If we take integration in R, we have
1
u(p) =
|BR (p)|
Hence, we have
Z
udx.
BR (p)
54
LI MA
Theorem 5. Given a harmonic function u in a domain D. We take
a ball BR (p) in the domain D. Then we have
Z
1
u(p) =
udθ.
|∂BR (p)| ∂BR (p)
and
1
u(p) =
|BR (p)|
Z
udx.
BR (p)
Mean value theorem is a basic property for harmonic functions. As
an application we prove the following result.
Theorem 6. (Liouville Theorem). If u is a bounded harmonic function 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
Z
1
u(x + y)dy.
u(x) =
|BR (0)| BR (0)
Hence,
Z
1
∂xi u(x) =
∇i u(x + y)dy
|BR (0)| BR (0)
Z
1
=
νi u(x + y)dσy ,
|BR (0)| ∂BR (0)
where νi = xi /R, and then,
M |∂BR (0)|
→0
|BR (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
Z
Z
1
(
udx −
udx)
u(x0 ) − u(x1 ) =
|BR (0)| BR (x0 )
BR (x1 )
Z
Z
1
=
(
udx −
udx).
|BR (0)| BR (x0 )−BR (x1 )
BR (x1 )−BR (x0 )
|∂xi u(x)| ≤
Hence,
|u(x0 ) − u(x1 ) ≤
Z
Z
1
(
|u|dx +
|u|dx)
|BR (0)| BR (x0 )−BR (x1 )
BR (x1 )−BR (x0 )
2M |BR (x1 ) − BR (x0 )|
|BR (0)|
→ 0
≤
EQUATIONS FROM MATHEMATICAL PHYSICS
55
as R → ∞ (since |BR (x1 ) − BR (x0 )| ≈ Rn−1 and |BR (0)| ≈ Rn ). That
is 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 and
parabolic equations. It is not required for the examination for students
even though it is really important for modern PDE. Recently I and
my students have found a monotonicity formula for solutions to a very
large class of elliptic and parabolic equations. We can not exhibit it
here, however, we would like to show the classical monotonicity formula
for harmonic functions.
Assume that u is a harmonic function in a ball BR := BR (p) with
center at the point p and of radius R. It is easy to see that the integral
Z
|∇u|2 dx
BR
is a non-decreasing function of R. To our surprise, so is
Z
2−n
R
|∇u|2 dx.
BR
Why? let us prove it. Recall that
∆u = 0
in the ball BR . Let X(x) = (Xj (x)) = x − p be a radial type vector
field in BR . Recall that on ∂BR , the outer unit normal is
ν = (x − p)/R.
We denote by
ui =
∂u
∂2u
=
∇
u,
u
=
.
i
ij
∂xi
∂xi xj
We now have
Z
Z
∆u∇u · X = −
0=−
Z
∇u · ν∇u · Xdσ +
Then we re-write it as
Z
Z
∇u · ν∇u · Xdσ =
∂BR
∇u · ∇(∇u · X)dx.
BR
∂BR
BR
∇u · ∇(∇u · X)dx.
BR
Note that
Z
Z
∇u · ν∇u · Xdσ = R
∂BR
and
Z
∂BR
Z
∇u·∇(∇u·X)dx =
BR
|∂r u|2 .
Z
∇i u∇i (∇j uXj )dx =
BR
BR
(|∇u|2 +uij ui Xj ).
EQUATIONS FROM MATHEMATICAL PHYSICS
57
Using integration by part to the last term, we have
Z
Z
Z
Z
1
R
1
2
2
uij ui Xj =
∇j |∇u| Xj =
n|∇u|2 .
|∇u| −
2
2
2
BR
∂BR
Then we
Z
Z
Z have
R
2−n
2
|∇u| +
|∇u|2 ,
∇u · ∇(∇u · X)dx =
2
2
BR
∂BR
BR
and
Z
(n − 2)
Z
2
(|∇u|2 − 2|∂r u|2 )dσ.
|∇u| = R
BR
∂BR
We now compute
Z
Z
Z
2−n
2
2−n
2
1−n
∂R [R
|∇u| ] = R
|∇u| + (2 − n)R
|∇u|2
BR
BR
Z∂BR
= 2R2−n
|∂r u|2
∂BR
≥ 0.
This implies that the function
R
2−n
Z
|∇u|2
BR
is a monotone non-decreasing function of R. More precisely, we have
proved the following result.
Theorem 7. Assume that u is a harmonic function in a ball BR . Then
Z
Z
Z
2−n
2
2−n
2
R
|∇u| − r
|∇u| = 2
|x − p|2 |∂r u|2
BR
BR −Br
Br
for all r < R. In particular, the function
Z
2−n
|∇u|2
r
Br
is a monotone non-decreasing function of r.
Monotonicity formulas play a very important role in the regularity theory of weak solutions to nonlinear parabolic and elliptic partial
differential equations/ systems.
58
LI MA
7. lecture seven
From the previous sections, we know that solving a PDE is not easy
matter. To handle PDE on unbounded domains, in particular, PDE on
whole space, people try to use integration type of transformations to
reduce ODE into algebraic equations and to reduce linear PDE into to
ODE. The famous Laplace and Fourier transforms are for this purpose.
In this section, we always assume that our functions have a good
decay at infinity such that the integrals converge.
7.1. Laplace transforms.
We define, for a function y = y(t) ∈ C 0 [0, ∞), the Laplace transform
by
Z ∞
y(t)e−st dt.
L(y)(s) =
0
Remark:
It is clear that if there is some constant M > 0 such that
|y(x)| ≤ eM x , f or 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
Z ∞
0
y 0 (t)e−st dt = −y(0) + sL(y)(s).
L(y )(s) =
0
By induction, we get
L(y
(k)
k
)(s) = s L(y)(s) −
k
X
y (j−1) (0)sk−j .
j=1
We can use this property to solve the following ODE:
00
y + ay 0 + by = g(t),
with initial conditions
y(0) = 0 = y 0 (0),
where a, b are constants and g(t) ∈ C 0 [0, ∞). In fact, let Y (s) =
L(y)(s). Then we have
s2 Y + asY + bY = L(g).
Hence, we have
Y =
s2
L(g)
.
+ as + b
EQUATIONS FROM MATHEMATICAL PHYSICS
59
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
Z a+i∞
1
−1
L (Y )(t) =
Y (s)est ds
2πi a−i∞
where a is the real part of s = a + ib, which is a complex variable in
the plane. Then we have
L(g)
y(t) = L−1 [ 2
].
s + as + b
In general, we have a list of Laplace transforms of special functions to
use. 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
Z
L(y)(s) =
∞
tn e−st dt.
0
We do the computation step by step.
Z
1 ∞ n −st
L(y)(s) = −
t e d(−st)
s 0
Z ∞
1
tn−1 e−st dt]
=
[n
s
0
= ...
n!
= n+1 .
s
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 function
on R. Note that C0∞ (R) is a subspace of S. The important example is
that the function
2
gA (x) = e−Ax
is in the space of Schwartz space, where A > 0 is a constant. We denote
g(x) = g1 (x).
Define the Fourier transform of a Schwartz function. Let f be a
smooth function in S. Define
Z
1
fˆ(ξ) = √
f (x)e−ixξ dx,
2π
which is called the Fourier transform (in short, F-transform) of f in
R. Set F (f ) = fˆ.
Remark: From the definition formula, we see that the Fourier transform makes sense for any function f ∈ L1 (Rn ), that is,
Z
|f (x)|dx < ∞.
Rn
From the definition of the F-transform, we can see that the Schwartz
space is the invariant set of F-transform, that is,
F : S → S.
Example 1: Let’s compute the Fourier transform of the function
gA (x) above. By definition,
Z
1
2
gc
e−Ax e−ixξ dx.
A (ξ) = √
2π
Note that
i 2
ξ2
2
,
−Ax − ixξ = −A(x +
ξ) −
2A
4A2
and
r
Z
π
−Ax2
e
dx =
.
A
Then we have
ξ2 Z
e− 4A2
1 − ξ22
−Ax2
√
√
gc
(ξ)
=
e
dx
=
e 4A .
A
2π
2A
EQUATIONS FROM MATHEMATICAL PHYSICS
61
From this we can see that, for A3 = 14 , gA is the eigenfunction of the
F-transform.
x2
For t > 0, let A = 4a12 t and let g(x, t) = a√12t e− 4a2 t . Then
F [g(x, t)](ξ) = e−a
2 ξ2 t
.
We shall use this expression.
Example 2: Let
qj (x) =
1
, f or |x| ≤ 1/j, and qj (x) = 0, f or |x| > 1/j.
2j
Then
1
qbj (ξ) = √
2 2πj
Z
1/j
1 sin jξ
e−ixξ dx = √
.
2π jξ
−1/j
Hence,
qbj (ξ) → 0, f or ξ 6= 0
as j → ∞, and
1
qbj (0) = √ .
2π
Given h, 0 6= λ ∈ R. Let
(τh f )(x) = f (x − h), (δλ f )(x) = f (λx).
Then by definition, we have
−i(hξ) ˆ
(τd
f (ξ)
h f )(ξ) = e
and
−1 ˆ
(δd
λ f )(ξ) = λ f (ξ/λ).
Hence, for
2 x2
(δλ g)(x) = e−λ
,
we have
1 − ξ22
−1
e 4λ .
δc
λ g(ξ) = λ ĝ(ξ/λ) = √
2λ
Given two functions f and g. Define their convolution by
Z
f ? g(x) = f (x − y)g(y)dy.
62
LI MA
Then we have
F (f ? g(x)) =
=
=
=
Z
1
√
f ? g(x)e−ixξ dx
2π
Z
Z
1
−ixξ
=√
e
dx f (x − y)g(y)dy
2π
Z
Z
1
−iyξ
√
g(y)e
dy f (x − y)e−i(x−y)ξ dx
2π
√
2π fˆ · ĝ.
The most important property is that the F-transform of the derivative
fx is a multiplication:
fˆx (ξ) = iξ fˆ(ξ).
This property gives us a chance to define the inverse operator ∂x−1 of
∂x by the definition
ˆ f (ξ) = (iξ)−1 fˆ(ξ).
∂x−1
Clearly, we have
∂x−1 ∂x f = f.
Remark: One may see more interesting properties about Fourier
transforms in the subject called ”Harmonic/Fourier Analysis”, which
is one of the beautiful theory of modern mathematics.
Let’s going back. By an elementary computation, we have the following symmetry formula:
Z
Z
u(x)ĝ(x)dx = û(y)g(y)dy.
Define
Z
1
ˇ
f (ξ) = √
f (x)eixξ dx,
2π
which is called the inverse Fourier transform of f . In short we write
it by F ? (f ).
By an elementary computation and the symmetry property, we have
Proposition 8.
fˆˇ = f
and
ˇ
fˆ = f.
Proof. Let’s verify the latter equality. In fact, note that
gA (x) → 1,
EQUATIONS FROM MATHEMATICAL PHYSICS
63
as A → 0 and
√
1 − ξ22
gc
e 4A → 2πδ0 ,
A (ξ) = √
2A
as A → 0. Then, by symmetry property,
Z
1
ˇˆ
f (0) = √
fˆ(ξ)dξ
2π
Z
1
= √ limA→0 gA (ξ)fˆ(ξ)dξ
2π
Z
1
= √ limA→0 gc
A (y)f (y)dy
2π
= f (0).
In general, we let f1 (y) = f (y + x). Then
Z
Z
1
1
b
f1 (ξ)dξ = √
fˆeixξ dξ.
f (x) = f1 (0) = √
2π
2π
Using this we can prove
Theorem 9. (Plancherel Theorem): Given u ∈ S, we have
Z
Z
2
|u| dx = |û|2 dξ
Proof. In fact, using the inverse Fourier transform, we have that
Z
1
b
u(x) = √
û(y)e−ixy dy = û(x).
2π
Let g = û in the above equation. So
ū = ĝ.
Using the symmetry formula. Then we have
Z
Z
Z
Z
Z
2
|û| = ûû = ûg = uĝ = |u|2 .
One can define n-dimensional Fourier transform of f in Rn by
Z
1
ˆ
f (ξ) = √
f (x)e−ix·ξ dx.
( 2π)n
64
LI MA
Here x · ξ is the inner product in Rn . The inverse Fourier transform of
f in Rn is defined by
Z
1
ˇ
f (ξ) = √
f (x)eix·ξ dx.
n
( 2π)
Similar properties in dimension n are also true for n-dimensional
Fourier transforms. In particular, Plancherel Theorem in Rn is true.
One important fact lying behind the Fourier transform is that for the
linear operator
00
A[u] = u (x),
we have the spectrum property:
A[eixξ ] = −|ξ|2 eixξ .
EQUATIONS FROM MATHEMATICAL PHYSICS
65
7.3. Heisenberg uncertainty principle.
For any smooth function u (maybe complex-valued) vanishing at infinity with
Z
|u|2 = 1,
we have
Z
Z
1
2
2
( x |u(x)| dx)( ξ 2 |û|2 dξ) ≥ ,
4
with its equality true only at
2
u(x) = Ae−βx ,
where β > 0 and A4 = 2β/π.
The inequality above is the so called Heisenberg uncertainty principle.
Proof. Integrating by part, we have
Z
Z
Z
0
0
2
1 = |u| = −2 xu · u ≤ 2 |x||u||u |.
Using the Cauchy-Schwartz inequality we have (H):
Z
Z
0
2 2
1 ≤ 4( x u )( |u |2 ),
with equality at
0
u = λxu.
The latter condition implies that
2 /2
u = Aeλx
R
.
2
Using u = 1 we have λ = −2β and the relation between A and β.
Note that
Z
Z
0 2
|u | = ξ 2 |û|2 .
Inserting this into (H), we then get what wanted.
Notice that, as a byproduct, we have
Z
Z
Z
1
0
2
2 2 1/2
u ≤ ( x u ) ( |u |2 )1/2 ,
2
1
for every u ∈ C0 (R).
There are a lot of applications of Fourier transforms in the various
branches in sciences and technologies. I hope our readers can read more
books on Fourier transforms and Fourier analysis. Please read as more
as 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
ûtt (ξ, t) = −|ξ|2 û(ξ, t).
Solving this ODE, we get
û = û(ξ, 0) cos(|ξ|t) + ubt (ξ, 0)
sin(|ξ|t)
.
|ξ|
Using the inverse F-transform we find the solution expression:
u(x, t) = F ? (û(ξ, 0) cos(|ξ|t) + ubt (ξ, 0)
sin(|ξ|t)
).
|ξ|
Set
u(x, 0) = f (x), ut (x, 0) = g(x).
The solution can be written as
Z
1
sin(|ξ|t) iξx
u(x, t) = √
(fˆ(ξ) cos(|ξ|t) + ĝ(ξ)
)e dξ.
n
|ξ|
( 2π)
Using the relations
1
cos(|ξ|t) = (ei|ξ|t + e−i|ξ|t )
2
and
sin(|ξ|t)
1
=
(ei|ξ|t − e−i|ξ|t ),
|ξ|
2i|ξ|
we have
1
u(x, t) = √
( 2π)n
where
Z
(fˆ+ (ξ)ei(ξx+|ξ|t) + fˆ− (ξ)ei(ξx−|ξ|t) )dξ,
1
1
ĝ(ξ))
fˆ+ (ξ) = (fˆ(ξ) +
2
i|ξ|
and
1
1
fˆ− (ξ) = (fˆ(ξ) −
ĝ(ξ)).
2
i|ξ|
Using the similar approach, we can use properties of F-transform to
solve the heat equation
ut = ∆u, on Rn × R+
EQUATIONS FROM MATHEMATICAL PHYSICS
67
Take F-transform for the variable x on both sides. Then we have
ût (ξ, t) = −|ξ|2 û(ξ, t).
This is a first order ode, and we can solve it to get
û = û(ξ, 0)exp(−|ξ|2 t).
Using the inverse F-transform we have
u(x, t) = F ? (û(ξ, 0)exp(−|ξ|2 t)).
Let
gt (x) := g(x, t) =
1
2
e−|x| /4t .
(2t)n/2
Then
ĝt (ξ) = exp(−|ξ|2 t).
Given the initial data u(x, 0) = f (x) in Rn . Let
1
f ? gt (x).
u(x, t) = √
( 2π)n
Let
1
2
Ht (x) := H(x, t) =
e−|x| /4t .
n/2
(4πt)
Then
u(x, t) = f ? Ht (x).
Hence, û(ξ, t) = fˆexp(−|ξ|2 t), and ût (ξ, t) = −|ξ|2 û(ξ, t). This shows
that u = f ? Ht is the solution to the heat equation with initial data
u(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 we
have
−|ξ|2 û(ξ, y) + ûyy (ξ, y) = 0.
Solve this and we find
û(ξ, y) = û(ξ, 0)exp(−|ξ|y) = fˆ(ξ)exp(−|ξ|y).
Define
Py (x) = F ? (exp(−| · |y))(x).
Then by an elementary computation, we have
Py (x) = F ? (exp(−| · |y))(x)
Z
1
= √
e−|ξ|y+ixξ dξ
2π
Z ∞
1
= √
e−yξ+ixξ dξ
2π 0
Z 0
1
eyξ+ixξ dξ
+ √
2π −∞
1
1
1
= √ (
+
).
2π y + ix y − ix
This implies that
1
2y
Py (x) = √
.
2
2π x + y 2
Hence
1
û = √ f\
? Py ,
2π
and
1
u(x, y) = √ f ? Py (x)
2π
is the solution wanted for the problem above.
EQUATIONS FROM MATHEMATICAL PHYSICS
69
8.3. Derivative bound. We ask the following question. Given a
smooth function u ∈ S(Rn ). Assume the bound M0 of |u| and the
bound M2 of |∆u| on Rn . Can we give a bound of |∇u| on Rn in term
of M0 and M2 ? We answer this question below.
Let
−∆u + u = f, in Rn .
Using the Fourier transform, we get
(|ξ|2 + 1)û = fˆ.
Then we have
û =
Define
1
fˆ.
|ξ|2 + 1
1
1
B2 (x) = √
F ?( 2
)(x).
|ξ| + 1
( 2π)n
Then we have
u(x) = B2 ? f (x)
and
|∇u(x)| = |∇B2 ? f (x)| ≤ |∇B2 | ? |f |(x).
Using the knowledge from calculus we can show that
Z
|∇B2 |(y)dy := B < ∞.
Rn
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| + r2 max|∆u|).
Since r > 0 can be any positive number, we get
|∇u(rx)|2 ≤ 4B 2 max|u|max|∆u|.
Therefore,
max|∇u| ≤ 2B
p
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 Maximum
principle to get uniform bound for a solution and its derivatives.
Consider a smooth solution u to the following elliptic partial differential 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 the
domain D. For otherwise, assume (x0 , y0 ) is and then we have
∇u(x0 , y0 ) = 0, DT DT u(x0 , y0 ) ≤ 0,
for any direction T ∈ R2 . The latter condition implies that the matrix
D2 u(x0 , y0 ) is non-positive, that is,
D2 u(x0 , y0 ) ≤ 0.
Hence,
(uxx + uyy )(x0 , y0 ) ≤ 0,
which contradicts to (IPDE). Therefore, we have
supD u = sup∂D u.
Assume that we only have
uxx + uyy ≥ 0
in the domain D. Then we choose
w = u + eβx
for small positive constant and large positive constant β. Note that
∆eβx > 0.
Then we have
∆w > 0.
Hence, we have
supD w = sup∂D w.
Sending →, we get that
supD u = sup∂D u.
Clearly, the above argument is true for general domains in Rn . This
is the maximum principle for elliptic pde of second order.
EQUATIONS FROM MATHEMATICAL PHYSICS
71
Theorem 10. Maximum Principle. Assume u is a smooth function
satisfying
∆u ≥ 0, in D.
Then we have
supD u = sup∂D u.
Remark: (1). It is clear from our argument that the function u ∈
C 2 (D) ∩ C(D̄) satisfying
uxx uyy < 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 we
have
uxx ≤ 0, uyy ≤ 0
at (x0 , y0 ). Hence, we have
uxx uyy (x0 , y0 ) ≥ 0,
a contradiction! This implies that
supD u = sup∂D u.
The same argument applies to the differential inequality:
det(D2 u) < 0, in D.
(2). The Maximum principle is true for many nonlinear elliptic and
parabolic partial differential equations of second order. It is a basic
tool 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
let
Q = {(x, t); x ∈ D, 0 < t ≤ T },
and let
∂p Q = {(x, t) ∈ Q̄; 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 ∈ C 2,1 (Q) ∩ C(Q̄) is a solution to (HE). Here
C 2,1 (Q) is the space of continuous functions which have continuous
x-derivatives up to second order and continuous t-derivative.
Theorem 11. Assume that u ∈ C 2,1 (Q) ∩ C(Q̄) satisfies Lu ≤ 0 on
Q. Then we have
maxQ̄ u(x, t) = max∂p Q u(x, t).
Proof. We first assume that Lu(x, t) < 0 in Q. Assume that there is
an interior point (x0 , t0 ) ∈ Q such that
u(x0 , t0 ) = maxQ̄ u(x, t).
Then we have at the point (x0 , t0 ), ∇u = 0, Dx2 u ≤ 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 and
then
maxQ̄ u(x, t) = max∂p Q u(x, t).
For general case when
Lu(x, t) ≤ 0, (x, t) ∈ Q,
we let, for > 0,
v(x, t) = u(x, t) − t.
Then
Lv(x, t) = Lu(x, t) − < 0, (x, t) ∈ Q.
Then we can use the result above to v(x, t) to get
maxQ̄ v(x, t) = max∂p Q v(x, t).
Note that
maxQ̄ u(x, t) ≤ maxQ̄ v(x, t)
and
max∂p Q v(x, t) ≤ max∂p Q u(x, t) + T.
EQUATIONS FROM MATHEMATICAL PHYSICS
73
Sending → 0, we get
maxQ̄ u(x, t) ≤ max∂p Q u(x, t).
However, since ∂p Q ⊂ Q̄, we have
maxQ̄ u(x, t) ≥ max∂p Q u(x, t).
Hence,
maxQ̄ u(x, t) = max∂p Q u(x, t).
We are done.
Let’s consider an easy application of the maximum principle.
Theorem 12. Given f ∈ C(Q̄) with F := maxQ̄ |f |. Assume that
u ∈ C 2,1 (Q) ∩ C(Q̄) satisfies
Lu = f (x, t), (x, t) ∈ Q
with u(x, t) = φ(x, t) on ∂p Q. Then we have
maxQ̄ u(x, t) = F T + max∂p Q |φ(x, t)|.
Proof. Denote by
B = max∂p Q |φ(x, t)|.
Consider the function w(x, t) defined by
w(x, t) = F t + B ± u(x, t).
Then
Lw = F ± f ≥ 0, on Q
and on ∂p Q,
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 smooth
bounded domain D ⊂ Rn . Let u = u(x, t) be a non-trivial smooth
solution in C(D̄) 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,
max |u(x, t)|
x∈D
is a monotone decreasing function of t.
Proof. Consider a new function defined by
Z
Jk (t) =
|u|2k dx.
D
Recall that, for each t,
M (t) := supx∈D |u(x, t)| = lim Jk (t)1/2k .
k→+∞
First, we note that
inf limJk (t)1/2k ≤ suplimJk (t)1/2k ≤ M (t).
So we only need to show that
M (t) ≤ inf limJk (t)1/2k .
In fact, For fixed t, since u is a bounded continuous function in D̄, we
have a sub-domain Q ⊂ D̄ 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) ≤ inf limJk (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 continuous function defined on a unbounded domain D.
EQUATIONS FROM MATHEMATICAL PHYSICS
75
Note that
Z
Z
dJk
2k−1
= 2k u
ut = 2k u2k−1 ∆u.
dt
By using integration by part, we have
Z
dJk
= −2k(2k − 1) u2k−2 |∇u|2 < 0.
dt
Hence Jk (t) is a monotone decreasing function of t. Therefore,
max |u(x, t)|
x∈D
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 that
M (t) ≤ M (0)e2at .
Remark 3. Note that for any bounded
continuous function f defined
R
p
on a unbounded domain D with D |f | dx < ∞ for some p > 0, we still
have
Z
M := supD |f (x)| = lim ( |f |k dx)1/k .
k→∞
D
According to the remark made in the proof of the theorem above, we
only need to prove that
Z
suplimk→∞ ( |f |k dx)1/k ≤ M.
D
However, it is almost trivial since
Z
Z
k
1/k
k−p
suplimk→∞ ( |f | dx) ≤ suplimk→∞ (M
|f |p dx)1/k = M.
D
D
With an easy modification, we can extend the above theorem into
unbounded 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 concept
of variational structure. We discuss the variational method for Laplace
equation in a region of dimension two.
Given a bounded domain D ⊂ Rn . We introduce the Dirichlet energy
(so called the variational structure)
Z
I(u) =
|∇u|2 dx,
D
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 ∈ A
such that
I(w) = infu∈A I(u).
Assume such a function w exists. Then for every ξ ∈ C01 (D), we have
w + tξ ∈ A
for every t ∈ R. So we have, for all t ∈ R,
I(w) ≤ I(w + tξ)
and then
d
I(w + tξ)|t=0 = 0.
dt
Note that
Z
I(w + tξ) =
[|∇w|2 + 2t∇w · ∇ξ + t2 |∇ξ|2 ].
Hence,
Z
d
I(w + tξ)|t=0 = 2 ∇w · ∇ξ = 0.
dt
Assume that w ∈ C 2 . Then using integration by part, we have
Z
(∆wξ) = 0.
Since ξ can be any testing function in C01 (D), we get the Euler-Lagrange
equation for the functional I(·), by the variational lemma (see our
textbook [1]), that
∆w = 0, in D,
EQUATIONS FROM MATHEMATICAL PHYSICS
77
with the boundary condition w = φ on ∂D. This is the famous Dirichlet
problem in the history. The recent study is about the obstacle problem
for harmonic functions.
Remark: We mention that the existence of the infimum w can be
proved by using the Hilbert space theory, which is a beautiful part of
the course Functional Analysis. It will be really helpful for readers
to know some knowledge about Sobolev spaces. Here are some simple
facts about it. Let I = [a, b]. We consider L2 (I) as the completion of
the space C0∞ (I) with respect to the norm
Z
|f | = ( |f (x)|2 dx)1/2 .
I
In the other word, the element f (which will be called a L2 -function)
in L2 (I) is defined by a Cauchy sequence {fk } ⊂ C0∞ (I), i.e.,
|fk − fj | → 0
as k, j → ∞. We say that the element f ∈ L2 (I) has derivative
g ∈ L2 (I) if there is a Cauchy sequence {fk } ⊂ C0∞ (I) such that
fk → f, and (fk )x → g
2
in L . If this is true, we often call g the strong derivative of f and
denote by g = fx . Then is easy to see that the integration by part
formula
Z
Z
f φx dx = −
gφdx,
I
I
for all φ ∈ C0∞ (I), is true. People often use this formula to define the
weak derivative of f . It can be showed that the strong derivative is the
same weak derivative. Furthermore, one can show that both definitions
are equivalent each other. Hence, we simply call g the derivative of f .
We denote by H01 (I) the Sobolev space which consists of L2 functions
with derivatives. For f, h ∈ H01 (I), we define
Z
(f, h) = fx hx dx.
I
Then one can show that (·, ·) is an inner product on H01 (I), and H01 (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 section
that readers know the Banach fixed point theorem. We recall the Fixed
Point Theorem: Let E be a Banach space and let B ⊂ E be a closed
convex subset. Let f : B → B be contraction mapping, i.e., there is a
positive 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 | · | is
the norm of E.
One often use Fixed Point Theorem to get the local existence of
solutions to non-linear evolution problem. Let’s see a little more. We
study the evolution problem of type
ut = ∆u + F (u)
or
utt = ∆u + F (u)
in R with nice initial data. Using the Duhamel principle, we can
reduce the problem into the form:
Z t
S(t − τ )F (u)dτ := f (u).
u = S(t)(u0 ) +
n
0
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, there
exists 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 to
take E to be defined by the energy method such that E is a Banach
algebra. Then we can use the interpolation inequality method to show
that f is in fact a contraction mapping.
The interesting feature in non-linear problems is that there is a blow
up 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
a
u(t) =
,
1 − ta
EQUATIONS FROM MATHEMATICAL PHYSICS
79
+
which goes to infinity as t → a1 . 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 the
integration like
p(t) = minD u(x, t)
or
Z
p(t) := ρ(x)u(x, t)k
or
Z
p(t) :=
ρ(x)|∇x u(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, which
makes p(t) go to negative as t → +∞. This is absurd since p(t) > 0
for all t. This gives us the blow up property.
Example Two: We consider the following nonlinear parabolic equation
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 have
0
p (t) ≥ p(t)2 .
This gives us that
a
1 − ta
for a = minx∈[0,2π] φ(x) > 0 and u has to blow up to +∞ in finite time
before a−1 .
Example Three: Given a constant β ≥ 14 . We consider the following
nonlinear parabolic equation
p(t) ≥
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 the
Maximum principle, we have
u(x, t) > 0, x ∈ [0, 2π] × {t > 0}.
80
LI MA
Choose C =
1
4
such that
Z
2π
C
0
Let
Z
p(t) = C
0
x
sin( ) = 1.
2
2π
x
sin( )u(x, t)dx.
2
Then, we have
p
0
Z
2π
x
sin( )ut
2
0
Z 2π
Z 2π
x
x
sin( )∆u + C
sin( )u2 + βp(t)
= C
2
2
0
0
Z 2π
1
x
= C
sin( )u2 + (β − )p(t)
2
4
0
2
≥ p(t) .
= C
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 Riemannian
manifolds.
EQUATIONS FROM MATHEMATICAL PHYSICS
81
10.3. Nonlinear PDE of first order.
This is a fun model for non-linear pde of first order.
Let us first consider a nice example:
u2x + u2y = 1,
which is called the eikonal equation. Introduce the curve (x(t), y(t))
by
dx
dy
= ux ,
= uy .
dt
dt
Then by the equation, we have
dx
dy
( )2 + ( )2 = 1.
dt
dt
This implies that t is the arc-length parameter for the curve. Doing
differentiation to the eikonal equation, we have
ux uxx + uy uyx = 0.
Hence along the curve,
d2 x
dx
dy
= uxx
+ uxy
= 0.
2
dt
dt
dt
Similarly, we have
d2 y
= 0.
dt2
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.
So
u = 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):
X
Ai (x, u)uxi = B(x, u).
i
We look at the graph
z = u(x).
Then the PDE says that the normal vector (∇u, −1) of the graph is
orthogonal 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
dz
= Ai (x, z),
= B(x, z).
dt
dt
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)), then
u = 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].
EQUATIONS FROM MATHEMATICAL PHYSICS
83
11. Appendix A
One may consider this section as a last one to our course. We have
three parts. One is for the variational lemma. Another is about the
method of separation of variables. The other is about some open problems in PDE’s. Our readers may do research for these problems in the
future.
11.1. Variational lemma.
Lemma 14. Let f ∈ C 0 (D) for a bounded domain in Rn . Assume
that
Z
f (x)ξ(x)dx = 0
D
for every ξ ∈
C0∞ (D).
Then f = 0.
Proof. We argue by contradiction. Assume that there is a point x0 ∈ D
such that f (x0 ) 6= 0. Then we may assume that f (x0 ) > 0 and there is
a small ball B (x0 ) such that f (x) 6= 0 for every x ∈ B (x0 ). Take
ξ(x) = ρ(|x − x0 |)
where
−
ρ(r) = e
1
1−r 2
, 0≤r<
and ρ(r) = 0 for all r ≥ . Note that ρ(|x − x0 |) ∈ C0∞ (D), and
f (x)ρ(|x − x0 |) > 0, in B (x0 ).
Hence,
Z
Z
f (x)ρ(|x − x0 |) > 0.
f (x)ξ(x)dx =
D
C0∞ (D)
This gives a contradiction to our assumption of our lemma. Therefore,
the lemma is true.
Example. Assume that u = u(t) ∈ C02 [0, L]. Define
Z L
1
1
F (u) =
( |u0 |2 − u3 − g(t)u(t))dt,
2
3
0
where g(t) is a given continuous function on [0, L]. Then, if δF (u) = 0,
u satisfies
00
−u = u(t)2 + g(t).
In fact, by definition,
Z
0 =< δF (u), φ >=
0
L
(u0 φ0 − u2 φ − g)dt
84
LI MA
for all φ ∈ C0∞ (0, L). Integrating by part, we get
Z L
(−u00 − u2 − g)φdt.
0=
0
By variational lemma, we know that
u00 + u2 + g = 0,
which is equivalent to our equation above.
EQUATIONS FROM MATHEMATICAL PHYSICS
85
11.2. Sturm-Liouville theory. In the old time before Fourier, people
often 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 be
expressed as the expansion of trigonometric functions. The answer is
yes since we know the beautiful theory of Fourier series.
In the method of separation of variables, we meet the problem if we
can use the eigenfunctions of the following eigenvalue problem

0
 (p(x)X )0 + [λ + q(x)]X = 0, 0 < x < L,
−a1 X 0 (0) + b1 X(0) = 0),

a2 X 0 (L) + b2 X(L) = 0,
where the constants satisfy ai ≥ 0, bi ≥ 0 and ai + bi 6= 0, and the
functions 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 an
increasing eigenvalues λn → ∞ and eigenfunctions Xn (corresponding
to λn ) with
Z L
Z L
Xn Xm dx = 0,
|Xn |2 dx = 1
0
0
for λn 6= λn , such that every smooth function f with the boundary
condition
−a1 f 0 (0) + b1 f (0) = 0),
a2 f 0 (L) + b2 f (L) = 0,
can be expressed as
X
Cn Xn (x).
f (x) =
n
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
∂
u
∂xn
> 0 on Rn , and
lim u(x0 , xn ) = ±1,
xn →±∞
for any x0 ∈ Rn−1 , then all level set of u are hyper-planes, at least for
n ≤ 8.
The case when n ≤ 8 was solved recently by Savin, who is a PhD
student of L. Caffarelli. For more reference, please see our paper:
Y.H.Du and Li Ma; Some remarks related to the De Giorgi conjecture, Proc. AMS, 131(2002)2415-2422.
2 Serrin conjecture
In 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 = v q ,
−∆v = up ,
where the indices p > 0, q > 0 satisfy
1
1
2
+
>1− .
p+1 q+1
n
We have some progress in this problem. Please see our paper
Li Ma and D.Z.Chen, A Liouville type Theorem for an integral System, CPAA, 5(2006)855-859.
3. A question from Yanyan Li and L.Nirenberg
It is not hard to show by using the Taylor expansion that, if a nonnegative function satisfying
00
|u (x)| ≤ M
on [−R, R], then we have
0
|u (0)| ≤
for M ≥
2u(0)
,
R2
0
2u(0)
.
R2
2u(0)M
and
|u (0)| ≤
for M <
p
u(0) R
+ M
R
2
EQUATIONS FROM MATHEMATICAL PHYSICS
87
Yanyan Li and L.Nirenberg (2004) asked what is the best constant C
in the following result in the ball BR ⊂ Rn :
Assume that u is a non-negative C 2 function in closure of the ball
BR satisfying:
max |∆u| = M.
BR
Then there is a constant C depending only on n such that
p
|∇u(x)| ≤ C u(0)M
q
≥ 2|x|,
if R ≥ u(0)
M
|∇u(x)| ≤ C(
q
.
if ≥ 2|x| ≥ R < u(0)
M
This is an open question.
u(0)
+ RM )
R
88
LI MA
12. Appendix B
This is a section for part of homework exercises in my courses in the
past 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 position
forms a triangle with the x-axis such that the middle point of the
string is the highest point and its height is b > 0. Find the determined
problem for this string.
Answer: Let u be the height position of the vibrating string at
(x, t). Then u satisfies
utt = a2 uxx , f or (x, t) ∈ (0, A) × (0, ∞)
with the initial data
2b
A
2b
A
u(x, 0) = x, x ∈ [0, ] and u(x, 0) = (A − x), x ∈ [ , A]
A
2
A
2
and the boundary data u(0, t) = 0 and u(A, t) = 0. The relation
between L, b, and A is
A2
L2
=
+ b2 .
4
4
Exercise 2. Assume 0 6= x ∈ Rn , n ≥ 2. Verify that
∆|x|β = (nβ + β(β − 2))|x|β−2 .
i
Answer: Note that ∂xi r = xr , for r = |x|.
Exercise 3. Assume 0 < u = u(x) ∈ C 2 . Denote by
X
|∇u|2 =
|∂i u|2 .
i
(1). Assume that f : R → R is in C 2 . Prove that
00
∆f (u) = f 0 (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
Z ∞
Γ(s) =
e−r rs−1 dr.
0
Show that the definition is well-defined, and Γ(s +1) = sΓ(s) for s > 0.
EQUATIONS FROM MATHEMATICAL PHYSICS
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, ∞)
2
with initial data u = 2xex /2 at t = 0.
Answer: Note that along the characteristic curve
t
x = − + A,
2
we have
du
t
= −xu = (− + A)u.
dt
2
Then,
2
u(x, t) = (2x + t)ex /2 .
Exercise 4. Solve the problem
y 2 uyy − x2 uxx = 0
with the initial data
u(x, 1) = φ(x), uy (x, 1) = ψ(x).
Answer:
y x
1
φ(xy) + φ( )
2
2 y
√ Z x
xy y − 3
+
s 2 φ(s)ds
4
xy
√ Z x
xy y − 3
−
s 2 ψ(s)ds.
2
xy
u(x, y) =
89
90
LI MA
12.3. Homework 3.
Exercise 1. Assume that Y is a solution to
00
−y + y = 0, x > 0
with the initial conditions y(0) = 0 and y 0 (0) = 1. Using the Duhamel
principle to find the expression for solution to the the following ODE
00
−y + y = g(x), x > 0
with the initial conditions y(0) = A and y 0 (0) = B.
Answer:
Z
0
x
g(τ )Y (x − τ )dτ.
y(x) = AY (x) + BY (x) −
0
Exercise 2. Given constants a1 , ..., ak . Assume Y (x) is a solution to
the the higher order ODE of the form
y (k) + a1 y (k−1) + ...ak y = 0
with the initial data
y(0) = 0, y 0 (0) = 0, ... y (k−1) (0) = 1.
Find the solution expression for the higher order ODE of the form
y (k) + a1 y (k−1) + ...ak y = g(x)
with the initial data
y(0) = b0 , y 0 (0) = b1 , ... y (k−1) (0) = bk−1 .
Answer: Define, for 0 ≤ j ≤ k − 1,
uj (x) = Y (j) (x) + a1 Y (j−1) (x) + ... + aj Y (x).
Then the solution expression is
Z
X
y(x) =
bj uj (x) +
j
0
x
g(τ )Y (x − τ )dτ.
EQUATIONS FROM MATHEMATICAL PHYSICS
91
12.4. Homework 4.
Exercise 1. Find a solution to the following Sturm-Liouville problem:
00
X + λX = 0, x ∈ (0, L),
and X(0) = X 0 (L) = 0.
Answer:
λk =
(k + 21 )2 π 2
L2
and
X = 0, or Xk (x) = B sin(
(k + 12 )π
x)
L
where k = 0, 1, 2.....
Exercise 2. 
 utt = a2 uxx , 0 < x < L, t > 0,
u(0, t) = 0 = ux (L, t),

u(x, 0) = sin2 πx
, ut (x, 0) = x(L − x).
L
Answer:
u(x, t) =
X
n≥1
(An cos(
anπ
anπ
nπ
t) + Bn sin(
t)) sin( x),
L
L
L
where
8
− 4)π
for n = 2k + 1, and An = 0 for n = 2k, and
8L3
Bn = 4 4
an π
for n = 2k, and Bn = 0 for n = 2k + 1,
An = −
n(n2
92
LI MA
12.5. Homework 5.
Exercise 1. Given a domain Ω ⊂ R2 . Consider the following problem:
−∆u = f (x, y), (x, y) ∈ Ω
with the boundary condition u = φ on ∂Ω. Find the Green function
for 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)) =
∞
X
G1 ((x, y), (ζ, 2kL + η)).
k=−∞
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 our
normal is ν = (0, −1).
? Exercise 3. Let B = BR (0) and assume φ ∈ C(∂B). Define
Z
R2 − |x|2
φ(y)
u(x) =
dσy ,
n
nωn R
∂B |x − y|
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 ∈
C (D) with
∆u(x) = f (x), ; x ∈ D.
2
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 I 2 = identity, that is, I(I(x)) = x for x ∈ D. This
implies that D = I(D? ).
EQUATIONS FROM MATHEMATICAL PHYSICS
93
(2). Let
w(x) = |x|2−n u(x/|x|2 ), x ∈ D?
which is called the Kelvin transform of the function u. Then
∆w(x) = |x|2−n f (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−n Γ(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:
r
2 sin(Bξ)
F (f )(ξ) =
.
B
ξ
Exercise 2. Given two constants a > 0, c and two bounded smooth
functions f (x, t) and φ(x). Using the Fourier transform to solve the
following problem
ut = a2 uxx + cu + f (x, t), −∞ < x < ∞, t > 0,
u(x, 0) = φ(x).
Answer:
Z
u(x, t) = e [ K(x − ξ, t)φ(ξ)dξ
Z
Z t
−cτ
e dτ K(x − ξ, t − τ )f (ξ, τ )dξ].
+
ct
0
We can obtain this result by two methods.
Method One: Let w = e−ct u and f1 (x, t) = e−ct f (x, t). Then we have
wt = a2 wxx + f1
with
w(x, t) = φ(x).
Method Two: Note that for û(ξ, t),
ût = (−a2 |ξ|2 + c)û + fˆ(ξ, t),
with
û(ξ, 0) = φ̂(ξ).
? Exercise 3. Define, for u ∈ S,
d2
+ x2 )u(x),
dx2
d
A[u](x) = ( + x)u(x),
dx
d
C[u](x) = (− + x)u(x).
dx
H[u](x) = (−
EQUATIONS FROM MATHEMATICAL PHYSICS
95
Show that
CA[u] = (H − 1)[u],
AC[u] = (H + 1)[u],
H(e−x
H(C k (e−x
2 /2
2 /2
) = e−x
2 /2
)) = (2k + 1)C k (e−x
2 /2
), k = 1, 2, ....
−x2 /2
We remark that C k (e
) are eigenfunctions of the operator H and
the polynomials defined by
hk (x) = ex
2 /2
C k (e−x
2 /2
)
are called the Hermite functions. H is called the Hermite operator,
which plays an important role in modern methods of mathematical
physics 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
∞
X
U (t) =
u(t + 2πn),
n=−∞
which is called the periodization of u. Clearly, U (t) is a periodic
function such that U (t + 2π) = U (t) on R. Denote by û the Fourier
transform of u. Using Fourier series theory, you show that
∞
√ −1 X
U (t) = ( 2π)
û(n)eint ,
n=−∞
which is called the Poisson summation formula. Then setting t = 0,
we get
∞
∞
X
√ −1 X
u(n) = ( 2π)
û(n).
−∞
−∞
? Exercise 5. Recall that the theta function ϑ(t), t > 0, is defined
by
∞
X
2
ϑ(t) =
e−πn t .
n=−∞
Using the Poisson summation, you show that
t−1/2 ϑ(1/t) = ϑ(t).
? Exercise 6. Assume that
B̂(ξ) =
1
.
1 + |ξ|2
96
LI MA
Find B(x).
Answer: First we write
Z ∞
1
2
e−s(1+|ξ| ) ds.
=
2
1 + |ξ|
0
Then we have
Z
Z ∞
1
2
−s
eix·ξ−s|ξ| dξ.
e ds
B(x) = √
( 2π)n/2 0
Rn
Recall that
Z
a2
π
2
eiay−sy dy = ( )1/2 e− 4s .
s
R
Then we have
Z
|x|2
π
2
eix·ξ−s|ξ| dξ = ( )n/2 e− 4s .
s
Rn
Hence, we have
Z ∞
|x|2
−n/2
B(x) = 2
s−n/2 e−s− 4s ds,
0
which is called the second Bessel potential.
? Exercise 7. Given β > 0. Assume that
1
B̂β (ξ) =
.
(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.
EQUATIONS FROM MATHEMATICAL PHYSICS
97
12.7. Homework 7.
Exercise 1. Let D be a bounded domain in Rn . Consider u ∈
C 2 (D) ∩ C(D) such that
−∆u = f (x), x ∈ D
u(x) = φ(x), x ∈ ∂D.
Then
maxD |u(x)| ≤ max∂D |φ(x)| +
d
sup|f (x)|,
2n
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 2
(d − |x − p|2 ) + M.
2n
Note that
ξ(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
d
maxD |u(x)| ≤ max∂D |φ(x)| + sup|f (x)|.
2n
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 ∈ C 2,1 (Q) satisfies

 ut = a2 uxx + f (x, t), 0 < x < L, 0 < t ≤ T,
u(0, t) = 0 = ux (L, t),
 u(x, 0) = φ(x)
Show that maxQ̄ |ut (x, t)| can be bounded by the suitable norm of f
and φ.
Answer: There is a uniform constant C = C(T, a) such that
00
maxQ̄ |ut | ≤ C[|f |C 1 (Q̄) + |φ |C[0,L] ].
Exercise 2. Given φ ∈ C(D̄) and a(x, t) ∈ C(Q̄) with A = maxQ̄ |a(x, t)|.
Assume that u ∈ C 2,1 (Q) ∩ C(Q̄) satisfies
Lu := ut − a2 ∆u = −u2 + a(x, t)u, (x, t) ∈ Q
and 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) ≤ M supD |φ(x)|
for some constant M > 0 depending only on n, T, and A.
Proof. Assume that u < 0 somewhere, saying,
u(x1 , t1 ) = −(t1 + 2 )2
for some arbitrary small and t1 > 0. Consider
w = u + t + 2 .
At (x1 , t1 ), we have
w(x1 , t1 ) = 0.
Note that
∆w = ∆u
and at t = 0, w = φ + > 0. Hence, there is a small 0 < t0 ≤ t1
such that w(x, t) ≥ 0 for (x, t)D̄ × (0, t0 ] and for some minimum point
x0 ∈ D for w such that
2
w(x0 , t0 ) = 0, ∆u(x0 , t0 ) = ∆w(x0 , t0 ) ≥ 0,
which implies that
u(x0 , t0 ) = −(t0 + 2 ).
Hence, we have
Lw(x0 , t0 ) ≤ 0,
EQUATIONS FROM MATHEMATICAL PHYSICS
99
which gives us that
Lu(x0 , t0 ) ≤ −,
Then, using the equation for u at (x0 , t0 ),
− ≥ (t0 + 2 )2 − a(x0 , t0 )(t0 + 2 ).
However, this is impossible for small and t1 . Hence, we have
u ≥ 0, on Q.
Once we have u ≥ 0, we can use the maximum principle 2 (see
Section 9.3) to conclude that
u(x, t) ≤ M supD φ(x).
Remark. One may try another method below. Assume that m :=
u(x2 , t2 ) = maxQ̄ u > 2supD φ. Then at (x2 , t2 ), we have
Lu ≥ 0.
Using the equation, we get
−m2 + a(x2 , t2 )m ≥ 0.
Hence, we have
m2 ≤ 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) ∈ C 1 (R × (0, T )) is a bounded function such that
limx→±∞ u(x, t) = 0
uniformly in t and
∂t F (u) + ∂x G(u) = 0.
Here ∂x G(u) = G (u)ux . Show that
Z
d
F (u)dx = 0.
dt R
R
We call the quantity R F (u)dx the conservation law.
Exercise 2. Let G be a vector-valued smooth function in R. Assume
that u(x, t) ∈ C 1 (Rn × (0, T )) is a bounded function such that
0
lim|x|→∞ u(x, t) = 0
uniformly in t and
∂t u + divx G(u) = 0.
Show that
d
dt
Z
uk dx = 0
Rn
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
Z
En (u) =
2
(n−1) 2
([u(n)
] − qn (u, ..., u(n−2)
)dx.
x ] + cn u[un
x
R
(n)
for suitable constants cn and polynomials qn . Here ux 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 heat
equation
wt − wxx = 0.
−cu
Answer: w = e , which is called the Cole-Hopf transformation.
? Exercise 5. Consider the Burgers’ equation
ut − uxx + uux = 0, x ∈ R
EQUATIONS FROM MATHEMATICAL PHYSICS
101
with initial data u = f (x). Let
Z
x
w(x, t) =
u(y, t)dy.
−∞
Show that w satisfies
1
wt − wxx + wx2 = 0
2
and use the conclusion of Exercise 4 to find an explicit expression of
u.
Answer:
R
f (y)
−|x−y|2
− 02
4t
dy
(x
−
y)e
R
u(x, t) =
,
R −|x−y|2 − f0 (y)
2 dy
t R e 4t
Rx
where f0 (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
Z p
A(u) =
1 + |∇u|2 dx
D
among the class A above.
Answer: From this variational structure, we know the famous
minimal surface equation:
−div( p
∇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. Assume u = u(x, y) = f (x) + g(y) satisfies
uy
ux
) + ∂y ( p
) = 0.
∂x ( p
2
2
1 + ux + uy
1 + u2x + u2y
Then we have
00
00
u(xu) =
1
cos(Ay)
log|
|.
A
cos(Ax)
f
g
+
= 0.
1 + f 0 (x)2 1 + g 0 (y)2
Prove that for some constant A > 0,
This solution is called the Sherk’s surface.
p Exercise 3. Find solution u(x, y) of the form u = f (r), r =
x2 + y 2 , to the equation
uy
ux
∂x ( p
) + ∂y ( p
) = 0.
2
2
1 + ux + uy
1 + u2x + u2y
Answer:
1
cosh(Ar + B).
A
Exercise 4. Find the Euler-Lagrange equation for the integral
Z
I(u) =
L(x, u, ux1 , ..., uxn )dx.
f (y) =
D
Answer:
∂u L −
n
X
j=1
∂xj (∂uxj L) = 0.
EQUATIONS FROM MATHEMATICAL PHYSICS
? Exercise 5. Find the Euler-Lagrange equation for the integral
Z
J(u) =
|∇u|2 dx
D
on the space
C02 (D)
with the constraint
Z
|u|2 dx = 1.
D
Answer:
−∆u = λu.
103
104
LI MA
Summary
This is a brief lecture note for the courses on ”Introduction to Equations from Mathematical Physics” or ”Methods for Mathematical Physics”.
Our motivation to write this note is to show the mainstream methods
used in solving some typical partial differential equations like wave
equation, heat equations, and the Laplace equation.
For students who would like to take the final examination of this
course, 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 Mathematical Physics,in Chinese, Higher education press, Beijing,1999.
[2] D.Hilbert, R.Courant, Methods of Mathematical Physics, Vol. 2. Interscience Publicaters, 1962.
EQUATIONS FROM MATHEMATICAL PHYSICS
105
Index
Bessel function, 28
blow up, 60
Characteristic curve method, 13
continuity equation, 9
D’alembert, 21
Dirac’s delta function, 39
divergence, 5
Duhamel principle, 17
Fourier, 24
fundamental solution, 35
Gaussian kernel, 36
Green, 5, 42
Gronwall, 18
harmonic function, 41
heat equation,11
Heisenberg, 51
Hermite operator, 90
Huygens, 30
Inversion transform, 87
Kelvin transform, 88
Kirchhoff, 29
Laplacian, 5, 11
Liouville, 67
maximum principle, 51
mean value theorem, 44
monotonicity formula, 47
partial differential operator, 10
Plancherel, 59
Poisson, 42
Schwartz, 48
string,7
Sturm, 67
support, 4
variational lemma, 65
wave equation, 7
Department of mathematical sciences, Tsinghua university, Beijing 100084,
China
E-mail address: [email protected]