MA3G1: Theory of PDEs
Dr. Manh Hong Duong
Mathematics Institute, University of Warwick
Email: [email protected]
January 4, 2017
Abstract
The important and pervasive role played by pdes in both pure and applied mathematics
is described in MA250 Introduction to Partial Differential Equations. In this module I
will introduce methods for solving (or at least establishing the existence of a solution!)
various types of pdes. Unlike odes, the domain on which a pde is to be solved plays an
important role. In the second year course MA250, most pdes were solved on domains
with symmetry (eg round disk or square) by using special methods (like separation of
variables) which are not applicable on general domains. You will see in this module the
essential role that much of the analysis you have been taught in the first two years plays
in the general theory of pdes. You will also see how advanced topics in analysis, such
as MA3G7 Functional Analysis I, grew out of an abstract formulation of pdes. Topics in
this module include:
• Method of characteristics for first order PDEs.
• Fundamental solution of Laplace equation, Green’s function.
• Harmonic functions and their properties, including compactness and regularity.
• Comparison and maximum principles.
• The Gaussian heat kernel, diffusion equations.
• Basics of wave equation (time permitting).
Aims: The aim of this course is to introduce students to general questions of existence, uniqueness and properties of solutions to partial differential equations.
Objectives: Students who have successfully taken this module should be aware of
several different types of pdes, have a knowledge of some of the methods that are used for
discussing existence and uniqueness of solutions to the Dirichlet problem for the Laplacian,
have a knowledge of properties of harmonic functions, have a rudimentary knowledge of
solutions of parabolic and wave equations.
Further reading. The contents of this course are basic and can be found in many
textbooks. The bibliography lists some books for further reading.
Contents
1 Introduction
3
1.1
What are partial differential equations . . . . . . . . . . . . . . . . . . . .
3
1.2
Examples of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3
Well-posedness of a PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.4
Classification of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2 Quasilinear first order PDEs
10
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2
The method of characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.1
The characteristic equations . . . . . . . . . . . . . . . . . . . . . . 13
2.2.2
Geometrical interpretation . . . . . . . . . . . . . . . . . . . . . . . 14
2.3
Compatibility of initial conditions . . . . . . . . . . . . . . . . . . . . . . . 18
2.4
Local existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . 19
2 Laplace’s equation and harmonic functions
23
2.1
Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2
Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.3
Mean value property for harmonic functions . . . . . . . . . . . . . . . . . 26
2.4
Smoothness and estimates on derivatives of harmonic functions . . . . . . . 28
2.5
The maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.5.1
Subharmonic and superharmonic functions . . . . . . . . . . . . . . 32
2.5.2
Some properties of subharmonic and superharmonic functions . . . 33
1
2
2.5.3
2.6
The maximum principle . . . . . . . . . . . . . . . . . . . . . . . . 33
Harnack’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3 The Dirichlet problem for harmonic functions
42
3.0.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.0.2
Representation formula based on the Green’s function . . . . . . . . 44
3.1
The Green function for a ball . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.2
C 0 -subharmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.3
Perron’s method for the Dirichlet problem on a general domain . . . . . . . 57
4 The theory of distributions and Poisson’s equation
63
4.1
The theory of distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.2
Convolutions and the fundamental solution . . . . . . . . . . . . . . . . . . 68
4.3
Poisson’s equation
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3.1
The fundamental solution . . . . . . . . . . . . . . . . . . . . . . . 74
4.3.2
Poisson’s equation in a bounded domain . . . . . . . . . . . . . . . 76
5 The heat equation
80
5.1
The fundamental solution of the heat equation . . . . . . . . . . . . . . . . 80
5.2
The maximum principle for the heat equation on a bounded domain . . . . 86
5.3
The maximum principle for the heat operator on the whole space . . . . . 88
Acknowledgements
94
Appendix
95
Chapter 1
Introduction
1.1
What are partial differential equations
A partial differential equation is an equation which involves partial derivatives of some
unknown function. PDEs are often used to describe a wide variety of phenomena such as
sound, heat, electrostatics, electrodynamics, fluid flow, elasticity, or quantum mechanics.
Definition 1 (PDE and classical solutions). Suppose that Ω ⊂
Rd
is some open set. A
partial differential equation (PDE) of order k is an equation of the form
h
i
F x, u(x), Du(x), D2 u(x), · · · , Dk u(x) = 0.
In the expression above, F : Ω × R × Rd × Rd × Rd is a given function; u : Ω →
2
k
(1.1)
R is
the unknown function and Dk u contains all the partial derivatives of u of order k, see
Appendix 5.3. A function u ∈ C k (Ω) is said to be a classical solution to the PDE on
the domain Ω if on substitution of u and its partial derivatives into (1.1) the relation is
identically satisfied on Ω.
Traditionally, PDEs are often derived by using conservation laws and constitutive laws.
Another method to derive PDEs, which have been developed recently, is to use operators
and functionals. The derivations of basic equations (transport equation, Laplace equation,
heat equation and wave equations) can be found in any textbook on the introduction of
PDEs (e.g., the the lecture notes of the course MA250 Introduction to Partial Differential
Equations). Therefore, we will not go into details on those in this module.
3
4
1.2
Examples of PDEs
PDEs are ubiquitous and appear in almost all fields in applied sciences. We list here
just some popular examples. Some of these will be treated in the subsequent chapters.
Example 1 (The continuity/transport equation). In general, a continuity equation describes the transport of some moveable quantity. For example in fluid dynamics, the
continuity equation reads
∂u
(x, t) + div(u(t, x)v(t, x)) = 0,
∂t
(1.2)
where u(x, t) is the fluid density at a point x ∈ R3 and at a time t, v(x, t) is the flow velocity
vector field and div denotes the divergence operator. Note that the above equation can
be written as
∂u
(x, t) + Du(t, x) · v(t, x) + u(t, x) div(v(t, x)) = 0,
∂t
Example 2 (The Laplace and Possion equations). The Laplace equation is a second-order
partial differential equation given by
∆u =
d
X
∂ 2u
i=1
∂x2i
= 0.
(1.3)
We often want to solve this equation in some bounded domain Ω ⊂ Rd and therefore, some
boundary conditions need to be provided. Now let us give an example of application (or
actually a derivation of this equation). We consider a solid body at thermal equilibrium
and denotes by u(x) its temperature. Let Q denotes the heat flux. Since the body is at
equilibrium, there is no flux through the boundary ∂Ω for any smooth domain Ω
ˆ
ˆ
0=
Q · dσ =
div(Q) dx,
∂Ω
Ω
where we have used the divergence theorem to obtain the second equality. This implies
that div Q must vanish identically
div Q = 0.
By Fourier’s law of heat conduction, the heat flux is proportional to the temperature
gradient, so that Q = k(x)Du(x), where k is the conductivity. If the body is homogeneous,
then k is constant and we deduce that
0 = div Q = div(kDu(x)) = k∆u(x).
5
Thus the temperature satisfies the Laplace equation.
In conclusion: the Laplace equation can be used to describe the temperature distribution of a solid body at thermal equilibrium.
The Poisson equation is the Laplace equation with an inhomogeneous right-hand side,
∆u(x) = f (x),
(1.4)
for some given function f . For instance, the Poisson equation appears in electrostatics.
For a static electric field E, Maxwell’s equations reduce to
curl E = 0,
div E = ρ,
where ρ is the charge density (in appropriate units). By Helmholtz decomposition, the first
equation implies that E = −Du, for some scalar electric potential field u. By substituting
this into the second equation, we obtain the Possion equation (1.4) with f = −ρ.
Example 3 (The heat/diffusion equation). In example 2, if the (homogeneous) body is
not in equilibrium, the the heat flux through the boundary of a region will equal the rate
of change of internal energy of the material inside that region. If no work is done by the
heat flow, then the rate of change of the internal energy is
ˆ
ˆ
ˆ
ˆ
dE
∂u
=
(t, x) dx =
Q · dσ =
div(Q) dx = k ∆u(t, x) dx.
dt
Ω ∂t
∂Ω
Ω
Ω
This implies that the temperature u satisfies the heat equation
∂u
(t, x) = k∆u(t, x).
∂t
(1.5)
This equation also can be used to describe other diffusive processes such as a diffusion
process (and called diffusion equation).
Example 4 (The wave equation). The wave equation is a second-order linear partial
differential equation
∂ 2u
(t, x) = c2 ∆u(t, x).
∂t2
(1.6)
Here u is used to model, for example, the mechanical displacement of a wave. The wave
equation arises in many fields like acoustics, electromagnetics, and fluid dynamics.
6
Example 5 (The Kramers equation). The Kramer equation (or kinetic Fokker-Planck
equation) describes the time evolution of the probability density of a particle moving
under the influence of an external potential, a friction and a stochastic noise given by
p
p
∂t ρ +
· ∇q ρ − ∇V · ∇q ρ = γ divp
ρ + β −1 ∇p ρ ,
m
m
where m is the mass of the particle, γ is the friction coefficient, β −1 is the temperature.
Another important equation of the same kind is the Boltzmann equation
∂t ρ +
p
· ∇q ρ − ∇V · ∇q ρ = Q(ρ, ρ)
m
where the right hand side describes the effect of collisions between particles, which is a
complicated formula.
Example 6 (A convection and nonlinear diffusion equation). The following equation
describes the time evolution of the probability density function, i.e., for each t ∈ [0, T ],
ρ(t) is a probability measure on
Rd , of some quantity,
∂t ρ = div ρ∇[U 0 (ρ) + V ] ,
where U , V are known as the internal and external energy functionals. In particular,
when U (ρ) = ρ log ρ, we obtain the Fokker-Planck equation, which is linear,
∂t ρ = ∆ρ + div(ρ∇V ).
When V = 0, U (ρ) =
1
ρm ,
m−1
we obtain the porous medium equation
∂t ρ = ∆ρm .
Example 7 (The Allen-Cahn equation). The Allen-Cahn euqation is a reaction-diffusion
equation and describes the process of phase separation
∂t ρ = D∆ρ − f 0 (ρ),
where f is some free energy functional. A typical example of f is f (ρ) = 41 ρ4 − 12 ρ2 .
Example 8 (The Cahn-Hilliard equation). A very closely related equation to the AllenCahn equation is the Cahn-Hilliard equation which is obtained by taking minus the Laplacian of the right-hand side of the former.
∂t ρ = −D∆2 u + ∆f 0 (ρ).
7
1.3
Well-posedness of a PDE
In all of the examples in the previous section, we need further information in order to
solve the problem. We broadly refer to this information as the data for the problem.
Ex. 1 We can specify the density distribution at some initial time t0 . This is an example
of Cauchy data (or an initial value problem).
Ex. 2 For the Laplace equation on a bounded domain, we need to prescribe the value of
u on the boundary of the domain. This is an example of Dirichlet data (a Dirichlet
problem). For the Poisson equation, we specify the function f and also the value
of u on the boundary.
Ex. 3 For the heat equation, we specify the initial temperature of the body, as well as
some boundary condition. For example, a Dirichlet boundary condition when the
temperature at the boundary is prescribed or a Neumann boundary condition when
the flux through the boundary is imposed.
An important aspect of the study of a PDE is about its wellposedness. A PDE
problem, consisting of a PDE together with some data is well-posed if
a) There exists a solution to the problem (existence).
b) For given data, the solution is unique (uniqueness).
c) The solution depends continuously on the data.
These are called Hadamard’s conditions. A problem is said to be ill-posed if any of the
condition above fails to hold.
Notice that existence and uniqueness involves boundary conditions. While continuous
dependence depends on considered metric/norm.
Wellposedness is important because for the vast majority of PDE problems that we
encounter, it is not possible to write down a solution explicitly. However, if well-posedness
is satisfied, we can often deduce properties of that solution directly from the PDE it
satisfies without ever needing an explicit solution.
While the issue of existence and uniqueness of solutions of ordinary differential equations has a very satisfactory answer with the Picard-Lindelöf theorem, that is far from
8
the case for partial differential equations. In general, the answers to the above questions depend heavily on the PDE itself as well as the domain Ω. Definition 1.1 is often
too general to provide any useful information. In this notes, we will be working with
much simpler examples of PDEs having explicit forms, and nevertheless exhibit lots of
interesting behaviour.
Example 9. Consider the following heat equation on (0, 1).




ut = uxx



u(0, t) = u(1, t) = 0,





u(x, 0) = u0 (x).
This PDE is well-posed.
Example 10. Consider the following backwards heat equation




ut = −uxx



u(0, t) = u(1, t),





u(x, 0) = u0 (x).
This PDE is ill-posed.
In general, it is not straightforward to see whether a PDE is well-posed or not. More
knowledge from subsequent chapters is needed.
1.4
Classification of PDEs
There are several ways to categorise PDEs. Following is a possibility.
Definition 2.
i) We say that (1.1) is linear if F is a linear function of u and its deriva-
tives, so that we can re-write (1.1) as (see Appendix 5.3 for the multi-index notation)
X
|α|≤k
aα (x)
∂ |α| u
= f (x).
∂xα
If f = 0, we say the equation is homogeneous.
ii) We say that (1.1) is semilinear if it is of the form
h
i
X
∂ |α| u
aα (x) α + a0 x, u(x), Du(x), D2 u(x), · · · , Dk−1 u(x) = 0,
∂x
|α|=k
9
so that the highest order derivatives of u appear linearly, with coefficients depending
only on x but not on u and its derivatives.
iii) We say that (1.1) is quasilinear if it is of the form
h
i ∂ |α| u
aα x, u(x), Du(x), D2 u(x), · · · , Dk−1 u(x)
∂xα
|α|=k
h
i
+ a0 x, u(x), Du(x), D2 u(x), · · · , Dk−1 u(x) = 0,
X
so that the highest order derivatives of u appear linearly, with coefficients possibly
depending only on the lower order derivatives of u.
iv) We say that (1.1) is fully nonlinear if is not linear, semilinear or quasilinear.
Exercise 1. Classify equations in the examples in Section 1.2.
Chapter 2
Quasilinear first order PDEs
In this chapter, we will study quasilinear first order PDEs. We will be able to apply
the method of characteristic to solve these equations.
2.1
Introduction
For simplicity we will consider functions defined on a subset of
R2 .
However, the
methods in this chapter are applicable for functions defined on higher dimensional sets.
In
R2 , a quasilinear first order PDE has the form
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u),
(2.1)
where a(x, y, u), b(x, y, u) and c(x, y, u) are coefficients; ux and uy denote the partial
derivatives of u with respect to x and y respectively. We often want to solve the above
equation in some bounded domain Ω ⊂ R2 .
Example 11. Let us start with the most simplest quasilinear first order PDE. Let Ω ⊂ R2 ,
we consider the following equation
∂u
(x, y) = 0 in Ω.
∂x
Solutions to this equation depend on the domain Ω. For instance, if Ω = [0, 1] × [0, 1]
then the value of u on {0} × [0, 1] will determine u everywhere in Ω. In fact, suppose
that u(0, y) = h(y) with h continuous on [0, 1], then using the fundamental theorem of
calculus we have,
ˆ
x
u(x, y) = u(0, y) +
0
10
∂u
(s, y) ds = h(y).
∂s
11
y
1
Ω
1
x
Figure 2.1: The domain Ω = [0, 1] × [0, 1]
Figure 2.2: [L] A domain on which ux = 0 is solvable with data on {0} × [0, 1]. [R] A
domain on which ux = 0 is not solvable with data on {0} × [0, 1].
This means that u does not change if we move along a curve of constant y and we say
that the value of u propagates along lines of constant y (Figure 2.1).
How about other domains? When is prescribing the value of u on {0} × [0, 1] enough
to determine u everywhere? Obviously for this approach to work, we must be able to
connect every point in Ω to {0} × [0, 1] by a horizontal line which lies in Ω. See Figure 2.1
for different situations. Note that while for a classical solution of a first order equation
we would require u ∈ C 1 (Ω), the procedure above in fact makes sense even if h(y) is only
continuous, so that uy need not to be C 0 . This suggests that we should consider a notion
of solutions for some PDEs which is weaker than classical solutions.
12
2.2
The method of characteristic
Let us take another example.
Example 12 (Transport equation with constant speed). Let c ∈
R and h : R → R be a
given function. Consider the following PDE
cux + uy = 0,
u(x, 0) = h(x).
(2.2)
Solution. The solution consists of three main steps
Step 1) Change the co-ordinate system (x, y)
(s, t) by writing


x = x(s, t)

y = y(s, t)
and define z(s, t) := u(x(s, t), y(s, t)).
Step 2) Solve the equation for z in the new co-ordinate system (s, t), which by construction will be easier to solve.
Step 3) Transform back to the original co-ordinate system


s = s(x, y)

t = t(x, y)
and u(x, y) = z(s(x, y), t(x, y)).
By the chain rule, we have
dz
dx
dy
=
ux + uy .
dt
dt
dt
We want that the right-hand side of the expression above is equal to cux + uy that appears
in the PDE. Therefore, we obtain the following system of ODEs
dx
=c
dt
dy
= 1.
dt
Solving this system, we obtain x(s, t) = ct + A(s),
y(s, t) = t + B(s). To find A(s), B(s)
we impose that
(t = 0) ⇒ (y = 0),
(t = 0) ⇒ (x = s),
13
from which we deduce that A(s) = s, B(s) = 0, hence, x(s, t) = ct + s and y(s, t) = t. In
the new co-ordinate system, we now have
dz
= 0,
dt
z(s, 0) = h(s).
Solving this equation gives
z(s, t) = h(s).
We now transform back to (x, y) and u(x, y). It is straightforward to find s, t in terms of
x, y
s = x − cy,
t = y.
Therefore, u(x, y) = z(s(x, y), t(x, y)) = h(x − cy). This is the solution of the transport
equation with constant speed.
Some observations:
1) To establish the new co-ordinate system, we need to solve a system of ODEs
dx
= c,
dt
dy
= 1,
dt
dz
= 0,
dt
x(s, 0) = s,
y(s, 0) = 0,
z(s, 0) = h(s).
These are called the characteristic equations associated to the PDE, and x(s, t), y(s, t), z(s, t)
are called the characteristic curves.
2) To transform back to the original co-ordinate system, we need to have the correspondence


x = x(s, t)


s = s(x, y)

y = y(s, t)

t = t(x, y).
3) If x − cy = k, then u(x, y) = u(k). In other words, u is constant along each line
x − cy = k.
2.2.1
The characteristic equations
In the general case
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u),
u(f (x), g(x)) = h(x),
(2.3)
14
where f, g, h are given functions. The method of characteristics to solve the above equation
above consists of three steps.
Step 1) Solve the following characteristic equations, which is a system of ODEs, for x, y, z
as functions of s, t
dx
= a(x, y, z), x(s, 0) = f (s),
dt
dy
= b(x, y, z), y(s, 0) = g(s),
dt
dz
= c(x, y, z), z(s, 0) = h(s).
dt
Step 2) Find the transformation


s = s(x, y)

t = t(x, y).
Step 3) Find the solution u(x, y) = z(s(x, y), t(x, y)).
The characteristic projection is defined by
σs (t) = x(s, t), y(s, t) .
A shock is formed when two characteristic projections meet and assign different values of
u, i.e., there exist (s, t), (s0 , t0 ) such that
x(s, t) = x(s0 , t0 ),
y(s, t) = y(s0 , t0 ),
but z(s, t) 6= z(s0 , t0 ).
2.2.2
Geometrical interpretation
We now give a geometrical interpretation (derivation) for the method of characteristics. Suppose that we are given a curve γ(s) = (f (s), g(s)), s ∈ (α, β). We will find a
function u satisfying
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u),
such that for s ∈ (a, b), we have [u ◦ γ](s) = u(f (s), g(s)) = h(s) for some given function
h : (α, β) → R. In other words, we specify the value of u along the curve γ.
15
The idea we shall pursue is to find the graph of u over some domain Ω by showing
that we can foliate it by curves along which the value of u is determined by ODEs. Recall
that the graph of u over Ω is the surface in
R3 given by
Graph(u) = {x, y, u(x, y) : (x, y) ∈ Ω}.
We know from the initial conditions that the curve Γ, given by (f (s), g(s), h(s)), s ∈ (α, β)
must belong to Graph(u). We want to write
[
Cs ⊂ Graph(u),
s∈(α,β)
where
Cs = {(x(s, t), y(s, t), z(s, t)) : t ∈ (−εs , εs )},
εs > 0,
with x(s, 0) = f (s), y(s, 0) = g(s) and z(s, 0) = h(s). In other words, through each line
of the curve (f (s), g(s), h(s)) we would like to find another curve which lies in the graph
of u.
Since the tangent vector to Cs , given by t :=
dx
(s, t), dy
(s, t), dz
(s, t)
dt
dt
dt
lies in the
surface Graph(u), it must be orthogonal to any vector which is normal to Graph(u).
Since the map
(x, y) → u(x, y) := (x, y, u(x, y))
is a parameterisation of the surface Graph(u), a vector normal to Graph(u) can be found
by
N := ux ∧ uy
   
0
1
   
   
=0∧1
   
ux
uy


−u
 x


= −uy  .


1
Therefore, Cs lies in the surface Graph(u) if and only if t · N = 0,i.e.,
dx
dy
dz
ux + uy −
= 0,
dt
dt
dt
16
should hold at each point on Cs . Recall that u satisfies the PDE
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u).
By comparing the two relationships above, we see that if x(t; s), y(t; s), z(t; s) is the unique
solution of the systems of ODEs
dx
= a(x, y, z),
dt
dy
= b(x, y, z),
dt
dz
= c(x, y, z),
dt
x(s, 0) = f (s)
y(s, 0) = g(s)
(2.4)
z(s, 0) = h(s)
then the curve Cs will lie in Graph(u). The Picard-Lindelöf theorem states that if a, b, c
are sufficiently well behaved, then there exists a unique solution to (2.4) on some interval
t ∈ (−εs , εs ) and that moreover the solution depends continuously on s. Later we will
state more precisely these conditions on a, b, c such that we can solve (2.4).
The equation (2.4) are known as the characteristic equations and the corresponding
solution curves (x(t; s), y(t; s), z(t; s)) are the characteristic curves or simply characteristics. The union of the characteristics forms a surface
[
Cs = {(x(s, t), y(s, t), z(s, t)) : t ∈ (−εs , εs ), s ∈ (α, β)}
s∈(α,β)
which is a portion of the graph of u that includes the curve Γ. We want to write this
as a graph of the form {(x, y, u(x, y)) : (x, y) ∈ Ω}. We can do this, provided we can
invert the map (s, t) 7→ (x(s, t), y(s, t)) to give s(x, y), t(x, y). Then we have u(x, y) =
z(s(x, y); t(x, y)).
Example 13. Solve the PDE
ux + xuy = u,
u(2, x) = h(x).
Solution. The characteristic equations are
dx
= 1,
dt
dy
= x,
dt
dz
= z,
dt
x(s, 0) = 2,
y(s, 0) = s,
z(s, 0) = h(s).
17
The first equation gives x(s, t) = t + 2. Substituting this into the second equation we get
y(s, t) =
t2
2
+ 2t + s. Finally, from the last equation, we have z(s, t) = et h(s). By writing
s, t in terms of x, y, we find
(x − 2)2
s=y−
− 2(x − 2), t = x − 2.
2
2
Therefore, u(x, y) = z(s, t) = ex−2 h y − (x−2)
−
2(x
−
2)
.
2
Note that the method of characteristic does not require that the PDEs are linear. We
consider the following quasilinear equation.
Example 14 (Burger’s equation). Solve the following Burger’s equation
uux + uy = 0,
x, y ∈ R, y ≥ 0.
u(x, 0) = h(x),
Solution. The characteristic equations reduce to
dx
= z,
dt
dy
= 1,
dt
dz
= 0,
dt
x(s, 0) = s,
y(s, 0) = 0,
z(s, 0) = h(s).
The last two equations are easily solved to get
y(s, t) = t,
z(s, t) = h(s).
Substituting these back to the first equation we get
x(s, t) = s + h(s)t.
In this case, we can not explicitly invert the relationship (t, s) 7→ (x, y). However, we
have
s(x, y) = x − h(s(x, y))y,
t(x, y) = y.
Therefore,
u(x, y) = z(s(x, y), t(x, y)) = h(s(x, y)) = h(x − h(s(x, y))y) = h(x − u(x, y)y),
i.e., u satisfies an implicit equation
u(x, y) = h(x − u(x, y)y).
18
In a particular case when h(x) =
√
x2 + 1, we have
u=
p
(x − uy)2 + 1,
which implies that
u(x, y) =
xy −
p
x2 − y 2 + 1
.
y2 − 1
Note that this solution is only valid for |y| < 1. At this point, u blows up. To explore
the blow-up phenomena in more generality, let us return to the characteristic equations
for the Burger’s equation
x(s, t) = h(s)t + s,
y(s, t) = t,
z(s, t) = h(s).
At fixed s the characteristic is simply a straight line parallel to the x − y plane
y=
x−s
,
h(s)
z = h(s).
This means that u(x, y) = z remains constant on each of the line y =
1
(x
h(s)
− s). The
meeting point of two characteristic projections is determined by
x − s1
x − s2
s2 − s1
=
=
.
h(s1 )
h(s2 )
h(s1 ) − h(s2 )
If h is increasing, then
s2 −s1
h(s1 )−h(s2 )
≤ 0. In this case, the lines of constant u are “fanning
out” and do not meet on y ≥ 0.
If h is decreasing, then
s2 −s1
h(s1 )−h(s2 )
≥ 0. The lines of constant u must meet on y ≥ 0. At
such a meeting point, the solution obtained by the method of characteristics is no longer
admissible since the two characteristic projections are trying to assign two different values
of u. This continuity is called a shock.
2.3
Compatibility of initial conditions
We recall that the initial condition is given by
u(f (x), g(x)) = h(x).
Differentiating this equation with respect to x, we find
f 0 (x)ux (f (x), g(x)) + g 0 (x)uy (f (x), g(x)) = h0 (x).
(2.5)
19
On the other hand, from the PDE we have
a(f (x), g(x), h(x))ux (f (x), g(x)) + b(f (x), g(x), h(x))uy (f (x), g(x)) = c(f (x), g(x), h(x)).
(2.6)
Equations (3.9) and (2.6) forms a linear system to determine ux (f (x), g(x)) and uy (f (x), g(x)).
• If a(f (x), g(x), h(x))g 0 (x) − b(f (x), g(x), h(x))f 0 (x) 6= 0, then this system has a
unique solution. In this case, we expect that the method of characteristics will
work. We say that the initial conditions are compatible with the PDE.
• If a(f (x), g(x), h(x))g 0 (x) − b(f (x), g(x), h(x))f 0 (x) = 0, the initial conditions are
incompatible with the PDE or they are redundant.
2.4
Local existence and uniqueness
Recall that we want to solve the PDE
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u) on Ω,
with initial data given along a curve γ : x 7→ (f (x), g(x)) ∈ Ω such that u(f (x), g(x)) =
h(x). We will show that the compatibility of the initial conditions is the only obstruction
(besides standard assumptions) to finding a solution in a neighbourhood of γ. Suppose
that:
• a, b, c ∈ C 1 (Ω × R),
• f, g, h ∈ C 1 ((α, β)) for some interval (α, β).
• The initial conditions are compatible with the PDE, i.e., a(f (x), g(x), h(x))g 0 (x) −
b(f (x), g(x), h(x))f 0 (x) 6= 0 for all x ∈ (α, β).
The the characteristic equations
dx
= a(x, y, z), x(s, 0) = f (s),
dt
dy
= b(x, y, z), y(s, 0) = g(s),
dt
dz
= c(x, y, z), z(s, 0) = h(s).
dt
20
can be solved uniquely for each s ∈ (α, β) and for t ∈ (−εs , εs ), and the solution will be of
class C 1 . To find the solution in terms of (x, y) one now to invert the map (s, t) 7→ (x, y).
Let’s define Ψ(s, t) = (x(s, t), y(s, t)). Then Ψ is of class C 1 in a neighbourhood of γ and




∂x
∂x
0
f (s) a(f (s), g(s), h(s))
.
DΨ(s, 0) =  ∂s ∂t  (s, 0) = 
∂y
∂y
0
g
(s)
b(f
(s),
g(s),
h(s))
∂s
∂t
According to the compatibility conditions, we have
det DΨ(s, 0) = a(f (s), g(s), h(s))g 0 (s) − b(f (s), g(s), h(s))f 0 (s) 6= 0,
for all s ∈ (α, β),
so that DΨ(s, 0) is invertible. By the inverse function theorem, Ψ−1 exists in a neighbourhood of (f (s), g(s)). Furthermore, we have


∂s
 ∂x
∂t
∂x
∂s
∂y 
∂t
∂y

(Ψ(s, t)) = (DΨ−1 )(Ψ(s, t)) = (DΨ(s, t))−1 =
b ∂x
∂s

b
−a
1


∂y
− a ∂s − ∂y ∂x
∂s
∂s
By the chain rule (remembering that u(x, y) = z(s(x, y), t(x, y)) )
∂z ∂s ∂z ∂t
+
∂s ∂x ∂t ∂x
∂z ∂s ∂z ∂t
uy =
+
∂s ∂y
∂t ∂y
ux =
or equivalently, in a matrix form

∂s
∂z  ∂x
ux uy = ∂z
∂s
∂t
∂t
∂x


∂s
∂y 
∂t
∂y
=
b ∂x
∂s
1
− a ∂y
∂s
∂z
∂s
∂z
∂t

b
−a
− ∂y
∂s
∂x
∂s


Hence
aux + buy = ux uy
 
a
 
b

=
b ∂x
∂s
1
− a ∂y
∂s
b ∂x
∂s
1
− a ∂y
∂s
∂z
∂s
∂z
∂t

b
− ∂y
∂s

=
∂z
∂s
0
c 
b ∂x
∂s
−
 
−a
a
 
∂x
b
∂s

a ∂y
∂s

= c,
so the equation is indeed satisfied. Uniqueness follows from the fact that the characteristic through a point (x0 , y0 , z0 ) is uniquely determined and moreover lies completely in
21
Graph(u) if (x0 , y0 , z0 ) lies in Graph(u). Thus any solution would have to include all of
the characteristics through (f (s), g(s), h(s)) and hence would have to locally agree with
the solution we construct above.
We summarise this section by the following theorem
Theorem 1. Suppose that a, b, c ∈ C 1 (Ω × R); f, g, h ∈ C 1 ((α, β)); and that s 7→ γ(s) =
(f (s), g(s)) is injective. Suppose further that a(f (x), g(x), h(x))g 0 (x)−b(f (x), g(x), h(x))f 0 (x) 6=
0 for all x ∈ (α, β). Then, given K ⊂ (α, β) a closed (and hence compact) interval, there
exists a neighbourhood U of γ(K) such that the equation
aux + buy = c
has a unique solution u of class C 1 such that u(γ(s)) = h(s).
Problem sheet 1
Exercise 2. Solve the problem
ux + uy = u,
u(x, 0) = cos x
Exercise 3. Solve the problem
ux + xuy = y,
u(0, y) = cos y
Exercise 4. Consider the Burger’s equation
uux + uy = 0,
u(x, 0) = h(x),
x, y ∈ R,
y≥0
Suppose that h ∈ C 1 (R), h bounded and decreasing, h0 < 0, and assume that h0 is
bounded. Show that no shock forms in the region
1
y<
and that for any y >
1
supR |h0 |
,
supR |h0 |
a shock has formed for some x.
Exercise 5. a) Solve the problem
ux + 3x2 uy = 1,
u(x, 0) = h(x),
(x, y) ∈ R2 ,
for h ∈ C 1 (R).
b) Show and explain why the problem
ux + 3x2 uy = 1,
u(s, s3 ) = 1,
(x, y) ∈ R2
has no solution u ∈ C 1 (R2 ).
c) Show and explain why the problem
ux + 3x2 uy = 1,
u(s, s3 ) = s − 1,
has infinitely many solutions u ∈ C 1 (R2 ).
22
(x, y) ∈ R2
Chapter 2
Laplace’s equation and harmonic
functions
In this chapter, we will study the Laplace’s equation, which is an important equation
both in physics and mathematics. Solutions of the Laplace’s equation are called harmonic
functions. We will study some important, both qualitative and quantitative, properties
of harmonic functions such as the mean value property, smoothness and estimates of the
derivatives, the maximum principle and Harnack’s inequality.
2.1
Laplace’s Equation
Let Ω ⊂ Rn , u : Ω → R be a C 2 function. The Laplacian operator is defined by
∆u(x) :=
n
X
∂ 2u
i=1
∂x2i
(x).
The Laplace’s equation is
∆u(x) = 0,
for all x ∈ Ω.
(2.1)
Remark 1 (Relations between the Laplacian, the gradient and the divergence operators).
Recall that the gradient of u, ∇u (or Du) is a vector field given by
∇u(x) = (∂x1 u, . . . , ∂xn u)T ,
and that if X = (X1 , . . . , Xn )T is a vector field then the divergence of X is
div(X) =
n
X
∂Xi
i=1
23
∂xi
.
24
We have the following relation between the Laplacian, the gradient and the divergence
operators
∆u = div(∇u).
This relation will be used throughout for example when applying the divergence theorem.
Example 15. 1) If a ∈ Rn , b ∈ R and u(x) = a · x + b (i.e., u is affine) then ∆u(x) = 0.
2) If u(x) =
Pn
i,j=1
aij xi xj , with aij = aji (i.e., the matrix a = (aij )i,j=1,...,n is symmetric),
then
(∇u)i = 2
n
X
aij xj ,
and ∆u(x) = 2
i=1
n
X
aii = 2 Tr(a).
i=1
So if a is trace-free (i.e., Tr(a) = 0) then ∆u = 0.
Example 16 (Laplacian operator of a radial function). Suppose u(x) = u(|x|) = u(r),
where r = |x|. Then we have
∆u(|x|) = ∆u(r) = u00 (r) +
n−1 0
u (r).
r
We will provide two proofs for this assertion. The first one is based on direct computations,
the second one is hinged on a property that the Laplacian operator is rotation invariant.
The first proof (direct computations). Since r = |x| =
pPn
i=1
x2i , we have
the chain rule we have
du ∂r
xi
= u0 (r) .
dr ∂xi
r
xi xi
∂ xi 00
0
+ u (r)
.
uxi xi = u (r)
r r
∂xi r
x2i
1 x2i
00
0
= u (r) 2 + u (r)
− 3 .
r
r
r
P
Therefore, recalling that r2 = ni=1 x2i ,
uxi =
∆u(x) =
n
X
uxi xi =
i=1
= u00 (r) +
n X
i=1
x2
u (r) 2i + u0 (r)
r
00
n−1 0
u (r).
r
1 x2i
− 3
r
r
∂r
∂xi
=
xi
.
r
By
25
The second proof.
First observation: the Laplacian is rotation invariant. Let Q be a constant n × n
matrix satisfying QQT = QT Q = I. Let v(x) = u(Qx). By the chain rule, we have
vxi (x) =
n
X
uxj (Qx)Qji ,
j=1
and similarly
n
X
vxi xi =
uxj xk (Qx)Qji Qki = (QT HQ)ii ,
j,k=1
where H = (Hessu)(Qx). Therefore
∆v(x) =
n
X
vxi xi
n
X
=
(QT HQ)ii = Tr(QT HQ) = Tr(QT QH) = Tr(H) = ∆u(Qx).
i=1
i=1
Here we use the fact that Q is orthogonal and the community of the trace T r(AB) =
T r(BA).
Note further that since Q is orthogonal, |x|2 = hx, xi = hQx, Qxi = |Qx|2 , it follows
that |x| = |Qx|.
Now if u is radial, i.e., u(x) = u(|x|) = u(|Qx|) = u(Qx), then v(x) = u(Qx) = u(x),
so that ∆v(x) = ∆u(Qx) = ∆u(x).
Second observation: By the divergence theorem, we have
ˆ
ˆ
ˆ
∇u · n dσ =
div(∇u) dx =
∆u(x) dx.
∂Br
Br
(2.2)
Br
x
Since u is radial, ∇u(x)·n = u0 (r) |x|
·n = u0 (r). The first integral in (2.2) can be computed
as
ˆ
ˆ
0
dσ = u0 (r)σn−1 rn−1 ,
∇u · n dσ = u (r)
∂Br
(2.3)
∂Br
where σn−1 is the area of the unit (n − 1)−sphere.
Now using the fact that ∆u(x) is radial, we can write the last integral in (2.2) as a
radial integral
ˆ
ˆ
Br
r
σn−1 sn−1 ∆u(s) ds.
∆u(x) dx =
0
From (2.3) and (2.4) we obtain that for any r1 < r2
ˆ r2
0
n−1
0
n−1
u (r2 )r2 − u (r1 )r1 =
sn−1 ∆u(s) ds.
r1
∆u(r) is simply the derivative of rn−1 u0 (r), thus
1 d n−1 0 n−1 0
r u (r) = u00 (r) +
u (r).
∆u(r) = n−1
r
dr
r
This equality implies that r
n−1
(2.4)
26
2.2
Harmonic functions
Definition 3. Let Ω ⊂ Rn be open. A function u ∈ C 2 (Ω) is said to be harmonic in Ω if
∆u(x) = 0
for all x ∈ Ω. In other words, harmonic functions solve Laplace’s equation.
Example 17. 1) We saw from Example 15 that affine functions u(x) = a · x + b, for
a ∈ Rn , b ∈ R, and u(x) = x21 + . . . + x2n−1 − (n − 1)x2n are harmonic in
Rn .
2) In polar coordinates, the function rk sin(kθ) is harmonic in R2 and the function r−k sin(kθ)
is harmonic in
R2 \ {0} for k ∈ N.
3) The function ex sin y is harmonic in
R2 ,
When Ω is a domain in
R2 .
an important class of harmonic functions come from
holomorphic functions.
Example 18. Suppose Ω ⊂ Rn ' C and that f (z) is holomorphic in Ω, i.e., the limit
f 0 (z0 ) := lim
z→z0
f (z) − f (z0 )
z − z0
exists for all z0 ∈ Ω. Suppose that f (x + iy) = u(x, y) + iv(x, y). Then u, v are harmonic
function in Ω. In fact, by the Cauchy-Riemann equations, we have


 ∂u = ∂v ,
∂x
∂y

 ∂u = − ∂v .
∂y
∂x
Therefore, ∆u(x, y) =
2.3
∂2u
∂x2
+
∂2u
∂y 2
=
∂2v
∂x∂y
−
∂2v
∂y∂x
= 0. Similarly, ∆v(x, y) = 0.
Mean value property for harmonic functions
Let us motivate this section by looking at a one-dimensional example. Suppose that
u : (a, b) →
R
is harmonic, i.e., u00 (x) = 0 in (a, b). It follows that u(x) = cx + d
for some constants c, d ∈
R.
Let x0 ∈ (a, b) and r > 0 is sufficiently small such that
[x0 − r, x0 + r] ⊂ (a, b). Let us compute
1
1
[u(x0 − r) + u(x0 + r)] = [c(x0 − r) + d + c(x0 + r) + d]
2
2
= cx0 + d,
27
and
1
2r
ˆ
x0 +r
x0 −r
ˆ x0 +r
1
u(y) dy =
(cy + d) dy
2r x0 −r
2
1
y
x0 +r
=
c + dy 2r
2
x0 −r
= cx0 + d.
Therefore,
1
1
u(x0 ) = [u(x0 − r) + u(x0 + r)] =
2
2r
ˆ
x0 +r
u(y) dy.
x0 −r
This property holds true for harmonic function in any dimensional space.
Theorem 2 (Mean value property). Suppose Ω ⊂
Rn
is open and that u ∈ C 2 (Ω) is
harmonic. Then if Br (x0 ) ⊂ Ω then
ˆ
1
u(x0 ) =
u(y) dσ(y)
|∂Br (x0 )| ∂Br (x0 )
ˆ
1
u(y) dy.
=
|(Br (x0 ))| Br (x0 )
(2.5)
(2.6)
Proof. We first prove (2.5). We perform two changes of variables, successively setting
y = x0 + z and z = rζ, which shifts Br (x0 ) to be centred at the origin and then scales it
to have unit radius. We obtain
ˆ
ˆ
u(y) dσ(y) =
∂Br (x0 )
ˆ
u(x0 + z) dσ(z) = r
n−1
∂Br (0)
u(x0 + rζ) dσ(ζ)
∂B1 (0)
Therefore, recalling that |∂Br (x0 )| = rn−1 σn−1 , where σn−1 is the area of the (n − 1)-unit
sphere,
1
|∂Br (x0 )|
ˆ
u(y) dσ(y) =
∂Br (x0 )
1
σn−1
ˆ
u(x0 + rζ) dσ(ζ).
(2.7)
∂B1 (0)
Let F (r) be defined as the right-hand side of the above equality. Let v(ζ) := u(x0 + rζ),
then ∇v(ζ) = r∇u(x0 + rζ) and ∆v(z) = r2 ∆u(x0 + rζ) = 0. We will show, by computing
28
the derivative of F (r), that F (r) is indeed a constant function.
ˆ
dF (r)
1 d
=
u(x0 + rζ) dσ(ζ)
dr
σn−1 dr ∂B1 (0)
ˆ
1
d
=
u(x0 + rζ) dσ(ζ)
σn−1 ∂B1 (0) dr
ˆ
1
=
ζ · ∇u(x0 + rζ) dσ(ζ)
σn−1 ∂B1 (0)
ˆ
1 1
=
ζ · ∇v(ζ) dσ(ζ)
σn−1 r ∂B1 (0)
ˆ
1 1
(∗)
=
∆v(ζ) dζ
σn−1 r B1 (0)
= 0,
where we have used the divergence theorem to obtain the equality (∗) above. Hence by
the continuity of u, we have
F (r) = lim F (r) =
r→0
ˆ
1
σn−1
u(x0 ) dσ(ζ) = u(x0 ).
(2.8)
∂B1 (0)
From (2.7) and (2.8), we obtain (2.5).
The equality (2.6) follows from (2.5). Indeed, we have
ˆ
ˆ r
ˆ
ˆ r
u(y) dy =
ds
u(y) dσ(y) =
|∂Bs (x0 )|u(x0 ) ds
Br (x0 )
0
∂Bs (x0 )
0
ˆ r
= u(x0 )
|∂Bs (x0 )| ds = u(x0 )|Br (x0 )|
0
Remark 2. The converse results are also true, namely that if u ∈ C(Ω) satisfies the mean
value property (either (2.5) or (2.6)) then u ∈ C 2 (Ω) and u is harmonic in Ω. Note that
it requires only that u ∈ C(Ω). See Exercises.
2.4
Smoothness and estimates on derivatives of harmonic functions
Proposition 1. Let Ω ⊂ Rn be open and let u be harmonic in Ω. Then u ∈ C ∞ (Ω) and
furthermore
α
|D u(x)| ≤
where d(x) = dist(x, ∂Ω).
n|α|
d(x)
|α|
sup |u(y)|,
y∈Ω
(2.9)
29
Proof. We first prove that u ∈ C ∞ (Ω). Let φ be satisfy the following property
(i) φ ∈ C ∞ (Rn ),
(ii) φ is radial,
(iii) φ is supported in Br for some r > 0,
´
(iv) Rn φ(x) dx = 1.
We will show that u(x) = (u ∗ φ)(x), where the convolution is defined by
ˆ
(u ∗ φ)(x) =
u(y)φ(x − y) dy.
Rn
Without loss of generality, set x = 0, and we take r small enough so that Br ⊂ Ω. Now
we have
ˆ
(u ∗ φ)(0) =
n
ˆR
u(y)φ(−y) dy
u(y)φ(−y) dy
=
ˆ
Br
u(y)φ(y) dy
=
B
ˆ rr ˆ
u(y) dS(y) φ(s) ds
=
ˆ0 r
∂Bs
|∂Bs |u(0)φ(s) ds
ˆ r
|∂Bs |φ(s) ds
= u(0)
0
ˆ rˆ
= u(0)
φ(s)dS ds
0
∂Bs
ˆ
= u(0)
φ(y) dy
=
0
Br
= u(0).
So u(x) = (u ∗ φ)(x). Since u ∗ φ ∈ C ∞ (Ω), it follows that u ∈ C ∞ (Ω) as claimed.
Step 1. Next we will prove the estimate (2.9) by induction on |α|. For |α| = 1, we
need to prove that
|Di u(x)| ≤
n
sup |u(y)|.
d(x) y∈Ω
Take r < d(x) then Br (x) ⊂ Ω. According to the mean value property
ˆ
ˆ
1
1
Di u(x) =
Di u(y) dy =
u(y)ni (y) dS(y),
|Br (x)| Br (x)
|Br (x)| ∂Br (x)
(2.10)
30
Since |ni | = 1, we can estimate
|Di u(x)| ≤
|∂Br (x)|
rn−1 σn−1
n
sup |u(y)| =
sup |u(y)| =
sup |u(y)|.
n
|Br (x)| y∈Br (x)
r ωn y∈Br (x)
r y∈Br (x)
Since this is true for any r < d(x) and Br (x) ⊂ Ω, it follows that
|Di u(x)| ≤
n
sup |u(y)|.
d(x) y∈Ω
Step 2. Suppose that the statement is true for |α| ≤ N .
Step 3. We need to prove that it is also true for |α| = N + 1. Write Dα u = Di Dβ u,
where β is a multi-index such that |β| = |α| − 1 = N . Now we fix some λ ∈ (0, 1). We
apply the technique in Step 1 to r = λd(x), to obtain
|Dα u(x)| = |Di Dβ u(x)| ≤
n
sup |Dβ u(y)|.
λd(x) y∈Bλd(x) (x)
By the triangle inequality if y ∈ Bλd(x) (x), then d(y) ≥ (1 − y)d(x), so applying the
induction assumption we have
β
|D (y)| ≤
n|β|
d(y)
|β|
sup |u(z)| ≤
z∈Ω
n|β|
(1 − λ)d(x)
|β|
sup |u(z)|,
z∈Ω
which implies that
sup |u(y)| ≤
y∈Bλd(x)
n|β|
(1 − λ)d(x)
|β|
sup |u(z)|,
z∈Ω
and that (recalling that |α| = |β| + 1)
n
|D u(x)| ≤
λd(x)
α
n|β|
(1 − λ)d(x)
|β|
sup |u(z)| =
z∈Ω
n |α| |β|
1
|β| sup |u(z)|
.
d(x)
λ(1 − λ)|β|
z∈Ω
(2.11)
To obtain the sharpest upper bound, we now minimizes the function λ 7→
1
,
λ(1−λ)|β|
is equivalent to maximises the function
λ 7→ g(λ) := λ(1 − λ)|β| ,
to obtain the optimal λ. The optimal λ satisfies the equation
0=
dg(λ)
= (1 − λ)|β| − λ|β|(1 − λ)|β|−1
dλ
= (1 − λ)|β|−1 [1 − λ − |β|λ] = (1 − λ)|β|−1 )(1 − |α|λ),
which
31
which implies that λ =
1
,
|α|
which is in the interval (0, 1) as required. Substituting this
value to g(λ), we get (recalling that |α| = |β| + 1)
|β|
1
|β||β|
1
g(λ) 1 =
1−
=
|α|
|α|
|α||α|
λ= |α|
Putting this back into (2.11) we obtain
|Dα u(x)| ≤
n|α| |α||α|
sup |u|,
d(x)|α| Ω
which is the reduction assumption for |α| = N + 1. Hence the result holds true for any
N.
Proposition 1 has two interesting consequences in the following theorems.
Theorem 3. Harmonic functions are analytic.
Proof. Let x0 , x ∈ Ω and suppose that the segment [x0 , x] = tx + (1 − t)x0 , t ∈ [0, 1] is
contained in Ω. Without loss of generality we assume that x0 = 0. Let u be a harmonic
function in Ω. By Taylor’s theorem we have
u(x) = Pm−1 (x) + Rm (x),
where Pm−1 (x) is a polynomial of degree m − 1 and
Rm (x) =
X Dα u(ξ)
xα ,
α!
|α|=m
for some ξ ∈ [x0 , x]. To show that u is analytic we need to show that Rm (x) → 0 as
m → ∞ uniformly on some small disc around x0 . By Proposition 1, we have
m
X |xα |
nm
|Rm (x)| ≤
sup |u|
d(ξ)
α!
Ω
|α|=m
m
X
1
m!
nm
=
sup |u|
|x1 |α1 . . . |xn |αn .
m! d(ξ)
α1 ! . . . αn !
Ω
|α|=m
Set ||x||1 :=
Pn
i=1
|xi |, then by the multinomial theorem we have
||x||m
1 =
X
|α|=m
m!
|x1 |α1 . . . |xn |αn .
α1 ! . . . αn !
Therefore, we obtain that
||x||m
1
|Rm (x)| ≤
m!
nm
d(ξ)
m
sup |u|.
Ω
(2.12)
32
We now use a weak version of the Stirling’s approximation, which holds true for all positive
integer m,
mm
≤ em .
m!
We get
m
ne||x||1
|Rm (x)| ≤
sup |u|.
d(ξ)
Ω
1
Taking |x| = |x − x0 | < 2 d(x0 ), by the triangular inequality, we have d(ξ) ≤ 12 d(x0 ), so
that
|Rm (x)| ≤
Finally, if we take x such that ||x||1 <
2ne||x||1
d(x0 )
d(x0 )
,
2ne
m
then
sup |u|.
(2.13)
Ω
2ne||x||1
d(x0 )
< 1 and
2ne||x||1
d(x0 )
m
→ 0 as
m → ∞, that implies that |Rm (x)| → 0 as m → ∞.
Theorem 4 (Liouville). Suppose u is a bounded harmonic function in
Rn ,
then u is a
constant.
Proof. Suppose |u| ≤ C in
Rn .
Let x ∈
Rn
and r > 0. Since u is harmonic in Br (x),
applying Proposition 1 in Br (x), we have
|Di u(x)| ≤
n
nC
sup |u| ≤
.
r Br (x)
r
Since r is arbitrary and C is independent of r, the above estimate implies that |∇u(x)| = 0.
So u is constant.
2.5
2.5.1
The maximum principle
Subharmonic and superharmonic functions
Definition 4. Let Ω ⊂ Rn be an open set, and let u ∈ C 2 (Ω). We say u is subharmonic
(respectively superhamornic) in Ω iff ∆u ≥ 0 (respectively ∆u ≤ 0).
Note that a function u ∈ C 2 (Ω) is harmonic if and only if it is both subharmonic and
superharmonic.
Example 19. 1) In one dimension, u is subharmonic iff u00 > 0, which means that u0 is
increasing and u is convex. Similarly u is superharmonic iff u is concave.
2) If u = xT Ax for some constant, symmetric matrix A. Then ∆u = 2Tr(A), so u is
subharmonic (respectively, superharmonic) iff Tr(A) ≥ 0 (respectively, Tr(A) ≤ 0).
33
2.5.2
Some properties of subharmonic and superharmonic functions
Recall that harmonic functions satisfy the mean value property, which means that for
any ball Br (x0 ) ⊂ Ω, we have
1
u(x0 ) =
|∂Br x0 |
ˆ
1
u(y)dσ(y) =
|Br (x0 )|
∂Br (x0 )
ˆ
u(y) dy.
Br (x0 )
Recall that to prove this in Theorem 2, we considered the function F (r) =
1
|∂Br x0 |
´
∂Br (x0 )
u(y)dσ(y)
and showed
(i) F (r) → u(x0 ) as r → 0. This relied on the continuity of u,
(ii) For all r > 0 we have
1
F (r) =
rσn−1
ˆ
0
∆u(x0 + rζ) dζ.
(2.14)
B1 (0)
It follows from (2.14) that if u is subharmonic (respectively, superharmonic) then F 0 (r) ≥
0 (respectively, F 0 (r) ≤ 0).
Proposition 2. If u is subharmonic in Ω and Br (x0 ) ⊂ Ω, then
ˆ
ˆ
1
1
u(x0 ) ≤
u(y)dσ(y) and u(x0 ) ≤
u(y) dy.
|∂Br x0 | ∂Br (x0 )
|Br (x0 )| Br (x0 )
(2.15)
If u is superharmonic then the reverse inequalities hold.
It follows from this Proposition and Theorem 2 that the value of a subharmonic
(respectively, superharmonic) function at the center of a ball is less (respectively, greater)
than or equal to the value of a harmonic function with the same values on the boundary.
Thus, the graphs of subharmonic functions lie below the graphs of harmonic functions
and the graphs of superharmonic functions lie above, which explains the terminology.
2.5.3
The maximum principle
An important feature of subharmonic functions is the following theorem.
Theorem 5 (Weak maximum principle). Let Ω be an open and bounded subset of
Rn .
Suppose that u ∈ C 2 (Ω) ∩ C 0 (Ω) be subharmonic in Ω. Then
max u = max u.
Ω
∂Ω
(2.16)
34
If, rather, u is superharmonic in Ω then
min u = min u.
(2.17)
∂Ω
Ω
Proof. Suppose u is subharmonic. We consider two cases.
Case 1: ∆u > 0 in Ω. Suppose that there exists x0 ∈ Ω such that u attains a local
maximum at x0 , then we must have ∇u(x0 ) = 0 and Hess u(x0 ) is negative definite, i.e.,
ξ T Hess u(x0 )ξ ≤ 0 for all ξ ∈ Rn , ξ 6= 0. It implies that
∆u(x0 ) = Tr(Hess u(x0 )) =
n
X
eTi Hess u(x0 )ei ≤ 0,
i=1
which is a contradiction. Therefore, u can not have maxima in the interior. Thus, the
maximum of u in Ω must be achieved on the boundary Ω, i.e., (2.16) holds.
Case 2: ∆u ≥ 0 in Ω. Define uε (x) := u(x)+ε|x|2 . Then ∆uε (x) = ∆u(x)+2εn > 0,
so that we can apply Case 1 above for uε (x) and obtain
max uε = max uε .
Ω
∂Ω
Since Ω is bounded, |x|2 is a bounded function in Ω, and uε → u uniformly on Ω̄ as ε → 0.
It follows that
max u = lim max uε = lim max uε = max u.
ε→0
Ω
Ω
ε→0 ∂Ω
∂Ω
The statement (2.17) for superhamornic function is obtained by applying (2.16) for −u,
which is subharmonic, and using the fact that max(−u) = − min u.
A
A
Since a harmonic function is both subharmonic and superhamornic, we obtain the
following corollary.
Corollary 1. 1) Let Ω ⊂ Rn be open and bounded and u ∈ C 2 (Ω) ∩ C 0 (Ω̄) is harmonic.
Then
max u = max u,
Ω
∂Ω
and
min u = min u.
Ω
∂Ω
This implies
max |u| = max |u|.
Ω
∂Ω
2) (The comparison principle) If Ω ⊂ Rn is open and bounded and if u, v ∈ C 2 (Ω)∩C 0 (Ω̄)
satisfying ∆u ≥ ∆v in Ω and u ≤ v on ∂Ω then u ≤ v in Ω.
35
The comparison follows by applying the weak maximum principle to the function
w := u − v, which is subharmonic and satisfies w ≤ 0 on ∂Ω, hence w ≤ 0 (i.e, u ≤ v) in
Ω.
Note that the weak maximum principle states that the that the maximum of a subharmonic function is to be found on the boundary, but may re-occur in the interior as
well. The following theorem provides a stronger statement.
Theorem 6 (Strong maximum principle). Let Ω be open and connected (possibly unbounded), and let u be subharmonic in Ω. If u attains a global maximum value in Ω, then
u is constant in Ω.
Proof. Define M := max u, and A := u−1 {M } = {x ∈ Ω : u(x) = M }. By the assumption
Ω
of the theorem, A is non-empty. Since u is continuous and {M } is a closed set in R, it
follows that A is relatively closed in Ω (i.e., there exists a closed F ⊂
Rn
such that
A = F ∩ Ω). We now show that A is also open. Indeed, let x ∈ Ω and let r be
sufficiently small that Br (x) ⊂ Ω. By the mean value property for subharmonic functions
in Proposition 2, we have
1
0 = u(x) − M ≤
|Br (x)|
ˆ
(u(y) − M ) dy.
Br (x)
Since M = max u, we have u(y) − M ≤ 0 for all y ∈ Br (x). This implies that
Ω
1
0 = u(x) − M ≤
|Br (x)|
ˆ
(u(y) − M ) dy ≤ 0.
Br (x)
Since u is continuous, this happens only if u = M in Br (x). So that Br (x) ⊂ A, which
implies that A is open. Since Ω is connected and A 6= ∅ is both open and closed, it follows
that A = Ω, and hence u ≡ M in Ω. This finishes the proof.
Again, since a harmonic function is both subharmonic and superharmonic, we get a
stronger result.
Corollary 2 (Strong maximum principle for harmonic functions). Let Ω be open and
connected and u : Ω → R a harmonic function. Then if u attains either a global maximum
or global minimum in Ω then u is constant in Ω.
36
2.6
Harnack’s inequality
The next result we shall derive is Harnack’s inequality for harmonic functions. To
state this theorem, we need to define the notion of a compactly contained subset. Let
Ω, Ω0 be two open subsets of
Rn ,
then we say that Ω0 is compactly contained in Ω and
write Ω0 b Ω if there exists a compact set K such that Ω0 ⊂ K ⊂ Ω.
Theorem 7 (Harnack’s inequality for harmonic functions). Let Ω ⊂
u:Ω→
R
Rn
be open and let
be a non-negative harmonic function. Then for any connected Ω0 b Ω there
exists a constant C depending only on Ω, Ω0 , n such that
sup ≤ C inf0 u.
Ω0
Ω
Proof. Let r > 0 satisfy that inf Ω0 d(x) > 4r where d(x) = dist(x, ∂Ω). Let x, y ∈ Ω0 be
such that |x − y| ≤ r. Applying the mean value property for u we have
ˆ
ˆ
ˆ
1
1
1
u(x) =
u(z) dz = n n
u(z) dz ≥ n n
u(z) dz,
|B2r (x)| B2r (x)
2 r ωn B2r (x)
2 r ωn Br (y)
where ωn = |B1 (0)| and we have used that u ≥ 0 and Br (y) ⊂ B2r (x) to obtain the above
inequality. Now using the mean value property again, we have
ˆ
ˆ
1
1
1
1
u(z) dz = n
u(z) dz = n u(y).
n
n
2 r ωn Br (y)
2 |Br (y)| Br (y)
2
It follows from these two inequalities that
1
u(y) ≤ u(x).
2n
By interchanging x and y, we obtain u(x) ≤ 2n u(y), so that for all |x − y| ≤ r we have
1
u(y) ≤ u(x) ≤ 2n u(y).
n
2
(2.18)
Now we will expand this estimate to any two points x0 , x1 in Ω0 . Since Ω0 is open and
connected, it is path connected. Hence there exists a curve γ : [0, 1] → Ω0 such that
γ(0) = x0 and γ(1) = x1 . Since Ω0 ⊂ K for some compact subset K, we can cover Ω0 with
at most N open balls of radius r, where N depends only on Ω, Ω0 . From these balls, we
select M ≤ N balls Bi , i = 1, . . . , M such that
x0 ∈ B1 ,
x1 ∈ BM ,
γ([0, 1]) ⊂
M
[
i=1
Bi ,
37
and Bi ∩ Bi+1 6= ∅,
i = 1, . . . , M − 1.
Applying the estimate (2.18) successively on these balls, we get
u(x1 ) ≤ 2M n u(x0 ) ≤ 2N n u(x0 ).
Since this inequality holds for any two points x0 , x1 ∈ Ω0 , it follows that
sup u ≤ 2nN inf0 u,
Ω0
Ω
which is the claim.
Harnack’s inequality is used to prove Harnack’s theorem about the convergence of
sequences of harmonic functions. Harnack’s inequality can also be used to show the
interior regularity of weak solutions of partial differential equations.
Corollary 3. Let Ω be connected and let uk be a sequence of harmonic functions in Ω
such that uk ≤ uk+1 for all k. If there exists x0 ∈ Ω such that uk (x0 ) converges (to a finite
value) then uk converges to some harmonic function u uniformly on the compact sets of
Ω.
Problem Sheet 2
Exercise 6 (Laplacian operator in the cylindrical and spherical coordinates).
1) Consider the cylindrical coordinates in R3 defined by




x1 (r, ϕ, z) = r cos ϕ,



x2 (r, ϕ, z) = r sin ϕ,
r ∈ [0, ∞), ϕ ∈ [0, 2π), z ∈ R,





x3 (r, ϕ, z) = z.
Prove that in these coordinates
1 ∂
∆u(r, ϕ, z) =
r ∂r
2) Consider the spherical coordinates in




x1 (r, θ, ϕ) = r sin θ cos ϕ,



x2 (r, θ, ϕ) = r sin θ sin ϕ,





x3 (r, ϕ, z) = z.
∂u
r
∂r
+
1 ∂ 2u ∂ 2u
+
.
r2 ∂ϕ2 ∂z 2
R3 defined by
r ∈ [0, ∞), θ ∈ [0, π), ϕ ∈ [0, 2π),
Prove that in these coordinates
1 ∂
1
∂ 2u
1
∂
∂u
2 ∂u
∆u(r, θ, ϕ) = 2
r
+ 2 2
+
sin θ
.
r ∂r
∂r
∂θ
r sin θ ∂ϕ2 r2 sin2 θ ∂θ
[Hint: write gradients and use integration by parts]
3) Show that if u, v are C 2 functions then
∆(uv)(x) = v(x)∆u(x) + u(x)∆v(x) + 2∇u(x) · ∇v(x).
Exercise 7.
38
39
1) Find the general form of radial harmonic functions in Rn \ {0}. Which of these extend
to
Rn as smooth functions?
2) Find all radial solutions of
∆u(x) =
1
(1 + |x|2 )2
in R2 \{0}. Which of these solutions extend smoothly to all of R2 (i.e., near the origin).
3) Let (z, ϕ) be polar coordinates in
R2 , and consider the set Ω = R2 \ {(x, 0) : x ≥ 0}.
Show that the functions log r, ϕ are harmonic in Ω.
Exercise 8 (Converse of the mean value property of harmonic functions in 1D).
1) Suppose u : (a, b) → R is continuous and satisfies
1
u(x0 ) = [u(x0 − r) + u(x0 + r)],
2
∀x0 , r such that [x0 − r, x0 + r] ⊂ (a, b).
Show that u is harmonic (i.e., affine).
2) Prove the same if u : (a, b) → R is continuous and satisfies
ˆ x0 +r
1
u(x) dx, ∀x0 , r such that [x0 − r, x0 + r] ⊂ (a, b).
u(x0 ) =
2r x0 −r
Exercise 9. 1) Prove that if u ∈ C 2 (Ω) satisfies the mean value property (either in the
spherical or bulk form), then ∆u = 0 in Ω.
2) Suppose Ω ⊂ Rn is open and such that Ω0 = Ω ∩ {xn = 0} =
6 ∅. Let Ω+ := Ω ∩ {xn >
0}, and define Ω̃ = Ω+ ∪ Ω0 ∪ Ω− , where
Ω− = {x = (x0 , xn ) ∈ Rn
Suppose that u ∈ C 2 (Ω+ ) is such that
ũ(x) =


u(x),
∂u
∂xn
such that (x, −xn ) ∈ Ω+ }.
= 0 on Ω0 , and define ũ : Ω̃ → R by
x ∈ Ω+ ∪ Ω0 ,

u(x0 , −xn ),
x = (x0 , xn ) ∈ Ω− .
Prove that ũ ∈ C 2 (Ω̃) and that ũ is harmonic in Ω̃. See the figure below for illustration
of the sets Ω, Ω0 , Ω± .
40
Figure 2.1: Illustration of the sets Ω, Ω0 , Ω±
3) Let Ω, Ω0 , Ω± , Ω̃ be as above (see also the figure), and suppose that v ∈ C 2 (Ω) satisfies
v = 0 on Ω0 . Define ṽ : Ω̃ → R by


v(x),
x ∈ Ω+ ∪ Ω0 ,
ṽ(x) =

−v(x0 , −xn ),
x = (x0 , xn ) ∈ Ω− .
Prove that ṽ ∈ C 2 (Ω̃) and that ṽ is harmonic in Ω̃.
Exercise 10. Recall the Azelà-Ascoli theorem: Let K ⊂ Rn be compact and let fi : K →
R be a sequence of functions that are uniformly bounded and equicontinuous.
Then there
exists a sequence fik which converges uniformly. [Recall that fi are equicontinuous if for
any ε > 0 there exists δ > 0 such that |fi (x) − fi (y)| < ε whenever |x − y| < δ].
Let Ω ⊂
Rn
be an open set and let ui : Ω →
R be a sequence of harmonic functions
which are uniformly bounded. Prove that for any multi-index α and for any compact
K ⊂ Ω there exists a subsequence uik such that Dα uik converges uniformly in K.
Exercise 11. 1) Suppose u : Ω → R is harmonic and that F : R → R is a convex function
of class C 2 . Show that F (u(x)) is subharmonic.
41
2) Suppose u : Rn → R is harmonic and that
ˆ
u2 (x) dx = M < ∞.
n
R
Prove that u ≡ 0 in
Rn .
3) If u is harmonic in Ω ⊂ Rn and if Br (x0 ) ⊂ Ω, prove that
d ´
2r ´
2
− u(y) dσ(y) =
− |∇u|2 (x) dx.
dr ∂Br (x0 )
n Br (x0 )
Chapter 3
The Dirichlet problem for harmonic
functions
In this chapter we will study the Direchlet problem for harmonic functions in a
bounded domain. We will establish a representation formula based on the Green’s function, find the Green’s function and solve the problem in a ball, and finally use the Perron’s
method to solve the problem in a general domain.
3.0.1
Introduction
In this chapter, we will study the following Dirichlet problem for harmonic functions.
The Dirichlet problem: Let Ω ⊂
Rn
be open and bounded. Let g ∈ C(∂Ω) be a
given function. Find a function u ∈ C 2 (Ω) ∩ C 0 (∂Ω) such that


∆u = 0, in Ω,

u = g,
(3.1)
on ∂Ω.
The schedule to solve this problem will be the following.
Step 1 Establish a representation formula for a solution to (3.1) based on Green’s function.
Step 2 Find the Green’s function and solve the problem (3.1) in a ball,
Step 3 Solve the problem (3.1) for a general domain using the Perron’s method.
42
43
To start with, we have the following lemma
Lemma 1. Let Ω ⊂
Rn
be open and bounded, let g1 , g2 ∈ C(∂Ω) be given. Suppose that
u1 , u2 solve the following problem for i = 1, 2 respectively


∆ui = 0, in Ω,

ui = gi ,
on ∂Ω.
Then
sup |u1 − u2 | = sup |g1 − g2 |.
Ω
(3.2)
Ω
As a consequence, a solution to (3.1), if it exists, is unique and depends continuously on
the data.
Proof. This is a direct consequence of the maximum principle for w := u1 − u2 , which
satisfies that


∆w = 0,
in Ω,

w = g1 − g2 ,
on ∂Ω.
In view of this Lemma, the main task is now to find a solution to (3.1). We first recall
the Green’s identities that will be used throughout this chapter. The starting point for
these identities is the divergence theorem.
ˆ
ˆ
→
−
div F dx =
Ω
→
− →
F ·−
n dS.
∂Ω
→
−
Given f, g ∈ C 2 (Ω). Applying the divergence theorem above for F = f ∇g, we obtain
ˆ
ˆ
−
div(f ∇g) dx =
f ∇g · →
n dS.
Ω
∂Ω
Note that
div(f ∇g) = f ∆g + ∇g · ∇f,
∂g
−
∇g · →
n = →
,
∂−
n
so that we can write the equality above as
ˆ
ˆ
ˆ
f ∆g dx = − ∇f · ∇g dx +
Ω
Ω
This relation is known as the first Green’s identity.
∂g
f →
− dS.
∂Ω ∂ n
(3.3)
44
By interchanging f and g in the first Green’s identity, we get
ˆ
ˆ
ˆ
∂f
g∆f dx = − ∇g · ∇f dx +
g →
− dS.
Ω
Ω
∂Ω ∂ n
This together with the first Green’s identity implies the following Green’s second identity
ˆ
ˆ ∂g
∂f (f ∆g − g∆f ) dx =
f →
−g →
dS.
(3.4)
∂−
n
∂−
n
Ω
∂Ω
Define the function
Φ(x) =



1
2π


1
|x|2−n ,
(2−n)σn−1
log |x|,
if n = 2,
(3.5)
if n = 3,
where σn−1 = |∂B1 (0)|. Then Φ ∈ C ∞ Rn \ {0}. For any x 6= 0, we have
1
xi |x|−n ,
σn−1
1
−nx2i
2
−n
∂xi xi Φ(x) =
+ |x|
.
σn−1 |x|n+2
∂xi Φ(x) =
Therefore,
∆Φ(x) =
n
X
∂x2i xi Φ(x)
i=1
n
X
1
−nx2i
−n
=
+ |x|
= 0.
n+2
σ
|x|
n−1
i=1
So
∆Φ(x) = 0,
3.0.2
∀x 6= 0,
lim Φ(x) = −∞.
x→0
(3.6)
Representation formula based on the Green’s function
We first provide a motivation for the Green’s function. Recall that we want to solve


∆u = 0, in Ω,

u = g,
in ∂Ω.
Suppose that for each x ∈ Ω, we can find a function G(x, y) such that


∆y G(x, y) = δx , in Ω,

G(x, y) = 0,
in ∂Ω.
45
Then formally for a solution u to (3.1), using the second Green’s identity, we have
ˆ
u(y)δx dy
u(x) =
Ω
ˆ
=
u(y)∆y G(x, y) dy
ˆΩ ˆ
∂G
∂u
=
u(y) →
(x, y) − G(x, y) →
(y) dS(y) + G(x, y)∆u(y) dy
∂−
n
∂−
n
∂Ω
Ω
ˆ
∂G
=
→
− (x, y)g(y) dS(y).
Ω ∂n
Therefore, we can represent u in terms of G and g. This suggests that to solve the
problem (3.1), one should find the function G(x, y). We observe that Φ(x − y) satisfies
that ∆y Φ(y − x) = δx , but it does not satisfy the boundary condition. The idea that
we shall follow is that we will construct G(x, y) from Φ(y − x) taking into account the
boundary conditions.
To do so, let u ∈ C 2 (Ω) and consider the following integral
ˆ
Φ(y − x)∆u(y) dy.
Ω
We want to integrate by parts this integral. However, note that Φ(y − x) has a singularity
at x = y. To integrate this, therefore, we proceed as follows. Take x ∈ Ω and ε > 0
sufficiently small such that Bε (x) ⊂ Ω. Define Vε := Ω \ Bε (x). By the Green’s second
identity, we have
ˆ
ˆ
ˆ
∂u
∆y Φ(y − x)u(y) dy +
Φ(y − x) →
Φ(y − x)∆u(y) dy =
(y) dS(y)
∂−
n
∂Vε
Vε
Vε
ˆ
∂Φ
−
u(y) →
(y − x) dS(y)
∂−
n
∂Vε
ˆ
ˆ
∂u
∂Φ
=
Φ(y − x) →
(y) dS(y) −
u(y) →
(y − x) dS(y),
−
∂n
∂−
n
∂Vε
∂Vε
where we have used that ∆y Φ(y − x) = 0. Next, we will pass to the limit ε → 0 this
relation. The LHS is straightforward by the dominated convergence theorem
ˆ
ˆ
lim
Φ(y − x)∆u(y) dy =
Φ(y − x)∆u(y) dy.
ε→0
Vε
Ω
For the boundary terms, we will show later that
Claim 1
ˆ
lim −
ε→0
ˆ
∂Φ
∂Φ
u(y) →
(y − x) dS(y) = −
u(y) →
(y − x) dS(y) + u(x).
−
∂n
∂−
n
∂Vε
∂Ω
(3.7)
46
Claim 2
ˆ
∂u
lim
Φ(y − x) →
(y) dS(y) =
ε→0 ∂V
∂−
n
ε
ˆ
∂u
Φ(y − x) →
(y) dS(y)
∂−
n
∂Ω
(3.8)
Assuming these claims at the moment, then we find that for any u ∈ C 2 (Ω), we have
ˆ
ˆ
ˆ
∂Φ
∂u
u(x) =
Φ(y−x)∆u(y) dy+
u(y) →
(y−x) dS(y)−
Φ(y−x) →
(y) dS(y). (3.9)
−
∂n
∂−
n
Ω
∂Ω
∂Ω
Note that in the Direchlet problem, we know ∆u in Ω and u on ∂Ω, but we do not know
∂u
→
∂−
n
on ∂Ω.
To overcome this, for each x ∈ Ω, we introduce a corrector function hx (y) such that


∆y hx (y) = 0, y ∈ Ω,

hx (y) = Φ(y − x),
y ∈ ∂Ω.
Then using the Green’s second identity, we have
ˆ ˆ
ˆ
∂u
∂hx
x
x
x
∆y h (y)u(y) dy +
h (y) →
h (y)∆u(y) dy =
(y) − u(y) →
(y) dS(y).
∂−
n
∂−
n
∂Ω
Ω
ˆΩ
ˆ
∂u
∂hx
=
Φ(y − x) →
(y) dS(y) −
u(y) →
(y) dS(y).
∂−
n
∂−
n
∂Ω
∂Ω
Hence,
ˆ
∂u
Φ(y − x) →
(y) dS(y) =
∂−
n
∂Ω
ˆ
∂hx
u(y) →
(y) dS(y) +
∂−
n
∂Ω
ˆ
hx (y)∆u(y) dy.
Ω
Substituting this equality back to (3.9), we obtain that
ˆ
ˆ ∂Φ
∂hx x
u(x) = [Φ(y − x) − h (y)]∆u(y) dy +
(y
−
x)
−
(y) u(y) dS(y). (3.10)
→
−
−
∂→
n
Ω
∂Ω ∂ n
By defining G(x, y) := Φ(y − x) − hx (y), then we get that for any u ∈ C 2 (Ω)
ˆ
ˆ
∂G
u(x) =
G(x, y)∆u(y) dy +
→
− (y − x) u(y) dS(y).
Ω
∂Ω ∂ n
(3.11)
The function G(x, y) is called the Green’s function for Ω. In particular, if u is harmonic
then
ˆ
u(x) =
∂Ω
∂G
(y − x) u(y) dS(y).
−
∂→
n
We now prove the two claims.
Proof of Claim 1. We have
ˆ
∂Φ
−
u(y) →
(y −x) dS(y) = −
∂−
n
∂Vε
ˆ
∂Φ
u(y) →
(y −x) dS(y)+
∂−
n
∂Ω
ˆ
∂Φ
u(y) →
(y −x) dS(y).
∂−
n
∂Bε (x)
47
The first term on the right-hand side does not depend on ε, so it is unchanged when
passing ε → 0. We consider the second term. We have
∂Φ
−
(y − x) = ∇y Φ(y − x) × →
n (y),
→
−
∂n
−
where →
n (y) is the outward normal vector on ∂ε (x). By direct computations
∇y Φ(y − x) =
1
1 y−x
∇y |y − x|2−n =
,
(2 − n)σn−1
σn−1 |y − x|n
y−x
→
−
,
n (y) =
|y − x|
so that
∂Φ
1 y−x y−x
1
1
(y − x) =
=
.
→
−
n
σn−1 |y − x| |y − x|
σn−1 |y − x|n−1
∂n
Therefore,
ˆ
ˆ
∂Φ
1
1
u(y) →
u(y) dS(y)
(y − x) dS(y) =
−
σn−1 ∂Bε (x) |y − x|n−1
∂n
∂Bε (x)
ˆ
1
=
u(y) dS(y)
σn−1 εn−1 ∂Bε (x)
ˆ
1
=
u(y) dS(y).
|∂Bε (x)| ∂Bε (x)
Since u is continuous, we have that
ˆ
ˆ
∂Φ
1
u(y) →
lim
u(y) dS(y) = u(x),
(y − x) dS(y) = lim
ε→0 ∂B (x)
ε→0 |∂Bε (x)| ∂B (x)
∂−
n
ε
ε
from which the assertion of the claim 1 is followed.
Proof of the claim 2. Similarly as in the proof of the claim 1, we have
ˆ
ˆ
ˆ
∂u
∂u
∂u
Φ(y − x) →
(y) dS(y) =
Φ(y − x) →
(y) dS(y) −
Φ(y − x) →
(y) dS(y).
−
−
∂n
∂n
∂−
n
∂Vε
∂Ω
∂Bε (x)
We now show the second term vanishes as ε → 0. In deed, using the explicit formula for
Φ for n ≥ 3 (the case n = 2 is similar), we have
ˆ
ˆ
∂u
∂u
1
1
(y) dS(y) ≤
(y) dS(y)
Φ(y − x) →
→
−
−
n−2
(n − 2)σn−1 ∂Bε (x) |y − x|
∂n
∂n
∂Bε (x)
ˆ
∂u 1
dS(y)
≤ →
∂−
n L∞ (B (x)) (n − 2)σ εn−2
ε
≤ Cε,
where to obtain the last step, we have used that
ˆ
dS(y) = σn−1 εn−1 .
∂Bε (x)
n−1
∂Bε (x)
48
Therefore
ˆ
lim
ε→0
∂u
Φ(y − x) →
(y) dS(y) = 0,
∂−
n
∂Bε (x)
and the claim 2 is followed.
To summarise, we have proved the following.
Theorem 8. If u ∈ C ( Ω) is a solution of


∆u = f,

u = g,
in Ω,
(3.12)
in
Ω,
where f, g are continuous, then for x ∈ Ω,
ˆ
ˆ
u(x) =
G(x, y)f (y) dy +
Ω
∂Ω
∂G
(x, y)g(y) dS(y).
−
∂→
n
(3.13)
In particular, if f ≡ 0, then
ˆ
u(x) =
∂Ω
3.1
∂G
(x, y)g(y) dS(y).
−
∂→
n
(3.14)
The Green function for a ball
Let Ω = BR (0). Recall that to solve the Dirichlet problem in Ω we need to find, for
each x ∈ Ω, the corrector function hx that satisfies


∆y hx (y) = 0, in Ω,

hx (y) = Φ(y − x),
(3.15)
on ∂Ω.
We will solve this problem by the method of image charge. Think of (3.15) as that we
want to find the electrostatic potential hx inside a spherical conductor of radius R de to
the point charge located at x.
We consider the image of x under the inversion in ∂BR ,
x∗ =
R2
x,
|x|2
which in particular, implies that
|x∗ | =
R2
,
|x|
x∗ · x = R2 .
49
Then, if y ∈ ∂BR , we have
|y − x|2 = |y|2 − 2x · y + |x|2
= R2 − 2x · y + |x|2
|x|2 R4
xR2
2
= 2
−2 2 ·y+R
R
|x|2
|x|
2
|x|
= 2 |x∗ |2 − 2x∗ · y + |y|2
R
|x|2
= 2 |x∗ − y|2 .
R
So we obtain that
Let
|y − x| =
|x|
|y − x∗ |
R



|x|
|y
R
∗
−
x
|
,
n = 2,
hx (y) :=
2−n

|x|
1

|y − x∗ |2−n ,
(2−n)σn−1 R
1
2π
log
∀y ∈ ∂BR .
n ≥ 3.
Note that since |x∗ | |x| = R2 and x ∈ BR , it follows that x∗ 6∈ BR , so that y 6= x∗ .
Therefore y 7→ hx (y) is harmonic in BR . In addition, by the computation above, it holds
that hx (y) = Φ(y − x) on ∂BR . Therefore, hx (y) is the unique corrector function.
Hence, the Green’s function for the ball is given by



1
R |y−x|
 2π
log |x| |y−x∗ | ,
n = 2,
GBR (x, y)
2−n
|x|

1
2−n
∗
2−n

− R
|y − x |
,
 (2−n)σn−1 |y − x|
By the representation formula (3.14), we have
ˆ
∂G
u(x) =
→
− (x, y)g(y) dS(y).
∂Ω ∂ n
Next, we will compute
∂G
(x, y).
→
∂−
n
We have
∂G
−
(x, y) = ∇y G(x, y) · →
n.
−
∂→
n
Here
y
→
−
n = ,
R
∇y G(x, y) =
1
σn−1
"
y−x
−
|y − x|n
|x|
R
2−n
#
y − x∗
.
|y − x∗ |n
n ≥ 3.
50
Note that this formula is true for n ≥ 2. Therefore, we obtain
"
#
2−n
∗
∂G
1
y−x
y−x
|x|
(x, y) =
·y
−
→
−
n
Rσn−1 |y − x|
R
|y − x∗ |n
∂n
2
1
|y| − x · y |x| 2−n |y|2 − x∗ · y
=
−
Rσn−1
|y − x|n
R
|y − x∗ |n
|x| −n |x|2
|x| −n
|y|2
1
1
1
=
−
−x·y
−
Rσn−1 |y − x|n
R
|y − x∗ |n
|y − x|n
R
|y − x∗ |n
1 R2 − |x|2
=
.
Rσn−1 |y − x|n
Set
KR (x, y) :=
1 R2 − |x|2
.
Rσn−1 |y − x|n
(3.16)
This function is called the Poisson kernel. We have proved that
Proposition 3. Let u ∈ C 2 (BR ) be harmonic in BR . Then for x ∈ BR , one has
ˆ
ˆ
1
R2 − |x|2
u(y) dS(y).
u(x) =
KR (x, y)u(y) dS(y) =
Rσn−1 ∂BR |y − x|n
∂BR
(3.17)
The formula above is called the Poisson’s integral formula.
Note that this proposition only tells us that if a solution exists, then it is given in
the interior by the Poisson’s integral formula. We need to show that for suitable g, the
Poisson’s integral formula indeed defines us a harmonic function.
We will need some properties of the Poisson’s kernel KR (x, y).
Lemma 2. The function KR (x, y) satisfies the following properties
(i) For any y ∈ ∂BR , the map
KRy : BR → R
x 7→ KR (x, y)
is harmonic.
(ii) If y ∈ ∂BR and x ∈ BR then KR (x, y) > 0.
(iii) For all x ∈ BR , we have
ˆ
KR (x, y) dS(y) = 1.
∂BR
51
(iv) For y 6= z with y, z ∈ ∂BR , we have
lim KR (x, y) = 0
x→z
x∈BR
and this convergence is uniform in y for |y − z| > δ > 0.
Proof.
(i) We notice that for each i = 1, . . . , n, the function
xi
|x|n
is harmonic on Rn \{0}.
Indeed, since
xi
∂
=
n
|x|
∂xi
|x|2−n
2−n
so
2−n 2−n xi
|x|
∂
|x|
∂
∆ n =∆
=
∆
= 0.
|x|
∂xi 2 − n
∂xi
2−n
It follows that |x − y|2−n and
Rσn−1 KR (x, y) =
(x−y)i
|x−y|n
are both harmonic. So that
2|y|2 − |x − y|2 − 2x · y
y−x
R2 − |x|2
2−n
=
=
−|x
−
y|
+
2y
·
|x − y|n
|x − y|n
|y − x|n
is also a harmonic function.
(ii) This is obvious since |x| < R for x ∈ BR .
(iii) Applying the Poisson’s integral formula for u ≡ 1 yields the result.
(iv) x→z
lim
R2 −|x|2
|y−x|n
=
0
|y−z|n
= 0. For |x − z| < 2δ , then |y − x| ≥ |y − z| − |x − z| ≥ δ − 2δ = 2δ .
x∈BR
Therefore,
R2 − |x|2
R2 − |x|2
≤
→ 0.
n
δ/2
|y−z|>δ |y − x|
sup
Theorem 9. Let g ∈ C 0 (∂BR ) and let us define the function u : BR → R by
ˆ
u(x) =
KR (x, y)g(y) dS(y).
BR
Then u is harmonic in BR , u ∈ C 0 (BR ) and for x0 ∈ ∂BR , we have
lim u(x) = g(x0 ).
x→x0
x∈BR
Proof. Since KR (x, y) is harmonic in x, it is also in C ∞ . Define
I(x, y) := KR (x, y)g(y).
52
Let U be compactly contained in BR . Then I : U × ∂BR → R is infinitely differentiable in
x and the derivative Dxα I(x, y) are continuous maps from U × ∂BR to R. Moreover, since
∂BR is compact, according to the results of the Appendix, we have that u is C ∞ and that
for each x ∈ U we have
ˆ
α
Dxα KR (x, y)g(y) dS(y).
D u(x) =
∂BR
Since any x ∈ BR belongs to some open set U b Ω, it follows that u is smooth and
ˆ
∆u(x) =
∆x KR (x, y)g(y) dS(y) = 0.
∂BR
Now we show the continuity at the boundary. Let x0 ∈ ∂BR . By Property (iii) in 2, we
have
ˆ
KR (x, y)(g(y) − g(x0 )) dS(y) = I1 + I2 ,
u(x) − g(x0 ) =
∂BR
where
ˆ
KR (x, y)(g(y) − g(x0 )) dS(y),
I1 :=
∂BR ∩Bδ (x0 )
and
ˆ
KR (x, y)(g(y) − g(x0 )) dS(y),
I2 :=
∂BR \Bδ (x0 )
for some δ > 0. Since g ∈ C 0 (∂BR ), for a given ε > 0, we can take δ > 0 small enough
that |g(y) − g(x0 )| ≤ ε for y ∈ ∂BR ∩ Bδ (x0 ). Therefore, the term I1 can be estimated by
ˆ
KR (x, y)|g(y) − g(x0 )| dS(y)
|I1 | ≤
∂BR ∩Bδ (x0 )
ˆ
≤ε
KR (x, y) dS(y)
∂BR ∩Bδ (x0 )
ˆ
≤ε
KR (x, y) dS(y) = ε.
(3.18)
∂BR
Since g is continuous (and hence bounded) on ∂BR from part (iv) of Lemma 2, we have
that
KR (x, y)(g(y) − g(x0 )) → 0 as x → x0
uniformly in y ∈ ∂BR \ Bδ (x0 ). Thus as x → x0 , we have I2 → 0.
Together, with the estimate (3.18), we have that limx→x0 u(x) = g(x0 ) as claimed.
As consequences of the Poisson’s integral formula, we obtain the following strengthened versions of the Harnack’s inequality and the Liouville theorem.
53
Theorem 10 (Harnack’s inequality for the ball). Let u ≥ 0 be harmonic in BR . Then
for any x ∈ BR we have the following estimate
|x|
|x|
1 − R u(0)
1 + R u(0)
n−1 ≤ u(x) ≤ n−1 .
|x|
1 + |x|
1
−
R
R
Proof. According to the Poisson’s integral formula we have
ˆ
R2 − |x|2
1
u(x) =
u(y) dS(y).
Rσn−1 ∂BR |y − x|n
By the triangle inequality, we have for any y ∈ ∂BR
R − |x| ≤ |y − x| ≤ |x| + R
So that
R2 − |x|2
σn−1 R(R + |x|)n
ˆ
∂BR
R2 − |x|2
u(y) dS(y) ≤ u(x) ≤
σn−1 R(R − |x|)n
ˆ
u(y) dS(y)
∂BR
By the mean value property
ˆ
u(y) dS(y) = |BR |u(0) = Rn−1 σn−1 u(0).
∂BR
Hence
R2 − |x|2
R2 − |x|2
n−1
R
σ
u(0)
≤
u(x)
≤
Rn−1 σn−1 u(0)
n−1
σn−1 R(R + |x|)n
σn−1 R(R − |x|)n
Simplifying this equality, we obtain
|x|
|x|
1 − R u(0)
1 + R u(0)
n−1 ≤ u(x) ≤ n−1
|x|
1 + |x|
1
−
R
R
as claimed.
Theorem 11 (Improved Liouville theorem). Suppose that u :
Rn → R is harmonic and
bounded from below. Then u must be a constant.
Proof. Without loss of generality, we assume u is harmonic in Rn and u ≥ 0. Take x ∈ Rn ,
according to Theorem 10 , for any R > 0, we have
|x|
u(0)
1
+
u(0)
1 − |x|
R
R
≤
u(x)
≤
n−1
n−1
|x|
1 + |x|
1
−
R
R
Sending R → ∞, we find that u(x) = u(0), ∀x ∈ Rn .
54
C 0-subharmonic functions
3.2
Suppose Ω ⊂ Rn .
Definition 5. We say that a function u ∈ C 0 (Ω) is C 0 -subharmonic in Ω if for any
Bρ (x) b Ω, we have the implication


h ∈ C 2 (Bρ (x)) ∩ C 0 (Bρ (x))



=⇒ u ≤ h in Bρ (x).
∆h = 0, in Bρ (x),





u ≤ h, on ∂Bρ (x)
Note that in this definition, there is no requirement that u ∈ C 2 (Ω).
Lemma 3. If u is C 0 -subharmonic, then for any Bρ (x) b Ω, we have
u(x) ≤
where
ffl
u(x) ≤
u(y) dS(y) and
∂Bρ (x)
u(y) dy,
Bρ (x)
denotes the average integral. And conversely, if u ∈ C 0 and satisfies either the
inequalities above for any ball Bρ (x) b Ω then u is C 0 -subharmonic.
Proof. By Theorem 9, there exists a function h solving



h ∈ C 2 (Bρ (x)) ∩ C 0 (Bρ (x))



∆h = 0, in Bρ (x),




u = h, on ∂Bρ (x)
Since u is C 0 -subharmonic, we have
u(x) ≤ h(x) =
h(y) dS(y) =
∂Bρ (x)
u(y) dS(y),
∂Bρ (x)
where the first equality follows from the mean value property for h.
The second estimate of the Lemma is similar.
Now suppose that


h ∈ C (Bρ (x)) ∩ C (Bρ (x))



∆h = 0, in Bρ (x),





u ≤ h, on ∂Bρ (x)
2
0
Then
(u − h)(x) ≤
(u − h)(y) dy,
Bρ (x)
55
i.e., u − h satisfies the mean value property. Recalling that to prove the strong maximum principle, we only need the mean value property and connectedness of the domain.
Therefore, it implies that u − h satisfies the strong maximum principle. Hence,
u − h ≤ sup (u − h) ≤ 0 in Bρ (x)
Bρ (x)
i.e., the implication in the definition holds.
Proposition 4 (Properties of C 0 -subharmonic functions).
(i) If u is C 0 -subharmonic function then the strong maximum principle holds. Moreover,
we have comparison principle with harmonic functions on every connected set Ω0 b
Ω, i.e., in the definition 5, one can replace the ball Bρ (x) by Ω0 .
(ii) Let u be C 0 -subharmonic in Ω and let Bρ (x) b Ω. We define ū to be the solution of
the problem


∆ū = 0,

ū = u,
and then define the function U by


ū(y),
U (y) =

u(y),
in Bρ (x),
on ∂Bρ (x).
if y ∈ Bρ (x),
if y ∈ Ω \ Bρ (x).
Then U is C 0 -subharmonic in Ω.
The function U is called the harmonic lifting of u in Bρ (x).
(iii) Let u, v be C 0 -subharmonic in Ω and such that u ≤ v. Suppose Bρ (x) b Ω and let
U, V respectively be the harmonic liftings of u and v in Bρ (x). Then U ≤ V .
(iv) If u1 , . . . , uN are C 0 -subharmonic in Ω, then
u(x) := max u1 (x), . . . , uN (x)
is also C 0 -subharmonic in Ω.
Proof.
1. From the proof of Theorem 6, the strong maximum principle follows from the
mean value property. According to Lemma 3 a C 0 -subharmonic function satisfies
the mean value property, so that it satisfies a strong maximum principle.
56
2. Let Br (z) b Ω and suppose that


h ∈ C (Br (z)) ∩ C (Br (z))



∆h = 0, in Br (z),





U ≤ h, on ∂Br (z).
2
0
First of all, note that since ū ≥ u in Bρ (x), then U ≥ u in in Bρ (x) and hence in
Ω. Hence u ≤ U ≤ h in ∂Br (z), it follows that u ≤ h in Br (z). Therefore, U ≤ h
in Br (z) \ Bρ (x). In addition, since ū = u on ∂Bρ (x), it follows that ū ≤ u ≤ h in
∂(Br (x)∩Bρ (x)). Since ū is harmonic in ∂(Br (x)∩Bρ (x)), we obtain that U = ū ≤ h
in Br (x)∩Bρ (x). So U ≤ h in Br (z, which means that U is C 0 -subharmonic function
in Ω.
3. On the set Ω \ Bρ (x), we have U = u ≤ v = V . In addition, we have
∆ū = 0,
ū = u,
∆v̄ = 0,
v̄ = v
in Bρ (x),
on ∂Bρ (x)
Since u ≤ v on ∂Bρ (x), by the comparison principle, we have that ū ≤ v̄ in Bρ (x),
so that also U ≤ V in Bρ (x).
In conclusion, we get U ≤ V in Ω.
4. For two functions f, g ∈ C 0 (Ω) it holds that
1
max{f, g} = [f + g + |f − g|],
2
which implies that max{f, g} ∈ C 0 (Ω). Repeating this argument, we have that if
u1 , . . . , uN ∈ C 0 (Ω) then u = max{u1 , . . . , uN } ∈ C 0 (Ω). Let h be the function
satisfies the conditions in the definition 5. Then ui ≤ h in ∂Bρ (x) and hence ui ≤ h
in Bρ (x) for all i = 1, . . . , N since ui are C 0 -subharmonic. Therefore u ≤ h in Bρ (x),
so u is also C 0 -subharmonic.
57
3.3
Perron’s method for the Dirichlet problem on a
general domain
We are now ready to introduce Perron’s method to solve the Dirichlet problem on a
general domain


∆u = 0,
in Ω

 u = g,
on ∂Ω.
Let g ∈ C 0 (Ω), define
S g := {v : Ω → R such that v is C 0 -subharmonic in Ω and v ≤ g on ∂Ω}.
Theorem 12. Suppose that Ω is open and bounded and that g ∈ C 0 (∂Ω). Define
u(x) := sup {v(x)}.
v∈S g
Then u is harmonic in Ω.
Proof. Step 1) First of all S g 6= ∅ since inf ∂Ω g ∈ S g . Moreover, for each v ∈ S g then
v ≤ sup∂Ω g. So that S g is non-empty, bounded above. Therefore, for each x,
u(x) is well-defined, finite number.
Step 2) Now pick y ∈ Ω. By definition of u, there exist vi ∈ S g such that vi (y) → u(y)
as i → ∞. Define
ṽi (x) = max{inf g, v1 (x), . . . , vi (x)}.
∂Ω
Then
• inf ∂Ω g ≤ ṽ1 ≤ ṽ2 ≤ . . . ≤ ṽi ≤ . . . ≤ sup∂Ω g.
• ṽi are subharmonic.
Step 3) Take ρ sufficiently small such that Bρ (y) b Ω. Let Vi be the harmonic lifting of
ṽi in Bρ (y). Then
• Vi ∈ S g , Vi (y) → u(y),
• Vi are harmonic in Bρ (y),
• inf ∂Ω g ≤ V1 ≤ V2 ≤ . . . ≤ Vi ≤ . . . ≤ sup∂Ω g.
58
Step 4) By Harnack’s theorem 3, Vi → V uniformly in Bρ0 (y) for any ρ0 < ρ, where V is
harmonic. Since Vi ∈ S g , it follows that V ≤ u in Ω.
Step 5) We now show that V = u in Bρ (y). Suppose this is not true. Then there exists
z ∈ Bρ (y) such that V (z) < u(z). By definition of u, there exists ṽ ∈ S g such
that V (z) < ṽ(z). We define
wi = max{ṽ, Vi },
and let Wi be the harmonic lifting of wi in Bρ (y). By similar arguments as in the
previous steps, there exists a new harmonic function W such that
• V ≤ W ≤ u in Bρ (y),
• V (y) = W (y) = u(y),
• W (z) ≥ ṽ(z).
The function W − V is harmonic in Bρ (y) satisfying
W − V ≥ 0 in Bρ (y) and (W − V )(y) = 0.
By the strong maximum principle it follows that V = W in Bρ (y). However, this
is a contradiction since
V (z) < ṽ(z) ≤ W (z).
Therefore, V = u in Bρ (y).
Step 6) Since y is arbitrary, we conclude that V = u in Ω, thus u is harmonic in Ω.
The function u constructed by Theorem 12 is called Perron’s solution. It is a good
candidate for a solution of the Dirichlet problem (3.1). It remains to show that it satisfies
the boundary condition. To achieve this, we need to make some assumptions on the
domain Ω.
Definition 6. We say that Ω satisfies the barrier postulate if for each y ∈ ∂Ω, there exists
a function (called a barrier function) Qy (x) ∈ C 0 (Ω̄) ∩ C 2 (Ω) satisfying
(i) Qy (x) is superharmonic in Ω,
59
(ii) Qy (x) > 0 in Ω̄ \ {y} and Qy (y) = 0.
Theorem 13. Suppose that Ω is open, bounded and satisfies the barrier postulate and
that g ∈ C 0 (∂Ω). Let u be the Perron’s solution. Then for any y ∈ ∂Ω, we have
lim u(x) = g(y).
x∈Ω
x→y
Proof. Let ε > 0 and set M = sup∂Ω |g|. Pick y ∈ ∂Ω and let Qy (x) be the barrier
function. By the continuity of g and the positivity of Qy (x) in Ω̄ \ {y}, there exist δ, K
depending on ε such that
|g(y) − g(z)| < ε for z ∈ ∂Ω, |y − z| < δ,
(3.19)
KQy (x) > 2M for z ∈ ∂Ω, |y − z| ≥ δ.
(3.20)
and
We consider the functions
uy,ε (x) = g(y) − ε − KQy (x),
ūy,ε (x) = g(y) + ε + KQy (x).
Since Qy (x) is superharmonic, uy,ε (x) is subharmonic. If z ∈ ∂Ω, then from (3.19) and
(3.20), we have
uy,ε (z) − g(z) = g(y) − g(z) − ε − KQy (x) ≤ 0,
which implies that uy,ε ∈ S g . Similarly, we get ūy,ε ∈ S̄g . Hence
uy,ε ≤ u(x) ≤ ūy,ε (x) ∀x ∈ Ω,
so that
g(y) − ε − KQy (x) ≤ u(x) ≤ g(y) + ε + KQy (x) ∀x ∈ Ω.
This implies that
−ε − KQy (x) ≤ u(x) − g(y) ≤ ε + KQy (x),
i.e.,
|u(x) − g(y)| ≤ ε + KQy (x).
Since Qy ∈ C 0 (Ω̄ and Qy (y) = 0, there exists δ 0 such that if |x−y| < δ 0 then Qy (x) ≤ K −1 ε
and
|u(x) − g(y)| ≤ 2ε,
which implies that limx→y u(x) = g(y) as claimed.
60
Corollary 4. Under the assumptions as in Theorem 13, the Dirichlet problem (3.1) is
well-posed.
To conclude this section, we provide a criterion that a barrier function exists on Ω for
the point y.
Lemma 4. Suppose that Ω ⊂
Rn
is open and bounded, and that y ∈ ∂Ω. If there exists
Br (z) such that Br (z) ∩ Ω̄ = {y}, then there exists a barrier function on Ω for y.
Proof. Consider the function


r2−n − |x − z|2−n , n ≥ 3,
Qy (x) =

log |x−z| , n = 2.
r
This function is a barrier function on Ω for y.
Problem Sheet 3
Exercise 12 (The Kelvin transform). Given a > 0 and consider the following map
Ta :
Rn \ {0} → Rn \ {0}
x 7→ a2
x
|x|2
which is known as inversion with respect to the sphere of radius a.
a) Show that the map is conformal, i.e., it preserves the angles between vectors
(v, w)
(DTa (x)[v], DTa (x)[w])
=
||DTa (x)[v]|| ||DTa (x)[w]||
||v|| ||w||
(3.21)
for all v, w ∈ Rn , x ∈ Rn \ {0}.
b) Show that in two dimensions Ta sends lines and circles into lines and circles.
c) If Ta is as above, define
va (x) =
a
|x|
n−2
a
|x|
n+2
Show that
∆va (x) =
u(Ta (x)).
∆u(Ta (x)).
What can we say if u is harmonic? va is called the the Kelvin transform of u.
Exercise 13 (Hopf Lemma. See Section 6.4.2 in Evans’s book with Lu = ∆u). Suppose Ω ⊂
Rn
is a smooth bounded domain (which implies the interior ball condition in
Evans’book). Suppose that u ∈ C 2 (Ω) ∩ C 1 (Ω) is subharmonic in Ω and suppose there
exists x0 ∈ ∂Ω with u(x0 ) > u(x) for all x ∈ Ω. Then
∂u
(x0 ) > 0,
∂ν
where ν is the outer unit normal vector.
61
62
Exercise 14 (Symmetry of the Green’s function). Let Ω ⊂ Rn be open, bounded and of
class C 1 . Let G(x, y) be the Green’s function for Ω. Prove that G is symmetric, i.e., for
all x, y ∈ Ω, x 6= y, we have
G(x, y) = G(y, x)
Exercise 15. Using the method of image charges, find the Green’s function of the half
space
Ω1 := {x ∈ Rn : xn > 0}
and of the half ball in
R3 :
Ω2 := {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1, z > 0}.
Exercise 16. a) Find an example of a C 0 -subharmonic function on (0, 1) that is not
everywhere differentiable.
b) Find an example of a C 1 -subharmonic function on (0, 1) which is C 1 -regular but not
everywhere of class C 2 .
Chapter 4
The theory of distributions and
Poisson’s equation
In this chapter, we will solve Poisson’s equation: given f : Rn → R, solve
∆u = f
for some u :
Rn
→
R.
(4.1)
The approach will be via the fundamental solution which is the
solution to the equation where the right-hand side is replaced by the so-called Dirac
distribution (Dirac delta function). The solution to Poisson’s equation then will be determined by a convolution of the fundamental solution with the function f . The main tool
of this chapter will be the theory of distributions. This theory plays an important role in
mathematics and is of independent interest.
4.1
The theory of distributions
As a motivation, let us consider the following simple problem: solve
x2 = a.
We knew that if a ≥ 0, this equation has real solutions x ∈
R.
However, if a < 0,
R to the set of
To extend R to C we not
operations on R, such as
it has no real solution. However, if we extend the set of real numbers
complex numbers C, then it has solutions z ∈
only enlarge
R
to a bigger set
R
⊂
C,
C even if a < 0.
but also generalise the
addition, subtraction and multiplication, to C. After extension to the complex numbers,
63
64
the problem of solving an algebraic equation (not only a quadratic one as the one above)
is in some ways simpler than the original problem: the number of complex roots in an
algebraic equation of order k is always k (counting multiplicities) while over the real line
the number can be any thing between 0 and k. Having found the complex roots, we then
can consider separately the problem of showing that some of them are in fact real.
Turing back to Poisson’s equation (4.1), the aim of this chapter is to treat the situation
where u and f need not be of class C 2 . Generally, we want to make sense of the PDE
equation of order k
Lu :=
X
aα Dα u = f,
|α|≤k
where aα ∈ C ∞ (Ω) for an open Ω ⊂ Rn , in the situation where u and f need not be of class
C k (Ω). To do so, similarly as in the motivating example above, we need to extend the
class of smooth functions to a larger set and generalise the operations on smooth functions
to elements of this set. Let us denote by D0 (Ω) the space to which our generalised solution
u and the right-hand side function f should belong. What properties do we require for
D0 (Ω) so that we can at least make sense of the PDE (4.1)? The list of desirable properties
should contain
1) The smooth functions C ∞ (Ω) should be contained in D0 (Ω) in such a way that we can
recover them, i.e., the inclusion map ι : C ∞ (Ω) ,→ D0 (Ω) should be injective.
2) We need to be able to add elements of D0 (Ω).
3) We need to be able to multiply elements of D0 (Ω) by elements of C ∞ (Ω).
4) We need to be able to differentiate elements of D0 (Ω). Moreover, the generalised
notion of differentiation should agree with the classical one when restricted to smooth
functions.
5) We need some topology on D0 (Ω) such that the above operations are continuous.
Before introducing D0 (Ω), we need to introduce another space, the space of test functions
D(Ω) := Cc∞ (Ω). A topology on D(Ω) is defined as follows.
Definition 7 (Convergence in D(Ω)). A sequence (φ)j∈N , φj ∈ D(Ω) = Cc∞ (Ω) converges
to φ ∈ D(Ω) if there exists a compact set K ⊂ Ω such that supp(φj ), supp(φ) ⊂ K for all
65
j ∈ N and that
Dα φj → Dα φ as j → ∞
uniformly in K for any multi-index α.
The space D(Ω) equipped with this topology becomes a topological vector space over
the real number, i.e, a vector space together with a topology such that addition of vectors
and scalar multiplication are continuous functions. For any topological vector space V ,
one can define the continuous dual space which consists of continuous linear maps
V 0 = {ω : V → R| ω linear, continuous}.
The space D0 (Ω) that we are seeking for will be the continuous dual space of D(Ω).
Definition 8. A distribution T ∈ D0 (Ω) is a linear functional on the space of test functions
T : D(Ω) → R
which is continuous with respect to the topology on D(Ω) defined in Definition 7. In other
words, if φj → φ in the sense of Definition 7 then
T φj → T φ as j → ∞.
We consider two important examples
Example 20.
1. If f : Ω →
R
is continuous, the distribution Tf associated to f is
define by
ˆ
f (x)φ(x) dx ∀φ ∈ Cc∞ (Ω).
Tf (φ) :=
Ω
In one-dimension, we can allow f to have a finite number of points of finite discontinuity. Because of this, a distribution is also called generalised function.
2. (The Dirac delta) For x ∈ Ω, the Dirac distribution δx is defined via
δx φ = φ(x).
We now show that the wish-list properties above are satisfied.
Proposition 5. Suppose that f ∈ C 0 (Ω). Let ρk,y , k = 1, 2, . . ., be a family of mollifiers
concentrating at y, i.e., ρk,y ∈ Cc∞ (Ω), ρk,y ≥ 0 and
ˆ
ρk,y (x) dx = 1, supp(ρk,y ) ⊂ B1/k (y).
Ω
66
Then
f (y) = lim Tf (ρk,y ).
k→∞
As a consequence, if f, g ∈ C 0 (Ω) and Tf = Tg then f = g.
Proof. Since
´
Ω
ρk,y (x) dx = 1, we have that
ˆ
f (y) − Tf (ρk,y ) = (f (y) − f (x)) ρk,y (x) dx.
Ω
Since f is continuous, given ε > 0, there exists K ∈
N such that |f (y) − f (x)| < ε for
x ∈ B1/k (y) whenever k ≥ K. For such a k, we have
ˆ
ˆ
|f (y) − Tf (ρk,y )| ≤
|f (y) − f (x)| ρk,y (x) dx ≤ ε ρk,y (x) dx = ε,
Ω
Ω
which implies that f (y) = limk→∞ Tf (ρk,y ).
Now suppose that f, g ∈ C 0 (Ω) and Tf = Tg . For any y ∈ Ω, let ρk,y be the mollifiers
concentrating at y. Then
f (y) = lim Tf (ρk,y ) = lim Tg (ρk,y ) = g(y),
k→∞
k→∞
so that f = g as claimed.
Next, we extend the operations on functions to distributions as follows.
1) Given T1 , T2 ∈ D0 (Ω). The distribution T1 + T2 is defined by
(T1 + T2 )(φ) := T1 (φ) + T2 (φ), ∀φ ∈ D(Rn ).
2) Given a ∈ C ∞ (Ω) and T ∈ D0 (Ω), the distribution aT is defined by
∀φ ∈ D(Rn ).
(aT )(φ) := T (aφ),
Note that the right-hand side makes sense since aφ ∈ D(Rn ).
3) First, note that if f ∈ C ∞ (Ω), then Di f ∈ C ∞ (Ω) and
ˆ
ˆ
TDi f (φ) =
Di f (x) φ(x) dx = − f (x) Di φ(x) dx = −Tf (Di φ),
Ω
Ω
Note that in the above computations, there are no boundary terms because φ vanishes
at the infinity. Motivated by this, we define the derivatives of a distribution as follows.
For any multi-index α, we define
(Dα T )(φ) := (−1)|α| T (Dα φ).
67
Let us do some examples
Example 21. 1) T = δx . Then
ˆ
(Di T )(φ) = −T (Di φ) = −
δx Di φ(y) dy = −(Di φ)(x),
Ω
2) Consider the Heaviside step function


1,
H(x) =

0,
x ≥ 0,
∀φ ∈ D(Rn ).
(x ∈ R).
x<0
Note that this function is not differentiable at x = 0. We consider the distribution TH
by
ˆ
TH (φ) =
R
H(x)φ(x) dx,
for φ ∈ D(Rn ).
Then we compute
(Dx TH )(φ) = −TH (Dx φ)
ˆ
= − H(x)Dx φ(x) dx
ˆR∞
Dx φ(x) dx
=−
0
= φ(0) = δ0 (φ).
Therefore, Dx TH = δ0 .
So the theory of distributions provides us some sort of meaning to the derivative of a
function whose classical derivatives do note exist.
Theorem 14. Suppose that f ∈ C 0 (Ω) and let Tf be the distribution associated to f .
Suppose further that for any multi-index α, with |α| ≤ k, there exist gα ∈ C 0 (Ω) such that
Dα Tf = Tgα
where Dα is the distributional derivative. Then f ∈ C k (Ω) and Dα f = gα in the classical
sense.
This theorem states that the distributional derivatives agree with the classical ones
when they are both defined.
68
We have addressed the four wish-lists that we asked in the introduction. It is straightforward to check that all the operations that we have defined are indeed continuous with
respect to the topology that we have introduced.
The advantage of introducing distributions is expressed in the following result, which
is followed from Theorem 14 and all the properties we have defined.
Proposition 6. Suppose that f ∈ C 0 (Ω) and that T ∈ D0 (Rn ) satisfies the distributional
equation
X
aα Dα T = Tf ,
|α|≤k
where aα ∈ C ∞ (Ω) and that there exist functions uα ∈ C 0 (Ω) such that Dα T = uα . Then
u := u0 is in C k (Ω) and is a classical solution of the PDE
X
aα Dα u = f.
(4.2)
|α|≤k
This proposition provides us a method to solve the PDE (4.2) that consists of two
steps.
1) Lift up the PDE (4.2) to the distributional PDE
X
aα Dα T = Tf
|α|≤k
and solve this distributional equation.
2) Check whether a distributional solution is a classical one.
In the next section, we will study how to solve the distributional equation above.
4.2
Convolutions and the fundamental solution
Recalling that for f ∈ C 0 (Rn ) and φ ∈ Cc∞ (Rn ), the convolution f ∗ φ is again a
function defined by
ˆ
(f ∗ φ)(x) :=
f (y)φ(x − y) dy.
Rn
How to extend this definition to T ∗ φ where T ∈ D0 (Rn ), φ ∈ Cc∞ (Rn )? To do so we define
τx φ : Rn → R
y 7→ φ(x − y).
69
If φ ∈ Cc∞ (Rn ) then so is τx φ. Now we can write
(f ∗ φ)(x) = Tf (τx φ).
The advantage of writing this way is that we can extend it to distributions: let T ∈
D0 (Ω), φ ∈ D(Rn ), we define
(T ∗ φ)(x) := T (τx φ).
Before stating some important properties of convolutions, we need to introduce the concept of support of a distribution.
Definition 9 (Support of a distribution). A distribution T ∈ D0 (Rn ) is supported in the
closed set K ⊂ Ω if
T φ = 0 ∀φ ∈ Cc∞ (Ω \ K),
i.e, T vanishes when applying to any test function that vanishes on K. The support of T
is defined by
suppT =
\
{K : T supported in K}.
Example if T = δx then suppT = {x}.
Proposition 7 (Properties of convolutions). Suppose T, Ti ∈ D0 (Rn ) and φ ∈ D(Rn ).
Then
(i) If T1 ∗ φ = T2 ∗ φ,
∀φ ∈ D(Rn ) then T1 = T2 .
(ii) T ∗ φ ∈ C ∞ (Rn ) and
Dα (T ∗ φ) = T ∗ Dα φ = Dα T ∗ φ.
(iii) If T has compact support, then T ∗ φ also has compact support; in particular T ∗ φ
is a test function.
Proof.
(i) We have (τx φ)(y) = φ(x − y), so that φ(−y) = (τ0 φ)(y), or equivalently,
φ = τ0 φ̃, where φ̃(y) := φ(−y). Then we have
T1 (φ) = T1 (τ0 φ̃) = (T1 ∗ φ̃)(0) = (T2 ∗ φ̃)(0) = T2 (φ),
which means that T1 = T2 as desired.
70
(ii) We first show that Di (T ∗ φ) = T ∗ Di φ. In fact, let ei be the i-th unit vector, we
calculate
i
(T ∗ φ)(x + εei ) − (T ∗ φ)(x)
1h
= T (τx+εei φ) − T (τx φ)
ε
ε
h1
i
= T (τx+εei φ − τx φ) ,
ε
where we have used the linearity of T . Since φ ∈ D(Rn ), we have
1
1
lim (τx+hei φ − τx φ)(y) = lim (φ(x − y + εei ) − φ(x − y))
ε→0 ε
ε→0 ε
∂
φ(x − y) = (τx Di φ)(y),
=
∂xi
with convergence in the topology of D(Rn ). By the continuity of distributions, we
have
(T ∗ φ)(x + εei ) − (T ∗ φ)(x)
= T (τx Di φ) = (T ∗ Di φ)(x),
ε→0
ε
lim
i.e., Di (T ∗ φ) = T ∗ Di φ. By repeating this arguments for higher derivatives, we
establish the first equality. To get the second equality, we calculate
Dα (τx φ)(y) =
∂ |α|
φ(x − y)
∂y α
= (−1)|α| (Dα φ)(x − y)
= (−1)|α| (τx Dα φ)(y),
so that Dα (τx φ) = (−1)|α| (τx Dα φ).
Therefore
(Dα T ∗ φ)(x) = Dα T [τx φ]
= (−1)|α| T (Dα τx φ)
= (−1)|α| T ((−1)|α| τx Dα φ)
= T (τx Dα φ)
= T ∗ Dα φ(x).
Hence Dα T ∗ φ = T ∗ Dα φ.
(iii) Suppose without loss of generality that T and φ are supported in BR (0) for some large
R. Let x ∈ Rn be such that |x| > 2R. Then if y ∈ BR (0), by the triangle inequality,
|y − x| > R. So that (τx φ)(y) = φ(x − y) = 0, i.e., τx φ ∈ Cc∞ (Rn \ BR (0)). By the
71
assumption that T is supported in BR (0), we have that T (τx φ) = (T ∗ φ)(x) = 0,
hence T ∗ φ(x) = 0 for all |x| > 2R, i.e., T ∗ φ has compact support.
We have extended addition, multiplication, derivatives and convolution of a distribution with a test function. Next, we want to extend the convolution of two distributions:
given T1 , T2 ∈ D0 (Rn ), how to define T1 ∗ T2 ? We recall that if f, g, h ∈ Cc0 (Rn ), then
(f ∗ g) ∗ h = f ∗ (g ∗ h).
In deed,
ˆ
[(f ∗ g) ∗ h](x) =
(f ∗ g)(y)h(x − y) dy
n
R
ˆ ˆ
f (z)g(y − z) dz h(x − y) dy
=
n
n
R
R
ˆ
ˆ
g(y − z)h(x − y) dy dz
=
f (z)
n
n
R
R
ˆ
ˆ
g(ỹ)h(x − ỹ − z) dỹ dz
=
f (z)
n
n
R
R
ˆ
=
f (z)(g ∗ h)(x − z) dz
Rn
= (f ∗ (g ∗ h))(x).
Motivated by this we define
Definition 10. Suppose that T1 , T2 ∈ D0 (Rn ) and that T2 has compact support. The
convolution T1 ∗ T2 is defined to be the unique distribution that satisfies
(T1 ∗ T2 ) ∗ φ = T1 ∗ (T2 ∗ φ),
∀φ ∈ D(Rn ).
Note that in the definition above, the requirement that T2 has compact support ensures
that T2 ∗ φ is a test function according to the last property in Proposition 7.
Theorem 15. Suppose that T1 , T2 ∈ D0 (Rn ) and that T2 has compact support. Then
Dα (T1 ∗ T2 ) = T1 ∗ Dα T2 = Dα T1 ∗ T2 .
72
Proof. We have
Dα (T1 ∗ T2 ) ∗ φ = (T1 ∗ T2 ) ∗ Dα φ
= T1 ∗ (T2 ∗ Dα φ)
= T1 ∗ (Dα T2 ∗ φ)
= (T1 ∗ Dα T2 ) ∗ φ
so that Dα (T1 ∗ T2 ) = (T1 ∗ Dα T2 ). Similarly,
(T1 ∗ Dα T2 ) ∗ φ = T1 ∗ (Dα T2 ∗ φ)
= T1 ∗ Dα (T2 ∗ φ)
= Dα T1 ∗ (T2 ∗ φ)
= (Dα T1 ∗ T2 ) ∗ φ,
so that (T1 ∗ Dα T2 ) = (Dα T1 ∗ T2 ) and we are done.
Example 22.
(i) If φ ∈ D(Rn ), then δ0 ∗ φ = φ.
(ii) If T ∈ D0 (Rn ) has compact support then δ0 ∗ T = T .
Definition 11 (Definition of the fundamental solution). We say that a distribution G is
a fundamental solution of the partial differential operator
L :=
X
aα D α
|α|≤k
where aα are constants, if it satisfies the distributional equation
LG = δ0 .
Lemma 5. Suppose that G ∈ D0 (Rn ) is a fundamental solution of L and that T0 is a
distribution with compact support. Then the distribution G ∗ T0 solves the distributional
equation
LT =
X
|α|≤k
aα Dα T = T0 .
73
Proof. We have
L(G ∗ T0 ) =
X
aα Dα (G ∗ T0 )
|α|≤k
=
X
aα (Dα G ∗ T0 )
|α|≤k
=
X
aα Dα G ∗ T0
|α|≤k
= (LG) ∗ T0
= δ0 ∗ T0 = T0 .
4.3
Poisson’s equation
We will apply the abstract framework in the previous section to solve Poisson’s equation: given f : Rn → R has compact support. Find u such that
∆u = f
in
Rn .
First of all, we notice that, for n ≥ 3, classical solutions to this problem are unique (if
they exist) if we assume that u → 0 as |x| → ∞. Indeed, suppose that u1 , u2 satisfy both
Poisson’s equation and the decay condition. Then w = u1 − u2 is harmonic in
Rn and is
bounded. By Liouville’s theorem, w must be a constant. Since w → 0 as |x| → ∞, the
constant must be 0. Hence, u1 = u2 .
Next, we will show that provided ρ ∈ Cc2 (Rn ), a classical solution satisfying the decay
condition indeed exists. The procedure follows from the abstract framework
1. We first show that a distributional solution exists using the fundamental solution.
2. Then we show that the distributional solution arises from a C 2 function. This
function will be the classical solution to Poisson’s equation.
74
4.3.1
The fundamental solution
Recall that in Chapter 3, we have defined the function


 1 log |x|, n = 2,
2π
Φ(x) =

1

|x|2−n , n ≥ 3.
(2−n)σn−1
Furthermore, for any φ ∈ Cc∞ (Rn ) we have that
ˆ
Φ(x − y)∆φ(y) dy.
φ(x) = lim
ε→0 Rn \B (x)
ε
By changing variable z = x − y, the above identity can be written as
ˆ
φ(x) = lim
Φ(z)∆φ(x − z) dz = (G ∗ ∆φ)(x),
ε→0 Rn \B (0)
ε
where the distribution G is defined by
ˆ
Φ(z)∆φ(z) dz.
Gφ = lim
ε→0 Rn \B (0)
ε
So we have φ = G ∗ ∆φ = ∆G ∗ φ, which means that ∆G = δ0 , i.e., G is the fundamental
solution to Poisson’s equation.
As a consequence, if f ∈ Cc0 (Rn ), then
∆(G ∗ Tf ) = Tf
i.e., G ∗ Tf is a solution to the distributional equation ∆T = Tf .
Lemma 6. Suppose f ∈ Cck (Rn ). Then
G ∗ Tf = Tu ,
where
ˆ
u(x) := lim
ε→0
Rn \Bε (x)
Φ(x − y) f (y) dy
and u ∈ C k (Rn ).
Proof. Let φ ∈ Cc∞ (Rn ). Let x ∈
Rn
and R > 0 sufficiently large that R > |x| and that
75
supp(φ) ∪ supp(f) ⊂ BR (0). We calculate
[(G ∗ Tf ) ∗ φ](x) = [G ∗ (Tf ∗ φ)](x)
= G[τx (Tf ∗ φ)]
ˆ
= lim
Φ(y)[τx (Tf ∗ φ)](y) dy
ε→0 Rn \B (0)
ε
ˆ
Φ(y)(Tf ∗ φ)(x − y) dy
= lim
ε→0 Rn \B (0)
ε
ˆ
hˆ
i
Φ(y)
f (z)φ(x − y − z) dz dy
= lim
ε→0 Rn \B (0)
Rnˆ
ε
ˆ
h
i
Φ(y)
f (z)φ(x − y − z) dz dy
= lim
ε→0 B (0)\B (0)
BR (0)
ε
ˆ 3R
hˆ
i
Φ(y)
f (w − y)φ(x − w) dw dy
= lim
ε→0 B (0)\B (0)
B4R (0)
ε
ˆ 3R h ˆ
i
= lim
Φ(y)f (w − y) dy φ(x − w) dw
ε→0 B (0)
B3R (0)\Bε (0)
4R
ˆ
hˆ
i
= lim
Φ(w − z)f (z) dz φ(x − w) dw
ε→0 B (0)
Rn \Bε (w)
2R
Define
ˆ
Φ(w − z)f (z) dz.
Rn \Bε (w)
Then due to the property of Φ, we have that uε → u uniformly in w, so that we can take
uε (w) =
the limit inside the integral above and obtain
ˆ
ˆ
h
i
lim
Φ(w − z)f (z) dz φ(x − w) dw
[(G ∗ Tf ) ∗ φ](x) =
B (0) ε→0 Rn \Bε (w)
ˆ 2R
=
u(w)φ(x − w) dw
B2R (0)
= (Tu ∗ φ)(x),
so that G ∗ Tf = Tu .
Next we show that u ∈ C k (Rn ). Since f ∈ Cck (Rn ), we have
ˆ
ˆ
uε (x) =
Φ(w − z)f (z) dz =
Φ(y)f (x − y) dy.
Rn \Bε (w)
BR (0)\Bε (0)
If |α| ≤ k, then
ˆ
α
Φ(y)Dα f (x − y) dy
D uε =
B2R (0)\Bε (0)
ˆ
Φ(x − y)Dα f (y) dy
=
BR (0)\Bε (0)
ˆ
=
Rn \Bε (0)
Φ(x − y)Dα f (y) dy
76
which converges uniformly to some limit by our previous results. Since uε and Dα uε
converge uniformly for |α| ≤ k, we conclude that u ∈ C k (Rn ).
Now we are ready to state the following theorem on the existence of a classical solution
satisfying the decay condition.
Theorem 16. If f ∈ Cc2 (Rn ) then
ˆ
u(x) = lim
ε→0
Rn \Bε (x)
Φ(x − y) f (y) dy
is a classical to Poisson’s equation. Furthermore, if n ≥ 3 then u → 0 as |x| → ∞.
Proof. By Lemma 6 u ∈ C 2 (Rn ) and Tu is a distributional solution of Poisson’s equation.
By Proposition 6, u is a classical solution of Poisson’s equation. Now let n ≥ 3. Let
R > 0 be such that supp(f) ⊂ BR (0). For |x| > 2R and y ∈ BR (0) then |y| ≤
|x − y| >
|x|
,
2
so that
|x|
.
2
We then can estimate
ˆ
n−2
2
Vol(B
(0))
R
|u(x)| = Φ(x − y) f (y) dy ≤
sup |f |,
(n − 2)σn−1 |x|
Rn
BR (0)
which implies that u(x) → 0 as |x| → ∞.
4.3.2
Poisson’s equation in a bounded domain
Let Ω ⊂ Rn be open, bounded and satisfies the barrier postulate. Poisson’s equation
in Ω is: given f ∈ Cc2 (Ω) and g ∈ C 0 (∂Ω), find u ∈ C 2 (Ω) ∩ C 0 (Ω) such that


∆u = f, in Ω,

u = g,
(4.3)
in ∂Ω.
This problem can be solved in two steps as follows. Each step is a problem that we know
how to solve.
Step 1) Find w ∈ C 2 (Rn ) solve ∆w = f in
Rn .
Step 2) Find v ∈ C 2 (Ω) ∩ C 0 (Ω) such that


∆u = 0,
in Ω,

u = g − w,
in ∂Ω.
Then u = w + v solve (4.3). Indeed, u ∈ C 2 (Ω) ∩ C 0 (Ω) and
∆u = ∆w + ∆v = f
in Ω, u = v + w = g − w + w = g
on ∂Ω.
77
Problem Sheet 4
Exercise 17. Consider the following domain of R2 , defined using polar coordinates (r, ϕ):
Ω0 := {(r, ϕ) : θ < ϕ < 2π − θ; 0 < r}
where θ ∈ (0, π).
a) Determine, with justification, for which values of θ the exterior sphere condition is
satisfied for every point of ∂Ω0 . (Recall: Suppose that Ω is open and bounded and
that y ∈ ∂Ω. If there exists Br (z) such that Br (z) ∩ Ω = {y} then we say that the
exterior sphere condition holds at y.)
b) Show that a barrier function exists at all points of ∂Ω0 , even when the exterior sphere
conditions does not hold. [Hint: Consider the expression for the Laplacian on
R2
in
polar coordinates.]
c) An open domain Ω of R2 is said to satisfy the exterior cone condition at a point ξ ∈ ∂Ω
if there exists a cone C with vertex ξ and a ball Br (ξ) such that
Ω ∩ Br (ξ) ∩ C = {ξ}
Prove using the idea for b) that if the exterior cone condition is satisfied at ξ the a
barrier function exists for ξ.
Exercise 18. a) Show that if f1 , f2 ∈ C 0 (Ω) and a ∈ C ∞ (Ω) then
aTf1 + Tf2 = Taf1 +f2 .
b) Show that if f ∈ C k (Ω) then
Dα Tf = TDα f
for |α| ≤ k.
78
79
c) Deduce that if f ∈ C k (Ω) then
X
aα Dα Tf = TLf ,
|α|≤k
where
Lf =
X
aα Dα f.
|α|≤k
Exercise 19. a) Show that if f, g, h ∈ Cc0 (Rn ) then
Tf ∗g = Tf ∗ Tg .
b) Show that convolution is linear in both of its arguments, i.e, if Ti ∈ D0 (Rn ) and T3 , T4
have compact support then
(T1 + aT2 ) ∗ T3 = T1 ∗ T3 + aT2 ∗ T3 ,
and
T1 ∗ (T3 + aT4 ) = T1 ∗ T3 + aT1 ∗ T4
where a ∈ R is a constant.
Chapter 5
The heat equation
In this chapter we will study the heat equation: given u0 : Rn → R. We want to find
a function u : Rn × [0, ∞) that satisfies the following heat equation
∂t u(x, t) = ∆u(x, t),
u(x, 0) = u0 (x),
x ∈ Rn , t > 0,
x ∈ Rn .
(5.1)
(5.2)
Note that ∆ is the Laplacian operator with respect to the spatial variable only. The heat
equation describes the evolution of the temperature of a homogeneous body given the
temperature profile at the initial time.
The procedure to solve this equation will be similar as in the previous chapter: first, we
construct the fundamental solution, and then taking the convolution of the fundamental
solution with the initial data.
5.1
The fundamental solution of the heat equation
We first have two observations.
Lemma 7. If u(x, t) satisfies (5.1) then so does v(x, t) := u(λx, λ2 t) for any λ > 0.
Proof. By the chain rule we have
∂t v(x, t) = λ2 ∂t u(λx, λ2 t) and ∆v(x, t) = λ2 ∆u(λx, λ2 t).
Thus ∂t v(x, t) = λ2 [∂t u − ∆u](λx, λ2 t) = 0 as desired.
80
81
Lemma 8. If u satisfies (5.1) and Du vanishes at the infinity then for any t > 0
ˆ
ˆ
u(x, t) dx =
u0 (x) dx.
Rn
Rn
Proof. We calculate
ˆ
ˆ
d
u(x, t) dx =
∂t u(x, t) dx
dt Rn
n
ˆR
=
=
n
ˆR
Rn
∆u(x, t) dx
div(Du(x, t)) dx
= 0,
where in the last equality we have applied the divergence theorem and the assumption
that Du vanishes at the infinity.
Motivated by these two observations, we will seek a self-similar solution of the heat
equation of the form
u(x, t) = h(t)f (ξ) where ξ =
|x|2
∈ R,
t
´
´
for some function h and f , such that Rn u(x, t) dx = Rn u0 (x) dx = 1. The last identity
is assumed without loss of generality.
We first find h. We have
ˆ
ˆ
|x|2 1=
u(x, t) dx = h(t)
f
dx
t
n
Rn
R
ˆ
ˆ ∞ 2
r
f
= h(t)
rn−1 dw dr.
t
n−1
S
0
By changing of variable y =
r2
,
t
1
we have r = (yt) 2 and dr =
t
2r
1
dy =
this back into the calculations above, we get
ˆ
n
n
σn−1 ∞
2
1 = t h(t)
f (y)y 2 −1 dy.
2
0
This implies that
−n
2
h(t) = t
σn−1
and 1 =
2
ˆ
∞
n
f (y)y 2 −1 dy.
0
t2
1
2y 2
dy. Substituting
82
It remains to find f . We have
|x|2 i
hn
2
n |x|
n
n n
−n
2
∂t u = ∂t t f
= − t− 2 −1 f (ξ) − t− 2 2 f 0 (ξ) = −t− 2 −1 f (ξ) + ξf 0 (ξ)
2
2
t
2
n xi
Di u = 2t− 2 f 0 (ξ),
t
n
2
2
n x
2t
Dii2 u =
f 0 (ξ) + 4t− 2 2i f 00 (ξ)
t
t
n
X
n
∆u =
Dii2 u = t− 2 −1 (2nf 0 (ξ) + 4ξf 00 (ξ)) .
i=1
Therefore, we obtain the following equation for f (ξ)
4ξf 00 (ξ) + ξf 0 (ξ) + 2nf 0 (ξ) +
n
f (ξ) = 0.
2
Set g(ξ) = f 0 (ξ) + 14 f (ξ), the above equation can be written as
ξg 0 (ξ) +
n
g(ξ) = 0,
2
which implies that
n
g(ξ) = Aξ − 2
for some constant A. Therefore,
n
1
f 0 (ξ) + f (ξ) = Aξ − 2 ,
4
i.e.,
n
1
A = ξ 2 (f 0 (ξ) + f (ξ)).
4
Assume that f and f 0 decay fast enough at the infinity, we obtain that the right-hand
side vanishes as |ξ| → ∞. Thus A = 0 and so
1
f 0 (ξ) + f (ξ) = 0.
4
ξ
By multiplying with the integrating factor e 4 , we obtain
d ξ
e 4 f (ξ) = 0
dξ
Solving this equation, we find
ξ
f (ξ) = Be− 4 .
(5.3)
83
We now find B. We have
σn−1
1=B
2
ˆ
∞
ξ
n
e− 4 ξ 2 −1 dξ
0
n
2
ˆ
∞
2
e−s sn−1 ds (change of variable ξ = 4s2 )
= Bσn−1 4
0
ˆ
n
2
= B4 2
e−|x| dx
n
n
ˆR ∞
√ n
−x2
e
dx = B(2 π) 2 .
=B 2
0
so B =
√1
.
(2 π)n
Conclusion
n
u(x, t) = t− 2
|x|2
1 − |x|2
1
− 4t
4t =
.
e
n e
(2π)n
(4πt) 2
To summarise, we have proved the following result
Proposition 8. The function
Γ(x, t) =
|x|2
1
− 4t
e
n
(4πt) 2
is a solution of the equation (5.1) and satisfies
ˆ
Γ(x, t) dx = 1,
Rn
∀t > 0.
The function Γ is called the heat kernel or the fundamental solution of the heat equation. It plays a similar role to the Green’s function for the Laplace operator, which can
be seen in the following theorem.
Theorem 17. Let u0 ∈ Cc0 (Rn ). Define u : Rn × (0, ∞) → R by
ˆ
ˆ
|x−y|2
1
u(x, t) :=
Γ(x − y, t)u0 (y) dy =
e− 4t u0 (y) dy.
n
(4πt) 2 Rn
Rn
Then u solves the heat equation (5.1)-(5.2). That is, u satisfies
∂t u(x, t) = ∆u(x, t)
∀(x, t) ∈ Rn × (0, ∞),
and
lim u(x, t) = u0 (x)
t→0
∀x ∈ Rn .
Proof. According to Proposition 8, Γ(x − y, t) satisfies
∂t Γ(x − y, t) = ∆Γ(x − y, t).
(5.4)
84
Furthermore, as a function of x and t, Γ(x − y, t) is smooth in
Rn × (0, ∞).
Suppose that
u0 is supported in BR (0). Then
ˆ
Γ(x − y, t)u0 (y) dy.
u(x, t) =
BR (0)
This implies that u(x, t) is also smooth in
integral and obtain that
ˆ
(∂t u − ∆u)(x, t) =
BR (0)
Rn × (0, ∞) and we can differentiate under the
(∂t − ∆x )Γ(x − y, t)u0 (y) dy = 0 ∀(x, t) ∈ Rn × (0, ∞).
We now verify the initial condition. Since u0 is continuous and compactly supported, it
is uniformly continuous. Therefore, given any ε > 0 there exists δ > 0 such that for any
x, z ∈ Rn with |x − z| < δ then |u0 (x) − u0 (z)| < ε. Since
ˆ
Γ(x − y, t) dy = 1,
Rn
we can write
ˆ
u(x, t) − u0 (x) =
Rn
Γ(x − y, t)(u0 (y) − u0 (x)) dy
and estimate
ˆ
|u(x, t) − u0 (x)| ≤
ˆR
Γ(x − y, t)|u0 (y) − u0 (x)| dy
ˆ
n
Γ(x − y, t)|u0 (y) − u0 (x)| dy +
=
Bδ (x)
Rn \Bδ (x)
Γ(x − y, t)|u0 (y) − u0 (x)| dy.
:= I1 + I2 .
Since for y ∈ Bδ (x), we have |u0 (y) − u0 (x)| < ε, the term I1 can be estimated
ˆ
ˆ
I1 ≤ ε
Γ(x − y, t) dy ≤ ε
Γ(x − y, t) dy = ε.
Bδ
Rn
The term I2 can be estimated as follows
ˆ
I2 ≤ 2 sup |u0 |
Γ(x − y, t) dy
Rn
Bδ (x)
ˆ
= 2 sup |u0 |
Γ(y, t) dy
Rn
Bδ (0)
ˆ ∞
s2
σn−1
= 2 sup |u0 |
e− 4t sn−1 ds
n
(4πt) 2 δ
Rn
ˆ
σn−1 ∞ −z2 n−1
= 2 sup |u0 |
dz → 0 as t → 0.
e z
n
(π) 2 2√δ t
Rn
85
Therefore, there exists δ 0 > 0 depending on δ (thus on ε but not on x) such that I2 < ε
for t < δ 0 . In conclusion, we have shown that for any ε > 0 there exists δ 0 > 0 such that
for t < δ 0 and for any x, we have
|u(x, t) − u0 (x)| < 2ε.
So limt→0 u(x, t) = u0 (x) as claimed.
We now show that the solution of the heat equation is always bounded.
Theorem 18. Let u ∈ Cc0 (Rn ) and let u(x, t) be defined as in (5.4). Then
|u(x, t)| ≤ max
|u |
Rn 0
∀(x, t) ∈ Rn × (0, ∞).
Proof. This estimate is straightforward:
ˆ
ˆ
|u(x, t)| ≤
Γ(x − y, t)|u0 (y)| dy ≤ max
|u |
Γ(x − y, t) dy = max
|u |.
Rn 0 Rn
Rn 0
Rn
Furthermore, we now show that the solution of the heat equation decays in time.
Theorem 19. Let u ∈ Cc0 (Rn ) and let u(x, t) be defined as in (5.4). Then there exists a
constant C depending on n and u0 such that
|u(x, t)| ≤
C
n
t2
∀(x, t) ∈ Rn × (0, ∞).
Proof. Suppose that u0 is supported in BR (0). Since Γ(x − y, t) ≤
1
n
(4πt) 2
, we can estimate
ˆ
|u(x, t)| ≤
Γ(x − y, t)|u0 (y)| dy
ˆ
≤ max
|u |
Γ(x − y, t) dy
Rn 0 BR (0)
1
C
≤ max
|u
|
n Vol(BR (0)) =
n ,
0
Rn
(4πt) 2
t2
BR (0)
where C = max
|u | 1 n Vol(BR (0).
Rn 0 (4π) 2
Theorem 17 only tells us that the function u defined by (5.4) is a solution to the heat
equation, but does not ensure it uniqueness. To obtain a statement about the uniqueness,
we will need to use the maximum principle in the next sections.
86
5.2
The maximum principle for the heat equation on
a bounded domain
We will show that the heat equation also satisfies a similar maximum principle as
harmonic functions.
Definition 12. Let Ω ⊂
Rn
be open and bounded and T > 0 be given. The space-time
domain ΩT is defined by
ΩT = Ω × (0, T ).
The parabolic boundary of ΩT is defined by
∂p ΩT = ∂(ΩT ) \ (Ω × {T }).
We denote by C 2,1 (ΩT ) the set of functions that are continuously differentiable twice
in space and once in time. For functions u ∈ C 2,1 (ΩT ), we define the heat operator L by
(Lu)(x, t) := ∂t u(x, t) − ∆u(x, t).
We have the following maximum principle.
Theorem 20 (The maximum principle for the heat operator on bounded domains).
Suppose that Ω ⊂
Rn
is open and bounded, T > 0 and that u ∈ C 2,1 (ΩT ) ∩ C 0 (ΩT )
satisfies
Lu ≤ 0
in ΩT .
Then
sup u = sup u.
ΩT
∂p ΩT
If instead Lu ≥ 0 in ΩT , then
inf u = inf u.
ΩT
∂p ΩT
Proof. Suppose that Lu ≤ 0. Let 0 < ε < T be arbitrary and let us define the function
v : ΩT −ε → R
(x, t) 7→ u(x, t) − εt.
87
The function v satisfies
v(x, t) ≤ u(x, t) ≤ v(x, t) + εT,
and
(5.5)
vt − ∆v ≤ −ε < 0 in Ω × (0, T − ε].
(5.6)
Since v is continuous, it attains a global maximum at some point (x0 , t0 ) ∈ ΩT −ε .
1. If (x0 , t0 ) ∈ ΩT −ε then we must have that ∂t v(x0 , t0 ) = 0 and Hessx v(x0 , t0 ) ≤ 0. So
(vt − ∆v)(x0 , t0 ) = −∆v(x0 , t0 ) ≥ 0,
which contradicts (5.6).
2. If (x0 , t0 ) ∈ Ω×{T −ε} then we must have that vt (x0 , t0 ) ≥ 0 (since v must decrease if
we fix x and allow t to decrease) and Hessx v(x0 , t0 ) ≤ 0 (since the function v(T −ε, x)
must have a local max at x0 ). Therefore,
(vt − ∆v)(x0 , t0 ) ≥ 0,
again contradicts (5.6).
Therefore, we have that (x0 , t0 ) ∈ ∂p ΩT −ε and hence
sup v = sup v ≤ sup u ≤ sup u.
ΩT −ε
∂p ΩT −ε
∂p ΩT −ε
∂p ΩT
On the other hand, since u ≤ v + εT , we have
sup u ≤ sup v + εT ≤ εT + sup u.
ΩT −ε
(5.7)
∂p ΩT
ΩT −ε
Since u is uniformly continuous on ΩT , we have that
lim sup u = sup u.
ε→0
ΩT −ε
ΩT
Thus taking ε → 0 in (5.7), we obtain that
sup u = lim sup u ≤ lim (εT + sup u) = sup u ≤ sup u.
ΩT
ε→0
ΩT −ε
ε→0
∂p ΩT
∂p ΩT
(5.8)
ΩT
Therefore, all of the inequalities in (5.8) can be replaced with inequalities. In particular,
we get
sup u = sup u
∂p ΩT
ΩT
as desired.
To obtain the result for Lu ≥ 0, we note that −u satisfies L(−u) ≤ 0 and apply the
first part of the theorem.
88
Corollary 5 (The maximum principle for solutions of the heat equation on bounded
domain). Suppose u ∈ C 2,1 (ΩT ) ∩ C 0 (ΩT ) solves the heat equation in ΩT . Then
max |u| = max |u|.
∂p ΩT
ΩT
Corollary 6. Let f ∈ C 0 (ΩT ) and g ∈ C 0 (∂p ΩT ). Then there is at most one solution
u ∈ C 2,1 (ΩT ) ∩ C 0 (ΩT ) of the boundary value problem


ut − ∆u = f in ΩT ,

u = g
on
∂p ΩT ,
Proof. Suppose there are two solutions u1 and u2 . Then u = u1 − u2 satisfies the equation


ut − ∆u = 0 in ΩT ,

u = 0 on ∂p ΩT ,
Apply Corollary 5 we obtain that
max |u| = max |u| = 0,
∂p ΩT
ΩT
which implies that u ≡ 0, i.e., u1 = u2 in ΩT .
Note that unlike the Dirichlet problem for harmonic functions, we are not free to
prescribe the value of u on all of ∂p ΩT . The restriction is because of the maximum
principle: we can not specify the boundary of u such that u achieves a value on Ω × {T }
that is strictly greater than the value it achieves on ∂p ΩT .
5.3
The maximum principle for the heat operator on
the whole space
In this section, we show a maximum principle for the heat operator in the whole
space. We need to make an assumption on the behaviour of u at the infinity. Denote by
ST = Rn × (0, T ) the space-time domain in this case.
Theorem 21 (The maximum principle for the heat operator on ST ). Let u ∈ C 2,1 (ST ) ∩
C 0 (S T ) satisfy Lu ≤ 0 in ST . In addition, assume that there exist constants C, α > 0
such that
2
u(x, t) ≤ Ceα|x| ,
∀x ∈ Rn , 0 ≤ t ≤ T.
89
Then
sup u = sup u(x, 0).
ST
x∈Rn
If instead, Lu ≥ 0 in ST and there exist constants C, α > 0 such that
∀x ∈ Rn , 0 ≤ t ≤ T.
2
u(x, t) ≥ −Ceα|x| ,
Then
inf u = infn u(x, 0).
ST
x∈R
Proof. We will prove the minimum principle. The maximum principle then follows by
changing u to −u. Suppose that Lu ≥ 0 in ST and there exist constants C, α > 0 such
that
2
u(x, t) ≥ −Ceα|x| ,
∀x ∈ Rn , 0 ≤ t ≤ T.
We need to show that
inf u = infn u(x, 0).
x∈R
ST
If u(x, 0) is not bounded from below then the assertion trivially holds. Therefore, we can
assume that u(x, 0) is bounded from below and define I := inf x∈Rn u(x, 0). Let β > α
such that T1 :=
1
8β
< T . The idea of the proof is as follows. We first show that u ≥ I in
ST1 and then by dividing the time interval [0, T ] into finitely many intervals of length less
than T1 and successively applying the result of the first step.
We define
|x|2
eβ 1−4βt
v(x, t) :=
n .
(1 − 4βt) 2
By direct computations, we have
−n
−1
2
∂t v = 4β(1 − 4βt)
n
n
β|x|2
+
2 1 − 4βt
β|x|2
e 1−4βt ,
β|x|2
∂xi v = (1 − 4βt)− 2 −1 2βxi e 1−4βt ,
β|x|2
1
βx2i
2
−n
−1
∂xi xi v = 4β(1 − 4βt) 2
+
e 1−4βt ,
2 1 − 4βt
n
X
β|x|2
n
β|x|2
2
−n
−1
2
∆v =
∂xi xi v = 4β(1 − 4βt)
+
e 1−4βt ,
2 1 − 4βt
i=1
which implies that v(x, t) satisfies the heat equation in ST1 . In addition, since (1−4βt)−1 >
1 in ST1 , we have that
2
v(x, t) ≥ eβ|x| .
90
Let ε > 0 and set w := u + εv − I. To prove u ≥ I in ST1 , it suffices to prove that
w ≥ 0 in ST1 for any ε > 0. The function w satisfies that Lw = Lu + εLv ≥ 0 in ST1 and
w(x, 0) = u(x, 0) + εv(x, 0) − I ≥ 0 for all x ∈ Rn . By the assumption on u we have that
2
2
w = u + εv − I ≥ εeβ|x| − Ceα|x| − I,
from where we deduce that
2
2
≥ εeβr − Ceαr − I.
inf
∂Br (0)×[0,T1 ]
2
Since β > α the term εeβr dominates for r large enough, i.e., there exists R ≥ 0 such
that w ≥ 0 on ∂Br (0) × [0, T1 ] for all r > R.
On the parabolic boundary of the cylinder Br (0) × (0, T1 ), we have w ≥ 0. Therefore,
applying Theorem 20 to this cylinder, we obtain that
inf
w=
Br (0)×(0,T1 )
inf
w ≥ 0,
∂p (Br (0)×(0,T1 ))
so that w ≥ 0 in Br (0) × (0, T1 ) for any r > R. Thus w ≥ 0 in ST1 for any ε > 0 and so
u ≥ I. Now dividing the interval [0, T ] into finitely many intervals of lengths less than T1
and successively applying the result on each interval, we obtain
u(x, t) ≥ infn u(y, 0)
y∈R
for all (x, t) ∈ ST . Taking the infimum over ST we get that
inf u ≥ infn u(x, 0).
ST
x∈R
On the other hand, it is obvious that
inf u ≤ infn u(x, 0).
ST
x∈R
Therefore, we have inf ST u = inf x∈Rn u(x, 0) as desired.
Corollary 7. Suppose that u ∈ C 2,1 (ST ) ∩ C 0 (S T ) and there exist constants C, α > 0
such that
2
|u(x, t)| ≤ Ceα|x| ,
∀x ∈ Rn , 0 ≤ t ≤ T.
Suppose further that u solves the heat equation
∂t u(x, t) = ∆u(x, t),
u(x, 0) = u0 (x),
x ∈ Rn , t > 0,
x ∈ Rn ,
where u0 ∈ C 0 (Rn ) is bounded. Then u is unique.
91
Proof. Suppose that there are two functions u1 and u2 satisfying the hypotheses of the
theorem. Define w := u1 − u2 . Then
2
0
2
|w(x, t)| ≤ |u1 (x, t)| + |u2 (x, t)| ≤ Ceα|x| + C 0 eα |x| ≤ C 00 eα
00 |x|2
,
where C 00 = 2 max{C, C 0 } and α00 = max{α, α0 }. In addition, w solves


wt − ∆w = 0 in Rn × (0, T ),

w = 0 on
Rn × {0}.
Applying Theorem 21 and noting that Lu = 0, we conclude that w = 0 and hence u1 = u2
as desired.
2
Note that if the growth assumption on u, |u(x, t)| ≤ Ceα|x| is removed then the
uniqueness statement above does not hold. A counter example for this was constructed
by Tychonoff in 1935.
Problem Sheet 5
Exercise 20. Suppose that n ≥ 3 and G is given by
G(x) =
1
|x|2−n .
σn−1 (2 − n)
Let ρ ∈ Cc2 (Rn ) be supported in BR (0) and define u by
ˆ
u(x) = lim
G(x − y)ρ(y) dy.
ε→0 Rn \B (x)
ε
Show that for |α| ≤ 2,
R2
sup |Dα u| ≤
sup |Dα ρ|.
n
2(n
−
2)
R
Rn
Deduce that the problem of finding u ∈ C 2 (Rn ) such that u → 0 as |x| → ∞ satisfying
∆u = ρ in
Rn
is well-posed.
Exercise 21 (The inhomogeneous heat equation). (a) Let φ ∈ Cc∞ (Rn × R). Show that
ˆ
lim
Γ(x, t)φ(x, t) dx = φ(0, 0),
t→0 Rn
where Γ denotes the fundamental solution of the heat equation.
Deduce that the function
ˆ
t 7→
Γ(x, t)φ(x, t) dx,
Rn
is uniformly continuous on (0, ∞) and vanishes for large values of t.
(b) Define the distribution T ∈ D0 (Rn × R) by
ˆ ∞ˆ
Tφ =
Γ(x, t)φ(x, t) dx dt,
0
Rn
∀φ ∈ Cc∞ (Rn × R).
Let δ > 0. Show that, for any φ ∈ Cc∞ (Rn × R), we have
ˆ δˆ
ˆ
ˆ ∞
Dt T (φ) = −
Γ(x, t)φt (x, t) dx dt+
Γ(x, δ)φ(x, δ)dx+
Γt (x, t)φ(x, t) dx dt,
0
Rn
Rn
δ
92
93
and
ˆ δˆ
Dxi xj T (φ) =
0
ˆ
Rn
∞
Γ(x, t)φxi xj (x, t) dx dt +
Γxi xj (x, t)φ(x, t) dx dt.
δ
(c) Deduce that for any δ > 0, we have
ˆ δˆ
ˆ
(Dt T − ∆T )(φ) = −
Γ(x, t)(φt + ∆φ)(x, t) dx dt +
Γ(x, δ)φ(x, δ) dx.
0
Rn
Rn
By bounding the first integral and making use of part (a), show that
(Dt T − ∆T )(φ) = φ(0, 0).
(d) Find a distributional solution to the inhomogeneous heat equation
ut − ∆u = f
in
Rn × R,
where f ∈ Cc∞ (Rn × R).
Exercise 22. Let Ω = {x ∈ Rn : xn > 0} be the half-space. Consider the boundary value
problem




ut = ∆u in Ω × (0, T )



u = 0 on ∂Ω × (0, T )





u = u0 on Ω × {0}.
Suppose that u0 ∈ Cc0 (Ω). By considering a reflection in the plane xn = 0, find a solution
u.
Acknowledgements
I would like to thank Dr. Claude Warnick, who lectured this course from 2013-2014,
for kindly sharing his notes1 , on which these notes (and exercise sheets) are largely based.
1
which was in turn based on notes of Andrea Malchiodi, who lectured this course from 2011-2013
94
Appendix
We review relevant knowledge about differentiation of functions of several variables.
Suppose Ω is an open subset of
Rn .
f : Ω → Rm is said to be differentiable at a point
x ∈ Ω if there exists a linear map, Df (x) from
Rn to Rm such that if h ∈ Rn , we have
f (x + h) = f (x) + Df (x) · h + o(h),
as h → 0.
(5.9)
i
We relate this definition to the usual partial derivatives. If f = (f i )m
i=1 , i.e., f is the
i-th component of f (x) with respect t the canonical basic of
derivative
∂f i
(x)
∂xj
Rm ,
we define the partial
to be the limit (if exists)
∂f i
f (x1 , . . . , xj + ε, . . . , xn ) − f (x1 , . . . , xj , . . . , xn )
.
(x) = lim
ε→0
∂xj
ε
In other words, the partial derivative of f i in the xj direction is simply the derivative of
f i when considering it as a function of xj alone. By setting h = εej in (5.9), we see that
if f is differentiable at x then
∂f i
(x)
∂xj
exists for all i = 1, . . . , m; j = 1, . . . , n. To obtain
the conserve, we need a bit stronger condition.
Lemma 9. Suppose that the partial derivative
∂f i
(x)
∂xj
exist and are continuous on an open
neighbourhood of x, then f is differentiable at x.
Since Df (x) is a linear map, we can represent it with respect to the canonical bases
for
Rn and Rm as a matrix.
matrix
Then the partial derivatives of f are the components of this

∂f1
(x)
 ∂x1
 ∂f2
 ∂x1 (x)


Df (x) = 

 ..
 .

∂fm
(x)
∂x1
∂f1
(x)
∂x2
...
∂f2
(x)
∂x2
...

∂f1
(x)
∂xn


∂f2
(x) 
∂xn




..
.
..
.. 
.

.

∂fm
∂fm
(x) . . . ∂xn (x)
∂x2
95
96
or in a compact way
[Df (x)]ij =
∂f i
.
∂xj
If f is differentiable at each point in Ω, we can think of Df as a map from Ω to the space
of m × n matrices which is identified to
Rm×n
Df : Ω → Rm×n
x 7→ Df (x).
If this map is continuous, the we say that f ∈ C 1 (Ω). Equivalently f ∈ C 1 (Ω) if all its
partial derivatives exist and are continuous in Ω.
Similarly, we say that f ∈ C 2 (Ω) if Df ∈ C 1 (Ω) and so on inductively. We can think
of D2 f as a map from Ω to
Rm × Rn × Rn , then its components are
[D2 f (x)]ij1 j2 =
∂ 2f i
(x).
∂xj1 ∂xj2
Lemma 10. If f ∈ C k (Ω) if and only if all the partial derivatives
∂kf i
:U →R
∂xj1 ∂xj2 . . . ∂xjk
exist and are continuous in Ω.
We now introduce multi-indices notations. Let α = (α1 , . . . , αm ) ∈ (Z≥ 0)n , we define
P
|α| := ni=1 αi and
∂ α
∂ α
∂ |α|f
=
.
.
.
,
∂xα
∂x1 1
∂xn n
i.e., we differentiate α1 times with respect to x1 , α2 times with respect to x2 and so on.
We also use more compact notation
Di :=
∂
,
∂xi
Dα :=
∂ |α|
.
∂xα
Similarly for h = (h1 , . . . , hn ), we denote
hα = hα1 1 . . . hαnn .
Finally, we introduce the multi-index factorial α! := α1 ! . . . αn !.
Theorem 22 (Multivariate Taylor’s theorem). Suppose that Ω is an open subset of
Rn ,
x ∈ Ω and f : U → R belongs to C k (Ω). Then writing h = (hj )nj=1 and using multi-indices
notation, we have
f (x + h) = f (x) +
X hα
Dα f (x) + Rk (x, h),
α!
|α|≤k
97
where the sum is taken over all multi-indices α = (α1 , . . . , αn ) with |α| ≤ k and the
remainder term satisfies
Rk (x, h)
→ 0 as
|h|k
h → 0.
Bibliography
[1] Fritz J. Partial Differential Equations. Springer, 1982.
[2] L.C. Evans. Partial Differential Equations. American Mathematical Society, 1998.
[3] E. Dibenedetto. Partial Differential Equations. Birkhauser, 2010.
[4] D. Gilbarg and N. Trudinger. Elliptic Partial Differential Equations of Second Order.
Springer, 2001.
98