Partial differential equations
International Journal for Numerical Methods in Engineering (2001)
Available from strathprints.strath.ac.uk
or
Abstract
Over the second half of the 20th century the subject area loosely referred to as numerical analysis of partial differential equations (PDEs) has undergone unprecedented development. At its practical end, the vigorous growth and steady diversification of the field were stimulated by the demand for accurate and reliable tools for computational modelling in physical sciences and engineering, and by the rapid development of computer hardware and architecture. At the more theoretical end, the analytical insight into the underlying stability and accuracy properties of computational algorithms for PDEs was deepened by building upon recent progress in mathematical analysis and in the theory of PDEs.
Author-supplied keywords
Available from strathprints.strath.ac.uk
Page 1
Partial differential equations
Partial Differential Equations
T. W. Ko¨rner after Joshi and Wassermann
October 12, 2002
Small print These notes are a digest of much more complete notes by M. S. Joshi and
A. J. Wassermann which are also being issued for this course. I should very much
appreciate being told of any corrections or possible improvements and might even part
with a small reward to the first finder of particular errors. This document is written
in LATEX2e and stored in the file labelled ~twk/IIB/PDE.tex on emu in (I hope) read
permitted form. My e-mail address is twk@dpmms.
Contents
1 Introduction 2
1.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 The symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Ordinary differential equations 6
2.1 The contraction mapping theorem . . . . . . . . . . . . . . . . 6
2.2 Vector fields, integral curves and flows . . . . . . . . . . . . . 8
2.3 First order linear and semi-linear PDEs . . . . . . . . . . . . . 10
2.4 First order quasi-linear PDEs . . . . . . . . . . . . . . . . . . 11
3 Distributions 13
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 The support of a distribution . . . . . . . . . . . . . . . . . . 17
3.3 Fourier transforms and the Schwartz space . . . . . . . . . . . 19
3.4 Tempered distributions . . . . . . . . . . . . . . . . . . . . . . 21
4 Convolution and fundamental solutions 23
4.1 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.2 Fundamental solutions . . . . . . . . . . . . . . . . . . . . . . 27
4.3 Our first fundamental solution . . . . . . . . . . . . . . . . . . 28
4.4 The parametrix . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1
T. W. Ko¨rner after Joshi and Wassermann
October 12, 2002
Small print These notes are a digest of much more complete notes by M. S. Joshi and
A. J. Wassermann which are also being issued for this course. I should very much
appreciate being told of any corrections or possible improvements and might even part
with a small reward to the first finder of particular errors. This document is written
in LATEX2e and stored in the file labelled ~twk/IIB/PDE.tex on emu in (I hope) read
permitted form. My e-mail address is twk@dpmms.
Contents
1 Introduction 2
1.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 The symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Ordinary differential equations 6
2.1 The contraction mapping theorem . . . . . . . . . . . . . . . . 6
2.2 Vector fields, integral curves and flows . . . . . . . . . . . . . 8
2.3 First order linear and semi-linear PDEs . . . . . . . . . . . . . 10
2.4 First order quasi-linear PDEs . . . . . . . . . . . . . . . . . . 11
3 Distributions 13
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 The support of a distribution . . . . . . . . . . . . . . . . . . 17
3.3 Fourier transforms and the Schwartz space . . . . . . . . . . . 19
3.4 Tempered distributions . . . . . . . . . . . . . . . . . . . . . . 21
4 Convolution and fundamental solutions 23
4.1 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.2 Fundamental solutions . . . . . . . . . . . . . . . . . . . . . . 27
4.3 Our first fundamental solution . . . . . . . . . . . . . . . . . . 28
4.4 The parametrix . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1
Page 2
4.5 Existence of the fundamental solution . . . . . . . . . . . . . . 30
5 The Laplacian 32
5.1 A fundamental solution . . . . . . . . . . . . . . . . . . . . . . 32
5.2 Identities and estimates . . . . . . . . . . . . . . . . . . . . . 33
5.3 The dual Dirichlet problem for the unit ball . . . . . . . . . . 36
6 Dirichlet’s problem for the ball and Poisson’s formula 38
7 The wave equation 40
8 The heat equation 45
8.1 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
8.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
9 References 48
1 Introduction
1.1 Generalities
When studying ordinary (i.e. not partial) differential equations we start
with linear differential equations with constant coefficients. We then study
linear differential equations and then plunge into a boundless ocean of non-
linear differential equations. The reader will therefore not be surprised if
most of a first course on the potentially much more complicated study of
partial differential equations should limit itself essentially to linear partial
differential equations.
A linear partial differential equation is an equation of the form Pu = f
where u and f are suitable functions from Rn to R and
P =
∑
|α|≤k
aα(x)∂α.
Our definition introduces some of the condensed notation conventions which
make the subject easier for the expert and harder for the beginner. The
multi-index
α = (α1, α2, . . . , αn) ∈ Nn.
We write
|α| =
n∑
j=1
αj
2
5 The Laplacian 32
5.1 A fundamental solution . . . . . . . . . . . . . . . . . . . . . . 32
5.2 Identities and estimates . . . . . . . . . . . . . . . . . . . . . 33
5.3 The dual Dirichlet problem for the unit ball . . . . . . . . . . 36
6 Dirichlet’s problem for the ball and Poisson’s formula 38
7 The wave equation 40
8 The heat equation 45
8.1 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
8.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
9 References 48
1 Introduction
1.1 Generalities
When studying ordinary (i.e. not partial) differential equations we start
with linear differential equations with constant coefficients. We then study
linear differential equations and then plunge into a boundless ocean of non-
linear differential equations. The reader will therefore not be surprised if
most of a first course on the potentially much more complicated study of
partial differential equations should limit itself essentially to linear partial
differential equations.
A linear partial differential equation is an equation of the form Pu = f
where u and f are suitable functions from Rn to R and
P =
∑
|α|≤k
aα(x)∂α.
Our definition introduces some of the condensed notation conventions which
make the subject easier for the expert and harder for the beginner. The
multi-index
α = (α1, α2, . . . , αn) ∈ Nn.
We write
|α| =
n∑
j=1
αj
2
Page 3
and
∂αu = ∂
|α|
∂xα11 ∂xα22 . . . ∂xαnn
.
Sometimes we write ∂αu = ∂|α|u∂x .
Although the first part of the course will deal with ‘first order’ linear
partial differential equations without restrictions on the coefficients (and in-
deed even with slightly more general partial differential equations) the main
part of the course will deal with linear partial differential equations with
constant coefficients. The main tools used will be Laurent Schwartz’s theory
of distributions and the Fourier transform.
The fact that we do not deal with non-linear equations does not mean that
they are not important. The equations of general relativity are non-linear
and at a more mundane level the Navier-Stokes equation of fluid dynamics
∂u
∂t −∆u+ u.∆u = f −∆p, ∆.u = 0
is non-linear.
We call P =∑|α|≤k aα(x)∂α a differential operator (or just an ‘operator’).
We look at three such operators in detail. The first which must be most
studied non-trivial differential operator in mathematics is the Laplacian ∆
known in more old fashioned texts as ∇2 and defined by
∆u =
n∑
j=1
∂2u
∂x2j
where u : Rn → R is a suitable function. The second is the wave operator 2
given by
2u(t,x) = ∂
2u
∂t2 (t,x)−∆xu(t,x)
where t ∈ R, x ∈ Rn and u : R × Rn → R is a suitable function. The third
is the heat operator J given by
Ju(t,x) = ∂u∂t (t,x)−∆xu(t,x).
The notations ∆ and 2 are standard but the notation J is not.
3
∂αu = ∂
|α|
∂xα11 ∂xα22 . . . ∂xαnn
.
Sometimes we write ∂αu = ∂|α|u∂x .
Although the first part of the course will deal with ‘first order’ linear
partial differential equations without restrictions on the coefficients (and in-
deed even with slightly more general partial differential equations) the main
part of the course will deal with linear partial differential equations with
constant coefficients. The main tools used will be Laurent Schwartz’s theory
of distributions and the Fourier transform.
The fact that we do not deal with non-linear equations does not mean that
they are not important. The equations of general relativity are non-linear
and at a more mundane level the Navier-Stokes equation of fluid dynamics
∂u
∂t −∆u+ u.∆u = f −∆p, ∆.u = 0
is non-linear.
We call P =∑|α|≤k aα(x)∂α a differential operator (or just an ‘operator’).
We look at three such operators in detail. The first which must be most
studied non-trivial differential operator in mathematics is the Laplacian ∆
known in more old fashioned texts as ∇2 and defined by
∆u =
n∑
j=1
∂2u
∂x2j
where u : Rn → R is a suitable function. The second is the wave operator 2
given by
2u(t,x) = ∂
2u
∂t2 (t,x)−∆xu(t,x)
where t ∈ R, x ∈ Rn and u : R × Rn → R is a suitable function. The third
is the heat operator J given by
Ju(t,x) = ∂u∂t (t,x)−∆xu(t,x).
The notations ∆ and 2 are standard but the notation J is not.
3
Page 4
1.2 The symbol
A key concept in more advanced work is the total symbol σ(P ) of a linear
partial differential operator
P (x, ∂) =
∑
|α|≤k
aα(x)∂α
obtained by replacing ∂∂xj by iξj so that
σ(P ) = p(x, ξ) =
∑
|α|≤k
aα(x)(iξ)α = e−ix.ξP (eix.ξ).
Note that x and ξ are n dimensional vectors and that we use the convention
yα =
n∏
j=1
yαjj .
To see where the symbol comes from, observe that taking Fourier transforms
P̂ u(ξ) = p(x, ξ)uˆ(ξ)
and so taking inverse Fourier transforms
P (x, ∂)u =
( 1
2pi
)n/2 ∫
Rn
eix.ξp(x, ξ)uˆ(ξ) dξ.
Applying the Leibnitz rule we get the following result.
Lemma 1.1 (Proposition 1). If P and Q are linear partial differential op-
erators
σ(PQ)(x, ξ) =
∑ (−i)|α|
α! ∂
α
ξ σ(P )(x, ξ)∂αxσ(Q)(x, ξ).
Here as one might expect α! = ∏nj=1 αj!.
We say that the differential operator
P (x, ∂) =
∑
|α|≤k
aα(x)∂α
has order k and that it has principal symbol
σk(P )(x, ξ) =
∑
|α|=k
aα(x)(iξ)α.
4
A key concept in more advanced work is the total symbol σ(P ) of a linear
partial differential operator
P (x, ∂) =
∑
|α|≤k
aα(x)∂α
obtained by replacing ∂∂xj by iξj so that
σ(P ) = p(x, ξ) =
∑
|α|≤k
aα(x)(iξ)α = e−ix.ξP (eix.ξ).
Note that x and ξ are n dimensional vectors and that we use the convention
yα =
n∏
j=1
yαjj .
To see where the symbol comes from, observe that taking Fourier transforms
P̂ u(ξ) = p(x, ξ)uˆ(ξ)
and so taking inverse Fourier transforms
P (x, ∂)u =
( 1
2pi
)n/2 ∫
Rn
eix.ξp(x, ξ)uˆ(ξ) dξ.
Applying the Leibnitz rule we get the following result.
Lemma 1.1 (Proposition 1). If P and Q are linear partial differential op-
erators
σ(PQ)(x, ξ) =
∑ (−i)|α|
α! ∂
α
ξ σ(P )(x, ξ)∂αxσ(Q)(x, ξ).
Here as one might expect α! = ∏nj=1 αj!.
We say that the differential operator
P (x, ∂) =
∑
|α|≤k
aα(x)∂α
has order k and that it has principal symbol
σk(P )(x, ξ) =
∑
|α|=k
aα(x)(iξ)α.
4
Page 5
Notice that if P has degree k and Q degree l then the principal symbol of
PQ is given by the simple formula
σl+k(PQ) = σk(P )σl(Q).
If the principal symbol is never zero or only vanishes to first order then
the operator is said to be of principal type. In more advanced work it is
shown that when the operator is of principal type the lower order terms have
little effect on the qualitative behaviour of the associated partial differential
equation.
We define the characteristic set to be the subset of Rn × Rn where the
principal symbol vanishes.
char(P ) = {(x, ξ) : σk(P )(x, ξ) = 0}.
We say that P is elliptic at x if the principle symbol σk(P )(x, ξ) 6= 0 for
ξ 6= 0. If P is elliptic at x for all x we say that P is elliptic.
The reason for the use of the word ‘elliptic’ may be traced to the symbols
of the three special operators.
σ(∆)(ξ) = −|ξ|2 (Laplacian)
σ(∆)(τ, ξ) = −τ 2 + |ξ|2 (Wave)
σ(J) = iτ + |ξ|2 (Heat)
Traditionally second order operators which behaved like the Laplacian were
called elliptic, those which behaved like the wave operator were called hy-
perbolic and those that behaved like the heat operator were called parabolic.
The distinction is very useful but the reference to conic sections is not.
Example 1.2 (Example 1). (i) We could have considered complex valued
u in place of real valued u. If we do this the operator
P = ∂∂x + i
∂
∂y
has principal symbol
σ1(P )(x, y : ξ, η) = iξ − η
and so is elliptic.
(ii) The Laplacian ∆ is elliptic but the wave operator 2 and the heat
operator J are not.
5
PQ is given by the simple formula
σl+k(PQ) = σk(P )σl(Q).
If the principal symbol is never zero or only vanishes to first order then
the operator is said to be of principal type. In more advanced work it is
shown that when the operator is of principal type the lower order terms have
little effect on the qualitative behaviour of the associated partial differential
equation.
We define the characteristic set to be the subset of Rn × Rn where the
principal symbol vanishes.
char(P ) = {(x, ξ) : σk(P )(x, ξ) = 0}.
We say that P is elliptic at x if the principle symbol σk(P )(x, ξ) 6= 0 for
ξ 6= 0. If P is elliptic at x for all x we say that P is elliptic.
The reason for the use of the word ‘elliptic’ may be traced to the symbols
of the three special operators.
σ(∆)(ξ) = −|ξ|2 (Laplacian)
σ(∆)(τ, ξ) = −τ 2 + |ξ|2 (Wave)
σ(J) = iτ + |ξ|2 (Heat)
Traditionally second order operators which behaved like the Laplacian were
called elliptic, those which behaved like the wave operator were called hy-
perbolic and those that behaved like the heat operator were called parabolic.
The distinction is very useful but the reference to conic sections is not.
Example 1.2 (Example 1). (i) We could have considered complex valued
u in place of real valued u. If we do this the operator
P = ∂∂x + i
∂
∂y
has principal symbol
σ1(P )(x, y : ξ, η) = iξ − η
and so is elliptic.
(ii) The Laplacian ∆ is elliptic but the wave operator 2 and the heat
operator J are not.
5
Page 6
char(2) = {(x, t, τ, ξ) ∈ R× Rn × R× Rn : τ 2 − |ξ|2 = 0}
char(J) = {(x, t, τ, ξ) ∈ R× Rn × R× Rn : τ − |ξ|2 = 0}
2 Ordinary differential equations
2.1 The contraction mapping theorem
Hadamard introduced the idea of a well posed problem into the study of
partial differential equations. According to Hadamard a well posed problem
must have a solution which exists, is unique and varies continuously on the
given data. Without going too deeply into the matter we may agree that
these are reasonable matters to investigate.
Thus given a partial differential equation with boundary conditions we
shall study the following problems.
Existence Can we show that there is a solution in a neighbourhood of a
given point? Can we show that there exists a solution everywhere?
Uniqueness Is the solution unique?
Continuity Does u depend continuously on the boundary conditions?
Does u depend continuously on on other elements of the problem?
Smoothness How many times is u differentiable? Does u have points of
singularity? Does the solution blow up in some way after a finite time?
We may illustrate these ideas in the case of ordinary differential equa-
tions. Some of the material in this section will be familiar from examination
questions on the first analysis course in 1B but you should consider it all
with the exception of the contraction mapping theorem (Theorem 2.1) itself
to be examinable.
In 1B we proved Banach’s contraction mapping theorem.
Theorem 2.1 (Theorem 1). Let (X, d) be a complete non-empty metric
space and T : X → X a map such that d(Tx, Ty) ≤ k(x, y) for all x, y ∈ X
and some k with 0 ≤ k < 1. Then there exists a unique x0 ∈ X such that
Tx0 = x0. If x ∈ X then T nx→ x0 as n→∞.
This can be strengthened as follows.
Theorem 2.2 (Corollary 1). Let (X, d) be a complete non-empty metric
space and T : X → X a map. Suppose further that there exists an integer
6
char(J) = {(x, t, τ, ξ) ∈ R× Rn × R× Rn : τ − |ξ|2 = 0}
2 Ordinary differential equations
2.1 The contraction mapping theorem
Hadamard introduced the idea of a well posed problem into the study of
partial differential equations. According to Hadamard a well posed problem
must have a solution which exists, is unique and varies continuously on the
given data. Without going too deeply into the matter we may agree that
these are reasonable matters to investigate.
Thus given a partial differential equation with boundary conditions we
shall study the following problems.
Existence Can we show that there is a solution in a neighbourhood of a
given point? Can we show that there exists a solution everywhere?
Uniqueness Is the solution unique?
Continuity Does u depend continuously on the boundary conditions?
Does u depend continuously on on other elements of the problem?
Smoothness How many times is u differentiable? Does u have points of
singularity? Does the solution blow up in some way after a finite time?
We may illustrate these ideas in the case of ordinary differential equa-
tions. Some of the material in this section will be familiar from examination
questions on the first analysis course in 1B but you should consider it all
with the exception of the contraction mapping theorem (Theorem 2.1) itself
to be examinable.
In 1B we proved Banach’s contraction mapping theorem.
Theorem 2.1 (Theorem 1). Let (X, d) be a complete non-empty metric
space and T : X → X a map such that d(Tx, Ty) ≤ k(x, y) for all x, y ∈ X
and some k with 0 ≤ k < 1. Then there exists a unique x0 ∈ X such that
Tx0 = x0. If x ∈ X then T nx→ x0 as n→∞.
This can be strengthened as follows.
Theorem 2.2 (Corollary 1). Let (X, d) be a complete non-empty metric
space and T : X → X a map. Suppose further that there exists an integer
6
Page 7
N ≥ 1 and a k with 0 ≤ k < 1 such that d(TNx, TNy) ≤ k(x, y) for all
x, y ∈ X. Then there exists a unique x0 ∈ X such that Tx0 = x0. If x ∈ X
then T nx→ x0 as n→∞.
Note that these theorems not only give a fixed point but also give a
method for finding it.
For the rest of section 2.1 f will be a function from R× Rn. Let
E = {t ∈ R : |t− t0| ≤ a} × {x ∈ Rn : ‖x− x0‖ ≤ b}.
We assume that f satisfies the Lipschitz condition
‖f(t, x1)− f(t, x2)‖ ≤ c‖x1 − x2‖
on E.
Exercise 2.3. (i) Show that a function that satisfies a Lipschitz condition
is continuous.
(ii) Show that any continuously differentiable function satisfies a Lipschitz
condition.
(iii) Show that a function that satisfies a Lipschitz condition need not be
differentiable everywhere.
(iv) Show that the function f considered above is bounded on E.
We set M = sup(t,x)∈E ‖f(t, x)‖ and h = min(a, bM−1).
Theorem 2.4. If f is as above the differential equation
dx
dt = f(t, x), x(t0, 0) = x0 F
has a unique solution for |t− t0| ≤ h.
Example 2.5. (Here x : R → R.)
(i) The differential equation
dx
dt = 0, x(0) = 0, x(1) = 1
has no solution.
(ii) Show that the differential equation
dx
dt = x
2
3 , x(0) = 0
has at least two solutions.
7
x, y ∈ X. Then there exists a unique x0 ∈ X such that Tx0 = x0. If x ∈ X
then T nx→ x0 as n→∞.
Note that these theorems not only give a fixed point but also give a
method for finding it.
For the rest of section 2.1 f will be a function from R× Rn. Let
E = {t ∈ R : |t− t0| ≤ a} × {x ∈ Rn : ‖x− x0‖ ≤ b}.
We assume that f satisfies the Lipschitz condition
‖f(t, x1)− f(t, x2)‖ ≤ c‖x1 − x2‖
on E.
Exercise 2.3. (i) Show that a function that satisfies a Lipschitz condition
is continuous.
(ii) Show that any continuously differentiable function satisfies a Lipschitz
condition.
(iii) Show that a function that satisfies a Lipschitz condition need not be
differentiable everywhere.
(iv) Show that the function f considered above is bounded on E.
We set M = sup(t,x)∈E ‖f(t, x)‖ and h = min(a, bM−1).
Theorem 2.4. If f is as above the differential equation
dx
dt = f(t, x), x(t0, 0) = x0 F
has a unique solution for |t− t0| ≤ h.
Example 2.5. (Here x : R → R.)
(i) The differential equation
dx
dt = 0, x(0) = 0, x(1) = 1
has no solution.
(ii) Show that the differential equation
dx
dt = x
2
3 , x(0) = 0
has at least two solutions.
7
Page 8
From now on we move away from 1B.
Theorem 2.6 (Theorem 3). The solution of F in Theorem 2.4 depends
continuously on x0. More formally, if we define T : R → C([t− h, t+ h]) by
taking Ty to be the solution of
dx
dt = f(t, x), x(t0) = y
and give C([t− h, t+ h]) the uniform norm then T is continuous.
In the special case of linear ordinary differential equations we can give a
rather strong perturbation result.
Theorem 2.7 (Theorem 4). We use the standard operator norm on the
space L = L(Rn,Rn) of linear maps. Suppose that A,B : R × Rn → L are
continuous and that M ≥ ‖A(t, x)‖, ‖B(t, x)‖ for all t and x.
dξ
dt (t, x) = A(t, x)ξ(t, x), ξ(t0, x) = a(x)
dη
dt (t, x) = B(t, x)η(t, x), η(t0, x) = b(x)
then if |t− t0| ≤ K
‖ξ(t, x)− η(t, x)‖ ≤ C(a,K,M)‖A−B‖(eM |t−t0| − 1) + ‖a− b‖eM |t−t0|
where C(a,K,M) depends only on a, M and K.
2.2 Vector fields, integral curves and flows
Let U be an open subset of Rn. A time dependent vector field on U is a map
f : (−², ²)× U → Rn
which associates a vector f(t, x) to each time t and point x.
Let x0 ∈ U . An integral curve for the vector field f with starting point
x0 is a map
φ : (−δ, δ)→ U
such that
dφ
dt = f(t, φ(t)) and φ(0) = x0
so that the tangent vectors to φ are just the values of the vector field at that
point at time.
8
Theorem 2.6 (Theorem 3). The solution of F in Theorem 2.4 depends
continuously on x0. More formally, if we define T : R → C([t− h, t+ h]) by
taking Ty to be the solution of
dx
dt = f(t, x), x(t0) = y
and give C([t− h, t+ h]) the uniform norm then T is continuous.
In the special case of linear ordinary differential equations we can give a
rather strong perturbation result.
Theorem 2.7 (Theorem 4). We use the standard operator norm on the
space L = L(Rn,Rn) of linear maps. Suppose that A,B : R × Rn → L are
continuous and that M ≥ ‖A(t, x)‖, ‖B(t, x)‖ for all t and x.
dξ
dt (t, x) = A(t, x)ξ(t, x), ξ(t0, x) = a(x)
dη
dt (t, x) = B(t, x)η(t, x), η(t0, x) = b(x)
then if |t− t0| ≤ K
‖ξ(t, x)− η(t, x)‖ ≤ C(a,K,M)‖A−B‖(eM |t−t0| − 1) + ‖a− b‖eM |t−t0|
where C(a,K,M) depends only on a, M and K.
2.2 Vector fields, integral curves and flows
Let U be an open subset of Rn. A time dependent vector field on U is a map
f : (−², ²)× U → Rn
which associates a vector f(t, x) to each time t and point x.
Let x0 ∈ U . An integral curve for the vector field f with starting point
x0 is a map
φ : (−δ, δ)→ U
such that
dφ
dt = f(t, φ(t)) and φ(0) = x0
so that the tangent vectors to φ are just the values of the vector field at that
point at time.
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Lemma 2.8. With the notation above, if f has a continuous derivative then
an integral curve always exists.
A local flow for f at x0 is a map
α : (−δ, δ)× U0 → U
where ² > δ > 0, U0 is open and x0 ∈ U0 ⊆ U , such that
d
dtα(t, x) = f(t, α(t, x)), α(0, x) = x.
Thus if x is fixed αx(t) = α(t.x) is an integral curve with starting point x.
In some sense a flow is a collection of integral curves.
We note but shall not prove that the smoother the vector field the smoother
the associated flow.
Theorem 2.9 (Theorem 5). If f is Ck and
d
dtα(t, x) = f(t, α(t, x)), α(0, x) = x.
then α ∈ Ck [1 ≤ k].
Now suppose f does not depend on t. Let αx(t) = α(t, x) as before.
Lemma 2.10. With the notation just established there exists an ² > 0 such
that
αt+s(u) = αt(αs(u))
for all |t|, |s|, |u| < ².
It is clear from the proof of Lemma 2.10 that the relation
αt+s = αt ◦ αs
will hold whenever it is reasonable for it to hold and in particular if the flow
is determined for all time we have a group action. This kind of group action
is called a dynamical system (see Course O6).
Definition 2.11. Let f : U → Rn be a vector field. We say that x0 ∈ U is
a critical point of f if f(x0) = 0.
Lemma 2.12. Let f : U → Rn be a (time independent) vector field.
(i) If φ is an integral curve passing through a critical point x0 then φ is
constant.
(ii) Suppose φ is an integral curve defined for all times t and φ(t)→ x0 ∈
U as t→∞. Then x0 is a critical point.
9
an integral curve always exists.
A local flow for f at x0 is a map
α : (−δ, δ)× U0 → U
where ² > δ > 0, U0 is open and x0 ∈ U0 ⊆ U , such that
d
dtα(t, x) = f(t, α(t, x)), α(0, x) = x.
Thus if x is fixed αx(t) = α(t.x) is an integral curve with starting point x.
In some sense a flow is a collection of integral curves.
We note but shall not prove that the smoother the vector field the smoother
the associated flow.
Theorem 2.9 (Theorem 5). If f is Ck and
d
dtα(t, x) = f(t, α(t, x)), α(0, x) = x.
then α ∈ Ck [1 ≤ k].
Now suppose f does not depend on t. Let αx(t) = α(t, x) as before.
Lemma 2.10. With the notation just established there exists an ² > 0 such
that
αt+s(u) = αt(αs(u))
for all |t|, |s|, |u| < ².
It is clear from the proof of Lemma 2.10 that the relation
αt+s = αt ◦ αs
will hold whenever it is reasonable for it to hold and in particular if the flow
is determined for all time we have a group action. This kind of group action
is called a dynamical system (see Course O6).
Definition 2.11. Let f : U → Rn be a vector field. We say that x0 ∈ U is
a critical point of f if f(x0) = 0.
Lemma 2.12. Let f : U → Rn be a (time independent) vector field.
(i) If φ is an integral curve passing through a critical point x0 then φ is
constant.
(ii) Suppose φ is an integral curve defined for all times t and φ(t)→ x0 ∈
U as t→∞. Then x0 is a critical point.
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2.3 First order linear and semi-linear PDEs
Integral curves give us a method for solving first order partial differential
equations
n∑
j=1
aj(x)
∂u
∂xj
= f(x).
In fact the method works for the slightly more general semi-linear first order
partial differential equation
n∑
j=1
aj(x)
∂u
∂xj
= f(x, u). (*)
Here we assume that aj is a real once continuously differentiable function
and that f is a real or complex continuously differentiable function.
The idea is to consider the vector field
A(x) = (a1(x), a2(x), . . . , an(x)).
If γ is an integral curve of A then
d
dtu(γ(t)) =
n∑
j=1
aj(γ(t))
∂u
∂xj
(γ(t)).
Thus solving (*) is equivalent to solving the ordinary differential equation
d
dtu(γ(t)) = f(γ(t), u(γ(t))) (**)
This not only gives us a method for solving (*) but tells us the appropriate
boundary conditions to use. For example, specifying u(0, y) along an integral
curve will, in general, give a problem with no solution. What we need to do
is to specify u(0, y) on a hypersurface S and solve along each integral curve.
Recall that that we defined the characteristic set of a linear differential
operator P (x, ∂) to be the subset of Rn × Rn where the principal symbol
vanishes.
char(P ) = {(x, ξ) : σk(P )(x, ξ) = 0}.
In the case of the linear differential operator
L(x, ∂) =
n∑
j=1
aj(x)
∂
∂xj
10
Integral curves give us a method for solving first order partial differential
equations
n∑
j=1
aj(x)
∂u
∂xj
= f(x).
In fact the method works for the slightly more general semi-linear first order
partial differential equation
n∑
j=1
aj(x)
∂u
∂xj
= f(x, u). (*)
Here we assume that aj is a real once continuously differentiable function
and that f is a real or complex continuously differentiable function.
The idea is to consider the vector field
A(x) = (a1(x), a2(x), . . . , an(x)).
If γ is an integral curve of A then
d
dtu(γ(t)) =
n∑
j=1
aj(γ(t))
∂u
∂xj
(γ(t)).
Thus solving (*) is equivalent to solving the ordinary differential equation
d
dtu(γ(t)) = f(γ(t), u(γ(t))) (**)
This not only gives us a method for solving (*) but tells us the appropriate
boundary conditions to use. For example, specifying u(0, y) along an integral
curve will, in general, give a problem with no solution. What we need to do
is to specify u(0, y) on a hypersurface S and solve along each integral curve.
Recall that that we defined the characteristic set of a linear differential
operator P (x, ∂) to be the subset of Rn × Rn where the principal symbol
vanishes.
char(P ) = {(x, ξ) : σk(P )(x, ξ) = 0}.
In the case of the linear differential operator
L(x, ∂) =
n∑
j=1
aj(x)
∂
∂xj
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Page 11
we have
char(L) =
{
(x, ξ) :
n∑
j=1
aj(x)ξj = 0
}
.
The hypersurface S is said to be characteristic for P at x if the normal
vector is a characteristic vector for P . The hypersurface S is called non-
characteristic if it is not characteristic at any point. In the case of the linear
differential operator L which we are considering, S is non-characteristic if at
each x ∈ S the normal ζ to S satisfies
n∑
j=1
ajζj 6= 0.
Retracing our steps it is clear that we have essentially (there is a slight
subtlety which we shall deal with when we prove Theorem 2.15) proved the
following theorem. (Note the use of the word ‘locally’.)
Theorem 2.13 (Theorem 6). Locally, there is a unique solution of (∗)
with u(x, 0) given on a non-characteristic hypersurface S.
Working through a few examples will make the theory much clearer.
Example 2.14 (Example 2). Solve the partial differential equation
∂u
∂x + 2x
∂u
∂y = u
2
subject to u(0, y) = f(y).
2.4 First order quasi-linear PDEs
An extension of the previous technique enables us to solve the slightly more
general first order semi-linear partial differential equation
n∑
j=1
aj(x, u)
∂u
∂xj
= b(x, u). (***)
subject to u|S = φ where S is a non-characteristic hypersurface. Here we
assume that the aj is a real once continuously differentiable function and
that b is a real continuously differentiable function. (Note that the technique
does not apply if B is complex valued and that a counter-example of Hans
Lewy shows that a solution may not exist.)
11
char(L) =
{
(x, ξ) :
n∑
j=1
aj(x)ξj = 0
}
.
The hypersurface S is said to be characteristic for P at x if the normal
vector is a characteristic vector for P . The hypersurface S is called non-
characteristic if it is not characteristic at any point. In the case of the linear
differential operator L which we are considering, S is non-characteristic if at
each x ∈ S the normal ζ to S satisfies
n∑
j=1
ajζj 6= 0.
Retracing our steps it is clear that we have essentially (there is a slight
subtlety which we shall deal with when we prove Theorem 2.15) proved the
following theorem. (Note the use of the word ‘locally’.)
Theorem 2.13 (Theorem 6). Locally, there is a unique solution of (∗)
with u(x, 0) given on a non-characteristic hypersurface S.
Working through a few examples will make the theory much clearer.
Example 2.14 (Example 2). Solve the partial differential equation
∂u
∂x + 2x
∂u
∂y = u
2
subject to u(0, y) = f(y).
2.4 First order quasi-linear PDEs
An extension of the previous technique enables us to solve the slightly more
general first order semi-linear partial differential equation
n∑
j=1
aj(x, u)
∂u
∂xj
= b(x, u). (***)
subject to u|S = φ where S is a non-characteristic hypersurface. Here we
assume that the aj is a real once continuously differentiable function and
that b is a real continuously differentiable function. (Note that the technique
does not apply if B is complex valued and that a counter-example of Hans
Lewy shows that a solution may not exist.)
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Page 12
The solution technique relies on regarding u as a variable just like x. We
give the technique as a recipe. Suppose that S is parameterised by a function
g so that
S = {x = g(s) : s ∈ Rn−1}.
We work with the vector field
(a1, a2, . . . , an, b)
on Rn+1 and solve for the integral curves
dxs
dt = a(xs(t), ys(t)) xs(0) = g(s)
dys
dt = b(xs(t), ys(t)) ys(0) = φ(s).
Our solution is basically y(s, t) but we want it as a function of x not (s, t).
The map
(s, t) 7→ x(s, t)
will have an invertible derivative at t = 0 provided that the vector
(a1(g(s), φ(s)), a2(g(s), φ(s)), . . . , an(g(s), φ(s)))
is not tangent to S and since we have specified that S is a non-characteristic
this will always be the case. A theorem of differential geometry (proved in
Course B4 Differential Manifolds, but in any case very plausible) called the
‘inverse function theorem’ asserts that under these circumstances (continuous
derivative invertible at a point) the map
(s, t) 7→ x(s, t)
is locally invertible. We may thus define
u(x) = y(s(x), t(x))
and direct calculation shows that it is a solution. Checking reversibility we
see that we have proved the following theorem.
Theorem 2.15 (Theorem 7). Equation (***) has a unique solution lo-
cally.
Once again, working through a few examples will make the theory much
clearer.
12
give the technique as a recipe. Suppose that S is parameterised by a function
g so that
S = {x = g(s) : s ∈ Rn−1}.
We work with the vector field
(a1, a2, . . . , an, b)
on Rn+1 and solve for the integral curves
dxs
dt = a(xs(t), ys(t)) xs(0) = g(s)
dys
dt = b(xs(t), ys(t)) ys(0) = φ(s).
Our solution is basically y(s, t) but we want it as a function of x not (s, t).
The map
(s, t) 7→ x(s, t)
will have an invertible derivative at t = 0 provided that the vector
(a1(g(s), φ(s)), a2(g(s), φ(s)), . . . , an(g(s), φ(s)))
is not tangent to S and since we have specified that S is a non-characteristic
this will always be the case. A theorem of differential geometry (proved in
Course B4 Differential Manifolds, but in any case very plausible) called the
‘inverse function theorem’ asserts that under these circumstances (continuous
derivative invertible at a point) the map
(s, t) 7→ x(s, t)
is locally invertible. We may thus define
u(x) = y(s(x), t(x))
and direct calculation shows that it is a solution. Checking reversibility we
see that we have proved the following theorem.
Theorem 2.15 (Theorem 7). Equation (***) has a unique solution lo-
cally.
Once again, working through a few examples will make the theory much
clearer.
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Page 13
Example 2.16 (Example 3). Solve the partial differential equation
u∂u∂x +
∂u
∂y = 1
subject to u = s/2 on x = y = s.
If a semi-linear equation (with real coefficients) has real right hand side
it can be solved by the technique of this subsection but if the right hand side
is complex we must use the previous method.
3 Distributions
3.1 Introduction
Distribution theory is a synthesis of ideas coming from partial differential
equations (e.g. weak solutions), physics (the Dirac delta function) and func-
tional analysis but the whole is much more than the sum of its parts.
We shall need the notion of the support of a continuous function.
Definition 3.1. If f : Rn → C is continuous we define the support supp f
of f by
supp f = closure{x : f(x) 6= 0}.
Lemma 3.2. If f : Rn → C is continuous the support of f is compact if and
only if there exists an R ≥ 0 such that f(x) = 0 if ‖x‖ > R.
Remember that, for Rn, a set is compact if and only if it is closed and
bounded.
We use the space of test functions D = C∞0 (Rn) consisting of smooth (that
is infinitely differentiable functions) functions of compact support. This is a
very restricted space of functions but fortunately non-trivial.
Lemma 3.3 (Lemma 1). (i) If we define E : R → R by
E(t) = exp(−1/t2) for t > 0
E(t) = 0 otherwise
then E is infinitely differentiable.
(ii) Given δ > η > 0 we can find F : R → R such that
1 ≥ F (t) ≥ 0 for all t
F (t) = 1 if |t| ≤ η
F (t) = 0 if |t| ≥ δ.
13
u∂u∂x +
∂u
∂y = 1
subject to u = s/2 on x = y = s.
If a semi-linear equation (with real coefficients) has real right hand side
it can be solved by the technique of this subsection but if the right hand side
is complex we must use the previous method.
3 Distributions
3.1 Introduction
Distribution theory is a synthesis of ideas coming from partial differential
equations (e.g. weak solutions), physics (the Dirac delta function) and func-
tional analysis but the whole is much more than the sum of its parts.
We shall need the notion of the support of a continuous function.
Definition 3.1. If f : Rn → C is continuous we define the support supp f
of f by
supp f = closure{x : f(x) 6= 0}.
Lemma 3.2. If f : Rn → C is continuous the support of f is compact if and
only if there exists an R ≥ 0 such that f(x) = 0 if ‖x‖ > R.
Remember that, for Rn, a set is compact if and only if it is closed and
bounded.
We use the space of test functions D = C∞0 (Rn) consisting of smooth (that
is infinitely differentiable functions) functions of compact support. This is a
very restricted space of functions but fortunately non-trivial.
Lemma 3.3 (Lemma 1). (i) If we define E : R → R by
E(t) = exp(−1/t2) for t > 0
E(t) = 0 otherwise
then E is infinitely differentiable.
(ii) Given δ > η > 0 we can find F : R → R such that
1 ≥ F (t) ≥ 0 for all t
F (t) = 1 if |t| ≤ η
F (t) = 0 if |t| ≥ δ.
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Page 14
(iii) Given δ > η > 0 we can find G : Rn → R such that
1 ≥ G(x) ≥ 1 for all x
G(x) = 1 if ‖x‖ ≤ η
G(x) = 0 if ‖x‖ ≥ δ.
Functions like G are called bump functions.
We equip D with a notion of convergence.
Definition 3.4. Suppose that fj ∈ D for each j and f ∈ D. We say that
fj →D f
if the following two conditions hold.
(i) There exists an R such that supp fj ⊆ B(0, R) for all j.
(ii) For each α we have ∂αfn → ∂αf uniformly as n→∞.
This is a definition that needs to be thought about carefully.
We can now define a distribution to be a linear map T : D → C which is
continuous in the sense that
fn →D f implies Tfn → Tf.
We write D′ for the set of distributions. we shall often write
Tf = 〈t, f〉.
Lemma 3.5. The set D′ is a vector space if we use the natural definitions
〈T + S, f〉 = 〈T, f〉+ 〈S, f〉, 〈λT, f〉 = λ〈T, f〉.
Our first key insight is the following.
Lemma 3.6. If g ∈ C(Rn) then if we define Tg by
〈Tg, f〉 =
∫
Rn
g(t)f(t) dt
we have Tg ∈ D′.
We usually write Tg = g, i.e.
〈Tg, f〉 =
∫
Rn
g(t)f(t) dt
and say that every continuous function is a distribution. (Hence the name
‘generalised function’ sometimes given to distributions.)
The second insight is more in the nature of a recipe than a theorem.
14
1 ≥ G(x) ≥ 1 for all x
G(x) = 1 if ‖x‖ ≤ η
G(x) = 0 if ‖x‖ ≥ δ.
Functions like G are called bump functions.
We equip D with a notion of convergence.
Definition 3.4. Suppose that fj ∈ D for each j and f ∈ D. We say that
fj →D f
if the following two conditions hold.
(i) There exists an R such that supp fj ⊆ B(0, R) for all j.
(ii) For each α we have ∂αfn → ∂αf uniformly as n→∞.
This is a definition that needs to be thought about carefully.
We can now define a distribution to be a linear map T : D → C which is
continuous in the sense that
fn →D f implies Tfn → Tf.
We write D′ for the set of distributions. we shall often write
Tf = 〈t, f〉.
Lemma 3.5. The set D′ is a vector space if we use the natural definitions
〈T + S, f〉 = 〈T, f〉+ 〈S, f〉, 〈λT, f〉 = λ〈T, f〉.
Our first key insight is the following.
Lemma 3.6. If g ∈ C(Rn) then if we define Tg by
〈Tg, f〉 =
∫
Rn
g(t)f(t) dt
we have Tg ∈ D′.
We usually write Tg = g, i.e.
〈Tg, f〉 =
∫
Rn
g(t)f(t) dt
and say that every continuous function is a distribution. (Hence the name
‘generalised function’ sometimes given to distributions.)
The second insight is more in the nature of a recipe than a theorem.
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Page 15
Lemma 3.7 (The standard recipe). Suppose that A : D → D is a linear
map which is continuous in the sense that
fn →D f implies Afn →D Af (1)
and that At : D → D is a linear map which is continuous in the sense that
fn →D f implies A
tfn →D A
tf. (2)
Suppose in addition
〈Ag, f〉 = 〈g, Atf〉 (3)
Then, if S is any distribution we can define a distribution TAS by the equation
〈TAS, f〉 = 〈S,Atf〉.
The mapping TA : D′ → D′ is linear and whenever g ∈ D we will have
Ag = TAg.
We usually write TA = A, so that
〈AT, f〉 = 〈T,Atf〉.
Exercise 3.8. Although we shall not use this in the course, convergence in
D′ is defined as follows. If Sj [j ≥ 1] are distributions we say that Tj →D′ S if
〈Sn, f〉 → 〈S, f〉
for all f ∈ D. Show that if A satisfies the conditions of Lemma 3.7, the map
TA : D′ → D′ is continuous in the sense that
Sj →D′ S implies TAS →D′ TAS.
Lemma 3.9. If T is a distribution we can define associated distributions by
following Lemma 3.7 as follows.
(i) (Multiplication by a smooth function.) If ψ ∈ C∞(Rn)
〈ψT, f〉 = 〈T, ψf〉
for all f ∈ D.
(ii) (Differentiation)
〈∂xjT, f〉 = −〈T, ∂xjf〉
15
map which is continuous in the sense that
fn →D f implies Afn →D Af (1)
and that At : D → D is a linear map which is continuous in the sense that
fn →D f implies A
tfn →D A
tf. (2)
Suppose in addition
〈Ag, f〉 = 〈g, Atf〉 (3)
Then, if S is any distribution we can define a distribution TAS by the equation
〈TAS, f〉 = 〈S,Atf〉.
The mapping TA : D′ → D′ is linear and whenever g ∈ D we will have
Ag = TAg.
We usually write TA = A, so that
〈AT, f〉 = 〈T,Atf〉.
Exercise 3.8. Although we shall not use this in the course, convergence in
D′ is defined as follows. If Sj [j ≥ 1] are distributions we say that Tj →D′ S if
〈Sn, f〉 → 〈S, f〉
for all f ∈ D. Show that if A satisfies the conditions of Lemma 3.7, the map
TA : D′ → D′ is continuous in the sense that
Sj →D′ S implies TAS →D′ TAS.
Lemma 3.9. If T is a distribution we can define associated distributions by
following Lemma 3.7 as follows.
(i) (Multiplication by a smooth function.) If ψ ∈ C∞(Rn)
〈ψT, f〉 = 〈T, ψf〉
for all f ∈ D.
(ii) (Differentiation)
〈∂xjT, f〉 = −〈T, ∂xjf〉
15
Page 16
for all f ∈ D.
(iii) (Reflection) If we write Rf(x) = f(−x) then
〈RT, f〉 = 〈T,Rf〉
for all f ∈ D.
(iv) (Translation) If we write τaf(x) = f(x− a) then
〈τaT, f〉 = 〈T, τ−af〉
for all f ∈ D.
Here is another example along the same lines.
Lemma 3.10. If A is a linear map on Rn then it induces a linear change of
coordinate map on functions A∗ by the formula
A∗f(x) = f(Ax).
If A is invertible we can define a transpose map (A∗)t by (A∗)t = | detA|−1(A−1)∗
i.e. by
((A∗)t(f))(x) = | detA|−1f(T−1x).
If T is a distribution we can define an associated distribution (i.e. obtain
a change of variables result for distributions) by following Lemma 3.7 to
obtain
〈A∗T, f〉 = 〈T, | detA|−1(A−1)∗f〉.
[Joshi and Wasserman think in terms of ‘signed areas’ and have detA
where we have | detA| in accordance with the conventions of applied math-
ematics.]
As an indication that we are on the right road, the next example shows
consistency with the Dirac deltas of mathematical methods in 1B.
Example 3.11. (i) If we write
〈δa, f〉 = f(a)
for all f ∈ D then δa is a distribution.
(ii) With the notation of Lemma 3.9 (iv),
τbδa = δa+b.
16
(iii) (Reflection) If we write Rf(x) = f(−x) then
〈RT, f〉 = 〈T,Rf〉
for all f ∈ D.
(iv) (Translation) If we write τaf(x) = f(x− a) then
〈τaT, f〉 = 〈T, τ−af〉
for all f ∈ D.
Here is another example along the same lines.
Lemma 3.10. If A is a linear map on Rn then it induces a linear change of
coordinate map on functions A∗ by the formula
A∗f(x) = f(Ax).
If A is invertible we can define a transpose map (A∗)t by (A∗)t = | detA|−1(A−1)∗
i.e. by
((A∗)t(f))(x) = | detA|−1f(T−1x).
If T is a distribution we can define an associated distribution (i.e. obtain
a change of variables result for distributions) by following Lemma 3.7 to
obtain
〈A∗T, f〉 = 〈T, | detA|−1(A−1)∗f〉.
[Joshi and Wasserman think in terms of ‘signed areas’ and have detA
where we have | detA| in accordance with the conventions of applied math-
ematics.]
As an indication that we are on the right road, the next example shows
consistency with the Dirac deltas of mathematical methods in 1B.
Example 3.11. (i) If we write
〈δa, f〉 = f(a)
for all f ∈ D then δa is a distribution.
(ii) With the notation of Lemma 3.9 (iv),
τbδa = δa+b.
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Page 17
(iii) We have
〈∂xjδa, f〉 = −(∂xjf)(a)
for all f ∈ D.
(iv) If we work on R and define the Heaviside function H : R → R by
H(t) = 1 for t ≥ 0,
H(t) = 0 otherwise
then, in the distributional sense, H ′ = δ.
3.2 The support of a distribution
We start with a definition.
Definition 3.12. If T ∈ D′ then the support suppT of T is defined as fol-
lows. A point x /∈ suppT if we can find an open set U such that x ∈ U and
whenever f ∈ D is such that supp f ⊆ U we have 〈T, f〉 = 0.
Exercise 3.13. (i) Check that if f is a continuous function its support as
a continuous function is the same as its support when it is considered as a
distribution.
(ii) Show that if T is a distribution suppT is closed.
The following theorem is extremely important as is the method (partition
of unity) by which we obtain it.
Theorem 3.14. If f ∈ D and T ∈ D′ then
suppT ∩ supp f = ∅ implies 〈T, f〉 = 0.
In my opinion the partition of unity is more an idea rather than a theorem
but here is one form expressed as a theorem.
Exercise 3.15 (Theorem 12). Let K be a compact subset of Rn and U1,
U2, . . . Um open sets in Rn such that K ⊆ ⋃Uj. Then we can find fj ∈ D
with 0 ≤ fi(x) ≤ 1 for all x, supp fj ⊆ Uj and
∑ fj(x) = 1 for x ∈ K,∑ fj(x) ≤ 1 everywhere.
When looking at Theorem 3.14 you should keep the following important
example in mind.
17
〈∂xjδa, f〉 = −(∂xjf)(a)
for all f ∈ D.
(iv) If we work on R and define the Heaviside function H : R → R by
H(t) = 1 for t ≥ 0,
H(t) = 0 otherwise
then, in the distributional sense, H ′ = δ.
3.2 The support of a distribution
We start with a definition.
Definition 3.12. If T ∈ D′ then the support suppT of T is defined as fol-
lows. A point x /∈ suppT if we can find an open set U such that x ∈ U and
whenever f ∈ D is such that supp f ⊆ U we have 〈T, f〉 = 0.
Exercise 3.13. (i) Check that if f is a continuous function its support as
a continuous function is the same as its support when it is considered as a
distribution.
(ii) Show that if T is a distribution suppT is closed.
The following theorem is extremely important as is the method (partition
of unity) by which we obtain it.
Theorem 3.14. If f ∈ D and T ∈ D′ then
suppT ∩ supp f = ∅ implies 〈T, f〉 = 0.
In my opinion the partition of unity is more an idea rather than a theorem
but here is one form expressed as a theorem.
Exercise 3.15 (Theorem 12). Let K be a compact subset of Rn and U1,
U2, . . . Um open sets in Rn such that K ⊆ ⋃Uj. Then we can find fj ∈ D
with 0 ≤ fi(x) ≤ 1 for all x, supp fj ⊆ Uj and
∑ fj(x) = 1 for x ∈ K,∑ fj(x) ≤ 1 everywhere.
When looking at Theorem 3.14 you should keep the following important
example in mind.
17
Page 18
Exercise 3.16. We work in R.
(i) Show that there exists an f ∈ D with f(0) = 0, f ′(0) = 1.
(ii) Observe that f(x) = 0 when x ∈ supp δ0 but 〈δ′0, f〉 = −1.
(iii) Why does this not contradict Theorem 3.14?
The support tells us where a distribution lives. A related concept is the
singular support.
Definition 3.17. If T ∈ D′ then the singular support singsuppT of T is
defined as follows. A point x /∈ singsuppT if we can find an f ∈ D such that
x /∈ suppT − f .
A careful argument using partitions of unity shows that if singsuppT = ∅
then T is in fact a smooth function.
When working with distributions it is helpful to recall the following re-
sults.
Lemma 3.18. If T is a distribution and ∂jT = 0 for all j then T = c where
c is a constant.
(We shall not prove this plausible result but like much else it may be
found in Friedlander’s beautiful little book, see section 2.1 [2].)
Lemma 3.19 (Theorem 28, A version of Taylor’s theorem). If f ∈ C∞(R)
and f(0) = f ′(0) = · · · = fm−1(0) = 0 then we can find g ∈ C∞(R) such that
f(x) = xmg(x)
for all x.
There is no difficulty (apart from notation) in extending Lemma 3.19 to
higher dimensions. In the next section we shall make use of the following
special case.
Lemma 3.20. If f ∈ C∞(Rn) and f(0) = 0 then we can find gj ∈ C∞(Rn)
such that
f(x) =
∑
gjxj
for all x.
As an example of the use of Lemma 3.19 we do the following exercise.
Exercise 3.21. Show that the general solution of
xT = 0
is T = cδ0.
18
(i) Show that there exists an f ∈ D with f(0) = 0, f ′(0) = 1.
(ii) Observe that f(x) = 0 when x ∈ supp δ0 but 〈δ′0, f〉 = −1.
(iii) Why does this not contradict Theorem 3.14?
The support tells us where a distribution lives. A related concept is the
singular support.
Definition 3.17. If T ∈ D′ then the singular support singsuppT of T is
defined as follows. A point x /∈ singsuppT if we can find an f ∈ D such that
x /∈ suppT − f .
A careful argument using partitions of unity shows that if singsuppT = ∅
then T is in fact a smooth function.
When working with distributions it is helpful to recall the following re-
sults.
Lemma 3.18. If T is a distribution and ∂jT = 0 for all j then T = c where
c is a constant.
(We shall not prove this plausible result but like much else it may be
found in Friedlander’s beautiful little book, see section 2.1 [2].)
Lemma 3.19 (Theorem 28, A version of Taylor’s theorem). If f ∈ C∞(R)
and f(0) = f ′(0) = · · · = fm−1(0) = 0 then we can find g ∈ C∞(R) such that
f(x) = xmg(x)
for all x.
There is no difficulty (apart from notation) in extending Lemma 3.19 to
higher dimensions. In the next section we shall make use of the following
special case.
Lemma 3.20. If f ∈ C∞(Rn) and f(0) = 0 then we can find gj ∈ C∞(Rn)
such that
f(x) =
∑
gjxj
for all x.
As an example of the use of Lemma 3.19 we do the following exercise.
Exercise 3.21. Show that the general solution of
xT = 0
is T = cδ0.
18
Page 19
3.3 Fourier transforms and the Schwartz space
We would like to take Fourier transforms of distributions but if we look at
the recipe of Lemma 3.7 we see that it fails since f ∈ D does not imply
fˆ ∈ D. (In fact, unless f = 0, f ∈ D implies fˆ /∈ D but we do not need this
result.)
Schwartz introduced another space of functions S(Rn which behaves ex-
traordinarily well under Fourier transforms.
S(Rn) = {f ∈ C∞(Rn) : sup
x
(1 + ‖x‖)m|∂αf(x)| <∞ for all m ≥ 0 and α}.
Thus S ‘consists of all infinitely differentiable functions all of whose derivative
tend to zero faster than polynomial towards infinity’. Since S ⊇ D we have
plentiful supply of such functions.
If f ∈ S(Rn) then the Fourier transform
fˆ(ξ) =
( 1
2pi
)n/2 ∫
Rn
e−ix.ξf(x) dx
We sometimes write Ff = fˆ . We also write
Djf = −i
∂
∂xj
and abusing notation take xjf to be the function
x 7→ xjf(x).
It is clear that
Dj : S(Rn)→ S(Rn)
xj : S(Rn)→ S(Rn)
Familiar computations now give our first result.
Lemma 3.22 (Lemma 2). The Fourier transform f 7→ fˆ takes S(Rn) to
S(Rn) and
D̂jf = ξj fˆ
x̂jf = −Dj fˆ
Lemma 3.22 can be used to give a neat proof of the Fourier inversion
formula for S. We use the following lemma.
19
We would like to take Fourier transforms of distributions but if we look at
the recipe of Lemma 3.7 we see that it fails since f ∈ D does not imply
fˆ ∈ D. (In fact, unless f = 0, f ∈ D implies fˆ /∈ D but we do not need this
result.)
Schwartz introduced another space of functions S(Rn which behaves ex-
traordinarily well under Fourier transforms.
S(Rn) = {f ∈ C∞(Rn) : sup
x
(1 + ‖x‖)m|∂αf(x)| <∞ for all m ≥ 0 and α}.
Thus S ‘consists of all infinitely differentiable functions all of whose derivative
tend to zero faster than polynomial towards infinity’. Since S ⊇ D we have
plentiful supply of such functions.
If f ∈ S(Rn) then the Fourier transform
fˆ(ξ) =
( 1
2pi
)n/2 ∫
Rn
e−ix.ξf(x) dx
We sometimes write Ff = fˆ . We also write
Djf = −i
∂
∂xj
and abusing notation take xjf to be the function
x 7→ xjf(x).
It is clear that
Dj : S(Rn)→ S(Rn)
xj : S(Rn)→ S(Rn)
Familiar computations now give our first result.
Lemma 3.22 (Lemma 2). The Fourier transform f 7→ fˆ takes S(Rn) to
S(Rn) and
D̂jf = ξj fˆ
x̂jf = −Dj fˆ
Lemma 3.22 can be used to give a neat proof of the Fourier inversion
formula for S. We use the following lemma.
19
Page 20
Lemma 3.23 (Lemma 4). If T : S(Rn) → S(Rn) is a linear map which
commutes with xj and Dj for all j then, writing I for the identity map,
T = cI
for some constant c.
Our proof of Lemma 3.23 depends on a slight improvement of the Taylor
series result Lemma 3.20.
Lemma 3.24 (Corollary 2). If f ∈ S(Rn) and f(a) = 0 then we can find
fj ∈ S(Rn) such that
f(x) =
n∑
j=1
(xj − aj)fj(x)
for all x.
We can now prove our inversion theorem.
Theorem 3.25. If R : S(Rn)→ S(Rn) is the map given by Rf(x) = f(−x)
then
F2 = R.
Observe that this means that F 4 = I so F is invertible. Stating our result
in a more traditional but equivalent manner we have the following theorem.
Theorem 3.26 (Theorem 9). The Fourier transform
f 7→ fˆ
is an isomorphism of S onto itself with inverse given by
f(x) =
( 1
2pi
)n/2 ∫
Rn
eix.ξfˆ(ξ) dx.
If f, g ∈ F , or more generally if the integral makes sense, we define the
convolution f ∗ g by
f ∗ g(x) =
∫
Rn
f(x− t)g(t) dt
The standard results that we recall from mathematical methods hold.
Lemma 3.27 (Theorem 10). If f, g ∈ F
(i)
∫
fgˆ =
∫
fˆ g,
(ii)
∫
fg∗ =
∫
fˆ(gˆ)∗, (Parseval)
(iii) f̂ ∗ g = fˆ gˆ,
(iv) f̂ g = fˆ ∗ gˆ,
(v) f ∗ g ∈ F .
20
commutes with xj and Dj for all j then, writing I for the identity map,
T = cI
for some constant c.
Our proof of Lemma 3.23 depends on a slight improvement of the Taylor
series result Lemma 3.20.
Lemma 3.24 (Corollary 2). If f ∈ S(Rn) and f(a) = 0 then we can find
fj ∈ S(Rn) such that
f(x) =
n∑
j=1
(xj − aj)fj(x)
for all x.
We can now prove our inversion theorem.
Theorem 3.25. If R : S(Rn)→ S(Rn) is the map given by Rf(x) = f(−x)
then
F2 = R.
Observe that this means that F 4 = I so F is invertible. Stating our result
in a more traditional but equivalent manner we have the following theorem.
Theorem 3.26 (Theorem 9). The Fourier transform
f 7→ fˆ
is an isomorphism of S onto itself with inverse given by
f(x) =
( 1
2pi
)n/2 ∫
Rn
eix.ξfˆ(ξ) dx.
If f, g ∈ F , or more generally if the integral makes sense, we define the
convolution f ∗ g by
f ∗ g(x) =
∫
Rn
f(x− t)g(t) dt
The standard results that we recall from mathematical methods hold.
Lemma 3.27 (Theorem 10). If f, g ∈ F
(i)
∫
fgˆ =
∫
fˆ g,
(ii)
∫
fg∗ =
∫
fˆ(gˆ)∗, (Parseval)
(iii) f̂ ∗ g = fˆ gˆ,
(iv) f̂ g = fˆ ∗ gˆ,
(v) f ∗ g ∈ F .
20
Page 21
3.4 Tempered distributions
Just as we constructed a space of distributions D′ using functions from D as
test functions so we can construct a space of distributions S ′ using functions
from S as test functions.
Of course, we need to equip S with a notion of convergence.
Definition 3.28. Suppose that fj ∈ S for each j and f ∈ S. We say that
fj →S f
if for each m ≥ 0 and α we have (1 + ‖x‖)m∂α(fj(x)− f(x))→ 0 uniformly
as n→∞.
We can now say that T ∈ S ′ to be a linear map T : D → C which is
continuous in the sense that
fn →S f impliesTfn → Tf.
As before we shall often write
Tf = 〈T, f〉.
We can now develop the theory of tempered distributions T ∈ S ′ just as
we did that of distributions T ∈ D′. Using results in the previous subsection
it is easy to check that F : S → S is a linear map which is continuous in the
sense that
fn →S f implies Ffn →S Ff (1)
and that
〈Fg, f〉 = 〈g,Ff〉 (3)
for any f, g ∈ S. (Thus, in the language of Lemma 3.7, F t = F .) Thus,
applying the result corresponding to Lemma 3.7 if T ∈ S ′ we can define FT
by the equation
〈FT, f〉 = 〈T,Ff〉,
or in the obvious corresponding notation
〈Tˆ , f〉 = 〈T, fˆ〉,
There is no difficulty in proving results corresponding to Lemma 3.22.
21
Just as we constructed a space of distributions D′ using functions from D as
test functions so we can construct a space of distributions S ′ using functions
from S as test functions.
Of course, we need to equip S with a notion of convergence.
Definition 3.28. Suppose that fj ∈ S for each j and f ∈ S. We say that
fj →S f
if for each m ≥ 0 and α we have (1 + ‖x‖)m∂α(fj(x)− f(x))→ 0 uniformly
as n→∞.
We can now say that T ∈ S ′ to be a linear map T : D → C which is
continuous in the sense that
fn →S f impliesTfn → Tf.
As before we shall often write
Tf = 〈T, f〉.
We can now develop the theory of tempered distributions T ∈ S ′ just as
we did that of distributions T ∈ D′. Using results in the previous subsection
it is easy to check that F : S → S is a linear map which is continuous in the
sense that
fn →S f implies Ffn →S Ff (1)
and that
〈Fg, f〉 = 〈g,Ff〉 (3)
for any f, g ∈ S. (Thus, in the language of Lemma 3.7, F t = F .) Thus,
applying the result corresponding to Lemma 3.7 if T ∈ S ′ we can define FT
by the equation
〈FT, f〉 = 〈T,Ff〉,
or in the obvious corresponding notation
〈Tˆ , f〉 = 〈T, fˆ〉,
There is no difficulty in proving results corresponding to Lemma 3.22.
21
Page 22
Lemma 3.29. (i) If R : S(Rn) → S ′(Rn) is the extension to S ′ of the
reflection function on S given by Rf(x) = f(−x) then, working in S ′
F2 = R.
(ii) The Fourier transform
f 7→ fˆ
is an isomorphism of S ′ onto itself with inverse given by F 3.
f(x) =
( 1
2pi
)n/2 ∫
Rn
eix.ξfˆ(ξ) dx
(iii) If T ∈ S ′ then, in an obvious notation,
D̂jT = ξjTˆ
x̂jT = −DjTˆ
However we do not have to develop the theory of S ′ and D′ separately
since D sits nicely in S and so S ′ sits nicely in D′.
Theorem 3.30. (i) We have D ⊆ S. Moreover if that fj ∈ D for each j
and f ∈ D then
fj →D f implies fj →S f
(ii) We may consider any element T ∈ S ′ as an element of D in a natural
manner.
Theorem 3.31 (Theorem 13). (i) Given any f ∈ S we may find a se-
quence fj ∈ D such that
fj →S f
as j →∞.
(ii) If T ∈ F and 〈T, f〉 = 0 whenever f ∈ D then 〈T, f〉 = 0 whenever
f ∈ S.
(iii) Let T, S ∈ F . If T and S are unequal when considered as members
of F they remain unequal when considered as members D.
Thus we may, and shall consider S ′ as a subspace of D′. When we talk
about distributions we shall mean members of D′. When we talk about
tempered distributions we shall mean members of S ′.
22
reflection function on S given by Rf(x) = f(−x) then, working in S ′
F2 = R.
(ii) The Fourier transform
f 7→ fˆ
is an isomorphism of S ′ onto itself with inverse given by F 3.
f(x) =
( 1
2pi
)n/2 ∫
Rn
eix.ξfˆ(ξ) dx
(iii) If T ∈ S ′ then, in an obvious notation,
D̂jT = ξjTˆ
x̂jT = −DjTˆ
However we do not have to develop the theory of S ′ and D′ separately
since D sits nicely in S and so S ′ sits nicely in D′.
Theorem 3.30. (i) We have D ⊆ S. Moreover if that fj ∈ D for each j
and f ∈ D then
fj →D f implies fj →S f
(ii) We may consider any element T ∈ S ′ as an element of D in a natural
manner.
Theorem 3.31 (Theorem 13). (i) Given any f ∈ S we may find a se-
quence fj ∈ D such that
fj →S f
as j →∞.
(ii) If T ∈ F and 〈T, f〉 = 0 whenever f ∈ D then 〈T, f〉 = 0 whenever
f ∈ S.
(iii) Let T, S ∈ F . If T and S are unequal when considered as members
of F they remain unequal when considered as members D.
Thus we may, and shall consider S ′ as a subspace of D′. When we talk
about distributions we shall mean members of D′. When we talk about
tempered distributions we shall mean members of S ′.
22
Page 23
Example 3.32 (Example 6). The delta function δ0 is a tempered distri-
bution. Its Fourier transform δˆ0 is the constant function (2pi)−n/2.
The constant function 1 is a tempered distribution. Its Fourier transform
1ˆ is (2pi)n/2δ0.
The way we have defined Tˆ by Lemma 3.7 ensures that
distributional Fourier transform(f) = classical Fourier transform(f)
whenever f ∈ F . In fact more is true.
Lemma 3.33 (Simple version of Proposition 2). If f is continuous and∫
|f(x)| dx <∞ then
distributional Fourier transform(f) = classical Fourier transform(f)
i.e.
fˆ(ξ) =
( 1
2pi
)n/2 ∫
Rn
e−ix.ξf(x) dx.
We shall also need the following simple result which the reader may al-
ready have met.
Lemma 3.34. If f is continuous and
∫
|f(x)| dx <∞ then fˆ is continuous
and bounded.
As a corollary we have the following useful lemma.
Lemma 3.35 (Corollary 5). If q is smooth and
|q(ξ)| ≤ C((1 + ‖ξ‖2)1/2)−n−l−²
for some ² > 0 then the Fourier transform and Fourier inverse transform of
q is C l.
4 Convolution and fundamental solutions
In this chapter we shall return at last to the topic of partial differential
equations but first we need to study convolution of distributions.
23
bution. Its Fourier transform δˆ0 is the constant function (2pi)−n/2.
The constant function 1 is a tempered distribution. Its Fourier transform
1ˆ is (2pi)n/2δ0.
The way we have defined Tˆ by Lemma 3.7 ensures that
distributional Fourier transform(f) = classical Fourier transform(f)
whenever f ∈ F . In fact more is true.
Lemma 3.33 (Simple version of Proposition 2). If f is continuous and∫
|f(x)| dx <∞ then
distributional Fourier transform(f) = classical Fourier transform(f)
i.e.
fˆ(ξ) =
( 1
2pi
)n/2 ∫
Rn
e−ix.ξf(x) dx.
We shall also need the following simple result which the reader may al-
ready have met.
Lemma 3.34. If f is continuous and
∫
|f(x)| dx <∞ then fˆ is continuous
and bounded.
As a corollary we have the following useful lemma.
Lemma 3.35 (Corollary 5). If q is smooth and
|q(ξ)| ≤ C((1 + ‖ξ‖2)1/2)−n−l−²
for some ² > 0 then the Fourier transform and Fourier inverse transform of
q is C l.
4 Convolution and fundamental solutions
In this chapter we shall return at last to the topic of partial differential
equations but first we need to study convolution of distributions.
23
Page 24
4.1 Convolution
Unfortunately the notion of convolution of distributions is not as ‘clean’ as
the rest of the theory but it is essential.
Recall that if f and g are ‘nice functions’ (e.g. if they are both Schwartz
functions) we define the convolution f ∗ g by
f ∗ g(x) =
∫
Rn
f(x− t)g(t) dt.
Convolution plays an essential role in probability theory, signal processing,
linear partial differential equations (as we shall see) and many other parts of
analysis. We proved in Lemma 3.27 (iii) that f̂ ∗ g = fˆ gˆ, and it can be argued
that importance of the Fourier transform is that it converts convolution into
multiplication.
However if we try out this formula on the tempered distribution 1 we get
1̂ ∗ 1 = 1ˆ1ˆ = (2pi)nδ0δ0
an ‘equation’ in which neither side makes sense (at least within distribution
theory as developed here). In general we can not multiply two distributions
and we can not convolve two distributions.
Having said this, there are many circumstances in which it is possible
to convolve two distributions. To see this observe that under favourable
circumstances, if f , g and h are all nice functions
〈f ∗ g, h〉 =
∫ (∫
Rn
f(s− y)g(y) dy
)
h(s) ds
=
∫ ∫
f(s− y)g(y)h(y) dy ds
=
∫ ∫
f(s− y)g(y)h(s) ds dy
=
∫ ∫
f(x)g(y)h(x+ y) dx, dy
= 〈f(x), 〈g(y), h(x+ y)〉〉.
We shall attempt to generalise this. Our treatment is not quite as general
as that given by Joshi and Wassermann in section 4.4 of their notes but
follows the treatment given by Friedlander in his book [2]. The underlying
idea is identical and either treatment can be used in examination questions.
Looking at the discussion of convolution in ‘favourable circumstances’ we
observe that we have made use of the ‘exterior product’ (f, g) 7→ f ⊗g where
f ⊗ g(x, y) = f(x)g(y).
24
Unfortunately the notion of convolution of distributions is not as ‘clean’ as
the rest of the theory but it is essential.
Recall that if f and g are ‘nice functions’ (e.g. if they are both Schwartz
functions) we define the convolution f ∗ g by
f ∗ g(x) =
∫
Rn
f(x− t)g(t) dt.
Convolution plays an essential role in probability theory, signal processing,
linear partial differential equations (as we shall see) and many other parts of
analysis. We proved in Lemma 3.27 (iii) that f̂ ∗ g = fˆ gˆ, and it can be argued
that importance of the Fourier transform is that it converts convolution into
multiplication.
However if we try out this formula on the tempered distribution 1 we get
1̂ ∗ 1 = 1ˆ1ˆ = (2pi)nδ0δ0
an ‘equation’ in which neither side makes sense (at least within distribution
theory as developed here). In general we can not multiply two distributions
and we can not convolve two distributions.
Having said this, there are many circumstances in which it is possible
to convolve two distributions. To see this observe that under favourable
circumstances, if f , g and h are all nice functions
〈f ∗ g, h〉 =
∫ (∫
Rn
f(s− y)g(y) dy
)
h(s) ds
=
∫ ∫
f(s− y)g(y)h(y) dy ds
=
∫ ∫
f(s− y)g(y)h(s) ds dy
=
∫ ∫
f(x)g(y)h(x+ y) dx, dy
= 〈f(x), 〈g(y), h(x+ y)〉〉.
We shall attempt to generalise this. Our treatment is not quite as general
as that given by Joshi and Wassermann in section 4.4 of their notes but
follows the treatment given by Friedlander in his book [2]. The underlying
idea is identical and either treatment can be used in examination questions.
Looking at the discussion of convolution in ‘favourable circumstances’ we
observe that we have made use of the ‘exterior product’ (f, g) 7→ f ⊗g where
f ⊗ g(x, y) = f(x)g(y).
24
Page 25
We observe that f , g and f ⊗ g are considered as distributions then
〈f ⊗ g, h〉 =
∫ ∫
f(x)g(y)h(x, y) dx dy. †
The exterior product carries over easily to distributions. If S ∈ D(Rn)
T ∈ D(Rm) we set
〈S ⊗ T, h〉 = 〈S, 〈T, hx〉〉
where hx(y) = h(x, y) and h ∈ Rn+m. Using dummy variables
〈S ⊗ T (x, y), h(x, y)〉 = 〈S(x), 〈T (y), h(x, y)〉〉.
Lemma 4.1. With the notation and hypotheses of the previous paragraph,
S ⊗ T is a well defined member of D(Rm+n).
Suppose now S, T ∈ D(Rn). If h ∈ D(Rn) the discussion of convolution
in ‘favourable circumstances’ suggests that we should look at the function
h˜ : R2n → C given by h˜(x, y) = h(x+y) and apply S⊗T to h˜. Unfortunately,
although h˜ ∈ C∞(R2n), h˜ is not of compact support unless h = 0.
Exercise 4.2. Prove the two statements made in the last sentence.
So far we have only allowed elements of D′ to operate on elements of D
so we appear to be stuck.
However, under certain circumstances, elements of D′ will operate on
more general C∞ functions in a natural manner.
Lemma 4.3. (i) Suppose T ∈ D′(Rn) and f ∈ C∞(Rn). If supp f ∩ suppT
is compact then there is a natural definition of 〈T, f〉.
(ii) Suppose T ∈ D′(Rn) and fj ∈ C∞(Rn). Suppose further that there
exists a compact set K such that K ⊇ supp fj ∩ suppT for all j. Then if
∂αfj → 0 uniformly on K for each α it follows that 〈T, fj〉 → 0 as j →∞.
We have arrived at our goal.
Theorem 4.4. If S, T ∈ D′(Rn) and suppS or suppT is compact then
we may define S ∗ T as follows. If h ∈ D(Rn) define h˜ : R2n → C by
h˜(x, y) = h(x + y), the expression 〈S ⊗ T, h˜〉 is well defined in the sense of
Lemma 4.3 so we set
〈S ∗ T, h〉 = 〈S ⊗ T, h˜〉
The map S ∗ T so defined is a distribution.
25
〈f ⊗ g, h〉 =
∫ ∫
f(x)g(y)h(x, y) dx dy. †
The exterior product carries over easily to distributions. If S ∈ D(Rn)
T ∈ D(Rm) we set
〈S ⊗ T, h〉 = 〈S, 〈T, hx〉〉
where hx(y) = h(x, y) and h ∈ Rn+m. Using dummy variables
〈S ⊗ T (x, y), h(x, y)〉 = 〈S(x), 〈T (y), h(x, y)〉〉.
Lemma 4.1. With the notation and hypotheses of the previous paragraph,
S ⊗ T is a well defined member of D(Rm+n).
Suppose now S, T ∈ D(Rn). If h ∈ D(Rn) the discussion of convolution
in ‘favourable circumstances’ suggests that we should look at the function
h˜ : R2n → C given by h˜(x, y) = h(x+y) and apply S⊗T to h˜. Unfortunately,
although h˜ ∈ C∞(R2n), h˜ is not of compact support unless h = 0.
Exercise 4.2. Prove the two statements made in the last sentence.
So far we have only allowed elements of D′ to operate on elements of D
so we appear to be stuck.
However, under certain circumstances, elements of D′ will operate on
more general C∞ functions in a natural manner.
Lemma 4.3. (i) Suppose T ∈ D′(Rn) and f ∈ C∞(Rn). If supp f ∩ suppT
is compact then there is a natural definition of 〈T, f〉.
(ii) Suppose T ∈ D′(Rn) and fj ∈ C∞(Rn). Suppose further that there
exists a compact set K such that K ⊇ supp fj ∩ suppT for all j. Then if
∂αfj → 0 uniformly on K for each α it follows that 〈T, fj〉 → 0 as j →∞.
We have arrived at our goal.
Theorem 4.4. If S, T ∈ D′(Rn) and suppS or suppT is compact then
we may define S ∗ T as follows. If h ∈ D(Rn) define h˜ : R2n → C by
h˜(x, y) = h(x + y), the expression 〈S ⊗ T, h˜〉 is well defined in the sense of
Lemma 4.3 so we set
〈S ∗ T, h〉 = 〈S ⊗ T, h˜〉
The map S ∗ T so defined is a distribution.
25
Page 26
Lemma 4.5. With the notation and hypotheses of Theorem 4.4
suppS ∗ T ⊆ suppS + suppT.
(Here suppS + suppT = {x+ y : x ∈ suppS, y ∈ suppT}.)
Note that if the hypotheses of Lemma 4.4 hold then using dummy vari-
ables we have the elegant formula
〈S ∗ T (t), f(t)〉 = 〈S(x)〈T (y), f(x+ y)〉.
The hypotheses of Theorem 4.4 can clearly be relaxed somewhat, so Joshi
and Wassermann refer to ‘convolvable distributions’.
We summarise some of the useful facts about convolutions.
Lemma 4.6. (i) If T and S are distributions at least one of which has com-
pact support then
T ∗ S = S ∗ T.
(ii) We recall that δ0 has compact support. If T is any distribution
δ0 ∗ T = T.
(iii) If R, T and S are distributions at least two of which have compact
support then
R ∗ (S ∗ T ) = (R ∗ S) ∗ T.
(iv) If T and S are distributions at least one of which has compact support
then
∂α(T ∗ S) = (∂αT ) ∗ S = T ∗ (∂αS).
Note that this gives ∂αu = ∂αδ0 ∗ u, so applying a differential operator is
equivalent to convolving with some distribution.
One of the key facts about convolution is that it smooths.
Lemma 4.7. If T is a distribution and f an m times continuously dif-
ferentiable function and at least one of the two has compact support then
〈Ty, f(x− y)〉 is a function and, as distributions,
〈Ty, f(x− y)〉 = T ∗ f(x).
Thus T ∗ f is an m times continuously differentiable function.
26
suppS ∗ T ⊆ suppS + suppT.
(Here suppS + suppT = {x+ y : x ∈ suppS, y ∈ suppT}.)
Note that if the hypotheses of Lemma 4.4 hold then using dummy vari-
ables we have the elegant formula
〈S ∗ T (t), f(t)〉 = 〈S(x)〈T (y), f(x+ y)〉.
The hypotheses of Theorem 4.4 can clearly be relaxed somewhat, so Joshi
and Wassermann refer to ‘convolvable distributions’.
We summarise some of the useful facts about convolutions.
Lemma 4.6. (i) If T and S are distributions at least one of which has com-
pact support then
T ∗ S = S ∗ T.
(ii) We recall that δ0 has compact support. If T is any distribution
δ0 ∗ T = T.
(iii) If R, T and S are distributions at least two of which have compact
support then
R ∗ (S ∗ T ) = (R ∗ S) ∗ T.
(iv) If T and S are distributions at least one of which has compact support
then
∂α(T ∗ S) = (∂αT ) ∗ S = T ∗ (∂αS).
Note that this gives ∂αu = ∂αδ0 ∗ u, so applying a differential operator is
equivalent to convolving with some distribution.
One of the key facts about convolution is that it smooths.
Lemma 4.7. If T is a distribution and f an m times continuously dif-
ferentiable function and at least one of the two has compact support then
〈Ty, f(x− y)〉 is a function and, as distributions,
〈Ty, f(x− y)〉 = T ∗ f(x).
Thus T ∗ f is an m times continuously differentiable function.
26
Page 27
We shall not need the next result so I leave it as an exercise but the
discussion would seem incomplete without it.
Exercise 4.8. (i) If T is a distribution of compact support then, using the
extension of Lemma 4.3,
Tˆ (ξ) = (2pi)−n/2〈T (x), exp(iξ.x)〉.
Further Tˆ is a continuous function.
(ii) If S and T are distributions of compact support
Ŝ ∗ T (ξ) = T̂ (ξ)Ŝ(ξ).
4.2 Fundamental solutions
We now have the tools to make substantial progress with the study of linear
partial differential equations with constant coefficients. The key notion is
that of a fundamental solution.
Definition 4.9. A fundamental solution of P (D) where P (ξ1, . . . , ξn) is a
polynomial with constant coefficients and P (D) = P (−i∂1, . . . , −i∂n) is a
distributional solution, T , of
P (D)T = δ0.
Note that we do not say that a fundamental solution is unique.
Example 4.10. If n = 1 and P (x) = ix then Ec = H + c is a fundamental
solution where H is the Heaviside function and c is any constant.
The interest of fundamental solutions is shown by the next lemma.
Lemma 4.11. We use the notation of Definition 4.9
(i) If T is a distribution of compact support and E is a fundamental
solution of p(D) then S = E ∗ T is a solution of P (D)S = T .
(ii) If f is an m times continuously differentiable function of compact
support and E is a fundamental solution of P (D) then S = E ∗ f is an m
times continuously differentiable solution of P (D)S = T .
Example 4.12. (i) We continue with Example 4.10. Suppose f is a contin-
uous function of compact support and we wish to solve
u′(x) = f(x).
27
discussion would seem incomplete without it.
Exercise 4.8. (i) If T is a distribution of compact support then, using the
extension of Lemma 4.3,
Tˆ (ξ) = (2pi)−n/2〈T (x), exp(iξ.x)〉.
Further Tˆ is a continuous function.
(ii) If S and T are distributions of compact support
Ŝ ∗ T (ξ) = T̂ (ξ)Ŝ(ξ).
4.2 Fundamental solutions
We now have the tools to make substantial progress with the study of linear
partial differential equations with constant coefficients. The key notion is
that of a fundamental solution.
Definition 4.9. A fundamental solution of P (D) where P (ξ1, . . . , ξn) is a
polynomial with constant coefficients and P (D) = P (−i∂1, . . . , −i∂n) is a
distributional solution, T , of
P (D)T = δ0.
Note that we do not say that a fundamental solution is unique.
Example 4.10. If n = 1 and P (x) = ix then Ec = H + c is a fundamental
solution where H is the Heaviside function and c is any constant.
The interest of fundamental solutions is shown by the next lemma.
Lemma 4.11. We use the notation of Definition 4.9
(i) If T is a distribution of compact support and E is a fundamental
solution of p(D) then S = E ∗ T is a solution of P (D)S = T .
(ii) If f is an m times continuously differentiable function of compact
support and E is a fundamental solution of P (D) then S = E ∗ f is an m
times continuously differentiable solution of P (D)S = T .
Example 4.12. (i) We continue with Example 4.10. Suppose f is a contin-
uous function of compact support and we wish to solve
u′(x) = f(x).
27
Page 28
Applying Lemma 4.11 we obtain
u(x) = c
∫ ∞
−∞
f(t) dt+
∫ x
−∞
f(t) dt.
(ii) Particular interest attaches to the two cases c = 0 when the solution
is
u(x) =
∫ x
−∞
f(t) dt (A)
and c = −1 when the solution is
u(x) = −
∫ ∞
x
f(t) dt (B)
since equation (A) produces a solution for u′(x) = f(x) valid whenever
f(x) = 0 for x large and positive and equation (B) produces a solution for
u′(x) = f(x) valid whenever f(x) = 0 for x large and negative.
One of the points to note about Example 4.12 (ii) is that it shows the
advantage of extending the notion of convolution as far as as we can.
Example 4.13. If S and T are distributions on R such that we can find a
real number R such that
suppS, suppT ⊆ (−R,∞).
Explain in as much detail as you consider desirable why we can define S ∗T .
It also shows that it may be desirable to consider different fundamental
solutions for different kinds of problems. We can think of fundamental so-
lutions as inverses on different classes of functions in accordance with the
formula
E ∗ P (D) = P (D) ∗ E = δ0.
4.3 Our first fundamental solution
If E is a fundamental solution of P (D) then by definition
P (D)E = δ0
so, provided that E is a tempered distribution, we may take Fourier trans-
forms to obtain
P (ξ)Eˆ(ξ) = (2pi)−n/2,
28
u(x) = c
∫ ∞
−∞
f(t) dt+
∫ x
−∞
f(t) dt.
(ii) Particular interest attaches to the two cases c = 0 when the solution
is
u(x) =
∫ x
−∞
f(t) dt (A)
and c = −1 when the solution is
u(x) = −
∫ ∞
x
f(t) dt (B)
since equation (A) produces a solution for u′(x) = f(x) valid whenever
f(x) = 0 for x large and positive and equation (B) produces a solution for
u′(x) = f(x) valid whenever f(x) = 0 for x large and negative.
One of the points to note about Example 4.12 (ii) is that it shows the
advantage of extending the notion of convolution as far as as we can.
Example 4.13. If S and T are distributions on R such that we can find a
real number R such that
suppS, suppT ⊆ (−R,∞).
Explain in as much detail as you consider desirable why we can define S ∗T .
It also shows that it may be desirable to consider different fundamental
solutions for different kinds of problems. We can think of fundamental so-
lutions as inverses on different classes of functions in accordance with the
formula
E ∗ P (D) = P (D) ∗ E = δ0.
4.3 Our first fundamental solution
If E is a fundamental solution of P (D) then by definition
P (D)E = δ0
so, provided that E is a tempered distribution, we may take Fourier trans-
forms to obtain
P (ξ)Eˆ(ξ) = (2pi)−n/2,
28
Page 29
so, at least formally,
Eˆ(ξ) = 1(2pi)n/2P (ξ) .
If P (ξ) is never zero the argument can easily be reversed in rigorous manner
to give the following theorem.
Theorem 4.14. If P is a polynomial with no real zeros then there is a fun-
damental solution given by the tempered distribution u with
uˆ(ξ) = 1(2pi)n/2P (ξ) .
There is only one fundamental solution which is also a tempered distribution.
Exercise 4.15. Show that Theorem 4.14 applies to the operator −∆ + 1.
Use Theorem 4.14 to solve the one dimensional case
−u′′ + u = δ0
and verify by direct calculation that you have indeed got a solution.
A large part of the theory of partial differential equations consists of
different ways of deal with the zeros of P (ξ).
4.4 The parametrix
If P (D) is an elliptic operator, P (ξ) may have real zeros but they must lie
in a bounded set. Using this we can obtain something ‘almost as good’ as a
fundamental solution.
Theorem 4.16 (Theorem 14). If P (D) is elliptic, φ a smooth function of
compact support identically 1 on a neighbourhood of the zero set of P (ξ) then
we can find a tempered distribution E with
Eˆ(ξ) =
( 1
2pi
)n/2 1
P (ξ)(1− φ)(ξ).
We have
P (D)E = δ0 + f
where f ∈ S(Rn). Also
singsuppE ⊆ {0}.
29
Eˆ(ξ) = 1(2pi)n/2P (ξ) .
If P (ξ) is never zero the argument can easily be reversed in rigorous manner
to give the following theorem.
Theorem 4.14. If P is a polynomial with no real zeros then there is a fun-
damental solution given by the tempered distribution u with
uˆ(ξ) = 1(2pi)n/2P (ξ) .
There is only one fundamental solution which is also a tempered distribution.
Exercise 4.15. Show that Theorem 4.14 applies to the operator −∆ + 1.
Use Theorem 4.14 to solve the one dimensional case
−u′′ + u = δ0
and verify by direct calculation that you have indeed got a solution.
A large part of the theory of partial differential equations consists of
different ways of deal with the zeros of P (ξ).
4.4 The parametrix
If P (D) is an elliptic operator, P (ξ) may have real zeros but they must lie
in a bounded set. Using this we can obtain something ‘almost as good’ as a
fundamental solution.
Theorem 4.16 (Theorem 14). If P (D) is elliptic, φ a smooth function of
compact support identically 1 on a neighbourhood of the zero set of P (ξ) then
we can find a tempered distribution E with
Eˆ(ξ) =
( 1
2pi
)n/2 1
P (ξ)(1− φ)(ξ).
We have
P (D)E = δ0 + f
where f ∈ S(Rn). Also
singsuppE ⊆ {0}.
29
Page 30
Lemma 4.17. If P (D) is elliptic and U a neighbourhood of 0 then we can
find a tempered distribution E with suppE ⊆ U , singsuppE ⊆ {0}, and
P (D)E = δ0 + f
where f ∈ S(Rn).
We call a distribution like E in Theorem 4.16 or Lemma 4.17 a parametrix.
It will, no doubt, have occured to the reader that if we solve a partial
differential equation
P (D)u = f
in the distributional sense what we get is a distribution u and not necessarily
a classical function even if f is a classical function. The next theorem removes
this problem in certain cases.
Theorem 4.18 (Theorem 15, special case of Weyl’s lemma). If u ∈
D′ and P (D) is an elliptic differential operator then u and P (D)u have the
same singular support. In particular if P (D)u is smooth so is u.
4.5 Existence of the fundamental solution
Except when we discussed first order equations, this course has been about
linear partial differential equations with constant coefficients. It is an im-
portant fact, which we shall not prove, that in this case there is always a
fundamental solution.
Theorem 4.19 (Theorem 16). If P (ξ1, . . . , ξn) is a polynomial with con-
stant coefficients and P (D) = P (−i∂1, . . . , −i∂n) then we can find a distri-
bution E such that
P (D)E = δ0.
A proof of Theorem 4.19 is given in section 10.4 of Friedlander’s book [2].
We shall prove an important special case.
Theorem 4.20. If P (ξ1, . . . , ξn) is a polynomial with constant coefficients
and there exists an a ∈ Rn and an η > 0 such that
|P (ξ + ia)| ≥ η
for all ξ ∈ Rn then we can find a distribution E such that
P (D)E = δ0.
30
find a tempered distribution E with suppE ⊆ U , singsuppE ⊆ {0}, and
P (D)E = δ0 + f
where f ∈ S(Rn).
We call a distribution like E in Theorem 4.16 or Lemma 4.17 a parametrix.
It will, no doubt, have occured to the reader that if we solve a partial
differential equation
P (D)u = f
in the distributional sense what we get is a distribution u and not necessarily
a classical function even if f is a classical function. The next theorem removes
this problem in certain cases.
Theorem 4.18 (Theorem 15, special case of Weyl’s lemma). If u ∈
D′ and P (D) is an elliptic differential operator then u and P (D)u have the
same singular support. In particular if P (D)u is smooth so is u.
4.5 Existence of the fundamental solution
Except when we discussed first order equations, this course has been about
linear partial differential equations with constant coefficients. It is an im-
portant fact, which we shall not prove, that in this case there is always a
fundamental solution.
Theorem 4.19 (Theorem 16). If P (ξ1, . . . , ξn) is a polynomial with con-
stant coefficients and P (D) = P (−i∂1, . . . , −i∂n) then we can find a distri-
bution E such that
P (D)E = δ0.
A proof of Theorem 4.19 is given in section 10.4 of Friedlander’s book [2].
We shall prove an important special case.
Theorem 4.20. If P (ξ1, . . . , ξn) is a polynomial with constant coefficients
and there exists an a ∈ Rn and an η > 0 such that
|P (ξ + ia)| ≥ η
for all ξ ∈ Rn then we can find a distribution E such that
P (D)E = δ0.
30
Page 31
We prove this result by contour pushing so we need a result on analyticity
and growthR
Lemma 4.21 (Proposition 5.5, the Paley-Wiener estimate). If f ∈ D(Rn)
then there is a natural extension of fˆ to a function on Cn which is analytic
in each variable. If supp f ⊆ B(0, r) then
|fˆ(z)| ≤ CN(1 + ‖z‖)−Ner|=z|
for all z and integers N ≥ 0. Here CN is a constant depending only on N
and f .
With the proof of Theorem 4.20 we come to the end of our general account
of linear constant coefficient partial differential equations. For the rest of the
course we shall study the Laplacian, the heat and the wave operator. We
close this section by looking at the heat and wave operator in the context of
Theorem 4.20.
Example 4.22 (Example 5.7). If
P (D) = ∂∂t −
∑ ∂2
∂x2j
(the heat operator) then if a < 0
|P (τ + ia, ξ)| ≥ −a
but if a ≥ 0 there exists a ξ ∈ Rn such that
|P (τ + ia, ξ)| = 0.
Example 4.23 (Example 5.8). If
P (D) = ∂
2
∂t2 −
∑ ∂2
∂x2j
(the wave operator) then if a 6= 0
|P (τ + ia, ξ)| ≥ a4 > 0
for τ ∈ R, ξ ∈ Rn.
Note that our shifting contour argument gives one fundamental solution
for the heat operator but two for the wave equation.
31
and growthR
Lemma 4.21 (Proposition 5.5, the Paley-Wiener estimate). If f ∈ D(Rn)
then there is a natural extension of fˆ to a function on Cn which is analytic
in each variable. If supp f ⊆ B(0, r) then
|fˆ(z)| ≤ CN(1 + ‖z‖)−Ner|=z|
for all z and integers N ≥ 0. Here CN is a constant depending only on N
and f .
With the proof of Theorem 4.20 we come to the end of our general account
of linear constant coefficient partial differential equations. For the rest of the
course we shall study the Laplacian, the heat and the wave operator. We
close this section by looking at the heat and wave operator in the context of
Theorem 4.20.
Example 4.22 (Example 5.7). If
P (D) = ∂∂t −
∑ ∂2
∂x2j
(the heat operator) then if a < 0
|P (τ + ia, ξ)| ≥ −a
but if a ≥ 0 there exists a ξ ∈ Rn such that
|P (τ + ia, ξ)| = 0.
Example 4.23 (Example 5.8). If
P (D) = ∂
2
∂t2 −
∑ ∂2
∂x2j
(the wave operator) then if a 6= 0
|P (τ + ia, ξ)| ≥ a4 > 0
for τ ∈ R, ξ ∈ Rn.
Note that our shifting contour argument gives one fundamental solution
for the heat operator but two for the wave equation.
31
Page 32
5 The Laplacian
5.1 A fundamental solution
The reader can probably guess a fundamental solution for the Laplacian.
We shall proceed more slowly in order to introduce some useful ideas. Ob-
serve first that by Theorem 4.18 any fundamental solution E will have
singsuppE ⊆ {0}. Since the Laplacian is rotationally invariant an averaging
argument shows (as one might fairly confidently guess) that there must be a
rotationally invariant fundamental solution.
Exercise 5.1. We could now proceed as follows. Suppose f(x) = g(r) with
r = ‖x‖. Show that
(∆f)(x) = 1rn−1
d
drr
n−1fg(r).
If f is smooth and ∆f = 0 on Rn \ {0} find f .
Instead of following the path set out in Exercise 5.1 we use another nice
idea — that of homogeneity. The following remark is due to Euler.
Lemma 5.2. Let f : Rn \ {0} → R be continuously differentiable and m be
an integer. The following statements are equivalent.
(i) f(λx) = λmf(x) for all λ > 0. (In other words, f is homogeneous of
degree m.)
(ii)
( n∑
j=1
xj
∂
∂xj
−m
)
f(x) = 0.
In view of Lemma 5.2 we make the following definition.
Definition 5.3. If u ∈ D′ then u is homogeneous of degree m ∈ C if
( n∑
j=1
xj
∂
∂xj
−m
)
f(x) = 0.
Lemma 5.4. (i) The delta function δ0 on Rn is homogeneous of degree −n.
(ii) If is u ∈ D′ is homogeneous of degree m then ∂xju is homogeneous of
degree m− 1.
Looking at Lemma 5.4 we guess that one fundamental solution En of the
Laplacian will be radially symmetric of degree 2 − n and we arrive at the
guess
En = Cn‖x‖2−n.
If n 6= 2 we shall now define En = Cn‖x‖2−n and verify that it is indeed a
fundamental solution for the appropriate choice of Cn.
32
5.1 A fundamental solution
The reader can probably guess a fundamental solution for the Laplacian.
We shall proceed more slowly in order to introduce some useful ideas. Ob-
serve first that by Theorem 4.18 any fundamental solution E will have
singsuppE ⊆ {0}. Since the Laplacian is rotationally invariant an averaging
argument shows (as one might fairly confidently guess) that there must be a
rotationally invariant fundamental solution.
Exercise 5.1. We could now proceed as follows. Suppose f(x) = g(r) with
r = ‖x‖. Show that
(∆f)(x) = 1rn−1
d
drr
n−1fg(r).
If f is smooth and ∆f = 0 on Rn \ {0} find f .
Instead of following the path set out in Exercise 5.1 we use another nice
idea — that of homogeneity. The following remark is due to Euler.
Lemma 5.2. Let f : Rn \ {0} → R be continuously differentiable and m be
an integer. The following statements are equivalent.
(i) f(λx) = λmf(x) for all λ > 0. (In other words, f is homogeneous of
degree m.)
(ii)
( n∑
j=1
xj
∂
∂xj
−m
)
f(x) = 0.
In view of Lemma 5.2 we make the following definition.
Definition 5.3. If u ∈ D′ then u is homogeneous of degree m ∈ C if
( n∑
j=1
xj
∂
∂xj
−m
)
f(x) = 0.
Lemma 5.4. (i) The delta function δ0 on Rn is homogeneous of degree −n.
(ii) If is u ∈ D′ is homogeneous of degree m then ∂xju is homogeneous of
degree m− 1.
Looking at Lemma 5.4 we guess that one fundamental solution En of the
Laplacian will be radially symmetric of degree 2 − n and we arrive at the
guess
En = Cn‖x‖2−n.
If n 6= 2 we shall now define En = Cn‖x‖2−n and verify that it is indeed a
fundamental solution for the appropriate choice of Cn.
32
Page 33
Lemma 5.5. Suppose n 6= 2.
(i) En is a well defined distribution according to our usual formula
〈En, f〉 =
∫
Rn
En(x)f(x) dx.
(ii) supp∆En ⊆ {0}.
(iii) ∆En is a multiple of δ0.
(iv) If Cn = ((2− n)ωn−1)−1 where ωn−1 is the area of the surface of the
unit sphere in Rn then ∆En = δ0.
If n = 2 we use previous knowledge to guess that
E2(x) = −
1
2pi log ‖x‖.
The verification that this is indeed a fundamental solution follows the pattern
of Lemma 5.5.
Theorem 5.6 (Theorem 5.1). The Laplacian on Rn has the fundamental
solution
E(x) = 1(2− n)ωn−1‖x‖2−n
when n 6= 2 and
E(x) = − 12pi log ‖x‖
for n = 2.
5.2 Identities and estimates
The reader will be familiar with the Gauss divergence theorem and the associ-
ated Green’s identities. There are problems (or at least apparent problems) in
proving this for complicated domains but the standard mathematical method
proof is rigorous when applied to the ball which is all we shall need.
Theorem 5.7 (Theorem 5.2, Gauss divergence theorem). Let g : B(0, r)→
Rn be continuous on B(0, r) and continuously differentiable on B(0, r). Then
∫
‖x‖<r
O.g dV =
∫
‖x‖=r
g.nx dS
where nx = ‖x‖−1x is the outward normal.
33
(i) En is a well defined distribution according to our usual formula
〈En, f〉 =
∫
Rn
En(x)f(x) dx.
(ii) supp∆En ⊆ {0}.
(iii) ∆En is a multiple of δ0.
(iv) If Cn = ((2− n)ωn−1)−1 where ωn−1 is the area of the surface of the
unit sphere in Rn then ∆En = δ0.
If n = 2 we use previous knowledge to guess that
E2(x) = −
1
2pi log ‖x‖.
The verification that this is indeed a fundamental solution follows the pattern
of Lemma 5.5.
Theorem 5.6 (Theorem 5.1). The Laplacian on Rn has the fundamental
solution
E(x) = 1(2− n)ωn−1‖x‖2−n
when n 6= 2 and
E(x) = − 12pi log ‖x‖
for n = 2.
5.2 Identities and estimates
The reader will be familiar with the Gauss divergence theorem and the associ-
ated Green’s identities. There are problems (or at least apparent problems) in
proving this for complicated domains but the standard mathematical method
proof is rigorous when applied to the ball which is all we shall need.
Theorem 5.7 (Theorem 5.2, Gauss divergence theorem). Let g : B(0, r)→
Rn be continuous on B(0, r) and continuously differentiable on B(0, r). Then
∫
‖x‖<r
O.g dV =
∫
‖x‖=r
g.nx dS
where nx = ‖x‖−1x is the outward normal.
33
Page 34
Green’s identities correspond to integrating by parts and are more of a
method than a theorem.
Lemma 5.8 (Proposition 5.1, Green’s identities). If u, v ∈ C2(B(0, r))
(a)
∫
‖x‖<r
v∆u dV = −
∫
‖x‖<r
Ov.Ou dV +
∫
‖x‖=r
vOu.nx dS.
(b)
∫
‖x‖<r
v.∆u− u.∆v dV =
∫
‖x‖=r
(vOu− uOv).nx dS.
Much of the early work on the Laplacian used the following observation.
Lemma 5.9 (Proposition 5.2, energy estimate). If u ∈ C2(B(0, r))
∫
‖x‖<r
‖Ou‖2 dV =
∫
‖x‖=r
uOu.nx dS −
∫
‖x‖<r
u∆u dV.
Theorem 5.10. (i) The Dirichlet problem:- find u ∈ C2(B(0, r)) subject to
the conditions
∆u = f on B(0, r)
u = g on {x : ‖x‖ = r}
has at most one solution.
(ii) The Neumann problem:- find u ∈ C2(B(0, r)) subject to the conditions
∆u = f on B(0, r)
Ou.nx = h on {x : ‖x‖ = r}
has at most one solution up to a constant.
We can do much better than Theorem 5.10 (i) by using the weak maxi-
mum principle.
Theorem 5.11 (Proposition 5.3, weak maximum principle). Let Ω be
a non-empty bounded open set in Rn with boundary ∂Ω. If u ∈ C(Ω) and u
is twice continuously differentiable in Ω with ∆u ≥ 0 in Ω then
sup
x∈Ω
u(x) = sup
x∈∂Ω
u(x).
Theorem 5.12 (Uniqueness for the Dirichlet problem). The Dirichlet
problem:- find u ∈ C(Ω) with u is twice continuously differentiable in Ω sub-
ject to the conditions
∆u = f on Ω
u = g on {x : ‖x‖ = r}
has at most one solution.
34
method than a theorem.
Lemma 5.8 (Proposition 5.1, Green’s identities). If u, v ∈ C2(B(0, r))
(a)
∫
‖x‖<r
v∆u dV = −
∫
‖x‖<r
Ov.Ou dV +
∫
‖x‖=r
vOu.nx dS.
(b)
∫
‖x‖<r
v.∆u− u.∆v dV =
∫
‖x‖=r
(vOu− uOv).nx dS.
Much of the early work on the Laplacian used the following observation.
Lemma 5.9 (Proposition 5.2, energy estimate). If u ∈ C2(B(0, r))
∫
‖x‖<r
‖Ou‖2 dV =
∫
‖x‖=r
uOu.nx dS −
∫
‖x‖<r
u∆u dV.
Theorem 5.10. (i) The Dirichlet problem:- find u ∈ C2(B(0, r)) subject to
the conditions
∆u = f on B(0, r)
u = g on {x : ‖x‖ = r}
has at most one solution.
(ii) The Neumann problem:- find u ∈ C2(B(0, r)) subject to the conditions
∆u = f on B(0, r)
Ou.nx = h on {x : ‖x‖ = r}
has at most one solution up to a constant.
We can do much better than Theorem 5.10 (i) by using the weak maxi-
mum principle.
Theorem 5.11 (Proposition 5.3, weak maximum principle). Let Ω be
a non-empty bounded open set in Rn with boundary ∂Ω. If u ∈ C(Ω) and u
is twice continuously differentiable in Ω with ∆u ≥ 0 in Ω then
sup
x∈Ω
u(x) = sup
x∈∂Ω
u(x).
Theorem 5.12 (Uniqueness for the Dirichlet problem). The Dirichlet
problem:- find u ∈ C(Ω) with u is twice continuously differentiable in Ω sub-
ject to the conditions
∆u = f on Ω
u = g on {x : ‖x‖ = r}
has at most one solution.
34
Page 35
The point here is not so much the generality of Ω as that we only need u
differentiable on Ω.
If u is twice continuously differentiable on some open set Ω with ∆u =
we say that u is a harmonic function. The reader is already aware of the
connection with analytic functions. As an example we give another proof of
a result from 1B.
Lemma 5.13 (Maximum principle for analytic functions). If Ω is a
bounded non-empty open set in C and f : Ω → C is a continuous function,
analytic on Ω then
sup
z∈Ω
|f(z)| = sup
z∈∂Ω
|f(z)|.
Lemma 5.14 (Gauss mean value theorem). (a) The value of a harmonic
function at a point is equal to its average over any sphere (boundary of a ball)
centred at that point.
(b) The value of a harmonic function at a point is equal to its average
over any ball centred at that point.
We shall see in Lemma 6.6 that condition (a) actually characterises har-
monic functions.
Lemma 5.15 (Proposition 5.5, the strong maximum principle). If Ω
is a bounded open connected set and u ∈ C(Ω) satisfies
u(ξ) ≤ 1Area sphere radius ρ
∫
‖x−ξ‖=ρ
u(x) dS
for all ρ sufficiently small (depending on ξ) and all ξ ∈ Ω then either u is
constant or u(ξ) < supx∈∂Ω u(x) for all ξ ∈ Ω.
We shall probably not have time to develop the matter much further but
the associated definition (Definition 5.16) is important in later work.
Definition 5.16. If Ω is open and u ∈ C(Ω) satisfies
u(ξ) ≤ 1Area sphere radius ρ
∫
‖x−ξ‖=ρ
u(x) dS
for all ρ sufficiently small (depending on ξ) and all ξ ∈ Ω then we say that
u is subharmonic.
Lemma 5.17. If Ω is open and u ∈ C(Ω) and u is twice differentiable on Ω
with ∆u ≥ 0 then u is subharmonic.
35
differentiable on Ω.
If u is twice continuously differentiable on some open set Ω with ∆u =
we say that u is a harmonic function. The reader is already aware of the
connection with analytic functions. As an example we give another proof of
a result from 1B.
Lemma 5.13 (Maximum principle for analytic functions). If Ω is a
bounded non-empty open set in C and f : Ω → C is a continuous function,
analytic on Ω then
sup
z∈Ω
|f(z)| = sup
z∈∂Ω
|f(z)|.
Lemma 5.14 (Gauss mean value theorem). (a) The value of a harmonic
function at a point is equal to its average over any sphere (boundary of a ball)
centred at that point.
(b) The value of a harmonic function at a point is equal to its average
over any ball centred at that point.
We shall see in Lemma 6.6 that condition (a) actually characterises har-
monic functions.
Lemma 5.15 (Proposition 5.5, the strong maximum principle). If Ω
is a bounded open connected set and u ∈ C(Ω) satisfies
u(ξ) ≤ 1Area sphere radius ρ
∫
‖x−ξ‖=ρ
u(x) dS
for all ρ sufficiently small (depending on ξ) and all ξ ∈ Ω then either u is
constant or u(ξ) < supx∈∂Ω u(x) for all ξ ∈ Ω.
We shall probably not have time to develop the matter much further but
the associated definition (Definition 5.16) is important in later work.
Definition 5.16. If Ω is open and u ∈ C(Ω) satisfies
u(ξ) ≤ 1Area sphere radius ρ
∫
‖x−ξ‖=ρ
u(x) dS
for all ρ sufficiently small (depending on ξ) and all ξ ∈ Ω then we say that
u is subharmonic.
Lemma 5.17. If Ω is open and u ∈ C(Ω) and u is twice differentiable on Ω
with ∆u ≥ 0 then u is subharmonic.
35
Page 36
5.3 The dual Dirichlet problem for the unit ball
Let Ω be a bounded open set with smooth boundary ∂Ω.
(a) The Dirichlet problem asks for a solution of ∆f = 0 in Ω with f = h
on ∂Ω.
(b) The dual Dirichlet problem asks for a solution of ∆f = 0 in Ω with
f = g on ∂Ω.
We have been deliberately vague about the exact nature of f , h and g.
Maintaining the same vagueness we see that solving one problem is ‘more or
less’ equivalent to solving the other.
Let us try to solve the dual Dirichlet problem for the unit ball B.
∆f = g for ‖x‖ < 1
f = 0 for ‖x‖ = 1
To make life easier for ourselves we shall initially assume g ∈ C(B) and
g infinitely differentiable on B. (In Lemma 6.3 we shall see that that the
formula obtained can be extended to a much wider class of functions.) We
shall work in Rn with n ≥ 3.
We start by defining
g˜ = g for ‖x‖ ≤ 1
g˜ = 0 for ‖x‖ > 1.
If E is the fundamental solution obtained earlier we start by looking at
f1 = E ∗ g˜. We know that, in a distributional sense,
∆E ∗ g˜ = g˜
but we wish to have a classical solution.
Lemma 5.18 (Lemma 5.1). (i) f1 ∈ C1(Rn) and ∂jf1 = ∂jE ∗ g˜.
(ii) There is a constant such that |f1(x)| ≤ A‖x‖2−n, |∂jf1(x)| ≤ A‖x‖1−n
for ‖x‖ large.
Lemma 5.19 (Lemma 5.2). We have f1 infinitely differentiable on Rn \
{x : ‖x‖ = 1} and
∆f1 = g for ‖x‖ < 1
∆f1 = 0 for ‖x‖ > 1.
We thus have ∆f1 = g in B but the boundary conditions are wrong.
To get round this we use Kelvin’s method of reflections. (If you have done
electrostatics you will recognise this for dimension n = 2.) We put
(Kf)(x) = ‖x‖2−nf
( x
‖x‖2
)
.
36
Let Ω be a bounded open set with smooth boundary ∂Ω.
(a) The Dirichlet problem asks for a solution of ∆f = 0 in Ω with f = h
on ∂Ω.
(b) The dual Dirichlet problem asks for a solution of ∆f = 0 in Ω with
f = g on ∂Ω.
We have been deliberately vague about the exact nature of f , h and g.
Maintaining the same vagueness we see that solving one problem is ‘more or
less’ equivalent to solving the other.
Let us try to solve the dual Dirichlet problem for the unit ball B.
∆f = g for ‖x‖ < 1
f = 0 for ‖x‖ = 1
To make life easier for ourselves we shall initially assume g ∈ C(B) and
g infinitely differentiable on B. (In Lemma 6.3 we shall see that that the
formula obtained can be extended to a much wider class of functions.) We
shall work in Rn with n ≥ 3.
We start by defining
g˜ = g for ‖x‖ ≤ 1
g˜ = 0 for ‖x‖ > 1.
If E is the fundamental solution obtained earlier we start by looking at
f1 = E ∗ g˜. We know that, in a distributional sense,
∆E ∗ g˜ = g˜
but we wish to have a classical solution.
Lemma 5.18 (Lemma 5.1). (i) f1 ∈ C1(Rn) and ∂jf1 = ∂jE ∗ g˜.
(ii) There is a constant such that |f1(x)| ≤ A‖x‖2−n, |∂jf1(x)| ≤ A‖x‖1−n
for ‖x‖ large.
Lemma 5.19 (Lemma 5.2). We have f1 infinitely differentiable on Rn \
{x : ‖x‖ = 1} and
∆f1 = g for ‖x‖ < 1
∆f1 = 0 for ‖x‖ > 1.
We thus have ∆f1 = g in B but the boundary conditions are wrong.
To get round this we use Kelvin’s method of reflections. (If you have done
electrostatics you will recognise this for dimension n = 2.) We put
(Kf)(x) = ‖x‖2−nf
( x
‖x‖2
)
.
36
Page 37
Lemma 5.20. If we set f2 = Kf1 and f = f1 − f2 then
(i) ∆f2 = 0 in B.
(ii) f1 = f2 on ∂B.
(iii) f solves our Dirichlet problem.
The proof of Lemma 5.20 (i) involves substantial calculation.
Lemma 5.21 (Lemma 5.3). (i) If f is twice differentiable
∆(Kf) = ‖x‖−4K(∆f).
(ii) If f is once differentiable and we write r ∂∂r =
∑ xj ∂∂xj we have
r ∂∂r = (−n+ 2)Kf −K
(
r ∂∂r
)
.
Lemma 5.21 shows us that f2 is harmonic in B \ {0} but a little more
work is needed to establish the full result.
Lemma 5.22 (Lemma 4). The function f2 is harmonic on B.
Doing the appropriate simple substitutions we obtain our final result.
Theorem 5.23 (Theorem 5.4). We work in Rn with n ≥ 3. If we set
G(x, y) = ‖x− y‖
2−n
ωn−1(n− 2)
− ‖x‖
2−n‖‖x‖−2x− y‖2
ωn−1(n− 2)
then
f(x) =
∫
‖y‖<1
G(x, y)g(y) dy
solves the dual Dirichlet problem.
The reader will recognise G(x, y) as a Green’s function.
We shall not have time to do the case n = 2 but the reader will find no
problem in redoing our arguments to obtain.
Theorem 5.24. We work in R2. If we set
G(x, y) = 12pi (log ‖x− y‖ − log ‖‖x‖
−2x− y‖)
then
f(x) =
∫
‖y‖<1
G(x, y)g(y) dy
solves the dual Dirichlet problem.
37
(i) ∆f2 = 0 in B.
(ii) f1 = f2 on ∂B.
(iii) f solves our Dirichlet problem.
The proof of Lemma 5.20 (i) involves substantial calculation.
Lemma 5.21 (Lemma 5.3). (i) If f is twice differentiable
∆(Kf) = ‖x‖−4K(∆f).
(ii) If f is once differentiable and we write r ∂∂r =
∑ xj ∂∂xj we have
r ∂∂r = (−n+ 2)Kf −K
(
r ∂∂r
)
.
Lemma 5.21 shows us that f2 is harmonic in B \ {0} but a little more
work is needed to establish the full result.
Lemma 5.22 (Lemma 4). The function f2 is harmonic on B.
Doing the appropriate simple substitutions we obtain our final result.
Theorem 5.23 (Theorem 5.4). We work in Rn with n ≥ 3. If we set
G(x, y) = ‖x− y‖
2−n
ωn−1(n− 2)
− ‖x‖
2−n‖‖x‖−2x− y‖2
ωn−1(n− 2)
then
f(x) =
∫
‖y‖<1
G(x, y)g(y) dy
solves the dual Dirichlet problem.
The reader will recognise G(x, y) as a Green’s function.
We shall not have time to do the case n = 2 but the reader will find no
problem in redoing our arguments to obtain.
Theorem 5.24. We work in R2. If we set
G(x, y) = 12pi (log ‖x− y‖ − log ‖‖x‖
−2x− y‖)
then
f(x) =
∫
‖y‖<1
G(x, y)g(y) dy
solves the dual Dirichlet problem.
37
Page 38
6 Dirichlet’s problem for the ball and Pois-
son’s formula
We noted earlier that a solution of the dual Dirichlet problem will give us a
solution for Dirichlet’s problem.
Lemma 6.1. Consider the Dirichlet problem for the unit ball B.
∆f = 0 for ‖x‖ < 1
f = h for ‖x‖ = 1
with h infinitely differentiable.
If n ≥ 3 the solution is given by
f(x) =
∫
‖y‖=1
P (x, y)h(y) dS(y)
where P (x, y) = ddnG(x, y) the directional derivative along the outward nor-
mal.
Lemma 6.2. Consider Let the Dirichlet problem for the unit ball B.
∆f = 0 for ‖x‖ < 1
f = h for ‖x‖ = 1
with h infinitely differentiable.
If n ≥ 2 the solution is given by
f(x) =
∫
‖y‖=1
P (x, y)h(y) dS(y)
where
P (x, y) = 1ωn−1
1− ‖x‖2
‖x− y‖n
We call P the Poisson kernel. We can improve Lemma 6.2 and provide
a proof of the improvement which is essentially independent of the work
already done.
Lemma 6.3. Consider the Dirichlet problem for the unit ball B.
∆f = 0 for ‖x‖ < 1
f = h for ‖x‖ = 1
38
son’s formula
We noted earlier that a solution of the dual Dirichlet problem will give us a
solution for Dirichlet’s problem.
Lemma 6.1. Consider the Dirichlet problem for the unit ball B.
∆f = 0 for ‖x‖ < 1
f = h for ‖x‖ = 1
with h infinitely differentiable.
If n ≥ 3 the solution is given by
f(x) =
∫
‖y‖=1
P (x, y)h(y) dS(y)
where P (x, y) = ddnG(x, y) the directional derivative along the outward nor-
mal.
Lemma 6.2. Consider Let the Dirichlet problem for the unit ball B.
∆f = 0 for ‖x‖ < 1
f = h for ‖x‖ = 1
with h infinitely differentiable.
If n ≥ 2 the solution is given by
f(x) =
∫
‖y‖=1
P (x, y)h(y) dS(y)
where
P (x, y) = 1ωn−1
1− ‖x‖2
‖x− y‖n
We call P the Poisson kernel. We can improve Lemma 6.2 and provide
a proof of the improvement which is essentially independent of the work
already done.
Lemma 6.3. Consider the Dirichlet problem for the unit ball B.
∆f = 0 for ‖x‖ < 1
f = h for ‖x‖ = 1
38
Page 39
with h continuous.
If n ≥ 2 the solution is given by
f(x) =
∫
‖y‖<1
P (x, y)h(y) dy
We need a preliminary lemma.
Lemma 6.4. (i) If y is fixed with ‖y‖ < 1 then P (x, y) is harmonic in x for
‖x‖ < 1.
(ii)
∫
‖y‖=1
P (x, y) dS(y) = 1
Dilation and translation gives the following result.
Lemma 6.5. The Dirichlet problem for any ball with continuous data has a
unique solution.
Lemma 6.5 is a very useful tool as the proof of the converse of Lemma 5.14 (a)
shows.
Lemma 6.6 (Proposition 5.7). Let Ω be open and u : Ω→ R continuous.
If the value of u at any point x ∈ Ω is equal to its average over any sufficiently
small sphere centred at that x then u is harmonic.
Lemma 6.7 (Corollary 5.2). The uniform limit of harmonic functions is
itself harmonic.
(Compare Morera’s theorem in complex variable theory.)
Poisson’s solution for the Dirichlet problem for the ball gives a useful
inequality.
Theorem 6.8 (Harnack’s inequality, Theorem 5.5). If f ∈ C(B(0, R))
is everywhere non-negative and f is harmonic on B(0, R) then if ‖x‖ = r < R
we have
Rn−2(R− r)
(R + r)n−1 f(0) ≤ f(x) ≤
Rn−2(R + r)
(R− r)n−1 f(0).
Lemma 6.9 (Harnack’s convergence theorem, Theorem 5.6). Let Ω
be an open set in Rn. Suppose that um is harmonic on Ω and that
(a) um(x) ≤ um+1(x) for all m ≥ 1
(b) um(x)→ u(x) as m→∞
for all x ∈ Ω. Then um → u uniformly on bounded sets and u is harmonic.
39
If n ≥ 2 the solution is given by
f(x) =
∫
‖y‖<1
P (x, y)h(y) dy
We need a preliminary lemma.
Lemma 6.4. (i) If y is fixed with ‖y‖ < 1 then P (x, y) is harmonic in x for
‖x‖ < 1.
(ii)
∫
‖y‖=1
P (x, y) dS(y) = 1
Dilation and translation gives the following result.
Lemma 6.5. The Dirichlet problem for any ball with continuous data has a
unique solution.
Lemma 6.5 is a very useful tool as the proof of the converse of Lemma 5.14 (a)
shows.
Lemma 6.6 (Proposition 5.7). Let Ω be open and u : Ω→ R continuous.
If the value of u at any point x ∈ Ω is equal to its average over any sufficiently
small sphere centred at that x then u is harmonic.
Lemma 6.7 (Corollary 5.2). The uniform limit of harmonic functions is
itself harmonic.
(Compare Morera’s theorem in complex variable theory.)
Poisson’s solution for the Dirichlet problem for the ball gives a useful
inequality.
Theorem 6.8 (Harnack’s inequality, Theorem 5.5). If f ∈ C(B(0, R))
is everywhere non-negative and f is harmonic on B(0, R) then if ‖x‖ = r < R
we have
Rn−2(R− r)
(R + r)n−1 f(0) ≤ f(x) ≤
Rn−2(R + r)
(R− r)n−1 f(0).
Lemma 6.9 (Harnack’s convergence theorem, Theorem 5.6). Let Ω
be an open set in Rn. Suppose that um is harmonic on Ω and that
(a) um(x) ≤ um+1(x) for all m ≥ 1
(b) um(x)→ u(x) as m→∞
for all x ∈ Ω. Then um → u uniformly on bounded sets and u is harmonic.
39
Page 40
7 The wave equation
Recall that the wave operator 2 given by
2u(t,x) = ∂
2u
∂t (t,x)−∆xu(t,x)
where t ∈ R, x ∈ Rn and u : R× Rn → R is a suitable function.
Guided by our knowledge of physics we concentrate on solving two prob-
lems. The first is the forcing problem.
2u = f subject to u(t, x) = 0 for t ≤ 0
where f(t, x) = 0 for t ≤ 0. (Later we shall replace t ≤ 0 by the slightly
more general t large and negative.) The second is the Cauchy problem
2u = f subject to u(x, 0) = u0(x) and
∂u
∂t (x, 0) = u1(x).
The two problems are, as one might expect, closely related and we shall solve
the second problem by reducing it to the first.
Looking at Theorem 4.20 and Example 4.23 we see that we know two
fundamental solutions E+ and E− are given by
〈E±, φ〉 =
( 1
2pi
)(n+1)/2 ∫
Rn
∫
R
φˆ(−(τ ± iα),−ξ)
‖ξ‖2 − (τ ± iα)2 dτ dξ,
where α is a strictly positive number. We can make a good guess at what
E+ and E− are though, unless we have a good background in physics, our
guess will be wrong when n = 2 (and indeed for all n ≥ 4)! We therefore
start with the case n = 3 when our guess turns out to be right and look at
E+. Since our guess suggests that E+ is not a function but a distribution we
look at 〈E±, φ〉 rather than E+ by itself.
Lemma 7.1. For all n have
〈E±, φ〉 =
( 1
2pi
)n/2 ∫
Rn
∫ ∞
0
φˆt(−ξ)
sin(‖ξ‖t)
‖ξ‖ dξ dt
We now see if our guess is appropriate.
Lemma 7.2. Let n = 3. The formula
〈u, f〉 =
∫
‖x‖=t
f(x) dS(x)
defines a distribution of compact support. We have
uˆ(ξ) = 4pit‖ξ‖ sin(‖ξ‖t)
40
Recall that the wave operator 2 given by
2u(t,x) = ∂
2u
∂t (t,x)−∆xu(t,x)
where t ∈ R, x ∈ Rn and u : R× Rn → R is a suitable function.
Guided by our knowledge of physics we concentrate on solving two prob-
lems. The first is the forcing problem.
2u = f subject to u(t, x) = 0 for t ≤ 0
where f(t, x) = 0 for t ≤ 0. (Later we shall replace t ≤ 0 by the slightly
more general t large and negative.) The second is the Cauchy problem
2u = f subject to u(x, 0) = u0(x) and
∂u
∂t (x, 0) = u1(x).
The two problems are, as one might expect, closely related and we shall solve
the second problem by reducing it to the first.
Looking at Theorem 4.20 and Example 4.23 we see that we know two
fundamental solutions E+ and E− are given by
〈E±, φ〉 =
( 1
2pi
)(n+1)/2 ∫
Rn
∫
R
φˆ(−(τ ± iα),−ξ)
‖ξ‖2 − (τ ± iα)2 dτ dξ,
where α is a strictly positive number. We can make a good guess at what
E+ and E− are though, unless we have a good background in physics, our
guess will be wrong when n = 2 (and indeed for all n ≥ 4)! We therefore
start with the case n = 3 when our guess turns out to be right and look at
E+. Since our guess suggests that E+ is not a function but a distribution we
look at 〈E±, φ〉 rather than E+ by itself.
Lemma 7.1. For all n have
〈E±, φ〉 =
( 1
2pi
)n/2 ∫
Rn
∫ ∞
0
φˆt(−ξ)
sin(‖ξ‖t)
‖ξ‖ dξ dt
We now see if our guess is appropriate.
Lemma 7.2. Let n = 3. The formula
〈u, f〉 =
∫
‖x‖=t
f(x) dS(x)
defines a distribution of compact support. We have
uˆ(ξ) = 4pit‖ξ‖ sin(‖ξ‖t)
40
Page 41
Putting everything together we obtain our solution.
Lemma 7.3. If n = 3 the the forward fundamental solution E+ is given by
〈E+, f〉 =
∫ ∞
0
1
4pit
∫
‖x‖=t
f(t, x) dS(x) dt.
The formula just given can be rewritten (we write E3,+ to emphasise that
n = 3) in various ways such as
〈E3,+, φ〉 =
∫ ∞
0
t
4pi
∫
‖x‖=1
f(xt, t) dS(x) dt = 14pi
∫
R3
f(y, ‖y‖)
‖y‖ dy.
We now turn our attention to the case n = 2. Observe that if a, b : R2 →
R are reasonable functions and we define a˜, b˜ : R3 → R by a˜(t, x1, x2, x3) =
a(t, x1, x2) and b˜(t, x1, x2, x3) = b(t, x1, x2) then 2a = b if and only if 2a˜ =
2b˜. This strongly suggests the guess that
〈²2,+, f〉 = 〈E3,+, f˜〉.
The main point at issue is whether this actually defines a distribution since
even when f ∈ D we will not (unless f = 0) have f˜ ∈ D. However the
argument of Lemma 4.3 applies since suppE3,+ ⊆ {(t, x) : ‖x‖ ≤ t} so ²2,+
is well defined. Now
〈2²2,+, f〉 = 〈²2,+,2f〉
= 〈E3,+, 2˜f〉
= 〈E3,+,2f˜〉
= 〈2E3,+, f˜〉
= 〈δ0, f˜〉
= 〈δ0, f〉
for all f ∈ D(R2) so 2²2,+ = δ0.
Thus ²2,+ is a fundamental solution. It is easily checked that supp ²2,+ ⊆
{(t, x) : ‖x‖ ≤ t}. We may thus apply a simple uniqueness result.
Lemma 7.4. If E and F are fundamental solutions with suppE, suppF ⊆
{(t, x) : t ≥ 0} then E = F .
Lemma 7.5. If n = 2 the the forward fundamental solution E2,+ is given by
〈E2,+, f〉 =
1
2pi
∫ ∞
0
∫
‖x‖≤t
f(t, x)
(t2 − ‖x‖2)1/2 dx dt.
41
Lemma 7.3. If n = 3 the the forward fundamental solution E+ is given by
〈E+, f〉 =
∫ ∞
0
1
4pit
∫
‖x‖=t
f(t, x) dS(x) dt.
The formula just given can be rewritten (we write E3,+ to emphasise that
n = 3) in various ways such as
〈E3,+, φ〉 =
∫ ∞
0
t
4pi
∫
‖x‖=1
f(xt, t) dS(x) dt = 14pi
∫
R3
f(y, ‖y‖)
‖y‖ dy.
We now turn our attention to the case n = 2. Observe that if a, b : R2 →
R are reasonable functions and we define a˜, b˜ : R3 → R by a˜(t, x1, x2, x3) =
a(t, x1, x2) and b˜(t, x1, x2, x3) = b(t, x1, x2) then 2a = b if and only if 2a˜ =
2b˜. This strongly suggests the guess that
〈²2,+, f〉 = 〈E3,+, f˜〉.
The main point at issue is whether this actually defines a distribution since
even when f ∈ D we will not (unless f = 0) have f˜ ∈ D. However the
argument of Lemma 4.3 applies since suppE3,+ ⊆ {(t, x) : ‖x‖ ≤ t} so ²2,+
is well defined. Now
〈2²2,+, f〉 = 〈²2,+,2f〉
= 〈E3,+, 2˜f〉
= 〈E3,+,2f˜〉
= 〈2E3,+, f˜〉
= 〈δ0, f˜〉
= 〈δ0, f〉
for all f ∈ D(R2) so 2²2,+ = δ0.
Thus ²2,+ is a fundamental solution. It is easily checked that supp ²2,+ ⊆
{(t, x) : ‖x‖ ≤ t}. We may thus apply a simple uniqueness result.
Lemma 7.4. If E and F are fundamental solutions with suppE, suppF ⊆
{(t, x) : t ≥ 0} then E = F .
Lemma 7.5. If n = 2 the the forward fundamental solution E2,+ is given by
〈E2,+, f〉 =
1
2pi
∫ ∞
0
∫
‖x‖≤t
f(t, x)
(t2 − ‖x‖2)1/2 dx dt.
41
Page 42
When we throw a pebble into a pond we do not get a single ripple but a
train of ripples!
It may be shown (see [1]) that in all dimensions the forward fundamental
solution has support in the ‘light cone’ {(t, x) : ‖x‖ ≤ t}, and that in odd
dimensions n > 1 the support lies on the surface {(t, x) : ‖x‖ ≤ t} (but
if n ≥ 5 the form of En,+ is more complicated than that of E3,+, involving
analogues of δ′0, δ′′0 and so on). Notice the contrast with ‘elliptic regularity’
as described in Theorem Weyl. If n is even the support of En,+ is spread out
over the whole light cone.
If n = 1 the matter follows an idea of D’Alembert familiar from 1B
mathematical methods.
Exercise 7.6. Let n = 1.
(i) Show that if we make the change of coordinates w = t+ x, y = t− x
the wave operator becomes 2 ∂
2
∂w∂y .
(ii) Show that the fundamental solution with support in {(w, y) : w ≥
0, y ≥ 0} is 12H(w)H(y).
A little thought shows that we have in fact a very general solution to our
forcing problem.
Lemma 7.7. Let A be a real number. If T is a distribution with suppT ⊆
{(t, x) ∈ R × Rn : t ≥ A} then there exists a unique distribution S with
suppS ⊆ {(t, x) ∈ R× Rn : t ≥ A} and
2S = T.
We have S = E+ ∗ T . If T is a k times differentiable function then so is S.
Reversing the sign of t we obtain E− the backwards fundamental solution.
We have the analogue of Lemma 7.7.
Lemma 7.8. Let A be a real number. If T is a distribution with suppT ⊆
{(t, x) ∈ R × Rn : t ≤ A} then there exists a unique distribution S with
suppS ⊆ {(t, x) ∈ R× Rn : t ≤ A} and
2S = T.
We have S = E− ∗ T . If T is a k times differentiable function then so is S.
If T ∈ D′ we can write T = T1 + T2 with
suppT1 ⊆ {(t, x) ∈ R× Rn : t ≥ A} and suppT2 ⊆ {(t, x) ∈ R× Rn : t ≤ B}
for some A and B.
42
train of ripples!
It may be shown (see [1]) that in all dimensions the forward fundamental
solution has support in the ‘light cone’ {(t, x) : ‖x‖ ≤ t}, and that in odd
dimensions n > 1 the support lies on the surface {(t, x) : ‖x‖ ≤ t} (but
if n ≥ 5 the form of En,+ is more complicated than that of E3,+, involving
analogues of δ′0, δ′′0 and so on). Notice the contrast with ‘elliptic regularity’
as described in Theorem Weyl. If n is even the support of En,+ is spread out
over the whole light cone.
If n = 1 the matter follows an idea of D’Alembert familiar from 1B
mathematical methods.
Exercise 7.6. Let n = 1.
(i) Show that if we make the change of coordinates w = t+ x, y = t− x
the wave operator becomes 2 ∂
2
∂w∂y .
(ii) Show that the fundamental solution with support in {(w, y) : w ≥
0, y ≥ 0} is 12H(w)H(y).
A little thought shows that we have in fact a very general solution to our
forcing problem.
Lemma 7.7. Let A be a real number. If T is a distribution with suppT ⊆
{(t, x) ∈ R × Rn : t ≥ A} then there exists a unique distribution S with
suppS ⊆ {(t, x) ∈ R× Rn : t ≥ A} and
2S = T.
We have S = E+ ∗ T . If T is a k times differentiable function then so is S.
Reversing the sign of t we obtain E− the backwards fundamental solution.
We have the analogue of Lemma 7.7.
Lemma 7.8. Let A be a real number. If T is a distribution with suppT ⊆
{(t, x) ∈ R × Rn : t ≤ A} then there exists a unique distribution S with
suppS ⊆ {(t, x) ∈ R× Rn : t ≤ A} and
2S = T.
We have S = E− ∗ T . If T is a k times differentiable function then so is S.
If T ∈ D′ we can write T = T1 + T2 with
suppT1 ⊆ {(t, x) ∈ R× Rn : t ≥ A} and suppT2 ⊆ {(t, x) ∈ R× Rn : t ≤ B}
for some A and B.
42
Page 43
Lemma 7.9. If T is a distribution then we can find a distribution S with
2S = T.
If T is a k times differentiable (respectively smooth) function then we can
choose S to be k times differentiable (respectively smooth).
It is pretty obvious that the decomposition T = T1+T2 and the resulting
solution S in Lemma 7.9 are not unique. A more direct way of seeing this is
to observe that if u is a twice differentiable function on R and ω ∈ Rn is a
vector of unit length then 2u(t− x.ω) = 0.
Exercise 7.10. (i) Verify this.
(ii) Why should you have guessed this from physical or other considera-
tions.
(iii) Why does the result not contradict Lemma 7.7.
The nature of the lack of uniqueness can be resolved if we can prove the
following theorem.
Theorem 7.11. Suppose that u0, u1 : Rn → R are smooth. Then if f is a
continuous function there is a unique solution of
2u = f
with
u(0, x) = u0(x),
∂u
∂t (0, x) = u1(x).
Clearly it is enough to prove a simpler version.
Theorem 7.12. Suppose that u0, u1 : Rn → R are smooth. Then there is a
unique solution of
2u = 0
with
u(0, x) = u0(x),
∂u
∂t (0, x) = u1(x).
It should be noted that if
2u = 0
43
2S = T.
If T is a k times differentiable (respectively smooth) function then we can
choose S to be k times differentiable (respectively smooth).
It is pretty obvious that the decomposition T = T1+T2 and the resulting
solution S in Lemma 7.9 are not unique. A more direct way of seeing this is
to observe that if u is a twice differentiable function on R and ω ∈ Rn is a
vector of unit length then 2u(t− x.ω) = 0.
Exercise 7.10. (i) Verify this.
(ii) Why should you have guessed this from physical or other considera-
tions.
(iii) Why does the result not contradict Lemma 7.7.
The nature of the lack of uniqueness can be resolved if we can prove the
following theorem.
Theorem 7.11. Suppose that u0, u1 : Rn → R are smooth. Then if f is a
continuous function there is a unique solution of
2u = f
with
u(0, x) = u0(x),
∂u
∂t (0, x) = u1(x).
Clearly it is enough to prove a simpler version.
Theorem 7.12. Suppose that u0, u1 : Rn → R are smooth. Then there is a
unique solution of
2u = 0
with
u(0, x) = u0(x),
∂u
∂t (0, x) = u1(x).
It should be noted that if
2u = 0
43
Page 44
with
u(0, x) = u0(x),
∂u
∂t (0, x) = u1(x),
then by the definition of 2
∂2u
∂t2 = ∆u
so in particular
∂2u
∂2t (0, x) = ∆u0(x).
Again
∂3u
∂t3 =
∂u
∂t∆u
so
∂3u
∂3t (0, x) = ∆u1(x)
and so on. Thus ∂
ju
∂jt (0, x) is specified to all orders. We shall need a converse
observation.
Lemma 7.13. If u0, u1 : Rn → R are smooth we can find smooth uj : Rn →
R [j ≥ 2] such that if u : R× Rn → R is smooth and
∂ju
∂jt (0, x) = uj(x)
for all j ≥ 0 then
(∂ku
∂kt2u
)
(0, x) = 0
for all x.
We also need a result of Borel.
Theorem 7.14 (Theorem 6.1, Borel’s lemma). Given smooth uj : Rn →
R we can find a smooth u : R× Rn → R such that
∂ju
∂jt (0, x) = uj(x)
for all x and all j ≥ 0.
44
u(0, x) = u0(x),
∂u
∂t (0, x) = u1(x),
then by the definition of 2
∂2u
∂t2 = ∆u
so in particular
∂2u
∂2t (0, x) = ∆u0(x).
Again
∂3u
∂t3 =
∂u
∂t∆u
so
∂3u
∂3t (0, x) = ∆u1(x)
and so on. Thus ∂
ju
∂jt (0, x) is specified to all orders. We shall need a converse
observation.
Lemma 7.13. If u0, u1 : Rn → R are smooth we can find smooth uj : Rn →
R [j ≥ 2] such that if u : R× Rn → R is smooth and
∂ju
∂jt (0, x) = uj(x)
for all j ≥ 0 then
(∂ku
∂kt2u
)
(0, x) = 0
for all x.
We also need a result of Borel.
Theorem 7.14 (Theorem 6.1, Borel’s lemma). Given smooth uj : Rn →
R we can find a smooth u : R× Rn → R such that
∂ju
∂jt (0, x) = uj(x)
for all x and all j ≥ 0.
44
Page 45
Borel’s lemma is plausible but not trivial. (The obvious Taylor series
argument fails because Taylor series need not converge except at the origin.)
The proof of Theorem 7.12 is now easy. By Theorem 7.14 and Lemma 7.13
we can find a smooth function v : R× Rn → R such that
(∂kv
∂kt2u
)
(0, x) = 0
and
v(0, x) = u0(x),
∂V
∂t (0, x) = u1(x).
If we set f = 2u then by our choice of u all the partial derivatives of f with
respect to t vanish when t = 0 and the functions f1, f2 : R×Rn → R defined
by
f1(t, x) = H(t)f(t, x), f2(t, x) = (1−H(t))F (t, x)
are smooth. Setting
u = v − E+ ∗ f1 − E− ∗ f2
we have a solution of the required form. (Since E+ ∗ f1 is infinitely differen-
tiable with support in t ≥ 0 all its derivatives must vanish at t = 0.)
To prove uniqueness, observe that by linearity we need only prove that if
2u = 0 and u and its first partial derivative in t vanish at t = 0 then u = 0.
But if 2u = 0 and u and its first partial derivative in t vanish at t = 0 then
all the partial derivatives of u with respect to t vanish when t = 0. We can
thus write
u = u+ + u0
with u± smooth, supported in ±t ≥ 0 and satisfying 2u± = 0. By Lemma 7.7
u+ = 0. Similarly u− = 0 and we are done.
8 The heat equation
8.1 Existence
Recall that the heat operator J is given by
Ju(t, x) =
(∂
∂t −∆
)
u(t, x)
where u : R× Rn → R.
45
argument fails because Taylor series need not converge except at the origin.)
The proof of Theorem 7.12 is now easy. By Theorem 7.14 and Lemma 7.13
we can find a smooth function v : R× Rn → R such that
(∂kv
∂kt2u
)
(0, x) = 0
and
v(0, x) = u0(x),
∂V
∂t (0, x) = u1(x).
If we set f = 2u then by our choice of u all the partial derivatives of f with
respect to t vanish when t = 0 and the functions f1, f2 : R×Rn → R defined
by
f1(t, x) = H(t)f(t, x), f2(t, x) = (1−H(t))F (t, x)
are smooth. Setting
u = v − E+ ∗ f1 − E− ∗ f2
we have a solution of the required form. (Since E+ ∗ f1 is infinitely differen-
tiable with support in t ≥ 0 all its derivatives must vanish at t = 0.)
To prove uniqueness, observe that by linearity we need only prove that if
2u = 0 and u and its first partial derivative in t vanish at t = 0 then u = 0.
But if 2u = 0 and u and its first partial derivative in t vanish at t = 0 then
all the partial derivatives of u with respect to t vanish when t = 0. We can
thus write
u = u+ + u0
with u± smooth, supported in ±t ≥ 0 and satisfying 2u± = 0. By Lemma 7.7
u+ = 0. Similarly u− = 0 and we are done.
8 The heat equation
8.1 Existence
Recall that the heat operator J is given by
Ju(t, x) =
(∂
∂t −∆
)
u(t, x)
where u : R× Rn → R.
45
Page 46
Lemma 8.1. A fundamental solution of the heat equation is given by
K(t, x) =
( 1
4pit
)n/2
exp(−‖x‖
2
4t ) if t > 0,
K(t, x) = 0 if t > 0.
Note that we do not find a ‘backward fundamental solution’ – the heat
equation has an arrow of time built in.
Since the only singularity of our fundamental solution is at (0, 0) we can
build a parametrix and apply the argument of Theorem 4.18 to obtain a
corresponding result.
Lemma 8.2 (Proposition 7.1). If u ∈ D′ and J is the heat operator then
u and Ju have the same singular support. In particular if Ju is smooth so is
u.
This property is sometimes called hypo-ellipticity — saying that although
the operator is not elliptic it behaves like an elliptic operator it behaves
similarly in this respect.
We prove the next result by direct verification.
Theorem 8.3. If u0 : Rn → R and g : R×Rn → R are continuous functions
then the system
(∂
∂t −∆
)
u(t, x) = g(t, x) for t > 0
subject to
lim
t→0+
u(t, x) = u0(x)
has a solution
u(t, x) = Kt ∗ u0 +
∫ t
0
Kt−s ∗ gs ds
where Kt(x) = K(t, x) and gs(x) = g(s, x).
If g = 0 we observe that u(t, x) is a smooth function of x for fixed t > 0
whatever the choice of u0 this is another indication of the arrow of time.
46
K(t, x) =
( 1
4pit
)n/2
exp(−‖x‖
2
4t ) if t > 0,
K(t, x) = 0 if t > 0.
Note that we do not find a ‘backward fundamental solution’ – the heat
equation has an arrow of time built in.
Since the only singularity of our fundamental solution is at (0, 0) we can
build a parametrix and apply the argument of Theorem 4.18 to obtain a
corresponding result.
Lemma 8.2 (Proposition 7.1). If u ∈ D′ and J is the heat operator then
u and Ju have the same singular support. In particular if Ju is smooth so is
u.
This property is sometimes called hypo-ellipticity — saying that although
the operator is not elliptic it behaves like an elliptic operator it behaves
similarly in this respect.
We prove the next result by direct verification.
Theorem 8.3. If u0 : Rn → R and g : R×Rn → R are continuous functions
then the system
(∂
∂t −∆
)
u(t, x) = g(t, x) for t > 0
subject to
lim
t→0+
u(t, x) = u0(x)
has a solution
u(t, x) = Kt ∗ u0 +
∫ t
0
Kt−s ∗ gs ds
where Kt(x) = K(t, x) and gs(x) = g(s, x).
If g = 0 we observe that u(t, x) is a smooth function of x for fixed t > 0
whatever the choice of u0 this is another indication of the arrow of time.
46
Page 47
8.2 Uniqueness
We did not claim uniqueness in Theorem 8.3 because the infinite propagation
speed of heat exhibited by our solution allows other (non-physical) solutions.
Exercise 8.4. (This is included for interest, it is non-examinable.) We work
with n = 1 for simplicity.
(i) Show formally (without worrying about rigour) that if g is smooth
u(t, x) =
∞∑
k=0
gk(t)x2k
(2k)!
satisfies Ju = 0.
(ii) If gk(0) = 0 for all k but g is not identically zero then u is a solution
of Ju = 0 with u(t, x)→ 0 as t→ 0+ but u is not the zero function.
(iii) We now need to choose g and make the argument rigorous. It is
up to the reader to choose whether to try this. My own preference would be
to manufacture g by hand but g(t) = exp(−1/t2) [t 6= 0], g(0) = 0 will do
(see [3], Section 7.1).
Because parabolic equations lie ‘on the boundary of elliptic equations’ we
try and use ideas about maxima of solutions similar to those of Theorems 5.11
and 5.12.
Lemma 8.5. Consider the bounded open set
Ω = {(t, x) : ‖x− y‖ < r, 0 < t < T}.
If u : Ω → R is continuous and u is twice differentiable with Ju ≤ 0 on Ω
then the maximum value of u occurs on
{(t, x) : ‖x− y‖ ≤ r, t = 0} ∪ {(t, x) : ‖x− y‖ = r, 0 ≤ t ≤ T}.
Theorem 8.6. Consider the open set
Ω = {(t, x) : 0 < t < T} = (0, T )× Rn.
If u : Ω → R is continuous and u is twice differentiable with Ju ≤ 0 on Ω
and further there exists an M such that
u(t, x) ≤M for all (t, x) ∈ Ω
then
u(t, x) ≤ sup
y∈Rn
u(0, y)
for all (t, x) ∈ Ω.
47
We did not claim uniqueness in Theorem 8.3 because the infinite propagation
speed of heat exhibited by our solution allows other (non-physical) solutions.
Exercise 8.4. (This is included for interest, it is non-examinable.) We work
with n = 1 for simplicity.
(i) Show formally (without worrying about rigour) that if g is smooth
u(t, x) =
∞∑
k=0
gk(t)x2k
(2k)!
satisfies Ju = 0.
(ii) If gk(0) = 0 for all k but g is not identically zero then u is a solution
of Ju = 0 with u(t, x)→ 0 as t→ 0+ but u is not the zero function.
(iii) We now need to choose g and make the argument rigorous. It is
up to the reader to choose whether to try this. My own preference would be
to manufacture g by hand but g(t) = exp(−1/t2) [t 6= 0], g(0) = 0 will do
(see [3], Section 7.1).
Because parabolic equations lie ‘on the boundary of elliptic equations’ we
try and use ideas about maxima of solutions similar to those of Theorems 5.11
and 5.12.
Lemma 8.5. Consider the bounded open set
Ω = {(t, x) : ‖x− y‖ < r, 0 < t < T}.
If u : Ω → R is continuous and u is twice differentiable with Ju ≤ 0 on Ω
then the maximum value of u occurs on
{(t, x) : ‖x− y‖ ≤ r, t = 0} ∪ {(t, x) : ‖x− y‖ = r, 0 ≤ t ≤ T}.
Theorem 8.6. Consider the open set
Ω = {(t, x) : 0 < t < T} = (0, T )× Rn.
If u : Ω → R is continuous and u is twice differentiable with Ju ≤ 0 on Ω
and further there exists an M such that
u(t, x) ≤M for all (t, x) ∈ Ω
then
u(t, x) ≤ sup
y∈Rn
u(0, y)
for all (t, x) ∈ Ω.
47
Page 48
We can now provide a uniqueness result to go with the existence result
of Theorem 8.3.
Theorem 8.7. If u0 : Rn → R and g : R×Rn → R are bounded continuous
functions then the system
(∂
∂t −∆
)
u(t, x) = g(t, x) for t > 0
subject to
lim
t→0+
u(t, x) = u0(x)
has exactly one solution.
We also have a result on continuous dependence on initial data.
Exercise 8.8. Suppose u1, u2 : Rn → R and g : R × Rn → R are bounded
continuous functions. If u˜j is the bounded solution of the system
(∂
∂t −∆
)
u˜j(t, x) = g(t, x) for t > 0
subject to
lim
t→0+
u˜j(t, x) = uj(x)
for j = 1, 2. Then
sup
t≥0, x∈Rn
‖u˜2(t, x)− u˜1(t, x)‖ = sup
x∈Rn
‖u2(x)− u1(t, x)‖.
9 References
The reader will be in doubt of my admiration for Friedlander’s little book [2].
The additions to the second edition do not concern this course but some nig-
gling misprints have been removed and you should tell your college library
to buy the second edition even it already has the first. Friedlander’s book
mainly concerns the ‘distributional’ side of the course. Specific partial differ-
ential equations are dealt with along with much else in Folland’s Introduction
to Partial Differential Equations [1]. The two books [2] and [1] are also ex-
cellent further reading. The book [3] is a classic by a major worker in the
field.
48
of Theorem 8.3.
Theorem 8.7. If u0 : Rn → R and g : R×Rn → R are bounded continuous
functions then the system
(∂
∂t −∆
)
u(t, x) = g(t, x) for t > 0
subject to
lim
t→0+
u(t, x) = u0(x)
has exactly one solution.
We also have a result on continuous dependence on initial data.
Exercise 8.8. Suppose u1, u2 : Rn → R and g : R × Rn → R are bounded
continuous functions. If u˜j is the bounded solution of the system
(∂
∂t −∆
)
u˜j(t, x) = g(t, x) for t > 0
subject to
lim
t→0+
u˜j(t, x) = uj(x)
for j = 1, 2. Then
sup
t≥0, x∈Rn
‖u˜2(t, x)− u˜1(t, x)‖ = sup
x∈Rn
‖u2(x)− u1(t, x)‖.
9 References
The reader will be in doubt of my admiration for Friedlander’s little book [2].
The additions to the second edition do not concern this course but some nig-
gling misprints have been removed and you should tell your college library
to buy the second edition even it already has the first. Friedlander’s book
mainly concerns the ‘distributional’ side of the course. Specific partial differ-
ential equations are dealt with along with much else in Folland’s Introduction
to Partial Differential Equations [1]. The two books [2] and [1] are also ex-
cellent further reading. The book [3] is a classic by a major worker in the
field.
48
Page 49
References
[1] G. B. Folland Introduction to Partial Differential Equations 2nd Edition,
Princeton University Press, 1995. QA324
[2] F. G. Friedlander and M. Joshi Introduction to the Theory of Distributions
2nd Edition, CUP, 1998. QA324
[3] F. John Partial Differential Equations 4th Edition, Springer-Verlag, 1982
QA1.A647
49
[1] G. B. Folland Introduction to Partial Differential Equations 2nd Edition,
Princeton University Press, 1995. QA324
[2] F. G. Friedlander and M. Joshi Introduction to the Theory of Distributions
2nd Edition, CUP, 1998. QA324
[3] F. John Partial Differential Equations 4th Edition, Springer-Verlag, 1982
QA1.A647
49
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