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MA3G1: Theory of PDEs Dr. Manh Hong Duong Mathematics Institute, University of Warwick Email: [email protected] January 4, 2017
MA3G1: Theory of PDEs
Dr. Manh Hong Duong
Mathematics Institute, University of Warwick
Email: [email protected]
January 4, 2017
Abstract
The important and pervasive role played by pdes in both pure and applied mathematics
is described in MA250 Introduction to Partial Differential Equations. In this module I
will introduce methods for solving (or at least establishing the existence of a solution!)
various types of pdes. Unlike odes, the domain on which a pde is to be solved plays an
important role. In the second year course MA250, most pdes were solved on domains
with symmetry (eg round disk or square) by using special methods (like separation of
variables) which are not applicable on general domains. You will see in this module the
essential role that much of the analysis you have been taught in the first two years plays
in the general theory of pdes. You will also see how advanced topics in analysis, such
as MA3G7 Functional Analysis I, grew out of an abstract formulation of pdes. Topics in
this module include:
• Method of characteristics for first order PDEs.
• Fundamental solution of Laplace equation, Green’s function.
• Harmonic functions and their properties, including compactness and regularity.
• Comparison and maximum principles.
• The Gaussian heat kernel, diffusion equations.
• Basics of wave equation (time permitting).
Aims: The aim of this course is to introduce students to general questions of exis-
tence, uniqueness and properties of solutions to partial differential equations.
Objectives: Students who have successfully taken this module should be aware of
several different types of pdes, have a knowledge of some of the methods that are used for
discussing existence and uniqueness of solutions to the Dirichlet problem for the Laplacian,
have a knowledge of properties of harmonic functions, have a rudimentary knowledge of
solutions of parabolic and wave equations.
Further reading. The contents of this course are basic and can be found in many
textbooks. The bibliography lists some books for further reading.
Contents
1 Introduction 3
1.1 What are partial differential equations . . . . . . . . . . . . . . . . . . . . 3
1.2 Examples of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Well-posedness of a PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4 Classification of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2 Quasilinear first order PDEs 10
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 The method of characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.1 The characteristic equations . . . . . . . . . . . . . . . . . . . . . . 13
2.2.2 Geometrical interpretation . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Compatibility of initial conditions . . . . . . . . . . . . . . . . . . . . . . . 18
2.4 Local existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . 19
2 Laplace’s equation and harmonic functions 23
2.1 Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.3 Mean value property for harmonic functions . . . . . . . . . . . . . . . . . 26
2.4 Smoothness and estimates on derivatives of harmonic functions . . . . . . . 28
2.5 The maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.5.1 Subharmonic and superharmonic functions . . . . . . . . . . . . . . 32
2.5.2 Some properties of subharmonic and superharmonic functions . . . 33
1
2
2.5.3 The maximum principle . . . . . . . . . . . . . . . . . . . . . . . . 33
2.6 Harnack’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3 The Dirichlet problem for harmonic functions 42
3.0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.0.2 Representation formula based on the Green’s function . . . . . . . . 44
3.1 The Green function for a ball . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.2 C0-subharmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.3 Perron’s method for the Dirichlet problem on a general domain . . . . . . . 57
4 The theory of distributions and Poisson’s equation 63
4.1 The theory of distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.2 Convolutions and the fundamental solution . . . . . . . . . . . . . . . . . . 68
4.3 Poisson’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.3.1 The fundamental solution . . . . . . . . . . . . . . . . . . . . . . . 74
4.3.2 Poisson’s equation in a bounded domain . . . . . . . . . . . . . . . 76
5 The heat equation 80
5.1 The fundamental solution of the heat equation . . . . . . . . . . . . . . . . 80
5.2 The maximum principle for the heat equation on a bounded domain . . . . 86
5.3 The maximum principle for the heat operator on the whole space . . . . . 88
Acknowledgements 94
Appendix 95
Chapter 1
Introduction
1.1 What are partial differential equations
A partial differential equation is an equation which involves partial derivatives of some
unknown function. PDEs are often used to describe a wide variety of phenomena such as
sound, heat, electrostatics, electrodynamics, fluid flow, elasticity, or quantum mechanics.
Definition 1 (PDE and classical solutions). Suppose that Ω ⊂ Rd is some open set. A
partial differential equation (PDE) of order k is an equation of the form
F[x, u(x), Du(x), D2u(x), · · · , Dku(x)
]= 0. (1.1)
In the expression above, F : Ω × R × Rd × Rd2 × Rdk is a given function; u : Ω → R is
the unknown function and Dku contains all the partial derivatives of u of order k, see
Appendix 5.3. A function u ∈ Ck(Ω) is said to be a classical solution to the PDE on
the domain Ω if on substitution of u and its partial derivatives into (1.1) the relation is
identically satisfied on Ω.
Traditionally, PDEs are often derived by using conservation laws and constitutive laws.
Another method to derive PDEs, which have been developed recently, is to use operators
and functionals. The derivations of basic equations (transport equation, Laplace equation,
heat equation and wave equations) can be found in any textbook on the introduction of
PDEs (e.g., the the lecture notes of the course MA250 Introduction to Partial Differential
Equations). Therefore, we will not go into details on those in this module.
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1.2 Examples of PDEs
PDEs are ubiquitous and appear in almost all fields in applied sciences. We list here
just some popular examples. Some of these will be treated in the subsequent chapters.
Example 1 (The continuity/transport equation). In general, a continuity equation de-
scribes the transport of some moveable quantity. For example in fluid dynamics, the
continuity equation reads
∂u
∂t(x, t) + div(u(t, x)v(t, x)) = 0, (1.2)
where u(x, t) is the fluid density at a point x ∈ R3 and at a time t, v(x, t) is the flow velocity
vector field and div denotes the divergence operator. Note that the above equation can
be written as∂u
∂t(x, t) +Du(t, x) · v(t, x) + u(t, x) div(v(t, x)) = 0,
Example 2 (The Laplace and Possion equations). The Laplace equation is a second-order
partial differential equation given by
∆u =d∑i=1
∂2u
∂x2i
= 0. (1.3)
We often want to solve this equation in some bounded domain Ω ⊂ Rd and therefore, some
boundary conditions need to be provided. Now let us give an example of application (or
actually a derivation of this equation). We consider a solid body at thermal equilibrium
and denotes by u(x) its temperature. Let Q denotes the heat flux. Since the body is at
equilibrium, there is no flux through the boundary ∂Ω for any smooth domain Ω
0 =
ˆ∂Ω
Q · dσ =
ˆΩ
div(Q) dx,
where we have used the divergence theorem to obtain the second equality. This implies
that div Q must vanish identically
div Q = 0.
By Fourier’s law of heat conduction, the heat flux is proportional to the temperature
gradient, so that Q = k(x)Du(x), where k is the conductivity. If the body is homogeneous,
then k is constant and we deduce that
0 = div Q = div(kDu(x)) = k∆u(x).
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Thus the temperature satisfies the Laplace equation.
In conclusion: the Laplace equation can be used to describe the temperature distri-
bution of a solid body at thermal equilibrium.
The Poisson equation is the Laplace equation with an inhomogeneous right-hand side,
∆u(x) = f(x), (1.4)
for some given function f . For instance, the Poisson equation appears in electrostatics.
For a static electric field E, Maxwell’s equations reduce to
curl E = 0, div E = ρ,
where ρ is the charge density (in appropriate units). By Helmholtz decomposition, the first
equation implies that E = −Du, for some scalar electric potential field u. By substituting
this into the second equation, we obtain the Possion equation (1.4) with f = −ρ.
Example 3 (The heat/diffusion equation). In example 2, if the (homogeneous) body is
not in equilibrium, the the heat flux through the boundary of a region will equal the rate
of change of internal energy of the material inside that region. If no work is done by the
heat flow, then the rate of change of the internal energy is
dE
dt=
ˆΩ
∂u
∂t(t, x) dx =
ˆ∂Ω
Q · dσ =
ˆΩ
div(Q) dx = k
ˆΩ
∆u(t, x) dx.
This implies that the temperature u satisfies the heat equation
∂u
∂t(t, x) = k∆u(t, x). (1.5)
This equation also can be used to describe other diffusive processes such as a diffusion
process (and called diffusion equation).
Example 4 (The wave equation). The wave equation is a second-order linear partial
differential equation∂2u
∂t2(t, x) = c2∆u(t, x). (1.6)
Here u is used to model, for example, the mechanical displacement of a wave. The wave
equation arises in many fields like acoustics, electromagnetics, and fluid dynamics.
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Example 5 (The Kramers equation). The Kramer equation (or kinetic Fokker-Planck
equation) describes the time evolution of the probability density of a particle moving
under the influence of an external potential, a friction and a stochastic noise given by
∂tρ+p
m· ∇qρ−∇V · ∇qρ = γ divp
( pmρ+ β−1∇pρ
),
where m is the mass of the particle, γ is the friction coefficient, β−1 is the temperature.
Another important equation of the same kind is the Boltzmann equation
∂tρ+p
m· ∇qρ−∇V · ∇qρ = Q(ρ, ρ)
where the right hand side describes the effect of collisions between particles, which is a
complicated formula.
Example 6 (A convection and nonlinear diffusion equation). The following equation
describes the time evolution of the probability density function, i.e., for each t ∈ [0, T ],
ρ(t) is a probability measure on Rd, of some quantity,
∂tρ = div(ρ∇[U ′(ρ) + V ]
),
where U , V are known as the internal and external energy functionals. In particular,
when U(ρ) = ρ log ρ, we obtain the Fokker-Planck equation, which is linear,
∂tρ = ∆ρ+ div(ρ∇V ).
When V = 0, U(ρ) = 1m−1
ρm, we obtain the porous medium equation
∂tρ = ∆ρm.
Example 7 (The Allen-Cahn equation). The Allen-Cahn euqation is a reaction-diffusion
equation and describes the process of phase separation
∂tρ = D∆ρ− f ′(ρ),
where f is some free energy functional. A typical example of f is f(ρ) = 14ρ4 − 1
2ρ2.
Example 8 (The Cahn-Hilliard equation). A very closely related equation to the Allen-
Cahn equation is the Cahn-Hilliard equation which is obtained by taking minus the Lapla-
cian of the right-hand side of the former.
∂tρ = −D∆2u+ ∆f ′(ρ).
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1.3 Well-posedness of a PDE
In all of the examples in the previous section, we need further information in order to
solve the problem. We broadly refer to this information as the data for the problem.
Ex. 1 We can specify the density distribution at some initial time t0. This is an example
of Cauchy data (or an initial value problem).
Ex. 2 For the Laplace equation on a bounded domain, we need to prescribe the value of
u on the boundary of the domain. This is an example of Dirichlet data (a Dirichlet
problem). For the Poisson equation, we specify the function f and also the value
of u on the boundary.
Ex. 3 For the heat equation, we specify the initial temperature of the body, as well as
some boundary condition. For example, a Dirichlet boundary condition when the
temperature at the boundary is prescribed or a Neumann boundary condition when
the flux through the boundary is imposed.
An important aspect of the study of a PDE is about its wellposedness. A PDE
problem, consisting of a PDE together with some data is well-posed if
a) There exists a solution to the problem (existence).
b) For given data, the solution is unique (uniqueness).
c) The solution depends continuously on the data.
These are called Hadamard’s conditions. A problem is said to be ill-posed if any of the
condition above fails to hold.
Notice that existence and uniqueness involves boundary conditions. While continuous
dependence depends on considered metric/norm.
Wellposedness is important because for the vast majority of PDE problems that we
encounter, it is not possible to write down a solution explicitly. However, if well-posedness
is satisfied, we can often deduce properties of that solution directly from the PDE it
satisfies without ever needing an explicit solution.
While the issue of existence and uniqueness of solutions of ordinary differential equa-
tions has a very satisfactory answer with the Picard-Lindelof theorem, that is far from
8
the case for partial differential equations. In general, the answers to the above ques-
tions depend heavily on the PDE itself as well as the domain Ω. Definition 1.1 is often
too general to provide any useful information. In this notes, we will be working with
much simpler examples of PDEs having explicit forms, and nevertheless exhibit lots of
interesting behaviour.
Example 9. Consider the following heat equation on (0, 1).ut = uxx
u(0, t) = u(1, t) = 0,
u(x, 0) = u0(x).
This PDE is well-posed.
Example 10. Consider the following backwards heat equationut = −uxx
u(0, t) = u(1, t),
u(x, 0) = u0(x).
This PDE is ill-posed.
In general, it is not straightforward to see whether a PDE is well-posed or not. More
knowledge from subsequent chapters is needed.
1.4 Classification of PDEs
There are several ways to categorise PDEs. Following is a possibility.
Definition 2. i) We say that (1.1) is linear if F is a linear function of u and its deriva-
tives, so that we can re-write (1.1) as (see Appendix 5.3 for the multi-index notation)∑|α|≤k
aα(x)∂|α|u
∂xα= f(x).
If f = 0, we say the equation is homogeneous.
ii) We say that (1.1) is semilinear if it is of the form∑|α|=k
aα(x)∂|α|u
∂xα+ a0
[x, u(x), Du(x), D2u(x), · · · , Dk−1u(x)
]= 0,
9
so that the highest order derivatives of u appear linearly, with coefficients depending
only on x but not on u and its derivatives.
iii) We say that (1.1) is quasilinear if it is of the form∑|α|=k
aα
[x, u(x), Du(x), D2u(x), · · · , Dk−1u(x)
]∂|α|u∂xα
+ a0
[x, u(x), Du(x), D2u(x), · · · , Dk−1u(x)
]= 0,
so that the highest order derivatives of u appear linearly, with coefficients possibly
depending only on the lower order derivatives of u.
iv) We say that (1.1) is fully nonlinear if is not linear, semilinear or quasilinear.
Exercise 1. Classify equations in the examples in Section 1.2.
Chapter 2
Quasilinear first order PDEs
In this chapter, we will study quasilinear first order PDEs. We will be able to apply
the method of characteristic to solve these equations.
2.1 Introduction
For simplicity we will consider functions defined on a subset of R2. However, the
methods in this chapter are applicable for functions defined on higher dimensional sets.
In R2, a quasilinear first order PDE has the form
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u), (2.1)
where a(x, y, u), b(x, y, u) and c(x, y, u) are coefficients; ux and uy denote the partial
derivatives of u with respect to x and y respectively. We often want to solve the above
equation in some bounded domain Ω ⊂ R2.
Example 11. Let us start with the most simplest quasilinear first order PDE. Let Ω ⊂ R2,
we consider the following equation
∂u
∂x(x, y) = 0 in Ω.
Solutions to this equation depend on the domain Ω. For instance, if Ω = [0, 1]× [0, 1]
then the value of u on 0 × [0, 1] will determine u everywhere in Ω. In fact, suppose
that u(0, y) = h(y) with h continuous on [0, 1], then using the fundamental theorem of
calculus we have,
u(x, y) = u(0, y) +
ˆ x
0
∂u
∂s(s, y) ds = h(y).
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11
x
y
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
1
1
Ω
Figure 2.1: The domain Ω = [0, 1]× [0, 1]
Figure 2.2: [L] A domain on which ux = 0 is solvable with data on 0 × [0, 1]. [R] A
domain on which ux = 0 is not solvable with data on 0 × [0, 1].
This means that u does not change if we move along a curve of constant y and we say
that the value of u propagates along lines of constant y (Figure 2.1).
How about other domains? When is prescribing the value of u on 0 × [0, 1] enough
to determine u everywhere? Obviously for this approach to work, we must be able to
connect every point in Ω to 0× [0, 1] by a horizontal line which lies in Ω. See Figure 2.1
for different situations. Note that while for a classical solution of a first order equation
we would require u ∈ C1(Ω), the procedure above in fact makes sense even if h(y) is only
continuous, so that uy need not to be C0. This suggests that we should consider a notion
of solutions for some PDEs which is weaker than classical solutions.
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2.2 The method of characteristic
Let us take another example.
Example 12 (Transport equation with constant speed). Let c ∈ R and h : R → R be a
given function. Consider the following PDE
cux + uy = 0, u(x, 0) = h(x). (2.2)
Solution. The solution consists of three main steps
Step 1) Change the co-ordinate system (x, y) (s, t) by writingx = x(s, t)
y = y(s, t)
and define z(s, t) := u(x(s, t), y(s, t)).
Step 2) Solve the equation for z in the new co-ordinate system (s, t), which by construc-
tion will be easier to solve.
Step 3) Transform back to the original co-ordinate systems = s(x, y)
t = t(x, y)
and u(x, y) = z(s(x, y), t(x, y)).
By the chain rule, we havedz
dt=dx
dtux +
dy
dtuy.
We want that the right-hand side of the expression above is equal to cux+uy that appears
in the PDE. Therefore, we obtain the following system of ODEs
dx
dt= c
dy
dt= 1.
Solving this system, we obtain x(s, t) = ct+A(s), y(s, t) = t+B(s). To find A(s), B(s)
we impose that
(t = 0) ⇒ (y = 0), (t = 0) ⇒ (x = s),
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from which we deduce that A(s) = s, B(s) = 0, hence, x(s, t) = ct+ s and y(s, t) = t. In
the new co-ordinate system, we now have
dz
dt= 0, z(s, 0) = h(s).
Solving this equation gives
z(s, t) = h(s).
We now transform back to (x, y) and u(x, y). It is straightforward to find s, t in terms of
x, y
s = x− cy, t = y.
Therefore, u(x, y) = z(s(x, y), t(x, y)) = h(x − cy). This is the solution of the transport
equation with constant speed.
Some observations:
1) To establish the new co-ordinate system, we need to solve a system of ODEs
dx
dt= c, x(s, 0) = s,
dy
dt= 1, y(s, 0) = 0,
dz
dt= 0, z(s, 0) = h(s).
These are called the characteristic equations associated to the PDE, and x(s, t), y(s, t), z(s, t)
are called the characteristic curves.
2) To transform back to the original co-ordinate system, we need to have the correspon-
dence x = x(s, t)
y = y(s, t)
s = s(x, y)
t = t(x, y).
3) If x − cy = k, then u(x, y) = u(k). In other words, u is constant along each line
x− cy = k.
2.2.1 The characteristic equations
In the general case
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u), u(f(x), g(x)) = h(x), (2.3)
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where f, g, h are given functions. The method of characteristics to solve the above equation
above consists of three steps.
Step 1) Solve the following characteristic equations, which is a system of ODEs, for x, y, z
as functions of s, t
dx
dt= a(x, y, z), x(s, 0) = f(s),
dy
dt= b(x, y, z), y(s, 0) = g(s),
dz
dt= c(x, y, z), z(s, 0) = h(s).
Step 2) Find the transformation s = s(x, y)
t = t(x, y).
Step 3) Find the solution u(x, y) = z(s(x, y), t(x, y)).
The characteristic projection is defined by
σs(t) =x(s, t), y(s, t)
.
A shock is formed when two characteristic projections meet and assign different values of
u, i.e., there exist (s, t), (s′, t′) such that
x(s, t) = x(s′, t′), y(s, t) = y(s′, t′),
but z(s, t) 6= z(s′, t′).
2.2.2 Geometrical interpretation
We now give a geometrical interpretation (derivation) for the method of characteris-
tics. Suppose that we are given a curve γ(s) = (f(s), g(s)), s ∈ (α, β). We will find a
function u satisfying
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u),
such that for s ∈ (a, b), we have [u γ](s) = u(f(s), g(s)) = h(s) for some given function
h : (α, β)→ R. In other words, we specify the value of u along the curve γ.
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The idea we shall pursue is to find the graph of u over some domain Ω by showing
that we can foliate it by curves along which the value of u is determined by ODEs. Recall
that the graph of u over Ω is the surface in R3 given by
Graph(u) = x, y, u(x, y) : (x, y) ∈ Ω.
We know from the initial conditions that the curve Γ, given by (f(s), g(s), h(s)), s ∈ (α, β)
must belong to Graph(u). We want to write⋃s∈(α,β)
Cs ⊂ Graph(u),
where
Cs = (x(s, t), y(s, t), z(s, t)) : t ∈ (−εs, εs), εs > 0,
with x(s, 0) = f(s), y(s, 0) = g(s) and z(s, 0) = h(s). In other words, through each line
of the curve (f(s), g(s), h(s)) we would like to find another curve which lies in the graph
of u.
Since the tangent vector to Cs, given by t :=(dxdt
(s, t), dydt
(s, t), dzdt
(s, t))
lies in the
surface Graph(u), it must be orthogonal to any vector which is normal to Graph(u).
Since the map
(x, y)→ u(x, y) := (x, y, u(x, y))
is a parameterisation of the surface Graph(u), a vector normal to Graph(u) can be found
by
N := ux ∧ uy
=
1
0
ux
∧
0
1
uy
=
−ux−uy
1
.
Therefore, Cs lies in the surface Graph(u) if and only if t · N = 0,i.e.,
dx
dtux +
dy
dtuy −
dz
dt= 0,
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should hold at each point on Cs. Recall that u satisfies the PDE
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u).
By comparing the two relationships above, we see that if x(t; s), y(t; s), z(t; s) is the unique
solution of the systems of ODEs
dx
dt= a(x, y, z), x(s, 0) = f(s)
dy
dt= b(x, y, z), y(s, 0) = g(s) (2.4)
dz
dt= c(x, y, z), z(s, 0) = h(s)
then the curve Cs will lie in Graph(u). The Picard-Lindelof theorem states that if a, b, c
are sufficiently well behaved, then there exists a unique solution to (2.4) on some interval
t ∈ (−εs, εs) and that moreover the solution depends continuously on s. Later we will
state more precisely these conditions on a, b, c such that we can solve (2.4).
The equation (2.4) are known as the characteristic equations and the corresponding
solution curves (x(t; s), y(t; s), z(t; s)) are the characteristic curves or simply characteris-
tics. The union of the characteristics forms a surface⋃s∈(α,β)
Cs = (x(s, t), y(s, t), z(s, t)) : t ∈ (−εs, εs), s ∈ (α, β)
which is a portion of the graph of u that includes the curve Γ. We want to write this
as a graph of the form (x, y, u(x, y)) : (x, y) ∈ Ω. We can do this, provided we can
invert the map (s, t) 7→ (x(s, t), y(s, t)) to give s(x, y), t(x, y). Then we have u(x, y) =
z(s(x, y); t(x, y)).
Example 13. Solve the PDE
ux + xuy = u, u(2, x) = h(x).
Solution. The characteristic equations are
dx
dt= 1, x(s, 0) = 2,
dy
dt= x, y(s, 0) = s,
dz
dt= z, z(s, 0) = h(s).
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The first equation gives x(s, t) = t+ 2. Substituting this into the second equation we get
y(s, t) = t2
2+ 2t+ s. Finally, from the last equation, we have z(s, t) = eth(s). By writing
s, t in terms of x, y, we find
s = y − (x− 2)2
2− 2(x− 2), t = x− 2.
Therefore, u(x, y) = z(s, t) = ex−2h(y − (x−2)2
2− 2(x− 2)
).
Note that the method of characteristic does not require that the PDEs are linear. We
consider the following quasilinear equation.
Example 14 (Burger’s equation). Solve the following Burger’s equation
uux + uy = 0, u(x, 0) = h(x), x, y ∈ R, y ≥ 0.
Solution. The characteristic equations reduce to
dx
dt= z, x(s, 0) = s,
dy
dt= 1, y(s, 0) = 0,
dz
dt= 0, z(s, 0) = h(s).
The last two equations are easily solved to get
y(s, t) = t, z(s, t) = h(s).
Substituting these back to the first equation we get
x(s, t) = s+ h(s)t.
In this case, we can not explicitly invert the relationship (t, s) 7→ (x, y). However, we
have
s(x, y) = x− h(s(x, y))y, t(x, y) = y.
Therefore,
u(x, y) = z(s(x, y), t(x, y)) = h(s(x, y)) = h(x− h(s(x, y))y) = h(x− u(x, y)y),
i.e., u satisfies an implicit equation
u(x, y) = h(x− u(x, y)y).
18
In a particular case when h(x) =√x2 + 1, we have
u =√
(x− uy)2 + 1,
which implies that
u(x, y) =xy −
√x2 − y2 + 1
y2 − 1.
Note that this solution is only valid for |y| < 1. At this point, u blows up. To explore
the blow-up phenomena in more generality, let us return to the characteristic equations
for the Burger’s equation
x(s, t) = h(s)t+ s, y(s, t) = t, z(s, t) = h(s).
At fixed s the characteristic is simply a straight line parallel to the x− y plane
y =x− sh(s)
, z = h(s).
This means that u(x, y) = z remains constant on each of the line y = 1h(s)
(x − s). The
meeting point of two characteristic projections is determined by
x− s1
h(s1)=x− s2
h(s2)=
s2 − s1
h(s1)− h(s2).
If h is increasing, then s2−s1h(s1)−h(s2)
≤ 0. In this case, the lines of constant u are “fanning
out” and do not meet on y ≥ 0.
If h is decreasing, then s2−s1h(s1)−h(s2)
≥ 0. The lines of constant u must meet on y ≥ 0. At
such a meeting point, the solution obtained by the method of characteristics is no longer
admissible since the two characteristic projections are trying to assign two different values
of u. This continuity is called a shock.
2.3 Compatibility of initial conditions
We recall that the initial condition is given by
u(f(x), g(x)) = h(x).
Differentiating this equation with respect to x, we find
f ′(x)ux(f(x), g(x)) + g′(x)uy(f(x), g(x)) = h′(x). (2.5)
19
On the other hand, from the PDE we have
a(f(x), g(x), h(x))ux(f(x), g(x)) + b(f(x), g(x), h(x))uy(f(x), g(x)) = c(f(x), g(x), h(x)).
(2.6)
Equations (3.9) and (2.6) forms a linear system to determine ux(f(x), g(x)) and uy(f(x), g(x)).
• If a(f(x), g(x), h(x))g′(x) − b(f(x), g(x), h(x))f ′(x) 6= 0, then this system has a
unique solution. In this case, we expect that the method of characteristics will
work. We say that the initial conditions are compatible with the PDE.
• If a(f(x), g(x), h(x))g′(x) − b(f(x), g(x), h(x))f ′(x) = 0, the initial conditions are
incompatible with the PDE or they are redundant.
2.4 Local existence and uniqueness
Recall that we want to solve the PDE
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u) on Ω,
with initial data given along a curve γ : x 7→ (f(x), g(x)) ∈ Ω such that u(f(x), g(x)) =
h(x). We will show that the compatibility of the initial conditions is the only obstruction
(besides standard assumptions) to finding a solution in a neighbourhood of γ. Suppose
that:
• a, b, c ∈ C1(Ω× R),
• f, g, h ∈ C1((α, β)) for some interval (α, β).
• The initial conditions are compatible with the PDE, i.e., a(f(x), g(x), h(x))g′(x)−
b(f(x), g(x), h(x))f ′(x) 6= 0 for all x ∈ (α, β).
The the characteristic equations
dx
dt= a(x, y, z), x(s, 0) = f(s),
dy
dt= b(x, y, z), y(s, 0) = g(s),
dz
dt= c(x, y, z), z(s, 0) = h(s).
20
can be solved uniquely for each s ∈ (α, β) and for t ∈ (−εs, εs), and the solution will be of
class C1. To find the solution in terms of (x, y) one now to invert the map (s, t) 7→ (x, y).
Let’s define Ψ(s, t) = (x(s, t), y(s, t)). Then Ψ is of class C1 in a neighbourhood of γ and
DΨ(s, 0) =
∂x∂s
∂x∂t
∂y∂s
∂y∂t
(s, 0) =
f ′(s) a(f(s), g(s), h(s))
g′(s) b(f(s), g(s), h(s))
.
According to the compatibility conditions, we have
detDΨ(s, 0) = a(f(s), g(s), h(s))g′(s)− b(f(s), g(s), h(s))f ′(s) 6= 0, for all s ∈ (α, β),
so that DΨ(s, 0) is invertible. By the inverse function theorem, Ψ−1 exists in a neigh-
bourhood of (f(s), g(s)). Furthermore, we have ∂s∂x
∂s∂y
∂t∂x
∂t∂y
(Ψ(s, t)) = (DΨ−1)(Ψ(s, t)) = (DΨ(s, t))−1 =1
b∂x∂s− a∂y
∂s
b −a
−∂y∂s
∂x∂s
By the chain rule (remembering that u(x, y) = z(s(x, y), t(x, y)) )
ux =∂z
∂s
∂s
∂x+∂z
∂t
∂t
∂x
uy =∂z
∂s
∂s
∂y+∂z
∂t
∂t
∂y
or equivalently, in a matrix form
(ux uy
)=(∂z∂s
∂z∂t
) ∂s∂x
∂s∂y
∂t∂x
∂t∂y
=1
b∂x∂s− a∂y
∂s
(∂z∂s
∂z∂t
) b −a
−∂y∂s
∂x∂s
Hence
aux + buy =(ux uy
)ab
=
1
b∂x∂s− a∂y
∂s
(∂z∂s
∂z∂t
) b −a
−∂y∂s
∂x∂s
ab
=
1
b∂x∂s− a∂y
∂s
(∂z∂s
c) 0
b∂x∂s− a∂y
∂s
= c,
so the equation is indeed satisfied. Uniqueness follows from the fact that the character-
istic through a point (x0, y0, z0) is uniquely determined and moreover lies completely in
21
Graph(u) if (x0, y0, z0) lies in Graph(u). Thus any solution would have to include all of
the characteristics through (f(s), g(s), h(s)) and hence would have to locally agree with
the solution we construct above.
We summarise this section by the following theorem
Theorem 1. Suppose that a, b, c ∈ C1(Ω× R); f, g, h ∈ C1((α, β)); and that s 7→ γ(s) =
(f(s), g(s)) is injective. Suppose further that a(f(x), g(x), h(x))g′(x)−b(f(x), g(x), h(x))f ′(x) 6=
0 for all x ∈ (α, β). Then, given K ⊂ (α, β) a closed (and hence compact) interval, there
exists a neighbourhood U of γ(K) such that the equation
aux + buy = c
has a unique solution u of class C1 such that u(γ(s)) = h(s).
Problem sheet 1
Exercise 2. Solve the problem
ux + uy = u, u(x, 0) = cos x
Exercise 3. Solve the problem
ux + xuy = y, u(0, y) = cos y
Exercise 4. Consider the Burger’s equation
uux + uy = 0, u(x, 0) = h(x), x, y ∈ R, y ≥ 0
Suppose that h ∈ C1(R), h bounded and decreasing, h′ < 0, and assume that h′ is
bounded. Show that no shock forms in the region
y <1
supR|h′|
,
and that for any y > 1sup
R|h′| a shock has formed for some x.
Exercise 5. a) Solve the problem
ux + 3x2uy = 1, u(x, 0) = h(x), (x, y) ∈ R2,
for h ∈ C1(R).
b) Show and explain why the problem
ux + 3x2uy = 1, u(s, s3) = 1, (x, y) ∈ R2
has no solution u ∈ C1(R2).
c) Show and explain why the problem
ux + 3x2uy = 1, u(s, s3) = s− 1, (x, y) ∈ R2
has infinitely many solutions u ∈ C1(R2).
22
Chapter 2
Laplace’s equation and harmonic
functions
In this chapter, we will study the Laplace’s equation, which is an important equation
both in physics and mathematics. Solutions of the Laplace’s equation are called harmonic
functions. We will study some important, both qualitative and quantitative, properties
of harmonic functions such as the mean value property, smoothness and estimates of the
derivatives, the maximum principle and Harnack’s inequality.
2.1 Laplace’s Equation
Let Ω ⊂ Rn, u : Ω→ R be a C2 function. The Laplacian operator is defined by
∆u(x) :=n∑i=1
∂2u
∂x2i
(x).
The Laplace’s equation is
∆u(x) = 0, for all x ∈ Ω. (2.1)
Remark 1 (Relations between the Laplacian, the gradient and the divergence operators).
Recall that the gradient of u, ∇u (or Du) is a vector field given by
∇u(x) = (∂x1u, . . . , ∂xnu)T ,
and that if X = (X1, . . . , Xn)T is a vector field then the divergence of X is
div(X) =n∑i=1
∂Xi
∂xi.
23
24
We have the following relation between the Laplacian, the gradient and the divergence
operators
∆u = div(∇u).
This relation will be used throughout for example when applying the divergence theorem.
Example 15. 1) If a ∈ Rn, b ∈ R and u(x) = a · x + b (i.e., u is affine) then ∆u(x) = 0.
2) If u(x) =∑n
i,j=1 aijxixj, with aij = aji (i.e., the matrix a = (aij)i,j=1,...,n is symmetric),
then
(∇u)i = 2n∑i=1
aijxj, and ∆u(x) = 2n∑i=1
aii = 2 Tr(a).
So if a is trace-free (i.e., Tr(a) = 0) then ∆u = 0.
Example 16 (Laplacian operator of a radial function). Suppose u(x) = u(|x|) = u(r),
where r = |x|. Then we have
∆u(|x|) = ∆u(r) = u′′(r) +n− 1
ru′(r).
We will provide two proofs for this assertion. The first one is based on direct computations,
the second one is hinged on a property that the Laplacian operator is rotation invariant.
The first proof (direct computations). Since r = |x| =√∑n
i=1 x2i , we have ∂r
∂xi= xi
r. By
the chain rule we have
uxi =du
dr
∂r
∂xi= u′(r)
xir.
uxixi = u′′(r)xir
xir
+ u′(r)∂
∂xi
(xir
).
= u′′(r)x2i
r2+ u′(r)
(1
r− x2
i
r3
).
Therefore, recalling that r2 =∑n
i=1 x2i ,
∆u(x) =n∑i=1
uxixi =n∑i=1
[u′′(r)
x2i
r2+ u′(r)
(1
r− x2
i
r3
)]= u′′(r) +
n− 1
ru′(r).
25
The second proof.
First observation: the Laplacian is rotation invariant. Let Q be a constant n × n
matrix satisfying QQT = QTQ = I. Let v(x) = u(Qx). By the chain rule, we have
vxi(x) =n∑j=1
uxj(Qx)Qji,
and similarly
vxixi =n∑
j,k=1
uxjxk(Qx)QjiQki = (QTHQ)ii,
where H = (Hessu)(Qx). Therefore
∆v(x) =n∑i=1
vxixi =n∑i=1
(QTHQ)ii = Tr(QTHQ) = Tr(QTQH) = Tr(H) = ∆u(Qx).
Here we use the fact that Q is orthogonal and the community of the trace Tr(AB) =
Tr(BA).
Note further that since Q is orthogonal, |x|2 = 〈x, x〉 = 〈Qx,Qx〉 = |Qx|2, it follows
that |x| = |Qx|.
Now if u is radial, i.e., u(x) = u(|x|) = u(|Qx|) = u(Qx), then v(x) = u(Qx) = u(x),
so that ∆v(x) = ∆u(Qx) = ∆u(x).
Second observation: By the divergence theorem, we haveˆ∂Br
∇u · n dσ =
ˆBr
div(∇u) dx =
ˆBr
∆u(x) dx. (2.2)
Since u is radial, ∇u(x)·n = u′(r) x|x| ·n = u′(r). The first integral in (2.2) can be computed
as ˆ∂Br
∇u · n dσ = u′(r)
ˆ∂Br
dσ = u′(r)σn−1rn−1, (2.3)
where σn−1 is the area of the unit (n− 1)−sphere.
Now using the fact that ∆u(x) is radial, we can write the last integral in (2.2) as a
radial integral ˆBr
∆u(x) dx =
ˆ r
0
σn−1sn−1∆u(s) ds. (2.4)
From (2.3) and (2.4) we obtain that for any r1 < r2
u′(r2)rn−12 − u′(r1)rn−1
1 =
ˆ r2
r1
sn−1∆u(s) ds.
This equality implies that rn−1∆u(r) is simply the derivative of rn−1u′(r), thus
∆u(r) =1
rn−1
d
dr
(rn−1u′(r)
)= u′′(r) +
n− 1
ru′(r).
26
2.2 Harmonic functions
Definition 3. Let Ω ⊂ Rn be open. A function u ∈ C2(Ω) is said to be harmonic in Ω if
∆u(x) = 0
for all x ∈ Ω. In other words, harmonic functions solve Laplace’s equation.
Example 17. 1) We saw from Example 15 that affine functions u(x) = a · x + b, for
a ∈ Rn, b ∈ R, and u(x) = x21 + . . .+ x2
n−1 − (n− 1)x2n are harmonic in Rn.
2) In polar coordinates, the function rk sin(kθ) is harmonic in R2 and the function r−k sin(kθ)
is harmonic in R2 \ 0 for k ∈ N.
3) The function ex sin y is harmonic in R2.
When Ω is a domain in R2, an important class of harmonic functions come from
holomorphic functions.
Example 18. Suppose Ω ⊂ Rn ' C and that f(z) is holomorphic in Ω, i.e., the limit
f ′(z0) := limz→z0
f(z)− f(z0)
z − z0
exists for all z0 ∈ Ω. Suppose that f(x+ iy) = u(x, y) + iv(x, y). Then u, v are harmonic
function in Ω. In fact, by the Cauchy-Riemann equations, we have∂u∂x
= ∂v∂y,
∂u∂y
= − ∂v∂x.
Therefore, ∆u(x, y) = ∂2u∂x2
+ ∂2u∂y2
= ∂2v∂x∂y− ∂2v
∂y∂x= 0. Similarly, ∆v(x, y) = 0.
2.3 Mean value property for harmonic functions
Let us motivate this section by looking at a one-dimensional example. Suppose that
u : (a, b) → R is harmonic, i.e., u′′(x) = 0 in (a, b). It follows that u(x) = cx + d
for some constants c, d ∈ R. Let x0 ∈ (a, b) and r > 0 is sufficiently small such that
[x0 − r, x0 + r] ⊂ (a, b). Let us compute
1
2[u(x0 − r) + u(x0 + r)] =
1
2[c(x0 − r) + d+ c(x0 + r) + d]
= cx0 + d,
27
and
1
2r
ˆ x0+r
x0−ru(y) dy =
1
2r
ˆ x0+r
x0−r(cy + d) dy
=1
2r
(cy2
2+ dy
) ∣∣∣x0+r
x0−r
= cx0 + d.
Therefore,
u(x0) =1
2[u(x0 − r) + u(x0 + r)] =
1
2r
ˆ x0+r
x0−ru(y) dy.
This property holds true for harmonic function in any dimensional space.
Theorem 2 (Mean value property). Suppose Ω ⊂ Rn is open and that u ∈ C2(Ω) is
harmonic. Then if Br(x0) ⊂ Ω then
u(x0) =1
|∂Br(x0)|
ˆ∂Br(x0)
u(y) dσ(y) (2.5)
=1
|(Br(x0))|
ˆBr(x0)
u(y) dy. (2.6)
Proof. We first prove (2.5). We perform two changes of variables, successively setting
y = x0 + z and z = rζ, which shifts Br(x0) to be centred at the origin and then scales it
to have unit radius. We obtain
ˆ∂Br(x0)
u(y) dσ(y) =
ˆ∂Br(0)
u(x0 + z) dσ(z) = rn−1
ˆ∂B1(0)
u(x0 + rζ) dσ(ζ)
Therefore, recalling that |∂Br(x0)| = rn−1σn−1, where σn−1 is the area of the (n− 1)-unit
sphere,
1
|∂Br(x0)|
ˆ∂Br(x0)
u(y) dσ(y) =1
σn−1
ˆ∂B1(0)
u(x0 + rζ) dσ(ζ). (2.7)
Let F (r) be defined as the right-hand side of the above equality. Let v(ζ) := u(x0 + rζ),
then ∇v(ζ) = r∇u(x0 +rζ) and ∆v(z) = r2∆u(x0 +rζ) = 0. We will show, by computing
28
the derivative of F (r), that F (r) is indeed a constant function.
dF (r)
dr=
1
σn−1
d
dr
ˆ∂B1(0)
u(x0 + rζ) dσ(ζ)
=1
σn−1
ˆ∂B1(0)
d
dru(x0 + rζ) dσ(ζ)
=1
σn−1
ˆ∂B1(0)
ζ · ∇u(x0 + rζ) dσ(ζ)
=1
σn−1
1
r
ˆ∂B1(0)
ζ · ∇v(ζ) dσ(ζ)
(∗)=
1
σn−1
1
r
ˆB1(0)
∆v(ζ) dζ
= 0,
where we have used the divergence theorem to obtain the equality (∗) above. Hence by
the continuity of u, we have
F (r) = limr→0
F (r) =1
σn−1
ˆ∂B1(0)
u(x0) dσ(ζ) = u(x0). (2.8)
From (2.7) and (2.8), we obtain (2.5).
The equality (2.6) follows from (2.5). Indeed, we haveˆBr(x0)
u(y) dy =
ˆ r
0
ds
ˆ∂Bs(x0)
u(y) dσ(y) =
ˆ r
0
|∂Bs(x0)|u(x0) ds
= u(x0)
ˆ r
0
|∂Bs(x0)| ds = u(x0)|Br(x0)|
Remark 2. The converse results are also true, namely that if u ∈ C(Ω) satisfies the mean
value property (either (2.5) or (2.6)) then u ∈ C2(Ω) and u is harmonic in Ω. Note that
it requires only that u ∈ C(Ω). See Exercises.
2.4 Smoothness and estimates on derivatives of har-
monic functions
Proposition 1. Let Ω ⊂ Rn be open and let u be harmonic in Ω. Then u ∈ C∞(Ω) and
furthermore
|Dαu(x)| ≤(n|α|d(x)
)|α|supy∈Ω|u(y)|, (2.9)
where d(x) = dist(x, ∂Ω).
29
Proof. We first prove that u ∈ C∞(Ω). Let φ be satisfy the following property
(i) φ ∈ C∞(Rn),
(ii) φ is radial,
(iii) φ is supported in Br for some r > 0,
(iv)´Rnφ(x) dx = 1.
We will show that u(x) = (u ∗ φ)(x), where the convolution is defined by
(u ∗ φ)(x) =
ˆRn
u(y)φ(x− y) dy.
Without loss of generality, set x = 0, and we take r small enough so that Br ⊂ Ω. Now
we have
(u ∗ φ)(0) =
ˆRn
u(y)φ(−y) dy
=
ˆBr
u(y)φ(−y) dy
=
ˆBr
u(y)φ(y) dy
=
ˆ r
0
(ˆ∂Bs
u(y) dS(y)
)φ(s) ds
=
ˆ r
0
|∂Bs|u(0)φ(s) ds
= u(0)
ˆ r
0
|∂Bs|φ(s) ds
= u(0)
ˆ r
0
ˆ∂Bs
φ(s)dS ds
= u(0)
ˆBr
φ(y) dy
= u(0).
So u(x) = (u ∗ φ)(x). Since u ∗ φ ∈ C∞(Ω), it follows that u ∈ C∞(Ω) as claimed.
Step 1. Next we will prove the estimate (2.9) by induction on |α|. For |α| = 1, we
need to prove that
|Diu(x)| ≤ n
d(x)supy∈Ω|u(y)|. (2.10)
Take r < d(x) then Br(x) ⊂ Ω. According to the mean value property
Diu(x) =1
|Br(x)|
ˆBr(x)
Diu(y) dy =1
|Br(x)|
ˆ∂Br(x)
u(y)ni(y) dS(y),
30
Since |ni| = 1, we can estimate
|Diu(x)| ≤ |∂Br(x)||Br(x)|
supy∈Br(x)
|u(y)| = rn−1σn−1
rnωnsup
y∈Br(x)
|u(y)| = n
rsup
y∈Br(x)
|u(y)|.
Since this is true for any r < d(x) and Br(x) ⊂ Ω, it follows that
|Diu(x)| ≤ n
d(x)supy∈Ω|u(y)|.
Step 2. Suppose that the statement is true for |α| ≤ N .
Step 3. We need to prove that it is also true for |α| = N + 1. Write Dαu = DiDβu,
where β is a multi-index such that |β| = |α| − 1 = N . Now we fix some λ ∈ (0, 1). We
apply the technique in Step 1 to r = λd(x), to obtain
|Dαu(x)| = |DiDβu(x)| ≤ n
λd(x)sup
y∈Bλd(x)(x)
|Dβu(y)|.
By the triangle inequality if y ∈ Bλd(x)(x), then d(y) ≥ (1 − y)d(x), so applying the
induction assumption we have
|Dβ(y)| ≤(n|β|d(y)
)|β|supz∈Ω|u(z)| ≤
(n|β|
(1− λ)d(x)
)|β|supz∈Ω|u(z)|,
which implies that
supy∈Bλd(x)
|u(y)| ≤(
n|β|(1− λ)d(x)
)|β|supz∈Ω|u(z)|,
and that (recalling that |α| = |β|+ 1)
|Dαu(x)| ≤ n
λd(x)
(n|β|
(1− λ)d(x)
)|β|supz∈Ω|u(z)| =
[( n
d(x)
)|α||β||β| sup
z∈Ω|u(z)|
]1
λ(1− λ)|β|.
(2.11)
To obtain the sharpest upper bound, we now minimizes the function λ 7→ 1λ(1−λ)|β|
, which
is equivalent to maximises the function
λ 7→ g(λ) := λ(1− λ)|β|,
to obtain the optimal λ. The optimal λ satisfies the equation
0 =dg(λ)
dλ= (1− λ)|β| − λ|β|(1− λ)|β|−1
= (1− λ)|β|−1[1− λ− |β|λ] = (1− λ)|β|−1)(1− |α|λ),
31
which implies that λ = 1|α| , which is in the interval (0, 1) as required. Substituting this
value to g(λ), we get (recalling that |α| = |β|+ 1)
g(λ)∣∣∣λ= 1|α|
=1
|α|
(1− 1
|α|
)|β|=|β||β|
|α||α|
Putting this back into (2.11) we obtain
|Dαu(x)| ≤ n|α||α||α|
d(x)|α|sup
Ω|u|,
which is the reduction assumption for |α| = N + 1. Hence the result holds true for any
N .
Proposition 1 has two interesting consequences in the following theorems.
Theorem 3. Harmonic functions are analytic.
Proof. Let x0, x ∈ Ω and suppose that the segment [x0, x] = tx+ (1− t)x0, t ∈ [0, 1] is
contained in Ω. Without loss of generality we assume that x0 = 0. Let u be a harmonic
function in Ω. By Taylor’s theorem we have
u(x) = Pm−1(x) +Rm(x),
where Pm−1(x) is a polynomial of degree m− 1 and
Rm(x) =∑|α|=m
Dαu(ξ)
α!xα,
for some ξ ∈ [x0, x]. To show that u is analytic we need to show that Rm(x) → 0 as
m→∞ uniformly on some small disc around x0. By Proposition 1, we have
|Rm(x)| ≤(nm
d(ξ)
)msup
Ω|u|∑|α|=m
|xα|α!
=1
m!
(nm
d(ξ)
)msup
Ω|u|∑|α|=m
m!
α1! . . . αn!|x1|α1 . . . |xn|αn .
Set ||x||1 :=∑n
i=1 |xi|, then by the multinomial theorem we have
||x||m1 =∑|α|=m
m!
α1! . . . αn!|x1|α1 . . . |xn|αn .
Therefore, we obtain that
|Rm(x)| ≤ ||x||m1
m!
(nm
d(ξ)
)msup
Ω|u|. (2.12)
32
We now use a weak version of the Stirling’s approximation, which holds true for all positive
integer m,mm
m!≤ em.
We get
|Rm(x)| ≤(ne||x||1d(ξ)
)msup
Ω|u|.
Taking |x| = |x − x0| < 12d(x0), by the triangular inequality, we have d(ξ) ≤ 1
2d(x0), so
that
|Rm(x)| ≤(
2ne||x||1d(x0)
)msup
Ω|u|. (2.13)
Finally, if we take x such that ||x||1 < d(x0)2ne
, then 2ne||x||1d(x0)
< 1 and(
2ne||x||1d(x0)
)m→ 0 as
m→∞, that implies that |Rm(x)| → 0 as m→∞.
Theorem 4 (Liouville). Suppose u is a bounded harmonic function in Rn, then u is a
constant.
Proof. Suppose |u| ≤ C in Rn. Let x ∈ R
n and r > 0. Since u is harmonic in Br(x),
applying Proposition 1 in Br(x), we have
|Diu(x)| ≤ n
rsupBr(x)
|u| ≤ nC
r.
Since r is arbitrary and C is independent of r, the above estimate implies that |∇u(x)| = 0.
So u is constant.
2.5 The maximum principle
2.5.1 Subharmonic and superharmonic functions
Definition 4. Let Ω ⊂ Rn be an open set, and let u ∈ C2(Ω). We say u is subharmonic
(respectively superhamornic) in Ω iff ∆u ≥ 0 (respectively ∆u ≤ 0).
Note that a function u ∈ C2(Ω) is harmonic if and only if it is both subharmonic and
superharmonic.
Example 19. 1) In one dimension, u is subharmonic iff u′′ > 0, which means that u′ is
increasing and u is convex. Similarly u is superharmonic iff u is concave.
2) If u = xTAx for some constant, symmetric matrix A. Then ∆u = 2Tr(A), so u is
subharmonic (respectively, superharmonic) iff Tr(A) ≥ 0 (respectively, Tr(A) ≤ 0).
33
2.5.2 Some properties of subharmonic and superharmonic func-
tions
Recall that harmonic functions satisfy the mean value property, which means that for
any ball Br(x0) ⊂ Ω, we have
u(x0) =1
|∂Brx0|
ˆ∂Br(x0)
u(y)dσ(y) =1
|Br(x0)|
ˆBr(x0)
u(y) dy.
Recall that to prove this in Theorem 2, we considered the function F (r) = 1|∂Brx0|
´∂Br(x0)
u(y)dσ(y)
and showed
(i) F (r)→ u(x0) as r → 0. This relied on the continuity of u,
(ii) For all r > 0 we have
F ′(r) =1
rσn−1
ˆB1(0)
∆u(x0 + rζ) dζ. (2.14)
It follows from (2.14) that if u is subharmonic (respectively, superharmonic) then F ′(r) ≥
0 (respectively, F ′(r) ≤ 0).
Proposition 2. If u is subharmonic in Ω and Br(x0) ⊂ Ω, then
u(x0) ≤ 1
|∂Brx0|
ˆ∂Br(x0)
u(y)dσ(y) and u(x0) ≤ 1
|Br(x0)|
ˆBr(x0)
u(y) dy. (2.15)
If u is superharmonic then the reverse inequalities hold.
It follows from this Proposition and Theorem 2 that the value of a subharmonic
(respectively, superharmonic) function at the center of a ball is less (respectively, greater)
than or equal to the value of a harmonic function with the same values on the boundary.
Thus, the graphs of subharmonic functions lie below the graphs of harmonic functions
and the graphs of superharmonic functions lie above, which explains the terminology.
2.5.3 The maximum principle
An important feature of subharmonic functions is the following theorem.
Theorem 5 (Weak maximum principle). Let Ω be an open and bounded subset of Rn.
Suppose that u ∈ C2(Ω) ∩ C0(Ω) be subharmonic in Ω. Then
maxΩ
u = max∂Ω
u. (2.16)
34
If, rather, u is superharmonic in Ω then
minΩu = min
∂Ωu. (2.17)
Proof. Suppose u is subharmonic. We consider two cases.
Case 1: ∆u > 0 in Ω. Suppose that there exists x0 ∈ Ω such that u attains a local
maximum at x0, then we must have ∇u(x0) = 0 and Hessu(x0) is negative definite, i.e.,
ξTHessu(x0)ξ ≤ 0 for all ξ ∈ Rn, ξ 6= 0. It implies that
∆u(x0) = Tr(Hessu(x0)) =n∑i=1
eTi Hessu(x0)ei ≤ 0,
which is a contradiction. Therefore, u can not have maxima in the interior. Thus, the
maximum of u in Ω must be achieved on the boundary Ω, i.e., (2.16) holds.
Case 2: ∆u ≥ 0 in Ω. Define uε(x) := u(x)+ε|x|2. Then ∆uε(x) = ∆u(x)+2εn > 0,
so that we can apply Case 1 above for uε(x) and obtain
maxΩ
uε = max∂Ω
uε.
Since Ω is bounded, |x|2 is a bounded function in Ω, and uε → u uniformly on Ω as ε→ 0.
It follows that
maxΩ
u = limε→0
maxΩ
uε = limε→0
max∂Ω
uε = max∂Ω
u.
The statement (2.17) for superhamornic function is obtained by applying (2.16) for −u,
which is subharmonic, and using the fact that maxA
(−u) = −minAu.
Since a harmonic function is both subharmonic and superhamornic, we obtain the
following corollary.
Corollary 1. 1) Let Ω ⊂ Rn be open and bounded and u ∈ C2(Ω) ∩ C0(Ω) is harmonic.
Then
maxΩ
u = max∂Ω
u, and minΩu = min
∂Ωu.
This implies
maxΩ|u| = max
∂Ω|u|.
2) (The comparison principle) If Ω ⊂ Rn is open and bounded and if u, v ∈ C2(Ω)∩C0(Ω)
satisfying ∆u ≥ ∆v in Ω and u ≤ v on ∂Ω then u ≤ v in Ω.
35
The comparison follows by applying the weak maximum principle to the function
w := u− v, which is subharmonic and satisfies w ≤ 0 on ∂Ω, hence w ≤ 0 (i.e, u ≤ v) in
Ω.
Note that the weak maximum principle states that the that the maximum of a sub-
harmonic function is to be found on the boundary, but may re-occur in the interior as
well. The following theorem provides a stronger statement.
Theorem 6 (Strong maximum principle). Let Ω be open and connected (possibly un-
bounded), and let u be subharmonic in Ω. If u attains a global maximum value in Ω, then
u is constant in Ω.
Proof. Define M := maxΩ
u, and A := u−1M = x ∈ Ω : u(x) = M. By the assumption
of the theorem, A is non-empty. Since u is continuous and M is a closed set in R, it
follows that A is relatively closed in Ω (i.e., there exists a closed F ⊂ Rn such that
A = F ∩ Ω). We now show that A is also open. Indeed, let x ∈ Ω and let r be
sufficiently small that Br(x) ⊂ Ω. By the mean value property for subharmonic functions
in Proposition 2, we have
0 = u(x)−M ≤ 1
|Br(x)|
ˆBr(x)
(u(y)−M) dy.
Since M = maxΩ
u, we have u(y)−M ≤ 0 for all y ∈ Br(x). This implies that
0 = u(x)−M ≤ 1
|Br(x)|
ˆBr(x)
(u(y)−M) dy ≤ 0.
Since u is continuous, this happens only if u = M in Br(x). So that Br(x) ⊂ A, which
implies that A is open. Since Ω is connected and A 6= ∅ is both open and closed, it follows
that A = Ω, and hence u ≡M in Ω. This finishes the proof.
Again, since a harmonic function is both subharmonic and superharmonic, we get a
stronger result.
Corollary 2 (Strong maximum principle for harmonic functions). Let Ω be open and
connected and u : Ω→ R a harmonic function. Then if u attains either a global maximum
or global minimum in Ω then u is constant in Ω.
36
2.6 Harnack’s inequality
The next result we shall derive is Harnack’s inequality for harmonic functions. To
state this theorem, we need to define the notion of a compactly contained subset. Let
Ω,Ω′ be two open subsets of Rn, then we say that Ω′ is compactly contained in Ω and
write Ω′ b Ω if there exists a compact set K such that Ω′ ⊂ K ⊂ Ω.
Theorem 7 (Harnack’s inequality for harmonic functions). Let Ω ⊂ Rn be open and let
u : Ω → R be a non-negative harmonic function. Then for any connected Ω′ b Ω there
exists a constant C depending only on Ω,Ω′, n such that
supΩ′≤ C inf
Ω′u.
Proof. Let r > 0 satisfy that infΩ′ d(x) > 4r where d(x) = dist(x, ∂Ω). Let x, y ∈ Ω′ be
such that |x− y| ≤ r. Applying the mean value property for u we have
u(x) =1
|B2r(x)|
ˆB2r(x)
u(z) dz =1
2nrnωn
ˆB2r(x)
u(z) dz ≥ 1
2nrnωn
ˆBr(y)
u(z) dz,
where ωn = |B1(0)| and we have used that u ≥ 0 and Br(y) ⊂ B2r(x) to obtain the above
inequality. Now using the mean value property again, we have
1
2nrnωn
ˆBr(y)
u(z) dz =1
2n1
|Br(y)|
ˆBr(y)
u(z) dz =1
2nu(y).
It follows from these two inequalities that
1
2nu(y) ≤ u(x).
By interchanging x and y, we obtain u(x) ≤ 2nu(y), so that for all |x− y| ≤ r we have
1
2nu(y) ≤ u(x) ≤ 2nu(y). (2.18)
Now we will expand this estimate to any two points x0, x1 in Ω′. Since Ω′ is open and
connected, it is path connected. Hence there exists a curve γ : [0, 1] → Ω′ such that
γ(0) = x0 and γ(1) = x1. Since Ω′ ⊂ K for some compact subset K, we can cover Ω′ with
at most N open balls of radius r, where N depends only on Ω,Ω′. From these balls, we
select M ≤ N balls Bi, i = 1, . . . ,M such that
x0 ∈ B1, x1 ∈ BM , γ([0, 1]) ⊂M⋃i=1
Bi,
37
and Bi ∩Bi+1 6= ∅, i = 1, . . . ,M − 1.
Applying the estimate (2.18) successively on these balls, we get
u(x1) ≤ 2Mnu(x0) ≤ 2Nnu(x0).
Since this inequality holds for any two points x0, x1 ∈ Ω′, it follows that
supΩ′u ≤ 2nN inf
Ω′u,
which is the claim.
Harnack’s inequality is used to prove Harnack’s theorem about the convergence of
sequences of harmonic functions. Harnack’s inequality can also be used to show the
interior regularity of weak solutions of partial differential equations.
Corollary 3. Let Ω be connected and let uk be a sequence of harmonic functions in Ω
such that uk ≤ uk+1 for all k. If there exists x0 ∈ Ω such that uk(x0) converges (to a finite
value) then uk converges to some harmonic function u uniformly on the compact sets of
Ω.
Problem Sheet 2
Exercise 6 (Laplacian operator in the cylindrical and spherical coordinates).
1) Consider the cylindrical coordinates in R3 defined byx1(r, ϕ, z) = r cosϕ,
x2(r, ϕ, z) = r sinϕ, r ∈ [0,∞), ϕ ∈ [0, 2π), z ∈ R,
x3(r, ϕ, z) = z.
Prove that in these coordinates
∆u(r, ϕ, z) =1
r
∂
∂r
(r∂u
∂r
)+
1
r2
∂2u
∂ϕ2+∂2u
∂z2.
2) Consider the spherical coordinates in R3 defined byx1(r, θ, ϕ) = r sin θ cosϕ,
x2(r, θ, ϕ) = r sin θ sinϕ, r ∈ [0,∞), θ ∈ [0, π), ϕ ∈ [0, 2π),
x3(r, ϕ, z) = z.
Prove that in these coordinates
∆u(r, θ, ϕ) =1
r2
∂
∂r
(r2∂u
∂r
)+
1
r2 sin2 θ
∂2u
∂ϕ2+
1
r2 sin2 θ
∂
∂θ
(sin θ
∂u
∂θ
).
[Hint: write gradients and use integration by parts]
3) Show that if u, v are C2 functions then
∆(uv)(x) = v(x)∆u(x) + u(x)∆v(x) + 2∇u(x) · ∇v(x).
Exercise 7.
38
39
1) Find the general form of radial harmonic functions in Rn \ 0. Which of these extend
to Rn as smooth functions?
2) Find all radial solutions of
∆u(x) =1
(1 + |x|2)2
in R2\0. Which of these solutions extend smoothly to all of R2 (i.e., near the origin).
3) Let (z, ϕ) be polar coordinates in R2, and consider the set Ω = R
2 \ (x, 0) : x ≥ 0.
Show that the functions log r, ϕ are harmonic in Ω.
Exercise 8 (Converse of the mean value property of harmonic functions in 1D).
1) Suppose u : (a, b)→ R is continuous and satisfies
u(x0) =1
2[u(x0 − r) + u(x0 + r)], ∀x0, r such that [x0 − r, x0 + r] ⊂ (a, b).
Show that u is harmonic (i.e., affine).
2) Prove the same if u : (a, b)→ R is continuous and satisfies
u(x0) =1
2r
ˆ x0+r
x0−ru(x) dx, ∀x0, r such that [x0 − r, x0 + r] ⊂ (a, b).
Exercise 9. 1) Prove that if u ∈ C2(Ω) satisfies the mean value property (either in the
spherical or bulk form), then ∆u = 0 in Ω.
2) Suppose Ω ⊂ Rn is open and such that Ω0 = Ω ∩ xn = 0 6= ∅. Let Ω+ := Ω ∩ xn >
0, and define Ω = Ω+ ∪ Ω0 ∪ Ω−, where
Ω− = x = (x′, xn) ∈ Rn such that (x,−xn) ∈ Ω+.
Suppose that u ∈ C2(Ω+) is such that ∂u∂xn
= 0 on Ω0, and define u : Ω→ R by
u(x) =
u(x), x ∈ Ω+ ∪ Ω0,
u(x′,−xn), x = (x′, xn) ∈ Ω−.
Prove that u ∈ C2(Ω) and that u is harmonic in Ω. See the figure below for illustration
of the sets Ω,Ω0,Ω±.
40
Figure 2.1: Illustration of the sets Ω,Ω0,Ω±
3) Let Ω,Ω0,Ω±, Ω be as above (see also the figure), and suppose that v ∈ C2(Ω) satisfies
v = 0 on Ω0. Define v : Ω→ R by
v(x) =
v(x), x ∈ Ω+ ∪ Ω0,
−v(x′,−xn), x = (x′, xn) ∈ Ω−.
Prove that v ∈ C2(Ω) and that v is harmonic in Ω.
Exercise 10. Recall the Azela-Ascoli theorem: Let K ⊂ Rn be compact and let fi : K →
R be a sequence of functions that are uniformly bounded and equicontinuous. Then there
exists a sequence fik which converges uniformly. [Recall that fi are equicontinuous if for
any ε > 0 there exists δ > 0 such that |fi(x)− fi(y)| < ε whenever |x− y| < δ].
Let Ω ⊂ Rn be an open set and let ui : Ω → R be a sequence of harmonic functions
which are uniformly bounded. Prove that for any multi-index α and for any compact
K ⊂ Ω there exists a subsequence uik such that Dαuik converges uniformly in K.
Exercise 11. 1) Suppose u : Ω→ R is harmonic and that F : R→ R is a convex function
of class C2. Show that F (u(x)) is subharmonic.
41
2) Suppose u : Rn → R is harmonic and that
ˆRn
u2(x) dx = M <∞.
Prove that u ≡ 0 in Rn.
3) If u is harmonic in Ω ⊂ Rn and if Br(x0) ⊂ Ω, prove that
d
dr−
∂Br(x0)u(y)2dσ(y) =
2r
n−
Br(x0)|∇u|2(x) dx.
Chapter 3
The Dirichlet problem for harmonic
functions
In this chapter we will study the Direchlet problem for harmonic functions in a
bounded domain. We will establish a representation formula based on the Green’s func-
tion, find the Green’s function and solve the problem in a ball, and finally use the Perron’s
method to solve the problem in a general domain.
3.0.1 Introduction
In this chapter, we will study the following Dirichlet problem for harmonic functions.
The Dirichlet problem: Let Ω ⊂ Rn be open and bounded. Let g ∈ C(∂Ω) be a
given function. Find a function u ∈ C2(Ω) ∩ C0(∂Ω) such that∆u = 0, in Ω,
u = g, on ∂Ω.
(3.1)
The schedule to solve this problem will be the following.
Step 1 Establish a representation formula for a solution to (3.1) based on Green’s func-
tion.
Step 2 Find the Green’s function and solve the problem (3.1) in a ball,
Step 3 Solve the problem (3.1) for a general domain using the Perron’s method.
42
43
To start with, we have the following lemma
Lemma 1. Let Ω ⊂ Rn be open and bounded, let g1, g2 ∈ C(∂Ω) be given. Suppose that
u1, u2 solve the following problem for i = 1, 2 respectively∆ui = 0, in Ω,
ui = gi, on ∂Ω.
Then
supΩ
|u1 − u2| = supΩ
|g1 − g2|. (3.2)
As a consequence, a solution to (3.1), if it exists, is unique and depends continuously on
the data.
Proof. This is a direct consequence of the maximum principle for w := u1 − u2, which
satisfies that ∆w = 0, in Ω,
w = g1 − g2, on ∂Ω.
In view of this Lemma, the main task is now to find a solution to (3.1). We first recall
the Green’s identities that will be used throughout this chapter. The starting point for
these identities is the divergence theorem.
ˆΩ
div−→F dx =
ˆ∂Ω
−→F · −→n dS.
Given f, g ∈ C2(Ω). Applying the divergence theorem above for−→F = f∇g, we obtain
ˆΩ
div(f∇g) dx =
ˆ∂Ω
f∇g · −→n dS.
Note that
div(f∇g) = f∆g +∇g · ∇f, ∇g · −→n =∂g
∂−→n,
so that we can write the equality above as
ˆΩ
f∆g dx = −ˆ
Ω
∇f · ∇g dx+
ˆ∂Ω
f∂g
∂−→ndS. (3.3)
This relation is known as the first Green’s identity.
44
By interchanging f and g in the first Green’s identity, we get
ˆΩ
g∆f dx = −ˆ
Ω
∇g · ∇f dx+
ˆ∂Ω
g∂f
∂−→ndS.
This together with the first Green’s identity implies the following Green’s second identity
ˆΩ
(f∆g − g∆f) dx =
ˆ∂Ω
(f∂g
∂−→n− g ∂f
∂−→n
)dS. (3.4)
Define the function
Φ(x) =
1
2πlog |x|, if n = 2,
1(2−n)σn−1
|x|2−n, if n = 3,
(3.5)
where σn−1 = |∂B1(0)|. Then Φ ∈ C∞Rn \ 0. For any x 6= 0, we have
∂xiΦ(x) =1
σn−1
xi|x|−n,
∂2xixi
Φ(x) =1
σn−1
[−nx2
i
|x|n+2+ |x|−n
].
Therefore,
∆Φ(x) =n∑i=1
∂2xixi
Φ(x) =n∑i=1
1
σn−1
[−nx2
i
|x|n+2+ |x|−n
]= 0.
So
∆Φ(x) = 0, ∀x 6= 0, limx→0
Φ(x) = −∞. (3.6)
3.0.2 Representation formula based on the Green’s function
We first provide a motivation for the Green’s function. Recall that we want to solve∆u = 0, in Ω,
u = g, in ∂Ω.
Suppose that for each x ∈ Ω, we can find a function G(x, y) such that∆yG(x, y) = δx, in Ω,
G(x, y) = 0, in ∂Ω.
45
Then formally for a solution u to (3.1), using the second Green’s identity, we have
u(x) =
ˆΩ
u(y)δx dy
=
ˆΩ
u(y)∆yG(x, y) dy
=
ˆ∂Ω
(u(y)
∂G
∂−→n(x, y)−G(x, y)
∂u
∂−→n(y))dS(y) +
ˆΩ
G(x, y)∆u(y) dy
=
ˆΩ
∂G
∂−→n(x, y)g(y) dS(y).
Therefore, we can represent u in terms of G and g. This suggests that to solve the
problem (3.1), one should find the function G(x, y). We observe that Φ(x − y) satisfies
that ∆yΦ(y − x) = δx, but it does not satisfy the boundary condition. The idea that
we shall follow is that we will construct G(x, y) from Φ(y − x) taking into account the
boundary conditions.
To do so, let u ∈ C2(Ω) and consider the following integral
ˆΩ
Φ(y − x)∆u(y) dy.
We want to integrate by parts this integral. However, note that Φ(y−x) has a singularity
at x = y. To integrate this, therefore, we proceed as follows. Take x ∈ Ω and ε > 0
sufficiently small such that Bε(x) ⊂ Ω. Define Vε := Ω \ Bε(x). By the Green’s second
identity, we have
ˆVε
Φ(y − x)∆u(y) dy =
ˆVε
∆yΦ(y − x)u(y) dy +
ˆ∂Vε
Φ(y − x)∂u
∂−→n(y) dS(y)
−ˆ∂Vε
u(y)∂Φ
∂−→n(y − x) dS(y)
=
ˆ∂Vε
Φ(y − x)∂u
∂−→n(y) dS(y)−
ˆ∂Vε
u(y)∂Φ
∂−→n(y − x) dS(y),
where we have used that ∆yΦ(y − x) = 0. Next, we will pass to the limit ε → 0 this
relation. The LHS is straightforward by the dominated convergence theorem
limε→0
ˆVε
Φ(y − x)∆u(y) dy =
ˆΩ
Φ(y − x)∆u(y) dy.
For the boundary terms, we will show later that
Claim 1
limε→0
[−ˆ∂Vε
u(y)∂Φ
∂−→n(y − x) dS(y)
]= −
ˆ∂Ω
u(y)∂Φ
∂−→n(y − x) dS(y) + u(x). (3.7)
46
Claim 2
limε→0
ˆ∂Vε
Φ(y − x)∂u
∂−→n(y) dS(y) =
ˆ∂Ω
Φ(y − x)∂u
∂−→n(y) dS(y) (3.8)
Assuming these claims at the moment, then we find that for any u ∈ C2(Ω), we have
u(x) =
ˆΩ
Φ(y−x)∆u(y) dy+
ˆ∂Ω
u(y)∂Φ
∂−→n(y−x) dS(y)−
ˆ∂Ω
Φ(y−x)∂u
∂−→n(y) dS(y). (3.9)
Note that in the Direchlet problem, we know ∆u in Ω and u on ∂Ω, but we do not know
∂u∂−→n on ∂Ω.
To overcome this, for each x ∈ Ω, we introduce a corrector function hx(y) such that∆yhx(y) = 0, y ∈ Ω,
hx(y) = Φ(y − x), y ∈ ∂Ω.
Then using the Green’s second identity, we have
ˆΩ
hx(y)∆u(y) dy =
ˆΩ
∆yhx(y)u(y) dy +
ˆ∂Ω
[hx(y)
∂u
∂−→n(y)− u(y)
∂hx
∂−→n(y)
]dS(y).
=
ˆ∂Ω
Φ(y − x)∂u
∂−→n(y) dS(y)−
ˆ∂Ω
u(y)∂hx
∂−→n(y) dS(y).
Hence,
ˆ∂Ω
Φ(y − x)∂u
∂−→n(y) dS(y) =
ˆ∂Ω
u(y)∂hx
∂−→n(y) dS(y) +
ˆΩ
hx(y)∆u(y) dy.
Substituting this equality back to (3.9), we obtain that
u(x) =
ˆΩ
[Φ(y − x)− hx(y)]∆u(y) dy +
ˆ∂Ω
( ∂Φ
∂−→n(y − x)− ∂hx
∂−→n(y))u(y) dS(y). (3.10)
By defining G(x, y) := Φ(y − x)− hx(y), then we get that for any u ∈ C2(Ω)
u(x) =
ˆΩ
G(x, y)∆u(y) dy +
ˆ∂Ω
∂G
∂−→n(y − x)u(y) dS(y). (3.11)
The function G(x, y) is called the Green’s function for Ω. In particular, if u is harmonic
then
u(x) =
ˆ∂Ω
∂G
∂−→n(y − x)u(y) dS(y).
We now prove the two claims.
Proof of Claim 1. We have
−ˆ∂Vε
u(y)∂Φ
∂−→n(y−x) dS(y) = −
ˆ∂Ω
u(y)∂Φ
∂−→n(y−x) dS(y)+
ˆ∂Bε(x)
u(y)∂Φ
∂−→n(y−x) dS(y).
47
The first term on the right-hand side does not depend on ε, so it is unchanged when
passing ε→ 0. We consider the second term. We have
∂Φ
∂−→n(y − x) = ∇yΦ(y − x)×−→n (y),
where −→n (y) is the outward normal vector on ∂ε(x). By direct computations
∇yΦ(y − x) =1
(2− n)σn−1
∇y |y − x|2−n =1
σn−1
y − x|y − x|n
,
−→n (y) =y − x|y − x|
,
so that∂Φ
∂−→n(y − x) =
1
σn−1
y − x|y − x|n
y − x|y − x|
=1
σn−1
1
|y − x|n−1.
Therefore,ˆ∂Bε(x)
u(y)∂Φ
∂−→n(y − x) dS(y) =
1
σn−1
ˆ∂Bε(x)
1
|y − x|n−1u(y) dS(y)
=1
σn−1εn−1
ˆ∂Bε(x)
u(y) dS(y)
=1
|∂Bε(x)|
ˆ∂Bε(x)
u(y) dS(y).
Since u is continuous, we have that
limε→0
ˆ∂Bε(x)
u(y)∂Φ
∂−→n(y − x) dS(y) = lim
ε→0
1
|∂Bε(x)|
ˆ∂Bε(x)
u(y) dS(y) = u(x),
from which the assertion of the claim 1 is followed.
Proof of the claim 2. Similarly as in the proof of the claim 1, we haveˆ∂Vε
Φ(y − x)∂u
∂−→n(y) dS(y) =
ˆ∂Ω
Φ(y − x)∂u
∂−→n(y) dS(y)−
ˆ∂Bε(x)
Φ(y − x)∂u
∂−→n(y) dS(y).
We now show the second term vanishes as ε→ 0. In deed, using the explicit formula for
Φ for n ≥ 3 (the case n = 2 is similar), we have∣∣∣∣ˆ∂Bε(x)
Φ(y − x)∂u
∂−→n(y) dS(y)
∣∣∣∣ ≤ 1
(n− 2)σn−1
ˆ∂Bε(x)
1
|y − x|n−2
∣∣∣ ∂u∂−→n
(y)∣∣∣ dS(y)
≤∥∥∥ ∂u∂−→n
∥∥∥L∞(Bε(x))
1
(n− 2)σn−1εn−2
ˆ∂Bε(x)
dS(y)
≤ Cε,
where to obtain the last step, we have used thatˆ∂Bε(x)
dS(y) = σn−1εn−1.
48
Therefore
limε→0
ˆ∂Bε(x)
Φ(y − x)∂u
∂−→n(y) dS(y) = 0,
and the claim 2 is followed.
To summarise, we have proved the following.
Theorem 8. If u ∈ C(Ω) is a solution of∆u = f, in Ω,
u = g, in Ω,
(3.12)
where f, g are continuous, then for x ∈ Ω,
u(x) =
ˆΩ
G(x, y)f(y) dy +
ˆ∂Ω
∂G
∂−→n(x, y)g(y) dS(y). (3.13)
In particular, if f ≡ 0, then
u(x) =
ˆ∂Ω
∂G
∂−→n(x, y)g(y) dS(y). (3.14)
3.1 The Green function for a ball
Let Ω = BR(0). Recall that to solve the Dirichlet problem in Ω we need to find, for
each x ∈ Ω, the corrector function hx that satisfies∆yhx(y) = 0, in Ω,
hx(y) = Φ(y − x), on ∂Ω.
(3.15)
We will solve this problem by the method of image charge. Think of (3.15) as that we
want to find the electrostatic potential hx inside a spherical conductor of radius R de to
the point charge located at x.
We consider the image of x under the inversion in ∂BR,
x∗ =R2
|x|2x,
which in particular, implies that
|x∗| = R2
|x|, x∗ · x = R2.
49
Then, if y ∈ ∂BR, we have
|y − x|2 = |y|2 − 2x · y + |x|2
= R2 − 2x · y + |x|2
=|x|2
R2
(R4
|x|2− 2
xR2
|x|2· y +R2
)=|x|2
R2
(|x∗|2 − 2x∗ · y + |y|2
)=|x|2
R2|x∗ − y|2.
So we obtain that
|y − x| = |x|R|y − x∗| ∀y ∈ ∂BR.
Let
hx(y) :=
1
2πlog(|x|R|y − x∗|
), n = 2,
1(2−n)σn−1
(|x|R
)2−n|y − x∗|2−n, n ≥ 3.
Note that since |x∗| |x| = R2 and x ∈ BR, it follows that x∗ 6∈ BR, so that y 6= x∗.
Therefore y 7→ hx(y) is harmonic in BR. In addition, by the computation above, it holds
that hx(y) = Φ(y − x) on ∂BR. Therefore, hx(y) is the unique corrector function.
Hence, the Green’s function for the ball is given by
GBR(x, y)
1
2πlog(R|x||y−x||y−x∗|
), n = 2,
1(2−n)σn−1
[|y − x|2−n −
(|x|R
)2−n|y − x∗|2−n
], n ≥ 3.
By the representation formula (3.14), we have
u(x) =
ˆ∂Ω
∂G
∂−→n(x, y)g(y) dS(y).
Next, we will compute ∂G∂−→n (x, y). We have
∂G
∂−→n(x, y) = ∇yG(x, y) · −→n .
Here
−→n =y
R,
∇yG(x, y) =1
σn−1
[y − x|y − x|n
−(|x|R
)2−ny − x∗
|y − x∗|n
].
50
Note that this formula is true for n ≥ 2. Therefore, we obtain
∂G
∂−→n(x, y) =
1
Rσn−1
[y − x|y − x|n
−(|x|R
)2−ny − x∗
|y − x∗|n
]· y
=1
Rσn−1
[|y|2 − x · y|y − x|n
−( |x|R
)2−n |y|2 − x∗ · y|y − x∗|n
]=
1
Rσn−1
[|y|2
|y − x|n−( |x|R
)−n |x|2
|y − x∗|n− x · y
( 1
|y − x|n−( |x|R
)−n 1
|y − x∗|n)]
=1
Rσn−1
R2 − |x|2
|y − x|n.
Set
KR(x, y) :=1
Rσn−1
R2 − |x|2
|y − x|n. (3.16)
This function is called the Poisson kernel. We have proved that
Proposition 3. Let u ∈ C2(BR) be harmonic in BR. Then for x ∈ BR, one has
u(x) =
ˆ∂BR
KR(x, y)u(y) dS(y) =1
Rσn−1
ˆ∂BR
R2 − |x|2
|y − x|nu(y) dS(y). (3.17)
The formula above is called the Poisson’s integral formula.
Note that this proposition only tells us that if a solution exists, then it is given in
the interior by the Poisson’s integral formula. We need to show that for suitable g, the
Poisson’s integral formula indeed defines us a harmonic function.
We will need some properties of the Poisson’s kernel KR(x, y).
Lemma 2. The function KR(x, y) satisfies the following properties
(i) For any y ∈ ∂BR, the map
KyR : BR → R
x 7→ KR(x, y)
is harmonic.
(ii) If y ∈ ∂BR and x ∈ BR then KR(x, y) > 0.
(iii) For all x ∈ BR, we have ˆ∂BR
KR(x, y) dS(y) = 1.
51
(iv) For y 6= z with y, z ∈ ∂BR, we have
limx→zx∈BR
KR(x, y) = 0
and this convergence is uniform in y for |y − z| > δ > 0.
Proof. (i) We notice that for each i = 1, . . . , n, the function xi|x|n is harmonic on Rn\0.
Indeed, sincexi|x|n
=∂
∂xi
(|x|2−n
2− n
)so
∆xi|x|n
= ∆
[∂
∂xi
(|x|2−n
2− n
)]=
∂
∂xi
[∆
(|x|2−n
2− n
)]= 0.
It follows that |x− y|2−n and (x−y)i|x−y|n are both harmonic. So that
Rσn−1KR(x, y) =R2 − |x|2
|x− y|n=
2|y|2 − |x− y|2 − 2x · y|x− y|n
= −|x− y|2−n + 2y · y − x|y − x|n
is also a harmonic function.
(ii) This is obvious since |x| < R for x ∈ BR.
(iii) Applying the Poisson’s integral formula for u ≡ 1 yields the result.
(iv) limx→zx∈BR
R2−|x|2|y−x|n = 0
|y−z|n = 0. For |x−z| < δ2, then |y−x| ≥ |y−z|− |x−z| ≥ δ− δ
2= δ
2.
Therefore,
sup|y−z|>δ
R2 − |x|2
|y − x|n≤ R2 − |x|2
δ/2→ 0.
Theorem 9. Let g ∈ C0(∂BR) and let us define the function u : BR → R by
u(x) =
ˆBR
KR(x, y)g(y) dS(y).
Then u is harmonic in BR, u ∈ C0(BR) and for x0 ∈ ∂BR, we have
limx→x0x∈BR
u(x) = g(x0).
Proof. Since KR(x, y) is harmonic in x, it is also in C∞. Define
I(x, y) := KR(x, y)g(y).
52
Let U be compactly contained in BR. Then I : U ×∂BR → R is infinitely differentiable in
x and the derivative DαxI(x, y) are continuous maps from U × ∂BR to R. Moreover, since
∂BR is compact, according to the results of the Appendix, we have that u is C∞ and that
for each x ∈ U we have
Dαu(x) =
ˆ∂BR
DαxKR(x, y)g(y) dS(y).
Since any x ∈ BR belongs to some open set U b Ω, it follows that u is smooth and
∆u(x) =
ˆ∂BR
∆xKR(x, y)g(y) dS(y) = 0.
Now we show the continuity at the boundary. Let x0 ∈ ∂BR. By Property (iii) in 2, we
have
u(x)− g(x0) =
ˆ∂BR
KR(x, y)(g(y)− g(x0)) dS(y) = I1 + I2,
where
I1 :=
ˆ∂BR∩Bδ(x0)
KR(x, y)(g(y)− g(x0)) dS(y),
and
I2 :=
ˆ∂BR\Bδ(x0)
KR(x, y)(g(y)− g(x0)) dS(y),
for some δ > 0. Since g ∈ C0(∂BR), for a given ε > 0, we can take δ > 0 small enough
that |g(y)− g(x0)| ≤ ε for y ∈ ∂BR ∩Bδ(x0). Therefore, the term I1 can be estimated by
|I1| ≤ˆ∂BR∩Bδ(x0)
KR(x, y)|g(y)− g(x0)| dS(y)
≤ ε
ˆ∂BR∩Bδ(x0)
KR(x, y) dS(y)
≤ ε
ˆ∂BR
KR(x, y) dS(y) = ε. (3.18)
Since g is continuous (and hence bounded) on ∂BR from part (iv) of Lemma 2, we have
that
KR(x, y)(g(y)− g(x0))→ 0 as x→ x0
uniformly in y ∈ ∂BR \Bδ(x0). Thus as x→ x0, we have I2 → 0.
Together, with the estimate (3.18), we have that limx→x0 u(x) = g(x0) as claimed.
As consequences of the Poisson’s integral formula, we obtain the following strength-
ened versions of the Harnack’s inequality and the Liouville theorem.
53
Theorem 10 (Harnack’s inequality for the ball). Let u ≥ 0 be harmonic in BR. Then
for any x ∈ BR we have the following estimate(1− |x|
R
)u(0)(
1 + |x|R
)n−1 ≤ u(x) ≤
(1 + |x|
R
)u(0)(
1− |x|R
)n−1 .
Proof. According to the Poisson’s integral formula we have
u(x) =1
Rσn−1
ˆ∂BR
R2 − |x|2
|y − x|nu(y) dS(y).
By the triangle inequality, we have for any y ∈ ∂BR
R− |x| ≤ |y − x| ≤ |x|+R
So that
R2 − |x|2
σn−1R(R + |x|)n
ˆ∂BR
u(y) dS(y) ≤ u(x) ≤ R2 − |x|2
σn−1R(R− |x|)n
ˆ∂BR
u(y) dS(y)
By the mean value property
ˆ∂BR
u(y) dS(y) = |BR|u(0) = Rn−1σn−1u(0).
Hence
R2 − |x|2
σn−1R(R + |x|)nRn−1σn−1u(0) ≤ u(x) ≤ R2 − |x|2
σn−1R(R− |x|)nRn−1σn−1u(0)
Simplifying this equality, we obtain(1− |x|
R
)u(0)(
1 + |x|R
)n−1 ≤ u(x) ≤
(1 + |x|
R
)u(0)(
1− |x|R
)n−1
as claimed.
Theorem 11 (Improved Liouville theorem). Suppose that u : Rn → R is harmonic and
bounded from below. Then u must be a constant.
Proof. Without loss of generality, we assume u is harmonic in Rn and u ≥ 0. Take x ∈ Rn,
according to Theorem 10 , for any R > 0, we have(1− |x|
R
)u(0)(
1 + |x|R
)n−1 ≤ u(x) ≤
(1 + |x|
R
)u(0)(
1− |x|R
)n−1
Sending R→∞, we find that u(x) = u(0),∀x ∈ Rn.
54
3.2 C0-subharmonic functions
Suppose Ω ⊂ Rn.
Definition 5. We say that a function u ∈ C0(Ω) is C0-subharmonic in Ω if for any
Bρ(x) b Ω, we have the implication
h ∈ C2(Bρ(x)) ∩ C0(Bρ(x))
∆h = 0, in Bρ(x),
u ≤ h, on ∂Bρ(x)
=⇒ u ≤ h in Bρ(x).
Note that in this definition, there is no requirement that u ∈ C2(Ω).
Lemma 3. If u is C0-subharmonic, then for any Bρ(x) b Ω, we have
u(x) ≤ ∂Bρ(x)
u(y) dS(y) and u(x) ≤ Bρ(x)
u(y) dy,
whereffl
denotes the average integral. And conversely, if u ∈ C0 and satisfies either the
inequalities above for any ball Bρ(x) b Ω then u is C0-subharmonic.
Proof. By Theorem 9, there exists a function h solvingh ∈ C2(Bρ(x)) ∩ C0(Bρ(x))
∆h = 0, in Bρ(x),
u = h, on ∂Bρ(x)
Since u is C0-subharmonic, we have
u(x) ≤ h(x) =
∂Bρ(x)
h(y) dS(y) =
∂Bρ(x)
u(y) dS(y),
where the first equality follows from the mean value property for h.
The second estimate of the Lemma is similar.
Now suppose that
h ∈ C2(Bρ(x)) ∩ C0(Bρ(x))
∆h = 0, in Bρ(x),
u ≤ h, on ∂Bρ(x)
Then
(u− h)(x) ≤ Bρ(x)
(u− h)(y) dy,
55
i.e., u − h satisfies the mean value property. Recalling that to prove the strong maxi-
mum principle, we only need the mean value property and connectedness of the domain.
Therefore, it implies that u− h satisfies the strong maximum principle. Hence,
u− h ≤ supBρ(x)
(u− h) ≤ 0 in Bρ(x)
i.e., the implication in the definition holds.
Proposition 4 (Properties of C0-subharmonic functions).
(i) If u is C0-subharmonic function then the strong maximum principle holds. Moreover,
we have comparison principle with harmonic functions on every connected set Ω′ b
Ω, i.e., in the definition 5, one can replace the ball Bρ(x) by Ω′.
(ii) Let u be C0-subharmonic in Ω and let Bρ(x) b Ω. We define u to be the solution of
the problem ∆u = 0, in Bρ(x),
u = u, on ∂Bρ(x).
and then define the function U by
U(y) =
u(y), if y ∈ Bρ(x),
u(y), if y ∈ Ω \Bρ(x).
Then U is C0-subharmonic in Ω.
The function U is called the harmonic lifting of u in Bρ(x).
(iii) Let u, v be C0-subharmonic in Ω and such that u ≤ v. Suppose Bρ(x) b Ω and let
U, V respectively be the harmonic liftings of u and v in Bρ(x). Then U ≤ V .
(iv) If u1, . . . , uN are C0-subharmonic in Ω, then
u(x) := maxu1(x), . . . , uN(x)
is also C0-subharmonic in Ω.
Proof. 1. From the proof of Theorem 6, the strong maximum principle follows from the
mean value property. According to Lemma 3 a C0-subharmonic function satisfies
the mean value property, so that it satisfies a strong maximum principle.
56
2. Let Br(z) b Ω and suppose that
h ∈ C2(Br(z)) ∩ C0(Br(z))
∆h = 0, in Br(z),
U ≤ h, on ∂Br(z).
First of all, note that since u ≥ u in Bρ(x), then U ≥ u in in Bρ(x) and hence in
Ω. Hence u ≤ U ≤ h in ∂Br(z), it follows that u ≤ h in Br(z). Therefore, U ≤ h
in Br(z) \ Bρ(x). In addition, since u = u on ∂Bρ(x), it follows that u ≤ u ≤ h in
∂(Br(x)∩Bρ(x)). Since u is harmonic in ∂(Br(x)∩Bρ(x)), we obtain that U = u ≤ h
in Br(x)∩Bρ(x). So U ≤ h in Br(z, which means that U is C0-subharmonic function
in Ω.
3. On the set Ω \Bρ(x), we have U = u ≤ v = V . In addition, we have
∆u = 0, ∆v = 0, in Bρ(x),
u = u, v = v on ∂Bρ(x)
Since u ≤ v on ∂Bρ(x), by the comparison principle, we have that u ≤ v in Bρ(x),
so that also U ≤ V in Bρ(x).
In conclusion, we get U ≤ V in Ω.
4. For two functions f, g ∈ C0(Ω) it holds that
maxf, g =1
2[f + g + |f − g|],
which implies that maxf, g ∈ C0(Ω). Repeating this argument, we have that if
u1, . . . , uN ∈ C0(Ω) then u = maxu1, . . . , uN ∈ C0(Ω). Let h be the function
satisfies the conditions in the definition 5. Then ui ≤ h in ∂Bρ(x) and hence ui ≤ h
in Bρ(x) for all i = 1, . . . , N since ui are C0-subharmonic. Therefore u ≤ h in Bρ(x),
so u is also C0-subharmonic.
57
3.3 Perron’s method for the Dirichlet problem on a
general domain
We are now ready to introduce Perron’s method to solve the Dirichlet problem on a
general domain ∆u = 0, in Ω
u = g, on ∂Ω.
Let g ∈ C0(Ω), define
Sg := v : Ω→ R such that v is C0-subharmonic in Ω and v ≤ g on ∂Ω.
Theorem 12. Suppose that Ω is open and bounded and that g ∈ C0(∂Ω). Define
u(x) := supv∈Sgv(x).
Then u is harmonic in Ω.
Proof. Step 1) First of all Sg 6= ∅ since inf∂Ω g ∈ Sg. Moreover, for each v ∈ Sg then
v ≤ sup∂Ω g. So that Sg is non-empty, bounded above. Therefore, for each x,
u(x) is well-defined, finite number.
Step 2) Now pick y ∈ Ω. By definition of u, there exist vi ∈ Sg such that vi(y) → u(y)
as i→∞. Define
vi(x) = maxinf∂Ωg, v1(x), . . . , vi(x).
Then
• inf∂Ω g ≤ v1 ≤ v2 ≤ . . . ≤ vi ≤ . . . ≤ sup∂Ω g.
• vi are subharmonic.
Step 3) Take ρ sufficiently small such that Bρ(y) b Ω. Let Vi be the harmonic lifting of
vi in Bρ(y). Then
• Vi ∈ Sg, Vi(y)→ u(y),
• Vi are harmonic in Bρ(y),
• inf∂Ω g ≤ V1 ≤ V2 ≤ . . . ≤ Vi ≤ . . . ≤ sup∂Ω g.
58
Step 4) By Harnack’s theorem 3, Vi → V uniformly in Bρ′(y) for any ρ′ < ρ, where V is
harmonic. Since Vi ∈ Sg, it follows that V ≤ u in Ω.
Step 5) We now show that V = u in Bρ(y). Suppose this is not true. Then there exists
z ∈ Bρ(y) such that V (z) < u(z). By definition of u, there exists v ∈ Sg such
that V (z) < v(z). We define
wi = maxv, Vi,
and let Wi be the harmonic lifting of wi in Bρ(y). By similar arguments as in the
previous steps, there exists a new harmonic function W such that
• V ≤ W ≤ u in Bρ(y),
• V (y) = W (y) = u(y),
• W (z) ≥ v(z).
The function W − V is harmonic in Bρ(y) satisfying
W − V ≥ 0 in Bρ(y) and (W − V )(y) = 0.
By the strong maximum principle it follows that V = W in Bρ(y). However, this
is a contradiction since
V (z) < v(z) ≤ W (z).
Therefore, V = u in Bρ(y).
Step 6) Since y is arbitrary, we conclude that V = u in Ω, thus u is harmonic in Ω.
The function u constructed by Theorem 12 is called Perron’s solution. It is a good
candidate for a solution of the Dirichlet problem (3.1). It remains to show that it satisfies
the boundary condition. To achieve this, we need to make some assumptions on the
domain Ω.
Definition 6. We say that Ω satisfies the barrier postulate if for each y ∈ ∂Ω, there exists
a function (called a barrier function) Qy(x) ∈ C0(Ω) ∩ C2(Ω) satisfying
(i) Qy(x) is superharmonic in Ω,
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(ii) Qy(x) > 0 in Ω \ y and Qy(y) = 0.
Theorem 13. Suppose that Ω is open, bounded and satisfies the barrier postulate and
that g ∈ C0(∂Ω). Let u be the Perron’s solution. Then for any y ∈ ∂Ω, we have
limx∈Ωx→y
u(x) = g(y).
Proof. Let ε > 0 and set M = sup∂Ω |g|. Pick y ∈ ∂Ω and let Qy(x) be the barrier
function. By the continuity of g and the positivity of Qy(x) in Ω \ y, there exist δ,K
depending on ε such that
|g(y)− g(z)| < ε for z ∈ ∂Ω, |y − z| < δ, (3.19)
and
KQy(x) > 2M for z ∈ ∂Ω, |y − z| ≥ δ. (3.20)
We consider the functions
uy,ε(x) = g(y)− ε−KQy(x),
uy,ε(x) = g(y) + ε+KQy(x).
Since Qy(x) is superharmonic, uy,ε(x) is subharmonic. If z ∈ ∂Ω, then from (3.19) and
(3.20), we have
uy,ε(z)− g(z) = g(y)− g(z)− ε−KQy(x) ≤ 0,
which implies that uy,ε ∈ Sg. Similarly, we get uy,ε ∈ Sg. Hence
uy,ε ≤ u(x) ≤ uy,ε(x) ∀x ∈ Ω,
so that
g(y)− ε−KQy(x) ≤ u(x) ≤ g(y) + ε+KQy(x) ∀x ∈ Ω.
This implies that
−ε−KQy(x) ≤ u(x)− g(y) ≤ ε+KQy(x),
i.e.,
|u(x)− g(y)| ≤ ε+KQy(x).
Since Qy ∈ C0(Ω and Qy(y) = 0, there exists δ′ such that if |x−y| < δ′ then Qy(x) ≤ K−1ε
and
|u(x)− g(y)| ≤ 2ε,
which implies that limx→y u(x) = g(y) as claimed.
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Corollary 4. Under the assumptions as in Theorem 13, the Dirichlet problem (3.1) is
well-posed.
To conclude this section, we provide a criterion that a barrier function exists on Ω for
the point y.
Lemma 4. Suppose that Ω ⊂ Rn is open and bounded, and that y ∈ ∂Ω. If there exists
Br(z) such that Br(z) ∩ Ω = y, then there exists a barrier function on Ω for y.
Proof. Consider the function
Qy(x) =
r2−n − |x− z|2−n, n ≥ 3,
log(|x−z|r
), n = 2.
This function is a barrier function on Ω for y.
Problem Sheet 3
Exercise 12 (The Kelvin transform). Given a > 0 and consider the following map
Ta : Rn \ 0 → R
n \ 0
x 7→ a2 x
|x|2
which is known as inversion with respect to the sphere of radius a.
a) Show that the map is conformal, i.e., it preserves the angles between vectors
(DTa(x)[v], DTa(x)[w])
||DTa(x)[v]|| ||DTa(x)[w]||=
(v, w)
||v|| ||w||(3.21)
for all v, w ∈ Rn, x ∈ Rn \ 0.
b) Show that in two dimensions Ta sends lines and circles into lines and circles.
c) If Ta is as above, define
va(x) =
(a
|x|
)n−2
u(Ta(x)).
Show that
∆va(x) =
(a
|x|
)n+2
∆u(Ta(x)).
What can we say if u is harmonic? va is called the the Kelvin transform of u.
Exercise 13 (Hopf Lemma. See Section 6.4.2 in Evans’s book with Lu = ∆u). Sup-
pose Ω ⊂ Rn is a smooth bounded domain (which implies the interior ball condition in
Evans’book). Suppose that u ∈ C2(Ω) ∩ C1(Ω) is subharmonic in Ω and suppose there
exists x0 ∈ ∂Ω with u(x0) > u(x) for all x ∈ Ω. Then
∂u
∂ν(x0) > 0,
where ν is the outer unit normal vector.
61
62
Exercise 14 (Symmetry of the Green’s function). Let Ω ⊂ Rn be open, bounded and of
class C1. Let G(x, y) be the Green’s function for Ω. Prove that G is symmetric, i.e., for
all x, y ∈ Ω, x 6= y, we have
G(x, y) = G(y, x)
Exercise 15. Using the method of image charges, find the Green’s function of the half
space
Ω1 := x ∈ Rn : xn > 0
and of the half ball in R3:
Ω2 := (x, y, z) ∈ R3 : x2 + y2 + z2 = 1, z > 0.
Exercise 16. a) Find an example of a C0-subharmonic function on (0, 1) that is not
everywhere differentiable.
b) Find an example of a C1-subharmonic function on (0, 1) which is C1-regular but not
everywhere of class C2.
Chapter 4
The theory of distributions and
Poisson’s equation
In this chapter, we will solve Poisson’s equation: given f : Rn → R, solve
∆u = f (4.1)
for some u : Rn → R. The approach will be via the fundamental solution which is the
solution to the equation where the right-hand side is replaced by the so-called Dirac
distribution (Dirac delta function). The solution to Poisson’s equation then will be deter-
mined by a convolution of the fundamental solution with the function f . The main tool
of this chapter will be the theory of distributions. This theory plays an important role in
mathematics and is of independent interest.
4.1 The theory of distributions
As a motivation, let us consider the following simple problem: solve
x2 = a.
We knew that if a ≥ 0, this equation has real solutions x ∈ R. However, if a < 0,
it has no real solution. However, if we extend the set of real numbers R to the set of
complex numbers C, then it has solutions z ∈ C even if a < 0. To extend R to C we not
only enlarge R to a bigger set R ⊂ C, but also generalise the operations on R, such as
addition, subtraction and multiplication, to C. After extension to the complex numbers,
63
64
the problem of solving an algebraic equation (not only a quadratic one as the one above)
is in some ways simpler than the original problem: the number of complex roots in an
algebraic equation of order k is always k (counting multiplicities) while over the real line
the number can be any thing between 0 and k. Having found the complex roots, we then
can consider separately the problem of showing that some of them are in fact real.
Turing back to Poisson’s equation (4.1), the aim of this chapter is to treat the situation
where u and f need not be of class C2. Generally, we want to make sense of the PDE
equation of order k
Lu :=∑|α|≤k
aαDαu = f,
where aα ∈ C∞(Ω) for an open Ω ⊂ Rn, in the situation where u and f need not be of class
Ck(Ω). To do so, similarly as in the motivating example above, we need to extend the
class of smooth functions to a larger set and generalise the operations on smooth functions
to elements of this set. Let us denote by D′(Ω) the space to which our generalised solution
u and the right-hand side function f should belong. What properties do we require for
D′(Ω) so that we can at least make sense of the PDE (4.1)? The list of desirable properties
should contain
1) The smooth functions C∞(Ω) should be contained in D′(Ω) in such a way that we can
recover them, i.e., the inclusion map ι : C∞(Ω) → D′(Ω) should be injective.
2) We need to be able to add elements of D′(Ω).
3) We need to be able to multiply elements of D′(Ω) by elements of C∞(Ω).
4) We need to be able to differentiate elements of D′(Ω). Moreover, the generalised
notion of differentiation should agree with the classical one when restricted to smooth
functions.
5) We need some topology on D′(Ω) such that the above operations are continuous.
Before introducing D′(Ω), we need to introduce another space, the space of test functions
D(Ω) := C∞c (Ω). A topology on D(Ω) is defined as follows.
Definition 7 (Convergence in D(Ω)). A sequence (φ)j∈N, φj ∈ D(Ω) = C∞c (Ω) converges
to φ ∈ D(Ω) if there exists a compact set K ⊂ Ω such that supp(φj), supp(φ) ⊂ K for all
65
j ∈ N and that
Dαφj → Dαφ as j →∞
uniformly in K for any multi-index α.
The space D(Ω) equipped with this topology becomes a topological vector space over
the real number, i.e, a vector space together with a topology such that addition of vectors
and scalar multiplication are continuous functions. For any topological vector space V ,
one can define the continuous dual space which consists of continuous linear maps
V ′ = ω : V → R| ω linear, continuous.
The space D′(Ω) that we are seeking for will be the continuous dual space of D(Ω).
Definition 8. A distribution T ∈ D′(Ω) is a linear functional on the space of test functions
T : D(Ω)→ R
which is continuous with respect to the topology on D(Ω) defined in Definition 7. In other
words, if φj → φ in the sense of Definition 7 then
Tφj → Tφ as j →∞.
We consider two important examples
Example 20. 1. If f : Ω → R is continuous, the distribution Tf associated to f is
define by
Tf (φ) :=
ˆΩ
f(x)φ(x) dx ∀φ ∈ C∞c (Ω).
In one-dimension, we can allow f to have a finite number of points of finite discon-
tinuity. Because of this, a distribution is also called generalised function.
2. (The Dirac delta) For x ∈ Ω, the Dirac distribution δx is defined via
δxφ = φ(x).
We now show that the wish-list properties above are satisfied.
Proposition 5. Suppose that f ∈ C0(Ω). Let ρk,y, k = 1, 2, . . ., be a family of mollifiers
concentrating at y, i.e., ρk,y ∈ C∞c (Ω), ρk,y ≥ 0 andˆ
Ω
ρk,y(x) dx = 1, supp(ρk,y) ⊂ B1/k(y).
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Then
f(y) = limk→∞
Tf (ρk,y).
As a consequence, if f, g ∈ C0(Ω) and Tf = Tg then f = g.
Proof. Since´
Ωρk,y(x) dx = 1, we have that
f(y)− Tf (ρk,y) =
ˆΩ
(f(y)− f(x)) ρk,y(x) dx.
Since f is continuous, given ε > 0, there exists K ∈ N such that |f(y) − f(x)| < ε for
x ∈ B1/k(y) whenever k ≥ K. For such a k, we have
|f(y)− Tf (ρk,y)| ≤ˆ
Ω
|f(y)− f(x)| ρk,y(x) dx ≤ ε
ˆΩ
ρk,y(x) dx = ε,
which implies that f(y) = limk→∞ Tf (ρk,y).
Now suppose that f, g ∈ C0(Ω) and Tf = Tg. For any y ∈ Ω, let ρk,y be the mollifiers
concentrating at y. Then
f(y) = limk→∞
Tf (ρk,y) = limk→∞
Tg(ρk,y) = g(y),
so that f = g as claimed.
Next, we extend the operations on functions to distributions as follows.
1) Given T1, T2 ∈ D′(Ω). The distribution T1 + T2 is defined by
(T1 + T2)(φ) := T1(φ) + T2(φ),∀φ ∈ D(Rn).
2) Given a ∈ C∞(Ω) and T ∈ D′(Ω), the distribution aT is defined by
(aT )(φ) := T (aφ), ∀φ ∈ D(Rn).
Note that the right-hand side makes sense since aφ ∈ D(Rn).
3) First, note that if f ∈ C∞(Ω), then Dif ∈ C∞(Ω) and
TDif (φ) =
ˆΩ
Dif(x)φ(x) dx = −ˆ
Ω
f(x)Diφ(x) dx = −Tf (Diφ),
Note that in the above computations, there are no boundary terms because φ vanishes
at the infinity. Motivated by this, we define the derivatives of a distribution as follows.
For any multi-index α, we define
(DαT )(φ) := (−1)|α|T (Dαφ).
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Let us do some examples
Example 21. 1) T = δx. Then
(DiT )(φ) = −T (Diφ) = −ˆ
Ω
δxDiφ(y) dy = −(Diφ)(x), ∀φ ∈ D(Rn).
2) Consider the Heaviside step function
H(x) =
1, x ≥ 0,
0, x < 0
(x ∈ R).
Note that this function is not differentiable at x = 0. We consider the distribution TH
by
TH(φ) =
ˆR
H(x)φ(x) dx, for φ ∈ D(Rn).
Then we compute
(DxTH)(φ) = −TH(Dxφ)
= −ˆR
H(x)Dxφ(x) dx
= −ˆ ∞
0
Dxφ(x) dx
= φ(0) = δ0(φ).
Therefore, DxTH = δ0.
So the theory of distributions provides us some sort of meaning to the derivative of a
function whose classical derivatives do note exist.
Theorem 14. Suppose that f ∈ C0(Ω) and let Tf be the distribution associated to f .
Suppose further that for any multi-index α, with |α| ≤ k, there exist gα ∈ C0(Ω) such that
DαTf = Tgα
where Dα is the distributional derivative. Then f ∈ Ck(Ω) and Dαf = gα in the classical
sense.
This theorem states that the distributional derivatives agree with the classical ones
when they are both defined.
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We have addressed the four wish-lists that we asked in the introduction. It is straight-
forward to check that all the operations that we have defined are indeed continuous with
respect to the topology that we have introduced.
The advantage of introducing distributions is expressed in the following result, which
is followed from Theorem 14 and all the properties we have defined.
Proposition 6. Suppose that f ∈ C0(Ω) and that T ∈ D′(Rn) satisfies the distributional
equation ∑|α|≤k
aαDαT = Tf ,
where aα ∈ C∞(Ω) and that there exist functions uα ∈ C0(Ω) such that DαT = uα. Then
u := u0 is in Ck(Ω) and is a classical solution of the PDE∑|α|≤k
aαDαu = f. (4.2)
This proposition provides us a method to solve the PDE (4.2) that consists of two
steps.
1) Lift up the PDE (4.2) to the distributional PDE∑|α|≤k
aαDαT = Tf
and solve this distributional equation.
2) Check whether a distributional solution is a classical one.
In the next section, we will study how to solve the distributional equation above.
4.2 Convolutions and the fundamental solution
Recalling that for f ∈ C0(Rn) and φ ∈ C∞c (Rn), the convolution f ∗ φ is again a
function defined by
(f ∗ φ)(x) :=
ˆRn
f(y)φ(x− y) dy.
How to extend this definition to T ∗φ where T ∈ D′(Rn), φ ∈ C∞c (Rn)? To do so we define
τxφ : Rn → R
y 7→ φ(x− y).
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If φ ∈ C∞c (Rn) then so is τxφ. Now we can write
(f ∗ φ)(x) = Tf (τxφ).
The advantage of writing this way is that we can extend it to distributions: let T ∈
D′(Ω), φ ∈ D(Rn), we define
(T ∗ φ)(x) := T (τxφ).
Before stating some important properties of convolutions, we need to introduce the con-
cept of support of a distribution.
Definition 9 (Support of a distribution). A distribution T ∈ D′(Rn) is supported in the
closed set K ⊂ Ω if
Tφ = 0 ∀φ ∈ C∞c (Ω \K),
i.e, T vanishes when applying to any test function that vanishes on K. The support of T
is defined by
suppT =⋂K : T supported in K.
Example if T = δx then suppT = x.
Proposition 7 (Properties of convolutions). Suppose T, Ti ∈ D′(Rn) and φ ∈ D(Rn).
Then
(i) If T1 ∗ φ = T2 ∗ φ, ∀φ ∈ D(Rn) then T1 = T2.
(ii) T ∗ φ ∈ C∞(Rn) and
Dα(T ∗ φ) = T ∗Dαφ = DαT ∗ φ.
(iii) If T has compact support, then T ∗ φ also has compact support; in particular T ∗ φ
is a test function.
Proof. (i) We have (τxφ)(y) = φ(x − y), so that φ(−y) = (τ0φ)(y), or equivalently,
φ = τ0φ, where φ(y) := φ(−y). Then we have
T1(φ) = T1(τ0φ) = (T1 ∗ φ)(0) = (T2 ∗ φ)(0) = T2(φ),
which means that T1 = T2 as desired.
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(ii) We first show that Di(T ∗ φ) = T ∗Diφ. In fact, let ei be the i-th unit vector, we
calculate
(T ∗ φ)(x+ εei)− (T ∗ φ)(x)
ε=
1
ε
[T (τx+εeiφ)− T (τxφ)
]= T
[1
ε(τx+εeiφ− τxφ)
],
where we have used the linearity of T . Since φ ∈ D(Rn), we have
limε→0
1
ε(τx+heiφ− τxφ)(y) = lim
ε→0
1
ε(φ(x− y + εei)− φ(x− y))
=∂
∂xiφ(x− y) = (τxDiφ)(y),
with convergence in the topology of D(Rn). By the continuity of distributions, we
have
limε→0
(T ∗ φ)(x+ εei)− (T ∗ φ)(x)
ε= T (τxDiφ) = (T ∗Diφ)(x),
i.e., Di(T ∗ φ) = T ∗ Diφ. By repeating this arguments for higher derivatives, we
establish the first equality. To get the second equality, we calculate
Dα(τxφ)(y) =∂|α|
∂yαφ(x− y)
= (−1)|α|(Dαφ)(x− y)
= (−1)|α|(τxDαφ)(y),
so that Dα(τxφ) = (−1)|α|(τxDαφ).
Therefore
(DαT ∗ φ)(x) = DαT [τxφ]
= (−1)|α|T (Dατxφ)
= (−1)|α|T ((−1)|α|τxDαφ)
= T (τxDαφ)
= T ∗Dαφ(x).
Hence DαT ∗ φ = T ∗Dαφ.
(iii) Suppose without loss of generality that T and φ are supported inBR(0) for some large
R. Let x ∈ Rn be such that |x| > 2R. Then if y ∈ BR(0), by the triangle inequality,
|y − x| > R. So that (τxφ)(y) = φ(x − y) = 0, i.e., τxφ ∈ C∞c (Rn \ BR(0)). By the
71
assumption that T is supported in BR(0), we have that T (τxφ) = (T ∗ φ)(x) = 0,
hence T ∗ φ(x) = 0 for all |x| > 2R, i.e., T ∗ φ has compact support.
We have extended addition, multiplication, derivatives and convolution of a distribu-
tion with a test function. Next, we want to extend the convolution of two distributions:
given T1, T2 ∈ D′(Rn), how to define T1 ∗ T2? We recall that if f, g, h ∈ C0c (Rn), then
(f ∗ g) ∗ h = f ∗ (g ∗ h).
In deed,
[(f ∗ g) ∗ h](x) =
ˆRn
(f ∗ g)(y)h(x− y) dy
=
ˆRn
[ˆRn
f(z)g(y − z) dz
]h(x− y) dy
=
ˆRn
f(z)
[ˆRn
g(y − z)h(x− y) dy
]dz
=
ˆRn
f(z)
[ˆRn
g(y)h(x− y − z) dy
]dz
=
ˆRn
f(z)(g ∗ h)(x− z) dz
= (f ∗ (g ∗ h))(x).
Motivated by this we define
Definition 10. Suppose that T1, T2 ∈ D′(Rn) and that T2 has compact support. The
convolution T1 ∗ T2 is defined to be the unique distribution that satisfies
(T1 ∗ T2) ∗ φ = T1 ∗ (T2 ∗ φ), ∀φ ∈ D(Rn).
Note that in the definition above, the requirement that T2 has compact support ensures
that T2 ∗ φ is a test function according to the last property in Proposition 7.
Theorem 15. Suppose that T1, T2 ∈ D′(Rn) and that T2 has compact support. Then
Dα(T1 ∗ T2) = T1 ∗DαT2 = DαT1 ∗ T2.
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Proof. We have
Dα(T1 ∗ T2) ∗ φ = (T1 ∗ T2) ∗Dαφ
= T1 ∗ (T2 ∗Dαφ)
= T1 ∗ (DαT2 ∗ φ)
= (T1 ∗DαT2) ∗ φ
so that Dα(T1 ∗ T2) = (T1 ∗DαT2). Similarly,
(T1 ∗DαT2) ∗ φ = T1 ∗ (DαT2 ∗ φ)
= T1 ∗Dα(T2 ∗ φ)
= DαT1 ∗ (T2 ∗ φ)
= (DαT1 ∗ T2) ∗ φ,
so that (T1 ∗DαT2) = (DαT1 ∗ T2) and we are done.
Example 22. (i) If φ ∈ D(Rn), then δ0 ∗ φ = φ.
(ii) If T ∈ D′(Rn) has compact support then δ0 ∗ T = T .
Definition 11 (Definition of the fundamental solution). We say that a distribution G is
a fundamental solution of the partial differential operator
L :=∑|α|≤k
aαDα
where aα are constants, if it satisfies the distributional equation
LG = δ0.
Lemma 5. Suppose that G ∈ D′(Rn) is a fundamental solution of L and that T0 is a
distribution with compact support. Then the distribution G ∗ T0 solves the distributional
equation
LT =∑|α|≤k
aαDαT = T0.
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Proof. We have
L(G ∗ T0) =∑|α|≤k
aαDα(G ∗ T0)
=∑|α|≤k
aα(DαG ∗ T0)
=( ∑|α|≤k
aαDαG)∗ T0
= (LG) ∗ T0
= δ0 ∗ T0 = T0.
4.3 Poisson’s equation
We will apply the abstract framework in the previous section to solve Poisson’s equa-
tion: given f : Rn → R has compact support. Find u such that
∆u = f in Rn.
First of all, we notice that, for n ≥ 3, classical solutions to this problem are unique (if
they exist) if we assume that u→ 0 as |x| → ∞. Indeed, suppose that u1, u2 satisfy both
Poisson’s equation and the decay condition. Then w = u1 − u2 is harmonic in Rn and is
bounded. By Liouville’s theorem, w must be a constant. Since w → 0 as |x| → ∞, the
constant must be 0. Hence, u1 = u2.
Next, we will show that provided ρ ∈ C2c (Rn), a classical solution satisfying the decay
condition indeed exists. The procedure follows from the abstract framework
1. We first show that a distributional solution exists using the fundamental solution.
2. Then we show that the distributional solution arises from a C2 function. This
function will be the classical solution to Poisson’s equation.
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4.3.1 The fundamental solution
Recall that in Chapter 3, we have defined the function
Φ(x) =
1
2πlog |x|, n = 2,
1(2−n)σn−1
|x|2−n, n ≥ 3.
Furthermore, for any φ ∈ C∞c (Rn) we have that
φ(x) = limε→0
ˆRn\Bε(x)
Φ(x− y)∆φ(y) dy.
By changing variable z = x− y, the above identity can be written as
φ(x) = limε→0
ˆRn\Bε(0)
Φ(z)∆φ(x− z) dz = (G ∗∆φ)(x),
where the distribution G is defined by
Gφ = limε→0
ˆRn\Bε(0)
Φ(z)∆φ(z) dz.
So we have φ = G ∗∆φ = ∆G ∗ φ, which means that ∆G = δ0, i.e., G is the fundamental
solution to Poisson’s equation.
As a consequence, if f ∈ C0c (Rn), then
∆(G ∗ Tf ) = Tf
i.e., G ∗ Tf is a solution to the distributional equation ∆T = Tf .
Lemma 6. Suppose f ∈ Ckc (Rn). Then
G ∗ Tf = Tu,
where
u(x) := limε→0
ˆRn\Bε(x)
Φ(x− y) f(y) dy
and u ∈ Ck(Rn).
Proof. Let φ ∈ C∞c (Rn). Let x ∈ Rn and R > 0 sufficiently large that R > |x| and that
75
supp(φ) ∪ supp(f) ⊂ BR(0). We calculate
[(G ∗ Tf ) ∗ φ](x) = [G ∗ (Tf ∗ φ)](x)
= G[τx(Tf ∗ φ)]
= limε→0
ˆRn\Bε(0)
Φ(y)[τx(Tf ∗ φ)](y) dy
= limε→0
ˆRn\Bε(0)
Φ(y)(Tf ∗ φ)(x− y) dy
= limε→0
ˆRn\Bε(0)
Φ(y)[ ˆ
Rn
f(z)φ(x− y − z) dz]dy
= limε→0
ˆB3R(0)\Bε(0)
Φ(y)[ ˆ
BR(0)
f(z)φ(x− y − z) dz]dy
= limε→0
ˆB3R(0)\Bε(0)
Φ(y)[ ˆ
B4R(0)
f(w − y)φ(x− w) dw]dy
= limε→0
ˆB4R(0)
[ ˆB3R(0)\Bε(0)
Φ(y)f(w − y) dy]φ(x− w) dw
= limε→0
ˆB2R(0)
[ ˆRn\Bε(w)
Φ(w − z)f(z) dz]φ(x− w) dw
Define
uε(w) =
ˆRn\Bε(w)
Φ(w − z)f(z) dz.
Then due to the property of Φ, we have that uε → u uniformly in w, so that we can take
the limit inside the integral above and obtain
[(G ∗ Tf ) ∗ φ](x) =
ˆB2R(0)
[limε→0
ˆRn\Bε(w)
Φ(w − z)f(z) dz]φ(x− w) dw
=
ˆB2R(0)
u(w)φ(x− w) dw
= (Tu ∗ φ)(x),
so that G ∗ Tf = Tu.
Next we show that u ∈ Ck(Rn). Since f ∈ Ckc (Rn), we have
uε(x) =
ˆRn\Bε(w)
Φ(w − z)f(z) dz =
ˆBR(0)\Bε(0)
Φ(y)f(x− y) dy.
If |α| ≤ k, then
Dαuε =
ˆB2R(0)\Bε(0)
Φ(y)Dαf(x− y) dy
=
ˆBR(0)\Bε(0)
Φ(x− y)Dαf(y) dy
=
ˆRn\Bε(0)
Φ(x− y)Dαf(y) dy
76
which converges uniformly to some limit by our previous results. Since uε and Dαuε
converge uniformly for |α| ≤ k, we conclude that u ∈ Ck(Rn).
Now we are ready to state the following theorem on the existence of a classical solution
satisfying the decay condition.
Theorem 16. If f ∈ C2c (Rn) then
u(x) = limε→0
ˆRn\Bε(x)
Φ(x− y) f(y) dy
is a classical to Poisson’s equation. Furthermore, if n ≥ 3 then u→ 0 as |x| → ∞.
Proof. By Lemma 6 u ∈ C2(Rn) and Tu is a distributional solution of Poisson’s equation.
By Proposition 6, u is a classical solution of Poisson’s equation. Now let n ≥ 3. Let
R > 0 be such that supp(f) ⊂ BR(0). For |x| > 2R and y ∈ BR(0) then |y| ≤ |x|2
, so that
|x− y| > |x|2
. We then can estimate
|u(x)| =∣∣∣∣ˆBR(0)
Φ(x− y) f(y) dy
∣∣∣∣ ≤ Vol(BR(0))
(n− 2)σn−1
(2
|x|
)n−2
supRn
|f |,
which implies that u(x)→ 0 as |x| → ∞.
4.3.2 Poisson’s equation in a bounded domain
Let Ω ⊂ Rn be open, bounded and satisfies the barrier postulate. Poisson’s equation
in Ω is: given f ∈ C2c (Ω) and g ∈ C0(∂Ω), find u ∈ C2(Ω) ∩ C0(Ω) such that∆u = f, in Ω,
u = g, in ∂Ω.
(4.3)
This problem can be solved in two steps as follows. Each step is a problem that we know
how to solve.
Step 1) Find w ∈ C2(Rn) solve ∆w = f in Rn.
Step 2) Find v ∈ C2(Ω) ∩ C0(Ω) such that∆u = 0, in Ω,
u = g − w, in ∂Ω.
Then u = w + v solve (4.3). Indeed, u ∈ C2(Ω) ∩ C0(Ω) and
∆u = ∆w + ∆v = f in Ω, u = v + w = g − w + w = g on ∂Ω.
77
Problem Sheet 4
Exercise 17. Consider the following domain of R2, defined using polar coordinates (r, ϕ):
Ω0 := (r, ϕ) : θ < ϕ < 2π − θ; 0 < r
where θ ∈ (0, π).
a) Determine, with justification, for which values of θ the exterior sphere condition is
satisfied for every point of ∂Ω0. (Recall: Suppose that Ω is open and bounded and
that y ∈ ∂Ω. If there exists Br(z) such that Br(z) ∩ Ω = y then we say that the
exterior sphere condition holds at y.)
b) Show that a barrier function exists at all points of ∂Ω0, even when the exterior sphere
conditions does not hold. [Hint: Consider the expression for the Laplacian on R2 in
polar coordinates.]
c) An open domain Ω of R2 is said to satisfy the exterior cone condition at a point ξ ∈ ∂Ω
if there exists a cone C with vertex ξ and a ball Br(ξ) such that
Ω ∩Br(ξ) ∩ C = ξ
Prove using the idea for b) that if the exterior cone condition is satisfied at ξ the a
barrier function exists for ξ.
Exercise 18. a) Show that if f1, f2 ∈ C0(Ω) and a ∈ C∞(Ω) then
aTf1 + Tf2 = Taf1+f2 .
b) Show that if f ∈ Ck(Ω) then
DαTf = TDαf
for |α| ≤ k.
78
79
c) Deduce that if f ∈ Ck(Ω) then ∑|α|≤k
aαDαTf = TLf ,
where
Lf =∑|α|≤k
aαDαf.
Exercise 19. a) Show that if f, g, h ∈ C0c (Rn) then
Tf∗g = Tf ∗ Tg.
b) Show that convolution is linear in both of its arguments, i.e, if Ti ∈ D′(Rn) and T3, T4
have compact support then
(T1 + aT2) ∗ T3 = T1 ∗ T3 + aT2 ∗ T3,
and
T1 ∗ (T3 + aT4) = T1 ∗ T3 + aT1 ∗ T4
where a ∈ R is a constant.
Chapter 5
The heat equation
In this chapter we will study the heat equation: given u0 : Rn → R. We want to find
a function u : Rn × [0,∞) that satisfies the following heat equation
∂tu(x, t) = ∆u(x, t), x ∈ Rn, t > 0, (5.1)
u(x, 0) = u0(x), x ∈ Rn. (5.2)
Note that ∆ is the Laplacian operator with respect to the spatial variable only. The heat
equation describes the evolution of the temperature of a homogeneous body given the
temperature profile at the initial time.
The procedure to solve this equation will be similar as in the previous chapter: first, we
construct the fundamental solution, and then taking the convolution of the fundamental
solution with the initial data.
5.1 The fundamental solution of the heat equation
We first have two observations.
Lemma 7. If u(x, t) satisfies (5.1) then so does v(x, t) := u(λx, λ2t) for any λ > 0.
Proof. By the chain rule we have
∂tv(x, t) = λ2∂tu(λx, λ2t) and ∆v(x, t) = λ2∆u(λx, λ2t).
Thus ∂tv(x, t) = λ2[∂tu−∆u](λx, λ2t) = 0 as desired.
80
81
Lemma 8. If u satisfies (5.1) and Du vanishes at the infinity then for any t > 0
ˆRn
u(x, t) dx =
ˆRn
u0(x) dx.
Proof. We calculate
d
dt
ˆRn
u(x, t) dx =
ˆRn
∂tu(x, t) dx
=
ˆRn
∆u(x, t) dx
=
ˆRn
div(Du(x, t)) dx
= 0,
where in the last equality we have applied the divergence theorem and the assumption
that Du vanishes at the infinity.
Motivated by these two observations, we will seek a self-similar solution of the heat
equation of the form
u(x, t) = h(t)f(ξ) where ξ =|x|2
t∈ R,
for some function h and f , such that´Rnu(x, t) dx =
´Rnu0(x) dx = 1. The last identity
is assumed without loss of generality.
We first find h. We have
1 =
ˆRn
u(x, t) dx = h(t)
ˆRn
f( |x|2
t
)dx
= h(t)
ˆSn−1
ˆ ∞0
f(r2
t
)rn−1 dw dr.
By changing of variable y = r2
t, we have r = (yt)
12 and dr = t
2rdy = t
12
2y12dy. Substituting
this back into the calculations above, we get
1 = tn2 h(t)
σn−1
2
ˆ ∞0
f(y)yn2−1 dy.
This implies that
h(t) = t−n2 and 1 =
σn−1
2
ˆ ∞0
f(y)yn2−1 dy.
82
It remains to find f . We have
∂tu = ∂t
(t−
n2 f( |x|2
2
))= −n
2t−
n2−1f(ξ)− t−
n2|x|2
t2f ′(ξ) = −t−
n2−1[n
2f(ξ) + ξf ′(ξ)
]Diu = 2t−
n2xitf ′(ξ),
D2iiu =
2tn2
tf ′(ξ) + 4t−
n2x2i
t2f ′′(ξ)
∆u =n∑i=1
D2iiu = t−
n2−1 (2nf ′(ξ) + 4ξf ′′(ξ)) .
Therefore, we obtain the following equation for f(ξ)
4ξf ′′(ξ) + ξf ′(ξ) + 2nf ′(ξ) +n
2f(ξ) = 0.
Set g(ξ) = f ′(ξ) + 14f(ξ), the above equation can be written as
ξg′(ξ) +n
2g(ξ) = 0,
which implies that
g(ξ) = Aξ−n2
for some constant A. Therefore,
f ′(ξ) +1
4f(ξ) = Aξ−
n2 ,
i.e.,
A = ξn2 (f ′(ξ) +
1
4f(ξ)).
Assume that f and f ′ decay fast enough at the infinity, we obtain that the right-hand
side vanishes as |ξ| → ∞. Thus A = 0 and so
f ′(ξ) +1
4f(ξ) = 0.
By multiplying with the integrating factor eξ4 , we obtain
d
dξ
(eξ4f(ξ)
)= 0
Solving this equation, we find
f(ξ) = Be−ξ4 . (5.3)
83
We now find B. We have
1 = Bσn−1
2
ˆ ∞0
e−ξ4 ξ
n2−1 dξ
= Bσn−14n2
ˆ ∞0
e−s2
sn−1 ds (change of variable ξ = 4s2)
= B4n2
ˆRn
e−|x|2
dx
= B
(2
ˆ ∞0
e−x2
dx
)n= B(2
√π)
n2 .
so B = 1(2√π)n
. Conclusion
u(x, t) = t−n2
1
(2π)ne−|x|24t =
1
(4πt)n2
e−|x|24t .
To summarise, we have proved the following result
Proposition 8. The function
Γ(x, t) =1
(4πt)n2
e−|x|24t
is a solution of the equation (5.1) and satisfies
ˆRn
Γ(x, t) dx = 1, ∀t > 0.
The function Γ is called the heat kernel or the fundamental solution of the heat equa-
tion. It plays a similar role to the Green’s function for the Laplace operator, which can
be seen in the following theorem.
Theorem 17. Let u0 ∈ C0c (Rn). Define u : Rn × (0,∞)→ R by
u(x, t) :=
ˆRn
Γ(x− y, t)u0(y) dy =1
(4πt)n2
ˆRn
e−|x−y|2
4t u0(y) dy. (5.4)
Then u solves the heat equation (5.1)-(5.2). That is, u satisfies
∂tu(x, t) = ∆u(x, t) ∀(x, t) ∈ Rn × (0,∞),
and
limt→0
u(x, t) = u0(x) ∀x ∈ Rn.
Proof. According to Proposition 8, Γ(x− y, t) satisfies
∂tΓ(x− y, t) = ∆Γ(x− y, t).
84
Furthermore, as a function of x and t, Γ(x− y, t) is smooth in Rn× (0,∞). Suppose that
u0 is supported in BR(0). Then
u(x, t) =
ˆBR(0)
Γ(x− y, t)u0(y) dy.
This implies that u(x, t) is also smooth in Rn× (0,∞) and we can differentiate under the
integral and obtain that
(∂tu−∆u)(x, t) =
ˆBR(0)
(∂t −∆x)Γ(x− y, t)u0(y) dy = 0 ∀(x, t) ∈ Rn × (0,∞).
We now verify the initial condition. Since u0 is continuous and compactly supported, it
is uniformly continuous. Therefore, given any ε > 0 there exists δ > 0 such that for any
x, z ∈ Rn with |x− z| < δ then |u0(x)− u0(z)| < ε. Since
ˆRn
Γ(x− y, t) dy = 1,
we can write
u(x, t)− u0(x) =
ˆRn
Γ(x− y, t)(u0(y)− u0(x)) dy
and estimate
|u(x, t)− u0(x)| ≤ˆRn
Γ(x− y, t)|u0(y)− u0(x)| dy
=
ˆBδ(x)
Γ(x− y, t)|u0(y)− u0(x)| dy +
ˆRn\Bδ(x)
Γ(x− y, t)|u0(y)− u0(x)| dy.
:= I1 + I2.
Since for y ∈ Bδ(x), we have |u0(y)− u0(x)| < ε, the term I1 can be estimated
I1 ≤ ε
ˆBδ
Γ(x− y, t) dy ≤ ε
ˆRn
Γ(x− y, t) dy = ε.
The term I2 can be estimated as follows
I2 ≤ 2 supRn
|u0|ˆBδ(x)
Γ(x− y, t) dy
= 2 supRn
|u0|ˆBδ(0)
Γ(y, t) dy
= 2 supRn
|u0|σn−1
(4πt)n2
ˆ ∞δ
e−s2
4t sn−1 ds
= 2 supRn
|u0|σn−1
(π)n2
ˆ ∞δ
2√t
e−z2
zn−1 dz → 0 as t→ 0.
85
Therefore, there exists δ′ > 0 depending on δ (thus on ε but not on x) such that I2 < ε
for t < δ′. In conclusion, we have shown that for any ε > 0 there exists δ′ > 0 such that
for t < δ′ and for any x, we have
|u(x, t)− u0(x)| < 2ε.
So limt→0 u(x, t) = u0(x) as claimed.
We now show that the solution of the heat equation is always bounded.
Theorem 18. Let u ∈ C0c (Rn) and let u(x, t) be defined as in (5.4). Then
|u(x, t)| ≤ maxRn|u0| ∀(x, t) ∈ Rn × (0,∞).
Proof. This estimate is straightforward:
|u(x, t)| ≤ˆRn
Γ(x− y, t)|u0(y)| dy ≤ maxRn|u0|
ˆRn
Γ(x− y, t) dy = maxRn|u0|.
Furthermore, we now show that the solution of the heat equation decays in time.
Theorem 19. Let u ∈ C0c (Rn) and let u(x, t) be defined as in (5.4). Then there exists a
constant C depending on n and u0 such that
|u(x, t)| ≤ C
tn2
∀(x, t) ∈ Rn × (0,∞).
Proof. Suppose that u0 is supported in BR(0). Since Γ(x−y, t) ≤ 1
(4πt)n2
, we can estimate
|u(x, t)| ≤ˆBR(0)
Γ(x− y, t)|u0(y)| dy
≤ maxRn|u0|
ˆBR(0)
Γ(x− y, t) dy
≤ maxRn|u0|
1
(4πt)n2
Vol(BR(0)) =C
tn2
,
where C = maxRn|u0| 1
(4π)n2
Vol(BR(0).
Theorem 17 only tells us that the function u defined by (5.4) is a solution to the heat
equation, but does not ensure it uniqueness. To obtain a statement about the uniqueness,
we will need to use the maximum principle in the next sections.
86
5.2 The maximum principle for the heat equation on
a bounded domain
We will show that the heat equation also satisfies a similar maximum principle as
harmonic functions.
Definition 12. Let Ω ⊂ Rn be open and bounded and T > 0 be given. The space-time
domain ΩT is defined by
ΩT = Ω× (0, T ).
The parabolic boundary of ΩT is defined by
∂pΩT = ∂(ΩT ) \ (Ω× T).
We denote by C2,1(ΩT ) the set of functions that are continuously differentiable twice
in space and once in time. For functions u ∈ C2,1(ΩT ), we define the heat operator L by
(Lu)(x, t) := ∂tu(x, t)−∆u(x, t).
We have the following maximum principle.
Theorem 20 (The maximum principle for the heat operator on bounded domains).
Suppose that Ω ⊂ Rn is open and bounded, T > 0 and that u ∈ C2,1(ΩT ) ∩ C0(ΩT )
satisfies
Lu ≤ 0 in ΩT .
Then
supΩT
u = sup∂pΩT
u.
If instead Lu ≥ 0 in ΩT , then
infΩTu = inf
∂pΩTu.
Proof. Suppose that Lu ≤ 0. Let 0 < ε < T be arbitrary and let us define the function
v : ΩT−ε → R
(x, t) 7→ u(x, t)− εt.
87
The function v satisfies
v(x, t) ≤ u(x, t) ≤ v(x, t) + εT, and (5.5)
vt −∆v ≤ −ε < 0 in Ω× (0, T − ε]. (5.6)
Since v is continuous, it attains a global maximum at some point (x0, t0) ∈ ΩT−ε.
1. If (x0, t0) ∈ ΩT−ε then we must have that ∂tv(x0, t0) = 0 and Hessxv(x0, t0) ≤ 0. So
(vt −∆v)(x0, t0) = −∆v(x0, t0) ≥ 0,
which contradicts (5.6).
2. If (x0, t0) ∈ Ω×T−ε then we must have that vt(x0, t0) ≥ 0 (since v must decrease if
we fix x and allow t to decrease) and Hessxv(x0, t0) ≤ 0 (since the function v(T−ε, x)
must have a local max at x0). Therefore,
(vt −∆v)(x0, t0) ≥ 0,
again contradicts (5.6).
Therefore, we have that (x0, t0) ∈ ∂pΩT−ε and hence
supΩT−ε
v = sup∂pΩT−ε
v ≤ sup∂pΩT−ε
u ≤ sup∂pΩT
u.
On the other hand, since u ≤ v + εT , we have
supΩT−ε
u ≤ supΩT−ε
v + εT ≤ εT + sup∂pΩT
u. (5.7)
Since u is uniformly continuous on ΩT , we have that
limε→0
supΩT−ε
u = supΩT
u.
Thus taking ε→ 0 in (5.7), we obtain that
supΩT
u = limε→0
supΩT−ε
u ≤ limε→0
(εT + sup∂pΩT
u) = sup∂pΩT
u ≤ supΩT
u. (5.8)
Therefore, all of the inequalities in (5.8) can be replaced with inequalities. In particular,
we get
sup∂pΩT
u = supΩT
u
as desired.
To obtain the result for Lu ≥ 0, we note that −u satisfies L(−u) ≤ 0 and apply the
first part of the theorem.
88
Corollary 5 (The maximum principle for solutions of the heat equation on bounded
domain). Suppose u ∈ C2,1(ΩT ) ∩ C0(ΩT ) solves the heat equation in ΩT . Then
maxΩT
|u| = max∂pΩT|u|.
Corollary 6. Let f ∈ C0(ΩT ) and g ∈ C0(∂pΩT ). Then there is at most one solution
u ∈ C2,1(ΩT ) ∩ C0(ΩT ) of the boundary value problemut −∆u = f in ΩT ,
u = g on ∂pΩT ,
Proof. Suppose there are two solutions u1 and u2. Then u = u1−u2 satisfies the equationut −∆u = 0 in ΩT ,
u = 0 on ∂pΩT ,
Apply Corollary 5 we obtain that
maxΩT
|u| = max∂pΩT|u| = 0,
which implies that u ≡ 0, i.e., u1 = u2 in ΩT .
Note that unlike the Dirichlet problem for harmonic functions, we are not free to
prescribe the value of u on all of ∂pΩT . The restriction is because of the maximum
principle: we can not specify the boundary of u such that u achieves a value on Ω× T
that is strictly greater than the value it achieves on ∂pΩT .
5.3 The maximum principle for the heat operator on
the whole space
In this section, we show a maximum principle for the heat operator in the whole
space. We need to make an assumption on the behaviour of u at the infinity. Denote by
ST = Rn × (0, T ) the space-time domain in this case.
Theorem 21 (The maximum principle for the heat operator on ST ). Let u ∈ C2,1(ST )∩
C0(ST ) satisfy Lu ≤ 0 in ST . In addition, assume that there exist constants C, α > 0
such that
u(x, t) ≤ Ceα|x|2
, ∀x ∈ Rn, 0 ≤ t ≤ T.
89
Then
supST
u = supx∈Rn
u(x, 0).
If instead, Lu ≥ 0 in ST and there exist constants C, α > 0 such that
u(x, t) ≥ −Ceα|x|2 , ∀x ∈ Rn, 0 ≤ t ≤ T.
Then
infSTu = inf
x∈Rnu(x, 0).
Proof. We will prove the minimum principle. The maximum principle then follows by
changing u to −u. Suppose that Lu ≥ 0 in ST and there exist constants C, α > 0 such
that
u(x, t) ≥ −Ceα|x|2 , ∀x ∈ Rn, 0 ≤ t ≤ T.
We need to show that
infSTu = inf
x∈Rnu(x, 0).
If u(x, 0) is not bounded from below then the assertion trivially holds. Therefore, we can
assume that u(x, 0) is bounded from below and define I := infx∈Rn u(x, 0). Let β > α
such that T1 := 18β< T . The idea of the proof is as follows. We first show that u ≥ I in
ST1 and then by dividing the time interval [0, T ] into finitely many intervals of length less
than T1 and successively applying the result of the first step.
We define
v(x, t) :=eβ
|x|21−4βt
(1− 4βt)n2
.
By direct computations, we have
∂tv = 4β(1− 4βt)−n2−1
(n
2+
β|x|2
1− 4βt
)eβ|x|21−4βt ,
∂xiv = (1− 4βt)−n2−12βxie
β|x|21−4βt ,
∂2xixi
v = 4β(1− 4βt)−n2−1
(1
2+
βx2i
1− 4βt
)eβ|x|21−4βt ,
∆v =n∑i=1
∂2xixi
v = 4β(1− 4βt)−n2−1
(n
2+
β|x|2
1− 4βt
)eβ|x|21−4βt ,
which implies that v(x, t) satisfies the heat equation in ST1 . In addition, since (1−4βt)−1 >
1 in ST1 , we have that
v(x, t) ≥ eβ|x|2
.
90
Let ε > 0 and set w := u + εv − I. To prove u ≥ I in ST1 , it suffices to prove that
w ≥ 0 in ST1 for any ε > 0. The function w satisfies that Lw = Lu+ εLv ≥ 0 in ST1 and
w(x, 0) = u(x, 0) + εv(x, 0)− I ≥ 0 for all x ∈ Rn. By the assumption on u we have that
w = u+ εv − I ≥ εeβ|x|2 − Ceα|x|2 − I,
from where we deduce that
inf∂Br(0)×[0,T1]
≥ εeβr2 − Ceαr2 − I.
Since β > α the term εeβr2
dominates for r large enough, i.e., there exists R ≥ 0 such
that w ≥ 0 on ∂Br(0)× [0, T1] for all r > R.
On the parabolic boundary of the cylinder Br(0)× (0, T1), we have w ≥ 0. Therefore,
applying Theorem 20 to this cylinder, we obtain that
infBr(0)×(0,T1)
w = inf∂p(Br(0)×(0,T1))
w ≥ 0,
so that w ≥ 0 in Br(0)× (0, T1) for any r > R. Thus w ≥ 0 in ST1 for any ε > 0 and so
u ≥ I. Now dividing the interval [0, T ] into finitely many intervals of lengths less than T1
and successively applying the result on each interval, we obtain
u(x, t) ≥ infy∈Rn
u(y, 0)
for all (x, t) ∈ ST . Taking the infimum over ST we get that
infSTu ≥ inf
x∈Rnu(x, 0).
On the other hand, it is obvious that
infSTu ≤ inf
x∈Rnu(x, 0).
Therefore, we have infST u = infx∈Rn u(x, 0) as desired.
Corollary 7. Suppose that u ∈ C2,1(ST ) ∩ C0(ST ) and there exist constants C, α > 0
such that
|u(x, t)| ≤ Ceα|x|2
, ∀x ∈ Rn, 0 ≤ t ≤ T.
Suppose further that u solves the heat equation
∂tu(x, t) = ∆u(x, t), x ∈ Rn, t > 0,
u(x, 0) = u0(x), x ∈ Rn,
where u0 ∈ C0(Rn) is bounded. Then u is unique.
91
Proof. Suppose that there are two functions u1 and u2 satisfying the hypotheses of the
theorem. Define w := u1 − u2. Then
|w(x, t)| ≤ |u1(x, t)|+ |u2(x, t)| ≤ Ceα|x|2
+ C ′eα′|x|2 ≤ C ′′eα
′′|x|2 ,
where C ′′ = 2 maxC,C ′ and α′′ = maxα, α′. In addition, w solveswt −∆w = 0 in Rn × (0, T ),
w = 0 on Rn × 0.
Applying Theorem 21 and noting that Lu = 0, we conclude that w = 0 and hence u1 = u2
as desired.
Note that if the growth assumption on u, |u(x, t)| ≤ Ceα|x|2
is removed then the
uniqueness statement above does not hold. A counter example for this was constructed
by Tychonoff in 1935.
Problem Sheet 5
Exercise 20. Suppose that n ≥ 3 and G is given by
G(x) =1
σn−1(2− n)|x|2−n.
Let ρ ∈ C2c (Rn) be supported in BR(0) and define u by
u(x) = limε→0
ˆRn\Bε(x)
G(x− y)ρ(y) dy.
Show that for |α| ≤ 2,
supRn
|Dαu| ≤ R2
2(n− 2)supRn
|Dαρ|.
Deduce that the problem of finding u ∈ C2(Rn) such that u→ 0 as |x| → ∞ satisfying
∆u = ρ in Rn
is well-posed.
Exercise 21 (The inhomogeneous heat equation). (a) Let φ ∈ C∞c (Rn × R). Show that
limt→0
ˆRn
Γ(x, t)φ(x, t) dx = φ(0, 0),
where Γ denotes the fundamental solution of the heat equation.
Deduce that the function
t 7→ˆRn
Γ(x, t)φ(x, t) dx,
is uniformly continuous on (0,∞) and vanishes for large values of t.
(b) Define the distribution T ∈ D′(Rn × R) by
Tφ =
ˆ ∞0
ˆRn
Γ(x, t)φ(x, t) dx dt, ∀φ ∈ C∞c (Rn × R).
Let δ > 0. Show that, for any φ ∈ C∞c (Rn × R), we have
DtT (φ) = −ˆ δ
0
ˆRn
Γ(x, t)φt(x, t) dx dt+
ˆRn
Γ(x, δ)φ(x, δ)dx+
ˆ ∞δ
Γt(x, t)φ(x, t) dx dt,
92
93
and
DxixjT (φ) =
ˆ δ
0
ˆRn
Γ(x, t)φxixj(x, t) dx dt+
ˆ ∞δ
Γxixj(x, t)φ(x, t) dx dt.
(c) Deduce that for any δ > 0, we have
(DtT −∆T )(φ) = −ˆ δ
0
ˆRn
Γ(x, t)(φt + ∆φ)(x, t) dx dt+
ˆRn
Γ(x, δ)φ(x, δ) dx.
By bounding the first integral and making use of part (a), show that
(DtT −∆T )(φ) = φ(0, 0).
(d) Find a distributional solution to the inhomogeneous heat equation
ut −∆u = f in Rn × R,
where f ∈ C∞c (Rn × R).
Exercise 22. Let Ω = x ∈ Rn : xn > 0 be the half-space. Consider the boundary value
problem ut = ∆u in Ω× (0, T )
u = 0 on ∂Ω× (0, T )
u = u0 on Ω× 0.
Suppose that u0 ∈ C0c (Ω). By considering a reflection in the plane xn = 0, find a solution
u.
Acknowledgements
I would like to thank Dr. Claude Warnick, who lectured this course from 2013-2014,
for kindly sharing his notes1, on which these notes (and exercise sheets) are largely based.
1which was in turn based on notes of Andrea Malchiodi, who lectured this course from 2011-2013
94
Appendix
We review relevant knowledge about differentiation of functions of several variables.
Suppose Ω is an open subset of Rn. f : Ω→ Rm is said to be differentiable at a point
x ∈ Ω if there exists a linear map, Df(x) from Rn to Rm such that if h ∈ Rn, we have
f(x+ h) = f(x) +Df(x) · h+ o(h), as h→ 0. (5.9)
We relate this definition to the usual partial derivatives. If f = (f i)mi=1, i.e., f i is the
i-th component of f(x) with respect t the canonical basic of Rm, we define the partial
derivative ∂f i
∂xj(x) to be the limit (if exists)
∂f i
∂xj(x) = lim
ε→0
f(x1, . . . , xj + ε, . . . , xn)− f(x1, . . . , xj, . . . , xn)
ε.
In other words, the partial derivative of f i in the xj direction is simply the derivative of
f i when considering it as a function of xj alone. By setting h = εej in (5.9), we see that
if f is differentiable at x then ∂f i
∂xj(x) exists for all i = 1, . . . ,m; j = 1, . . . , n. To obtain
the conserve, we need a bit stronger condition.
Lemma 9. Suppose that the partial derivative ∂f i
∂xj(x) exist and are continuous on an open
neighbourhood of x, then f is differentiable at x.
Since Df(x) is a linear map, we can represent it with respect to the canonical bases
for Rn and Rm as a matrix. Then the partial derivatives of f are the components of this
matrix
Df(x) =
∂f1∂x1
(x) ∂f1∂x2
(x) . . . ∂f1∂xn
(x)
∂f2∂x1
(x) ∂f2∂x2
(x) . . . ∂f2∂xn
(x)
......
. . ....
∂fm∂x1
(x) ∂fm∂x2
(x) . . . ∂fm∂xn
(x)
95
96
or in a compact way
[Df(x)]ij =∂f i
∂xj.
If f is differentiable at each point in Ω, we can think of Df as a map from Ω to the space
of m× n matrices which is identified to Rm×n
Df : Ω→ Rm×n
x 7→ Df(x).
If this map is continuous, the we say that f ∈ C1(Ω). Equivalently f ∈ C1(Ω) if all its
partial derivatives exist and are continuous in Ω.
Similarly, we say that f ∈ C2(Ω) if Df ∈ C1(Ω) and so on inductively. We can think
of D2f as a map from Ω to Rm × Rn × Rn, then its components are
[D2f(x)]ij1j2 =∂2f i
∂xj1∂xj2(x).
Lemma 10. If f ∈ Ck(Ω) if and only if all the partial derivatives
∂kf i
∂xj1∂xj2 . . . ∂xjk: U → R
exist and are continuous in Ω.
We now introduce multi-indices notations. Let α = (α1, . . . , αm) ∈ (Z≥0)n, we define
|α| :=∑n
i=1 αi and∂|α|f
∂xα=( ∂
∂x1
)α1. . .( ∂
∂xn
)αn,
i.e., we differentiate α1 times with respect to x1, α2 times with respect to x2 and so on.
We also use more compact notation
Di :=∂
∂xi, Dα :=
∂|α|
∂xα.
Similarly for h = (h1, . . . , hn), we denote
hα = hα11 . . . hαnn .
Finally, we introduce the multi-index factorial α! := α1! . . . αn!.
Theorem 22 (Multivariate Taylor’s theorem). Suppose that Ω is an open subset of Rn,
x ∈ Ω and f : U → R belongs to Ck(Ω). Then writing h = (hj)nj=1 and using multi-indices
notation, we have
f(x+ h) = f(x) +∑|α|≤k
hα
α!Dαf(x) +Rk(x, h),
97
where the sum is taken over all multi-indices α = (α1, . . . , αn) with |α| ≤ k and the
remainder term satisfiesRk(x, h)
|h|k→ 0 as h→ 0.
Bibliography
[1] Fritz J. Partial Differential Equations. Springer, 1982.
[2] L.C. Evans. Partial Differential Equations. American Mathematical Society, 1998.
[3] E. Dibenedetto. Partial Differential Equations. Birkhauser, 2010.
[4] D. Gilbarg and N. Trudinger. Elliptic Partial Differential Equations of Second Order.
Springer, 2001.
98
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