MA3G1: Theory of Pdes

MA3G1: Theory of PDEs

Dr. Manh Hong Duong Mathematics Institute, University of Warwick Email: [email protected]

January 4, 2017 Abstract

The important and pervasive role played by pdes in both pure and applied mathematics is described in MA250 Introduction to Partial Diﬀerential Equations. In this module I will introduce methods for solving (or at least establishing the existence of a solution!) various types of pdes. Unlike odes, the domain on which a pde is to be solved plays an important role. In the second year course MA250, most pdes were solved on domains with symmetry (eg round disk or square) by using special methods (like separation of variables) which are not applicable on general domains. You will see in this module the essential role that much of the analysis you have been taught in the ﬁrst two years plays in the general theory of pdes. You will also see how advanced topics in analysis, such as MA3G7 Functional Analysis I, grew out of an abstract formulation of pdes. Topics in this module include:

• Method of characteristics for ﬁrst order PDEs.

• Fundamental solution of Laplace equation, Green’s function.

• Harmonic functions and their properties, including compactness and regularity.

• Comparison and maximum principles.

• The Gaussian heat kernel, diﬀusion equations.

• Basics of wave equation (time permitting).

Aims: The aim of this course is to introduce students to general questions of existence, uniqueness and properties of solutions to partial diﬀerential equations.

Objectives: Students who have successfully taken this module should be aware of several diﬀerent types of pdes, have a knowledge of some of the methods that are used for discussing existence and uniqueness of solutions to the Dirichlet problem for the Laplacian, have a knowledge of properties of harmonic functions, have a rudimentary knowledge of solutions of parabolic and wave equations.

Further reading. The contents of this course are basic and can be found in many textbooks. The bibliography lists some books for further reading. Contents

1 Introduction 3

1.1 What are partial diﬀerential equations ...... 3

1.2 Examples of PDEs ...... 4

1.3 Well-posedness of a PDE ...... 7

1.4 Classiﬁcation of PDEs ...... 8

2 Quasilinear ﬁrst order PDEs 10

2.1 Introduction ...... 10

2.2 The method of characteristic ...... 12

2.2.1 The characteristic equations ...... 13

2.2.2 Geometrical interpretation ...... 14

2.3 Compatibility of initial conditions ...... 18

2.4 Local existence and uniqueness ...... 19

2 Laplace’s equation and harmonic functions 23

2.1 Laplace’s Equation ...... 23

2.2 Harmonic functions ...... 26

2.3 Mean value property for harmonic functions ...... 26

2.4 Smoothness and estimates on derivatives of harmonic functions ...... 28

2.5 The maximum principle ...... 32

2.5.1 Subharmonic and superharmonic functions ...... 32

2.5.2 Some properties of subharmonic and superharmonic functions . . . 33

1 2

2.5.3 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

h i 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.

3 4

1.2 Examples of PDEs

PDEs are ubiquitous and appear in almost all ﬁelds in applied sciences. We list here just some popular examples. Some of these will be treated in the subsequent chapters.

Example 1 (The continuity/transport equation). In general, a continuity equation describes the transport of some moveable quantity. For example in ﬂuid dynamics, the continuity equation reads

∂u (x, t) + div(u(t, x)v(t, x)) = 0, (1.2) ∂t where u(x, t) is the fluid density at a point x ∈ R3 and at a time t, v(x, t) is the flow velocity vector field and div denotes the divergence operator. Note that the above equation can be written as ∂u (x, t) + Du(t, x) · v(t, x) + u(t, x) div(v(t, x)) = 0, ∂t Example 2 (The Laplace and Possion equations). The Laplace equation is a second-order partial differential equation given by

d X ∂2u ∆u = = 0. (1.3) ∂x2 i=1 i 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 = Q · dσ = div(Q) dx, ˆ∂Ω ˆΩ where we have used the divergence theorem to obtain the second equality. This implies that div Q must vanish identically div Q = 0.

By Fourier’s law of heat conduction, the heat ﬂux is proportional to the temperature gradient, so that Q = k(x)Du(x), where k is the conductivity. If the body is homogeneous, then k is constant and we deduce that

0 = div Q = div(kDu(x)) = k∆u(x). 5

Thus the temperature satisﬁes the Laplace equation.

In conclusion: the Laplace equation can be used to describe the temperature distribution of a solid body at thermal equilibrium.

The Poisson equation is the Laplace equation with an inhomogeneous right-hand side,

∆u(x) = f(x), (1.4) for some given function f. For instance, the Poisson equation appears in electrostatics. For a static electric ﬁeld E, Maxwell’s equations reduce to

curl E = 0, div E = ρ, where ρ is the charge density (in appropriate units). By Helmholtz decomposition, the ﬁrst equation implies that E = −Du, for some scalar electric potential ﬁeld u. By substituting this into the second equation, we obtain the Possion equation (1.4) with f = −ρ.

Example 3 (The heat/diffusion equation). In example 2, if the (homogeneous) body is not in equilibrium, the the heat flux through the boundary of a region will equal the rate of change of internal energy of the material inside that region. If no work is done by the heat flow, then the rate of change of the internal energy is

dE ∂u = (t, x) dx = Q · dσ = div(Q) dx = k ∆u(t, x) dx. dt ˆΩ ∂t ˆ∂Ω ˆΩ ˆΩ This implies that the temperature u satisﬁes the heat equation

∂u (t, x) = k∆u(t, x). (1.5) ∂t

This equation also can be used to describe other diffusive processes such as a diffusion process (and called diffusion equation).

Example 4 (The wave equation). The wave equation is a second-order linear partial differential equation ∂2u (t, x) = c2∆u(t, x). (1.6) ∂t2 Here u is used to model, for example, the mechanical displacement of a wave. The wave equation arises in many fields like acoustics, electromagnetics, and fluid dynamics. 6

Example 5 (The Kramers equation). The Kramer equation (or kinetic Fokker-Planck equation) describes the time evolution of the probability density of a particle moving under the inﬂuence of an external potential, a friction and a stochastic noise given by p p ∂ ρ + · ∇ ρ − ∇V · ∇ ρ = γ div ρ + β−1∇ ρ , t m q q p m p 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 p ∂ ρ + · ∇ ρ − ∇V · ∇ ρ = Q(ρ, ρ) t m q q 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,

0 ∂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 ).

1 m When V = 0,U(ρ) = m−1 ρ , we obtain the porous medium equation

m ∂tρ = ∆ρ .

Example 7 (The Allen-Cahn equation). The Allen-Cahn euqation is a reaction-diﬀusion equation and describes the process of phase separation

0 ∂tρ = D∆ρ − f (ρ),

1 4 1 2 where f is some free energy functional. A typical example of f is f(ρ) = 4 ρ − 2 ρ .

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.

2 0 ∂tρ = −D∆ u + ∆f (ρ). 7

1.3 Well-posedness of a PDE

In all of the examples in the previous section, we need further information in order to solve the problem. We broadly refer to this information as the data for the problem.

Ex. 1 We can specify the density distribution at some initial time t0. This is an example of Cauchy data (or an initial value problem).

Ex. 2 For the Laplace equation on a bounded domain, we need to prescribe the value of u on the boundary of the domain. This is an example of Dirichlet data (a Dirichlet problem). For the Poisson equation, we specify the function f and also the value of u on the boundary.

Ex. 3 For the heat equation, we specify the initial temperature of the body, as well as some boundary condition. For example, a Dirichlet boundary condition when the temperature at the boundary is prescribed or a Neumann boundary condition when the ﬂux through the boundary is imposed.

An important aspect of the study of a PDE is about its wellposedness. A PDE problem, consisting of a PDE together with some data is well-posed if a) There exists a solution to the problem (existence). b) For given data, the solution is unique (uniqueness). c) The solution depends continuously on the data.

These are called Hadamard’s conditions. A problem is said to be ill-posed if any of the condition above fails to hold.

Notice that existence and uniqueness involves boundary conditions. While continuous dependence depends on considered metric/norm.

Wellposedness is important because for the vast majority of PDE problems that we encounter, it is not possible to write down a solution explicitly. However, if well-posedness is satisﬁed, we can often deduce properties of that solution directly from the PDE it satisﬁes without ever needing an explicit solution.

While the issue of existence and uniqueness of solutions of ordinary differential equations has a very satisfactory answer with the Picard-Lindelöftheorem, that is far from 8 the case for partial differential equations. In general, the answers to the above questions depend heavily on the PDE itself as well as the domain Ω. Definition 1.1 is often too general to provide any useful information. In this notes, we will be working with much simpler examples of PDEs having explicit forms, and nevertheless exhibit lots of interesting behaviour.

Example 9. Consider the following heat equation on (0, 1).  u = u  t xx  u(0, t) = u(1, t) = 0,   u(x, 0) = u0(x). This PDE is well-posed.

Example 10. Consider the following backwards heat equation  u = −u  t xx  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 derivatives, so that we can re-write (1.1) as (see Appendix 5.3 for the multi-index notation) X ∂|α|u a (x) = f(x). α ∂xα |α|≤k If f = 0, we say the equation is homogeneous.

ii) We say that (1.1) is semilinear if it is of the form X ∂|α|u h i a (x) + a x, u(x), Du(x),D2u(x), ··· ,Dk−1u(x) = 0, α ∂xα 0 |α|=k 9

so that the highest order derivatives of u appear linearly, with coeﬃcients depending only on x but not on u and its derivatives.

iii) We say that (1.1) is quasilinear if it is of the form

X h i∂|α|u a x, u(x), Du(x),D2u(x), ··· ,Dk−1u(x) α ∂xα |α|=k

h 2 k−1 i + a0 x, u(x), Du(x),D u(x), ··· ,D u(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, y) = 0 in Ω. ∂x Solutions to this equation depend on the domain Ω. For instance, if Ω = [0, 1] × [0, 1] then the value of u on {0} × [0, 1] will determine u everywhere in Ω. In fact, suppose that u(0, y) = h(y) with h continuous on [0, 1], then using the fundamental theorem of calculus we have, x ∂u u(x, y) = u(0, y) + (s, y) ds = h(y). ˆ0 ∂s

10 11

1 Ω

1 x

Figure 2.1: The domain Ω = [0, 1] × [0, 1]

Figure 2.2: [L] A domain on which ux = 0 is solvable with data on {0} × [0, 1]. [R] A domain on which ux = 0 is not solvable with data on {0} × [0, 1].

This means that u does not change if we move along a curve of constant y and we say that the value of u propagates along lines of constant y (Figure 2.1).

How about other domains? When is prescribing the value of u on {0} × [0, 1] enough to determine u everywhere? Obviously for this approach to work, we must be able to connect every point in Ω to {0} × [0, 1] by a horizontal line which lies in Ω. See Figure 2.1 for diﬀerent situations. Note that while for a classical solution of a ﬁrst order equation we would require u ∈ C1(Ω), the procedure above in fact makes sense even if h(y) is only

0 continuous, so that uy need not to be C . This suggests that we should consider a notion of solutions for some PDEs which is weaker than classical solutions. 12

2.2 The method of characteristic

Let us take another example.

Example 12 (Transport equation with constant speed). Let c ∈ R and h : R → R be a given function. Consider the following PDE

cux + uy = 0, u(x, 0) = h(x). (2.2)

Solution. The solution consists of three main steps

Step 1) Change the co-ordinate system (x, y) (s, t) by writing  x = x(s, t)

y = y(s, t)

and deﬁne z(s, t) := u(x(s, t), y(s, t)).

Step 2) Solve the equation for z in the new co-ordinate system (s, t), which by construc- tion will be easier to solve.

Step 3) Transform back to the original co-ordinate system  s = s(x, y)

t = t(x, y)

and u(x, y) = z(s(x, y), t(x, y)).

By the chain rule, we have dz dx dy = u + u . dt dt x dt y

We want that the right-hand side of the expression above is equal to cux +uy that appears in the PDE. Therefore, we obtain the following system of ODEs

dx = c dt dy = 1. dt

Solving this system, we obtain x(s, t) = ct + A(s), y(s, t) = t + B(s). To ﬁnd A(s),B(s) we impose that (t = 0) ⇒ (y = 0), (t = 0) ⇒ (x = s), 13 from which we deduce that A(s) = s, B(s) = 0, hence, x(s, t) = ct + s and y(s, t) = t. In the new co-ordinate system, we now have dz = 0, z(s, 0) = h(s). dt Solving this equation gives z(s, t) = h(s).

We now transform back to (x, y) and u(x, y). It is straightforward to ﬁnd s, t in terms of x, y s = x − cy, t = y.

Therefore, u(x, y) = z(s(x, y), t(x, y)) = h(x − cy). This is the solution of the transport equation with constant speed.

Some observations:

1) To establish the new co-ordinate system, we need to solve a system of ODEs dx = c, x(s, 0) = s, dt dy = 1, y(s, 0) = 0, dt dz = 0, z(s, 0) = h(s). dt 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) s = s(x, y)

y = y(s, t) t = t(x, y).

3) If x − cy = k, then u(x, y) = u(k). In other words, u is constant along each line x − cy = k.

2.2.1 The characteristic equations

In the general case

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u), u(f(x), g(x)) = h(x), (2.3) 14 where f, g, h are given functions. The method of characteristics to solve the above equation above consists of three steps.

Step 1) Solve the following characteristic equations, which is a system of ODEs, for x, y, z as functions of s, t

dx = a(x, y, z), x(s, 0) = f(s), dt dy = b(x, y, z), y(s, 0) = g(s), dt dz = c(x, y, z), z(s, 0) = h(s). dt

Step 2) Find the transformation  s = s(x, y)

t = t(x, y).

Step 3) Find the solution u(x, y) = z(s(x, y), t(x, y)).

The characteristic projection is deﬁned by

σs(t) = x(s, t), y(s, t) .

A shock is formed when two characteristic projections meet and assign diﬀerent values of u, i.e., there exist (s, t), (s0, t0) such that

x(s, t) = x(s0, t0), y(s, t) = y(s0, t0), but z(s, t) 6= z(s0, t0).

2.2.2 Geometrical interpretation

We now give a geometrical interpretation (derivation) for the method of characteristics. Suppose that we are given a curve γ(s) = (f(s), g(s)), s ∈ (α, β). We will ﬁnd a function u satisfying

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u), such that for s ∈ (a, b), we have [u ◦ γ](s) = u(f(s), g(s)) = h(s) for some given function h :(α, β) → R. In other words, we specify the value of u along the curve γ. 15

The idea we shall pursue is to ﬁnd the graph of u over some domain Ω by showing that we can foliate it by curves along which the value of u is determined by ODEs. Recall that the graph of u over Ω is the surface in R3 given by

Graph(u) = {x, y, u(x, y):(x, y) ∈ Ω}.

We know from the initial conditions that the curve Γ, given by (f(s), g(s), h(s)), s ∈ (α, β) must belong to Graph(u). We want to write

[ Cs ⊂ Graph(u), s∈(α,β) where

Cs = {(x(s, t), y(s, t), z(s, t)) : t ∈ (−εs, εs)}, εs > 0, with x(s, 0) = f(s), y(s, 0) = g(s) and z(s, 0) = h(s). In other words, through each line of the curve (f(s), g(s), h(s)) we would like to ﬁnd another curve which lies in the graph of u.

dx dy dz Since the tangent vector to Cs, given by t := dt (s, t), dt (s, t), dt (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         =  0  ∧  1      ux uy   −ux     = −uy .   1

Therefore, Cs lies in the surface Graph(u) if and only if t · N = 0,i.e.,

dx dy dz u + u − = 0, dt x dt y dt 16

should hold at each point on Cs. Recall that u satisﬁes the PDE

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u).

By comparing the two relationships above, we see that if x(t; s), y(t; s), z(t; s) is the unique solution of the systems of ODEs

dx = a(x, y, z), x(s, 0) = f(s) dt dy = b(x, y, z), y(s, 0) = g(s) (2.4) dt dz = c(x, y, z), z(s, 0) = h(s) dt then the curve Cs will lie in Graph(u). The Picard-Lindel¨oftheorem states that if a, b, c are suﬃciently well behaved, then there exists a unique solution to (2.4) on some interval t ∈ (−εs, εs) and that moreover the solution depends continuously on s. Later we will state more precisely these conditions on a, b, c such that we can solve (2.4).

The equation (2.4) are known as the characteristic equations and the corresponding solution curves (x(t; s), y(t; s), z(t; s)) are the characteristic curves or simply characteristics. The union of the characteristics forms a surface

[ Cs = {(x(s, t), y(s, t), z(s, t)) : t ∈ (−εs, εs), s ∈ (α, β)} s∈(α,β) which is a portion of the graph of u that includes the curve Γ. We want to write this as a graph of the form {(x, y, u(x, y)) : (x, y) ∈ Ω}. We can do this, provided we can invert the map (s, t) 7→ (x(s, t), y(s, t)) to give s(x, y), t(x, y). Then we have u(x, y) = z(s(x, y); t(x, y)).

Example 13. Solve the PDE

ux + xuy = u, u(2, x) = h(x).

Solution. The characteristic equations are

dx = 1, x(s, 0) = 2, dt dy = x, y(s, 0) = s, dt dz = z, z(s, 0) = h(s). dt 17

The ﬁrst equation gives x(s, t) = t + 2. Substituting this into the second equation we get

t2 t y(s, t) = 2 + 2t + s. Finally, from the last equation, we have z(s, t) = e h(s). By writing s, t in terms of x, y, we ﬁnd

(x − 2)2 s = y − − 2(x − 2), t = x − 2. 2

x−2 (x−2)2 Therefore, u(x, y) = z(s, t) = e h y − 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 = z, x(s, 0) = s, dt dy = 1, y(s, 0) = 0, dt dz = 0, z(s, 0) = h(s). dt

The last two equations are easily solved to get

y(s, t) = t, z(s, t) = h(s).

Substituting these back to the ﬁrst equation we get

x(s, t) = s + h(s)t.

In this case, we can not explicitly invert the relationship (t, s) 7→ (x, y). However, we have s(x, y) = x − h(s(x, y))y, t(x, y) = y.

Therefore,

u(x, y) = z(s(x, y), t(x, y)) = h(s(x, y)) = h(x − h(s(x, y))y) = h(x − u(x, y)y), i.e., u satisﬁes an implicit equation

u(x, y) = h(x − u(x, y)y). 18

√ In a particular case when h(x) = x2 + 1, we have

u = p(x − uy)2 + 1, which implies that xy − px2 − y2 + 1 u(x, y) = . 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

x − s y = , z = h(s). h(s)

1 This means that u(x, y) = z remains constant on each of the line y = h(s) (x − s). The meeting point of two characteristic projections is determined by

x − s x − s s − s 1 = 2 = 2 1 . h(s1) h(s2) h(s1) − h(s2)

If h is increasing, then s2−s1 ≤ 0. In this case, the lines of constant u are “fanning h(s1)−h(s2) out” and do not meet on y ≥ 0.

If h is decreasing, then s2−s1 ≥ 0. The lines of constant u must meet on y ≥ 0. At h(s1)−h(s2) 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

0 0 0 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))g0(x) − b(f(x), g(x), h(x))f 0(x) 6= 0, then this system has a unique solution. In this case, we expect that the method of characteristics will work. We say that the initial conditions are compatible with the PDE.

• If a(f(x), g(x), h(x))g0(x) − b(f(x), g(x), h(x))f 0(x) = 0, the initial conditions are incompatible with the PDE or they are redundant.

2.4 Local existence and uniqueness

Recall that we want to solve the PDE

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u) on Ω, with initial data given along a curve γ : x 7→ (f(x), g(x)) ∈ Ω such that u(f(x), g(x)) = h(x). We will show that the compatibility of the initial conditions is the only obstruction (besides standard assumptions) to ﬁnding a solution in a neighbourhood of γ. Suppose that:

• a, b, c ∈ 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))g0(x) − b(f(x), g(x), h(x))f 0(x) 6= 0 for all x ∈ (α, β).

The the characteristic equations

dx = a(x, y, z), x(s, 0) = f(s), dt dy = b(x, y, z), y(s, 0) = g(s), dt dz = c(x, y, z), z(s, 0) = h(s). dt 20

can be solved uniquely for each s ∈ (α, β) and for t ∈ (−εs, εs), and the solution will be of class 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     ∂x ∂x 0 ∂s ∂t f (s) a(f(s), g(s), h(s)) DΨ(s, 0) =   (s, 0) =   . ∂y ∂y 0 ∂s ∂t g (s) b(f(s), g(s), h(s)) According to the compatibility conditions, we have

det DΨ(s, 0) = a(f(s), g(s), h(s))g0(s) − b(f(s), g(s), h(s))f 0(s) 6= 0, for all s ∈ (α, β), so that DΨ(s, 0) is invertible. By the inverse function theorem, Ψ−1 exists in a neighbourhood of (f(s), g(s)). Furthermore, we have     ∂s ∂s ∂x ∂y −1 −1 1 b −a   (Ψ(s, t)) = (DΨ )(Ψ(s, t)) = (DΨ(s, t)) =   ∂t ∂t b ∂x − a ∂y ∂y ∂x ∂x ∂y ∂s ∂s − ∂s ∂s

By the chain rule (remembering that u(x, y) = z(s(x, y), t(x, y)) )

∂z ∂s ∂z ∂t u = + x ∂s ∂x ∂t ∂x ∂z ∂s ∂z ∂t u = + y ∂s ∂y ∂t ∂y or equivalently, in a matrix form     ∂s ∂s b −a ∂z ∂z ∂x ∂y 1 ∂z ∂z ux uy =   =   ∂s ∂t ∂t ∂t b ∂x − a ∂y ∂s ∂t ∂y ∂x ∂x ∂y ∂s ∂s − ∂s ∂s Hence   a aux + buy = ux uy   b     1 b −a a = ∂z ∂z     b ∂x − a ∂y ∂s ∂t ∂y ∂x ∂s ∂s − ∂s ∂s b   1 0 = ∂z c   b ∂x − a ∂y ∂s ∂x ∂y ∂s ∂s b ∂s − a ∂s = c, so the equation is indeed satisﬁed. Uniqueness follows from the fact that the characteristic through a point (x0, y0, z0) is uniquely determined and moreover lies completely in 21

Graph(u) if (x0, y0, z0) lies in Graph(u). Thus any solution would have to include all of the characteristics through (f(s), g(s), h(s)) and hence would have to locally agree with the solution we construct above.

We summarise this section by the following theorem

Theorem 1. Suppose that a, b, c ∈ 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))g0(x)−b(f(x), g(x), h(x))f 0(x) 6= 0 for all x ∈ (α, β). Then, given K ⊂ (α, β) a closed (and hence compact) interval, there exists a neighbourhood U of γ(K) such that the equation

aux + buy = c has a unique solution u of class 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, h0 < 0, and assume that h0 is bounded. Show that no shock forms in the region 1 y < 0 , supR |h | 1 and that for any y > 0 a shock has formed for some x. supR |h | Exercise 5. a) Solve the problem

2 2 ux + 3x uy = 1, u(x, 0) = h(x), (x, y) ∈ R ,

for h ∈ C1(R). b) Show and explain why the problem

2 3 2 ux + 3x uy = 1, u(s, s ) = 1, (x, y) ∈ R

has no solution u ∈ C1(R2). c) Show and explain why the problem

2 3 2 ux + 3x uy = 1, u(s, s ) = s − 1, (x, y) ∈ R

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 n X ∂2u ∆u(x) := (x). ∂x2 i=1 i 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

T ∇u(x) = (∂x1 u, . . . , ∂xn u) ,

T and that if X = (X1,...,Xn) is a vector ﬁeld then the divergence of X is n X ∂Xi div(X) = . ∂x i=1 i

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.

Pn 2) If u(x) = i,j=1 aijxixj, with aij = aji (i.e., the matrix a = (aij)i,j=1,...,n is symmetric), then n n X X (∇u)i = 2 aijxj, and ∆u(x) = 2 aii = 2 Tr(a). i=1 i=1 So if a is trace-free (i.e., Tr(a) = 0) then ∆u = 0.

Example 16 (Laplacian operator of a radial function). Suppose u(x) = u(|x|) = u(r), where r = |x|. Then we have

n − 1 ∆u(|x|) = ∆u(r) = u00(r) + u0(r). r

We will provide two proofs for this assertion. The ﬁrst one is based on direct computations, the second one is hinged on a property that the Laplacian operator is rotation invariant.

The ﬁrst proof (direct computations). Since r = |x| = pPn x2, we have ∂r = xi . By i=1 i ∂xi r the chain rule we have

du ∂r 0 xi uxi = = u (r) . dr ∂xi r 00 xi xi 0 ∂ xi uxixi = u (r) + u (r) . r r ∂xi r x2 1 x2 = u00(r) i + u0(r) − i . r2 r r3

2 Pn 2 Therefore, recalling that r = i=1 xi ,

n n X X x2 1 x2 ∆u(x) = u = u00(r) i + u0(r) − i xixi r2 r r3 i=1 i=1 n − 1 = u00(r) + u0(r). r 25

The second proof.

First observation: the Laplacian is rotation invariant. Let Q be a constant n × n matrix satisfying QQT = QT Q = I. Let v(x) = u(Qx). By the chain rule, we have n X vxi (x) = uxj (Qx)Qji, j=1 and similarly n X T vxixi = uxj xk (Qx)QjiQki = (Q HQ)ii, j,k=1 where H = (Hessu)(Qx). Therefore n n X X T T T ∆v(x) = vxixi = (Q HQ)ii = Tr(Q HQ) = Tr(Q QH) = Tr(H) = ∆u(Qx). i=1 i=1 Here we use the fact that Q is orthogonal and the community of the trace T r(AB) = T r(BA).

Note further that since Q is orthogonal, |x|2 = hx, xi = hQx, Qxi = |Qx|2, it follows that |x| = |Qx|.

Now if u is radial, i.e., u(x) = u(|x|) = u(|Qx|) = u(Qx), then v(x) = u(Qx) = u(x), so that ∆v(x) = ∆u(Qx) = ∆u(x).

Second observation: By the divergence theorem, we have

∇u · n dσ = div(∇u) dx = ∆u(x) dx. (2.2) ˆ∂Br ˆBr ˆBr 0 x 0 Since u is radial, ∇u(x)·n = u (r) |x| ·n = u (r). The ﬁrst integral in (2.2) can be computed as

0 0 n−1 ∇u · n dσ = u (r) dσ = u (r)σn−1r , (2.3) ˆ∂Br ˆ∂Br where σn−1 is the area of the unit (n − 1)−sphere. Now using the fact that ∆u(x) is radial, we can write the last integral in (2.2) as a radial integral r n−1 ∆u(x) dx = σn−1s ∆u(s) ds. (2.4) ˆBr ˆ0 From (2.3) and (2.4) we obtain that for any r1 < r2

r2 0 n−1 0 n−1 n−1 u (r2)r2 − u (r1)r1 = s ∆u(s) ds. ˆr1 This equality implies that rn−1∆u(r) is simply the derivative of rn−1u0(r), thus 1 d n − 1 ∆u(r) = rn−1u0(r) = u00(r) + u0(r). rn−1 dr 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 Rn 2 2 2 Rn a ∈ , b ∈ R, and u(x) = x1 + ... + xn−1 − (n − 1)xn are harmonic in .

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

0 f(z) − f(z0) f (z0) := lim z→z0 z − z0 exists for all z0 ∈ Ω. Suppose that f(x + iy) = u(x, y) + iv(x, y). Then u, v are harmonic function in Ω. In fact, by the Cauchy-Riemann equations, we have   ∂u ∂v  ∂x = ∂y ,  ∂u ∂v  ∂y = − ∂x .

∂2u ∂2u ∂2v ∂2v Therefore, ∆u(x, y) = ∂x2 + ∂y2 = ∂x∂y − ∂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., u00(x) = 0 in (a, b). It follows that u(x) = cx + d for some constants c, d ∈ R. Let x0 ∈ (a, b) and r > 0 is suﬃciently small such that

[x0 − r, x0 + r] ⊂ (a, b). Let us compute 1 1 [u(x − r) + u(x + r)] = [c(x − r) + d + c(x + r) + d] 2 0 0 2 0 0

= cx0 + d, 27 and

1 x0+r 1 x0+r u(y) dy = (cy + d) dy 2r ˆx0−r 2r ˆx0−r 2 1 y x0+r = c + dy 2r 2 x0−r

= cx0 + d.

Therefore, 1 1 x0+r u(x0) = [u(x0 − r) + u(x0 + r)] = u(y) dy. 2 2r ˆx0−r This property holds true for harmonic function in any dimensional space.

Theorem 2 (Mean value property). Suppose Ω ⊂ Rn is open and that u ∈ C2(Ω) is harmonic. Then if Br(x0) ⊂ Ω then

1 u(x0) = u(y) dσ(y) (2.5) |∂Br(x0)| ˆ∂Br(x0) 1 = u(y) dy. (2.6) |(Br(x0))| ˆBr(x0)

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

n−1 u(y) dσ(y) = u(x0 + z) dσ(z) = r u(x0 + rζ) dσ(ζ) ˆ∂Br(x0) ˆ∂Br(0) ˆ∂B1(0)

n−1 Therefore, recalling that |∂Br(x0)| = r σn−1, where σn−1 is the area of the (n − 1)-unit sphere,

1 1 u(y) dσ(y) = u(x0 + rζ) dσ(ζ). (2.7) |∂Br(x0)| ˆ∂Br(x0) σn−1 ˆ∂B1(0)

Let F (r) be deﬁned as the right-hand side of the above equality. Let v(ζ) := u(x0 + rζ), 2 then ∇v(ζ) = r∇u(x0 +rζ) and ∆v(z) = r ∆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) 1 d = u(x0 + rζ) dσ(ζ) dr σn−1 dr ˆ∂B1(0) 1 d = u(x0 + rζ) dσ(ζ) σn−1 ˆ∂B1(0) dr 1 = ζ · ∇u(x0 + rζ) dσ(ζ) σn−1 ˆ∂B1(0) 1 1 = ζ · ∇v(ζ) dσ(ζ) σn−1 r ˆ∂B1(0) (∗) 1 1 = ∆v(ζ) dζ σn−1 r ˆB1(0) = 0, where we have used the divergence theorem to obtain the equality (∗) above. Hence by the continuity of u, we have 1 F (r) = lim F (r) = u(x0) dσ(ζ) = u(x0). (2.8) r→0 σn−1 ˆ∂B1(0) From (2.7) and (2.8), we obtain (2.5).

The equality (2.6) follows from (2.5). Indeed, we have r r u(y) dy = ds u(y) dσ(y) = |∂Bs(x0)|u(x0) ds ˆBr(x0) ˆ0 ˆ∂Bs(x0) ˆ0 r = u(x0) |∂Bs(x0)| ds = u(x0)|Br(x0)| ˆ0

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 harmonic functions

Proposition 1. Let Ω ⊂ Rn be open and let u be harmonic in Ω. Then u ∈ C∞(Ω) and furthermore n|α||α| |Dαu(x)| ≤ sup |u(y)|, (2.9) d(x) y∈Ω 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) Rn φ(x) dx = 1. ´ We will show that u(x) = (u ∗ φ)(x), where the convolution is deﬁned by

(u ∗ φ)(x) = u(y)φ(x − y) dy. ˆRn

Without loss of generality, set x = 0, and we take r small enough so that Br ⊂ Ω. Now we have

(u ∗ φ)(0) = u(y)φ(−y) dy ˆRn = u(y)φ(−y) dy ˆBr = u(y)φ(y) dy ˆBr r = u(y) dS(y) φ(s) ds ˆ0 ˆ∂Bs r = |∂Bs|u(0)φ(s) ds ˆ0 r = u(0) |∂Bs|φ(s) ds ˆ0 r = u(0) φ(s)dS ds ˆ0 ˆ∂Bs = u(0) φ(y) dy ˆBr = 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 n |Diu(x)| ≤ sup |u(y)|. (2.10) d(x) y∈Ω

Take r < d(x) then Br(x) ⊂ Ω. According to the mean value property 1 1 Diu(x) = Diu(y) dy = u(y)ni(y) dS(y), |Br(x)| ˆBr(x) |Br(x)| ˆ∂Br(x) 30

Since |ni| = 1, we can estimate

n−1 |∂Br(x)| r σn−1 n |Diu(x)| ≤ sup |u(y)| = n sup |u(y)| = sup |u(y)|. |Br(x)| y∈Br(x) r ωn y∈Br(x) r y∈Br(x)

Since this is true for any r < d(x) and Br(x) ⊂ Ω, it follows that

n |Diu(x)| ≤ sup |u(y)|. d(x) y∈Ω

Step 2. Suppose that the statement is true for |α| ≤ N.

α β Step 3. We need to prove that it is also true for |α| = N + 1. Write D u = DiD u, where β is a multi-index such that |β| = |α| − 1 = N. Now we ﬁx some λ ∈ (0, 1). We apply the technique in Step 1 to r = λd(x), to obtain

α β n β |D u(x)| = |DiD u(x)| ≤ sup |D u(y)|. λd(x) y∈Bλd(x)(x)

By the triangle inequality if y ∈ Bλd(x)(x), then d(y) ≥ (1 − y)d(x), so applying the induction assumption we have

n|β||β| n|β| |β| |Dβ(y)| ≤ sup |u(z)| ≤ sup |u(z)|, d(y) z∈Ω (1 − λ)d(x) z∈Ω which implies that

n|β| |β| sup |u(y)| ≤ sup |u(z)|, y∈Bλd(x) (1 − λ)d(x) z∈Ω and that (recalling that |α| = |β| + 1)

|β| |α| α n n|β| n |β| 1 |D u(x)| ≤ sup |u(z)| = |β| sup |u(z)| |β| . λd(x) (1 − λ)d(x) z∈Ω d(x) z∈Ω λ(1 − λ) (2.11)

1 To obtain the sharpest upper bound, we now minimizes the function λ 7→ λ(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

dg(λ) 0 = = (1 − λ)|β| − λ|β|(1 − λ)|β|−1 dλ = (1 − λ)|β|−1[1 − λ − |β|λ] = (1 − λ)|β|−1)(1 − |α|λ), 31

1 which implies that λ = |α| , which is in the interval (0, 1) as required. Substituting this value to g(λ), we get (recalling that |α| = |β| + 1)

1 1 |β| |β||β| g(λ) = 1 − = 1 |α| λ= |α| |α| |α| |α| Putting this back into (2.11) we obtain

|α| |α| α n |α| |D u(x)| ≤ |α| sup |u|, d(x) Ω which is the reduction assumption for |α| = N + 1. Hence the result holds true for any N.

Proposition 1 has two interesting consequences in the following theorems.

Theorem 3. Harmonic functions are analytic.

Proof. Let x0, x ∈ Ω and suppose that the segment [x0, x] = tx + (1 − t)x0, t ∈ [0, 1] is contained in Ω. Without loss of generality we assume that x0 = 0. Let u be a harmonic function in Ω. By Taylor’s theorem we have

u(x) = Pm−1(x) + Rm(x),

where Pm−1(x) is a polynomial of degree m − 1 and

X Dαu(ξ) R (x) = xα, m α! |α|=m for some ξ ∈ [x0, x]. To show that u is analytic we need to show that Rm(x) → 0 as m → ∞ uniformly on some small disc around x0. By Proposition 1, we have

m nm X |xα| |Rm(x)| ≤ sup |u| d(ξ) Ω α! |α|=m 1 nm m m! X α1 αn = sup |u| |x1| ... |xn| . m! d(ξ) Ω α1! . . . αn! |α|=m Pn Set ||x||1 := i=1 |xi|, then by the multinomial theorem we have m! m X α1 αn ||x||1 = |x1| ... |xn| . α1! . . . αn! |α|=m Therefore, we obtain that

m m ||x||1 nm |Rm(x)| ≤ sup |u|. (2.12) m! d(ξ) Ω 32

We now use a weak version of the Stirling’s approximation, which holds true for all positive integer m, mm ≤ em. m! We get m ne||x||1 |Rm(x)| ≤ sup |u|. d(ξ) Ω 1 1 Taking |x| = |x − x0| < 2 d(x0), by the triangular inequality, we have d(ξ) ≤ 2 d(x0), so that m 2ne||x||1 |Rm(x)| ≤ sup |u|. (2.13) d(x0) Ω m d(x0) 2ne||x||1 2ne||x||1 Finally, if we take x such that ||x||1 < , then < 1 and → 0 as 2ne d(x0) d(x0) 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.

n n Proof. Suppose |u| ≤ C in R . Let x ∈ R and r > 0. Since u is harmonic in Br(x), applying Proposition 1 in Br(x), we have n nC |Diu(x)| ≤ sup |u| ≤ . r Br(x) r Since r is arbitrary and C is independent of r, the above estimate implies that |∇u(x)| = 0. So u is constant.

2.5 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 u00 > 0, which means that u0 is increasing and u is convex. Similarly u is superharmonic iﬀ u is concave.

2) If u = xT Ax for some constant, symmetric matrix A. Then ∆u = 2Tr(A), so u is subharmonic (respectively, superharmonic) iﬀ Tr(A) ≥ 0 (respectively, Tr(A) ≤ 0). 33

2.5.2 Some properties of subharmonic and superharmonic functions

Recall that harmonic functions satisfy the mean value property, which means that for any ball Br(x0) ⊂ Ω, we have 1 1 u(x0) = u(y)dσ(y) = u(y) dy. |∂Brx0| ˆ∂Br(x0) |Br(x0)| ˆBr(x0) Recall that to prove this in Theorem 2, we considered the function F (r) = 1 u(y)dσ(y) |∂Brx0| ∂Br(x0) ´ and showed

(i) F (r) → u(x0) as r → 0. This relied on the continuity of u,

(ii) For all r > 0 we have

0 1 F (r) = ∆u(x0 + rζ) dζ. (2.14) rσn−1 ˆB1(0)

It follows from (2.14) that if u is subharmonic (respectively, superharmonic) then F 0(r) ≥ 0 (respectively, F 0(r) ≤ 0).

Proposition 2. If u is subharmonic in Ω and Br(x0) ⊂ Ω, then 1 1 u(x0) ≤ u(y)dσ(y) and u(x0) ≤ u(y) dy. (2.15) |∂Brx0| ˆ∂Br(x0) |Br(x0)| ˆBr(x0) 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

max u = max u. (2.16) Ω ∂Ω 34

If, rather, u is superharmonic in Ω then

min u = min u. (2.17) Ω ∂Ω

Proof. Suppose u is subharmonic. We consider two cases.

Case 1: ∆u > 0 in Ω. Suppose that there exists x0 ∈ Ω such that u attains a local maximum at x0, then we must have ∇u(x0) = 0 and Hess u(x0) is negative deﬁnite, i.e., T n ξ Hess u(x0)ξ ≤ 0 for all ξ ∈ R , ξ 6= 0. It implies that

n X T ∆u(x0) = Tr(Hess u(x0)) = ei Hess u(x0)ei ≤ 0, i=1 which is a contradiction. Therefore, u can not have maxima in the interior. Thus, the maximum of u in Ω must be achieved on the boundary Ω, i.e., (2.16) holds.

2 Case 2: ∆u ≥ 0 in Ω. Deﬁne uε(x) := u(x)+ε|x| . Then ∆uε(x) = ∆u(x)+2εn > 0, so that we can apply Case 1 above for uε(x) and obtain

max uε = max uε. Ω ∂Ω

2 ¯ Since Ω is bounded, |x| is a bounded function in Ω, and uε → u uniformly on Ω as ε → 0. It follows that

max u = lim max uε = lim max uε = max u. Ω ε→0 Ω ε→0 ∂Ω ∂Ω The statement (2.17) for superhamornic function is obtained by applying (2.16) for −u, which is subharmonic, and using the fact that max(−u) = − min u. A A

Since a harmonic function is both subharmonic and superhamornic, we obtain the following corollary.

Corollary 1. 1) Let Ω ⊂ Rn be open and bounded and u ∈ C2(Ω) ∩ C0(Ω)¯ is harmonic. Then max u = max u, and min u = min u. Ω ∂Ω Ω ∂Ω This implies max |u| = max |u|. Ω ∂Ω

2) (The comparison principle) If Ω ⊂ Rn is open and bounded and if u, v ∈ 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 subharmonic function is to be found on the boundary, but may re-occur in the interior as well. The following theorem provides a stronger statement.

Theorem 6 (Strong maximum principle). Let Ω be open and connected (possibly 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 := max u, and A := u−1{M} = {x ∈ Ω: u(x) = M}. By the assumption Ω of the theorem, A is non-empty. Since u is continuous and {M} is a closed set in R, it follows that A is relatively closed in Ω (i.e., there exists a closed F ⊂ Rn such that A = F ∩ Ω). We now show that A is also open. Indeed, let x ∈ Ω and let r be suﬃciently small that Br(x) ⊂ Ω. By the mean value property for subharmonic functions in Proposition 2, we have

1 0 = u(x) − M ≤ (u(y) − M) dy. |Br(x)| ˆBr(x)

Since M = max u, we have u(y) − M ≤ 0 for all y ∈ Br(x). This implies that Ω 1 0 = u(x) − M ≤ (u(y) − M) dy ≤ 0. |Br(x)| ˆBr(x)

Since u is continuous, this happens only if u = M in Br(x). So that Br(x) ⊂ A, which implies that A is open. Since Ω is connected and A 6= ∅ is both open and closed, it follows that A = Ω, and hence u ≡ M in Ω. This ﬁnishes the proof.

Again, since a harmonic function is both subharmonic and superharmonic, we get a stronger result.

Corollary 2 (Strong maximum principle for harmonic functions). Let Ω be open and connected and u :Ω → R a harmonic function. Then if u attains either a global maximum or global minimum in Ω then u is constant in Ω. 36

2.6 Harnack’s inequality

The next result we shall derive is Harnack’s inequality for harmonic functions. To state this theorem, we need to deﬁne the notion of a compactly contained subset. Let Ω, Ω0 be two open subsets of Rn, then we say that Ω0 is compactly contained in Ω and

0 0 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

0 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, Ω0, n such that

sup ≤ C inf u. Ω0 Ω0

0 Proof. Let r > 0 satisfy that infΩ0 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

1 1 1 u(x) = u(z) dz = n n u(z) dz ≥ n n u(z) dz, |B2r(x)| ˆB2r(x) 2 r ωn ˆB2r(x) 2 r ωn ˆBr(y) where ωn = |B1(0)| and we have used that u ≥ 0 and Br(y) ⊂ B2r(x) to obtain the above inequality. Now using the mean value property again, we have

1 1 1 1 n n u(z) dz = n u(z) dz = n u(y). 2 r ωn ˆBr(y) 2 |Br(y)| ˆBr(y) 2 It follows from these two inequalities that

1 u(y) ≤ u(x). 2n

By interchanging x and y, we obtain u(x) ≤ 2nu(y), so that for all |x − y| ≤ r we have

1 u(y) ≤ u(x) ≤ 2nu(y). (2.18) 2n

0 0 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] → Ω0 such that

0 0 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, Ω0. From these balls, we select M ≤ N balls Bi, i = 1,...,M such that

M [ x0 ∈ B1, x1 ∈ BM , γ([0, 1]) ⊂ Bi, i=1 37

and Bi ∩ Bi+1 6= ∅, i = 1,...,M − 1. Applying the estimate (2.18) successively on these balls, we get

Mn Nn u(x1) ≤ 2 u(x0) ≤ 2 u(x0).

0 Since this inequality holds for any two points x0, x1 ∈ Ω , it follows that

sup u ≤ 2nN inf u, Ω0 Ω0 which is the claim.

Harnack’s inequality is used to prove Harnack’s theorem about the convergence of sequences of harmonic functions. Harnack’s inequality can also be used to show the interior regularity of weak solutions of partial diﬀerential equations.

Corollary 3. Let Ω be connected and let uk be a sequence of harmonic functions in Ω such that uk ≤ uk+1 for all k. If there exists x0 ∈ Ω such that uk(x0) converges (to a ﬁnite value) then uk converges to some harmonic function u uniformly on the compact sets of Ω. Problem Sheet 2

Exercise 6 (Laplacian operator in the cylindrical and spherical coordinates).

1) Consider the cylindrical coordinates in R3 deﬁned by  x (r, ϕ, z) = r cos ϕ,  1  x2(r, ϕ, z) = r sin ϕ, r ∈ [0, ∞), ϕ ∈ [0, 2π), z ∈ R,   x3(r, ϕ, z) = z.

Prove that in these coordinates

1 ∂ ∂u 1 ∂2u ∂2u ∆u(r, ϕ, z) = r + + . r ∂r ∂r r2 ∂ϕ2 ∂z2

2) Consider the spherical coordinates in R3 deﬁned by  x (r, θ, ϕ) = r sin θ cos ϕ,  1  x2(r, θ, ϕ) = r sin θ sin ϕ, r ∈ [0, ∞), θ ∈ [0, π), ϕ ∈ [0, 2π),   x3(r, ϕ, z) = z.

Prove that in these coordinates

1 ∂ ∂u 1 ∂2u 1 ∂ ∂u ∆u(r, θ, ϕ) = r2 + + sin θ . r2 ∂r ∂r r2 sin2 θ ∂ϕ2 r2 sin2 θ ∂θ ∂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 1 ∆u(x) = (1 + |x|2)2 in R2 \{0}. Which of these solutions extend smoothly to all of R2 (i.e., near the origin).

3) Let (z, ϕ) be polar coordinates in R2, and consider the set Ω = R2 \{(x, 0) : x ≥ 0}. Show that the functions log r, ϕ are harmonic in Ω.

Exercise 8 (Converse of the mean value property of harmonic functions in 1D).

1) Suppose u :(a, b) → R is continuous and satisﬁes

1 u(x ) = [u(x − r) + u(x + r)], ∀x , r such that [x − r, x + r] ⊂ (a, b). 0 2 0 0 0 0 0

Show that u is harmonic (i.e., aﬃne).

2) Prove the same if u :(a, b) → R is continuous and satisﬁes

1 x0+r u(x0) = u(x) dx, ∀x0, r such that [x0 − r, x0 + r] ⊂ (a, b). 2r ˆx0−r

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.

n 0 + 2) Suppose Ω ⊂ R is open and such that Ω = Ω ∩ {xn = 0}= 6 ∅. Let Ω := Ω ∩ {xn > 0}, and deﬁne Ω˜ = Ω+ ∪ Ω0 ∪ Ω−, where

− 0 n + Ω = {x = (x , xn) ∈ R such that (x, −xn) ∈ Ω }.

Suppose that u ∈ C2(Ω+) is such that ∂u = 0 on Ω0, and deﬁneu ˜ : Ω˜ → R by ∂xn  u(x), x ∈ Ω+ ∪ Ω0, u˜(x) = 0 0 − u(x , −xn), x = (x , xn) ∈ Ω .

Prove thatu ˜ ∈ C2(Ω)˜ and thatu ˜ 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. Definev ˜ : Ω˜ → R by  v(x), x ∈ Ω+ ∪ Ω0, v˜(x) = 0 0 − −v(x , −xn), x = (x , xn) ∈ Ω .

Prove thatv ˜ ∈ C2(Ω)˜ and thatv ˜ is harmonic in Ω.˜

n Exercise 10. Recall the Azel`a-Ascolitheorem: Let K ⊂ R 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].

n Let Ω ⊂ R 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

α K ⊂ Ω there exists a subsequence uik such that D uik converges uniformly in K.

Exercise 11. 1) Suppose u :Ω → R is harmonic and that F : R → R is a convex function of class C2. Show that F (u(x)) is subharmonic. 41

2) Suppose u : Rn → R is harmonic and that

u2(x) dx = M < ∞. ˆRn

Prove that u ≡ 0 in Rn.

n 3) If u is harmonic in Ω ⊂ R and if Br(x0) ⊂ Ω, prove that

d 2r − u(y)2dσ(y) = − |∇u|2(x) dx. dr ∂B (x ) n B (x ) ´r 0 r´ 0 Chapter 3

The Dirichlet problem for harmonic functions

In this chapter we will study the Direchlet problem for harmonic functions in a bounded domain. We will establish a representation formula based on the Green’s function, ﬁnd the Green’s function and solve the problem in a ball, and ﬁnally use the Perron’s method to solve the problem in a general domain.

3.0.1 Introduction

In this chapter, we will study the following Dirichlet problem for harmonic functions.

The Dirichlet problem: Let Ω ⊂ Rn be open and bounded. Let g ∈ C(∂Ω) be a given function. Find a function u ∈ C2(Ω) ∩ C0(∂Ω) such that  ∆u = 0, in Ω, (3.1) u = g, on ∂Ω.

The schedule to solve this problem will be the following.

Step 1 Establish a representation formula for a solution to (3.1) based on Green’s function.

Step 2 Find the Green’s function and solve the problem (3.1) in a ball,

Step 3 Solve the problem (3.1) for a general domain using the Perron’s method.

42 43

To start with, we have the following lemma

n Lemma 1. Let Ω ⊂ R be open and bounded, let g1, g2 ∈ C(∂Ω) be given. Suppose that u1, u2 solve the following problem for i = 1, 2 respectively  ∆ui = 0, in Ω,

ui = gi, on ∂Ω.

Then

sup |u1 − u2| = sup |g1 − g2|. (3.2) Ω Ω As a consequence, a solution to (3.1), if it exists, is unique and depends continuously on the data.

Proof. This is a direct consequence of the maximum principle for w := u1 − u2, which satisﬁes that  ∆w = 0, in Ω,

w = g1 − g2, on ∂Ω.

In view of this Lemma, the main task is now to ﬁnd a solution to (3.1). We ﬁrst recall the Green’s identities that will be used throughout this chapter. The starting point for these identities is the divergence theorem.

−→ −→ div F dx = F · −→n dS. ˆΩ ˆ∂Ω −→ Given f, g ∈ C2(Ω). Applying the divergence theorem above for F = f∇g, we obtain

div(f∇g) dx = f∇g · −→n dS. ˆΩ ˆ∂Ω Note that ∂g div(f∇g) = f∆g + ∇g · ∇f, ∇g · −→n = , ∂−→n so that we can write the equality above as

∂g f∆g dx = − ∇f · ∇g dx + f −→ dS. (3.3) ˆΩ ˆΩ ˆ∂Ω ∂ n This relation is known as the ﬁrst Green’s identity. 44

By interchanging f and g in the ﬁrst Green’s identity, we get

∂f g∆f dx = − ∇g · ∇f dx + g −→ dS. ˆΩ ˆΩ ˆ∂Ω ∂ n This together with the ﬁrst Green’s identity implies the following Green’s second identity

∂g ∂f (f∆g − g∆f) dx = f −→ − g −→ dS. (3.4) ˆΩ ˆ∂Ω ∂ n ∂ n Deﬁne the function   1 log |x|, if n = 2, Φ(x) = 2π (3.5)  1 |x|2−n, if n = 3, (2−n)σn−1 ∞ n where σn−1 = |∂B1(0)|. Then Φ ∈ C R \{0}. For any x 6= 0, we have

1 −n ∂xi Φ(x) = xi|x| , σn−1 1 −nx2 ∂2 Φ(x) = i + |x|−n . xixi n+2 σn−1 |x|

Therefore, n n X X 1 −nx2 ∆Φ(x) = ∂2 Φ(x) = i + |x|−n = 0. xixi σ |x|n+2 i=1 i=1 n−1 So ∆Φ(x) = 0, ∀x 6= 0, lim Φ(x) = −∞. (3.6) x→0

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) = u(y)δx dy ˆΩ

= u(y)∆yG(x, y) dy ˆΩ ∂G ∂u = u(y) −→(x, y) − G(x, y) −→(y) dS(y) + G(x, y)∆u(y) dy ˆ∂Ω ∂ n ∂ n ˆΩ ∂G = −→(x, y)g(y) dS(y). ˆΩ ∂ n Therefore, we can represent u in terms of G and g. This suggests that to solve the problem (3.1), one should ﬁnd the function G(x, y). We observe that Φ(x − y) satisﬁes that ∆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

Φ(y − x)∆u(y) dy. ˆΩ 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

∂u Φ(y − x)∆u(y) dy = ∆yΦ(y − x)u(y) dy + Φ(y − x) −→(y) dS(y) ˆVε ˆVε ˆ∂Vε ∂ n ∂Φ − u(y) −→(y − x) dS(y) ˆ∂Vε ∂ n ∂u ∂Φ = Φ(y − x) −→(y) dS(y) − u(y) −→(y − x) dS(y), ˆ∂Vε ∂ n ˆ∂Vε ∂ n 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

lim Φ(y − x)∆u(y) dy = Φ(y − x)∆u(y) dy. ε→0 ˆVε ˆΩ For the boundary terms, we will show later that

Claim 1

∂Φ ∂Φ lim − u(y) (y − x) dS(y) = − u(y) (y − x) dS(y) + u(x). (3.7) ε→0 −→ −→ ˆ∂Vε ∂ n ˆ∂Ω ∂ n 46

Claim 2 ∂u ∂u lim Φ(y − x) (y) dS(y) = Φ(y − x) (y) dS(y) (3.8) ε→0 −→ −→ ˆ∂Vε ∂ n ˆ∂Ω ∂ n

Assuming these claims at the moment, then we ﬁnd that for any u ∈ C2(Ω), we have ∂Φ ∂u u(x) = Φ(y−x)∆u(y) dy+ u(y) −→(y−x) dS(y)− Φ(y−x) −→(y) dS(y). (3.9) ˆΩ ˆ∂Ω ∂ n ˆ∂Ω ∂ n 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  x ∆yh (y) = 0, y ∈ Ω,

hx(y) = Φ(y − x), y ∈ ∂Ω.

Then using the Green’s second identity, we have

x x x x ∂u ∂h h (y)∆u(y) dy = ∆yh (y)u(y) dy + h (y) −→(y) − u(y) −→(y) dS(y). ˆΩ ˆΩ ˆ∂Ω ∂ n ∂ n ∂u ∂hx = Φ(y − x) −→(y) dS(y) − u(y) −→(y) dS(y). ˆ∂Ω ∂ n ˆ∂Ω ∂ n Hence,

x ∂u ∂h x Φ(y − x) −→(y) dS(y) = u(y) −→(y) dS(y) + h (y)∆u(y) dy. ˆ∂Ω ∂ n ˆ∂Ω ∂ n ˆΩ Substituting this equality back to (3.9), we obtain that

x x ∂Φ ∂h u(x) = [Φ(y − x) − h (y)]∆u(y) dy + −→(y − x) − −→(y) u(y) dS(y). (3.10) ˆΩ ˆ∂Ω ∂ n ∂ n By deﬁning G(x, y) := Φ(y − x) − hx(y), then we get that for any u ∈ C2(Ω) ∂G u(x) = G(x, y)∆u(y) dy + −→(y − x) u(y) dS(y). (3.11) ˆΩ ˆ∂Ω ∂ n The function G(x, y) is called the Green’s function for Ω. In particular, if u is harmonic then ∂G u(x) = −→(y − x) u(y) dS(y). ˆ∂Ω ∂ n We now prove the two claims.

Proof of Claim 1. We have ∂Φ ∂Φ ∂Φ − u(y) −→(y−x) dS(y) = − u(y) −→(y−x) dS(y)+ u(y) −→(y−x) dS(y). ˆ∂Vε ∂ n ˆ∂Ω ∂ n ˆ∂Bε(x) ∂ n 47

The ﬁrst term on the right-hand side does not depend on ε, so it is unchanged when passing ε → 0. We consider the second term. We have ∂Φ (y − x) = ∇ Φ(y − x) × −→n (y), ∂−→n y −→ where n (y) is the outward normal vector on ∂ε(x). By direct computations

1 2−n 1 y − x ∇yΦ(y − x) = ∇y |y − x| = n , (2 − n)σn−1 σn−1 |y − x| y − x −→n (y) = , |y − x| so that ∂Φ 1 y − x y − x 1 1 −→(y − x) = n = n−1 . ∂ n σn−1 |y − x| |y − x| σn−1 |y − x| Therefore, ∂Φ 1 1 u(y) −→(y − x) dS(y) = n−1 u(y) dS(y) ˆ∂Bε(x) ∂ n σn−1 ˆ∂Bε(x) |y − x| 1 = n−1 u(y) dS(y) σn−1ε ˆ∂Bε(x) 1 = u(y) dS(y). |∂Bε(x)| ˆ∂Bε(x) Since u is continuous, we have that ∂Φ 1 lim u(y) (y − x) dS(y) = lim u(y) dS(y) = u(x), ε→0 −→ ε→0 ˆ∂Bε(x) ∂ n |∂Bε(x)| ˆ∂Bε(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 have ∂u ∂u ∂u Φ(y − x) −→(y) dS(y) = Φ(y − x) −→(y) dS(y) − Φ(y − x) −→(y) dS(y). ˆ∂Vε ∂ n ˆ∂Ω ∂ n ˆ∂Bε(x) ∂ n 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 ∂u 1 1 ∂u Φ(y − x) (y) dS(y) ≤ (y) dS(y) −→ n−2 −→ ˆ∂Bε(x) ∂ n (n − 2)σn−1 ˆ∂Bε(x) |y − x| ∂ n ∂u 1 ≤ −→ dS(y) L∞(B (x)) n−2 ∂ n ε (n − 2)σn−1ε ˆ∂Bε(x) ≤ Cε, where to obtain the last step, we have used that

n−1 dS(y) = σn−1ε . ˆ∂Bε(x) 48

Therefore ∂u lim Φ(y − x) (y) dS(y) = 0, ε→0 −→ ˆ∂Bε(x) ∂ n 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, (3.12) u = g, in Ω, where f, g are continuous, then for x ∈ Ω,

∂G u(x) = G(x, y)f(y) dy + −→(x, y)g(y) dS(y). (3.13) ˆΩ ˆ∂Ω ∂ n In particular, if f ≡ 0, then

∂G u(x) = −→(x, y)g(y) dS(y). (3.14) ˆ∂Ω ∂ n

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  x ∆yh (y) = 0, in Ω, (3.15) hx(y) = Φ(y − x), on ∂Ω.

We will solve this problem by the method of image charge. Think of (3.15) as that we want to ﬁnd the electrostatic potential hx inside a spherical conductor of radius R de to the point charge located at x.

We consider the image of x under the inversion in ∂BR,

R2 x∗ = x, |x|2 which in particular, implies that

R2 |x∗| = , x∗ · x = R2. |x| 49

Then, if y ∈ ∂BR, we have

|y − x|2 = |y|2 − 2x · y + |x|2

= R2 − 2x · y + |x|2 |x|2 R4 xR2 = − 2 · y + R2 R2 |x|2 |x|2 |x|2 = |x∗|2 − 2x∗ · y + |y|2 R2 |x|2 = |x∗ − y|2. R2

So we obtain that |x| |y − x| = |y − x∗| ∀y ∈ ∂B . R R Let  |x|  1 log |y − x∗| , n = 2, hx(y) := 2π R 2−n  1 |x| |y − x∗|2−n, n ≥ 3. (2−n)σn−1 R ∗ 2 ∗ ∗ Note that since |x | |x| = R and x ∈ BR, it follows that x 6∈ BR, so that y 6= x . x Therefore y 7→ h (y) is harmonic in BR. In addition, by the computation above, it holds x x that h (y) = Φ(y − x) on ∂BR. Therefore, h (y) is the unique corrector function. Hence, the Green’s function for the ball is given by   1 R |y−x|  2π log |x| |y−x∗| , n = 2, G (x, y) BR 2−n  1 |y − x|2−n − |x| |y − x∗|2−n , n ≥ 3.  (2−n)σn−1 R

By the representation formula (3.14), we have

∂G u(x) = −→(x, y)g(y) dS(y). ˆ∂Ω ∂ n

∂G Next, we will compute ∂−→n (x, y). We have ∂G (x, y) = ∇ G(x, y) · −→n . ∂−→n y

Here

y −→n = , R " # 1 y − x |x|2−n y − x∗ ∇yG(x, y) = n − ∗ n . σn−1 |y − x| R |y − x | 50

Note that this formula is true for n ≥ 2. Therefore, we obtain " # ∂G 1 y − x |x|2−n y − x∗ −→(x, y) = n − ∗ n · y ∂ n Rσn−1 |y − x| R |y − x | 1 |y|2 − x · y |x|2−n |y|2 − x∗ · y = n − ∗ n Rσn−1 |y − x| R |y − x | 1 |y|2 |x|−n |x|2 1 |x|−n 1 = n − ∗ n − x · y n − ∗ n Rσn−1 |y − x| R |y − x | |y − x| R |y − x | 1 R2 − |x|2 = n . Rσn−1 |y − x|

Set 1 R2 − |x|2 KR(x, y) := n . (3.16) Rσn−1 |y − x| This function is called the Poisson kernel. We have proved that

2 Proposition 3. Let u ∈ C (BR) be harmonic in BR. Then for x ∈ BR, one has

1 R2 − |x|2 u(x) = KR(x, y)u(y) dS(y) = n u(y) dS(y). (3.17) ˆ∂BR Rσn−1 ˆ∂BR |y − x|

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

y R KR : BR →

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

KR(x, y) dS(y) = 1. ˆ∂BR 51

(iv) For y 6= z with y, z ∈ ∂BR, we have

lim K (x, y) = 0 x→z R x∈BR

and this convergence is uniform in y for |y − z| > δ > 0.

xi Rn Proof. (i) We notice that for each i = 1, . . . , n, the function |x|n is harmonic on \{0}. Indeed, since 2−n xi ∂ |x| n = |x| ∂xi 2 − n so 2−n 2−n xi ∂ |x| ∂ |x| ∆ n = ∆ = ∆ = 0. |x| ∂xi 2 − n ∂xi 2 − n

2−n (x−y)i It follows that |x − y| and |x−y|n are both harmonic. So that

R2 − |x|2 2|y|2 − |x − y|2 − 2x · y y − x Rσ K (x, y) = = = −|x − y|2−n + 2y · n−1 R |x − y|n |x − y|n |y − x|n

is also a harmonic function.

(ii) This is obvious since |x| < R for x ∈ BR.

(iii) Applying the Poisson’s integral formula for u ≡ 1 yields the result.

2 2 (iv) lim R −|x| = 0 = 0. For |x−z| < δ , then |y −x| ≥ |y −z|−|x−z| ≥ δ − δ = δ . x→z |y−x|n |y−z|n 2 2 2 x∈BR Therefore, R2 − |x|2 R2 − |x|2 sup n ≤ → 0. |y−z|>δ |y − x| δ/2

0 Theorem 9. Let g ∈ C (∂BR) and let us deﬁne the function u : BR → R by

u(x) = KR(x, y)g(y) dS(y). ˆBR

0 Then u is harmonic in BR, u ∈ C (BR) and for x0 ∈ ∂BR, we have

lim u(x) = g(x0). x→x0 x∈BR

∞ Proof. Since KR(x, y) is harmonic in x, it is also in C . Deﬁne

I(x, y) := KR(x, y)g(y). 52

Let U be compactly contained in BR. Then I : U × ∂BR → R is inﬁnitely diﬀerentiable in α R x and the derivative Dx I(x, y) are continuous maps from U × ∂BR to . 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

α α D u(x) = Dx KR(x, y)g(y) dS(y). ˆ∂BR

Since any x ∈ BR belongs to some open set U b Ω, it follows that u is smooth and

∆u(x) = ∆xKR(x, y)g(y) dS(y) = 0. ˆ∂BR

Now we show the continuity at the boundary. Let x0 ∈ ∂BR. By Property (iii) in 2, we have

u(x) − g(x0) = KR(x, y)(g(y) − g(x0)) dS(y) = I1 + I2, ˆ∂BR where I := K (x, y)(g(y) − g(x )) dS(y), 1 ˆ R 0 ∂BR∩Bδ(x0) and I := K (x, y)(g(y) − g(x )) dS(y), 2 ˆ R 0 ∂BR\Bδ(x0) 0 for some δ > 0. Since g ∈ C (∂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

|I | ≤ K (x, y)|g(y) − g(x )| dS(y) 1 ˆ R 0 ∂BR∩Bδ(x0) ≤ ε K (x, y) dS(y) ˆ R ∂BR∩Bδ(x0)

≤ ε KR(x, y) dS(y) = ε. (3.18) ˆ∂BR

Since g is continuous (and hence bounded) on ∂BR from part (iv) of Lemma 2, we have that

KR(x, y)(g(y) − g(x0)) → 0 as x → x0 uniformly in y ∈ ∂BR \ Bδ(x0). Thus as x → x0, we have I2 → 0.

Together, with the estimate (3.18), we have that limx→x0 u(x) = g(x0) as claimed.

As consequences of the Poisson’s integral formula, we obtain the following 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

|x| |x| 1 − R u(0) 1 + R u(0) n−1 ≤ u(x) ≤ n−1 . |x| |x| 1 + R 1 − R Proof. According to the Poisson’s integral formula we have 1 R2 − |x|2 u(x) = n u(y) dS(y). Rσn−1 ˆ∂BR |y − x|

By the triangle inequality, we have for any y ∈ ∂BR

R − |x| ≤ |y − x| ≤ |x| + R

So that R2 − |x|2 R2 − |x|2 n u(y) dS(y) ≤ u(x) ≤ n u(y) dS(y) σn−1R(R + |x|) ˆ∂BR σn−1R(R − |x|) ˆ∂BR By the mean value property

n−1 u(y) dS(y) = |BR|u(0) = R σn−1u(0). ˆ∂BR Hence

2 2 2 2 R − |x| n−1 R − |x| n−1 n R σn−1u(0) ≤ u(x) ≤ n R σn−1u(0) σn−1R(R + |x|) σn−1R(R − |x|) Simplifying this equality, we obtain

|x| |x| 1 − R u(0) 1 + R u(0) n−1 ≤ u(x) ≤ n−1 |x| |x| 1 + R 1 − R as claimed.

Theorem 11 (Improved Liouville theorem). Suppose that u : Rn → R is harmonic and bounded from below. Then u must be a constant.

Proof. Without loss of generality, we assume u is harmonic in Rn and u ≥ 0. Take x ∈ Rn, according to Theorem 10 , for any R > 0, we have

|x| |x| 1 − R u(0) 1 + R u(0) n−1 ≤ u(x) ≤ n−1 |x| |x| 1 + R 1 − R

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(B (x)) ∩ C0(B (x)) ρ ρ   ∆h = 0, in Bρ(x), =⇒ u ≤ h in Bρ(x).   u ≤ h, on ∂Bρ(x) 

Note that in this deﬁnition, there is no requirement that u ∈ C2(Ω).

0 Lemma 3. If u is C -subharmonic, then for any Bρ(x) b Ω, we have

u(x) ≤ u(y) dS(y) and u(x) ≤ u(y) dy, ∂Bρ(x) Bρ(x) where denotes the average integral. And conversely, if u ∈ C0 and satisﬁes either the

ﬄ 0 inequalities above for any ball Bρ(x) b Ω then u is C -subharmonic.

Proof. By Theorem 9, there exists a function h solving  h ∈ C2(B (x)) ∩ C0(B (x))  ρ ρ  ∆h = 0, in Bρ(x),   u = h, on ∂Bρ(x)

Since u is C0-subharmonic, we have

u(x) ≤ h(x) = h(y) dS(y) = u(y) dS(y), ∂Bρ(x) ∂Bρ(x) where the ﬁrst equality follows from the mean value property for h.

The second estimate of the Lemma is similar.

Now suppose that  h ∈ C2(B (x)) ∩ C0(B (x)) ρ ρ   ∆h = 0, in Bρ(x),   u ≤ h, on ∂Bρ(x)  Then (u − h)(x) ≤ (u − h)(y) dy, Bρ(x) 55 i.e., u − h satisﬁes the mean value property. Recalling that to prove the strong maximum principle, we only need the mean value property and connectedness of the domain. Therefore, it implies that u − h satisﬁes the strong maximum principle. Hence,

u − h ≤ sup (u − h) ≤ 0 in Bρ(x) Bρ(x) i.e., the implication in the deﬁnition holds.

Proposition 4 (Properties of C0-subharmonic functions).

(i) If u is C0-subharmonic function then the strong maximum principle holds. Moreover,

0 we have comparison principle with harmonic functions on every connected set Ω b 0 Ω, i.e., in the deﬁnition 5, one can replace the ball Bρ(x) by Ω .

0 (ii) Let u be C -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), if y ∈ Bρ(x), U(y) =

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).

0 (iii) Let u, v be C -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 .

0 (iv) If u1, . . . , uN are C -subharmonic in Ω, then

u(x) := max u1(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(B (z)) ∩ C0(B (z)) r r   ∆h = 0, in Br(z),   U ≤ h, on ∂Br(z). 

First of all, note that sinceu ¯ ≥ 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, sinceu ¯ = u on ∂Bρ(x), it follows thatu ¯ ≤ u ≤ h in

∂(Br(x)∩Bρ(x)). Sinceu ¯ is harmonic in ∂(Br(x)∩Bρ(x)), we obtain that U =u ¯ ≤ h 0 in Br(x)∩Bρ(x). So U ≤ h in Br(z, which means that U is C -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 thatu ¯ ≤ 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

1 max{f, g} = [f + g + |f − g|], 2

which implies that max{f, g} ∈ C0(Ω). Repeating this argument, we have that if

0 0 u1, . . . , uN ∈ C (Ω) then u = max{u1, . . . , uN } ∈ C (Ω). Let h be the function

satisﬁes the conditions in the deﬁnition 5. Then ui ≤ h in ∂Bρ(x) and hence ui ≤ h 0 in Bρ(x) for all i = 1,...,N since ui are C -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

R 0 Sg := {v :Ω → such that v is C -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) := sup{v(x)}. v∈Sg

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

v˜i(x) = max{inf g, v1(x), . . . , vi(x)}. ∂Ω Then

• inf∂Ω g ≤ v˜1 ≤ v˜2 ≤ ... ≤ v˜i ≤ ... ≤ sup∂Ω g.

• v˜i are subharmonic.

Step 3) Take ρ suﬃciently small such that Bρ(y) b Ω. Let Vi be the harmonic lifting of

v˜i 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

0 Step 4) By Harnack’s theorem 3, Vi → V uniformly in Bρ0(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 existsv ˜ ∈ Sg such that V (z) < v˜(z). We deﬁne

wi = max{v,˜ 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 0 ¯ 2 a function (called a barrier function) Qy(x) ∈ C (Ω) ∩ C (Ω) satisfying

(i) Qy(x) is superharmonic in Ω, 59

¯ (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

lim u(x) = g(y). x∈Ω x→y

Proof. Let ε > 0 and set M = sup∂Ω |g|. Pick y ∈ ∂Ω and let Qy(x) be the barrier ¯ function. By the continuity of g and the positivity of Qy(x) in 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),

u¯y,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 getu ¯y,ε ∈ Sg. Hence

uy,ε ≤ u(x) ≤ u¯y,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).

0 ¯ 0 0 −1 Since Qy ∈ C (Ω and Qy(y) = 0, there exists δ such that if |x−y| < δ then Qy(x) ≤ K ε and |u(x) − g(y)| ≤ 2ε, which implies that limx→y u(x) = g(y) as claimed. 60

Corollary 4. Under the assumptions as in Theorem 13, the Dirichlet problem (3.1) is well-posed.

To conclude this section, we provide a criterion that a barrier function exists on Ω for the point y.

Lemma 4. Suppose that Ω ⊂ Rn is open and bounded, and that y ∈ ∂Ω. If there exists ¯ Br(z) such that Br(z) ∩ Ω = {y}, then there exists a barrier function on Ω for y.

Proof. Consider the function  r2−n − |x − z|2−n, n ≥ 3, Qy(x) =  |x−z| log 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

n n Ta : R \{0} → R \{0} x x 7→ a2 |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

(DT (x)[v],DT (x)[w]) (v, w) a a = (3.21) ||DTa(x)[v]|| ||DTa(x)[w]|| ||v|| ||w||

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 a n−2 v (x) = u(T (x)). a |x| a Show that a n+2 ∆v (x) = ∆u(T (x)). a |x| a

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 (x ) > 0, ∂ν 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

n Ω1 := {x ∈ R : xn > 0} and of the half ball in R3:

3 2 2 2 Ω2 := {(x, y, z) ∈ R : x + y + z = 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 determined by a convolution of the fundamental solution with the function f. The main tool of this chapter will be the theory of distributions. This theory plays an important role in mathematics and is of independent interest.

4.1 The theory of distributions

As a motivation, let us consider the following simple problem: solve

x2 = a.

We knew that if a ≥ 0, this equation has real solutions x ∈ R. However, if a < 0, 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

X α Lu := aαD u = f, |α|≤k ∞ n where aα ∈ C (Ω) for an open Ω ⊂ R , 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 D0(Ω) the space to which our generalised solution u and the right-hand side function f should belong. What properties do we require for D0(Ω) so that we can at least make sense of the PDE (4.1)? The list of desirable properties should contain

1) The smooth functions C∞(Ω) should be contained in D0(Ω) in such a way that we can recover them, i.e., the inclusion map ι : C∞(Ω) ,→ D0(Ω) should be injective.

2) We need to be able to add elements of D0(Ω).

3) We need to be able to multiply elements of D0(Ω) by elements of C∞(Ω).

4) We need to be able to diﬀerentiate elements of D0(Ω). Moreover, the generalised notion of diﬀerentiation should agree with the classical one when restricted to smooth functions.

5) We need some topology on D0(Ω) such that the above operations are continuous.

Before introducing D0(Ω), we need to introduce another space, the space of test functions

∞ D(Ω) := Cc (Ω). A topology on D(Ω) is deﬁned as follows.

∞ Deﬁnition 7 (Convergence in D(Ω)). A sequence (φ)j∈N, φj ∈ D(Ω) = Cc (Ω) converges to φ ∈ D(Ω) if there exists a compact set K ⊂ Ω such that supp(φj), supp(φ) ⊂ K for all 65 j ∈ N and that

α α D φj → D φ as j → ∞ uniformly in K for any multi-index α.

The space D(Ω) equipped with this topology becomes a topological vector space over the real number, i.e, a vector space together with a topology such that addition of vectors and scalar multiplication are continuous functions. For any topological vector space V , one can deﬁne the continuous dual space which consists of continuous linear maps

V 0 = {ω : V → R| ω linear, continuous}.

The space D0(Ω) that we are seeking for will be the continuous dual space of D(Ω).

Deﬁnition 8. A distribution T ∈ D0(Ω) is a linear functional on the space of test functions

T : D(Ω) → R which is continuous with respect to the topology on D(Ω) defined in Definition 7. In other words, if φj → φ in the sense of Definition 7 then

T φj → T φ as j → ∞.

We consider two important examples

Example 20. 1. If f :Ω → R is continuous, the distribution Tf associated to f is deﬁne by

∞ Tf (φ) := f(x)φ(x) dx ∀φ ∈ Cc (Ω). ˆΩ 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.

0 Proposition 5. Suppose that f ∈ C (Ω). Let ρk,y, k = 1, 2,..., be a family of molliﬁers ∞ concentrating at y, i.e., ρk,y ∈ Cc (Ω), ρk,y ≥ 0 and

ρk,y(x) dx = 1, supp(ρk,y) ⊂ B1/k(y). ˆΩ 66

Then

f(y) = lim Tf (ρk,y). k→∞ 0 As a consequence, if f, g ∈ C (Ω) and Tf = Tg then f = g.

Proof. Since Ω ρk,y(x) dx = 1, we have that ´

f(y) − Tf (ρk,y) = (f(y) − f(x)) ρk,y(x) dx. ˆΩ 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)| ≤ |f(y) − f(x)| ρk,y(x) dx ≤ ε ρk,y(x) dx = ε, ˆΩ ˆΩ which implies that f(y) = limk→∞ Tf (ρk,y).

0 Now suppose that f, g ∈ C (Ω) and Tf = Tg. For any y ∈ Ω, let ρk,y be the molliﬁers concentrating at y. Then

f(y) = lim Tf (ρk,y) = lim Tg(ρk,y) = g(y), k→∞ k→∞ so that f = g as claimed.

Next, we extend the operations on functions to distributions as follows.

0 1) Given T1,T2 ∈ D (Ω). The distribution T1 + T2 is deﬁned by

n (T1 + T2)(φ) := T1(φ) + T2(φ), ∀φ ∈ D(R ).

2) Given a ∈ C∞(Ω) and T ∈ D0(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) = Dif(x) φ(x) dx = − f(x) Diφ(x) dx = −Tf (Diφ), ˆΩ ˆΩ 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). 67

Let us do some examples

Example 21. 1) T = δx. Then

n (DiT )(φ) = −T (Diφ) = − δxDiφ(y) dy = −(Diφ)(x), ∀φ ∈ D(R ). ˆΩ

2) Consider the Heaviside step function  1, x ≥ 0, H(x) = (x ∈ R). 0, x < 0

Note that this function is not diﬀerentiable at x = 0. We consider the distribution TH by

n TH (φ) = H(x)φ(x) dx, for φ ∈ D(R ). ˆR Then we compute

(DxTH )(φ) = −TH (Dxφ)

= − H(x)Dxφ(x) dx ˆR ∞ = − Dxφ(x) dx ˆ0

= 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.

0 Theorem 14. Suppose that f ∈ C (Ω) and let Tf be the distribution associated to f. 0 Suppose further that for any multi-index α, with |α| ≤ k, there exist gα ∈ C (Ω) such that

α D Tf = Tgα

α k α where D is the distributional derivative. Then f ∈ C (Ω) and D f = gα in the classical sense.

This theorem states that the distributional derivatives agree with the classical ones when they are both deﬁned. 68

We have addressed the four wish-lists that we asked in the introduction. It is straightforward to check that all the operations that we have deﬁned are indeed continuous with respect to the topology that we have introduced.

The advantage of introducing distributions is expressed in the following result, which is followed from Theorem 14 and all the properties we have deﬁned.

Proposition 6. Suppose that f ∈ C0(Ω) and that T ∈ D0(Rn) satisﬁes the distributional equation

X α aαD T = Tf , |α|≤k ∞ 0 α where aα ∈ C (Ω) and that there exist functions uα ∈ C (Ω) such that D T = uα. Then k u := u0 is in C (Ω) and is a classical solution of the PDE

X α aαD u = f. (4.2) |α|≤k

This proposition provides us a method to solve the PDE (4.2) that consists of two steps.

1) Lift up the PDE (4.2) to the distributional PDE

X α aαD T = Tf |α|≤k and solve this distributional equation.

2) Check whether a distributional solution is a classical one.

In the next section, we will study how to solve the distributional equation above.

4.2 Convolutions and the fundamental solution

0 Rn ∞ Rn Recalling that for f ∈ C ( ) and φ ∈ Cc ( ), the convolution f ∗ φ is again a function defined by (f ∗ φ)(x) := f(y)φ(x − y) dy. ˆRn 0 Rn ∞ Rn How to extend this definition to T ∗φ where T ∈ D ( ), φ ∈ Cc ( )? To do so we define

n τxφ : R → R y 7→ φ(x − y). 69

∞ Rn If φ ∈ Cc ( ) 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 ∈ D0(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 ∈ D0(Rn) is supported in the closed set K ⊂ Ω if

∞ T φ = 0 ∀φ ∈ Cc (Ω \ K), i.e, T vanishes when applying to any test function that vanishes on K. The support of T is deﬁned by \ suppT = {K : T supported in K}.

Example if T = δx then suppT = {x}.

0 n n Proposition 7 (Properties of convolutions). Suppose T,Ti ∈ D (R ) and φ ∈ D(R ). Then

n (i) If T1 ∗ φ = T2 ∗ φ, ∀φ ∈ D(R ) 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. 70

(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 + εe ) − (T ∗ φ)(x) 1h i i = T (τ φ) − T (τ φ) ε ε x+εei x h1 i = T (τ φ − τ φ) , ε x+εei x

where we have used the linearity of T . Since φ ∈ D(Rn), we have

1 1 lim (τx+he φ − τxφ)(y) = lim (φ(x − y + εei) − φ(x − y)) ε→0 ε i ε→0 ε ∂ = φ(x − y) = (τxDiφ)(y), ∂xi with convergence in the topology of D(Rn). By the continuity of distributions, we have (T ∗ φ)(x + εei) − (T ∗ φ)(x) lim = T (τxDiφ) = (T ∗ Diφ)(x), ε→0 ε

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

∂|α| Dα(τ φ)(y) = φ(x − y) x ∂yα = (−1)|α|(Dαφ)(x − y)

|α| α = (−1) (τxD φ)(y),

α |α| α so that D (τxφ) = (−1) (τxD φ).

Therefore

α α (D T ∗ φ)(x) = D T [τxφ]

|α| α = (−1) T (D τxφ)

|α| |α| α = (−1) T ((−1) τxD φ)

α = T (τxD φ) = T ∗ Dαφ(x).

Hence DαT ∗ φ = T ∗ Dαφ.

(iii) Suppose without loss of generality that T and φ are supported in BR(0) for some large n R. Let x ∈ R be such that |x| > 2R. Then if y ∈ BR(0), by the triangle inequality, ∞ Rn |y − x| > R. So that (τxφ)(y) = φ(x − y) = 0, i.e., τxφ ∈ Cc ( \ 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 distribution with a test function. Next, we want to extend the convolution of two distributions: 0 Rn 0 Rn given T1,T2 ∈ D ( ), how to deﬁne T1 ∗ T2? We recall that if f, g, h ∈ Cc ( ), then

(f ∗ g) ∗ h = f ∗ (g ∗ h).

In deed,

[(f ∗ g) ∗ h](x) = (f ∗ g)(y)h(x − y) dy ˆRn = f(z)g(y − z) dz h(x − y) dy ˆRn ˆRn = f(z) g(y − z)h(x − y) dy dz ˆRn ˆRn = f(z) g(˜y)h(x − y˜ − z) dy˜ dz ˆRn ˆRn = f(z)(g ∗ h)(x − z) dz ˆRn = (f ∗ (g ∗ h))(x).

Motivated by this we deﬁne

0 n Definition 10. Suppose that T1,T2 ∈ D (R ) and that T2 has compact support. The convolution T1 ∗ T2 is defined to be the unique distribution that satisfies

n (T1 ∗ T2) ∗ φ = T1 ∗ (T2 ∗ φ), ∀φ ∈ D(R ).

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.

0 n Theorem 15. Suppose that T1,T2 ∈ D (R ) and that T2 has compact support. Then

α α α D (T1 ∗ T2) = T1 ∗ D T2 = D T1 ∗ T2. 72

Proof. We have

α α D (T1 ∗ T2) ∗ φ = (T1 ∗ T2) ∗ D φ

α = T1 ∗ (T2 ∗ D φ)

α = T1 ∗ (D T2 ∗ φ)

α = (T1 ∗ D T2) ∗ φ

α α so that D (T1 ∗ T2) = (T1 ∗ D T2). Similarly,

α α (T1 ∗ D T2) ∗ φ = T1 ∗ (D T2 ∗ φ)

α = T1 ∗ D (T2 ∗ φ)

α = D T1 ∗ (T2 ∗ φ)

α = (D T1 ∗ T2) ∗ φ,

α α so that (T1 ∗ D T2) = (D T1 ∗ T2) and we are done.

n Example 22. (i) If φ ∈ D(R ), then δ0 ∗ φ = φ.

0 n (ii) If T ∈ D (R ) 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

X α L := aαD |α|≤k where aα are constants, if it satisﬁes the distributional equation

LG = δ0.

0 n Lemma 5. Suppose that G ∈ D (R ) is a fundamental solution of L and that T0 is a distribution with compact support. Then the distribution G ∗ T0 solves the distributional equation

X α LT = aαD T = T0. |α|≤k 73

Proof. We have

X α L(G ∗ T0) = aαD (G ∗ T0) |α|≤k

X α = aα(D G ∗ T0) |α|≤k

X α = aαD G ∗ T0 |α|≤k

= (LG) ∗ T0

= δ0 ∗ T0 = T0.

4.3 Poisson’s equation

We will apply the abstract framework in the previous section to solve Poisson’s equation: given f : Rn → R has compact support. Find u such that

∆u = f in Rn.

First of all, we notice that, for n ≥ 3, classical solutions to this problem are unique (if they exist) if we assume that u → 0 as |x| → ∞. Indeed, suppose that u1, u2 satisfy both n Poisson’s equation and the decay condition. Then w = u1 − u2 is harmonic in R 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. 2 Rn Next, we will show that provided ρ ∈ Cc ( ), 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. 74

4.3.1 The fundamental solution

Recall that in Chapter 3, we have deﬁned the function   1 log |x|, n = 2, Φ(x) = 2π  1 |x|2−n, n ≥ 3. (2−n)σn−1

∞ Rn Furthermore, for any φ ∈ Cc ( ) we have that

φ(x) = lim Φ(x − y)∆φ(y) dy. ε→0 n ˆR \Bε(x) By changing variable z = x − y, the above identity can be written as

φ(x) = lim Φ(z)∆φ(x − z) dz = (G ∗ ∆φ)(x), ε→0 n ˆR \Bε(0) where the distribution G is deﬁned by

Gφ = lim Φ(z)∆φ(z) dz. ε→0 n ˆR \Bε(0)

So we have φ = G ∗ ∆φ = ∆G ∗ φ, which means that ∆G = δ0, i.e., G is the fundamental solution to Poisson’s equation.

0 Rn As a consequence, if f ∈ Cc ( ), then

∆(G ∗ Tf ) = Tf

i.e., G ∗ Tf is a solution to the distributional equation ∆T = Tf .

k Rn Lemma 6. Suppose f ∈ Cc ( ). Then

G ∗ Tf = Tu, where u(x) := lim Φ(x − y) f(y) dy ε→0 n ˆR \Bε(x) and u ∈ Ck(Rn).

∞ Rn Rn Proof. Let φ ∈ Cc ( ). Let x ∈ 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)]

= lim Φ(y)[τx(Tf ∗ φ)](y) dy ε→0 n ˆR \Bε(0)

= lim Φ(y)(Tf ∗ φ)(x − y) dy ε→0 n ˆR \Bε(0) h i = lim Φ(y) f(z)φ(x − y − z) dz dy ε→0 n n ˆR \Bε(0) ˆR h i = lim Φ(y) f(z)φ(x − y − z) dz dy ε→0 ˆB3R(0)\Bε(0) ˆBR(0) h i = lim Φ(y) f(w − y)φ(x − w) dw dy ε→0 ˆB3R(0)\Bε(0) ˆB4R(0) h i = lim Φ(y)f(w − y) dy φ(x − w) dw ε→0 ˆB4R(0) ˆB3R(0)\Bε(0) h i = lim Φ(w − z)f(z) dz φ(x − w) dw ε→0 n ˆB2R(0) ˆR \Bε(w) Deﬁne

uε(w) = Φ(w − z)f(z) dz. n ˆR \Bε(w)

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 h i [(G ∗ Tf ) ∗ φ](x) = lim Φ(w − z)f(z) dz φ(x − w) dw ε→0 n ˆB2R(0) ˆR \Bε(w) = u(w)φ(x − w) dw ˆB2R(0)

= (Tu ∗ φ)(x), so that G ∗ Tf = Tu. k Rn k Rn Next we show that u ∈ C ( ). Since f ∈ Cc ( ), we have

uε(x) = Φ(w − z)f(z) dz = Φ(y)f(x − y) dy. n ˆR \Bε(w) ˆBR(0)\Bε(0) If |α| ≤ k, then

α α D uε = Φ(y)D f(x − y) dy ˆB2R(0)\Bε(0) = Φ(x − y)Dαf(y) dy ˆBR(0)\Bε(0) = Φ(x − y)Dαf(y) dy n ˆR \Bε(0) 76

α which converges uniformly to some limit by our previous results. Since uε and D uε 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.

2 Rn Theorem 16. If f ∈ Cc ( ) then

u(x) = lim Φ(x − y) f(y) dy ε→0 n ˆR \Bε(x) is a classical to Poisson’s equation. Furthermore, if n ≥ 3 then u → 0 as |x| → ∞.

2 n Proof. By Lemma 6 u ∈ C (R ) 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 |x| R > 0 be such that supp(f) ⊂ BR(0). For |x| > 2R and y ∈ BR(0) then |y| ≤ 2 , so that |x| |x − y| > 2 . We then can estimate n−2 Vol(BR(0)) 2 |u(x)| = Φ(x − y) f(y) dy ≤ sup |f|, Rn ˆBR(0) (n − 2)σn−1 |x| 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

2 0 2 0 in Ω is: given f ∈ Cc (Ω) and g ∈ C (∂Ω), ﬁnd u ∈ C (Ω) ∩ C (Ω) such that  ∆u = f, in Ω, (4.3) u = g, in ∂Ω. This problem can be solved in two steps as follows. Each step is a problem that we know how to solve.

Step 1) Find w ∈ 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.

0 ∞ Exercise 18. a) Show that if f1, f2 ∈ C (Ω) and a ∈ C (Ω) then

aTf1 + Tf2 = Taf1+f2 . b) Show that if f ∈ Ck(Ω) then

α D Tf = TDαf

for |α| ≤ k.

78 79 c) Deduce that if f ∈ Ck(Ω) then

X α aαD Tf = TLf , |α|≤k

where

X α Lf = aαD f. |α|≤k

0 Rn Exercise 19. a) Show that if f, g, h ∈ Cc ( ) then

Tf∗g = Tf ∗ Tg.

0 n b) Show that convolution is linear in both of its arguments, i.e, if Ti ∈ D (R ) 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

n In this chapter we will study the heat equation: given u0 : R → R. We want to ﬁnd a function u : Rn × [0, ∞) that satisﬁes the following heat equation

n ∂tu(x, t) = ∆u(x, t), x ∈ R , t > 0, (5.1)

n u(x, 0) = u0(x), x ∈ R . (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

2 2 2 2 ∂tv(x, t) = λ ∂tu(λx, λ t) and ∆v(x, t) = λ ∆u(λx, λ t).

2 2 Thus ∂tv(x, t) = λ [∂tu − ∆u](λx, λ t) = 0 as desired.

80 81

Lemma 8. If u satisﬁes (5.1) and Du vanishes at the inﬁnity then for any t > 0

u(x, t) dx = u0(x) dx. ˆRn ˆRn

Proof. We calculate

d u(x, t) dx = ∂tu(x, t) dx dt ˆRn ˆRn = ∆u(x, t) dx ˆRn = div(Du(x, t)) dx ˆRn = 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

|x|2 u(x, t) = h(t)f(ξ) where ξ = ∈ R, t for some function h and f, such that Rn u(x, t) dx = Rn u0(x) dx = 1. The last identity ´ ´ is assumed without loss of generality.

We ﬁrst ﬁnd h. We have

|x|2 1 = u(x, t) dx = h(t) f dx ˆRn ˆRn t ∞ r2 = h(t) f rn−1 dw dr. ˆSn−1 ˆ0 t

1 r2 1 t t 2 By changing of variable y = , we have r = (yt) 2 and dr = dy = 1 dy. Substituting t 2r 2y 2 this back into the calculations above, we get

∞ n σn−1 n −1 1 = t 2 h(t) f(y)y 2 dy. 2 ˆ0 This implies that

∞ − n σn−1 n −1 h(t) = t 2 and 1 = f(y)y 2 dy. 2 ˆ0 82

It remains to ﬁnd f. We have

2 2 − n |x| n − n −1 − n |x| 0 − n −1 hn 0 i ∂ u = ∂ t 2 f = − t 2 f(ξ) − t 2 f (ξ) = −t 2 f(ξ) + ξf (ξ) t t 2 2 t2 2

− n xi 0 D u = 2t 2 f (ξ), i t n 2 2 2t 2 0 − n xi 00 D u = f (ξ) + 4t 2 f (ξ) ii t t2 n X n 2 − 2 −1 0 00 ∆u = Diiu = t (2nf (ξ) + 4ξf (ξ)) . i=1 Therefore, we obtain the following equation for f(ξ)

n 4ξf 00(ξ) + ξf 0(ξ) + 2nf 0(ξ) + f(ξ) = 0. 2

0 1 Set g(ξ) = f (ξ) + 4 f(ξ), the above equation can be written as n ξg0(ξ) + g(ξ) = 0, 2 which implies that

− n g(ξ) = Aξ 2 for some constant A. Therefore,

0 1 − n f (ξ) + f(ξ) = Aξ 2 , 4 i.e.,

n 0 1 A = ξ 2 (f (ξ) + f(ξ)). 4 Assume that f and f 0 decay fast enough at the inﬁnity, we obtain that the right-hand side vanishes as |ξ| → ∞. Thus A = 0 and so

1 f 0(ξ) + f(ξ) = 0. 4

ξ By multiplying with the integrating factor e 4 , we obtain

d ξ e 4 f(ξ) = 0 dξ

Solving this equation, we ﬁnd

− ξ f(ξ) = Be 4 . (5.3) 83

We now ﬁnd B. We have

∞ σn−1 − ξ n −1 1 = B e 4 ξ 2 dξ 2 ˆ0 ∞ n −s2 n−1 2 = Bσn−14 2 e s ds (change of variable ξ = 4s ) ˆ0 n −|x|2 = B4 2 e dx ˆRn ∞ n −x2 √ n = B 2 e dx = B(2 π) 2 . ˆ0

√1 so B = (2 π)n . Conclusion

2 2 n 1 |x| 1 |x| − 2 − 4t − 4t u(x, t) = t n e = n e . (2π) (4πt) 2 To summarise, we have proved the following result

Proposition 8. The function

1 |x|2 − 4t Γ(x, t) = n e (4πt) 2 is a solution of the equation (5.1) and satisﬁes

Γ(x, t) dx = 1, ∀t > 0. ˆRn

The function Γ is called the heat kernel or the fundamental solution of the heat equation. It plays a similar role to the Green’s function for the Laplace operator, which can be seen in the following theorem.

0 Rn Rn R Theorem 17. Let u0 ∈ Cc ( ). Deﬁne u : × (0, ∞) → by

1 |x−y|2 − 4t u(x, t) := Γ(x − y, t)u0(y) dy = n e u0(y) dy. (5.4) ˆRn (4πt) 2 ˆRn Then u solves the heat equation (5.1)-(5.2). That is, u satisﬁes

n ∂tu(x, t) = ∆u(x, t) ∀(x, t) ∈ R × (0, ∞), and

n lim u(x, t) = u0(x) ∀x ∈ R . t→0

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) = Γ(x − y, t)u0(y) dy. ˆBR(0) This implies that u(x, t) is also smooth in Rn × (0, ∞) and we can diﬀerentiate under the integral and obtain that

n (∂tu − ∆u)(x, t) = (∂t − ∆x)Γ(x − y, t)u0(y) dy = 0 ∀(x, t) ∈ R × (0, ∞). ˆBR(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

n x, z ∈ R with |x − z| < δ then |u0(x) − u0(z)| < ε. Since

Γ(x − y, t) dy = 1, ˆRn we can write

u(x, t) − u0(x) = Γ(x − y, t)(u0(y) − u0(x)) dy ˆRn and estimate

|u(x, t) − u0(x)| ≤ Γ(x − y, t)|u0(y) − u0(x)| dy ˆRn

= Γ(x − y, t)|u0(y) − u0(x)| dy + Γ(x − y, t)|u0(y) − u0(x)| dy. ˆ ˆ n Bδ(x) R \Bδ(x)

:= I1 + I2.

Since for y ∈ Bδ(x), we have |u0(y) − u0(x)| < ε, the term I1 can be estimated

I1 ≤ ε Γ(x − y, t) dy ≤ ε Γ(x − y, t) dy = ε. ˆ ˆ n Bδ R

The term I2 can be estimated as follows

I2 ≤ 2 sup |u0| Γ(x − y, t) dy Rn ˆ Bδ(x)

= 2 sup |u0| Γ(y, t) dy Rn ˆ Bδ(0) ∞ σn−1 s2 − 4t n−1 = 2 sup |u0| n e s ds Rn (4πt) 2 ˆδ ∞ σn−1 −z2 n−1 = 2 sup |u0| n e z dz → 0 as t → 0. Rn (π) 2 ˆ √δ 2 t 85

0 Therefore, there exists δ > 0 depending on δ (thus on ε but not on x) such that I2 < ε for t < δ0. In conclusion, we have shown that for any ε > 0 there exists δ0 > 0 such that for t < δ0 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.

0 Rn Theorem 18. Let u ∈ Cc ( ) and let u(x, t) be deﬁned as in (5.4). Then

n |u(x, t)| ≤ max |u0| ∀(x, t) ∈ R × (0, ∞). Rn

Proof. This estimate is straightforward:

|u(x, t)| ≤ Γ(x − y, t)|u0(y)| dy ≤ max |u0| Γ(x − y, t) dy = max |u0|. n n ˆRn R ˆRn R

Furthermore, we now show that the solution of the heat equation decays in time.

0 Rn Theorem 19. Let u ∈ Cc ( ) and let u(x, t) be deﬁned as in (5.4). Then there exists a constant C depending on n and u0 such that

C n |u(x, t)| ≤ n ∀(x, t) ∈ R × (0, ∞). t 2

1 Proof. Suppose that u0 is supported in BR(0). Since Γ(x − y, t) ≤ n , we can estimate (4πt) 2

|u(x, t)| ≤ Γ(x − y, t)|u0(y)| dy ˆBR(0)

≤ max |u0| Γ(x − y, t) dy Rn ˆBR(0) 1 C ≤ max |u0| n Vol(BR(0)) = n , Rn (4πt) 2 t 2

1 where C = max |u0| n Vol(BR(0). Rn (4π) 2

Theorem 17 only tells us that the function u deﬁned by (5.4) is a solution to the heat equation, but does not ensure it uniqueness. To obtain a statement about the uniqueness, we will need to use the maximum principle in the next sections. 86

5.2 The maximum principle for the heat equation on a bounded domain

We will show that the heat equation also satisﬁes a similar maximum principle as harmonic functions.

Deﬁnition 12. Let Ω ⊂ Rn be open and bounded and T > 0 be given. The space-time domain ΩT is deﬁned by

ΩT = Ω × (0,T ).

The parabolic boundary of ΩT is deﬁned by

∂pΩT = ∂(ΩT ) \ (Ω × {T }).

2,1 We denote by C (ΩT ) the set of functions that are continuously diﬀerentiable twice 2,1 in space and once in time. For functions u ∈ C (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).

n 2,1 0 Suppose that Ω ⊂ R is open and bounded, T > 0 and that u ∈ C (ΩT ) ∩ C (ΩT ) satisﬁes

Lu ≤ 0 in ΩT .

Then sup u = sup u. ΩT ∂pΩT

If instead Lu ≥ 0 in ΩT , then inf u = inf u. ΩT ∂pΩT

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

sup v = sup v ≤ sup u ≤ sup u. ΩT −ε ∂pΩT −ε ∂pΩT −ε ∂pΩT On the other hand, since u ≤ v + εT , we have

sup u ≤ sup v + εT ≤ εT + sup u. (5.7) ΩT −ε ΩT −ε ∂pΩT

Since u is uniformly continuous on ΩT , we have that

lim sup u = sup u. ε→0 ΩT −ε ΩT Thus taking ε → 0 in (5.7), we obtain that

sup u = lim sup u ≤ lim (εT + sup u) = sup u ≤ sup u. (5.8) ε→0 ε→0 ΩT ΩT −ε ∂pΩT ∂pΩT ΩT Therefore, all of the inequalities in (5.8) can be replaced with inequalities. In particular, we get sup u = sup u ∂pΩT ΩT as desired.

To obtain the result for Lu ≥ 0, we note that −u satisﬁes L(−u) ≤ 0 and apply the ﬁrst part of the theorem. 88

Corollary 5 (The maximum principle for solutions of the heat equation on bounded

2,1 0 domain). Suppose u ∈ C (ΩT ) ∩ C (ΩT ) solves the heat equation in ΩT . Then

max |u| = max |u|. ΩT ∂pΩT 0 0 Corollary 6. Let f ∈ C (ΩT ) and g ∈ C (∂pΩT ). Then there is at most one solution 2,1 0 u ∈ C (ΩT ) ∩ C (ΩT ) of the boundary value problem  ut − ∆u = f in ΩT ,

u = g on ∂pΩT ,

Proof. Suppose there are two solutions u1 and u2. Then u = u1 −u2 satisﬁes the equation  ut − ∆u = 0 in ΩT ,

u = 0 on ∂pΩT ,

Apply Corollary 5 we obtain that

max |u| = max |u| = 0, ΩT ∂pΩT 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

n ST = R × (0,T ) the space-time domain in this case.

2,1 Theorem 21 (The maximum principle for the heat operator on ST ). Let u ∈ C (ST ) ∩ 0 C (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 sup u = sup u(x, 0). n ST x∈R

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 inf u = inf u(x, 0). n ST x∈R

Proof. We will prove the minimum principle. The maximum principle then follows by changing u to −u. Suppose that Lu ≥ 0 in ST and there exist constants C, α > 0 such that u(x, t) ≥ −Ceα|x|2 , ∀x ∈ Rn, 0 ≤ t ≤ T.

We need to show that inf u = inf u(x, 0). n ST x∈R 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 1 such that T1 := 8β < T . The idea of the proof is as follows. We ﬁrst show that u ≥ I in

ST1 and then by dividing the time interval [0,T ] into finitely many intervals of length less than T1 and successively applying the result of the first step. We define 2 β |x| e 1−4βt v(x, t) := n . (1 − 4βt) 2 By direct computations, we have

2 β|x|2 − n −1 n β|x| ∂ v = 4β(1 − 4βt) 2 + e 1−4βt , t 2 1 − 4βt 2 n β|x| − 2 −1 1−4βt ∂xi v = (1 − 4βt) 2βxie ,

2 β|x|2 2 − n −1 1 βxi ∂ v = 4β(1 − 4βt) 2 + e 1−4βt , xixi 2 1 − 4βt n 2 β|x|2 X 2 − n −1 n β|x| ∆v = ∂ v = 4β(1 − 4βt) 2 + e 1−4βt , xixi 2 1 − 4βt i=1

−1 which implies that v(x, t) satisﬁes the heat equation in ST1 . In addition, since (1−4βt) >

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 ≥ εeβr2 − Ceαr2 − I. ∂Br(0)×[0,T1] 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

inf w = inf w ≥ 0, Br(0)×(0,T1) ∂p(Br(0)×(0,T1)) so that w ≥ 0 in Br(0) × (0,T1) for any r > R. Thus w ≥ 0 in ST1 for any ε > 0 and so u ≥ I. Now dividing the interval [0,T ] into ﬁnitely many intervals of lengths less than T1 and successively applying the result on each interval, we obtain

u(x, t) ≥ inf u(y, 0) y∈Rn for all (x, t) ∈ ST . Taking the inﬁmum over ST we get that

inf u ≥ inf u(x, 0). n ST x∈R On the other hand, it is obvious that

inf u ≤ inf u(x, 0). n ST x∈R

n Therefore, we have infST u = infx∈R u(x, 0) as desired.

2,1 0 Corollary 7. Suppose that u ∈ C (ST ) ∩ C (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

n ∂tu(x, t) = ∆u(x, t), x ∈ R , t > 0,

n u(x, 0) = u0(x), x ∈ R ,

0 n where u0 ∈ C (R ) 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

α|x|2 0 α0|x|2 00 α00|x|2 |w(x, t)| ≤ |u1(x, t)| + |u2(x, t)| ≤ Ce + C e ≤ C e , where C00 = 2 max{C,C0} and α00 = max{α, α0}. In addition, w solves  n wt − ∆w = 0 in R × (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 1 G(x) = |x|2−n. σn−1(2 − n) 2 Rn Let ρ ∈ Cc ( ) be supported in BR(0) and deﬁne u by

u(x) = lim G(x − y)ρ(y) dy. ε→0 n ˆR \Bε(x) Show that for |α| ≤ 2, R2 sup |Dαu| ≤ sup |Dαρ|. Rn 2(n − 2) Rn Deduce that the problem of ﬁnding u ∈ C2(Rn) such that u → 0 as |x| → ∞ satisfying

∆u = ρ in Rn is well-posed.

∞ Rn R Exercise 21 (The inhomogeneous heat equation). (a) Let φ ∈ Cc ( × ). Show that

lim Γ(x, t)φ(x, t) dx = φ(0, 0), t→0 ˆRn where Γ denotes the fundamental solution of the heat equation.

Deduce that the function t 7→ Γ(x, t)φ(x, t) dx, ˆRn is uniformly continuous on (0, ∞) and vanishes for large values of t.

(b) Deﬁne the distribution T ∈ D0(Rn × R) by

∞ ∞ Rn R T φ = Γ(x, t)φ(x, t) dx dt, ∀φ ∈ Cc ( × ). ˆ0 ˆRn ∞ Rn R Let δ > 0. Show that, for any φ ∈ Cc ( × ), we have δ ∞ DtT (φ) = − Γ(x, t)φt(x, t) dx dt+ Γ(x, δ)φ(x, δ)dx+ Γt(x, t)φ(x, t) dx dt, ˆ0 ˆRn ˆRn ˆδ

92 93

and

δ ∞

Dxixj T (φ) = Γ(x, t)φxixj (x, t) dx dt + Γxixj (x, t)φ(x, t) dx dt. ˆ0 ˆRn ˆδ

δ (DtT − ∆T )(φ) = − Γ(x, t)(φt + ∆φ)(x, t) dx dt + Γ(x, δ)φ(x, δ) dx. ˆ0 ˆRn ˆRn 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

n ut − ∆u = f in R × R,

∞ Rn R where f ∈ Cc ( × ).

n Exercise 22. Let Ω = {x ∈ R : xn > 0} be the half-space. Consider the boundary value problem  u = ∆u in Ω × (0,T )  t  u = 0 on ∂Ω × (0,T )   u = u0 on Ω × {0}.

0 Suppose that u0 ∈ Cc (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)

i m i We relate this deﬁnition to the usual partial derivatives. If f = (f )i=1, i.e., f is the i-th component of f(x) with respect t the canonical basic of Rm, we deﬁne the partial

i derivative ∂f (x) to be the limit (if exists) ∂xj

∂f i f(x , . . . , x + ε, . . . , x ) − f(x , . . . , x , . . . , x ) (x) = lim 1 j n 1 j n . ∂xj ε→0 ε

i In other words, the partial derivative of f in the xj direction is simply the derivative of i f when considering it as a function of xj alone. By setting h = εej in (5.9), we see that i if f is diﬀerentiable at x then ∂f (x) exists for all i = 1, . . . , m; j = 1, . . . , n. To obtain ∂xj the conserve, we need a bit stronger condition.

i Lemma 9. Suppose that the partial derivative ∂f (x) exist and are continuous on an open ∂xj 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   ∂f1 ∂f1 ∂f1 ∂x (x) ∂x (x) ... ∂x (x)  1 2 n   ∂f2 (x) ∂f2 (x) ... ∂f2 (x)  ∂x1 ∂x2 ∂xn      Df(x) =      . . .. .   . . . .    ∂fm (x) ∂fm (x) ... ∂fm (x) ∂x1 ∂x2 ∂xn

95 96 or in a compact way i i ∂f [Df(x)]j = . ∂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

2 i 2 i ∂ f [D f(x)]j1j2 = (x). ∂xj1 ∂xj2 Lemma 10. If f ∈ Ck(Ω) if and only if all the partial derivatives ∂kf i : U → R ∂xj1 ∂xj2 . . . ∂xjk exist and are continuous in Ω.

n We now introduce multi-indices notations. Let α = (α1, . . . , αm) ∈ (Z≥0) , we deﬁne Pn |α| := i=1 αi and ∂|α|f ∂ α ∂ α α = ... , ∂x ∂x1 1 ∂xn n i.e., we diﬀerentiate α1 times with respect to x1, α2 times with respect to x2 and so on. We also use more compact notation

|α| ∂ α ∂ Di := ,D := α . ∂xi ∂x

Similarly for h = (h1, . . . , hn), we denote

α α1 αn h = h1 . . . hn .

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, R k n x ∈ Ω and f : U → belongs to C (Ω). Then writing h = (hj)j=1 and using multi-indices notation, we have X hα f(x + h) = f(x) + Dαf(x) + R (x, h), α! k |α|≤k 97

where the sum is taken over all multi-indices α = (α1, . . . , αn) with |α| ≤ k and the remainder term satisﬁes R (x, h) k → 0 as h → 0. |h|k 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.