EXAMENSARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET

EXAMENSARBETEN I MATEMATIK

MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET

The Heat Equation

av

Myrto Barrdahl

2008 - No 4

The Heat Equation

Myrto Barrdahl

Examensarbete i matematik 30 h¨ogskolepo¨ang, f¨ordjupningskurs Handledare: Andrzej Szulkin

Abstract

The aim of this thesis is to investigate existence and uniqueness of solutions to the heat conduction equation in Rⁿ and in bounded domains in Rⁿ. Two different types of solutions to a non-linear diffusion equation, known as the Porous Medium Equation, are also presented.

Contents

1 Introduction 3

2 Preliminaries 6

2.1 Definitions . . . 6

2.2 Distributions and weak solutions . . . 9

3 The Heat Equation in a Bounded Domain 15 3.1 Eigenvalues . . . 17

3.2 Separation of variables . . . 23

4 The Pure Initial Value Problem 34 4.1 The Fourier Transform . . . 34

4.2 Regularity . . . 41

4.3 The non-homogeneous problem . . . 42

4.4 Scaling . . . 47

5 The Porous Medium Equation 49 5.1 Separation of Variables . . . 49

5.2 Similarity under Scaling . . . 50

Chapter 1 Introduction

Heat is distributed in a body the same way that a fluid or gas would be flowing from parts of lower concentration to parts of higher concentration - a process known as diffusion. The transfer of thermal energy in a domain Ω ⊂ Rⁿ is carried out by means of conduction and can be described by the following partial differential equation

K4u = ut, x ∈ Ω, t > 0

which is known as the homogeneous Diffusion or Heat Equation. Here u(x, t) represents the temperature of the body or the density of the fluid or gas at a given point and time instant x and t respectively and K is a constant of thermal conductivity that we will set equal to 1. The equation is normally stated together with an initial condition and (when not defined in all of Rⁿ) also a boundary condition such as

 u(x, t) = 0, x ∈ ∂Ω, t > 0 u(x, 0) = g(x), x ∈ Ω.

We shall assume that g(x) is smooth. The boundary condition is referred to as the homogeneous Dirichlet condition if the temperature is kept at zero on the boundary. If the flow of heat across the boundary is controlled, the boundary condition is called the Neumann condition and if the heat flow obeys Newton’s law of cooling it is known as the Robin condition. This thesis however, is concerned only with the homogeneous Dirichlet condition, both for the homogeneous heat equation and for the non-homogeneous case

u_t = 4u + f (x, t), x ∈ Ω, t > 0

where we shall assume that f ∈ L²(Ω).

The heat equation was first presented by J.B. Fourier (1768-1830) (al- though in a slightly different form from the one given here) in 1807 and he solved many important special cases of it through a combination of expan- sions in Fourier series and Fourier integrals and separation of variables. Even though Euler, D’Alembert and Lagrange, all former contributors of work on solutions in the form of trigonometric series viewed this approach with great suspicion, Fourier’s methods greatly influenced the solutions of other prob- lems of mathematical physics during the next nearly two hundred years. In 1855 Adolf Fick (1829-1901) who was studying the diffusion of solutes in liquids and was aware of the analogy between heat conduction and diffusion formulated the partial differential equation describing this process. Later on, different scientists have introduced several other applications for the heat equation like diffusion of gases and Brownian motion [5].

The non-linear diffusion equation

u_t− 4(u^γ) = 0 in Rⁿ× (0, ∞) (1.0.1) where γ > 1 is a constant and u ≥ 0, is known as the Porous Medium Equa- tion since it can be used to describe the flow of a liquid in a homogenous, porous medium. It was first presented for this purpose by the French scien- tist J. Boussinesq (1842-1929) who used the special case of γ = 2 to model groundwater infiltration. Although the equation might look like just a sim- ple non-linear version of the heat equation it took many years before it was solved. Two engineers, L. Leibenzon and M. Muskat used it in the 1930s to study gases in porous media, which corresponds to the case where γ ≥ 2 and which is applicable on the extraction of oil. Later, in 1948, the Rus- sian scientist Polubarinova-Kochina contributed to the knowledge of special solutions to it by considering groundwater infiltration into a porous layer of rock or soil. In the 1950s Ya. Zel’dovich and collaborators used the equation to describe heat propagation in ionized gases, an application of this could take place at the first stage after a nuclear explosion when heat is traveling through gas that is practically motionless. Together with Kompaneets and Barenblatt he also found a solution that respresents release of heat from a point source [6].

This thesis begins with an introductory chapter on weak solutions and some very basic functional analysis along with all necessary definitions and theorems. Chapter 3, in which we look for solutions to the equation in a

bounded domain in Rⁿ, relies heavily on the theory presented in [1]. In the next chapter, where the pure initial value problem is treated a lot of information has been gathered both from [4] and [1]. Finally, for the last chapter that deals with the porous medium equation the main sources are [1] and [6].

Chapter 2

Preliminaries

In this chapter all necessary definitions are made together with a presentation of some important concepts. The first section is focused around definitions and results from functional analysis and the second deals with weak solutions and some basic theory of distributions.

2.1 Definitions

Before starting out with the definitions it must be stated that throughout this paper Ω denotes a domain in Rⁿ. Many of the functions that we work with here are members of various Hibert spaces and therefore we begin with the following definitions:

Definition 2.1.1. A Banach space X is a complete, normed, linear space and the norm is denoted k · k_X.

Definition 2.1.2. A Hilbert space X is a complete, normed, linear space with an inner product that generates the norm. The inner product is denoted by (·, ·)_X or (·, ·) when it is clear from the context what space is being used.

In order for some notation that appears in the next chapter to make sense we also state:

Definition 2.1.3. A function u(x, t) defined on Ω × (0, ∞) ⊂ Rⁿ⁺¹ belongs to the space C^2;1(Ω × (0, ∞)) if u_t and u_xixj are all continuous in Ω × (0, ∞).

Together with the function spaces are the norms and there is one in particular that is important to us, namely the L^p-norm for a certain kind of functions:

Definition 2.1.4. Let X denote a real Hilbert space. The space L^p(0, T ; X), where 1 ≤ p < ∞, contains all functions u : [0, T ] → X such that

kuk_L^p_{(0,T ;X)}:=

Z T 0

ku(t)k^p dt

^1/p

< ∞

and the space L^∞(0, T ; X) contains all functions u : [0, T ] → X such that kuk_L^∞_{(0,T ;X)} = ess sup

0≤t≤T

ku(t)k < ∞.

The space L^p(0, T ; X) is a Banach space and for p = 2 it is a Hilbert space.

Theory from functional analysis is vital for the solution of the heat equa- tion and so the very basic concepts are introduced here.

Definition 2.1.5. Let H be a Hilbert space. An operator from H into R, is called a functional. The collection of all bounded, linear functionals on H constitute its dual space H^∗.

Definition 2.1.6. Consider a measure space X. A property of say, a func- tion f defined on this space, holds a.e. (almost everywhere) if it holds every- where on X except on a set of measure 0.

Weak convergence of a sequence is used frequently in this paper and so we give the definition below.

Definition 2.1.7. Let X be a real Hilbert space. The sequence {u_m} ⊂ X converges to u ∈ X in the weak sense if hu^∗, u_mi → hu^∗, ui for every bounded, linear functional u^∗ ∈ X^∗. Here hu^∗, u_mi denotes the action of u^∗ on u, u^∗(u_m). Weak convergence is denoted u_m * u.

Note that |hu^∗, u_mi − hu^∗, vi| = |hu^∗, u_m− vi| ≤ ku^∗k_X^∗ku_m − vk_X and thus it is clear that strong convergence implies weak convergence. Moreover, any weakly convergent sequence is bounded.

Definition 2.1.8. Let X, Y be real Hilbert spaces. A bounded, linear operator K : X → Y is said to be compact if every bounded sequence {x_k} ⊂ X has a subsequence {x_kj} such that {K(x_kj)} converges in Y .

It can be shown that if K : X → Y is a compact, linear operator and there is a sequence {x_m} ⊂ X such that x_m * x then Kx_m → Kx.

Definition 2.1.9. The support of a function f : Ω → R that is continuous on Ω is defined as suppf = {x ∈ Ω : f (x) 6= 0} and we write f ∈ C₀(Ω) when suppf is compact.

Definition 2.1.10. A multiindex is a vector of the form α = (α₁, . . . , α_n) where α_i ≥ 0 are integers for all i, and its order is defined as |α| = α₁ + . . . + α_n. We also define D^αv = (_∂x^∂

1)^α1· · · (_∂x^∂

n)^αnv.

Definition 2.1.11. The function space H¹(Ω) is a Hilbert space that contains all locally integrable functions f , such that for every multiindex |α| ≤ 1, D^αf exists in the weak sense and belongs to L²(Ω). Its norm is kf k_H¹_(Ω) =

R

Ω(|∇f |²+ |f |²) dx
¹₂

. Consider the space C₀^∞(Ω), which consists of all infinitely differentiable functions on Ω whose support is compact. The closure of this space in H¹(Ω) is denoted H₀¹(Ω). The dual space of H₀¹(Ω) is denoted H⁻¹(Ω).

The following Lemma that is used several times in the report takes ad- vantage of a well known result from integration theory.

Lemma 2.1.1. Let E be a measurable set and {f_k} a sequence of nonnega- tive, measurable functions such that f (x) = P∞

k=1f_k(x) then Z

E

f (x) dµ =

∞

X

k=1

Z

E

f_k(x) dµ. (2.1.1)

Proof. Since the partial sums of (2.1.1) form a monotonically increasing se- quence the result follows from the Lebesgue monotone convergence theorem [2, p. 58].

Lemma 2.1.2. Let A, B be Hilbert spaces. A linear operator L : A → B is bounded if and only if it is continuous.

Proof. Assume that u → v ∈ A and that L is bounded. Then we have kLu − LvkB = kL(u − v)kB ≤ kLkku − vkA.

This implies that Lu → Lv in B and L is thereby continuous.

On the other hand, if we assume that L is not bounded then for every n ∈ N there exist u⁰n∈ A such that

kL(u⁰_n)k > nku⁰_nk.

If we put u_n = _nku^u0n0

nk then u_n → 0 but kL(u_n)k > 1 and so L cannot be continuous at u = 0.

Lemma 2.1.3. i) If u_m * u then kuk ≤ lim inf ku_mk.

ii) If u_m * u and lim ku_mk exists then kuk = lim ku_mk if and only if u_m → u.

Proof. i) We have

0 ≤ ku_m− uk² = ku_mk²− 2(u_m, u) + kuk². (2.1.2) From this it follows that 0 ≤ lim inf kumk²− 2kuk²+ kuk² and consequently kuk² ≤ lim inf ku_mk².

ii) Taking the limit of both sides of (2.1.2) we get lim ku_m− uk² = lim ku_mk²− kuk² which shows that u_m → u if and only if ku_mk → kuk.

Finally there are two other important theorems that will be used.

Theorem 2.1.4. [1, p. 639] (Riesz representation theorem) For every u^∗ ∈ H^∗, there exists a unique element u ∈ H such that hu^∗, vi = (u, v) for all v ∈ H.

Theorem 2.1.5. [1, p. 639] (Weak Compactness) Let X be a Hilbert space and assume the sequence {u_k} ⊂ X is bounded. Then there exists a subse- quence {u_kj} ⊂ {u_k} and u ∈ X such that

u_kj * u.

2.2 Distributions and weak solutions

When searching for a solution to a PDE it is not always convenient to demand from the beginning that it be continuously differentiable, or even continuous.

It is therefore advantageous to commence by taking a wider class of solutions into consideration: weak solutions. The following definitions are needed:

Definition 2.2.1. Take u, v ∈ L¹_loc(Ω) and let α be a multiindex. We say that D^αu = v in the weak sense if

Z

Ω

uD^αφ dx = (−1)^|α|

Z

Ω

vφ dx for all φ ∈ C₀^∞(Ω).

In what follows, two definitions of the concept of weak solution are given.

Definition 2.2.2. A function u ∈ H₀¹(Ω) is a weak solution of

 −4u = f (x) in Ω u = 0 on ∂Ω

where f ∈ L²(Ω), if for every v ∈ H₀¹(Ω) the following holds:

Z

Ω

∇u · ∇v dx = Z

Ω

f v dx.

Definition 2.2.3. Consider a function u ∈ L²(0, T ; H₀¹(Ω)) such that u⁰ ∈ L²(0, T ; H⁻¹(Ω)). We call u a weak solution of







4u = ut, x ∈ Ω, t > 0 u(x, t) = 0, x ∈ ∂Ω, t > 0 u(x, 0) = g(x), x ∈ Ω

if

i) hu⁰, vi +R

Ω∇u · ∇v dx = 0 for all v ∈ H₀¹(Ω) and a.e. 0 ≤ t ≤ T ii) u(0) = g.

We emphasize that hu⁰, vi denotes the action of the functional u⁰(t) ∈ H⁻¹(Ω) = H₀¹(Ω)^∗ on v for a.e. fixed t.

The method used to solve the heat equation rests upon first finding a weak solution by means of functional analysis and then proving this solution to be regular.

In what follows we give the definition of a distribution and present some of its properties. These concepts will then be used in Chapter 3 and 4.

Definition 2.2.4. A distribution F in Ω is a linear mapping F : C₀^∞(Ω) → R such that F (v_j) → 0 for every sequence {v_j} ⊂ C₀^∞(Ω) with support in a fixed compact set K ⊂ Ω and whose derivatives D^αv_j → 0 uniformly in K for all

|α| ≥ 0, as j → ∞. The support of a distribution F in Ω is the smallest, closed set K ⊂ Ω such that F (v) = 0 whenever v ≡ 0 in a neighborhood of K [4, p. 65].

To denote the action of F on v we shall usually write hF, vi instead of F (v). Furthermore, if F ∈ L¹_loc(Ω) the notation F (v) = R

ΩF v dx is used.

This can be done since if f ∈ L¹_loc(Ω) then it can be shown that Ff as defined by hF_f, vi =R

Ωf v dx is a distribution. However, we shall make no distinction between F_f and f .

Definition 2.2.5. Two functions u and f satisfy 4u = f in Ω, in the sense of distributions if for all v ∈ C₀^∞(Ω)

hu, 4vi = hf, vi.

Distributions have the advantage that they may always be differentiated.

Distributional derivative is defined through the formula below:

hD^αF, vi = (−1)^|α|hF, D^αvi

for all v ∈ C₀^∞(Rⁿ). Here hD^αF, vi denotes the action of D^αF on v. Note in particular that if u ∈ L¹_loc(Ω) is a weak solution of 4u = f , then it is a distribution solution and the condition hu, 4vi = hf, vi may be expressed as R

Ωu4v dx = R

Ωf v dx. It is worth noting that a weak solution is a distri- bution solution that belongs to an appropriate function space, for example H₀¹(Ω).

The Dirac delta function, δ and the distribution F_δ will be presented below. We begin by considering H : R → R, defined as

H(x) = 1 if x ≥ 0 0 if x < 0.

This function is discontinuous at x = 0, formally however, we can define H⁰(x) by

hH⁰, vi = −hH, v⁰i = − Z ∞

−∞

H(x)v⁰(x) dx = − Z ∞

0

v⁰(x) dx = v(0) for every v ∈ C₀¹(R), provided that the weak derivative H⁰(x), exists. More- over, if v is smooth enough it is possible to obtain higher order weak deriva- tives of H through iterated formal partial integration.

Now set H⁰(x) = δ(x). Then δ is defined as δ(x) such that for every v ∈ C₀^∞(Rⁿ) we formally have

hδ, vi = v(0) (2.2.1)

which is usually written F_δ(v) =R δ(x)v(x) dx = v(0). It is clear that F_δ is linear. Take any sequence {v_j} ⊂ C₀^∞(Ω) with support in a fixed compact set K ⊂ Ω, and suppose the derivatives D^αvj → 0 uniformly in K, as j → ∞.

From the definition of F_δwe now see that F_δ(v_j) → 0 as j → ∞ and therefore F_δ indeed is a distribution.

Definition 2.2.6. The convolution in Rⁿ of two functions f and g, where at least one has compact support is a new function f ? g on Rⁿ, defined by

(f ? g)(x) = Z

Rⁿ

f (x − y)g(y) dy provided the integral exists.

We analogously make the following definition.

Definition 2.2.7. The convolution of two distributions f and g is defined as hf ? g, vi =

Z

Rⁿ

Z

Rⁿ

f (z)g(y)v(z + y) dydz for all v ∈ C₀^∞(Rⁿ).

Convolution is commutative (both for functions and distributions), that is f ? g = g ? f . This can be shown by the variable substitution z = x − y.

Moreover, taking distributional derivatives of a convolution f ? g where v ∈ C₀^∞(Rⁿ), gives

h4(f ? g), vi = h(f ? g), 4vi = Z

Rⁿ

Z

Rⁿ

f (x − y)g(y) dy



4v(x) dx

= Z

Rⁿ

Z

Rⁿ

f (z)g(y) dy



4_zv(z + y) dz

= Z

Rⁿ

Z

Rⁿ

f (z)g(y)4_yv(z + y) dydz.

Note that since 4 is operating only on the y or z-variable we may take advantage of distributional derivatives once more to find that

Z

Rⁿ

Z

Rⁿ

f (z)g(y)4_yv(z + y) dydz = Z

Rⁿ

Z

Rⁿ

f (z)4_yg(y)v(z + y) dydz

and that Z

Rⁿ

Z

Rⁿ

f (z)g(y)4_zv(z + y) dydz = Z

Rⁿ

Z

Rⁿ

4_zf (z)g(y)v(z + y) dydz which allows us to conclude

4(f ? g) = (4f ) ? g = f ? (4g). (2.2.2) Lemma 2.2.1. If f ∈ C₀(Rⁿ) and g ∈ L¹_loc(Rⁿ), then f ? g ∈ C(Rⁿ).

Proof. Let K denote the support of f and recall that f is uniformly continu- ous on this set and hence on Rⁿ. For a fixed x let eK = {y−z : |x−y| ≤ 1, z ∈ K}. Then eK is compact and so g ∈ L¹_loc(Rⁿ) means that R

Ke |g| dx = C for some constant C. By uniform continuity we have that, given any ε > 0, there exists δ ∈ (0, 1) such that |f (x − z) − f (y − z)| < ε/C whenever |x − y| < δ and z ∈ Rⁿ. This implies that

|(f ? g)(x) − (f ? g)(y)| =     Z

Ke

(f (x − z) − f (y − z))g(z) dz    

≤ Z

Ke

|f (x − z) − f (y − z)||g(z)| dz

< ε C

Z

Ke

|g(z)| dz ≤ ε and so the continuity of f ? g follows.

Lemma 2.2.2. If w ∈ L¹_loc(Rⁿ) then δ ? w = w.

Proof. Consider (2.2.1) and note that it implies Z

Rⁿ

δ(x)v(x + y) dx = v(y).

For φ ∈ C₀^∞(Rⁿ) we then have hδ ? w, φi =

Z

Rⁿ

Z

Rⁿ

δ(z)w(y)φ(z + y) dydz

= Z

Rⁿ

Z

Rⁿ

δ(z)φ(z + y) dz



w(y) dy

= Z

Rⁿ

φ(y)w(y) dy

= hw, φi.

Hence δ ? w = w.

As we shall see, convolution of distributions will be used to solve the non- homogeneous heat equation. This is done by first showing that the special case

u_t− 4u = δ

can be solved. A solution to this equation is called a fundamental solution of the heat operator and solves the equation in the sense of distributions. It is however, not unique since it is possible to add solutions of the homogenous problem to it. The necessary concepts of distribution theory have now been presented and we are ready to start working on a solution to our problem.

Chapter 3

The Heat Equation in a Bounded Domain

In this chapter we establish the existence of solutions to the following initial- boundary value problems:







4u = u_t, x ∈ Ω, t > 0 u(x, t) = 0, x ∈ ∂Ω, t > 0 u(x, 0) = g(x), x ∈ Ω

(3.0.1)

and







u_t− 4u = f (x, t), x ∈ Ω, t > 0 u(x, t) = 0, x ∈ ∂Ω, t > 0 u(x, 0) = g(x), x ∈ Ω

(3.0.2)

where, in both problems, it is assumed that g ∈ L²(Ω) and in the second problem that f ∈ L²(Ω × (0, ∞)).

The two basic ideas used in the first section of this chapter are the method known as separation of variables and the concept of weak solution. Separation of variables leads to a new boundary value problem that requires existence of eigenvalues and an orthonormal basis of eigenfunctions to be solved. In many books on the topic, this is taken care of using somewhat advanced functional analysis and in order to get around that we instead begin by constructing a new norm, equivalent to that of H₀¹(Ω). From this we proceed to define a linear functional on the same space, that according to Riesz theorem has a counterpart in the form of a function defined on H₀¹(Ω). Using this argument

the problem proves to have a weak solution. The main result of the section is a theorem that proves existence of eigenvalues and an orthonormal basis of eigenfunctions.

The second part of this chapter starts over by actually performing the method of separation of variables and proceeds by taking advantage of the conclusions from the first section, namely the existence of eigenvalues and orthonormal basis of eigenfunctions. The idea is to begin by considering an approximation of the problem in an eigenfunction space of finite dimension and then pass to limits, a method known as Galerkin approximation. The section ends with a theorem that deals with the regularity of the solution and at the end of this chapter uniqueness of the solution is investigated.

The following important theorems shall be needed:

Theorem 3.0.3. [1, p. 272] (Special case of the Rellich-Kondrachov com- pactness theorem) Let Ω ⊂ Rⁿ be a bounded domain and let {u_k} denote a sequence of functions in H₀¹(Ω). If u_k * u in H₀¹(Ω) then u_k→ u in L²(Ω).

Theorem 3.0.4. [1, p. 265] (Special case of the Poincar´e inequality) Let Ω be a bounded domain in Rⁿ. If u ∈ H₀¹(Ω), then there exists a constant C, depending only on n and Ω, such that

kuk_L²_(Ω) ≤ Ck∇uk_L²_(Ω).

Proposition 3.0.5. [1, p. 622] (H¨older’s inequality). Suppose 1 ≤ p, q ≤ ∞,

1

p + ¹_q = 1. If u ∈ L^p(Ω), v ∈ L^q(Ω), we have Z

Ω

|uv| dx ≤ kuk_L^p_(Ω)kvk_L^q_(Ω).

Lemma 3.0.6. [1, p. 624] (Gronwall’s inequality) Let η be a nonnegative, differentiable function on [0, T ] that satisfies η⁰(t) ≤ φ(t)η(t) + ψ(t) where φ(t), ψ(t) are nonnegative, integrable functions on [0, T ]. Then for all 0 ≤ t ≤ T we have

η(t) ≤ e^{R0tφ(s) ds}

 η(0) +

Z t 0

ψ(s) ds

 .

Theorem 3.0.7. [4, p. 144] (Weak Maximum Principle) Take u ∈ C^2;1(Ω × (0, T )) ∩ C(Ω × (0, T )) and suppose that it satisfies 4u ≥ u_t in Ω × (0, T ).

Then u assumes its maximum on the parabolic boundary Γ = {(x, t) ∈ ΩT : x ∈ ∂Ω or t = 0}, where Ω_T = Ω × (0, T ], that is

max

(x,t)∈ ¯ΩT

u(x, t) = max

(x,t)∈Γu(x, t)

Proof. We begin by assuming that 4u > u_t in Ω_T. Let τ be such that 0 < τ < T and consider

Ω_τ = Ω × (0, τ ), Γ_τ = {(x, t) ∈ Ω_τ : x ∈ ∂Ω or t = 0}.

Suppose max_Ωτ u = u(x, τ ) for some x ∈ Ω. Then we must have that u_t(x, τ ) ≥ 0 due to the fact that the maximum is taken for t = τ and 4u(x, τ ) ≤ 0 since the second derivative (if it exists) of any function in its maximum point is non-positive. But this contradicts our assumption and by the same argument we see that it is impossible for u to attain its maximum inside Ω_τ. Therefore the theorem holds for Ω_τ. However since u is continuous we have that

τ →Tlimmax

Ωτ

u(x, t) = max

ΩT

u(x, t) and thus the conclusion holds.

Suppose now that 4u ≥ u_t in Ω_T. Set v = u − kt where k > 0. This implies that v ≤ u on Ω_T and 4v − v_t = 4u − u_t+ k > 0 in Ω_T. By the previous case we now get

max

ΩT

u = max

ΩT

(v + kt) ≤ max

ΩT

v + kT = max

Γ v + kT ≤ max

Γ u + kT which, if we let k → 0 in turn gives max_Ω

T u ≤ max_Γu. Since the reverse inequality clearly must hold we finally get

max

Ω_T

u = max

Γ u.

3.1 Eigenvalues

Throughout this chapter we assume that ∂Ω is C¹. It is actually possible to prove that this assumption is not necessary to solve the equations that we will be working with, but this will not be done here.

The method of separation of variables rests on constructing solutions to the boundary value problem

 −4u = u_t x ∈ Ω, t > 0 u(x, t) = 0 x ∈ ∂Ω, t > 0

which are of the form u(x, t) = v(t)w(x). We will conduct the process in detail in the following section but for now we just state that this method requires us to find all the eigenvalues to the following problem:

 −4w(x) = λw(x) in Ω w(x) = 0 on ∂Ω.

A generalization of this would be to find all eigenvalues to a compact, linear, symmetric operator K in H₀¹(Ω). Before this is taken care of though we have to make sure that K actually exists. So, suppose that we have the following problem

 −4u = λu in Ω

u = 0 on ∂Ω (3.1.1)

and recall that if u ∈ H₀¹(Ω) andR

Ω∇u·∇v dx = λR

Ωuv dx for all v ∈ C₀^∞(Ω) then u is a weak solution to the above equation. Using Green’s formula and the boundary condition we find that

Z

Ω

λuv dx = Z

Ω

(−4u)v dx = Z

Ω

∇u · ∇v dx

for u ∈ C²(Ω) and v ∈ C₀^∞(Ω) which means that u ∈ C²(Ω) is a weak solution to (3.1.1). It is also clear that

(u, v) = Z

Ω

∇u · ∇v dx (3.1.2)

satisfies the conditions for an inner product.

Recall that two norms k · k_a and k · k_b are equivalent if there exist positive numbers α, β such that αkuk_a ≤ kuk_b ≤ βkuk_a. Denote the norm of u in H₀¹(Ω) as defined above, by kuk_H(Ω) and note that kuk_H(Ω) = k∇uk_L²_(Ω). Now the H-norm is equivalent to the H₀¹-norm kuk²_H1

0(Ω) =R

Ω(|∇u|²+ u²) dx, for

kuk²_H(Ω) ≤ kuk²_H1

0(Ω)= k∇uk²_L2(Ω)+ kuk²_L2(Ω) ≤

k∇uk²_L2(Ω)+ Ck∇uk²_L2(Ω) = βk∇uk²_L2(Ω) = βkuk²_H(Ω) where the second inequality follows from Poincar´e’s inequality. As shall be seen this new norm plays a crucial role when it comes to restating problem (3.1.1).

Now we will define the functional that provides the link that makes it possible for us to find the operator that we are looking for. Let u^∗ ∈ H⁻¹(Ω) be such that

hu^∗, vi = Z

Ω

uv dx

where v ∈ H₀¹(Ω) and u is a given function in L²(Ω). It follows that hu^∗, αv+βwi =

Z

Ω

u(αv+βw) dx = Z

Ω

αuv dx+

Z

Ω

βuv dx = αhu^∗, vi+βhu^∗, vi so u^∗ is linear and again, according to Poincar´e’s inequality

| Z

Ω

uv dx| ≤ Z

Ω

|u||v| dx ≤ kuk_L²_(Ω)kvk_L²_(Ω)

≤ C₀kuk_L²_(Ω)k∇vk_L²_(Ω) = C₀kuk_L²_(Ω)kvk_H(Ω) = Ckvk_H(Ω) for constants C₀, C₁, C, and so u^∗ is bounded.

According to Theorem 2.1.4 there exists an element Ku ∈ H₀¹(Ω) such that

hu^∗, vi = (Ku, v) = Z

Ω

uv dx for all v ∈ H₀¹(Ω). (3.1.3) This is the operator that we need. Apparently K is linear and from the equation above we see that it must be bounded too, since u^∗ is.

Now since u is a weak solution of problem (3.1.1) we get from (3.1.2) and (3.1.3) that

Z

Ω

λuv dx = Z

Ω

∇u · ∇v dx = (u, v), and (Ku, v) =

Z

Ω

uv dx = 1 λ

Z

Ω

∇u · ∇v dx which allows us to write

1

λ(u, v) = (Ku, v).

It is now possible to restate problem (3.1.1) and for µ = 1/λ it becomes

Ku = µu, u ∈ H₀¹(Ω). (3.1.4)

Note that (Ku, v) =R

Ωuv dx = (u, Kv) which means that K is symmet- ric. In what follows we also prove that it is compact.

Let {u_k} be a bounded sequence of functions in H₀¹(Ω). According to Theorem 2.1.5 this sequence contains a subsequence {u_kj} with the property that ukj * u, that is (ukj, v) → (u, v) ∀ v ∈ H₀¹(Ω). With the aid of the inequalities of H¨older and Poincar´e and Theorem 3.0.3 we have

sup

kvk_H(Ω)≤1

|(K(ukj − u), v)| ≤ sup

kvk_H(Ω)≤1

Z

Ω

|ukj − u||v| dx

≤ cku_kj − uk_L²_(Ω) → 0 as j → ∞ and hence K is compact.

We are now ready to state and prove the following theorem that is crucial for the method of separation of variables.

Theorem 3.1.1. Consider problem (3.1.4). There exists a sequence {µk} of eigenvalues such that µ_k → 0 as k → ∞, and an orthonormal basis {u_k} in H₀¹(Ω) of eigenfunctions corresponding to these.

Remark : This theorem holds in any Hilbert space, provided K is sym- metric and compact. Moreover, since the eigenfunctions make up a basis all existing eigenvalues must be included in the sequence.

Proof. 1. Set

µ1 = sup

u∈H(Ω) u6=0

(Ku, u)

(u, u) = sup

kuk_H(Ω)=1

(Ku, u) and suppose

(Ku_m, u_m) → µ₁, as m → ∞

for some bounded sequence of functions, {u_m} ⊂ H₀¹(Ω) such that ku_mk_H(Ω) = 1 for all m. Our purpose now is to prove that the above supremum is as- sumed. Recall that we have shown K to be a compact, symmetric, linear operator.

By Theorem 2.1.5 {u_m} has a subsequence, {u_ml} such that u_ml * u in H₀¹(Ω). The compactness of K now implies Ku_ml → Ku. For the sake of notation we write u_m instead of u_ml. From this it then follows that (Ku_m, u_m) → (Ku, u) as m → ∞.

Now

(Kum, um) = (Ku_m, u_m) (u_m, u_m)

and since (Ku, u) > 0 if u 6= 0 it is clear that µ₁ > 0 and hence u_m * u 6= 0.

By the first part of Lemma 2.1.3 we also see that kuk ≤ ku_mk = 1 which gives

(Ku, u)

(u, u) ≥ (Ku, u) = lim

m→∞(Ku_m, u_m) = µ₁ and so

(Ku, u) (u, u) ≥ µ₁. But from the definition of µ1 we have that

(Ku, u) (u, u) ≤ µ1

so therefore

(Ku, u) (u, u) = µ1.

We get kuk = 1 and thus conclude that the maximum is assumed for this u.

2. Orthogonality is now to be shown, that is if µ_k 6= µ_m then (u_k, u_m) = 0.

We have

 Ku_k= µ_ku_k Ku_m = µ_mu_m

and (Ku_k, u_m) = (Ku_m, u_k), due to the symmetry of K and commutativity of the inner product. The equalities

 (Ku_k, u_m) = µ_k(u_k, u_m) (Kum, uk) = µm(um, uk)

then imply µ_k(u_k, u_m) = µ_m(u_m, u_k) which means that (u_k, u_m) = 0 must hold.

3. We now show that µ₁ is an eigenvalue.

Let u₁ be such that ku₁k = 1 and the supremum in part 1 is assumed for it and take φ ∈ H₀¹(Ω). Again, for notational convenience, we write u instead of u₁. We have

f (t) := (K(u + tφ), u + tφ)

ku + tφk² = (Ku, u) + 2t(Ku, φ) + t²(Kφ, φ) kuk²+ 2t(u, φ) + t²kφk²

and

f⁰(t) = [2(Ku, φ) + 2t(Kφ, φ)][ku + tφk²] ku + tφk⁴

−[2(u, φ) + 2tkφk²][(Ku, u) + 2t(Ku, φ) + t²(Kφ, φ)]

ku + tφk⁴ .

Now since f (t) attains its maximum for t = 0 and f⁰(0) = 2(Ku, φ)kuk²− 2(u, φ)(Ku, u)

kuk⁴ ,

we get f⁰(0) = 2(Ku, φ) − 2µ₁(u, φ) = 0 because of kuk = 1 and (Ku, u) = µ₁ and this implies

(Ku, φ) = µ₁(u, φ) for all φ ∈ H₀¹(Ω), so Ku = µ₁u.

Hence µ₁ is an eigenvalue of the operator K.

4. The existence of an orthonormal basis of eigenfunctions is shown here.

Define H₁ = (span{u₁})^⊥ and set µ₂ = sup

u∈H1u6=0

(Ku, u) (u, u) .

Taking φ ∈ H1 and proceeding as in part 3 we conclude that (Ku2, v) = µ₂(u₂, v) for all v ∈ H₁. Apart from that we have (Ku₂, u₁) = (u₂, Ku₁) = µ₁(u₂, u₁) = 0 = µ₂(u₂, u₁) since u₁⊥H₁ and so (Ku, φ) = µ₂(u, φ), ∀φ ∈ H₀¹(Ω) and thus Ku2 = µ2u2. Note that µ2 ≤ µ1 since this supremum is taken over a smaller set. Iterating the procedure by putting H₂ := (span{u₁, u₂})^⊥ etc generates eigenvalues µ₁ ≥ µ₂ ≥ µ₃. . . such that µ_m → µ ≥ 0.

Let H = span{u1, u2, u3, . . .} and consider H^⊥. We now claim that H^⊥ = {0}. Suppose dim H^⊥ > 0. Letting

µ = sup

u∈H⊥

u6=0

(Ku, u) (u, u)

generates a new, positive eigenvalue but this is impossible since, as we shall see, µ_m → 0 as m → ∞. Now u₁, u₂, . . . are all orthogonal and ku_ik = 1 for

i = 1, 2, 3, . . . and since dim H^⊥ = 0 they constitute an orthonormal basis in H₀¹(Ω).

5. Finally it is proved that µ_m → 0.

Consider the orthonormal eigenfunctions {u_m}. We know that u_m * 0 in H(Ω).

For assume um * u 6= 0. For every ε > 0 it is then possible to find an integer k > 0 and constants α₁, . . . , α_k such that

u −

k

X

i=1

αiui

< ε.

Moreover, (u_m,Pk

i=1α_iu_i) = 0 if m > k, by orthogonality of {u_m}. But if u_m * u then

0 = lim

m→∞(u_m,

k

X

i=1

α_iu_i) = (u,

k

X

i=1

α_iu_i) which enables us to write kuk² = (u, u−Pk

i=1αiui) < εkuk and so accordingly we must have u = 0.

Due to the compactness of K, the fact that lim_m→∞(Ku_m, u_m) = (Ku, u) = 0 and kumk = 1 give

m→∞lim µ_m = lim

m→∞µ_m(u_m, u_m) = lim

m→∞(Ku_m, u_m) = 0.

Now the necessary work has been done and we are ready to proceed to separation of variables where the usefulness of Theorem 3.1.1 shall become evident.

3.2 Separation of variables

Let us now return to the Dirichlet problem (3.0.1) and recall that we begin by looking for solutions u(x, t) that can be separated in the following way:

u(x, t) = v(t)w(x).

Now

u_t(x, t) = v⁰(t)w(x), 4u(x, t) = v(t)4w(x)

and so for all x ∈ Ω and t > 0 such that v(t), w(x) 6= 0 we have that u_t(x, t) = 4u(x, t) implies v⁰(t)w(x) = v(t)4w(x) which in turn leads to

v⁰(t)

v(t) = 4w(x) w(x) .

Since both the left and the right hand side of the above equation depend on only one, distinct from the opposite side, variable, we must have

v⁰(t) = −µv(t), 4w(x) = −µw(x) and w(x) = 0 on ∂Ω where µ is a constant. If µ is known, the solution to the first equation is

v(t) = C₀e^−µt

for some constant C0. Thus only the second equation needs to be investi- gated. Problem (3.1.1) and the theory from the previous section give the existence of real eigenvalues and an orthonormal basis of eigenfunctions.

Set

u(x, t) = v(t)w(x). (3.2.1)

Since there for every eigenvalue µk exists an eigenfunction wk(x) we have that v_k(t) = e^−µkt and also that u_k(x, t) = Ce^−µktw_k(x) solves

 (uk)t− 4uk= 0 in Ω × (0, ∞)

u_k(x, t) = 0 x ∈ ∂Ω, t > 0 (3.2.2) for some constant C that can be set equal to 1.

What we now wish to do is to use the superposition principle to find solutions to (3.0.1) and (3.0.2) as infinite sums involving all the u_k. The natural way to do this would be to start with a finite sum and then with the aid of some limit argument establish the infinite case.

Assume that u(x, t) is a solution of equation (3.0.2) and associate with it a mapping u : [0, T ] → H₀¹(Ω) such that [u(t)](x) := u(x, t). Analogously, define f : [0, T ] → L²(Ω) through [f(t)](x) := f (x, t). Now if we can find u(t) our problem is solved. The idea is to start by looking for approximate solutions of equation (3.0.2) in the subspace spanned by the eigenfunctions {wk}^m_k=1 and then use a limit argument to solve the actual problem.

It is possible to show that the functions w_k(x) (k = 1, . . .) are smooth [1, p. 335]. Moreover, through multiplication with a suitable constant they can

be orthonormalized in L²(Ω). This follows since (w_k, w_l)_H(Ω) = 0 for k 6= l implies

0 = Z

Ω

∇w_l· ∇w_k dx = µ_l Z

Ω

w_lw_k dx.

and so (wk, wl)_L²_(Ω) = 0 for k 6= l. Recall also, that {wk} is an orthogonal basis in H₀¹(Ω).

For a fixed integer m let u_m(t) :=

m

X

l=1

C_l^m(t)e^−µltw_l. (3.2.3)

Then u_m : [0, T ] → H₀¹(Ω) and by the superposition principle u_m solves (3.2.2) in Ω × (0, T ) for all T . We wish that C_l^m(t) be such that

(u_m(0), w_k) = (g, w_k) (3.2.4) holds for every k = 1, . . . , m. Now u_m(t) satisfies the boundary condition in (3.2.2). Multiplying by w_l and using Green’s formula we get

Z

Ω

(u_m)_tw_l dx + Z

Ω

∇u_m· ∇w_l dx = Z

Ω

fw_ldx which is the same as

(u⁰_m, w_l)_L²_(Ω)+ (u_m, w_l)_H(Ω) = (f, w_k)_L²_(Ω), (3.2.5) where the derivative is taken with respect to t.

Due to {w_k} being orthonormal in L²(Ω) we have that R

Ωw_lw_k dx = 0 if l 6= k and R

Ωwlwk dx = 1 when l = k. This implies (u⁰_m, w_k)_L²_(Ω)= (C_k^m(t)⁰− µ_kC_k^m(t)) e^−µkt and since

Z

Ω

∇w_l· ∇w_k dx = Z

Ω

(−4w_l)w_k dx = λ_l Z

Ω

w_lw_k dx (3.2.6) we have that

(u_m, w_k)_H(Ω) = Z

Ω m

X

l=1

C_l^m(t)e^−µlt∇w_l· ∇w_kdx = C_k^m(t)e^−µktµ_k.

Adding these two last inner products together yields

C_k^m(t)⁰e^−µkt= (f, w_k)_L²_(Ω). (3.2.7) From the orthonormality of {w_k} in L²(Ω) we have that

(um(0), wk)_L²_(Ω) =

m

X

l=1

C_l^m(0) Z

Ω

wl(x)wk(x) dx = C_k^m(0) (3.2.8)

and so from (3.2.4) it is clear that C_k^m(t) should satisfy

C_k^m(0) = (g, w_k)_L²_(Ω). (3.2.9) From (3.2.7) and (3.2.9) we now deduce that C_k^m(t) = C_k(t) for every fixed integer m and thus

u_m(t) =

m

X

l=1

C_l(t)e^−µltw_l where C_l(t) =Rt

0 e^µls(f, w_l)_L²_(Ω)ds + (g, w_l)_L²_(Ω).

The purpose now is to show that, as m → ∞, the sequence of approximate solutions {u_m} has a subsequence that converges to a weak solution of (3.0.2).

This is done with the help of two theorems. The first one is an energy estimate and states the following:

Theorem 3.2.1. There exists a constant C that depends only on Ω and T such that

max

0≤t≤Tku_m(t)k_L²_(Ω)+ ku_mk_L²(0,T ;H(Ω)) + ku⁰_mk_L²_{(0,T ;H}⁻¹_(Ω))

≤ C kf k_L²_{(0,T ;L}²_(Ω))+ kgk_L²_(Ω)
. Proof. 1. Multiply equation (3.2.5) by C_l(t)e^−µlt for l = 1, . . . , m and add the equations together so as to obtain

(u⁰_m, um)_L²_(Ω)+ kumk²_H(Ω) = (f, um)_L²_(Ω), (3.2.10) for 0 ≤ t ≤ T . We also have that

(u⁰_m, u_m)_L²_(Ω) = d dt

 1

2ku_mk²_L2(Ω)

 .

From this it follows that d

dt

 1

2ku_mk²_L2(Ω)



+ ku_mk²_H(Ω) = (f, u_m)_L²_(Ω). (3.2.11) Since

(f, u_m)_L²_(Ω) ≤ kfk_L²_(Ω)ku_mk_L²_(Ω) ≤ 1

2(kfk²_L2(Ω)+ ku_mk²_L2(Ω)), we obtain

d dt

 1

2ku_mk²_L2(Ω)



≤ d

dt

 1

2ku_mk²_L2(Ω)



+ ku_mk²_H(Ω)

≤ 1 2

kfk²_L2(Ω)+ ku_mk²_L2(Ω)



. (3.2.12) 2. Observe that g =P∞

l=1(g, wl)_L²_(Ω)wl and that kgk²_L2(Ω) =

Z

Ω

∞

X

l=1

(g, w_l)_L²_(Ω)w_l

!2

dx

=

∞

X

l=1

(g, wl)²_L2(Ω)

due to [2, p. 214]. Note also that ku_m(0)k²_L2(Ω) =

Z

Ω m

X

l=1

C_l(0)w_l dx

!2

=

m

X

l=1

Z

Ω

(C_l(0)w_l)² dx

=

m

X

l=1

C_l²(0) Z

Ω

w²_l dx

=

m

X

l=1

(g, w_l)²_L2(Ω). (3.2.13) From this we see that

ku_m(0)k²_L2(Ω) =

m

X

l=1

(g, w_l)²_L2(Ω)

≤

∞

X

l=1

(g, w_l)²_L2(Ω) = kgk²_L2(Ω).

3. Now we set η(t) := ku_mk²_L2(Ω) and ξ(t) := kf(t)k²_L2(Ω). From (3.2.12) it follows that

η⁰(t) ≤ η(t) + ξ(t) and so lemma 3.0.6 gives that

η(t) ≤ e^t

 η(0) +

Z t 0

ξ(s) ds

 .

Since by definition, η(0) = kum(0)k²_L2(Ω) ≤ kgk²_L2(Ω), we have

0≤t≤Tmax ku_m(t)k²_L2(Ω) ≤ D₀

kgk²_L2(Ω)+ kfk²_L2(0,T ;L²(Ω))



(3.2.14) for some constant D₀.

4. Now, using the second inequality in (3.2.12) it is possible to integrate from 0 to T and take advantage of (3.2.14) to obtain, for some constant D₁, ku_mk²_L2(0,T ;H(Ω)) =

Z T 0

ku_m(t)k²_H(Ω)dt

≤ 1

2 Z T

0

kfk²_L2(Ω)+ ku_mk²_L2(Ω)dt + 1

2ku_m(0)k²_L2(Ω)−1

2ku_m(T )k²_L2(Ω)

≤ D₁

kfk²_L2(0,T ;L²(Ω))+ kgk²_L2(Ω)



. (3.2.15)

5. Choose a function v ∈ H(Ω) with the property: kvk_H(Ω) ≤ 1 and set v = v¹ + v² where v¹ ∈ span{w_l}^m_l=1 and v² belongs to the orthogonal complement of this set in H(Ω). We know that {w_l} are orthogonal in H(Ω) and therefore kv¹k_H(Ω) ≤ kvk_H(Ω) ≤ 1. From (3.2.5) we now get

(u⁰_m, v¹)_L²_(Ω)+ (u_m, v¹)_H(Ω) = (f, v¹)_L²_(Ω) (3.2.16) where the fact that u⁰_m is smooth allows us to write hu⁰_m, vi = (u⁰_m, v)_L²_(Ω). With the use of the definition of u_m(t) we now have, for every fixed t

hu⁰_m, vi = (u⁰_m, v)_L²_(Ω)= (u⁰_m, v¹)_L²_(Ω) = (f, v¹)_L²_(Ω)− (u_m, v¹)_H(Ω). (3.2.17) Thus, due to kv¹k_H(Ω) ≤ 1 and Poincar´e’s inequality we have

|hu⁰_m, vi| ≤ D₂(kfk_L²_(Ω)+ ku_mk_H(Ω)).