One-Dimensional Fractal Wave Equations

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Electronic Theses and Dissertations Graduate Studies, Jack N. Averitt College of

Spring 2011

Fun Choi Chan

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ONE-DIMENSIONAL FRACTAL WAVE EQUATIONS

by F. CHAN

(Under the Direction of Dr. Sze-Man Ngai)

ABSTRACT

We study one-dimensional wave equations deﬁned by a class of fractal Laplacians.

These Laplacians are defined by fractal measures generated by iterated function systems with overlaps, such as the well-know infinite Bernoulli convolution associated with golden ratio and the 3-fold convolution of the Cantor measure. The iterated function systems defining these measures do not satisfy the open set condition or the post-critically finite condition, and therefore the existing theory, introduced by Kigami and developed by many other mathematicians, cannot be applied. First, by using a weak formulation of the problem, we prove the existence, uniqueness and regularity of weak solutions of these wave equations. Second, we study numerical computations of the solutions. By using the second-order self-similar identities introduced by Strichartz et al., we discretize the equation and use the finite element method and central difference method to obtain numerical solutions. Last, we also prove that the numerical solutions converge to the weak solution, and obtain estimates for the convergence of this approximation scheme.

INDEX WORDS: Fractal, wave equation, iterated function system, second-order self-similar identities, weak solution, ﬁnite element method

ii ONE-DIMENSIONAL FRACTAL WAVE EQUATIONS

by F. CHAN

B.S. in Mathematics

A Thesis Submitted to the Graduate Faculty of Georgia Southern University in Partial Fulﬁllment of the Requirement for the Degree

MASTER OF SCIENCE

STATESBORO, GEORGIA

2011 c 2011

iv ONE-DIMENSIONAL FRACTAL WAVE EQUATIONS

by F. CHAN

Major Professor: Dr. Sze-Man Ngai

Committee: Dr. Scott Kersey

Dr. Frederic Mynard Dr. Shijun Zheng

Electronic Version Approved: March 2011

v ACKNOWLEDGMENTS

I would like to express my gratitude to Dr. Sze-Man Ngai for his advice and guidance during my graduate study, and especially while conducting this research.

I gratefully appreciate my thesis committee members, Dr. Scott Kersey, Dr. Frederic Mynard and Dr. Shijun Zheng for their advice and comments.

I also appreciate Dr. Alexander Teplyaev for his suggestions and comments.

Last, I would like to thank Dr. Yan Wu for his attention to my graduation procedures. I would also like to thank my friend Yin-Tat Lee for his help. I am also grateful to everyone who has played a role in this thesis. My apologies to anybody whom I did not mention personally one by one.

vi TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS ...... vi

LIST OF FIGURES ...... x

CHAPTER

0.1 Notation ...... 1

1 Introduction ...... 3

1.1 Preliminaries ...... 3

1.1.1 Fractal measures ...... 4

1.1.2 Operator ∆µ ...... 5

1.2 Main problems in our project ...... 7

1.3 Main result ...... 8

2 Existence and uniqueness of weak solution ...... 11

3 The ﬁnite element method ...... 23

4 The central diﬀerence method ...... 29

5 Fractal measures deﬁned by iterated function systems ...... 33

vii 5.0.1 Weighted Lebesgue measure ...... 33

5.1 Inﬁnite Bernoulli convolution associated with the golden ratio ...... 34

5.2 3-fold convolution of the Cantor measure ...... 37

6 Convergence of the approximation scheme ...... 39

7 Numerical results ...... 45

BIBLIOGRAPHY ...... 51

APPENDICES ...... 54

A FRACTAL MEASURES ...... 54

A.1 BERNOULLI CONVOLUTION ASSOCIATED WITH THE GOLDEN RATIO ...... 54

A.2 THREE-FOLD CONVOLUTION OF THE CANTOR MEASURE ...... 62

A.3 INVERTIBILITY OF M ...... 68

B NORMED SPACES INVOLVING TIME ...... 71

B.1 COMPLETENESS OF L2([0,T ]; X) ...... 71

C ABSOLUTE CONTINUITY ...... 75

viii 2 C.1 ABSOLUTE CONTINUITY OF (um(x,t))t µ AND (u (x,t),u (x,t))...... k k 75 E m m

D SEPARABILITY ...... 77

E WEAK DERIVATIVE ...... 78

E.1 BANACH SPACE VALUED FUNCTIONS ...... 78

1 2 F EMBEDDING OF H0 IN Lµ ...... 86

G DIFFERENTIABILITY OF DISTRIBUTION AND BANACH SPACE VALUED FUNCTIONS ...... 88

ix LIST OF FIGURES

Figure Page

Figure1 Dirichlet boundary condition for the weighted Lebesque measure associated with the weight p =2 √3 and 1 p = √3 1. . 46 − − −

Figure 2 Dirichlet boundary conditions for the inﬁnite Bernoulli convolution associated with the golden ratio...... 48

Figure 3 Dirichlet boundary c ondition for the 3-fold convolution of the Cantor measure...... 50

x 1

0.1 Notation

Let us introduce some notation and symbols used in this thesis.

Let ∆µ be Laplacian with respect to µ.

Let Cc∞(a,b) be the set of inﬁnitely diﬀerentiable functions with compact sup- ports [a,b]. Let X be a Banach space and denote the corresponding norm. k·kX

Let Lp([0,T ]; X) be the space of all measurable functions u : [0,T ] X. →

Let Xbe a Banach space. The space C([0,T ]; X) comprises all continuous functions u : [0,T ] X. The space C1([0,T ]; X) comprises all C1 functions u : [0,T ] → → X. The space C2([0,T ]; X) comprises all C2 functions u : [0,T ] X. →

Let Dom( ) := H1(a,b) be the Sobolev space of all functions f : [a,b] R in E 0 → L2([a,b],dx) with f(a)= f(b)=0.

Let (Dom( ))′ denote the dual space of Dom( ). Then each u (Dom( ))′ E E ∈ E deﬁnes a continuous linear functional := (u,v)µ.

(u,v)= b u vdx with domain Dom( ). E a ∇ ∇ E R (u,v) := b uv dµ denotes the inner product in L2 [a,b] and let u denote the µ a µ k kµ R corresponding norm.

Let u X, where X is Dom( ), L2([a,b],dx) or L2 [a,b]. Then u is the ∈ E µ ∇ distributional derivative with respect to x.

Let u : [0,T ] X. Thenu ˙ denotes the strong derivative. See Appendix G. →

Let α : [0,T ] R. Then α′ denotes the strong derivative. See Appendix G. → 2

Let u : [0,T ] X, where X be Dom( ) or L2 [a,b]. Then ∂u denotes the partial → E µ ∂t derivative. See Appendix G.

Let u : [0,T ] Dom( ) or Let u : [0,T ] L2 [a,b]. Then u and (u ) denote → E → µ t m t the weak derivative. CHAPTER 1

INTRODUCTION

In the real world, many geometric objects are best modeled by fractals, such as trees, coastlines, mountains, and clouds. In 1975, Mandelbrot[19] coined the term of fractals and argued the theory of fractals.

Analysis on fractals [13] actors of Laplacian on the Sierpinski gasket, which is first introduced by J. Kigami in [12], and eigenvalues and eigenfunctions of Laplacian on post critically finite self-similar sets. In this work, Kigami also explain how to construct Dirichlet forms, harmonic functions, Green’s functions and Laplacians on the post critically finite self-similar sets.

Dalrymple et al. [6] studied analogues of some classical diﬀerential equations, such as heat and wave equations on the Sierpinski gasket.

In this thesis, we study one-dimensional wave equations deﬁned by a class of fractal Laplacians. These Laplacians are deﬁned by fractal measures generated by iterated function systems with overlaps. We proved the existence and uniqueness of the weak solution of the hyperbolic initial/boundary value problem of the non- homogeneous wave equation. (see chapter 2.) We also solve the homogeneous wave equation numerically. Finally, we prove that the numerical solutions converge to the weak solution.

1.1 Preliminaries

In this section, we will introduce known results related to our projects. 4

1.1.1 Fractal measures

Let D be a non-empty compact subset of Rd. A function S : D D is called → contraction on D if there is a number c with 0

Si(x)= ρiRix + bi, i =0, 1,...,m, (1.1.1)

where 0 <ρ < 1,R is an orthogonal transformation and b Rd (see [8]). i i i ∈

To each set of probability weights p m , where p 0 and m p = 1, there { i}i=0 i ≥ i=0 i P exists a unique probability measure, called a self-similar measure, satisfying the identity m 1 µ = p µ S− (1.1.2) i ◦ i Xi=0 (see [10]).

We say that S m satisﬁes the open set condition (OSC) if the exist a nonempty { i}i=0 bounded open set U such that m S (U) U and S (U) S (U) = for all i = j ∪i=0 i ⊂ i ∩ j ∅ 6 (see [8]). Let S m and µ be of the form (1.1.1) and (1.1.2), respectively. Also, we { i}i=0 assume that supp(µ)=[a,b]. Deﬁne

nj Tj(x)= ρ x + dj, j =1, 2,...,N, (1.1.3) where n N and d Rd. µ is said to satisfy a family of second-order self-similar j ∈ j ∈ identities (or simply second-order identities) with respect to T N (see [17]) if { j}j=1

(i) supp(µ) N T (supp(µ)), and ⊆ j=1 j S 5

(ii) for each Borel set A supp(µ) and 0 i, j N, µ(T T A) can be expressed ⊆ ≤ ≤ i ◦ j as a linear combination of µ(T A): k =1,...,L as { k } m µ(T T A)= c µ(T A), (1.1.4) i ◦ j k k Xk=0

where ck = ck(i, j) are independent of A.

For our purposes, T N has to satisfy the OSC. { j}i=0

1.1.2 Operator ∆µ

Let µ be a continuous positive ﬁnite Borel measure on R with supp(µ) [a,b]. ⊆ 2 It is well known (see e.g., [3, 9]) that µ deﬁnes a Dirichlet Laplacian ∆µ on Lµ[a,b], described as follows. Let H1(a,b) be the Sobolev space of all functions in L2([a,b],dx)

whose distributional derivatives (see Appendix G) belong to L2[a,b], with the inner product b b (u,v)H1(a,b) := uv dx + u v dx. Za Za ∇ ∇ 1 1 1 Let H0 (a,b) be the completion of Cc∞(a,b) in H (a,b). H0 (a,b) is a dense subspace

2 2 of Lµ[a,b]. Deﬁne a quadratic form on Lµ[a,b],

b (u,v)= u v dx, (1.1.5) E Za ∇ ∇

֒ (with domain equal to H1(a,b) in the Dirichlet case. Since the embeddings H1(a,b 0 0 → -L2 [a,b] and H1(a,b) ֒ L2 [a,b] are compact, is closed. Thus there exists a non µ → µ E negative self-adjoint operator T on L2 [a,b] such that Dom( ) = Dom(T 1/2) and µ E

E(u,v)=(T 1/2u,T 1/2v) for all u,v Dom( ), µ ∈ E 6

where

(u,v)µ := uv dµ Z denotes the inner product in L2([a,b],dµ). Let denote the corresponding norm. k · kµ

We deﬁne ∆ := T and call it the Dirichlet Laplacian with respect to µ if µ − Dom( )= H1(a,b). E 0

Let u Dom( ) and f L2 [a,b]. It is known that u Dom(∆ ) and ∆ u = f ∈ E ∈ µ ∈ µ µ if and only if ∆u = fdµ in the sense of distribution, i.e.,

b b b b u vdx = ( ∆u)vdx = ( fu)vdµ = ( ∆µu)vdµ Za ∇ ∇ Za − Za − Za − for all v C∞(a,b). ∈ c

It is known (see, e.g., [3, 9]) that there exists an orthonormal basis of eigenfunctions of ∆ and the eigenvalues λ are discrete and satisfy 0 λ <λ < with µ { n} ≤ 1 2 ···

limn λn = . →∞ ∞

The eigenvalues λ and eigenfunctions u are deﬁned by the following equation:

u(x) v(x) dx = λ u(x) v(x) dµ(x), (1.1.6) Z ∇ ∇ Z

where the equation holds for all v C∞(a,b). ∈ 0

The operators ∆µ and their generalizations have been studied in connection with spectral functions of the string and diﬀusion processes . More recently, they have been

studied in connection with fractal measures (see [3, 9, 20]). 7

1.2 Main problems in our project

We ﬁrst give a heuristic derivation of the general wave equation of a vibrating string with non-homogeneous mass density.(see e.g. [25]) In later chapters, we will study the properties of one-dimensional fractal wave equations, such as weak solutions and

their regularity. Moreover, we will use numerical methods to ﬁnd approximations to the weak solution of this type of equations.

We use the following notation: Let u(x,t) be vertical displacement from equilib- rium position at position x and time t.

Let T (x,t) be magnitude of tension.

Let µ be measure representing mass density (non-constant).

u(x,t)

Applying Newton’s second law of motion, we get

T (x,t) T (x,t) = 0, (1.2.7) 1+ u2 x=x1 − 1+ u2 x=x0 x x and p p x1 T (x,t)ux T (x,t)ux = utt dµ. (1.2.8) 2 x=x1 2 x=x0 1+ u − 1+ u Zx0 x x p p 8

Assume that the vertical displacement is small, i.e. u << 1. Then 1+ u2 = | x| x p 1+ O(u2). Therefore, (1.2.7) implies T (x ,t) T (x ,t), i.e., T constant, and x 1 ≈ 0 ≈ (1.2.8) implies x1 T ux T ux = utt dµ. · x=x1 − · x=x0 Z x0

x1 x1 T uxx dx = utt dµ. Zx0 Zx0 Therefore,

Tuxx = utt dµ.

Normalize so that T =1. Suppose we let ∆ u = f ∆u = fdµ µ ⇔

Then,

∆µu = utt .

If µ is Lebesgue measure, this reduces to the standard wave equation ∆u = utt.

1.3 Main result

The main purpose of this thesis is to study the one-dimensional wave equations deﬁned by a class of fractal Laplacians.

In Chapter 2, we proved that the existent and uniqueness of solution of these kind equations. The most important result of Chapter 2 is as follows:

Theorem 1.3.1. There exists a constant C > 0, depending only on U and T , such 9

that for all m N, ∈

max um(x,t) Dom ( ) + (um(x,t))t µ + (um(x,t))tt L2([0,T ],(Dom ( ))′) 0 t T k k E k k k k E ≤ ≤ (1.3.9) 2 2 C( f L2([0,T ],L2 [a,b]) + g Dom ( ) + h µ). ≤ k k µ k k E k k

This helps us prove the existent and uniqueness of solution. In Chapter 3, we used the ﬁnite element method to approximate our weak solution. There is a main result in chapter 3:

If um(x,t) be deﬁned by

N m m u (x,t) := βj(t)φj(x), where βj(t) := βm,j(t) and φj(x) := φm,j(x) (1.3.10) Xj=0 where φ0,φ1,...,φN m are the standard piecewise linear ﬁnite element basis functions deﬁned as:

x xk−1 m − if x x x , k =1,...,N x x −1 k 1 k  k− k − ≤ ≤   x xk+1 m φk(x)=  − if x x x , k =0,...,N 1 (1.3.11)  x x +1 k k+1  k− k ≤ ≤ −  0 otherwise.    and satisfy:

b b m m m u (x,t) φi(x) dx = u tt(x,t)φi(x) dµ, for i =0, ,N . (1.3.12) − Za ∇ ∇ Za ···

Moreover, d2w M = Kw, t> 0  dt2 −   (1.3.13) 

w(0) = w0, w′(0) = w0′ .   Theorem 1.3.2. Let µ be deﬁned by (1.1.2) on R with supp[µ]=[a,b]and satis- ﬁes a family of second-order self-similar identities. Assume M is invertible. Then, 10

(1.3.12) could be discretized into a systme of second order ordinary diﬀerential equations (1.3.13). Thus, it could be solved numerically by the ﬁnite element method.

In Chapter 6, we proved the convergence of the approximation solution. The main result is as follows:

Theorem 1.3.3. Let v be a absolute continuous function on [a,b]. Let π = x Nm m { i}i=0 be any partition of [a,b]. Then, v x) v(x) C π 1/2 for all x [a,b]. ( −Pm ≤ k mk ∈

m 2 This helps us to show: u deﬁned in(1.3.10), converges in Lµ[a,b] to u. CHAPTER 2

EXISTENCE AND UNIQUENESS OF WEAK SOLUTION

In this section, we modify the proof in [7], and replace the standard Laplacian ∆

by the µ-Laplacian ∆µ. Then, Our goal is to prove the existence and uniqueness of a weak solution of the following non-homogeneous hyperbolic initial/boundary value problem (IVBP):

u ∆ u = f on U =[a,b] [0,T ],  tt − µ T ×   u =0 on a,b [0,T ], (2.0.1)  { }× u = g, u = h on U t =0 .  t × { }   Let (u,v) be the quadratic form deﬁned in (1.1.5), with domain H1(a,b). First E 0 of all, we will discuss the solution of abstract homogeneous wave equations.

Theorem 2.0.4. (Shinbrot [22]) Let H be a complex Hilbert space. Deﬁne u : R H, → and let A be a self-adjoint operator on H satisfying

(Au(t),u(t)) 0 for each t such that u(t) Dom(A). (2.0.2) ≥ ∈

1 Let g Dom(A),h Dom(A 2 ). Then the initial value problem ∈ ∈

u¨(t)+ Au(t)=0, (2.0.3)

u(0) = g, u˙(0) = h, (2.0.4)

sin(t√λ) has a unique solution, given by u(t) = ∞ cos(t√λ)dE g + ∞ dE h, where 0 λ 0 √λ λ R R Eλ λ R is the spectral family associated with A . { } ∈

Proof. See [22]. For completeness we also include some details in Appendix G. 12

1 Proposition 2.0.5. If g Dom( ∆ ) and h Dom(( ∆ ) 2 ) then (2.0.1) has a ∈ − µ ∈ − µ sin(t√λ) unique solution, given by u(t)= ∞ cos(t√λ)dEλg + ∞ dEλh, where Eλ λ R 0 0 √λ { } ∈ R R is the spectral family associated with ∆ . − µ

Proof. Since g and h are supported on [a,b] and vanish at the end points a and b.

The proposition follows from theorem 2.0.4.

Remark: In general, the classical wave equation may not have a solution unless f has

the proper Darboux structure. (See [4]).

Deﬁnition 2.0.1. (i) A function s : [0,T ] X is called simple if it has the form → N s(t)= χ (t)u for t [0,T ], (2.0.5) Em m ∈ mX=1 where each E is a Lebesgue measurable subset of [0,T ] and u X(m = m m ∈ 1,...,N).

(ii) A function u : [0,T ] X is strongly measurable if there exist simple function → s : [0,T ] X such that N →

s (t) u(t) for Labesgue a.e. (0 t T ). N → ≤ ≤

(iii) A function u : [0,T ] X is weakly measurable if for each u∗ X∗, the → ∈ mapping t u∗,u(t) is Lebesgue measurable. 7→ h i

Deﬁnition 2.0.2. We say u : [0,T ] X is almost separably valued if there exists a → subset E [0,T ], with (E)=0, such that the set u(t) t [0,T ] E is separable. ⊂ L { | ∈ \ }

Theorem 2.0.6. (Pettis [21]). The mapping u : [0,T ] X is strongly measurable if → and only if u is weakly measurable and is almost separably valued. 13

Deﬁnition 2.0.3. Let X be a separable Banach space with norm . Then deﬁne k kX Lp([0,T ]; X) to be the space of all measurable functions u : [0,T ] X satisfying →

T 1 p p (i) u p := ( u( ,t) dt) < for 1 p< , and k kL ([0,T ];X) 0 k · k ∞ ≤ ∞ R

(ii) u L∞([0,T ];X) := esssup0 t T u(t) X < . k k ≤ ≤ k k ∞

Remark: For 1 p , Lp([0,T ]; X) are Banach spaces. (See Appendix B.) ≤ ≤ ∞

Deﬁnition 2.0.4. g,h Dom( ),f L2([0,T ];Dom( )), A function u L2([0,T ];Dom( )), ∈ E ∈ E ∈ E 2 2 2 with u L ([0,T ]; L [a,b]) and u L ([0,T ];(Dom( ))′) is a weak solution of t ∈ µ tt ∈ E IBVP (2.0.1) if the following conditions are satisﬁed:

(i) u ,v + (u,v)=(f,v) for each v Dom( ), f L2 [a,b] and Lebesgue a.e. h tt i E µ ∈ E ∈ µ t [0,T ]; ∈

(ii) u(x, 0) = g(x) and u (x, 0) = h(x) for all x [a,b]. t ∈

1 2 Let w ∞ C [a,b] be an orthonormal basis of L [a,b] consisting of the { k}k=1 ⊂ µ eigenfunctions of ∆ with eigenvalues λ ∞ . Then − µ { k}k=1 b b wk vdx = λk wk v dµ, v Dom( ). Za ∇ ∇ Za ∀ ∈ E

The existence of such an orthonormal basis of eigenfunctionsof∆µ can be found in [3].

Fix a positive integer m, and deﬁne

um(x,t) := αm,k(t)wk(x), (2.0.6) Xk=1 14

where we will show that the coeﬃcients α (t) m can be chosen to belong to { m,k }k=1 C2(0,T ) and satisfy

αm,k(0) = (g,wk)µ, k =1,...,m (2.0.7)

αm,k′ (0) = (h,wk)µ, k =1,...,m (2.0.8)

and

(um)tt,wk + (um,wk)=(f,wk)µ, 0 t T, k =1,...,m. (2.0.9) µ E ≤ ≤

Theorem 2.0.7. For each m N, there exists a unique function u of the form ∈ m (2.0.6) satisfying (2.0.7)-(2.0.9).

Proof. Let um(x,t) be deﬁned in (2.0.6). Then, we have

(um)tt,wk = αm,k′′ (t). (2.0.10) µ

Moreover, (u ,w ) = m (w ,w )α (t), j, k = 1,...,m, and f := (f,w ) . E m k j=1 E j k m,j k k µ P Consequently, (2.0.9) is discretized into the linear system of ODEs:

m α′′ (t) (w ,w )α (t)= f , 0 t T, k =1,...,m, m,k − E j k m,j k ≤ ≤ Xj=1 or

α′′ (t) λ α (t)= f , 0 t T, k =1,...,m, (2.0.11) m,k − k m,k k ≤ ≤ with the initial conditions (2.0.7) and (2.0.8). There exists a unique C2 vector-valued

function αm′′ (t)=(αm,′′ 1(t),...,αm,m′′ (t)) satisfying (2.0.7), (2.0.8) and (2.0.11) for 0 t T, namely ≤ ≤

αm,k(t)=(g,wk)cos( λkt)+ Pm,k(t), p where Pm,k(t) is the particular solution that depends on f(t,x). 15

We now need to take the limit as m . To this end we need the following → ∞

estimates for um, together with its partial derivatives with respect to time, that are uniform in m.

Theorem 2.0.8. There exists a constant C > 0, depending only on U and T , such that for all m N, ∈

max um(x,t) + (um(x,t))t + (um(x,t))tt 2 ′ 0 t T Dom ( ) µ L ([0,T ],(Dom ( )) ) ≤ ≤ k k E k k k k E 2 2 C( f L2([0,T ],L2 [a,b]) + g Dom ( ) + h µ). ≤ k k µ k k E k k (2.0.12)

Proof. Multiplying equality (2.0.9) by αm,k′ (t), we have

(um)tt,αm,k′ (t)wk + um,αm,k′ (t)wk = f,αm,k′ (t)wk , k =1,...,m. µ E µ

Summing the k equations up, we have

m m m

(um)tt, αm,k′ (t)wk + um, αm,k′ (t)wk = f, αm,k′ (t)wk . µ E µ Xk=1 Xk=1 Xk=1

That is,

(um)tt, (um)t + um, (um)t = f, (um)t . (2.0.13) µ E µ Since d 1 ( (u ) 2 )=((u ) , (u ) ) dt 2 k m tkµ m tt m t µ and d 1 ( (u ,u )) = (u , (u ) ), dt 2E m m E m m t from (2.0.13) we have

d 1 2 1 1 2 1 2 (um)t µ+ (um,um) = f, (um)t (um)t µ f µ (um)t µ+ f µ . dt2 k k 2E µ ≤ k k k k ≤ 2 k k 2 k k 16

Consuquently,

d 2 2 2 (um)t µ + (um,um) (um)t µ + (um,um)+ f µ . dt k k E ≤ k k E k k

Since (u ) 2 and (u ,u (x,t)) are absolutely continuous functions (see Ap- k m tkµ E m m pdeix C1), Gronwall’s inequality, yields the estimate:

(u (x,t)) 2 + (u (x,t),u (x,t)) k m tkµ E m m t (2.0.14) t 2 2 e (um(x, 0))t µ + (um(x, 0),um(x, 0)) + f(s) µ ds . ≤ k k E Z0 k k

Moreover, for the ﬁrst term on right hand side of (2.0.14), we have m m b 2 b 2 2 2 (um(x, 0))t = α′ (0)wk(x) dµ = (α′ (0)) (wk(x)) dµ k kµ Z m,k Z m,k a Xk=1 Xk=1 a m 2 = (αm,k′ (0)) . Xk=1

By expressing h(x)= ∞ α′ (0)w (x), we have for all m N, k=1 m,k k ∈ n P n n 2 2 h = lim αn,k′ (0)wk(x), αn,k′ (0)wk(x) = lim (αn,k′ (0)) k kµ n n →∞ Xk=1 Xk=1 →∞ Xk=1 (2.0.15) (u (x, 0)) 2 . ≥ k m tkµ

Also, for the second term on right hand side of (2.0.14), we have m m (u (x, 0),u (x, 0)) = α (0)w , α (0)w E m m E m,k k m,k k Xk=1 Xk=1 m m (2.0.16) = (α (0)w ,α (0)w )= (α (0))2λ . E m,k k m,k k m,k k Xk=1 Xk=1

By expressing g(x)= k∞=1 αm,k(0)wk(x), we obtain Pn n n 2 (g,g)= lim αn,k(0)wk, αn,k(0)wk = lim (αn,k(0)) (wk,wk) E n E n E →∞ Xk=1 Xk=1 →∞ Xk=1 n 2 = lim (αn,k(0)) λk. n →∞ Xk=1 (2.0.17) 17

Combining (2.0.16) and (2.0.17), we have, for all m N, ∈

(u (x, 0),u (x, 0)) (g,g). (2.0.18) E m m ≤E

Now, combining (2.0.14), (2.0.15) and (2.0.18), we get, for all m N, ∈

2 t 2 2 (um(x,t))t + um(x,t),um(x,t) e h + (g,g)+ f(t) 2 2 . k kµ E ≤ k kµ E k kL (0,T,Lµ[a,b]) Since 0 t T was arbitrary, we see from this estimate that ≤ ≤

2 t 2 2 max (um(x,t))t + (um(x,t),um(x,t)) e ( h + (g,g)+ f(t) 2 2 ). 0 t T µ µ L (0,T,Lµ[a,b]) ≤ ≤ k k E ≤ k k E k k (2.0.19)

Next, we ﬁx any v Dom( ), with v Dom ( ) 1, and write v = v1 + v2, where ∈ E k k E ≤ m v1 span wk k=1 and (v2,wk)µ =0(k =1, 2,...,m). Then, v1 Dom ( ) 1, (2.0.6) ∈ { } k k E ≤ and (2.0.9) imply

(um)tt,v := (um)tt,v = (um)tt,v1 =(f,v1)µ (um,v1). h i µ µ −E

So, (u ) ,v (f,v ) (u ,v ) , which implies that |h m tt i|≤| 1 µ −E m 1 |

(um)tt,v Dom ( ))′ C f µ + (um,um). kh ik E ≤ k k pE Thus, we have

2 2 (um)tt (Dom ( ))′ C( f µ + (um,um)). k k E ≤ k k E From (2.0.19), this implies

T 2 2 (u ) ′ dt C( f 2 2 + h + (g,g)). (2.0.20) m tt (Dom ( )) L ([0,T ],Lµ[a,b]) µ Z0 k k E ≤ k k k k E

Remark

Deﬁnition 2.0.5. Let Xbe a Banach space. 18

(i) The space C([0,T ]; X) comprises all continuous functions u : [0,T ] Xwith →

u := max u < , C([0,T ];X) 0 t T X k k ≤ ≤ k k ∞

(ii) The space C1([0,T ]; X) comprises all C1 functions u : [0,T ] Xwith →

u˙ := max u˙ < , C([0,T ];X) 0 t T X k k ≤ ≤ k k ∞

(iii) The space C2([0,T ]; X) comprises all C2 functions u : [0,T ] Xwith →

u¨ := max u¨ < . C([0,T ];X) 0 t T X k k ≤ ≤ k k ∞ Deﬁnition 2.0.6. Let u L1([0,T ]; X). We say v L1([0,T ]; X) is the weak ∈ ∈ derivative of u, written

ut = v,

if T T φt(t)u(t)dt = φ(t)ut(t)dt. Z0 − Z0

for all scalar test function φ(t) C∞(0,T ). ∈ c

Deﬁnition 2.0.7. Let Xbe a Banach space. We say a sequence u ∞ X { m}m=1 ⊂ converges weakly to u X, written ∈

um ⇀ u,

u∗,u u∗,u h mi→h i for each bounded linear functional u∗ X∗. ∈

Theorem 2.0.9. Assume g Dom( ),h L2 [a,b] and f L2([0,T ]; L2 [a,b]). ∈ E ∈ µ ∈ µ Then the IBVP (2.0.1) has a weak solution. 19

Proof. From the energy estimate (2.0.12), we know that u ∞ is bounded in { m}m=1 2 2 2 L ([0,T ];Dom( )), (u ) ∞ is bounded in L ([0,T ]; L [a,b]), and (u ) ∞ E { m t}m=1 µ { m tt}m=1 2 is bounded in L (([0,T ];Dom( ))′). E

So, by Banach Alaoglu theorem, there exists a subsequence u (x,t) ∞ and { ml }l=1 2 2 2 2 u L ([0,T ];Dom( )), with u L ([0,T ]; L [a,b]), and u L ([0,T ];(Dom( ))′) ∈ E t ∈ µ tt ∈ E such that

2 uml (x,t) ⇀ u(x,t) in L ([0,T ];Dom( )),  E     2 2 (2.0.21)  (uml (x,t))t ⇀ utt(x,t) in L ([0,T ]; Lµ[a,b]),     (u (x,t)) ⇀ u (x,t) in L2([0,T ];(Dom( )) ).  ml tt tt ′  E 

∂uml (x,t) 2 2 Remark: Originally, ∂t weakly converge to some γ in L ([0,T ]; Lµ[a,b]) and 2 ∂ uml (x,t) 2 2 weakly converge to some σ in L ([0,T ];(Dom( ))′). It can be proved that ∂t E ∂u(x,t) ∂2u(x,t) ∂t = γ and ∂t2 = σ. (See Appendix E).

Now, we ﬁx an integer N and choose a function v C1([0,T ];Dom( )) of the ∈ E form N v(x,t)= α (t)w (x), where α N C2[0,T ]. (2.0.22) k k { k}k=1 ⊂ Xk=1

We select m N, multiply (2.0.9) by α(t), sum k =1,...,N, and then integrate ≥ with respect to t to get

T T (um(x,t))tt,v(x,t) + (um(x,t),v(x,t)) dt = (f(x,t),v(x,t))µ dt. Z0 h i E Z0 (2.0.23) Setting m = m , letting l tend to , and using (2.0.21), we have l ∞ T T (utt(x,t)),v(x,t) + (u(x,t),v(x,t)) dt = (f(x,t),v(x,t))µ dt. (2.0.24) Z0 h i E Z0 20

Since w ∞ is a basis of Dom( ), the set of functions of the form (2.0.22) is { k}k=1 E dense in L2([0,T ];Dom( )), and thus (2.0.24) holds for all v L2([0,T ];Dom( )). E ∈ E Then, (2.0.24) implies

u (x,t),v(x,t) + (u(x,t),v(x,t))=(f(x,t),v(x,t)) h tt i E µ

for all v Dom( ) and a.e. t [0,T ]. ∈ E ∈

Next we will verify

u(x, 0) = g(x) and ut(x, 0) = h(x). (2.0.25)

For this, in (2.0.24), we choose any function v C2([0,T ];Dom( )), with v(T ) = ∈ E vt(T ) = 0. Integrating by parts twice with respect to t for the ﬁrst term of (2.0.24), we get T u(x,t),vtt(x,t) + (u(x,t),v(x,t)) dt Z0 h i E T = (f(x,t),v(x,t))µ dt (u(x, 0),vt(x, 0))+(ut(x, 0),v(x, 0)). (2.0.26) Z0 −

Similar to (2.0.23),

T ( um(x,t),vt(x,t) + (um(x,t),v(x,t))) dt Z0 h i E T = (f(x,t),v(x,t))µ dt (um(x, 0),vt(x, 0))+(um,t(x, 0),v(x, 0)). Z0 −

Setting m = ml and combining (2.0.7), (2.0.8) and (2.0.21), we have

T ∂2v(x,t) u(x,t), 2 + (u(x,t),v(x,t)) dt Z0 h ∂t i E T = (f,v(x,t))µ dt (g(x),vt(x, 0))+(h(x),v(x, 0)). (2.0.27) Z0 −

Comparing (2.0.26) and (2.0.27), and noting that v(x, 0) and vt(x, 0) were arbitrary, we conclude that (2.0.25) holds. Therefore, u(x,t) is a weak solution of (2.0.1). 21

Theorem 2.0.10. Assume the same hypotheses of Theorem 2.0.9. Then the weak solution of the IBVP (2.0.1) is unique.

Proof. To show this, it suﬃces to show that the only weak solution of (2.0.1) with g(x)= h(x)= f(x,t)=0is u 0 in L2([0,T ];Dom( )). To show this, ﬁx 0 r T ≡ E ≤ ≤ and set s u(x,τ)dτ if 0 t s,  t ≤ ≤ v(x,t) := R  0 if s t T. ≤ ≤ Then v(x,t) Dom( ) for each t [0,T ] and so ∈ E ∈ s utt(x,t),v(x,t) + (u(x,t),v(x,t)) dt =0. (2.0.28) Z0 h i E

We have ut(x, 0) = 0 and by the deﬁnition of v(x,t), we have v(x,s)=0. Integrating by parts in the ﬁrst term of (2.0.28), we obtain:

s utt(x,t),vt(x,t) + (u(x,t),v(x,t)) dt =0. − Z0 h i E

Moreover, we have v (x,t)= u(x,t) for 0 t s. Hence t − ≤ ≤ s ut(x,t),u(x,t) (vt(x,t),v(x,t)) dt =0 Z0 h i−E s ∂ 1 2 1 u(x,t) µ (v(x,t),v(x,t))) dt =0 ⇒ Z0 ∂t2 k k − 2E s 1 2 1 u(x,t) µ (v(x,t),v(x,t)) =0 fora.e. t [0,T ]. ⇒ h2 k k − 2E i0 ∈

u(x,s) 2 (v(x,s),v(x,s))) = u(x, 0) 2 (v(x, 0),v(x, 0)). k kµ −E k kµ −E Since u(x, 0) = v(x,s)=0, u(x,s) 2 + (v(x, 0),v(x, 0)) = 0. This implies k kµ E

u(x,s)=0 fora.e. x [a,b] and for all s [0,T ] ∈ ∈ 22 and u(x,t) is continuous in Dom( ). Thus u(x,t) 0. E ≡ CHAPTER 3

THE FINITE ELEMENT METHOD

In this section, we use the ﬁnite element method to solve the homogeneous IBVP (2.0.1).

Multiplying the ﬁrst equation in (2.0.1), where f(x,t) = 0, by v Dom( ), ∈ E integrating both sides, and then using integration by parts, we obtain b b u(x,t) v(x,t) dx = utt(x,t) v(x,t) dµ. (3.0.1) − Za ∇ ∇ Za Next, we apply the ﬁnite element method to approximate u(x,t) by N m m u (x,t)= βj(t)φj(x), where βj(t) := βm,j(t) and φj(x) := φm,j(x) (3.0.2) Xj=0

and φ0,φ1,...,φN m are the standard piecewise linear ﬁnite element basis functions deﬁned as:

x x1 − if x x x x0 x1 0 1 φ (x) =  − ≤ ≤ 0   0 otherwise,   x xk−1 m − if x x x , 1 k N 1 x x −1 k 1 k  k− k − ≤ ≤ ≤ ≤ −   x xk+1 φk(x) =  − if x x x  x x +1 k k+1  k− k ≤ ≤

 0 otherwise,    x xNm−1 − if x m x x m x m x m−1 N 1 N N − N − ≤ ≤ φ m (x) =  N   0 otherwise.  Equivalently, 

x xk−1 m − if x x x , k =1,...,N x x −1 k 1 k  k− k − ≤ ≤   x xk+1 m φk(x)=  − if x x x , k =0,...,N 1 (3.0.3)  x x +1 k k+1  k− k ≤ ≤ −  0 otherwise.    24

We require um(x,t) to satisfy:

b b m m m u (x,t) φi(x) dx = utt (x,t)φi(x) dµ, for i =0, ,N . (3.0.4) − Za ∇ ∇ Za ···

Substituting (3.0.2) into (3.0.4) and using u(a,t)= u(b,t)=0, we get

N m 1 N m 1 − b − b βj(t) φj(x) φi(x) dx = β′′(t) φj(x)φi(x) dµ . (3.0.5) − Z ∇ ∇ j Z Xj=1 a Xj=1 a

We deﬁne the matrices M and K (the mass and stiﬀness matrices, respectively) by b b dφj dφi Mij = φj(x)φi(x) dµ, Kij = (x) (x) dx, (3.0.6) Za − Za dx dx and the vector-valued function w(t) by

β1(t)   . w(t)= . .       βN m 1(t)  −  Then (3.0.5) can be put in a matrix form as

Mw′′ + Kw =0. (3.0.7)

Equivalently,

Mw′′ = Kw. (3.0.8) −

We have a system of second-order linear ODEs (3.0.8) with constant coeﬃcients.

We need two initial conditions, and they are obtained from the initial conditions in (2.0.1). We have

u(x, 0) = g(x), 0

and g(x) can be approximated by its linear interpolant:

N m 1 − g˜(x)= g(xi)φi(x). Xi=1 Therefore, we set

wi(0) = g(xi).

Similarly, we set

wi′(0) = h(xi).

These lead to the initial conditions

g(x1) h(x1)     . . w(0) = w0 = . , w′(0) = w′ = .   0           g(xN m 1) h(xN m 1)  −   −  Therefore, we get the IVP

d2w M = Kw, t> 0  dt2 −   (3.0.9) 

w(0) = w0, w′(0) = w0′ .    m Theorem 3.0.11. Deﬁne M by (3.0.6). If for i = 1,...,N 1, Mi,i Mi,i 1 − − − − M 0, then M is strictly diagonally dominant matrices. Thus, (3.0.9) has a i,i+1 ≥ unique solution w(t).

In the case of the inﬁnite Bernoulli convolution associated with the golden ratio, and the 3-fold convolutions of the Cantor measure, we prove M is strictly diagonally dominant. (See [27].) The proof is shown in Appendix A.3.

Remarks: From this theorem, we know that for j = 1, ,N m 1, β (t) are C2 ··· − j functions on (0,T ). 26

Remarks: will move to below of the following paragraph.

In the rest of this thesis we will consider self-similar measures µ deﬁned by IFSs of the form S N as given in (1.1.1) and (1.1.2). We assume that µ satisﬁes a family { i}i=1 of second-order self-similar identities with respect to T N . { j}j=1

Since TJ [a,b] can represent each m-level interval TJ [a,b], where J =(j1, . . . , jm) and jk 1,...,N , can be written as [xi 1,xi], where the index k can be obtained ∈ { } − directly from J as follows (see [5]):

m 1 m 2 0 i = i(J):=(j 1)N − +(j 1)N − + +(j 1)N +1. . 1 − 2 − ··· m −

For example, if J = (1,..., 1), then i(J) = 1, andif J =(N,...,N), then i(J)= N m. It follows that

x a TJi := TJ [a,b]=[xi 1,xi], or TJi (x) := TJ (x)=(xi xi 1) − + xi 1. (3.0.10) − − − b a − − j j m We deﬁne ci := cJ for j =1,...,N and i =1,...,N .

By assumption, we can evaluate the measure of each 1-level interval, i.e., µ(Tj[a,b]) = dµ T , and the integrals of xdµ T and x2 dµ T , j =1,...,N. ◦ j ◦ j ◦ j R R R Theorem 3.0.12. Let µ be deﬁned by (1.1.2) on R with supp[µ]=[a,b] and assume that µ satisﬁes a family of second-order self-similar identities. Assume the mass

matrix M is invertible. Then (3.0.4) can be discretized into a system of second order ordinary diﬀerential equations0 (3.0.9). Thus, it can be solved numerically by the

ﬁnite element method.

We remark that b Mii = φi(x)φi(x) dµ, Za 27

while,

xi xi+1 x xi 1 x xi 1 x xi+1 x xi+1 Mii = − − − − dµ + − − dµ Zx − xi xi 1 xi xi 1 Zx xi xi+1 xi xi+1 i 1 − − − − i − − b b x a 2 x a 2 = ( − ) dµ(TJ (x)) + ( − + 1) dµ(TJ (x)). Z b a i Z − b a i+1 a − a − For change of variable, formula, see Appendeix A.3.

Method 1. We let w := w(t ), n 1 and use the central diﬀerence method n n ≥ − (CDM) to solve the IVP (3.0.9).

Using the approximations

2 d w(tn) wn+1 2wn + wn 1 wn+1 wn 1 − − and w′(t ) − − . (3.0.11) dt2 ≈ (∆t)2 n ≈ 2∆t

Substituting (3.0.11) into (3.0.8) yields

wn+1 2wn + wn 1 1 − − = M− Kw (∆t)2 − n

2 1 wn+1 2wn + wn 1 = (∆t) M− Kwn − − − 2 1 wn+1 = (2I (∆t) M− K)wn wn 1. − − −

Moreover, we have

2 1 w1 w 1 w1 = (2I (∆t) M− K)w0 w 1, w0′ = − − − − − 2∆t or

2 1 w1 = (2I (∆t) M− K)w0 w 1, w 1 = w1 2∆tw0′ − − − − −

From the last two equations we get

2 (∆t) 1 w1 = I M− K w0 +∆tw0′ . − 2 28

Therefore, the IVP becomes, by the CDM approximation:

2 1 wn+1 = (2I (∆t) M− K)wn wn 1, n =1, 2,...  − − −   w(t0)= w0   2 (3.0.12)  (∆t) 1 w(t1)= w1 = I M− K w0 +∆tw0′ − 2   t = n∆t.  n   Method 2. We transform the second-order system of ODEs to an equivalent ﬁrst- order system.

Let y(t)= w′(t), and thus y′(t)= w′′(t), and let

y(t) Y(t)=   .  y′(t)    Then (3.0.9) becomes the following equivalent ﬁrst-order system:

w(t0) 0 I Y′(t)= AY(t), Y(t0)=   , A =   . (3.0.13) 1 w′(t0)  M− K 0   −  This system can be solved by using standard ODE theory. CHAPTER 4

THE CENTRAL DIFFERENCE METHOD

In this section, we use the central diﬀerence method to solve the wave equation:

u = u dµ  xx tt   u(a,t) = u(b,t)=0 (4.0.1)  u(x, 0) = g(x), u (x, 0) = h(x).  t   Let = ( x J , t N ) be a partition of the rectangle [a,b] [0,T ], i.e., P P { j}j=1 { n}n=0 ×

a = x

Also, let

∆xj = xj xj 1 and ∆tn = tn tn 1. − − − −

Deﬁneu ¯x =u ¯x( ) on ((xj 1 + xj)/2,tn):1 j J, 0 n N by P { − ≤ ≤ ≤ ≤ }

xj 1 + xj u(xj,tn) u(xj 1,tn) u¯x − ,tn := − − . 2 ∆xj

u¯x approximates the partial derivative ux at the points ((xj 1 + xj)/2,tn). −

Next, deﬁneu ¯ =u ¯ ( ) on x ,t ):1 j J 1, 0 n N by xx xx P { j n ≤ ≤ − ≤ ≤ } u¯x((xj + xj+1)/2,tn) u¯x((xj 1 + xj)/2,tn) u¯xx(xj,tn):= − − (∆xj +∆j+1)/2 (4.0.2) 2 u(xj+1,tn) 1 + 1 u(x ,t )+ 1 u(x ,t ) ∆x +1 ∆x ∆x +1 j n ∆x j 1 n = h j − j j j − i. ∆xj +∆xj+1

Thus,u ¯xx(xj,tn) approximates the second-order partial derivative uxx at (xj,tn).

In a similar fashion, we deﬁneu ¯t =u ¯t( ) on (xj, (tn 1 +tn)/2):0 j J, 1 P { − ≤ ≤ ≤ n N by ≤ } tn 1 + tn u(xj,tn) u(xj,tn 1) u¯t xj, − := − − , 2 ∆tn 30

and deﬁneu ¯ =u ¯ ( ) on (x ,t ):0 j J, 1 n N 1 by tt tt P { j n ≤ ≤ ≤ ≤ − } u¯t(xj, (tn + tn+1)/2) u¯t(xj, (tn + tn 1)/2) u¯tt := − − (∆tn +∆n+1)/2 (4.0.3) 2 1 u(x ,t ) 1 + 1 u(x ,t )+ 1 u(x ,t ) ∆t +1 j n+1 ∆tn ∆t +1 j n ∆tn j n 1 = h n − n − i. ∆tn +∆tn+1

Obviously,u ¯t andu ¯tt approximate the partial derivatives ut and utt, respectively.

In all of our computations, we set ∆tn = ∆t for all n = 1,...,N. In this case, equation (4.0.3) simpliﬁes to

u(xj,tn+1) 2u(xj,tn)+ u(xj,tn 1) u¯ (x ,t )= − − . (4.0.4) tt j n (∆t)2

Substituting (4.0.2) and (4.0.4) into (4.0.1) leads to the following discretized wave

equation:

1 1 1 1 2( u˜(xj+1,tn) ( + )˜u(xj,tn)+ u˜(xj 1,tn)) ∆xj+1 − ∆xj ∆xj+1 ∆xj − ∆x +∆x j+1 j (4.0.5) u˜(xj,tn+1) 2˜u(xj,tn)+˜u(xj,tn 1) µ([xj 1,xj+1]) − − = − 2 . (∆t) (∆xj +∆xj+1)

We consider the two cases with the same or diﬀerent ∆xj.

Case 1: ∆x =∆x for all 1 j J. In this case equation (4.0.5) becomes j ≤ ≤ u˜(xj+1,tn) 2˜u(xj,tn)+˜u(xj 1,tn) − − (∆x)2

u˜(xj,tn+1) 2˜u(xj,tn)+˜u(xj,tn 1 µ([xj 1,xj+1]) = − − − . (∆t)2 2(∆x) Let 2(∆t)2 sj = sj( ) := , j =1,...,J 1. P ∆x(µ[xj 1,xj+1]) − − Then we get

sj u˜(xj+1,tn) 2˜u(xj,tn)+˜u(xj 1,tn) =u ˜(xj,tn+1) 2˜u(xj,tn)+˜u(xj,tn 1), − − − − 31

u˜(xj,tn+1)+˜u(xj,tn 1)= sju˜(xj+1,tn) 2(sj 1)˜u(xj,tn)+ sju˜(xj 1,tn). (4.0.6) − − − −

Moreover, we have

2 u˜(xj,t1) u˜(xj,t 1)=˜ut(xj,t0) (2∆t)+ (∆t) ut(xj,t0)(2∆t) hj(2∆t). (4.0.7) − − · O ≈ ≈

Therefore, adding (4.0.6) (for n = 0) and (4.0.7), we get

sj sj u˜(xj,t0+1) u˜(xj+1, 0) (sj 1) u˜(xj, 0) + u˜(xj 1, 0) + hj (∆t). (4.0.8) ≈ 2 · − − · 2 − ·

Finally,

u˜(xj,t0)=˜u(xj,t0)= g(xj)= gj, u˜(x0,t)=˜u(a,t) andu ˜(xJ ,t)=˜u(b,t)=0. (4.0.9)

Therefore, using the central diﬀerence method, we can approximate the solution of the system (4.0.1) by the following formulas:

u˜(xj,tn+1) sj u˜(xj+1,tn)+2(sj 1)˜u(xj,tn)+ sju˜(xj 1,tn) u˜(xj,tn 1)  ≈ · − − − −  sj sj  u˜(xj,t1) u˜(xj+1,t0)+(sj 1) u˜(xj,t0)+ u˜(xj 1,t0)+ hj (∆t)  ≈ 2 · − · 2 − ·   u˜(xj,t0)=˜u(xj, 0) = g(xj)= gj   u˜(x ,t)=˜u(a,t)=0, u˜(x ,t)=˜u(b,t)=0.  0 J   (4.0.10)

Case 2. ∆xj is not constant.

If xj+1 xj = xj xj 1 and let ∆xj = xj xj 1, where j =1, 2,... . Then from − 6 − − − − (4.0.5), we have

1 1 1 1 2( u˜(xj+1,tn) ( + )˜u(xj,tn)+ u˜(xj 1,tn)) ∆xj+1 − ∆xj ∆xj+1 ∆xj − ∆xj+1 +∆xj n+1 uj 2˜u(xj,tn)+˜u(xj,tn 1) µ([xj 1,xj+1]) − − − = 2 , (∆t) (∆xj +∆xj+1) 32

which is simpliﬁed to 1 1 1 1 sj u˜(xj+1,tn) ( + )˜u(xj,tn)+ u˜(xj 1,tn) ∆xj+1 − ∆xj ∆xj+1 ∆xj −

=˜u(xj,tn+1) 2˜u(xj,tn)+˜u(xj,tn 1), − − or

u˜(xj,tn+1)+˜u(xj,tn 1) − (4.0.11) sj 1 1 s = u˜(xj+1,tn) (sj( + ) 2)˜u(xj,tn)+ u˜(xj 1,tn). ∆xj+1 · − ∆xj ∆xj+1 − ∆xj − Moreover, from (2.12), if n = 0, we have

2 u˜(xj,t1) u˜(xj,t 1)=˜ut(xj,t0) (2∆t)+ (∆t) u˜t(xj,t0)(2∆t) − − · O ≈ (4.0.12)

u˜(xj,t1) u˜(xj,t 1) hj(2∆t). ≈ − − ≈

By summing (4.0.11) (for n = 0) and (4.0.12), we get s 1 1 u˜(x ,t ) u˜(x , 0) (s( + )/2 1)˜u(x ,t ) j 1 ≈ 2∆x · j+1 − ∆x ∆x − j 0 j+1 j j+1 (4.0.13) sj + u˜(xj 1,t0)+ hj (∆t). 2∆xj − · Moreover,

u˜(xj,t0)= u(xj, 0) = g(xj)= gj, u˜(x0,t)= u(a,t)=0, u˜(xJ ,t)= u(b,t)=0. (4.0.14)

Therefore, we can approximate the solution of the system(4.0.1) by the following scheme:

sj 1 1 u˜(xj,tn+1)= u˜(xj+1,tn) (sj( + ) 2)˜u(xj,tn)  ∆xj+1 · − ∆xj ∆xj+1 −  sj  + u˜(xj 1,tn) u˜(xj,tn 1)  ∆xj  − − −  sj 1 1  u˜(xj,t1)= u˜(xj+1, 0) (sj( + )/2 1)˜u(xj,t0)  2∆xj+1 ∆xj ∆xj+1  · − − (4.0.15)  sj  + u˜(xj 1,t0)+ hj (∆t) 2∆xj − ·   u˜(xj, 0) = g(xj)= gj    u˜(a,t)=0, u˜(b,t)=0.    CHAPTER 5

FRACTAL MEASURES DEFINED BY ITERATED FUNCTION SYSTEMS

In this chapter, we solve the homogeneous IBVP (2.0.1) numerically for three diﬀerent

measures namely, the weighted Lebesgue measure, the inﬁnite Bernoulli convolution associated with the golden ratio, and the 3-fold convolutions of the Cantor measure.

Let S N be an IFS of contractive similitudes on R of the form { i}i=1

Si(x)= ρx + bi, i =1,...,N, (5.0.1) and let µ be an associated self-similar measure with supp[µ]=[a,b] satisfying the following self-similar identity:

N 1 µ = p µ S− , (5.0.2) i ◦ i Xi=1

N where 0

5.0.1 Weighted Lebesgue measure

The weighted Lebesgue measure is deﬁned by the system of

1 1 1 S (x)= x, S (x)= x + 1 2 2 2 2

and

1 1 µ = pµ S− + (1 p)µ S− . ◦ 1 − ◦ 2 In view of [3], we will choose the weight p =2 √3. − 34

0 1 S1 ¡ A S2 ¡ AU

01 1 ¡A 2 ¡A ¡ AUS2S1 ¡ AU S2S2 S1S1 S1S2

01 1 3 1 4 2 4

For any Borel subset A [0, 1], we have: ⊆

µ(S1TiA) µ(S1A)   = Mi   , i =1, 2,  µ(S2TiA)   µ(S2A)      where p 0 1 p 0 M1 =   ,M2 =  −  ,  0 p   0 (1 p)     − 

Let J = j1j2 . . . jm, ji =1, 2 or 3. Then

µ(S1A) µ(SJ A)= cJ   ,  µ(S2A)    where

c = e M M =(c1 ,c2 ). J j1 j2 ··· jm J J

5.1 Inﬁnite Bernoulli convolution associated with the golden ratio

The infinite Bernoulli convolution associated with the golden ratio satisfies a family of second-order self-similar identities. This was first pointed out by Strichartz et al.

[26]. We can make use of this to calculate the measure of suitable subintervals. 35

The inﬁnite Bernoulli convolution associated with the golden ratio is deﬁned the by the IFS

√5 1 S (x)= ρx, S (x)= ρx + (1 ρ), ρ = − . 1 2 − 2

0 1 S1 ¡ A S2 ¡ AU

01 ρ ρ 1 −

For each 0

1 1 µ(A)= pµ S− (A)+(1 p)µ S− (A). ◦ 1 − ◦ 2 the weighted inﬁnite Bernoulli convolution associated with the golden ratio. If p =1/2, we get the classical one.

Deﬁne

2 3 2 2 T1(x)= ρ x. T2(x)= ρ x + ρ , T3(x)= ρ x + ρ.

Then µ satisﬁes the following second-order self-similar identities (see [17]): for any

Borel subset A [0, 1], we have: ⊆

µ(T1TiA) µ(T1A)     µ(T T A) = Mi µ(T A) , i =1, 2, 3,  2 i   2           µ(T3TiA)   µ(T3A)      where

p2 0 0 0 p2 0     2 M1 = (1 p)p (1 p)p 0 ,M2 = 0 (1 p)p 0 ,      − −   −     2   0 1 p 0   0 (1 p) 0   −   −  36

0 p 0   2 M3 = 0 (1 p)p (1 p) p .    − −   2   0 0 (1 p)   −  If p =1/2, we obtain

2 0 0 0 1 0 0 4 0 1   1   1   M1 = 1 2 0 ,M2 = 0 1 0 ,M3 = 0 2 1 . 8   4   8                0 4 0   0 1 0   0 0 2       

Let J = j1j2 . . . jm, ji =1, 2 or 3. Then

µ(T1A)   µ(TJ A)= cJ µ(T A) ,  2       µ(T3A)    where c = e M M =(c1 ,c2 ,c3 ). J j1 j2 ··· jm J J J

Moreover, we have

b b b f(x) dµ = p f(S1(x)) dµ + (1 p) f(S2(x)) dµ. (5.1.3) Za Za − Za

Then, we use (5.1.3) to evaluate the measure of each interval µ(T [0, 1]) = 1 dµ j 0 ◦ R T , 1 xdµ T and 1 x2 dµ T , j = 1, 2, 3 for any probability weights p and 1 p. j 0 ◦ j 0 ◦ j − R R 1 The following result is the matrix with p = 2 :

1 1 1 1 1 1 dµ T1 dµ T2 dµ T3  0 ◦ 0 ◦ 0 ◦   3 3 3  R1 R1 R1 1 1 1 xdµ T xdµ T xdµ T = 2 .  0 1 0 2 0 3   6(3ρ 1) 6 6(3ρ +3)   ◦ ◦ ◦   −  R1 2 R1 2 R1 2 5ρ+4 ρ+5 2 ρ  x dµ T x dµ T x dµ T   −   0 ◦ 1 0 ◦ 2 0 ◦ 3   6(ρ+8) 6(ρ+8) 6(ρ+8)   R R R   (5.1.4) 37

We will apply this matrix to calculate the value of the entries of M in (3.0.12), which is very important for ﬁnding out the measure of each interval on the level m of the ﬁnite element method.

5.2 3-fold convolution of the Cantor measure

The 3-fold convolutions of the Cantor measure satisﬁes a family of second-order identities. It is deﬁned by the IFS

1 2 S (x)= x + (i 1), for i =1, 2, 3, 4, i 3 3 − does not satisfy the OSC.

0 3 S1¡ S2£ C S3 A S4 ¡ ¡ ££ CCW AAU

01 2 3

Its corresponding self-similar measures

1 1 3 1 3 1 1 1 µ = µ S− + µ S− + µ S− + µ S− 8 ◦ 1 8 ◦ 2 8 ◦ 3 8 ◦ 4

Deﬁne, 1 1 1 T (x)= x, T (x)= x +1, T (x)= x +2 1 3 2 3 3 3 Then µ satisﬁes the following second-order self-similar identities (see [17]): for any Borel subset A [0, 3], ⊆

µ(T1jA) µ(T1A)     µ(T A) = Mj µ(T A) , j =1, 2, 3,  2j   2           µ(T3jA)   µ(T3A)      38

where the coeﬃcient matrices Mj are given by

1 0 0 0 1 0 3 0 1 1   1   1   M1 = 0 3 0 ,M2 = 3 0 3 ,M3 = 0 3 0 . 8   8   8                1 0 3   0 1 0   0 0 1       

Let J = j1j2 . . . jm, ji =1, 2 or 3. Then

µ(T1A)   µ(TJ A)= cJ µ(T A) ,  2       µ(T3A)    where

c =e M M =(c1 ,c2 ,c3 ). J j1 j2 ··· jm J J J

We can also evaluate the measure of each interval TJ . They are given by

3 3 3 1 3 1 dµ T1 dµ T2 dµ T3  0 ◦ 0 ◦ 0 ◦   5 5 5  R R R          3 3 3   27 9 3   0 xdµ T1 0 xdµ T2 0 xdµ T3   70 10 14   ◦ ◦ ◦  =   . (5.2.5)  R R R                 3 x2dµ T 3 x2dµ T 3 x2dµ T   5517 11943 63   0 1 0 2 0 3   6440 6440 184   ◦ ◦ ◦     R R R            We will apply this matrix to calculate the value of entries of M in (3.0.12), that is very important for ﬁnding out the measure of each interval on level m of the ﬁnite

element method. CHAPTER 6

CONVERGENCE OF THE APPROXIMATION SCHEME

In this chapter, we proved the convergence for the approximation scheme for the homogeneous IBVP (2.0.1). Some of our results are obtained by modifying similarly

ones in [24].

Let Vm be the set of end-points of all the m-level intervals, and rearrange its elements so that V = x : k =0, 1,...,N m with x

Sm := u Sm : u(a)= u(b)=0 D { ∈ } be the subspace of Sm consisting of functions satisfying the Dirichlet boundary conditions. Then

dim Sm =#V 2= N m 1. D m − −

We will choose the basis of Sm consisting of the tent functions deﬁned in (3.0.3):

Then ﬁx t [0,T ], any um(x,t) Sm deﬁned in(3.0.2) ∈ ∈

Deﬁnition 6.0.1. The linear map : Dom( ) Sm is deﬁned by Pm E → D N m 1 − := u(x )φ (x) for all u Dom( ). Pm i i ∈ E Xi=0 is called the Rayleigh-Ritz projection.

Lemma 6.0.1. Let v be a absolutely continuous function on [a,b]. Then v(x) v(y) | − |≤ 1/2 x y v 1 for all x,y [a,b]. | − | k kH0 ∈ 40

Proof. Since v is a absolutely continuous function on [a,b], and for all x,y [a,b], ∈ x y x v(x) v(y) = v(s) ds v(s) ds = v(s) ds | − | Z ∇ − Z ∇ Z ∇ 0 0 y x (6.0.1) 1/2 v(s) ds x y v 1 .(Seee.g[7].) H0 ≤ Zy |∇ | ≤| − | k k

Lemma 6.0.2. Let be the Rayleigh-Ritz projection,and u Dom( ). Then, Pm ∈ E u is its component in the subspace Sm, u u vanishes on the boundary and Pm D −Pm

(u u, u¯(m))=0 for all u¯(m) Sm.(See[24]). E −Pm ∈ D

Proof.

b n (u mu,φm,i(x)) = (u u(xi)φm,i(x)) φm,i(x) dx E −P Z ∇ − ∇ a Xi=1 xi = (u u(xi 1)φm,i 1(x) u(xi)φm,i(x)) dx − − Zxi−1 ∇ − − xi+1 (u u(xi)φm,i(x) u(xi+1)φm,i+1(x)) dx − Zxi ∇ − −

xi−1 = u u(xi 1)φm,i 1(x) u(xi)φm,i(x) xi − − − − | xi u u(xi)φm,i(x) u(xi+1)φm,i+1(x) xi+1 − − − |

= u(xi) u(xi 1)φm,i 1(xi) u(xi)φm,i(xi) − − − −

u(xi 1) u(xi 1)φm,i 1(xi 1) u(xi)φm,i(xi 1) − − − − − − − −

u(xi+1) u(xi)φm,i(xi+1) u(xi+1)φm,i+1(xi+1) − − −

+ u(xi) u(xi)φm,i(xi) u(xi+1)φm,i+1(xi) − −

=(u(xi) u(xi)) (u(xi 1) u(xi 1)) (u(xi+1) u(xi+1))+(u(xi) u(xi)) − − − − − − − − =0.

N m 1 m (m) Since φ − is a basis of S , (u u, u¯ )=0. { m,i}i=1 D E −Pm 41

Lemma 6.0.3. Let v be in Dom( ), and let v be the Rayleigh-Ritz projection of E Pm v to the subspace Sm of piecewise linear functions with any partition π = x Nm . D m { i}i=0 Then

v = v . |πm Pm |πm

Proof. Similar to that of [5, Lemma 5.3].

Theorem 6.0.4. Let v be a absolute continuous function on [a,b]. Let π = x Nm m { i}i=0 be any partition of [a,b]. Then, v x) v(x) C π 1/2 for all x [a,b]. ( −Pm ≤ k mk ∈

Proof. Since v(x) be a absolute continuous function on [a,b], and x [a,b], there ∈ m exists i 1,...,N such that x [xi 1,xi] and lemma 6.0.1 and 6.0.3 . Thus, ∈ { } ∈ −

v(x) mv(x) v(x) v(xi 1)+ v(xi 1) mv(x) | −P |≤| − − − −P |

v(x) v(xi 1) + v(xi 1) mv(x) ≤| − − | | − −P | x 1/2 1/2 xi 1/2 1/2 2 2 v dx x xi 1 + v dx xi xi 1 − − ≤ Zxi−1 |∇ | − Zxi−1 |∇ | − 1/2 2M xi xi 1 , where M = v Dom ( ) . ≤ − − k k E Let h = max xi xi 1 . Then, { − − } v(x) v(x) < 2Mh1/2 2M π 1/2 , for all x [a,b]. (6.0.2) | −Pm | ≤ k mk ∈

Let g,h Dom( ),f L2([0,T ];Dom( )), and u be deﬁned by in the IVBP ∈ E ∈ E (2.0.1). Then we have,

+ (u,v)=(f,v) for all v Dom( ) (6.0.3) tt E µ ∈ E Lemma 6.0.5. Fix m. Let u L2([0,T ], Dom( )) be a weak solution of IBVP ∈ E (2.0.1). Let um be deﬁned in (3.0.2). We can choose it to satisfy: 42

(i) (um,vm) + (um,vm)=(f,vm) for all vm Sm, (6.0.4) tt µ E µ ∈ D

m m m N 1 m N 1 (ii) u (x, 0) = i=0− g(xi)φi(x), and ut (x, 0) = i=0− h(xi)φi(x). P P

Deﬁne e(t) := Pu(t) um(t). Then, (e ,e ) + (e,e )=([Pu u] ,e ) . − tt t µ E t − tt t µ

Proof. Since e Sm, substituting e for v in (6.0.3) and e for vm in (6.0.4) respec- t ∈ D t t tively. Using the deﬁnition of pairing and subtracting these equations, we get

(u um,e ) + (u um,e )=0. tt − tt t µ E − t

Equivalently,

(u ( u) +( u) um,e ) + (u u + u um,e )=0, tt − Pm tt Pm tt − tt t µ E −Pm Pm − t which imply

(( u) um,e ) + ( u um,e ) = (( u) u ,e ) , because (u Pm tt − tt t µ E Pm − t Pm tt − tt t µ E − u,e ) = 0. By the deﬁnition of e(t), this becomes Pm t

(e ,e ) + (e,e )=([ u u] ,e ) . (6.0.5) tt t µ E t Pm − tt t µ

m 2 Proposition 6.0.6. Fix t. Then u converges in Lµ[a,b] to u.

1 1 1 2 1 2 Proof. Let E(t) := 2 (et,et)µ + 2 (e,e)= 2 et(t) µ + 2 e(t) Dom . Then E k k k k E

2 √ et µ 2 E(t), (6.0.6) k k ≤ p 2 √ e Dom 2 E(t), (6.0.7) k k E ≤ p 43

1 2 E(t) et µ + e Dom . (6.0.8) ≤ 2 k k k k E

1 2 1 2 Left hand side of (6.0.5) is equal to ( 2 et µ)t +( 2 e Dom )t = Et(t). k k k k E

For the right hand side of (6.0.5), it follows Cauchy Swartz inequality and (6.0.6), that’s

√ ([ mu u]tt,et)µ [ mu u]tt µ et µ [ mu u]tt µ 2 E(t). P − ≤ k P − k k k ≤ k P − k p So combine (6.0.7) and (6.0.8),

√ Et(t) [ mu u]tt µ 2 E(t), ≤ k P − k p Et(t) [ mu u]tt √2, E(t) ≤ k P − kµ (6.0.9) p s √ 2 E(s) 2 E(0) 2 [ mu u]tt µ dt, − ≤ Z0 k P − k p p s √ 2 E(s) 2+ 2 [ mu u]tt µ dt, because E(0) = 0. p ≤ Z0 k P − k p From (6.0.7) and (6.0.8), we have 2 s √ e(s) Dom 2 E(s) 2 [ mu u]tt µ dt, √2 k k E ≤ ≤ Z0 k P − k p s √ e(s) Dom 2 [ mu u]tt µ dt, (6.0.10) k k E ≤ Z0 k P − k 2 s √ e(s) µ 2 [ mu u]tt µ dt. √2 k k ≤ Z0 k P − k s T

e(s) µ Cs [ mu u]tt µ dt Cs [ mu u]tt µ dt k k ≤ Z0 k P − k ≤ Z0 k P − k T T 2 1 2 1 2 √ m 2 √ (6.0.11) Cs( [ mu u]tt µ dt) T Cs( 2 utt µ ρ dt) T ≤ Z0 k P − k ≤ Z0 k k 1 √ m 2 Cs 2T (ρ ) utt L2([0,T ],Dom ( ) . ≤ k k E Therefore, ﬁx t,

um u um u + u u k − kµ ≤ k −Pm kµ kPm − kµ √ m/2 m/2 Cs 2T (ρ ) utt L2([0,T ],Dom ( )) + 2(ρ ) u Dom ( ) 0 as m . ≤ k k E k k E → ← ∞ (6.0.12) 44 CHAPTER 7

NUMERICAL RESULTS 46

Timestep :0 Timestep :900 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :100 Timestep :1000 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :200 Timestep :1100 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :300 Timestep :1200 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :400 Timestep :1300 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :500 Timestep :1400 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :600 Timestep :1500 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :700 Timestep :1600 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :800 Timestep :1700 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 1: Dirichlet boundary condition for the weighted Lebesque measure associated with the weight p =2 √3 and 1 p = √3 1. − − − 47

Timestep :1800 Timestep :2700 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :1900 Timestep :2800 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2000 Timestep :2900 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2100 Timestep :3000 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2200 Timestep :3100 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2300 Timestep :3200 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2400 Timestep :3300 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2500 Timestep :3400 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :2600 Timestep :3500 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Continuation of Fig. 1 48

Timestep :0 Timestep :500 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :100 Timestep :600 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :200 Timestep :700 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :300 Timestep :800 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :400 Timestep :900 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Figure 2: Dirichlet boundary conditions for the inﬁnite Bernoulli convolution associated with the golden ratio. 49

Timestep :1000 Timestep :1500 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :1100 Timestep :1600 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :1200 Timestep :1700 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :1300 Timestep :1800 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Timestep :1400 Timestep :1900 1 1 0 0 −1 −1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

Continuation of Fig. 2 50

Timestep :0 Timestep :500 1 1 0 0 −1 −1 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Timestep :100 Timestep :600 1 1 0 0 −1 −1 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Timestep :200 Timestep :700 1 1 0 0 −1 −1 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Timestep :300 Timestep :800 1 1 0 0 −1 −1 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Timestep :400 Timestep :900 1 1 0 0 −1 −1 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3

Figure 3: Dirichlet boundary c ondition for the 3-fold convolution of the Cantor measure. 51

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FRACTAL MEASURES

A.1 BERNOULLI CONVOLUTION ASSOCIATED WITH THE GOLDEN RATIO

1 1 S (x)= ρx = S− (x)= ρ− x 1 ⇒ 1 2 1 1 S (x)= ρx + (1 ρ)= ρx + ρ = S− (x)= ρ− x ρ 2 − ⇒ 2 − 2 1 2 T (x)= ρ x = T − (x)= ρ− x 1 ⇒ 1 3 2 1 3 T (x)= ρ x + ρ = T − (x)= ρ− x (1 + ρ) 2 ⇒ 2 − 2 1 2 T (x)= ρ x + ρ = T − (x)= ρ− x (1 + ρ) 3 ⇒ 3 − 1 1 Supp(µ)=[0, 1]&µ = pµ S− + (1 p)µ S− ◦ 1 − ◦ 2

f(x)dµ = p f(S1x)dµ + (1 p) f(S2x)dµ −1 −1 ZA ZS1 [A] − ZS2 [A] and

1 f(x)dµ Ti = f(Ti− x)dµ ZA ◦ ZTi[A]

ρ2 =1 ρ, ρ3 =2ρ 1, ρ4 =2 3ρ, ρ5 =5ρ 3, ρ6 =5 8ρ. − − − − − 55

Let p =1/2. Then, our goal is to claim that (5.1.4) is true.

Proof. Consider the ﬁrst row:

ρ2 dµ =1/2 dµ +1/2 dµ −1 2 −1 2 Z0 ZS1 [0,ρ ] ZS2 [0,ρ ] ρ 0 =1/2 dµ +1/2 dµ Z0 Z ρ ρ − =1/2 dµ Z0

Therefore,

ρ2 ρ2 ρ ρ2 ρ dµ =1/2 dµ +1/2 dµ dµ = dµ. (A.1.1) Z0 Z0 Zρ2 ⇒ Z0 Z0 Moreover,

ρ dµ = dµ +1/2 dµ 2 −1 2 −1 2 Zρ ZS1 [ρ ,ρ] ZS2 [ρ ,ρ] 1 ρ2 =1/2 dµ +1/2 dµ Zρ Z0 1 ρ =1/2 dµ +1/2 dµ by(A.1.1) Zρ Zρ2

Then, by (A.1.1)

ρ 1 ρ2 1 dµ = dµ = dµ and dµ =1 Zρ2 Zρ Z0 Z0

ρ2 ρ 1 1 dµ = dµ = dµ =1/3 dµ =1/3 Z0 Zρ2 Zρ Z0 Therefore, ρ2 ρ2 dµ T1 = dµ = dµ =1/3 Z0 ◦ ZT1[0,1] Z0 56

ρ2 ρ dµ T2 = dµ = dµ =1/3 2 Z0 ◦ ZT2[0,1] Zρ ρ2 1 dµ T3 = dµ = dµ =1/3 Z0 ◦ ZT3[0,1] Zρ

Now consider the 2nd row

1 xdµ = S1(x)dµ +1/2 S2(x)dµ −1 −1 Z0 ZS1 [0,1] ZS2 [0,1] 1+ρ 1 =1/2 ρxdµ +1/2 (ρx + ρ2)dµ Z0 Z ρ − 1 1 1 =1/2 ρxdµ +1/2 ρxdµ +1/2 ρ2dµ. Z0 Z0 Z0 Therefore,

1 1 1 1 (1 ρ) xdµ =1/2ρ2 dµ xdµ =1/2 dµ =1/2. − Z0 Z0 ⇒ Z0 Z0

ρ2 xdµ =1/2 S1(x)dµ +1/2 S2(x)dµ −1 2 −1 2 Z0 ZS1 [0,ρ ] ZS2 [0,ρ ] ρ 0 =1/2 ρxdµ +1/2 (ρx + ρ2)dµ Z0 Z ρ − ρ2 ρ =1/2ρ xdµ +1/2ρ xdµ. Z0 Zρ2

ρ2 1 ρ ρ 2 ρ ρ xdµ = 1 xdµ = xdµ. (A.1.2) Z 1 ρ Z 2 2 ρ Z 2 0 − 2 ρ − ρ ρ xdµ =1/2 S1(x)dµ +1/2 S2(x)dµ 2 −1 2 −1 2 Zρ ZS1 [ρ ,ρ] ZS2 [ρ ,ρ] 1 ρ2 =1/2 ρxdµ +1/2 (ρx + ρ2)dµ Zρ Z0 2 2 1 1 1 ρ 1 ρ = ρ xdµ + ρ xdµ + ρ2 dµ. 2 Zρ 2 Z0 2 Z0 57

Thus,

2 2+ ρ ρ 1 1 1 ρ 1 1 1 1 1 1 xdµ = ρ xdµ + ρ2 xdµ = ρ + ρ2 = ρ + ρ2. 2 Zρ2 2 Z0 2 Z0 2 2 2 3 4 6

ρ 2 1 1 1 1 1 xdµ = ( ρ + ρ2)= ( ρ + ρ2) Z 2 2+ ρ 4 6 2+ ρ 2 3 ρ (A.1.3) 1 3ρ +2ρ2 1 2+ ρ2 1 = ( )= = . 2+ ρ 6 2+ ρ 6 6

Therefore, (A.1.2) becomes 2 ρ ρ xdµ = . (A.1.4) Z 6(2 ρ) 0 − 1 xdµ =1/2 S1(x)dµ +1/2 S2(x)dµ −1 −1 Zρ ZS1 [ρ,1] ZS2 [ρ,1] 1+ρ 1 =1/2 ρxdµ +1/2 (ρx + ρ2)dµ Z1 Zρ2 1 1 1 1 = ρ xdµ + ρ2 xdµ. 2 Zρ2 2 Zρ2 1 1 1 ρ ρ2 (1 ρ) xdµ = ρxdµ + . ⇒ − 2 Zρ 2 Zρ2 3 1 2 ρ ρ2 2 ρ +4ρ2 1+3ρ2 4 3ρ xdµ = ( + )= ( )= = − ⇒ Z 2 ρ 12 3 2 ρ 12 6(2 ρ) 6(2 ρ) ρ − − − −

Therefore,

1 ρ2 1 2 2 ρ xdµ T1 = T − (x)dµ = ρ− dµ = ρ− Z ◦ Z 1 Z 6(2 ρ) 0 T1[0,1] 0 − 1 1 = = 6ρ(2 ρ) 6(3ρ 1) − −

1 ρ ρ ρ 1 3 3 xdµ T2 = T2− (x)dµ = [ρ− x (1+ ρ)]dµ = ρ− xdµ (1+ ρ)dµ 2 2 2 Z0 ◦ ZT2[0,1] Zρ − Zρ − Zρ 58

1 3 1 1 2+ ρ 1+ ρ 1 xdµ T2 = ρ− (1 + ρ) = = ⇒ Z0 ◦ 6 − 3 6ρ − 3 6

1 1 1 2 xdµ T3 = T3− (x)dµ = [ρ− x (1 + ρ)]dµ Z0 ◦ ZT3[0 1] Zρ − 1 1 2 = ρ− xdµ (1 + ρ)dµ Zρ − Zρ

1 2 4 3ρ 1 5+ ρ 1+ ρ xdµ T3 =ρ− − (1 + ρ) = ⇒ Z ◦ 6(2 ρ) − 3 6(2 ρ) − 3 0 − − 3ρ2 ρ 1 = = = 6(2 ρ) 2(1+2ρ) 2(3 + ρ) −

Finally, we consider the third row

1 2 1 2 1 2 x dµ = (S1(x)) dµ + (S2(x)) dµ −1 −1 Z0 2 ZS1 [0,1] 2 ZS2 [0,1] 1 1 1 1 = ρ2x2dµ + (ρx + ρ2)2dµ 2 Z0 2 Z0 1 1 1 1 1 1 1 = ρ2x2dµ + ρ2x2dµ + ρ3 xdµ + ρ4 dµ 2 Z0 2 Z0 Z0 2 Z0 1 1 1 1 ρ (1 ρ2) x2dµ = (2ρ 1) + (2 3ρ)1 = − ⇒ − Z0 2 − 2 − 2 1 1 ρ ρ x2dµ = − = ⇒ Z 2(1 ρ2) 2 0 − Moreover,

ρ2 2 1 2 1 2 x dµ = (S1(x)) dµ + (S2(x)) dµ −1 2 −1 2 Z0 2 ZS1 [0,ρ ] 2 ZS2 [0,ρ ] 2 1 ρ 1 0 1 ρ 1 ρ = ρ2x2dµ + (ρx + ρ2)2dµ = ρ2x2dµ + ρ2x2dµ 2 Z0 2 Z ρ 2 Z0 2 Zρ2 − ρ2 ρ (2 ρ2) x2dµ = ρ2 x2dµ ⇒ − Z0 Zρ2 59

ρ2 2 ρ ρ 2 ρ 2 2 x dµ = 2 x dµ = (2ρ 1) x dµ ⇒ Z 2 ρ Z 2 − Z 2 0 − ρ ρ Moreover,

ρ 2 1 2 1 2 x dµ = (S1(x)) dµ + (S2(x)) dµ 2 −1 2 −1 2 Zρ 2 ZS1 [ρ ,ρ] 2 ZS2 [ρ ,ρ] 2 1 1 1 ρ = ρ2x2dµ + (ρx + ρ2)2dµ 2 Zρ 2 Z0 2 2 2 1 1 1 ρ ρ 1 ρ = ρ2x2dµ + ρx2dµ + ρ3 xdµ + ρ4 dµ 2 Zρ 2 Z0 Z0 2 Z0

ρ 1 1 1 (2 + ρ2) x2dµ = ρ2 x2dµ +2ρ3 xdµ + ρ4 dµ ⇒ Zρ2 Z0 Z0 Z0 ρ ρ ρ 2 3ρ (2 + ρ2) x2dµ = ρ2 + 2(2ρ 1) + − ⇒ Z 2 2 − 6(2 ρ) 3 ρ − ρ 2ρ 1 4ρ2 2ρ 2 3ρ (2 + ρ2) x2dµ = − + − + − ⇒ Z 2 2 6(2 ρ) 3 ρ − ρ 3(2ρ 1)(2 ρ)+4ρ2 2ρ + 2(2 3ρ)(2 ρ) (2 + ρ2) x2dµ = − − − − − ⇒ Z 2 6(2 ρ) ρ − 2 3ρ +4ρ2 = − 6(2 ρ) ρ − 2 6 7ρ 6 7ρ 6 7ρ x = −2 = − = − ⇒ Z 2 6(2 + ρ )(2 ρ) 6(3 ρ)(2 ρ) 6(7 6ρ) ρ − − − − Therefore,

ρ2 ρ 2 2 1 ρ 2 1 ρ 6 7ρ 6 7ρ 6ρ +7rho 13 20ρ x = − x = − − = − − 2 = − Z 1+ ρ Z 2 1+ ρ 6(3 ρ)(2 ρ) 6(7 6ρ +7ρ 6ρ ) 6(1+7ρ) 0 ρ − − − −

And,

1 2 1 2 1 2 x dµ = (S1(x)) dµ + (S2(x)) dµ −1 −1 Zρ 2 ZS1 [ρ,1] 2 ZS2 [ρ,1] 1 1+ρ 1 1 = ρ2x2dµ + (ρx + ρ2)2dµ 2 Z1 2 Zρ2 1 1 1 1 ρ 1 1 1 x2dµ = ρ2x2dµ + ρ2x2dµ + ρ3 xdµ + ρ4 dµ ⇒ Zρ 2 Zρ 2 Zρ2 Zρ2 2 Zρ2 60

2 1 1 1 1 ρ 1 ρ 1 x2dµ = ρ2x2dµ + ρ2x2dµ + ρ3( xdµ xdµ)+ ρ4 ⇒ Zρ 2 Zρ 2 Zρ2 Z0 − Z0 3 1 1 1 6 7ρ 1 ρ 1 (1 ρ2) x2dµ = ρ2 − + ρ3( )+ ρ4 ⇒ − 2 Z 2 6(7 6ρ) 2 − 6(2 ρ) 3 ρ − − 1 1 1 6 7ρ 1 2 3ρ 1 (1 ρ2) x2dµ = ρ2 − + (2ρ 1) − )+ (2 3ρ) ⇒ − 2 Z 2 6(7 6ρ) 2 − − 6(2 ρ) 3 − ρ − − 1 1 1 6 7ρ 1 2 3ρ (1 ρ2) x2dµ = ρ2 − + − ) ⇒ − 2 Z 2 6(7 6ρ) 6 − 6(2 ρ) ρ − − 1 1 ρ2(6 7ρ)(2 ρ)+2(7 6ρ)(2 ρ) 2(2 3ρ)(7 6ρ) (1 ρ2) x2dµ = − − − − − − − ⇒ − 2 Z 12(7 6ρ)(2 ρ) ρ − − 1 1 (19 46ρ + 27ρ2)+( 14ρ + 12ρ2 + 42ρ 36ρ2) (1 ρ2) x2dµ = − − − ⇒ − 2 Z 12(14 19ρ +6ρ2) ρ − 1 1 19 18ρ 3ρ2 22 21ρ (1 ρ2) x2dµ = − − = − ⇒ − 2 Z 12(20 25ρ) 12(20 25ρ) ρ − − 1 22 21ρ 22 21ρ x2dµ = − = − ⇒ Z 6(20 25ρ)(2 ρ2) 6(20 25ρ)(1 + ρ) ρ − − − 1 22 21ρ 22 21ρ x2dµ = − = − ⇒ Z 6(20 5ρ 25ρ2) 6(20ρ 5) ρ − − − Therefore,

1 ρ2 2 1 2 4 2 1 13 20ρ x dµ T1 = (T − (x)) dµ = ρ− x dµ = − Z ◦ Z 1 Z 2 3ρ 6(1+7ρ) 0 T1[0,1] 0 −

1 2 13 20ρ 13 20ρ 6 7ρ x dµ T1 = − = − = − ⇒ Z ◦ 6(2 + 11ρ 21ρ2) 6(32ρ 19) 6(13ρ 6) 0 − − − 1 2 1+6ρ 5ρ +4 x dµ T1 = − = ⇒ Z ◦ 6(7 6ρ) 6(8 + ρ) 0 −

1 ρ 2 1 2 3 2 x dµ T2 = (T2− (x)) dµ = (ρ− x (1 + ρ) dµ 2 Z0 ◦ ZT2[0,1] Zρ − ρ 6 2 3 2 = (ρ− x 2ρ (1 + ρ)x +(1+ ρ) dµ Zρ2 − 61

1 2 11 6 7ρ 2(1 + ρ) 1 2 1 x dµ T2 = − +(1+ ρ) ⇒ Z ◦ 5 8ρ 6(7 6ρ) − 2ρ 1 6 3 0 − − − 1 2 6 7ρ (1 + ρ) 2+ ρ x dµ T2 = − + ⇒ Z ◦ 6(35 86ρ + 48ρ2) − 3(2ρ 1) 3 0 − − 1 2 6 7ρ (1 + ρ) (2 ρ)(2ρ 1) x dµ T2 = − − − − ⇒ Z ◦ 6(83 134ρ) − 3(2ρ 1) 0 − − 1+6ρ 1 = − 6( 51+83ρ) − 3(2ρ 1) − − 1 2 5 ρ 1 (5 ρ)(2ρ 1) 2(32 51ρ) x dµ T2 = − = − − − − ⇒ Z ◦ 6(32 51ρ) − 3(2ρ 1) 6(32 51ρ)(2ρ 1) 0 − − − − 1 2 2 69 + 113ρ 2ρ 71 + 115ρ x dµ T2 = − − = − ⇒ Z ◦ 6( 32 + 115ρ 102ρ2) 6( 134 + 217ρ) 0 − − − 1 2 44 71ρ 27+44ρ 17 27ρ x dµ T2 = − = − = − ⇒ Z ◦ 6(83 134ρ) 6( 51+83ρ) 6(32 51ρ) 0 − − − 1 2 10+17ρ 7 10ρ 3+7ρ x dµ T2 = − = − = − ⇒ Z ◦ 6( 19+32ρ) 6(13 19ρ) 6( 6+13ρ) 0 − − − 1 2 4 3ρ 1+4ρ 5+ ρ x dµ T2 = − = = ⇒ Z ◦ 6(7 6ρ) 6(1+7ρ) 6(8 + ρ) 0 −

1 1 2 1 2 2 2 x dµ T3 = (T3− (x)) dµ = (ρ− x (1 + ρ) dµ Z0 ◦ ZT3[0,1] Zρ − 1 1 2 4 2 2 2 x dµ T3 = (ρ− x 2ρ− (1 + ρ)x +(1+ ρ) dµ ⇒ Z0 ◦ Zρ − 1 2 1 22 21ρ 2(1 + ρ) 4 3ρ 2 1 x dµ T3 = − − +(1+ ρ) ⇒ Z ◦ 2 3ρ 6(20ρ 5) − 1 ρ 6(2 ρ) 3 0 − − − − 1 2 22 21ρ (1 + ρ)(4 3ρ) 2+ ρ x dµ T3 = − − + ⇒ Z ◦ 6( 10+55ρ 60ρ2) − 3(1 ρ)(2 ρ) 3 0 − − − − 1 2 22 21ρ (1+4ρ) 2+ ρ x dµ T3 = − + ⇒ Z ◦ 6( 70 + 115ρ) − 3((2 3ρ + ρ2) 3 0 − − 1 2 22 21ρ (1+4ρ)+(2+ ρ)(3 4ρ) x dµ T3 = − − ⇒ Z ◦ 6( 70 + 115ρ) − 3(3 4ρ) 0 − − 1 2 2 1+22ρ 5+9ρ +4ρ x dµ T3 = − ⇒ Z ◦ 6(45 70ρ) − 3(3 4ρ) 0 − − 62

1 2 1+22ρ 1+5ρ x dµ T3 = − ⇒ Z ◦ 6(45 70ρ) − 3(3 4ρ) 0 − − 1 2 (1+22ρ)(3 4ρ) 2( 1+5ρ)(45 70ρ) x dµ T3 = − − − − ⇒ Z ◦ 6(45 70ρ)(3 4ρ) 0 − − 1 2 2 2 (3 62ρ 88ρ ) 2( 45 + 295ρ 350ρ ) x dµ T3 = − − − − − ⇒ Z ◦ 6(135 390ρ + 280ρ2) 0 − 1 2 2 5 440ρ 700ρ x dµ T3 = − − ⇒ Z ◦ 6(415 670ρ) 0 − 1 2 705 1140ρ 435 + 705ρ x dµ T3 = − = − ⇒ Z ◦ 6(415 670ρ) 6( 255 + 415ρ) 0 − − 1 2 270 435ρ 54 87ρ 33+54ρ x dµ T3 = − = − = − ⇒ Z ◦ 6(160 255ρ) 6(32 51ρ) 6( 19+32ρ) 0 − − − 1 2 21 33ρ 12+21ρ 9 12ρ x dµ T3 = − = − = − ⇒ Z ◦ 6(13 19ρ) 6( 6+13ρ) 6(7 6ρ) 0 − − − 1 2 3+9ρ 6 3ρ 2 ρ x dµ T3 = − = − = − ⇒ Z0 ◦ 6(1+7ρ) 6(8 + ρ) 2(8 + ρ)

This is to complete of proof of the claim of (5.1.4).

A.2 THREE-FOLD CONVOLUTION OF THE CANTOR MEASURE

1 1 2 1 4 1 S (x)= x, S (x)= x + , S (x)= x + , S (x)= x +2 1 3 2 3 3 3 3 3 4 3

1 1 1 1 S− (x)=3x, S− (x)=3x 2, S− (x)=3x 4, S− (x)=3x 6 1 2 − 3 − 4 − 1 1 1 T (x)= x, T (x)= x +1, T (x)= x +2 1 3 2 3 3 3

1 1 1 T − (x)=3x, T − (x)=3x 3, T − (x)=3x 6 1 2 − 3 − 1 3 f(x)dµ = f(S1x)dµ + f(S2x)dµ −1 −1 ZA 8 ZS1 [A] 8 ZS2 [A] 3 1 + f(S3x)dµ + f(S4x)dµ −1 −1 8 ZS3 [A] 8 ZS4 [A] 63

1 f(x)dµ Ti = f(Ti− x)dµ ZA ◦ ZTi[A]

But, we have

1 f(Tx)dµ = f(y)dµ T − (y) ′ ZΩ ZΩ ◦

Our goal is to claim that (5.2.5) is true.

Proof.

3 3 3 1 3 1 dµ T1 dµ T2 dµ T3  0 ◦ 0 ◦ 0 ◦   5 5 5  R R R          3 3 3   27 9 3   0 xdµ T1 0 xdµ T2 0 xdµ T3   70 10 14   ◦ ◦ ◦  =    R R R                 3 x2dµ T 3 x2dµ T 3 x2dµ T   5517 11943 63   0 1 0 2 0 3   6440 6440 184   ◦ ◦ ◦     R R R           

Proof : Consider the ﬁrst row:

1 1 3 3 1 dµ = dµ + dµ + dµ + dµ −1 −1 −1 −1 Z0 8 ZS1 [0 1] 8 ZS2 [0 1] 8 ZS3 [0 1] 8 ZS4 [0 1]

1 3 1 1 3 1 3 3 − 1 − dµ = dµ + dµ + dµ + dµ ⇒ Z0 8 Z0 8 Z 2 8 Z 4 8 Z 6 − − − 5 1 1 3 dµ = dµ ⇒ 8 Z0 8 Z0 1 1 3 1 dµ = dµ = ⇒ Z0 5 Z0 5

2 1 3 3 1 dµ = dµ + dµ + dµ + dµ −1 −1 −1 −1 Z1 8 ZS1 [1 2] 8 ZS2 [1 2] 8 ZS3 [1 2] 8 ZS4 [1 2] 64

2 1 6 3 4 3 2 1 0 dµ = dµ + dµ + dµ + dµ ⇒ Z1 8 Z3 8 Z1 8 Z 1 8 Z 3 − − 2 3 1 3 1 2 dµ = (1 )+ ( + dµ) ⇒ Z1 8 − 5 8 5 Z1 3 2 3 3 15 (1 ) dµ = + = ⇒ − 8 Z1 10 40 40 2 15 8 3 dµ =( )( )= ⇒ Z1 40 5 5

3 1 3 3 1 dµ = dµ + dµ + dµ + dµ −1 −1 −1 −1 Z2 8 ZS1 [2,3] 8 ZS2 [2,3] 8 ZS3 [2,3] 8 ZS4 [2,3] 3 1 9 3 7 3 5 1 3 dµ = dµ + dµ + dµ + dµ ⇒ Z2 8 Z6 8 Z4 8 Z2 8 Z0 3 3 3 1 dµ = dµ + ⇒ Z2 8 Z2 8 3 1 dµ = ⇒ Z2 5

Therefore, 3 1 1 dµ T1 = dµ = dµ = Z0 ◦ ZT1[0,3] Z0 5

3 2 3 dµ T2 = dµ = dµ = Z0 ◦ ZT2[0,3] Z1 5

3 3 1 dµ T3 = dµ = dµ = Z0 ◦ ZT3[0,3] Z2 5

Now consider the 2nd row: 3 1 3 xdµ = S1(x)dµ + S2(x)dµ −1 −1 Z0 8 ZS1 [0,3] 8 ZS2 [0,3] 3 1 + S3(x)dµ + S4(x)dµ −1 −1 8 ZS3 [0,3] 8 ZS4 [0,3] 65

3 1 9 1 3 7 1 2 xdµ = xdµ + ( x + )dµ ⇒ Z0 8 Z0 3 8 Z 2 3 3 − 3 5 1 4 1 3 1 + ( x + )dµ + ( x + 2)dµ 8 Z 4 3 3 8 Z 6 3 − − 3 1 3 1 3 1 1 3 1 1 3 1 xdµ = xdµ + xdµ + xdµ + + xdµ + ⇒ Z0 24 Z0 8 Z0 4 8 Z0 2 24 Z0 4 3 3 xdµ = ⇒ Z0 2 1 1 3 1 3 1 1 2 xdµ = xdµ + ( x + )dµ +0 Z0 8 Z0 3 8 Z 2 3 3 − 7 1 1 3 1 1 xdµ = xdµ + dµ ⇒ 8 Z0 24 Z0 4 Z0 1 1 3 4 1 1 3 4 1 9 xdµ = xdµ +( )( )=( )( )+( )( )= ⇒ Z0 21 Z0 14 5 21 2 14 5 70

Moreover,

2 3 4 1 1 3 2 1 4 xdµ =0+ ( x + )dµ + ( x + )dµ +0 Z1 8 Z1 3 3 8 Z 1 3 3 − 2 1 3 1 3 1 2 1 2 xdµ = xdµ + dµ + xdµ + dµ Z1 8 Z1 4 Z1 8 Z0 2 Z0 7 2 1 3 1 1 1 1 2 xdµ = xdµ + + dµ + dµ ⇒ 8 Z1 8 Z0 4 4 Z0 2 Z1 7 2 1 3 1 1 1 1 3 63 xdµ =( )( )+( )+( )( )+( )( )= ⇒ 8 Z1 8 2 4 4 5 2 5 80 2 9 xdµ = ⇒ Z1 10

Moreover,

3 3 5 1 4 1 3 1 xdµ =0+ ( x + )dµ + ( x + 2)dµ +0 Z2 8 Z2 3 3 8 Z0 3

3 1 3 1 3 1 3 1 3 xdµ = xdµ + dµ + xdµ + dµ ⇒ Z2 8 Z2 2 Z2 24 Z0 4 Z0 66

7 3 1 1 1 3 1 33 xdµ = + + 1= ⇒ 8 Z2 2 5 24 2 4 80 3 33 xdµ = ⇒ Z2 70

Therefore,

3 1 1 27 xdµ T1 = T1− (x)dµ = 3xdµ == Z0 ◦ ZT1[0,3] Z0 70

3 2 1 9 3 9 xdµ T2 = T2− (x)dµ = (3x 3)dµ =3( )= Z0 ◦ ZT2[0,3] Z1 − 10 − 5 10 3 3 1 33 1 3 xdµ T3 = T3− (x)dµ = (3x 6)dµ =3 6 = Z0 ◦ ZT3[0,3] Z2 − 70 − 5 14

Consider the third row :

3 2 1 2 3 2 x dµ = (S1(x)) dµ + (S2(x)) dµ −1 −1 Z0 8 ZS1 [0,3] 8 ZS2 [0,3]

3 2 1 2 + (S3(x)) dµ + (S4(x)) dµ −1 −1 8 ZS3 [0,3] 8 ZS4 [0,3]

3 1 9 1 2 3 7 1 2 x2dµ = x dµ + ( x + )2dµ ⇒ Z0 8 Z0 3 8 Z 2 3 3 − 3 5 1 4 1 3 1 + ( x + )2dµ + ( x + 2)2dµ 8 Z 4 3 3 8 Z 6 3 − − 3 1 3 3 3 12 3 12 3 3 3 x2dµ = x2dµ + x2dµ + xdµ + dµ + x2dµ Z0 72 Z0 72 Z0 72 Z0 72 Z0 72 Z0 ⇒ 24 3 48 3 1 3 12 3 36 3 + xdµ + dµ + x2dµ + xdµ + dµ 72 Z0 72 Z0 72 Z0 72 Z0 72 Z0 3 1 3 48 3 12 48 36 x2dµ = x2dµ + xdµ + + + ⇒ Z0 9 Z0 72 Z0 72 72 72 8 3 48 3 4 7 x2dµ =( )( )+ = ⇒ 9 Z0 72 2 3 3 3 21 x2dµ = ⇒ Z0 8 67

1 1 3 1 3 1 1 2 x2dµ = ( x)2dµ + ( x + )2dµ Z0 8 Z0 3 8 Z 2 3 3 − 1 1 3 3 1 12 1 12 1 x2dµ = x2dµ + x2dµ + xdµ + dµ ⇒ Z0 72 Z0 72 Z0 72 Z0 72 Z0 69 1 1 21 12 9 12 1 613 x2dµ =( )( )+( )( )+( )( )= ⇒ 72 Z0 72 8 72 10 72 5 6720 1 613 x2dµ = ⇒ Z0 6440

2 3 4 1 2 3 1 1 4 x2dµ =0+ ( x + )2dµ + ( x + )2dµ +0 Z1 8 Z1 3 3 8 Z 2 3 3 − 2 3 3 12 3 x2dµ = x2dµ + xdµ ⇒ Z1 72 Z1 72 Z1 12 3 3 2 24 2 48 2 + dµ + x2 + xdµ + dµ 72 Z1 72 Z0 72 Z0 72 Z0 2 3 3 12 3 12 4 3 2 24 2 48 4 x2dµ = x2dµ + xdµ + + x2 + xdµ + ⇒ Z1 72 Z1 72 Z1 72 5 72 Z0 72 Z0 72 5 69 2 3 21 12 3 9 24 9 9 12 4 48 4 1811 x2dµ =( )( )+ ( )+ ( + )+( )( )+( )( )= ⇒ 72 Z1 72 8 72 2 − 70 72 70 10 72 5 72 5 1344 2 1811 x2dµ = ⇒ Z1 1288

3 3 5 1 4 1 3 1 x2dµ =0+ ( x + )2dµ + ( x + 2)2dµ Z2 8 Z2 3 3 8 Z0 3 Thus, we have

3 3 3 24 3 x2dµ = x2dµ + xdµ Z2 72 Z2 72 Z2 48 3 1 3 12 3 36 3 + dµ + x2dµ + xdµ + dµ 72 Z2 72 Z0 72 Z0 72 Z0

69 3 24 33 48 1 1 21 12 3 36 7237 x2dµ =( )( )+( )( )+( )( )+( )( )+( )1 = ⇒ 72 Z2 72 70 72 5 72 8 72 2 72 6720 68

3 7237 x2dµ = ⇒ Z2 6440

Therefore,

3 1 1 2 1 2 2 2 613 5517 x dµ T1 = (T1− (x)) dµ = (3x) dµ =9 x dµ = 9( )= Z0 ◦ ZT1[0,3] Z0 Z0 6440 6440 3 2 2 1 2 2 x dµ T2 = (T2− (x)) dµ = (3x 3) dµ Z0 ◦ ZT2[0,3] Z1 − 3 2 2 2 2 2 2 2 x dµ T2 =9 (x 2x + 1)dµ =9 x dµ 18 xdµ +9 dµ ⇒ Z0 ◦ Z1 − Z1 − Z1 Z2 3 2 1811 9 3 11943 x dµ T2 = 9( ) 18( )+9( )= ⇒ Z0 ◦ 1288 − 10 5 6440

3 3 2 1 2 2 x dµ T3 = (T3− (x)) dµ = (3x 6) dµ Z0 ◦ ZT3[0,3] Z2 − 3 3 3 3 3 2 2 2 x dµ T3 =9 (x 4x + 4)dµ =9 x dµ 36 xdµ + 36 dµ ⇒ Z0 ◦ Z2 − Z2 − Z2 Z2 3 2 7237 33 1 63 x dµ T3 = 9( ) 36( )+36( )= ⇒ Z0 ◦ 6440 − 70 5 184 This is to complete of proof of the claim of (5.2.5).

A.3 INVERTIBILITY OF M

Theorem A.3.1. (see [2]) If f is nonnegative, then

1 f(T ω)µ(dω)= f(ω′)µT − (dω′). (A.3.5) ZΩ ZΩ′ 1 A function f (not necessarily nonnegative) is integrable with respect to µT − if and only if fT is integrable with respect to µ, in which case (A.3.5) and

1 f(T ω)µ(dω)= f(ω′)µT − (dω′) (A.3.6) ZT −1A′ ZA′ 69

holds. For nonnegative f, (A.3.6) always holds.

Lemma A.3.2. From the weighted Lebesgue measure, the inﬁnite Bernoulli convolution associated with the golden ratio, we have show that (5.1.4) is true. For

i = 1, 2, 3, 1(2x2 x) dµ T > 0 and 1(2x2 3x + 1) dµ T > 0. Thus, 0 − ◦ i 0 − ◦ i R R 1(2x2 x) dµ T > 0 and 1(2x2 3x + 1) dµ T > 0. Then, M is invertible in 0 − ◦ J 0 − ◦ J R R this case.

Proof.

1 2 2 2(5ρ + 4) 1 1 13ρ 16+30ρ (2x x) dµ T1 = = − > 0 Z − ◦ 6(ρ + 8) − 6(3ρ 1) 6 (3ρ 1)(ρ + 8) 0 − − and

1 2 2 2(5ρ + 4) 3 1 1 12ρ + 19ρ 16 (2x 3x + 1) dµ T1 = + = − > 0. Z − ◦ 6(ρ + 8) − 6(3ρ 1) 3 2 (3ρ 1)(ρ + 8) 0 − −

1 2 2(ρ + 5) 1 1 ρ +2 (2x x) dµ T2 = = > 0 Z0 − ◦ 6(ρ + 8) − 6 6 ρ +8 and 1 2 2(ρ + 5) 1 1 1 ρ +2 (2x 3x + 1) dµ T2 = 3( )+ = > 0. Z0 − ◦ 6(ρ + 8) − 6 3 6 ρ +8

1 2 3 2 2(2 ρ) 1 1 7ρ 4 12ρ +6ρ (2x x) dµ T3 = − 2 = −2 − > 0 Z0 − ◦ 6(ρ + 8) − 6(3ρ + 3) −18 (ρ + 1)(ρ + 8) and

1 2 2 2(2 ρ) 3 1 1 20ρ ρ + 12 (2x 3x + 1) dµ T3 = − 2 + = 2 − > 0. Z0 − ◦ 6(ρ + 8) − 6(3ρ + 3) 3 6 (ρ + 1)(ρ + 8) 70

Lemma A.3.3. From the 3-fold convolutions of the Cantor measure, we have show that (5.2.5) is true. For i = 1, 2, 3, 3( 2 x2 1 x) dµ T > 0 and 3( 2 x2 x + 0 9 − 3 ◦ i 0 9 − R R 1) dµ T > 0. Thus, 3( 2 x2 1 x) dµ T > 0 and 3( 2 x2 x + 1) dµ T > 0. ◦ i 0 9 − 3 ◦ J 0 9 − ◦ J R R Then, M is invertible in this case.

Proof. 3 2 2 1 2 5517 1 27 199 ( x x) dµ T1 = = > 0, Z0 9 − 3 ◦ 9 6440 − 3 70 3220 and 3 2 2 2 5517 27 1 3 ( x x + 1) dµ T1 = + = > 0. Z0 9 − ◦ 9 6440 − 70 5 644

3 2 2 1 2 11943 1 9 361 ( x x) dµ T2 = = > 0, Z0 9 − 3 ◦ 9 6440 − 3 10 3220 and 3 2 2 2 11943 9 3 361 ( x x + 1) dµ T2 = + = > 0. Z0 9 − ◦ 9 6440 − 10 5 3220

3 2 2 1 2 63 1 3 3 ( x x) dµ T3 = = > 0, Z0 9 − 3 ◦ 9 184 − 3 14 644 and 3 2 2 2 63 3 1 199 ( x x + 1) dµ T3 = + = > 0. Z0 9 − ◦ 9 184 − 14 5 3220 APPENDIX B

NORMED SPACES INVOLVING TIME

B.1 COMPLETENESS OF L2([0,T ]; X)

Theorem B.1.1. Let X be a Banach space and for 1 p . Then Lp([0,T ]; X) ≤ ≤ ∞ is a Banach space.

Proof. For 1 p< , ≤ ∞

(1) For all f Lp([0,T ]; X), we have := T f p dt 0. = 0, if and only if ∈ If 0 k kX ≥ If R f = 0 for Lebesgue a.e. t [0,T ]. ∈

(2) For all f Lp([0,T ]; X) and α R, we have T αf p dt = T α p f p dt = ∈ ∈ 0 k kX 0 | | k kX R R α p . | | If

(3) For all f,g Lp([0,T ]; X), we have T f + g p dt T ( f + g )p dt ∈ 0 k kX ≤ 0 k kX k kX ≤ R R + . If Ig

These three conditions hold for p = with its norm. Thus, for 1 p , ∞ ≤ ≤ ∞ Lp([0,T ]; X) are normed spaces.

We modify the proof in [18]. Let u Lp([0,T ]; X) be a Cauchy sequence. It { n}⊂ suﬃces to show that u has a convergent subsequence. Let n N such that { n} 1 ∈

u u p 1/2 for all n n . k n − n1 kL ([0,T ];X) ≤ ≥ 1

Let n N such that n n and 2 ∈ 2 ≥ 1

2 u u p 1/2 for all n n . k n − n2 kL ([0,T ];X) ≤ ≥ 2 72

In general, for each k N, let nk N such that nk nk 1 and ∈ ∈ ≥ −

2 u u p 1/2 for all n n . k n − nk kL ([0,T ];X) ≤ ≥ k

Now for each m N, deﬁne S : [0,T ] [0, ] by ∈ m → ∞ m S (t) := u (t) + u (t) u (t) ,t [0,T ]. (B.1.1) m k n1 kX nk+1 − nk X ∈ Xk=1

Then S is a monotone increasing sequence of real-valued functions on [0,T ]. Ap- { m} plying the triangle inequality in the space Lp([0,T ],dt) to (B.1.1), yields

S p = u p + u (t) u (t) k mkL ([0,T ];X) k n1 kL ([0,T ];X) nk+1 − nk Lp([0,T ];X) Xk=1 m (B.1.2) k u p + 1/2 = u p +1 < . ≤ k n1 kL ([0,T ];X) k n1 kL ([0,T ];X) ∞ Xk=1 Deﬁne S : [0,T ] [0, ] by → ∞

S(t):= lim Sm : [0,T ]. (B.1.3) m →∞ Then for 1 p< , by the monotone convergence theorem and (B.1.2), ≤ ∞ T T p p S dt = lim Sm dt < . m Z0 | | →∞ Z0 | | ∞ Hence S Lp([0,T ],dt), and thus S(t) < for Lebesgue a.e. t [0,T ]. The same ∈ ∞ ∈ holds by (B.1.2) if p = . Next, we note that for each t [0,T ], ∞ ∈ u (t)=u (t)+(u (t) u (t)) + + ((u (t) u (t)) nm+1 n1 n2 − n1 ··· nm+1 − nm m (B.1.4) =u (t)+ ((u (t) u (t)). n1 nk+1 − nk Xk=1

Moreover, by (B.1.3), the series u (t)+ ∞ ((u (t) u (t)) is absolutely summable n1 k=1 nk+1 − nk P in X, provided S(t) < . Since X is a Banach space, the series is summable in X ∞ for such t. Hence for all t such that S(t) < , we could deﬁne ∞ 73

∞ u(t) := u (t)+ (u (t) u (t)). (B.1.5) n1 nk+1 − nk Xk=1 Equations (B.1.4) and (B.1.5) imply that

u(t)= lim unm+1 (t) in X, provided S(t) < infty. (B.1.6) m →∞ Moreover, for all such t,

u(t) = lim unm+1 (t) S(t) < infty. k kX m X ≤ →∞

Hence, u(t) Lp([0,T ],dt). This means u Lp([0,T ],X). Lastly, we notice that k kX ∈ ∈ for all t such that S(t) < infty,

u (t) u(t) u (t) + u(t) nm+1 − X ≤ nm+1 X k kX (B.1.7) S (t)+ u( t) Lp([0,T ],dt). ≤ k kX ∈ Combining (B.1.6), (B.1.6) and using the dominated convergence theorem, we get

lim unm+1 (t) u(t) dt = 0. m Z − X →∞ 0

That is , u converges to u in Lp([0,T ],X). Thus, Lp([0,t]; X) is a Banach { nm+1 } space.

Lemma B.1.2. If X is a Hilbert space, and for all f,g,h L2([0,T ]; X) and ∈ T := 0 (f(t),g(t))X dt. Then, R

(1) 0,=0 if and only if f =0 for a.e. t [0,T ]; ≥ ∈

(2) =;

(3) <αf,g >= α for all α R; ∈

(4) = + . 74

Thus, L2([0,T ]; X) is a Hilbert space.

Proof.

(1) = T (f(t),f(t)) dt = T f(t) 2 dt 0; 0 X 0 k kX ≥ R R =0 T f(t) 2 dt = 0 f(t) = 0 for a.e. [0,T ] f(t)=0 ⇔ 0 k kX ⇔ k kX ∈ ⇔ R for a.e. t [0,T ]; ∈

T T (2) = 0 (f(t),g(t))X dt = 0 (g(t),f(t))X dt =; R R T T (3) <αf,g >= 0 (αf(t),g(t))X dt = 0 α(f(t),g(t))X dt = α; R R T T (4) = 0 (f(t)+ h(t),g(t))X dt = 0 (f(t),g(t))X +(h(t),g(t))X dt =< R R f,g > + .

Thus, L2([0,T ]; X) is a Hilbert space. APPENDIX C

ABSOLUTE CONTINUITY

C.1 ABSOLUTE CONTINUITY OF (u (x,t)) 2 AND k m tkµ (u (x,t),u (x,t)). E m m

The following lemma is used in the proof of the theorem 2.0.8 in Chapter 2.

Lemma C.1.1. (u (x,t)) 2 and (u (x,t),u (x,t)) are absolute continuous func- k m tkµ E m m tions on t [0,T ] for any x R. ∈ ∈

Proof. Since d 1 ( (u (x,t)) 2 )=((u (x,t)) , (u (x,t)) ) dt 2 k m tkµ m tt m t µ and d 1 ( (u (x,t),u (x,t))) = (u (x,t), (u (x,t)) ), dt 2E m m E m m t it suﬃces to show that

((um(x,t))tt, (um(x,t))t)µ and

(u (x,t), (u (x,t)) ) E m m t are continuous.

By using (2.0.6), (2.0.10) and (2), we have

b b m m (um(x,t))tt(um(x,t))t dµ = α′′ (t)wk(x) α′ (t)wj(x) dµ Z Z m,k m,j a a Xk=1 Xj=1 m

= αm,k′′ (t)αm,k′ (t). Xk=1 76

b m b um(x,t) (um(x,t))t dx = αm,k(t)λk wk(x) (um(x,t))t dµ Z ∇ ∇ Z ∇ a Xk=1 a m

= λkαm,k(t)αm,k′ (t). Xk=1

Thus,

(u (x,t)) 2 = (u (x, 0)) 2 + 2((u (x,t)) , (u (x,t)) ) k m tkµ k m tkµ m tt m t µ and

(um(x,t),um(x,t)) = (um(x, 0),um(x, 0))+2 um(x,t), (um(x,t))t . E E E

Lemma C.1.2 (Gronwall’s inequality). Let η( ) be a non-negative, absolutely con- · tinuous function on [0,T ], which satisﬁes for a.e. t the diﬀerential inequality

η′(t) φ(t)η(t)+ ψ(t), ≤

where φ(t) and ψ(t) are non-negative, summable functions on [0,T ]. Then

t t η(t) eR0 φ(s)ds(η(0) + ψ(s)ds) ≤ Z0

for all t [0,T ]. ∈ APPENDIX D

SEPARABILITY

The following lemma is used in the proof of the theorem 2.0.6 in Chapter 2.

Lemma D.0.3. Any subset of separable Banach Space X, with norm , is separable. k k

Proof. Let Y = y ∞ be a countable dense subset of X. Let n N. Then for each { k}k=1 ∈ a A, let (a, 1 ) be the open ball in (X, , with center a and radius 1 . For each ∈ B n k·kX n 1 (m) 1 such open ball, there exists y Y (a, ). Let yn m∞=1 = Y ( a A (a, )). For ∈ ∩B n { } ∩ ∪ ∈ B n (m) each m, let xn A such that ∈ 1 x(m) y(m) . n − n X ≤ n

Let ǫ> 0, be arbitrary and for all x A be arbitrary. Let N := 2 +1. Then there ∈ ǫ ǫ exists y(m0) (x, 1 ) B(x,ǫ). Then, there exists x(m0) A such that Nǫ ∈B Nǫ ≤ Nǫ ∈

(m0) (m0) 1 xNǫ yNǫ . − X ≤ Nǫ

Therefore,

(m0) (m0) (m0) 1 (m0) 1 1 2 xNǫ x xNǫ yNǫ + yNǫ x + = ǫ. − X ≤ − X ≤ Nǫ − X ≤ Nǫ Nǫ Nǫ ≤

APPENDIX E

WEAK DERIVATIVE

E.1 BANACH SPACE VALUED FUNCTIONS

Let X be a seperable Banach space, with norm . k k

m Deﬁnition E.1.1. (i) If s(t)= i=1 χEi (t)ui is simple, we deﬁne P T m s(t) dt := Ei ui. (E.1.1) Z | | 0 Xi=1

(ii) We say f : [0,T ] X is summable if there exists a sequence s ∞ of simple → { k}k=1 functions such that

T sk(t) f(t) dt 0 as k . (E.1.2) Z0 k − k → → ∞

(iii) If f is summable, we deﬁne

T T f(t) dt = lim sk(t) dt. (E.1.3) k Z0 →∞ Z0 Theorem E.1.1 (Bochner). A strongly measurable function u : [0,T ] X is → summable if and only if t u(t) is summable. In this case 7→ k kX T T u(t) dt u(t) dt, Z0 ≤ Z0 k k

and T T u∗, u(t) dt u∗,f(t) dt, D Z0 E ≤ Z0 h i

for each u∗ X∗. ∈

Lemma E.1.2. If α(t) := m χ (t) c is simple on [0,T ] and u X, then i=1 Ei i ∈ T T P 0 α(t)udt = u 0 α(t) dt. R R 79

Proof.

T T m m α(t)udt = χE (t) ci udt = Ei ci u Z Z i | | 0 0 Xi=1 Xi=1 m T = u Ei ci = u α(t) dt. | | Z Xi=1 0

Lemma E.1.3. If f : [0,T ] R is a Lebesgue integrable function on [0,T ] and → u X, then T f(t)udt = u T f(t) dt. ∈ 0 0 R R

Proof. Since f(t) is a Lebesgue integrable function on [0,T ], there exists a sequence

of simple functions αn n∞=1 on [0,T ] such that f(t) = limn αn(t). Hence, { } →∞ T T αn(t) u f(t) u dt = αn(t) f(t) u dt Z0 k − k Z0 | − | k k T = u αn(t) f(t) dt k k Z0 | − | 0 as n . → → ∞

Therefore, f(t)u is summable. Thus, we have

T T T T f(t)udt = lim αn(t) udt = u lim αn(t) dt = u f(t) dt. n n Z0 →∞ Z0 →∞ Z0 Z0

Lemma E.1.4. If u ⇀u in L2([0,T ];Dom( )), then u ⇀u in L2([0,T ]; L2 [a,b]). m E m µ

2 2 Proof. Suppose for v∗ in the dual space of L ([0,T ]; Lµ[a,b]) and v is the correspond- 80

ing Reisz’s representation of v∗. Then,

v∗,um L2([0,T ];L2 [a,b]) v∗,u L2([0,T ];L2 [a,b]) h i µ −h i µ T T T

= (v,um)µ dt (v,u)µ dt = (v,um u)µ dt Z − Z Z − 0 0 0 T T

(v,um u)µ dt v µ um u µ dt ≤ Z0 | − | ≤ Z0 k k k − k T 1 T 1 2 2 2 2 v µ dt um u µ dt ≤ Z0 k k Z0 k − k 0 as m . → → ∞

Lemma E.1.5. For all u L2([0,T ];Dom( )), we have ∈ E T T d u(x,t) dt = u(x,t) dt, where = . ∇ Z0 Z0 ∇ ∇ dx

2 Proof. We know that v(x) L [a,b]. In fact, for all v(x) C∞(a,b), we have ∈ ∈ c b T b T v(x) u(x,t) dt dx = v′(x) u(x,t) dt dx Za ∇ Z0 − Za Z0 T b = v′(x) u(x,t) dxdt − Z0 Za T b = v(x) u(x,t) dxdt Z0 Za ∇ b T = v(x) u(x,t) dt dx. Za Z0 ∇ Then, we have T u(x,t) dt = T u(x,t) dt. ∇ 0 0 ∇ R R Lemma E.1.6. Let u L2([0,T ]; X), where X is Dom( ) or L2 [a,b]. Deﬁne v( ,t) := ∈ E µ · t u( ,τ) dτ. Then v(x,t) L2([0,T ];Dom( )), and v ( ,t)= u( ,t). 0 · ∈ E t · · R 81

Proof. Fix t [0,T ].Then, ∈ b b t 2 v(x,t) 2 dx u(x,τ) dτ dx Za |∇ | ≤ Za Z0 |∇ | b t 2 2 1 1 ( u(x,τ) dτ) 2 T 2 dx ≤ Za Z0 |∇ | b t T u(x,τ) 2 dτ dx ≤ Za Z0 |∇ | t b T u(x,τ) 2 dxdτ ≤ Z0 Za |∇ | t 2 T u( ,τ) Dom ( ) dτ TM. ≤ Z0 k · k E ≤ Thus, for each t [0,T ], v(x,t) Dom( ). ∈ ∈ E

Moreover, let ϕ(t) C∞(0,T ). Then, ∈ C T T t ϕt(t)v(x,t) dt = ϕt(t) u( ,τ) dτ dt Z0 Z0 Z0 · T T T T = ϕt(t) u( ,τ) dtdτ = ( ϕt(t) dt)u( ,τ) dτ Z0 Zt=τ · Z0 Zt=τ · T = ϕ(t)u(x,t) dt. Z0 − So, v ( ,t)= u( ,t). t · ·

Lemma E.1.7. The set ϕψ : ϕ C∞(0,T ),ψ(x) C∞(a,b) is dense in { ∈ c ∈ c } 2 2 L ([0,T ]; Lµ[a,b]).

Proof. Suppose u L2([0,T ];Dom( )). For each ﬁx t, u( ,t) Dom( ). Thus, there ∈ E · ∈ E n exists ψn(x) Cc∞[a,b] such that ψn(x) u Dom ( ) < ǫ/2 . Now, ϕn(t) := 1 for t ∈ k − k E ∈ 82

E [0,T ],ϕ (t):< 1 for t E¯ , where E¯ = [0,T ] E , and ϕ (t) C∞(0,T ). n ⊂ n ∈ n n \ n n ∈ c T 2 2 ϕn(t)ψn(x) u L2([0,T ];Dom ( )) = ϕn(t)ψn(x) u Dom ( ) dt k − k E Z0 k − k E 2 2 = ϕn(t)ψn(x) u Dom ( ) dt + ϕn(t)ψn(x) u Dom ( ) dt ZEn k − k E ZE¯n k − k E ǫ 2 2 2 n (En)+2 ϕ(t)n ψn(x) Dom ( ) + u Dom ( ) dt ¯ ≤2 L ZEn | | k k E k k E ǫ 2 2 n (En)+2 ψn(x) Dom ( ) + u Dom ( ) dt ¯ ≤2 L ZEn k k E k k E ǫ ǫ 2 2 n (En)+2 ( n + u )Dom ( ) + u Dom ( ) dt ¯ E ≤2 L ZEn 2 k k k k E ǫ ǫ 2 2 n (En)+2 (2( n ) +3 u Dom ( ) dt ¯ ≤2 L ZEn 2 k k E ǫ 2 2 5( n ) (En)+6 u Dom ( ) dt 0. ≤ 2 L ZE¯n k k E →

1 n Remark: (En) 0 and any f L (X),X R . Then, limn fdx =0. L → ∈ ⊂ →∞ En R Proposition E.1.8. Suppose u ⇀ u in L2([0,T ];Dom( )), and (u ) ⇀ γ in m E m t 2 2 2 2 L ([0,T ]; Lµ[a,b]). Then (um)t ⇀ut in L ([0,T ]; Lµ[a,b]).

T 2 2 Proof. Suppose β := γdt, v∗ L ([0,T ]; L [a,b]). Then v∗, (u ) v∗,γ 0 ∈ µ h m ti→h i R T b T b v∗ (um)t dµ dt v∗ γ dµ dt. ⇔ Z0 Za → Z0 Za

Choosing ϕ(t) C∞(0,T ) and ψ(x) C∞(a,b) and replacing v∗by ϕ(t)ψ(x) in the ∈ C ∈ C above equation. We get

T b L.H.S = lim (um)t ϕ(t)ψ(x) dµ dt m →∞ Z0 Za b T b T = lim ϕ(t)ψ(x)(um)t dt dµ = lim ψ(x) ϕ(t)(um)t dt dµ m m →∞ Za Z0 →∞ Za Z0 b T = lim ψ(x) ϕt(t) um dt dµ m →∞ Za Z0 − T b T b = lim ϕt(t)ψ(x) um dµ dt = ϕt(t)ψ(x) udµ dt. m →∞ Z0 Za − Z0 Za − 83

T b b T R.H.S = γϕ(t)ψ(x) dµ dt = ϕ(t)ψ(x) γ dt dµ Z0 Za Za Z0 b T b T = ψ(x) ϕ(t) γ dt dµ = ψ(x) ϕt(t) β dt dµ Za Z0 Za Z0 − T b = ϕt(t)ψ(x) β dµ dt. Z0 Za −

So, we have T b T b ϕt(t)ψ(x) udµ dt = ϕt(t)ψ(x) β dµ dt. Z0 Za − Z0 Za − Since ϕ (t)ψ(x) is dense in L2([0,T ]; L2 [a,b]), we get u = β. This implies u = { t } µ t γ.

2 2 Lemma E.1.9. Fix v L ([0,T ], Dom( )), and for all w L ([0,T ], Dom( )′), ∈ E ∈ E T 2 deﬁne l(w) := w,v dt. Then, l (L ([0,T ], Dom( )′))′. 0 h i ∈ E R

Proof. T T T

l(w) = w,v dt w,v dt w Dom ( )′ v Dom ( ) dt | | Z h i ≤ Z |h i| ≤ Z k k E k k E 0 0 0 T T 2 1/2 2 1/2 ( w Dom ( )′ dt) ( v Dom ( ) dt) ≤ Z0 k k E Z0 k k E 2 2 w L2([0,T ],Dom ( )′) v L2([0,T ],Dom ( )) . ≤ k k E k k E Thus, l(w) is bounded.

2 Moreover, if α ,α R and w ,w L ([0,T ], Dom( )′) 1 2 ∈ 1 2 ∈ E T T l(α1w1 + α2w2)= α1w1 + α2w2,v dt = α1 w1,v + α2 w2,v dt Z0 h i Z0 h i h i T T = α1(˜w1,v)µ dt + α2(˜w2,v)µ dt Z0 Z0 T T =α1 w1,v dt + α2 w2,v dt Z0 h i Z0 h i

=α1l(w1)+ α2l(w2). 84

2 2 Proposition E.1.10. Suppose (um)t ⇀ut in L ([0,T ]; Lµ[a,b]) , and l(um)tt ⇀lσ in

2 2 L ([0,T ];Dom( )′). Then l ⇀l in L ([0,T ];Dom( )′) . E (um)tt (u)tt E

Proof. Suppose σ L2([0,T ]; L2 [a,b]) is the representative of l , and deﬁne ∈ µ σ T 2 2 γ := σ dt, for any v L ([0,T ];Dom ). Then by l ⇀l in L ([0,T ];Dom( )′) 0 ∈ E (um)tt σ E R and Lemma E.1.9, we have

T T l(u ) ,v dt lσ,v dt. Z m tt → Z h i 0 0

i.e. T b T b (um)tt vdµ dt lσ vdµ dt. Z0 Za → Z0 Za

Choosing ϕ(t) C∞(0,T ) and ψ(x) C∞(a,b) and replacing v by ϕ(t)ψ(x) in the ∈ C ∈ C above equation.

T b Lefthandside = lim (um)tt ϕ(t)ψ(x) dµ dt m →∞ Z0 Za b T b T = lim ϕ(t)ψ(x)(um)tt dt dµ = lim ψ(x) ϕ(t)(um)tt dt dµ m m →∞ Za Z0 →∞ Za Z0 b T = lim ψ(x) ϕ(t)′ (um)t dt dµ m →∞ Za Z0 − T b T b = lim ϕt(t)ψ(x)(um)t dµ dt = ϕt(t)ψ(x) ut dµ dt. m →∞ Z0 Za − Z0 Za −

T b b T Righthandside = σϕ(t)ψ(x) dµ dt = ϕ(t)ψ(x) σdt dµ Z0 Za Za Z0 b T b T = ψ(x) ϕ(t) σdt dµ = ψ(x) ϕt(t) γ dt dµ Za Z0 Za Z0 − T b = ϕt(t)ψ(x) γ dµ dt. Z0 Za − So, we have

T b T b ϕt(t)ψ(x) ut dµ dt = ϕt(t)ψ(x) γ dµ dt. Z0 Za − Z0 Za − 85

Since ϕ (t)ψ(x) is dense in L2([0,T ]; L2 [a,b]), we get u = γ. This implies u = { t } µ t tt σ.

Lemma E.1.11. Let u, v L2([0,T ];Dom( )). Then, all Lebesgue a.e. x [a,b] , ∈ E ∈

(u(x,t)v(x,t))t = u(x,t)tv(x,t)+ u(x,t)vt(x,t).

T T Proof. Let ϕ(t) Cc∞. Then, 0 u(x,t)v(t) ϕ(t) dt = 0 u(x,t)v(t) ϕt(t)dt. ∈ t − R R T ut(x,t)v(t)+ u(x,t)vt(t) ϕ(t)dt Z0 T T = u(x,t) v(t)ϕ(t) dt + u(x,t)vtϕ(t)dt − Z0 t Z0 T T T = u(x,t)v(t)ϕt(t) dt u(x,t)vt(t)ϕ(t) dt + u(x,t)vt(t)ϕ(t)dt − Z0 − Z0 Z0 T = u(x,t)v(t) ϕt(t)dt. − Z0

1 Now, let v (x, ) C∞(0,T ) such that v (x, ) v(x, ) in H (0,T ). Then, n · ∈ c n · → · 0 T T u(x,t)v(x,t) ϕ(t) dt = u(x,t)v(x,t) ϕt(t)dt Z0 t − Z0 T = lim u(x,t)vn(x,t)ϕt(t)dt n − →∞ Z0 T = lim ut(x,t)vn(x,t)+ u(x,t)(vn(x,t))t ϕ(t)dt n →∞ Z0 T = ut(x,t)v(x,t)+ u(x,t)vt(x,t) ϕ(t)dt. Z0 For the last equality, we use the fact v (x, ) v(x, ) in H1(0,T ). n · → · 0 APPENDIX F

1 2 EMBEDDING OF H0 IN Lµ

1 Proposition F.0.12. Let u H (a,b) and let φ C∞(a,b) such that φ u ∈ 0 { n} ⊂ c n → in H1(a,b). Then, there exists a subsequence φ such that φ u everywhere 0 { nk } nk → c

in [a,b], where uc is the continuous representative of the equivalence class of u in

1 H0 (a,b).

Proof. Since φ u in H1(a,b), there exists a subsequence φ converging point- n → 0 { nk } wise Lebesgue a.e. to u on (a,b). Now let x (a,b), and let ǫ> 0 be arbitrary. We c ∈

ﬁrst notice that since φn is convergent, there exists C > 0 such that

N φn Dom C for all n . (F.0.1) k k E ≤ ∈

Next, by the continuity of u , there exists 0 <δ < ( ǫ )2 such that for all y [a,b], c ǫ 3+C ∈ with y x <δ , we have | − | ǫ u (x) u (y) < ǫ/3. (F.0.2) | c − c | Hence,

φ (x) u (x) φ (x) φ (y) + φ (y) u (y) + u (y) u (x) . (F.0.3) | nk − c |≤| nk − nk | | nk − c | | c − c |

The ﬁrst term can be estimated as follows,

x x 1/2 2 1/2 φn (x) φn (y) = φn (s) ds φn (s) ds x y | k − k | Z ∇ k ≤ Z |∇ k | | − | y y (F.0.4)

1/2 ǫ 2 φnk Dom x y C( ) ǫ/3. ≤ k k E | − | ≤ 3+ C ≤ Substituting (F.0.2) and (F.0.4) into (F.0.3), we get

φ (x) u (x) ǫ/3+ φ (y) u (y) + ǫ/3, | nk − c |≤ | nk − c | for all y (x δ ,x + δ ). ∈ − ǫ ǫ 87

Let y (a,b)satisfy limk φn (y) = uc(y). Then, for all k suﬃcient large, ∈ →∞ k

φn (y) uc(y) < ǫ/3, and hence φn (x) uc(x) < ǫ. Thus, limk φn (x)= uc(x) | k − | | k − | →∞ k for al x [a,b]. ∈

Corollary F.0.13. Let u H1(a,b) and let u¯ be its unique L2 [a,b] representative. ∈ 0 µ

Then we can take u¯ to be uc.

Corollary F.0.14. If supp(u)=[a,b],a

Proof. Let u H1(a,b) such that I(u) = 0. Then we haveu ¯ = u = 0 in L2 [a,b] . ∈ 0 c µ Since supp(u)=[a,b], we have u 0on[a,b]. Thus, u = 0 Lebesgue a.e on [a,b] c ≡ APPENDIX G

DIFFERENTIABILITY OF DISTRIBUTION AND BANACH SPACE VALUED FUNCTIONS

Let denote the collection of all test functions on [a,b]. D

Deﬁnition G.0.2. (see e.g.[25]) A distribution f is a functional : R which is D 7→ linear and continuous in the following sense. If φ is a test function, then we ∈ D denote the corresponding real number by (f,φ).

By linearity we mean that

(f,αφ + βψ)= α(f,φ)+ β(f,ψ) for all constants α, β and all test functions φ,ψ.

By continuity we mean following. If φ is a sequence of test functions that { n} vanish outside a common interval and converge uniformly to a test function φ, and if all their derivatives do as well, then

(f,φ ) (f,φ) as n . n → → ∞

Deﬁnition G.0.3. (see e.g.[25]) For any distribution f, we deﬁne its derivative f ∇ by the formula

( f,φ)=(f, φ) for all test functions φ. ∇ ∇

Deﬁnition G.0.4 (Strong derivative). (see e.g. [11]) If f : (a,b) R X, where ⊂ → f(t+h) f(t) X is a Banach space, then it is diﬀerentiable at t (a,b) if L := limh 0 − ∈ → h exists in X. The limit L, if it exists, will be denoted by f˙(t).

Definition G.0.5 ( Fréchet derivative). (see e.g.[15]) Let V and W be Banach spaces, and U V be an open subset of V . A function f : U W is called Fréchet ⊂ → 89

differentiable at t U if there exists a bounded linear operator A : V W such that ∈ t → f(t + h) f(t) A (h) lim k − − t kW =0 h 0 h → k kV Definition G.0.6 (Gâteaux derivative). (see e.g.[23]) Suppose X and Y are locally convex topological vector spaces (for example, Banach spaces), U X is open, and ⊂ F : X Y . The Gâteaux differential dF (u;Ψ) of F at u U in the direction → ∈ Ψ X is defined as → F (u + hΨ) F (u) dF (u;Ψ) = − . h

Remark: In the case, f : (a,b) R X, where X is a Banach space, if f has a ⊂ → strong derivative at t (a,b), then f is Fréchet differentiable at t and also Gâteaux ∈ derivative at t.

Proposition G.0.15. If f is strongly differentiable at t (a,b), and L be the strong ∈ derivative of f, then f is Frećhet differentiable at t.

Proof. f(t + h) f(t) lim − L =0. h 0 h − → X f(t + h) f(t) hL ) lim − − =0. ⇔ h 0 h → X f(t + h) f(t) hL) lim k − − k X =0. ⇔ h 0 h → | | Define A (h)= hL. Then A (h):(a,b) X is a linear operator. Hence f is Fréchet t t → differentiable at t.

∂ u(x,t+h) u(x,t) Deﬁnition G.0.7 (Partial derivative). u(x,t) = limh 0 − exists in R. ∂t → h

u(t+h) u(t) Proposition G.0.16. Suppose for t (0,T ). u˙(t) := limh 0 − , the limit ∈ → h exists in Dom( ). Then the partial derivative ∂ u(x,t)=u ˙(x,t) for Lebesgue a.e. E ∂t x [a,b]. Consequently, we can regard u =u ˙ in Dom( )= H1(a,b). ∈ t E 0 90

Proof. u(t + h) u(t) lim − u˙(t) =0. h 0 h − → Dom ( ) E b u(t + h) u(t) 2 lim ( − u˙(x,t)) dx =0. ⇔ h 0 Z ∇ h − → a

b u(t + h) u(t) 2 lim − u˙(x,t) dx =0. ⇔ h 0 Z h − → a

u(t + h) u(t) for a.e x [a,b], lim − u˙(x,t) =0. ⇒ ∈ h 0 h − → ∂ for a.e x [a,b ], u(x,t)=u ˙(x,t). ⇒ ∈ ∂t

Deﬁnition G.0.8 (Spectral Family). (see e.g.[14]) A real spectral family is a one-

parameter family Eλ λ R of projections Eλ deﬁned on a Hilbert space H (of any { } ∈ dimension) which depends on a real parameter λ and is such that

(i) E E , hence E E = E E = E if λ η, λ ≤ η λ η η λ λ ≤

(ii) limλ Eλx =0, limλ + Eλx = x for any x H, →−∞ → ∞ ∈

(iii) Eλ+0x = limη λ+0 Eηx = Eλx. →

Proof of theorem 2.0.4. The uniqueness of (2.0.3) is well known.

Let v(t) := ∞ √λ sin(t√λ)dE g + ∞ cos(t√λ)dE h. Fix t R. Then, for 0 − λ 0 λ ∈ R R any h R 0 , ∈ \{ }

u(t + h) u(t) ∞ cos((t + h)√λ) cos(t√λ) − v(t)= − + √λ sin(t√λ)dEλg h − Z0 h ∞ sin((t + h)√λ) sin(t√λ) + − cos(t√λ)dEλh. Z0 h − 91

Hence,

u(t + h) u(t) − v(t) h −

√ √ ∞ cos((t + h) λ) cos(t λ) √ √ − + λ sin(t λ)dEλg ≤ Z0 h

√ √ ∞ sin((t + h) λ) sin(t λ) √ + − cos(t λ)dEλh Z0 h − (G.0.1) 2 √ √ 1/2 ∞ cos((t + h) λ) cos(t λ) √ √ − + λ sin(t λ) dEλg ≤ Z0 h

2 √ √ 1/2 ∞ sin((t + h) λ) sin(t λ) √ + − cos(t λ) dEλh . Z0 h −

(see, e.g.,Yosida [28], p312, Corollary 2)

Note that for the ﬁrst term in (G.0.1), we have,

√ √ cos((t + h) λ) cos(t λ) √ √ − + λ sin(t λ) h

√ √ √ √ √ cos(t λ)cos(h λ) sin(t λ)sin(h λ) cos(t λ) √ √ = − − + λ sin(t λ) h

√ √ √ √ √ cos(t λ)(cos(h λ) 1) λ sin(t λ)sin(h λ) √ √ = − + λ sin(t λ) h − h√λ (G.0.2)

√ √ √ √ (cos(h λ) 1) √ √ sin(h λ) √ cos(t λ) λ − + λ sin(t λ) + λ ≤ h√λ h√λ

(1)(1)(√λ)+( √λ)2 + √λ = 4( √λ), ≤ for all h suﬃcient small. 92

Also for the second term in (G.0.1), we have, √ √ sin((t + h) λ) sin(t λ) √ − cos(t λ) h√λ −

√ √ √ √ √ sin(t λ)cos(h λ) + cos(t λ)sin(h λ ) sin(t λ) √ = − cos(t λ) h√λ −

√ √ √ sin(t λ)(cos(h λ) 1) √ sin(h λ) √ = − cos(t λ) cos(t λ) h√λ − h√λ − (G.0.3)

√ √ √ cos(h λ 1) √ sin(h λ) √ sin(t λ) − + cos(t λ) + cos(h λ) ≤ h√λ h√λ

(1)(1)+(1)(2)+1 = 4, ≤ for all h suﬃcient small.

Combining (G.0.1),(G.0.2), (G.0.3) and by the dominated convergence theorem, we

u(t+h) u(t) − ˙ have limh 0 h v(t) =0. Hence u(t)= v(t). Let → −

∞ ∞ w(t) := λ cos(t√λ)dEλg √λ sin(t√λ)dEλh Z0 − − Z0 . Then for any h R 0 , ∈ \{ } u˙(t + h) u˙(t) − w(t) h −

√ √ √ √ ∞ λ sin((t + h) λ)+ λ sin(t λ) √ − + λ cos(t λ)dEλg ≤ Z0 h

√ √ √ √ ∞ λ cos((t + h) λ) λ cos(t λ) √ √ + − + λ sin(t λ)dEλh (G.0.4) Z0 h√λ

2 √ √ √ √ 1/2 ∞ λ sin((t + h) λ)+ λ sin(t λ) √ − + λ cos(t λ) dEλg ≤ Z0 h

2 √ √ 1/2 ∞ cos((t + h) λ) cos(t λ) √ √ + − + λ sin(t λ) dEλh . Z0 h

Using (G.0.3), for the ﬁrst term of (G.0.4), we have

2 √ √ √ √ λ sin((t + h) λ)+ λ sin(t λ) √ 2 − + λ cos(t λ) (4λ) (G.0.5) h ≤

Similarly, by using (G.0.2), for the second integral of (G.0.4), satisﬁes

2 √ √ cos((t + h) λ) cos(t λ) √ √ √ 2 − + λ sin(t λ) (4 λ) = 16λ. (G.0.6) h ≤

Combining (G.0.4),(G.0.5), (G.0.6), by the dominated convergence theorem and letting h tend to 0, we getu ¨(t)= w(t).

Finally,

∞ ∞ Au(t),x = λd Eλu(t),x = λd u(t),Eλx h i Z0 h i Z0 h i ∞ ∞ ∞ sin(t√η) = λd cos(t√η)dEηg + dEηh,Eλx Z0 Z0 Z0 √η ∞ ∞ ∞ sin(t√η) = λd cos(t√η)d Eηg,Eλx + d Eηh,Eλx Z0 Z0 h i Z0 √η h i ∞ ∞ ∞ sin(t√η) = λd cos(t√η)d EλEηg,x + d EλEηh,x (G.0.7) Z0 Z0 h i Z0 √η h i λ λ ∞ sin(t√η) = λd cos(t√η)d Eηg,x + d Eηh,x Z0 Z0 h i Z0 √η h i λ ∞ sin(t√λ) = λ cos(t√λ)d Eλg,x + λ d Eλh,x Z0 h i Z0 √λ h i = u¨(t),x . −h i

Since E h,x is of bounded variation on [0,λ], , we get the last equality from h λ i the second equality. (See e.g. Apostal [1].)

The initial conditions are obvious.