.32). 1...
.....xi 2..

5.3,.“
. 4....

r; .....rx..:,....$..

.- Oval-.2117:
7 us. A .:
.

& x'
.13 \
3;” I V;

I .v ..
.- 21.)». v
.1311:- 7.
1.13.571)?!
1

q’

pup-E; 4

e
.....ri.
5.3%....
. 113‘-
e 1;! .
...?» $3 ,

£33.... a;
:95; Ru... 3}
....3.

“an?! .. ... in

. ...: ml... .
$1.... . I": «1.....11.
... ..

"(I o»:
- 41,

 

 

 

 

 

astiggZTQt'am mlljlfljjjljlllllllzill/ill
University 17 2211

 

 

 

This is to certify that the

dissertation entitled

Wavelet-Based Numerical
Methods for Some Boundary Value Problems

presented by

Xiaodi Wang

has been accepted towards fulfillment
of the requirements for

 

 

Ph .0. degree in Applied Mathematics

I? amt 1/ id 9. {A

Major professor

Date 5% 0&1"qu 2‘ [995‘
- , , , ,

0

MS U is an AMrmatiw Action/Equal Opportunity Institution 0-12771

PLACE ll RETURN BOX to remove thb checkout from your record.
TO AVOID FINES return on or before date duo.

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TWP—1T1

MSU In An Affirmative ActionIEqunl Opportunity Irutltwon
Wm:

 

Wavelet—Based Numerical
Methods for Some Boundary Value Problems

By

Xiaodi Wang

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1995

ABSTRACT

Wavelet—Based Numerical Methods for Some Boundary Value Problems

By

Xiaodi Wang

Elliptic and their related parabolic boundary value problems for partial differ-
ential equations have wide applications in sciences and engineering. Wavelet-based
numerical methods have been developed in recent years for solving such problems and
may lead to a major computational breakthrough in this field.

This work estibalishes some unconditional bases for some Sobolev spaces by means
of antiderivatives of wavelets and investigates the use of the Galerkin method with
such bases combining with domain decomposition technique and finite difference time
discretization scheme in solving elliptic and their relate parabolic singular boundary
value problems.

We first construct some unconditional bases for some Sobolev spaces and derive
some approximation properties of those bases. We then consider self-adjoint elliptic
boundary value problems with no singularities throughout their domains. By the
weighted wavelet-Galerkin method we come up with well behaved diagonal matrices.
Thus the number of operations needed to solve the problem is greatly reduced in

comparison with other methods.

Next we look at problems with singularities caused by the vanishing of leading
coefficients at the boundary. In such cases problems are difficult to solve efficiently.
However, by using a weighted wavelet—based domain decomposition method solutions
of such problems become more manageable.

Finally we study singular parabolic boundary value problems. We use a combi-
nation of the finite difference time discretization scheme and wavelet-based domain
decomposition methods. Such methods have been applied to some Fokker-Planck
equations. The correspoding numerical results indicate that wavelet based methods

are excellent candidates for tackling such problems.

To my wife and my parents

iv

ACKNOWLEDGMENTS

I am most indebted to my dissertation advisor, Professor David H. Y.
Yen, for his constant encouragement and support during my graduate study
at Michigan State University. I would like to thank him for suggesting the

problem and the helpful directions which make this dissertation a reality.

I also would like to thank my dissertation committee members Profes-
sor Chichia Chiu, Professor Qiang Du, Professor Michael W. Frazier, and

Professor Zhengfang Zhou for their valuable suggestions and precious time.

| wish to express my gratitude to my wife, Chundong Fu, my parents,
Rouhuai Wang and Qizhao Luo for their patience, encouragement, and
support in daily life.

Contents

1 Introduction 1
1.1 Preliminaries and Notations ....................... 1
1.2 Overview of The Thesis ......................... 7

2 Wavelets and Corresponding Basis Emctions for L2 and Some Sobolev

Spaces 9
2.1 Properties of Compactly Supported Wavelets on R1 ........

2.2 Construction of Multidimensional Compactly Supported Wavelets . . 13
2.3 Wavelet Bases for L2(a,b) and L2([a,b] x [c,d]) ............ 15
2.4 Construction of Weighted Wavelet Basis for Lila, b) .......... 23
2.5 Approximation Properties of Wavelet Bases .............. 25

2.6 Wavelet Related Bases for Some Sobolev Spaces and Their Approxi-
mation Properties ............................. 27
2.6.1 Frames and Bases of Some Sobolev Spaces by Antiderivatives

of Wavelets ............................ 27
2.6.2 “Pseudo—Antiderivatives” of Weighted Wavelets and Their Prop-
erties ............................... 49

3 Wavelet Approximation Solutions for Some Linear Elliptic Boundary
Value Problems 55
3.1 Some Classical Results Concerning the Approximation of Linear Vari-

ational Problems ............................. 55

vi

3.2 Wavelet Based Solutions of Regular Linear Sturm—Liouville Bound—

ary Value Problems ............................ 59
3.2.1 Wavelet Approximation Solutions of Dirichlet Problems . . . . 60
3.2.2 Wavelet Approximation Solutions of Neumann Problems . . . 64

3.2.3 Wavelet Approximation Solutions of Mixed Boundary Value
Problems ............................. 66

3.3 Wavelet Based Domain Decomposition Methods for Singular Sturm-
Liouville Boundary Value Problems ................... 67
3.4 Steady-State Solutions of Some Fokker-Planck Equations ....... 69

4 Wavelet Approximation Solutions Some for Parabolic Initial Bound-

ary Value Problems 84
4.1 Wavelet-Galerkin Approximation of Some Linear Parabolic Initial-Boundary
Value Problems .............................. 84

4.2 Wavelet Based Domain Decomposition Methods for Some Singular
Fokker-Planck Equations ......................... 90
5 Concluding Remarks and Discussions 100
5.1 Concluding Remarks ........................... 100
5.2 Discussions ................................ 101

vii

List of Figures

3.1
3.2
3.3
3.4
3.5
3.6

4.1
4.2

(8q,8r) = (1,1) ............................... 81
(8q,8r) = (1.5,1.5) ............................. 81
(8q,8r) = (1,2) ............................... 82
(8q,8r) = (2,1) ............................... 82
(8q,8r) = (0.7,0.4) ............................. 83
(8q,8r) = (O.3,0.2) ............................. 83
Numerical solutions at t = 1,1.5,2. ................... 99
Numerical solutions from t = 0 to t = 2. ................ 99

viii

Chapter 1

Introduction

Elliptic and their related parabolic boundary value problems for partial differential
equations have wide applications in sciences and engineering. In recent years, a
number of numerical methods involving the use of wavelets have been developed for
solving such problems(See [3], [6], [35], [37], [38], [48], [54], [60] and [62]). This
thesis investigates the application of wavelets in numerical boundary value problems
with or without singularities (caused by the vanishing of the leading coefficients) at

the boundary.

1.1 Preliminaries and Notations

We introduce first the following preliminaries.

Let us start with some notational conventions. Throughout this thesis we use N,
Z, R, C for the sets of naturals, integers, real and complex numbers, respectively,
while R" stands for n-dimensional Euclidean spaces for n 2 1, n E N.

On R",n Z 1, we shall use the Lebesgue measure only. If E C R" is Lebesgue
measurable, then we denote the measure of E by [E]; in particular, []a,b[] = b — a,
where b > a and ]a,b[ stands for one of (a,b), (a,b], [a,b),[a,b]. We shall often use
Banach spaces LP(RN) for 1 S p S 00, defined as

L”(R") = [f l urn... = (f... |f(w)|pdx)% < co}, p < oo (1.1)

and
L°°<R”) = {f l ”in... = esssup..R~If(z)i < oo}. (12)

When p = 2, L2(RN) is a Hilbert space with the inner product

(L9) = f..~ fédx for Lg e HR”), (13)

where g denotes the complex conjugate of g, and (f, f) = llfllip We shall use H to

denote a general Hilbert space and

(uvv7H’ [lullil = lain)” (1'4)

to represent the corresponding inner product and norm of H. Similar to L2(RN)

space, we often consider

[2(Z) = {(Ck)keZ[Ck E C and 2 [ct]2 < +00} (1.5)

If we denote C = (Ck)k€Za d = (dk)keZ, then
+00 _ 2
(0,6012 = Z dek, ||Cl|12 = (0, CM (1-6)
k=—oo
We shall often drop the limits on the integration or the summation index when we
integrate over the whole space or sum over all integers. The most important property

about the Hilbert spaces is that they always have orthonormal bases, namely, there

exist a set of elements {(1%} C H such that

<¢ka ¢kl>H = 51w (1-7)
and
||f||iq = Z llf,¢k>H|2, for all f E H, (1-8)
I:
1, if k = k’,
where (Sky 2
0, otherwise.

If in addition the set {gbk} is countable, i.e. k E Z, then H is called a separable

Hilbert space. We shall consider separable Hilbert spaces only. The most important

inequality for a Hilbert space H, which we shall frequently use in this thesis, is the
Cauchy—Schwarz inequality
|<uav>|H S IIUIlllvll- (1-9)
If H = L2(R"), we obtain
[/ fgda: g (/|f|2da:)l/2 (/[g|2da:)l/2, for f,g e L2(R“) (1.10)

and if H = 12(Z), ( 1.9 ) becomes

3 (Zia/42) (Zld:|2)m. (1.11)

We also have the Holder inequality for LP(R"),1 1<p

we

The next best thing to an orthonormal basis for a Hilbert space H is unconditional

 

      

 

1 1
S ||f||,,]|g||q, for f E L", g E L", ;+ E = 1. (1.12)

 

bases which are also called Riesz bases. A basis {qbk} of H is called unconditional if

it satisfies one of the following three equivalent properties:

2cm 6 H => 5: law. 6 H, (1.13)
k k

20k¢k E H => 261,045]; 6 H, (1.14)
k k

where 6;, 2 21:1 are chosen at random for each k, or there exist 0 < A S B < 00 such

that
A Milli. s ; mom 3 B ”fut. (1.15)

Note that if {cm} is an orthonormal basis of H, than A = B = 1 in (1.15). This means
that any orthonormal basis of H is an unconditional basis. However, the converse is
not true.

We often use C8°(R")(or C8°(Q)) to denote the collection of infinitely differentiable

functions with compact supports.

We define the H61der spaces CO’”(R) by

 

Co’r(R) = {f E L°°(R);s;1’p [f(:1: + (2|,— f($)[ < 00} (1.16)

for0<r<1,and
00"(12) = [f e 0W ); d —-,;df 6 0°” (3)] (1.17)

for r = k + r", 0 < r’ < 1, where Ck(R) consists of the functions that are k times

continuously differentiable.

We also define Sobolev spaces Hm(R"), Hm(Q), H3), Hi(a, b) by

Hm(R") = {sz—>C|D’fEL2(R"),|r|§m}, (1.18)
Hmm) = {f=fl—>C|D'f€l>2(9) mm}, (1.19)
113(5)) 2 {feH1(n)|f=00nan}, (1.20)
H1(a,b) = {feH1((a,b)|fa)-_—o}, (1.21)

where Q is a bounded domain in R", (n 2 1), and D" stands for derivatives in the
weak sense, that is, for r = (r1,r2,-- - ,rn), [r] = r1+ 1'; + - -- +73, 3 m E N
arl+r2+m+rn

D” = . 1.22
Bx'ilax? - - - 8:13;" ( )

 

Iff E L1(R), (or f E L1(Q))

]fqude =(—11)"r1+',.+'"/g cbda: (1.23)

for all g!) E C§°(R"), (or ()3 E CS°(Q)), then we denote D'f 2 9., a generalized
derivative of f. If m = 0, we define H°(R") = L2(R") (or H°(fl) = L2(Q)). Sobolev
spaces H "‘(R"), H ”(51), H362) and H 3(0) are also Hilbert spaces, with corresponding

inner products and norms as follows

< u,v >m-E Z((u,v))k, or (u,v)m,g E Z((u,v))k,g, (1.24)
k=0 [13:0
and
m % m %
uunm :< u 1» >342: (21111:) or llullmn =< u u >22: (2 111.12..) ,
k=0 k=0
(1.25)
where
((u, v) )k— —— Z [DruDrvdx, k = O,1,2,---, (1.26)

lrl=k

OI'

((u,v))k,g = Z ]SzDruDrvdx, k = 0,1,2,- - - , (1.27)
Irl=k
and
(14,. E< (11,11) >,‘,”, or (111,... E ((u,u));{;‘;. (1.28)

It is well-known that the semi-norm | - I1 is an equivalent norm of H362) and H362)

if (2 is a finite open interval of R, that is, there exist two constants A, B > 0 such

that for any 11 6 H662) (or H}62))
Alulu!) S ”Hill 3 Blulm- (1°29)

Quite often in our work here, operators that are continuous linear maps from a

Hilbert space H to itself are employed. The norm of an operator A is defined as

“All = sup HAUIIH (130)

u ”:1
By a continuous linear map A we mean that A is linear, i.e.,
A(au + 62)) = aAu + ,BA'U, (1.31)
for any (1,5 6 C and u, v E H, and there exists c > 0 such that

IIAUIIH S CllullH (1-32)

for any 11 E H. Because of (1.32) we also call'it a bounded linear operator (or map).
If (1.32) is valid, then the smallest such number c is ”A”.

For each operator A if there is an operator A“ corresponding to it such that
< Au,v >H=< u,A"v >H (1.33)
for all u, v E H, then A“ is said to be the adjoint operator of A. The following is true
”4"” = IIAII, HIV/1H: llAllz- (1-34)

In the case A" = A, A is said to be self adjoint.

The kernel of an operator A is given by
kerA = {u E HIAu = 0}, (1.35)

5

and its range by
rangeA = {Aulu E H}. (1.36)

If kerA = {0} and if there is a bounded operator B such that AB 2 BA = I(I is the
identity operator, i.e. [U = u for all u E H), then A is said to be invertible and B is
said to be the inverse of A and denoted by A-l.

The spectrum 0(A) of an operator A from H to H consists of all the /\ E C
such that A — AI does not have a bounded inverse. If H is finite—dimensional, 0(A)
contains only the eigenvalues of A. On the other hand if H is infinitely dimensional,
0(A) consists of all eigenvalues, which constitute the point spectrum 00(A), but often

contains other /\ as well, which form the continuous spectrum 01(A). So in this case

0(A) = 00(A) U 01(A). The spectral radius p(A) is defined as
p(A) = supml | A e own}. (137)

The condition number n, which is a crucial point in the computational aspect, is
defined by
K = ||A||||A_1||- (1-38)

A is said to be a linear functional if A is a linear operator from H to C. The Riesz
representation theorem states that: If A is a bounded linear functional on H, then

there is a unique 11,. e H such that
A(u) =< 11,224 >H, for all u E H. (1.39)
We use the following convention for the Fourier transform for f E L2(R")
(me) = fie = (21¢? ] f(:v)e"“"d€. (1.40)
The inverse Fourier transform is then given by
fftv) = me) = (270-? / hoards. (1.41)

It follows from a simple computation that

nine = llfllo- (1.42)

6

One can also show that

k
[$93.11)] (5) = (it)k(ff)(€). (1.43)

With the help of the Fourier transform we may define the equivalent norm for the

Sobolev space H "‘( R")

“11113.. = [mu +|€|)"‘I1‘1(€)|2d€- (144)

Therefore for any number s > 0, we may extend our definition for Sobolev spaces to

H’(R") = {u e L2(R")|/(1+ IEI)‘|&(€)l2dE < oo} (1.45)

where s is not necessarily an integer.

In (1.44) and (1.45) R" can be replaced by (2.

1.2 Overview of The Thesis

This thesis is organized as follows.

In Chapter 2, we first introduce the basic idea of wavelets on R", which can be
found in [9] and [23]. Next we discuss wavelet bases on intervals and rectangles,
specially periodized wavelet bases on [0,1] (see [23], [24], [51] and [55]). We find
the explicit form of such functions in terms of the wavelet basis on the entire line,
and obtain some properties of such functions which are the same as nonperiodized
wavelets except vanishing moments. Then we construct the weighted wavelet basis
for L3,,(a, b) = L2((a, b), wdrr)[21], which is orthonormal with respect to weight w, and
extend the construction to L3,,((a, b) x (c, d)). Then we state that the characterization
of some Sobolev spaces by means of wavelet coefficients [23], [45], [46], and establish
some approximation properties of wavelets. Finally, we construct various bases of
H162), H362) and H362) by antiderivatives ( or pseudo antiderivatives) of (weighted)
wavelets, and prove that in fact they are unconditional bases of these Sobolev spaces,
and derive some approximation properties for both (weighted) wavelets and (pseudo)

antiderivatives of (weighted) wavelets.

In Chapter 3, we use (weighted) wavelets and their (pseudo) antiderivatives to ob-
tain a better treatment of elliptic boundary value problems. We first study (weighted)
wavelet-Galerkin (and domain decomposition) methods for (singular) two—point bound
ary value problems. Then we apply this general method to some steady state of
Fokker-Planck equations with singularities at the boundary. We also discuss the
stability and the error estimate of corresponding problems.

Finally, in Chapert 5 we draw some conclusions and remarks about wavelet based
methods for ODE and PDE problems, and discuss the possible directions of the future
research.

In Chapter 4 we introduce some efficient methods which combine finite difference
time discretization schemes and wavelet-based (domain decomposition) spatial pro-
jections for solving parabolic initial (singular) boundary value problems. In particular

we illustrate how these methods may be applied to some Fokker-Planck equations.

Chapter 2

Wavelets and Corresponding Basis
Functions for L2 and Some Sobolev

Spaces

2.1 Properties of Compactly Supported Wavelets

on R1

Daubechies gives the detailed construction of compactly supported wavelets on R1
and their properties in her book [23]. In this section, we shall present a brief review
of this construction and the properties of such wavelets.

For an integer p 2 1, construct a finite sequence {dflfifil satisfying

Ed). = 2. (2.1)
k
decfk_2m = 260,m for every m E Z, (2.2)
k
where if k — 2m ¢ {0,1,---,2p —1},dk_2m = 0,

Z(—1)kk"‘d_k+1 = 0 for 0 g m g p — 1 (2.3)
1:

Solutions of dk for 1 S p S 10 can be found in [23] , where d). = \/2h(k). To find

compactly supported wavelets, we define

= 12 their. (2.4)
2 k

We then define the so—called scaling function d) by its Fourier transformation

at) = mom/mien)
_\/1——7r- fi 7720(2- jé).

j=1

So ¢(a:) = 2:1: qub(2:r — It). Once (f) is found we can construct the wavelet function w
by

use) = Zena—mew — k). (2.5)
It now follows from a lemma in Deslauriers and Dubuc [27](We discuss this lemma

below.) that ()5 has compact support and supp¢ Q [0, 2p — 1].

Lemma 2.1.1 Ifm(€)= Elm—1:0 ake‘ké, with Zak = 1, then H311 m(2'j§) is an entire
function of exponential type. In particular it is the Fourier transforms of a distribution

with support in [k0, k1].

Proof. It follows from the Paley-Wiener theorem for distributions that it suffices to

prove that IT}; m(2"j§) is an entire function of exponential type with bounds

 

 

firm—J's) s 01(1 +0“) eXP(—k‘01m§) for 1m 5 s o, (2.6)
i=1
fiMZ < C2(1 + OM” exp(kllm§) for 1mg 2 0, (2,7)

 

 

for some constants Cl, Cg, M1 and M2.

Define
m1(€) = 642%“) = Z ak+koeik£

Then

00

H m(2"j€) = e“c015 fi m1(2-j§).

i=1 j=l
So to prove (2.6), it suffices to show that

fi m1(2-j€) s 01(1+I€I)M* for 1m: s 0 (2.8)

i=1

 

 

10

Note that for Imé S 0

1: —ko .
lm1(€)—1| S E lak+ko||€'k£—1l
k=0

ki-ko
2 Z lak+ko l min(19 klél)
[:20

< Cmin(1, IEI).

|/\

For an arbitrary 6 with Imf S 0, if [f] S 1, then

 

 

fim1(2-j§) S fi(1+C2—j)
j=l j=l
S E]: exp(2'jC) S 60. (2.9)

If [é] 2 1, then there is some jo Z 0 such that 2j° S [5] < 215“, and

 

 

 

 

00 jo+l oo
Ema—"6) S H(1+C) Hm1(2_j2—j°_l€)
i=1 j=l j=l

S (1 + C)jo+leC
S 60(1+C)exp[ln(1+C)ln[§I/ln2]
S (1 + C)eC|€|ln(l+C)/ln2. (210)

Now combining (2.9) for [E] S 1 and (2.10) for [6] 2 1 gives us (2.8).
Similarly we can prove (2.7). #
The construction of 21) now tells us that supprb g [1 — p, p]. We define

so that suppi/j (_I [0, 2p — 1] _D_ suppgb. We still use w to denote iii. If we define
¢j,k($) = 2j/2¢(2jx — k) and I/Jj,k($) = 2172214212 — k) for any pair of integers (j,k),
and define

V, = span{qb,,,)c : k E Z}, (2.11)

and

W" = span{ib,,,k : k E Z}, (2.12)

11

then V, and Wu are subspaces of L2(R). Also we have

V. C Vn+1 (2.13)
UV. = L2(R) (2.14)
full/n = {0} (2.15)
{n¢n,k}k€z is an orthonormal basis of V,

and V, contains polynomials of degree S p — 1,

l¢n,k}kez is an orthonormal basis of W", (2.16)
WnLVn, WkLWn if k 75 n and Vn+1 = Vn@Wn = V0 69 EBWk, (2.17)
k=0
+00 .
/ ¢j,kdl‘ = 2-1/2, (2.18)
+00
/ era-131: = 0, 0 g m g p — 1. (2.19)

Definition 2.1.1 If the wavelet 2,!) satisfies ff; xmw,,kd:r = 0 for 0 S m S p — 1,

then we say that it has p vanishing moments.

By virtue of (2.14) and (2.17), we know that W,- contains the “detailed” information
needed to go from the approximation at resolution j to an approximation at resolution

j + 1. Consequently

L2(R) = 69W, 2 V0€B$Wka (2.20)
.162 k=0

a decomposition of L2(R) into mutually orthogonal subspaces, i.e. {212M} je 3ch 2 con-
stitutes an orthonormal basis of L2(R).

Moreover, for suppzl) g [0,2p — 1] 2 suppd) with p > 1 the following property is
satisfied:

PM) 112,3), 6 C ON”) 2 {Hélder continuous functions with exponent /\(p)} (2.21)

where

y
E
22

0.55,
2(3) e 1.087833,

>’
E
22

1.617926,

12

)1(p) z 0.34851) for large p.

It follows from the above that for each f 6 L2(R), f can be expressed as

=chk¢2tvk )+Zd0k¢0k( :v) (2.22)

1:62

where OJ"); = (f, w”), d0,k = (f, (bo’k). If using notations 1b-”, = 00,). and c-” = do’k,
(2.22) then becomes

2:) = -:16j’kwj’k($). (2.23)
This means that ]—
lim []f— Z c,- “1),-Alp: 0 for all f E L2(R). (2.24)
j=—l
Therefore
llfllz.2 - Z1 [0.1]2 Vf E 142(19)- (2-25)
,__

For p = 1, we obtain the simplest example of a wavelet function

 

1 0 £2: < 3,
We) = < —1 3 S2: < 1, (2.26)
1 0 otherwise,

where 112(2) is the well-known Haar function, and {112“} is the corresponding Haar

basis for L2(R).

2.2 Construction of Multidimensional Compactly
Supported Wavelets

In Section 2.1 we focused our attention on the construction and the properties of
one—dimensional compactly supported wavelets. By using tensor products, we can

construct higher-dimensional wavelets. To demonstrate the idea, let us take a look

13

at the two—dimensional case. Let p Z l, and let (1) and d) be corresponding scaling

function and wavelet function in Section 2.1 with suppczS g [0, 2p— 1] 2 suppgb. Define

(PM 31) = ¢(~’r) ° (My), (2-27)

and

me) = mom). (2.28)

Since for each j E Z, {qSJ-‘khez is an orthonormal set in L2(R), (DJ-,khkz is an orthonor-

mal basis of V32, where
V32 2 span{<I>j,k,,k2|k1, ’92 E Z} (2.29)

with

‘91 k1,=k2 ¢j k1($)¢j.k2(y)° (2-30)
Hence V32 Q Vfil,‘v’j E Z. Let WJ-2_LV-2 such that Vial — V-2 EB W2. Then it turns out
from the properties in Section 2.1 and the definitions (2.27), (2.28), (2.29) and (2.30)
that

2 3
W? = span{\ll(l ),kga‘l’iiil,k2a ‘I’iflglkzlklvhezl

NJki

v}, = VJEBEBW}, (2.31)

L2(R2) = UV-=2 V2®®W2= 69 W}

j=0 jz-oo

where w;,‘2,,.,(x,y) = ¢.,..<x>w..k.(y)w $2., ..(x y): ¢.,.,(x)¢,-,..(y), w§22,,.,(x,y) =
¢J.k1($)¢j.k2 (y). If we denote $0,]: by 1b-},k, then @0,k1.k2 = ‘II—1,k1,k2-

Consequently,
{‘113,c1 kzlj E N for l = 1,3;j E NU {—1} for l = 2; [91,162 E Z} (2.32)

forms an orthonormal basis for L2(R2). Since 45 and 1% are compactly supported,

(2.32) is the compactly supported wavelet basis for L2(R2).

l4

There is an even more straightforward way to construct wavelet basis for L2(R2).

Let f E L2(R2), then for each a: E R, f(:I:,) E L2(R). Hence

=2 cj k( :r)tpj( kg for each fixed a: E R (2.33)

1‘62
]=—l

where cj,k(a:) = (f(:r,-),¢j,k)L2(R). Notice that f E L2(R2) and (by, E L2(R) for
j Z —1,k E Z. Therefore cj,k(1:) E L2(R) for a.e. :c, and j Z -—l,k E Z, and

Cj,(k$ :2 di ,I¢i,(lx (234)
i=—l
where
611,: = (Cj,k,¢i,z)L2(R)
= ((f($, '), ¢j,k)L2(R), ¢i,l)L2(R)
= (f($ay)awi,l($)¢j.k(y))L2(R2)° (2-35)

It follows from (2.33),(2.34) and (2.35) that

=2 2( (fat/H.439 )W( ))¢i.z($)¢j.k(y) (2-36)

i,j=—1k,I€Z

for f E L2(R2). It turns out that {¢§’[($)1/Jj,k(y)} constitutes a compactly supported
orthonormal basis of L2(R2).

Remark: In the first case, the compact supports of the wavelet basis are squares
while in second case they are rectangles. There are also several constructions of non-
separable wavelets. For example Cohen and Daubechies [18] constructed smooth,
compactly supported, biorthogonal wavelets using ideas from the univariate construc-

tion.

2.3 Wavelet Bases for L2(a., b) and L2([a, b] x [c,d])

In applications for numerical PDE and ODE problems we often need bases for L2(a, b)
and L2([a, b] x [c, d]). So far all wavelets that we have discussed in previous sections

lead to bases for L2(R). Let us start with a compactly supported scaling function 45

15

and a wavelet function #2 with supp¢,supp¢ Q [0, 2p — 1]. Without loss of generality,
wavelets on a special interval [0,1] are being constructed as more general intervals can
be transformed into [0,1] by a simple linear map.

In the case p = 1, we have the Haar basis {tbj,k},kz—Zi of L2(R) where
E

1 ifxE [0,1/2),

t/J—uc = 450,1” (15(53) = 450.0(1') = X[0.1]($), W53) = { ,
—l 1f:L' E [l/2,1).

Since each of these functions has a support either within [0,1] or completely outside
of [0,1], the collection {¢j,k|supp¢j,k 0 [0,1] aé {D} = {¢j,k|j Z —1,k 6 Ij} is an
orthonormal basis of L2(0,1), where I]- : {klO S k S 23 — 1, where 23 = 1 ifj =
—1,23'= 21'1sz 0}.

The situation is somewhat more complicated when one starts from wavelet bases
with p > 1. Several solutions are proposed for this case below.

The simplest solution seems to be the following. For any function f E L2(0, 1) we
can extend it to the entire line by setting f (x) E 0 outside of [0,1]. This function can
then be expressed in terms of a wavelet basis for L2(R).

Another approach is to consider

Mac) = 2 Mac +l)X[o,1]($)a (2.37)
152

$1.1M = 12: ¢j.k($ +1)X[0,1]($), (2.38)
62

with j 2 jo Z O and k E 11-, where 43 and 1b are scaling and wavelet functions with
support g [0, 2p— 1], jo is chosen such that 2].0 2 2p— 1, and I J- is the same as before.
Clearly for p = l, (15“ = d) and d)" = 11). Hence it suffices to discuss the cases for
p > 1. By the construction 4);),(0) = 2,62 ¢j,k(l) = 2,62 ¢j,k(1 +1) = 31,,(1) and

;,k(0) = 2121,,(1). So we can easily extend such functions to the entire line as periodic

functions. We call them periodized wavelets.
Now define
V" = span{¢3,k|k E 1]}. (2.39)

.7

We claim that

16

Theorem 2.3.1 Forj Z jo, {¢;,k}k61, forms an orthonormal basis of V].

Proof. Let k,k’ E 13-. Then

in which we have used

/1¢j,(kx ¢j,(k’$
[0 (2:53.143? + Z) (Z ¢j,k'($ + l’)) (133

IEZ 1’62
22/ m. ()x+t )¢,,'d,..(.+z).
IEZI’EZ
22/, ¢..(’1y+l—t)¢..ydy.:()
IEZI’EZ

Z j,” (Z ¢j.k(y +1 — l’)) ¢j.k'(y)dy
1’62 162
Z/ll-H (12¢ij y+l))¢1k’()

(’62 162
ZZv/Im ¢j,()ky+l¢j,ydyk’()
lEZI’EZ

E]: $ij y+l)¢jk’(y )dy

lEZ
Z(¢j,k—2m WM)
162

0 Hk#H,
{1 szk,

«busty +1) = 22¢(2’1y +1) — k)—- - 22¢(2y —(k— 221)) = (251.1421.

This means that {$131,611. constitutes an orthonormal basis of Vn“. #

For later applications, we need to observe some properties of (153$ and 1113*. First

we denote

suppcblk = 11.k(¢’), 811121212311. = war).

and define

It is easy to check that

={k|0§kg2i—2p+1},

. . (2.40)
Sf={k|2J—2p+1<kg22 —1}.

17

Proposition 2.3.1

k e S} 4:) 1),). C [0,1],

(2.41)
ICES]R 4:161?) and0¢1j,k,

where [M = suppgbik = [2'jk, (2p — 1 + k)2‘j].

A careful investigation now leads to explicit expressions for $32k and $3,: respec-

tively.
Proposition 2.3.2

$31,. = 45.... 11);. = 4.... ifk e S}, (2.42)
and

963,100): 921.1(3) + 45,-).(1; +1),

2.43
¢i.k($) = $1405) + $1.143 +1) 1fit 6 5,3- ( )

Proof. We need only to show that (2.42) and (2.43) for qblk, since proofs for $39,): are
exactly the same as those for (1533*. Let k 6 S}. Then suppquJ-Jc = I 3;. C [0,1]. Therefore
for any 1 ¢ 0 and a: 6 [0,1], we have x+l ¢ [0,1] and hence x+l ¢ I”. It follows that
¢j,k($+l) E O for a: 6 [0,1] and l 75 0. Consequently 9133,,‘(23 = Zz¢j,k($+1) = ¢j,k($).

To prove (2.43), let I: E 533. Then k > 2j—2p+l > 0. Hence, forl < 0, x+l < 0 <
2"jk for every :1: 6 [0,1], which implies ¢j’k($ +1) 5 0 ifl < 0. Ifl Z 2, then a: +1 2 2
for a: 6 [0,1]. However, (2p — 1+ k)2"j = (2p — l)/2j + k/2j < 1+1 = 2. It turns
out that :1: +1 ¢ 1,3,. for l 2 2 Va: 6 [0,1], that is, ¢j,k(:r +1) E 0 for a: E [0,1],l Z 2.
Since ¢j,k(a: +1) gé 0 on [0,1], we therefore arrive at (b3,k(:r) = ¢j,k($) + ¢j,k(a: +1). #

As consequences of Proposition 2.3.2, we have the following.
Proposition 2.3.3
Ij.k(¢i) = [MU/2*) 2 [£17 (2.44)
1;), = 1,), ifk e S}, (2.45)
= [1.2—4,11.) 10.(2p — 1+ 102-2 — 11
= (1,). m [0,1]) u (1,-1-2.- 0 [0,1]) ifk e Sf. (2.46)

18

Next we define
W; = span{z/J;’k}kelj. (2.47)

The proof of the following theorem is nearly identical to that of Theorem 2.3.1.

Theorem 2.3.2 Forj Z jo, {ibis/chef, constitutes an orthonormal basis of W3“.

Since
2p—l 1 2P—1
¢j,k($) _ —23/2 2 d r1(¢22(2j$ "' M)’ n =\/§ Z dn¢j+1,2k+n($ )
11:0 11:0
1 22‘
z1).)",(k3'572-‘(X)ind—n-Jrlq5.7'.2}c-{-71(:l7)3
fn:0
we have

¢§,k($) = 245,-).(2—1)

leZ
2p-l
= 22/22 2d 3(2 2(2J‘( (:r—l)— k)—n)
IEZ 71:0
_1_ 22‘
= J22?) dn 2451'“, 2k+n($ _1) (2.48)
n_0 162
1 212-1
= Cl” 71(‘1: )°
7 “2:; j+l, 2k+

Notice that k E Ij does not imply 2k + n E 11-“ for n 6 {0,1, - . ~ ,2p — 1}. However,
if 2k + n > 2J-+1 — 1, then 2k + n — 2i 3 2"+1 — 1. Hence by the definition of the

periodized wavelets

¢;+l,2k+n—21 ((13) = (Z ¢j+l,k+n—2J($ + 1)) X[o'1]($)

IEZ
: (Z ¢J+l,k+n($ + 1+ 1)) X[O1I]($)
IEZ

: ¢i+1,2k+n($)-

Hence ¢§+1,2k+n($) E 1611(9) and therefore $33,}: 6 1631(0).

Similarly
_1_ 22’

$331417 T2? 1rid-n+1 j+l,2k+n(x) (2'49)

\/—n=0

19

implies $32k E V111.
Consequences of Theorems 2.3.1 and 2.3.2 and the scaling relations (2.48), (2.49)

are:
* t
Vn C Viz-+1,

V1:+l 3 V7: U W1:

(2.50)

The same as for nonperiodized wavelets, we have

Proposition 2.3.4 For n 2 jo,
i) V’: C v;+1,
ii) W; J. Vn“,
iii) Vn"‘+1 = V; 63 W;

Proof: We need only to prove ii) and iii).
Let k,k’ E In, Then
fl ¢n,( k 11bn, k'( (Edit)

= 22/: ¢.,()kx+l¢.,xkl( +z')dx

IEZI’EZ
= 22f“ ¢.,(ky+I—l')¢.,ydyki()
lEZl’EZ
l'+1
: Z] (Z¢n,(ky+1_l))¢n,ydyk’()
I’EZ IEZ
l’ 1
= Z] i (Z¢..y+1)w.kl()
(’62 16Z
= 22/7“ <f>nk 31+ )¢nk'(y)dy
IEZI’EZ
= Z]: ¢n,(),ydky+l¢nk’()y
162
: Z(¢n,k—2"la¢n,k’)
162
= 0

It follows that {¢;,k}k61n _L {wz‘k1}k’eln. That is W; _L Vn“.
To prove iii), we notice that dimV,',"+1 : 2"+1 = 2" + 2” =dimV,f+dimW,: and
W; .1. VJ. Hence V,:‘+1 : V; 63 W5. #

The main result of the periodized wavelet basis is as follows.

20

Theorem 2.3.3 {¢;k};210_1 forms an orthonormal basis of L2(0,1), that is,
’ he!)

11,14:— — jge 69W ,;+,,_ — L2(0,1), (2.51)

n:0

where ¢;0_1,k 2 95:0,). for k 6 13-0.

Proof. Let f E L2(0,1). We extend f to the entire line by

m) = { flat) as e (0.1). (2.52)
0 :16 ¢ (0,1).

Then f“3 6 L2(R). Since {¢j,k}j2jo_1’kez(¢jo_1’k = ¢jo,k) is an orthonormal basis of
L2(R), we have f" : 21-21-04 zkez(fe,¢j’k)¢lj’k. Letj Z jo. We define Sf : {k 6
Ile 631;”,c and 1 ¢ 1]- k} and let 5" — Si”). Then we have

Jo-l—

f”: 2 [2(fea¢},k)¢},k+2(f Mime (n+2 A011: 1421:, (2-53)

ijo-1 1:63} kesf IcesR
where 5'3 and 5‘ are the same as in (2.40) for j Z jo. For j : jo — 1 we define

SR = S” and 5’

Jo_ _1 Jo_ ——1 - 5310. By a simple computation and Proposition 2.3.2 we have
156wa —2p+2gkg—1,
keS}<=>0_<_k_<_2i—2p+1, (2.54)
Ices}? ¢=> 2j—2p+2Sk_<_2j—1.

Hence 511’ and 513 have the same number of elements, namely 2p— 2 elements. For any
h E 5?, $13,451: +1) : 2j/2¢(2j(:c +1)— k) : 2j/2¢(2ja: — (k — 21)) = 1PM-” (2:). Since
2i—2p+2 g k g Zj—1,—2p+2 3 15—215 —1. So we havek e Sf ¢=> k—2J' e Sf.
With this relation if we periodize both sides of (2.53) and use the fact that

thin—21(33): (:1: ¢j,k—2J($ + 1))(X[o 11(1'3 )=(Z( ¢j( k3? $+ l + 1)) MO 11(37 37): $3,]:

then

f"(=v) = (Z f‘3(n +1))X[0,1]($)

IeZ

= (2 [EU-faith” Z¢1k($+1)+ 2(fe’z’b-7’k) 21:45“ (3+1)

ijo-1 1:68! keSL

21

keSR

+ 2 (“if $15k);¢2k($+l)])X[0,1](~’F)

P

= Z 2(fe’¢;,k)¢;,k+2(f ahbjk) @bM‘l‘Z 5%,)”11’1]

ijo-l L_kESJ’ kESJL kESR

= Z Z (fe,¢},k)¢;,k + Z (fca¢j,k_2:)¢§,k_2: + (fc,t/)j.k)¢},k]

J'Zio—l Ices} kESJR

 

: Z 2 (f6, ¢;k)w;k + 2 (f6, ¢j,k—2J + ¢j,k)1/’;k] -

ijo-1 1:68} 1:6sz

since for k E SE, ibj,k_2,(x) : i/2j,k(a: + 1), 1,0“ + gig-$-21 = wzk. We finally arrive at

f(l')=fe($)= Z Z (fa ;,k)¢;,k- (255)

1215-1 kesfusf

Remark: The periodized wavelets preserve almost all the desired properties of
the original wavelets except the vanishing moments. To see this, let p Z 2 and k E 313.

Then fol ¢;,kd$ : 0. However,
1 1 1
f0 :czbzkda: = A xibj,k(:r)da: +/0 xibj,k_2,(:r)da:
. , 21—1: , , 21+1—k
23/2 [2" [k (a: + k)2"2b(:c)d:c +2"J/, k
_ 21-
. 21+1—k .
2'3”2 []k (at + k)z/)(:r)d:c — 2”]
- 2

_ _2—5j/2/2J+1—k¢(x)dx 75 0
_ 2 ')

J—k

(:r + k — 2j)2—j¢(m)dx]
21+1-k

wedge]

J—k

since suppd) : [0,2p — 1] and 2i — (2i — 1) : 1.
To construct wavelet bases for L2([a,b], [c,d]), we can use the methods already

discussed above and the tensor product in Section 2.2.

22

2.4 Construction of Weighted Wavelet Basis for
L2..(a,b)

First we denote

I“. : [a + 2’jk(b — a),a + 2"j(k +1)(b — a)], [0,0 : 1-1.0 : [a, b],
1;). = [a + 2-1‘ k(b —— a), a + 2-i(k + 2-1)(b —- 5)],
57), = [a + 2‘J'(k + 2‘1)(b — a)), a + 2"J'(k +1)(b — (1)],

where -1 S j and 0 S k S 2i — 1. It follows from the definition above that

I“, : I 11,]: U I J" k and [131: Q [a,b] for any j and I: under consideration. Now we define

3—1 "’(Ii'k) % wU’.) %
h,,k(a:)=w2(I,-,k) (mfg) XI;.,($)-(w—(1j:3) X1;_k($) .

h—1.0(~”17) = X[0.1]($)w(1—1,0)- ,

(2.56)

NIF‘

where w E L°°(a,b),w(a:) > c > 0 a.e., and w(E) = wadx. Let L?”(a,b) :

{f= llfllw < 00}, where < f,g >w= fffgwdw, llfllw = \/< 7,7 >w- Hence Lamb)
is a weighted L2 space with weight w. It is easy to check that

1, .f .,k : .3 3
<hj,k,h.,m >w={ 1” ) (. m) (2.57)

0, otherwise.

Thus {hj,k} is an orthonormal set of L30(a,b). The most important result we shall

use later about hi). is the following.
Theorem 2.4.1 {hj.k} constitutes an orthonormal basis of Li(a,b).
To prove this theorem we need the following lemmas.

Lemma 2.4.1 Let
(M) = mourn-1 f, f w dt. (258)
Then
(24% + Auk — AIM) f =< f, hi). >,,, h,,,.. (2.59)

23

Proof. It follows from (2.56) and (2.59) that

(141;, +11; -A1,)(f)

= 2.1211) In, 110111.21. w< >11,,,fwdt
—XI,,,.w w([j.k) lz,_,,fwdt

= X1}, {w(1},.)-1 11;, 1w dt — w(1..1)-11,,, f w 11}
+24, {w( },.)-l 11;, fw dt — w(1}..)-111,, f w 11}

= x.;,,w<1.-,k)-lw(1},.rl {11050111, fw «11- w(1},.)11,,, f w 11}
+x.;,w(1.-..)-1w( gin-1 {110.611,}; fwdt — w(1;,.)1,,,fwdt}

1 11
-1 w(1‘- ) w“ 1
-w(15,k) ’{(:w(i;,:)) X'h‘ w(w(1,l',:))(5 Trill"

: < f, hj‘k >w hj’k.

#

Lemma 2.4.2 Suppose that f E L2w(a,b). Then Zj__122301 < f, hJ-Jc >u, h“, con-

verges to f pointwise a.e..

Proof. To prove this lemma, we define the “conditional expectation” operator 8,, by

2"-1
= Z Az.,.(f). (2.60)
k:O
We also define
Dn = n+1 —' an. (2.61)
Then by Lemma 2.4.1 and (2.5)
2n—1
Dn(f) = Z < Na... >.. h... (2.62)
k=0
Notice that
211—1
=12) xi...{w1,,,} f1” fwdt. (2.63)

It follows from the Lebesgue point theorem that En( f) ——) f a.e. pointwise. Thus by
(2.61), (2.62), and (2.63)

20:0 Dn(f) = limN—ioo 5N(f) — 50(f)

(2.64)
= f— X[a,b]w([a,b])—lf[a’b] fwdt a.e.

24

01‘
00

f = Z Z < f,h..,k >.. h... (2.65)

pointwise a.e.. #

Now we are ready to prove Theorem 2.4.1.
Proof of Theorem 2.4.1: We need only to show that {h,-,k} is dense in Li(a,b).
Let f E L3,,(a,b) such that < f, h“, >w= 0 for 0 Sj and 0 S k S 25 —— 1. Then by
Lemma 2.4.2 f = 0 a.e.. This completes the proof of the theorem. #

2.5 Approximation Properties of Wavelet Bases

In their papers, Jaffard and Meyer [45, 46] proved that for Sobolev spaces H ‘( R)
f 6 [13(3) ¢¢ Z I(f,¢’j.k)|2(1 + 22”) < 00 (2-65)
jk

for p sufficiently large. This means that we can define an equivalent norm on H ’(R)

by means of wavelet coefficients

A
2

||f|lH;, = ()3 (1.1.6120 + 2218)) Vf 6 H302). (2.67)
j,k

In what follows, we shall prove that if 45, 211 E H ”((1), then we have following approx-

imation property.

Proposition 2.5.1 Let p be suficiently large and let ()5 and 2b be the corresponding
scaling and wavelet functions. If 6b,zb E H’(Q), then we can find error bounds in the

H” norm form S s.

Proof. Let Pn denote the projection operator onto Vn. Then for m S s

llf-Pnfllfir = (2|(f.t/1j,k)|2(1+22jm))

J)"
kEZ

l

_ _ 2 2js (1+22jm) 5
— (2|(f,z/1.,k)l(1+2 )——(1 +2212)

#62

25

s (1 + 222118-111)? (Z(f.¢...)2(1+ 2215))

M

< 2‘(”+1)("m)||f| H5). (2.68)

 

Jaffard and Meyer also proved that

Theorem 2.5.1 If (Ail! E C” and if («MAJ-24 is an orthonormal basis of L261),
then {¢j,k}j2-1 also forms an unconditional basis of H1(Q), where r > 1 and Q is
a smooth region in R". In addition, for f E H1(Q), the following condition on its
wavelet coeflicients 6ch = (f, 1b“) holds:

C [256,42 g If f3 0’ (216,42 with 0 < C s C" < oo. 2.69
.7 J

This theorem provides the mathematical justification for wavelet—based methods for
boundary value problems discussed in later sections. As a consequence of above

theorem, we have:
Corollary 2.5.1 Forp Z 3, {¢;,k}kel,,j2—1 forms an unconditional basis of Hl(0, 1).

We end this section by establishing estimates on wavelet coefficients of functions

in 11”.
Proposition 2.5.2 Forj 2 0 andp 2 1, let f e H”. Then
leJcl s C2'J’mlflm,z,,., 0 s m s p. (270)

Proof. Since w has p vanishing moments, for any polynomial of degree S m — 1 S

19.1,

 

|c,-,k| = M:(f-€I)I/)j.kdx
/ (f - (Illbmdx

),1:
Hf — QIIO.I,,kll¢j,kl|0
= llf— gll0.1,,k° (2-71)

Notice that |Ij,k| = 2‘j(2p — 1). The Bramble-Hilbert theorem shows that

 

 

|/\

leJcl S iglfllf - qllo.1,,. S 02_j’|f|s.I,-,.-

26

2.6 Wavelet Related Bases for Some Sobolev Spaces
and Their Approximation Properties

In this section we shall construct unconditional bases for H1(fl), H361) and H362)
respectively by using antiderivatives or pseudo-antiderivatives. Also we shall establish

their approximation properties.

2.6.1 Frames and Bases of Some Sobolev Spaces by Antideriva-

tives of Wavelets

Let us consider 0 = (0, 1). Let p Z 1 and let (b and #2 be the corresponding scaling and
wavelet functions. We begin our discussion by introducing the following definitions

and theorems.

Definition 2.6.1 A sequence {qbn} in a Hilbert space H is said to be a frame if there
exist constants A and B > 0 such that for every f E H

Allfllz E El < f, 111.. > I2 S Bllfllz- (2.72)

From the Plancheral theorem we see that every orthonormal basis of H is a frame
with A = 1 = B. It follows from Definition 2.6.1 that W = H, that is, {gbn} is
a “basis” of H in the weak sense, i.e., {cbn} is not necessarily a linearly independent
set. On the other hand W = H does not imply that {451,} is a frame. For our

applications, we need the following more general definition.

Definition 2.6.2 Let Man}? be subset of a Hilbert space H. Let span {an} = V be
the set of all elements chqbn(cn E R) which converges strongly in H. Then {457:} is
said to be a generalized frame of H ifV = H.

Now let us denote dzzi—a f: cbdx for (b E L1(a,b). The following theorem will
provide a solid base for establishing unconditional wavelet related bases of H1(Sl),

H} (9) and H6 (9).

27

Theorem 2.6.1 If {¢,,}:°=l is a generalized frame of L2(a,b), then

{251(6) = [andt — (.7.- — axis... :1: 6 [a,b]} (2.73)
{6;(6) = fawn, :1: 6 [a,b]} (2.74)
{1,<I>,,(:r) = 02(2), :1: E [a,b]} (2.75)

are generalized frames of H6(a,b), H}(a,b) and Hl(a,b) respectively.

Proof. Let us prove the theorem for H}(a, b) first.

We note that (I); E H}(a,b) and that the semi—norm | - II is an equivalent norm
in H}(a,b). Let f E Hf(a,b) and let g = f’. Since g E L2(a,b) there exist {on} C I2
such that N

0 =N1i_r,n,ug— 2cm“: = )ggoflg—givrdx, (2.76)

n=1

where gN = 277:1 cnqbn. Thus if we take fN 2 ZS; cud); then fN E H}(a,b) and

gig, If — f~l1 = gig, Hg - 9~|Io = 0. (277)

Thus we conclude that W = H}(a, b).

Now we show that {(1)2} is a generalized frame of H6(a, b). Obviously, <99, 6
H3(a,b) for n E N. Let f E H&(a,b), and let g = f’. Then ffgdsc = f(b) — f(a) = 0.
Since g E L2(a,b), 3{cn} E I2 such that (2.76) is true. Notice that {m = 22;, and)”,

we obtain

1 b
w = ml. g~(w)d11|
1 b

= all (gN—gldib‘l
l
mllg—QNHO ——) 0

S

 

as N —> 00. Hence
N _
H9 — Z Cn(¢n($) — ¢nlll0 S “9 — gNllo + Vb — al§N| “—1 0
n=1
as N —> 00. Therefore fN = 2L, cn<l>2 satisfies fN E H3(a,b) and
gigolf —le1 = 0-

28

Since I - I; is also an equivalent norm for Hé(a, b), we complete the proof for H6(a, b).

~

Finally, we prove the case for H1(a,b). Let f E H1(a,b). Then f(zc) = f(:r) —
f(a) E Hl(a,b), and f’(a:) = f’(:r) = g(:r). So by the proof for H}(a,b)

N
lf_le1=lf_ch(pnll—‘)O

11:1

as N —> 00. Thus if we take fN = 221:1 c,,<l>n + f(a)-1, then

llf-fNIIO = llf—f(a)-ch<l>nllo
= llf—lelo
S le— lel —‘> 0

asN—>oo,and

lf—fzvll = lf-f(a)-ch‘1’n|1

: lf—le1—>0

as N —+ 00. The reason for the first inequality is that f E H}(a, b) and I ~ I1 is an
equivalent norm of H}(a, b).

Now we complete the proof of the theorem. #

With the above theorem, we now examine the wavelets {¢jo’k,¢’j,kl j 2 jo,2j° 2
2p —1,k E [J’- = {2 — 2p S k S 2i — 1}} with p 21, which form a generalized frame
of L262). It then follows from theorem 2.6.1 that

Theorem 2.6.2
(Dim/.973) = hi ¢jo.kdt - $4202
Wale”) = hi ‘Pjacdt - CPI/3311:
q’jo.k.($) = l: <l>jo,kdt
‘I’j,k.(-’L‘) = his $51.6”
<I>jo_1,o(:r)51,<I>J-o,k(:r)= (150,,“
‘I’j.k($) = ‘I’joJc.

mews), 3'2}... keI} (2.80)

form generalized frames of HMQ), Hf(Q) and H1(Q) respectively.

29

In particular, if p = 1, we have the surprising conclusion.

Theorem 2.6.3 . For the Haar basis {1%)in Z —1,k E 1,} of L2(Q), the corre-
sponding antiderivatives {1, \I'Mlj _>_ —1, k E 11-} constitutes an unconditional basis of

H1(Q), where ‘Iljgc is defined as in (2.79) and ‘11-”) = (1)0,0.

Proof. It follows from Theorem 2.6.2 that {1, \IIM} forms a generalized frame of
H1(Q). Hence to show that it is an unconditional basis of H 1((2), we need only to
show that {1, \II 13k} is a linearly independent set, and satisfies (2.72). Let
C ° 1+ 2: Z Cj,k‘pj,k = 0. (2.81)
jZ-l k6],

Taking derivatives of both sides, we obtain

2 Cj,k1l)j,k = 0.
Since {7/5ij 2 —1, k E 1,} is an orthonormal basis of L262), we conclude that 6ch = 0
for j Z —1,k E Ij. Substituting CM = 0 back in (2.81) leads to c = 0.
Let 1 = ‘11-“). To prove that {WM} 2.2 is an unconditional basis, we also need

to prove that (2.72) holds. First of all, notice that supp‘IIJ-Jc Q suppzbJ-Jc Q [i.k- Hence

I(f w,.)|2= |f1,,($folbjkd8)dxl

3.,(11. lflm)l1. |¢},m|dsdm) 2

:0, ,,fl (l(m> (f§dS)%(fJ|1/1}.k|2ds)idx)

(”If (x)|\/' 1mm) (2.82)
:b2

2111:) Ilfll31,.
4*—-b—||f||o, 1,.
S (1)]: — aj,k)llfllo,l,,k
= 2_jllfll(i,1,,k-

Therefore

23:4 21:61, l(fa‘1’j.k)l2

S 23:-12k61, 2—jllfllli,1,,,,
= 2332-12‘jllfllfin

= 4|lfllbn-

(2.83)

30

Next, if f = 29313113,), + c- 1, then f’ = Egg/13*. So
I(f'.¢j.k)|2 = lcj.kl2 and Ill2 = Z ICj,k|2- (2-84)

Also
I((f, ‘P_2,o)>1 l2 = I(f,1)|2 S llf||3,n- (2-85)

Combining (2.83), (2.84) and (2.85) establishes

Z I(f,‘1’j.k)|2 + Z I(f’, (PM)?
S 4|lf|l3,n + If li,n (2-86)

S 4llfll1,n-

Z l ((f, @2011 |2

Now let us examine the left hand side of (2.72).

llflli = llfllg + lfli‘

= lfzdiv + Z I(f’a¢j.kll2

= I“: Cj.k‘1’j.k)d:r + Z I(f'. 1PM”2

= 23 Cch I f Wilda: + Z ICJ'Jcl2

= ch,k(f,‘1’j,k) + Z ICj,lcl2 (2-87)
.<. (z ch.kl2)‘/2(Z I(f. ‘I’j.1=)|2)‘/2 + 2: lcjmlz

S. E I(f, \11,-,k)|2/2 + Z |Cj.1=|2/2 + Z ICJ'Jcl2

S 3 (Z |(f,‘1’j,1c)|2 + Z I(f’. moi”)

=%2l<(f1\pj.k))1l2'

The relation (2.72) then follows immediately from (2.86) and (2.87). #
It follows from the proof of Theorem 2.6.3 that the following approximation prop-

erty of antiderivatives of the Haar basis functions holds.

Corollary 2.6.1 Let {Wj,k}k€[j be the antiderivatives of the Haar basis functions.
Then for any f E H1(Q),j 2 0, and k E I,-

|(f,‘1’j,k)| S c2‘jllfllo

for some constant c.

31

Remark. From the above theorem we see the advantages of the antiderivatives
of wavelets: They are smoother than the original functions, and make the “useless”
Haar basis very useful in applications. A simple calculation shows that {\II M} ,20 are
the “hat” functions as in finite element methods.

Recall that for p 2 1, {7/1},ka E Ij,j 2 jo — 1,660,), = $324,} is an orthonormal
basis of L262). In particular, for p = 1, {tbs-3,} = {21“,} is the Haar basis of L262).

Hence it is natural to establish the following more general theorem.

Theorem 2.6.4 For p Z 1, let {$35k} be a periodized wavelet basis of L262). Then
{1, ‘1’;k}j2jo-lkelj is an unconditional basis of H162) where 35,,(22 2:): for ”(s sds)

:1: 6 (0,1) and \II;O_ 1),—“(1)330 k with ”bio k— -f6" @530,de-

Proof. The proof is almost the same as that of Theorem 2.6.3. #

With similar proofs we are ready to establish analogous theorems for H662) and
H362)-
Theorem 2.6.5 For p Z 1, let {(121),},21014 be a periodized wavelet basis of L262).
he 1'
The" {wg} k*($ )= f: $33:de” Z jO _ 112]” Z 2}? _11k 6 1]" [Jo = [Jo-11 and ¢fo—l,k =

obj-M} constitutes an unconditional basis for H662).

Proof. Clearly {\Ilj k,
is linearly independent, let 2631\ka = 0. Then ch,kib-

3,1“

} forms a generalized frame of H162). To prove that {‘11} k}
= 0. This again implies
cJ-Jc = 0.

To establish (2.72), the equivalent norm I - I1 of H662) is used. Since {\II;,),.} is
an orthonormal basis of H662) with respect to I - I1, (2.72) is automatically satisfied.

This means that {‘Il;’k.} is not only an unconditional basis of H662), but also an

orthonormal basis of H662). #

Theorem 2.6.6 Forp > 1, let {$332k} be the periodized wavelet basis of L262). Then
{\Il'i‘ok},>,oI—1 \{\Il;'o_10} forms an unconditional basis of H662), where

‘I’joflkml: [if—(1501.613 x¢jmkk61j0

For p = 1, {$353320 forms an orthonormal basis of H662) with respect to the semi-

norm I ° ll.“

32

Before proving Theorem 2.6.6 we shall investigate some approximation properties

of antiderivatives of wavelets considered thus far.

For p Z 1, define

0171(9)

5636(1)}, = [may — mmm 6 me e 13},
span{\I13,k =/0x1/)J,kdy '7 3357.133? E 9’19 6 lb},
span{<l>}?k = [453,111.11 — 637332: e S2,k e 11},
span{‘§[l:",?,c = for l/Jhdyll‘ E 91"? E {J},
565661,,“ = [61,166 6 11,1: e 1}},
span{‘llJ.k.. = [6},kdylm 6 9,1: 6 13},
spams}, = [madam 6 9.1m 6 1m},
spmnm... = [madam e 11.1 e 1.},
span{1,<I>J,k.I:r 6 52,1: 6 [3},

spam... = [0 . 41.1199 6 mm e 13},
span{1,<I>}’k,I:r 6 S2,]: 6 1.1},

span{\ll},k, 2/0 (blkdylx 6 (2,11: 6 IJ}.

(2.88)
(2.89)
(2.90)
(2.91)
(2.92)
(2.93)
(2.94)

(2.95)
(2.96)
(2.97)
(2.98)
(2.99)

Then it is natural to have the following propositions which are similar to those

relating to wavelets as in Section 2.1.

Proposition 2.6.1 Forp Z 1 andJ satisfying 2" Z 4p—4, {9%,} we and {\IlgkhEpJ
1 kEI’J 1

constitutes bases for 017,162) and 0WJ(Q) respectively.

Proof. Let 35‘, S} and S 5% be defined as before.
1) Show that {(13,} .... is a basis of 012(9).
kel’

1
We shall show that {(16.1}161', is a linearly dependent set, but {63),} k¢0 is a
’ hers

linearly independent set.

Let f(as) = 21:61.1, Q3, Then f(0) = f(1) = O and f E 017,].
Claim 1. f(cr) :- 0, that is, {(Eakhepj is a linearly dependent set.

33

NOLlCC that k E Sf ¢==> k — 2J 6 $52,453,]: = ¢JJ¢ + éJ,k_2J and Elk + Elk—2’ =

$1,), = 2"”2 for k 6 5?, hence we have

f’($)

2: (95M — 5J3.)

k6];

Z ($1.1m — 2—1/2) + 2 (451,11 — 331,11) + 2 (95M - $33.1)
kesj Hes; was?

2 (953,1: — 24/2) "I“ Z {(QJJc + QbJJc—ZJ) — (3J9. + aJ,k—2J)}

kES§ kesf

2 (m3). — 2"“) + 2 (m3... -— 2‘“)

1:655 1:6352

2 (453,1: — 24/2)-

kelg

Therefore for m E I 3

(fl? $3,"; — 2_J/2) : Z (¢3,k — 2—J/2’ ¢3,m _ 2_J/2)

kelg

: 26153.1.) 953,...) — Til/2E.“ _ 24/224572". + 24/2)
kerg

=1—Z2-J
kelg
=1—1=0.

It follows from the above that f’ _L H =span{¢}’m —2‘J/2Im E 1",} with respect to the
inner product (--.,) Therefore f’ E 0. This implies f(x) :— c. But f(0) = f(1) = 0,
thus c = 0.

Claim 2. {Q9 k} kaéo is a linearly independent set.
’ kel’

Let 2,5,5 CMQSJ, = 0. We shall show that cu, = 0 for k E 1",. Taking derivatives
k 0

¢
of both sides of the above equality, we obtain

(3r

Hence

2 CJ,k(¢>J,k — $19.) = 0 a

I
kEIJ
k¢0

Z CJ,k¢>J,k = Z 6.1.1.521: -

I I
keIJ keIJ
k¢o k¢o

—/0-1( 2 CJ,k¢J,k)¢J,od:L‘ = [01(2 CJ,k$J’k)¢J’Od$ ,

ks]; kels
k¢o k¢o

34

()1

Z CJ,k [:0 ¢J,k¢J,od-’B = (2 0.1231024” -

he]; he];
k¢o k¢o

Since fie? ¢J’k¢‘]'0d$ = 0 for k 75 0, the above equality gives us

( Z CJ.k$J,k)2—J/2 = 0

kelg
k¢o
()I
Z CJ’kaJ’k = 0.
k6];
k¢o
Therefore
2 Cj,k¢J’k($) = 0 for IL‘ 6 [0,1]. (2.100)
kg];
k¢o

FormGS} andm¢0
1
f0 ( Z CJ,k¢J,k)¢J,mdiv = 0

I
keIJ
k¢o

implies cf", 2 0. Thus (2.100) becomes

2 6.154511. + Z CJ,k'¢J,k1 = 0

hes? H655
()r
E (CJJccfiJJC + CJ’k_2J¢J’k_2J) = 0. (2.101)
keSf

Periodizing both sides of (2.101) leads
{ Z (CJJ: Z ¢J,k($ +1) + CJ’k_2J Z ¢J,k_2J(IB + 1)) } X[0,1]($) = 0
1:15:552 IeZ leZ

This is the same as

Z (cJ,,,¢f,, + c,,,,_2.¢3,_,_,) = 0. (2.102)
keSf

Since (bi), = ¢}’k_21, (2.102) becomes

2: (CJJc + CJ,k—2J)¢},k = 0- (2-103)

keSf

35

Multiplying both sides of (2.103) by (153m for m E S? and then integrating from 0 to

1 we obtain

CJ,m : _CJ,m—2J'

Hence (2.101) is equivalent to

Z Clix-$2655) = Z CJ,k¢J,k—2J(CC).

Let
IL : [0’1]n(UkES§IJ,k—2J)m
IR = [0111 O (UkeslflJJc).
Since 2J Z 4P — 41 IL 0 IR = 0,“: turns out
Zkesf 611.491.1013) = 0
Zkesf CJ,k¢J,k—2J(:c) = 0

Hence for m E 55?

1
Z CJ,k /0 4511953,... = 0,

keSf

1
Z CJJc/O ¢J,k—2J¢},m = 0.

kesf

Adding the above two equalities and noting that 051,], + ¢J,k_2.l = 051,, results

1
Z c}, [0 63,453,, = 0 for m e 55'.

keSf

for :1: E [0,1].

(2.104)

(2.105)

(2.106)

(2.107)

This implies cf", 2 0 for m E 5?. Therefore we complete the proof for Claim 2.

Consequently, {Q3336 is a basis of 01762).
2) Show that {\Ilg'khepjjis a basis if 0WJ.
Set

2 CJJc‘IIS'k = 0.

k6];

Taking derivatives of both sides of (2.108) gives

2 611.110,]. = Z GMT/711..

kel; 161;

36

(2.108)

(2.109)

If m E 35, one has at": = 0. Multiplying both sides of (2.109) by 2b,,” and then

integrating from 0 to 1 one arrives at cf", = 0 for m E 55. Therefore (2.109) becomes

2 (CJch’JJc + CJ,k—2J¢J,k—2J) = Z (CHEM + CJJc—ZJJJJc—ZJ)‘

kesf kesf

(2.110)

Taking inner products of both sides of (2.110) with respect to 05,10 and noting that
m = fol ¢J,o = 2‘J/2 and f6 wJ,k¢J,odat 2 f6 ¢J’k_2J¢J’0d$ = O for k E 552, one

obtains
Z (Wk-1;”: + CJ,k—2Jll'-J,k—2J) 2-J/2 Z 01 (2-111)
keSf
i.e.,
Z: (CJ,kK—l)-J,k + CJ,k_2J$J'k_2J) = 0. (2.112)
keSf
So
2 (CJJchJJc + CJ,k—2“/’J,k-2J) = 0- (2-113)
keSf
Periodizing (2.113) we have
2 (Cu: + CJ,k—2J)¢3,k = 0, (2.114)
keSf
in which we have used the result: (bi), = ([23,,‘41.
Now (2.114) implies CJJc = —CJ,k_2J for k E 5",}. Thus one can rewrite (2.113) as

Z CJ.k(’l/)J,k — zl’J,k—2J) = 0.

keSf

Notice that [L 0 IR = 0. Hence

2 CJMbJJ. = 0. and Z CJ,k¢J,k—2J = 0-

kesf kesf

Therefore, for m E S?

1
Z Cum/0 ¢J,k¢},md$ = 0,

keSf

1
Z elk/0 ¢J.k—2Jll)3,md$ = 0.

keSf

37

Adding the above two equalities results

1
Z CJJc/O $3,123,}ndrv = 0-

165?

This implies CJ.m = 0 for m E 5"}. #

Proposition 2.6.2 For p 2 1 and J 2 j0(Zj° 2 2p — 1), {Q}?k}::lo and {Wither}

form bases of 017} and 01717} respectively.

Proof. 1) Show that {(1)30k}:¢10 is a basis of 017;.
. E J
First, we show that {Qficheh is linearly dependent. Let
f = 2 Q3?!”
kEIJ

then
f’ = Elms... —2-’/2) = 2 m3}. 4’”.

kEIJ kEIJ

By the same argument as in the proof of Proposition 2.6.1, we have
f' _L H = span{qb},,c — 2"J/2Ik E IJ}.

Thus f’(m) E 0. Hence f(x) = c. Since f(0) = 0 = f(1), c = 0.

Now we prove that {Qfl} f”? is linearly independent.
. e ,

Set
2 CJvk¢33c : 01
her]
k¢0
we have
2: 611.053,), = Z (Mk-42:11.-
kGIJ kEIJ
k¢o k¢o
Thus
1 J 1
2 01$] ¢3.k¢3,0d$ = ( Z Cum—7] 653,061.13,
kEIJ 0 keIJ 0
k¢o k¢o
i.e.,

0 = ( Z cJ,,,)2-%2-%,

kEIJ
k¢0

38

(2.115)

i.e.,

E: 6,1,;c = 0.

kEIJ
k¢o

Hence (2.115) becomes
2 Q1453, = 0. (2.116)

RGIJ
k¢0

Taking inner products of both sides of (2.116) with respect to 433,", for m E I] and
m 75 0, we arrive at cl", = 0 for m # 0,m E 11, that is, {Q}?k}ku:IoJ is linearly
independent.
2) Show that {01333161, is a basis of 01/17}.
It is sufficient to show that if
2 c1111}, = 0, (2.117)
kEIJ
then cm, 2 0 for all k E IJ. So we have
2 61,163,, = 0. (2.118)
kEIJ

Since {1&2},th is an orthonormal basis of WJ, we have of), = 0 for k E [1. #

Proposition 2.6.3 Forp 2 1 and 2" 2 4p—4, {QJ,k.}k6pJ and {\IIJ,k.}kEpJ constitute

bases of .17; and .1717.) respectively.

Proof. 1) Show that {QJ,),,..}kEpJ is a basis of .17.].
Set

2 CJJCQJJn : 0-
kEIf,

Then we have

0 = Z CJ,k¢J,k = 2 611.9511: + Z (CJ,k1<J5J,k' + CJ,k'—2J¢J,k'—2J)- (2-119)

“’3 1:635 MES}2

FormE S}

1 1 1
0 = 2 6.1,]: f0 ¢J,k¢J,md$ + Z (CJJC’ f0 QJ,k'¢J,md$ +CJ,k'—2J f0 ¢J,k'—2J¢J,md$) 1

1:655 k'esj‘
0 = Cfm.

39

Thus (2.119) becomes
2 (CJ.k¢J,k + CJ,k—2J(/>J,k—2J) = 0- (2.120)
keSf
Now the exact same argument as in the proof of Proposition 2.6.1 shows that CM =
CJ,k_2J = 0 for all k E Sf}. Hence {<I>J,k.}kepj is a basis of .17].

2) Show that {‘Iluflhepj forms a basis of .1717}.

Let
Z CJ,k‘I’J,k. = 0-
kezg
Then
2 CJMZ’JJ: = 0. (2.121)
kel}

For any m E Sf,

1
0 = 2 61,1: ]0 ¢J,k1/)J,mdir

kelg
= elm.

Hence (2.121) becomes

2 (CJJc'vbJJc + CJ.k—2J¢J,k_21) = O.

keS§
Again the same argument as in the proof of Proposition 2.6.1 shows that CM =

CJ,k_2J = 0 for all k E Sf}. It follows that {‘IJJ,k.}kepJ is a basis of .147]. #

Proposition 2.6.4 For p Z 1 and J 2 jo, {11¢3,k*}k611 and {\Ilihhelj form or-

thonormal bases of .17; and .W} with respect to ((-, -))1,g respectively.
Proof. Let k,m E IJ. Then
(( 3:)“, ¢;,m*))lvfl : (¢;,Ikt$ Q:ilfmax) = (¢;,k3 ¢;,m) : 6km;

and

(( 21:11:? \p;’m*))lafl = (¢;,ki¢;,m) : (Sk’m.

40

Proposition 2.6.5 For p Z 1 and 2J 2 4p — 4, {1,¢J,k*}kel; and {\IIJ,k.}kEpJ con-
stitute bases of 117.1(0) and 1WJ(Q) respectively.

Proof. Since 1W] = .1371 as in Proposition 2.6.3, we need only to prove the above

proposition for 1 l7].

Set
C ' 1 + Z C.I,Ic(I)J,knl = 0,
kel}
then
2 CJ,k<I5J,k. = 0-
her;

The proof of Proposition 2.6.3 shows that cu. = 0 for all k E 1",. So c + 0 = 0, thus
c = 0. #

Proposition 2.6.6 Forp 2 1 andJ Z jo(23 Z 2p—1), {1,<I>},k*}kelj and{\I'§,k,}k61J
form bases of 1Vj(fl) and IWflQ) = iii/3(9) respectively.

Proof. It is the consequence of Proposition 2.6.4. #

Now we are ready to establish the following.

Proposition 2.6.7 For p Z 1 and 2" 2 4p — 4, we have

017.“ = 0% + OWJ, (2.122)
..VJ+1 = .17.] + .WJ, (2.123)
IVJ+1 = 117.] + 1W], (2.124)

where “+” stands for the direct sum.

Proof. Since the proofs of (2.122)—(2.124) are similar, we shall prove (2.122) only.

Claim 1. {(9%} no U {\IIS k'}k'613 is linearly independent.
kelf] ’

In fact,
0 0
Z cJJc‘I’JJc + Z dJ.k"I’J.k' = 0
k¢0 klell
tel; J

41

implies

Z 0.1145“ + Z dJ,k'1/)J,k' = 261131,]. + dJJc—IZJJC) (2-125)

k¢o [yep] um

I I
kEIJ k6]!

since 212m = 0. Hence

1 1
Z 6.1,}. f0 ¢J,k¢J,odIr + Z dJ,k' f0 ¢J,k'¢J,od$

k¢o k’ I’
he]; 6 J

_ _ 1
2(CJ.k¢J,k+dJ,k¢J,k) [0 ¢J,odx.

ago

I
keIJ

Since fol ¢J.k¢J,0d$ = fol ¢J,m¢J,oda: : 0 for all m 75 0, k,m E I", and (,1qude = 2-1/2,
we have

2 (CHEM + dJ,k—J;J,k) = 0.

Ho
kEIG
Therefore (2.125) becomes
2 CJ,kQ5J,k + Z dJ,kI'</JJ,kI) = 0. (2.126)
k¢o k161i]

I
kEIJ

For m 75 0,m E 55 and m’ E 5'5, we then obtain

1 1
Z 61,]: ]0 ¢J,k¢J,deE + 2 (1.1.1:! [0 ¢J,k'¢J,md$ = 0,

kgo k’ I!
keI’J E J
1 1
2 61,1: / ¢J,k¢J,m'dIL‘ + 2 due / ¢J,k'¢J,m'd$ = 0,
k¢0 0 klel! 0
k6], J

J

that is, cJJn = 0 and dJJnI = 0 for m aé 0,m E I", and m’ E 1",. It turns out

2 (CJ,k¢J,k + CJ,k_2J¢J,k_2J + dJJci/JJJc + dJ’k_2J¢J,k_2./) = 0.
hes?

Since 2" 2 4p — 4, IL 0 [R = 6. Thus we actually have

Z (6mm + aim) = 0. (2.127)
kesf
Z (CJ,k—2J¢J,k—2J + dJ,k—2J¢J,k—2J) = 0- (2-128)
hes?

42

Periodizing (2.127) and (2.128) respectively results

2 (CJ.k¢3,k + dJJclL'lk) = 0,
kesf

Z (CJak_2J¢3,k—2J + dJ.k—2J¢3,k-2J) = 0.
keSf

Notice that (123 k_2J = 963,1: and whq, = wik for k E 55*, we then obtain

2: (CJ,k¢3,k + CUM/13,1.) = 01 (2-129)
keSf
Z (CJ,k—2J¢},k + dJ,k—2Jbb3,k) = 0- (2-130)
keSf

Adding (2.129) and (2.130) gives

2 {(CJJ: + CJ,k-2J)¢},k + (dJJc + (111.4015)} = 0 (2.131)

keSf

Multiplying both sides of (2.131) by (film and 2,123.", respectively for m E 552 and then

integrating from 0 to 1, we have
CJJTI + CJ,m—2J : 0, dJ,m + dJ,m—2J : 0'

So it is sufficient to show that c“, = d“, = 0 for k E 552. This is true because

I
0 = ]0 { Z (CJ,k¢},k + dud/’31)} Q51";

keSf

1 1
= )2 {ka / (savanna. / v3,.¢3,mdx}

keSf
: CJ,m

for m E 5"}. Similarly dJM = O for m E 5"}. Now we complete the proof of Claim 1.
Claim 2. 01714.1 3 017,] + OWJ.
Let u E 017.] + ()Wj. Then u can be written as

0 O
u = Z: CJJc‘PJJC'i' Z eJ’k’WJ’kI

k¢0 I I
keI’J k 61"
a: _ 1‘ _
= Z CJJ: (/ ¢J,kd77 — $21.33) + Z 6M! (/ $214177 — 1121,13)
k¢o 0 k'EI' 0
kelg J

43

PM

1 2P:

Z CJJc dn(f (PJ-H, 2k+nd77_ ¢J+1 2k+n$ (I?)

k¥0 [71:0
her;
3 —

+ 2: m— fifl— 1)"d_.+1 (/ wJ+1,2k+ndn—w...,2k+.x) (2.132)

klel’ 0

219—1
= £2 E CJkdnq>J+12k+n 7152 Z eJk’d-n' '(+1 ‘1”) \pJ+l,2k’+n’
”‘0 n:0\/§k'61’n'=2—2p

I
fikelj

Notice that for k 75 0,k,k’ E 13, n E {0,1,---,2p- 1} and n’ E {2 — 2p,---,1},
if 21: + n > 2"+1 — 1 or 219’ + n’ < 2 — 2p, then supquJHan 0 [0,1] = (b and
supp¢J+mkv+nI F) [0,1] = 0. Therefore (1)3 +1..” M E 0 E ‘1’3+1,2k'+n'- In other words, if
2k+n,2k’+n’ ¢ 13“, then ¢3+L2k+n E O E \IIgHflkIHI. This implies that u E 09]“,
i.e., 017.)“ D 017,) + 0W1.

Claim 3. 01711.1 2 0171 + OWJ.

It follows from Proposition 2.6.1 that dim 0171,11 = 2J+1 — 1 = (2" — 1) + 2" =dim
0171+dim OWJ- #

Proposition 2.6.8 For p Z 1 and J 2 j0(23 2 2p — 1), we have

017;... = 017; +027}, (2.133)
.17]+1 = .17} EB .1717} with respect to ((-,-))1,9, (2.134)
117;... = .17; +127}. (2.135)

Proof. Since the proofs of (2.133)-(2.135) are similar to the proof of Proposition
2.6.7, we shall prove (2.134) only.

Claim 1. .17} _L .Wj with respect to ((o, -))1,9.

In fact, if k E I’J and m E 1",, then

(( 2k!) w;,mt))119 : (¢3,kt3¢;,mt) = 0'
Claim 2. 37;“ 3 .17; 69.67;.
Let u E .17] EB .1717}. Then there exists {cJ,k, 6,],ka E 1]} such that

u = Z(CJ,k‘I’},k.+dJ,k‘I’3,k*)

k6];
2p—l l
= 27‘ CJJ¢ Z dn¢3+1,2k+n + 6.1,); 2 (‘1)n¢3+1,2k+n -
1:6ij n=0 n=2—2p

44

Sure

F1
WEHE
Theo;
theon
isnot

N1
So{¢

ht

lhhi

fince

There
bans

lb

The C1
01‘ e.

Since ¢3+1,2k+n = (,bDH’zHMzJ = ¢3+l,2k+n+2-” we have 2k + n — 2J E [1+1 or 2k +
n + 2" E [1+1 for 2k +n ¢ IJ+1. It then follows that u E .VjH. Hence Claim 2 holds.

Claim 3. dimJ7j+1 =dim.l7f+dim.W].

In fact, dim.i7;+, = (11,4 = 21+l = 21 + 2’ =dim.17;+dim.W;.

The theorem now follows from Claim 1-Claim 3. #

We are then ready to prove Theorem 2.6.6.

Proof of Theorem 2.6.6:

For p = 1,1113?) = ‘pj.ka¢;,k = quJ, are Haar basis functions. In this case jo = 0. So
we need to prove that {@311} 13.2210, is an unconditional basis of H661). It follows from
Theorem 2.6.2 that 811333th is a generalized frame of H661). Hence to prove the
theorem we need only to show that {\Ilfi} 2-1 is a linear dependent set but {111:1} jzo
is not, and (2.72) is true.

Notice that ab 2 60,0 2 1, :1: E [0,1), hence $101,061) 2 f6” gbopds—xd—Jop = x—z = 0.
So {\II'J'fSC}J-Z_1, is linearly dependent. To see that {@3196th is linearly independent,
let

2: 01.11133. = 0
This implies

E 0111/51 = 0
Since {$33k} is linearly independent, OJ"): = 0.

To establish (2.72) we use the equivalent norm | - ll of H661).

Z l((f, 23%))62 = Z I(f’,t/)J'.zc)|2 = llf’llon = lflm

Therefore {\Ilfi} is not only an unconditional basis for H661) but also an orthonormal
basis for H661) with respect to the equivalent norm I - I13.

For p > 1, it follows from Theorem 2.6.2 that

 

H661) = span{<I>;:’k,\II;-‘gn|j _>_ jo,k E IJ-o,m E 1,}.

The consequence of Proposition 2.6.1-2.6.8 now gives that {@fik, \Ilfinlj _>_ jo,k ¢
0, k E I jmm E Ij} is linearly independent, and

 

H661) = span{<I>3-‘£k,\ll;gn|j 2 j0,k # 0,1: E Ijo,m E 1,}

45

 

= 0173,; + (+310 0117501,). (2.136)

To establish (2.72) we use the equivalent norm | - ll of H661). We notice that
(fl, 1) 1" fol f'dx = 0a and $30,), = 2_j°/2. Hence

2 |((f.‘1'}-‘,‘3.))1|2

JZJo-l
= Z I(f’,¢}‘,k)|2+ Z |(f’.¢},,k—<E},,k)|2
jZJo-l HO
“6’10
= Z I(fzwzvr + 23 mag...) — (f',1)2"'°/2|2 (2.137)
j>jo k¢0
" helm
= Z I(f’,¢;,,.)l"’+ Z |(f',¢;,,,.)|2 g mm.
JZJo-l :50

Since \II’O =<I>‘0

~11: ' 1110 ° ' ~11-
Jo_1 Jo 0 E Ovjo and Since {(pjo’k}kk€130 IS a baSIS of 0V]-

#0
q>Jo.— 0“ Z CJO k (1)10 k

k¢0
“5110
Hence
11: _Lo. ,, ...-1Q
(1510.0 — 2 2 — Z Cjo.k(¢jo,k — 2 2 )1
##0
“€110
i.e.,
_ ' -111
¢j00: ZCJOkW jok k—2 lQ)+2 2
kgo
“6110
Thus

I(f',¢}o,o)|2 = I: Cjo.kl(f’a¢;o,k)_(f,12—%)l+(f,32_%)l2

Ho
“6110
= I Z Cjo.k(f’v¢;o,k)l2
k¢o
1:510
S (002201001): I(f io.k)|
k¢o k¢o
“6110 "6110
A
= A Z «132...»? (2138)
Ho
"EIJo

46

It follows from (2.137) and (2.138) that

lflm = Z I(f’,¢},,k)|2+ Z I(f',ib}'-',m)l2

Ice]Jo 1210
m6!)
= Z I” (”¢jo,)k ”2+ 2f I(f (I’F‘Vbj,)m )’3|2+|(f ¢jo,20)|
Ice!Jo J>Jo
k¢0 me!)
3 Z I(f’.¢;.,.)l2+ Z I(f’,w;,m)l2+A Z I(f’,¢>;.,o)I2
1:650 1210 "Elio
Ho "16111:;10
< ()(I’1+A{Z|f¢jo,)(fk|2+2| (lizvbj,)m|2}
he! 0 J>Jo
k¢0 "'61;
= (1 +A){ Z “(L ;O,k))1|2+ Z I((faW§,m))1|2}-
M5110 1210
k¢o "‘61;

Therefore
'2 <
1_1_+A|fIWQ<ZI(fa ))l| Ifllfi

We conclude that {\Il; klj > jo — 1, k 6 1,0, Ij0_1 = [jo}\{‘1l;o_l’0} is an unconditional
basis for HMQ). #

Now let us go back to consider non-periodic wavelets. let p > 1, and let 45 and
11) be the corresponding scaling and wavelet functions. With Propositions 2.6.1-2.6.3
and the similar argument as in the proof of Theorem 2.6.6 we obtain the following

theorem.

Theorem 2. 6. 7 For 2’0 > 4p — 4, {(1)00 ”All kll E 1301 75 0,j Z Jo,k E 1;},
{<I>Jo,1..., \Ilj,k...|l 6 IJo,k E 1;,j 2 Jo} and {1,<I>JOJ, \Ilj,k|l E I§O,j Z Jo,k E I,’} consti-
tute unconditional bases for H662), H}(Q) and HI(Q) respectively, where (1)30,“ W9,“
QJO,I:,\I’j,k.,¢JO,(,‘IIj,k are defined in {2.78), (2.79) and (2.80).

In the following, we shall explore further approximation properties of antideriva-

tives of wavelets.
Theorem 2.6.8 For any u E H6(Q)flHm+l(Q) {or u E H:(Q)0Hm+l(fl)), we have

inf Iu — 'UlLQ S CZ‘JquImHQ, O S m S p, (2.139)
‘UEoVJ.

47

07‘

Pr

wh

Le?

It.‘

Si]

fol

01‘

inf lu — vllfl S 02-Jmlulp+l.fla 0 S m S p)
v6.V;

where C is a constant.

Proof. Let u 6 H362) fl HP+I(Q). Then by Theorem 2.6.6
u = Z X c,,,.\11;3,
ijO—l k6};
where
If : Ii ifj 2 Jo
[jo\{0} ifj = jo — 1.
Let

J
UJ : Z Z \I’Rk.

j=jo-1 1:67;

It then follows from Proposition 2.5.2 that

111—qu”; = | Z ch.k‘1’§,k|1.n

jZJ+l kEIJ

= ( 2: Z 1c.,.12)%

j=J+1 keIJ

Np—

S ( E Z C2—2jmlu’13n,l,,k) 0 S m S p

j=J+l kEIJ
S 82—Jmlullmfi

'—_- B2_Jm |U|m+1,fl.

Therefore

inf |u — Ull'fl S In — lulu; S B2"Jm|u|m+1,g.
11601,;

Similarly we can establish (2.140). #

(2.140)

(2.141)

(2.142)

For p > 1, let {112“} be the corresponding wavelet basis of L2(R). Then we have

following estimates.

Theorem 2.6.9

inf Iu — Ull’fl S C2_J8|U|3+1’Q 0 S s < 1)
onVJ

48

(2.143)

[0:

Le

11:

We

11 SI

for any 11 E H3(fl) fl H’+1(Q),
inf Iu — '0an S B2'J’|u|3+1,g 0 S s < p (2.144)
UE.VJ

for any 11 E Hi(fl) fl H‘+1(Q).

Proof. Since proofs of (2.143) and (2.144) are similar, we present only that of (2.144).
Let u 6 H301) fl Hm+l(Q) and in 2 23-1404 2kg; cj,k\Il,-,k., where \Iljo_1,k. = (1),-0,)",
110—1 = [jo-

1/2

1 l/2 1
S (/ wfkdx) (/ wlzmdx) S 1, k,l E I}.
o ’ o '

Then it follows from the above, Theorem 2.5.2 and Schwarz’s inequality that

]1 IPj,k1/)1,mdw
o

 

 

 

lu — “Jlin = I: Z gull)... in
j>Jkeq
= || 2 Z Cj.k¢1.k||3,a
j>Jkeg
1
s (E E 14111:; )3 101ml / 21:11,".de
j>J he]; j,l>J mEIJ’ 0
S (E Z 2’j‘|u’|,,1,,)2
j>Jkeg
S (C E 2_Js|u’|s,fl)2
j>J

= (02—Jsiuis+l.fl)2-

#

2.6.2 “Pseudo—Antiderivatives” of Weighted Wavelets and Their

Properties

In this subsection, we shall first construct the “pseudo—antiderivatives” of the weighted
wavelets discussed in the Section 2.4, and then study their properties which will be
used later.

Let

H_ = rh_ (11—32 .h_ dt,
{ 1.0(x) fa. 110w b—a fa 1,010 (2.145)

Hj.k($) = If hm: wdt,

49

for a: 6 [a,b]. Notice that ifa: < a+k2‘j(b—-a), then supij,kfl[a,a:] = ,,kfl[a,:r] = (ll.
Hence Hj,k(:c) = 0. If :r > a + (k + 1)2'j(b — a), then H,,k(:c) = f: hj’kwdt =<
hj,k,¢ >w= 0. It is easy to check then that H_1,o(a) = H_1,o(b) = 0. Therefore
supijJ, Q 1,31,. We shall call HM “pseudo—antiderivative” of h”. We are now in a

position to discuss properties of the “pseudo-antiderivatives”.

Let f 6 H6(Q). Then f’ E L2(Q). Since w E L°°(Q), and w 2 c,

 

 

 

I 2 < w 00 I 2 = w 00 2 ,
{ ||f 11 _ 11 11 Ilf 11. II 11 mm (2.146)
llf'lli Z Cllf’ll3 = lelino
So f' 6 L362). Thus
llf'lli, = Z: l (f'1hj.k)w l2 + I(f',h—1.o>wl2
j=0
2
2 :1: — a b
= Z | (f, 111.0191 + f,H_1,0 + b / h_1,0wdt . (2.147)
’ —a a 1,9
Notice that < f,l >15): 0 for a linear function 1. Hence (2.147) becomes
2| < f» Hue >1.n l2 + | < f, H—1.o >1.n> l2 = llf'lli. (2.148)
A combination of (2.146) and (2.148) now gives us
leli,n S 2 l < flHch >111 l2 S llwlloolflin- (2-149)
j=-1
If we denote B = c, A = IIwIIOO, then (2.149) becomes
Blflixz S 2| < f, Hi1]? >1.n l2 S Alflin' (2-150)

We have thus proved the following theorem.

Theorem 2.6.10 {Hugh}? constitutes a frame of HM”).
E J

Let NJ(Q) = span {HJ-Jc : —1 Sj S J,k E 1,}. Then by Theorem 2.6.10 NJ(Q) is
a finite dimensional subspace of H6(fl) and U3°NJ(Q) is dense in H362). Hence any

f E HMQ) can be approximated by elements of NJ(Q). A consequence of Theorem
2.6.10 is the following.

50

Theorem 2.6.11 {Hj,k}0<j<J constitutes a basis of NJ, and {Hj,k}0<j constitutes

an unconditional basis of H362).

Proof. It suffices to show that {Hj’k}0<j<J is linearly independent and {HM} -1519
_ _ kEIJ'
is not.

Let
Z c,,,.H,-,,. = 0. (2.151)

OSiSJ
Taking derivatives of both sides of (2.151), we find

2 chChJ-ch = 0. (2.152)
OSiSJ
We now take inner products of both sides of (2.152) with respect to him, and get
em, = 0 for 0 S i S J. It then follows that {H 13k}o<j < J are linearly independent.

Now we show that {Hid—15,31 is linearly dependent.
kEIJ

Let
f(JI) = Z dj,kHj,k + H-1,o,
0515*
1:6],
where
w a,b b
dj’k 2 HA hj,kd$.
Then
f' = Z d,,,.h,-,kw + (ll-1,0w - w((01, b)l/(b - 0))-
0<J<k
F67,

It is easy to check that (f’,h1,m) = O for —1 S l < +00 and m E I). So it turns out
that f’ E 0, i.e., f(zc) E c. But f(0) = f(1) = 0 gives c = 0.
The conclusion above and Theorem 2.6.10 now guarantee that {H 13k} 120 is an
ker,

unconditional basis of H362). #
With the similar proof as in that of Theorem 2.6.10 we have

Corollary 25.2 {fill—1.010%, Hchlj Z 0,k 6 11'} and {lalii (14.010477, Hchlj 2
OJ: 6 1,} constitute unconditional bases of H362) and H162) respectively.

The following theorems give us some approximation properties of NJ62).

51

Theorem 2.6.12 For any f E H362)flH262) or H362)flH262) , we have
£151.15)” — 51m 3 C2-J|f|m (2.153)
where C does not depend on f and J.
We need the following lemma to prove the above theorem.
Lemma 2.6.1 Let u 6 L362), and let u“, =< u,h,-,1c >,,,. Then forj 2 O
|u,-,,.| g B2'J’|u|,,,,', (2.154)
where B does not depend on u,j and k.
Proof. Let (a,,k,b,~,k) = Ij.k- Since < h_1,0, h“. >w= 0 forj 2 0,
IuMI = I < u,h,-,k >,,, I = I < u — q,h,-,1c >w I, (2.155)
where q is an arbitrary constant. By the Schwarz inequality
luchl S gggllu — (Ill0.1,,.llhj.kwll0- (2-155)
Notice that ||hj,kw||0 S leléo. Thus (2.156) becomes
14.11 s 11w”: 331,124 — qua... (2.157)
Notice also that

inquR ll“ — Qll0.I,-,1. 3 ll“ — “(ai.k)ll0’11.k
. 2 1
(1,], (fan: u’dt) d5)
(f1), (:1: — a“) (ffjk u’2 dt) dx)
2-é- (5,). — 5),.) |u|1.1,,.

C 2-jlul1911,k’

NI”

(2.158)

l/\

”D |/\

where (bJ-Jc — a“) = IIMI = 2"j(b — a) and C = 2‘i‘(b — a). If we now denote
1

B = HwIISoC, (2.157) and (2.158) then yield the desired inequality in the lemma. #

Proof of Theorem 2.6.12. Let f E H362)flH262). Then by Theorem 2.6.11

f = ij,kHj,ka (2.159)
i=0

52

in the sense of | - I”). Let f] = 231:0 fchHch. We have

lfj,kl = l(f,i hj1k)l S lelIJJ‘JU (2160)

where C 2 Ilthcllo for anyj and k. For any givenj and 16 taking (2.150) into account,

we have

lflT,IJ'k S 2 l < f, Hivm >lej,k '2' (2'161)

1131;91ch

Now Lemma 2.6.1 gives us

213mg!” l < f7 Hi,m >191],k l2 S B2 Zhungljfi 2_2£lf’ {Itm

24 _2_ 2 (2.162)
: B 32 Jlfl2,lj'k'

It follows from (2.159), (2.160), (2.161) and (2.162) that

lf — lei,n lzj>J 2051:5214 fj,kHj.kli,n

ll 21» 20991—1 fj.khj.kw||3,n

l/\

Zj>J ll 20991—1 fj.khj.kwll(21,n

Zj>J Z ffik l hi1: "’2 d3

C1 232.] 2051:3214 fj2,k (2'163)
Cz Zj>J Zogkng—l lfli,1,,,.

Ca 212.] 2031:3214 ZI.,.,,§I,~,,. l < fa HM >1.I,,k l2

C4 23>.) 20991-1 24]. lfl2,1,,,.

052-2Jlfl2n-

|/\ l/\ |/\ |/\ ||

|/\

Finally,

/\

infveNJ(Q) If“ Ullfl __ lf‘le1,Q

(2.164)
S 02-Jlfl2,n-

A consequence of Theorem 2.6.12 is the following.
Theorem 2.6.13 For any f 6 H362)flH262),we have

'f — OO<BZ"’ 2.1
veglflmllf 12“ _ |f|2,n ( 65)

where B does not depend on f and J.

53

Proof.

infvENJ(Q) “f " ””00 : supueNAQ) infrEfl ”(53) "' ”(93”
= suvaNJ(Q) infxen If: (f' — '0') dtl (2 166)
S infveN,(n)(b - allf — vlm
S BQ'JIfIw
#

We are now in a position to investigate wavelet based solution methods for nu-

merical PDE and ODE problems in the following chapters.

54

Chapter 3

Wavelet Approximation Solutions
for Some Linear Elliptic Boundary

Value Problems

3.1 Some Classical Results Concerning the Ap-
proximation of Linear Variational Problems

The Galerkin methods for the solution of boundary value problems are often based on
the approximate solution of related weak problems. The bilinear forms in the weak
problems play the role of the differential operator of the boundary value problem.

Hence, to begin with, we introduce the following definitions.

Definition 3.1.1 Let H be a real Hilbert space equipped with a norm || - II. A bilinear
forms (-, ) : H x H —> R is said to be continuous if there is a constant C > 0 such
that

(u,v) g C “all Hi)”, Vu,v E H. (3.1)

Definition 3.1.2 Let V be a closed subspace of a real Hilbert space H with norm
|| - ||. A bilinear forms (-, ) : H -—> R is said to be V-elliptic if there is a constant

55

C > 0 such that
(m) 2 C M2, Vv e v. (3.2)

The fundamental linear variational problem under consideration is formulated as

follows:

Find u E V such that
(3.3)

a(u,v) = L(v), V1) 6 V.

We consider:

(i) V is a suitable real Hilbert space for the problem with scalar product (., -) and

associated norm II” 1..
(ii) a (~, -) : V x V —-> R is a bilinear form and V-elliptic over V X V.
(iii) L : V ——> R is a continuous linear functional.

The following theorem is known as the Lax-Milgram theorem.

Theorem 3.1.1 Assume (i), (ii) and (iii). The problem (3.3) has a unique solution.

Let V“ denote the dual space of V and let (~, ) : V X V —> R denote the canonical
duality pairing. Then the Riesz Representation Theorem guarantees that there exists

a unique continuous linear operator A EIsom( V, V') and a unique f E V'“ such that

a(u,v) = < Au,v >, Vu,v E V (3.4)
L(v) = < f,v > Vv E V (3.5)

It turns out that (3.1) is equivalent to the following “linear operator equation”:
Au = f. (3.6)
It follows from (3.4), (3.5) and (3.6) that

|a(u,v)l S IIAIIIIUIIIIUII, WW 6 V C”)

56

|a(v,v)| 2 IIA“lll'lllv||2, W E V- (3-8)

Where ”A” and “14‘1“" are the standard operator norms. Indeed, “A” and ||A"1||‘1
are the smallest and the largest constant in (3.1) and (3.2) respectively.

We now demonstrate the variational formulation of some elliptic boundary value
problems for second order differential equations. First, let us consider the following

homogeneous Dirichlet problem on [a, b]:

—(PU')' + qu = f, (3-9)

u(a) = u(b) = 0. (3.10)

We also let 9 = (a, b). We assume f E L2(Q),p and q are smooth in 6 with
0 < po 3 p(:z:) and 0 S q(:c). (3.11)

It is well known that solving problem (3.9) and (3.10) in H3(a,b) is equivalent to

solving a linear variational problem of (3.3) with

a(u,v) = [f(pu'v’ + quv) dx, (3.12)
V = H}, (3.13)
and
b
L(v) = / fv dz. (3.14)

It is easy to see that the hypotheses of the Lax-Milgram Theorem are satisfied. There-
fore (3.3) has a unique solution in H362) which is also the unique solution of (3.9)
and (3.10). In this case V" = H‘1(Q), and A : 113(0) —+ H'1(Q) is defined by
A = —ad;pfi + q.

Now we consider (3.9) with mixed boundary condition:
u(a) = 0,u'(b) = c. (3.15)
If we let

v = Him) (3.16)

57

a(-, ) is the same as before, and L(v) = (f, v)+cp(b)v(b), then again the hypotheses of
the Lax-Milgram Theorem are satisfied here, and (3.3) therefore has a unique solution
in H.162) which is also the unique solution of (3.9) and (3.15). Here V" = H'1 (9) x R,
and A is defined by

Au = {—%pdixu + qu, {(u — u0)'(b)}} , (3.17)

where uo is the unique solution in HMO) of the Dirichlet problem

-(PUB)' + (1710 = f, (3-18)

u0(a) = u0(b) = 0. (3.19)

—u'(a) = 0, u’(b) = d. (3.20)

Now

V = H‘(Q) (3.21)

a(-, ) is still the same as before, but L(v) = (f, v)+dp(b)v(b) —cp(a)v(a). Once again
we have a unique solution (3.3) in H1(Q). And now V" = H’1(Q) x R2, and A is
defined by

Au = f-(PU'Y + qu, {(u - uo)'(b), -(u - uo)'(a)}} (322)

where no is the unique solution of (3.18) and (3.19).
The Galerkin method now is to consider a family Vn of approximation closed
subspaces of V with Vn being finite dimensional. We approximate problem (3.3) on

V7, by

Find un E Vn such that
(3.23)

a(u,,,v) = L(v), V?) E Vn
It follows from the Lax-Milgram Theorem that (3.23) has a unique solution if a(-, )
and L() satisfy (ii) and (iii). We then have the following approximation property:

nu—u..|| s ||A||||A-1|| “6—6”, W e v... (3.24)

58

This implies
llu — an” s IIAII llA“|| infwu — vll =v 6 Va}. (325)

In addition, if a(-, ) is symmetric, then (3.24) can be improved to yield

Ilu - an“ s (HAIIIIA'1 )% uu - vn, Vv e v... (3.26)

 

 

Hence

”a — u,“ g (HANNA-1 )iinmlu — 6|) : v e vn}. (3.27)

 

 

Consequently, if lim,HOO inf{||u — v|| : v E Vn} = 0, then

lim ||u — 21,,” = 0. (3.28)

fl—FOO

In the following sections we shall use Vn constructed in Chapter 2 as our approx-
imation closed subspaces of H1(Q),H}(Q) and H6(fl) respectively. Since Vn C
V,.,...1,U,,Vn = H1(Q) (or H3, H6), (3.28) is automatically satisfied.

3.2 Wavelet Based Solutions of Regular Linear
Sturm—Liouville Boundary Value Problems

In this section, we shall study the numerical solution of the following regular Sturm-
Liouville boundary value problems by means of pseudo—antiderivatives of weighted

wavelets. We consider
—(pu')' + qu = f(x) for a: e (o, 1) = n, (3.29)

with boundary data

—au(0) + Bu’(0) : 0; a 2 0,fi Z 0 and 02 + H” > 0. (3 30)
au(1)+bu’(1) =0; a20,b20and a2+b2 >0. .
We assume that f E L2(Q), and p,q are smooth in H with
0 < p S p($) and 0 S q($), (3-31)

59

where p is a constant. We shall first derive the solution of problem (3.29)+(3.30)
with a = a = 1 and ,8 = b = 0, i.e., the Dirichlet boundary value problem. Then we
shall consider the Neumann problem, that is, problem (3.29)+(3.30) with a = a = 0,

B = b = 1. Finally, we shall consider the mixed boundary value problem, i.e.,

(3.29)+(3.30) with a = b = 1, 6 = a = 0.

3.2.1 Wavelet Approximation Solutions of Dirichlet Prob-

lems

From the Section 3.1 the variational form of (3.29) with the homogeneous Dirichlet

boundary condition is

Find u E H561) such that
(3.32)
a(u,v) = L(v), for all '0 E HMO)
where a(-, ) is given by
1
a(u,v) 2] (pu'v' + quv) dx, (3.33)
o
and L is defined by
L(v) = (f,v), for v E H3362). (3.34)

The condition (3.31) on p and q guarantees that a(-, ) is continuous and Hé-elliptic.
Therefore it follows from the Lax-Milgram theorem that (3.33) has a unique solution
u E Han), which is the weak solution of the problem (3.29) with the Dirichlet
boundary condition.

Let us describe the weighted wavelet-Galerkin method as follows. Construct a
weighted Haar basis functions as in Section 2.4 and their pseudo-antiderivatives as
in Section 2.6.2 with (a,b) = (0,1) = 0,m = l/p. Let NJ(Q) be defined as in
Section 2.6.2. Then base on approximation properties of the pseudo—antiderivatives
we may choose NJ(Q) as Galerkin approximation spaces. To approximate (3.32) it
is then quite natural to use the following Galerkin formulation of the above Dirichlet

problem:

Findu E N (I such that
{ " ’( ) (3.35)

a(uJ,v) = L(v) for every 12 E NJ(Q).

60

It follows from theorem 2.6.8 and (3.28) that

lim HUJ — u||H1 S c liminf HvJ -— u||H1— _ 0.
J->1VJEN

For any given J E N, the solution UJ E NJ(Q) of (3.35) has the forms

uJ(:c) = :1: uj,kHJ-,k(:r) :1: E [0,1], (3.36)
i=0,k61,
where H 3"]: are pseudo—antiderivatives of weighted Haar basis functions hj,k. Substi-
tuting (3.36) back in (3.33) and choosing v = Ham, 0 S i S J, m E I.- lead to the
following system of linear equations

J
E uj,ka(HJ-,k, Hm”) = L(H;,m), 0 S 2 S J, m E I". (3.37)

j=0,kEIJ'

In matrix notation, (3.37) may be rewritten as
AU 2 F (3.38)

For later convenience, let I : (j,k) 1—> i be a mapping from N x N —> N such that

l(j,m) S l(i,k) ifj S 2' or j = i and m S 11:. Then we get

= (cu-,1), 013k = a(Hj. H1). (339)
= (11,-) (3.40)
= (f1), f1 = L(Hj) (3-41)

In the case q = 0, knowing that p = 1/w, we have

1 1
(1,-,1, = ]pH§H,2d:1:=/ phJ-whkwda:
o o

= [01 hjhkwda: =< 15.6,. >..,= j). (3.42)
This implies
1 0
A = (3.43)
0 1
So (3.38) becomes
U = F, i.e. =fj— — (f,H ). (3.44)

61

an

be

C0

H1

CC

af

If we compare the matrix A in (3.43) with corresponding matrices in finite element
methods or other wavelet based methods we see that the weighted wavelet based
method makes the problem surprisingly simple! In the case q(.7:) E 0 and p(:c) E 1,
we have w = 1. So HM are antiderivatives of Haar basis hJ-Jc. We may also replace
H M by \Ilh, antiderivatives of periodized wavelets it?“ and resulting matrix A is still
an unit matrix with p = 1, q = 0. Notice that a(-, -) is symmetric and NASH-elliptic,
hence A is symmetric and positive definite in the general case.

Taking the facts suppHJ-Jc = Ich C [0,1] and for k’ 75 k, I“. 0 [MI = (ll or {QM}, a
common boundary point of 1]";c and 1331.1, into account, we know that even for p and

q being not constant we still come up with very nice matrix

A A M 0
A: “ ‘2 , D1: (3.45)
Ai2 DA
0 Am

where A11 is a positive definite matrix with order O(logN), N =dimNJ = 2’ — 1.

Hence we can find a permutation matrix B such that

CO T
A=B B an)
0 0,,

By applying the Courant-Fischer Minimax Theorem, one can show the following [62].

Proposition 3.2.1 For general coefiicients p(:r) and q(:c) satisfying (3.31), all but
O(logN) of the eigenvalues of A are uniformly bounded below and above.

This excellent feature allows us to apply the conjugate gradient method to solve
the system with fast speed of convergence even if the condition number is large.

The next proposition combining with Proposition 3.2.1 will guarantee that the
convergence of the conjugate gradient method for solving AU = F will be uniform

after O(logN) steps.

Proposition 3.2.2 Let U" be the n-th approximated solution of AU 2 F by the
conjugate gradient method. Suppose the spectrum ofA 2 0'0 U 01. [fl = loll and

maxdo /\
k = -—.——,
mmao )1

62

then

 

“£11"

where H - ”3, = (., -) and M is a constant depending on the spectrum of A.

Iw—UwASM(

By the preconditioning technique we can improve the performance of the conjugate

gradient method. It is natural to choose the preconditioner P for A defined by

(”l
P=B , an)
01

where LLT = C‘1 in (3.46). (since C and C"1 are symmetric and positive definite,
there exists the Cholesky decomposition LLT = C '1.) So instead of solving AU = F,

we solve the equivalent problem
PTAPV = STF, U = 51/, (3.43)

where PTAP is a well-conditioned matrix. The action of B can be easily derived
during the computation, so we only have to compute and store the lower triangular
matrix L which has O(log2 N) entries. Theoretically the condition number of PTAP
is bounded above by maxxeg p(:r) + max q/ p. The preconditioner P is made possible
by the orthogonality property of wavelets.

Before we derive the error estimate, we shall also discuss some computational
aspects of agj and fj in (3.39) and (3.40) required to compute the solution of the
approximation problem (3.34). The evaluation of agj and fj can be performed by
using the Gaussian quadrature method. By their special structure, we may start
from the highest scale 2". Once a(HJ,K, Him) and (f, HJ,K) for k,k +1 E IJ, m E I,-
have been computed, we may find a(HJ_1‘kI, H -,m) and (f, H j‘m). Therefore the total
number of computation needed reduced greatly.

To get error estimate for u E H6(Q)0H2(Q), theorem 2.6.11, 2.6.12 and a standard

duality argument give

IU-‘Ujll’fl S cinfveNJ lit—”Ulla S C2-JIUI2,Q (3 49)
In — UJloo’Q S ClnfveNJ |u — vloog S B2_J|u|2,g

Also we have the following.

63

WE

d1

F1

Theorem 3.2.1 Let u and UJ be the solutions of (3.32) and {3.34) respectively. If
u E HMO) fl H"+1(O),

llu - UJllon + 2_Jllu - UJllm S 02—J(’+1’HUHs+1.n (3-50)

For inhomogeneous Dirichlet problems, that is (3.29) plus boundary condition
u(0) = c, u(1)= d, (3.51)

we introduce new function w(:z:) = u(x) — (1 — :r)c — d2). Then w(0) = w(1) = 0.

Therefore (3.29)+(3.51) is equivalent to the following problem.
(3.52)

where F(3:) = f(zr) + (d — c)p’ -— [(1 — :r)c + dat]q. Then by the formulation for
homogeneous Dirichlet problems and the discussion above the variational problem of
(3.29)+(3.51) is

Find u E H1(O) such that

a(u,v) = L(v), Vv E HMO)
where a(u,v) = a(w,v) + a(qb0(:r),v), ¢0(:c) = (1 — 2:)c + dz. Then the scheme we

(3.53)

discussed before can be applied to this case.

3.2.2 Wavelet Approximation Solutions of Neumann Prob-

lems

For the homogeneous Neumann problem (3.29)+(3.30) with [3 = b = 1, oz 2 a = 0,

the corresponding variational form is

Find u E HI(O) such that
(3.54)
a(u,v) = L(v), Vv E H1(O)
where a(-, ) is given by
1
a(u,v) 2] (pu'v' + quv)da:, Vu,v E H1(O) (3.55)
o

64

L6

(16

1‘95

for

and L is defined as
1
L(v) = / fvdx v11 6 111(9).
0

Again the hypothesis of the Lax-Milgram Theorem are satisfied. So (3.29)+(3.30)
with H = b = 1, a = a = 0 has a unique solution in H1(O) which is also the unique
solution of (3.54).

In this case we choose N}(O) =span{1,fg" h_1,0wdn,HJ-,k|k E 11,0 S j S J} as
our approximation subspaces of H1(O). Hence the Galerkin approximation of (3.54)
is
{ Find 111 E N}(O) such that (3.56)

a(uJ,v) = L(v), Vv E N}(O).
Let “J = Zj=—2,kEIJ Uj,kHj,k, where H_2,0 = 1. Substituting 11,; and v = Hl,m into
(3.56) gives the following linear system.

J

2 uj,ka(HJ-,k, Hum) = L(Hzm). (3.57)

J=-2
Ice!J

(3.57) may be written in matrix form
AU = F (3.58)

where A = (a(Hj,k,Hl,m)), F = (L(H1,m))T, U = [u_2,0,u_1'0,UI,0,"',UJ’2J_1]T.
Since any constant function belongs to H1(O), sufficient conditions for a(-, ) to be
H1(O)-elliptic are (3.31) with

0 < qo S q(a:). (3.59)

If (3.31) and (3.59) are satisfied, the resulting matrix A is symmetric and positive
definite. We may use Preconditioned Conjugate Gradient or Gauss-Seidel method to
solve (3.58).

If 117]“, IV)“ are used as approximation subspaces of H1(O), we have similar
resulting matrix A. Intuitively for q =constant, the resulting matrix has a very simple

form when we use {1, wik, (blkhe 1, as basis functions,

A = ( 7;” Z ) (3.60)

where

D: , 2i:(a1,a2,---,a21,0,---,0).
0 A2J+l
Therefore we may even get the “exact” solution of (3.58) in 2" +1 operations!

For the inhomogeneous Neumann problem (3.29) with boundary conditions

(3.61)
P(1)U'(1) = 41
our variational formulation will be slightly different from (3.55):
Find it E H1(O) such that
(3.62)
a(u, v) = L(v)

where a(-, ) is the same as in (3.55) and L(v) = (f,v) + cv(0) + dv(1).

3.2.3 Wavelet Approximation Solutions of Mixed Boundary
Value Problems

In this subsection we consider the mixed boundary value problem (3.29)+(3.30) with

a = b = 1, fl = a = 0. Then we have the corresponding variational formulation:

Find u E H}(O) such that
(3.63)
a(u,v) = L(v), Vv e H.110)
where a(-, ) is the same as before and
L(v) = (12v) + 40(1) — cum). (3.64)

We may use NJ. 2 span”; h_1,owdn,HJ-,k|j 2 0,11: E Ij} as our approximation
1 0
subspaces of H}(O).If q E 0, then the resulting matrix is A = . We

0 1

may also use ...Vj'H and .VJ+1 as approximation subspaces of H}(O).

66

If‘

(it.

511

If)

[6

9’1

3.3 Wavelet Based Domain Decomposition Meth-
ods for Singular Sturm—Liouville Boundary Value
Problems

In this section we consider the singular Sturm-Liouville boundary value problem as

follows.

i — (p(a:)U’) + q($)U($) = f0”), ‘5 6 [0’1) (365)

u(l) = d

where p(0) = 0, p’(0) > 0, p(a:) > 0 for :1: E (0.1], q($) Z 0, and q(0) > 0. We also
require that u(0) and u’ (0) are finite.

Let us give a brief introduction about how to solve the problem first.

We solve the problem (3.65) by first decomposing [0, 1] into two subintervals [0, 6],
[6, 1] for small enough <5, 0 < 6 < 1. Over [5, l] we solve the problem by using method
described in Section 3.2. Over [0,6] we solve the problem by the asymptotic expan-
sions. Finally, we use the domain decomposition technique to derive the matching
inner boundary condition and solve the problem over [0, 1].

Now let us describe the method in details.

First, we note that u’ (0) is bounded and p’(0) > 0. Hence integrating both sides
of (3.65) results

—par( )u ’(x)— 0)+/0c1(s s)ds: f0 f(s) (3.66)

This implies

u'() =—/ [f(8 )— )us( )lds/p(a= ) (3.67)

Since p(0) = 0 and u’(0) lS bounded, p(0)u’ (0 ) = 0. Taking limit on both sides of
(3.67) gives

u'(0) = -[f(0) - q(0)U(0)l/p’(0)-
Let u(0) = /\ (note that u(0) is bounded, so /\ is finite). Then u’(0) = —[f(0) —
q(0)/\] / p’ (0) By Taylor expansion
11(4) a 2401+ 41011: = ,1 — 11(0) — «1101114240101 é um). (3.68)

67

He:

and

The

Th1
p16
fun)
Call

tha

Th1

W111

and

So
{ «46) z 1 — 11(0) — 41011131710) = ...(1)
17(6) z 4(0) = ~00) — «0111/1401 = 14(6).
Hence from (3.69)
121011405) = 41(0) — «am
and

A = [f(0) + P’(0)U'(5)l/q(0)-

Therefore

 

(3.72) then implies

 

1.1161110) — 1(0).

“9(5) = q<016+pl<0>

(3.69)

(3.70)

(3.71)

(3.72)

(3.73)

Thus over [0, 5] we use u1(a:) = u(0)+u’(0)x to approximate u. But before we solve the

problem over [6, 1], we do not know u(O) and u’(0), therefore u1(:r) is still an unknown

function. However, we have the relationship (3.73) between 111(6) and u’1(6), hence we

can use it as the starting point, the matching inner boundary condition. This means

that we need to solve the problem over [5, 1]

-(pU’2) + qwz = f
112(6) = u1(6)

“9(5) = “3(5) = 21311631127610
112(1) d

The corresponding variational form of (3.74) is given by

{ Find 11. e H(6, 1) = {f e H1(6,1)|f(1) = 0} such that
a(u2,v) = L(v)

where a(, ) is defined by

 

 

_ 1 I I p(5)u2(d)q(0)
a(u,v) _fa (puzv + quzv)da: + q(0)6 +p’(0) 72(5)
and
_ _ ,0 WWW) v

68

(3.74)

(3.75)

(3.76)

(3.77)

It can be shown that a(-,-) is continuous and H-elliptic. So (3.75) has a unique
solution. Now one can approximate the (3.75) by the methods discussed in Section
3.3. Once we find the approximation solution for (3.75), then u1(6), u’1(6) are known.
Hence the asymptotic solution of the problem over [0,6] u1(:r) = u1(6) + u’l(6).’r
is obtained. Finally, 17(5):) = u1(:1:)X[0,5](:r) + u2(a:)x(5,1](2:) gives the approximation
solution for (3.65).

3.4 Steady-State Solutions of Some Fokker-Planck
Equations

In Chapter 4 we shall discuss the the following Fokker-Planck equation
3 _ 132]ap] _ M)

2

)

Q3
H
to

8‘” (at) e (0.1) x (0,+oo) (3.78)

with boundary conditions at :1: = 0 and a: = 1.
Before tackling the complete Fokker-Planck equation in Chapter 4, we shall solve

a simplified version called the steady-state or equilibrium problem of (3.78)

14120172) _ d(bp)
2 d:1:"2 da:

 

=0 xEmJ) (3m)

with the same boundary conditions at :c = 0 and a: = 1.
At present, we assume that a(x) and b(:1:) are smooth and that a(O) = a(l) =
0,a’(0) > 0,a’(1) < 0 and b(0) 2 0,b(1) S 0, the following corresponding boundary

conditions are imposed (see [59])

311$ a($)p(:r) = 0 if 6(0) = 0, (3.30)
.1331- a(at)p(a:) = 0 if b(1) = 0, (3.81)
xgrg+{-;-[a(x)p(x)l’ — 111-112(4)} = o if 12(0) 75 o. (3.82)
x§q1_{%[a(w)p(x)l’ — b(:v)p(:r)} = o 1111(1) 79 o. (3.83)
In particular we take a(cr) and b(a:) as in [59] to be
a(a) = $141 — :13),

69

3111

is 1
the

01]

111

Le

aP

b(m) = 3:3(1— 2:)[h +(1— 2h);r] — ra: + q(1— :r),

where s, h, r, q are some non—negative numbers.

Let u = ap. Then (3.79)-(3.83) become

1 ,, b ’_
é—u — (Eu) — 0, :1: E (0,1), (3.84)
lim u(x) = 0 if b(0) = 0, (3.85)
x—+0+

lirp u(:1:) = 0. if b(l) = 0, (3.86)
x—+ '7

. 1 , b _ .
lerglJ-iu — gu) — 0 1f b(0) 74 0, (3.87)
lim (lu' — 9u) = 0 if b(l) # 0. (3.88)
17—H- 2 a

We shall consider the following cases.

Case 1. r=q=0,s;£0.

In this case b(O) = 0 = b(1), and b(r)/a(a:) = 4s[h + (1 — 2h):r] is smooth on
[0,1]. So the corresponding Dirichlet boundary value problem (3.84)+(3.85)+(3.86)
is regular. Since u E 0 is a solution of this problem, it follows from the uniqueness
theorem that u E 0 is the only solution of the problem. Thus ap E 0. Since a(m) > 0
on (0,1), p(:c) E 0 on (0, 1).The numerical treatment of the problem is given below.

As in Section 3.2.1 the weak form of the above problem is

F'nduE H1O s hthat
( ' °( ) “C (3.89)

a(u,v) = 0 for all v E HMO)

where a(-, ) is defined by

1 1
a(u,v) = —%/0 u'v'da: — 43/ (h +(1— 2h)a:)uv'd:c.

0

Let {z/Jj,k},371 be the Haar basis of L2(O). Let us choose 0VJ(O) as the Galerkin
6 J
approximation subspace of HMO). Then the approximation problem of (3.89) is

Find u E V such that
{ J ° J (3.90)

a(uJ,v) = 0 for every 2) E oVJ.
Since w; E OVJ, there exist {uj,k}og,lgj C R, such that
k6 J
J
U] = Z Uikwak.
i=0

70

F1
he

Then the corresponding linear system of (3.90) is
AU = 0, (3.91)

where A = (a(‘II‘JlJc, \Ilfim)), U = (Uj,k)T.
Numerical experiments show that for our data (3, h) = (2, 0.5), (2,1), (4.8, 1.3), (2,2),
A is nonsingular with the real parts of all eigenvalues of A being greater than 0.4.
Therefore (3.91) has the unique solution U E 0 and (3.90) has the unique solution
u J E 0. Hence we reach the same conclusion as the theoretical one.
Case 2. s 75 0,r7é0,q5£0.
In this case, b(0) = 0,b(1) 75 0, and
b(r) 45(1— 2:)(h + (1 -- 2h):c) — 4r
a(517) (1 - 1‘) '
For the data (3, h) of interest, one can show that the problem (3.84)+(3.85)+(3.88)

 

 

has the unique solution u E 0.
Since the problem now has singularity at x = 1, we apply the domain decomposi-
tion technique to solve the problem. Notice that b(x)/a(;r) z —4r/ (1 — x) for :1: near

1. Suppose that u(x) = c(1 — :13)" for a: near 1. Then

0 = 1u" — (Eu) = écoda — 1)(1 — :1:)O"2 + C(a —1)(-—4r(1 — :r)a — 2).

2 a
So
ca(a — 1) — 8rc(a — 1) = 0,
or
ca2 — (l + 8r)ca + 8rc = 0,
or

(1:1 or 8r.

To satisfy the boundary condition (3.88), we must have

0 = lim {lu’— éu}
:r—tl- 2 a

= lim {__c_2€1_(1_ :1:)°"l + 4cr(1— mad}.

33—H—

71

Si

111

OI

11']

8' is a solution of the problem near :1: = 1.

This implies a = 81'. Therefore 11 = c(1 — 3:)

Now we choose 6 small, and decompose [0, 1] into two subintervals [0,1 — 6] and
[1 — 6,1]. Over [1 — 6,1], we choose u2(:c) = c(1 — s)s” as the asymptotic solution of
the problem. It leads to the following relation

u'2(:1:) = — 8r u2(m). (3.92)

l—x

 

Suppose ul is a solution of the problem over [0,1 — 6]. We have
u1(1— 6) = 112(1— 6), (3.93)
u'1(1— 6) = u'2(l — 6), (3.94)

Multiplying both sides of (3.93) by u’2(1 — 6) and those of (3.94) by u2(1 — 6) and

then subtracting both sides of two resulting equations, we have
u2(1— 6)u'1(1- 6) = u'2(1— 6)u1(1— 6)

01‘

21111—6): —8—"u2(1—6).

«111—6): “'2‘“ 6

”2(1 — 5)
Therefore over [0, 1 — 6], the problem becomes
%u'1'— (3111) = 0
111111404. u1(:1:) = 0 (3.95)
u’1(1— 6) = ——u1(1 — 6).
The weak form of (3.95) is

Find u.) E ...Vj' such that
~ (3.96)
a(uJ,v) = 0 for every ’11 E ...Vj‘,
where
1 1—6 1— 6 b
a(u,v) = —1/ u'v'dzr—f —uv 'da:
2 0 a

b(l — 6)
.(1 — 6)

1—6 1— 6 b
= —/01u'v’d:1: — f0 —uv 'da:
2 a

(4% + 436(h + (1 — 21;)(1— 6)) —— 4r)u(1— 6)v(1_ 6)

[01—6 a'v’da _ ,/01-6 fiuv'dx 43(h + (1 — 2h)(1— 6))u(1_ 5)v(1_ 5).

a

—-u'(1— 6)v(1—6)+

 

u(l — 6)v(1— 6)

 

+

[\DIH

72

Again our numerical experiment shows that the corresponding matrix A of (3.96) is
nonsingular. In fact, all eigenvalues of A are bounded away from 0. So the problem
(3.96) has the unique solution 111 E 0. Thus 0 = u1(1 — 6) = 112(1 -— 6) = C68”, that
is,c=0and u2E0.

Case 3. s 75 0,q#0,r = 0.

We can reduce Case 3 to Case 2 by the simple linear transformation t = 1 — x.
The case then proceeds in the same way as in Case 2.

Case 4. s=r=0,q7é0.

In this case, b/a = 4q/x, and the associated problem is (3.84)+(3.86)+(3.87). By
the same technique as in Case 2 near :1: = 0, we choose the asymptotic solution of the

problem over [0, 6] by u1 = cxsq. So the weak form of the problem over [6,1] is given

by

(3.97)

Find 17“ = U2J(1 — 3:) E ...Vj' such that
a(iIgJ,v) = 0 for every 11 E ..Vj',

where

 

1—6 1—6 _
a(u,v) : év/ou [Hid +/ b(l $)u 'U’dx

a()l—xu

1 1 (5(5)
—§u (1 — 6)v(1— 6) — m

1 1—6 1—6 4
= -2-/ u'v'dx—j 1 q uv'dm.
0 0 —:1:

A same reasoning as before shows that the problem has the unique solution u E 0.

Case 5. s=q=0,r7é0.

u(l — 6)v(1—6)

 

This case is similar to Case 4 and the corresponding problem has the unique
solution 11 E 0.

Finally we consider the following case.

Case 6. s = 0,r,q 75 0.

An exact nontrivial solution is
u(x) = 6689(1 — a)8'. (3.98)
Since p = u/a and since 1,1 pda: = 1,
p(:1:) = :68‘1—1(1 — :r)8r"l/B(8r,8q). (3.99)

73

1111

iii)

of

C0

11

So c = 1/B(8r, 8q), where B(-, ) is the Beta function. The numerical formulation of
the problem is as follows.

Since b(O), b(1) 75 0, as in Case 2 and Case 4, we choose the asymptotic solutions
of the problem over [0, 6] and [1 — 6,1] as u1(:1:) = c338" and u3(:c) = ca:(1- s)s”. Over

[6,1 — 6], we derive the matching inner boundary conditions as in Case 2 and Case 4

, , _ 8_q _ 89,
111(6) - “2(6) (5 ”2(6) _ (S 1(6)?
ug(1— 6) = u'2(1— 6) = —§(S:u2(1 —— 6) = ~¥u3(6).

Therefore over [6, 1 — 6], the weak formulation is given by

{ Find 112.1 6 1171* such that (3.100)

a(u2J,v) = 0 for every 11 E 1V},
where

1 1—6 11 1-6b ’
a(u,v) = 2/0 uvdrc—jo —uvd:1:

4q 4r
1_ 6u(1—— 6)v(1-— 6) + 1 _ 6u(6)v(6).

 

 

+

We have used the periodized Daubechies wavelets and their antiderivatives for
p = 2 to solve the problem. The numerical problem reduces to the matrix equation
of the form (3.91). Using Matlab we look for the smallest eigenvalues of A and their
corresponding eigenvectors. The smallest eigenvalues in the case we considered are of
the order 0(10‘4) to 0(10‘5) with mesh size 2‘7. The eigenvectors are normalized so
that resulting p’satisfies fol p“(:1:)d:1: = 1.The numerical results ,for (8g, 8r) = (1,1),
(1,2), (2,1), (1.5,1.5), (0.7,0.4), (O.3,0.2) are as in the following tables and figures,
where :1: denotes the grid points on [0,1], p(:1:) and p*(a:) denote the values of the

exact and the numerical solutions respectively.

74

 

Table 3.1:

(8q, 8r) = (1,1).

 

1'

19(3)

11"(33)

10(93) - p“(z)

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000
1.000000000000

 

0.999999048097
0.996912165532
0.998508113060
0.999313895649
0.999477914686
0.999589025905
0.999732506081
0.999782719459
0.999824579798
0.999892435566
0.999921362870
0.999948297705
0.999999088097
1.000024186901
1.000049884918
1.000105768717
1.000137639638
1.000173670418
1.000265839001
1.000328357564
1.000409538191
1.000685323810
1.0009564023 78
1.001494784564
0.999999048097

 

0000000951903
0003087834468
0001491886940
0000686104351
0000522085314
0000410974095
0000267493919
0000217280541
0000175420202
0000107564434
0000078637130
0000051702295
0.000000911903
-0000024186901
-0000049884918
-0.000105768717
-0000137639638
-0.000173670418
-0000265839001
-0000328357564
-0000409538191
-0000685323810
-0000956402378
-0001494784564
0000000951903

 

75

 

Table 3.2: (8q,8r) = (1.5,1.5).

 

27

19(3)

11"(x)

19(3) - 11"(55)

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

0000000000000
0.445670291151
0620023946075
0.847113979886
0.930035919112
0999758572674
1.109132552881
1.151643366362
1.187255466860
1.240047892149
1.258001949822
1.270670969829
1.280715955945
1.278212110039
1.270670969829
1.240047892149
1.216574831553
1.187255466860
1.109132552881
1.058892474857
0.999758572674
0.847113979886
0746607698777
0.620023946075
0000000000000

 

0000000000000
0444294320090
0619099198269
0846533123834
0929550748125
0.999348113975
1 . 108836328438
1.151393616162
1 . 187047692609
1.239915023458
1.257903548009
1.270605802363
1280715321406
1278243558349
1270734886340
1240179566892
1.216742787180
1.187462152528
1.109427865588
1.059240611328
1.000168428508
0847694880286
0747322067376
0620951006891
0000000000000

 

0000000000000
0001375971062
0000924747806
0000580856052
0000485170987
0000410458699
0000296224442
0000249750200
0000207774251
0000132868691
0000098401813
0000065167466
0000000634539

—0000031448310
-0000063916511
-0000131674743
-0000167955627
-0000206685668
-0000295312708
-0000348136471
-0000409855834
-0000580900400
-0000714368598
-0000927060817

0000000000000

 

76

 

Table 3.3:

(8q, 8r) = (1,2).

 

37

11(1“)

11*(16)

p($) - 19*(8')

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

2.000000000000
1.937500000000
1.875000000000
1.750000000000
1.687500000000
1.625000000000
1.500000000000
1.437500000000
1.375000000000
1250000000000
1.187500000000
1.125000000000
1.000000000000
0937500000000
0875000000000
0750000000000
0687500000000
0625000000000
0500000000000
0437500000000
0375000000000
0250000000000
0.187500000000
0125000000000
0000000000000

 

2.000834346526
1.932324937488
1.872985527803
1.749530535087
1.687324199545
1.625011341900
1.500225779699
1.437788584123
1.375333619021
1.250388145193
1.187903103471
1.125412201858
1.000417213280
0937914677417
0875409527423
0750392953879
0687882128142
0625369917874
0500342037860
0437826646435
0375310438123
0250275933981
0187757798970
0125239191892
0000000000000

 

-0000834346526
0005175062512
0002014472197
0000469464913
0000175800455
—0000011341900
-0000225779699
-0000288584123
-0000333619021
-0000388145193
-0000403103471
-0000412201858
-0000417213280
-0000414677417
-0000409527423
-0000392953879
-0000382128142
-0000369917874
-0000342037860
-0000326646435
-0000310438123
-0000275933981
-0000257798970
-0000239191892
0000000000000

 

77

 

Table 3.4:

(861.87“) = (2.1)-

 

23

11(3)

11"(x)

p($) - P‘(~'v)

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

0000000000000
0062500000000
0.125000000000
0250000000000
0.312500000000
0375000000000
0500000000000
0562500000000
0625000000000
0 750000000000
0812500000000
0875000000000
1.000000000000
1.062500000000
1.125000000000
1250000000000
1.312500000000
1.375000000000
1.500000000000
1.562500000000
1.625000000000
1.750000000000
1.812500000000
1.875000000000
2.000000000000

 

0000000000000
0062280979960
0124761370017
0249724101522
0312206361438
0374689283026
0.499657420818
0562142831645
0624629298054
0749606028229
0812096690864
0.874589225218
0999581312279
1.062081800635
1 . 124586098864
1249609935413
1.312132245292
1.374664254789
1.499771927613
1.562360069040
1.624986335208
1.750467707274
1.813475535744
1.877018219774
1.999162544592

 

0.000000000000
0.000219020040

0.000238629983
0.000275898478
0000293638562
0000310716974
0.000342579182
0000357168355

0.000370701946
0000393971771

0000403309136
0000410774782
0000418687721

0000418199365

0000413901136
0000390064587
0000367754708

0000335745211

0000228072387
0000139930960
0000013664792
0.000467707274
0.000975535744
0.002018219774
0000837455408

 

78

 

Table 3.5: (8q,8r) = (0.7,0.4).

 

13

10(3)

11"(x)

11(3) - P‘(-’v)

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

00
0952518817015
0789057820735
0668002001943
0638530020478
0618388203538
0595164253755
0589360636368
0586458076652
0587919965668
0591909641217
0597982764286
0616570292573
0629362914569
0644810361093
0685287537496
0.711523389074
0742957166075
0827509932981
0885624767572
0960080890596
1.197587284027
1.408310534817
1.778018217454

00

 

00
0955675312689
0790406646515
0668613244350
0639014676473
0618794345875
0595481812296
0589652299931
0.586731146022
0588170893487
0592155109312
0598225828568
0616816994699
0629615723208
0645072459562
0685580022387
0.711838908619
0743303105286
0827951462937
0886143236105
0.960711448382
1.198683113279
1.409969919664
1.781057585099

OO

 

-0003156495674
-0001348825781
-0000611242407
—0000484655995
-0000406142337
-0000317558541
-0000291663563
-0000273069369
-0000250927819
-0000245468095
-0000243064282
-0000246702125
-0000252808638
-0000262098470
-0.000292484891
-0000315519546
-0000345939212
-0000441529956
-0000518468533
-0000630557786
-0001095829253
-0001659384847
-0003039367645

 

79

 

Table 3.6: (8q,8r) = (O.3,0.2).

 

 

 

3: 11(4) 11*(93) 11(73) - 11"(33)

0.000000 00 00

(1031250 1 497684264712 10502647351865 41004963087152
0.062500 0946436806064 0948054657545 -0001617851481
0.125000 0615660375678 0616223723885 -0000563348207
0.156250 0.542175149189 0542586670070 -0000411520881
0.187500 0491841384533 0492164413996 —0000323029463
0.250000 0428723843745 0428952595585 -0000228751840
0.281250 0408467583973 0408669726945 -0000202142972
0.312500 0393161947228 0393345013143 -0000183065916
0.375000 0373472607006 0373632007388 -0000159400382
0.406250 0367914068236 0368066644169 -0000152575933
0.437500 0364755620760 0364903884336 -0000148263576
0.500000 0.365030915819 0365176972001 -0000146056182
0.531250 0368402331290 0368550314734 -0000147983444
0.562500 0374038614724 0374190651593 -0000152036869
0.625000 0393046223858 0393213978378 -0000167754520
0.656250 0407232065961 0407412649872 -0000180583911
0.687500 0425415927154 0.425614011320 -0000198084166
0.750000 0478508617264 0478763932494 -0000255315230
0.781250 0517455710332 0.517758642807 -0000302932475
0.812500 0569517768304 0569891813728 -0000374045423
0.875000 0747912870732 0748597234046 -0000684363314
0.906250 0918616030089 0919698417448 —0001082387359
(1937500 1J240797035430 10242918070145 41002121034715

1.000000

 

 

OO

 

OO

 

 

80

 

 

 

 

 

 

1.25 . . r ' T ' T ‘ '
1.2 - l
1.15 - i
1.1 ~ A
1.05 - i
1E E
0.95 - i
0.9 - i
0.85 » i
0.8 - d
0'750 01 oi2 031 0:4 ois ole oi7 ole 0:9 1

X

Figure 3.1: (8q,8r) = (1,1).

 

1.4 I I I T I I I I I

p‘(x)

 

 

 

 

Figure 3.2: (8q, 8r) = (1.5,1.5).

81

2.5

P'(X)

0.5

1.8

1.6

P'(X)

0.6

0.4

0.2

 

 

 

0.1

0.2 03 0.4 0.5 0.6 0.7

Figure 3.3: (8q, 8r) = (1,2).

0.8

0.9

 

 

 

Figure 3.4: (8q, 8r) = (2,1).

82

 

 

2.2

0.6

04

2.5

P'(X)

0.5

 

 

 

02
0

 

 

 

Figure 3.6: (8q, 8r) = (O.3,0.2).

 

 

Chapter 4

Wavelet Approximation Solutions
Some for Parabolic Initial

Boundary Value Problems

In this chapter we study initial parabolic boundary value problems of the following

form:

ut=Au+f t>0,a:E[0,1]
u(zr,0) = u0(a:)
with appropriate Dirichlet, Neumann, or flux condition, where A is an elliptic operator

in the space variable 2:.

4.1 Wavelet-Galerkin Approximation of Some Lin-
ear Parabolic Initial—Boundary Value Problems

In this section we shall focus on the following representative parabolic equation, which

describes, for example, the time-dependent flow of heat:

pcut—V'kVuzf, :rE0,0StST<oo, (4.1)

84

where u is the temperature, p the density, c the specific heat, It the thermal conduc-
tivity, and f a heat source. Here we assume that O is a bounded in R", and that p, c,
and k are positive functions in O that are independent of t.

In addition to (4.1), the temperature u(x,t) is required to satisfy an initial con-

dition:
u(r,0) = u0(:r) :1: E O (4.2)
and boundary conditions:
u($,t) = b(at), :1: E 8521, 0 < t < T, (4.3)
-Vu-7‘2‘=0, xEBQ;,O<t<T, (4.4)
—kVu 1 ii = h0(u — u,), :1: E 8O3, 0 < t < T, (4.5)

where the boundary of O,8O = BOI U BOg U (9O3, ii is the unit outward normal to
80, ho is the coefficient of heat transfer (ho > 0), and u, if the specified temperature
of the surrounding medium, we also assume that 0O is piecewise smooth. To solve
the problem numerically we shall give approaches. For the first approach, the initial-
boundary value problem 4.1 —> 4.5 is replaced by the following equivalent variational

problem.

(4.6)

Find u(x,t) E S, such that
(pout, vZ) = a(u,v) + L(u, v), Vv E 50

where

S(O) : {u | for each0 < t S T, u(o,t), ut(-,t) E H1(O); u(:1:,t) = b(x)on0O1}

(4.7)

and
50(0) = {v 6 111(0) | 6(a) = 0, 60.} (4.8)

and
a(u,v) = “/9 kVu - v6 dx + 893 110(11, — u)v d8 (4.9)
umth mm

85

We consider the case R" = R1,O = (0,1),0O1 = 80, and b(x) = 0 first. In this case
SHUthuhflEH$mhflEH%m} (4n)

a=mm) (um

We choose one of OVJ(O),0 Vj‘, or NJ(O) as defined in §2.6.1, 2.6.2 as our finite dimen-
sional approximation subspace of 30(O) = HMO). Then the so called semi-discrete

Galerkin approximation uJ(-,t) to u(-,t) is a one-parameter family of elements in

NJ(O) (or 0VJ(O),0 Vj‘) of the form

J
=X:ZfiflM%Afl mm)

j=-l kEIJ'

In case we choose OVJ(O) or 0V}, (4.13) is replaced by

”JG” 0 =2 ECO or (4.14)

j=jo—l kESJ

222 EC" j \If'k(x) (4.15)

1:10-11:61,-
where the coefficients Cj,k(t), C2,,(t), and C;k(t) are continuously differentiable func-

tions of t on [0, T]. These coefficients are determined by the system of equations
(10614]. ‘16,...) = a(uJ1 ‘16,...) + 14016,...) (4-16)
Hence the approximation problem of (4.6) is given by

Find uJ E SJ, such that
(4.17)

(pcuf, Hl,m) = a(u", H1,m)+ L(Hzm), [_>_ 0,m E 11.

Substituting (4.13) into (4.17) gives a system of ordinary differential equations:

d _8 _L
M7%=_KC+& mm)
where
M E (]QPCHj’kHl’mdx)’ (4.19)

86

K E (fa kH;’kH,"md:1:) = (fa khj,kwh1,mwd:r)

= (]n hj,kh1,mwd:1:) = (5j15km)

and the N x 1 source vector

(1
Thus (4.18) becomes
d C _1 _s

where 5 (t) is N x 1 vector unknowns
5(1) 2 [c.1111]?
In addition, the initial condition (4.2) requires that
E (0) =50: 14.110117".

where

J
uJ(IL‘, 0) = Z: Cj,k(0)Hj,k($) z 210(3)).

(4.20)

(4.21)

(4.22)

(4.23)

(4.24)

(4.25)

Since {1111:105ng is an unconditional basis of HMO) and p,c > 0, M is nonsingular.
RE 1

To see this let 2: [Zj’k]T E R214, E750, and V(:r) = 231:0 21,61]. zj,kHJ-,k, then

__sT __L
Z M Z=/flpcV2(:1:)d:r > 0

(4.26)

Hence M is not only nonsingular, but also positive definite. So if one can solve

the system of ordinary differential equations (4.22) subject to (4.24) numerically,

one obtains the approximation solution U] of u, the exact solution. The numerical

solution of the matrix differential equation (4.22) with initial condition (4.24) can be

discretized in time by a variety of techniques. Here, we discuss in detail the general

weighted implicit finite difference approximation, with parameter 0.

 

5111+. — 511
M( a.

87

)=_[461..+(1_41a]+§

(4.27)

where 0 S 0 S 1. The implicit character of the scheme becomes clear when it is
written as

511.11: T(At) 5N +At(M + (mm-1 5 (4.28)

where

T(At) = I — At(M + 6At1)'1. (4.29)

We obtain the forward difference, Crank-Nicolson, and backward difference approxi-
mations for 0 = 0, 1 / 2 and 1, respectively.

Now we turn our attention to the convergence of the semi-discrete Galerkin ap-
proximation u"(-,t), which is in (4.13) (or (4.14), (4.15)), to the true solution u(-,t)
for sequences of finite-dimensional subspaces NJ (or OVJ(O), on(O)). Then Theorem

2.6.12 can be used to prove the following theorem.

Theorem 4.1.1 Let u(x,t) be the solution of (4.3)-(4.5) with R" = R, O = (0,1),
80; = 3O and 6(1) 2 0, and for each t let uJ(-,t) be its semi-discrete Galerkin
approximation (4.13) in NJ. Assume that for each t, u(-,t) and ut(-,t) E H2(O).
Then 0 S t S T

[u — ullln S C2_J|u|2,9, (4.30)

where C depends on T and u.

Proof. Since for each t E [0, T], u(-,t) E HMO), and since {Hj,k}j20 is an uncondi-
tional basis of HMO), u(:1:,t) = ch,k(t)HJ-,k(:r). Hence u — UJ = 23-” Cj’k(t)Hj’k($).
Therefore it follows from Theorem 2.6.12 that

I’d - “J 1.0 = || 2 Cj.k(t)hj.k($)w($)||3

j>J
g C(t)2-J|u|m.

We claim that C may be chosen so that it does not depend on t. In fact, from

the proof of Theorem 2.6.12, C(t) = B(t)]lwlloo, where B(t) satisfies [c,-,k(t)|2 S

Bz(t)2‘2‘]|u|§,.k. Hence to show that C(t) is independent of t it suffices to show
’ Jo

that B(t) is independent of t. To see this we notice that C: (c131,)T is the solution

of (4.18) and M and K are nonsingular matrices. Hence it follows from the well

88

known theorem that C(t) is bounded for t Z 0 if ReAj < 0, where Aj,1 S j S 2"
are eigenvalues of —M‘1K = —M’l. Since M is positive definite, M "1 is positive
definite, thus ReAJ- < 0. Therefore Cj,k(t) are uniformly bounded. So B(t) does not
depend on t. #

Higher convergence rates can be obtained by replacing NJ by UV; or 0171.

Theorem 4.1.2 Let p > 1 and 45,111 be corresponding scaling and wavelet functions.
Then for each u(-,t),ut(-,t) E HMO) r1 H’+1(O), one has

|u —— UJILQ S C2'J’|u|3+1,9, 0 S s S p (4.31)
where 11.] is defined as in (4.14) or (4.15).

Now we return to the initial value problem (4.22), (4.24). From the discussion in
Theorem 4.1.1, 5 (t) is bounded for t 2 0

We are interested in the similar situation for the approximations 5N given by
(4.28), that is, we look for such conditions that the vectors 5N, N = 0,1,- - -, are

bounded.
By repeated application of (4.28) we arrive at

_k

511: (T(At))N(56 — §)+ s. (4.32)
Therefore, for C N to be bounded, it suffices to have
11714011 5 1. (4.33)

Consider the three weighted finite difference approximations as in (4.27), (4.28) for
0:0,1/2 and 1. For 0:0

T(At) = I — AtM‘l. (4.34)

Since M is symmetric positive definite, M '1 is symmetric positive definite. So T(At)
is symmetric. Thus if we choose I] - II to be the Euclidean matrix norm, then (4.33)
becomes p(T(At)) S 1, where p(T(At)) denotes the spectral radius of T(At)), that
is, the condition p(T(At)) S 1 is just the definition of matrix stability condition. So
||T(At))ll S 1 4: p(AtM’1)S 2 <==> At S 2/P(M’l).

89

4.2 Wavelet Based Domain Decomposition Meth-
ods for Some Singular Fokker-Planck Equa-
tions

In this section, we apply the method discussed in the previous section and combination

with the domain decomposition technique to the following singular Fokker-Planck

equation:
5%.?) :_;_a_<5§121 _ @1332), (x,t) 6 (0,1) x (0.T)
(4.35)
110.0) = 9(3)

with boundary conditions to be discussed below.

The above equation arises in probability theory, where p(:r, t) stands for the prob-
ability density distribution of a stochastic process driven by a random Brownian
motion in space :1: and time t. The coefficients a and b in the above equation are
related to the “diffusion” and “drift” of the stochastic process.

We shall be concerned with the case where :3 lies in a finite interval which we
assume without loss of generality to be (0,1). we shall further assume that the
leading coefficient a(rc) in (4.35) vanishes at :1: = 0 and :1: = 1 so that we have a
“singular” boundary value problem.

Before discussing appropriate boundary conditions we mention that the problem
above has been studied in [41] by a Gauss-Galerkin method consisting in approxima-
tions the probability density function p(:r, t) by an n-point discrete measure with time
dependent nodes and weights. See also [41] for related literature on this problem.

In what follows we shall for the most part restrict ourselves to the case where
equation (4.35) arises in the context of population genetics [59] where the problem
of a population of fixed size N, with 2N alleles of types A1 and A2, segregating at a
particular locus is considered. In this context :1: denotes the fraction of the A1 allele
at time t. The coefficients a and b depend upon the particular mechanisms that affect
the population and various special cases are considered in [59].

In [59] the “ray method”, or the method of asymptotic expansion, is employed

to obtain solutions for early t and compared with known solutions whenever possible.

90

Another aspect is the existence of nontrivial steady-state solution of the related elliptic
boundary value problems which we addressed in Chapter 3.

We consider the case of a(:1:),b(:r) being smooth and a(:1:) > 0 in (0,1), a(0) =
a(1) = 0, a’(0) > 0, a’(1) < 0, 6(0) S 0, b(1) S 0. We assume the following conditions

atm=0andx=1z

{ limxno a(z)p(:1:,t) = 0 if b(0) = 0, (4.36)

limx.“ a(x)p(:1:,t) = 0 if b(0) = 1,

{ 11m,_,0{(2egx_0)z _ b(:1:)p(a:,t)} = 0 if 6(0) 74 0, (4.37)
11111,,_.l “Mix—”1). — b(x)p(:1:,t)} = 0 if b(1) 94 0.

Note that a’(0) > 0, a’(1) < 0 imply that a(:1:) has a simple zero at :1: = 0 and :c = 1.
To be specific, we take a(at) and 6(2) as in Section 3.4.
We study (4.35) with the following different boundary conditions.
Case 1. b(0) = 0, b(1) 74 0
In this case we consider (4.35) with corresponding boundary conditions described
in (4.36) and (4.37). To make the problem simpler as in Chapter 3, we let u(ac,t) =

a(x)p(:c, t). Then the problem becomes

s=aa—80,
u(x. 0) = 9(4)
u(0,t) = 0

(4.38)

b(0) = 0, b(1) 75 0 implies q = 0 and both 3 and r are nonzero. Thus
b(z) 4s(1— :13)(h +(1— 2h):c) — 4r

 

a(:1:) 1 — :1:
Hence the problem has a singularity at :1: = 1.
To solve the above singular problem, we decompose [0, 1] into two subintervals:
[0, 1—6], [1 —6, 1]. Over [1 —6, 1], we use asymptotic solution for small 6 to approximate
the solution of (4.38) as in Section 3.4. Over [0,1 — 6] we use the method described
in Section 4.1. Then we use domain decomposition technique to derive the matching

inner boundary conditions and solve the problem over O.

91

Intuitively, as in Section 3.4, for a small 6 > 0, b(x)/a(:1:) z —4r/(1 — :13). Since

limxnl Gut — Eu) 2 0, it behaves like

u(x,t) z (1 — a)8"H(t)

on [1 —6, 1] for some function H. Let us take u2(:1:, t) = (1 -:1:)8"H(t) as an asymptotic

solution of the problem over [1 — 6,1]. Then

87'

—.’L'

 

u2$(:1:,t)= —8r(1 — :1:)8r_lH(t) = —1 u2(:1:,t).
Over [0, 1 — 6], let u1(:1:, t) be the solution of the problem, then we should have

u1(1—6,t) = u2(1—-6,t),

11;,(1— 6,t) —_- u23(1— 6,1).

The relations above give us

 

1 _ _ u23(1— 6,t) 1 _
ux(1 6,t) — u2(1— 6,t) u (1 6,t), (4.39)
= —§6Cu1(1 — 6,t). (4.40)

The weak form of the problem over [0,1 — 6] now is given by the following

 

Find “41(31) 6 .Vj‘ for t > 0 such that (4 41)
(23‘1”): a(u1,v) for every v E 1.17.71 .
where
1—6 1-5
a(u, v) = —% uxv'dx +/ gut/d1:
0 0
1 b(1 - <5)
+2ux(1— 6)?)(1— 6) " a(1 _ 6)u(1— (5)1)“ — 6)
1

2 —§ 01-6 (ax — Eu) v’dx + 4s(h + (1 — 2h)(1— 6))u(1— 6)v(1— 6).

Once we find out ul(:c,t), we then get u2(a:,t) since u1(1— 6,t) = u2(1 — 6,t) =
68’H(t).
Case 2. b(1) 2 06(0) ¢ 0.

92

This implies that s, q 75 0 and r :: 0 So

 

 

b(:1:) _ 4s:r(1 + (1 — 2h):1:) + 4q
a(:1:) :1: i
In this case the problem has a singularity at :1: = 0 However, it can be reduced to
Case 1 by a linear map g = 1 - :13.
Case 3. 6(0) 2 b(1) = 0.
There are two situations: b(m) E 0 or b(a') E 0

If b(:1:) E 0, we are facing

_ a
“t "" aura:

“(5510): 9(3) (4-42)
u(0,t) = 11(10): 0.

Hence the semi-discrete Galerkin approximation solution is obtained by solving the

following weak form

{ Find uJ(-,t) E 6171“ for t > 0 “Ch that (4.43)

(151,10) 2 a(uh'v) for every '0 E 0171*,

where

a(u(.,t),v) = —%/01 ux(av)'d:1:.

Since the problem (4.42) has no singularity, we do not need the domain decom-
position. If g(:1:) E a(:1:), then the exact solution of the problem is u = ae‘zt, that is,
p(:1:,t) = e‘2‘.

If b(m) E 0, we must have s 75 0r = q = 0 So b/a = 4s(h + (1 — 2h):1:) is a linear

function. Hence

u(a,0) = g(:c) (4.44)

The corresponding bilinear form is given by

1 1 , 1 b ,
a(u,v) = —§/0 u,;(av) dzr + —u(av) d:1:.

0 a

Finally, we discuss the following case.

93

Case 4. 0(0) ¢ 0, b(1) 75 0

We have
ut/a = §um — (£1016 ,
1106.0) = a(w),

limx—yo (lax _ Eu) : 0, (445)

2

lim,,_,1 Gum — Eu) = 0
We let ul be the same as in Case 2, and let 1.12 be the same as in Case 1. Then we
can solve the problem similarly.

The numerical solutions of the problem in Case 3 for b E O are as in the following
tables and figures. In Table 4.1 to Table 4.3, u(at,t) is the exact solution, u*(:1:,t) is
the numerical solution. In Table 4.4, p and p“ denote exact solution and numerical
solution respectively (notice that p(:r, t) is actually independent of :6).

Some explanation of the details in solving the numerical problem is in order. The
Daubechies wavelets and their antiderivatives are used as in Section 3.4 with mesh
size 2'7, and the problem is reduced to the form (4.18). Explicit time schemes are
used to solve the time matching problem with At = 0.01, which make these schemes
stable. The solutions are computed to values t = 10 at which these solutions well

approach the steady state solution u E 0

94

Table 4.1: Solution at t = 1.

 

(l?

u(x,t)

u*(:1:,t)

u—u"

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

0000000000000
0001024266060
0001982450438
0003700574151
0004460513486
0.005154371139
0006343841402
0006839454011
0007268984939
0.007929801752
0008161087637
0008326291840
0008458455202
0008425414362
0.008326291840
0.007929801752
0007632434186
0.007268984939
0006343841402
0.005782147111
0.005154371139
0.003700574151
0002874553135
0.001982450438
0000000000000

 

0000000000000
0001024601912
0001983100475
0003701787553
0004461976068
0005156061234
0006345921519
0006841696638
0007271368407
0007932401899
0008163763621
0008329021994
0008461228692
0008428177018
0.008329021994
0.007932401899
0007634936828
0.007271368407
0.006345921519
0005784043051
0.005156061234
0.003701787553
0002875495688
0.001983100475
0000000000000

 

0000000000000
-0000000335852
-0000000650037
-0000001213402
-0000001462583
-0000001690096
-0000002080118
-0000002242627
-0000002383468
-0000002600147
-0000002675985
-0.000002730154
-0000002773490
-0.000002762656
-0000002730154
-0000002600147
-0000002502641
-0000002383468
-0000002080118
—0000001895940
-0000001690096
-0000001213402
~0000000942553
-0000000650037
0000000000000

 

95

 

Table 4.2: Solution at t = 1.5.

 

III

u(z, t)

u“(:1:,t)

u—u“

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

0000000000000
0000376806426
0000729302759
0001361365151
0001640931208
0001896187174
0002333768830
0002516094520
0.002674110117
0.002917211037
0003002296359
0003063071589
0.003111691773
0003099536727
0.003063071589
0.002917211037
0002807815623
0.002674110117
0002333768830
0002127133048
0.001896187174
0.001361365151
0001057489001
0000729302759
0000000000000

 

0000000000000
0000376929979
0000729541894
0.001361811536
0001641469263
0001896808926
0002334534062
0002516919536
0002674986946
0002918167578
0003003280799
0003064075957
0.003112712083
0003100553051
0.003064075957
0.002918167578
0002808736294
0.002674986946
0.002334534062
0002127830525
0.001896808926
0.001361811536
0001057835747
0000729541894
0000000000000

 

0000000000000
-0000000123553
-0000000239135
-0000000446 386
-0000000538054
—0000000621751
-0000000765 232
-0000000825016
-0000000876829
-0000000956541
-0000000984440
—0000001004368
-0000001020310
-0000001016324
-0000001004368
—0000000956541
—0000000920670
-0000000876829
-0000000765 232
-0000000697478
-0000000621751
-0000000446386
—0000000346 746
-0000000239135
0000000000000

 

96

 

Table 4.3: Solution at t = 2.

 

(l7

u(z, t)

u*(:r,t)

u—u"

 

 

0.000000
0.031250
0.062500
0.125000
0.156250
0.187500
0.250000
0.281250
0.312500
0.375000
0.406250
0.437500
0.500000
0.531250
0.562500
0.625000
0.656250
0.687500
0.750000
0.781250
0.812500
0.875000
0.906250
0.937500
1.000000

 

 

0000000000000
0000138619337
0000268295492
0000500818251
0000603664856
0000697568278
0000858545573
0000925619446
0000983750136
0001073181966
0001104483107
0.001126841064
0.001144727431
0.001140255839
0.001126841064
0.001073181966
0001032937642
0000983750136
0000858545573
0.000782528517
0000697568278
0000500818251
0000389028463
0000268295492
0000000000000

 

0000000000000
0000138664790
0000268383464
0000500982467
0000603862795
0000697797008
0000858827086
0000925922952
0000984072703
0001073533858
0.001104845262
0.001127210551
0.001145102782
0.001140629724
0.001127210551
0.001073533858
0001033276338
0000984072703
0000858827086
0000782785105
0000697797008
0000500982467
0000389156023
0000268383464
0000000000000

 

0000000000000
-0000000045453
-0000000087973
-0000000164216
-0000000197939
-0000000228 730
-0000000281513
-0000000303507
~0000000322567
-0000000351892
-0000000362155
—0000000369486
—0000000375351
-0000000373885
-0000000369486
-0000000351892
-0000000338696
-0000000322567
-0000000281513
-0000000256588
—0000000228 730
-0000000164216
-0000000127561
-0000000087973

0000000000000

 

97

 

Table 4.4: Solutions for t = 0 through t = 5.

 

 

 

‘ P p“ 2-2‘

0.00 1000000000000 0.999351012867 0000648987133
0.12 0778800783071 0778295351384 0000505431687
0.25 0.606530659713 0.606137029119 0000393630594
0.50 0.367879441171 0.367640692148 0000238749024
0.62 0.286504796860 0.286318858934 0000185937927
0.75 0.223130160148 0.222985351546 0.000144808603
1.00 0.135335283237 0.135247452379 0000087830857
1.12 0.105399224562 0.105330821821 0.000068402741
1.25 0082084998624 0082031726516 0000053272108
1.50 0049787068368 0049754757201 0000032311167
1.62 0038774207832 0038749043870 0000025163962
1.75 0030197383422 0030177785709 0000019597713
2.00 0018315638889 0018303752275 0000011886614
2.12 0014264233909 0014254976605 0000009257304
2.25 0011108996538 0011101786942 0000007209596
2.50 0006737946999 0006733574158 0000004372841
2.62 0005247518399 0005244112827 0000003405572
2.75 0004086771438 0004084119176 -0000002652262
3.00 0002478752177 0002477143498 0000001608678
3.12 0001930454136 0001929201296 0000001252840
3.25 0001503439193 0001502463480 0000000975713
3.50 0000911881966 0000911290166 0000000591800
3.62 0000710174389 0000709713495 0000000460894
3.75 0000553084370 0000552725426 0000000358945
4.00 0000335462628 0000335244917 0000000217711
4.12 0000261258557 0000261089004 0000000169553
4.25 0000203468369 0000203336321 0000000132048
4.50 0000123409804 0000123329713 0000000080091
4.62 0000096111652 0000096049277 0000000062375
4.75 0000074851830 0000074803252 0000000048578
5.00 0000045399930 0000045370466 0000000029464

 

 

 

 

 

98

 

 

x10
9 T I T I I Y I I Y
8" i=1 _.
7'- -4
6” 4
AS- 4
25:
a“ ‘
1:15
3- 4
2* 4
1:2
1’ 4

 

 

Figure 4.1: Numerical solutions at t = 1,1.5,2.

 

 

 

 

 

Figure 4.2: Numerical solutions from t = 0 to t = 2.

99

Chapter 5

Concluding Remarks and

Discussions

5.1 Concluding Remarks

In this section we shall make some observations. As one can see, we have been explored
in this thesis the potential applicability of wavelet based methods for the numerical
solutions of some boundary value and initial value problems, especially for problems
with singularities at boundaries. The numerical results strongly indicate that wavelet
based methods (using antiderivatives of wavelets as basis functions) share some of the
computational properties of finite element method and are well suited for multi-level
solution methods. They should be excellent candidates for solving ODE and PDE
problems in the future. As the theoretical base of such numerical methods, we have
established that antiderivatives of Daubechies wavelets with p Z 1 actually provide
unconditional bases (the next best thing to orthonormal wavelets) for some Sobolev
spaces under our consideration. We have also been derived some approximation
properties for both wavelets and their antiderivatives, which support wavelet based
numerical methods we have discussed in Chapter 3 and Chapter 4. We observe that
similar to a multi-resolution analysis for L2(R), for the given Daubechies wavelets

with p _>_ 1, we have the corresponding “multi-resolution analysis” for H1(Q) (or

HIM), 0r 171(9))-

100

5.2 Discussions

As the up and coming algorithms, wavelet based methods are suitable for various
numerical ODE and PDE problems and computing architectures. The results ob—
tained in this thesis are just one small step in the whole evolution of the application
of wavelets and their antiderivatives. There is a tremendous amount of work ahead
in the developments and analyses of the algorithms in different cases, as well as for
various applications. There is also a large amount of work ahead in the development
of wavelet analysis itself to suit different computational needs.

Finally we shall say a few words about possible immediate future research. The
first direction of the future research involves the further investigation and imple-
mentation on wavelet based methods for other elliptic and parabolic boundary value
problems, and in particular for other Fokker-Planck equations with singularities at
boundaries that we have not reported here. There is also a work ahead involving new
constructions of other (weighted) wavelets and their antiderivtives on bounded one

and higher dimensional domains.

101

Bibliography

[1] W. F. Ames, Numerical Methods for Partial Diflerential Equations, Academic
Press, 1992.

[2] O. Axelsson and G. Lindskog, 0n the rate of convergence of the preconditioned
conjugate gradient method, Numer. Math., 48(1986), 499-523.

[3] E. Bacry, S. Mallat and G. Papanicolaou A wavelet based space-time adaptive
numerical method for partial diflerential equations, Mathematical Modeling and

Numerical Analysis, 26(1992), no.6, 793-834.

[4] J. Benedetto and M. Frazier, eds., Wavelets: Mathematics and Applications,
CRC Press, 1993.

[5] G. Beylkin, Wavelets, Multiresolution Analysis and Fast Numerical Algorithms,
A draft of IN RIA Lecture Notes(1991).

[6] G. Beylkin, 0n wavelet-based algorithms for solving differential equations,
preprint( 1992).

[7] G. Beylkin, Wavelets and fast numerical algorithms, in Different Perspectives on
Wavelets (I. Daubechies ed.), Proceedings of Symposia in Applied Mathematics,
47(1993), 89-117.

[8] G. Beylkin, R. R. Coifman, and V. Rokhlin, Fast wavelet transforms and nu-
merical algorithms I, Comm. Pure and Appl. Math., 44(1991), 141-183.

[9] C. K. Chui, An Introduction to Wavelets, Academic Press, New York, 1992.

102

[10] C. K. Chui (ed.), Wavelets: A Tutorial in Theory and Applications, Academic
Press, New York, 1992.

[11] C. K. Chui and J. Z. Wang, A general framework of compactly supported splines
and wavelets, J. Approx. Theory, 71(1992), 263-304.

[12] C. K. Chui and J. Z. Wang, A cardinal spline approach to wavelets, Proc. Amer.
Math. 800, 113(1991), 785-793.

[13] C. K. Chui and J. Z. Wang, 0n compactly supported spline wavelets and a
duality principle, Trans. Amer. Math. 800, 330(1992), 903—916.

[14] P. G. Ciarlet, Finite Element Methods for Elliptic Problems, North-Holland,
Amsterdam, 1978.

[15] A. Cohen, Ondelettes, analysis multire’solutions et filtres miroir en quadrature,

Ann. Inst. H. Poincaré, Anal. non Linéaire, 7(1990), 439-459.

[16] A. Cohen and J. P. Conze, Re’gularite’ des bases d’ondelettes et measures er-

godiques, Rev. Math. Iberoamer, to appear.

[17] A. Cohen and I. Daubechies, Orthonormal bases of compactly supported wavelets
III: Better frequency localization, SIAM J. Math. Anal., to appear.

[18] A. Cohen, 1. Daubechies, J. Feauveau, Bi-orthogonal bases of compactly sup-
ported wavelets, Comm. Pure Appl. Matyh., 45(1992), 485-560.

[19] A. Cohen, 1. Daubechies, B. Jawerth and P. Vial, Multiresolution analysis,
wavelets and fast algorithms on an interval, Comptes Rendus Acad. Sc. Paris,

Série 1, 316(1992), 417—421.

[20] A. Cohen, I. Daubechies and P. Vial, Wavelets on the interval and fast wavelet
transforms, preprint (1993).

[21] R. R. Coifman, P.W. Jones, and S. Semmes, Two elementary proofs of the
L2 boundedness of Cauchy integrals on Lipschitz curves, J. Amer. Math. Soc.,

2(1989), 553-564.

103

[22]

[23]

[24]

[25]

[26]

[27]

[28]

[29]

[30]

[31]

[32]

I. Daubechies, Orthonormal bases of compactly supported wavelets, Comm. Pure

and Appl. Math., 41(1988), 909-996.

I. Daubechies, Ten Lectures on Wavelets, CBMS-N SF Series in Applied Math-
ematics , SIAM, 1992.

I. Daubechies, Two recent results on wavelets: Wavelet bases for the interval,
and biorthogonal wavelets diagonalizing the derivative operator, in Recent Ad-
vances in Wavelet Analysis (L. L. Schumaker and G. Webb eds.), Academic
Press, New York, 1993, 237-258.

I. Daubechies, editor, Different Perspectives on Wavelets, Amer. Math. Soc.,
1993.

I. Daubechies, Orthonormal bases of compactly supported wavelets II: Variations

on a theme, SIAM J. Math. Anal., 24(1993), 499-519.

G. Deslauriers and S. Dubug, Interpolation dyadique, In Fractals, Dimeions non

Entieres et Applications, G. Cherbit ed., Masson, Paris, 1987, 44-55.

R. J. Duffin and A. C. Schaeffer, A class of nonharmonic Fourier series, Trans.

Am. Math. Soc., 72(1952), 341—366.

J. Epperson and M. Frazier, An almost orthogonal radial wavelet expansion for

radial distributions, preprint, 1992.

J. Epperson and M. Frazier, Polar wavelets and associated Littlewood-Paley

theory, preprint, 1994.

M. Frazier and B. Jawerth, The cb-transform and applications to distribution

spaces, in Function Spaces and Applications (M. Cwikel et at. eds.), Lecture

Notes in Math., 1302(1988), Springer, 223-246.

M. Frazier and B. Jawerth, Applications of the d) and wavelet transforms to
the theory of function, in Wavelets and Their Applications (Ruskai et al. eds.),
Jones and Bartlett, Boston, 1992, 337-417.

104

[33] M. Frazier and B. Jawerth, and G. Weiss, Littlewood—Paley Theory and the
Study of Function Spaces, CBMS Regional Conference Series 79(1991), A.M.S.,

Providence, RI.

[34] R. Glowinski, Numerical Methods for Nonlinear Variational Problems, Springer-
Verlag, 1993.

[35] R. Glowinski, W. Lawton, M. Ravachol, and E. Tenenbaum, Wavelet solutions
of linear and nonlinear elliptic, parabolic and hyperbolic problems in one space

dimension, in Computing Methods in Applied Sciences and Engineering (R.

Glowinski et al. eds.)(1992), SIAM, 55—120.

[36] R. Glowinski, J. L. Lions and R. Treémolieres, Numerical Analysis of Varia-
tional Inequalities, North-Holland, 1976.

[37] R. Glowinski, T. W. Pan, R. 0. Wells, X. Zhou, Wavelet and finite element

solutions for the Neumann problem using fictitious domains, preprint(1994).

[38] R. Glowinski, A. Rieder, R. 0. Wells, X. Zhou, A preconditioned CG-method
for wavelet-Galerkin discretizations of elliptic problems, preprint(1995).

[39] G. H. Golub and C. F. Van Loan, Matrix Computations, 2nd ed., The Johns
Hopkins University Press, 1989.

[40] A. Haar, Zur theorie der orthogonalen funktionensysteme, Mathemetische An-
nalen (1910), 331-371.

[41] A. Hajjafar, H. Salehi, and D.H.Y. Yen, On a class of Gauss-Galerkin methods
for Markov processes, in Nonstationary Stochastic Stochastic Processes and

Their Applications (A. G. Miamee ed.), 1992, 126-146.

[42] C. A. Hall, T. A. Porsching, Numerical Analysis of Partial Differential Equa-
tions, Pretice Hall, 1990.

[43] C. Heil, Wavelets and frames, in Signal Processing, 2nd ed., Springer—Verlag,
New York, 1990, 147-160.

105

[44] E. Issacson and H. B. Keller, Analysis of Numerical Methods, John Wiley &
Sons,1966.

[45] P. S. Jaffard, Wavelet methods for fast resolution of elliptic problems, SIAM J.
on Numer. Anal., 29(1992), 965-986.

[46] P. S. Jaffard et Y. Meyer, Bases d’ondelettes dans des ouverts de R”, J. Math.
Pures et Appl., 68(1989), 95-108.

[47] F. John, Partial Differential Equations, 4th ed., Springer-Verlag.

[48] S. Mallat, Multiresolution approximation and wavelet orthonormal bases of

L2(R), Trans. Am. Math. 800, 315(1989), 69-88.

[49] Y. Meyer, Principe d ’incertitude, bases hilbertiennes et algebres d ’operateurs,

Seminaire Bourbaki (1985-1986), no. 662.

[50] Y. Meyer, Wavelets and Operators, Cambridge Studies in Advanced Mathemat-
ics, 37, 1992.

[51] Y. Meyer, Ondelettes sur l’intervalle, Revista Mathematica lberoamericana,

7(1991), 115-133.

[52] D. O’regan, Theory of Singular Boundary Value Problems, World Scientific
Publishing, Singapore, 1994.

[53] D. L. Powers, Boundary Value Problems, Academic Press, New York, 1979.

[54] J. M. Restrepo, G. K. Leaf, Wavelet-Galerkin discretization of hyperbolic equa-
tions, preprint(1994).

[55] J. M. Restrepo, G. K. Leaf, G. Schlossnagle, Periodized Daubechies wavelets,
preprint(1994).

[56] G. D. Smith, Numerical Solution of Partial Differential Equations: Finite Dif-
ference Methods, Oxford University Press, Oxford, 1990.

106

[57] E. M. Stein, Singular Integrals and Difierentiability Properties of Functions,

Princeton University Press, 1970.

[58] J. Stoer and R. Bulirsch, Introduction to Numerical Analysis, 2nd ed., Springer-
Verlag, 1993.

[59] R. Voronka, and J. B. Keller, Asymptotic analysis of stochastic models in pop-
ulation genetics, Mathematical Biosciences 25(1975), 331-362.

[60] R. 0. Wells and X. Zhou, Wavelet solutions for the Dirichlet problem,
preprint(1993).

[61] J. Xu, Iterative methods by space decomposition and subspace correction, SIAM

Review, 34(1992), 581-613.

[62] J. Xu, and W. Shann, Galerkin-wavelet methods for two-point boundary value
problems, Numerische Mathematik, Springer-Verlag, 63, 123-144.

[63] O. C. Ziencicwicz, D. W. Kelly, J. Gago, and I. Babu§ka, Hierarchical finite ele-
ment approaches, error estimates and adaptive refinement, in The Mathematics
of Finite Elements and Applications IV(1982), Academic Press, London, 313-
346.

[64] A. Zygmand, Trigonometric Series, 2nd ed., Cambridge University Press, 1968.

107