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The Discretized Korteweg-de Vries Equation
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The Discretized Korteweg—de Vries Equation

By

Michael J. Nixon

A Dissertation

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1997

i‘ll' ’ it'll! l

ABSTRACT

The Discretized Korteweg—de Vries Equation
By

Michael J. Nixon

In this dissertation, we study a discretized version of the (generalized) Korteweg—
de Vries equation, Btu + Bin + u4axu = 0. After a number of estimates, we utilize the
Contraction Mapping Principle to prove the global well-posedness of this equation
in a certain discrete Banach space. Our results are analogous to those of Kenig,
Ponce, and Vega in the continuous setting. However, due to the nature of the Fourier
multipliers, the proofs of several of these estimates in the discrete setting require
new techniques. Our results yield a numerical procedure for computing the solution.
We present a numerical algorithm which is based on successive iterations to obtain
a fixed point guaranteed by the Contraction Mapping Principle. This fixed point is

the desired solution which we demonstrate with several numerical experiments.

To my family

iii

ACKNOWLEDGMENTS

Several individuals deserve my gratitude. First and foremost, I wish to express my
deep appreciation to my advisor, Professor Michael Frazier, for his excellent guidance,
encouragement, and commitment to all aspects of my graduate studies. For his efforts,
I am extremely indebted.

I would like to thank the other members of my Thesis Committee: Professors
Patricia Lamm, Joel Shapiro, William Sledd, and Zhengfang Zhou. Each of them
has contributed significantly to my graduate experience. Furthermore, I would like
to thank Professor Bruce Sagan for bringing a very helpful combinatorial identity to
our attention.

I would also like to convey my sincere appreciation to my fellow graduate student,
Richard Hensh, for his effort and thoughtful suggestions in regards to the numerical
examples in my thesis.

iv

TABLE OF CONTENTS

INTRODUCTION

1 The Main Result
1.1 Preliminaries ...............................
1.2 Discretization of (K dV)4 .........................
1.3 A Discretized Version of Duhamel’s Principle ..............
1.4 The Operator (Pm .............................
1.5 The Space Xh ...............................
1.6 Contraction Mapping Theorem .....................
1.7 Well-Posedness of (K dV): in X h .....................

2 The Estimates
2.1 A Discrete Version of Fractional Integration ..............
2.2 A Lemma Involving Fractional Differentiation .............
2.3 Proof of Lemma 1.17 ...........................
2.4 Discrete Complex Interpolation .....................

3 Numerical Results
3.1 The Cutoff Function 7,,(n, m) ......................
3.2 The Coefficients Q[n, m] .........................
3.3 A Few Examples .............................

BIBLIOGRAPHY

14
14
18
2O
21
22
25
34

36
36
38
47
83

85
85
90
92

99

Introduction

Let 8' (R) be the set of tempered distributions on the real line. Let [3 denote a Banach
space such that B C 8' (R) The following initial value problems which arise in the
study of water waves are called the (generalized) K orteweg-de Vries equations,

8tu+82u+ukaxu=0 xER,tER,k€N
(KdV)k

U|t=0 2 U0 E B.
We say that (K dV),c is globally well-posed in B if there exists a unique u = u(:c, t)
such that the following holds:

0 u E C((—oo, +00); 8), i.e., u(-,t) E B for all t E R and t —+ u(-,t) is continuous

from R to B.

o u(-,0) = no.

(-, t + h) — u(-, t)

 

o For t E R, Ilia?) it exists as an element of 8’ (R). Define
—)

amp, t) = ’12:?) u(., t + h}: — u(., t).

 

o The nonlinear term uk(-, t)8$u(o, t) E 8' (R) for all t E R.
o u is a solution to (K dV),c in the distributional sense.

0 The mapping no —> u is continuous from B to C((—oo, +00); 8).

We say that (K dV),c is locally well-posed if the above holds with C((—oo, +00); 8)
replaced by C((Ao, A1); 8) for some A0 < O and A1 > 0.

1

For F E 8' (R), let F be the Fourier Transform of F, defined by

- 1 .
mg) = Zr' A F(:c)e"x‘€da:.

For 3 E R, the Sobolev Space of order s, denoted H ”(R), is the set of tempered

distributions F for which F is a function and

llFllip = l. |F(€)I2(1 + was < 00.

The homogeneous Sobolev Space of order s, denoted H‘(R), is the set of tempered

distributions F for which F is a function and

I|F||2-. = [R |F(€)|2€2‘d{ < 00.

Obviously, H ’ (R) C H 3 (R) for s 2 0. It is well known that H ‘ (R) and H 3 (R) are

Hilbert spaces with inner product in H ‘(R) defined by

< HG >= AF(£)5(£)(1+€2)Sd£

and in H 3 (R) by
< F,G >= [R F(€)5(£)§2’d€-

In their 1976 paper [1], using energy methods, J. Bona and R. Scott proved that
(K dV),c is locally well-posed in H ’(R) for s > 3/ 2, for all k E N. By the Sobolev
Imbedding Theorem for s > 3/ 2, H ‘(R) C Cl(R), the set of all continuously dif-
ferentiable functions on R. Hence, this result is restrictive in the sense that it can
not guarantee a solution to (K dV),c if the initial data no has, say, a jump discontinu-
ity. However, the homogeneity of the nonlinear term suggests that well-posedness in

H ‘(R) may hold for smaller values of 3. That is, if u solves (K dV)k, then for A > 0,

so does

u,\(:c, t) = A2/ku()\:r, A3t)

with initial data

u,\(a:, 0) = Az/kuO‘m, 0).

Since we are using a scaling argument, let us assume that these solutions are in the

homogeneous Sobolev space, H"(R). Note that

lluA($,0)llip = AQ/kA’A‘l/2IIU($,0)IIH»

This suggests that the optimal s for (K dV)k is

NIH
|
Prlto

8);:

Note that 3;, Z 0 if and only if k 2 4.

In their 1993 paper [5], Kenig, Ponce, and Vega proved that (K dV),c is globally
well-posed in H” (R) for k 2 4 provided that the initial data is small, i.e., Iluol I gm 3 6
for some 6 > 0. If the initial data is not small, they proved local well-posedness in
H 3" (R) Note that by Plancherel’s Theorem, H°(R) = L2(R), the set of square
integrable functions on R. Thus, for k = 4, the smallest value of k for which Kenig,
Ponce, and Vega obtained well-posedness in H ”k (R), we have global well-posedness
in the familiar function space L2 (R) Because of this, the remainder of this thesis
will focus on the Korteweg-de Vries equation with k = 4 and with B = L2(R), i.e.,

6tu+82u+u40xu=0 a: E R,t€R
“{th

. Ult=o = no G L2(R).

We now give a brief outline of the proof in [5] for k = 4. Consider the associated

homogeneous linear equation

wltzO = 100 E L2(R)

and the corresponding inhomogeneous equation

6v+0§v=
(IH) t .9

'U]t:0 = 0.

One can easily verify that (H) is globally well-posed in L2(R) and the solution is

given by
wait) = W(t)wo($) = Ae‘<$‘+“3>wo<5)d€. (1)

Note that W(t) is a Fourier multiplier operator with multiplier ems, i.e.,

(W(t)wo)“(€) = e“‘3<120(€)-

Using Duhamel’s Principle, one can obtain that (I H ) is globally well-posed in L2(R)

for 9 such that g(-, t) E L2(R) for all t E R with the solution given by
v(:r, t) = [0‘ we — t')g(-, t')(a:)dt'. (2)
To see the connection with (K dV)4, if
Su(:c, t) = W(t)u0(:z:) — [0‘ we — t')(u46$u)(o, t')(:z:)dt', (3)

then (1) and (2) imply that u solves (K dV)4 if and only if u = Su, provided Sn

makes sense as a distribution. In other words, a is a solution of (K dV)4 if and only

if u is a fixed point of the nonlinear Operator S.

More precisely, for T E (0, +00], define

2T = {veC((—T,T);L2(R))nL°°((-T,T);L2(R))=

||v||L2Lpo < 00 and ”mung; < 00},

where

T We l/p
IIvIILm = (f, (LT Iv(m.t)l"dt) dx]

forlgp,q§00. Ifq=00, then

1/10
may“ sup |v(:v,t)l"drr)

tel—737"]

with a similar definition for p = 00. The norm for these Banach spaces is

”UHZT =max{ SUP llv('at)l]L2;llUlngL10;llaxvlngotzl-
t _ T] T T

1

By proving various estimates with these mixed norms involving the terms
W(t)uo($) (4)

and

[we — t’><u4a.u)(-,t’)<x)dt’. (5)

Kenig, Ponce, and Vega were able to show that given small initial data, S is a con-
traction mapping on B, = {v 6 Zoo : ||v||zm S r} for some r > O. For arbitrary initial
data, they were able to find a T > 0 for which S is a contraction on B, C ZT. Hence,

the Contraction Mapping Principle guarantees a solution to u = So in this ball in

ZT.

Contraction mappings and fixed point iterations are useful tools in the numerical
study of differential equations. Can the method developed in [5] be adapted to show
that a discrete version of (K dV)4 is well-posed? Is the solution a fixed point of some
operator in some space? If so, does this lead to a method that can be implemented
numerically? In this dissertation, we answer these questions in the affirmative.

The general outline of the proof for the discrete setting is analogous to that of
the continuous setting. However, there are major differences that make the discrete
proofs different and in some cases quite a bit more involved than their continuous
counterparts.

The first step in this process is to discretize (K dV)4 in a natural way. To his end,
fix h > O and let us first consider the associated linear equation (H). Let w(n, m)
be a discrete function defined on Z x Z. The obvious replacement for 63u(x,t) is
aghwm, m) (see Definition 1.6). However, the replacement of am is not so clear.

Assuming m > O, on the one hand, we could replace 6tu(:r, t) with

w(n, m + 1) — w(n, m)

h3

 

and (H) becomes

w(n, m +1) - w(n, m)
h3

 

+ 83,,,w(n, m) = 0. (6)

Note, for homogeneity purposes we are letting h be the step size in the LII-direction,
while h3 is the step size in the t-direction. On the other hand, we could replace

Btu(a', t) with
w(n, m) — w(n,m — 1)

h3

 

and (H) becomes

w(n, m) — w(n, m — 1)

h3

 

+ 8,3,,hw(n, m) = O. (7)

The scheme in (6) is referred to as an explicit scheme, because, given the values
w(n, m) at height m, one can solve explicitly for the values of w(n, m + 1) at height
m+ 1. The other is called an implicit scheme since the values w(n, m) at height m are
determined implicitly by the values w(n, m — 1) at height m — 1. It may seem that the
explicit scheme is the more convenient of the two. However, considering the Fourier
multiplier which corresponds to ems, we will see that the implicit scheme is the better
choice. Here, the Fourier Transform takes functions defined on Z to functions defined
on [-7r/h,7r/h] (see Definition 1.2).

Let wo be the initial data. If we discretize explicitly, taking the Fourier Transform

of both sides in the first variable, (6) becomes

w(1)(g m+ 1)- (”(0, m) isin3(h9)a,<1)
h3 + ’13

 

(0, m)=0,

which implies

am, m +1): (1 — <h6)>ai”(0,m)

and hence,

afflw, m) = (1 — isin3 (ham, <1 (0).

Notice the multiplier (1 — isin3(h9))’" has magnitude (1 + sin6(h0))'"/2, which blows
up for 0 < l0] < 1r/h as m —) 00. Hence, by Plancherel’s Theorem (Proposition 1.3),
the norms of w(-, m) in the discrete space 1,2,(Z) (see Definition 1.1) go to infinity with

m. This is undesirable both mathematically and physically.

Applying the same argument to (7 ), it follows that

will?)
(1 + isin3(h6))m

 

aw, m) =

(for the proof of this see §1.2). Note that the multiplier (1 + i sin3(h0))"" has magni-
tude which is bounded by one for all 0 6 [—7r / h, 7r / h] and hence, the norm of w(, m)
in l2 h(Z) will be bounded by ||wo||,2 for every m. Therefore, we choose the implicit

scheme 1n (7) and the resulting multiplier

I
1 + isin3(h0)' (8)

 

The fact that the multiplier is not a pure exponential, like eitfa, causes difficulties in
the proofs of the discrete case that do not arise in the continuous case. For example,
consider the following estimate proved in [5], which is used in establishing that S is

a contraction mapping.

Lemma 0.1 Let mo 6 L2(R). Then
llaxW(t)wol|LgoL§ S Cllwolltz,

where

1/2
anrm = sup (/ WW) .
16R R

Proof: Recall, W(t)wo(:r) = [R ei($£+‘€3)wo(€)d€. It follows, after the change of vari-

ables 53 = 7, that

azW(t)’LUo($) = i/ée'($€+t€3)co( £)d€

: fife itr eixrl flT—2/3+1/31D0(T1/3)dT.

Thus, using Plancherel’s Theorem in the t-variable,

- 1/3 _ .. 2
6131 7. 1/3w0(T1/3)l d7"

 

1
Hawumonig = 5 [R

c (1. words

again using 5 = 71/3, which finishes the proof.

This proof is simplified by the fact that the multiplier eit’i3 has magnitude one for
all 5 6 R, and the fact that after the substitution 7 = 63, the term e“T is just what is
needed to regard the integral as a Fourier transform in the t-variable. The multiplier
(1 + isin3(h0))"" is not exactly of this form, so when we proceed as in the proof of
Lemma 0.1 we encounter an “error” term after the change of variables. The proof

will be reduced to the following inequality (see Proposition 2.9):

W .
2 / e“'""r cos'mlrg(r)dr
mEZ ’1'

2 s c f; Igmrczr, (9)

 

 

where the cosIml r is the anticipated extra term. If the proof was completely analogous
to the proof of Lemma 0.1, we would not have this term and (9) would become
a statement of Plancherel’s Theorem involving Fourier series. Instead, the proof
requires some additional techniques including a combinatorial identity (see the proof
of Claim 2.11 following Proposition 2.9) pointed out to us by Bruce Sagan.

Next, consider m < 0. The implicit scheme in this case is

w(n, m +123- w(n, m) + fi’hwm, m) = 0. (10)

 

10
After applying the argument in §1.2, (10) becomes

cow)
(1 — isin3(h0))|ml

 

«rifle, m) =

and the resulting multiplier is
1
1 — isin3(h0)° (11)

 

Notice that because of our need to control the size of the multiplier, we now have two
multipliers, (8) for m > 0 and (11) for m < 0.

Recall that the Fourier multiplier in the continuous case associated with W(t) is
eitia. Since

- 3 - 3 - 3
et(t+s)£ : eit£ €136 ,

the family of operators {W(t)}teR forms a group under composition, i.e.,

W(t + s)wo = W(t)W(s)w0 for all s,t E R.

This plays an important role in the proof of the key estimate

 

‘ t I I I

la: / W(t—t)g(-.t><x)dt ]| s angina? (12)
0 Lg°L§

needed in [5]. If

(770)1109)
(1 + Sgn(m)i sin3(h0))|m|’

 

(Hh(m)no)‘(9) =

then it follows for m1, m2 6 Z that

Hh(m1+ m2)170 ?'é Hh(m1)Hh(m2)7lo

if m1 . m2 < O. This forces the corresponding proof in the discrete setting (see Lemma

1.17(ii)) to proceed along different lines than the proof of (12) in [5]. This necessitates

11

the technically difficult proof of Pr0position 2.19 in addition to the relatively simple
proof of PrOposition 2.17 (see Remark 2.20).

Despite the additional difficulties resulting from the multiplier in the discrete
formulation, we are still able to obtain the full analog of the results in [5]. We now
turn to the formal statement of the problem and our results. Since the nonlinear term

of (K dV)4 can be written as $83,053), we discretize (K dV)4 as follows:

"(72,771) — ”(71,772. _ 1)

 

1
+ 33mm”, m) + gan,h(775)(na m) = 0 m > 0

 

h3
(KW): n(n, 0) = new m z 0
1 _ 1
770% m + h)3 r)(n, m) + 3131,1100”: m) + gan,,,(n5)(n, m) = O m < 0’

where 770 is the discrete initial data. If

Hh(m)no(n)

is the analog of (4) and solves the discrete version of (H) and

1
gAhan,h(715)(n, m)

is the analog of (5) (see Definitions 1.12 and 1.14), then the analog of the Operator S
in (3) is

mm. m) = Hh(m>no(n> — §Ah6.,h(n5>(n. m).

It will follow (see §1.4) that 17 solves (K dV)2 if and only if

n = q’non-

Analogous to the space Zco from [5], we will define a discrete space X h (see Definition

1.16). The primary conclusion of this thesis is the following:

12

Main Result: Suppose the initial data 770 is small, i.e.,

nEZ

1/2
llnollzg = (h 2 WWW) S 60

for some do > 0. Then there exists r > 0 such that the operator (1),,0 is a contraction
mapping in

Br = {77 E Xh 3 ”77]th S T},

with 60 and r independent of h > 0. Consequently, by the Contraction Mapping Prin-
ciple, there exists a unique solution 17 E B, of (K dVfl. Moreover, n E l°°(l,2,(Z); Z)

and the map

770—”)

is continuous from lflZ) to Xh. Hence, (K dV)f,i is well-posed in Xh.

This result lends itself to a numerical procedure for computing the solution to
(K dV)Z. To do this, we first numerically implement the operator <I>,,0. This will be
done by using the coefficients Q[n, m] which have been precomputed and stored (see
§3.2). Then, by picking the initial guess to be the zero function on Z x Z, we run
the fixed point iteration. If the initial data is small enough, then the main result
guarantees that the iterates (bglm) converge to the solution of (K dV): as n —> 00.
Also, because the contraction mapping constant is less than one, the convergence is
exponential. The fact that the operator (Pm, and the iteration scheme are easy to
implement is the motivation for this thesis.

This approach differs from the standard approach in several ways. The standard

approach is to solve the difference equations iteratively going up one level at a time,

13

while we work on the entire space Z x Z for each iteration. A disadvantage of our
approach is the amount of computer storage required. Each global iteration has to
be stored in order to define the next iteration, whereas when solving the difference
equations level by level, one can discard the data at each level. However, the advan-
tages of our method include the rapid computation of the entire solution on the entire
grid, and the fact that we avoid problems with ill-conditioning and accumulation of
round-off error. In addition, our approach guarantees certain size estimates on the
iterations, namely the three norms of the space Xh (see Definition 1.16) including a
bound on the l2 norms of each level. Finally, note that our result holds for all h > 0
not just, for example, small h. This distinguishes our approach from another possible
approach in which one tries to compare <I>,,o, for small h, with S, which is known to
be contraction map in the continuous setting. Thus what we have obtained is a true
difference equation result.

In Chapter 1, we state several definitions which are used throughout this the-
sis. After this, we show that our choice of discretization yields a nonlinear discrete
operator whose fixed point solves (K dV)f,‘. Then, we state the crucial estimates and
define the discrete Banach space X h in which we find our solution. Finally, with these
estimates, we establish that we have a contraction mapping, which quickly leads to a
solution of (K dV)j,‘. This part of the proof follows the same general outline as in [5].
Chapter 2 contains the proofs of the main estimates. It is here that techniques dif-
ferent from those in [5] are required to overcome the difficulties noted above related
to the the multipliers in the discrete setting. Finally, results of various numerical
experiments are presented in Chapter 3. An explanation of the numerical algorithm

is included there as well.

CHAPTER 1

The Main Result

1 . 1 Preliminaries

Let h > 0. In the following definitions, let a and 9 denote functions defined on Z and

w denote a function defined on Z x Z.

Definition 1.1 o E 1,2,(Z) if

Hallzg = (h 2 |0(n)|2)1/2 < 00.

7162

If h represents the step size and 0(n) = f (nh) for, say, f E C (R) 0L2 (R), then Hallfg

represents a Riemann sum of I f |2. Hence, ”Gilli ~ I] f H L2 for small h.

Definition 1.2 For 0 E [—7r/h,7r/h], let

6,,(0) = h z a(k)e"‘ho.

kEZ

14

15

Note that by Fourier inversion,

_ 1 "/h . -inh0
0(n) — 27r [Tr/h ah(0)e d0.

Proposition 1. 3 If a E l2 ,,,(Z) then

1 “M 1/2
Halli; = (5; lo. (more) .

Proof:

1 "/h . 2 _ 1 "/h ikho tho
2n/_./1."”‘(9)'d9 _ 27: /_W/h(hZa(k)e) ()(hzao )eJ

kEZ jEZ

= 72/1r/hh /h h2 Z Z( 009 —j)ei(k-j)h0d6

kEZjEZ

1 vr/h
= -— h2 k 2d
2.1,. EM 9

kEZ

= hEIOUCV

kEZ
= Halli;-

 

Definition 1.4 For a and g E 1,2,(Z), let

(o*g)(n =(lc)h£a

kEZ

be the convolution of or and 9.

Remark 1.5 One can see as in the standard case with h = 1 that

nEZ kEZ

(0 =1: 9),, 2h E (h 2 o(k)g( n-—k)) einho

16

= h 2: 0(k)eikh0h 2 9(7), _ k)ei(n—k)h0
lcEZ nEZ

= 5h(9)§h(9)

for 0 E [—7r/h,7r/h].

Definition 1.6 Let
0(n +1) — 0(n —- 1)

6ho(n) 2 2h

 

For k E N, define 6§o(n) inductively, i.e.,

8§o(n) = chat-10(7),).

o(n+3) —3o(n+1)+30(n— 1) —o(n—3).

 

For example, 03,,0(n) 2

8h3
Proposition 1.7 Let a E l§(Z). Then
A —-i sin h0 .
(6h0)h(0) = ——h(—)-oh(0).

Proof: By definition,

(0h0)7,(6) = h26h0(k)eikh9
keZ

= h: 0(k‘l' 1) ”JUV- 1)eikh0
1Icez 2h

2 _ 220( [9+ 1)€ ikh0__ 5W )eikhe.
2k€Z 2kEZ

 

17

By setting j = k + 1 in the first sum and setting j = k — 1 in the second sum, we

have

(aloha) = $2; 0(j)e"'"”(e""‘” — em)
singhfl) 5;; (0).

=—i

 

By induction on k, it follows that

 

Definition 1.8 For a E lfi(Z), fl 6 C, let

sin(h0)

l9
h &h(0)e"""9d0.

 

1 1r/h
D5001) : 57; ./;1r/h

 

 

Remark 1.9 Due to the homogeneity of the associated linear equation, if h is the
step size in the x-direction, then h3 will be the step size in the t-direction. This

motivates the following definition.

Definition 1.10 For 1 g p, q s 00, w E lfilfn,h(Z x Z) if

nEZ mez

p/q llp
(bum:(hzhazlwmmn‘?) ] <oo.

Up = 00, then

1/a
llwllzgozjnh = SUP (ha: |w(n,m)|q)
' nEZ

mEZ

18

and if q = 00, then

1/r
”canto, = (h : sup Iw(n,m)l”) .

Definition 1.11 Let

w(n+1,m) —- w(n — 1,m)
2h

an,hw(n, m) =

 

and define aghwm, m) inductively as above.

1.2 Discretization of (K dV)4

Recall our discretization of (K dV)4,

77(n, m) - 17(n, m - 1)

 

1
+ 82,,,17(n, m) + gan,h(n5)(n, m) = 0 m > O

 

h3
(Ken/)2 170210) = now) m z 0
770% m +123- "(na 7’” + aghna, m) + gamrxn, m) = 0 m < 0-

We discretize the linear equation implicitly as well,

w(n, m) -— w(n, m — 1)

 

+ 83,hw(n,m) = 0 m > 0

 

h3

w(n,0) = 01001) m = O (1.1)
1 _

W(n,m 'l" ’33 W(nam) + ag,hw(n’ m) = 0 m < 0.

Assume m > 0 and let w(',m) 6 lf,(Z) for all m 6 Z be a solution to (1.1). By

taking ‘ of both sides of (1.1) in the first variable, we have

affix, m) - affix, m — 1) isin3(h6)
h3 + h3

 

(12,9)(9, m) = 0,

19

with

(22,9)(6, m) = h E w(n, m)ei"h9.
7162
This implies
wile. m — 1)
1 + isin3(h6) ’

 

cow, m) =

and hence,

 

.. (1) _ (L00)h(0)
to), (01m) ‘ (1 + isin3(h0))'"'

If m < 0, reasoning as above, it follows that

.(1) __ (030)}:(9)
wh (0’ m) — (1 —— isin3(h0))lml'

 

This leads to the following definition.

Definition 1.12 For m,n 6 Z and wo E 1,2,(Z), let

Hh(m)wo(n)

 

1 [m e“""”<eo)h(0)
= — . , 3 d6.
27r -—1r/h (1 + sgn(m)z s1n (h0))|m|

Remark 1.13 Note,

A _ (Leona)
(Hh(m)w0) (9) - (1+sgn(m)isin3(h0))'m"

 

Hence, by Plancherel’s Theorem, Hh(m) : l§(Z) —-> l,2,(Z) is bounded with operator
norm less than or equal to 1. Also, Hh(0) = I where I is the identity operator on

l,2,(Z). Moreover, by the argument above and Fourier inversion, if

w(n, m) = Hh(m)wo(n),

then (12 is the unique solution to (1.1) for w(-, m) 6 1?,(Z) for all m E Z.

20
1.3 A Discretized Version of Duhamel’s Principle

Proceeding along the same path as [5], we now consider a discretized version of the

inhomogeneous linear equation.

Definition 1.14 For (1) such that w(-,m) E lflZ) for all m E Z, let

h3ZHh(m +1 —j)w(',j)(n) m > 0
j=l
Ahw(n, m) = 0 m = O

-—h3 i: Hh(m —1-j)w(-,j)(n) m < 0.

Proposition 1.15 Let n be such that n(-, m) E 12 ,,(Z) for all m E Z. Let w(n, m)— —

Ahn(n, m). Then w solves

w(n, m) — w(n, m — 1)

 

+ aghwm, m) = n(n, m) m > O

 

h3
wmm+1-wmm
( ’33 ( ) + 33,,hw(n, m) = 17(n, m) m < 0.

Moreover, w is unique for w(-, m) E lf,(Z) for all m E Z.

Proof: The case m = 0 is trivial by definition of Am. Assume m > 0. Then

 

(“’"ml 3:,“ m " 1) + 63,hw(-,m))i(6>

= (ZHh(m+1—j)n(,j)- -EH Hh(m— ])77(' J)

+h3§233,hHh( m(+1-J')n ( 9)) (9)

 

 

j=1
___ 2m: 1 fl(1)(j)0 _mE—f 1 fi(1)(0 )
j=l( (1+isin3(h0))m+1‘3 h ’ i=1 (1+isin3(h0))m'j h "7
+isin3(h9) Z offlw, j)

 

j=1((1+isin3(h0))'"+1 J

21

_ m 1(1)
— Z (1+isin3 (h6))m+1—Jnh (0’ j)

j=l(

 

1(1)
(1+isin3 (h6))"‘+1 17711 (6 j)

 

—(1+isin3(h(6))))zl1

 

m 1 1(1) .
+ZSin3( ()h6 )1; (1+isin 3(h6))m+1‘jnh (9’3)

«(1)
_ 77h (6,m) 3 6(1)
— 1+ isin3(h6) —isin (h6) 12:1(1 + isin 3(h6))m+1‘Jnh (0 j)

 

 

 

m 1 ~(1)
+isin3 ()h6 )E=:1( 1+isin3 (h6))m+1‘ _J-77h (91])

H.

1+ isin 3(h6) ,(1)
" 1+isin3(h6) " (9’7")

= 17%, m).

 

Thus,

 

(whm) 3:0,?” -1) + 63,hw(',m)) (9) — 6i1)(91m)1

with n£1)(6,m) = h 2 n(k,m)ei"h9. The proposition follows after integrating both
kEZ
sides against em”. The proof for m < 0 is very similar.

1.4 The Operator (1)170

Our goal is the following: given no E lf,(Z), find 17 such that n(-,m) E lf,(Z) for all

m E Z and
776». m) = Hh(m)no(n) - éAhan,h(775)(n1m)

”(n1 0) = 770 (n)

(1.2)

for m, n E Z. Combining Remark 1.13 and Proposition 1.15, such an n solves (K dVfl.

To find an n solving (1.2), we will prove that the operator

nun, m) = Hh(m)no(n> — §A16.,1(n5>(n,m>

22

is a contraction mapping on a ball centered at the origin in a particular Banach space.

1.5 The Space X),

We now define the discrete Banach space upon which we will show (Pm, is a contraction.

Definition 1.16 Let
X), = {W(mm) E l°°(l?.(Z); Z) 3 llwllzgzgg‘h < 001||3n,hwllzgozghh < 00},

with

ilwiixh = max{81ég ||w(-, m)llz§; llwlll5llo illan,h“1llt°<>z2 }
m

n m.h n m.h

Obviously, X h is a Banach space with the above norm. Using the estimates below,
we can find the solution to (1.2) in this Banach space X h. This will be done by proving
that (Pm, is a contraction map from a small ball around the origin in the space X), to

itself. The estimates needed are the following:

Lemma 1.17 For h > 0,

2') ||3n,hHh(m)Tlollt:°l3n,h S cllnollt:

and

ii) ||63,,,,Ahwlllgozgn.h S Cllwllzgtfmi

with c independent of h > 0.

23

Lemma 1.18 For h > 0,

2') lth(m)770llzgz;gh s cunt,

and

12') ”mum, s cunning),

with c independent of h > 0.

Before proceeding, we need the following proposition.

Pr0position 1.19 If (.1) E Xh, then 6",),(105) E 12/41253, with

ilafl,h(w5)iilz/4l;fl/h9 S 5||wllih

Proof: By definition,

5 _ 5 _
anh(w5)=w(n+1,m) w(n 1,m)
’ 2h
= w(n + 1,m)2—hw(n —1,m) (w4(n +1,m) + w3(n +1,m)w(n — 1,m)

+w2(n + 1, m)w2(n — 1,m) + w(n +1,m)w3(n — 1,m) + w4(n — 1,m))

 

 

= 8n,hw(n, m) (w4(n +1,m)+ w3(n +1,m)w(n —1,m)

+w2(n +1,m)w2(n --1,m)+ w(n +1,m)w3(n - 1,m) + w4(n - 1,m)).

Set V(n, m) = w(n + 1, m) and u(n, m) = w(n — 1, m). It follows that

4

]]6n,h(w5)]]l:/4110/9 S Z“(amhwlu4—jl‘jii15/4110/9
m,h j=0 n 771,):

4 9/8 4/5
2: h: (h3 Z Ian,hw|10/9|V4—j#j|10/9)

i=0 nEZ mE Z

24

24: (h 2 (ha 2 lan’hwlz) 5/8 (h3 E lu4‘jw'l5/2)1/2)4/5

S
j=0 nEZ mEZ mEZ
1/2 4 1/2 4/5
3 Sup h32|0n,hw|2 2 fix h32:|1/4"1uj|"’/2
"Ez mEZ j=0 nEZ mEZ
4
S iiwllthAj’
:0

with

nEZ mEZ

1/2 4/5
Aj : (h 2 (hit 2 lV4—j/J‘ji5/2) ) .

Note that the second inequality above follows from an application of Hblder’s Inequal-

. . . 9
1ty on the sum 1n m w1th p = -5- and q = 4'

Claim 1.20 For V,u E X), and 0 g j S 4,

A: s llVllx,’llull’x,,-

 

Proof:(of Claim 1.20) Fix j. Set p = 4 4

4
and q = 3 (p or q = 00 when dividing by

zero). Then using Hblder’s Inequality on each sum, it follows that

4—3-1 t- 4/5
Aj S It: h3 Z lVllo ha 2: l/Jllo
nEZ mEZ mEZ

. (h;(hs;a.ullo)l/2)*(h;(ha;a.1.m)‘”)i

4—“ 2'
llVllthlll‘llxh-

|/\

///

25

Since Ill/Hxh = ||u|lxh = lelxh, it follows that A,- g llwll‘fiq. Therefore,
“albino/m9 3 sum,

which concludes the proof of Proposition 1.19.

1.6 Contraction Mapping Theorem

In this section, assuming Lemmas 1.17 and 1.18, we prove (Pm, is a contraction map-

ping on a ball centered about the origin in the Banach Space X h.

Theorem 1.21 There exists do > 0 and r > 0 such that, if B, = {w E X), :

Iiwilxh S 7'} and ”Halli; _<. 50; then
i) (Pm, : B, ——> B, continuously and

ii) ||<I>,,0(1/) — <I>,,o(u)llxh 5 HIV — ullxh, for some )1 < 1 and V,u E B,,

with A, do, and r independent of h > 0.

Proof(i): Recall, the operator (Pm, has two terms, Hh(m)no(n) and Ahan,h(n5)(n, m).
In order to establish the existence of an r > 0 such that (1),", : B, —+ B,, we need six
estimates, three involving each term.

First, we focus on Hh(m)no(n). Since the multiplier associated with the operator

Hh(m) is bounded by one, we have

SUP lth(m)770llz§ S ”Wilt; S 50- (1-3)
mEZ

26

By Lemma 1.17(i),

llan,hHh(m)7iollz;-,ozfmh S Cllflollzg S C<50 (1-4)
and by Lemma 1.18(i),
lth(m)770|lzgz},9,h S Cllnolltg S 050, (1-5)

with 0 independent of h > 0. Combining (1.3), (1.4), and (1.5), we can conclude
Hh(m)770 E X}; and

lth(m)770||X1. S 050, (1-6)

with c independent of h > 0.
Next, consider the nonlinear term Ahan,h(n5)(n, m). Fix m > 0. For a sequence
770, let

SinnoWaJ') = Hh(.l — m — 1)3n,h770(n)Xlgjgm-

Again, by Lemma 1.17(i),

llshllouzgozffi S Cllnollzgi

with c independent of m > 0. By duality, Sm : 1,1,13,,,(Z x Z) -> l§(Z) is bounded with

bounds independent of m > 0 where

Smw(n) = h3 i Hh(m +1 —j)(9n,hw(-,j)(n)

J—l
= Ahan,hw(n, m).

Therefore,

llAhaann m)|lz§ S Cllwllzizihi (1-7)

27

with c independent of m > 0 and h > 0. Letting w = n5 in (1.7) and taking the

supremum over m > 0, it follows that

sup iiAhan,h(775)('il’fnlillfI S Cllnsllzgzzh
m>0 J.
= Cllnlliw,

The same proof can be used for m < 0 (by definition, Ahan,h(n5)(n, 0) = 0). Hence,
s13; ”Ataturxummo s Cllnll‘},- (1.8)

Next, by Lemma 1.17(ii),

llan,hAhan,h(775)lil;',°lg1,h '3 lla§,hAh(775)llla°lfn,h

|/\

curling",
= Cllfllli’gtfiffi

Cllnllio- (1-9)

|/\

Finally, by Lemma 1.18(ii) and PrOposition 1.19,

l/\

llAhang (775) l hang, cilan,h(775)||1:/4z’1:/h9

< 5c]|n||§(h. (1.10)

Thus, ifn E Xh, combining (1.8), (1.9), and (1.10), it follows Ahan,h(n5) E X), as well.

Furthermore,

llAth,h(n5)llx,. S Cllnllio, (1-11)

28

with c independent of h > 0. To conclude the proof of (i), if n E X), with ”77]th S r,

then by (1.6) and (1.11) there exists c1 and e; such that

“(1)1707“th < 0160 + 627‘s.

If (So and r are chosen small enough, then

H‘I’noflllxh S 6150 + C275 S 7'

which implies

(13,70:B, —>B,.

Proof(ii): Let u and u be in B, C Xh. To prove (1),,0 is a contraction mapping on

B, for some 1' > 0, we need three estimates, one for each norm in the definition of

H ' Hxh. First,
1
SUP ”(pilot/('1 m) - q’noManII}, = g SUP llAhan.h(V5 " #5)(‘1’m)llz§
m>0 m>0
1 m . .
= g SUP “’13 Z: Hh(m + 1 " J)3n,h(V5 — #5)(',J)llz§
m>0 1'21
S 0|le - milling“,

where the last inequality follows from (1.7) with w replaced with v5 — ’15. As before,

this can be proven for m g 0 as well. Thus,

Sig; i|@00V(°im) — @flo”('1m)lllfi S Cii(V — ”)(V4 + V3” + V2l1'2 + VH3 + ”finial?" h
m .

1/2
= ch 2: (h3 2 |(u — [1)(1/4 + V311 + uzuz + 1413 + #4)|2)

nEZ mEZ

29

1/10 2/5
S— chZ (ha 2 IV-ul‘“) - (123 Z |u4 +V3u+u2p2+ufl3+fl4|5fl>

nEZ mEZ mEZ

c (h; (113:; (u — ”(10) W)

1/2 4/5
. (h E (h3 2 lu" + V311. + V2112 + 1413 + “4'5/2) )

1/5

l/\

nEZ mEZ
S CHV — pllxh(Ao + A1 + A2 + A3 + A4),

with

1/2 4/5
A) = (’1 2: (ha 2 IV4‘jujl5/2) )

nEZ mEZ

for 0 g j g 4. By Claim 1.20, A,- 3 r4. Therefore,
311g; ll‘PnoI/(w m) — ¢,.u(-, mm), s 07‘4”!» — .Ullxh- (1.12)
Note,
3”,,(as)(n, m) = 8n,ha(n, m)6(n + 1, m) + oz(n — 1, m)6,,,hfl(n, m).
Thus,

1
”@170” " q’ml‘llzszlo = gllAhan.h(V5 — #Ullzgzgfm

7‘ m,h

S Cilan,h(V5 "' ”SHIP/411°”
n "uh

nEZ mEZ

9/3
C (h 2 (’13 2 i071). [(11 - 160/“ + 1231‘ + V213 + 1413 + #4)] Im/g) )

|/\

c (h 2 (h3 2 I8",h(1/ — u)(n, m)

nEZ mEZ

4 3 2 2 3 4 10/9 9/8 4/5
-(v +vu+uu +vu +u)(n+1,m)| )

30

1162 mEZ

+c (h 2 (h3 2“]! — u)(n —1,m)

4 3 2 2 3 4 10/9 9/8 4/5
4%MV+V#+VH+WM+M)MmM

= 1+11,

where the first inequality follows from Lemma 1.18(ii). By applying Hiilder’s inequal-

9 9
ity to the sum on m with p = E and q = 4’ it follows that

5/8
I g c (h 2 (h3 2 Ian,h(V — u)(n.m)|2)

nEZ mEZ

1/2 4/5
5 2
. (I13 2 ](1/4 +V3u+u2u2 +1413 +u4)(n+1,m)] / ) )

mEZ

1/2
3 csup (h3 2 lan,h(V — H)(n:m)l2)

nEZ mEZ

1/2 4/5
2
. (h E (h3 E ](1/4 + V311 + V2112 + VH3 + u4)(n, m)]5/ ) )

l/\

CHI/ - [.tllxh(Ao +A1 + A2 + A3 + A4)

3 emu—mu, min

with the last inequality following from Claim 1.20. Next, we consider II. By applying
H61der’s Inequality to both the sum in m with p = 9 and q = g and to the the sum

4
in n withp=4 and q: 3’ we have

1/2
II g h: (h3 Ely—[1]”)
nEZ mEZ

4/3
- (h 2 (h3 2 lamb/4 + V3lt + 112/12 + V113 + 114)]5/4) )

1/5

3/5

7162 mEZ

31

4 f 4/3 3/5
S IIV-MIIth hZ(h3:I3n.h(I/4_’ WW”) )

j=0 \ nEZ mEZ
3/5

4 I 54 4/3
s Ilv-ulli: (h:(ha:lanturjxmmmum1,m)l’) ]

 

j=0 nEZ mEZ
- 4/3 3/5
+(h:(is:(um-1,m)o,,(,o(,,m)]5“) ) - (1.14)
7162 mEZ

Claim 1.22 Let V and u be in Xh. Then for an integer j such that O S j S 4,

4/3 3/5
(h: (ha 2]” 4 3&1th ’)5] H) ) SJIIVIIx,jII/1IIx,

nEZ mEZ

Proof: Fix j such that 1 S j g 4 (the case j = 0 is trivial). Using Héilder’s Inequality

 

I 8 C O. I I
on the sum in m w1th p1 = , and q, = —+4’ then usmg Holder’s Inequality agam
— .7

 

on the sum in n with p, = 4_ j and q2= -—_—1-(p,-, q,= 00 for i = 1, 2 when dividing

by zero), it follows that

(z (W >)

nEZ mEZ

4—;1 1— “5
s (1263:2110) -,<(h3:|a.,> 1%
1162 mEZ mEZ

1/2 71 10 51%, 5
s (h: (It: >314”) ) (h: (13: law, )7) )
nEZ mEZ nEZ mEZ

4-
S IIVIIX:

II

 

mEZ k=0

j—l . 3% 21—2
h 2: (ha 2 (Z Ian,h.u'(na m)||u-7_1-k(n +12mIII/‘k(n —1,m)|) )
nEZ

32

4_
S IIVIIX,J
- 1 1— it?“
‘7_ log—14:2 ’—
2 h: ’33: Ian,h#(n,m)IJ—I+%I#(n+1,m)| 1"“ INN—1 m)I‘Lgi“)
k=0 nEZ mEZ
: IIVIIX—hjZBjJCa
with

17—1
5

4
21—2
_151, log—14:2 _1_0_
BM = (h: (’13 Z lan,hu(n, mMIMI/102+ 1,m)l 1+4 lu(n- 1,m)|1+'i) )

nEZ mEZ

 

' 4
Fix It. Again, apply Holder’s Inequality to the sum on m with p, = J: and

 

. ,_ _ 1
q1 = j _ 1 and to the sum on n with p2 = 75—1—3 and q2 = 316—. This gives
7.53 1/2 “I?"
10 '—1-—k 1_0k_
Bch S (h: (’13 Z lamb/42) (ha 2 W(n +1,m)| 1" W(n — 1 m)| 1)
nEZ mEZ mEZ
1/2 1/2 31
10 '—l—k 10’:
.<. sup (M 2 law?) h: (,3 z: |u(n +1,m)I4L—ZJ—l lu(n —1 mu;- )
"EZ mez nEZ mEZ
-_1 I; ig—l
20— 56715
S llullx, h 2 (’13 Z lu(n+ 1,m)lm) (h3 E |u(n —1m)|1°)
neZ mEZ mEZ
1/2 t—ii 1/2 i
S llullx, h: (’13 Z lull") h: (’13 Z lull”)
nEZ mEZ nEZ \ mEZ
S Willi},
Therefore,

54
IhZIhszI van,“ ')| ) ) <jllV||x,’l|u|lx,,

nEZ mEZ

33

which concludes Claim 1.22.

///

Notice the second term of (1.14) can be bounded as above by switching the roles

of l/ and ,u. Hence,

II S CT4||V_' IU’HXh'

Combining this with (1.13), it follows that

”(1)770” _ Qflo/‘Hlfllo S 67.4“” — HHXh'

n m,h

Finally,

1
HangM’noV — (I’ml‘mzooz? = gllagmAMI/S - Hsmzmz?

VI. 173.}; n m,h

< 6|le "' Hslltufn'h

1/2
= ch 2 (h3 E |(u — p)(z/4 + 113/; + u2p2 + 1413 + #4)|2)

nEZ mEZ

|/\

1/10
Ch 2 (h3 E I” _ ”'10) ' (h3 2 IV4 + 113;! + U2l12 + 1413 + ”4'5/2

nEZ mEZ mEZ

1/2 1/5
c h: (h3 XIV—M10)
nEZ mEZ

l/\

7162 mEZ

|/\

CHV - flllxh(Ao +141 + A2 + A3 + A4)

|/\

67‘4HV_/—"HXM

(1.15)

:)2/5

1/2 4/5
- (h 2 (123 2 ll/4 + 113;» + u2p2 + up3 + [fr/2) )

(1.16)

where the first inequality follows from Lemma 1.17(ii) and the last follows from Claim

1.20. Finally, (1.12), (1.15), and (1.16) imply that for V and ,u 6 Br,

“(1)7101! — (Pm/ilk}; S 07‘4“” — ”llxh'

34

By picking r small enough, i.e., cr4 = A < 1, it follows that (1)00 is a contraction map

on B,, which concludes the proof of Theorem 1.21(ii).

1.7 Well-Posedness of (K dV)ff in X 7,

If the initial data is small, the Contraction Mapping Principle guarantees the existence
of a unique 17 E B, C X}. such that 17 = <I>,,017. By definition of (Pm, 17 solves (1.2).
Also, by definition of the space X h, 77 E l°°(l2(Z) ; Z). Moreover, we have the following

theorem.

Theorem 1.23 Let 60 and r be as in Theorem 1.21. Let 170 E lfi(Z) such that ”170”,: g
60 and let 17 6 B, be the unique solution of (K dV)flI guaranteed by Theorem 1.21. Then

the map

’70—’77

is continuous from B(0; 60) to B, with B(0;60) = {o : ||0||7§ S 60}.

Proof: Fix h > 0. Let am, 770 6 B(0; 60). Then the unique solutions to (K (1V): in B,

with initial data we and 170 are w and 17, respectively, which satisfy
1 5
w(n, m) = tho(n) — gAhBhQu )(n, m)

and

m m) = Hh770(n) — éAhamSXn, m).

Thus,

w(n, m) — n(n, m) = Hmong — noxn) — §Ahah<w5 — n5)(n, m),

35

which implies

llw - nllxh S Ith(m)(wo - no)||x,. + ell/ham)5 - 775)||x,.-

= I + II.
It follows from the proof of Theorem 1.21 that
I S 0|le — Delhi,

and

11 g cr4||w — nllx, S Allw -77||xh,

with A < 1 (by the choice of r). Hence,

llw — Ullxh S Cllwo - Uon; + All“! - nllxh

which implies

6
“cu ‘ 77'th S ___}:Hwo — "allig-

This concludes the proof of Theorem 1.23.

Remark 1.24 By definition of X h, the map 770 —+ 17 is continuous from B (0; 60) to
l°°(12(Z) ;Z)-

CHAPTER 2

The Estimates

In this chapter, we will prove the estimates stated in the previous chapter, namely

Lemmas 1.17 and 1.18.

2.1 A Discrete Version of Fractional Integration
Before proceeding, we need the following lemma.

Lemma 2.1 Let k be a sequence such that

C

k < —.
I (n)l—|n|1/2+1

Let
T(b) = b a: k

for a sequence b. Then

T: W(Z) —, 10(2)

1 1
isboundedif1<p<q<ooand6= 5.

"SIP“

36

37

Proof: Fix A > 0. Let p > 0 which is yet to be determined. Set k1 = k . X{OSInL<.u},
k2 = k - X{lnl>u}’ and let b E l”(Z) with ”bllzp = 1. Then b * k = b * k1+ b * k2, both

of which are well defined. This leads to

#{n: |(b* k)(n)| > 2A} 3 #{n2 |(b* k1)(n)| > A} + #{n: |(b* k2)(n)| > A}

 

 

 

2' IA+IIA- (2.1)
Now,
“(1* kllli’p llklllil Clip/2
[AS-TS ,, 3 AP. (2.2)
Let p, = p 1 be the conjugate exponent of p. Then
“12* k2||z°° S Ilkzllw S mil/q,
since p < 2 hence, p’ > 2. Pick [1 such that curl/‘1 = A, i.e., u = ch‘q. Thus,
#{nz |(b * k2)(n) > A} = 0, (2.3)

which implies I I ,\ = 0. Therefore, combining (2.1), (2.2), and (2.3), it follows that

 

. cup/2 c __ c
#{n.|(b*k)(n)| >2A} 3 AP +E— E.
. . 1 1 1
Hence, T 18 weak-type (p, q) for 1 S p < q < oo w1th 6:1-7— 5. By the
Marcinkiewicz Interpolation Theorem, T is strong-type (p, q) for 1 < p < 2 and q

1 1 1
such that - = — — —.
q p 2

38
2.2 A Lemma Involving Fractional Differentiation

We now proceed with proving the main estimates. To this end, we make use of the
operators D3 and complex interpolation. The following lemma provides half of the

necessary results.

Lemma 2.2 For 7 E R,
- —1 4 i
.) |th ” 7Hh(m)770||zgz;gh s Cllflollzg

and

ii) l |D,:1/2+”A,,w| luv»

11 m.h

S C7l|wllly3lrlmha
with C, = C(I’yl + 1) where c is independent ofcy and h > 0.

Remark 2.3 Before proving (i), we check to see if the homogeneity is correct. As-

sume (i) is true for h = 1. Then

1/4
Ithl/HWHMmMOHIgg-g, = (h 2(3111) ngl/4+'7Hh(m)7)o(n)l)4)

nEZ mEZ
1 /h 4 1/4
= h sup — I/

. . 1/4
1 I 1r/h . . (h: 770(k)ezkh0)e-mh0 4
= h1/2 _/ h6 —1/4+:7 kez d0
("26, (3:22 21r -./hls‘“( )' (1+Sgn(m)isin3(h0))|m|

>4)

"“7 (fi0)h(0)e“"""
(1 + sgn(m)i sin3(h0))lm|

sin(h6)

h d0

 

 

 

 

 

 

 

 

 

1 " - _ - (mane-W
= hl/2 _ f 6 l/4+ry
("26:2 (78:21; 27f -1r I sm I (1 + sgn(m)i sin3 0)|m| d6

 

s ch1/2(Z |17o(n)|2)1/2

n62

= 00% Z Ino(n)12)‘/2-

nEZ

39

Proof(i): By Remark 2.3, the case h = 1 implies all other cases with h > 0. Hence,

we can assume h = 1 and drop the subscript notation. Note that

Z Z D’1/4+i7H(m)770(n)w(n, m)

 

mEZnEZ
1r . * (9 _—
= Z 217;] lsingl—1/4+n(1+s 770( ..) 3 lmla(1)(g,m)d9
mez —1r gn(m)ism 0)

 

 

7r “(1)
= 2%] 70(o)(|sin6I-1/4-‘7 ” (0’7") )d9

mez (1 -— sgn(m)i sin3 0)|”‘|

__— z Z 170(n)D‘1/4“7H(-m)w('am)(n)-

mEZ nEZ

 

To prove (i), by duality, we need to show that

ll 2 D‘1/4+‘7H(m)W(nm)llzg S Clleg/sun, (2-4)

mEZ

with c independent of ’y. The square of the left hand side of (2.4) equals

 

Z Z ZD—1/4+i7H(m)w(nam)D'1/4+i7H(u)w(n,1/)
nEZmEZ uEZ
__ 1 1r - —1/2 7;,(1)(g, m)<IJ(1)(0, u)
_ Z Z 271' f... |sm0| (1 +sgn(m)isin36)|ml(1 — sgn(u)isin30)|V|d0

 

 

mEZ 1162
= Z Z -1- [1r w‘”(0,m)lsin0|‘1/2T(m, V)(0)122(1)(0,V)d0
mezuez 27f ”'
= Z Z Z ”(n’m)D-l/2Tm.vw(niV)
mEZ uEZ nEZ
= Z Z “)(n, m) 2: D-l/sz,uw(nv V), (2.5)
mEZ 7162 1162

where the operator Tm,” is the operator with Fourier multiplier T (m, u)(6) on the first

variable with

T(m, 10(9) = (1 + sgn(m)isin3 0)"'"'(1 — sgn(y)isin3 (9)—le
1 1
(1 :1: isin3 0y (1 + sin6 9)r’

 

40

where p Z O and r _>_ 0 depend on m and V. By (2.5) and duality, we need to show

 

 

 

 

 

 

that
H 2 D_1/2Tm,l’w(ni V)||l;‘,l$,? S Cllwllzi/flgn' (2'6)
VEZ
Now,
2 D-l/ZTm,,,w(n,z/) = 22—1-e-W1sin01 1/2T(m, u)(0)a<1>(0,u)00
V62 V62 —7I
= 22—1]: e””9| Sln 6| 1/2T(m,) V) (0) (Z w(k,V)eIk0d0
VEZ 2’” 1:62
5 Z Z 271/; a“: ">91sin01 1/2T(m, u) )(0) 00] 1100. u) )1.
kEZVEZ
Claim 2.4
C.,
in9 1/2+i7
l/tre |sin0| T(m, V) (0)<d0| _ ——————|n|1/2, + 1

with C, = c(|'y| + 1) and c independent of m, V, and n E Z.
Remark 2.5 Assume Claim 2.4 for the moment. It follows that

C

 

sup E D 1/2‘1m' ”w n, V < 2| E w k,V .
"‘6z I VEZ ( )I kEZI -III1/2 I 1 VEZI ( )I
By Lemma 2.1,

II sup 2 D l/2'-"m,uw(n v)||z4 < Cll Z M n V)| |l14/s — Cllwllwsp,

"‘52 uez uez

which is (2.6). Hence, we can conclude the proof of Lemma 2.2(i) once we have proven

Claim 2.4.

Proof:(of Claim 2.4) Obviously, the left hand side of the claim is bounded indepen-

dent of '7, m, V, and n. Hence, we can assume n 31$ 0. Let A1(0) E C8°(R) supported

41

in [—7r/4,1r/4] and A1 E 1 on [—7r/8,7r/8]. Set A3(6) = A1(0—7r) and A2 =1-A1—A3.
Then

1]" 0"“91 sin 0|‘I/2+i7T(m, V)(0)d6|

<

 

1r/4 , .
f /4 emol sin 6|"1/2+"’A1(0)T(m, V)(0)d0l

+ 81110. sin 0|‘1/2+i7A2(9)T(m, V)(0)d0l

[r/8<|9|<7n/8

 

51r/4 , .
+ /3 /4 emolsin0|“1/2+'7A3(6)T(m,V)(0)d0l

= 1”,, + 11",, + 111,,

First, consrder 11,”. Since @I srn0|‘1/2+” g c(1 + |71)| srn0|"3/2, after integrating

 

 

by parts, we obtain
[In ___ _1- / _d_(ein9)| Singl-1/2+i7A2(6)T(m V) (0)d0
I7 Inl "/8<l9|<71r/8 d0 1

mo— . —1/2+z'7
lnl ffl/8<l01<71r/8 e 9 (I 8m 0| A2(0)T(m, 10(9)) (10

|/\

 

 

— 1119— . ‘1/2+!7
lnl A/8<l0|<71r/8 6 d9 (|sm01 ) /\2(9)T(m, V)(0)d6

 

1 in - - i ’
+I—TTI /1r/8<101<71r/8e alsmdl 1/2+7A2(0)T(m,V)(0)d0|
_1_ ind - —l/2+i7 d
+lni [r/4<|9|<3w/4e |sm0| d0A2I0)T(mIV)(0)d0
07/ . _3 C C d
__ 19 fidg _ __ — T 0 d0
lnl "/kwKM/Slsm I + W + W 1r/8<|0|<71r/8 ddI (m,V)() a

 

where C, = c(|7|+1). Thus,

 

 

 

 

C c d 1
11,, —1 — f — .
’7 _ 1n| + |n| 1r/8<|9|<71r/8 d0 ((1iisin30)/’) d6 (2 7)
d 1
'—c— _ . 6 d0.
|n| 1r/8<10|<71r/8 d0 (1+srn 0)r

 

 

42

The second integral in (2.7) is easily bounded independent of r by a bounded variation

argument. For the first integral in (2.7),

c c 3psin2 0| cos 0|

 

 

 

 

 

 

 

 

_/ i( I ) d9 = — f 00
lnl w/8<101<7«/8 d6 (liisin36)0 |n| n/8<101<71r/8(1+sin60)(P+1)/2
< __C_ 6(£;—£)|sin50cosd|d6
— lnl w/8<101<7«/s (1+sin60)(p+1)/2
= ‘6“ i( 1 ) d0
lnl 1r/8<10|<71r/8 do (1+sin60)(p—1)/2
< _C_
_ lnl

for p at 1 (with a similar proof for p = 1) which is again bounded independent of p

using a bounded variation argument. Thus,

C.

Next,

eiip arctan(sin3 9)

 

1r/4 , .
m0 - —l/2+ry
[4/4 e |s1n BI AI (6) (1 + sin6 0)r+p/2

_ [II/4 ei(n0:i:p arctan(sin3 9))A1(0) I Sin 0I_1/2+i7
——1r/4 (1 + sin6 0)?

I... 00|

 

 

 

with q = r + p/2. Let (pg be a C°° function such that cpo E 1 on |0| Z 2 and

suppgpo C {IBI Z 1}. Then

 

Inn S

 

1r/4 I 3111 0| 1/2+‘7
Biz/1.140)
(.04 (MD 9), 11610000101014 (2.9)

 

+

 

"/4 |sin9| ”2+”
Lie“ (1+8... 9, ()< 0011111)) ,

with

z/Jp,,,(0) 2 n0 :t parctan(sin3 6).

43

Since sind ~ 0 for 6 E [~7r/4, 7r/4], the second integral in (2.9) is easily seen to be

less than IT—lI—fig—fi. We now focus on the first integral in (2.9). Set

Apn = {92 |¢,,n(9)l S VII/2}

and

3,... = {0: I¢;,..(6)I s InI/3}-

Let (1, (2 E C°° with (1 + C2 = 1, (1 E 1 on BM and suppCl C AM. Note, (1 and (2
can be chosen such that C; and (5 change sign only a finite number of times. If p = 0,

then (1 E 0 and (2 :- 1. The first integral in (2.9) equals

 

 

11/4 lsin 0'“ l/Z‘H'Y
e11 <9) 0 0 d0
/_«/46 (1+sin 60)? AI( )I'OIIIIIII ) I

1r/4 sing 1/2‘I‘i7A
[01.011 'A<6><po(1n161c1<01d6|

<
-1r/4 (1 + sin 60)?

 

 

 

 

"/4 |sin0| 1/2+i7/\
610,10) 0 0 0
+ [tr/46 (1+sin6 0)? [III )<P0(Inl )C2(0)d
= Imus.

Note, ¢1Io,n(0) = n :1: (p arctan(sin3 0)) with (parctan(sin3 (9))I ~ p62 and w;,,(0) =
(p arctan(sin3 0))" ~ p0 for 0 E [—7r/4,7r/4], both of which can be easily verified.
Inl

Since 11%| 3 F on the set AM, it follows p62 ~ |n| on suppCl. Hence,

191~_I1I
fl

for 0 E suppCl. Since ¢;,n(0) ~ p0, there exists c > 0 such that

|¢Z,n(0)l Z c Inlp (2-10)

44

suppC1C{0:c1‘lI—:—I S 1913 ”III—ELI} (2.11)

Before proceeding, we need the following lemma [10, pg. 342].

OD

Lemma(van der Corput) Suppose 45 is C°° and compactly supported. Let

1/1 be a real-valued function so that, for some I: E N, k 2 2,
10%| 2 C,c > 0
throughout the support of 01. Then

00 , C
l/;0061¢($)¢($)d$ S WI(II¢IIL°°+IID¢IIL1)

Applying van der Corput’s Lemma to [5,12, with k = 2 and appealing to (2.10) and
(2.11), it follows that

 

I“) < ___—9— F .. f F’ 0 d0 , 2.12
"17-1n11/4p1/4 (ll 11. (supp(1)+ suppcll <11 ( 1
with
0(0) z Isin01‘1’2+"’/\1(0)<po(lnl0)cl(0)
(1+sin60)‘1 °
Note that
1/4
1F(0)1< p (2.13)

-— cInII/4

45

on suppCl. Furthermore,

1F’(611 s camel-3”

 

 

 

11(0100111116113101 . _,
(1+sin60)q l+lsm0| 12

 

(11(0100110101<1(o1)’l

(1 + sin6 0)?

 

 

 

 

 

 

 

“4 A101 annex (01 ’

< —3/2 P 1 $00 1

- C7|0| +c|n|1/4 ( (1+sin60)‘1
for 0 e suppCl. Therefore,

1 ”4 A(9)<P0(I7II9)C1(9) '

F 0 00 < C/ 0-3/200 " f 1 d0
fsuppclI I )I " 7 “£99” I +C|n|1/4 1019/4 (1+sin60)‘1

p1/4
S 07W (2.14)

again using a bounded variation argument. Finally, combining (2.12), (2.13), and

(2.14), we have

[(11 < ___91__,

Next, we need to show

0
2) 7
171,7 <

—— 2.1

with C7 = c(l'yl + 1) and c independent of p, q, and n. It follows that

 

 

1r/4 1 d |sin6| ”2+”
(2) = _ 6210.101
1 [4,4 0' (0) 00( I (1+sin6 0)a A1(6)¢°(I"I0)C2I0)d6

"/4 d 1 Isin 0| 1/2+"7)‘
: 61111,, n(9)_
./-1r/4ed0 (WWW) (1 .1. Sin60)q10)(900()InI0 (2(9 d0

[_m eitban) ¢0n(6) lsin9| ”2+”

 

 

 

|/\

A (01,011.10) <2 (0)) “I

 

1r/4 (1/1,,,,,(6I))2 (1 + sin 60)?
71/4 9 0 3/2+i’7
+01 fw/‘te ei'l’pm (9) ¢:OS(0) |(31i1: linGA 69)q A16“ )900(|n|0)C2(0))d6|

 

+

 

[II/4 610p,.(01|8in9| ”2+” *1“) )¢°(I"I6)C2(9) d0
—«/4 111;,(0) (1 + sins 0)“ '

 

 

46

Recall, l/JIMIQI = n :1: p(arctan(sin3 0)), with p(arctan(sin3 6)) ~ p92 on {—71/4, 11/4].
Since Iz/JL,,,,(9)| Z I3£I on suppcg, it follows that 11/);m(0)| _>_ cp02 on supng. This and

the fact that |w;,n(0)| S cp|0| implies

 

 

 

C plgl _ C —
1512) < 1/2dO 0+____ 3/2d9
’7 " lnl l/ln |<101p02I0I Inl 1/lnl<|9|I0I
+ c [04 1.1010011n101czw1 0,,
[ml/2 _fl/4 (1+sin60)q
C.
S InII/2

Since |I,(,2,),| g c, this finishes the proof of (2.15) which implies

C
In <—7—-.
’7 _ |n|1/2 + 1

The estimate for I I Inn can be obtained from the estimate for In” using a change of
variable, namely r = 0 — 71. These facts and (2.8) conclude the proof of Claim 2.4

and hence concludes the proof of Lemma 2.2(i).

Proof:(ii) We need to show that

11D;1/2+‘7A,,w111.,..

umh

S CIVIIwIIlz/i’l'ln'hv
with C, = c(I'yl + 1). Suppose m > 0. Then, by Claim 2.4,

|D;I/2+i7Ahw1(n, m)|

h3 —inh0
h:_ 271' 1./—://he

_ hl/2

“”7 0111110 01
(1 + sgn(m +1 - j)isin3(h0))lm+1-J'I

~ 0|—1/2+i’7
h3(k_2}1_r_/:I i(k— n)0 I SlIl d9
2 Z( w j) e (1 + 0sin3(0))m+1-J

kEZ j=1

sin(h0)

h d0

 

 

 

 

 

 

 

 

47

 

 

m ' 0|-1/2+i'7
< hl/2h3 (k ei(lc— n)0 ISln d0
- £32,212”); (1+0s1n3(0))m+1—J
1
g C, 111/2113 |w( (k ,j)| ,
BI: — 01/2 +1

with C, = c(I'yl + 1) and independent of m. The same proof works for m S 0.
Therefore,

 

1/2 +1”) 7/2 1
SUP ID), Ahw(n, m)| < C h Iw( (k III .
mEZ £3.26; lk _ ”ll/2 +1
Hence, by Lemma 2.1,
IID;I/2+I7Ahwlllgl:h = (h(z:(sup lD-1/2+i7Ahwl)4)1/4
I n62 mez
S 0101/42/28: |w( 71021112313“
1162 mEZ

‘7 C7IIwIIgg/31'1n h

and this concludes the proof of Lemma 2.2(ii).

2.3 Proof of Lemma 1.17

The following lemma implies Lemma 1.17.

Lemma 2.6 For 7 E R,

1) IIDh IIIIHh(m)770II1°°12, < CIIIIOIII}:

and

i2) IIDIIIIAIIWII100:2 S CVIIWII11121

"111,11 ’1th

with C, = c(IyI + 1) and c independent of”) and h > 0.

48

Remark 2.7 To see how this implies Lemma 1.17, first consider Lemma 1.17(ii)
which says

IIaimAthIzg-ezfm S CvIIwIIIngn‘h-
The multiplier associated with 6,2”, is

— sin2(h0)
(1.2 I

which is a constant multiple of the multiplier

| sin(h0) |2
7,2

associated with D,2,. Hence, Lemma 1.17 (ii) reduces to Lemma 2.6(ii) with 'y = 0.

As for Lemma 1.17(i) which says

IIan,hHh(m)770IIz;-,ozfmh S CIIUOIIzg,

the multiplier associated with 0",), is

—i sin(h0)
h

while the multiplier associated with D}, is

| sin(h0)|
_h .

Let M(h19) = —isgn(h0) for 0 6 [—7r/h,7r/h]. Hence,

—i sin(h6)
h

|sin(h0)|

-_— M(h0) h

49

which implies for m E Z

3n,hHh(m)no = DIHhThflo,

where T), is the Fourier multiplier on [—1r/h,1r/h] associated with M (h0). Since

T), : 1,2,(Z)—1 lflZ) is bounded (with norm less than one for all h > 0), Lemma 2.6(i)

implies Lemma 1.17(i).

Remark 2.8 Before proving Lemma 2.6(i), we again check to see if the homogeneity

is correct. Assume (i) is true for h = 1. Then

IIDIIIIIIHMmIUOIIu-szghh

|/\

 

 

 

 

 

. 2 1/2
1 “/1 ... sin<h61 (0011(0)
3 __ znh0
:IeIIzI (II "a; 27? f-r/he h (1 +sgn(m)isin3(h0))lmld0

 

 

 

1 7' _- - |sin0|1+’7
h1/2 _/ 1710 [6 :k0 do
i‘é‘z’ (0.26:2 27r _..e (5’7“ Ie )(1+sgn(m)isin3(0))ImI
ChI/2(Z I77o(77I)II)I/2 = CIIUOIIzfi-

nEZ

Proof(i): As before, assume h = 1 and drop the notation. For m, n E Z,

 

. 11 _ ' 1+i'7"
DI+'7H(m)flo(n) 1 / em0 ISlngI 770(0) (10

% -1r (1 + sgn(m)i sin3 0)I"‘|
_ _1_ [m 61.1 1011012200161
271 -1r/4 (1 + sgn(m)i sin3 0)|m|
+_I_ / ein0 ISinoIIIIII’IOIOI
27, 104510133104 (1 + sgn(m)i sin3 0)Iml
+1. [SI/4 ein0 |sin0|1+t7770(0) d0
27" 32/4 (1 + sgn(m)i s1n3 0)|m|
= I(n, m) + [I(n, m) + III(n,m).

d0

 

 

d0

 

50

First,

 

”2 1* (0112 "I
II , 2 < f ’7” 00
i‘é‘z’(ZI (n m)|) — C(mez «143109.14 (1+sin601lml I

mez
1 In
2 c (fw/4310Isavr/4 IIIOIIIIII ("g (1+ sin60)lml) do)
3 c (j; 110(011200)“
= c(Z 10010112112. (2.161

nez

 

Next, we need to show

1/2 1/2
sup (2: |I(n,m)|2) g c (E |170(n)|2) . (2.17)

nEZ mEZ nEZ
2 1/2
d0

2 1/2
) , (2.18)

Note that

 

[II/4 fnrr(0)

—1r/4 (1 + sgn(m)i sin3 0)I’"|

 

 

"62 mez "62 mez

1/2
sup (2 |I(n,m)|2) = csup (Z

6

[II/II —im arctan(sir13 0) fun (0) d0
-1r/4 (1 + sin6 0)I"‘I/2

 

 

= csup (Z
nEZ mEZ
with
0.401 = emol Sinall‘I'I’fio(9)-
To motivate the proof of (2.17), consider Lemma 0.1 of [5] which is the continuous
analog of Lemma 2.6(i) (with 6,, replaced with D1+ i7). The next logical step would

be to follow the proof of Lemma 0.1 by transforming the Fourier series variable from

n to m via a change in variables. Hence, by setting r = arctan(sin3 0), i.e.,

0 = arcsin(tan1/3 r) = h(r)

51

and
sec2 7

d0 =
tan2/3 r \/1 — tan”3 7

dr = hI (T)d7’,

OOH—I

(2.18) becomes

b 2 1/2
[be—I'm cosImIrfn,,(r)hI(r)dr ) , (2.19)

.(z

mEZ

 

 

with

inn (T) = fn,'7(h(T))

and b = arctan(sin3 (7r /4)) Since hI(r) ~ 7‘2/3, it follows that

 

1/2 III
(f: Ifn,(’YT) ”(TIII‘Id’r) -<- c(f-II,’ Ifn'IITI (IIIT —2/3IIII( IIII I
1/2
: c/(II/ /4 Ifm(9 III(arCta'n(SIn3 0)) —2/3d0)
sin2 9 III
: c/(I: IIIIIIIIII 2(arctan(Sin3 9II2/3 d0)
/ 2 1/2
S C(/_::II7 0w )1 d0) . (2.20)

Thus, (2.17) has been reduced to showing the following proposition.

Proposition 2.9

11' . 2
2 / e“""T cosImIrg(r)dr

mEZ

 

 

_<_ 441901120.

Remark 2.10 By setting g(r) 2 fn,,(r)hI (TIM—1:519} in (2.19), with (2.20), Propo—

sition 2.9 implies (2.17).

52

Proof: Suppose 9(7) 6 L2[—7r,7r]. Let a), be the k“ Fourier coefficient of g, i.e.,

l .
9(7) = — Z ake‘I". Assume m 2 0. Then
27’ kez

[II e‘ImT cosmrg(0)d0 = [II 6"” (II—LL) g(r)dr

_, -1; 2

= 2-m/ (1—0-‘2T)mg(7—)0T
= 2"" IZ(II,I)e_I2jTg(r)dr
-"j=0 J
m m
=m<1
i=0 3
: ZBmJaZj)
.20

1
witthJ=§;1-(I;I) formZOandOSjSm.

Claim 2.11 For all m 2 1,
i) in: Bm’k = 1
1:20
and for all k 2 O,
0.) f: B... = 2.

m=k

Proof:(of Claim 2.11) To prove (i), observe

°° 1 m m 1
B... =— 1*1m-kz—1 1m=1.
z. ,. 02(1) .1 +1

k=0

 

gImI( ) = ("Wk-1)

no a 1 k
To prove (11), COIlSldeI‘ gk(x) — (___—E) . It follows that m! k _ 1

1
This implies that for |a:| < 1,

 

53

Therefore,
IEk co (m) m
—— = a: .
(1 _ x)k+l 77;]: 1:

Setting a; := 1 / 2 completes the proof of (ii).

///

By Schur’s Lemma (see e.g., [2, pg. 394]),

(2 )1” = (22%“)

mZO j=0

 

1r .
/ e‘mT cos'm'T g(7')d7'
-—1r

 

1/2
S ‘5 (Z lajlz)

jEZ

= V5(/_:lg(r)l2dv')1/2

After making the change of variables (p = —0 we obtain a similar estimate with Z

mZO

replaced by Z and this concludes the proof of Proposition 2.9 and hence, concludes
mSO

the proof of (2.17). Notice that the same estimate works with (1 + sgn(m)i sin3 0)""‘I

replaced by (1 — sgn(m)i sin3 0)""", setting 43 = 6 — 7r yields
5119(2 Ill-’1(n,m)|2)1/2 S C(Z Ino(n)|2)l/2-

"62 mEZ nEZ

This fact along with (2.16) and (2.17) conclude the proof of Lemma 2.6(i).

The following is a useful corollary of the proof of (2.17).

Corollary 2. 12

(.2

 

[”4 f (9)

—1r/4 (1 + sgn(m)i sin3 0)|m|

 

 

2 1/2 1/2
"/4 |f(‘9)|2
< —— .
d0 ) _ (far/4 sin20 d6

54

The following proposition is a consequence of Lemma 2.6(i) via duality.

Proposition 2.13

llh3 Z Di+i7Hh(-m)w(-.m)llz,2, S Cllwllzngm,
mEZ

with c independent of ’7 E R and h > 0.

Before we prove Lemma 2.6(ii), we define three functions on Z x Z.

Definition 2.14

h3 )3 Hh(m+1—j)w(-,j)(n) ifm>0

j¢m+1
Ahw(n, m) = 0 ifm = 0
-h3 2: Hum —1—j)w(-,j)(n) ifm < 0.
#m—1

Definition 2.15

h3 2 sgn(m +1 —j)Hh(m + 1 — j)w(-,j)(n) ifm > 0

j¢m+l
Bhw(n,m) = 0 ifm = 0
ha 2 sgn(m —1—j>H,.(m —1—j>w(-.j)<n> ifm < o.

j¢m-1

Definition 2.16

h3 2 Hh(m+1—j)w(-,j)(n) ifm>0

11 jz-oo

Ehw(n, m) = 0 ifm = 0

—h3 :0: Hh(m — 1 — j)w(-,j)(n) ifm < 0.

It follows that

Bhw = 2Ahw — Ahw + 2Ehw,

55

which implies

DfimB w— — zpfimA w— D§+‘7A,.w + 2D§+i7Eh u). (2.21)

Proposition 2.17

HDZHVEIIW I [13°12

m, S 0110011511an

with c independent of 7 and h > 0.

Remark 2.18 For the last time, we check the homogeneity. Assume Proposition

2)1/2

2.17 is true for h = 1 and let m > 0. Then

2+” wil’wJ)
(1 + isin3(h0))m+1*j

h49)

sin

d0

 

(
h

h3 i 1 lfl/he ~inh0
27f —1r/he

j=-oo

sup(h3 Z

 

 

 

 

1

Z27r

jz-oo

 

A (1) 0 .
—in9 2+i'y w ( 3 .7)
A. ‘3 |s1n0| (1 + isin3 0)m+1-J' d0

 

m>0

= h3/2h3h—2 (Z

 

1/2
S chm/2: Z:|(.u(n,m)|2
nEZ mEZ
= own...“

The same proof works for m < 0. Combining these and the case h = 1 implies all

other cases for h > 0.

Proof: Assume h = 1 and assume m > 0. Then

D2+i7Ew(n, m) D1+17H(m)[H(1)(Z(D1H(—j)[x{jso}w(,j)])] .

jEZ

56

This implies

 

 

 

 

||X{m20}Dz+i7Ew||z;-.oen S C 11(1)(ZDlm-J'qusowhfll) (2-22)
jEZ (’2‘
S CHZD1H(—j)[X{jgo}w(°,jlllllg (2°23)
jez
S C||X{mgo}w(n,m)||z,1,zz,, (2-24)
—<— cllwlllhlgn:

where (2.22) follows from Lemma 2.6(i), (2.23) follows from fact that H (1) is bounded
on l2(Z), and finally (2.24) follows from Proposition 2.13. The case m < O is similar.
Combining these two cases and applying the triangle inequality finishes the proof of

Proposition 2.17.
Proposition 2.19
“DizzflvAhWngozfmh S Cllwllzngn'h,

with 6 independent of ”y E R and h > 0.

Remark 2.20 If the set of operators {Hh(m)}m€z did form a group under composi-
tion, i.e.,

Hh(m1 + m2) = Hh(m1)Hh(m2) (2.25)

for all m1, m2 6 Z, then the proof of Proposition 2.19 would proceed exactly as the
proof of Proposition 2.17, which used the fact that (2.25) is true if m1 and m2 are
the same sign. However, because of our choice of discretization, we do not have the

group structure in general. Hence, our proof will require different methods.

57

Proof: By a similar homogeneity argument as in Remark 2.18, we can assume h = 1.

Suppose m > 0. Then

D2+'7Aw( (,n m)

A (1) .
w (0,2) d0

 

j¢m+1

OO

1

—in0 2+i
0 7 .
:2 [22:6 (Sin I (1 + sgn(m + 1 — j)isin3 0)Im+1-2I

= lim—/ If?! e—mol sin 6|2+i7(1 - isin3 0) < Z (1 — isin3 6)m—jcb(1)(0,j)) d0
< <1r—c J

c—>0 271’ c

-in0 1 Sin 0|2+i7

'=m+2

WW 2')

 

+ lim / e
e—>0 27r <|9|<1r—€

| sin 0|2+i7

m
_fi—
1+is1n 0 (J.__ ”(1+isin3 0)”

| sin 0|2+'Ve" ‘2‘P

J)do

 

 

= 1 —

. 1r .
m 2 6""0/ e‘mup . , 3 . +
c—>0 47r e<|9|<1r-e —1r 1 + i s1n l9 — e"?

1— isin36 — e“
I Sill 0|2+i7e—i2<p

up) (21(0, (p)dgod0

 

 

' 2+i’7
= lim i 1r "6—imw/ —m0 1 Slngl
6—)0 47r2— c<|9|<1r—c 1 + isin3 0 - eW’

 

1 7' _J
= {giggle "W’ZKnm- j,so)w (2)0 de,
jEZ
with
1 |sin0|2+i7
K6, m, _/ —im0 . , ,
(m (p): 27r SlOlSw—ee (p(l+ism30—e“f’
w(2)(J =Zw(j,k We”,
keZ
and

0(0, v2) = 2 0(2’0, w)?”

jEZ

1— isin3 0 — e‘W’

. )(D(0, <p)d6d<p

| sin 0|2+i7e"2“’

 

. d0,
1— isin3 9 — e‘W’)

Before proceeding with the proof of Proposition 2.19, we need the following defi-

nition and a few claims concerning it.

Definition 2.21 For 1;) ¢ 0, let

I sin 0|“i7

 

 

1 «r .
K , = _/ —im0 _
7(m ‘pl 21r -26 (1+isin30 — ew +

' 2+i7 —i2(p
|s1n0| e ) d0.

1— isin3 0 - e"‘P

58

Note, 1131(1) KEJ.,(m,cp) = K,(m,<p) for all m E Z and for all (p E [—7r,1r] with <p 75 0.
Since K.J,(m, 0) = 0 for all e > 0, m 6 Z, and 7 E R, set K,(m, 0) = 0. Hence,

£216 Ke,7(m1 (10) : K7(m’ (p)

for all m E Z and (,0 6 [—7r, 7r].

Claim 2.22 There exists C such that
||K€,7(m1')HL2[—1r,1r] S Ca

with C independent of e, m, and '7.

Proof:(of Claim 2.22) By definition,

. 1 7’ _,-
K.,.(m, -) (z) = gJ-J [_J K.,.<m, we "de

1 7' . sin 6 2+” sin 0 “he—‘2” -
=_// .w (J .+(L .dWW¢
47r2 —1r £<|0|<1r-e 1 + 2 sm 0 — eW’ 1 — 1 sm 0 — e'W’

1 1r _. J .
= _J / f e 'm9|sm0|2+”
471’ —1r £<|0|<1r—c

 

 

 

1 ( 1 )
. . 3 81¢
1+ism 0 l—m

+ 6—1.2“, 1 d0 e-“V’d
1 — isin3 0 1 — -e—”33— (p

 

 

l—isin 9
1 f” / . - 1 °° e‘k‘P
_ —:m0 - 2+1'7
_ — 6 sm 6 —-.-.— E . .
47r2 —1r £<|0|<W—e I I 1 + ism3 0 k=0 (1 + ism3 0)"
e—i2cp oo e—ilup

 

—— 41¢de
+1—isin30k:0(1—isin30)"]e cp

1 _imo , 6 2+.in an 00 ei(k-l)cp e-i(k+2+l)<p d d6
_ 4?]. e Ism l /— z:((1+isin3l9)"+1Jr (1-isin30)"+1) (p

<|0|<1r—c 1r k=0

1 —' a |sin 912+” .
— 'm d0 fl > 0
27f /€<|9|<7r—e e (1 + isin3 9)l+1 ‘ —

0 ifl=—1

1 _- 9 |sin0|2+i7 ,
-— "n d0 fl<—2.
27r £<|0|<1r—ee (1—isin3l9)(‘(“1 l —

 

 

 

 

59

By Plancherel’s Theorem,

1/2
(lKe.7(m1')(|L2[—1r,1r] = (Z lKe.~y(m, -)‘(l)l2)

(62

|/\

1/2 1/2
(J2; (KW )(lzfl)+ (;IKW )(l) l 2) (2-26)

First, consider the second sum in (2.26). Assuming e is small,

1/2 1 Isinl9|2H7 2 1/2
, .. 2 : _ —im9
(21mm, > ml) (; 2.12mi (1 ”2.3121“ 1

130

 

 

 

 

 

 

 

 

 

 

 

 

 

. 1/2
1 ‘ 024-17 2
S X -/ e ""9 (3191, may
120 277 6<|9|<7r/4 (1 +ism 0)
. 1/2
1 - |sin0|2+'7 2
+ — f "m" . . d0
(5 271' 1r/4<|9|<31r/4 e (1 + is1n3 0)1+1
. 1/2
1 _' 9 lSln0|2+t7 2
+ — / "" . . d6
(g5 27’ e<Il9-1r|<1r/4 e (1 + 1 sm3 (9)1+1
= I + II + III.
By Jensen’s Inequality,
1/2 1/2
3W4 I sin 0|4 \ 31/42 | sin 9'4
H < I - d9 ‘ = fJJJJZ . d0
__ C ((220 1r/4 (1 + 81116 0)l+l )( C( (>0 (1 + Sln6 0)(+1
31r/4 1 d6 1/2
— C (fr/4 sin20 )
S C,

with C independent of m, c,and 7. Next, turning to I, it follows that

2 1/2
120

 

1r/4 1
f . . 3 ,f.,m,.(0)d6

—7r/4 (1 + ism 0)

 

 

60

e—imOI sin 0l2+i7

1+isin36

"/4 If... ((9):? ”2 «x4 0 1 1/2
I < / —’—’7-—d0 /
- C ( -1r/4 sin2 6 C 4/4 1 + sin6 6 sin2 00w

—n/4 1 +sm 0

C,

with f€.m,.,(6) =

 

X{¢<|o|<7r/4}. By Corollary 2.12,

|/\

 

l/\

with C independent of m, e, and 7. Furthermore, III can be reduced to I (but with
1 — z'sin3 0 in place of 1 + z'sin3 0 which can be handled similarly to I). The first sum
in (2.26) can be bounded by C independently of e, m, and 7 using similar arguments
as above.

///
Claim 2.23

lim Km(m, ) = K1(m, )

c—iO
in L2[—7r, 7r], uniformly in m.

Remark 2.24 Assume Claim 2.23 for the moment. If w E lll2 then

nm’

lim
6—H)

 

f... 64'” (2mm - 33 so) — K702 — 3', WW 90)) ds0

jEZ

 

€—)0

5 lim / :3 IK.,.,(n — j, .p) — K.(n — j. (p)||o(2)(j, sandy)
'"jez

£401.62

S £136: H Ke,7(n “ j: ) " K‘r(n " j, ') ”L2[—1r,1r] ' H 03(2)“) ) ”UH—war]
jEZ

=0.

= lim 2 j... IK.,7(n - j. w) - K702 - j, w)ll<51(2)(j, w)ld<p

61

Therefore,

11'

. 1 _,.,,, . . .
D2+'”’Aw(n. m) = fl / e ‘p 2: K702 — J. 90)w(2)(3. <p)dso- (2-27)
_, jez

Proof:(of Claim 2.23) Let 62 > 61 > 0 with 62 < 7r /4 and assume I _>_ 0. Then

(K62.’7(m’ ) _ K61,'7(m7 ))A(l)

1 _- 9 |sin6|2+i7
;— — "n d6. 2.28
277 (v/61<|9l<€2 + v/17—62<|9I<1r—61) 6 (1 + 2181113 0)H'l ( )

By Corollary 2.12,

2 1/2 . 2 1/2
2 d0 5 c / .3;ng
120 €1<I9|<£2 1 + $111 6

which goes to O as 61,62 -—) 0. Again, a change of variables gives us the same result for

 

 

/ e—imfll sin 0|2+i7 1
a<|0|<52 1 + z'sin3 0 (1 + isin3 0)‘

 

 

the second integral of (2.28). It follows that {Km(m, -)“(l)},20 is uniformly Cauchy
in l2(Z+) as e —+ 0 independent of m E Z. A similar argument would give us the same
result for l < 0. Therefore, {Km(m, -)‘(j)}j€z is uniformly Cauchy in l2(Z) indepen-
dent of m E Z. Hence, by Plancherel’s Theorem, K6,,(m, ) converges in L2[—-7r, 7r] as
e -—> 0, uniformly in 177.. Since Km(m, .) —-) K7(m, ) pointwise, K6,,(m, ) —) K7(m, )

in L2[—7r, 7r] uniformly in m. Furthermore, it follows that

 

( i f" —im9 ISin0|2+i7 d0 fl >
I 21r —1r e (1+z'sin319)’+1 l " 0

Kv(m. -)‘(l) = 0 ifl = —1 (2.29)

1 7' _- 0 |sin0|2+i7 .
— ”’1 d0 fl<—2.
27r/_1re (l—z'sin30)|‘|—1 ‘ —

 

///

62

Claim 2.25 The Fourier coefficients of K0(0, ) are absolutely summable, i.e.,

Z|K0(0,)1)| < C.
(EZ

Proof: Suppose *y = 0 and l 2 0. Let 1/11 E 08° with 1/21 E 1 on (—7r/8, 71/8) and
suppwl C (—7T/4,7T/4). Let ’l/J2(0) = 1,!)1(0 — 7T) and $3 =1- 1&1 — 1/12. From (2.29) it
follows that

"/4 sin20

Ko(0,‘)A(l) = 0 121(0) . . 3 d9

—1r/4 (1 + z Sln 6)‘+1

 

sin2 0
0 d0
) (1 + isin3 19)1+1

 

+C/ 1P3
11/8<|0|<71r/8
51r/4sin2(5l
116
+6 3.. /. M0 )(1+z'sin3 11)!+1

= 1(1) + 11(1) + 111(1).

 

First,

sin2 0
0 d9
) (1 + z'sin3 0)‘+1
1

d6
/1r/8<|0|<71r/8: _0 (1 + sin 60)”2

 

:Imm = i

(=0

 

3
[11/8<I0|<77r/8

|/\
n

 

|/\
Q

(2.30)

By the same change of variables as before, namely

0 = arcsin(tan1/3 r) = h(r),

it follows that

 

oo °° “/4 2121(6) sin20 1
II = /
12%| ()| 12:; 4/4 1+isin36(1+1’sin30)’

 

63

 

17/4 e—il a.rcta.n(sin3 0) w1(9) 81112 0
/_«/4 (1+ sin6 6)”2 1 + isin3 6

 

 

2
—/_:6 —ilr COS'T $1,“ (7.)) SEC T TdT
3 1 + i tan 1' ‘ /1 —2/tan

3/bb12: T(cos Tg( 'r'r)d

3 —/:e T'cos T Z ak e'deT,

kEZ

 

 

M2 1M8 1M8 EM8

(2.31)

 

 

N
II
o

 

with
b = arctan(sin3(7r/4)),
we» T
9(T) : 1+ itanr m,
and

1 11 —i T
ak = -2—7r/_1rg(r)e k dT.

Before proceeding with the proof of Claim 2.25, we need the following definition,

proposition, corollary, and a theorem of Bernstein.

Definition 2.26 A function f defined on II is in Lip°(II) for 0 < a S 1 if there

exists 0 > 0 such that for :c, y E H,

lflfl-fWMSCW-m2

with distance measured on the unit circle.

Proposition 2.27 Let 0 < a S 1 and suppose f is a continuous function defined on

H which is differentiable for :c E H\{O}. If there exists C such that
IfWHSCW”5

then f E Lip°(H).

64

Proof: We need only concern ourselves with a: and y near the origin, say, 2,3; 6

[—7r/4, 71/4]. Suppose |:1:| < lyl. If neither :1: nor y equal 0 and they are on the same
y 1

side of the origin, then f (y) — f (z) = :1: f f (t)dt, which implies

|f(y) -f(-’r)|

|/\

y 1
if 111mm
:tC [y ltla‘ldt
0.01/1“ — 1x10)

Caly — ail".

|/\

|/\

The last inequality follows from the fact that the function no, (2:) = lea E Lip"(H).

Now, if :1: = 0, then

f(y) — 1(0) = 21/07th

since f' has an integrable singularity at the origin and f is continuous there as well.
Hence, the argument above works when either a: or y are zero. Finally, suppose :1: and

y are on Opposite sides of the origin with :1: negative. Then

lf(y)—f($)| S |f(y)—f(0)|+|f(0)—f($)|
g o [0” (110-111 + c [0 (1101-1111
S Ca(|y|a+|$|a)
S éa|y+lrvlla
= Caly-xl“,

where the last inequality comes from the equivalence of the l1 and 11/“ norms on R”.

This concludes the proof of Proposition 2.27.

65

Recall,
h 1' secz'r 1
gm = (“f _f l)
2 tan 7’ \/1 _ tan2/3 7.

 

= 91(7) '92”),

 

with

h(r) = arcsin(tanl/3 7').

Since 1121 is smooth, compactly supported, and 1111 E 1 in neighborhood of the origin,

it follows that 91 E 08°(I'I). Next,

1
92(7) = —§(1—tan2/3 T)—3/2 tan‘l/3 7 sec2 ’7',

which implies

193(T)| S CITI‘”3

0n SUPP¢1(h(T))-
Corollary 2.28
g E Lip2/3(l'l).
Proof: Since 91 is smooth and |g'2(r)| g Clrl‘1/3, we have
lg'(T)| S CITI'l/B-

Since 9 is continuous, Proposition 2.27 implies that g E Lip2/3(l'l).

///

The following is a theorem of Bernstein [4, pg. 32].

66

Theorem 2.29 If
feLme

for some a E (1 / 2, 1], then the Fourier coefiicients of f are absolutely summable, i.e.,

Z If(n)l s G.

1162

Corollary 2.28 implies that

Zlakl <00

kEZ

where ak is the kth Fourier coefficient of

h 7' sec2 1 1
flfl=wfip
2 tan 7' \/1 _ tan2/3 7.

 

= 91(T) '92(T)o

 

By the proof of Proposition 2.9, (2.31) becomes

1 , (2.32)

m
2 31.1021
(:0 0

j:

 

 

with Bid = $6). After interchanging the order of summation and applying Claim

2.11(ii), (2.32) becomes

2 NW _<_ 2 lilsz Z 31.1
i=0 1:1

(:0
w
2 2 Mil
i=0

C 2 lajl
jez

c. (2.33)

l/\

|/\

67

Again, by a change of variables, 2: II I I (l)| S C. With (2.30) and (2.33), this implies
(=0

2 moo->101 < ca

120

The proof showing 2 |(K0(0, ))“(l)| < 00 is nearly identical. Therefore,
(<0

:22 |Ko(0. -)‘(l)l < oo. (2.34)

which concludes the proof of Claim 2.25.

///

Finally, to finish the proof of Proposition 2.19, observe that (2.34) implies K 0(0, (p)

is a continuous function of p, which implies that there exists C such that

”(0(0: 90)] S C- (235)
Claim 2.30 There exists C such that

|K7(m. <19)| S C.

with C independent of m, (,0, and 7.

Remark 2.31 Claim 2.30 will finish the proof of Proposition 2.19. To see why this

is the case, assume there exists a C such that |K,(m, 90)] g C independently of m, (p,

and '7. Hence, by (2.27),

I... 8—... 2 mm — 1,2)a<2>(1.<p)d2

nEZ mEZ jEZ

llD2+hAwlllgozg = csup (Z

 

 

68

2 Km - 1.220911, so)

jez

2 1/2
dcp)

 

 

71
= csup /
nEZ -fl

 

 

1r 2 1/2
S csupZ (/ [Km-1' ww (”(7. 11)] dr)
nEZ jEz '—
1/2
S 02 ( / Iw(2’( (1M¢)|2dsp)
jEZ
= 02(2 M M 2?”
jEZ kEZ
= Cllwllzgzzn-
Proof:(of Claim 2.30) Recall,
1 7' sin2 0 sin2 Oe‘flw
K , = —— f , . .
0(0 (’0) 271 —« (1+isin30—e'¢+1—isin30-e‘W’) d6

 

 

1 1' sin20 sin20
= _/ ..3 . + ..3 _. d9
271 —7r 1+ism6—ew’ l-ismd—e"?

 

 

 

1 7' sin2 0(e —12¢_ 1)
+%/_1r1—isin3 0—e“‘-0d]9
7' — 7r ' 2 -i2‘P_
= if (1 coscp.):in2 0 . d0+—1—/ 8111.0.(63 1.)d0
71 —71 (1—coscp)2+(sm 9—smcp)2 271 —7r 1 —ism 0—e“‘P
= I+II.

First, for 71/4 3 lgol S 71, II is trivially bounded. If |cp| < 71/4,

 

 

lHl < _1_ f" |e“2"’ —1|sin20
" —1r|1—cos<,0+i(sin30-sin<,p)|
< / Igol sm 0 / |<p| sm 0
_ {0:lsin3 0|g2|sin1p|} |1 — cos <p| {0:181:13 9|>2|singp|} lsin3 0|
2 0 __ 2
g C/ 0—d0+C/ ( 7r>111)
{0:10:3chva Icpl {ale—«191211131 lrl
/‘ lgo|s1n 0
{0:I sin3 0|>2| sin gal} ] sin [p]
02
g C / d0 + C
{9 101<c1<le/3} M

g C. (2.36)

69

Notice
(1 - cos (,0) sin2 0
(1 - cos 1p)2 + (sin3 0 — sin 1p)2

 

2 0 (2.37)

for all 0,4,0 6 [—71,71]. Combining (2.35) and (2.36), it follows I = |I| S C. Hence,
(2.37) implies

 

7' (1 — cos cp) sin2 6
d0 < .
[.11 (1 — cos cp)2 + (sin3 0 — sin (p)2 ‘ C (2 38)

Finally, (2.36) and (2.38) imply

 

 

 

 

 

 

 

 

1 7' - |sin0|2+"1 |sin0|2+i7e“2‘P
K, =-—/-""9 .. . .. .do
I 7(m WI 271 — e (1+iSIn30—e“l’+1—231n30—e"‘P
1 W . - sin20 sin20
< _ —:m9 - 0 :7 . . d0
“ 271/46 lsm I (1+isin30—e"P+1—isin30—e“"P) ]
1 7' _- 9 I sin 0|2+‘7(e“2"’ — 1)
— "n . d6
+ 271/— e 1—isin30—e"‘P
_1_ 1' (1—coscp):in20. d0+C
271 -—7r(]. — cos (p)2 + (sm 0 — sm p)2
S C,

with C independent of m, p, and 7. This concludes the proof of Claim 2.30 which

concludes the proof of Proposition 2.19.
Proposition 2.32
llD121+hBhwllzgozgh _<. Cvllwlllglfn‘h:

with C7 = c(1 + WI) and c independent ofh > 0.

Proof: By a similar homogeneity argument as in Remark 2.18, we can assume h = 1.

Suppose m > 0. Then

 

_ . A(1) .
D2+"’Bw( (,n m) =2 —/: e""9|sin0]2+iv sgn(m+1 le £911) 1 .d6
J';r£rrz+127r (1+ sgn(m +1 — ])ism 0)lm+ “3|

70

1 . ' 62+” m “(1)0 '
= lim—/ e—zn0_l_31_n._l.__3__ w. .(373) . d9
HO 27r £<|0|<1r-—c 1+zsm 0 j=_°o (1+zsm 0 "“1

 

 

 

 

 

 

 

, 1 - . - . . °° . . . A .
—£1_1+16 -2—7r/€<l0|<1rf€—m9|sm 0|2+’7(1 — 2 sm3 0) (_Z (1 — 2 sm3 0)m—Jw(1)(0, 3)) d9
J-m+2
1 - 7" - | sin 0|“i7 sin 6|2+i7e"2"’
= lim — f ”"9 f """P . —— . “ 0, d d6
6—)0 4772 £<l0l<7rf€ _"8 (1 + iSin3 0 _ 6W; 1 _ iSin3 0 __ 6-,‘p w( ()0) (,0
1 7r . . I sin 9|2+i7 l sin 0|2+i78—i2¢p
= 1' _ We / -m9 . — . * 0, d0d
€136 471'2 —1r 6 £<l0|<1r—ee (I + isina 6 - e"? 1 — z'sin3 0 - e‘W’ w( (p) (’0
= lim -1- 1r 6“” Z I? (n -J' <p)u‘1(2’(j <p) d<p
€—)0 271' —1r 'Ez 6’7 , , ’
with
~ 1 . l sin 9|2+i7 l sin 6|2+i78—i2¢
K , = — f “'m” . — . d6.
W(m (p) 27F e<|0|<1r-e e (I + z'sin3 0 — eW’ 1 - z'sin3 0 — e‘W’

Now, as with K6,,(m, (p) in Claim 2.22, it follows that

1.66:7 (m, .)A(l)
1 ei(k—l)cp e—i(k+2+l)¢p

= — —im0 ' 2+i’7 W 00 _
4W2 /e<|o|<1r—ee |sm0| / Z ((1 + isin3 6)k+1 (1 _ isin3 g)k+1) d‘Pdo

_W Ic=0
= sgn<l+1)K.,.(m,-)“(z).

 

 

It follows, as in Claim 2.23, that {Km(m, -)‘(j)}j€z is uniformly bounded and uni-
formly Cauchy in l2(Z) as e —) 0 independent of m E Z. Again, by Plancherel’s

Theorem, R6,,(m, ) converges in L2[—7r, 7r] uniformly in m.

Definition 2.33 Let

in L2[—7r, 7r].

71

It follows, as in Remark 2.24, that

i 1 71’ ——im ~ . A .
02+ le(n, m) = 5; /_ n e "’ 2 K701 — J, cp)w‘2)(3, W190

 

 

jez
where, for (,0 79 0,
~ 1 7* . |sin0|2+i7 |sin0|2+i7e“‘2¢
K , = __ / -zm9 . — . d6
7(m (p) 27r -1re (1+isin36—e'v’ 1—z'sin30—e‘W’

_ _1_ / (gs-mo ( I smell?” _ | sin 0|?” ) d9
_ 27r —‘n' 1+isin30—e‘V’ 1—isin30—e‘W’

_1_ /" 6-...9 I smoking — M) d,

21r -1r 1 — zsm 0 — e‘W’

_ _ 7' (my I sin 0|2+""(sin3 9 — sin go) d6

_ 7r -7r 2—2coscp—sin2<,o+(sin30—sincp)2

L f" 8-.....) I sin0|2+"’(1 - e-W)
27r _« 1 — z'sin3 0 — e-w

= W7(mv (P) + 177(m,<p).

 

 

 

+

 

+ d6

 

To finish the proof of the proposition, we need to show that there exists C such
that

|1?7(m,so)l S C(l'rl + 1)

independently of m E Z, for almost every (p 6 [—7r, 7r] and ’7 (see Remark 2.31). Note

that

At

lV‘Y(mv 90” S C:

by absolute value estimates and (2.36). Thus, we need to show

|W7(m, son 3 C(l’rl + 1), (2.39)

with C independent of 'y, m,and for almost every 90. Note, the integrand of Wo(0, (p)

is not of constant sign. Hence, we can not use the same methods used in the proof of

 

72

Proposition 2.19.

Definition 2.34 Let

 

—i _i sin 6’ 2+” sin3 0 — sin
W€,7(m, $0) = _/ 6 m0 l l ( (,0)

d0.
71' {0:6<Isin 9—sinl/3 SP” 2 _' 2 cos 90 _ Sinz (P + (Sin3 0 — Sin $0)2

Obviously, £13 W€,.,(m, (,0) = W7(m, (,0) for (,0 75 0.

Claim 2.35 There exists C, such that
IWm(m, w)| S Cw

for all m E Z and (0 ¢ 0, where C, = C(I'yl + 1) with C independent of c, m, 7, and
<0 75 0-

Claim 2.35 implies (2.39).

Proof:(of Claim 2.35) We can assume both 6 and |(p| are small. Fix (,0 75 0.
Set 7' = sing0 and r = 2 — 2cos<,0 — sin2 (0. It follows that IT] ~ |<,0| and

r = (1 — cos (0)2 ~ ccp". Thus,

 

 

 

 

- sin 0 2+” sin3 0 — r
7r|Wen(m, (PM S /|-0|<1r/4,€<|sin g_11/3l thl r +l (Sing 0 _ T)2 )d0
. / I...
"/4<l9|<31r/4,£<|sin 0—1-1/3| 'r + (sin3 0 _. T)2
+ / e_,-m9 I sin 0|2+‘7(sin3 0 — 7') d0
37r/4<0<5‘n'/4,e<|sin 0—71/31 ,- + (31113 9 _ 7.)2

 

 

= 16,7(m, (,0) + 116,7(m, (p) + 1116,7(m, (,0).

73

For small (0 (recall 7' is small), clearly
116,7(m) (p) S 0) (2'40)

with C independent of e, (0, m, and 7. Next, consider 16,,(m, (,0). By setting é = sin 0,

we have

e—im arcsin£€2l€li7(€3 — T)

/|‘arcsinE|<1r/4,e<|£—Tl/3| T + (£3 — T)2 V1 — dgégl

 

16:7 (m, (p) =

 

By a second change of variable, namely 5 = 74/33:,

1/3$)x2|x|‘7(a:3 — 1) dz

r'+(x3—1)2\/1—W’

—im arcsin(1

 

8

 

 

It (m 90) z /
’7 , |$|<J¥T_1/3,6’<I$-ll

with e' = 1‘1/36 and 1" = T—27‘ ~ C(02. Note that r' > 0. Let (0 E C8°(R) be an even

function so that Co E 1 for |a:| _<_ 2 and suppCo C {le S 3}. Also, let (1 = 1 - Co-

 

 

Then
I (m (.0) < / e-‘marcsin<r”3m)$21xl£7<x3 " 1) C0“ — 1) :1:
"7 ’ " |x|<:,@r-1/3,e’<|x—1| r' + ($3 - 1)2 x/l — 72/332

 

 

+ / e—imarcsmws.)wZIxIW — 1) cum — 1) ..
|$|<J§ZT“’3»6'<lx—ll r' + (2:3 — 1)2 ‘/1 _ 72/31,?
: A6,,7(m’ T) + BE’,7(m’ T).

First, consider Aer,7(m, 7'). It follows that

/ e—imarcsin(71/3x) lexli7($3 _ 1)
|x|£4,e'<|x—l| T, + ($3 — 1)2
_ 3lfivl‘7($ - 1) Co(rv - 1) d2:
r' + 9(2: — 1)? \/_—1 _ 00.2
+ / e"'"arcsin(r”3x) 3'34”“ " 1) (c(1? - 1) a:
|$|S4.e'<|a:-1| 1" + 9(a: — 1)2 ‘/1 _ 72/3132
= Ag) (m,7') + AG) (m, 7'),
5 :7 C ,‘Y

 

A n(m, T) g

 

 

 

 

 

74

r’(:).- — 1)2(:z:3 + 2:):2 + 3x + 3) + 3(a: — 1)4(2a:3 + 3::2 + 3:): + 1)
(r’)2 + r'(a: — 1)2(:1:2 + :1: +1)2 + 9r'(:1: — 1)2 + 9(1): -— 1)“!(11:2 + :1: + 1)2

 

S C /|$|S4

 

 

 

|x3+2m2+35c+3| |2x3+3$2+3x+1|
< C dx
" |m|<4 (:1:2 + a: +1)2 3(5):2 + a: + 1)2
S C. (2.41)
Next, let a E R such that a = — arcsin(¢1/ 3), which implies |a| ~ |T|1/ 3. Then

 

A:.2l1(m,'r) _<_

9

/ (e—imarcsin(7-1/3x) _. eixmm 3l$li7($ — 1) (0(5): - 1) d1:
I$IS4,6’<Ix-ll<Fn-l—a-[ 1J+ 9(1: _ 1)2 m

 

3|$|‘7($ - 1) Co(-’L‘ - 1)

ixma

 

 

 

 

 

+ f e I dill.
|$|S4.6'<lx—ll<]—,;1;I 'r + 9(2: — 1)2 ‘/1 _ 71.2/31:2

+ / e—imarcsin(TI/3a:) 3I$Ii7(x — 1) (0(33 — 1) dz
|x|S4.maX(Tfipe')<|x—1| 7" + 9(a) - 1)2 fi_ 7.2/31;

2 A9)7(m, 7') + Aif)7(m, T) + AS’Mm, 7').

After applying the Mean Value Theorem to the real and imaginary parts of 3(x) =

e“’" main(Tl/3z) — e‘i‘m“, it follows that

 

(3) _
A6,,7(m, 7’) .—

 

/ m71/3 sin(im arcsin(T1/3z ))— ma sin( 2 ma)
|x|S4.6'<lx—1I<T;‘;T ,/1 _ 72/323 9’ 2
'(IL‘ _1) 3l$li7($ — 1) (0(3) _ 1)
T-I- 9(a: — 1)2‘/1_7-2/3$2d$

m'rl/3 _ , 1 3
cos(zm arcs1n(T / 313)) + ma cos(y$ma)

+ /
|x|g4,e’<|x—1|<WIJ(\/1 _ 72/33)},

 

 

 

 

.(x _1), 3W2: - 1) Co( as — \/__d$|1)d

7' + 9(13 - 1)21 —- 72/3332

|/\

clma|/ 1 d2:

lx-ll<m
< C, (2.42)

75

where .2; and ya; are between 1 and it. Let 1/2 6 C80 with 1/2 E 1 for |:z:| g 1/4 and

SUpp’l/J C {|er S 1/2}. By changing variables,

 

 

 

 

 

 

 

 

 

Ag) (m 7.) = / ixmale +1li7$ (0(1)) d3?
‘ 1’7 ’ |a.)|g3,e’<|ac|<r;fa r' + 9x2 \/1 __ 7.2/3“ + 1)2
. 1'7
_<_ / ezzma 3$|$ +,1l 1M; + 1) (C(32) (1117
|z|g.°.,.g<|:)c|<ml,TI r -+- 9x \/1 _ r2/3(a: + 1)2
/ m 31:11 + 11211 — we + 1)] coca) d.
|z|§3,e'<|z|<l—r—nl:[ r' + 92:2 \/1 _ 72/30” + 1)2
i a:
S C + 0 [Re m (r’ + 9x2X16'<lxl<rnfi1}(x))

 

 

. ()3.- + 1|”[1 — (12(1: + 1)]Co(a:)) dx
\/1— T2/3($ + 1)2

 

= C + (1)/R «swig... (x)a.,.(x)das

 

= C + Cl(fe',m,a,r' * 97,7)(ma)l1 (243)
with
" _ (I:
fc',m,a,r'($) — WX{£'<III<T;IH}(23)
and

= Irv + 1|”[1 — «M1 + 1)]<o(:v)
\/1 — r2/3(a: + 1)2

 

 

gmr (37)

Claim 2.36 There exists C such that

er’,m,a,r' llL°° S C)

with C independent of 6', m, a, and r'.

Proof: By Fourier inversion,

 

lfe'.m.a,r'(y)l : C [Reizy c',m,a,r'($)d$

76

/ ixy (13
I
c’<[:n|<];la 7‘ + 911:2

x
= c sinx T————dx
/e’<|x|< 1 ( y)? +9232 ,

lm—aT

2
x
0 —,———dx
/| lylr + 9x2

IKHIIT
c d x
/ ——(cos(xy)) , ,da:
|m|>1tr1€'<|x|<1..+..( d2: r +92:

+—
lyl
r, — 9x2
2dx

1
C _
+ M (r' + 9x2)

dx

 

 

 

|/\

 

 

|/\

cos(xy) + c

 

 

[mbfig'defia
c 1
WI Izl>¢1 1172

C,

|/\

|/\

where C and c are independent of e', m, a, and r' (c is the boundary term).

///

Claim 2.37 There exists C such that

llgwllu S C(I7I + 1),

with C independent of 7 and T(small).

Remark 2.38 Since

llfe’m,a,r' * g7,TllL°° S llfe'm,a,r'llL°° ° llg'YflllLlr

Claims 2.36 and 2.37 imply that

lfe'm,a,r' * 9717(ma’)l S C(l’Yl +1)

With (2.43) this implies
Agmmfl) g C(|7| +1). (2.44)

77

Proof:(of Claim 2.37) Assume r is small. Recall,

g (x) z Ix + 1|“’[1 — «)(n: +1)1<o(a:)
,/1_.2/a(.+1)2 ’

 

 

which is smooth and compactly supported for all 7. By Fourier inversion,

97,7(31) = Leixyg7,r(x)dxa

which implies there exists C which is independent of 7 and small r such that

I 97,1(y)l g C for all y E R. Also, one can check that there exists C such that

d2

397.1(1) < C

 

 

for small |7| (|7| g 1) and that

d2
I s 01))?

35521.10”)

 

for large |7I (|7I > 1). Thus,

lg'm (y) Id?!

1 . d2

_ ’33!— "

y. [Re d$2(g%.,)(x)dx
1

CmaX(|7|. 1) + 011180107th / —dy

lyIZmaX(l7l.1) y2

g ,T y dy = /
[RI 7 ( )I IyISmaxmm)

dy

 

 

+/
lyIZmaX(|7|.1)

|/\

|/\

CmaXUVI. 1)

|/\

C(|7| + 1).

and this concludes the claim.

///

78

Finally, assuming m 75 0 and by change of variables, A:.5L(m, 7') becomes

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

/ e—imarcsin(rl/3[x+1])3lx + 1|i7x (0(5):) d2?
max(e',1#[)<|x|<3 r' + 9:1:2 \/1 _ 7.2/3“, + D2
= / e—im arcsin(rl/3[x+1])3$l$ + ”in/)0: + 1) (0(3) d2?
max(e’,m)<|x|<3 r' + 9x2 \/1 _ 72/3“: + 1)2
+ / e—imarcsin(rl/3[x+1])3$lx +1li7l1 _ WI” + 1)] (OCT) d$
max(e',I;10—r)<|x|<3 r' + 9x2 \/1 _ 7.2/30; + U2
1 - r2/3 x +1 2

S C+c / \/ 1(3 )

max(e',]—mi;[)<|x|<3 mr /

i (e—imarcsin(rl/3[x+1])) 113 III: + ”fill _ ¢($ + 1)]C0($) d2?
dx r' + 9x2 \/1 _ 7.2/3($ + 1)2
< C + ___C_ / e—imarcsin(rl/3[x+1])
_ ImT1/3I max(e’.I.—.‘;I)<le<s
d x - cImaI
-— 1 '7 1 — 1 ) d
dx (7" + 9x2I$ + I I (0(2): + ”(006) x + |m11/3I

c d x
< c —/ — (:——) .
_ + |mr1/3I $14“ dx r + 9x2 x

c :1; d .
—— —— — 1 '7 1 — 1 d

+|mrl/3| fm<lxl<3 r' + 9x2 Idxux + I I 10(93 + )lCo)($) x + C

5 0+ c / _1_d c(I)I+1)ImaI
Imrl/3I ($1411 x2 Imrl/3I

S C(|7|+1)- (2-45)

Therefore, combining (2.42), (2.44), and (2.45), it follows that
4334mm) 3 0(1)) +1).

With (2.41), this implies
4.1,.(m. 1) s 0(1)) +1). (246)

79

Next, we need to show

B.),.(m,r) .<_ 0(1)) + 1). (2.47)

with C independent of 6', 7, m, and 7'. Using similar methods as above,

 

 

 

 

- - ~’172|.’17|"7(953 - 1)
B ’ , < / —zmarcsm(‘rl/3x) I
‘ ”(m T) ‘ lesér-l/aeklz—u e 1‘ + (It?3 -1)2
_ 1412(1- — 1) (1(4 — 1) d.
7" + (5’3 - 1)2 V1 — “r2/3x2

+ / e—im arcsin(rl/3x) |x|i7($ — 1) <1 (22 —' 1) d3}
lxlslzér—l/fle’qz-u r' + (x — 1)2 ‘/1 _ 72/33;

= B:.1L(m,r) + B:?L(m, r).

 

 

 

Since r' ~ ccp2 is small, \/1 — 7'2/32:2 2 c for x g lax-V3, and suppCl x — 1 C
2

{le > 2}, it follows that

dx

 

(1) m T c |7"11=2(II=3 - 1) - r'(II: -1)|
B" ( ’ ) S [Imlzz (1‘ - 102(1133 - 1)2

|$2($ -1)2($3 - 1) — (a? - 1X51?3 -1)2|
“($.22 (x — 1)2(x3 — 1)2 dz

5 6
x x

c/ I—I-dx + c/ —8dx
IIIZ2 17 le22 517

C. (2.48)

 

|/\

|/\

/ (e—imarcsin(rl/3x) _ eixma\ I$Ii7($ _' 1) C1($ — 1)

I2I5391-1/3.c'<lz—11<r.—3;I ’r’ + (:2: — 1)2\/1 - T2/3x2

/ eixma lehh’ — 1) <1 (11: — 1) d2?

IxI391-1/3,e'<Ix-II<W 7" + (a; _. 1)? ‘ /1 _ T2/3III2

+ / e-imarcsin(‘rl/3x) I$|i7(x _ 1) (1(3) —' 1) dd:
lxls91-1/3,max(e’,I,—nla)<|x—1| r' + (x _ 1)2 ( f——1 _ 72/3332

_ (3) (4) (5)
_ B6,,7(m, r) + B6,,7(m, 7') + B£,,7(m, r).

 

 

+

 

 

 

 

80

It can be shown

3911(7)"), r) g C (2.49)

using similar techniques as those used in (2.42). By changing variables,

 

 

 

 

 

BS4) (m 7.) = / ixma I113 — llWm (1(3) dIE
‘3 ’7 , I$+1IS-?T_l/3,€’<I$I<TKIJ 7" + $2 \/1 _ 7-2/3($ _ 1)2
< / ixma I17 — 1|i7$ (1(3)) d2! + C
— |a):|51§511-1/3,c’<|:z:|<]"f1 r' + :1:2 \/1 _ 7.2/3” _ 1)2 ’

 

 

with C arising as a result of the change in region of integration (recall 7' is small).

Hence,

 

 

 

 

 

 

 

 

 

_ 1 1')
B(,4) (m, r) S / cos(xma) Ix , I x C1 (x) dx
5 ’7 lxlgAQT-l/3,e’<|x|<I—n-f—J r + x2 \/1 _ 7.2/3“; _ 1)2
_ 1 1")
/ sin(xma) Ix, I22: C1 (x) dx + C
legagr—r/s,g<lx|<1ml_d r + x \[1 _ 7.2/3” _ D2
= I + II + C.
First, assuming 7' is small,
2
II 3 clmaI ,—x—2dx S C. (2.50)
|x|<Tml—a. ’l" + at
Next, using the fact that (1 is even,
_ 1 1")
I :: / cos(xma) Ix, I 2x C1 (x)
d<zgmin(1§r-1/3,]—ml—GT) r + x \/1 _ 72/3” _ 1)2

 

dx

 

_ 1 1'7
+ / f , cos(xma) Ix, I 2x C1 (x)
max(:-2—3|T—1/3I,-]_"}_a[)gx<—e r + x \/1 _ 72/3($ _ 1)2

81

 

 

 

 

 

 

x — 1 ‘7 x + 1 ‘7
- I I - I I C1($)d?
\/1 — 72/3(x — 1)? 1/1 - r2/3(x + 1)?

U

 

 

 

(I:
l’<x$min(3§T-ll3, 1 )COS($ma) 7.’ + $2 (f7($ + 1) — 12(2) _ 1))(l($)d$

ImOI

 

with
_ lcvli7
f7($) — r————1 _ 72/3332
and

I _ i7|x|i7 —2Ix|‘77'2/3x

f’r( )_ mm + (1 __ T2/3x2)3/2'

Using the fact that for a complex-valued C1 function f and x1, x2 6 R,

 

 

|f(=l?1)--f(?32)|S2 SUP If'(y)||$1-$2|)

yEIx1,x2]
I is bounded by
1 I
2/ — d
2325151974” le ”filial |f7(y)| :1:

since suppCl C {le Z 2}. For 2 g |x| g #74” and y such that Ix — yI S 1, we have

IyI ~ |x|. Hence

I C ”Y
sup 11.14)) s —'—I + 0.2/1.1.
ly—xlsl |$|

Therefore

1
I _<_ c|7I ——dx + 09”

d < C 1 , 2.51
2S|x| £132 Asxng—l/a IE —' (|7| + ) ( )

since T is small. It follows by (2.50) and (2.51) that

B:?L(m,r) g C(|7|+1). (2.52)

 

82

Finally, by changing variables once more, assuming m 75 0, and integrating by parts

(as in (2.45)), B:?L(m,r) becomes

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 d _' 3C8. ( l/3[ +11) :13 '
_ 1m 1' 1117' :1: 137 d
C [x+1|53?r‘1/3,mu(e',m)<IxI mrl/3 d1!) (8 ) 7" +$2Ix + I (1(3)) (13
c f d ( x ) dx
Imrl/3I Find“ dx r’+x2
c x d - crl/3
__ __ _ 117 d
+Im'rl/3I filaqxl r'+:1:2 dx(Im+ I CHI”) 3+ ImaI
c|7I m
C —/ d
" + |m71/3I max(2,m)<Ix| 1" +x2 |x+1|I<l<$ $)I x
d
+Im'rC1/3I/1 (In: | r +x-_—_m—2 Cl($)I x
' 12
_ 171mm / AM) +0
Imrl/3I
S C(l’)|+1)- (2-53)

Together, (2.48), (2.49), (2.52), and (2.53) imply (2.47). Combining this with (2.46),
we conclude that

1.,.(m,<p) s 0(1)) + 1). (2.54)

with C independent of 6,7, m, and (,0. Setting 0' = 0 + 7r, it follows
1116,,(m,—(p) = 15,7(m,(0), which implies 1116,7(m, (,0) S C(|7| + 1) with C
independent of e, 7, m, and (0. This, (2.40), and (2.54) conclude the proof of Claim

2.35 which concludes the proof of PrOposition 2.32.

Recall,
Dfi+‘?B,,w = 2D§+‘?A,.w — D§+‘?A,.w + 2D§+I7Ehw.

83

Combining Pr0positions 2.17, 2.19, and 2.32 finishes the proof of Lemma 2.6(ii), and

hence completes the proof of Lemma 1.17(ii).

2.4 Discrete Complex Interpolation

We are now in a position to prove the final crucial estimate, namely Lemma 1.18.

Recall the statement of Lemma 1.18(i),

lth(m)no|Izgz;,gh S Cllflolllg, (2-55)

with c independent of h > 0.

To prove (2.55), consider the analytic family of operators
rznom) = D;‘/4D§,‘—Z>Hh(m)no(n), o g Rez g 1,
with 770 E lflZ). If 2 = i7, then

IszUOIllgOIEn'h = IIDl—iS/Mflollzooz? < Cllflollzg: (2-56)

nm,h_

by Lemma 2.6(i). If 2 = 1 + i7, then

IITz770||141°° = IID_l/4-i5/47Hh(m)flolItglsgm S CIIflong, (257)

n m.h

by Lemma 2.2(i). Combining (2.56) and (2.57) with Stein’s analytic interpolation
theorem [7] and letting x = 4/5 implies (2.55) which finishes the proof of Lemma
1.18(i).

84

Next, recall the statement of Lemma 1.18(ii),
llAhwllzgzggh S Cllwllgmwf) (2-58)

with c independent of h > 0. Similar to the proof of (2.55), consider the analytic

family of operators

TM = Dgz/Zpfill‘zlAhw, 0 g Rez g 1,

 

with (.0 defined on Z x Z and compactly supported. If 2 = i7, then
“1.421143”, = ”Dr‘s/24141141,, s 0.111111%“ (2.59)
by Lemma 2.6(ii). If z = 1 + 27, then
IITzwIIIgzsgm = IID-lfl—iszthzgzgh S Cvllwlllg/fi'lmh, (2-60)

by Lemma 2.2(ii). In both (2.59) and (2.60), C, = c(1 + |7|) with 0 independent of 7
and h > 0. Again, applying Stein’s analytic interpolation theorem finishes the proof

of (2.58) which concludes the proof of Lemma 1.18(ii).

CHAPTER 3

Numerical Results

In this chapter, we discuss the numerical implementation of the fixed point iteration

with the operator (Pm.

3.1 The Cutoff Function 7h(n,rn)

The iterates of the contraction mapping (Pm are defined on the entire grid, Z x Z.

To go from one iteration to the next, the entire iteration is needed. This is obviously

not feasible numerically. Hence, we need to introduce a “smooth” cutoff function

which is zero for n or m large. One may think we can simply set the iterates equal

to zero for large n or m. However, this may adversely affect the norm estimates of

the contraction mapping. This is due to the fact that if 02 is defined on Z x Z and
1ifIn| SN,|m| 3M

XN,M(n1m) :
0 else,

then it is not necessarily the case that

llan.h(XN.Mw)II(;-,°zghh S Cllanmwlllgwgha

85

86

with c independent of h > 0. In fact, it is quite the contrary as h —) 0. Thus, we
need to introduce a cutoff function which is one for Inl S N and |m| S M, but decays
slowly to zero for large n.

Let h > 0 be small. Suppose we would like a solution of (K dV): for (n, m) E Z x Z
such that |nh| S N and Imh3| S M where h is chosen so that N is an integer multiple
of h and M is an integer multiple of h3. Let 7;,(n, m) be a piecewise linear function
such that

7,, E 1 for InhI S N and |mh3I S M

and

supp7h = {(n, m) E Z x Z: Inhl S 2Nh"1/5, Imh3| S M}.

Obviously, 7), can be chosen so that

chl/S
sup lamh'th S N
n,meZ

 

(3.1)

for small h. Note that we do not need any decay in the m—direction since none of the

three norms of X h involves differences in the m—direction.

Definition 3.1 For a discrete function 77 defined on Z x Z, let

~

877(77’1 m) : 711(7),, m)"(") m)

Recall the definition of (Pm,

1412(1), m) = Hh(m)7?o(n) — éAhanmmn, m).

Wlth 770 E 1,2;(Z).

 

87

Theorem 3.2 There exists ()0 > 0 and r > 0 such that, if B, = {w E X), :

lelx. < r} and “none, 3 60, then

i) S‘Dm : B, —+ B, continuously and

ii) |IS<I>,,O(V) — S<I>m(u)IIxh < AIIV — ,uIIXh, for some A < 1 and 14/1 6 B,,

with A, 60, and 7‘ independent of h > 0.

Proof(i): First, we would like to determine a bound for the operator norm of S on

X h. Obviously,

llgflllzgzfih S Il’lllzgzggh (3-2)
and
sup ||§17(-.m)llzg S sup ||n(-.m)|lz,2,- (3-3)
mEZ mEZ
Note that
0n,h(7hn)(n, m) = 7;,(n + 1, m)3n,h17(n, m) + 8n,h7h(n, m)n(n — 1, m).
Then

1/2
llan,h(nn)llzgo)2 < llamhnllzsgzgn, + 8161121 (ha 2 (6.171(1). m)|2|n(n -1.m)l2)

nah "-
mEZ

, 2/5 1/10
E IlanmnlI123)?“+ 51113013 Z Ian,h7h(n) m)|”) (ha 2 WW — 1) m)|10) (3-4)

"6 mlSMh-3 mez
1/10
ch1/5M2/5
S Ilan,h77llz;,-o13n h + _—N—_ SUP ’13 Z W71 — limlllo (3-5)
' nEZ mez

nEZ mEZ

1/2 1/5

S(1+5)||?7||xh,

88

with 5 depending on N and M. Note that (3.4) follows from H61der’s inequality and
(3.5) follows from (3.1). By this estimate, (3.2), and (3.3), we have

lls‘nllxh S (1+ é)||17||x,.- (3-6)
Recall that from the proof of Theorem 1.21, we were able to show that
I|<I>,,077||Xh S C5,, + Cr5
for 17 E B,. By choosing r and (50 small enough, we have

1.
c60+Cr5S .

 

 

1+ 6
Hence,
(1511.47))...s(1+&)1+E =r
for 77 6 B,, which implies
30,, :B, —) 3,.

Proof(ii): Since S is linear, (3.6) implies that if [1,12 6 Xh, then

”59%)” — S‘I’nol/llx). S (1+ 5)||‘I’nou - ‘I’nonlxh- (3-7)

From the proof of Theorem 1.21(ii), if 11,11 6 B,, then

ll‘I’nou - (Paul/Hm. S C(“’llu - Vllxh- (3-8)

89

1
If we choose r small enough so that Cr4 < Till—5’ then (3.7) and (3.8) imply
llgéflof‘ _ 3(1)?)onth 5 All” — VIIXM

with A = Cr4(1 + E) < 1.

Remark 3.3 In the definition of 7),, if we fix m E Z such that Imh3| S M, then
7h(n, m) = 0 only if InhI > 2N h‘1/5. This was necessary to insure (3.1), which led
to (3.6) independent of h > 0. If we assume h is bounded below, then the h‘l/5 can

be discarded, i.e., we can select 7,, with 7,,(n, m) = 0 only if |nh| > 2N.
Hence, S<I>,,0 is a contraction mapping on B, C X), for some 7‘ > 0. By the
Contraction Mapping Principle, there exists a unique 17 E B, such that
S (1)1107? = 77
which implies
VhQnofl = 77

Let A = N/h E N. If (n, m) E Z x Z such that InI S A and |mh3| S M, then
c11.401111) = 7701.171)- (3.9)

By definition of the Operator (Pm, (3.9) implies that 77 solves (KdV): for —A + 3 S
n S A — 3 and |mh3I S M. Since the definition of 62,,,w(n, m) includes w(n + 3, m)
and w(n — 3, m), it is necessary to have n satisfy —A + 3 S n S A — 3 so that when

the linear operator (as in (1.1)) is applied to both sides of (3.9), we maintain equality.

90

3.2 The Coefficients Q[n,m]

In this section, we now turn to the actual implementation of the fixed point iteration.

Fix h > 0 and let 770 E 1,2,(Z). Recall the definition of the operator Hh(m),

 

Hh(m)no(n) = 1 [Wk e“"h9(fio)h(9) d0

2_7r —1r/h (1 + sgn(m)i sin3(h0))l’"|

 

6

1 [WM e—inhO -imarctan(sin3(h9)) thez 770(kleikh9

 

 

 

 

‘ 2; ./). (1 + sin6(h0))ImI/?
h w/1 . . . 3
: k _/ —i(n—k)h0 —imarctan(sm (h0)) (10
£0“ )21r —1r/h e e (1 + sin6(h0))l”"l/2
1 1f . . - 3 1
= k _/ -:(n—k)0—zmarctan(sm (0)) d
262’“ I27. .. e (1 + sin60)lml/2
1 7' cos((n - k)6 + m arctan(sin3(0)))
= Z 770095;] (1 + sin6 6)ImI/2
kez ‘”
Definition 3.4 For n, m E Z, let
_ 1 1' cos(n0 + marctan(sin3(0)))
an, ml - Er- _W (1 + sins 0)Im|/2 d6.
It follows that
Hh(m)770(n) = : 110(11)an — k, m). (3.10)

kEZ

We can represent Ahw(n, m) using these coefficients as well. Note that QIn, m] does

not depend on h. These coefficients can be computed and stored for future use.

To begin, we choose the zero function as our starting point. Independent of 1',

this guarantees that we are starting in 8,. With Q[n, m] precomputed for n and m

91

such that Inhl S 2N h"1/ 5 and Imh3| S M, we compute the first iteration, (’01. Let

wl (n, m) = (1),,0 (0) (n, m)

= Hh(m)770(n) - 14.6....(oxn. m)

5
= Hh(m)770(”)
= £7700“)an - k) m].

It follows that (01 solves the associated discrete linear equation. To obtain the first

iteration, we need to multiply by the cutoff, i.e.,

601 (n, m) = 7,,(n, m) -w1(n, m) = 3(1),“, (0)(n, m),

which still solves the linear equation for n and m such that —A + 3 S n S A —— 3 and
|mh3I S M. Since (721 is the solution to the associated linear equation, we check the

computation’s accuracy by means of maximum relative error,

 

 

max max
InISA—3 0<mSMh‘3

(3.11)

01(71, m)

 

(In n,m —(Dl n,m—-1 3 ~
( Ih3( J +6n,,,w1(n,m)l

and similarly for m < 0. We can also check how well it solves (K dV): by considering

 

01(fl1m)‘$1("1mfl + aihfijl (n, m) 'l' éan,h(af)(n’ m)

max ..
“’1 (n) m)

max
InISA—3 0<mSMh‘3

 

 

I (3.12)

and similarly for m < 0.

The second iteration (112 is computed as above, i.e.,

(7)2(n, m) = 7;,(n, m) - <I>,,0(IJI(n, m) = SQmw1(n, m).

 

92

By induction,
(I'Jk(n, m) = 7h(n, m) - <I>,,0(12k_1(n,m) = $0,012,-de

for k 6 N. At each stage, we compute the maximum relative error by replacing (221
with a), in (3.12) for m > 0 and m < 0. Finally, we compute the Xh-norm of each

iteration. Recall, this involves three size estimates,

1/5

1/2
Ilaklll=llakllzgqgh= h 2 (’13 Z l@k(n,m)llo) ,

In|S2Nh—3/5 ImISMh3

1/2
ll‘IJkll2 = llamhaklllg'flfim = lnlgmgm (’13 Z lan,h0k(n,m)l2) ,

lmlSMh"

and

1/2
lldklls = SUP H5110 m)||,2 = max ’1 land" m)|2 -
mEZ , h ImISMha Inls2§1—6/5 ,

3.3 A Few Examples

In this section, we give a few examples of the convergence and divergence of the
operator <I>,,O. Using Mathematica, we have computed Q[n, m] for |n| S 32 and
Iml S 32. Hence, our iterates will be defined on InI S 16 and ImI S 32. In the
following examples, the step size h will be no smaller than 0.1. By Remark 3.3, we

can set
1 lnl s 8. Iml _<. 32

3+2 —16SnS —8,|m| 332
7h(n,m)=

—Lg-+2 8SnS16,ImIS32

0 else.

 

93

This implies that we only consider the points {(n, m) : In| S 5 and ImI S 32} for the
solution of (K dV)51". Since we are limited in the number of coefiicients available, we

will choose our initial data in each example to be compactly supported near the origin.

Example 3.5 Suppose our initial data is given by the function

0.5x + 0.5 —0.5 S x S 0
110(1) = -0.5x + 0.5 0 g x g 0.5
0 IxI 2 0.5.

Note that an is an even function. It is easy to see that if w(x, t) solves the associated
linear problem (H) (see pg. 4), then so does w(—x, —t). Furthermore, if u(x, t) solves
(KdV)4, then u(—x, —t) does as well. By uniqueness, u(—x, —t) = u(x,t). In fact,
this symmetry prOperty should be true for all the iterates of the operator 5' in (3),
since the right-hand side of (I H ) (see pg. 4), namely g(x, t), will have the property
that —g(x, t) = g(—x, —t). Recall that we choose the zero function as our starting
point. It follows that

5(")(0)(x, t) = SI")(0)(—x, —t) (3.13)

for all x,t E R and n E N.
Let h = 1. To begin, we need to discretize our initial data. We do so by choosing

the values of uo at the integers. Thus, our discrete initial data is

0.5 n = 0
1M") =
0 n 75 0.
In this example, we run four successive iterations. Recall that the first iteration

solves the associated linear equation. To measure the accuracy of the computation,

we compute the the maximum relative error of the first iteration with respect to the

94

linear equation using (3.11) with A = 8 and M = 32. The result is 8.62601 x 10‘“.
All three norms of each iteration and the maximum relative error (M RE) of each
iteration with respect to the (K dV)§I equation are listed in Figure 3.6. For the last
three iterations, we include the distance of the current iteration from the previous
one measured in the Xh-norm under the column with heading “distance”. As in the

continuous setting, we have

(3%)(I‘I(0)(n,m) = (§¢no)I’°I(0)(-n. -m) (3-14)

for all n and m on our grid and k 6 N. One can see from the norms that these
iterations are converging very rapidly. Since the relative error is becoming extremely

small, we can conclude that the iterations are converging to a solution of (K dV)j,‘I.

 

 

 

iteration M RE norm one norm two norm three distance
1 0.00094209 0.516778 0.409208 0.5
2 0.0000615788 0.516887 0.408461 0.5 0.03331
3 3.87118 x 10‘7 0.51688 0.408455 0.5 0.0000105811
4 2.43126 x 10‘9 0.51688 0.408455 0.5 5.08566 x 10‘8

 

 

 

 

Figure 3.6

Example 3.7 Let the initial data no be as in Example 3.5. However, in this example,

we set h = 0.5. If no(n) = uo(nh), then

0.5 n = 0
77001) =
0 n 75 0,

which is the same discrete initial data as in Example 3.5. One might think that

we should get the same iterations as well. However, recall that the operator A), is

 

95

dependent upon h. This influences each iteration.

As in the previous example, we compute the first four iterations (starting from
the zero function) and their relevant data (see Figure 3.8). Once again, these
iterations have the property mentioned in (3.14). By considering the distance
estimates between successive iterations and the size of the maximum relative error,
one can see that the iterations are converging exponentially to the solution of (K dV);d

guaranteed by Theorem 3.2. See Figure 3.8.

 

 

iteration M RE norm one norm two norm three distance
1 0.0188406 0.365417 0.289354 0.353553
2 0.0000305815 0.365431 0.289222 0.353553 0.0231647
3 4.78587 x 10’8 0.365431 0.289222 0.353553 4.6644 x 10‘7
4 6.90206 x 10"10 0.365431 0.289222 0.353553 7.68505 x 10-10

 

 

 

 

Figure 3.8

Example 3.9 Suppose our initial data is given by the function

1 —1SxS0
110(23): —1 0SxS1
0 IxIZl.

Unlike the previous example, this function is discontinuous. However, it does have
finite left and right-hand limits for all x 6 R. Assuming h = 1, let

1

170(7)) = 5 (5133» 110(1). + a) + 5133+ u0(n — 5)) ,

96

i.e.,
0 5 n = —1
0 n = 0
770(71) =
—0.5 n = 1
0 other

for n E Z. Also, note that uo is an odd function. In this case, if w(x, t) solves the
associated linear equation (H), then so does —w(—x, —t) and if u(x, t) solves (K W )4,
then so does —u(—x,—t). Hence, u(—x, —t) = u(x, t). By a similar argument as
before, this is also true of all the iterates of the operator S.

As in the previous example, our initial guess is the zero function. However,
in this example, we compute the first seven iterations. This data illustrates the
odd reflexivity noted above. Notice that for iterations five through seven, the only
column which changes is the “distance” column which becomes extremely small (see
Figure 3.10). This implies that by the fourth iteration, we are very close to the fixed

point of our operator. Apparently, the best we can hope for in this example in terms

of the relative error is 1.04304 x 10-8.

 

 

 

 

 

iteration M RE norm one norm two norm three distance
1 0.0099275 0.587994 0.737931 0.707107
2 0.000044212 0.588019 0.738561 0.707107 0.036962
3 1.35837 x 10‘7 0.588019 0.738563 0.707107 5.61944 x 10‘6
4 1.04304 x 10‘8 0.588019 0.738563 0.707107 2.61774 x 10"8
5 1.04309 x 10‘8 0.588019 0.738563 0.707107 1.37671 x 10"10
6 1.04309 x 10“8 0.588019 0.738563 0.707107 6.09476 x 10’13
7 1.04309 x 10“8 0.588019 0.738563 0.707107 2.2745 x 10‘15

 

Figure 3.10

 

97

Example 3.11 Again consider the initial data from Example 3.9, but with h = 0.1.

Hence,

r
05 n = —10
1 ——9 S n S —1
0 = 0

770(71): I

—1 1S n S 9
—05 n: 10

1 0 other.

 

Again, we compute seven iterations. Unlike the previous examples, the maximum
relative error of the first iteration is quite large, approximately 5.9. This is mainly due
to the larger data which can be seen from the norms in Figure 3.12 and the smaller h.

However, by the seventh iteration the maximum relative error is once again very small.

 

 

 

 

 

 

iteration M RE norm one norm two norm three distance
1 5.90663 0.93733 0.834393 1.33463
2 2.00782 0.932334 0.838427 1.31646 0.186061
3 0.0562563 0.932087 0.838494 1.31646 0.00255853
4 0.00457199 0.932068 0.838484 1.31646 0.0000613827
5 0.000312503 0.932067 0.838483 1.31646 4.0051 x 10’6
6 9.41672 x 10‘6 0.932067 0.838483 1.31646 2.27996 x 10"7
7 2.4938 x 10‘6 0.932067 0.838483 1.31646 1.31978 x 10'8

 

 

Figure 3.12

Example 3.13 In the previous example, the initial data was larger than in previous
examples. This seemed to lead to slower convergence of the iterations. Can the initial

data be too large to yield convergence? With h = 1, we consider the following initial

98

data,

0 InI > 5
770(71) =
n —5 S n S 5.

After two iterations, it is evident that the iterations in this example are diverging in
the space X), (see Figure 3.14). This suggests that the condition in Theorem 1.21

that the initial data be small is necessary.

 

iteration M RE norm one norm two norm three distance

 

1 128.373 8.07398 6.46922 10.4881

 

 

 

2 1.44893 x 1012 850.296 1000.54 1071.73 1070.25

 

Figure 3.14

BIBLIOGRAPHY

BIBLIOGRAPHY

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[3] I. I. Hirschman, Jr., A convexity theorem for certain groups of transformations,
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[4] Y. Katznelson, An Introduction to Harmonic Analysis, Dover Publications Inc.,
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[5] C. Kenig, G. Ponce, and L. Vega, Well-posedness and scattering results for the
generalized Korteweg-de Vries equation via the contraction mapping principle,
Comm. in Pure and Appl. Math. 66, 1993, 527-620.

[6] W. Rudin, Real and Complex Analysis, third ed., McGraw Hill, New York, 1987.

[7] E. Stein, Interpolation of Linear Operators, Trans. Amer. Math. Soc. 83, 1956,
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[8] E. Stein, Singular Integrals and Difierentiability Properties of Functions. Prince-
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[9] E. Stein and C. Weiss, Introduction to Fourier Analysis on Euclidean Spaces.
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99

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