一種擾動區塊循環矩陣種類的特徵曲線問題

國 立 交 通 大 學

應用數學系

碩 士 論 文

一種擾動區塊循環矩陣種類的特徵曲線問題

Eigencurve Problems For A Class Of Perturbed Block

Circulant Matrices

研 究 生：張靖尉

指導老師：莊 重 教授

一種擾動區塊循環矩陣種類的特徵曲線問題

Eigencurve Problems For A Class Of Perturbed Block Circulant Matrices

　 　 研 究 生：張靖尉 Student：Jin-Wei Chang

指導教授：莊 重 Advisor：Dr. Jonq Juang

國 立 交 通 大 學

應用數學 學 系

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics

College of Science

National Chiao Tung University

in partial Fulfillment of the Requirements

for the Degree of

Master

in

Applied Mathematics

June 2006

Hsinchu, Taiwan, Republic of China

一種擾動區塊循環矩陣種類的特徵曲線問題

學生：

張靖尉

指導教授

：莊 重

國立交通大學

應用數學系﹙研究所﹚

摘

要

　 關心的是，對於

β

任意固定，擾動區塊循環矩陣

C

( , )

α β

種類的特徵曲線問題。

這裡

α

>

0

是(微波)純量積因子且

β

∈ \

表示混合邊界常數。

C

( , )

α β

是一個區塊循

環矩陣只有在

β

=

1

。對於每個

α

，

C

( ,1)

α

的特徵值包含它的區塊矩陣作線性組合

後的特徵值這件事是已經被知道的。這樣的結果被稱作對於

C

( ,1)

α

的降低特徵值

問題。在這篇論文裡，我們得到二個主要結果。首先，對於

C

( , 0)

α

的降低特徵值

問題被完全解決了。

C

( , )

α β

的降低特徵值問題也得到一些部分結果。第二，對於

( , 0)

C

α

和

C

( ,1)

α

利用微波方法控制混沌，扮演必要角色的第二大特徵曲線問題將

被討論。

Eigencurve Problems For A Class Of

Perturbed Block Circulant Matrices

Student: Jing-Wei Chang

Advisor: Jonq Juang

Department of Applied Mathematics

National Chiao Tung University

Hsinchu, Taiwan, R.O.C.

ABSTRACT

Of concern is the eigencurve problems for a class of ”perturbed” block circulant matrices C(α, β) with β arbitrary fixed. Here α > 0 is a (wavelet) scalar factor and β ∈ R represents a mixed boundary constant. C(α, β) is a block circulant matrix only if β = 1. It is well-known that for each α the eigenvalues of C(α, 1) consists of eigenvalues of a certain linear combinations of its block matrices. Such results are called the reduced eigenvalue problem for C(α, 1). In this thesis, we obtain two main results. First, the reduced eigenvalue problem for C(α, 0) is completely solved. Some partial results for the reduced eigenvalue problem of C(α, β) are also obtained. Second, the second eigencurve problem, which plays essential role for wavelet method for controlling chaos, for C(α, 0) and C(α, 1) are discussed.

誌

謝

這篇論文首先必須感謝我的指導教授莊重老師，這二年

來細心及耐心的教導與勉勵，在我論文遇到瓶頸時，總是給

予幫助與想法，使我度過難關，更有勇氣地往前走。與老師

研究過程當中，老師的研究熱誠讓我了解做研究應有的態度

及方法，使我獲益良多。

其次，感謝金龍學長、郁泉學姊適時給我意見與指導，

並陪我度過許多充實的時光，讓我更有動力做研究。還有感

謝奐勛的陪伴與幫助。

最後，要謝謝研究所的同學，這二年來，一起玩樂、聊

天，每當我遇到挫折時能給我扶持，讓我重拾信心，這將是

我最美的回憶

當然，家人的支持與鼓勵更是讓我感到溫馨與感動，並

辛苦的工作讓我無憂無慮的完成學業，在這由衷地感謝爸、

媽的栽培和姐姐關懷。

目

錄

1.

Introduction………….……… …1

2.

Reduced Eigenvalue Problems………..4

3.

The Second Eigencurve Of

C

( , 0)

α

And ( ,1)

C

α

………9

1. Introduction

Of concern here is the eigencurve problem for a class of ”perturbed” block circulant matrices.

C(α, β)b = λ(α, β)b. (1.1a)

Here C(α, β) is an n × n block matrix of the following form.

C(α, β) =            C1(α, β) C2(α, 1) 0 · · · 0 C2T(α, β) C₂T(α, 1) C1(α, 1) C2(α, 1) · · · 0 0 .. . . .. . .. . .. ... .. . . .. . .. . .. ... 0 0 · · · CT 2(α, 1) C1(α, 1) C2(α, 1) C2(α, β) 0 · · · 0 C2T(α, 1) ICˆ 1(α, β) ˆI            n×n (1.1b) Here C1(α, β) =           −1 − β 1 0 · · · 0 1 −2 1 0 · · · 0 0 1 −2 1 · · · 0 .. . . .. . .. . .. . .. ... 0 · · · 0 1 −2 1 0 · · · 0 1 −2           2j_×2j −α(1 + β) 22j ee T =: A1(β, 2j) − α(1 + β) 22j ee T_{, (1.1c)}

where e = (1, 1, ..., 1)T_{, j is a positive integer, α > 0 is a (wavelet) scalar factor and}

β ∈ R represents a mixed boundary constant. Moreover,

C2(α, β) =      0 0 · · · 0 .. . ... 0 0 β 0 · · · 0      +αβ 22jee T =: A2(β, 2j) + αβ 22jee T _(1.1d) 1

ˆ I =            0 0 · · · 0 1 0 0 · · · 0 1 0 .. . · · · ... .. . · · · ... 0 1 0 · · · 0 0 1 0 · · · 0 0            . (1.1e)

C(α, β) is a block circulant matrix (see e.g., [1]) only if β = 1. It is well-known, see e.g., Theorem 5.6.4 of [1], that for each α the eigenvalues of C(α, 1) consists of eigenvalues of a certain linear combinations of its block matrices. Such results are called the reduced eigenvalue problem for C(α, 1). In this thesis, we obtain two main results. First, the reduced eigenvalue problem for C(α, 0) is completely solved. Some partial results for the reduced eigenvalue problem of C(α, β) are also obtained. The second problem in question is to describe the second eigencurve λ2(α), which plays essential role for wavelet method for controlling chaos, of (1.1a)

for fixed β. By the second largest eigencurve λ2(α) of C(α, β) for fixed β, we mean

that for given α > 0, λ2(α, β) is the second largest eigenvalue of C(α, β). We

re-mark that 0 is the largest eigenvalue of C(α, β) for any α > 0 and β ∈ R. This is to say for fixed β, 0 is the first eigencurve of C(α, β). A nontrivial upper bound is conjuctured for the second eigencurve λ2(α, 0) (resp., λ2(α, 1)) when α is large is

obtained for any j and n ∈ N. The remainder of this introductory section is devoted to a brief description about how this eigencurve problem arises and its related work.

This problem arises in the wavelet method for chaotic control ([4]). It is found there that the modification of a tiny fraction of wavelet subspaces of a coupling matrix could lead to a dramatic change in chaos synchronizing properties. We begin with describing their work. Let there be N nodes (oscillators). Assume ui

is the m-dimensional vector of dynamical variables of the ith node. Let the iso-lated (uncoupling) dynamics be ˙ui = f (ui) for each node. Used in the coupling,

h : Rm→ Rm_{is an arbitrary function of each node’s variables. Thus, the dynamics}

of the ith node are

˙ ui= f (ui) +  N X j=1 aijh(uj), i = 1, 2, ..., N, (1.2a)

where  is a coupling strength. The sum

N

X

j=1

aij = 0. Let u = (u1, u2, ..., uN)T,

F (u) = (f (u1), f (u2), ..., f (uN))T, H(u) = (h(u1), h(u2), ..., h(uN))T, and A =

(aij). We may write (1.1a) as

˙

u = F (u) + A × H(u). (1.2b)

Here × is the direct product of two matrices B and C defined as follows. Let B = (bij)k1×k2 be a k1× k2 matrix and C = (Cij)k2×k3 be a k2× k3 block matrix,

where each of Cij, 1 ≤ i ≤ k2, 1 ≤ j ≤ k3, is a k4× k5 matrix. Then

B × C = (

k2

X

l=1

bilClj)k1×k3.

Many coupling schemes are covered by Equation(1.2b). For example, if the Lorenz system is used and the coupling is through its three components x, y, and z, then the function h is just the matrix

I3=   1 0 0 0 1 0 0 0 1   . (1.3)

The choice of A will provide the connectivity of nodes. For instance, the nearest neighbor coupling with periodic, Neumann boundary conditions and mixed bound-ary conditions are, respectively, given as A = A1(1, N ) + A2(1, N ) + AT2(1, N ), A =

A1(0, N ) + A2(1, N ) ˆI and A = A1(β, N ) + A2(β, N ) + AT2(β, N ) + (1 − β)A2(1, N ) ˆI,

where those A0s are defined in (1.1c,d).

Mathematical speaking ([2]), the second largest eigenvalue λ2 of A is dominant in

controlling the stability of chaotic synchronization, and the critical strength c for

synchronization can be determined in term of λ2,

c=

Lmax

−λ2

. (1.4)

The eigenvalues of A = A1(1, N ) are given by λi = −4 sin2 π(i−1)_N , i=1,2,...,N. In

general, a larger number of nodes gives a smaller nonzero eigenvalue λ2 in

magni-tude and, hence, a larger c. In controlling a given system, it is desirable to reduce

the critical coupling strength c. The wavelet method in [4] would transform A

into C(α, β). Consequently, it is of great interest to study the second eigencurve of C(α, β) for each β. A numerical simulation of a coupled system of N = 512 Lorenz oscillators in [4] shows that with h = I3 and A = A(1, N ), the critical coupling

strength c decreases linearly with respect to the increase of α up to a critical value

αc. The smallest c is about 6, which is about 103 times smaller than the original

critical coupling strength, indicating the efficiency of the proposed approach. The mathematical verification of such phenomena is first achieved by Shieh, Wei, Wang and Lai [3]. Specifically, they solved the second eigencurve problem for C(α, 1) with n being a multiple of 4 and j being any positive integer. Subse-quently, in [5], the second eigencurve problem for C(α, 0) and C(α, 1) with n being any positive integer and j = 1 is solved.

2. Reduced Eigenvalue problems

Writing the eigenvalue problem C(α, β)b = λb, where b = (b1, b2, ..., bn)T and

bi∈ C2

j

, in block component form, we get

C₂T(α, 1)bi−1+ C1(α, 1)bi+ C2(α, 1)bi+1= λbi, 1 ≤ i ≤ n. (2.1a)

Mixed boundary conditions would yield that

C2T(α, 1)b0+C1(α, 1)b1+C2(α, 1)b2= λb1= C1(α, β)b1+C2(α, 1)b2+C2T(α, β)bn, and C₂T(α, 1)bn−1+C1(α, 1)bn+C2(α, 1)bn+1= λbn= C2(α, β)b1+C2T(α, 1)bn−1+ ˆIC1(α, β) ˆIbn, or, equivalently, C₂T(α, 1)b0= (C1(α, β) − C1(α, 1))b1+ C2T(α, β)bn = [      1 − β 0 · · · 0 0 0 · · · 0 .. . ... . .. ... 0 0 · · · 0      +α(1 − β) 22j ee T_]b 1+ [      0 · · · 0 β 0 · · · 0 0 .. . . .. ... ... 0 · · · 0 0      +αβ 22jee T_]b n = (1 − β)C2T(α, 1) ˆIb1+ βC2T(α, 1)bn, (2.1b) and C2(α, 1)bn+1= ( ˆIC1(α, β) ˆI − C1(α, 1))bn+ C2(α, β)b1 = (1 − β)C2(α, 1) ˆIbn+ βC2(α, 1)b1. (2.1c)

To study the block difference equation (2.1), we set

bj= δjv, (2.2)

where v ∈ C2j and δ ∈ C.

Substituting (2.2) into (2.1a), we have

[C₂T(α, 1) + δ(C1(α, 1) − λI) + δ2C2(α, 1)]v = 0. (2.3)

To have a nontrivial solution v satisfying (2.3), we need to have

det[C₂T(α, 1) + δ(C1(α, 1) − λI) + δ2C2(α, 1)] = 0. (2.4)

Definition 2.1. Equation (2.4) is to be called the characteristic equation of the block difference equation (2.1a). Let δk= δk(λ) 6= 0 and vk= vk(λ) 6= 0 be complex

numbers and vectors, respectively, satisfying (2.3). Here k = 1, 2, ..., m and m ≤ 2j_.

Assume that there exists a λ ∈ C, such that bj = Σmk=1ckδ j

k(λ)vk(λ), j=0,1,...,n+1,

satisfy equation (2.1b,c), where ck ∈ C. If, in addition, bj, j = 1, 2, ..., n, are not

all zero vectors. then such δk(λ) is called the characteristic value of equation (2.1)

or (1.1a) with respect to λ and vk(λ) its corresponding characteristic vector.

Remark 2.1. Clearly, for each α and β, λ in the Definition of 2.1 is an eigenvalue of C(α, β).

Should no ambiguity arises, we will write C2T(α, 1) = C2T, C1(α, 1) = C1 and

C2(α, 1) = C2. Likewise, we will write A2(β, 2j) = A2(β) and A1(β, 2j) = A1(β).

Proposition 2.1. Let ρ(λ) = {δi(λ) : δi(λ) is a root of equation (2.4)}, and let

ρ0(λ) = {_δ 1

i(λ) : δi(λ) is a root of equation (2.4)}. Then ρ(λ) = ρ

0_{(λ). Let δ}

iand δk

be in ρ(λ). We further assume that δi and vi =

   vi1 .. . vi2j    satisfy (2.3). Suppose δi· δk = 1. Then δk and vk =         vi2j vi2j₋₁ .. . vi2 vi1         =: vs

i also satisfy (2.3). Conversely, if

δi· δk 6= 1, then vk6= vsi.

Proof. To proof ρ(λ) = ρ0(λ), we see that det[C₂T + δ(C1− λI) + δ2C2] = δ2det[

1 δ2C T 2 + 1 δ(C1− λI) + C2] = δ2det[ 1 δ2C T 2 + 1 δ(C1− λI) + C2] T = δ2det[C2T + 1 δ(C1− λI) + 1 δ2C2].

Thus, if δ is a root of equation (2.4), then so is 1

δ. To see the last assertion of the 5

proposition, we write equation (2.3) with δ = δiand v = vi in component form. 2j

X

m=1

[(C₂T)lmvim+ δi( ¯C1)lmvim+ δ2i(C2)lmvim] = 0, l = 1, 2, ..., 2j. (2.5)

Here ¯C1= C1− λI. Now the right hand side of (2.5) becomes

(1 δk )2{ 2j X m=1 [(C2)l(2j+1−m)vi(2j+1−m)+ δk( ¯C1)l(2j+1−m)vi(2j+1−m) +δ2_k(C₂T)l(2j_+1−m)v_i(2j_+1−m)]} = (1 δk )2{ 2j X m=1 [(C2T)(2j_+1−l)mv_i(2j_+1−m)+ δk( ¯C1)(2j_+1−l)mv_i(2j_+1−m) +δ2_k(C2)(2j_+1−l)mv_i(2j_+1−m)]}, l = 1, 2, ..., 2j. (2.6)

We have used the fact that

(A)(2j_+1−l)m= (AT)_l(2j_+1−m), (2.7)

where A = CT

2 or ¯C1 or C2 to justify the equality in (2.6). However, (2.7) follows

from (1.1c) and (1.1d). Letting vi(2j_+1−m) = vkm, we have that the pair (δk, vk)

satisfies (2.3). Suppose vk = vsi, we see, similarly, that the pair (δ1i, vk) also satisfy

(2.3). Thus _δ1

i = δk.

 Definition 2.2. We shall call vsand −vs, the symmetric vector and antisymmetric vector of v, respectively. A vector v is symmetric (resp., antisymmetric) if v = vs

(resp., v = −vs_). Theorem 2.1. Let δk = e πk ni, k is an integer and i = √ −1, then δ2k, k=0,1,...,n-1,

are characteristic values of equation (2.1) with β = 1. For each α, if λ ∈ C satisfies

det[C2T+ δ2k(C1− λI) + δ2k2 C2] = 0,

for some k ∈ Z, 0 ≤ k ≤ n − 1, then λ is an eigenvalue of C(α, 1). Proof. Let λ be as assumed. Then there exists a v ∈ C2j_{, v 6= 0 such that}

[C₂T + δ2k(C1− λI) + δ2k2 C2]v = 0. 6

Let bj = δ j

2kv, 0 ≤ j ≤ n + 1. Then such b 0

js satisfy (2.1a), (2.1b), and (2.1c). We

just proved the assertion of the theorem. _ Corollary 2.1. Set

Γk = C1+ δ2n−kC2T+ δkC2. (2.8)

Then the eigenvalues of C(α, 1), for each α, consists of eigenvalues of Γk, k =

0, 2, 4, ..., 2(n − 1). That is ρ(C(α, 1)) =

n−1

[

k=0

ρ(Γ2k). Here ρ(A) = the spectrum of

the matrix A.

Remark 2.2. C(α, 1) is a block circulant matrix. The assertion of Corollary 2.1 is not new (see e.g., Theorem 5.6.4 of [3]). Here we mere gave a different proof. To study the eigenvalue of C(α, 0) for each α, we begin with considering the eigen-values and eigenvectors of CT

2 + C1+ C2 and C2T − C1+ C2.

Proposition 2.2. Let T1(C) (resp., T2(C)) be the set of linearly independent

eigenvectors of the matrix C that are symmetric (resp., antisymmetric). Then |T1(C2T+C1+C2)| = |T2(C2T+C1+C2)| = |T1(C2T−C1+C2)| = |T2(C2T−C1+C2)| =

2j−1_{. Here |A| denote the cardinality of the set A.}

Proof. We will only illustrate the case for CT

2 − C1+ C2 =: C. We first

ob-serve that |T1(C)| is less than or equal to 2j−1. So is |T2(C)|. We also remark

the cardinality of the set of all linearly independent eigenvectors of C is 2j_{. If}

0 < |T1(C)| < 2j−1, there must exist an eigenvector v for which v 6= vs, v 6= −vs

and v /∈ span{T1(C), T2(C)}, the span of the vectors in T1(C) and T2(C). It then

follows from Proposition 2.1 that v + vs, a symmetric vector, is in the span{T1(C)}.

Moreover, v − vs _{is in span{T}

2(C)}. Hence v ∈ span{T1(C), T2(C)}, a

contradic-tion. Hence, |T1(C)| = 2j−1. Similarly, we conclude that |T2(C)| = 2j−1. 

Theorem 2.2. Let δk = e πk ni, k is an integer, i = √ −1. For each α, if λ ∈ C satisfies det[C₂T+ δk(C1− λI) + δk2C2] = 0,

for some k ∈ Z, 1 ≤ k ≤ n − 1, then λ is an eigenvalue of C(α, 0). Let λ be the eigenvalue of CT

2 + C1+ C2 (resp., −C2T + C1− C2) for which its associated

eigenvector v satisfies ˆIv = v (resp., ˆIv = −v), then λ is also an eigenvalue of C(α, 0).

Proof. For any 1 ≤ k ≤ n − 1, let δk be as assumed. Let λk and vk be a number

and a nonzero vector, respectively, satisfying

[C₂T+ δk(C1− λkI) + δk2C2]vk = 0. (2.9) 7

Using Proposition 2.1, we see that λk satisfies

det[C₂T + δ2n−k(C1− λkI) + δ22n−kC2] = 0. (2.10)

Let v2n−kbe a nonzero vector satisfying [C2T+δ2n−k(C1−λkI)+δ22n−kC2]v2n−k= 0.

Letting

bi= δikvk+ δkδ2n−ki v2n−k, i = 0, 1, ..., n + 1,

we conclude, via (2.9) and (2.10), that bi satisfy (2.1a) with λ = λk. Moreover,

ˆ

Ib1= δkIvˆ k+ ˆIv2n−k= δkv2n−k+ vk= b0.

We have used Proposition 2.1 to justify the second equality above. Similarly, bn+1= ˆIbn. To see λ = λk, 1 ≤ k ≤ n − 1, is indeed an eigenvalue of C(α, 0) for

each α, it remains to show that bi6= 0 for some i. Using Proposition 2.1, we have

that there exists an m, 1 ≤ m ≤ 2j _{such that v}

km = v(2n−k)(2j_−m+1)6= 0. We first

show that b06= 0. Let m be the index for which vkm 6= 0. Suppose b0= 0. Then

vkm+ δkv(2n−k)m= 0

and

vk(2j_−m+1)+ δkv(2n−k)(2j_−m+1)= v(2n−k)m+ δkvkm= 0.

And so, vkm = δk2vkm, a contradiction. Let λ and v be as assumed in the last

assertion of theorem. Letting bi= v (resp., bi= (−1)iv), we conclude that λ is an

eigenvalue of C(α, 0) with corresponding eigenvector      b1 b2 .. . bn      .

Thus, λk is an eigenvalue of C(α, 0) for each α. 

Corollary 2.2. Let δk = e

πk

ni, k is an integer, i =

√

−1. Then, for each α, ρ(C(α, 0)) = n−1 [ k=1 ρ(Γk) [ ρS(Γ0) [

ρAS(Γn), where ρS(A) (resp., ρAS(A)) the set

of eigenvalues of A for which their corresponding eigenvectors are symmetric (resp., antisymmetric).

We next consider the eigenvalues of C(α, β).

Theorem 2.3. Let δk= e

πk

ni, k is an integer, i =

√

−1. Then, for each α,

ρ(C(α, β)) ⊃              [n 2] [ k=1 ρ(Γ2k) [ ρS(Γ0), n is odd, n 2−1 [ k=1 ρ(Γ2k) [ ρS(Γ0) [ ρAS(Γn), n is even.

Here [n₂] is the greatest integer that is less than or equal to n₂.

Proof. We illustrate only the case that n is even. Let k be such that 1 ≤ k ≤ n₂− 1. Let bi = δ2ki v2k+ δ2kδi2n−2kv2n−2k, we see clearly that such bi, i = 0, 1, n, n + 1,

satisfy both Neumann and periodic boundary conditions, respectively. And so

b0= (1 − β)b0+ βb0= (1 − β) ˆIb1+ βbn,

and

bn+1= (1 − β)bn+1+ βbn+1= (1 − β) ˆIbn+ βb1.

Here, δ2k, 1 ≤ k ≤ n₂ − 1, are characteristic values of equation of (2.1). Thus, if

λ ∈ ρ(Γ2k), then λ is an eigenvalue of C(α, β). The assertions for Γ0 and Γn can

be done similarly. _

Remark 2.3. If n is an even number, for each α and β, half of the eigenvalues of C(α, β) are independent of the choice of β. The other characteristic values of (1) seem to depend on β. It is of interest to find them.

3. The Second Eigencurve of C(α, 0) and C(α, 1) We begin with considering the eigencurves of Γk, as given in (2.8). Clearly,

Γk=           −2 1 0 · · · δ2n−k 1 −2 1 0 · · · 0 0 1 −2 1 · · · 0 .. . . .. . .. . .. . .. ... 0 · · · 0 1 −2 1 δk · · · 0 1 −2           m×m −α(2 − 2 cos πk n ) m ee T =: D1(k) − α(k)eeT, (3.1)

where m = 2j_{. We next find a unitary matrix to diagonalize D} 1(k).

Remark 3.1. Let (λ(k), v(k)) be the eigenpair of D1(k). If eTv(k) = 0, then λ(k)

is also an eigenvalue of Γk.

Proposition 3.1. Let θl,k= 2lπ m + kπ nm, l = 0, 1, ..., m − 1, (3.2a) p_l(k) =      eiθl,k ei2θl,k .. . eimθl,k      (3.2b) and P (k) = p_√0(k) m · · · pm−1_√ (k) m  . (3.2c)

(i) Then P (k) is a unitary matrix and PH(k)D1(k)P (k) = Diag(λ0,k· · · λm−1,k),

where PH _{is the conjugate transpose of P , and}

λl,k= 2 cos θl,k− 2, l = 0, 1, .., m − 1. (3.2d)

(ii) Moreover, for 0 ≤ k ≤ 2n, the eigenvalues of D1(k) are distinct if and only if

k 6= 0, n or 2n.

Proof. Let b = (b1, ..., bm)T. Writing the eigenvalue problem D1(k)b = λb in

com-ponent form, we get

bj−1− (2 + λ)bj+ bj+1= 0, j = 2, 3, ..., m − 1, (3.3a)

−(2 + λ)b1+ b2+ δ2n−kbm= 0, (3.3b)

δkb1+ bm−1− (2 + λ)bm= 0. (3.3c)

Set bj= δj, where δ satisfies the characteristic equation 1 − (2 + λ)δ + δ2= 0 of the

system D1(k)b = λb. Then the boundary conditions (3.3b) and (3.3c) are reduced

to

δm= δk. (3.4)

Thus, the solutions eiθl,k_{, l = 0, 1, ..., m − 1, of (3.4) are the candidates for the}

characteristic values of (3.3). Substituting eiθl,k _{into (3.3a) and solving for λ,}

we see that λ = λl,k are the candidates for the eigenvalues of D1(k). Clearly,

(λ, b) = (λl,k, pl(k)) satisfies D1(k)b = λb and b = pl(k) 6= 0. To complete the 10

proof of the proposition, it suffices to show that P (k) is unitary. To this end, we have that pHl (k) · pl0(k) =  m l = l0 0 l 6= l0 _, Clearly, pH

l (k) · pl(k) = m. Now, let l 6= l0, we have that

pH_l (k) · p_l0(k) = m X j=1 eij(θl,k−θl0 ,k)₌ m X j=1 eij(2(l−l0 )m π)₌r(1 − r m₎ 1 − r = 0,

where r = ei(2(l−l0 )_m π)_{. The last assertion of the proposition is obvious.}

 Remark 3.2. The eigenvalues and eigenvectors of D1(k) were first given in the

Proposition 2.3 of [3]. For completeness, we give an easier proof above.

To prove the main results in this section, we also need the following proposition. Proposition 3.2. Suppose D = diag(d1, ..., dm) ∈ Rm×m and that the diagonal

entries satisfy d1 > · · · > dm. Let γ 6= 0 and z = (z1, ..., zm)T ∈ Rn. Assume that

(λi(γ), vi(γ)) are the eigenpairs of D + γzzT with λ1(γ) ≥ λ(γ) ≥ ... ≥ λm(γ). (i)

Let A = {k : 1 ≤ k ≤ m, zk = 0}, Ac = {1, ..., m} − A. If k ∈ A, then dk= λk. (ii)

Assume γ > 0. Then the following interlacing relations hold λ1(γ) ≥ d1≥ λ2(γ) ≥

d2 ≥ ... ≥ λm(γ) ≥ dm. Moreover, the strict inequality holds for these indexes

i ∈ Ac_{. (iii) Let i ∈ A}c_{, λ}

i(γ) are strictly increasing in γ and lim

γ→∞λi(γ) = ¯λi for

all i, where ¯λi are the roots of g(λ) =

X

k∈Ac

z_i2 dk− λ

with ¯λi∈ (di, di−1). In case that

1 ∈ Ac_{, d} 0= ∞.

Proof. The proof of interlacing relations in (ii) and the assertion in (i) can be found in Theorem 8.6.2 of [4]. We only prove the remaining assertions of the propo-sition. Rearranging z so that zT _{= (0, 0, ..., 0, z}

i1, ..., zik) =: (0, ..., 0, ¯z

T_{), where}

i1 < i2 < ... < ik and ij ∈ Ac, j = 1, ..., k. The diagonal matrix D is rearranged

accordingly. Let D = diag(D1, D2), where D2 = diag(di1, ..., dik). Following

The-orem 8.6.2 of [4], we see that λij(γ) are the roots of the scalar equation fγ(λ), where

fγ(λij(γ)) = 1 + γ k X j=1 z2 ij dij− λij(γ) = 0. (3.5)

Differentiate the equation above with respect to γ, we get

k X j=1 z2 ij dij− λij(γ) + (γ k X j=1 z2 ij (dij− λik(γ)) 2) dλij(γ) dγ = 0. Thus, 11

dλij(γ) dγ = 1 γ2 k X j=1 z2 ij (dij − λij(γ)) 2 > 0.

Clearly, the limit of λij(γ) as γ → ∞ exists, say ¯λij. Since, for dij < λij < dij−1,

k X j=1 z2 ij dij − λij(γ) = −1 γ.

Taking the limit as γ → ∞ on both side of the equation above, we get

k X j=1 z2_ij dij − ¯λij = 0 (3.6) as desired. _

We are now in the position to state the following theorems.

Theorem 3.1. For each k and α, denote by λl,k(α) ,l = 0, 1, ..., 2j−1 =: m−1, the

eigenvalues of Γk. For k = 1, 2, ..., n − 1, let (λl,k, ul,k) be the eigenpairs of D1(k),

as defined in (3.1), then there exist λ∗_l,k such that lim

α→∞λl.k(α) = λ ∗ l,k. Moreover, gk(λ∗l,k) = 0, where gk(λ) = m X l=1 1 (λl−1,k)(λl−1,k+ λ) . (3.7)

Proof. Let k be as assumed. Set, for l = 0, 1, ..., m − 1,

zl+1= pHl (k)e = m X j=1 eijθl,k₌ e −iθl,k_{(1 − e}−imθl,k₎ 1 − e−iθl,k = e−iθl,k_{(1 − e}−ikπn) 1 − e−iθl,k . Then ¯ zl+1zl+1= 2 − 2 cos mθl,k 2 − 2 cos θl,k = 2 cos kπ n − 2 λl,k 6= 0. (3.8)

Let P (k) be as given in (3.2c). Then

−PH(k) · Γk· P (k) = Diag(−λ0,k, ..., −λm−1,k) + α(k)PlH(k)e(P H l (k)e)

H_.

Note that if k is as assumed, it follows from Proposition 3.1-(ii) that λl,k, l =

0, ..., m − 1, are distinct. Thus, we are in the position to apply Proposition 3.2.

Specifically, by noting Ac = φ, we see that λ∗_l,ksatisfies gk(λ) = 0, where gk(λ) = m X l=1 1 (λl−1,k)(λl−1,k+ λ) .  Conjecture 3.1. We conjecture that

λ∗_0,k≤ λ0,n, k = 1, 2, ..., n − 1. (3.9)

The calculation via computer seems to confirm that (3.9) holds. Note that if gk(−λ0,n) < 0, then (3.9) holds true.

Remark 3.3. For m = 2, we have that

λ∗_0,k= λ 2 0,k+ λ 2 1,k λ0,k+ λ1,k . (3.10)

Treating k as a real parameter, and differentiating (3.9) with respect to k, we get

(λ0,k+ λ1,k)2 dλ∗_0,k dk = 2(λ0,k λ0,k dk + λ1,k λ1,k dk )(λ0,k+ λ1,k) − (λ 2 0,k+ λ 2 1,k)( λ0,k dk + λ1,k dk ) = λ2_0,kλ0,k dk + λ 2 1,k λ1,k dk + (λ0,kλ1,k)(λ0,k λ0,k dk + λ1,k λ1,k dk ) < 0. Consequently λ∗_0,k≤ λ∗

0,1. A direct computation would yield that λ∗0,1 ≤ λ0,n.

References

[1] P.J. Davis, Circulant Matrices, John Wiley, New York, 1979.

[2] L.M. Pecora and T.L. Carroll, Master stability Functions for Synchronized Coupled Systems, Physical Review Letters 80(1998), 2109-2112.

[3] S.F. Shieh, G.W. Wei, Y.Q. Wang, and C.-H. Lai, Mathematical proof for wavelet method of chaos control, Journal of Mathematical Physics, J. Math. Phy., to appear.

[4] G.W. Wei, M. Zhan and C.-H. Lai, Tailoring wavelets for chaos control, Physical Review Letters 89(2002), 284103.

[5] Jong Juang and Chin-Lung Li, Eigenvalue problem aand their application to the wavelet method of chaotic control, J. Math. Phy., to appear.