chaotic control
Jonq Juang, Chin-Lung Li, and Jing-Wei Chang
Citation: Journal of Mathematical Physics 47, 122702 (2006); doi: 10.1063/1.2400828 View online: http://dx.doi.org/10.1063/1.2400828
View Table of Contents: http://scitation.aip.org/content/aip/journal/jmp/47/12?ver=pdfcov Published by the AIP Publishing
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Perturbed block circulant matrices and their application
to the wavelet method of chaotic control
Jonq Juang,a兲 Chin-Lung Li,b兲and Jing-Wei Changc兲
Department of Applied Mathematics, National Chiao Tung University, Hsinchu, Taiwan 300, Republic of China
共Received 1 August 2006; accepted 23 October 2006; published online 26 December 2006兲 Controlling chaos via wavelet transform was proposed by Wei et al.关Phys. Rev. Lett. 89, 284103.1–284103.4 共2002兲兴. It was reported there that by modifying a tiny fraction of the wavelet subspace of a coupling matrix, the transverse stability of the synchronous manifold of a coupled chaotic system could be dramatically enhanced. The stability of chaotic synchronization is actually controlled by the second largest eigenvalue 2共␣,兲 of the 共wavelet兲 transformed coupling matrix C共␣,兲 for each␣ and. Hereis a mixed boundary constant and ␣is a scalar factor. In particular, = 1 共0兲 gives the nearest neighbor coupling with periodic 共Neumann兲 boundary conditions. In this paper, we obtain two main results. First, the reduced eigenvalue problem for C共␣, 0兲 is completely solved. Some partial results for the reduced eigenvalue problem of C共␣,兲 are also obtained. Second, we are then able to understand behavior of2共␣, 0兲 and 2共␣, 1兲 for any wavelet dimension j苸N and block dimension n苸N. Our results complete and strengthen the work of Shieh et al.关J. Math. Phys. 47, 082701.1–082701.10 共2006兲兴 and Juang and Li关J. Math. Phys. 47, 072704.1–072704.16 共2006兲兴. © 2006 American Insti-tute of Physics.关DOI:10.1063/1.2400828兴
I. INTRODUCTION
Of concern here is the eigencurve problem for a class of “perturbed” block circulant matrices.
C共␣,兲b = 共␣,兲b. 共1.1a兲
Here C共␣,兲 is an n⫻n block matrix of the following form:
C共␣,兲 =
冢
C1共␣,兲 C2共␣,1兲 0 ¯ 0 C2 T共␣ ,兲 C2T共␣,1兲 C1共␣,1兲 C2共␣,1兲 ¯ 0 0 ⯗ ⯗ ⯗ ⯗ 0 0 ¯ C2T共␣,1兲 C1共␣,1兲 C2共␣,1兲 C2共␣,兲 0 ¯ 0 C2 T共 ␣,1兲 IˆC1共␣,兲Iˆ冣
n⫻n . 共1.1b兲 Herea兲Electronic mail: jjuang@math.nctu.edu.tw b兲Electronic mail: presidentf.am92g@nctu.edu.tw c兲Electronic mail: jof12789@yahoo.com.tw
47, 122702-1
C1共␣,兲 =
冢
− 1 − 1 0 ¯ ¯ 0 1 − 2 1 0 ¯ 0 0 1 − 2 1 ¯ 0 ⯗ ⯗ 0 ¯ 0 1 − 2 1 0 ¯ ¯ 0 1 − 2冣
2j⫻2j −␣共1 +兲 22j ee T ¬ A1共,2j兲 − ␣共1 +兲 22j ee T , 共1.1c兲 where e =共1,1, ... ,1兲T, j is a positive integer,␣⬎0 is a 共wavelet兲 scalar factor, and苸R repre-sents a mixed boundary constant. Moreover,C2共␣,兲 =
冢
0 0 ¯ 0 ⯗ ⯗ 0 0  0 ¯ 0冣
+␣ 22jee T ¬ A2共,2j兲 + ␣ 22jee T , 共1.1d兲 Iˆ =冢
0 0 ¯ ¯ 0 1 0 0 ¯ 0 1 0 ⯗ · · · ⯗ ⯗ · · · ⯗ 0 1 0 ¯ 0 0 1 0 ¯ ¯ 0 0冣
. 共1.1e兲The dimension of C共␣,兲 is n2j⫻n2j. From here on, we shall call n and j the block and the wavelet dimensions of C共␣,兲, respectively. C共␣,兲 is a block circulant matrix 共see, e.g., Ref.1兲
only if= 1. It is well known, see, e.g., Theorem 5.6.4 of Ref.1, that for each␣the eigenvalues of C共␣, 1兲 consist of eigenvalues of a certain linear combinations of its block matrices. Such results are called the reduced eigenvalue problem for C共␣, 1兲.
This problem arises in the wavelet method for a chaotic control.7 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共oscillators兲. Assume uiis the m-dimensional vector of dynamical variables of the ith node. Let the isolated 共uncoupling兲 dynamics be u˙i= f共ui兲 for each node. Used in the coupling, h:Rm →Rmis an arbitrary function of each node’s variables. Thus, the dynamics of the ith node is
u˙i= f共ui兲 +⑀
兺
j=1 Naijh共uj兲, i = 1,2, ... ,N, 共1.2a兲
where ⑀ is a coupling strength. The sum 兺j=1N aij= 0. Let u =共u1, u2, . . . , uN兲T, F共u兲 =共f共u1兲, f共u2兲, ... , f共uN兲兲T, H共u兲=共h共u1兲,h共u2兲, ... ,h共uN兲兲T, and A =共aij兲. We may write Eq.
共1.1a兲as
u˙ = F共u兲 +⑀A⫻ H共u兲. 共1.2b兲
Here⫻ is the direct product of two matrices B and C defined as follows. Let B=共bij兲k
1⫻k2 be a k1⫻k2matrix and C =共Cij兲k
B⫻ C =
冉
兺
l=1 k2 bilClj冊
k1⫻k3 .Many coupling schemes are covered by Eq.共1.2b兲. 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=
冢
1 0 0 0 1 0 0 0 1冣
. 共1.3兲
The choice of A will provide the connectivity of nodes. For instance, the nearest neighbor coupling with periodic, Neumann boundary conditions and mixed boundary conditions are, respectively, given as A = A1共1,N兲+A2共1,N兲+A2
T共1,N兲¬AP, A = A
1共0,N兲+A2共1,N兲Iˆ¬AN and A = A1共, N兲 + A2共, N兲+A2
T共, N兲+共1−兲A
2共1,N兲Iˆ¬AM, where those Ai’s, i = 1 , 2, are defined in Eqs.共1.1c兲 and共1.1d兲.
Mathematically speaking,5the second largest eigenvalue2of A is dominant in controlling the stability of chaotic synchronization, and the critical strength⑀cfor synchronization can be deter-mined in terms of2,
⑀c=
Lmax −2
. 共1.4兲
The eigenvalues of A = APare given byi= −4 sin2关共i−1兲/N兴, i=1,2, ... ,N. In general, a larger number of nodes give a smaller nonzero eigenvalue2 in magnitude and, hence, a larger⑀c. In controlling a given system, it is desirable to reduce the critical coupling strength⑀c. The wavelet method in Ref.7will, in essence, transform A into C共␣,兲. Consequently, it is of great interest to study the second eigencurve of C共␣,兲 for each. By the second largest eigencurve2共␣,兲 of C共␣,兲 for fixed, we mean that for given␣⬎0, 2共␣,兲 is the second largest eigenvalue of C共␣,兲. We remark that 0 is the largest eigenvalue of C共␣,兲 for any␣⬎0 and苸R. This is to say that for fixed,2共␣,兲=0 is the first eigencurve of C共␣,兲. A numerical simulation7of a coupled system of N = 512 Lorenz oscillators shows that with h = I3 and A = AP, the critical cou-pling strength⑀cdecreases linearly with respect to the increase of␣up to a critical value␣c. The smallest⑀cis about 6, which is about 103times 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 et al.6 Specifi-cally, they solved the second eigencurve problem of C共␣, 1兲 with n being a multiple of 4 and j being any positive integer. Subsequently, in Ref.4the second eigencurve problem for C共␣, 0兲 and
C共␣, 1兲 with n being any positive integer and j=1 are solved without touching on the reduced eigenvalue problem. In this paper, we obtain two main results. First, the reduced eigenvalue problem for C共␣, 0兲 is completely solved. Some partial results for the reduced eigenvalue problem of C共␣,兲 are also obtained. Second, we are then able to understand the behavior of 2共␣, 0兲 and 2共␣, 1兲 for any j and n苸N.
II. REDUCED EIGENVALUE PROBLEMS
Writing the eigenvalue problem C共␣,兲b=b, where b=共b1, b2, . . . , bn兲T and bi苸C2
j
, in block component form, we get
C2T共␣,1兲bi−1+ C1共␣,1兲bi+ C2共␣,1兲bi+1=bi, 1艋 i 艋 n. 共2.1a兲 Mixed boundary conditions would yield that
C2T共␣,1兲b0+ C1共␣,1兲b1+ C2共␣,1兲b2=b1= C1共␣,兲b1+ C2共␣,1兲b2+ C2 T共␣
,兲bn and
C2T共␣,1兲bn−1+ C1共␣,1兲bn+ C2共␣,1兲bn+1=bn= C2共␣,兲b1+ C2 T共␣,1兲bn−1 + IˆC1共␣,兲Iˆbn, or, equivalently, C2T共␣,1兲b0=共C1共␣,兲 − C1共␣,1兲兲b1+ C2T共␣,兲bn =
冤冢
1 − 0 ¯ 0 0 0 ¯ 0 ⯗ ⯗ ⯗ 0 0 ¯ 0冣
+␣共1 −兲 22j ee T冥
b 1+冤冢
0 ¯ 0  0 ¯ 0 0 ⯗ ⯗ ⯗ 0 ¯ 0 0冣
+␣ 22jee T冥
b n =共1 −兲C2T共␣,1兲Iˆb1+C2共␣,1兲bn 共2.1b兲 and C2共␣,1兲bn+1=„IˆC1共␣,兲Iˆ − C1共␣,1兲…bn+ C2共␣,兲b1=共1 −兲C2 T共␣,1兲Iˆbn +C2共␣,1兲b1. 共2.1c兲 To study the block difference equation关Eq. 共2.1兲兴, we setbj=␦jv, 共2.2兲
wherev苸C2j and␦苸C.
Substituting Eq.共2.2兲into Eq.共2.1a兲, we have 关C2
T共␣
,1兲 +␦„C1共␣,1兲 − I… +␦2C2共␣,1兲兴v = 0. 共2.3兲 To have a nontrivial solutionv satisfying Eq.共2.3兲, we need to have
det关C2
T共␣,1兲 +␦共C
1共␣,1兲 − I兲 +␦2C2共␣,1兲兴 = 0. 共2.4兲 Definition 2.1: Equation共2.4兲is to be called the characteristic equation of the block differ-ence equation关Eq. 共2.1a兲兴. Let␦k=␦k共兲⫽0 and vk=vk共兲⫽0 be complex numbers and vectors,
respectively, satisfying Eq. 共2.3兲. Here k = 1 , 2 , . . . , m and m艋2j. Assume that there exists a 苸C, such that bj=⌺k=1m ck␦k
j共兲vk共兲, j=0,1, ... ,n+1, satisfy Eqs. 共2.1b兲
and 共2.1c兲, where ck 苸C. If, in addition, bj, j = 1 , 2 , . . . , n, are not all zero vectors, then such␦k共兲 is called a charac-teristic value of Eq.共2.1a兲,共2.1b兲, and共2.1c兲or共1.1a兲with respect to and vk共兲 its correspond-ing characteristic vector.
Remark 2.1: Clearly, for each␣and, in Definition 2.1 is an eigenvalue of C共␣,兲. Should no ambiguity arises, we will write C2T共␣, 1兲=C2T, C1共␣, 1兲=C1, and C2共␣, 1兲=C2. Likewise, we will write A2共, 2j兲=A2共兲 and A1共, 2j兲=A1共兲.
Proposition 2.1: Let共兲= 兵␦i共兲:␦i共兲 is a root of Eq.(2.4)其, and let¯共兲= 兵1/␦i共兲:␦i共兲 is a root of Eq. (2.4)其. Then 共兲=¯共兲. Let ␦i and ␦kbe in 共兲. We further assume that␦i and
vi=共vi1, . . . ,vi2j兲T satisfy Eq. (2.3). Suppose ␦i·␦k= 1. Then ␦k and vk
=共vi2j,vi2j−1, . . . ,vi2,vi1兲T¬vis also satisfy Eq.(2.3). Conversely, if␦i·␦k⫽1, then vk⫽vis.
Proof: To prove共兲=¯共兲, we see that det关C2 T +␦共C1−I兲 +␦2C2兴 =␦2det
冋
1 ␦2C2 T +1 ␦共C1−I兲 + C2册
=␦2det冋
1 ␦2C2 T +1 ␦共C1−I兲 + C2册
T =␦2det冋
C2 T +1 ␦共C1−I兲 + 1 ␦2C2册
.Thus, if␦is a root of Eq.共2.4兲, then so is 1 /␦. To see the last assertion of the proposition, we write Eq.共2.3兲with␦=␦i andv = viin component form.
兺
m=1 2j 关共C2 T兲lm vim+␦i共C¯1兲lmvim+␦i2共C2兲lmvim兴 = 0, l = 1,2, ... ,2 j . 共2.5兲Here C¯1= C1−I. Now the right hand side of Eq.共2.5兲becomes
冉
1 ␦k冊
2再
兺
m=1 2j 关共C2兲l共2j+1−m兲vi共2j+1−m兲+␦k共C¯1兲l共2j+1−m兲vi共2j+1−m兲+␦k2共C2T兲l共2j+1−m兲vi共2j+1−m兲兴冎
=冉
1 ␦k冊
2再
兺
m=1 2j 关共C2 T兲 共2j+1−l兲mvi共2j+1−m兲+␦k共C¯1兲共2j+1−l兲mvi共2j+1−m兲+␦k2共C2兲共2j+1−l兲m兲vi共2j+1−m兲兴冎
, l = 1,2, . . . ,2j. 共2.6兲We have used the fact that
共A兲共2j+1−l兲m=共AT兲l共2j+1−m兲, 共2.7兲
where A = C2Tor C¯1or C2to justify the equality in Eq.共2.6兲. However, Eq.共2.7兲follows from Eqs.
共1.1c兲and共1.1d兲. Lettingvi共2j+1−m兲=vkm, we have that the pair共␦k,vk兲 satisfies Eq.共2.3兲. Suppose
vk=vi s
, we see, similarly, that the pair共1/␦i,vk兲 also satisfies Eq. 共2.3兲. Thus 1 /␦i=␦k. 䊐 Remark 2.2: Equation共2.4兲is a palindromic equation. That is, for each,␦and␦−1are both the roots of Eq. 共2.4兲. However, the eigenvalue problem discussed here is not a palindromic eigenvalue problem.3
Definition 2.2: We shall callvsand −vs, the symmetric vector and antisymmetric vector ofv, respectively. A vectorv is symmetric共antisymmetric兲 if v=vs共v=−vs兲.
Theorem 2.1: Let ␦k= e共k/n兲i, k is an integer and i =
冑
−1, then ␦2k, k = 0 , 1 , . . . , n − 1, are characteristic values of Eq.共2.1a兲,共2.1b兲, and 共2.1c兲with= 1. For each␣, if苸C satisfiesdet关C2T+␦2k共C1−I兲 +␦2k2C2兴 = 0, for some k苸Z, 0艋k艋n−1, then is an eigenvalue of C共␣, 1兲.
Proof: Let be as assumed. Then there exists a v苸C2j,v⫽0 such that 关C2
T
+␦2k共C1−I兲 +␦2k2C2兴v = 0. Let bj=␦2k
j
v, 0艋 j艋n+1. Then such bj’s satisfy Eqs.共2.1a兲,共2.1b兲, and共2.1c兲. We just proved the
assertion of the theorem. 䊐
Corollary 2.1: Set
⌫k= C1+␦2n−kC2 T
+␦kC2. 共2.8兲
Then the eigenvalues of C共␣, 1兲, for each␣, consist of eigenvalues of⌫k, k = 0 , 2 , 4 , . . . , 2共n−1兲. That is,共C共␣, 1兲兲=艛k=0n−1共⌫2k兲. Here共A兲= the spectrum of the matrix A.
Remark 2.3: C共␣, 1兲 is a block circulant matrix. The assertion of Corollary 2.1 is not new 共see, e.g., Theorem 5.6.4 of Ref.1兲. Here we merely gave a different proof.
To study the eigenvalue of C共␣, 0兲 for each␣, we begin with considering the eigenvalues and eigenvectors of C2T+ C1+ C2 and C2
T
− C1+ C2.
Proposition 2.2: Let T1共C兲 共T2共C兲兲 be the set of linearly independent eigenvectors of the matrix C that are symmetric (antisymmetric). Then兩T1共C2
T
+ C1+ C2兲兩 =兩T2共C2 T
+ C1+ C2兲兩 =兩T1共C2 T
− C1+ C2兲兩 =兩T2共CT2− C1+ C2兲兩 =2j−1. Here兩A兩 denote the cardinality of the set A.
Proof: We will only illustrate the case for C2T− C1+ C2= : C. We first observe that兩T1共C兲兩 is less than or equal to 2j−1. So is兩T
independent eigenvectors of C is 2j. If 0⬍兩T
1共C兲兩 ⬍2j−1, there must exist an eigenvectorv for whichv⫽vs,v⫽−vs, andv苸span兵T
1共C兲,T2共C兲其, the span of the vectors in T1共C兲 and T2共C兲. It then follows from Proposition 2.1 thatv + vs, a symmetric vector, is in the span兵T
1共C兲其. Moreover,
v − vs is in span兵T
2共C兲其. Hence v苸span兵T1共C兲,T2共C兲其, a contradiction. Hence, 兩T1共C兲兩 =2j−1.
Similarly, we conclude that兩T2共C兲兩 =2j−1. 䊐
Theorem 2.2: Let␦k= e共k/n兲i, where k is an integer and i =
冑
−1. For each␣, if苸C satisfies det关C2T
+␦k共C1−I兲 +␦k 2
C2兴 = 0,
for some k苸Z, 1艋k艋n−1, then is an eigenvalue of C共␣, 0兲. Let be the eigenvalue of C2T + C1+ C2共−C2
T
+ C1− C2兲 for which its associated eigenvector v satisfies Iˆv=v 共Iˆv=−v兲, then is also an eigenvalue of C共␣, 0兲.
Proof: For any 1艋k艋n−1, let ␦kbe as assumed. Letkandkbe a number and a nonzero vector, respectively, satisfying
关C2 T
+␦k共C1−kI兲 +␦k 2
C2兴vk= 0. 共2.9兲
Using Proposition 2.1, we see thatksatisfies
det关C2T+␦2n−k共C1−kI兲 +␦2n−k2 C2兴 = 0. 共2.10兲 Letv2n−kbe a nonzero vector satisfying关C2
T +␦2n−k共C1−kI兲+␦2n−k2 C2兴v2n−k= 0. Letting bi=␦k i vk+␦k␦2n−k i v2n−k, i = 0,1, . . . ,n + 1,
we conclude, via Eqs.共2.9兲and共2.10兲, that bisatisfy Eq.共2.1a兲 with=k. Moreover, Iˆb1=␦kIˆvk+ Iˆv2n−k=␦kv2n−k+vk= b0.
We have used Proposition 2.1 to justify the second equality above. Similarly, bn+1= Iˆbn. To see =k, 1艋k艋n−1, is indeed an eigenvalue of C共␣, 0兲 for each ␣, it remains to show that bi ⫽0 for some i. Using Proposition 2.1, we have that there exists an m, 1艋m艋2j
such that vkm =v共2n−k兲共2j−m+1兲⫽0. We first show that b0⫽0. Let m be the index for which vkm⫽0. Suppose b0
= 0. Then vkm+␦kv共2n−k兲m= 0 and vk共2j−m+1兲+␦kv共2n−k兲共2j−m+1兲=v共2n−k兲m+␦kvkm= 0. And so,vkm=␦k 2
vkm, a contradiction. Let and v be as assumed in the last assertion of theorem. Letting bi=v (bi=共−1兲iv), we conclude that is an eigenvalue of C共␣, 0兲 with corresponding eigenvector共b1, b2, . . . , bn兲T. Thus,kis an eigenvalue of C共␣, 0兲 for each␣. 䊐 Corollary 2.2: Let ␦k= e共k/n兲i, where k is an integer and i =
冑
−1. Then, for each ␣,(C共␣, 0兲)=艛k=1n−1共⌫k兲艛S共⌫
0兲艛AS共⌫n兲, where S共A兲 (AS共A兲) the set of eigenvalues of A for which their corresponding eigenvectors are symmetric (antisymmetric).
We next consider the eigenvalues of C共␣,兲.
Theorem 2.3: Let␦k= e共k/n兲i, where k is an integer and i =
冑
−1. Then, for each␣,„C共␣,兲… 傻
冦
艛 k=1 关n/2兴 共⌫2k兲 艛S共⌫0兲, n is odd 艛 k=1 共n/2兲−1 共⌫2k兲 艛S共⌫0兲 艛AS共⌫n兲, n is even.冧
Here关n/2兴 is the greatest integer that is less than or equal to n/2.Proof: We illustrate only the case that n is even. Assume that k is such that 1艋k艋n/2−1. Let bi=␦2ki v2k+␦2k␦2n−2k
i
v2n−2k, we see clearly that such bi, i = 0 , 1 , n , n + 1, satisfy both Neumann and periodic boundary conditions, respectively. And so
b0=共1 −兲b0+b0=共1 −兲Iˆb1+bn and
bn+1=共1 −兲bn+1+bn+1=共1 −兲Iˆbn+b1.
Here,␦2k, 1艋k艋共n/2兲−1, are characteristic values of Eq.共2.1a兲, 共2.1b兲, and共2.1c兲. Thus, if 苸共⌫2k兲, then is an eigenvalue of C共␣,兲. The assertions for ⌫0and⌫ncan be done similarly.䊐 Remark 2.4: If n is an even number, for each␣and, half of the eigenvalues of C共␣,兲 are independent of the choice of. The other characteristic values of Eq.共2.1兲 seem to depend on. It is of interest to find them.
III. THE SECOND EIGENCURVE OF C„␣,0… AND C„␣,1…
We begin with considering the eigencurves of ⌫k, as given in Eq.共2.8兲. Clearly,
⌫k=
冢
− 2 1 0 ¯ ¯ ␦2n−k 1 − 2 1 0 ¯ 0 0 1 − 2 1 ¯ 0 ] ] 0 ¯ 0 1 − 2 1 ␦k ¯ ¯ 0 1 − 2冣
m⫻m −␣共2 − 2 cos共k/n兲兲 m ee T ¬ D1共k兲 −␣共k兲eeT, 共3.1兲 where m = 2j. We next find a unitary matrix to diagonalize D1共k兲.
Remark 3.1: Let 共共k兲,v共k兲兲 be the eigenpair of D1共k兲. If eTv共k兲=0, then 共k兲 is also an eigenvalue of⌫k. Proposition 3.1: Let l,k= 2l m + k nm, l = 0,1, . . . ,m − 1, 共3.2a兲 pl共k兲 = 共eil,k,ei2l,k,¯ ,eiml,k兲T, 共3.2b兲 and P共k兲 =
冉
p冑
0共k兲 m , . . . , pm−1共k兲冑
m冊
. 共3.2c兲共i兲 Then P共k兲 is a unitary matrix and PH共k兲D
1共k兲P共k兲=diag共0,k¯m−1,k兲, where PHis the con-jugate transpose of P, and
l,k= 2 cosl,k− 2, l = 0,1, . . . ,m − . 共3.2d兲 共ii兲 Moreover, for 0艋k艋2n, the eigenvalues of D1共k兲 are distinct if and only if k⫽0, n, or 2n.
Proof: Let b =共b1, . . . , bm兲T. Writing the eigenvalue problem D1共k兲b=b in component form, we get
bj−1−共2 + 兲bj+ bj+1= 0, j = 2,3, . . . ,m − 1, 共3.3a兲 −共2 + 兲b1+ b2+␦2n−kbm= 0, 共3.3b兲
␦kb1+ bm−1−共2 + 兲bm= 0. 共3.3c兲 Set bj=␦j, where ␦ satisfies the characteristic equation 1 −共2+兲␦+␦2= 0 of the system D1共k兲b=b. Then the boundary conditions 共3.3b兲and共3.3c兲 are reduced to
␦m=␦
k. 共3.4兲
Thus, the solutions eil,k, l = 0 , 1 , . . . , m − 1, of Eq. 共3.4兲are the candidates for the characteristic
values of Eq.共3.3兲. Substituting eil,kinto Eq.共3.3a兲and solving for, we see that =l,kare the
candidates for the eigenvalues of D1共k兲. Clearly, 共,b兲=共l,k, pl共k兲兲 satisfies D1共k兲b=b and b = pl共k兲⫽0. Thus, =l,k are, indeed, the eigenvalues of D1共k兲. To complete the proof of the proposition, it suffices to show that P共k兲 is unitary. To this end, we need to compute plH共k兲·pl
⬘共k兲. Clearly, pl H共k兲·pl共k兲=m. Now, let l⫽l
⬘
, we have that pl H共k兲 · pl ⬘共k兲 =兺
j=1 m eij共l,k−l⬘,k兲=兺
j=1 m eij共关2共l−l⬘兲/m兴兲=r共1 − r m兲 1 − r = 0,where r = ei共关2共l−l⬘兲/m兴兲. Hence, P共k兲 is unitary. The last assertion of the proposition is obvious. 䊐 To prove the main results in this section, we also need the following proposition. Some assertions of the proposition are from Theorem 8.6.2 of Ref.2.
Proposition 3.2: Suppose D = diag共d1, . . . , dm兲苸Rm⫻m and that the diagonal entries satisfy d1⬎ ¯ ⬎dm. Let␥⫽0 and z=共z1, . . . , zm兲T苸Rn. Assume that (i共␥兲,vi共␥兲) are the eigenpairs of D +␥zzT with 1共␥兲艌共␥兲艌 ¯ 艌m共␥兲. (i) Let A=兵k:1艋k艋m,zk= 0其, Ac=兵1, ... ,m其−A. If k 苸A, then dk=k. (ii) Assume ␣⬎0. Then the following interlacing relations hold 1共␥兲艌d1 艌2共␥兲艌d2艌 ¯ 艌m共␥兲艌dm. Moreover, the strict inequality holds for these indices i苸Ac. (iii) Let i苸Ac,i共␥兲 are strictly increasing in␥ and lim
␣→⬁i共␥兲=¯ifor all i, where¯iare the roots of g共兲=兺k苸Aczi2/共dk−兲 with ¯i苸共di, di−1兲. In the case that 1苸Ac, d0=⬁.
Proof: The proof of interlacing relations in 共ii兲 and the assertion in 共i兲 can be found in Theorem 8.6.2 of Ref.2. We only prove the remaining assertions of the proposition. Rearranging
z so that zT=共0,0, ... ,0,zi
1, . . . , zik兲ª共0, ... ,0,zគ
T兲, where i
1⬍i2⬍ ¯ ⬍ik and ij苸Ac, j = 1 , . . . , k. The diagonal matrix D is rearranged accordingly. Let D = diag共D1, D2兲, where D2 = diag共di
1, . . . , dik兲. Following Theorem 8.6.2 of Ref.2, we see thatij共␥兲 are the roots of the scalar
equation f␥共兲, where f␥共i j共␥兲兲 = 1 +␥
兺
j=1 k zj 2 dij−ij共␥兲 = 0. 共3.5兲Differentiating the equation above with respect to␥, we get
兺
j=1 k z ij 2 dij−ij共␥兲 +冉
␥兺
j=1 k zi2j 共dij−i k共␥兲兲 2冊
dij共␥兲 d␥ = 0. Thus, dij共␥兲 d␥ = 1 ␥2兺
j=1 k z ij 2 共dij−i j共␥兲兲 2⬎ 0.Clearly, for each ij, the limit ofij共␥兲 as␥→⬁ exists, say, ¯ij. Since, for dij⬍⬍dij−1,
兺
j=1 k zij 2 dij−ij共␥兲 =1 ␥.兺
j=1 k zij 2 dij−¯ij = 0. 共3.6兲 as desired. 䊐We are now in the position to state the following theorems.
Theorem 3.1: Let n and m = 2jbe given positive integers. For each k, k = 1 , 2 , . . . , n − 1, and␣, we denote by l,k共␣兲, l=0,1, ... ,2j− 1, the eigenvalues of ⌫k. For k = 1 , 2 , . . . , n − 1, we let 共l,k, ul,k兲, l=0,1, ... ,2j− 1, be the eigenpairs of D
1共k兲, as defined in Eq.共3.1兲. Then the following hold true.
(i) l,k共␣兲 is strictly decreasing in␣, l = 0 , 1 , . . . , 2j− 1 and k = 1 , 2 , . . . , n − 1. (ii) There existl,k* such that lim␣→⬁l,k共␣兲=l,k*. Moreover, gk共l,k* 兲=0, where
gk共兲 =
兺
l=1 m1
共l−1,k兲共l−1,k+兲. 共3.7兲
Proof: The first assertion of the theorem follows from Proposition 3.2 共iii兲. Let k be as assumed. Set, for l = 0 , 1 , . . . , m − 1,
zl+1= pl H共k兲e =
兺
j=1 m eijl,k=e −l,k共1 − e−iml,k兲 1 − e−l,k = e−l,k共1 − e−ik共/n兲兲 1 − e−l,k . Then z ¯l+1zl+1= 2 − 2 cos ml,k 2 − 2 cosl,k =2 cos共k/n兲 − 2 l,k ⫽ 0. 共3.8兲Let P共k兲 be as given in Eq.共3.2c兲. Then
− PH共k兲 · ⌫k· P共k兲 = diag共− 0,k, . . . ,−m−1,k兲 +␣共k兲Pl
H共k兲e共PlH共k兲e兲H .
Note that if k is as assumed, it follows from Proposition 3.1共ii兲 that l,k, l = 0 , . . . , m − 1, are distinct. Thus, we are in the position to apply Proposition 3.2. Specifically, by noting Ac=, we see that0,k* satisfies g共兲=0, where
g共兲 =
兺
l=1 m1
共l−1,k兲共l−1,k+兲.
We have used Eqs.共3.2d兲,共3.6兲, and共3.8兲, to find g共兲. 䊐 We next give an upper bound for0,k* , k = 1 , 2 , . . . , n − 1.
Theorem 3.2: The following inequalities hold true:
0,k* ⬍ 0,n, k = 1,2, . . . ,n − 1. 共3.9兲 Proof: To complete the proof of Eq.共3.9兲, it suffices to show that gk共−0,n兲⬍0. Now, gk共− 0,n兲
=
兺
l=1 m1
关2 cos共关2共l − 1兲/m兴 + 共k/nm兲兲 − 2兴关2 cos„关2共l − 1兲/m兴 + 共k/nm兲… − 2 cos共/m兲兴
¬ h共m,n,k兲 = h共2j,n,k兲. 共3.10兲
We shall prove that h共2j, n , k兲⬍0 by the induction on j. For j=1, h共2,n,k兲=1
2关关1/cos2共k/ 2n兲 − 1兴兴⬍0, k=1,2, ... ,n−1. Assume h共2j, n , k兲⬍0. Here, n苸N and k=1,2, ... ,n−1. We first note that
cos
冉
2共2 j+ i − 1兲 2j+1 + k 2j+1n冊
= − cos冉
2共i − 1兲 2j+1 + k 2j+1n冊
¬ − cosi−1,k,j+1, i = 1,2, . . . ,2 j . 共3.11兲 Moreover, upon using Eq.共3.11兲, we get that1
共cosi−1,k,j+1− 1兲共cosi−1,k,j+1− cos0,n,j+1兲
+ 1
共cos2j+i−1,k,j+1− 1兲共cos2j+i−1,k,j+1− cos0,n,j+1兲
= 1
共cosi−1,k,j+1− 1兲共cosi−1,k,j+1− cos0,n,j+1兲
+ 1
共cosi−1,k,j+1+ 1兲共cosi−1,k,j+1+ cos0,n,j+1兲
= 2 cos
2
i−1,k,j+1+ 2 cos0,n,j+1 共cos2
i−1,k,j+1− 1兲共cos2i−1,k,j+1− cos20,n,j+1兲
= 8共cos
2
i−1,k,j+1+ cos0,n,j+1兲
共cos 2i−1,k,j+1− 1兲共cos 2i−1,k,j+1− cos 20,n,j+1兲
= 2共cos
2
i−1,k,j+1+ cos0,n,j+1兲 共cosi−1,k,j− 1兲共cosi−1,k,j− cos0,n,j兲
. 共3.12兲 We are now in a position to compute h共2j+1, n , k兲. Using Eq.共3.12兲, we get that
h共2j+1,n,k兲 =
兺
l=1 2j+11
4共cosl−1,k,j+1− 1兲共cosl−1,k,j+1− cos0,n,j+1兲 =
兺
l=1 2j
2共cos2
l−1,k,j+1+ cos0,n,j+1兲 共cosl−1,k,j− 1兲共cosi−1,k,j− cos0,n,j兲
艋 8共cos2
0,k,j+1+ cos0,n,j+1兲h共2j,n,k兲. 共3.13兲 We have used the facts that cos2
0,k,j+1⬎cos2i−1,k,j+1, i = 2 , . . . , 2j, and that the first term共i=1兲 of the summation in Eq.共3.13兲is negative while all the others are positive to justify the inequality in Eq.共3.13兲. It then follows from Eq.共3.13兲that h共2j+1, n , k兲⬍0. We just complete the proof of the
theorem. 䊐
Theorem 3.3: Let n and j be the block and wavelet dimensions of C共␣, 1兲, respectively. Assume n and j are any positive integers. Let2共␣兲 be the second eigencurve of C共␣, 1兲. Then the following hold.
共i兲 2共␣兲 is a nonincreasing function of␣.
共ii兲 If n is an even number, then2共␣兲=0,nwhenever ␣艌␣*for some ␣*⬎0. 共iii兲 If n is an odd number, then2共␣兲⬍0,nwhenever ␣艌␣¯ for some␣¯⬎0.
Proof: We first remark that in the case of = 1, the set of the indices k’s in Eq. 共3.1兲 is 兵0,2,4, ... ,2共n−1兲其ªIn. Suppose n is an even number. Then n苸In. Thus,␦n= −1,0,n=/ m, and
p0共n兲=共ei共/m兲, ei共2/m兲, . . . , ei兲T. Applying Proposition 3.1, we see that p0共n兲−p0
s共n兲, an antisym-metric vector, is also an eigenvector of D1共n兲. And so eT(p0共n兲−p0
s共n兲)=0. It then follows from Remark 3.1 that 0,n is an eigenvalue of ⌫n= D1共n兲−共n兲eeT for all ␣. The first and second assertions of the theorem now follow from Theorems 3.1 and 3.2. Let n be an odd number. Then
␦i·␦i⫽1 for any i苸In. Thus, if the pair共␦i,vi兲 satisfy Eq.共2.3兲, thenvi⫽−vi s
. Otherwise, the pair 共␦i,vi−共−vi兲s兲=共␦i,vi+vi
s兲 also satisfy Eq. 共2.3兲
. This is a contradiction to the last assertion in Proposition 2.1. Thus, viH· e⫽0 for any i苸In. We then conclude, via Proposition 3.2 共iii兲 and Theorem 3.2, that the last assertion of the theorem holds. 䊐 Remark 3.2: 共i兲 Let the number of uncoupled 共chaotic兲 oscillators be N=2jn. If n is an odd number, then the wavelet method for controlling the coupling chaotic oscillators work even better in the sense that the critical coupling strength ⑀ can be made even smaller. 共ii兲 For n being a
multiple of 4 and j苸N, the assertions in Theorem 3.3 were first proved in Ref. 6by a different method.
Theorem 3.4: Let n and j be the block and wavelet dimensions of C共␣, 0兲, respectively. Assume n and j are any positive integers. Let2共␣兲 be the second eigencurve of C共␣, 0兲. Then for any n, there exists a␣˜ such that2共␣兲=0,nwhenever␣艌␣˜ .
Remark 3.3: For n苸N and j=1, the explicit formulas for the eigenvalues of C共␣, 0兲 were obtained in. Ref.4Such results are possible due to the fact that the dimension of the matrices in Eq.共2.4兲is 2⫻2.
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