Applied Mathematics E-Notes, 9(2009), 109-145 c
Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Synchronization of Strongly Coupled Neural Networks∗
Sui Sun Cheng and Yi Feng Wu†
Received 1 May 2008
Abstract
Two identical discrete time cellular neural networks are coupled and sharp
conditions are found so that some or all neural units will eventually synchronize.
In deriving these criteria, we make use of symmetry (invariance) principles, Banach contraction technique and spectral properties of several band matrices with
block components.
1
Introduction
The fact that various parts of a biological system operate in harmony is taken to
be an important indication of normal functioning of the system. One concept that
is essential in describing such harmonious operations is ‘synchronization’. In order
to design computing machines that simulate harmonious operations, it is therefore
necessary to find mathematical models that are capable of generating synchronized
outputs. Synchronization can occur in a number of continuous dynamical systems.
Synchronization can also occur in artificial neural network models where time and
space are both assumed to be discrete. Such an example is studied in [1, 2]. For
the sake of convenience, we briefly recall this network model here. Let x1 , ..., xn be
(t)
n (n ≥ 2) neuron units placed at the vertices of a regular polygon. Let xi be the
state values of the neuron unit xi in the time period t. During the time period t,
(t)
(t)
if the state value x1 is larger than x2 , information will “flow” from the unit x1 to
(t+1)
(t)
the unit x2 . The subsequent change of state value of x2 is x2
− x2 and under a
first order approximation assumption,
it is reasonable that it is proportional to the
(t)
(t)
(t)
(t)
difference x1 − x2 , say, γ x1 − x2 , where γ is a positive rate constant. Similarly,
(t)
(t)
information will flow from the point x3 to the point x2 if x3 > x2 . Thus, it is
reasonable that the total effect is
(t+1)
(t)
(t)
(t)
(t)
(t)
(t)
(t)
(t)
x2
− x2 = γ x1 − x2 + γ x3 − x2 = γ x1 − 2x2 + x3 .
By similar considerations, we may then obtain the following dynamic system of equations
x(t+1) = (1 − 2γ)x(t) + γAn x(t) , t = 0, 1, 2, ...,
∗ Mathematics
† Department
Subject Classifications: 37N25, 39A10, 92B20.
of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043, R. O. China
109
110
Synchronization of Strongly Coupled Dynamical Networks
†
(t)
(t)
where x(t) = x1 , ..., xn
and An is the circulant matrix defined by





An = 



0 1 0
1 0 1
0 1 0
.. .. ..
. . .
0 0 0
1 0 0

··· 0 1
··· 0 0 

··· 0 0 

.. .. 
..
. . . 

··· 0 1 
· · · 1 0 n×n
(1)
when n ≥ 2.
Now suppose there is another identical neural network with neuron units denoted by
y1 , ..., yn. Suppose further that the neuron pairs xi and yi , i = 1, 2, ..., n, are
“strongly
(t+1)
(t)
(t)
(t)
connected” so that the change, say, x2
− x2 is also proportional to 2γ y2 − x2
(note the factor 2 here), then the subsequent equation for the neuron x2 is
(t+1)
x2
(t)
(t)
(t)
(t)
(t)
(t)
(t)
− x2 = γ x1 − x2 + γ x3 − x2 + 2γ y2 − x2 .
The complete set of equations for the neurons x1 , ..., xn and y1 , ..., yn is of the form
x(t+1)
y(t+1)
=
=
(1 − 4γt )x(t) + γt An x(t) + 2γt y(t),
(1 − 4γt )y(t) + γt An y(t) + 2γt x(t) ,
which describes the evolution process of the state values as t tends to infinity. In
Fig. 1, we illustrate a coupled network with 6 neuron units x1 , ..., x6 and 6 neuron
units y1 , ..., y6 in two manners. The one on the right is a planar graph, which is more
convenient for illustrating later results.
Figure 1
There are many questions of interests which we can raise regarding the above evolutionary system. An interesting one is whether some or all neuron units are ultimately
in synchronization, so that, as time evolves, they differ from each other only by infinitesimal values.
111
S. S. Cheng and Y. F. Wu
In this paper, we will consider such a problem for a slightly more general nonlinear
system of equations of the form
x(t+1) = (1 − 4γt )F x(t) + γt An F x(t) + 2γt F y(t) ,
(2)
y(t+1) = (1 − 4γt )F y(t) + γt An F y(t) + 2γt F x(t) ,
†
for t = 0, 1, 2, . . . , where {γt } is a real sequence, F(u) = (f(u1 ), f(u2 ), ..., f(un)) for
†
u = (u1 , ..., un) and f : R → R is a Lipschitz function1 which satisfies
|f (x) − f (y)| ≤ Γ |x − y| , x, y ∈ R,
for some fixed positive constant Γ. †
x(t)
(t)
(t)
Note that if we denote the vector
by z(t), where x(t) = x1 , ..., xn
and
(t)
y
†
(t)
(t)
y(t) = y1 , ..., yn
, then our system can be written in the form z(t+1) = F t, z(t) .
Thus, given an initial distribution z(0) = z, it is easily seen that we can calculate z(1) , z(2) , . . . successively and in a unique manner from (2). Such a sequence
z(0) , z(1), z(2), . . . is said to be a solution of (2).
An important property of our system (2) is its ‘invariance’ under ‘rotations’. To be
†
more precise, let us call the vector (un , u1 , u2 , ..., un−1
the(forward) rotation of the
) (t)
x
is a solution of (2), then
vector u = (u1 , u2 , ..., un)† and denote it by θ(u). If
y(t)
θ(x(t))
it is easily checked that
is also a solution since we are simply rearranging
θ(y(t))
the equations
in (2).
∞
Let z(t) t=0 be a solution sequence of (2). Given any two distinct neuron units u, v
∞
∞
in {x1 , . . . , xn, y1 , . . . , yn } , let u(t) t=0 and v(t) t=0 be the corresponding component
∞
∞
sequences in the solution sequence z(t) t=0 . We say that the solution z(t) t=0 is
{u, v} synchronized if
lim u(t) − v(t) = 0.
t→∞
∞
More generally, let Ω be a subset of {x1 , . . . , xn , y1 , . . . , yn }, if z(t) t=0 is {u, v} syn
∞
chronized for each pair of distinct neuron units in Ω, then we say that z(t) t=0 is Ω
(partially)
synchronized. In case Ω = {x1 , . . . , xn , y1 , . . . , yn }, then it is natural to say
∞
that z(t) t=0 is (fully) synchronized. In Fig. 12, a fully synchronized neural network
(n = 5) can be found in which any two (distinct) units are connected with a dash line
to show synchronization.
Two neuron units from {x1 , .., xn} or from {y1 , ..., yn} are said to be of the same
type, while a neuron unit u from {x1 , ..., xn} and v from {y1 , ..., yn} are said to be
of different type. In this paper, we are interested in finding various synchronization
phenomena that involves neuron units of the same or different types.
1 For example, the tent map g defined by g(x) = 2x for 0 ≤ x ≤ 1/2, 2(1 − x) for 1/2 ≤ x ≤ 1, and
0 elsewhere is a Lipschitz function with Lipschitz constant 2.
112
Synchronization of Strongly Coupled Dynamical Networks
To motivate the main results that follow, let us consider the case where n = 2. For
the sake of convenience, we use ρ (W ) to denote the spectral radius of a square matrix
W.
When n = 2, our system is

(t+1)
(t)
(t)
(t)

x1
= (1 − 4γt ) f x1 + 2γt f x2 + 2γt f y1 ,





 x(t+1) = 2γt f x(t) + (1 − 4γt ) f x(t) + 2γt f y(t) ,
2
1 2 2 (3)
(t+1)
(t)
(t)
(t)

+
(1
−
4γ
)
f
y
+
2γ
f
y2 ,
y
=
2γ
f
x

t
t
t
1
1
1




 y(t+1) = 2γt f x(t) + 2γt f y(t) + (1 − 4γt ) f y(t) .
2
2
1
2
We assert that if
lim sup |1 − 4γt | <
t→∞
1
,
Γ
(4)
then every solution of (3) is {x2 , y1 } synchronized. Indeed, note that
(t+1)
(t+1)
(t)
(t)
x2
− y1
= (1 − 4γt ) f x2 − f y1
.
(5)
If (4) holds, then there exist constant d ∈ (0, 1) and positive integer T such that (cf.
Banach contraction principle)
Γ |1 − 4γt | < d < 1, t ≥ T,
so that by (5),
(t+1)
x2
(t+1)
− y1
and hence
(T +n)
x2
(t)
(t)
(t)
(t)
= |1 − 4γt | f x2 − f y1
≤ Γ |1 − 4γt | x2 − y1 ,
(T +n)
− y1
(T +n−1)
< d x2
(T +n−1)
− y1
(T )
< · · · < d n x2
(T )
− y1
for n = 1, 2, . . . . If we now let n → ∞, we see that
(t)
(t)
lim x2 − y1
t→∞
= 0.
This shows that {x2 , y1 } are synchronized. By considering the absolute difference
(t)
(t)
(t)
(t)
x1 − y2 , we may proceed in a similar manner to show that limt→∞ x1 − y2
That is, every solution of (3) is {x1 , y2 } synchronized.
Similarly, note that
 
! (t)
(t)
(t+1)
(t+1)
f
x
−
f
x
1
2
x1
− x2
1 − 6γt
2γt
 .
=
(t+1)
(t+1)
(t)
(t)
2γt
1 − 6γt
y1
− y2
f y
−f y
1
If
lim sup ρ
t→∞
1 − 6γt
2γt
2γt
1 − 6γt
<
1
,
Γ
= 0.
(6)
2
(7)
113
S. S. Cheng and Y. F. Wu
there exist constant d ∈ (0, 1) and positive integer T such that
 
!
(t)
(t)
(t+1)
(t+1)
f
x
−
f
x
x1
− x2
1 − 6γt
2γt
 1 2 
≤
(t+1)
(t+1)
(t)
(t)
2γt
1 − 6γt
y1
− y2
f
y
−
f
y2
2
1
2
!
(t)
(t)
1 − 6γt
2γt
x1 − x2
≤ Γρ
(t)
(t)
2γt
1 − 6γt
y1 − y2
2
!
(t)
(t)
x1 − x2
≤ d
, t ≥ T.
(t)
(t)
y1 − y2
2
2
so that
(T +n)
(T +n)
x1
− x2
(T +n)
(T +n)
y1
− y2
!
(T )
(T )
x1 − x2
(T )
(T )
y1 − y2
< · · · < dn
2
!
2
for n = 1, 2, . . . . As n → ∞, we see that
(t)
lim
t→∞
(t)
x1 − x2
(t)
(t)
y1 − y2
!
= 0.
That is, every solution of (3) is {x1 , x2 } and {y1 , y2 } synchronized. Similarly, we may
show that every solution of (2) is {x1 , y1 } and {x2 , y2 } synchronized.
We remark that when γt = γ for all t and Γ = 1, condition (4) holds if, and only if,
|1 − 4γ| < 1 or 0 < γ < 1/2. Note that the eigenvalues of
1 − 6γ
2γ
2γ
1 − 6γ
are 1 − 8γ and 1 − 4γ. Thus, condition (7) becomes
max {|1 − 8γ| , |1 − 4γ|} < 1,
or 0 < γ < 1/4. Note that this condition is sharp. Indeed, when γ = 1/4 and f is the
identity function, (6) becomes
!
!
(t+1)
(t+1)
(t)
(t)
1
x1
− x2
−1 1
x1 − x2
.
=
(t+1)
(t+1)
(t)
(t)
1 −1
2
y1
y1 − y2
− y2
If we let
(0)
(0)
(0)
(0)
x1 = −1, y1 = 1, x2 = y2 = 0,
then since
t t 1
−1 1
−1
−1
t
= (−1)
,
1 −1
1
1
2
n
o∞
n
o∞
(t)
(t)
(t)
(t)
we see that neither x1 − x2
nor y1 − y2
converge to zero.
t=0
t=0
114
Synchronization of Strongly Coupled Dynamical Networks
The condition (4) is also sharp since when f is the identity function,
(t+1)
(t+1)
(t)
(t)
t+2
x2
− y1
= − x2 − y1
= (−1)
,
we see that
n
(t)
(t)
x2 − y1
o∞
t=0
does not converge to zero. Thus every solution of (2),
where n = 2, is synchronized when 0 < γ < 1/4 and this condition is sharp.
To facilitate later discussions, we need to introduce some simple notations. The n
by n identity matrix is denoted by In . We let J, U and V be respectively the matrices
0 1
1 0
0 1
J=
, U =
and V =
.
(8)
1 0
−1 0
0 −1
Note that the matrix J has eigenvalues −1 and 1 with the corresponding eigenvectors
†
†
u
b = (−1, 1) and vb = (1, 1) respectively. Note further that u
b and vb are linearly
independent. The matrix U has eigenvalues 0 and 1 with corresponding eigenvectors
(0, 1)† and u
b respectively. The matrix V has eigenvalues 0 and −1 with eigenvectors
(1, 0)† and u
b respectively. The matrix U † has eigenvalues 0 and 1 with corresponding
eigenvectors b
v and (1, 0)† respectively. The matrix V † has eigenvalues 0 and −1 with
corresponding eigenvectors b
v and (0, 1)† respectively.
The Case where n = 3
2
From (2), we see that



(t+1)
x1
(t+1)
x2
(t+1)
x3
−
−
−
(t+1)
y1
(t+1)
y2
(t+1)
y3


 
=
1 − 6γt
γt
γt
γt
1 − 6γt
γt

γt


γt


1 − 6γt
and
(t+1)
(t+1)
x1
− x2
(t+1)
(t+1)
y1
− y2
!
=
1 − 5γt
2γt
2γt
1 − 5γt
as well as



(t+1)
x1
(t+1)
y1
(t+1)
x3
(t+1)
− y2
(t+1)
− x2
(t+1)
− y3


 
=
1 − 4γt
γt
γt
γt
1 − 4γt
−γt

(t)
(t)
f x1 − f y1
(t)
(t)
f x2 − f y2
(t)
(t)
f x3 − f y3

(t)
(t)
f x1 − f x2

 (t)
(t)
f y1 − f y2



γt


−γt  

1 − 6γt
(t)
(t)
f x1 − f y2
(t)
(t)
f y1 − f x2
(t)
(t)
f x3 − f y3
and by the rotational invariance of (2), we also have

 ! (t)
(t)
(t+1)
(t+1)
−
f
x
f
x
1 − 5γt
2γt
x2
− x3
3 ,
 2 =
(t+1)
(t+1)
(t)
(t)
2γt
1 − 5γt
y2
− y3
f y2 − f y3



,

(9)
(10)



,

(11)
(12)
115
S. S. Cheng and Y. F. Wu
(t+1)
x1
(t+1)
y1
−
−
(t+1)
x3
(t+1)
y3
!
=
1 − 5γt
2γt
2γt
1 − 5γt
and






(t+1)
x1
(t+1)
y1
(t+1)
x2
(t+1)
− y3
(t+1)
− x3
(t+1)
− y2
(t+1)
x2
(t+1)
y2
(t+1)
x1
(t+1)
− y3
(t+1)
− x3
(t+1)
− y1




 
=
 
=
1 − 4γt
γt
γt


(t)
(t)
f x1 − f x3
 ,
(t)
(t)
f y1 − f y3

γt
1 − 4γt
−γt
1 − 4γt
γt
γt

γt


−γt  

1 − 6γt

γt
1 − 4γt
−γt

γt


−γt  

1 − 6γt
(t)
(t)
f x1 − f y3
(t)
(t)
f y1 − f x3
(t)
(t)
f x2 − f y2
(t)
(t)
f x2 − f y3
(t)
(t)
f y2 − f x3
(t)
(t)
f x1 − f y1
(13)



,




.

For the same reason as used in the case where n = 2, we may now see that if
1
1 − 5γt
2γt
< ,
lim sup ρ
(14)
2γt
1 − 5γt
Γ
t→∞
then every solution of (2) is {x1 , x2 } , {x2 , x3 } , {x1 , x3 } , {y1 , y2 } , {y2 , y3 } , and {y1 , y3 }
synchronized. Moreover, every solution of (2) is {x1 , x2 , x3} and {y1 , y2 , y3 } synchronized. If


1 − 6γt
γt
γt
 < 1,
γt
1 − 6γt
γt
(15)
lim sup ρ 
Γ
t→∞
γt
γt
1 − 6γt
then every solution of (2) is {x1 , y1 }, {x2 , y2 } and {x3 , y3 } synchronized; and that if


1 − 4γt
γt
γt
1
lim sup ρ 
γt
1 − 4γt
−γt  < ,
(16)
Γ
t→∞
γt
−γt
1 − 6γt
then every solution of (2) is {x1 , y2 }, {x2 , y1 }, {x3 , y3 }, {x2 , y3 }, {x3 , y2 }, {x1 , y1 },
{x1 , y3 }, {x3 , y1 } and {x2 , y2 } synchronized.
We remark that when γt = γ for all t and Γ = 1, the condition (14) becomes
max {|1 − 7γ| , |1 − 3γ|} < 1,
or 0 < γ < 2/7. When γ = 2/7 and f is the identity function, (10) becomes
!
!
(t)
(t)
(t+1)
(t+1)
1
−3 4
x1 − x2
x1
− x2
=
.
(t+1)
(t+1)
(t)
(t)
4 −3
7
y1
− y2
y1 − y2
If we let
(0)
(0)
(0)
(0)
x1 = y2 = 1, x2 = y1 = 0,
then since
t 1
−3
4
7
4
−3
t 1
−1
t
= (−1)
1
−1
,
116
Synchronization of Strongly Coupled Dynamical Networks
we see that there is a solution of (2) which is not {x1 , x2 } nor {y1 , y2 } synchronized.
Next, the matrix


1 − 6γt
γt
γt


γt
1 − 6γt
γt
γt
γt
1 − 6γt
has the eigenvalues 1 − 7γ, 1 − 7γ and 1 − 4γ. The condition (15) becomes
max {|1 − 7γ| , |1 − 4γ|} < 1,
or 0 < γ < 2/7. When γ = 2/7 and f is the identity function, (9) becomes

If we let


(t+1)
(t+1)
x1
− y1
−5 2
1
 (t+1)
(t+1) 
 x2
 =  2 −5
− y2
7
(t+1)
(t+1)
2
2
x3
− y3
(0)
(0)
(0)
(0)

  (t)
(t)
x1 − y1
2

(t) 
2   x(t)
.
2 − y2
(t)
(t)
−5
x3 − y3
(0)
(0)
x1 = x2 = x3 = 1, y1 = y2 = y3 = 0,
then since

t
−5 2
1 
2 −5
7
2
2
n
o∞
(t)
(t)
we see that xi − yi
does
Finally, the matrix
t=0

t 

 
t
2
1
1
−1
 1 ,
2   1 =
7
−5
1
1
not converge to zero for i = 1, 2, 3.
1 − 4γt

γt
γt
γt
1 − 4γt
−γt

γt
−γt 
1 − 6γt
has the eigenvalues 1 − 7γ, 1 − 4γ and 1 − 3γ. By similar reasonings, we see that the
condition (16) is equivalent to 0 < γ < 2/7 which is also sharp. When γ = 2/7 and f
is the identity function, (11) becomes

Since


(t+1)
(t+1)
x1
− y2
−1 2
1
 (t+1)

(t+1)
 y1
− x2
 =  2 −1
7
(t+1)
(t+1)
2 −2
x3
− y3

t
−1 2
1 
2 −1
7
2 −2

  (t)
(t)
x1 − y2
2


−2   y1(t) − x(t)
.
2
(t)
(t)
−5
x3 − y3
t 



t
2
1
1
−1 
−2   −1  =
−1  ,
7
−5
1
1
we see that there is a solution of (2) which is not {x1 , y2 }, {x2 , y1 } nor {x3 , y3 } synchronized. Thus every solution of (2), where n = 3, is synchronized when 0 < γ < 2/7
and the condition is sharp.
117
S. S. Cheng and Y. F. Wu
3
The Case where n = 4
In view of our previous discussions for the cases n = 2 and n = 3, it is reasonable to
proceed to the general case. However, there are enough difference in the case n = 4
from the general case to warrant a sketch of the various synchronization phenomena.
• If
lim sup ρ (1 − 6γt )I4 + γt
t→∞
J
J
J
J
< 1/Γ,
(17)
then every solution of (2) is {xi , yi } synchronized for i = 1, 2, 3, 4. Indeed, from
(2), we see that





(t+1)
x1
(t+1)
x2
(t+1)
x3
(t+1)
x4
−
−
−
−

(t)
(t)
f x1 − f y1
 

(t)
(t)

 
 f x2 − f y2
J J

.

 = (1 − 6γt )I4 + γt


(t)
(t)
J J


 f x3 − f y3

 (t)
(t)
f x4 − f y4
(18)
Banach contraction technique described previously, the condition
(t+1)
y1
(t+1)
y2
(t+1)
y3
(t+1)
y4
Then by the


J
lim sup ρ (1 − 6γt )I4 + γt
J
t→∞
J
J
< 1/Γ
will imply
(t+1)
lim xi
t→∞
(t+1)
− yi
= 0 for i = 1, 2, 3, 4.
• If
lim sup ρ ((1 − 4γt )I2 + 2γt J) < 1/Γ,
(19)
t→∞
then every solution of (2) is {x1 , x3 } and {y1 , y3 } synchronized. Indeed, from (2),
we may show that
(t+1)
(t+1)
x1
− x3
(t+1)
(t+1)
y1
− y3
!


(t)
(t)
f x1 − f x3
.
= ((1 − 4γt )I2 + 2γt J)  (t)
(t)
f y1 − f y3
(20)
Then the same argument which we used before leads to the desired conclusion.
Furthermore, by the rotation invariance of our system (2), we may also conclude
that every solution is {x2 , x4 } and {y2 , y4 } synchronized.
• If
lim ρ (1 − 4γt )I2 + γt
t→∞
2J
I2 − J
I2 − J
2J
< 1/Γ,
(21)
118
Synchronization of Strongly Coupled Dynamical Networks
then every solution of (2) is {x1 , y3 } , {x3 , y1 } , {x2 , y4 } and {x4 , y2 } synchronized.
Indeed, from (2), we may show that
 (t+1)

(t+1)
x1
− y3
 (t+1)
(t+1) 
− x3
2J
I2 − J
 y1

(1 − 4γt )I2 + γt
 (t+1)
(t+1)  =
I2 − J
2J
 x2

− y4
(t+1)
(t+1)
y2
− x4
 
(t)
(t)
f x1 − f y3
 


(t)
(t)

 f y1 − f x3
.

(22)
×

(t)
(t)

 f x2 − f y4

 (t)
(t)
f y2 − f x4
• If
2J
lim ρ (1 − 5γt )I2 + γt
I2
t→∞
I2
2J
J
lim ρ (1 − 4γt )I2 + γt
t→∞
I2
I2
J
< 1/Γ,
< 1/Γ,
(23)
then every solution of (2) is {x1 , x2 } , {x4 , x3 } , {y1 , y2 } and {y4 , y3 } synchronized.
Indeed, from (2), we may show that
 
(t)
(t)
 (t+1)

f x1 − f x2
(t+1)
 x1
− x2



(t)
(t)
 (t+1)
(t+1) 
f
y
−
f
y2


1
− y2
2J I2
 y1


.
=
(1
−
5γ
)I
+
γ
 (t+1)
t 2
t
(t+1) 


(t)
(t)
I2 2J
 x4

− x3
 f x4 − f x3



(t+1)
(t+1)
(t)
(t)
y4
− y3
f y4 − f y3
(24)
Then the same argument which we used before leads to the desired conclusion.
Furthermore, by the rotation invariance of our system (2), we may also conclude
that every solution is {x1 , x4 } , {x2 , x3 } , {y1 , y4 } and {y2 , y3 } synchronized.
• If
(25)
then every solution of (2) is {x1 , y2 } , {x2 , y1 } , {x4 , y3 } and {x3 , y4 } synchronized.
Indeed, from (2), we may show that
 
(t)
(t)
 (t+1)

f x1 − f y2
(t+1)
 x1
− y2



(t)
(t)
 (t+1)
(t+1) 
 f y1 − f x2

− x2
J I2
 y1


.
 (t+1)
(t+1)  = (1 − 4γt )I2 + γt


(t)
(t)
I
J
 x4

2
− y3
 f x4 − f y3

 
(t+1)
(t+1)
(t)
(t)
y4
− x3
f y4 − f x3
(26)
Then the same argument which we used before leads to the desired conclusion.
Furthermore, by the rotation invariance of our system (2), we may also conclude
that every solution is {x1 , y4 } , {x4 , y1 } , {x2 , y3 } and {x3 , y2 } synchronized.
119
S. S. Cheng and Y. F. Wu
The conditions derived above are sharp. To see this, we assume that γt = γ for all
t, f is the identity function and Γ = 1.
• Then the matrix
(1 − 6γ)I4 + γt
J
J
J
J
has eigenvalues 1 − 8γ, 1 − 6γ, 1 − 6γ and 1 − 4γ. The condition (17) becomes
max{|1 − 8γ| , |1 − 6γ| , |1 − 4γ|} < 1
or 0 < γ < 1/4. When γ = 1/4, (18) becomes





(t+1)
(t+1)
x1
− y1
(t+1)
(t+1)
x2
− y2
(t+1)
(t+1)
x3
− y3
(t+1)
(t+1)
x4
− y4

 −1
1
J

I4 +
=
J

2
4





J
J
(t)
(t)
x1 − y1
(t)
(t)
x2 − y2
(t)
(t)
x3 − y3
(t)
(t)
x4 − y4



.

If we let
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
x2 = x4 = y1 = y3 = 1, x1 = x3 = y2 = y4 = 0,
then since
we see that
n
1
−1
I4 + A4
2
4
(t)
(t)
xi − yi
o∞
t=0
t



−1
 1 
t



 −1  = (−1) 
1

−1
1 
,
−1 
1
does not converge to zero for i = 1, 2, 3, 4.
• Next, the condition (19) becomes max{|1 − 6γ| , |1 − 2γ|} < 1 or 0 < γ < 1/3.
When γ = 1/3, (20) becomes
! !
(t+1)
(t+1)
(t)
(t)
−1
2
x1
− x3
x1 − x3
=
I2 + J
.
(t+1)
(t+1)
(t)
(t)
3
3
y1
− y3
y1 − y3
If we let
(0)
(0)
(0)
(0)
x3 = y1 = 1, x1 = y3 = 0,
then since
−1
2
I2 + J
3
3
t −1
1
t
= (−1)
−1
1
,
we see that there is a solution of (2) which is not {x1 x3 } nor {y1 , y3 } synchronized.
• Next, the matrix
(1 − 4γt )I2 + γt
2J
I2 − J
I2 − J
2J
120
Synchronization of Strongly Coupled Dynamical Networks
has eigenvalues 1 − 8γ, 1 − 4γ and 1 − 2γ, 1 − 2γ. The condition (21) becomes
max{|1 − 8γ| , |1 − 4γ| , |1 − 2γ|} < 1
or 0 < γ < 1/4. When γ = 1/4, (22) becomes
 (t+1)

(t+1)
x1
− y3
 (t+1)
(t+1) 
− x3
2J
 y1

=
(1
−
4γ
)I
+
γ
 (t+1)

t 2
t
(t+1)
I2 − J
 x2

− y4
(t+1)
(t+1)
y2
− x4
I2 − J
2J
If we let
(0)
(0)
(0)
(0)
(0)
(0)
(0)





(t)
(t)
x1 − y3
(t)
(t)
y1 − x3
(t)
(t)
x2 − y4
(t)
(t)
y2 − x4



.

(0)
x1 = x3 = y2 = y4 = 1, x2 = x4 = y1 = y3 = 0,
then since
t 1
2J
I2 − J
4
I2 − J
2J
t



1
 −1 

t


 −1  = (−1) 
1

1
−1 
,
−1 
1
we see that there is a solution of (2) which is not {x1 , y3 } , {x3 , y1 } , {x2 , y4 } nor
{x4 , y2 } synchronized.
• Next, the matrix
(1 − 5γt )I2 + γt
2J
I2
I2
2J
has eigenvalues 1 −8γ, 1 −6γ, 1 −4γ and 1 −2γ. The condition (23) then becomes
max{|1 − 8γ| , |1 − 6γ| , |1 − 4γ| , |1 − 2γ|} < 1
or 0 < γ < 1/4. When γ = 1/4, then (24) becomes
 (t+1)

 (t)
(t+1)
(t)
x1
− x2
x1 − x2
 (t+1)
 (t)
(t+1) 
(t)
− y2
2J I2
 y1

 y1 − y2
 (t+1)
 (t)
(t+1)  = (1 − 5γt )I2 + γt
I2 2J
 x4

 x4 − x3(t)
− x3
(t+1)
(t+1)
(t)
(t)
y4
− y3
y4 − y3
If we let
(0)
(0)
(0)
(0)
(0)
(0)
(0)



.

(0)
x1 = x3 = y2 = y4 = 1, x2 = x4 = y1 = y3 = 0,
then since
−1
1
I2 +
4
4
2J
I2
I2
2J
t



1
 −1 

t


 −1  = (−1) 
1

1
−1 
,
−1 
1
we see that there is a solution of (2) which is not {x1 , x2} , {x4 , x3 } , {y1 , y2 } nor
{y4 , y3 } synchronized.
121
S. S. Cheng and Y. F. Wu
• Finally, the matrix
(1 − 4γt )I2 + γt
J
I2
I2
J
has eigenvalues 1 −6γ, 1 −4γ, 1 −4γ and 1 −2γ. The condition (25) then becomes
max{|1 − 6γ| , |1 − 4γ| , |1 − 2γ|} < 1
or 0 < γ < 1/3. When γ = 1/3, then (26) becomes





(t+1)
(t+1)
x1
− y2
(t+1)
(t+1)
y1
− x2
(t+1)
(t+1)
x4
− y3
(t+1)
(t+1)
y4
− x3
If we let
(0)

 J

 = (1 − 4γt )I2 + γt
I

2
(0)
(0)
(0)
(0)
I2
J
(0)





(0)
(t)
(t)
x1 − y2
(t)
(t)
y1 − x2
(t)
(t)
x4 − y3
(t)
(t)
y4 − x3



.

(0)
x1 = x2 = y3 = y4 = 1, x3 = x4 = y1 = y2 = 0,
then since
1
−1
I2 +
3
3
J
I2
I2
J
t



1
 −1 

t


 −1  = (−1) 
1

1
−1 
,
−1 
1
we see that there is a solution of (2) which is not {x1 , y4 } , {x4 , y1 } , {x2 , y3 } nor
{x3 , y2 } synchronized.
Hence we concluded that every solution of (2), where n = 4, is synchronized when
0 < γ < 1/4 and the condition is sharp.
4
Synchronization Criteria for n ≥ 5
The same principles used in the previous discussions can be used again for the case
where n ≥ 2, although there are some other details.
THEOREM 4.1. Suppose n ≥ 5. Let
bt = (1 − 6γt )In + γt An
A
(27)
bt < 1/Γ, then every solution of (2) is
where An is defined by (1). If lim supt→∞ ρ A
{xi , yi } synchronized for i = 1, 2, . . . , n.
Indeed, from (2), we see that
(t+1)
xi
(t+1)
− yi
=
n X
k=1
bt
A
(t)
(t)
f xk − f yk
for i = 1, 2, . . ., n.
ik
(28)
122
Synchronization of Strongly Coupled Dynamical Networks
Then by the same
arguments used in the derivations in the last section, the condition
bt < 1/Γ will imply
lim supt→∞ ρ A
(t+1)
lim xi
t→∞
(t+1)
− yi
= 0 for i = 1, 2, . . ., n.
THEOREM 4.2. Suppose n = 2m, where m ≥ 3. For t ≥ 0, let


2J I2
0


..


.
bt = (1 − 4γt )I2m−2 + γt  I2 2J

.
(29)
B


..
..

.
. I2 
0
I2 2J (2m−2)×(2m−2)
bt < 1/Γ, then every solution of (2) is {x1 , x3 , . . . , x2m−1 }, {x2 , x4 , . . . ,
If lim supt→∞ ρ B
x2m }, {y1 , y3 , . . . , y2m−1 } and {y2 , y4 , . . . , y2m} synchronized.
Indeed, from (2) we have


(t)
(t)
f x1 − f x3
 (t+1)

(t+1)


x1
− x3


(t)
(t)
 (t+1)
 f y1 − f y3

(t+1) 
 y1


− y3

 (t+1)



(t)
(t)
(t+1) 
 x
 f x2m − f x4

 2m − x4



bt 
 y(t+1) − y(t+1)  = B
.
(t)
(t)
 2m

 f y2m − f y4

4




···




·
·
·
 (t+1)



(t+1)


(t)
(t)
 xm+3 − xm+1 
 f xm+3 − f xm+1 
(t+1)
(t+1)


ym+3 − ym+1
(t)
(t)
f ym+3 − f ym+1
Then by reasonings we have explained before, we see that
(t+1)
(t+1)
(t+1)
(t+1)
(t+1)
(t+1)
lim x1
− x3
, x2m − x4
, . . . , xm+3 − xm+1
=0
t→∞
and
lim
t→∞
(t+1)
y1
(t+1)
− y3
(t+1)
, y2m
(t+1)
− y4
(t+1)
(t+1)
, . . . , ym+3 − ym+1
= 0.
(30)
(31)
By the rotational invariance of (2), we see further that every solution of (2) is
{x2 , x4 } , {y2 , y4 } , {x1 , x5 } , {y1 , y5 } , {x2m+6−j , xj } , and {y2m+6−j , yj } synchronized
for j = 6, 7, . . ., m + 2 as well. To complete our explanation, let us consider a neural
network in which n = 6. To illustrate (30) and (31), we draw dash lines connecting x1
and x3 , x4 and x6 , y1 and y3 , as well as y4 and y6 to show synchronization. Then we
add new dash lines to connect x2 and x4 , y2 and y4 , etc. These are illustrated in Fig.
2. From this Figure, we may easily see that every solution is {x1 , x3 , x5}, {y1 , y3 , y5 },
{x2 , x4 , x6 } and {y2 , y4 , y6 } synchronized (See Fig. 3). The same reasoning shows
that every solution of the general system (2) is {x1 , x3 , . . . , x2m−1 }, {x2 , x4 , . . . , x2m },
{y1 , y3 , . . . , y2m−1 }, and {y2 , y4 , . . . , y2m } synchronized2 .
2 The formal approach is to consider a graph with vertices {x , ..., x , y , .., y } and synchronized
n 1
n
1
vertices as edges. Then the proof reduces to finding the connected components which is an easy matter.
123
S. S. Cheng and Y. F. Wu
Figure 2
Figure 3
In the above result, we see that there are four groups of synchronized units. These
groups may or may not be distinct. For example, consider a neural network where n =
6. We consider the special case where γt = γ for all t, f is the tent map function,
and the
bt < 1/Γ
Lipschitz constant Γ = 2. From the Appendix, the condition lim supt→∞ ρ B
in Theorem 4.2 can be replaced by 1/6 < γ < 3/14. Then choosing γ = 0.2 and
(0)
(0)
(0) (0)
(0)
(0) (0) (0) (0) (0) (0) (0)
x1 , x2 , x3 , x4 , x5 , x6 , y1 , y2 , y3 , y4 , y5 , y6
=
1
(3, 1, 3, 1, 3, 1, 2, 5, 2, 5, 2, 5),
10
(∞)
(∞)
(∞)
(∞)
(∞)
(∞)
we may compute x1 = x3 = x5 = y2
= y4
= y6 =
(∞)
(∞)
(∞)
(∞)
(∞)
x4 = x6 = y1 = y3 = y5 = 0.4747.
THEOREM 4.3. Suppose n = 2m, where m ≥ 3. For t ≥ 0, let

2J I2
U

..
 I2 2J
. 0
0


..
.
.
.
.
.
.
.
.
.
.
bt = (1 − 4γt )I2m + γt 
C


.
.
..
.. I

0
0
2


I2 2J
V
U † 0 · · · 0 V † −2I2
(∞)
0.3323 and x2











2m×2m
.
=
124
Synchronization of Strongly Coupled Dynamical Networks
bt < 1/Γ, then every solution of (2) is {x1 , x3 , . . . , x2m−1 , y1 , y3 , . . . ,
If lim supt→∞ ρ C
y2m−1 } and {x2 , x4 , . . . , x2m, y2 , y4 , . . . , y2m } synchronized.
As in the proof of Theorem 4.2, we may first show that

(t)
(t)
f x1 − f y3

 (t+1)
(t+1) 

(t)
(t)
x1
− y3
 f y1 − f x3
 y(t+1) − x(t+1) 

 1


(t)
(t)
3
 (t+1)
 f x2m − f y4
(t+1) 
 x2m − y4


 (t+1)

(t)
(t)
(t+1) 
 y

 f y2m − f x4
−
x
4
 2m



 = Cbt 
···


 · · ·  x(t+1) − y(t+1) 

(t)
(t)
 m+3
 f xm+3 − f ym+1
m+1 
 (t+1)
 (t+1) 
 ym+3 − xm+1


(t)
(t)
 (t+1)

 f ym+3 − f xm+1
(t+1)

 x2

− y2

(t)
(t)
(t+1)
(t+1)
f x2 − f y2

xm+2 − ym+2
 (t)
(t)
f xm+2 − f ym+2











.










Figure 4
Figure 5
Then the condition lim supt→∞ ρ Cbt < 1/Γ shows that every solution is {x1 , y3 },
{x3 , y1 } , {x2 , y2 } , {xm+2 , ym+2 } , {x2m+4−j , yj }, and {xj , y2m+4−j } synchronized for
125
S. S. Cheng and Y. F. Wu
j = 4, 5, . . . , m+1. Then the rotation invariance of (2) shows further that every solution
of (2) is {x2 , y4 } , {x4 , y2 } , {x1 , y5 } , {x5 , y1 } , {x3 , y3 } , {xm+3 , ym+3 } , {x2m+6−j , yj } ,
and {xj , y2m+6−j } synchronized for j = 6, 7, . . ., m + 2. By carefully inspecting the
connections, we may then conclude the proof of our theorem. These are illustrated in
Figures 4 and 5.
THEOREM 4.4. Suppose n = 2m, where m ≥ 3. For t ≥ 0, let




b t = (1 − 4γt )I2m + γt 
D




bt
If lim supt→∞ ρ D
y2m } synchronized.
−I2 + 2J
I2
I2
2J
..
.
0
..
.
..
.
..
.

0
..
.
2J
I2
I2
−I2 + 2J








.
(32)
2m×2m
< 1/Γ, then every solution of (2) is {x1 , . . . , x2m} and {y1 , . . . ,
As in the proof of Theorem 4.2, we may first show that

(t)
(t)
f x1 − f x2
 (t+1)

(t+1)

x1
− x2

(t)
(t)
 (t+1)
 f y1 − f y2
(t+1) 
 y1


− y2
 (t+1)

(t)
(t)
(t+1) 
f
x
−
f
x
 x


 2m − x3


2m
3  y(t+1) − y(t+1)  = D
bt 
(t)
(t)
 2m
 f y2m − f y3

3



···



 (t+1)
 · · · (t+1) 

 xm+2 − xm+1 
(t)
(t)
f
x
−
f
x

(t+1)
(t+1)
 m+2 m+1
ym+2 − ym+1
(t)
(t)
f ym+2 − f ym+1








.







b t < 1/Γ shows that every solution is {x1 , x2 },
Then the condition lim supt→∞ ρ D
{y1 , y2 }, {x2m+3−j , xj }, and {y2m+3−j , yj } synchronized for j = 3, 4, . . . , m + 1. Then
the rotation invariance of (2) shows further that every solution of (2) is {x2 .x3 } ,
{y2 .y3 } , {x1 .x4 } , {y1 .y4 } , {xj , x2m+5−j } and {yj , y2m+5−j } synchronized for j =
5, 6, . . . , m + 3. By carefully inspecting the connections, we may then conclude the
proof of our theorem. These are illustrated in Figure 6.
THEOREM 4.5. Suppose n = 2m, where m ≥ 3. For t ≥ 0, let

J

 I2

bt = (1 − 4γt )I2m + γt 
E




I2
2J
..
.
0
..
.
..
.
..
.
0
..
.
2J
I2





.



I2 
J
2m×2m
126
Synchronization of Strongly Coupled Dynamical Networks
Figure 6
bt < 1/Γ, then every solution of (2) is {x1 , x3 , . . . , x2m−1, y2 , y4 , . . . , y2m }
If lim supt→∞ ρ E
and {x2 , x4 , . . . , x2m , y1 , y3 , . . . , y2m−1 } synchronized.
As in the proof of Theorem 4.2, we may first show that


(t)
(t)
f x1 − f y2
 (t+1)

(t+1)


x1
− y2


(t)
(t)

 (t+1)
 f y1 − f x2
(t+1) 
 y1


− x2


 (t+1)


(t)
(t)
(t+1) 

 x
 f x2m − f y3

 2m − y3


bt 
.
 y(t+1) − x(t+1)  = E
(t)
(t)

 2m

 f y2m − f x3
3




···




·
·
·

 (t+1)


(t+1)


(t)
(t)
 xm+2 − ym+1 
f
x
−
f
y


(t+1)
(t+1)
m+1 
 m+2
ym+2 − xm+1
(t)
(t)
f ym+2 − f xm+1
bt < 1/Γ shows that every solution is {x1 , y2 },
Then the condition lim supt→∞ ρ E
{x2 , y1 }, {x2m+3−j , yj }, and {xj , y2m+3−j } synchronized for j = 3, 4, . . ., m + 1. Then
the rotation invariance of (2) shows further that every solution of (2) is {x2 , y3 } ,
{x3 , y2 } , {x1 , y4 } , {x4 , y1 } , {x2m+5−j , yj } , and {xj , y2m+5−j } synchronized for j =
5, 6, . . . , m+2. By carefully inspecting the connections, we may then conclude the proof
of our theorem. These are illustrated in Figures 7 and 8.
In the above result, we see that there are two groups of synchronized units. These
groups may or may not be distinct. For example, consider a neural network where n =
6. We consider the special case where γt = γ for all t, f is the tent map function,
and the
bt < 1/Γ
Lipschitz constant Γ = 2. From the Appendix, the condition lim supt→∞ ρ B
in Theorem 4.2 can be replaced by 1/6 < γ < 3/14. Then choosing γ = 0.21 and
(0)
(0)
(0) (0)
(0)
(0) (0) (0) (0) (0) (0) (0)
x1 , x2 , x3 , x4 , x5 , x6 , y1 , y2 , y3 , y4 , y5 , y6
=
1
(2, 1, 2, 1, 2, 1, 3, 5, 3, 5, 3, 5),
10
(∞)
(∞)
(∞)
(∞)
we may compute x1 = x3 = x5 = y2
(∞)
(∞)
(∞)
(∞)
(∞)
x4 = x6 = y1 = y3 = y5 = 0.5167.
(∞)
= y4
(∞)
= y6
(∞)
= 0.8105 and x2
=
127
S. S. Cheng and Y. F. Wu
Figure 7
Figure 8
THEOREM 4.6. Suppose n = 2m, where m ≥ 3. For t ≥ 0, let

2J

 I2


b
Ft = (1 − 4γt )I2m + γt 



−I2
I2
2J
..
.
0
−I2
..
.
..
.
..
.
0
..
.
2J
I2
I2
2J









.
(33)
2m×2m
If lim supt→∞ ρ Fbt < 1/Γ, then every solution of (2) is {xj , xm+j } and {yj , ym+j }
synchronized for j = 1, 2, . . . , m.
128
Synchronization of Strongly Coupled Dynamical Networks
Figure 9
As in the proof of Theorem 4.2, we may first show that












(t+1)
(t+1)
x1
− xm+1
(t+1)
(t+1)
y1
− ym+1
(t+1)
(t+1)
x2
− xm+2
(t+1)
(t+1)
y2
− ym+2
···
(t+1)
(t+1)
xm − x2m
(t+1)
(t+1)
ym
− y2m


(t)
(t)
f x1 − f xm+1
(t)
(t)
f y1 − f ym+1
(t)
(t)
f x2 − f xm+2
(t)
(t)
f y2 − f ym+2
· · · 











 = Fbt 









(t)
(t)

 f xm − f x2m

(t)
(t)
f ym − f y2m








,







and employing the Banach contracting technique to conclude our proof.
In Figure 9, we consider a neural network where n = 6. To illustrate Theorem 4.6,
we draw dash lines connecting x1 and x4 , y1 and y4 , etc. to show synchronization. The
rotational invariance of (2), however, leads to no new information.
As an example, we consider a neural network where n = 6. We consider the special
case where γt = γ for all t, f is the tent map function,
the Lipschitz constant
and
Γ = 2. From the Appendix, the condition lim supt→∞ ρ Fbt < 1/Γ in Theorem 4.6 is
replaced by 1/8 < γ < 3/16. By choosing γ = 0.16 and
=
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(∞)
= x5
(0)
x1 , x2 , x3 , x4 , x5 , x6 , y1 , y2 , y3 , y4 , y5 , y6
1
(2, 1, 6, 2, 1, 6, 7, 4, 5, 7, 4, 5),
10
(∞)
(∞)
(∞)
(∞)
we may compute x1 = x4 = y1 = y4 = 0.3409, x2
(∞)
(∞)
(∞)
(∞)
0.3651, x3 = x6 = y3 = y6 = 0.3351.
(∞)
(∞)
= y2
(∞)
= y5
=
129
S. S. Cheng and Y. F. Wu
THEOREM 4.7. Suppose n = 2m, where m ≥ 3. For t ≥ 0, let

2J
I2

 I2


b
Gt = (1 − 4γt )I2m + γt 



2J
..
.
0
−J
−J
..
.
..
.
..
.
0
..
.
2J
I2
I2
2J









.
2m×2m
b t < 1/Γ, then every solution of (2) is {xj , ym+j } and {xm+j , yj }
If lim supt→∞ ρ G
synchronized for j = 1, 2, . . . , m.
This follows easily from the fact that












(t+1)
(t+1)
x1
− ym+1
(t+1)
(t+1)
y1
− xm+1
(t+1)
(t+1)
x2
− ym+2
(t+1)
(t+1)
y2
− xm+2
···
(t+1)
(t+1)
xm − y2m
(t+1)
(t+1)
ym
− x2m














=G
bt 












(t)
(t)
f x1 − f ym+1
(t)
(t)
f y1 − f xm+1
(t)
(t)
f x2 − f ym+2
(t)
(t)
f y2 − f xm+2
· · · (t)
(t)
f xm − f y2m
(t)
(t)
f ym − f x2m








.







In Figure 10, we consider a neural network where n = 6. To illustrate Theorem 4.7,
we draw dash lines connecting x1 and y4 , x2 and y5 , etc. to show synchronization. The
rotation invariance of (2), however, leads to no new information.
Figure 10
130
Synchronization of Strongly Coupled Dynamical Networks
THEOREM 4.8. Suppose n = 2m + 1, where m ≥ 2. For t ≥ 0, let

2J

 I2

b t = (1 − 4γt )I2m + γt 
H





I2
2J
..
.
0
..
.
..
.
..
.
0
..
.
2J
I2
I2
−I2 + 2J








.
(34)
2m×2m
b t < 1/Γ, then every solution of (2) is {x1 , x2 , . . . , x2m+1 } and
If lim supt→∞ ρ H
{y1 , y2 , . . . , y2m+1 } synchronized.
As in the proof of Theorem 4.2, we may first show that












(t+1)
(t+1)
x1
− x3
(t+1)
(t+1)
y1
− y3
(t+1)
(t+1)
x2m+1 − x4
(t+1)
(t+1)
y2m+1 − y4
···
(t+1)
(t+1)
xm+3 − xm+2
(t+1)
(t+1)
ym+3 − ym+2














=H
bt 












(t)
(t)
f x1 − f x3
(t)
(t)
f y1 − f y3
(t)
(t)
f x2m+1 − f x4
(t)
(t)
f y2m+1 − f y4
· · · (t)
(t)
f xm+3 − f xm+2
(t)
(t)
f ym+3 − f ym+2








.







b t < 1/Γ implies that every solution is {x1 , x3 },
Then the condition lim supt→∞ ρ H
{y1 , y3 }, and {x2m+5−j , xj }, {x2m+5−j , xj } , where j = 4, 5, . . . , m + 2, synchronized.
The rotational invariance of (2) shows further that every solution of (2) is {x2 , x4 },
{y2 , y4 }, {x1 , x5 }, {y1 , y5 }, {x2m+7−j , xj }, and {y2m+7−j , yj } synchronized for j =
6, 7, . . . , m + 3. Then careful inspection of the connections completes our proof. These
are illustrated in Figure 11.
Figure 11
131
S. S. Cheng and Y. F. Wu
THEOREM 4.9. Suppose n = 2m + 1, where

2J
I2

..
 I2 2J
.


.
.
..
..
b t = (1 − 4γt )I2m+1 + γt 
K


..

.
0


−b
u† 0 · · ·
m ≥ 2. For t ≥ 0, let

−b
u

0
0 

.. 
..
.
. 
.

.. 
2J I2
. 

I2 J
0 
· · · 0 −2 (2m+1)×(2m+1)
b t < 1/Γ, then every solution of (2) is (fully) synchronized.
If lim supt→∞ ρ K
As in the proof of Theorem 4.2, we may first show that

(t)
(t)
f
x
−
f
y
 (t+1)


1 3 (t+1)

x1
− y3
(t)
(t)
 f y1 − f x3
 (t+1)
(t+1) 

− x3
 y1


(t)
(t)
 (t+1)
(t+1) 
f
x
−
f
y

 x2m+1 − y4

 2m+1
4  (t+1)


(t)
(t)
(t+1) 
 y
f y2m+1 − f x4
bt 
 2m+1 − x4
=K




···


 · · ·  (t+1)
(t+1) 

(t)
(t)
x
−
y
 m+3
m+2 
f
x
−
f
y

 (t+1)
(t+1) 
 m+3
m+2  ym+3 − xm+2


(t)
(t)
 f ym+3 − f xm+2
(t+1)
(t+1)

x2
− y2
(t)
(t)
f x2 − f y2










.








b t < 1/Γ shows that every solution of (2) is
Then the condition lim supt→∞ ρ K
{x1 , y3 }, {x3 , y1 }, {x2 , y2 } , {x2m+5−j , yj }, and {xj , y2m+5−j } synchronized for j =
4, 5, . . . , m + 2. The rotation invariance of (2) then shows that every solution of (2) is
{x2 , y4 }, {y2 , x4 }, {x1 , y5 }, {y1 , x5 }, {x3 , y3 } , {x2m+7−j , yj }, and {xj , y2m+7−j } synchronized for j = 6, 7, . . . , m + 3. Careful inspection of the connections then completes
our proof. These are illustrated in Figure 12.
Figure 12
132
5
Synchronization of Strongly Coupled Dynamical Networks
Sharpness
We will now consider the special case where γt = γ for all t, f is the identity function,
and the Lipschitz constant Γ = 1.
• Since
8
if n is even
2kπ
max 6 − 2 cos
=
,
1≤k≤n
n
6 − 2 cos (n−1)π
if n is odd
n
bt < 1/Γ in Theorem 4.1 is replaced by
the condition lim supt→∞ ρ A
0<γ<
2
6 − 2 cos
max1≤k≤n
=
2kπ
n
(
1/4
1/ 3 − cos
(n−1)π
n
if n is even
.
if n is odd
First note that this condition is sharp when n is even. Indeed, take f to be the
identity function. When γ = 1/4, (28) becomes
(t+1)
xi
(t+1)
− yi
=
n X
−1
k=1
1
In + An
2
4
(t)
(t)
xk − yk
ik
for i = 1, 2, . . ., n.
1
The number −1 is an eigenvalue of −1
2 In + 4 An and the corresponding eigenvector
†
is u
e = (−1, 1, −1, . . ., 1) . If we let
(0)
(0)
(0)
(0)
x2k = y2k−1 = 1, x2k−1 = y2k = 0 for k = 1, 2, . . ., n/2,
then since


(t)
(t)
x1 − y1


···

 =
(t)
(t)
xn − yn
=
we see that
n
(t)
(t)
o∞
−1
1
In + An
2
4
t
−1
1
In + An
2
4
t


(0)
(0)
x1 − y1


···


(0)
(0)
xn − yn
t
u
e = (−1) u
e,
does not converge to zero for i = 1, . . . , n. Next,
when n is odd, this condition is also sharp since if γ = 1/ 3 − cos (n−1)π
, the
n
number −1 is an eigenvalue of the matrix and the corresponding eigenvector is
ve =
c1 cos
xi − yi
t=0
(n − 1) π
(n − 1) π
+ c2 sin
, . . . , c1 cos (n − 1) π + c2 sin (n − 1) π
n
n
†
,
where c1 and c2 are not both equal to zero. If we take f to be the identity
function and let
(0)
xk = c1 cos
(n − 1) kπ (0)
(n − 1) kπ
, yk = −c2 sin
, for k = 1, . . . , n,
n
n
133
S. S. Cheng and Y. F. Wu
then since
we see that
• Since
n
(t)
(t)
xi − yi
o∞
t=0
3 − 2 cos
bt
A
t
t
v = (−1) e
e
v,
does not converge to zero for i = 1, . . . , n.
(m − 1) π
(m − 1) π
< 6 − 2 cos
,
m
m
we have
kπ
kπ
(m − 1) π
6 − 2 cos
, 2 − 2 cos
= 6 − 2 cos
.
1≤k≤m−1
m
m
m
bt < 1/Γ in Theorem 4.2 is replaced by
Thus, the condition lim supt→∞ ρ B
0 < γ < 1/ 3 − cos (m−1)π
. This condition is sharp since when γ is equal to
m
bt and the
, the number −1 is an eigenvalue of the matrix B
1/ 3 − cos (m−1)π
m
corresponding eigenvector is
max
2
u
e=
(m − 1) π †
2 (m − 1) π †
(m − 1) π †
sin
u
b , sin
u
b , . . . , sin
u
b
m
m
m
!†
.
If we take f to be the identity function and let
(0)
y1 = sin
(m − 1) π
(0)
(0)
(0)
(0)
, x1 = y3 = x2m+4−k = yk = 0, k = 4, 5, . . . , m − 1,
m
(0)
xk = sin
(k − 2) (m − 1) π
, k = 3, 4, . . . , m + 1,
m
and
(0)
y2m+2−k = sin
then since

(t)
(t)
x1 − x3

 y1(t) − y3(t)

 x(t) − x(t)
2m
4

 y(t) − y(t)

2m
4

···

 (t)
(t)
 xm+3 − xm+1
(t)
(t)
ym+3 − ym+1

(m − 1) kπ
, k = 2, 3, . . . , m − 1,
m







 
t

= B
bt 










(0)
(0)
x1 − x3
(0)
(0)
y1 − y3
(0)
(0)
x2m − x4
(0)
(0)
y2m − y4
···
(0)
(0)
xm+3 − xm+1
(0)
(0)
ym+3 − ym+1




 t

t
= B
bt u
e = (−1) u
e,





n
o∞ n
o∞ n
o∞
(t)
(t)
(t)
(t)
(t)
(t)
we see that neither x1 − x3
, y1 − y3
, xj − x2m+4−j
t=0
t=0
t=0
n
o∞
(t)
(t)
converges to zero for j = 4, 5, . . . , m + 1.
nor yj − y2m+4−j
t=0
134
Synchronization of Strongly Coupled Dynamical Networks
• Since
2 − 2 cos
(m − 1) π
(m − 1) π
< 4 < 6 − 2 cos
< 8,
m
m
we have
kπ
kπ
4, 8, 6 − 2 cos
, 2 − 2 cos
= 8.
1≤k≤m−1
m
m
bt < 1/Γ in Theorem 4.3 is replaced by
Thus, the condition lim supt→∞ ρ C
0 < γ < 1/4. This condition is sharp since when γ = 1/4, the number −1 is an
bt and the corresponding eigenvector is
eigenvalue of the matrix C
(
†
−b
u† , u
b†, . . . , −b
u† , u
b†, u
b†
if m is odd
u
e=
.
†
†
†
†
† † †
u
b , −b
u , . . ., u
b , −b
u ,u
b ,b
v
if m is even
max
If we take f to be the identity function and let
(
(0)
(0)
(0)
(0)
x2k−1 = y2k = 1, x2k = y2k−1 = 0 for k = 1, . . . , m if m is odd
,
(0)
(0)
(0)
(0)
x2k = y2k−1 = 1, x2k−1 = y2k = 0 for k = 1, . . . , m if m is even
then since

(t)
(t)
x1 − y3
 y(t) − x(t)

1
3

(t)
(t)
 x2m
− y4

 y(t) − x(t)
2m
4


···

 x(t) − y(t)
 m+3
m+1
 (t)
 ym+3 − x(t)
m+1

 x2(t) − y2(t)
(t)
(t)
xm+2 − ym+2












 
t

= C
bt 














(0)
(0)
x1 − y3
(0)
(0)
y1 − x3
(0)
(0)
x2m − y4
(0)
(0)
y2m − x4
···
(0)
(0)
xm+3 − ym+1
(0)
(0)
ym+3 − xm+1
(0)
(0)
x2 − y2
(0)
(0)
xm+2 − ym+2






 t

t
 = Cbt u
e = (−1) u
e,







o∞
o∞ n
o∞ n
n
(t)
(t)
(t)
(t)
(t)
(t)
, nor
, x2 − y2
, x3 − y1
we see that neither x1 − y3
t=0
t=0
t=0
n
o∞ n
o∞ n
o∞
(t)
(t)
(t)
(t)
(t)
(t)
xm+2 − ym+2
, x2m+4−j − yj
xj − y2m+4−j
, where j =
t=0
t=0
t=0
4, 5, . . . , m + 1, can converge to zero.
• Since
kπ
kπ
, 2 − 2 cos
= max {8, 4} = 8,
6 − 2 cos
1≤k≤m
m
m
b t < 1/Γ in Theorem 4.4 is replaced by
Thus, the condition lim supt→∞ ρ D
0 < γ < 1/4. This condition is sharp since when γ = 1/4, the number −1 is an
b t and the corresponding eigenvector is
eigenvalue of the matrix D
max
†
u
e= u
b† , −b
u† , u
b†, −b
u† , . . . , (−1)m+1 u
b† .
135
S. S. Cheng and Y. F. Wu
If we take f to be the identity function and let
(0)
(0)
(0)


(0)
x2k = y2k−1 = 1, x2k−1 = y2k = 0 for k = 1, . . . , m,
then since

(t)
(t)
x1 − x2

(t)
(t)
 y1 − y2

 x(t) − x(t)
2m
3

 y(t) − y(t)

2m
3

···

 (t)
(t)
 xm+2 − xm+1
(t)
(t)
ym+2 − ym+1






 
t

bt 
= D










(0)
(0)
x1 − x2
(0)
(0)
y1 − y2
(0)
(0)
x2m − x3
(0)
(0)
y2m − y3
···
(0)
(0)
xm+2 − xm+1
(0)
(0)
ym+2 − ym+1




 t

t
bt u
= D
e = (−1) u
e,





n
o∞ n
o∞ n
o∞
(t)
(t)
(t)
(t)
(t)
(t)
we see that neither x1 − x2
, y1 − y2
, x2m+3−j − xj
t=0
t=0
t=0
n
o∞
(t)
(t)
nor y2m+3−j − yj
can converge to zero for j = 3, 4, . . ., m + 1.
t=0
• Since
4 < 6 − 2 cos
(m − 1) π
< 8,
m
we have
(k − 1) π
kπ
(m − 1) π
max 6 − 2 cos
, 2 − 2 cos
= 6 − 2 cos
.
1≤k≤m
m
m
m
bt < 1/Γ in Theorem 4.5 is replaced by
Thus, the condition lim supt→∞ ρ E
(m−1)π
. This condition is sharp since when γ is equal to
0 < γ < 1/ 3 − cos m
(m−1)π
bt and the
, the number −1 is an eigenvalue of the matrix E
1/ 3 − cos m
corresponding eigenvector is
u
e=
(m − 1) π †
3 (m − 1) π †
(2m − 1) (m − 1) π †
cos
u
b , cos
u
b , . . . , cos
u
b
m
m
m
†
If we take f to be the identity function and let
(0)
(0)
−x1 = y1 = cos
(0)
(m − 1) π
3 (m − 1) π
(0)
(0)
, − x2m = y2m = cos
,
m
m
(0)
−x2m+2−k = y2m+2−k = cos
(m − 1) (2k − 1) π
, for k = 3, 4, . . . , m,
m
and
(0)
(0)
xk = yk = 0, for k = 2, 3, . . . , m + 1,
.
136
Synchronization of Strongly Coupled Dynamical Networks
then since




(0)
(0)
(t)
(t)
x1 − y2
x1 − y2




 y1(t) − x(t)

 y1(0) − x(0)

2
2




 x(t) − y(t)   x(0) − y(0)  2m
3
2m
3
t
t



t
 y(t) − x(t)  = E
bt  y(0) − x(0)  = E
bt u
e = (−1) u
e,




2m
3
2m
3




···
···




 (t)

 (0)

(t)
(0)
 xm+2 − ym+1 
 xm+2 − ym+1 
(t)
(t)
(0)
(0)
ym+2 − xm+1
ym+2 − xm+1
n
o∞ n
o∞ n
o∞
(t)
(t)
(t)
(t)
(t)
(t)
we see that neither x1 − y2
, x2 − y1
, x2m+3−j − yj
t=0
t=0
t=0
n
o∞
(t)
(t)
nor xj − y2m+3−j
converges to zero for j = 3, 4, . . . , m + 1.
t=0
• Since
8
(2k − 1) π
max 6 − 2 cos
=
1≤k≤m
m
6 − 2 cos (m−1)π
m
if m is odd
,
if m is even
4
(2k − 1) π
max 2 − 2 cos
=
1≤k≤m
2 − 2 cos (m−1)π
m
m
if m is odd
,
if m is even
and
we have
(2k − 1) π
(2k − 1) π
, 2 − 2 cos
6 − 2 cos
= 8.
1≤k≤m
m
m
Thus, the condition lim supt→∞ ρ Fbt < 1/Γ in Theorem 4.6 is replaced by
0 < γ < 1/4. This condition is sharp since when γ = 1/4, the number −1 is an
eigenvalue of the matrix Fbt and the corresponding eigenvector is
†
u
e = −b
u† , u
b† , . . . , −b
u† , u
b† , −b
u† .
max
If we take f to be the identity function and let
(0)
(0)
(0)
(0)
x2k−1 = y2k = 1, x2k = y2k−1 = 0, for k = 1, 2, . . . , m,
then since
 (t)
(t)
x1 − xm+1
 (t)
(t)
 y1 − ym+1
 (t)
 x − x(t)
m+2
 2
 y(t) − y(t)
 2
m+2

···

 (t)
(t)
 xm − x2m
(t)
(t)
ym − y2m








 
t

 = Fbt 










(0)
(0)
x1 − xm+1
(0)
(0)
y1 − ym+1
(0)
(0)
x2 − xm+2
(0)
(0)
y2 − ym+2
···
(0)
(0)
xm − x2m
(0)
(0)
ym − y2m




 t

t
 = Fbt u
e = (−1) u
e,





137
S. S. Cheng and Y. F. Wu
we see that neither
for j = 1, 2, . . . , m.
n
(t)
(t)
xj − xm+j
o∞
t=0
nor
n
(t)
(t)
yj − ym+j
o∞
t=0
converges to zero
• Since
max
1≤k≤m
6 − 2 cos
2kπ
m
=
6 − 2 cos (m−1)π
m
8
if m is odd
if m is even
and
4
(2k − 1) π
max 2 − 2 cos
=
1≤k≤m
m
2 − 2 cos (m−1)π
m
if m is odd
,
if m is even
we have
2kπ
(2k − 1) π
max 6 − 2 cos
, 2 − 2 cos
= 8.
1≤k≤m
m
m
b t < 1/Γ in Theorem 4.7 is replaced by
Thus, the condition lim supt→∞ ρ G
0 < γ < 1/4. This condition is sharp since when γ = 1/4, the number −1 is an
b t and the corresponding eigenvector is
eigenvalue of the matrix G
u
e = −b
u† , u
b†, . . . , −b
u† , u
b†
If we take f to be the identity function and let
(0)
(0)
(0)
†
.
(0)
x2k−1 = y2k = 1, x2k = y2k−1 = 0, for k = 1, 2, . . . , m,
then since
 (t)
(t)
x1 − ym+1
 (t)
(t)
 y1 − xm+1
 (t)
 x − y(t)
m+2
 2
 y(t) − x(t)
 2
m+2

···

 (t)
(t)
 xm − y2m
(t)
(t)
ym − x2m
we see that neither
for j = 1, 2, . . . , m.
n








 
t

= G
bt 










(t)
(t)
xj − ym+j
• Since
2 − 2 cos
(0)
(0)
x1 − ym+1
(0)
(0)
y1 − xm+1
(0)
(0)
x2 − ym+2
(0)
(0)
y2 − xm+2
···
(0)
(0)
xm − y2m
(0)
(0)
ym − x2m
o∞
t=0
nor
n




 t

t
= G
bt u
e = (−1) u
e,





(t)
(t)
xm+j − yj
o∞
t=0
converges to zero
2mπ
2mπ
< 6 − 2 cos
,
2m + 1
2m + 1
we have
max 6 − 2 cos
1≤k≤m
2kπ
2kπ
, 2 − 2 cos
2m + 1
2m + 1
= 6 − 2 cos
2mπ
.
2m + 1
138
Synchronization of Strongly Coupled Dynamical Networks
b t < 1/Γ in Theorem 4.8 is replaced by 0 <
Thus, the condition lim supt→∞ ρ H
2mπ
2mπ
γ < 1/ 3 − cos 2m+1
. This condition is sharp since when γ = 1/ 3 − cos 2m+1
,
b
the number −1 is an eigenvalue of the matrix Ht and the corresponding eigenvector is
†
2mπ †
4mπ †
2m2 π †
u
e = sin
u
b , sin
u
b , . . . , sin
u
b
.
2m + 1
2m + 1
2m + 1
If we take f to be the identity function and let
y1 = sin
(0)
2mπ
4mπ
, y2m+1 = sin
,
2m + 1
2m + 1
xk+2 = sin
2mkπ
, for k = 1, 2, . . . , m,
2m + 1
and
(0)
y2m+3−k = sin
2mkπ
, for k = 3, 4, . . . , m,
2m + 1
then since




(0)
(0)
(t)
(t)
x1 − x3
x1 − x3




(0)
(0)
(t)
(t)
 y1 − y3

 y1 − y3




 (t)
(t) 
 x(0) − x(0)   x
−
x
4
2m+1
4
t
t


 2m+1
t
 y(t) − y(t)  = H
b t  y(0) − y(0)  = H
bt u
e = (−1) u
e,
 2m+1

 2m+1

4
4




···
···




 (t)

 (0)

(0)
(t)
 xm+3 − xm+2 
 xm+3 − xm+2 
(t)
(t)
(0)
(0)
ym+3 − ym+2
ym+3 − ym+2
n
o∞ n
o∞ n
o∞
(t)
(t)
(t)
(t)
(t)
(t)
we see that neither x1 − x3
, y1 − y3
, x2m+5−j − xj
t=0
t=0
t=0
n
o∞
(t)
(t)
nor y2m+5−j − yj
converges to zero for j = 4, 5, . . . , m + 2.
t=0
• Since
2 − 2 cos
we have
2kπ
2kπ
2mπ
, 4, 2 − 2 cos
= 6 − 2 cos
.
1≤k≤m
2m + 1
2m + 1
2m + 1
b t < 1/Γ in Theorem 4.9 is replaced by 0 <
Thus, the condition lim supt→∞ ρ K
2mπ
2mπ
γ < 1/ 3 − cos 2m+1
. This condition is sharp since when γ = 1/ 3 − cos 2m+1
,
b t and the corresponding eigenthe number −1 is an eigenvalue of the matrix K
vector is
†
2mπ †
4mπ †
2m2 π †
u
e = − cos
u
b , − cos
u
b , . . . , − cos
u
b ,1 .
2m + 1
2m + 1
2m + 1
max
6 − 2 cos
2mπ
2mπ
< 4 < 6 − 2 cos
,
2m + 1
2m + 1
139
S. S. Cheng and Y. F. Wu
If we take f to be the identity function and let
(0)
(0)
x1 = −y1 = cos
2mπ
4mπ
(0)
(0)
(0)
(0)
, x
= −y2m+1 = cos
, x = 1, y2 = 0,
2m + 1 2m+1
2m + 1 2
(0)
2mkπ
, for k = 3, 4, . . . , m,
2m + 1
(0)
x2m+3−k = −y2m+3−k = cos
and
(0)
(0)
xk = yk = 0, for k = 3, 4, . . . , m + 2,
then since

(t)
(t)
x1 − y3

(t)
(t)
 y1 − x3
 (t)
 x2m+1 − y4(t)
 (t)
(t)
 y
 2m+1 − x4

···

 (t)
(t)
x
 m+3 − ym+2
 (t)
(t)
 ym+3 − xm+2
(t)
(t)
x2 − y2










 
t

= K
bt 












(0)
(0)
x1 − y3
(0)
(0)
y1 − x3
(0)
(0)
x2m+1 − y4
(0)
(0)
y2m+1 − x4
···
(0)
(0)
xm+3 − ym+2
(0)
(0)
ym+3 − xm+2
(0)
(0)
x2 − y2





 t

t
= K
bt u
e = (−1) u
e,






n
o∞ n
o∞ n
o∞
(t)
(t)
(t)
(t)
(t)
(t)
, x3 − y1
, x2 − y2
nor
we see that neither x1 − y3
t=0
t=0
t=0
n
o∞ n
o∞
(t)
(t)
(t)
(t)
, xj − y2m+5−j
, where j = 4, 5, . . . , m + 2, can
x2m+5−j − yj
t=0
t=0
converge to zero.
We conclude our investigation with the following remark: Suppose n = 2m where
m ≥ 3. When 0 < γ < 1/4, every solution of (2) is (fully) synchronized; and when 1/4 ≤
γ < 2/7, every solution of (2) is {x1 , x3 , . . . , x2m−1 , y2 , y4 , . . . , y2m } and {x2 , x4 , . . . , x2m ,
y1 , y3 , . . . , y2m−1 } synchronized. But in general, (2) is not {x1 , x3 , . . . , x2m−1, y1 , y3 , . . . ,
y2m−1 }, {x2 , x4 , . . . , x2m ,ny2 , y4 , . . . , yo
2m}, {x1 , x2 , . . . , x2m}, nor {y1 , y2 , . . . , y2m } syn(t)
(t)
xi − yi
chronized. In general,
does not converge to 0 for all i = 1, . . . , n;
n
o
n
o
(t)
(t)
(t)
(t)
xj − xm+j does not converge to 0 for all j = 1, . . . , m (if m is odd); yj − ym+j
n
o
(t)
(t)
does not converge to zero for all j = 1, 2, . . . , m (if m is odd); xj − ym+j does not
o
n
(t)
(t)
does not converge
converge to zero for j = 1, 2, . . ., m (if m is even); xm+j − yj
to zero for j = 1, 2, . . . , m (if
which n = 6:
γ = 0.25
t=0
t=1
t=2
t = 10
t = 100
m is even). We give some data for a neural network in
(t)
x1
1
2
2
2
2
(t)
x2
3
3
3
3
3
(t)
x3
1
2
2
2
2
(t)
x4
3
3
3
3
3
(t)
x5
1
2
2
2
2
(t)
x6
3
3
3
3
3
140
Synchronization of Strongly Coupled Dynamical Networks
γ = 0.25
t=0
t=1
t=2
t = 10
t = 100
(∞)
(t)
(t)
y1
1
3
3
3
3
(t)
y2
5
2
2
2
2
(∞)
y3
1
3
3
3
3
(∞)
(t)
y4
5
2
2
2
2
(∞)
(t)
y5
1
3
3
3
3
(t)
y6
5
2
2
2
2
(∞)
(∞)
(∞)
This is an example for x1 = x3 = x5 = x2 = x4 = x6 = 2 and x2 =
(∞)
(∞)
(∞)
(∞)
(∞)
(∞)
(∞)
x4
= x6
= x1
= x3
= x5
= 3. But xi
6= yi
for i = 1, 2, . . ., 6,
(∞)
(∞)
(∞)
(∞)
xj 6= x3+j , and yj 6= y3+j for j = 1, 2, 3.
If n = 2m +1, where m ≥ 2, every solution of (2) is synchronized if 0 < γ <
2mπ
1/ 3 − cos 2m+1
. If we let m → ∞, then whether n is even or odd, when 0 < γ < 1/4,
every solution of (2) is (fully) synchronized. Furthermore, the conditions are sharp.
6
Appendix
We collect here the eigenvalues and eigenvectors of matrices used in the previous discussions. They can be verified in a straightforward manner and details can be found
in [6]. We denote the n by n identity matrix by In , J, U and V are respectively
0 1
1 0
0 1
J=
, U =
and V =
.
1 0
−1 0
0 −1
The matrix J has eigenvalues −1 and 1 with the corresponding (independent) eigenvec†
†
tors u
b = (−1, 1) and vb = (1, 1) respectively. The matrix U has eigenvalue 0 and 1 with
corresponding eigenvectors (0, 1)† and u
b respectively. The matrix V has eigenvalues 0
and −1 with eigenvectors (1, 0)† and u
b respectively. The matrix U † has eigenvalues 0
and 1 with corresponding eigenvectors b
v and (1, 0)† respectively. The matrix V † has
eigenvalues 0 and −1 with corresponding eigenvectors vb and (0, 1)† respectively.
• First of all, the eigenvalues and the

0 1 0
 1 0 1

 0 1 0

An =  . . .
 .. .. ..

 0 0 0
1 0 0
corresponding eigenvectors of the matrix

··· 0 1
··· 0 0 

··· 0 0 

, n ≥ 3,
. . 
..
. .. .. 

··· 0 1 
· · · 1 0 n×n
are well known [3, 4, 5] and given respectively by
λk (An ) = 2 cos
2kπ
, k = 1, 2, . . . , n
n
†
(k)
(k)
and u(k) = u1 , . . . , un
, 1 ≤ k ≤ n, where
(k)
uj
= c1 cos
2jkπ
2jkπ
+ c2 sin
, j = 1, 2, . . . , n,
n
n
141
S. S. Cheng and Y. F. Wu
where c1 and c2 are not both equal to zero.
• The matrix

2J
I2

 I2



0
0
2J
..
.
..
.
..
.
I2
2J
I2
has the eigenvalues
−2 + 2 cos






, m ≥ 3,
(2m−2)×(2m−2)
kπ
kπ
and 2 + 2 cos
, k = 1, . . . , m − 1
m
m
with the corresponding eigenvectors
†
†
(k) †
(k)
(k) †
(k)
u(k) = u1 u
b , . . . , um−1 u
b†
and v(k) = v1 b
v , . . . , vm−1 vb†
respectively, where
(k)
uj
(k)
= vj
= sin
kjπ
, j = 1, . . . , m − 1,
m
for k = 1, . . . , m − 1.
• The matrix

2J

 I2








U†
I2
2J
..
.

U
..
.
..
.
0
..
.
.
..
0
..
0
···
.
I2
0
0
..
.
I2
2J
V†
0
V
−2I2










, m ≥ 3,
2m×2m
has the eigenvalues
−2 + 2 cos
kπ
kπ
and 2 cos
, k = 1, . . . , m
m
m
with the corresponding eigenvectors
 †

(k)
†
†
†
 u(k)u
b
,
.
.
.
,
u
u
b
,
u
b
1
m−1
u(k) =
†
(k)

 u(k)
b† , . . . , um−1 u
b† , b
v†
1 u
and
if m is odd
, 1 ≤ k ≤ m,
if m is even
†
(k)
(k)
v(k) = v1 vb† , . . . , vm−1 b
v† , 0b
v† , 1 ≤ k ≤ m − 1,
142
Synchronization of Strongly Coupled Dynamical Networks
v(m) = −b
u† , . . . , −b
u† , b
v†
respectively, where
(k)
uj
=
cos kjπ
m
− cos kjπ
m
†
if m is odd
, j = 1, . . . , m,
if m is even
for k = 1, 2, . . ., m, and
(k)
vj
= sin
kjπ
, j = 1, . . . , m − 1,
m
for k = 1, 2, . . ., m − 1.
• Then the matrix

−I2 + 2J


I2







I2
2J
..
.
0
..
.
..
.
..
.
0
..
.
2J
I2
I2
−I2 + 2J
has the eigenvalues
−2 + 2 cos








, m ≥ 3,
2m×2m
kπ
kπ
and 2 + 2 cos
, k = 1, . . . , m
m
m
with the corresponding eigenvectors
†
†
(k) †
(k)
(k) †
u(k) = u1 u
b , . . . , u(k)
b†
and v(k) = v1 vb† , . . . , vm
vb
m u
respectively, where
(k)
uj
(k)
= vj
= sin
k (2j − 1) π
, j = 1, . . . , m,
2m
for k = 1, 2, . . ., m.
• The matrix

J

 I2






has the eigenvalues
−2 + 2 cos
I2
2J
..
.
0
..
.
..
.
..
.
0
..
.
2J
I2





, m ≥ 3,



I2 
J
2m×2m
(k − 1) π
kπ
and 2 + 2 cos
, k = 1, . . . , m,
m
m
143
S. S. Cheng and Y. F. Wu
with the corresponding eigenvectors
†
†
(k) †
(k) †
(k) †
u(k) = u1 u
b , . . . , u(k)
b†
and v(k) = v1 b
v , . . . , vm
v
b
,
m u
where
(k)
uj
= cos
k (2j − 1) π
(k − 1) (2j − 1) π
(k)
and vj = sin
, j = 1, . . . , m
m
m
for k = 1, 2, . . ., m.
• The matrix

2J

 I2






−I2
I2
2J
..
.
0
−I2
..
.
..
.
..
.
0
..
.
2J
I2
I2
2J
has the eigenvalues
−2 + 2 cos









, m ≥ 3,
2m×2m
(2k − 1) π
(2k − 1) π
and 2 + 2 cos
, k = 1, . . . , m
m
m
with the corresponding eigenvectors
†
†
(k) †
(k)
(k) †
u(k) = u1 u
b , . . . , u(k)
b†
and v(k) = v1 vb† , . . . , vm
vb
, 1≤k≤m
m u
where
(k)
uj
(k)
= vj
= c1 cos
(2k − 1) jπ
(2k − 1) jπ
+ c2 sin
, j = 1, . . . , m,
m
m
where c1 and c2 are not both equal to zero.
• The matrix

2J

 I2






−J
has the eigenvalues
−2 + 2 cos
I2
2J
..
.
0
−J
..
.
..
.
..
.
0
..
.
2J
I2
I2
2J









, m ≥ 3,
2m×2m
2kπ
(2k − 1) π
and 2 + 2 cos
, k = 1, . . . , m
m
m
with the corresponding eigenvectors
†
†
(k) †
(k)
(k) †
u(k) = u1 u
b , . . . , u(k)
b†
and v(k) = v1 vb† , . . . , vm
vb
, 1≤k≤m
m u
144
Synchronization of Strongly Coupled Dynamical Networks
where
(k)
uj
= c1 cos
2jkπ
2jkπ
+ c2 sin
, j = 1, . . . , m,
m
m
and
(2k − 1) jπ
(2k − 1) jπ
+ c2 sin
, j = 1, ..., m,
m
m
where c1 and c2 are not both equal to zero.
(k)
vj
= c1 cos
• The matrix

2J

I2

 I2






2J
..
.
0
..
.
..
.
..
.
0
..
.
2J
I2
I2
−I2 + 2J








, m ≥ 2,
2m×2m
has eigenvalues
−2 + 2 cos
2kπ
2kπ
and 2 + 2 cos
, k = 1, . . . , m
2m + 1
2m + 1
with the corresponding eigenvectors
†
†
(k) †
(k)
(k) †
u(k) = u1 u
b , . . . , u(k)
b†
and v(k) = v1 vb† , . . . , vm
vb
, 1≤k≤m
m u
where
(k)
uj
(k)
= vj
= sin
2jkπ
, j = 1, . . . , m,
2m + 1
for k = 1, 2, ..., m.
• The matrix

2J

 I2








−b
u†
I2
−b
u
..
.
..
.
0
..
.
0
···
2J
..
.
0
..
.
2J
I2
···
has eigenvalues
−2 + 2 cos
and
2 + 2 cos
I2
J
0
0
..
.
..
.
0
−2











(2m+1)×(2m+1)
2kπ
, k = 0, 1, . . . , m,
2m + 1
2kπ
, k = 1, . . . , m,
2m + 1
, m ≥ 2,
145
S. S. Cheng and Y. F. Wu
with the corresponding eigenvectors
†
(k) †
†
u(k) = u1 u
b , . . . , u(k)
u
b
,
1
, 0 ≤ k ≤ m,
m
and
where
(k)
uj
†
(k)
(k) †
v(k) = v1 vb† , . . . , vm
vb , 0 , 1 ≤ k ≤ m
= − cos
2kjπ
2kjπ
(k)
and vj = sin
, j = 1, . . . , m.
2m + 1
2m + 1
References
[1] S. S. Cheng, C. J. Tian and M. Gil’, Synchronization in a discrete circular network,
Proceedings of the Sixth International Conference on Difference Equations, 61-73,
CRC, Boca Raton, FL, 2004.
[2] W. C. Yueh and S. S. Cheng, Synchronization in an artificial neural network, Chaos
Solutions & Fractals, 30(2006), 734–747.
[3] W. C. Yueh and S. S. Cheng, Explicit eigenvalues and inverse of several Toeplitz
matrices, Anziam J., 48(2006) 73-97.
[4] R. T. Gregory and D. Karney, A Collection of Matrices for Testing Computational
Algorithms, Wiley-Interscience, New York, 1969.
[5] W. C. Yueh, Eigenvalues of several tridiagonal matrices, Appl. Math. E-Notes,
5(2005), 66–4.
[6] Y. F. Wu, Synchronization in a class of strongly coupled discrete time neural networks, Master Thesis, Tsing Hua University, 2007.