Multiple steady state solutions of a circular neural network

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Chaos, Solitons and Fractals 37 (2008) 1228–1232 www.elsevier.com/locate/chaos

Multiple steady state solutions of a circular neural network Gen-Qiang Wang a, Sui Sun Cheng a

b,*

Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, PR China b Department of Mathematics, Tsing Hua University, Hsinchu 30043, Taiwan, ROC Accepted 9 October 2006

Abstract A three term discrete system arising from seeking steady state solutions of a discrete time neural network is considered and multiple periodic solutions are found by means of Clark’s theorem in the critical point theory.  2006 Elsevier Ltd. All rights reserved.

1. Introduction To motivate what follows, let us consider x neuron units placed on the vertices of a regular x-polygon. Let xðtÞ n denote the state value of the nth neuron unit during the time period t 2 {0, 1, 2, . . .}. Assume that each neuron unit is activated by its two neighbors so that the change of state values between two consecutive time periods is given by ðtÞ

ðtÞ

ðtÞ xðtþ1Þ  xðtÞ n n ¼ xn1 þ xnþ1  f ðn; xn Þ;

where f stands for the bias mechanism inherent in the nth neuron unit. Then we have an evolutionary system (discrete time dynamical system DTDS) of the form ðtÞ

ðtÞ

ðtÞ  xðtÞ xðtþ1Þ n n ¼ xn1 þ xnþ1  f ðn; xn Þ; ðtÞ x0 ðtÞ x1

¼ ¼

n 2 f1; . . . ; xg;

xðtÞ x ; ðtÞ xxþ1

for t = 0, 1, 2, . . .. In order to understand the dynamics n of the neural o1network modeled by the above evolutionary sysðtÞ y tem, it is of interest to seek ‘steady state’ solutions ðx1 ; . . . ; xðtÞ Þ such that xðtÞ x n ¼ xn for n 2 {1, . . . , x} and t P 0. t¼0

One of the reasons is that these steady state solutions correspond to the fixed points of the DTDS and fixed points are candidates for storing information facets. Replacing xðtÞ n by xn in the DTDS leads us to solutions of the steady state system: xn1 þ xnþ1  f ðn; xn Þ ¼ 0;

n 2 f1; . . . ; xg;

x0 ¼ xx ; x1 ¼ xxþ1 ; *

Corresponding author. Tel.: +88 635713784; fax: +88 635723888. E-mail address: [email protected] (S.S. Cheng).

0960-0779/$ - see front matter  2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2006.10.024

G.-Q. Wang, S.S. Cheng / Chaos, Solitons and Fractals 37 (2008) 1228–1232

1229

or equivalently, x-periodic solutions fxn g1 n¼1 of xn1 þ xnþ1  f ðn; xn Þ ¼ 0;

n 2 Z ¼ f0; 1; 2; . . .g:

In [2], a slightly more general system of the form X nþ1 þ X n1  f ðn; X n Þ ¼ 0;

n2Z

ð1Þ  

k

k

is studied where Xn is a k-vector for each n 2 Z (instead of a scalar variable), and f = (f1, f2, . . . , fk) 2 C(Z · R , R ) such that there is a fixed positive integer x such that f(n + x, U) = f(n, U) for all (n, U) 2 Z · Rk. As usual, a solution {Xn}n2Z of (1) is a real vector sequence that renders (1) into an identity after substitution. It is said to be x-periodic if Xn+x = Xn for all n 2 Z. Under the assumption that there exists a continuously differentiable function F 2 C1(R · Rk, R) such that $UF(n, U) = f(n, U) and F(n + x, U) = F(n, U) for all (n, U) 2 R · Rk, where $U denotes the gradient operator in U, sufficient conditions for (1) to have least one nontrivial x-periodic solution of (1) is found by means of a mountain pass theorem in the critical point theory. However, for artificial neural networks to be useful in real applications, it is of great interest to find more than one steady state solutions. For this reason, in this paper, we consider (1) again under the same condition on f and provide sufficient conditions for the existence of 2kx x-periodic solutions of (1). Our technique is based on Clark’s theorem in the critical point theory instead of the mountain pass theorem, and hence this study is independent from the results in [2]. For general information on critical point theory, we refer to [13,14]. For additional information on discrete systems, we refer to [1–12]. In particular, we remark that in [1], the authors consider the discrete time second order dynamical systems X nþ1  2X n þ X n1  gðn; X n Þ ¼ 0;  

k

n 2 Z;

ð2Þ

k

k

where g = (g1, g2, . . . , gk) 2 C(Z · R , R ) and g(n + x, U) = g(n, U) for any (n, U) 2 Z · R and some positive integer x P 3. Note that if we let I: Rk ! Rk be the identity map and let f(n, Xn) = g(n, Xn)  2I(Xn), then (1) and (2) seem to be ‘equivalent’. However, as pointed out in [2], an existence condition for (2) in terms of g may not be applicable to yield existence of periodic solutions of (1). Also, we will see later (see Example 1) that our conditions may guarantee more periodic solutions.

2. Existence criteria The main result of this paper is the following. Theorem 1. Suppose F(n, U) P 0 for any (n, U) 2 Z · Rk and F(n, U) is even in U. Suppose further that (G1) there are constants d > 0 and a 2 (0, 1) such that for any n 2 Z and U 2 Rk satisfying jUj 6 d, we have F(n, U) 6 ajUj2 and (G2) there are constants q > 0, c > 0 and b 2 (1, 1) such that for any n 2 Z and U 2 Rk satisfyingjUj P q, we have F(n, U) P bjUj2  c. Then (1) possesses at least 2kx nontrivial x-periodic solutions. Corollary 1. Suppose F(n, U) P 0 for any (n, U) 2 Z · Rk and F(n, U) is even in U. Suppose further that (G3) F(n, U) = o(jUj2) as U ! 0 and (G4) there are constants R1 > 0 and a1 > 2 such that for any n 2 Z and U 2 Rk satisfying jUj P R1, hU ; rU F ðn; U Þi P a1 F ðn; UÞ > 0: Then (1) possesses at least 2kx nontrivial x-periodic solutions. Corollary 2. Suppose k = 1 and for each n 2 Z;

Rx

Rx (G5) 0 f ðn; tÞ dt P 0 for any n 2 Z and x 2 R, (G6) f(n, x) = o(x) as x ! 0 and

0

f ðn; uÞ du is even in x. Suppose further that

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(G7) there are constants R1 > 0 and a1 > 2 such that for any n 2 Z and x 2 R satisfying jxj P R1, Z x f ðn; uÞ du > 0: xf ðn; xÞ P a1 0

Then (1) possesses at least 2x nontrivial x-periodic solutions. Before turning to the proof of our results, let us first show that our results are truly different from those in [1]. Example 1. Let k = 1 and x = 3. Set f ðn; xÞ ¼ 2að2 þ sin 23 npÞðx  sin xÞ where a > 1, and set F ðn; xÞ ¼ 2 2að2 þ sin 23 npÞðx2 þ cos x  1Þ: Then $xF(n, x) = f(n, x), F(n, x) P 0 for any (n, x) 2 Z · R and F(n, x) is even in x . It is easy to see from limx!0 maxn2RjF(t, x)/x2j = 0 that (G1) is satisfied. Since F ðn; xÞ ¼ 2að2 þ sin 23 npÞ 2 ðx2 þ cos x  1Þ P ax2  6a for any (n, x) 2 Z · R, if we take q > 0, b = a and c = 6a, then (G2) is satisfied. Thus by Theorem 1, (1) possesses at least 6 nontrivial 3-periodic solutions. It is easy to see that in this case if we use the main theorem in [1], then we may only infer 2 nontrivial 3-periodic solutions at best. Example 2. Consider a scalar equation of the form xnþ1 þ xn1 

24 ðxn Þ3 ¼ 0; n 5

2 Z:

ð3Þ

We assert that (3) possesses at least 8 nontrivial 4-periodic solutions. Indeed, here f ðn; xÞ ¼ 245 x3 and F ðn; xÞ ¼ 65 x4 : Then $xF(n, x) = f(n, x), F(n, x) P 0 for any (n, x) 2 Z · R and F(n, x) is even in x. It is easy to see that limx!0 maxn2ZjF(n, x)/ x2j = 0 and that (G1) is satisfied. If we take q > 1, b = 6/5 and c > 0, then (G2) is satisfied. Thus by Theorem 1, (3) possesses at least 8 nontrivial 4-periodic solutions. However, if we use the main theorem in [2], then we may only infer at most 1 nontrivial 4-periodic solution at best. Actually, if we rewrite (3) in the form (2), then gðn; xÞ ¼ 245 x3  2x and the conditions of the main Theorem in [1] are not satisfied. Indeed, otherwise there is G, such that $UG(n, U) = g(n, U), G(n, x) P 0 for 23 and G(n, 0) = 0 for n 2 Z. Thus Gðn; xÞ ¼ 65 x4  x2 ; which is contrary to the fact that G(n, x) P 0 for n 2 Z and x 2 R. The spirit of the proof of Theorem 1 is similar to that of the main theorem in [1,2] but we use Clark’s theorem instead of a linking theorem or a mountain pass theorem. For the sake of completeness, we give Pa complete proof as follows. First, for any U = (U1, . . . , Uk) , V = (V1, . . . , Vk)  2 Rk, their inner product is hU ; V i ¼ ki¼1 U i V i and the norm of V is jVj = hV, Vi1/2. Let S be the set of all real vector sequences X = {Xn}n2Z where Xn = (Xn1, Xn2, . . . , Xnk)  2 Rk. For any X, Y 2 S and a, b 2 R, aX + bY is defined by aX + bY = {aXn + bYn}n2Z. Then S is a linear space. Let Ex be the set of all x-periodic vector sequences in S. When endowed with the norm k  kEx and inner product h; iEx defined by !1=2 x X 2 jX n j ; ð4Þ kX kEx ¼ n¼1

and hX ; Y iEx ¼

x X hX n ; Y n i n¼1

for any X = {Xn}n2Z, Y = {Yn}n2Z in Ex, the pair ðEx ; h; iEx Þ is a Hilbert space. We now formulate our problem as a critical point problem. Consider the functional I defined on Ex by IðX Þ ¼

x X fX nþ1 X n þ F ðn; X n Þg;

X 2 Ex :

ð5Þ

n¼1

Since Ex is linearly homeomorphic to Rxk, I can be viewed as a continuously differentiable functional defined on the finite dimensional Hilbert space Rxk by taking X0 = Xx and Xx+1 = X1. In particular, the Frechet derivative I 0 (X) is Þ zero if, and only if, oIðX ¼ 0 for all n 2 {1, . . . , x} and l 2 {1, . . . , k}. Since oX n;l oIðX Þ ¼ fX n;l  X n1;l þ fl ðn; X n Þg; oX n;l

1 6 n 6 x; 1 6 l 6 k:

ð6Þ

We see that I 0 (X) = 0 if, and only if, X nþ1 þ X n1  f ðn; X n Þ ¼ 0;

n 2 f1; . . . ; xg: 0

That is, X 2 Ex is a critical point of I (i.e. I (X) = 0) if, and only if, X is an x-periodic solution of (1).

ð7Þ

G.-Q. Wang, S.S. Cheng / Chaos, Solitons and Fractals 37 (2008) 1228–1232

1231

Let H be a real Banach space. A continuously differentiable functional J 2 C1(H, R) is said to satisfy the Palais– Smale condition (P–S condition) if any sequence {un}  H for which {J(un)} is bounded and J 0 (un) ! 0 as n ! 1 possesses a convergent subsequence in H. Lemma 1 [14, Clark’s Theorem 9.1]. Let E be a real Banach space and I a functional in C1(E, R) which is even, bounded from below, and satisfies the P–S condition. Suppose further that I(0) = 0, there is a set K  E such that K is homeomorphic to Sj1 by an odd map, and supk I < 0. Then I possesses at least j distinct pairs of nontrivial critical points. Lemma 2. If (G2) holds, then the functional I defined by (5) is bounded from below in Ex. Proof. According to (G2), if we let n o  c1 ¼ max F ðn; U Þ  bjU j2 þ c : n 2 Z; jU j 6 q ; and c 0 = c + c1. Then for any n 2 Z and U 2 Rk, we have F ðn; U Þ P bjU j2  c0 :

ð8Þ

By (5) and Cauchy’s inequality we see that for any X 2 Ex, I ðX Þ ¼

x X

fX nþ1 X n þ F ðn; X n Þg P 

n¼1

x X

jX nþ1 X n j þ

n¼1

¼ kX k2Ex þ

x  X

x X

F ðn; X n Þ P kX k2Ex þ

n¼1

x  X

bjX n j2  c0



n¼1



bjX n j2  c0 ¼ ðb  1ÞkX k2Ex  c0 x:

n¼1

Since b > 1 by (G2), for any X 2 Ex, we have I(X) P  c 0 x. The proof is complete.

h

Lemma 3. If (G2) holds, then the functional I defined by (5) satisfies the P–S condition. Proof. Let fIðX ðiÞ Þg1 i¼1 be a sequence bounded from above, that is, there exists a positive constant M such that  ðiÞ  6 M; i ¼ 1; 2; . . . I X In view of the proof of Lemma 2, it is easy to see that  2   M P I X ðiÞ P ðb  1ÞX ðiÞ Ex  c0 x; which implies that  ðiÞ 2 X  6 ðb  1Þ1 ðM þ c0 xÞ: Ex That is, {X(i)} is a bounded sequence in the finite dimensional space Ex. Hence {X(i)} has a convergent subsequence. The proof is complete. h Proof of Theorem 1. We now turn to the proof of Theorem 1. It suffices to find 2kx nontrivial critical points of the functional I defined by (5). First of all, by (G1) and the fact that F(n, U) P 0 for any (n, U) 2 Z · Rk, we see that F(n, 0) = 0 for n 2 Z. It follows that I(0) = 0. Since F(n, U) is even in U, we know that I 2 C1(E, R) and is even. Furthermore, by Lemmas 2 and 3, I is bounded from below in Ex and satisfies the P–S condition. Let K ¼ fX 2 Ex jkX kEx ¼ dg  Ex . It is easy to see from dim Ex = kx that K is homeomorphic to Skx1 by an odd map, and that for any X 2 K satisfying jX jEx ¼ d, I ðX Þ ¼

x X

fX nþ1 X n þ F ðn; X n Þg 6 kX k2Ex þ

n¼1

þa

x X

x X

F ðn; X n Þ 6 kX k2Ex

n¼1 2

j X n j ¼  ð1 

aÞkX k2Ex

¼ r < 0;

n¼1

where r = (1  a)d2 > 0. Thus by Lemma 1, we know that I possesses at least 2kx nontrivial critical points. That is, (1) possesses at least 2kx nontrivial x-periodic solution. The proof is complete. h

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G.-Q. Wang, S.S. Cheng / Chaos, Solitons and Fractals 37 (2008) 1228–1232

Proof of Corollary 1. We now turn to the proof of Corollary 1. It is easy to see that if (G3) holds, then the condition (G1) of Theorem 1 is true. From (G4), we have  U rU F ðn; U Þ a1 ; P ð9Þ for n 2 Z and jU j P R1 : jU j F ðn; U Þ jU j Thus d ln F ðn; U Þ a1 ; P djU j jU j which implies d ðln F ðn; U Þ  a1 ln jU jÞ P 0 for n 2 Z d jU j

and

jU j P R1 :

ð10Þ

Let / ¼ min fln F ðn; U Þ  a1 ln jU j : n 2 Z and jU j P R1 g: By (9), ln F ðn; U Þ  a1 ln jU j P / for n 2 Z

and jU j P R1 :

That is F ðn; U Þ P b1 jU ja1

for n 2 Z

and

jU j P R1 ;

where b1 = exp(/). Let q1 P R1 satisfying b1 qa11 2 > 1: Then for n 2 Z and jUj P q1, F ðn; U Þ P b1 jU ja1 2 jU j2 P b; where b ¼ b1 qa11 2 > 1: Thus the condition (G2) of Theorem 1 holds. The proof is complete. h Rx Finally, the proof of Corollary 2 follows from Corollary 1 by taking F ðt; xÞ ¼ 0 f ðt; xÞ dx:

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