International Journal of Bifurcation and Chaos, Vol. 14, No. 7 (2004) 2223–2228 c
World Scientific Publishing Company
PIECEWISE TWO-DIMENSIONAL MAPS AND
APPLICATIONS TO CELLULAR NEURAL NETWORKS
HSIN-MEI CHANG and JONG JUANG
Department of Applied Mathematics, National Chiao Tung University, Hsin-Chu 30050, Taiwan
Received March 13, 2003; Revised September 9, 2003
Of concern is a two-dimensional map T of the form T (x, y) = (y, F (y) − bx). Here F is a three-piece linear map. In this paper, we first prove a theorem which states that a semiconjugate condition for T implies the existence of Smale horseshoe. Second, the theorem is applied to show the spatial chaos of one-dimensional Cellular Neural Networks. We improve a result of Hsu [2000].
Keywords: Cellular Neural Networks; Smale horseshoe; piecewise two-dimensional map.
1. Introduction
We consider a piecewise two-dimensional map of the form T (x, y) = (y, F (y) − bx) , (1) where F (y) = a1y + a0− a1+ c1 y ≥ 1, a0y + c1 |y| ≤ 1, a−1y + a−1− a0+ c1 y ≤ −1. (2) Here a0 < 0, a1, a−1 > 1, b > 0, and c1 ∈ R is a biased term. The graph of F is given in Fig. 1.
The motivation for studying such a map is, in part, due to the form of the map is a gen-eralized version of Lozi map [Lozi, 1978]. More importantly, the map arises in the study of com-plexity of a set of bounded stable stationary solu-tions of one-dimensional Cellular Neural Networks (CNNs) (see e.g. [Chua, 1998; Chua & Yang, 1998a, 1998b]). In this paper, we first prove a theorem which states that a semiconjugate condition for T implies the existence of Smale horseshoe. Second, we apply the theorem to show the spatial chaos of one-dimensional Cellular Neural Networks. Such CNNs are of the form (e.g. [Ban et al., 2002, 2001;
Hsu, 2000]). dxi
dt = −xi+ z + αf (xi−1) + af (xi)
+ βf (xi+1) , i ∈ Z (3a) where f (x) is a piecewise-linear output function defined by f (x) = rx + 1 − r x ≥ 1 x |x| ≤ 1 lx + l − 1 x ≤ −1, (3b)
where r and l are positive constants. The quantity z is called threshold or bias term, related to indepen-dent voltage sources in electric circuits. The con-stants α, a and β are the interaction weights be-tween neighboring cells. The study of problems for the case of r = l = 0 and α = β has been estab-lished in [Chua, 1998; Chua & Yang, 1998a; Juang & Lin, 2000]. Here we consider r > 0 and l > 0. Then the main results are the following. Given α and β, if (z, a) is in a certain parameter region Σα,β (see Theorem 3.1), then there exist r and l suffi-ciently small for which Λl,r (see Theorem 3.1) is a hyperbolic invariant set. Consequently, the spatial entropy of the corresponding set of bounded, stable stationary solutions is ln 2.
2223
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y F( )y
Fig. 1. a1= 1.2, a0= −0.5, a−1= 1.5, c1= 0.2.
2. Main Results
We first introduce some notations. Let
S = {(x, y) ∈ R2: |x| ≤ p, |y| ≤ p} . (4) Here p > 1. Let the four corners of S be labeled as
K = (p, p) , L = (p, −p) ,
M = (−p, −p) , N = (−p, p) . (5a) Set
K = (p, 1) , L = (p, −1) ,
M = (−p, −1) , N = (−p, 1) . (5b) The x and y coordinates of K are denoted, respec-tively, by Kx and Ky.
We next number the following conditions. K1y ≥ p > 1 , (6a)
Ny1 ≤ −p , (6b)
Ly1 ≥ p , (6c)
and
M1y≤ −p . (6d)
Here the subscript denotes the iteration index un-der the map T . For instance, K1y denotes the y co-ordinate of T (K) = K1. Suppose (6) holds. Then T (S) ∩ S has three vertical strips. See Fig. 2. Sim-ilarly, T−1
(S)T S has three horizontal strips, and T−1(S)T S T T (S) has 9 components. By induction Tn
j=−n Tj(S) has 9n components. With this infor-mation we can define a semiconjugate
h : Λ → {0, 1, 2}2 (7) which is onto. Here Λ =T∞
j=−∞(Tj(S)T S). If the components of Λ are points, then Λ is a Cantor set.
N1 K1 K( p, p) K( P, 1) L( p, -1) L( p, -p) N1 K1 L1 M1 N(-p, p) N(-p, 1) M(-p, -1) M(-p, -p) M1 L1 V1 U1 S1 Fig. 2.
This, in turn, implies that the semiconjugacy h is one to one and so is a conjugacy. This motivates the following definition.
Definition 1.1. Conditions on b, a−1, a0, and a1 so that there exists a p > 1 for which (6) holds are called a semiconjugate condition for T .
To prove the main theorem, we need to intro-duce more notations. Now, T (S)T S, has three ver-tical strips, say S1, U1and V1. The one on the right, see Fig. 2, is labeled as S1. Clearly, T (S1)T S also has three vertical strips. The strip of T (S1)T S1 is to be denoted by S2. We then define Sninductively. Note that Sn, n ∈ N, are all parallelograms. Us and Vn are defined similarly.
The parallelogram N1K1K1N1, see Fig. 2, is to be denoted by S1. Likewise, Sndenotes the parallel-ogram NnKnKnNn. The length of the shorter side of the parallelogram Sn (resp. Sn) is to be denoted by
dn(resp. cn) . (8a) The slope of the longer side of the parallelogram Sn is to be denoted by
mn. (8b)
Lemma 2.1. The following recursive relations
hold.
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c
i LK
( , )
p p
l
i LK
( , )
p 1
K
iK
id
i Ld
i l Fig. 3. (i) di = mcii, ci+1= bdi, (ii) mi+1= a1−mb i, m1 = a1.Proof. The first recursive relation is obvious. To see (ii), let li be given as in Fig. 3. We then see that Ki = (p − (li/mi), p) and Ki = (p − (li + p − 1)/mi, 1). Now, the slope mi+1 = the slope of T (Ki)T (Ki) = Ki+1Ki+1 = F (p) − F (1) + b((1 − p)/mi)/(p − 1) = a1− (b/mi).
Lemma 2.2. If b > 0 and a1 ≥ 2(1 + b), then
limn→∞ cn= 0.
Proof. We first prove that limn→∞ mn = (a1 + pa2
1− 4b)/2. To this end, we see that an induc-tion would yield that mi ≥ 1 for all i ∈ N and that mi is decreasing in i. Suppose x is the limit of {mn}. Then x must satisfy equation x = a1− (b/x). Upon using the fact that m1 = a1, we conclude that x = a1+pa21− 4b/2 as asserted. Now, using Lemma 2.1(i), we get dn= bn−1d1/Qni=2mi. Thus,
dn≤ 2b a1+pa21− 4b !n−1 d1 ≤ 2b a1 n−1 d1 ≤ b 1 + b n−1 d1.
We have just completed the proof of the lemma. Similarly, we have the following lemma.
Lemma 2.3. If b > 0 and a−1 > 2(1 + b), then
the length of the shorter side of the parallelogram Vn shrinks to zero as n → ∞.
Using Lemmas 2.2 and 2.3, we have the follow-ing lemma.
Lemma 2.4. If b > 0, min{a1, a−1} > 2(1 + b), then the length of the shorter side of the parallelo-gram Un shrinks to zero as n → ∞.
Remark. The assumptions on Lemmas 2.2–2.4 would also yield thatT−∞
j=0(Tj(S)T S) are pairwise disjoint horizontal line segments.
We are now ready to state our main results.
Theorem 2.1. Let F be a piecewise linear map
defined as in (2) and the bias term c1 satisfy the inequality
max{−1 − b, a0+ 1 + b}
< c1< min{1 + b, −a0− 1 − b} , (9) then a semiconjugate condition for T implies the conjugate of h.
Proof. Note that K1y ≥ p, (6b) and (6d) are equiv-alent to the following inequalities.
p(a1− 1 − b) ≥ a1− a0− c1, (10a) −a0+ c1 ≥ p(1 + b) , (10b) −a0− c1≥ p(1 + b) , (10c) and
p(a−1− 1 − b) ≥ a−1− a0+ c1, (10d) respectively. We remark (10b) and (10c) to ensure that −a0 − 1 − b > 0, as a result, inequality (9) makes sense. Using (10a) and (10b), we see imme-diately that −a0+ c1 b + 1 ≥ p ≥ a1− a0− c1 a1− b − 1 . (11)
Note that a1 − b − 1 being positive is guaranteed by the fact that p > 1 and the assumptions on c1. Using (10), we get that
a1 ≥ −2a0(b + 1) c1− a0− 1 − b = 2(b + 1) 1 +1 + b − c1 a0 ≥ 2(b + 1) . (12a)
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The last inequality is justified by the assumptions on c1. Similarly, we see that
a−1≥ 2a0(b + 1) c1+ a0+ 1 + b = 2(b + 1) 1+1 + b + c1 a0 ≥ 2(b + 1) . (12b) It then follows from Lemmas 2.2–2.4 that T∞
j=−∞(Tj(S)T S) is a Cantor set. We thus com-plete the proof of the main theorem.
Remarks
(1) If F (y), as defined in 2, is such that a0 > 0, and a1, a−1< −1, then a similar result can also be obtained.
(2) The theorem holds true in general for F being a finitely many piecewise linear map. Specifically, if the bias term c1 is not “too biased”, then a semiconjugate condition for T implies the exis-tence of Smale horseshoe.
In the following, we give conditions on a0, a1, a−1, b and c for which T has a semiconjugate condition.
Theorem 2.2. Let a0 < 0, a1, a−1 > 1 and b > 0. Suppose a0+ 1 + b < 0, min{a1, a−1} > 2(1 + b). Let the bias term c1 satisfy (9), and that
a1 ≥ −2a0(b + 1) c1− a0− 1 − b (13a) and a−1 ≥ 2a0(b + 1) c1+ a0+ 1 + b . (13b)
then there exists a p > 1 such that T has a semi-conjugate condition.
3. Applications to CNNs
A basic and important class of solutions of (1) is the bounded, stable stationary solutions. In the case that r = l = 0 and α = β, the corresponding stable stationary solutions have been studied in [Chua & Yang, 1998a; Juang & Lin, 2000]. The case that r and l are positive is considered in [Ban et al., 2002, 2001; Hsu, 2000]. The techniques in these two cases are quite different. Specifically, in the latter case, the question of complexity of a set of stable station-ary solutions is converted to asking how chaotic is a map. If α or β = 0, then the resulting map is one-dimensional [Ban et al., 2002, 2001]. If α, β 6= 0,
then the resulting map is a two-dimensional of the following form [Hsu, 2000]
T (x, y) = y, 1 β (F (y) − ay − z) − α βx
=: (y, F (y) − bx) . (14a) Here, F (y) = 1 r y − 1 r + 1 y ≥ 1 y |y| ≤ 1 1 l y − 1 + 1 l y ≤ −1. (14b)
Hsu [2000] used a theorem of Afraimovich (see e.g. [Afraimovich, 1993]) as well as a semicon-jugate condition to show that in certain param-eters’ region, the map T has Smale horseshoe structure. However, Afraimovich’s Theorem is not needed in this case. Only a semiconjugate condition is required.
To apply Theorem 2.2, we first note that a−1= 1 β( 1 l − a), a0 = 1 β(1 − a), a1 = 1 β( 1 r − a), c1 = −z β , b = αβ. With the above identifications, we immedi-ately have the following results concerning the com-plexity of a set of bounded, stable stationary mosaic solutions of (3). Here the stationary mosaic solu-tions (xi)∞i=−∞ means that (xi)∞i=−∞ is a stationary solution of (3) and that |xi| > 1 for all i ∈ Z. More-over, the mosaic solutions obtained in the following theorem are bounded and stable (see e.g. [Chua & Yang, 1998a; Hsu, 2000]).
Define s = α + a + β. Assume the bias term z satisfies the following inequality.
max{−s + a, s − 2a + 1}
< z < min{s − a, 2a − 1 − s} . (15) Define, respectively, the regions Σα,β and Σα,β,l,r as follows. Σα,β = {(z, a) ∈ R2| (15) holds} , (16) and Σα,β,l,r = {(z, a) ∈ R2|r < r+, and l < l+} . (17) Here, r+z,α,a,β = 2a − s − 1 − z a(1 + s − z) − 2s, (18a) and l+z,α,a,β = 2a − s − 1 + z a(1 + s + z) − 2s. (18b) We are now in a position to state the following results.
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Theorem 3.1. Let α and β be positive numbers
and let a > 1 + α + β. Suppose (z, a) ∈ P α,β. Then there exist r and l sufficiently small, more precisely 0 < r < r+ = r+
z,α,a,β and 0 < l < l+= l+z,α,a,β for which T has a hyperbolic invariant
set Λl,r(z, α, a, β) = Λl,r in the (x, y) plane such that T |Λl,r is topologically conjugate to a two-side Bernoulli shift of two symbols. Hence, the spatial entropy of the corresponding set of stationary solu-tions equals ln 2.
l
1p
0p
0r
-1 z aFig. 4. = 13, l1 : −z + a(1 − 2) = 1, p0 : z = 2a,
r−1: z + a(1 − 2) = 1, p0: z = −2a.
l
1r
-1p
0p
0 z aFig. 5. = 16, l1 : −z + a(1 − 2) = 1, p0 : z = 2a,
r−1: z + a(1 − 2) = 1, p0: z = −2a. (I) (II) (III) (IV) (V) (VI) (VII) (VIII) (IX) Fig. 6.
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Remarks
(1) Note that if (z, a) ∈ Σα,β, then −2s + a(1 + s − z) = a(−z − 1 − s + 2a) + 2(a − 1)(s − a) > 0 and −2s + a(1 + s + z) = a(z − 1 − s + 2a) + 2(a − 1)(s − a) > 0. Consequently, those r+and l+ are positive.
(2) Adapting the notations in [Juang & Lin, 2000] we let α = β = a. Then the set Σα,β = Σ is given in the following figure.
Note that for 0 < ε < 14, Σ ( [3, 3] (see Fig. 5.1 of [Juang & Lin, 2000] for the definition of [3, 3]), and for 14 ≤ < 12, Σ = [3, 3] (see Figs. 4 and 5). Applying Theorem 3.1, we conclude that let α = β = a, 14 ≤ ε < 12, and if (z, a) ∈ Σ= [3, 3], then there exist r and l sufficiently small for which Λl,r is a hyperbolic invariant set. This result gener-alized those in [Chua, 1998; Chua & Yang, 1998a; Juang & Lin, 2000]. For 0 < < 1
4, if (z, a) ∈ Σ and r, l > 0 is sufficiently small, then the corre-sponding set of stable, bounded stationary solutions also has spatial entropy ln 2.
(3) To get a feel of how small r and l are required to be, set = 14 and z = 0. We see easily that r+= l+ has a maximum 1
16 for 2 < a < ∞. (4) Figure 6 is a collection of a computer
sim-ulation with a set of parameters, satisfying a > 1 + α + β, 0 < r < r+ = r+z,α,a,β and 0 < l < l+ = l+
z,α,a,β. Specifically, we choose α = β = 1, r = l = 0.005, z = 0, a = 4. Each collection in Fig. 6 contains two arrays of col-ors. The first array is the initial outputs. The second array represents the final outputs. If the state xj of a cell cj is such that |xj| < 1, then
we color it green. If the state xj of a cell cj is less than −1 (greater than 1, respectively), then we color it blue (red, respectively).
Acknowledgment
We thank Dr. C. J. Yu for providing the simulation work in the paper.
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4. JONQ JUANG, CHIN-LUNG LI, MING-HUANG LIU. 2006. CELLULAR NEURAL NETWORKS: MOSAIC PATTERNS, BIFURCATION AND COMPLEXITY. International Journal of Bifurcation and Chaos 16:01, 47-57. [Abstract] [References] [PDF] [PDF Plus]
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