World Scientific Publishing Company
SYNCHRONIZATION AND ASYNCHRONIZATION IN
A LATTICE OF COUPLED LORENZ-TYPE MAPS
WEN-WEI LIN∗ and SHIH-FENG SHIEH†
Department of Mathematics, National Tsing Hua University, Hsinchu, 300, Taiwan
∗[email protected] †[email protected]
YI-QIAN WANG
Department of Mathematics, Nanjing University, Nanjing, 210008, P. R. China
Received June 30, 2003; Revised March 1, 2005
In this paper, we study synchronization and asynchronization in a Coupled Lorenz-type Map Lattice (CLML). Lorenz-type map forms a chaotic system with an appropriate discontinuous function. We prove that in a CLML with suitable coupling strength, there is a subset of full measure in the phase space such that chaotic synchronization occurs for any orbit starting from this subset and there is a dense subset of measure zero in the phase space such that synchronization will never occur. We also provide numerical observations to explain our results. Keywords: Chaotic synchronization; Lorenz-type map.
1. Introduction
Synchronization is a fundamental phenomenon in Coupled Map Lattices (CMLs). Experimen-tal observation shows that maps manifest similar behavior in discrete time, provided they are cou-pled with suitable coupling strengths and the lat-tice size. The behavior of periodic synchronization in CMLs has been well studied and used in many practical applications (see [Amritkar et al., 1991; Wu & Chua, 1994]). However, one of the most excit-ing recent developments is to study chaotic synchro-nization in CMLs. Since 1990, people found ways to exploit “synchronized chaos” for practical applica-tions in signal processing and secure communica-tion (see e.g. [Cuomo & Oppenheim, 1992, 1993; Heagy, et al., 1995; Pecora & Carroll, 1990; Pec-ora et al., 1997; Vohra et al., 1992; Wu & Chua, 1994]). Thus, the problem of coming up with a rigorous mathematical description of synchronized chaotic behavior of CMLs appears to be attractive
and important from both theoretical and practical points of view.
The simplest type of chaotic synchronization of CMLs occurs in stable spatially homogeneous regimes corresponding to the existence of attractive spatially homogeneous solutions. In other words, in such cases there is a large (open) set of initial con-ditions such that a solution starting from an ini-tial condition in the set becomes spaini-tially homoge-neous as discrete time n becomes very large, i.e. the coordinates of the individual maps become equal to each other n→ ∞. In established regimes, individ-ual maps become indistinguishable and we observe exact perfect synchronization. Thus, it is possible that a suitable coupling strength permits the exis-tence of a spatially homogeneous solution provided all individual maps are identical. Recently, synchro-nization in lattices of maps with various types has been studied in [Jost & Joy, 2002; Lin et al., 1999; Lin et al., 2002; Lin & Wang, 2001].
In the last decade, studies aimed at understand-ing collective behavior of nerve systems have been more concerned with synchronization behavior of neuron ensembles. For example, people found that epilepsy was caused by an abnormal synchronized discharge of cortical neurons in the central nervous system [Sirven & Varrato, 1999]. Biologists have proposed many discontinuous dynamical models to describe neuronal impulses. Some models of cou-pled piecewise continuous maps lattices have also been used for modeling synchronization behavior of neuron ensembles [Andreev & Krasichkov, 2003; Hayakawa & Sawada, 2000; Freeman, 2000]. In this paper, we mainly study the dynamics of Coupled Lorenz-type Map Lattice (CLML) in the following form:
x = (1− c)f(x) + cf(y),
y = cf (x) + (1− c)f(y), (1a) where c is the coupling strength and f (x) is a piecewise linear Lorenz-type map defined by (see Fig. 1) f (x) = l + 2(1− l)x, 0≤ x < 1 2, l + 2(1− l) x−1 2 , 1 2 ≤ x ≤ 1 (1b) with 0≤ l < 1/2.
Lorenz-type map is a class of important dynam-ical systems and has been widely studied (see e.g. [Afraimovich & Hsu, 2002; Malkin, 1989; Milnor & Thurston, 1988]). It can be regarded as a dis-cretized form of the famous Lorenz equation and exhibits similar chaotic behaviors as the Lorenz
Fig. 1. A piecewise linear Lorenz-type map.
equation. In recent years, some results of chaotic synchronization in lattices of coupled Lorenz equa-tions have been obtained (see [Chiu et al., 2000; Lin & Peng, 2002]). However, to our knowledge, synchronization rarely results in CLMLs.
Our numerical experiments show that the behaviors of synchronization and asynchronization in CLMLs are very complex. On the one hand, an orbit starting from a randomly chosen initial point will tend to the spatially homogeneous regime {x = y}; on the other hand, the occurrence of synchronization is not uniform, that is, after any number of iterations, there exists an orbit of a pos-itive distance to the spatially homogeneous regime {x = y}, though it converges to {x = y} as the number of iterations becomes sufficiently large. This kind of synchronization is quite different from the synchronization of examples in that occur uniformly for any initial points.
Let
a = 2(1− l)(1 − 2c), (2)
where c and l are defined in (1a) and (1b), respec-tively. Denote
Sa= the “synchronization region”, (3) i.e. the subset of the phase space {0 ≤ x, y ≤ 1} from which the orginating iterations (1) will syn-chronize. If a ≥ 1, one can easily verify that syn-chronization will not occur. Hence, we only consider the case of 0 ≤ a < 1. In accordance with our numerical observations, in this paper, we will prove the following properties:
(I) For any 0 ≤ a < 1 in (2), there is a dense subset of measure zero in {0 ≤ x, y ≤ 1} such that the system (1) will never synchronize. (II) For any 0 ≤ a < 1 in (2), the measure of
the subset Sa as in (3) for the system (1) is positive.
(III) For any 0≤ a ≤ 1/2 in (2), the synchroniza-tion region Sa as in (3) is of full measure. By the result of (I), it is difficult to use the tradi-tional Liapunov method to prove (II) and (III). Our new proof technique is based on the construction of infinitely many Cantor sets that satisfy: (a) points in any such Cantor set will uniformly synchronize; (b) the “speeds” of synchronization are different for any two such Cantor sets; (c) the union of these Cantor sets is of full measure. For 1/2 < a < 1, numerical results show that synchronization occurs
much more slowly than 0 < a≤ 1/2. A more careful analysis is necessary for this case.
In the above, the result of (II) can be gener-alized to Lorenz-type maps. However, it is difficult to prove the result of (III) for general Lorenz-type maps by using our technique. We will dwell on this in a future paper.
This paper is organized as follows. We prove statements (I)–(III) in Secs. 2–4, respectively. Numerical results are given in the last section. 2. Asynchronization in a Dense Set Let
Φ(x, y) = ((1− c)f(x) + cf(y), cf(x) + (1− c)f(y)),
where f (x) is given in (1b) with l = 0, i.e.
f (x) = 2x, 0≤ x < 1 2, 2 x−1 2 , 1 2 ≤ x ≤ 1. (4)
Note that the proof for the case of 0 < l < 1/2 is similar and omitted below. The iteration (1a) can be written as
(x, y) = Φ(x, y). (5)
The Jacobi matrix for Φ is denoted by J and is equal to J = 2 1− c c c 1− c .
The eigenvalues and their associated eigenvectors of J are{2, 2(1 − 2c)} and {(1, 1), (1, −1)}, respec-tively. It holds that Φ is expanding in the direc-tion (1, 1) and is contracting in the direcdirec-tion (1,−1) provided a = 2(1 − 2c) < 1. Using the change of variables X = (x + y)/2 and Y = (x− y)/2, we thus obtain an equivalent system to (1) (X, Y ) = Ψa(X, Y ), (6) where Ψa(X, Y ) = (2X, aY ), X + Y < 1 2, X− Y < 1 2, (2X− 1, aY ), X + Y > 1 2, X− Y > 1 2, 2X−1 2, aY + a 4 , X + Y < 1 2, X− Y > 1 2, 2X−1 2, aY − a 4 , X + Y > 1 2, X− Y < 1 2. Thus our phase space is
Ω ={(x, y)| 0 ≤ x, y ≤ 1}
={(X, Y )| 0 ≤ X + Y ≤ 1, 0 ≤ X − Y ≤ 1}.
Theorem 2.1. The line {X = 1/2} is invariant under the iteration of Ψa. Moreover for 0≤ a < 1, every point on the line {X = 1/2} tends to the points {(1/2, Y∗), (1/2,−Y∗)} of period 2, where Y∗ = a/(4(a + 1)).
Proof. From the definition of Ψa(X, Y ), for any point (X, Y ) ∈ {X = 1/2} ⊂ {X + Y < 1/2, X − Y > 1/2} ∪ {X + Y > 1/2, X − Y < 1/2}, we have X = (2X − 1/2)|X=1/2 = 1/2. Hence {X = 1/2} is an invariant subspace. Thus the restriction of Ψa onto {X = 1/2} can be reduced to
g(Y ) = aY + a 4, Y ≤ 0, aY −a 4, Y > 0.
By a basic calculation, we obtain that the points {±Y∗ = ±a/(4(a + 1))} are of period 2. Since g is piecewise linear and 0 ≤ g(Y ) = a < 1, the points of period 2 are globally contracting except for Y = 0. This completes the proof.
By Theorem 2.1, all points on {X = 1/2} except (X, Y ) = (1/2, 0) will never synchronize.
Theorem 2.2. There exists a dense subset of mea-sure zero in the phase space Ω such that any orbit starting from this subset will never synchronize.
Proof. We claim that any point (X, Y ) ∈ b∈AVb will arrive at {X = 1/2} in finite time, where Vb ={X = b} and A = {b ∈ (0, 1)|b =kk=1b ik/2k, ik = 0 or 1, ikb = 1, kb ∈ N}. In fact, for any point (X, Y ) ∈ b∈AVb, from the definition of A follows that X is of the form X = kk=1X ik/2k with ik= 0 or 1, and ikX = 1 kX ∈ N. From the definition of the map Ψa, we then have X = kk=1X−1ik/2k with ik= 0 or 1, that is, the number of terms in the sum is shortened by 1. After kX − 1 iterations, the ini-tial X becomes i/2 with i = 0 or 1. Again by the definition of Ψa, one can verify that i = 1. Thus we prove that two points in b∈AVb will eventually arrive at {X = 1/2}. Hence by Theorem 2.1 they cannot synchronize. On the other hand, b∈AVb is obviously dense in Ω. Thus we complete the proof.
In Theorem 2.2, we see that a dense subset of measure zero will never synchronize even though a = 0. We will see in Secs. 3 and 4, for a range of specified a, CLML (1) will synchronize in the remainder of this dense set. Due to the zero mea-sure of this dense subset, for any random choice of initial point in Ω, synchronization will occur.
3. Synchronization on a Positive Measure Set
In this section, we show that for 0 ≤ a < 1, there is a subset of positive measure such that two orbits which start from this subset will synchronize. To this end, we denote that
Ω = (X, Y )∈ Ω, −a 4 ≤ Y ≤ a 4 .
The set Ωis a trapping region of (6) since Ψa(Ω)⊂ Ω. We now let Z =|Y | and denote JR the “jump region” J R = J R(a) = (X, Z) X − Z ≥ 1 2, X + Z≤ 1 2, 0≤ Z ≤ a 4 . Here the jump region represents the cases when either x > 1/2, y < 1/2 or x < 1/2, y > 1/2. If (X, Y ) ∈ Ω\JR, then we have Y = aY < Y . Consequently, if the initial point (X(0), Y (0)) and the first n iteration points are in Ω\JR, then Y (n + 1) = an+1Y (0). Thus we have Y (n) → 0 as n → ∞ if Y (n) ∈ Ω\JR for all n = 1, 2, . . . . So
we prove the following result:
Proposition 3.1. If the initial point (X, Y ) = ((x + y)/2, (x− y)/2) and its orbits Ψna(X, Y ), n = 1, 2, . . . , stay in Ω\JR, then synchronization occurs for (x, y), i.e. Φn(x, y)→ {x = y}, as n → ∞.
However, if (x, y)∈ JR and |x − y| 1, then |x − y|/|x − y| 1. We see a jump occurs. This is the reason why we call J R a “jump” region. We now extend the upper half domain of Ω to
˜ Ω = ˜Ω(a) = 0≤ X ≤ 1, 0 ≤ Z ≤ a 4 , and consider (X, Z) = Fa(X, Z), where Fa(X, Z) = (f (X), aZ), (X, Z)∈ ˜Ω\JR 2X−1 2,−aZ + a 4 , (X, Z)∈ JR. Without loss of generality, in the following, we restrict all statements in ˜Ω. We note that the synchronization problem for (1), i.e.
lim n→∞Φn(x, y)→ {x = y}, is now equivalent to lim n→∞F n a(X, Z)→ {Z = 0}. Given n ∈ N ∪ {0}, 0 ≤ a < 1 and > 0. We first introduce some notations. Let x(n)i = i/2n, i = 0, . . . , 2n, be some nodes on [0, 1]. Let y(n)i = (x(n)i + x(n)i+1)/2 = (2i + 1)/2n+1 be the middle point of x(n)i and x(n)i+1, i = 0, . . . , 2n−1. Let rn= rn(a, ) = (a/2)n, and Ii(n)= [y(n)i − rn, yi(n)+ rn]. Define
Λ(n)= Λ(n)(a, ) = 0≤j≤n 1≤i≤2j Ii(j), Λ(∞)= Λ(∞)(a, ) = 0≤j<∞ 1≤i≤2j Ii(j). Later we will see that the map Fa behaves like a Lorenz-type map on the first n + 1 iterations when it is restricted to Λ(n)× [0, a/4].
Proposition 3.2
(i) µ1(Λ(∞)) < 2/(1− a), where µ1 denotes the Lebesgue measure on R1.
(ii) For n ≥ 0, Ii(n)(a, ) ⊃ Ii(n)(a, ) provided a ≤ a and ≤ .
Proof. Note that µ1(Λ(n))≤ 0≤j≤n 0≤i≤2j µ1(Ii(j)) = 0≤j≤n aj(2)→ 2 1− a, as n→ ∞. The strict inequality in the first assertion holds, since one can always find some Ii(n) and Ij(m) such that Ii(n)∩ Ij(m) = ∅. The second assertion directly follows from rn(a, ) ≤ rn(a, ) for a ≤ a and ≤ .
Remark 3.1. Actually [0, 1]\Λ(∞) defines a Cantor set with measure larger than 1− 2(/(1 − a)) pro-vided 1− 2(/(1 − a)) > 0.
The next proposition states the dynamics of Fa on those Ii(n)’s defined above. Let ˜Λ(n) =
˜
Λ(n)(a, ) =1≤i≤2nIi(n).
Proposition 3.3. Let 0 < ≤ a/4 and p = (X, ) ∈ ˜
Ω(a). Assume p ∈ Λ(n)× {}, then it holds (i) Fai(p) = (fi(X), ai), for 0≤ i ≤ n + 1. (ii) Fai(p) /∈ JR, for 0 ≤ i ≤ n.
(iii) Fan+1(p)∈ JR if and only if p ∈ ˜Λ(n+1)(a, )× {}.
Proof. For n = 0, Λ(0)(a, ) = [(1/2)− , (1/2) + ]. Thus if p /∈ Λ(0) × {}, then Fai(p) = (fi(X), ai) for i = 0, 1. Assume the assertions (i)–(iii) hold for n = k. Using the fact that Λ(j)(a, ) ⊂ Λ(j+1)(a, ) for each j ∈ N, it suffices to prove the assertion (i) whenever i = k + 2 and assertion (ii) whenever i = k + 1. To see this, we first note that for each j ∈ N, fj is continuous on (x(j)i , x(j)i+1), 0 ≤ i ≤ 2j − 1. Furthermore,
fj(X) = 2jX− 2jx(j)i (7) for X ∈ [x(j)i , x(j)i+1]. Assume X ∈ [x(k+1)i , x(k+1)i+1 ] for some 0≤ i ≤ 2k+1− 1. Since p /∈ Λ(k+1)× {},
we thus have
X∈ [x(k+1)i , yi(k+1)− rk+1]∪ [y(k+1)i + rk+1, x(k+1)i+1 ]. (8) Moreover, since Fak(p) /∈ JR, we have
Fak+1(p) = Fa(Fak(p)) = Fa(fk(X), ak)
= (fk+1(X), ak+1). (9) Combining (7)–(9), Fak+1(p) /∈ JR. This proves assertion (ii) and thus assertion (i), i.e.
Fk+2(p) = (fk+2(X), ak+2).
The assertion (iii) directly follows from (8). The proof is thus given inductively.
Proposition 3.3 states the dynamics of Fa restricted on Λ(n) × [0, a/4]. For any p ∈ Λ(n)× [0, a/4], it guarantees that the first n points of the orbit {Fak(p)}∞k=1 behave like a Lorenz-type map and stay in Ω\JR. Proposition 3.3(iii) gives the cri-teria of Fan+1(p) ∈ JR. Moreover, if Ii(n) satisfies Ii(n)∩ Ij(m)=∅ for all m < n and 1 ≤ j ≤ 2m, then
Fan(Ii(n)× {}) = 1 2 − a n,1 2 + a n× {an} which lies in the jump region, see Fig. 2. We are now ready to state the main theorem of this section.
Theorem 3.1. Let 0 ≤ a < 1 and Sa ⊂ {0 ≤ x, y ≤ 1} be the subset such that any two orbits starting from Sa will synchronize. Then µ2(Sa) > 0, where µ2 is the Lebesgue measure on the plane R2. Furthermore, for sufficiently small , the following holds µ2(Sa∩ {|x − y| < }) ≥ 2 − 1 + 1 1− a 2 .
Proof. Given sufficiently small satisfying ≥ δ ≥ 0. From the definition of JR and Λ(∞)(a, δ) follows that synchronization occurs for every point
p ∈ ([0, 1]\Λ(∞)(a, δ))× {δ}. By Propositions 3.1, 3.3 and 3.2(i) we have
Fan(([0, 1]\Λ(∞)(a, δ))×{δ}) → {Z = 0} as n → ∞ and
µ1([0, 1]\Λ(∞)(a, δ))≥ 1 − 2δ
1− a. (10)
Integrating both sides of (10) with respect to δ, we get µ2 0≤δ≤ ([0, 1]\Λ(∞)(a, δ))× {δ} ≥ 0 1− 2δ 1− a dδ = − 2 1− a. (11) Here we note that since we extend the upper half domain of ˜Ω to Ω = {0 ≤ X ≤ 1, 0 ≤ Z ≤ a/4} (see Fig. 3), it follows that
Sa∩{|x−y| ≤ } ⊂ 0≤δ≤ (([0, 1]\Λ(∞)(a, δ))×{δ}). Thus µ2(Sa∩ {|x − y| ≤ }) ≥ 2 − 1 + 1 1− a 2 . This completes the proof.
Remark 3.2. (i) Let p = (X, Z) ∈ 0≤δ≤([0, 1]\Λ(∞))× {δ}. Since Fan(p) /∈ JR for all n ≥ 0, from Proposition 3.3(ii), we thus have
dist(Fan(p),{Z = 0}) = O(an).
On the other hand, (11) gives the estimates for Sa ∩ {|x − y| ≤ } ⊂ 0≤δ≤(([0, 1]\Λ(∞)(a, δ))× {δ}). The ratio between the measures of the
Fig. 3. The regions Ω and ˜Ω on the phase space.
synchronization region and that of the strip {|x − y| < } is given as µ2(Sa∩ {|x − y| ≤ }) µ2({|x − y| < }) ≥ 1 − 1 + 1 1− a . We may roughly say that for the initial condition (x, y) with|x − y| 1, the iterations will possibly uniformly synchronize.
(ii) Following a similar proof as above, Theorem 3.1 also holds for a general Lorenz-type map of a small C1 perturbation of a piecewise linear Lorenz-type map.
4. Synchronization of Full Measure In this section, we will show the behavior of syn-chronization of full measure. To this end, we need more investigation on Ii(n)’s introduced in Sec. 3 and on the dynamics of Farestricted to the jump region.
Proposition 4.1. Let a = 1/2 and = a/4 = 1/8. For given Ii(n)(a, ) = [y(n)i − rn, y(n)i + rn] and Ij(m)(a, ) = [yj(m) − rm, yj(m) + rm], n < m, if Ii(n)∩ Ij(m) = ∅, then either Ij(m)⊂ Ii(n) (12a) or Ij(m)∩ Ii(n) = [yj(m)− rm, yj(m)] or [y(m)j , y(m)j + rm]. (12b) Proof. We introduce the transformation h : I = [0, 1]→ Σ2 ={0, 1}N defined by h(x) = (i1, i2, i3, i4, . . .), (13a) where x = ∞ j=1 ij 1 2 j . (13b)
Note that h is multiple-valued. Without loss of gen-erality, we only need to check that for m > n, Ij(m) = [yj(m)− rm, y(m)j + rm] satisfies (12), where y(m)j is the nearest to y(n)i − rn among all y(m)k ’s. Since y(n)i = (2i + 1)/2n+1 and rn = (a/2)n(2) = 2−2n−2, y(n)i − rn can be written as a binary code. By (13) we have
h(y(n)i − rn) = (i1, i2, . . . , in+1, 1, 1, . . . , 1, 0, 0, . . .), (14)
where the adjacent 1s in (14) is from the (n + 2)th digit to the (2n + 2)th digit. Then
(a) For n + 2≤ m ≤ 2n + 1,
h(y(m)) = (i1, i2, . . . , in+1, 1, 1, . . . , 1, 0, 0, . . . , 0, . . .), where the adjacent 1s is from the (n + 2)th digit to the (m + 1)th digit. It follows that
|yi(n)− rn− y(m)| = 2m+11 +· · · + 22n+31 ≥ rm
= 1
22m+3. Thus (12a) holds for this case. (b) For m = 2n + 2,
h(y(m)) = (i1, i2, . . . , in, 1, . . . , 1, 0, 0, . . .), where the 1s end at the (2n + 2)th digit. Hence (12b) holds for this case.
(c) For m ≥ 2n + 3, this case is the same as case (a). The proof is now completed.
Now we let Ji(n)= [yi(n)− rn, y(n)i ] and Ki(n) = [y(n)i , y(n)i + rn].
Proposition 4.2. Let a = 1/2 and = a/4. If Ji(n)= Ji(n)(a, ) ⊂ Ij(m) (respectively, Ki(n) ⊂ Ij(m)), for all m < n, then for all k < n the following hold (i) Fak(Ji(n)× {}) ⊂ JR (respectively, Fak(Ki(n)× {}) ⊂ JR). (ii) Fan(Ji(n) × {}) = [(1/2) − (1/2n) · 1/23, 1/2]× {1/2n · 1/23} (respectively, Fan(Ki(n)× {}) = [1/2, (1/2) + (1/2n) · 1/23] × {1/2n· 1/23}).
Proof. Proposition 4.1 indicates that the condition of the proposition is well posed. The rest of the proof follows from Proposition 3.3.
The next proposition gives the properties of Fak-images of Ji(n) × {} and Ki(n) × {} with
a≤ a = 1/2 and ≤ = a/4.
Proposition 4.3. Let a ≤ a = 1/2 and ≤ = a/4. If Ji(n) = Ji(n)(a, ) ⊂ Λ(n−1)(a, ) (respec-tively, Ki(n)= Ki(n)(a, ) ⊂ Λ(n−1)(a, )), then
(i) For all k < n, Fak(Ji(n)× {}) ⊂ JR
(respec-tively, Fak(Ki(n)× {}) ⊂ JR). (ii) Fan(Ji(n)× {}) = 1 2− 2n, 1 2 × {(a)n} respectively, Fan(Ki(n){}) = 1 2, 1 2 + 2n ×{(a)n}. (15) Proof. Let n, a, be given in the assumption. From Proposition 3.2(ii) and Proposition 3.3 follows that Ji(n)(a, )⊂ Ji(n) 1 2, 1 8 respectively, Ki(n)(a, )⊂ Ki(n) 1 2, 1 8 . for all a ≤ 1/2 and ≤ 1/8. The assertion (i) holds. We note that
Fan(Ji(n)(a, )) = 1 2− (a )n,1 2 × {(a)n} respectively, Fan(Ki(n)(a, )) = 1 2, 1 2 + (a )n ×{(a)n}. (16) Assume that p = (X, )∈ (Ji(n)(a, )\Ji(n)(a, ))× {} (respectively, p ∈ (K(n)
i (a, )\Ki(n)(a, )) × {}). From Proposition 3.2(ii) again follows that p ∈ ˜Λ(m) for all m < n, and
Fan(p) = (fn(X), (a)n). (17)
Combining (16) and (17), we obtain (15). This com-pletes the proof.
Remark 4.1. (i) We remark that Fan with a ≤ 1/2
homeomorphically maps the rectangle Ji(n)(1/2, 1/8) × [, 1/8] to the rectangle [(1/2) − (1/2n) · 1/8, 1/2]× [(a)n, (a)n/8].
(ii) Proposition 4.3 implies that for all a≤ 1/2, the orbits starting from p∈ [0, 1]\Λ(∞)(1/2, 1/8)× {} under the map Fa will never enter the jump region,
Let Ln, L∗n, Rn, R∗n ⊂ ˜Ω(1/8), n = 1, 2, 3, . . ., be the rectangle regions defined as
Ln= 1 2n+2 ≤ X ≤ 1 2n+1, 0≤ Z ≤ 1 8 , L∗n= 1 2 + 1 2n+2 ≤ X ≤ 1 2+ 1 2n+1, 0≤ Z ≤ 1 8 , Rn= 1− 1 2n+1 ≤ X ≤ 1 − 1 2n+2, 0≤ Z ≤ 1 8 , and Rn∗ = 1 2− 1 2n+1 ≤ X ≤ 1 2 − 1 2n+2, 0≤ Z ≤ 1 8 . See Fig. 4 for the illustration.
Here L∗n and R∗n may cross the jump regions. We also define the projection P : ˜Ω(1/8)→ [0, 1/2] as follows: P (X, Z) = X, 0≤ X ≤ 1 2, X− 1 2, 1 2 < X≤ 1.
Proposition 4.4. Let l1, . . . , lk be k horizontal line segments in Ln∪ L∗n (respectively, Rn∪ R∗n), n≥ 2, i.e.
li ={xLi ≤ x ≤ xRi } × {i}.
Assume that each two P (li) are disjoint (here, the two intervals can only be allowed to intersect at the end point) and
1≤i≤k P (li) = 1 2n+2, 1 2n+1 = P (Ln) respectively = 1− 1 2n+1, 1− 1 2n+2 = P (Rn) . Then the following hold
(i) each two P (Fa(li)) are disjoint (ii) 1≤i≤k P (Fa(li)) = 1 2n+1, 1 2n respectively = 1− 1 2n, 1− 1 2n+1 . (18)
Proof. We only show the proof for a part of Ln ∪ L∗n. Let p = (X, Z) and p = (X, Z) be in Ln∪ L∗n. We claim P (Fa(p)) = P (Fa(p)) which gives the assertion (i). It suffices to consider the case when p ∈ JR and p ∈ JR. Since p ∈ JR, we have p ∈ L∗n, and hence X > 1/2. It thus follows that P (Fa(p)) = 2X − 1/2. If X > 1/2, then P (Fa(p)) = 2X − 1. If X ≤ 1/2, then P (Fa(p)) = 2X. Suppose P (Fa(p)) = P (Fa(p)). Then either X = X+ 1/4 or X + 1/4 = X, i.e. P (p) = P (p). This contradicts the assumption P (p) = P (p). Assertion (i) is proved. To show (ii), let X ∈ [1/2n+1, 1/2n]∪ [(1/2) + (1/2n+1), (1/2) + (1/2n)]. Since P (X, Z) ∈ [1/2n+1, 1/2n], we may, without loss of generality, assume that X ∈ [1/2n+1, 1/2n]. If p = (X, Z) satisfying P (Fa(p)) = X, then (a) X = X + 1 2 , if p ∈ JR and X > 1 2, (b) X = X + 1 2 2 , if p∈ JR, (c) X = X 2, if p ∈ JR and X ≤ 1 2. If there does not exist p = (X, Z) ∈ li such that P (Fa(p)) = X, (b) and (c) indicate that there is no such p satisfying X = (X + 1/2)/2 ∈ [(1/4) + (1/2n+2), (1/24) + (1/2n+1)] or X = X/2∈ [1/2n+2, 1/2n+1]. This contradicts that iP (li) = [1/2n+2, 1/2n+1] and completes the proof.
Note that if p = (X, Z), and p = (X, Z) ∈ Ln∪ L∗n(respectively, Rn∪ R∗n) with P (p)≤ P (p), then
P (Fa(p))≤ P (Fa(p)). (19) Furthermore, if those line segments l1, . . . , lk given in Proposition 4.4 are in L1 ∪ L∗1 (respectively, R1∪ R∗1), then Fa(li)⊂ 1 4, 1 2 × 0,1 8 respectively 1 2, 3 4 × 0, 1 8 (20a)
Fig. 5. The first four images underFa ofQ0⊂ I1(1)(1/2, 1/8) for a = 1/2. and ∪P (Fa(li)) = 1 4, 1 2 . (20b)
Also, the order of each li is preserved as in (19). Now we are in a position to give the main theorem of this section.
Theorem 4.1. For 0 ≤ a ≤ 1/2 (i.e. 3/8 ≤ c ≤ 1/2), the subset such that the orginating system (1 ) (with l = 0) will synchronize is of full measure, i.e. µ2(Sa) = 1.
Proof. Since ˜Ω(1/2) ={0 ≤ X ≤ 1, 0 ≤ Z ≤ 1/8} is a trapping region for Fa, for 0 ≤ a ≤ 1/2, it suffices to show that
lim
n→∞dist(F n
a(p),{Z = 0}) = 0
for almost all p ∈ ˜Ω(1/2). For simplicity, we write Ki(n) = Ki(n)(1/2, 1/8) and Jj(n) = Jj(n)(1/2, 1/8). Now, let n ≥ 1 and 1 ≤ i ≤ 2n. From Proposi-tion 4.1, we assume that Ki(n) ⊂ Ij(m) for all m < n and 1 ≤ j ≤ 2m. From Proposition 4.3, it follows that Fan|K(n) i ×{}(X, Z) = (f n(X), an). (21) We now partition Ki(n)by Ki(n)= [yi(n), yi(n)+ rn] = ∞ j=0 Qj, where Qj = [yi(n)+ rn/2j+1, y(n)i + rn/2j]. From Proposition 4.3(ii) and (21), it follows that
Fan(Qj× {}) = 1 2+ 1 2n+j+2, 1 2+ 1 2n+j+1 × {an}.
Since for each j, P (Fan(Qj × {}) = [1/2n+j+2, 1/2n+j+1], thus, by Proposition 4.4 and (20), we have Fan+(n+j+2)(Qj× {}) = l1∪ l2∪ · · · ∪ lk ⊂ 1 4, 1 2 × 0,1 8 , where the li’s are some horizontal line segments such that each two P (li) are disjoint andP (li) = [1/4, 1/2]. See Fig. 5 for the illustration.
As mentioned in Remark 4.1 the orbit of p∈[1/4, 1/2]\Λ(∞)(1/2, 1/8)×[0, 1/8] will never enter the jump region. Thus those p ∈ Qj × {} with Fan+(n+j+2)(p) ∈ ([1/4, 1/2]\Λ(∞))× [0, 1/8] will never enter the jump region after the (2n + j + 2)th iteration. Since µ1([1/4, 1/2]\Λ(∞)(1/2, 1/8)) > 0, we see that a subset of positive measure in Ji(n) will enter the jump region at most one time. (Actu-ally, the measure = µ1(Ji(n))µ1([1/4, 1/2]\Λ(∞)(1/2, 1/8))/µ1([1/4, 1/8].)
Now, let ρ = µ1([1/4, 1/2]\Λ∞(1/2, 1/8))/ µ1([1/4, 1/8]) and Ki(n )⊂ [1/4, 1/2] satisfying
Kin ⊂ Ijm for all m < n and 1≤ j ≤ 2m. Here,
we note that those (Ki(n ) × [0, 1/8]) ∩ F(2n+j+2)
(Qj× {}) are the (2n + j + 2)th images of points in Qj× {} which may re-enter the jump region. We also similarly partition Ki(n ) by
Ki(n )= ∞ j=0 Qj where Qj = [yi(n )+ rn/2j+1, yi(n )+ rn/2j]. Let Q = F−(2n+j+2)((Qj × [0, 1/8]) ∩ F(2n+j+2)(Qj×
as on Qj, we alternatively obtain F(2n+j+2+2n+j+2)(Q) = l1∪ · · · ∪ lk ⊂ 1 4, 1 2 × 0,1 8
and P (li) = [1/4, 1/2]. We see that a positive measure set in Q of measure µ1(Q)ρ will never enter the jump region after the (2n + j + 2 + 2n+ j+ 2)th iteration. An inductive process thus com-pletes the proof.
Remark 4.2. Following in a way similar to the above proof, Theorem 4.1 also holds for (1) with 0 < l < 1/2 by carefully choosing the rectangle regions Ln, L∗n, Rn and R∗n.
5. Numerical Experiments
In this section, we will exhibit the behavior of nonuniform synchronization of (1) by numerical experiments. Theoretically by Theorem 2.2, syn-chronization for the system (1) with 0 < a < 1 is nonuniform. Due to truncation errors of com-puters, it is not easy to find synchronization for (1) with 0 < a < 1/2 in numerical computa-tions. Hence our numerical experiments will focus on (1) with a 1. In Figs. 8–10, the horizon-tal axis denotes the iteration time and the vertical axis denotes the absolute value of difference of the states x and y, |x − y|. We show that the speeds of synchronization are quite different for various
Fig. 6. Synchronization for CLML with l = 0, c = 0.28, a = 0.88 and the initial point (x, y) = (0.849987498779205, 0.964226163269648).
parameters and different initial points. In Figs. 6–8 we simulate the CLMLs with a = 0.88 for various initial points. In Fig. 6, we see the orbit never enters the jump region and hence CLML occurs in synchronization rapidly. In Figs. 7 and 8, these two orbits re-enter the jump region for a huge number of times, thus these two CLMLs occur in the synchronization at about n = 1300 and n = 1600, respectively. In Figs. 9 and 10, we set a 1, the synchronization speed is very slow, approximately n = 8× 105.
Fig. 7. Synchronization for CLML with l = 0, c = 0.28, a = 0.88 and the initial point (x, y) = (0.4474544783526994, 1.654624747929506).
Fig. 8. Synchronization for CLML with l = 0, c = 0.28, a = 0.88 and the initial point (x, y) = (0.1315429151518945, 0.4728897370785985).
Fig. 9. Synchronization for CLML with l = 0.05, c = 0.2370, a = 0.9994 and the initial point (x, y) = (0.1762661444946180, 0.4057062130620954).
Fig. 10. Synchronization behavior for CLML withl = 0.125, c = 0.21435, a = 0.999775 and the initial point (x, y) = (0.1527392702903627, 0.7467856765644292).
Acknowledgments
The third author is currently a postdoc at the National Center for Theoretical Sciences, Taiwan. He would like to thank the hospitality of the insti-tute and Professor Chang-Shou Lin of National Chung Cheng University, Taiwan and Professor Song-Sun Lin of National Chiao Tung University, Taiwan.
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![Fig. 2. If I i (n) ∩ I j (m) = ∅ for all m < n and 1 ≤ j ≤ 2 m , then F a n ( I i (n) × {}) = [1/2 − a n , 1/2 + a n ] × {a n }.](https://thumb-ap.123doks.com/thumbv2/9libinfo/7491516.115199/5.918.89.467.112.288/fig-if-all-lt-and-then-f-n.webp)
