# Periodicity in a Class of Systems of Delay Difference Equations

Authors: Zhijian Wei
Journal: Journal of Applied Mathematics
 (2011)

---

## Abstract

We study a system of delay difference equations modeling four-dimensional discrete-time delayed neural networks with no internal decay. Such a discrete-time system can be regarded as the discrete analog of a differential equation with piecewise constant argument. By using semicycle analysis method, it is shown that every bounded solution of this discrete-time system is eventually periodic. The obtained results are new, and they complement previously known results.


---

## Body

## 1. Introduction

Over the past decades, there has been increasing interest in the potential applications of the dynamics of artificial neural networks in signal and image processing. Among the most popular models in the literature of artificial neural networks is the following well-known Hopfield's model [1, 2]: (1.1)dui(t)dt=μiui(t)+∑j≠iTijfj(uj(t-τij)),1≤i≤n,
where μi≥0 is the internal decay rate, τij≥0 is the delay incorporated by Marcus and Westervelt [3] to account for the finite switching speed of amplifiers (neurons), fj:R→R (the set of all real numbers) is the signal functions, and Tij represents the connection strengths between neurons and if the output from neuron j excites (resp., inhibits) neuron i, then Tij>0 (resp., <0), 1≤i,j≤n.Much work on Hopfield-type neural networks with constant time delays has been carried out. However, from the point of view of electronic implementation of neural networks and control, the time-varying delay case is more suitable for practical neural networks, and we can achieve this by using piecewise constant arguments. Because of the wide application of differential equations with piecewise constant argument in certain biomedical models (see, e.g., [4]), much progress has been made in the study of differential equations with piecewise constant arguments since the pioneering work of Cooke and Wiener [5] and Shah and Wiener [6]. For more details and references on this subject, interested readers may refer to a survey of Cooke and Wiener [7]. Motivated by these discussions, we propose (1.2)dui(t)dt=μiui(t)+∑j≠iTijfj(uj([t-τij])),1≤i≤n,
as a new neural network model, where [·] denotes the greatest integer function and fj (1≤j≤n) are the McCulloch-Pitts nonlinear function given by (1.3)f(u)={1ifu>σ,-1ifu≤σ.
Here, σ∈R is referred as the threshold. The McCulloch-Pitts nonlinearity reflects the fact that the signal transmission is of digital nature: a neuron is either fully active or completely inactive.Neural networks are complex and large-scale nonlinear dynamical systems, while the dynamics of the delayed neural networks is even richer and more complicated [8]. In order to obtain a deep and clear understanding of the dynamics of neural networks, the dynamics of delayed neural networks consisting of a few neurons has received increasing attention over the past years [9–14]. The studies on these neural networks of a few neurons are potentially useful because the complexities found in such models often provide promising information for the studies on more complicated neural networks with a large number of neurons [15]. So, we are inspired to consider a special case of (1.2), (1.4)x1′(t)=f(x2([t-k])),x2′(t)=f(x3([t-k])),x3′(t)=-f(x4([t-k])),x4′(t)=-f(x1([t-k])),
where f is the activation function with McCulloch-Pitts nonlinearity (1.3), k is a nonnegative integer and the negative (resp., positive) sign on the right indicates negative (resp., positive) feedback. (1.4) describes the dynamics of four interacting neurons with no internal decays.To study (1.4), it is convenient to consider the following difference system: (1.5)x1(n+1)=x1(n)+f(x2(n-k)),x2(n+1)=x2(n)+f(x3(n-k)),x3(n+1)=x3(n)-f(x4(n-k)),x4(n+1)=x4(n)-f(x1(n-k)),
where k is a nonnegative integer and f satisfies (1.3). In fact, integrate (1.4) from n to t∈[n,n+1) to get (1.6)x1(t)=x1(n)+∫ntf(x2([s-k]))ds=x1(n)+f(x2(n-k))(t-n),x2(t)=x2(n)+∫ntf(x3([s-k]))ds=x2(n)+f(x3(n-k))(t-n),x3(t)=x3(n)-∫ntf(x4([s-k]))ds=x3(n)-f(x4(n-k))(t-n),x4(t)=x4(n)-∫ntf(x1([s-k]))ds=x4(n)-f(x1(n-k))(t-n).
Letting t→n+1, we get a special case of (1.5) with k=0. Hence, system (1.5) can be regarded as a more general neural network model of four neurons with transmission delay.Our study is inspired by the work of Chen [16–18], who considered (1.7)x(n+1)=x(n)-f(x(n-k)),x(n+1)=x(n)+f(x(n-k)),x(n+1)=x(n)+f(y(n-k)),y(n+1)=y(n)-f(x(n-k)),
where f:R→R is given by (1.3).The main objective of this paper is to study (1.5) with f satisfying (1.3). The discontinuity of f makes it difficult to apply directly the theory of discrete dynamical systems to system (1.5). However, by using semicycle analysis method, we obtain some interesting results for the periodicity of (1.5). It is shown that every bounded solution of system (1.5) is eventually periodic.The rest of the paper is organized as follows. In Section2, we introduce some necessary notations and lemmas that will be used later. In Section 3, we state and prove our main result.


## 2. Notations and Preliminaries

For the sake of simplicity, we introduce some notations. LetZ denote the set of all integers. For a, b∈Z, define Z(a)={a,a+1,…} and Z(a,b)={a,a+1,…,b} whenever a≤b.By a solution of (1.5), we mean a sequence {(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ that satisfies (1.5). Clearly, for given (x¯1(-k),x¯2(-k),x¯3(-k),x¯4(-k)),(x¯1(-k+1),x¯2(-k+1), x¯3(-k+1),x¯4(-k+1)),…,(x¯1(0),x¯2(0), x¯3(0), x¯4(0))∈R4, system (1.5) has a unique solution {(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ satisfying the initial condition: (2.1)(x1(i),x2(i),x3(i),x4(i))=(x¯1(i),x¯2(i),x¯3(i),x¯4(i))fori∈Z(-k,0).
If all terms of the sequence {vn}n=-k+∞ are integers, we say that it is an integer sequence. If {xi(n)}n=-k∞ for all i∈Z(1,4) are integer sequences and also satisfy (1.5) and (1.3), then {(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ is said to be an integer sequence solution of (1.5) and (1.3).Let{(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ be a solution of (1.5). Define (2.2)τ(x)={x-[x]forx≠0,1forx=0.
Clearly, τ(x)≥0. From (1.5) and (1.3), we have τ(xi(n)-σ)=τ(xi(0)-σ), i∈Z(1,4). Denote xĩ(n)=xi(n)-σ-τ(xi(0)-σ)+1, i∈Z(1,4). Then xĩ(n)∈Z for all n∈Z(-k), i∈Z(1,4). Moreover, if xi(n)>σ, then xĩ(n)=xi(n)-σ-τ(xi(0)-σ)+1≥xi(n)-σ>0, and if xi(n)≤σ, then xĩ(n)=xi(n)-σ-τ(xi(0)-σ)+1≤-1+1=0, i∈Z(1,4). Hence, {(x1̃(n),…,x4̃(n))}n=-k+∞ is an integer sequence solution of (1.5) and (1.3) with σ=0.So, without loss of generality, in the paper we always assume thatσ=0, and a solution of (1.5) means an integer sequence solution.A solution{(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ is said to be eventually periodic if there exist n0∈{-k,-k+1,-k+2,…} and ω∈{1,2,3,…} such that (2.3)(x1(n+n0+ω),x2(n+n0+ω),…,x4(n+n0+ω))=(x1(n+n0),x2(n+n0),…,x4(n+n0)),
for n=0,1,2,…, and ω is called a period. The smallest such ω is called the minimal period of {(x1(n),x2(n),x3(n),x4(n))}n=-k+∞.Definition 2.1.
For a sequence{un}n=-k+∞, we say that {un1,un1+1,…,un1+l-1} with l≥1 is a positive semicycle with the length l if un>0 for n∈Z(n1,n1+l-1), and n1 and l satisfy
(2.4)eithern1=-korun1-1≤0,eitherl=+∞orun1+l≤0.
We call l the length of the positive semicycle. A negative semicycle is defined similarly (replace > by ≤ and vice versa).
To simplify the following arguments, for an integer sequence, we denoteAr+ the positive semicycle {vn1,…,vn1+2r} with vn1+r±m=r-m+1 for m∈Z(0,r) and Ar- the negative semicycle {vn1,…,vn1+2r} with vn1+r±m=m-r for m∈Z(0,r).Definition 2.2.
An integer sequence{vn}n=n1n1+L-1 (may be infinite terms) is said to be an r-normal cycle if it possesses one of the representations as follows: {Ar+,Ar-,…,Ar+,Ar-}, {Ar+,Ar-,…,Ar-,Ar+}, {Ar-,Ar+,…,Ar-,Ar+}, and {Ar-,Ar+,…,Ar+,Ar-}. In addition, if L=+∞, then it is said to be a total r-normal cycle.Remark 2.3.
For ar-normal cycle {vn}n=n1n1+L-1, from Definition 2.2, it is easy to see that there must exist i∈Z(0) such that L=(2r+1)i. If L=+∞, that is, {vn}n=n1n1+L-1 is a total r-normal cycle, then i=+∞.
Throughout this section, we tacitly assume that(S)={(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ is a periodic solution of system (1.5) satisfying the initial values:
(2.5)(x1(i),x2(i),x3(i),x4(i))∈Z4fori∈Z(-k,0).
In view of (1.5) and (1.3), it is easy to see that all terms x1(n), x2(n), x3(n), and x4(n) of (S) are integers, that is, (S) is an integer sequence solution of (1.5).
For a periodic solution(S) of (1.5), denote
(2.6)Pxi+=sup⁡{xi(n):n∈Z(-k)}-1,Pxi-=-inf⁡{xi(n):n∈Z(-k)},
and let Lxi+ and Lxi- be the length of the longest positive semicycle of {xi(n)}n=-k+∞ and the length of the longest negative semicycle of {xi(n)}n=-k+∞, respectively, where i=1,2,…,4. Note that Lxi+,Lxi-∈Z(1) for i=1,2,…,4.
The following several lemmas will be useful to the proof of the main result in this paper.Lemma 2.4.
(a)
EitherLxi+>Lxj+ or Lxi->Lxj+ or min{Lxi+,Lxi-}≥Lxj+ for i,j=1,2,…,4 and i≠j.
(b)
EitherLxi+>Lxj- or Lxi->Lxj- or min{Lxi+,Lxi-}≥Lxj- for i,j=1,2,…,4 and i≠j.Proof.
We only prove the conclusion: eitherLx1+>Lx2+ or Lx1->Lx2+ or min{Lx1+,Lx1-}≥Lx2+. The other conclusions can be proved similarly. Now, we distinguish two cases to finish the proof.
Case 1.
Lx2+=2r for some r>0. Without loss of generality, we assume that x2(0)≤0,x2(1)>0,x2(2)>0,…,x2(2r)>0,x2(2r+1)≤0,…. Then, from (1.5) and (1.3), we have
(2.7)x1(2r+k+1)=x1(r+k+1)+∑i=r+12rf(x2(i))=x1(r+k+1)+r,(2.8)x1(r+k+1)=x1(k+1)+∑i=1rf(x2(i))=x1(k+1)+r.
(i)
Ifx1(r+k+1)≥1, it follows from (2.7) that x1(2r+k+1)≥r+1. Since the distance between consecutive elements of {x1(n)}n=0∞ is 1, we have Lx1+≥2r+1>Lx2+.
(ii)
Ifx1(r+k+1)≤0, it follows from (2.8) that x1(k+1)≤-r. Thus, Lx1-≥2r+1>Lx2+.
Case 2.
Lx2+=2r+1 for some r>0. Without loss of generality, we assume that x2(0)≤0,x2(1)>0,x2(2)>0,…,x2(2r+1)>0, and x2(2r+2)≤0,…. Then, from (1.5) and (1.3), we have
(2.9)x1(k+2r+2)=x1(r+k+1)+∑i=1+r2r+1f(x2(i))=x1(r+k+1)+(r+1),(2.10)x1(r+k+1)=x1(k+1)+∑i=1rf(x2(i))=x1(k+1)+r.(i)
Ifx1(r+k+1)>0, it follows from (2.9) that x1(k+2r+2)>r+1. Thus, Lx1+>2r+1=Lx2+.(ii)
Ifx1(r+k+1)<0, it follows from (2.10) that x1(k+1)<-r. Thus, Lx1->2r+1=Lx2+.(iii)
Ifx1(r+k+1)=0, it follows from (2.9) and (2.10) that x1(k+2r+2)=r+1 and x1(k+1)=-r. Thus, min{Lx1+,Lx1-}≥2r+1=Lx2+.
Therefore, we have eitherLx1+>Lx2+ or Lx1->Lx2+ or min{Lx1+,Lx1-}≥Lx2+. This completes the proof.Lemma 2.5.
Lxi+=Lxj- for i,j=1,…,4. Moreover, Lx1+ is odd.Proof.
The lemma follows immediately from Lemma2.4 together with its proof.Lemma 2.6.
Pxi+=Pxj- for i,j=1,…,4.Proof.
First, we claim thatPx2+≤Px1+. Otherwise, Px2+>Px1+. Since the distance between consecutive elements of {x2(n)}n=0∞ is 1, it follows easily from the definitions of Px2+ and Lx2+ that Lx2+≥2Px2++1. By Lemma 2.5, there exists r≥0 such that Lx2+=2r+1≥2Px2++1, which implies that r≥Px2+. Without loss of generality, we assume that x2(0)≤0,x2(1)>0,x2(2)>0,…,x2(2r+1)>0, and x2(2r+2)≤0,…. Note that r≥Px2+ and Px2+>Px1+. It follows from (2.9) and the definition of Px1+ that
(2.11)x1(k+1+r)=x1(k+2r+2)-(r+1)≤x1(k+2r+2)-Px2+-1<x1(k+2r+2)-Px1+-1≤0.
Thus, x1(k+1+r)≤-1, which, together with (2.10), implies that
(2.12)x1(k+1)=x1(k+1+r)-r≤-(r+1).
It follows that Px1-≥r+1 and hence, Lx1-≥2r+3>Lx2+, a contradiction to Lemma 2.5. This proves the claim. Repeating the same argument as that in the proof of the above claim, we have Pxi+≤Pxj+, Pxi-≤Pxj-, and Pxi+≤Pxj- for i,j=1,…,4 and i≠j. Thus, we get Pxi+=Pxj- for i,j=1,…,4. This completes the proof.Lemma 2.7.
Lx1+=2Px1++1.Proof.
It follows from the definitions ofPx1+ and Lx1+ that Lx1+≥2Px1++1. We show Lx1+=2Px1++1 by way of contradiction. Assume that Lx1+>2Px1++1. By Lemma 2.5, there exists r≥0 such that Lx1+=2r+1≥2Px1++1, which implies that r≥Px1+. Without loss of generality, we assume that x1(0)≤0,x1(1)>0,x1(2)>0,…,x1(2r+1)>0, and x1(2r+2)≤0,…. Now, we distinguish two cases to finish the proof.
Case 1.
x4(r+k+1)≥1. For this case, from (1.5) and (1.3), we have
(2.13)x4(k+1)=x4(k+r+1)+∑i=1rf(x1(i))≥1+r>1+Px1+,
which implies that Px4+>Px1+, a contradiction to Lemma 2.6.
Case 2.
x4(r+k+1)≤0. For this case, from (1.5) and (1.3), we have
(2.14)x4(k+2r+2)=x4(r+k+1)-∑i=1+r2r+1f(x1(i))≤-(1+r),
which implies that Px4-≥1+r>Px1+, a contradiction to Lemma 2.6. This completes the proof.Throughout the remaining part of this paper, we denotePx1+=p.Lemma 2.8.
(a)
x1(n1)=p+1 for some n1∈Z(-k). Moreover, x4(n1+k)=1 and x4(n1+k+p+1)=-p.
(b)
x4(n2)=p+1 for some n2∈Z(-k). Moreover, x3(n2+k)=1 and x3(n2+k+p+1)=-p.
(c)
x2(n3)=p+1 for some n3∈Z(-k). Moreover, x1(n3+k)=0 and x1(n3+k+p+1)=p+1.
(d)
x3(n4)=p+1 for some n4∈Z(-k). Moreover, x2(n4+k)=0 and x2(n4+k+p+1)=p+1.Proof.
We only give the proof of the conclusion (c). The other conclusions can be proved similarly. By the definition ofPx2+, there must exist n3∈Z(-k) such that x2(n3)=Px2++1=p+1. Since the distance between consecutive elements of {x2(n)}n=0∞ is 1, it follows that x2(n3±j)>0 for all j∈Z(1,p). We claim that x1(n3+k)=0. Otherwise, x1(n3+k)≤-1 or x1(n3+k)≥1. If x1(n3+k)≥1, it follows from (1.5) and (1.3) that
(2.15)x1(n3+k+p+1)=x1(n3+k)+∑j=n3n3+pf(x2(j))=x1(n3+k)+(p+1).
Thus, x1(n3+k+p+1)≥p+2, which implies that Px1+≥p+2>Px2+, a contradiction to Lemma 2.6. If x1(n3+k)≤-1, it follows from (1.5) and (1.3) that
(2.16)x1(n3+k)=x1(n3+k-p)+∑j=n3-pn3-1f(x2(j))=x1(n3+k-p)+p.
Thus, x1(n3+k-p)=x1(n3+k)-p≤-1-p, which implies that Px1-≥p+1>Px2+, a contradiction to Lemma 2.6. This proves the claim. By (1.5) and (1.3), we must have
(2.17)x1(n3+k+p+1)=x1(n3+k)+∑j=n3n3+pf(x2(j))=p+1.
This completes the proof.Arguing as in the proof of Lemma2.8, we can get the following result.Lemma 2.9.
(a)
x1(n1)=-p for some n1∈Z(-k). Moreover, x4(n1+k)=0 and x4(n1+k+p+1)=p+1.
(b)
x4(n2)=-p for some n2∈Z(-k). Moreover, x3(n2+k)=0 and x3(n2+k+p+1)=p+1.
(c)
x2(n3)=-p for some n3∈Z(-k). Moreover, x1(n3+k)=1 and x1(n3+k+p+1)=-p.
(d)
x3(n4)=-p for some n4∈Z(-k). Moreover, x2(n4+k)=1 and x2(n4+k+p+1)=-p.Lemma 2.10.
p≤2k.Proof.
Without loss of generality, we assume thatx1(0)=0,x1(1)=1,…,x1(p)=p,x1(p+1)=p+1,x1(p+2)=p,…,x1(2p)=2, x1(2p+1)=1, and x1(2p+2)=0. It follows from Lemma 2.8 that x4(p+k+1)=1. By (1.5) and (1.3), we have
(2.18)x4(p+k+1)=x4(k+1)-∑i=1pf(x1(i)).
Thus, x4(k+1)=p+1. It follows from Lemma 2.8 that x3(p+2k+2)=-p, and hence, by Lemma 2.9, we have x2(p+3k+2)=1. From (1.5) and (1.3), we get
(2.19)x2(p+3k+2)=x2(3k+2)+∑i=2k+2p+2k+1f(x3(i))=x2(3k+2)-p.
Thus, x2(3k+2)=p+1. Therefore, by Lemma 2.8, we have x1(4k+2)=0. Since x1(2p+2)=0, it follows that 4k+2≥2p+2, and hence 2k≥p. This completes the proof.We hereafter useMxi to denote the length of the longest normal cycle of {xi(n)}n=-k+∞, where i=1,…,4.Lemma 2.11.
Mxi=+∞, i=1,…,4.Proof.
We show thatMxi=+∞ for i∈Z(1,4) by way of contradiction. Without loss of generality, we assume that Mx1<+∞. We will distinguish several cases to finish the proof.
Case 1.
Mxi=+∞, i=2,3,4. Without loss of generality, we assume that {x2(i)}n=1+∞ is a total p-normal cycle and x2(1)=1. Then, x2(2i(2p+1)+p+1)=p+1 for i∈Z(0), and x2(2i(2p+1)+3p+2)=-p for i∈Z(0). By Lemmas 2.8 and 2.9, we have
(2.20)x1(2i(2p+1)+(p+1)+(k+p+1))=p+1,x1(2i(2p+1)+(3p+2)+(k+p+1))=-p,
for i∈Z(0). Therefore, {x1(n)}n=p+k+2+∞ is a total p-normal cycle, which is a contradiction to Mx1<+∞.
Case 2.
Mxi<+∞, i=2,3,4. Without loss of generality, we assume that Mx2≥max{Mx1,Mx3,Mx4}. Then, by the definition of Mx2, there exists L≥1 such that Mx2=(2p+1)L is the length of the longest normal cycle of {x2(n)}n=-k+∞. Without loss of generality, we assume that {x2(n)}n=1Mx2 is a p-normal cycle and xMx2-p=p+1,xMx2-p+1=p,…,xMx2-1=2,xMx2=1. To finish the proof of Case 2, we will distinguish two subcases.
Subcase 1.
x2(1)=1. Then there exists q≥0 such that L=2q+1. Thus,
(2.21)x2(2i(2p+1)+p+1)=p+1fori∈Z(0,q),x2(2i(2p+1)+3p+2)=-pfori∈Z(0,q-1).
By Lemmas 2.8 and 2.9, we have x1(2i(2p+1)+(p+1)+(k+p+1))=p+1 for i∈Z(0,q) and x1(2i(2p+1)+(3p+2)+(k+p+1))=-p for i∈Z(0,q-1). It follows from x2(p+1)=p+1 that x1(p+k+1)=0. This, combined with
(2.22)x1(p+k+1)=x1(k+1)+∑i=1pf(x2(i))=x1(k+1)+p,
gives x1(k+1)=-p. By Lemma 2.10, we have k-p+1≥k-2k+1>-k. Therefore, the total number of p+1 and -p contained in {x1(i)}i=k+1-p(2q+1)(2p+1)+k+1+p is 2q+2. Since
(2.23)(2q+1)(2p+1)+k+1+p-(k-p+1)+1=(2q+1)(2p+1)+2p+1,
it follows that {x1(i)}i=k+1-p(2q+1)(2p+1)+k+1+p is a normal cycle with the length of Mx2+2p+1, and hence Mx1>Mx2, which leads to a contradiction.
Subcase 2.
x2(1)=0. Then, using a similar argument to that in the proof of Subcase 1, we can show that this is also a contradiction.
IfMx2<+∞, Mx3=+∞, Mx4=+∞ or Mx2<+∞, Mx3<+∞, Mx4=+∞, then we have the same contradiction by similar arguments. This completes the proof.


## 3. Main Result and Proof

Theorem 3.1.
Fork∈Z(0), every bounded solution of (1.5) is eventually periodic with the minimal period 4r+2 for some r∈Ω(k), where Ω(k)={r∈Z(0):(2k+1)/(2r+1)∈Z(0)}.Proof.
Let{(x1(n),x2(n),x3(n),x4(n))}n=-k+∞ be a bounded solution of (1.5). Then there exists a positive integer M0>0 such that |x1(n)|+⋯+|x4(n)|<M0 for all n∈Z(-k). Denote xjn(i)=xj(n-i) for i∈Z(0,k) and j∈Z(1,…,4), and
(3.1)Xn=(x1n(0),x1n(1),…,x1n(k),x2n(0),x2n(1),…,x2n(k),…,x4n(0),…,x4n(k)).
Define Pi:Z4k+4→Z by setting Pi(v1,v2,…,v4k+4)=vi for vi∈Z and i∈Z(1,4k+4), and define F=(F1,F2,…,F4k+4) as
(3.2)Fi(X)={P1X+f(P2k+2X)fori=1,Pi-1Xfor2≤i≤k+1,Pk+2X+f(P3k+3X)fori=k+2,Pi-1Xfork+3≤i≤2k+2,P2k+3X-f(P4k+4X)fori=2k+3,Pi-1Xfor2k+4≤i≤3k+3,P3k+4X-f(Pk+1X)fori=3k+4,Pi-1Xfor3k+5≤i≤4k+4.
Then, Xn+1=F(Xn). Since |x1(n)|+⋯+|x4(n)|<M0 for all n∈Z(-k) and xj(n)∈Z for all n∈Z(-k) and j∈Z(1,…,4), it follows that {Xn}n=-k+∞ is a finite set. Therefore, there exist n1>n2≥1 such that Xn2=Xn1=Xn2+n1-n2. This, combined with Xn+1=F(Xn), gives Xn=Xn+n1-n2 for n≥n2+1. Hence, {Xn}n=-k+∞ is eventually periodic with the period (n1-n2). It follows that {PjXn}n=-k+∞ are also eventually periodic, j=1,k+2,2k+3,3k+4. Therefore, {(x1(n),…,x4(n))}n=-k+∞ is eventually periodic with the minimal period T∈Z(1). By Lemma 2.11, there exists some r∈Z(0) such that T=4r+2. Without loss of generality, we assume that {x1(n)}n=1+∞ is a total r-normal cycle, and x1(1)=1. Then, x1(r+1)=r+1. By Lemma 2.8, we can get that x4(r+k+1)=1 and x4(2r+2+k)=-r. It follows from Lemma 2.9 that x3(2r+2k+2)=0 and x3(3r+2k+3)=r+1, and so, again by Lemma 2.8, x2(3r+3k+3)=0 and x2(4r+3k+4)=r+1. By Lemma 2.8, we have x1(4r+4k+4)=0 and x1(5r+4k+5)=r+1. From (1.5) and (1.3), we can get x1(4r+4k+5)=x1(4r+4k+4)+f(x2(4r+3k+4)). Thus, x1(4r+4k+5)=0+1=1. Therefore, {x1(i)}i=14r+4k+4 consists of Ar+,Ar-,…,Ar+,Ar-, consecutively, in total of N semicycles, in which N is a positive even integer. Consequently, there exists an integer M≥0 such that 4r+4k+4=(2r+1)·(2M+2), that is, (2k+1)/(2r+1)=M, which implies that r∈Ω(k). This completes the proof.



---

*Source: 735825-2011-10-27.xml*