# Periodicities of a System of Difference Equations

Authors: Weizhen Quan; Miaoqiao Pan; Xiaopei Li
Journal: Journal of Function Spaces
 (2017)

---

## Abstract

We study the periodicities of a system of difference equationsxn+1=max1/xn,An/yn-k,yn+1=max1/yn,Bn/xn-k, where initial values (x-k,y-k),…,(x0,y0)∈(0,+∞)×(0,+∞). We show that if An,Bn∈(0,1) are two periodic sequences, then every solution of the above system is eventually periodic with period 2. If k is even, there must be one in xn and yn converges to period two solution.


---

## Body

## 1. Introduction

Difference equations are powerful tool that describe the law of nature. Recently, the theory of difference equations has received extensive attention because it can be applied to many areas of science and technology, such as the fields of information and e-science. Moreover, people pay more attention to the dynamics of max-type difference equations; they are concerned about the research of the period character of solutions and the convergence of positive solutions. Please refer to [1–7].Szalkai [8] studied the max-type difference equation (1)xn=max⁡A0xn,A1xn-1,…,Akxn-k,n=0,1,2,…,where Ai(i=1,2,…,k)∈R, and at least one of them is nonzero. The initial values x0,x-1,…,x-k are nonzero real numbers, so every solution of this equation is periodic solution. Chen [9] studied the dynamics of max-type difference equation(2)xn+1=max⁡1xn,Anxn-1,n=0,1,2,…,where {An}n=0∞=…,α,β,α,β,… is a two-cycle sequence, β>α>1. The initial values x0, x-1 are nonzero real numbers; then every positive solution of this equation is bounded and persists, and xn∈[α/β,βα] for all n≥0. Furthermore, every solution of this equation is periodic with period 4. Yang [10] studied the max-type difference equation(3)xn=max⁡1xn-1α,Axn-2,n=0,1,2,…,where 0<α<1, A>0; then every solution of this equation tends to be x-=1 or is eventually periodic with period 4. Sun et al. [11] studied the dynamics of max-type difference equation (4)xn+1=max⁡1xn,Anxn-1,n=0,1,2,…,where {An}n=0∞ is a k-cycle sequence with k≥2. The initial values x0, x-1 are positive real numbers. Let limsupn→∞⁡xn=S; if there exist infinitely many n such that xn≥S and xn+1≥S, then {xn} is eventually equal to 1. every positive solution of this equation is eventually periodic with period 2 provided An∈(0,1).Motivated by the work of above max-type difference equations, the aim of this paper is to investigate the periodicities of a system of max-type difference equations(5)xn+1=max⁡1xn,Anyn-k,yn+1=max⁡1yn,Bnxn-k,where initial values (x-k,y-k),(x-k+1,y-k+1),…,(x0,y0)∈(0,+∞)×(0,+∞),  k∈N+, and  An,Bn are positive sequences. More precisely, we will prove that if An,Bn∈(0,1) are two periodic sequences, then every solution of (5) is eventually periodic with period 2. if k is even, then one of xn and yn converges to a period two solution.


## 2. Main Results

Before proceeding with the proof of our main results in this section, we shall need the following lemmas.Lemma 1.
Let{(xn,yn)}n=-k+∞ be a solution of system (5); then xn+1xn≥1, yn+1yn≥1, and xn+1+xn≥2,  yn+1+yn≥2 for all n≥0.Proof.
By system (5), we have(6)xn+1≥1xn,yn+1≥1ynfor all n≥0. Hence xn+1xn≥1, yn+1yn≥1, and xn+1+xn≥2xn+1xn≥2, yn+1+yn≥2yn+1yn≥2. This completes the proof of the lemma.Lemma 2.
Let{(xn,yn)}n=-k+∞ be a solution of system (5). If n>k, then(7)xn+1≤max⁡xn-1,Anyn-k-1≤max⁡xn-1,yn-k-1,yn+1≤max⁡yn-1,Bnxn-k-1≤max⁡yn-1,xn-k-1.Proof.
By Lemma1 and An<1, we obtain that, for any n>k,(8)xn+1=max⁡xn-1xnxn-1,Anyn-k-1yn-kyn-k-1≤max⁡xn-1,Anyn-k-1≤max⁡xn-1,yn-k-1.Similarly, we also obtain that, for any n>k,(9)yn+1=max⁡yn-1ynyn-1,Bnxn-k-1xn-kxn-k-1≤max⁡yn-1,Bnxn-k-1≤max⁡yn-1,xn-k-1.This completes the proof of the lemma.Theorem 3.
Let{(xn,yn)}n=-k+∞ be a solution of system (5) and rn+1=max⁡xn,xn-1,…,xn-k-1,yn,yn-1,…,yn-k-1. Then(1)
rn is eventually monotonically decreasing. Moreover, if T=limn→∞⁡rn, then T≥1;(2)
max⁡{limsupn→∞⁡xn,limsupn→∞⁡yn}=T.Proof.
(
1
) By Lemma 2, if n>k, we have(10)xn+1≤max⁡xn-1,yn-k-1≤max⁡xn,xn-1,…,xn-k-1,yn,yn-1,…,yn-k-1=rn+1.Similarly, we obtain yn+1≤rn+1. Hence(11)rn+2=max⁡xn+1,xn,…,xn-k,yn+1,yn,…,yn-k≤rn+1,so {rn} is eventually monotonically decreasing. On the other hand, by Lemma 1, we know xn+1xn≥1, yn+1yn≥1, so rn≥1. Hence T≥1.
(
2
) It is easy to know that there are infinitely many n such that xn=rn+1 or yn=rn+1, so limsupn→∞⁡xn≥T, or limsupn→∞⁡yn≥T. On the other hand, xn≤rn+1 and yn≤rn+1. Hence limsupn→∞⁡xn≤T and limsupn→∞⁡yn≤T. Thus(12)max⁡limsupn→∞xn,limsupn→∞yn=T.This completes the proof of the theorem.Lemma 4.
Leta=max⁡{An,Bn:n∈N+} and η=1-a/2T. Then there exists N∈N+ such that(13)xn+1≤max⁡xn-1,T-η,yn+1≤max⁡yn-1,T-η,for all n≥N.Proof.
Letη1=1-a/2aT. By Theorem 3, there exists N∈N+ such that n-k-1≥N:(14)yn-k-1<T+η1=1+a2aT.Then Anyn-k-1<a1+a/2aT=T-1-a/2T=T-η. By Lemma 2, it follows that(15)xn+1≤max⁡xn-1,T-η.Similarly, we obtain yn+1≤max⁡{yn-1,T-η}. This completes the proof of the lemma.Lemma 5.
There exists a positive integerN1∈N+ such that, for all t≥0, xN1+2t≥xN1+2t+2≥T (if  limsupn→∞⁡xn=T), or yN1+2t≥yN1+2t+2≥T (if  limsupn→∞⁡yn=T).Proof.
LetN and η be as Lemma 4. By Theorem 3, if limsupn→∞⁡xn=T, then there exists a monotonically increasing sequence nt→+∞ such that xnt≥T. On the other hand, there exists a subsequence {ntj} of {nt} such that ntj=2ltj+r(r∈0,1). Without loss of generality, we may assume that ∃N∈N+, nt=2lt+r≥N (for all nt). By Lemma 4, we get(16)T≤xnt=x2lt+r≤max⁡x2lt-1+r,T-η=x2lt-1+r≤max⁡x2lt-2+r,T-η=x2lt-2+r≤max⁡x2lt-3+r,T-η=⋯=x2lt-1+r.Write N1=2l1+r. Then xN1+2t≥xN1+2t+2≥T for all t≥0.
Similarly, we obtain that iflimsupn→∞⁡yn=T, then yN1+2t≥yN1+2t+2≥T. This completes the proof of the Lemma.Lemma 6.
(1) Iflimsupn→∞⁡xn=T, then there exists an integer N∈N+ such that both the number sequences {xN+2t} and {yN+2t+k+1} are monotonically decreasing, and xN+2txN+2t-1=1, yN+2t+kyN+2t+k+1=1 for all t≥0.
(2) Iflimsupn→∞⁡yn=T, then there exists an integer N∈N+ such that both the number sequences {yN+2t} and {xN+2t+k+1} are monotonically decreasing, and yN+2tyN+2t-1=1, xN+2t+kxN+2t+k+1=1 for all t≥0.Proof.
Here, we only prove(1). By Lemma 5, if limsupn→∞⁡xn=T and limsupn→∞⁡yn=M≤T, then there exists N∈N+ such that xN+2t≥xN+2t+2≥T for all t≥0. So {xN+2t} is monotonically decreasing for all t≥0. There exists a constant ε>0 such that ayN+2t-k≤T-ε. Hence,(17)1xN+2t-1≤xN+2t=max⁡1xN+2t-1,AN+2t-1yN+2t-k-1≤max⁡1xN+2t-1,byN+2t-k≤max⁡1xN+2t-1,T-ε≤1xN+2t-1.This implies xN+2txN+2t-1=1. Without loss of generality, we assume yN+2t+k<T/a, so(18)1yN+2t+k≤yN+2t+k+1=max⁡1yN+2t+k,BN+2t+kxN+2t=1yN+2t+k≤yN+2t+k-1.Hence, yN+2t+kyN+2t+k+1=1, and {yN+2t+k+1} is monotonically decreasing for all t≥0. This completes the proof of the Lemma.Theorem 7.
Let{(xn,yn)}n=-k+∞ be a solution of system (5). If k is even, then one of the {xn} and {yn} is eventually periodic with period 2.Proof.
By Lemma6, if limsupn→∞⁡xn=T, then there exists N∈N+ such that both {xN+2t} and {yN+2t+k+1} are monotonically decreasing. We can easily get xN+2txN+2t-1=1 and yN+2t+kyN+2t+k+1=1 for all t≥0.
We will prove that{yn} is eventually periodic with period 2. Otherwise, there will be infinitely many kj such that(19)yN+2kj+1=max⁡1yN+2kj,BN+2kjxN+2kj-k=BN+2kjxN+2kj-k>1yN+2kj.We take a subsequence {BN+2kj} and assume BN+2kj→B≤a, yN+2kj→L. Since -k is even, xN+2kj-k→T. Take the limit on both sides of the inequality BN+2kj/xN+2kj-k>1/yN+2kj; hence BL>T. Since T≥L, 1>a≥B, we have aT≥BL, so T>aT≥BL>T. This is a contradiction. Therefore {yn} is eventually periodic with period 2.
Similarly, we obtain that iflimsupn→∞⁡yn=T, then {xn} is eventually periodic with period 2. This completes the proof of the theorem.Theorem 8.
If bothAn and Bn are periodic sequences, then the solution {(xn,yn)}n=-k+∞ of system (5) converges to a period two solution.Proof.
We will only prove the case oflimsupn→∞⁡xn=T. By Lemma 6, we know that if limsupn→∞⁡xn=T, then there exists N∈N+ such that both {xN+2t} and {yN+2t+k+1} are monotonically decreasing. We can get xN+2txN+2t-1=1 and yN+2t+kyN+2t+k+1=1 for all t≥0 easily. So both {xN+2t-1} and {yN+2t+k} are monotonically increasing.
Now, we prove that{xn} is eventually periodic with period 2. Assume that {xn} is not eventually periodic with period 2; then {xN+2t-1} is not an eventual constant, and there exist infinitely many kj→+∞ such that(20)xN+2kj-1=max⁡1xN+2kj-2,AN+2kj-2yN+2kj-k-2=AN+2kj-2yN+2kj-k-2>1xN+2kj-2.Therefore(21)AN+2kj-2=xN+2kj-1yN+2kj-k-2.By Lemma 6, we know that {xN+2kj-1yN+2kj-k-2} is monotonically increasing, not an eventual constant, so {AN+2kj-2} is monotonically increasing, not an eventual constant. This is a contradiction. So {xn} is eventually periodic with period 2.
Similarly, we obtain that{yn} is eventually periodic with period 2. So {(xn,yn)}n=-k+∞ converges to a period two solution. This completes the proof of the theorem.



---

*Source: 2095805-2017-12-24.xml*