# Global Behavior of the Max-Type Difference Equationxn+1=max{1/xn,An/xn−1}

Authors: Taixiang Sun; Bin Qin; Hongjian Xi; Caihong Han
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We study global behavior of the following max-type difference equationxn+1=max{1/xn,An/xn−1}, n=0,1,…, where {An}n=0∞ is a sequence of positive real numbers with 0≤infAn≤supAn<1. The special case when {An}n=0∞ is a periodic sequence with period k and An∈(0,1) for every n≥0 has been completely investigated by Y. Chen. Here we extend his results to the general case.


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## Body

## 1. Introduction

In the recent years, there has been a lot of  interest in studying the global behavior of, the so-called, max-type difference equations; see, for example, [1–17] (see also references therein). In [1, 3–5, 7, 8], the second order max-type difference equation
(1.1)xn+1=max{1xn,Anxn-1},n=0,1,…
has been studied for positive coefficients An, which are periodic with period k. The case k=1 was studied in [1], the case k=2 was studied in [3], the case k=3 was studied in [4, 8], and the more difficult case k=4 was studied in [7]. Chen [5] found that every positive solution of (1.1) is eventually periodic with period 2 when {An}n=0∞ is a periodic sequence of positive real numbers with period k≥2 and An∈(0,1) for all n≥0. These results were also included in the recent monograph [9] along with other related references. In this paper, we study global behavior of (1.1) when {An}n=0∞ is a sequence of positive real numbers with 0≤infAn≤supAn<1.


## 2. Main Results

The main results of this paper are established through the following lemmas.Lemma 2.1.
Let{xn}n=-1∞ be a positive solution of (1.1), then(1)
xn+1xn≥1 for all n≥0;(2)
ifxk+1xk>1 for some k≥1, then xk+2xk+1=1.Proof.
(1) is obvious sincexn+1≥1/xn for all n≥0.(2)
Ifxk+1xk>1 for some k≥1, then xk+1xk-1=Ak. Suppose for the sake of contradiction that xk+2xk+1>1, then similarly we get xk+2xk=Ak+1 and
(2.1)Ak+1Ak=xk+1xk-1xk+2xk≥1.This is a contradiction since Ak+1<1 and Ak<1. The proof is complete.Lemma 2.2.
Let{xn}n=-1∞ be a positive solution of (1.1) and Pn=max{xn,xn-1} for all n≥1. Then(1)
xn+1≤Pn and Pn is nonincreasing;(2)
xn is bounded, and moreover 1/P1≤xn≤P1 for any n≥1.Proof.
By Lemma2.1(1) and the assumption An<1, we obtain that for any n≥1,
(2.2)xn+1=max{xn-1xnxn-1,Anxnxnxn-1}≤max{xn-1,xn}=Pn.
Hence
(2.3)Pn+1=max{xn+1,xn}≤Pn,
which implies that for all n≥1,(2.4)xn≤P1.
Furthermore, it follows that for all n≥1,(2.5)xn+1=max{1xn,Anxn-1}≥1xn≥1P1.
The proof is complete.Remark 2.3.
Note that from the proof of Lemma2.2 we have that P1≥1.Remark 2.4.
Various sequences which satisfy inequality in Lemma2.2(1), that is, xn+1≤Pn have been studied, for example, in [18–24].Lemma 2.5.
Let{xn}n=-1∞ be a positive solution of (1.1) and limn→∞Pn=S. Then S=limsupn→∞xn.Proof.
SincePn is a subsequence of xn, it follows that
(2.6)S≤limsupn→∞xn.
On the other hand, by xn+1≤Pn for all n≥1, we obtain
(2.7)limsupn→∞xn≤limsupn→∞Pn=S.
The proof is complete.Remark 2.6.
Let{xn}n=-1∞ be a positive solution of (1.1). By Lemma 2.2, we see that if S=limsupn→∞xn and xN<S for some N>0, then xN-1,xN+1∈[S,+∞). For example, if it were xN-1<S, then it would be PN<S, which would imply limsupn→∞xn<S.Lemma 2.7.
Suppose that{xn}n=-1∞ is a positive solution of (1.1) and S=limsupn→∞xn. Write
(2.8)ω(xn)={x:thereexist-1≤k1<k2<⋯<kn<⋯suchthatlimn→∞xkn=x}.
Then ω(xn)={S,1/S}.Proof.
Ifω(xn) contains only one point, we may assume by taking a subsequence that Ank→μ(<1). By taking the limit in the following relationship:
(2.9)xnk+1=max{1xnk,Ankxnk-1},
as k→∞, we obtain
(2.10)S=max{1S,μS}=1S,
which implies that S=1.
If ω(xn) contains at least two points, let L∈ω(xn)-{S}, then there exists a subsequence xnk of xn such that
(2.11)xnk→L<S.
By Remark 2.6, we see that there exists N>0 such that for every nk>N,
(2.12)xnk<S,xnk+1,xnk-1∈[S,+∞),
from which it follows that
(2.13)xnk+1→S,xnk-1→S.
By taking a subsequence we may assume that Ank→μ(<1). By taking the limit in the following relationship:
(2.14)xnk+1=max{1xnk,Ankxnk-1},
as k→∞, we obtain
(2.15)S=max{1L,μS}=1L,
which implies
(2.16)L=1S.
The proof is complete.Theorem 2.8.
Let{xn}n=-1∞ be a positive solution of (1.1) and S=limsupn→∞xn. Then one of the following two statements is true.(1)
If there exist infinitely manyn such that xn≥S and xn+1≥S, then {xn}n=-1∞ is eventually equal to 1.(2)
If there existsN such that xN+2k<S and xN+2k-1≥S for all k≥0, then xN+2k→1/S and xN+2k-1→S.Proof.
(1) We assume that there exists an infinite sequencen1<n2<n3<⋯<nk<⋯ such that
(2.17)xnk≥S,xnk+1≥S.
By taking a subsequence we may assume from Lemma 2.7 that
(2.18)Ank→μ<1,xnk-1→l∈{S,1S}.
By taking the limit in the following relationship:
(2.19)xnk+1xnk=max{1,Ankxnkxnk-1},
as k→∞, we get
(2.20)S2=max{1,Sμl}.
Since Sμ/l∈{μ,μS2} and μ<1, it follows that S2=1 and ω(xn)={1}.In the following, we show that {xn}n=-1∞ is eventually equal to 1. It only needs to prove that there exists N≥0 such that for all n≥N,(2.21)1xn>Anxn-1.
Indeed, if there exist infinitely many nk such that
(2.22)xnk+1=Ankxnk-1,
by taking a subsequence we may assume that Ank→μ<1, then it follows that
(2.23)1=μ1,μ=1,
which is a contradiction. Therefore there exists N such that for all n≥N,(2.24)xn+1=1xn.
Thus
(2.25)xn=xN,forn=N+2k,xn=xN+1,forn=N+2k+1.
Since xn→1, we have xN+1=xN=1.
(2)  If S=1, then the result follows from Lemma 2.7. In the following, we assume S≠1. Suppose for the sake of contradiction that there exists a subsequence xN+2ki of xN+2k such that
(2.26)xN+2ki→S.
By taking a subsequence we may assume that
(2.27)AN+2ki→μ.
By taking the limit in the following relationship:
(2.28)xN+2ki+1=max{1xN+2ki,AN+2kixN+2ki-1},
as ki→∞, we get
(2.29)S=max{1S,μS},
which implies
(2.30)S=1.
This is a contradiction. The proof is complete.Corollary 2.9.
Let{An}n=0∞ be a periodic sequence of positive real numbers, then every positive solution of (1.1) is eventually periodic with period 2.Proof.
Let{xn}n=-1∞ be a positive solution of (1.1) and S=limsupn→∞xn. By Remark 2.6 and Theorem 2.8, we may assume without loss of generality that x2k<S, x2k-1≥S≥1 for all k≥0. Suppose for the sake of contradiction that there exists a sequence m1<m2<⋯<mk<⋯ such that(1)
xmk+1xmk-1=Amk, and xmk+1xmk>1;(2)
xn+1xn=1, for n≠mk.Then mk is odd for every k≥1. Let mk=2nk+1, then it follows from Lemma 2.1 that
(2.31)x2nk+2x2nk=A2nk+1<1=x2nk+1x2nk<x2nk+1x2nk+2.
From this and by (2) it follows that
(2.32)A2nk+1x2nk+2=x2nk<x2nk+2=x2nk+4=⋯=x2nk+1<x2nk+1+2=A2nk+1+1x2nk+1.
Therefore for every k≥1,
(2.33)A2nk+1<x2nk+22=x2nk+12<A2nk+1+1,
which is a contradiction since {An}n=0∞ is a periodic sequence. The proof is complete.Remark 2.10.
Corollary2.9 is the main result of [5].


## 3. Example

In this section, we give an example for{An}n=0∞ to be no periodic sequence.Example 3.1.
Consider(3.1)xn+1=max{1xn,Anxn-1},n=0,1,…,
where A2n=A2n+1=(2-1/2n)(2-1/2n+1)/16 for any n≥0. Then solution {xn}n=-1∞ of (3.1) with the initial values x-1=1/4 and x0=4 satisfies the following.(1)
x2p-1x2p=1, for any p≥0.(2)
x2p-1<x2p+1=A2px2p-1<12<2<x2p+2<x2p,for anyp≥0.Proof.
By simple computation, we have(3.2)A2p=(2-1/2p)(2-1/2p+1)16>{x-12,ifp=0,(A0x-1)2,ifp=1,(A2p-2A2p-6⋯A2A2p-4A2p-8⋯A0x-1)2,ifp≥2iseven,(A2p-2A2p-6⋯A4A0A2p-4A2p-8⋯A2x-1)2,ifp≥2isodd.
It follows from (3.1) and (3.2) that
(3.3)x1x-1=max{x-1x0,A0}=max{x-12,A0}=A0,x2x1=max{1,x1A1x0}=max{1,A0A1x-1x0}=1,x3x1=max{x1x2,A2}=max{x12x2x1,A2}=max{(A0x-1)2,A2}=A2,x4x3=max{1,x3A3x2}=max{1,A2A3x2x1}=1,x5x3=max{x3x4,A4}=max{x32x4x3,A4}=max{(x3x1x1x-1x-1)2,A4}=max{(A2A0x-1)2,A4}=A4,x6x5=max{1,x5A5x4}=max{1,A4A5x4x3}=1,x7x5=max{x5x6,A6}=max{x52x6x5,A6}=max{(x5x3x1x-1x3x1x-1)2,A6}=max{(A4A0A2x-1)2,A6}=A6,x8x7=max{1,x7A7x6}=max{1,A6A7x6x5}=1.
By induction, we have from (3.1) and (3.2) that for any p≥1,(3.4)x4p+1x4p-1=max{x4p-1x4p,A4p}=max{x4p-12x4px4p-1,A4p}=max{x4p-12,A4p}=max{(x4p-1x4p-3x4p-5⋯x1x4p-3x4p-5x4p-7⋯x-1x-1)2,A4p}=max{(A4p-2A4p-6⋯A2A4p-4A4p-8⋯A0x-1)2,A4p}=A4p,x4p+2x4p+1=max{1,x4p+1A4p+1x4p}=max{1,A4pA4p+1x4px4p-1}=1,x4p+3x4p+1=max{x4p+1x4p+2,A4p+2}=max{x4p+12x4p+2x4p+1,A4p+2}=max{x4p+12,A4p+2}=max{(x4p+1x4p-1x4p-3x4p-5⋯x1x-1x4p-1x4p-3x4p-5x4p-7⋯x1x-1)2,A4p+2}=max{(A4pA4p-4⋯A4A0A4p-2A4p-6⋯A2x-1)2,A4p+2}=A4p+2,x4p+4x4p+3=max{1,x4p+3A4p+3x4p+2}=max{1,A4p+2A4p+3x4p+2x4p+1}=1.
from which the result follows. The proof is   complete.



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