AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation15296410.1155/2009/152964152964Research ArticleGlobal Behavior of the Max-Type Difference Equation xn+1=max{1/xn,An/xn1}SunTaixiang1QinBin2XiHongjian2HanCaihong1StevicStevo1College of Mathematics and Information ScienceGuangxi UniversityNanning 530004Chinagxu.edu.cn2Department of MathematicsGuangxi College of Finance and EconomicsNanning 530003China20093103200920090402200902032009080320092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study global behavior of the following max-type difference equation xn+1=max{1/xn,An/xn1}, n=0,1,, where {An}n=0 is a sequence of positive real numbers with 0infAnsupAn<1. The special case when {An}n=0 is a periodic sequence with period k and An(0,1) for every n0 has been completely investigated by Y. Chen. Here we extend his results to the general case.

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,  (see also references therein). In [1, 35, 7, 8], the second order max-type difference equation 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 , the case k=2 was studied in , the case k=3 was studied in [4, 8], and the more difficult case k=4 was studied in . Chen  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 k2 and An(0,1) for all n0. These results were also included in the recent monograph  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 0infAnsupAn<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

xn+1xn1 for all n0;

if xk+1xk>1 for some k1, then xk+2xk+1=1.

Proof.

(1) is obvious since xn+11/xn for all n0.

If xk+1xk>1 for some k1, 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 Ak+1Ak=xk+1xk-1xk+2xk1.

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 n1. Then

xn+1Pn and Pn is nonincreasing;

xn is bounded, and moreover 1/P1xnP1 for any n1.

Proof.

By Lemma 2.1(1) and the assumption An<1, we obtain that for any n1, xn+1=max{xn-1xnxn-1,Anxnxnxn-1}max{xn-1,xn}=Pn. Hence Pn+1=max{xn+1,xn}Pn, which implies that for all n1,xnP1. Furthermore, it follows that for all n1,xn+1=max{1xn,Anxn-1}1xn1P1. The proof is complete.

Remark 2.3.

Note that from the proof of Lemma 2.2 we have that P11.

Remark 2.4.

Various sequences which satisfy inequality in Lemma 2.2(1), that is, xn+1Pn have been studied, for example, in .

Lemma 2.5.

Let {xn}n=-1 be a positive solution of (1.1) and limnPn=S. Then S=limsupnxn.

Proof.

Since Pn is a subsequence of xn, it follows that Slimsupnxn. On the other hand, by xn+1Pn for all n1, we obtain limsupnxnlimsupnPn=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=limsupnxn 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 limsupnxn<S.

Lemma 2.7.

Suppose that {xn}n=-1 is a positive solution of (1.1) and S=limsupnxn. Write ω(xn)={x:there  exist  -1k1<k2<<kn<  such  that  limnxkn=x}. Then ω(xn)={S,1/S}.

Proof.

If ω(xn) contains only one point, we may assume by taking a subsequence that Ankμ(<1). By taking the limit in the following relationship: xnk+1=max{1xnk,Ankxnk-1}, as k, we obtain S=max{1S,μS}=1S, which implies that S=1. If ω(xn) contains at least two points, let Lω(xn)-{S}, then there exists a subsequence xnk of xn such that xnkL<S. By Remark 2.6, we see that there exists N>0 such that for every nk>N, xnk<S,xnk+1,xnk-1[S,+), from which it follows that xnk+1S,xnk-1S. By taking a subsequence we may assume that Ankμ(<1). By taking the limit in the following relationship: xnk+1=max{1xnk,Ankxnk-1}, as k, we obtain S=max{1L,μS}=1L, which implies L=1S. The proof is complete.

Theorem 2.8.

Let {xn}n=-1 be a positive solution of (1.1) and S=limsupnxn. Then one of the following two statements is true.

If there exist infinitely many n such that xnS and xn+1S, then {xn}n=-1 is eventually equal to 1.

If there exists N such that xN+2k<S and xN+2k-1S for all k0, then xN+2k1/S and xN+2k-1S.

Proof.

(1) We assume that there exists an infinite sequence n1<n2<n3<<nk< such that xnkS,xnk+1S. By taking a subsequence we may assume from Lemma 2.7 that Ankμ<1,xnk-1l{S,1S}. By taking the limit in the following relationship: xnk+1xnk=max{1,Ankxnkxnk-1}, as k, we get 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 N0 such that for all nN,1xn>Anxn-1. Indeed, if there exist infinitely many nk such that xnk+1=Ankxnk-1, by taking a subsequence we may assume that Ankμ<1, then it follows that 1=μ1,μ=1, which is a contradiction. Therefore there exists N such that for all nN,xn+1=1xn. Thus xn=xN,for  n=N+2k,xn=xN+1,for  n=N+2k+1. Since xn1, we have xN+1=xN=1. (2) If S=1, then the result follows from Lemma 2.7. In the following, we assume S1. Suppose for the sake of contradiction that there exists a subsequence xN+2ki of xN+2k such that xN+2kiS. By taking a subsequence we may assume that AN+2kiμ. By taking the limit in the following relationship: xN+2ki+1=max{1xN+2ki,AN+2kixN+2ki-1}, as ki, we get S=max{1S,μS}, which implies 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=limsupnxn. By Remark 2.6 and Theorem 2.8, we may assume without loss of generality that x2k<S, x2k-1S1 for all k0. Suppose for the sake of contradiction that there exists a sequence m1<m2<<mk< such that

xmk+1xmk-1=Amk, and xmk+1xmk>1;

xn+1xn=1, for nmk.

Then mk is odd for every k1. Let mk=2nk+1, then it follows from Lemma 2.1 that x2nk+2x2nk=A2nk+1<1=x2nk+1x2nk<x2nk+1x2nk+2. From this and by (2) it follows that A2nk+1x2nk+2=x2nk<x2nk+2=x2nk+4==x2nk+1<x2nk+1+2=A2nk+1+1x2nk+1. Therefore for every k1, 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.

Corollary 2.9 is the main result of .

3. Example

In this section, we give an example for {An}n=0 to be no periodic sequence.

Example 3.1.

Consider xn+1=max{1xn,Anxn-1},n=0,1,, where A2n=A2n+1=(2-1/2n)(2-1/2n+1)/16 for any n0. Then solution {xn}n=-1 of (3.1) with the initial values x-1=1/4 and x0=4 satisfies the following.

x2p-1x2p=1, for any p0.

x2p-1<x2p+1=A2px2p-1<12<2<x2p+2<x2p,  for any  p0.

Proof.

By simple computation, we have A2p=(2-1/2p)(2-1/2p+1)16>{x-12,if  p=0,(A0x-1)2,if  p=1,(A2p-2A2p-6A2A2p-4A2p-8A0x-1)2,if  p2  is  even,(A2p-2A2p-6A4A0A2p-4A2p-8A2x-1)2,if  p2  is  odd.   It follows from (3.1) and (3.2) that 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 p1,x4p+1x4p-1=max{x4p-1x4p,A4p}=max{x4p-12x4px4p-1,A4p}=max{x4p-12,A4p}=max{(x4p-1x4p-3x4p-5x1x4p-3x4p-5x4p-7x-1x-1)2,A4p}=max{(A4p-2A4p-6A2A4p-4A4p-8A0x-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-5x1x-1x4p-1x4p-3x4p-5x4p-7x1x-1)2,A4p+2}=max{(A4pA4p-4A4A0A4p-2A4p-6A2x-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.

Acknowledgments

The project is supported by NNSF of China(10861002), NSF of Guangxi (0640205,0728002), and Innovation Project of Guangxi Graduate Education(2008105930701M43).

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