We study global behavior of the following max-type difference equation xn+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.

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
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

xn+1xn≥1 for all n≥0;

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

Proof.

(1) is obvious since xn+1≥1/xn for all n≥0.

If xk+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
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

xn+1≤Pn and Pn is nonincreasing;

xn is bounded, and moreover 1/P1≤xn≤P1 for any n≥1.

Proof.

By Lemma 2.1(1) and the assumption An<1, we obtain that for any n≥1,
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 n≥1,xn≤P1.
Furthermore, it follows that for all n≥1,xn+1=max{1xn,Anxn-1}≥1xn≥1P1.
The proof is complete.

Remark 2.3.

Note that from the proof of Lemma 2.2 we have that P1≥1.

Remark 2.4.

Various sequences which satisfy inequality in Lemma 2.2(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.

Since Pn is a subsequence of xn, it follows that
S≤limsupn→∞xn.
On the other hand, by xn+1≤Pn for all n≥1, we obtain
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
ω(xn)={x:thereexist-1≤k1<k2<⋯<kn<⋯suchthatlimn→∞xkn=x}.
Then ω(xn)={S,1/S}.

Proof.

If ω(xn) contains only one point, we may assume by taking a subsequence that Ank→μ(<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 L∈ω(xn)-{S}, then there exists a subsequence xnk of xn such that
xnk→L<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+1→S,xnk-1→S.
By taking a subsequence we may assume that Ank→μ(<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=limsupn→∞xn. Then one of the following two statements is true.

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

If there exists N 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 sequence n1<n2<n3<⋯<nk<⋯ such that
xnk≥S,xnk+1≥S.
By taking a subsequence we may assume from Lemma 2.7 that
Ank→μ<1,xnk-1→l∈{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 N≥0 such that for all n≥N,1xn>Anxn-1.
Indeed, if there exist infinitely many nk such that
xnk+1=Ankxnk-1,
by taking a subsequence we may assume that Ank→μ<1, then it follows that
1=μ1,μ=1,
which is a contradiction. Therefore there exists N such that for all n≥N,xn+1=1xn.
Thus
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
xN+2ki→S.
By taking a subsequence we may assume that
AN+2ki→μ.
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=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

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

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
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 k≥1,
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 [5].

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 n≥0. 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 p≥0.

x2p-1<x2p+1=A2px2p-1<12<2<x2p+2<x2p,for anyp≥0.

Proof.

By simple computation, we have
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
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,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.

Acknowledgments

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

AmlehA. M.HoagJ.LadasG.A difference equation with eventually periodic solutionsBerenhautK. S.FoleyJ. D.StevićS.Boundedness character of positive solutions of a max difference equationBridenW. J.GroveE. A.LadasG.McGrathL. C.On the nonautonomous equation xn+1=max{An/xn,Bn/xn−1}Proceedings of the 3rd International Conference on Difference EquationsSeptember 1997Taipei, TaiwanGordon and Breach4973MR1735533ZBL0938.39012BridenW. J.GroveE. A.LadasG.KentC. M.Eventually periodic solutions of xn+1=max{1/xn,An/xn−1}ChenY.Eventual periodicity of xn+1=max{1/xn,An/xn−1} with periodic coefficientsÇinarC.StevićS.YalçinkayaI.On positive solutions of a reciprocal difference equation with minimumFeuerJ.On the eventual periodicity of xn+1=max{1/xn,An/xn−1} with a period-four parameterGroveE. A.KentC.LadasG.RadinM. A.On xn+1=max{1/xn,An/xn−1} with a period 3 parameterGroveE. A.LadasG.KentC. M.RadinM. A.On the boundedness nature of positive solutions of the difference equation xn+1=max{An/xn,Bn/xn−1} with periodic parametersPatulaW. T.VoulovH. D.On a max type recurrence relation with periodic coefficientsStevićS.On the recursive sequence xn+1=A+(xnp/xn−1r)StevićS.On the recursive sequence xn+1=max{c,xnp/xn−1p}SunF.On the asymptotic behavior of a difference equation with maximumSzalkaiI.On the periodicity of the sequence xn+1=max{A0/xn,⋯,Ak/xn−k}VoulovH. D.Periodic solutions to a difference equation with maximumVoulovH. D.On the periodic nature of the solutions of the reciprocal difference equation with maximumBibbyJ.Axiomatisations of the average and a further generalisation of monotonic sequencesCopsonE. T.On a generalisation of monotonic sequencesStevićS.A note on bounded sequences satisfying linear inequalitiesStevićS.A generalization of the Copson's theorem concerning sequences which satisfy a linear inequalityStevićS.A global convergence resultStevićS.Asymptotic behavior of a sequence defined by iteration with applicationsStevićS.Asymptotic behaviour of a nonlinear difference equation