On the Global Attractivity of a Max-Type Difference Equation

Abstract

We investigate asymptotic behavior and periodic nature of positive solutions of the difference equation 𝑥_𝑛=max{𝐴/𝑥_𝑛−1,1/𝑥^𝛼_𝑛−3},𝑛=0,1,…, where 𝐴>0 and 0<𝛼<1. We prove that every positive solution of this difference equation approaches 𝑥=1 or is eventually periodic with period 2.

1. Introduction

In the recent years, there has been a lot of interest in studying the global attractivity and the periodic nature of, so-called, max-type difference equations (see, e.g., [1–17] and references therein).

In [10], the following difference equation was proposed by Ladas:𝑥_𝑛𝐴=max{¹𝑥_𝑛−1,𝐴₂𝑥_𝑛−2𝐴,…,^𝑝𝑥_𝑛−𝑝},𝑛=0,1,…, (1.1) where 𝐴₁,𝐴₂,…,𝐴_𝑝 are real numbers and initial conditions are nonzero real numbers.

In [17], asymptotic behavior of positive solutions of the difference equation was investigated𝑥_𝑛1=max{𝑥^𝛼_𝑛−1,𝐴𝑥_𝑛−2},𝑛=0,1,…, (1.2)where 0<𝛼<1 and 0<𝐴. It was showed that every positive solution of this difference equation approaches 𝑥=1 or is eventually periodic with period 4.

In [14], it was proved that every positive solution of the difference equation,𝑥_𝑛𝐴=max{𝑥^𝛼_𝑛−1,𝐵𝑥^𝛽_𝑛−2},𝑛=0,1,…, (1.3) where 0<𝛼,𝛽<1,0<𝐴, and 0<𝐵, converges to 𝑥=max{𝐴^1/(𝛼+1),𝐵^1/(𝛽+1)}.

In this paper, we investigate the difference equation𝑥_𝑛𝐴=max{𝑥_𝑛−1,1𝑥^𝛼_𝑛−3},𝑛=0,1,…, (1.4)where 0<𝐴,0<𝛼<1 and initial conditions are positive real numbers. We prove that every positive solution of this difference equation approaches 𝑥=1 or is eventually periodic with period 2.

2. The Case 𝐴=1

In this section, we consider the difference equation𝑥_𝑛1=max{𝑥_𝑛−1,1𝑥^𝛼_𝑛−3},𝑛=0,1,…, (2.1)where 0<𝛼<1.Theorem 2.1. Let 𝑥_𝑛 be a solution of (2.1). Then 𝑥_𝑛 approaches 𝑥=1.Proof. Choose a number 𝐵 such that 0<𝐵<1, let 𝑥_𝑛=𝐵^𝑦𝑛 for 𝑛≥−3. Then (2.1) implies the difference equation𝑦_𝑛=min{−𝑦_𝑛−1,−𝛼𝑦_𝑛−3},𝑛=0,1,…, (2.2) where 0<𝛼<1 and initial conditions are real numbers.
Let 𝑦_𝑛 be a solution of (2.2). Then it suffices to prove 𝑦_𝑛→0. Observe that there exists a positive integer 𝑁 such that𝑦_𝑁=−𝑦_𝑁−1,𝑦_𝑁+1=−𝛼𝑦_𝑁−2for𝑁≥0. (2.3)By computation, we get that 𝑦_𝑁+2=−𝑦_𝑁+1,𝑦_𝑁+3=−𝛼𝑦_𝑁,𝑦_𝑁+4=−𝑦_𝑁+3 (2.4)and then𝑦_4𝑛+𝑁=𝛼^𝑛𝑦_𝑁,𝑦_4𝑛+𝑁+1=𝛼^𝑛𝑦_𝑁+1,𝑦_4𝑛+𝑁+2=−𝑦_4𝑛+𝑁+1,𝑦_4𝑛+𝑁+3=−𝛼𝑦_4𝑛+𝑁∀𝑛≥0. (2.5) So, 𝑦_4𝑛+𝑁→0,𝑦_4𝑛+𝑁+1→0,𝑦_4𝑛+𝑁+2→0,𝑦_4𝑛+𝑁+3→0. This implies 𝑦_𝑛→0.

3. The Case 0<𝐴<1

In this section, we consider (1.4), where 0<𝛼<1.

Let 𝑥_𝑛=𝐴^𝑦𝑛,𝑛≥−3. Equation (1.4) implies the difference equation𝑦_𝑛=min{1−𝑦_𝑛−1,−𝛼𝑦_𝑛−3},𝑛=0,1,…, (3.1)where initial conditions are real numbers.Lemma 3.1. Let 𝑦_𝑛 be a solution of (3.1). Then for all 𝑛≥0,|𝑦_𝑛|≤max{|𝑦_𝑛−1|−1,𝛼|𝑦_𝑛−3|}. (3.2)Proof. From (3.1), we have the following statements:
 if 𝑦_𝑛−1≥0 and 𝑦_𝑛−3≥0, then |𝑦_𝑛|≤max{|𝑦_𝑛−1|−1,𝛼|𝑦_𝑛−3|}; if 𝑦_𝑛−1≤0 and 𝑦_𝑛−3≤0, then |𝑦_𝑛|≤𝛼|𝑦_𝑛−3|; if 𝑦_𝑛−1≥0 and 𝑦_𝑛−3≤0, then |𝑦_𝑛|≤max{|𝑦_𝑛−1|−1,𝛼|𝑦_𝑛−3|}; if 𝑦_𝑛−1≤0 and 𝑦_𝑛−3≥0, then |𝑦_𝑛|=𝛼|𝑦_𝑛−3|.
In general, we have |𝑦_𝑛|≤max{|𝑦_𝑛−1|−1,𝛼|𝑦_𝑛−3|}forall𝑛≥0.Theorem 3.2. if𝑥_𝑛 is a solution of (1.4), 𝑥_𝑛 approaches 𝑥=1.Proof. Let 𝑦_𝑛 be a solution of (3.1). To prove 𝑥_𝑛→1, it suffices to prove 𝑦_𝑛→0.
Choose a number 𝛽 such that 0<|𝑦_𝑛−1|−1≤𝛽|𝑦_𝑛|. Then from inequality (3.2), we get that|𝑦_𝑛|≤max{𝛽|𝑦_𝑛−1|,𝛼|𝑦_𝑛−3|},𝑛≥0. (3.3)
Let 𝛾=max{𝛽,𝛼}, then 0<𝛾<1 and|𝑦_𝑛|≤𝛾max{|𝑦_𝑛−1|,|𝑦_𝑛−3|},𝑛≥0. (3.4)
From (3.4) and by induction, we get that|𝑦_𝑛|≤𝛾^[𝑛/3]+1max{|𝑦₋₁|,|𝑦₋₂|,|𝑦₋₃|},𝑛≥0. (3.5)
This implies 𝑦_𝑛→0.

4. The Case 𝐴>1

In this section, we consider (1.4). Let 𝑥_𝑛=𝐴^𝑦𝑛 for 𝑛≥−3. Equation (1.4) implies the difference equation𝑦_𝑛=max{1−𝑦_𝑛−1,−𝛼𝑦_𝑛−3},𝑛=0,1,…, (4.1) where 0<𝛼<1 and initial conditions are real numbers.Theorem 4.1. If 𝑥_𝑛 is a solution of (1.4), then the following statements are true:
(a)𝑥_𝑛 approaches 𝑥=1, if there is an integer 𝑁 such that𝑥_𝑁=𝐴𝑥_𝑁−1,𝑥_𝑁+1=1𝑥^𝛼_𝑁−2for0≤𝑛≤𝑁. (4.2)(b)𝑥_𝑛 is eventually periodic with period 2, if there is an integer 𝑁 such that𝑥_𝑁=𝐴𝑥_𝑁−1,𝑥_𝑁+1=𝑥_𝑁−1,𝐴^{−𝛼/(1−𝛼)}≤𝑥_𝑁−1≤𝐴^1/(1−𝛼)for0≤𝑛≤𝑁. (4.3)Proof. (a) Change of variables 𝑥_𝑛=𝐴^𝑦𝑛,𝑛≥−3. If 𝑥_𝑁=𝐴/𝑥_𝑁−1 and 𝑥_𝑁+1=1/𝑥^𝛼_𝑁−2 for 0≤𝑛≤𝑁, then 𝑦_𝑁=1−𝑦_𝑁−1 and 𝑦_𝑁+1=−𝛼𝑦_𝑁−2. Let 𝑦_𝑛 be a solution of (4.1). So, to prove 𝑥_𝑛→1, it suffices to prove 𝑦_𝑛→0. From (4.1), there is at least an integer 𝑁 such that 𝑦_𝑁=1−𝑦_𝑁−1>0 and 𝑦_𝑁+1=−𝛼𝑦_𝑁−2 for 0≤𝑛≤𝑁. By computation from (4.1), we get that 𝑦_𝑁+1<0, 𝑦_𝑁−2≥𝑦_𝑁>0 and then𝑦_𝑁+2=1+𝛼𝑦_𝑁−2,𝑦_𝑁>𝑦_𝑁+2𝑦>0,_𝑁+3=−𝛼𝑦_𝑁,𝑦_𝑁+1<𝑦_𝑁+3𝑦<0,_𝑁+4=1+𝛼𝑦_𝑁,𝑦_𝑁>𝑦_𝑁+2≥𝑦_𝑁+4𝑦>0,_𝑁+5=−𝛼𝑦_𝑁+2,𝑦_𝑁+1≤𝑦_𝑁+3<𝑦_𝑁+5<0. (4.4)
So, we have𝑦_4𝑛+𝑁>𝑦_4𝑛+𝑁+2≥𝑦_4𝑛+𝑁+4>0,𝑦_4𝑛+𝑁+1≤𝑦_4𝑛+𝑁+3<𝑦_4𝑛+𝑁+5<0 (4.5)for all 𝑛≥0. This implies 𝑦_𝑛→0.
(b) Change of variables 𝑥_𝑛=𝐴^𝑦𝑛, 𝑛≥−3. Let 𝑦_𝑛 be a solution of (4.1).
If 𝑥_𝑁=𝐴/𝑥_𝑁−1,𝑥_𝑁+1=𝑥_𝑁−1, and 𝐴^{−𝛼/(1−𝛼)}≤𝑥_𝑁−1≤𝐴^1/(1−𝛼) for 0≤𝑛≤𝑁, then𝑦_𝑁=1−𝑦_𝑁−1,𝑦_𝑁+1=𝑦_𝑁−1,−𝛼1−𝛼≤𝑦_𝑁−1≤11−𝛼. (4.6)Clearly, there is at least an integer 𝑁 such that 𝑦_𝑁=1−𝑦_𝑁−1 for 0≤𝑛≤𝑁.
Suppose that 𝑦_𝑁+1=𝑦_𝑁−1 and 𝑦_𝑁−1∉[−𝛼/(1−𝛼),1/(1−𝛼)].
If 𝑦_𝑁−1>1/(1−𝛼) then from (4.1), we have 𝑦_𝑁+2=−𝛼𝑦_𝑁−1. So, from (a) we get immediately that 𝑦_𝑛→0.
If 𝑦_𝑁−1<−𝛼/(1−𝛼), then we have 𝑦_𝑁+2=1−𝑦_𝑁+1 and 𝑦_𝑁+3=−𝛼𝑦_𝑁. Then we get that 𝑦_𝑛→0, from (a).
We assume that 𝑦_𝑁+1=𝑦_𝑁−1 and −𝛼/(1−𝛼)≤𝑦_𝑁−1≤1/(1−𝛼). To prove the desired result, it suffices to show that 𝑦_𝑛 is eventually periodic with period 2. By computation from (4.1), we get immediately 𝑦_𝑁=1−𝑦_𝑁−1 for all 0≤𝑛≤𝑁. This is the desired result.