# On the Global Attractivity of a Max-Type Difference Equation

Authors: Ali Gelişken; Cengiz Çinar
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We investigate asymptotic behavior and periodic nature of positive solutions of the difference equationxn=max{A/xn−1,1/xn−3α},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.


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## Body

## 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:(1.1)xn=max{A1xn−1,A2xn−2,…,Apxn−p},n=0,1,…, where A1,A2,…,Ap are real
numbers and initial conditions are nonzero real
numbers.In [17], asymptotic behavior of positive solutions of
the difference equation was investigated(1.2)xn=max{1xn−1α,Axn−2},n=0,1,…,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,(1.3)xn=max{Axn−1α,Bxn−2β},n=0,1,…, where 0<α,β<1,0<A, and 0<B, converges
to  x¯=max{A1/(α+1),B1/(β+1)}.In this paper, we investigate the difference
equation(1.4)xn=max{Axn−1,1xn−3α},n=0,1,…,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 CaseA=1

In this section, we
consider the difference equation(2.1)xn=max{1xn−1,1xn−3α},n=0,1,…,where 0<α<1.Theorem 2.1.
Letxn be a solution
of (2.1). Then xn approaches x¯=1.Proof.
Choose a numberB such that 0<B<1, let xn=Byn for n≥−3. Then (2.1)
implies the difference equation(2.2)yn=min{−yn−1,−αyn−3},n=0,1,…, where 0<α<1   and initial
conditions are real numbers.
Letyn be a solution
of (2.2). Then it suffices to prove yn→0. Observe that
there exists a positive integer N such
that(2.3)yN=−yN−1,yN+1=−αyN−2forN≥0.By computation, we get that
(2.4)yN+2=−yN+1,yN+3=−αyN,yN+4=−yN+3and then(2.5)y4n+N=αnyN,y4n+N+1=αnyN+1,y4n+N+2=−y4n+N+1,y4n+N+3=−αy4n+N∀n≥0. So, y4n+N→0,y4n+N+1→0,y4n+N+2→0,y4n+N+3→0.     This implies yn→0.


## 3. The Case0<A<1

In this section, we consider (1.4), where 0<α<1.Letxn=Ayn,n≥−3. Equation (1.4)
implies the difference equation(3.1)yn=min{1−yn−1,−αyn−3},n=0,1,…,where initial conditions are real
numbers.Lemma 3.1.
Letyn be a solution
of (3.1). Then for all n≥0,(3.2)|yn|≤max{|yn−1|−1,α|yn−3|}.Proof.
From (3.1), we have the following
statements:
ifyn−1≥0 and yn−3≥0, then |yn|≤max{|yn−1|−1,α|yn−3|};ifyn−1≤0 and yn−3≤0, then |yn|≤α|yn−3|;ifyn−1≥0 and yn−3≤0, then |yn|≤max{|yn−1|−1,α|yn−3|};ifyn−1≤0 and yn−3≥0, then |yn|=α|yn−3|.
In general, we have|yn|≤max{|yn−1|−1,α|yn−3|}foralln≥0.Theorem 3.2.
ifxn is a solution
of (1.4), xn approaches x¯=1.Proof.
Letyn  be a solution of
(3.1). To prove xn→1, it suffices to
prove yn→0.
Choose a numberβ such that 0<|yn−1|−1≤β|yn|. Then from
inequality (3.2), we get that(3.3)|yn|≤max{β|yn−1|,α|yn−3|},n≥0.
Letγ=max{β,α}, then 0<γ<1 and(3.4)|yn|≤γmax{|yn−1|,|yn−3|},n≥0.
From (3.4) and by induction, we get
that(3.5)|yn|≤γ[n/3]+1max{|y−1|,|y−2|,|y−3|},n≥0.
This impliesyn→0.


## 4. The CaseA>1

In this section, we consider (1.4).  Let xn=Ayn for n≥−3. Equation
(1.4) implies the difference
equation(4.1)yn=max{1−yn−1,−αyn−3},n=0,1,…, where 0<α<1 and initial
conditions are real numbers.Theorem 4.1.
Ifxn is a solution
of (1.4), then the following statements are true:
(a)
xn approaches x¯=1, if there is an
integer N such
that(4.2)xN=AxN−1,xN+1=1xN−2αfor0≤n≤N.(b)
xn is eventually
periodic with period 2, if there is an
integer N such
that(4.3)xN=AxN−1,xN+1=xN−1,A−α/(1−α)≤xN−1≤A1/(1−α)for0≤n≤N.Proof.
(a)  Change of
variablesxn=Ayn,n≥−3. If xN=A/xN−1 and xN+1=1/xN−2α for 0≤n≤N, then yN=1−yN−1 and yN+1=−αyN−2.
Let yn be a solution
of (4.1). So, to prove xn→1, it suffices to
prove yn→0. From (4.1),
there is at least an integer N such that yN=1−yN−1>0 and yN+1=−αyN−2 for 0≤n≤N. By computation
from (4.1), we get that yN+1<0, yN−2≥yN>0 and
then(4.4)yN+2=1+αyN−2,yN>yN+2>0,yN+3=−αyN,yN+1<yN+3<0,yN+4=1+αyN,yN>yN+2≥yN+4>0,yN+5=−αyN+2,yN+1≤yN+3<yN+5<0.
So, we have(4.5)y4n+N>y4n+N+2≥y4n+N+4>0,y4n+N+1≤y4n+N+3<y4n+N+5<0for all n≥0. This implies yn→0.
(b)  Change of
variablesxn=Ayn, n≥−3. Let yn be a solution
of (4.1).
IfxN=A/xN−1,xN+1=xN−1, and A−α/(1−α)≤xN−1≤A1/(1−α)   for 0≤n≤N, then(4.6)yN=1−yN−1,yN+1=yN−1,−α1−α≤yN−1≤11−α.Clearly, there is at least an
integer N such that yN=1−yN−1 for 0≤n≤N.
Suppose thatyN+1=yN−1 and yN−1∉[−α/(1−α),1/(1−α)].
IfyN−1>1/(1−α) then from
(4.1), we have yN+2=−αyN−1. So, from (a) we get
immediately that yn→0.
IfyN−1<−α/(1−α), then we have yN+2=1−yN+1 and yN+3=−αyN. Then we get
that yn→0, from (a).
We assume thatyN+1=yN−1 and −α/(1−α)≤yN−1≤1/(1−α). To prove the
desired result, it suffices to show that yn is eventually
periodic with period 2. By computation from (4.1), we get immediately yN=1−yN−1 for all 0≤n≤N. This is the
desired result.



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*Source: 812674-2009-04-26.xml*