# On the Asymptotic Behavior of a Difference Equation with Maximum

Authors: Fangkuan Sun
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

We study the asymptotic behavior of positive solutions to the difference equationxn=max{A/xn-1α,B/xn−2β}, n=0,1,…,
where 0<α,β<1,A,B>0. We prove that every positive solution to this equation converges to x∗=max{A1/(α+1),B1/(β+1)}.


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## Body

## 1. Introduction

Recently, there has been a considerable interest in
studying, the so-called, max-type difference
equations, see for example, [1–21] and the references cited therein. The max-type operators arise naturally in certain models in automatic control theory
(see [9, 11]). The investigation of the
difference equation(1.1)xn=max{A1xn−1,A2xn−2,…,Apxn−p},n=0,1,…,where p∈ℕ,Ai,i=1,…,p,
are real numbers such that at least one of them is different from zero and the
initial values x−1,…,x−p are different from zero was proposed in
[6]. Some results
about (1.1) and its generalizations can be found in [1, 3–5, 7, 8, 10, 12, 17, 18, 19] (see also the references therein). The study of
max-type equations whose some terms contain nonconstant numerators was
initiated by Stević,
see for example, [2, 14–16]. For some closely related papers, see also [20, 21].Motivated by the aforementioned papers and by computer
simulations, in this paper we study the asymptotic behavior of positive
solutions to the difference equation(1.2)xn=max{Axn−1α,Bxn−2β},n=0,1,…,where 0<α,β<1,A,B>0.
We prove that every positive solution of this equation converges to x*=max{A1/(α+1),B1/(β+1)}.


## 2. Main Results

In this section, we will prove the following result concerning (1.2).Theorem 2.1.
Let(xn) be a positive solution to (1.2).
Then(2.1)xn→max{A1/(α+1),B1/(β+1)}asn→∞.In order to establish Theorem2.1, we need the following
lemma and its corollary which can be found in [13].Lemma 2.2.
Let(an)n∈ℕ be a sequence of positive numbers which
satisfies the inequality(2.2)an+k≤qmax{an+k−1,an+k−2,…,an},forn∈ℕ,where q>0 and k∈ℕ are fixed. Then there exist L∈ℝ+ such that(2.3)akm+r≤Lqm∀m∈ℕ0,1≤r≤k.Corollary 2.3.
Let(an)n∈ℕ be a sequence of positive numbers as in Lemma 2.2. Then there exists M>0 such that(2.4)an≤M(qk)n,n∈ℕ.Now, we are in a position to prove Theorem2.1.Proof.
We proceed by distinguishing two
possible cases.
Case (A1/(α+1)≥B1/(β+1)).
We provexn→A1/(α+1) as n→∞.
Setxn=ynA1/(α+1),
then (1.2) becomes(2.5)yn=max{1yn−1α,Cyn−2β},n=0,1,…,where C=B/A(α+1)/(β+1).
Since A1/(α+1)≥B1/(β+1),
we have C≤1.
To prove xn→A1/(α+1) as n→∞,
it suffices to prove yn→1 as n→∞.
We proceed by two cases:C=1 and 0<C<1.
CaseC=1. In this case (2.5) is reduced to(2.6)yn=max{1yn−1α,1yn−2β},n=0,1,…,where 0<α,β<1.
Choose a number D so that 0<D<1.
Let yn=Dzn,n≥−2.
Then, (zn) is a solution to the difference
equation(2.7)zn=min{−αzn−1,−βzn−2},n=0,1,….To prove yn→1 as n→∞,
it suffices to prove zn→0 as n→∞.
It can be easily proved that there is a positive
integerN such that for all n≥0,(2.8)z3n+N≥0,z3n+N+1≤0,z3n+N+2≤0.By simple computation, we get
that, for all n≥0,(2.9)z3n+N+2=min{−αz3n+N+1,−βz3n+N}=−βz3n+N,(2.10)0≤z3n+N+3=min{−αz3n+N+2,−βz3n+N+1}=min{αβz3n+N,−βz3n+N+1}≤αβz3n+N,(2.11)z3n+N+4=min{−αz3n+N+3,−βz3n+N+2}=−αz3n+N+3.
Since0<αβ<1,
(2.10) implies z3n+N→0 as n→∞.
From (2.9) and (2.11), it follows that z3n+N+1→0,z3n+N+2→0 as n→∞.
This implies zn→0.
Case0<C<1. Let yn=Czn,
then (zn) is a solution to the difference
equation(2.12)zn=min{−αzn−1,1−βzn−2},n=0,1,….To prove yn→1 as n→∞,
it suffices to prove zn→0 as n→∞.
If z−1=0,z−2=0,
then we have zn=0 for all n≥−2.
Next, we assume either z−1≠0 or z−2≠0.
Then the following four claims are obviously true.
Claim 1.
Ifzn−1≥0 and zn−2≥0 for some n,
then(2.13)|zn|≤max{α|zn−1|,β|zn−2|−1}.
Claim 2.
Ifzn−1≤0 and zn−2≤0 for some n,
then |zn|≤α|zn−1|.
Claim 3.
Ifzn−1≥0 and zn−2≤0 for some n,
then |zn|=α|zn−1|.
Claim 4.
Ifzn−1≤0 and zn−2≥0 for some n,
then(2.14)|zn|≤max{α|zn−1|,β|zn−2|−1}.
In general, we have(2.15)|zn|≤max{α|zn−1|,β|zn−2|−1}≤max{α|zn−1|,β|zn−2|}≤γmax{|zn−1|,|zn−2|},where 0<γ=max{α,β}<1.
From (2.15) and Corollary 2.3, there exists M>0 such that(2.16)|zn|≤M(γ)n.This implies zn→0 as n→∞.
Case (A1/(α+1)<B1/(β+1)).
We provexn→B1/(β+1) as n→∞.
Similar to the proof of Case1, we set xn=ynB1/(β+1),
then (1.2) becomes(2.17)yn=max{Cyn−1α,1yn−2β},n=0,1,…,where C=A/B(α+1)/(β+1)<1.
To prove xn→B1/(β+1) as n→∞,
it suffices to prove yn→1 as n→∞.
Let yn=Czn,
then (zn) is a solution to the difference
equation(2.18)zn=min{1−αzn−1,−βzn−2},n=0,1,….To prove yn→1 as n→∞,
it suffices to prove zn→0 as n→∞.
If z−1=0,z−2=0,
then we have zn=0 for all n≥−2.
Next, we assume either z−1≠0 or z−2≠0,
then the following four claims are obviously true.
Claim 1.
Ifzn−1≥0 and zn−2≥0 for some n,
then(2.19)|zn|≤max{α|zn−1|−1,β|zn−2|}.
Claim 2.
Ifzn−1≤0 and zn−2≤0 for some n,
then |zn|≤β|zn−2|.
Claim 3.
Ifzn−1≥0 and zn−2≤0 for some n,
then(2.20)|zn|≤max{α|zn−1|−1,β|zn−2|}.
Claim 4.
Ifzn−1≤0 and zn−2≥0 for some n,
then |zn|=β|zn−2|.
In general, we have(2.21)|zn|≤max{α|zn−1|−1,β|zn−2|}≤max{α|zn−1|,β|zn−2|}≤γmax{|zn−1|,|zn−2|},where 0<γ=max{α,β}<1.
Then the rest of the proof is similar to the proof of Case 1 and will be omitted. The proof is complete.Theorem 2.4.
Every solution to the difference
equationxn=A/xn−mα,0<α<1,A>0 converges to x*=A1/(α+1).Proof.
Letxn=ynA1/(α+1),
then the equation becomes(2.22)yn=1yn−mα=yn−2mα2=yn−4mα4=⋯=yn−2⌈n/2m⌉mα2⌈n/2m⌉.From this and the condition 0<α<1,
it follows that yn→1 as n→∞ which implies xn→A1/(α+1) as n→∞.


## 3. Conclusions and Remarks

This paper
examines the asymptotic behavior of positive solutions to the difference
equation (1.2) with 0<α,β<1,A,B>0.
The method used in this work may provide insight into the asymptotic behavior
of positive solutions to the generic difference equation(3.1)xn=max{A1xn−1α1,A2xn−2α2,…,Apxn−pαp},n=0,1,…,where 0<αi<1,Ai>0,i=1,…,p.
We close this work by proposing the following conjecture.Conjecture 3.1.
Assume that(xn) is a positive solution to (3.1). Then xn→max1≤i≤p{Ai1/(αi+1)} as n→∞.



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*Source: 243291-2008-07-28.xml*