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Discrete Dynamics in Nature and Society
VolumeΒ 2008Β (2008), Article IDΒ 243291, 6 pages
http://dx.doi.org/10.1155/2008/243291
Research Article

On the Asymptotic Behavior of a Difference Equation with Maximum

College of Computer Science, Chongqing University, Chongqing 400044, China

Received 25 May 2008; Revised 6 June 2008; Accepted 18 June 2008

Academic Editor: StevoΒ Stevic

Copyright Β© 2008 Fangkuan Sun. This 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.

Abstract

We study the asymptotic behavior of positive solutions to the difference equation π‘₯𝑛=max{A/x𝛼n-1,𝐡/π‘₯π›½π‘›βˆ’2}, 𝑛=0,1,…, where 0<𝛼,𝛽<1,𝐴,𝐡>0. We prove that every positive solution to this equation converges to π‘₯βˆ—=max{A1/(𝛼+1),𝐡1/(𝛽+1)}.

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 equationπ‘₯𝑛𝐴=max1π‘₯π‘›βˆ’1,𝐴2π‘₯π‘›βˆ’2𝐴,…,𝑝π‘₯π‘›βˆ’π‘ξ‚‡,𝑛=0,1,…,(1.1)where π‘βˆˆβ„•,𝐴𝑖,𝑖=1,…,𝑝, are real numbers such that at least one of them is different from zero and the initial values π‘₯βˆ’1,…,π‘₯βˆ’π‘ 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 SteviΔ‡, 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 equationπ‘₯𝑛𝐴=maxπ‘₯π›Όπ‘›βˆ’1,𝐡π‘₯π›½π‘›βˆ’2,𝑛=0,1,…,(1.2)where 0<𝛼,𝛽<1,𝐴,𝐡>0. We prove that every positive solution of this equation converges to π‘₯βˆ—=max{𝐴1/(𝛼+1),𝐡1/(𝛽+1)}.

2. Main Results

In this section, we will prove the following result concerning (1.2).

Theorem 2.1. Let (π‘₯𝑛) be a positive solution to (1.2).
Then π‘₯𝑛𝐴→max1/(𝛼+1),𝐡1/(𝛽+1)asπ‘›β†’βˆž.(2.1)

In order to establish Theorem 2.1, we need the following lemma and its corollary which can be found in [13].

Lemma 2.2. Let (π‘Žπ‘›)π‘›βˆˆβ„• be a sequence of positive numbers which satisfies the inequality π‘Žπ‘›+π‘˜ξ‚†π‘Žβ‰€π‘žmax𝑛+π‘˜βˆ’1,π‘Žπ‘›+π‘˜βˆ’2,…,π‘Žπ‘›ξ‚‡,forπ‘›βˆˆβ„•,(2.2)where π‘ž>0 and π‘˜βˆˆβ„• are fixed. Then there exist πΏβˆˆβ„+ such that π‘Žπ‘˜π‘š+π‘Ÿβ‰€πΏπ‘žπ‘šβˆ€π‘šβˆˆβ„•0,1β‰€π‘Ÿβ‰€π‘˜.(2.3)

Corollary 2.3. Let (π‘Žπ‘›)π‘›βˆˆβ„• be a sequence of positive numbers as in Lemma 2.2. Then there exists 𝑀>0 such that π‘Žπ‘›ξ‚€β‰€π‘€π‘˜βˆšπ‘žξ‚π‘›,π‘›βˆˆβ„•.(2.4)

Now, we are in a position to prove Theorem 2.1.

Proof. We proceed by distinguishing two possible cases.
Case (𝐴1/(𝛼+1)β‰₯𝐡1/(𝛽+1)). We prove π‘₯𝑛→𝐴1/(𝛼+1) as π‘›β†’βˆž.
Set π‘₯𝑛=𝑦𝑛𝐴1/(𝛼+1), then (1.2) becomes𝑦𝑛1=maxπ‘¦π›Όπ‘›βˆ’1,πΆπ‘¦π›½π‘›βˆ’2,𝑛=0,1,…,(2.5)where 𝐢=𝐡/𝐴(𝛼+1)/(𝛽+1). Since 𝐴1/(𝛼+1)β‰₯𝐡1/(𝛽+1), we have 𝐢≀1. To prove π‘₯𝑛→𝐴1/(𝛼+1) as π‘›β†’βˆž, it suffices to prove 𝑦𝑛→1 as π‘›β†’βˆž.
We proceed by two cases: 𝐢=1 and 0<𝐢<1.

Case 𝐢=1. In this case (2.5) is reduced to𝑦𝑛1=maxπ‘¦π›Όπ‘›βˆ’1,1π‘¦π›½π‘›βˆ’2,𝑛=0,1,…,(2.6)where 0<𝛼,𝛽<1. Choose a number 𝐷 so that 0<𝐷<1. Let 𝑦𝑛=𝐷𝑧𝑛,𝑛β‰₯βˆ’2. Then, (𝑧𝑛) is a solution to the difference equation𝑧𝑛=minβˆ’π›Όπ‘§π‘›βˆ’1,βˆ’π›½π‘§π‘›βˆ’2,𝑛=0,1,….(2.7)To prove 𝑦𝑛→1 as π‘›β†’βˆž, it suffices to prove 𝑧𝑛→0 as π‘›β†’βˆž.
It can be easily proved that there is a positive integer 𝑁 such that for all 𝑛β‰₯0,𝑧3𝑛+𝑁β‰₯0,𝑧3𝑛+𝑁+1≀0,𝑧3𝑛+𝑁+2≀0.(2.8)By simple computation, we get that, for all 𝑛β‰₯0,𝑧3𝑛+𝑁+2=minβˆ’π›Όπ‘§3𝑛+𝑁+1,βˆ’π›½π‘§3𝑛+𝑁=βˆ’π›½π‘§3𝑛+𝑁,(2.9)0≀𝑧3𝑛+𝑁+3=minβˆ’π›Όπ‘§3𝑛+𝑁+2,βˆ’π›½π‘§3𝑛+𝑁+1=min𝛼𝛽𝑧3𝑛+𝑁,βˆ’π›½π‘§3𝑛+𝑁+1≀𝛼𝛽𝑧3𝑛+𝑁𝑧,(2.10)3𝑛+𝑁+4=minβˆ’π›Όπ‘§3𝑛+𝑁+3,βˆ’π›½π‘§3𝑛+𝑁+2=βˆ’π›Όπ‘§3𝑛+𝑁+3.(2.11)
Since 0<𝛼𝛽<1, (2.10) implies 𝑧3𝑛+𝑁→0 as π‘›β†’βˆž. From (2.9) and (2.11), it follows that 𝑧3𝑛+𝑁+1β†’0,𝑧3𝑛+𝑁+2β†’0 as π‘›β†’βˆž. This implies 𝑧𝑛→0.
Case 0<𝐢<1. Let 𝑦𝑛=𝐢𝑧𝑛, then (𝑧𝑛) is a solution to the difference equation𝑧𝑛=minβˆ’π›Όπ‘§π‘›βˆ’1,1βˆ’π›½π‘§π‘›βˆ’2,𝑛=0,1,….(2.12)To prove 𝑦𝑛→1 as π‘›β†’βˆž, it suffices to prove 𝑧𝑛→0 as π‘›β†’βˆž. If π‘§βˆ’1=0,π‘§βˆ’2=0, then we have 𝑧𝑛=0 for all 𝑛β‰₯βˆ’2. Next, we assume either π‘§βˆ’1β‰ 0 or π‘§βˆ’2β‰ 0. Then the following four claims are obviously true.
Claim 1. If π‘§π‘›βˆ’1β‰₯0 and π‘§π‘›βˆ’2β‰₯0 for some 𝑛, then|||𝑧𝑛|||𝛼|||𝑧≀maxπ‘›βˆ’1||||||𝑧,π›½π‘›βˆ’2|||ξ‚‡βˆ’1.(2.13)
Claim 2. If π‘§π‘›βˆ’1≀0 and π‘§π‘›βˆ’2≀0 for some 𝑛, then |𝑧𝑛|≀𝛼|π‘§π‘›βˆ’1|.
Claim 3. If π‘§π‘›βˆ’1β‰₯0 and π‘§π‘›βˆ’2≀0 for some 𝑛, then |𝑧𝑛|=𝛼|π‘§π‘›βˆ’1|.
Claim 4. If π‘§π‘›βˆ’1≀0 and π‘§π‘›βˆ’2β‰₯0 for some 𝑛, then
In general, we havewhere . From (2.15) and Corollary 2.3, there exists such thatThis implies as .
Case (). We prove as .
Similar to the proof of Case 1, we set π‘₯𝑛=𝑦𝑛𝐡1/(𝛽+1), then (1.2) becomes𝑦𝑛𝐢=maxπ‘¦π›Όπ‘›βˆ’1,1π‘¦π›½π‘›βˆ’2,𝑛=0,1,…,(2.17)where 𝐢=𝐴/𝐡(𝛼+1)/(𝛽+1)<1. To prove π‘₯𝑛→𝐡1/(𝛽+1) as π‘›β†’βˆž, it suffices to prove 𝑦𝑛→1 as π‘›β†’βˆž. Let 𝑦𝑛=𝐢𝑧𝑛, then (𝑧𝑛) is a solution to the difference equation𝑧𝑛=min1βˆ’π›Όπ‘§π‘›βˆ’1,βˆ’π›½π‘§π‘›βˆ’2,𝑛=0,1,….(2.18)To prove 𝑦𝑛→1 as π‘›β†’βˆž, it suffices to prove 𝑧𝑛→0 as π‘›β†’βˆž. If π‘§βˆ’1=0,π‘§βˆ’2=0, then we have 𝑧𝑛=0 for all 𝑛β‰₯βˆ’2. Next, we assume either π‘§βˆ’1β‰ 0 or π‘§βˆ’2β‰ 0, then the following four claims are obviously true.
Claim 1. If π‘§π‘›βˆ’1β‰₯0 and π‘§π‘›βˆ’2β‰₯0 for some 𝑛, then|||𝑧𝑛|||𝛼|||𝑧≀maxπ‘›βˆ’1||||||π‘§βˆ’1,π›½π‘›βˆ’2|||.(2.19)
Claim 2. If π‘§π‘›βˆ’1≀0 and π‘§π‘›βˆ’2≀0 for some 𝑛, then |𝑧𝑛|≀𝛽|π‘§π‘›βˆ’2|.
Claim 3. If π‘§π‘›βˆ’1β‰₯0 and π‘§π‘›βˆ’2≀0 for some 𝑛, then|||𝑧𝑛|||𝛼|||𝑧≀maxπ‘›βˆ’1||||||π‘§βˆ’1,π›½π‘›βˆ’2|||.(2.20)
Claim 4. If π‘§π‘›βˆ’1≀0 and π‘§π‘›βˆ’2β‰₯0 for some 𝑛, then |𝑧𝑛|=𝛽|π‘§π‘›βˆ’2|.
In general, we have|||𝑧𝑛|||𝛼|||𝑧≀maxπ‘›βˆ’1||||||π‘§βˆ’1,π›½π‘›βˆ’2|||𝛼|||𝑧≀maxπ‘›βˆ’1||||||𝑧,π›½π‘›βˆ’2||||||𝑧≀𝛾maxπ‘›βˆ’1|||,|||π‘§π‘›βˆ’2|||,(2.21)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 equation π‘₯𝑛=𝐴/π‘₯π›Όπ‘›βˆ’π‘š,0<𝛼<1,𝐴>0 converges to π‘₯βˆ—=𝐴1/(𝛼+1).

Proof. Let π‘₯𝑛=𝑦𝑛𝐴1/(𝛼+1), then the equation becomes𝑦𝑛=1π‘¦π›Όπ‘›βˆ’π‘š=𝑦𝛼2π‘›βˆ’2π‘š=𝑦𝛼4π‘›βˆ’4π‘š=β‹―=𝑦𝛼2βŒˆπ‘›/2π‘šβŒ‰π‘›βˆ’2βŒˆπ‘›/2π‘šβŒ‰π‘š.(2.22)From this and the condition 0<𝛼<1, it follows that 𝑦𝑛→1 as π‘›β†’βˆž which implies π‘₯𝑛→𝐴1/(𝛼+1) as π‘›β†’βˆž.

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,𝐡>0. The method used in this work may provide insight into the asymptotic behavior of positive solutions to the generic difference equationπ‘₯𝑛𝐴=max1π‘₯𝛼1π‘›βˆ’1,𝐴2π‘₯𝛼2π‘›βˆ’2𝐴,…,𝑝π‘₯π›Όπ‘π‘›βˆ’π‘ξ‚‡,𝑛=0,1,…,(3.1)where 0<𝛼𝑖<1,𝐴𝑖>0,𝑖=1,…,𝑝. We close this work by proposing the following conjecture.

Conjecture 3.1.Assume that (π‘₯𝑛) is a positive solution to (3.1). Then π‘₯𝑛→max1≀𝑖≀𝑝{𝐴1/(𝛼𝑖𝑖+1)} as π‘›β†’βˆž.

Acknowledgments

The author is grateful to the anonymous referees for their huge number of valuable comments and suggestions, which considerably improved the paper. This work is supported by Natural Science Foundation of China (10771227).

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