Research Article | Open Access

# Global Behavior of the Max-Type Difference Equation

**Academic Editor:**Stevo Stevic

#### Abstract

We study global behavior of the following max-type difference equation , where is a sequence of positive real numbers with . The special case when is a periodic sequence with period and for every 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 has been studied for positive coefficients , which are periodic with period . The case was studied in [1], the case was studied in [3], the case was studied in [4, 8], and the more difficult case was studied in [7]. Chen [5] found that every positive solution of (1.1) is eventually periodic with period 2 when is a periodic sequence of positive real numbers with period and for all . 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 is a sequence of positive real numbers with .

#### 2. Main Results

The main results of this paper are established through the following lemmas.

Lemma 2.1. *Let be a positive solution of (1.1), then*(1)* for all ;*(2)*if for some , then .*

*Proof. *(1) is obvious since for all .(2)If for some , then . Suppose for the sake of contradiction that , then similarly we get and
This is a contradiction since and . The proof is complete.

Lemma 2.2. *Let be a positive solution of (1.1) and for all . Then*(1)* and is nonincreasing;*(2)* is bounded, and moreover for any .*

*Proof. *By Lemma 2.1(1) and the assumption , we obtain that for any ,
Hence
which implies that for all
Furthermore, it follows that for all
The proof is complete.

*Remark 2.3. *Note that from the proof of Lemma 2.2 we have that .

*Remark 2.4. *Various sequences which satisfy inequality in Lemma 2.2(1), that is, have been studied, for example, in [18–24].

Lemma 2.5. *Let be a positive solution of (1.1) and . Then .*

*Proof. *Since is a subsequence of , it follows that
On the other hand, by for all , we obtain
The proof is complete.

*Remark 2.6. *Let be a positive solution of (1.1). By Lemma 2.2, we see that if and for some , then For example, if it were , then it would be , which would imply .

Lemma 2.7. *Suppose that is a positive solution of (1.1) and . Write
**
Then .*

*Proof. *If contains only one point, we may assume by taking a subsequence that By taking the limit in the following relationship:
as , we obtain
which implies that
If contains at least two points, let , then there exists a subsequence of such that
By Remark 2.6, we see that there exists such that for every ,
from which it follows that
By taking a subsequence we may assume that By taking the limit in the following relationship:
as , we obtain
which implies
The proof is complete.

Theorem 2.8. *Let be a positive solution of (1.1) and . Then one of the following two statements is true.*(1)*If there exist infinitely many such that and , then is eventually equal to 1.*(2)*If there exists such that and for all then and *

*Proof. *(1) We assume that there exists an infinite sequence such that
By taking a subsequence we may assume from Lemma 2.7 that
By taking the limit in the following relationship:
as , we get
Since and , it follows that and .In the following, we show that is eventually equal to 1. It only needs to prove that there exists such that for all
Indeed, if there exist infinitely many such that
by taking a subsequence we may assume that , then it follows that
which is a contradiction. Therefore there exists such that for all
Thus
Since , we have
(2) If , then the result follows from Lemma 2.7. In the following, we assume . Suppose for the sake of contradiction that there exists a subsequence of such that
By taking a subsequence we may assume that
By taking the limit in the following relationship:
as we get
which implies
This is a contradiction. The proof is complete.

Corollary 2.9. *Let be a periodic sequence of positive real numbers, then every positive solution of (1.1) is eventually periodic with period 2.*

*Proof. *Let be a positive solution of (1.1) and . By Remark 2.6 and Theorem 2.8, we may assume without loss of generality that , for all Suppose for the sake of contradiction that there exists a sequence such that(1) and (2) for .Then is odd for every . Let , then it follows from Lemma 2.1 that
From this and by (2) it follows that
Therefore for every ,
which is a contradiction since 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 to be no periodic sequence.

*Example 3.1. *Consider
where for any Then solution of (3.1) with the initial values and satisfies the following.(1) for any (2)

*Proof. *By simple computation, we have
It follows from (3.1) and (3.2) that
By induction, we have from (3.1) and (3.2) that for any
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).

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#### Copyright

Copyright © 2009 Taixiang Sun et al. 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.