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Research Article | Open Access

Volume 2009 |Article ID 152964 | https://doi.org/10.1155/2009/152964

Taixiang Sun, Bin Qin, Hongjian Xi, Caihong Han, "Global Behavior of the Max-Type Difference Equation ", Abstract and Applied Analysis, vol. 2009, Article ID 152964, 10 pages, 2009. https://doi.org/10.1155/2009/152964

# Global Behavior of the Max-Type Difference Equation

Revised02 Mar 2009
Accepted08 Mar 2009
Published01 Jun 2009

#### 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,  (see also references therein). In [1, 35, 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 , the case was studied in , the case was studied in [4, 8], and the more difficult case was studied in . Chen  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  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 .

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 .

#### 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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