Abstract

We consider the following difference equation , where initial values and is a strictly decreasing continuous surjective function. We show the following. (1) Every positive solution of this equation converges to or for some . (2) Assume . Then the set of initial conditions such that the positive solutions of this equation converge to or is a unique strictly increasing continuous function or an empty set.

1. Introduction

Recently there have been published quite a lot of works concerning global behavior of the difference equations [18]. These results are not only valuable in their own right, but they can provide insight into their differential counterparts.

In [9], Kulenović and Ladas considered the positive solutions for difference equation with . They gave some partial results on the convergence of this equation.

Kalikow et al. [10] studied the following difference equation: where initial values and is in a certain class of increasing continuous functions. They showed that the set of initial conditions of in the first quadrant that converge to any given boundary point of the first quadrant forms a unique strictly increasing continuous function.

Motivated by the above studies, in this paper, we consider the following difference equation: where initial values and is a strictly decreasing continuous surjective function. Our main result is the following theorem.

Theorem 1. Every positive solution of (2) converges to for some .
Assume . Then the set of initial conditions such that the positive solutions of (2) converge to is a unique strictly increasing continuous function or an empty set.

2. The Main Result

Proof of Theorem 1. Let be a positive solution of (2). Then and are decreasing sequences since . Let and . Then we have which implies or . If , then since is a strictly decreasing continuous surjective function with . This completes proof of Theorem 1.

Write and define by for all . It is easy to see that if is a solution of (2), then for any . In the following, let for some .

Lemma 2. The following statements are true:(i)is a homeomorphism;(ii);(iii) and .

Proof. (i) Since for any with and is continuous for any , is a homeomorphism.
(ii) Let and . Then , , and which implies .
Let . Then and . Choose . Then . Thus .
The proof of (iii) is similar to that of (ii). This completes the proof of Lemma 2.

In order to show Theorem 1, we will construct two families of strictly increasing functions and () as follows. Set Then is a strictly increasing function which maps onto . Set Then is a strictly increasing function which maps onto .

Assume that, for some positive integer , we already define strictly increasing functions and such that both and map onto . Set Then both and are strictly increasing functions which map onto . In such a way, we construct two families of strictly increasing functions and ().

Set and . For any , write

Let . Since and , it follows that Thus and .

Let . Since and , it follows that Thus and . Using induction, one can easily show that, for any , Since is a homeomorphism and with is the boundary of , we have that, for any , In a similar fashion, we may show that Since , , and is a homeomorphism, we have that and , which implies that, for any , It follows from (12) and (18) that, for , and for , Noting (19) and (20), we may assume that, for every , and for every ,

Set

Lemma 3. The following statements are true:(i) and ;(ii)both and are increasing continuous functions which map onto .

Proof. (i)  Let . Then we have , which follows that Since , we have Let . It follows from (24) and (25) that so we have It follows from (25) and (27) that Thus we have .
Let . Then we have , which follows that Since , we have Let . It follows from (29) and (30) that so we have It follows from (31) and (32) that Thus we have . In a similar fashion, we can show that .
(ii) Since are strictly increasing functions, we have that is an increasing function. For any , let then .
Now we claim that . Indeed, if , then it follows from (6) that So we have that It follows from (34) and (36) that there exist such that It follows from Lemma 3(i) and (37) that and this is a contradiction. The claim is proven.
In a similar fashion, we may show that Thus is an increasing continuous function. In a similar fashion, we may show that is an increasing continuous function. Lemma 3 is proven.

Let where and . It follows from Lemma 2(iii) and Lemma 3(ii) that

Proof of Theorem 1. Noting (40), we consider the following two cases.
Case  1 . It follows from (40) that Let and be a solution of (2) with initial value ; it follows from Lemma 3(i) that which implies that . It follows from (42) and Theorem 1 that Next we claim that is a strictly increasing function. Indeed, if there exists , such that and , then there exist such that . Set Then we have Using induction, one can show that, for any , It follows from (44) and (47) that This is a contradiction. The claim is proven.
Now let with and be a solution of (2) with initial value .
If , then it follows from Theorem 1 and (2) that which implies .
If and , then there exists such that from which it follows that Then we have , which implies .
If and , then let and , and there exists such that . We can show that, for any , which implies From all abovementioned, the set of initial conditions such that the positive solutions of (2) converge to is .
In a similar fashion, we also may show that the set of initial conditions such that the positive solutions of (2) converge to is .
Case  2 (). It follows from (41) and Case 1 that the set of initial conditions such that the positive solutions of (2) converge to is an empty set. This completes the proof of Theorem 1.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This project was supported by NNSF of China (11261005), NSF of Guangxi (2012GXNSFDA276040), and SF of ED of Guangxi (2013ZD061).