Abstract and Applied Analysis
Volume 2011, Article ID 815285, 15 pages
http://dx.doi.org/10.1155/2011/815285
Research Article

On the Difference Equation 𝑥𝑛+1=𝑥𝑛𝑥𝑛−2−1

1Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Richmond, VA 23284, USA
2Department of Mathematical Sciences, Appalachian State University, Boone, NC 28608, USA
3Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia

Received 24 October 2010; Accepted 19 January 2011

Academic Editor: Yong Zhou

Copyright © 2011 Candace M. Kent 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.

Abstract

The long-term behavior of solutions of the following difference equation: , , where the initial values , , are real numbers, is investigated in the paper.

1. Introduction

Recently there has been a huge interest in studying nonlinear difference equations which do not stem from differential equations (see, e.g.,  and the references therein). Usual properties which have been studied are the boundedness character [8, 13, 15, 2830, 33, 35, 36], the periodicity [8, 13], asymptotic periodicity [1619, 21], local and global stability [1, 8, 13, 15, 16, 2834], as well as the existence of specific solutions such as monotone or nontrivial [2, 3, 5, 9, 10, 15, 18, 20, 2227].

In this paper we will study solutions of the following difference equation:

The difference equation (1.1) belongs to the class of equations of the form with particular choices of and , where . Although (1.2) looks simple, it is fascinating how its behavior changes for different choices of and . The cases and , and have been correspondingly investigated in papers [11, 12]. This paper can be regarded as a continuation of our systematic investigation of (1.2).

Note that (1.1) has two equilibria:

2. Periodic Solutions

In this section we prove some results regarding periodicity of solutions of (1.1). The first result concerns periodic solutions with prime period two which will play an important role in studying the equation.

Theorem 2.1. Equation (1.1) has prime period-two solutions if and only if the initial conditions are , , or , , .

Proof. Let be a prime period-two solution of (1.1). Then and , for every and for some such that . We have and . From these two equations we obtain or equivalently We have four cases to be considered.
Case 1. If , then , and we obtain the first prime period-two solution.
Case 2. If , then , and we obtain the second prime period-two solution.
Case 3. If , then , which is an equilibrium solution.
Case 4. If , then , which is the second equilibrium solution. Thus, the result holds.

Theorem 2.2. Equation (1.1) has no prime period-three solutions.

Proof. Let be a prime period-three solution of (1.1). Then ,  ,  , for every and some such that at least two of them are different. We have From (2.2) we easily see that , , and , so that From (2.3) we obtain which implies . Hence or . From this and (2.3) it follows that or , from which the result follows.

Theorem 2.3. Equation (1.1) has no prime period-four solutions.

Proof. Let be a prime period-four solution of (1.1) and , , . Then we have Thus, from (2.6) and (2.8), we have that . This along with (2.7) gives while from (2.8) we get or equivalently
Case 1. Suppose . Then and , which, by Theorem 2.1, yields a period-two solution.
Suppose . If , then from (2.10) we get a contradiction. If , then , so that . Hence , , or .

Case 2. Suppose . Then and which results in a period-two solution as proved in Theorem 2.1.
Case 3. Suppose . Then and which is an equilibrium solution.
Case 4. Suppose . Then and which is the second equilibrium solution. Proof is complete.

3. Local Stability

Here we study the local stability at the equilibrium points and .

Theorem 3.1. The negative equilibrium of (1.1), , is unstable. Moreover, it is a hyperbolic equilibrium.

Proof. The linearized equation associated with the equilibrium is Its characteristic polynomial is Hence Since there is a zero of .
On the other hand, from and (3.3), it follows that is a unique real zero of . Hence, the other two roots are conjugate complex.
Since we obtain . From this, the theorem follows.

Theorem 3.2. The positive equilibrium of (1.1), , is unstable. Moreover, it is also a hyperbolic equilibrium.

Proof. The linearized equation associated with the equilibrium is Its characteristic polynomial is
We have Since there is a zero of .
From this and since , we have that is a unique real zero of . Thus, the other two roots are conjugate complex.
Since we obtain . From this, the theorem follows.

4. Case 𝑥−2,𝑥−1,𝑥0∈(−1,0)

This section considers the solutions of (1.1) with . Before we formulate the main result in this section we need some auxiliary results.

Lemma 4.1. Suppose that . Then the solution of (1.1) is such that for .

Proof. We have which implies . Assume that we have proved for , for some . Then we have , finishing an inductive proof of the lemma.

Remark 4.2. We would like to say here that a similar argument gives the following extension of Lemma 4.1.
Suppose , , , where . Then the solution of the difference equation is such that for .

We now find an equation which is satisfied for the even terms of a solution of (1.1) as well as for the odd terms of the solution.

From (1.1) we have Then we have the following:

and similarly

Hence, the subsequences and satisfy the difference equation and .

For convenience, we make another change of variable . Then (4.5) becomes Note also that .

It is easy to see that (4.6) has the following four equilibria:

If we let where , then we find the following:(1), (2), (3).

Thus, the function is strictly increasing in each argument.

Lemma 4.3. Let be a solution of (1.1) which is not equal to the equilibrium solution of the equation. Suppose that and that one of the following conditions holds: (H1), , with at least one of the inequalities strict,(H2), , with at least one of the inequalities strict.
Then , for every and
(a)if (H1) holds, then there is an such that (b)if (H2) holds, then there is an such that

Proof. By Lemma 4.1 we have that , for . We will prove only (a). The proof of (b) is dual and is omitted. Since , we have From this and since , we have If or , then inequality (4.13) is strict and, consequently, inequality (4.14) is strict too. If , then , from which it follows that inequality (4.14) is strict. In this case we have which is a strict inequality. Hence and are the obvious candidates, depending on which of the two cases, just described, holds.
Assume that we have proved (4.11) for and that . The case is proved similarly and so is omitted. Then we have From this and since , we have Hence by induction the lemma follows.

Theorem 4.4. Let be a solution of (1.1) which is not equal to the equilibrium solution of the equation. Suppose that and that one of the conditions, (H1) or (H2), holds.
Then , for every , and converges to a two-cycle.

Proof. First of all , for , by Lemma 4.1. We next show that converges to a two-cycle. To this end we show that one of the subsequences, or , converges to 0 and the other one to . Showing this, in turn, is equivalent to showing the following:(a)the corresponding solution of (4.6) converges to if for some , , for , where for ,(b)the corresponding solution of (4.6) converges to if for some , for , where for .
We prove (a). The proof of (b) is similar and will be omitted. We have Let Then we have
First assume . From (4.20) it follows that there is an such that . By the monotonicity of and since we have From this and by induction we obtain , , which implies , which is a contradiction.
Now assume . Let be a subsequence of such that . Then there is a subsequence of , which we may denote the same, such that there are the following limits: , , and , which we denote, respectively, by , . From this and by (4.23) we have that Hence there is an such that . Otherwise, for and by the monotonicity of we would get which is a contradiction. On the other hand, contradicts the choice of . Hence cannot be in the interval .
From all of the above we have that . Therefore, , as desired.

Theorem 4.5. Assume that for a solution of (1.1) there is an such that Then the solution converges to a two-cycle or to the equilibrium .

Proof. First note that by Lemma 4.1 we have , . From (1.1) we obtain the identity Applying (4.28) for and using the fact , we get . Hence Using induction along with identity (4.28) it is shown that for every . Hence, there are finite limits and , say and . Letting in the relations we get and . Hence From this we have , or if , then so that and .

Remark 4.6. Let , and with . For , (4.27) will be Hence, under the conditions we have that (4.27) is satisfied for . It is easy to show that there are some such that the set in (4.34) is nonempty.

Note also that in the proof of Theorem 4.5 the relation (4.28) plays an important role. Relations of this type have been successfully used also in [17, 21].

It is a natural question if there are nontrivial solutions of (1.1) converging to the negative equilibrium . The next theorem, which is a product of an E-mail communication between Stević and Professor Berg , gives a positive answer to the question. In the proof of the result we use an asymptotic method from Proposition 3.3 in . Some asymptotic methods for solving similar problems have been also used, for example, in the following papers: [25, 20, 2227]. For related results, see also [9, 10, 15, 18] and the references therein.

Theorem 4.7. There are nontrivial solutions of (1.1) converging to the negative equilibrium .

Proof. In order to find a solution tending to , we make the substitution , yielding the equation and for we make the ansatz with , where and are the conjugate complex zeros of the characteristic polynomial Note that .
Replacing (4.36) into (4.35) and comparing the coefficients, we find with
Equation (4.38) is satisfied for , where and are arbitrary, so that it suffices to consider (4.38) for and such that . If and are chosen to be conjugate complex numbers, then according to (4.38) all are conjugate complex numbers to and consequently series (4.36) is real. We look for a solution (4.36) with , , and determine a positive constant such that Since the inequality is valid for , by induction, we get from (4.38)
Note that
It is not difficult to check that
Hence (4.40) holds with
For such chosen the series in (4.36) converges if , which implies . We have , so that , and therefore we have the convergence of the series for . In this way for , we obtain a solution of (4.35) converging to a real solution of (4.35) as , that is, a solution of (1.1) converging to , as desired.

Remark 4.8. For , the first coefficients in (4.38) are

Remark 4.9. If we replace by in (4.35) with an arbitrary , then we can choose in such a way that we get arbitrary first coefficients (not only of modulus 1).

5. Unbounded Solutions of (1.1)

In this section we find sets of initial values of (1.1) for which unbounded solutions exist. For related results, see, for example, [8, 13, 15, 2830, 33, 35, 36] and the references therein.

The next theorem shows the existence of unbounded solutions of (1.1).

Theorem 5.1. Assume that Then

Proof. From the hypothesis we have that , and so . Therefore, , and so . On the other hand, we have Combining the last two inequalities, we have that . Assume that we have proved for some . We have , which implies , or equivalently . From this and (1.1), we get finishing the inductive proof of the theorem.

Corollary 5.2. Assume that the initial values of a solution of (1.1) satisfy the condition Then the solution tends to .

Proof. Assume to the contrary that the sequence does not tend to plus infinity. Since the sequence is increasing and bounded, then it must converge. But (1.1) has only two equilibria, and they are both less than . We have a contradiction. The proof is complete.

For our next result, we need to introduce the following definition.

Definition 5.3. Let be a solution of (1.1), and let . Then we say that the solution has the eventual semicycle pattern (or ) if there exists such that, for , and (or, resp., and ).

Remark 5.4. Note that the eventual semicycle pattern can be extended to for .

Theorem 5.5. Assume that is a solution of (1.1) such that and that at least one of , , is negative. Then and the solution is separated into seven unbounded eventually increasing subsequences such that the solution has the eventual semicycle pattern

Proof. Assume that . Then we have Hence , for and . An inductive argument shows that for each , from which the first part of the result follows in this case. The other six cases follow from the above case by shifting indices for 1, 2, 3, 4, 5, or 6 places forward.
From this and Theorem 5.1, we see that the sequences monotonically tend to or with the aforementioned eventual semicycle pattern.

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