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Abstract and Applied Analysis

VolumeÂ 2010Â (2010), Article IDÂ 469683, 13 pages

http://dx.doi.org/10.1155/2010/469683

## Solutions of the Difference Equation

^{1}Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Richmond, VA 23284, USA^{2}Department of Mathematical Sciences, Appalachian State University, Boone, NC 28608, USA^{3}School of Mathematical Sciences, Rochester Institute of Technology, Rochester, NY 14623, USA^{4}Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia

Received 28 April 2010; Accepted 9 June 2010

Academic Editor: AğacikÂ Zafer

Copyright Â© 2010 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

Our goal in this paper is to investigate the long-term behavior of solutions of the following difference equation: , where the initial conditions and are real numbers. We examine the boundedness of solutions, periodicity of solutions, and existence of unbounded solutions and how these behaviors depend on initial conditions.

#### 1. Introduction and Preliminaries

We investigate the long-term behavior of solutions of the second-order difference equation where the initial conditions and are real numbers. In particular, we examine the boundedness and periodic behaviors of solutions and the dependence of such behaviors on initial conditions.

Over the last decade, a number of rational second-order difference equations have been extensively studied due to their unique and diversified behavior of solutions. See, for example, [1–10] and the related references therein. Several of the rational difference equations have been used in mathematical biology by Beverton and Holt [11] and Pielou [12]. In addition, some rational difference equations exhibit a trichotomy behavior of solutions relative to the relationships between the parameters: convergence of solutions, periodicity of solutions, and existence of unbounded solutions. See [13] for one such example. The difference equation (1.1) belongs to the class of equations of the form and a particular choice of , where

#### 2. The Equilibria and Periodic Solutions of (1.1)

In this section, we show that (1.1) possesses exactly two equilibria and three periodic solutions with minimal period three.

After solving the equation , we find that (1.1) has exactly two equilibria, one positive and one negative, which we denote by and , respectively:

Note that there are no solutions which are eventually constant. Indeed, if for some then from , it follows that Repeating this procedure, we obtain , for as claimed.

Also note that there are no solutions that are eventually periodic with minimal period two. Indeed, if and , , then we have so that , which is a contradiction.

The following result shows that there exist exactly three periodic solutions of (1.1) with minimal period three and gives a description of each.

Theorem 2.1. *There exist exactly three periodic solutions of (1.1) with minimal period three. They are given by the three pairs of initial conditions , ; , ; and , . *

*Proof. *We can write terms of a period-three solution of (1.1) as
Therefore, this is indeed a solution of period three if the system below is satisfied.
Hence we see that . Thus, this is true if , or if , or if when and . Now it suffices to consider the following three cases.*Case 1. *Suppose that . Then implies that and hence gives us . Therefore, a period-three solution exists if and .*Case 2. *Suppose that . Then implies that and hence gives us . Therefore, a period-three solution exists if and .*Case 3. *Suppose that such that and . Then implies that , which gives us . Therefore, a period-three solution exists if and . (Note that are also solutions of , which are the two equilibria of (1.1).)Hence, there exist exactly three periodic solutions with minimal period three of (1.1) given by
as claimed.

In the sequel, we will refer to any one of these three periodic solutions of (1.1) as

The following theorem demonstrates, in a similar way as in [14, 15], that (1.1) has solutions that are eventually periodic with minimal period three.

Theorem 2.2. *All eventually periodic solutions with minimal period three are of the form
**
Where , , , and, if , for *

*Proof. *If is an eventually periodic solution with minimal period three, then by Theorem 2.1 there is an such that and Hence and consequently and . Let (which implies that ). From (1.1), if , we get for as desired.

*Remark 2.3. *If, in Theorem 2.2, , then, among some others, we obtain the solutions in Theorem 2.1.

The next theorem shows that no periodic or eventually periodic solution of (1.1) converges to a minimal period-three solution.

Theorem 2.4. *Let be a solution of (1.1) that is neither periodic nor eventually periodic with minimal period three. Then does not converge to the minimal period-three solution
*

*Proof. *Let be a solution of (1.1) that is neither periodic nor eventually periodic with minimal period three. For the sake of contradiction, assume that converges to the minimal period-three solution
Then we can choose such that there exist
with
that is, for ,

Now, first observe that
We claim that . Otherwise, . Then, by (1.1), , which gives us a solution that is eventually periodic with minimal period three, a contradiction. This leaves us with or . If , then since by (2.11), we have and so . If , then since and by (2.11), we have and so . Therefore, we need only consider the following two cases.*Case 1. *If , , then we obtain the following inequalities from (2.11) and (2.12):

We next compute the following:
which is true. Hence, and . Therefore, this case applies to , and by induction and (2.11) and (2.12), we obtain
Hence, cannot converge to as in this case.*Case 2. *If , , then we obtain the following:
We next compute the following:
which is true by (2.11) and (2.12). Hence, and . Therefore, this case applies to , and by induction and (2.11) and (2.12), we obtain the following inequalities:

Hence, cannot converge to as in this case, and we are done with our proof by contradiction.

Finally we show that the interval is invariant.

Theorem 2.5. *If , , then for all . *

*Proof. *If , , then . From (1.1) and by induction, we then have that for all .

*Remark 2.6. *Notice, if the solution is not periodic or eventually periodic with minimal period three, then as we see from Theorem 2.5 and as we shall see from Theorems 4.1–4.5, either the solution is bounded, while inside the invariant interval , or the solution becomes unbounded.

#### 3. Stability and Convergence of Solutions of (1.1)

In this section, we determine the stability nature of the two equilibria of (1.1) and leave open for the reader the possibility of convergence of solutions to the negative equilibrium.

Lemma 3.1. *The negative equilibrium solution of (1.1), , is locally asymptotically stable. The positive equilibrium solution of (1.1), , is unstable. *

*Proof. *The characteristic equation of the equilibria of (1.1) is the following:
with eigenvalues
Therefore, it is easy to show that the eigenvalues for are complex with , and for , and .

*Open Problem 3. *If , show whether or not every solution of (1.1) converges to (see Theorem 2.5 as well as Lemma 3.1).

#### 4. Unbounded Solutions of (1.1)

In this section, we find sets of initial conditions of (1.1) for which unbounded solutions exist.

First observe that when the initial conditions or , then existence of unbounded solutions appears. Specifically, the following two theorems will show existence of unbounded solutions relative to the set of these initial conditions. The theorem below is included to make this paper self-contained. One can see that the theorem below overlaps with [1, Theorem .1].

Theorem 4.1. *If , then the solution is eventually strictly increasing and tends to . *

*Proof. *Since , we have . Thus,
Therefore, . Thus . Rewriting, . Hence, . It follows by induction that the solution is eventually strictly increasing.

To prove that the sequence tends to , we assume otherwise. Since the sequence is increasing, it must be bounded and thus must converge. But, the equation has only two equilibria, and they are both less than . We have a contradiction. The proof is complete.

Recently there has been considerable interest in showing the existence of monotonic solutions of nonlinear difference equations. Various methods for this can be found, for example, in [6, 16–23].

Theorem 4.2. *Suppose . Then
**
and the subsequences and tend to and tends to .*

*Proof. *We first have
Then obviously
But we will prove that .

Since , we have that . Thus, . Since , we must have . Thus . Therefore, . Hence . From here we have , and the result that follows.

Next, we show that is not only less than , but it is less than . To this end, since , and , we have . This gives , and hence, .

To show that , we start by observing that , , and so . Thus, since , we get, with ,

By induction, it can be proved that
that is, is a positive increasing subsequence and and are negative decreasing subsequences.

Next we verify that these subsequences are unbounded and thus that our solution is unbounded. Assume that the two decreasing sequences, and , are bounded from below. Then they each must converge to a finite limit (which is the same finite limit and less than ). But by (1.1), the third increasing subsequence, , must also converge to a finite limit (which is positive), where
This is impossible because there are no periodic solutions with minimal period two (see Section 2). So and are unbounded (they tend to ) and thus is also unbounded (it tends to ) by (1.1) again.

Theorem 4.3. *Suppose that and . Then the solution of (1.1) has three subsequences , , and , where two of them tend to and one tends to .*

*Proof. *Note that and . Then the result follows from Theorem 4.2.

It might be interesting to note that if the order of the initial conditions in the above theorem is changed, the behavior of the solution can be drastically different. For example, if we let and then the solution does what the above theorem guarantees. On the other hand, if and , then the solution enters the interval , remains there, and thus is bounded. Furthermore, it should be pointed out that if the initial conditions are such that and , then the solution in certain cases is bounded and in other cases is unbounded.

In addition, observe that if , then the solutions of (1.1) exhibit somewhat chaotic behavior relative to the initial conditions. A little change in the initial conditions can cause a drastic difference in the long-term behavior of the solutions. For instance, if and , then the solution enters and then remains in the interval , and hence is bounded. Whereas if and , then the solution has three unbounded subsequences: one tending to and the other two tending to .

Theorem 4.4. *Let be a solution of (1.1). Suppose that*(i)*, ;*(ii)*.**Then is a positive increasing subsequence; is a negative decreasing subsequence; is a negative decreasing subsequence. Consequently, there does not exist an such that for all . Furthermore, all three subsequences are unbounded and so the solution is unbounded.*

*Proof. *Let be a solution of (1.1). Suppose that(i), ;(ii).Since by Hypothesis (i), then we have
Then, since , we have

We compute the sign of :
which is true by Hypotheses (i) and (ii). We next have
since , . Then, also,
since , .

With , replacing in Theorem 4.2, we have that

Hence, is a positive increasing subsequence, is a negative decreasing subsequence, and is a negative decreasing subsequence, and so there does not exist an such that for all . Furthermore, by Theorem 4.2, tends to and , and both tend to .

In conclusion, we show that, for a certain range of initial conditions, the solution of (1.1) is either unbounded or eventually enters the interval and remains there.

Theorem 4.5. *Let be a solution of (1.1). Suppose that . Then one of the following occurs.*(i)*The solution is unbounded.*(ii)*There exists such that for all .*

*Proof. *Let be a solution of (1.1). Assume that
We will show that the solution is decreasing while the terms are positive. Since , we have
Thus
which, since , implies that

Observe then that we have a decreasing sequence of terms as long as the terms are positive. Since our decreasing sequence is bounded above by and since is less than zero, this sequence is not bounded below by zero and thus crosses over to negative values.

Let be the last positive term and the first negative term of our solution. We now determine ranges of values of terms of our solution beginning with and , using “” in place of “” for convenience.

(1) and : observe that
which is true. Hence, . Note that this, in turn, implies that . Then at least since .(2): by Statement (), and so .(3)*Case 1 (). *Then from Statement (), and thus . By Theorem 4.2, the solution is unbounded.*Case 2 (). *We wish to show that :
which is true since, by Statements () and (), .We next show that . First we show that :
which is true since, by Statements () and (), and .

Second, we show that : notice that
which is true since and by Statement (). Because , it follows from Theorem 2.5 that there exists such that for all .

#### 5. Conclusions and Future Work

It is of interest to continue the investigation of the monotonicity, periodicity, and boundedness nature of solutions of (1.1). It is of further interest to extend our study of solutions of (1.1) to an equation with an arbitrary constant parameter or a nonautonomous parameter, or to an equation with arbitrary delays.(i), where is any real.(ii), where is a periodic sequence of real numbers.(iii), where .

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