`Abstract and Applied AnalysisVolume 2010, Article ID 152378, 17 pageshttp://dx.doi.org/10.1155/2010/152378`
Research Article

## Long-Term Behavior of Solutions of the Difference Equation

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 2 August 2010; Revised 24 October 2010; Accepted 28 October 2010

Copyright © 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

We investigate the long-term behavior of solutions of the following difference equation: , , where the initial values , , and are real numbers. Numerous fascinating properties of the solutions of the equation are presented.

#### 1. Introduction and Preliminaries

Recently there has been great interest in studying nonlinear difference equations which do not stem from differential equations (see, e.g., [128] and the references therein). Standard properties which have been studied are boundedness [5, 9, 2325], periodicity [2, 5, 9, 10, 27], asymptotic periodicity [3, 4, 8, 1114, 16, 17, 19, 20, 23], and local and global stability [5, 911, 2326], as well as existence of specific solutions such as monotone or nontrivial solutions [1, 6, 7, 13, 15, 1822].

In this paper, we investigate the long-term behavior of solutions of the third-order difference equation where the initial values are real numbers.

The difference equation (1.1) belongs to the class of equations of the form where , , and . The case , has been recently investigated in [8].

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

This section is devoted to the study of the equilibria and periodic solutions of (1.1).

##### 2.1. Equilibria of (1.1)

If is an equilibrium of (1.1), then it satisfies the equation Hence, (1.1) has exactly two equilibria, one positive and one negative, which we denote by and , respectively: (the golden number and its conjugate).

##### 2.2. Periodic Solutions of (1.1)

Here, we study the existence of periodic solutions of (1.1). For related results, see, for example, [2, 5, 9, 10, 27] and the references therein. The first two results are simple, but we will prove them for the completeness, the benefit of the reader, and since we use them in the sequel.

Theorem 2.1. There are no eventually constant solutions of difference equation (1.1).

Proof. If is an eventually constant solution of (1.1), then , for some , where is an equilibrium point. In this case, (1.1) gives , which implies Repeating this procedure, we obtain for . Hence, there are no eventually constant solutions.

Theorem 2.2. Difference equation (1.1) has no nontrivial period two solutions nor eventually period two solutions.

Proof. Assume that and , for every , and some , with . Then, we have From this and since , we obtain a contradiction, finishing the proof of the result.

Theorem 2.3. There are no periodic or eventually periodic solutions of (1.1) with prime period three.

Proof. If for some , we have If , , or , then from (2.6) we easily obtain contradictions in all these cases. Hence, we may assume that , , and . Equalities in (2.6) also imply that From (2.7), we get which implies that , that is, . From this and (2.7) we obtain , implying ,, from which the result follows.

Theorem 2.4. There are no periodic or eventually periodic solutions of (1.1) with prime period four.

Proof. Assume for some . Then we have If for some , then from (2.10) we easily obtain a contradiction. For example, if , then from (2.10) we get . This implies . From this and since we would get , a contradiction. The other cases are proved analogously.
Hence we may assume that , . From (2.10) we have that is, From (2.10) and (2.12), we obtain Since the relations in (2.12) are cyclic, we also obtain that Equalities (2.13) and (2.14) imply that the expressions , have the same sign. Assume that they are all positive (the case when they are all negative is considered similarly so it is omitted). We have from which easily follows that . From this, (2.13), and (2.14), it follows that , , which implies the result.

The following result shows that there exist periodic solutions of (1.1) with prime period five.

Theorem 2.5. A solution of (1.1) is of period five if and only if (i), , , and , or(ii), or (iii), , and .

Proof. If the initial conditions are as given, then by some calculations it is easy to see that these solutions are of period five. Now we assume that a solution is of period five. Then, we can write terms of the solution of (1.1) as
Note that the expressions for and both yield the same condition, namely, . The condition for gives , so that . Hence, either or , which was a restriction coming from and . If , then . The second restriction can be rewritten as , provided . If , then .

Remark 2.6. If , where , then starts a 5-cycle (in particular, in view of the above theorem, we have a 5 cycle if the initial conditions are ,,, where and are any real numbers). If , then we have from which the statement follows in this case.
If , then by direct calculation we have from which the statement follows in this case.

Remark 2.7. There are period-five solutions of (1.1) that do not have a zero term. It is enough to use and such that . For example, we can choose , ,.

Remark 2.8. There exist solutions that are eventually of period five; for example, choose , , , or , , .

#### 3. Solutions in the Interval

Here, we study the solutions of (1.1) with initial values in the interval or for which there are three subsequent terms that are eventually in the interval.

The next result shows that the interval is an invariant interval for (1.1).

Theorem 3.1. If , then for all .

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

Remark 3.2. There are solutions that eventually enter the interval . One example of such a solution is one with the initial conditions , , and .

##### 3.1. Convergence to Period-Five Solutions

The next theorem is devoted to the convergence of solutions of (1.1) with initial conditions in the interval to period-five solutions. For related results on the asymptotic periodicity of difference equations, see [3, 4, 8, 1014, 16, 17, 19, 20, 23].

Note that if , then , and we have that . Indeed, since , then clearly , and so that , from which it follows that . Hence, by Theorems 2.5 and 3.1 it follows that these solutions are periodic of period five belonging to the interval .

Before we formulate and prove the main result in this section, we need an auxiliary result. Relations of this type were first discovered and used by Stević in [12] and then subsequently used in several papers (e.g., in [16]).

Lemma 3.3. Any solution of (1.1) satisfies the following equality

Proof. If , we have
Using (3.3) and then (3.2), we get which is equality (3.1).

Theorem 3.4. Consider the difference equation (1.1) with initial conditions . Then, this solution converges to a period-five solution.

Proof. Consider five subsequences , , , , and . By Lemma 3.3, we know that our solution satisfies (3.1) for all . Thus, we have Therefore, and, since the values of the sequence are in the interval , we have, Without loss of generality, we will consider the subsequence . Let By inequality (3.7), we have Since the sequence is positive, nonincreasing, and bounded below by 0, it must converge to, say, . We will prove that . Assume to the contrary that . Observe that from (3.6) we have Since the sequence converges to a nonzero value, it follows that converges to a nonzero value. Hence, . Therefore, is close to −1 for large. Suppose is large. Then, is close to 0, which contradicts the fact that as . Hence, , as desired.

Remark 3.5. As it has been already mentioned, there are solutions that have initial values outside the interval and enter the interval. Such solutions also converge to a period five solution by the previous theorem.

Corollary 3.6. Assume that a solution of (1.1) has initial conditions ,. Then, , all future terms are in the interval , and thus this solution converges to a periodic solution with period five.

Proof. Clearly . Since and , we have , and so . Furthermore, we have . Similarly, . The rest of the proof is the same as in Theorem 3.1, and so is omitted.

#### 4. 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 4.1. The positive equilibrium of (1.1),  , is unstable.

Proof. The characteristic equation of the equilibrium is the following: Let . Since and , it follows that there is a such that , from which the result follows for the equilibrium .

Remark 4.2. Consider the characteristic equation of the negative equilibrium of (1.1), , and the function . Observe the following: (i). (ii) for all . Hence, we have one negative eigenvalue, and a complex conjugate pair of eigenvalues, and , such that It follows that , and so is a nonhyperbolic equilibrium where the roots of the characteristic equation have absolute values less than or equal to one.

Open Problem 4.3. Determine the stability nature of the negative equilibrium of (1.1), .

#### 5. More on Invariant Intervals

In this section, we discuss invariant intervals regarding the subsequences , , , , and of a solution of (1.1).

Theorem 5.1. Assume that a solution of (1.1) has initial conditions , and . Then, , , and for and also .

Proof. Since , we have . Now, we prove the above statement for . From and , we have . From this and since , we get .
Furthermore, we have that Since and , it follows that . Now, we verify that must be in a more restrictive interval . We prove this by contradiction. Assume that . Since with and , we must have that which is equivalent to On the other hand, since and , we have . However, since , it follows that , which is a contradiction. Hence, , as claimed.
Next, since and , we have , and so, , finishing the proof for the case .
Now assume that , , and for .
Since and , we have
Since and , we have Assume that . Then, we have which is equivalent to Now note that and since and . Hence, which along with implies that , which is a contradiction. Hence,
Since , we have Assume that .
Since by Lemma 3.3 and , we have . Therefore, . From this and since , we have that . We then have Since , we have . Furthermore we have Since , we have .
By Lemma 3.3, for each , we have By the inductive hypothesis, we have that so that the sign of the difference is the same as the sign of , that is, On the other hand, , which is a contradiction. Hence,
Since and , we have Assume that . Then, we have which is equivalent to On the other hand, since and since from and we obtain , it follows that This fact along with implies which is a contradiction. Hence,
Finally, since and , we obtain From (5.4)–(5.22) and by the method of induction, the proof follows.

Theorem 5.2. Any solution of (1.1) with initial values satisfying the following conditions ,, and converges to a period-five solution.

Proof. By Theorem 5.1, we have that , , , , , , and for .
We prove by induction that all the subsequences , , , , and are monotone.
Assume . Then, by Lemma 3.3 and above comments, we have
Assume that we have proved for .
Since for each and , and since by Theorem 5.1, we obtain , , we have that the differences , have the same sign for each . From this and by Theorem 5.1, the claim follows.
Since all the subsequences , , , and are monotone and bounded, they are convergent, and consequently the solution converges to a period-five solution, as claimed.

#### 6. 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 values 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.

Theorem 6.1. If , then the following statements hold true:(a) and ;(b)the solution tends to .

Proof. (a) Since , we have . Thus, Therefore, . Thus, . Rewriting, . Hence, . One can follow the same steps to prove that , and the rest of the proof goes by a simple inductive argument.
(b) Suppose, on the contrary, that one of these subsequences given in part (a) is bounded. Then, by the relationship it would follow that both subsequences and converge. Hence the whole solution either converges to a period-two solution or to an equilibrium. However, (1.1) does not have any nontrivial period-two solution. Thus, it must converge to an equilibrium. But, this is not possible because the largest equilibrium point is smaller than and . This is a contradiction. Hence, the proof is complete.

Theorem 6.2. Assume that and . Then, each solution with the initial conditions , , and tends to plus infinity.

Proof. We have , , and By Lemma 3.3, we have Using (6.4) and the fact , by induction it easily follows that that is, the subsequences , , , and are increasing.
We also have Note that Using this fact and (6.6) we obtain from which it follows that as .
From (6.4), the monotonicity of those five subsequences, and (6.8), we get for , from which it follows that as for each , finishing the proof of the theorem.

Remark 6.3. Note that the last theorem shows that if initial values are moved to the right with respect to the initial values of a positive solution of period five then such solutions go to plus infinity.

#### 7. Case

Here, we consider the case . The next theorem shows that there is a large class of eventually nondecreasing solutions of (1.1) converging to . For some results of this type, see, for example, [1, 6, 7, 10, 13, 15, 1822] and the related references therein.

Theorem 7.1. Assume and . Then, every solution with such initial values is eventually nondecreasing and converges to .

Proof. Multiplying the assumption , by we obtain . From this and since , we obtain , that is, . Hence, . Now assume for some . Multiplying the inequality by and using (7.1), it follows that that is, . Hence, by induction we have proved that the sequence is nondecreasing and bounded above by , from which the result easily follows.

Theorem 7.2. Assume , , , and . Then, for every solution with such initial values there is an , such that

Proof. According to the assumptions we have
If , then by multiplying by , we obtain
If , then since , we obtain
Multiplying the inequality (see (7.4)) by , we obtain which, together with inequality (7.6), implies again .
Note that now we cannot guarantee the positivity of . Similarly, from the inequality (see (7.5)), we obtain and from the inequality , we obtain
Now assume that and . Then, by multiplying the inequality by and subtracting 1, we obtain . Hence, we proved by induction that (7.11) holds as far as is positive.
If for all , the sequence is convergent and its limit is nonnegative. However, this is not possible since the only nonnegative equilibrium of (1.1) is . From this, the result follows.

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