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

On the Difference Equation π‘₯𝑛+1=π‘₯𝑛π‘₯π‘›βˆ’2βˆ’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: YongΒ Zhou

Copyright Β© 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: π‘₯𝑛+1=π‘₯𝑛π‘₯π‘›βˆ’2βˆ’1, π‘›βˆˆβ„•0, where the initial values π‘₯βˆ’2, π‘₯βˆ’1, π‘₯0 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., [1–36] and the references therein). Usual properties which have been studied are the boundedness character [8, 13, 15, 28–30, 33, 35, 36], the periodicity [8, 13], asymptotic periodicity [16–19, 21], local and global stability [1, 8, 13, 15, 16, 28–34], as well as the existence of specific solutions such as monotone or nontrivial [2, 3, 5, 9, 10, 15, 18, 20, 22–27].

In this paper we will study solutions of the following difference equation: π‘₯𝑛+1=π‘₯𝑛π‘₯π‘›βˆ’2βˆ’1,π‘›βˆˆβ„•0.(1.1)

The difference equation (1.1) belongs to the class of equations of the form π‘₯𝑛+1=π‘₯π‘›βˆ’π‘˜π‘₯π‘›βˆ’π‘™βˆ’1,π‘›βˆˆβ„•0,(1.2) with particular choices of π‘˜ and 𝑙, where π‘˜,π‘™βˆˆβ„•0. Although (1.2) looks simple, it is fascinating how its behavior changes for different choices of π‘˜ and 𝑙. The cases π‘˜=0 and 𝑙=1, π‘˜=1 and 𝑙=2 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: π‘₯1=√1βˆ’52,π‘₯2=√1+52.(1.3)

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 π‘₯βˆ’2=0, π‘₯βˆ’1=βˆ’1, π‘₯0=0 or π‘₯βˆ’2=βˆ’1, π‘₯βˆ’1=0, π‘₯0=βˆ’1.

Proof. Let (π‘₯𝑛)βˆžπ‘›=βˆ’2 be a prime period-two solution of (1.1). Then π‘₯2π‘›βˆ’2=π‘Ž and π‘₯2π‘›βˆ’1=𝑏, for every π‘›βˆˆβ„•0 and for some π‘Ž,π‘βˆˆβ„ such that π‘Žβ‰ π‘. We have π‘₯1=π‘₯0π‘₯βˆ’2βˆ’1=π‘Ž2βˆ’1=𝑏 and π‘₯2=π‘₯1π‘₯βˆ’1βˆ’1=𝑏2βˆ’1=π‘Ž. From these two equations we obtain (π‘Ž2βˆ’1)2βˆ’1=π‘Ž or equivalently ξ€·π‘Žπ‘Ž(π‘Ž+1)2ξ€Έβˆ’π‘Žβˆ’1=0.(2.1) We have four cases to be considered.
Case 1. If π‘Ž=0, then 𝑏=βˆ’1, and we obtain the first prime period-two solution.
Case 2. If π‘Ž=βˆ’1, then 𝑏=0, and we obtain the second prime period-two solution.
Case 3. If π‘Ž=π‘₯1, then 𝑏=π‘Ž2βˆ’1=π‘₯1, which is an equilibrium solution.
Case 4. If π‘Ž=π‘₯2, then 𝑏=π‘Ž2βˆ’1=π‘₯2, which is the second equilibrium solution. Thus, the result holds.

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

Proof. Let (π‘₯𝑛)βˆžπ‘›=βˆ’2 be a prime period-three solution of (1.1). Then π‘₯3π‘›βˆ’2=π‘Ž,  π‘₯3π‘›βˆ’1=𝑏,  π‘₯3𝑛=𝑐, for every π‘›βˆˆβ„•0 and some π‘Ž,𝑏,π‘βˆˆβ„ such that at least two of them are different. We have π‘₯1=π‘₯0π‘₯βˆ’2π‘₯βˆ’1=π‘Žπ‘βˆ’1=π‘Ž,2=π‘₯1π‘₯βˆ’1π‘₯βˆ’1=π‘Žπ‘βˆ’1=𝑏,3=π‘₯2π‘₯0βˆ’1=π‘π‘βˆ’1=𝑐.(2.2) From (2.2) we easily see that π‘Žβ‰ 0, 𝑏≠0, and 𝑐≠0, so that 1𝑐=1+π‘Ž1,π‘Ž=1+𝑏1,𝑏=1+𝑐.(2.3) From (2.3) we obtain 𝑏𝑐=1+=𝑏+12𝑏+1𝑏+1βŸΉπ‘=1+𝑏+1=2𝑏+13𝑏+2,2𝑏+1(2.4) which implies 𝑏2βˆ’π‘βˆ’1=0. Hence 𝑏=π‘₯1 or 𝑏=π‘₯2. From this and (2.3) it follows that π‘Ž=𝑏=𝑐=π‘₯1 or π‘Ž=𝑏=𝑐=π‘₯2, from which the result follows.

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

Proof. Let (π‘₯𝑛)βˆžπ‘›=βˆ’2 be a prime period-four solution of (1.1) and π‘₯βˆ’2=π‘Ž, π‘₯βˆ’1=𝑏, π‘₯0=𝑐. Then we have π‘₯1=π‘₯0π‘₯βˆ’2βˆ’1=π‘Žπ‘βˆ’1,(2.5)π‘₯2=π‘₯1π‘₯βˆ’1βˆ’1=(π‘Žπ‘βˆ’1)π‘βˆ’1=π‘Ž,(2.6)π‘₯3=π‘₯2π‘₯0βˆ’1=π‘Žπ‘βˆ’1=𝑏,(2.7)π‘₯4=π‘₯3π‘₯1βˆ’1=𝑏(π‘Žπ‘βˆ’1)βˆ’1=𝑐.(2.8) Thus, from (2.6) and (2.8), we have that π‘Ž=𝑐. This along with (2.7) gives π‘Ž2βˆ’1=𝑏,(2.9) while from (2.8) we get 𝑏(π‘Ž2βˆ’1)βˆ’1=π‘Ž or equivalently (π‘Ž+1)(𝑏(π‘Žβˆ’1)βˆ’1)=0.(2.10)
Case 1. Suppose π‘Ž=βˆ’1. Then 𝑏=π‘Ž2βˆ’1=0 and 𝑐=π‘Ž=βˆ’1, which, by Theorem 2.1, yields a period-two solution.
Suppose π‘Žβ‰ βˆ’1. If π‘Ž=1, then from (2.10) we get a contradiction. If π‘Žβ‰ 1, then π‘Ž2βˆ’1=𝑏=1/(π‘Žβˆ’1), so that π‘Ž(π‘Ž2βˆ’π‘Žβˆ’1)=0. Hence π‘Ž=0, π‘Ž=π‘₯1, or π‘Ž=π‘₯2.

Case 2. Suppose π‘Ž=0. Then 𝑏=π‘Ž2βˆ’1=βˆ’1 and 𝑐=π‘Ž=0 which results in a period-two solution as proved in Theorem 2.1.
Case 3. Suppose π‘Ž=π‘₯1. Then 𝑏=π‘Ž2βˆ’1=π‘₯1 and 𝑐=π‘₯1 which is an equilibrium solution.
Case 4. Suppose π‘Ž=π‘₯2. Then 𝑏=π‘Ž2βˆ’1=π‘₯2 and 𝑐=π‘₯2 which is the second equilibrium solution. Proof is complete.

3. Local Stability

Here we study the local stability at the equilibrium points π‘₯1 and π‘₯2.

Theorem 3.1. The negative equilibrium of (1.1), π‘₯1, is unstable. Moreover, it is a hyperbolic equilibrium.

Proof. The linearized equation associated with the equilibrium π‘₯1√=(1βˆ’5)/2∈(βˆ’1,0) is π‘₯𝑛+1βˆ’π‘₯1π‘₯π‘›βˆ’π‘₯1π‘₯π‘›βˆ’2=0.(3.1) Its characteristic polynomial is 𝑃π‘₯1(πœ†)=πœ†3βˆ’π‘₯1πœ†2βˆ’π‘₯1.(3.2) Hence π‘ƒξ…žπ‘₯1(πœ†)=3πœ†2βˆ’2π‘₯1ξ€·πœ†=πœ†3πœ†βˆ’2π‘₯1ξ€Έ.(3.3) Since 𝑃π‘₯1(βˆ’2)=βˆ’8βˆ’5π‘₯1<0,𝑃π‘₯1(βˆ’1)=βˆ’1βˆ’2π‘₯1=√5βˆ’2>0,(3.4) there is a zero πœ†1∈(βˆ’2,βˆ’1) of 𝑃π‘₯1.
On the other hand, from 𝑃π‘₯1(0)=βˆ’π‘₯1>0 and (3.3), it follows that πœ†1 is a unique real zero of 𝑃π‘₯1. Hence, the other two roots πœ†2,3 are conjugate complex.
Since πœ†1||πœ†2||2=π‘₯1,(3.5) we obtain |πœ†2|=|πœ†3|<1. From this, the theorem follows.

Theorem 3.2. The positive equilibrium of (1.1), π‘₯2, is unstable. Moreover, it is also a hyperbolic equilibrium.

Proof. The linearized equation associated with the equilibrium π‘₯2√=(1+5)/2∈(1,2) is π‘₯𝑛+1βˆ’π‘₯2π‘₯π‘›βˆ’π‘₯2π‘₯π‘›βˆ’2=0.(3.6) Its characteristic polynomial is 𝑃π‘₯2(πœ†)=πœ†3βˆ’π‘₯2πœ†2βˆ’π‘₯2.(3.7)
We have π‘ƒξ…žπ‘₯2(πœ†)=3πœ†2βˆ’2π‘₯2ξ€·πœ†=πœ†3πœ†βˆ’2π‘₯2ξ€Έ.(3.8) Since 𝑃π‘₯2(2)=8βˆ’5π‘₯2=ξ‚€βˆš11βˆ’552<0,𝑃π‘₯2(3)=27βˆ’10π‘₯2√=22βˆ’55>0,(3.9) there is a zero πœ†1∈(2,3) of 𝑃π‘₯2.
From this and since 𝑃π‘₯2(0)=βˆ’π‘₯2<0, we have that πœ†1 is a unique real zero of 𝑃π‘₯2. Thus, the other two roots πœ†2,3 are conjugate complex.
Since πœ†1||πœ†2||2=π‘₯2,(3.10) we obtain |πœ†2|=|πœ†3|<1. From this, the theorem follows.

4. Case π‘₯βˆ’2,π‘₯βˆ’1,π‘₯0∈(βˆ’1,0)

This section considers the solutions of (1.1) with π‘₯βˆ’2,π‘₯βˆ’1,π‘₯0∈(βˆ’1,0). Before we formulate the main result in this section we need some auxiliary results.

Lemma 4.1. Suppose that π‘₯βˆ’2,π‘₯βˆ’1,π‘₯0∈(βˆ’1,0). Then the solution (π‘₯𝑛)βˆžπ‘›=βˆ’2 of (1.1) is such that π‘₯π‘›βˆˆ(βˆ’1,0) for 𝑛β‰₯βˆ’2.

Proof. We have π‘₯βˆ’2,π‘₯0∈(βˆ’1,0) which implies π‘₯1=π‘₯0π‘₯βˆ’2βˆ’1∈(βˆ’1,0). Assume that we have proved π‘₯π‘›βˆˆ(βˆ’1,0) for βˆ’2β‰€π‘›β‰€π‘˜, for some π‘˜β‰₯2. Then we have π‘₯π‘˜+1=π‘₯π‘˜π‘₯π‘˜βˆ’2βˆ’1∈(βˆ’1,0), 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 π‘˜π‘–βˆˆβ„•, 1≀𝑖≀2π‘š, π‘₯βˆ’π‘ ,…,π‘₯βˆ’1,π‘₯0∈(βˆ’1,0), where 𝑠=max{π‘˜1,…,π‘˜2π‘š}. Then the solution (π‘₯𝑛)βˆžπ‘›=βˆ’π‘  of the difference equation π‘₯𝑛+1=2π‘šξ‘π‘–=1π‘₯π‘›βˆ’π‘˜π‘–βˆ’1,π‘›βˆˆβ„•0,(4.1) is such that π‘₯π‘›βˆˆ(βˆ’1,0) for 𝑛β‰₯βˆ’π‘ .

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 π‘₯2𝑛+3=π‘₯2𝑛+2π‘₯2π‘›βˆ’1,π‘₯2𝑛+2=π‘₯2𝑛+1π‘₯2π‘›βˆ’1βˆ’1,π‘›βˆˆβ„•0.(4.2) Then we have the following:π‘₯2𝑛+3=ξ€·π‘₯2𝑛+1π‘₯2π‘›βˆ’1π‘₯βˆ’1ξ€Έξ€·2π‘›βˆ’1π‘₯2π‘›βˆ’3ξ€Έβˆ’1βˆ’1=π‘₯2π‘›βˆ’1ξ€·π‘₯2𝑛+1π‘₯2π‘›βˆ’1π‘₯2π‘›βˆ’3βˆ’π‘₯2𝑛+1βˆ’π‘₯2π‘›βˆ’3ξ€Έ,π‘›βˆˆβ„•,(4.3)

and similarly π‘₯2𝑛+4=π‘₯2𝑛π‘₯2𝑛+2π‘₯2𝑛π‘₯2π‘›βˆ’2βˆ’π‘₯2𝑛+2βˆ’π‘₯2π‘›βˆ’2ξ€Έ,π‘›βˆˆβ„•.(4.4)

Hence, the subsequences 𝑦𝑛=π‘₯2𝑛+1 and 𝑧𝑛=π‘₯2𝑛+2 satisfy the difference equation 𝑒𝑛+1=π‘’π‘›βˆ’1ξ€·π‘’π‘›π‘’π‘›βˆ’1π‘’π‘›βˆ’2βˆ’π‘’π‘›βˆ’π‘’π‘›βˆ’2ξ€Έ,π‘›βˆˆβ„•,(4.5) and π‘’π‘›βˆˆ(βˆ’1,0).

For convenience, we make another change of variable 𝑣𝑛=βˆ’π‘’π‘›. Then (4.5) becomes 𝑣𝑛+1=π‘£π‘›βˆ’1𝑣𝑛+π‘£π‘›βˆ’2βˆ’π‘£π‘›π‘£π‘›βˆ’1π‘£π‘›βˆ’2ξ€Έ,π‘›βˆˆβ„•.(4.6) Note also that π‘£π‘›βˆˆ(0,1).

It is easy to see that (4.6) has the following four equilibria: 𝑣0√=βˆ’1+52,𝑣1=0,𝑣2=√5βˆ’12,𝑣3=1.(4.7)

If we let 𝑓(𝑒,𝑣,𝑀)=𝑣(𝑒+π‘€βˆ’π‘’π‘£π‘€),(4.8) where 𝑒,𝑣,π‘€βˆˆ(0,1), then we find the following:(1)𝑓𝑒=π‘£βˆ’π‘£2𝑀=𝑣(1βˆ’π‘£π‘€)>0, (2)𝑓𝑣=𝑒+π‘€βˆ’2𝑣𝑒𝑀=𝑒(1βˆ’π‘£π‘€)+𝑀(1βˆ’π‘£π‘’)>0, (3)𝑓𝑀=π‘£βˆ’π‘£2𝑒=𝑣(1βˆ’π‘£π‘’)>0.

Thus, the function 𝑓(𝑒,𝑣,𝑀) is strictly increasing in each argument.

Lemma 4.3. Let (π‘₯𝑛)βˆžπ‘›=βˆ’2 be a solution of (1.1) which is not equal to the equilibrium solution π‘₯1=√1βˆ’52(4.9) of the equation. Suppose that π‘₯βˆ’2,π‘₯βˆ’1,π‘₯0∈(βˆ’1,0)(4.10) and that one of the following conditions holds: (H1)π‘₯βˆ’2≀π‘₯1, π‘₯βˆ’1β‰₯π‘₯1, π‘₯0≀π‘₯1 with at least one of the inequalities strict,(H2)π‘₯βˆ’2β‰₯π‘₯1, π‘₯βˆ’1≀π‘₯1, π‘₯0β‰₯π‘₯1 with at least one of the inequalities strict.
Then π‘₯π‘›βˆˆ(βˆ’1,0), for every 𝑛β‰₯βˆ’2 and
(a)if (H1) holds, then there is an π‘βˆˆβ„•0 such that π‘₯2𝑛+1>π‘₯1,π‘₯2𝑛+2<π‘₯1,𝑛β‰₯𝑁,(4.11)(b)if (H2) holds, then there is an π‘βˆˆβ„•0 such that π‘₯2𝑛+1<π‘₯1,π‘₯2𝑛+2>π‘₯1,𝑛β‰₯𝑁.(4.12)

Proof. By Lemma 4.1 we have that π‘₯π‘›βˆˆ(βˆ’1,0), for 𝑛β‰₯βˆ’2. We will prove only (a). The proof of (b) is dual and is omitted. Since βˆ’1<π‘₯0,π‘₯βˆ’2≀π‘₯1, we have π‘₯1=π‘₯0π‘₯βˆ’2βˆ’1β‰₯π‘₯21βˆ’1=π‘₯1.(4.13) From this and since π‘₯1≀π‘₯βˆ’1<0, we have π‘₯2=π‘₯1π‘₯βˆ’1βˆ’1≀π‘₯21βˆ’1=π‘₯1.(4.14) If π‘₯βˆ’2<π‘₯1 or π‘₯0<π‘₯1, then inequality (4.13) is strict and, consequently, inequality (4.14) is strict too. If π‘₯βˆ’2=π‘₯0=π‘₯1, then π‘₯βˆ’1>π‘₯1, from which it follows that inequality (4.14) is strict. In this case we have π‘₯3=π‘₯2π‘₯0βˆ’1>π‘₯21βˆ’1=π‘₯1,(4.15) which is a strict inequality. Hence 𝑁=0 and 𝑁=1 are the obvious candidates, depending on which of the two cases, just described, holds.
Assume that we have proved (4.11) for π‘β‰€π‘›β‰€π‘˜ and that 𝑁=0. The case 𝑁=1 is proved similarly and so is omitted. Then we have π‘₯2π‘˜+3=π‘₯2π‘˜+2π‘₯2π‘˜βˆ’1>π‘₯21βˆ’1=π‘₯1.(4.16) From this and since π‘₯1<π‘₯2π‘˜+1<0, we have π‘₯2π‘˜+4=π‘₯2π‘˜+3π‘₯2π‘˜+1βˆ’1<π‘₯21βˆ’1=π‘₯1.(4.17) Hence by induction the lemma follows.

Theorem 4.4. Let (π‘₯𝑛)βˆžπ‘›=βˆ’2 be a solution of (1.1) which is not equal to the equilibrium solution π‘₯1=√1βˆ’52(4.18) of the equation. Suppose that π‘₯βˆ’2,π‘₯βˆ’1,π‘₯0∈(βˆ’1,0)(4.19) and that one of the conditions, (H1) or (H2), holds.
Then π‘₯π‘›βˆˆ(βˆ’1,0), for every 𝑛β‰₯βˆ’2, and (π‘₯𝑛)βˆžπ‘›=βˆ’2 converges to a two-cycle{βˆ’1,0}.

Proof. First of all π‘₯π‘›βˆˆ(βˆ’1,0), for 𝑛β‰₯βˆ’2, by Lemma 4.1. We next show that (π‘₯𝑛)βˆžπ‘›=βˆ’2 converges to a two-cycle{βˆ’1,0}. To this end we show that one of the subsequences, (π‘₯2𝑛)βˆžπ‘›=βˆ’1 or (π‘₯2𝑛+1)βˆžπ‘›=βˆ’1, converges to 0 and the other one to βˆ’1. Showing this, in turn, is equivalent to showing the following:(a)the corresponding solution (𝑣𝑛)βˆžπ‘›=βˆ’1 of (4.6) converges to 𝑣3 if for some 𝑁β‰₯βˆ’1, 𝑣𝑛>𝑣2, for 𝑛β‰₯𝑁, where π‘£π‘›βˆˆ(0,1)=(𝑣1,𝑣3) for 𝑛β‰₯βˆ’1,(b)the corresponding solution (𝑣𝑛)βˆžπ‘›=βˆ’1 of (4.6) converges to 𝑣1 if for some 𝑁β‰₯βˆ’1,𝑣𝑛<𝑣2, for 𝑛β‰₯𝑁, where π‘£π‘›βˆˆ(0,1)=(𝑣1,𝑣3) for 𝑛β‰₯βˆ’1.
We prove (a). The proof of (b) is similar and will be omitted. We have π‘£π‘›βˆˆξ€·π‘£2,𝑣3ξ€Έ,𝑛β‰₯𝑁.(4.20) Let 𝐼=liminfπ‘›β†’βˆžπ‘£π‘›,𝑆=limsupπ‘›β†’βˆžπ‘£π‘›.(4.21) Then we have 𝑣2≀𝐼≀𝑆≀𝑣3.(4.22)
First assume 𝐼=𝑣2. From (4.20) it follows that there is an πœ€>0 such that 𝐼+πœ€<𝑣𝑁,𝑣𝑁+1,𝑣𝑁+2<𝑣3. By the monotonicity of 𝑓 and since 𝑓(π‘₯,π‘₯,π‘₯)>π‘₯forξ€·π‘₯βˆˆπ‘£2,𝑣3ξ€Έ,(4.23) we have 𝑣𝑁+3𝑣=𝑓𝑁+2,𝑣𝑁+1,𝑣𝑁>𝑓(𝐼+πœ€,𝐼+πœ€,𝐼+πœ€)>𝐼+πœ€.(4.24) From this and by induction we obtain 𝑣𝑛>𝐼+πœ€, 𝑛β‰₯𝑁, which implies liminfπ‘›β†’βˆžπ‘£π‘›β‰₯𝐼+πœ€, which is a contradiction.
Now assume 𝐼∈(𝑣2,𝑣3). Let (π‘£π‘›π‘˜)π‘˜βˆˆβ„• be a subsequence of (𝑣𝑛)βˆžπ‘›=βˆ’1 such that limπ‘˜β†’βˆžπ‘£π‘›π‘˜=𝐼. Then there is a subsequence of (π‘£π‘›π‘˜)π‘˜βˆˆβ„•, which we may denote the same, such that there are the following limits: limπ‘˜β†’βˆžπ‘£π‘›π‘˜βˆ’1, limπ‘˜β†’βˆžπ‘£π‘›π‘˜βˆ’2, and limπ‘˜β†’βˆžπ‘£π‘›π‘˜βˆ’3, which we denote, respectively, by πΎβˆ’π‘–, 𝑖=1,2,3. From this and by (4.23) we have that π‘“ξ€·πΎβˆ’1,πΎβˆ’2,πΎβˆ’3ξ€Έ=𝐼<𝑓(𝐼,𝐼,𝐼).(4.25) Hence there is an 𝑖0∈{1,2,3} such that πΎβˆ’π‘–0<𝐼. Otherwise, πΎβˆ’π‘–β‰₯𝐼 for 𝑖=1,2,3 and by the monotonicity of 𝑓 we would get 𝑓𝐾(𝐼,𝐼,𝐼)β‰€π‘“βˆ’1,πΎβˆ’2,πΎβˆ’3ξ€Έ=𝐼<𝑓(𝐼,𝐼,𝐼),(4.26) which is a contradiction. On the other hand, πΎβˆ’π‘–0<𝐼 contradicts the choice of 𝐼. Hence 𝐼 cannot be in the interval (𝑣2,𝑣3).
From all of the above we have that 𝑣3=𝐼≀𝑆≀𝑣3. Therefore, limπ‘›β†’βˆžπ‘£π‘›=𝑣3, as desired.

Theorem 4.5. Assume that for a solution (π‘₯𝑛)βˆžπ‘›=βˆ’2 of (1.1) there is an 𝑁β‰₯βˆ’1 such that βˆ’1<π‘₯𝑁<π‘₯𝑁+2<0,0>π‘₯π‘βˆ’1>π‘₯𝑁+1>π‘₯𝑁+3>βˆ’1.(4.27) Then the solution converges to a two-cycle{βˆ’1,0} or to the equilibrium π‘₯1.

Proof. First note that by Lemma 4.1 we have π‘₯π‘›βˆˆ(βˆ’1,0), 𝑛β‰₯𝑁. From (1.1) we obtain the identity π‘₯𝑛+4βˆ’π‘₯𝑛+2=π‘₯𝑛+1ξ€·π‘₯𝑛+3βˆ’π‘₯π‘›βˆ’1ξ€Έ.(4.28) Applying (4.28) for 𝑛=𝑁 and using the fact π‘₯𝑁+1∈(βˆ’1,0), we get 0>π‘₯𝑁+4>π‘₯𝑁+2. Hence π‘₯𝑁<π‘₯𝑁+2<π‘₯𝑁+4<0,π‘₯π‘βˆ’1>π‘₯𝑁+1>π‘₯𝑁+3>βˆ’1.(4.29) Using induction along with identity (4.28) it is shown that π‘₯𝑁<π‘₯𝑁+2<β‹―<π‘₯𝑁+2π‘˜<0,π‘₯π‘βˆ’1>π‘₯𝑁+1>β‹―>π‘₯𝑁+2π‘˜+1>βˆ’1,(4.30) for every π‘˜βˆˆβ„•. Hence, there are finite limits limπ‘˜β†’βˆžπ‘₯𝑁+2π‘˜ and limπ‘˜β†’βˆžπ‘₯𝑁+2π‘˜+1, say 𝑙1 and 𝑙2. Letting π‘˜β†’βˆž in the relations π‘₯𝑁+2π‘˜+2=π‘₯𝑁+2π‘˜+1π‘₯𝑁+2π‘˜βˆ’1βˆ’1,π‘₯𝑁+2π‘˜+3=π‘₯𝑁+2π‘˜+2π‘₯𝑁+2π‘˜βˆ’1,(4.31) we get 𝑙1=𝑙22βˆ’1 and 𝑙2=𝑙21βˆ’1. Hence 𝑙1βˆ’π‘™2𝑙1+𝑙2ξ€Έ+1=0.(4.32) From this we have 𝑙1=𝑙2=π‘₯1, or if 𝑙1≠𝑙2, then 𝑙1+𝑙2=βˆ’1 so that 𝑙1=0 and 𝑙2=βˆ’1.

Remark 4.6. Let π‘₯βˆ’2=π‘Ž, π‘₯βˆ’1=𝑏 and π‘₯0=𝑐 with π‘Ž,𝑏,π‘βˆˆ(βˆ’1,0). For 𝑁=βˆ’1, (4.27) will be π‘₯βˆ’2>π‘₯0π‘₯βŸΊπ‘Ž>𝑐,1βˆ’π‘₯βˆ’1=π‘₯0π‘₯βˆ’2βˆ’1βˆ’π‘₯βˆ’1π‘₯=π‘Žπ‘βˆ’1βˆ’π‘>0,2βˆ’π‘₯0=π‘₯1π‘₯βˆ’1βˆ’1βˆ’π‘₯0=π‘₯0π‘₯βˆ’1π‘₯βˆ’2βˆ’π‘₯βˆ’1βˆ’π‘₯0βˆ’1=π‘Žπ‘π‘βˆ’π‘βˆ’π‘βˆ’1<0.(4.33)Hence, under the conditions π‘Ž>𝑐,π‘Žπ‘>𝑏+1,π‘Žπ‘π‘<𝑏+𝑐+1,(4.34) we have that (4.27) is satisfied for 𝑁=βˆ’1. It is easy to show that there are some π‘Ž,𝑏,π‘βˆˆ(βˆ’1,0) 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 π‘₯1. The next theorem, which is a product of an E-mail communication between SteviΔ‡ and Professor Berg [6], gives a positive answer to the question. In the proof of the result we use an asymptotic method from Proposition 3.3 in [3]. Some asymptotic methods for solving similar problems have been also used, for example, in the following papers: [2–5, 20, 22–27]. 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 π‘₯1.

Proof. In order to find a solution tending to π‘₯1, we make the substitution π‘₯𝑛=𝑦𝑛+π‘₯1, yielding the equation 𝑦𝑛+3βˆ’π‘₯1𝑦𝑛+2+𝑦𝑛=𝑦𝑛+2𝑦𝑛,𝑛β‰₯βˆ’2,(4.35) and for π‘›βˆˆβ„•0 we make the ansatz 𝑦𝑛=βˆžξ“βˆžπ‘˜=0𝑙=0π‘Žπ‘˜π‘™π‘π‘›π‘˜π‘žπ‘›π‘™,(4.36) with π‘Ž00=0, where 𝑝 and π‘ž are the conjugate complex zeros of the characteristic polynomial 𝑃(πœ†)=πœ†3βˆ’π‘₯1ξ€·πœ†2ξ€Έ.+1(4.37) Note that |𝑝|=|π‘ž|=π‘Ÿβ‰ˆ0.74448.
Replacing (4.36) into (4.35) and comparing the coefficients, we find π‘‘π‘˜π‘™π‘Žπ‘˜π‘™=π‘˜ξ“π‘™π‘–=0𝑗=0,𝑗+𝑖≠0π‘Žπ‘–π‘—π‘2π‘–π‘ž2π‘—π‘Žπ‘˜βˆ’π‘–,π‘™βˆ’π‘—,(4.38) with π‘‘π‘˜π‘™=𝑝3π‘˜π‘ž3π‘™βˆ’π‘₯1𝑝2π‘˜π‘ž2𝑙.+1(4.39)
Equation (4.38) is satisfied for π‘˜+𝑙≀1, where π‘Ž10 and π‘Ž01 are arbitrary, so that it suffices to consider (4.38) for π‘˜ and 𝑙 such that π‘˜+𝑙>1. If π‘Ž10 and π‘Ž01 are chosen to be conjugate complex numbers, then according to (4.38) all π‘Žπ‘˜π‘™ are conjugate complex numbers to π‘Žπ‘™π‘˜ and consequently series (4.36) is real. We look for a solution (4.36) with π‘Ž10=π‘Ž01, |π‘Ž10|=1, and determine a positive constant πœ† such that ||π‘Žπ‘˜π‘™||β‰€πœ†π‘˜+π‘™βˆ’1.(4.40) Since the inequality is valid for π‘˜+𝑙≀1, by induction, we get from (4.38) ||π‘Žπ‘˜π‘™||β‰€πœ†π‘˜+π‘™βˆ’21||π‘‘π‘˜π‘™||π‘˜ξ“π‘™π‘–=0𝑗=0,𝑗+𝑖≠0π‘Ÿ2(𝑖+𝑗).(4.41)
Note that π‘˜ξ“π‘™π‘–=0𝑗=0,𝑗+𝑖≠0π‘Ÿ2(𝑖+𝑗)<1ξ€·1βˆ’π‘Ÿ2ξ€Έ2βˆ’1=2π‘Ÿ2βˆ’π‘Ÿ4ξ€·1βˆ’π‘Ÿ2ξ€Έ2.(4.42)
It is not difficult to check that 𝐷∢=supπ‘˜+𝑙β‰₯21||π‘‘π‘˜π‘™||=1||𝑑21||β‰ˆ2.095.(4.43)
Hence (4.40) holds with πœ†=𝐷2π‘Ÿ2βˆ’π‘Ÿ4ξ€·1βˆ’π‘Ÿ2ξ€Έ2.(4.44)
For such chosen πœ† the series in (4.36) converges if πœ†π‘Ÿπ‘›<1, which implies 𝑛>lnπœ†/ln(1/π‘Ÿ). We have πœ†β‰ˆ8.450, so that lnπœ†/ln(1/π‘Ÿ)β‰ˆ7.233, and therefore we have the convergence of the series for 𝑛>7. In this way for 𝑛>7, we obtain a solution of (4.35) converging to a real solution of (4.35) as π‘›β†’βˆž, that is, a solution of (1.1) converging to π‘₯1, as desired.

Remark 4.8. For π‘Ž10=π‘Ž01=1, the first coefficients in (4.38) are π‘Ž20=𝑝2𝑑20,π‘Ž11=𝑝2+π‘ž2𝑑11,π‘Ž02=π‘ž2𝑑02,π‘Ž30=𝑝2𝑝2ξ€Έπ‘Ž+120𝑑30,π‘Ž21=𝑝4+π‘ž2ξ€Έπ‘Ž20+𝑝2ξ€·π‘ž2ξ€Έπ‘Ž+111𝑑21.(4.45)

Remark 4.9. If we replace 𝑛 by 𝑛+𝑐 in (4.35) with an arbitrary π‘βˆˆβ„, then we can choose 𝑐 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, 28–30, 33, 35, 36] and the references therein.

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

Theorem 5.1. Assume that ξ€½||π‘₯minβˆ’2||,||π‘₯βˆ’1||,||π‘₯0||ξ€Ύ>π‘₯2=√1+52.(5.1) Then π‘₯2<||π‘₯0||<||π‘₯1||<||π‘₯2||||π‘₯<β‹―<𝑛||<β‹―.(5.2)

Proof. From the hypothesis we have that |π‘₯βˆ’2|βˆ’1>π‘₯2βˆ’1, and so |π‘₯0|(|π‘₯βˆ’2|βˆ’1)>π‘₯2(π‘₯2βˆ’1)=1. Therefore, |π‘₯0||π‘₯βˆ’2|βˆ’|π‘₯0|>1, and so |π‘₯0||π‘₯βˆ’2|βˆ’1>|π‘₯0|. On the other hand, we have ||π‘₯1||=||π‘₯0π‘₯βˆ’2||>||π‘₯βˆ’10||||π‘₯βˆ’2||βˆ’1.(5.3) Combining the last two inequalities, we have that |π‘₯1|>|π‘₯0|>π‘₯2. Assume that we have proved π‘₯2<||π‘₯0||<||π‘₯1||<||π‘₯2||||π‘₯<β‹―<π‘˜||,(5.4) for some π‘˜βˆˆβ„•. We have |π‘₯π‘˜βˆ’2|βˆ’1>π‘₯2βˆ’1, which implies |π‘₯π‘˜|(|π‘₯π‘˜βˆ’2|βˆ’1)>π‘₯2(π‘₯2βˆ’1)=1, or equivalently |π‘₯π‘˜||π‘₯π‘˜βˆ’2|βˆ’1>|π‘₯π‘˜|. From this and (1.1), we get ||π‘₯π‘˜+1||=||π‘₯π‘˜π‘₯π‘˜βˆ’2||>||π‘₯βˆ’1π‘˜||||π‘₯π‘˜βˆ’2||||π‘₯βˆ’1>π‘˜||>π‘₯2,(5.5) finishing the inductive proof of the theorem.

Corollary 5.2. Assume that the initial values of a solution (π‘₯𝑛)βˆžπ‘›=βˆ’2 of (1.1) satisfy the condition ξ€½π‘₯minβˆ’2,π‘₯βˆ’1,π‘₯0ξ€Ύ>π‘₯2=√1+52.(5.6) 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 π‘₯0. We have a contradiction. The proof is complete.

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

Definition 5.3. Let (π‘₯𝑛)βˆžπ‘›=βˆ’2 be a solution of (1.1), and let π‘–βˆˆ{1,2}. Then we say that the solution has the eventual semicycle pattern π‘˜+,π‘™βˆ’ (or π‘˜βˆ’,𝑙+) if there exists π‘βˆˆβ„• such that, for π‘›βˆˆβ„•0, π‘₯𝑁+𝑛(π‘˜+𝑙)+1,…,π‘₯𝑁+𝑛(π‘˜+𝑙)+π‘˜β‰₯π‘₯𝑖 and π‘₯𝑁+𝑛(π‘˜+𝑙)+π‘˜+1,…,π‘₯𝑁+𝑛(π‘˜+𝑙)+π‘˜+𝑙<π‘₯𝑖 (or, resp., π‘₯𝑁+𝑛(π‘˜+𝑙)+1,…,π‘₯𝑁+𝑛(π‘˜+𝑙)+π‘˜<π‘₯𝑖 and π‘₯𝑁+𝑛(π‘˜+𝑙)+π‘˜+1,…,π‘₯𝑁+𝑛(π‘˜+𝑙)+π‘˜+𝑙β‰₯π‘₯𝑖).

Remark 5.4. Note that the eventual semicycle pattern can be extended to π‘˜Β±1,π‘˜βˆ“2,π‘˜Β±3,…,π‘˜βˆ“π‘€ for 𝑀>2.

Theorem 5.5. Assume that (π‘₯𝑛)βˆžπ‘›=βˆ’2 is a solution of (1.1) such that ξ€½||π‘₯minβˆ’2||,||π‘₯βˆ’1||,||π‘₯0||ξ€Ύ>π‘₯2=√1+52(5.7) and that at least one of π‘₯βˆ’2, π‘₯βˆ’1, π‘₯0 is negative. Then ||π‘₯𝑛||β‰₯π‘₯2,𝑛β‰₯βˆ’2,(5.8) and the solution is separated into seven unbounded eventually increasing subsequences such that the solution has the eventual semicycle pattern 1+,1βˆ’,2+,3βˆ’.(5.9)

Proof. Assume that π‘₯βˆ’2,π‘₯βˆ’1,π‘₯0<βˆ’π‘₯2. Then we have π‘₯1=π‘₯0π‘₯βˆ’2βˆ’1>π‘₯22βˆ’1=π‘₯2,π‘₯2=π‘₯1π‘₯βˆ’1βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,π‘₯3=π‘₯2π‘₯0βˆ’1>π‘₯22βˆ’1=π‘₯2,π‘₯4=π‘₯3π‘₯1βˆ’1>π‘₯22βˆ’1=π‘₯2,π‘₯5=π‘₯4π‘₯2βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,π‘₯6=π‘₯5π‘₯3βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,π‘₯7=π‘₯6π‘₯4βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2.(5.10) Hence |π‘₯𝑖|>π‘₯2, for βˆ’2≀𝑖≀7 and π‘₯5,π‘₯6,π‘₯7<βˆ’π‘₯2βˆ’2<βˆ’π‘₯2<0. An inductive argument shows that π‘₯7π‘˜+1=π‘₯7π‘˜π‘₯7π‘˜βˆ’2βˆ’1>π‘₯22βˆ’1=π‘₯2,π‘₯7π‘˜+2=π‘₯7π‘˜+1π‘₯7π‘˜βˆ’1βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,π‘₯7π‘˜+3=π‘₯7π‘˜+2π‘₯7π‘˜βˆ’1>π‘₯22βˆ’1=π‘₯2,π‘₯7π‘˜+4=π‘₯7π‘˜+3π‘₯7π‘˜+1βˆ’1>π‘₯22βˆ’1=π‘₯2,π‘₯7π‘˜+5=π‘₯7π‘˜+4π‘₯7π‘˜+2βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,π‘₯7π‘˜+6=π‘₯7π‘˜+5π‘₯7π‘˜+3βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,π‘₯7π‘˜+7=π‘₯7π‘˜+6π‘₯7π‘˜+4βˆ’1<βˆ’π‘₯22βˆ’1=βˆ’π‘₯2βˆ’2<βˆ’π‘₯2,(5.11) for each π‘˜βˆˆβ„•0, 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 (π‘₯7π‘˜+𝑖)βˆžπ‘˜=0 monotonically tend to βˆ’βˆž or +∞ with the aforementioned eventual semicycle pattern.

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