Long-Term Behavior of Solutions of the Difference Equation 𝑥_𝑛+1=𝑥_𝑛−1𝑥_𝑛−2−1

Abstract

We investigate the long-term behavior of solutions of the following difference equation: 𝑥_𝑛+1=𝑥_𝑛−1𝑥_𝑛−2−1, 𝑛∈ℕ₀, where the initial values 𝑥₋₂, 𝑥₋₁, and 𝑥₀ 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., [1–28] and the references therein). Standard properties which have been studied are boundedness [5, 9, 23–25], periodicity [2, 5, 9, 10, 27], asymptotic periodicity [3, 4, 8, 11–14, 16, 17, 19, 20, 23], and local and global stability [5, 9–11, 23–26], as well as existence of specific solutions such as monotone or nontrivial solutions [1, 6, 7, 13, 15, 18–22].

In this paper, we investigate the long-term behavior of solutions of the third-order difference equation𝑥_𝑛+1=𝑥_𝑛−1𝑥_𝑛−2−1,𝑛∈ℕ₀, (1.1) where the initial values 𝑥₋₂,𝑥₋₁,𝑥₀ are real numbers.

The difference equation (1.1) belongs to the class of equations of the form 𝑥_𝑛+1=𝑥_𝑛−𝑘𝑥_𝑛−𝑙−1,𝑛∈ℕ₀, (1.2) where 𝑘,𝑙∈ℕ₀, 𝑘<𝑙, and gcd(𝑘;𝑙)=1. The case 𝑘=0, 𝑙=1 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 𝑥 is an equilibrium of (1.1), then it satisfies the equation 𝑥²−𝑥−1=0. (2.1) Hence, (1.1) has exactly two equilibria, one positive and one negative, which we denote by 𝑥₁ and 𝑥₂, respectively:𝑥₁√=∶1+52,𝑥₂√=∶1−52 (2.2) (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 {𝑥_𝑛}^∞_𝑛=−2 is an eventually constant solution of (1.1), then 𝑥_𝑁=𝑥_𝑁+1=𝑥_𝑁+2=𝑥, for some 𝑁∈ℕ₀, where 𝑥 is an equilibrium point. In this case, (1.1) gives 𝑥_𝑁+2=𝑥_𝑁𝑥_𝑁−1−1, which implies 𝑥_𝑁−1=𝑥_𝑁+2+1𝑥_𝑁=𝑥+1𝑥=𝑥. (2.3) Repeating this procedure, we obtain 𝑥_𝑛=𝑥 for −2≤𝑛≤𝑁+2. 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 𝑥_𝑁=𝑥_𝑁+2𝑘 and 𝑥_𝑁+1=𝑥_𝑁+2𝑘+1, for every 𝑘∈ℕ₀, and some 𝑁≥−2, with 𝑥_𝑁≠𝑥_𝑁+1. Then, we have 𝑥_𝑁+4=𝑥_𝑁+2𝑥_𝑁+1−1=𝑥_𝑁𝑥_𝑁+1−1=𝑥_𝑁+3=𝑥_𝑁+1. (2.4) From this and since 𝑥_𝑁+4=𝑥_𝑁, 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 𝑥_𝑁=𝑥_𝑁+3𝑘,𝑥_𝑁+1=𝑥_𝑁+3𝑘+1,𝑥_𝑁+2=𝑥_𝑁+3𝑘+2,𝑘≥0, (2.5) for some 𝑁≥−2, we have 𝑥_𝑁+3=𝑥_𝑁+1𝑥_𝑁−1=𝑥_𝑁,𝑥_𝑁+4=𝑥_𝑁+2𝑥_𝑁+1−1=𝑥_𝑁+1,𝑥_𝑁+5=𝑥_𝑁+3𝑥_𝑁+2−1=𝑥_𝑁𝑥_𝑁+2−1=𝑥_𝑁+2. (2.6) If 𝑥_𝑁=0, 𝑥_𝑁+1=0, or 𝑥_𝑁+2=0, then from (2.6) we easily obtain contradictions in all these cases. Hence, we may assume that 𝑥_𝑁≠0, 𝑥_𝑁+1≠0, and 𝑥_𝑁+2≠0. Equalities in (2.6) also imply that 𝑥_𝑁+1=𝑥_𝑁+1𝑥_𝑁,𝑥_𝑁+2=𝑥_𝑁+1+1𝑥_𝑁+1,𝑥_𝑁=𝑥_𝑁+2+1𝑥_𝑁+2. (2.7) From (2.7), we get 𝑥_𝑁=𝑥_𝑁+2+1𝑥_𝑁+2=2𝑥_𝑁+1+1𝑥_𝑁+1=+13𝑥_𝑁+22𝑥_𝑁,+1 (2.8) which implies that 𝑥²_𝑁−𝑥_𝑁−1=0, that is, 𝑥_𝑁=(1±√5)/2. From this and (2.7) we obtain 𝑥_𝑁+1=𝑥_𝑁+2=(1±√5)/2, implying 𝑥_𝑛=(1±√5)/2,𝑛≥𝑁, 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 𝑥_𝑁=𝑥_𝑁+4𝑘,𝑥_𝑁+1=𝑥_𝑁+4𝑘+1,𝑥_𝑁+2=𝑥_𝑁+4𝑘+2,𝑥_𝑁+3=𝑥_𝑁+4𝑘+3,𝑘≥0, (2.9) for some 𝑁≥−2. Then we have 𝑥_𝑁+4=𝑥_𝑁+2𝑥_𝑁+1−1=𝑥_𝑁,𝑥_𝑁+5=𝑥_𝑁+3𝑥_𝑁+2−1=𝑥_𝑁+1,𝑥_𝑁+6=𝑥_𝑁+4𝑥_𝑁+3−1=𝑥_𝑁𝑥_𝑁+3−1=𝑥_𝑁+2,𝑥_𝑁+7=𝑥_𝑁+5𝑥_𝑁+4−1=𝑥_𝑁+1𝑥_𝑁−1=𝑥_𝑁+3. (2.10) If 𝑥_𝑁+𝑖=0 for some 𝑖∈{0,1,2,3}, then from (2.10) we easily obtain a contradiction. For example, if 𝑥_𝑁=0, then from (2.10) we get 𝑥_𝑁+2=−1=𝑥_𝑁+3. This implies 𝑥_𝑁+1=𝑥_𝑁+3𝑥_𝑁+2−1=0. From this and since 𝑥_𝑁+2𝑥_𝑁+1−1=𝑥_𝑁 we would get 𝑥_𝑁=−1, a contradiction. The other cases are proved analogously.
Hence we may assume that 𝑥_𝑁+𝑖≠0, 𝑖∈{0,1,2,3}. From (2.10) we have 𝑥_𝑁=𝑥_𝑁+3+1𝑥_𝑁+1=𝑥_𝑁+2+1/𝑥_𝑁+1𝑥_𝑁+1/𝑥_𝑁+2=𝑥_𝑁+2𝑥_𝑁+2+𝑥_𝑁+1𝑥_𝑁𝑥_𝑁=𝑥+1^𝑁+1𝑥_𝑁+1+1𝑥_𝑁𝑥²_𝑁+1, (2.11) that is, 𝑥_𝑁𝑥_𝑁+1²=𝑥_𝑁+1𝑥_𝑁+1+1. (2.12) From (2.10) and (2.12), we obtain 𝑥_𝑁+3+1²=𝑥_𝑁+1𝑥_𝑁+1+1. (2.13) Since the relations in (2.12) are cyclic, we also obtain that 𝑥_𝑁+1²=𝑥_𝑁+1+1𝑥_𝑁+2,𝑥+1_𝑁+1+1²=𝑥_𝑁+2+1𝑥_𝑁+3,𝑥+1_𝑁+2+1²=𝑥_𝑁+3+1𝑥_𝑁+1. (2.14) Equalities (2.13) and (2.14) imply that the expressions 𝑥_𝑁+𝑖+1,𝑖∈{0,1,2,3} 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 𝑥_𝑁+3+1²=𝑥_𝑁+1𝑥_𝑁+1+1=𝑥_𝑁+1𝑥_𝑁+3+1𝑥_𝑁+2=𝑥+1_𝑁+1𝑥_𝑁+3+1𝑥_𝑁+3+1𝑥_𝑁+1, (2.15) from which easily follows that 𝑥_𝑁+3=𝑥_𝑁. From this, (2.13), and (2.14), it follows that 𝑥_𝑁+𝑖=𝑥_𝑁, 𝑖∈{1,2,3}, 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)𝑥₋₂=𝑎, 𝑥₋₁=𝑏, 𝑎𝑏≠1, and 𝑥₀=(1+𝑎)/(𝑎𝑏−1), or(ii)𝑥₋₂=𝑥₋₁=−1, or (iii)𝑥₋₂=𝑎, 𝑥₋₁=0, and 𝑥₀=−𝑎−1.

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 {𝑥_𝑛}^∞_𝑛=−2 is of period five. Then, we can write terms of the solution of (1.1) as 𝑥₋₂𝑥=𝑎,₋₁𝑥=𝑏,₀𝑥=𝑐,₁𝑥=𝑑=𝑎𝑏−1,₂𝑥=𝑒=𝑏𝑐−1,₃()𝑥=𝑓=𝑐𝑑−1=𝑐𝑎𝑏−1−1=𝑎,₄()𝑥=𝑔=𝑎𝑏−1)(𝑏𝑐−1−1=𝑏,₅=ℎ=(𝑏𝑐−1)𝑓−1=(𝑏𝑐−1)𝑎−1=𝑐. (2.16)
Note that the expressions for 𝑓 and ℎ both yield the same condition, namely, 𝑎𝑏𝑐=𝑎+𝑐+1. The condition for 𝑔 gives 𝑎𝑏²𝑐−𝑎𝑏−𝑏𝑐=𝑏, so that 𝑏(𝑎𝑏𝑐−𝑎−𝑐−1)=0. Hence, either 𝑏=0 or 𝑎𝑏𝑐=𝑎+𝑐+1, which was a restriction coming from 𝑓 and ℎ. If 𝑏=0, then 𝑓=𝑎=−𝑐−1. The second restriction can be rewritten as 𝑐=(1+𝑎)/(𝑎𝑏−1), provided 𝑎𝑏≠1. If 𝑎𝑏=1, then 𝑓=𝑎=−1=𝑏.

Remark 2.6. If 𝑥_𝑘=0, where 𝑘≥−1, then 𝑥_𝑘 starts a 5-cycle (in particular, in view of the above theorem, we have a 5 cycle if the initial conditions are 𝑥₋₂=𝑎,𝑥₋₁=0,𝑥₀=𝑐, where 𝑎 and 𝑐 are any real numbers). If 𝑘=−1, then we have 𝑥₋₁𝑥=0,₀=𝑥₀,𝑥₁=𝑥₋₁𝑥₋₂𝑥−1=−1,₂=𝑥₀𝑥₋₁𝑥−1=−1,₃=𝑥₁𝑥₀−1=−𝑥₀𝑥−1,₄=𝑥₂𝑥₁𝑥−1=0,₅=𝑥₃𝑥₂−1=−−𝑥₀−1−1=𝑥₀,𝑥₆=𝑥₄𝑥₃−1=−1, (2.17) from which the statement follows in this case.
If 𝑘≥0, then by direct calculation we have 𝑥_𝑘𝑥=0,_𝑘+1=𝑥_𝑘−1𝑥_𝑘−2𝑥−1,_𝑘+2=𝑥_𝑘𝑥_𝑘−1𝑥−1=−1,_𝑘+3=𝑥_𝑘+1𝑥_𝑘𝑥−1=−1,_𝑘+4=𝑥_𝑘+2𝑥_𝑘+1−1=−𝑥_𝑘−1𝑥_𝑘−2−1−1=−𝑥_𝑘−1𝑥_𝑘−2,𝑥_𝑘+5=𝑥_𝑘+3𝑥_𝑘+2𝑥−1=0,_𝑘+6=𝑥_𝑘+4𝑥_𝑘+3−1=−−𝑥_𝑘−1𝑥_𝑘−2−1=𝑥_𝑘−1𝑥_𝑘−2𝑥−1,_𝑘+7=𝑥_𝑘+5𝑥_𝑘+4−1=−1, (2.18) 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 𝑎≠−1 and 𝑏∈ℝ⧵{−1,0} such that 𝑎𝑏≠1. For example, we can choose 𝑥₋₂=2, 𝑥₋₁=2,𝑥₀=1.

Remark 2.8. There exist solutions that are eventually of period five; for example, choose 𝑥₋₂=4, 𝑥₋₁=0.5, 𝑥₀=1, or 𝑥₋₂=0, 𝑥₋₁=4, 𝑥₀=−0.5.

3. Solutions in the Interval (−1,0)

Here, we study the solutions of (1.1) with initial values in the interval (−1,0) or for which there are three subsequent terms that are eventually in the interval.

The next result shows that the interval (−1,0) is an invariant interval for (1.1).

Theorem 3.1. If −1<𝑥₋₂,𝑥₋₁,𝑥₀<0, then −1<𝑥_𝑛<0 for all 𝑛≥−2.

Proof. If −1<𝑥₋₂,𝑥₋₁,𝑥₀<0, then −1<𝑥₁=𝑥₋₁𝑥₋₂−1<0. From (1.1) and by induction, we then have that −1<𝑥_𝑛<0 for all 𝑛≥−2.

Remark 3.2. There are solutions that eventually enter the interval (−1,0). One example of such a solution is one with the initial conditions 𝑥₋₂=1.3, 𝑥₋₁=1.5, and 𝑥₀=1.6.

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 (−1,0) to period-five solutions. For related results on the asymptotic periodicity of difference equations, see [3, 4, 8, 10–14, 16, 17, 19, 20, 23].

Note that if 𝑎,𝑏∈(−1,0), then 𝑎𝑏≠1, and we have that (1+𝑎)/(𝑎𝑏−1)∈(−1,0). Indeed, since 𝑎,𝑏∈(−1,0), then clearly (1+𝑎)/(𝑎𝑏−1)<0, and 𝑎𝑏<−𝑎 so that 1+𝑎<1−𝑎𝑏, from which it follows that (1+𝑎)/(𝑎𝑏−1)>−1. Hence, by Theorems 2.5 and 3.1 it follows that these solutions are periodic of period five belonging to the interval (−1,0).

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 𝑥_𝑛+5−𝑥_𝑛=𝑥_𝑛𝑥_𝑛+4−𝑥_𝑛−1,for𝑛≥−1. (3.1)

Proof. If 𝑛≥−1, we have 𝑥_𝑛+2=𝑥_𝑛𝑥_𝑛−1𝑥−1,_𝑛+3=𝑥_𝑛+1𝑥_𝑛𝑥−1,_𝑛+4=𝑥_𝑛+2𝑥_𝑛+1−1=𝑥_𝑛𝑥_𝑛−1−1𝑥_𝑛+1−1=𝑥_𝑛+1𝑥_𝑛𝑥_𝑛−1−𝑥_𝑛+1−1,𝑥_𝑛+5=𝑥_𝑛+3𝑥_𝑛+2−1=𝑥_𝑛+1𝑥_𝑛−1𝑥_𝑛𝑥_𝑛−1−1−1=𝑥_𝑛+1𝑥_𝑛²𝑥_𝑛−1−𝑥_𝑛+1𝑥_𝑛−𝑥_𝑛𝑥_𝑛−1=𝑥_𝑛𝑥_𝑛+1𝑥_𝑛𝑥_𝑛−1−𝑥_𝑛+1−𝑥_𝑛−1. (3.2)(3.3)
Using (3.3) and then (3.2), we get 𝑥_𝑛+5−𝑥_𝑛=𝑥_𝑛𝑥_𝑛+1𝑥_𝑛𝑥_𝑛−1−𝑥_𝑛+1−𝑥_𝑛−1−𝑥_𝑛=𝑥_𝑛𝑥_𝑛+1𝑥_𝑛𝑥_𝑛−1−𝑥_𝑛+1−𝑥_𝑛−1−1=𝑥_𝑛𝑥_𝑛+1𝑥_𝑛𝑥_𝑛−1−𝑥_𝑛+1−1−𝑥_𝑛−1=𝑥_𝑛𝑥_𝑛+4−𝑥_𝑛−1, (3.4) which is equality (3.1).

Theorem 3.4. Consider the difference equation (1.1) with initial conditions 𝑥₋₂,𝑥₋₁,𝑥₀∈(−1,0). Then, this solution converges to a period-five solution.

Proof. Consider five subsequences {𝑥_5𝑛}^∞_𝑛=0, {𝑥_5𝑛+1}^∞_𝑛=0, {𝑥_5𝑛+2}^∞_𝑛=0, {𝑥_5𝑛+3}^∞_𝑛=0, and {𝑥_5𝑛+4}^∞_𝑛=0. By Lemma 3.3, we know that our solution satisfies (3.1) for all 𝑛≥−1. Thus, we have 𝑥_𝑛+10−𝑥_𝑛+5=𝑥_𝑛+5𝑥_𝑛+9−𝑥_𝑛+4=𝑥_𝑛+5𝑥_𝑛+4𝑥_𝑛+8−𝑥_𝑛+3=𝑥_𝑛+5𝑥_𝑛+4𝑥_𝑛+3𝑥_𝑛+7−𝑥_𝑛+2=𝑥_𝑛+5𝑥_𝑛+4𝑥_𝑛+3𝑥_𝑛+2𝑥_𝑛+6−𝑥_𝑛+1=𝑥_𝑛+5𝑥_𝑛+4𝑥_𝑛+3𝑥_𝑛+2𝑥_𝑛+1𝑥_𝑛+5−𝑥_𝑛. (3.5) Therefore, 𝑥_𝑛+10−𝑥_𝑛+5=𝑥_𝑛+5𝑥_𝑛+4𝑥_𝑛+3𝑥_𝑛+2𝑥_𝑛+1𝑥_𝑛+5−𝑥_𝑛, (3.6) and, since the values of the sequence are in the interval (−1,0), we have, 𝑥_𝑛+10−𝑥_𝑛+5≤𝑥_𝑛+5−𝑥_𝑛. (3.7) Without loss of generality, we will consider the subsequence {𝑥_5𝑛}^∞_𝑛=0. Let 𝑎_𝑛=𝑥_5(𝑛+1)−𝑥_5𝑛,𝑛∈ℕ₀. (3.8) By inequality (3.7), we have 𝑎₀≥𝑎₁≥𝑎₂≥⋯. (3.9) Since the sequence {|𝑎_𝑛|}^∞_𝑛=0 is positive, nonincreasing, and bounded below by 0, it must converge to, say, 𝐴. We will prove that 𝐴=0. Assume to the contrary that 𝐴≠0. Observe that from (3.6) we have 𝑥_5(𝑛+1)−𝑥_5𝑛=^5𝑛_𝑘=1𝑥_𝑘𝑥₅−𝑥₀. (3.10) Since the sequence {|𝑎_𝑛|}^∞_𝑛=0 converges to a nonzero value, it follows that ∏^∞_𝑘=1|𝑥_𝑘| converges to a nonzero value. Hence, lim_𝑘→∞|𝑥_𝑘|=1. Therefore, 𝑥_𝑘 is close to −1 for 𝑘 large. Suppose 𝑘 is large. Then, 𝑥_𝑘+3=𝑥_𝑘+1𝑥_𝑘−1 is close to 0, which contradicts the fact that |𝑥_𝑘+3|→1 as 𝑘→∞. Hence, 𝐴=0, as desired.

Remark 3.5. As it has been already mentioned, there are solutions that have initial values outside the interval (−1,0) 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 𝑥₋₂=0,𝑥₋₁,𝑥₀∈(−1,0). Then, 𝑥₁=−1, all future terms are in the interval (−1,0), and thus this solution converges to a periodic solution with period five.

Proof. Clearly 𝑥₁=𝑥₋₁𝑥₋₂−1=−1. Since 0<−𝑥₀<1 and 0<−𝑥₋₁<1, we have 0<𝑥₀𝑥₋₁<1, and so −1<𝑥₂<0. Furthermore, we have 𝑥₃=𝑥₁𝑥₀−1=−𝑥₀−1∈(−1,0). Similarly, 𝑥₄=𝑥₂𝑥₁−1=−𝑥₂−1∈(−1,0). 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),  𝑥₁, is unstable.

Proof. The characteristic equation of the equilibrium 𝑥₁ is the following: 𝜆³−𝑥₁𝜆−𝑥₁=0. (4.1) Let 𝑃₁(𝑥)=𝑥³−𝑥₁𝑥−𝑥₁. Since 𝑃₁(1)=1−2𝑥₁=−√5<0 and lim_𝑥→+∞𝑃₁(𝑥)=+∞, it follows that there is a 𝜆₀>1 such that 𝑃₁(𝜆₀)=0, from which the result follows for the equilibrium 𝑥₁.

Remark 4.2. Consider the characteristic equation of the negative equilibrium of (1.1), 𝑥₂, 𝜆³−𝑥₂𝜆−𝑥₂=0 (4.2) and the function 𝑃₂(𝑥)=𝑥³−𝑥₂𝑥−𝑥₂. Observe the following: (i)𝑃₂(𝑥₂)=0. (ii)𝑃^′₂(𝑥)=3𝑥²−𝑥₂>0 for all 𝑥∈ℝ. Hence, we have one negative eigenvalue, 𝜆₁=𝑥₂∈(−1,0) and a complex conjugate pair of eigenvalues, 𝜆₂ and 𝜆₂, such that 𝜆₁𝜆₂²=𝑥₂. (4.3) It follows that |𝜆₂|=|𝜆₂|=1, and so 𝑥₂ 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), 𝑥₂.

5. More on Invariant Intervals

In this section, we discuss invariant intervals regarding the subsequences {𝑥_5𝑛}^∞_𝑛=0, {𝑥_5𝑛+1}^∞_𝑛=0, {𝑥_5𝑛+2}^∞_𝑛=0, {𝑥_5𝑛+3}^∞_𝑛=0, and {𝑥_5𝑛+4}^∞_𝑛=0 of a solution {𝑥_𝑛}^∞_𝑛=−2 of (1.1).

Theorem 5.1. Assume that a solution of (1.1) has initial conditions 𝑥₋₂=0,𝑥₋₁∈(0,1) and 𝑥₀∈(−1,0). Then, 𝑥_5𝑛+2,𝑥_5𝑛+6∈(−2,−1), 𝑥_5𝑛+3,𝑥_5𝑛+5∈(−1,0), and 𝑥_5𝑛+4∈(0,1) for 𝑛∈ℕ₀ and also 𝑥₁=−1.

Proof. Since 𝑥₋₂=0, we have 𝑥₁=𝑥₋₁𝑥₋₂−1=−1. Now, we prove the above statement for 𝑛=0. From 0<𝑥₋₁<1 and −1<𝑥₀<0, we have −1<𝑥₀𝑥₋₁<0. From this and since 𝑥₂=𝑥₀𝑥₋₁−1, we get 𝑥₂∈(−2,−1).
Furthermore, we have that 𝑥₃=𝑥₁𝑥₀−1=−𝑥₀(),𝑥−1∈−1,0₄=𝑥₂𝑥₁−1=−𝑥₂−1∈(0,1). (5.1) Since 0<−𝑥₃<1 and 1<−𝑥₂<2, it follows that 𝑥₅=𝑥₃𝑥₂−1∈(−1,1). Now, we verify that 𝑥₅ must be in a more restrictive interval (−1,0). We prove this by contradiction. Assume that 𝑥₅∈[0,1). Since 0≤𝑥₅=𝑥₃𝑥₂−1<1 with 𝑥₃=𝑥₁𝑥₀−1=−𝑥₀−1 and 𝑥₂=𝑥₀𝑥₋₁−1, we must have that 0≤−𝑥₀−1𝑥₀𝑥₋₁−1−1<1⟺0≤𝑥₀1−𝑥₋₁−𝑥₀𝑥₋₁<1, (5.2) which is equivalent to 0≤𝑥₀−𝑥₂−𝑥₋₁<1. (5.3) On the other hand, since 1<−𝑥₂<2 and −1<−𝑥₋₁<0, we have 0<−𝑥₂−𝑥₋₁<2. However, since 𝑥₀<0, it follows that 𝑥₀(−𝑥₂−𝑥₋₁)<0, which is a contradiction. Hence, 𝑥₅∈(−1,0), as claimed.
Next, since 0<𝑥₄<1 and 0<−𝑥₃<1, we have −1<𝑥₄𝑥₃<0, and so, −2<𝑥₆<−1, finishing the proof for the case 𝑛=0.
Now assume that 𝑥_5𝑛+2,𝑥_5𝑛+6∈(−2,−1), 𝑥_5𝑛+3,𝑥_5𝑛+5∈(−1,0), and 𝑥_5𝑛+4∈(0,1) for 0≤𝑛≤𝑘.
Since 𝑥_5𝑘+4∈(0,1) and 𝑥_5𝑘+5∈(−1,0), we have 𝑥_5(𝑘+1)+2=𝑥_5𝑘+5𝑥_5𝑘+4−1∈(−2,−1). (5.4)
Since 𝑥_5𝑘+6∈(−2,−1) and 𝑥_5𝑘+5∈(−1,0), we have 𝑥_5(𝑘+1)+3=𝑥_5𝑘+6𝑥_5𝑘+5−1∈(−1,1). (5.5) Assume that 𝑥_5(𝑘+1)+3∈[0,1). Then, we have 0≤𝑥_5(𝑘+1)+3=𝑥_5𝑘+6𝑥_5𝑘+5−1=𝑥_5𝑘+4𝑥_5𝑘+3−1𝑥_5𝑘+3𝑥_5𝑘+2−1−1<1 (5.6) which is equivalent to 0≤𝑥_5𝑘+3𝑥_5𝑘+4𝑥_5𝑘+3𝑥_5𝑘+2−𝑥_5𝑘+4−𝑥_5𝑘+2<1. (5.7) Now note that 𝑥_5𝑘+4𝑥_5𝑘+3𝑥_5𝑘+2>0 and −𝑥_5𝑘+4−𝑥_5𝑘+2>0 since −1<−𝑥_5𝑘+4<0 and 1<−𝑥_5𝑘+2<2. Hence, 𝑥_5𝑘+4𝑥_5𝑘+3𝑥_5𝑘+2−𝑥_5𝑘+4−𝑥_5𝑘+2>0 which along with 𝑥_5𝑘+3<0 implies that 𝑥_5𝑘+3(𝑥_5𝑘+4𝑥_5𝑘+3𝑥_5𝑘+2−𝑥_5𝑘+4−𝑥_5𝑘+2)<0, which is a contradiction. Hence, 𝑥_5(𝑘+1)+3∈(−1,0). (5.8)
Since −2<𝑥_5(𝑘+1)+2,𝑥_5𝑘+6<−1, we have 𝑥_5(𝑘+1)+4=𝑥_5(𝑘+1)+2𝑥_5𝑘+6−1∈(0,3). (5.9) Assume that 𝑥_5(𝑘+1)+4∈[1,3).
Since by Lemma 3.3𝑥_5(𝑘+1)+4−𝑥_5𝑘+4=𝑥_5𝑘+4𝑥_5(𝑘+1)+3−𝑥_5𝑘+3 (5.10) and 0<𝑥_5𝑘+4<1≤𝑥_5(𝑘+1)+4, we have 𝑥_5(𝑘+1)+4−𝑥_5𝑘+4>0. Therefore, 𝑥_5𝑘+4(𝑥_5(𝑘+1)+3−𝑥_5𝑘+3)>0. From this and since 𝑥_5𝑘+4>0, we have that 𝑥_5(𝑘+1)+3−𝑥_5𝑘+3>0. We then have 𝑥_5(𝑘+1)+3−𝑥_5𝑘+3=𝑥_5𝑘+3𝑥_5(𝑘+1)+2−𝑥_5𝑘+2>0. (5.11) Since 𝑥_5𝑘+3<0, we have 𝑥_5(𝑘+1)+2−𝑥_5𝑘+2<0. Furthermore we have 𝑥_5(𝑘+1)+2−𝑥_5𝑘+2=𝑥_5𝑘+2𝑥_5(𝑘+1)+1−𝑥_5𝑘+1<0. (5.12) Since 𝑥_5𝑘+2<0, we have 𝑥_5(𝑘+1)+1−𝑥_5𝑘+1>0.
By Lemma 3.3, for each 1≤𝑛≤𝑘, we have 𝑥_5(𝑛+1)+1−𝑥_5𝑛+1=𝑥_5𝑛+1𝑥_5𝑛𝑥_5𝑛−1𝑥_5𝑛−2𝑥_5𝑛−3𝑥_5𝑛+1−𝑥_{5(𝑛−1)+1}. (5.13) By the inductive hypothesis, we have that 𝑥_5𝑛+1𝑥_5𝑛𝑥_5𝑛−1𝑥_5𝑛−2𝑥_5𝑛−3>0 so that the sign of the difference is the same as the sign of 𝑥_5(𝑘+1)+1−𝑥_5𝑘+1, that is, 𝑥_5(𝑛+1)+1−𝑥_5𝑛+1>0,0≤𝑛≤𝑘. (5.14) On the other hand, 𝑥₆−𝑥₁=𝑥₆+1<0, which is a contradiction. Hence, 𝑥_5(𝑘+1)+4∈(0,1). (5.15)
Since −2<𝑥_5(𝑘+1)+2<−1 and 𝑥_5(𝑘+1)+3∈(−1,0), we have 𝑥_5(𝑘+1)+5=𝑥_5(𝑘+1)+3𝑥_5(𝑘+1)+2−1∈(−1,1). (5.16) Assume that 𝑥_5(𝑘+1)+5∈[0,1). Then, we have 0≤𝑥_5𝑘+6𝑥_5𝑘+5−1𝑥_5𝑘+5𝑥_5𝑘+4−1−1<1 (5.17) which is equivalent to 0≤𝑥_5𝑘+5𝑥_5𝑘+6𝑥_5𝑘+5𝑥_5𝑘+4−𝑥_5𝑘+6−𝑥_5𝑘+4<1. (5.18) On the other hand, since 𝑥_5𝑘+6𝑥_5𝑘+5𝑥_5𝑘+4>0 and since from 1<−𝑥_5𝑘+6<2 and −1<−𝑥_5𝑘+4<0 we obtain −𝑥_5𝑘+6−𝑥_5𝑘+4>0, it follows that 𝑥_5𝑘+6𝑥_5𝑘+5𝑥_5𝑘+4−𝑥_5𝑘+6−𝑥_5𝑘+4>0. (5.19) This fact along with 𝑥_5𝑘+5<0 implies 𝑥_5𝑘+5𝑥_5𝑘+6𝑥_5𝑘+5𝑥_5𝑘+4−𝑥_5𝑘+6−𝑥_5𝑘+4<0 (5.20) which is a contradiction. Hence, 𝑥_5(𝑘+1)+5∈(−1,0). (5.21)
Finally, since 0<𝑥_5(𝑘+1)+4<1 and −1<𝑥_5(𝑘+1)+3<0, we obtain 𝑥_5(𝑘+1)+6=𝑥_5(𝑘+1)+4𝑥_5(𝑘+1)+3−1∈(−2,−1). (5.22) 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 𝑥₋₂=0,𝑥₋₁∈(0,1), and 𝑥₀∈(−1,0) converges to a period-five solution.

Proof. By Theorem 5.1, we have that 𝑥₋₂=0, 𝑥₋₁∈(0,1), 𝑥₀∈(−1,0), 𝑥₁=−1, 𝑥_5𝑛+2,𝑥_5𝑛+6∈(−2,−1), 𝑥_5𝑛+3,𝑥_5𝑛+5∈(−1,0), and 𝑥_5𝑛+4∈(0,1) for 𝑛∈ℕ₀.
We prove by induction that all the subsequences {𝑥_5𝑛+3}^∞_𝑛=−1, {𝑥_5𝑛+4}^∞_𝑛=−1, {𝑥_5𝑛+5}^∞_𝑛=−1, {𝑥_5𝑛+6}^∞_𝑛=−1, and {𝑥_5𝑛+7}^∞_𝑛=−1 are monotone.
Assume 𝑛=−1. Then, by Lemma 3.3 and above comments, we have 𝑥₃−𝑥₋₂=𝑥₃𝑥<0,₄−𝑥₋₁=𝑥₋₁𝑥₃−𝑥₋₂=𝑥₋₁𝑥₃𝑥<0,₅−𝑥₀=𝑥₀𝑥₄−𝑥₋₁𝑥>0,₆−𝑥₁=𝑥₁𝑥₅−𝑥₀𝑥<0,₇−𝑥₂=𝑥₂𝑥₆−𝑥₁>0. (5.23)
Assume that we have proved −1<𝑥_5𝑛+3<𝑥_{5(𝑛−1)+3}<0=𝑥₋₂,0<𝑥_5𝑛+4<𝑥_{5(𝑛−1)+4}<1,0>𝑥_5𝑛+5>𝑥_{5(𝑛−1)+5}>−1,−2<𝑥_5𝑛+6<𝑥_{5(𝑛−1)+6}<−1,−1>𝑥_5𝑛+7>𝑥_{5(𝑛−1)+7}>−2, (5.24) for 0≤𝑛≤𝑘.
Since 𝑥_{5(𝑘+1)+𝑗}−𝑥_5𝑘+𝑗=𝑥_5𝑘+𝑗𝑥_{5𝑘+𝑗−1}𝑥_{5𝑘+𝑗−2}𝑥_{5𝑘+𝑗−3}𝑥_{5𝑘+𝑗−4}𝑥_5𝑘+𝑗−𝑥_{5(𝑘−1)+𝑗}, (5.25) for each 𝑘∈ℕ₀ and 𝑗=3,4,5,6,7, and since by Theorem 5.1, we obtain 𝑥_5𝑘+𝑗𝑥_{5𝑘+𝑗−1}𝑥_{5𝑘+𝑗−2}𝑥_5k+𝑗−3𝑥_{5𝑘+𝑗−4}>0, (5.26)𝑘∈ℕ₀, 𝑗=3,4,5,6,7, we have that the differences 𝑥_{5(𝑘+1)+𝑗}−𝑥_5𝑘+𝑗, 𝑘∈ℕ₀ have the same sign for each 𝑗∈{3,4,5,6,7}. From this and by Theorem 5.1, the claim follows.
Since all the subsequences {𝑥_5𝑛+3}^∞_𝑛=−1, {𝑥_5𝑛+4}^∞_𝑛=−1, {𝑥_5𝑛+5}^∞_𝑛=−1, {𝑥_5𝑛+6}^∞_𝑛=−1 and {𝑥_5𝑛+7}^∞_𝑛=−1 are monotone and bounded, they are convergent, and consequently the solution {𝑥_𝑛}^∞_𝑛=−2 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 𝑥₋₂,𝑥₋₁,𝑥₀>𝑥₁ or 𝑥₋₂,𝑥₋₁,𝑥₀<−1, 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 𝑥₋₂,𝑥₋₁,𝑥₀>𝑥₁=(1+√5)/2, then the following statements hold true:(a)𝑥₋₁<𝑥₁<x₃<⋯ and 𝑥₀<𝑥₂<𝑥₄<⋯;(b)the solution tends to +∞.

Proof. (a) Since 𝑥₋₁>(1+√5)/2, we have 1/𝑥₋₁<2/(1+√5)=(√5−1)/2. Thus, 11+𝑥₋₁√<1+5−12=1+√52<𝑥₋₂. (6.1) Therefore, 𝑥₋₂>1+1/𝑥₋₁. Thus, 𝑥₋₁𝑥₋₂>𝑥₋₁+1. Rewriting, 𝑥₋₁𝑥₋₂−1>𝑥₋₁. Hence, 𝑥₁>𝑥₋₁. One can follow the same steps to prove that 𝑥₂>𝑥₀, 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 𝑥_𝑛−2=1+𝑥_𝑛+1𝑥_𝑛−1,𝑛∈ℕ₀, (6.2) it would follow that both subsequences {𝑥_2𝑛}^∞_𝑛=0 and {𝑥_2𝑛−1}^∞_𝑛=0 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 𝑥₋₁ and 𝑥₀. This is a contradiction. Hence, the proof is complete.

Theorem 6.2. Assume that 𝑎,𝑏>0 and 𝑎𝑏>1. Then, each solution with the initial conditions 𝑥₋₂=𝑎, 𝑥₋₁=𝑏, and 𝑥₀>(1+𝑎)/(𝑎𝑏−1) tends to plus infinity.

Proof. We have 𝑥₁=𝑎𝑏−1>0, 𝑥₂=𝑥₀𝑥₋₁−1>(𝑏+1)/(𝑎𝑏−1)>0, and 𝑥₃=𝑥₁𝑥₀−1>(𝑎𝑏−1)1+𝑎𝑎𝑏−1−1=𝑎=𝑥₋₂>0. (6.3) By Lemma 3.3, we have 𝑥_𝑛−𝑥_𝑛−5=𝑥₃−𝑥₋₂^𝑛−5_𝑗=−1𝑥_𝑗,𝑛≥4. (6.4) Using (6.4) and the fact 𝑥₃−𝑥₋₂>0, by induction it easily follows that 𝑥_5𝑘+𝑗>𝑥_{5(𝑘−1)+𝑗}>0,𝑘∈ℕ₀,𝑗=3,4,5,6,7, (6.5) that is, the subsequences {𝑥_5𝑘+3}^∞_𝑘=−1, {𝑥_5𝑘+4}^∞_𝑘=−1, {𝑥_5𝑘+5}^∞_𝑘=−1, {𝑥_5𝑘+6}^∞_𝑘=−1 and {𝑥_5𝑘+7}^∞_𝑘=−1 are increasing.
We also have 𝑥_5𝑘+3−𝑥_{5(𝑘−1)+3}=𝑥₃−𝑥₋₂^𝑘_𝑗=1𝑥_{5(𝑗−1)+3}𝑥_{5(𝑗−1)+2}𝑥_{5(𝑗−1)+1}𝑥_5(𝑗−1)𝑥_{5(𝑗−1)−1}≥𝑥₃−𝑥₋₂𝑥₃𝑥₂𝑥₁𝑥₀𝑥₋₁^𝑘≥𝑥₃−𝑥₋₂𝑎𝑏(𝑎+1)(𝑏+1)𝑎𝑏−1^𝑘. (6.6) Note that ()𝑞∶=𝑎𝑏𝑎+1)(𝑏+1𝑎𝑏−1>1. (6.7) Using this fact and (6.6) we obtain 𝑥_5𝑘+3>𝑥_{5(𝑘−1)+3}+𝑥₃−𝑥₋₂𝑞^𝑘⟹𝑥_5𝑘+3>𝑥₋₂+𝑥₃−𝑥₋₂^𝑘_𝑗=0𝑞^𝑗, (6.8) from which it follows that 𝑥_5𝑘+3→∞ as 𝑘→∞.
From (6.4), the monotonicity of those five subsequences, and (6.8), we get 𝑥_5𝑘+3+𝑖−𝑥_{5(𝑘−1)+3+𝑖}=𝑥_{5(𝑘−1)+3+𝑖}⋯𝑥_{5(𝑘−1)+3+1}𝑥_5𝑘+3−𝑥_{5(𝑘−1)+3}>𝑥_𝑖−2⋯𝑥₋₁𝑥_5𝑘+3−𝑥_{5(𝑘−1)+3}>𝑥_𝑖−2⋯𝑥₋₁𝑥₃−𝑥₋₂𝑞^𝑘, (6.9) for 𝑖=1,2,3,4, from which it follows that 𝑥_5𝑘+3+𝑖→∞ as 𝑘→∞ for each 𝑖∈{1,2,3,4}, 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 𝑥₋₂,𝑥₋₁,𝑥₀∈(1,𝑥₁)

Here, we consider the case 𝑥₋₂,𝑥₋₁,𝑥₀∈(1,𝑥₁). The next theorem shows that there is a large class of eventually nondecreasing solutions of (1.1) converging to 𝑥₁. For some results of this type, see, for example, [1, 6, 7, 10, 13, 15, 18–22] and the related references therein.

Theorem 7.1. Assume 𝑥₋₂,𝑥₋₁,𝑥₀∈(1,𝑥₁) and 𝑥₋₂,𝑥₋₁≤𝑥₀≤𝑥₋₁𝑥₋₂−1. Then, every solution with such initial values is eventually nondecreasing and converges to 𝑥₁.

Proof. Multiplying the assumption 𝑥₋₂≤𝑥₀, by 𝑥₋₁ we obtain 𝑥₋₁𝑥₋₂≤𝑥₋₁𝑥₀. From this and since 1<𝑥₋₂,𝑥₋₁,𝑥₀<𝑥₁, we obtain 0<𝑥₋₁𝑥₋₂−1≤𝑥₋₁𝑥₀−1<𝑥²₁−1=𝑥₁, that is, 0<𝑥₁≤𝑥₂<𝑥₁. Hence, 1<𝑥₋₂,𝑥₋₁≤𝑥₀≤𝑥₁≤𝑥₂<𝑥₁. Now assume 1<𝑥₋₂,𝑥₋₁≤𝑥₀≤𝑥₁≤𝑥₂≤⋯≤𝑥_𝑛<𝑥₁, (7.1) for some 𝑛≥2. Multiplying the inequality 𝑥_𝑛−3≤𝑥_𝑛−1 by 𝑥_𝑛−2 and using (7.1), it follows that 𝑥_𝑛−3𝑥_𝑛−2−1≤𝑥_𝑛−1𝑥_𝑛−2−1<𝑥²₁−1=𝑥₁, (7.2) that is, 𝑥_𝑛≤𝑥_𝑛+1<𝑥₁. Hence, by induction we have proved that the sequence {𝑥_𝑛}^∞_𝑛=−1 is nondecreasing and bounded above by 𝑥₁, from which the result easily follows.

Theorem 7.2. Assume 𝑥₋₂,𝑥₋₁,𝑥₀∈(1,𝑥₁), 𝑥₀≤max{𝑥₋₁,𝑥₋₂}, 𝑥₋₁𝑥₋₂−1≤min{𝑥₋₂,𝑥₋₁,𝑥₀}, and 𝑥₀𝑥₋₁−1≤𝑥₀. Then, for every solution with such initial values there is an 𝑁≥2, such that 𝑥₂≥𝑥₃≥⋯≥𝑥_𝑁≥0>𝑥_𝑁+1. (7.3)

Proof. According to the assumptions we have 0=1⋅1−1<𝑥₁=𝑥₋₁𝑥₋₂−1≤𝑥₋₁<𝑥₁,0=1⋅1−1<𝑥₂=𝑥₀𝑥₋₁−1≤𝑥₀<𝑥₁. (7.4)(7.5)
If 𝑥₀≤𝑥₋₂, then by multiplying by 𝑥₋₁, we obtain 𝑥₂=𝑥₀𝑥₋₁−1≤𝑥₋₁𝑥₋₂−1=𝑥₁. (7.6)
If 𝑥₀≤𝑥₋₁, then since 𝑥₁=𝑥₋₁𝑥₋₂−1≤𝑥₋₂, we obtain 𝑥₃=𝑥₁𝑥₀−1≤𝑥₋₁𝑥₋₂−1=𝑥₁. (7.7)
Multiplying the inequality 𝑥₁≤𝑥₋₁ (see (7.4)) by 𝑥₀, we obtain 𝑥₃≤𝑥₂, (7.8) which, together with inequality (7.6), implies again 𝑥₃≤𝑥₁.
Note that now we cannot guarantee the positivity of 𝑥₃. Similarly, from the inequality 𝑥₂≤𝑥₀ (see (7.5)), we obtain 𝑥₄≤𝑥₃, (7.9) and from the inequality 𝑥₃≤𝑥₁, we obtain 𝑥₅≤𝑥₄. (7.10)
Now assume that 0<𝑥_𝑛≤𝑥_𝑛−1≤⋯𝑥₃≤𝑥₂<𝑥₁ (7.11) and 𝑥_𝑛+1>0. Then, by multiplying the inequality 𝑥_𝑛−1≤𝑥_𝑛−3 by 𝑥_𝑛−2 and subtracting 1, we obtain 𝑥_𝑛+1≤𝑥_𝑛. Hence, we proved by induction that (7.11) holds as far as 𝑥_𝑛 is positive.
If 𝑥_𝑛>0 for all 𝑛≥2, the sequence {𝑥_𝑛}^∞_𝑛=−2 is convergent and its limit is nonnegative. However, this is not possible since the only nonnegative equilibrium of (1.1) is 𝑥₁. From this, the result follows.