Solutions of the Difference Equation 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−1−1

Abstract

Our goal in this paper is to investigate the long-term behavior of solutions of the following difference equation: 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−1−1, 𝑛=0,1,2,…, where the initial conditions 𝑥₋₁ and 𝑥₀ 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 equation𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−1−1,𝑛=0,1,2,…,(1.1) where the initial conditions 𝑥₋₁ and 𝑥₀ 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 form𝑥_𝑛+1=𝑥_𝑛−𝑘𝑥_{𝑛−𝑘−1}−1,for𝑛=0,1,…,(1.2) and a particular choice of 𝑘, where 𝑘=0,1,….

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 𝑥²−𝑥−1=0, we find that (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.(2.1)

Note that there are no solutions which are eventually constant. Indeed, if 𝑥_𝑁=𝑥_𝑁+1=𝑥 for some 𝑁≥0, then from 𝑥_𝑁+1=𝑥_𝑁𝑥_𝑁−1−1, it follows that 𝑥_𝑁−1=𝑥_𝑁+1+1𝑥_𝑁=𝑥+1𝑥=𝑥.(2.2) Repeating this procedure, we obtain 𝑥_𝑛=𝑥, for −1≤𝑛≤𝑁+1, as claimed.

Also note that there are no solutions that are eventually periodic with minimal period two. Indeed, if 𝑥_𝑁=𝑥_𝑁+2𝑘 and 𝑥_𝑁+1=𝑥_𝑁+2𝑘+1, 𝑘≥0, then we have 𝑥_𝑁+3=𝑥_𝑁+2𝑥_𝑁+1−1=𝑥_𝑁𝑥_𝑁+1−1=𝑥_𝑁+2=𝑥_𝑁,(2.3) so that 𝑥_𝑛=𝑥_𝑁, 𝑛≥𝑁, 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 𝑥₋₁=−1, 𝑥₀=−1; 𝑥₋₁=−1, 𝑥₀=0; and 𝑥₋₁=0, 𝑥₀=−1.

Proof. We can write terms of a period-three solution of (1.1) as 𝑥₋₁=𝑎,𝑥₀=𝑏,𝑥₁𝑥=𝑎𝑏−1,₂=𝑏(𝑎𝑏−1)−1=𝑎⇔𝑎𝑏²−𝑏−1=𝑎⇔𝑎𝑏²𝑥=𝑎+𝑏+1,₃=𝑎(𝑎𝑏−1)−1=𝑏⇔𝑎²𝑏−𝑎−1=𝑏⇔𝑎²𝑏=𝑎+𝑏+1.(2.4) Therefore, this is indeed a solution of period three if the system below is satisfied. 𝑎𝑏²𝑎=𝑎+𝑏+1,₂𝑏=𝑎+𝑏+1.(2.5) Hence we see that 𝑎𝑏²=𝑎²𝑏. Thus, this is true if 𝑎=0, or if 𝑏=0, or if 𝑎=𝑏 when 𝑎≠0 and 𝑏≠0. Now it suffices to consider the following three cases.Case 1. Suppose that 𝑎=0. Then 𝑎²𝑏=𝑎+𝑏+1 implies that 0=𝑏+1 and hence gives us 𝑏=−1. Therefore, a period-three solution exists if 𝑎=0 and 𝑏=−1.Case 2. Suppose that 𝑏=0. Then 𝑎²𝑏=𝑎+𝑏+1 implies that 0=𝑎+1 and hence gives us 𝑎=−1. Therefore, a period-three solution exists if 𝑎=−1 and 𝑏=0.Case 3. Suppose that 𝑎=𝑏 such that 𝑎≠0 and 𝑏≠0. Then 𝑎²𝑏=𝑎+𝑏+1 implies that 𝑎³=2𝑎+1, which gives us 𝑎=−1. Therefore, a period-three solution exists if 𝑎=−1 and 𝑏=−1. (Note that 𝑎=(1±√5)/2 are also solutions of 𝑎³=2𝑎+1, which are the two equilibria of (1.1).)Hence, there exist exactly three periodic solutions with minimal period three of (1.1) given by 𝑥₋₁=−1,𝑥₀=−1,𝑥₁=0,…;𝑥₋₁=−1,𝑥₀=0,𝑥₁𝑥=−1,…;₋₁=0,𝑥₀=−1,𝑥₁=−1,…,(2.6) as claimed.

In the sequel, we will refer to any one of these three periodic solutions of (1.1) as…,0,−1,−1,0,−1,−1,….(2.7)

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 (𝑥₋₁,𝑥₀,…,𝑥_𝑁,𝑥_𝑁+1,0,−1,−1,0,−1,−1,0,−1,−1,…),(2.8) Where 𝑁≥−1, 𝑥_𝑁+1=𝑎∈𝑅⧵{0}, 𝑥_𝑁=1/𝑎, and, if 𝑁≠−1, 𝑥_𝑛−1=(𝑥_𝑛+1+1)/𝑥_𝑛 for 0≤𝑛≤𝑁.

Proof. If {𝑥_𝑛}^∞_𝑛=−1 is an eventually periodic solution with minimal period three, then by Theorem 2.1 there is an 𝑁≥−1 such that 𝑥_𝑁+2=0 and 𝑥_𝑁+3=−1. Hence 0=𝑥_𝑁+2=𝑥_𝑁+1𝑥_𝑁−1 and consequently 𝑥_𝑁≠0≠𝑥_𝑁+1 and 𝑥_𝑁+1=1/𝑥_𝑁. Let 𝑥_𝑁+1=𝑎 (which implies that 𝑥_𝑁=1/𝑎). From (1.1), if 𝑁≠−1, we get 𝑥_𝑛−1=(𝑥_𝑛+1+1)/𝑥_𝑛, for 0≤𝑛≤𝑁, as desired.

Remark 2.3. If, in Theorem 2.2, 𝑎=−1, 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 {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1) that is neither periodic nor eventually periodic with minimal period three. Then {𝑥_𝑛}^∞_𝑛=−1 does not converge to the minimal period-three solution …,0,−1,−1,0,−1,−1,….(2.9)

Proof. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1) that is neither periodic nor eventually periodic with minimal period three. For the sake of contradiction, assume that {𝑥_𝑛}^∞_𝑛=−1 converges to the minimal period-three solution …,0,−1,−1,0,−1,−1,….(2.10) Then we can choose 𝑁≥0 such that there exist −12<𝛿₁,𝛿₂,𝛿₃1,…<2,−1<𝜖₀,𝜖₁,𝜖₂,𝜖₃,…<1,(2.11) with 𝑥_𝑁−2=−1+𝜖₀,𝑥_𝑁−1=𝛿₁,𝑥_𝑁=−1+𝜖₁,𝑥_𝑁+1=−1+𝜖₂,𝑥_𝑁+2=𝛿₂,𝑥_𝑁+3=−1+𝜖₃,𝑥_𝑁+4=−1+𝜖₄,𝑥_𝑁+5=𝛿₃,𝑥_𝑁+6=−1+𝜖₅,…,(2.12) that is, for 𝑛=0,1,…, 𝑥_{𝑁+(3𝑛−2)}=−1+𝜖_2𝑛,𝑥_{𝑁+(3𝑛−1)}=𝛿_𝑛+1,𝑥_𝑁+3𝑛=−1+𝜖_2𝑛+1.(2.13)
Now, first observe that 𝑥_𝑁−2=−1+𝜖₀,𝑥_𝑁−1=𝛿₁,𝑥_𝑁=−1+𝜖₁=−1+𝛿₁(−1+𝜖₀).(2.14) We claim that 𝛿₁≠0. Otherwise, 𝑥_𝑁−1=𝛿₁=0. Then, by (1.1), 𝑥_𝑁=𝑥_𝑁+1=−1, which gives us a solution that is eventually periodic with minimal period three, a contradiction. This leaves us with 𝛿₁>0 or 𝛿₁<0. If 𝛿₁>0, then since 𝑥_𝑁−2=−1+𝜖₀<0 by (2.11), we have 𝑥_𝑁<−1 and so 𝜖₁<0. If 𝛿₁<0, then since 𝑥_𝑁−2=−1+𝜖₀<0 and 𝑥_𝑁=−1+𝜖₁<0 by (2.11), we have 𝑥_𝑁∈(−1,0) and so 𝜖₁>0. Therefore, we need only consider the following two cases.
Case 1. If 𝛿₁>0, 𝜖₁<0, then we obtain the following inequalities from (2.11) and (2.12): 𝑥_𝑁−1=𝛿₁𝑥>0,_𝑁=−1+𝜖₁𝑥<−1,_𝑁+1=−1+𝜖₂=−1+𝛿₁(−1+𝜖₁𝑥)<−1,_𝑁+2=𝛿₂=(−1+𝜖₁)[−1+𝛿₁(−1+𝜖₁𝑥)]−1>0,_𝑁+3=−1+𝜖₃=𝛿₂[−1+𝛿₁(−1+𝜖₁)]−1<−1.(2.15)
We next compute the following: 𝑥_𝑁+2=𝛿₂=(−1+𝜖₁)[−1+𝛿₁(−1+𝜖₁)]−1>𝛿₁⇔(−1+𝜖₁)[−1+𝛿₁(−1+𝜖₁)]>𝛿₁+1⇔1+𝛿₁−𝛿₁𝜖₁−𝜖₁−𝛿₁𝜖₁+𝛿₁𝜖²₁>𝛿₁+1⇔−2𝛿₁𝜖₁−𝜖₁+𝛿₁𝜖²₁>0⇔𝛿₁𝜖²₁>𝜖₁(2𝛿₁+1)⇔𝛿₁𝜖₁<2𝛿₁+1,(2.16) which is true. Hence, 𝛿₂>𝛿₁>0 and 𝜖₃<0. Therefore, this case applies to 𝛿₂,𝜖₃, and by induction and (2.11) and (2.12), we obtain 0<𝛿₁<𝛿₂<⋯<𝛿_𝑛<⋯.(2.17) Hence, 𝛿_𝑛 cannot converge to 0 as 𝑛→∞ in this case.Case 2. If 𝛿₁<0, 𝜖₁>0, then we obtain the following: 𝑥_𝑁−1=𝛿₁𝑥∈(−1,0),_𝑁=−1+𝜖₁𝑥∈(−1,0),_𝑁+1=−1+𝜖₂=−1+𝛿₁(−1+𝜖₁𝑥)∈(−1,0),_𝑁+2=𝛿₂=(−1+𝜖₁)[−1+𝛿₁(−1+𝜖₁𝑥)]−1∈(−1,0),_𝑁+3=−1+𝜖₃=𝛿₂[−1+𝛿₁(−1+𝜖₁)]−1∈(−1,0).(2.18) We next compute the following: 𝑥_𝑁+2=𝛿₂=(−1+𝜖₁)[−1+𝛿₁(−1+𝜖₁)]−1<𝛿₁⇔(−1+𝜖₁)[−1+𝛿₁(−1+𝜖₁)]<𝛿₁+1⇔1+𝛿₁−𝛿₁𝜖₁−𝜖₁−𝛿₁𝜖₁+𝛿₁𝜖²₁<𝛿₁+1⇔−2𝛿₁𝜖₁−𝜖₁+𝛿₁𝜖²₁<0⇔𝛿₁𝜖²₁<𝜖₁(2𝛿₁+1)⇔𝛿₁𝜖₁<2𝛿₁+1,(2.19) which is true by (2.11) and (2.12). Hence, 𝛿₂<𝛿₁<0 and 𝜖₃>0. Therefore, this case applies to 𝜖₃,𝛿₂, and by induction and (2.11) and (2.12), we obtain the following inequalities: 0>𝛿₁>𝛿₂>⋯>𝛿_𝑛>⋯.(2.20)
Hence, 𝛿_𝑛 cannot converge to 0 as 𝑛→∞ in this case, and we are done with our proof by contradiction.

Finally we show that the interval (−1,0) is invariant.

Theorem 2.5. If −1<𝑥₋₁, 𝑥₀<0, then −1<𝑥_𝑛<0 for all 𝑛≥−1.

Proof. If −1<𝑥₋₁, 𝑥₀<0, then −1<𝑥₁=𝑥₀𝑥₋₁−1<0. From (1.1) and by induction, we then have that −1<𝑥_𝑛<0 for all 𝑛≥−1.

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 (−1,0), 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), 𝑥₂, is locally asymptotically stable. The positive equilibrium solution of (1.1), 𝑥₁, is unstable.

Proof. The characteristic equation of the equilibria of (1.1) is the following: 𝜆²−𝑥_1,2𝜆−𝑥_1,2=0,(3.1) with eigenvalues 𝜆_±=𝑥_1,2±𝑥²_1,2+4𝑥_1,22.(3.2) Therefore, it is easy to show that the eigenvalues for 𝑥₂ are complex with |𝜆_±|<1, and for 𝑥₁, |𝜆₊|>1 and |𝜆₋|<1.

Open Problem 3. If −1<𝑥₋₁, 𝑥₀<0 show whether or not every solution {𝑥_𝑛}^∞_𝑛=−1 of (1.1) converges to 𝑥₂ (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 𝑥₋₁,𝑥₀>𝑥₁ or 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. The theorem below is included to make this paper self-contained. One can see that the theorem below overlaps with [1, Theorem 9.1].

Theorem 4.1. If 𝑥₋₁,𝑥₀>𝑥₁=(1+√5)/2, then the solution {𝑥_𝑛}^∞_𝑛=0 is eventually strictly increasing and tends to +∞.

Proof. Since 𝑥₀>(1+√5)/2, we have 1/𝑥₀<2/(1+√5)=(√5−1)/2. Thus, 11+𝑥₀√<1+5−12=1+√52<𝑥₋₁.(4.1) Therefore, 𝑥₋₁>1+1/𝑥₀. Thus 𝑥₋₁𝑥₀>𝑥₀+1. Rewriting, 𝑥₀𝑥₋₁−1>𝑥₀. Hence, 𝑥₁>𝑥₀. It follows by induction that the solution {𝑥_𝑛}^∞_𝑛=0 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 𝑥₀. 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 𝑥₋₁,𝑥₀<−1. Then 0<𝑥₁<𝑥₄<𝑥₇<⋯,⋯<𝑥₈<𝑥₆<𝑥₅<𝑥₃<𝑥₂<𝑥₀<−1,(4.2) and the subsequences {𝑥_3𝑛}^∞_𝑛=0 and {𝑥_3𝑛+2}^∞_𝑛=0 tend to −∞ and {𝑥_3𝑛+1}^∞_𝑛=0 tends to +∞.

Proof. We first have 𝑥₁=𝑥₀𝑥₋₁−1>0.(4.3) Then obviously 𝑥₂=𝑥₁𝑥₀−1<−1.(4.4) But we will prove that 𝑥₂<𝑥₀<−1.
Since (𝑥₀+1)²>0, we have that 𝑥²₀+2𝑥₀+1>0. Thus, −1<2/𝑥₀+1/𝑥²₀. Since 𝑥₋₁<−1, we must have 𝑥₋₁<2/𝑥₀+1/𝑥²₀. Thus 𝑥₋₁𝑥₀>2+1/𝑥₀. Therefore, 𝑥₋₁𝑥₀−1>1+1/𝑥₀. Hence 𝑥₁>1+1/𝑥₀. From here we have 𝑥₁𝑥₀−1<𝑥₀, and the result that 𝑥₂<𝑥₀ follows.
Next, we show that 𝑥₃ is not only less than −1, but it is less than 𝑥₂. To this end, since 𝑥₂<𝑥₀, and 𝑥₁=𝑥₀𝑥₋₁−1>0, we have 𝑥₁𝑥₂<𝑥₁𝑥₀. This gives 𝑥₂𝑥₁−1<𝑥₁𝑥₀−1, and hence, 𝑥₃<𝑥₂.
To show that 𝑥₄>𝑥₁, we start by observing that 𝑥₃<−1, 𝑥₁>0, and so 𝑥₃<𝑥₁. Thus, since 𝑥₃=𝑥₂𝑥₁−1<𝑥₁, we get, with 𝑥₂+1<0, 𝑥₁𝑥₂−𝑥₁−1<0⇒𝑥₁(𝑥₂−1)<1⇒𝑥₁(𝑥₂−1)(𝑥₂+1)>𝑥₂+1⇒𝑥₁𝑥²₂−𝑥₁>𝑥₂+1⇒𝑥₁𝑥²₂−𝑥₂−1>𝑥₁⇒𝑥₂(𝑥₂𝑥₁−1)−1>𝑥₁⇒𝑥₃𝑥₂−1>𝑥₁⇒𝑥₄>𝑥₁.(4.5)
By induction, it can be proved that 0<𝑥₁<𝑥₄<𝑥₇<⋯,⋯<𝑥₈<𝑥₆<𝑥₅<𝑥₃<𝑥₂<𝑥₀<−1,(4.6) that is, {𝑥_3𝑛+1}^∞_𝑛=0 is a positive increasing subsequence and {𝑥_3𝑛}^∞_𝑛=0 and {𝑥_3𝑛+2}^∞_𝑛=0 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, {𝑥_3𝑛}^∞_𝑛=0 and {𝑥_3𝑛+2}^∞_𝑛=0, are bounded from below. Then they each must converge to a finite limit (which is the same finite limit and less than 𝑥₂). But by (1.1), the third increasing subsequence, {𝑥_3𝑛+1}^∞_𝑛=0, must also converge to a finite limit (which is positive), where 𝑥_3𝑛+1=𝑥_3𝑛𝑥_3𝑛−1−1=|𝑥_3𝑛|⋅|𝑥_{3(𝑛−1)+2}|−1.(4.7) This is impossible because there are no periodic solutions with minimal period two (see Section 2). So {𝑥_3𝑛}^∞_𝑛=0 and {𝑥_3𝑛+2}^∞_𝑛=0 are unbounded (they tend to −∞) and thus {𝑥_3𝑛+1}^∞_𝑛=0 is also unbounded (it tends to +∞) by (1.1) again.

Theorem 4.3. Suppose that 𝑥₋₁<0 and 𝑥₀>0. Then the solution of (1.1) has three subsequences {𝑥_3𝑛}^∞_𝑛=1, {𝑥_3𝑛+1}^∞_𝑛=1, and {𝑥_3𝑛+2}^∞_𝑛=1, where two of them tend to −∞ and one tends to +∞.

Proof. Note that 𝑥₁=𝑥₋₁𝑥₀−1<−1 and 𝑥₂=𝑥₀𝑥₁−1<−1. 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 𝑥₋₁=−0.58 and 𝑥₀=0.618 then the solution does what the above theorem guarantees. On the other hand, if 𝑥₋₁=0.618 and 𝑥₀=−0.58, then the solution enters the interval (−1,0), remains there, and thus is bounded. Furthermore, it should be pointed out that if the initial conditions are such that 𝑥₋₁>0 and 𝑥₀<0, then the solution in certain cases is bounded and in other cases is unbounded.

In addition, observe that if 0<𝑥₋₁,𝑥₀<𝑥₁, 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 𝑥₋₁=1.5 and 𝑥₀=1.6, then the solution enters and then remains in the interval (−1,0), and hence is bounded. Whereas if 𝑥₋₁=1.5 and 𝑥₀=1.61, then the solution has three unbounded subsequences: one tending to +∞ and the other two tending to −∞.

Theorem 4.4. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). Suppose that(i)0<𝑥₋₁, 𝑥₀<1;(ii)𝑥²₀𝑥²₋₁−2𝑥₀𝑥₋₁+1−𝑥₋₁>0.Then {𝑥_3𝑛}^∞_𝑛=2 is a positive increasing subsequence; {𝑥_3𝑛+2}^∞_𝑛=1 is a negative decreasing subsequence; {𝑥_3𝑛+1}^∞_𝑛=2 is a negative decreasing subsequence. Consequently, there does not exist an 𝑁≥−1 such that −1<𝑥_𝑛<0 for all 𝑛≥𝑁. Furthermore, all three subsequences are unbounded and so the solution is unbounded.

Proof. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). Suppose that(i)0<𝑥₋₁, 𝑥₀<1;(ii)𝑥²₀𝑥²₋₁−2𝑥₀𝑥₋₁+1−𝑥₋₁>0.Since 𝑥₀𝑥₋₁<1 by Hypothesis (i), then we have 𝑥₁=𝑥₀𝑥₋₁−1<0.(4.8) Then, since 𝑥₁𝑥₀<0, we have 𝑥₂=𝑥₁𝑥₀−1<−1.(4.9)
We compute the sign of 𝑥₃=𝑥₂𝑥₁−1: 𝑥₃>0⇔𝑥₂𝑥₁−1>0⇔(𝑥₁𝑥₀−1)𝑥₁−1>0⇔𝑥²₁𝑥₀−𝑥₁−1>0⇔(𝑥₀𝑥₋₁−1)²𝑥₀−(𝑥₀𝑥₋₁−1)−1>0⇔𝑥³₀𝑥²₋₁−2𝑥²₀𝑥₋₁+𝑥₀−𝑥₀𝑥₋₁+1−1>0⇔𝑥²₀𝑥²₋₁−2𝑥₀𝑥₋₁+1−𝑥₋₁>0,(4.10) which is true by Hypotheses (i) and (ii). We next have 𝑥₄=𝑥₃𝑥₂−1<−1(4.11) since 𝑥₃>0, 𝑥₂<0. Then, also, 𝑥₅=𝑥₄𝑥₃−1<−1(4.12) since 𝑥₄<0, 𝑥₃>0.
With 𝑥₄, 𝑥₅<−1 replacing 𝑥₋₁,𝑥₀<−1 in Theorem 4.2, we have that 0<𝑥₆<𝑥₉<𝑥₁₂<⋯,⋯<𝑥₁₃<𝑥₁₁<𝑥₁₀<𝑥₈<𝑥₇<𝑥₅<−1.(4.13)
Hence, {𝑥_3𝑛}^∞_𝑛=2 is a positive increasing subsequence, {𝑥_3𝑛+1}^∞_𝑛=2 is a negative decreasing subsequence, and {𝑥_3𝑛+2}^∞_𝑛=1 is a negative decreasing subsequence, and so there does not exist an 𝑁≥−1 such that −1<𝑥_𝑛<0 for all 𝑛≥𝑁. Furthermore, by Theorem 4.2, {𝑥_3𝑛}^∞_𝑛=2 tends to +∞ and {𝑥_3𝑛+1}^∞_𝑛=2, and {𝑥_3𝑛+2}^∞_𝑛=1 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 (−1,0) and remains there.

Theorem 4.5. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). Suppose that 1<𝑥₋₁,𝑥₀<𝑥₁. Then one of the following occurs.(i)The solution is unbounded.(ii)There exists 𝑛₀≥1 such that 𝑥_𝑛∈(−1,0) for all 𝑛≥𝑛₀.

Proof. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). Assume that 1<𝑥₋₁,𝑥₀<𝑥₁=1+√52.(4.14) We will show that the solution is decreasing while the terms are positive. Since 1<𝑥₀<(1+√5)/2, we have 1𝑥₀>21+√5=√5−12.(4.15) Thus 11+𝑥₀2>1+1+√5√=1+5−12>𝑥₋₁,(4.16) which, since 𝑥₀>0, implies that 𝑥₀>𝑥₀𝑥₋₁−1=𝑥₁.(4.17)
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 𝑥₁ and since 𝑥₂ is less than zero, this sequence is not bounded below by zero and thus crosses over to negative values.
Let 𝑥_𝑁 be the last positive term and 𝑥_𝑁+1 the first negative term of our solution. We now determine ranges of values of terms of our solution beginning with 𝑥_𝑁 and 𝑥_𝑁+1, using “𝑛” in place of “𝑁” for convenience.
(1)𝑥_𝑛 and 𝑥_𝑛+1: observe that 𝑥_𝑛𝑥_𝑛−1−1>−1⇔𝑥_𝑛𝑥_𝑛−1>0,(4.18) which is true. Hence, −1<𝑥_𝑛+1<0. Note that this, in turn, implies that 0<𝑥_𝑛𝑥_𝑛−1<1. Then at least 0<𝑥_𝑛<1 since 𝑥_𝑛<𝑥_𝑛−1.(2)𝑥_𝑛+2: by Statement (1), −1<𝑥_𝑛+1𝑥_𝑛<0 and so −2<𝑥_𝑛+2<−1.(3)𝑥_𝑛+3,𝑥_𝑛+4,….Case 1 (𝑥_𝑛+3>0). Then from Statement (2), 𝑥_𝑛+4=𝑥_𝑛+3𝑥_𝑛+2−1<−1 and thus 𝑥_𝑛+5=𝑥_𝑛+4𝑥_𝑛+3−1<−1. By Theorem 4.2, the solution is unbounded.Case 2 (𝑥_𝑛+3<0). We wish to show that 𝑥_𝑛+3>−1: 𝑥_𝑛+3=𝑥_𝑛+2𝑥_𝑛+1−1>−1⇔𝑥_𝑛+2𝑥_𝑛+1>0,(4.19) which is true since, by Statements (1) and (2), 𝑥_𝑛+1,𝑥_𝑛+2<0.We next show that −1<𝑥_𝑛+4<0. First we show that 𝑥_𝑛+4<0: 𝑥_𝑛+4=𝑥_𝑛+3𝑥_𝑛+2−1<0⇔(𝑥_𝑛+2𝑥_𝑛+1−1)𝑥_𝑛+2−1<0⇔(𝑥²_𝑛+1𝑥_𝑛−𝑥_𝑛+1−1)(𝑥_𝑛+1𝑥_𝑛−1)−1<0⇔𝑥³_𝑛+1𝑥²_𝑛−2𝑥²_𝑛+1𝑥_𝑛+𝑥_𝑛+1−𝑥_𝑛+1𝑥_𝑛+1−1<0⇔𝑥³_𝑛+1𝑥²_𝑛−2𝑥²_𝑛+1𝑥_𝑛+𝑥_𝑛+1−𝑥_𝑛+1𝑥_𝑛<0⇔𝑥²_𝑛+1𝑥²_𝑛−2𝑥_𝑛+1𝑥_𝑛+1−𝑥_𝑛>0⇔𝑥_𝑛+1𝑥_𝑛(𝑥_𝑛+1𝑥_𝑛−2)−(𝑥_𝑛−1)>0⇔(𝑥²_𝑛+2−1)−(𝑥_𝑛−1)>0⇔𝑥²_𝑛+2>𝑥_𝑛,(4.20) which is true since, by Statements (1) and (2), 𝑥²_𝑛+2>1 and 0<𝑥_𝑛<1.
Second, we show that 𝑥_𝑛+4>−1: notice that 𝑥_𝑛+4=𝑥_𝑛+3𝑥_𝑛+2−1>−1⇔𝑥_𝑛+3𝑥_𝑛+2>0,(4.21) which is true since 𝑥_𝑛+3<0 and 𝑥_𝑛+2<0 by Statement (2). Because −1<𝑥_𝑛+3,𝑥_𝑛+4<0, it follows from Theorem 2.5 that there exists 𝑛₀≥1 such that 𝑥_𝑛∈(−1,0) for all 𝑛≥𝑛₀.

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)𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−1−𝑐, 𝑛=0,1,…, where 𝑐 is any real.(ii)𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−1−𝑐_𝑛, 𝑛=0,1,…, where {𝑐_𝑛}^∞_𝑛=0 is a periodic sequence of real numbers.(iii)𝑥_𝑛+1=𝑥_𝑛−ℓ𝑥_𝑛−𝑘−1, 𝑛=0,1,…, where ℓ,𝑘∈{0,1,…}.