# Long-Term Behavior of Solutions of   the Difference Equationxn+1=xn−1xn−2−1

Authors: Candace M. Kent; Witold Kosmala; Stevo Stević
Journal: Abstract and Applied Analysis
 (2010)

---

## Abstract

We investigate the long-term behavior of solutions of the following difference equation:xn+1=xn−1xn−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.


---

## Body

## 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 equation(1.1)xn+1=xn-1xn-2-1,n∈N0,
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 (1.2)xn+1=xn-kxn-l-1,n∈N0,
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)

Ifx¯ is an equilibrium of (1.1), then it satisfies the equation (2.1)x¯2-x¯-1=0.
Hence, (1.1) has exactly two equilibria, one positive and one negative, which we denote by x¯1 and x¯2, respectively:(2.2)x¯1=:1+52,x¯2=:1-52
(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{xn}n=-2∞ is an eventually constant solution of (1.1), then xN=xN+1=xN+2=x¯, for some N∈ℕ0, where x¯ is an equilibrium point. In this case, (1.1) gives xN+2=xNxN-1-1, which implies
(2.3)xN-1=xN+2+1xN=x¯+1x¯=x¯.
Repeating this procedure, we obtain xn=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 thatxN=xN+2k and xN+1=xN+2k+1, for every k∈ℕ0, and some N≥-2, with xN≠xN+1. Then, we have
(2.4)xN+4=xN+2xN+1-1=xNxN+1-1=xN+3=xN+1.
From this and since xN+4=xN, 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(2.5)xN=xN+3k,xN+1=xN+3k+1,xN+2=xN+3k+2,k≥0,
for some N≥-2,  we have
(2.6)xN+3=xN+1xN-1=xN,xN+4=xN+2xN+1-1=xN+1,xN+5=xN+3xN+2-1=xNxN+2-1=xN+2.
If xN=0, xN+1=0, or xN+2=0, then from (2.6) we easily obtain contradictions in all these cases. Hence, we may assume that xN≠0, xN+1≠0, and xN+2≠0. Equalities in (2.6) also imply that
(2.7)xN+1=xN+1xN,xN+2=xN+1+1xN+1,xN=xN+2+1xN+2.
From (2.7), we get
(2.8)xN=xN+2+1xN+2=2xN+1+1xN+1+1=3xN+22xN+1,
which implies that xN2-xN-1=0, that is, xN=(1±5)/2. From this and (2.7) we obtain xN+1=xN+2=(1±5)/2, implying xn=(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(2.9)xN=xN+4k,xN+1=xN+4k+1,xN+2=xN+4k+2,xN+3=xN+4k+3,k≥0,
for some N≥-2. Then we have
(2.10)xN+4=xN+2xN+1-1=xN,xN+5=xN+3xN+2-1=xN+1,xN+6=xN+4xN+3-1=xNxN+3-1=xN+2,xN+7=xN+5xN+4-1=xN+1xN-1=xN+3.
If xN+i=0 for some i∈{0,1,2,3}, then from (2.10) we easily obtain a contradiction. For example, if xN=0, then from (2.10) we get xN+2=-1=xN+3. This implies xN+1=xN+3xN+2-1=0. From this and since xN+2xN+1-1=xN we would get xN=-1, a contradiction. The other cases are proved analogously.
Hence we may assume thatxN+i≠0, i∈{0,1,2,3}. From (2.10) we have
(2.11)xN=xN+3+1xN+1=(xN+2+1)/xN+1(xN+1)/xN+2=xN+2(xN+2+xN+1)xN(xN+1)=(xN+1)(xN+1+1)xNxN+12,
that is,
(2.12)(xNxN+1)2=(xN+1)(xN+1+1).
From (2.10) and (2.12), we obtain
(2.13)(xN+3+1)2=(xN+1)(xN+1+1).
Since the relations in (2.12) are cyclic, we also obtain that
(2.14)(xN+1)2=(xN+1+1)(xN+2+1),(xN+1+1)2=(xN+2+1)(xN+3+1),(xN+2+1)2=(xN+3+1)(xN+1).
Equalities (2.13) and (2.14) imply that the expressions xN+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
(2.15)(xN+3+1)2=(xN+1)(xN+1+1)=(xN+1)(xN+3+1)(xN+2+1)=(xN+1)(xN+3+1)(xN+3+1)(xN+1),
from which easily follows that xN+3=xN. From this, (2.13), and (2.14), it follows that xN+i=xN, 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{xn}n=-2∞ is of period five. Then, we can write terms of the solution of (1.1) as
(2.16)x-2=a,x-1=b,x0=c,x1=d=ab-1,x2=e=bc-1,x3=f=cd-1=c(ab-1)-1=a,x4=g=(ab-1)(bc-1)-1=b,x5=h=(bc-1)f-1=(bc-1)a-1=c.
Note that the expressions forf 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.
Ifxk=0, where k≥-1, then xk 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
(2.17)x-1=0,x0=x0,x1=x-1x-2-1=-1,x2=x0x-1-1=-1,x3=x1x0-1=-x0-1,x4=x2x1-1=0,x5=x3x2-1=-(-x0-1)-1=x0,x6=x4x3-1=-1,
from which the statement follows in this case.
Ifk≥0, then by direct calculation we have
(2.18)xk=0,xk+1=xk-1xk-2-1,xk+2=xkxk-1-1=-1,xk+3=xk+1xk-1=-1,xk+4=xk+2xk+1-1=-(xk-1xk-2-1)-1=-xk-1xk-2,xk+5=xk+3xk+2-1=0,xk+6=xk+4xk+3-1=-(-xk-1xk-2)-1=xk-1xk-2-1,xk+7=xk+5xk+4-1=-1,
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, choosex-2=4,  x-1=0.5,  x0=1, or x-2=0,  x-1=4,  x0=-0.5.


## 2.1. Equilibria of (1.1)

Ifx¯ is an equilibrium of (1.1), then it satisfies the equation (2.1)x¯2-x¯-1=0.
Hence, (1.1) has exactly two equilibria, one positive and one negative, which we denote by x¯1 and x¯2, respectively:(2.2)x¯1=:1+52,x¯2=:1-52
(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{xn}n=-2∞ is an eventually constant solution of (1.1), then xN=xN+1=xN+2=x¯, for some N∈ℕ0, where x¯ is an equilibrium point. In this case, (1.1) gives xN+2=xNxN-1-1, which implies
(2.3)xN-1=xN+2+1xN=x¯+1x¯=x¯.
Repeating this procedure, we obtain xn=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 thatxN=xN+2k and xN+1=xN+2k+1, for every k∈ℕ0, and some N≥-2, with xN≠xN+1. Then, we have
(2.4)xN+4=xN+2xN+1-1=xNxN+1-1=xN+3=xN+1.
From this and since xN+4=xN, 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(2.5)xN=xN+3k,xN+1=xN+3k+1,xN+2=xN+3k+2,k≥0,
for some N≥-2,  we have
(2.6)xN+3=xN+1xN-1=xN,xN+4=xN+2xN+1-1=xN+1,xN+5=xN+3xN+2-1=xNxN+2-1=xN+2.
If xN=0, xN+1=0, or xN+2=0, then from (2.6) we easily obtain contradictions in all these cases. Hence, we may assume that xN≠0, xN+1≠0, and xN+2≠0. Equalities in (2.6) also imply that
(2.7)xN+1=xN+1xN,xN+2=xN+1+1xN+1,xN=xN+2+1xN+2.
From (2.7), we get
(2.8)xN=xN+2+1xN+2=2xN+1+1xN+1+1=3xN+22xN+1,
which implies that xN2-xN-1=0, that is, xN=(1±5)/2. From this and (2.7) we obtain xN+1=xN+2=(1±5)/2, implying xn=(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(2.9)xN=xN+4k,xN+1=xN+4k+1,xN+2=xN+4k+2,xN+3=xN+4k+3,k≥0,
for some N≥-2. Then we have
(2.10)xN+4=xN+2xN+1-1=xN,xN+5=xN+3xN+2-1=xN+1,xN+6=xN+4xN+3-1=xNxN+3-1=xN+2,xN+7=xN+5xN+4-1=xN+1xN-1=xN+3.
If xN+i=0 for some i∈{0,1,2,3}, then from (2.10) we easily obtain a contradiction. For example, if xN=0, then from (2.10) we get xN+2=-1=xN+3. This implies xN+1=xN+3xN+2-1=0. From this and since xN+2xN+1-1=xN we would get xN=-1, a contradiction. The other cases are proved analogously.
Hence we may assume thatxN+i≠0, i∈{0,1,2,3}. From (2.10) we have
(2.11)xN=xN+3+1xN+1=(xN+2+1)/xN+1(xN+1)/xN+2=xN+2(xN+2+xN+1)xN(xN+1)=(xN+1)(xN+1+1)xNxN+12,
that is,
(2.12)(xNxN+1)2=(xN+1)(xN+1+1).
From (2.10) and (2.12), we obtain
(2.13)(xN+3+1)2=(xN+1)(xN+1+1).
Since the relations in (2.12) are cyclic, we also obtain that
(2.14)(xN+1)2=(xN+1+1)(xN+2+1),(xN+1+1)2=(xN+2+1)(xN+3+1),(xN+2+1)2=(xN+3+1)(xN+1).
Equalities (2.13) and (2.14) imply that the expressions xN+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
(2.15)(xN+3+1)2=(xN+1)(xN+1+1)=(xN+1)(xN+3+1)(xN+2+1)=(xN+1)(xN+3+1)(xN+3+1)(xN+1),
from which easily follows that xN+3=xN. From this, (2.13), and (2.14), it follows that xN+i=xN, 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{xn}n=-2∞ is of period five. Then, we can write terms of the solution of (1.1) as
(2.16)x-2=a,x-1=b,x0=c,x1=d=ab-1,x2=e=bc-1,x3=f=cd-1=c(ab-1)-1=a,x4=g=(ab-1)(bc-1)-1=b,x5=h=(bc-1)f-1=(bc-1)a-1=c.
Note that the expressions forf 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.
Ifxk=0, where k≥-1, then xk 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
(2.17)x-1=0,x0=x0,x1=x-1x-2-1=-1,x2=x0x-1-1=-1,x3=x1x0-1=-x0-1,x4=x2x1-1=0,x5=x3x2-1=-(-x0-1)-1=x0,x6=x4x3-1=-1,
from which the statement follows in this case.
Ifk≥0, then by direct calculation we have
(2.18)xk=0,xk+1=xk-1xk-2-1,xk+2=xkxk-1-1=-1,xk+3=xk+1xk-1=-1,xk+4=xk+2xk+1-1=-(xk-1xk-2-1)-1=-xk-1xk-2,xk+5=xk+3xk+2-1=0,xk+6=xk+4xk+3-1=-(-xk-1xk-2)-1=xk-1xk-2-1,xk+7=xk+5xk+4-1=-1,
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, choosex-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<xn<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<xn<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 ifa,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
(3.1)xn+5-xn=xn(xn+4-xn-1),forn≥-1.Proof.
Ifn≥-1, we have
(3.2)xn+2=xnxn-1-1,xn+3=xn+1xn-1,xn+4=xn+2xn+1-1=(xnxn-1-1)xn+1-1=xn+1xnxn-1-xn+1-1,(3.3)xn+5=xn+3xn+2-1=(xn+1xn-1)(xnxn-1-1)-1=xn+1(xn)2xn-1-xn+1xn-xnxn-1=xn(xn+1xnxn-1-xn+1-xn-1).
Using (3.3) and then (3.2), we get
(3.4)xn+5-xn=xn(xn+1xnxn-1-xn+1-xn-1)-xn=xn(xn+1xnxn-1-xn+1-xn-1-1)=xn[(xn+1xnxn-1-xn+1-1)-xn-1]=xn(xn+4-xn-1),
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{x5n}n=0∞, {x5n+1}n=0∞, {x5n+2}n=0∞, {x5n+3}n=0∞, and {x5n+4}n=0∞. By Lemma 3.3, we know that our solution satisfies (3.1) for all n≥-1. Thus, we have
(3.5)xn+10-xn+5=xn+5(xn+9-xn+4)=xn+5xn+4(xn+8-xn+3)=xn+5xn+4xn+3(xn+7-xn+2)=xn+5xn+4xn+3xn+2(xn+6-xn+1)=xn+5xn+4xn+3xn+2xn+1(xn+5-xn).
Therefore,
(3.6)|xn+10-xn+5|=|xn+5||xn+4||xn+3||xn+2||xn+1||xn+5-xn|,
and, since the values of the sequence are in the interval (-1,0), we have,
(3.7)|xn+10-xn+5|≤|xn+5-xn|.
Without loss of generality, we will consider the subsequence {x5n}n=0∞. Let
(3.8)an=x5(n+1)-x5n,n∈N0.
By inequality (3.7), we have
(3.9)|a0|≥|a1|≥|a2|≥⋯.
Since the sequence {|an|}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
(3.10)|x5(n+1)-x5n|=(∏k=15n|xk|)|x5-x0|.
Since the sequence {|an|}n=0∞ converges to a nonzero value, it follows that ∏k=1∞|xk| converges to a nonzero value. Hence, limk→∞|xk|=1. Therefore, xk is close to −1 for k large. Suppose k is large. Then, xk+3=xk+1xk-1 is close to 0, which contradicts the fact that |xk+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.
Clearlyx1=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.


## 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 ifa,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
(3.1)xn+5-xn=xn(xn+4-xn-1),forn≥-1.Proof.
Ifn≥-1, we have
(3.2)xn+2=xnxn-1-1,xn+3=xn+1xn-1,xn+4=xn+2xn+1-1=(xnxn-1-1)xn+1-1=xn+1xnxn-1-xn+1-1,(3.3)xn+5=xn+3xn+2-1=(xn+1xn-1)(xnxn-1-1)-1=xn+1(xn)2xn-1-xn+1xn-xnxn-1=xn(xn+1xnxn-1-xn+1-xn-1).
Using (3.3) and then (3.2), we get
(3.4)xn+5-xn=xn(xn+1xnxn-1-xn+1-xn-1)-xn=xn(xn+1xnxn-1-xn+1-xn-1-1)=xn[(xn+1xnxn-1-xn+1-1)-xn-1]=xn(xn+4-xn-1),
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{x5n}n=0∞, {x5n+1}n=0∞, {x5n+2}n=0∞, {x5n+3}n=0∞, and {x5n+4}n=0∞. By Lemma 3.3, we know that our solution satisfies (3.1) for all n≥-1. Thus, we have
(3.5)xn+10-xn+5=xn+5(xn+9-xn+4)=xn+5xn+4(xn+8-xn+3)=xn+5xn+4xn+3(xn+7-xn+2)=xn+5xn+4xn+3xn+2(xn+6-xn+1)=xn+5xn+4xn+3xn+2xn+1(xn+5-xn).
Therefore,
(3.6)|xn+10-xn+5|=|xn+5||xn+4||xn+3||xn+2||xn+1||xn+5-xn|,
and, since the values of the sequence are in the interval (-1,0), we have,
(3.7)|xn+10-xn+5|≤|xn+5-xn|.
Without loss of generality, we will consider the subsequence {x5n}n=0∞. Let
(3.8)an=x5(n+1)-x5n,n∈N0.
By inequality (3.7), we have
(3.9)|a0|≥|a1|≥|a2|≥⋯.
Since the sequence {|an|}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
(3.10)|x5(n+1)-x5n|=(∏k=15n|xk|)|x5-x0|.
Since the sequence {|an|}n=0∞ converges to a nonzero value, it follows that ∏k=1∞|xk| converges to a nonzero value. Hence, limk→∞|xk|=1. Therefore, xk is close to −1 for k large. Suppose k is large. Then, xk+3=xk+1xk-1 is close to 0, which contradicts the fact that |xk+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.
Clearlyx1=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),  x¯1, is unstable.Proof.
The characteristic equation of the equilibriumx¯1 is the following:
(4.1)λ3-x¯1λ-x¯1=0.
Let P1(x)=x3-x¯1x-x¯1. Since P1(1)=1-2x¯1=-5<0 and limx→+∞P1(x)=+∞, it follows that there is a λ0>1 such that P1(λ0)=0, from which the result follows for the equilibrium x¯1.Remark 4.2.
Consider the characteristic equation of the negative equilibrium of (1.1), x¯2,
(4.2)λ3-x¯2λ-x¯2=0
and the function P2(x)=x3-x¯2x-x¯2. Observe the following: (i)
P2(x¯2)=0.(ii)
P2′(x)=3x2-x¯2>0 for all x∈ℝ. Hence, we have one negative eigenvalue, λ1=x¯2∈(-1,0) and a complex conjugate pair of eigenvalues, λ2 and λ¯2, such that
(4.3)λ1|λ2|2=x¯2.
It follows that |λ2|=|λ¯2|=1, and so x¯2 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), x¯2.


## 5. More on Invariant Intervals

In this section, we discuss invariant intervals regarding the subsequences{x5n}n=0∞, {x5n+1}n=0∞, {x5n+2}n=0∞, {x5n+3}n=0∞, and {x5n+4}n=0∞ of a solution {xn}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, x5n+2,x5n+6∈(-2,-1),  x5n+3,x5n+5∈(-1,0), and x5n+4∈(0,1) for n∈ℕ0 and also x1=-1.Proof.
Sincex-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(5.1)x3=x1x0-1=-x0-1∈(-1,0),x4=x2x1-1=-x2-1∈(0,1).
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
(5.2)0≤(-x0-1)(x0x-1-1)-1<1⟺0≤x0(1-x-1-x0x-1)<1,
which is equivalent to
(5.3)0≤x0(-x2-x-1)<1.
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, since0<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 thatx5n+2,x5n+6∈(-2,-1), x5n+3,x5n+5∈(-1,0), and x5n+4∈(0,1) for 0≤n≤k.
Sincex5k+4∈(0,1) and x5k+5∈(-1,0), we have
(5.4)x5(k+1)+2=x5k+5x5k+4-1∈(-2,-1).
Sincex5k+6∈(-2,-1) and x5k+5∈(-1,0), we have
(5.5)x5(k+1)+3=x5k+6x5k+5-1∈(-1,1).
Assume that x5(k+1)+3∈[0,1). Then, we have
(5.6)0≤x5(k+1)+3=x5k+6x5k+5-1=(x5k+4x5k+3-1)(x5k+3x5k+2-1)-1<1
which is equivalent to
(5.7)0≤x5k+3(x5k+4x5k+3x5k+2-x5k+4-x5k+2)<1.
Now note that x5k+4x5k+3x5k+2>0 and -x5k+4-x5k+2>0 since -1<-x5k+4<0 and 1<-x5k+2<2. Hence, x5k+4x5k+3x5k+2-x5k+4-x5k+2>0 which along with x5k+3<0 implies that x5k+3(x5k+4x5k+3x5k+2-x5k+4-x5k+2)<0, which is a contradiction. Hence,
(5.8)x5(k+1)+3∈(-1,0).
Since-2<x5(k+1)+2,x5k+6<-1, we have
(5.9)x5(k+1)+4=x5(k+1)+2x5k+6-1∈(0,3).
Assume that x5(k+1)+4∈[1,3).
Since by Lemma3.3(5.10)x5(k+1)+4-x5k+4=x5k+4(x5(k+1)+3-x5k+3)
and 0<x5k+4<1≤x5(k+1)+4, we have x5(k+1)+4-x5k+4>0. Therefore, x5k+4(x5(k+1)+3-x5k+3)>0. From this and since x5k+4>0, we have that x5(k+1)+3-x5k+3>0. We then have
(5.11)x5(k+1)+3-x5k+3=x5k+3(x5(k+1)+2-x5k+2)>0.
Since x5k+3<0, we have x5(k+1)+2-x5k+2<0. Furthermore we have
(5.12)x5(k+1)+2-x5k+2=x5k+2(x5(k+1)+1-x5k+1)<0.
Since x5k+2<0, we have x5(k+1)+1-x5k+1>0.
By Lemma3.3, for each 1≤n≤k, we have
(5.13)x5(n+1)+1-x5n+1=x5n+1x5nx5n-1x5n-2x5n-3(x5n+1-x5(n-1)+1).
By the inductive hypothesis, we have that x5n+1x5nx5n-1x5n-2x5n-3>0 so that the sign of the difference is the same as the sign of x5(k+1)+1-x5k+1, that is,
(5.14)x5(n+1)+1-x5n+1>0,0≤n≤k.
On the other hand, x6-x1=x6+1<0, which is a contradiction. Hence,
(5.15)x5(k+1)+4∈(0,1).
Since-2<x5(k+1)+2<-1 and x5(k+1)+3∈(-1,0), we have
(5.16)x5(k+1)+5=x5(k+1)+3x5(k+1)+2-1∈(-1,1).
Assume that x5(k+1)+5∈[0,1). Then, we have
(5.17)0≤(x5k+6x5k+5-1)(x5k+5x5k+4-1)-1<1
which is equivalent to
(5.18)0≤x5k+5(x5k+6x5k+5x5k+4-x5k+6-x5k+4)<1.
On the other hand, since x5k+6x5k+5x5k+4>0 and since from 1<-x5k+6<2 and -1<-x5k+4<0 we obtain -x5k+6-x5k+4>0, it follows that
(5.19)x5k+6x5k+5x5k+4-x5k+6-x5k+4>0.
This fact along with x5k+5<0 implies
(5.20)x5k+5(x5k+6x5k+5x5k+4-x5k+6-x5k+4)<0
which is a contradiction. Hence,
(5.21)x5(k+1)+5∈(-1,0).
Finally, since0<x5(k+1)+4<1 and -1<x5(k+1)+3<0, we obtain
(5.22)x5(k+1)+6=x5(k+1)+4x5(k+1)+3-1∈(-2,-1).
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 Theorem5.1, we have that x-2=0,  x-1∈(0,1),  x0∈(-1,0),  x1=-1,  x5n+2,x5n+6∈(-2,-1),    x5n+3,x5n+5∈(-1,0),  and x5n+4∈(0,1) for n∈ℕ0.
We prove by induction that all the subsequences{x5n+3}n=-1∞, {x5n+4}n=-1∞, {x5n+5}n=-1∞, {x5n+6}n=-1∞, and {x5n+7}n=-1∞ are monotone.
Assumen=-1. Then, by Lemma 3.3 and above comments, we have
(5.23)x3-x-2=x3<0,x4-x-1=x-1(x3-x-2)=x-1x3<0,x5-x0=x0(x4-x-1)>0,x6-x1=x1(x5-x0)<0,x7-x2=x2(x6-x1)>0.
Assume that we have proved(5.24)-1<x5n+3<x5(n-1)+3<0=x-2,0<x5n+4<x5(n-1)+4<1,0>x5n+5>x5(n-1)+5>-1,-2<x5n+6<x5(n-1)+6<-1,-1>x5n+7>x5(n-1)+7>-2,
for 0≤n≤k.
Since(5.25)x5(k+1)+j-x5k+j=x5k+jx5k+j-1x5k+j-2x5k+j-3x5k+j-4(x5k+j-x5(k-1)+j),
for each k∈ℕ0 and j=3,4,5,6,7, and since by Theorem 5.1, we obtain
(5.26)x5k+jx5k+j-1x5k+j-2x5k+j-3x5k+j-4>0,k∈ℕ0, j=3,4,5,6,7, we have that the differences x5(k+1)+j-x5k+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{x5n+3}n=-1∞, {x5n+4}n=-1∞, {x5n+5}n=-1∞, {x5n+6}n=-1∞ and {x5n+7}n=-1∞ are monotone and bounded, they are convergent, and consequently the solution {xn}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 valuesx-2,x-1,x0>x¯1 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.
Ifx-2,x-1,x0>x¯1=(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) Sincex-1>(1+5)/2, we have 1/x-1<2/(1+5)=(5-1)/2. Thus,
(6.1)1+1x-1<1+5-12=1+52<x-2.
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(6.2)xn-2=1+xn+1xn-1,n∈N0,
it would follow that both subsequences {x2n}n=0∞ and {x2n-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 thata,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 havex1=ab-1>0, x2=x0x-1-1>(b+1)/(ab-1)>0, and
(6.3)x3=x1x0-1>(ab-1)1+aab-1-1=a=x-2>0.
By Lemma 3.3, we have
(6.4)xn-xn-5=(x3-x-2)∏j=-1n-5xj,n≥4.
Using (6.4) and the fact x3-x-2>0, by induction it easily follows that
(6.5)x5k+j>x5(k-1)+j>0,k∈N0,j=3,4,5,6,7,
that is, the subsequences {x5k+3}k=-1∞, {x5k+4}k=-1∞, {x5k+5}k=-1∞, {x5k+6}k=-1∞ and {x5k+7}k=-1∞ are increasing.
We also have(6.6)x5k+3-x5(k-1)+3=(x3-x-2)∏j=1k(x5(j-1)+3x5(j-1)+2x5(j-1)+1x5(j-1)x5(j-1)-1)≥(x3-x-2)(x3x2x1x0x-1)k≥(x3-x-2)(ab(a+1)(b+1)ab-1)k.
Note that
(6.7)q∶=ab(a+1)(b+1)ab-1>1.
Using this fact and (6.6) we obtain
(6.8)x5k+3>x5(k-1)+3+(x3-x-2)qk⟹x5k+3>x-2+(x3-x-2)∑j=0kqj,
from which it follows that x5k+3→∞ as k→∞.
From (6.4), the monotonicity of those five subsequences, and (6.8), we get
(6.9)x5k+3+i-x5(k-1)+3+i=x5(k-1)+3+i⋯x5(k-1)+3+1(x5k+3-x5(k-1)+3)>xi-2⋯x-1(x5k+3-x5(k-1)+3)>xi-2⋯x-1(x3-x-2)qk,
for i=1,2,3,4, from which it follows that x5k+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. Casex-2,x-1,x0∈(1,x¯1)

Here, we consider the casex-2,x-1,x0∈(1,x¯1). The next theorem shows that there is a large class of eventually nondecreasing solutions of (1.1) converging to x¯1. 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.
Assumex-2,x-1,x0∈(1,x¯1) and x-2,x-1≤x0≤x-1x-2-1. Then, every solution with such initial values is eventually nondecreasing and converges to x¯1.Proof.
Multiplying the assumptionx-2≤x0, by x-1 we obtain x-1x-2≤x-1x0. From this and since 1<x-2,x-1,x0<x¯1, we obtain 0<x-1x-2-1≤x-1x0-1<x¯12-1=x¯1, that is, 0<x1≤x2<x¯1. Hence, 1<x-2,x-1≤x0≤x1≤x2<x¯1. Now assume
(7.1)1<x-2,x-1≤x0≤x1≤x2≤⋯≤xn<x¯1,
for some n≥2. Multiplying the inequality xn-3≤xn-1 by xn-2 and using (7.1), it follows that
(7.2)xn-3xn-2-1≤xn-1xn-2-1<x¯12-1=x¯1,
that is, xn≤xn+1<x¯1. Hence, by induction we have proved that the sequence {xn}n=-1∞ is nondecreasing and bounded above by x¯1, from which the result easily follows.Theorem 7.2.
Assumex-2,x-1,x0∈(1,x¯1), 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
(7.3)x2≥x3≥⋯≥xN≥0>xN+1.Proof.
According to the assumptions we have(7.4)0=1⋅1-1<x1=x-1x-2-1≤x-1<x¯1,(7.5)0=1⋅1-1<x2=x0x-1-1≤x0<x¯1.
Ifx0≤x-2, then by multiplying by x-1, we obtain
(7.6)x2=x0x-1-1≤x-1x-2-1=x1.
Ifx0≤x-1, then since x1=x-1x-2-1≤x-2, we obtain
(7.7)x3=x1x0-1≤x-1x-2-1=x1.
Multiplying the inequalityx1≤x-1 (see (7.4)) by x0, we obtain
(7.8)x3≤x2,
which, together with inequality (7.6), implies again x3≤x1.
Note that now we cannot guarantee the positivity ofx3. Similarly, from the inequality x2≤x0 (see (7.5)), we obtain
(7.9)x4≤x3,
and from the inequality x3≤x1, we obtain
(7.10)x5≤x4.
Now assume that(7.11)0<xn≤xn-1≤⋯x3≤x2<x¯1
and xn+1>0. Then, by multiplying the inequality xn-1≤xn-3 by xn-2 and subtracting 1, we obtain xn+1≤xn. Hence, we proved by induction that (7.11) holds as far as xn is positive.
Ifxn>0 for all n≥2, the sequence {xn}n=-2∞ is convergent and its limit is nonnegative. However, this is not possible since the only nonnegative equilibrium of (1.1) is x¯1. From this, the result follows.



---

*Source: 152378-2010-12-29.xml*