# Solutions of the Difference Equationxn+1=xnxn−1−1

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

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## Abstract

Our goal in this paper is to investigate the long-term behavior of
solutions of the following difference equation:xn+1=xnxn−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.


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## Body

## 1. Introduction and Preliminaries

We investigate the long-term behavior of solutions of the second-order difference equation(1.1)xn+1=xnxn-1-1,n=0,1,2,…,
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 form(1.2)xn+1=xn-kxn-k-1-1,forn=0,1,…,
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 equationx¯2-x¯-1=0, we find that (1.1) has exactly two equilibria, one positive and one negative, which we denote by x¯1 and x¯2, respectively:(2.1)x¯1=:1+52,x¯2=:1-52.Note that there are no solutions which are eventually constant. Indeed, ifxN=xN+1=x¯ for some N≥0, then from xN+1=xNxN-1-1, it follows that (2.2)xN-1=xN+1+1xN=x¯+1x¯=x¯.
Repeating this procedure, we obtain xn=x¯, for -1≤n≤N+1, as claimed.Also note that there are no solutions that are eventually periodic with minimal period two. Indeed, ifxN=xN+2k and xN+1=xN+2k+1, k≥0, then we have (2.3)xN+3=xN+2xN+1-1=xNxN+1-1=xN+2=xN,
so that xn=xN, 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
(2.4)x-1=a,x0=b,x1=ab-1,x2=b(ab-1)-1=a⇔ab2-b-1=a⇔ab2=a+b+1,x3=a(ab-1)-1=b⇔a2b-a-1=b⇔a2b=a+b+1.
Therefore, this is indeed a solution of period three if the system below is satisfied. 							(2.5)ab2=a+b+1,a2b=a+b+1.
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 thata=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 thatb=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 thata=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
(2.6)x-1=-1,x0=-1,x1=0,…;x-1=-1,x0=0,x1=-1,…;x-1=0,x0=-1,x1=-1,…,
as claimed.In the sequel, we will refer to any one of these three periodic solutions of (1.1) as(2.7)…,0,-1,-1,0,-1,-1,….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(2.8)(x-1,x0,…,xN,xN+1,0,-1,-1,0,-1,-1,0,-1,-1,…),
Where N≥-1, xN+1=a∈R∖{0}, xN=1/a, and, if N≠-1, xn-1=(xn+1+1)/xn for 0≤n≤N.Proof.
If{xn}n=-1∞ is an eventually periodic solution with minimal period three, then by Theorem 2.1 there is an N≥-1 such that xN+2=0 and xN+3=-1. Hence 0=xN+2=xN+1xN-1 and consequently xN≠0≠xN+1 and xN+1=1/xN. Let xN+1=a (which implies that xN=1/a). From (1.1), if N≠-1, we get xn-1=(xn+1+1)/xn, for 0≤n≤N, as desired.Remark 2.3.
If, in Theorem2.2, 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{xn}n=-1∞ be a solution of (1.1) that is neither periodic nor eventually periodic with minimal period three. Then {xn}n=-1∞ does not converge to the minimal period-three solution
(2.9)…,0,-1,-1,0,-1,-1,….Proof.
Let{xn}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 {xn}n=-1∞ converges to the minimal period-three solution
(2.10)…,0,-1,-1,0,-1,-1,….
Then we can choose N≥0 such that there exist
(2.11)-12<δ1,δ2,δ3,…<12,-1<ϵ0,ϵ1,ϵ2,ϵ3,…<1,
with
(2.12)xN-2=-1+ϵ0,xN-1=δ1,xN=-1+ϵ1,xN+1=-1+ϵ2,xN+2=δ2,xN+3=-1+ϵ3,xN+4=-1+ϵ4,xN+5=δ3,xN+6=-1+ϵ5,…,
that is, for n=0,1,…,
(2.13)xN+(3n-2)=-1+ϵ2n,xN+(3n-1)=δn+1,xN+3n=-1+ϵ2n+1.
Now, first observe that(2.14)xN-2=-1+ϵ0,xN-1=δ1,xN=-1+ϵ1=-1+δ1(-1+ϵ0).
We claim that δ1≠0. Otherwise, xN-1=δ1=0. Then, by (1.1), xN=xN+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 xN-2=-1+ϵ0<0 by (2.11), we have xN<-1 and so ϵ1<0. If δ1<0, then since xN-2=-1+ϵ0<0 and xN=-1+ϵ1<0 by (2.11), we have xN∈(-1,0) and so ϵ1>0. Therefore, we need only consider the following two cases.
Case 1.
Ifδ1>0, ϵ1<0, then we obtain the following inequalities from (2.11) and (2.12):
(2.15)xN-1=δ1>0,xN=-1+ϵ1<-1,xN+1=-1+ϵ2=-1+δ1(-1+ϵ1)<-1,xN+2=δ2=(-1+ϵ1)[-1+δ1(-1+ϵ1)]-1>0,xN+3=-1+ϵ3=δ2[-1+δ1(-1+ϵ1)]-1<-1.
We next compute the following:(2.16)xN+2=δ2=(-1+ϵ1)[-1+δ1(-1+ϵ1)]-1>δ1⇔(-1+ϵ1)[-1+δ1(-1+ϵ1)]>δ1+1⇔1+δ1-δ1ϵ1-ϵ1-δ1ϵ1+δ1ϵ12>δ1+1⇔-2δ1ϵ1-ϵ1+δ1ϵ12>0⇔δ1ϵ12>ϵ1(2δ1+1)⇔δ1ϵ1<2δ1+1,
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
(2.17)0<δ1<δ2<⋯<δn<⋯.
Hence, δn cannot converge to 0 as n→∞ in this case.Case 2.
Ifδ1<0, ϵ1>0, then we obtain the following:
(2.18)xN-1=δ1∈(-1,0),xN=-1+ϵ1∈(-1,0),xN+1=-1+ϵ2=-1+δ1(-1+ϵ1)∈(-1,0),xN+2=δ2=(-1+ϵ1)[-1+δ1(-1+ϵ1)]-1∈(-1,0),xN+3=-1+ϵ3=δ2[-1+δ1(-1+ϵ1)]-1∈(-1,0).
We next compute the following:
(2.19)xN+2=δ2=(-1+ϵ1)[-1+δ1(-1+ϵ1)]-1<δ1⇔(-1+ϵ1)[-1+δ1(-1+ϵ1)]<δ1+1⇔1+δ1-δ1ϵ1-ϵ1-δ1ϵ1+δ1ϵ12<δ1+1⇔-2δ1ϵ1-ϵ1+δ1ϵ12<0⇔δ1ϵ12<ϵ1(2δ1+1)⇔δ1ϵ1<2δ1+1,
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:
(2.20)0>δ1>δ2>⋯>δn>⋯.
Hence,δn 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<xn<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<xn<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 Theorem2.5 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), x¯2, is locally asymptotically stable. The positive equilibrium solution of (1.1), x¯1, is unstable.Proof.
The characteristic equation of the equilibria of (1.1) is the following:
(3.1)λ2-x¯1,2λ-x¯1,2=0,
with eigenvalues
(3.2)λ±=x¯1,2±x¯1,22+4x¯1,22.
Therefore, it is easy to show that the eigenvalues for x¯2 are complex with |λ±|<1, and for x¯1, |λ+|>1 and |λ-|<1.Open Problem 3.
If-1<x-1, x0<0 show whether or not every solution {xn}n=-1∞ of (1.1) converges to x¯2 (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 conditionsx-1,x0>x¯1 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.
Ifx-1,x0>x¯1=(1+5)/2, then the solution {xn}n=0∞ is eventually strictly increasing and tends to +∞.Proof.
Sincex0>(1+5)/2, we have 1/x0<2/(1+5)=(5-1)/2. Thus,
(4.1)1+1x0<1+5-12=1+52<x-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 {xn}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.
Supposex-1,x0<-1. Then
(4.2)0<x1<x4<x7<⋯,⋯<x8<x6<x5<x3<x2<x0<-1,
and the subsequences {x3n}n=0∞ and {x3n+2}n=0∞ tend to -∞ and {x3n+1}n=0∞ tends to +∞.Proof.
We first have(4.3)x1=x0x-1-1>0.
Then obviously
(4.4)x2=x1x0-1<-1.
But we will prove that x2<x0<-1.
Since(x0+1)2>0, we have that x02+2x0+1>0. Thus, -1<2/x0+1/x02. Since x-1<-1, we must have x-1<2/x0+1/x02. 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 thatx3 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 thatx4>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,
(4.5)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.
By induction, it can be proved that(4.6)0<x1<x4<x7<⋯,⋯<x8<x6<x5<x3<x2<x0<-1,
that is, {x3n+1}n=0∞ is a positive increasing subsequence and {x3n}n=0∞ and {x3n+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,{x3n}n=0∞ and {x3n+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 x¯2). But by (1.1), the third increasing subsequence, {x3n+1}n=0∞, must also converge to a finite limit (which is positive), where
(4.7)x3n+1=x3nx3n-1-1=|x3n|·|x3(n-1)+2|-1.
This is impossible because there are no periodic solutions with minimal period two (see Section 2). So {x3n}n=0∞ and {x3n+2}n=0∞ are unbounded (they tend to -∞) and thus {x3n+1}n=0∞ is also unbounded (it tends to +∞) by (1.1) again.Theorem 4.3.
Suppose thatx-1<0 and x0>0. Then the solution of (1.1) has three subsequences {x3n}n=1∞, {x3n+1}n=1∞, and {x3n+2}n=1∞, where two of them tend to -∞ and one tends to +∞.Proof.
Note thatx1=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 letx-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 if0<x-1,x0<x¯1, 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{xn}n=-1∞ be a solution of (1.1). Suppose that(i)
0<x-1, x0<1;(ii)
x02x-12-2x0x-1+1-x-1>0.Then {x3n}n=2∞ is a positive increasing subsequence; {x3n+2}n=1∞ is a negative decreasing subsequence; {x3n+1}n=2∞ is a negative decreasing subsequence. Consequently, there does not exist an N≥-1 such that -1<xn<0 for all n≥N. Furthermore, all three subsequences are unbounded and so the solution is unbounded.Proof.
Let{xn}n=-1∞ be a solution of (1.1). Suppose that(i)
0<x-1, x0<1;(ii)
x02x-12-2x0x-1+1-x-1>0.Since x0x-1<1 by Hypothesis (i), then we have
(4.8)x1=x0x-1-1<0.
Then, since x1x0<0, we have
(4.9)x2=x1x0-1<-1.
We compute the sign ofx3=x2x1-1:
(4.10)x3>0⇔x2x1-1>0⇔(x1x0-1)x1-1>0⇔x12x0-x1-1>0⇔(x0x-1-1)2x0-(x0x-1-1)-1>0⇔x03x-12-2x02x-1+x0-x0x-1+1-1>0⇔x02x-12-2x0x-1+1-x-1>0,
which is true by Hypotheses (i) and (ii). We next have
(4.11)x4=x3x2-1<-1
since x3>0, x2<0. Then, also,
(4.12)x5=x4x3-1<-1
since x4<0, x3>0.
Withx4, x5<-1 replacing x-1,x0<-1 in Theorem 4.2, we have that
(4.13)0<x6<x9<x12<⋯,⋯<x13<x11<x10<x8<x7<x5<-1.
Hence,{x3n}n=2∞ is a positive increasing subsequence, {x3n+1}n=2∞ is a negative decreasing subsequence, and {x3n+2}n=1∞ is a negative decreasing subsequence, and so there does not exist an N≥-1 such that -1<xn<0 for all n≥N. Furthermore, by Theorem 4.2, {x3n}n=2∞ tends to +∞ and {x3n+1}n=2∞, and {x3n+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{xn}n=-1∞ be a solution of (1.1). Suppose that 1<x-1,x0<x¯1. Then one of the following occurs.(i)
The solution is unbounded.(ii)
There existsn0≥1 such that xn∈(-1,0) for all n≥n0.Proof.
Let{xn}n=-1∞ be a solution of (1.1). Assume that
(4.14)1<x-1,x0<x¯1=1+52.
We will show that the solution is decreasing while the terms are positive. Since 1<x0<(1+5)/2, we have
(4.15)1x0>21+5=5-12.
Thus
(4.16)1+1x0>1+21+5=1+5-12>x-1,
which, since x0>0, implies that
(4.17)x0>x0x-1-1=x1.
Observe then that we have a decreasing sequence of terms as long as the terms are positive. Since our decreasing sequence is bounded above byx¯1 and since x¯2 is less than zero, this sequence is not bounded below by zero and thus crosses over to negative values.
LetxN be the last positive term and xN+1 the first negative term of our solution. We now determine ranges of values of terms of our solution beginning with xN and xN+1, using “n” in place of “N” for convenience.
(1)
xn and xn+1: observe that
(4.18)xnxn-1-1>-1⇔xnxn-1>0,
which is true. Hence, -1<xn+1<0. Note that this, in turn, implies that 0<xnxn-1<1. Then at least 0<xn<1 since xn<xn-1.(2)
xn+2: by Statement (1), -1<xn+1xn<0 and so -2<xn+2<-1.(3)
xn+3,xn+4,….Case 1 (xn+3>0).
Then from Statement (2), xn+4=xn+3xn+2-1<-1 and thus xn+5=xn+4xn+3-1<-1. By Theorem 4.2, the solution is unbounded.Case 2 (xn+3<0).
We wish to show thatxn+3>-1:
(4.19)xn+3=xn+2xn+1-1>-1⇔xn+2xn+1>0,
which is true since, by Statements (1) and (2), xn+1,xn+2<0.We next show that -1<xn+4<0. First we show that xn+4<0:
(4.20)xn+4=xn+3xn+2-1<0⇔(xn+2xn+1-1)xn+2-1<0⇔(xn+12xn-xn+1-1)(xn+1xn-1)-1<0⇔xn+13xn2-2xn+12xn+xn+1-xn+1xn+1-1<0⇔xn+13xn2-2xn+12xn+xn+1-xn+1xn<0⇔xn+12xn2-2xn+1xn+1-xn>0⇔xn+1xn(xn+1xn-2)-(xn-1)>0⇔(xn+22-1)-(xn-1)>0⇔xn+22>xn,
which is true since, by Statements (1) and (2), xn+22>1 and 0<xn<1.
Second, we show thatxn+4>-1: notice that
(4.21)xn+4=xn+3xn+2-1>-1⇔xn+3xn+2>0,
which is true since xn+3<0 and xn+2<0 by Statement (2). Because -1<xn+3,xn+4<0, it follows from Theorem 2.5 that there exists n0≥1 such that xn∈(-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)
xn+1=xnxn-1-c, n=0,1,…,  where c is any real.(ii)
xn+1=xnxn-1-cn, n=0,1,…,  where {cn}n=0∞ is a periodic sequence of real numbers.(iii)
xn+1=xn-ℓxn-k-1, n=0,1,…,  where ℓ,k∈{0,1,…}.



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