# On the Difference Equationxn+1=xnxn−2−1

Authors: Candace M. Kent; Witold Kosmala; Stevo Stević
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The long-term behavior of solutions of the following difference
equation:xn+1=xnxn−2−1, n∈ℕ0, where the initial values x−2, x−1, x0 are real numbers, is investigated in the paper.


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## Body

## 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:(1.1)xn+1=xnxn-2-1,n∈N0.The difference equation (1.1) belongs to the class of equations of the form  (1.2)xn+1=xn-kxn-l-1,n∈N0,
with particular choices of k and l, where k,l∈ℕ0. Although (1.2) looks simple, it is fascinating how its behavior changes for different choices of k and l. The cases k=0 and l=1, k=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: (1.3)x¯1=1-52,x¯2=1+52.


## 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 x-2=0, x-1=-1, x0=0 or x-2=-1, x-1=0, x0=-1.Proof.
Let(xn)n=-2∞ be a prime period-two solution of (1.1). Then x2n-2=a and x2n-1=b, for every n∈ℕ0 and for some a,b∈ℝ such that a≠b. We have x1=x0x-2-1=a2-1=b and x2=x1x-1-1=b2-1=a. From these two equations we obtain (a2-1)2-1=a or equivalently
(2.1)a(a+1)(a2-a-1)=0.
We have four cases to be considered.
Case 1.
Ifa=0, then b=-1, and we obtain the first prime period-two solution.
Case 2.
Ifa=-1, then b=0, and we obtain the second prime period-two solution.
Case 3.
Ifa=x¯1, then b=a2-1=x¯1, which is an equilibrium solution.
Case 4.
Ifa=x¯2, then b=a2-1=x¯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(xn)n=-2∞ be a prime period-three solution of (1.1). Then x3n-2=a,  x3n-1=b,  x3n=c, for every n∈ℕ0 and some a,b,c∈ℝ such that at least two of them are different. We have
(2.2)x1=x0x-2-1=ac-1=a,x2=x1x-1-1=ab-1=b,x3=x2x0-1=bc-1=c.
From (2.2) we easily see that a≠0, b≠0, and c≠0, so that
(2.3)c=1+1a,a=1+1b,b=1+1c.
From (2.3) we obtain
(2.4)c=1+bb+1=2b+1b+1⟹b=1+b+12b+1=3b+22b+1,
which implies b2-b-1=0. Hence b=x¯1 or b=x¯2. From this and (2.3) it follows that a=b=c=x¯1 or a=b=c=x¯2, from which the result follows.Theorem 2.3.
Equation (1.1) has no prime period-four solutions.Proof.
Let(xn)n=-2∞ be a prime period-four solution of (1.1) and x-2=a, x-1=b, x0=c. Then we have
(2.5)x1=x0x-2-1=ac-1,(2.6)x2=x1x-1-1=(ac-1)b-1=a,(2.7)x3=x2x0-1=ac-1=b,(2.8)x4=x3x1-1=b(ac-1)-1=c.
Thus, from (2.6) and (2.8), we have that a=c. This along with (2.7) gives
(2.9)a2-1=b,
while from (2.8) we get b(a2-1)-1=a or equivalently
(2.10)(a+1)(b(a-1)-1)=0.
Case 1.
Supposea=-1. Then b=a2-1=0 and c=a=-1, which, by Theorem 2.1, yields a period-two solution.
Supposea≠-1. If a=1, then from (2.10) we get a contradiction. If a≠1, then a2-1=b=1/(a-1), so that a(a2-a-1)=0. Hence a=0, a=x¯1, or a=x¯2.
Case 2.
Supposea=0. Then b=a2-1=-1 and c=a=0 which results in a period-two solution as proved in Theorem 2.1.
Case 3.
Supposea=x¯1. Then b=a2-1=x¯1 and c=x¯1 which is an equilibrium solution.
Case 4.
Supposea=x¯2. Then b=a2-1=x¯2 and c=x¯2 which is the second equilibrium solution. Proof is complete.


## 3. Local Stability

Here we study the local stability at the equilibrium pointsx¯1 and x¯2.Theorem 3.1.
The negative equilibrium of (1.1), x¯1, is unstable. Moreover, it is a hyperbolic equilibrium.Proof.
The linearized equation associated with the equilibriumx¯1=(1-5)/2∈(-1,0) is
(3.1)xn+1-x¯1xn-x¯1xn-2=0.
Its characteristic polynomial is
(3.2)Px¯1(λ)=λ3-x¯1λ2-x¯1.
Hence
(3.3)Px¯1′(λ)=3λ2-2x¯1λ=λ(3λ-2x¯1).
Since
(3.4)Px¯1(-2)=-8-5x¯1<0,Px¯1(-1)=-1-2x¯1=5-2>0,
there is a zero λ1∈(-2,-1) of Px¯1.
On the other hand, fromPx¯1(0)=-x¯1>0 and (3.3), it follows that λ1 is a unique real zero of Px¯1. Hence, the other two roots λ2,3 are conjugate complex.
Since(3.5)λ1|λ2|2=x¯1,
we obtain |λ2|=|λ3|<1. From this, the theorem follows.Theorem 3.2.
The positive equilibrium of (1.1), x¯2, is unstable. Moreover, it is also a hyperbolic equilibrium.Proof.
The linearized equation associated with the equilibriumx¯2=(1+5)/2∈(1,2) is
(3.6)xn+1-x¯2xn-x¯2xn-2=0.
Its characteristic polynomial is
(3.7)Px¯2(λ)=λ3-x¯2λ2-x¯2.
We have(3.8)Px¯2′(λ)=3λ2-2x¯2λ=λ(3λ-2x¯2).
Since
(3.9)Px¯2(2)=8-5x¯2=(11-55)2<0,Px¯2(3)=27-10x¯2=22-55>0,
there is a zero λ1∈(2,3) of Px¯2.
From this and sincePx¯2(0)=-x¯2<0, we have that λ1 is a unique real zero of Px¯2. Thus, the other two roots λ2,3 are conjugate complex.
Since(3.10)λ1|λ2|2=x¯2,
we obtain |λ2|=|λ3|<1. From this, the theorem follows.


## 4. Casex-2,x-1,x0∈(-1,0)

This section considers the solutions of (1.1) with x-2,x-1,x0∈(-1,0). Before we formulate the main result in this section we need some auxiliary results.Lemma 4.1.
Suppose thatx-2,x-1,x0∈(-1,0). Then the solution  (xn)n=-2∞ of  (1.1) is such that  xn∈(-1,0) for  n≥-2.Proof.
We havex-2,x0∈(-1,0) which implies x1=x0x-2-1∈(-1,0). Assume that we have proved xn∈(-1,0) for -2≤n≤k, for some k≥2. Then we have xk+1=xkxk-2-1∈(-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 Lemma4.1.
Supposeki∈ℕ, 1≤i≤2m, x-s,…,x-1,x0∈(-1,0), where s=max{k1,…,k2m}. Then the solution (xn)n=-s∞ of the difference equation
(4.1)xn+1=∏i=12mxn-ki-1,n∈N0,
is such that xn∈(-1,0) for n≥-s.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 (4.2)x2n+3=x2n+2x2n-1,x2n+2=x2n+1x2n-1-1,n∈N0.
Then we have the following:(4.3)x2n+3=(x2n+1x2n-1-1)(x2n-1x2n-3-1)-1=x2n-1(x2n+1x2n-1x2n-3-x2n+1-x2n-3),n∈N,and similarly(4.4)x2n+4=x2n(x2n+2x2nx2n-2-x2n+2-x2n-2),n∈N.Hence, the subsequencesyn=x2n+1 and zn=x2n+2 satisfy the difference equation  (4.5)un+1=un-1(unun-1un-2-un-un-2),n∈N,
and un∈(-1,0).For convenience, we make another change of variablevn=-un. Then (4.5) becomes  (4.6)vn+1=vn-1(vn+vn-2-vnvn-1vn-2),n∈N.
Note also that vn∈(0,1).It is easy to see that (4.6) has the following four equilibria: (4.7)v¯0=-1+52,v¯1=0,v¯2=5-12,v¯3=1.If we let(4.8)f(u,v,w)=v(u+w-uvw),
where u,v,w∈(0,1), then we find the following:(1)
fu=v-v2w=v(1-vw)>0,(2)
fv=u+w-2vuw=u(1-vw)+w(1-vu)>0,(3)
fw=v-v2u=v(1-vu)>0.Thus, the functionf(u,v,w) is strictly increasing in each argument.Lemma 4.3.
Let(xn)n=-2∞ be a solution of  (1.1) which is not equal to the equilibrium solution
(4.9)x¯1=1-52
of the equation. Suppose that
(4.10)x-2,x-1,x0∈(-1,0)
and that one of the following conditions holds:  (H1)
x-2≤x¯1, x-1≥x¯1, x0≤x¯1 with at least one of the inequalities strict,(H2)
x-2≥x¯1, x-1≤x¯1, x0≥x¯1 with at least one of the inequalities strict.
Thenxn∈(-1,0), for every n≥-2 and
(a)
if(H1) holds, then there is an N∈ℕ0 such that(4.11)x2n+1>x¯1,x2n+2<x¯1,n≥N,(b)
if(H2) holds, then there is an N∈ℕ0 such that(4.12)x2n+1<x¯1,x2n+2>x¯1,n≥N.Proof.
By Lemma4.1 we have that xn∈(-1,0), for n≥-2. We will prove only (a). The proof of (b) is dual and is omitted. Since -1<x0,x-2≤x¯1, we have
(4.13)x1=x0x-2-1≥x¯12-1=x¯1.
From this and since x¯1≤x-1<0, we have
(4.14)x2=x1x-1-1≤x¯12-1=x¯1.
If x-2<x¯1 or x0<x¯1, then inequality (4.13) is strict and, consequently, inequality (4.14) is strict too. If x-2=x0=x¯1, then x-1>x¯1, from which it follows that inequality (4.14) is strict. In this case we have
(4.15)x3=x2x0-1>x¯12-1=x¯1,
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 N≤n≤k and that N=0. The case N=1 is proved similarly and so is omitted. Then we have
(4.16)x2k+3=x2k+2x2k-1>x¯12-1=x¯1.
From this and since x¯1<x2k+1<0, we have
(4.17)x2k+4=x2k+3x2k+1-1<x¯12-1=x¯1.
Hence by induction the lemma follows.Theorem 4.4.
Let(xn)n=-2∞ be a solution of (1.1) which is not equal to the equilibrium solution
(4.18)x¯1=1-52
of the equation. Suppose that
(4.19)x-2,x-1,x0∈(-1,0)
and that one of the conditions, (H1) or (H2), holds.
Thenxn∈(-1,0), for every n≥-2, and (xn)n=-2∞ converges to a two-cycle{-1,0}.Proof.
First of allxn∈(-1,0), for n≥-2, by Lemma 4.1. We next show that (xn)n=-2∞ converges to a two-cycle{-1,0}. To this end we show that one of the subsequences, (x2n)n=-1∞ or (x2n+1)n=-1∞, converges to 0 and the other one to -1. Showing this, in turn, is equivalent to showing the following:(a)
the corresponding solution(vn)n=-1∞ of (4.6) converges to v¯3 if for some N≥-1, vn>v¯2, for n≥N, where vn∈(0,1)=(v¯1,v¯3) for n≥-1,(b)
the corresponding solution(vn)n=-1∞ of (4.6) converges to v¯1 if for some N≥-1,vn<v¯2, for n≥N, where vn∈(0,1)=(v¯1,v¯3) for n≥-1.
We prove (a). The proof of (b) is similar and will be omitted. We have(4.20)vn∈(v¯2,v¯3),n≥N.
Let
(4.21)I=lim⁡inf⁡n→∞vn,S=limsup⁡n→∞vn.
Then we have
(4.22)v¯2≤I≤S≤v¯3.
First assumeI=v¯2. From (4.20) it follows that there is an ɛ>0 such that I+ɛ<vN,vN+1,vN+2<v¯3. By the monotonicity of f and since
(4.23)f(x,x,x)>xforx∈(v¯2,v¯3),
we have
(4.24)vN+3=f(vN+2,vN+1,vN)>f(I+ɛ,I+ɛ,I+ɛ)>I+ɛ.
From this and by induction we obtain vn>I+ɛ, n≥N, which implies liminfn→∞vn≥I+ɛ, which is a contradiction.
Now assumeI∈(v¯2,v¯3). Let (vnk)k∈ℕ be a subsequence of (vn)n=-1∞ such that limk→∞vnk=I. Then there is a subsequence of (vnk)k∈ℕ, which we may denote the same, such that there are the following limits: limk→∞vnk-1, limk→∞vnk-2, and limk→∞vnk-3, which we denote, respectively, by K-i, i=1,2,3. From this and by (4.23) we have that
(4.25)f(K-1,K-2,K-3)=I<f(I,I,I).
Hence there is an i0∈{1,2,3} such that K-i0<I. Otherwise, K-i≥I for i=1,2,3 and by the monotonicity of f we would get
(4.26)f(I,I,I)≤f(K-1,K-2,K-3)=I<f(I,I,I),
which is a contradiction. On the other hand, K-i0<I contradicts the choice of I. Hence I cannot be in the interval (v¯2,v¯3).
From all of the above we have thatv¯3=I≤S≤v¯3. Therefore, limn→∞vn=v¯3, as desired.Theorem 4.5.
Assume that for a solution(xn)n=-2∞ of (1.1) there is an N≥-1 such that
(4.27)-1<xN<xN+2<0,0>xN-1>xN+1>xN+3>-1.
Then the solution converges to a two-cycle{-1,0} or to the equilibrium x¯1.Proof.
First note that by Lemma4.1 we have xn∈(-1,0), n≥N. From (1.1) we obtain the identity
(4.28)xn+4-xn+2=xn+1(xn+3-xn-1).
Applying (4.28) for n=N and using the fact xN+1∈(-1,0), we get 0>xN+4>xN+2. Hence
(4.29)xN<xN+2<xN+4<0,xN-1>xN+1>xN+3>-1.
Using induction along with identity (4.28) it is shown that
(4.30)xN<xN+2<⋯<xN+2k<0,xN-1>xN+1>⋯>xN+2k+1>-1,
for every k∈ℕ. Hence, there are finite limits limk→∞xN+2k and limk→∞xN+2k+1, say l1 and l2. Letting k→∞ in the relations
(4.31)xN+2k+2=xN+2k+1xN+2k-1-1,xN+2k+3=xN+2k+2xN+2k-1,
we get l1=l22-1 and l2=l12-1. Hence
(4.32)(l1-l2)(l1+l2+1)=0.
From this we have l1=l2=x¯1, or if l1≠l2, then l1+l2=-1 so that l1=0 and l2=-1.Remark 4.6.
Letx-2=a, x-1=b and x0=c with a,b,c∈(-1,0). For N=-1, (4.27) will be
(4.33)x-2>x0⟺a>c,x1-x-1=x0x-2-1-x-1=ac-1-b>0,x2-x0=x1x-1-1-x0=x0x-1x-2-x-1-x0-1=abc-b-c-1<0.Hence, under the conditions
(4.34)a>c,ac>b+1,abc<b+c+1,
we have that (4.27) is satisfied for N=-1. It is easy to show that there are some a,b,c∈(-1,0) such that the set in (4.34) is nonempty.Note also that in the proof of Theorem4.5 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 x¯1. The next theorem, which is a product of an E-mail communication between Stević 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 x¯1.Proof.
In order to find a solution tending tox¯1, we make the substitution xn=yn+x¯1, yielding the equation
(4.35)yn+3-x¯1(yn+2+yn)=yn+2yn,n≥-2,
and for n∈ℕ0 we make the ansatz
(4.36)yn=∑k=0∞∑l=0∞aklpnkqnl,
with a00=0, where p and q are the conjugate complex zeros of the characteristic polynomial
(4.37)P(λ)=λ3-x¯1(λ2+1).
Note that |p|=|q|=r≈0.74448.
Replacing (4.36) into (4.35) and comparing the coefficients, we find
(4.38)dklakl=∑i=0k∑j=0,j+i≠0laijp2iq2jak-i,l-j,
with
(4.39)dkl=p3kq3l-x¯1(p2kq2l+1).
Equation (4.38) is satisfied for k+l≤1, where a10 and a01 are arbitrary, so that it suffices to consider (4.38) for k and l such that k+l>1. If a10 and a01 are chosen to be conjugate complex numbers, then according to (4.38) all akl are conjugate complex numbers to alk and consequently series (4.36) is real. We look for a solution (4.36) with a10=a01¯, |a10|=1, and determine a positive constant λ such that
(4.40)|akl|≤λk+l-1.
Since the inequality is valid for k+l≤1, by induction, we get from (4.38)
(4.41)|akl|≤λk+l-21|dkl|∑i=0k∑j=0,j+i≠0lr2(i+j).
Note that(4.42)∑i=0k∑j=0,j+i≠0lr2(i+j)<1(1-r2)2-1=2r2-r4(1-r2)2.
It is not difficult to check that(4.43)D∶=sup⁡k+l≥21|dkl|=1|d21|≈2.095.
Hence (4.40) holds with
(4.44)λ=D2r2-r4(1-r2)2.
For such chosenλ the series in (4.36) converges if λrn<1, which implies n>lnλ/ln(1/r). We have λ≈8.450, so that lnλ/ln(1/r)≈7.233, 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 n→∞, that is, a solution of (1.1) converging to x¯1, as desired.Remark 4.8.
Fora10=a01=1, the first coefficients in (4.38) are
(4.45)a20=p2d20,a11=p2+q2d11,a02=q2d02,a30=p2(p2+1)a20d30,a21=(p4+q2)a20+p2(q2+1)a11d21.Remark 4.9.
If we replacen by n+c in (4.35) with an arbitrary c∈ℝ, 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(5.1)min⁡{|x-2|,|x-1|,|x0|}>x¯2=1+52.
Then
(5.2)x¯2<|x0|<|x1|<|x2|<⋯<|xn|<⋯.Proof.
From the hypothesis we have that|x-2|-1>x¯2-1, and so |x0|(|x-2|-1)>x¯2(x¯2-1)=1. Therefore, |x0||x-2|-|x0|>1, and so |x0||x-2|-1>|x0|. On the other hand, we have
(5.3)|x1|=|x0x-2-1|>|x0||x-2|-1.
Combining the last two inequalities, we have that |x1|>|x0|>x¯2. Assume that we have proved
(5.4)x¯2<|x0|<|x1|<|x2|<⋯<|xk|,
for some k∈ℕ. We have |xk-2|-1>x¯2-1, which implies |xk|(|xk-2|-1)>x¯2(x¯2-1)=1, or equivalently |xk||xk-2|-1>|xk|. From this and (1.1), we get
(5.5)|xk+1|=|xkxk-2-1|>|xk||xk-2|-1>|xk|>x¯2,
finishing the inductive proof of the theorem.Corollary 5.2.
Assume that the initial values of a solution(xn)n=-2∞ of (1.1) satisfy the condition
(5.6)min⁡{x-2,x-1,x0}>x¯2=1+52.
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 x0. We have a contradiction. The proof is complete.For our next result, we need to introduce the following definition.Definition 5.3.
Let(xn)n=-2∞be a solution of (1.1), and leti∈{1,2}.Then we say that the solution has the eventual semicycle pattern k+,l- (or k-,l+) if there existsN∈ℕsuch that, forn∈ℕ0,xN+n(k+l)+1,…,xN+n(k+l)+k≥x¯iandxN+n(k+l)+k+1,…,xN+n(k+l)+k+l<x¯i(or, resp.,xN+n(k+l)+1,…,xN+n(k+l)+k<x¯iandxN+n(k+l)+k+1,…,xN+n(k+l)+k+l≥x¯i).Remark 5.4.
Note that the eventual semicycle pattern can be extended tok1±,k2∓,k3±,…,kM∓ for M>2.Theorem 5.5.
Assume that(xn)n=-2∞ is a solution of (1.1) such that
(5.7)min⁡{|x-2|,|x-1|,|x0|}>x¯2=1+52
and that at least one of x-2, x-1, x0 is negative. Then
(5.8)|xn|≥x¯2,n≥-2,
and the solution is separated into seven unbounded eventually increasing subsequences such that the solution has the eventual semicycle pattern
(5.9)1+,1-,2+,3-.Proof.
Assume thatx-2,x-1,x0<-x¯2. Then we have
(5.10)x1=x0x-2-1>x¯22-1=x¯2,x2=x1x-1-1<-x¯22-1=-x¯2-2<-x¯2,x3=x2x0-1>x¯22-1=x¯2,x4=x3x1-1>x¯22-1=x¯2,x5=x4x2-1<-x¯22-1=-x¯2-2<-x¯2,x6=x5x3-1<-x¯22-1=-x¯2-2<-x¯2,x7=x6x4-1<-x¯22-1=-x¯2-2<-x¯2.
Hence |xi|>x¯2, for -2≤i≤7 and x5,x6,x7<-x¯2-2<-x¯2<0. An inductive argument shows that
(5.11)x7k+1=x7kx7k-2-1>x¯22-1=x¯2,x7k+2=x7k+1x7k-1-1<-x¯22-1=-x¯2-2<-x¯2,x7k+3=x7k+2x7k-1>x¯22-1=x¯2,x7k+4=x7k+3x7k+1-1>x¯22-1=x¯2,x7k+5=x7k+4x7k+2-1<-x¯22-1=-x¯2-2<-x¯2,x7k+6=x7k+5x7k+3-1<-x¯22-1=-x¯2-2<-x¯2,x7k+7=x7k+6x7k+4-1<-x¯22-1=-x¯2-2<-x¯2,
for each k∈ℕ0, 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 Theorem5.1, we see that the sequences (x7k+i)k=0∞ monotonically tend to -∞ or +∞ with the aforementioned eventual semicycle pattern.



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*Source: 815285-2011-03-30.xml*