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.

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: xn+1=xnxn-2-1,n∈N0.

The difference equation (1.1) belongs to the class of equations of the form 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: 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
a(a+1)(a2-a-1)=0.
We have four cases to be considered.

Case 1.

If a=0, then b=-1, and we obtain the first prime period-two solution.

Case 2.

If a=-1, then b=0, and we obtain the second prime period-two solution.

Case 3.

If a=x¯1, then b=a2-1=x¯1, which is an equilibrium solution.

Case 4.

If a=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
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
c=1+1a,a=1+1b,b=1+1c.
From (2.3) we obtain
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
x1=x0x-2-1=ac-1,x2=x1x-1-1=(ac-1)b-1=a,x3=x2x0-1=ac-1=b,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
a2-1=b,
while from (2.8) we get b(a2-1)-1=a or equivalently
(a+1)(b(a-1)-1)=0.

Case 1.

Suppose a=-1. Then b=a2-1=0 and c=a=-1, which, by Theorem 2.1, yields a period-two solution.

Suppose a≠-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.

Suppose a=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.

Suppose a=x¯1. Then b=a2-1=x¯1 and c=x¯1 which is an equilibrium solution.

Case 4.

Suppose a=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 points x¯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 equilibrium x¯1=(1-5)/2∈(-1,0) is
xn+1-x¯1xn-x¯1xn-2=0.
Its characteristic polynomial is
Px¯1(λ)=λ3-x¯1λ2-x¯1.
Hence
Px¯1′(λ)=3λ2-2x¯1λ=λ(3λ-2x¯1).
Since
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, from Px¯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
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 equilibrium x¯2=(1+5)/2∈(1,2) is
xn+1-x¯2xn-x¯2xn-2=0.
Its characteristic polynomial is
Px¯2(λ)=λ3-x¯2λ2-x¯2.

We have
Px¯2′(λ)=3λ2-2x¯2λ=λ(3λ-2x¯2).
Since
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 since Px¯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
λ1|λ2|2=x¯2,
we obtain |λ2|=|λ3|<1. From this, the theorem follows.

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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 that x-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 have x-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 Lemma 4.1.

Suppose ki∈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
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 x2n+3=x2n+2x2n-1,x2n+2=x2n+1x2n-1-1,n∈N0.
Then we have the following: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 x2n+4=x2n(x2n+2x2nx2n-2-x2n+2-x2n-2),n∈N.

Hence, the subsequences yn=x2n+1 and zn=x2n+2 satisfy the difference equation un+1=un-1(unun-1un-2-un-un-2),n∈N,
and un∈(-1,0).

For convenience, we make another change of variable vn=-un. Then (4.5) becomes 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: v¯0=-1+52,v¯1=0,v¯2=5-12,v¯3=1.

If we let f(u,v,w)=v(u+w-uvw),
where u,v,w∈(0,1), then we find the following:

fu=v-v2w=v(1-vw)>0,

fv=u+w-2vuw=u(1-vw)+w(1-vu)>0,

fw=v-v2u=v(1-vu)>0.

Thus, the function f(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
x¯1=1-52
of the equation. Suppose that
x-2,x-1,x0∈(-1,0)
and that one of the following conditions holds:

x-2≤x¯1, x-1≥x¯1, x0≤x¯1 with at least one of the inequalities strict,

x-2≥x¯1, x-1≤x¯1, x0≥x¯1 with at least one of the inequalities strict.

Then xn∈(-1,0), for every n≥-2 and

if (H1) holds, then there is an N∈ℕ0 such that

x2n+1>x¯1,x2n+2<x¯1,n≥N,

if (H2) holds, then there is an N∈ℕ0 such that

x2n+1<x¯1,x2n+2>x¯1,n≥N.Proof.

By Lemma 4.1 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
x1=x0x-2-1≥x¯12-1=x¯1.
From this and since x¯1≤x-1<0, we have
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
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
x2k+3=x2k+2x2k-1>x¯12-1=x¯1.
From this and since x¯1<x2k+1<0, we have
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
x¯1=1-52
of the equation. Suppose that
x-2,x-1,x0∈(-1,0)
and that one of the conditions, (H1) or (H2), holds.

Then xn∈(-1,0), for every n≥-2, and (xn)n=-2∞ converges to a two-cycle{-1,0}.

Proof.

First of all xn∈(-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:

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,

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
vn∈(v¯2,v¯3),n≥N.
Let
I=liminfn→∞vn,S=limsupn→∞vn.
Then we have
v¯2≤I≤S≤v¯3.

First assume I=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
f(x,x,x)>xforx∈(v¯2,v¯3),
we have
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 assume I∈(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
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
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 that v¯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
-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 Lemma 4.1 we have xn∈(-1,0), n≥N. From (1.1) we obtain the identity
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
xN<xN+2<xN+4<0,xN-1>xN+1>xN+3>-1.
Using induction along with identity (4.28) it is shown that
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
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
(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.

Let x-2=a, x-1=b and x0=c with a,b,c∈(-1,0). For N=-1, (4.27) will be
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
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 Theorem 4.5 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 to x¯1, we make the substitution xn=yn+x¯1, yielding the equation
yn+3-x¯1(yn+2+yn)=yn+2yn,n≥-2,
and for n∈ℕ0 we make the ansatz
yn=∑k=0∞∑l=0∞aklpnkqnl,
with a00=0, where p and q are the conjugate complex zeros of the characteristic polynomial
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
dklakl=∑i=0k∑j=0,j+i≠0laijp2iq2jak-i,l-j,
with
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
|akl|≤λk+l-1.
Since the inequality is valid for k+l≤1, by induction, we get from (4.38)
|akl|≤λk+l-21|dkl|∑i=0k∑j=0,j+i≠0lr2(i+j).

Note that
∑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
D∶=supk+l≥21|dkl|=1|d21|≈2.095.

Hence (4.40) holds with
λ=D2r2-r4(1-r2)2.

For such chosen λ 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.

For a10=a01=1, the first coefficients in (4.38) are
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 replace n 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 (<xref ref-type="disp-formula" rid="EEq1">1.1</xref>)

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
min{|x-2|,|x-1|,|x0|}>x¯2=1+52.
Then
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
|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
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
|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
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 let i∈{1,2}. Then we say that the solution has the eventual semicycle pattern k+,l- (or k-,l+) if there exists N∈ℕ such that, for n∈ℕ0, xN+n(k+l)+1,…,xN+n(k+l)+k≥x¯i and xN+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¯i and xN+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 to k1±,k2∓,k3±,…,kM∓ for M>2.

Theorem 5.5.

Assume that (xn)n=-2∞ is a solution of (1.1) such that
min{|x-2|,|x-1|,|x0|}>x¯2=1+52
and that at least one of x-2, x-1, x0 is negative. Then
|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
1+,1-,2+,3-.

Proof.

Assume that x-2,x-1,x0<-x¯2. Then we have
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
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 Theorem 5.1, we see that the sequences (x7k+i)k=0∞ monotonically tend to -∞ or +∞ with the aforementioned eventual semicycle pattern.

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