# Global Behavior ofxn+1=(α+βxn-k)/(γ+xn)

Authors: Chenquan Gan; Xiaofan Yang; Wanping Liu
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

This paper aims to investigate the global stability of negative solutions of the difference equationxn+1=(α+βxn-k)/(γ+xn), n=0,1,2,…, where the initial conditions x-k,…,x0∈-∞,0,k is a positive integer, and the parameters β,γ<0,α>0. By utilizing the invariant interval and periodic character of solutions, it is found that the unique negative equilibrium is globally asymptotically stable under some parameter conditions. Additionally, two examples are given to illustrate the main results in the end.


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## Body

## 1.  Introduction

The study of nonlinear difference equations has always attracted a considerable attention (see, e.g., [1–30] and the references cited therein). In particular, some references investigated the dynamical behavior of positive solutions of difference equations (see, e.g., [3–5, 11]), and some references examined the dynamical behavior of negative solutions of some difference equations (see, e.g., [6, 7]).Gibbons et al. [4] studied the behavior of nonnegative solutions to the recursive sequence
(1)xn+1=α+βxn-1γ+xn,n=0,1,…,
with α,β,γ≥0, and also presented an open problem, which had been solved by Stevic´ in [8]. Kulenovic´ and Ladas, in addition, considered (1) in their book [9].Stevi
c
´ [10] considered the behavior of nonnegative solutions of the following second-order difference equation
(2)xn+1=α+βxn-11+g(xn),n=0,1,…,
where g:ℝ+⋃‍{0}→ℝ is a nonnegative increasing mapping.Douraki et al. [11] studied the qualitative behavior of positive solutions of the difference equation
(3)xn+1=p+qxn-k1+xn,n=0,1,…,
where the initial values x-k,…,x-1,x0∈(0,+∞), k is a positive integer, and p,q≥0. Moreover, (3) is a special case of the following open problem (see also in [11]), which was proposed by Kulenovic´ and Ladas in [9].Open Problem (equation(6.97) in [9]).
Assume thatp,q∈[0,∞). Investigate the global behavior of positive solutions of (3).Stevi
c
´ [12] considered the boundedness, oscillatory behavior, and global stability of nonnegative solutions of the difference equation
(4)xn+1=α+βxn-kf(xn,…,xn-k+1),n=0,1,…,
where k∈ℕ,α,β≥0, and f:ℝ+k→ℝ+ is a continuous function nondecreasing in each variable such that f(0,…,0)>0.It is worthwhile to note that the above mentioned references ([4, 10–12]), especially [4, 11], only discussed the dynamical behavior of positive solutions of difference equation. Furthermore, inspired by the above work and [6, 7], the main goal of this paper is to study the global behavior of negative solutions of the difference equation
(5)xn+1=α+βxn-kγ+xn,n=0,1,…,
where k is a positive integer, α>0,β,γ<0, and the initial conditions x-k,…,x0∈(-∞,0).In fact, it is easy to see that (5) is an extension of an open problem introduced by Kulenovic´ and Ladas in [9] and also is a special case of (4) by a simple change. However, here we establish some results regarding the global stability, invariant interval, and periodic character of negative solutions of (5).


## 2. Linearized Stability and Period 2 Solutions

The aim of this section is to discuss the local stability of the unique negative equilibrium of (5). The period 2 solutions of (5), in addition, will be verified.In this section, we need the following lemma.Lemma 1 (see [3]).
Assume thata,b∈ℝ and k∈{1,2,…}. Then
(6)|a|+|b|<1
is a sufficient condition for the asymptotic stability of the difference equation
(7)xn+1-axn+bxn-k=0,n=0,1,….
Suppose in addition that one of the following two cases holds.(a)
kis odd andb<0.(b)
kis even andab<0.Then (6) is also a necessary condition for the asymptotic stability of (7).Assume thatx¯ is an equilibrium of (5). Then it satisfies the equation x¯2+(γ-β)x¯-α=0, which implies that the unique negative equilibrium of (5) is
(8)x¯=β-γ-(β-γ)2+4α2.Let(9)xn+1=f(xn,xn-1,…,xn-k)=α+βxn-kγ+xn.
Then the linearized equation of (5) at x¯ is
(10)un+1=p0un+p1un-1+⋯+pkun-k,
where
(11)pi=∂f∂xn-i(x¯,x¯,…,x¯).
Straightforward calculations yield
(12)un+1+x¯γ+x¯un-βγ+x¯un-k=0.From here and Lemma1, we obtain the following result.Theorem 2.
The unique negative equilibriumx¯ of (5) is locally asymptotically stable if γ<β.Proof.
Ifγ<β, then
(13)|x¯γ+x¯|+|-βγ+x¯|=β+x¯γ+x¯<1.
It follows from Lemma 1 that x¯ is locally asymptotically stable. Thus the proof is complete.Now, we will examine the existence of period 2 solutions of (5).Theorem 3.
Let{xn}n=-k∞ be a negative solution of (5). Then the following statements hold.(a)
Assume thatk is odd. Then,
(i)
equation (5) has a negative prime period 2 solution …,w,r,w,r,… if and only if β=γ;(ii)
ifβ=γ, then the values w and r of all negative prime period 2 solutions are given by {w,r∈(-∞,0):wr=α};(iii)
ifβ≠γ, then (5) has no negative solutions with prime period 2.(b)
Assume thatk is even. Then (5) has no negative solutions with prime period 2.Proof.
Let{xn}n=-k∞ be a negative solution of (5).  (a)
Assume thatk is odd. Then,
(i)
suppose that…,w,r,w,r,… is a negative prime period 2 solution. Then from (5), we have
(14)w=α+βwγ+r,r=α+βrγ+w.
From the above relations, we derive that
(15)(w-r)(β-γ)=0.
Since w≠r, then β=γ.The reverse part is clear by a simple computation. So it is omitted.(ii)
Applying the relations (14), we get that w(γ+r)=α+βw and r(γ+w)=α+βr; namely, w(γ-β)+wr=α and r(γ-β)+wr=α. As β=γ, then wr=α.(iii)
Ifβ≠γ, then it follows from (15) that w=r.(b)
Assume thatk is even. Let …,w,r,w,r,…(w≠r) be a negative prime period 2 solution; then
(16)w=α+βrγ+r,r=α+βwγ+w.
It follows from the above equations that (w-r)(β+γ)=0. As β+γ<0, then w=r, which contradicts the hypothesis that w≠r.


## 3. Invariant Interval

In this section, we will consider the invariant interval of negative solutions of (5).Let(17)f(u,v)=α+βvγ+u.Lemma 4.
The following statements are true.(a)
Assume thatγ∈(-∞,(β-β2+4α)/2]. Then,
(i)
γ<β<0;(ii)
ifβ<-α/2, then βγ>α and x¯∈(β,0)⊂[γ,0].(b)
Assume thatf(u,v) is defined by (17) and u,v∈(-∞,0]. Then f(u,v) is nonincreasing in u and nondecreasing in v.Proof.
The proofs of (a) and (b) are as follows:(a)
ifγ∈(-∞,(β-β2+4α)/2], then
(ii)
γ-β≤(β-β2+4α)/2-β=-(β+β2+4α)/2<0, clearly, γ<β<0;(iii)
the conditionβ<-α/2 implies that α/β2<2. Then
(18)|(β-β2+4α)/2||α/β|=β(β-β2+4α)2α=-2ββ+β2+4α=1((-1/2)+(1/4)+(α/β2))>1(-1/2+1/4+2)=1,
which leads to γ≤(β-β2+4α)/2<α/β. Note that γ<β<0; thus βγ>α. Also
(19)x¯-β=2(βγ-α)(β-γ)2+4α-(β+γ)>0.
Clearly, x¯∈(β,0)⊂[γ,0].(b)
Note that(20)∂f∂u=-α+βv(γ+u)2<0,∂f∂v=βγ+u>0.
So the result holds.Theorem 5.
Assume thatγ∈(-∞,(β-β2+4α)/2]. Then [γ,0] is an invariant interval of (5).Proof.
Suppose that{xn}n=-k∞ is a solution to (5) with initial conditions x-k,…,x0∈[γ,0]. Since γ∈(-∞,(β-β2+4α)/2], it follows by a direct computation that α+βγ-γ2≤0. By Lemma 4(b), we immediately get that the function f(u,v) is nonincreasing in u and nondecreasing in v with u,v∈(-∞,0). Then
(21)x1=α+βx-kγ+x0=f(x0,x-k)<f(x0,0)<f(γ,0)=α2γ<0,x1=α+βx-kγ+x0=f(x0,x-k)>f(0,x-k)>f(0,γ)=α+βγγ≥γ,
which implies that x1∈[γ,0]. It follows by induction that xn∈[γ,0] for all n≥1. Thus, the proof is complete.


## 4. Global Stability

Recall that (5) has a unique negative equilibrium x¯, which is locally asymptotically stable if γ<β by Theorem 2. In this section, we will show that (a) x¯ is globally asymptotically stable under the conditions γ∈(-∞,(β-β2+4α)/2] and β<-α/2; (b) every negative solution converges to x¯ when βγ=α.Lemma 6 (see [3]).
Consider the difference equation(22)yn+1=f(yn,yn-k),n=0,1,…,
where k∈{1,2,…}. Let I=[a,b] be some interval of real numbers and assume that
(23)f:[a,b]×[a,b]⟶[a,b]
is a continuous function satisfying the following properties:  (a)
f(u,v) is nonincreasing in u and nondecreasing in v;(b)
if(m,M)∈[a,b] is a solution of the system
(24)m=f(M,m),M=f(m,M),thenm=M.Then (22) has a unique equilibrium y¯ and every solution of (22) converges to y¯.Theorem 7.
Assume thatγ∈(-∞,(β-β2+4α)/2] and β<-α/2. Then the following statements are true:  (i)
the unique negative equilibriumx¯=(β-γ-(β-γ)2+4α)/2 of (5) is a global attractor with a basin S=[γ,0]k+1;(ii)
the unique negative equilibriumx¯=(β-γ-(β-γ)2+4α)/2 of (5) is globally asymptotically stable.Proof.
Suppose that{xn}n=-k∞ is a solution to (5) with initial values x-k,…,x0∈S. By Lemma 4(b), we immediately get that the function f(u,v) is continuous and nonincreasing in u and nondecreasing in v on the invariant interval [γ,0]. Furthermore, let m,M∈I be a solution of the system m=f(m,M), M=f(M,m); then by Lemma 4(a)(i) and (ii), we have γ<β<0 and x¯∈(β,0)⊂[γ,0], and by Theorem 3, we obtain m=M. Finally, it follows by Lemma 6 that limn→∞xn=x¯. Hence, the proof is complete.Lemma 8.
Letβγ=α. Then (5) has no nontrivial negative periodic solutions of (not necessarily prime) period k.Proof.
Ifβγ=α, then substituting (8), it becomes that x¯=β. Suppose that {xn}n=-k∞ is a negative solution to (5) satisfying xn=xn-k for all n≥0; then
(25)xn+1=α+βxnγ+xn.
Simplifying the above equation, it follows that (γ+xn)(xn+1-β)=0. Clearly, xn+1=β=x¯ for all n≥0. The proof is complete.Theorem 9.
Assume thatβγ=α. Then every negative solution to (5) converges to the unique negative equilibrium x¯.Proof.
Assume thatβγ=α. It follows from Lemma 8 that x¯=β. Then
(26)xn+1-x¯=α+βxn-kγ+xn-β=β(xn-k-xn)γ+xn.
Case  1. If xn-k-xn<0, then xn+1<x¯, so xn-k<xn<x¯.
Case  2. If xn-k-xn≥0, then xn+1≥x¯, so x¯≤xn≤xn-k.
Next, we consider Case 1 (Case 2 is similar and thus it is omitted). Ifxn-k<xn<x¯ for all n≥0, then it is clear that for i∈{0,1,…,k-1} there exists αi such that
(27)limm→∞xmk+i=αi.
But then α0,α1,…,αk-1 is a periodic solution of (not necessarily prime) period k. By Lemma 8 the result holds.


## 5. Examples

To illustrate the main results in Section4, here we present two examples.Example 1.
Consider (1) with α=8,β=-4,γ=-6, and k=3. Then x¯=-2. As γ∈(-∞,(β-β2+4α)/2] and β<-α/2, it follows from Theorem 7(ii) that the unique negative equilibrium is globally asymptotically stable. For the initial conditions x-i=-10,i=0,1,2,3, Figure 1 exhibits how xn+1 evolves with n.Figure 1
Evolution ofxn+1 for (1) with parameters and initial condition given in Example 1.Example 2.
Consider (1) with α=12,β=-3,γ=-4, and k=3. Then x¯=-3. As βγ=α, it follows from Theorem 9 that the unique negative equilibrium is globally attractive. For the initial conditions x-i=-4,i=0,1,2,3, Figure 2 exhibits how xn+1 evolves with n.Figure 2
Evolution ofxn+1 for (1) with parameters and initial condition given in Example 2.



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