# On the Difference Equationxn+1=α+xn−m/xnk

Authors: İbrahim Yalçinkaya
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

We investigate the global behaviour of the difference
equation of higher orderxn+1=α+xn−m/xnk, n=0,1,…, where the parameters α,k∈(0,∞) and the initial values x−m,x−(m−1),…,x−2,x−1, and x0 are arbitrary positive real numbers.


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## Body

## 1. Introduction

Although
difference equations  are relatively simple in form, it is, unfortunately,
extremely difficult to understand thoroughly the global behavior of their
solutions. See, for example, [1–12] and the relevant references cited therein. Difference equations appear naturally as discrete analogues and as
numerical solutions of differential and delay differential equations having
applications in various scientific branches, such
as in ecology, economy, physics, technics, sociology, and biology.  Hamza and Morsy in [5] investigated the global behavior of the difference
equation(1.1)xn+1=α+xn−1xnk,n=0,1,…,
where the
parameters α,k∈(0,∞) and the initial values x−1 and x0 are arbitrary positive real numbers.Equation (1.1) was investigated when k=1 where α∈(0,∞) (see [1, 3]). There are some other examples of the research regarding (1.1) (e.g., [4, 8]).Yalçinkaya in [11] investigated the
global behavior of the difference equation(1.2)xn+1=α+xn−2xnk,n=0,1,…,where the parameters α,k∈(0,∞) and the initial values are arbitrary positive
real numbers.Also, in [12], we investigated the global behavior of the
difference equation(1.3)xn+1=α+xn−3xnk,n=0,1,…,where the parameters α,k∈(0,∞) and the initial values are arbitrary positive
real numbers.In this paper, we consider the following difference
equation of higher order(1.4)xn+1=α+xn−mxnk,n=0,1,…,where the parameters α,k∈(0,∞) and the initial values are arbitrary positive
real numbers.Here, we review some results which will be useful in
our investigation of the behavior of (1.4) solutions (cf. [10]).Definition 1.1.
LetI be an interval of real numbers and let f:Ik+1→I be a continuously differentiable function
where k is a nonnegative integer. Consider the
difference equation(1.5)xn+1=f(xn,xn−1,…,xn−k),n=0,1,…,with the
initial values x−k,…,x0∈I.
A point x¯ is called an equilibrium point of (1.5)
if(1.6)x¯=f(x¯,x¯,…,x¯).Definition 1.2.
Letx¯ be an equilibrium point of (1.5).
(a) The
equilibriumx¯ is called locally stable if for every ε>0,
there exists δ>0 such that x0,…,x−k∈I and |x0−x¯|+⋯+|x−k−x¯|<δ,
then(1.7)|xn−x¯|<ε,∀n≥−k.
(b) The
equilibriumx¯ is called locally asymptotically stable if it is locally stable and if there exists γ>0 such that if x0,…,x−k∈I and |x0−x¯|+⋯+|x−k−x¯|<γ,
then(1.8)limn→∞xn=x¯.
(c) The
equilibriumx¯ is called global attractor if for
every x0,…,x−k∈I,(1.9)limn→∞xn=x¯.
(d) The
equilibriumx¯ is called globally asymptotically stable if it is locally stable and is a global attractor.
(e) The
equilibriumx¯ is called unstable if is not stable.Definition 1.3.
Letai=(∂f/∂ui)(x¯,…,x¯) for each i=0,1,…,k denote the partial derivatives of f(u0,u1,…,uk) evaluated at an equilibrium x¯ of (1.5).
Then(1.10)zn+1=a0zn+a1zn−1+⋯+akzn−k,n=0,1,…is called the linearized
equation of (1.5) about the equilibrium point x¯.Theorem 1.4 (Clark's theorem).
Consider the difference equation (1.10).
Then(1.11)∑i=0k|ai|<1is a sufficient
condition for the locally asymptotically stability of (1.5).Definition 1.5.
The sequence{xn} is said to be periodic with period p if xn+p=xn for n=0,1,… (cf. [2]).


## 2. Main Results

In this
section, we investigate the global behavior, the boundedness, and some
periodicity of (1.4).A pointx¯∈ℝ is an equilibrium point of (1.4) if and only if
it is a zero for the function(2.1)g(x)=x−x1−k−α,that is,(2.2)x¯−x¯1−k−α=0.Lemma 2.1.
Equation (1.4) has a unique equilibrium point x¯>1.Proof.
Case 1.
Assume thatk=1,
then (1.4) has a unique equilibrium point x¯=α+1>1.Case 2.
Assume that0<k<1.
The function g defined by (2.1) is decreasing on [0,(1−k)1/k] and increasing on [(1−k)1/k,∞). Since g(1)=−α and limx→∞g(x)=∞,
then g has a unique zero x¯>1.Case 3.
Assume
that1<k. Since g is increasing on [0,∞),g(1)=−α and limx→∞g(x)=∞,
then g has a unique zero x¯>1.
Therefore, the proof is complete.Theorem 2.2.
Assume thatx¯ is the equilibrium point of (1.4). If k(k+1)(1−k)/k<α, then x¯ is locally asymptotically stable.Proof.
From (1.5) and (1.10), we see
that(2.3)f(u0,u1,…,um)=α+u0−kum,then(2.4)a0=−kx¯k,ai=0∀i∈{1,2,…,m−1},am=1x¯k.
By using Clark's theorem, we get thatx¯ is locally asymptotically stable if and only
if x¯k>k+1.
Letk(k+1)(1−k)/k<α, a simple calculations shows
that(2.5)g((k+1)1/k)=k(k+1)(1−k)/k−α<0,where g is defined by (2.1). Then, since limx→∞g(x)=∞,x¯>(k+1)1/k and x¯k>k+1. Therefore, the proof is complete.Lemma 2.3.
Ifα≠1, then every solution of (1.4) is bounded.Proof.
We get that(2.6)α<xn+1<α+βxn−m,∀n=1,2,…,where β=1/αk.
By induction we obtain(2.7)α<x(m+1)n+p<α1−βn1−β+βnxp,∀p∈{−(m−1),−(m−2),…,−1,0,1}.
Also, we see that ifα>1, (2.8)α<x(m+1)n+p<α11−β+xp,∀p∈{−(m−1),(−m−2),…,−1,0,1}.
Therefore, the proof is complete.Theorem 2.4.
Assume
thatx¯ is the equilibrium point of (1.4). If α>k1/k≥1, then x¯ is globally asymptotically stable.Proof.
We must show that the equilibrium
pointx¯ of (1.4) is both locally asymptotically stable and limx→∞xn=x¯.
Firstly, sincek≥1, then k≥k(k+1)(1−k)/k and since α>k1/k,
we get α>k(k+1)(1−k)/k. By Theorem 2.2, x¯ is locally asymptotically stable.
Let{xn}n=−m∞ be a solution of (1.4). By Lemma 2.3, {xn}n=−m∞ is bounded.
Let us introduce(2.9)Λ1=limn→∞infxn,Λ2=limn→∞supxn.
Then, for allε∈(0,Λ1) there exists n0∈ℕ such that for all n≥n0,
we get(2.10)Λ1−ε≤xn≤Λ2+ε.
This implies that(2.11)α+Λ1−ε(Λ2+ε)k≤xn+1≤α+Λ2+ε(Λ1−ε)kforn≥n0+1.Then, we obtain(2.12)α+Λ1−ε(Λ2+ε)k≤Λ1≤Λ2≤α+Λ2+ε(Λ1−ε)k,and from the above
inequality(2.13)α+Λ1Λ2k≤Λ1≤Λ2≤α+Λ2Λ1k,which implies
that(2.14)(αΛ2kΛ1k−1+Λ1k)≤Λ1kΛ2k≤(αΛ2k−1Λ1k+Λ2k).
Consequently, we obtain(2.15)αΛ2k−1Λ1k−1(Λ2−Λ1)≤(Λ2k−Λ1k).
Suppose thatΛ1≠Λ2,
we get that(2.16)αΛ2k−1Λ1k−1≤Λ2k−Λ1kΛ2−Λ1.
There existsγ∈(Λ1,Λ2) such that(2.17)Λ2k−Λ1kΛ2−Λ1=kγk−1≤kΛ2k−1.
This implies thatαk≤k, which is a contradiction. Hence, Λ1=Λ2=x¯. So, we have shown that(2.18)limn→∞xn=x¯.Therefore, the proof is complete.Theorem 2.5.
Suppose that m is odd, then let{xn}n=−m∞ be a positive solution of (1.4) which consists of at least two semicycles.
Then {xn}n=−m∞ is oscillatory and, except possibly for the
first semicycle, every semicycle is of length one.Proof.
Assume
thatxn−2a<x¯≤xn−(2a+1),∀a∈{0,1,2,…,(m−1)/2} for some n≥0,
then(2.19)xn+1>α+x¯xk¯=x¯,xn+2<α+x¯xk¯=x¯,xn+3>α+x¯xk¯=x¯,⋮xn+m+1<α+x¯xk¯=x¯.
Second, considerxn−(2a+1)<x¯<xn−2a,∀a∈{0,1,2,…,(m−1)/2},then(2.20)xn+1<α+x¯xk¯=x¯,xn+2>α+x¯xk¯=x¯,xn+3<α+x¯xk¯=x¯,⋮xn+m+1>α+x¯xk¯=x¯,which ends the proof.Theorem 2.6.
Equation (1.4) has a period (m+1) solution (not necessary
prime) {xn}n=−m∞ if and only if (x−m,x−m+1,…,x−2,x−1,x0) is a solution of the system(2.21)at=α+atat−1k,∀t∈{1,2,…,m+1}.
Moreover, if at least one of the initial values of (1.4) is different from the others, then {xn}n=−m∞ has a prime period (m+1) solution.Proof.
First, assume that{xn}n=−m∞ is a prime period (m+1) solution of (1.4),
then(2.22)x−m=x1=α+x−mx0k,and for all t∈{2,3,4,…,m+1},(2.23)xt−(m+1)=xt=α+xt−(1+m)xt−1k=α+xtxt−1k.
Then,(x−m,x−m+1,…,x−2,x−1,x0) is a solution of the system (2.21).
Second, assume that(x−m,x−m+1,…,x−2,x−1,x0) is a solution of the system (2.21),
then(2.24)x−m=α+x−mx0k=x1,x−(m−1)=α+x−(m−1)x−mk=α+x−(m−1)x1k=x2,x−(m−2)=α+x−(m−2)x−(m−1)k=α+x−(m−2)x2k=x3,⋮x−1=α+x−1x−2k=α+x−1xm−1k=xm,x0=α+x0x−1k=α+x0xmk=xm+1.
By induction we see that(2.25)xn+m+1=xn∀n≥−m.
In the case where at least one of the initial values
of (1.4) is different from the others, clearly {xn}n=−m∞ is a prime period (m+1) solution.



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