# Global Attractivity of a Higher-Order Difference Equation

Authors: R. Abo-Zeid
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

The aim of this work is to investigate the global stability, periodic nature, oscillation, and the boundedness of all admissible solutions of the difference equationxn+1=Axn-2r-1/(B-C∏i=lkxn-2i),n=0,1,2,… where A,B,C are positive real numbers and l,r,k are nonnegative integers, such that l≤k.


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## Body

## 1. Introduction and Preliminaries

Although some difference equations look very simple, it is extremely difficult to understand thoroughly the global behaviors of their solutions. One can refer to [1, 2]. The study of nonlinear rational difference equations of higher order is of paramount importance, since we still know so little about such equations. It is worthwhile to point out that although several approaches have been developed for finding the global character of difference equations [2–4], relatively a large number of difference equations have not been thoroughly understood yet [5–8].Aloqeili in [9] discussed the stability properties and semicycle behavior of the solutions of the difference equation:
(1.1)xn+1=xn-1a-xnxn-1,n=0,1,2,…,
with real initial conditions and positive real number a.In [10], the authors investigated the global asymptotic stability of the difference equation:
(1.2)xn+1=Axn-1B+C∏i=lkxn-2i,n=0,1,2,…,
where A, B, C are nonnegative real numbers and l,k are nonnegative integers, such that l≤k.Also in [11], they discussed the existence of unbounded solutions under certain conditions of the difference equation:
(1.3)xn+1=A∏i=lkxn-2i-1B+C∏i=lk-1xn-2i,n=0,1,2,…,
where A, B, C are nonnegative real numbers and l,k are nonnegative integers, l<kIn [12], the global asymptotic stability of the difference equation:
(1.4)xn+1=Axn-2r-1B+Cxn-2lxn-2k,n=0,1,2,…
was discussed, where A, B, C are nonnegative real numbers and r, l, k are nonnegative integers such that l≤k and r≤k.In [13], the global stability and periodic nature of the solutions of the difference equations:
(1.5)xn+1=xn-2±1+xnxn-1xn-2,n=0,1,2,…
were discussed, where the initial conditions x-2,x-1,x0 are real numbers.In [14], we discussed the oscillation, boundedness, and the global behavior of all admissible solutions of the difference equation:
(1.6)xn+1=Axn-1B-Cxnxn-2,n=0,1,2,…,
where A,B,C are positive real numbers.In this paper, we study the global asymptotic stability of the difference equation(1.7)xn+1=Axn-2r-1B-C∏i=lkxn-2i,n=0,1,2,…,
where A,B,C are nonnegative real numbers and l,r,k are nonnegative integers, such that l≤k.Consider the difference equation:(1.8)xn+1=f(xn,xn-1,…,xn-k),n=0,1,2…,
where f:Rk+1→R. An equilibrium point for (1.8) is a point x-∈R such that x-=f(x-,x-,…,x-).(1)
An equilibrium pointx- for (1.8) is called locally stable if for every ϵ>0,∃δ>0 such that every solution {xn} with initial conditions x-k,x-k+1,…,x0∈]x--δ,x-+δ[ is such that xn∈]x--ϵ,x-+ϵ[, ∀n∈N. Otherwise x- is said to be unstable.(2)
The equilibrium pointx- of (1.8) is called locally asymptotically stable if it is locally stable and there exists γ>0 such that for any initial conditions x-k,x-k+1,…,x0∈]x--γ,x-+γ[, the corresponding solution {xn} tends to x-.(3)
An equilibrium pointx- for (1.8) is called global attractor if every solution {xn} converges to x- as n→∞.(4)
The equilibrium pointx- for (1.8) is called globally asymptotically stable if it is locally asymptotically stable and global attractor.The linearized equation associated with (1.8) is
(1.9)yn+1=∑i=0k‍∂f∂xn-i(x-,…,x-)yn-i,n=0,1,2,….
The characteristic equation associated with (1.9) is
(1.10)λk+1-∑i=0k‍∂f∂xn-i(x-,…,x-)λk-i=0.Theorem 1.1 (see [2]).
Assume thatf is a C1 function and let x- be an equilibrium point of (1.8). Then the following statements are true.(1)
If all roots of (1.10) lie in the open disk |λ|<1, then x- is locally asymptotically stable.(2)
If at least one root of (1.10) has absolute value greater than one, then x- is unstable.


## 2. Linearized Stability Analysis

Consider the difference equation:(2.1)xn+1=Axn-2r-1B-C∏i=lkxn-2i,n=0,1,2,…,
where A,B,C are nonnegative real numbers and l,r,k are nonnegative integers, such that l≤k.The change of variablesxn=B/Ck-l+1yn reduces (1.7) to the difference equation:
(2.2)yn+1=γyn-2r-11-∏i=lkyn-2i,n=0,1,2,…,
where γ=A/B.Now we determine the equilibrium points of (2.2) and discuss their local asymptotic behavior. It is clear that the values of the equilibrium points depend on whether k-l is even or odd.Whenk-l is odd, we have the equilibrium points y-=0 and y-=±1-γk-l+1 if γ<1 and y-=0 only if γ≥1.Whenk-l is even, we have the equilibrium points y-=0 and y-=1-γk-l+1.Now assume that𝔎=max{2k,2r+1}.The linearized equation associated with (2.2) about y- is
(2.3)zn+1-γ1-y-k-l+1zn-2r-1-γy-k-l+1(1-y-k-l+1)2∑i=lk‍zn-2i=0,n=0,1,2,….
The characteristic equation associated with this equation is
(2.4)λK+1-γ1-y-k-l+1λK-2r-1-γy-k-l+1(1-y-k-l+1)2∑i=lk‍λK-2i=0.
We summarize the results of this section in the following two theorems.Theorem 2.1.
Assume that𝔎=2k>2r+1. Then the following statements are true.  (1)
The zero equilibrium point is locally asymptotically stable ifγ<1 and unstable (saddle point) if γ>1.(2)
Whenk-l is even, the equilibrium point y-=1-γk-l+1 is unstable if γ<1 and unstable (saddle point) if γ>1.(3)
Whenk-l is odd, the equilibrium points y-=±1-γk-l+1 are unstable.Proof.
(
1
) The linearized equation (2.3) about y-=0 is
(2.5)zn+1-γzn-2r-1=0,n=0,1,2,….
The characteristic equation associated with this equation is(2.6)λ2k+1-γλ2k-2r-1=0.
So λ=0, λ=±γ2r+2. Therefore the result follows.
(
2
) Suppose that k-l is even. The linearized equation (2.3) about y-=1-γk-l+1 is
(2.7)zn+1-zn-2r-1-1γ(1-γ)∑i=lk‍zn-2i=0,n=0,1,2,….
The associated characteristic equation (2.4) becomes
(2.8)λ2k+1-λ2k-2r-1-1γ(1-γ)∑i=lk‍λ2k-2i=0.
Let(2.9)f(λ)=λ2k+1-λ2k-2r-1-1γ(1-γ)∑i=lk‍λ2k-2i=0.
We can see thatf(λ) has a real root in (1,∞) if γ<1 and when γ>1, f(λ) has a root in (1,∞) and some roots with |λ|<1. Therefore the result follows.
(
3
) When k-l is odd, f(λ) has a root in (1,∞) and some roots with |λ|<1, if γ<1. Therefore y-=±1-γ are unstable.Theorem 2.2.
Assume that𝔎=2r+1>2k. Then the following statements are true.  (1)
The zero equilibrium point is locally asymptotically stable ifγ<1 and a source if γ>1.(2)
Ifk-l is even, then the equilibrium point y-=1-γk-l+1 is unstable (saddle point).(3)
Ifk-l is odd, then the equilibrium points y-=±1-γk-l+1 are unstable (saddle points).Proof.
It is sufficient to consider the linearized equation abouty-:
(2.10)zn+1-γ1-y-k-l+1zn-2r-1-γy-k-l+1(1-y-k-l+1)2∑i=lk‍zn-2i=0,n=0,1,2,…
and its associated characteristic equation:
(2.11)λ2r+2-γ1-y-k-l+1-γy-k-l+1(1-y-k-l+1)2∑i=lk‍λ2r+1-2i=0.


## 3. Oscillation

Lett be the largest nonnegative integer such that 0<2t+1≤𝔎 and let s be the largest nonnegative integer such that 0≤2s≤𝔎.Theorem 3.1.
Assume thatγ<1. Then the interval (-1-γk-l+1,1-γk-l+1) is an invariant interval for (2.2).Proof.
The proof is by induction. Suppose thaty-i∈(-1-γk-l+1,1-γk-l+1), i=0,1,…,𝔎. Hence |y-i|<1-γk-l+1, i=0,1,…,𝔎.
This implies that|∏i=lky-2i|<1-γ. Then
(3.1)|y1|=γ|y-2r-1||1-∏i=lky-2i|≤γ|y-2r-1||1-|∏i=lky-2i||<|y-2r-1|,|y2|=γ|y-2r||1-∏i=lky-2i+1|≤γ|y-2r||1-|∏i=lky-2i+1||<|y-2r|.
If for a certain n0∈ℕ we have yn0-𝔎,yn0-𝔎+1,…,yn0∈(-1-γk-l+1,1-γk-l+1), then
(3.2)|yn0+1|=γ|yn0-2r-1||1-∏i=lky-2i|≤γ|yn-2r-1||1-|∏i=lky-2i||<|yn0-2r-1|<1-γk-l+1.
This completes the proof.Corollary 3.2.
Assume that{yn}n=-𝔎∞ be a solution of (2.2) such that either y-𝔎,y-𝔎+1,…,y-1,y0∈(0,1-γk-l+1)(or(-1-γk-l+1,0)). Then {yn}n=-𝔎∞ is positive (or negative). Moreover, {yn}n=-𝔎∞ converges to the zero equilibrium point.Theorem 3.3.
Let{yn}n=-𝔎∞ be a nontrivial solution of (2.2) such that either(C1)
-1-γk-l+1<y-2t-1, y-2t+1,…,y-1<0<y-2s, y-2s+2,…,y0<1-γk-l+1or(C2)
-1-γk-l+1<y-2s, y-2s+2,…,y0<0<y-2t-1, y-2t+1,…,y-1<1-γk-l+1 is satisfied. Then {yn}n=-𝔎∞ oscillates about y-=0 with semicycles of length one. Moreover y2(r+1)n+2j<(or>)y2(r+1)(n-1)+2j and y2(r+1)n+2j+1>(or<)y2(r+1)(n-1)+2j+1, j=1,2,…,r+1 and n=0,1,2,….Proof.
Assume that condition(C1) is satisfied. Then we have y1=γy-2r-1/(1-∏i=lk‍y-2i)>y-2r-1, and y2=γy-2r/(1-∏i=lk‍y-2i+1)<y-2r.
By induction we get0>y2(r+1)n+2j+1>y2(r+1)(n-1)+2j+1, and 0<y2(r+1)n+2j<y2(r+1)(n-1)+2j,n=0,1,2,….
If condition(C2) is satisfied, the result is similar and will be omitted.


## 4. Global Behavior of (2.2)

Theorem 4.1.
The following statements are true.(1)
Ifγ<1, then the zero equilibrium point is a global attractor with basin (-1-γk-l+1,1-γk-l+1)𝔎+1.(2)
Ifγ=1, then (2.2) has prime period two solutions of the form …,0,φ,0,φ,0,…, where φ∈ℝ.(3)
Ifγ>1, then there exist solutions which are neither bounded nor persist.Proof.
(
1
) Suppose that y-𝔎,y-𝔎+1,…,y-1,y0∈(-1-γk-l+1,1-γk-l+1). Then using Theorem 3.1, we have that yn∈(-1-γk-l+1,1-γk-l+1), n≥1.
Moreover, we have|yn+1|<|yn-2r-1|, n=0,1,2,….
That is, the subsequences{|y2(r+1)n+j|}n=-1∞, j=1,2,…,2r+2 are decreasing.
From (2.2) we have
(4.1)|y2(r+1)n+j|=γ|y2(r+1)(n-1)+j||1-∏i=lky2(r+1)n+j-2i-1|≤γ|y2(r+1)(n-1)+j||1-|∏i=lky2(r+1)n+j-2i-1||.
Now suppose that|y2(r+1)n+j|→Lj as n→∞, j=1,2,…,2r+2. Then the last inequality implies that
(4.2)Lj≤γLj|1-∏i=lkLj-2i-1|,j=1,2,…,2r+2.
If for a certain j∈{1,2,…,2r+2} we have Lj≠0, then |1-∏i=lk‍Lj-2i-1|≤γ. This implies that
(4.3)1+γ≥∏i=lk‍Lj-2i-1≥1-γ.
This is a contradiction as the the subsequences{|y2(r+1)n+j|}n=-1∞, j=1,2,…,2r+2 are decreasing. Therefore, Lj=0, j=1,2,…,2r+2, and {yn}n=-𝔎∞ converges to zero.
(
2
) Clear!
(
3
) Let {yn}n=-𝔎∞ be a solution of (2.2) with initial conditions, |y-i|<γ-1k-l+1(>γ+1k-l+1), i=2s,2s-2,…,2,0, and |y-i|>γ+1k-l+1(<γ-1k-l+1), i=2t-1,2t-3,…,1.
We consider only the case|y-i|<γ-1k-l+1, i=2s,2s-2,…2,0, and |y-i|>γ+1k-l+1, i=2t-1,2t-3,…,1.
It follows that|∏i=lk‍y-2i|=|y-2k||y-2k+2|⋯|y0|<γ-1.
That is,(4.4)-γ+1<∏i=lk‍y-2i<γ-1.
This implies that-γ+2<1-∏i=lk‍y-2i<γ and so |1-∏i=lk‍y-2i|<γ. Hence we have
(4.5)|y1|=|γy-2r-1||1-∏i=lky-2i|>γ|y-2r-1|γ=|y-2r-1|>γ+1k-l+1.
Also|∏i=lk‍y-2i+1|>γ+1 implies that γ+2<1-∏i=lk‍y-2i+1<-γ and so |1-∏i=lk‍y-2i+1|>γ.
Hence we have(4.6)|y2|=|γy-2r||1-∏i=lky-2i+1|<γ|y-2r|γ=|y-2r|<γ-1k-l+1.
By induction we get(4.7)|y2(r+1)n+2j+1|>|y2(r+1)(n-1)+2j+1|>γ+1k-l+1,|y2(r+1)n+2j|<|y2(r+1)(n-1)+2j|<γ-1k-l+1,n≥0 and j=0,1,…,r. Now suppose that
(4.8)|y2(r+1)(n-1)+2j+1|⟶L2j+1∈(γ+1k-l+1,∞],|y2(r+1)(n-1)+2j|⟶L2j∈[0,γ-1k-l+1),
as n→∞, j=0,1,…,r.
But as(4.9)|y2(r+1)n+2j|=γ|y2(r+1)(n-1)+2j||1-∏i=lky2(r+1)n+2j-2i-1|≤γ|y2(r+1)(n-1)+2j||1-|∏i=lky2(r+1)n+j-2i-1||,
then
(4.10)L2j≤γL2j|1-∏i=lkL2j-2i-1|,j=0,1,…,r.
We claim that for each j=0,1,…,r, L2j=0.
For the sake of contradiction suppose that there existsj∈{0,1,…,r} with L2j∈(0,γ-1k-l+1).
Then (4.10) gives
(4.11)|1-∏i=lk‍L2j-2i-1|≤γ.
This implies that
(4.12)-γ+1≤∏i=lk‍L2j-2i-1≤γ+1.
As L2j+1∈(γ+1k-l+1,∞], j=0,1,…,r, we have a contradiction.
Thus it is true that for eachj=0,1,…,r we have L2j=0 and so limn→∞y2n=0.
We now claim that for eachj=0,1,…,r, L2j=∞.
For the sake of contradiction, suppose that there existsj∈{0,1,…,r} with L2j+1∈(γ+1k-l+1,∞). Then
(4.13)L2j+1=lim⁡n→∞|y2(r+1)(n-1)+2j+1|=γlim⁡n→∞⁡|y2(r+1)(n-1)+2j+1||1-∏i=lklim⁡n→∞⁡y2(r+1)n+2j-2i|≥γlim⁡n→∞⁡|y2(r+1)(n-1)+2j+1|1+∏i=lklim⁡n→∞⁡|y2(r+1)n+2j-2i|=γL2j+11+∏i=lkL2j=γL2j+1.
This is a contradiction. Therefore for each j=0,1,…,r we have L2j+1=∞ and so limn→∞y2n+1=∞.
The case when|y-i|>γ+1k-l+1, i=2s,2s-2,…,2,0 and |y-i|<γ-1k-l+1, i=2t-1,2t-3,…,1 is similar and will be omitted.


## 5. Numerical Examples

Example 5.1.
Figure1 shows that if r=0, l=0, k=1 (𝔎=max{2k,2r+1}=2) and γ=0.6, then the solution {yn}n=-2∞ with initial conditions y-2=-1, y-1=1.3, y0=1.1 converges to zero.Figure 1
The difference equationyn+1=0.6yn-1/(1-ynyn-2).Example 5.2.
Figure2 shows that if r=1, l=0, k=1 (𝔎=max{2k,2r+1}=3) and γ=0.7, then the solution {yn}n=-3∞ with initial conditions y-3=-1,y-2=-1.2, y-1=1.3, y0=1.1 converges to zero.Figure 2
The difference equationyn+1=0.7yn-3/(1-ynyn-2).Example 5.3.
Figure3 shows that if r=0, l=0, k=1 (𝔎=max{2k,2r+1}=2) and γ=2, then the solution {yn}n=-3∞ with initial conditions y-2=2,y-1=0.4, y0=2.1 is unbounded.Figure 3
The difference equationyn+1=2yn-1/(1-ynyn-2).Example 5.4.
Figure4 shows that if r=1, l=0, k=1 (𝔎=max{2k,2r+1}=3) and γ=2, then the solution {yn}n=-3∞ with initial conditions y-3=0.5, y-2=2, y-1=0.4, y0=2.1 is unbounded.Figure 4
The difference equationyn+1=2yn-3/(1-ynyn-2).



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