# Global Asymptotic Stability for a Fourth-Order Rational Difference Equation

Authors: Meseret Tuba Gülpinar; Mustafa Bayram
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

Our aim is to
investigate the global behavior of the following
fourth-order rational difference equation:xn+1=(xnxn−2xn−3+xn+xn−2+xn−3+a)/(xnxn−2+xnxn−3+xn−2xn−3+1+a), n=0,1,2,… where a∈[0,∞) and the initial values x−3,x−2,x−1,x0∈(0,∞). To verify that the positive equilibrium point of the
equation is globally asymptotically stable, we used the rule of
the successive lengths of positive and negative semicycles of
nontrivial solutions of the aforementioned equation.


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## Body

## 1. Introduction

There has been a great interest in studying global behaviors of rational difference equations. One can easily see that it is hard to understand thoroughly the global behaviors of solutions of rational difference equations although they have simple forms. And there has not been any general method to identify the global behaviors of rational difference equations of order greater than one and so far [1–3].Let us consider the following fourth-order difference equation:(1.1)xn+1=xnxn-2xn-3+xn+xn-2+xn-3+axnxn-2+xnxn-3+xn-2xn-3+1+a,n=0,1,2,…,
where a∈[0,∞) and the initial values x-3,x-2,x-1,x0∈(0,∞) in this paper.  By determining the rule for the positive and negative semicycles, we assigned the global behavior of the positive equilibrium point. The unique positive equilibrium point x̅ of (1.1)  is obtained x̅=1 by solving
(1.2)x̅=x̅3+3x̅+a3x̅2+1+a.Definition 1.1.
A solution{xn}n=-3∞ of  (1.1) is said to be  eventually trivial  if xn is eventually equal to x̅=1; otherwise, the solution is said to be  nontrivial  [4–6].Definition 1.2.
A solution{xn}n=-3∞ of  (1.1) is said to be  eventually positive  if xn is eventually greater than x̅=1.
A solution{xn}n=-3∞ of  (1.1) is said to be  eventually negative  if xn is eventually less than x̅=1 [1, 2, 4–6].Definition 1.3.
Apositive semicycle  of a solution {xn}n=-3∞ of  (1.1) consists of a “string” of terms {xl,xl+1,…,xm}, all greater than or equal to the equilibrium point x̅, with l≥-3 and m<∞ such that
(1.3)eitherl=-3orl>-3,xl-1<x̅,eitherm=∞orm<∞,xm+1<x̅.
A  negative semicycle  of a solution {xn}n=-3∞ of  (1.1) consists of a “string” of terms {xl,xl+1,…,xm}, all greater than or equal to the equilibrium point x̅, with l≥-3 and m<∞ such that
(1.4)eitherl=-3orl>-3,xl-1≥x̅,eitherm=∞orm<∞,xm+1≥x̅.
And also the  lengths of a semicycle  is (m-l+1), the number of the terms contained in it. And we denote that the lengths of a positive semicycle are by (m-l+1)+ and the lengths of a negative semicycle are by (m-l+1)-   [1, 2, 4–6].Definition 1.4.
A solution{xn}n=-3∞ of  (1.1) is called nonoscillatory about x̅, or simply  nonoscillatory,  if there exists N≥-3 such that either
(1.5)xn≥x̅,∀n≥N
or
(1.6)xn<x̅,∀n≥N.
Otherwise, the solution {xn}n=-3∞ is called oscillatory about x̅, or simply  oscillatory  [1, 2].Lemma 1.5.
A positive solution{xn}n=-3∞ of  (1.1) is eventually trivial if and only if
(1.7)(x-3-1)(x-2-1)(x-1-1)(x0-1)=0.Proof.
To prove the lemma, first we assume that(x-3-1)(x-2-1)(x-1-1)(x0-1)≠0 and then we must show that xn≠1 for any n≥1.
Assume that for someN≥1,  xN=1 and that xn≠1 for -3≤n≤N-1. So
(1.8)xN=xN-1xN-3xN-4+xN-1+xN-3+xN-4+axN-1xN-3+xN-1xN-4+xN-3xN-4+1+a=1,
and we obtain (xN-4-1)(xN-3-1)(xN-1-1)=0; hence  xN-4=1,xN-3=1,xN-1=1, when we solve the equation above. It is easy to see that xN-4=1,  xN-3=1, or  xN-1=1 contradicts with xn≠1 for -3≤n≤N-1.
If  (1.7)  holds, it is clear that the following conclusions hold:(i)
ifx-3=1,  xn=1 for n≥1,(ii)
ifx-2=1,  xn=1 for n≥1,(iii)
ifx-1=1,  xn=1 for n≥2,(iv)
ifx0=1,  xn=1 for n≥1.It is obvious that if the initial conditions do not satisfy  (1.7), then the positive solution {xn}n=-3∞ of  (1.1) is eventually nontrivial.Lemma 1.6.
If{xn}n=-3∞ is a nontrivial positive solution of  (1.1), then the following conclusions are satisfied:(i)
(xn+1-1)(xn-1)(xn-2-1)(xn-3-1)>0,(ii)
(xn+1-xn)(xn-1)<0,(iii)
(xn+1-xn-2)(xn-2-1)<0,(iv)
(xn+1-xn-3)(xn-3-1)<0.Proof.
(i) The proof of the inequality (i) is obtained by subtracting 1 from (1.1)
(1.9)xn+1-1=xnxn-2xn-3+xn+xn-2+xn-3+axnxn-2+xnxn-3+xn-2xn-3+1+a-1=(xn-1)(xn-2-1)(xn-3-1)xnxn-2+xnxn-3+xn-2xn-3+1+a.
The dominator of this fraction is positive so (xn+1-1)(xn-1)(xn-2-1)(xn-3-1)>0 also.
(ii) If we subtractxn from (1.1), we obtain
(1.10)xn+1-xn=xnxn-2xn-3+xn+xn-2+xn-3+axnxn-2+xnxn-3+xn-2xn-3+1+a-xn=-(xn-1)[xn-2(xn+1)+xn-3(xn+1)+a]xnxn-2+xnxn-3+xn-2xn-3+1+a.
The expression [xn-2(xn+1)+xn-3(xn+1)+a]/(xnxn-2+xnxn-3+xn-2xn-3+1+a) is positive, and so we get
(1.11)(xn+1-xn)(xn-1)<0.
The proofs for inequalities (iii) and (iv) are similar to the one for (ii).


## 2. Main Results and Their Proofs

The trajectory of (1.1) and global asymptotic stability of the positive solution are considered in this part of the paper.Theorem 2.1.
Let{xn}n=-3∞ be a strictly oscillatory solution of (1.1). Then the positive and negative semicycles of (1.1)  are …,3+,3-,3+,3-,3+,3-,3+,3-,… or  …,2+,1-,2+,1-,2+,1-,2+,1-,…  or  …,2-,1+,2-,1+,2-,1+,2-,1+,… or  …,1+,1-,1+,1-,1+,1-,1+,1-,….Proof.
Assume that{xn}n=-3∞ is a strictly oscillatory solution of  (1.1), then the initial values must satisfy one of the following four cases: (i)
xp-3>1,  xp-2<1,  xp-1>1,  xp>1,(ii)
xp-3>1,  xp-2<1,  xp-1>1,  xp<1,(iii)
xp-3>1,  xp-2<1,  xp-1<1,  xp>1,(iv)
xp-3>1,  xp-2<1,  xp-1<1,  xp<1.If (i) occurs, it follows from Lemma 1.6(i) that
(2.1)xp-3>1,xp-2<1,xp-1>1,xp>1,xp+1<1,xp+2>1,xp+3>1,xp+4<1,xp+5>1,xp+6>1,xp+7<1,xp+8>1,xp+9>1,xp+10<1,xp+11>1,xp+12>1,xp+13<1,xp+14>1,xp+15>1,xp+16<1,xp+17>1,xp+18>1,xp+19<1,xp+20>1,xp+21>1,….
It means that the rule for the lengths of positive and negative semicycles of the solution of  (1.1) occurs successively as …,2+,1-,2+,1-,2+,1-,2+,1-,….
If (ii) happens, the positive and negative semicycles are(2.2)xp-3>1,xp-2<1,xp-1>1,xp<1,xp+1>1,xp+2<1,xp+3>1,xp+4<1,xp+5>1,xp+6<1,xp+7>1,xp+8<1,xp+9>1,xp+10<1,xp+11>1,xp+12<1,xp+13>1,xp+14<1,xp+15>1,xp+16<1,xp+17>1,xp+18<1,xp+19>1,xp+20<1,xp+21>1,xp+22<1,xp+23>1,….
The regulation for the lengths of positive and negative semicycles which occur successively is …,1+,1-,1+,1-,1+,1-,1+,1-,….
The other cases can be shown similarly.Theorem 2.2.
The positive equilibrium point of (1.1) is globally asymptotically stable.Proof.
Let us show that the positive equilibrium pointx̅=1 of (1.1) is locally asymptotically stable and also globally attractive, then it  is globally asymptotically stable.  The linearized equation of  (1.1) about the positive equilibrium  x̅=1 is
(2.3)yn+1=q0yn+q1yn-1+q2yn-2+q3yn-3,
where qi=(∂F/∂ui)(x̅,x̅,x̅,x̅) and F(u0,u1,u2,u3)=(u0u2u3+u0+u2+u3+a)/(u0u2+u0u3+u2u3+1+a).
And we obtain(2.4)yn+1=q0yn+q1yn-1+q2yn-2+q3yn-3=0·yn+0·yn-1+0·yn-2+0·yn-3=0,
thereby x̅=1 is locally asymptotically stable.
Now we must show thatlimn→∞xn=x̅=1. The proof is as follows.(1)
If initial values of (1.1) satisfy (1.7), then xn=1 according to Lemma 1.5, so limn→∞xn=x̅=1.(2)
If the initial values of (1.1) do not satisfy (1.7), then for any solution of (1.1), xn≠1 for  n≥-3.
(i)
If the solution is nonoscillatory about the positive equilibrium point of (1.1), then {xn} is monotonic and bounded because of Lemma 1.6. So, the limit
(2.5)limn→∞xn=L
exists and is finite. If we take limits on both sides of (1.1), then we obtain L=(L3+3L+a)/(3L2+1+a) and thereby L=1. So limn→∞xn=L=1.(ii)
If the solution is strictly oscillatory, then trajectory structure of nontrivial solutions of (1.1) is …,3+,3-,3+,3-,3+,3-,3+,3-,… or  …,2+,1-,2+,1-,2+,1-,2+,1-,…  or  …,2-,1+,2-,1+,2-,1+,2-,1+,…  or …,1+,1-,1+,1-,1+,1-,1+,1-,….First, we investigate the case where the rule of the trajectory structure is …,3+,3-,3+,3-,3+,3-,3+,3-,… in a period. We denote positive semicycles by {xp,xp+1,xp+2}+ and negative semicycles by {xp+3,xp+4,xp+5}-. The rule for the positive and negative semicycles can be “periodically’’ expressed as follows:
(2.6){xp+6n,xp+6n+1,xp+6n+2}+,{xp+6n+3,xp+6n+4,xp+6n+5}-,n=0,1,….
By using Lemma 1.6 we obtain(i)
xp+6n+2<xp+6n+1<xp+6n; xp+6n+3<xp+6n+4<xp+6n+5,(ii)
xp+6n+6<xp+6n+2;  xp+6n+9>xp+6n+5.This relations give rise to
(2.7)xp+6n+6<xp+6n+2<xp+6n+1<xp+6n,xp+6n+3<xp+6n+4<xp+6n+5<xp+6n+9.
It is easy to see that {xp+6n}n=0∞  is decreasing with its lower bound 1 because of the inequality xp+6n+6<xp+6n+2<xp+6n+1<xp+6n. Hence the limit exists and is finite.  The limit is then
(2.8)limn→∞xp+6n=limn→∞xp+6n+1=limn→∞xp+6n+2=L.
Similarly {xp+6n+3}n=0∞ is increasing and its upper bound 1 because of the inequality xp+6n+3<xp+6n+4<xp+6n+5<xp+6n+9. So, the limit limn→∞xp+6n+3=M exists and is finite. And we derive limn→∞xp+6n+4=limn→∞xp+6n+5=M.
Now, we must show thatL=M=1. To do this, let us take
(2.9)xp+6n+6=xp+6n+5xp+6n+3xp+6n+2+xp+6n+5+xp+6n+3+xp+6n+2+axp+6n+5xp+6n+3+xp+6n+5xp+6n+2+xp+6n+3xp+6n+2+1+a,
and taking the limit on both sides of the above equality, we obtain
(2.10)L=M·M·L+M+M+L+aM·M+M·L+M·L+1+a.
By solving this equation we have
(2.11)(L-1)[2M(L+1)+a]=0,
and then L=1.  Similarly to obtain M by using 1≤xp+6n+2<xp+6n+1 we are taking
(2.12)xp+6n+5=xp+6n+4xp+6n+2xp+6n+1+xp+6n+4+xp+6n+2+xp+6n+1+axp+6n+4xp+6n+2+xp+6n+4xp+6n+1+xp+6n+2xp+6n+1+1+a.
And taking the limit on both sides of the above equality we get
(2.13)M=M·L·L+M+L+L+aM·L+M·L+L·L+1+a.
By solving this equation we have (M-1)[2L(M+1)+a]=0 and  M=1. Hence, we derive L=M=1, so
(2.14)limn→∞xn=1.
It can be shown that limn→∞xn=1 for the other rules of the trajectory structures with the same manner. Therefore, the positive equilibrium point is globally asymptotically stable.



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*Source: 616982-2009-08-06.xml*