# Global Attractivity and Periodic Character of Difference Equation of Order Four

Authors: M. A. Obaid; E. M. Elsayed; M. M. El-Dessoky
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We investigate the global convergence result, boundedness, and periodicity of solutions of the recursive sequencexn+1=axn+((bxn-1+cxn-2+dxn-3)/(αxn-1+βxn-2+γxn-3)),n=0,1,…, where the parameters a,b,c,d,α,β, andγare positive real numbers and the initial conditions x-3,x-2,x-1, and x0are positive real numbers.


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## Body

## 1. Introduction

Our goal in this paper is to investigate the global stability character, boundedness, and the periodicity of solutions of the recursive sequence(1.1)xn+1=axn+bxn-1+cxn-2+dxn-3αxn-1+βxn-2+γxn-3,
where the parameters a,b,c,d,α,β,andγare positive real numbers and the initial conditions x-3,x-2,x-1, and x0are positive real numbers.Recently there has been a lot of interest in studying the global attractivity, the boundedness character and the periodicity nature of nonlinear difference equations, see for example [1–15].The study of the nonlinear rational difference equations of a higher order is quite challenging and rewarding, and the results about these equations offer prototypes towards the development of the basic theory of the global behavior of nonlinear difference equations of a big order, recently many researchers have investigated the behavior of the solution of difference equations—for example, in [3] Elabbasy et al. investigated the global stability, periodicity character and gave the solution of special case of the following recursive sequence:
(1.2)xn+1=axn-bxncxn-dxn-1.
In [5] Elabbasy and Elsayed investigated the global stability character and boundedness of solutions of the recursive sequence
(1.3)xn+1=axnp+b∏r=1pxn-rcxnp+d∏r=1pxn-r.
Elsayed [11] investigated the global character of solutions of the nonlinear, fourth-order, rational difference equation
(1.4)xn+1=axn-2+bxnxn-2cxn+dxn-3.
Saleh and Aloqeili [16] investigated the difference equation
(1.5)yn+1=A+ynyn-k.
Yang et al. [17] investigated the invariant intervals, the global attractivity of equilibrium points, and the asymptotic behavior of the solutions of the recursive sequence
(1.6)xn+1=axn-1+bxn-2c+dxn-1xn-2.
For some related work see [16–26].Here, we recall some basic definitions and some theorems that we need in the sequel.LetI be some interval of real numbers and let
(1.7)F:Ik+1→I
be a continuously differentiable function. Then for every set of initial conditions x-k,x-k+1,…,x0∈I, the difference equation
(1.8)xn+1=F(xn,xn-1,…,xn-k),n=0,1,…,
has a unique solution {xn}n=-k∞.Definition 1.1   (Equilibrium Point).
A pointx¯∈I is called an equilibrium point of (1.8) if
(1.9)x¯=F(x¯,x¯,…,x¯).
That is, xn=x¯ for n≥0is a solution of (1.8), or equivalently, x¯ is a fixed point of f.Definition 1.2   (Periodicity).
A sequence{xn}n=-k∞is said to be periodic with period p if  xn+p=xnfor all n≥-k.Definition 1.3   (Stability).
(i) The equilibrium pointx¯of (1.8) is locally stable if for every  ϵ>0,  there exists δ>0  such that for all x-k,  x-k+1,…,  x-1,x0∈Iwith
(1.10)|x-k-x¯|+|x-k+1-x¯|+⋯+|x0-x¯|<δ,
we have
(1.11)|xn-x¯|<ϵ∀n≥-k.
(ii) The equilibrium pointx¯of (1.8) is locally asymptotically stable if x¯is locally stable solution of (1.8) and there exists γ>0, such that for all x-k,x-k+1,…,x-1, x0∈Iwith
(1.12)|x-k-x¯|+|x-k+1-x¯|+⋯+|x0-x¯|<γ,
we have
(1.13)limn→∞xn=x¯.
(iii) The equilibrium pointx¯of (1.8) is a global attractor if for all x-k,x-k+1,…,x-1,x0∈I,we have
(1.14)limn→∞xn=x¯.
(iv) The equilibrium pointx¯of (1.8) is globally asymptotically stable if x¯is locally stable, and x¯is also a global attractor of (1.8).
(v) The equilibrium pointx¯of (1.8) is unstable if x¯is not locally stable.
The linearized equation of (1.8) about the equilibrium x¯is the linear difference equation
(1.15)yn+1=∑i=0k‍∂F(x¯,x¯,…,x¯)∂xn-iyn-i.Theorem A (see [18]).
Assume thatpi∈R,i=1,2,…,kandk∈{0,1,2,…}. Then
(1.16)∑i=1k‍|pi|<1
is a sufficient condition for the asymptotic stability of the difference equation
(1.17)xn+k+p1xn+k-1+⋯+pkxn=0n=0,1,….Theorem B (see [19]).
Letg:[a,b]k+1→[a,b] be a continuous function, where k is a positive integer, and where [a,b] is an interval of real numbers. Consider the difference equation
(1.18)xn+1=g(xn,xn-1,…,xn-k),n=0,1,….
Suppose that g satisfies the following conditions.(1)
For each integeri with 1≤i≤k+1; the function g(z1,z2,…,zk+1) is weakly monotonic in zi for fixed z1,z2,…,zi-1,zi+1,…,zk+1.(2)
If(m,M) is a solution of thmhje system
(1.19)m=g(m1,m2,…,mk+1),M=g(M1,M2,…,Mk+1),then m=M, where for each i=1,2,…,k+1, we set
(1.20)mi={mifgisnondecreasinginzi,Mifgisnonincreasinginzi,Mi={Mifgisnondecreasinginzi,mifgisnonincreasinginzi.
Then there exists exactly one equilibrium x¯ of (1.18), and every solution of (1.18) converges to x¯.
The paper proceeds as follows. In Section2, we show that when
(1.21)(α+β+γ)(b+c+d)>max{|2α(c+d)-2b(β+γ)|,|2γ(b+c)-2d(α+β)|,|2β(b+d)-2c(α+γ)|}
then the equilibrium point of (1.1) is locally asymptotically stable. In Section 3 we prove that the solution is bounded and persists when a<1 and the solution of (1.1) is unbounded if a>1. In Section 4 we prove that there exists a period two solution of (1.1). In Section 5 we prove that the equilibrium point of (1.1) is global attractor. Finally we give a numerical examples of some special cases of (1.1) and draw it by using Matlab.


## 2.   Local Stability of the Equilibrium Point of (1.1)

This section deals with study of the local stability character of the equilibrium point of  (1.1).Equation (1.1) has equilibrium point and is given by
(2.1)x¯=ax¯+b+c+dα+β+γ.
If a<1, then the only positive equilibrium point of (1.1) is given by
(2.2)x¯=b+c+d(α+β+γ)(1-a).
Let f:(0,∞)4→(0,∞) be a continuously differentiable function defined by
(2.3)f(u,v,w,t)=au+bv+cw+dtαv+βw+γt.
Therefore, it follows that
(2.4)∂f(u,v,w,t)∂u=a,∂f(u,v,w,t)∂v=(bβ-cα)w+(bγ-dα)t(αv+βw+γt)2,∂f(u,v,w,t)∂w=-(bβ-cα)v+(cγ-dβ)t(αv+βw+γt)2,∂f(u,v,w,t)∂t=-(bγ-dα)v-(cγ-dβ)w(αv+βw+γt)2.
Then we see that
(2.5)∂f(x¯,x¯,x¯,x¯)∂u=a=-a3,∂f(x¯,x¯,x¯,x¯)∂v=(bβ-cα)+(bβ-dα)(α+β+γ)2x¯=[(bβ-cα)+(bγ-dα)](1-a)(α+β+γ)(b+c+d)=-a2,∂f(x¯,x¯,x¯,x¯)∂w=-(bβ-cα)+(cγ-dβ)(α+β+γ)2x¯=[-(bβ-cα)+(cγ-dβ)](1-a)(α+β+γ)(b+c+d)=-a1,∂f(x¯,x¯,x¯,x¯)∂t=-(bγ-dα)-(cγ-dβ)(α+β+γ)2x¯=[-(bγ-dα)-(cγ-dβ)](1-a)(α+β+γ)(b+c+d)=-a0.
Then the linearized equation of (1.1) about x¯ is
(2.6)yn+1+a3yn+a2yn-1+a1yn-2+a0yn-3=0,
whose characteristic equation is
(2.7)λ4+a3λ3+a2λ2+a1λ+a0=0.Theorem 2.1.
Assume that(2.8)(α+β+γ)(b+c+d)>max{|2α(c+d)-2b(β+γ)|,|2γ(b+c)-2d(α+β)|,|2β(b+d)-2c(α+γ)|}.
Then the positive equilibrium point of (1.1) is locally asymptotically stable.Proof.
It is follows by Theorem A that  (2.6) is asymptotically stable if all roots of (2.7) lie in the open disc |λ|<1, that is, if
(2.9)|a3|+|a2|+|a1|+|a0|<1.|a|+|[(bβ-cα)+(bγ-dα)](1-a)(α+β+γ)(b+c+d)|+|[-(bβ-cα)+(cγ-dβ)](1-a)(α+β+γ)(b+c+d)|+|[-(bγ-dα)-(cγ-dβ)](1-a)(α+β+γ)(b+c+d)|<1,
and so
(2.10)|[(bβ-cα)+(bγ-dα)](1-a)|+|[-(bβ-cα)+(cγ-dβ)](1-a)|+|[-(bγ-dα)-(cγ-dβ)](1-a)|<[(α+β+γ)(b+c+d)](1-a).
Dividing the denominator and numerator by (1-a) gives
(2.11)|(bβ-cα)+(bγ-dα)|+|-(bβ-cα)+(cγ-dβ)|+|-(bγ-dα)-(cγ-dβ)|<(α+β+γ)(b+c+d).
Suppose that
(2.12)B1=(bβ-cα)+(bγ-dα),B2=-(bβ-cα)+(cγ-dβ),B3=-(bγ-dα)-(cγ-dβ).
We consider the following cases.
(
1
)
B
1
>
0,  B2>0,  and  B3>0. In this case we see from (2.11) that
(2.13)(bβ-cα)+(bγ-dα)-(bβ-cα)+(cγ-dβ)-(bγ-dα)-(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.14)(α+β+γ)(b+c+d)>0,
which is always true.
(
2
)
B
1
>
0,  B2>0,  and  B3<0.It follows from (2.11) that
(2.15)(bβ-cα)+(bγ-dα)-(bβ-cα)+(cγ-dβ)+(bγ-dα)+(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.16)2γ(b+c)-2d(α+β)<(α+β+γ)(b+c+d),
which is satisfied by (2.8).
(
3
)
B
1
>
0,  B2<0,  and  B3>0.  We see from (2.11) that
(2.17)(bβ-cα)+(bγ-dα)+(bβ-cα)-(cγ-dβ)-(bγ-dα)-(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.18)2β(b+d)-2c(α+γ)<(α+β+γ)(b+c+d),
which is satisfied by (2.8).
(
4
)
B
1
>
0,  B2<0,  and  B3<0. It follows from (2.11) that
(2.19)(bβ-cα)+(bγ-dα)+(bβ-cα)-(cγ-dβ)+(bγ-dα)+(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.20)2b(β+γ)-2α(c+d)<(α+β+γ)(b+c+d),
which is satisfied by (2.8).
(
5
)
B
1
<
0,  B2>0,  and  B3>0.  We see from (2.11) that
(2.21)-(bβ-cα)-(bγ-dα)-(bβ-cα)+(cγ-dβ)-(bγ-dα)-(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.22)2α(c+d)-2b(β+γ)<(α+β+γ)(b+c+d),
which is satisfied by (2.8).
(
6
)
B
1
<
0,  B2>0, and  B3<0.  It follows from (2.11) that
(2.23)-(bβ-cα)-(bγ-dα)-(bβ-cα)+(cγ-dβ)+(bγ-dα)+(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.24)2c(α+γ)-2β(b+d)<(α+β+γ)(b+c+d),
which is satisfied by (2.8).
(
7
)
B
1
<
0,  B2<0, and  B3>0.  We see from (2.11) that
(2.25)-(bβ-cα)-(bγ-dα)+(bβ-cα)-(cγ-dβ)-(bγ-dα)-(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.26)2d(α+β)-2γ(b+c)<(α+β+γ)(b+c+d),
which is satisfied by (2.8).
(
8
)
B
1
<
0,  B2<0,  and  B3<0.  It follows from (2.11) that
(2.27)-(bβ-cα)-(bγ-dα)+(bβ-cα)-(cγ-dβ)+(bγ-dα)+(cγ-dβ)<(α+β+γ)(b+c+d),
if and only if
(2.28)(α+β+γ)(b+c+d)>0,
which is always true. The proof is complete.


## 3. Boundedness of Solutions of (1.1)

Here we study the boundedness nature and persistence of solutions of (1.1).Theorem 3.1.
Every solution of (1.1) is bounded and persists if a<1.Proof.
Let{xn}n=-3∞ be a solution of (1.1). It follows from (1.1) that
(3.1)xn+1=axn+bxn-1+cxn-2+dxn-3αxn-1+βxn-2+γxn-3=axn+bxn-1αxn-1+βxn-2+γxn-3+cxn-2αxn-1+βxn-2+γxn-3+dxn-3αxn-1+βxn-2+γxn-3.
Then
(3.2)xn+1≤axn+bxn-1αxn-1+cxn-2βxn-2+dxn-3γxn-3=axn+bα+cβ+dγ∀n≥1.
By using a comparison, we see that
(3.3)limsupn→∞xn≤bβγ+cαγ+dαβαβγ(1-a)=M.
Thus the solution is bounded.
Now we wish to show that there existsm>0 such that
(3.4)xn≥m∀n≥1.
The transformation
(3.5)xn=1yn
will reduce (1.1) to the equivalent form
(3.6)1yn+1=a(αyn-2yn-3+βyn-1yn-3+γyn-1yn-2)+yn(byn-2yn-3+cyn-1yn-3+dyn-1yn-2)yn(αyn-2yn-3+βyn-1yn-3+γyn-1yn-2),
or
(3.7)yn+1=yn(αyn-2yn-3+βyn-1yn-3+γyn-1yn-2)a(αyn-2yn-3+βyn-1yn-3+γyn-1yn-2)+yn(byn-2yn-3+cyn-1yn-3+dyn-1yn-2).
It follows that
(3.8)yn+1≤yn(αyn-2yn-3+βyn-1yn-3+γyn-1yn-2)yn(byn-2yn-3+cyn-1yn-3+dyn-1yn-2)=αyn-2yn-3byn-2yn-3+cyn-1yn-3+dyn-1yn-2+βyn-1yn-3byn-2yn-3+cyn-1yn-3+dyn-1yn-2+γyn-1yn-2byn-2yn-3+cyn-1yn-3+dyn-1yn-2≤αyn-2yn-3byn-2yn-3+βyn-1yn-3cyn-1yn-3+γyn-1yn-2dyn-1yn-2=αb+βc+γd=αcd+βbd+γbcbcd=H∀n≥1.
Thus we obtain
(3.9)xn=1yn≥1H=bcdαcd+βbd+γbc=m∀n≥1.
From (3.3) and (3.9), we see that
(3.10)m≤xn≤M∀n≥1.
Therefore, every solution of (1.1) is bounded and persists.Theorem 3.2.
Every solution of (1.1) is unbounded if a>1.Proof.
Let{xn}n=-3∞ be a solution of (1.1). Then from (1.1) we see that
(3.11)xn+1=axn+bxn-1+cxn-2+dxn-3αxn-1+βxn-2+γxn-3>axn∀n≥1.
We see that the right-hand side can be written as follows:
(3.12)yn+1=ayn⇒yn=any0,
and this equation is unstable because a>1, and limn→∞yn=∞. Then by using ratio test {xn}n=-3∞ is unbounded from above.


## 4. Existence of Periodic Solutions

In this section we study the existence of periodic solutions of (1.1). The following theorem states the necessary and sufficient conditions that this equation has periodic solutions of prime period two.Theorem 4.1.
Equation (1.1) has positive prime period two solutions if and only if (i)
(b+d-c)(α+γ-β)(1+a)+4(aβ(b+d)+c(α+γ))>0,α+γ>β,b+d>c.Proof.
First suppose that there exists a prime period two solution(4.1)…,p,q,p,q,…,
of (1.1). We will prove that condition (i) holds.
We see from (1.1) that
(4.2)p=aq+bp+cq+dpαp+βq+γp=aq+ep+cqfp+βq,wheree=b+d,f=α+γ,q=ap+bq+cp+dqαq+βp+γq=ap+eq+cpfq+βp.
Then
(4.3)fp2+βpq=afpq+aβq2+ep+cq,(4.4)fq2+βpq=afpq+aβp2+eq+cp.
Subtracting (4.3) from (4.4) gives
(4.5)f(p2-q2)=-aβ(p2-q2)+(e-c)(p-q).
Since  p≠q, it follows that
(4.6)p+q=(e-c)(f+aβ).
Again, adding (4.3) and (4.4) yields
(4.7)f(p2+q2)+2βpq=2afpq+aβ(p2+q2)+(e+c)(p+q)
or
(4.8)(f-aβ)(p2+q2)+2(β-af)pq=(e+c)(p+q).
It follows by (4.6), (4.8) and the relation
(4.9)p2+q2=(p+q)2-2pq∀p,q∈R
that
(4.10)2(β-f)(1+a)pq=2(eaβ+cf)(e-c)(f+aβ)2.
Thus
(4.11)pq=(eaβ+cf)(e-c)(f+aβ)2(β-f)(1+a).
Now it is clear from (4.6) and (4.11) that p and q are the two distinct roots of the quadratic equation
(4.12)t2-((e-c)(f+aβ))t+((eaβ+cf)(e-c)(f+aβ)2(β-f)(1+a))=0,(4.13)(f+aβ)t2-(e-c)t+((eaβ+cf)(e-c)(f+aβ)(β-f)(1+a))=0,
and so
(4.14)[e-c]2-4(eaβ+cf)(e-c)(β-f)(1+a)>0,
or
(4.15)[e-c]2+4(eaβ+cf)(e-c)(f-β)(1+a)>0,(e-c)(f-β)(1+a)+4(eaβ+cf)>0.
Therefore, inequality (i) holds.
Second, suppose that inequality (i) is true. We will show that (1.1) has a prime period two solution.
Assume that(4.16)p=e-c+ζ2(f+aβ),q=e-c-ζ2(f+aβ),
where ζ=[e-c]^2-4(eaβ+cf)(e-c)/(β-f)(1+a).
We see from inequality (i) that(4.17)(e-c)(f-β)(1+a)+4(eaβ+cf)>0,e>c,f>β,
which is equivalent to
(4.18)(e-c)2>4(eaβ+cf)(e-c)(β-f)(1+a).
Therefore,  p  and q are distinct real numbers.
Set(4.19)x-3=p,x-2=q,x-1=p,x0=q.
We wish to show that
(4.20)x1=x-1=p,x2=x0=q.
It follows from (1.1) that
(4.21)x1=aq+bp+cq+dpαp+βq+γp=aq+ep+cqfp+βq=a(e-c-ζ2(f+aβ))+e((e-c+ζ)/2(f+aβ))+c((e-c-ζ)/2(f+aβ))f((e-c+ζ)/2(f+aβ))+β((e-c-ζ)/2(f+aβ)).
Dividing the denominator and numerator by 2(f+aβ)gives
(4.22)x1=ae-ac-aζ2(f+aβ)+e(e-c+ζ)+c(e-c-ζ)f(e-c+ζ)+β(e-c-ζ)=ae-ac-aζ2(f+aβ)+(e-c)[(e+c)+ζ](f+β)(e-c)+(f-β)ζ.
Multiplying the denominator and numerator of the right side by (f+β)(e-c)-(f-β)ζ gives
(4.23)x1=ae-ac-aζ2(f+aβ)+(e-c)[(e+c)+ζ][(f+β)(e-c)-(f-β)ζ][(f+β)(e-c)+(f-β)ζ][(f+β)(e-c)-(f-β)ζ]=ae-ac-aζ2(f+aβ)+(e-c){(f+β)(e2-c2)+ζ[(f+β)(e-c)-(f-β)(e+c)]-(f-β)ζ2}(f+β)2(e-c)2-(f-β)2ζ2=ae-ac-aζ2(f+aβ)+(e-c){(f+β)(e2-c2)+2ζ(βe-cf)-(f-β)𝒜}(f+β)2(e-c)2-(f-β)2𝒜=ae-ac-aζ2(f+aβ)+(e-c){(f+β)(e2-c2)+2ζ(βe-cf)-ℬ}(f+β)2(e-c)2-(f-β)2𝒜=ae-ac-aζ2(f+aβ)+(e-c){2(e-c)[fc+βe-2(eaβ+cf)/(1+a)]+2ζ(βe-cf)}4(e-c)[βf(e-c)+(β-f)(eaβ+cf)/(1+a)],
where 𝒜  denotes to  [e-c]2-4(eaβ+cf)(e-c)/(β-f)(1+a) and ℬ  denotes to (f-β)(e-c)2+4(eaβ+cf)(e-c)/(1+a).
Multiplying the denominator and numerator of the right side by(1+a), we obtain(4.24)x1=ae-ac-aζ2(f+aβ)+(e-c)[(fc+βe)(1+a)-2(eaβ+cf)]+ζ(1+a)(βe-cf)2[βf(e-c)(1+a)+(β-f)(eaβ+cf)]=ae-ac-aζ2(f+aβ)+(e-c)(βe-fc)(1-a)+ζ(1+a)(βe-cf)2[βf(e-c)(1+a)+(β-f)(eaβ+cf)]=ae-ac-aζ2(f+aβ)+(βe-fc){(e-c)(1-a)+ζ(1+a)}2(βe-cf)(f+aβ)=ae-ac-aζ2(f+aβ)+(e-c)(1-a)+ζ(1+a)2(f+aβ)=ae-ac-aζ+(e-c)(1-a)+ζ(1+a)2(f+aβ)=e-c+ζ2(f+aβ)=p.
Similarly as before one can easily show that
(4.25)x2=q.
Then it follows by induction that
(4.26)x2n=q,x2n+1=p∀n≥-1.
Thus (1.1) has the prime period two solution
(4.27)…,p,q,p,q,…,
where p and q are the distinct roots of the quadratic equation (4.13) and the proof is complete.


## 5. Global Attractor of the Equilibrium Point of (1.1)

In this section, we investigate the global asymptotic stability of (1.1).Lemma 5.1.
For any values of the quotientb/α,c/β,  andd/γ,the function f(u,v,w,t)defined by (2.3) has the monotonicity behavior in its three arguments.Proof.
The proof follows by some computations and it will be omitted.Remark 5.2.
It follows from (1.1), whenb/α=c/β=d/γ, that
(5.1)xn+1=axn+λ∀n≥-3andforsomeconstantλ.
Whenever the quotientsα/A,  β/B,andγ/C are not equal, we get the following result.Theorem 5.3.
The equilibrium pointx¯ is a global attractor of (1.1) if one of the following statements holds:
(5.2)(1)bα≥cβ≥dγ,d≥b+c,(2)bα≥dγ≥cβ,c≥b+d,(3)cβ≥bα≥dγ,d≥b+c,(4)cβ≥dγ≥bα,b≥c+d,(5)dγ≥cβ≥bα,b≥c+d,(6)dγ≥bα≥cβ,c≥b+d.Proof.
Let{xn}n=-3∞ be a solution of (1.1) and again let fbe a function defined by (2.3).
We will prove the theorem when case (1.1) is true, and the proof of the other cases is similar and so will be omitted.Assume that (5.2) is true, then it is easy from the equations after (2.3) to see that the function f(u,v,w,t)is nondecreasing in u,v and nonincreasing in t and it is not clear what is going on with w. So we consider the following two cases.Case 1.
Assume that the functionf(u,v,w,t)is nondecreasing in w.
Suppose that(m,M)is a solution of the system  M=f(M,M,M,m)  and  m=g(m,m,m,M). Then from (1.1), we see that
(5.3)M=aM+bM+cM+dmαM+βM+γm,m=am+bm+cm+dMαm+βm+γM,
or
(5.4)M(1-a)=bM+cM+dmαM+βM+γm,m(1-a)=bm+cm+dMαm+βm+γM.
Then
(5.5)(α+β)(1-a)M2+γ(1-a)Mm=(b+c)M+dm,(α+β)(1-a)m2+γ(1-a)Mm=(b+c)m+dM.
Subtracting this two equations, we obtain
(5.6)(M-m){(α+β)(1-a)(M+m)+(d-b-c)}=0,
under the conditions d≥b+c,  a<1,we see that
(5.7)M=m.
It follows by Theorem B that x¯ is a global attractor of (1.1) and then the proof is complete.Case 2.
Assumes that the functionf(u,v,w,t)is nonincreasing in w.
Suppose that(m,M)is a solution of the system  M=f(M,M,m,m)  and  m=g(m,m,M,M). Then from (1.1), we see that
(5.8)M=aM+bM+cM+dmαM+βM+γm,m=am+bm+cm+dMαm+βm+γM,
or
(5.9)M(1-a)=bM+cm+dmαM+βm+γm,m(1-a)=bm+cM+dMαm+βM+γM.
Then
(5.10)α(1-a)M2+(β+γ)(1-a)Mm=bM+(c+d)m,α(1-a)m2+(β+γ)(1-a)Mm=bm+(c+d)M.
Subtracting these two equations we obtain
(5.11)(M-m){α(1-a)(M+m)+(c+d-b)}=0.
Under the conditions d≥b+c,  a<1,we see that
(5.12)M=m.
It follows by Theorem B that x¯ is a global attractor of (1.1) and then the proof is complete.


## 6.   Numerical Examples

For confirming the results of this paper, we consider numerical examples which represent different types of solutions to (1.1).Example 6.1.
We assumex-3=2,x-2=3,x-1=8,x0=5,a=0.6,b=7,c=3,d=9,  α=3.8,β=0.2,  and  γ=1.2. See Figure 1.Figure 1Example 6.2 (see [Figure2]).
Sincex-3=5,x-2=6,x-1=3,x0=4,a=1.3,b=5,c=3,d=8,  α=6,β=4,  and   γ=7.Figure 2Example 6.3.
We considerx-3=5,x-2=7,x-1=11,x0=3,  a=5,b=2,  c=3,  d=4,α=3,  β=2,  and  γ=7. See Figure 3.Figure 3Example 6.4 (see [Figure4]).
Sincex-3=7,x-2=2,x-1=3,x0=5,a=0.1,b=2,c=5,d=1.4,α=1.3,β=2,andγ=3.Figure 4Example 6.5 (see [Figure5]).
It shows the solutions whenx-3=x-1=p,x-2=x0=q,a=0.6,b=7,c=3,d=9,α=3.8,β=0.2,γ=1.2(6.1)Sincep,q=((e-c)±[e-c]2-4(eaβ+cf)(e-c)/(β-f)(1+a)2(f+aβ)).Figure 5



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*Source: 746738-2012-10-31.xml*