# Solution and Attractivity for a Rational Recursive Sequence

Authors: E. M. Elsayed
Journal: Discrete Dynamics in Nature and Society
 (2011)

---

## Abstract

This paper is concerned with the behavior of solution of the nonlinear difference
equationxn+1=axn−1+bxnxn−1/(cxn+dxn−2), n=0,1,…, where the initial conditions x−2, x−1, x0 are arbitrary positive real numbers
and a,b,c,d are positive constants. Also, we give specific form of the solution of four special cases of this equation.


---

## Body

## 1. Introduction

In this paper we deal with the behavior of the solution of the following difference equation:(1.1)xn+1=axn-1+bxnxn-1cxn+dxn-2,n=0,1,…,
where the initial conditions x-2,x-1,x0 are arbitrary positive real numbers and a,b,c,d are positive constants. Also, we obtain the solution of some special cases of (1.1).Let us introduce some basic definitions and some theorems that we need in the sequel.LetI be some interval of real numbers and let(1.2)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(1.3)xn+1=f(xn,xn-1,…,xn-k),n=0,1,…,
has a unique solution {xn}n=-k∞ [1].Definition 1.1 (equilibrium point).
A pointx¯∈I is called an equilibrium point of (1.3) if
(1.4)x¯=f(x¯,x¯,…,x¯).
That is, xn=x¯ for n≥0,is a solution of (1.3), or equivalently, x¯ is a fixed point of f.Definition 1.2 (stability).
(i) The equilibrium pointx¯of (1.3) is locally stable if for every ϵ>0, there exists δ>0such that for all x-k,x-k+1,…,x-1,x0∈I with
(1.5)|x-k-x¯|+|x-k+1-x¯|+⋯+|x0-x¯|<δ,
we have
(1.6)|xn-x¯|<ϵ∀n≥-k.
(ii) The equilibrium pointx¯of (1.3) is locally asymptotically stable if x¯is locally stable solution of (1.3) and there exists γ>0, such that for all x-k,x-k+1,…,x-1,x0∈I with
(1.7)|x-k-x¯|+|x-k+1-x¯|+⋯+|x0-x¯|<γ,
we have
(1.8)lim⁡n→∞xn=x¯.
(iii) The equilibrium pointx¯ of (1.3) is global attractor if for all x-k,x-k+1,…,x-1,x0∈I, we have
(1.9)lim⁡n→∞xn=x¯.
(iv) The equilibrium pointx¯of (1.3) is globally asymptotically stable if x¯is locally stable, and x¯is also a global attractor of (1.3).
(v) The equilibrium pointx¯of (1.3) is unstable if x¯is not locally stable.The linearized equation of (1.3) about the equilibrium x¯is the linear difference equation
(1.10)yn+1=∑i=0k∂f(x¯,x¯,…,x¯)∂xn-iyn-i.Theorem A (see [2]).
Assume thatp,q∈R and k∈{0,1,2,…}. Then
(1.11)|p|+|q|<1,
is a sufficient condition for the asymptotic stability of the difference equation
(1.12)xn+1+pxn+qxn-k=0,n=0,1,….Remark 1.3.
Theorem A can be easily extended to a general linear equations of the form(1.13)xn+k+p1xn+k-1+⋯+pkxn=0,n=0,1,…,
where p1,p2,…,pk∈R and k∈{1,2,…}. Then (1.13) is asymptotically stable provided that
(1.14)∑i=1k|pi|<1.Consider the following equation(1.15)xn+1=g(xn,xn-1,xn-2).
The following theorem will be useful for the proof of our results in this paper.Theorem B (see [1]).
Let[a,b] be an interval of real numbers and assume that
(1.16)g:[a,b]3⟶[a,b],
is a continuous function satisfying the following properties:(a)
g(x,y,z) is nondecreasing in x and y in [a,b] for each z∈[a,b], and is nonincreasing in z∈[a,b] for each x and y in [a,b];(b)
if(m,M)∈[a,b]×[a,b] is a solution of the system(1.17)M=g(M,M,m),m=g(m,m,M),
then
(1.18)m=M.
Then (1.15) has a unique equilibrium x¯∈[a,b] and every solution of (1.15) converges to x¯.Definition 1.4 (periodicity).
A sequence{xn}n=-k∞is said to be periodic with period p if  xn+p=xnfor all n≥-k.Definition 1.5 (Fibonacci sequence).
The sequence{Fm}m=0∞={1,2,3,5,8,13,…}, that is,  Fm=Fm-1+Fm-2,m≥0,F-2=0,F-1=1 is called Fibonacci sequence.Recently there has been a great interest in studying the qualitative properties of rational difference equations. Some prototypes for the development of the basic theory of the global behavior of nonlinear difference equations of order greater than one come from the results for rational difference equations.However, there have not been any effective general methods to deal with the global behavior of rational difference equations of order greater than one so far. From the known work, one can see that it is extremely difficult to understand thoroughly the global behaviors of solutions of rational difference equations although they have simple forms (or expressions). One can refer to [3–23] for examples to illustrate this. Therefore, the study of rational difference equations of order greater than one is worth further consideration.Many researchers have investigated the behavior of the solution of difference equations, for example, Aloqeili [24] has obtained the solutions of the difference equation(1.19)xn+1=xn-1a-xnxn-1.
Amleh et al. [25] studied the dynamics of the difference equation(1.20)xn+1=a+bxn-1A+Bxn-2.Çinar [26, 27] got the solutions of the following difference equation(1.21)xn+1=xn-1±1+axnxn-1.
In [28], Elabbasy et al. investigated the global stability and periodicity character and gave the solution of special case of the following recursive sequence(1.22)xn+1=axn-bxncxn-dxn-1.
Elabbasy et al. [29] investigated the global stability, boundedness, and periodicity character and gave the solution of some special cases of the difference equation(1.23)xn+1=αxn-kβ+γ∏i=0kxn-i.
In [30], Ibrahim got the form of the solution of  the rational difference equation (1.24)xn+1=xnxn-2xn-1(a+bxnxn-2).
Karatas et al. [31] got the solution of the difference equation(1.25)xn+1=xn-51+xn-2xn-5.
Yalçinkaya and Çinar [32] considered the dynamics of the difference equation(1.26)xn+1=axn-kb+cxnp.
Yang [33] investigated the global asymptotic stability of the difference equation (1.27)xn+1=xn-1xn-2+xn-3+axn-1+xn-2xn-3+a.
See also [1, 2, 30, 31, 34–40]. Other related results on rational difference equations can be found in [32, 33, 41–48].


## 2. Local Stability of (1.1)

In this section we investigate the local stability character of the solutions of (1.1). Equation (1.1) has a unique equilibrium point and is given by(2.1)x¯=ax¯+bx¯2cx¯+dx¯,
or(2.2)x¯2(1-a)(c+d)=bx¯2,
if (c+d)(1-a)≠b, then the unique equilibrium point is x¯=0.Letf:(0,∞)3→(0,∞) be a function defined by(2.3)f(u,v,w)=av+buvcu+dw.
Therefore it follows that (2.4)fu(u,v,w)=bdvw(cu+dw)2,fv(u,v,w)=a+bucu+dw,fw(u,v,w)=-bduv(cu+dw)2,
we see that(2.5)fu(x¯,x¯,x¯)=bd(c+d)2,fv(x¯,x¯,x¯)=a+bc+d,fw(x¯,x¯,x¯)=-bd(c+d)2.
The linearized equation of (1.1) about x¯ is(2.6)yn+1-bd(c+d)2yn-(a+bc+d)yn-1+bd(c+d)2yn-1=0.Theorem 2.1.
Assume that(2.7)b(c+3d)<(1-a)(c+d)2.
Then the equilibrium point of (1.1) is locally asymptotically stable.Proof.
It follows from Theorem A that (2.6) is asymptotically stable if
(2.8)|bd(c+d)2|+|a+bc+d|+|bd(c+d)2|<1,
or
(2.9)a+bc+d+2bd(c+d)2<1,
and so,
(2.10)b(c+3d)(c+d)2<(1-a).
The proof is complete.


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

In this section we investigate the global attractivity character of solutions of (1.1).Theorem 3.1.
The equilibrium pointx¯ of (1.1) is global attractor if c(1-a)≠b.Proof.
Letp,q be real numbers and assume thatg:[p,q]3→[p,q] is a function defined by g(u,v,w)=av+buv/(cu+dw), then we can easily see that the function g(u,v,w) is increasing in u,v and decreasing in w.Suppose that (m,M)is a solution of the system
(3.1)M=g(M,M,m),m=g(m,m,M).
Then from (1.1), we see that
(3.2)M=aM+bM2cM+dm,m=am+bm2cm+dM,
or
(3.3)M(1-a)=bM2cM+dm,m(1-a)=bm2cm+dM,
then
(3.4)d(1-a)Mm+c(1-a)M2=bM2,d(1-a)Mm+c(1-a)m2=bm2,
subtracting, we obtain
(3.5)c(1-a)(M2-m2)=b(M2-m2),c(1-a)≠b.
Thus
(3.6)M=m.
It follows from Theorem B that x¯ is a global attractor of (1.1), and then the proof is complete.


## 4. Boundedness of Solutions of (1.1)

In this section we study the boundedness of solutions of (1.1).Theorem 4.1.
Every solution of (1.1) is bounded if (a+b/c)<1.Proof.
Let{xn}n=-2∞ be a solution of (1.1). It follows from (1.1) that
(4.1)xn+1=axn-1+bxnxn-1cxn+dxn-2≤axn-1+bxnxn-1cxn=(a+bc)xn-1.
Then
(4.2)xn+1≤xn-1∀n≥0.
Then the subsequences {x2n-1}n=0∞, {x2n}n=0∞ are decreasing and so are bounded from above by M=max{x-2,x-1,x0}.


## 5. Special Cases of (1.1)

Our goal in this section is to find a specific form of the solutions of some special cases of (1.1) when a,b,c, and d are integers and give numerical examples of each case and draw it by using MATLAB 6.5.


### 5.1. On the Difference Equationxn+1=xn-1+xnxn-1/(xn+xn-2)

In this subsection we study the following special case of (1.1):
(5.1)xn+1=xn-1+xnxn-1xn+xn-2,
where the initial conditions x-2,x-1,x0are arbitrary positive real numbers.Theorem 5.1.
Let{xn}n=-2∞ be a solution of (5.1). Then for n=0,1,2,…(5.2)x2n-1=k∏i=0n-1(F4i+3h+F4i+2rF4i+2h+F4i+1r),x2n=h∏i=0n-1(F4i+5h+F4i+4rF4i+4h+F4i+3r),
wherex-2=r,x-1=k,x0=h,{Fm}m=0∞={0,1,1,2,3,5,8,13,…}.Proof.
Forn=0 the result holds. Now suppose that n>0 and that our assumption holds for n-1,n-2. That is,
(5.3)x2n-3=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r),x2n-2=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r),x2n-4=h∏i=0n-3(F4i+5h+F4i+4rF4i+4h+F4i+3r).
Now, it follows from (5.1) that
(5.4)x2n-1=x2n-3+x2n-2x2n-3x2n-2+x2n-4=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))+h∏i=0n-3((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+((F4n-3h+F4n-4r)/(F4n-4h+F4n-5r))k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))((F4n-3h+F4n-4r)/(F4n-4h+F4n-5r))+1=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+(F4n-3h+F4n-4r)k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))F4n-3h+F4n-4r+F4n-4h+F4n-5r=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+(F4n-3h+F4n-4r)k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))F4n-2h+F4n-3r=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)(1+F4n-3h+F4n-4rF4n-2h+F4n-3r)=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)(F4n-1h+F4n-2rF4n-2h+F4n-3r).
Therefore
(5.5)x2n-1=k∏i=0n-1(F4i+3h+F4i+2rF4i+2h+F4i+1r).
Also, we see from (5.1) that
(5.6)x2n=x2n-2+x2n-1x2n-2x2n-1+x2n-3=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+k∏i=0n-1((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))k∏i=0n-1((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))+k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+((F4n-1h+F4n-2r)/(F4n-2h+F4n-3r))h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))(F4n-1h+F4n-2r/F4n-2h+F4n-3r)+1=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+(F4n-1h+F4n-2r)h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))F4n-1h+F4n-2r+F4n-2h+F4n-3r=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+(F4n-1h+F4n-2r)h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))F4nh+F4n-1r=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)(1+F4n-1h+F4n-2rF4nh+F4n-1r)=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)(F4n+1h+F4nrF4nh+F4n-1r).
Thus
(5.7)x2n=h∏i=0n-1(F4i+5h+F4i+4rF4i+4h+F4i+3r).
Hence, the proof is completed.For confirming the results of this section, we consider numerical example forx-2=7,x-1=6,x0=3.(See Figure 1).Figure 1


### 5.2. On the Difference Equationxn+1=xn-1+xnxn-1/(xn-xn-2)

In this subsection we give a specific form of the solutions of the difference equation(5.8)xn+1=xn-1+xnxn-1xn-xn-2,
where the initial conditions x-2,x-1,x0are arbitrary positive real numbers with x-2≠x0.Theorem 5.2.
Let{xn}n=-2∞ be a solution of (5.8). Then for n=0,1,2,…(5.9)x2n-1=k∏i=0n-1(F2i+3h-F2i+1rF2i+1h-F2i-1r),x2n=h∏i=0n-1(F2i+4h-F2i+2rF2i+2h-F2ir),
where x-2=r,x-1=k,x0=h,{Fm}m=-1∞={1,0,1,1,2,3,5,8,13,…}.Proof.
Forn=0 the result holds. Now suppose that n>0 and that our assumption holds for n-1,n-2. That is,
(5.10)x2n-3=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r),x2n-2=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir),x2n-4=h∏i=0n-3(F2i+4h-F2i+2rF2i+2h-F2ir).
Now, it follows from (5.8) that
(5.11)x2n-1=x2n-3+x2n-2x2n-3x2n-2-x2n-4=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)+h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))-h∏i=0n-3((F2i+4h-F2i+2r)/(F2i+2h-F2ir))=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)+((F2nh-F2n-2r)/(F2n-2h-F2n-4r))k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))(F2nh-F2n-2r/F2n-2h-F2n-4r)-1=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)+(F2nh-F2n-2r)k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))F2n-1h-F2n-3r=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)(1+F2nh-F2n-2rF2n-1h-F2n-3r)=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)(F2n-1h-F2n-3r+F2nh-F2n-2rF2n-1h-F2n-3r)=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)(F2n+1h-F2n-1rF2n-1h-F2n-3r).
Therefore
(5.12)x2n-1=k∏i=0n-1(F2i+3h-F2i+1rF2i+1h-F2i-1r).
Also, we see from (5.8) that
(5.13)x2n=x2n-2+x2n-1x2n-2x2n-1-x2n-3=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+k∏i=0n-1((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))k∏i=0n-1((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))-k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+((F2n+1h-F2n-1r)/(F2n-1h-F2n-3r))h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))((F2n+1h-F2n-1r)/(F2n-1h-F2n-3r))-1=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+(F2n+1h-F2n-1r)h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))F2n+1h-F2n-1r-F2n-1h+F2n-3r=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+(F2n+1h-F2n-1r)h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))F2nh-F2n-2r=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)(1+F2n+1h-F2n-1rF2nh-F2n-2r)=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)(F2n+2h-F2nrF2nh-F2n-2r).
Thus
(5.14)x2n=h∏i=0n-1(F2i+4h-F2i+2rF2i+2h-F2ir).
Hence, the proof is completed.Assume thatx-2=0.7,x-1=1.6,x0=13. (See Figure 2), and for x-2=9,x-1=5,x0=2. (See Figure 3).Figure 2Figure 3The following cases can be treated similarly.


### 5.3. On the Difference Equationxn+1=xn-1-xnxn-1/(xn+xn-2)

In this subsection we obtain the solution of the following difference equation(5.15)xn+1=xn-1-xnxn-1xn+xn-2,
where the initial conditions x-2,x-1,x0 are arbitrary positive real numbers.Theorem 5.3.
Let{xn}n=-2∞ be a solution of (5.15). Then for n=0,1,2,…(5.16)x2n-1=k∏i=0n-1(F2ih+F2i+1rF2i+1h+F2i+2r),x2n=h∏i=0n-1(F2i+1h+F2i+2rF2i+2h+F2i+3r),
where x-2=r,x-1=k,x0=h,{Fm}m=0∞={0,1,1,2,3,5,8,13,…}.
Figure4 shows the solution when x-2=3,x-1=7,x0=12.Figure 4


### 5.4. On the Difference Equationxn+1=xn-1-xnxn-1/(xn-xn-2)

In this subsection we give the solution of the following special case of (1.1)
(5.17)xn+1=xn-1-xnxn-1xn-xn-2
where the initial conditions x-2,x-1,x0 are arbitrary real numbers. with x-2≠x0,x-2,x-1,x0≠0.Theorem 5.4.
Let{xn}n=-2∞ be a solution of (5.17). Then every solution of (5.17) is periodic with period 12. Moreover {xn}n=-2∞ takes the form
(5.18){r,k,h,krr-h,h-r,hkr-h,-r,-k,-h,krh-r,-(h-r),hkh-r,r,k,h,…},
where x-2=r,x-1=k,x0=h.
Figure5 shows the solution when x-2=3,x-1=7,x0=2.Figure 5


## 5.1. On the Difference Equationxn+1=xn-1+xnxn-1/(xn+xn-2)

In this subsection we study the following special case of (1.1):
(5.1)xn+1=xn-1+xnxn-1xn+xn-2,
where the initial conditions x-2,x-1,x0are arbitrary positive real numbers.Theorem 5.1.
Let{xn}n=-2∞ be a solution of (5.1). Then for n=0,1,2,…(5.2)x2n-1=k∏i=0n-1(F4i+3h+F4i+2rF4i+2h+F4i+1r),x2n=h∏i=0n-1(F4i+5h+F4i+4rF4i+4h+F4i+3r),
wherex-2=r,x-1=k,x0=h,{Fm}m=0∞={0,1,1,2,3,5,8,13,…}.Proof.
Forn=0 the result holds. Now suppose that n>0 and that our assumption holds for n-1,n-2. That is,
(5.3)x2n-3=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r),x2n-2=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r),x2n-4=h∏i=0n-3(F4i+5h+F4i+4rF4i+4h+F4i+3r).
Now, it follows from (5.1) that
(5.4)x2n-1=x2n-3+x2n-2x2n-3x2n-2+x2n-4=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))+h∏i=0n-3((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+((F4n-3h+F4n-4r)/(F4n-4h+F4n-5r))k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))((F4n-3h+F4n-4r)/(F4n-4h+F4n-5r))+1=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+(F4n-3h+F4n-4r)k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))F4n-3h+F4n-4r+F4n-4h+F4n-5r=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)+(F4n-3h+F4n-4r)k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))F4n-2h+F4n-3r=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)(1+F4n-3h+F4n-4rF4n-2h+F4n-3r)=k∏i=0n-2(F4i+3h+F4i+2rF4i+2h+F4i+1r)(F4n-1h+F4n-2rF4n-2h+F4n-3r).
Therefore
(5.5)x2n-1=k∏i=0n-1(F4i+3h+F4i+2rF4i+2h+F4i+1r).
Also, we see from (5.1) that
(5.6)x2n=x2n-2+x2n-1x2n-2x2n-1+x2n-3=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+k∏i=0n-1((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))k∏i=0n-1((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))+k∏i=0n-2((F4i+3h+F4i+2r)/(F4i+2h+F4i+1r))=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+((F4n-1h+F4n-2r)/(F4n-2h+F4n-3r))h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))(F4n-1h+F4n-2r/F4n-2h+F4n-3r)+1=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+(F4n-1h+F4n-2r)h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))F4n-1h+F4n-2r+F4n-2h+F4n-3r=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)+(F4n-1h+F4n-2r)h∏i=0n-2((F4i+5h+F4i+4r)/(F4i+4h+F4i+3r))F4nh+F4n-1r=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)(1+F4n-1h+F4n-2rF4nh+F4n-1r)=h∏i=0n-2(F4i+5h+F4i+4rF4i+4h+F4i+3r)(F4n+1h+F4nrF4nh+F4n-1r).
Thus
(5.7)x2n=h∏i=0n-1(F4i+5h+F4i+4rF4i+4h+F4i+3r).
Hence, the proof is completed.For confirming the results of this section, we consider numerical example forx-2=7,x-1=6,x0=3.(See Figure 1).Figure 1


## 5.2. On the Difference Equationxn+1=xn-1+xnxn-1/(xn-xn-2)

In this subsection we give a specific form of the solutions of the difference equation(5.8)xn+1=xn-1+xnxn-1xn-xn-2,
where the initial conditions x-2,x-1,x0are arbitrary positive real numbers with x-2≠x0.Theorem 5.2.
Let{xn}n=-2∞ be a solution of (5.8). Then for n=0,1,2,…(5.9)x2n-1=k∏i=0n-1(F2i+3h-F2i+1rF2i+1h-F2i-1r),x2n=h∏i=0n-1(F2i+4h-F2i+2rF2i+2h-F2ir),
where x-2=r,x-1=k,x0=h,{Fm}m=-1∞={1,0,1,1,2,3,5,8,13,…}.Proof.
Forn=0 the result holds. Now suppose that n>0 and that our assumption holds for n-1,n-2. That is,
(5.10)x2n-3=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r),x2n-2=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir),x2n-4=h∏i=0n-3(F2i+4h-F2i+2rF2i+2h-F2ir).
Now, it follows from (5.8) that
(5.11)x2n-1=x2n-3+x2n-2x2n-3x2n-2-x2n-4=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)+h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))-h∏i=0n-3((F2i+4h-F2i+2r)/(F2i+2h-F2ir))=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)+((F2nh-F2n-2r)/(F2n-2h-F2n-4r))k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))(F2nh-F2n-2r/F2n-2h-F2n-4r)-1=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)+(F2nh-F2n-2r)k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))F2n-1h-F2n-3r=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)(1+F2nh-F2n-2rF2n-1h-F2n-3r)=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)(F2n-1h-F2n-3r+F2nh-F2n-2rF2n-1h-F2n-3r)=k∏i=0n-2(F2i+3h-F2i+1rF2i+1h-F2i-1r)(F2n+1h-F2n-1rF2n-1h-F2n-3r).
Therefore
(5.12)x2n-1=k∏i=0n-1(F2i+3h-F2i+1rF2i+1h-F2i-1r).
Also, we see from (5.8) that
(5.13)x2n=x2n-2+x2n-1x2n-2x2n-1-x2n-3=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+k∏i=0n-1((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))k∏i=0n-1((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))-k∏i=0n-2((F2i+3h-F2i+1r)/(F2i+1h-F2i-1r))=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+((F2n+1h-F2n-1r)/(F2n-1h-F2n-3r))h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))((F2n+1h-F2n-1r)/(F2n-1h-F2n-3r))-1=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+(F2n+1h-F2n-1r)h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))F2n+1h-F2n-1r-F2n-1h+F2n-3r=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)+(F2n+1h-F2n-1r)h∏i=0n-2((F2i+4h-F2i+2r)/(F2i+2h-F2ir))F2nh-F2n-2r=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)(1+F2n+1h-F2n-1rF2nh-F2n-2r)=h∏i=0n-2(F2i+4h-F2i+2rF2i+2h-F2ir)(F2n+2h-F2nrF2nh-F2n-2r).
Thus
(5.14)x2n=h∏i=0n-1(F2i+4h-F2i+2rF2i+2h-F2ir).
Hence, the proof is completed.Assume thatx-2=0.7,x-1=1.6,x0=13. (See Figure 2), and for x-2=9,x-1=5,x0=2. (See Figure 3).Figure 2Figure 3The following cases can be treated similarly.


## 5.3. On the Difference Equationxn+1=xn-1-xnxn-1/(xn+xn-2)

In this subsection we obtain the solution of the following difference equation(5.15)xn+1=xn-1-xnxn-1xn+xn-2,
where the initial conditions x-2,x-1,x0 are arbitrary positive real numbers.Theorem 5.3.
Let{xn}n=-2∞ be a solution of (5.15). Then for n=0,1,2,…(5.16)x2n-1=k∏i=0n-1(F2ih+F2i+1rF2i+1h+F2i+2r),x2n=h∏i=0n-1(F2i+1h+F2i+2rF2i+2h+F2i+3r),
where x-2=r,x-1=k,x0=h,{Fm}m=0∞={0,1,1,2,3,5,8,13,…}.
Figure4 shows the solution when x-2=3,x-1=7,x0=12.Figure 4


## 5.4. On the Difference Equationxn+1=xn-1-xnxn-1/(xn-xn-2)

In this subsection we give the solution of the following special case of (1.1)
(5.17)xn+1=xn-1-xnxn-1xn-xn-2
where the initial conditions x-2,x-1,x0 are arbitrary real numbers. with x-2≠x0,x-2,x-1,x0≠0.Theorem 5.4.
Let{xn}n=-2∞ be a solution of (5.17). Then every solution of (5.17) is periodic with period 12. Moreover {xn}n=-2∞ takes the form
(5.18){r,k,h,krr-h,h-r,hkr-h,-r,-k,-h,krh-r,-(h-r),hkh-r,r,k,h,…},
where x-2=r,x-1=k,x0=h.
Figure5 shows the solution when x-2=3,x-1=7,x0=2.Figure 5


## 6. Conclusion

This paper discussed global stability, boundedness, and the solutions of some special cases of (1.1). In Section 2 we proved whenb(c+3d)<(1-a)(c+d)2, (1.1) local stability.In Section 3 we showed that the unique equilibrium of (1.1) is globally asymptotically stable if c(1-a)≠b. In Section 4 we proved that the solution of (1.1) is bounded if (a+b/c)<1. In Section 5 we gave the form of the solution of four special cases of (1.1) and gave numerical examples of each case and drew them by using Matlab 6.5.



---

*Source: 982309-2011-06-08.xml*