# The Dynamics of the Solutions of Some Difference Equations

Authors: H. El-Metwally; R. Alsaedi; E. M. Elsayed
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

This paper is devoted to investigate the global behavior of the following rational difference equation:yn+1=αyn-t/(β+γ∑i=0kyn-(2i+1)p∏i=0kyn-(2i+1)q),n=0,1,2,…, where α,β,γ,p,q∈(0,∞) and k,t∈{0,1,2,…} with the initial conditions x0,x-1,…,x-2k,x-2max{k,t}-1∈ (0,∞). We will find and classify the equilibrium points of the equations under studying and then investigate their local and global stability. Also, we will study the oscillation and the permanence of the considered equations.


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## Body

## 1. Introduction

The aim of this paper is to study the dynamics of the solutions of the following recursive sequence:(1.1)yn+1=αyn-tβ+γ∑i=0kyn-(2i+1)p∏i=0kyn-(2i+1)q,n=0,1,2,…,
where α,β,γ,p,q∈(0,∞) and K∈{0,1,2,…}, where K=max{k,t}, with the initial conditions x0,  x-1,…,  x-2k,x-2K-1∈(0,∞). We deal with the classification of the equilibrium points of (1.1) in terms of being stable or unstable, where we investigate the global attractor of the solutions of (1.1) as well as the permanence of the equation. Also, we establish some appropriate conditions, which grantee the oscillation of the solutions of (1.1). For more results in the direction of this study, see, for example, [1–23] and the papers therein.In the sequel, we present some well-known results and definition that will be useful in our investigation of (1.1). Let I be some interval of real numbers and let 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
(1.2)xn+1=f(xn,xn-1,…,xn-k),n=0,1,…,
has a unique solution {xn}n=-k∞.Definition 1.1 (permanence).
The difference equation (1.2) is said to be permanent if there exist numbers m and M with 0<m≤M<∞ such that, for any initial conditions x-k,x-k+1,…,x-1,x0∈(0,∞), there exists a positive integer N which depends on the initial conditions such that m≤xn≤M for all n≥N.Definition 1.2 (periodicity).
A sequence{xn}n=-k∞ is said to be periodic with period p if xn+p=xn for all n≥-k.Definition 1.3 (semicycles).
A positive semicycle of a sequence{xn}n=-k∞ consists of a “string” of terms {xl,xl+1,…,xm} all greater than or equal to the equilibrium point x¯, with l≥-k and m≤∞ such that either l=-k or l>-k and xl-1<x¯; either m=∞ or m<∞ and xm+1<x¯. A negative semicycle of a sequence {xn}n=-k∞ consists of a “string” of terms {xl,xl+1,…,xm} all less than the equilibrium point x¯, with l≥-k and m≤∞ such that either l=-k or l>-k and xl-1≥x¯; either m=∞ or m<∞ and xm+1≥x¯.Definition 1.4 (oscillation).
A sequence{xn}n=-k∞ is called nonoscillatory about the point x¯ if there exists N≥-k such that either xn>x¯ for all n≥N or xn<x¯ for all n≥N. Otherwise, {xn}n=-k∞ is called oscillatory about x¯.


## 2. Dynamics of (1.1)

The change of variablesyn=(β/γ)1/[p+(k+1)q]xn reduces (1.1) to the following difference equation
(2.1)xn+1=rxn-t1+∑i=0kxn-(2i+1)p∏i=0kxn-(2i+1)q,n=0,1,2,…,
where r=α/β.In this section, we study the local stability character and the global stability of the equilibrium points of the solutions of (2.1). Also, we give some results about the oscillation and the permanence of (2.1).Recall that the equilibrium points of (2.1) are given by
(2.2)x¯=rx¯1+(k+1)x¯p+(k+1)q.
Then, whenever r≤1, (2.1) has the only equilibrium point x¯=0, and, while at r>1, (2.1) possesses the unique positive equilibrium point x¯=((r-1)/(k+1))1/(p+(k+1)q).The following theorem deals with the local stability of the equilibrium pointx¯=0 of (2.1).Theorem 2.1.
The following statements are true.(i)
Ifr<1, then the equilibrium point x¯=0 of (2.1) is locally asymptotically stable.(ii)
Ifr>1, then the equilibrium point x¯=0 of (2.1) is a saddle point.(iii)
Ifr=1, then the equilibrium point x¯=0 of (2.1) is nonhyperbolic with λ=0<1 and λ=1.Proof.
The linearized equation of (2.1) about x¯=0 is un+1-run-t=0. Then, the associated eigenvalues are λ=0 and λ=r. Then, the proof is complete.Theorem 2.2.
Assume thatr<1, then the equilibrium point x¯=0 of (2.1) is globally asymptotically stable.Proof.
Let{xn}n=-2k-1∞ be a solution of (2.1). It was shown in Theorem 2.1 that the equilibrium point  x¯=0 of (2.1) is locally asymptotically stable. So it suffices to show that
(2.3)lim⁡n→∞⁡xn=0.
Now, it follows from (2.1) that
(2.4)xn+1=rxn-t1+xn-1p+qxn-3q⋯xn-(2k+1)q+xn-1qxn-3p+q⋯xn-(2k+1)q+⋯+xn-1qxn-3q⋯xn-(2k+1)p+q≤rxn-2t.
Now, assume that X=max{x-2t,x-2t+1,…,x-1,x0/r}. Then, it follows from (1.1) and after some simple computations are achieved that x2(n-1)t+i≤rnX, i=0,1,…,2t. Therefore xn→0 as n→∞, and this completes the proof.Theorem 2.3.
Assume thatr>1.  Then, every solution of (2.1) is either oscillatory or tends to the equilibrium point  x-=((r-1)/(k+1))1/(p+(k+1)q).Proof.
Let{xn}n=-2K-1∞ be a solution of (2.1). Without loss of generality, assume that {xn}n=-2K-1∞ is a nonoscillatory solution of (2.1), then it is suffices to show that limn→∞xn=x¯. Assume that xn≥x¯ for n≥n0 (the case where xn≤x¯ for n≥n0 is similar and will be omitted). It follows from (2.1) that
(2.5)xn+1=rxn-t1+xn-1p+qxn-3q⋯xn-(2k+1)q+xn-1qxn-3p+q⋯xn-(2k+1)q+⋯+xn-1qxn-3q⋯xn-(2k+1)p+q≤xn-t(r1+(k+1)x¯p+(k+1)q)=xn-t.
Hence, each subsequence {x(t+1)n+i},  j=0,1,…,t, of {xn}n=0∞ is decreasing sequence and therefore it has a limit. Let for some j=0,1,…,t, limn→∞x(t+1)n+j=μ, and, for the sake of contradiction, assume that μ>x¯. Then, by taking the limit of both sides of
(2.6)x(t+1)n+j+1=rx(t+1)n+j-t1+∑s=0kx(t+1)n+j-(2s+1)p∏s=0kx(t+1)n+j-(2s+1)q,n=0,1,2,…,
we obtain μ=rμ/(1+(k+1)μp+(k+1)q), which contradicts the hypothesis that x¯=((r-1)/(k+1))1/(p+(k+1)q) is the only positive solution of (2.2). Therefore, limn→∞x(t+1)n+j=μ, for all j=0,1,…,t. This means that all the subsequences {x(t+1)n+i},  j=0,1,…,t, of {xn}n=0∞ have the same limit, x¯, and therefore limn→∞xn=x¯, which completes the proof.Theorem 2.4.
Assume thatt=0, r>1, and let {xn}n=-2k-1∞ be a solution of (2.1) which is strictly oscillatory about the positive equilibrium point x¯=((r-1)/(k+1))1/(p+(k+1)q) of (2.1). Then, the extreme point in any semicycle occurs in one of the first 2(k+1) terms of the semicycle.Proof.
Assume that{xn}n=-2k-1∞ is a strictly oscillatory solution of (2.1). Let L≥M≥N≥2(k+1), and let {xN,xN+1,…,xM} be a positive semicycle followed by the negative semicycle {xM,xM+1,…,xL}. Now, it follows from (2.1) that
(2.7)xN+2(k+1)-xN=rxN+2k+11+xN+2kp+qxN+2k-2q⋯xNq+xN+2kqxN+2k-2p+q⋯xNq+⋯+xN+2kq⋯xNp+q-xN≤xN+2k+1(r1+(k+1)x¯p+(k+1)q)-xN=xN+2k+1-xN=rxN+2(k+t)1+xN+2k-1p+qxN+2k-3q⋯xN-1q+⋯+xN+2k-1q⋯xN-1p+q-xN≤xN+2k(r1+(k+1)x¯p+(k+1)q)-xN=xN+2k-xN≤xN+2k-1-xN≤⋯≤xN+1-xN≤xN-xN=0.
Then, xN≥xN+2(k+1) for all N≥2(k+1).
Similarly, we see from (2.1) that
(2.8)xM+2(k+1)-xM=rxM+2k+11+xM+2kp+qxM+2k-2q⋯xMq+xM+2kqxM+2k-2p+q⋯xMq+⋯+xM+2kq⋯xMp+q-xM≥xM+2k+1(r1+(k+1)x¯p+(k+1)q)-xM=xM+2k+1-xM=rxM+2k1+xM+2k-1p+qxM+2k-3q⋯xM-1q+⋯+xM+2k-1q⋯xM-1p+q-xM≥xM+2k(r1+(k+1)x¯p+(k+1)q)-xM=xM+2k-xM≤xM+2k-1-xM≥⋯≥xM+1-xM≥xM-xM=0.
Therefore,xM+2(k+1)≥xM for all M≥2(k+1). The proof is complete.Theorem 2.5.
Assume thatt=2k,r>1, and let {xn}n=-2k-1∞ be a solution of (2.1) which is strictly oscillatory about the positive equilibrium point x¯=((r-1)/(k+1))1/(p+(k+1)q) of (2.1). Then, the extreme point in any semicycle occurs in one of the first 2k terms of the semicycle.Proof.
Assume that{xn}n=-2k-1∞ is a strictly oscillatory solution of (2.1). Let L≥M≥N≥2k+1, and  let {xN,xN+1,…,xM} be a positive semicycle followed by the negative semicycle {xM,xM+1,…,xL}. Now, it follows from (2.1) that
(2.9)xN+2k+1-xN=rxN1+xN+2kp+qxN+2k-2q⋯xNq+xN+2kqxN+2k-2p+q⋯xNq+⋯+xN+2kq⋯xNp+q-xN≤xN(r1+(k+1)x¯p+(k+1)q)-xN=xN-xN=0.
Then, xN≥xN+2k for all N≥2k+1.
Similarly, we see from (2.1) that
(2.10)xM+2k+1-xM=rxM1+xM+2kp+qxM+2k-2q⋯xMq+xM+2kqxM+2k-2p+q⋯xMq+⋯+xM+2kq⋯xMp+q-xM≥xM(r1+(k+1)x¯p+(k+1)q)-xM=xM-xM=0.
Therefore, xM+2k≥xM for all M≥2k+1. The proof is complete.Theorem 2.6.
Assume thatt=2k+1, r>1, and let {xn}n=-2k-1∞ be a solution of (2.1) which is strictly oscillatory about the positive equilibrium point x¯=((r-1)/(k+1))1/(p+(k+1)q) of (2.1). Then, the extreme point in any semicycle occurs in one of the first 2k+1 terms of the semicycle.Proof.
Assume that{xn}n=-2k-1∞ be a strictly oscillatory solution of (2.1). Let L≥M≥N≥2k+1 and let {xN,xN+1,…,xM} be a positive semicycle followed by the negative semicycle {xM,xM+1,…,xL}. Now it follows from (2.1) that
(2.11)xN+2(k+1)-xN=rxN1+xN+2k+1p+qxN+2k-1q⋯xNq+xN+2k+1qxN+2k-1p+q⋯xNq+⋯+xN+2k+1q⋯xNp+q-xN≤xN(r1+(k+1)x¯p+(k+1)q)-xN=xN-xN=0.
Then, xN≥xN+2k for all N≥2k+1.
Similarly, we see from (2.1) that
(2.12)xM+2(k+1)-xM=rxM1+xM+2k+1p+qxM+2k-1q⋯xMq+xM+2k+1qxM+2k-1p+q⋯xMq+⋯+xM+2k+1q⋯xMp+q-xM≥xM(r1+(k+1)x¯p+(k+1)q)-xM=xM-xM=0.
Therefore xM+2k≥xM for all M≥2k+1. The proof is complete.Theorem 2.7.
Equation (2.1) is permanent.Proof.
Let{xn}n=-2k-1∞ be a solution of (2.1). There are two cases to consider.(i)
{xn}n=-2k-1∞ is a nonoscillatory solution of (2.1). Then, it follows from Theorem 2.3 that
(2.13)lim⁡n→∞⁡xn=x¯,
that is, there is a sufficiently large positive integer N such that |xn-x¯|<ɛ for all n≥N and for some ɛ>0. So x¯-ɛ<xn<x¯+ɛ, this means that there are two positive real numbers, say C and D, such that C≤xn≤D.(ii)
{xn}n=-2k-1∞ is strictly oscillatory about x¯=((r-1)/(k+1))1/(p+(k+1)q).
Now, let{xs+1,xs+2,…,xt} be a positive semicycle followed by the negative semicycle {xt+1,xt+2,…,xu}. If xV and xW are the extreme values in these positive and negative semicycles, respectively, with the smallest possible indices V and W, then by Theorem 2.4 we see that V-s≤2(k+1) and W-u≤2(k+1). Now, for any positive indices μ and L with μ<L, it follows from (2.1) for n=μ,μ+1,…,L-1 that
(2.14)xL=xL-1(r1+xL-2p+qxL-4q⋯xL-2k-2q+xL-2qxL-4p+q⋯xL-2k-2q+⋯+xL-2qxL-4q⋯xL-2k-2p+q)=r2xL-2(1+xL-3p+q⋯xL-2k-3q+⋯+xL-3q⋯xL-2k-3p+q)(1+xL-2p+q⋯xL-2k-2q+⋯+xL-2q⋯xL-2k-2p+q)⋮=xL-ζrζ∏η=1ζ(11+∑i=0kxL-(2i+1)-ηp∏i=0kxL-(2i+1)-ηq)=xμrL-μ∏η=μL-1(11+∑i=0kxη-(2i+1)p∏i=0kxη-(2i+1)q).
Therefore, for V=L and s=μ, we obtain
(2.15)xV=xsrV-s∏η=sV-1(11+∑i=0kxη-(2i+1)p∏i=0kxη-(2i+1)q)≤x¯r2k+1=H.
Again whenever W=L and μ=t, we see that
(2.16)xW=xtrW-t∏η=tW-1(11+∑i=0kxη-(2i+1)p∏i=0kxη-(2i+1)q)≥x¯rW-t∏η=tW-1(11+(k+1)Hp+(k+1)q)=x¯rW-t(11+(k+1)Hp+(k+1)q)W-t-1≥x¯(11+(k+1)Hp+(k+1)q)2k+1=G.
That is G≤xn≤H. It follows from (i) and (ii) that min{C,G}≤xn≤max{D,H}. Then, the proof is complete.



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