# The Periodicity of Positive Solutions of the Nonlinear Difference Equationxn+1=α+(xn−kp)/(xnq)

Authors: Mehmet Gümüş
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We give a remark about the periodic character of positive
solutions of the difference equationxn+1=α+xn−kp/xnq, n=0,1,…, where k>1 is an odd integer, α,p, q∈(0,∞), and the initial conditions x−k,…,x0 are arbitrary positive numbers.


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## Body

## 1. Introduction

In this paper, we consider the difference equation(1)xn+1=α+xn-kpxnq,n=0,1,…,
where k>1 is an odd integer, α is positive, p,q∈(0,∞), and the initial conditions x-k,…,x0 are arbitrary positive numbers. Equation (1) was studied by many authors for different cases of k,p,q.In [1], the authors studied the special case of k=1 of (1), that is, the recursive sequence
(2)xn+1=α+xn-1pxnq,n=0,1,…,
where α,p,q are positive and the initial values x-1,x0 are positive numbers.Recently, in [2], the authors obtained the periodicity results of positive solutions of (1). They investigated the existence of a prime periodic solution of (1). But they did not investigate the positive solutions of (1) which converge to a prime two-periodic solution.There exist many other papers related with (1) and on its extensions (see [3–8]).Our aim in this paper is to give a remark about the periodic character of all positive solutions of (1). We show that all positive solutions of (1), for k is odd, converge to a prime two-periodic solution.We believe that difference equations, also referred to as recursive sequence, are a hot topic. There has been an increasing interest in the study of qualitative analysis of difference equations and systems of difference equations. Difference equations appear naturally as discrete analogues and as numerical solutions of differential and delay differential equations having applications in biology, ecology, economics, physics, computer sciences, and so on.Here, we recall some notations and results which will be useful in our proofs.LetI be some interval of real numbers and let F be continuous function defined on Ik+1. Then, for initial conditions x-k,…,x0∈I, it is easy to see that the difference equation
(3)xn+1=F(xn,xn-1,…,xn-k),n=0,1,…,
has a unique solution {x}n=-k∞.We say that the equilibrium pointx¯ of (3) is the point that satisfies the condition
(4)x¯=F(x¯,x¯,…,x¯).
That is, xn=x¯ for n≥0 is a solution of (3), or equivalently, x¯ is a fixed point of F.A solution{x}n=-k∞ of (3) is said to be periodic with period p if xn+p=xn for all n≥-k(5)xn+p=xn,∀n≥-k.A solution{x}n=-k∞ of (3) is said to be periodic with prime period p, or a p-cycle if it is periodic with period p and p is the least positive integer for which (5) holds.The linearized equation for (1) about the positive equilibrium x¯ is
(6)yn+1+qx¯p-q-1yn-px¯p-q-1yn-k=0,n=0,1,….


## 2. Main Results

In this section, we find solutions, fork is odd of (1), which converge to a prime two-periodic solution. The following three results are essentially proved in [1, 2]. Hence, we omit their proofs.Lemma 1.
Suppose that(7)p∈(0,1),
then every positive solution of (1) is bounded.Lemma 2.
Assume that k is odd. Then, (1) has prime two-periodic solutions if and only if
(8)0<p<1<q
and there exists a sufficient small positive number ε1, such that
(9)1(α+ε1)q-p>ε1,(α+ε1)p/qε1-1/q<α+ε1-p/q(α+ε1)p2-q2/q.Lemma 3.
If either(10)0<q<p<1,
or
(11)0<p<q,q↛∞
hold, then (1) has a unique equilibrium point x¯.Theorem 4.
Consider (1) where k is odd, for every positive solution {xn} of (1) which satisfies any of the following initial conditions: (i)
x-k+2l≥x-k+2l-2 for  all l=1,2,…,(k-1)/2 and x-k+2s+1≤x-k+2s-1 for  all s=1,2,…,(k+1)/2.(ii)
x-k+2l≤x-k+2l-2 for  all l=1,2,…,(k-1)/2 and x-k+2s+1≥x-k+2s-1for  alls=1,2,…,(k+1)/2.(iii)
x-k+2l≥x-k+2l-2for  alll=1,2,…,(k-1)/2 and x-k+2s+1≥x-k+2s-1for  alls=1,2,…,(k+1)/2.(iv)
x-k+2l≤x-k+2l-2for  alll=1,2,…,(k-1)/2 and x-k+2s+1≤x-k+2s-1for  alls=1,2,…,(k+1)/2.Then, the sequences{x2n} and {x2n+1} are eventually monotone.Proof.
We have(12)x2n+1-x2n-1=x2n-2q(x2n-kp-x2n-k-2p)+x2n-k-2p(x2n-2q-x2nq)(x2nx2n-2)q,(13)x2n+2-x2n=x2n-1q(x2n-k+1p-x2n-k-1p)+x2n-k-1p(x2n-1q-x2n+1q)(x2n+1x2n-1)q.
If x-k+2l≥x-k+2l-2for  alll=1,2,…,(k-1)/2 and x-k+2s+1≤x-k+2s-1 for  alls=1,2,…,(k+1)/2, from (12)  and (13) we obtain x3≥x1 and consequently x4≤x2. By induction we obtain
(14)x0≥x2≥⋯≥x2n≥⋯and⋯≥x2n+1≥x2n-1≥⋯≥x1.
Similarly if x-k+2l≤x-k+2l-2for  alll=1,2,…,(k-1)/2 and x-k+2s+1≥x-k+2s-1  for  alls=1,2,…,(k+1)/2, from (12) and (13) using induction we obtain
(15)x0≤x2≤⋯≤x2n≤⋯and⋯≤x2n+1≤x2n-1≤⋯≤x1.
If x-k+2l≥x-k+2l-2for  alll=1,2,…,(k-1)/2 and x-k+2s+1≥x-k+2s-1for  alls=1,2,…,(k+1)/2 and x1≥x3 we can obtain from (12) and (13)
(16)x0≤x2≤⋯≤x2n≤⋯and⋯≤x2n+1≤x2n-1≤⋯≤x1.
Therefore, the result follows immediately.The following result is essentially proved in [1] for k=1 of (1). We obtain the same result, for k>1 is an odd integer.Lemma 5.
Consider (1) where (8) and (9) hold and k is odd. Let {xn}n=-k∞ be a solution of (1) such that either
(17)α<x-k,x-k+2,…,x-1<α+ε1,(α+ε1)p/qε1-1/q<x-k+1,x-k+3,…,x0
or
(18)α<x-k+1,x-k+3,…,x0<α+ε1,(α+ε1)p/qε1-1/q<x-k,x-k+2,…,x-1.
Then if (17) holds, we have
(19)α<x2n-1<α+ε1,x2n>(α+ε1)p/qε1-1/q,n=0,1,…,.
Also if (18) holds, we have
(20)α<x2n<α+ε1,x2n-1>(α+ε1)p/qε1-1/q,n=0,1,…,.Proof.
Firstly, assume that (17) holds. Then, from (9) we have
(21)α<x1=α+x-kpx0q<α+ε1(α+ε1)p(α+ε1)p=α+ε1,x2=α+x-k+1px1q>α+(α+ε1)(p2-q2)/qε1-p/q>(α+ε1)p/qε1-1/q.
Working inductively, we can get
(22)α<xk=α+x-1pxk-1q<α+ε1(α+ε1)p(α+ε1)p=α+ε1,xk+1=α+x0pxkq>α+(α+ε1)(p2-q2)/qε1-p/q>(α+ε1)p/qε1-1/q.
Therefore, we can easily prove relations (19). Similiarly if (18) holds, we can prove that (20) is satisfied.Lemma 6.
Ifα>1, then every positive solution {xn}n=-k∞ of (1) satisfies the following inequalities:
(23)α<x2n+k<α1-βn1-β+βnx2n-1p,n=1,2,…,α<x2n+k+1<α1-βn1-β+βnx2np,n=1,2,…
and here the cases
(24)α<x2n<α1-βn1-β+βnx-k+1p,n=1,2,…,α<x2n-1<α1-βn1-β+βnx-kp,n=1,2,…,(25)β=1αq
hold.Proof.
We haveα<xn+1<α+βxn-kp for  alln=1,2,…. By induction, we obtain (23) and (24). If here α>1, we also see that
(26)α<x2n<α1-β+x-k+1p,n=1,2,…α<x2n-1<α1-β+x-kp,n=1,2,…Now, we are ready for the main result of this paper.Theorem 7.
Consider (1) where (8) and (9) hold and k is odd. Suppose that
(27)α+ε1<1,
then every positive solution {xn}n=-k∞ with initial values x-k,x-k+1,…,x-1,x0, which satisfy conditions of Theorem 4 and either (17) or (18), converges to a prime two-periodic solution.Proof.
Let{xn}n=-k∞ be a solution with initial values x-k,x-k+1,…,x-1,x0, which satisfy conditions of Theorem 4, and either (17) or (18). Using Lemma 1 and Theorem 4, we have that there exist
(28)limn→∞x2n+1=L,limn→∞x2n=S.
Besides, from Lemma 5 we have that either L or S belongs to the interval (α,α+ε1). Furthermore, from Lemma 3 we have that (1) has a unique equilibrium x¯ such that 1<x¯<∞. Therefore, from (27) we have that L≠S. So {xn}n=-k∞ converges to a prime two-periodic solution. The proof is complete.



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*Source: 742912-2013-03-26.xml*