The Periodicity of Positive Solutions of the Nonlinear Difference Equation 𝑥_𝑛+1=𝛼+(𝑥^𝑝_𝑛−𝑘)/(𝑥^𝑞_𝑛)

Abstract

We give a remark about the periodic character of positive solutions of the difference equation 𝑥_𝑛+1=𝛼+𝑥^𝑝_𝑛−𝑘/𝑥^𝑞_𝑛, 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.

1. Introduction

In this paper, we consider the difference equation 𝑥_𝑛+1𝑥=𝛼+^{𝑝𝑛−𝑘}𝑥^𝑞_𝑛,𝑛=0,1,…, (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 𝑥_𝑛+1𝑥=𝛼+^𝑝𝑛−1𝑥^𝑞_𝑛,𝑛=0,1,…, (2) 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.

Let 𝐼 be some interval of real numbers and let 𝐹 be continuous function defined on 𝐼^𝑘+1. Then, for initial conditions 𝑥_−𝑘,…,𝑥₀∈𝐼, it is easy to see that the difference equation 𝑥_𝑛+1=𝐹𝑥_𝑛,𝑥_𝑛−1,…,𝑥_𝑛−𝑘,𝑛=0,1,…, (3) has a unique solution {𝑥}^∞_𝑛=−𝑘.

We say that the equilibrium point 𝑥 of (3) is the point that satisfies the condition 𝑥=𝐹(𝑥,𝑥,…,𝑥). (4) That is, 𝑥_𝑛=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 𝑥_𝑛+𝑝=𝑥_𝑛 for all 𝑛≥−𝑘𝑥_𝑛+𝑝=𝑥_𝑛,∀𝑛≥−𝑘. (5)

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 𝑦_𝑛+1+𝑞𝑥^{𝑝−𝑞−1}𝑦_𝑛−𝑝𝑥^{𝑝−𝑞−1}𝑦_𝑛−𝑘=0,𝑛=0,1,…. (6)

2. Main Results

In this section, we find solutions, for 𝑘 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 𝑝∈(0,1), (7) 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 0<𝑝<1<𝑞 (8) and there exists a sufficient small positive number 𝜀₁, such that 1𝛼+𝜀₁^𝑞−𝑝>𝜀₁,𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞<𝛼+𝜀₁^−𝑝/𝑞𝛼+𝜀₁^{𝑝2−𝑞2/𝑞}. (9)

Lemma 3. If either 0<𝑞<𝑝<1, (10) or 0<𝑝<𝑞,𝑞↛∞ (11) hold, then (1) has a unique equilibrium point 𝑥.

Theorem 4. Consider (1) where 𝑘 is odd, for every positive solution {𝑥_𝑛} 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−1}for  all𝑠=1,2,…,(𝑘+1)/2.(iii)𝑥_{−𝑘+2𝑙}≥𝑥_{−𝑘+2𝑙−2}for  all𝑙=1,2,…,(𝑘−1)/2 and 𝑥_{−𝑘+2𝑠+1}≥𝑥_{−𝑘+2𝑠−1}for  all𝑠=1,2,…,(𝑘+1)/2.(iv)𝑥_{−𝑘+2𝑙}≤𝑥_{−𝑘+2𝑙−2}for  all𝑙=1,2,…,(𝑘−1)/2 and 𝑥_{−𝑘+2𝑠+1}≤𝑥_{−𝑘+2𝑠−1}for  all𝑠=1,2,…,(𝑘+1)/2.

Then, the sequences {𝑥_2𝑛} and {𝑥_2𝑛+1} are eventually monotone.

Proof. We have 𝑥_2𝑛+1−𝑥_2𝑛−1=𝑥^𝑞_2𝑛−2𝑥^𝑝_2𝑛−𝑘−𝑥^𝑝_{2𝑛−𝑘−2}+𝑥^𝑝_{2𝑛−𝑘−2}𝑥^𝑞_2𝑛−2−𝑥^𝑞_2𝑛𝑥_2𝑛𝑥_2𝑛−2^𝑞,𝑥_2𝑛+2−𝑥_2𝑛=𝑥^𝑞_2𝑛−1𝑥^𝑝_{2𝑛−𝑘+1}−𝑥^𝑝_{2𝑛−𝑘−1}+𝑥^𝑝_{2𝑛−𝑘−1}𝑥^𝑞_2𝑛−1−𝑥^𝑞_2𝑛+1𝑥_2𝑛+1𝑥_2𝑛−1^𝑞. (12)(13) If 𝑥_{−𝑘+2𝑙}≥𝑥_{−𝑘+2𝑙−2}for  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 𝑥₀≥𝑥₂≥⋯≥𝑥_2𝑛≥⋯and⋯≥𝑥_2𝑛+1≥𝑥_2𝑛−1≥⋯≥𝑥₁. (14) Similarly if 𝑥_{−𝑘+2𝑙}≤𝑥_{−𝑘+2𝑙−2}for  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 𝑥₀≤𝑥₂≤⋯≤𝑥_2𝑛≤⋯and⋯≤𝑥_2𝑛+1≤𝑥_2𝑛−1≤⋯≤𝑥₁. (15) If 𝑥_{−𝑘+2𝑙}≥𝑥_{−𝑘+2𝑙−2}for  all𝑙=1,2,…,(𝑘−1)/2 and 𝑥_{−𝑘+2𝑠+1}≥𝑥_{−𝑘+2𝑠−1}for  all𝑠=1,2,…,(𝑘+1)/2 and 𝑥₁≥𝑥₃ we can obtain from (12) and (13) 𝑥₀≤𝑥₂≤⋯≤𝑥_2𝑛≤⋯and⋯≤𝑥_2𝑛+1≤𝑥_2𝑛−1≤⋯≤𝑥₁. (16) 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 {𝑥_𝑛}^∞_𝑛=−𝑘 be a solution of (1) such that either 𝛼<𝑥_−𝑘,𝑥_−𝑘+2,…,𝑥₋₁<𝛼+𝜀₁,𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞<𝑥_−𝑘+1,𝑥_−𝑘+3,…,𝑥₀ (17) or 𝛼<𝑥_−𝑘+1,𝑥_−𝑘+3,…,𝑥₀<𝛼+𝜀₁,𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞<𝑥_−𝑘,𝑥_−𝑘+2,…,𝑥₋₁. (18) Then if (17) holds, we have 𝛼<𝑥_2𝑛−1<𝛼+𝜀₁,𝑥_2𝑛>𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞,𝑛=0,1,…,. (19) Also if (18) holds, we have 𝛼<𝑥_2𝑛<𝛼+𝜀₁,𝑥_2𝑛−1>𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞,𝑛=0,1,…,. (20)

Proof. Firstly, assume that (17) holds. Then, from (9) we have 𝛼<𝑥₁𝑥=𝛼+^𝑝−𝑘𝑥^𝑞₀<𝛼+𝜀₁𝛼+𝜀₁^𝑝𝛼+𝜀₁^𝑝=𝛼+𝜀₁,𝑥₂𝑥=𝛼+^{𝑝−𝑘+1}𝑥^𝑞₁>𝛼+𝛼+𝜀₁^{(𝑝2−𝑞2)/𝑞}𝜀₁^−𝑝/𝑞>𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞. (21) Working inductively, we can get 𝛼<𝑥_𝑘𝑥=𝛼+^𝑝−1𝑥^𝑞_𝑘−1<𝛼+𝜀₁𝛼+𝜀₁^𝑝𝛼+𝜀₁^𝑝=𝛼+𝜀₁,𝑥_𝑘+1𝑥=𝛼+^𝑝0𝑥^𝑞_𝑘>𝛼+𝛼+𝜀₁^{(𝑝2−𝑞2)/𝑞}𝜀₁^−𝑝/𝑞>𝛼+𝜀₁^𝑝/𝑞𝜀₁^−1/𝑞. (22) Therefore, we can easily prove relations (19). Similiarly if (18) holds, we can prove that (20) is satisfied.

Lemma 6. If 𝛼>1, then every positive solution {𝑥_𝑛}^∞_𝑛=−𝑘 of (1) satisfies the following inequalities: 𝛼<𝑥_2𝑛+𝑘<𝛼1−𝛽^𝑛1−𝛽+𝛽^𝑛𝑥^𝑝_2𝑛−1,𝑛=1,2,…,𝛼<𝑥_2𝑛+𝑘+1<𝛼1−𝛽^𝑛1−𝛽+𝛽^𝑛𝑥^𝑝_2𝑛,𝑛=1,2,… (23) and here the cases 𝛼<𝑥_2𝑛<𝛼1−𝛽^𝑛1−𝛽+𝛽^𝑛𝑥^𝑝_−𝑘+1,𝑛=1,2,…,𝛼<𝑥_2𝑛−1<𝛼1−𝛽^𝑛1−𝛽+𝛽^𝑛𝑥^𝑝_−𝑘,𝑛=1,2,…,1𝛽=𝛼^𝑞 (24)(25) hold.

Proof. We have 𝛼<𝑥_𝑛+1<𝛼+𝛽𝑥^𝑝_𝑛−𝑘 for  all𝑛=1,2,…. By induction, we obtain (23) and (24). If here 𝛼>1, we also see that 𝛼<𝑥_2𝑛<𝛼1−𝛽+𝑥^𝑝_−𝑘+1,𝑛=1,2,…𝛼<𝑥_2𝑛−1<𝛼1−𝛽+𝑥^𝑝_−𝑘,𝑛=1,2,… (26)

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 𝛼+𝜀₁<1, (27) then every positive solution {𝑥_𝑛}^∞_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 {𝑥_𝑛}^∞_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 lim_𝑛→∞𝑥_2𝑛+1=𝐿,lim_𝑛→∞𝑥_2𝑛=𝑆. (28) 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 {𝑥_𝑛}^∞_𝑛=−𝑘 converges to a prime two-periodic solution. The proof is complete.

Acknowledgments

The authors would like to thank the referees for their helpful suggestions. This research is supported by TUBITAK and Bulent Ecevit University Research Project Coordinatorship.