Some Notes on the Difference Equation 𝑥_𝑛+1=𝛼+(𝑥_𝑛−1/𝑥^𝑘_𝑛)

Abstract

We investigate the behavior of the solutions of the recursive sequence𝑥_𝑛+1=𝛼+𝑥_𝑛−1/𝑥^𝑘_𝑛,𝑛=0,1,…, where 𝛼∈[0,∞), 𝑘∈(0,∞), and the initial conditions 𝑥₋₁,𝑥₀ are arbitrary positive numbers. Included are results that considerably improve those in the recently published paper by Hamza and Morsy (2009).

1. Introduction

Our aim in this paper is to give some remarks for the positive solutions of the difference equation𝑥_𝑛+1𝑥=𝛼+^𝑛−1𝑥^𝑘_𝑛,𝑛=0,1,…, (1.1) where 𝛼∈[0,∞), 𝑘∈(0,∞), and the initial conditions 𝑥₋₁,𝑥₀ are arbitrary positive numbers. Amleh et al. in [1] obtained important results for the difference equation 𝑥_𝑛+1𝑥=𝛼+^𝑛−1𝑥_𝑛, (1.2) which guide many authors. It was proved in [1] that, when 𝛼>1, the equilibrium 𝑥=𝛼+1 of (1.2) is globally asymptotically stable. When 𝛼=1, every positive solution of (1.2) converges to a period-two solution. Every positive solution of (1.2) is bounded if and only if 𝛼≥1. Finally, when 0<𝛼<1, the equilibrium is an unstable saddle point. Closely related equations to (1.1) are investigated by many authors, for example, [2–10].

In [4] the authors investigated the behavior of positive solutions of (1.1). It was proved in [4] that, when 𝛼≠1, every positive solution of (1.1) is bounded and when 𝛼>𝑘^1/𝑘≥1, the equilibrium 𝑥 of (1.1) is globally asymptotically stable. But in [4] the authors obtain some incorrect results for the boundedness character and the global stability of solutions of (1.1), and it is not shown that (1.1) has periodic solutions with conditions of 𝛼 and 𝑘.

Our aim here is to improve and correct these results and extend some of the results in [4].

We say that the equilibrium point 𝑥 of the equation 𝑥_𝑛+1=𝐹𝑥_𝑛,𝑥_𝑛−1,…,𝑥_𝑛−𝑘,𝑛=0,1,…, (1.3) is the point that satisfies the condition 𝑥=𝐹(𝑥,𝑥,…,𝑥). (1.4) A positive semicycle of a solution {𝑥_𝑛}^∞_𝑛=−1 of (1.1) consists of a “string” of terms {𝑥_𝑙,𝑥_𝑙+1,…,𝑥_𝑚} all greater than or equal to 𝑥, with 𝑙≥−1 and 𝑚≤∞, such that either𝑙=−1or𝑙>−1,𝑥_𝑙−1<𝑥,either𝑚=∞or𝑚<∞,𝑥_𝑚+1<𝑥. (1.5) A negative semicycle of a solution {𝑥_𝑛}^∞_𝑛=−1 of (1.1) consists of a “string” of terms {𝑥_𝑙,𝑥_𝑙+1,…,𝑥_𝑚} all less than 𝑥, with 𝑙≥−1 and 𝑚≤∞, such that either𝑙=−1or𝑙>−1,𝑥_𝑙−1≥𝑥,either𝑚=∞or𝑚<∞,𝑥_𝑚+1≥𝑥. (1.6) A solution {𝑥_𝑛}^∞_𝑛=−𝑘 of (1.1) is called nonoscillatory if there exists 𝑁≥−1 such that either 𝑥_𝑛>𝑥∀𝑛≥𝑁 (1.7) or 𝑥_𝑛<𝑥∀𝑛≥𝑁. (1.8){𝑥_𝑛}^∞_𝑛=−1 of (1.1) is called oscillatory if it is not nonoscillatory. We say that a solution {𝑥_𝑛}^∞_𝑛=−1 of (1.1) is bounded and persists if there exist positive constants 𝑃 and 𝑄 such that 𝑃≤𝑥_𝑛≤𝑄 for 𝑛=−1,0,….

The linearized equation for (1.1) about the positive equilibrium 𝑥 is 𝑦_𝑛+1+𝑘𝑥^−𝑘𝑦_𝑛−𝑥^−𝑘𝑦_𝑛−1=0,𝑛=0,1,…. (1.9) We need the following lemmas, which were given in [4].

Lemma 1.1. Let 𝑥 be the equilibrium point of (1.1).(i)If 𝑘(1+𝑘)^{(1−𝑘)/𝑘}<𝛼, then the equilibrium point 𝑥 of (1.1), is locally asymptotically stable.(ii)If 𝑘(1+𝑘)^{(1−𝑘)/𝑘}>𝛼, then the equilibrium point 𝑥 of (1.1) is unstable.

Lemma 1.2. The following statements are true.(i)If 𝑘=1, then (1.1) has a unique equilibrium point 𝑥=𝛼+1.(ii)If 𝑘≠1, then (1.1) has a unique equilibrium point 𝑥>1.

Lemma 1.3. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1), which consists of at least two semicycles. Then, {𝑥_𝑛}^∞_𝑛=−1 is oscillatory and, except possibly for the first semicycle, every semicycle is of length one.

The paper is organized as follows. In Section 2 we investigate the boundedness character of positive solutions of (1.1). We prove that if 0≤𝛼<1, then there exist unbounded solutions of (1.1) and when the cases either 𝛼∈(0,∞) and 𝑘→0⁺ or 𝛼>1 and 𝑘→∞, then every positive solution of (1.1) is unbounded. We show that when 𝛼>1 and 𝑘↛0⁺,𝑘↛∞, then every positive solution of (1.1) is bounded. Also we show that if 𝛼>1, 𝛼>𝑘(1+𝑘)^{(1−𝑘)/𝑘} and 𝑘↛0⁺,𝑘↛∞, then the equilibrium point 𝑥 of (1.1) is globally asymptotically stable. Section 3 is devoted to the periodic character of the positive solutions of (1.1). Finally we show that a sufficient condition that every positive solution of (1.1) converges to a prime two periodic solution.

2. Boundedness and Global Stability of (1.1)

In this section, we present some results for the boundedness character of positive solutions and global stability of the equilibrium point of (1.1).

Theorem 2.1. Consider (1.1). Then, the following statements are true.(a)If 𝛼∈(0,∞) and 𝑘→0⁺, then every positive solution of (1.1) is unbounded.(b)If 𝛼>1 and 𝑘→∞, then every positive solution of (1.1) is unbounded.

Proof. (a) On the contrary, we assume that {𝑥_𝑛}^∞_𝑛=−1 is a positive bounded solution of (1.1). Then, we have 𝑠=liminf𝑥_𝑛,𝑆=limsup𝑥_𝑛<∞. (2.1) Thus, from (1.1) we get 𝑠𝑠≥𝛼+𝑆^𝑘. (2.2) Let 𝑘→0⁺; then we obtain 𝑠≥𝛼+𝑠,𝛼≤0, (2.3) which contradicts 𝛼>0, so the proof is complete.
(b) Again we assume that {𝑥_𝑛}^∞_𝑛=−1 is a positive bounded solution of (1.1). Then, we have 𝛼<𝑠=liminf𝑥_𝑛,𝑆=limsup𝑥_𝑛<∞. (2.4) Thus, from (1.1), we have 𝑆𝑆≤𝛼+𝑠^𝑘. (2.5) Let 𝑘→∞; then 𝑆≤𝛼, (2.6) which contradicts 𝛼<𝑠, so the proof is complete.

Now, we show that if 0≤𝛼<1, then there exist positive solutions of (1.1) that are unbounded.

Theorem 2.2. One has 0≤𝛼<1. (2.7) Then there exist positive solutions of (1.1) that are unbounded.

Proof. Assume that 𝛼∈(0,1). Choose 𝛿∈(0,1−𝛼), and let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1) with the initial conditions such that 𝑥₋₁>𝛼+𝛿𝛿^1/𝑘−𝛼(𝛼+𝛿),𝛼<𝑥₀<𝛼+𝛿. (2.8) Then, 𝑥₁𝑥=𝛼+⁻¹𝑥^𝑘₀𝑥>𝛼+⁻¹(𝛼+𝛿)^𝑘,𝑥₂𝑥=𝛼+⁰𝑥^𝑘₁<𝛼+𝛼+𝛿𝛼+𝑥₋₁/(𝛼+𝛿)^𝑘𝑘<𝛼+𝛿. (2.9) Further we have 𝑥₃𝑥=𝛼+¹𝑥^𝑘₂𝑥>𝛼+¹(𝛼+𝛿)^𝑘1>𝛼+(𝛼+𝛿)^𝑘𝑥𝛼+⁻¹(𝛼+𝛿)^𝑘𝑥>𝛼+⁻¹(𝛼+𝛿)^𝑘,𝑥₄𝑥=𝛼+²𝑥^𝑘₃<𝛼+𝛼+𝛿𝛼+𝑥₋₁/(𝛼+𝛿)^𝑘𝑘<𝛼+𝛿. (2.10) Therefore, working inductively we can prove that for 𝑛=0,1,…𝑥_2𝑛+1𝑥>𝛼+^2𝑛−1(𝛼+𝛿)^𝑘𝑥>𝛼+⁻¹(𝛼+𝛿)^𝑘,𝛼<𝑥_2𝑛<𝛼+𝛿. (2.11) Hence, 𝑥_2𝑛+1>𝑥₋₁(𝛼+𝛿)^𝑘𝑛+11+𝛼1+(𝛼+𝛿)^𝑘1+⋯+(𝛼+𝛿)^𝑘𝑛. (2.12) Since 1(𝛼+𝛿)^𝑘>1, (2.13) which implies that lim_𝑛→∞𝑥_2𝑛+1=∞ (2.14){𝑥_𝑛}^∞_𝑛=−1 is unbounded. For 𝛼∈(0,1), the proof is complete.
Now, we assume that 𝛼=0 and choose the initial conditions such that 0<𝑥₋₁𝑥≤1,₀>1(1−𝜀)^1/𝑘forsome0<𝜀<1. (2.15) So, we have 0<𝑥₁=𝑥₋₁𝑥^𝑘₀≤1𝑥^𝑘₀𝑥<1−𝜀,₂=𝑥₀𝑥^𝑘₁>𝑥₀(1−𝜀)^𝑘. (2.16) Further we have 𝑥₃=𝑥₁𝑥^𝑘₂<1𝑥^𝑘₀𝑥<1−𝜀,₄=𝑥₂𝑥^𝑘₃>𝑥₀/(1−𝜀)^𝑘(1−𝜀)^𝑘=𝑥₀(1−𝜀)^2𝑘. (2.17) By induction we have 𝑥_2𝑛+1∈(0,1−𝜀),𝑥_2𝑛>𝑥₀(1−𝜀)^𝑘𝑛∀𝑛=0,1,…. (2.18) Thus, lim_𝑛→∞𝑥_2𝑛=∞,lim_𝑛→∞𝑥_2𝑛+1=0. (2.19) This completes the proof.

The following theorem is given in [4].

Theorem 2.3. Suppose that 𝛼≠1; then; every positive solution of (1.1) is bounded.

In [4] this result is not correct. So, we give the following theorem for the boundedness of (1.1).

Theorem 2.4. Suppose that 𝛼>1, 𝑘↛0⁺, and 𝑘↛∞; then every positive solution of (1.1) is bounded.

Proof . From (1.1), 𝑥_𝑛≥𝛼 for 𝑛≥0. Thus, from (1.1), without loss of generality, we obtain for 𝑛≥0𝑥_2𝑛+1𝑥=𝛼+^2𝑛−1𝑥^𝑘_2𝑛𝑥≤𝛼+^2𝑛−1𝛼^𝑘,𝑥_2𝑛𝑥=𝛼+^2𝑛−2𝑥^𝑘_2𝑛−1𝑥≤𝛼+^2𝑛−2𝛼^𝑘. (2.20) From (2.20) using induction, we obtain 𝑥_2𝑛+11≤𝛼1+𝛼^𝑘+1𝛼^𝑘21+⋯+𝛼^𝑘𝑛+𝑥₋₁𝛼^𝑘𝑛+1≤𝛼^𝑘𝛼^𝑘+𝑥−1𝛼^{−1𝑘𝑛+1},𝑥_2𝑛1≤𝛼1+𝛼^𝑘+1𝛼^𝑘21+⋯+𝛼^𝑘𝑛−1+𝑥₀𝛼^𝑘𝑛≤𝛼^𝑘𝛼^𝑘+𝑥−1𝛼^0𝑘𝑛. (2.21) From which the proof follows.

Actually, Hamza and Morsy in [4] obtained global stability of the equilibrium point of (1.1). But the result does not include the case 𝑘∈(0,1) and some parts of its proof are incorrect. So, here we will obtain global stability of the equilibrium point of (1.1) when 𝑘∈(0,1).

Theorem 2.5. Consider (1.1). Let 𝑘↛0⁺ and 𝑘↛∞. Suppose that 𝛼>1,𝛼>𝑘(1+𝑘)^{(1−𝑘)/𝑘} (2.22) hold. Then, the unique positive equilibrium 𝑥 of (1.1) is globally asymptotically stable.

Proof. By Lemma 1.1, 𝑥 is locally asymptotically stable. Thus, it is enough for the proof that every positive solution of (1.1) tends to the unique positive equilibrium 𝑥. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). By Theorem 2.4, {𝑥_𝑛}^∞_𝑛=−1 is bounded. Thus, we have 1<𝑠=liminf𝑥_𝑛,𝑆=limsup𝑥_𝑛<∞. (2.23) Then, from (2.23), we get 𝑆𝑆≤𝛼+𝑠^𝑘𝑠,𝑠≥𝛼+𝑆^𝑘. (2.24) We claim that 𝑠=𝑆, otherwise 𝑆>𝑠. From (2.24), we obtain 𝑆𝑠^𝑘≤𝛼𝑠^𝑘+𝑆,𝑆^𝑘𝑠≥𝛼𝑆^𝑘+𝑠, (2.25) And, from (2.25), 𝛼𝑆^𝑘𝑠^𝑘−1+𝑠^𝑘≤𝛼𝑠^𝑘𝑆^𝑘−1+𝑆^𝑘. (2.26) Thus, 𝛼𝑆^𝑘−1𝑠^𝑘−1(𝑆−𝑠)≤𝑆^𝑘−𝑠^𝑘. (2.27) Assume that 𝑘≥1. We consider 𝑓(𝑥)=𝑥^𝑘 with 𝑥∈(0,∞); then there exists 𝑎,𝑐∈(𝑠,𝑆) such that 𝑆^𝑘−𝑠^𝑘𝑆−𝑠=𝑘𝑐^𝑘−1≤𝑘𝑆^𝑘−1. (2.28) From (2.27) and (2.28), we obtain 𝛼𝑆^𝑘−1𝑠^𝑘−1≤𝑘𝑆^𝑘−1, (2.29) which is equivalent to 𝛼𝑠^𝑘−1≤𝑘. (2.30) Since we have 𝑆,𝑠≥𝛼, we get 𝛼^𝑘≤𝑘. (2.31) Since 𝛼>1 and 𝛼>𝑘(1+𝑘)^{(1−𝑘)/𝑘}, for some values 𝛼 and 𝑘, (2.31) is not satisfied. This is a contradiction. Thus, we find 𝑆=𝑠.
Now, assume that 𝑘<1. Then, from (2.27) and arguing as above, we get 𝛼𝑆^𝑘−1𝑠^𝑘−1≤𝑘𝑠^𝑘−1. (2.32) Furthermore, we have 𝑥_2𝑛+1𝑥<𝛼+^2𝑛−1𝛼^𝑘. (2.33) We consider the following difference equation: 𝑦_𝑚+1𝑦=𝛼+^𝑚𝛼^𝑘. (2.34) Every positive solution of the previous equation converges to 𝛼^𝑘+1/(𝛼^𝑘−1). It follows that 𝑆=limsup𝑥_𝑛≤𝛼^𝑘+1/(𝛼^𝑘−1). Then, we obtain that 𝛼^𝑘+1𝛼^𝑘−1^𝑘−1≤𝑆^𝑘−1≤𝑘𝛼. (2.35) Thus, 𝛼𝛼^𝑘+1𝛼^𝑘−1^𝑘−1≤𝑘. (2.36) Since 𝛼>1 and 𝛼>𝑘(1+𝑘)^{(1−𝑘)/𝑘}, for some values 𝛼 and 𝑘, (2.36) is not satisfied. So 𝑠=𝑆, which implies that {𝑥_𝑛}^∞_𝑛=−1 tends to the unique positive equilibrium. From which the proof follows.

Remark 2.6. Consider (1.1), where 0≤𝛼<1. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). If {𝑥_𝑛}^∞_𝑛=−1 is bounded, then it is stable too.

3. Periodicity of the Solutions of (1.1)

In this section we investigate the periodicity of (1.1) when 𝛼>1 and 𝑘>1,𝑘↛∞.

We need the following lemma whose proof follows by simple computation and thus it will be omitted.

Lemma 3.1. Let {𝑥_𝑛}^∞_𝑛=−𝑘 be a solution of (1.1), and let 𝐿>𝛼. Then, the following statements are satisfied.(i)lim_𝑛→∞𝑥_2𝑛=𝐿 if and only if lim_𝑛→∞𝑥_2𝑛+1=𝐿^1/𝑘/(𝐿−𝛼)^1/𝑘,(ii)lim_𝑛→∞𝑥_2𝑛+1=𝐿 if and only if lim_𝑛→∞𝑥_2𝑛=𝐿^1/𝑘/(𝐿−𝛼)^1/𝑘.

Theorem 3.2. Consider (1.1), where 𝛼>1,𝑘>1,𝑘↛∞. (3.1) Assume that there exists a sufficient small positive number 𝜀₁ such that 𝛼+𝜀₁^1/𝑘−𝛼+𝜀₁^{(1−𝑘2)/𝑘}<𝛼𝜀₁^1/𝑘,𝜀₁<𝛼+𝜀₁^1−𝑘. (3.2)(3.3) Then, (1.1) has a periodic solution of prime period two.

Proof. Let {𝑥_𝑛}^∞_𝑛=−1 be a solution of (1.1). It is obvious that if 𝑥₋₁𝑥=𝛼+⁻¹𝑥^𝑘₀,𝑥₀𝑥=𝛼+⁰𝑥^𝑘₋₁ (3.4) hold, then {𝑥_𝑛}^∞_𝑛=−1 is a periodic solution of period two. Consider the system𝑥𝑥=𝛼+𝑦^𝑘𝑦,𝑦=𝛼+𝑥^𝑘. (3.5) Then this system is equivalent to 𝑦𝑦−𝛼−𝑥^𝑘𝑥=0,𝑦=𝑥−𝛼^1/𝑘, (3.6) and so we get the equation 𝐹(𝑥)=𝑥𝑥−𝛼^1/𝑘𝑥−𝛼−^{(1−𝑘2)/𝑘}(𝑥−𝛼)^1/𝑘 (3.7) and obtain 𝐹(𝑥)=1(𝑥−𝛼)^1/𝑘𝑥^1/𝑘−𝑥^{(1−𝑘2)/𝑘}−𝛼. (3.8) Thus, lim_𝑥→𝛼+𝐹(𝑥)=∞. (3.9) Moreover, from (3.2) we can show that 𝐹𝛼+𝜀₁<0. (3.10) Therefore, the equation 𝐹(𝑥)=0 has a solution 𝑥=𝛼+𝜀₀, where 0<𝜀₀<𝜀₁, in the interval (𝛼,𝛼+𝜀₁). So, we have 𝑦=𝑥𝑥−𝛼^1/𝑘. (3.11) We now consider the function 𝐾(𝜀)=(𝛼+𝜀)^1−𝑘−𝜀. (3.12) Since from (3.1) 𝐾^′(𝜀)=(1−𝑘)(𝛼+𝜀)^−𝑘−1<0, (3.13) we have 𝐾𝜀₀>𝐾𝜀₁. (3.14) From (3.3), we have 𝐾(𝜀₁)>0, and thus 𝐾(𝜀₀)>0, which implies that 𝑥=𝛼+𝜀₀<𝛼+𝜀₀𝜀₀^1/𝑘=𝑦. (3.15) Hence, if 𝑥₋₁=𝑥,  𝑥₀=𝑦, then the solution {𝑥_𝑛}^∞_𝑛=−1 with initial values 𝑥₋₁,𝑥₀ is a prime 2-periodic solution. This completes the proof.

In the following theorem we will generalize the result due to Stević [6, Theorem 2].

Theorem 3.3. For every positive solution {𝑥_𝑛} of (1.1), the sequences {𝑥_2𝑛} and {𝑥_2𝑛+1} are eventually monotone.

Proof. We have 𝑥_2𝑛+1−𝑥_2𝑛−1=𝑥^𝑘_2𝑛−2𝑥_2𝑛−1−𝑥_2𝑛−3+𝑥_2𝑛−3𝑥^𝑘_2𝑛−2−𝑥^𝑘_2𝑛𝑥_2𝑛𝑥_2𝑛−2^𝑘,𝑥_2𝑛+2−𝑥_2𝑛=𝑥^𝑘_2𝑛−1𝑥_2𝑛−𝑥_2𝑛−2+𝑥_2𝑛−2𝑥^𝑘_2𝑛−1−𝑥^𝑘_2𝑛+1𝑥_2𝑛+1𝑥_2𝑛−1^𝑘. (3.16) If 𝑥₁≥𝑥₋₁ and 𝑥₂≤𝑥₀, we obtain from (3.16) 𝑥₃≥𝑥₁ and consequently 𝑥₄≤𝑥₂. By induction we obtain 𝑥₀≥𝑥₂≥⋯≥𝑥_2𝑛≥⋯,⋯≥𝑥_2𝑛+1≥𝑥_2𝑛−1≥⋯≥𝑥₁≥𝑥₋₁. (3.17) Similarly if 𝑥₁≤𝑥₋₁ and 𝑥₂≥𝑥₀, using induction we obtain from (3.16) 𝑥₀≤𝑥₂≤⋯≤𝑥_2𝑛≤⋯,⋯≤𝑥_2𝑛+1≤𝑥_2𝑛−1≤⋯≤𝑥₁≤𝑥₋₁. (3.18) If 𝑥₁≥𝑥₋₁, 𝑥₂≥𝑥₀ and 𝑥₁≥𝑥₃, we can obtain from (3.16) 𝑥₀≤𝑥₂≤⋯≤𝑥_2𝑛≤⋯,⋯≤𝑥_2𝑛+1≤𝑥_2𝑛−1≤⋯≤𝑥₁. (3.19) Hence, we may assume that 𝑥₃≥𝑥₁≥𝑥₋₁ and 𝑥₀≤𝑥₂. If further 𝑥₂≥𝑥₄, then 𝑥₂≥⋯≥𝑥_2𝑛≥⋯,⋯≥𝑥_2𝑛+1≥𝑥_2𝑛−1≥⋯≥𝑥₁≥𝑥₋₁. (3.20) So we may assume that 𝑥₃≥𝑥₁≥𝑥₋₁ and 𝑥₀≤𝑥₂≤𝑥₄. By induction we obtain the result in this case.
The cases 𝑥₁≤𝑥₋₁ and 𝑥₂≤𝑥₀ can be treated similarly.

Theorem 3.4. Consider (1.1) where (3.1), (3.2), and (3.3) hold. Then, every positive solution of (1.1) converges to a prime two periodic solution.

Proof. By Theorem 3.3, for every positive solution of (1.1) the sequences {𝑥_2𝑛} and {𝑥_2𝑛+1} are eventually monotone. By Theorem 2.4, the sequences {𝑥_2𝑛} and {𝑥_2𝑛+1} are bounded. Hence, the sequences {𝑥_2𝑛} and {𝑥_2𝑛+1} are convergent. Using Lemmas 1.3 and 3.1 the result follows.