On the Recursive Sequence 𝑥_𝑛=𝐴+𝑥^𝑝_𝑛−𝑘/𝑥^𝑟_𝑛−1

Abstract

This paper studies the dynamic behavior of the positive solutions to the difference equation 𝑥_𝑛=𝐴+𝑥^𝑝_𝑛−𝑘/𝑥^𝑟_𝑛−1, 𝑛=1,2,…, where 𝐴,𝑝, and 𝑟 are positive real numbers, and the initial conditions are arbitrary positive numbers. We establish some results regarding the stability and oscillation character of this equation for 𝑝∈(0,1).

1. Introduction

In recent years, there has been intense interest in the dynamic behavior of the positive solutions to a class of difference equations of the form 𝑥_𝑛𝑥=𝐴+^{𝑝𝑛−𝑘}𝑥^𝑝_𝑛−1,𝑛∈ℕ, (1.1) where 𝐴 and 𝑝 are positive real numbers. Now, let us make a brief review on the advances in this class of difference equations.

In 1999, Amleh et al. [1] studied the second-order rational difference equation 𝑥_𝑛𝑥=𝐴+^𝑛−2𝑥_𝑛−1,𝑛∈ℕ. (1.2)

Later, Berenhaut and Stević [2], Stević [3], and El-Owaidy et al. [4] extended this work to the following more general second-order difference equation: 𝑥_𝑛𝑥=𝐴+^𝑝𝑛−2𝑥^𝑝_𝑛−1,𝑛∈ℕ. (1.3)

On the other hand, DeVault et al. [5] investigated the following higher-order version of (1.2): 𝑥_𝑛𝑥=𝐴+^𝑛−𝑘𝑥_𝑛−1,𝑛∈ℕ. (1.4)

By combining (1.3) and (1.4), Berenhaut and Stević [6] examined a larger class of difference equations, which are of the form 𝑥_𝑛𝑥=𝐴+^{𝑝𝑛−𝑘}𝑥^𝑝_𝑛−1,𝑛∈ℕ. (1.5)

Very recently, Berenhaut et al. [7] studied the following generalization of (1.5): 𝑥_𝑛𝑥=𝐴+^{𝑝𝑛−𝑘}𝑥^𝑝_𝑛−𝑚,𝑛∈ℕ. (1.6)

For some related work, the interested reader is referred to [1, 3, 8–19].

Inspired by the previous work and by the work owing to Stević [15], this paper studies the behavior of the recursive equation 𝑥_𝑛𝑥=𝐴+^{𝑝𝑛−𝑘}𝑥^𝑟_𝑛−1,𝑛∈ℕ. (1.7) We establish some interesting results regarding the stability and oscillation character of this equation for 𝑝∈(0,1).

2. Stability Character

In this section we investigate the stability character of the positive solutions to (1.7).

A point 𝑥∈ℝ is an equilibrium point of (1.7) if and only if it is a root for the function 𝑔(𝑥)=𝑥−𝑥^𝑝−𝑟−𝐴, (2.1) that is, 𝑥=𝑥^𝑝−𝑟+𝐴. (2.2)

Lemma 2.1. Let 0<𝑝<𝑟+1, then (1.7) has a unique equilibrium point 𝑥>1.

Proof
Case 1. 𝑝=𝑟. Then 𝑥=𝐴+1>1.Case 2. 𝑟<𝑝<𝑟+1. Then 𝑔 defined by (2.1) is decreasing on [0,(𝑝−𝑟)^{1/(𝑟−𝑝+1)}] and increasing on [(𝑝−𝑟)^{1/(𝑟−𝑝+1)},∞). Since 𝑔(1)=−𝐴 and lim_𝑥→∞𝑔(𝑥)=∞, then 𝑔 has a unique zero 𝑥>1.Case 3. 0<𝑝<𝑟. Since 𝑔 is increasing on [0,∞),𝑔(1)=−𝐴 and lim_𝑥→∞𝑔(𝑥)=∞, then 𝑔 has a unique zero 𝑥>1.

Lemma 2.2. Let 0<𝑝<𝑟+1. Assume that 𝑥 is the equilibrium point of (1.7). If (𝑝+𝑟)^{(𝑝−𝑟)/(𝑟+1−𝑝)}(𝑝+𝑟−1)<𝐴, then 𝑥 is locally asymptotically stable.

Proof. By the Linearized Stability Theorem [11], 𝑥 is locally asymptotically stable if and only if 𝑥^{𝑟+1−𝑝}>𝑝+𝑟. A simple calculations shows that 𝑔((𝑝+𝑟)^{1/(𝑟+1−𝑝)})=(𝑝+𝑟)^{(𝑝−𝑟)/(𝑟+1−𝑝)}(𝑝+𝑟−1)−𝐴<0, (2.3) where 𝑔 is defined by (2.1). Then since lim_𝑥→∞𝑔(𝑥)=∞, we have 𝑥>(𝑝+𝑟)^{1/(𝑟+1−𝑝)} and 𝑥^{𝑟+1−𝑝}>𝑝+𝑟. The proof is complete.

Lemma 2.3. If 𝑝∈(0,1), then every positive solution to (1.7) is bounded.

Proof. Note that each 𝑛∈ℕ can be written in the form 𝑙𝑘+𝑖 for some 𝑙∈ℕ₀ and 𝑖∈{0,1,…,𝑘−1}. From (1.7) and since 𝑥_𝑛>𝐴 for every 𝑛≥0, we have that 𝑥_{𝑙𝑘+𝑖}𝑥=𝐴+^{𝑝(𝑙−1)𝑘+𝑖}𝑥^𝑟_{𝑙𝑘+𝑖−1}𝑥<𝐴+^{𝑝(𝑙−1)𝑘+𝑖}𝐴^𝑟, (2.4) for every 𝑙∈ℕ₀ and 𝑖∈{0,1,…,𝑘−1}. Let (𝑢_𝑙^(𝑖))_𝑙∈ℕ0 be the solution to the difference equation 𝑢_𝑙^(𝑖)=𝐴+(𝑢^(𝑖)_𝑙−1)^𝑝𝐴^𝑟,𝑢₀^(𝑖)=𝑥_−𝑘+𝑖. (2.5)
From (2.4) and by induction we see that 𝑥_{(𝑙−1)𝑘+𝑖}≤𝑢_𝑙^(𝑖),𝑙∈ℕ₀. Hence it is enough to prove that the sequences (𝑢_𝑙^(𝑖))_𝑙≥0,𝑖∈{0,1,…,𝑘−1} are bounded.
Since the function 𝑓(𝑥)=𝐴+𝑥^𝑝/𝐴^𝑟,𝑥∈(0,∞) is increasing and concave for 𝑝∈(0,1), it follows that there is a unique fixed point 𝑥 of the equation 𝑓(𝑥)=𝑥 and that the function 𝑓 satisfies (𝑓(𝑥)−𝑥)(𝑥−𝑥)<0,𝑥∈(0,∞). (2.6)
Using this fact it is easy to see that if 𝑢_𝑙^(𝑖)∈(0,𝑥], the sequence is nondecreasing and bounded from above by 𝑥, and if 𝑢_𝑙^(𝑖)≥𝑥, it is nonincreasing and bounded from below by 𝑥. Hence for every 𝑢₀^(𝑖)∈(0,∞), each of the sequences 𝑢_𝑙^(𝑖),𝑖∈{0,1,…,𝑘−1} is bounded. The claimed result follows.

Lemma 2.4 (see [18]). Let 𝑠,𝑡 be distinct nonnegative integers. Consider the difference equation 𝑥_𝑛=𝑓(𝑥_𝑛−𝑠,𝑥_𝑛−𝑡𝑥),𝑛=1,2,3,…,_{1−max(𝑠,𝑡)},𝑥_{2−max(𝑠,𝑡)},…,𝑥₀∈[𝑎,𝑏]. (2.7) Suppose 𝑓 satisfies the following conditions.(H₁)𝑓∶[𝑎,𝑏]²→[𝑎,𝑏] is a continuous function that is nondecreasing in the first argument and is nonincreasing in the second argument.(H₂)The system 𝑥=𝑓(𝑥,𝑦),𝑦=𝑓(𝑦,𝑥) (2.8) has a unique solution (𝑥,𝑥)∈[𝑎,𝑏]×[𝑎,𝑏].Then 𝑥 is the global attractor of all solutions to (2.7).

Theorem 2.5. Let 𝑝+𝑟≤1, then the unique equilibrium 𝑥 to (1.7) is globally asymptotically stable.

Proof. By Lemma 2.3, there must exist positive constants 𝑃 and 𝑄 such that 𝑃≤𝑥_𝑛≤𝑄. Let 𝑓(𝑢,𝑣)=𝐴+𝑢^𝑝/𝑣^𝑟,𝑢,𝑣∈[𝑃,𝑄], it is easy to verify that (H₁) holds. In addition, if 𝑥𝑥=𝐴+^𝑝𝑦^𝑟,𝑦𝑦=𝐴+^𝑝𝑥^𝑟, (2.9) then 𝑥−𝐴=𝑥𝑦−𝐴𝑦^{𝑝+𝑟𝑝+𝑟}. (2.10) Assume that 𝑥≠𝑦, then 𝑥>𝑦 or 𝑥<𝑦.
In case 𝑥>𝑦, we have (𝑥−𝐴)/(𝑦−𝐴)>𝑥/𝑦≥𝑥^𝑝+𝑟/𝑦^𝑝+𝑟, which contradicts with (2.10).
In case 𝑥<𝑦, we have (𝑥−𝐴)/(𝑦−𝐴)<𝑥/𝑦≤𝑥^𝑝+𝑟/𝑦^𝑝+𝑟, again a contradiction.
Thus 𝑥=𝑦=𝑥. By Lemma 2.4, the required result follows.

Theorem 2.6. Let 0<𝑝≤𝑟<1 and 𝐴^{𝑟−𝑝+1}≥𝑝/𝑟. Then every positive solution to (1.7) converges to the unique equilibrium 𝑥.

Proof. By Lemma 2.3, every positive solution {𝑥_𝑛} to (1.7) is bounded, which implies that there are finite liminf𝑥_𝑛=𝐼 and limsup𝑥_𝑛=𝑆. Assume that 𝐼≠𝑆(𝐼<𝑆). Taking the liminf and limsup in (1.7), it follows that 𝐼𝐴+^𝑝𝑆^𝑟𝑆≤𝐼<𝑆≤𝐴+^𝑝𝐼^𝑟. (2.11) From this and 𝑟∈(0,1), it follows that 𝐴𝑆^𝑟+𝐼^𝑝≤𝐼𝑆^𝑟<𝑆𝐼^𝑟≤𝐴𝐼^𝑟+𝑆^𝑝, (2.12) yielding 𝐴𝑆^𝑟−𝑆^𝑝<𝐴𝐼^𝑟−𝐼^𝑝. (2.13)
Define function 𝑓(𝑥)=𝐴𝑥^𝑟−𝑥^𝑝,𝑥∈(𝐴,∞). Since 𝑓^′(𝑥)=𝐴𝑟𝑥^𝑟−1−𝑝𝑥^𝑝−1=𝑥^𝑝−1(𝐴𝑟𝑥^𝑟−𝑝−𝑝)>𝑥^𝑝−1(𝑟𝐴^{𝑟−𝑝+1}−𝑝)≥0, (2.14) we deduce that 𝑓 is increasing, and thus (2.13) cannot hold. Therefore we have 𝐼=𝑆, which implies the result.

Theorem 2.7. Let 0<𝑝<1,𝑟≥1, and 𝐴^{𝑟−𝑝+1}≥𝑟+𝑝−1. Then every positive solution to (1.7) converges to the unique equilibrium 𝑥.

Proof. From (2.11) we have 𝐴𝐼^𝑟−1𝑆^𝑟+𝐼^{𝑝+𝑟−1}≤𝐼^𝑟𝑆^𝑟≤𝐴𝐼^𝑟𝑆^𝑟−1+𝑆^{𝑝+𝑟−1}. (2.15) Consequently, we obtain (𝐴𝐼^𝑟−1𝑆^𝑟−1)(𝑆−𝐼)≤(𝑆^{𝑟+𝑝−1}−𝐼^{𝑟+𝑝−1}). Suppose that 𝐼≠𝑆, we get 𝐴𝐼^𝑟−1𝑆^𝑟−1≤𝑆^{𝑟+𝑝−1}−𝐼^{𝑟+𝑝−1}𝑆−𝐼=(𝑟+𝑝−1)𝛾^{𝑝+𝑟−2}, (2.16) where 𝛾∈(𝐼,𝑆), leading to 𝐴^𝑟𝑆^𝑟−1≤𝐴𝐼^𝑟−1𝑆^𝑟−1≤(𝑟+𝑝−1)𝛾^{𝑝+𝑟−2}<(𝑟+𝑝−1)𝐴^𝑝−1𝑆^𝑟−1. (2.17)
This implies that 𝐴^{𝑟−𝑝+1}<𝑟+𝑝−1, which is a contradiction. Hence, 𝐼=𝑆=𝑥.

3. Oscillation Character

In this section we investigate the oscillation character of the positive solutions to (1.7).

Theorem 3.1. Let {𝑥_𝑛}^∞_𝑛=−𝑘 be a positive solution to (1.7). Then either {𝑥_𝑛}^∞_𝑛=−𝑘 consists of a single semicycle or {𝑥_𝑛}^∞_𝑛=−𝑘 oscillates about the equilibrium 𝑥 with semicycles having at most 𝑘−1 terms.

Proof. Suppose that {𝑥_𝑛}^∞_𝑛=−𝑘 has at least two semicycles. Then there exists 𝑁≥−𝑘 such that either 𝑥_𝑁<𝑥≤𝑥_𝑁+1 or 𝑥_𝑁+1<𝑥≤𝑥_𝑁. Assume that 𝑥_𝑁<𝑥≤𝑥_𝑁+1. (The argument for the case 𝑥_𝑁+1<𝑥≤𝑥_𝑁 is similar and is omitted). Now suppose that the positive semicycle beginning with the term 𝑥_𝑁+1 has 𝑘−1 terms. Then 𝑥_𝑁<𝑥≤𝑥_{𝑁+𝑘−1} and so 𝑥_𝑁+𝑘𝑥=𝐴+^𝑝𝑁𝑥^𝑟_{𝑁+𝑘−1}<𝐴+𝑥^𝑝𝑥^𝑟=𝐴+𝑥^𝑝−𝑟=𝑥. (3.1) This completes the proof.

Theorem 3.2. Suppose that 𝑘 is even and let {𝑥_𝑛}^∞_𝑛=−𝑘 be a solution to (1.7), which has 𝑘−1 consecutive semicycles of length one, then every semicycle after this point is of length one.

Proof. There exists 𝑁≥−𝑘 such that either 𝑥_𝑁,𝑥_𝑁+2,…,𝑥_{𝑁+𝑘−2}<𝑥≤𝑥_𝑁+1,𝑥_𝑁+3,…,𝑥_{𝑁+𝑘−1} (3.2) or 𝑥_𝑁+1,𝑥_𝑁+3,…,𝑥_{𝑁+𝑘−1}<𝑥≤𝑥_𝑁,𝑥_𝑁+2,…,𝑥_{𝑁+𝑘−2}. (3.3) We prove the former case. The proof for the latter is similar and is omitted. Now, we have 𝑥_𝑁+𝑘𝑥=𝐴+^𝑝𝑁𝑥^𝑟_{𝑁+𝑘−1}<𝐴+𝑥^𝑝𝑥^𝑟=𝐴+𝑥^𝑝−𝑟=𝑥𝑥,_𝑁+𝑘+1𝑥=𝐴+^𝑝𝑁+1𝑥^𝑟_𝑁+𝑘>𝐴+𝑥^𝑝𝑥^𝑟=𝐴+𝑥^𝑝−𝑟=𝑥. (3.4)
The result then follows by induction.

Lemma 3.3. Let 0<𝑝<𝑟+1. Then (1.7) has no nontrivial periodic solutions of (not necessarily prime) period 𝑘−1.

Proof. Suppose that {𝑥_𝑛}^∞_𝑛=−𝑘 is a positive solution to (1.7) satisfying 𝑥_𝑛−1=𝑥_𝑛−𝑘 for all 𝑛≥1, then 𝑥_𝑛=𝐴+𝑥^𝑝_𝑛−𝑘/𝑥^𝑟_𝑛−1=𝐴+𝑥^𝑝−𝑟_𝑛−1 implies that 𝑥_𝑛−1=𝑥_𝑛=𝑥 for all 𝑛>−𝑘. The proof is complete.

Theorem 3.4. Assume that 𝑝≤𝑟. Let {𝑥_𝑛}^∞_𝑛=−𝑘 be a positive solution to (1.7), which consists of a single semicycle, then{𝑥_𝑛}^∞_𝑛=−𝑘 converges to the equilibrium 𝑥.

Proof. Suppose 𝑥_𝑛≥𝑥 (the case for 𝑥_𝑛≤𝑥 is similar and is omitted) for all 𝑛≥−𝑘, then 𝑥_𝑛+1𝑥=𝐴+^{𝑝𝑛−(𝑘−1)}𝑥^𝑟_𝑛≥𝑥=𝐴+𝑥^𝑝−𝑟, (3.5) implying that 𝑥_{𝑛−(𝑘−1)}≥𝑥^{(𝑝−𝑟)/𝑝}𝑥_𝑛^𝑟/𝑝≥𝑥_𝑛^{(𝑝−𝑟)/𝑝}𝑥_𝑛^𝑟/𝑝=𝑥_𝑛, (3.6) and so 𝑥_{𝑛−(𝑘−1)}≥𝑥_𝑛≥𝑥for𝑛=1,2,…. (3.7) From here it is clear that for 𝑖=0,…,𝑘−2 there exists 𝛼_i such that lim_𝑛→∞𝑥_{𝑛(𝑘−1)+𝑖}=𝛼_𝑖. (3.8) But then 𝛼₀,𝛼₁,…,𝛼_𝑘−2 is a periodic solution of (not necessarily prime) period 𝑘−1. By Lemma 3.3 the result holds.

Acknowledgments

The author is grateful to the anonymous referees for their valuable comments and suggestions. This work was supported by the Natural Science Foundation of China (no. 10771227) and the Project for New Century Excellent Talents of Educational Ministry of China (no. NCET-05-0759).