On Global Attractivity of a Class of Nonautonomous Difference Equations

Abstract

We mainly investigate the global behavior to the family of higher-order nonautonomous recursive equations given by 𝑦_𝑛=(𝑝+𝑟𝑦_𝑛−𝑠)/(𝑞+𝜙_𝑛(𝑦_𝑛−1,𝑦_𝑛−2,…,𝑦_𝑛−𝑚)+𝑦_𝑛−𝑠), 𝑛∈ℕ₀, with 𝑝≥0,𝑟,𝑞>0,𝑠,𝑚∈ℕ and positive initial values, and present some sufficient conditions for the parameters and maps 𝜙_𝑛∶(ℝ⁺)^𝑚→ℝ⁺,𝑛∈ℕ₀, under which every positive solution to the equation converges to zero or a unique positive equilibrium. Our main result in the paper extends some related results from the work of Gibbons et al. (2000), Iričanin (2007), and Stević (vol. 33, no. 12, pages 1767–1774, 2002; vol. 6, no. 3, pages 405–414, 2002; vol. 9, no. 4, pages 583–593, 2005). Besides, several examples and open problems are presented in the end.

1. Introduction

There has been a great interest in studying classes of nonlinear difference equations and systems, particularly those which model real situations in engineering and science, for example, [1–15]. On the other hand, non-autonomous difference equations also have a ubiquitous presence in applications from automatic controlling, ecology, economics, biology, population dynamics and so forth. Thus the main task when dealing them is to know the asymptotical behaviour of their solutions. For some recent advances in this area see [1, 16–24] and the references cited therein.

Gibbons et al. [25] discussed the behavior of nonnegative solutions to the rational recursive equation𝑥_𝑛+1=𝛼+𝛽𝑥^𝑛−1𝛾+𝑥_𝑛,𝑛∈ℕ₀,(1.1) with 𝛼,𝛽,𝛾≥0, and also proposed an open problem, which had been solved by Stević in [4], concerning the particular case 𝛼=0,𝛾=𝛽 in (1.1) (see also [26, 27] for the case of some related higher-order difference equations, as well as [28–30]).

In [3], Stević studied the behavior of nonnegative solutions of the following second-order difference equation𝑥_𝑛+1=𝛼+𝛽𝑥^𝑛−11+𝑔𝑥_𝑛,𝑛∈ℕ⁰,(1.2) where 𝑔:ℝ⁺⋃{0}→ℝ is a nonnegative increasing mapping. Obviously (1.2) is a generalization of (1.1).

Later, Stević [6] extended (1.1) and (1.2) to the following more general equation𝑥_𝑛+1=𝛼+𝛽𝑥^𝑛−𝑘𝑓𝑥_𝑛,…,𝑥_{𝑛−𝑘+1},𝑛∈ℕ⁰,(1.3) where 𝑘∈ℕ,𝛼,𝛽≥0 and 𝑓:ℝ^𝑘+→ℝ₊ is a continuous function nondecreasing in each variable such that 𝑓(0,0,…,0)>0; and investigated the oscillatory behavior, the boundedness character and the global stability of nonnegative solutions to the equation.

Recently, Iričanin [2] studied the asymptotic behavior of the following class of autonomous difference equations:𝑥_𝑛=𝛼𝑥^𝑛−𝑘1+𝑥_𝑛−𝑘+𝑓𝑥_𝑛−1,…,𝑥_𝑛−𝑚,𝑛∈ℕ⁰,(1.4) where 𝛼>0,𝑘,𝑚∈ℕ and 𝑓 is a continuous mapping satisfying the condition𝛽min𝑢₁,…,𝑢_𝑚≤𝑓𝑢₁,𝑢₂,…,𝑢_𝑚≤𝛽max𝑢₁,…,𝑢_𝑚,(1.5) for certain 𝛽∈(0,1). In [2] he adopted the approach of frame sequences (a discrete analog of the method of frame curves used in the theory of differential equations), which has been used in the literature for many times, for example, [26–28, 30–38]; and showed that all positive solutions converge to zero if 0<𝛼≤1 and converge to the unique positive equilibrium if 𝛼>1.

Motivated by the above works, especially [2, 5], our aim in this paper is to study the global attractivity in the following family of non-autonomous difference equations:𝑦_𝑛=𝑝+𝑟𝑦^𝑛−𝑠𝑞+𝜙_𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚+𝑦_𝑛−𝑠,𝑛∈ℕ₀,(1.6) where 𝑝≥0,𝑟,𝑞>0,𝑠,𝑚∈ℕ, and 𝜙_𝑛∶(ℝ⁺)^𝑚→ℝ⁺,𝑛∈ℕ₀ are mappings satisfying the following condition𝛽min𝑥₁,…,𝑥_𝑚≤𝜙_𝑛𝑥₁,𝑥₂,…,𝑥_𝑚≤𝛽max𝑥₁,…,𝑥_𝑚,(1.7) for some fixed 𝛽∈(0,+∞).

Through careful analysis, we find that the results in [2] also persist if the function 𝑓 in (1.4) is replaced by variable functions such as 𝜙_𝑛 satisfying condition (1.7). If 𝑝=0, then (1.6) can be transformed into the following form𝑥_𝑛=(𝑟/𝑞)𝑥^𝑛−𝑠1+𝐺_𝑛𝑥_𝑛−1,…,𝑥_𝑛−𝑚+𝑥_𝑛−𝑠,𝑛∈ℕ₀,(1.8) where 𝐺_𝑛(𝑥₁,…,𝑥_𝑚)=𝜙_𝑛(𝑞𝑥₁,𝑞𝑥₂,…,𝑞𝑥_𝑚)/𝑞, by setting 𝑦_𝑛=𝑞𝑥_𝑛. Then according to the results in [2], we have that if 𝑟≤𝑞, then lim_𝑛→∞𝑦_𝑛=0; and if 𝑟>𝑞, then lim_𝑛→∞𝑦_𝑛=𝑞lim_𝑛→∞𝑥_𝑛=𝑞((𝑟/𝑞−1)/(1+𝛽))=(𝑟−𝑞)/(1+𝛽), for some 𝛽∈(0,1). Thus it suffices to consider the case when 𝑝>0 in the following.

Note that if 𝑝>0, then by relation (1.7), 𝑦=(√(𝑞−𝑟)²+4𝑝(1+𝛽)+𝑟−𝑞)/2(1+𝛽) is the unique positive equilibrium of (1.6). And in Section 3, we will prove the following main theorem.

Theorem 1.1. Consider (1.6), where 𝑠,𝑚∈ℕ,𝑝,𝑟,𝑞>0 with 𝑟𝑞≥𝑝, and 𝜙_𝑛∶(ℝ⁺)^𝑚→ℝ⁺ are functions satisfying the condition 𝛽min𝑥₁,…,𝑥_𝑚≤𝜙_𝑛𝑥₁,𝑥₂,…,𝑥_𝑚≤𝛽max𝑥₁,…,𝑥_𝑚,(1.9) for some fixed 𝛽∈(0,+∞). If 𝑞≥𝑟,𝛽∈(0,+∞) or 𝑞<𝑟,𝛽∈(0,𝛽₀], where 𝛽₀=4𝑝/(𝑞−𝑟)²+1, then the unique positive equilibrium 𝑦 of (1.6) is a global attractor.

2. Auxiliary Results

Before proving the main result of this paper, in this section we first confirm two preliminary lemmas.

Let Φ∶ℝ⁺⋃{0}×ℝ⁺→ℝ be the mapping Φ(𝑥,𝑤)=(𝑝+𝑟𝑤)/(𝑞+𝛽𝑥+𝑤), where 𝑝,𝑞,𝑟,𝛽>0 and 𝑟𝑞≥𝑝, so as Φ is decreasing in the first variable and increasing in the second one. Then (1.6) can be simplified to the following form:𝑦_𝑛=Φ𝜙^𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚𝛽,𝑦^𝑛−𝑠,𝑛∈ℕ⁰.(2.1)

Lemma 2.1. Consider the following higher-order rational difference equation: 𝑤_𝑛=Φ𝑥,𝑤_𝑛−𝑠,𝑛∈ℕ₀,(2.2) where 𝑝,𝑟,𝑞,𝛽∈ℝ⁺,𝑠∈ℕ, the parameter 𝑥≥0 and initial values 𝑤_𝑘,𝑘∈{−1,…,−𝑠} are arbitrary nonnegative numbers. Then every positive solution (𝑤_𝑛)^{∞𝑛=−𝑠} to (2.2) converges to the unique positive equilibrium point 𝑆(𝑥)=12(𝑞−𝑟+𝛽𝑥)²+4𝑝−(𝑞−𝑟+𝛽𝑥).(2.3)

Proof. First we show that (2.2) has a unique positive equilibrium. Assume that 𝑤>0 is an equilibrium point of (2.2), then 𝑤=(𝑝+𝑟𝑤)/(𝑞+𝛽𝑥+𝑤) which implies only one positive root 𝑤=𝑆(𝑥)=12(𝑞−𝑟+𝛽𝑥)²+4𝑝−(𝑞−𝑟+𝛽𝑥).(2.4) If 𝑠≥2, then (2.2) can be separated into 𝑠 analogous first-order difference equations of the form 𝑤^𝑛(𝑘)=𝑝+𝑟𝑤^{(𝑘)𝑛−1}𝑞+𝛽𝑥+𝑤^{(𝑘)𝑛−1},𝑛∈ℕ₀,(2.5) with different initial values 𝑤^(𝑘)−1=𝑤_−(𝑘+1), where 𝑘∈{0,1,…,𝑠−1}. Note that the equation is Riccati, so it can be solved and the convergence of its solutions can be proved (see, e.g., [39] or a recent comment in [40]).
Let the symbol [] symbolize the greatest integer function and define a sequence 𝑃(𝑛)≡𝑛(mod𝑠),𝑛∈ℕ₀. Obviously, for each positive solution (𝑤_𝑛)^{∞𝑛=−𝑠} to (2.2) we have𝑤_𝑛=𝑤^{[](𝑠−𝑃(𝑛)−1)𝑛/𝑠},𝑛≥−𝑠.(2.6)
From the above analysis, it suffices to prove the case when 𝑠=1. Suppose that 𝑠=1 for (2.2), then for all 𝑛∈ℕ₀, we have𝑤_𝑛+1−𝑤_𝑛=𝑤⁰−𝑤₋₁(𝑟(𝑞+𝛽𝑥)−𝑝)^𝑛+1𝑤𝑞+𝛽𝑥+𝑤_𝑛∏^{𝑛−1𝑖=0}𝑞+𝛽𝑥+𝑤_𝑖²𝑞+𝛽𝑥+𝑤₋₁,(2.7)_2𝑘+2−𝑤_2𝑘=(𝑟(𝑞+𝛽𝑥)−𝑝)𝑤^2𝑘+1−𝑤_2𝑘−1𝑞+𝛽𝑥+𝑤_2𝑘+1𝑞+𝛽𝑥+𝑤_2𝑘−1,𝑘∈ℕ⁰𝑤,(2.8)_2𝑘+3−𝑤_2𝑘+1=(𝑟(𝑞+𝛽𝑥)−𝑝)𝑤^2𝑘+2−𝑤_2𝑘𝑞+𝛽𝑥+𝑤_2𝑘+2𝑞+𝛽𝑥+𝑤_2𝑘,𝑘∈ℕ⁰.(2.9)Case 1. If 𝑟(𝑞+𝛽𝑥)≥𝑝, then by (2.7) (𝑤_𝑛)^∞𝑛=−1 is either nonincreasing or nondecreasing. On the other hand, we have that min{𝑝,𝑟}max{𝑞+𝛽𝑥,1}≤𝑤^𝑛≤max{𝑝,𝑟}min{𝑞+𝛽𝑥,1}(2.10) for all 𝑛≥0. Therefore, the limit of (𝑤_𝑛)^∞𝑛=−1 exists, and through simple calculations, we get lim_𝑛→∞𝑤_𝑛=𝑆(𝑥).Case 2. If 𝑟(𝑞+𝛽𝑥)<𝑝, then by (2.8) and (2.9) and inductively we have that (𝑤_2𝑘) is nonincreasing and (𝑤_2𝑘−1) nondecreasing, or (𝑤_2𝑘) is nondecreasing and (𝑤_2𝑘−1) nonincreasing. Again by (2.10), the limits of (𝑤_2𝑘) and (𝑤_2𝑘−1) exist, denoted by lim_𝑘→∞𝑤_2𝑘=𝛼 and lim_𝑘→∞𝑤_2𝑘−1=𝛾. From (2.2) we have 𝛼=𝑝+𝑟𝛾𝑞+𝛽𝑥+𝛾,𝛾=𝑞+𝑝𝛽+𝑥𝑟+𝛼𝛼,(2.11) which imply that 𝛼=𝛾=𝑤. Hence lim_𝑛→∞𝑤_𝑛=𝑆(𝑥).
The proof of Lemma 2.1 is complete.

Lemma 2.2. Suppose that the parameters, in (2.3), satisfy 𝑝,𝑟,𝑞,𝛽>0 with 𝑡=𝑞−𝑟. Define two sequences (𝑀_𝑘)^∞𝑘=1 and (𝑚_𝑘)^∞𝑘=1 as follows: 𝑚_𝑘=𝑆𝑀_𝑘+𝜀𝑀𝑘,𝑘=1,2,…,_𝑘=𝑆𝑚_𝑘−1−𝜀𝑘−1,𝑘=2,3,…,(2.12) where the initial value 𝑀₁=𝑆(0), and 𝜀∈(0,𝜆), 𝜆=12(1+𝛽)𝑡+𝛽𝑀₁²+4𝑝(1+𝛽)−𝑡+𝛽𝑀₁.(2.13) If 𝑞≥𝑟,𝛽∈(0,+∞) or 𝑞<𝑟,𝛽∈(0,𝛽₀], where 𝛽₀=4𝑝/(𝑞−𝑟)²+1, then lim_𝑘→∞𝑀_𝑘=lim_𝑘→∞𝑚_𝑘.(2.14)

Proof. By simple calculations, we have 𝑀₂−𝑀₁==𝑆1𝑚₁−𝜀−𝑆(0)2𝑡+𝛽𝑚₁=1−𝜀²+4𝑝−√𝑡²+4𝑝−𝛽𝑚₁−𝜀2⎛⎜⎜⎜⎝𝛽²𝑚₁−𝜀²+2𝛽𝑡𝑚₁−𝜀𝑡+𝛽𝑚₁−𝜀²+4𝑝+√𝑡²+4𝑝−𝛽𝑚¹=−𝛽𝑚₁−𝜀⎞⎟⎟⎟⎠−𝜀⎛⎜⎜⎜⎝𝑆𝑚¹−𝜀+𝑆(0)𝑡+𝛽𝑚₁−𝜀²+4𝑝+√𝑡²+4𝑝⎞⎟⎟⎟⎠.(2.15) Obviously, 𝑆(𝑚₁−𝜀)+𝑆(0)>0.Claim 1. 𝑚₁−𝜀>0.Proof. Define a function 𝑓(𝑥)=2(𝑆(𝑀₁+𝑥)−𝑥). It suffices to prove that 𝑓(𝑥)>0 for all 𝑥∈(0,𝜆). The derivative of 𝑓(𝑥) is 𝑑𝑓𝑑𝑥=𝛽𝑡+𝛽𝑀¹+𝑥𝑡+𝛽(𝑀₁+𝑥)²+4𝑝−𝛽−2<0.(2.16) Since 𝑓(𝜆)=0 and 𝑓(0)>0, thus 𝑓(𝑥)>0 for 𝑥∈(0,𝜆).Therefore, it follows from (2.15) and Claim 1 that 𝑀₂−𝑀₁<0.(2.17) Denote 𝐿_𝑘=2𝑡+𝛽𝑀^𝑘+1+𝑀_𝑘+𝜀/𝑘+𝜀/(𝑘+1)𝑄𝑡+𝛽𝑀_𝑘+1+𝜀/(𝑘+1)²+4𝑝+𝑡+𝛽𝑀_𝑘+𝜀/𝑘²+4𝑝_𝑘,=2𝑡+𝛽𝑚^𝑘+𝑚_𝑘−1−𝜀/𝑘−𝜀/(𝑘−1)𝑡+𝛽𝑚_𝑘−𝜀/𝑘²+4𝑝+𝑡+𝛽𝑚_𝑘−1−𝜀/(𝑘−1)²+4𝑝.(2.18) Simply, we obtain that |𝐿_𝑘|<1 and |𝑄_𝑘|<1.
Observe that2𝑚_𝑘+1−𝑚_𝑘=𝛽1−𝐿_𝑘𝑀_𝑘−𝑀_𝑘+1+𝜀2𝑀𝑘(𝑘+1),𝑘=1,2,…,_𝑘+1−𝑀_𝑘=𝛽𝑄_𝑘−1𝑚_𝑘−𝑚_𝑘−1+𝜀𝑘(𝑘+1),𝑘=2,3,….(2.19) With (2.17) and (2.19), it follows by induction that (𝑚_𝑘)^∞𝑘=1,(𝑀_𝑘)^∞𝑘=1 are strictly increasing and decreasing, respectively. In addition, 𝑀_𝑘>0,𝑘=1,2,…, hence (𝑀_𝑘)^∞𝑘=1 possesses a finite limit denoted by 𝜑=lim_𝑘→∞𝑀_𝑘. From (2.12), we know that the limit of (𝑚_𝑘)^∞𝑘=1 (denoted by 𝜇=lim_𝑘→∞𝑚_𝑘) also exists. Therefore, taking limits on both sides of (2.12), we have 𝜇𝜑==𝑆𝑆((𝜑𝜇)),(2.20) which imply that 𝜑𝜇²+𝑡𝜇+𝛽𝜇𝜑=𝑝,₂(𝜇−𝜑+)𝑡(𝜑𝜇++𝛽𝜑𝜇+𝜑𝑡=)=𝑝,0.((22..2212))Claim 2. If 𝑞≥𝑟,𝛽∈(0,+∞) or 𝑞<𝑟,𝛽∈(0,𝛽₀], then 𝜇=𝜑.Proof. Suppose that 𝜇≠𝜑, then it follows from (2.22) that 𝜇=−𝜑−𝑡. By substituting 𝜇=−𝜑−𝑡 into the second identity of (2.21), we get (1−𝛽)𝜑²+𝑡(1−𝛽)𝜑−𝑝=0.(2.23)(i)If 𝛽=1, then 𝑝=0 which is a contradiction to 𝑝>0,(ii)If 𝛽∈(0,1), then the unique positive root of (2.23) is 𝜑=√𝑡²+4𝑝/(1−𝛽)−𝑡2>𝑀¹.(2.24) However, 𝜑<𝑀₁ since (𝑀_𝑘) is strictly decreasing.(iii)If 𝑞=𝑟,𝛽∈(1,+∞), then (2.23) reduces to 0>(1−𝛽)𝜑²=𝑝>0.(iv)If 𝑞≠𝑟,𝛽∈(1,𝛽₀), then for (2.23), Δ=𝑡²(1−𝛽)²+4𝑝(1−𝛽)<0 which implies that (2.23) has no real roots.(v)For 𝑞>𝑟,𝛽=𝛽₀, we have Δ=0. So, 𝜑=(𝑟−𝑞)/2<0 which is contradictive to 𝜑≥0.(vi)For 𝑞>𝑟,𝛽∈(𝛽₀,+∞), (2.23) has two negative roots.(vii)For 𝑞<𝑟,𝛽=𝛽₀. Solving (2.23), we get 𝜑=(𝑟−𝑞)/2 implying 𝜇=(𝑟−𝑞)/2. Hence 𝜇=𝜑, which contradicts the assumption.
Obviously Claim 2 follows directly from (i)–(vii).
Applying Claim 2 and (2.21), we conclude thatlim_𝑘→∞𝑀_𝑘=lim_𝑘→∞𝑚_𝑘=√𝑡²+4𝑝(1+𝛽)−𝑡2(1+𝛽).(2.25) Hence the lemma is complete.

3. Main Results

Obviously, condition (1.7) in Section 1 guarantees the fact that (1.6) possesses a unique equilibrium point 𝑦=(√𝑡²+4𝑝(1+𝛽)−𝑡))/(2(1+𝛽)), where 𝑡=𝑞−𝑟.

First, we present a proposition concerning the boundedness of all positive solutions to (1.6).

Proposition 3.1. Consider (1.6) with condition (1.7) and 𝑝,𝑞,𝑟∈ℝ⁺, then every positive solution to (1.6) has permanent bounds.

Proof. Let (𝑦_𝑛) be a solution to (1.6) with positive initial values. Then, we have 𝑦_𝑛=𝑝+𝑟𝑦^𝑛−𝑠𝑞+𝜙_𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚+𝑦_𝑛−𝑠≤max{𝑝,𝑟}≤max{𝑝,𝑟}min𝑞+𝜙_𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚,1𝑦min{𝑞,1}=𝑈,∀𝑛≥0,_𝑛=𝑝+𝑟𝑦^𝑛−𝑠𝑞+𝜙_𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚+𝑦_𝑛−𝑠≥min{𝑝,𝑟}≥min{𝑝,𝑟}max𝑞+𝜙_𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚,1max{𝑞+𝛽𝑈,1}=𝐿,∀𝑛≥𝑚.(3.1) Thus we have 𝐿≤𝑦_𝑛≤𝑈, for all 𝑛≥𝑚.

In the following, we will give the proof of the main result (i.e., Theorem 1.1) presented in Section 1.

Proof of Theorem 1.1. Let 𝜀∈ℝ⁺ be an arbitrary fixed number satisfying 0<𝜀<𝜆 (𝜆 defined by (2.13) in Lemma 2.2). Define two sequences (𝑀_𝑘)^∞𝑘=1,(𝑚_𝑘)^∞𝑘=1 as shown by (2.12) in Lemma 2.2. Let (𝑦_𝑛) be any positive solution to (1.6). In the following, we proceed by presenting two claims.Claim 1. There exists 𝑁₁∈ℕ, such that 𝑚₁−𝜀≤𝑦_𝑛≤𝑀₁+𝜀 for all 𝑛≥ℕ₁.Proof. From (2.1), we have that 𝑦_𝑛=Φ𝜙^𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚𝛽,𝑦^𝑛−𝑠≤Φ0,𝑦^𝑛−𝑠.(3.2) Suppose that (𝑥_𝑛) is a solution to the following difference equation 𝑥_𝑛=Φ0,𝑥_𝑛−𝑠,𝑛∈ℕ₀,(3.3) with initial values 𝑥_−𝑠=𝑦_−𝑠,…,𝑥₋₁=𝑦₋₁. From this and in view of the monotonicity of the function 𝑓(𝑥)=(𝑝+𝑟𝑥)/(𝑞+𝑥),𝑥∈ℝ⁺, by induction we can easily get that 𝑦_𝑛≤𝑥_𝑛 for 𝑛≥−𝑠.
By Lemma 2.1, lim_𝑛→∞𝑥_𝑛=𝑀₁. Hence, there exists 𝑏₁∈ℕ such that 𝑥_𝑛≤𝑀₁+𝜀 for 𝑛≥𝑏₁, then𝑦_𝑛≤𝑀₁+𝜀(3.4) for all 𝑛≥𝑏₁.
From (2.1), (1.7), and (3.4), it follows that𝑦_𝑛≥Φmax𝑦_𝑛−1,…,𝑦_𝑛−𝑚,𝑦_𝑛−𝑠≥Φ𝑀₁+𝜀,𝑦_𝑛−𝑠(3.5) for all 𝑛≥𝑏₁+𝑚.
Suppose that (𝑢_𝑛) is a solution to the following difference equation:𝑢_𝑛=Φ𝑀₁+𝜀,𝑢_𝑛−𝑠,𝑛∈ℕ₀,(3.6) with initial values 𝑢_{𝑏1+𝑚−𝑠}=𝑦_{𝑏1+𝑚−𝑠},…,𝑢_{𝑏1+𝑚−1}=𝑦_{𝑏1+𝑚−1}.
Since the function ℎ(𝑥)=Φ(𝑀₁+𝜀,𝑥) is increasing on the interval (0,+∞), we can easily get by induction that 𝑦_𝑛≥𝑢_𝑛 for 𝑛≥𝑏₁+𝑚−𝑠, and by Lemma 2.1, lim_𝑛→∞𝑢_𝑛=𝑚₁. Hence there exists a natural number 𝑁₁≥𝑏₁ such that 𝑢_𝑛≥𝑚₁−𝜀 for 𝑛≥𝑁₁, then 𝑚₁−𝜀≤𝑦_𝑛≤𝑀₁+𝜀 for 𝑛≥𝑁₁.
Working inductively, we will eventually reach the following claim.
Claim 2. For each 𝑘∈ℕ, there exists 𝑁_𝑘∈ℕ such that 𝑚_𝑘−𝜀/𝑘≤𝑦_𝑛≤𝑀_𝑘+𝜀/𝑘 for all 𝑛≥𝑁_𝑘.Proof. By Claim 1, if 𝑘=1, we have 𝑁₁∈ℕ such that 𝑚₁−𝜀≤𝑦_𝑛≤𝑀₁+𝜀 for all 𝑛≥𝑁₁. Then by the method of induction, we can assume that for 𝑘∈ℕ fixed, there exists 𝑁_𝑘∈ℕ such that 𝑚_𝑘−𝜀/𝑘≤𝑦_𝑛≤𝑀_𝑘+𝜀/𝑘 for all 𝑛≥𝑁_𝑘. Thus, it suffices to show that there exists 𝑁_𝑘+1∈ℕ such that 𝑚_𝑘+1−𝜀/(𝑘+1)≤𝑦_𝑛≤𝑀_𝑘+1+𝜀/(𝑘+1) for all 𝑛≥𝑁_𝑘+1.
Let 𝑧=max{𝑠,𝑚}. Define a sequence (𝑥^𝑛(𝑘+1)) as follows𝑥^𝑛(𝑘+1)=Φ𝑆⁻¹𝑀_𝑘+1,𝑥^{(𝑘+1)𝑛−𝑠},𝑛≥𝑁^𝑘+𝑧,(3.7) with 𝑥^𝑛(𝑘+1)=𝑦_𝑛, for 𝑛=𝑁_𝑘,…,𝑁_𝑘+𝑧−1.
By reasoning inductively on 𝑛≥𝑁_𝑘+𝑧, one has𝑦_𝑛≤≤ΦΦm𝑆in₋₁𝑦_𝑛−1,…,𝑦_𝑛−𝑚,𝑦_𝑛−𝑠≤Φ𝑆⁻¹𝑀_𝑘+1,𝑦_𝑛−𝑠𝑀_𝑘+1,𝑥^{(𝑘+1)𝑛−𝑠}=𝑥^𝑛(𝑘+1),∀𝑛≥𝑁_𝑘+𝑧.(3.8) By Lemma 2.1, lim_𝑛→∞𝑥^𝑛(𝑘+1)=𝑀_𝑘+1. Therefore, there is 𝑏_𝑘+1≥𝑁_𝑘 such that 𝑦_𝑛≤𝑀_𝑘+1+𝜀/(𝑘+1)forall𝑛≥𝑏_𝑘+1.
Define the other sequence (𝑢^𝑛(𝑘+1)) as follows:𝑢^𝑛(𝑘+1)=Φ𝑆⁻¹𝑚_𝑘+1,𝑢^{(𝑘+1)𝑛−𝑠},for𝑛≥𝑏^𝑘+1,(3.9) where 𝑢^𝑛(𝑘+1)=𝑦_𝑛,for𝑛=𝑏_𝑘+1,…,𝑏_𝑘+1+𝑧−1.
Once more, by induction on 𝑛≥𝑏_𝑘+1+𝑧,𝑦_𝑛≥Φmax𝑦_𝑛−1,…,𝑦_𝑛−𝑚,𝑦_𝑛−𝑠≥Φ𝑀_𝑘+1+𝜀≥Φ𝑆₋₁𝑘+1,𝑦^𝑛−𝑠𝑚_𝑘+1,𝑢^{(𝑘+1)𝑛−𝑠}=𝑢^𝑛(𝑘+1),∀𝑛≥𝑏_𝑘+1+𝑧.(3.10) By Lemma 2.1, lim_𝑛→∞𝑢^𝑛(𝑘+1)=𝑚_𝑘+1. Thus, let 𝑁_𝑘+1≥𝑏_𝑘+1 be greater enough so as 𝑦_𝑛≥𝑚_𝑘+1−(𝜀/(𝑘+1))=𝑆⁻¹(𝑀_𝑘+2) for all 𝑛≥𝑁_𝑘+1.
Therefore, we get that there exists 𝑁_𝑘+1∈ℕ such that𝑚_𝑘+1−𝜀𝑘+1≤𝑦^𝑛≤𝑀_𝑘+1+𝜀𝑘+1(3.11) for all 𝑛≥𝑁_𝑘+1.By Claim 2, we have lim_𝑘→∞𝑚_𝑘=lim_𝑘→∞𝑚^𝑘−𝜀𝑘≤liminf^𝑛→∞𝑦_𝑛≤limsup_𝑛→∞𝑦_𝑛≤lim_𝑘→∞𝑀^𝑘+𝜀𝑘=lim^𝑘→∞𝑀_𝑘.(3.12) This plus Lemma 2.2 leads to lim_𝑛→∞𝑦_𝑛=𝑦=√𝑡²+4𝑝(1+𝛽)−𝑡2(1+𝛽).(3.13) The proof is complete.

4. Applications and Future Work

Next, several examples are presented.

Example 4.1. Let 𝑝_𝑛∈(0,+∞) for all 𝑛∈ℕ₀, and 𝜙_𝑛𝑥₁,𝑥₂,…,𝑥_𝑚=𝛽^𝑝𝑛∑^𝑚𝑖=1𝑥^𝑝𝑛_𝑖𝑚,𝑛∈ℕ⁰(4.1) for some 𝛽>0. If 𝑟𝑞>𝑝 and 𝑞≥𝑟,𝛽∈(0,+∞) or 𝑞<𝑟,𝛽∈(0,𝛽₀], where 𝛽₀=4𝑝/(𝑞−𝑟)²+1, then by Theorem 1.1 we conclude that every positive solution to the following non-autonomous difference equation: 𝑦_𝑛=𝑝+𝑟𝑦^𝑛−𝑠𝑞+𝛽^𝑝𝑛∑^𝑚𝑖=1𝑦^𝑝𝑛_𝑛−𝑖/𝑚+𝑦_𝑛−𝑠,𝑛∈ℕ₀,(4.2) converges to the unique positive equilibrium point 𝑦=(√(𝑞−𝑟)²+4𝑝(1+𝛽)+𝑟−𝑞)/2(1+𝛽).

Example 4.1 extends Example 2.4 in [2].

Example 4.2. Let 𝜙_𝑛(𝑥₁,𝑥₂,…,𝑥_𝑚)=𝛽max{𝑥₁,𝑥₂,…,𝑥_𝑚} for all 𝑛∈ℕ₀, then under the conditions of Theorem 1.1, all positive solutions to the recursive equation 𝑦_𝑛=𝑝+𝑟𝑦^𝑛−𝑠𝑞+𝛽max𝑦_𝑛−1,𝑦_𝑛−2,…,𝑦_𝑛−𝑚+𝑦_𝑛−𝑠,𝑛∈ℕ₀,(4.3) converge to the unique positive equilibrium 𝑦=(√(𝑞−𝑟)²+4𝑝(1+𝛽)+𝑟−𝑞)/2(1+𝛽).

In this paper, the behavior of positive solutions to the case when 𝑟𝑞≥𝑝,𝑞<𝑟,𝛽∈(𝛽₀,+∞), where 𝛽₀=4𝑝/(𝑞−𝑟)²+1, isn't investigated, since we have no further new ideas for the particular case. Through certain calculations, easily we know that the equation 𝑆∘𝑆(𝑥)=𝑥 has two different positive roots, if 𝑞<𝑟,𝛽∈(𝛽₀,+∞), which implies lim_𝑘→∞𝑀_𝑘>lim_𝑘→∞𝑚_𝑘. From this we propose the following open problem.

Open Problem 4. Is there a positive solution (𝑦_𝑛) to (1.6) with condition (1.7) when 𝑟𝑞≥𝑝,𝑞<𝑟,𝛽∈(𝛽₀,+∞), where 𝛽₀=4𝑝/(𝑞−𝑟)²+1, such that (𝑦_𝑛) eventually converges to a periodic solution?

Furthermore, the case 𝑟𝑞<𝑝 for (1.6) is also of extreme value to study.

Acknowledgments

The authors are grateful to the referees for their huge number of valuable suggestions, which considerably improved the presentation in the paper. Besides, the authors thank Professor. Iričanin for very valuable comments regarding this subject. This work was financially supported by National Natural Science Foundation of China (no. 10771227).