Dynamics of a Higher-Order Nonlinear Difference Equation

Abstract

We consider the higher-order nonlinear difference equation 𝑥_𝑛+1=(𝛼+𝑥_𝑛)/(𝐴+𝐵𝑥_𝑛+𝑥_𝑛−𝑘), 𝑛=0,1,…, where parameters are positive real numbers and initial conditions 𝑥_−𝑘,…,𝑥₀ are nonnegative real numbers, 𝑘≥2. We investigate the periodic character, the invariant intervals, and the global asymptotic stability of all positive solutions of the abovementioned equation. We show that the unique equilibrium of the equation is globally asymptotically stable under certain conditions.

1. Introduction and Preliminaries

In this paper, we will investigate the global behavior of solutions of the following nonlinear difference equation:𝑥_𝑛+1=𝛼+𝑥^𝑛𝐴+𝐵𝑥_𝑛+𝑥_𝑛−𝑘,𝑛=0,1,…,(1.1) where parameters are positive real numbers and initial conditions 𝑥_−𝑘,…,𝑥₀ are nonnegative real numbers, 𝑘≥2.

In 2003, the authors in [1] considered the difference equation𝑥_𝑛+1=𝛼+𝛽𝑥^𝑛𝐴+𝐵𝑥_𝑛+𝐶𝑥_𝑛−1,𝑛=0,1,…,(1.2) with nonnegative parameters 𝛼,𝛽,𝐴,𝐵,𝐶 and nonnegative initial conditions 𝑥₋₁,𝑥₀. They obtained some global asymptotic stability results for the solutions of (1.2). For (1.2), we can also see [2–4].

For the global behavior of solutions of some related equations, see [5–13]. Other related results can be found in [14–20]. For the sake of convenience, we recall some definitions and theorems which will be useful in the sequel.

Definition 1.1. Let 𝐼 be some interval of real numbers, and let 𝑓∶𝐼^𝑚+1⟶𝐼(1.3) be a continuously differential function. Then for every set of initial conditions 𝑦_−𝑘,…,𝑦₋₁,𝑦₀∈𝐼, the difference equation 𝑦_𝑛+1=𝑓𝑦_𝑛,𝑦_𝑛−1,…,𝑦_𝑛−𝑘,𝑛=0,1,2,…,(1.4) has a unique solution {𝑦_𝑛}^{∞𝑛=−𝑘}.
A point 𝑦 is called an equilibrium point of (1.4) if 𝑦=𝑓𝑦,𝑦,…,𝑦.(1.5) That is, 𝑦_𝑛=𝑦,for𝑛≥0,(1.6) is a solution of (1.4), or equivalently 𝑦 is a fixed point of 𝑓.

Definition 1.2. Let 𝑦 be an equilibrium point of (1.4). Then the following are considered.(i)The equilibrium 𝑦 is called locally stable (or stable) if, for every 𝜀>0, there exists 𝛿>0 such that, for all 𝑦_−𝑘,…,𝑦₋₁,𝑦₀∈𝐼 with ∑^{𝑖=0𝑖=−𝑘}|𝑦_𝑖−𝑦|<𝛿, we have |𝑦_𝑛−𝑦|<𝜀 for all 𝑛≥𝑘.(ii)The equilibrium 𝑦 of (1.4) is called locally asymptotically stable (asymptotic stable) if it is locally stable and if there exists 𝛾>0 such that, for all 𝑦_−𝑘,…,𝑦₋₁,𝑦₀∈𝐼 with ∑^{𝑖=0𝑖=−𝑘}|𝑦_𝑖−𝑦|<𝛾, we have lim_𝑛→∞𝑦_𝑛=𝑦.(iii)The equilibrium 𝑦 of (1.4) is called a global attractor if, for every 𝑦_−𝑘,…,𝑦₋₁,𝑦₀∈𝐼, we have lim_𝑛→∞𝑦_𝑛=𝑦.(iv)The equilibrium 𝑦 of (1.4) is globally asymptotically stable if it is locally stable and is a global attractor. (v)The equilibrium 𝑦 of (1.4) is called unstable if it is not stable. (vi)The equilibrium 𝑦 of (1.4) is called a source, or a repeller, if there exists 𝑟>0 such that, for all 𝑦_−𝑘,…,𝑦₋₁,𝑦₀∈𝐼 with ∑^{𝑖=0𝑖=−𝑘}|𝑦_𝑖−𝑦|<𝛾, there exists 𝑁≥1 such that |𝑦_𝑁−𝑦|≥𝑟.

An interval 𝐽⊆𝐼 is called an invariant interval for (1.4) if 𝑦_−𝑘,…,𝑦₀∈𝐽⟹𝑦_𝑛∈𝐽∀𝑛>0.(1.7) That is, every solution of (1.4) with initial conditions in 𝐽 remains in 𝐽.

The linearized equation associated with (1.4) about the equilibrium 𝑦 is 𝑦_𝑛+1=^𝑘_𝑖=𝑜𝜕𝑓𝜕𝑢_𝑖𝑦,…,𝑦𝑦_𝑛−𝑖,𝑛=0,1,2,….(1.8) Its characteristic equation is 𝜆^𝑘+1=^𝑘_𝑖=𝑜𝜕𝑓𝜕𝑢_𝑖𝑦,…,𝑦𝜆^𝑘−𝑖.(1.9)

Theorem 1.3 (see [10]). Assume that 𝑓 is a 𝐶¹ function, and let 𝑦 be an equilibrium of (1.4). Then the following statements are true. (i)If all the roots of (1.9) lie in the open unit disk |𝜆|<1, then the equilibrium 𝑦 of (1.4) is asymptotically stable.(ii)If at least one root of (1.9) has absolute value greater than one, then the equilibrium 𝑦 of (1.4) is unstable.

Theorem 1.4 (see [10]). Assume that 𝑃,𝑄∈𝑅 and 𝑘∈{1,2,…}. Then ||𝑃||+||𝑄||<1(1.10) is a sufficient condition for the asymptotic stability of the difference equation 𝑦_𝑛+1=𝑃𝑦_𝑛+𝑄𝑦_𝑛−𝑘,𝑛=0,1,….(1.11)

Lemma 1.5 (see [8]). Consider the difference equation 𝑦_𝑛+1=𝑓𝑦_𝑛,𝑦_𝑛−𝑘,𝑛=0,1,…,(1.12) where 𝑘∈{1,2,…}. Let 𝐼=[𝑎,𝑏] be some interval of real numbers and assume that 𝑓∶[𝑎,𝑏]×[𝑎,𝑏]⟶[𝑎,𝑏](1.13) is a continuous function satisfying the following properties. (a)𝑓(𝑥,𝑦) is nondecreasing in 𝑥∈[𝑎,𝑏] for each 𝑦∈[𝑎,𝑏], and 𝑓(𝑥,𝑦) is nonincreasing in 𝑦∈[𝑎,𝑏] for each 𝑥∈[𝑎,𝑏]. (b)If (𝑚,𝑀)∈[𝑎,𝑏]×[𝑎,𝑏] is a solution of the system 𝑓(𝑚,𝑀)=𝑚,𝑓(𝑀,𝑚)=𝑀,(1.14) then 𝑚=𝑀.
Then (1.12) has a unique equilibrium 𝑥∈[𝑎,𝑏] and every solution of (1.12) converges to 𝑥.

Lemma 1.6 (see [8]). Consider the difference equation 𝑦_𝑛+1=𝑓𝑦_𝑛,𝑦_𝑛−𝑘,𝑛=0,1,…,(1.15) where 𝑘∈{1,2,…}. Let 𝐼=[𝑎,𝑏] be some interval of real numbers and assume that 𝑓∶[𝑎,𝑏]×[𝑎,𝑏]⟶[𝑎,𝑏](1.16) is a continuous function satisfying the following properties. (a)𝑓(𝑥,𝑦) is nonincreasing in each of its arguments. (b)If (𝑚,M)∈[𝑎,𝑏]×[𝑎,𝑏] is a solution of the system 𝑚=𝑓(𝑀,𝑀),𝑀=𝑓(𝑚,𝑚),(1.17) then 𝑚=𝑀.
Then (1.15) has a unique equilibrium 𝑦∈[𝑎,𝑏] and every solution of (1.15) converges to 𝑦.

2. Local Stability and Period-Two Solutions

The equilibria of (1.1) are the solutions of the equation 𝑥=𝛼+𝑥𝐴+𝐵𝑥+𝑥.(2.1) So (1.1) possesses the unique positive equilibrium 𝑥=1−𝐴+√(1−𝐴)²+4𝛼(𝐵+1)2(𝐵+1).(2.2) The linearized equation associated with (1.1) about the positive equilibrium is 𝑧_𝑛+1+𝐵𝑥−1𝐴+𝐵𝑥+𝑥𝑧^𝑛+𝑥𝐴+𝐵𝑥+𝑥𝑧^𝑛−𝑘=0.(2.3)

The next result follows from Theorem 1.4.

Theorem 2.1. Assume that either𝐴≥1or𝐴<1,(𝐵−1)(1−𝐴)²+4𝐵²𝛼>0.(2.4) Then the positive equilibrium 𝑥 of (1.1) is locally asymptotically stable.

Theorem 2.2. Equation (1.1) has no nonnegative prime period-two solution.

Proof. Assume for the sake of contradiction that there exist distinct nonnegative real numbers 𝜙 and 𝜓 such that …,𝜙,𝜓,𝜙,𝜓,…(2.5) is a prime period-two solution of (1.1). (a)Assume that 𝑘 is odd. Then 𝑥_𝑛+1=𝑥_𝑛−𝑘 and 𝜙,𝜓 satisfy the following system: 𝜙=𝛼+𝜓𝐴+𝐵𝜓+𝜙,𝜓=𝐴+𝛼𝐵+𝜙𝜙+𝜓.(2.6) Subtracting both sides of the above two equations, we obtain (𝜙−𝜓)[𝜙+𝜓+(𝐴+1)]=0.(2.7) If 𝜙≠𝜓, then 𝜙+𝜓=−(𝐴+1); this contradicts the hypothesis that 𝜙,𝜓≥0.(b)Assume that 𝑘 is even. Then 𝑥_𝑛=𝑥_𝑛−𝑘 and 𝜙,𝜓 satisfy the following system: 𝜙=𝛼+𝜓𝐴+𝐵𝜓+𝜓,𝜓=𝐴+𝛼𝐵+𝜙𝜙+𝜙.(2.8) Subtracting both sides of the above two equations, we obtain (𝜙−𝜓)(𝐴+1)=0.(2.9) If 𝜙≠𝜓, then 𝐴=−1; this contradicts the hypothesis that 𝐴≥0.
The proof is complete.

3. Boundedness and Invariant Interval

In this section, we will investigate the boundedness and invariant interval of (1.1).

Theorem 3.1. Every solution of (1.1) is bounded from above and from below by positive constants.

Proof. Let {𝑥_𝑛}^{∞𝑛=−𝑘} be a positive solution of (1.1). Clearly, if the solution is bounded from above by a constant 𝑀, then 𝑥_𝑛+1≥𝛼𝐴+(𝐵+1)𝑀,for𝑛≥−𝑘,(3.1) and so it is also bounded from below. Now for the sake of contradiction assume that the solution is not bounded from above. Then there exists a subsequence {𝑥_𝑛𝑚+1}^∞𝑚=0 such that lim_𝑚→∞𝑛_𝑚=∞,lim_𝑚→∞𝑥_𝑛𝑚+1=∞(3.2) and also 𝑥_𝑛𝑚+1=max𝑥_𝑛∶𝑛≤𝑛_𝑚for𝑚≥0.(3.3) From (1.1) we see that 𝑥_𝑛+1<𝛼𝐴+𝐴1𝑥^𝑛for𝑛≥0,(3.4) and so lim_𝑚→∞𝑥_𝑛𝑚+1=lim_𝑚→∞𝑥_𝑛𝑚=∞.(3.5) Hence, for sufficiently large 𝑚, 0≤𝑥_𝑛𝑚+1−𝑥_𝑛𝑚=𝛼+𝑥^𝑛𝑚𝐴+𝐵𝑥_𝑛𝑚+𝑥_{𝑛𝑚−𝑘}−𝑥_𝑛𝑚=𝛼+(1−𝐴)−𝐵𝑥^𝑛𝑚−𝑥_{𝑛𝑚−𝑘}𝑥_𝑛𝑚𝐴+𝐵𝑥_𝑛𝑚+𝑥_{𝑛𝑚−𝑘}<0,(3.6) which is a contradiction.
The proof is complete.

Let 𝑓(𝑥,𝑦)=𝛼+𝑥𝐴+𝐵𝑥+𝑦.(3.7) Then the following statements are true.

Lemma 3.2. (a) Assume that 𝐴≥𝐵𝛼. Then 𝑓(𝑥,𝑦) is increasing in 𝑥 for each 𝑦 and decreasing in 𝑦 for each 𝑥.
(b) Assume that 𝐴<𝐵𝛼. Then 𝑓(𝑥,𝑦) is decreasing in 𝑦 for each 𝑥, decreasing in 𝑥 for 𝑦∈[0,𝐵𝛼−𝐴], and increasing in 𝑥 for 𝑦∈[𝐵𝛼−𝐴,∞].

Proof. The proofs of (a) and (b) are simple and will be omitted.

Theorem 3.3. Equation (1.1) possesses the following invariant intervals: (a)[0,1/𝐵], when 𝐵𝛼≤𝐴;(b)[𝐵𝛼−𝐴,1/𝐵], when 𝐴<𝐵𝛼<𝐴+1/𝐵;(c)[0,𝛼/𝐴], when 𝐵𝛼=𝐴+1/𝐵;(d)[1/𝐵,𝐵𝛼−𝐴], when 𝐴+1/𝐵<𝐵𝛼<𝐴+𝛼/𝐴;(e)[1/𝐵,𝛼/𝐴], when 𝐵𝛼≥𝐴+𝛼/𝐴.

Proof. (a) Set 𝑔(𝑥)=(𝛼+𝑥)/(𝐴+𝐵𝑥), so 𝑔(𝑥) is nondecreasing for 𝑥 and 𝑔(1/𝐵)≤1/𝐵 if 𝐵𝛼≤𝐴, when 𝑥_−𝑘,…,𝑥₀∈[0,1/𝐵]; then we have 𝑥₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘≤𝛼+𝑥⁰𝐴+𝐵𝑥₀≤𝑔1𝐵≤𝐵1.(3.8)
The proof follows by induction.
(b) In view of Lemma 3.2(b), by using the monotonic character of the function 𝑓(𝑥,𝑦) and the condition 𝐴<𝐵𝛼<𝐴+(1/𝐵), when 𝑥_−𝑘,…,𝑥₀∈[𝐵𝛼−𝐴,1/𝐵], we can get 𝑥₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘=𝑓𝑥_𝑜,𝑥_−𝑘≥𝑓𝐵𝛼−𝐴,1𝑥𝐵>𝐵𝛼−𝐴,₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘=𝑓𝑥_𝑜,𝑥_−𝑘≤𝑓1𝐵,𝐵𝛼−A=𝐵1.(3.9)
The proof follows by induction.
(c) Set ℎ(𝑥)=(𝛼+𝑥)/(𝐴+𝐵𝑥+𝛼/𝐴) and 𝑔(𝑥)=(𝛼+𝑥)/(𝐴+𝐵𝑥), so ℎ(𝑥) is increasing and 𝑔(𝑥) is decreasing for 𝑥 if 𝐵𝛼=𝐴+1/𝐵. In view of Lemma 3.2(b), by using the monotonic character of the function 𝑓(𝑥,𝑦),   when 𝑥_−𝑘,…,𝑥₀∈[0,𝛼/𝐴], we have 𝑥₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘≥𝛼+𝑥⁰𝑥𝐴+𝐵𝑥₀+(𝛼/𝐴)≥ℎ(0)>0,₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘≤𝛼+𝑥⁰𝐴+𝐵𝑥₀≤𝑔(0)=𝛼𝐴.(3.10)
The proof follows by induction.
(d) In view of Lemma 3.2(b), by using the monotonic character of the function 𝑓(𝑥,𝑦) and the condition 𝐴+1/𝐵<𝐵𝛼<𝐴+𝛼/𝐴, when 𝑥_−𝑘,…,𝑥₀∈[1/𝐵,𝐵𝛼−𝐴], we obtain 𝑥₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘=𝑓𝑥_𝑜,𝑥_−𝑘≥𝑓(𝐵𝛼−𝐴,𝐵𝛼−𝐴)>1𝑥₁𝐵,=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘=𝑓𝑥_𝑜,𝑥_−𝑘≤𝑓1𝐵,𝐵1=𝐴𝐵𝐵+𝛼+𝐵1+1<𝐵𝛼−𝐴.(3.11)
The proof follows by induction.
(e) In view of the condition 𝐵𝛼≥𝐴+𝛼/𝐴, we can get 𝐵𝛼−𝐴≥𝛼/𝐴; by using the monotonic character of the function 𝑓(𝑥,𝑦) and the condition 𝐵𝛼≥𝐴+𝛼/𝐴, when 𝑥_−𝑘,…,𝑥₀∈[1/𝐵,𝛼/𝐴], we have 𝑥₁=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘=𝑓𝑥_𝑜,𝑥_−𝑘≥𝑓𝛼𝑥₁𝐴,𝐴𝛼=𝐴²𝐴+𝛼𝐵+𝛼𝛼+𝛼≥𝐵1,=𝛼+𝑥⁰𝐴+𝐵𝑥₀+𝑥_−𝑘=𝑓𝑥_𝑜,𝑥_−𝑘≤𝑓1𝐵,𝐵1=𝐴𝐵𝐵+𝛼+𝐵1+1<𝐴𝛼.(3.12)
The proof follows by induction.
The proof is complete.

4. Semicycles Analysis

We now give the definitions of positive and negative semicycles of a solution of (1.4) relative to an equilibrium point 𝑥.

A positive semicycle of a solution {𝑥_𝑛} of (1.4) consists of a string of terms {𝑥_𝑙,𝑥_𝑙+1,…,𝑥_𝑚}, all greater than or equal to the equilibrium 𝑥, with 𝑙≥−𝑘 and 𝑚≤∞ and such that eeitihtheerr𝑚𝑙==−∞𝑘oorr𝑚𝑙><−∞𝑘,,𝑥𝑥_{𝑙−1𝑚+1}<𝑥,<𝑥.(4.1)

A negative semicycle of a solution {𝑥_𝑛} of (1.4) consists of a string of terms {𝑥_𝑙,𝑥_𝑙+1,…,𝑥_𝑚}, all less than the equilibrium 𝑥, with 𝑙≥−𝑘 and 𝑚≤∞ and such that eeiitthheerr𝑚𝑙==−∞𝑘oorr𝑚𝑙><−∞𝑘,,x𝑥_{𝑙−1𝑚+1}≥𝑥,≥𝑥.(4.2)

Theorem 4.1 (see [12]). Assume that 𝑓∈𝐶[(0,∞)×(0,∞),(0,∞)] is such that 𝑓(𝑥,𝑦) is increasing in 𝑥 for each fixed 𝑦 and is decreasing in 𝑦 for each fixed 𝑥. Let 𝑥 be a positive equilibrium of (1.12). Then the following are considered. (a)If 𝑘=1, then every solution of (1.12) has semicycles of length at least two. (b)If 𝑘≥2, then every solution of (1.12) has semicycles that are either of length at least 𝑘+1 or of length at most 𝑘−1.

Let {𝑥_𝑛} be a positive solution of (1.1). Then one has the following identities:𝑥_𝑛+1−1𝐵=𝐵1(𝐴𝐵𝛼+−𝐵𝑥𝐴_𝑛)+−𝑥𝑥^𝑛−𝑘_𝑛−𝑘𝑥,𝑛∈ℕ₀,(4.3)_𝑛+1−𝛼𝐴=−𝐴1(𝐵𝛼−𝐴)𝑥^𝑛+𝛼𝑥_𝑛−𝑘𝐴+𝐵𝑥_𝑛+𝑥_𝑛−𝑘𝑥,𝑛∈ℕ₀,(4.4)_𝑛+1−(𝐵𝛼−𝐴)=𝐵[1/𝐵−(𝐵𝛼−𝐴)]𝑥^𝑛+𝐵𝛼1/𝐵−𝑥_𝑛−𝑘+𝐴𝑥_𝑛−𝑘−(𝐵𝛼−𝐴)𝐴+𝐵𝑥_𝑛+𝑥_𝑛−𝑘𝑥,𝑛∈ℕ₀,(4.5)_𝑛+1−𝑥=𝑥𝑥−𝑥_𝑛−𝑘+𝐵𝑥−1/𝐵𝑥−𝑥_𝑛𝐴+𝐵𝑥_𝑛+𝑥_𝑛−𝑘,𝑛∈ℕ₀,(4.6)𝑥_{2(𝑘+1)(𝑛+1)}−𝑥_{2(𝑘+1)𝑛}=𝐵𝑥^{2(𝑘+1)(𝑛+1)−1}1/𝐵−𝑥_{2(𝑘+1)𝑛}𝐴+𝑥_{2(𝑘+1)𝑛}+𝐵𝑥_{2(𝑘+1)𝑛+𝑘}𝐴+𝐵𝑥_{2(𝑘+1)(𝑛+1)−1}𝐴+𝐵𝑥_𝑛+1+𝑥_{2(𝑘+1)𝑛}+𝛼+𝑥_{2(𝑘+1)𝑛+𝑘}+(1+𝐴𝐵)𝑥^{2(𝑘+1)𝑛+𝑘}𝛼𝐵/(1+𝐴𝐵)−𝑥_{2(𝑘+1)𝑛}+𝐴𝛼−𝐴𝑥_{2(𝑘+1)𝑛}−𝑥^{22(𝑘+1)𝑛}𝐴+𝐵𝑥_{2(𝑘+1)(𝑛+1)−1}𝐴+𝐵𝑥_{2(𝑘+1)𝑛+𝑘}+𝑥_{2(𝑘+1)𝑛}+𝛼+𝑥_{2(𝑘+1)𝑛+𝑘},𝑛∈ℕ₀.(4.7)1/𝐵=𝐵𝛼−𝐴𝑥=1/𝐵𝑥_𝑛+1−1𝐵=𝐵1𝐴+1/𝐵𝐵𝑥−_𝑛𝑥+𝑥^𝑛−𝑘_𝑛−𝑘,𝑛∈ℕ₀,(4.8)

If 𝑥_{2(𝑘+1)(𝑛+1)}−𝑥_{2(𝑘+1)𝑛}=1/𝐵−𝑥^{2(𝑘+1)𝑛}𝐴𝐵𝑥_{2(𝑘+1)(𝑛+1)−1}+𝐵𝑥_{2(𝑘+1)𝑛}𝑥_{2(𝑘+1)(𝑛+1)−1}𝐴+𝐵𝑥_{2(𝑘+1)(𝑛+1)−1}𝐴+𝐵𝑥_{2(𝑘+1)𝑛+𝑘}+𝑥_{2(𝑘+1)𝑛}+𝛼+𝑥_{2(𝑘+1)𝑛+𝑘}+1/𝐵−𝑥^{2(𝑘+1)𝑛}𝐵²𝑥_{2(𝑘+1)𝑛+𝑘}𝑥_{2(𝑘+1)(𝑛+1)−1}+𝐵²𝛼𝑥_{2(𝑘+1)𝑛+𝑘}+𝐴𝑥_{2(𝑘+1)𝑛}+𝐴𝐵𝛼𝐴+𝐵𝑥_{2(𝑘+1)(𝑛+1)−1}𝐴+𝐵𝑥_{2(𝑘+1)𝑛+𝑘}+𝑥_{2(𝑘+1)𝑛}+𝛼+𝑥_{2(𝑘+1)𝑛+𝑘}𝑛∈ℕ₀,.(4.9), then 𝐵𝛼≤𝐴,(4.10) and (4.3), (4.7) change into{𝑥_𝑛}   𝑥_𝑛≤1/𝐵

The following lemmas are straightforward consequences of identities (4.3)–(4.9).

Lemma 4.2. Assume that 𝑛≥1 and let 𝑁≥0 be a solution of (1.1). Then the following statements are true. (i)𝑥_𝑁−𝑘≤𝑥 for all 𝑥_𝑁≥𝑥. (ii)If for some 𝑥_𝑁+1≥𝑥, 𝑁≥0 and 𝑥_𝑁−𝑘>𝑥, then 𝑥_𝑁<𝑥. (iii)If for some 𝑥_𝑁+1<𝑥, 0<𝑥<1/𝐵 and 𝐴<𝐵𝛼<𝐴+1𝐵,(4.11), then {𝑥_𝑛}. (iv)𝑁≥0.

Lemma 4.3. Assume that 𝑥_𝑁<𝐵𝛼−𝐴 and let 𝑥_𝑁+𝑘+1>1/𝐵 be a solution of (1.1). Then the following statements are true. (i)If for some 𝑁≥0, 𝑥_𝑁=𝐵𝛼−𝐴, then 𝑥_𝑁+𝑘+1=1/𝐵.(ii)If for some 𝑁≥0, 𝑥_𝑁>𝐵𝛼−𝐴, then 𝑥_𝑁+𝑘+1<1/𝐵.(iii)If for some 𝑁≥0, 𝐵𝛼−𝐴<𝑥_𝑁<1/𝐵, then 𝐵𝛼−𝐴<𝑥_𝑁+𝑘+1<1/𝐵. (iv)If for some 𝑁≥0, 𝑥_𝑁−𝑘≤𝑥, then 𝑥_𝑁≥𝑥.(v)If for some 𝑥_𝑁+1≥𝑥, 𝑁≥0 and 𝑥_𝑁−𝑘>𝑥, then 𝑥_𝑁<𝑥. (vi)If for some 𝑥_𝑁+1<𝑥, 𝑁≥0 and 𝑥_{2(𝑘+1)𝑁}<𝐵𝛼−𝐴, then 𝑥_{2(𝑘+1)(𝑁+1)}>𝑥_{2(𝑘+1)𝑁}. (vii)If for some 𝑁≥0, 𝑥_{2(𝑘+1)𝑁}>1/𝐵, then 𝑥_{2(𝑘+1)(𝑁+1)}<𝑥_{2(𝑘+1)𝑁}.(viii)If for some 𝐵𝛼−𝐴<𝑥<1/𝐵, 𝐵𝛼=𝐴+1𝐵,(4.12), then {𝑥_𝑛}. (ix)𝑁≥0.

Lemma 4.4. Assume that 𝑥_𝑁>1/𝐵 and let 𝑥_𝑁+𝑘+1<1/𝐵 be a solution of (1.1). Then the following statements are true. (i)If for some 𝑁≥0, 𝑥_𝑁=1/𝐵, then 𝑥_𝑁+𝑘+1=1/𝐵. (ii)If for some 𝑁≥0, 𝑥_𝑁<1/𝐵, then 𝑥_𝑁+𝑘+1>1/𝐵. (iii)If for some 𝑁≥0, 𝑥_{2(𝑘+1)𝑁}>1/𝐵, then 𝑥_{2(𝑘+1)(𝑁+1)}<𝑥_{2(𝑘+1)𝑁}. (iv)If for some 𝑁≥0, 𝑥_{2(𝑘+1)𝑁}<1/𝐵, then 𝑥_{2(𝑘+1)(𝑁+1)}>𝑥_{2(𝑘+1)𝑁}. (v)If for some 𝑥=1/𝐵, 𝐴+1𝐵<𝐵𝛼<𝐴+𝐴𝛼,(4.13), then {𝑥_𝑛}. (vi)𝑁≥0.

Lemma 4.5. Assume that 𝑥_𝑁<𝐵𝛼−𝐴 and let 𝑥_𝑁+𝑘+1>1/𝐵 be a solution of (1.1). Then the following statements are true. (i)If for some 𝑁≥0, 𝑥_𝑁=𝐵𝛼−𝐴, then 𝑥_𝑁+𝑘+1=1/𝐵. (ii)If for some 𝑁≥0, 𝑥_𝑁>𝐵𝛼−𝐴, then 𝑥_𝑁+𝑘+1<1/𝐵. (iii)If for some 𝑁≥0, 1/𝐵<𝑥_𝑁<𝐵𝛼−𝐴, then 1/𝐵<𝑥_𝑁+𝑘+1<𝐵𝛼−𝐴. (iv)If for some 𝑁≥0, 𝑥_𝑁−𝑘≤𝑥, then 𝑥_𝑁≤𝑥. (v)If for some 𝑥_𝑁+1≥𝑥, 𝑁≥0 and 𝑥_𝑁−𝑘>𝑥, then 𝑥_𝑁>𝑥. (vi)If for some 𝑥_𝑁+1<𝑥, 𝑁≥0 and 𝑥_{2(𝑘+1)𝑁}<1/𝐵, then 𝑥_{2(𝑘+1)(𝑁+1)}>𝑥_{2(𝑘+1)𝑁}. (vii)If for some 𝑁≥0, 𝑥_{2(𝑘+1)𝑁}>𝐵𝛼−𝐴, then 𝑥_{2(𝑘+1)(𝑁+1)}<𝑥_2(k+1)𝑁. (viii)If for some 1/𝐵<𝑥<𝐵𝛼−𝐴, 𝐵𝛼≥𝐴+𝛼𝐴,(4.14), then {𝑥_𝑛}. (ix)𝑥_𝑛<𝛼/𝐴.

Lemma 4.6. Assume that 𝑛≥1 and let 𝑁≥0 be a solution of (1.1). Then the following statements are true. (i)𝑥_𝑁<𝛼/𝐴 for all 𝑥_𝑁+𝑘+1>1/𝐵. (ii)If for some 𝑁≥0, 1/𝐵<𝑥_𝑁<𝛼/𝐴, then 1/𝐵<𝑥_𝑁+𝑘+1<𝛼/𝐴. (iii)If for some 𝑁≥0, 𝑥_𝑁−𝑘≤𝑥, then 𝑥_𝑁≤𝑥. (iv)If for some 𝑥_𝑁+1≥𝑥, 𝑁≥0 and 𝑥_𝑁−𝑘>𝑥, then 𝑥_𝑁>𝑥. (v)If for some 𝑥_𝑁+1<𝑥, 𝑁≥0 and 𝑥_{2(𝑘+1)𝑁}<1/𝐵, then 𝑥_{2(𝑘+1)(𝑁+1)}>𝑥_{2(𝑘+1)𝑁}. (vi)If for some 𝑁≥0, 𝑥_{2(𝑘+1)𝑁}>𝛼/𝐴, then 𝑥_{2(𝑘+1)(𝑁+1)}<𝑥_{2(𝑘+1)𝑁}. (vii)If for some 1/𝐵<𝑥<𝛼/𝐴, {𝑥_𝑛}^{∞𝑛=−𝑘}, then 𝐵𝛼≤𝐴.. (viii)𝑘+1.

The following result is a consequence of Theorem 4.1 and Lemmas 4.2–4.6.

Theorem 4.7. Let 𝑘−1 be a nontrivial solution of (1.1). Then the following statements are true. (a)Assume that 𝐴<𝐵𝛼<𝐴+1/𝐵. Then, except possibly for the first semicycle, every oscillatory solution of (1.1) has semicycles that are either of length at least [𝐵𝛼−𝐴,1/𝐵], or of length at most 𝑘+1.(b) Assume that 𝑘−1 Then, except possibly for the first semicycle, every oscillatory solution of (1.1) which lies in the invariant interval 𝐵𝛼=𝐴+1/𝐵. has semicycles that are either of length at least {𝑥_𝑛}^{∞𝑛=−𝑘}, or of length at most 2(𝑘+1).(c)Assume that 𝐴+1/𝐵<𝐵𝛼<𝐴+𝛼/𝐴. Then, except possibly for the first semicycle, [1/𝐵,𝐵𝛼−𝐴] is oscillatory and the sum of the lengths of two consecutive semicycles is equal to 𝑘+1.(d)Assume that 𝐵𝛼≥𝐴+𝛼/𝐴. Then, except possibly for the first semicycle, every oscillatory solution of (1.1) which lies in the invariant interval [1/𝐵,𝛼/𝐴] has semicycles at most 𝑘+1.(e)Assume that {𝑥_𝑛}^{∞𝑛=−𝑘} Then, except possibly for the first semicycle, every oscillatory solution of (1.1) which lies in the invariant interval 𝐵𝛼≤𝐴. has semicycles at most [0,1/𝐵].

5. Global Stability Proof

In this section, we will investigate the global stability of all positive solutions of (1.1).

Theorem 5.1. Let 𝐴<𝐵𝛼<𝐴+1/𝐵. be a positive solution of (1.1). Then the following statements are true. (a)Assume that [𝐵𝛼−𝐴,1/𝐵] Then every solution of (1.1) eventually enters the interval 𝐴+1/𝐵<𝐵𝛼<𝐴+𝛼/𝐴..(b)Assume that [1/𝐵,𝐵𝛼−𝐴] Then every solution of (1.1) eventually enters the interval 𝐵𝛼≥𝐴+𝛼/𝐴..(c)Assume that [1/𝐵,𝛼/𝐴] Then every solution of (1.1) eventually enters the interval 𝑥_𝑛≤1/𝐵.(d)Assume that 𝑛≥1 Then every solution of (1.1) eventually enters the interval 𝑥∈[0,1/𝐵].

Proof. (a) In view of Lemma 4.2, we know that [0,1/𝐵] for all 𝑥_−𝑘,…,𝑥₀∈[𝐵𝛼−𝐴,1/𝐵] and 𝑥_𝑛∈[𝐵𝛼−𝐴,1/𝐵],forall𝑛≥0; that is, all solutions of (1.1) eventually enter the interval [𝐵𝛼−𝐴,1/𝐵].
(b) If 2(𝑘+1), by Theorem 3.3(b), then we have {𝑥_{2(𝑘+1)𝑛+𝑗}}^{2𝑘+1𝑗=0}. If the initial conditions are not in the interval {𝑥_𝑛}, then we consider the {𝑥_{2(𝑘+1)𝑛}}th subsequences 𝑁 of the solution 𝑥_{2(𝑘+1)𝑁}<𝐵𝛼−𝐴. We will give the proof for the subsequence 𝑥_{2(𝑘+1)𝑁}>1/𝐵. The proof for all the other subsequences is similar and will be omitted. Without loss of generality, we assume that there exists 𝑥_{(𝑘+1)(2𝑁+1)}>1/𝐵>𝐵𝛼−𝐴 sufficiently large such that 𝑥_{2(𝑘+1)(𝑁+1)}<1/𝐵 if (𝑥_{2(𝑘+1)(𝑁+1)}≥𝐵𝛼−𝐴, then the proof is similar and will be omitted); then in view of Lemmas 4.3(ii) and (iv), we know that 𝑥_{2(𝑘+1)(𝑁+1)}<𝐵𝛼−𝐴 and 𝑥_{2(𝑘+1)(𝑁+1)}>𝑥_{2(𝑘+1)𝑁}. If {𝑥_{2(𝑘+1)(𝑁+𝑚)}}^∞𝑚=0, then, by induction, we know that the former assertion implies that the result is true. If 𝑥_{2(𝑘+1)(𝑁+𝑚)}<𝐵𝛼−𝐴, by Lemma 4.3(viii), then we can get lim_𝑚→∞𝑥_{2(𝑘+1)(𝑁+𝑚)}. It follows by induction that the subsequence lim_𝑚→∞𝑥_{2(𝑘+1)(𝑁+𝑚)}≤𝐵𝛼−𝐴 is increasing, and because 𝑥_𝑛<𝛼/𝐴, so 𝑛≥1 exists and [0,𝛼/𝐴]. However, taking limits by (4.7), we get a contradiction.
(c) The proof is similar to (b), so will be omitted.
(d) In view of Lemma 4.6, we know that [1/𝐵,𝛼/𝐴] for all [1/𝐵,𝛼/𝐴]; that is, all solutions of (1.1) eventually enter the interval {𝑥_{2(𝑘+1)(𝑁+𝑚)}}^∞𝑚=0. Furthermore, by Theorem 3.3, [0,1/𝐵] is an invariant interval of (1.1). Now, assume for the sake of contradiction that all solutions never enter the interval 𝑥_{2(𝑘+1)𝑁}≤1/𝐵, then the subsequence 𝐵𝛼≥𝐴+𝛼/𝐴 enters the interval 𝑥_{2(𝑘+1)(𝑁+1)}>𝑥_{2(𝑘+1)𝑁}. Because {𝑥_{2(𝑘+1)(𝑁+𝑚)}}^∞𝑚=0 and [0,1/𝐵], then, by Lemma 4.6, we know that lim_𝑚→∞𝑥_{2(𝑘+1)(𝑁+𝑚)}; it follows by induction that the subsequence lim_𝑚→∞𝑥_{2(𝑘+1)(𝑁+𝑚)}≤1/𝐵 is increasing in the interval [0,1/𝐵]. So 𝑥 exists and 𝐵𝛼≤𝐴, which is a contradiction because (1.1) has no equilibrium point in the interval [0,1/𝐵].
The proof is complete.

Theorem 5.2. Assume that (2.4) holds. Then the positive equilibrium [0,1/𝐵] of (1.1) is a global attractor of all positive solutions of (1.1).

We consider the following five cases.

Case 1. Assume that 𝑓(𝑥,𝑦). By Theorems 3.3(a) and 5.1(a), we know that (1.1) possesses an invariant interval 𝑥 and every solution of (1.1) eventually enters the interval 𝑦. Further, it is easy to see that [0,1/𝐵] increases in 𝑚,𝑀∈[0,1/𝐵] and decreases in 𝛼+𝑚𝐴+𝐵𝑚+𝑀=𝑚,𝐴+𝛼𝐵+𝑀𝑀+𝑚=𝑀,(5.1) in 𝛼+𝑚=𝐴𝑚+𝐵𝑚²+𝑀𝑚,𝛼+𝑀=𝐴𝑀+𝐵𝑀²+𝑀𝑚.(5.2).
Let (𝑚−𝑀)[1−𝐴−𝐵(𝑀+𝑚)]=0.(5.3) be a solution of the system 𝑚+𝑀≠(1−𝐴)/𝐵 which is equivalent to 𝑚=𝑀 Hence 𝐴≥1 Now if 𝑚+𝑀=(1−𝐴)/𝐵, then 𝑚. For instance, this is the case if 𝑀 is satisfied.
If 𝑚+𝑀=1−𝐴𝐵,𝑚𝑀=1−𝛼𝐵(5.4), then 𝐵(𝐵−1)𝑚²+(𝐵−1)(𝐴−1)𝑚−𝐵𝛼=0,(5.5) and Δ=(𝐵−1)(𝐵−1)(𝐴−1)²+4𝐵²𝛼.(5.6) satisfy the system 𝐵<1 and the equation Δ<0 whose discriminant is 𝑚=𝑀 Clearly, in this case, 𝑥∈[0,1/𝐵], and in view of condition (2.4), we have 𝑥, from which it follows that 𝐴<𝐵𝛼<𝐴+1/𝐵. In view of Lemma 1.5, (1.1) has a unique equilibrium [𝐵𝛼−𝐴,1/𝐵] and every solution of (1.1) converges to [𝐵𝛼−𝐴,1/𝐵].

Case 2. Assume that 𝑓(𝑥,𝑦). By Theorems 3.3(b) and 5.1(b), we know that (1.1) possesses an invariant interval 𝑥 and every solution of (1.1) eventually enters the interval 𝑦. Further, it is easy to see that [𝐵𝛼−𝐴,1/𝐵] increases in 𝑥∈[𝐵𝛼−𝐴,1/𝐵] and decreases in 𝑥 in 𝐵𝛼=𝐴+1/𝐵. Then using the same argument in Case 1, (1.1) has a unique equilibrium 2(𝑘+1) and every solution of (1.1) converges to {𝑥_{2(𝑘+1)𝑛+𝑗}}^∞𝑛=0.

Case 3. Assume that (𝑗∈{0,1,…,2𝑘+1}). Considering the 𝑛≥0,th subsequences 2(𝑘+1)1/𝐵, 1/𝐵 then by Lemma 4.4, we know that each one of the 1/𝐵th subsequences is above2(𝑘+1), belowforall𝑛∈ℕ, or identically equal to 𝑥_{2(𝑘+1)(𝑛+1)}=𝑥_{2(𝑘+1)𝑛}iff𝑥_{2(𝑘+1)𝑛}=1𝐵.(5.7). Furthermore, by the identity (4.9), we can get that all 2(𝑘+1)th subsequences converge monotonically to limits, and {𝑥_{2(𝑘+1)(𝑛+1)+𝑗}}^∞𝑛=0, (𝑗∈{0,1,…,2𝑘+1}) So all the 1/𝐵th subsequences 𝑥=1/𝐵𝐴+1/𝐵<𝐵𝛼<𝐴+𝛼/𝐴. converge to [1/𝐵,𝐵𝛼−𝐴]. That is, [1/𝐵,𝐵𝛼−𝐴] is a global attractor of (1.1).

Case 4. Assume that 𝑓(𝑥,𝑦) By Theorems 3.3(d) and 5.1(c), we know that (1.1) possesses an invariant interval [1/𝐵,𝐵𝛼−𝐴] and every solution of (1.1) eventually enters the interval 𝑚,𝑀∈[1/𝐵,𝐵𝛼−𝐴]. Furthermore, it is easy to see that the function 𝛼+𝑚𝐴+𝐵𝑚+𝑚=𝑀,𝐴+𝛼𝐵+𝑀𝑀+𝑀=𝑚,(5.8) decreases in each of its arguments in the interval 𝛼+𝑚=𝐴𝑀+(𝐵+1)𝑚𝑀,𝛼+𝑀=𝐴𝑚+(𝐵+1)𝑚𝑀.(5.9). Let (𝑚−𝑀)(𝐴+1)=0 be a solution of the system 𝑚=𝑀 that is, the solution of the system 𝑥∈[1/𝐵,𝐵𝛼−𝐴] Then 𝑥, which implies that 𝐵𝛼≥𝐴+𝛼/𝐴. Employing Lemma 1.6, we see that (1.1) has a unique equilibrium [1/𝐵,𝛼/𝐴] and every solution of (1.1) converges to [1/𝐵,𝛼/𝐴].

Case 5. Assume that 𝑓(𝑥,𝑦). By Theorems 3.3(e) and 5.1(d), we know that (1.1) possesses an invariant interval [1/𝐵,𝛼/𝐴] and every solution of (1.1) eventually enters the interval 𝑥∈[1/𝐵,𝛼/𝐴]. Further, it is clear to see that the function 𝑥 decreases in each of its arguments in the interval 𝑥. Then, using the same argument as in Case 4, (1.1) has a unique equilibrium 𝑥_𝑛+1=(𝛼+𝛽𝑥_𝑛)/(𝐴+𝐵𝑥_𝑛+𝐶𝑥_𝑛−1) and every solution of (1.1) converges to 𝑦_𝑛+1=(𝑝+𝑦_𝑛−𝑘)/(𝑞𝑦_𝑛+𝑦_𝑛−𝑘).
The proof is complete.

In view of Theorems 2.1 and 5.2, we have the following result.

Theorem 5.3. Assume that (2.4) holds. Then the unique positive equilibrium 𝑦_𝑛+1=(𝑝+𝑦_𝑛−1)/(𝑞𝑦_𝑛+𝑦_𝑛−1) of (1.1) is globally asymptotically stable.