Abstract

We investigate the local stability, prime period-two solutions, boundedness, invariant intervals, and global attractivity of all positive solutions of the following difference equation: 𝑦𝑛+1=(𝑟+𝑝𝑦𝑛+𝑦𝑛−𝑘)/(𝑞𝑦𝑛+𝑦𝑛−𝑘), 𝑛∈ℕ0, where the parameters 𝑝,𝑞,𝑟∈(0,∞),𝑘∈{1,2,3,…} and the initial conditions 𝑦−𝑘,…,𝑦0∈(0,∞). We show that the unique positive equilibrium of this equation is a global attractor under certain conditions.

1. Introduction and Preliminaries

Our aim in this paper is to study the dynamical behavior of the following rational difference equation

𝑦𝑛+1=𝑟+𝑝𝑦𝑛+𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0,(1.1) where 𝑝,𝑞,𝑟∈(0,∞), ℕ0∈{0,1,…}, 𝑘∈{1,2,3,…} and the initial conditions 𝑦−𝑘,…,𝑦0∈(0,∞).

When 𝑘=1, (1.1) reduces to

𝑦𝑛+1=𝑟+𝑝𝑦𝑛+𝑦𝑛−1𝑞𝑦𝑛+𝑦𝑛−1,𝑛∈ℕ0.(1.2)

In [1] (see also [2]), the authors investigated the global convergence of solutions to (1.2) and they obtained the following result.

Theorem 1.1. Let 𝑝, 𝑞 and 𝑟 be positive numbers. Then every solution of (1.2) converges to the unique equilibrium or to a prime-two solution.

The main purpose of this paper is to further consider the global attractivity of all positive solutions of (1.1). That is to say, we will prove that the unique positive equilibrium of (1.1) is a global attractor under certain conditions (see Theorem 4.10).

For the general theory of difference equations, one can refer to the monographes [3] and [2]. For other related results on nonlinear difference equations, see, for example, [1–18].

For the sake of convenience, we firstly present some definitions and known results which will be useful in the sequel.

Let 𝐼 be some interval of real numbers and let 𝑓∶𝐼×𝐼→𝐼 be a continuously differentiable function. Then for initial conditions 𝑥−𝑘,…,𝑥0∈𝐼, the difference equation

𝑥𝑛+1𝑥=𝑓𝑛,𝑥𝑛−𝑘,𝑛∈ℕ0(1.3) has a unique solution {𝑥𝑛}∞𝑛=−𝑘.

A point 𝑥 is called an equilibrium of (1.3) if

𝑥=𝑓𝑥,𝑥.(1.4)

That is, 𝑥𝑛=𝑥 for 𝑛≥0 is a solution of (1.3), or equivalently, 𝑥 is a fixed point of 𝑓.

An interval 𝐽⊆𝐼 is called an invariant interval of (1.3) if

𝑥−𝑘,…,𝑥0∈𝐽⟹𝑥𝑛∈𝐽∀𝑛∈ℕ0.(1.5)

That is, every solution of (1.3) with initial conditions in J remains in J.

Let

𝑃=𝜕𝑓𝜕𝑢𝑥,𝑥,𝑄=𝜕𝑓𝜕𝑣𝑥,𝑥(1.6)

denote the partial derivatives of 𝑓(𝑢,𝑣) evaluated at an equilibrium 𝑥 of (1.3). Then the linearized equation associated with (1.3) about the equilibrium 𝑥 is

𝑧𝑛+1=𝑃𝑧𝑛+𝑄𝑧𝑛−𝑘,𝑛=0,1,…,(1.7) and its characteristic equation is

𝜆𝑘+1−𝑃𝜆𝑘−𝑄=0.(1.8)

Lemma 1.2 (see [3]). Assume that 𝑃,𝑄∈𝑅 and 𝑘∈{1,2,…}. Then ||𝑃||+||𝑄||<1(1.9) is a sufficient condition for asymptotic stability of the difference equation (1.7). Suppose in addition that one of the following two cases holds: (i)𝑘 odd and 𝑄>0,(ii)𝑘 even and 𝑃𝑄>0.Then (1.9) is also a necessary condition for the asymptotic stability of the difference equation (1.7).

The following result will be useful in establishing the global attractivity character of the equilibrium of (1.1), and it is a reformulation of [2, 7].

Lemma 1.3. Suppose that a continuous function 𝑓∶[𝑎,𝑏]×[𝑎,𝑏]→[𝑎,𝑏] satisfies one of (i)–(iii):

(i)𝑓(𝑥,𝑦) is nonincreasing in 𝑥,𝑦, and []×[]∀(𝑚,𝑀)∈𝑎,𝑏𝑎,𝑏,(𝑓(𝑚,𝑚)=𝑀,𝑓(𝑀,𝑀)=𝑚)⟹𝑚=𝑀,(1.10)(ii)𝑓(𝑥,𝑦) is nondecreasing in 𝑥 and nonincreasing in 𝑦, and []×[]∀(𝑚,𝑀)∈𝑎,𝑏𝑎,𝑏,(𝑓(𝑚,𝑀)=𝑚,𝑓(𝑀,𝑚)=𝑀)⟹𝑚=𝑀,(1.11)(iii)𝑓(𝑥,𝑦) is nonincreasing in 𝑥 and nondecreasing in 𝑦, and []×[]∀(𝑚,𝑀)∈𝑎,𝑏𝑎,𝑏,(𝑓(𝑀,𝑚)=𝑚,𝑓(𝑚,𝑀)=𝑀)⟹𝑚=𝑀.(1.12)

(Note that for 𝑘 odd this is equivalent to (1.3) having no prime period-two solution)

Then (1.3) has a unique equilibrium in [𝑎,𝑏] and every solution with initial values in [𝑎,𝑏] converges to the equilibrium.

This work is organized as follows. In Section 2, the local stability and periodic character are discussed. In Section 3, the boundedness, invariant intervals of (1.1) are presented. Our main results are formulated and proved in Section 4, where the global attractivity of (1.1) is investigated.

2. Local Stability and Period-Two Solutions

The unique positive equilibrium of (1.1) is

√𝑦=(1+𝑝)+(1+𝑝)2+4𝑟(1+𝑞)2.(1+𝑞)(2.1)

The linearized equation associated with (1.1) about 𝑦 is

𝑧𝑛+1−(𝑝−𝑞)𝑦−𝑞𝑟(𝑞+1)𝑟+(𝑝+1)𝑦𝑧𝑛+(𝑝−𝑞)𝑦+𝑟(𝑞+1)𝑟+(𝑝+1)𝑦𝑧𝑛−𝑘=0,(2.2)

and its characteristic equation is

𝜆𝑘+1−(𝑝−𝑞)𝑦−𝑞𝑟(𝑞+1)𝑟+(𝑝+1)𝑦𝜆𝑘+(𝑝−𝑞)𝑦+𝑟(𝑞+1)𝑟+(𝑝+1)𝑦=0.(2.3)

From this and Lemma 1.2, we have the following result.

Theorem 2.1. Assume that 𝑝,𝑞,𝑟∈(0,∞) and initial conditions 𝑦−𝑘,…,𝑦0∈(0,∞). Then the following stataments are true. (i)If (𝑝−𝑞)𝑦−𝑞𝑟≥0,(𝑝−3𝑞−𝑝𝑞−1)𝑦<2𝑞𝑟,(2.4) then the unique positive equilibrium 𝑦 of (1.1) is locally asymptotically stable;(ii) If (𝑝−𝑞)𝑦−𝑞𝑟<0<(𝑝−𝑞)𝑦+𝑟,(2.5) then the unique positive equilibrium 𝑦 of (1.1) is locally asymptotically stable. In particular, if 𝑘 is even, then the equilibrium 𝑦 is locally asymptotically stable if and only if (2.5) holds;(iii)If (𝑝−𝑞)𝑦+𝑟≤0,(2.6) then the unique positive equilibrium 𝑦 of (1.1) is locally asymptotically stable.

In the following, we will consider the period-two solutions of (1.1).

Let

…,𝜙,𝜓,𝜙,𝜓,…(2.7)

be a period-two solution of (1.1), where 𝜙 and 𝜓 are two arbitrary positive real numbers.

If 𝑘 is even, then 𝑦𝑛=𝑦𝑛−𝑘, and 𝜙 and 𝜓 satisfy the following system:

𝜙=𝑟+𝑝𝜓+𝜓𝑞𝜓+𝜓,𝜓=𝑟+𝑝𝜙+𝜙,𝑞𝜙+𝜙(2.8)

then (𝜙−𝜓)(𝑝+1)=0, we have 𝜙=𝜓, which is a contradiction.

If 𝑘 is odd, then 𝑦𝑛+1=𝑦𝑛−𝑘, and 𝜙 and 𝜓 satisfy the following system:

𝜙=𝑟+𝑝𝜓+𝜙𝑞𝜓+𝜙,𝜓=𝑟+𝑝𝜙+𝜓,𝑞𝜙+𝜓(2.9)

then 𝜙+𝜓=1−𝑝, 𝜙𝜓=𝑝(1−𝑝)/(𝑞−1). By calculating, (1.1) has prime period-two solution if and only if

𝑝<1,𝑞>1,4𝑟<(1−𝑝)(𝑞−1−𝑝𝑞−3𝑝).(2.10)

From the above discussion, we have the following result.

Theorem 2.2. Equation (1.1) has a positive prime period-two solution …,𝜙,𝜓,𝜙,𝜓,…(2.11) if and only if 𝑘isodd,𝑝<1,𝑞>1,4𝑟<(1−𝑝)(𝑞−𝑝𝑞−3𝑝−1).(2.12) Furthermore, if (2.12) holds, then the prime period-two solution of (1.1) is “unique” and the values of 𝜙 and 𝜓 are the positive roots of the quadratic equation 𝑡2−(1−𝑝)𝑡+𝑟+𝑝(1−𝑝)𝑞−1=0.(2.13)

3. Boundedness and Invariant Intervals

In this section, we discuss the boundedness, invariant intervals of (1.1).

3.1. Boundedness

Theorem 3.1. All positive solutions of (1.1) are bounded.

Proof. Equation (1.1) can be written as 𝑦𝑛+1=𝑟+𝑝𝑦𝑛+𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘≥𝑝𝑦𝑛+𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘≥min{(𝑝/𝑞),1}𝑞𝑦𝑛+𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘𝑝=min𝑞,1(3.1) for all 𝑛≥0. We denote 𝑝𝐾=min𝑞,1.(3.2) Then 𝑦𝑛+1=𝑟+𝑝𝑦𝑛+𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘≤𝑟+𝑝𝑦𝑛+𝑦𝑛−𝑘(𝑞/2)𝐾+(𝐾/2)+(𝑞/2)𝑦𝑛+(1/2)𝑦𝑛−𝑘≤max{𝑟,𝑝,1}1+𝑦𝑛+𝑦𝑛−𝑘min{(𝑞/2)𝐾+(𝐾/2),(𝑞/2),(1/2)}1+𝑦𝑛+𝑦𝑛−𝑘=max{𝑟,𝑝,1}min{(𝑞/2)𝐾+(𝐾/2),(𝑞/2),(1/2)}(3.3) for all 𝑛>𝑘. The proof is complete.

Let {𝑦𝑛}∞𝑛=−𝑘 be a positive solution of (1.1). Then the following identities are easily established:

𝑦𝑛+1−1=(𝑞−𝑝)(𝑟/(𝑞−𝑝))−𝑦𝑛𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0,𝑦(3.4)𝑛+1−𝑝𝑞=((𝑝−𝑞)/𝑞)(𝑞𝑟/(𝑝−𝑞))−𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0,𝑦(3.5)𝑛+1−𝑞𝑟=𝑝𝑝−𝑞2−𝑝𝑞−𝑞2𝑟𝑦/(𝑝−𝑞)𝑛+(1/𝑞)𝑞𝑦𝑛+𝑦𝑛−𝑘+𝑦((𝑝−𝑞−𝑞𝑟)/(𝑝−𝑞))𝑛−𝑘−(𝑝/𝑞)𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0,𝑦(3.6)𝑛+1−𝑝+𝑟𝑞=𝑟1−𝑦𝑛+((𝑞−𝑝−𝑟)/𝑞)𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0,𝑦(3.7)𝑛−𝑦𝑛+2(𝑘+1)=𝑦𝑛𝑞−(𝑝/𝑞)2𝑦𝑛+𝑘𝑦𝑛+2𝑘+1+𝑞𝑦𝑛𝑦𝑛+2𝑘+1𝑞𝑦𝑛+2𝑘+1𝑞𝑦𝑛+𝑘+𝑦𝑛+𝑟+𝑝𝑦𝑛+𝑘+𝑦𝑛+𝑝𝑦𝑛+𝑘𝑦𝑛+𝑦−((𝑝+𝑞𝑟)/𝑝)2𝑛−𝑦𝑛−𝑟𝑞𝑦𝑛+2𝑘+1𝑞𝑦𝑛+𝑘+𝑦𝑛+𝑟+𝑝𝑦𝑛+𝑘+𝑦𝑛,𝑛∈ℕ0.(3.8)

When 𝑞=𝑝+𝑟, the unique positive equilibrium of (1.1) is 𝑦=1, (3.4) becomes

𝑦𝑛+1𝑟−1=1−𝑦𝑛𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0.(3.9)

When √𝑝=𝑞(1+1+4𝑟)/2, the unique positive equilibrium is 𝑦=𝑝/𝑞, (3.5) becomes

𝑦𝑛+1−𝑝𝑞=(𝑝−𝑞)/𝑞(𝑝/𝑞)−𝑦𝑛−𝑘𝑞𝑦𝑛+𝑦𝑛−𝑘,𝑛∈ℕ0,(3.10) and (3.8) becomes

𝑦𝑛−𝑦𝑛+2(𝑘+1)=𝑦𝑛𝑞−(𝑝/𝑞)2𝑦𝑛+𝑘𝑦𝑛+2𝑘+1+𝑞𝑦𝑛𝑦𝑛+2𝑘+1+𝑝𝑦𝑛+𝑘+𝑦𝑛+(𝑝−𝑞)/𝑞𝑞𝑦𝑛+2𝑘+1𝑞𝑦𝑛+𝑘+𝑦𝑛+𝑟+𝑝𝑦𝑛+𝑘+𝑦𝑛,𝑛∈ℕ0.(3.11)

Set

𝑓(𝑥,𝑦)=𝑟+𝑝𝑥+𝑦.𝑞𝑥+𝑦(3.12)

Lemma 3.2. Assume that 𝑓(𝑥,𝑦) is defined in (3.12). Then the following statements are true: (i)assume 𝑝<𝑞. Then 𝑓(𝑥,𝑦) is strictly decreasing in 𝑥 and increasing in 𝑦 for 𝑥≥𝑟/(𝑞−𝑝); and it is strictly decreasing in each of its arguments for 𝑥<𝑟/(𝑞−𝑝);(ii)assume 𝑝>𝑞. Then 𝑓(𝑥,𝑦) is increasing in 𝑥 and strictly decreasing in 𝑦 for 𝑦≥𝑞𝑟/(𝑝−𝑞); and it is strictly decreasing in each of its arguments for 𝑦<𝑞𝑟/(𝑝−𝑞).

Proof. By calculating the partial derivatives of the function 𝑓(𝑥,𝑦), we have 𝑓𝑥(𝑥,𝑦)=(𝑝−𝑞)𝑦−𝑞𝑟(𝑞𝑥+𝑦)2,𝑓𝑦(𝑥,𝑦)=(𝑞−𝑝)𝑥−𝑟(𝑞𝑥+𝑦)2,(3.13) from which these statements easily follow.

3.2. Invariant Interval

In this subsection, we discuss the invariant interval of (1.1).

3.2.1. The Case 𝑝<𝑞

Lemma 3.3. Assume that 𝑝<𝑞, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)𝑦𝑛>𝑝/𝑞 for all 𝑛≥1;(ii)If for some 𝑁≥0, 𝑦𝑁>𝑟/(𝑞−𝑝), then 𝑦𝑁+1<1;(iii)If for some 𝑁≥0, 𝑦𝑁=𝑟/(𝑞−𝑝), then 𝑦𝑁+1=1;(iv)If for some 𝑁≥0, 𝑦𝑁<𝑟/(𝑞−𝑝), then 𝑦𝑁+1>1;(v)If 𝑝<𝑞<𝑝+𝑟, then (1.1) possesses an invariant interval [1,𝑟/(𝑞−𝑝)] and 𝑦∈(1,𝑟/(𝑞−𝑝));(vi)If 𝑝+𝑟<𝑞<𝑝+𝑞𝑟/𝑝, then (1.1) possesses an invariant interval [𝑟/(𝑞−𝑝),1] and 𝑦∈(𝑟/(𝑞−𝑝),1);(vii)If 𝑞≥𝑝+𝑞𝑟/𝑝, then (1.1) possesses an invariant interval [𝑝/𝑞,1] and 𝑦∈(𝑝/𝑞,1).

Proof. The proofs of (i)–(iv) are straightforward consequences of the identities (3.5) and (3.4). So we only prove (v)–(vii). By the condition (i) of Lemma 3.2, the function 𝑓(𝑥,𝑦) is strictly decreasing in 𝑥 and increasing in 𝑦 for 𝑥≥𝑟/(𝑞−𝑝); and it is strictly decreasing in both arguments for 𝑥<𝑟/(𝑞−𝑝).
(v) Using the decreasing character of 𝑓, we obtain
𝑟1=𝑓,𝑟𝑞−𝑝𝑞−𝑝<𝑓(𝑥,𝑦)<𝑓(1,1)=𝑟+𝑝+1<𝑟𝑞+1𝑞−𝑝.(3.14) The inequalities 𝑟1<,𝑞−𝑝𝑟+𝑝+1<𝑟𝑞+1𝑞−𝑝(3.15) are equivalent to the inequality 𝑞<𝑝+𝑟.
On the other hand, 𝑦 is the unique positive root of quadratic equation
(𝑞+1)𝑦2−(𝑝+1)𝑦−𝑟=0.(3.16) Since 𝑟(𝑞+1)𝑞−𝑝2𝑟−(𝑝+1)𝑞−𝑝−𝑟=𝑟(𝑞+1)(𝑝+𝑟−𝑞)(𝑞−𝑝)2(>0,𝑞+1)−(𝑝+1)−𝑟=𝑞−𝑝−𝑟<0,(3.17) then we have that 𝑦∈(1,𝑟/(𝑞−𝑝)).
(vi) By using the monotonic character of 𝑓, we obtain
(𝑞−𝑝)(𝑝+𝑟)+𝑟𝑞2𝑟−𝑝𝑞+𝑟=𝑓1,𝑟𝑞−𝑝≤𝑓(𝑥,𝑦)≤𝑓𝑞−𝑝,1=1.(3.18) The inequalities (𝑞−𝑝)(𝑝+𝑟)+𝑟𝑞2>𝑟−𝑝𝑞+𝑟,𝑟𝑞−𝑝𝑞−𝑝<1(3.19) follow from the inequality 𝑞>𝑝+𝑟.
On the other hand, similar to (v) it can be proved that 𝑦∈(𝑟/(𝑞−𝑝),1).
(vii) In this case note that 𝑟/(𝑞−𝑝)≤𝑝/𝑞<1 holds, and using the monotonic character of 𝑓, we obtain
𝑝𝑞<𝑞𝑟+𝑝𝑞+𝑝𝑞2𝑝+𝑝=𝑓1,𝑞𝑝≤𝑓(𝑥,𝑦)≤𝑓𝑞=,1𝑞𝑟+𝑝2+𝑞𝑞(𝑝+1)≤1.(3.20) Furthermore, similar to (v) it follows 𝑦∈(𝑝/𝑞,1). The proof is complete.

When 𝑞=𝑝+𝑟, (3.9) implies that the following result holds.

Lemma 3.4. Assume that 𝑞=𝑝+𝑟, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)If for some 𝑁≥0, 𝑦𝑁>1, then 𝑦𝑁+1<1;(ii)If for some 𝑁≥0, 𝑦𝑁=1, then 𝑦𝑁+1=1;(iii)If for some 𝑁≥0, 𝑦𝑁<1, then 𝑦𝑁+1>1.

3.2.2. The Case 𝑝>𝑞

Lemma 3.5. Assume that 𝑝>𝑞, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)𝑦𝑛>1 for all 𝑛≥1;(ii)If for some 𝑁≥0, 𝑦𝑁<𝑞𝑟/(𝑝−𝑞), then 𝑦𝑁+𝑘+1>𝑝/𝑞;(iii)If for some 𝑁≥0, 𝑦𝑁=𝑞𝑟/(𝑝−𝑞), then 𝑦𝑁+𝑘+1=𝑝/𝑞;(iv)If for some 𝑁≥0, 𝑦𝑁>𝑞𝑟/(𝑝−𝑞), then 𝑦𝑁+𝑘+1<𝑝/𝑞;(v)If √𝑞<𝑝<𝑞(1+1+4𝑟)/2, then (1.1) possesses an invariant interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)] and 𝑦∈(𝑝/𝑞,𝑞𝑟/(𝑝−𝑞));(vi)If √𝑞(1+1+4𝑟)/2<𝑝<𝑞+𝑞𝑟, then (1.1) possesses an invariant interval [𝑞𝑟/(𝑝−𝑞),𝑝/𝑞] and 𝑦∈(𝑞𝑟/(𝑝−𝑞),𝑝/𝑞);(vii)If 𝑝≥𝑞+𝑞𝑟, then (1.1) possesses an invariant interval [1,𝑝/𝑞] and 𝑦∈(1,𝑝/𝑞).

Proof. The proofs of (i)–(iv) are direct consequences of the identities (3.4) and (3.5). So we only give the proofs (v)–(vii). By Lemma 3.2 (ii), the function 𝑓(𝑥,𝑦) is increasing in 𝑥 and strictly decreasing in 𝑦 for 𝑦≥𝑞𝑟/(𝑝−𝑞); and it is strictly decreasing in each of its arguments for 𝑦<𝑞𝑟/(𝑝−𝑞).
(v) Using the decreasing character of 𝑓, we obtain
𝑝𝑞=𝑓𝑞𝑟,𝑝−𝑞𝑞𝑟𝑝𝑝−𝑞≤𝑓(𝑥,𝑦)≤𝑓𝑞,𝑝𝑞=𝑞𝑟+𝑝(𝑝+1)≤𝑝(𝑞+1)𝑞𝑟𝑝−𝑞.(3.21) The inequalities 𝑞𝑟+𝑝(𝑝+1)≤𝑝(𝑞+1)𝑞𝑟,𝑝𝑝−𝑞𝑞<𝑞𝑟𝑝−𝑞(3.22) are equivalent to the inequality √𝑝<𝑞(1+1+4𝑟)/2. That is, [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)] is an invariant interval of (1.1).
On the other hand, similar to Lemma 3.3 (v), it can be proved that 𝑦∈(𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)).
(vi) By using the monotonic character of 𝑓, we obtain
𝑞𝑟≤𝑝−𝑞(𝑞𝑟+𝑝)(𝑝−𝑞)+𝑝𝑞2𝑟𝑞3𝑟+𝑝(𝑝−𝑞)=𝑓𝑞𝑟,𝑝𝑝−𝑞𝑞𝑝≤𝑓(𝑥,𝑦)≤𝑓𝑞,𝑞𝑟=𝑝𝑝−𝑞𝑞.(3.23) The inequalities (𝑞𝑟+𝑝)(𝑝−𝑞)+𝑝𝑞2𝑟𝑞3≥𝑟+𝑝(𝑝−𝑞)𝑞𝑟,𝑝−𝑞𝑞𝑟≤𝑝𝑝−𝑞𝑞(3.24) are equivalent to the inequality √𝑝>𝑞(1+1+4𝑟)/2.
On the other hand, similar to Lemma 3.3 (v) it can be proved that 𝑦∈(𝑞𝑟/(𝑝−𝑞),𝑝/𝑞).
(vii) In this case note that 𝑞𝑟/(𝑝−𝑞)≤1<𝑝/𝑞 holds. By the monotonic character of 𝑓, we have
1<𝑞𝑟+𝑝𝑞+𝑝𝑞2𝑝+𝑝=𝑓1,𝑞𝑝≤𝑓(𝑥,𝑦)≤𝑓𝑞=,1𝑞𝑟+𝑝2+𝑞≤𝑝𝑞(𝑝+1)𝑞.(3.25) The inequalities 𝑞𝑟+𝑝𝑞+𝑝𝑞2+𝑝>1,𝑞𝑟+𝑝2+𝑞≤𝑝𝑞(𝑝+1)𝑞(3.26) are equivalent to the inequality 𝑝≥𝑞+𝑞𝑟.
Furthermore, similar to Lemma 3.3 (v), it follows 𝑦∈(1,𝑝/𝑞). The proof is complete.

Lemma 3.6. Assume that √𝑝=𝑞(1+1+4𝑟)/2, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)If for some 𝑁≥0, 𝑦𝑁<𝑝/𝑞, then 𝑦𝑁+𝑘+1>𝑝/𝑞;(ii)If for some 𝑁≥0, 𝑦𝑁=𝑝/𝑞, then 𝑦𝑁+𝑘+1=𝑝/𝑞;(iii)If for some 𝑁≥0, 𝑦𝑁>𝑝/𝑞, then 𝑦𝑁+𝑘+1<𝑝/𝑞;(iv)If for some 𝑁≥0, 𝑦𝑁>𝑝/𝑞, then 𝑦𝑁>𝑦𝑁+2(𝑘+1)>𝑝/𝑞;(v)If for some 𝑁≥0, 𝑦𝑁<𝑝/𝑞, then 𝑦𝑁<𝑦𝑁+2(𝑘+1)<𝑝/𝑞.

Proof. In this case, we have that 𝑞𝑟/(𝑝−𝑞)=𝑝/𝑞. These results follow from the identities (3.10) and (3.11) and the details are omitted.

4. Global Attractivity

In this section, we discuss the global attractivity of the positive equilibrium of (1.1). We show that the unique positive equilibrium 𝑦 of (1.1) is a global attractor when 𝑝=𝑞 or 𝑝<1 and 𝑝<𝑞≤𝑝𝑞+1+3𝑝 or 𝑞<𝑝≤1.

4.1. The Case 𝑝=𝑞

In this subsection, we discuss the behavior of positive solutions of (1.1) when 𝑝=𝑞.

Theorem 4.1. Assume that 𝑝=𝑞 holds, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the unique positive equilibrium 𝑦 of (1.1) is a global attractor.

Proof. By the change of variables 𝑦𝑛𝑟=1+𝑢𝑝+1𝑛,(4.1) Equation (1.1) reduces to the difference equation 𝑢𝑛+1=11+𝑝𝑟/(𝑝+1)2𝑢𝑛+𝑟/(𝑝+1)2𝑢𝑛−𝑘,𝑛∈ℕ0.(4.2) The unique positive equilibrium 𝑢 of (4.2) is √𝑢=−(𝑝+1)+(𝑝+1)2+4𝑟(𝑝+1).2𝑟(4.3) Applying Lemma 1.3 in interval [0,1], then every positive solution of (1.1) converges to 𝑢. That is, 𝑢 is a global attractor. So, 𝑦 is a global attractor.

4.2. The Case 𝑝<𝑞

In this subsection, we present global attractivity of (1.1) when 𝑝<𝑞.

The following result is straightforward consequence of the identity (3.7).

Lemma 4.2. Assume that 𝑝<𝑞 holds, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)Suppose that 𝑞<𝑝+𝑟. If for some 𝑁≥0, 𝑦𝑁>1, then 𝑦𝑁+1<(𝑝+𝑟)/𝑞;(ii)Suppose that 𝑞>𝑝+𝑟. If for some 𝑁≥0, 𝑦𝑁<1, then 𝑦𝑁+1>(𝑝+𝑟)/𝑞.

Theorem 4.3. Assume that 𝑝<𝑞, 𝑝<1 and 𝑞≤𝑝𝑞+1+3𝑝 hold. Let {𝑦𝑛}∞𝑛=−𝑘 be a positive solution of (1.1). Then the following statements hold true: (i)Suppose 𝑞<𝑝+𝑟. If 𝑦0∈[1,𝑟/(𝑞−𝑝)], then 𝑦𝑛∈[1,𝑟/(𝑞−𝑝)] for 𝑛≥1. Furthermore, every positive solution of (1.1) lies eventually in the interval [1,𝑟/(𝑞−𝑝)].(ii)Suppose 𝑞>𝑝+𝑟. If 𝑦0∈[𝑟/(𝑞−𝑝),1], then 𝑦𝑛∈[𝑟/(𝑞−𝑝),1] for 𝑛≥1. Furthermore, every positive solution of (1.1) lies eventually in the interval [𝑟/(𝑞−𝑝),1].

Proof. We only give the proof of (i), the proof of (ii) is similar and will be omitted. First, note that in this case 𝑝/𝑞<1<(𝑝+𝑟)/𝑞<𝑟/(𝑞−𝑝) holds.
If 𝑦0∈[1,𝑟/(𝑞−𝑝)], then by Lemma 3.3 (iv), we have that 𝑦1>1, and by Lemma 4.2 (i), we obtain that 𝑦1<(𝑝+𝑟)/𝑞<𝑟/(𝑞−𝑝), which implies that 𝑦1∈[1,𝑟/(𝑞−𝑝)], by induction, we have 𝑦𝑛∈[1,𝑟/(𝑞−𝑝)], for 𝑛≥1.
Now, to complete the proof it remains to show that when 𝑦0∉[1,𝑟/(𝑞−𝑝)], there exists 𝑁>0 such that 𝑦𝑁∈[1,𝑟/(𝑞−𝑝)].
If 𝑦0∉[1,𝑟/(𝑞−𝑝)], then we have the following two cases to be considered:
(a)𝑦0>𝑟/(𝑞−𝑝);(b)𝑦0<1.
Case (a). From Lemma 3.3 (ii), we see that 𝑦1<1. Thus, in the sequel, we only consider case (b).
If 𝑦0<1, then by Lemma 3.3 (iv), we have 𝑦1>1, and from Lemma 4.2 (i), we have 𝑦2<(𝑝+𝑟)/𝑞<𝑟/(𝑞−𝑝). So 𝑦3>1 and 𝑦4<𝑟/(𝑞−𝑝). By induction, there exists exactly one term greater than 1, which is followed by exactly one term less than 𝑟/(𝑞−𝑝), which is followed by exactly one term greater than 1, and so on. If for some 𝑁>0, 1≤𝑦𝑁≤𝑟/(𝑞−𝑝), then the former assertion implies that the result is true.
So assume for the sake of contradiction, that for all 𝑛≥1, 𝑦𝑛 never enter the interval [1,𝑟/(𝑞−𝑝)], then the sequence {𝑦𝑛}∞𝑛=1 will oscillate relative to the interval [1,𝑟/(𝑞−𝑝)] with semicycles of length one. Consider the subsequence {𝑦2𝑛}∞𝑛=1 and {𝑦2𝑛+1}∞𝑛=1 of solution {𝑦𝑛}∞𝑛=−𝑘, we have
𝑦2𝑛<1,𝑦2𝑛+1>𝑟𝑞−𝑝for𝑛≥1.(4.4) Let 𝐿=lim𝑛→∞sup𝑦2𝑛,𝑙=lim𝑛→∞inf𝑦2𝑛,𝑀=lim𝑛→∞sup𝑦2𝑛+1,𝑚=lim𝑛→∞inf𝑦2𝑛+1,(4.5) which in view of Theorem 3.1 exist as finite numbers, such that 𝐿≤𝑟+𝑝𝑚+𝑙𝑞𝑚+𝑙,𝑙≥𝑟+𝑝𝑀+𝐿,𝑞𝑀+𝐿(4.6)𝑀≤𝑟+𝑝𝑙+𝑚𝑞𝑙+𝑚,𝑚≥𝑟+𝑝𝐿+𝑀𝑞𝐿+𝑀.(4.7) From (4.6), we have 𝑞(𝐿𝑚−𝑙𝑀)≤𝑝(𝑚−𝑀)+(𝑙−𝐿), which implies that 𝐿𝑚−𝑙𝑀≤0. Also, from (4.7), we have 𝑞(𝑙𝑀−𝐿𝑚)≤𝑝(𝑙−𝐿)+(𝑚−𝑀), which implies that 𝑙𝑀−𝐿𝑚≤0. Thus 𝑙𝑀−𝐿𝑚=0 and 𝐿=𝑙, 𝑀=𝑚 hold, from which it follows that lim𝑛→∞𝑦2𝑛 and lim𝑛→∞𝑦2𝑛+1 exist.
Set
lim𝑛→∞𝑦2𝑛=𝐿,lim𝑛→∞𝑦2𝑛+1=𝑀,(4.8) then 𝐿≤1, 𝑀≥𝑟/(𝑞−𝑝) and 𝐿, 𝑀 satisfies the system 𝐿=𝑟+𝑝𝑀+𝐿𝑞𝑀+𝐿,𝑀=𝑟+𝑝𝐿+𝑀,𝑞𝐿+𝑀(4.9) which implies that 𝐿, 𝑀 is a period-two solution of (1.1). Furthermore, in view of Theorem 2.2, (1.1) has no period-two solution when 𝑝<1 and 𝑞≤𝑝𝑞+1+3𝑝 hold. This is a contradiction, as desired. The proof is complete.

Theorem 4.4. Assume that 𝑝<𝑞, 𝑝<1 and 𝑞≤𝑝𝑞+1+3𝑝 hold. Then the unique positive equilibrium 𝑦 of (1.1) is a global attractor.

Proof. To complete the proof, there are four cases to be considered.
Case (i). 𝑞<𝑝+𝑟.
By Theorem 4.3 (i), we know that all solutions of (1.1) lies eventually in the invariant interval [1,𝑟/(𝑞−𝑝)]. Furthermore, the function 𝑓(𝑥,𝑦) is non-increasing in each of its arguments in the interval [1,𝑟/(𝑞−𝑝)]. Thus, applying Lemma 1.3, every solution of (1.1) converges to 𝑦, that is, 𝑦 is a global attractor.
Case (ii). 𝑞=𝑝+𝑟.
In this case, the only positive equilibrium is 𝑦=1. In view of Lemma 3.4, we see that, after the first semicycle, the nontrivial solution oscillates about 𝑦 with semicycles of length one. Consider the subsequences {𝑦2𝑛}∞𝑛=1 and {𝑦2𝑛+1}∞𝑛=1 of any nontrivial solution {𝑦𝑛}∞𝑛=−𝑘 of (1.1). We have
𝑦2𝑛<1,𝑦2𝑛+1>1,for𝑛≥1,(4.10) or vice versa. Here, we may assume, without loss of generality, that 𝑦2𝑛<1 and 𝑦2𝑛+1>1, for 𝑛≥1.
Let
𝐿=lim𝑛→∞sup𝑦2𝑛,𝑙=lim𝑛→∞inf𝑦2𝑛,𝑀=lim𝑛→∞sup𝑦2𝑛+1,𝑚=lim𝑛→∞inf𝑦2𝑛+1,(4.11) which, in view of Theorem 3.1, exist. Then as the same argument in Theorem 4.3, we can see that lim𝑛→∞𝑦2𝑛 and lim𝑛→∞𝑦2𝑛+1 exist.
Set
lim𝑛→∞𝑦2𝑛=𝐿,lim𝑛→∞𝑦2𝑛+1=𝑀,(4.12) then 𝐿≤1, 𝑀≥1. If 𝐿≠𝑀, then, also as the same argument in Theorem 4.3, we can see that 𝐿, 𝑀 is a period-two solution of (1.1), which contradicts Theorem 2.2. Thus 𝐿=𝑀, from which it follows that lim𝑛→∞𝑦𝑛=1, which implies that 𝑦=1 is a global attractor.
Case (iii). 𝑝+𝑟<𝑞<𝑝+(𝑞/𝑝)𝑟.
By Theorem 4.3 (ii), we know that all solutions of (1.1) lies eventually in the invariant interval [𝑟/(𝑞−𝑝),1]. Furthermore, the function 𝑓(𝑥,𝑦) decreases in 𝑥 and increases in 𝑦 in the interval [𝑟/(𝑞−𝑝),1]. Thus, applying Lemma 1.3, every solution of converges to 𝑦, that is, 𝑦 is a global attractor.
Case (iv). 𝑞≥𝑝+(𝑞/𝑝)𝑟.
In this case, we note that 𝑟/(𝑞−𝑝)≤𝑝/𝑞<1 holds. From Theorem 4.3 (ii) and Lemma 3.3 (i), we know that all solutions of (1.1) eventually enter the invariant interval [𝑝/𝑞,1]. Hence, by using the same argument in (iii), 𝑦 is a global attractor. The proof is complete.

4.3. The Case 𝑝>𝑞

In this subsection, we discuss the global behavior of (1.1) when 𝑝>𝑞.

The following three results are the direct consequences of equations (3.4), (3.5), (3.6), and (3.8).

Lemma 4.5. Assume that √𝑞<𝑝<𝑞(1+1+4𝑟)/2, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)If for some 𝑁≥0, 𝑦𝑁<𝑝/𝑞, then 𝑦𝑁<𝑦𝑁+2(𝑘+1)<𝑞𝑟/(𝑝−𝑞);(ii)If for some 𝑁≥0, 𝑦𝑁>𝑞𝑟/(𝑝−𝑞), then 𝑝/𝑞<𝑦𝑁+2(𝑘+1)<𝑦𝑁;(iii)If for some 𝑁≥0, 𝑝/𝑞≤𝑦𝑁≤𝑞𝑟/(𝑝−𝑞), then 𝑝/𝑞≤𝑦𝑁+2(𝑘+1)≤𝑞𝑟/(𝑝−𝑞).

Lemma 4.6. Assume that √𝑞(1+1+4𝑟)/2<𝑝<𝑞+𝑞𝑟, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)If for some 𝑁≥0, 𝑦𝑁<𝑞𝑟/(𝑝−𝑞), then 𝑦𝑁<𝑦𝑁+2(𝑘+1)<𝑝/𝑞;(ii)If for some 𝑁≥0, 𝑦𝑁>𝑝/𝑞, then 𝑞𝑟/(𝑝−𝑞)<𝑦𝑁+2(𝑘+1)<𝑦𝑁;(iii)If for some 𝑁≥0, 𝑞𝑟/(𝑝−𝑞)≤𝑦𝑁≤𝑝/𝑞, then 𝑞𝑟/(𝑝−𝑞)≤𝑦𝑁+2(𝑘+1)≤𝑝/𝑞.

Lemma 4.7. Assume that 𝑝≥𝑞+𝑞𝑟, and {𝑦𝑛}∞𝑛=−𝑘 is a positive solution of (1.1). Then the following statements are true: (i)𝑦𝑛>1 for 𝑛≥1;(ii)If for some 𝑁≥0, 𝑦𝑁>𝑝/𝑞, then 1<𝑦𝑁+2(𝑘+1)<𝑝/𝑞 and 𝑦𝑁+2(𝑘+1)<𝑦𝑁;(iii)If for some 𝑁≥0, 1<𝑦𝑁≤𝑝/𝑞, then 1<𝑦𝑁+2(𝑘+1)≤𝑝/𝑞.

Theorem 4.8. Assume that 𝑝>𝑞 holds, and let {𝑦𝑛}∞𝑛=−𝑘 be a positive solution of (1.1). Then the following statements hold true: (i)If √𝑝<𝑞(1+1+4𝑟)/2, then every positive solution of (1.1) lies eventually in the interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)].(ii)If √𝑞(1+1+4𝑟)/2<𝑝<𝑞+𝑞𝑟, then every positive solution of (1.1) lies eventually in the interval [𝑞𝑟/(𝑝−𝑞),𝑝/𝑞].(iii)If 𝑝≥𝑞+𝑞𝑟, then every positive solution of (1.1) lies eventually in the interval [1,𝑝/𝑞].

Proof. We only give the proof of (i), the proofs of (ii) and (iii) are similar and will be omitted.
When √𝑞<𝑝<𝑞(1+1+4𝑟)/2, recall that from Lemma 3.5, [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)] is an invariant interval and so it follows that every solution of (1.1) with 𝑘+1 consecutive values in [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)], lies eventually in this interval. If the solution is not eventually in [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)], there are three cases to be considered.
Case (i). If for some 𝑁≥0, 𝑦𝑁>𝑞𝑟/(𝑝−𝑞), then there are two cases to be considered. If 𝑦𝑁+2(𝑘+1)𝑛≥𝑞𝑟/(𝑝−𝑞) for every 𝑛∈𝑁, then by Lemma 4.5, we have
𝑦𝑁+2(𝑘+1)(𝑛−1)>𝑦𝑁+2(𝑘+1)𝑛>𝑝𝑞,(4.13) hence, the subsequence {𝑦𝑁+2(𝑘+1)𝑛} is strictly monotonically decreasing convergent and its limit 𝑆 satisfies 𝑆≥𝑞𝑟/(𝑝−𝑞). Taking limit on both sides of (3.8), we obtain a contradiction. If for some 𝑛0, 𝑦𝑁+2(𝑘+1)𝑛0<𝑞𝑟/(𝑝−𝑞), then by Lemma 4.5 we obtain that {𝑦𝑁+2(𝑘+1)𝑛} is eventually in the interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)].
Case (ii). If for some 𝑁≥0, 𝑦𝑁<𝑝/𝑞, then there are two cases to be considered. If 𝑦𝑁+2(𝑘+1)𝑛<𝑝/𝑞 for every 𝑛∈𝑁, then by Lemma 4.5 we obtain
𝑦𝑁+2(𝑘+1)(𝑛−1)<𝑦𝑁+2(𝑘+1)𝑛<𝑞𝑟,𝑝−𝑞(4.14) which implies that the subsequence {𝑦𝑁+2(𝑘+1)𝑛} is convergent. Then as the same argument in case (i), obtain a contradiction. If for some 𝑛0, 𝑦𝑁+2(𝑘+1)𝑛0>𝑝/𝑞, then by Lemma 4.5 we have that {𝑦𝑁+2(𝑘+1)𝑛} is eventually in the interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)].
Case (iii). If for some 𝑁≥0,𝑝/𝑞≤𝑦𝑁≤𝑞𝑟/(𝑝−𝑞), then by Lemma 4.5 it follows that 𝑝/𝑞≤𝑦𝑁+2(𝑘+1)𝑛≤𝑞𝑟/(𝑝−𝑞). Assume that there is a subsequence {𝑦𝑁0+2(𝑘+1)𝑛} such that 𝑦𝑁0+2(𝑘+1)𝑛≥𝑞𝑟/(𝑝−𝑞), or 𝑦𝑁0+2(𝑘+1)𝑛≤𝑝/𝑞, for every 𝑛∈𝑁. Then its limit 𝑆 satisfies 𝑆≥𝑞𝑟/(𝑝−𝑞), or 𝑆≤𝑝/𝑞. Taking limit on both sides of (3.8), obtain a contradiction. Hence, for all 𝑁∈{1,2,…,2(𝑘+1)} the subsequences {𝑦𝑁+2(𝑘+1)𝑛} are eventually in the interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)].

Theorem 4.9. Assume that 𝑝>𝑞 and 𝑝≤1 hold. Then the unique positive equilibrium 𝑦 of (1.1) is a global attractor.

Proof. The proof will be accomplished by considering the following four cases.
Case (i). √𝑝<𝑞(1+1+4𝑟)/2.
By part (i) of Theorem 4.8, we know that all positive solutions of (1.1) lie eventually in the invariant interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)]. Furthermore, the function 𝑓(𝑥,𝑦) is nonincreasing in each of its arguments in the interval [𝑝/𝑞,𝑞𝑟/(𝑝−𝑞)]. Thus, applying Lemma 1.3, every solution of (1.1) converges to 𝑦, that is, 𝑦 is a global attractor.
Case (ii). √𝑝=𝑞(1+1+4𝑟)/2.
In this case, the only positive equilibrium of (1.1) is 𝑦=𝑝/𝑞. From Lemma 3.6 and (3.11), we know that each of the 2(𝑘+1) subsequences
𝑦2(𝑘+1)𝑛+𝑖∞𝑛=0for𝑖=1,2,…,2(𝑘+1)(4.15) of any solution {𝑦𝑛}∞𝑛=−𝑘 of (1.1) is either identically equal to 𝑝/𝑞 or strictly monotonically convergent and its limit is greater than zero. Set 𝐿𝑖=lim𝑛→∞𝑦2(𝑘+1)𝑛+𝑖for𝑖=1,2,…,2(𝑘+1).(4.16) Then, clearly, …,𝐿1,𝐿2,…,𝐿2(𝑘+1),…(4.17) is a period solution of (1.1) with period 2(𝑘+1). By applying (3.11) to the solution (4.17) and using the fact 𝐿𝑖>0 for 𝑖=1,2,…,2(𝑘+1), we see that 𝐿𝑖=𝑝𝑞for𝑖=1,2,…,2(𝑘+1),(4.18) and so lim𝑛→∞𝑦𝑛=𝑝𝑞,(4.19) which implies that 𝑦=𝑝/𝑞 is a global attractor.
Case (iii). √𝑞(1+1+4𝑟)/2<𝑝<𝑞+𝑞𝑟.
By Theorem 4.8 (ii), all positive solutions of (1.1) eventually enter the invariant interval [𝑞𝑟/(𝑝−𝑞),𝑝/𝑞]. Furthermore, the function 𝑓(𝑥,𝑦) increases in 𝑥 and decreases in 𝑦 in the interval [𝑞𝑟/(𝑝−𝑞),𝑝/𝑞]. Thus, applying Lemma 1.3 and assumption 𝑝≤1, every solution of (1.1) converges to 𝑦. So, 𝑦 is a global attractor.
Case (iv). 𝑝≥𝑞+𝑞𝑟.
In this case, we note that 𝑞𝑟/(𝑝−𝑞)≤1<𝑝/𝑞 holds. In view of Theorem 4.8 (iii), we obtain that all solutions of (1.1) eventually enter the invariant interval [1,𝑝/𝑞]. Furthermore, the function 𝑓(𝑥,𝑦) increases in 𝑥 and decreases in 𝑦 in the interval [1,𝑝/𝑞]. Then using the same argument in case (iii), every solution of (1.1) converges to 𝑦. Thus the equilibrium 𝑦 is a global attractor. The proof is complete.

Finally, we summarize our results and obtain the following theorem, which shows that 𝑦 is a global attractor in three cases.

Theorem 4.10. The unique positive equilibrium 𝑦 of (1.1) is a global attractor, when one of the following three cases holds: (i)𝑝=𝑞;(ii)𝑝<1 and 𝑝<𝑞≤𝑝𝑞+1+3𝑝;(iii)𝑞<𝑝≤1.