Discrete Dynamics in Nature and Society
Volume 2010 (2010), Article ID 610467, 17 pages
doi:10.1155/2010/610467
Research Article

Boundedness and Global Attractivity of a Higher-Order Nonlinear Difference Equation

Xiu-Mei Jia1,2 and Wan-Tong Li2

1Department of Mathematics, Hexi University, Zhangye, Gansu 734000, China
2School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, China

Received 5 November 2009; Accepted 4 February 2010

Academic Editor: Guang Zhang

Copyright © 2010 Xiu-Mei Jia and Wan-Tong Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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: 𝑦 𝑛 + 1 = ( 𝑟 + 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 ) / ( 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 ) , 𝑛 ∈ ℕ 0 , where the parameters 𝑝 , 𝑞 , 𝑟 ∈ ( 0 , ∞ ) , 𝑘 ∈ { 1 , 2 , 3 , … } and the initial conditions 𝑦 − 𝑘 , … , 𝑦 0 ∈ ( 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

𝑦 𝑛 + 1 = 𝑟 + 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , ( 1 . 1 ) where 𝑝 , 𝑞 , 𝑟 ∈ ( 0 , ∞ ) , ℕ 0 ∈ { 0 , 1 , … } , 𝑘 ∈ { 1 , 2 , 3 , … } and the initial conditions 𝑦 − 𝑘 , … , 𝑦 0 ∈ ( 0 , ∞ ) .

When 𝑘 = 1 , (1.1) reduces to

𝑦 𝑛 + 1 = 𝑟 + 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 1 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 1 , 𝑛 ∈ ℕ 0 . ( 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 , … , 𝑥 0 ∈ 𝐼 , the difference equation

𝑥 𝑛 + 1 𝑥 = 𝑓 𝑛 , 𝑥 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 ( 1 . 3 ) has a unique solution { 𝑥 𝑛 } ∞ 𝑛 = − 𝑘 .

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

𝑥 = 𝑓 𝑥 , 𝑥 . ( 1 . 4 )

That is, 𝑥 𝑛 = 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 , … , 𝑥 0 ∈ 𝐽 ⟹ 𝑥 𝑛 ∈ 𝐽 ∀ 𝑛 ∈ ℕ 0 . ( 1 . 5 )

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

Let

𝑃 = 𝜕 𝑓 𝜕 𝑢 𝑥 , 𝑥 , 𝑄 = 𝜕 𝑓 𝜕 𝑣 𝑥 , 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

𝑧 𝑛 + 1 = 𝑃 𝑧 𝑛 + 𝑄 𝑧 𝑛 − 𝑘 , 𝑛 = 0 , 1 , … , ( 1 . 7 ) and its characteristic equation is

𝜆 𝑘 + 1 − 𝑃 𝜆 𝑘 − 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 𝑃 𝑄 > 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 , 𝑏 𝑎 , 𝑏 , ( 𝑓 ( 𝑚 , 𝑚 ) = 𝑀 , 𝑓 ( 𝑀 , 𝑀 ) = 𝑚 ) ⟹ 𝑚 = 𝑀 , ( 1 . 1 0 ) (ii) 𝑓 ( 𝑥 , 𝑦 ) is nondecreasing in 𝑥 and nonincreasing in 𝑦 , and [ ] × [ ] ∀ ( 𝑚 , 𝑀 ) ∈ 𝑎 , 𝑏 𝑎 , 𝑏 , ( 𝑓 ( 𝑚 , 𝑀 ) = 𝑚 , 𝑓 ( 𝑀 , 𝑚 ) = 𝑀 ) ⟹ 𝑚 = 𝑀 , ( 1 . 1 1 ) (iii) 𝑓 ( 𝑥 , 𝑦 ) is nonincreasing in 𝑥 and nondecreasing in 𝑦 , and [ ] × [ ] ∀ ( 𝑚 , 𝑀 ) ∈ 𝑎 , 𝑏 𝑎 , 𝑏 , ( 𝑓 ( 𝑀 , 𝑚 ) = 𝑚 , 𝑓 ( 𝑚 , 𝑀 ) = 𝑀 ) ⟹ 𝑚 = 𝑀 . ( 1 . 1 2 )

(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 + 4 𝑟 ( 1 + 𝑞 ) 2 . ( 1 + 𝑞 ) ( 2 . 1 )

The linearized equation associated with (1.1) about 𝑦 is

𝑧 𝑛 + 1 − ( 𝑝 − 𝑞 ) 𝑦 − 𝑞 𝑟 ( 𝑞 + 1 ) 𝑟 + ( 𝑝 + 1 ) 𝑦 𝑧 𝑛 + ( 𝑝 − 𝑞 ) 𝑦 + 𝑟 ( 𝑞 + 1 ) 𝑟 + ( 𝑝 + 1 ) 𝑦 𝑧 𝑛 − 𝑘 = 0 , ( 2 . 2 )

and its characteristic equation is

𝜆 𝑘 + 1 − ( 𝑝 − 𝑞 ) 𝑦 − 𝑞 𝑟 ( 𝑞 + 1 ) 𝑟 + ( 𝑝 + 1 ) 𝑦 𝜆 𝑘 + ( 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 , … , 𝑦 0 ∈ ( 0 , ∞ ) . Then the following stataments are true. (i)If ( 𝑝 − 𝑞 ) 𝑦 − 𝑞 𝑟 ≥ 0 , ( 𝑝 − 3 𝑞 − 𝑝 𝑞 − 1 ) 𝑦 < 2 𝑞 𝑟 , ( 2 . 4 ) then the unique positive equilibrium 𝑦 of (1.1) is locally asymptotically stable;(ii) If ( 𝑝 − 𝑞 ) 𝑦 − 𝑞 𝑟 < 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 𝑦 𝑛 = 𝑦 𝑛 − 𝑘 , and 𝜙 and 𝜓 satisfy the following system:

𝜙 = 𝑟 + 𝑝 𝜓 + 𝜓 𝑞 𝜓 + 𝜓 , 𝜓 = 𝑟 + 𝑝 𝜙 + 𝜙 , 𝑞 𝜙 + 𝜙 ( 2 . 8 )

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

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

𝜙 = 𝑟 + 𝑝 𝜓 + 𝜙 𝑞 𝜓 + 𝜙 , 𝜓 = 𝑟 + 𝑝 𝜙 + 𝜓 , 𝑞 𝜙 + 𝜓 ( 2 . 9 )

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

𝑝 < 1 , 𝑞 > 1 , 4 𝑟 < ( 1 − 𝑝 ) ( 𝑞 − 1 − 𝑝 𝑞 − 3 𝑝 ) . ( 2 . 1 0 )

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 . 1 1 ) if and only if 𝑘 i s o d d , 𝑝 < 1 , 𝑞 > 1 , 4 𝑟 < ( 1 − 𝑝 ) ( 𝑞 − 𝑝 𝑞 − 3 𝑝 − 1 ) . ( 2 . 1 2 ) 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 𝑡 2 − ( 1 − 𝑝 ) 𝑡 + 𝑟 + 𝑝 ( 1 − 𝑝 ) 𝑞 − 1 = 0 . ( 2 . 1 3 )

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 𝑦 𝑛 + 1 = 𝑟 + 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 ≥ 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 ≥ m i n { ( 𝑝 / 𝑞 ) , 1 } 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 𝑝 = m i n 𝑞 , 1 ( 3 . 1 ) for all 𝑛 ≥ 0 . We denote 𝑝 𝐾 = m i n 𝑞 , 1 . ( 3 . 2 ) Then 𝑦 𝑛 + 1 = 𝑟 + 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 ≤ 𝑟 + 𝑝 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 ( 𝑞 / 2 ) 𝐾 + ( 𝐾 / 2 ) + ( 𝑞 / 2 ) 𝑦 𝑛 + ( 1 / 2 ) 𝑦 𝑛 − 𝑘 ≤ m a x { 𝑟 , 𝑝 , 1 } 1 + 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 m i n { ( 𝑞 / 2 ) 𝐾 + ( 𝐾 / 2 ) , ( 𝑞 / 2 ) , ( 1 / 2 ) } 1 + 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 = m a x { 𝑟 , 𝑝 , 1 } m i n { ( 𝑞 / 2 ) 𝐾 + ( 𝐾 / 2 ) , ( 𝑞 / 2 ) , ( 1 / 2 ) } ( 3 . 3 ) for all 𝑛 > 𝑘 . The proof is complete.

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

𝑦 𝑛 + 1 − 1 = ( 𝑞 − 𝑝 ) ( 𝑟 / ( 𝑞 − 𝑝 ) ) − 𝑦 𝑛 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , 𝑦 ( 3 . 4 ) 𝑛 + 1 − 𝑝 𝑞 = ( ( 𝑝 − 𝑞 ) / 𝑞 ) ( 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ) − 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , 𝑦 ( 3 . 5 ) 𝑛 + 1 − 𝑞 𝑟 = 𝑝 𝑝 − 𝑞 2 − 𝑝 𝑞 − 𝑞 2 𝑟 𝑦 / ( 𝑝 − 𝑞 ) 𝑛 + ( 1 / 𝑞 ) 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 + 𝑦 ( ( 𝑝 − 𝑞 − 𝑞 𝑟 ) / ( 𝑝 − 𝑞 ) ) 𝑛 − 𝑘 − ( 𝑝 / 𝑞 ) 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , 𝑦 ( 3 . 6 ) 𝑛 + 1 − 𝑝 + 𝑟 𝑞 = 𝑟 1 − 𝑦 𝑛 + ( ( 𝑞 − 𝑝 − 𝑟 ) / 𝑞 ) 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , 𝑦 ( 3 . 7 ) 𝑛 − 𝑦 𝑛 + 2 ( 𝑘 + 1 ) = 𝑦 𝑛 𝑞 − ( 𝑝 / 𝑞 ) 2 𝑦 𝑛 + 𝑘 𝑦 𝑛 + 2 𝑘 + 1 + 𝑞 𝑦 𝑛 𝑦 𝑛 + 2 𝑘 + 1 𝑞 𝑦 𝑛 + 2 𝑘 + 1 𝑞 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 + 𝑟 + 𝑝 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 + 𝑝 𝑦 𝑛 + 𝑘 𝑦 𝑛 + 𝑦 − ( ( 𝑝 + 𝑞 𝑟 ) / 𝑝 ) 2 𝑛 − 𝑦 𝑛 − 𝑟 𝑞 𝑦 𝑛 + 2 𝑘 + 1 𝑞 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 + 𝑟 + 𝑝 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 , 𝑛 ∈ ℕ 0 . ( 3 . 8 )

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

𝑦 𝑛 + 1 𝑟 − 1 = 1 − 𝑦 𝑛 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 . ( 3 . 9 )

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

𝑦 𝑛 + 1 − 𝑝 𝑞 = ( 𝑝 − 𝑞 ) / 𝑞 ( 𝑝 / 𝑞 ) − 𝑦 𝑛 − 𝑘 𝑞 𝑦 𝑛 + 𝑦 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , ( 3 . 1 0 ) and (3.8) becomes

𝑦 𝑛 − 𝑦 𝑛 + 2 ( 𝑘 + 1 ) = 𝑦 𝑛 𝑞 − ( 𝑝 / 𝑞 ) 2 𝑦 𝑛 + 𝑘 𝑦 𝑛 + 2 𝑘 + 1 + 𝑞 𝑦 𝑛 𝑦 𝑛 + 2 𝑘 + 1 + 𝑝 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 + ( 𝑝 − 𝑞 ) / 𝑞 𝑞 𝑦 𝑛 + 2 𝑘 + 1 𝑞 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 + 𝑟 + 𝑝 𝑦 𝑛 + 𝑘 + 𝑦 𝑛 , 𝑛 ∈ ℕ 0 . ( 3 . 1 1 )

Set

𝑓 ( 𝑥 , 𝑦 ) = 𝑟 + 𝑝 𝑥 + 𝑦 . 𝑞 𝑥 + 𝑦 ( 3 . 1 2 )

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 ≥ 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ; and it is strictly decreasing in each of its arguments for 𝑦 < 𝑞 𝑟 / ( 𝑝 − 𝑞 ) .

Proof. By calculating the partial derivatives of the function 𝑓 ( 𝑥 , 𝑦 ) , we have 𝑓 𝑥 ( 𝑥 , 𝑦 ) = ( 𝑝 − 𝑞 ) 𝑦 − 𝑞 𝑟 ( 𝑞 𝑥 + 𝑦 ) 2 , 𝑓 𝑦 ( 𝑥 , 𝑦 ) = ( 𝑞 − 𝑝 ) 𝑥 − 𝑟 ( 𝑞 𝑥 + 𝑦 ) 2 , ( 3 . 1 3 ) 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 { 𝑦 𝑛 } ∞ 𝑛 = − 𝑘 is a positive solution of (1.1). Then the following statements are true: (i) 𝑦 𝑛 > 𝑝 / 𝑞 for all 𝑛 ≥ 1 ;(ii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 > 𝑟 / ( 𝑞 − 𝑝 ) , then 𝑦 𝑁 + 1 < 1 ;(iii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 = 𝑟 / ( 𝑞 − 𝑝 ) , then 𝑦 𝑁 + 1 = 1 ;(iv)If for some 𝑁 ≥ 0 , 𝑦 𝑁 < 𝑟 / ( 𝑞 − 𝑝 ) , then 𝑦 𝑁 + 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 + 𝑞 𝑟 / 𝑝 , then (1.1) possesses an invariant interval [ 𝑟 / ( 𝑞 − 𝑝 ) , 1 ] and 𝑦 ∈ ( 𝑟 / ( 𝑞 − 𝑝 ) , 1 ) ;(vii)If 𝑞 ≥ 𝑝 + 𝑞 𝑟 / 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
𝑟 1 = 𝑓 , 𝑟 𝑞 − 𝑝 𝑞 − 𝑝 < 𝑓 ( 𝑥 , 𝑦 ) < 𝑓 ( 1 , 1 ) = 𝑟 + 𝑝 + 1 < 𝑟 𝑞 + 1 𝑞 − 𝑝 . ( 3 . 1 4 ) The inequalities 𝑟 1 < , 𝑞 − 𝑝 𝑟 + 𝑝 + 1 < 𝑟 𝑞 + 1 𝑞 − 𝑝 ( 3 . 1 5 ) are equivalent to the inequality 𝑞 < 𝑝 + 𝑟 .
On the other hand, 𝑦 is the unique positive root of quadratic equation
( 𝑞 + 1 ) 𝑦 2 − ( 𝑝 + 1 ) 𝑦 − 𝑟 = 0 . ( 3 . 1 6 ) Since 𝑟 ( 𝑞 + 1 ) 𝑞 − 𝑝 2 𝑟 − ( 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 . 1 7 ) then we have that 𝑦 ∈ ( 1 , 𝑟 / ( 𝑞 − 𝑝 ) ) .
(vi) By using the monotonic character of 𝑓 , we obtain
( 𝑞 − 𝑝 ) ( 𝑝 + 𝑟 ) + 𝑟 𝑞 2 𝑟 − 𝑝 𝑞 + 𝑟 = 𝑓 1 , 𝑟 𝑞 − 𝑝 ≤ 𝑓 ( 𝑥 , 𝑦 ) ≤ 𝑓 𝑞 − 𝑝 , 1 = 1 . ( 3 . 1 8 ) The inequalities ( 𝑞 − 𝑝 ) ( 𝑝 + 𝑟 ) + 𝑟 𝑞 2 > 𝑟 − 𝑝 𝑞 + 𝑟 , 𝑟 𝑞 − 𝑝 𝑞 − 𝑝 < 1 ( 3 . 1 9 ) 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
𝑝 𝑞 < 𝑞 𝑟 + 𝑝 𝑞 + 𝑝 𝑞 2 𝑝 + 𝑝 = 𝑓 1 , 𝑞 𝑝 ≤ 𝑓 ( 𝑥 , 𝑦 ) ≤ 𝑓 𝑞 = , 1 𝑞 𝑟 + 𝑝 2 + 𝑞 𝑞 ( 𝑝 + 1 ) ≤ 1 . ( 3 . 2 0 ) 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 { 𝑦 𝑛 } ∞ 𝑛 = − 𝑘 is a positive solution of (1.1). Then the following statements are true: (i)If for some 𝑁 ≥ 0 , 𝑦 𝑁 > 1 , then 𝑦 𝑁 + 1 < 1 ;(ii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 = 1 , then 𝑦 𝑁 + 1 = 1 ;(iii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 < 1 , then 𝑦 𝑁 + 1 > 1 .

3.2.2. The Case 𝑝 > 𝑞

Lemma 3.5. Assume that 𝑝 > 𝑞 , and { 𝑦 𝑛 } ∞ 𝑛 = − 𝑘 is a positive solution of (1.1). Then the following statements are true: (i) 𝑦 𝑛 > 1 for all 𝑛 ≥ 1 ;(ii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 < 𝑞 𝑟 / ( 𝑝 − 𝑞 ) , then 𝑦 𝑁 + 𝑘 + 1 > 𝑝 / 𝑞 ;(iii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 = 𝑞 𝑟 / ( 𝑝 − 𝑞 ) , then 𝑦 𝑁 + 𝑘 + 1 = 𝑝 / 𝑞 ;(iv)If for some 𝑁 ≥ 0 , 𝑦 𝑁 > 𝑞 𝑟 / ( 𝑝 − 𝑞 ) , then 𝑦 𝑁 + 𝑘 + 1 < 𝑝 / 𝑞 ;(v)If √ 𝑞 < 𝑝 < 𝑞 ( 1 + 1 + 4 𝑟 ) / 2 , then (1.1) possesses an invariant interval [ 𝑝 / 𝑞 , 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ] and 𝑦 ∈ ( 𝑝 / 𝑞 , 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ) ;(vi)If √ 𝑞 ( 1 + 1 + 4 𝑟 ) / 2 < 𝑝 < 𝑞 + 𝑞 𝑟 , then (1.1) possesses an invariant interval [ 𝑞 𝑟 / ( 𝑝 − 𝑞 ) , 𝑝 / 𝑞 ] and 𝑦 ∈ ( 𝑞 𝑟 / ( 𝑝 − 𝑞 ) , 𝑝 / 𝑞 ) ;(vii)If 𝑝 ≥ 𝑞 + 𝑞 𝑟 , 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 ≥ 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ; and it is strictly decreasing in each of its arguments for 𝑦 < 𝑞 𝑟 / ( 𝑝 − 𝑞 ) .
(v) Using the decreasing character of 𝑓 , we obtain
𝑝 𝑞 = 𝑓 𝑞 𝑟 , 𝑝 − 𝑞 𝑞 𝑟 𝑝 𝑝 − 𝑞 ≤ 𝑓 ( 𝑥 , 𝑦 ) ≤ 𝑓 𝑞 , 𝑝 𝑞 = 𝑞 𝑟 + 𝑝 ( 𝑝 + 1 ) ≤ 𝑝 ( 𝑞 + 1 ) 𝑞 𝑟 𝑝 − 𝑞 . ( 3 . 2 1 ) The inequalities 𝑞 𝑟 + 𝑝 ( 𝑝 + 1 ) ≤ 𝑝 ( 𝑞 + 1 ) 𝑞 𝑟 , 𝑝 𝑝 − 𝑞 𝑞 < 𝑞 𝑟 𝑝 − 𝑞 ( 3 . 2 2 ) are equivalent to the inequality √ 𝑝 < 𝑞 ( 1 + 1 + 4 𝑟 ) / 2 . That is, [ 𝑝 / 𝑞 , 𝑞 𝑟 / ( 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 , 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ) .
(vi) By using the monotonic character of 𝑓 , we obtain
𝑞 𝑟 ≤ 𝑝 − 𝑞 ( 𝑞 𝑟 + 𝑝 ) ( 𝑝 − 𝑞 ) + 𝑝 𝑞 2 𝑟 𝑞 3 𝑟 + 𝑝 ( 𝑝 − 𝑞 ) = 𝑓 𝑞 𝑟 , 𝑝 𝑝 − 𝑞 𝑞 𝑝 ≤ 𝑓 ( 𝑥 , 𝑦 ) ≤ 𝑓 𝑞 , 𝑞 𝑟 = 𝑝 𝑝 − 𝑞 𝑞 . ( 3 . 2 3 ) The inequalities ( 𝑞 𝑟 + 𝑝 ) ( 𝑝 − 𝑞 ) + 𝑝 𝑞 2 𝑟 𝑞 3 ≥ 𝑟 + 𝑝 ( 𝑝 − 𝑞 ) 𝑞 𝑟 , 𝑝 − 𝑞 𝑞 𝑟 ≤ 𝑝 𝑝 − 𝑞 𝑞 ( 3 . 2 4 ) are equivalent to the inequality √ 𝑝 > 𝑞 ( 1 + 1 + 4 𝑟 ) / 2 .
On the other hand, similar to Lemma 3.3 (v) it can be proved that 𝑦 ∈ ( 𝑞 𝑟 / ( 𝑝 − 𝑞 ) , 𝑝 / 𝑞 ) .
(vii) In this case note that 𝑞 𝑟 / ( 𝑝 − 𝑞 ) ≤ 1 < 𝑝 / 𝑞 holds. By the monotonic character of 𝑓 , we have
1 < 𝑞 𝑟 + 𝑝 𝑞 + 𝑝 𝑞 2 𝑝 + 𝑝 = 𝑓 1 , 𝑞 𝑝 ≤ 𝑓 ( 𝑥 , 𝑦 ) ≤ 𝑓 𝑞 = , 1 𝑞 𝑟 + 𝑝 2 + 𝑞 ≤ 𝑝 𝑞 ( 𝑝 + 1 ) 𝑞 . ( 3 . 2 5 ) The inequalities 𝑞 𝑟 + 𝑝 𝑞 + 𝑝 𝑞 2 + 𝑝 > 1 , 𝑞 𝑟 + 𝑝 2 + 𝑞 ≤ 𝑝 𝑞 ( 𝑝 + 1 ) 𝑞 ( 3 . 2 6 ) are equivalent to the inequality 𝑝 ≥ 𝑞 + 𝑞 𝑟 .
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 + 4 𝑟 ) / 2 , and { 𝑦 𝑛 } ∞ 𝑛 = − 𝑘 is a positive solution of (1.1). Then the following statements are true: (i)If for some 𝑁 ≥ 0 , 𝑦 𝑁 < 𝑝 / 𝑞 , then 𝑦 𝑁 + 𝑘 + 1 > 𝑝 / 𝑞 ;(ii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 = 𝑝 / 𝑞 , then 𝑦 𝑁 + 𝑘 + 1 = 𝑝 / 𝑞 ;(iii)If for some 𝑁 ≥ 0 , 𝑦 𝑁 > 𝑝 / 𝑞 , then 𝑦 𝑁 + 𝑘 + 1 < 𝑝 / 𝑞 ;(iv)If for some 𝑁 ≥ 0 , 𝑦 𝑁 > 𝑝 / 𝑞 , then 𝑦 𝑁 > 𝑦 𝑁 + 2 ( 𝑘 + 1 ) > 𝑝 / 𝑞 ;(v)If for some 𝑁 ≥ 0 , 𝑦 𝑁 < 𝑝 / 𝑞 , then 𝑦 𝑁 < 𝑦 𝑁 + 2 ( 𝑘 + 1 ) < 𝑝 / 𝑞 .

Proof. In this case, we have that 𝑞 𝑟 / ( 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 ≤ 𝑝 𝑞 + 1 + 3 𝑝 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 { 𝑦 𝑛 } ∞ 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 𝑦 𝑛 𝑟 = 1 + 𝑢 𝑝 + 1 𝑛 , ( 4 . 1 ) Equation (1.1) reduces to the difference equation 𝑢 𝑛 + 1 = 1 1 + 𝑝 𝑟 / ( 𝑝 + 1 ) 2 𝑢 𝑛 + 𝑟 / ( 𝑝 + 1 ) 2 𝑢 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 . ( 4 . 2 ) The unique positive equilibrium 𝑢 of (4.2) is √ 𝑢 = − ( 𝑝 + 1 ) + ( 𝑝 + 1 ) 2 + 4 𝑟 ( 𝑝 + 1 ) . 2 𝑟 ( 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 { 𝑦 𝑛 } ∞ 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 , 𝑦 𝑁 > 1 , then 𝑦 𝑁 + 1 < ( 𝑝 + 𝑟 ) / 𝑞 ;(ii)Suppose that 𝑞 > 𝑝 + 𝑟 . If for some 𝑁 ≥ 0 , 𝑦 𝑁 < 1 , then 𝑦 𝑁 + 1 > ( 𝑝 + 𝑟 ) / 𝑞 .

Theorem 4.3. Assume that 𝑝 < 𝑞 , 𝑝 < 1 and 𝑞 ≤ 𝑝 𝑞 + 1 + 3 𝑝 hold. Let { 𝑦 𝑛 } ∞ 𝑛 = − 𝑘 be a positive solution of (1.1). Then the following statements hold true: (i)Suppose 𝑞 < 𝑝 + 𝑟 . If 𝑦 0 ∈ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] , then 𝑦 𝑛 ∈ [ 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 𝑦 0 ∈ [ 𝑟 / ( 𝑞 − 𝑝 ) , 1 ] , then 𝑦 𝑛 ∈ [ 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 𝑦 0 ∈ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] , then by Lemma 3.3 (iv), we have that 𝑦 1 > 1 , and by Lemma 4.2 (i), we obtain that 𝑦 1 < ( 𝑝 + 𝑟 ) / 𝑞 < 𝑟 / ( 𝑞 − 𝑝 ) , which implies that 𝑦 1 ∈ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] , by induction, we have 𝑦 𝑛 ∈ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] , for 𝑛 ≥ 1 .
Now, to complete the proof it remains to show that when 𝑦 0 ∉ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] , there exists 𝑁 > 0 such that 𝑦 𝑁 ∈ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] .
If 𝑦 0 ∉ [ 1 , 𝑟 / ( 𝑞 − 𝑝 ) ] , then we have the following two cases to be considered:
(a) 𝑦 0 > 𝑟 / ( 𝑞 − 𝑝 ) ;(b) 𝑦 0 < 1 .
Case (a). From Lemma 3.3 (ii), we see that 𝑦 1 < 1 . Thus, in the sequel, we only consider case (b).
If 𝑦 0 < 1 , then by Lemma 3.3 (iv), we have 𝑦 1 > 1 , and from Lemma 4.2 (i), we have 𝑦 2 < ( 𝑝 + 𝑟 ) / 𝑞 < 𝑟 / ( 𝑞 − 𝑝 ) . So 𝑦 3 > 1 and