On the Rational Recursive Sequence 𝑥_𝑛+1=(𝛼−𝛽𝑥_𝑛)/(𝛾−𝛿𝑥_𝑛−𝑥_𝑛−𝑘)

Abstract

We study the global stability, the periodic character, and the boundedness character of the positive solutions of the difference equation 𝑥_𝑛+1=(𝛼−𝛽𝑥_𝑛)/(𝛾−𝛿𝑥_𝑛−𝑥_𝑛−𝑘),𝑛=0,1,2,…,𝑘∈{1,2,…}, in the two cases: (i) 𝛿≥0,𝛼>0,𝛾>𝛽>0; (ii) 𝛿≥0,𝛼=0,𝛾,𝛽>0, where the coefficients 𝛼,𝛽,𝛾,and𝛿, and the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀ are real numbers. We show that the positive equilibrium of this equation is a global attractor with a basin that depends on certain conditions posed on the coefficients of this equation.

1. Introduction

The asymptotic stability of the rational recursive sequence, 𝑥_𝑛+1=𝛼+𝛽𝑥_𝑛𝛾+∑^𝑘_𝑖=0𝛾_𝑖𝑥_𝑛−𝑖,𝑛=0,1,2,…, (1.1) was investigated when the coefficients 𝛼,𝛽,𝛾,and𝛾_𝑖 are nonnegative real numbers (see [1–3]). Studying the asymptotic behavior of the rational sequence (1.1) when some of the coefficients are negative was suggested in [3]. Recently, Aboutaleb et al. [4] studied the rational recursive sequence, 𝑥_𝑛+1=𝛼−𝛽𝑥_𝑛𝛾+𝑥_𝑛−1,𝑛=0,1,2,…, (1.2)where 𝛼,𝛽,and𝛾 are nonnegative real numbers and obtained sufficient conditions for the global attractivity of the positive equilibria. Yan et al. [5] studied recently the rational recursive sequence, 𝑥_𝑛+1=𝛼−𝛽𝑥_𝑛𝛾−𝑥_𝑛−𝑘,𝑛=0,1,2,…, (1.3) where 𝛼≥0,𝛾,𝛽>0 are real numbers while 𝑘≥1 is an integer number, and the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀ are arbitrary real numbers. They proved that the positive equilibrium 𝑥 of (1.3) is a global attractor with a basin that depends on certain conditions of the coefficients. He et al. [6] studied recently the rational recursive sequence, 𝑥_𝑛+1=𝑎−𝑏𝑥_𝑛−𝑘𝐴+𝑥_𝑛,𝑛=0,1,2,…, (1.4) where 𝑎≥0,𝐴,𝑏>0 are real numbers while 𝑘≥1 is an integer number and the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀ are arbitrary real numbers. They proved the global attractivity and periodic character of the positive solution of (1.4). Stević [7] studied recently the rational recursive sequence, 𝑥_𝑛+1=𝛼+𝛽𝑥_𝑛𝛾−𝑥_𝑛−𝑘,𝑛=0,1,2,…, (1.5) where the parameters 𝛼,𝛽,and𝛾 are nonnegative real numbers and 𝑘≥1 is an integer number while the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀ are arbitrary real numbers. Other related results can be found in [8–19].

Our aim in this paper is to study the global attractivity, the periodicity, and the boundedness of the positive solution of the following rational recursive sequence: 𝑥_𝑛+1=𝛼−𝛽𝑥_𝑛𝛾−𝛿𝑥_𝑛−𝑥_𝑛−𝑘,𝑛=0,1,2,…, (1.6) in the two cases (i) 𝛿≥0,𝛼>0,𝛾>𝛽>0, (ii) 𝛿≥0,𝛼=0,𝛾,𝛽>0, where the coefficients 𝛼,𝛽,𝛾,and𝛿 are real numbers and 𝑘≥1 is an integer number, while the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀ are arbitrary real numbers. We will prove that the positive equilibrium 𝑥 of (1.6) is a global attractor with a basin that depends on certain conditions of these coefficients.

2. Local Stability and Permanence

We first recall some results which will be useful in the sequel. Let 𝐈 be some real interval and let 𝐹 be a continuous function defined on 𝐈^𝑘+1. Then, for initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀∈𝐈, it is easy to see that the difference equation,𝑥_𝑛+1=𝐹(𝑥_𝑛,𝑥_𝑛−1,…,𝑥_𝑛−𝑘),𝑛=0,1,2,…,𝑘≥1, (2.1) has a unique solution {𝑥_𝑛}.

Definition 2.1. A point 𝑥 is called an equilibrium of (2.1), if 𝑥=𝐹(𝑥,…,𝑥). That is, 𝑥_𝑛=𝑥 for 𝑛≥0 is a solution of (2.1), or equivalently, is a fixed point of 𝐹.

Definition 2.2. An interval 𝐉⊂𝐈 is called an invariant interval of (2.1) if the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀∈𝐉 imply that the solution 𝑥_𝑛∈𝐉 for 𝑛>0. That is, every solution of (2.1) with initial conditions in 𝐉 remains in 𝐉.

Definition 2.3. The difference equation (2.1) is said to be permanent if there exist numbers 𝑃 and 𝑄 with 0<𝑃≤𝑄<∞ such that for any initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀ there exists a positive integer 𝑁 which depends on the initial conditions such that 𝑃≤𝑥_𝑛≤𝑄, for all 𝑛≥𝑁.

The linearized equation associated with (2.1) about the equilibrium 𝑥 is 𝑦_𝑛+1=^𝑘_𝑖=0𝜕𝐹(𝑥,…,𝑥)𝜕𝑢_𝑖𝑦_𝑛−𝑖,𝑛=0,1,2,…. (2.2) Its characteristic equation is 𝜆^𝑛+1=^𝑘_𝑖=0𝜕𝐹(𝑥,…,𝑥)𝜕𝑢_𝑖𝜆^𝑛−𝑖,𝑛=0,1,2,…. (2.3)

Theorem 2.4 (see [3]). Assume that 𝐹 is a 𝐶¹-function and let 𝑥 be an equilibrium of (2.1). Then, the following statements are true:
(a) if all the roots of (2.3) lie in the open unit disk |𝜆|<1, then the equilibrium 𝑥 of (2.1) is locally asymptotically stable; (b) if at least one root of (2.3) has absolute value greater than one, then the equilibrium 𝑥 of (2.1) is unstable.

Theorem 2.5 (see [3, 8]). Assume that 𝑝,𝑞∈ℝ, and 𝑘∈{1,2,…}. Then, |𝑝|+|𝑞|<1 (2.4) is a sufficient condition for the asymptotic stability of the difference equation 𝑥_𝑛+1−𝑝𝑥_𝑛+𝑞𝑥_𝑛−𝑘=0,𝑛=0,1,2,…. (2.5) Suppose in addition that one of the following two cases holds: (i) k is odd and 𝑞<0, or (ii) 𝑘 is even and 𝑝𝑞<0. Then, (2.4) is also a necessary condition for the asymptotic stability of (2.5) (see [6]).

First, we study the rational recursive sequence 𝑥_𝑛+1=𝛼−𝛽𝑥_𝑛𝛾−𝛿𝑥_𝑛−𝑥_𝑛−𝑘,𝑛=0,1,2,…, (2.6) together with the conditions 𝛿≥0,𝛼>0,𝛾>𝛽>0,𝑘∈{1,2,…}. (2.7)

The unique positive equilibrium point 𝑥 of (2.6) is the solution of the equation 𝑥=𝛼−𝛽𝑥𝛾−(𝛿+1)𝑥, (2.8) which is given by √𝑥=(𝛾+𝛽)±𝑇,2(𝛿+1) (2.9) where 𝑇=(𝛾+𝛽)²−4𝛼(𝛿+1). (2.10) If (2.7) holds and 𝛼=(𝛾+𝛽)²/4(𝛿+1), then (2.6) has a unique positive equilibrium 𝑥₀=(𝛾+𝛽)/2(𝛿+1). If (2.7) holds and 𝛼<(𝛾+𝛽)²/4(𝛿+1) then (2.6) has two positive equilibria 𝑥_1,2 given by (2.9).

The linearized equation of (2.6) about the equilibrium 𝑥_𝑖(𝑖=0,1,2) is given by 𝑦_𝑛+1+𝛽−𝛿𝑥_𝑖[𝛾−(𝛿+1)𝑥_𝑖]𝑦_𝑛−𝑥_𝑖[𝛾−(𝛿+1)𝑥_𝑖]𝑦_𝑛−𝑘=0. (2.11) The characteristic equation associated with (2.6) about 𝑥₀ is 𝜆^𝑘+1+[2𝛽−𝛾−𝛽𝛿(𝛾+𝛽)(𝛿+1)(𝛾−𝛽)]𝜆^𝑘−𝛾+𝛽(𝛿+1)(𝛾−𝛽)=0. (2.12) Now, we have the following results:

(a)if 0≤𝛿<2𝛽/(𝛾−𝛽), then (𝛾+𝛽)/(𝛿+1)(𝛾−𝛽)>1 and hence the equilibrium 𝑥₀ of (2.6) is unstable (see Figure 1);(b)if 𝛿>2𝛽/(𝛾−𝛽), then |2𝛽−𝛾−𝛽𝛿(𝛾+𝛽)(𝛿+1)(𝛾−𝛽)|+|𝛾+𝛽(𝛿+1)(𝛾−𝛽)|=1. (2.13) Thus, the linearized stability analysis fails. On the other hand, the characteristic equation associated with (2.6) about 𝑥₁ is 𝜆^𝑘+1+[2𝛽𝛾−𝛽−√𝑇−𝛿(𝛾+𝛽+√𝑇)(𝛿+1)(𝛾−𝛽−√𝑇)]𝜆^𝑘−𝛾+𝛽+√𝑇(𝛿+1)(𝛾−𝛽−√𝑇)=0. (2.14) Now, we have the following results:

Figure 1 

(a)if 0≤𝛿<2𝛽/(𝛾−𝛽), then it is obvious that |𝛾+𝛽+√𝑇(𝛿+1)(𝛾−𝛽−√𝑇)|≥(𝛾−𝛽)√𝛾+𝛽[𝛾+𝛽+𝑇][𝛾−𝛽−√𝑇]>1, (2.15) hence the equilibrium 𝑥₁ of (2.6) is unstable;(b)if 𝛿≥2𝛽/(𝛾−𝛽), then it is easy to see that 2𝛽(𝛿+1)<𝛿[𝛾+𝛽+√𝑇], and consequently, we have |2𝛽𝛾−𝛽−√𝑇−𝛿[𝛾+𝛽+√𝑇](𝛿+1)[𝛾−𝛽−√√𝑇]|+|𝛾+𝛽+𝑇(𝛿+1)[𝛾−𝛽−√√𝑇]|=𝛾−𝛽+𝑇𝛾−𝛽−√𝑇>1, (2.16) and hence the equilibrium 𝑥₁ of (2.6) is unstable.

For the positive equilibrium 𝑥₂, in view of conditions (2.7) and 𝛼<(𝛾+𝛽)²/4(𝛿+1), we have 𝑥₂=𝛾+𝛽−√𝑇<2(𝛿+1)𝛾+𝛽<𝛾2(𝛿+1).𝛿+1 (2.17) Hence, if 0<𝛼≤𝛽(𝛾−𝛽),𝛿+1 (2.18) then √𝑇≥(𝛾+𝛽)²−4𝛽(𝛾−𝛽)>(𝛾+𝛽)²−(𝛾+3𝛽)(𝛾−𝛽)=(𝛾+𝛽)²−(𝛾+𝛽)²+4𝛽²=2𝛽. (2.19) Consequently, we have |𝛽−𝛿𝑥₂𝛾−(𝛿+1)𝑥₂|+|𝑥₂𝛾−(𝛿+1)𝑥₂|<𝛽+(𝛿+1)𝑥₂𝛾−(𝛿+1)𝑥₂=3𝛽+𝛾−√𝑇𝛾−𝛽+√𝑇<3𝛽+𝛾−2𝛽𝛾−𝛽+2𝛽=1, (2.20) which by Theorem 2.5 implies that 𝑥₂ is locally asymptotically stable (see Figure 2).

Figure 2 

Lemma 2.6. Let 𝑓(𝑢,𝑣)=(𝛼−𝛽𝑢)/(𝛾−𝛿𝑢−𝑣) and assume that conditions (2.7) and (2.18) hold. Then, the following statements are true:
(a) 0<𝑥₂<𝛼/𝛽,𝛼/𝛽<𝑥₁<∞; (b) 𝑓(𝑥,𝑥) is a strictly decreasing function in (−∞,𝛼/𝛽); (c) let 𝑢,𝑣∈(−∞,𝛼/𝛽), then the function 𝑓(𝑢,𝑣) is a strictly decreasing function in 𝑢 and 𝑎 strictly increasing function in 𝑣.

Proof. We prove (a) only. The proofs of (b) and (c) are omitted here. In view of (2.7) and (2.18), we have 𝑥₂=𝛾+𝛽−√𝑇<2(𝛿+1)𝛾+𝛽<𝛾2(𝛿+1).𝛿+1 (2.21) From (2.8) and (2.21), we have 𝛼−𝛽𝑥₂>0 and so 𝑥₂<𝛼/𝛽. Also, in view of (2.7) and (2.18), we have 0<𝛼−𝛽𝑥₁𝛾−(𝛿+1)𝑥₁=𝑥₁=𝛾+𝛽+√𝑇≥2(𝛿+1)𝛾+𝛽+(𝛾+𝛽)²−4𝛽(𝛾−𝛽)=2(𝛿+1)𝛾+𝛽+(𝛾−𝛽)²+4𝛽²>2(𝛿+1)𝛾+𝛽+(𝛾−𝛽)²=𝛾2(𝛿+1),𝛿+1 (2.22) and so 𝛾−𝑥₁(𝛿+1)<0. Consequently, 𝛼−𝛽𝑥₁<0 which implies that 𝑥₁>𝛼/𝛽. The proof is completed.

Theorem 2.7. Assume that the conditions (2.7) and (2.18) hold. Let {𝑥_𝑛} be any solution of (2.6). If 𝑥_𝑖∈(−∞,𝛼/𝛽],for 𝑖=−𝑘,−𝑘+1,…,−1 and if 𝑥₀∈[0,𝛼/𝛽], then 0≤𝑥_𝑛≤𝛼𝛽,𝑛=1,2,…. (2.23) That is the solution {𝑥_𝑛} is bounded.

Proof. By part (c) of Lemma 2.6, we have 0=𝛼−𝛽⋅(𝛼/𝛽)𝛾−𝛿𝑥₀−𝑥_−𝑘≤𝑥₁=𝛼−𝛽𝑥₀𝛾−𝛿𝑥₀−𝑥_−𝑘≤𝛼−𝛽⋅0=𝛼𝛾−𝛿⋅0−𝛼/𝛽=𝛾−𝛼/𝛽𝛽𝛼.𝛾𝛽−𝛼 (2.24) From (2.18), we deduce that 𝛾𝛽−𝛼>𝛽², and then we have 0≤𝑥₁≤𝛼𝛽. (2.25) Also, we have 0=𝛼−𝛽⋅(𝛼/𝛽)𝛾−𝛿𝑥₁−𝑥_−𝑘+1≤𝑥₂=𝛼−𝛽𝑥₁𝛾−𝛿𝑥₁−𝑥_−𝑘+1≤𝛼−𝛽⋅0<𝛼𝛾−𝛿⋅0−𝛼/𝛽𝛽. (2.26) Thus, 0≤𝑥₂≤𝛼𝛽. (2.27) The result (2.23) now follows by induction. The proof is completed.

3. Global Attractivity

In this section, we will study the global attractivity of positive solutions of (2.6). We show that the positive equilibrium 𝑥 of (2.6) is a global attractor with a basin that depends on certain conditions imposed on the coefficients.

Theorem 3.1. Assume that the conditions (2.7) and (2.18) hold. Then, the equilibrium point 𝑥₂ of (2.6) is globally asymptotically stable.

Proof. In Section 2, we have shown under the assumptions (2.7) and (2.18) that the equilibrium 𝑥₂ is locally asymptotically stable. It remains to prove that the equilibrium 𝑥₂ is a global attractor. To this end, set 𝐼=lim_𝑛→∞inf𝑥_𝑛 and 𝑆=lim_𝑛→∞sup𝑥_𝑛 which by Theorem 2.7 exist and are positive numbers. Then, from (2.6) we deduce that 𝑆≤𝛼−𝛽S𝛾−(𝛿+1)𝐼,𝐼≥𝛼−𝛽𝐼.𝛾−(𝛿+1)𝑆 (3.1) Consequently, we have −𝛼+(𝛾+𝛽)𝑆≤(𝛿+1)𝐼𝑆≤−𝛼+(𝛾+𝛽)𝐼, (3.2) from which it follows that 𝐼=𝑆. Thus, the the proof of Theorem 3.1 is completed.

Lemma 3.2 (see [8]). Consider the difference equation 𝑥_𝑛+1=𝑓(𝑥_𝑛,𝑥_𝑛−𝑘),𝑘≥1,𝑛=0,1,2,…. (3.3) Let [𝑎,𝑏] be some interval of real numbers, and assume that 𝑓:[𝑎,𝑏]×[𝑎,𝑏]→[𝑎,𝑏] is a continuous function satisfying the following properties:
(a) 𝑓(𝑢,𝑣) is a nonincreasing function in u, and a nondecreasing function in 𝑣; (b) if (𝑚,𝑀)∈[𝑎,𝑏]×[𝑎,𝑏] is a solution of the system 𝑚=𝑓(𝑀,𝑚),𝑀=𝑓(𝑚,𝑀), (3.4) then, 𝑚=𝑀. Then, (3.3) has a unique equilibrium point 𝑥 and every solution of (3.3) converges to 𝑥.

Theorem 3.3. Assume that the conditions (2.7) and (2.18) hold. Then, the positive equilibrium 𝑥 of (2.6) is a global attractor with a basin 𝑆^∗=[0,𝛼/𝛽]^𝑘+1.

Proof. For 𝑢,𝑣∈[0,𝛼/𝛽], set 𝑓(𝑢,𝑣)=𝛼−𝛽𝑢.𝛾−𝛿𝑢−𝑣 (3.5) We claim that 𝑓:[0,𝛼/𝛽]×[0,𝛼/𝛽]→[0,𝛼/𝛽]. In fact, if we set 𝑎=0,𝑏=𝛼/𝛽, then 𝑓(𝑏,𝑎)=𝛼−𝛽𝑏=𝛾−𝛿𝑏−𝑎𝛼−𝛼𝛾−𝛿(𝛼/𝛽)=0=𝑎, (3.6) and in view of the condition (2.18), we have 𝑓(𝑎,𝑏)=𝛼−𝛽𝑎=𝛼𝛾−𝛿𝑎−𝑏=𝛾−𝛼/𝛽𝛽𝛼<𝛼𝛾𝛽−𝛼𝛽=𝑏. (3.7) Since 𝑓(𝑢,𝑣) is decreasing in 𝑢 and increasing in 𝑣, it follows that 𝑎≤𝑓(𝑢,𝑣)≤𝑏, for all 𝑢,𝑣∈[𝑎,𝑏], which implies that our assertion is true. On the other hand, conditions (a) and (b) of Lemma 3.2 are clearly true. Let {𝑥_𝑛} be a solution of (2.6) with the initial conditions (𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀)∈S. By Lemma 3.2, we have lim_𝑛→∞𝑥_𝑛=𝑥. The proof is completed.

Theorem 3.4. Assume that conditions (2.7) and (2.18) hold. Then, the positive equilibrium 𝑥 of (2.6) is a global attractor with a basin 𝑆^∗=(−∞,𝛼/𝛽]^𝑘×[0,𝛼/𝛽].

Proof. Let {𝑥_𝑛} be a solution of (2.6) with the initial conditions 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑥₀∈𝑆^∗. Then, by Theorem 2.7, we have 𝑥_𝑛𝛼∈[0,𝛽],𝑛=1,2,…. (3.8) By Theorem 3.3, we have lim_𝑛→∞𝑥_𝑛+𝑘=𝑥 and so lim_𝑛→∞𝑥_𝑛=𝑥. The proof is completed.

Theorem 3.5. Assume that conditions (2.7) hold with 0≤𝛿<1. Also, assume that k is an odd positive integer. Then, the necessary and sufficient condition for (2.6) to have positive solutions of prime period two is that 𝛽(𝛾−𝛽)<𝛼<(𝛾−𝛽)4[𝛾+3𝛽−𝛿(𝛾−𝛽)]. (3.9)

Proof. First, suppose that there exist distinctive positive solutions of prime period two, …,𝑃,𝑄,𝑃,𝑄,…, (3.10) of the difference equation (2.6).
If 𝑘 is odd, then 𝑥_𝑛+1=𝑥_𝑛−𝑘. It follows from the difference equation (2.6) that 𝑃=𝛼−𝛽𝑄𝛾−𝛿𝑄−𝑃,𝑄=𝛼−𝛽𝑃.𝛾−𝛿𝑃−𝑄 (3.11) Consequently, we obtain 𝑃+𝑄=𝛾−𝛽,𝑃𝑄=𝛼−𝛽(𝛾−𝛽).1−𝛿 (3.12) Thus, we deduce that 𝛼>𝛽(𝛾−𝛽),0≤𝛿<1. (3.13) Now it is clear that 𝑃,𝑄 are two positive distinct real roots of the quadratic equation 𝑡²−(𝑃+𝑄)𝑡+𝑃𝑄=0. (3.14) Therefore, we have (𝛾−𝛽)²>4[𝛼−𝛽(𝛾−𝛽)].1−𝛿 (3.15) From (3.13) and (3.15) we obtain condition (3.9). Conversely, suppose that the condition (3.9) is valid. Then, we deduce that (3.13) and (3.15) hold. Consequently, there exists two positive distinct real numbers 𝑃 and 𝑄 such that 𝑃=𝛾−𝛽2−√𝐾2,𝑄=𝛾−𝛽2+√𝐾2, (3.16)(3.17) where 𝐾>0 is given by 𝐾=(𝛾−𝛽)²−4[𝛼−𝛽(𝛾−𝛽)1−𝛿],0≤𝛿<1. (3.18) Thus, 𝑃 and 𝑄 given by (3.16) and (3.17) represent two positive distinct real roots of the quadratic equation (3.14). Now, we are going to prove that 𝑃 and 𝑄 given by (3.16) and (3.17) are positive solutions of prime period two of the difference equation (2.6). To this end, we assume that 𝑥_−𝑘=𝑃,𝑥_−𝑘+1=𝑄,…,𝑥₋₁=𝑃,𝑥₀=𝑄. We wish to prove that 𝑥₁=𝑃 and 𝑥₂=𝑄.
It follows from the difference equation (2.6) and the formulas (3.16) and (3.17) that 𝑥₁=𝛼−𝛽𝑥₀𝛾−𝛿𝑥₀−𝑥_−𝑘=𝛼−𝛽𝑄=√𝛾−𝛿𝑄−𝑃2𝛼−𝛽[𝛾−𝛽+𝐾]2𝛾−𝛿[𝛾−𝛽+√𝐾]−[𝛾−𝛽−√=√𝐾]𝛽[2𝛼/𝛽−(𝛾−𝛽)−𝐾]2𝛾−(1+𝛿)(𝛾−𝛽)+(1−𝛿)√𝐾𝛽=()√1−𝛿{2𝛼/𝛽−(𝛾−𝛽)−𝐾}{2𝛾/(1−𝛿)−((1+𝛿)/(1−𝛿))(𝛾−𝛽)−√𝐾}{2𝛾/(1−𝛿)−((1+𝛿)/(1−𝛿))(𝛾−𝛽)+√𝐾}{2𝛾/(1−𝛿)−((1+𝛿)/(1−𝛿))(𝛾−𝛽)−√.𝐾} (3.19) After some reduction, we deduce that 𝑥₁𝛽=()1−𝛿[𝛼(1−𝛿)+𝛽²][𝛾−𝛽−√𝐾]/𝛽(1−𝛿)2[𝛼(1−𝛿)+𝛽²]/(1−𝛿)²=𝛾−𝛽−√𝐾2=𝑃. (3.20) Similarly, we can show that, 𝑥₂=𝛼−𝛽𝑥₁𝛾−𝛿𝑥₁−𝑥_−𝑘+1=𝛼−𝛽𝑃𝛾−𝛿𝑃−𝑄=𝑄. (3.21) By using the induction, we have 𝑥_𝑛=𝑃,𝑥_𝑛+1=𝑄,∀𝑛≥−𝑘. (3.22) Thus, the difference equation (2.6) has positive solutions of prime period two. Hence, the proof of Theorem 3.5 is completed.

Theorem 3.6. Assume that conditions (2.7) hold. If k is even, then (2.6) has no positive solutions of prime period two.

Proof. Suppose that there exists distinctive positive solutions of prime period two, …,𝑃,𝑄,𝑃,𝑄,…, (3.23) of the difference equation (2.6).
If 𝑘 is even, then 𝑥_𝑛=𝑥_𝑛−𝑘. It follows from the difference equation (2.6) that 𝑃=𝛼−𝛽𝑄𝛾−(𝛿+1)𝑄,𝑄=𝛼−𝛽𝑃.𝛾−(𝛿+1)𝑃 (3.24) From which we have (𝛾−𝛽)(𝑃−𝑄)=0 and by using (2.7), we deduce that 𝑃=𝑄. This is a contradiction. Thus, the proof of Theorem 3.6 is completed.

4. The Case 𝛼=0

Secondly, we study the rational recursive sequence 𝑥_𝑛+1=−𝛽𝑥_𝑛𝛾−𝛿𝑥_𝑛−𝑥_𝑛−𝑘,𝑛=0,1,2,…, (4.1) where 𝛿≥0,𝛾,𝛽>0 are real numbers and 𝑘∈{1,2,…}. By putting 𝑥_𝑛=𝛽𝑦_𝑛, (4.1) yields 𝑦_𝑛+1=−𝑦_𝑛𝐴−𝛿𝑦_𝑛−𝑦_𝑛−𝑘,𝑛=0,1,2,…, (4.2) where 𝐴=𝛾/𝛽. Equation (4.2) has two equilibrium points 𝑦₁=0,𝑦₂=1+𝐴.1+𝛿 (4.3) The linearized equation associated with (4.2) about the equilibria 𝑦_𝑖,(𝑖=1,2) is 𝑧_𝑛+1+1−𝛿𝑦_𝑖𝐴−(𝛿+1)𝑦_𝑖𝑧_𝑛−𝑦_𝑖𝐴−(𝛿+1)𝑦_𝑖𝑧_𝑛−𝑘=0. (4.4) The characteristic equation of (4.4) about the equilibrium 𝑦₂=(1+𝐴)/(1+𝛿) is 𝜆^𝑘+1+[𝛿𝐴−1𝛿+1]𝜆^𝑘+𝐴+1𝛿+1=0. (4.5) Now, we deduce from (4.5) the following results:

(a)if 𝛿=0, and since 𝐴+1>1, then the equilibrium 𝑦₂ is unstable (see [7]);(b)if 𝐴>𝛿>0, and since (𝐴+1)/(𝛿+1)>1, then the equilibrium 𝑦₂ is unstable;(c)if 𝐴=𝛿, then |𝛿𝐴−1𝛿+1|+|𝛿+1𝛿+1|=|𝛿−1|+1. (4.6) Now, we have the following results from case (c): (i) if 𝐴=𝛿>1, then the equilibrium 𝑦₂ is unstable; (ii) if 0<𝐴=𝛿<1, then the equilibrium 𝑦₂ is unstable; (iii) if 𝐴=𝛿=1, then the linearized stability analysis fails;

(d)if 1<𝐴<𝛿, |𝛿𝐴−1𝛿+1|+|𝐴+1𝛿+1|=𝛿𝐴−1+𝛿+1𝐴+1=𝛿+1𝐴(𝛿+1)𝛿+1=𝐴>1, (4.7) and hence the equilibrium 𝑦₂ is unstable;(e)if 𝐴<𝛿≤1,|𝛿𝐴−1𝛿+1|+|𝐴+1𝛿+1|=1−𝛿𝐴+1+𝐴𝐴𝛿+1≥1+2(1−𝛿)≥1, (4.8) and hence the equilibrium 𝑦₂ is unstable. The characteristic equation of (4.4) about the equilibrium 𝑦₁=0 is 𝜆^𝑘+1+1𝐴𝜆^𝑘=0. (4.9) This equation has two roots 𝜆₁=0,𝜆₂1=−𝐴. (4.10) Now, we deduce from (4.10) the following results:

(i)if 𝐴>1, then the equilibrium 𝑦₁=0 is locally asymptotically stable (see Figure 3);(ii)if 0<𝐴<1, then the equilibrium 𝑦₁=0 is unstable (see Figure 4);(iii)if 𝐴=1, then the linearized stability analysis fails.

Figure 3 

Figure 4 

In the following results, we assume that 𝐴≥𝛿+2, where 𝛿≥0.

Lemma 4.1. Assume that the initial conditions 𝑦_−𝑖∈[−1,1], for 𝑖=1,2,…,𝑘 and 𝑦₀∈[−1,0]. Then, {𝑦_2𝑛−1} is nonnegative and monotonically decreasing to zero, while {𝑦_2𝑛} is nonpositive and monotonically increasing to zero.

Proof. Suppose that 𝑦_−𝑖∈[−1,1], for 𝑖=1,2,…,𝑘 and 𝑦₀∈[−1,0]. Clearly, 0≤𝑦₁≤1 and −1≤𝑦₂≤0. By induction, we can see that 0≤𝑦_2𝑛−1≤1 and −1≤𝑦_2𝑛≤0 for 𝑛≥1.
If 𝐴≥𝛿+2,𝛿≥0 we have 𝑦_2𝑛−1𝑦_2𝑛+1=(𝐴−𝛿𝑦_2𝑛−𝑦_2𝑛−𝑘)(𝐴−𝛿𝑦_2𝑛−1−𝑦_{2𝑛−𝑘−1})>1, (4.11) and hence 𝑦_2𝑛−1>𝑦_2𝑛+1,𝑛=1,2,…. (4.12) Similarly, we can show that 𝑦_2𝑛<𝑦_2𝑛+2,𝑛=1,2,…. The proof of Lemma 4.1 is completed.

On using arguments similar to that used in Lemma 4.1, we can easily prove the following lemma.

Lemma 4.2. Assume that the initial conditions 𝑦_−𝑖∈[−1,1], for 𝑖=1,2,…,𝑘 and 𝑦₀∈[0,1]. Then, {𝑦_2𝑛−1} is nonpositive and monotonically increasing to zero, while {𝑦_2𝑛} is nonnegative and monotonically decreasing to zero.

Corollary 4.3. The equilibrium point 𝑦₁=0 of (4.1) is a global attractor with a basin 𝑆^∗=[−1,1]^𝑘+1.

Theorem 4.4. The equilibrium point 𝑦₁=0 of (4.1) is a global attractor with a basin 𝑆^∗=(−∞,1]^𝑘×[(−𝐴+1)/(𝛿+1),(𝐴−1)/(𝛿+1)], where 𝛿≥0.

Proof. Assuming that the initial conditions 𝑦_−𝑘,𝑦_−𝑘+1,…,𝑦₋₁,𝑦₀∈S^∗. If 𝐴≥𝛿+2, with 𝛿≥0, then we deduce that −1≤(1−𝐴)/(1+𝛿)𝐴−𝛿𝑦₀−𝑦_−𝑘≤𝑦₁=−𝑦₀𝐴−𝛿𝑦₀−𝑦_−𝑘≤(𝐴−1)/(𝛿+1)(𝐴−1)/(𝛿+1)=1,−1≤−1𝐴−𝛿𝑦₁−𝑦_−𝑘+1≤𝑦₂=−𝑦₁𝐴−𝛿𝑦₁−𝑦_−𝑘+1≤1𝐴−𝛿−1≤1. (4.13) By induction, it follows that 𝑦_𝑖∈[−1,1] for 𝑖≥1. Thus, the proof of Theorem 4.4 follows from Corollary 4.3.

Theorem 4.5. If 𝐴>1, then the equilibrium point 𝑦₁=0 of (4.2) is globally asymptotically stable.

Finally, on using arguments similar to that used in Theorems 3.5 and 3.6, we can prove easily the following results.

Theorem 4.6. Assume that 𝛿𝑎𝑛𝑑𝐴>1. If k is an odd positive integer, then the necessary and sufficient condition for (4.2) to have positive solutions of prime period two is that (see Figure 5) (𝐴−1)𝛿>𝐴+3. (4.14)

Figure 5 

Theorem 4.7. If k is an even positive integer, then (4.2) has no positive solutions of prime period two (see Figure 6).

Figure 6