On the Period-Two Cycles of 𝑥_𝑛+1=(𝛼+𝛽𝑥_𝑛+𝛾𝑥_𝑛−𝑘)/(𝐴+𝐵𝑥_𝑛+𝐶𝑥_𝑛−𝑘)

Abstract

We consider the higher order nonlinear rational difference equation 𝑥_𝑛+1=(𝛼+𝛽𝑥_𝑛+𝛾𝑥_𝑛−𝑘)/(𝐴+𝐵𝑥_𝑛+𝐶𝑥_𝑛−𝑘),𝑛=0,1,2,…, where the parameters 𝛼,𝛽,𝛾,𝐴,𝐵,𝐶 are positive real numbers and the initial conditions 𝑥_−𝑘,…,𝑥₋₁,𝑥₀ are nonnegative real numbers, 𝑘∈{1,2,…}. We give a necessary and sufficient condition for the equation to have a prime period-two solution. We show that the period-two solution of the equation is locally asymptotically stable.

1. Introduction

Recently, dynamics of nonnegative solutions of higher order rational difference equation has been an area of intense interest. Related to this subject, researches are done by Dehghan et al. [1–4], Zayed [5–7], Huang and Knopf [8, 9], Karatas [10, 11], and others. For the general theory of difference equations, one can refer to the monographes of Kocic and Ladas [12], Elaydi [13], Agarwal [14], Kulenovic and Ladas [15], and Camouzis and Ladas [16]. Other related results can be found in [17–24].

Our aim in this paper is to study the higher order nonlinear rational difference equation 𝑥_𝑛+1=𝛼+𝛽𝑥_𝑛+𝛾𝑥_𝑛−𝑘𝐴+𝐵𝑥_𝑛+𝐶𝑥_𝑛−𝑘,𝑛=0,1,2,…, (1) where the parameters 𝛼,𝛽,𝛾,𝐴,𝐵,𝐶 are positive real numbers and the initial conditions 𝑥_−𝑘,…,𝑥₋₁,𝑥₀ are nonnegative real numbers, 𝑘∈{1,2,…}. Our concentration is on the periodic character of all positive solutions of (1).

The periodic character of positive solutions of (1) for 𝑘=1 has been investigated by the authors in [25]. They showed that the period-two solution of (1) for 𝑘=1 is locally asymptotically stable if it exists.

Motivated by the above results, our interest is now to study and generalize the previous results to the general case depicted in (1).

The change of variable 𝑥_𝑛=𝛾𝐶𝑦_𝑛 (2) reduces (1) to 𝑦_𝑛+1=𝑟+𝑝𝑦_𝑛+𝑦_𝑛−𝑘𝑧+𝑞𝑦_𝑛+𝑦_𝑛−𝑘,𝑛=0,1,2,…, (3) where 𝑟=𝛼𝐶𝛾²𝛽,𝑝=𝛾,𝐴𝑧=𝛾𝐵,𝑞=𝐶 (4) are positive real numbers and the initial conditions 𝑦_−𝑘,…,𝑦₋₁,𝑦₀ are nonnegative real numbers.

This paper is organized besides this introduction in three sections. In Section 2, we present some preliminaries and some results which can be mainly deduced from the general situation studied in [12–16, 26]. Our main results are presented in Section 3; we give a necessary and sufficient condition for the equation to have a prime period-two solution, in addition to providing a necessary and sufficient conditions for the prime period-two solution of the equation to be locally asymptotically stable. In order to illustrate the results of the previous section and to support our theoretical discussion, we consider several numerical examples in Section 4; we use MATLAB to see how the behaviors of (1) look like. Finally, we conclude in Section 5 with suggestions for future research.

2. Preliminaries

For the sake of self-containment and convenience, we recall the following definitions and results from [16].

Let 𝐼 be some interval of real numbers and let 𝑓∶𝐼^𝑘+1⟶𝐼 (5) be a continuously differentiable function. Then for every set of initial conditions 𝑥_−𝑘,…,𝑥₋₁,𝑥₀∈𝐼, the difference equation 𝑥_𝑛+1=𝑓𝑥_𝑛,𝑥_𝑛−1,…,𝑥_𝑛−𝑘,𝑛=0,1,2,…, (6) has a unique solution {𝑥_𝑛}^∞_𝑛=−𝑘.

A solution of (6) that is constant for all 𝑛≥−𝑘 is called an equilibrium solution of (6). If 𝑥_𝑛=𝑥for𝑛≥−𝑘 (7) is an equilibrium solution of (6), then 𝑥 is called an equilibrium point or simply an equilibrium of (6).

Definition 1. (i) The equilibrium point 𝑥 of (6) is called locally stable if for every 𝜖>0, there exists 𝛿>0 such that if 𝑥_−𝑘,…,𝑥₋₁,𝑥₀∈𝐼, and |𝑥_−𝑘−𝑥|+⋯,|𝑥₋₁−𝑥|+|𝑥₀−𝑥|<𝛿, we have 𝑥_𝑛−𝑥<𝜖,∀𝑛≥−𝑘. (8)
(ii) The equilibrium point 𝑥 of (6) is called locally asymptotically stable if it is locally stable, and if there exists 𝛾>0, if 𝑥_−𝑘,…,𝑥₋₁,𝑥₀∈𝐼, and |𝑥_−𝑘−𝑥|+⋯,|𝑥₋₁−𝑥|+|𝑥₀−𝑥|<𝛿, we have lim_𝑛→∞𝑥_𝑛=𝑥. (9)
(iii) The equilibrium point 𝑥 of (6) is called a global attractor if for every 𝑥_−𝑘,…,𝑥₋₁,𝑥₀∈𝐼, we have lim_𝑛→∞𝑥_𝑛=𝑥. (10)
(iv) The equilibrium point 𝑥 of (6) is called globally asymptotically stable if it is locally stable and a global attractor.
(v) The equilibrium point 𝑥 of (6) is called unstable if it is not stable.

Definition 2. (i) A solution {𝑥_𝑛} of (6) is said to be periodic with period 𝑃 if 𝑥_𝑛+𝑃=𝑥_𝑛,∀𝑛≥−𝑘. (11)
(ii) A solution {𝑥_𝑛} of (6) is said to be periodic with prime period 𝑃 if 𝑃 is the least positive integer for which (11) holds.

Definition 3. Let 𝑞_𝑖=𝜕𝑓𝜕𝑢_𝑖(𝑥,𝑥,…,𝑥) (12) denote the partial derivatives of 𝑓(𝑢₀,𝑢₁,…,𝑢_𝑘) evaluated at the equilibrium 𝑥 of (6). Then the equation 𝑦_𝑛+1=𝑞₀𝑦_𝑛+𝑞₁𝑦_𝑛−1+⋯+𝑞_𝑘𝑦_𝑛−𝑘,𝑛=0,1,2,…, (13) is called the linearized equation associated with (6) about the equilibrium point 𝑥. Its characteristic equation is 𝜆^𝑘+1−𝑞₀𝜆^𝑘−⋯−𝑞_𝑘−1𝜆−𝑞_𝑘=0. (14)

Theorem 4 (linearized stability). (a) If all roots of (14) lie in the open unit disk |𝜆|<1, then the equilibrium 𝑥 of (6) is locally asymptotically stable.
(b) If at least one of the roots of (14) has absolute value greater than one, then 𝑥 is unstable.

The following result from [26] will become handy in the sequel.

Lemma 5. If 𝐹_𝑛=𝑎₁𝑏₁𝑐₁𝑎₂𝑏₂𝑐₂⋱⋱⋱⋱𝑏_𝑛−1𝑐_𝑛−1𝑎_𝑛, (15) then 𝐹_𝑛 satisfies the following recursive formula: 𝐹_𝑛=𝑎_𝑛𝐹_𝑛−1−𝑏_𝑛−1𝑐_𝑛−1𝐹_𝑛−2. (16)

The aforementioned lemma leads to the following conclusion.

Corollary 6. If 𝐷_𝑛=,01−𝜆01−𝜆⋱⋱⋱⋱1−𝜆0 (17) then 𝐷_𝑛=𝜆𝐷_𝑛−2=𝜆0𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑,_𝑛/2𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛. (18)

3. Main Result

In this section, we give a necessary and sufficient condition for (1) to have a prime period-two solution. We show that the period-two solution of (1) is locally asymptotically stable.

Equation (3) has a unique positive equilibrium given by 𝑦=1+𝑝−𝑧+1+𝑝−𝑧²+4𝑟𝑞+12.𝑞+1 (19) The linearized equation associated with (3) about the equilibrium is given by 𝑧_𝑛+1=𝑝−𝑞𝑦𝑧+𝑞+1𝑦𝑧_𝑛+1−𝑦𝑧+𝑞+1𝑦𝑧_𝑛−𝑘, (20) and its characteristic equation is 𝜆^𝑘+1−𝑝−𝑞𝑦𝑧+𝑞+1𝑦𝜆^𝑘−1−𝑦𝑧+𝑞+1𝑦=0. (21)

Theorem 7. (a) If 𝑝+𝑧≥1, (22) ?then (3) has no nonnegative prime period-two solution.
(b) If 𝑝+𝑧<1, (23) ?then (3) has prime period-two solution …,Φ,Ψ,Φ,Ψ,… (24) ?if and only if 𝑘 is odd and 𝑞−𝑟<1−𝑝−𝑧1−𝑝−𝑧1+3𝑝−𝑧4, (25) ?where the values of Φ and Ψ are the positive and distinct solutions of the quadratic equation 𝑡²−𝑝1−𝑧−𝑝𝑡+1−𝑧−𝑝+𝑟𝑞−1=0,𝑞>1. (26)

Proof. Assume that there exist distinct nonnegative real numbers Φ and Ψ, such that …,Φ,Ψ,Φ,Ψ,… (27) is a prime period-two solution of (3); there are two cases to be considered.
Case??1??(k is even). In this case Φ and Ψ satisfy Φ=𝑟+𝑝Ψ+Ψ𝑧+𝑞Ψ+Ψ,Ψ=𝑟+𝑝Φ+Φ.𝑧+𝑞Φ+Φ (28) Furthermore, 𝑧Φ+𝑞+1ΨΦ=𝑟+𝑝+1Ψ,𝑧Ψ+𝑞+1ΦΨ=𝑟+𝑝+1Φ. (29)(30) Subtracting (30) from (29), we have 𝑧(Φ−Ψ)=𝑝+1(Ψ−Φ), (31) so (Φ−Ψ)𝑧+𝑝+1=0⟹Φ=Ψor𝑧+𝑝=−1. (32) This contradicts the hypothesis that Φ and Ψ are distinct nonnegative real numbers. Also, 𝑧+𝑝=−1 contradicts the hypothesis that 𝑧 and 𝑝 are positive real numbers.
Case??2??(k is odd). (a) If 𝑝+𝑧≥1, (33) then in this case Φ and Ψ satisfy Φ=𝑟+𝑝Ψ+Φ𝑧+𝑞Ψ+Φ,Ψ=𝑟+𝑝Φ+Ψ.𝑧+𝑞Φ+Ψ (34) Furthermore, Φ𝑧+𝑞Ψ+Φ=𝑟+𝑝Ψ+Φ,Ψ𝑧+𝑞Φ+Ψ=𝑟+𝑝Φ+Ψ. (35)(36) Subtracting (35) from (36), we have (Φ+Ψ)=1−𝑧−𝑝. (37) But, 𝑝+𝑧≥1, this implies that Φ+Ψ≤0 which contradicts the hypothesis that Φ,Ψ are distinct positive real numbers.
(b) If 𝑝+𝑧<1, (38) then in this case Φ and Ψ satisfy Φ=𝑟+𝑝Ψ+Φ,𝑧+𝑞Ψ+ΦΨ=𝑟+𝑝Φ+Ψ.𝑧+𝑞Φ+Ψ (39)(40) Moreover, Φ𝑧+𝑞Ψ+Φ=𝑟+𝑝Ψ+Φ,Ψ𝑧+𝑞Φ+Ψ=𝑟+𝑝Φ+Ψ. (41)(42) Subtracting (41) from (42), we have (Φ+Ψ)=1−𝑧−𝑝. (43) Furthermore, one adds (41) to (42), makes use of (43), and then does some elementary algebraic manipulation; we have 𝑝ΦΨ=1−𝑧−𝑝+𝑟.𝑞−1 (44) Equation (44) leads to the following conclusion: 𝑞>1, (45) that follows from the facts that ΦΨ>0,1−𝑝−𝑧>0. (46) Notice that when 𝑞=1 then adding (41) to (42) gives 2𝑟+2𝑝(1−𝑝−𝑧)=0, which is impossible.
Construct the quadratic equation 𝑡²−𝑝1−𝑧−𝑝𝑡+1−𝑧−𝑝+𝑟𝑞−1=0,𝑞>1. (47) So Φ and Ψ are the positive and distinct solutions of the above quadratic equation, that is, ±𝑡=1−𝑧−𝑝1−𝑧−𝑝²−4𝑝1−𝑧−𝑝+𝑟/𝑞−12,𝑞>1. (48)

Theorem 8. Suppose (3) has a prime period-two solution. Then, the period-two solution is locally asymptotically stable.

Proof. To investigate the local stability of the two cycles …,Φ,Ψ,Φ,Ψ,… (49) we first vectorize (3) by introducing the following change of variables: 𝑧₁(𝑛)=𝑦_𝑛−𝑘,𝑧₂(𝑛)=𝑦_{𝑛−𝑘+1},𝑧₃(𝑛)=𝑦_{𝑛−𝑘+2},⋮𝑧_𝑘(𝑛)=𝑦_𝑛−1,𝑧_𝑘+1(𝑛)=𝑦_𝑛, (50) and write (3) in the equivalent form: 𝑧₁()𝑧𝑛+1₂()⋮𝑧𝑛+1_𝑘()𝑧𝑛+1_𝑘+1()𝑧𝑛+1=𝑇¹(𝑛)𝑧₂(𝑛)⋮𝑧_𝑘(𝑛)𝑧_𝑘+1(𝑛),𝑛=0,1,2,…, (51) where 𝑇𝑧₁𝑧₂⋮𝑧_𝑘𝑧_𝑘+1=𝑧₂𝑧₃⋮𝑧_𝑘+1𝑟+𝑝𝑧_𝑘+1+𝑧₁𝑧+𝑞𝑧_𝑘+1+𝑧₁. (52) Now Φ and Ψ generate a period-two solution of (3) only if ΦΨ⋮ΦΨ (53) is a fixed point of 𝑇², the second iterate of 𝑇. Furthermore, 𝑇²𝑧₁𝑧₂⋮𝑧_𝑘𝑧_𝑘+1=𝑓₁𝑧₁,𝑧₂,…,𝑧_𝑘+1𝑓₂𝑧₁,𝑧₂,…,𝑧_𝑘+1⋮𝑓_𝑘𝑧₁,𝑧₂,…,𝑧_𝑘+1𝑓_𝑘+1𝑧₁,𝑧₂,…,𝑧_𝑘+1, (54) where 𝑓₁𝑧₁,𝑧₂,…,𝑧_𝑘+1=𝑧₃,𝑓₂𝑧₁,𝑧₂,…,𝑧_𝑘+1=𝑧₄,⋮𝑓_𝑘𝑧₁,𝑧₂,…,𝑧_𝑘+1=𝑟+𝑝𝑧_𝑘+1+𝑧₁𝑧+𝑞𝑧_𝑘+1+𝑧₁𝑓_𝑘+1𝑧₁,𝑧₂,…,𝑧_𝑘+1=𝑟+𝑝𝑟+𝑝𝑧_𝑘+1+𝑧₁/𝑧+𝑞𝑧_𝑘+1+𝑧₁+𝑧₂𝑧+𝑞𝑟+𝑝𝑧_𝑘+1+𝑧₁/𝑧+𝑞𝑧_𝑘+1+𝑧₁+𝑧₂. (55) The prime period-two solution of (3) is asymptotically stable if the eigenvalues of the Jacobian matrix 𝐽_𝑇², evaluated at ΦΨ⋮ΦΨ lie inside the unit disk.
We have𝐽_𝑇²ΦΨ⋮ΦΨ=0010...00001...000001..0................0000...11−Φ𝑧+𝑞Ψ+Φ000...𝑝−𝑞Φ()𝑧+𝑞Ψ+Φ𝑝−𝑞Ψ1−Φ𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ1−Ψ𝑧+𝑞Φ+Ψ00...𝑝−𝑞Φ𝑝−𝑞Ψ.𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ (56)
Now let 𝑃(𝜆)=det(𝐽_𝑇²−𝜆𝐼) be the characteristic polynomial of 𝐽_𝑇². Then, by the Laplace expansion in the (𝑘+1) row, 𝑃(𝜆)=−𝜆010⋯⋯00−𝜆01⋯⋯000−𝜆01⋯0⋮⋮⋮⋮⋮⋱⋮0000−𝜆⋯11−Φ𝑧+𝑞Ψ+Φ000⋯−𝜆𝑝−𝑞Φ()𝑧+𝑞Ψ+Φ𝑝−𝑞Ψ1−Φ𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ1−Ψ𝑧+𝑞Φ+Ψ0000𝑝−𝑞Ψ𝑝−𝑞Φ()𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ−𝜆=−𝑝−𝑞Ψ1−Φ𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ010⋯⋯0−𝜆01⋯⋯00−𝜆01⋯0⋮⋮⋮⋮⋱⋮000−𝜆⋯1000⋯−𝜆𝑝−𝑞Φ𝑧+𝑞Ψ+Φ_𝐴𝑘+1−Ψ𝑧+𝑞Φ+Ψ−𝜆10⋯⋯0001⋯⋯00−𝜆01⋯0⋮⋮⋮⋮⋱⋮000−𝜆⋯11−Φ𝑧+𝑞Ψ+Φ00⋯−𝜆𝑝−𝑞Φ𝑧+𝑞Ψ+Φ_𝐵𝑘+𝑝−𝑞Ψ𝑝−𝑞Φ𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ−𝜆−𝜆010⋯⋯00−𝜆01⋯⋯000−𝜆01⋯⋮⋮⋮⋮⋮⋱10000−𝜆⋯01−Φ.𝑧+𝑞Ψ+Φ000⋯⋯−𝜆_𝐶𝑘 (57)
However; by Lemma 5, Corollary 6, and the fact that 𝑘 is odd, 𝐴_𝑘=𝑝−𝑞Φ𝐷𝑧+𝑞Ψ+Φ^𝑘−1+𝜆𝐷_𝑘−2=𝑝−𝑞Φ𝜆𝑧+𝑞Ψ+Φ^(𝑘−1)/2,𝐵_𝑘=−𝜆𝐴_𝑘−1+1−Φ𝑧+𝑞Ψ+Φ=−𝜆𝑝−𝑞Φ𝐷𝑧+𝑞Ψ+Φ^𝑘−2+𝜆𝐷_𝑘−3+1−Φ𝑧+𝑞Ψ+Φ=−𝜆^(𝑘+1)/2+1−Φ,𝐶𝑧+𝑞Ψ+Φ_𝑘=(−𝜆)^𝑘+1−Φ𝐷𝑧+𝑞Ψ+Φ^𝑘−1=−𝜆^𝑘+1−Φ𝜆𝑧+𝑞Ψ+Φ^(𝑘−1)/2. (58) Therefore, 𝑃(𝜆)()=−𝑝−𝑞Ψ1−Φ×𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ𝑝−𝑞Φ𝜆𝑧+𝑞Ψ+Φ^(𝑘−1)/2+1−Ψ𝑧+𝑞Φ+Ψ−𝜆^(𝑘+1)/2+1−Φ+𝑧+𝑞Ψ+Φ𝑝−𝑞Ψ𝑝−𝑞Φ×𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ−𝜆−𝜆^𝑘+1−Φ𝜆𝑧+𝑞Ψ+Φ^(𝑘−1)/2=𝜆^𝑘+1−𝑝−𝑞Ψ𝑝−𝑞Φ𝜆𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ^𝑘−1−Ψ+𝑧+𝑞Φ+Ψ1−Φ𝜆𝑧+𝑞Ψ+Φ^(𝑘+1)/2+(1−Ψ)(1−Φ).𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ (59)
Hence, the characteristic polynomial is given by 𝑓(𝜆)=𝜆^𝑘+1−𝑄𝜆^𝑘−𝐿𝜆^(𝑘+1)/2+𝜇=0, (60) where 𝑄=𝑝−𝑞Φ𝑝−𝑞Ψ,𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ𝐿=1−Φ𝑧+𝑞Ψ+Φ_𝑎+1−Ψ𝑧+𝑞Φ+Ψ_𝑏,()𝜇=1−Φ)(1−Ψ𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ=𝑎𝑏. (61)
Assume that 0<Φ<Ψ. Then, by (39), ()()1=𝑟/Φ+𝑝Ψ/Φ+1>1𝑧+𝑞Ψ+Φ.𝑧+𝑞Ψ+Φ (62) Hence, 𝑧+𝑞Ψ+Φ>1. (63) Similarly, we observe that 𝑧+𝑞Φ+Ψ>1. (64) Furthermore, since 𝑝+𝑧<1, (43) implies the sum of Φ, Ψ is less than 1 and, a fortiori, each is less than 1. Indeed, we have 10<Φ<minΨ,2<1. (65)
With that in mind, it is clear that 0<𝑎,𝑏<1. (66) In addition, with understanding that ()=𝑝Φ+Ψ1−𝑧−𝑝>0,ΦΨ=1−𝑧−𝑝+𝑟𝑞−1 (67) and the fact that 𝑞>1, (68) we have to establish 𝑄>0,𝑄+𝐿<1+𝜇. (69)(70) First, we will establish inequality (69). To this end, observe that inequality (69) is equivalent to 𝑝−𝑞Φ𝑝−𝑞Ψ𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ>0, (71) which is true if and only if 𝑝²−𝑝𝑞(Φ+Ψ)+𝑞²ΦΨ𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ>0, (72) which is true if and only if 𝑝²−𝑝𝑞1−𝑝−𝑧+𝑞²𝑝1−𝑝−𝑧+𝑟/𝑞−1𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ>0, (73) which is true if and only if 𝑝²𝑞−1−𝑝𝑞𝑞−11−𝑝−𝑧+𝑝𝑞²1−𝑝−𝑧+𝑟𝑞²𝑞−1𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ>0, (74) which is true if and only if 𝑝²𝑞−1−𝑝𝑞1−𝑝−𝑧𝑞−1−𝑞+𝑟𝑞²𝑞−1𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ>0, (75) which is true if and only if 𝑝²𝑞−1+𝑝𝑞1−𝑝−𝑧+𝑟𝑞²𝑞−1𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ>0, (76) which is clearly satisfied.
Next we will establish inequality (70). Observe that inequality (70) is equivalent to 1−Φ+𝑧+𝑞Ψ+Φ1−Ψ+−()𝑧+𝑞Φ+Ψ𝑝−𝑞Φ𝑝−𝑞Ψ1−Φ)(1−Ψ𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ<1, (77) which is true if and only if 1−Φ+𝑧+𝑞Ψ+Φ1−Ψ+𝑝𝑧+𝑞Φ+Ψ₂−1+𝑞²−1ΦΨ+1−𝑝𝑞(Φ+Ψ)𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ<1, (78) which is true if and only if 1−Φ+𝑧+𝑞Ψ+Φ1−Ψ+𝑝𝑧+𝑞Φ+Ψ₂×𝑝/+×−1+𝑞−1𝑞+11−𝑝−𝑧+𝑟𝑞−11−𝑝𝑞1−𝑝−𝑧𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ⁻¹<1, (79) which is true if and only if 1−Φ+𝑧+𝑞Ψ+Φ1−Ψ+𝑝𝑧+𝑞Φ+Ψ₂−1+𝑟𝑞+1+1−𝑝−𝑧𝑝+1𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ<1, (80) which is true if and only if 1−Φ+𝑧+𝑞Ψ+Φ1−Ψ+𝑟𝑧+𝑞Φ+Ψ𝑞+1−𝑧𝑝+1𝑧+𝑞Φ+Ψ𝑧+𝑞Ψ+Φ<1, (81) which is true if and only if ()+()<.1−Φ𝑧+𝑞Φ+Ψ1−Ψ𝑧+𝑞Ψ+Φ+𝑟𝑞+1−𝑧𝑝+1𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ (82) Now observe that the righthand side of (82) is 𝐼=𝑧+𝑞Ψ+Φ𝑧+𝑞Φ+Ψ=𝑧²)Ψ+𝑧𝑞+1(Ψ+Φ+𝑞²+Φ²+1+𝑞²ΨΦ=𝑧²()()+𝑧𝑞+1Ψ+Φ+𝑞Ψ+Φ²−2ΨΦ+1+𝑞²ΨΦ=𝑧²()()+𝑧𝑞+1Ψ+Φ+𝑞Ψ+Φ²+𝑞−1²ΨΦ=𝑧²+𝑧𝑞+11−𝑝−𝑧+𝑞1−𝑝−𝑧²+𝑞−1²𝑝1−𝑝−𝑧+𝑟𝑞−1=𝑧²+×𝑧+𝑟𝑞−11−𝑝−𝑧𝑞+1+𝑞1−𝑝−𝑧+𝑝𝑞−1=𝑧²+−=()++𝑟𝑞−11−𝑝−𝑧𝑞−𝑝+𝑧=𝑟𝑞−1−𝑧𝑝−1𝑝−𝑞1−𝑝−𝑧=𝑟𝑞−1+𝑧1−𝑞+𝑞1−𝑝+𝑝1−𝑝𝑧−𝑟1−𝑞𝑝−𝑞𝑝−1=𝑧−𝑟−𝑞𝑧+𝑞𝑟+𝑝²−𝑝−𝑞𝑝+𝑞. (83) The lefthand side of (82) is ()+()()+()()Ψ𝐼𝐼=1−Φ𝑧+𝑞Φ+Ψ1−Ψ𝑧+𝑞Ψ+Φ+𝑟𝑞+1−𝑧𝑝+1=2𝑧+𝑞Φ+ΨΦ+Ψ−𝑧Φ+Ψ−𝑞²+Φ²()Ψ−2ΦΨ+𝑟𝑞+1−𝑧𝑝+1=2𝑧+𝑞+1−𝑧Φ+Ψ−𝑞²+Φ²())−2ΦΨ+𝑟𝑞+1−𝑧𝑝+1=2𝑧+𝑞+1−𝑧Φ+Ψ−𝑞(Ψ+Φ²()()−2ΦΨ−2ΦΨ+𝑟𝑞+1−𝑧𝑝+1=2𝑧+𝑞+1−𝑧Φ+Ψ−𝑞Ψ+Φ²()()+2𝑞−1ΦΨ+𝑟𝑞+1−𝑧𝑝+1=2𝑧+𝑞+1−𝑧Φ+Ψ−𝑞Ψ+Φ²𝑝+2𝑞−11−𝑝−𝑧+𝑟(()𝑞−1+𝑟𝑞+1−𝑧𝑝+1=2𝑧+2𝑟+Φ+Ψ)1−𝑧+𝑞−𝑞Ψ+Φ+2𝑝+𝑟𝑞+1−𝑧𝑝+1=2𝑧+2𝑟+1−𝑝−𝑧1−𝑧+𝑞−𝑞1−𝑝−𝑧+2𝑝+𝑟𝑞+1−𝑧𝑝+1=2𝑧+2𝑟+1−𝑝−𝑧1−𝑧+𝑞𝑝+𝑞𝑧+2𝑝+𝑟𝑞+1−𝑧𝑝+1=3𝑟+1+𝑝−𝑧+𝑟𝑞+𝑝𝑞+𝑧𝑞−2𝑝𝑧−2𝑝𝑧𝑞+𝑧²−𝑞𝑧²−𝑞𝑝²−2𝑝²−2𝑝𝑧. (84) Hence, inequality (70) is true if and only if 3𝑟+1+𝑝−𝑧+𝑟𝑞+𝑝𝑞+𝑧𝑞−2𝑝𝑧−2𝑝𝑧𝑞+𝑧²−𝑞𝑧²−𝑞𝑝²−2𝑝²−2𝑝𝑧<𝑧−𝑟−𝑞𝑧+𝑞𝑟+𝑝²−𝑝−𝑞𝑝+𝑞 (85) or equivalently 4𝑟<𝑞−𝑞𝑝−𝑞𝑧−1−3𝑝+𝑧−𝑝𝑞+𝑞𝑝²+𝑞𝑧𝑝+𝑝+3𝑝²−𝑧𝑝−𝑧𝑞+𝑞𝑧𝑝+𝑞𝑧²+𝑧+3𝑝𝑧−𝑧²𝑞−𝑞−⟺4𝑟<1−𝑝−𝑧𝑞−𝑞𝑝−𝑞𝑧−1−3𝑝+𝑧⟺4𝑟<1−𝑝−𝑧1−𝑝−𝑧1+3𝑝−𝑧⟺𝑟<1−𝑝−𝑧1−𝑝−𝑧1+3𝑝−𝑧4, (86) which is clearly satisfied (condition (25)).
Now, by applying Theorem 4 we shall show that the zeros of 𝑓 in (60) lie in the open unit disk |𝜆|<1. To do so, suppose to the contrary that 𝑓 has a zero 𝜆 such that |𝜆|≥1. Then, by the triangle inequality, |𝜆|^(𝑘+1)/2−𝑎|𝜆|^(𝑘+1)/2≤𝜆−𝑏_(𝑘+1)/2−𝑎𝜆^(𝑘+1)/2−𝑏=𝑄|𝜆|^𝑘. (87) Thus, 𝑓(|𝜆|)=|𝜆|^𝑘+1−𝑄|𝜆|^𝑘−𝐿|𝜆|^(𝑘+1)/2+𝜇≤0. (88)
However, by the Descartes’ Rule of Signs 𝑓 has either two or no positive zeros. Furthermore, 𝑓(0)𝑓=𝜇>0,_()𝑘+1/2√𝑎𝑓(1)<0,=1+𝜇−𝑄−𝐿>0, (89) and so, by the Intermediate Value Theorem, 𝑓(𝑥) has two positive zeros in the open interval (0,1). Moreover, since 𝑓(1)>0, we conclude that 𝑓(𝑥)>0 for all 𝑥≥1 which contradicts inequality (88).
The proof is complete.

Remark 9. The characteristic equation of the linearized equation at the equilibrium solution is given by 𝜆^𝑘+1−𝑝−𝑞𝑦𝑧+𝑞+1𝑦𝜆^𝑘−1−𝑦𝑧+𝑞+1𝑦=0. (90) Since the magnitude of the constant term is less than 1, the equation has at least one root inside the unit disk. As such, by the Stable Manifold Theorem, there is a manifold of solutions, of dimension bigger than or equal to 1, that converge to the equilibrium solution. Hence, the period-two solution cannot be globally asymptotically stable.

4. Numerical Examples

In order to illustrate the results of the previous section and to support our theoretical discussion, we consider several numerical examples generated by MATLAB.

Case??1??(k is even). For this case we consider the following example: 𝑦_𝑛+1=0.0001+0.4𝑦_𝑛+𝑦_𝑛−20.5+20𝑦_𝑛+𝑦_𝑛−2. (91) The dynamics of (91) is shown in Figure 1, no prime period-two solution.

Figure 1 

Dynamics of 𝑦𝑛+1=(0.0001+0.4𝑦𝑛+𝑦𝑛−2)/(0.5+20𝑦𝑛+𝑦𝑛−2).

Case??2??(k is odd). There are two cases to be considered.

Subcase??2.1?? (𝑝+𝑧≥1). For this case we consider the following example: 𝑦_𝑛+1=0.01+0.5𝑦_𝑛+𝑦_𝑛−30.7+2𝑦_𝑛+𝑦_𝑛−3. (92) The dynamics of (92) is shown in Figure 2, no prime period-two solution.

Figure 2 

Dynamics of 𝑦𝑛+1=(0.01+0.5𝑦𝑛+𝑦𝑛−3)/(0.7+2𝑦𝑛+𝑦𝑛−3).

Subcase??2.2?? (𝑝+𝑧<1 and 𝑟<((1−𝑝−𝑧)[𝑞(1−𝑝−𝑧)−(1+3𝑝−𝑧)]/4)). For this case we consider the following example: 𝑦_𝑛+1=0.01+0.4𝑦_𝑛+𝑦_𝑛−30.3+10𝑦_𝑛+𝑦_𝑛−3. (93) The dynamics of (93) is shown in Figure 3; it has prime period-two solution.

Figure 3 

Dynamics of 𝑦𝑛+1=(0.01+0.4𝑦𝑛+𝑦𝑛−3)/(0.3+10𝑦𝑛+𝑦𝑛−3).

5. Conclusion

In this paper, we showed that the period-two solution of the higher order nonlinear rational difference equation 𝑥_𝑛+1=𝛼+𝛽𝑥_𝑛+𝛾𝑥_𝑛−𝑘𝐴+𝐵𝑥_𝑛+𝐶𝑥_𝑛−𝑘,𝑛=0,1,2,…, (94) where the parameters 𝛼,𝛽,𝛾,𝐴,𝐵,𝐶 are positive real numbers and the initial conditions 𝑥_−𝑘,…,𝑥₋₁,𝑥₀ are nonnegative real numbers, 𝑘∈{1,2,…}, is locally asymptotically stable if it exists.

We consider the aforementioned result as a step forward in investigating bigger classes of difference equations which afford the ELAS property; that is, the existence of a periodic solution implies its local asymptotic stability.

Acknowledgment

The authors are very grateful to the anonymous referees for carefully reading the paper and for their comments and valuable suggestions that lead to an improvement in the paper.