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

Definition 1. (i) The equilibrium point $$ of (6) is called locally stable if for every ϵ > 0, there exists δ > 0 such that if x_−k, …, x₋₁, x₀ ∈ I, and $$, we have

(ii) The equilibrium point $$ of (6) is called locally asymptotically stable if it is locally stable, and if there exists γ > 0, if x_−k, …, x₋₁, x₀ ∈ I, and $$, we have

(iii) The equilibrium point $$ of (6) is called a global attractor if for every x_−k, …, x₋₁, x₀ ∈ I, we have

(iv) The equilibrium point $$ of (6) is called globally asymptotically stable if it is locally stable and a global attractor.

(v) The equilibrium point $$ 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

(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

denote the partial derivatives of f(u₀, u₁, …, u_k) evaluated at the equilibrium $$ of (6). Then the equation is called the linearized equation associated with (6) about the equilibrium point $$. Its characteristic equation is

Theorem 4 (linearized stability). (a) If all roots of (14) lie in the open unit disk |λ | < 1, then the equilibrium $$ of (6) is locally asymptotically stable.

(b) If at least one of the roots of (14) has absolute value greater than one, then $$ is unstable.

Lemma 5. If

then F_n satisfies the following recursive formula:

Corollary 6. If

then

Theorem 7. (a) If

 then (3) has no nonnegative prime period-two solution.

(b) If

 then (3) has prime period-two solution  if and only if k is odd and  where the values of Φ and Ψ are the positive and distinct solutions of the quadratic equation

Proof. Assume that there exist distinct nonnegative real numbers Φ and Ψ, such that

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

Furthermore, Subtracting (30) from (29), we have so 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

then in this case Φ and Ψ satisfy Furthermore, Subtracting (35) from (36), we have But, p + z ≥ 1, this implies that Φ + Ψ ≤ 0 which contradicts the hypothesis that Φ, Ψ are distinct positive real numbers.

(b) If

then in this case Φ and Ψ satisfy Moreover, Subtracting (41) from (42), we have Furthermore, one adds (41) to (42), makes use of (43), and then does some elementary algebraic manipulation; we have Equation (44) leads to the following conclusion: that follows from the facts that Notice that when q = 1 then adding (41) to (42) gives 2r + 2p(1 − p − z) = 0, which is impossible.

Construct the quadratic equation

So Φ and Ψ are the positive and distinct solutions of the above quadratic equation, that is,

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

we first vectorize (3) by introducing the following change of variables: and write (3) in the equivalent form: where Now Φ and Ψ generate a period-two solution of (3) only if is a fixed point of T², the second iterate of T. Furthermore, where The prime period-two solution of (3) is asymptotically stable if the eigenvalues of the Jacobian matrix $$, evaluated at $$ lie inside the unit disk.

We have

Now let $$ be the characteristic polynomial of $$. Then, by the Laplace expansion in the (k + 1) row,

However; by Lemma 5, Corollary 6, and the fact that k is odd,

Therefore,

Hence, the characteristic polynomial is given by

where

Assume that 0 < Φ < Ψ. Then, by (39),

Hence, Similarly, we observe that 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

With that in mind, it is clear that

In addition, with understanding that and the fact that we have to establish First, we will establish inequality (69). To this end, observe that inequality (69) is equivalent to which is true if and only if which is true if and only if which is true if and only if which is true if and only if which is true if and only if which is clearly satisfied.

Next we will establish inequality (70). Observe that inequality (70) is equivalent to

which is true if and only if which is true if and only if which is true if and only if which is true if and only if which is true if and only if Now observe that the righthand side of (82) is The lefthand side of (82) is Hence, inequality (70) is true if and only if or equivalently 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,

Thus,

However, by the Descartes’ Rule of Signs f has either two or no positive zeros. Furthermore,

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

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.