- Open Access
© Qian and Qi-hong; licensee Springer. 2011
- Received: 10 February 2011
- Accepted: 3 June 2011
- Published: 3 June 2011
This article is concerned with the following rational difference equation y n+1= (y n + y n-1)/(p + y n y n-1) with the initial conditions; y -1, y 0 are arbitrary positive real numbers, and p is positive constant. Locally asymptotical stability and global attractivity of the equilibrium point of the equation are investigated, and non-negative solution with prime period two cannot be found. Moreover, simulation is shown to support the results.
- Global stability
- solution with prime period two
- numerical simulation
with initial conditions y -1, y 0 ∈ (0, + ∞), p ∈ R +.
Let us introduce some basic definitions and some theorems that we need in what follows.
- (1)The equilibrium point of Equation 2 is locally stable if for every ε > 0, there exists δ > 0, such that for any initial data x -k , x -k+1, ..., x 0 ∈ I, with
- (2)The equilibrium point of Equation 2 is locally asymptotically stable if is locally stable solution of Equation 2, and there exists γ > 0, such that for all x -k , x -k+1, ..., x 0 ∈ I, with
The equilibrium point of Equation 2 is a global attractor if for all x -k , x -k+1, ..., x 0 ∈ I, we have .
The equilibrium point of Equation 2 is unstable if is not locally stable.
Moreover, suppose p 2 > 0, then, |p 1| + |p 2| < 1 is also a necessary condition for the asymptotic stability of Equation 4.
g(x, y) is non-decreasing in x ∈ [p, q] for each fixed y ∈ [p, q], and g(x, y) is non-increasing in y ∈ [p, q] for each fixed x ∈ [p, q].
If (m, M) is a solution of system
M = g(M, m) and m = g(m, M),
then M = m.
It follows by Lemma 2, Equation 7 is asymptotically stable, if p > 2.
Theorem 2. Assume that , the equilibrium point and of Equation 1 is a global attractor.
Suppose that (m, M) is a solution of system
M = g(M, m) and m = g(m, M).
Lemma 3 suggests that is a global attractor of Equation 1 and then, the proof is completed.
Theorem 3. (1) has no non-negative solution with prime period two for all p ∈ R +.
is a prime period-two solution of (1).
so φ = ψ, which contradicts the hypothesis φ ≠ ψ. The proof is complete.
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