ON DISCRETE-TIME DYNAMICAL SYSTEMS UNDER PERIODIC PERTURBATION

We present some results on the existence
and the minimum period of periodic orbits for discrete-time
dynamical systems under periodic perturbation. Some examples are
presented to illustrate these results.


Introduction
Consider the following discrete-time dynamical systems: where f is a continuous function, y(n) is a perturbation function.
We are interested in the following question.
Question 1. 1.Under what conditions system (1.1) has a periodic orbit when the perturbation is a periodic signal?

Main results
In this section, we present some results on the existence and the minimum period of periodic orbits of system (1.1) under the following assumptions.2 Discrete-time systems under periodic perturbation Assumption 2.1.The perturbation y(n) is assumed to be a periodic signal of minimum period k satisfying y i = y i+k , that is, the periodic perturbation is {y 0 , y 1 ,..., y k−1 }, where i is a nonnegative integer and k is a positive integer.Assumption 2.2.f (0, y i ) = 0 provided that y i = 0, i = 0,1,...,k − 1.
Assumption 2.3.We have lim Before stating our results, we first recall a fixed-point theorem due to Brower, which is useful in the following arguments.
where C is a nonempty, bounded, close and convex set.Then F has a fixed point.Its proof can be found in [2].Now we present the following theorems.
Proof.According to Assumption 2.1 we know that the periodic signal is {y 0 , y 1 ,..., y k−1 } with minimum period k satisfying y i = y i+k , where i is a nonnegative integer and k is a positive integer.
Since lim x →∞ ( f (x, y i ) / x ) < 1, for every y i , i = 0,1,...,k − 1, there exist corresponding positive numbers m i , i = 0,1,...,k − 1 such that f (x, y i ) < x when x > m i . Let is a positive invariant set of the following systems: (2.2) Then F : U → U is a continuous mapping.By Brower fixed-point theorem, the mapping F has a fixed point x 0 , that is, is a periodic orbit.This implies that o(x 0 ) contains a periodic orbit whose minimum period j divides k, where j is a positive integer.
In addition, according to Assumption 2.2, we have that there exists i 0 ∈ {0, 1,...,k−1} such that x i0 = 0.This shows that o(x 0 ) contains a nontrivial periodic orbit.The proof is complete.
H. Li and X. Yang 3 It seems that the period of o(x 0 ) should not be less than that of the perturbation y(n), which is often the case when dynamical systems described by ordinary differential equations are forced by periodic signals.However, the following example shows that this can not be the case in discrete-time systems.
Consider the following example.
Example 2.6.We have where f (x, y) = x/(2 It is easy to see that for Example 2.6 the conditions stated in Theorem 2.5 are satisfied.Thus according to Theorem 2.5, system (2.3) has a periodic orbit.In fact, it can be verified that the orbit {5/13, 6/13} is a periodic orbit of (2.3).
Remark 2.7.In Example 2.6, we have the following facts.
(a) The system x(n + 1)= f (x(n),0) = (1/2)x(n) is globally asymptotically stable which implies that the system does not have nontrivial periodic orbits.However, its perturbed system has a nontrivial periodic orbit.
(b) The functions f i = f (x, y i ), i = 0,1,2,3, satisfy the following condition: which shows that and f 1 • f 0 (x) have the same fixed points.(c) The minimum period of the periodic orbit of the perturbation is 4, but the minimum period of its corresponding periodic orbit is 2, which defies our intuition.
Considering the above fact (c), we want to know under what conditions (1.1) has a nontrivial k-period orbit.This will be given in Theorem 2.8.Before presenting Theorem 2.8, we assume that the integer k (in Assumption 2.1) has the following factors: k 1 ,...,k r and 0 < k 1 < k 2 < ••• < k r < k, where r and k 1 ,...,k r are positive integers and k l divides k, l = 1,2,...,r.Moreover, we let Theorem 2.8.Suppose that Assumptions 2.1, 2.2, and 2.3 hold, and for any k l ∈ {k 1 ,...,k r }, F kl and F 2kl do not have the same fixed points.Then system (1.1) also has a periodic orbit of minimum period k.
Proof.In terms of the proof of Theorem 2.5, we have that system (1.1) has a periodic orbit {x 0 ,x 1 ,...,x j−1 } of minimum period j, where j ∈ {k 1 ,...,k r ,k}.Below, our purpose is to prove that the minimum period of the above periodic orbit is k, that is, j = k.For this purpose, we assume that j ∈ {k 1 ,...,k r }.Since {x 0 ,x 1 ,...,x j−1 } is a periodic orbit, it is evident that x 0 is the fixed point of F j and F 2 j , which is a contradiction to the assumption of Theorem 2.8.Therefore the minimum period of the periodic orbit {x 0 ,x 1 ,...,x j−1 } is also k.The proof is complete.
It is easy to verify that in Example 2.9 the conditions stated in Theorem 2.8 are satisfied.Thus by Theorem 2.8, we know that system (2.5)only has a periodic orbit of period 4. It is easy to verify that the period orbit is {1/17, 6/17,−1/17,−6/17}.