# Dynamical Systems - Chaos: Stability condition for a 2-cycle system

Master1022
Homework Statement:
Find the values of the parameter ##\lambda## such that the 2-cycle is stable
Relevant Equations:
None
Hi,

(This question is part of the same example as a previous post of mine, but I have a question about a different part of it)

I was looking at a question from an exam for a course I am self-teaching. There is a sub-question which asks us to find the values of a parameter for which the 2-cycle system is stable. However, when looking at the solution I don't understand the condition they have used (and I can't see any reference to it in the materials I am using).

Question & Attempt:
Let us imagine we have a system: ## x_{n + 1} = F(x_n) = \lambda ( 1 - \alpha x_n ^2) ##. Then there are some sub-questions about finding the 2-cycle system and the fixed points of it, before the parts I specifically I don't understand.

(i) Find the 2-cycle system
We can do this by doing ## F(F(x_n)) = \lambda - \alpha \lambda ^3 - 2 \alpha ^2 \lambda ^3 x^2 - \alpha ^3 \lambda ^3 x ^4 ##

(ii) Then we find the fixed points of the 2-cycle system
Then we look for the fixed points by setting ## F(F(x)) = x ## which results in the following:
$$(\lambda - \alpha \lambda x^2 - x) (1 - \alpha x \lambda - \alpha \lambda ^2 + \alpha ^2 x^2 \lambda ^2) = 0$$

Then the second term is the condition for ##F(F(x)) = x ##, which I thought was also a fixed point.

$$\hat x_{1, 2} = \frac{\alpha \lambda \pm \alpha \lambda \sqrt{ 4 \alpha \lambda ^2 - 3}}{2 \alpha ^2 \lambda ^2}$$

We are told to use ## \alpha = \frac{1}{2} ## from here onwards.
Thus the solutions reduce to: ## \hat x_{1, 2} = \frac{1}{\lambda} \left( 1 \pm \sqrt{ 2 \lambda ^2 - 3} \right) ##

Below are the parts I don't understand the theory for:
(iii) Find the values of ##\lambda## for which the 2-cycle exists (subject of previous question)

The solution exists when ## 2 \lambda ^2 - 3 > 0 \rightarrow \lambda > \sqrt{\frac{3}{2}} ##

(iv) Find the values of ## \lambda ## for which the 2-cycle is stable

The solution scheme says that we do:
$$\left| \frac{dF(F(x))}{dx} |_{\hat x_{1, 2}} \right| = \left| F'(\hat x_1) F'(\hat x_2) \right| < 1$$

and then it arrives at the answer: ## -\sqrt{\frac{3}{2}} < \lambda < \sqrt{\frac{5}{2}} ##

I am not sure where this condition comes from... Any help would be greatly appreciated.

Homework Helper
2022 Award
(iii): The fixed points of $F \circ F$ must be real for a 2-cycle to exist.

(iv): Which do you not understand: why $|f'(x)| < 1$ is the condition for $x_{n+1} = f(x_n)$ to be stable, or why the chain rule gives $(F \circ F)'(x_1) = (F \circ F)'(x_2) = F'(x_1)F'(x_2)$?

Master1022
Thanks @pasmith !

(iii): The fixed points of $F \circ F$ must be real for a 2-cycle to exist.
Sure thing!

(iv): Which do you not understand: why $|f'(x)| < 1$ is the condition for $x_{n+1} = f(x_n)$ to be stable, or why the chain rule gives $(F \circ F)'(x_1) = (F \circ F)'(x_2) = F'(x_1)F'(x_2)$?
Neither unfortunately.