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.

This leads to two solutions:
$$\hat x_{1, 2} = \frac{\alpha \lambda \pm \alpha \lambda \sqrt{ 4 \alpha \lambda ^2 - 3}}{2 \alpha ^2 \lambda ^2}$$

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.

 

[pasmith]

(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.