# Homework Help: Help with a proof with discrete dynamical sysmtes / chaos theory.

1. Oct 22, 2008

### goosefrabbas

1. The problem statement, all variables and given/known data
Consider the families of iterating functions Fλ(x) = λ(x3 - x). Fλ(x) undergoes a bifurcation at λ=1, about the fixed point x=0. Figure out what ilk of bifurcation is occurring for Fλ(x) and prove your assertion rigorously.

2. Relevant equations
My book says this about period-doubling bifurcations, and I need to prove all four of these to prove the problem.
Definition: A one-parameter family of functions Fλ undergoes a period-doubling bifurcation at the parameter value λ=λ0 if there is an open interval and an ε such that:
1. For each λ in the interval [λ0 - ε, λ0 + ε], there is a unique fixed point pλ for Fλ in I.
2.For λ0 - ε < λ < λ0, Fλ has no cycles of period 2 in I and pλ is attracting (resp. repelling).
3.For λ0 < λ < λ0 + ε, there is a unique 2-cycle q1λ,q2λ in I with Fλ(q1λ)=q2λ. This 2-cycle is attracting (resp. repelling). Meanwhile, the fixed point pλ is repelling (resp. attracting).
4.As λ -> λ0, we have qiλ -> pλ0
Also, these theorems are necessary (I think).
Chain Rule Along A Cycle: Suppose x0, x2, ..., xn-1 lie on a cycle of period n for F with xi = Fi(x0). Then
(Fn)'(x0) = F'(xn-1) * ... * F'(x1) * F'(x0).
The corollary for this is:
Suppose x0, x1, ..., xn-1 lie on an n-cycle for F. Then
(Fn)'(x0) = (Fn)'(x1) = ... = (Fn)'(xn-1)

3. The attempt at a solution\
I have almost no idea where to start the proof. So far I have:
The interval I can = (-1, 1).

To answer 1., I know that there are fixed points at 0, +sqrt(2), -sqrt(2). So to have a unique fixed point pλ in the interval [λ0 - ε, λ0 + ε], 0<ε<sqrt(2). Can I just choose an arbitrary ε, like ε=1?

To answer 2., if ε=1, λ0 - ε < λ < λ0, so 0 - 1 < λ < 0, so -1<λ<0. I used algebra and got 2-cycles for x = -1, -.7548777, 0, 1 1.4655712, but all but x=1.4655712 lie in the interval I. Could someone check this? I may have made a mistake in my algebra. I don't see why else I would get this.

I really appreciate the help!!

Last edited: Oct 23, 2008
2. Oct 23, 2008

### goosefrabbas

I changed my interval I to (-.5, .5) so that part 2 would work.

And for part 3, I got this far.
Fλ(x)=λ(x3-x) is an odd function, so the second iteration of Fλ(x0) = x0. Using algebra I got x = (1 +- sqrt(1 - 4λ))/(2λ), so this is the 2-cycle for Fλ, but only for 0<λ<1/4.

For part 4, As λ -> λ0, we have qiλ -> pλ0. So substituting, I get as λ -> 0, qiλ -> 0.

Is everything right so far?
I'm not sure how to prove part 4.