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Help with a proof with discrete dynamical sysmtes / chaos theory.

  1. Oct 22, 2008 #1
    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. jcsd
  3. Oct 23, 2008 #2
    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.
     
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