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Lyapunov Exponent

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the Lyapunov exponent for the linear map xn+1= rxn.

    2. Relevant equations

    λ = Lyapunov Exponent

    λ = [itex]\lim_{n \rightarrow \infty} \begin{bmatrix}\dfrac{1}{n} \sum_{i = 0}^{n - 1} ln|f'(x_i)| \end{bmatrix} [/itex]

    3. The attempt at a solution

    f'(x) = r.

    λ = [itex]\lim_{n \rightarrow \infty} \begin{bmatrix}\dfrac{1}{n} \sum_{i = 0}^{n - 1} ln|r| \end{bmatrix} [/itex]

    = [itex]\lim_{n \rightarrow \infty} \begin{bmatrix}\dfrac{ln(r)}{n} \end{bmatrix} [/itex]



    This is where I'm a bit lost. Is λ = ∞, or is λ = ln(r)?

    In another example, with the tent function [itex] f(x) =\begin{cases}rx, \hspace{4mm} 0 \leq x \leq \dfrac{1}{2} \\ r - rx, \hspace{4mm} \dfrac{1}{2} \leq x \leq 1 \end{cases} [/itex]

    λ = ln(r).

    Is the Lyapunov exponent for both of these systems the same?
     
  2. jcsd
  3. Apr 29, 2013 #2
  4. Apr 29, 2013 #3

    fzero

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    You haven't properly accounted for the sum here.
     
  5. Apr 29, 2013 #4
    I have no idea how to really go about doing that. Any advice for starting it?
     
  6. Apr 29, 2013 #5

    fzero

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    Since ##\ln |r|## is independent of ##n##, it can be factored out of the sum. What is left over? How many terms appear in the sum?
     
  7. Apr 29, 2013 #6
    If it's factored out isn't there nothing remaining?

    As of today in class the professor mentioned something about n being the period, and a particular function he gave had period 2, so it was 1/2 times the sum. I'm just completely lost on this and the professor blows through everything at lightning speed as if we've all had the same 30+ years of experience in the topic as he has.

    My brain is completely fried at this point.
     
  8. May 1, 2013 #7

    fzero

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    You had

    Factoring out the ##\ln|r|##, we can write this as

    $$ \lambda = \lim_{n \rightarrow \infty} \frac{\ln |r|}{n} \sum_{i = 0}^{n - 1} 1.$$

    You're left with computing the sum, which depends on ##n##.

    The Lyapunov exponent is defined for any orbit ##\{x_0,\cdots\}##. A periodic orbit is an orbit ##\{z_0,\cdots, z_{k-1}\}## such that ##z_{i+k-1}=z_i## for all ##i\geq 0## (##k-1## looks a bit weird here, but that's because we started counting at 0). This means that the orbit goes from

    $$z_0\rightarrow z_1 \rightarrow \cdots \rightarrow z_{k-1} \rightarrow z_0 \rightarrow \cdots,$$

    i.e. it is a loop of sorts. This is why we call the orbit periodic and the integer ##k## is the period. If the prof used ##n## for the period, I can see how that can be confusing.

    For a periodic orbit, the Lyapunov exponent reduces because of the periodicity. We can write

    $$\lambda = \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^{n-1} \ln | f'(z_i)|,$$

    as you have in the OP. Now let ##n = Nk## and we take ##n\rightarrow \infty## by taking the limit ##N\rightarrow \infty##. The convenience of this is that, because of the periodicity,

    $$\sum_{i=0}^{n-1} \ln | f'(z_i)| = N \sum_{i=0}^{k-1} \ln | f'(z_i)|.$$

    What's happened here is that, for ##n## much, much larger than ##k##, we essentially have an integer number ##N## of repeats of the periodic orbit. The "round-off" error should vanish in the limit that ##n## and ##N## go to infinity.

    Going back to the formula for the exponent, we then find that

    $$\lambda = \lim_{N\rightarrow \infty} \frac{1}{Nk} N \sum_{i=0}^{k-1} \ln | f'(z_i)| =\frac{1}{k} \sum_{i=0}^{k-1} \ln | f'(z_i)| .$$

    Which is equivalent to the formula you describe, but perhaps written in a less confusing way by using ##k## instead of ##n##.
     
  9. May 2, 2013 #8
    That drastically cleared things up for me. Thank you!
     
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