Re: Local Lyapunov Exponents
From: Luca Mihai (luca_at_univ-brest.fr)
Date: 09/14/04
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Date: 14 Sep 2004 15:43:53 -0700
Lou Pecora <pecora@anvil.nrl.navy.mil> wrote in message news:<pecora-EDED4F.11371713092004@ra.nrl.navy.mil>...
> In article <9baf5e12.0409110236.633da19f@posting.google.com>,
> luca@univ-brest.fr (Luca Mihai) wrote:
>
> > I'm interested in calculating the local Lyapunov exponents for example
> > for the zero mean logistic map.
> > If I use the definitions provided by Abarbanel and some methods on
> > numerical integration I obtain an equality between all the local
> > Lyapunov exponents, with the global one.
> > Moreover for the first Local Lyapunov exponents (L=1), it can be
> > proven that it's value is ln(2).
> > Please any advice, or new method will be helpfull.
> > Thank you
>
> How many time steps are you using for the local exponents? Can you
> provide the form of the logistic map and parameter values?
>
> -- Lou Pecora (my views are my own)
Sorry I should have them specified the first time.
The logistic map I use is given by the recurence:
x(n+1) = 1-2*x(n)^2
Now the global Lyapunov exponent is ln(2)
The first local one I calculated using
L(1) = int(log(abs(4*x)*ro(x)),x=-1..1)
where ro(x)=1/pi/sqrt(1-x^2)
I obtain L(1) = ln(2)
For the others with a numerical integration method is the same
Thank you
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