Lyapunov exponent of the logistic map (Mathematica Code)

In a previous post I’d shown a way to get the Lyapunov exponent from the time series data of any map. In this quick tutorial, I’ll show you a cleaner way to get the Lyapunov exponent for the specific case of the logistic map, and then using a really short script in Mathematica, plot it against r.

First the mathematical analysis that simplifies the expression for the Lyapunov exponent for a map, and particularly the logistic map. This discussion follows this article.

Suppose the initial infinitesimal perturbation is $\delta x_{0}$ . Then we have, for $n\rightarrow\infty$ :

$\left|\delta x_{n}\right|=\left|\delta x_{0}\right|e^{\lambda n}\Rightarrow e^{\lambda n}=\underset{\delta x_{0}\rightarrow0}{\lim}\left|\frac{\delta x_{n}}{\delta x_{0}}\right|=\left|\frac{dx_{n}}{dx_{0}}\right|$ .

But we can write:

$\frac{dx_{n}}{dx_{0}}=\frac{dx_{n}}{dx_{n-1}}\frac{dx_{n-1}}{dx_{n-2}}\ldots\frac{dx_{1}}{dx_{0}},$

where each $x_{i}=f(x_{i-1})$ . So we have:

$\frac{dx_{n}}{dx_{0}}=f'(x_{n-1})f'(x_{n-2})\ldots f'(x_{0}).$

Therefore,

$\;e^{\lambda n}=\left|f'(x_{n-1})f'(x_{n-2})\ldots f'(x_{0})\right|$

$\Rightarrow\lambda(r;x_{0})=\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\ln\left|f'(x_{n-1})f'(x_{n-2})\ldots f'(x_{0})\right|$

$=\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\sum\limits_{k=0}^{n-1}\ln\left|f'(x_{k})\right|.$

For the logistic map, $f'(x)=r(1-2x)$ .
So we have:

$\lambda(r;x_{0})=\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\sum\limits_{k=0}^{n-1}\ln\left|r(1-2x_{k})\right|$ .

We can put the above formula in a short Mathematica script to obtain $\lambda$ as a function of $r$ and plot it. The following is the code:

\[Lambda][r_] := Module[{f, l},
   f[x_] := r x (1 - x);
   l[x_] := Log[Abs[r (1 - 2 x)]];
   Mean[l[NestList[f, 0.1, 1*^2]]]];
Plot[\[Lambda][r], {r, 0, 4}, PlotStyle -> Thickness[.0001], 
 AxesLabel -> {"r", "\[Lambda](r)"}]

And the following is the output:
In the line that uses Nestlist, we specify the starting point of the trajectories. However, I noticed that the output does not depend on the starting point.

5 thoughts on “Lyapunov exponent of the logistic map (Mathematica Code)”

Pingback: Calculating the Lyapunov Exponent of a Time Series (with python code) | One Life

Hi, very good. But, and of system dynamics, i. e., flow?

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November 6, 2017 at 12:34 pm Reply

Is it l or f ? in the function

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November 12, 2018 at 10:21 am Reply

if the function is sin(x) then can you draw it graphically its lyaponuv exponent?

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April 5, 2022 at 7:43 am Reply

Hey Neel!

Tanks a lot for your help respective the code. I have the problem, that this code isn’t working in Mathematica. It just shows the axes but without the graphics. Does this code is still working on your program?

I`d be very thankful for your help!

Best regards

David Reiter

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October 28, 2022 at 4:10 am Reply