# Discrete-time Lyapunov exponent plot

I am trying to create a Lyapunov exponent plot as a function of $$\alpha$$ for two functions:

$$f(x) = (\alpha + 1)x - \alpha x^{3}$$

and a piecewise function given by following Mathematica code:

f[x] = Piecewise[{{-1, x < -1}, {1, x > 1}, {x * (1 - alfa) + alfa, 1 >= x > alfa}, {x * (1 - alfa) - alfa, -alfa > x >= -1}, {x * (2 - alfa), alfa >= x >= -alfa}}]


Here you have an appropriate manipulation plot for second function:

Manipulate[Plot[Piecewise[{{-1, x < -1}, {1, x > 1}, {alfa + (1 - alfa) x,  1 >= x > alfa}, {-alfa + (1 - alfa) x, -alfa > x >= -1}, {(2 - alfa) x, alfa >= x >= -alfa}}, 0], {x, -2, 2}], {alfa, 0, 1}]


There is already a similar post here but I don't know how to apply that solution to my case. Is it even possible for discontinuous functions like my second piecewise function?

I would like to achieve a plot similar to those made for logistic map found on the internet: Something like this?

g[x_, alfa_] := (alfa + 1) x - alfa x^3;

p[x_, alfa_] := Piecewise[
{{-1, x < -1},
{x (1 - alfa) - alfa, -alfa > x >= -1},
{x (2 - alfa), alfa >= x >= -alfa},
{x (1 - alfa) + alfa, 1 >= x > alfa},
{1, x > 1}}
];

lyapunov[f_, x0_, alfa_, n_, tr_: 0] := Module[
{df, xi},
df = Derivative[1, 0][f];
xi = NestList[f[#, alfa] &, x0, n - 1];
(1/n) Total[Log[Abs[df[#, alfa]]] & /@ Drop[xi, tr]]
];


The function lyapunov computes the Lyapunov exponent of the function f as $$\lambda = \frac{1}{n} \sum_{i=0}^{n-1} \ln \left| \, f^\prime (x_i) \right|$$ where $x_{n+1} = f(x_n)$. The starting position is given as x0 and n steps are taken, with the first tr steps excluded.

Applying this to g and p, we can make plots by varying alfa:

gtable = Table[{alfa, lyapunov[g, 0.5, alfa, 10000, 5000]}, {alfa, 0, 2, 0.01}];

ptable = Table[{alfa, lyapunov[f, 0.2, alfa, 10000, 5000]}, {alfa, 0, 1, 0.01}];

ListPlot[#,
Frame -> True,
FrameLabel -> {"\[Alpha]", "\[Lambda]"},
Joined -> True] & /@ {gtable, ptable} This first is the plot for g, and looks like the map you posted. But this second, the plot for p, is more unusual.

• That is exactly what I needed. Thank you very much for a quick response. Apr 30, 2015 at 16:05
• @JarekMazur: best of luck! Apr 30, 2015 at 18:53