Skip to main content

plotting - 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:





Answer



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}


gplot


This first is the plot for g, and looks like the map you posted.


pplot


But this second, the plot for p, is more unusual.


Comments

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...