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I’m relatively new to Mathematica and also new to this forum. In fact, this is my first question so please apologise if I do some formatting errors. I want to calculate the Lyapunov exponent of a driven and damped spherical pendulum. For this, I tried to use the code provided by Chris K. for my problem. Which is not working properly. To see if I did a general mistake, I calculated the Lyapunov exponent for a simple damped and driven pendulum and the code from Chris K. works perfectly. Which brings me to my four questions for my problem. If you can only answer one question, please do so every help is very much appreciated.

  1. The code is only working if the damped pendulum which is not driven. After I switch on the excitation (change excitation amplitude from $0$ to e.g. $0.05 m$) the code is not working anymore. According to the bifurcation logistic map of this spherical pendulum, there should be a chaotic behaviour for excitation amplitudes $U_0$ in areas from $0.01-0.055 m$ and from $0.8-0.1 m$. Why is it that way that Chris K. code is not working anymore after the excitation is included?
  2. I cannot seem to find the option to switch on the axis labels? Answered by ChrisK:LyapunovExponents[eqns, ics, ShowPlot -> True, PlotOpts -> {AxesLabel -> {"iteration", "exponent"}}]
  3. I want to plot a graph that shows the Lyapunov exponent over the bifurcation parameter x-axis: $U_0$ , y-axis: $\theta(t)$ or $y(t)$ (after state space for) like in this question by Jarek Mazur. Is there a way to do this for my problem preferably without using AUTO-07p?
  4. Even though the code works for the unforced spherical pendulum, loads of error messages are produced. Is that normal?

The ODEs for the spherical pendulum are as follows: $$ \theta ''(t) +2 \zeta _{\theta } \omega _n \theta '(t)+ \frac{g \sin (\theta (t))}{l} - \sin (\theta(t))\cos (\theta (t)) \phi '(t)^2 =- \frac{U _0 \Omega _u^2 \cos (\theta (t)) \sin (\phi (t))\cos(t \Omega _u)}{l}; \phi ''(t)+\frac{2 \zeta _{\phi } \omega _n}{\sin^2 (\theta (t))} \phi '(t)+\frac{2 \theta '(t) \cos(\theta (t)) \phi '(t)}{\sin(\theta (t))}=-\frac{U_0 \Omega _u^2 \cos (\phi (t)) \cos (t \Omega_u)}{l\sin (\theta (t))} $$

The ODEs are converted into the state space form that is required for the code from Chris K.

$$ x'(t)=-2. \zeta _{\theta } \omega _n x(t)-\frac{ g \sin (y(t))}{l}+0.5 z(t)^2 \sin (2 y(t))-\frac{U_0 \Omega _u^2 \sin (w(t)) \cos (y(t)) \cos (t \Omega_u)}{l}; y'(t)=x(t); z'(t)= -\frac{2 \zeta _{\phi } \omega _n}{\sin^2 (y (t))} z(t)-\frac{2 x(t) \cos (y(t)) z(t)}{\sin(y(t))}-\frac{U_0 \Omega _u^2 \cos (w(t)) \cos (t \Omega _u)}{l\sin (y(t))}; w'(t) = z(t) $$

As mentioned before I used Chris K. GramaSchmidt and LyapunovExponent function and added my code and variables which are as follows:

l = 0.5                                   
g = 9.81                               
Subscript[\[Omega], n] = Sqrt[g/l]      
Subscript[\[CapitalOmega], u] = Subscript[\[Omega], n]  
Subscript[U, 0] = 0.05                     
Subscript[\[Zeta], \[Theta]] = 0.0025      
Subscript[\[Zeta], \[Phi]] = 0.0025        

Equations for the spherical pendulum in state-space form

steq1 = Derivative[1][y][t] == x[t]   
steq2 = Derivative[1][x][t] == -((1.*g*Sin[y[t]])/l) - (Cos[y[t]]*1.*Cos[t*Subscript[\[CapitalOmega],u]]*Sin[w[t]]*Subscript[U, 0]*Subscript[\[CapitalOmega], u]^2)/l - 2.*Subscript[\[Zeta], \[Theta]]*Subscript[\[Omega], n]*x[t] + 0.5*Sin[2.*y[t]]*z[t]^2  
steq3 = Derivative[1][w][t] == z[t]    
steq4 = Derivative[1][z][t] == (1/(0.5 - 0.5*Cos[2.*y[t]]))*(-((1.*Cos[t*Subscript[\[CapitalOmega],u]]*Cos[w[t]]*Sin[y[t]]*Subscript[U, 0]*Subscript[\[CapitalOmega], u]^2)/l) - (2.*Subscript[\[Zeta], \[Phi]]*Subscript[\[Omega], n] + 1.*Sin[2.*y[t]]*x[t])*z[t])    
eqns = {steq2, steq1, steq4, steq3}  
  
ics = {x -> 0, y -> 0.78, z -> 0., w -> 0.78}    
  
LyapunovExponents[eqns, ics, ShowPlot -> True]  

Thank you very much for your help.

Edit: After some consideration, I realised, that the proposed parameters for the pendulum make the pendulum instable. This is why I choose to increase the damping ratio and decrease the excitation frequency as follows.

l = 0.5                                   
g = 9.81                               
Subscript[\[Omega], n] = Sqrt[g/l]      
Subscript[\[CapitalOmega], u] = 3 
Subscript[U, 0] = 0.05                     
Subscript[\[Zeta], \[Theta]] = 0.05     
Subscript[\[Zeta], \[Phi]] = 0.05 

I also updated the initial conditions:

steq1 = Derivative[1][y][t] == x[t]   
steq2 = Derivative[1][x][t] == -((1.*g*Sin[y[t]])/l) - (Cos[y[t]]*1.*Cos[t*Subscript[\[CapitalOmega],u]]*Sin[w[t]]*Subscript[U, 0]*Subscript[\[CapitalOmega], u]^2)/l - 2.*Subscript[\[Zeta], \[Theta]]*Subscript[\[Omega], n]*x[t] + 0.5*Sin[2.*y[t]]*z[t]^2  
steq3 = Derivative[1][w][t] == z[t]    
steq4 = Derivative[1][z][t] == (1/(0.5 - 0.5*Cos[2.*y[t]]))*(-((1.*Cos[t*Subscript[\[CapitalOmega],u]]*Cos[w[t]]*Sin[y[t]]*Subscript[U, 0]*Subscript[\[CapitalOmega], u]^2)/l) - (2.*Subscript[\[Zeta], \[Phi]]*Subscript[\[Omega], n] + 1.*Sin[2.*y[t]]*x[t])*z[t])    
eqns = {steq2, steq1, steq4, steq3}  
  
ics = {x -> 0.78, y -> 0.78, z -> 0.78, w -> 0.78}    
  
LyapunovExponents[eqns, ics, ShowPlot -> True] 

This gives me the following results:

{-0.0850468, -0.213523, -0.213502, Indeterminate}

However, the last Lyapunov exponent can't be calculated. Did anyone else had a similar issue?

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  • $\begingroup$ Is LyapunovExponents a new Mathematica function (>v12) ? $\endgroup$ Commented Jul 1, 2020 at 11:15
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    $\begingroup$ @UlrichNeumann no, not in v12.1.1 - LyapunovSolve exists but it's unrelated. $\endgroup$
    – flinty
    Commented Jul 1, 2020 at 11:34
  • $\begingroup$ For calculating Lyapunov exponents there are some answers here which may be helpful: mathematica.stackexchange.com/a/17608/72682 mathematica.stackexchange.com/a/179470/72682 $\endgroup$
    – flinty
    Commented Jul 1, 2020 at 11:38
  • $\begingroup$ @UlrichNeumann no the LyapunovExconents in not a new function in Mathematica. I used the prior work ofChris K $\endgroup$
    – Philipp
    Commented Jul 1, 2020 at 12:05

1 Answer 1

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Not an answer, merely some observations. It seems that the problem might come from NDSolve not LyapunovExponents. If you simulate the system long enough, NDSolve runs into trouble:

tmax = 10000;
sol = NDSolve[Join[eqns, {x[0] == 0, y[0] == 0.78, z[0] == 0, w[0] == 0.78}],
  {x, y, z, w}, {t, 0, tmax}];
(* NDSolve::ndcf -- Repeated convergence test failure at t == 940.4341901984399`; unable to continue. *)

Running for a shorter time gives some clues:

tmax = 20;
sol = NDSolve[Join[
   eqns, {x[0] == 0, y[0] == 0.78, z[0] == 0, w[0] == 0.78}], {x, y, 
   z, w}, {t, 0, tmax}];

GraphicsColumn[{
  Plot[Evaluate[x[t] /. sol], {t, 0, tmax}, PlotRange -> All],
  Plot[Evaluate[y[t] /. sol], {t, 0, tmax}, PlotRange -> All],
  Plot[Evaluate[z[t] /. sol], {t, 0, tmax}, PlotRange -> All],
  Plot[Evaluate[w[t] /. sol], {t, 0, tmax}, PlotRange -> All]
}]

Mathematica graphics

Notice that when y[t] goes through zero, z[t] takes a rapid excursion. I suppose that's due to the denominator of z'[t] having being zero when y[t]==0.

Hopefully someone with more knowledge of spherical pendulums or NDSolve issues can weigh in.

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  • 1
    $\begingroup$ on geneal question to your LyapunovExponents code, is it possible to include AxesLabel or FrameLabel in the plots? $\endgroup$
    – Philipp
    Commented Jul 1, 2020 at 13:10
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    $\begingroup$ @Philipp Oops, I meant to add that option last time I edited the code, but forgot part. I updated it so that it should work now. Try e.g. LyapunovExponents[eqns, ics, ShowPlot -> True, PlotOpts -> {AxesLabel -> {"iteration", "exponent"}}] $\endgroup$
    – Chris K
    Commented Jul 1, 2020 at 15:22
  • $\begingroup$ Re: looping over a parameter, right now you'll need to do that manually. But I suppose you've got it get it working for a single value first. $\endgroup$
    – Chris K
    Commented Jul 1, 2020 at 15:24
  • $\begingroup$ Thank you @Chris K $\endgroup$
    – Philipp
    Commented Jul 1, 2020 at 17:20
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    $\begingroup$ @Philipp The computation is in error when the denominator of the expression for z'[t], namely, (0.5 - 0.5 Cos[2. y[t]]), becomes very close to zero. $\endgroup$
    – bbgodfrey
    Commented Jul 3, 2020 at 2:18

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