Finding the Period of a Limit Cycle

Question

I'm interested in the periods of limit cycles of the Wilson-Cowan equations which have the form $$x'(t) = -x + S(ax(t) - by(t) +e)$$ $$y'(t) = -y + S(cx(t) - dy(t) + f)$$

where $$S(x) = 1 + \frac{\tanh(\frac{x}{2})}{2}$$

You can observe a limit cycle with the parameters seen in the following code:

    s[x_] := (1  + Tanh[x/2]/2);
a = 10;
b = 10;
c = 10;
d = -5;
e = -0.75;
f = -15;

wc = {-x + s[(a*x) - (b*y) + e], -y + s[(c*x) - (d*y) + f]};
T = 40;
point = {0.77, 0.29};
LimCyc = ParametricPlot[
   Evaluate[
    First[{x[t], y[t]} /. 
      NDSolve[{x'[t] == -x[t] + s[(a*x[t]) - (b*y[t]) + e], 
        y'[t] == -y[t] + s[(c*x[t]) - (d*y[t]) + f], 
        Thread[{x[0], y[0]} == point]}, {x, y}, {t, 0, T}]]], {t, 0, 
    T}, PlotStyle -> Red];

Show[StreamPlot[wc, {x, 0, 2}, {y, 0, 2}, PlotRangePadding -> 0, 
  ImageSize -> {500, 500}], LimCyc]

Is there an easy way to numerically compute the period of a limit cycle for a given set of parameters?

Answer 1

Although it's primarily designed for ecological models, my EcoEvo package can help. First, you need to install it with

PacletInstall["EcoEvo", "Site" -> "http://raw.githubusercontent.com/cklausme/EcoEvo/master"]

Then, load the package and define your model:

<< EcoEvo`;
S[z_] := 1 + Tanh[z/2]/2;
SetModel[{
  Aux[x] -> {Equation :> -x + S[a x - b y + e]},
  Aux[y] -> {Equation :> -y + S[c x - d y + f]}
}]

Doublecheck that it matches your results:

a = 10; b = 10; c = 10; d = -5; e = -0.75; f = -15;
sol = EcoSim[{x -> 0.75, y -> 0.25}, 20];
Show[
 PlotEcoStreams[{x, 0, 2}, {y, 0, 2}],
 RuleListPlot[sol, PlotStyle -> Pink]
]

Now use the final result from the simulation as an initial guess for FindEcoCycle:

ec = FindEcoCycle[FinalSlice[sol]];
PlotDynamics[ec]

The period can be found as the final time of ec:

FinalTime[ec]
(* 5.27899 *)

As a bonus, you can calculate Floquet multipliers with EcoEigenvalues:

EcoEigenvalues[ec]
(* {3.6338*10^-7, -0.71155} *)

If you want to avoid the package, the idea is to warm up the simulation, look for a maximum in one variable (say x), take a tiny step further, then use WhenEvent to look for when you return to that point. There's also a Method using FindRoot.

Answer 2

Here is a simple approach to get the period of the unknown limit cycle. The idea is to approximate the limitcycle by a circle (1st harmonic) around the mean of the limitcycle:

solution NDSolve

XY = NDSolveValue[{x'[t] == -x[t] + s[(a*x[t]) - (b*y[t]) + e],y'[t] == -y[t] + s[(c*x[t]) - (d*y[t]) + f],Thread[{x[0], y[0]} == point]}, {x, y}, {t, 0, T}]

some data of the last points

txy = Table[ { t , Norm[ Through[XY[t]]] } , {t,Subdivide[T/2, T, 100]}];

Fit the circle

{m1, m2} = NIntegrate[Through[XY[t]], {t, T/2, T}]/(T/2);
mod = NonlinearModelFit[txy, {Norm[{m1, m2} +r {Cos[2 Pi t/T1 - \[Alpha]1], Sin[2 Pi t/T1 - \[Alpha]1]}],r > 0}, { r, T1, \[Alpha]1}, t, Method -> "NMinimize"]
mod["BestFitParameters"]
(*{r -> 0.406525, T1 -> 5.28612, \[Alpha]1 -> 2.39255}*)

the period of the limitcycle T1 -> 5.28612

check result

Plot[ Evaluate[Through[XY[t]]] , {t, T/2, T},GridLines ->Evaluate[{{T - T1, T}, None} /. mod["BestFitParameters"]]]

Answer 3

I'll elaborate on the method indicated by @ChrisK involving use of WhenEvent to find a pair of maxima. Here I find a bunch of such pairs and take differences. It will be clear that they converge.

s[x_] := (1 + Tanh[x/2]/2);
{a, b, c, d, e, f} = {10, 10, 10, -5, -0.75, -15};
T = 40;
point = {0.77, 0.29};

We find both max and min values un y[t] (could also do this for x[t] but one suffices). THis is done by recording values t for which y'[t] vanishes.

extrema = 
  Reap[NDSolveValue[{x'[t] == -x[t] + s[(a*x[t]) - (b*y[t]) + e], 
      y'[t] == -y[t] + s[(c*x[t]) - (d*y[t]) + f], 
      Thread[{x[0], y[0]} == point], 
      WhenEvent[y'[t] == 0, Sow[t]]}, {x[t], y[t]}, {t, 0, 3 T}]][[2, 
    1]];

We want to go from peaks to peaks and valleys to valleys so we find time differences between extrema located two apart.

Differences[Partition[extrema, 2]]
(* Out[457]= {{5.38632, 5.29292}, {5.2813, 5.27931}, {5.27904, 
  5.27899}, {5.27899, 5.27899}, {5.27899, 5.27899}, {5.27899, 
  5.27899}, {5.27899, 5.27899}, {5.27899, 5.27899}, {5.27899, 
  5.27899}, {5.27899, 5.27899}, {5.27899, 5.27899}, {5.27899, 
  5.27899}, {5.27899, 5.27899}, {5.27899, 5.27899}, {5.27899, 
  5.27899}, {5.27899, 5.27899}, {5.27899, 5.27899}, {5.27899, 
  5.27899}, {5.27899, 5.27899}, {5.27899, 5.27899}, {5.27899, 
  5.27899}} *)

And 5.27899 falls out as the period.