The model is a nonautonomous system
$x'=(2+cos(2 π t))x -0.5 x^2-0.5$
and it can be transformed to the autonomous form by

eqns = {
  x'[t] == (2 + y[t]) x[t] - 0.5 x[t]^2 - 0.5,
  y'[t] == z[t],
  z'[t] == -4 π^2 y[t]}

The equilibria for the autonomous system are caculated as (0.268,0,0) and (3.732,0,0).
I need to analysis the system by a vectorfield to determine the type of equlibria and if limet cycles exist.
Any suggestions would be much appreciated!


I don't think transforming the non-autonomous system into that autonomous system is the right way to go. Notice that the equilibria you mention both have {y, z} == {0, 0}, which contradicts the given periodic forcing.

Instead, I suggest transforming it by adding $dt/dt=1$ (see e.g. Strogatz 2015, p. 8). Plotting that vector field using myStreamPlot (original by @Rahul) from here:

sp = myStreamPlot[{1, (2 + Cos[2 π t]) x - 0.5 x^2 - 0.5},
  {t, 0, 4}, {x, -10, 10}, AspectRatio -> 0.3, ImageSize -> 600,
  FrameLabel -> {"t", "x"}]

Mathematica graphics

You can already make out a stable limit cycle oscillating around $x \approx 4$, but to highlight it with a trajectory from NDSolve:

sol = NDSolve[{x'[t] == (2 + Cos[2 π t]) x[t] - 0.5 x[t]^2 - 0.5,
  x[0] == 8}, x, {t, 0, 4}][[1]];
Show[sp, Plot[x[t] /. sol, {t, 0, 4}, PlotStyle -> Pink]]

Mathematica graphics


Strogatz SH. 2015. Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering. 2nd ed.

  • $\begingroup$ Just to complete this helpful answer: There is a second limit cycle near x~0.27 $\endgroup$ – Ulrich Neumann Jul 23 '19 at 10:44
  • $\begingroup$ @UlrichNeumann Thanks, I overlooked that one. I'd assume it's unstable. $\endgroup$ – Chris K Jul 23 '19 at 10:57
  • $\begingroup$ @Chris K It's a good answer. As pointed out by @UlrichNeumann, the second limit cycle is near 0.27 and it is unstable. But I'm more familiar with the phase diagram with Axes x and x', and the stable limit cycle will converge to a closed curve. Is it possible to plot in this way within the vector field? $\endgroup$ – keanhy14 Jul 23 '19 at 11:16
  • $\begingroup$ @keanhy14 I don't think so. Usually such plots make sense for second-order systems involving x''. Only thing I can think of is wrapping it around a cylinder as done here but I don't think that will be particularly insightful. $\endgroup$ – Chris K Jul 23 '19 at 11:44
  • $\begingroup$ Thanks, and I love the answer in the way with some references, that' great! $\endgroup$ – keanhy14 Jul 23 '19 at 11:51

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