I am curious to know how to recover the differential equation that goes with a phase portrait. I have seen the following posts but the first one was a $y'$ (and easy enough to "do in my head") and the second one lacks the figure.

As an example, I solved (in Mathematica) the non-linear pendulum equation and plotted the phase portrait as below:

$$y''(t)+\frac{y'(t)}{q}+\sin \left(y(t)\right)=g \cos (\omega t)$$

Where $q$ is damping, $g$ is the forcing term while $\omega$ is the frequency of forcing. The phase portrait ($y'$ vs $y$) is plotted:

enter image description here

So, now if I were given this phase portrait, how do I derive the autonomous differential equation for it? I ask because I do not know the "language" to search for online.

The reason I posted this question here instead of mathematica.SE is because this is more of a Mathematics question and not so much a Mathematica question.

My Mathematica code, in case interested:

g = 1/10; q = 8; \[Omega] = -0.04; TMax = 40;
pSolNL = NDSolveValue[{y''[t] + (1/q) y'[t] + Sin[y[t]] == 
    g  Cos[\[Omega]  t], y[0] == 0, y'[0] == 0}, y, {t, 0, TMax}];

ParametricPlot[{pSolNL[\[Tau]] /. \[Tau] -> t, 
  D[pSolNL[\[Tau]], \[Tau]] /. \[Tau] -> t}, {t, 0, TMax}, 
 AxesLabel -> {"y(t)", "y'(t)"}]
  • $\begingroup$ The system that produced this portrait is nonautonomous, so it's hard to see how to derive an autonomous equation for it. The phase portrait involves a loss of information (time is implicit). You'd be better off fitting y[t] and y'[t] rather than y'[y]. That said, check this out: science.sciencemag.org/content/324/5923/81.abstract $\endgroup$ – Chris K Feb 3 '16 at 15:23

There is no autonomous differential equation on the plane with this phase portrait. Notice that at some points you would need to have at least two tangents to the curve, which is impossible in autonomous systems.

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  • $\begingroup$ So there is no way I can "reverse engineer" the pendulum equation from this? $\endgroup$ – dearN Feb 3 '16 at 0:27
  • $\begingroup$ Never from a single trajectory. But you could try to find the equation by adding $t'=1$ and thus rediscovering yours plus $t'=1$, provided that you had many trajectories and looked at the phase portrait as if the curve evolved also along a third direction (according to $t'=1$). Tortuous, and difficult to be rigorous, never seen it really, but feasible in theory. :-) $\endgroup$ – John B Feb 3 '16 at 0:30
  • $\begingroup$ Converse to my question, is there a way I can change my pendulum differential equation towards a certain phase portrait (without a manual parametric study)? For example, I would like to decrease the number of spirals. Clearly, I would need to find a balance between the damping and frequency and forcing. What kind of stability analysis would I be performing? $\endgroup$ – dearN Feb 3 '16 at 1:04
  • $\begingroup$ In principle, you should increase $q$ and decrease $g$, but you surely know that already. Other than this, I would plot many other solutions. Really, to do something like you are asking for a nonautonomous equation is very much equation dependent (and for the same reason parameter dependent). $\endgroup$ – John B Feb 3 '16 at 1:11

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