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Does the curve traced out by the tip of a double pendulum have cusps?

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  • $\begingroup$ I don't know, I would need the expression of the curve to tell. But speaking qualitatively from the animation, and complete guesswork, it looks like it is possible for the pendulum to be left shortly in some extreme position, only to be jerked back in a non-smooth manner. $\endgroup$ – Mankind Feb 26 '16 at 1:40
  • $\begingroup$ The path is smooth in $\mathbb R^4$, simply because of the right-hand sides of the differential equations. But in general it is not smooth in the plane (or would you in fact consider smooth the trajectory of a simple pendulum inside the heteroclinics?). $\endgroup$ – John B Feb 26 '16 at 1:50
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Yes, the curve may have cusps: If you release the pendulum in a state of rest, the resulting motion can be mirrored in time to obtain a trajectory on which the pendulum retraces its path, which thus "ends" at the point at which the pendulum was released. Here the time derivative is zero and changes sign, thus satisfying the definition of a cusp.

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  • $\begingroup$ I'll edit the question to be more precise. $\endgroup$ – Mark B Feb 26 '16 at 2:04
  • $\begingroup$ @MarkB: OK, I've edited the answer in response. $\endgroup$ – joriki Feb 26 '16 at 2:11
  • $\begingroup$ I guess this supposes (for a given set of IC) the existence of a solution for all time. I assume this is true, but why? $\endgroup$ – Mark B Feb 26 '16 at 2:22
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    $\begingroup$ @MarkB: I'm sure there's a mathematical theorem to prove this, but the most direct proof is physical: If you release a double pendulum, it does something :-) $\endgroup$ – joriki Feb 26 '16 at 2:23
  • $\begingroup$ I guess my main hangup is that in order for the derivative to change sign, it needs to be defined for $t<0$, and I'm not sure how that's true from a physical point of view. $\endgroup$ – Mark B Feb 26 '16 at 2:27

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