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It's said in elementary classical mechanics texts that the phase trajectories of an isolated system can't cross. But clearly they can, for example for the pendulum, the trajectories look like this:

The points where they cross correspond to unstable equilibrium. Is there some sort of general result about this, when can phase trajectories indeed cross?

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    $\begingroup$ Arnold's? :) love that book! $\endgroup$ – Phoenix87 Jan 31 '15 at 11:37
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Look carefully at it, especially at the arrows. You should think of the "intersections" as limits of the two different situations (the loops and the waves). Trajectories that actually reach the unstable equilibrium will remain there indefinitely, so there is no crossing over the potential peak. If you miss that peak by a tiny amount it will look like you have reached it, but you then fall back to your steps again. So trajectories really don't intersect in Hamiltonian mechanics!

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Comments to the question (v1):

  1. The property of non-crossing trajectories holds more generally for (not necessarily Hamiltonian) autonomous$^1$ 1st order ODEs $$\dot{z}^I~=~f^I(z),$$ where the functions $f^I$ are assumed to be sufficiently regular in order to ensure local uniqueness of the solutions. A sufficient condition for local uniqueness is if $f$ is Lipschitz continuous.

  2. Crossing trajectories could possibly happen for sufficiently singular (Hamiltonian) systems, such as, e.g. Norton's Dome $$ H(q,p)~=~\frac{1}{2}p^2 - \frac{2K}{3}|q|^{3/2}, $$ cf. e.g. this Phys.SE post.

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$^1$ Autonomous means no explicit time dependence.

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  • $\begingroup$ Thanks, that's an interesting example with Norton's dome! $\endgroup$ – Spine Feast Feb 1 '15 at 14:49

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