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For a dynamical system $\dot{x} = f(x)$, I understand the Poincare map is defined by successive intersections of an (n-1) dimensional surface $\Sigma$ with trajectories in n dimensional phase space.

I can see that a fixed point must correspond to a closed orbit (because trajectories in phase space cannot intersect). Is there also a simple interpretation for how a n-cycle of the Poincare map corresponds to orbits?

I can imagine something like a figure 8 orbit would give us a three cycle however I know this is not possible as trajectories cannot intersect! I cannot see anyother way in which a trajectory could intersect 3 (or more) times with $\Sigma$.

Wooster
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1 Answers1

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In simple terms n-cycle corresponds to a periodic orbit as well. It is easy to see because n-cycle gives you n fixed points of n-th power of the Poincare map.

Pictures like this are very common (here you have 2-cycle) 2-cycle

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demitau
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  • Okay great, thanks for your answer. So in your picture, where would the surface $\Sigma$ be? – Wooster Apr 22 '15 at 13:33
  • You may use the following 2D-square, for example $0\times [-5,-2] \times [-1,2]$. – demitau Apr 22 '15 at 13:48
  • Ah I see - thanks! Is this not perhaps a four cycle actually? I was originally trying to come up with an example in 2d phase space which was making it harder - is this in fact impossible? – Wooster Apr 22 '15 at 19:26
  • If you take the cross-section I specified, you Poincare map will have a periodic orbit of period 2 (of the Poincare map) that is given by intersection of this curve with the cross-section. If you are allowed to take a very large square (which is not always the case) as a cross-section, you get period 4 orbit of the Poincare map. With the same picture you can add as many dimensions you want -- just add several equations describing constant motion -- in those coordinates dynamics would be trivial and in tree frist ones -- like the one on the picture. – demitau Apr 23 '15 at 08:48