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What are the conditions for classical motion to be periodic? In one dimension, if the motion is bounded, then it is also periodic. However, I don't think this generalizes to higher dimensions.

I am interested in some general condition that does not require obtaining the exact solution. This would be useful, for example, in applying the action-angle coordinate method, which does not require solving the equations of motions, but assumes that the motion is periodic in the first place.

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  • $\begingroup$ Check KAM theorem. All systems with enough number of constants of motion are integrable and from the KAM theorem follows they are periodic in some chosen coordinate system. On the other side, I don't know if there is any other requirement which guarantee periodic motion. $\endgroup$ – kakaz Dec 11 '17 at 19:01
  • $\begingroup$ I am surprised as to why noone said this so far, but the statement "In one dimension, if the motion is bounded, then it is also periodic" is false. Simple example: 1D motion that ends up at a fixed point. Clearly "bounded" but also "not periodic". $\endgroup$ – George Datseris Dec 12 '17 at 8:38
  • $\begingroup$ @GeorgeDatseris, a fixed point is a degenerate case of periodicity, especially in the classification regular vs. chaotic motion. In a discrete-time system (i.e., a map), that's perhaps more natural, as a fixed point is a period-1 orbit (an orbit that "returns" to the same state at every map iteration / time step). $\endgroup$ – stafusa Dec 13 '17 at 10:02
  • $\begingroup$ @stafusa I did not talk about motion on a periodic point, but motion that goes to a periodic point (i.e. tends there for $t\to\infty$). E.g. consider dissipative pendulum. I would never call this motion periodic, yet it is bounded. You are correct about fixed points being a degenerate case of periodicity, but my comment has nothing to do with it. $\endgroup$ – George Datseris Dec 13 '17 at 13:57
  • $\begingroup$ @GeorgeDatseris You're correct. It's more common to care about the asymptotic states, forgetting about the 'transient' ones, so I misread your comment. The orbit you describe is regular (non-chaotic), but indeed not periodic. $\endgroup$ – stafusa Dec 13 '17 at 14:01
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The sufficient, but not necessary condition for a system to present regular motion is that it's integrable. An integrable system has enough constants of motion to restrict the system orbits in phase space $-$ specifically, its phase space is foliated by invariant regular tori. Regular means non-chaotic, i.e., periodic or quasi-periodic motion. As for how to find constants of motion, there are a couple of methods, in particular using Poisson brackets (see also this answer).

Now, there can be plenty of periodic motion in non-integrable systems. In particular, as kakaz pointed out in their comment, the KAM theorem is very relevant here: it states, loosely speaking, that the periodic orbits of a perturbed integrable system will remain regular, as long as they do not resonate with the perturbation and the latter is not too strong, which means that large fractions of the perturbed phase space typically remain regular under small perturbations.

Note that 2D systems are also regular (if they are smooth and otherwise well behaved), and that the question Are there necessary and sufficient conditions for ergodicity? is in some sense a complement of the present question.

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sufficiency condition for a motion to be periodic

I) one generalized coordinate $q$

the equation of motion (all constants are equal to one)

$$\frac{d^2 q}{dt^2}+F=0$$

The conditions that $q(t)$ periodic $(q(t)=q(t+T))$ are:

the force $F$ must be only a function of the generalized coordinate $q$

$F= F(q)$ is equal to:

$F_1=q\,,q^3\,,q^7\,\ldots=q^{2n+1}\quad n=0\,,1\,,2\ldots$ or

$F_2=\sin^{2n+1}(q)$ or

$F_3=\cos^{2n+1}(q)$ or

the potential energy is: $U_i=\int F_i(q)\,dq$

$ U_1=\int F_1(q)\,dq=\frac{1}{2}\frac{q^{2n+2}}{n+1}$

II) two generalized coordinate $q_1\,,q_2$

the equations of motion (all constants are equal to one$

$$\frac{d^2q_1}{dt^2}-q_1\,\dot{q}_2^2+F=0$$

$$\frac{d q_2}{dt}\,q_1^2=\text{const}$$

The conditions that q_1(t) periodic are:

the force F must be only a function of the generalized coordinate $q_1$

$ F=F(q_1)$ is equal to:

$\frac{1}{q_1^2}\,,\frac{1}{q_1^4}\,,\frac{1}{q_1^6}\,,\ldots$

the potential energy is:

$U=\int F(q_1)\,dq_1=\frac{1}{2n-1}\,q_1^{1-2 n}\,,\quad n=1\,,2\,,\ldots$

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