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Sure! In classical mechanics it's nice when you have a problem for which the solution is integrable. Oftentimes these solutions will also be periodic. One very useful example of this is an elliptical orbit---the object will return to the same position in the orbit at equal intervals. Unfortunately, there aren't very many problems that are simple enough to have periodic solutions in the real world. While Mars's orbit is approximately periodic, perturbations from Jupiter cause it to vary a little bit, so it's not strictly periodic. Now, we know from observations that the orbit of Mars is almost an ellipse, so these perturbations from Jupiter don't blow up and cause Mars to go careening through the Solar System. Intuitively, it makes sense that this should be possible because if you have some infinitesimally small perturbation, the orbit should remain approximately the same as the unperturbed orbit. The question the KAM theorem answers is "Under what kinds of perturbations do the perturbed orbits remain approximately the same as the unperturbed orbits?" The answer is that if the perturbations are (1) small, (2) smooth, and (3) non-resonant, then the perturbed orbits will be close to the unperturbed orbits. Now, condition (1) doesn't tell you exactly how small those perturbations need to be---only that they exist if they're small enough. Condition (2) will be met for any realistic system---it just prevents you from choosing some pathological perturbation function like the Weierstrass function. Condition (3) basically states that if the perturbation is periodic, then the ratio between its frequency and the frequency of the orbit must be an irrational number. If you had, for example, a 1:1 resonance, so the two frequencies are the same, then the perturbation will be applied at the same spot in every orbit. It will then build up over time until it will eventually have a large effect and the perturbation will no longer be small. The KAM theorem goes further, too. If you look at the unperturbed orbit in phase space, it will just trace out a nice, simple ellipse (or, in general, some closed loop). Since the perturbed orbit will be close to the unperturbed orbit, the KAM theorem also tells you what shape it will have in phase space. By energetic arguments, the KAM theorem tells you that the perturbed orbit is restricted to a torus centered around the unperturbed orbit. This is what's called the invariant torus. Not only is the perturbed orbit restricted to this torus, but given enough time, the perturbed orbit fills out the entire torus. The perturbed orbit will eventually come arbitrarily close to any point on the torus. (I believe this is due to the non-resonance condition---that is, if the perturbed orbit was strictly periodic then there would be a resonance between the perturbation and the orbit---but I'm not sure on this point.) These orbits are known as quasiperiodic because they are confined to a torus centered around a periodic orbit, but they never strictly repeat themselves. |