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I have a question regarding KAM curves.

KAM theory says that if we add a very small non-integrable Hamiltonian

perturbation to an integrable Hamiltonian system, then most of the

invariant tori are preserved if the perturbation is very small and the

measure of the destroyed invariant tori scales with the magnitude of

the perturbation.

Now, are the KAM curves for perturbed Hamiltonian systems "exact

invariants" or "invariants upto all orders". Since the perturbation is

non-integrable, I would believe that the KAM curves would only be

"invariants upto all orders" and not really be "exact invariants". Is

this right?

If the above is right, then, in addition to Arnold Diffusion, should

not there be an exponentially slow diffusion across the KAM curves

also since they are not exact invariants? And if the above statement

is wrong, then how does a non-integrable perturbation happen to

conserve the integrability of the Hamiltonian in certain regions in

phase-space?

Thanks,

Kushal.

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# KAM curves

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