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KAM curves

  1. Jul 21, 2008 #1

    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

  2. jcsd
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