Is any Hamiltonian system integrable?

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The discussion centers on the integrability of Hamiltonian systems in classical mechanics, specifically questioning the assumption that any Hamiltonian system is integrable according to Liouville's theorem. The argument presented claims that the inverse function provides 2n constants of motion, suggesting functional independence of the coordinates. However, this reasoning overlooks the necessity of the existence of sufficient independent constants of motion for integrability, which is not guaranteed for all Hamiltonian systems. The fallacy lies in assuming that the complexity of the functions does not affect their integrability.

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LLSM
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This is related to classical Hamiltonian mechanics. There is something wrong
in the following argument but I cannot pinpoint where exactly the pitfall is:
Consider an arbitrary (smooth) Hamiltonian (let us assume conservative) and 2n
phase space coordinates (q,p). The Hamiltonian flow gives the evolution
starting the initial conditions (q(q0,p0,t),p(q0,p0,t)). Now, clearly the
inverse function (q0(q,p,t),p0(q,p,t)) provides 2n constants of motion: for
any point (q,p) in the orbit and t one can reconstruct the initial values of
the coordinates, obtaining always the same values. In addition the
transformation from (q,p) to (q0,p0) is a canonical one, so the Poisson
brackets between the various q0 vanish. I would say that obviously the various
q0(q,p,t) are functionally independent. So apparently, any Hamiltonian system
is integrable in the sense of Liouville (the fact that these n functions
q0(q,p,t) are very complicated is a different matter). Can anyone point out
to me where is the fallacy? Thanks.
 
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Is there a reason you think there must be a fallacy? Do you think not all Hamiltonian systems are intrgrable?
 

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