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Suppose the Hamiltonian is time-dependent, $H(t)$. The condition on a time-dependent invariant is

$$ 0 = \frac{d I}{d t } = \frac{\partial I }{\partial t} +\{ I, H\} . $$

We can rewrite it as

$$ \frac{\partial I}{\partial t} = - \{ I, H \} . $$

This is a first order differential equation. For given value of $I(t = 0)$, generally there will be a unique solution, right?

But if so, there would be no chaos.

So, where is wrong?

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  • $\begingroup$ (a) What do you mean by "there would be no chaos"? (b) I think you already got this, but to be sure: you say "For given value of $I(t=0)$", but $I(t=0)$ is still a function of $\mathbf{p}$ and $\mathbb{q}$, so "value" isn't probably the right word. $\endgroup$ – JiK Sep 13 '16 at 8:31
  • $\begingroup$ (a) If $I$ is an invariant, and if the system has only one degree-of-freedom, then the system is confined to the loop of $I = const $. (b) Yes, you can give $I(0)$ as an arbitrary function of $p$ and $q$. $\endgroup$ – J.Bates Sep 13 '16 at 11:38

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