# Existence of time-dependent invariants for a classical Hamiltonian system

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?

• (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. – JiK Sep 13 '16 at 8:31
• (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 $q$. – J.Bates Sep 13 '16 at 11:38