Are Hamilton's equations reversible?

Question

Say I define a time dependent vector field $\Psi(t):\mathbb{R}^d\to \mathbb{R}^d$ as reversible (also here) if, for $f(x,y)=(x,-y)$, we have:

$$ f\circ \Psi \circ f =\Psi(-t)=\Psi^{-1}(t).$$

Just to clarify $\Psi:[0,T]\times \mathbb{R}^d\to \mathbb{R}^d$. Now let $\Psi$ be some Hamiltonian dynamics $\Psi=(q,p)$, where:

$$ \frac{d q}{dt}=\nabla_p H (q,p), \ \ \ ~q(0)=x, $$ $$ \frac{d p}{dt}=-\nabla_q H (q,p), \ \ \ ~p(0)=y, $$ where $H(q,p)=H(q,-p)$. Is it obvious that $\Psi$ is reversible? Can I use Reversibility of Hamiltonian dynamics somehow?

(Note the 2nd equality $\Psi(-t)=\Psi^{-t}(t)$ follows since $(q,p)$ is a flow. https://en.wikipedia.org/wiki/Flow_(mathematics) )

Answer 1

1. A dynamical equation system with dynamical variables $z$ is called reversible if it is invariant under the combination $(t\to -t, z\to I(z))$ where $I$ is an involution$^1$, cf. Ref. 1.

2. Main example: Any autonomous Hamiltonian $H(q,p)=H(q,-p)$ that is even in momenta describes a reversible system. Here the involution is $I(q,p)=(q,-p)$.

3. Concerning OP's title question, it is possible to find non-reversible non-autonomous Hamiltonian systems. Think e.g. on a Hamiltonian of the form $H(q,p,t)=f(t)q+g(t)p$ for 2 appropriate functions $f,g$.

4. More interestingly, Ref. 1 claims [e.g. around eq. (1.29)] that there exist non-reversible autonomous Hamiltonian systems.

References:

1. J.A.G. Roberts & G.R.W. Quispel, Chaos and time-reversal symmetry. Order and chaos in reversible dynamical systems. Phys. Rep. 216 (1992) 63.

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$^1$ More generally in a geometric language: If $M$ denotes the manifold of dynamical variables (sans time $t$), then the map $I:\Gamma(TM)\to \Gamma(TM)$ is a (possibly time-dependent, not necessarily integrable) mixed tensor field, such that it is pointwise an involution $\forall z\in M: I_z^2={\bf 1}_{TM}$. This is often called an almost product structure in the literature.