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?

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) )

Say I define a time dependent vector field $\Psi(t):\mathbb{R}^d\to \mathbb{R}^d$ as reversible 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?

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?

Say I define a time dependent vector field $\Psi(t):\mathbb{R}^d\to \mathbb{R}^d$ as reversible 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?

Say I define a time dependent vector field $\Psi(t):\mathbb{R}^d\to \mathbb{R}^d$ as reversible 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?