Suppose I have a Hamiltonian $H=H(p,q)=H(p,q+1)$ defined on a cylinder $\mathbb{T} \times \mathbb{R}^{+}$, such that $$H(p,q) = \sqrt{p}f(q)$$ where $f(q)=f(q+1)>0$ is a periodic function of $q$, so that we have $p \in \mathbb{R}^{+}$ and $q \in \mathbb{T}$. I want to convert the system to action-angle variables and verify that the time-$1$ map of the flow is twist in those variables.
Hamilton's equations are \begin{equation} \dot{q} = \frac{f(q)}{2\sqrt{p}}, \qquad \dot{p}=-\sqrt{p}f'(q) \label{hamiltonsinneq} \end{equation}
Due to the form of $H$, all its orbits are closed curves and we may define the action-angle variables $(I, \vartheta)$. By definition we have $$I = \int^{1}_{0} p dq = \int^{1}_{0} \frac{H^{2}}{f^{2}(q)} dq = H^{2}\int^{1}_{0}\frac{dq}{ f^{2}(q)} .$$ From this we express $H$ in terms of $(I, \vartheta)$:
$$H = \sqrt{\frac{I}{\int^{1}_{0}\frac{dq}{ f^{2}(q)} }}$$ Since $H$ depends nonlinearly on $I$, and $$\dot{\vartheta} = \omega(I)= \frac{\partial H}{\partial I},$$ the time-$1$ map of $H$ is an integrable map satisfying the twist condition $\frac{d\omega}{dI} \neq 0$.
Can anyone confirm if the above derivation is correct?
