Suppose I have the following Hamiltonian $$H(q, p, x,y) = \frac{p^{2}}{2} - \frac{q^{2}}{2} \left(\left(\frac{y^{2}}{2} + \omega^{2}x^{2}\right)^{2} - \frac{q^{2}}{2}\right)$$
Where $(q, p, x,y) \in \mathbb{R}^{4}$. The system is integrable, with two integrals $H$ and $J = y^{2}/2 + \omega^{2}x^{2}$. We have Hamilton's equations
$$\dot{q} = p$$ $$\dot{p} = q\left(\left(\frac{y^{2}}{2} + \omega^{2}(\tau)x^{2}\right)^{2} - q^{2}\right)$$ $$\dot{x} = -q^{2}\left(y\left(\frac{y^{2}}{2} + \omega ^{2}(\tau)x^{2}\right)\right)$$ $$\dot{y} = 2\omega^{2}(\tau)xq^{2}\left(\frac{y^{2}}{2} + \omega ^{2}(\tau)x^{2}\right)$$
The fixed point $(0,0,0,0)$ is degenerate with $4$ zero eigenvalues, hence it has a $4$ dimensional center manifold. However, in terms of $(q, p)$ (considering $(x,y)$ fixed), the point $(0,0)$ is saddle with homoclinic connection. In terms of $(x,y)$ variables, for fixed $(q, p)$ we have a center equilibrium at $(x,y) = (0,0)$ surrounded by periodic orbits.
I would like to analyse the global topology of the phase space of this system. To this end, I hope to find an appropriate change of coordinates.
My question is, can we replace at least one pair of canonically conjugated variables $(x,y)$ and $(q,p)$ by $\textbf{global}$ action-angle variables in this system?
Considering that the system is a center in $(x,y)$ only, is it correct to introduce action-angle variables $(I, \theta)$ through the transformation $(q,p,x,y) \mapsto (q,p, I(x,y), \theta(x,y))$ ? Or must action - angle variables also depend on $(q,p)$ ?