Changing to action-angle variables to reduce the degrees of freedom

I have a Hamiltonian with two degrees of freedom, and when I change to action angle variables, one of the action variables does not appear in the final Hamiltonian. The reason seems to be because the redundent action variable shifts the area of the second variable and therefore does not appear in the final energy expression.

I have something of the form

$$H(x_1,x_2,p_1,p_2)=2-cosleft(frac{2pi}{L_1}p_1+frac{2pialpha x_2}{L_2}right)-cosleft(frac{2pi }{L_1}p_2right)$$ where $p_iin[0,L_1)$, $x_iin[0,L_2).$ Since $x_1$ is cyclic $p_1=kappa$ is constant therefore the first action variable is therefore
$$J_1=frac{1}{2pi}oint p_1dx_1=frac{kappa L_2}{2pi}.$$ Because of the periodic nature of the problem the orbits hit the boundary and continue to form a loop. The turning points are
$$x_m^{pm}=-frac{L_2kappa}{a L_1}pmfrac{L_2}{2pi a}arccos(1-E)+frac{L_2 m}{a}.$$ The left and right turning points for an orbit with energy $E=0.44$ are $x_0^-$ and $x_0^+$ respectively. Calculating the remaining action variable we find
$$J_2=frac{1}{2pi}oint p_2dx_2=2L_1(x_0^+-x_0^-)-int_{x_0^-}^{x_0^+}p’_2dx_2$$
where $p’_2=frac{L_1}{2pi}arccos(2-E-cos(frac{2pi}{L_1}kappa+frac{2pi a}{L_2}x_2)).$ On then see that changing variables, $kappa$ is eliminated from $J_2$. Therefore, $$H(J_1,J_2)=H(J_2)=E.$$ Do I then treat this as a one dimensional classical system? From $$omega_1=frac{partial H}{partial J_1}=0,$$ this does seem that the motion is indeed around the $J_2$ torus. Is this correct? Does anybody know of a similar problem in classical mechanics? I am also interested in the quantum spectra of such a system, this seems to imply that the energy contour has no curvature, this would affect the eigenvalues available, and cause degeneracy… Any feedback would be greatly appreciated.