 Solution

Take the separate oscillator energiesThey depend on disjoint canonical coordinate pairs, soThus they are first integrals in involution. Their differentialsare nonzero precisely when $(x_i,y_i)\ne (0,0)$ and have disjoint supports. Hence they are functionally independent on the specified open set.

To understand all further global integrals of motion, use the actionsDirect differentiation gives $d{x}_i\wedge d{y}_i=d{\theta }_i\wedge d{R}_i$, so these are action-angle variables withWhen ${\alpha }_2/{\alpha }_1$ is irrational, every orbit is dense in its fixed-action two-dimensional flat torus. One elementary proof samples the orbit at times $2\pi j/{\alpha }_1$: the first angle returns, and the second undergoes an irrational rotation. These samples are dense in the second circle. Allowing a fixed additional time supplies any desired first angle, proving density in the whole flat torus.

A continuous, globally defined integral of motion must have the same value on an orbit and its closure. Therefore it is constant on every fixed-action flat torus. A $C^1$ such integral has the form $F=\varphi (R_1,R_2)$, with $\varphi$ $C^1$, by evaluating $F$ at a fixed choice of angles. HenceIt cannot be independent of $E_1,E_2$. This proves there are at most two globally independent differentiable integrals. If a third global integral were independent at a point on an excluded coordinate plane, independence would persist in a neighbourhood and hence at nearby points of the regular open set, again a contradiction.

The word global matters: on an angular chart, ${\alpha }_2{\theta }_1-{\alpha }_1{\theta }_2$ is a local integral of motion independent of the actions. It is not single-valued on the flat torus when the frequency ratio is irrational. Thus the assertion would be false for arbitrary local, rather than global, integrals.