Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 20 / 2

A function $F$ is an integral of motion of $H$ when it is constant along every trajectory of the Hamiltonian flow. Infinitesimally,The antisymmetry of a Lie bracket of the Poisson bracket gives $X_{F}H=\{H,F\}=0$. Hence $H$ is also an integral of motion of $F$.

For the commuting assertion, ${\mathcal{L}}_{X_F}\omega =d({\iota }_{X_F}\omega )=d^{2}F=0$, because $\omega$ is closed. The contraction identity for a Lie bracket of vector fields then givesSince the symplectic form is a nondegenerate bilinear form at every point, the Hamiltonian vector fields commute:With the chosen convention the general identity is $[X_F,X_H]=X_{\{H,F\}}$. Equivalently, it follows by applying the Jacobi identity for the Poisson bracket to an arbitrary test function. Local flows commute whenever both compositions are defined; no global completeness assumption is needed.

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.

The Hamiltonian perturbation preserving one regular level is explicit:Linearity of the Hamiltonian vector field in the differential of a smooth map givesIt vanishes on ${\Sigma }_0$. At a point of ${\Sigma }_c$, its contraction with the symplectic form isSince the differentials are independent there, this vanishes exactly when $c=0$. Thus on every nonempty nonzero regular fibre the two Hamiltonian vector fields differ at every point, not merely as functions somewhere:Mutual Poisson commutation of the $G_j$ is unnecessary; the given first-integral property does not by itself make the perturbed flow preserve every common level, nor is that requested.

There is a genuine missing nonemptiness hypothesis in the literal statement. Independent differentials on an empty set is a vacuous condition, and any two vector-field restrictions to an empty set are equal. For example, on ${\mathbb{R}}^2$ let $r=1$, $G_1=H=e^q$, and $\rho =1$. Every nonempty ${\Sigma }_c$, $0<c<1$, is regular, while ${\Sigma }_c=\varnothing$ for $-1<c\le 0$. In particular, the restrictions at $c=-1/2$ can never be unequal for any $K$. Therefore the printed universal inequality requires either all the compared fibres to be nonempty, or the pointwise formulation displayed above. The explicit $K$ fully proves either intended qualification.