Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 32C (Integrable Systems)

In canonical coordinates, Hamilton's equations areA function $F(q,p)$ is an integral of motion if it is constant along every Hamiltonian flow, equivalentlyfor a time-independent $F$. The system is an integrable Hamiltonian system when it has $n$ first integrals $F_1,\dots ,F_n$ whose differentials are linearly independent almost everywhere and which are in mutual Poisson involution: $\{F_i,F_j\}=0$ for all $i,j$.

For one degree of freedom, take $F_1=H$. Antisymmetry of the Poisson bracket gives $\{H,H\}=0$, so energy is conserved. The stated condition is exactly $dH\ne 0$, which supplies the required functional independence. Hence the system is integrable at every such regular point.

Let $F_1,\dots ,F_n$ be commuting independent integrals for $(M,H)$ and let $G_1,\dots ,G_m$ be such integrals for $(\tilde{M},\tilde{H})$. Regard each as a function on the product phase space. ThenThe $F_i$ commute among themselves, the $G_a$ commute among themselves, and every cross bracket $\{F_i,G_a\}$ vanishes because the functions depend on disjoint canonical coordinates. Their differentials form two independent coordinate blocks, so all $n+m$ are independent on the product of the two regular sets. Thus the combined system is integrable, an instance of the direct product of integrable Hamiltonian systems.

Write the Hamiltonian as the sum of separated one-dimensional energiesEach $(q_i,p_i)$ system is a simple harmonic oscillator and is integrable away from $(q_i,p_i)=(0,0)$ by part (b). The separated energies satisfyand their differentials are independent whenever every oscillator pair is nonzero. The Poisson-commuting separated energies therefore make the $n$-dimensional oscillator integrable. The exceptional coordinate strata have measure zero, so this proves Liouville integrability in its usual almost-everywhere sense; the origin is the only point where $dH$ itself vanishes.

No. Consider two uncoupled oscillators with nonzero amplitudes:The orbit is bounded. If it had a period $T>0$, then ${\omega }_{1}T$ and ${\omega }_{2}T$ would both be integer multiples of $2\pi$, forcing ${\omega }_1/{\omega }_2\in \mathbb{Q}$. Choosing an irrational frequency ratio gives an incommensurate harmonic-oscillator orbit, which is quasiperiodic on an invariant torus and not periodic.