Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 31C / Solution

A Hamiltonian system with $n$ degrees of freedom is Liouville integrable if it has $n$ functionally independent first integrals, including the Hamiltonian, whose pairwise Poisson brackets vanish. Functional independence is required on a regular open set. The Arnold-Liouville theorem says that a connected compact regular common level of these integrals is an $n$-torus, with a neighborhood admitting action-angle coordinates; the Hamiltonian flow there has constant angular velocities and fixed actions. Compactness is essential to the torus conclusion.

Use $\{f,g\}={\sum }_j(f_{q_j}{g}_{p_j}-f_{p_j}{g}_{q_j})$. On $r>0$, for differentiable $F$, Hamilton's equations give $\dot{q}=p$, $\dot{p}=-F^{\prime }(r)q/r$. Hence the angular momentum $M=q\times p$ obeysIn particular $M_1$ and $M_2$ are first integrals. The Jacobi identity for the Poisson bracket isTaking $f=H,g=M_1,h=M_2$ shows $\{H,\{M_1,M_2\}\}=0$. Direct differentiation gives $M_3=\{M_1,M_2\}=q_1{p}_2-q_2{p}_1$ and, cyclically, $\{M_i,M_j\}={ϵ}_{ijk}{M}_k$.

Set $L^2=M_1^2+M_2^2+M_3^2$. Then $\{L^2,M_3\}=2M_1(-M_2)+2M_2{M}_1=0$, and the brackets of $H$ with both expressions vanish. ThusFor functional independence, at $q=(r,0,0)$ with $r>0$ the Jacobian matrix of $(H,L^2,M_3)$ with respect to $(p_1,p_2,p_3)$ iswhose determinant is $-2r^3{p}_1{p}_3$. It is nonzero when $p_1{p}_3\ne 0$. More generally this minor is a nonzero polynomial in $q,p$, independent of $F$, so its zero set has empty interior and the integrals are independent on a dense open regular set. This proves Liouville integrability on the regular set. The arbitrary potential in the question does not ensure compact common levels: $F=0$ has unbounded free trajectories. The full torus conclusion applies to those regular connected common levels which are compact, rather than to every choice of $F$ and every level.