Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 15B / c

For $H=H(p)$, Hamilton's equations giveThus $P=p$ and $Q=q-t{H}^{\prime }(p)$. The Poisson bracket is $\{Q,P\}=1$, so the transformation is canonical for every $t$. A type-2 generator isIndeed $p={\partial }_q{F}_2=P$, $Q={\partial }_P{F}_2=q-t{H}^{\prime }(P)$, and the transformed Hamiltonian is $K=0$, as expected because $Q,P$ are constants of motion.

For the harmonic oscillator Hamiltonian $H=(q^2+p^2)/2$,This rotation preserves the symplectic form $dq\wedge dp$, so it is a canonical transformation for every $t$. Where $\cos t\ne 0$, solving for the initial variables givesand hence the type-2 generatorIts derivatives give the displayed transformation and $H+{\partial }_t{F}_2=0$. At $\cos t=0$ the canonical rotation remains perfectly regular, but its projection to $(q,P)$ is singular, so no type-2 generating function in those variables exists there; another generating-function chart covers those isolated times.