Past exam of the topics in applied mathematics course of the University of Oxford / 2010 / B / 4 / v / Solution

First assume ${\omega }_0\ne 0$ and regard $\theta$ modulo $2\pi$. The section $\theta =0$ has return time $T=2\pi /|{\omega }_0|$. Solving the resonantly forced oscillator from this section givesAt $t=T$, the sine vanishes and the cosine is one, while differentiating the secular term adds $\gamma T/2$ to $y$. Hence the Poincaré map isFor $\gamma >0$, every iterate has $y_n=y_0+n\gamma T/2\to \infty$. Thus every oscillator trajectory is unbounded, regardless of its initial amplitude. For an arbitrary section phase ${\theta }_0$, the return map translates by $(\gamma T\sin {\theta }_0/(2{\omega }_0),\gamma T\cos {\theta }_0/2)$, which is still nonzero. At $\gamma =0$ the map is the identity, and all $(x,y)$ motions are bounded harmonic oscillations, including the zero-amplitude equilibrium.

The zero-frequency case allowed by the printed parameter range is different. If ${\omega }_0=0$, then $\theta ={\theta }_0$ is constant and $x=x_0+y_{0}t+\frac{1}{2}\gamma \cos {\theta }_0{t}^2$. There is no phase-crossing Poincaré section. For $\gamma >0$, choosing $\cos {\theta }_0=0$ and $y_0=0$ gives bounded equilibria, contradicting an unqualified “all trajectories” statement. For $\gamma =0$ at zero frequency, free-particle motion is unbounded unless $y_0=0$. The resonance conclusion therefore requires the natural nonzero-frequency assumption.