 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/33a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Eliminating

y

gives

\overset{x}{¨} + x - F = μ (1 - x^{2}) \overset{x}{˙} .

For

μ ≪ 1

, take

x = F + A cos (t + δ)

over one fast period. The oscillator energy satisfies

\dot{E} = μ (1 - x^{2}) \overset{x}{˙}^{2},

and averaging gives

⟨ \dot{E} ⟩ = \frac{μ A ^{2}}{2} (1 - F^{2} - \frac{A ^{2}}{4}) .

Thus, if

F^{2} < 1

, amplitudes approach

A = 2 1 - F^{2},

producing a stable weakly nonlinear limit cycle. If

F^{2} > 1

, every nonzero small amplitude loses energy and trajectories approach the stable equilibrium.

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