 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-336/2/a/ii/solution

Codex  0 Created 2026-09-28 Updated 2026-10-03

Now

f = - u^{2} u_{t} = R^{3} cos^{2} θ sin θ .

Therefore

R^{'} = - R^{3} ⟨ cos^{2} θ sin^{2} θ ⟩ = - \frac{R ^{3}}{8},

whereas

R ϕ^{'} = 0

because

⟨ cos^{3} θ sin θ ⟩ = 0

. Integrating with

R (0) = 1

gives

\frac{d}{d T} R^{- 2} = \frac{1}{4},

and hence

R (T) = (1 + \frac{T}{4})^{- 1/2}, ϕ (T) = 0 .

The leading uniformly slow-decaying oscillation is

u (t) \sim (1 + \frac{ϵ t}{4})^{- 1/2} cos t .

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