 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4/38b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

For

0 < V < β / (8 ℓ^{2})

there are two positive solutions

k_{1} \in (ℓ, 3 ℓ)

and

k_{2} > 3 ℓ

, together with their negatives. The second derivative is

ω^{''} (k) = \frac{2 β k ( 3 ℓ ^{2} - k ^{2} )}{( k ^{2} + ℓ ^{2} ) ^{3}} .

Each negative stationary point has the conjugate contribution of its positive partner, because

A

is real even and

Φ_{V}

is odd. Therefore

φ (V t, t) \sim 2 \frac{2 π}{t} [\frac{A ( k _{1} )}{∣ ω ^{''} ( k _{1} ) ∣} cos (t Φ_{V} (k_{1}) - \frac{π}{4}) + \frac{A ( k _{2} )}{∣ ω ^{''} ( k _{2} ) ∣} cos (t Φ_{V} (k_{2}) + \frac{π}{4})] .

If a leading amplitude vanishes, this gives a zero coefficient and a higher-order term supplies that branch's first nonzero contribution.

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