 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-331/1/b/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

For a possibly unstable normal mode,

c_{i} \neq = 0

ensures

V = U - c

never vanishes, so choose one continuous branch of

V^{a}

. Substitute

w = V^{a} q

into the Taylor–Goldstein equation and collect terms:

q^{''} + 2 a \frac{U ^{'}}{V} q^{'} + [- k^{2} + (a - 1) \frac{U ^{''}}{V} + \frac{N ^{2} + a ( a - 1 ) ( U ^{'} ) ^{2}}{V ^{2}}] q = 0.

V^{2 a}

puts its first two terms in divergence form:

(V^{2 a} q^{'})^{'} - k^{2} V^{2 a} q + [(a - 1) U^{''} V^{2 a - 1} + {N^{2} + a (a - 1) (U^{'})^{2}} V^{2 a - 2}] q = 0.

\overline{q}

q (\pm L) = 0

\int_{- L}^{L} V^{2 a} (∣ q^{'} ∣^{2} + k^{2} ∣ q ∣^{2}) d z = \int_{- L}^{L} [{N^{2} + a (a - 1) (U^{'})^{2}} V^{2 a - 2} + (a - 1) U^{''} V^{2 a - 1}] ∣ q ∣^{2} d z .

The weight is

V^{2 a}

, not

∣ V ∣^{2 a}

: preserving its complex phase is essential for the subsequent stability proofs.

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