 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-74/3/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Let

K^{2} = ∣ k ∣^{2} = k_{x}^{2} + k_{z}^{2} + Ω^{2} t^{2} k_{x}^{2}

. For the complex wave before taking its real part,

(\partial_{t} + Ω y \partial_{x}) (v e^{i k \cdot x}) = [\dot{v} + i (\dot{k} \cdot x + Ω y k_{x}) v] e^{i k \cdot x} = \dot{v} e^{i k \cdot x},

since

\dot{k}_{y} = - Ω k_{x}

. The asserted velocity equation therefore follows directly from

\dot{v} + Ω v_{y} x = - i k q

. To verify the pressure amplitude, solenoidality is

k \cdot v = 0

. Its derivative requires

0 = \dot{k} \cdot v + k \cdot \dot{v} = - 2Ω k_{x} v_{y} - i K^{2} q,

and hence

q = 2 i Ω k_{x} v_{y} / K^{2}

, as claimed.

\overset{v}{˙}_{y} = a v_{y}, \overset{v}{˙}_{z} = b v_{y}, a = \frac{2Ω k _{x} k _{y}}{K ^{2}}, b = \frac{2Ω k _{x} k _{z}}{K ^{2}} .

Both coefficients are real. Differentiating

v_{y} v_{z}^{*} - v_{z} v_{y}^{*}

makes the terms proportional to

b ∣ v_{y} ∣^{2}

cancel, while the remaining terms give

a (v_{y} v_{z}^{*} - v_{z} v_{y}^{*})

\dot{H} = \frac{2Ω k _{x} k _{y}}{K ^{2}} H = - \frac{2 Ω ^{2} t k _{x}^{2}}{k _{x}^{2} + k _{z}^{2} + Ω ^{2} t ^{2} k _{x}^{2}} H, H (t) = \frac{H _{0} ( k _{x}^{2} + k _{z}^{2} )}{k _{x}^{2} + k _{z}^{2} + Ω ^{2} t ^{2} k _{x}^{2}} .

Here

H

is real because the difference in its definition is purely imaginary. Its sign is preserved, and shear reduces its magnitude away from

t = 0

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