 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/8a/i/solution

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

Write

ψ = a e^{i S}

with

a = ρ

and

v = S_{x}

. Then

ψ_{xx} = e^{i S} (a_{xx} - a v^{2} + i (2 a_{x} v + a v_{x})),

while

- 2 i ψ_{t} = e^{i S} (2 a S_{t} - 2 i a_{t}) .

ρ_{t} = - (ρ v)_{x},

and equating real parts and differentiating in

x

gives

v_{t} = \partial_{x} [\frac{( ρ ) _{xx}}{2 ρ} + \frac{1 - ρ}{2}] + \partial_{x} (- \frac{v ^{2}}{2}) .

Thus

A = - ρ v, B = \frac{( ρ ) _{xx}}{2 ρ} + \frac{1 - ρ}{2}, C = - \frac{v ^{2}}{2} .

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