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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/32d/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 32D / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Direct substitution gives

g^{s} g^{r} = g^{s + r}

. Differentiating at

s = 0

gives

t \partial_{x} + i x (ψ \partial_{ψ} - ψ^{*} \partial_{ψ^{*}})

.

Writing the transformed field in transformed coordinates gives

ψ (t, x) = e^{i (s x - s^{2} t /2)} ψ (t, x - s t) .

Consequently

ψ_{t} = e^{i Φ} (ψ_{t} - s ψ_{x} - i s^{2} ψ /2),

ψ_{x} = e^{i Φ} (ψ_{x} + i s ψ), ψ_{xx} = e^{i Φ} (ψ_{xx} + 2 i s ψ_{x} - s^{2} ψ),

where

Φ = s x - s^{2} t /2

. These formulas, together with their complex conjugates and the analogous mixed derivatives, give the second prolongation. Differentiating at zero yields

η^{0} = i x ψ_{t} - ψ_{x}, η^{11} = i x ψ_{xx} + 2 i ψ_{x} .

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