 Solution

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

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

ξ = x - U t

gives

ψ_{0}^{''} - 2 i U ψ_{0}^{'} + (1 - ∣ ψ_{0} ∣^{2}) ψ_{0} = 0 .

Write

ψ_{0} (ξ) = u (ξ) + i 2 U .

Its real part satisfies

u^{''} + (1 - 2 U^{2} - u^{2}) u = 0 .

For

U > 0

, the imaginary part also gives

2 u^{'} = 1 - 2 U^{2} - u^{2} .

With

α = 1 - 2 U^{2}

and a translation chosen so that the notch is centred at zero,

ψ_{0} (ξ) = α tanh (\frac{α ξ}{2}) + i 2 U .

Its limiting modulus is

α^{2} + 2 U^{2} = 1 = d

, as required. The same expression at

U = 0

is obtained directly from the second-order equation.

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