 Solution

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

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

∣ ψ_{0} (ξ) ∣^{2} = 1 - (1 - 2 U^{2}) sech^{2} (\frac{1 - 2 U ^{2}}{2} ξ) .

For

U = 0

∣ ψ_{0} (ξ) ∣^{2} = tanh^{2} (ξ / 2),

a stationary notch reaching zero at its centre. For

U = 1/2

∣ ψ_{0} (ξ) ∣^{2} = 1 - \frac{1}{2} sech^{2} (ξ /2),

a shallower notch with minimum

1/2

. Since

ξ = x - U t

, the first profile remains fixed and the second translates rigidly to the right at speed

1/2

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