 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/39c/b/i/solution

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

The Fourier modes evolve according to the dispersion relation found in part (a), so

ϕ (V t, t) = \int_{- \infty}^{\infty} A (k) e^{i t ψ (k)} d k, ψ (k) = (V - U) k - \frac{k ^{9}}{9} .

For

V > U

ψ^{'} (k) = V - U - k^{8} = 0.

It has the two real solutions

k = \pm k_{0}, k_{0} = (V - U)^{1/8} .

Ω = ψ (k_{0}) = \frac{8}{9} k_{0}^{9}, ∣ ψ^{''} (\pm k_{0}) ∣ = 8 k_{0}^{7} .

ϕ (V t, t) \sim \frac{π}{4 t k _{0}^{7}} [A (k_{0}) e^{i (Ω t - π /4)} + A (- k_{0}) e^{- i (Ω t - π /4)}] .

Because the initial field is real, its Fourier transform has the conjugate symmetry

A (- k) = A (k)^{*}

. Hence the same result can be written

ϕ (V t, t) \sim \frac{π}{t k _{0}^{7}} Re [A (k_{0}) e^{i (Ω t - π /4)}] .

 New to topics? Read the docs here!