 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/33c/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Differentiate the spatial equation with respect to

t

and the temporal equation with respect to

x

. Substitution of

ψ_{x} = ik ψ + u, ψ_{t} = - i k^{2} ψ + i u_{x} - k u

into

ψ_{x t} - ik ψ_{t}

cancels every term involving

ψ

u

, and

u_{x}

, leaving

ψ_{x t} - ik ψ_{t} = i u_{xx} .

The left side is

u_{t}

by differentiating

ψ_{x} - ik ψ = u

, so consistency is exactly

u_{t} = i u_{xx}

. This is a scalar Lax pair formulation.

Taking the half-line Fourier transform and integrating twice by parts gives

\partial_{t} u (k, t) + i k^{2} u (k, t) = kh (t) - i u_{x} (0, t) .

e^{i k^{2} t}

and integration in time yields

u (k, t) e^{i k^{2} t} - u_{0} (k) = \int_{0}^{t} e^{i k^{2} τ} [kh (τ) - i u_{x} (0, τ)] d τ .

Replace

k

- k

and subtract. The unknown Neumann boundary value cancels:

e^{i k^{2} t} [u (k, t) - u (- k, t)] = u_{0} (k) - u_{0} (- k) + 2 k \int_{0}^{t} e^{i k^{2} τ} h (τ) d τ .

For

x > 0

, Fourier inversion of the zero extension gives

\int_{R} e^{ik x} u (- k, t) d k = 0.

Inverting the preceding identity therefore gives

u (x, t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- i k^{2} t + ik x} [u_{0} (k) - u_{0} (- k)] d k + \frac{1}{π} \int_{- \infty}^{\infty} \int_{0}^{t} e^{- i k^{2} (t - τ) + ik x} kh (τ) d τ d k .

Thus

G (k, τ) = kh (τ) .

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