 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-61/2/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For the specified time operator,

N_{k} E_{k} = 3 ik q E_{k}

, because the derivatives of

e^{i (k x + k^{3} t)}

cancel between

\partial_{t}

and

\partial_{x}^{3}

. Expanding the derivatives of

E_{k} A_{k}

gives

N_{k} (E_{k} A_{k}) = E_{k} [\int_{L} \frac{N _{l} φ ( l )}{k + l} d λ (l) + 3 ik \int_{L} \frac{M _{l} φ ( l )}{k + l} d λ (l) - 3 k I_{x}] .

For completeness, the extra derivative terms before simplification are

3 ik A_{k, xx} - 3 k^{2} A_{k, x} + 3 ik q A_{k}

. Substituting

φ_{xx} + qφ = M_{l} φ + i l φ_{x}

turns their derivative contribution into

- 3 k \int_{L} φ_{x} d λ = - 3 k I_{x}

Applying

N_{k}

to the original linear integral equation and using

I_{x} = - q

cancels its right side. Thus

F (Nφ) = 3 k E_{k} \int_{L} \frac{M _{l} φ ( l )}{k + l} d λ (l) .

Part (a) gives

E_{k} \int_{L} M_{l} φ / (k + l) d λ = i M_{k} φ

. Therefore the exact requested identity is

F (Nφ) = 3 ik Mφ .

Since

Mφ = 0

, this becomes homogeneous too. The same uniqueness argument gives

Nφ = 0

; it is not necessary to assume a separate uniqueness theorem for the differential time equation.

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