 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3/32a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Differentiating the two equations for a fundamental matrix of solutions gives

ψ_{ρ τ} = (U_{τ} + U V) ψ, ψ_{τ ρ} = (V_{ρ} + V U) ψ .

Equality for all independent initial vectors is equivalent to the zero-curvature condition

U_{τ} - V_{ρ} + [U, V] = 0.

A single possibly zero solution vector alone would not imply a matrix identity; compatibility is understood for the complete linear system.

Write

q = ϕ_{ρ}

s = sin ϕ

c = cos ϕ

U = iλ σ_{3} + \frac{i q}{2} σ_{1}, V = \frac{1}{4 iλ} (c σ_{3} + s σ_{2}), λ \neq = 0.

[U, V] = - \frac{i s}{2} σ_{1} - \frac{i q c}{4 λ} σ_{2} + \frac{i q s}{4 λ} σ_{3},

whereas

V_{ρ} = - (i q c / (4 λ)) σ_{2} + (i q s / (4 λ)) σ_{3}

. Thus these last two components cancel, and the remaining curvature is

(i /2) (ϕ_{ρ τ} - sin ϕ) σ_{1}

. Therefore

ϕ_{ρ τ} = sin ϕ,

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