 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2/33a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Lax's equation in the convention used here is

\dot{L} = [L, A] = L A - A L .

Let

L ψ = λ ψ

with

(ψ, ψ) = 1

. Since

L

A

is anti-self-adjoint, differentiating

λ = (ψ, L ψ)

gives

\dot{λ} = (\dot{ψ}, L ψ) + (ψ, \dot{L} ψ) + (ψ, L \dot{ψ}) .

The first and third terms equal

λ, d (ψ, ψ) / d t = 0

. For the middle term,

(ψ, [L, A] ψ) = (L ψ, A ψ) - (ψ, A L ψ) = 0.

Thus

\dot{λ} = 0

, proving the Isospectral Lax equation property: every eigenvalue is independent of $t$ .

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