 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/33d/c/solution

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

When

u

is independent of

y

and

z = x + i y

\partial_{z} = \partial_{\overset{z}{ˉ}} = \frac{1}{2} \frac{d}{d x} .

Put

L = U - V, M = U + V .

The zero-curvature equation becomes

\frac{1}{2} L_{x} + [U, V] = 0.

Since

[M, L] = [U + V, U - V] = - 2 [U, V],

L_{x} = [M, L] .

Therefore, for every positive integer

n

, the cyclic property of the matrix trace gives

\frac{d}{d x} tr (L^{n}) = n tr (L^{n - 1} [M, L]) = n tr (L^{n - 1} M L - M L^{n}) = 0.

The PDE from part (b) reduces to

\frac{1}{4} u_{xx} = \frac{1}{2} sinh (2 u) .

With

ϕ = 2 u

, this is

ϕ^{''} = 4 sinh ϕ .

To extract a nontrivial invariant, use

n = 2

p = u_{x}

, direct multiplication gives

tr (L^{2}) = \frac{1}{2} [p^{2} + λ^{2} + λ^{- 2} - 2 cosh (2 u)] .

The terms involving

λ

are constant, and

p = ϕ^{'} /2

. Hence

\frac{1}{2} (ϕ^{'})^{2} - 4 cosh ϕ = C

is a first integral. Direct differentiation verifies it:

\frac{d}{d x} [\frac{1}{2} (ϕ^{'})^{2} - 4 cosh ϕ] = ϕ^{'} (ϕ^{''} - 4 sinh ϕ) = 0.

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