 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/4d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

λ

for the normalization and vary

J [u] = \int_{D} (∣\nabla u ∣^{2} - λ u^{2}) d x d y .

For a smooth variation

u + ε η

with

η = 0

\partial D

\frac{d}{d ε} J [u + ε η]_{ε = 0} = 2 \int_{D} (\nabla u \cdot \nabla η - λ u η) = - 2 \int_{D} (\nabla^{2} u + λ u) η .

\nabla^{2} u + λ u = 0 .

Multiplying by

u

and integrating, while using

\nabla \cdot (u \nabla u) = ∣\nabla u ∣^{2} + u \nabla^{2} u,

u = 0

on the boundary give

0 = I [u] + \int_{D} u \nabla^{2} u = I [u] - λ \int_{D} u^{2} .

The normalization is one, so the multiplier equals the stationary value:

λ = I [u] .

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