 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-3/11b/a/solution

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

The identity follows immediately from the product rule:

\nabla \cdot (κ ψ \nabla ϕ) = ψ \nabla \cdot (κ \nabla ϕ) + κ \nabla ψ \cdot \nabla ϕ .

ϕ_{1}, ϕ_{2}

have the same boundary data, put

ψ = ϕ_{1} - ϕ_{2}

. Integrating the identity with

ϕ = ψ

gives

\int_{V} κ ∣\nabla ψ ∣^{2} d V = 0

because the volume equation and boundary term vanish. Positivity of

κ

makes

ψ

constant, and its boundary value makes it zero, proving uniqueness.

For any

w = ϕ + ψ

with

ψ = 0

on the boundary, expansion gives

\int κ ∣\nabla w ∣^{2} = \int κ ∣\nabla ϕ ∣^{2} + in t κ ∣\nabla ψ ∣^{2},

because the cross term vanishes by the same integration by parts. This proves the inequality and the Dirichlet principle.

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