 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3/11a/ii/solution

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

Put

η = ψ - ϕ

η = 0

S

E [ψ] - E [ϕ] = \int_{V} (∣\nabla η ∣^{2} + m^{2} η^{2}) d V + 2 \int_{V} (\nabla ϕ \cdot \nabla η + m^{2} ϕ η) d V .

Integration by parts and the field equation make the cross term zero:

\int_{V} (\nabla ϕ \cdot \nabla η + m^{2} ϕ η) d V = \int_{S} η \frac{\partial ϕ}{\partial n} d S - \int_{V} η (\nabla^{2} ϕ - m^{2} ϕ) d V = 0.

Consequently

E [ψ] - E [ϕ] = \int_{V} (∣\nabla η ∣^{2} + m^{2} η^{2}) d V \geq 0 .

For

m > 0

, equality holds exactly when

η = 0

. For

m = 0

, equality holds exactly when

\nabla η = 0

, so

η

is constant on the connected region; its zero boundary value then again forces

η = 0

. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:

ψ = ϕ is the equality condition for every m \geq 0 .

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