 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-3/12b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

\int_{V} ∣\nabla u ∣^{2} d V = \int_{S} u \frac{\partial u}{\partial n} d S - \int_{V} u \nabla^{2} u d V = 0.

Thus

\nabla u = 0

, so

u

is constant on each connected component of

V

; its zero boundary value forces

u = 0

Since

w - v = 0

S

and

\nabla^{2} v = 0

, another application of Green's first identity gives

\int_{V} \nabla v \cdot \nabla (w - v) d V = 0.

Consequently

\int_{V} \nabla v \cdot \nabla w d V = \int_{V} ∣\nabla v ∣^{2} d V .

Expanding

\nabla w = \nabla v + \nabla (w - v)

and using the vanishing cross term yields the Dirichlet principle

\int_{V} ∣\nabla w ∣^{2} d V = \int_{V} (∣\nabla v ∣^{2} + ∣\nabla (w - v) ∣^{2}) d V \geq \int_{V} ∣\nabla v ∣^{2} d V .

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