 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/1/f/solution

Choose the symmetric tensor field

Σ_{kj} = (\partial_{k} ψ) (\partial_{j} ψ) - \frac{1}{2} δ_{kj} ∣\nabla ψ ∣^{2} .

The product rule and equality of mixed partial derivatives give

\partial_{k} Σ_{kj} = (\nabla^{2} ψ) \partial_{j} ψ + (\partial_{k} ψ) \partial_{k} \partial_{j} ψ - \frac{1}{2} \partial_{j} ∣\nabla ψ ∣^{2} = (\nabla^{2} ψ) \partial_{j} ψ .

This is the Euclidean scalar-gradient stress identity. Substitute it into part (e), and use

Σ_{ki} \overset{n}{^}_{k} = (\partial_{n} ψ) \partial_{i} ψ - \frac{1}{2} ∣\nabla ψ ∣^{2} \overset{n}{^}_{i}

, where

\partial_{n} ψ = \hat{n} \cdot \nabla ψ

. Reassembling the components gives the required pure surface integral:

\int_{V} r \nabla \cdot [(\nabla^{2} ψ) \nabla ψ] d V = \int_{S} [r (\nabla^{2} ψ) \partial_{n} ψ - (\partial_{n} ψ) \nabla ψ + \frac{1}{2} ∣\nabla ψ ∣^{2} \hat{n}] d S .

Sufficient smoothness of

ψ

is understood for the displayed derivatives and applications of the divergence theorem.

 New to topics? Read the docs here!