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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-3/11c/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 11C / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

The product rule for the divergence gives

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

Subtracting cancels the two gradient products. Therefore

\nabla \cdot (u \nabla v - v \nabla u) = u \nabla^{2} v - v \nabla^{2} u .

To obtain the requested order of terms, apply this identity with

u, v

interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is

r

; on the inner sphere it is

- r

. Write

\partial_{r} = r \cdot \nabla

on both spheres. Then the resulting Green second identity is

\int_{ρ \leq ∣ x ∣ \leq r} (v \nabla^{2} u - u \nabla^{2} v) d V = \int_{∣ x ∣ = r} (v \partial_{r} u - u \partial_{r} v) d S - \int_{∣ x ∣ = ρ} (v \partial_{r} u - u \partial_{r} v) d S .

The minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

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