 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v/2/a/ii/solution

F = u \nabla v - v \nabla u

\nabla \cdot F = u \nabla^{2} v - v \nabla^{2} u,

because the two gradient inner products cancel. Also

F \cdot n = u \partial_{n} v - v \partial_{n} u

. Thus

∭_{R} (u \nabla^{2} v - v \nabla^{2} u) d V = \iint_{\partial R} (u \partial_{n} v - v \partial_{n} u) d S .

This is Green's second identity, under the usual regularity assumptions permitting these derivatives and boundary integrals.

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