 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-v-resit/question-2/b/solution

Write

\partial_{i} = \partial / \partial x_{i}

and use

\times

for the paper's wedge notation, which denotes the cross product. The given expression for the curl and the product rule give

\nabla \times (v \times w) = \sum_{i} e_{i} \times ((\partial_{i} v) \times w + v \times (\partial_{i} w)) .

a \times (b \times c) = b (a \cdot c) - c (a \cdot b)

separately to the two terms:

e_{i} \times ((\partial_{i} v) \times w) e_{i} \times (v \times \partial_{i} w) = w_{i} \partial_{i} v - w \partial_{i} v_{i}, = v \partial_{i} w_{i} - v_{i} \partial_{i} w .

Summing over the Cartesian basis vectors identifies the two directional derivatives and the two divergences:

\nabla \times (v \times w) = (w \cdot \nabla) v + v (\nabla \cdot w) - (v \cdot \nabla) w - w (\nabla \cdot v) .

Here, for example,

((w \cdot \nabla) v)_{j} = \sum_{i} w_{i} \partial_{i} v_{j}

. This proves the curl of a cross product directly from the formula supplied.

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