Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 1 / i

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- Solution i

Solution

Work in a Cartesian coordinate system, write

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

, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols gives

[\nabla \times (\nabla \times u)]_{i} = ϵ_{ijk} ϵ_{k l m} \partial_{j} \partial_{l} u_{m} = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) \partial_{j} \partial_{l} u_{m} = \partial_{i} (\partial_{j} u_{j}) - \partial_{j} \partial_{j} u_{i} .

Here the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identity

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

For the second identity, the same contraction followed by the product rule yields

[\nabla \times (u \times v)]_{i} = \partial_{j} (u_{i} v_{j} - u_{j} v_{i}) = u_{i} \partial_{j} v_{j} - v_{i} \partial_{j} u_{j} + v_{j} \partial_{j} u_{i} - u_{j} \partial_{j} v_{i} .

Thus the curl of a cross product is

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

The directional derivatives in the last two terms act on the vector field immediately to their right.

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