 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/1/a/solution

[u \times (\nabla \times u)]_{i} = ϵ_{ijk} ϵ_{k l m} u_{j} \partial_{l} u_{m} = u_{j} \partial_{i} u_{j} - u_{j} \partial_{j} u_{i} = \frac{1}{2} \partial_{i} (u_{j} u_{j}) - [(u \cdot \nabla) u]_{i} .

The contraction of the two symbols uses the Kronecker delta, and the final step is the product rule. Thus the convective acceleration identity is

(u \cdot \nabla) u = \frac{1}{2} \nabla∣ u ∣^{2} - u \times (\nabla \times u) .

Put

ω = \nabla \times u

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

Since the divergence of a curl is zero, taking the curl of the first identity yields

\nabla \times [(u \cdot \nabla) u] = (\nabla \cdot u) ω + (u \cdot \nabla) ω - (ω \cdot \nabla) u .

Twice continuous differentiability suffices for the mixed-derivative identities used here.

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