 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ia/paper-3/4b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For Cartesian components,

\partial_{j} x_{i} = δ_{ij}

, the Kronecker delta. Differentiating

r^{2} = x_{i} x_{i}

gives

\partial_{j} r = \frac{x _{j}}{r} (r > 0) .

Put

s = k_{ℓ} x_{ℓ}

. Using the product rule and

\partial_{j} r^{- q} = - q x_{j} r^{- q - 2}

d_{ij} = - \frac{k _{i} x _{j}}{r ^{3}} + \frac{k _{j} x _{i}}{r ^{3}} + \frac{s δ _{ij}}{r ^{3}} - \frac{3 s x _{i} x _{j}}{r ^{5}} .

E_{ij} = \frac{d _{ij} + d _{ji}}{2} = \frac{s δ _{ij}}{r ^{3}} - \frac{3 s x _{i} x _{j}}{r ^{5}}, W_{ij} = \frac{d _{ij} - d _{ji}}{2} = \frac{k _{j} x _{i} - k _{i} x _{j}}{r ^{3}} .

Here

E_{ji} = E_{ij}

and

W_{ji} = - W_{ij}

, and

d = E + W

. Taking the trace gives the divergence

\nabla \cdot u = d_{ii} = \frac{3 s}{r ^{3}} - \frac{3 s r ^{2}}{r ^{5}} = 0.

For the curl, use the Levi-Civita symbol. Because

d_{kj} = \partial_{j} u_{k}

, the symmetric part makes no contribution, and

(\nabla \times u)_{i} = ϵ_{ijk} d_{kj} = \frac{ϵ _{ijk} ( k _{j} x _{k} - k _{k} x _{j} )}{r ^{3}} = \frac{2 ϵ _{ijk} k _{j} x _{k}}{r ^{3}} .

Thus

\nabla \times u = 2 k \times x / r^{3}

. All derivative identities hold away from the singular origin.

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