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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-3/4c/solution

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Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 4C / Solution by

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

In Cartesian coordinates, independent coordinate variables satisfy

\frac{\partial x _{i}}{\partial x _{j}} = δ_{ij}, \frac{\partial r}{\partial x _{j}} = \frac{x _{j}}{r} (r > 0),

where

δ_{ij}

is the Kronecker delta. Applying the product rule and chain rule to

u_{i} = r^{α} x_{i}

gives the Jacobian matrix

d_{ij} = r^{α} δ_{ij} + α r^{α - 2} x_{i} x_{j} .

This tensor is symmetric in

i, j

. Hence its contraction with the antisymmetric Levi-Civita symbol vanishes:

(\nabla \times u)_{i} = ϵ_{ijk} d_{kj} = 0.

Similarly,

v_{i} = ϵ_{ijk} k_{j} u_{k}

and the constancy of

k

give

\nabla \cdot v = ϵ_{ijk} k_{j} d_{ki} = 0.

The trace of the Jacobian matrix is

\nabla \cdot u = (3 + α) r^{α},

which vanishes for

α = - 3

. The curl of a cross product identity, with

k

constant, now gives

\nabla \times (k \times u) = k (\nabla \cdot u) - (k \cdot \nabla) u .

For

α = - 3

,

(k \cdot \nabla) u = r^{- 3} k - 3 r^{- 5} (k \cdot x) x,

so

\nabla \times v = \frac{3 ( k \cdot x ) x - r ^{2} k}{r ^{5}} .

All identities for this singular case hold on

r > 0

. In particular, zero divergence here is a pointwise assertion away from the origin, not an assertion that the flux through a sphere enclosing the singularity is zero.

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