 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/7a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

R = x^{2} + y^{2}

, direct differentiation gives

\frac{\partial u}{\partial x} = \frac{- 2 x ( x ^{2} - 3 y ^{2} )}{R ^{3}}, \frac{\partial v}{\partial y} = \frac{2 x ( x ^{2} - 3 y ^{2} )}{R ^{3}},

\nabla \cdot u = 0

. Likewise,

\frac{\partial v}{\partial x} = \frac{- 2 y ( 3 x ^{2} - y ^{2} )}{R ^{3}} = \frac{\partial u}{\partial y},

so the two-dimensional vorticity

\partial_{x} v - \partial_{y} u

vanishes. The flow is therefore incompressible and irrotational away from the origin.

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