 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4/38c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Curling Navier-Stokes and using incompressibility gives

ω_{t} + (u \cdot \nabla) ω = (ω \cdot \nabla) u + ν \nabla^{2} ω

. For pure swirl,

ω = r^{- 1} (r v)_{r}

, so regularity gives

v (r, t) = r^{- 1} \int_{0}^{r} s ω (s, t) d s

. Adding strain

(- α r, v, 2 α z)

yields

ω_{t} - α r ω_{r} = 2 α ω + ν (ω_{rr} + r^{- 1} ω_{r}) .

2 π r

and integration proves

\dot{Γ} = 0

. The decaying steady Burgers vortex is

ω_{s} (r) = \frac{Γ α}{2 π ν} e^{- α r^{2} / (2 ν)} .

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