 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/16c/b/i/solution

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

A radially symmetric velocity potential has

u_{r} = \partial_{r} ϕ

. The incompressible flow condition in two-dimensional polar coordinates is

\frac{1}{r} \frac{\partial}{\partial r} (r u_{r}) = 0,

r u_{r} = A (t)

u_{r} (a, t) = \overset{a}{˙}

, hence

u_{r} (r, t) = \frac{a a ˙}{r}, ϕ (r, t) = a \overset{a}{˙} lo g r + C (t) .

At the outer interface,

\dot{b} = u_{r} (b, t) = a \overset{a}{˙} / b

. Therefore

\frac{d}{d t} (b^{2} - a^{2}) = 2 b \dot{b} - 2 a \overset{a}{˙} = 0,

and

b (t)^{2} - a (t)^{2} = b_{0}^{2} - a_{0}^{2} .

π (b^{2} - a^{2})

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