 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/39a/b/solution

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

q = - \frac{h _{0}^{3}}{12 μ} \nabla p = \frac{h _{0}^{3} G}{12 μ} e_{x},

ψ \sim A r sin θ, A = \frac{h _{0}^{3} G}{12 μ} .

Continuity of

q_{r} = r^{- 1} ψ_{θ}

gives

[ψ_{θ}]_{-}^{+} = 0

. Since

\nabla p = - 12 μ h^{- 3} q

, continuity of tangential pressure derivative gives

[h^{- 3} ψ_{r}]_{-}^{+} = 0

The harmonic solutions regular at zero and with the prescribed far field are

ψ_{>} = (A r + \frac{B}{r}) sin θ, ψ_{<} = C r sin θ .

The interface conditions give

B = A a^{2} \frac{h _{1}^{3} - h _{0}^{3}}{h _{0}^{3} + h _{1}^{3}}, C = \frac{2 A h _{1}^{3}}{h _{0}^{3} + h _{1}^{3}} .

Integrating the outward flux across the appropriate semicircle gives

2 a C = \frac{4 A a h _{1}^{3}}{h _{0}^{3} + h _{1}^{3}} .

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