 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/39b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

With

B = 0

ψ = A sin^{2} θ + C r sin θ + \frac{D s i n ^{3} θ}{r} .

Requiring it to vanish at

r = 2 a sin θ

and

r = 4 a sin θ

gives

A + 2 a C + \frac{D}{2 a} = 0, A + 4 a C + \frac{D}{4 a} = 0.

Thus

A = - 6 a C

and

D = 8 a^{2} C

C = α

yields

ψ = α (r sin θ - 6 a sin^{2} θ + \frac{8 a ^{2} sin ^{3} θ}{r}) = \frac{α sin θ}{r} (r - 2 a sin θ) (r - 4 a sin θ) .

Using

u_{r} = ψ_{θ} / r

and

u_{θ} = - ψ_{r}

, differentiation shows that

u_{r} cos θ + u_{θ} sin θ

equals

α

on the inner boundary and

- α /2

on the outer boundary. Both wall conditions hold precisely when

α = a Ω.

 New to topics? Read the docs here!