Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 39B / iii

Table of contents
- Solution iii

Solution

With

B = 0

, the stated biharmonic stream function for planar Stokes flow is

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.

This is the Stokes flow between touching counter-rotating cylinders.

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