Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 37A / c

Table of contents
- Solution c

Solution

Let the sphere have radius

a

a = 1

for the unit sphere implicit in the question. Since

Φ = \frac{α U}{r}, χ = β U \cdot \nabla \frac{1}{r} = - β \frac{U \cdot x}{r ^{3}},

one finds

u = (- \frac{α}{r} - \frac{β}{r ^{3}}) U + (- \frac{α}{r ^{3}} + \frac{3 β}{r ^{5}}) (U \cdot x) x .

On

r = a

, the no-slip boundary condition requires

u = U

for every surface normal. The coefficient of

(U \cdot n) n

must vanish, while the coefficient of

U

must equal one:

- \frac{α}{a} + \frac{3 β}{a ^{3}} = 0, - \frac{α}{a} - \frac{β}{a ^{3}} = 1.

Solving,

α = - \frac{3 a}{4}, β = - \frac{a ^{3}}{4} .

For a unit sphere,

α = - \frac{3}{4}, β = - \frac{1}{4} .

Substitution reproduces the standard translating sphere in Stokes flow.

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