 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/39d/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

At zero Reynolds number, an incompressible Newtonian fluid obeys the Stokes equations

- \nabla p + μ \nabla^{2} u = 0, \nabla \cdot u = 0.

Taking the divergence gives

\nabla^{2} p = 0

. Taking the curl and writing

ω = \nabla \times u

gives

μ \nabla^{2} ω = 0.

Thus both pressure and vorticity are harmonic in the fluid.

u \to 0 (r \to \infty),

and, on

r = a

u \cdot n = U \cdot n, (I - n n^{T}) u = (I - n n^{T}) U .

The first is impermeability of the solid surface and the second is viscous no slip. Together they say

u = U

on the sphere.

For the stated solution, put

A (r) = \frac{3 a}{4 r} + \frac{a ^{3}}{4 r ^{3}}, B (r) = \frac{3 a}{4 r ^{3}} - \frac{3 a ^{3}}{4 r ^{5}}, s = U \cdot x .

Then

u_{j} = A U_{j} + B s x_{j}

, and

\frac{\partial u _{j}}{\partial x _{i}} = \frac{A ^{'}}{r} x_{i} U_{j} + \frac{B ^{'}}{r} s x_{i} x_{j} + B U_{i} x_{j} + B s δ_{ij},

where

A^{'} = - \frac{3 a}{4 r ^{2}} - \frac{3 a ^{3}}{4 r ^{4}}, B^{'} = - \frac{9 a}{4 r ^{4}} + \frac{15 a ^{3}}{4 r ^{6}} .

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