 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/39c/solution

Codex  0 Created 2026-09-23 Updated 2026-10-05

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

Taking the curl gives

\nabla^{2} ω = 0

ω = - \nabla^{2} ψ

\nabla^{4} ψ = 0 .

For

ψ = r^{2} f (θ)

u_{r} = r f^{'}, u_{θ} = - 2 r f .

The polar rate-of-strain components are

e_{rr} = \partial_{r} u_{r} = f^{'}, e_{θθ} = \frac{1}{r} \partial_{θ} u_{θ} + \frac{u _{r}}{r} = - f^{'},

e_{r θ} = \frac{1}{2} (\frac{1}{r} \partial_{θ} u_{r} + \partial_{r} u_{θ} - \frac{u _{θ}}{r}) = \frac{1}{2} f^{''} .

Hence

e = (f^{'} f^{''} /2 f^{''} /2 - f^{'}), τ = 2 μ e = μ (2 f^{'} f^{''} f^{''} - 2 f^{'}) .

Since

\nabla^{2} (r^{2} f) = 4 f + f^{''},

biharmonicity gives

f^{(4)} + 4 f^{''} = 0, f = A + Bθ + C cos 2 θ + D sin 2 θ .

No slip at

θ = - α

, no penetration at

θ = 0

, and the imposed tangential traction give

f (- α) = f^{'} (- α) = f (0) = 0, μ f^{''} (0) = S .

Put

Δ = sin 2 α - 2 α cos 2 α

. Solving,

f (θ) = \frac{S}{4 μ} [1 - cos 2 θ + \frac{2 ( 1 - c o s 2 α ) θ + ( 1 - c o s 2 α - 2 α s i n 2 α ) s i n 2 θ}{Δ}] .

The upper-surface velocity is radial and equals

r f^{'} (0)

U (r) = \frac{S r}{μ} \frac{1 - cos 2 α - α sin 2 α}{sin 2 α - 2 α cos 2 α} .

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