Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 39A / a

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- Solution a

Solution

The lubrication approximation requires

H / L ≪ 1

, small wall slope, and negligible inertia, for example

ρ U H / μ ≪ L / H

. To leading order,

p_{z} = 0, μ u_{zz} = \nabla p, \nabla \cdot u + w_{z} = 0.

With no slip at

z = 0, h

,

u = U (1 - \frac{z}{h}) + \frac{z ( z - h )}{2 μ} \nabla p .

Integration gives

q = \frac{U h}{2} - \frac{h ^{3}}{12 μ} \nabla p .

Integrated incompressibility gives

\nabla \cdot q = 0

, so locally

q = \nabla \land (0, 0, ψ)

. Solving the flux law for

\nabla p

and taking its curl yields

\nabla \cdot (h^{- 3} \nabla ψ) = - U h^{- 3} h_{y} .

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