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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 38C / c / Solution

Integrating the incompressible continuity equation

u_{x} + v_{y} = 0

across the moving gap, with

v (x, 0) = 0

and

v (x, h) = - V

, gives the Reynolds lubrication equation

\frac{d Q}{d x} = V .

Hence

Q = V x + C

for a constant

C

, and the flux formula from part (b) gives

p_{x} = \frac{6 μU}{h ^{2}} - \frac{12 μ ( V x + C )}{h ^{3}} .

Define the spatial mean

⟨ f ⟩ = \frac{1}{L} \int_{0}^{L} f (x) d x .

Because both ends meet fluid at pressure

p_{0}

, the pressure recovery condition in lubrication flow is

0 = p (L) - p (0) = \int_{0}^{L} p_{x} d x .

It determines

C = \frac{U}{2} \frac{⟨ h ^{- 2} ⟩}{⟨ h ^{- 3} ⟩} - V \frac{⟨ x h ^{- 3} ⟩}{⟨ h ^{- 3} ⟩} .

Therefore the requested pressure gradient in a translating and squeezing finite gap is

p_{x} = \frac{6 μU}{h ^{2}} - \frac{12 μ}{h ^{3}} [V x + \frac{U}{2} \frac{⟨ h ^{- 2} ⟩}{⟨ h ^{- 3} ⟩} - V \frac{⟨ x h ^{- 3} ⟩}{⟨ h ^{- 3} ⟩}] .

Solved by gpt-5.6-sol high.

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