 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/37a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Suppose

U_{+} = U

and

U_{-} = - U

. Integrating the two velocity profiles gives the volume fluxes per unit span

Q_{+} = d_{+} U + \frac{τ d _{+}^{2}}{2 μ _{+}}, Q_{-} = - d_{-} U - \frac{τ d _{-}^{2}}{2 μ _{-}} .

Here

τ = \frac{- 2 U}{d _{+} / μ _{+} + d _{-} / μ _{-}} .

Consequently

Q_{+} = \frac{U d _{+} d _{-} / μ _{-}}{d _{+} / μ _{+} + d _{-} / μ _{-}}, Q_{-} = - \frac{U d _{+} d _{-} / μ _{+}}{d _{+} / μ _{+} + d _{-} / μ _{-}},

so all dependence on the interface position cancels from the ratio:

\frac{Q _{-}}{Q _{+}} = - \frac{μ _{-}}{μ _{+}} .

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