 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/38a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The similarity form gives

u = - \frac{A}{r ^{2}} F^{'}, w = \frac{( A ν ) ^{1/2}}{2 r ^{3/2}} (F - 3 η F^{'}) .

ρ^{- 1} d P / d r = 2 A^{2} / r^{5}

in the correctly written radial boundary-layer equation

u u_{r} + w u_{z} = - ρ^{- 1} P_{r} + ν u_{zz}

gives

F^{(3)} + 2 (1 - F^{'2}) - \frac{1}{2} F F^{''} = 0.

No slip and matching to the sink require

F (0) = F^{'} (0) = 0

and

F^{'} (\infty) = 1

. The boundary-layer flux is of order

r U δ

, relative to the sink flux

A

it is

\frac{r U δ}{A} \sim (ν r / A)^{1/2} .

Mass conservation therefore requires an outer correction of this relative size, which is small in the assumed region.

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