 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/38b/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

ϵ = \frac{ν}{Q}, η = \frac{y}{ϵ x}, ψ = ν Q f (η) = Q ϵ f (η) .

u = ψ_{y}

v = - ψ_{x}

gives

u = \frac{Q}{x} f^{'} (η), v = \frac{Q ϵ}{x} η f^{'} (η) .

At fixed

y

u_{x} = - \frac{Q}{x ^{2}} [f^{'} + η f^{''}], u_{y} = \frac{Q}{ϵ x ^{2}} f^{''}, u_{yy} = \frac{Q}{ϵ ^{2} x ^{3}} f^{'''} .

The terms proportional to

η f^{'} f^{''}

cancel, leaving

u u_{x} + v u_{y} = - \frac{Q ^{2}}{x ^{3}} (f^{'})^{2} .

Since

U = - Q / x

U U^{'} = - Q^{2} / x^{3}

, while

ν u_{yy} = Q^{2} f^{'''} / x^{3}

. The boundary-layer equation therefore becomes

f^{'''} = 1 - (f^{'})^{2} .

Choosing the wall value of the streamfunction as zero, imposing no slip, and matching to

U = - Q / x

give

f (0) = 0, f^{'} (0) = 0, f^{'} (η) \to - 1 (η \to \infty) .

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