 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/16a/a/solution

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

For the parallel flow

u = (u (y, t), 0)

\partial_{t} u = ν \partial_{y}^{2} u .

Substitution of

u = Re [U_{0} f (y) e^{iω t}]

gives

ν f^{''} = iω f .

The boundary velocities require

f (L_{0}) = 1

and

f (- L_{0}) = - 1

. The problem is antisymmetric under

y \mapsto - y

, so

f

is odd. With

k = \frac{iω}{ν} = (1 + i) \frac{ω}{2 ν} = \frac{( 1 + i ) Δ}{L _{0}},

the unique odd solution is

f (y) = \frac{sinh [( 1 + i ) Δ y ]}{sinh [( 1 + i ) Δ ]}, y = \frac{y}{L _{0}}, Δ = (\frac{ω L _{0}^{2}}{2 ν})^{1/2} .

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