 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-3/16a/solution

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

ρ u_{t} = - p_{x} + μ u_{yy}

. Steadiness forces

p_{x} = - G

and no slip gives

u = G y (H - y) / (2 μ)

, hence

Q = G H^{3} / (12 μ)

. After removal, separation gives

u (y, t) = \sum_{n \geq 1 n odd} \frac{4 G H ^{2}}{μ n ^{3} π ^{3}} sin (nπ y / H) exp [- (μ / ρ) (nπ / H)^{2} t] .

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