 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-2/5d/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

For incompressible inviscid flow without body force,

\nabla \cdot u = 0, \partial_{t} u + (u \cdot \nabla) u = - ρ^{- 1} \nabla p .

Taking curl gives

\partial_{t} ω + (u \cdot \nabla) ω = (ω \cdot \nabla) u, ω = \nabla \times u,

equivalently

\partial_{t} ω = \nabla \times (u \times ω)

Direct differentiation of the stated field gives

ω = u

. Hence

u \times ω = 0

\partial_{t} ω = 0

and the field is steady. Finally

(u \cdot \nabla) u = \nabla (∣ u ∣^{2} /2) - u \times ω = \nabla (∣ u ∣^{2} /2) .

Euler's equation therefore implies

\nabla (p + \frac{1}{2} ρ ∣ u ∣^{2}) = 0,

H

is spatially uniform.

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