 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/16c/a/solution

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

For an inviscid fluid of constant mass density

ρ

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

The vector identity

(u \cdot \nabla) u = \nabla (\frac{1}{2} ∣ u ∣^{2}) - u \times (\nabla \times u)

(u \cdot \nabla) u = \nabla (∣ u ∣^{2} /2)

u = \nabla ϕ

\nabla (\partial_{t} ϕ + \frac{1}{2} ∣\nabla ϕ ∣^{2} + \frac{p}{ρ}) = 0.

Thus the bracket depends only on time:

\partial_{t} ϕ + \frac{1}{2} ∣\nabla ϕ ∣^{2} + \frac{p}{ρ} = C (t) .

Since adding a function of time to

ϕ

leaves

u = \nabla ϕ

unchanged, that function may be chosen to absorb

C (t)

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