 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/13c/solution

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

Write

u = \nabla ϕ + β \nabla α, L = - β α_{t} - \frac{1}{2} u \cdot u .

ϕ

gives

\nabla \cdot u = 0

α

gives

β_{t} + \nabla \cdot (β u) = 0,

β

gives

α_{t} + u \cdot \nabla α = 0.

Using incompressibility, the middle equation becomes

β_{t} + u \cdot \nabla β = 0.

These are the three required Euler-Lagrange equations.

Let

D_{t} = \partial_{t} + u \cdot \nabla

. Since

D_{t} α = D_{t} β = 0

, differentiating

u_{i} = \partial_{i} ϕ + β \partial_{i} α

gives

D_{t} u_{i} = \partial_{i} (D_{t} ϕ) - u_{j} \partial_{i} u_{j} = \partial_{i} (D_{t} ϕ - \frac{1}{2} u^{2}) .

Now

D_{t} ϕ = ϕ_{t} + u \cdot \nabla ϕ = ϕ_{t} + u^{2} + β α_{t},

because

u \cdot \nabla α = - α_{t}

. Hence

D_{t} u_{i} = \partial_{i} (ϕ_{t} + β α_{t} + \frac{1}{2} u^{2}) = - \partial_{i} p,

where

p = - \frac{1}{2} u^{2} - ϕ_{t} - β α_{t}, f (ϕ_{t}, α_{t}, β) = - ϕ_{t} - β α_{t} .

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