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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-65/3/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 65 / 3 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Interpret the two-dimensional dependence as independence of

z

, allowing the velocity to have a vertical component

U = u_{z} (x, y)

. Write

u_{h} = (u_{x}, u_{y})

and

B_{h} = (A_{y}, - A_{x})

. The resistive induction equation, with a gauge fixing the potential's zero mean, gives

A_{t} + u_{h} \cdot \nabla A = η \nabla^{2} A, B_{t} + u_{h} \cdot \nabla B = B_{h} \cdot \nabla U + η \nabla^{2} B .

The first is a homogeneous advection-diffusion equation. Multiply it by

A

, average over the periodic box and use integration by parts. Advection contributes zero because

\nabla \cdot u_{h} = 0

; there are no surface terms. Thus

\frac{1}{2} \frac{d}{d t} ⟨ A^{2} ⟩ = - η ⟨ ∣\nabla A ∣^{2} ⟩ \leq - η k^{2} ⟨ A^{2} ⟩ .

Multiplying by

e^{2 η k^{2} t}

makes its derivative nonpositive. With the initial norm convention

⟨ A^{2} (0)⟩ = A_{0}^{2}

, this proves

⟨ A^{2} (t)⟩ \leq A_{0}^{2} e^{- 2 η k^{2} t} .

The Poincaré inequality uses the zero-mean assumption; a spatially constant flux potential would otherwise have no gradient. Zero mean remains preserved by this incompressible periodic equation.

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