Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 2 / b / i

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- Solution i

Solution

Dot the momentum equation with

u

, multiply the temperature equation by

σ Ra θ

, and integrate over

V

. The pressure term vanishes by incompressibility, while skew-symmetry of incompressible transport removes both nonlinear advection terms. Integration by parts and the boundary conditions give

\frac{d E}{d t} = σ \int_{V} [2 Ra wθ - ∣\nabla u ∣^{2} - Ra ∣\nabla θ ∣^{2}], d V,

where

E = \frac{1}{2} \int_{V} (∣ u ∣^{2} + σ Ra θ^{2}) d V .

For

Ra = 0

, this reduces to

\frac{d E}{d t} = - σ \int_{V} ∣\nabla u ∣^{2} d V \leq 0,

so viscosity monotonically dissipates kinetic energy.

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