Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 2 / a

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- Solution a

Solution

Let

D / D t = \partial_{t} - S x \partial_{y}

be the material derivative along the background linear shear flow

u_{0} = - S x e_{y}

. Linearizing about constant

Σ

gives

\frac{D Σ ^{'}}{D t} = - Σ (\partial_{x} v_{x} + \partial_{y} v_{y}),

\frac{D v _{x}}{D t} - 2Ω v_{y} = - \partial_{x} Ψ - \frac{v _{s}^{2}}{Σ} \partial_{x} Σ^{'},

\frac{D v _{y}}{D t} + (2Ω - S) v_{x} = - \partial_{y} Ψ - \frac{v _{s}^{2}}{Σ} \partial_{y} Σ^{'} .

The coefficient

2Ω - S

includes the contribution

v \cdot \nabla u_{0} = - S v_{x} e_{y}

from perturbation advection of the background shear.

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