Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 333 / 3 / b

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- Solution b

Solution

Differentiate horizontal momentum with respect to

z

and vertical momentum with respect to

x

. When the two equations are subtracted, the terms proportional to

\overline{u}_{z} (u_{x}^{'} + w_{z}^{'})

vanish by incompressible flow. The pressure derivatives also cancel, leaving

D_{t} (u_{z}^{'} - w_{x}^{'}) + \overline{u}_{zz} w^{'} + σ_{x}^{'} = 0 .

Differentiate this equation with respect to

x

. Since continuity gives

(u_{z}^{'} - w_{x}^{'})_{x} = - (w_{xx}^{'} + w_{zz}^{'}),

one obtains

- D_{t} \nabla^{2} w^{'} + \overline{u}_{zz} w_{x}^{'} + σ_{xx}^{'} = 0.

Apply

D_{t}

and use the buoyancy equation

D_{t} σ^{'} = - N^{2} w^{'}

. Multiplication by

- 1

then gives

D_{t}^{2} (w_{xx}^{'} + w_{zz}^{'}) - \overline{u}_{zz} D_{t} w_{x}^{'} + N^{2} w_{xx}^{'} = 0 .

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