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Waves and compressible flow course of the University of Oxford

Past exam of the waves and compressible flow course of the University of Oxford

2023

For $w=e^{\alpha z/2}W$, direct <differentiation> gives
$$
w_{zz}-\alpha w_z=e^{\alpha z/2}\left(W_{zz}-\frac{\alpha^2}{4}W\right),\qquad w_{xx}=e^{\alpha z/2}W_{xx}.
$$
Thus the transformed <partial differential equation> and its impermeable-wall <boundary conditions> are
$$
\boxed{\partial_t^2\left(\nabla^2W-\frac{\alpha^2}{4}W\right)+\alpha g W_{xx}=0,\qquad W(x,0,t)=W(x,h,t)=0}.
$$
The positive exponential factor does not change where the vertical <velocity> vanishes.

Solution

Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 2 / a / ii / Solution

Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 2 / a / ii / Solution

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Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 2 / a / ii / Solution

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