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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2010

Paper 71

The <internal-wave ray tracing> equations preserve the horizontal <wavenumber> $k$ and laboratory frequency. For the increasing wind, the upward branch satisfies
$$
m(z)=\frac{N_0}{U(z)}\sqrt{1-q^2(z)},\qquad
q(z)=\frac{kU(z)}{N_0}=q_0(1+z/H),\qquad q_0=\frac{kU_0}{N_0}<1.
$$
The group-velocity ratio therefore gives
$$
\frac{dx}{dz}=\frac{k}{m}=\frac{q}{\sqrt{1-q^2}}.
$$
Integrating from $(0,0)$ yields
$$
\boxed{x(z)=\frac{H}{q_0}\left[\sqrt{1-q_0^2}-\sqrt{1-q_0^2(1+z/H)^2}\right].}
$$
This is the ascending branch of the <circular mountain-wave ray in linear shear>:
$$
(x-x_t)^2+(z+H)^2=\left(\frac{H}{q_0}\right)^2,\qquad
x_t=\frac{H}{q_0}\sqrt{1-q_0^2}.
$$
It reaches $z_t=H(q_0^{-1}-1)$ at $x=x_t$, where the energy direction is horizontal. After reflection, $m<0$ and the ray continues downwind and downward along the other branch
$$
x(z)=\frac{H}{q_0}\left[\sqrt{1-q_0^2}+\sqrt{1-q_0^2(1+z/H)^2}\right],
$$
returning to $z=0$ at $x=2x_t$. The last panel of the figure in part (b) shows this path. These are WKB energy rays, with the short neighborhood of the turning height supplied by its transition solution.

Solution

Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 71 / 2 / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 71 / 2 / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 71 / 2 / d / Solution

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