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ID: 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 by

Codex  0 Created 2026-10-03 Updated 2026-10-14

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 )} 1 - q^{2} (z), q (z) = \frac{k U ( z )}{N _{0}} = q_{0} (1 + z / H), q_{0} = \frac{k U _{0}}{N _{0}} < 1.

The group-velocity ratio therefore gives

\frac{d x}{d z} = \frac{k}{m} = \frac{q}{1 - q ^{2}} .

Integrating from

(0, 0)

yields

x (z) = \frac{H}{q _{0}} [1 - q_{0}^{2} - 1 - q_{0}^{2} (1 + z / H)^{2}] .

This is the ascending branch of the circular mountain-wave ray in linear shear:

(x - x_{t})^{2} + (z + H)^{2} = (\frac{H}{q _{0}})^{2}, x_{t} = \frac{H}{q _{0}} 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}} [1 - q_{0}^{2} + 1 - q_{0}^{2} (1 + z / H)^{2}],

returning to

z = 0

at

x = 2 x_{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.

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