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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-333/4/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 4 / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

Put

v = 0

and use amplitudes proportional to

e^{i (k x + m z - ω t)}

. Zonal momentum gives

\overset{u}{^} = \frac{k}{ω} \hat{ϕ},

while incompressibility, buoyancy evolution, and hydrostatic balance give

\overset{w}{^} = - \frac{k}{m} \overset{u}{^} = - \frac{ωm}{N ^{2}} \hat{ϕ} .

Hence

ω^{2} = \frac{N ^{2} k ^{2}}{m ^{2}} .

Meridional geostrophic balance requires

\frac{d ϕ ^}{d y} = - \frac{β k}{ω} y \hat{ϕ},

so

\hat{ϕ} (y) = Φ_{0} exp (- \frac{β k}{2 ω} y^{2}) .

Decay as

∣ y ∣ \to \infty

, together with

β > 0

and

k > 0

, requires

ω > 0

. Therefore

ω = \frac{N k}{∣ m ∣}, \hat{ϕ} (y) = Φ_{0} exp (- \frac{β ∣ m ∣}{2 N} y^{2}) .

The negative-frequency root makes the Gaussian exponent positive and the solution diverge away from the equator. The acceptable branch is the eastward equatorial Kelvin wave.

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