Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 3

In linear quasi-geostrophic approximation, buoyancy and vertical velocity satisfyImpermeability means $w^{\prime }=0$ at $z=-H$ and $z=0$. For a normal mode with nonzero frequency this is equivalent to the Neumann conditionswhere the primes on $\hat{\psi }$ denote derivatives with respect to $z$. Equivalently, ${\psi }_z^{\prime }=0$ at both rigid boundaries.

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Linearizing conservation of three-dimensional quasi-geostrophic potential vorticity about rest givesFor $K^2=k^2+l^2$, the plane-wave ansatz yieldsThus the vertical structure equation isThe rigid-boundary eigenfunctions areSubstitution gives the Baroclinic Rossby wave dispersion relationsThe $n=0$ member is the Barotropic Rossby wave; $n\ge 1$ are baroclinic vertical modes.

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At the free surface, the linear kinematic boundary condition isHydrostatic pressure and geostrophic-streamfunction normalization giveCombining this withgivesFor an oscillatory disturbance with no time-independent boundary offset, the free-surface boundary condition is thereforeThe lower rigid boundary retains ${\psi }_z^{\prime }(-H)=0$.

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Forthe surface value is simply ${\psi }^{\prime }(0)=A$. The dynamic free-surface condition from part i givesThe cosine is the leading small-surface-displacement approximation to the exact quasi-geostrophic vertical mode; its derivative vanishes at both rigid-boundary locations.

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The isopycnal displacement isFor the $n$th cosine mode,Its maximum magnitude is thereforeFor $H=3000\mathrm{m}$, $N=10^{-3}{\mathrm{s}}^{-1}$, and $|\eta {|}_{\max }=10^{-2}\mathrm{m}$,The first baroclinic mode therefore displaces interior density surfaces by about $10^2\mathrm{m}$ even though the surface moves only one centimetre.

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