Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 333 (Fluid Dynamics of Climate)

Let $\mathbf{u}=(u,v)$ and let the layer thickness be $H=H_0+\eta$. The inviscid shallow water equations areThey follow from a homogeneous incompressible fluid with small aspect ratio, hydrostatic pressure, negligible vertical acceleration, horizontal velocity nearly uniform through the depth, a material free surface, and a rigid stationary bottom. Rotation may be represented by an f-plane or beta plane. The model is useful because many atmospheric and oceanic motions are horizontally much broader than their depth, while the free surface or an internal density interface still supports waves and potential vorticity dynamics.

Let $\zeta =v_x-u_y$. Taking the vertical curl of momentum giveswhere $Df/Dt=\beta v$ supplies the planetary-vorticity term when $f$ varies. Continuity gives $DH/Dt=-H\nabla \cdot \mathbf{u}$. Combining the two equations yields material conservation of shallow-water potential vorticity:

With no $y$ dependence, the linearized shallow water equations areTheir conserved linear potential-vorticity anomaly isbecause the initial velocity jump has derivative $2v_0\delta (x)$. The final steady state is in geostrophic balance, so $u=0$ and $fv=g{\eta }_x$. HenceTaking $f>0$ for definiteness and requiring decay at infinity gives

Initially, the potential energy is zero and, per unit length in $y$,For $L\gg R_d$, the final kinetic and potential energies are equal:Thus $E_{msteady}=\rho H_0{v}_0^2{R}_d$. During geostrophic adjustment, inertia-gravity waves carry the excess energy out of $|x|<L$; energy in that finite region is therefore not conserved. Their long-wave speed is $c=\sqrt{g{H}_0}$, so the adjustment of the stated region takes

Sverdrup balance applies to a steady, large-scale, small-Rossby number, hydrostatic and nearly geostrophic ocean interior on a beta plane. The flow is depth-integrated, relative-vorticity advection and interior friction are negligible, density is treated as constant for the barotropic balance, and the principal vorticity source is the curl of wind stress. The balancesays that wind input of vertical vorticity is balanced by meridional advection of planetary vorticity, or equivalently by the stretching needed to conserve potential vorticity as parcels move across latitude circles.

Use $u=-{\psi }_y$, $v={\psi }_x$. Since $v=W(y)/\beta$ and the eastern boundary may be chosen as the zero streamline,The lower half-basin has southward interior flow and a northward western return current, giving a clockwise ocean gyre. The upper half has northward interior flow and a southward western return current, giving a counterclockwise gyre. The interior solution cannot satisfy no slip at every wall, so viscous layers complete the circulation.

With constant lateral viscosity $\nu$, the vorticity equation contains $\nu {\nabla }^4\psi$. At the western and eastern walls, balancing $\beta {\psi }_x$ against four $x$ derivatives gives the Munk boundary layer scaleAt the southern and northern walls, $x$ variation remains on the basin scale $L_X$ while four $y$ derivatives occur across the layer, so

At small Rossby number, $q=(\zeta +f_0)/H\simeq f_0/H$. For steady flow with $H=y^2$, the forced potential-vorticity equation becomessoDefine a transport streamfunction by $Hu=-{\psi }_y$ and $Hv={\psi }_x$. Choosing the eastern wall as $\psi =0$ gives the topographic Sverdrup balance solutionand thereforeIn each half-basin, the transport streamlines are the level curves $(L_X-x)y^3=\text{constant}$. They move westward and toward $y=0$. This interior solution treats the two sides of the degenerate line $H=0$ separately and requires boundary layers to enforce solid-wall conditions.

For both $y>0$ and $y<0$, the velocity $v=-W_{0}y/(2f_0)$ carries parcels toward $y=0$, where $H=y^2$ decreases. Their leading potential vorticity $f_0/H$ therefore increases along the path, exactly as required by the positive source $W_0/H$. The explicit interior velocity has zero relative vorticity, so the leading change comes from vortex-column stretching rather than from $\zeta /H$. Relative vorticity and friction become important only where boundary layers close the circulation or where the singular depth profile near $y=0$ invalidates the approximation.

The quasi-geostrophic approximation requires small Rossby number, nearly horizontal geostrophic balance, hydrostatic vertical balance, small interface or density displacements, stable background stratification, and horizontal scales much larger than the vertical scale. The Boussinesq approximation and a beta plane are used, ageostrophic motion enters only at the order needed to evolve potential vorticity, and here the buoyancy frequency $N$ is constant. Under these assumptions the materially conserved three-dimensional quasi-geostrophic potential vorticity is

The QG thermodynamic relation makes vertical velocity proportional to the material derivative of ${\psi }_z$. At a flat rigid boundary, no normal flow therefore givesor ${\hat{\psi }}^{\prime }(0)=0$ for a nonzero-frequency mode. Let $K^2=k^2+l^2$ and take the vertical structure $\hat{\psi }={\psi }_0\cos (mz)$, which satisfies this condition. Linearizing $q_t+\beta {\psi }_x=0$ gives the Baroclinic Rossby wave dispersion relationFor long horizontal waves, $K{R}_d\ll 1$, so $\omega \simeq -\beta k{R}_d^2$: the vertical-mode deformation term controls the response. For short waves, $K{R}_d\gg 1$, so $\omega \simeq -\beta k/K^2$, the Barotropic Rossby wave form. In both limits the zonal phase propagation is westward relative to the mean flow.

Linearization about $U\hat{\mathbf{x}}$ changes the frequency to the intrinsic value $\omega -Uk$. ThusA mountain-fixed disturbance has $\omega =0$, so its vertical wavenumber satisfiesThe quasi-geostrophic mountain wave propagates vertically only when $m^2>0$. For $U>0$, this requiresOtherwise the disturbance is vertically evanescent.

At the topographic boundary, no normal flow gives $w^{\prime }=U{h}_x$. Since QG buoyancy is $b^{\prime }=f_0{\psi }_z^{\prime }$ and $U{b}_x^{\prime }+N^2{w}^{\prime }=0$, the linearized lower condition isFor the specified ridge $h=h_0\cos kx$, matching its horizontal structure sets $l=0$; retaining $K^2=k^2+l^2$ also covers a sinusoidal ridge with that horizontal wavevector. When $U>\beta /K^2$, defineThe decaying solution and its boundary amplitude areIncreasing $N$ raises the streamfunction amplitude in proportion to $N$ while reducing the vertical e-folding length ${\gamma }^{-1}$ in proportion to $N^{-1}$: stronger stratification produces a stronger but more shallowly trapped disturbance.

Define the transformed Eulerian mean circulation byThe added terms form a nondivergent eddy-induced circulation, so ${\partial }_y{\overline{v}}_a^{\ast }+{\partial }_z{\overline{w}}_a^{\ast }=0$. The buoyancy equation becomesSubstituting ${\overline{v}}_a={\overline{v}}_a^{\ast }+N^{-2}{\partial }_z\overline{v^{\prime }{\sigma }^{\prime }}$ into the momentum equation giveswhere the components of the Eliassen–Palm flux are

For an Equatorial Kelvin wave, set $\hat{v}=0$. The zonal and meridional momentum equations givesoHydrostatic balance and buoyancy evolution give $\hat{\sigma }=im\hat{\varphi }$ and $\hat{w}=-\omega m\hat{\varphi }/N^2$. Continuity then givesBecause the mode must decay as $|y|\to \infty$, one needs $k/\omega >0$. With the stipulated $\omega >0$, the acceptable branch is thereforeThe branch $\omega =-Nk/|m|$ would require $k<0$ when $\omega >0$, making the Gaussian exponent positive and the mode unbounded; it is not an equatorially trapped solution.

For real $k,m,\omega$,Upward group propagation has $m<0$, while a Kelvin wave has $k>0$, so $\overline{u^{\prime }{w}^{\prime }}>0$. Dissipation makes this eastward momentum flux decrease with height; therefore $-{\partial }_z\overline{u^{\prime }{w}^{\prime }}>0$ and the wave exerts an eastward force on the mean flow.

Choose the phase so that the trapped meridional structure $\hat{v}(y)$ is real. The linear wave equations then make $\hat{\varphi }$ and $\hat{u}$ purely imaginary relative to $\hat{v}$, soFor every equatorial mode, zonal momentum, hydrostatic balance, and buoyancy evolution giveSubstitution into the transformed flux shows that the terms involving $\hat{v}{\hat{\varphi }}^{\ast }$ cancel between momentum and buoyancy transport. Hence

For upward-propagating waves $m<0$, so $F^{(z)}$ has the opposite sign to $k$. Where a wave dissipates, its flux tends toward zero with height: eastward modes with $k>0$ have $F^{(z)}<0$ and exert an eastward force, while westward modes with $k<0$ have $F^{(z)}>0$ and exert a westward force. Thus dissipation deposits each wave's signed zonal pseudomomentum in the stratospheric mean flow. The formula also shows why $F^{(z)}$ is a wave-activity or pseudomomentum flux rather than merely $-\overline{u^{\prime }{w}^{\prime }}$: the eddy buoyancy transport supplies the correction required by the transformed circulation.