Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 333 (Fluid Dynamics of Climate)

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.

In the quasi-geostrophic approximation, replace $f_0^2-{\omega }^2$ by $f_0^2$. The resulting explicit frequency, and then its leading small-$d/H_0$ form, areFor $d>0$, the phase velocity has the sign of $f_0$. Thus propagation is eastward in the Northern Hemisphere, with the shallower side on the right. Differentiating the leading dispersion relation givesIn the long-wave limit the group velocity and phase velocity coincide:

To find the slow amplitude, integrate the low-frequency height equation across the step. This gives ${\partial }_t[H{\eta }_y]=f_0[H]{\eta }_x$, where brackets mean the upper-side minus lower-side value. For the given exponentially trapped outer profile, $[H{\eta }_y]=2H_{0}A/R_D$ and ${\eta }_x(x,0)=2{\eta }_0\delta (x)-A_x$. Hence long-wave transport along a depth step obeysThe method of characteristics, interpreted for the Dirac delta distribution source, givesFor $c_T>0$ this is $2{\eta }_0$ in $0<x<c_{T}t$ and zero outside: a surface depression is left in the wake of the wave. The sharp $x$-jump is a long-wave outer approximation. The fast adjustment region of width $O(R_D)$ near the origin, and the detailed wavefront, are unresolved. Keeping a smoothed background ${\eta }_s(x)$ instead would give $A_t+c_T{A}_x=c_T{\eta }_s^{\prime }(x)$ and $A={\eta }_s(x)-{\eta }_s(x-c_{T}t)$ at this order.

The along-step geostrophic flow is $u=-g{\eta }_y/f_0$. Integrating separately over the two sides givesIn the wake, $T_{-}=Q_{-}$ and $T_{+}=-Q_{+}$. The shallow-side transport is eastward and the deep-side transport westward, with magnitudes equal to the respective far-field currents. The net along-step transport is $T_{-}+T_{+}=-4g{\eta }_{0}d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.

Adopt the Boussinesq approximation, hydrostatic approximation, traditional approximation, and inviscid quasi-geostrophic approximation. Assume stable constant $N_0^2>0$, nonzero Coriolis parameter $f_0$, a beta plane, small Rossby number, $\beta L/|f_0|$ of that small order, and small displacement of the background stratification. Vertical advection of the buoyancy perturbation is then higher order. Define the quasi-geostrophic streamfunction byThe buoyancy and vertical vorticity equations at the retained order areDifferentiate the first in $z$. Since $J({\psi }_z,{\psi }_z)=0$, its material-derivative term becomes $f_0{D}_g{\psi }_{zz}$. Elimination of $w_z$ gives diabatically forced quasi-geostrophic potential vorticity:Thus it is the height gradient of heating, rather than heating alone, that produces potential vorticity at this order.

For the steady response linearized about rest, assume $\beta \ne 0$. The quasi-geostrophic potential-vorticity equation reduces to $\beta {\psi }_x=f_0{R}_z/N_0^2$. Choosing the horizontally uniform pressure gauge so that $\psi \to 0$ in the east gives the steady quasi-geostrophic response to localized heatingSetUsing the error function to integrate the Gaussian function givesThese velocities vanish as $x\to +\infty$. A depth-dependent but horizontally uniform addition to $\psi$ is dynamically irrelevant for the requested horizontal velocity. At $z=0$, streamlines are contours of $e^{-y^2/L^2}I(x)$. For $f_0,\beta ,R_0>0$, flow approaches the heating region from the west on its northern side, turns southward, and returns westward on its southern side. The contours are open, not closed gyres: $I$ tends to a nonzero constant in the west.

For $-H<z<0$, $R_z<0$ and the forcing decreases potential vorticity; steady balance requires $\beta v_g=f_0{R}_z/N_0^2<0$, giving southward flow. Below $z=-H$, $R_z>0$ and the meridional flow reverses. A constant heating rate independent of $z$ would have no such interior potential vorticity source.

For the nonlinear diagnostic quasi-geostrophic omega equation, write $\zeta ={\nabla }_h^2\psi$. Apply ${\nabla }_h^2$ to buoyancy evolution and $f_0{\partial }_z$ to vorticity evolution:Subtracting eliminates the pressure tendency and yieldsThe beta term is linear and must remain when the nonlinear terms are neglected. For the steady weak-forcing solution, $f_0\beta {\psi }_{xz}=f_0^2{R}_{zz}/N_0^2$, soThis also follows immediately from the steady linear buoyancy equation. It satisfies $w=0$ at $z=0$, decay at depth and horizontal infinity, and the linear quasi-geostrophic omega equation; with these homogeneous conditions the elliptic problem has no extra decaying homogeneous solution. The rigid-lid interpretation is an additional boundary approximation for this diagnostic response.

At $x=0$, the displayed $v_g$ and $w$ provide the requested meridional/vertical arrows. For positive heating there is upwelling concentrated near $y=0$, maximal in depth at $z=-H$, with northward flow below that depth and southward flow above. The vertical circulation tends to zero far from the forcing. A western zonal wake persists, but its meridional and vertical components vanish as $x\to -\infty$. Horizontal ageostrophic flow supplies the divergence associated with the upwelling; the projected $(v_g,w)$ slice is not itself a two-dimensional incompressible flow and need not have closed streamlines.

The steady calculation fixes an interior circulation, not its attainability from every possible initial and boundary state. For example, a rigid lid with initially uniform boundary buoyancy conserves that buoyancy at linear order because $R(0)=w(0)=0$. The formal steady profile instead generally has nonuniform $b^{\prime }(0)=f_0{\psi }_z(0)$. Such additional initial boundary data require a time-dependent adjustment and cannot simply be imposed on this particular steady solution. No boundary buoyancy data are supplied for the requested steady problem.

In an adiabatic quasi-geostrophic approximation, hydrostatic approximation and geostrophic balance give $b^{\prime }=f_0{\psi }_z$. The leading buoyancy equation is $D_g{b}^{\prime }+N^{2}w=0$; hence $w=-D_g(f_0{\psi }_z/N^2)$. At the rigid surface $z=\alpha y$, the no-normal-flow condition is $D_g(z-\alpha y)=0$, so $w=\alpha v_g=\alpha {\psi }_x$. The small boundary slope allows the disturbance condition to be evaluated at the flattened boundary $z=0$ to the retained order.

For thermal wind $U=\Lambda z$, choose the basic quasi-geostrophic streamfunction $\overline{\psi }=-\Lambda zy$. Its buoyancy anomaly is $\overline{b}=-f_0\Lambda y$, and its interior quasi-geostrophic potential vorticity is constant. Therefore the linear disturbance equations areThe term $-\Lambda {\psi }_x^{\prime }$ is the advection of the basic meridional buoyancy gradient. Since $q^{\prime }$ is simply advected at each fixed height, $q^{\prime }=0$ initially implies $q^{\prime }=0$ forever. At the lower boundary define $a=\alpha N^2/f_0$; thenThe PDF uses $\alpha$ for the lower-boundary slope throughout; the TeX's isolated $z=cy$ is a transcription error.

For nonzero zonal wavenumber let $\mu =N|k|/|f_0|>0$. The zero-interior-potential vorticity condition gives ${\hat{\psi }}_{zz}-{\mu }^2\hat{\psi }=0$. In the semi-infinite domain choose decay as $z\to \infty$. Writing $B_0(t)={\hat{\psi }}_z(0,t)$, potential-vorticity inversion then givesThis sloping-boundary Eady edge wave has all its time dependence in the boundary buoyancy equation; the interior responds instantaneously through elliptic potential-vorticity inversion. A single boundary wave oscillates without exponential growth. Vertical shear and topography contribute oppositely to its propagation. The contributions cancel when $\alpha =f_0\Lambda /N^2$, the slope of a background isopycnal. For $f_0>0$, positive $\Lambda$ without slope gives eastward propagation, while a positive slope without shear gives westward propagation. The phase velocity varies as $1/|k|$, and the trapping depth is $1/\mu$. Its frequency is independent of $|k|$ on each signed branch, so its zonal group velocity is zero in this ideal semi-infinite model.

With a rigid horizontal upper boundary, put $B_D(t)={\hat{\psi }}_z(D,t)$ and $M=\mu D$. Solving the same second-order inversion problem with both derivative data givesDifferentiation at $0$ and $D$ checks the prescribed values. At the upper boundary $w^{\prime }=0$, so $({\partial }_t+ik\Lambda D)B_D=ik\Lambda \hat{\psi }(D)$. Combining this with the lower-boundary condition givesThus two Boundary Rossby waves interact across the layer. For normal modes proportional to $e^{-ikct}$ their dimensional wave-speed matrix isFor $\Lambda \ne 0$, define $C=c/(\Lambda D)$ and $\tilde{\alpha }=a/\Lambda$. Its characteristic polynomial is the Eady model with a sloping lower boundary dispersion relationThe source writes $\tilde{\mu }=NkD/f_0$; for $k,f_0>0$ this is $M$, and the dispersion relation is unchanged by replacing that signed quantity by its absolute value. Its two roots areIf $\tilde{\alpha }>1$, the second term is positive, so $\Delta >0$ for every $M>0$: all modes are spectrally stable. If $\tilde{\alpha }<1$, the increasing function $M\tanh M$ runs from zero to infinity, so there is a unique $M$ withThere the squared term is zero and $\Delta <0$, giving an exponentially growing member of the complex conjugate pair. By continuity it lies in an unstable band. This proves both the requested instability for $0\le \tilde{\alpha }<1$ and instability for negative slopes in the sloping-boundary Eady model. It does not assert that every wavelength is unstable. Strongly negative slopes move the resonant band to large $M$, where the coupling is exponentially weak.

At $\tilde{\alpha }=1$ the speeds are real, but their crossing at $M=\coth M$ can give a defective neutral mode and algebraic growth; absence of exponential instability is weaker than boundedness of every initial disturbance. The nondimensional slope is undefined when $\Lambda =0$; the dimensional matrix remains valid and then has two real speeds. The instability for $\tilde{\alpha }<1$ is the counterpropagating wave instability mechanism: lower and upper boundary waves can match their laboratory phase speeds and exchange energy with the vertical shear.

Use the corrected PDF continuity equation ${\overline{v}}_{a,y}+{\overline{w}}_{a,z}=0$. Set $S=d{\rho }_s/dz=-{\rho }_0{N}^2/g<0$, $M=\overline{u^{\prime }{v}^{\prime }}$, and $B=\overline{{\rho }^{\prime }{v}^{\prime }}/S$. The transformed Eulerian mean replaces the Eulerian mean meridional circulation byThese substitutions preserve nondivergence, while density evolution becomes ${\overline{\rho }}_t+S{\overline{w}}_a^{\ast }=0$. Substituting ${\overline{v}}_a={\overline{v}}_a^{\ast }+B_z$ into the momentum equation givesThe geostrophic balance and hydrostatic approximation retain their original form. The vector $\mathbf{F}$ is the Eliassen–Palm flux. It gathers both eddy stresses into one mean-flow forcing rather than treating the eddy density flux as direct heating of the mean density.

For small-amplitude Rossby waves, the quasi-geostrophic wave-activity conservation law has the form ${\mathcal{A}}_t+\nabla \cdot \mathbf{F}=\mathcal{D}$, with the consistent signed wave activity convention and $\mathcal{D}$ representing sources or dissipation. Locally for a slowly varying wave packet, its Eliassen–Palm flux follows the group velocity multiplied by the signed wave-activity density. In particular, a wave can have westward phase propagation and upward wave-activity transport. Flux convergence during dissipation transfers wave momentum to the mean flow.

The non-acceleration theorem for quasi-geostrophic waves states that steady conservative waves have $\nabla \cdot \mathbf{F}=0$, and, with no independent boundary forcing or residual circulation, cannot accelerate the balanced mean zonal flow. The residual elliptic problem then has the zero solution under homogeneous impermeable/decaying conditions. The Eulerian mean circulation need not vanish: its eddy correction may be nonzero even when the residual mean circulation is zero. Time-dependent wave activity, dissipation and nonhomogeneous boundary conditions are the ways to evade the theorem. This distinction and the flux convention agree with Haynes's wave–mean-flow discussion, sections 6.2–6.3.

For the specified forcing let $\ell =\pi /L$ and $m=N\ell /|f_0|>0$. ThenIntroduce a streamfunction for the residual flow with the explicit sign conventionThe impermeable sidewalls require $v_a=0$. Since $B$ and $B_z$ vanish there, they also require $v_a^{\ast }=0$, giving this sine dependence. Differentiating geostrophic balance and hydrostatic approximation yields thermal wind, $f_0{\overline{u}}_z=(g/{\rho }_0){\overline{\rho }}_y$. Its time derivative, together with the transformed density equation, gives $f_0{\overline{u}}_{tz}=N^2{\overline{w}}_{a,y}^{\ast }$. The momentum equation consequently reduces to the Eliassen equation for residual circulation:Choose decay of the residual response at both vertical infinities, eliminating independently imposed remote circulation and exponentially growing homogeneous solutions. The wave flux itself need not vanish below the dissipating layer. These are conditions on the residual response, not an assumption that both Eulerian velocities vanish at both infinities.

Because ${\mathcal{F}}^{\prime \prime }=[\delta (z+D)-\delta (z-D)]/(2D)$, the decaying Green function $-e^{-m|z|}/(2m)$ gives the localized wave-drag response in a stratified channel:The derivative coefficient is explicitlyThe streamfunction is continuous; its derivative jumps by $+G_0/(2f_{0}D)$ at $z=-D$ and $-G_0/(2f_{0}D)$ at $z=D$ in accordance with the delta sources. The sharp change in dissipation rate permits the residual velocity jumps. The Eulerian velocity and zonal acceleration remain continuous.

Using the momentum equation givesHere $U(z)<0$ at every finite height, and it tends to zero at both infinities. Dissipation therefore produces westward acceleration throughout the channel, with exponentially decaying tails outside the directly forced layer. The geostrophic balance relation fixes the pressure tendency, up to a spatially uniform gauge:Set that gauge to zero by requiring pressure tendency to decay at both vertical infinities. For $f_0>0$, it is negative on $0<y<L/2$, zero at $L/2$, and positive on $L/2<y<L$.

The density tendency needed for the PDF's sketch follows from the transformed buoyancy equation:For $f_0>0$, it is negative in the upper southern quadrant and positive in the upper northern quadrant; below $z=0$ those signs reverse. It vanishes along $y=L/2$, along $z=0$, and at both vertical infinities. The TeX has misread this density-tendency request as a pressure tendency; the pressure formula above is also provided, but its signs do not reverse across $z=0$.

The corresponding Eulerian mean streamfunction isFor $f_0>0$ its bracket is positive and decreases with $z$. Thus the Eulerian meridional flow is southward everywhere; the vertical flow is downward in the southern half and upward in the northern half. Above the wave region it vanishes, but below itThis nonzero remote Eulerian vertical flow is cancelled by the eddy contribution in the residual circulation. Imposing zero Eulerian vertical velocity there would be incompatible with the prescribed undamped wave density flux.

The residual circulation consists of two oppositely directed cells: $C$ is positive above $z=0$ and negative below. Inside the dissipation layer $v_a^{\ast }>0$; outside it $v_a^{\ast }<0$. The upper cell moves northward on its lower flank, upward on the northern side, southward aloft and downward on the southern side; the lower cell has the reverse vertical sense.

Finally, integrate the momentum equation in height. Since $C$ vanishes at both ends and $\mathcal{F}(+\infty )-\mathcal{F}(-\infty )=1$, vertically integrated wave-drag acceleration isPutting $s=mD$ and integrating the interior expression givesThe full vertical integral equals the deposited wave momentum independently of stratification. The fraction within the dissipating layer increases from zero to one: for $s\ll 1$ it is $s+O(s^2)$, while for $s\gg 1$ it is $1-1/(2s)+O(e^{-2s}/s)$. Weak stratification or a thin layer spreads acceleration over the remote vertical scale $1/m$; strong stratification or a thick layer confines most acceleration to the dissipation region.