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The hydrostatic approximation gives $p=\rho g(h-z)$ above the horizontal floor, up to atmospheric pressure. The horizontal Stokes flow equation is $\mu {\mathbf{u}}_{zz}=\rho g\nabla h$, with the no-slip boundary condition at $z=0$ and zero tangential stress on the free surface. Integrating twice gives $\mathbf{u}=(\rho g/\mu )\nabla h(z^2/2-hz)$ and the lubrication gravity-current fluxFor the squeegee lubrication model, use $h_0$ vertically, $\ell =\rho g{h}_0^3/(\mu U)$ horizontally, and $\ell /U$ in time. If $X$ is the laboratory coordinate in the direction of travel, set $x=(X-Ut)/\ell$, $y=Y/\ell$, $t=U{t}_{\mathrm{lab}}/\ell$, and $h=h_{\mathrm{lab}}/h_0$. The blade half-length is $\alpha =L/\ell$. The moving-frame flux is $\mathbf{q}=-h{\mathbf{e}}_x-h^3\nabla h/3$. Conservation of volume gives the dimensionless lubrication equationUsing these scales as representative of the actual flow, lubrication theory requires a small aspect ratio and negligible inertia relative to vertical viscous resistance:The second is the Reynolds number based on $h_0$, multiplied by $h_0/\ell$. Surface tension is neglected as assumed. Large piles, rounded gaps, and narrow end regions need their own local slope and inertia checks if these representative scales cease to describe them.

Let $z$ increase upwards, so the bed is at $z=-b(x)$, the surface is at $z=h(x,t)$, and $H=h+b$. In the shallow-shelf approximation the incompressible flow has nearly depth-uniform horizontal speed $u$ and vertical strain $w_z=-u_x$. The Newtonian fluid stress tensor, together with the hydrostatic approximation, then givesThe factor four includes both horizontal extension and the pressure correction required by vertical compression. Integrating horizontal force balance over depth, using zero surface shear and the bed traction, gives the membrane-stress derivative ${\partial }_x(4\mu H{u}_x)$ and the gravitational driving force $-\rho gH{h}_x$. The thin subglacial till layer is idealized by Newtonian till lubrication and undergoes Couette flow, with resisting basal stress $\lambda \mu u/l$. Hence force balance and mass conservation yieldSince $b$ is fixed, $H_t=h_t$. No accumulation or ablation is included.

At the grounding line, the first boundary condition is ice-sheet flotation:For an unbuttressed ice shelf, the second is the extensional stress required to balance the difference between the integrated ice and seawater hydrostatic pressures. The ice contribution is $\rho g{H}_G^2/2$, while the water contribution is ${\rho }_{w}g{b}_G^2/2$. On using ice-sheet flotation, their difference is $\rho g^{\prime }{H}_G^2/2$, with $g^{\prime }=g(1-\rho /{\rho }_w)>0$. Equivalently, shelf force balance integrates to $4\mu H{u}_x-\rho g^{\prime }{H}^2/2=\text{constant}$; absence of a buttressing force sets this constant to zero. Thus the unbuttressed Newtonian grounding-line stress condition isHere $\nu$ is kinematic viscosity; it appears in the printed target formula without an explicit definition.

The ratio of longitudinal stress divergence to till drag is of order $4lH/(\lambda L^2)$. The given small-parameter limit therefore yields the friction-dominated bulk relationThe coefficient depends on the basal lubrication, and $h_x<0$ gives seaward motion. This is a bulk reduction: longitudinal stress has been neglected in the differential equation but its boundary traction remains specified. A complete uniformly valid approximation close to the grounding line could require a membrane-stress boundary layer. In the remainder, use the pointwise friction closure and retain the shelf stress condition as the boundary data, as requested in the paper.

Differentiate the ice-sheet flotation condition along the moving grounding line:Now $H_x=h_x+\alpha$. Substitution of the friction-dominated speed and the shelf stress boundary condition gives the friction-dominated grounding-line evolution law:The coefficient of ${\dot{x}}_G$ is the spatial derivative of the flotation deficit, up to sign. If it vanishes, the implicit equation remains the correct compatibility condition but division to determine a finite speed is invalid; tangential contact needs separate analysis.

In a steady state, volume flux per unit width is constant: $Hu=q_0$. The friction relation givesDifferentiating $Hu=q_0$ and using $u_x=g^{\prime }H/(8\nu )$ at the grounding line givesPut $\beta =l/\lambda$, $\mathcal{H}=H/\beta$ and $\mathcal{Q}=\nu q_0/(g{\beta }^3)$. Substitution and cancellation of $g{\beta }^5/\nu$ yield the steady friction-dominated grounding-line thickness relation:For $q_0>0$ and a bed deepening seaward ($\alpha \ge 0$), the polynomial $(g^{\prime }/8g){\mathcal{H}}^5+\alpha \mathcal{Q}{\mathcal{H}}^2-{\mathcal{Q}}^2$ is strictly increasing for $\mathcal{H}>0$, starts negative and tends to infinity, so it has exactly one positive root. At that root $\mathcal{Q}>\alpha {\mathcal{H}}^2$ and $H_x<0$, as required by steady outward extension. The relation belongs to the specified friction closure; it is not a flux law for arbitrary till rheology or arbitrary membrane-stress matching.

For slow cooling of an astrophysical disk, let $t_{\mathrm{cool}}\gg {\Omega }^{-1}$ and assume a vertically symmetric laminar column with no significant radial mass exchange. Its vertical velocity is of order $H/t_{\mathrm{cool}}$, so the vertical material derivative of that velocity is of order $H/t_{\mathrm{cool}}^2$. Relative to gravity or pressure acceleration $\sim {\Omega }^{2}H$, the inertial correction is $O[(\Omega t_{\mathrm{cool}}{)}^{-2}]$.

The hydrostatic approximation therefore holds to leading order while the column slowly cools and contracts. Non-turbulent evolution removes the prescribed turbulent heating, but not the compressional work in the ideal gas energy equation:The continuity equation must still determine the slow vertical flow; $D_t$ is not generally ${\partial }_t$ at fixed height. This is a quasi-static approximation for a well-prepared, mechanically stable column, with fast free oscillations or growing convection excluded from the assumed slow solution.

At fixed cylindrical radius, let $w=u_z$ and use the continuity equation ${\partial }_t\rho +{\partial }_z(\rho w)=0$. For the similarity solution $z=\xi \eta (t)$ and $\rho =\tilde{\rho }(\xi )/\eta$, the continuity equation reduces toHere $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:The hydrostatic approximation and ideal gas law becomeThe material derivatives are $D_{t}P=\tilde{P}\dot{\eta }$ and $D_t\rho =-\tilde{\rho }\dot{\eta }/{\eta }^2$, so the energy equation givesSeparation of variables requires a positive constant $K$ such thatIntegrating with $\eta (0)=1$ gives the slow cooling of an astrophysical disk similarity factorA closed set of profile equations is thereforeThese have the same polytropic vertical structure in stellar gravity as part (b). Writingwith $P_0=\mathcal{R}{\rho }_0{T}_0$, fixesReturning to physical height, the explicit profiles and evolving semi-thickness for $|z|\le H(t)$ areandOutside this moving surface $\rho =P=0$. The surface density of a disk is conserved because $\rho dz=\tilde{\rho }d\xi$. The column cools and contracts without changing its scaled profile: its midplane temperature falls as ${\eta }^2$, its mass density rises as ${\eta }^{-1}$ and its pressure falls as $\eta$.

The value $T_0$ determines $H_0$, but the mass density normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating givesOtherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is $\ddot{\eta }\xi$, with $\ddot{\eta }/(\eta {\Omega }^2)=2(1+\beta )C^2/[(1+2\beta {)}^2{\Omega }^2(1+Ct{)}^2]$, which remains small and decreases if $C\ll \Omega$.

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.

The constant $f=2\Omega \sin \phi$ is the positive Coriolis parameter at northern latitude $\phi$, $g$ is the gravitational acceleration, and $h$ is the undisturbed mean depth of the layer.

The linearized shallow water equations assume a homogeneous, incompressible, inviscid layer over a flat bed; a horizontal scale much larger than the depth, so the shallow-water approximation and hydrostatic approximation apply; depth-independent horizontal velocity; a free-surface displacement $|\eta |\ll h$ and sufficiently small velocity to neglect nonlinear products; and an f-plane on which $f$ is constant. The equations also omit viscosity, forcing, and bottom topography.

At the displaced free surface, constant atmospheric pressure and the hydrostatic approximation give the dynamic conditionThe linearized kinematic condition is $w=D_g\eta /Dt$. Combining these relations withgives

Use a shallow, homogeneous active upper layer of mean thickness $h_0$ and mass density ${\rho }_1$, over a much deeper, effectively stationary layer with mass density ${\rho }_2>{\rho }_1$. The Boussinesq approximation keeps the small density contrast only in buoyancy, giving reduced gravity $g^{\prime }=g({\rho }_2-{\rho }_1)/{\rho }_0$. The hydrostatic approximation and shallow-water approximation make horizontal velocity nearly depth independent and turn the interface slope into the horizontal force $-g^{\prime }\nabla h$.

Linearize about rest and uniform $h_0$, with $|h-h_0|\ll h_0$, neglecting products of perturbations and convective acceleration. The f-plane approximation sets the Coriolis parameter to constant $f_0$. Neglect wind forcing, viscosity, entrainment and bottom drag. The two horizontal momentum equations then balance local acceleration, the Coriolis force and the pressure gradient. The signs $-f_{0}v$ and $+f_{0}u$ follow from eastward and northward coordinates.

The thickness equation is the linearized conservation of mass law: horizontal divergence decreases $h$, while horizontal convergence increases it. Here $h$ is active-layer thickness, not total ocean depth, and $g^{\prime }$ rather than $g$ gives the restoring force because the motion principally displaces the thermocline.

Rotation rate, oblateness and measured external gravitational multipoles provide constraints beyond mass and radius; spin-axis precession/nutation and satellite dynamics can add moment-related information. For a giant planet the deep rotation rate and hydrostatic approximation matter. These measurements require an interior model or supplementary constraints for an accurate moment: a spherically symmetric body's exterior monopole alone cannot reveal its radial density distribution. An exact moment cannot generally be inferred just by labeling a measured $J_2$ as $C$.