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Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.

Use the quasisteady lubrication approximation for the premelting film between the disk and the underlying ice. The ice is anchored to the lower plane; new ice formed from water supplied through this upper film raises the disk. Let $U$ be its upward separation rate, $d(r)$ the premelted thickness, $\mu$ the liquid dynamic viscosity, $\rho$ a common ice/water mass density, $L$ the specific latent heat of fusion, and $T_m$ the absolute bulk melting temperature. As usual in this leading model, neglect the ice–water density difference, pressure-dependent changes in $T_m$ due to a common phase pressure, inertia and edge corrections. The imposed thermal field is taken as maintained by an external heat supply/removal system, so no additional heat-conduction bottleneck is introduced.

Write $\Delta =T_m-T_0>0$, $K_T=\rho L/T_m$ and $\Delta (r)=\Delta (1-r^2/a^2)$. The Clapeyron pressure relation for equal-density phases makes the pressure difference across the solid–liquid interfaceThis thermomolecular pressure is the repulsive stress transmitted through the film. To calculate a definite rate one also needs a film-thickness constitutive law. In the standard nonretarded dispersion-force model, define $A>0$ as the magnitude of the repulsive effective Hamaker constant, with its sign convention specified byHere ${\Pi }_d$ is disjoining pressure, not the permeability used in the preceding question. The sign of an effective Hamaker coefficient depends on the three materials; the positive parameter here denotes an interaction that stabilizes a premelted film. The problem does not specify $A$ numerically, so it must remain a defined material parameter.

Take the reservoir pressure at the rim to be $p_{\infty }$. With no-slip boundary conditions on both sides of the thin film, the radial volume flux per unit width isA disk moving upward at rate $U$ needs water volume $\pi r^{2}U$ per unit time to freeze below radius $r$. Regularity on the axis and mass conservation therefore give $2\pi r{q}_r=-\pi r^{2}U$, or $q_r=-Ur/2$. ThusDefine $B=A/(6\pi K_T)$, so $d^3=B/[\Delta (1-r^2/a^2)]$. Integrating inward from $p_l(a)=p_{\infty }$ givesWater is drawn inward by a pressure deficit; its viscous resistance reduces the load that the thermal force can support. The net normal stress on the disk is $p_l-p_{\infty }+P_T$. Let $W$ be the downward apparent load after subtracting ordinary buoyancy. For an ideal thin disk with negligible buoyant volume, $W=Mg$, where $g$ is gravitational acceleration. If a finite displaced volume is known, use its submerged weight instead; the stated mass and radius alone do not determine that volume.

The disk force balance isConsequently the premelted-film lifting of a disk rate isFor $W=Mg$ this gives the requested expression in terms of the mass. Every additional symbol in this formula has been defined above. The growing-ice lifting branch exists only when the bracket is positive.

At the threshold the required water flux and viscous pressure deficit vanish. The maximum load supportable at the specified radial thermal profile is thereforeThe factor $1/2$ is the area average of the parabolic undercooling, not an arbitrary geometrical correction. At equality the lift rate is zero; strictly positive lift requires $W<W_{\max }$. There is no radius-only, temperature-independent maximum load in this ideal model: one must specify the imposed undercooling or a physically admissible range of it.

For a fixed positive load define ${\Delta }_c=2W{T}_m/(\pi a^2\rho L)$ and $U_{\infty }=A/(6\pi \mu a^2)$. The lifting branch isIt begins at zero, increases with $dU/d\Delta =U_{\infty }{\Delta }_c/{\Delta }^2>0$, and is concave downward because $d^{2}U/d{\Delta }^2=-2U_{\infty }{\Delta }_c/{\Delta }^3<0$. Below threshold no positive steady heave is possible: the thermomolecular force is too weak to support the load. The mathematical continuation to negative $U$ is not a prediction of upward ice-driven separation there.

Increasing undercooling increases the available thermal pressure, but it also thins the premelting film. In the inverse-cube disjoining pressure model, $d^3\propto {\Delta }^{-1}$, so the hydraulic conductance decreases in exact inverse proportion to the driving scale. The remaining load correction tends to zero and the rate approaches a plateau:For positive load this is approached from below, not attained at finite undercooling. With zero load the reduced model gives the same constant rate wherever its thin-film assumptions hold.

Two limits of the calculation are worth making explicit. First, the formal $d(r)$ diverges at the warm rim, so a narrow outer connection to bulk water replaces the thin-film solution there; this leading model assumes that its contribution to the integrals is negligible. Second, very large undercooling produces a molecularly thin film and eventually invalidates continuum lubrication theory and the simple dispersion-force law. Thus the plateau is the maximum within the stated ideal constitutive model, not an unrestricted extrapolation for real water.

If a different molecular interaction is intended, the result is not universal. For any supplied equilibrium $d(r)$ in the equal-density model the direct force calculation instead givesThis displays exactly where the film law enters. The closed formula and plateau above use the conventional repulsive inverse-cube law; premelting alone, without that material law, would not determine a unique rate curve.

Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. Absorb the reference hydrostatic pressure into a kinematic pressure $p$. The ideal nonrotating Boussinesq equations areFor a resting density profile ${\rho }_b(z)$, the background satisfies $p_b^{\prime }(z)={\sigma }_b(z)$ and the squared buoyancy frequency isStable stratification has $N^2>0$. The Boussinesq approximation requires fractional density variations to be small over the domain and parcel excursions, so density can be replaced by ${\rho }_0$ in inertia while its small variation is retained in the gravitational force. For an approximately uniform density gradient over height $H$, this requires $N^{2}H/g\ll 1$. The stipulated incompressibility alone does not justify neglecting density variation in inertia.

Put $b=\sigma -{\sigma }_b$ and $p^{\prime }=p-p_b$. To linear order the two-dimensional internal gravity wave equations areThe exact impermeability condition on $z=\eta (x,t)$ is the kinematic boundary conditionTaylor expansion about $z=0$ permits $w(x,0,t)={\eta }_t$ at leading order if the boundary slope and its displacement relative to the vertical wave scale are small: $|kϵ|\ll 1$ and $|mϵ|\ll 1$. The second condition controls both the omitted evaluation correction $\eta w_z$ and the product $u{\eta }_x$, since continuity gives $u\sim (m/k)w$. Hence for the harmonic forcing,The propagating calculation assumes $k\ne 0$ and $0<|\omega |<N$; a time-independent boundary does no oscillatory work on fluid initially at rest. If $k=0$ but $\omega \ne 0$, incompressibility forces the vertical velocity to be independent of height, so the unbounded column has no finite-energy radiating internal-wave solution. The propagating wave-generation problem therefore needs the stated nonzero horizontal wavenumber.

For a plane mode with phase $\theta =kx+mz-\omega t$, let $U,W,P,B$ be its complex amplitudes. The linear equations giveSubstituting into $-i\omega W=-imP+B$ gives the dispersion relationNo waves enter from above. The appropriate radiation condition therefore selects upward vertical group velocity, not upward phase propagation:The boundary-forced internal gravity wave satisfying the boundary condition isThe displacement is $ϵ\sin \theta$, consistent with $b=-N^2$ times the displacement. For a right-moving boundary choose $k>0$, $\omega >0$ without loss of generality. Then $m<0$, and the constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m>0$: the crests rise toward the right. Their acute angle with the horizontal isThe group velocity is along these crests in the upward-right direction, whereas the wavevector and phase propagation point downward-right.

The physical pressure perturbation is ${\rho }_0{p}^{\prime }$. Pressure work at the lower boundary, to second order in amplitude, is its period-mean pressure times vertical boundary velocity. Static hydrostatic terms average to zero over a cycle. ThereforeAs an independent energy check, the wave's mean kinetic-plus-buoyancy energy density is ${\rho }_0{ϵ}^2{N}^2/2$, and multiplication by $c_{gz}$ gives exactly this upward energy flux.

For the exact horizontal mean-momentum equation, incompressibility writes the horizontal momentum equation asAverage over a horizontal period. Require the pressure and velocity fluxes to be periodic, or more generally their horizontal derivative averages to vanish; in particular there must be no imposed mean pressure gradient. ThenHorizontal averaging also makes ${\partial }_z\overline{w}=0$; the zero mean boundary displacement rate gives $\overline{w}=0$. These conditions exclude an independent mean mass flux or horizontal force. The mean equation applies in the fluid, with flattening of the small moving boundary understood at the order used above.

For slow switch-on, define $A(z,t)=ϵ(t-z/c_{gz})$, using $ϵ(s)=0$ for $s<0$. The outgoing internal-wave envelope radiation condition is $A_t+c_{gz}{A}_z=0$. Replace $ϵ$ in the oscillatory fields by $A$. A modulation time $T$ must satisfy $|\omega |T\gg 1$ and $|m|c_{gz}T\gg 1$, with a frequency bandwidth sufficiently narrow to linearize the dispersion relation. The exact boundary derivative also contains $\dot{ϵ}\sin (kx-\omega t)$; it is smaller by the slow-modulation parameter and is beyond this leading envelope approximation.

To leading slow-modulation order, the wave momentum flux isSince ${\partial }_z{A}^2=-c_{gz}^{-1}{\partial }_t{A}^2$, integrating the mean-momentum equation from rest gives the internal-wave momentum depositionThis is second order in wave amplitude, with smaller relative slow-modulation corrections. Mean-flow feedback on the leading wave is higher order. The mean current is created as the wave envelope arrives; an already uniform wavetrain has no local flux divergence.

Finally move horizontally with speed $c_p=\omega /k$. The initial uniform relative flow is $-c_p$, with infinite total energy in the unbounded column. Its finite mean-energy change is insteadAt finite time the envelope is absent sufficiently far above, so integrating the momentum equation givesThis is the boundary work and mean-flow energy of an internal wave relation. In the moving frame the boundary pattern is stationary to leading modulation order. It extracts energy from the incident mean flow and converts it to outgoing internal-wave energy; in the laboratory frame the same energy is supplied as boundary work.

Here write $\sigma$ for the buoyancy perturbation about the resting constant-$N$ stratification; the total buoyancy also includes its background gradient $N^2$. Use kinematic perturbation pressure $p$. The weak-forcing Linearized Boussinesq equations areWeakness means that perturbation advection is small compared with the retained linear terms. With the specified streamfunction convention, the transverse vorticity isDifferentiate the horizontal momentum equation in $z$ and subtract the $x$ derivative of vertical momentum. The pressure derivatives cancel, givingThe second equation uses ${\sigma }_t=N^2{\psi }_x$ and is the forced internal-wave equation.

For free waves at frequency $|\omega |\ll N$, the internal gravity wave dispersion relation gives $|k|/|m|\simeq |\omega |/N\ll 1$. Thus the horizontal response scale is much larger than the vertical scale. Vertical acceleration is small in the vertical momentum equation, leaving the hydrostatic approximation $p_z=\sigma$. Correspondingly the transverse vorticity is predominantly $u_z$: neglect $-w_x={\psi }_{xx}$ relative to ${\psi }_{zz}$ in the inertial vorticity term. The restoring term $N^2{\psi }_{xx}$ must still be retained, since it balances the small time derivatives.

For $m\ne 0$ put $\psi =e^{imz}P(x,t)$, $\sigma =e^{imz}B(x,t)$ and interpret physical fields as real parts. The reduced equation isOne can solve it while obtaining the buoyancy simultaneously. Define $U=imP$ and $H=iB/(cm)$. The first-order hydrostatic equations becomeHence $R_{+}=U+H$ and $R_{-}=U-H$ obeyApply the method of characteristics with zero fields in the past. Withthe solution is $R_{\pm }=I_{\pm }$. Consequently the hydrostatic response to a localized horizontal force isThese are exactly the required two characteristic integrals, derived with their relative sign fixed by the buoyancy equation. Quiet-past initial conditions exclude additional freely propagating solutions.

For the consistency check, set $\tau =s-t$, so both integrals run over the fixed range $(-\infty ,0)$. Differentiation at fixed $x,z$ now givesThis change of variable is important: it includes the time dependence of both the sampled forcing and the moving integration geometry without omitting a term. Each straight path intersects $|x|\le a$ for a time interval of length at most $2a/c$. If $L=\sup |f_{xt}|$, thenThe characteristic size of the hydrostatic buoyancy and pressure-gradient balance is $|m|a{f}_{\max }$, as follows from the analogous bound on $B$ in its integral formula. Thus the omitted acceleration has relative scale at most a constant timesTogether with the low-frequency assumption, this verifies the requested hydrostatic consistency. It is a comparison of characteristic magnitudes, rather than division by a buoyancy value at a node where both terms may vanish. The capital $N$ here is the buoyancy frequency, as printed in the PDF.

For $t\ge 0$, only $0\le s\le t$ contributes. If $|x|>ct+a$, every point $x\mp c(s-t)$ lies outside the source interval, so both integrals vanish:This finite propagation cone belongs to the reduced hydrostatic wave equation.

The hypotheses do not require an even forcing, so no single symmetric graph follows for every permitted $f$. For an illustrative smooth even nonnegative profile, $U$ is even and $H$ is odd. At late time and fixed $x$, if $A_f={\int }_{-a}^{a}f(r,\infty )dr$, thenThus the streamfunction develops a broad nearly level interior response with outgoing fronts, while the buoyancy changes sign across the source and approaches opposite interior plateaus to its left and right. Both vanish beyond the fronts. For $m>0$, a vertical phase with $\sin (mz)=1$ gives real fields $\psi =U/m$ and $\sigma =cmH$. The figure plots those profiles up to positive normalization factors, for an even compact forcing increasing slowly from zero.

Choose a fixed vertical datum so the physical free surface is at $H+\zeta$, with $H$ constant; its depth above the bottom is $h=H+\zeta -b$. The gradient of the constant datum does not enter the pressure force. With $D_H={\partial }_t+u{\partial }_x+v{\partial }_y$, the rotating shallow water equations areThe fluid is hydrostatic with depth-independent horizontal velocity at this order. The pressure-gradient force uses the free-surface slope, not the layer-depth slope: $\nabla \zeta =\nabla h+\nabla b$.

Let $q=v_x-u_y$ be relative vertical vorticity, $q_a=f+q$ the absolute vorticity, and $\delta =u_x+v_y$. Taking the horizontal curl of the two momentum equations and expanding the advective derivatives givesMass conservation is $D_{H}h=-h\delta$. The two stretching factors cancel, proving the exact shallow-water potential vorticity conservation lawThe denominator is $h$, the actual column thickness that changes under horizontal convergence; an arbitrary surface elevation $\zeta$ does not obey that column mass equation. There is no $w{\partial }_z$ term because the reduced layer quantities have no vertical dependence. The vertical motion has already supplied the layer-thickness equation, rather than remaining an independent coordinate of this model.

Now take $H=h_{00}$ and assume $|q|\ll |f|$, $|\zeta |,|b|\ll H$. To first order,For a small Rossby number, leading geostrophic flow satisfies $fv=g{\zeta }_x$ and $fu=-g{\zeta }_y$. ThusFor $f\ne 0$, the barotropic deformation radius is $L_R=\sqrt{gH}/|f|$. Multiplication of the first-order PV by $H$ givesThis is the required shallow-water quasi-geostrophic potential vorticity, including the bottom contribution.

Write $S=fϵ/H$ and $L=L_R$. The initial state has $\psi =0$, so ${\mathcal{Q}}_i=f+(f/H)b$. The specified homogenization sets ${\mathcal{Q}}_f=f$ in $|y|<a$ and leaves it unchanged outside. Hence the requested change isFor $f>0$ this is a descending straight segment from $+Sa$ to $-Sa$, with zero exterior values and jumps at the two edges. Endpoint values do not affect the inversion. It is the anomaly, not the full final PV profile, that appears as the source below.

For the zonal mean, $v=0$, $u=-{\psi }^{\prime }(y)$, and potential-vorticity inversion requireswith decaying disturbance at both infinities. The odd source makes the unique decaying solution odd. A particular interior solution is $S{L}^{2}y$, so writeBoth $\psi$ and ${\psi }^{\prime }$ must be continuous at $y=\pm a$: a jump would introduce a delta function or its derivative into a source that has only finite jumps. Matching at $a$ givesand thereforeDifferentiate the interior streamfunction to obtainThe exterior profiles needed for the sketches are $u=(D/L)e^{-(|y|-a)/L}$ and $\zeta =(f/g)\mathrm{sgn}(y)D{e}^{-(|y|-a)/L}$.

For $L\gg a$, put $A=a/L$, $Y=y/L$, so $|Y|\le A$. Uniform Taylor expansion givesConsequentlyFor $f>0$ the leading interior current is westward and approximately parabolic.

For $L\ll a$ and away from the edges, $u\simeq -S{L}^2$ and $\zeta =(f/g)\psi \simeq ϵy$: the surface follows the bottom slope centrally. Near $|y|=a$, the exponentially thin corrections give positive edge jets of width $L$ and peak scale $SLa/2$ when $f>0$. Those jets decay in the exterior. The elevation is odd, has opposite-sign extrema just inside the edges, and decays to zero outside; at the positive edge $\zeta (a)\simeq ϵ(a-L)/2$. The following sketch uses the full matched solution with $L/a=0.08$. For negative $f$, the current and PV-change signs reverse, while this elevation profile does not.

Potential-vorticity inversion is the diagnostic recovery of the balanced flow from the instantaneous PV distribution, together with appropriate boundary or far-field data. In the unbounded constant-$N$ case the mathematical problem is the Poisson equationRequire a regular solution, with interface continuity in a piecewise-smooth case. For a compactly supported bounded anomaly, the unique decaying solution is the Newtonian-potential integralThe sign follows from ${\nabla }_{\ast }^2[-1/(4\pi r)]={\delta }_0$. A nonzero total anomaly gives a $1/r$ far-field streamfunction, which is allowed by the stated decay condition. The measure is in stretched coordinates; converting to physical coordinates introduces their volume Jacobian.

Two regular solutions with these data differ by a harmonic function tending to zero, which must vanish by the maximum principle. This also excludes imposed uniform currents or shear fields encoded by nondecaying harmonic additions and fixes the irrelevant additive pressure gauge. Once $\psi$ is found, ${\mathbf{u}}_g=(-{\psi }_y,{\psi }_x)$, $p^{\prime }=f\psi$ and $b=f{\psi }_z$ follow. The smaller vertical and ageostrophic velocities require the next-order balance equations.

PV is not by itself an inversion variable for arbitrary unbalanced fluid motion: the quasi-geostrophic approximation supplies the diagnostic geostrophic and hydrostatic relations, filtering the independent inertia-gravity-wave motions. In bounded domains, boundary buoyancy or equivalent pressure data are additional indispensable inputs.

Potential-vorticity conservation makes the evolving PV distribution a set of parcel-carried labels. In the ideal model it can be rearranged into filaments and complicated geometry, but it is not created or destroyed by advection. Apparent homogenization, such as that assumed for the shallow-water experiment, describes a coarse-grained mixed distribution rather than pointwise destruction of materially conserved PV.

Within a balanced regime, potential-vorticity inversion turns those labels and the boundary data into a velocity field. This makes understanding a flow a coupled geometric problem: determine how PV is rearranged, invert the resulting distribution, and use the recovered velocity to find the next rearrangement. Inversion is nonlocal, so a compact PV anomaly influences fluid well outside its support, as the spherical example explicitly demonstrates.

The quasi-geostrophic potential-vorticity equation has only one time derivative:An initial PV field and the inversion data determine the balanced evolution; an independent initial acceleration is not needed. Pressure, buoyancy and the leading velocity are diagnosed from that field at each instant. This reduced initial-data requirement reflects the removal of freely oscillating inertia-gravity waves, which in the full fluid equations require additional initial information. The principles are consequently powerful for slow large-scale flow, but do not recover every unbalanced motion or remove the need to state boundary conditions and the balance approximation.