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A local stellar relaxation time depends on the number density $n$ of scatterers. Calling $N$ a total number, as the PDF does, leaves the displayed expression dimensionally wrong: ${\sigma }^3/(G^2{m}^{2}N)$ has units of time per volume. The intended local expression must use $n$, or $N/V$ for a homogeneous sample of volume $V$. We shall also make the normalization of the stellar relaxation time explicit, since an unspecified representative one-dimensional velocity dispersion does not fix an exact numerical coefficient.

In a small-angle gravitational encounter, approximate the relative trajectory by a straight line of speed $u$ and impact parameter $b$. The perpendicular acceleration of one star due to the other is $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. Its impulse is thereforeA star encounters $2\pi bdbnu$ scatterers per unit time in this interval of impact parameters. Independent deflections have random directions, so their variances add as in a random walk. Hence the accumulated squared velocity change has rateThe Coulomb logarithm in stellar dynamics arises because each logarithmic interval of impact parameter contributes equally. The lower cutoff is approximately the strong-deflection scale, $b_{\min }\simeq G(2m)/u^2$ for equal masses, or the physical collision scale if larger. The upper cutoff is a scale on which the local density and straight-line encounter approximation cease to apply, such as the local disk thickness. Thus $\Lambda$ is dimensionless; it is not a number density or an arbitrary dimensional argument of a logarithm.

To display the normalization issue, use the representative relative speed $u=\sqrt{2}\sigma$ and define $T(\kappa )$ by $D_{v}T=\kappa {\sigma }^2$. This givesChoosing $\kappa =3/4$ recovers precisely the printed coefficient after repairing $N$ to $n$:This is an explicitly defined convention, not a uniquely derivable normalization from the supplied data. If instead relaxation means a mean-square change equal to the full initial random-speed variance, $3{\sigma }^2$, the same encounter approximation gives $T_R=4T_R^{\mathrm{printed}\mathrm{convention}}$. Averaging the encounter speed distribution instead of using one representative $u$ changes the coefficient again. The robust result is the scaling $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$.

For a transparent solar-neighbourhood estimate, take $n=0.1{\mathrm{pc}}^{-3}$, $m=M_{\odot }$, $\sigma =30{\mathrm{km}\mathrm{s}}^{-1}$ and $b_{\max }=300\mathrm{pc}$. With $G=4.3009\times 10^{-3}\mathrm{pc}{M}_{\odot }^{-1}({\mathrm{km}\mathrm{s}}^{-1}{)}^2$, one obtains $b_{\min }=4.78\times 10^{-6}\mathrm{pc}$, $\ln \Lambda =17.96$, andThese illustrative parameters give an order $10^{13}$--$10^{14}$ year stellar relaxation time, much longer than a galactic lifetime of order $10^{10}$ years. The solar neighbourhood is consequently a collisionless stellar system with respect to encounters between individual stars. Scattering by massive clouds and collective structure is a different process; the estimate does not assert that every mechanism of disk heating is negligible.

Take $z$ upward from the unperturbed interface in the apparatus frame, with liquid at $z>0$ and pulling velocity $-V{\mathbf{e}}_z$. The primary ice contains no salt. We use the dilute, equal-density approximation and neglect fluid motion relative to the mould. The initial salt inventory per unit horizontal area, with a common density factor suppressed, is $N=h{C}_0$. Its conservation is essential: the far liquid has zero concentration, so an imposed nonzero far-field composition would describe a different experiment.

The steady advection-diffusion equation is $-V{C}^{\prime }=D{C}^{\prime \prime }$. Decay into the pure water and interfacial salt rejection giveSince ${\int }_0^{\infty }C(z)dz=D{C}_I/V=N$, the finite-inventory directional salt rejection solution isThe decay length is $D/V$. Local liquidus equilibrium givesThus the thermal apparatus selects the interface position at this isotherm. The stated speed inequality is precisely $C_I<C_E$, so the interface is warmer than the eutectic temperature, $T_I>T_E=-m{C}_E$. These are steady values after the initial finite saline layer has rearranged; the final exponential distribution is not confined to the original depth $h$.

For $Vh{C}_0/D>C_E$, a steady pure-ice front would require an interfacial composition beyond the eutectic composition. That branch is unavailable. During the transient, rejected brine concentrates until it reaches $C_E$; secondary salt-bearing crystals can then form and the residual brine freezes as an eutectic system. Salt is incorporated into the product rather than remaining indefinitely in a super-eutectic exponential layer. The detailed pattern can include brine pockets and a nonplanar freezing region; it depends on the thermal history and the morphological stability. The speed inequality alone does not decide that latter stability.

For the stability calculation use an interface displacement $\eta =\hat{\eta }{e}^{ikx+st}$ and a liquid perturbation $c=\hat{c}{e}^{ikx+st-qz}$. Here $k>0$ is the wavenumber, $s$ is the temporal rate and $\Re q>0$ gives spatial decay. The frozen thermal field is $T_I+Gz$. Define $\Gamma ={\gamma }_{sl}{T}_m/(\rho L)>0$, where ${\gamma }_{sl}$ is solid-liquid surface energy, $T_m$ is absolute pure-solvent melting temperature, $\rho$ is mass density and $L$ is specific latent heat. For a solid protrusion into liquid the curvature is $\mathcal{K}=-{\eta }_{xx}=k^2\eta$, and the Gibbs-Thomson effect gives local equilibrium $T=-mC-\Gamma \mathcal{K}$. We assume isotropic surface energy and instantaneous interface kinetics, so no kinetic undercooling term is present.

Linearizing the equilibrium condition at the displaced interface givesThe perturbation advection-diffusion equation givesThe local solute balance at a moving graph, to first order, isIts perturbation is $(V-Dq)\hat{c}+s{C}_I\hat{\eta }=0$, because the terms $D\eta C^{\prime \prime }(0)$ and $V\eta C^{\prime }(0)$ cancel. Consequently the full frozen-temperature solutal interface dispersion isNo quasistatic approximation to solute diffusion was made here. For a growing mode the unique admissible liquid root isWriting $b=(G+\Gamma k^2)/(m{C}_I)$ and eliminating $s$ gives $(q-V/D)(q-V/D+b)=k^2$. The root continuously connected to the growing interface branch is thereforeIn particular, unstable wavelengths exist exactly when $mV{C}_I/D>G$, and their band isThis is the constitutional supercooling criterion: immediately above the plane, the local liquidus rises at rate $-m{C}^{\prime }(0)=mV{C}_I/D$, so it overtakes the imposed thermal gradient when that rate exceeds $G$. The Gibbs-Thomson effect cuts off short waves. At long wavelength $s\sim D(V/D-b)k^2/b$ for positive $G$. The exactly uniform inventory-changing displacement is not an instability of a fixed-inventory experiment. On stable branches the full diffusion problem also admits relaxation transients; the root selection above is sufficient to identify every exponentially growing interface mode.

If one additionally neglects the perturbation time derivative in the liquid diffusion equation, the alternative reduced rate isIt has the same onset and cutoff, but need not have the same rate away from the slow-growth regime. The frozen-temperature approximation by itself does not justify this extra reduction.

After onset, protrusions reach more weakly saline liquid and advance faster, while grooves accumulate rejected brine. The planar interface develops cells and, as growth becomes sufficiently strong, crystal dendrites with concentrated liquid between them. Nonlinear competition changes the selected spacing; linear rates alone cannot fix the final geometry. The interdendritic liquid forms a mushy layer, concentrates toward the eutectic composition, and ultimately freezes into a salt-bearing eutectic product when it reaches $T_E$. This replaces the pure planar salt-rejection picture with trapping and freezing of brine between crystals.

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.

Take any fixed material subvolume $B_0$ of the reference configuration, with outward normal vector $N$, current particle velocity $v_i={\dot{x}}_i$, and positive temperature $\theta$. The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because $N_I{q}_I^0$ is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.

Apply the divergence theorem and use the reference momentum conservation equation ${\rho }_0{\dot{v}}_i=P_{Ii,I}+{\rho }_0{g}_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_i{P}_{Ii,I}$ term. Since $v_{i,I}={\dot{F}}_{iI}$, the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because $B_0$ was arbitrary. In particular,

The specific Helmholtz free energy is $\psi =u-\theta \eta$. Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor $\psi =\psi (F,\theta ,{\xi }_r)$, expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of $\dot{F}$ and $\dot{\theta }$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor ${\rho }_0$ on its internal-variable term. The PDF defines $f_r=-{\psi }_{,{\xi }_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ${\hat{f}}_r=-{\rho }_0{\psi }_{,{\xi }_r}$; the term is then ${\hat{f}}_r{\dot{\xi }}_r$ with no extra density. The two conventions must not be mixed.

For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.

For linear thermoelasticity with prescribed thermal volume, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta =\theta -{\theta }_0$. Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at $(I,{\theta }_0)$ and zero reaction. The stress-free reference state has ${\psi }_{,F}=0$. Since $\partial \det F/\partial F=I$ there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is $u_{i,i}=h^{\prime }({\theta }_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to $h^{\prime }=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is ${\theta }_0{C}_e$.

Together with ${\rho }_0{\ddot{u}}_i={\sigma }_{ji,j}+{\rho }_0{g}_i$, the first-order thermal balance is ${\theta }_0(C_e\dot{\vartheta }-{\beta }_{ji}{\dot{u}}_{i,j}+h^{\prime }({\theta }_0)\dot{q})={\rho }_{0}r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.

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.

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.