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Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. For any specific quantity $f$, conservation of mass givesDotting the momentum equation with $\Sigma \mathbf{u}$ produces the kinetic energy balanceThe Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Since the shearing-sheet tidal potential is time-independent, its advected potential-energy density obeysFor the isothermal equation of state, define the barotropic energy density $U=c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})$, with fixed positive reference surface density ${\Sigma }_{\mathrm{ref}}$. The continuity equation impliesAdding these balances combines the two pressure terms into $-\nabla \cdot (P\mathbf{u})$. This establishes isothermal shearing-sheet energy conservation. Thus the conserved energy density and its flux areChoosing the density unit so that ${\Sigma }_{\mathrm{ref}}=1$ reproduces the printed logarithm. Changing the reference adds a multiple of the conserved mass to $E$ and its advective energy flux. The barotropic energy density is the mathematical energy of this fixed-isothermal sound speed closure; it is not the microscopic thermal energy of a thermally isolated gas. Maintaining an isothermal process can require heat exchange.

On a sphere of radius $R$ rotating at angular speed $\Omega$, the Coriolis parameter is $f=2\Omega \sin \varphi$. Near latitude ${\varphi }_0$, write $y=R(\varphi -{\varphi }_0)$, use locally eastward and northward Cartesian coordinates, and expandThe beta plane retains the first northward variation of planetary vorticity while neglecting higher latitude dependence and metric curvature. A midlatitude local calculation assumes $L/R\ll 1$ and $|\beta |L/|f_0|\ll 1$; near the equator the distinct equatorial beta plane has $f_0=0$, so the midlatitude low-frequency reduction used below does not apply.

For a homogeneous shallow layer, the hydrostatic approximation gives $p=p_{\mathrm{atm}}+\rho g(\eta -z)$ and hence $p_x/\rho =g{\eta }_x$, $p_y/\rho =g{\eta }_y$, independently of depth. The approximation follows from the small aspect ratio $H/L$ and neglect of vertical acceleration. Differentiate the linear horizontal momentum equations in $z$. Their depth derivatives obeyStarting from rest gives $u_z=v_z=0$ initially, and the unique solution remains zero. This establishes depth independence of hydrostatic shallow-water flow; an arbitrary pre-existing shear would not be removed merely by taking the hydrostatic approximation. Integrating the continuity equation between the rigid bottom and the moving surface gives ${\eta }_t+H(u_x+v_y)=0$ at linear order. With the depth-integrated shallow-water transports $U=Hu$, $V=Hv$, one obtains

It is useful to make the coefficient approximation in the height reduction explicit. Put ${\mathcal{F}}_y={\partial }_t^2+f(y{)}^2$. Differentiating the two momentum equations in time and eliminating the other transport givesSince ${\partial }_y({\mathcal{F}}_{y}V)={\mathcal{F}}_y{V}_y+2f\beta V$, taking their horizontal divergence and using the continuity equation givesApplying ${\mathcal{F}}_y$ proves the exact variable-Coriolis shallow-water height equationAt the reference latitude, or after the usual local freezing of undifferentiated $f$ factors to $f_0$, this gives the height relation written with ${\mathcal{F}}_0={\partial }_t^2+f_0^2$. With $f=f_0+\beta y$ over a finite region, that constant-coefficient version is a local approximation, not an exact identity. Keeping ${\mathcal{F}}_y$ as above avoids silently commuting a variable Coriolis parameter through a spatial derivative.

For the slow Rossby wave branch, take $|\omega |/|f_0|\ll 1$, approximate ${\mathcal{F}}_0$ by $f_0^2$, and discard the two time-derivative terms on the right compared with $-f_0^2{\eta }_x$. This is the regular midlatitude long-time ordering, with nondegenerate zonal variation. If $a=f_0/c$, the result isThe printed low-frequency equation has the opposite right-hand sign. The minus sign follows directly from the preceding eliminated equation and is also the sign needed for the printed isofrequency-circle centre. An independent check uses geostrophic balance: $\psi =g\eta /f_0$, $(u,v)=(-{\psi }_y,{\psi }_x)$, and the shallow-water quasi-geostrophic potential vorticity is ${\nabla }_h^2\psi -a^2\psi +\beta y$. Linear potential-vorticity conservation gives $({\nabla }_h^2\psi -a^2\psi {)}_t+\beta {\psi }_x=0$.

For $\eta =A{e}^{i(kx+ly-\omega t)}$, the shallow-water Rossby-wave dispersion relation isFor fixed $l$ and $\beta >0$, it is odd in $k$, zero at $k=0$, negative for $k>0$, and tends to zero from below as $k\to +\infty$. Its minimum occurs at $k=\sqrt{l^2+a^2}$ with ${\omega }_{\min }=-\beta /[2\sqrt{l^2+a^2}]$. Thus both long and short waves have small frequency. The zonal phase velocity is westward, $\omega /k<0$, whereas the zonal group velocity isIt changes sign at the frequency minimum.

At fixed nonzero frequency, completing the square yields the Rossby-wave isofrequency circleThe radius is real only if $|\omega |\le |\beta |/(2|a|)$. For $\beta >0$, the branch with $k>0$ has $\omega <0$ and a circle centred on the positive $k$-axis. If instead the printed plus-sign wave equation is taken literally with this same Fourier convention, its dispersion is $\omega =+\beta k/(k^2+l^2+a^2)$ and its circle is centred at $(+\beta /(2\omega ),0)$. These two conventions cannot be mixed.

For reflection of a Rossby wave at a meridional wall, the stationary wall preserves frequency, and its translation invariance in $y$ preserves the tangential wavenumber. Thus ${\omega }_r={\omega }_i$, $l_r=l_i$. Both $k$ values solveTheir sum and product giveAn incident wave in $x>0$ travels toward the wall in group velocity, not necessarily in phase velocity. For $\beta >0$, $0<k_i<\sqrt{l_i^2+a^2}$ has $c_{gx,i}<0$, while its partner $k_r>\sqrt{l_i^2+a^2}$ has $c_{gx,r}>0$. The double-root case has zero normal group velocity and does not describe a wave packet incident on the wall.

At leading quasi-geostrophic approximation, the impermeability condition is $u=-g{\eta }_y/f_0=0$. For $l_i\ne 0$, the boundary condition givesThe reflected height therefore has equal amplitude and a phase change of $\pi$. Equality here concerns the height or quasi-geostrophic streamfunction amplitudes in the reduced model. At the degenerate $l_i=0$, geostrophic no-normal-flow is automatically satisfied and alone does not determine their ratio; the usual homogeneous wall-streamfunction condition supplies $A_r=-A_i$ if imposed. Retaining the small ageostrophic transport changes the boundary condition towhich tends to $-1$ in the regular low-frequency ordering but is not generally of unit modulus. In particular, $l_i=0$ gives $A_r/A_i=-k_i/k_r$ in that more complete boundary relation. Thus the printed equal-amplitude assertion requires the nondegenerate leading quasi-geostrophic approximation, or an explicit homogeneous wall condition; it is not a general exact shallow-water reflection law.

For a particle of mass $m$, the probability density and one-dimensional probability current areThe Schrodinger equation and its complex conjugate, for real $V$, giveThe potential terms cancel precisely because $V$ is real. Thus ${\rho }_t+j_x=0$, the continuity equation for probability. Integrating over an interval says that its probability changes by the incoming current minus the outgoing current.

Let $\mathbf{u}=-(3/2)\Omega x{\mathbf{e}}_y+\mathbf{v}$ and $\rho ={\rho }_0(1+\delta )$, with $\delta$ and $\mathbf{v}$ small. Define the sound speed by $c^2=(dp/d\rho {)}_0$, assumed positive. The background Keplerian shear balances the radial Coriolis force and the shearing-sheet tidal potential. For disturbances preserving axisymmetry ${\partial }_y=0$, its advective derivative ${\mathbf{u}}_0\cdot \nabla$ vanishes, but $\mathbf{v}\cdot \nabla {\mathbf{u}}_0=-(3/2)\Omega v_x{\mathbf{e}}_y$ remains. The linearized Euler equations and continuity equation areThe azimuthal coefficient is $(2-3/2)\Omega =\Omega /2$, not $2\Omega$: it combines the background velocity gradient with the Coriolis force. The uniform background has no buoyancy or vertical stratification in this stated model.

For the polytropic equation of state $p=K{\rho }^{1+1/n}$, take $K>0$ and $n>0$ for the physical model. The specific enthalpy relative to vacuum isIts gradient obeys $\nabla Q={\rho }^{-1}\nabla p$, so the Euler equations becomeFor the material derivative $D_t={\partial }_t+\mathbf{u}\cdot \nabla$, $D_{t}Q=(Q/n\rho )D_t\rho$. The continuity equation thus becomesThe choice $Q=0$ at zero mass density fixes the additive convention in this equation. Although the momentum equation alone is unchanged by adding a constant to $Q$, the polytropic relation and enthalpy evolution must use the same vacuum reference.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

The vertical component $q_z=\zeta -f\eta /H$ is $H$ times the first-order anomaly in shallow-water potential vorticity about rest:Thus $\mathbf{q}=q_z\hat{\mathbf{z}}$ records the combination of relative vorticity and column stretching that is conserved by the unforced linearized shallow water equations. Indeed their curl and continuity equation give $(q_z{)}_t=0$. It is the persistent forcing of the balanced component during geostrophic adjustment, while inertia-gravity waves carry away the unbalanced component.

Initially $q_z=-f{\eta }_0\mathrm{sgn}(x)/H$. In the final geostrophic flow, $u=0$, $v=g{\eta }_x/f$, and $\zeta =g{\eta }_{xx}/f$. Conservation of the linearized shallow-water potential vorticity therefore givesBoundedness on both sides and continuity of $\eta ,{\eta }_x$ at the origin determine the constants:The solution is the locally attained balanced state after outgoing inertia-gravity waves have departed; the full inviscid system need not lose its globally conserved wave energy. The signed volume flux in the $y$ direction is

For double Kelvin-wave adjustment at a depth step, put $c_{\pm }=\sqrt{g{H}^{\pm }}$ and use $R_D^{\pm }=c_{\pm }/f$ on the corresponding side of the step. The PDF's far-field wording must mean $\eta \to {\eta }_0\mathrm{sgn}(x)$, rather than ${\eta }_0$ for every $x$. In the exponentially trapped outer response, cross-step geostrophic balance gives, at $y=0$,To retain the propagation of this slowly varying trapped disturbance, the along-step acceleration in $u_t-fv=-g{\eta }_x$ must also be retained. It supplies the small cross-step ageostrophic flow. Thus the two cross-step volume fluxes areHere $d\mathrm{sgn}(x)/dx=2\delta (x)$ as a distributional derivative. Equating the fluxes and dividing by $c_{+}+c_{-}$ givesThe printed forcing is missing a factor of two. It cannot follow from the printed sign-function surface profile with the usual Dirac delta distribution normalization. Pure geostrophic balance in both horizontal directions would also omit $A_t$ and could not determine the wave evolution.

The method of characteristics with zero initial amplitude gives the corrected weak solutionBehind the eastward front, $0<x<\Delta ct$, the surface elevation at the step becomes $-{\eta }_0$, matching the upstream side. All the original cross-step transport is diverted into flow along the step once the wave has passed a fixed location: $v=0$ there, with opposing along-step velocities above and below the step. This is the complete barrier in the adjusted region, and in the long-time limit at every fixed $x$. At finite time a front still carries cross-step flux: the formulas above give $H^{\pm }{v}^{\pm }=(2{\eta }_0{c}_{+}{c}_{-}/f)\delta (x-\Delta ct)$. The discontinuous outer representation does not resolve the short-scale fast adjustment at either front.

For completeness, solving the equation literally as printed instead gives $A={\eta }_0[\Theta (x)-\Theta (x-\Delta ct)]$. Its amplitude is half that required; $\eta (x,0,t)$ retains a jump ${\eta }_0$ at the origin after passage, so that literal equation does not imply complete blocking. The corrected equation and the literal printed equation are distinguished rather than asserting an inconsistent barrier result.

For steady depth and no sources of fluid mass, the continuity equation gives ${\nabla }_h\cdot (H\mathbf{u})=0$. In a simply connected region this defines an ocean transport streamfunction:A multiply connected basin additionally needs its circulation or boundary-constant data. Taking the vertical curl of $\mathbf{u}=(-{\psi }_y,{\psi }_x)/H$ gives the exact relative vorticity from a variable-depth transport streamfunction relationNo small-depth-variation expansion has been made. No normal flow at a basin wall means that $\psi$ is constant along that connected wall.

For an inclined viscous film with opposing surface shear, the steady parallel velocity automatically satisfies the continuity equation and the advective acceleration vanishes. The two components of the Navier-Stokes equations reduce toThe uniform thickness and constant atmospheric pressure make $p_x=0$. The boundary conditions areThese are respectively no-slip boundary condition, imposed tangential traction on the surface with outward normal $+{\mathbf{e}}_z$, and the normal stress condition on the flat free surface. The normal viscous stress is zero because the normal velocity vanishes. Integrating givesIn particular $\mu u^{\prime }(0)=\rho gh\sin \alpha -S$ is the downslope tangential traction exerted by the fluid on the wall. The wall's tangential traction on the fluid has the opposite sign, because the outward normal of the fluid at its bottom is $-{\mathbf{e}}_z$. This fixes the potentially confusing wall sign convention.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

Let ${\Sigma }_{d0}$ be the uniform dust background and put ${\delta }_d={\Sigma }_d^{\prime }/{\Sigma }_{d0}$. The linearized continuity equation is ${\partial }_t{\delta }_d=-{\partial }_x{v}_x^{\prime }$ because background shear has no divergence or axisymmetric advection. Using the late entrained velocity $v_x^{\prime }=u\cos \theta$ givesThus the coherent travelling wave has maximum fractional excess $|ku|/\Omega$, and its crest density is ${\Sigma }_{d0}[1+|ku|/\Omega ]$ when the stationary part is absent. This is the density contrast of an entrained dust wave, and requires $|ku|/\Omega \ll 1$ in the present linear approximation. Its fractional oscillating density amplitude is the same as the gas's long-wavelength amplitude, so the coherent wave produces no additional dust-to-gas concentration ratio at this order.

Velocity damping alone does not erase the time-independent $C(x)$; initial density and the history of the forcing are needed to fix it. The printed data therefore determine the travelling-wave contrast, rather than a unique total late-time concentration for arbitrary initial histories. As an explicit example, switch on the prescribed gas epicycle at $t=0$ with initially uniform dust and ${\mathbf{v}}^{\prime }=0$. The corrected velocity pair gives $v_x^{\prime }=u(1-e^{-ϵ\Omega t})\cos \theta$. Integrating continuity then leavesat late times. It is smaller by order $ϵ$ in the stipulated weak-coupling regime, and confirms the stated leading wave-crest estimate while showing why arbitrary stationary density structure is not removed by the velocity transient.

The field $\varphi$ is a compositional order parameter in a binary fluid mixture. Its spatial integral is conserved in a closed sample, so advection and diffusion enter a continuity equation; the material derivative follows the incompressible fluid. The diffusive current $\mathbf{J}=-M\nabla \mu$, with positive order-parameter mobility $M$, moves material down the chemical potential gradient. The chemical potential of a composition field is the functional derivative of the symmetric quartic Landau free energyso $\mu =a\varphi +b{\varphi }^3-\kappa {\nabla }^2\varphi$. Positive $b,\kappa$ stabilize large amplitudes and penalize sharp interfaces; $a<0$ permits phase separation.