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Write $c=\sqrt{gH}$ and $R_d=c/|f|$, with $f\ne 0$. For the profiles below take $f>0$, so $\alpha =L/R_d$ agrees with the stated parameter; changing the sign of $f$ reverses the balanced meridional velocity but not the height profile. The linearized shallow water equations areTaking the horizontal curl gives $(v_x-u_y{)}_t=-f(u_x+v_y)=f{\eta }_t/H$. Hence the signed linear shallow-water potential vorticity anomaly is conserved:The initial data are independent of $y$, so the solution remains independent of $y$ and $v_x=f(\eta -{\eta }_0)/H$. Differentiating continuity in time and using the zonal momentum equation gives

For a horizontal Fourier mode with wavenumber $k$, put ${\Omega }_k^2=f^2+c^2{k}^2$. The solution splits into a balanced component and inertia-gravity waves:These waves have frequencies $|\omega |\ge |f|$, phase velocity $\omega /k$, and group velocity $c^{2}k/\omega$, whose magnitude is less than $c$. They have zero perturbation potential vorticity and carry the unbalanced part away. On the unbounded line, the localized difference between the initial and balanced profiles disperses, leaving a local steady limit. No damping is needed for this local geostrophic adjustment; in a closed periodic domain, undamped waves would persist and a literal pointwise steady limit would generally not exist.

The bounded steady height satisfies the modified Helmholtz equationOddness removes the even homogeneous solution in the central region. The exterior solution is bounded and tends to the appropriate plateau. Matching both height and its first derivative at $x=\pm L$ gives the geostrophic adjustment of a finite-width height ramp:For example, writing the interior solution as $hx/L+B\sinh (x/R_d)$, matching its slope to the decaying exterior at $L$ gives $B=-h{e}^{-\alpha }/\alpha$.

The steady meridional momentum equation gives $u_s=0$, and the zonal equation gives the geostrophic balance $f{v}_s=g{\eta }_s^{\prime }$. ThusThe height is odd and monotone, and the meridional current is even, with maximum $v_s(0)=gh(1-e^{-\alpha })/(fL)$ for $h>0$.

For $\alpha \ll 1$, the initial ramp is narrow compared with the Rossby deformation radius. To leading order,Near the origin the height rises with slope $h/R_d$, rather than $h/L$; the current has width $R_d=L/\alpha$ and peak $gh\alpha /(fL)$. For $\alpha \gg 1$, the central height remains nearly $hx/L$ and the central velocity nearly $gh/(fL)$. Around each end of the ramp there is a matching layer of width $R_d=L/\alpha$, with height correction of order $h/\alpha$; the exterior current decays over the same width. At $x=\pm L$ the current is approximately half its central value. In both limits the zonal current is zero. These are the broad rotational adjustment of a narrow disturbance and the weak edge adjustment of a broad disturbance, respectively.

Use the energy normalization specified for the height energy: the linear total energy density is $(u^2+v^2)/2+g{\eta }^2/(2H)$. The linear equations implyFor $\alpha \ll 1$, let the potential energy loss mean the positive difference between the initial and final values, per unit transverse length. Although both absolute potential energies diverge on the infinite line, their difference is finite. The leading step approximation givesThe finite ramp changes this leading expression by a relative $O(\alpha )$ correction. The initial kinetic energy is zero and the final gain isThe remainder $\Delta V-\Delta T\simeq g{h}^2{R}_d/H$ is energy carried away by the outgoing inertia-gravity waves. This explains why the ratio is less than one without invoking dissipation: conservation of energy includes the waves, not only the final balanced flow.

Let $S(z)=d{\rho }_s/dz$, $N^2=-gS/{\rho }_0>0$, and define the quasi-geostrophic streamfunction by $\psi =\tilde{p}/({\rho }_0{f}_0)$. Use horizontal scale $L$, vertical scale $D$, velocity scale $U$ and advective time $L/U$. The required quasi-geostrophic approximation has small Rossby number $\epsilon =U/(|f_0|L)$, $\beta L/|f_0|=O(\epsilon )$, and Burger number $N^2{D}^2/(f_0^2{L}^2)=O(1)$. Hydrostatic, thin-layer scaling has $D/L\ll 1$; density variations are small compared with ${\rho }_0$, as required by the Boussinesq approximation. The leading horizontal flow is nondivergent, with vertical velocity $w=O(\epsilon UD/L)$. Retain forcing of the same slow order as density advection, $r=O(U\tilde{\rho }/L)$; stronger forcing need not preserve this balanced scaling.

Leading geostrophic balance and the hydrostatic approximation giveWrite $D_g/Dt={\partial }_t-{\psi }_y{\partial }_x+{\psi }_x{\partial }_y$. The density equation becomesTaking the curl of the horizontal momentum equations at the next order and using continuity gives the stretching equationSubstitute the expression for $w$. Vertical differentiation commutes with the geostrophic material derivative acting on ${\psi }_z$ here: the extra Jacobian is $J({\psi }_z,{\psi }_z)=0$. Since $N$ depends only on height, this yields the diabatically forced quasi-geostrophic potential vorticity equationThe minus sign is appropriate to the specified density forcing; a positive upward buoyancy forcing is $R=-gr/{\rho }_0$ and gives $D_{g}q/Dt=f_0{\partial }_z(R/N^2)$.

Now take constant $N$ and linearize about rest. Put $s=f_0^2/N^2$ and $S_0=g{f}_0\mu r_0/({\rho }_0{N}^2)$. The forced equation isFor the harmonic heating response of quasi-geostrophic flow, represent the two traveling Fourier components with frequencies ${\omega }_{+}={\omega }_0$ and ${\omega }_{-}=-{\omega }_0$:Each component of the forcing has amplitude $S_0/2$, soEquivalently its vertical structure obeysThese are the time-periodic forced responses; unspecified initial data may add free Rossby waves. Selecting no incoming waves describes forcing switched on causally and then its persistent harmonic part.

For $\beta >0$, the eastward component always has ${\lambda }_{+}^2>0$. Let ${\lambda }_{+}>0$. The growing exponential would introduce an unbounded velocity and pressure disturbance at infinity, so discard it. The bottom condition ${\varphi }_{+}(0)=0$ then givesIf $\mu ={\lambda }_{+}$, this expression has the regular limit $-C_{+}z{e}^{-\mu z}/(2\mu )$, rather than a physical singularity.

For ${\omega }_0>\beta /k$, the westward component also has positive ${\lambda }_{-}^2$. Choose ${\lambda }_{-}>0$ and use the same decaying expression with minus subscripts, including the same coincident-exponent limit. Thus both components are vertically evanescent above the heating.

For $0<{\omega }_0<\beta /k$, putBoth homogeneous solutions $e^{\pm imz}$ are bounded, so boundedness is insufficient. A free Rossby wave with phase $kx+mz-\omega t$ hasThe radiation condition demands energy propagation away from the heating, upward into $z>0$. For $\beta ,k>0$ that selects positive $m$, hence $e^{imz}$ in the westward component. The resulting solution isIt satisfies ${\varphi }_{-}(0)=0$ and, away from the decaying forcing, ${\varphi }_{-}^{\prime }\sim im{\varphi }_{-}$. Its far-field amplitude does not decay in this nondissipative model: continuous forcing supplies an outgoing wave train. Thus the low-frequency westward component propagates vertically while the eastward component remains evanescent.

At the threshold ${\omega }_0=\beta /k$, the westward vertical wavenumber is zero. Excluding the linearly growing homogeneous solution and retaining the bottom condition gives ${\varphi }_{-}=C_{-}(e^{-\mu z}-1)/{\mu }^2$. This bounded limiting response tends to a constant and has zero vertical group velocity; demanding both bottom zero and decay to zero at infinity would overdetermine this threshold problem. For $\beta \le 0$, the branch labels and propagation inequality must be reconsidered; the comparison stated here assumes the usual positive beta-plane gradient.

The hydrostatic buoyancy anomaly is $b^{\prime }=f_0{\psi }_z$, and its leading adiabatic evolution is $D_g{b}^{\prime }/Dt+N_0^{2}w=0$. The small vertical velocity advects the background stratification; vertical advection of the perturbation buoyancy is higher order. ConsequentlyThe impermeability condition at a stationary lower boundary $z=\alpha y$ is $w=\alpha v$. Using $v_g={\psi }_x$ gives the topographic buoyancy boundary condition for quasi-geostrophic flowFor application at the reference plane, the actual small-height condition is $|\alpha |L/D\ll 1$. Slopes scaled by $D/L$ are of Rossby-number order, so this topographic contribution is retained alongside the small ageostrophic vertical velocity. The dimensional slope should not be compared to the Rossby number without also specifying the aspect ratio.

The basic quasi-geostrophic streamfunction is $\overline{\psi }=-\Lambda zy$. It gives $U=\Lambda z$, basic buoyancy gradient ${\overline{b}}_y=-f_0\Lambda$, and spatially constant interior quasi-geostrophic potential vorticity. Linearizing its conservation law therefore givesThe printed equation omits the shear-advection term. This cannot be omitted for general disturbances: an interior anomaly proportional to $\cos [k(x-\Lambda zt)]$ is materially conserved but not independent of time at a fixed location. The intended zero-interior-PV Eady model modes do have $q^{\prime }=0$ and satisfy both expressions; the ensuing mode calculation is valid in that subspace.

At either boundary, linearizing the geostrophic material derivative includes $v^{\prime }{\overline{\psi }}_{zy}=-\Lambda {\psi }_x^{\prime }$. LetThe two boundary equations areThese are exactly the respective bottom and top slope corrections to the rigid-boundary buoyancy condition for quasi-geostrophic waves.

Assume periodicity horizontally and first consider $q^{\prime }=0$. Multiply its defining equation by ${\psi }_x^{\prime }$ and integrate over the layer. The two horizontal Laplacian terms vanish by integration by parts; the remaining volume term after one vertical integration is an $x$-derivative. ThereforeFor nonzero $A_0,A_D$, the boundary equations giveThe top advection term integrates to zero. Subtracting proves conservation of the boundary pseudomomentum of a sloping Eady layer:The original PDF uses squared vertical derivatives and the upper slope $\gamma$ in the first denominator; these details are corrupted in the converted TeX. For arbitrary nonzero interior PV the same calculation instead gives $d\mathcal{P}/dt=(f_{0}s{)}^{-1}\int {\psi }_x^{\prime }{q}^{\prime }dV$, which need not vanish. For example, at an instant take ${\psi }^{\prime }=\cos kx+(z/D{)}^2\sin kx$, independent of $y$; the zonal average of ${\psi }_x^{\prime }{q}^{\prime }$ is $-ks/D^2$. Thus the boundary-only conservation statement also needs the zero-interior-PV qualification.

For an exponentially growing smooth normal mode, this qualification is automatic: the interior equation is $ik(\Lambda z-c)\hat{q}=0$, and a phase speed with nonzero imaginary part cannot equal the real basic velocity. Therefore $\hat{q}=0$. If $A_0{A}_D<0$, the conserved boundary quadratic form is definite, so a nonzero exponentially growing boundary amplitude is impossible. Instability consequently requiresfor $\Lambda \ne 0$. When one $A$ vanishes, the corresponding boundary derivative is simply advected and must vanish for a growing mode. The other boundary then yields a real single-edge-wave speed; if both vanish, the zero-interior-PV Neumann problem has no nonzero growing mode. These degenerate cases are neutral and should not be handled by dividing by zero.

For equal slopes, put $\zeta =z/D$, $C=c/(\Lambda D)$, $M=N_0|k|D/|f_0|$, and $a=N_0^2\alpha /(f_0\Lambda )-1$. The normal-mode interior equation is ${\Phi }_{\zeta \zeta }-M^2\Phi =0$ for the zero-interior-PV modes. ThusThe coefficient equations areSetting their determinant to zero, without dividing by $C$ or $a$, givesso the Eady model with parallel sloping boundaries hasFor positive $k,f_0$, $M$ is the printed $\mu$, and $a$ is the printed $\tilde{\alpha }$. The expression is even in the sign of that vertical scale. For $a\ge 0$ the radicand is at least $1/4$, so both speeds are real and these modes have no exponential growth. For equal slopes the earlier necessary condition is merely $a^2>0$, which is satisfied in many stable cases and is not sufficient. At $a=0$ the speeds are $0$ and $\Lambda D$; the invariant with nonzero denominators cannot be used, but the determinant still correctly gives neutral modes. No claim excluding transient amplification is needed.

Use the meridional overturning streamfunction conventionDifferentiating geostrophic and hydrostatic balance gives the thermal wind relation $f_0{\overline{u}}_z=(g/{\rho }_0){\overline{\rho }}_y$. Differentiate it in time and use the mean momentum and density equations. The resulting diagnostic elliptic boundary value problem isFor the transformed circulation it becomes the Eliassen equation for residual circulation, with signs appropriate to the convention above:At the rigid meridional walls, impermeability requires each streamfunction to be constant along each wall. Set those constants to zero, choosing the solution with no imposed net vertical throughflow. Since the prescribed density flux vanishes at both walls, the same conditions apply to ${\mathcal{X}}^{\ast }$.

There is a genuine data limitation: prescribing $F$ alone does not prescribe $R$. The term $-f_0{R}_{yz}$ remains in both circulation equations. To give the usual explicit step-flux circulation, assume vanishing meridional momentum-flux divergence, $R_y=0$, with no independently imposed circulation. The formulas below are conditional on that assumption, not a unique consequence of the density flux alone.

Let $l=\pi /L$, $\kappa =N_{0}l/|f_0|$ and $A_0=-1/S>0$, using the flux unit in which its lower plateau is $-1$. The bounded solution has $\mathcal{X}=\sin (ly)\chi (z)$. Its vertical equation isAt vertical infinity require bounded Eulerian flow and vanishing residual flow. Requiring the Eulerian vertical velocity to vanish also at lower infinity would be inconsistent with the continuing lower-region flux divergence. At $z=0$, continuity of $\chi$ and ${\chi }^{\prime }$ follows because the forcing has a step but no delta function. The step-flux residual circulation in a stratified channel is thereforeHere ${\mathcal{X}}^{\ast }=\sin (ly){\chi }^{\ast }$, because $F/S=A_0\sin (ly)$ below zero and vanishes above it. The residual streamfunction decays at both infinities, whereas the Eulerian streamfunction tends to $-A_0\sin (ly)$ below and to zero above.

For $z\ne 0$, the Eulerian velocity isThe flow descends near the $y=0$ wall and rises near $y=L$, with a negative meridional return flow concentrated near $z=0$. Below the dissipation level its vertical flow balances the continuing divergence of eddy density transport. There is no independently added throughflow, but opposing vertical transports at lower infinity are required by this idealized forcing.

Away from the discontinuity the residual meridional velocity equals the Eulerian value, whileThus the residual mean circulation forms two oppositely rotating cells, confined within vertical distance $O(1/\kappa )$ of the termination of the wave flux. Its streamfunction jumps by $-A_0\sin (ly)$ at zero. Distributionally this supplies the horizontal return transportThe Dirac delta sheet is a consequence of the discontinuous imposed flux, not an extra boundary condition. Smoothing the flux termination replaces it by a thin finite return current. Streamlines must not be drawn as continuously crossing the jump without that transport.

The physical Eulerian mean flow is continuous. In the transformed momentum equation, the sheet contribution to $f_0{v}^{\ast }$ cancels the sheet in the flux divergence, leaving the smooth mean accelerationThus termination of the vertical Eliassen–Palm flux produces a finite column-integrated wave drag, redistributed over depth by the residual mean circulation.

For completeness, the general bounded response can be calculated without setting $R_y$ to zero. Expand $R_{yz}$ in wall-compatible sine modes, with coefficients $B_n(z)$, $l_n=n\pi /L$ and ${\kappa }_n=N_0{l}_n/|f_0|$. Add to either displayed streamfunction the same correctionfor decaying momentum-flux forcing. The Green function sign follows from $({\partial }_z^2-{\kappa }_n^2)e^{-{\kappa }_n|z-z^{\prime }|}=-2{\kappa }_n\delta (z-z^{\prime })$.

An explicit counterexample to uniqueness is the additional momentum flux $R=ϵ[1-\cos (2ly)]e^{-a|z|}$, with $a>0$. It vanishes at both walls and leaves the prescribed density flux unchanged. For $a\ne 2\kappa$ it addsto both circulations. This nonzero bounded, wall-impermeable solution verifies that the original mean equations need an additional momentum-flux specification to select a unique sketch.

Put $r=|\mathbf{x}|$ and $s=\mathbf{F}\cdot \mathbf{x}$. In the Unscaled Papkovich–Neuber representation, take the harmonic potentials $\mathbf{\Phi }=-\mathbf{F}/(8\pi \mu r)$ and $\chi =0$. Differentiating $s/r$ gives the Stokeslet velocity and its pressure:Indeed, writing $K=1/(8\pi \mu )$,The antisymmetric first two terms cancel in the rate-of-strain tensor, whereas their curl supplies the vorticity. ThusThe rate-of-strain tensor has zero trace, as required by incompressibility. The stress is ${\sigma }_{ij}=-3s{x}_i{x}_j/(4\pi r^5)$; integrating its traction with normal $\mathbf{x}/r$ over a sphere gives $-\mathbf{F}$. This fixes the force normalization of the Stokeslet.

For a sphere in a uniform straining Stokes flow, write the ambient velocity as $\mathbf{Ex}$, where $\mathbf{E}$ is symmetric and trace-free, and set $q=\mathbf{x}\cdot \mathbf{Ex}$. Symmetry gives zero translation and rotation of the centred force-free, torque-free sphere. Seek a decaying disturbance using the Unscaled Papkovich–Neuber representation withThe components of $\mathbf{Ex}/r^3$ are derivatives of $1/r$, and $q/r^5$ is a trace-free harmonic quadrupole, so both potentials satisfy Laplace's equation outside the sphere. Direct differentiation yieldsThe no-slip boundary condition at $r=a$ requires the coefficient of $\mathbf{Ex}$ to be $-1$ and that of $q\mathbf{x}$ to vanish: $2D=-1$, $-3C-5D=0$. Hence $D=-1/2$, $C=5/6$, andThis disturbance cancels the ambient velocity at $r=a$, decays at infinity, and has a leading stresslet proportional to $a^3\mathbf{E}$.

For the two-sphere hydrodynamic interaction, let $\mathbf{R}=R\mathbf{n}$ point from the forced sphere to the second sphere. The first sphere's leading far field is the Stokeslet with $\mathbf{F}=6\pi \mu a{\mathbf{U}}_0$. Its incident rate-of-strain tensor at the second sphere isThe force-free sphere translates with the local uniform incident flow, and the torque-free sphere rotates with half the incident vorticity. Those parts cause no leading Stokeslet or rotlet disturbance. The remaining incident strain produces the stresslet just calculated. At displacement $-\mathbf{R}$ from the second sphere its velocity isThe first sphere translates in this returned incident flow. Finite-size sampling and higher incident multipoles enter only beyond the retained reflection order. Thus the stresslet reflection between two force-free and forced spheres givesThe factor $|{\mathbf{U}}_0|$ makes explicit the velocity scale implicit in the printed remainder.

The resulting input power, equal to the viscous dissipation, isIt decreases at fixed force. The Minimum-dissipation theorem for Stokes flow instead compares admissible fields at fixed first-sphere velocity: extending a two-sphere field rigidly into the second sphere gives an admissible one-sphere field, so the extra rigid inclusion can only increase the minimum resistance. Allowing the second sphere to choose its translation and rotation minimizes over its rigid motions, producing zero force and couple there. An increase of this effective hydrodynamic resistance matrix decreases its inverse hydrodynamic mobility matrix, and hence decreases fixed-force power. The negative mobility correction is therefore consistent with minimum dissipation.

If the second sphere is prevented from rotating, its angular velocity relative to the incident fluid becomesUse the rotating sphere in Stokes flow at $-\mathbf{R}$. The additional reflected rotlet givesAdding this rotation-clamped reflection between two spheres to the strain reflection,The holding couple supplies no power because the held sphere's angular velocity is zero.

Take $x$ downslope and $z$ normal to the plane. To leading order in the lubrication approximation, Newtonian gravity gives the hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. The downslope Stokes equation is therefore $\mu u_{zz}=\rho g(h_x-\theta )$. Imposing the no-slip boundary condition at $z=0$ and the stress-free boundary condition at $z=h$ givesHere $\nu =\mu /\rho$ is the kinematic viscosity; $u$ is the downslope velocity. Applying mass conservation to this lubrication gravity-current flux,For typical thickness $H=\hat{h}$ and length $L=\hat{x}$, the transverse velocity is $O(UH/L)$ by incompressibility. Both convective terms are then $O(U^2/L)$, while transverse viscous acceleration is $O(\nu U/H^2)$. Thus the inertia criterion for an inclined viscous film is $U{H}^2/(\nu L)\ll 1$, in addition to $H/L\ll 1$. With evolution time $T\sim L/U$ the same condition controls unsteady inertia. If a faster time is independently imposed, also require $H^2/(\nu T)\ll 1$. Estimating $U$ from the velocity profile gives the two requested limits:

At long times, $h\sim A/x_N$ and the ratio of hydrostatic-gradient forcing to downslope Newtonian gravity in the bulk is $h_x/\theta =O(A/(\theta x_N^2))\ll 1$. Put $K=g\theta /(3\nu )$; the bulk conservation law is $h_t+(K{h}^3{)}_x=0$. The finite-volume gravity-driven thin-film current scales as $x\sim t^{1/3}$, $h\sim t^{-1/3}$. More explicitly, writing $h=t^{-1/3}f(\xi )$, $\xi =x/t^{1/3}$, givesThe constant of integration is zero for the zero-flux dry tail. ConsequentlyThe reduced bulk solution is zero outside this interval. Its thin nose region is resolved below. Integrating to impose the conserved area,The bulk front height is $h_N=3A/(2x_N)$, with $h_N^2=x_N/(3Kt)$. Differentiation of the front law givesEquivalently, this is the Rankine-Hugoniot condition for the bulk shock wave, whose speed is flux divided by height.

In moving coordinate $y=x-x_N(t)$, the full thin-film equation becomes $h_t{|}_y-{\dot{x}}_N{h}_y+q_y=0$. A nose thickness $h_N$ and slope $\theta$ give axial scale ${\ell }_N=h_N/\theta$. Since ${\ell }_N/x_N=O(A/(\theta x_N^2))\ll 1$, the explicit time derivative is smaller than translation by that ratio. The travelling front of a gravity-driven thin film is therefore quasisteady. Integrate once with dry-tip flux zero to obtain $q={\dot{x}}_{N}h$, orWith $f=h/h_N$, separation of variables and the convention $y=0$ at $f=0$ giveThe nose approaches $h_N$ exponentially far behind its tip, then falls to zero with a vertical tangent in this small-slope model. The requested sketch shows this monotone profile:

Expanding the inverse hyperbolic tangent, $f-\mathrm{atanh}f=-f^3/3+O(f^5)$, so $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. Its slope has magnitude $\theta /f^2$ near the tip. The small-slope breakdown at an inclined-current nose occurs at $f=O(\sqrt{\theta })$, which givesFinally, surface tension produces a pressure-gradient scale $\gamma h_N/{\ell }_N^3$, whereas downslope Newtonian gravity gives $\rho g\theta$. Their ratio is $\gamma {\theta }^2/(\rho g{h}_N^2)$. Thus surface tension changes the principal nose length whenThis is a bulk-nose crossover; capillarity and contact-line physics can already affect a smaller region very close to the dry tip even when the bulk nose is Newtonian gravity dominated.