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The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.

Linearizing the vertical velocity requires retaining advection of ${\bar{\psi }}_z=-\Lambda y$ by $v^{\prime }={\psi }_x^{\prime }$. ThereforeThe impermeable boundary at $z=0$ imposesThis is conservation of boundary buoyancy, not merely a condition that ${\psi }_z^{\prime }$ vanish.

For zero interior potential vorticity, write ${\psi }^{\prime }=\mathrm{Re}\{\varphi (z,t)e^{i(kx+ly)}\}$. With $K_h=\sqrt{k^2+l^2}>0$, stratified quasi-geostrophic inversion givesThe decaying solution isNo time dependence has been assumed to obtain this vertical structure. Applying the boundary condition gives $-\mu \dot{A}-i\Lambda kA=0$, henceThus the Eady edge wave travels eastward with horizontal phase velocity $\omega /k=\Lambda /\mu$ for $k\ne 0$.

The propagation mechanism can be made explicit using boundary parcels. Thermal wind gives the background buoyancy gradient ${\bar{b}}_y=f{\bar{\psi }}_{zy}=-f\Lambda$. A small meridional parcel displacement $d$ produces boundary buoyancy perturbation $b^{\prime }=-d{\bar{b}}_y=f\Lambda d$. Meanwhile the decaying potential-vorticity inversion gives $b^{\prime }=f{\psi }_z^{\prime }=-f\mu {\psi }^{\prime }$ at the boundary. Consequently ${\psi }^{\prime }=-\Lambda d/\mu$ there, and the parcel equation $d_t=v^{\prime }={\psi }_x^{\prime }$ becomesA displaced buoyancy contour therefore induces a velocity field that advances that displacement eastward. No interior potential-vorticity gradient is needed because the boundary buoyancy supplies the invertibility data.

Vortex stretching is what sets the induced flow's vertical reach. In the interior, the zero-potential vorticity condition requiresBoundary isopycnal displacement decreases with height as $e^{-\mu z}$, so fluid columns are stretched or compressed over depth ${\mu }^{-1}=f/(N{K}_h)$. Conservation of potential vorticity makes this stretching balance the relative vorticity. The resulting inversion factor $b^{\prime }=-f\mu {\psi }^{\prime }$ gives the propagation speed: stronger density stratification reduces penetration and speed, while larger $f$ increases them. Zero interior potential vorticity therefore does not imply zero interior flow.

For imposed boundary vertical velocity, retain the same decaying structure and writeThe boundary relation becomes the forced Eady edge wave equationThe integrating factor $e^{i\omega t}$ gives, for ${\omega }_0\ne \omega$,The arbitrary homogeneous coefficient $B$ sets the initial disturbance. With $A(0)=0$, this is $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$.

At resonance, ${\omega }_0=\omega$, integration instead yieldsAn initially undisturbed fluid has ${\psi }^{\prime }=Ct{e}^{-\mu z}\cos (kx+ly-\omega t)$. The imposed pumping reinforces the same boundary normal mode every cycle, with no dissipative loss, so the linear amplitude grows secularly. The divergence describes resonance of the undamped linear model; sufficiently large wave amplitude invalidates the assumed small-disturbance approximation.

Potential-vorticity inversion is the diagnostic recovery of the flow from its advected vortical scalar, using the balance relation and specified boundary conditions. Its usefulness is a separation of tasks: transport the scalar by the current velocity, invert to find the new velocity, and repeat. The inverse is spatially nonlocal even though the transport law is local. The four models differ chiefly in the operator being inverted and the information needed at boundaries.

In two-dimensional vortex dynamics, planar incompressibility permits $\mathbf{u}=(-{\psi }_y,{\psi }_x)$, and vertical vorticity is $\zeta ={\nabla }^2\psi$. The curl of inviscid momentum givesThere is no vortex stretching in strictly planar motion, so vorticity is materially conserved. The two-dimensional potential-vorticity inversion is the Poisson equation ${\nabla }^2\psi =\zeta$. On the plane, with suitable integrability and far-field behavior, its Green's function is $(2\pi {)}^{-1}\log |\mathbf{x}-{\mathbf{x}}^{\prime }|$, givingDifferentiation gives the planar Biot-Savart kernel. Here ${\psi }_h$ is a harmonic function fixed by boundary, circulation and far-field data. A rigid impermeable wall is a streamline, so $\psi$ is constant on it; constants or circulations around distinct holes require specification. The vorticity alone cannot distinguish two flows differing by a permitted harmonic contribution, such as a uniform current on an unbounded plane. A constant added to $\psi$ has no physical effect.

In nonrotating layerwise-two-dimensional vortex dynamics, strong stable density stratification constrains motion to almost horizontal layers. In the ideal independent-layer limit, vertical velocity vanishes at leading order, $(u,v)=(-{\psi }_y,{\psi }_x)$ andhold separately at each fixed $z$. Height is a parameter: the horizontal inverse Laplacian contains no vertical derivative. With background buoyancy $N^{2}z$, the leading Ertel potential vorticity is $Q=\mathbf{\omega }\cdot \nabla b\simeq N^2\zeta$, so the materially advected normalized scalar is $Q/N^2\simeq \zeta$. The same horizontal Poisson equation is inverted layer by layer, using each layer's boundary data. This is an asymptotic constrained model, not arbitrary three-dimensional Euler motion with vertical velocity set to zero: its validity requires small Froude number and controlled vertical scales. At finite buoyancy frequency, the weak vertical motion and tilted isopycnals restore coupling. The inversion may be independent by layer while the reconstructed fields vary with height. The horizontal Euler equations for an inviscid fluid determine the pressure gradient, while hydrostatic balance relates its vertical derivative to buoyancy.

In shallow-water quasi-geostrophic inversion, begin with the exact shallow-water potential vorticity $(f+\zeta )/h$. Linearizing its anomaly about depth $H$ gives, after multiplying by $H$, $\zeta -f_0\eta /H$. For small Rossby number, geostrophic balance gives $\psi =g\eta /f_0$, while $\zeta ={\nabla }_h^2\psi$. Keeping the beta plane background gradient givesThus $q-\beta y$ is inverted with the modified Helmholtz equation $({\nabla }_h^2-L_D^{-2})\psi =q-\beta y$, under the specified boundary and far-field conditions. The velocity follows from $\psi$ and surface elevation from $\eta =f_0\psi /g$. A horizontal Fourier mode of the anomaly satisfiesThe Rossby deformation radius therefore limits the reach of the balanced response. Layer stretching, represented by $-\psi /L_D^2$, is absent from ordinary planar vorticity inversion. Unlike the pure Poisson equation, the finite-deformation operator also responds to the constant part of $\psi$, which specifies mean elevation rather than an arbitrary streamfunction offset. Specified bottom topography adds a known source to $q$; it is subtracted along with the background before inversion.

In stratified quasi-geostrophic inversion, continuously varying vertical displacement replaces the shallow layer's single stretching term. For positive $N^2(z)$ and constant nonzero $f_0$, the three-dimensional quasi-geostrophic potential vorticity isGeostrophic balance and hydrostatic balance give velocity $(-{\psi }_y,{\psi }_x)$ and buoyancy perturbation $b=f_0{\psi }_z$. Invert the anisotropic elliptic boundary value problem using interior $q$, prescribed top and bottom buoyancy as Neumann boundary conditions, and the lateral, far-field and global-flow data. Boundary buoyancy evolves by its own material conservation when the boundary is impermeable. It cannot be discarded: the Eady edge wave in the preceding solution has zero interior potential vorticity anomaly but nonzero boundary buoyancy and a nonzero induced flow.

For constant $N,f_0$, the stretched coordinates for quasi-geostrophic inversion $Z=Nz/|f_0|$ turn the inversion operator into the three-dimensional Laplacian. Its infinite-space Green's function is $-1/(4\pi |\mathbf{X}-{\mathbf{X}}^{\prime }|)$ in $(x,y,Z)$ coordinates. Hence a horizontal disturbance of scale $L$ communicates vertically over scale $|f_0|L/N$, explicitly coupling layers. In a bounded domain with homogeneous difference data, multiplying the difference of two inversions by its streamfunction and applying integration by parts givesThis proves uniqueness up to a constant when the relevant boundary terms vanish; a mean or pressure reference fixes the remaining constant. Compatible boundary fluxes are needed for a purely Neumann boundary-value problem.

Across all four models, inversion provides the induced flow, while scalar transport supplies time evolution. The independent horizontal inverse in the first two models becomes a finite-deformation inverse in shallow water and a vertically coupled elliptic inverse in continuous stratification. The principle applies within the stated balance or constrained model. In the unrestricted Boussinesq equations or shallow water equations, gravity waves and divergent motion can share the same potential vorticity, so the vortical scalar alone does not reconstruct every possible flow.

Let $\ell$ and $\mathcal{T}$ denote longitudinal length and time scales, and let $U$ be a typical streamwise velocity. The shallow water approximation requires $h/\ell \ll 1$, slow changes in the channel geometry $h|B^{\prime }/B|\ll 1$, and small longitudinal slopes. The vertical velocity is then of order $Uh/\ell$; its acceleration must be small compared with the restoring acceleration, for example $Uh/(\mathcal{T}\ell )+U^{2}h/{\ell }^2\ll g^{\prime }$ in a reduced-gravity layer. The leading pressure is hydrostatic pressure, and cross-sectional horizontal velocity is approximated by $u(x,t)$. High Reynolds number permits negligible interior viscous terms but does not itself guarantee the geometric or hydrostatic assumptions. We also require $B>0$ and $h>0$ in the wetted interior. The dimensionless $B$ need not be small: it measures the cross-section's opening angle, while the smallness condition concerns streamwise variation.

The cross-sectional area and volume transport areFor an impermeable bed and no fluid entrainment, volume conservation givesor, equivalently,The geometrical source term expresses the depth change needed when a moving layer encounters a wider or narrower triangular section.

Write $g^{\prime }=g({\rho }_f-{\rho }_a)/{\rho }_f>0$. The Saint-Venant dry-front condition describes a liquid advancing over a dry bed, or into an ambient whose inertia is negligible, as when ${\rho }_a/{\rho }_f\ll 1$. The edge has $h_f=0$; its velocity is obtained by continuing the appropriate shallow water Riemann invariant to zero depth. There is no independent universal finite-depth head Froude number. In the triangular geometry a release of a resting layer gives $u_f=2\sqrt{2g^{\prime }{h}_0}$, whereas a rectangular release gives $2\sqrt{g^{\prime }{h}_0}$.

The Benjamin deep-ambient front condition includes the inertia of an ambient that must turn around a finite-depth gravity current head. In the ideal deep-ambient Boussinesq approximation, ${\rho }_a/{\rho }_f$ is close to one andHere $h_f$ is the current depth just behind its head, $u_f$ its speed relative to the far ambient, and $F$ the ideal head Froude number. The ambient is much deeper than $h_f$; a finite upper-layer depth changes the closure. The lower density must exceed the ambient density for a bottom current. A non-Boussinesq head requires an appropriate density-ratio correction rather than using $F=\sqrt{2}$ at every ratio. The Saint-Venant and Benjamin models therefore describe different front physics, even though both use the same long-wave equations in the interior.

Use $G=g\beta (T-T_0(z))$ for the plume reduced gravity, and take $z$ upwards. For a symmetric line plume define fluxes per unit source length,The integrals span the plume; its edges can move horizontally with $z$. The fluid entrainment rate at one moving edge is the inward normal transport relative to that edge. For an edge $x=b(z)$ it is $E=w{b}^{\prime }-u$; symmetry supplies equal transport through the other edge. The Batchelor entrainment hypothesis closes its mean value as $E=\alpha w_c$, where $w_c$ is a representative vertical velocity and $\alpha$ is the entrainment coefficient. The integrated volume conservation equation is therefore $q^{\prime }=2\alpha w_c$.

Using volume conservation, the vertical acceleration equation has conservative form ${\partial }_x(uw)+{\partial }_z(w^2)=G$. Incoming ambient has zero vertical momentum; the integrated momentum conservation equation is consequently $m^{\prime }=\int Gdx$. For the temperature anomaly, $T_0$ depends on height andAmbient fluid enters with $G=0$, giving the buoyancy flux balance. The integral model isHere $N$ is the buoyancy frequency when $N^2>0$; the same algebra permits unstable temperature gradients with $N^2<0$.

These are mean integral balances. The advective equations printed in the question omit explicit turbulent stresses and heat fluxes; a turbulent plume interpretation requires the usual entrainment closure for that unresolved transverse transport. Assuming vanishing edge stresses and heat fluxes, and negligible additional axial turbulent-flux contributions, gives the displayed bulk model. It would be inconsistent to treat a discontinuous top-hat profile as a smooth exact pointwise solution of the inviscid temperature-advection equation.

In the top-hat plume model, vertical velocity $w(z)$ and reduced gravity $G(z)$ are uniform over $-b<x<b$. The ambient has zero vertical velocity, negligible ambient turbulence and a prescribed $T_0(z)$. Thenand the equations becomeEquivalently, the ordinary differential equations for the three unknowns areThe last equation displays dilution by fluid entrainment as well as loss of relative buoyancy while rising through a stably stratified ambient. The model assumes a steady slender two-dimensional turbulent plume, large Reynolds number, the Boussinesq approximation, constant $\beta$ and $\alpha$, negligible heat loss, negligible imposed crossflow, and negligible pressure departure from ambient hydrostatic pressure. The factors of two count both entraining edges; $b$ is a half-width. A finite source also requires its volume, momentum and buoyancy fluxes as boundary data.

Let ${\mathcal{B}}_0>0$ be the source buoyancy flux per unit length. The conversion from the heat-source strength depends on its units: if $F$ is a kinematic temperature-volume flux, ${\mathcal{B}}_0=g\beta F$; if it is a physical heat power per unit length, ${\mathcal{B}}_0=g\beta F/({\rho }_0{c}_p)$, where $c_p$ is specific heat capacity at constant pressure. This explicit convention avoids treating heat power as a buoyancy flux without conversion.

For uniform ambient $N^2=0$, so $\mathcal{B}={\mathcal{B}}_0$. The ideal pure line-source limit has negligible initial volume and momentum transport. The flux equations admit $w=w_{\ast }$ constant, $q=2\alpha w_{\ast }z$ and $b=\alpha z$. Substituting into the momentum equation gives $2\alpha w_{\ast }^2=2bG$, and the conserved buoyancy flux then fixes $2\alpha w_{\ast }^3={\mathcal{B}}_0$. ThusThis verifies all three differential equations, not just their dimensional exponents. The plume grows in width while maintaining constant upward velocity; its temperature anomaly decreases like $z^{-1}$. The zero-height singularity is the ideal point-source limit. A finite source or nonzero momentum input introduces a plume virtual origin or a forced near-source region; the heat strength alone cannot uniquely determine that finite-source region.

The filling box model requires the first outflow to spread across the enclosure and remain above unmodified fluid, while a narrow nearly vertical plume rises below it. Several mechanisms can defeat those assumptions. A tall narrow box makes the plume a substantial fraction of the width, so its compensating return velocity is no longer small and modifies both fluid entrainment and momentum flux. A fast ceiling current can descend the sidewalls and overturn the initially stable upper region. In both cases a horizontally uniform slowly descending filling-box first front is inappropriate. Strong plume displacement from the centre, imposed crossflow, large source volume, vents or substantial wall cooling likewise alter the volume and heat budgets.

A useful diagnostic is whether the ceiling outflow's inertial acceleration, of order $U_c^2/{\ell }_c$, overwhelms the restoring buoyancy of the nascent layer. If $U_c^2/(G_H{d}_c)$ is not small for its depth $d_c$, the outflow can penetrate or overturn the stratification rather than remain trapped above it. This is a diagnostic Froude number, not a universal threshold without specifying the wall-current geometry and mixing. Low Reynolds number invalidates a constant turbulent entrainment coefficient, and thermal diffusion or wall heat loss can erase the sharp first-front jump. Thus large aspect ratio, strong return circulation, overturning, ventilation or diffusion require a model beyond the standard filling-box approximation.

Use the Boussinesq approximation and a top-hat plume model. Let $v(z)<0$ be the uniform ambient return velocity outside the plume. Negligible source volume flux and an impermeable cylindrical wall require zero net transport through every horizontal section:The entrainment speed must be specified relative to that moving ambient. Taking it to be $\alpha (w-v)$ givesThe surrounding unmodified fluid has zero buoyancy relative to the initial fluid. Neglecting thermal or compositional loss therefore givesLet $B_s$ denote the conserved specific source buoyancy flux: $B_s=B_o/\pi$ if $B_o$ is the actual integrated source flux, or $B_s=B_o$ if the source strength has already been normalized by $\pi$.

To obtain the usual integral-model singularity one must also state a pressure closure. Here take the whole-section mean pressure to have only its reference ambient hydrostatic pressure, and neglect wall force and extra vertical turbulent-stress fluxes in the integral budget. Internal momentum transfers between plume and ambient cancel in that whole-section balance. The remaining upward force is the plume buoyancy, givingThe two momentum fluxes add, even though $v$ is negative: vertical momentum transported downward also contributes $v^2$ to the flux through an upward-oriented horizontal section. These three flux balances, with the algebraic return-velocity constraint, govern $b,w,g^{\prime }$ in the confined plume with compensating return flow.

The pressure-neglecting closure is an integral approximation, not a claim that the return fluid has zero acceleration. If a mean pressure departure $p_{\ast }(z)$ is retained, the whole-section momentum balance instead contains $-(R^2/{\rho }_0)p_{\ast }^{\prime }(z)$, along with any wall-force or stress contribution. The entrainment hypothesis alone does not specify that function. The singularity calculated below belongs to the stated pressure-neglecting model; it is not forced by continuity alone or asserted to be an infinite velocity in the real fluid.

It is helpful to solve using the nonsingular total flux $J$ and volume flux $Q$ before recovering $M$. The relation $J=M/(1-Q^2/(R^{2}M))$ is quadratic:The plus sign selects the small-area pure-source branch. Its discriminant vanishes when $J=4Q^2/R^2$, givingThereforeThe specific upward plume and ambient momentum-flux contributions are equal, not opposite. The ambient's transported momentum has the opposite sign from its transported volume, so their product remains positive.

To show that the pure branch actually reaches this fold in finite height and that it is a genuine pole, introduceThen $s=q^2/m$ andThe pure-source condition is $j\sim (5/8{)}^{2/5}{q}^{4/5}$. Before the fold, $s<1/2$ and $m>2q^2$. Thus for every fixed positive $q_1$, the derivative $dj/dq$ is bounded above by a constant times $q^{-2}$ on $q\ge q_1$. If the branch continued for arbitrarily large $q$, $j$ would remain bounded, contradicting $j\ge 4q^2$. Hence it reaches $j=4q^2$ at a finite $q_o>0$. Its height is finite because $d\zeta /dq=(1-s)/(2\sqrt{m})$ is finite away from the source and behaves like a constant times $q^{-2/5}$ at the source, an integrable singularity.

For the numerator, write $m=j(1-s)$. ThenAt the fold $j=4q_o^2$, so $q_o^3\ge 5/256$. On the other hand the dimensionless numerator in the $m$ equation at $s=1/2$ isIt is strictly negative because $5/256>1/(64\sqrt{2})$. The denominator $1-2s$ approaches zero from above. Consequently $dM/dz\to -\infty$ at finite $z_o$, although the fluxes and velocities themselves remain finite. This proves the confined-plume momentum fold rather than merely noticing a possible zero denominator.

The dimensionless pure-source branch uniquely determines the remaining amplitudes. They can be specified without an elementary antiderivative by integrating the displayed scalar equation for $j(q)$ until $j=4q^2$, and settingA direct numerical integration gives $q_o\simeq 0.281610$, ${\zeta }_o\simeq 0.400049$, so the dimensional results for this specified entrainment convention areA convention using $\alpha w$ instead of $\alpha (w-v)$ would produce different numerical amplitudes; the relative-speed convention was stated in part (a).

Physically $z_o$ is a turnover or breakdown height for a slender steady plume surrounded by uniform counterflow. The opposing streams then occupy equal areas and have equal speeds; the single-valued small-area branch cannot continue smoothly. Above it one expects lateral spreading, stronger recirculation, mixing and unsteady buoyant structures rather than a continuation of the same slender top-hat column. A detailed flow or upper-layer evolution requires restoring pressure and stress dynamics and, eventually, the modified ambient stratification. The model does not predict an infinite physical velocity, nor does it require that all buoyant fluid forever remain below $z_o$.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.