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Let $T_0=T_i-T_{\infty }>0$, $S_0=S_i-S_{\infty }>0$ and $R_0=\beta S_0/(\alpha T_0)$. At steady volume, the weir outflow equals the inflow $Q$. The well-mixed heat and salt balances giveWith $\theta =(T_i-T)/T_0$ and $\eta =(S_i-S)/S_0$, the flux ratio implies ${\eta }_s=q{\theta }_s$, where $q=R_F/R_0$. The actual interface ratios are $\Delta T=T_0(1-\theta )$, $\Delta S=S_0(1-\eta )$ and $R_{\rho }=R_0(1-\eta )/(1-\theta )$.

The original PDF uses $R_{\rho }^{-2}$ in the heat-flux law; the converted TeX's exponent $-3$ is a transcription error. Using the PDF yieldsSubstitute this into the heat balance to obtain the double-diffusive overflow reservoir relationsFor the usual heat-dominated positive upward buoyancy flux, $0\le R_F<1$ and $R_0>1$ give $0\le q<1$. Then the right-hand side after substituting ${\eta }_s=q{\theta }_s$ decreases from $C$ to zero as ${\theta }_s$ runs from zero to one; there is a unique physical steady root. The net upward buoyancy flux is $B_0=g\alpha F_T(1-R_F)$.

Write $z_i=H-h$ for the height of the warm-layer interface above the heat source. A negligible-volume point heat source supplies a pure plume with conserved buoyancy flux $B_0$. The given plume buoyancy implies its volume flux at the interface isAt steady displacement ventilation, the plume feeds the warm layer at the same rate it empties through the upper vent: $q_v=Q_p(z_i)$. The upper-layer buoyancy balance gives $q_v{g}^{\prime }=B_0$, soEliminate $q_v$ and $g^{\prime }$ to getThe height equation has one solution with $0<h<H$. Its height is independent of source strength in this ideal pure-plume model, while its warm-layer density contrast increases as $B_0^{2/3}$. The sketch shows cool replacement air below, an entraining plume through that region, and the upper mixed layer feeding the ceiling vent.

The room's nearly motionless upper reservoir has interface $z_i=H-h$. A doorway roof at height $D$ has available light-layer head $\delta =D-z_i=h-(H-D)>0$. Evaluate the Bernoulli equation in the room and at a freely controlled doorway: $E=-g^{\prime }{z}_i$ upstream and $E=3g^{\prime }{h}_c/2-g^{\prime }D$ at its critical section. Therefore $h_c=2\delta /3$ andThis is the ideal broad-crested weir law, not a small-orifice law with $C_d$. A nonideal doorway would require its own measured loss correction.

When the floor heat source is on, volume conservation and buoyancy flux conservation still give $Q=B_0^{1/3}{z}_i^{5/3}/\gamma$ and $g^{\prime }=\gamma B_0^{2/3}{z}_i^{-5/3}$. Combine these with the doorway relation to obtain the plume-fed doorway ventilation equilibrium:There is one interface-height root in the indicated interval. The result is physically appropriate only if its upper outflow is thin enough for the one-layer doorway model to remain valid.

If $D<H$, air in the volume above the doorway is part of the upper thermal reservoir and can store heat and alter transients. But its extra depth $H-D$ does not add available head at the doorway: only $D-z_i$ enters the critical condition. In the steady ideal well-mixed model it shares the same density as the outflow. Treating the whole $h$ as the doorway head would overestimate discharge.

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.

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.

Normalize all fluxes by $\pi$, rather than including that factor in only some of them. The specific plume volume flux, momentum flux and buoyancy flux areThe ambient transport is $-Q$ and its specific momentum flux isPut $s=b^2/R^2=Q^2/(R^{2}M)$, with $0<s<1$. The total specific momentum flux isThe first differential equation follows immediately from relative-speed fluid entrainment:For the second, differentiate $J=M/(1-s)$ and use $s^{\prime }=2s{Q}^{\prime }/Q-s{M}^{\prime }/M$. Since $J^{\prime }=b^2{g}^{\prime }=B_{s}Q/M$, this yieldsThis is the requested two-equation system, with $B=B_s$ and zero total volume flux as the additional conservation conditions. Near the pure source, $Q,M\to 0$ and $s\to 0$. Eliminating height in that limit givesThe zero integration constant selects a pure plume; a nonzero imposed source momentum would select a different branch and generally a different turning height. The leading radius is $b\sim 6\alpha z/5$, which recovers the unconfined pure-plume limit near the source.

The mass and heat entrainment equations immediately giveUse the specified relation $\rho T={\rho }_0{T}_0=\beta$. ThenMeanwhile the kinematic buoyancy flux isSince $T_0$, ${\rho }_0$ and $c_p$ are constant,Thus conservation is a consequence of conserved excess heat in this ideal-gas, uniform-ambient model, not merely an assumption that arbitrary non-Boussinesq buoyancy fluxes are conserved in every environment.

Let ${\rho }_r$ be a constant reference density close to the ambient density, and let ${\rho }_p$ denote the mean plume density. The Boussinesq approximation permits full-cross-section kinematic fluxesThus $Q$ is the specific mass flux, equal to volume flux at this order; $M$ is the specific vertical momentum flux; and $F$ is the buoyancy flux. Physical mass and momentum fluxes are ${\rho }_{r}Q$ and ${\rho }_{r}M$. These definitions include the full circular area. If factors of $\pi$ are removed in a course's flux convention, divide $Q,M,F$ by $\pi$ throughout rather than mixing normalizations. The PDF's source density is ${\rho }_0$, distinct from ambient ${\rho }_a(z)$; the TeX changes the source symbol incorrectly.

For the top-hat plume model, let the plume radius be $r(z)$ and let $w,b$ be uniform inside it. ThenThe Batchelor entrainment hypothesis supplies the necessary closure: the inward edge velocity is $v_e={\alpha }_{e}w$, with constant entrainment coefficient ${\alpha }_e>0$. Entrainment across a cylindrical slice gives $Q^{\prime }=2\pi r{\alpha }_{e}w=2{\alpha }_e\sqrt{\pi }\sqrt{M}$. The entrained ambient has negligible upward momentum. The vertical buoyancy force on the slice is $\pi r^{2}b=F/w$, hence $M^{\prime }=FQ/M$.

For the buoyancy balance, let $R={\int }_A{\rho }_{p}wdA$ be the transported density scalar. Entrainment gives $R^{\prime }={\rho }_a{Q}^{\prime }$. Differentiate $F=g({\rho }_{a}Q-R)/{\rho }_r$ to get $F^{\prime }=g{\rho }_a^{\prime }Q/{\rho }_r$. Therefore the closed flux equations areHere $N$ is the buoyancy frequency when the ambient is stably stratified. These are the Boussinesq top-hat plume in a stratified ambient equations in the explicitly stated full-area convention. Stable stratification consumes positive buoyancy flux as the plume rises. The top-hat assumption alone would not determine $Q^{\prime }$ without an entrainment law.

For uniform ambient density, $F=F_0$ is constant. The source volume $Q_0$ becomes negligible when accumulated entrainment ${\int }_0^{z}c\sqrt{M(s)}ds$ is much greater than $Q_0$. In a jet-like region, $M\simeq M_0$, so the criterion is $z\gg Q_0/(c\sqrt{M_0})$. In a buoyancy-dominated far field with $Q\simeq C_Q{z}^{5/3}$, it is $z\gg (Q_0/C_Q{)}^{3/5}$, with $C_Q$ determined below. These are the two source-volume lengths of a turbulent plume. They are not the jet length, which measures the persistence of imposed source momentum. A nonzero source volume can also appear as a plume virtual origin; it is not valid to discard it at every distance simply because the source is geometrically small.

For $Q_0=M_0=0$, $F_0>0$, seek $Q=C_Q{z}^{5/3}$, $M=C_M{z}^{4/3}$. The two flux equations give $C_Q=(3c/5)\sqrt{C_M}$ and $C_M^{3/2}=9c{F}_0/20$. Thus the pure-plume solution isFor completeness its directly measurable top-hat fields areThis is an axisymmetric pure plume. In this regime the explicit source-volume criterion isFor $Q_0=F_0=0$, $M_0>0$, the buoyancy vanishes and momentum is constant. The pure-jet solution isThis is the jet limit of the top-hat plume model. Both ideal zero-volume sources have singular speeds at $z=0$; they model regions outside a finite source where entrainment has already made the source-volume correction small.

When $M_0$ and $F_0$ are both positive, use the jet solution near the source and the pure-plume solution far above it, with an appropriate virtual-origin correction. To quantify this transition, eliminate $z$:The conserved quantity is the plume flux-balance invariant. With negligible $Q_0$, comparison of its two terms and $Q\simeq c\sqrt{M_0}z$ gives a transition height of order $M_0^{3/4}/(\sqrt{c}\sqrt{F_0})$. The exact connection, including finite $Q_0$, is the finite-source forced-plume quadraturewith $M$ then recovered from the invariant. A positive integrand gives the physical increasing-volume branch. As $Q$ grows, the source constant is negligible relative to $Q^2$, proving attraction to pure plume similarity. One matches the two asymptotic regimes or uses this quadrature; the jet and plume solutions cannot be linearly superposed because their governing flux equations are nonlinear.

Use the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.

Let $z>0$ measure height from the lower edge and $y>0$ distance from one face of the vertical plate. Use the steady Boussinesq approximation, homogeneous isotropic permeability of a porous medium $K$, reference fluid mass density ${\rho }_0$, dynamic viscosity $\mu$, thermal expansion coefficient $\alpha$, effective thermal conductivity $k_e$ and liquid volumetric heat capacity $C_f={\rho }_0{c}_f$. Define $B={\rho }_{0}g\alpha K/\mu$ and $\kappa =k_e/C_f$, consistently with the thermal convention used above. Write $\theta =T-T_a$, where $T_a$ is the uniform ambient temperature.

Interpret $F$ as uniform wall heat-flux density: a plate segment of height $dz$ inputs heat $Fdz$ per unit horizontal span. The cumulative heat input up to height $z$ is therefore $Fz$. If $F$ instead denotes the sum of flux densities into two identically heated faces, replace it by $F/2$ in each face formula before summing. A specification only of total power per span, without its distribution in height, would not determine this uniformly heated-wall problem. We calculate one heated face; identical heating on both faces doubles the integrated fluxes.

Away from the leading edge, suppose the boundary layer is slender, $\delta (z)\ll z$, and porous inertia and longitudinal heat conduction are negligible. Lateral pressure adjustment leaves the vertical Darcy velocity proportional to buoyancy:At the plate, $v=0$ and $-k_e{\theta }_y=F$; in the far field $w,\theta \to 0$. Darcy law permits tangential discharge at an impermeable plate, so imposing a further no-slip condition $w=0$ would be inappropriate in this leading porous model.

Balance $\theta \sim F\delta /k_e$, $w\sim B\theta$ and $w\theta /z\sim \kappa \theta /{\delta }^2$ to obtainThus the warmed layer thickens and accelerates upward. Define a streamfunction by $w={\psi }_y$, $v=-{\psi }_z$ and setThe last two expressions agree because ${\delta }^3=\kappa k_{e}z/(BF)$. Differentiation gives $v=-(\kappa /3\delta )(2f-\eta f^{\prime })$. Substituting in the thermal equation leaves the single nonlinear equation for the uniformly heated vertical plate in a porous medium:The first condition is impermeability; the second fixes the wall heat flux; the third fixes ambient temperature and vertical velocity. Seek the positive-temperature branch with $f^{\prime }\ge 0$ and finite $f(\infty )$. Decay also gives $f^{\prime \prime }(\infty )=0$. The far-field transverse inflow need not vanish: $v_{\infty }=-(2\kappa /3\delta )f(\infty )$ supplies the increasing discharge.

An approximate solution can be obtained without a numerical shooting calculation. Integrate the differential equation over $0<\eta <\infty$. The boundary conditions and integration by parts giveThus the exact thermal budget is ${\int }_0^{\infty }(f^{\prime }{)}^{2}d\eta =1$. Assume an exponential thermal shape, $f=C(1-e^{-\eta /d})$. The wall derivative requires $C/d^2=1$, and the integral condition requires $C^2/(2d)=1$. Hence $d=2^{1/3}$, $C=2^{2/3}$ and the integral exponential profile for porous wall convection isIt satisfies all boundary conditions, has positive decaying temperature, and matches the exact integrated heat budget. It is an integral approximation, not an exact pointwise solution of the differential equation; its mass-flux coefficient should be read accordingly.

Define the volume discharge, kinematic Darcy momentum flux and buoyancy flux per horizontal span byWith these standard discharge conventions the mass flux is $\dot{m}={\rho }_{0}Q$. Their similarity expressions areThe integral identity evaluates two of them exactly:Equivalently $C_f\int w\theta dy=Fz$: all the heat supplied below height $z$ is carried upward through that section. Although these fluxes are exact within the boundary-layer equations, porous drag means that the momentum flux itself is not conserved with height.

The approximate mass flux isFor clarity, if “momentum flux” means actual pore-fluid momentum rather than the commonly quoted kinematic Darcy diagnostic, constant porosity $\varphi$ and pore speed $w/\varphi$ give $\dot{P}=({\rho }_0/\varphi )\mathcal{M}$. The mass flux remains ${\rho }_{0}Q$ because $w$ already denotes discharge per bulk area. Stating both prevents an implicit porosity error. Validity also requires $\alpha \theta \ll 1$ and a sufficiently slender, low-inertia porous boundary layer; the approximation is not the immediate leading-edge conduction field.

Use the conventional kinematic plume fluxesThen $\pi Q$ is the volume flux, $\pi {\rho }_{r}Q$ the dimensional mass flux, $\pi {\rho }_{r}M$ the dimensional momentum flux, and $\pi F$ the kinematic buoyancy flux. Calling $\pi Q$ a mass flux uses a density-normalized convention: the reference density has been suppressed. If density-weighted momentum and buoyancy fluxes are desired, multiply $M$ and $F$ by the same constant ${\rho }_r$ throughout. Here the conventional kinematic definitions make the Boussinesq approximation explicit.

For an upward plume with $Q,M>0$, the inverse relations areThe volume conservation, vertical momentum conservation, and buoyancy balances from part a becomeHere the vertical momentum conservation equation uses ${\rho }_r$ in the inertia and retains $({\rho }_0-\rho )gA={\rho }_{r}B$ in the buoyancy force. The ambient has no imposed vertical momentum at entrainment.

To classify the resulting partial differential equations, work in the nonsingular primitive variables $(A,w,g^{\prime })$ with $A>0$. Subtract $w$ times the volume equation from the momentum equation and $g^{\prime }$ times the volume equation from the buoyancy equation. This givesThe principal matrix is thereforeAll three characteristic speeds are real and equal to $w$, so the model is a hyperbolic system if that term means only reality of the characteristic roots. However, for $A>0$, $(K-wI)v=0$ requires $v_w=0$, leaving only two independent eigenvectors. HenceThis weak hyperbolicity of the unsteady top-hat plume prevents a complete set of independent characteristic fields. For example, freezing $A,w$ in the homogeneous principal part gives $D_t\delta w=0$ and $D_t\delta A=-A{\partial }_z\delta w$. A Fourier perturbation of wave number $k$ therefore produces a term proportional to $kt$ in $\delta A$. Real characteristic roots alone do not provide uniform control of high-frequency perturbations. The flux-variable transformation above is invertible in the stated region, so changing to $(Q,M,F)$ does not remove this degeneracy.

For a steady upward plume with prescribed $N^2=-G^2<0$, the kinematic plume fluxes satisfyThe positive sign in the last equation expresses the gain of buoyancy flux as the plume rises through the prescribed unstable ambient. Seek a positive power law solution $Q=q{z}^p$, $M=m{z}^r$, $F=f{z}^s$, with $z>0$. Matching powers givesThe last relation gives $s=p+1$ and the first $r=2p-2$. Substitution into the middle relation yields $4p-5=2p+1$, henceThe amplitude equations areFrom the first relation $m=9q^2/(4{\alpha }^2)$, while the last two give $m^2=G^2{q}^2/16$. Taking positive amplitudes gives $m=Gq/4$, and consequentlyReconstructing the radius, speed and reduced gravity gives the top-hat plume in constant unstable stratification:As a direct check, $M_z={\alpha }^2{G}^2{z}^3/9=FQ/M$ and $F_z={\alpha }^2{G}^3{z}^3/9=G^{2}Q$, while $Q_z={\alpha }^{2}G{z}^2/3=2\alpha \sqrt{M}$.