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Let $z_f(t)$ be the filling-box first front, with $z_f(0)=H$. Below it the original fluid has not yet been through the plume. In the filling box model, the turbulent plume adjusts rapidly and occupies a negligible fraction of the room's area. A point source with negligible source volume flux supplies $Q(z)=C{F}_s^{1/3}{z}^{5/3}$. All its volume crossing the front grows the processed region, so volume conservation givesPut $a_p=\pi C{F}_s^{1/3}/A_c$. Integrating $d(z_f^{-2/3})/dt=2a_p/3$ yieldsThe first front approaches the floor only as $t\to \infty$. A nonzero source volume flux would require an equal exhaust to keep the room volume fixed; it is treated in the following parts.

Let $\rho$ be the liquid density and neglect atmospheric shear. The shallow-channel lubrication approximation makes the surface nearly horizontal in each cross-section, at $z=h(x,t)$, with hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. At fixed $y$, the floor is $z_b=\alpha |y|$ and the local liquid thickness is $H_y=h-z_b$. Axial momentum and boundary conditions areWriting $\zeta =z-z_b$, integration givesThe wetted interval is $|y|<h/\alpha$. Its cross-sectional area and total axial flux areThe shallow V-channel lubrication flux and volume conservation $A_t+Q_x=0$ therefore giveThe slope $\alpha \ll 1$ makes transverse shear small relative to vertical shear; the longitudinal current must also be slender for this reduction.

For typical height $H$ and axial length $L$, fixed volume gives $H^{2}L\sim \alpha V$, while the equation gives $L^2\sim C{H}^{3}t$. Eliminating either scale yieldsThus the current grows in length like $t^{2/7}$ and thins like $t^{-1/7}$. Equivalently, the porous-medium transformation of V-channel spreading sets $w=h^2$ and gives $w_t=(C/5)(w^{5/2}{)}_{xx}$ with constant mass $\alpha V$.

Seek a symmetric solution $h=t^{-1/7}f(\xi )$ with $\xi =x/t^{2/7}$. Substitution into the conservative depth equation givesThe left side is $-(2/7)(\xi f^2{)}^{\prime }$. Symmetry sets the integration constant to zero, so $C{f}^4{f}^{\prime }=-(2/7)\xi f^2$ within the support. Integrating $f^2{f}^{\prime }=-2\xi /(7C)$ gives $f^3=3({\xi }_0^2-{\xi }^2)/(7C)$. In physical variables the constant-volume similarity in a V-shaped channel iswhere $X(t)$ is the half-length of its support and the positive part makes $h=0$ outside it.

Define the positive normalizationThe first substitution is $s=\cos \theta$; the gamma form follows from the beta function integral. Conservation of volume now givesand consequentlyThe full current length is $2X(t)$. Its flux vanishes at the two noses, since the integrated similarity equation gives $C{h}^4{h}_x=-2x{h}^2/(7t)$ there, so no volume is lost at the support boundary. A shifted origin or a positive shift of time gives the corresponding translated similarity family.

In an intended trigonometric $I(\beta )$ notation, $J=I(7/3)$ requires either cosine integrated over $[-\pi /2,\pi /2]$ or absolute cosine over $[0,\pi ]$. The printed ${\cos }^{\beta }\theta$ on $[0,\pi ]$ is not a positive real normalization for the required fractional exponent: with a real cube-root interpretation at $\beta =7/3$, its two halves cancel, while the principal complex power is not real. The similarity normalization uses the positive sine integral $J$ above.

Assume the shallow-water approximation: the horizontal scale is long compared with the depth, the hydrostatic approximation applies, and the cross-section-averaged velocity is approximately uniform. Neglect viscosity and drag for this part. Volume conservation gives the constant discharge $Q=bhu$. The steady horizontal momentum equation isso its integral is the Bernoulli equation head $\mathcal{H}=H+h+u^2/(2g)$. The shallow-water specific energy measured above the bed is thereforeThe Froude number compares the flow velocity with the long-wave speed $\sqrt{gh}$:At fixed $Q,b$, the shallow-water specific energy tends to infinity as $h$ tends to zero or infinity. Its unique minimum occurs at the critical depthThe two depth branches above that minimum are subcritical flow, $F<1$ with $h>h_c$, and supercritical flow, $F>1$ with $h<h_c$. The long-wave characteristic speeds relative to the bed are $u\pm \sqrt{gh}$. Thus a subcritical flow can receive information from downstream, whereas both characteristic curves travel downstream in a positive supercritical flow.

A hydraulic control is a critical section where one characteristic speed vanishes and a smooth flow can pass between the two branches. Regularity there supplies a condition relating $Q$, the depth and the geometry; together with reservoir or inlet data, it can determine the discharge. Merely finding $F=1$ is not enough: a finite depth gradient also requires the geometric forcing to vanish in the critical momentum equation.

The occupied cross-sectional area and interface width areThe incoming volume flux per unit channel length is $h{w}_e$. Volume conservation therefore gives $A_t+(Au{)}_x=h{w}_e$. Ambient fluid carries no excess mass density, so integrating the excess-mass density balance gives $(A{g}^{\prime }{)}_t+(Au{g}^{\prime }{)}_x=0$. These two equations imply dilution of the reduced gravity, even though total integrated buoyancy is conserved.

The integrated horizontal hydrostatic pressure force, divided by the reference mass density, isThe prismatic walls contribute no axial component of pressure force. Since ambient fluid enters with zero horizontal velocity, it supplies no horizontal momentum flux. Thus the entraining shallow-water current in a triangular channel obeys the conservative equationsTo put them in primitive form, let $D={\partial }_t+u{\partial }_x$ be the material derivative. The volume equation gives $Dh+(h/2)u_x=w_e$. Subtracting $u$ times the volume equation from the momentum equation, and $g^{\prime }$ times it from the buoyancy equation, yieldsThe momentum source is the acceleration of newly entrained ambient fluid, not bed drag. The source in the reduced gravity equation describes mixing with ambient fluid of zero excess mass density. In particular, the rectangular-channel coefficients cannot be used: the triangular area produces both $h/2$ in the depth equation and $h/3$ in the buoyancy-gradient force.

Hindered settling is the reduction of the particle settling velocity by displaced-fluid backflow and interactions with other particles. In the specified model, the downward speed is $V_s$ at vanishing concentration and falls to zero at the maximum packing fraction ${\varphi }_{\max }$. The characteristic speed of a concentration disturbance will differ from this individual-particle speed.

Take $z$ positive upwards. For a locally uniform horizontal suspension without diffusion or bulk fluid motion, the upward particle flux isParticle volume conservation gives the scalar conservation lawThis is the quadratic hindered-settling flux. Along its characteristic curves,For a smooth initial profile ${\varphi }_0(\zeta )$, the map is $z=\zeta +c({\varphi }_0(\zeta ))t$. Its Jacobian is $1+(2V_s/{\varphi }_{\max }){\varphi }_0^{\prime }(\zeta )t$. A shock develops where this first reaches zero: the compressive condition is ${\varphi }_0^{\prime }<0$, andwhen the minimum is negative and no earlier boundary discontinuity is imposed. A discontinuous compressive initial state can instead contain a shock from $t=0$.

Let $z=s(t)$ separate states ${\varphi }_L$ below the shock and ${\varphi }_R$ above it. Integrating the conservation law across a moving thin interval gives the Rankine-Hugoniot conditionEquivalently, the flux relative to the shock agrees on both sides. For this quadratic flux,Since $j^{\prime \prime }=2V_s/{\varphi }_{\max }>0$, a physical compression shock has ${\varphi }_L>{\varphi }_R$ and satisfies $c({\varphi }_L)>\dot{s}>c({\varphi }_R)$: characteristic curves enter it from both sides. These conditions select the shock over a nonphysical expansive discontinuity.

The entrainment coefficient $\alpha$ is the ratio of the inward ambient-fluid velocity normal to a buoyant plume edge to the characteristic buoyant plume velocity. For an upward buoyant plume, take $u_e=\alpha w>0$. There are two edges at half-width $b$, so the full-section volume source is $2u_e$. Dividing every integral balance by two gives consistent half-section equations.

For top-hat plume profiles, volume conservation and mass conservation areThe mass entering the buoyant plume is ambient mass with local mass density ${\rho }_0(z)$. The Boussinesq approximation uses a common reference mass density in inertia while retaining the mass density deficit in buoyancy. The mass equation may be kept to the first order in that deficit to derive its transport.

Multiply the volume equation by the time-independent ambient mass density. The product rule givesSubtract the mass equation and multiply by $g$. The entrainment sources cancel exactly, leavingwhereFor stable stratification, ${\rho }_0^{\prime }<0$ and the buoyancy frequency $N$ is real. Rising fluid moves into lighter ambient fluid and loses its mass density deficit relative to that local ambient. Thus environmental stratification reduces the buoyant plume's buoyancy flux, even though the ambient fluid entering by entrainment contributes no mass density deficit at its own height.

Apply differentiation under the integral sign on the moving drop:The moving-boundary term vanishes because $h(R,t)=0$. At the origin, symmetry and regularity give $h_r(0,t)=0$. At the front, the stated $h\propto (R-r{)}^{1/3}$ gives $h^3{h}_r=O((R-r{)}^{1/3})\to 0$. Thus $\boxed{dM/dt=0}$, expressing volume conservation for this axisymmetric viscous gravity current.

With $\eta =r/t^{\alpha }$ and $R={\eta }_0{t}^{\alpha }$, substitute the self-similar solution into the conserved integral:For a nonzero finite drop volume, the power of $t$ must vanish, so $\boxed{\alpha =1/8}$. Balancing powers in the evolution equation gives the same result: its left side scales as $t^{-5/4}$ and its right side as $t^{-1-2\alpha }$. The front-edge exponent alone would not determine this spreading exponent without the conserved volume.

At a fixed time, the complete spatial derivative isThe first term describes geometric variation along the channel at a fixed surface elevation. The second describes the extra wetted area produced by a varying water level; $S_h$ is the surface width.

Assume a fixed channel, negligible lateral inflow, uniform cross-sectional velocity, incompressible water and the hydrostatic long-wave approximation. Integrating volume conservation over a short channel segment givesInviscid horizontal momentum, with $h$ measured from the fixed datum, is $u_t+u{u}_x+g{h}_x=0$. The coefficient matrix acting on $(h_x,u_x)$ has eigenvaluesThese are the variable-section channel characteristics. Put $D_{\pm }={\partial }_t+(u\pm c){\partial }_x$. Adding the corresponding linear combinations of the two equations givesThe opposite signs in the two combinations cancel both derivative terms in the source calculation.

If $c=c(h)$, then $D_{\pm }c=c_h{D}_{\pm }h$ and $d{c}^2/dh=2c{c}_h$. Taking $\xi =t$ along a characteristic gives the requested formA sufficient geometry assumption is $S(x,h)=b(x)A(h)$, for which $S/S_h=A/A^{\prime }$ is independent of $x$ at fixed height.

For a completely general $S(x,h)$, the printed form omits an explicit wave-speed geometry term. The correct conversion is insteadFor example, a lake at rest with constant $h$ but position-dependent $c$ has $D_{\pm }c=\pm c{c}_x$. The printed left side is then nonzero while its right side is zero. The $h$-based compatibility equation remains valid, including when ${\partial }_h(c^2)=0$ makes the speed-based expression singular.

Let $\mathbf{U}=(u,v)$, $\zeta =v_x-u_y$ and $D/Dt={\partial }_t+\mathbf{U}\cdot \nabla$. The inviscid shallow water equations areTaking the vertical curl of the momentum equation, with constant $f$, gives $D\zeta /Dt=-(f+\zeta )\nabla \cdot \mathbf{U}$. Combining this with the depth equation provesThus shallow-water potential vorticity is materially conserved: changes in absolute vorticity are exactly compensated by stretching or compression of the layer depth.

The relation to a Bernoulli function presupposes steady flow. In that case define $B=g(h+H)+\frac{1}{2}|\mathbf{U}{|}^2$. The vector identity $(\mathbf{U}\cdot \nabla )\mathbf{U}=\nabla (\frac{1}{2}|\mathbf{U}{|}^2)+\zeta \hat{\mathbf{z}}\times \mathbf{U}$ givesSteady volume conservation allows a transport stream function with $h\mathbf{U}=({\psi }_y,-{\psi }_x)$. Hence $\hat{\mathbf{z}}\times (h\mathbf{U})=\nabla \psi$, soIn particular, $\mathbf{U}\cdot \nabla B=0$: the Bernoulli function is constant along each streamline. Where the flow permits a single-valued dependence on the stream function, the displayed derivative describes how that constant changes between neighbouring streamlines. For an unsteady flow, shallow-water potential vorticity remains conserved but the stated steady Bernoulli function relation does not generally hold.

Assume the ambient fluid has zero downslope velocity, no bed drag, and an entrainment coefficient $\alpha >0$. The entraining shallow-water layer receives volume at rate $\alpha |u|$, but neither excess buoyancy nor downslope momentum from that ambient fluid. Consequently volume conservation, conservation of excess buoyancy, and momentum conservation giveThe pressure contribution is the integral of the hydrostatic pressure from part (a). The last source is the downslope component of the gravitational force.

With $D={\partial }_t+u{\partial }_x$, the equivalent equations areThe last term is the momentum required to accelerate newly entrained fluid. For variables $(h,u,b)$, the matrix of the spatial derivatives isThus it is a strictly hyperbolic system for $h,b>0$ and $0\le \theta <\pi /2$. At $\theta =\pi /2$ all three characteristic speeds coincide and the matrix is not diagonalizable. The exclusion of that angle presumes the physical range with $\cos \theta \ge 0$: allowing an overhanging channel with $\cos \theta <0$ would also destroy hyperbolicity.

The channel is prismatic: its width depends on height, not on $x$. A layer of depth $h$ has cross-sectional area $A={\int }_0^h\beta zdz=\beta h^2/2$. Volume conservation gives $A_t+(Au{)}_x=0$. With $g^{\prime }=G\varphi$, the integrated excess hydrostatic pressure force isThus the streamwise momentum balance is $(Au{)}_t+(A{u}^2+G\varphi \beta h^3/6{)}_x=0$. Under the specified vertical-settling approximation, the horizontal projection of the depositional boundary has width $\beta h$, so the particle balance is $(A\varphi {)}_t+(Au\varphi {)}_x=-\beta hw\varphi$, where $w=w_{0}f(\varphi )>0$ is the downward speed magnitude. The prismatic triangular-channel shallow water equations are thereforeThe pressure coefficient $h/3$ is the section-weighted mean of $h-z$. The factor two in deposition comes from top width divided by area. There is no streamwise widening term, because $\beta$ is constant along the channel.

Define the positive characteristic speed scale $c=\sqrt{G\varphi h/2}$. In variables $(u,\varphi ,c)$ the equations becomeThe coefficient matrix of this hyperbolic system has eigenvalues $u,u+c,u-c$. Hence the three characteristic families areAlong $dx/dt=u$, the concentration equation is the ordinary differential equation $d\varphi /dt=-2w\varphi /h$. Along $dx/dt=u\pm c$, the left eigenvectors $(1,\mp 4c/(3\varphi ),\pm 4)$ give the sedimenting triangular-channel characteristic compatibility equationsHere every derivative in a given equation follows that characteristic family, and $h=2c^2/(G\varphi )$. These three compatibility equations form the characteristic description; they are not three independent closed equations for all fields on any one curve. In particular, $u\pm 4c$ are not conserved Riemann invariants when the concentration varies: its differential and the deposition source must both be retained. The description assumes $h,\varphi >0$; the dry or zero-buoyancy limit is degenerate.

Apply volume conservation, mass conservation and vertical momentum flux balance to a slice of the triangular-profile line plume. Ambient ingestion contributes $2e$ of volume and $2{\rho }_{0}e$ of mass per unit height and source length. The ambient is quiescent, so it supplies no leading vertical momentum. The integrated driving force is the stored buoyancy $\mathcal{B}$. Using the quantities from part (a), the non-Boussinesq triangular-profile line-plume balances areTogether with Batchelor entrainment, these are three equations for $W,\rho ,b$. They neglect viscous boundary stresses and streamwise pressure-force corrections within the integral-plume approximation.

Multiplying volume conservation by $g{\rho }_0$ and subtracting $g$ times mass conservation cancels the ambient sources exactly, givingThis is buoyancy conservation in the homogeneous ambient; it is a consequence of the first two balances, not an additional independent equation.

Now take the Boussinesq approximation in inertia and entrainment, retaining the small density deficit in buoyancy. Then $Q={\rho }_{0}bW$, $M=(2/3){\rho }_{0}b{W}^2$ and $F=(2/3)gbW({\rho }_0-\rho )$. ConsequentlyThe mass source is $2\alpha {\rho }_{0}W=3\alpha {\rho }_{0}M/Q$. Substitution into the mass conservation, momentum and buoyancy balances gives the unsteady Boussinesq triangular-profile line-plume balancesThese use density-weighted mass and momentum fluxes. Removing ${\rho }_0$ from some flux definitions while retaining it in the source would mix incompatible conventions.

Use the steady Boussinesq top-hat plume in a stratified ambient, with negligible ambient vertical momentum and no externally added heat or particles. The Batchelor entrainment hypothesis supplies inward velocity $\alpha w$ across circumference $2\pi b$. Volume conservation gives $\pi Q_z=2\pi b\alpha w$, and $bw=\sqrt{M}$, henceThe momentum flux increases by the integrated buoyancy force. Writing $g^{\prime }=g\Delta \rho /{\rho }_o$, the force per reference density and per $\pi$ is $g^{\prime }{b}^2$. Since $F=g^{\prime }Q$ and $b^2=Q^2/M$, this givesFor the buoyancy flux, entrained material has the local environmental mass density. Let ${\rho }_p$ be the plume's averaged density. Its scalar mass conservation is $({\rho }_{p}Q{)}_z={\rho }_a{Q}_z$. Since $F=(g/{\rho }_o)({\rho }_a-{\rho }_p)Q$, differentiating cancels both entrainment terms and givesIn stable density stratification, ${\rho }_a^{\prime }<0$ and a rising plume loses buoyancy flux. These are kinematic fluxes with the common factor $\pi {\rho }_o$ removed, so no additional factor of $\pi$ belongs in the three differential equations.

Seek a steady downslope solution with $u>0$ constant. The volume conservation equation reduces to $h_x=E$, so the constant-speed entraining gravity current on a slope hasThe conserved buoyancy flux per unit width is $B=u{g}^{\prime }h$, givingSubstitute these expressions into the momentum equation. Its advective term is $E{u}^2$ and its hydrostatic term is $\frac{1}{2}E(B/u)\cos \theta$. The balance is thereforewhich givesThe assumed slope inequality makes ${\beta }_{\ast }>0$. The increasing thickness dilutes reduced gravity while leaving $g^{\prime }h$ and the buoyancy flux constant. The denominator includes both bottom drag and the momentum needed to accelerate newly entrained ambient fluid.

Choose the positive $x$ direction along the background flow, so $U>0$; reverse $x$ if necessary. Let $z$ measure depth below the impermeable roof, and set $\Delta \rho ={\rho }_w-{\rho }_c>0$. The deep ambient water has $p_w=p_0(x)+{\rho }_{w}gz$. Continuity of pressure at $z=h$ and hydrostatic pressure inside the buoyant CO2 giveHere $k$ is permeability of a porous medium and $\mu$ is dynamic viscosity. Since $h/H\ll 1$, the ambient return-flow correction to the imposed pressure gradient is higher order. The small aspect ratio supplies the hydrostatic pressure approximation; capillary trapping, dissolution, compressibility and dispersion are omitted in this sharp-interface model.

The depth-integrated Darcy flux is $q=Uh-K_{b}h{h}_x$. The actual stored fluid volume is $\varphi h$ per unit horizontal area, where $\varphi$ is the aquifer porosity. Thus the porous gravity current with background flow satisfieswith $u=U/\varphi$, $a=K_b/\varphi$ and $J=Q/\varphi$. Both $Q$ and the later $V$ denote actual injected volume per unit transverse width. Away from injection, the equation is an advected porous medium equation. At each finite front impose $h=0$ and vanishing outgoing volume flux per unit width. At the injection point $h$ is continuous and $q(0^{+})-q(0^{-})=Q$. For an initially empty aquifer, $\varphi \int hdx=Qt$. The advective speed is $U/\varphi$, not $U$.

For a constant-flux porous gravity current, write characteristic depth $H_c$ and extent $L_c$ temporarily. At early times volume conservation gives $H_c{L}_c\sim Jt$; gravity spreading gives $L_c^2/t\sim a{H}_c$. HenceBackground advection becomes comparable when $L_c\sim ut$, so the transition scales areFor $t\ll t_c$, the leading similarity solution is symmetric:For $0<\eta <{\eta }_N$, its model ordinary differential equation and boundary conditions areChoose the nonnegative finite-front branch, reflect evenly in $x$, and require $F{F}^{\prime }\to 0$ at the front. The factor $1/2$ divides the injection equally between the two halves. Integration of the ordinary differential equation gives $2{\int }_0^{{\eta }_N}Fd\eta =1$; the limiting front slope is $F^{\prime }({\eta }_N^{-})=-2{\eta }_N/3$. This boundary-value problem determines the initially symmetric shape without assuming an unjustified elementary formula. Independent numerical integration gives ${\eta }_N\simeq 1.17619$ in this normalization.

For $t\gg t_c$, upstream storage approaches a finite steady value. With no upstream leakage, $q=0$ for $x<0$, giving $h_x=U/K_b$ wherever $h>0$. Downstream the uniform interior carries $q=Q$, soThe upstream volume flux vanishes even though local Darcy velocity contributions from background flow and buoyancy cancel. The upstream stored volume is $\varphi h_{\mathrm{int}}|x_{-}|/2$, a finite constant. The downstream plateau therefore extends a distance $ut+O({\ell }_c)$, and its front advances at $u$ to leading order.

The diffusive nose of an advected porous gravity current has widthIn a frame translating at $u$, balance the time derivative against the nonlinear gravity diffusion. With $\eta =(x-x_c(t))/\sqrt{a{h}_{\mathrm{int}}t}$ and $x_c(t)=ut+O({\ell }_c)$, the leading transition is $h=h_{\mathrm{int}}F(\eta )$, withIts finite front has $F\sim ({\eta }_N/2)({\eta }_N-\eta )$ as $\eta \uparrow {\eta }_N$. Independent numerical integration of this boundary-value problem gives ${\eta }_N\simeq 1.23849$, so a more accurate leading nose position is $x_N\simeq x_c+1.23849\sqrt{a{h}_{\mathrm{int}}t}$ and its local shape is $F\simeq 0.619247(1.23849-\eta )$. The entire transition has no elementary closed form. A useful explicit, approximate shape is the volume-preserving linear rampThis gives $x_N\simeq x_c+\sqrt{a{h}_{\mathrm{int}}t}$. Its ramp width is $2\sqrt{a{h}_{\mathrm{int}}t}$, and its edge obeys the exact kinematic boundary condition ${\dot{x}}_N=u-a{h}_x(x_N^{-},t)$ to the displayed order. It also has the same area as a sharp step at $x_c$. This is an approximate ramp closure, not an exact solution of the nonlinear nose ordinary differential equation. A constant phase shift $x_c=ut-|x_{-}|/2$ balances the finite upstream storage within this ramp approximation; the full solution's subleading matching may alter constant offsets.

For a rapid injection of total volume $V$, remove the source for $t>0$ and introduce $\xi =x-ut$. The advected constant-volume porous gravity current obeys $h_t=a(h{h}_{\xi }{)}_{\xi }$ and $\int hd\xi =M=V/\varphi$. A similarity solution has depth proportional to $t^{-1/3}$ and radius proportional to $t^{1/3}$. Direct substitution gives the compactly supported Barenblatt solutionIndeed ${\int }_{-R}^R(R^2-{\xi }^2)/(6at)d\xi =2R^3/(9at)=V/\varphi$. Its centre translates at the pore speed $u$, its two edges are $ut\pm R(t)$, and its maximum depth decays as $t^{-1/3}$. The expression is exact for a point release within this reduced model; finite-width initial data approach this similarity solution at late times. If $U=0$, the constant-rate current remains in the symmetric spreading regime and the constant-volume solution has a stationary centre.