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With exponentially decreasing injection, the advective approximation $\varphi h_t+u{h}_x=0$ and inlet condition $uh(0,t)=Q{e}^{-t/\tau }$ give $h_0=Q/u$ and the displayed solution along characteristic curves. It applies behind the advective nose $X=ut/\varphi$. The relative size of the discarded nonlinear spreading term for this profile is $2\varphi h\cot \theta /(u\tau )$. Smallness of that quantity, and matching through a narrow leading layer, are necessary qualifications: elapsed time alone does not make the approximation uniform at the nose. In a translating frame the full equation remains a porous medium equation with coefficient $u\cot \theta /\varphi$.

The Barenblatt solution is a mass-preserving similarity solution of the porous medium equation. For $u_t=\beta (u{u}_x{)}_x$ in one dimension it has the form

For a steady point-source shear-driven viscous gravity current far downstream, $h\ll y_N\ll x$ permits dropping the downstream gravity flux. With $w=h^2$, the equation is $w_x=D(w^2{)}_{yy}/(4A)$, a quadratic porous medium equation. Conserving $A\int wdy=Q$ yields $y_N=(3\mu \Delta \rho gQx/{\tau }^2{)}^{1/3}$ and $h^2=\tau (y_N^2-y^2)/(2\Delta \rho gx)$ inside the current. The central height decays as $x^{-1/6}$. Near the source, balancing shear and gravity gives an upstream reach of order $\sqrt{\mu \Delta \rho gQ}/\tau$, whose numerical coefficient is not determined by scaling.

The porous medium equation $h_t=\lambda (h{h}_x{)}_x-\Gamma h$ with constant $\Gamma >0$ becomes $H_{\tau }=\lambda (H{H}_x{)}_x$ under the displayed transformation. Substitution gives $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $\lambda (h{h}_x{)}_x=\lambda e^{-2\Gamma t}(H{H}_x{)}_x$, so ${\tau }^{\prime }=e^{-\Gamma t}$. This transformation applies to every solution with appropriately transformed initial and boundary conditions, not just a similarity solution. A one-sided constant-volume porous gravity current therefore has remaining physical volume $V{e}^{-\Gamma t}$ and a limiting front $[9\lambda V/(\varphi \Gamma ){]}^{1/3}$. The finite transformed time $\tau \to 1/\Gamma$ means that arbitrary initial profiles need not reach the universal large-time source shape before drainage removes them.

In a porous medium equation with exponent $m>1$, the effective diffusion coefficient vanishes at zero density. A nonnegative compactly supported source profile can therefore retain a finite front for every finite positive time. For example, the radial cubic-diffusion source profile has support radius proportional to $t^{1/6}$. This notion concerns a moving diffusion front; it does not require the constant characteristic cone used for a wave equation.

A wide, shallow drop with cross-sectional area $A$ has the gravity-driven volume flux $q=-g{h}^3{h}_x/(3\nu )$ when substrate no-slip and zero surface shear are imposed. Mass conservation yields a porous medium equation, $h_t=[g/(3\nu )](h^3{h}_x{)}_x$. Its symmetric source-type similarity solution has half-width $\ell$ in the display and height $h=[9\nu ({\ell }^2-x^2)/(10gt){]}_{+}^{1/3}$, where $I={\int }_0^{\pi /2}{\cos }^{5/3}\theta d\theta$. Integrating the profile recovers $A$. The contact-edge slope diverges, so the bulk lubrication approximation requires an inner edge regularization if contact-line microphysics or surface tension is retained.

For constant porosity and permeability, the equation is $h_T+h_X={\nabla }_{X,Y}\cdot (h{\nabla }_{X,Y}h)$. Under $\tau =T$, $\xi =X-T$, $\eta =Y$, the chain rule gives $h_T=h_{\tau }-h_{\xi }$ and $h_X=h_{\xi }$. Thus the moving-frame porous medium equation isIts conserved mass is $M=\iint hd\xi d\eta$.

There is a source inconsistency in the printed volume formula. Actual liquid volume is $V=P\iint hdxdy$. The stated coordinates have Jacobian determinantso the volume Jacobian for downslope stretched coordinates requiresThe printed factor ${\tan }^2\theta$ in the first expression is its reciprocal and cannot be proved with these coordinates. For example, at $\theta =\pi /6$ an integral equal to one corresponds to actual volume $3P$, whereas the printed formula gives $P/3$. The physical-volume normalization below uses the corrected Jacobian.

Let $r=\sqrt{{\xi }^2+{\eta }^2}$. For the source-type axisymmetric similarity solution, putMass conservation imposes $a=2b$, and balancing $h_{\tau }$ against $\nabla \cdot (h\nabla h)$ gives $a+1=2a+2b$. Hence $a=1/2$ and $b=1/4$. The profile equation isMultiplying by $s$, the left side is $-\frac{1}{4}(s^{2}f{)}^{\prime }$. Integration and regularity at the origin giveWhere $f>0$, divide by $sf$ to obtain $f^{\prime }=-s/4$, soSet the profile to zero beyond this front. The mass normalization isTherefore the two-dimensional Barenblatt profile for quadratic porous-medium diffusion iswhere $[q{]}_{+}=\max (q,0)$. Its radius and central thickness areThe profile is regular at the center, its flux $-h{h}_r$ vanishes at the dry edge, and the front speed $r_N^{\prime }=r_N/(4\tau )$ equals the limiting advective speed $-h_r$ there. Its support shrinks to the release point as $\tau \downarrow 0$ while its mass stays $M$, giving the concentrated initial release in the weak sense.

In physical coordinates its center travels atand $r=\tan \theta \sqrt{(x-x_c{)}^2+y^2}$ with $\tau =\rho gK\sin \theta \tan \theta t/(\mu P)$. If the printed volume expression were instead adopted merely as a definition of a formal parameter $\tilde{V}$, the same profile would use $M=\tilde{V}/(P{\tan }^2\theta )$; that parameter would equal $V{\tan }^4\theta$, not the actual released fluid volume.

Let $z$ measure height above the substrate, and write $\mu =\rho \nu$. In the lubrication approximation, the horizontal scale $\ell$ is much larger than the typical height $H_0$. Vertical momentum balance is hydrostatic to leading order, while the horizontal balance retains the dominant vertical shear:The no-slip boundary condition gives $u(0)=0$; negligible air viscosity and constant surface tension give $u_z(h)=0$. Integrating twice and then through the depth yieldsThe free-surface kinematic boundary condition and incompressibility imply $h_t+q_x=0$. ThereforeThe cross-sectional area $A=\int hdx$ is conserved when the endpoint volume flux vanishes.

The relevant small groups follow from the same balances. Put ${ϵ}_h=H_0/\ell$ and $U_{\ast }=g{H}_0^3/(3\nu \ell )$. Small ${ϵ}_h$ suppresses horizontal viscous derivatives and makes vertical viscous corrections to hydrostatic pressure smaller by $O({ϵ}_h^2)$. The ratio of horizontal inertia to vertical viscous stress isThus a small full-length Reynolds number is sufficient but not necessary: thin-layer geometry reduces the inertial ratio. The unsteady viscous ratio $H_0^2/(\nu t_{\ast })$ must likewise be small; on the spreading timescale $t_{\ast }=\ell /U_{\ast }$ it equals the displayed group. If surface tension is retained, its pressure-gradient correction relative to gravity isThe Bond number here uses the horizontal scale, because the capillary pressure is $O(\gamma H_0/{\ell }^2)$ whereas the hydrostatic pressure is $O(\rho g{H}_0)$. We also neglect air shear and evaporation and assume a Newtonian fluid with constant kinematic viscosity.

To find the similarity solution, put $K=g/(3\nu )$. Conservation of $A$ gives $h\sim A/\ell$, and balancing the time derivative against the nonlinear diffusion gives ${\ell }^5\sim K{A}^{3}t$. A normalized ansatz isSubstitution gives $(f^3{f}^{\prime }{)}^{\prime }=-(f+\xi f^{\prime })/5$. For the symmetric zero-flux profile, one integration gives $f^3{f}^{\prime }=-\xi f/5$, and on its positive support a second integration givesThus the solution is a compactly supported Barenblatt solution of the porous medium equation. Let $\ell (t)$ denote its half-width. In dimensional variables,Here the positive-part notation sets the height to zero outside $|x-x_c|<\ell$. The area normalization fixes the coefficient, not merely the exponent. With $x-x_c=\ell \sin \theta$,Writing the remaining integral as $I$, the final widths areThe height falls like $t^{-1/5}$ and the width grows like $t^{1/5}$. The volume flux tends to zero at the front, even though the profile slope diverges; this gives a valid compactly supported weak solution of the reduced equation. The diverging edge slope means the lubrication approximation is not uniform at a microscopic contact line. Surface-tension or contact-line physics supplies an inner regularization without changing the gravity-dominated bulk similarity law in its validity regime. This is an exact source-type solution for $t>0$; an arbitrary finite initial drop approaches an appropriate late-time profile rather than being exactly represented by this singular initial profile.

Write $G=\Delta \rho g>0$ and take $\tau >0$ in the positive $x$ direction. Removing the ambient hydrostatic pressure leaves excess pressure $G(h-z)$, so the horizontal pressure gradient in the dense gravity current is $G\nabla h$. The lubrication approximation givesThe no-slip boundary condition at $z=0$ and the imposed upper shear stress require $u(0)=v(0)=0$, $\mu u_z(h)=\tau$ and $\mu v_z(h)=0$. Integration yieldsConsequently the depth-integrated volume flux isUsing conservation of mass, $h_t+\nabla \cdot \mathbf{q}=0$, proves the evolution equation for a shear-driven viscous gravity current:The gravity term smooths thickness gradients, while the imposed shear transports the current downstream.

The approximations need more than a small ordinary Reynolds number. With thickness $H$, horizontal scale $L$, dense-fluid density ${\rho }_c=\rho +\Delta \rho$ and typical horizontal speed $V$, sufficient small parameters areThe first neglects horizontal viscous derivatives relative to vertical ones. The second compares convective inertia with vertical viscous stress: it is the reduced Reynolds number, ${\mathrm{Re}}_H(H/L)$, rather than a requirement that ${\mathrm{Re}}_H$ itself be small. The third keeps viscous normal stress small relative to the excess hydrostatic pressure $GH$. Here $V$ is of order $\tau H/\mu +G{H}^3/(\mu L)$, so this last condition requires $\tau /(GL)\ll 1$ together with $(H/L{)}^2\ll 1$. Explicit sufficient inertia conditions are ${\rho }_c\tau H^3/({\mu }^{2}L)\ll 1$ and ${\rho }_{c}G{H}^5/({\mu }^2{L}^2)\ll 1$. If an independently imposed unsteady time scale is $T$, also require ${\rho }_c{H}^2/(\mu T)\ll 1$; for $T\sim L/V$ this is the same reduced-inertia condition.

Set $A=\tau /(2\mu )$ and $D=G/(3\mu )$. Far downstream, the transverse scale $Y=y_N$ is much smaller than $x$, so the downstream gravity-flux divergence is smaller than the transverse one by order $(Y/x{)}^2$. For the steady flow, away from the source,The latter expresses conservation of the prescribed point-source volume flux. In scaling form, $Q\sim A{H}^{2}Y$ and $A{H}^2/x\sim D{H}^4/Y^2$. Eliminating $H$ gives $Y^3\sim DQx/A^2$, proving $y_N\propto x^{1/3}$ and $H\propto x^{-1/6}$.

For the exact downstream similarity solution, put $w=h^2$. Since $h^3{h}_y=(w^2{)}_y/4$, the reduced equation is the porous medium equationSeek $w=x^{-1/3}F(\eta )$ with $\eta =y/x^{1/3}$. Substitution givesIntegrate once, using symmetry and zero transverse flux at $\eta =0$, to obtain $-A\eta F/3=DF{F}^{\prime }/2$. Where $F>0$, this gives $F^{\prime }=-2A\eta /(3D)$, hence $F=A({\eta }_N^2-{\eta }^2)/(3D)$. Vanishing thickness fixes its support. The volume-flux condition then readsThus the downstream profile and half-width arewith $h=0$ outside that interval. This is the downstream similarity of a shear-driven gravity current. The square-root edge is the formal outer thin-layer solution; arbitrarily close to that edge its slope is not small, so the local front requires a separate description. This does not change the downstream scaling and flux calculation.

Near the source the two horizontal scales are comparable, say $\ell$, so the far-downstream reduction cannot locate the upstream nose. In the upstream direction the downstream shear flux $A{H}^2$ must balance the upstream gravity flux of order $D{H}^4/\ell$. Hence $\ell \sim (D/A)H^2$. The total discharge through a region of width $\ell$ is $Q\sim A{H}^2\ell$, giving $H^4\sim Q/D$. Combining these estimates gives the upstream reach scaleThe shear advects liquid toward positive $x$, while the hydrostatic thickness gradient drives liquid back toward negative $x$; their balance stops further upstream spreading. A numerical prefactor requires the full two-dimensional near-source free-boundary solution and cannot be fixed by these scaling arguments alone.

Choose transverse $x$ along the cusp from the line of cylinder contact, and $y$ across its narrow gap. Expanding the two circular boundaries for $x\ll a$ gives $y=\pm x^2/(2a)$, so the local full gap is $d(x)=x^2/a$. At the meniscus, $x\simeq w$, a semicircle fits across this gap with radius $d(w)/2=w^2/(2a)$. Thus the leading area and transverse curvature in the cusp flow between touching cylinders areThe meniscus area correction is $O(w^4/a^2)$ and is small relative to $w^3/a$. The curvature is stated as a positive suction magnitude; the liquid pressure relative to air is $p=-2\gamma a/w^2$. Axial curvature is negligible compared with this transverse curvature under the slow-variation approximation.

At fixed $x$, axial lubrication theory gives a local slit Poiseuille flow between rigid walls $y=\pm d/2$:Integrating across the cusp givesThe local meniscus edge correction occupies a transverse width small compared with $w$ and does not change this leading flux. Since $p_z=4\gamma a{w}^{-3}{w}_z$, the axial volume flux is $Q=-\gamma w^4{w}_z/(21\mu a^2)$. Area conservation ${\mathcal{A}}_t+Q_z=0$ now givesWriting $u=w^3$ also puts this in porous medium equation form $u_t=\gamma (u^{5/3}{)}_{zz}/(35\mu a)$, explaining the finite-support spreading profile.

Set $K=\gamma /(7\mu a)$. Fixed volume means $\int w^{3}dz=3aV$. If $w\propto t^{-\beta }$ and axial extent is proportional to $t^{\alpha }$, this conservation gives $\alpha =3\beta$. The evolution equation gives $3\beta +1=5\beta +2\alpha$, hence $\beta =1/8$, $\alpha =3/8$.

For the symmetric source-type similarity solution, write $w=t^{-1/8}f(\eta )$, $\eta =z/t^{3/8}$. The differential equation becomesIntegrate from the centre using zero flux and symmetry. This gives $K{f}^4{f}^{\prime }=-3\eta f^3/8$, so on the positive support $f{f}^{\prime }=-3\eta /(8K)$. With $f({\eta }_N)=0$,Therefore the fixed-volume capillary spreading in a cylindrical cusp profile isLet $B=21\mu a/(8\gamma t)$. Its volume isWith $s=\sin \theta$, the last integral is ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\theta d\theta =3\pi /8$, equal to the supplied sine integral by symmetry. Consequently $V=\pi B^{3/2}{z}_N^4/(8a)$, yieldingThe other tip is at $-z_N$. The maximum width decays as $t^{-1/8}$. This is the leading slender outer solution: the ideal point injection and the steep microscopic nose require local descriptions outside its uniform lubrication range.

Finally, for vertical cylinders let $\rho$ denote the liquid density, and take the bath pressure at $z=0$ as the air-pressure reference. At rest the liquid pressure is $p(z)=-\rho gz$. Equating hydrostatic suction with the wetting-meniscus capillary pressure gives the hydrostatic rise in a cylindrical cusp:The cusp can therefore support an indefinitely high, increasingly narrow ideal wetting tail rather than a fixed terminal rise height. Large $z$ makes $w\ll a$ and strengthens the transverse cusp approximation, until microscopic physics limits the continuum description.

For a concentrated line release, take $V$ to be actual fluid volume per unit source length and allow symmetric spreading to both sides of $x=0$. PutSubstitution in the porous medium equation gives the exponential drainage transform for porous-medium diffusion:The Barenblatt solution has conserved area ${\int }_{-\infty }^{\infty }Hdx=V/\varphi$. A quadratic cap satisfies this equation:Indeed, inside the current $H_x=-x/(3Ds)$ and $R\propto s^{1/3}$; differentiating verifies $H_s=D(H{H}_x{)}_x$. Its area is $\int Hdx=2R^3/(9Ds)=V/\varphi$, fixing the prefactor. The flux vanishes at the two moving noses because $H=0$ there. Hence the remaining physical volume and occupied width areFor a one-sided current on a reflecting half-line, replace the $2$ in $R^3$ by $1$ when $V$ denotes the total release into that side. Volume alone does not select this exact solution for an arbitrary finite initial patch; it describes an impulsive line release. Its initial central height is singular, so a finite injection region and the constraint $h\ll w$ regularize the near-source early-time behavior.

Take the boundary to be horizontal, put $z=0$ on it, and let $z=h(x,t)$ be the interface. Write $\Delta \rho =\rho -{\rho }_2>0$. Under the long-thin hydrostatic approximation, the deep ambient fluid has pressure $p_2=p_{\ast }-{\rho }_{2}gz$ to leading order, with negligible horizontal pressure gradient. Matching pressure at the interface gives, inside the dense porous gravity current,The Darcy velocity is superficial volume flux per total area, not pore speed. The horizontal Darcy law and depth-integrated flux therefore giveThe porosity $\varphi$ multiplies storage, so mass conservation is $\varphi h_t+q_x=0$. ConsequentlyThe deep-ambient and long-thin assumptions are what permit neglecting the ambient return-flow pressure variation, vertical inertia, and nonhydrostatic pressure corrections.

For a localized finite release, first use the reflecting half-line $x\ge 0$, so $h_x(0,t)=0$. Let $S=V/\varphi$ be the conserved cross-sectional area of the current; $V$ here is actual pore-fluid volume per unit cross-flow distance. A similarity solution of the porous medium equation has thickness proportional to $t^{-\alpha }$ and extent proportional to $t^{\beta }$. Conservation gives $\alpha =\beta$, and balancing $h_t$ against $(h{h}_x{)}_x$ gives $\alpha +2\beta =1$. Thus $\alpha =\beta =1/3$. PutSubstitution gives $-(F+\xi F^{\prime })/3=(F{F}^{\prime }{)}^{\prime }$. Zero flux at $\xi =0$ allows one integration: $F{F}^{\prime }=-\xi F/3$. In the positive-current region, $F^{\prime }=-\xi /3$, so $F=({\xi }_f^2-{\xi }^2)/6$. The normalization ${\int }_0^{{\xi }_f}Fd\xi =1$ gives ${\xi }_f^3=9$. The one-sided constant-volume porous gravity current is thereforeHere $[a{]}_{+}=\max (a,0)$. Its Darcy flux vanishes at the moving front because $h=0$ there, and its physical volume is $\varphi {\int }_0^{L}hdx=V$. This is the exact concentrated-release Barenblatt solution of the depth equation, physically used after the short initial adjustment to a long-thin current; an arbitrary finite initial patch requires its initial profile as well as its volume, although the same profile describes the usual late spreading regime. If a line release spreads symmetrically in both directions and $V$ denotes the total volume, use the same parabola on $-L<x<L$ with $L^3=9\lambda Vt/(2\varphi )$. If instead volume has already been divided by porosity, replace $V/\varphi$ by that geometric volume.

Drainage at superficial rate $\varphi \Gamma h$ changes mass conservation to $\varphi h_t+q_x=-\varphi \Gamma h$, soFor constant $\Gamma >0$, use the exponential drainage transform for porous-medium diffusionIndeed, $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $(h{h}_x{)}_x=e^{-2\Gamma t}(H{H}_x{)}_x$. Choosing ${\tau }^{\prime }=e^{-\Gamma t}$ leaves $H_{\tau }=\lambda (H{H}_x{)}_x$. Substituting the undrained Barenblatt solution givesUse the factor $1/2$ in $L_d^3$ for a total-volume symmetric release, as above. For general initial data the same transform is exact, but $H$ is the corresponding undrained solution rather than automatically a source parabola.

The remaining physical volume is $V(t)=V{e}^{-\Gamma t}$. Early times have $\tau \sim t$, recovering undrained spreading. Late times have $\tau \to 1/\Gamma$, so the front approaches a finite runout while the current drains exponentially:The front slows because drainage reduces the hydrostatic driving head. The finite transformed time also explains why drainage does not guarantee that an arbitrary initial profile reaches its undrained universal large-time shape.

Use the positive transport coefficient $D$ from the preceding solution. The porous medium equation here is $h_t=D(h^2{h}_x{)}_x=(D/3)(h^3{)}_{xx}$. A planar pulse conserves the water cross-sectional area $\mathcal{V}=\int hdx$. If the supplied lake volume is a three-dimensional volume $V$, introduce the constant out-of-plane width $B_y$ and take $\mathcal{V}=V/B_y$; alternatively $V$ may be understood as volume per unit span. A two-dimensional model cannot determine an absolute extent from an unspecified three-dimensional volume alone.

For a symmetric localized release, write $h=t^{-a}f(x/t^b)$. Area conservation requires $a=b$, and balancing the PDE powers gives $a+1=3a+2b$, hence $a=b=1/4$. With $\eta =x/t^{1/4}$, the profile equation isIntegrate once, using symmetry and zero central flux: $D{f}^2{f}^{\prime }=-\eta f/4$. Inside the wet region this gives $f^2=C-{\eta }^2/(4D)$. The dry continuation is zero. Write the result asThe normalization follows from ${\int }_{-L}^{L}hdx=\pi HL/2=\mathcal{V}$. This cubic diffusion pulse has finite support $-L<x<L$, total extent $2L$, and spreading speed $\dot{L}=L/(4t)$. Although $h_x$ is singular at the ideal nose, the flux vanishes there and $u=-Dh{h}_x=x/(4t)$ tends to the finite front speed.

For a one-sided pulse on $0<x<L$ with a reflecting boundary at zero and the same area $\mathcal{V}$, replace $\mathcal{V}$ in the displayed full-line formulas by $2\mathcal{V}$. The power laws are unchanged. These are source-type Barenblatt solutions for an instantaneous localized release; a finite initial lake footprint approaches the profile at long times and may require a virtual time origin, rather than matching this singular initial condition exactly.

Assume $D_0,n_0,Q>0$. The density-dependent diffusion coefficient makes this a cubic porous medium equation, since its two-dimensional form is $n_t=[D_0/(3n_0^2)]\Delta (n^3)$. The mass-preserving similarity solution is a radial cubic-diffusion source profile. At fixed $r$, the proposed scaled density givesSeparation is possible whenwhere $\gamma$ is a constant. The resulting ordinary differential equation isRecognize the second term as $\gamma ({\xi }^{2}F{)}^{\prime }$. Regularity and zero radial flux at the origin set the integration constant to zero:In the region $F>0$, this reduces to $F{F}^{\prime }=-\gamma \xi$, giving $F^2=1-\gamma {\xi }^2$ from the central normalization. For a nonnegative profile whose front is precisely at $\xi =1$, the first zero fixes $\gamma =1$. Continuity of the density excludes truncating a positive value at the front. HenceHere integrating the scale equation gives ${\lambda }^6=6D_{0}t/r_0^2+\text{constant}$, and the point-source initial condition requires $\lambda (0)=0$.

Use the planar area element, not the one-dimensional area under the drawn profile. Mass conservation givesThus the scale and the complete density arewhere $[u{]}_{+}=\max (u,0)$. The factors of $n_0$ are necessary on dimensional grounds; the length scale is not a function of $Q$ alone.

The front and peak obeyEach radial profile starts with horizontal tangent at the origin, decreases to a square-root edge, and is identically zero beyond $R(t)$. Later profiles are wider and lower. This is finite propagation in porous-medium diffusion, because the diffusion coefficient vanishes at zero density. The sketch uses dimensionless time $\tau =6D_{0}t/r_0^2$; it preserves the radial integral $2\pi \int rndr$, not the unweighted area under each curve.

At the front $F^{\prime }$ diverges, so the profile is a weak solution, not a globally smooth classical solution. Nevertheless $F^2{F}^{\prime }=-\xi \sqrt{1-{\xi }^2}$ tends to zero there. The density and flux both match their zero exterior values, so extending the interior solution by zero creates no spurious boundary source in the conservation law. The behavior at the origin is regular because $F^{\prime }(0)=0$.

Finally, the point release is recovered as a distributional initial condition. For any continuous compactly supported test function $\phi$ on the plane, the total density remains $Q$ and all its support lies in the shrinking disk $r\le R(t)$. ThereforeThus $n(\cdot ,t)\to Q{\delta }^{(2)}$ as $t\downarrow 0$, in the sense of the Dirac delta function.

Take $\Delta \rho >0$ and $\tau >0$, with $x$ pointing along the imposed shear stress, and put $G=\Delta \rho g$. Relative to the upper fluid's hydrostatic reference, the hydrostatic pressure in the current is $p=G(h-z)$. Thus the horizontal pressure gradient is $G\nabla h$. The lubrication theory momentum balance, lower no-slip boundary condition, and imposed upper shear stress giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe two terms have opposite roles: imposed shear stress carries fluid downstream, whereas the hydrostatic pressure gradient spreads it from thick regions towards thin ones.

For the approximation, a representative horizontal velocity is $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis condition controls viscous normal stress, vertical viscous corrections and the normal projection of shear at a slightly tilted interface relative to $GH$. With it and the previous inertia bounds, vertical inertia is small as well. Time variations here are on the transport timescale $L/U$; separately imposed rapid forcing would need its own unsteady inertia bound. A small ordinary Reynolds number is a stronger sufficient restriction, but the reduced inertia ratios above are what the thin-layer momentum balance directly requires.