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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.

In the steady far-downstream regime, cross-stream derivatives dominate gravity-driven spreading. Dropping the smaller downstream gravity flux leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, with $h=0$ outside. The squared height is a parabolic Barenblatt solution; the height cross-section is a semicircular profile after rescaling its axes. The cross-stream volume flux vanishes at the edges even though the height slope becomes singular there. The profile describes the outer lubrication theory region, not a resolved microscopic front.

Near the point source the two horizontal dimensions are comparable, say $\ell$. Upstream spreading arrests where outward gravity-driven volume flux balances downstream shear-driven volume flux:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. The same horizontal scale is obtained by setting $y_N\sim x$ in the downstream similarity solution, confirming where that solution fails.

For the line source the steady volume flux per unit width is $q=A{h}^2-D{h}^3{h}_x$. On the downstream constant-height branch, $q=Q_{2d}$ gives $h_d^2=Q_{2d}/A$. Upstream there is no net flux through the finite nose, so $q=0$. Thus $h{h}_x=A/D$, and continuity of height at the source givesThe line-source upstream reach under imposed shear is thereforeIn the upstream part, shear-driven and gravity-driven volume fluxes cancel. The volume flux per unit width jumps by exactly $Q_{2d}$ at the source. As in the point-source profile, the square-root nose is a formal outer solution with an unresolved steep front; its finite upstream reach follows from flux balance, not from assuming a small slope all the way to the nose.

Near the closest point, in $X=x-Ut$, the gap is $h(X)\simeq h_0+X^2/(2a)$. The horizontal length scale is $\ell =\sqrt{2a{h}_0}$, up to an inessential constant. Thus the aspect ratio $h_0/\ell \ll 1$ follows from $h_0\ll a$. The lubrication approximation assumes an incompressible Newtonian fluid, no slip on the surfaces, slow variation of the gap, and negligible inertia relative to transverse viscous stress:This is the reduced Reynolds-number condition; it is weaker than requiring a Reynolds number based on $\ell$ to be small. In the cylinder frame the geometry and flow are steady. Pressure is uniform across the narrow gap to leading order, and equal ambient pressures far to either side close the model. The calculation assumes a full liquid film; cavitation would change that pressure condition.

In a steady axisymmetric flow without swirl, time derivatives and azimuthal derivatives vanish. For a slender high-Reynolds number jet of width $\delta \ll z$, radial viscous derivatives dominate axial ones by $(z/\delta {)}^2$. The transverse momentum balance makes pressure nearly constant across the jet, matching the constant ambient pressure, so the leading axial pressure gradient vanishes. Retaining axial convection and radial diffusion in the Navier-Stokes equation gives the boundary-layer equationThe radial velocity is small but its convective term is of the same order as axial convection because radial gradients are larger.

The hydrostatic approximation gives $p=\rho g(h-z)$ above the horizontal floor, up to atmospheric pressure. The horizontal Stokes flow equation is $\mu {\mathbf{u}}_{zz}=\rho g\nabla h$, with the no-slip boundary condition at $z=0$ and zero tangential stress on the free surface. Integrating twice gives $\mathbf{u}=(\rho g/\mu )\nabla h(z^2/2-hz)$ and the lubrication gravity-current fluxFor the squeegee lubrication model, use $h_0$ vertically, $\ell =\rho g{h}_0^3/(\mu U)$ horizontally, and $\ell /U$ in time. If $X$ is the laboratory coordinate in the direction of travel, set $x=(X-Ut)/\ell$, $y=Y/\ell$, $t=U{t}_{\mathrm{lab}}/\ell$, and $h=h_{\mathrm{lab}}/h_0$. The blade half-length is $\alpha =L/\ell$. The moving-frame flux is $\mathbf{q}=-h{\mathbf{e}}_x-h^3\nabla h/3$. Conservation of volume gives the dimensionless lubrication equationUsing these scales as representative of the actual flow, lubrication theory requires a small aspect ratio and negligible inertia relative to vertical viscous resistance:The second is the Reynolds number based on $h_0$, multiplied by $h_0/\ell$. Surface tension is neglected as assumed. Large piles, rounded gaps, and narrow end regions need their own local slope and inertia checks if these representative scales cease to describe them.

Both eigenvalues of the linear generator, $-1/R$ and $-1/(4R)$, are negative. Nevertheless the non-normal matrix transfers the slowly decaying second component into the first before viscous decay wins. Modal decay coexists with energy amplification of order $R^2$ over a time of order $R$.

This illustrates transient growth from non-normal modes: large Reynolds number permits much greater disturbance amplification even though the linear stability analysis finds no unstable eigenvalue. The energy-neutral rotational nonlinearity can redirect the amplified state and affect its subsequent evolution. The linear gain alone establishes neither sustained nonlinear instability nor a nonlinear transition threshold.

For $f(u)=C{u}^p$, the dimensional expression is again $L=C{\eta }^{2-3p}{\rho }^{p-1}{\sigma }^{2p-1}{t}^p$. Independence of dynamic viscosity requires $2-3p=0$, giving inertial hydrodynamic coarsening:For this law, both inertial terms and the capillary term scale as $u^{-4/3}$, whereas the viscous term scales as $u^{-5/3}$. It is consequently subleading at large $u$. Direct substitution gives the leading amplitude conditionA growing branch requires the indicated ratio to be positive. The inertial combination must be nonzero: if $2\beta =\alpha$ its leading acceleration and convective contributions cancel, so this particular schematic equation cannot realize the generic inertial-capillary balance with that pure power law. The requested scaling assumes no such geometrical degeneracy. This also shows why the coefficients cannot determine a universal numerical prefactor without a morphology calculation. The relative viscous contribution decays as $u^{-1/3}$, consistently with an increasingly large domain-scale Reynolds number.

For a power law $f(u)=C{u}^p$, restoring dimensions givesIndependence of mass density requires $p-1=0$. Hence viscous hydrodynamic coarsening hasAt small $u$, the capillary and viscous terms are both of order $u^{-2}$, while the convective inertial term is of order $u^{-1}$ and the acceleration of the leading linear law is zero. Thus the leading force balance is viscous resistance against capillary driving, giving $C_v=c_s/c_v=-\delta /\gamma >0$ in the signed convention of the preceding part.

Equivalently, the domain-scale Reynolds number is $\rho L\dot{L}/\eta \sim C_v^{2}u$, so inertia is negligible as $u\to 0$. “Small” here refers to time compared with $t_0$ within a late-stage hydrodynamic window: it does not extend the model through the microscopic initial stage, where diffusion or interface formation can dominate. An initial-time offset changes the intercept, not this asymptotic growth exponent.

Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

Put $q=10/Re>0$. The frequency formula becomesFor large positive Reynolds number, $\sqrt{q-q^2/4}=\sqrt{q}[1-q/8+O(q^2)]$. ThereforeMore precisely the real correction is $-5/Re$, and the next imaginary correction is $O(R{e}^{-3/2})$. The growing branch has a rate tending to zero as $Re\to \infty$, despite being unstable at every sufficiently large finite $Re$.

Let $w$ be downward Darcy flux, measured per unit total horizontal area. Neglect the resistance of the displaced phase, capillary pressure, and horizontal subsurface flow. Taking atmospheric pressure as zero, the upper end of the saturated column has pressure $\rho gh$ and its advancing lower end has pressure zero. Across a column of depth $l$, Darcy law therefore givesThe pressure head from the overlying layer supplies $h/l$, and gravity supplies one. Because the new liquid fills a pore volume $\varphi dl$ per unit area, the vertical imbibition under a draining liquid layer obeys $\varphi \dot{l}=w$. Volume conservation gives $h+\varphi l=h_0$, so, until the surface layer is exhausted,The initial condition is understood as a limiting solution of this singular equation. For $0<\varphi <1$, separation of variables givesAt small penetration, expansion of the logarithm yields $t\sim \varphi \nu l^2/(2gk{h}_0)$, henceA sufficient dimensional early-time condition is $t\ll \varphi \nu h_0/(gk)$, for which $l\ll h_0$. The printed $t\ll 1$ needs a chosen time unit or nondimensionalization. In the limiting case $\varphi =1$, $l^2=2gk{h}_{0}t/\nu$ holds exactly until $h=0$. The divergent initial Darcy flux is an idealization of sudden contact with a dry continuum; it cannot describe penetration below the pore scale.

For the spreading surface layer, measure $z$ upwards from the substrate. A thin, slow gravity current has hydrostatic pressure $p=\rho g(h-z)$ and a leading horizontal viscous balance $\nu u_{zz}=g{h}_x$. A no-slip boundary condition at $z=0$ and a stress-free boundary condition at $z=h$ giveThis lubrication gravity-current flux drives horizontal spreading down the surface-pressure gradient. Surface volume conservation is $h_t+q_x=-w$, while conservation in the pores is $\varphi l_t=w$. Combining these balances gives the deep-substrate drainage of a gravity current modelDrainage transfers liquid from the surface layer into the substrate. Indeed $(h+\varphi l{)}_t+q_x=0$.

For definiteness take a one-sided current on $0<x<x_N(t)$, supplied at $x=0$. The imposed volume is per unit span, so $V_0$ has dimensions of area. The boundary conditions and volume conservation constraint areThe source starts from zero injected volume. Initially dry points acquire $l>0$ only when the advancing current reaches them; the formula $h/l$ is not applied outside that region. At the advancing nose these are limiting conditions for a singular drainage equation. For a source feeding two identical sides, each side has half the volume and inlet flux; only the corresponding numerical prefactors change. The integral and inlet-flux conditions are equivalent once the local balance and nose conditions hold.

The physical approximations are lubrication theory, negligible inertia, slowly varying surface height, small pore-scale Reynolds number, and negligible capillary entry pressure. Treating the substrate as vertically draining columns also requires its lateral flow to be negligible and its depth to exceed the penetration depth. The model is valid while surface liquid remains available above each draining column.

For a similarity solution, let $h$ and $l$ scale as $t^a$, and let the current length scale as $t^b$. Since $h/l$ then has no time dependence, drainage requires $a-1=0$. Balancing $h_t$ with the horizontal spreading term gives $a-1=4a-2b$, so $a=1$ and $b=2$. These exponents also give injected volume $t^{a+b}=t^3$, explaining the particular supply law.

Define scaleswhich satisfy $CD=V_0/{\tau }^3$ and $D^2=(g/\nu )C^3$. The cubic-input similarity for a draining gravity current isSubstitution into the two conservation equations giveswhereThus only $K$ and $\varphi$ remain in the dimensionless problem. Its constraints areIntegrating the sum of the two ordinary differential equations verifies that the inlet flux and integral normalization agree. The dimensionless endpoint ${\xi }_N={\xi }_N(K,\varphi )$ is the undetermined multiplicative constant in the length law; solving for it is not required.

The exact gap between the plane and the circle isIn the narrow region relevant to lubrication theory,so changes occur over the streamwise scale $x=O(\sqrt{a{h}_0})$; below use $x=\sqrt{2a{h}_0}X$.

Choose the positive $x$ direction to be the direction of the cylinder surface velocity $U=a\Omega$ in the gap, and write $\mu =\rho \nu$ for the dynamic viscosity. The lubrication velocity profile and volume flux per unit axial length areSince $Q$ is constant and the pressure returns to the same ambient value as $x\to \pm \infty$, ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$. ThereforeUsing $h=h_0(1+X^2)$ and the supplied integrals givesin the direction of the cylinder surface motion.

The shear stress exerted by the fluid on the cylinder is opposite toEliminating $p_x$ and inserting $Q=2U{h}_0/3$ giveswhere the sign is relative to $U$. Sincethe resisting torque per unit axial length is

Finally, the ratio of streamwise fluid inertia $U^2/\sqrt{a{h}_0}$ to transverse viscous acceleration $\nu U/h_0^2$ is of orderThe stated restriction makes this lubrication-scale Reynolds number small, justifying neglect of inertia.

The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

The shallow-water approximation requires a small depth-to-horizontal-scale ratio, slow streamwise variation of the channel and interface, small vertical acceleration, and rapid enough cross-sectional mixing for $u$ and $\varphi$ to be representative averages. Near the channel apex, or during the initial collapse at a sharp front, these conditions need not hold. The high Reynolds number permits neglect of leading viscous resistance; weak stirring mixes the particles without appreciable ambient fluid entrainment.

Put $\lambda =\beta /H$. At elevation $z$, the channel width is $2\lambda xz$, so the occupied area and interface width areVolume conservation and suspended-particle conservation giveThe excess hydrostatic pressure is ${\rho }_{0}d\varphi (h-z)$. Its streamwise gradient averaged over the triangular section gives acceleration $-d\varphi h_x-(dh/3){\varphi }_x$: the weighted mean of $h-z$ is $h/3$. Thus the triangular-channel shallow water equations areThe quasilinear coefficient matrix in $(h,u,\varphi )$ isIts three distinct real eigenvalues, for $h,d\varphi >0$, establish strict hyperbolicity:On the material characteristic curve, $d\varphi /dt=-2W_s\varphi /h$. On each of the other two characteristic curves, multiplying by the left eigenvector $(\pm 2c/h,1,\pm dh/(3c))$ gives the compatibility ordinary differential equationThe derivatives are along the indicated characteristic, with $g^{\prime }=d\varphi$. At vanishing concentration, the gravity-wave speeds coalesce and strict hyperbolicity is lost.

The sink speed scales as $U\sim A/R^2$, so the Reynolds number is $Re\sim UR/\nu =A/(\nu R)$. The approximation applies in the high-Reynolds-number region $R\ll A/\nu$. Radial inertia scales as $UdU/dR\sim A^2/R^5$, so the radial pressure gradient has the same $R^{-5}$ dependence.