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The Benney equation is a long-wave approximation for a gravity-driven thin film that expresses its volume flux per unit width locally in terms of height and its derivatives. For surface-velocity scaling on a plane inclined by $\alpha$, one normalization isHere $\epsilon$ is the film aspect ratio, $\mathrm{Re}$ the thickness-based Reynolds number, and $S$ the capillary parameter for a falling film. The leading velocity is parabolic. Its first inertial correction uses the leading mass conservation equation to eliminate the time derivative of the height. This local closure requires the correction to remain small; it does not describe arbitrary large-amplitude short-wave dynamics.

The Dupuit approximation treats flow in a shallow unconfined aquifer as predominantly horizontal, with hydrostatic pressure $p=\rho g(h-z)$ and horizontal pressure gradient independent of depth. Integrating Darcy's law over the saturated depth then gives a volume flux per unit width depending only on the groundwater height and its slope.

Consider a nonlinear diffusion equation with uniform supply, $\varphi h_t=D_m(h^m{h}_x{)}_x+R$, on $x>0$, with an absorbing boundary $h(0,t)=0$ and initially $h=0$. Far from the boundary, $h=Rt/\varphi$. Balancing the time derivative, supply, and diffusion givesThe similarity solution satisfiesThe outward boundary volume flux per unit width is $Q=R\ell c_m$, where $c_m={\lim }_{\eta \downarrow 0}{F}^m{F}^{\prime }$. Integrating the equation givesThus the growing region of depleted storage fixes the discharge prefactor, and $Q\propto t^{(m+1)/2}$.

For a thin Newtonian fluid film on a conical slope at angle $\alpha$, neglecting the thickness-gradient contribution to the driving pressure gives volume flux per unit width $q=D{h}^3$, where $D=g\sin \alpha /(3\nu )$. Radial mass conservation isThe distance $x$ is measured along the slope; the horizontal radius is $x\cos \alpha$, so total volume is $2\pi \cos \alpha \int xhdx$.

For a thin buoyant layer beneath an inclined impermeable roof, the hydrostatic approximation and Darcy's law give $u=kg\Delta \rho \sin \theta /\mu$ and the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. The first term transports fluid upslope; the second spreads the layer through its hydrostatic pressure gradient. The porosity multiplies the stored fluid volume, so the advection speed is $u/\varphi$, not the Darcy velocity $u$.

An inverted falling film flows on the underside of an inclined solid surface. Taking the normal coordinate from the solid into the liquid reverses the normal component of gravity relative to an upper-surface film. In the Benney equation, the hydrostatic contribution to the volume flux per unit width therefore changes from $-2\epsilon h^3{h}_x\cot \alpha /3$ to its negative. Normal gravity then amplifies disturbances of long wavelengths, while surface tension resists short wavelengths.

For a steady line-source shear-driven viscous gravity current with downstream constant thickness, $h_d^2=2\mu Q_{2d}/\tau$. Zero upstream volume flux per unit width gives $h{h}_x=3\tau /(2\Delta \rho g)$, so $h^2=h_d^2+3\tau x/(\Delta \rho g)$ until it vanishes at $x=-x_N$. Unlike the point-source scaling estimate, this reduced line-source model fixes the exact coefficient $2/3$. The square-root nose has an unresolved steep edge.

Even when most of an unconfined aquifer with depth-dependent permeability is deep, its height vanishes at an absorbing outlet. Near that outlet, finite outward volume flux per unit width $Q$ gives $u_{b}h{h}_x\simeq Q$ and hence $h\simeq (2Qx/u_b{)}^{1/2}$. The high-depth outer approximation instead has $h\simeq [6Qx/(u_b\beta ){]}^{1/3}$. Their crossover lies at height of order $1/\beta$ and distance of order $u_b/({\beta }^{2}Q)$; the thin inner boundary layer transmits the same leading discharge.

Take outward radial volume flux per unit width as positive. The streamfunction in polar coordinates immediately givesFor spanwise width $W$, the volumetric flow rate is $Q=Wq$. The wedge inside radius $r$ has cross-sectional area $\alpha r^2/2$, whose rate of increase is $r^2\dot{\alpha }/2$. Opening therefore draws in exactly the volume needed to fill the expanding wedge; closing expels it. This is also a direct check of conservation of mass for the moving control volume. It follows from the wall boundary conditions and remains true even though the printed interior streamfunction is erroneous.

Integrating the radial velocity over the thickness gives the radial volume flux per unit width:The full volumetric flow rate across a cylindrical surface, positive outward, isAn annulus of radial thickness $dr$ contains volume $2\pi rhdr$ to leading order. Its rate of gain equals incoming minus outgoing volume flux, soDividing and taking $dr\to 0$ yields the axisymmetric thin-film equation:The equation assumes no evaporation or injection through the boundaries. Integrating it over the whole drop also verifies conservation of mass when the edge carries zero volume flux.

Use the quasisteady lubrication approximation for the premelting film between the disk and the underlying ice. The ice is anchored to the lower plane; new ice formed from water supplied through this upper film raises the disk. Let $U$ be its upward separation rate, $d(r)$ the premelted thickness, $\mu$ the liquid dynamic viscosity, $\rho$ a common ice/water mass density, $L$ the specific latent heat of fusion, and $T_m$ the absolute bulk melting temperature. As usual in this leading model, neglect the ice–water density difference, pressure-dependent changes in $T_m$ due to a common phase pressure, inertia and edge corrections. The imposed thermal field is taken as maintained by an external heat supply/removal system, so no additional heat-conduction bottleneck is introduced.

Write $\Delta =T_m-T_0>0$, $K_T=\rho L/T_m$ and $\Delta (r)=\Delta (1-r^2/a^2)$. The Clapeyron pressure relation for equal-density phases makes the pressure difference across the solid–liquid interfaceThis thermomolecular pressure is the repulsive stress transmitted through the film. To calculate a definite rate one also needs a film-thickness constitutive law. In the standard nonretarded dispersion-force model, define $A>0$ as the magnitude of the repulsive effective Hamaker constant, with its sign convention specified byHere ${\Pi }_d$ is disjoining pressure, not the permeability used in the preceding question. The sign of an effective Hamaker coefficient depends on the three materials; the positive parameter here denotes an interaction that stabilizes a premelted film. The problem does not specify $A$ numerically, so it must remain a defined material parameter.

Take the reservoir pressure at the rim to be $p_{\infty }$. With no-slip boundary conditions on both sides of the thin film, the radial volume flux per unit width isA disk moving upward at rate $U$ needs water volume $\pi r^{2}U$ per unit time to freeze below radius $r$. Regularity on the axis and mass conservation therefore give $2\pi r{q}_r=-\pi r^{2}U$, or $q_r=-Ur/2$. ThusDefine $B=A/(6\pi K_T)$, so $d^3=B/[\Delta (1-r^2/a^2)]$. Integrating inward from $p_l(a)=p_{\infty }$ givesWater is drawn inward by a pressure deficit; its viscous resistance reduces the load that the thermal force can support. The net normal stress on the disk is $p_l-p_{\infty }+P_T$. Let $W$ be the downward apparent load after subtracting ordinary buoyancy. For an ideal thin disk with negligible buoyant volume, $W=Mg$, where $g$ is gravitational acceleration. If a finite displaced volume is known, use its submerged weight instead; the stated mass and radius alone do not determine that volume.

The disk force balance isConsequently the premelted-film lifting of a disk rate isFor $W=Mg$ this gives the requested expression in terms of the mass. Every additional symbol in this formula has been defined above. The growing-ice lifting branch exists only when the bracket is positive.

At the threshold the required water flux and viscous pressure deficit vanish. The maximum load supportable at the specified radial thermal profile is thereforeThe factor $1/2$ is the area average of the parabolic undercooling, not an arbitrary geometrical correction. At equality the lift rate is zero; strictly positive lift requires $W<W_{\max }$. There is no radius-only, temperature-independent maximum load in this ideal model: one must specify the imposed undercooling or a physically admissible range of it.

For a fixed positive load define ${\Delta }_c=2W{T}_m/(\pi a^2\rho L)$ and $U_{\infty }=A/(6\pi \mu a^2)$. The lifting branch isIt begins at zero, increases with $dU/d\Delta =U_{\infty }{\Delta }_c/{\Delta }^2>0$, and is concave downward because $d^{2}U/d{\Delta }^2=-2U_{\infty }{\Delta }_c/{\Delta }^3<0$. Below threshold no positive steady heave is possible: the thermomolecular force is too weak to support the load. The mathematical continuation to negative $U$ is not a prediction of upward ice-driven separation there.

Increasing undercooling increases the available thermal pressure, but it also thins the premelting film. In the inverse-cube disjoining pressure model, $d^3\propto {\Delta }^{-1}$, so the hydraulic conductance decreases in exact inverse proportion to the driving scale. The remaining load correction tends to zero and the rate approaches a plateau:For positive load this is approached from below, not attained at finite undercooling. With zero load the reduced model gives the same constant rate wherever its thin-film assumptions hold.

Two limits of the calculation are worth making explicit. First, the formal $d(r)$ diverges at the warm rim, so a narrow outer connection to bulk water replaces the thin-film solution there; this leading model assumes that its contribution to the integrals is negligible. Second, very large undercooling produces a molecularly thin film and eventually invalidates continuum lubrication theory and the simple dispersion-force law. Thus the plateau is the maximum within the stated ideal constitutive model, not an unrestricted extrapolation for real water.

If a different molecular interaction is intended, the result is not universal. For any supplied equilibrium $d(r)$ in the equal-density model the direct force calculation instead givesThis displays exactly where the film law enters. The closed formula and plateau above use the conventional repulsive inverse-cube law; premelting alone, without that material law, would not determine a unique rate curve.

Choose $x$ upslope and $z$ normally downwards from the impermeable roof, so the buoyant layer occupies $0<z<h(x,t)$. Use the hydrostatic approximation and neglect the ambient return-flow pressure correction of relative order $h/H$. Pressure continuity at the interface gives the pressure inside the buoyant layer asThe along-slope Darcy velocity is thereforeDefine $u=kg\Delta \rho \sin \theta /\mu$. Integrating over depth gives the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. Local fluid-volume conservation, including porosity, is $\varphi h_t+F_x=0$. Hence the inclined porous gravity current satisfiesThe first term is upslope advection driven by the mass density difference; the second is nonlinear hydrostatic spreading. The Darcy velocity $u$ and the corresponding pore velocity $u/\varphi$ must be distinguished.

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.

At leading order in the lubrication approximation, surface arclength is horizontal distance. An insoluble surfactant is transported by the surface velocity, without bulk exchange or diffusion. Conservation of insoluble surfactant on a moving interface therefore becomesThe local hydrostatic pressure and capillary normal stress giveSolve $\mu u_{zz}=p_x$ with no-slip boundary condition at $z=0$ and Marangoni stress $\mu u_z(h)={\gamma }_x=-A{C}_x$. The velocity, surface velocity and volume flux per unit width areMass conservation, $h_t+q_x=0$, yields the thin-film equations with insoluble surfactant:and the corresponding surfactant equation isThe hydrostatic and capillary terms thus affect both the film flux and the surface transport, with different coefficients.

For finite-mass Marangoni spreading on a liquid film, first neglect both pressure-gradient terms. If the spread has size $\ell$, then $C\sim M/\ell$ and $C_x\sim M/{\ell }^2$, giving surface speed $u_s\sim AM{h}_0/(\mu {\ell }^2)$. Equating $\ell /t$ to this speed givesSet $\ell =(AM{h}_{0}t/\mu {)}^{1/3}$ exactly as a similarity scale, and writeOn the positive half of the pool, substitution produces two ordinary differential equations:Symmetry fixes the flux integration constant in the second equation to zero, giving $H{\Gamma }^{\prime }=-\eta /3$ wherever $\Gamma >0$. Substitution into the first equation gives $\eta H^{\prime }=H$. Thus the linear similarity profiles for surfactant spreading have the formThe concentration vanishes at the moving front. The two integral constraints are total surfactant mass and conservation of the fluid volume relative to the undisturbed layer:The first gives $k={\eta }_N^2/3$; the second gives $k{\eta }_N=2$. Therefore ${\eta }_N^3=6$. In physical variables, the solution isOutside the pool the leading outer solution has $h=h_0$, $C=0$. In particular, $h(x_{N-},t)=2h_0$: the subscript means the one-sided limit just behind the front. As a consistency check, the surface speed there is $x_N/(3t)={\dot{x}}_N$ and the fluid jump satisfies the moving-front mass conservation condition. This ideal outer solution has a height jump from $2h_0$ to $h_0$.

That jump cannot persist in a physical interface with nonzero gravity or surface tension. Across a smoothing region of width $\Delta$, $h$ changes by $O(h_0)$. The Marangoni flux scales as $AM{h}_0^2/(\mu x_N^2)$. At $g=0$, its balance with the capillary flux ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$ gives capillary smoothing of a Marangoni front:Here the surface tension approaches ${\gamma }_0$ at the surfactant-free edge. The assumed concentration-gradient scale is $M/x_N^2$ even inside the smoothing region.

When gravity dominates capillarity, the hydrostatic flux is $\rho g{h}_0^4/(\mu \Delta )$. The same balance gives gravity smoothing of a Marangoni front:The gravity-dominated condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$, the square of the capillary length. The printed condition omits $\rho$; the expression above restores dimensional consistency.

This width grows faster than $x_N$. It becomes comparable with the pool size whenThe time is a scaling estimate, so numerical factors cannot be fixed by the width balance. For $t\gg t^{\ast }$, the sharp-front similarity profile no longer describes the film: hydrostatic pressure levels the layer over the whole pool, and its thickness tends towards $h_0$ with an increasingly small depression beneath the surfactant. The surface can still spread by Marangoni stress while a hydrostatic-pressure-driven return flow nearly cancels the net film flux. In this gravity-levelled surfactant film, $q\simeq 0$ gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$, so the fractional height variation is of order $AM/(\rho g{h}_0^2{x}_N)\ll 1$.

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.

Let $z$ increase upwards, so the bed is at $z=-b(x)$, the surface is at $z=h(x,t)$, and $H=h+b$. In the shallow-shelf approximation the incompressible flow has nearly depth-uniform horizontal speed $u$ and vertical strain $w_z=-u_x$. The Newtonian fluid stress tensor, together with the hydrostatic approximation, then givesThe factor four includes both horizontal extension and the pressure correction required by vertical compression. Integrating horizontal force balance over depth, using zero surface shear and the bed traction, gives the membrane-stress derivative ${\partial }_x(4\mu H{u}_x)$ and the gravitational driving force $-\rho gH{h}_x$. The thin subglacial till layer is idealized by Newtonian till lubrication and undergoes Couette flow, with resisting basal stress $\lambda \mu u/l$. Hence force balance and mass conservation yieldSince $b$ is fixed, $H_t=h_t$. No accumulation or ablation is included.

At the grounding line, the first boundary condition is ice-sheet flotation:For an unbuttressed ice shelf, the second is the extensional stress required to balance the difference between the integrated ice and seawater hydrostatic pressures. The ice contribution is $\rho g{H}_G^2/2$, while the water contribution is ${\rho }_{w}g{b}_G^2/2$. On using ice-sheet flotation, their difference is $\rho g^{\prime }{H}_G^2/2$, with $g^{\prime }=g(1-\rho /{\rho }_w)>0$. Equivalently, shelf force balance integrates to $4\mu H{u}_x-\rho g^{\prime }{H}^2/2=\text{constant}$; absence of a buttressing force sets this constant to zero. Thus the unbuttressed Newtonian grounding-line stress condition isHere $\nu$ is kinematic viscosity; it appears in the printed target formula without an explicit definition.

The ratio of longitudinal stress divergence to till drag is of order $4lH/(\lambda L^2)$. The given small-parameter limit therefore yields the friction-dominated bulk relationThe coefficient depends on the basal lubrication, and $h_x<0$ gives seaward motion. This is a bulk reduction: longitudinal stress has been neglected in the differential equation but its boundary traction remains specified. A complete uniformly valid approximation close to the grounding line could require a membrane-stress boundary layer. In the remainder, use the pointwise friction closure and retain the shelf stress condition as the boundary data, as requested in the paper.

Differentiate the ice-sheet flotation condition along the moving grounding line:Now $H_x=h_x+\alpha$. Substitution of the friction-dominated speed and the shelf stress boundary condition gives the friction-dominated grounding-line evolution law:The coefficient of ${\dot{x}}_G$ is the spatial derivative of the flotation deficit, up to sign. If it vanishes, the implicit equation remains the correct compatibility condition but division to determine a finite speed is invalid; tangential contact needs separate analysis.

In a steady state, volume flux per unit width is constant: $Hu=q_0$. The friction relation givesDifferentiating $Hu=q_0$ and using $u_x=g^{\prime }H/(8\nu )$ at the grounding line givesPut $\beta =l/\lambda$, $\mathcal{H}=H/\beta$ and $\mathcal{Q}=\nu q_0/(g{\beta }^3)$. Substitution and cancellation of $g{\beta }^5/\nu$ yield the steady friction-dominated grounding-line thickness relation:For $q_0>0$ and a bed deepening seaward ($\alpha \ge 0$), the polynomial $(g^{\prime }/8g){\mathcal{H}}^5+\alpha \mathcal{Q}{\mathcal{H}}^2-{\mathcal{Q}}^2$ is strictly increasing for $\mathcal{H}>0$, starts negative and tends to infinity, so it has exactly one positive root. At that root $\mathcal{Q}>\alpha {\mathcal{H}}^2$ and $H_x<0$, as required by steady outward extension. The relation belongs to the specified friction closure; it is not a flux law for arbitrary till rheology or arbitrary membrane-stress matching.