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Use the Dupuit approximation: the saturated region is shallow enough that hydrostatic pressure is $p=\rho g(h-z)$ and flow is predominantly horizontal. Darcy's law then gives the horizontal Darcy velocity $u_x=-u_b(1+\beta z)h_x$, where $u_b=k_0\rho g/\mu$. Integrating through the saturated depth gives the volume flux per unit widthThe stored water volume per unit area is $\varphi h$, so mass conservation yields the unconfined aquifer with depth-dependent permeability equationDefine the positive river discharge by $Q(t)=-q(0,t)$. A useful check on all subsequent results is the integrated mass conservation law

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.

The hydrostatic pressure gives $p_x=\rho g{h}_x$, independent of depth. Thus Darcy's law gives $u=-K{h}_x$ and the horizontal volume flux per unit width isThe coefficient $K$ is the hydraulic conductivity, with units of velocity. For a column of unit horizontal area, the stored water volume is $\varphi h$, and rainfall supplies volume flux $R$. The local mass conservation equation is $\varphi h_t+q_x=R$. Therefore the Boussinesq equation for an unconfined aquifer isThis assumes rainfall is entirely converted into recharge and neglects unsaturated storage and losses before the water reaches the water table.

Write $H(x^{\prime })=h^{\prime }(x^{\prime })$ and $p_x^{\prime }=d{p}^{\prime }/d{x}^{\prime }$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x^{\prime }$. The vertical pressure gradient is zero, and integrating $u_{z^{\prime }{z}^{\prime }}^{\prime }=p_x^{\prime }$ twice with the wall boundary conditions gives the local Couette-Poiseuille flowThe first two terms are Couette flow due to the moving plate; the last is plane Poiseuille flow due to the local pressure gradient. Its dimensionless volume flux per unit width isUse the Leibniz integral rule and incompressibility to differentiate this integral:Thus the flux is independent of position, andThe normal velocity can then be recovered from incompressibility:It is zero on the lower wall by construction and zero on the upper wall by the constant-volume flux relation. This supplies the remaining component of the leading velocity field.

The endpoint pressure conditions are matching conditions, not an assertion that the lubrication approximation remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous pressure variation is of size $\mu U/R$, whose ratio to the inner scale isEven an outer inertial pressure scale $\rho U^2$ has ratioConsequently the gap pressure is much larger than either outer variation. Matching to fluid with a common ambient pressure at both ends, and taking that value as zero gauge pressure, givesThere is an overlap region $1\ll |x^{\prime }|\ll {\delta }^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed pressure difference comparable with $P$ would change these conditions and the resulting volume flux; equal outer pressure is the physical assumption being used here.

Let the constant in the integrated Reynolds equation be $C$:Both endpoint pressures are zero, so the integral of $d{p}^{\prime }/d{x}^{\prime }$ on the real line vanishes. DefineThe substitution $x^{\prime }=\sqrt{2}\tan \theta$ givesThe second value follows by integrating the supplied identity for ${\cos }^4\theta$; the nonconstant cosine terms have zero integral over this interval. Hence $12C{I}_3-6I_2=0$, givingThe pressure gradient is nowDirect differentiation showsThe candidate tends to zero at both ends, so the additive constant is zero andTogether with the local Couette-Poiseuille flow profile, this solves the sliding-plate flow through a parabolic gap.

To restore dimensions, integrate $u=U{u}^{\prime }$ across $dz=h_{0}d{z}^{\prime }$. The signed dimensional volume flux per unit width, positive in the $+x$ direction, isThus the magnitude drawn in the direction of the moving plate is $2U{h}_0/3$. For a plate width $B$ large enough to neglect side-wall effects, the total volumetric flow rate is $-2BU{h}_0/3$; without a specified width the answer is per unit width.

The Couette flow and plane Poiseuille flow contributions to the dimensional volume flux per unit width are respectivelyThey point in the same direction precisely when $Q_P<0$, givingAt the endpoints the plane Poiseuille flow contribution is zero. Outside that interval it opposes the Couette flow contribution, while their sum remains the constant negative volume flux per unit width $-2U{h}_0/3$. Thus pressure assists the moving plate near the narrowest gap and drives a compensating return contribution farther away.

For $\varphi h_t=D_m(h^m{h}_x{)}_x$ on $0<x<L$, with $h(0,t)=0$ and zero right-hand volume flux, a separation of variables giveswhere the positive profile satisfiesIts boundary volume flux per unit width isThese profiles are exact separated solutions and describe the leading long-time discharge for a broad class of positive initial data. The virtual origin $t_0$ depends on the initial profile; it does not make the separated solution an exact representation of every initial condition.

A stationary upper wall $z=h_0+x^2/(2R)$ and a flat lower wall moving at velocity $-U$ generate a Couette-Poiseuille flow in the lubrication approximation. Write $L=\sqrt{R{h}_0}$, $X=x/L$, $H=1+X^2/2$ and $p=(\mu UL/h_0^2)\Pi$. Equal endpoint gauge pressures givewhere $Q$ is the signed volume flux per unit width. To determine its constant value, the Reynolds equation gives $H^3{\Pi }_X/12+H/2=C$. Integrating ${\Pi }_X$ over the real line gives $C=I_2/(2I_3)=2/3$, where $I_m={\int }_{-\infty }^{\infty }{H}^{-m}dX$ and the substitution $X=\sqrt{2}\tan \theta$ evaluates the ratio. The plane Poiseuille flow and Couette flow contributions both point towards negative $x$ exactly for $|x|<\sqrt{2R{h}_0/3}$. This is a full-film model with matched equal outer pressures.

For a fixed wall at $\theta =0$ and a wall rotating at $\dot{\alpha }$ at $\theta =\alpha$, the quasi-steady Stokes approximation with the no-slip boundary condition givesIndeed the biharmonic equation for $r^{2}G$ is $G^{(4)}+4G^{\prime \prime }=0$; the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$. The four wall conditions are $G(0)=G^{\prime }(0)=G^{\prime }(\alpha )=0$, $G(\alpha )=1$. The outward radial volume flux per unit width is $-r^2\dot{\alpha }/2$, precisely the negative rate of increase of wedge area. The associated local pressure is $p=\mu \dot{\alpha }[2\cos \alpha /(\sin \alpha -\alpha \cos \alpha )]\log (r/r_{\ast })+C$. Its logarithmic singularity is integrable, although the ideal hinge requires a microscopic regularization.

A zero-height outlet can carry finite nonzero volume flux per unit width in a nonlinear Darcy flow model with $q=-Kh{h}_x$. If $h{h}_x\to J>0$ as $x\downarrow 0$, then $(h^2{)}^{\prime }\to 2J$, and integration gives $h\sim \sqrt{2Jx}$ and $h_x\sim \sqrt{J/(2x)}$. Thus the height vanishes while its slope diverges; the flux must be evaluated as a limit of the product. The divergent slope also identifies a region where the outer Dupuit approximation eventually needs a local correction.

With constant volume flux per unit width $Hu=q_0>0$, define $\beta =l/\lambda$, $\mathcal{H}=H/\beta$, and $\mathcal{Q}=\nu q_0/(g{\beta }^3)$. The friction closure gives $H_x=\alpha -q_0\nu /(\beta g{H}^2)$. Differentiating constant flux and applying the unbuttressed Newtonian grounding-line stress condition gives $q_0{H}_x/H+g^{\prime }{H}^2/(8\nu )=0$. Substitution proves the displayed relation. For $\alpha \ge 0$, $g^{\prime }>0$ and $\mathcal{Q}>0$, the polynomial $(g^{\prime }/8g){\mathcal{H}}^5+\alpha \mathcal{Q}{\mathcal{H}}^2-{\mathcal{Q}}^2$ increases strictly on the positive half-line, begins negative and tends to infinity; hence it has exactly one positive root. Its root satisfies $\mathcal{Q}>\alpha {\mathcal{H}}^2$, so the thickness decreases seaward at that steady grounding line.