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Take $x$ downslope, $y$ across the slope, and $z$ normal to the plane. The leading normal momentum balance givesThe tangential lubrication theory equations, with no slip at $z=0$ and zero tangential stress at $z=h$, then give the depth-integrated fluxFor $\alpha \ll 1$, defineDropping tildes, steady mass conservation $\nabla \cdot \mathbf{q}=0$ becomeswith $h\to 1$ as $x\to -\infty$.

The leading streamwise momentum equation in lubrication theory iswith $p=p(x)$. The no-slip boundary condition isTwice integrating in $y$ gives the local Couette-Poiseuille flow in a thin gapIts volume flux per unit span is

Seek a similarity solution in the dimensionally completed formwhere the fixed scales absorb $A,D$, and $Q_0$. Its inlet volume flux scales asso the prescribed source requiresThe time derivative, horizontal lubrication theory term, and pressure-dependent Darcy drainage term scale respectively asEquating the three exponents givesSolving these linear equations together with the inlet condition yields the similarity exponents of a draining gravity currentEquivalently,and, up to constant dimensional scales,

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

Let $d(z,t)$ be the local water-film thickness. To first order in the interface amplitudes,The lubrication theory flux down the vertical surface, with a no-slip boundary condition at the ice and a stress-free boundary condition at the water-air interface, isFor the unperturbed film $p_z=0$, so

The linearized Young–Laplace equation gives the capillary-pressure perturbationBecause the prescribed volume flux is uniform, its first-order perturbation must vanish. Linearization of the flux law givesThus, with $\Gamma =\gamma /(3\rho g)$,and henceThe complex amplitude ratio records the phase shift caused by surface tension in the long-wave approximation.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.

A thin dense gravity current on a rigid no-slip substrate can be carried downstream by imposed upper shear stress $\tau$. For density contrast $\Delta \rho$, lubrication theory gives $\mathbf{q}=A{h}^2{\mathbf{e}}_x-D{h}^3\nabla h$, where $A=\tau /(2\mu )$ and $D=\Delta \rho g/(3\mu )$. The continuity equation gives the displayed equation. Besides small slopes and reduced inertia, the hydrostatic reduction requires $\tau /(\Delta \rho gL)\ll 1$ to keep shear-related normal stress small. Finite zero-thickness fronts are formal outer profiles and need a separate local description when their slopes are large.

If a draining gravity current is supplied with volume flux proportional to $t^{\alpha }$, a first-kind similarity solution hasprovidedThese relations follow by balancing the time derivative, horizontal lubrication theory flux, vertical drainage, and imposed inlet flux.

For a no-slip planar liquid film with an insoluble surfactant, lubrication theory gives $q=-h^3{p}_x/(3\mu )+h^2{\gamma }_x/(2\mu )$ and surface velocity $u_s=-h^2{p}_x/(2\mu )+h{\gamma }_x/\mu$. Mass conservation and conservation of insoluble surfactant on a moving interface give $h_t+q_x=0$ and $C_t+(C{u}_s{)}_x=0$. With $\gamma ={\gamma }_0-AC$ and small slopes, $p_x=\rho g{h}_x-(\gamma h_{xx}{)}_x$. The different coefficients in $q$ and $u_s$ are essential: Marangoni stress and hydrostatic return flow transport fluid and surfactant differently.

Let $x$ be slant distance from an upward apex, and let the surface slope down at angle $\alpha$ to the horizontal. Axisymmetric lubrication theory givesFor conserved volume $V$ and $x_N^3\gg V/\sin \alpha$, the leading similarity solution has $h=\sqrt{x/(5K\sin \alpha t)}$ in $0<x<x_N$, withThe outer front is a discontinuity in this reduced conservation law; the hydrostatic-gradient term and the full local flow resolve the front and apex regions where the outer approximation fails.

A fluid advancing under an elastic plate can be resisted mainly by lubrication theory flow near its front. For front gap $h_f$ and peeling length $l_p$, the elastic pressure gradient is of order $B{h}_f/l_p^5$, while the gap volume flux is of order $B{h}_f^4/(\mu l_p^5)$. Mass balance therefore gives peeling speed $U_p\sim B{h}_f^3/(\mu l_p^5)$.