Lubrication theory

Lubrication theory describes viscous flow through a gap whose thickness is much smaller than its streamwise length; pressure is nearly uniform across the gap and controls a locally parabolic velocity profile.

In a dimensionless thin bearing with lower-wall speed one, stationary upper wall and gap $h(x)$, lubrication theory gives $u=1-Y/h+(p^{\prime }/2)Y(Y-h)$ and constant volume flux $q=h/2-h^3{p}^{\prime }/12$. Equal endpoint pressures, chosen as zero at $x=0,1$, imply $(h^3{p}^{\prime }-6h{)}^{\prime }=0$.

For a converging exponential gap $h=e^{-\alpha x}$, $\alpha >0$, integration gives $p^{\prime }=6e^{3\alpha x}(e^{-\alpha x}-M)$ with $M=3(e^{2\alpha }-1)/(2(e^{3\alpha }-1))$. This constant is the positive weighted average $\int h{h}^{-3}dx/\int h^{-3}dx$, so $e^{-\alpha }<M<1$. It follows that the pressure first rises, reaches a unique maximum at the displayed $x_m$, and falls back to zero. The interior pressure is positive and supports a load. Expansion gives $x_m=1/2+5\alpha /24+O({\alpha }^3)$ as $\alpha \to 0$.

For liquid between spatially uniform parallel plates with gap $h(t)$ and circular material rim $R(t)$, lubrication theory gives ${\overline{u}}_r=-h^2{p}_r/(12\mu )$ and $\dot{h}+r^{-1}(rh{\overline{u}}_r{)}_r=0$. Bounded pressure at zero excludes the logarithmic solution, and $p(R)=0$ gives the displayed pressure. Then ${\overline{u}}_r=-r\dot{h}/(2h)$ and rim advection implies $(R^{2}h{)}^{\prime }=0$, expressing volume conservation. Closing plates have positive gauge pressure; opening requires suction, whose validity can be limited by cavitation or air invasion. Negative gauge pressure is not in itself a contradiction.

Take the downward coordinate along a vertical wall that moves upward with speed $U$. For a stress-free thin Newtonian film with negligible inertia and surface-pressure gradients, the lubrication approximation gives $\mu u_{zz}=-\rho g$, $u(0)=-U$, and $u_z(h)=0$. Thus $u=-U+(\rho g/\mu )(hz-z^2/2)$ and the volume flux per unit width is $q=-Uh+\rho g{h}^3/(3\mu )$. Integrating incompressibility across the material film gives the displayed conservation law. Gravity drives fluid downward, while the moving wall entrains it upward; the local characteristic speed is $q^{\prime }(h)=-U+\rho g{h}^2/\mu$.

In the leading steady coating model, $Q=Uh-g{h}^3\cos \theta /(3\nu )$ and $H=Uh/Q$ obey $(H-1)/H^3=\alpha \cos \theta$, where $\alpha =g{Q}^2/(3\nu U^3)$. The left side reaches its positive maximum $4/27$ at $H=3/2$. This gives the largest globally continuous finite wall-entrained branch, with critical thickness $3Q/(2U)$. It is an algebraic transport-capacity result; full temporal stability needs the corresponding time-dependent free-surface model.

For gap $h\ll \ell$ and speed $U$, shear stress scales as $\mu U/h$ and viscoelastic normal stresses as ${\psi }_i{U}^2/h^2$. Their force contributions scale respectively as $\mu U/h^2$ and ${\psi }_i{U}^2/(h^2\ell )$. Thus the displayed criterion permits neglecting the normal-stress corrections in the leading lubrication approximation. It is weaker than requiring the normal stresses themselves to be smaller than the shear stress. Rapid time dependence can impose an additional material-time criterion.

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.

A thin axisymmetric coating of physical thickness $d=ϵah$ on a rigid cylinder of radius $a$ has leading curvature $a^{-1}-d/a^2-d_{\bar{z}\bar{z}}$. With constant surface tension $\gamma$, negligible Newtonian gravity, no-slip boundary condition at the wall and negligible exterior stress, lubrication theory gives $q=-d^3{p}_{\bar{z}}/(3\mu )$. Set $z=\bar{z}/a$ and $t=\gamma {ϵ}^3\bar{t}/(3\mu a)$ to obtain the displayed thin-film equation. A perturbation of a uniform film $h_0$ proportional to $e^{st+ikz}$ hasThus the fastest Rayleigh–Plateau instability has $|k|=1/\sqrt{2}$, or physical wavelength $2\pi \sqrt{2}a$.

Expand the area of an axisymmetric cylindrical free surface to second order in film thickness and eliminate the first-order area term using conservation of the exact liquid volume. Up to positive factors and constants this gives the displayed surface energy. Its variation is $\delta \mathcal{E}/\delta h=-h-h_{zz}$. The thin-film equation is the conserved gradient flow $h_t={\partial }_z[h^3{\partial }_z(\delta \mathcal{E}/\delta h)]$. For periodic boundaries, or boundaries with vanishing energy flux,The decrease in capillary surface energy supplies viscous dissipation. On an infinite coating with unequal far films, use a local energy balance including boundary fluxes rather than an unsubtracted infinite energy integral.

A large collar translating slowly at speed $c$ over a film $h_{+}$ and leaving a thinner film $h_{-}$ must grow: its leading excess volume increases at rate $c(h_{+}-h_{-})$. An exact travelling wave connecting unequal stationary films is impossible, since integration of the thin-film equation would give a common constant $q-ch=-c{h}_{+}=-c{h}_{-}$. In a slowly evolving collar core the leading pressure gradient vanishes. Therefore $h^{\prime }+h^{\prime \prime \prime }=0$, and a centred, reflection-symmetric collar has $h=A(1+\cos x)+B$ on $-\pi <x<\pi$. Its leading volume is $2\pi A$, so for constant small $B$, $2\pi \dot{A}\simeq c(h_{+}-h_{-})$. The speed is supplied by collar-to-film capillary matching.

Near a collar edge adjoining a film of thickness $h_f$, local mass conservation gives $h^3{h}^{\prime \prime \prime }=c(h-h_f)$. The scaling $h=h_{f}H$, $x-x_f=h_{f}X/c^{1/3}$ yields the displayed equation. Its linearization at $H=1$ has exponents $1$ and $-1/2\pm i\sqrt{3}/2$. The positive growing rear branch approaching one as $X\to -\infty$ has one amplitude, removable by translation; the front's decaying oscillatory branches have two amplitudes, leaving one parameter after translation. This dimension count describes local branches; global positivity and selection require matching. If their core asymptotes, after removing linear terms by translation, are $H_{-}\sim a_{-}{X}^2+b_{-}$ and $H_{+}\sim a_{+}(\lambda )X^2+b_{+}(\lambda )$, matching to the collar gives $A/2=a_{\pm }{c}^{2/3}/h_{\pm }$ and $B=b_{\pm }{h}_{\pm }$. Consequently $a_{+}{b}_{+}=a_{-}{b}_{-}$; a selected root with $\alpha =a_{+}({\lambda }_0)>a_{-}$ yields $c=(A{h}_{+}/(2\alpha ){)}^{3/2}$, $B=h_{+}{b}_{+}({\lambda }_0)$ and $h_{-}/h_{+}=a_{-}/\alpha$. Local linearization alone does not establish uniqueness of this global matching root.

For a nonrotating cylinder of radius $a$ translating parallel to a stationary wall at minimum gap $b\ll a$, the leading gap is $h=b+x^2/(2a)$. In the cylinder frame, the constant flux is $-2Vb/3$ and the pressure gradient is $2\mu V(4b-3h)/h^3$. The pressure is $2\mu Vx/h^2$. Integrating the wall shear $4\mu V(1/h-b/h^2)$ gives the displayed leading force per unit axial length. Full wetting and negligible inertia are assumed; cavitation would change the pressure boundary conditions.

On one side of two touching cylinders of radius $a$, the leading gap is $d(x)=x^2/a$. A completely wetting semicircular meniscus at $x=w\ll a$ has radius $w^2/(2a)$ and suction $p=-2\gamma a/w^2$. The area is $\mathcal{A}={\int }_0^{w}d(x)dx=w^3/(3a)$ to leading order. Integrating the local slit Poiseuille flow gives $Q=-w^7{p}_z/(84\mu a^3)=-\gamma w^4{w}_z/(21\mu a^2)$. Thus $(w^3{)}_t=\gamma (w^4{w}_z{)}_z/(7\mu a)$.

For vertical touching cylinders in a bath, the hydrostatic suction at height $z$ is $\rho gz$. Equating it to the meniscus suction $2\gamma a/w^2$ gives the displayed large-height width. Ideal perfect wetting permits an arbitrarily high thin cusp tail, unlike a uniform finite-gap capillary. The prediction assumes the continuum model remains valid as the gap narrows.

For fixed volume $V$ and symmetric spreading, conservation of mass and the cusp flow between touching cylinders equation give width proportional to $t^{-1/8}$ and support length proportional to $t^{3/8}$. The source-type similarity solution has the displayed profile and $z_N=(8Va/\pi {)}^{1/4}(8\gamma t/(21\mu a){)}^{3/8}$. It is a formal outer lubrication solution; its microscopic nose requires a local description when the axial slope becomes large.

For a periodic thin mucus gap $h(x)$ translating as a backward wave relative to a force-free foot, put $I_j={\int }_0^L{h}^{-j}dx$. In the wave frame, the plane and foot have horizontal velocities $V-U$ and $V$. A Couette-Poiseuille flow with periodic pressure has flux $q=(V-U/2)I_2/I_3$. Zero net plane force gives the displayed crawling speed. The leading viscous dissipation per wavelength is $\Phi =2\mu UV{I}_1$.

For $h=h_0[1-A\sin (2\pi x/L)]$, the crawling speed is $U/V=3A^2/(1+2A^2)$. At prescribed $U$, eliminate $V$ to obtain $\Phi =2\mu L{U}^2(1+2A^2)/(3h_0{A}^2\sqrt{1-A^2})$. The derivative in $A^2$ vanishes only at $A^2=1/2$, and the dissipation diverges at both endpoints. The displayed optimum still dissipates $8\sqrt{2}/3$ times the simple Couette flow dissipation of an externally driven flat foot moving at speed $U$ across a uniform film of thickness $h_0$.

A rotating roller with top speed $U$ and gap $d(x)=b+x^2/(2a)$ drives a shear-free lubrication approximation flow. No slip below and zero shear above give $u=U+p_x(y^2-d^2)/(2\mu )$ and $Q=Ud-d^3{p}_x/(3\mu )$. The pressure recovery condition in lubrication flow then gives $Q=U\int d^{-2}dx/\int d^{-3}dx=4Ub/3$. With $s=x/\sqrt{2ab}$, the excess pressure is $-\mu U\sqrt{2a}{b}^{-3/2}s/(1+s^2{)}^2$. Gravity supports it through the small surface displacement $h=(p-p_0)/(\rho g)$; the flat leading geometry is consistent when $\mu |U|\sqrt{a}/(\rho g{b}^{5/2})\ll 1$.

Lubrication theory predicts large viscous resistance when fluid must be squeezed through a narrow gap. For normal approach of smooth curved surfaces with curvature scale $a$ and minimum separation $h\ll a$, the resisting force scales as $\mu a^{2}U/h$. Under a bounded normal force this gives $|\dot{h}|=O(h)$, so ideal smooth surfaces do not make contact in finite time. Tangential and rolling motions have different resistance laws; the normal squeeze law should not be applied to them.

Rotation about the offset axis has azimuthal Couette flow and no pressure variation at leading lubrication order. Integrating the tangential traction gives torque $-{\mathcal{R}}_{\Omega }\Omega$ on the inner sphere. The concentric limit is $8\pi \mu a^4/(3\Delta )$ and the contact limit is finite, $4\pi \mu a^4/\Delta$, because the wall velocity vanishes at the closest pole.

For an inner sphere of radius $a$, outer radius $a+\Delta$, and signed offset $\alpha \Delta$ along a fixed axis, the leading lubrication pressure generated by offset speed $V=\Delta \dot{\alpha }$ is $p=3\mu a^{2}V/[\alpha {\Delta }^3(1-\alpha \cos \theta {)}^2]+\text{constant}$. Its force on the inner sphere is $-\mathcal{R}V$ along that axis. Remove the spatially constant divergent term before taking $\alpha \to 0$; the resistance then tends to $8\pi \mu a^4/{\Delta }^3$. As $\alpha \to 1$, it is $6\pi \mu a^4/[{\Delta }^3(1-\alpha )]$ to leading order.

For smooth no-slip spheres of radius scale $a$ with minimum gap $\delta \ll a$, the lateral gap scale is $(a\delta {)}^{1/2}$. Squeeze-film lubrication theory gives pressure $p\sim \mu a(-\dot{\delta })/{\delta }^2$ and the displayed normal resistance. Bounded forcing therefore gives $-\dot{\delta }=O(\delta )$, so the gap cannot vanish in finite time. The conclusion uses smooth surfaces, no slip and continuum creeping flow; it is not a contact model for rough particles.

For a Newtonian layer with density $\rho$, dynamic viscosity $\mu$, thickness $h$ and a plate moving at $U$, a stress-free roof and no-slip base give $q=Uh-\rho g{h}^3{h}_x/(3\mu )$. Zero net mountain-material volume flux produces $h^3=9\mu Ux/(\rho g)$ when the contact height vanishes. Very near its idealized contact line the slope violates the lubrication approximation.

A lower-viscosity basal sediment layer changes the plate coupling in viscous mountain spreading over an underthrust plate. With mountain thickness $H$ and sediment depth $d$, equal densities give one hydrostatic pressure gradient $G=\rho g{\partial }_x(H+d)$. Fluxes are $q_s=Ud-G(d^{2}H/2+d^3/3)/{\mu }_s$ and $q_m=UH-G(d{H}^2+d^{2}H/2)/{\mu }_s-G{H}^3/(3\mu )$. Separate conservation of mass laws determine each layer. Steady zero mountain flux and sediment influx $U{d}_{\infty }$ give contact height $3d_{\infty }$ and surface slope $2{\mu }_{s}U/(9\rho g{d}_{\infty }^2)$.

In front of the contact line in the sediment-lubricated viscous mountain model, $h=d$ and $d_x=3{\mu }_{s}U(d-d_{\infty })/(\rho g{d}^3)$. For $\delta =d/d_{\infty }$ and $\ell =\rho g{d}_{\infty }^3/(3{\mu }_{s}U)$, integration gives $x/\ell =P(\delta )-P(3)$, with $P(\delta )={\delta }^3/3+{\delta }^2/2+\delta +\ln (\delta -1)$. The branch $1<\delta \le 3$ is unique and tends exponentially to the uniform far-field depth.

A blade moving over a thin gravity-driven viscous film produces a moving-frame flux $\mathbf{q}=-h{\mathbf{e}}_x-h^3\nabla h/3$. Scaling film thickness by $h_0$ and horizontal distance by $\rho g{h}_0^3/(\mu U)$ yields this dimensionless volume-conservation equation. Fluid can escape beneath the blade or around its ends.

A long blade with negligible under-blade leakage diverts incoming fluid laterally. A deep-pile profile $h^3=9[x_N(y)-x]$ and integrated flux conservation give $d(\int h^{4}dx)/dy=-12y$, hence $x_N=[126({\alpha }^2-y^2){]}^{3/7}/9$. This leading approximation excludes the nose and blade-tip transition regions.

For a rounded blade with local parabolic lubrication gap $d=ϵ+x^2/(2\delta )$, $ϵ\ll \delta$, the Couette-Poiseuille flow in a thin gap yields a floor-driven flux $2ϵ/3$ and pressure-driven leakage proportional to ${ϵ}^{5/2}\Delta p/\sqrt{\delta }$. Positive $J$ is from the upstream pile towards the rear.

The steady upstream film of an infinite blade has $h_x=-3(h-1)/h^3$. Thus $X(h)=h^3/9+h^2/6+h/3+\log (h-1)/3$. The head is fixed by under-blade leakage balancing the incoming film. The deep pile has a cubic profile, while the actual upstream transition has an exponential tail towards $h=1$.

A journal bearing supports a rotating axle inside a surrounding cylindrical sleeve with a thin lubricating film. An eccentric gap generates lubrication pressure as rotation transports fluid into and out of narrower regions. A full-film journal bearing model imposes periodic pressure around the entire circumference; a model with cavitation needs additional free-boundary conditions. The two models can predict different loads and dynamics.

A full-film journal bearing model assumes that liquid fills the entire annular gap and that pressure is single-valued around a complete circumference. No cavitation boundary is imposed. The full-film eccentric journal-bearing rotation and full-film journal-bearing whirl formulas use this assumption; their predicted motion depends on keeping it.

A freely translating inner axle with prescribed positive angular velocity $\Omega$ and downward load $F>0$ has full-film journal bearing force balance equationswhere $R_r,R_t$ are the squeeze resistance of an eccentric journal bearing. Rotation supplies tangential fluid force $R_t\Omega r/2$. The equilibrium lies horizontally to the right and satisfies $F=R_t\Omega r/2$. Linearization there has imaginary eigenvalues: the prescribed rotating axle continuously supplies energy, so viscous dissipation does not imply settling. For an initially concentric axle, putting $e=r/\Delta$ and $K=6\pi \mu \Omega a^3/{\Delta }^2$ gives the closed orbitThis follows by eliminating time and integrating $x_e+3ex/[2(1-e^2)]=\Omega {\Delta }^2{R}_{0}e/[2F(1-e^2{)}^{3/2}]$. It applies to the leading inertia-free full-film model; adding cavitation, constraints or finite-length leakage changes the prediction.

Let the inner-centre displacement from a fixed outer centre have magnitude $r=\Delta e$, with $0\le e<1$. The leading full-film journal bearing resistance to translation in the displacement direction and perpendicular to it isFor $h=\Delta (1-e\cos \theta )$, the Reynolds lubrication equation gives radial squeezing flux $a{v}_r\sin \theta$ and tangential squeezing flux $a{v}_t[-\cos \theta +3e/(2+e^2)]$. The additive constant enforces periodic pressure. Integrating the pressure traction gives these coefficients; at $e=0$ they coincide, as required by rotational symmetry.

For inner radius $a$, mean gap $\Delta \ll a$, fixed eccentricity parameter $|\alpha |<1$, gap $h=\Delta (1+\alpha \sin \theta )$, and inner angular velocity $\Omega$, the constant circumferential flux isIt follows by integrating the lubrication theory relation $p_s=6\mu \Omega a/h^2-12\mu q/h^3$ over a full period. The leading holding force is horizontal:Integration by parts computes this force without a closed-form pressure. The fluid couples about the separate cylinder axes obey $C_i+C_o=\alpha \Delta F_x$; moments about a common axis balance exactly at this order.

For a thin viscous gravity current with hydrostatic pressure $p=\rho g(h-z)$, a no-slip boundary condition at its base and a stress-free boundary condition at its top, horizontal momentum balance gives $u=(g/\nu )h_x(z^2/2-hz)$. Integrating across the depth gives the volume flux $q=-(g/3\nu )h^3{h}_x$. Horizontal spreading follows $h_t+q_x=0$ if the substrate is impermeable.

With $K=\rho g/(3\mu )$ and formal edge conditions $h(\pm L,t)=0$, the lubrication theory equation is $h_t=K(h^3{h}_x{)}_x$. Its positive separated long-time profile hasWriting $C=B(4/5,1/2)$ in terms of the beta function, the centre height is $h_0^3=10\mu L^2/(\rho g{C}^{2}t)$. The edge region of a draining viscous film determines how the singular outer profile matches the flow over the physical edge.

The outer similarity solution approaches a plate edge as $h/h_0\sim [(4C/5)(1-x/L){]}^{1/4}$, where $C=B(4/5,1/2)$. Its slope becomes order one atThis region lies outside lubrication theory because its vertical and horizontal scales are comparable. The outer zero-thickness boundary condition is consistent at leading order since the matching height is $O(\ell )\ll h_0$, while the actual edge flow requires an inner solution. A finite outward flux survives even though the outer thickness vanishes, because the slope diverges.

Lubrication pressure is the pressure variation required to drive viscous flow through a thin gap. Its characteristic scale is $P\sim \mu UL/H^2$ for velocity scale $U$, gap thickness $H$, and streamwise length $L$.

Hele–Shaw flow is slow flow between closely spaced parallel plates. Averaging across a gap of thickness $h$ gives the Darcy-type lawrelative to stationary plates.

For viscous fluid inside a circular fluid free boundary between parallel plates of separation $h$, the depth-averaged Hele-Shaw flow satisfies $\bar{u}=-K\nabla p$, $K=h^2/(12\mu )$. With volume injection rate $Q$, ${\bar{u}}_r=Q/(2\pi hr)$ and $R^2=R_0^2+Qt/(\pi h)$. Neglecting surface tension and exterior viscosity, pressure is constant at the interface. Perturb its radius by $a_n\cos n\theta$ and solve the interior Laplace equation with a regular pressure term proportional to $r^n\cos n\theta$. The linearized pressure boundary condition gives $\delta p(R)=a_{n}Q/(2\pi hKR)$, hence velocity perturbation $-n{a}_{n}Q/(2\pi h{R}^2)$. The displaced base velocity adds another $-a_{n}Q/(2\pi h{R}^2)$, proving the displayed growth law. Injection stabilizes the circular blob; suction destabilizes it. An exact radial solution can therefore exist while giving a fragile experimental prediction. These signs concern viscous fluid inside the blob, rather than the opposite geometry of a low-viscosity injected bubble.

A long-wave approximation assumes that a layer thickness $H$ is much smaller than its streamwise length $L$. The small aspect ratio $H/L$ makes transverse gradients dominate longitudinal gradients and is the geometric basis of lubrication theory.

For velocity scale $U$, gap thickness $H$, and streamwise length $L\gg H$, lubrication balance gives $P\sim \mu UL/H^2$ and $\tau \sim \mu U/H$. Hence

For gap scale $H$, length $L$, lower-wall tangential speed $U$, upper-wall normal speed $V$, and induced horizontal scale $W=|U|+VL/H$, the lubrication limit requiresThe last condition includes both $\rho |U|H^2/(\mu L)\ll 1$ and $\rho VH/\mu \ll 1$.

Uniform normal injection of speed $V$ through the lower plate of a stationary axisymmetric gap of thickness $h$ and radius $R$ producesWith ambient pressure at the rim, the lubrication pressure is $p-p_0=3\mu V(R^2-r^2)/h^3$, and its integral supports the load $3\pi \mu V{R}^4/(2h^3)$.

For a film of thickness $h(x,t)$ flowing down a plane inclined by $\alpha$, lubrication theory gives the fluxwhen surface tension is neglected. Mass conservation gives the thin-film equation $h_t+q_x=0$.

For a film of thickness $H$, axial length $L\gg H$, and velocity $U$, the ratio of convective inertia to transverse viscous stress in lubrication theory is $U{H}^2/(\nu L)$. On a small incline of slope $\theta$, lubrication gravity-current flux gives $U\sim g\theta H^2/\nu$ when $H\ll \theta L$, and $U\sim g{H}^3/(\nu L)$ when $H\gg \theta L$. The respective criteria are $g\theta H^4/({\nu }^{2}L)\ll 1$ and $g{H}^5/({\nu }^2{L}^2)\ll 1$. Unsteady inertia also requires $H^2/(\nu T)\ll 1$ if the evolution timescale $T$ is independent of $L/U$.

For $h_t={\partial }_x(-A{h}^3+B{h}^3{h}_x)$, $A,B>0$, a front $h=F(x-ct)$ joining $h_0$ behind it to zero at its tip has $c=A{h}_0^2$. Integrating once gives $F^{\prime }=(A/B)(1-h_0^2/F^2)$ andNear the tip, $F\sim (-3c\xi /B{)}^{1/3}$. Its unbounded slope signals the failure of the small-slope lubrication approximation sufficiently close to the contact line.

For a gravity-current nose of height $h_N$ on a small incline of slope $\theta$, the travelling front of a gravity-driven thin film satisfies $h_y=\theta (1-h_N^2/h^2)$. Close to its dry tip, $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. The slope becomes order one at $h\sim \sqrt{\theta }{h}_N$, giving an $O(\sqrt{\theta }{h}_N)$ region where lubrication theory fails. The main nose length is $h_N/\theta$. Comparing capillary pressure gradients $\gamma h_N/(h_N/\theta {)}^3$ with gravitational forcing $\rho g\theta$ gives the crossover $h_N\sim \theta \sqrt{\gamma /(\rho g)}$. This concerns the main nose length; capillarity may already matter in a much smaller tip region.

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.

Fora current of conserved area $V$ has the first-kind similarity solutionterminated by a shock atIts front height is $h_N(t)=(V/(2Ct){)}^{1/3}$.

The scaling $x\sim C{h}^{2}t$ and $V\sim hx$ gives $V\sim C{h}^{3}t$. Maintaining an asymptotically constant front height therefore requires constant volume flux of order $C{h}^3$.

Near the closest approach of a circle of radius $a$ to a wall, the gap is $h(x)=h_0+x^2/(2a)+O(x^4/a^3)$ and has streamwise scale $\sqrt{a{h}_0}$.

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.