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.

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.

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.

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 $h_t+q_x=0$.

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 sphere almost filling a tube has local parabolic lubrication gap $h\simeq a[ϵ+(z/a{)}^2]/2$. The resistance integrals are $I_1=2\pi {ϵ}^{-1/2}$, $I_2=(2\pi /a){ϵ}^{-3/2}$ and $I_3=(3\pi /a^2){ϵ}^{-5/2}$, obtained with $z=a\sqrt{ϵ}\xi$ and $\int (1+{\xi }^2{)}^{-n}d\xi$. Moving-boundary lubrication flux weights shear-driven and pressure-driven transport by different powers of the gap, explaining the three distinct divergence rates.

Take a fluid control volume spanning a nearly occluding sphere. The pressure difference contributes $\pi a^2\Delta P$ upwards, while the outer-wall shear in the sphere frame contributes $2\pi a\mu \int u_y(h)dz$. Since sphere-frame flux in a tube gives $u_y(h)=4U/h-6q/h^2$, the drag on the sphere is $F=\pi a^2\Delta P+2\pi a\mu (4U{I}_1-6q{I}_2)$. Using the wall shear avoids integrating traction over the strongly curved sphere directly. The remaining tube pressure drop is fixed by Hagen-Poiseuille flow.

Use $F^{\ast }=F/(6\pi \mu aU)$ and $L^{\ast }=4L/(3a)$. The control-volume drag formula for a sphere in a tube gives the displayed uniform leading drag. If $L^{\ast }\ll {ϵ}^{-1/2}$, $F^{\ast }\sim (4\pi /3){ϵ}^{-1/2}$ and local annular shear dominates. If ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$, $F^{\ast }\sim L^{\ast }$ and the Hagen-Poiseuille flow resistance of the long tube dominates. If $L^{\ast }\gg {ϵ}^{-5/2}$, $F^{\ast }\sim 3\pi {ϵ}^{-5/2}$ and pressure-driven bypass through the narrow gap dominates. The laboratory flow is piston-like in the first two limits and negligible in the third.

Two identical well-separated spheres falling at the same speed share one laboratory volume flux, rather than producing twice the single-sphere piston flux. In the middle of the three drag regimes for a nearly occluding sphere, the same overall Hagen-Poiseuille flow pressure drop is divided equally between them, halving each sphere's drag. In the local-shear and local-gap-pressure regimes each sphere retains its own leading resistance, so the drag per sphere is unchanged. This conclusion assumes nonoverlapping lubrication regions and sufficient separation from the tube ends.

Between surfaces of velocities $U_0,U_1$ separated by $h$, the local lubrication profile is

For $U_0=0$, $U_1=U$, and constant $G=p_x$, zero shear at $y=0$ requiresThe velocity becomes $u=U{y}^2/h^2$, the flux per unit width is $Uh/3$, and the fluid exerts shear $-2\mu U/h$ on the moving upper wall.

The flux of the local Couette-Poiseuille profile is

For a sphere falling with speed $U$ through a tube of radius $a$, let $Q$ be the downward laboratory volume flux. In the sphere frame the wall moves upwards and the narrow annulus carries upward flux per unit circumference $q=(\pi a^{2}U-Q)/(2\pi a)$. Couette-Poiseuille flow in a thin gap gives $q=Uh/2-h^3{p}_z/(12\mu )$. Thus the pressure below minus above the sphere is $\Delta P=6\mu (2q{I}_3-U{I}_2)$, using lubrication resistance integrals for a nearly occluding sphere. A nearly piston-like laboratory flux can coexist with a nonzero relative gap flux.

For a full Newtonian liquid film in the lubrication approximation, a cylinder translating parallel to a stationary plane with minimum gap $h_0\ll a$ exerts tangential force $2\pi \mu U\sqrt{2a/h_0}$ per unit axial length on the plane. The leading normal force vanishes by the odd symmetry of excess pressure.

Incompressible mass conservation in a moving gap givesCombined with the local Couette-Poiseuille flux, this is the Reynolds lubrication equation.

For a thin gap $h$ with tangential wall velocities ${\mathbf{v}}_0,{\mathbf{v}}_h$, lubrication theory and the no-slip boundary condition give the integrated volume fluxConservation of mass gives $h_t+{\nabla }_{\parallel }\cdot \mathbf{q}=0$. This is the Reynolds equation; it determines the lubrication pressure from wall motion and gap geometry.

A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

For a fixed outer radius $b\ll a$ and $h=aϵ+r^2/(2a)$,Indeed, for $a\sqrt{ϵ}\ll r\ll b$ the integrand is $(2a{)}^n/r$; integrating from a scale proportional to $a\sqrt{ϵ}$ to $b$ produces half of $\log (1/ϵ)$. The inner region contributes a bounded term after rescaling $r=a\sqrt{ϵ}s$. In particular, $\int h^{-1}dxdy$ and $\int x^2/(a{h}^2)dxdy$ have the same leading coefficient $2\pi a$.

For lower speed $U$, stationary upper tangential speed, upper normal speed $-V$, and equal endpoint pressures, define $⟨f⟩=L^{-1}{\int }_0^{L}f(x)dx$. Thenwhere

The horizontal force on the lower wall vanishes when

For a cylinder of radius $a$ in a coaxial container with gap $h\ll a$, global displacement forces reverse flow through the gap and createsThe pressure drag exceeds side shear by order $a/h$. Balancing it with excess weight gives $U\simeq \Delta \rho ,g{h}^3/(6\mu a)$.

If pressure approaches the same ambient value at both ends of a lubrication region, then ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$, which determines the conserved flux.

For a sphere of radius $(1-ϵ)a$ displaced by $ϵ\lambda a$ from the axis of a tube of radius $a$, the narrow gap isIts thickness is $O(ϵa)$ and its axial length is $O(a\sqrt{ϵ})$, so lubrication pressure dominates shear stress by $O({ϵ}^{-1/2})$.

An axial force balance on fluid between two remote cross-sections of a cylindrical tube giveswhen every immersed body between the sections is force-free.

A thin-film equation combines a lubrication flux with mass conservation to evolve a free-surface thickness. Gravity, capillary pressure, wall motion, and intermolecular forces may contribute to the flux.

For a liquid film heated below and cooled above, the quasistatic temperature of a conductively cooled film makes a thicker region colder at the surface. If surface tension decreases with temperature, Marangoni stress draws fluid towards thicker regions, competing with smoothing by capillary pressure. Writing ${\gamma }_s={\gamma }_0+bh$ gives $a_c={\gamma }_0/(3\mu )$, $a_m=b/(2\mu )$. A balanced long-wave scaling gives $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The dispersion relation $s=k^2-k^4$ has unstable band $0<|k|<1$ and fastest normal mode at $|k|=1/\sqrt{2}$. This thermal mechanism does not require surfactant.

Positive steady solutions of the balanced thermocapillary thin-film equation with zero volume flux satisfy $H_{XX}+\ln H=0$ after choosing the thickness scale. Their first integral has potential $V(H)=H(\ln H-1)$, with minimum $V(1)=-1$ and limiting value $V(0^{+})=0$. At $E=-1$ the film is uniform; for $-1<E<0$ it is positive and periodic. At $E=0$ a limiting drop has peak $e$, half-width $\sqrt{\pi e}$, zero limiting contact angle and divergent curvature at zero thickness. Small periodic profiles have period $2\pi$ and approach the neutral, rather than fastest-growing, normal mode of the thermocapillary film instability.

For a film with fixed lower temperature $T_0$ and upper condition $-\kappa T_z=\alpha (T_s-T_0+\Delta T)$, negligible horizontal thermal diffusion and thermal advection reduce the heat equation to $T_{zz}=0$. Thus $T=T_0-\alpha \Delta Tz/(\kappa +\alpha h)$. The local approximation requires $(h/L{)}^2\ll 1$ and $U{h}^2/(\kappa L)\ll 1$, together with slow time variation relative to $h^2/\kappa$. For $\alpha h/\kappa \ll 1$, $T_s\simeq T_0-\Delta T\alpha h/\kappa$. The coefficient $\kappa$ is thermal diffusivity, and the boundary coefficient $\alpha$ is normalized accordingly.

For an exterior thin film, circumferential and axial curvature together give $h_t+(h^3[h_{zzz}+h_z+G]{)}_z/3=0$ in appropriate dimensionless variables. The $h_z$ term destabilizes long waves through the Rayleigh–Plateau instability, whereas $h_{zzz}$ damps short waves. The linear growth rate is $(k^2-k^4)/3-iGk$, with fastest growth at $|k|=1/\sqrt{2}$. The gravity parameter is $G=\rho g{a}^2/(\gamma ϵ)$ when film thickness is scaled as $ϵah$ and axial distance by cylinder radius $a$.

For a travelling wave of large positive speed in the cylindrical film equation, the leading core is proportional to $1-\cos x$ over an axial interval of length $2\pi$. Edge coordinates $X=(3c{)}^{1/3}(x-x_e)$ reduce the transition equation to $H^3{H}_{XXX}=H-1$. Its real positive exponent gives a monotone trailing approach and its complex stable exponents give damped capillary waves ahead, with physical wavelength $4\pi a/(3^{5/6}{c}^{1/3})$. Matching the two quadratic edge expansions selects the uniform film thickness; this is a matched asymptotic expansion, valid while the pulse itself remains thin relative to the cylinder.

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.

In the same gravity–Marangoni reduction, zero liquid flux gives $\Gamma =1+(H_0^2-H^2)/3$ and $6JX=(1+H_0^2/3)(H^3-H_0^3)/3-(H^5-H_0^5)/15$. The lower part of the liquid flows backwards while the surface and upper third flow forwards. For zero initial height and concentration drop $\delta$, $J={\delta }^{3/2}(1-3\delta /5)/(2\sqrt{3})$ is largest formally at complete depletion. Since surface velocity is $J/\Gamma$, that endpoint is singular and requires physical regularization by diffusion, capillarity or an edge model. The algebraic maximum is not by itself a regular zero-concentration outflow.

In the steady nondiffusive gravity–Marangoni model with capillarity neglected, zero surfactant flux and positive concentration make the surface velocity zero. With $Q=-H^2{\Gamma }_X/2-H^3{H}_X/3$ and $J=-H\Gamma {\Gamma }_X-H^2\Gamma H_X/2$, integration gives $H^4=H_0^4-48QX$ and $\Gamma =1+(H_0^2-H^2)/4$. A concentration drop $\delta$ over unit length therefore selects $Q=-\delta (H_0^2+2\delta )/6$. The liquid flows opposite to the imposed Marangoni traction, with velocity proportional to $H_{X}Y(Y-H)/2$ and an immobilized surface. This reduction is singular at a dry endpoint; its small-slope assumption does not resolve that edge.

A localized amount $M=\int Cdx$ of insoluble surfactant spreads over a planar film of depth scale $h_0$. Neglect gravity, capillarity and diffusion. Then $C\sim M/\ell$ and Marangoni stress drives $u_s\sim AM{h}_0/(\mu {\ell }^2)$, giving $\ell \sim (AM{h}_{0}t/\mu {)}^{1/3}$. The thin-film equations with insoluble surfactant admit a similarity solution in which the film height and concentration depend on $x/\ell$, conserving total surfactant and fluid volume relative to the undisturbed layer.

When the smoothing width exceeds the capillary length, gravity dominates capillarity. Balancing hydrostatic flux $\rho g{h}_0^4/(\mu \Delta )$ against the Marangoni stress flux gives $\Delta \sim \rho g{h}_0^2{x}_N^2/(AM)\propto t^{2/3}$. The condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$ in physical units. The smoothing region catches up with the spreading pool at $x_N\sim AM/(\rho g{h}_0^2)$ and $t\sim \mu (AM{)}^2/[(\rho g{)}^3{h}_0^7]$, after which the sharp-front outer similarity solution ceases to apply.

After gravity smoothing of a Marangoni front reaches the whole spreading pool, the film approaches uniform depth while surface transport continues. A small height gradient creates a hydrostatic pressure return flow that almost cancels the net Marangoni stress flux. Setting $q\simeq 0$ in the thin-film equations with insoluble surfactant gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$ and fractional depth variation $O(AM/(\rho g{h}_0^2{x}_N))\ll 1$. The depressed region under the surfactant therefore becomes progressively shallower.

The height jump in linear similarity profiles for surfactant spreading is smoothed by surface tension. If $C_x\sim M/x_N^2$ through a region of width $\Delta$, the Marangoni stress flux is $AM{h}_0^2/(\mu x_N^2)$ and the capillary flux is ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$. Their balance gives $\Delta \sim ({\gamma }_0{h}_0^2{x}_N^2/(AM){)}^{1/3}\propto t^{2/9}$. This estimate assumes gravity is negligible and the smoothing region remains narrow compared with the pool.

For symmetric finite-mass Marangoni spreading on a liquid film, the leading outer profiles are $h=2h_0|x|/x_N$ and $C=(M/x_N)(1-|x|/x_N)$ inside $|x|<x_N$, where $x_N=(6AM{h}_{0}t/\mu {)}^{1/3}$. Outside, $h=h_0$ and $C=0$. The similarity solution follows from $H{\Gamma }^{\prime }=-\eta /3$, $\eta H^{\prime }=H$, $2\int \Gamma d\eta =1$ and $\int (H-1)d\eta =0$. The film reaches height $2h_0$ immediately behind the front, leaving an outer height jump that gravity or surface tension must smooth. These are ideal leading outer profiles, not a description of the microscopic central dry region or the smoothed edge.

Disjoining pressure is the thickness-dependent normal stress caused by intermolecular interactions between nearby interfaces. An attractive law $\Pi (h)=V/h^3$ grows as a film thins and can drive rupture.

For a viscous sheet of half-thickness $h_0$, attractive disjoining pressure $V/h^3$, and surface tension $\gamma$, a symmetric Fourier mode of dimensionless wavenumber $K=k{h}_0$ has a destabilizing molecular contribution and a stabilizing capillary contribution. The ratiosets the cutoff $K={\Gamma }^{-1/2}$.

The thin-film mass flux is the layer thickness times its depth-averaged velocity. In a quasi-steady transition region, mass conservation makes this flux approximately constant along the layer.

For a long axisymmetric bubble with slowly varying radius $a(z,t)$, the leading normal-stress balance and internal lubrication flux couple capillary pressure to conservation of the cross-sectional area.

Near a finite-time viscous-capillary pinch-off, balancing time dependence, viscous normal stress, capillary pressure, and axial lubrication flow can select self-similar radius and axial scales.

A slender viscous thread has a cross-sectional scale much smaller than its axial variation scale. For an axisymmetric Newtonian thread of area $A$ and nearly uniform axial velocity $w$, mass conservation and the Trouton ratio givewhen inertia and surface tension are negligible and $z$ points downward.

Uniform axial strain rate $E$ in an annular Newtonian fluid gives radial velocity $u_r=-Er/2+C/r$. For radii $R_i,R_e$, area parameter $A=R_e^2-R_i^2$, and constant gas pressures,The opposite signs of inner and outer curvature in the Young–Laplace equation give these expressions. The kinematic boundary condition implies $\dot{A}=-EA$ and