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For a cylinder falling parallel to a wall through a minimum gap $ϵa/2$, the parabolic lubrication gap has length scale $a\sqrt{ϵ}$. The leading tangential drag scales as $\mu U/\sqrt{ϵ}$, while the leading rotational resistance scales as $\mu \Omega a/\sqrt{ϵ}$.

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.

For an infinite blade, the steady moving-frame flux is constant and its upstream value is $-1$. The infinite-blade gravity pile therefore satisfies $-h-h^3{h}_x/3=-1$, or $h_x=-3(h-1)/h^3$. Integrating givesAn additive constant in $X$ simply shifts $x_0$. The head at the blade, and hence $x_0$, must supply the pressure difference that makes the flux beneath the blade equal to the incoming flux. For $x_0\gg 1$ the deep part has $h\simeq [9(x_0-x){]}^{1/3}$, followed by a smooth transition to $h=1$ with $h-1$ exponentially small, proportional to $e^{-3x}$. Thus $x_0$ is an approximate nose coordinate, not a sharp front of the exact solution.

Near the rounded edge the parabolic lubrication gap is $d(x)=ϵ+\delta -\sqrt{{\delta }^2-x^2}\simeq ϵ+x^2/(2\delta )$. In the blade frame the floor moves at $-1$ and the blade is stationary. The Couette-Poiseuille flow in a thin gap has flux $q_x=-d/2-d^3{p}_x/12$. Let $J=-q_x$ be positive leakage towards the rear, and let $\Delta p=p_{\mathrm{ahead}}-p_{\mathrm{rear}}>0$. The narrow-gap integrals areIntegrating the pressure gradient gives $\Delta p=12J{I}_3-6I_2$, so the squeegee gap leakage flux isThe first term is the floor-driven contribution. For the steady infinite blade, $J=1$, so $\Delta p\simeq 9\pi \sqrt{2\delta }/(2{ϵ}^{5/2})$. The hydrostatic pressure difference is the height difference at the two sides. A fully developed rear layer of thickness $h_{-}$ has moving-frame flux $-h_{-}$; hence $h_{-}=1$, andThe large head eventually forces all the arriving film beneath an infinite blade. Taking a perfectly closed gap instead would eliminate this steady state.

Matching the flat film to a cylindrical interface gives the local parabolic lubrication gapThe thickness changes by $O(h_0)$ across the transition, soIn the extensional-force equation, viscous and capillary terms have scalesTheir balance gives

The transition adjusts on time $\delta /U$, whereas the flat film drains on time $L/U$. Since $\delta \ll L$, the transition is quasi-steady. Its thin-film mass flux therefore satisfies

In the cylinder frame the wall moves at velocity $-U$. The parabolic lubrication gap and its natural stretched coordinate areThe wall values are $u(0)=-U$ and $u(h)=\Omega a$ to leading order. The Couette-Poiseuille flow in a thin gap is thereforeand its constant flux isThe pressure recovery condition in lubrication flow gives ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$. Sinceit follows that

Differentiating the velocity profile gives the two surface shear stressesUsing $\int h^{-1}dx=2\pi /\sqrt{ϵ}$ and $\int h^{-2}dx=2\pi /(a{ϵ}^{3/2})$, the shear force exerted by the fluid on the wall isand that exerted by the fluid on the cylinder isThey are not equal and opposite because pressure acting on the sloping cylinder surface also transfers tangential momentum. Indeed, integration by parts gives the cylinder's pressure forceso its total leading hydrodynamic force is $-4\pi \mu U/\sqrt{ϵ}$.

The cylinder's excess weight per unit axial length is $\pi a^2\Delta \rho g$ in the falling direction, and it has no gravitational couple about its axis. Force and couple balance therefore giveThe second result follows because the leading viscous couple is $-4\pi \mu \Omega a^2/\sqrt{ϵ}$.

For $\Omega =0$, write $h=(ϵa/2)(1+{\xi }^2)$. ThenThus $p$ is an odd pressure disturbance that vanishes at $x=0$ and at both infinities; its extrema occur at $\xi =\pm 1/\sqrt{3}$. The cylinder shear is positive near the narrowest point, negative in the outer parts of the gap, and vanishes atThe streamlines pass through the gap in the wall's direction overall. Pressure-driven backflow bends the interior streamlines and creates the two shear-reversal locations on the cylinder; the streamline sketch is symmetric under a half-turn combined with reversal of the flow direction.

For the final Couette flow, the lower and upper minimum gaps are $ϵa(1+\lambda )/2$ and $ϵa(1-\lambda )/2$. If the cylinder translates at speed $V$, the two leading lubrication drags are proportional toThe force-free condition givesFor $\lambda =0$, $V=0$ and the streamlines in the two equal gaps are mirror images with opposite directions. The $O(\mu Ua)$ subleading wall-driven couple must balance the leading rotational resistance $O(\mu \Omega a^2/\sqrt{ϵ})$, so

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.