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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 $\mu \Omega a/\sqrt{ϵ}$.

For a nonrotating force-free sphere moving with speed $U$ in a stationary tube, use its moving frame: the tube wall has velocity $-U$. In the parabolic lubrication gap $h=h_0+x^2/(2a)$, $h_0=ϵa(1+\lambda \cos \theta )$, moving-boundary lubrication flux gives $q=-h^3{p}_x/(12\mu )-Uh/2$. The wall-shear and pressure-drop balance in a tube rules out an $O({ϵ}^{-3/2})$ net pressure jump. Applying this common pressure recovery to each angular strip gives $q=-2U{h}_0/3$. The next balance yields the displayed positive localized pressure drop. Meanwhile the integrated leading shear stress on the sphere vanishes in each strip; this is consistent with negligible leading torque, and does not mean the local shear or pressure force vanishes.

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 $p$ as excess pressure relative to the fluid far from the gap, and let ${\dot{h}}_0<0$ for approach. The lower surface of the sphere is $h_0+a-\sqrt{a^2-r^2}$ above the plane. In the dominant region $r=O(\sqrt{a{h}_0})\ll a$, this is the parabolic lubrication gapIts slope is $O(\sqrt{h_0/a})\ll 1$, justifying lubrication theory. Both rigid surfaces have zero radial velocity, so the no-slip boundary condition and radial Stokes equation give the radial volume flux per unit circumferenceAxisymmetric conservation of mass gives $h_t+r^{-1}(rq{)}_r=0$. Symmetry at $r=0$ therefore yields $q=-r{\dot{h}}_0/2$, hence $p_r=6\mu {\dot{h}}_{0}r/h^3$. Matching to $p\to 0$ outside the pressure-bearing region givesExtending the parabolic profile to infinity in this integral is a leading-order matching device; the main pressure and force come from $r=O(\sqrt{a{h}_0})$.

The radial scale $\sqrt{a\hat{h}}$ balances the two contributions to the gap. The time scale $\hat{h}/\hat{v}$ is the time for a typical gap change, and the lubrication pressure scale $\mu a\hat{v}/{\hat{h}}^2$ follows either from the formula above or from $p\sim \mu \hat{v}{\ell }^2/{\hat{h}}^3$ with ${\ell }^2=a\hat{h}$. The elastic surface's upward displacement is $-\eta p$, so the actual gap is increased by $+\eta p$. Thus the dimensionless gap in weakly compliant sphere-plane squeeze flow isA dot now means differentiation with respect to $T$. The dimensionless Reynolds lubrication equation isEven without fluid inertia, the elastic surface has a pressure-dependent normal velocity, so ${\mathcal{H}}_T$ includes $\epsilon P_T$; omitting it would wrongly discard the acceleration term below.

Write $d=H_0+R^2/2$ and abbreviate $H=H_0$, $v={\dot{H}}_0$, $b={\ddot{H}}_0$ during the calculation. The rigid term isAt first order the Reynolds equation givesBecause $RdR=dd$, one integration givesSet $\xi =R^2/(2H)$, so $d=H(1+\xi )$. Dividing and changing variables yieldsThe first three terms are derivatives of $I_3,I_4,I_5$, respectively. Integrating from infinity to enforce $P_1\to 0$ givesThe rational expansion of $I_n$ follows by differentiating $\log [\xi /(1+\xi )]+{\sum }_{j=1}^{n-1}[j(1+\xi {)}^j{]}^{-1}$, obtaining $1/[\xi (1+\xi {)}^n]$, and fixing its value to zero at infinity.

Near $\xi =0$ every $I_n$ has the same singular term $\log \xi$. Its coefficient must vanish to keep the pressure regular, soEquivalently, the integrated radial flux above must vanish at $R=0$; otherwise $P_{1R}$ would have a $1/R$ singularity. Using $I_4=I_3+[3(1+\xi {)}^3{]}^{-1}$ and $I_5=I_4+[4(1+\xi {)}^4{]}^{-1}$ cancels all logarithms and gives the simplified regular expression

The upward hydrodynamic force has scale $\mu a^2\hat{v}/\hat{h}$. Its leading contribution is pressure integrated over horizontal projected area; vertical shear forces are smaller in the lubrication limit. Thus $F=2\pi {\int }_0^{\infty }(P_0+\epsilon P_1+\cdots )RdR$. Use $RdR=H_{0}d\xi$ and ${\int }_0^{\infty }(1+\xi {)}^{-n}d\xi =1/(n-1)$. The two contributions areTherefore

For weight-controlled sedimentation, let the fixed effective load be $F=6\pi A$, $A>0$, and neglect particle inertia as well as fluid inertia. At leading order ${\dot{H}}_0=-A{H}_0$ and ${\ddot{H}}_0=A^2{H}_0$. The first-order bracket evaluated on this trajectory is $-(7/5)A^2/H_0^2$. Solving the force balance perturbatively therefore givesHence weak compliance initially increases the sedimentation speed in the regime where the perturbation is valid. The squeeze pressure indents the coating away from the sphere, widens the flow passage and lowers resistance. This is a load-controlled conclusion on the slow sedimenting branch, not a claim about an arbitrary externally imposed acceleration at startup.

The load-controlled breakdown of weak squeeze-flow compliance occurs when the relative correction $O(\epsilon A/H_0^2)$ becomes order one:The central rigid pressure has size $P_0(0)\sim A/H_0$, so the relative elastic gap change is also $\epsilon P_0(0)/H_0\sim \epsilon A/H_0^2$. Thus breakdown means that elastic deflection is comparable with the nominal undeformed gap. The regular expansion about a rigid plane then ceases to be uniform and the coupled elastic-lubrication problem must be retained. It does not by itself imply contact, fluid inertia or failure of the thin-gap geometry.

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

The flat plate moves tangentially and the curved plate is stationary. The no-slip boundary condition and no-penetration boundary condition giveAt the curved plate, impermeability can also be written $w=u{h}_x$; together with zero tangential velocity, this gives both components zero. The native PDF writes $v^{\prime }=0$ in the scaled wall conditions although the normal component is defined as $w^{\prime }$. We use $w^{\prime }$ consistently.

The parabolic lubrication gap changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is thereforeThe proposed normal velocity scale $h_{0}U/L$ then makes the two terms of the continuity equation comparable. Balancing the horizontal pressure gradient against the transverse viscous term in the Navier-Stokes equation requires $P/L=\mu U/h_0^2$, soHere pressure is measured from a common reference value, which is chosen at the outer fluid in part (b).

For clarity, define the dimensionless material derivativeSubstitution of all the scalings gives the exact dimensionless equationsFor example, the factor in the last left side is obtained by dividing the vertical acceleration scale $\rho h_0{U}^2/L^2$ by the vertical pressure gradient scale $P/h_0$.

Taking $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$ leaves the lubrication approximationThe scaled gap and boundary conditions areThe Reynolds number itself need not be small: it is ${\delta }^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.

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.

A sphere approaching a locally compliant elastic coating has a parabolic lubrication gap increased by pressure-induced downward surface displacement. With a local compliance $\eta$, the dimensionless compliance is $\epsilon =\eta \mu a\hat{v}/{\hat{h}}^3$. The gap is $H=H_0+R^2/2+\epsilon P$ and its Reynolds lubrication equation is $12H_T=R^{-1}(R{H}^3{P}_R{)}_R$. The rigid term is $P_0=-3{\dot{H}}_0/(H_0+R^2/2{)}^2$. The first elastic correction contains both ${\dot{H}}_0^2$ and ${\ddot{H}}_0$: elastic gap evolution retains the time derivative of pressure even when fluid inertia is negligible.