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In this continuation $\lambda$ measures eccentricity, not the viscosity ratio used in Question 2. Take $ϵ\to 0$ with $0\le \lambda <1$ fixed. For the nearly occluding sphere in a cylindrical tube, the narrow gap has thickness $H\sim ϵa$ and axial length $\ell \sim a\sqrt{ϵ}$, obtained by balancing $h_0$ and $x^2/(2a)$. In the sphere frame the tangential velocity scale is $U$. The lubrication approximation then gives the gap scalesThe large local pressure excursion is distinct from the eventual net pressure jump. Away from the sphere the velocity varies across a radius $a$, so the far Hagen-Poiseuille flow hasThese scales also show why the gap pressure can be much larger than its shear stress while the leading net jump can be smaller than that pressure.

Use $y=a-r$, so $y=0$ is the moving tube wall and $y=h$ is the locally stationary sphere. The no-slip boundary conditions are $u(0)=-U$, $u(h)=0$. Integrating $\mu u_{yy}=p_x$, with pressure independent of $y$ to leading order, gives the Couette-Poiseuille flow in a thin gapDifferentiate at both surfaces and eliminate $p_x$ using this moving-boundary lubrication flux:The coordinate reversal means ${\sigma }_{rx}=-{\sigma }_{xy}$ at the tube wall.

At leading order $q$ is independent of $x$ in each angular strip, though it may depend on $\theta$. Both boundaries are stationary as shapes in the sphere frame and impermeable, so the integrated continuity equation is $q_x+a^{-1}(q_{\theta }^{\mathrm{circ}}{)}_{\theta }=0$. The pressure changes along $x$ over $a\sqrt{ϵ}$ but along the circumference over $a$. Thus circumferential velocity is only $O(U\sqrt{ϵ})$, and its flux is $O(Ua{ϵ}^{3/2})$. Circumferential flux divergence is smaller than the possible leading axial divergence by a factor $ϵ$. Consequently $q_x=0$ to the order needed. This is not an assumption of equal flux at different angular positions.

For fixed $\theta$ define the lubrication resistance integrals for a nearly occluding sphereIn particular, $J_1=\pi \sqrt{2a/h_0}$, $J_2=J_1/(2h_0)$ and $J_3=3J_1/(8h_0^2)$. The local pressure relation givesA putative generic value $q=O(U{h}_0)$ makes this jump $O(\mu U/(a{ϵ}^{3/2}))$. However, the integrated wall shear in part (ii) is only $O(\mu Ua/\sqrt{ϵ})$, so the force-free global balance permits a common net pressure jump only of order $\mu U/(a\sqrt{ϵ})$. The outer flow has circumferentially uniform pressures on both sides. Therefore the $O({ϵ}^{-3/2})$ pressure jump must cancel in every angular strip, not merely after angular averaging. The pressure recovery condition in lubrication flow determinesThis explains how the leading-order part of the global force balance selects the local flux.

The total sphere-frame volume flux is nowTransforming back to the laboratory adds $\pi a^{2}U$, so $Q_{\mathrm{lab}}=\pi a^{2}U(1-4ϵ/3)$ to this order. For physical pressure, Hagen-Poiseuille flow gives $Q_{\mathrm{lab}}=-\pi a^4{p}_x/(8\mu )$. ThusThe printed positive expression is the pressure-drop gradient in the direction of the laboratory flow; the Poiseuille hint omits the minus sign when $p_x$ is read as the derivative of physical pressure. With the specified wall velocity $-U$, the laboratory flow is positive and its physical pressure gradient is negative.

Next insert the leading flux into the wall shear stress:Let $\Delta p>0$ be the localized extra pressure drop, after subtracting the regular far Poiseuille gradient. Contributions outside the neck are only $O(\mu U/a)$ and are smaller than the $O({ϵ}^{-1/2})$ jump. Since ${\sigma }_{rx}=-{\sigma }_{xy}$, part (ii) gives $\pi a^2\Delta p=a\int \int {\sigma }_{xy}(0)dxd\theta$. Hence the leading pressure drop isFor example, the concentric limit is $\Delta p\sim 4\pi \mu U\sqrt{2/ϵ}/a$. The integral grows as the minimum gap shrinks; the fixed-$\lambda <1$ qualification prevents treating a further near-contact limit as uniformly covered.

On the sphere the leading shear stress is ${\sigma }_{xy}(h)=-2\mu U/h+4\mu U{h}_0/h^2$, soPositive and negative tangential stresses cancel within each angular strip. Thus the leading tangential resultant, and its leading torque contribution, vanish: negligible rotation is consistent with the torque-free condition. The shear is not zero pointwise, and the pressure forces are not zero. Pressure on a sphere cannot produce a torque about its centre because its normal is radial; its force contributions and the matched far pressure jump remain essential to the force-free transport of a nearly occluding sphere.

The nearly occluding sphere in a cylindrical tube has gap thickness $H=O(ϵa)$ and axial length $L=O(\sqrt{aH})=O(a\sqrt{ϵ})$. In the gap, lubrication theory therefore givesFar ahead of and behind the sphere, the length and width scales are both $a$, so

In the sphere frame the tube wall at $y=0$ moves with velocity $-U$ and the sphere surface at $y=h$ is stationary. The local Couette-Poiseuille flow in a thin gap isIts axial volume flux per unit circumferential width isDifferentiating the profile and eliminating $p_x$ gives the two wall stressesThe continuity equation integrated across the gap says that changes of $q$ along $x$ are balanced by circumferential flux divergence. Circumferential variations occur on scale $a$, much longer than the axial scale $a\sqrt{ϵ}$, so $q=q(\theta )$ is independent of $x$ at leading order.

Put $h=h_0(1+{\xi }^2)$ with $x=\sqrt{2a{h}_0}\xi$. The leading $O({ϵ}^{-3/2})$ pressure jump must vanish in the global balance from part ii. Equivalently, the pressure recovery condition in lubrication flow givesUsing $I_2=\pi /2$ and $I_3=3\pi /8$,

The total flux through the gap is consequentlyFar from the sphere, translation of the tube contributes $-\pi a^{2}U$, while Hagen-Poiseuille flow contributes $\pi a^4{p}_x/(8\mu )$. Equating these fluxes gives

At the next order, substitute $q=-2U{h}_0/3$ into the tube-wall shear and integrate through the gap:The balance in part ii then gives the leading pressure drop

Finally, the shear on the sphere integrates tobecause $I_1=\pi$ and $I_2=\pi /2$. Thus the narrow gap exerts no net leading axial shear force on the sphere even though its local shear is nonzero; the leading hydrodynamic force transfer to the sphere is through the large lubrication pressure acting on its gently sloping surface.

Integrating the incompressible continuity equation $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the Reynolds lubrication equationHencefor a constant $C$, and the flux formula from part (b) givesDefine the spatial meanBecause both ends meet fluid at pressure $p_0$, the pressure recovery condition in lubrication flow isIt determinesTherefore the requested pressure gradient in a translating and squeezing finite gap is

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

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 $\mu |U|\sqrt{a}/(\rho g{b}^{5/2})\ll 1$.