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In a closed-base cylindrical container, mass conservation makes the annular volume flux opposite to the core flux. With $h=R-a\ll a$ and annular viscosity $\lambda \mu$, Couette-Poiseuille flow in a thin gap gives $q_g=wh/2-h^3{P}_z/(12\lambda \mu )\simeq -aw/2$, where $P$ removes the annular hydrostatic pressure. Hence $P_z=6\lambda \mu aw/h^3[1+O(h/a)]$. The resulting column equation is $(a^2{w}_z{)}_z/a^2=\Delta \rho g/(3\mu )+2\lambda aw/h^3$. Curvature and the Couette flux change only the neglected relative order $h/a$.

In dimensionless Couette-Poiseuille flow in a thin gap with lower-wall velocity one and stationary upper wall, the conserved volume flux satisfies $Q=h/2-h^3{p}_x/12$. Equal pressures at the two ends imply $\int p_{x}dx=0$ and give the displayed expression. It is a weighted average of the positive gap height with weight $h^{-3}$, so $\min h\le 2Q\le \max h$. The local pressure gradient has the sign of $h-2Q$, making the flux constraint useful for locating pressure extrema without solving for the pressure explicitly.

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.

Take positive $\Omega$ counterclockwise in the $(x,y)$ plane and put $W=\Omega a$, $d=ϵa$. The bottom cylinder surface moves locally at $W$; the wall moves at $-U$. In the lubrication approximation,The leading equations are $p_y=0$, $\mu u_{yy}=p_x$ and incompressibility. No slip gives the Couette-Poiseuille flow in a thin gapThe steady gap flux is independent of $x$. Equal leading pressures at the two matched outer ends require $\int p_{x}dx=0$. Let $J_n=\int h^{-n}dx=\ell d^{-n}{I}_n$. Then $J_2/J_3=4d/3$, and $\boxed{q=\frac{2}{3}d(W-U)}$. The velocity field and flux determine the normal velocity by incompressibility and no penetration.

At the wall and cylinder, respectively,Here $\tau ={\sigma }_{xy}$; the fluid force on the cylinder has the opposite traction sign from that on the wall. Since $J_2=J_1/(2d)$,The shear and pressure force partition in cylinder lubrication explains why these two shear forces need not balance: pressure on the sloping cylinder surface supplies an equally important tangential component. Its contribution isBoundary terms vanish under outer matching. Thus the total cylinder force is $-2\mu U{J}_1$, and it balances the wall force once pressure is included. Pressure acts normally through the circular cylinder's center and gives no torque; leading shear torque is $a{F}_{\mathrm{cylinder}}^{\mathrm{shear}}=-2\mu \Omega a^2{J}_1$.

The effective weight per unit length is $+\pi a^2\Delta \rho g$ in the falling direction, with zero applied couple. Force and couple balance give $\boxed{U=\Delta \rho g{a}^2\sqrt{ϵ}/(2\sqrt{2}\mu ),\Omega =0\text{ at leading order}}$. Finite outer-region torque can generate smaller rotation; the leading singular gap resistance is what fixes this result.

For $\Omega =0$, $q=-2dU/3$ andThe pressure is odd, negative for $x<0$ and positive for $x>0$ with the wall moving toward decreasing $x$, and tends to the common reference pressure at either end. The cylinder shear changes sign at $\boxed{x=\pm \ell =\pm a\sqrt{2ϵ}}$. For $\zeta =y/h$, the velocity isFlow is entirely wall-directed near the neck, while a reversed layer next to the cylinder exists for $|\xi |>1$. These changes are the separation and reattachment points shown in the stream-function sketch.

For the final Couette flow, the printed center $(\lambda ϵa,0)$ is an offset parallel to the walls. Translational invariance removes it; the centered two-gap symmetry gives the literal answer $\boxed{V=0}$ for every $\lambda$. A dependence on $\lambda$ requires a transverse position, evidently the intended $(0,\lambda ϵa)$. With that repaired interpretation, the lower and upper gaps are $d_b=ϵa(1+\lambda )$ and $d_t=ϵa(1-\lambda )$.

In the translating cylinder frame the wall velocities are $-U-V$ and $U-V$. Each leading tangential drag is $2\mu J_1$ times its wall velocity, independent of the rotation. Hence zero net force requires $(U-V)J_t-(U+V)J_b=0$, where $J_{t,b}=\pi \sqrt{2a/d_{t,b}}$. For the intended transverse offset,This formula assumes both gaps remain lubrication gaps, with fixed $|\lambda |<1$. It gives $V=0$ at $\lambda =0$ and motion toward the faster nearby wall otherwise.

At $\lambda =0$, the lower-gap flow is the previous $-U$ flow and the upper gap its reversed mirror, with fluxes $\mp 2dU/3$. Their leading translational shear torques vanish. Any finite outer-fluid Couette torque is $O(\mu Ua)$, while the sum of the gap rotation torques is $O(\mu \Omega a^2/\sqrt{ϵ})$. Balancing them explains $\boxed{\Omega =O(\sqrt{ϵ}U/a)}$. Leading gap analysis alone does not determine its prefactor.

Use pressure with the ambient hydrostatic pressure removed, and let $\Delta P$ be the pressure below the falling sphere minus that above it. Away from the sphere the tube flow is Hagen-Poiseuille flow. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their pressure drops add:The large exterior reservoir returns this flux with negligible pressure loss compared with the long narrow tube. The sphere is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.

Now use a frame translating downwards with the sphere, and take $z$ positive upwards from its center. The sphere is stationary and the tube wall moves upwards with speed $U$. The narrow gap isIt has axial length scale $a\sqrt{ϵ}$, making lubrication theory appropriate. The sphere-frame flux in a tube is constant: the upward relative flux through the whole section is $\pi a^{2}U-Q$, so its flux per unit circumference isWith $y=0$ on the sphere and $y=h$ on the wall, Couette-Poiseuille flow in a thin gap givesSince $\Delta P=-\int p_{z}dz$, the pressure-jump relation isThe parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt{ϵ})$. The lubrication resistance integrals for a nearly occluding sphere areSetting $\xi =\tan \theta$ evaluates $J_1=\pi$, $J_2=\pi /2$, $J_3=3\pi /8$, hence

For the force, enclose the fluid between sections just above and below the sphere, including the annular layer. The net upward pressure force is $\pi a^2\Delta P$. The tube wall exerts upward shear traction $\mu u_y(h)$ on this fluid; the sphere exerts a downward force equal to the upward hydrodynamic force $F$ on itself. From the gap solution,The axial control volume force balance therefore gives the control-volume drag formula for a sphere in a tube:Using a control volume avoids having to integrate the curved-sphere normal pressure and tangential shear separately.

With $Q^{\ast }=Q/(\pi a^{2}U)$, $\Delta P^{\ast }=a\Delta P/(6\mu U)$, $q^{\ast }=2q/(aU)$, $F^{\ast }=F/(6\pi \mu aU)$, $I_n^{\ast }=a^{n-1}{I}_n$ and $L^{\ast }=4L/(3a)$, the three flux/pressure equations becomeSolving these linear relations givesThe force relation is $F^{\ast }=\Delta P^{\ast }+\frac{4}{3}{I}_1^{\ast }-q^{\ast }{I}_2^{\ast }$. Substitution first yieldsBecause $I_2^{\ast }/I_3^{\ast }=2ϵ/3$ and $I_1^{\ast }/I_3^{\ast }=2{ϵ}^2/3$, the terms $-2L^{\ast }{I}_2^{\ast }$ and $\frac{4}{3}{L}^{\ast }{I}_1^{\ast }$ are lower order than $L^{\ast }{I}_3^{\ast }$. Keeping the uniformly relevant leading terms givesThe squared $I_2^{\ast }$ is essential; replacing it by $I_2^{\ast }{I}_3^{\ast }$ would give the wrong sign and scaling. This establishes the three drag regimes for a nearly occluding sphere.

For $L^{\ast }\ll {ϵ}^{-1/2}$,The sphere nearly carries the tube's fluid down with it: the far-field flux is close to the piston displacement rate. The relative annular leakage is small in total volume, and the dominant force comes from the local annular shear term. The pressure force associated with the overall tube resistance is smaller.

For ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$,The flow remains approximately piston-like, but the dominant force is now the pressure needed to drive Hagen-Poiseuille flow through the long clear tube. Both displayed terms in $q^{\ast }$ are retained because their relative size changes inside this regime without producing an additional leading drag regime.

For $L^{\ast }\gg {ϵ}^{-5/2}$,The large tube resistance makes its net through-flow negligible. The sphere's displaced fluid instead passes upwards relative to it through the thin annular gap, with gap velocity of order $U/ϵ$. Its pressure drop and resulting pressure force dominate.

Finally consider two identical co-moving spheres, with nonoverlapping lubrication regions and the same total clear-tube length to leading order. The common through-flux is not doubled. In the middle regime it is still approximately $\pi a^{2}U$, so the same total tube pressure drop is shared equally between the two identical spheres. The load sharing between co-moving spheres in a tube gives half the single-sphere force on each sphere in regime (ii). In regime (i), drag is set locally by the nearly pressure-balanced annular shear and is unchanged on each sphere. In regime (iii), each sphere must pass essentially its own displacement flux through its annular gap, giving the same local pressure drag as before; adding another pressure jump changes the already small tube flux but not either leading drag. The force is unchanged in regimes (i) and (iii).

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.

Near the closest point, the cylinder–plane gap isFor translation parallel to the cylinder axis, the leading flow is Couette flow. Its shear traction is $\mu U/h$, soHence $A_{yy}=2\pi {\epsilon }^{-1/2}\mu L$.

For transverse translation, the local Couette-Poiseuille flow in a thin gap and mass conservation give the Reynolds lubrication equation. Writing $x=a\sqrt{\epsilon }X$ and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce towhere the final term is the direct Couette shear contribution. Therefore

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.

For steady flow the momentum equation isApplying no slip, $u(0)=0$ and $u(h)=U$, gives the Couette-Poiseuille flow in a thin gap

The leading streamwise momentum equation in lubrication theory iswith $p=p(x)$. The no-slip boundary condition isTwice integrating in $y$ gives the local Couette-Poiseuille flow in a thin gapIts volume flux per unit span 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

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 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.