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A microswimmer is a microscopic body that propels itself through a fluid. At small Reynolds number its motion is governed by Stokes flow, and self-propulsion must respect the force-free and torque-free conditions.

For two zero-body-force Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\hat{\mathbf{u}},\hat{\mathbf{\sigma }})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow statesApply it first with the auxiliary translating-sphere solution and then with the auxiliary rotating-sphere solution. The swimmer is force-free and torque-free, while the auxiliary surface tractions are known. The resulting surface slip velocity formulas are

On $r=a$, the first part of the prescribed slip isIts surface average is $-2\mathbf{A}/3$, whereas the $\mathbf{B}$ term has zero average by oddness. ThusThe $\mathbf{A}$ term contributes no rotation. For the other term, the isotropic second and fourth surface moments giveand hence

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