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Linearity of Stokes flow and Uniqueness of Stokes flow identify the ambient resistance to an expanding thin plug with the reversed perturbation around a stationary disc in an axisymmetric straining flow. The imposed straining flow has zero tangential stress on the disc plane. Each face then resists radial velocity $Er$ with traction $-8\mu Er/[\pi \sqrt{R^2-r^2}]$.

For a fixed geometry, the steady Stokes equation maps $(\mathbf{u},p^{\prime },\mathbf{f})$ to $(-\mathbf{u},-p^{\prime },-\mathbf{f})$ when all imposed boundary velocities and disturbance forces are reversed; $p^{\prime }$ is pressure relative to any unchanged hydrostatic pressure. Linearity of Stokes flow and Uniqueness of Stokes flow therefore reverse the entire velocity field. If the driving schedule and evolving geometry are retraced, fluid elements retrace their paths too. Reversing just one drive while leaving another applied force unchanged does not meet these conditions.

This reversal concerns the paths, while viscous dissipation remains positive: for a sphere, the required power $6\pi \mu a{U}^2$ is unchanged by $U\mapsto -U$. A reversed experiment must supply fresh energy. Inertia, Brownian motion and irreversible contact can prevent perfect retracing. Edward M. Purcell illustrated the mechanism with a viscous fluid flow demonstration in Life at low Reynolds number, American Journal of Physics 45, 3–11 (1977).

Put $r=|\mathbf{x}|$. In the Unscaled Papkovich–Neuber representation, choose the componentwise harmonic vector fieldBoth $\mathbf{\Omega }\times \mathbf{x}$ and $(\mathbf{\Omega }\times \mathbf{x})/r^3$ are harmonic away from the origin: the latter is a linear combination of derivatives of the harmonic function $1/r$. Also $\mathbf{x}\cdot \mathbf{\Phi }=0$, so the representation gives $\mathbf{u}=f(r)\mathbf{\Omega }\times \mathbf{x}$. The no-slip boundary condition requires $f(a)=1$ and $f(b)=0$, givingThus the Rotational Stokes flow between concentric spheres isThe divergence of $\mathbf{\Phi }$ is zero: the cross-product field itself has zero divergence, and $\nabla f$ is radial and perpendicular to $\mathbf{\Omega }\times \mathbf{x}$. Hence $p=2\mu \nabla \cdot \mathbf{\Phi }=0$ in this pressure gauge.

There is also a symmetry argument. Take the rotation axis as the polar axis. Linearity of Stokes flow makes velocity and pressure perturbations change sign when the rotation reverses. Reflection in any plane containing that axis reverses the axial rotation vector. Combined with axial symmetry, this excludes meridional velocity components and forces the pressure perturbation to be both unchanged and sign-reversed by that reflection. Equivalently a purely azimuthal axisymmetric Stokes flow has no meridional component of its vector Laplacian, so its meridional pressure gradient vanishes. Thus the pressure is constant; setting that arbitrary constant to zero gives the same result without solving the radial coefficients.

For the Newtonian fluid stress tensor, let $\mathbf{q}=\mathbf{\Omega }\times \mathbf{x}$. The symmetric part of the velocity gradient isbecause the gradient of rigid rotation is antisymmetric. Consequently the full stress tensor, with $p=0$, isFor $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, its only nonzero spherical-coordinate components are ${\sigma }_{r\varphi }={\sigma }_{\varphi r}=-3\mu B\Omega \sin \vartheta /r^3$. On the inner sphere the traction exerted by the fluid, using the sphere's outward normal $\mathbf{n}$, isIts torque isusing $\int n_i{n}_{j}d\omega =4\pi {\delta }_{ij}/3$. Therefore the applied Torque in rotational Stokes flow between concentric spheres isFor $b\gg a$, the relevant speed and length are $\Omega a$ and $a$, so negligible inertia requires the rotational Reynolds numberFor a narrow gap $d=b-a\ll a$, the viscous gradients are controlled by $d$, not $a$. A centrifugal acceleration of scale ${\Omega }^{2}a$ is then compared with viscous scale $\mu \Omega a/d^2$, rather than $\mu \Omega /a$. Their ratio is $\rho |\Omega |d^2/\mu$. This explains why the large-gap condition is not a necessary narrow-gap criterion; a thin gap greatly strengthens viscous resistance. This base-flow scaling is distinct from a criterion for stability to arbitrary short-scale disturbances.

The Minimum-dissipation theorem for Stokes flow compares a Stokes solution with all sufficiently regular incompressible trial velocities in the same fluid domain that have the same prescribed boundary velocities. With no body-force work, the dissipation functional isIf $\mathbf{v}=\mathbf{u}+\mathbf{w}$ and $\mathbf{w}=0$ on the prescribed boundary, integration by parts with $\nabla \cdot \mathbf{\sigma }(\mathbf{u})=0$ and $\nabla \cdot \mathbf{w}=0$ makes the cross term vanish. ThusThe comparison fixes velocities, rather than applied torques or forces.

Now let the particles take their actual rigid translations and rotations in the particle-containing flow. Extend the fluid velocity through each particle by its rigid velocity. No-slip boundary conditions make this extension continuous; rigid motion has zero rate-of-strain tensor and zero divergence. The extension is therefore an admissible trial field in the original particle-free annulus, and its dissipation equals that of the fluid outside the particles. Hence $D\ge D_0$.

The Stokes boundary-work identity gives $D=\mathbf{G}\cdot \mathbf{\Omega }$. The outer sphere is stationary, the maintained inner sphere has zero translational velocity, and every added particle has zero total force and torque, so none of those other boundaries contributes power. Likewise $D_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor nonzero rotation and a particle of positive volume the inequality is strict: equality in the dissipation identity would force the extension to be the unique original flow, but that original flow has nonzero strain on every open particle region, whereas a rigid interior has zero strain. This proves that force-free inclusions increase rotational resistance:No dilute-suspension approximation or assumption about the particle shapes was used.

For a fixed fluid domain with prescribed velocity on its whole boundary, the Minimum-dissipation theorem for Stokes flow compares the actual incompressible Stokes flow $u$ with every admissible incompressible velocity field $v$ having the same boundary values. The trial fields need not themselves satisfy the momentum equation. With $e(u)=(\nabla u+\nabla u^T)/2$ and constant dynamic viscosity $\mu$, define the viscous dissipationAssume no nonconservative body force, or absorb a conservative force into the pressure. Set $w=v-u$, so $\nabla \cdot w=0$ and $w=0$ on the boundary. The Stokes stress tensor $\sigma =-pI+2\mu e(u)$ obeys $\nabla \cdot \sigma =0$. Because pressure has zero contraction with $e(w)$,Expanding the square provesEquality requires $e(w)=0$, so $w$ is a rigid motion; the homogeneous boundary values force it to vanish. The theorem concerns fixed boundary velocities, not fixed applied forces or arbitrary comparisons between different fluid domains.

For the concentric-sphere motion, seek $u=F(r)\mathbf{\Omega }\times x$. Incompressibility is automatic. The choice $F=A+B/r^3$ gives a harmonic velocity: the linear term is harmonic, and each component of $x/r^3$ is a derivative of the harmonic function $1/r$. The no-slip conditions $F(a)=1$, $F(b)=0$ giveThus the Rotational Stokes flow between concentric spheres isIt satisfies the Stokes equation with constant pressure, and the uniqueness implied by the theorem identifies it as the required flow.

In the Unscaled Papkovich–Neuber representation, take $\Phi =-u/2$ and $\chi =0$. Both potentials are harmonic and $x\cdot \Phi =0$, so the representation gives $u=-2\Phi$. Alsoand therefore $p=0$, after choosing the arbitrary pressure constant.

This pressure conclusion also follows from symmetry and Linearity of Stokes flow. Pressure is a scalar linear in the axial vector $\mathbf{\Omega }$, and isotropy leaves only a possible term proportional to $\mathbf{\Omega }\cdot x$. That is a pseudoscalar, forbidden by reflection symmetry of the concentric spherical domain. Hence the pressure perturbation vanishes. Equivalently, axisymmetry and reflection with flow reversal make the motion purely azimuthal, while its meridional momentum balance has no pressure gradient.

Let $v=\mathbf{\Omega }\times x$ in the following stress calculation. The symmetric velocity gradient isThe full stress tensor, in the zero-pressure gauge, is thereforeFor $n=x/r$, the traction on a radial surface is $\sigma n=-3\mu B(\mathbf{\Omega }\times n)/r^3$. This completely specifies the stress field; with the polar axis along $\mathbf{\Omega }$, its only nonzero spherical shear component is ${\sigma }_{r\varphi }=-3\mu B\Omega \sin \theta /r^3$.

The fluid couple on the inner sphere is the integral of $x\times \sigma n$ with the normal pointing from the solid into the fluid. Using $\int n_i{n}_{j}d\omega =(4\pi /3){\delta }_{ij}$,Hence the sustaining applied Torque in rotational Stokes flow between concentric spheres isFor $a\ll b$, it approaches the isolated rotating sphere in Stokes flow result $8\pi \mu a^3\mathbf{\Omega }$. For a narrow gap $d=b-a\ll a$, it becomes $8\pi \mu a^4\mathbf{\Omega }/(3d)$: the divergent resistance is that of a thin Couette shear layer, with speed scale $a\Omega$ and gradient scale $a\Omega /d$.

Now add the force-free, couple-free rigid particles and denote the resulting velocity by $u_p$. Extend $u_p$ through each particle as its rigid translational and rotational velocity. No slip makes this extension continuous across every particle surface; it is incompressible and has zero rate-of-strain tensor inside each particle. It therefore gives an admissible trial velocity in the original particle-free annulus, with exactly the same sphere boundary velocities. Its full-annulus dissipation is the actual fluid dissipation ${\mathcal{D}}_p$, since the added interior integrals are zero. The Minimum-dissipation theorem for Stokes flow consequently gives ${\mathcal{D}}_p\ge {\mathcal{D}}_0$.

The boundary-work identity follows by integrating $\nabla \cdot (\sigma u)$: dissipation equals the mechanical power supplied at all rigid boundaries. The outer sphere is stationary; the inner sphere has no translation, so any force needed to hold its centre does no work; and each added particle contributes zero power because its net force and couple are zero. Thus ${\mathcal{D}}_p={\mathbf{G}}_p\cdot \mathbf{\Omega }$, while ${\mathcal{D}}_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor $\mathbf{\Omega }\ne 0$ and at least one particle of positive volume in the open annulus, the inequality is strict. Equality would make the extended velocity identical to the particle-free solution, yet that solution has nonzero strain on every open region away from the rotation axis and cannot be rigid throughout any particle interior. Hence the applied couple component along the imposed angular velocity increases. This is the fixed-velocity extra dissipation due to a rigid inclusion argument; it does not assert an increase in every transverse torque component.

The Minimum-dissipation theorem for Stokes flow compares fields with fixed geometry and boundary velocities. Moving the sphere changes the geometry, while sedimentation prescribes its effective weight rather than its velocity. It therefore supplies no rule for migrating toward a position of least dissipation. Indeed, at fixed force $W$, the power is $WV$, and a lower drag resistance gives a larger settling speed and larger power, rather than a smaller power. This is a fixed-force comparison of minimum viscous dissipation issue.

The proposed migration is itself false in ideal Stokes flow. Reflect the uniform vertical tube in a horizontal plane through the sphere centre. The sphere's lateral position is unchanged, but the applied axial force reverses. Reflection leaves a transverse translational velocity unchanged. By Linearity of Stokes flow, however, reversing the force reverses that transverse velocity. It must therefore be zero. This proves no lateral migration of a single sphere in a uniform tube in Stokes flow:Rotation and a wall-dependent vertical speed are allowed. Inertial lift, particle deformation or tube-end effects belong to different problems and do not follow from the dissipation theorem.

Use the Papkovich–Neuber representation in the conventionIt gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=-\mu \nabla (\nabla \cdot \mathbf{\Phi })=\nabla p$. Additive pressure constants are immaterial.

Let $q=\mathbf{x}\cdot \mathbf{Ex}$, with $\mathbf{E}$ symmetric and traceless. The harmonic vector $\mathbf{Ex}/r^3$ is a derivative of $1/r$. The scalar $q/r^5$ is a decaying degree-two harmonic because $q$ is a harmonic homogeneous quadratic. These Papkovich potentials for a strained sphere have precisely the strain symmetry and decay required for the disturbance, so trySubstitution givesAt $r=a$, require ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$. The independent vector structures give $B=a^5$ and $A=-5a^3/3$. The disturbance for a sphere in a uniform straining Stokes flow isIt decays at infinity and satisfies no slip exactly at the sphere. Its leading far field is a stresslet, of velocity scale $E{a}^3/r^2$.

A force-free sphere in a linear background translates with ${\mathbf{U}}_0$, by Faxén's first law, and a couple-free sphere rotates with the background rigid-body rotation $\mathbf{\Omega }$, by Faxén's rotational law. Uniform translation and solid-body rotation can then be matched without disturbance. The remaining relative boundary velocity is exactly $-\mathbf{Ex}$. The Linearity of Stokes flow and Uniqueness of Stokes flow therefore give the same disturbance. The pure strain produces no resultant force or couple, by its symmetric degree-two angular structure.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.

Write $\mathbf{n}=\mathbf{R}/R$. At fixed geometry, Linearity of Stokes flow makes the translational velocities linear in the couples. With zero forces and couples perpendicular to $\mathbf{R}$, isotropy and reflection symmetry allow only terms proportional to ${\mathbf{G}}_j\times \mathbf{n}$ in each ${\mathbf{U}}_i$. There is no longitudinal term: reflect in the plane containing $\mathbf{R}$ and the common torque axis, taking account of the axial-vector transformation of torque, and use linearity under torque reversal. HenceThe direction of $\mathbf{R}$ can still change; constant separation does not mean stationary centers.

When the couples are equal, a rotation through $\pi$ about their common axis exchanges the identical spheres and reverses the in-plane translational velocities while leaving the axial angular velocities unchanged. Uniqueness of Stokes flow then givesThe midpoint is stationary and the pair can orbit it. The body spin need not equal the angular rate of that orbit. These symmetry statements hold at arbitrary noncontact separation, not just in a distant rotlet approximation.

Write $d_1={\rho }_0-{\rho }_1>0$, $d_2={\rho }_2-{\rho }_0>0$, and choose the common undisturbed interface pressure $P$. The hydrostatic pressure is $p_1=P-{\rho }_{1}gz$ in the upper fluid and $p_2=P-{\rho }_{2}gz$ in the lower fluid. Continuity of hydrostatic pressure across both current interfaces givesThus $h_1=d_{2}h/(d_1+d_2)$ and $h_2=-d_{1}h/(d_1+d_2)$. This isostatic depth partition of an interfacial current expresses local vertical buoyancy balance: a displaced column with an imbalance would move vertically until the partition was restored. With the hydrostatic pressure $P-{\rho }_{0}gz$ subtracted, the pressure within the current is $\Delta \rho gh$, whereThe equivalent body force is outward for a profile decreasing with $r$. Its magnitude is $\Delta \rho g|h_r|$; the derivative without an absolute value is a signed quantity.

For an ambient-controlled interfacial plug current, integrating the body force through depth gives radial force per area $O(\Delta \rho g{H}^2/R)$. The ambient Stokes flow varies over distance $R$, exerting traction $O(\mu u/R)$, so $u\sim \Delta \rho g{H}^2/\mu$. The current's internal shear supports the ambient traction with a velocity difference $\Delta u\sim (\mu u/R)H/(\lambda \mu )$; $\Delta u/u\sim H/(\lambda R)\ll 1$ requires $\lambda \gg H/R$. The force per area from its in-plane viscous extension is $O(\lambda \mu uH/R^2)$, negligible against ambient traction when $\lambda \ll R/H$. Together these give the viscosity window for an interfacial plug currentFixed volume gives $V\sim H{R}^2$, and the spreading velocity has the scale $\dot{R}$. Thus $\dot{R}\sim \Delta \rho g{V}^2/(\mu R^4)$ andChoosing $A$ so that $R=A{t}^{1/5}$ exactly, the similarity solution has $h=(V/A^2)t^{-2/5}F(\eta )$, $\eta =r/(A{t}^{1/5})$, and $u=A{t}^{-4/5}U(\eta )$. In radial conservation $h_t+r^{-1}(rhu{)}_r=0$, substitution gives $(\eta FU{)}^{\prime }=({\eta }^{2}F{)}^{\prime }/5$. Regularity and no source at the centre remove the integration constant. Therefore the radial plug velocity is linear:

For the disc-traction analogy for an interfacial current, subtract the imposed straining flow from the stationary-disc solution. The perturbation has velocity $-Er{\mathbf{e}}_r$ on the disc and zero far-field velocity. Reverse its sign: the resulting Stokes flow is driven by a radially expanding disc with velocity $Er{\mathbf{e}}_r$ in otherwise quiescent fluid. The background straining flow contributes no radial shear traction on $z=0$, so the given disc traction changes sign. This step uses Linearity of Stokes flow, Uniqueness of Stokes flow, and the no-slip boundary condition. The thin plug current has the same leading boundary velocity and hence the same ambient resistance. Each face contributes $-8\mu Er/[\pi \sqrt{R^2-r^2}]$; both faces must be counted. With $E=1/(5t)$, radial force balance isIntegrate with $h(R,t)=0$. The interfacial plug-current similarity profile isVolume integration gives $V=2\pi {\int }_0^{R}rhdr=4\pi H{R}^2/5$. Eliminating $H$ gives the full prefactor:The formal edge slope diverges, so a narrow edge region lies outside the thin-current approximation. Its traction and volume contributions are integrable, and this is a leading thin-current solution rather than a uniformly valid edge description.

For a diffusion-controlled solute gravity current, excess solute mass, not enriched-fluid volume, is conserved. At late times the effective mass density excess is proportional to solute concentration, so $\Delta \rho (t)R^{2}H\sim \Delta {\rho }_0{V}_0$, and $H\sim (Dt{)}^{1/2}$. The same ambient resistance scaling givesThis scaling does not determine a universal numerical prefactor. The mass diffusivity $D$ controls the growing depth, while dilution weakens buoyancy; assuming a fixed current volume would incorrectly retain the $t^{1/5}$ law. The thin-layer requirement becomes $(\mu D/(\Delta {\rho }_{0}g{V}_0){)}^{1/3}\ll 1$; when only dissolved solute changes the upper fluid, its viscosity is of order the ambient viscosity, consistent with the plug-current window.

Linearity of Stokes flow and rotational symmetry, including reflection in mathematics invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a polar vector; its linear dependence on a single vector $\mathbf{A}$ must be $c\mathbf{A}$. A traceless second-rank tensor $\mathbf{B}$ cannot produce a polar vector by an isotropic linear map: tensor contraction with the identity matrix vanishes and tensor contraction with the totally antisymmetric tensor vanishes because $\mathbf{B}$ is a symmetric second-rank tensor. Angular velocity is an axial vector. Neither $\mathbf{A}$ nor a symmetric $\mathbf{B}$ can produce one by an isotropic linear map. Products such as $\mathbf{A}\times \mathbf{BA}$ are excluded by Linearity of Stokes flow. Thus $\mathbf{\Omega }=0$.

Use the Unscaled Papkovich–Neuber representation, with harmonic functions $\mathbf{\Phi }$ and $\chi$:Indeed $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$ because each potential is a harmonic function. Put $r=|\mathbf{x}|$ and $S=\mathbf{x}\cdot \mathbf{Bx}$. The decaying harmonic functions with the necessary angular dependence are $\mathbf{A}\cdot \mathbf{x}/r^3$, $\mathbf{Bx}/r^3$ and $S/r^5$. The last is a harmonic function precisely because $\mathrm{tr}\mathbf{B}=0$. A $1/r$ Stokeslet is excluded by the force-free condition, and a rotlet by the torque-free condition. TryThe resulting Stokes flow isAt $r=a$, the coefficient of $\mathbf{n}(\mathbf{A}\cdot \mathbf{n})$ fixes $c=a^3/3$; the remaining constant vector gives $\mathbf{U}=-2\mathbf{A}/3$. In the $\mathbf{B}$ mode, matching $\mathbf{Bn}-(\mathbf{n}\cdot \mathbf{Bn})\mathbf{n}$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete two-mode tensorial squirmer flow isThe boundary condition is satisfied in the laboratory frame, so the Stokes flow tends to zero at infinity. The potential dipole in the $\mathbf{A}$ mode gives an irrotational flow and decays as $r^{-3}$; the leading $\mathbf{B}$ mode is a stresslet, decaying as $r^{-2}$. Neither carries a net force or torque. Direct surface integral averaging with the surface slip velocity formula gives the same translation, providing an independent check. Uniqueness of Stokes flow then identifies this decaying solution.

First supply the introductory scaling and geometry. The leading tangential Stokes equation in a thin gap balances $P/L$ against $\mu U/H^2$. The shear stress scale is $\tau \sim \mu U/H$, so lubrication pressure dominates shear stress:The outer circle, viewed along a ray $r(\cos \theta ,\sin \theta )$ from the inner centre, satisfies $r^2-2\alpha \Delta r\sin \theta +{\alpha }^2{\Delta }^2=(a+\Delta {)}^2$. Expanding its positive root to first order in $\Delta /a$ gives $r=a+\Delta +\alpha \Delta \sin \theta +O({\Delta }^2/a)$, hence $h=\Delta (1+\alpha \sin \theta )$.

Take positive $\Omega$ to mean increasing $\theta$, and put $s=a\theta$, $y=r-a$, $U=\Omega a$. With no-slip boundary conditions at $y=0,h$, lubrication theory givesSteady mass conservation makes $q$ independent of $\theta$. Single-valued pressure requires ${\int }_0^{2\pi }{p}_{\theta }d\theta =0$. Write $I_n={\int }_0^{2\pi }(1+\alpha \sin \theta {)}^{-n}d\theta$ and $D_{\alpha }=1+{\alpha }^2/2$. It follows thatThe tangential fluid tractions on the inner and outer solids, measured in the same increasing-$\theta$ direction, are respectivelyThis sign convention distinguishes the traction on a wall from the corresponding stress component in the fluid.

To leading order the holding force balances the pressure traction on the inner circle. Integration by parts givesUse $p_{\theta }=a{p}_s$ and $\int \sin \theta /(1+\alpha \sin \theta {)}^{n}d\theta =(I_{n-1}-I_n)/\alpha$, with the continuous limit at $\alpha =0$. Substituting the boxed flux yieldsThe zero vertical component also follows from reflection in mathematics in the vertical line: this leaves the geometry and vertical force unchanged, reverses $\Omega$, and Linearity of Stokes flow then requires $F_z=0$.

Pressure gives no moment about either circular wall's own centre. Integrating the viscous shear torque with lever arm $a$ to leading order gives the fluid couplesIndeed $C_i=a^2\int {\tau }_{i}d\theta$ and $C_o=a^2\int {\tau }_{o}d\theta$ at this order. Applied holding or driving couples are the negatives of these fluid couples. They need not cancel because they are moments about different axes. The fluid force on the outer solid equals the holding force $F_x$ on the inner one. Transferring its moment to the inner axis adds $-\alpha \Delta F_x$, so angular momentum balance about a common axis isFor $\alpha =0$ the usual equal and opposite concentric couples are recovered. All these formulas are leading thin-gap expressions at fixed $|\alpha |<1$; taking the contact limit requires maintaining the local slender-gap assumptions.

Put the fixed outer centre at the origin, and let the moving inner centre be $\mathbf{d}=(x,z)$. This reverses the sign of the vertical offset used in the previous parts: now $h=\Delta -\mathbf{d}\cdot \mathbf{n}$. At concentricity rotation produces no force and translation has isotropic resistance $R_0=12\pi \mu a^3/{\Delta }^3$. Balancing the external load givesThere is no inertial acceleration in this Stokes flow model; velocity is determined instantaneously by force balance.

For the subsequent motion, the full-film journal bearing assumption matters: pressure is single-valued around the whole annulus, with no specified cavitation boundary. Let $r=|\mathbf{d}|$, $e=r/\Delta$, $\varphi =\arg (x+iz)$, and ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi }$ be its radial and tangential unit vectors. The squeeze resistance of an eccentric journal bearing has radial and tangential coefficientsFor completeness, rotate coordinates until $\mathbf{d}$ points along $x$, so $h=\Delta (1-e\cos \theta )$. Radial translation gives squeezing flux $q=a{v}_r\sin \theta$ and resistance proportional to $\int {\sin }^2\theta /(1-e\cos \theta {)}^3=\pi /(1-e^2{)}^{3/2}$. Tangential translation gives $q=a{v}_t[-\cos \theta +3e/(2+e^2)]$; the constant enforces periodic pressure. Its resistance usesThese expressions supply both translation directions; retaining only the vertical squeezing formula would miss the moving axle's horizontal dynamics.

The rotational fluid force from the earlier calculation points along ${\mathbf{e}}_{\varphi }$, with magnitude $R_t\Omega r/2$. Superposing rotation and translation by Linearity of Stokes flow, and resolving the load, givesFor positive $\Omega$ the axle initially moves down, then to the right because the rotation-generated force is perpendicular to the downward eccentricity. The equilibrium is displaced horizontally to the right, at the same height as the outer axis. Its eccentricity $0<e_{\ast }<1$ is the unique solutionThe rotational force then points upwards and balances the load. For negative $\Omega$ the displacement is to the left, using $|\Omega |$ in the magnitude equation. Uniqueness follows because the logarithmic derivative of $e/[\sqrt{1-e^2}(1+e^2/2)]$ has positive numerator $1-e^2/2+e^4$ after a common positive denominator. If $\Omega =0$ and $F>0$, no separated static equilibrium exists in this ideal model. If $F=0$, concentricity remains a solution.

One should not assert that this full-film model necessarily settles to the equilibrium. It predicts full-film journal-bearing whirl. For $F,\Omega >0$, eliminate time from the two boxed evolution equations and put $K=6\pi \mu \Omega a^3/{\Delta }^2$. The orbit starting at concentricity obeysso, with its zero initial horizontal displacement,The orbit begins on the lower branch, turns at $z=0$ before contact, and returns along the upper branch, forming a closed orbit through concentricity. The ratio of the right-hand side to $e$ increases from zero to infinity, giving a unique turning eccentricity below one. Linearizing at equilibrium likewise gives $\delta \dot{r}=-(F/R_{r,\ast })\delta \varphi$ and $\delta \dot{\varphi }=c_{\ast }\delta r$ with $c_{\ast }>0$: the eigenvalues are imaginary, not damped. Thus the equilibrium position exists, but the initially concentric axle circulates around it in the stated ideal full-film approximation. Cavitation or additional mechanical constraints would change the model; none were specified.

The normalized centre trajectory below illustrates the full-film prediction for $K/F=1$. The dashed circle is the contact locus of the axle centre, not either cylinder wall.