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For each incompressible Stokes flow without body force, ${\partial }_j{\sigma }_{ij}=0$ andThe divergence theorem givesIncompressibility removes the pressure term, antisymmetric velocity-gradient parts do not contract with the symmetric stress, and the remaining integrand is $2\mu e_{ij}^{(1)}{e}_{ij}^{(2)}$, symmetric under $1\leftrightarrow 2$. This proves the Lorentz reciprocal theorem for Stokes flow:

By Linearity of Stokes flow, the drag on a translating body is linear in its velocity, so $F_i=A_{ij}{U}_j$. Apply the reciprocal theorem to two translation solutions with velocities ${\mathbf{U}}^{(1)}$ and ${\mathbf{U}}^{(2)}$. Their surface velocities are constant and the far-field contribution vanishes, yieldingSince the two vectors are arbitrary,Thus the hydrodynamic resistance matrix is symmetric and, for fixed fluid viscosity, depends only on body geometry.

A straight rod has fore-aft symmetry about its centre. Gravity acts through that centre, so the translational-rotational coupling and applied torque vanish. Kinematic reversibility of Stokes flow and Uniqueness of Stokes flow then rule out spontaneous rotation: the inclination $\theta$ remains constant.

Let the speed for perpendicular translation be $U_0$. The given factor of two makes the parallel speed $2U_0$ for the same force component. Resolving gravity parallel and perpendicular to the rod givesResolving these velocities horizontally and vertically givesTherefore

The material derivative follows the interfacial surfactant. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ describes dilution by tangential expansion and concentration by compression. The term $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing surface area due to normal motion of a curved interface. Finally, $D_s{\nabla }_s^{2}C$ describes surface diffusion. Together these give conservation of insoluble surfactant on a moving interface.

The relevant surface Péclet number isWrite $S=\mathbf{n}\cdot \mathbf{A}\cdot \mathbf{n}$ and ${\mathbf{u}}_s=\mathbf{Ax}-S\mathbf{x}$. Since $\mathbf{A}$ has zero trace and ${\nabla }_s\mathbf{n}={\mathbf{I}}_s/a$, the surface divergence isThere is no normal motion. To first order in ${\mathrm{Pe}}_s$, the steady surfactant balance becomes $D_s{\nabla }_s^2{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. The traceless quadratic $S$ is a degree-two spherical harmonic, so ${\nabla }_s^{2}S=-6S/a^2$. The mean of $C^{\prime }$ is zero by total surfactant conservation; consequentlyThe discarded advective term ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$ is smaller by ${\mathrm{Pe}}_s$. This is the quadrupolar surfactant distribution on a spherical interface.

With the unit normal directed from the bubble into the exterior, the interfacial stress balance with variable surface tension isUsing the specified first-order curvature and $\gamma ={\gamma }_0-K{\gamma }_{1}S$, together with ${\nabla }_{s}S=2{\mathbf{I}}_s\mathbf{An}/a$, givesThe first term is the spherical capillary pressure; the last contains the normal tension correction and the tangential Marangoni stress.

The ambient Stokes flow has the symmetry of the symmetric traceless tensor $\mathbf{E}$. In the Unscaled Papkovich–Neuber representation, ${\chi }_{\infty }=\mathbf{x}\cdot \mathbf{E}\cdot \mathbf{x}/2$ is harmonic and yields $\mathbf{Ex}$. The decaying vector potential must have the form $\mathbf{E}\nabla (1/r)$, and the scalar disturbance must have the degree-two form $\mathbf{E}:\nabla \nabla (1/r)$. They are harmonic outside the bubble and have precisely the required rotational covariance. Dimensionless amplitudes may therefore be writtenFor a steady bubble, the no-penetration boundary condition gives $1+P-3Q=0$. Its tangential velocity gives $\alpha =1+2Q$. The tangential stress boundary condition then yieldsHence $\alpha =5/(5+2M)$, $Q=-M/(5+2M)$ and $P=-(5+5M)/(5+2M)$. The inviscid interior supplies only the constant pressure balancing $2{\gamma }_0/a$. Matching the remaining normal stress givesEliminating $P,Q,\alpha$ givesAs $M\to \infty$, $\alpha \to 0$: the Marangoni stress suppresses tangential motion and effectively immobilizes the interface. This limit must retain the small surface Péclet number and small-deformation assumptions; $M$ can grow through increasing ${\gamma }_1$ without invalidating the linear concentration approximation.

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

Let $\mathbf{u}$ be the Stokes flow and let $\mathbf{v}$ be any other admissible incompressible flow with the same no-slip boundary condition on the sphere and the same decay at infinity. PutThen $\nabla \cdot \mathbf{w}=0$ and $\mathbf{w}=0$ on the sphere. The viscous dissipation satisfiesFor the Stokes stressincompressibility of $\mathbf{w}$ givesThe divergence theorem, the homogeneous boundary data, decay at infinity, and $\nabla \cdot \mathbf{\sigma }=0$ giveThereforeThis proves the Minimum-dissipation theorem for Stokes flow.

More surfactant is encountered on the side toward which the far-field concentration $\Gamma$ increases. Adsorption lowers the surface tension there, so the resulting Marangoni stress drives interfacial flow toward the cleaner, higher-tension side. The reaction force propels the bubble along the concentration gradient; this is chemophoresis.

The bulk concentration obeys the advection-diffusion equationAt $r=a$,equates the outward bulk diffusive flux to minus the net adsorption rate: $C>b\Gamma$ favors desorption into the bulk, while $C<b\Gamma$ favors adsorption onto the interface.

With advection neglected, $\Gamma$ is harmonic. Rotational symmetry about $\mathbf{G}$ and the imposed far-field gradient select the dipolar formWhen $kab\ll D$, the boundary flux is smaller than the characteristic diffusive flux, so the leading boundary condition is ${\partial }_r\Gamma =0$ at $r=a$. This gives $B=a^3/2$, and hence

For an interface with unit normal directed from the bubble into the exterior liquid, the interfacial stress balance with variable surface tension may be writtenwith signs tied to the stated curvature convention. Put $C^{\prime }=A\mathbf{G}\cdot \mathbf{n}$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa =2/a$ andthe remaining exterior traction isIts resultant vanishes becauseand therefore the integrals of the two terms cancel. This is required because the bubble and its interfacial stresses exert no external body force on the combined bubble–fluid system.

The traction is a first spherical harmonic, so the decaying, force-free Stokes flow has no Stokeslet and is generated by the indicated Papkovich–Neuber representation. Comparing the supplied tractionwith the capillary traction givesIn the convention for these potentials, the normal velocity at $r=a$ is $(\beta /\mu )\mathbf{G}\cdot \mathbf{n}$. The kinematic boundary condition for a translating sphere is $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so

Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C^{\prime }$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential isUsing the identities supplied in the question,The bulk result gives ${\Gamma }^{\prime }=\frac{3}{2}a\mathbf{G}\cdot \mathbf{n}$ on the surface. The linearized equationthen yieldsThus

Increasing $k$ strengthens exchange with the imposed bulk gradient, so $A$ and $U$ increase toward a saturation value. Increasing $D_s$ smooths surface-concentration differences and decreases $U$. Increasing $C_0$ strengthens advective redistribution of the background surfactant; the resulting feedback opposes the imposed dipole, so $U$ decreases.

For an incompressible Newtonian fluid with negligible inertia and no body force, the Stokes equations areThey are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sendsFluid particles therefore retrace their paths under reversal. This is kinematic reversibility of Stokes flow.

Choose the vertical plane through the rod's axis. Reflection in that plane leaves the horizontal rod, the vertical gravitational force, and the Stokes equations unchanged, but reverses any velocity normal to the plane. A second reflection in the vertical plane perpendicular to the rod reverses velocity along the rod. By uniqueness of Stokes flow, both horizontal velocity components vanish. The rod therefore translates vertically.

Equivalently, the hydrodynamic resistance matrix of an axisymmetric rod has one coefficient parallel to its axis and another perpendicular to it. A vertical force on a horizontal rod lies entirely in the perpendicular eigenspace, so its velocity is vertical.

At negligible Reynolds number, the Stokes equations and their no-slip boundary conditions are linear in velocity. Scaling or superposing a prescribed rigid translation therefore scales or superposes the resulting stress and total hydrodynamic force. Hence, for a fixed rod orientation,The minus sign records that viscous drag opposes motion, and $\mathbf{R}$ is the translational hydrodynamic resistance matrix.

The Papkovich–Neuber representation writes a homogeneous incompressible Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and harmonic scalar $\chi$:with ${\nabla }^2\mathbf{\Phi }=0$ and ${\nabla }^2\chi =0$. The representation satisfies incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and substitution then verifies the Stokes equation.

For a sphere translating with constant vector velocity $\mathbf{U}$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:Substitution gives the translating sphere in Stokes flowAt $r=a$ the radial tensor terms cancel and $\mathbf{u}=\mathbf{U}$, while $\mathbf{u}\to 0$ as $r\to \infty$, so the no-slip boundary condition and far-field condition hold. The resulting traction integrates to the Stokes drag law $6\pi \mu a\mathbf{U}$ in magnitude.

The steady incompressible Stokes flow equations areTaking the curl eliminates the pressure gradient and commutes with the Laplacian, so for the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$,This is the vorticity part of Harmonic pressure and vorticity in Stokes flow.

For a two-dimensional incompressible flow, choose the stream function conventionThenApplying the harmonic-vorticity equation givesthe Biharmonic stream function for planar Stokes flow equation.

Let $H$ be a representative positive gap thickness and setthe larger-order horizontal speed after including the squeeze-induced flux. The lubrication-limit scaling for a moving thin gap requiresEquivalently, the inertial condition containsThe gap must also remain open, $h(x)>0$. These conditions make streamwise derivatives small relative to transverse derivatives, keep the pressure uniform across the gap to leading order, and reduce the Navier-Stokes equation to Stokes flow locally.

A rotation by $\pi$ about the vertical symmetry axis maps a horizontal angular velocity ${\mathbf{\Omega }}^{\prime }$ to $-{\mathbf{\Omega }}^{\prime }$ but leaves a possible vertical force unchanged. Linearity and uniqueness of Stokes flow therefore require that vertical force to equal its own negative, so it vanishes. A horizontal force and a horizontal couple are allowed by the same symmetry.

Write $\lambda =1-ϵ$ and examine the lower pole, where $\theta \ll 1$. Since $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$,Thus $h=O(ϵ\Delta )$ for $\theta =O({ϵ}^{1/2})$, a circular patch of radius

In the broad region, the surface speed is $O({\Omega }^{\prime }a)$, the shear is $O(\mu {\Omega }^{\prime }a/\Delta )$, the area is $O(a^2)$, and the moment arm is $O(a)$. HenceIn the narrow patch, the shear rises to $O(\mu {\Omega }^{\prime }a/(ϵ\Delta ))$, while its area falls to $O(ϵa^2)$; the moment arm remains $O(a)$. Therefore

Dropping the moment arm gives the shear-force scale. Pressure produces the same horizontal order after multiplication by the small surface slope. Both regions consequently contribute

Put $\delta =H/L\ll 1$. In a thin gap, the streamwise Stokes flow balance isso $P\sim \mu UL/H^2$. The characteristic shear stress is $\tau \sim \mu U/H$, and thereforeThis is the pressure-dominance scaling of lubrication theory.

Let the centre displacement be $\lambda \Delta$ in the downward direction. To first order in $\Delta /a$, projecting this displacement onto the radial direction at polar angle $\theta$ changes the concentric gap $\Delta$ by $-\lambda \Delta \cos \theta$. Hence

For two body-force-free Stokes flows $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow isIndeed, the difference of the two volume integrands isbecause both flows are incompressible and the Newtonian fluid stress tensor is symmetric. The divergence theorem proves the boundary identity.

On a rigid body, ${\mathbf{u}}^{(r)}={\mathbf{U}}^{(r)}+{\mathbf{\Omega }}^{(r)}\times \mathbf{x}$. The reciprocal theorem becomesWriting $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}(\mathbf{U},\mathbf{\Omega }{)}^T$ and choosing arbitrary pairs of rigid velocities shows thatThus the hydrodynamic resistance matrix is symmetric.

For an axisymmetric surface $r=a(z,t)$, twice the mean curvature with the outward-normal convention isWriting $a=a_0+\eta$ and retaining linear terms gives

At the unperturbed boundary $r=a_0$, the linearized kinematic boundary condition, tangential-stress condition, and Young–Laplace equation areFor a normal mode $e^{ikz+st}$ these become $u(a_0)=s\eta$, ${\sigma }_{rz}(a_0)=0$, and

The Papkovich–Neuber representation of body-force-free Stokes flow iswhere $\mathbf{\Phi }$ and $\chi$ are harmonic.

Let $x=kr$ and $E=e^{ikz+st}$. Substitution of the given radial vector potential and scalar potential givesUsing the modified Bessel function identity $(x{I}_1{)}^{\prime }=x{I}_0$, the tangential stress simplifies toThe stress-free condition at $r=a$ therefore yields

The Papkovich–Neuber representation writes a homogeneous Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and a harmonic scalar $\chi$ asFor translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic $1/r$ and its directional derivatives contracted with $\mathbf{U}$. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to $\mathbf{\Omega }\times \mathbf{x}/r^3$. Matching the no-slip boundary condition $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at $r=a$ gives the superposition of the translating sphere in Stokes flow and the rotating sphere in Stokes flow:Each term decays at infinity, and direct substitution at $r=a$ gives the prescribed rigid velocity.

When $\mathbf{U}=0$, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the Newtonian fluid stress tensor gives

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

Because the wavelength is much smaller than the shelf thickness, the ice-shelf corrugation relaxation may be treated as Stokes flow in the half-space $z>0$, with the mean ice-ocean boundary at $z=0$. Letso $u=-{\psi }_z$ and $w={\psi }_x$. Taking the curl of the Stokes equation gives the biharmonic stream function for planar Stokes flow equation ${\nabla }^4\psi =0$. Decay as $z\to \infty$ removes the growing modes, leavingThusand the perturbation pressure obtained from the Stokes equation is

The linearized zero-shear stress condition at $z=0$ isso $B=kA$. The water is hydrostatic, and displacement of the density interface gives the normal-stress conditionAfter $B=kA$, one has $w_z(0)=0$, soThe kinematic boundary condition ${\eta }_t=w(0)$ gives $\sigma \hat{\eta }=ikA$. Eliminating $A$ yieldsHence a corrugation of wavelength $\lambda =2\pi /k$ decays on the timescaleHydrostatic buoyancy supplies the restoring stress, while viscous deformation over depth $O(k^{-1})$ supplies the resistance. Shorter wavelengths deform a shallower but more strongly sheared layer and therefore have a longer decay time in this gravity-only model. As ice is advected away from the grounding line, long corrugations should disappear first, leaving progressively shorter-wavelength structure farther downstream.

Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration velocity is unchanged, and the axial angular velocity changes sign because it is an axial vector.

On the other hand, reversing the original Stokes flow makes all three velocities change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal velocities have opposite signs. Uniqueness of Stokes flow forces that velocity to equal its own negative, so it is zero. The rotational velocities already agree under the two operations, so this argument places no restriction on rotation.