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Write $D=R_e-R_i$, $S=R_e+R_i$, $A=R_e^2-R_i^2=DS$, and $\Delta p=p_i-p_e$. Incompressible flow with axial strain rate $E$ givesIts radial Stokes equation gives $p_r=0$, because $u_{r,rr}+u_{r,r}/r-u_r/r^2=0$. The radial Newtonian fluid stress tensor component is ${\sigma }_{rr}=-p-\mu E-2\mu C/r^2$. The outer and inner Young–Laplace equation conditions have opposite curvature signs:Subtracting and eliminating $C$ givesIn particular, the boxed pressure is equivalent to the printed identity for $(-p-\mu E)A$. For equal gas pressures, $C=-\gamma R_i{R}_e/(2\mu D)$, so surface tension draws both interfaces inward in addition to the imposed extension.

The kinematic boundary condition is ${\dot{R}}_j=-E{R}_j/2+C/R_j$ at either interface. Taking the difference, and then the difference of squared radii, givesThe cancellation of $C$ in the area equation is mass conservation: axial stretching reduces the liquid area while radial redistribution changes the hole size. For $E=0$, equal pressures, and $R_i(0)=b$, $R_e(0)=2b$, one has $A=3b^2$ and $D=b+\gamma t/(2\mu )$. Since $S=A/D$,Closure occurs at $D=\sqrt{A}=\sqrt{3}b$, soThis capillary collapse of an annular viscous cylinder has finite closure time for $\gamma >0$ and $b>0$. For $\gamma =0$ it does not close in this case. The annular formula is used only up to closure; the singular cylindrical inner curvature is not a boundary condition on the subsequent solid cylinder.

Let $z$ increase upwards, so the bed is at $z=-b(x)$, the surface is at $z=h(x,t)$, and $H=h+b$. In the shallow-shelf approximation the incompressible flow has nearly depth-uniform horizontal speed $u$ and vertical strain $w_z=-u_x$. The Newtonian fluid stress tensor, together with the hydrostatic approximation, then givesThe factor four includes both horizontal extension and the pressure correction required by vertical compression. Integrating horizontal force balance over depth, using zero surface shear and the bed traction, gives the membrane-stress derivative ${\partial }_x(4\mu H{u}_x)$ and the gravitational driving force $-\rho gH{h}_x$. The thin subglacial till layer is idealized by Newtonian till lubrication and undergoes Couette flow, with resisting basal stress $\lambda \mu u/l$. Hence force balance and mass conservation yieldSince $b$ is fixed, $H_t=h_t$. No accumulation or ablation is included.

At the grounding line, the first boundary condition is ice-sheet flotation:For an unbuttressed ice shelf, the second is the extensional stress required to balance the difference between the integrated ice and seawater hydrostatic pressures. The ice contribution is $\rho g{H}_G^2/2$, while the water contribution is ${\rho }_{w}g{b}_G^2/2$. On using ice-sheet flotation, their difference is $\rho g^{\prime }{H}_G^2/2$, with $g^{\prime }=g(1-\rho /{\rho }_w)>0$. Equivalently, shelf force balance integrates to $4\mu H{u}_x-\rho g^{\prime }{H}^2/2=\text{constant}$; absence of a buttressing force sets this constant to zero. Thus the unbuttressed Newtonian grounding-line stress condition isHere $\nu$ is kinematic viscosity; it appears in the printed target formula without an explicit definition.

The ratio of longitudinal stress divergence to till drag is of order $4lH/(\lambda L^2)$. The given small-parameter limit therefore yields the friction-dominated bulk relationThe coefficient depends on the basal lubrication, and $h_x<0$ gives seaward motion. This is a bulk reduction: longitudinal stress has been neglected in the differential equation but its boundary traction remains specified. A complete uniformly valid approximation close to the grounding line could require a membrane-stress boundary layer. In the remainder, use the pointwise friction closure and retain the shelf stress condition as the boundary data, as requested in the paper.

Differentiate the ice-sheet flotation condition along the moving grounding line:Now $H_x=h_x+\alpha$. Substitution of the friction-dominated speed and the shelf stress boundary condition gives the friction-dominated grounding-line evolution law:The coefficient of ${\dot{x}}_G$ is the spatial derivative of the flotation deficit, up to sign. If it vanishes, the implicit equation remains the correct compatibility condition but division to determine a finite speed is invalid; tangential contact needs separate analysis.

In a steady state, volume flux per unit width is constant: $Hu=q_0$. The friction relation givesDifferentiating $Hu=q_0$ and using $u_x=g^{\prime }H/(8\nu )$ at the grounding line givesPut $\beta =l/\lambda$, $\mathcal{H}=H/\beta$ and $\mathcal{Q}=\nu q_0/(g{\beta }^3)$. Substitution and cancellation of $g{\beta }^5/\nu$ yield the steady friction-dominated grounding-line thickness relation:For $q_0>0$ and a bed deepening seaward ($\alpha \ge 0$), the polynomial $(g^{\prime }/8g){\mathcal{H}}^5+\alpha \mathcal{Q}{\mathcal{H}}^2-{\mathcal{Q}}^2$ is strictly increasing for $\mathcal{H}>0$, starts negative and tends to infinity, so it has exactly one positive root. At that root $\mathcal{Q}>\alpha {\mathcal{H}}^2$ and $H_x<0$, as required by steady outward extension. The relation belongs to the specified friction closure; it is not a flux law for arbitrary till rheology or arbitrary membrane-stress matching.

Write $r=|\mathbf{x}|$ and take $\mu$ to be the dynamic viscosity. The Papkovich–Neuber representation of a homogeneous Stokes flow uses a harmonic vector potential $\mathbf{\Phi }$ and harmonic scalar potential $\chi$:These formulas give incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and give $\mu {\nabla }^2\mathbf{u}=\nabla p$, the Stokes equation away from forcing.

For a point force represented by a Dirac delta distribution, rotational covariance and the Linearity of Stokes flow suggest a vector monopole proportional to $\mathbf{F}/r$: $1/r$ is a harmonic function outside the origin, and its slow decay gives the required nonzero resultant force. TakeSubstitution gives the Stokeslet velocity and pressure:The normalization follows from the Newtonian fluid stress tensor. Differentiating the velocity gives the rate of strain and vorticity of a Stokeslet:Here $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$. Thus ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}=-3(\mathbf{F}\cdot \mathbf{x})x_i{x}_j/(4\pi r^5)$. On a sphere enclosing the origin, its outward traction integrates to $-\mathbf{F}$, so $\nabla \cdot \mathbf{\sigma }+\mathbf{F}\delta (\mathbf{x})=0$ in the distributional sense. All displayed fields away from the point force have $r>0$.

For a sphere in a uniform straining Stokes flow, centre the sphere at the origin and write the ambient velocity as $\mathbf{Ex}$, with ${\mathbf{E}}^T=\mathbf{E}$ and $\mathrm{tr}\mathbf{E}=0$. Faxén's first law and Faxén's rotational law give zero translation and rotation. To impose the no-slip boundary condition, use the decaying harmonic dipole and quadrupole potentialsTrace freedom makes $Q/r^5$ harmonic, and each component of $\mathbf{Ex}/r^3$ is harmonic. The Papkovich–Neuber representation givesAt $r=a$, the coefficient of $\mathbf{Ex}$ must give ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$, while the extra radial term must vanish. Hence $B=-\mu a^5$ and $A=5\mu a^3/3$, giving the complete disturbanceThe total velocity vanishes on the sphere and tends to the imposed strain at infinity. This field has no Stokeslet or rotlet contribution, consistent with a force-free, torque-free sphere. Its leading far field is the radial stresslet $-5a^{3}Q\mathbf{x}/(2r^5)$.

Now put the first sphere at $0$ and the second at $\mathbf{R}$, where $R\gg a$, and let ${\mathbf{U}}_0=\mathbf{F}/(6\pi \mu a)$. The first sphere's leading disturbance is its Stokeslet. The rate-of-strain tensor and vorticity incident on the second sphere areThe second sphere translates with the local flow and rotates with half its local vorticity. Its force-free and torque-free conditions therefore remove any reflected force or torque monopole. The incident strain produces the stresslet just obtained. Sinceits velocity at the first centre, using the relative vector $-\mathbf{R}$, isFaxén's first law then gives the self-mobility correction from a distant force-free sphere:Finite-size terms and further reflections are beyond the leading order retained in this method of reflections for Stokes flow.

Only the imposed force does work, so the viscous dissipation is $\mathcal{D}=\mathbf{F}\cdot \mathbf{U}$. With $\zeta =6\pi \mu a$, define ${\epsilon }_R=a/R$ and $\gamma =15a^4/(4R^6)$. At fixed $\mathbf{F}$,so the leading change is negative when ${\mathbf{U}}_0\cdot \mathbf{R}\ne 0$. The direction is held fixed in taking ${\epsilon }_R\to 0$.

The fixed-force comparison of minimum viscous dissipation distinguishes the prescribed force from prescribed boundary velocity. Indeed,and henceAt the actual reduced velocity, adding the second sphere increases the dissipation relative to an isolated translating sphere at that same velocity, exactly as required by the Minimum-dissipation theorem for Stokes flow. One can extend the two-sphere velocity rigidly into the second sphere, adding zero strain and producing an admissible trial field in the domain without it. The reflected angular velocity of the first sphere contributes only a higher-order squared rotational power to this comparison.

Finally hold the second sphere's angular velocity at zero. Faxén's rotational law requires the applied coupleThe resulting rotlet adds at the first centreThis is the additional rotational constraint correction to sphere mobility. Including the strain reflection, the total leading change relative to the isolated velocity isFor motion parallel to the line of centres the constraint adds nothing at this order; for transverse motion it supplies an extra resisting correction.

Take $\mathbf{F},\mathbf{G}$ to be the force and couple exerted by the body on the fluid, equivalently the external force and couple needed to maintain its motion. This fixes the sign for a positive hydrodynamic resistance matrix. If one uses the fluid's force on the body, both resultants have the opposite sign.

For two Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\tilde{\mathbf{u}},\tilde{\mathbf{\sigma }})$ in the same fluid domain $\mathcal{D}$, with zero body force, the Lorentz reciprocal theorem for Stokes flow isHere $\mathbf{n}$ points out of the fluid. To see the identity, take the divergence of the difference of the two cross-work fluxes. The stress divergences vanish, while incompressibility and symmetry of the Newtonian fluid stress tensor reduce the remaining terms to $2\mu (\mathbf{e}:\tilde{\mathbf{e}}-\tilde{\mathbf{e}}:\mathbf{e})=0$. The divergence theorem proves the result.

On the body, the no-slip boundary condition is $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{r}$, and the far-field contribution vanishes for decaying exterior flows. Reciprocity becomesWith $q=(\mathbf{U},\mathbf{\Omega }{)}^T$ and $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}q$, this is $q^T\mathsf{R}\tilde{q}={\tilde{q}}^T\mathsf{R}q$ for all pairs, so ${\mathsf{R}}^T=\mathsf{R}$. The power identity givesEquality would force zero strain throughout the connected exterior domain, hence a rigid fluid motion. Decay at infinity makes that motion zero, and the no-slip boundary condition then gives $q=0$. This proves the symmetry and positivity of a rigid-body resistance matrix:

For the helix, assume $0<\varphi <\pi /2$ and use cylindrical unit vectors. Its arclength element and tangent areIntegrating to $\theta =L\cos \varphi /b$ gives $\boxed{\text{total arc length}=L}$. In a combined axial translation and rotation, $\mathbf{V}=U{\mathbf{e}}_z+b\Omega {\mathbf{e}}_{\theta }$. The slender-body force density, or local resistive-force theory, givesWriting $s_{\varphi }=\sin \varphi$, $c_{\varphi }=\cos \varphi$, its axial and azimuthal components areThe axial couple density is $b{f}_{\theta }$. Integration over arclength gives the axial resistance matrix of a slender helix:Thus pure translation gives $(F_z,G_z)=(AU,BU)$, and pure rotation gives $(F_z,G_z)=(B\Omega ,D\Omega )$. The mixed coefficients agree, as reciprocity requires, and $AD-B^2=C^2{L}^2{b}^2/2>0$. The sign of $B$ follows the handedness in the parametrization; reversing handedness reverses propulsion.

For the helical microswimmer with a spherical head, the diagram's rotation relation is ${\Omega }_0=\Omega -\omega$. Neutral buoyancy and the absence of external forcing make the whole swimmer force-free and torque-free. Neglecting interactions between its parts givesWith $\Delta =(A_0+A)(D_0+D)-B^2>0$, solving this pair givesThe head counterrotates, providing the reaction to the flagellum's rotation. The figure fixes this relative rotation convention; it is duplicated and corrupted in the local TeX.

For a very large head, the coefficient regime is $A_0\gg A$ and $D_0\gg D$. ConsequentlyThe large translational resistance $A_0$ makes the microswimmer slow even though the flagellum rotates almost at the motor rate.

For a very small head, $A_0\ll A$ and $D_0\ll D$. The appropriate denominator retains the translation-rotation coupling:Both tend to zero with $D_0\propto a^3$, while ${\Omega }_0\simeq -\omega$. The motor mainly rotates the low-resistance head, and produces little flagellar motion relative to the fluid.

In the intermediate coefficient regime $A_0\ll A$ and $D_0\gg D$, the head supplies a strong rotational reaction with little translational penalty. ThenThe head radius has dropped out. To find the optimal pitch of a helical microswimmer, put $z=\tan \varphi$. The dimensionless speed becomes $z/(z^2+2)$ and has derivative $(2-z^2)/(z^2+2{)}^2$. Hence, for $\omega >0$ with the chosen handedness,The angle is measured from the horizontal plane. An intermediate pitch combines axial and transverse tangent directions, allowing anisotropic drag to convert rotation into translation; either extreme removes that coupling.

The geometric head-size regimes quoted in the paper hold with the logarithmic slenderness factor treated as fixed. More precisely, $A_0\ll A$ requires $a\ll L(1+{\cos }^2\varphi )/(3|\log ϵ|)$, while $D_0\gg D$ requires $a^3\gg L{b}^2(1+{\sin }^2\varphi )/(4|\log ϵ|)$. These coefficient inequalities state the validity of the intermediate approximation when the logarithm is quantitatively important. The many-turn and local-drag assumptions of slender-body theory also remain in force.

For the stated axisymmetric flow, the diagonal components of the rate-of-strain tensor areTheir sum is $u_r+u/r+w_z=0$, the incompressibility condition. In the leading thin-sheet approximation, vanishing tangential traction makes $u$ independent of $z$. The normal stress boundary condition is ${\sigma }_{zz}=-p_{\mathrm{ext}}$, so the Newtonian fluid stress tensor givesThereforeFor the small annular sector, the inner and outer radial faces contribute $2\delta \theta {\partial }_r(rh{\sigma }_{rr})\delta r$ in the radial direction. The two azimuthal faces contribute $-2\delta \theta h{\sigma }_{\theta \theta }\delta r$: their hoop tractions have inward radial components. The combined radial force of the external pressure on the sloping upper and lower surfaces is $2\delta \theta r{p}_{\mathrm{ext}}{h}_r\delta r$.

Neglecting inertia, force balance is thusSubstituting the two stresses cancels the terms involving $p_{\mathrm{ext}}{h}_r$ and gives the axisymmetric viscous-sheet stretching equations:Finally, conservation of mass in the sector gives

WriteThe velocity gradient of the solid-rotation part is antisymmetric, so it drops out of the Newtonian fluid stress tensor. Sinceand $p=0$, the stress isEquivalently, with $\mathbf{n}=\mathbf{x}/r$,

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

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.

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

The broad faces have curvature zero to leading order, so their stress boundary condition gives ${\sigma }_{xx}=-p_e$. The semicircular edges have curvature $1/h$, giving ${\sigma }_{yy}=-p_e-\gamma /h$. Incompressible flow gives $u_x+v_y+w_z=0$; uniform transverse normal stresses imply uniform transverse extension rates, and eliminating them from the Newtonian fluid stress tensor yields

At an edge, the kinematic boundary condition balances axial advection of $b$, lateral strain, and capillary retraction. This givesThe equation $hbw=Q$ expresses conservation of volume flux, whilestates that the total axial tension is constant. DefineThen $[\Gamma ]=L^{-2}$ and $[T]=L^{-1}$, and substitution of $h=Q/(bw)$ givesSubtracting these logarithmic-derivative equations givesThe remaining width equation isThus, for $T\ne 0$,while the continuous $T=0$ limit is $1/b=1/b_0+\Gamma z/6$.

If $\Gamma =0$, then $b/b_0=e^{-Tz/6}$, $w/w(0)=e^{Tz/3}$, and mass conservation gives $h/h(0)=e^{-Tz/6}$. Therefore a prescribed thinning ratio $R=h(L)/h(0)$ requires

The momentum equation isTaking its scalar product with $u$ givesFor the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, its contraction with the velocity gradient isbecause incompressibility removes the pressure term and the symmetric rate-of-strain tensor is orthogonal to the antisymmetric part of $\nabla u$. Integration over $\mathcal{D}$ and the divergence theorem therefore give the kinetic-energy balance for an incompressible Newtonian fluid:The terms are respectively the rate of change of kinetic energy, outward advective flux of kinetic energy, power supplied by the body force, power supplied by surface traction, and irreversible viscous dissipation. Since $f=-\nabla \psi$, the body-force power may also be written $-\nabla \cdot (\psi u)$ and interpreted as exchange with potential energy.

Differentiating the Stokeslet gives its rate-of-strain tensor and vorticity, displayed above for $r=|\mathbf{x}|>0$. The pressure is $p=\mathbf{F}\cdot \mathbf{x}/(4\pi r^3)$. Integrating the Newtonian fluid stress tensor over a sphere enclosing the origin gives traction resultant $-\mathbf{F}$, fixing the point-force normalization.

A spherical inviscid bubble moving slowly through a viscous liquid has a dipolar surfactant perturbation. The associated Marangoni stress reduces the tangential velocity from its clean-bubble value towards zero. In the linear regime, with ${\mathbf{u}}_s=A(I-\mathbf{nn})\mathbf{U}$, the consistent Newtonian fluid stress tensor gives $A=-1/[2(1+2\lambda )]$.