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Uniform axial strain rate $E$ in an annular Newtonian fluid gives radial velocity $u_r=-Er/2+C/r$. For radii $R_i,R_e$, area parameter $A=R_e^2-R_i^2$, and constant gas pressures,The opposite signs of inner and outer curvature in the Young–Laplace equation give these expressions. The kinematic boundary condition implies $\dot{A}=-EA$ and

A slender axisymmetric column of a Newtonian fluid, with radius $a(z,t)$, axial velocity $w$, external pressure $p_e$ and axial body-force density $f$, obeys $(a^2{)}_t+(a^{2}w{)}_z=0$ and $3\mu (a^2{w}_z{)}_z/a^2=p_{e,z}-f$ when inertia, surface tension and leading external shear are negligible. The factor three is the Trouton ratio. Radial mass conservation gives $u_r=-r{w}_z/2$; the Newtonian fluid stress tensor gives $p=p_e-\mu w_z$ and axial excess stress $3\mu w_z$. This reduces a three-dimensional Stokes flow to one-dimensional evolving geometry.

For a thin Newtonian fluid film on a conical slope at angle $\alpha$, neglecting the thickness-gradient contribution to the driving pressure gives volume flux per unit width $q=D{h}^3$, where $D=g\sin \alpha /(3\nu )$. Radial mass conservation isThe distance $x$ is measured along the slope; the horizontal radius is $x\cos \alpha$, so total volume is $2\pi \cos \alpha \int xhdx$.

A linear basal drag law has resisting traction $-\mu \beta u_b$, with $\beta >0$ of dimension inverse length. It idealizes lubrication by subglacial till and is equivalent to a Navier slip boundary condition of slip length $1/\beta$ when the ice is a Newtonian fluid. It is a constitutive idealization, not a universal sliding law.

Mantle convection is the slow buoyancy-driven motion of a rocky planetary mantle. On geological timescales the solid mantle behaves as a highly viscous Newtonian fluid to first approximation.

A Navier slip boundary condition relates fluid velocity relative to a stationary flat wall to the wall-normal tangential velocity gradient, $u_b={\ell }_s{u}_z$, with positive slip length ${\ell }_s$. For a Newtonian fluid, it is equivalent to linear resisting wall traction of magnitude $\mu u_b/{\ell }_s$. The normal points from the wall into the fluid; this fixes the sign convention.

For isotropic incompressible flow of a Newtonian fluid, the stress tensor is the displayed expression, with pressure $p$, dynamic viscosity $\mu$, velocity $\mathbf{u}$, and identity matrix $I$. In Couette flow, its off-diagonal component is the shear stress.

A thin layer of subglacial till idealized as a Newtonian fluid of dynamic viscosity $\lambda \mu$ and thickness $l$ undergoes Couette flow between stationary bedrock and ice moving at speed $u$. Its resisting basal shear stress has magnitude ${\tau }_b=\lambda \mu u/l$ and acts opposite to the sliding motion. Combining this drag with the hydrostatic driving force gives the bulk approximation used in the friction-dominated grounding-line evolution law.

Near the closest point, in $X=x-Ut$, the gap is $h(X)\simeq h_0+X^2/(2a)$. The horizontal length scale is $\ell =\sqrt{2a{h}_0}$, up to an inessential constant. Thus the aspect ratio $h_0/\ell \ll 1$ follows from $h_0\ll a$. The lubrication approximation assumes an incompressible Newtonian fluid, no slip on the surfaces, slow variation of the gap, and negligible inertia relative to transverse viscous stress:This is the reduced Reynolds-number condition; it is weaker than requiring a Reynolds number based on $\ell$ to be small. In the cylinder frame the geometry and flow are steady. Pressure is uniform across the narrow gap to leading order, and equal ambient pressures far to either side close the model. The calculation assumes a full liquid film; cavitation would change that pressure condition.

Take $z$ upward from the horizontal bed and measure $x$ away from an ice divide. In the shallow-ice approximation, the hydrostatic pressure is $p=p_a+\rho g(h-z)$ and the horizontal balance for a Newtonian fluid isThe upper stress-free boundary condition gives $u_z(h)=0$. The resisting basal traction is $-\mu \beta u_b$, so the fluid-side shear stress satisfies $\mu u_z(0)=\mu \beta u_b$. This is a Navier slip boundary condition with slip length ${\ell }_s=1/\beta$. Integrating twice givesThe first term is the uniform basal sliding contribution, while the quadratic term is internal viscous deformation. Both are positive where the thickness decreases downstream.

The shallow-ice flux with linear basal drag isNo ice accumulation or ice ablation appears. For a symmetric glacier, work on one half, $0\le x\le x_N(t)$, with $q(0,t)=0$, $h(x_N,t)=0$ and no outgoing volume flux. Its conserved half-volume per unit width is $V_g={\int }_0^{x_N}hdx$; the full glacier has volume $2V_g$. This specifies the volume convention rather than silently supplying an unspecified value. For a general finite release on the whole line, the late similarity solution is centred on its conserved centre of mass and uses half its total volume for $V_g$.

The internal-shear and linear basal drag law mobilities are equal at $H_{\ast }=3/\beta$. Suitable vertical, horizontal and temporal scales areWrite $h=H_{\ast }\mathcal{H}$, $x=L_{\ast }X$, $t=T_{\ast }\tau$. ThenThe initial typical thickness is $\mathcal{H}\simeq 10$, the initial extent is of order $L_{\ast }/10$, and its shear-controlled spreading time is of order $T_{\ast }/10^5$. The shallow-ice approximation also requires small aspect ratio: in particular the initial thickness and extent must satisfy approximately $100H_{\ast }/L_{\ast }\ll 1$. The equations model the broad shallow bulk; the very tip can require physics beyond that approximation.

The two limiting equations are porous medium equations of the form ${\mathcal{H}}_{\tau }=({\mathcal{H}}^n{\mathcal{H}}_X{)}_X$, with $n=3$ for a thick shear-dominated glacier and $n=2$ for a thin sliding-dominated glacier. Volume conservation and balance of the time derivative give exponent $\alpha =1/(n+2)$. Set $\mathcal{H}={\tau }^{-\alpha }f(\eta )$, $\eta =X{\tau }^{-\alpha }$. One integration, using zero volume flux at the ice divide, gives $f^n{f}^{\prime }=-\alpha \eta f$, henceThe planar volume-conserving nonlinear-diffusion similarity is thereforeThe Beta function fixes its constants:Consequently the explicit early shear and late sliding limits areTheir centre depths are $A_3{\tau }^{-1/5}$ with $A_3\simeq 0.842252$, and $A_2{\tau }^{-1/4}$ with $A_2=\sqrt{2/\pi }$. Thus the early extent grows as $t^{1/5}$ and the late extent as $t^{1/4}$. These are asymptotic regime profiles; they solve the respective limiting equations, not the full sum of both mobilities. For arbitrary finite-width initial data they are appropriate after the corresponding spatial relaxation, with a virtual time origin depending on that data.

The shear-to-slip transition in a shallow ice current occurs at thickness of order $H_{\ast }$. Using equality at the centre of the early similarity solution givesIf the initial profile is already approximated by that early similarity solution and its centre depth is 10, the virtual age is ${\tau }_0=A_3^5/10^5\simeq 4.2385\times 10^{-6}$, and the elapsed estimate is ${\tau }_{\mathrm{tr}}-{\tau }_0$. Other reasonable definitions of typical thickness change the order-one coefficient; the initial thickness alone does not uniquely determine an exact transition time or an exact initial profile. The robust nondimensional estimate is a transition time of order one in the $T_{\ast }$ scale.

Finally, let a local nose translate steadily at positive speed $v$ and write $s=x_N-x\ge 0$. This is a local traveling nose of a glacier with linear basal drag; the speed of the globally spreading glacier changes slowly with time. The local mass conservation equation integrates to $q=vh$, since both $q$ and $h$ vanish at the front. Therefore

The right side is strictly increasing, giving a unique positive thickness at every $s>0$. With $S=X_N-X$ and $c=d{X}_N/d\tau$, the same relation reads $cS={\mathcal{H}}^3/3+{\mathcal{H}}^2/2$. Its limiting profiles areThe square-root tip reflects basal sliding; the cube-root thicker region reflects internal viscous deformation. Even during the early thick regime, the very nose is thin and has the square-root inner form. At late times that sliding balance governs nearly the whole profile. The dimensionless crossover thickness is $\mathcal{H}=1$ and its distance is $S=5/(6c)$; the early thick cube-root outer profile matches this smaller sliding region. The diverging geometric slope at the idealized tip also marks the local limit of the long-wave shallow-ice approximation.

Let $\mu$ be the dynamic viscosity, $u_i={\dot{x}}_i$, and $F_i$ the signed force exerted by sphere $i$ on the fluid. For negligible sphere inertia, this also equals the externally or internally applied force on that sphere; its hydrodynamic force is $-F_i$. The Stokes drag law and the axial velocity of a Stokeslet give the leading hydrodynamic mobility matrixThe factor $1/(4\pi \mu \ell )$ follows from $(I+\hat{\mathbf{r}}\hat{\mathbf{r}})\mathbf{F}/(8\pi \mu \ell )$ when $\mathbf{F}$ is parallel to the line of centres. The longitudinal two-sphere mobility retains the first interaction in $a_i/\ell$; finite-size and repeated-reflection corrections are higher order.

For the linked force-free pair, $F_2=-F_1$. Put $X=(x_1+x_2)/2$ and $S_m=m_1+m_2$. ThensoThe coefficient depends only on the current separation. For any period $T$, $\ell (T)=\ell (0)$, and thereforeThis is a closed integral of a single-valued function of one real shape coordinate. Since $x_1=X-\ell /2$ and $x_2=X+\ell /2$, both spheres also have zero net displacement. Equal radii make $\dot{X}=0$ instantaneously; unequal radii generally permit oscillatory translation, but still no mean motion. This explicit force-free two-sphere stroke is the scallop theorem: a single-parameter reciprocal deformation in a Newtonian fluid at zero inertia cannot produce net free swimming. The timing of extension and contraction cannot change that conclusion, because Stokes flow has no inertial memory.

For the externally prescribed pair, put $q=\omega t$, ${\zeta }_i=1/m_i=6\pi \mu a_i$, andInvert the hydrodynamic mobility matrix:In a large-${\ell }_0$ expansion with $a_i,\delta$ fixed,The isolated-drag term and the constant-$h$ interaction have zero mean. The order-${\ell }_0^{-2}$ term from $h^2{u}_1$ also has zero mean, since its coefficient is constant at that order. Meanwhile,Consequently the externally driven two-sphere pump has the following phase-dependent mean force of an externally driven sphere pair:Here $⟨\cdot {⟩}_t$ denotes averaging over one period. The analogous calculation gives $⟨d{u}_1{⟩}_t={\delta }^2\omega \sin \varphi /2$, so $⟨F_2{⟩}_t=⟨F_1{⟩}_t$ at leading order, andThus the mean fluid forcing is nonzero except at $\varphi =0$ or $\pi$ modulo $2\pi$. The sign changes when the phase lag is reversed. Those exceptional strokes are reciprocal: the position-space loop collapses to a line, and their mean force vanishes by reversibility, not just by this leading expansion.

There is no contradiction with the scallop theorem. The externally imposed motion is not force-free, and the actuators prescribe two phase-shifted motions; their combined motion is not reciprocal for a generic phase. The spheres return to their prescribed positions while transferring a mean force to the fluid. This is pumping by external forcing, rather than propulsion of the freely linked one-shape-coordinate system.

The asymptotic calculation requires $\delta /{\ell }_0\ll 1$ and persistent large separation. The minimum separation of phase-shifted sphere oscillations isThe printed condition ${\ell }_0\ge \delta$ alone does not guarantee nonoverlap or the stipulated far-separated regime. The prescribed trajectories must additionally keep this minimum well above both radii. This is a compatibility qualification on the data, not a change of the phase-dependent force calculation.

In resistive-force theory (RFT), the hydrodynamic force per unit arclength exerted by the fluid on a slender filament isHere $\mathbf{t}$ is the unit tangent and $\mathbf{u}$ the local velocity relative to the background fluid. The parallel and perpendicular drag coefficients of a slender filament are positive and generally satisfy ${\xi }_{\perp }>{\xi }_{\parallel }$; their leading logarithmic ratio is about two. This local approximation assumes a small filament radius compared with length and curvature scales, negligible fluid inertia, and a Newtonian fluid. It represents drag by the local tangent direction and neglects nonlocal hydrodynamic interactions between separated filament segments. Boundaries, close approaches and end corrections may require slender-body theory or a more complete flow calculation. In this problem the background fluid is at rest, and the drag coefficients are taken uniform along the filament.

The rigid-body velocity in a deforming swimmer frame is the sum of translation, rotation and material deformation. With $\mathbf{r}=x{\mathbf{e}}_x+ϵg{\mathbf{e}}_y$,so, in instantaneous swimmer-frame components,The reference-frame conditions $g(0,t)=g_x(0,t)=0$ attach the origin and orientation to the filament's base and base tangent. They prevent arbitrary shape translations or tilts from being absorbed into the definition of $g$.

The exact tangent and arclength element areSince $U,V,\Omega =O(ϵ)$, the local velocity is $O(ϵ)$. Changing the drag tensor by its $O(ϵ)$ tangent correction therefore changes $\mathbf{f}$ only at $O({ϵ}^2)$. The arclength correction is smaller still at this order. Also $-ϵ\Omega g=O({ϵ}^2)$. Thus $\mathbf{t}={\mathbf{e}}_x$ is sufficient at order $ϵ$. This assumes the small-slope expansion is uniform, $ϵ|g_x|\ll 1$.

Put $G_0(t)=⟨g_t{⟩}_x$ and $G_1(t)=⟨x{g}_t{⟩}_x$, with $⟨h{⟩}_x=L^{-1}{\int }_0^{L}hdx$. The first-order local force isIts total hydrodynamic forces areTaking the moment about the swimmer-frame origin, $M_z=\int (x{f}_y-ϵg{f}_x)ds$, the second term is beyond first order. HenceFor force-free and torque-free motion, the three leading coefficients vanish. The longitudinal force immediately gives $U_1=0$. The two transverse equations areTheir determinant is $L^2/12>0$. Solving yields the first-order free swimming of a planar filament:The resistive-force theory coefficient cancels because both equations use the same transverse drag. Another interpretation is the least-squares projection of filament deformation velocity: $-V_1-{\Omega }_{1}x$ is the best affine approximation to $g_t$ on $[0,L]$. Zero transverse force and torque mean that the residual is orthogonal to $1$ and $x$.

If $g$ is sufficiently differentiable and periodic with period $T$, each $g_t(x,t)$ has zero temporal mean. More explicitly,Both bracketed quantities return to their initial values over a period. Thus the first-order periodic transverse swimming velocity satisfiesThese results concern first order and swimmer-frame components. Changes of orientation can affect laboratory displacements at second order; higher-order net swimming is not ruled out.

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ are the symmetric rate-of-strain tensors. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.

In components, with derivatives ${\partial }_j{u}_i$, use the product rule to compute the divergence of the cross-work flux:The pressure term vanishes by incompressibility; the antisymmetric part of the velocity gradient has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. ThusIntegrate over $V$ and apply the divergence theorem. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the Lorentz reciprocal theorem for Stokes flow:This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.

Let $\ell =(H_m-H_0)/\sin \alpha$ be the along-slope distance to the snowline, so $H(x)=H_m-x\sin \alpha$ and $a(x)=A(1-x/\ell )$. Define $D=g\sin \alpha /(3\nu )$.

Treat ice as an incompressible Newtonian fluid, neglect inertia, and use lubrication theory with thickness measured normal to the slope. The bed has a no-slip boundary condition, the free surface has zero tangential stress, and normal pressure is hydrostatic. The small thickness slope allows its pressure-gradient contribution to be neglected against gravity along the mountain. With normal coordinate $y$, the tangential equation and boundary conditions areIntegration givesThe horizontal radius of a ring is $x\cos \alpha$, so mass conservation gives the gravity-driven ice flow on a conical slope equationNegative accumulation is ice ablation and applies only where ice exists; the ice-free region has $h=0$.

In a steady state, regularity and zero total flux at the apex require $xD{h}^3\to 0$ as $x\to 0$. Integrating givesRequiring a continuous zero-thickness steady terminus yields the steady conical ice cap with a linear accumulation gradient:The terminus lies below the snowline, allowing the ablation region to balance snowfall. The maximum thickness occurs at $x=3\ell /4$.

The volume follows from integrating the ring areas. With $s=x/\ell$,Substituting $D$ and $\ell =\Delta H/\sin \alpha$ givesThe ideal outer profile has steep slopes very close to the apex and terminus. Those small regions require local corrections to the assumed slope balance, while the bulk profile and leading volume follow from the stated approximation.

Murray's law minimizes the sum of the power needed to pump a Newtonian fluid and the metabolic power needed to maintain blood volume. For a cylindrical vessel of radius $R$, length $\ell$, and prescribed volume flux $Q$, Hagen-Poiseuille flow givesIf maintenance costs $\alpha$ per unit volume,Setting the derivative of ${\mathcal{P}}_{\mathrm{pump}}+{\mathcal{P}}_{\mathrm{met}}$ with respect to $R$ to zero givesThus $Q\propto R^3$. Conservation of volume flux at a bifurcation gives

A force-free swimmer with a reciprocal shape stroke cannot achieve net locomotion in an inertia-free Newtonian fluid. A single real shape coordinate that retraces an interval gives a reciprocal cycle. The conclusion concerns free swimming; externally imposed translations can instead force and pump the fluid.

Subglacial till is a mixture of rock fragments and fine sediment beneath ice. A wet deformable layer can lubricate basal sliding. Its constitutive equation must be specified; Newtonian till lubrication is one idealization, rather than a claim that all subglacial till is a Newtonian fluid.

For a two-dimensional Newtonian fluid ice sheet attached to an unbuttressed ice shelf, ice-sheet flotation gives $\rho H={\rho }_{w}b$. The difference between the depth-integrated ice and ocean hydrostatic pressures is $\rho g{H}^2/2-{\rho }_{w}g{b}^2/2=\rho g^{\prime }{H}^2/2$, where $g^{\prime }=g(1-\rho /{\rho }_w)$. Shelf force balance integrates $4\mu H{u}_x-\rho g^{\prime }{H}^2/2$ to a constant; absence of a buttressing force makes that constant zero. The shallow-shelf approximation therefore gives the displayed boundary condition, or $u_x=g^{\prime }H/(8\nu )$ with kinematic viscosity $\nu =\mu /\rho$.

A velocity profile records the variation of velocity across a spatial coordinate, for example $u(z)$ across a parallel viscous layer. Its derivative sets the tangential shear stress in a Newtonian fluid, while its integral gives the volume flux.