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A nearly confined cylinder in Couette flow is driven by the unequal lubrication resistances of its two gaps. If their minimum widths are proportional to $1+\lambda$ and $1-\lambda$, force balance weights the relative wall velocities by the inverse square roots of those widths.

For inviscid plane Couette flow $U=z$ between impermeable walls at $z=\pm 1$, the nonzero real streamwise wavenumber $\alpha$ gives no nontrivial smooth discrete vertical-velocity modes. Instead, the vorticity variable $q=(D^2-{\alpha }^2)w$ obeys $({\partial }_t+i\alpha z)q=0$. Localized vorticity sheets give generalized eigenfunctions with real frequencies ${\omega }_{\xi }=\alpha \xi$. Their closure forms a neutral continuous spectrum.

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.

For this steady unidirectional flow, the convective acceleration $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanishes because $u$ depends only on $y$. The axial Navier-Stokes equation reduces to $0=-p_x+\mu u^{\prime \prime }$, with $p_x=-G$. The no-slip boundary conditions therefore giveThis is a superposition of Couette flow and pressure-driven Poiseuille flow.

Integrate the axial ordinary differential equation twice: $u=-G{y}^2/(2\mu )+C_{1}y+C_2$. The no-slip boundary condition at zero gives $C_2=0$, and the one at $h$ gives $C_1=U/h+Gh/(2\mu )$. HenceThe first term is Couette flow; the second is Poiseuille flow driven in positive $x$ when $G>0$. Their addition is valid because the reduced Stokes equation is linear.

This is a squeezing motion, not the rotational motion of the previous part. The gap changes at rate $h_t=V\sin \theta$. The Reynolds lubrication equation therefore givesThe outer wall also has tangential speed $V\cos \theta$, but its local Couette flow flux is $O(V\Delta )$ compared with the squeezing flux $O(Va)$ and does not affect the leading force. Thus $p_{\theta }=-12\mu aq/h^3$ at leading order. Periodic pressure gives $C=0$, because the integral of $\cos \theta /h^3$ vanishes over a full circle. Balancing pressure traction on the fixed inner cylinder and using integration by parts,The supplied integral gives the squeeze resistance of an eccentric journal bearing:The fluid tends to pull the inner cylinder upwards with the moving outer wall, so a downward holding force is required when $V>0$. At concentricity the translation resistance is $R_0=12\pi \mu a^3/{\Delta }^3$. The omitted wall-tangential flux and shear traction give smaller thin-gap contributions.

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.

Let the upper plate in flow A move at speed $U$, with the lower plate fixed. The Couette flow profile isso its mid-plane speed is $U/2$ and its total viscous dissipation per unit plate area is

Write $G=-dp/dx>0$ for the constant pressure-gradient magnitude in flow B. The plane Poiseuille flow profile isEquality of the mid-plane speeds givesSince $u_B^{\prime }(y)=G(h-2y)/(2\mu )$,Hence

When $\Delta \ll 1$, the small-argument expansion $\sinh z=z+O(z^3)$ givesThereforeThe viscous diffusion time $L_0^2/\nu$ is then short compared with the oscillation period. Viscosity communicates the wall motion across the entire layer almost instantaneously, producing the linear profile of quasi-steady Couette flow.

Near the closest point, the cylinder–plane gap isFor translation parallel to the cylinder axis, the leading flow is Couette flow. Its shear traction is $\mu U/h$, soHence $A_{yy}=2\pi {\epsilon }^{-1/2}\mu L$.

For transverse translation, the local Couette-Poiseuille flow in a thin gap and mass conservation give the Reynolds lubrication equation. Writing $x=a\sqrt{\epsilon }X$ and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce towhere the final term is the direct Couette shear contribution. Therefore

On $r=c$, the traction $\mathbf{\sigma }\mathbf{n}$ exerted by the outer fluid on the inner fluid isThe inner fluid exerts the opposite traction on the outer fluid. Its couple is thereforeUsingand the supplied spherical integral,we obtainThus the radius cancels:This is the Torque in rotational Stokes flow between concentric spheres.

When $a\ll b$,the rotational drag torque for a sphere in an effectively unbounded fluid. If $h=b-a\ll a$, thensothe inverse-gap growth expected from a thin Couette flow.