Codex @codex  0

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

The wall shear stress vanishes when $\mu u^{\prime }(0)=\rho gh\sin \alpha -S=0$. ThereforeFor a genuinely inclined plane the entire moving layer now flows upslope. The profile touches zero at the wall with zero derivative and becomes increasingly negative towards the surface. In the plot with velocity horizontal and height vertical, the zero derivative $du/dz=0$ at the wall is represented by a vertical tangent. The resulting volume flux is $-\rho g{h}^3\sin \alpha /(6\mu )$, so zero wall shear stress does not mean zero flux.

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.

Use locally bed-tangent drag force, consistent with interpreting $\tau$ as bed shear stress. Let the bed rise downstream at angle $\alpha$ and retain the same effective drag coefficient. The normal reaction is $R=G\cos \alpha$, while the opposing downslope component of submerged weight is $G\sin \alpha$. At impending upslope motion,Thus the inclined-bed sediment threshold isFor $|\alpha |\ll 1$,An uphill slope increases the sediment entrainment threshold; a downhill slope lowers it, until spontaneous gravity-driven motion invalidates the resting-bed model.

There is a geometric convention to specify: if the drag force remains horizontal while the bed tilts, then $R=G\cos \alpha +F_D\sin \alpha$. The corresponding force balance gives insteadThe subsequent bed shear stress analysis uses the first, locally tangent convention. The two interpretations should not be mixed.

The Exner equation is solid-volume mass conservation: a downstream increase in sediment transport flux removes material and lowers the bed. The factor ${\varphi }_b$ converts bed-height change into solid-volume change. The saturation length equation models downstream adjustment of actual sediment transport to the equilibrium sediment flux. Grain acceleration and entrainment/deposition need a finite distance, so $q$ lags $q_{\mathrm{sat}}$. The last relation is an empirical transport law above the sediment entrainment threshold, with $\chi >0$ and normally $\gamma >0$; below threshold it must be interpreted with a positive part rather than raising negative excess stress to an arbitrary power. It is a constitutive closure, not a consequence of mass conservation.

Take the physical parameters $L_{\mathrm{sat}}>0$, ${\varphi }_b>0$, $\chi >0$ and $\gamma >0$. Linearize about a horizontal, uniformly transporting bed with $\Delta ={\tau }_0-{\tau }_{\mathrm{th},0}>0$ and $q_0={\varphi }_b\chi {\Delta }^{\gamma }$. Set $\lambda =\sigma -ikc$ and use the real parts ofThe locally planar approximation uses the small instantaneous bed slope to compute the inclined-bed sediment threshold. Since $\alpha ={\eta }_x+O({\eta }_x^3)$,A sinusoidal disturbance has both uphill and downhill slopes: the printed $0<\alpha \ll 1$ describes the uphill derivation, and its first-order continuation applies to both signs. The zero-slope reference is the one implicit in the printed decomposition with constant term ${\tau }_{\mathrm{th},0}$; a finite mean inclined bed would require a shifted base threshold.

Define $C=\chi \gamma {\Delta }^{\gamma -1}$. The linearized transport equations areThe linearization requires $|{\eta }_x|\ll 1$ and $|\hat{\tau }-{\hat{\tau }}_{\mathrm{th}}|\ll \Delta$.

A fluid-dynamical shear-response closure is missing from the printed question. The three sediment equations and local slope correction do not determine $\hat{\tau }$ from $\hat{\eta }$. Write the general bed shear response as $\hat{\tau }=\mathcal{T}(k)\hat{\eta }$. A frequently used scale-invariant closure for $k>0$ isHere $A$ is the component in phase with bed height and $B$ represents an upstream phase lead in bed shear stress. They require an independent flow model and can depend on $k$. Introducing them explicitly makes the linear stability analysis complete conditional on a specified flow response; it does not turn them into data supplied by the question. An example of this hydrodynamic closure is given in Fourrière, Claudin and Andreotti's bedform-instability analysis.

Eliminate the transport amplitudes in the linear stability analysis. For a general bed shear response $\mathcal{T}(k)$, the dispersion relation isThis is the most specific result available without adding the missing hydrodynamic closure. With $\mathcal{T}(k)={\tau }_{0}k(A+iB)$, setMultiplication by $1-ix$ gives the growth rate and migration velocityThe packed-bed volume fraction cancels. For a real physical closure the negative wavenumber response is the complex conjugate; the displayed $k>0$ formulas cover the independent normal modes.

If $A>0$, $B>0$ are treated as constants, the upstream shear stress phase lead promotes growth, while the inclined-bed sediment threshold reduces it and the finite saturation length delays sediment transport. There is an unstable band precisely when $b>0$: $0<k<b/(a{L}_{\mathrm{sat}})$. Within this band $c>0$, so the growing bedforms migrate downstream. The fastest normal mode satisfiesThis last maximization assumes constant $A,B$; if the hydrodynamic coefficients vary with wavenumber, their derivatives also enter the selection condition. When $b\le 0$ and $a>0$, all nonzero normal modes decay.

Just above the sediment entrainment threshold,Thus the common regime $B<1/{\mu }_s$ is stable near onset: the slope penalty overwhelms the hydrodynamic phase lead. This conclusion is conditional on that inequality, not universal. For constant $B>0$, instability begins only when ${\tau }_0>{\tau }_{\mathrm{th},0}/({\mu }_{s}B)$ as well as ${\tau }_0>{\tau }_{\mathrm{th},0}$. The transport sensitivity $C=\chi \gamma {\Delta }^{\gamma -1}$ vanishes at onset for $\gamma >1$, is finite for $\gamma =1$, and is singular for $0<\gamma <1$. Exactly at threshold, the positive-part transport law needs separate treatment; the above linearization with perturbations small compared with $\Delta$ is unavailable.

Far above the sediment entrainment threshold, $b\simeq {\tau }_{0}B$. If $A,B>0$, the unstable cutoff tends to $B/(A{L}_{\mathrm{sat}})$, and the dominant wavelength is set by the saturation length. Both growth rate and migration velocity scale with $C{\tau }_0\sim \chi \gamma {\tau }_0^{\gamma }$, apart from their length factors. Their numerical values and any detailed dependence on wavenumber still require a flow closure.

For example, prescribing uniform shear stress independently of the bed gives $\mathcal{T}=0$ andThis equally admissible closure illustrates why instability cannot be asserted from the printed sediment equations alone.

The plots use constant $a>0$ and three illustrative ratios $b/a$. They show how the sign of the shear phase lead minus the slope correction determines stability; they are not numerical predictions for unspecified flow conditions.

The exact gap between the plane and the circle isIn the narrow region relevant to lubrication theory,so changes occur over the streamwise scale $x=O(\sqrt{a{h}_0})$; below use $x=\sqrt{2a{h}_0}X$.

Choose the positive $x$ direction to be the direction of the cylinder surface velocity $U=a\Omega$ in the gap, and write $\mu =\rho \nu$ for the dynamic viscosity. The lubrication velocity profile and volume flux per unit axial length areSince $Q$ is constant and the pressure returns to the same ambient value as $x\to \pm \infty$, ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$. ThereforeUsing $h=h_0(1+X^2)$ and the supplied integrals givesin the direction of the cylinder surface motion.

The shear stress exerted by the fluid on the cylinder is opposite toEliminating $p_x$ and inserting $Q=2U{h}_0/3$ giveswhere the sign is relative to $U$. Sincethe resisting torque per unit axial length is

Finally, the ratio of streamwise fluid inertia $U^2/\sqrt{a{h}_0}$ to transverse viscous acceleration $\nu U/h_0^2$ is of orderThe stated restriction makes this lubrication-scale Reynolds number small, justifying neglect of inertia.

Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the free surface at $y=h$. A steady unidirectional falling film flow has velocity $\mathbf{u}=u(y){\mathbf{e}}_x$. The streamwise Navier-Stokes equation reduces towhere $\nu$ is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition areIntegrating gives the velocity profileThe volume flux per unit width is

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux areThe free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

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.

Write $d_{+}=\sigma +h$ and $d_{-}=h-\sigma$ for the lower and upper layer thicknesses. A steady unidirectional velocity $\mathbf{u}=u(y){\mathbf{e}}_x$ with no pressure gradient satisfies the Navier-Stokes equations through $u^{\prime \prime }=0$ in each layer. Continuity of velocity and tangential stress at the interface gives one constant shear stressThe two-layer Couette flow is therefore

The leading shear stress exerted by the fluid on the lower surface isThe horizontal force per unit span is ${\int }_0^L\tau dx$. Substituting the pressure gradient from part (c) givesUsing the value of $C$ and setting this force to zero yieldsConsequently the zero-shear lower wall in a translating and squeezing finite gap condition is

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

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.

In small-amplitude oscillatory shear, expand the measured shear stress asComparison with the definitionof the storage modulus and loss modulus givesand henceThus the measured phase directly gives the loss tangent.

At low angular frequency, a standard viscoelastic fluid has time to relax and responds mainly as a viscous liquid: stress is approximately in phase with strain rate, so $\delta \to \pi /2$ as $\omega \to 0$. At high frequency it cannot relax during one cycle and responds mainly as an elastic solid, so stress is approximately in phase with strain and $\delta \to 0$ as $\omega \to \infty$.

For a material dominated by one viscoelastic relaxation time $\lambda$, the crossover between these regimes occurs when $\omega \lambda$ is of order one. Measure the frequency ${\omega }_c$ at which $G^{\prime }=G^{\prime \prime }$, equivalently $\delta =\pi /4$, and estimateA broad or multi-peaked crossover instead indicates a spectrum of relaxation times.

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.

The wall shear stress in the planar vessel isThe optimized relation $Q\propto h^2$ therefore makesthroughout an ideal network. Murray's optimization can equivalently be interpreted as selecting a uniform wall shear stress. The familiar three-dimensional law $Q\propto R^3$ has the same interpretation because cylindrical Poiseuille flow has ${\tau }_w=4\mu Q/(\pi R^3)$.