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Rotation about the offset axis has azimuthal Couette flow and no pressure variation at leading lubrication order. Integrating the tangential traction gives torque $-{\mathcal{R}}_{\Omega }\Omega$ on the inner sphere. The concentric limit is $8\pi \mu a^4/(3\Delta )$ and the contact limit is finite, $4\pi \mu a^4/\Delta$, because the wall velocity vanishes at the closest pole.

Consider steady flow of a Newtonian fluid in incompressible flow with constant dynamic viscosity $\mu >0$ between parallel plates at $z=0$ and $z=H$. The bottom plate is stationary and obeys the Navier slip boundary condition with slip length $b$. The top plate moves at velocity $U>0$ and obeys the no-slip boundary condition. There is no streamwise pressure gradient or body force. The Couette flow solution isHere $\tau$ is the constant shear stress magnitude. The lower wall's Navier slip boundary condition holds exactly because $b{u}^{\prime }(0)=bU/(H+b)=u(0)$. At the top, $u(H)=U$ is exactly the prescribed no-slip boundary condition; no finite upper-wall slip length is imposed in this example.

For a numerical example, stipulate $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $U=1\mathrm{mm}{\mathrm{s}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. The bottom velocity is $0.167\mathrm{mm}{\mathrm{s}}^{-1}$ and the shear stress is $0.0833\mathrm{Pa}$, compared with $0.100\mathrm{Pa}$ when both plates obey the no-slip boundary condition. The extrapolated profile reaches zero at $z=-2\mu \mathrm{m}$.

For fixed $H$, $U$ and $\mu$, $b\to 0$ recovers $u=Uz/H$. As $b\to \infty$, $u\to U$ throughout the gap and $\tau \to 0$: the moving upper wall selects the limiting uniform velocity, while the lower wall approaches a stress-free boundary condition. The dynamic viscosity stays positive; the disappearing shear stress comes from the disappearing velocity gradient.

A short derivation uses unidirectional flow: the convective acceleration vanishes and the Navier-Stokes equation reduces to $\mu u^{\prime \prime }=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.

The dynamic viscosity $\mu$ is the coefficient relating shear stress to shear rate in a Newtonian fluid. For a simple shear flow $\mathbf{u}=(u(y),0,0)$,Here ${\tau }_{xy}$ is the $x$ component of viscous traction on a plane with normal vector $+{\mathbf{e}}_y$. The shear rate $\dot{\gamma }=du/dy$ has units ${\mathrm{s}}^{-1}$. Thus dynamic viscosity measures the stress required to sustain a given rate of deformation. It is the shear coefficient; the separate coefficient for volume change is volume viscosity. ^[ref]

For a worked Couette flow example, place a Newtonian fluid of constant dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$ between parallel plates separated by $h=0.0010\mathrm{m}$. Hold the lower plate fixed and move the upper plate at $U=0.020\mathrm{m}{\mathrm{s}}^{-1}$. Assume steady flow, no streamwise pressure gradient, and the no-slip boundary condition. ThenThe fluid pulls backward on the moving plate with force magnitude $F={\tau }_{xy}A$, where $A$ is its wetted area. For $A=0.010{\mathrm{m}}^2$ this is $0.020\mathrm{N}$. Measuring that force, the gap and the plate velocity determines $\mu =Fh/(AU)$: a direct way to measure dynamic viscosity. The linear profile and opposing wall tractions are the standard plane Couette flow solution. ^[ref]

The velocity gradient matters, rather than the velocity itself. Adding the same constant velocity to both plates and the fluid leaves $du/dy$, and hence shear stress, unchanged. In uniform flow, every layer translates together and $du/dy=0$. More generally, even a nonzero velocity gradient need not generate viscous stress: solid-body rotation has zero strain-rate tensor, as the viscous stress tensor example shows. Thin gaps amplify shear stress for a fixed plate-speed difference, making dynamic viscosity useful for predicting the force needed to move lubricated surfaces.

A rotating inner cylinder of radius $R$ and height $H$ has a uniform gap $h$ from a fixed housing at its side and both ends. For $h\ll R,H$, no-slip boundary condition and laminar Couette flow give side torque $2\pi \eta \omega R^{3}H/h$. Each end contributes ${\int }_0^R(\eta \omega s/h)(2\pi sds)s=\pi \eta \omega R^4/(2h)$. Summing both end faces proves the displayed expression. Shaft area, corner regions, gas slip and viscous heating are neglected.

A fluid is a material that flows under sustained shear stress and cannot maintain a static shear stress in its fluid state. Liquids and gases are the usual examples. A Newtonian fluid has deviatoric stress proportional to its rate-of-strain tensor, whereas more general fluid constitutive equations can be nonlinear or have memory. In a porous medium, the fluid occupies connected voids within a load-bearing solid framework. The important distinction is the response to a continuing tangential force: an ordinary elastic solid can settle into a fixed deformation, while an ordinary fluid continues to deform. A fluid can nevertheless be at rest when no unbalanced shear stress drives it. ^[ref]

For a worked example, place a Newtonian fluid between parallel plates separated by $h=1\mathrm{mm}$, keep the lower plate fixed and move the upper one at $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$. With no-slip boundary conditions, steady flow, no imposed pressure gradient and negligible edge effects, Couette flow has $u(y)=Uy/h$. For dynamic viscosity $\mu =0.001\mathrm{Pa}\mathrm{s}$, the shear stress magnitude isThe velocity gradient describes neighbouring layers sliding at different speeds. Doubling $U$ doubles the required shear stress; maintaining the same velocity keeps producing deformation, rather than fixing a final shape. This gives viscosity a concrete meaning: resistance to a rate of motion, not a prohibition on motion.

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.

A gliding bacterium moves along a surface without swimming through the bulk fluid. In a thin-fluid-film model, its relative motion dissipates energy through Couette flow, although the actuation mechanism can involve additional dissipation.

For circular Couette flow with $v=Ar+B/r$, the rotating-frame reduction of circular viscoelastic shear gives shear rate $g=-2B/r^2$. The total shear stress is $\mu g$, and the azimuthal minus radial normal-stress difference is $2(\mu -{\mu }_r)\tau g^2$ in the stress convention where ${\sigma }_{rr}=-p$. Radial force balance without inertia is $p^{\prime }=-2(\mu -{\mu }_r)\tau g^2/r$, which integrates to the displayed pressure. The pressure constant is undetermined by these equations.

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.

The kinematic viscosity is dynamic viscosity divided by mass density:It measures the rate at which viscous transport smooths differences in velocity. For constant $\mu$ and $\rho$ in incompressible flow, dividing the Navier-Stokes equation by mass density makes the viscous acceleration $\nu {\nabla }^2\mathbf{u}$, where ${\nabla }^2$ is the Laplacian. ^[ref]

For example, stipulate two Newtonian fluids with the same dynamic viscosity $0.10\mathrm{Pa}\mathrm{s}$ but different mass densities. In the same steady Couette flow as above, both have shear stress $2.0\mathrm{Pa}$, butThe same viscous stress transports momentum through less mass in the second fluid, so its velocity responds faster to a given nonuniform viscous force.

To see the diffusion equation directly, consider an unsteady parallel velocity field $\mathbf{u}=(u(y,t),0,0)$ with constant material coefficients and no streamwise pressure gradient or body force. Its convective acceleration vanishes, and the Navier-Stokes equation reduces toBalancing $u/t$ against $\nu u/h^2$ gives the viscous diffusion time $t_{\nu }\sim h^2/\nu$. For the $1.0\mathrm{mm}$ gap, the two fluids have scales $0.010\mathrm{s}$ and $0.0050\mathrm{s}$. This is a characteristic scale, not an exact settling time. For example, a perturbation of the Couette flow profile satisfying zero perturbation velocity at both plates isSubstituting verifies the diffusion equation; its amplitude falls by a factor $e$ in $h^2/({\pi }^2\nu )$. Kinematic viscosity is therefore useful for predicting how quickly a moving wall influences the interior, while dynamic viscosity gives the shear stress at a prescribed shear rate.

In unbounded Couette flow with velocity $(y,0)$, linear vorticity is transported exactly as $\omega (x,y,t)={\omega }_0(x-yt,y)$. For a nonzero streamwise Fourier mode $\alpha$ and sufficiently localized smooth initial data, inversion by the one-dimensional modified Helmholtz Green function gives the displayed pointwise decay of the streamfunction and velocity, although vorticity does not decay in magnitude. The mechanism is phase mixing: shearing produces increasingly rapid transverse oscillations, whose elliptic velocity reconstruction suppresses them. Smoothness and decay of the initial streamfunction without derivative control do not alone imply these algebraic rates.

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 $h=h_0[1-A\sin (2\pi x/L)]$, the crawling speed is $U/V=3A^2/(1+2A^2)$. At prescribed $U$, eliminate $V$ to obtain $\Phi =2\mu L{U}^2(1+2A^2)/(3h_0{A}^2\sqrt{1-A^2})$. The derivative in $A^2$ vanishes only at $A^2=1/2$, and the dissipation diverges at both endpoints. The displayed optimum still dissipates $8\sqrt{2}/3$ times the simple Couette flow dissipation of an externally driven flat foot moving at speed $U$ across a uniform film of thickness $h_0$.

Approximate the gap by steady Couette flow, neglecting edge effects and additional losses from membrane deformation. Its velocity gradient is $U/h$, so the shear stress is $\eta U/h$. The force on the gliding bacterium and the rate of viscous dissipation areThe same power follows by integrating $\eta (U/h{)}^2$ over gap volume $Ah$. In SI units, $A=10^{-10}{\mathrm{m}}^2$, $h=10^{-7}\mathrm{m}$, and $U=10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$. ThusAn energy needs a duration or distance: $E=10^{-18}\Delta t\mathrm{J}$ for $\Delta t$ in seconds, equivalently $E=F_{d}s$ for travel distance $s$. Moving one micrometre takes one second and dissipates $10^{-18}\mathrm{J}$. This is the mechanical cost of the sheared film. A numerical biological efficiency would additionally require the motor's input power and the losses associated with its peristaltic actuation.

A generalized Newtonian fluid can approximate the shear response of a fluid with viscoelasticity when the flow varies slowly relative to its stress-memory time scales and elastic stresses are negligible, or in locally steady unidirectional flow where a fitted steady shear viscosity captures the relevant stress and elastic normal stresses do not significantly influence the motion. It is not a general replacement in rapidly changing flows or in flows controlled by elastic stresses. Two specific failures are stress relaxation after deformation stops, since this law has no memory, and nonzero normal-stress differences in steady simple shear flow, since the only nonzero nonpressure components are off diagonal. A rate-dependent viscosity alone cannot repair either limitation.

Take the usual physical power-law fluid range $k>0$, $n>0$, and $G>0$. The steady unidirectional Couette-Poiseuille flow satisfies $0=G+{\partial }_y{\sigma }_{xy}$. The stated upper-wall stress thus givesPut $Y=y/h$. Inverting this monotone stress law gives $u_y=(Gh/k{)}^{1/n}\mathrm{sgn}(\alpha -Y)|\alpha -Y{|}^{1/n}$. Integration from the lower no-slip boundary condition $u(-h)=-U$ gives the power-law Couette-Poiseuille flow profileThe upper no-slip boundary condition requiresThe bracket is a strictly increasing function of $\alpha$ with value zero at $\alpha =0$, so the equation determines a unique positive $\alpha$. The absolute values are essential: for sufficiently strong pressure driving, the shear can reverse inside the channel. For a Newtonian fluid, $n=1$, $\mu =k$, and the bracket reduces to $4\alpha$, yieldingFor zero pressure gradient the limit is pure Couette flow; the parameterization of wall stress by $Gh\alpha$ then becomes singular rather than indicating zero Couette stress.

For weak pressure forcing, let $s_0=U/h$ and $s_c=k{s}_0^n$. Write the center-plane shear stress as $s_c+\delta s$. Linearization of the power-law relation about the positive base shear givesIntegrating $u_y$ across the channel must still give $2U$. The Gy term has zero mean, so $\delta s=0$ at first order. Hence the weak-pressure Couette-Poiseuille expansion isIt is a small positive parabolic displacement of the straight Couette profile, preserving both wall velocities. For fixed positive n the stated $Gh\ll k(U/h{)}^n$ controls the expansion; when n is very small one needs the stronger incremental condition $Gh\ll nk(U/h{)}^n$.

The volume flux per unit spanwise width is $Q={\int }_{-h}^{h}u(y)dy$. The odd Couette part contributes zero and the parabola contributes $4h^3/3$ times its coefficient, soThis correctly reduces to the exact Newtonian flux $2G{h}^3/(3k)$ at $n=1$.

The extension to a general smooth generalized Newtonian fluid uses its differential shear viscosity, rather than its secant viscosity. Let $\mathcal{T}(s)=s\mu (s)$ for positive shear rate, and assumeLinearizing the stress law repeats the preceding calculation with $nk{s}_0^{n-1}$ replaced by ${\eta }_d$. ThusThe expansion requires small rate changes, $Gh\ll s_0|{\eta }_d|$, and a smooth constitutive law on the perturbed rates. At zero differential viscosity the regular first-order expansion fails. A negative differential viscosity is still locally invertible mathematically, so the formula is formally applicable, but belongs to a decreasing, mechanically unstable constitutive branch. The usual stable-flow interpretation requires positive differential viscosity. Thus “any” generalized-Newtonian law needs these local qualifications.

Axisymmetric rotation about the offset axis gives wall velocity $v_{\phi }=a\Omega \sin \theta$. Its azimuthal Couette flow has no pressure-driving divergence, so pressure remains constant. The fluid-on-inner-sphere tangential traction is $-\mu a\Omega \sin \theta /h$. Multiply it by the moment arm $a\sin \theta$ and integrate:Putting $t=\alpha z$ and using the second supplied identity with its generic parameter equal to $\alpha$ gives the axial rotational resistance of eccentric nested spheres:The continuous limit at $\alpha =0$ is $-8\pi \mu \Omega a^4/(3\Delta )$, consistent with a concentric narrow spherical shell. At $\alpha \to 1$, the torque tends to $-4\pi \mu \Omega a^4/\Delta$: it remains finite because the rotational wall speed vanishes at the closest pole.

Write $\mathbf{\tau }$ for the Kirchhoff stress tensor and $S$ for the second Piola-Kirchhoff stress tensor. Since $\mathbf{\tau }=FS{F}^T$ and $\dot{F}=LF$ for the velocity gradient,Therefore the upper-convected rate as a stress push-forward is

For the simple shear flow, put $\gamma =f_{,2}$ and $g=\dot{\gamma }$. The velocity gradient and rate-of-strain tensor areTheir upper-convected derivative givesDenote the symmetric extra stress by $E={\sigma }^d$, and the relaxation time by ${\tau }_r$ to distinguish it from $\mathbf{\tau }$. The Oldroyd-B model in the question becomesFor the steady constitutive response, $E_{22}=E_{33}=E_{13}=E_{23}=0$, $E_{12}=\mu g$ and $E_{11}=2(\mu -{\mu }_r){\tau }_r{g}^2$. HenceA constant imposed shear rate can still have startup transients; these values describe the steady response.

For circular Couette flow $v(r)e_{\theta }$, the velocity gradient in the orthonormal polar basis $(e_r,e_{\theta })$ isFor completeness, in fixed Cartesian coordinates let $\omega (r)=v(r)/r$ and $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Then $v=\omega Jx$ and $L=\omega J+({\omega }^{\prime }/r)(Jx)\otimes x$. Along a particle, the polar basis rotates at angular speed $\omega$. Its contribution to the material derivative cancels the rigid-rotation part of $L$ in the upper-convected derivative. The rotating-frame reduction of circular viscoelastic shear therefore leavesThe flow direction is now $e_{\theta }$ and the gradient direction is $e_r$. With $g=v^{\prime }-v/r$, the same simple-shear constitutive calculation gives

Angular force balance implies $r^2\mu g$ is constant. Integrating $v^{\prime }-v/r\propto r^{-2}$ gives $v=Ar+B/r$. With $v(a)=V_a$ and $v(b)=0$,The radial force balance is $-p^{\prime }-2(\mu -{\mu }_r){\tau }_r{g}^2/r=0$. Thus the inertialess Oldroyd-B circular Couette pressure isThe constant $p_0$ is fixed by a pressure or normal-traction boundary condition.

Use incompressible Euler equations with pressure divided by the constant reference density. Writing $L={\partial }_t+U(y){\partial }_x$, their linearization about the parallel shear flow isTaking the divergence requires the commutator $[{\partial }_y,L]=U^{\prime }{\partial }_x$. Both this commutator and the divergence of $U^{\prime }v$ contribute, so ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying the Laplacian to $Lv=-p_y$ givesHence the wall-normal velocity equation isTaking ${\partial }_z$ of the streamwise momentum equation minus ${\partial }_x$ of the spanwise equation cancels pressure and yields the inviscid Squire equation,This is the vorticity component in the original PDF; the converted TeX corrupts its definition. Impermeable inviscid walls impose $v(\pm 1,t)=0$. They do not impose a viscous no-slip condition on the tangential perturbation velocity.

For a Fourier mode, set $k^2={\alpha }^2+{\beta }^2$, $D=d/dy$ and $q=(D^2-k^2)\tilde{v}$. The equations becomeFor Couette flow $U=\lambda y$, $U^{\prime \prime }=0$, so $q(y,t)=e^{-i\alpha \lambda yt}{q}_0(y)$ with $q_0=(D^2-k^2){\tilde{v}}_0$. Invert this using the Dirichlet Green function of $D^2-k^2$ on $[-1,1]$. If $y_{<}=\min (y,s)$ and $y_{>}=\max (y,s)$, it isIt vanishes at both walls and its first derivative has jump one at $y=s$. ThusAt $t=0$, uniqueness of the Dirichlet inversion recovers the compatible initial velocity. This inviscid Couette initial-value Green function directly solves the initial-value problem without a Laplace transform.

The Squire equation is now a first-order forced equation in time. Its integrating factor givesFor nonzero $\alpha$, the velocity formula contains shearing phases and the inviscid Couette continuous spectrum. Velocity cancellation can occur even though the transported $q$ has constant magnitude; it must not be confused with exponential modal instability.

For the separate case $\alpha =0$, $\beta \ne 0$, the Dirichlet operator $D^2-{\beta }^2$ is invertible and $q_t=0$. Hence $\tilde{v}={\tilde{v}}_0$ andSince $\tilde{\eta }=i\beta \tilde{u}$ in this case, $\tilde{u}={\tilde{u}}_0-\lambda t{\tilde{v}}_0$. This is the lift-up effect: a transverse disturbance transports base-flow momentum and creates a streamwise perturbation growing linearly in time, with perturbation kinetic energy generically growing as $t^2$. The growth is algebraic and is possible without an exponentially unstable normal mode. For nonzero initial normal velocity and nonzero shear, nonlinear effects eventually limit the validity of the linearized calculation. The same streamwise-independent result holds with $\lambda$ replaced by $U^{\prime }(y)$ for a general parallel profile.

Take positive $x$ in the snail's direction of travel, and let $y$ measure height above the rigid plane. The wave moves at laboratory velocity $U-V$. In the wave frame the plane's horizontal velocity is $V-U$ and the foot's is $V$. The foot is stationary as a shape, so its material velocity is tangent to $y=h(x)$: to leading lubrication theory order its components are $(V,V{h}_x)$. Apply the no-slip boundary condition to both surfaces.

The leading Stokes flow equations are $p_y=0$ and $\mu u_{yy}=p_x$. Their solution is the local Couette-Poiseuille flowThe normal-velocity condition on each steady surface and incompressibility make the wave-frame flux independent of $x$. Integrating the velocity givesPut $C=V-U/2$. Periodicity of pressure gives $0={\int }_0^L{p}_{x}dx=12\mu (C{I}_2-q{I}_3)$, and henceThe shear traction exerted by the fluid on the plane, positive in the $x$ direction, is

The snail is force-free horizontally: its gait consists of internal motion and it has no externally applied horizontal drive. Uniform locomotion therefore requires zero net horizontal fluid force on the foot per wavelength. The stresses on opposite periodic fluid faces cancel, so horizontal momentum balance also makes the net force on the plane zero. ThusSubstitution of $q$ proves the lubrication model of a crawling snail:The Cauchy-Schwarz inequality applied to $h^{-1/2}$ and $h^{-3/2}$ gives $0<\alpha \le 1$, with equality exactly for a uniform gap. Thus the denominator is positive, and a nonuniform waveform gives forward motion; the flat gap gives $U=0$.

For the sinusoidal shape, the given integrals yieldTherefore $\alpha =1/(1+A^2/2)$ andFor small amplitude, $U\sim 3A^{2}V$: the leading locomotion is quadratic, and there is no propulsion from a flat foot. As $A\to 1^{-}$, $U\to V$, so the wave becomes nearly stationary in the laboratory frame while the snail advances relative to it. However, the minimum gap tends to zero; finite speed in this limit does not imply finite dissipation, and $A=1$ itself is excluded by the noncontact model.

For the general waveform, use the mechanical work identity for viscous dissipation. Pressure does no boundary work because the boundary motion is tangential in the wave frame. The plane's velocity is constant, and its total force is zero, so it contributes zero net work. The upper boundary supplies, at leading lubrication order,The zero plane-force relation is $\mu U{I}_1=\frac{1}{2}\int h{p}_{x}dx$. ConsequentlyAs a direct check, integrating the squared shear rate gives $\Phi ={\int }_0^L[\mu U^2/h+h^3{p}_x^2/(12\mu )]dx=\mu I_1[U^2+12C^2(1-\alpha )]$, which reduces to the same result using the force-free speed relation. Zero horizontal force on the foot does not mean zero gait work: the tangential material velocity along its sloping surface includes vertical motion.

For prescribed speed $U>0$ and sinusoidal shape, eliminate $V=U(1+2A^2)/(3A^2)$ to obtainWriting $y=A^2$, the logarithmic derivative of the amplitude-dependent factor isThe only root in $0<y<1$ is $y=1/2$. The dissipation diverges at both endpoints, so this is the global minimum and gives the optimal sinusoidal waveform for viscous snail locomotion:The magically driven flat glider has simple Couette flow $u=Uy/h_0$ in the laboratory frame, and dissipation ${\Phi }_{\mathrm{glide}}=\mu L{U}^2/h_0$ per wavelength. HenceThe comparison is at the same $U$, mean thickness $h_0$ and wavelength $L$; the flat glider is externally propelled, whereas the snail must power its internal waveform.