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In linear viscoelasticity, stress is a linear functional of past strain increments: scaling the deformation scales the response, and responses to small deformations add. For a time-translation-invariant material, Boltzmann superposition writes $\sigma (t)={\int }_{-\infty }^{t}G(t-s)\dot{ϵ}(s)ds$, where $G$ is a relaxation modulus. The material combines reversible storage with delayed or dissipative response; the description is restricted to small deformations about the chosen state.

The Maxwell model places an elastic spring and viscous dashpot in series. They carry the same shear stress $\sigma$, and their shear strains add:Differentiating givesHere $ϵ$ is dimensionless shear strain, $\sigma$ shear stress in pascals, $G_0$ the spring shear modulus, $\eta$ dashpot viscosity in $\mathrm{Pa}\mathrm{s}$, and $\tau$ the relaxation time. The first term describes changing elastic strain; the second is viscous flow. In this model the material flows at long times and has zero static shear modulus.

Use the Monge representation $\mathbf{r}(x,y)=(x,y,h(x,y))$ for a single-valued surface without overhangs. Its area element is $dA=\sqrt{1+|\nabla h{|}^2}dxdy$; at small slopes, $dA\simeq [1+|\nabla h{|}^2/2]dxdy$ and twice the mean curvature is $2\mathcal{H}\simeq {\nabla }^{2}h$, up to orientation sign. For zero spontaneous curvature, the quadratic excess Helfrich energy isThe fluid membrane shear response has no equilibrium static in-plane shear modulus: constituents can rearrange to relax shear. Bending and isotropic tension still cost energy; viscous shear stresses can matter during motion. A Gaussian curvature term is omitted under the stated approximation; for fixed closed-surface topology its integral is constant, but boundaries and topology changes require care. The small-slope Monge expression describes weak deformations, not an entire fully wrapped sphere.^[ref]

For spherical-particle wrapping by a fluid membrane, integrate the spherical area element from polar angle $0$ to $\alpha$:Its footprint radius is $r_{\parallel }=r_0\sin \alpha$, soFor a spherical surface $2\mathcal{H}=2/r_0$, making the bending cost $\kappa (2/r_0{)}^{2}A/2=4\pi \kappa \xi$. adhesion contributes $-2\pi w{r}_0^2\xi$, and tension contributes $\sigma \Delta A$. ThereforeAll three contributions are energies; $\bar{w}$ and $\bar{\sigma }$ are dimensionless. This is the specified bound-patch model: deformation of the free membrane and the neck is omitted.

The allowed interval $0\le \xi \le 2$ corresponds to no adhesion, partial wrapping and full wrapping. At $\xi =2$, the cap covers a whole sphere and its footprint shrinks to zero. This is a closed neck in the geometric limit. It does not by itself prove membrane scission: creating a detached enclosure changes topology and involves neck physics absent from this energy.

For positive membrane tension, $E^{\prime \prime }/(\pi \kappa )=2\bar{\sigma }>0$. Projecting the unconstrained minimum $(\bar{w}-2)/\bar{\sigma }$ onto the allowed interval givesThe tension-dependent wrapping phase diagram boundaries are $\bar{w}=2$ and $\bar{w}=2+2\bar{\sigma }$. At zero membrane tension, the energy is linear: the minimum jumps from $0$ to $2$ as $\bar{w}$ crosses $2$, while all indentation values are degenerate at $\bar{w}=2$. Negative membrane tension is outside the stable-flat-membrane interpretation used here. Increasing membrane tension makes the excess area required for wrapping more expensive, broadening the partially wrapped region.

TODO: user-deferred biological application. The final request for biological relevance is intentionally left unsolved; all preceding generic geometry and membrane mechanics are included.

A Couette rheometer measures the torque required to rotate one coaxial cylinder relative to another. With inner radius $R_i$, outer radius $R_o$, wetted length $L$, inner angular velocity $\Omega$ and stationary outer wall, laminar Newtonian flow hasThe shear stress at the inner wall is ${\tau }_i=\mathcal{T}/(2\pi L{R}_i^2)$. In a narrow gap $d=R_o-R_i\ll R_i$, $\dot{\gamma }\simeq \Omega R_i/d$, so $\eta \simeq \mathcal{T}d/(2\pi L\Omega R_i^3)$.

Measuring torque over a range of speeds gives a flow curve: constant viscosity identifies a Newtonian regime; decreasing or increasing apparent viscosity indicates shear thinning or thickening. A yield-stress fit needs a genuine low-rate regime and must not be confused with bearing drag or wall slip. Small oscillatory strain also probes viscoelasticity, with $\tau ={\gamma }_0[G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t]$ for imposed $\gamma ={\gamma }_0\cos \omega t$.

Control temperature, calibrate instrument torque and end effects, and check gap dependence or roughened surfaces for slip. Keep inertia small enough to avoid Taylor vortices. The exact Newtonian profile is a calibration model; a non-Newtonian material needs its own stress–rate relation.

Use a rheometer, for example a cone-and-plate rheometer or a parallel-plate rheometer, to impose a small deformation and measure the required torque, or impose a torque and measure the deformation. The apparatus geometry converts these into shear stress and strain. First check that reducing the imposed amplitude does not change the normalized response; this identifies the linear regime and avoids yielding or structural rearrangement.

Three complementary protocols characterize the material's memory. In a stress relaxation experiment impose a small step ${ϵ}_{0}H(t)$ and hold the deformation fixed. The output is the relaxation modulus $G(t)=\sigma (t)/{ϵ}_0$, revealing how stored stress decays and whether an elastic plateau remains. In a creep compliance measurement impose a small stress step ${\sigma }_{0}H(t)$ and record $J(t)=ϵ(t)/{\sigma }_0$. Removing the load also distinguishes recoverable deformation from permanent flow.

In small-amplitude oscillatory shear, impose sinusoidal strain over a range of frequencies and measure both stress amplitude and phase. With $ϵ={ϵ}_0\cos \omega t$, fitThe outputs are the storage modulus $G^{\prime }$ and loss modulus $G^{\prime \prime }$, or equivalently amplitude ratio and loss tangent $\tan \delta =G^{\prime \prime }/G^{\prime }$. These measure elastic storage and dissipative response, respectively. The step experiments give time-domain information and the oscillatory experiment gives the corresponding frequency-domain response. All comparisons must use the same deformation convention and controlled temperature.

Multiplying the deformation gradient by its matrix transpose gives the left Cauchy-Green deformation tensor and its trace:The derivative of the strain energy density isFor brevity put $g(r)={\mu }_1+{\mu }_2(3+{\tau }^2{r}^2{)}^n$. Part (a) gives the Cauchy stress tensor, in the current cylindrical coordinate system orthonormal basis, asIn particular, the end-face shear stress is $T_{\theta z}=2g\tau r$. The undetermined pressure $p$ affects the normal stresses but does not affect this shear stress. No normal traction conditions have been specified to determine $p$.

Using $a=A$ and the end-face shear stress from part (c), the printed torque normalization givesThe binomial theorem expands the remaining integral into powers of $r$:Integrating each term gives an expression valid also at $\tau =0$:For example, $n=1$ gives $M=({\mu }_1+3{\mu }_2)A^4\tau /2+{\mu }_2{A}^6{\tau }^3/3$. The torque is an odd polynomial in the twist per unit reference length, as expected when reversing the twist reverses the end-face shear stress.

For a physical end-face torque $N$, integration over the azimuth normally givesThe original PDF explicitly defines $M$ without this $2\pi$, so the boxed answer retains its convention. Since $\mathbf{T}$ is the Cauchy stress tensor, its shear component here is spatial $T_{\theta z}$; the printed capital $Z$ in $T_{\theta Z}$ does not require a change to a nominal stress tensor.

Let the gas have a small bulk velocity $u_x(y)$, an approximately isotropic local thermal velocity distribution, and a mean free path $\lambda$ short compared with the flow-gradient scale. A molecule transports the mean streamwise momentum acquired near its previous collision. To first order, its nonequilibrium momentum relative to the destination layer is approximately $-m{\tau }_c{v}_{y}d{u}_x/dy$. The resulting transverse momentum flux isThus the dynamic viscosity is $\eta \simeq \rho {\tau }_c⟨v_y^2⟩$. With isotropy, a representative collision time ${\tau }_c\sim \lambda /⟨v⟩$, and a chosen thermal speed average, this has the formThe dimensionless $K$ depends on how speeds and collision times are averaged; a simple one-speed estimate gives $K=1/3$. Assumptions include dilute binary collisions, a locally equilibrated distribution, weak shear, and negligible boundary slip. The kinetic theory identity $p=\rho ⟨v^2⟩/3$ then givesFor the Maxwell speed distribution, $⟨v^2⟩/⟨v{⟩}^2=3\pi /8$, soThis explains both the units of inverse time and the relation to molecular collision frequency, up to the averaging convention.

In the cylinder apparatus, distinguish the vector mean molecular velocity, $⟨\mathbf{v}⟩=\mathbf{u}$, from the mean thermal speed $⟨|\mathbf{v}-\mathbf{u}|⟩$. For a slow nearly isothermal Couette flow, the thermal speed is approximately position-independent, whereas the bulk mean is azimuthal and changes from the local rotating-wall speed to zero at the fixed wall. In the side gap, with $y$ measured radially from the inner wall,Across either end gap, at cylindrical radius $s$, the corresponding profile is $u_{\phi }(s,z)\simeq \omega s(1-z/h)$, with $z$ measured away from the moving end face. The no-slip boundary condition and small $h/R,h/H$ underlie these linear profiles.

The side-wall shear stress has magnitude $\eta \omega R/h$, producing resisting torque $2\pi \eta \omega R^{3}H/h$. The original diagram also has a gap at both circular end faces. At either face, the stress is $\eta \omega s/h$ and its ring contribution isAdding the side and both ends gives the end-face torque in narrow-gap cylindrical viscometry:The motor supplies the opposite of this viscous resisting torque. Corner flow and the small axle area are neglected. Retaining only the cylindrical wall would omit the $\pi R^4/h$ term shown by the actual closed-ended geometry.

For hard-sphere Ar atoms of collision diameter $d_{\mathrm{Ar}}$, $\lambda =1/(\sqrt{2}n\pi d_{\mathrm{Ar}}^2)$ and $\rho =n{m}_{\mathrm{Ar}}$. ThereforeUsing $m_{\mathrm{Ar}}=40m_u$, $T=273.15\mathrm{K}$ and the simple estimate $K=1/3$ givesThis is a kinetic estimate with a model-dependent numerical prefactor, rather than a precise atomic size measurement. Since the density cancels and the hard-sphere collision diameter is taken constant, the prediction is $\eta \propto \sqrt{T}$. The collision cross-section of a real atom can vary with speed, so its exact temperature dependence need not follow this power law.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

With $u=x{f}^{\prime }(\zeta )$ and $W=-f(\zeta )$, incompressibility is automatic because $u_x=f^{\prime }$ and $W_{\zeta }=-f^{\prime }$. In the streamwise boundary-layer equation,Equating coefficients of $x$ givesThis coefficient comparison determines the equation even though the streamwise equation vanishes identically at the stagnation point $x=0$.

The no-penetration boundary condition gives $f(0)=0$, the surface shear stress gives $x{f}^{\prime \prime }(0)=\gamma x$, and matching the tangential velocity to the outer solution gives $f^{\prime }\to 1$. ThusIf the correction $f^{\prime }-1$ is integrable at the far end, then $f=\zeta +O(1)$ there, which is consistent with $W\sim -\zeta$. The small-shear stress solution below has precisely this property.

In a thin boundary layer, take streamwise velocity and distance to be order one, set $z=\delta \zeta$, and use incompressibility to set $w=\delta W$. The streamwise convective acceleration is order one, whereas its dominant viscous contribution after division by the Reynolds number is ${\mathrm{Re}}^{-1}{\delta }^{-2}{u}_{\zeta \zeta }$. Their dominant balance selectsWith this choice, the exact rescaled Navier-Stokes equations areThus the leading pressure is independent of $\zeta$. Matching it to the outer solution from part (a), evaluated near $z=0$, gives $p=p_{\ast }-x^2/2$, hence $-p_x=x$. The leading boundary-layer equation is thereforeThe exact surface shear stress condition rescales to $u_{\zeta }+{\delta }^2{W}_x=\delta \beta x=\gamma x$. Since $W(x,0)=0$ for all $x$, its tangential derivative also vanishes there. The leading surface boundary conditions areMatching to the outer solution requiresThe second matching condition specifies the leading growth of $W$, allowing a constant displacement correction. Requiring $W+\zeta \to 0$ would impose an extra condition not supplied by the leading outer solution. This is the stress-driven stagnation-point boundary layer.

The held-flat interface imposes the no-penetration boundary condition, while its specified tangential traction gives a shear stress condition:A differentiable boundary value $w(x,0)\equiv 0$ implies $w_x(x,0)=0$, so the second boundary condition is also $u_z(x,0)=\beta x$. There is no imposed no-slip boundary condition at this interface; tangential velocity is determined by the shear stress and the fluid equations.

The solution in part (a) has zero tangential traction, and so remains valid for the whole interface if and only if $\beta =0$. At the single point $x=0$ the imposed shear stress vanishes for any $\beta$, but this does not satisfy a boundary condition that must hold for every $x$.

At zero surface shear stress, the solution connected to part (a) isIndeed, it has $f_0^{\prime }=1$, $f_0^{\prime \prime }=f_0^{\prime \prime \prime }=0$ and $f_0(0)=0$. Insert the regular perturbation $f=\zeta +\gamma f_1+O({\gamma }^2)$ into the nonlinear ordinary differential equation. The terms areThe coefficient of $\gamma$ therefore givesExpanding the surface boundary conditions and the far-field matching condition givesThe printed $+\infty$ limit is a sign error. It is present in the native PDF on paper page 3, not introduced by transcription. The fluid occupies $z<0$, the inner variable is $\zeta =z/\delta <0$, and part (b)(iii) correctly matches at $-\infty$. We retain that physical domain throughout. A change to the positive coordinate $\eta =-\zeta$ would also change the differential equation and surface derivative signs; it cannot be made by changing only the far-field sign.

The inequality ${\delta }^2\ll \gamma$ ensures that effects discarded at order ${\delta }^2$ in the boundary-layer scaling are smaller than the retained order-$\gamma$ perturbation. The inequality $\gamma \ll 1$ justifies this regular perturbation. The two conditions control different approximations.

A numerical answer needs an assumption about fault orientation. For a cohesionless fault optimally oriented for Mohr-Coulomb failure criterion, with vertical ${\sigma }_1$, write $\phi$ for the internal friction angle and $\delta$ for the fault dip. The angle between the plane and ${\sigma }_1$ is $90^{\circ }-\delta$, so the optimal normal-fault dip relation givesHence ${\mu }_f=\tan 40^{\circ }=0.839\simeq 0.84$ under that model.

For the Mohr circle, plot effective compression ${\sigma }_n^{\prime }$ horizontally and shear stress $\tau$ vertically. Draw a circle with diameter $[{\sigma }_3^{\prime },{\sigma }_1^{\prime }]$ tangent to $\tau ={\mu }_f{\sigma }_n^{\prime }$. Its centre and radius obey $R/C=\sin \phi$. One illustrative normalization is ${\sigma }_3^{\prime }=1$, ${\sigma }_1^{\prime }=4.59891$, with tangency at the resolved stresses on the $65^{\circ }$ plane. Pore fluid pressure shifts the total-stress circle horizontally; the friction law uses effective stress.

The dip does not determine the actual sliding coefficient of friction without the optimal-orientation assumption. A reactivated inherited plane may be nonoptimal, cohesion may matter, and the stress ratio and pore fluid pressure are not measured. For an arbitrary compression-positive stress state,at cohesionless sliding. Thus the requested single value is a conditional inference, not a friction measurement.

Effective stress and frictional strength generally increase with depth through the shallow crust undergoing brittle fracture. At $1\mathrm{km}$, small overburden and often weak, fractured near-surface materials limit the shear stress and elastic energy that can accumulate. Some shallow materials also undergo stable sliding rather than unstable earthquake nucleation. Near $15\mathrm{km}$, a locked fault prone to brittle fracture can sustain a much larger differential stress before frictional instability releases stored elastic energy.

This depth is plausibly near the lower, strong part of the seismogenic layer: below it, increasing temperature eventually promotes material creep across the brittle-ductile transition. Once nucleated at depth, a dynamic finite-source rupture can propagate into shallow regions that could not independently nucleate the same event. Depth alone does not determine the nucleation point: local pore pressure, frictional stability and stress heterogeneity also matter.

Use compression-positive effective stress normal to the plane ${\sigma }_n^{\prime }$ on the horizontal axis and shear stress $\tau$ on the vertical axis. For any chosen pair of principal stresses, Mohr's circle has centre $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and radius $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$.

For the reverse fault example, $C=60\mathrm{MPa}$ and $R=30\mathrm{MPa}$. Draw the two Mohr-Coulomb failure criterion lines $\tau =\pm {\sigma }_n^{\prime }/\sqrt{3}$. The tangencies areThese are shear failure points with nonzero shear stress. The two smaller three-dimensional Mohr circles use $({\sigma }_1^{\prime },{\sigma }_2^{\prime })$ and $({\sigma }_2^{\prime },{\sigma }_3^{\prime })$; they lie inside the outer circle and do not touch the envelope in this example.

For the opening geological joints, use the tensile failure criterion ${\sigma }_n^{\prime }=-T_0$. The example gives $C=3.5\mathrm{MPa}$, $R=6.5\mathrm{MPa}$ and $T_0=3\mathrm{MPa}$. The circle first touches this vertical cutoff at $({\sigma }_n^{\prime },\tau )=(-3,0)\mathrm{MPa}$. Thus the geological joints form on planes normal to ${\sigma }_3$ and have zero resolved shear stress. A complete brittle envelope can combine this tensile cutoff with compressive Mohr-Coulomb failure criterion, using sufficient cohesion that shear failure does not occur first. The failure-mechanics figure displays these two distinct constructions.

Trace the strike-slip fault along the southwest–northeast valley, approximately through $(u,v)=(0.12,0.83),(0.28,0.75),(0.44,0.68),(0.53,0.62),(0.65,0.42),(0.83,0.20),(0.97,0.04)$. The long oblique strand is interpreted as left-lateral strike-slip faulting: draw the southeastern side moving northeast and the northwestern side moving southwest, parallel to the trace.

This sense is consistent with horizontal north–south maximum compression: for an approximately northeast-striking fault, the resolved shear stress moves the southeastern side northeastward relative to the northwestern side. The deflected central drainage supports that interpretation. The regional stress direction supplies a consistency check rather than a uniquely precise trace location; follow the geomorphic break rather than drawing a single straight line through every coordinate.

Use compression-positive principal stresses ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$. The two-dimensional Mohr's circle shown is the ${\sigma }_1$–${\sigma }_3$ circle: its horizontal axis is compressive normal stress ${\sigma }_n$, its vertical axis signed shear stress $\tau$, its centre is $C=({\sigma }_1+{\sigma }_3)/2$, and its radius is $R=({\sigma }_1-{\sigma }_3)/2$. The intercepts $({\sigma }_1,0)$ and $({\sigma }_3,0)$ represent principal stress planes, on which shear stress vanishes. The top and bottom points have $|\tau |=R$, the maximum shear stress.

In the rock sketch, $\theta$ is the counterclockwise angle from the ${\sigma }_1$ direction to the plane's unit normal vector $\mathbf{n}=(\cos \theta ,\sin \theta )$. Choose the clockwise unit tangent $\mathbf{t}=(\sin \theta ,-\cos \theta )$ and define $\tau =\mathbf{t}\cdot {\mathbf{\sigma }}^c\mathbf{n}$, with ${\mathbf{\sigma }}^c=\mathrm{diag}({\sigma }_1,{\sigma }_3)$ compression positive. Thus a positive $\theta$ produces a positive $2\theta$ rotation on the plotted circle. The actual traction on the positive-normal face is $-{\mathbf{\sigma }}^c\mathbf{n}=-{\sigma }_n\mathbf{n}-\tau \mathbf{t}$. Reversing the tangent convention reverses the plotted shear sign, not the physical prediction.

In three dimensions there are circles for each pair of principal stresses. The largest is the ${\sigma }_1$–${\sigma }_3$ circle; a general three-dimensional plane requires all three principal stresses and its normal, rather than a single angle on this circle.

Draw a failure envelope in the $({\sigma }_n^{\prime },\tau )$ plane. A useful brittle shear-failure approximation is the Mohr-Coulomb failure criterionwhere $c$ is cohesion, $\phi$ the internal friction angle, and ${\sigma }_n^{\prime }$ the effective compressive normal stress. A circle wholly inside the envelope is below this ideal failure threshold. Increasing the differential stress expands it until it first touches the envelope: the tangent points identify the failing plane orientations. For a linear envelope, tangency impliesWith the angle convention above, the upper tangent point is at $2\theta =90^{\circ }+\phi$. Its plane therefore makes $45^{\circ }-\phi /2$ with the maximum-compression direction; there is a conjugate plane on the other side. For $\phi =30^{\circ }$, this is $30^{\circ }$, rather than the $45^{\circ }$ plane of maximum shear stress. An existing weak fault can reactivate first, with its own lower cohesion and friction, and need not have this optimum new-fracture orientation.

For tensile brittle fracture, add a tensile strength cutoff: failure occurs when ${\sigma }_3^{\prime }\le -T_0$, where $T_0>0$. The opening fracture is approximately perpendicular to the minimum-compression direction. Real rock envelopes curve, especially at small confinement; a straight shear envelope must not be extrapolated as an exact tensile law. High confinement can instead favor distributed or ductile deformation, so merely drawing a larger circle does not establish that brittle failure is the active mechanism.^[ref]

For a plane whose normal lies in the ${\sigma }_1$–${\sigma }_3$ plane, rotate through $2\theta$ from $({\sigma }_1,0)$ in the stated sign convention. The circle point givesThe normal-to-plane angle is essential: if the angle supplied is between the plane itself and ${\sigma }_1$, use its complement for $\theta$. The principal stress planes have zero shear stress, and planes at $45^{\circ }$ to those principal stress planes have maximum $|\tau |$.

For an arbitrary plane in a fully three-dimensional stress field, let $n_i$ be its direction cosines in the principal stress basis. ThenThese follow by resolving the traction into its normal and tangential components; the compression-positive tensor is the negative of the usual tension-positive Cauchy stress tensor. The family of Mohr circles visualizes the possible combinations, but the two-dimensional $2\theta$ rule alone does not specify an arbitrary three-dimensional orientation.^[ref]

Now both plates at $z=0$ and $z=H$ are stationary and both obey the Navier slip boundary condition with the same slip length $b$. Drive steady flow of a Newtonian fluid in incompressible flow by the constant pressure gradient $G=-dp/dx>0$, with constant dynamic viscosity $\mu >0$ and no other streamwise body force. The planar Poiseuille flow profile and its volumetric flow rate per unit width areThe full volumetric flow rate for channel width $W$ is $Q=Wq$ when side-wall effects are negligible. Thus $q$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, while $Q$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$. Denote the value at $b=0$ by $q_0=G{H}^3/(12\mu )$.

The two wall checks expose the normal signs:Both Navier slip boundary conditions therefore hold exactly. The two walls exert backward tangential tractions of magnitude $GH/2$ on the fluid. Increasing $b$ at fixed $G$ adds a uniform velocity offset, leaving the velocity gradient and wall shear stress unchanged. This differs from Couette flow with a slipping lower wall, where upper-wall velocity is fixed and shear stress decreases.

For $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $G=10^3\mathrm{Pa}{\mathrm{m}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, the wall velocity is $10\mu \mathrm{m}{\mathrm{s}}^{-1}$, the centre velocity is $22.5\mu \mathrm{m}{\mathrm{s}}^{-1}$, and $q=1.83\times 10^{-10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. The enhancement is $q/q_0=2.2$. Equivalently, a prescribed volumetric flow rate needs only $1/2.2$ of the original pressure gradient. This makes slip length useful for estimating pumping requirements in small channels.

At fixed $H$, $G>0$ and $\mu$, $b\to 0$ recovers ordinary planar Poiseuille flow. As $b\to \infty$, the velocity and $q$ grow without bound in this model, although the varying part of the profile stays the same. At exactly zero tangential wall traction, no steady solution can balance this nonzero pressure gradient: $u^{\prime }(0)=u^{\prime }(H)=0$ conflicts with $\mu [u^{\prime }(H)-u^{\prime }(0)]=-GH$. Unlimited enhancement is consequently not a prediction for a real channel; finite wall friction, finite driving capacity and the validity of the assumed steady flow must be assessed. At fixed finite $q$ instead, $G\to 0$ and the profile approaches uniform velocity $q/H$.

A short derivation integrates the reduced Navier-Stokes equation $\mu u^{\prime \prime }=-G$ twice. Equal slip lengths and the opposite inward unit normals give $u=G[z(H-z)+bH]/(2\mu )$. Its integral over the gap supplies the enhancement factor $1+6b/H$.