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Let $X=x_{\dim }/L$, $Y=y/h_0$, scale horizontal and vertical velocities by $U$ and $U{h}_0/L$, and define $P=(p_{\dim }-p_0)h_0^2/(\mu UL)$. The small aspect ratio makes $P$ independent of $Y$ at leading order. The second small parameter, $U{h}_0^2/(\nu L)$, makes inertia negligible compared with transverse viscous stress. Thus the lubrication equations are $u_{YY}=P_X$ and $u_X+V_Y=0$.

No slip gives $u(X,0)=1$, $u(X,h)=0$, soNo penetration at the two solid walls and continuity make $q$ constant. Therefore, writing the dimensionless variables again as $x,p$,The pressure scale is set by balancing the longitudinal pressure gradient with shear across the thin gap, not by the inertial dynamic-pressure scale.

Let $\delta =R{e}^{-1/2}$ and use the inner variables $\boxed{Y=y/\delta =R{e}^{1/2}y,V=v/\delta =R{e}^{1/2}v}$. Continuity balances $u_x$ with $V_Y$. In the tangential momentum balance, advection is order one while the normal viscous term is $R{e}^{-1}{\delta }^{-2}{u}_{YY}=u_{YY}$. Streamwise viscosity is smaller by $R{e}^{-1}$. The normal momentum balance forces the leading pressure to be independent of $Y$. ThusOn the stationary impermeable plate, no-slip boundary condition and no-penetration boundary condition give $u=V=0$ at $Y=0$, $x>0$. Matching to the uniform outer stream gives $u\to 1$ as $Y\to \infty$, and the leading-edge/upstream data are $u\to 1$ as $x\downarrow 0$ at fixed $Y>0$. There is no requirement that the scaled transverse velocity $V$ vanish at infinity; its physical value $v=\delta V$ matches a smaller outer correction. The outer streamwise Euler equation with constant speed one gives $p_x=0$, and $p_Y=0$ carries this pressure gradient throughout the layer.

Scale $x,y$ by $L$, $z$ by $\delta L$, tangential velocities by $U$, normal velocity by $\delta U$, and time by $L/U$. Continuity balances all three derivatives. Tangential viscous forces scale as $\mu U/({\delta }^2{L}^2)$, hence their pressure balance requiresThe inertial/normal-viscous ratio is ${\delta }^2\rho LU/\mu$, so it is negligible under the given smallness condition. Tangential viscous derivatives are smaller by ${\delta }^2$; the normal pressure balance gives $p_z=0$ at leading order. In dimensional variables the remaining equations are continuity, $\mu u_{zz}=p_x$, $\mu v_{zz}=p_y$, $p_z=0$.

No-slip boundary condition at each plate imposes $u=v=0$, while their normal velocities are $w(0)=0$ and $w(h)=\dot{h}$. Integrating the two tangential equations twice givesIntegrate continuity through the gap. Since $h$ is independent of in-plane position and the tangential velocity vanishes at the upper wall, the flux divergence is $\nabla \cdot (h\overline{\mathbf{u}})$ and the normal contribution is $\dot{h}$. Therefore

Rigid impermeable walls require zero normal fluid velocity, so the potential satisfies a homogeneous Neumann condition. In the cylinder,It is also single-valued and periodic in $\theta$, and regular on the axis. No tangential no-slip boundary condition is imposed on this inviscid acoustic potential.

For equilateral triangular pipe flow, the polynomial vanishes on all three walls, so it satisfies no-slip boundary conditions. Direct differentiation givesThusFor $G<0$, $A<0$ and the velocity is positive in the triangular interior.

At the wall $x=a$, the outward normal of the fluid is ${\mathbf{e}}_x$. The axial traction exerted by the wall on the fluid is $\mu w_x$, so the fluid exerts the opposite axial traction on the wall. Since $w_x(a,y)=A(a^2-3y^2)$, the force per unit axial length isIt points with the flow for a negative axial pressure gradient. As a check, the triangular area is $a^2/\sqrt{3}$, so the total pressure driving force per unit length is $-G{a}^2/\sqrt{3}$; each of the three equivalent walls receives one third.

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.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

Stretch $z=\delta Z$ and write $f=\delta h$, $w=\delta W$, with horizontal velocity and time left order one. The longitudinal viscous term is ${\delta }^{-2}{u}_{ZZ}$, so set $P={\delta }^{2}p$. For $Re{\delta }^2\ll 1$, the leading bulk equations are $P_Z=0$, $u_{ZZ}=P_x$ and $u_x+W_Z=0$.

The normal-stress condition gives $P=-{\delta }^3\hat{C}{h}_{xx}$ at leading order, and the tangential condition gives $u_Z(h)=0$. HenceNo slip at the bed, $u(0)=0$, yields $u=P_x(Z^2/2-hZ)$. Averaging through the depth givesThe kinematic condition becomes $h_t+u(h)h_x-W(h)=-\widehat{ϵ}/\delta$ at leading order. Integration of continuity from the impermeable bed then gives

A nonzero contact speed or evaporative flux is singular in the strictly no-slip sharp-contact-line model with mobility proportional to $h^3$. A finite nonzero limiting $\overline{u}$ would require $h_{xxx}=3\overline{u}/(C{h}^2)$, which is singular for a linearly vanishing height and generally incompatible with a prescribed finite slope at the exact line. Thus a dynamic finite-contact-angle condition needs microscopic regularization, such as slip or a precursor film. In the slow-drying limit a fixed contact angle can instead be used for the leading quasistatic outer profile; it is not an exact regular finite-rate contact-line solution.

At leading quasistatic order $h_{xxx}=0$, so the symmetric profile satisfying the imposed outer angle isIts cross-sectional volume is $V={\int }_{-s}^{s}hdx=2\theta s^2/3$. The moving endpoints contribute no boundary-height term, and the end flux vanishes, giving $V_t=-2ϵs$. ConsequentlyThese are leading slow-time results up to extinction; the approximation need not remain uniform at the final vanishing scale.

To recover the small velocity, differentiate the quasistatic profile using $s_t=-3ϵ/(2\theta )$:Symmetry makes the flux zero at the centre, so $h\overline{u}=-ϵx(s^2-x^2)/(4s^2)$. Division by the height givesThe flow is inward on both sides, suggesting transport of suspended matter toward a central deposit rather than a perimeter ring. Coffee commonly leaves the opposite, ring-like pattern. A possible missing ingredient is contact-line pinning: holding the perimeter fixed makes evaporation require outward replenishment, and stronger evaporation near the edge reinforces that transport. Receding constant-angle and pinned contact lines therefore have qualitatively different deposit mechanisms.

For comparison, the same uniform-evaporation parabolic model with fixed $s$ gives ${\theta }_t=-3ϵ/s$ and $\overline{u}=ϵx/(s\theta )$, an outward velocity. This comparison isolates the boundary assumption even before adding edge-enhanced evaporation.

The receding solution is the quasistatic evaporation of a receding parabolic drop.

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 $2U_0/3$.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

For a profile depending only on $y$, the Poisson equation becomes $w_1^{\prime \prime }=-G/\mu$. Integrating twice givesThe no-slip boundary conditions $w_1(b)=w_1(-b)=0$ imply $C_1=0$ and $C_2=G{b}^2/(2\mu )$. ThusThis is plane Poiseuille flow between the two horizontal walls. It is positive in the interior for $G>0$, so the velocity points down the pressure gradient. It does not satisfy the side-wall no-slip boundary conditions, which is precisely what the correction in part (iii) repairs.

For the unidirectional flow $\mathbf{u}=(0,0,w)$, incompressibility gives ${\partial }_{z}w=0$. Hence $w=w(x,y,t)$, and the convective acceleration $w{\partial }_{z}w$ vanishes. The transverse components of the Navier-Stokes equation give $p_x=p_y=0$, while the axial component givesA steady-flow assumption is implicit in the question. Neither unidirectional flow nor a time-independent pressure gradient alone eliminates ${\partial }_{t}w$: a decaying transient can be superposed on the final pipe profile. The native PDF does not explicitly say steady flow in this part. For the intended steady flow, $w_t=0$, and thereforeTaking the pipe walls to be stationary, the no-slip boundary condition isThe no-penetration boundary condition is already satisfied by the assumed velocity field. This is the Poisson equation defining rectangular duct flow.

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 $x$.

The flat plate moves tangentially and the curved plate is stationary. The no-slip boundary condition and no-penetration boundary condition giveAt the curved plate, impermeability can also be written $w=u{h}_x$; together with zero tangential velocity, this gives both components zero. The native PDF writes $v^{\prime }=0$ in the scaled wall conditions although the normal component is defined as $w^{\prime }$. We use $w^{\prime }$ consistently.

The parabolic lubrication gap changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is thereforeThe proposed normal velocity scale $h_{0}U/L$ then makes the two terms of the continuity equation comparable. Balancing the horizontal pressure gradient against the transverse viscous term in the Navier-Stokes equation requires $P/L=\mu U/h_0^2$, soHere pressure is measured from a common reference value, which is chosen at the outer fluid in part (b).

For clarity, define the dimensionless material derivativeSubstitution of all the scalings gives the exact dimensionless equationsFor example, the factor in the last left side is obtained by dividing the vertical acceleration scale $\rho h_0{U}^2/L^2$ by the vertical pressure gradient scale $P/h_0$.

Taking $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$ leaves the lubrication approximationThe scaled gap and boundary conditions areThe Reynolds number itself need not be small: it is ${\delta }^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

For a smooth incompressible Navier-Stokes equation trajectory with zero velocity at a fixed flat wall, the normal momentum equation determines the normal pressure gradient. In a perturbation model with buoyancy term ${\mathrm{Ri}}_B\theta \hat{\mathbf{y}}$, it gives ${\partial }_{n}p=\nu \mathbf{n}\cdot \Delta \mathbf{u}-{\mathrm{Ri}}_B\theta \mathbf{n}\cdot \hat{\mathbf{y}}$. An extra homogeneous pressure Neumann boundary condition is valid only when the right side vanishes; it does not follow from the no-slip boundary condition.

For an incompressible parallel flow with base velocity $U(y)$ and perturbation proportional to $e^{i\alpha (x-ct)}$, divergence of the linearized momentum equations gives $p\prime \prime -{\alpha }^{2}p=-2i\alpha U\prime v$. At a no-slip boundary condition, streamwise momentum gives $p(0)=u\prime \prime (0)/(i\alpha R)$. The equation encodes how base shear generates perturbation fluid pressure.

A steady two-dimensional purely radial incompressible Newtonian fluid flow has $u_r=g(\theta )/r$ by the continuity equation. The polar Navier-Stokes equation givesEquality of the mixed pressure derivatives yields the displayed third-order equation. On fixed walls $\theta =\pm \alpha$, no-slip boundary conditions require $g(\pm \alpha )=0$. A source with volume flux $Q$ per unit out-of-plane length requires ${\int }_{-\alpha }^{\alpha }gd\theta =Q$. Consequently $g=QG$ leaves the single inertial parameter $ϵ=Q/\nu$ in $G^{(3)}+4G^{\prime }+2ϵG{G}^{\prime }=0$. The origin is a source singularity; the radial ansatz describes the wedge away from that point, not the detailed source apparatus.

For $\mathrm{Kn}\ll 1$, frequent gas–gas collisions can establish a local Maxwellian before macroscopic conditions change appreciably. Under this additional near-equilibrium assumption, the hydrodynamic limit of the Boltzmann equation motivates a continuum approximation with Navier-Stokes equations and heat conduction. A no-slip boundary condition is a leading wall approximation, not a statement about individual molecular velocities.

With a small but finite Knudsen number, a useful slip approximation retains continuum bulk equations and corrects wall velocity and temperature. A Navier slip boundary condition can represent the leading tangential velocity correction, but its slip length must come from the relevant gas–wall interaction model. A nearby Knudsen layer still requires kinetic treatment if its detailed structure is wanted. Slip corrections are perturbations; extending them into order-one Knudsen numbers has no general accuracy guarantee. At order-one Knudsen number, wall and intermolecular collisions both matter; this transitional situation calls for a kinetic theory model. At $\mathrm{Kn}\gg 1$, free molecular flow becomes a useful limiting approximation. These are scale comparisons, with accuracy depending on geometry, flow conditions and the observable, rather than universal sharp numerical boundaries. ^[ref]

Incompressible mass conservation in a moving gap givesCombined with the local Couette-Poiseuille flux, this is the Reynolds lubrication equation.

For a thin gap $h$ with tangential wall velocities ${\mathbf{v}}_0,{\mathbf{v}}_h$, lubrication theory and the no-slip boundary condition give the integrated volume fluxConservation of mass gives $h_t+{\nabla }_{\parallel }\cdot \mathbf{q}=0$. This is the Reynolds equation; it determines the lubrication pressure from wall motion and gap geometry.