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Meaning and microscopic origin. For a Newtonian fluid, the dynamic viscosity $\eta$ measures tangential momentum transfer between fluid layers moving at different speeds. For the Newtonian shear considered here, ${\tau }_{xz}=\eta d{v}_z/dx$, with units $\mathrm{Pa}\mathrm{s}$ for $\eta$. Molecular motion and collisions transport momentum across layers in a gas; intermolecular interactions contribute strongly in liquids. The resulting viscous forces oppose relative shear motion and convert organized kinetic energy to heat. The constitutive law is Newtonian and isotropic; incompressibility alone does not establish it.

Vertical force balance. Choose $z$ downward, as in the original diagram, and $x$ outward normal to the plate. In a uniform steady film, $\mathbf{v}=v_z(x)\hat{\mathbf{z}}$ has zero advective acceleration and no along-plate pressure gradient. On a slice of thickness $dx$, the net shear force per unit face area is $\tau (x+dx)-\tau (x)$, while its weight per unit area is $\rho gdx$. ThereforeHere $\tau =\eta d{v}_z/dx$ denotes the downward traction on a face with positive $x$ normal. The opposite face has the opposite traction orientation, accounting for the sign in the force difference.

Boundary conditions. At the fixed solid wall, the no-slip boundary condition is $v_z(0)=0$, with no normal penetration. At the free surface $x=d$, neglect air shear, so $\tau (d)=0$ and hence $v_z^{\prime }(d)=0$. The flat free surface has normal pressure $p_{\mathrm{atm}}$, neglecting surface-curvature effects. These are a velocity condition at the wall and a stress condition at the free surface, not zero velocity at both boundaries.

The parabolic speed profile. Integrating ${\tau }^{\prime }=-\rho g$ from the shear-free surface gives $\tau (x)=\rho g(d-x)$. Thusand the wall condition yieldsIt is zero at the wall and largest at the free surface, where its gradient vanishes. The wall shear $\rho gd$ supports the downslope weight of the film per unit plate area.

Volume flux per unit width. Integrate through the depth:This is volume flow per unit length in $y$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Inclining the plate. Keep $x$ normal outward from the plate and use $s$ downslope. The gravitational components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plate. For a uniform film of normal thickness $d$, the steady slice equations areThe free-surface pressure condition givesUniform $d$ makes ${\partial }_{s}p=0$. The same no-slip and zero-surface-shear conditions then give the uniform gravity-driven Newtonian film,The component normal to the plate changes the hydrostatic pressure, while the tangential component drives the flow. The thickness is measured normal to the tilted plate.

The inclined-film volume flux. Integrating the new profile givesIt recovers the vertical result at $\theta =\pi /2$ and vanishes on a horizontal plate at fixed uniform depth. A varying depth, imposed pressure gradient or significant air shear would require additional terms; the solution here uses the uniform film specified by the simple draining-plate model.

Flow visualization and the idealization. A streamline is everywhere tangent to the instantaneous velocity field at one fixed time. A streakline is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a pathline. These coincide geometrically in steady flow, but need not do so in an unsteady draining flow.

The intended ideal liquid is approximately incompressible and inviscid: viscosity and associated shear stresses are negligible on the scales and times being considered, the Mach number is small, and the density is effectively constant. This does not require zero vorticity; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.

Spiralling flow and angular momentum. In an axisymmetric inviscid flow, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial specific angular momentum,The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The slow-drain angular-momentum amplification then givesThe assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.

A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_{\varphi }=C/r$ with positive constants $D,C$, giving $\varphi ={\varphi }_0-(C/D)\ln (r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.

The free-surface depression. Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal centripetal acceleration. The radial Euler equation and vertical hydrostatic equilibrium giveThus $g{H}^{\prime }(r)=u_{\varphi }^2/r=C^2/r^3$, with $C={\omega }_0{R}^2$. Integrating from $r$ to the rim givesFor $r\ll R$ this becomesThe surface rises outward because the inward pressure gradient force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.

The terrestrial estimate. Use the vertical component of Earth's rotation as the ideal initial angular speed, ${\omega }_0\simeq {\Omega }_E\sin \lambda$. At Cambridge, taking $\lambda \simeq 52.2^{\circ }$ and ${\Omega }_E=7.292\times 10^{-5}{\mathrm{s}}^{-1}$, this gives ${\omega }_0\simeq 5.76\times 10^{-5}{\mathrm{s}}^{-1}$. Hence, for $R=1\mathrm{m}$, $r=0.01\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\mu \mathrm{m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.

Curl of the azimuthal velocity. For axisymmetric $u_{\varphi }=r\omega (r)$ with no vertical dependence, the rotational part of the vorticity is vertical:The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.

The two limiting states. For initial uniform rotation, $\omega ={\omega }_0$, whereas the late approximation has $\omega =C/r^2$. ThereforeThe late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.

What is actually conserved. Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the vorticity equation isVertical compression of a material column changes its vorticity through vortex stretching, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, vorticity dilution in a shallow draining layer givesThus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.

Kelvin's circulation theorem conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi {\omega }_0{R}^2$, consistent with the late $u_{\varphi }=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding shallow-water potential vorticity is $(\zeta +f)/H$, with $f=2{\Omega }_E\sin \lambda$; the conserved quantity involves absolute vorticity.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

Signed radial balance. Let $r$ increase outward from the vortex centre and $U$ be signed azimuthal velocity, positive anticlockwise. At $45^{\circ }$N, $f=2{\Omega }_{\oplus }\sin 45^{\circ }>0$. The positive speed is $V=|U|$, $f$ is the Coriolis parameter, the density is $\rho$, and $p_r=\partial p/\partial r$. In circular motion the inward centripetal acceleration is $-U^2/r$, while the outward radial acceleration from the Coriolis force is $fU$. The pressure gradient acceleration is $-p_r/\rho$. HenceThe term $U^2/r$ is an acceleration, not an additional physical force in the rotating-frame momentum equation. It can be moved to the force-balance side and depicted as outward centrifugal force in a frame following the circular parcel.

For a cyclone, $U=+V$ and $p_r/\rho =V^2/r+fV>0$: pressure increases outward and the centre is a low. Its inward pressure force balances the outward Coriolis and centrifugal terms. For a conventional anticyclone, $U=-V$ and $G=-p_r/\rho >0$; the balance isThe inward Coriolis force balances the outward pressure and centrifugal terms, and pressure has a maximum at the centre.

Anticyclonic branches and near-centre regularity. The anticyclonic gradient-wind branches follow from the quadratic balance. Solving givesThe minus root is the normal branch connected continuously to geostrophic flow for small forcing. The plus root is an anomalous high-speed branch and must not be discarded algebraically, even though it often fails the stability conditions relevant to large-scale anticyclones. Reality requiresA finite smooth central pressure therefore has zero radial gradient at the origin, with $-p_r/\rho$ no larger than order $r$. For $p=p_c-\rho C{r}^2/2+\cdots$, the limiting curvature obeys $C\le f^2/4$. A pressure maximum with a finite nonzero limiting outward pressure force cannot support the required circular anticyclonic balance arbitrarily close to its centre.

For positive $\mathrm{Ro}=V/(fr)$ and geostrophic magnitudes,The cyclonic gradient wind is subgeostrophic and the normal anticyclonic wind supergeostrophic. On the latter branch $0\le \mathrm{Ro}\le 1/2$. At a point on the eastern side of a cyclone, geostrophic and actual winds point north but the actual arrow is shorter; the ageostrophic difference points south. On the eastern side of a normal anticyclone both geostrophic and actual winds point south, with the actual arrow longer and its ageostrophic addition also south.

At lower latitude, the same $V$ and radius give a larger Rossby number and larger differences from geostrophic wind. The contour topology remains a central low or a conventional central high, but for fixed $V,r$ the required radial pressure gradients change with $f$: the cyclonic gradient is $\rho (V^2/r+fV)$ and the anticyclonic outward pressure force is $fV-V^2/r$. The latter conventional branch cannot be maintained once the reduced Coriolis contribution is too weak. Actual spatial gradients, rather than an arbitrary spacing of contour values, determine how tightly drawn contours should be.

The high-centre conclusion is conditional on the conventional anticyclonic branch. Gradient balance alone also allows a sufficiently rapid clockwise circulation about a low: if $V>fr$, then $p_r/\rho =V^2/r-fV>0$. For example, $V=2fr$ gives $p_r/\rho =2f^{2}r>0$. Such a configuration is outside the usual geostrophic anticyclone regime. Tornado-scale vortices have very large Rossby number, so the dominant balance is nearly cyclostrophic, $p_r/\rho \simeq V^2/r>0$. Their centres are low pressure irrespective of the sign of rotation; low pressure supplies the inward acceleration needed to bend the fast flow.

Balances and density gradients. The first two relations describe horizontal geostrophic balance between pressure gradient and Coriolis force accelerations. The vertical relation is hydrostatic balance between the vertical pressure force and gravity. They describe a slowly varying large-scale interior circulation, excluding the frictional surface/bottom layers and large ageostrophic accelerations.

Differentiate the horizontal balances vertically and use $p_z=-\rho g$, treating $f,{\rho }_0$ as vertically constant:These are the oceanic thermal wind relations. With no motion at $z=-H$, the exact surface relations involve the depth-integrated density gradients. If the measured density is representative of the vertical mean, or if those gradients are approximately uniform with depth,Equivalently these hold exactly for the horizontal gradients of $\bar{\rho }=H^{-1}{\int }_{-H}^0\rho dz$ in this fixed-depth reference model. A single arbitrary surface-density gradient does not otherwise determine a vertical integral. In the Northern Hemisphere the inferred surface current runs along horizontal isopycnals with denser water on its left when the no-motion reference and this vertical-gradient approximation apply.

For a steady density pattern and a ship whose actual track velocity is $v_s(\cos \theta ,\sin \theta )$,If the field changes during the survey an Eulerian time derivative must also be included. Here $H=\Delta z>0$ is the upward separation from reference depth to surface, so its sign must not be reversed.

Inversion and velocity estimate. Let $C=f{\rho }_0{v}_s/(gH)$ and denote the two ship measurements by $y_{\theta },y_0$, with the east-going ship assigned the subscript zero. The observation system and its inverse areIt is invertible for $\sin \theta \ne 0$. For two measurements and two unknowns this inverse is also the stated least-squares estimate; the nearly parallel-track geometry explains its poor conditioning at small angle.

The supplied parameters give $C=5.654\times 10^{-5}$ in units converting velocity to the measured density tendency. With $y_{\theta }=1.9\times 10^{-5}$ and $y_0=1.8\times 10^{-5}\mathrm{kg}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$,The current therefore has eastward and southward components. Signs follow from the positive-upward, east–north coordinate convention and the specified no-motion reference.

Uncertainty and track design. The two-track thermal-wind covariance exposes the geometrical limitation. The stated error already applies to the difference of successive density observations. Dividing by the sampling interval, rather than adding an extra factor of $\sqrt{2}$, gives ${\sigma }_y=7.2\times 10^{-4}/3600=2.0\times 10^{-7}\mathrm{kg}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$. Assume independent errors of that size for the two ships, as in the supplied covariance model. ThenAt $5^{\circ }$,The correlation is positive,Both estimates depend on the east-going ship; its positive measurement error lowers $v_g$ and also lowers $u_g$, explaining the sign of the covariance. If the other ship goes north instead, $\theta =90^{\circ }$ gives orthogonal observations: $u_g=y_{90}/C$ and $v_g=-y_0/C$. Both standard errors become $0.00354\mathrm{m}{\mathrm{s}}^{-1}$ and their covariance vanishes. The north-going track improves the poorly constrained zonal component by about a factor of sixteen, while leaving the meridional precision unchanged under the same noise assumptions.

Use a shallow, homogeneous active upper layer of mean thickness $h_0$ and mass density ${\rho }_1$, over a much deeper, effectively stationary layer with mass density ${\rho }_2>{\rho }_1$. The Boussinesq approximation keeps the small density contrast only in buoyancy, giving reduced gravity $g^{\prime }=g({\rho }_2-{\rho }_1)/{\rho }_0$. The hydrostatic approximation and shallow-water approximation make horizontal velocity nearly depth independent and turn the interface slope into the horizontal force $-g^{\prime }\nabla h$.

Linearize about rest and uniform $h_0$, with $|h-h_0|\ll h_0$, neglecting products of perturbations and convective acceleration. The f-plane approximation sets the Coriolis parameter to constant $f_0$. Neglect wind forcing, viscosity, entrainment and bottom drag. The two horizontal momentum equations then balance local acceleration, the Coriolis force and the pressure gradient. The signs $-f_{0}v$ and $+f_{0}u$ follow from eastward and northward coordinates.

The thickness equation is the linearized conservation of mass law: horizontal divergence decreases $h$, while horizontal convergence increases it. Here $h$ is active-layer thickness, not total ocean depth, and $g^{\prime }$ rather than $g$ gives the restoring force because the motion principally displaces the thermocline.

At latitude $\ell$, the Earth's angular velocity in east–north–up coordinates is $\mathbf{\Omega }=(0,\Omega \cos \ell ,\Omega \sin \ell )$. Assume a small horizontal patch, constant $\Omega$, negligible vertical velocity, and a vertical constraint balancing effective gravity. Absorb the steady centrifugal force into that effective gravity. With $f=2\Omega \sin \ell$ and physical horizontal force per unit mass $F_x^{\prime },F_y^{\prime }$, the Coriolis force givesLarge-scale atmospheric and oceanic flows deflect to their right in the northern hemisphere and left in the southern hemisphere; their balance with a horizontal pressure gradient produces geostrophic flow. A Foucault pendulum has an oscillation plane that precesses relative to the ground at angular rate $-\Omega \sin \ell$. Both effects vanish at the equator within this horizontal approximation.

Hydrostatic equilibrium balances the vertical pressure gradient against weight, $dP/dz=-\rho g$. For a dry ideal gas parcel displaced slowly enough to maintain pressure balance but rapidly enough to exchange negligible heat, the first law givesCombining the two equations yields the dry adiabatic lapse rateusing $c_p\simeq 1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Here temperature in the thermodynamic differential is absolute, $g$ and $c_p$ are treated as constant, composition is fixed, and condensation is excluded. Hydrostatic balance alone does not force the atmosphere to be adiabatic; the parcel assumption supplies the extra relation.

Water condensation releases latent heat, reducing the cooling of a rising saturated parcel and giving a smaller moist adiabatic lapse rate. Radiation, mixing, large-scale motion and incomplete convective adjustment also shape the environmental lapse rate. Thus an observed $6$--$7\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ need not equal the dry parcel value and is broadly consistent with a moist, radiatively influenced atmosphere.

Let $z$ increase down the pipe, $a=0.015\mathrm{m}$ be its radius and $r$ the distance from its axis. For steady, fully developed laminar flow of a Newtonian fluid, the axial Navier-Stokes equations reduce toWrite $K=-dp/dz+\rho g\sin 20^{\circ }$. Integrating twice, regularity at $r=0$ excludes a logarithmic term, and the no-slip boundary condition $u(a)=0$ gives the Hagen-Poiseuille flow profileThe volumetric flow rate follows by integrating over the circular cross-section:For the intended gravity-driven case with no imposed axial pressure gradient, $dp/dz=0$. Substitution givesandThe corresponding mean speed is about $0.1047\mathrm{m}{\mathrm{s}}^{-1}$ and the diameter-based Reynolds number is $\rho \overline{u}(2a)/\eta \simeq 3.49$, consistent with laminar flow. The original PDF does not specify inlet and outlet pressures. If they impose an additional pressure gradient, the general result with $K$ above applies; gravity alone does not determine that extra contribution.

Take $\mathbf{u}=(u(y),0)$ and no imposed streamwise pressure gradient. The normal momentum balance gives $p=p_{\ast }-\rho gy\cos \alpha$. The streamwise equation is $\mu u^{\prime \prime }+\rho g\sin \alpha =0$, since the convective acceleration vanishes. With no slip at the lower wall and signed upper-wall traction $\tau$ in the positive $x$ direction, the conditions areIntegration givesThe required upper-wall velocity is its endpoint value $u(h)=\rho g{h}^2\sin \alpha /(2\mu )+\tau h/\mu$. The unfactored formula remains meaningful when $\sin \alpha =0$.

The fluid exerts shear $\boxed{\mu u^{\prime }(0)=\rho gh\sin \alpha +\tau }$ per unit length on the lower plane in the positive $x$ direction. The force of that plane on the fluid has the opposite sign. The total wall force on the fluid balances its downslope weight, providing an independent sign check.

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.

Introduce $\mathcal{R}=\rho UL/\mu$. Balancing the streamwise pressure gradient $\Pi /L$ with cross-layer viscosity $\mu U/({\delta }^2{L}^2)$ givesAfter division by that viscous scale, the streamwise momentum equation isThe transverse equation, divided by its pressure-gradient scale, isContinuity scales exactly to ${\hat{u}}_{\hat{x}}+{\hat{w}}_{\hat{z}}=0$. With $\delta \ll 1$, ${\delta }^2\mathcal{R}\ll 1$ and $S=O(1)$, the leading equations are

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.

The only velocity component is axial, so incompressibility gives ${\partial }_{z}w=0$. Thus $w=w(x,y)$, and the steady convective acceleration is $w{\partial }_z(w\mathbf{k})=0$. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial momentum equation is thereforeIts left side depends only on $x,y$, while its right side depends only on $z$. Equality throughout the pipe forces both to be one constant, say $G$. Consequently the axial pressure gradient is constant, and $p(z)=Gz+p_{\ast }$. This is absence of dependence on a coordinate, rather than probabilistic independence.

An attached similarity profile must approach $f^{\prime }\to A$ as $\eta \to \infty$. Linearizing about $f\sim A\eta +\text{constant}$ gives, for the small velocity correction $g=f^{\prime }-A$,When $\alpha A>0$, one independent far-field correction decays with a Gaussian factor; when $\alpha A<0$, the two nontrivial asymptotic modes grow, one algebraically and one exponentially, so a nonconstant wall profile cannot decay into the prescribed outer flow. Thus the attached exponential similarity construction requiresEquivalently the outer speed increases along the direction of motion, giving a favorable pressure gradient. In the intended flow away from the leading edge along positive $x$, this means $A>0$ and $\alpha >0$. The sign-reversed choice $A<0,\alpha <0$ describes locally favorable flow in the opposite direction; it is not a layer developing downstream from the stated leading edge.

The asymptotic approximation also requires a geometrically thin layer, away from the leading-edge region. Comparing its thickness with the streamwise variation length $1/|\alpha |$ gives the local condition $\mathrm{Re}|A|e^{\alpha x}/|\alpha |\gg 1$. An exponentially decaying outer speed eventually violates this condition on an infinite downstream interval. These scale and orientation requirements distinguish formal similarity reduction from a uniformly valid attached boundary layer.

Let $\mathrm{Re}=\rho UL/\mu$ and use the given anisotropic scales. Continuity makes the transverse velocity scale $\delta U$ consistent with the gap height $\delta L$. The streamwise viscous stress divergence has scale $\mu U/({\delta }^2{L}^2)$, while the streamwise pressure gradient has scale $\Pi /L$. Their balance selectsThe streamwise equation then has the dimensionless formThe normal equation becomesWhen $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$, neglecting the indicated smaller terms givesThe bottom conditions express no slip on a plate moving at speed $U$ and no penetration. The top conditions express no slip and no penetration on the stationary wall. The leading pressure is a function of the streamwise coordinate alone.

For pressure lobes in a symmetric translating lubrication gap, the height integrals areHenceThe denominator of the pressure gradient is positive. Its numerator changes sign when $\sqrt{1+{\hat{x}}^2}=\pi /(2\sqrt{2})$, that is at $\hat{x}=\pm x_{\ast }$ with $x_{\ast }=({\pi }^2/8-1{)}^{1/2}$. The gradient is positive for $|\hat{x}|>x_{\ast }$, zero at those two points, and negative for $|\hat{x}|<x_{\ast }$. Integration, with either endpoint condition, givesThis function is odd and vanishes at $-1,0,1$. It has a positive maximum at $-x_{\ast }$ and a negative minimum at $x_{\ast }$. Thus the sketch has a positive pressure lobe in the converging left half of the gap and a negative gauge-pressure lobe in the diverging right half.

The flux average for a translating lubrication gap follows by integrating the pressure gradient between the endpoints:If $2Q$ were smaller than the minimum height, the integrand would be strictly positive everywhere, contradicting zero total pressure change. If it were larger than the maximum, the integrand would be strictly negative. ThereforeEquivalently,is a weighted average of the height with strictly positive weight ${\hat{h}}^{-3}$. This also identifies the constant flux uniquely.

Define the aspect ratio and the thickness-based Reynolds number byUse the nondimensionalizationThe small transverse velocity scale follows from incompressibility. Dropping stars, the base fields become $\bar{u}=2z-z^2$ and $\bar{p}=(1-z)\cot \alpha$. Before subtracting the base state, the dimensionless Navier-Stokes equation readsFor example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0{h}_0^2/(\mu L)=\mathrm{Re}\epsilon$ and the pressure gradient coefficient $2\epsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller velocity scale; this gives the displayed $\mathrm{Re}{\epsilon }^2$ coefficient.

Now substitute $u=\bar{u}+\hat{u}$, $w=\hat{w}$, $p=\bar{p}+\hat{p}$. Since ${\bar{u}}_{zz}+2=0$, ${\bar{p}}_z=-\cot \alpha$, and the base fields have no $x$ or $t$ dependence, subtraction givesThese are nonlinear equations for the deviations; small amplitude has not been assumed. The later asymptotic expansion takes $\mathrm{Re}=O(1)$ as $\epsilon \to 0$, or more generally requires the retained inertial correction to remain small.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.