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Take $\Delta \rho >0$ and $\tau >0$, with $x$ pointing along the imposed shear stress, and put $G=\Delta \rho g$. Relative to the upper fluid's hydrostatic reference, the hydrostatic pressure in the current is $p=G(h-z)$. Thus the horizontal pressure gradient is $G\nabla h$. The lubrication theory momentum balance, lower no-slip boundary condition, and imposed upper shear stress giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe two terms have opposite roles: imposed shear stress carries fluid downstream, whereas the hydrostatic pressure gradient spreads it from thick regions towards thin ones.

For the approximation, a representative horizontal velocity is $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis condition controls viscous normal stress, vertical viscous corrections and the normal projection of shear at a slightly tilted interface relative to $GH$. With it and the previous inertia bounds, vertical inertia is small as well. Time variations here are on the transport timescale $L/U$; separately imposed rapid forcing would need its own unsteady inertia bound. A small ordinary Reynolds number is a stronger sufficient restriction, but the reduced inertia ratios above are what the thin-layer momentum balance directly requires.

In the steady far-downstream regime, cross-stream derivatives dominate gravity-driven spreading. Dropping the smaller downstream gravity flux leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, with $h=0$ outside. The squared height is a parabolic Barenblatt solution; the height cross-section is a semicircular profile after rescaling its axes. The cross-stream volume flux vanishes at the edges even though the height slope becomes singular there. The profile describes the outer lubrication theory region, not a resolved microscopic front.

Near the point source the two horizontal dimensions are comparable, say $\ell$. Upstream spreading arrests where outward gravity-driven volume flux balances downstream shear-driven volume flux:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. The same horizontal scale is obtained by setting $y_N\sim x$ in the downstream similarity solution, confirming where that solution fails.

For the line source the steady volume flux per unit width is $q=A{h}^2-D{h}^3{h}_x$. On the downstream constant-height branch, $q=Q_{2d}$ gives $h_d^2=Q_{2d}/A$. Upstream there is no net flux through the finite nose, so $q=0$. Thus $h{h}_x=A/D$, and continuity of height at the source givesThe line-source upstream reach under imposed shear is thereforeIn the upstream part, shear-driven and gravity-driven volume fluxes cancel. The volume flux per unit width jumps by exactly $Q_{2d}$ at the source. As in the point-source profile, the square-root nose is a formal outer solution with an unresolved steep front; its finite upstream reach follows from flux balance, not from assuming a small slope all the way to the nose.

Use the well-mixed natural ventilation model for a distributed floor source. Let the building height be $H_b$, its volume be $\mathcal{V}$, its temperature excess be $\theta$, and the exterior mass density be $\rho$. Let ${\alpha }_T$ be the coefficient of thermal expansion, so reduced gravity is $g^{\prime }=g{\alpha }_T\theta$. Convert the total heat input $Q$ to the temperature-volume flux $\mathcal{H}=Q/(\rho c_p)$, where $c_p$ is specific heat capacity. If the quoted heat flux is per floor area, first multiply it by that area.

For effective opening area for pressure-driven ventilation, define the equal-opening hydraulic coefficient $K_A$ such that the signed throughflow $q$ obeys $q|q|=K_A\Pi$, with $\Pi$ the total driving pressure divided by mass density. In the effective-area convention of question 6, each opening has discharge $A\sqrt{\delta p/\rho }$, so $K_A=A^2/2$. The usual physical-area orifice law instead gives $K_A=C_d^2{A}^2$; that prefactor does not affect the stability conclusions.

Set $W=\Delta P/\rho >0$ and $G_T=g{\alpha }_T{H}_b$. Positive $q$ denotes entry at the floor and exit at the roof. hydrostatic pressure and wind combine to giveBecause the inflowing air is at exterior temperature in either direction, the heat budget isThe well-mixed ventilation temperature balance at steady state is thereforeAll roots must satisfy $\theta >0$ and their appropriate flow-direction inequality; squaring a pressure relation without imposing those conditions could add spurious branches.

The assisting case has a unique positive solution and it is stable. For opposing wind, define ${\theta }_w=W/G_T$. There is always one solution with $\theta >{\theta }_w$, a buoyancy-dominated upward flow. Below ${\theta }_w$, the heat-removal curve is $R(\theta )=\sqrt{K_A}\theta \sqrt{W-G_T\theta }$. It is zero at both endpoints, and its maximum occurs at $\theta =2{\theta }_w/3$:For $0<\mathcal{H}<R_{\max }$ there are two downward-flow roots as well as the upward-flow root. Equivalently, setting $y=G_T\theta /W$ and $\epsilon =G_T\mathcal{H}/(\sqrt{K_A}{W}^{3/2})$, the equilibrium curve isFor linear stability analysis of the opposing-wind ventilation bistability, linearize the heat budget: the eigenvalue is $-R^{\prime }(\theta )/\mathcal{V}$. The cooler downward root has $0<\theta <2{\theta }_w/3$ and $R^{\prime }>0$, so it is stable. The warmer downward root has $2{\theta }_w/3<\theta <{\theta }_w$ and $R^{\prime }<0$, so it is unstable and separates the two basins. The upward root has $R^{\prime }>0$ and is stable. At $\mathcal{H}=R_{\max }$ the two downward roots merge at a saddle-node bifurcation; above that value only the upward equilibrium remains.

For ventilation switching under wind and heating changes, increasing wind pressure or decreasing heat input both reduce $\epsilon$, so both can create the pair of wind-driven equilibria. They nevertheless have different physical and transient effects. On the hot upward branch, implicit differentiation of $\mathcal{H}=\sqrt{K_A}\theta \sqrt{G_T\theta -W}$ shows that increasing $W$ increases $\theta$, while decreasing $\mathcal{H}$ decreases $\theta$. Both reduce the upward ventilation rate, but only a wind change shifts ${\theta }_w$.

Under slow parameter variation, a state follows its current stable branch while that branch persists; the upward branch does not disappear when wind increases or heating decreases. A sufficiently large abrupt wind increase can raise ${\theta }_w$ beyond the current temperature and place that state below the new unstable threshold, causing flow reversal and cooling towards the wind-dominated equilibrium. A jump insufficient to cross its basin boundary returns to the upward equilibrium.

At fixed $W$, reducing a still-positive heat input cannot force an initially upward-flow state through ${\theta }_w$: at that boundary the temperature balance has $\dot{\theta }=\mathcal{H}/\mathcal{V}>0$. Thus heat reduction alone does not cause that reversal in this model. Conversely, decreasing wind or increasing heat from the stable downward branch eventually destroys it at the fold and forces a switch to the upward branch, giving history dependence. Wind strengthening and heat reduction have the same effect on the steady dimensionless control parameter, but not the same temperature change or switching dynamics. These stability statements use quasi-steady opening flow and a single well-mixed temperature; other stratification or airflow-inertia models require their own stability analysis.

Let $q_b$ be floor inflow and $q_t$ roof outflow. Use effective-area discharge $A\sqrt{\delta p/\rho }$, matching the factors in the supplied relations. With negligible source volume, both equal $q$. The lower region has exterior mass density while the warm upper layer has reduced gravity $g^{\prime }$ and depth $H-h$. hydrostatic pressure therefore supplies a total opening pressure head $g^{\prime }(H-h)$. Equal openings sharing equal discharge take half each, givingSteady lower-layer volume conservation sets this ventilation rate equal to the entrained turbulent plume flux at the interface, $q=\lambda B^{1/3}{h}^{5/3}$. Upper-layer buoyancy conservation gives $B=q{g}^{\prime }$. These are the required balances for displacement ventilation:Eliminating $g^{\prime }$ and $q$ gives the displacement-ventilation interface height equationThe left side is strictly increasing from zero to infinity, so there is one interface height. It depends on opening geometry and plume entrainment, not on $B$; the resulting $g^{\prime }=B^{2/3}/(\lambda h^{5/3})$ and throughflow do depend on $B$.

For finite source volume $Q$, conservation instead gives $q_t=q_b+Q$. The two pressure drops are no longer equal:Let $q_p(h;Q)$ be a physically entraining source plume: $q_p(0;Q)=Q$ and $q_p(h;Q)>Q$ for $h>0$. Its interface balance is $q_p=q_t=Q+q_b$. Therefore, as $q_b$ reaches zero, a positive-depth lower layer has no replenishment and cannot remain steady: this is no steady displacement layer without ambient supply. The limiting interface is $h=0$. At the threshold $q_t=Q$ and $g^{\prime }=B/Q$, and the entire hydrostatic head drives the roof discharge. HenceThis is source-volume blocking of displacement ventilation. The half-head factor of the negligible-source state must not be kept after the floor inflow vanishes. In the usual physical-area convention $q=A_{\mathrm{phys}}{C}_d\sqrt{2\delta p/\rho }$, the equivalent formula is $Q_c=(2C_d^2{A}_{\mathrm{phys}}^{2}BH{)}^{1/3}$.

The quoted pure-plume law cannot be used unmodified down to the origin for a source with nonzero volume: it would incorrectly give zero flux there. A compatible finite-source example uses a virtual origin,It recovers the supplied pure-plume limit when $Q\to 0$, and gives the same blocking threshold. The complete finite-$Q$ interface trajectory depends on that plume model, but the threshold needs only source-volume conservation and positive entrainment. For $Q>Q_c$, the assumed steady regime with a lower inflow no longer exists.

Take $A>0$ for the repulsive molecular interaction, $p_0$ as the common reference pressure, and let $T_i=T_m-\delta T$ be the ice-film temperature. All ratios involving $T_m$ use its absolute value, approximately $273.15\mathrm{K}$, rather than the numerical Celsius value zero. The premelting film permits water to reach and freeze at the bottom of the ice while the upper surface is displaced upward without freezing there.

For equal phase mass densities, equality of the solid and liquid chemical potentials gives the Clapeyron pressure relation for equal-density phases. Indeed, $d\mu =-sdT+dp/\rho$ and $s_l-s_s=L/T_m$, so expansion around coexistence at $(T_m,p_0)$ yieldsThis is a pressure difference between phases, not a same-pressure Clausius-Clapeyron slope. The molecular disjoining pressure supports this difference:Treat the normal solid load as its effective pressure in this planar model. The gravitational load corresponding to hydrostatic pressure is $p_s=p_0+\rho gh$, so $p_l=p_0+\rho gh-p_T$. Refer the bath pressure to the sheet's upper gravitational datum, or neglect the sheet-scale hydrostatic head. The Darcy flow law then gives the upward supply and the ice growth rateHere $\Pi$ is permeability of a porous medium, $\mu$ is dynamic viscosity, and equal densities identify supplied liquid volume with added ice volume to leading order in $a/h$.

To recover the printed law, use the usual hydraulic control of frost heave idealization: the sheet's top is at $T_m$, the liquid film and ice have a common thermal conductivity $K$, and the flow is slow enough that latent-heat production is small compared with the conductive heat passing through the layer. In addition to a quasi-steady ice temperature field, this needsThe leading heat flux is consequently continuous through the film and ice. Their thermal resistances giveLet $B_T=\rho L\Delta T/T_m$, a pressure scale. The thermomolecular pressure and the molecular law then giveSubstitute this pressure into the Darcy flow law:The first term draws water toward the undercooled ice; the increasing gravitational load eventually cancels it.

Quasi-steady temperature alone is not sufficient to fix this exact prefactor. With film conductivity $K_l$, ice conductivity $K_s$, and nonnegligible latent heat, the appropriate additional balance isTogether with the Darcy equation, this is the implicit growth model. Even in the slow-flow limit, a conductivity ratio $K_s/K_l=r$ changes the leading driving pressure by $r^{3/4}$. Taking $r=2$ is a counterexample to obtaining the printed prefactor from quasi-steadiness alone. Thus the boxed equation is the intended equal-conductivity, thin-film, hydraulically limited model, rather than a consequence of only the stated quasi-steady assumption. This distinction is the heat-balance correction to a premelted-film growth model.

For that reduced equation defineThe scale $h_{\ast }$ balances $F{h}_{\ast }^{-3/4}$ against $\rho g{h}_{\ast }$; $F$ has units of pressure times length to the power $3/4$, so $h_{\ast }$ is a length and $t_{\ast }$ is a time. They yield

Set $u={\eta }^{7/4}$. The equation becomes linear:For initial thickness $\eta (0)={\eta }_0\ge 0$, the explicit solution of gravity-limited frost heave isThe usual zero-initial-thickness sketch usesIts early and late behaviour areFor $0<\eta <1$, the derivative is positive and ${\eta }^{\prime \prime }=(-\frac{3}{4}{\eta }^{-7/4}-1){\eta }^{\prime }<0$, so the curve is increasing and concave downward. It approaches the stable thickness $h_{\ast }$ exponentially. Initial thickness above $h_{\ast }$ instead relaxes downward; at $h_{\ast }$ it is stationary.

The singular slope at zero belongs to the formal reduced solution. Since $a/h\propto h^{-3/4}$, the thin-film assumption eventually fails as $h\to 0$. The early power law therefore describes an intermediate continuum regime after any microscopic startup, not arbitrarily early physical times. A finite positive ${\eta }_0$ supplies a regular initial condition when that regime begins. Gravity limits the ultimate thickness; molecular suction and the Darcy resistance set the growth toward it.

Apply mass conservation to a fixed horizontal region $D$. The layer mass is $\rho {\int }_{D}hdxdy$ and the outward horizontal flux is $\rho {\int }_{\partial D}h(u,v)\cdot \mathbf{n}ds$. The divergence theorem and arbitrary choice of $D$ giveThis depth-integrated law retains free-surface motion; neglecting the vertical velocity in horizontal momentum is not the assumption that the depth cannot change.

Take the free-surface pressure to be constant. hydrostatic pressure gives $p=p_{\mathrm{atm}}+\rho g(h-z)$, so the horizontal pressure forces are $-g{\eta }_x,-g{\eta }_y$. Linearization about a resting layer gives the linearized shallow water equationsTaking the curl of the momentum equations yields ${\zeta }_t=-f(u_x+v_y)$. Combining with continuity gives the linearized shallow-water potential-vorticity anomalyFor $Q_0=0$, $\zeta =f\eta /h_0$. Taking the divergence of momentum gives ${\partial }_t(u_x+v_y)-f\zeta =-g{\nabla }_h^2\eta$. Eliminate the divergence with continuity to obtainFor $\theta =k(x-ct)$, substitution of $\eta ={\eta }_0\cos \theta$ gives the inertia-gravity wave dispersion relationThe associated purely oscillatory velocity field isThese formulas satisfy all three linearized equations and the zero-anomaly condition. A separately added spatially uniform inertial oscillation is not part of this monochromatic wave.

To first order in amplitude evaluate the wave at a particle's equilibrium coordinate $x_{\ast }$, so ${\theta }_{\ast }=k{x}_{\ast }-\omega t$. Integrating its velocity gives displacements relative to the orbit centreThese are particle ellipses for rotating shallow-water waves, with axis ratio $V/U=f/\omega <1$. For the stipulated positive signs, the velocity at phases $0,\pi /2,\pi ,3\pi /2$ points respectively right, up, left, down, with magnitudes $U,V,U,V$. The particle orbits are clockwise as time advances, since the phase decreases.

Take $z$ positive upwards and let $f$ denote axial body force per unit volume. With inertia and surface tension omitted, the extensional equations for a slender Newtonian column areMass conservation gives the first equation. The second uses the Trouton ratio three: the leading radial velocity is $u_r=-r{w}_z/2$, the radial normal traction gives $p=p_{\mathrm{ext}}-\mu w_z$, and the axial stress is $-p_{\mathrm{ext}}+3\mu w_z$. An axial control volume balance then gives the displayed equation. The approximation requires a slender column and slowly varying radius.

Let $h=R-a$. Since the container base is closed, the total volume flux at every section is zero. The column carries flux $\pi a^{2}w$, so the annulus carries $-\pi a^{2}w$ and has mean speed of order $a|w|/h$. Its interfacial shear stress is therefore of order $\lambda \mu a|w|/h^2$. Transmitting this stress through the core produces an axial velocity variation of order $\lambda a^2|w|/h^2$. The plug-flow criterion for a column in a low-viscosity annulus is consequently

Use modified pressure $P=p_{\mathrm{ext}}+\rho gz$, removing the annular fluid's hydrostatic pressure. In a local planar gap, with $y=0$ at the column and $y=h$ at the container wall, lubrication theory givesThe required return flux per unit circumference is $q_g=-aw/2$ to leading order in $h/a$. Its pressure-driven component is larger than the Couette component by $a/h$, so the annular return flow around a slumping column givesThe annular shear contribution to the integrated axial force is smaller than this pressure-gradient term by $h/a$. Retaining the leading terms and substituting $f=-(\rho +\Delta \rho )g$ gives the dimensional equationsThe boundary conditions are $w(0,t)=0$ at the base and $w_z(L(t),t)=0$ at the free upper end; the latter expresses zero excess axial extensional stress.

For $\lambda =0$ and initially constant $a=a_0$, put $G=\Delta \rho g/(3\mu )$. Then $w_{zz}=G$, so

For $\lambda >0$, defineThis function records the leading thin-annulus resistance; differences from curvature and the Couette term enter at the already neglected relative order $h/a$. Choose the extensional screening length of a slumping column and the associated scales asHere $h_0=R-a_0$ denotes the initial annular gap in this question. With $Z=z/\hat{z}$, $T=t/\hat{t}$ and $W=w/\hat{w}$, the dimensional equations reduce to

Initially $\varphi ={\varphi }_0$ and ${\Lambda }_0=L_0/\hat{z}$, so $W_{ZZ}=1+W$, with $W(0)=0$ and $W_Z({\Lambda }_0)=0$. The initial velocity profile of an annularly confined column isThe column thickens everywhere below its stress-free top, while that initial thickening rate is zero at the top.

For a tall column, ${\Lambda }_0\gg 1$, the limiting forms away from exponentially small end corrections areThe bulk falls at approximately $-\hat{w}$, where excess weight balances annular hydraulic resistance. Extensional stresses adjust this velocity to zero in a basal boundary layer of thickness $\hat{z}$; almost all thickening occurs there.

For a short column on this scale, ${\Lambda }_0\ll 1$, while still slender, the limits areThe annular-resistance term $W$ is small, so excess weight is balanced predominantly by extensional viscous stress. The top speed is $-\hat{w}{\Lambda }_0^2/2$, recovering the zero-annular-viscosity result. Thus $\hat{z}$ is the vertical distance over which extensional stress communicates the basal constraint before annular resistance screens it. The sketches below use separate natural normalizations for the two limits.

At leading order in the lubrication approximation, surface arclength is horizontal distance. An insoluble surfactant is transported by the surface velocity, without bulk exchange or diffusion. Conservation of insoluble surfactant on a moving interface therefore becomesThe local hydrostatic pressure and capillary normal stress giveSolve $\mu u_{zz}=p_x$ with no-slip boundary condition at $z=0$ and Marangoni stress $\mu u_z(h)={\gamma }_x=-A{C}_x$. The velocity, surface velocity and volume flux per unit width areMass conservation, $h_t+q_x=0$, yields the thin-film equations with insoluble surfactant:and the corresponding surfactant equation isThe hydrostatic and capillary terms thus affect both the film flux and the surface transport, with different coefficients.

For finite-mass Marangoni spreading on a liquid film, first neglect both pressure-gradient terms. If the spread has size $\ell$, then $C\sim M/\ell$ and $C_x\sim M/{\ell }^2$, giving surface speed $u_s\sim AM{h}_0/(\mu {\ell }^2)$. Equating $\ell /t$ to this speed givesSet $\ell =(AM{h}_{0}t/\mu {)}^{1/3}$ exactly as a similarity scale, and writeOn the positive half of the pool, substitution produces two ordinary differential equations:Symmetry fixes the flux integration constant in the second equation to zero, giving $H{\Gamma }^{\prime }=-\eta /3$ wherever $\Gamma >0$. Substitution into the first equation gives $\eta H^{\prime }=H$. Thus the linear similarity profiles for surfactant spreading have the formThe concentration vanishes at the moving front. The two integral constraints are total surfactant mass and conservation of the fluid volume relative to the undisturbed layer:The first gives $k={\eta }_N^2/3$; the second gives $k{\eta }_N=2$. Therefore ${\eta }_N^3=6$. In physical variables, the solution isOutside the pool the leading outer solution has $h=h_0$, $C=0$. In particular, $h(x_{N-},t)=2h_0$: the subscript means the one-sided limit just behind the front. As a consistency check, the surface speed there is $x_N/(3t)={\dot{x}}_N$ and the fluid jump satisfies the moving-front mass conservation condition. This ideal outer solution has a height jump from $2h_0$ to $h_0$.

That jump cannot persist in a physical interface with nonzero gravity or surface tension. Across a smoothing region of width $\Delta$, $h$ changes by $O(h_0)$. The Marangoni flux scales as $AM{h}_0^2/(\mu x_N^2)$. At $g=0$, its balance with the capillary flux ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$ gives capillary smoothing of a Marangoni front:Here the surface tension approaches ${\gamma }_0$ at the surfactant-free edge. The assumed concentration-gradient scale is $M/x_N^2$ even inside the smoothing region.

When gravity dominates capillarity, the hydrostatic flux is $\rho g{h}_0^4/(\mu \Delta )$. The same balance gives gravity smoothing of a Marangoni front:The gravity-dominated condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$, the square of the capillary length. The printed condition omits $\rho$; the expression above restores dimensional consistency.

This width grows faster than $x_N$. It becomes comparable with the pool size whenThe time is a scaling estimate, so numerical factors cannot be fixed by the width balance. For $t\gg t^{\ast }$, the sharp-front similarity profile no longer describes the film: hydrostatic pressure levels the layer over the whole pool, and its thickness tends towards $h_0$ with an increasingly small depression beneath the surfactant. The surface can still spread by Marangoni stress while a hydrostatic-pressure-driven return flow nearly cancels the net film flux. In this gravity-levelled surfactant film, $q\simeq 0$ gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$, so the fractional height variation is of order $AM/(\rho g{h}_0^2{x}_N)\ll 1$.

Use pressure with the ambient hydrostatic pressure removed, and let $\Delta P$ be the pressure below the falling sphere minus that above it. Away from the sphere the tube flow is Hagen-Poiseuille flow. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their pressure drops add:The large exterior reservoir returns this flux with negligible pressure loss compared with the long narrow tube. The sphere is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.

Now use a frame translating downwards with the sphere, and take $z$ positive upwards from its center. The sphere is stationary and the tube wall moves upwards with speed $U$. The narrow gap isIt has axial length scale $a\sqrt{ϵ}$, making lubrication theory appropriate. The sphere-frame flux in a tube is constant: the upward relative flux through the whole section is $\pi a^{2}U-Q$, so its flux per unit circumference isWith $y=0$ on the sphere and $y=h$ on the wall, Couette-Poiseuille flow in a thin gap givesSince $\Delta P=-\int p_{z}dz$, the pressure-jump relation isThe parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt{ϵ})$. The lubrication resistance integrals for a nearly occluding sphere areSetting $\xi =\tan \theta$ evaluates $J_1=\pi$, $J_2=\pi /2$, $J_3=3\pi /8$, hence