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Use a fixed representative volume $V$ containing many pores. Let $\phi$ be the porosity and let $\mathbf{q}$ be the Darcy velocity, defined per unit total area. The conservation of mass isThe solid matrix is rigid and fixed, and the liquid is fully saturating and has constant mass density, so the stored mass on the left is constant. Microscopic internal liquid-solid boundaries contribute no through-flux by the no-penetration boundary condition. Applying the divergence theorem to the external flux givesSince the representative volume is arbitrary, the macroscopic continuity equation isThere is no extra factor of $\phi$ in the flux: that factor is already incorporated in the definition of the Darcy velocity. A compressible liquid or deformable matrix would require retaining the storage term.

Choose the stream function convention $u=-{\psi }_y$, $v={\psi }_x$, so that this is an incoming stagnation point flow rather than the reversed flow. The velocity field isAt the plate $v(x,0)=0$, satisfying the no-penetration boundary condition. The nonzero tangential velocity $u(x,0)=Ax$ is permitted for an inviscid flow; the no-slip boundary condition is not imposed on the Euler equations for an inviscid fluid.

The steady convective acceleration is $(A^{2}x,A^{2}y)$. Thus the Euler equations for an inviscid fluid requireIntegrating givesThe printed formula omits $\rho$ from its quadratic term. The source uses physical pressure, not pressure divided by mass density, as is clear from its $\rho gy$ term and its Navier-Stokes equation. The omitted factor is also required by dimensional analysis. For a concrete substitution check, the printed expression gives $-p_x/\rho =A^{2}x/\rho$ rather than the required $A^{2}x$ when $\rho \ne 1$ in the chosen numerical units.

Use the same stream function convention and denote differentiation with respect to the dimensionless variable $\eta =y/\delta$ by a prime. The proposed similarity solution givesIts continuity equation holds identically. Substitution in the tangential Prandtl boundary-layer equation givesChoose the constant scale $\delta =\sqrt{\nu /A}$. The Hiemenz flow equation and its boundary conditions areThe first two conditions impose the no-penetration boundary condition and no-slip boundary condition. The last matches the tangential outer velocity. One also has $f^{\prime \prime }\to 0$ and $f(\eta )/\eta \to 1$. It is too strong to impose $f-\eta \to 0$: the displacement thickness generally permits a constant offset, $f\sim \eta -B$.

The normal Navier-Stokes equation makes the approximation in part (c) explicit. With this similarity solution, it givesAt the wall this differs from $(p_{\infty }{)}_y=-\rho g$ by $-\rho A^2\delta f^{\prime \prime }(0)$. This is a concrete next-order counterexample to interpreting the printed normal pressure gradient equality exactly; the tangential Hiemenz flow equation remains valid at its stated order.

Each layer has constant mass density, initially negligible velocity and hence zero vorticity. Its Euler equations for an inviscid fluid have a conservative gravitational body force, so Kelvin's circulation theorem preserves zero bulk vorticity. In a simply connected layer, the irrotational flow therefore has ${\mathbf{v}}_i=\nabla {\varphi }_i$. The incompressible flow condition givesThis applies while the ideal layerwise solution is smooth. It allows an interfacial vortex sheet; it does not assert that the whole two-fluid flow has zero vorticity or that a smooth interface must exist for arbitrarily long times.

The rigid wall no-penetration boundary conditions arewith boundedness and ${\partial }_r{\varphi }_i=0$ at $r=0$ for regular axisymmetric fields. For a small interface displacement and small slope, linearization of the kinematic boundary condition about $z=0$ givesThe small-slope condition is an additional requirement for high-wavenumber modes: $|\alpha \eta |\ll 1$, not just $|\eta |\ll h,a$.

Choose the unperturbed hydrostatic pressures as ${\bar{p}}_i(z)=p_{\ast }-{\rho }_{i}gz$. The linearized Bernoulli equation gives the pressure perturbation $p_i^{\prime }=-{\rho }_i{\partial }_t{\varphi }_i$, after absorbing spatially constant gauges into ${\varphi }_i$. Since surface tension and viscous normal stress vanish, pressure continuity on $z=\eta$ givesorThe initial velocity is zero, so ${\eta }_t(r,0)=0$. Separate conservation of mass in each closed layer also requires ${\int }_0^a\eta (r,t)rdr=0$ relative to the initially horizontal interface.

Write $\eta =H{e}^{\sigma t}{J}_0(\alpha r)$ using the Bessel function of the first kind. The wall boundary conditions at $z=\pm h$ selectBecause $J_0^{\prime }(s)=-J_1(s)$, the sidewall no-penetration boundary condition requireswhere $j_{1,n}$ are the positive zeros of $J_1$. These radial eigenfunctions automatically preserve layer volume becauseThe constant radial mode $\alpha =0$ cannot change the interface height in a closed tank: the Laplace equation and the impermeable lids then give zero vertical velocity, and a uniform displacement would change each layer's volume.

The kinematic boundary condition fixesThus the interface values are ${\varphi }_1(0)=\sigma \eta \coth (\alpha h)/\alpha$ and ${\varphi }_2(0)=-\sigma \eta \coth (\alpha h)/\alpha$. Substitution in the normal stress condition gives the finite-depth Rayleigh-Taylor dispersion relation:soHere $\mathcal{A}$ is the Atwood number. Every allowed mode has a positive growth rate and is subject to Rayleigh-Taylor instability. Since $d[\alpha \tanh (\alpha h)]/d\alpha =\tanh (\alpha h)+\alpha h{\mathrm{sech}}^2(\alpha h)>0$, shorter radial scales grow faster. The limits areThere is no fastest finite wavenumber in the ideal sharp-interface model without surface tension, viscosity or diffusion. The zero-initial-velocity condition uses both exponential branches: an initial mode $H{J}_0({\alpha }_{n}r)$ evolves as $H\cosh ({\sigma }_{n,+}t)J_0({\alpha }_{n}r)$. If the mass density ordering were reversed the same calculation would give oscillatory stable modes; equal mass densities give neutral modes.

Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.

Let the perturbed interface be $z=\eta (x,y,t)$. For inviscid disturbances about rest, the initially irrotational velocities remain potential flows,in the upper layer $0<z<h$ and lower layer $-h<z<0$. The rigid walls impose no-penetration boundary conditionand zero normal derivative at $x,y=0,2h$. Linearizing the kinematic and pressure-continuity conditions at $z=0$ gives

The horizontal Neumann eigenfunctions arewhere $m,n$ are nonnegative integers, not both zero. For time dependence $e^{-i\omega t}$, the vertical factors satisfying the rigid lids areThe kinematic condition gives their interface valuesSubstitution into the dynamic condition yields the interfacial gravity-wave dispersion relationFor ${\rho }_2>{\rho }_1$, the lowest nontrivial wavenumber is $k=\pi /(2h)$, with the two degenerate modes $(m,n)=(1,0)$ and $(0,1)$. If ${\rho }_1\ll {\rho }_2$, the relation reduces to the finite-depth free-surface result ${\omega }^2\simeq gk\tanh (kh)$. If ${\rho }_1>{\rho }_2$, then ${\omega }^2<0$ and disturbances grow exponentially: this is the Rayleigh-Taylor instability.

The material derivative follows the interfacial surfactant. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ describes dilution by tangential expansion and concentration by compression. The term $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing surface area due to normal motion of a curved interface. Finally, $D_s{\nabla }_s^{2}C$ describes surface diffusion. Together these give conservation of insoluble surfactant on a moving interface.

The relevant surface Péclet number isWrite $S=\mathbf{n}\cdot \mathbf{A}\cdot \mathbf{n}$ and ${\mathbf{u}}_s=\mathbf{Ax}-S\mathbf{x}$. Since $\mathbf{A}$ has zero trace and ${\nabla }_s\mathbf{n}={\mathbf{I}}_s/a$, the surface divergence isThere is no normal motion. To first order in ${\mathrm{Pe}}_s$, the steady surfactant balance becomes $D_s{\nabla }_s^2{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. The traceless quadratic $S$ is a degree-two spherical harmonic, so ${\nabla }_s^{2}S=-6S/a^2$. The mean of $C^{\prime }$ is zero by total surfactant conservation; consequentlyThe discarded advective term ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$ is smaller by ${\mathrm{Pe}}_s$. This is the quadrupolar surfactant distribution on a spherical interface.

With the unit normal directed from the bubble into the exterior, the interfacial stress balance with variable surface tension isUsing the specified first-order curvature and $\gamma ={\gamma }_0-K{\gamma }_{1}S$, together with ${\nabla }_{s}S=2{\mathbf{I}}_s\mathbf{An}/a$, givesThe first term is the spherical capillary pressure; the last contains the normal tension correction and the tangential Marangoni stress.

The ambient Stokes flow has the symmetry of the symmetric traceless tensor $\mathbf{E}$. In the Unscaled Papkovich–Neuber representation, ${\chi }_{\infty }=\mathbf{x}\cdot \mathbf{E}\cdot \mathbf{x}/2$ is harmonic and yields $\mathbf{Ex}$. The decaying vector potential must have the form $\mathbf{E}\nabla (1/r)$, and the scalar disturbance must have the degree-two form $\mathbf{E}:\nabla \nabla (1/r)$. They are harmonic outside the bubble and have precisely the required rotational covariance. Dimensionless amplitudes may therefore be writtenFor a steady bubble, the no-penetration boundary condition gives $1+P-3Q=0$. Its tangential velocity gives $\alpha =1+2Q$. The tangential stress boundary condition then yieldsHence $\alpha =5/(5+2M)$, $Q=-M/(5+2M)$ and $P=-(5+5M)/(5+2M)$. The inviscid interior supplies only the constant pressure balancing $2{\gamma }_0/a$. Matching the remaining normal stress givesEliminating $P,Q,\alpha$ givesAs $M\to \infty$, $\alpha \to 0$: the Marangoni stress suppresses tangential motion and effectively immobilizes the interface. This limit must retain the small surface Péclet number and small-deformation assumptions; $M$ can grow through increasing ${\gamma }_1$ without invalidating the linear concentration approximation.

For a velocity potential, $\mathbf{u}=\nabla \varphi$. The incompressible flow condition $\nabla \cdot \mathbf{u}=0$ therefore becomes Laplace equation ${\nabla }^2\varphi =0$. The Laplace equation in polar coordinates givesso positivity of $\gamma$ and $\lambda$ requiresThe velocity components areThe no-penetration boundary condition on both walls says $u_{\theta }=0$ at $\theta =0$ and $\theta =\alpha$, hence $\lambda \alpha =n\pi$. If $n\ge 2$, $\sin (n\pi \theta /\alpha )$ changes sign inside the wedge. The stated sign condition therefore selects $n=1$, and

Equation (1) is Laplace equation for the velocity potential. It expresses incompressibility, $\nabla \cdot \mathbf{u}=0$, together with $\mathbf{u}=\nabla \varphi$. Equation (2) is the linearized dynamic free-surface condition: the unsteady Bernoulli equation and constant atmospheric pressure balance the potential acceleration against the gravitational restoring term $g\zeta$. Equation (3), with the printed $c$ understood as $\zeta$, is the kinematic free-surface condition: the surface moves with the vertical fluid velocity. Equations (4) and (5) are no-penetration boundary conditions at the rigid bottom and cylindrical sidewall, respectively.

Conservation of volume flux through a circular arc of radius $r$ givesso the outer flow is the sink flow in a sectorThis is a potential flow and already satisfies the no-penetration boundary condition on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the no-slip boundary condition. Thin vorticity-bearing boundary layers are needed to reduce that speed to zero at each wall.

Let the unit normal $\mathbf{n}$ point from the fluid toward the free surface, let $\mathbf{t}$ lie along the surface, and write the swimmer direction asReflecting the swimmer position across the plane and reflecting its orientation toconstructs the free-surface image of a force dipole. At the surface the two dipoles have equal tangential velocity and opposite normal velocity. Their sum therefore satisfies the no-penetration boundary condition; tangential velocity is even across the plane and normal velocity is odd, so the tangential traction vanishes and the stress-free boundary condition is also satisfied.

The vector from the image to the swimmer is $\mathbf{r}=-2h\mathbf{n}$, for which $({\mathbf{p}}^{\ast }\cdot \mathbf{r}{)}^2=4h^2{\sin }^2\theta$. Evaluating the image force-dipole flow at the swimmer givesThere is no tangential image velocity at the swimmer in this point-dipole approximation. With the convention that $\mathcal{P}>0$ is an extensile pusher microswimmer, a nearly parallel pusher is attracted toward the surface, whereas a nearly parallel contractile puller microswimmer is repelled. For $|\sin \theta |>1/\sqrt{3}$ the normal drift reverses because the image samples the axial rather than equatorial part of the dipolar flow.

Use $u={\psi }_Y$, $V=-{\psi }_x$, and put $\eta =Y/x^{\gamma }$, $\psi =x^{\beta }f(\eta )$. ThenMatching $u\sim x^m$ requires $\beta -\gamma =m$. Equating the powers in streamwise inertia $x^{2m-1}$ and transverse diffusion $x^{m-2\gamma }$ givesThe two convective terms combine to $x^{2m-1}[m(f^{\prime }{)}^2-\beta f{f}^{\prime \prime }]$, while $U{U}^{\prime }=m{x}^{2m-1}$. Hence the Falkner-Skan equation in this scaling isThe additive stream-function constant is fixed by $f(0)=0$; when $\beta \ne 0$ this also encodes no penetration. For $\beta =0$ it is a harmless choice of additive constant choice. Existence of an attached physical branch for a given $m$ is a separate issue from deriving the similarity reduction.

On an attached branch with $w=f^{\prime }>0$ away from the wall, use $f$ itself as the independent variable. The chain rule gives $f^{\prime \prime }=w{w}_f$ and $f^{(3)}=w(w_f^2+w{w}_{ff})$. ThereforeIf the wall shear $s=f^{\prime \prime }(0)>0$, then $f\sim s{\eta }^2/2$, so $w(f)\sim \sqrt{2sf}$. Thus $w_f$ is singular at zero while $w{w}_f\to s$ is finite. Imposing a finite arbitrary $w_f(0)$ would remove the usual attached branch. This is the velocity as dependent variable in the Falkner-Skan equation reduction.

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.

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 $2U_0/3$.

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.

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.

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.