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The Navier slip boundary condition allows a fluid to slide tangentially past a solid wall while resisting that motion through wall friction in the Navier slip boundary condition. For parallel flow past a flat, impermeable wall translating uniformly with tangential velocity $U_w$, write the fluid's tangential velocity as $u$ and its relative tangential velocity as $u_{\mathrm{rel}}=u-U_w$. Thenwhere $b$ is the slip length, also written ${\ell }_s$. The unit normal $\mathbf{n}$ points from the wall into the fluid, and ${\partial }_{n}u=\mathbf{n}\cdot \nabla u$ is the normal derivative. The separate no-penetration boundary condition requires $(\mathbf{u}-{\mathbf{U}}_w)\cdot \mathbf{n}=0$: tangential sliding does not allow flow through the wall.

The sign convention matters. In a channel $0<z<H$, the lower wall has inward unit normal $+{\mathbf{e}}_z$ and the upper wall has inward unit normal $-{\mathbf{e}}_z$. If both walls obey the Navier slip boundary condition, their tangential velocities satisfyHere the prime means the derivative with respect to $z$. Using $+b_H{u}^{\prime }(H)$ at the upper wall would reverse the friction law. The examples Couette flow with a slipping lower wall and planar Poiseuille flow with equal wall slip use different upper-wall boundary conditions.

This flat-wall formula is a local linear model. On a curved wall the Newtonian fluid stress tensor must supply the tangential traction; the normal derivative alone generally misses curvature terms. The author-written review by Lauga, Brenner and Stone states the full strain-based condition with the unit normal directed into the liquid. ^[ref]

The same review credits Navier's 1823 treatise with the linear slip law. Its history is a reminder that the no-slip boundary condition is an experimentally successful model, whose validity depends on the interface and scale. The history here follows that review rather than a new reading of Navier's original memoir. ^[ref]

Use the thin-annulus approximation stated in the source, so $u_r^{\prime }+ik{u}_z=0$. LetHere ${\kappa }_{\mathrm{ep}}^2$ is the conventional Rayleigh discriminant, also the squared local radial epicyclic frequency when positive. For $k\ne 0$ and $\omega \ne 0$, the azimuthal and axial mode equations giveThe thin-gap incompressible flow condition then gives $\pi =i\omega u_r^{\prime }/k^2$. Substitute these into the radial equation:ThusTo put this into exactly the form printed in the paper, defineThe narrow-gap centrifugal stability equation is consequentlyThe two rigid cylinders are impermeable, so the no-penetration boundary condition givesThere is no inviscid no-slip condition on the tangential or axial components.

The sign deserves emphasis: the source's printed plus sign makes its $\Phi$ the negative of the conventional Rayleigh discriminant. Using the conventional positive discriminant instead requires a minus sign in front of that potential term. These are equivalent conventions; replacing one without adjusting the other would falsely predict instability of solid-body rotation. The $k=0$ and marginal $\omega =0$ cases, excluded by these eliminations, must be treated separately or by limits.

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.