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Consider a steady incompressible flow with zero vorticity past a gently varying bottom $z=\eta (x)$, write the velocity potential as $\Phi =Ux+\varphi$, where $U$ is the uniform background speed and $|\nabla \varphi |\ll U$. Take $z$ upward and let $p_{\mathrm{ref}}$ be the background pressure at $z=0$. The Bernoulli equation isKeeping first-order terms and evaluating at the bottom givesThe first term is the hydrostatic pressure change due to elevation; the second comes from the perturbed kinetic energy per unit volume. The symbol $\eta$ denotes bottom height, while $\varphi$ denotes the perturbation velocity potential.

The linearized no-penetration boundary condition is ${\varphi }_z(x,0)=U{\eta }_x$. Thus the bottom geometry first determines a perturbation flow, whose tangential velocity supplies the dynamic pressure correction. The approximation neglects both the squared perturbation velocity and products of the small height with perturbation derivatives.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in a two-dimensional potential flow. Suppose an incident complex potential of a two-dimensional potential flow $f$ is analytic on and inside $|z|=a$, with its physical singularities outside that disk. Then the displayed $W$ has constant stream function on $|z|=a$. The proof is the identity $a^2/\bar{z}=z$ on the circle: there $W=f(z)+\overline{f(z)}$ is real. Hence the circle is a streamline and satisfies the no-penetration boundary condition. The reflected term has its image singularities inside the obstacle. Additional imposed circulation must be specified separately. The theorem is named after Louis Melville Milne-Thomson.

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]

The no-penetration boundary condition sets the relative normal velocity of a fluid and an impermeable boundary to zero. Tangential slip may still be permitted.

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.

Use instantaneous polar coordinates centered on the moving circular cylinder, but measure the fluid velocity in the laboratory frame. Impermeability equates the normal components of body and fluid velocity:Since ${\mathbf{e}}_x\cdot {\mathbf{e}}_r=\cos \theta$ and $u_r={\varphi }_r$, the no-penetration boundary condition isThe velocity potential obeys the exterior Laplace equation, by incompressibility. The boundary data have only the first cosine harmonic. A decaying harmonic potential of this form is $\varphi =C\cos \theta /r$. Matching its radial derivative gives $-C/a^2=U$, so $C=-U{a}^2$. The zero circulation condition rules out an angular potential term, and the integrated boundary condition gives zero net volume flux, ruling out a radial logarithmic source term. The remaining decaying harmonics have zero coefficients by the boundary data. HenceUsing the streamfunction in polar coordinates, $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$, givesOne may also obtain both functions as the real and imaginary parts of $w(z)=-U{a}^2/z$, a two-dimensional potential doublet. The velocity decays as $r^{-2}$ and the source of this singular representation lies inside the excluded body. These are the fields used to calculate the added mass of a circular cylinder.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in planar potential flow. Suppose $f(\zeta )$ is an incident complex potential of a two-dimensional potential flow with no singularities on or inside $|\zeta |=R$. Thenis a complex potential of a two-dimensional potential flow outside the circle, with the same exterior singularities and their strengths as the incident flow; the added image singularities lie inside the excluded disk. On $|\zeta |=R$, $R^2/\bar{\zeta }=\zeta$, so $W=f+\bar{f}$ is real. Its imaginary part, the streamfunction, is constant on the circle, proving the no-penetration boundary condition. The theorem gives the usual zero-circulation construction; a separately prescribed circulation can be added by its logarithmic vortex potential.

Put $s=a^2/r$ and define the image term $G(r,\theta )=(r/a)\Psi (s,\theta )$. The key step is to verify that inversion preserves the Stokes streamfunction equation for irrotational flow. Direct differentiation giveswhere the quantities on the right are evaluated at $(s,\theta )$. HenceBy linearity, $\tilde{\Psi }=\Psi -G$ obeys the same equation and describes an incompressible flow and an irrotational flow wherever the original exterior flow is regular.

On the sphere, $s=a$ and the two terms cancel identically:Thus the sphere is a stream surface and satisfies the no-penetration boundary condition. Tangential slip is allowed in this potential flow model. For every finite $r>a$, the image argument satisfies $0<s<a$, where $\Psi$ has no singularities. The prefactor $r/a$ is regular there. Therefore the construction creates no new finite exterior singularities; any original exterior singularities remain those of the prescribed incident flow. This is the sphere theorem for a Stokes streamfunction.

For the far-field claim, the stated bound givesOne must compare velocities, rather than require the correction to the Stokes streamfunction to tend to zero. Smoothness at the original origin, together with $\Psi (0,\theta )=0$, gives ${\Psi }_s=O(1)$ and ${\Psi }_{\theta }=O(s)$ there. Thus $G_r=O(r^{-1})$ and $G_{\theta }=O(1)$, and the correction hasThese estimates hold away from the coordinate poles, with the regular-axis limits supplied by the original smooth physical flow. ConsequentlyThe far-field velocity field is unchanged. For a fully smooth incident velocity field at the origin, the usual regular expansion is stronger and makes the sphere disturbance decay as $r^{-3}$.

The method of images places an equal line source at $-ia$, outside the physical upper half-plane. A convenient complex potential of a two-dimensional potential flow isThe two logarithms can be combined locally as $(Q/(2\pi ))\log (z^2+a^2)$; their branches of the complex logarithm affect only the potential convention, not the single-valued complex velocity. The point source in this two-dimensional problem has the line source normalization: its outward volume flux is $Q$ per unit span.

Writing $R_{-}^2=x^2+(y-a{)}^2$, $R_{+}^2=x^2+(y+a{)}^2$, and using $w^{\prime }=u-iv$, the velocity field isOn the wall $y=0$, the vertical components cancel:This is the required no-penetration boundary condition. Tangential slip is permitted. At large distance the velocity decays as $1/r$, and near $z=ia$ its singular part is exactly that of the specified source; the reflected source is regular throughout the fluid. Away from the injection point, this is an incompressible flow and an irrotational flow.

The fixed upper and lower walls have zero normal velocity. Since $v={\varphi }_y$, their no-penetration boundary conditions areFor the moving end wall, use its level-set equation $F(x,y,t)=x-X(y,t)=0$. The kinematic boundary condition $DF/Dt=0$ gives the exact impermeability relationBoth $v$ and $X_y$ are first order, so their product is second order. A Taylor expansion from $x=X$ to $x=0$ changes the first-order perturbation velocity only at second order as well. ThusAt infinity, select a bounded response with no waves incident from $x=+\infty$. A propagating acoustic waveguide mode must satisfy the outgoing radiation condition for a waveguide mode, while a nonpropagating mode must decay. For a time dependence $e^{-i\omega t}$, a propagating axial factor has the form $e^{iqx}$, $q>0$, and obeys ${\Phi }_x-iq\Phi \to 0$. An evanescent factor is $e^{-\alpha x}$, $\alpha >0$.

Requiring $\varphi \to 0$ for all frequencies would incorrectly exclude the radiated traveling waves. In this lossless channel they retain finite amplitude arbitrarily far downstream.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.

Take positive $\Gamma$ to mean counterclockwise circulation. A line vortex at ${\zeta }_0$ has complex potential of a two-dimensional potential flow $\Gamma \log (\zeta -{\zeta }_0)/(2\pi i)$. The method of images imposes the no-penetration boundary condition on the real axis by placing an image line vortex of strength $-\Gamma$ at ${\bar{\zeta }}_0$. ThusFor real $\zeta$, the numerator and denominator have equal modulus, so the streamfunction isThe real axis is therefore a streamline. The image line vortex lies outside the mapped fluid region, while the complex velocitydecays as $|\zeta |\to \infty$. This selects the flow due to the line vortex and the boundary, without superposing an imposed background velocity field.

For an irrotational flow, write $\mathbf{u}=\nabla \varphi$. The incompressible flow condition then gives the Laplace equationAt leading order in a small disturbance this equation holds on $-h<y<0$. Including gravity, the Unsteady Bernoulli equation isChoose the time-dependent additive freedom in the velocity potential so that $B(t)=p_0/\rho$. Neglecting surface tension, pressure continuity at the free surface gives the exact dynamic conditionThe free surface is a material boundary: applying the material derivative to $y-\eta (x,t)=0$ yields the kinematic boundary condition for a free-surface graphFor a small disturbance, the Taylor expansion from $y=\eta$ to $y=0$ changes the first-order velocity potential and its partial derivatives only at second order. Both $|\nabla \varphi {|}^2$ and ${\varphi }_x{\eta }_x$ are also second-order terms. The linearized free-surface boundary conditions are thereforeThe fixed bed has the no-penetration boundary conditionThe required approximation is small surface elevation, small wave steepness, and small velocity perturbations. The stated $p_1\ll p_0$ alone does not ensure this approximation near resonance.

Seek the travelling forced component of the surface elevation in the form $\eta =A\cos \chi$, where $\chi =x-ct$ and $A$ is a signed coefficient. The forcing has wavenumber one. Separation of variables in the Laplace equation, followed by the bed no-penetration boundary condition, givesSince ${\eta }_t=Ac\sin \chi$, the surface kinematic boundary condition requires $B\sinh h=Ac$, soAt $y=0$, ${\varphi }_t=-A{c}^2\coth h\cos \chi$. The dynamic member of the linearized free-surface boundary conditions therefore becomesConsequently the signed response and the nonnegative amplitude areThe travelling-pressure response of finite-depth surface gravity waves is opposite in phase to the imposed pressure below the resonant speed and in phase above it. These expressions describe the travelling forced component; freely propagating surface gravity waves can be superposed to meet additional initial conditions.

Use $W=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so $W^{\prime }(z)=u-iv$. The Milne-Thomson circle theorem says that if $f(z)$ is an incident complex potential of a two-dimensional potential flow whose singularities lie outside $|z|=a$, thenmakes $|z|=a$ a streamline. Indeed, on that circle $a^2/\bar{z}=z$, so $W=f+\bar{f}$ is real and the stream function is constant. This imposes the no-penetration boundary condition.

For the two-dimensional potential doublet, put $M=m{e}^{i\alpha }$ and $f=M/(z-b)$, where $b>a$ is real. The theorem givesUp to an additive constant, this is equivalentlyThe method of images has placed the image two-dimensional potential doublet at $a^2/b$, inside the obstacle. We use the solution with zero added circulation and no imposed far-field flow; the original doublet is the only singularity in the liquid.

At the rigid bed, the no-penetration boundary condition givesThe displaced surface is material, so taking the material derivative of $z-\eta (x,t)$ gives the exact kinematic boundary condition for a free-surface graph:With upward vertical coordinate $z$, the gravitational potential per unit mass is $gz$. The Unsteady Bernoulli equation, evaluated on the surface with the prescribed pressure $p=B{\eta }_{xxxx}$, givesHere the spatially uniform time-dependent Bernoulli term has been absorbed into the velocity potential. The pressure is measured relative to the constant exterior pressure. This dynamic boundary condition uses the specified massless bending-cover model, with no ice inertia or tension.

Let $\chi =kx-\omega t$, with $k>0$. The surface elevation is $\eta =\alpha \cos \chi$, so ${\eta }_t=\alpha \omega \sin \chi$. Use separation of variables in the interior Laplace equation:The $\sinh (k(z+h))$ term is excluded by the bed no-penetration boundary condition. The linear surface kinematic boundary condition determinesAn irrelevant additive constant is omitted. At $z=0$,The linear dynamic boundary condition then reduces toConsequently the dispersion relation of a floating elastic plate with zero plate inertia isThe negative-frequency branch gives propagation in the opposite direction. For signed wavenumber, replace $k$ by $|k|$ in the squared-frequency formula. Setting $B=0$ recovers the finite-depth surface gravity wave relation, a useful sign check.

Write ${\omega }_h=\dot{\alpha }$, positive on opening. The no-slip boundary condition equates the fluid velocity with each plate's velocity:The upper condition is a moving-wall no-penetration boundary condition, not zero normal velocity in the laboratory frame. At the hinge the velocity tends to zero. With the arbitrary constant of the streamfunction chosen so that $\psi (r,0)=0$, these conditions becomeA changing angle cannot give a literally steady flow in fixed coordinates. We interpret the stated steadiness as the quasi-steady Stokes approximation at each instantaneous $\alpha$.

The printed expression is not biharmonic. To test it without reproducing the source formula, write its bracket as $G_p(\theta )$ and its streamfunction as ${\psi }_p=-\Omega r^2{G}_p/2$. For any angular function $G$, the Laplacian givesAt the concrete angle $\alpha =\pi /4$, differentiation of the printed bracket givesFor example, at the interior point $\theta =\pi /8$,The printed bracket does satisfy $G_p(0)=G_p^{\prime }(0)=G_p^{\prime }(\alpha )=0$ and $G_p(\alpha )=1$, so its boundary velocities would require $\Omega ={\omega }_h$. Satisfying these boundary conditions does not cure its nonzero biharmonic operator. The stated series is likewise not a general biharmonic basis: its $r^2\theta \sin 2\theta$ term already fails the displayed test. The separated polar solutions of the biharmonic equation contain additional radial powers; for the required $r^2$ dependence, the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$.

The corrected slowly opening Stokes wedge similarity solution follows by setting $\psi =-{\omega }_h{r}^{2}G(\theta )/2$. The biharmonic equation gives $G^{(4)}+4G^{\prime \prime }=0$. The two lower-wall conditions reduce its general solution toThe upper-wall condition $G^{\prime }(\alpha )=0$ fixes the ratio of $A$ and $B$, and $G(\alpha )=1$ fixes their scale. An expression that also remains finite at $\alpha =\pi /2$ isFor $0<\alpha <\pi$, its denominator is positive: the derivative of $\sin \alpha -\alpha \cos \alpha$ is $\alpha \sin \alpha >0$. Direct differentiation verifies all four boundary conditions and the biharmonic equation. The resulting velocity components are $v_r=-\dot{\alpha }r{G}^{\prime }/2$ and $v_{\theta }=\dot{\alpha }rG$.

For a fixed angle of order one, $U(r)\sim |\dot{\alpha }|r$, so the local Reynolds number is ${\mathrm{Re}}_r\sim |\dot{\alpha }|r^2/\nu$. One breakdown length is thereforeA second is the geometric length $l_2\sim \min (L,W)$, where $L$ is the distance to the radial ends and $W$ the spanwise width. At $r\sim l_2$, matching to the surrounding bath or three-dimensional end flow matters. Thus $r\ll \min (l_1,l_2)$ is required for this local two-dimensional Stokes flow, with angle-dependent factors in the inertial estimate. Ideal infinite plates remove the geometric cutoff but not the inertial one.

There is also a temporal restriction: the relevant viscous diffusion time must be short compared with the time $T_s$ over which the shape and its velocity field change appreciably. Using $r$ as a conservative diffusion length gives $r\ll \sqrt{\nu T_s}$. For order-one angles and constant $\dot{\alpha }$, $T_s\sim |\dot{\alpha }{|}^{-1}$ gives the same scale as $l_1$; narrow angles require using their actual gap and angular factors rather than extrapolating this order-one-angle estimate.

The coefficient of the growing term is fixed by the far-field complex potential: $A=\beta e^{i\alpha }$. On $z=a{e}^{i\theta }$, the no-penetration boundary condition makes the cylinder a streamline, so the imaginary part of $A{a}^2{e}^{2i\theta }+B{a}^{-2}{e}^{-2i\theta }$ must be constant. Choosing the two terms to be complex conjugates makes that constant zero, givingThis is the zero-circulation solution, also obtained by the Milne-Thomson circle theorem. Impermeability and the leading far field alone permit an additional circulation term $-i\Gamma \log z/(2\pi )$; the requested algebraic form implicitly selects $\Gamma =0$. No no-slip boundary condition is imposed on an inviscid flow.

The streamfunction in polar coordinates isIt gives $v_r(a,\theta )=0$ andThe liquid still exerts normal pressure on each surface element even though tangential viscous traction is absent. For this steady flow with zero vorticity, the Bernoulli equation has one constant $C$ throughout the connected fluid region:There is no finite uniform far-field pressure to substitute for $C$, because this straining velocity field grows with radius. Taking $\mathbf{n}=(\cos \theta ,\sin \theta )$ outward from the cylinder, the fluid force per unit axial width isIndeed $p(\theta +\pi )=p(\theta )$ while $\mathbf{n}(\theta +\pi )=-\mathbf{n}(\theta )$: diametrically opposite elements cancel exactly. Zero resultant force does not mean zero local pressure or zero deformation tendency.

Take $x$ along a horizontal gap and $s$ perpendicular to its plates, with $0<s<b$. Assume a Newtonian fluid, stationary plates, a steady flow, a laminar flow and a fully developed flow away from the entrances and side edges. The no-slip boundary condition holds on both plates. Any transverse gravity is absorbed in the hydrostatic pressure; there is no imposed transverse through-flow. Translational symmetry and the no-penetration boundary condition then give $\mathbf{u}=(u(s),0,0)$. The Navier-Stokes equation reduces toTwice integrating gives the plane Poiseuille flow:The volumetric flow rate through one gap isThere are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the Darcy velocity isThus the permeability of a porous medium has the dimensions of squared length. If finite plate thickness were included, the Darcy velocity averaged over the whole manifold would also contain its porosity.

Both the local acceleration and the convective acceleration vanish: ${\partial }_t\mathbf{u}=0$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=u_x{\partial }_x\mathbf{u}=0$. The dominant balance is therefore between the pressure gradient and viscous stress, with no inertia contribution in this fully developed solution. This exact cancellation does not guarantee that arbitrary large-Reynolds number realizations remain laminar flows; entrances, disturbances and turbulence can invalidate the assumptions.