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Uniform axial strain rate $E$ in an annular Newtonian fluid gives radial velocity $u_r=-Er/2+C/r$. For radii $R_i,R_e$, area parameter $A=R_e^2-R_i^2$, and constant gas pressures,The opposite signs of inner and outer curvature in the Young–Laplace equation give these expressions. The kinematic boundary condition implies $\dot{A}=-EA$ and

An inviscid uniform-density self-gravitating cylinder in vacuum, with no rotation or surface tension, has the displayed axisymmetric dispersion relation, where $\kappa =|k_z|R$. To derive it, let $W=\psi +\pi /\rho =ϵ{\Pi }_0{I}_0(kr)e^{i(kz-\omega t)}$. The linear kinematic boundary condition gives ${\Pi }_0={\omega }^2/[k{I}_0^{\prime }(\kappa )]$. The Lagrangian pressure perturbation vanishes on the moving free surface, giving $\pi (R)=2\pi G{\rho }^{2}R\eta$. Substitute the axisymmetric gravitational potential of a displaced cylindrical interface and use $I_0^{\prime }=I_1$ to obtain the relation. Its nonzero neutral root is ${\kappa }_c\simeq 1.0667$: wavelengths longer than $2\pi R/{\kappa }_c\simeq 5.89R$ have ${\omega }^2<0$ and grow. The fastest-growing mode has $\kappa \simeq 0.58$, so its wavelength is about $11R$. Negative ${\omega }^2$ must be interpreted through an imaginary frequency, not through a real oscillatory ansatz.

A moving boundary excites an internal gravity wave through its kinematic boundary condition. For wavenumber $k$ and angular frequency $\omega$, the vertical velocity field satisfies $W^{\prime \prime }+k^2(N^2/{\omega }^2-1)W=0$. The radiation condition selects outward group velocity; when $\omega >N$, the bounded response is an evanescent wave.

Let fluid 1 lie above fluid 2, with mass densities ${\rho }_1<{\rho }_2$ and dynamic viscosities ${\mu }_1<{\mu }_2$. Define $\Delta \rho ={\rho }_2-{\rho }_1>0$, $\Delta \mu ={\mu }_2-{\mu }_1>0$, porosity $\varphi$, permeability of a porous medium $k$, and downward interface speed $W$. With Darcy flux $\varphi W$, no capillarity, and two unbounded layers, Darcy's law givesFor a Fourier mode of interface displacement, the two perturbation pressures decay exponentially into the respective layers. The kinematic boundary condition makes their amplitudes proportional to $\varphi {\mu }_1\sigma /(k|\alpha |)$ and $-\varphi {\mu }_2\sigma /(k|\alpha |)$. Linearized pressure continuity adds the base-gradient difference $\varphi W\Delta \mu /k-g\Delta \rho$. Eliminating the amplitudes proves the formula. The critical interface speed is $kg\Delta \rho /(\varphi \Delta \mu )$; the critical Darcy velocity is $kg\Delta \rho /\Delta \mu$. Without surface tension this model has unbounded short-wave growth rate on its unstable side.

For steady equal-pressure flow, mass conservation gives $Aw=A_0{w}_0$ and the kinematic boundary condition gives $(D\sqrt{w}{)}_z=\gamma /(2\mu \sqrt{w})$, where $D=R_e-R_i$. Therefore finite-height closure occurs exactly whenFor positive velocity the integral is increasing. If it diverges, closure is finite; if its limiting value lies below the right-hand threshold, there is no finite closure. Quadratic freely falling profiles have logarithmically divergent integrals, while sufficiently strongly drawn exponentially accelerating profiles can have finite integrals. Thus nozzle speed alone is insufficient to guarantee closure: the axial tension or downstream condition matters.

For bending stiffness $D$ and mass per area $m$, a plane wave of surface displacement $\eta$ satisfies $m{\eta }_{tt}+D{\nabla }_H^4\eta =p$. An inviscid deep-water velocity potential has vertical dependence $e^{kz}$; the kinematic boundary condition and linear Bernoulli equation give $p=({\rho }_w{\omega }^2/k-{\rho }_{w}g)\eta$. Eliminating $p$ gives the displayed dispersion relation. With water depth $d$, replace it by ${\omega }^2=(gk+D{k}^5/{\rho }_w)\tanh (kd)/[1+(m/{\rho }_w)k\tanh (kd)]$.

A flexural-gravity wave bends a floating elastic plate while displacing the underlying water. For a uniform plate with areal mass $m$ and bending stiffness $D$ above inviscid deep water of mass density ${\rho }_w$, put $\alpha =m/{\rho }_w$ and $\beta =D/{\rho }_w$. Combining the plate equation with the water's kinematic boundary condition gives the displayed dispersion relation. Both gravitational acceleration and plate bending restore displacement; water and plate inertia resist acceleration. It assumes small amplitude, no prestress and negligible shear deformation of the plate.

A free surface is a fluid boundary whose position is part of the solution and whose traction is prescribed by the adjoining medium. Its motion obeys a kinematic boundary condition, while its pressure or stress supplies a dynamic boundary condition.

At a material interface without surface tension, continuity of displacement and of the pressure evaluated on the displaced interface givesThe first follows from the kinematic boundary condition $ik(U-c)\hat{\eta }=\hat{w}$. The second uses $\hat{p}={\rho }_0[(U-c){\hat{w}}^{\prime }-U^{\prime }\hat{w}]/(ik)$ and hydrostatic displacement $[\hat{p}-g\rho \hat{\eta }]=0$. These conditions apply to density, velocity and vorticity jumps within the Boussinesq approximation.

Two deep layers of incompressible flow with upper mass density ${\rho }_1$, lower mass density ${\rho }_2$, and tangential velocities $U_1,U_2$ have this dispersion relation when surface tension is absent. Decaying velocity potentials and the linearized kinematic boundary condition determine their amplitudes; the dynamic boundary condition for an inviscid interface gives the displayed relation. For $k>0$ the squared departure of the phase velocity $c$ from the density-weighted mean velocity is $g({\rho }_2-{\rho }_1)/[k({\rho }_1+{\rho }_2)]-{\rho }_1{\rho }_2(U_1-U_2{)}^2/({\rho }_1+{\rho }_2{)}^2$. A negative value gives Kelvin-Helmholtz instability despite stable density stratification.

Use complex amplitudes and take their real parts at the end. Write the radial surface displacement as $\eta =ϵe^{i(kz-\omega t)}$ and the combined potential asThe linear kinematic boundary condition is $u_r(R)={\partial }_t\eta$. Differentiate it once more in time and use the radial momentum equation, ${\partial }_t{u}_r=-{\partial }_{r}W$, evaluated at the unperturbed radius $R$:With $\kappa =kR$, the mode amplitudes therefore satisfyThe prime differentiates the Modified Bessel function of the first kind with respect to its dimensionless argument, rather than with respect to $r$. The result has units $\mathrm{m}{\mathrm{s}}^{-2}$, as required because $ϵ{\Pi }_0$ is a potential. It applies to $k>0$; the uniform $k=0$ limit requires separate treatment.

For planar incompressible flow, $u_x+w_z=0$. Locally this permits a streamfunction with $u={\psi }_z$, $w=-{\psi }_x$, because equality of the mixed derivatives enforces incompressibility. The remaining component of the curl is $u_z-w_x={\psi }_{zz}+{\psi }_{xx}$, so irrotational flow requires Laplace's equation for $\psi$.

Put $K=\pi /b$. The sinusoidal contribution to the supplied streamfunction has ${\psi }_{xx}=U_{0}a{K}^2\cos (Kx)e^{-Kz}$ and ${\psi }_{zz}=-U_{0}a{K}^2\cos (Kx)e^{-Kz}$; the linear term is also harmonic. HenceAs $z\to \infty$, these approach the specified uniform flow, $(u,w)=(U_0,0)$.

The impermeability kinematic boundary condition is $w=u{z}_s^{\prime }(x)$ on $z=z_s(x)$, equivalently $\psi$ is constant on the wall. For $aK\ll 1$, evaluating the perturbed velocities at $z=0$ gives $w=-U_{0}aK\sin (Kx)=U_0{z}_s^{\prime }$ to first order. The residual is of order $U_0(aK{)}^2$. Equivalently $\psi (x,z_s)=U_{0}a\cos (Kx)[1-e^{-Ka\cos (Kx)}]=O(U_{0}K{a}^2)$. The surface velocity isThe first expression is the supplied approximate field evaluated exactly at the wall; its accuracy as a wall-flow solution is still only first order in $a/b$.

The interface is material. Applying the kinematic boundary condition to $y-\eta (x,t)=0$ gives, exactly,The disturbance velocities and $\eta$ are small. Products of two disturbance quantities are second order, and a Taylor expansion moves their evaluation to $y=0$ at linear order. HenceThe extra upper-fluid term is advection of the displaced interface by the base uniform flow.

Use the Unsteady Bernoulli equation separately on each side, with constant mass density $\rho$. There are no restoring terms specified at this internal interface, and its pressure is continuous. Choose the time-dependent potential gauges so that the undisturbed pressure is $p_0$ on both sides. Subtracting the base values from Bernoulli's principle gives, at linear order,The quadratic disturbance-velocity terms have been discarded, while the cross term with the base velocity is first order and must remain. Equating the perturbation pressures yields the third condition:

Use complex normal modes and take their real parts at the end. Write $\eta =A{e}^{i(kx-\omega t)}$, with $k>0$. The Laplace equation and the wall Neumann boundary conditions give the vertical formsAt $y=0$, the two kinematic boundary conditions becomeThusThe dynamic condition isSubstituting the two amplitudes and cancelling the common nonzero depth factor gives the dispersion relation for a vortex sheet in a symmetric rigid channel:Equivalently,For the specified $U>0$, the plus sign gives positive temporal growth rate $\gamma =kU/2$ in the convention $e^{-i\omega t}$. A real growing interface displacement is thereforeThe other branch decays at the same rate. Consequently the interface is linearly unstable, exhibiting Kelvin-Helmholtz instability. There is no nonzero-amplitude mode with a purely real temporal frequency for $k>0$ and $U\ne 0$; the paper's sinusoidal notation must allow a complex frequency, with the physical displacement obtained by a real part.

The wave pattern is advected at speed $U/2$, the mean of the two base velocities. The growth rate is independent of $h$ here because both layers have the same mass density and the same depth, making their factors $\coth (kh)$ cancel. In an asymmetric channel those inertia factors would differ. In the zero-shear limit $U=0$, the two roots merge at zero and there is no exponential growth.

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.

Linearize about fluid at rest, with mean free surface $y=0$ and mean bottom $y=-d$. Regard $a$, $\eta$ and the perturbation velocity potential as first-order quantities. A Taylor expansion from $y=\eta$ to $y=0$ produces products of first-order quantities, as do ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$. Discarding these second-order terms gives the linearized free-surface boundary conditionsEquivalently, differentiate the second condition in time and substitute the first to obtain ${\varphi }_{tt}+g{\varphi }_y=0$ at $y=0$.

The bottom is also impermeable and moving. Its exact kinematic boundary condition follows from $D(y-h)/Dt=0$:Here $h_t=-akc\cos (k(x-ct))$, while ${\varphi }_x{h}_x$ and the change from evaluating ${\varphi }_y$ at $h$ to evaluating it at $-d$ are second order. HenceThe interior Laplace equation remains ${\varphi }_{xx}+{\varphi }_{yy}=0$ on $-d<y<0$. This linearization requires small slopes and small resulting excursions as well as $a\ll d$; close to resonance a small prescribed bottom displacement can produce a response too large for it.

Treat the free surface as the material level set $F(x,y,t)=y-\eta (x,t)=0$. The kinematic boundary condition $DF/Dt=0$, with $\mathbf{u}=\nabla \varphi$, givesThis says that fluid initially on the free surface remains on it.

For constant mass density $\rho$, the Euler momentum equation under gravity is $D\mathbf{u}/Dt=-\nabla p/\rho -g{\mathbf{e}}_y$. Since the flow is irrotational, $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)$. Integration in space therefore gives the Unsteady Bernoulli equationAt the free surface, $p=P_a$ and $y=\eta$. Adding a function $G(t)$ to the velocity potential leaves the velocity field unchanged; choose $G^{\prime }=P_a/\rho -B(t)$. With this choice the surface dynamic boundary condition for an inviscid interface becomesThe constant atmospheric pressure has been absorbed into this choice of potential.

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 the spherical symmetry of the expansion, with fluid at rest at infinity and no superposed imposed flow. The radial Laplace equation givesThe moving sphere is a material boundary: its kinematic boundary condition is $u_r(R,t)=\dot{R}$. Consequently $A(t)=R^2\dot{R}$, and the spherically symmetric incompressible radial flow hasEquivalently, its volumetric flow rate through any concentric sphere is $4\pi r^2{u}_r=4\pi R^2\dot{R}$, the rate of change of the enclosed volume.

Linearize about the resting state $\eta =0$, $\varphi =0$. Introduce a dimensionless small parameter $ϵ$ and writeholding the depth, bending stiffness and variation scales fixed. Thus both boundary displacement and induced velocity are small. For any first-order derivative $D\varphi$, a Taylor expansion at the moving surface givesThe displacement correction to the leading field is ${ϵ}^2{\eta }_1{\partial }_z(D{\varphi }_1)(x,0,t)$, so evaluating it on $z=0$ changes it only at second order. Also ${\varphi }_x{\eta }_x=O({ϵ}^2)$ and $|\nabla \varphi {|}^2=O({ϵ}^2)$. The first-order kinematic boundary condition and dynamic boundary condition are thereforeDropping the subscript, the linearized boundary condition pair isThe fixed-bed condition remains ${\varphi }_z(x,-h,t)=0$. The source's shorthand $|\eta |\ll 1$ requires a length scale: for the subsequent plane wave, small surface slope $k|\alpha |\ll 1$ and displacement small compared with depth are sufficient conditions. Small height alone, with arbitrarily rapid spatial variation or a finite background disturbance, does not justify discarding the products.