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For irrotational flow, ${\mathbf{u}}_j=\nabla {\varphi }_j$. An impermeable stationary horizontal wall has zero normal velocity, so the kinematic boundary condition isNo tangential no-slip condition is imposed on these inviscid fluids.

Take $x$ in the injection direction and $y$ along the flat interface. The specified $U$ must be a Darcy velocity, so the pore-fluid velocity and unperturbed interface speed are $U/\varphi$. Write the perturbed interface as $x=X(t)+\eta (y,t)$, with $\dot{X}=U/\varphi$, and choose the normal from fluid 1 into fluid 2. The background pressure gradients from Darcy law are ${\partial }_x{p}_i^{(0)}=-{\mu }_{i}U/k$.

By incompressibility and uniform permeability of a porous medium, pressure perturbations are harmonic functions. A transverse Fourier mode with positive wavenumber $\alpha$ has the decaying formsThe kinematic boundary condition and equal normal Darcy flux implyWithout capillarity the pressure is continuous at the displaced interface. Its linear perturbation is thereforeConsequently the planar viscous-fingering dispersion relation reduces toThis is the Saffman–Taylor instability: less viscous fluid advances preferentially through protrusions. Without a short-wave regularizing mechanism the idealized rate has no finite maximum.

For the prescribed apparent capillary pressure, the chosen normal is $\mathbf{n}=(1,-{\eta }_y)/\sqrt{1+{\eta }_y^2}$, so $\nabla \cdot \mathbf{n}=-{\eta }_{yy}+O({\eta }^2)={\alpha }^2\eta +O({\eta }^2)$. Subtract the flat-front pressure jump $\gamma (1+\beta X){\kappa }_0$. Linearization of the jump at $x=X+\eta$ givesThere is no first-order product of the surface-tension perturbation and the perturbed macroscopic curvature. ThusUsing the capillary number convention $\mathrm{Ca}={\mu }_{2}U/\gamma$ and ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$, this isIf $x$ is measured from the unperturbed front and $\gamma$ denotes its local apparent tension, set $X=0$ to obtain the usual fixed-coefficient expression. For a physically fixed spatial gradient, $X=Ut/\varphi$ and the displayed rate is an instantaneous, generally time-dependent growth rate; the amplitude solves $\dot{a}=\sigma (\alpha ;X(t))a$, rather than one constant exponential law for the entire displacement. The parameter $\beta$ has dimensions of inverse length; the absolute tension gradient is $\gamma \beta$.

Assume $1+\beta X>0$, as required for positive apparent surface tension, and define $A=({\mu }_2-{\mu }_1)U-k\gamma \beta {\kappa }_0$. If $A>0$, unstable modes satisfy $0<{\alpha }^2<A/[k\gamma (1+\beta X)]$. Differentiating the cubic dispersion relation gives the fastest-growing modeWhen $A\le 0$, every nonzero mode decays, with a neutral zero-wavenumber translation in the infinite-domain limit. For $\beta {\kappa }_0>0$, surface-tension-gradient stabilization of viscous fingering suppresses the instability entirely at positive injection speeds satisfyingIf $\beta {\kappa }_0\le 0$, no positive injection speed eliminates every long-wave unstable mode in an unbounded interface. The prescribed wetting variation is treated as an effective normal pressure law for porous-media flow.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

At the change in buoyancy frequency, continuity of vertical velocity and fluid pressure givesThe background mass density is continuous, so there is no interfacial density jump or separate surface restoring force. Since $P=i\omega W^{\prime }/k^2$, these internal-wave transmission across a stratification step conditions are continuity of $W$ and $W^{\prime }$. The buoyancy perturbation need not be continuous, because the background density gradient jumps.

Let $W_0=-i\omega {\eta }_0$ and $\ell =k\sqrt{N_2^2/{\omega }^2-1}$. The lower unstratified layer obeys $W^{\prime \prime }-k^{2}W=0$ and the upper radiation condition givesPropagating these matching data downward givesThe kinematic boundary condition at the moving lower boundary therefore givesThe upper vertical-displacement complex amplitude at $H$ isIn particular, $|{\eta }_H/{\eta }_0|=[{\cosh }^2(kH)+(\ell /k{)}^2{\sinh }^2(kH){]}^{-1/2}$. The unstratified layer attenuates and phase-shifts the transmitted internal gravity wave. The matching height is $H$ in the original PDF; the TeX transcription's $H^{\prime }$ is a typo.

At the rigid surface, the no-slip boundary condition and no penetration giveAt the horizontal free surface, the kinematic boundary condition and the stress boundary condition areHere surface tension is absent and the inviscid air exerts no tangential traction.

The incompressible Stokes equation isUse the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ and setThenand the wall and zero-shear conditions becomeThe two momentum equations readEquality of mixed derivatives gives $f^{(4)}=0$, so the boundary conditions implyIntegrating for the pressure and imposing the normal-stress condition at $y=h$ determinesThe complete instantaneous velocity and pressure fields are thereforeIndeed, with the viscous stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$, these expressions satisfy both traction conditions. Evaluating the vertical velocity at the material surface gives the pressure-driven thinning of a uniform Stokes layer:Thus

Write the density and pressure as ${\rho }_0+{\rho }^{\prime }$ and $p_0+p^{\prime }$, and let the small fluid velocity be the gradient of an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$. The linear homentropic acoustic equations arewhere the adiabatic sound speed of a perfect gas is $c_0^2=\gamma p_0/{\rho }_0$. Absorbing a function of time into $\varphi$, the momentum equation givesEliminating $p^{\prime }$ and ${\rho }^{\prime }$ gives the acoustic wave equation

Let the shell displacement be $a(t)-a_0=\mathrm{Re}(\eta e^{-i\omega t})$ and put $k=\omega /c_0$, $\theta =k{a}_0$. The spherically symmetric solution regular at the centre isThe kinematic boundary condition ${\varphi }_r(a_0,t)=\dot{a}(t)$ givesSince $p^{\prime }=-{\rho }_0{\varphi }_t$, the pressure amplitude on the shell is thereforeThe radial Newton's second law per unit area isso its complex amplitude satisfiesFor a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(m{c}_0^2)$ yield the frequency equation for spherically symmetric vibration of a gas-filled elastic shell:The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta \cot \theta -1$ runs from $+\infty$ to $-\infty$ on $(n\pi ,(n+1)\pi )$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.

Take $k>0$; reflection in $x$ supplies the result for negative wavenumber. Since each region has uniform base velocity, the perturbation is an incompressible, irrotational flow with a velocity potential satisfying the Laplace equation. Its decaying outer forms areFor a varicose mode of a planar jet, reflection symmetry makes the inner potential proportional to $\cosh (ky)$; for a sinuous mode of a planar jet, it is proportional to $\sinh (ky)$.

At the upper interface, the linearized kinematic boundary condition isThe first equation gives ${\varphi }_{+}(h)=-\sigma \eta /k$. The second givesThe linearized Unsteady Bernoulli equation gives $p^{\prime }=-\rho ({\partial }_t+U{\partial }_x)\varphi$. Applying pressure continuity therefore produces the equal-density top-hat planar-jet dispersion relationSolving the quadratic equation givesEquivalently,The plus sign is the unstable Kelvin-Helmholtz instability branch. In particular, the two modes have exactly the same growth rate,Its limiting forms areThus neither type grows faster in this equal-density idealization, although their phase speeds differ. The unbounded large-$k$ growth is the familiar short-wave pathology of an inviscid vortex sheet.

Let $\hat{\eta }$ be the displacement amplitude of an interface. Its kinematic boundary condition is $ik(U-c)\hat{\eta }=\hat{w}$, on either side, so the first of the jump conditions for stratified inviscid shear flow isFor zero surface tension, pressure is continuous at the displaced interface. Since the background hydrostatic pressure has derivative $-g\rho$, linearization gives $[\hat{p}-g\rho \hat{\eta }{]}_{-}^{+}=0$. Using $\hat{p}={\rho }_0[(U-c){\hat{w}}^{\prime }-U^{\prime }\hat{w}]/(ik)$ and the kinematic condition givesThis derivation also works when $U$ jumps: the displacement is shared, while the vertical velocity need not be. It avoids ambiguous products of distributions at a discontinuous velocity profile.

Use a common factor $e^{i(kx-\omega t)}$, with $k=k_0>0$ and $\omega ={\omega }_0$. For the upward propagating branches with $0<{\hat{\omega }}_2=\omega -k{U}_2<N_2$, defineThese are the inclinations of upward-sloping constant-phase lines of an internal gravity wave to the horizontal. Write the incident, reflected and transmitted complex velocity amplitudes as $A_0,A_r,A_t$. Incompressibility gives $(u_i,w_i)=A_0(\cos {\theta }_1,\sin {\theta }_1)$, $(u_r,w_r)=A_r(\cos {\theta }_1,-\sin {\theta }_1)$, and $(u_t,w_t)=A_t(\cos {\theta }_2,\sin {\theta }_2)$.

The jump conditions for stratified inviscid shear flow require a common interface displacement, not a common vertical velocity: $w=-i\hat{\omega }\eta$ on each side. Since the background mass density is continuous, pressure is continuous without a hydrostatic jump. The horizontal momentum equation gives $p/{\rho }_0=(\hat{\omega }/k)u$. HenceThese two matching equations determine internal-wave transmission across a velocity jump:The amplitude formula printed in the PDF is inconsistent with these material-interface matching conditions. In particular, identical layers have $A_t/A_0=1$ and $A_r/A_0=0$, whereas its expression labelled “reflected” equals one. The displayed results above distinguish reflection from transmission and retain the intrinsic-frequency factors required by the kinematic boundary condition and pressure balance.

For equal buoyancy frequencies and the specified opposing current, ${\theta }_1=\pi /6$, ${\theta }_2=\arcsin (3/4)$, andThe incident phase lines have slope $1/\sqrt{3}$, the reflected lines slope $-1/\sqrt{3}$, and the transmitted lines slope $3/\sqrt{7}$: they steepen above the interface. For the other specified current, ${\hat{\omega }}_2=0$. The transmitted vertical wavenumber then diverges and there is no regular propagating upper-layer wave of that frequency: this is the critical level of an internal gravity wave limit. Approaching it from ${\hat{\omega }}_2>0$ gives $A_r/A_0\to -1$ and vanishing transmitted vertical energy flux; setting the intrinsic frequency to zero directly is outside the regular matching calculation.

Let ${\varphi }_{+}$ and $p_{+}^{\prime }$ denote the fields in $y>0$, and ${\varphi }_{-}$ and $p_{-}^{\prime }$ those in $y<0$. The linearized kinematic boundary condition equates the normal fluid velocity on both faces to the membrane velocity:The linear acoustic pressure relation isPressure below pushes upward and pressure above pushes downward. The transverse force balance for a membrane with mass per area $m$ and tension $T$ is consequentlyor, in terms of the velocity potentials,

Let $x$ point down the plane and $z$ normal to it, with velocity $(u,w)$. The two-dimensional incompressible Navier-Stokes equation isThe lubrication theory limit has film thickness much smaller than its streamwise length. Transverse viscous derivatives dominate, inertia is negligible at late times, and the leading equations areAt the wall, $u=w=0$ by the no-slip boundary condition. At the free surface $z=h(x,t)$, take $p=p_{\mathrm{atm}}$ and $u_z=0$, neglecting surface tension. The normal balance gives the hydrostatic pressureIf the current has thickness scale $H$ and length scale $L$, thenThus the streamwise pressure gradient is asymptotically smaller than the downslope body force for a fixed nonzero inclination. The streamwise equation becomesUsing $u(0)=0$ and $u_z(h)=0$ givesThe volume flux isIntegrating incompressibility across the film and using the kinematic boundary condition gives $h_t+q_x=0$, hence

Let the disturbed sheet be $y=h+\eta$ and take a normal mode proportional to $e^{i(kx-\omega t)}$, with $K=|k|$. Irrotational flow and the rigid-wall condition at $y=0$ give perturbation velocity potentialsThe two linearized kinematic boundary conditions at $y=h$ areThe Unsteady Bernoulli equation gives $p_{-}^{\prime }=i\rho (\omega -kU)A\cosh (Kh)$ and $p_{+}^{\prime }=i\rho \omega B$. Continuity of pressure therefore yields the finite-depth vortex-sheet dispersion relationor, with $T=\tanh (Kh)$,For $k>0$ its two roots areOne root has positive temporal growth for every $k\ne 0$, so every finite wavelength is subject to the Kelvin-Helmholtz instability.

The radial velocity and velocity potential areChoose the gauge in the Unsteady Bernoulli equation by evaluating it at infinity, where $u=0$, $\varphi =0$, and $p=p_0$:At the cavity surface the vacuum pressure is zero, and the kinematic boundary condition gives $u(a,t)=\dot{a}$. SinceBernoulli's equation becomesTherefore

More surfactant is encountered on the side toward which the far-field concentration $\Gamma$ increases. Adsorption lowers the surface tension there, so the resulting Marangoni stress drives interfacial flow toward the cleaner, higher-tension side. The reaction force propels the bubble along the concentration gradient; this is chemophoresis.

The bulk concentration obeys the advection-diffusion equationAt $r=a$,equates the outward bulk diffusive flux to minus the net adsorption rate: $C>b\Gamma$ favors desorption into the bulk, while $C<b\Gamma$ favors adsorption onto the interface.

With advection neglected, $\Gamma$ is harmonic. Rotational symmetry about $\mathbf{G}$ and the imposed far-field gradient select the dipolar formWhen $kab\ll D$, the boundary flux is smaller than the characteristic diffusive flux, so the leading boundary condition is ${\partial }_r\Gamma =0$ at $r=a$. This gives $B=a^3/2$, and hence

For an interface with unit normal directed from the bubble into the exterior liquid, the interfacial stress balance with variable surface tension may be writtenwith signs tied to the stated curvature convention. Put $C^{\prime }=A\mathbf{G}\cdot \mathbf{n}$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa =2/a$ andthe remaining exterior traction isIts resultant vanishes becauseand therefore the integrals of the two terms cancel. This is required because the bubble and its interfacial stresses exert no external body force on the combined bubble–fluid system.

The traction is a first spherical harmonic, so the decaying, force-free Stokes flow has no Stokeslet and is generated by the indicated Papkovich–Neuber representation. Comparing the supplied tractionwith the capillary traction givesIn the convention for these potentials, the normal velocity at $r=a$ is $(\beta /\mu )\mathbf{G}\cdot \mathbf{n}$. The kinematic boundary condition for a translating sphere is $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so

Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C^{\prime }$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential isUsing the identities supplied in the question,The bulk result gives ${\Gamma }^{\prime }=\frac{3}{2}a\mathbf{G}\cdot \mathbf{n}$ on the surface. The linearized equationthen yieldsThus

Increasing $k$ strengthens exchange with the imposed bulk gradient, so $A$ and $U$ increase toward a saturation value. Increasing $D_s$ smooths surface-concentration differences and decreases $U$. Increasing $C_0$ strengthens advective redistribution of the background surfactant; the resulting feedback opposes the imposed dipole, so $U$ decreases.

No normal flow at a rigid wall gives $\boxed{\varphi (\pm 1)=0}$. At a free surface $y=1+\eta$, the linearized kinematic boundary condition isso $(U-c)\eta =-\varphi$. The linearized tangential momentum equation gives the pressure amplitude $p=(U-c){\varphi }^{\prime }-U^{\prime }\varphi$. Constant surface pressure therefore requires

Let the unperturbed interface rise at pore velocity $U/\varphi$, and write its displacement asIn each fluid, Darcy's law and incompressible flow implyThe decaying pressure perturbations are proportional to $e^{-\alpha |z|}$. Continuity of normal velocity, the kinematic boundary condition $\varphi {\eta }_t=u_n^{\prime }$, and continuity of pressure giveThus a less mobile displaced fluid, ${\mu }_2>{\mu }_1$, and a denser fluid above a lighter one both drive the Saffman–Taylor instability.

For immiscible fluids, the Young–Laplace equation adds the pressure jump $-\gamma {\eta }_{xx}$. The dispersion relation becomeswhereIf $A>0$, the unstable band is $0<\alpha <\sqrt{A/C}$ and differentiation gives

Define the signed characteristic buoyancy velocitySince ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$ and $Ca={\mu }_{2}U/\gamma$,This expression applies when the quantity in the final parentheses is positive.