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Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

Separate the uniform stream and an arbitrary time-dependent gauge from each velocity potential: ${\varphi }_j=U_{j}x+f_j(t)+{\varphi }_j^{\prime }$. Only gradients of the velocity potential determine the velocity field. The remote fluid must remain in its unperturbed uniform stream, soAfter fixing the harmless gauge, the nonzero-wavenumber perturbation velocity potentials themselves vanish at the corresponding infinities. Incompressible flow and irrotational flow give the Laplace equation for each perturbation. For a normal mode with $k>0$, the vertical amplitude solves ${\Phi }_j^{\prime \prime }-k^2{\Phi }_j=0$, henceThe opposite exponentials are excluded because they make the perturbation grow at infinity. It is the perturbation, rather than the total potential $U_{j}x+f_j(t)$, that decays.

Expanding the kinetic helicity conservation law separates advection from the other flux:Thus material conservation of kinetic helicity density holds precisely whenAn especially useful sufficient pair of conditions is incompressible flow, $\nabla \cdot \mathbf{u}=0$, and constancy of $w-u^2/2$ along integral curves of the vorticity, $\mathbf{\omega }\cdot \nabla (w-u^2/2)=0$. Together these conditions make both terms vanish without cancellation. They are not necessary individually, because the two terms in the displayed balance can cancel. An irrotational flow is a trivial conserved case with $q=0$.

For contrast, combining the balance with mass conservation givesConsequently constancy of $w-u^2/2$ along integral curves of the vorticity makes $q/\rho$ an advected scalar even in compressible flow; it does not generally make $q$ itself constant.

Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.

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.

Write the density and pressure as $\rho ={\rho }_0+{\rho }^{\prime }$ and $p=p_0+p^{\prime }$, and let $\mathbf{u}$ be the small fluid velocity. The linear homentropic acoustic equations arewhere the adiabatic sound speed isIf the flow is irrotational, write $\mathbf{u}=\nabla \varphi$. The momentum equation integrates spatially toa purely time-dependent integration term having been absorbed into $\varphi$. Substitution into continuity gives the wave equationso the wave speed is $c_0$.

To obtain energy conservation, dot the momentum equation with $\mathbf{u}$ and use the product rule:The continuity and pressure-density relations giveTherefore the acoustic energy conservation law iswith acoustic energy density and intensity

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.

Writing $R=x^2+y^2$, direct differentiation givesso $\nabla \cdot \mathbf{u}=0$. Likewise,so the two-dimensional vorticity ${\partial }_{x}v-{\partial }_{y}u$ vanishes. The flow is therefore incompressible and irrotational away from the origin.

For an inviscid fluid of constant mass density $\rho$ with no body force, the Euler equations for an inviscid fluid areThe vector identityand irrotational flow give $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Writing $u=\nabla \varphi$ in terms of a velocity potential, we obtainThus the bracket depends only on time:Since adding a function of time to $\varphi$ leaves $u=\nabla \varphi$ unchanged, that function may be chosen to absorb $C(t)$. This is the Unsteady Bernoulli equation.

For irrotational flow, $\mathbf{u}=\nabla \varphi$. Incompressibility gives the Laplace equationThe rigid side walls impose no penetration:and decay in the infinitely deep fluid requires $\nabla \varphi \to 0$ as $z\to -\infty$.

Linearizing the kinematic condition from part (a) about $z=0$ givesThe Unsteady Bernoulli equation, evaluated at the surface and linearized about hydrostatic equilibrium, gives the dynamic conditionThese are the linearized free-surface boundary conditions. Initial values for $\eta$ and $\varphi$ specify the transient; below we also record the initially quiescent solution.

Let $a$ denote the bubble size, $U$ its speed, and $\mathrm{Re}=Ua/\nu \gg 1$ the Reynolds number. Outside a thin boundary layer, inertia dominates and the flow is an irrotational flow with velocity scale $U$ and strain scale $U/a$. Its viscous dissipation has orderin three dimensions.

The boundary-layer thickness isFor a clean bubble the stress-free boundary condition prescribes tangential stress rather than tangential velocity. The outer flow already has tangential strain of order $U/a$, so cancelling that stress requires a velocity correction of onlynot a correction of order $U$. Its gradient remains $O(U/a)$, and its boundary-layer dissipation iswhich is asymptotically smaller.

In steady translation, the drag power $DU$ balances viscous dissipation. The leading estimate therefore follows from the known outer flow alone:in three dimensions. This is the dissipation estimate for high-Reynolds-number bubble drag.

Write the velocity potentials in the upper, middle and lower domains as ${\varphi }_{+}$, ${\varphi }_0$ and ${\varphi }_{-}$. Irrotational flow and incompressible flow imply the Laplace equationThe far-field boundary conditions are

Every shear layer is a material interface. The exact kinematic boundary condition at $y=h+\eta$ isand at $y=-h-\eta$ it isHere each equation is evaluated on the indicated moving interface.

Finally, pressure continuity is the dynamic boundary condition for an inviscid interface. The Unsteady Bernoulli equation givesafter absorbing spatially constant functions of time into the potentials.

For time-dependent irrotational flow with $u=\nabla \varphi$, constant mass density, and no body force, the Euler equations for an inviscid fluid integrate toA time-dependent change of gauge in $\varphi$ may set $C(t)=0$.