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At an interface between two inviscid fluids without surface tension, the pressure is continuous. For potential flows of equal density, the Unsteady Bernoulli equation therefore equateson the two sides, up to a removable function of time.

For inviscid flow with a conservative body-force potential $\Phi$, the Euler momentum equation and the identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ apply. In an irrotational region write $\mathbf{u}=\nabla \varphi$. For constant density they giveThis is the Unsteady Bernoulli equation; its constant is spatially uniform, with time-dependent gauge absorbed into $\varphi$ if desired.

For the intended linearly elastic balloon discharge model, take an initially stationary, inviscid plug of water in the constant-area tube. Put $B=\pi a^2$ and $p_{\mathrm{in}}-p_{\mathrm{atm}}=KV$ with $K>0$. At the two tube ends the speeds are equal, so their kinetic terms cancel in the Bernoulli difference. Since $\varphi (L)-\varphi (0)=Lu$, this givesHence $\ddot{V}+{\Omega }^{2}V=0$, where ${\Omega }^2=KB/(\rho L)$. With $V(0)=V_0$ and $u(0)=0$,ThusAcceleration stays positive until emptying, so the maximum speed is reached at that instant. The time is independent of initial volume, while the maximum speed is not.

The initial-rest condition is implicit in the claimed volume independence, rather than specified in the text. For an independently prescribed positive initial speed $u_0$, the first zero instead occurs at ${\Omega }^{-1}\arctan [\Omega V_0/(B{u}_0)]$. The calculation also takes the given balloon pressure as the pressure directly at the tube entrance; a reservoir-to-exit kinetic-head or entrance-loss model would add other terms and is not this equal-speed tube model.

The free surface is a material surface, so its kinematic boundary condition isWith constant atmospheric pressure $p_a$ and no surface tension, the dynamic boundary condition for an inviscid interface follows from Unsteady Bernoulli equation:Write $\varphi =Ux+\phi$, where $\phi$ and $\eta$ are small. Choose the time-dependent potential gauge so the undisturbed right-hand side is $U^2/2$. Retaining first-order terms, evaluated at $y=0$, givesThe incompressibility condition also gives Laplace's equation for $\phi$.

Set $\Omega =\omega -kU$. Substitution of the specified normal mode into the two linearized boundary conditions yields $-i\Omega a=ikb$ and $\Omega b+ga=0$. Eliminating $b$ gives the dispersion relationThe decay factor $e^{ky}$ in a fluid extending to $y=-\infty$ uses $k>0$. With signed wavenumber the decaying factor is $e^{|k|y}$ and the right-hand side is $g|k|$. The frequency shift is the advection of a deep-water gravity wave by the uniform current.

The interface is material: fluid particles on $F=z-\eta (x,t)=0$ remain there. Its material derivative therefore vanishes, giving $w_j={\eta }_t+u_j{\eta }_x=D_j\eta /Dt$ on the interface. Since $w_j={\partial }_z{\varphi }_j$, this is the kinematic boundary condition for a free-surface graph on each side. The material derivatives use the respective tangential fluid velocities; they enforce the same moving normal boundary.

The Unsteady Bernoulli equation gives $p_j+{\rho }_j[{\partial }_t{\varphi }_j+|\nabla {\varphi }_j{|}^2/2+gz]=C_j(t)$. Absorb the spatially constant functions $C_j$ into the gauges of the velocity potentials. Evaluating at $z=\eta$ and subtracting the equations yields the stated dynamic boundary condition for an inviscid interface. Surface tension supplies the pressure jump, rather than pressure continuity. Thus the first condition enforces material motion, and the second enforces the Bernoulli pressure balance.

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 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

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.