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An infinitely long circular cylinder of radius $a$ translating across its axis through otherwise stationary, inviscid incompressible fluid has added mass $m_a=\rho \pi a^2$ per unit length. The decaying, circulation-free velocity potential is $\varphi =-U{a}^2\cos \theta /r$. In moving centre coordinates, the fixed-laboratory derivative is ${\varphi }_t=-\dot{U}{a}^2\cos \theta /r-U^2{a}^2\cos (2\theta )/r^2$. The Unsteady Bernoulli equation gives $p-p_{\infty }=\rho a^2\dot{U}\cos \theta /r+\rho U^2{a}^2\cos (2\theta )/r^2-\rho U^2{a}^4/(2r^4)$. Integrating $-pa\cos \theta$ around $r=a$ yields $F_x=-\rho \pi a^2\dot{U}$; the transverse force vanishes. Taking a derivative at fixed centre coordinates would omit the necessary convective correction.

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.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

For a slender tube of length $L$ supplied from a large reservoir with excess pressure $P$, negligible gravity and entrance inertia, the Unsteady Bernoulli equation gives $L\dot{U}+U^2/2=P/\rho$. A stationary initial liquid column has the displayed speed and volume flux $\pi a^{2}U$ for tube radius $a$. The inertial term controls the early linear acceleration; the kinetic-energy term sets the terminal inviscid speed. This is a leading slender-tube model while the reservoir remains large and the tube full.

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:

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.

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.

An irrotational flow has $\nabla \times \mathbf{u}=0$. Hence a velocity potential $\varphi$ exists locally, and globally on a simply connected domain, with $\mathbf{u}=\nabla \varphi$. The exterior of a sphere in three dimensions has this property. Constant mass density reduces the continuity equation to $\nabla \cdot \mathbf{u}=0$, soThus the velocity potential is a harmonic function. For the Euler equations, use the vorticity cross-product identityThe last term vanishes. Since ${\partial }_t\mathbf{u}=\nabla {\varphi }_t$, the Euler momentum equation becomesOn each connected fluid region, a function with zero gradient is spatially constant. Therefore the Unsteady Bernoulli equation isThe freedom to add a function of time to the velocity potential changes $F(t)$ but leaves the velocity field unchanged. On a multiply connected region, zero vorticity alone guarantees only local potentials; a global single-valued velocity potential additionally requires zero circulation around every closed path.

Put $A=R^2\dot{R}$. The partial derivative in the Unsteady Bernoulli equation is taken at fixed spatial position, soAt infinity the velocity vanishes and $p\to p_{\infty }$, giving $F(t)=\dot{G}+p_{\infty }/\rho$. Substitution cancels the arbitrary $G$ and yields the pressure field of a pulsating sphere:The distinction between ${\partial }_t\varphi$ at fixed $r$ and the material derivative along the moving boundary is essential here.

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.

For inviscid flow that is also incompressible flow and irrotational flow with velocity potential $\mathbf{u}=\nabla \varphi$, the Euler equations integrate to the Unsteady Bernoulli equationThe same $F(t)$ applies throughout a connected irrotational fluid region; an additive time-dependent potential can absorb it. With gravity absent, fluid at rest at infinity and $\varphi \to 0$, it is $p_{\infty }(t)/\rho$.

Take the cylinder centre at laboratory position $(X(t),0)$ with $\dot{X}=U(t)$, and define $r,\theta$ relative to this moving centre. The circulation-free potential flow satisfies Laplace's equation, decays at infinity, and has surface condition ${\varphi }_r(a,\theta )=U\cos \theta$. The $\cos \theta$ harmonic solution is consequentlyThe velocities are $u_r=U{a}^2\cos \theta /r^2$ and $u_{\theta }=U{a}^2\sin \theta /r^2$, hence $|\mathbf{u}{|}^2=U^2{a}^4/r^4$.

The time derivative must be taken at fixed laboratory coordinates, not fixed $r,\theta$. At a fixed fluid point $\dot{r}=-U\cos \theta$, $\dot{\theta }=U\sin \theta /r$, soSubstituting into the Unsteady Bernoulli equation gives the entire pressure fieldAt the cylinder surface this is $p_{\infty }+\rho a\dot{U}\cos \theta +\rho U^2(\cos 2\theta -1/2)$. Pressure acts against the outward radial normal of the cylinder, so the force per unit length isAll steady terms cancel by trigonometric orthogonality. ThereforeThis identifies the added mass of a circular cylinder as the mass of displaced fluid per unit length. The solution assumes the ambient fluid is initially at rest and has no separately prescribed circulation; these select the usual pure translating-cylinder flow.

For an inviscid fluid of constant density, with irrotational velocity $\mathbf{u}=\nabla \Phi$ and conservative body-force potential $\Psi$, the Unsteady Bernoulli equation isIt holds throughout a connected irrotational fluid region, up to the time-dependent gauge of $\Phi$. Indeed the Euler equations and $\nabla \times \mathbf{u}=0$ imply that the gradient of the left side is zero. In the bubble problem there is no body-force contribution compatible with the assumed spherical symmetry.

Spherical incompressibility gives ${\partial }_r(r^2{u}_r)=0$. The moving interface imposes $u_r(R,t)=\dot{R}$, soThis potential tends to zero at infinity, where the velocity tends to zero and $C(t)=p_{\infty }/\rho$. Differentiate $\Phi$ at fixed spatial $r$, not along the moving interface: ${\Phi }_t=-(2R{\dot{R}}^2+R^2\ddot{R})/r$. At $r=R$, pressure continuity without surface tension gives $p=p_{\infty }-\Delta p$. Bernoulli therefore yields $-2{\dot{R}}^2-R\ddot{R}+{\dot{R}}^2/2-\Delta p/\rho =0$, or the Rayleigh equation for an inviscid spherical bubble

Integrating the kinetic energy over the exterior fluid givesDifferentiate and use the radius equation:With $R(0)=R_0$ and $\dot{R}(0)=0$, the pressure-work energy balance for a spherical bubble integrates toorThe collapsing branch takes the negative square root for $\dot{R}$. The energy gain is exactly the pressure difference times the lost bubble volume, providing a check on the sign and numerical factor.

For an inviscid flow that is an incompressible flow and an irrotational flow, write $\mathbf{u}=\nabla \varphi$. The Euler equations for an inviscid fluid, without gravity, integrate spatially to the Unsteady Bernoulli equationA time-dependent addition to $\varphi$ can absorb $C(t)$.

In the ideal uniform tube-plug model, let $u(t)$ be the axial speed toward the outlet. Incompressibility and fixed cross-section make the speed the same at both ends. With $\varphi =u(t)x$, the kinetic terms cancel when the Unsteady Bernoulli equation is subtracted between entrance and exit, givingConservation of volume in the spherical balloon gives $4\pi R^2\dot{R}=-\pi a^{2}u$, or $u=-4R^2\dot{R}/a^2$. Differentiating and eliminating $u$ yieldsThis is the constant-tension balloon discharge equation; entrance losses and a reservoir-to-tube kinetic-head correction are excluded by this ideal plug model.

Multiplication by $2R\dot{R}$ turns the equation intoThe printed time formula additionally assumes the water is initially at rest, $\dot{R}(0)=0$, although only $R_0$ is explicitly specified in the question. Under that assumption $C=K{R}_0^2$. On the shrinking branch,Set $r=R_0\sin \vartheta$ and $\theta =\arcsin (R/R_0)$. The integral becomes $R_0^2{K}^{-1/2}{\int }_{\theta }^{\pi /2}{\sin }^2\vartheta d\vartheta$, hencePutting $R=0$, or $\theta =0$, gives the ideal emptying timeFor an independently prescribed initial rate $v_0=\dot{R}(0)$, the first integral instead has $C=K{R}_0^2+R_0^4{v}_0^2$, so radius alone does not determine that emptying time. This explicitly identifies the extra initial datum needed for the displayed answer.

For an inviscid flow of constant density $\rho$, driven by a conservative body force $-\nabla \Phi$, the Euler equation isThe vector identity $(u\cdot \nabla )u=\nabla (|u{|}^2/2)-u\times (\nabla \times u)$ and irrotationality give the Unsteady Bernoulli equation. Specifically, wherever a single-valued potential $u=\nabla \varphi$ is available,On a connected potential-flow region this meansThe integration function is spatially constant but need not be constant in time; it can be removed by changing the potential's time-dependent gauge. A simply connected irrotational region provides the requisite global potential. For a barotropic fluid the pressure term is replaced by $\int dp/\rho (p)$; here the liquid density is constant.

For the tube, let $U(t)$ be its approximately uniform axial velocity. The large reservoir's surface velocity is negligible and gravity is neglected as stipulated. Entrance-region inertia contributes an effective length of order $a$, negligible against $L$. The potential difference between the outlet and the reservoir is consequently $LU$ to leading order. Subtract Bernoulli at the reservoir surface, where pressure is $P+p_a$, from its value at the outlet, where pressure is $p_a$ and speed is $U$. This givesThe inertial term cannot be dropped during startup. With $U_{\infty }=\sqrt{2P/\rho }$, separate variables:Hence the inviscid startup in a pressure-driven tube givesInitially $U\sim Pt/(\rho L)$, the acceleration of the stationary liquid column. At long times $U\to \sqrt{2P/\rho }$, the inviscid pressure-to-kinetic-energy limit. The result applies while the tube remains full and reservoir dimensions are large enough that surface motion and the neglected entrance inertia remain small.

Take the undisturbed free surface as $z=0$, the bottom as $z=-h$, and the walls as $x,y=0,a$. For an incompressible flow which is an irrotational flow, write $\mathbf{u}=\nabla \varphi$ with a velocity potential satisfying ${\nabla }^2\varphi =0$. Impermeability gives ${\varphi }_z=0$ at the bottom, ${\varphi }_x=0$ at the $x$ walls, and ${\varphi }_y=0$ at the $y$ walls.

For the actual surface $z=\eta (x,y,t)$, the exact kinematic boundary condition for a free-surface graph isThe Unsteady Bernoulli equation, with atmospheric pressure at the surface and its time-dependent constant absorbed into $\varphi$, givesFor small disturbances about fluid at rest, $\varphi$ and $\eta$ are first order. Products of first-order quantities and Taylor corrections from evaluating at $z=\eta$ are second order. The linear surface conditions are thereforeor, after eliminating the elevation,

The separated normal modes of surface gravity waves in a rectangular tank havewhere $m,n$ are nonnegative integers not both zero. The horizontal factors satisfy the wall conditions, and the vertical factor satisfies both the Laplace equation and the bottom condition. At the top, ${\varphi }_z=k_{mn}\tanh (k_{mn}h)\varphi$; substitution into the surface condition givesThe elevation isReal physical standing waves are the real parts or linear combinations of these complex modes. The uniform $m=n=0$ mode has zero frequency and would change the mean depth; conservation of the fixed water volume excludes it from the disturbances. Modes with equal $m^2+n^2$ are degenerate and can be superposed.

The free surface is a material surface. For $F(x,y,t)=y-\eta (x,t)$, the material derivative vanishes on $F=0$. Since the velocity is $\nabla \varphi$, this gives the exact kinematic boundary condition for a free-surface graphAt the surface the pressure equals the constant atmospheric pressure. Absorbing that constant into the time-dependent potential gauge in Unsteady Bernoulli equation yields the exact dynamic boundary condition for an inviscid interfaceThere is no surface-tension term in this model.

Write $\varphi =Ux+\phi$ and linearize in the small perturbation $\phi$ and the surface displacement. The base value is $F=U^2/2$, and evaluation at the displaced surface can be replaced by evaluation at zero to first order. ThusIncompressibility requires Laplace's equation ${\phi }_{xx}+{\phi }_{yy}=0$. For $k>0$, the complex perturbation $b{e}^{ky}{e}^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta =\mathrm{Re}\{a{e}^{i(kx-\omega t)}\}$ and $\Omega =\omega -kU$. The linear boundary conditions becomeThey are consistent precisely whenConsequently the two laboratory phase velocities of a deep-water gravity wave areIf propagation of a wave packet is intended, its group velocities are instead $d\omega /dk=U\pm \frac{1}{2}\sqrt{g/k}$.

At the initial instant the circulation around every closed curve is zero because the fluid is at rest. A smooth material-flow map carries closed curves bijectively between times. Conservation of circulation around every material curve therefore makes the line integral of the velocity zero around every closed curve at the current time, not just small contractible loops. Fix a reference point in a connected fluid region and defineThe vanishing closed-curve integrals make this independent of the path. Thus $\mathbf{u}=\nabla \varphi$: the fluid has a globally defined velocity potential. Its incompressible flow condition is $\nabla \cdot \mathbf{u}=0$, hence ${\nabla }^2\varphi =0$. This argument establishes the global potential without silently assuming the fluid region is simply connected.

Use the laboratory frame, with the sphere instantaneously centred at the origin and moving in the $\theta =0$ direction. Impermeability equates fluid and sphere normal velocities, while the fluid is at rest at infinity:Choose the additive constant so that $\varphi \to 0$ at infinity. Substitution of $\varphi =f(r)\cos \theta$ into the Laplace equation in spherical polar coordinates givesThe far-field condition gives $A=0$, and $f^{\prime }(a)=U$ gives $B=-U{a}^3/2$. ThereforeThe radial and polar velocity components areFor fluid density $\rho$, its kinetic energy isThus $T_f=\frac{1}{2}{m}_a{U}^2$ with added mass of a sphere $m_a=2\pi \rho a^3/3$, half the mass of displaced fluid.

For the initial acceleration after release, let $\dot{U}$ denote the upward acceleration, $\mathcal{V}=4\pi a^3/3$, and $m_b={\rho }_b\mathcal{V}$. At the instant $U=0$, the velocity-squared term in the Unsteady Bernoulli equation vanishes. The time derivative of the potential at the sphere is ${\varphi }_t(a,\theta )=-a\dot{U}\cos \theta /2$; centre-motion corrections also vanish at this instant. Relative to hydrostatic pressure, the additional pressure is consequentlyIts upward force on the sphere isAdding buoyancy and weight, Newton's second law gives $(m_b+m_a)\dot{U}=(\rho -{\rho }_b)\mathcal{V}g$. ThereforeThis is the acceleration of a freely falling sphere with added mass, with upward chosen positive. If ${\rho }_b>\rho$, the negative value represents a downward acceleration.

Without surface tension, the exact dynamic boundary condition for an inviscid interface is equality of pressure, $p_1=p_2$, at $z=\zeta$. The Unsteady Bernoulli equation in each irrotational fluid isChoose the potential gauge so that $C_i=p_0/{\rho }_i+U_i^2/2$, its unperturbed value. Expanding the kinetic term as $U_i^2/2+U_i{\partial }_x{\phi }_i$ and evaluating gravity at the displaced interface givesto first order. Equality of these pressures gives

For constant density $\rho$, the inviscid Euler equations with conservative body force areThe vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ simplifies for potential flow $\mathbf{u}=\nabla \varphi$, since its vorticity vanishes. Also ${\partial }_t\mathbf{u}=\nabla {\varphi }_t$. ThereforeOn each connected fluid region the expression is spatially constant but may depend on time, giving the Unsteady Bernoulli equationThe arbitrariness of $f$ corresponds to the freedom to add a function of time to the velocity potential without changing $\mathbf{u}$. Replacing $\varphi$ by $\varphi -{\int }^{t}f(s)ds$ sets the right-hand side to zero.

The Euler equations for an inviscid fluid with no body force is ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho$. For irrotational flow $\mathbf{u}=\nabla \varphi$, the identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ removes the last term. Thus the gradient of ${\varphi }_t+|\nabla \varphi {|}^2/2+p/\rho$ is zero, giving the Unsteady Bernoulli equationThe function is spatially constant on the connected fluid region; adding a time-dependent constant to the potential changes its gauge.

For spherically symmetric incompressible radial flow, $r^2{u}_r$ is independent of $r$. The bubble boundary moves with the fluid, so $u_r(R)=\dot{R}$, giving $u_r=R^2\dot{R}/r^2$. Fix $\varphi \to 0$ at infinity, obtaining $\varphi =-R^2\dot{R}/r$ and $F=p_{\infty }/\rho$. The time derivative is taken at fixed $r$, then evaluated at the moving boundary: ${\varphi }_t(R)=-(2{\dot{R}}^2+R\ddot{R})$. With pressure continuity at the interface and no capillary or viscous correction, Bernoulli gives the Rayleigh equation for an inviscid spherical bubble

The gas law and initial pressure fix $C{M}^2=p_{\infty }{R}_0^6/2$. Multiply the radius equation by $2R^2\dot{R}$; its left side is $d(R^3{\dot{R}}^2)/dt$. Integrating from the zero-speed initial state gives the first integral of polytropic spherical-bubble motionTo find the permitted radii, put $s=R^3/R_0^3$. The bracket becomes $R_0^3(1-s)(2s-1)/(3s)$, nonnegative exactly for $1/2\le s\le 1$. ThereforeInitially $\ddot{R}=-p_{\infty }/(2\rho R_0)<0$, so the radius decreases from the upper endpoint. At the lower endpoint the gas pressure is $2p_{\infty }$, giving positive acceleration and an outward reversal. The first-integral zeros are simple, so the travel-time integrals near either endpoint behave as an integrable inverse square root. The bubble reaches both turning radii in finite time, and the autonomous undamped equation repeats this motion periodically. Thus the stated interval describes oscillation, rather than an irreversible approach to one endpoint.