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For gas pressure depending only on radius and initial speed zero, multiplying the Rayleigh equation for an inviscid spherical bubble by $2R^2\dot{R}$ integrates to the displayed identity. A polytropic pressure $p_g=K/R^{3\kappa }$ gives elementary power integrals when $\kappa \ne 1$, and a logarithm when $\kappa =1$. Turning radii are the positive zeros of this first integral. At $\kappa =2$ and $p_g(R_0)=p_{\infty }/2$, they are $R_0$ and $R_0/2^{1/3}$; the inward and outward accelerations at them have opposite signs, giving bounded undamped oscillation in the ideal model.

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 potential flow $\mathbf{u}=\nabla \phi$ in an inviscid fluid of constant density under the conservative force per unit mass $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by inserting the velocity potential into the Euler equations and integrating the resulting spatial gradient. A time-only change of the velocity potential changes $C(t)$ without altering the velocity field.

For spherical motion in an unbounded liquid, incompressibility gives $r^2{u}_r=A(t)$; the kinematic condition $u_r(R)=\dot{R}$ gives $A=R^2\dot{R}$. Choose $\phi =-R^2\dot{R}/r$, vanishing at infinity. To obtain the displayed bubble relation, take no spatial body-force contribution, neglect viscosity and surface tension, and equate the liquid-side interface pressure to $p_b$. With the far-field pressure $p_{\infty }$, Bernoulli equation evaluated at the surface uses the partial derivative at fixed $r$:Hence the Rayleigh equation for an inviscid spherical bubble is

The exterior kinetic energy isDifferentiating it and using $\dot{V}=4\pi R^2\dot{R}$ gives the pressure-work energy balance for a spherical bubble:This derivation also holds at a turning point $\dot{R}=0$, without dividing by $\dot{R}$.

For constant $p_{\infty }$ and the stated inverse-volume bubble pressure, integrate this exact time derivative:The relation is valid on physically accessible portions of the motion, where the resulting kinetic energy is nonnegative. In particular $\log q\le q-1$ makes the bracket nonpositive, as expected for the restoring pressure about $V_0$.

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.

For spherical expansion in an unbounded incompressible flow, continuity gives $u_r=R^2\dot{R}/r^2$. Integrating the kinetic energy outside the bubble gives $K=2\pi \rho R^3{\dot{R}}^2$. The Rayleigh equation for an inviscid spherical bubble then yields $\dot{K}=(p_b-p_{\infty })\dot{V}$, where $V=4\pi R^3/3$. With $p_b=p_{\infty }{V}_0/V$ and constant $p_{\infty }$, integration gives $K-K_0=p_{\infty }[V_0\log (V/V_0)-V+V_0]$. This neglects viscosity, surface tension and external body-force differences, which contribute additional terms in other bubble models.