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Use the spherical symmetry of the expansion, with fluid at rest at infinity and no superposed imposed flow. The radial Laplace equation givesThe moving sphere is a material boundary: its kinematic boundary condition is $u_r(R,t)=\dot{R}$. Consequently $A(t)=R^2\dot{R}$, and the spherically symmetric incompressible radial flow hasEquivalently, its volumetric flow rate through any concentric sphere is $4\pi r^2{u}_r=4\pi R^2\dot{R}$, the rate of change of the enclosed volume.

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.

In spherical symmetry, the velocity is radial and constant over each sphere. HenceApply the divergence theorem to the fluid shell between radii $R_1$ and $R_2$. Since the flow is incompressible,Thus $Q$ is independent of $R$ and is a function of time alone. This is the flux law for spherically symmetric incompressible radial flow.