Past exam of the fluids and waves course of the University of Oxford / 2023 / 1

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.

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.

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.

With zero internal pressure and negligible surface tension, pressure continuity sets the liquid-side boundary value to $p(R,t)=0$. Evaluating the pressure field of a pulsating sphere there givesThus the Rayleigh equation for an inviscid spherical bubble, together with its initial conditions, isFor $p_{\infty }>0$, initially $\ddot{R}(0)=-p_{\infty }/(\rho a)<0$, so the physically relevant branch describes Rayleigh collapse of a spherical cavity.

Write $P=p_{\infty }/\rho >0$. Multiplying the radius ordinary differential equation by $2R^2\dot{R}$ gives an exact derivative:The initial conditions therefore give the first integralChoose the negative square root because the cavity starts accelerating inward. It cannot turn around while $0<R<a$, since the squared speed is then strictly positive. Separating variables, with $s=R/a$, proves the collapse time of a spherical cavity isHenceThe improper integral is finite: its integrand behaves as $s^{3/2}$ near zero and as $[3(1-s){]}^{-1/2}$ near one. The divergent inward speed as $R\to 0$ signals a singular endpoint of this ideal incompressible flow model; the finite collapse time follows within that model.