Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 16C / b

A radially symmetric velocity potential has $u_r={\partial }_r\varphi$. The incompressible flow condition in two-dimensional polar coordinates isso $r{u}_r=A(t)$. The kinematic boundary condition at the bubble surface gives $u_r(a,t)=\dot{a}$, henceAt the outer interface, $\dot{b}=u_r(b,t)=a\dot{a}/b$. ThereforeandThe area $\pi (b^2-a^2)$ of the incompressible liquid is conserved. These formulas are the radially symmetric incompressible flow in a planar annulus.

Apply the Unsteady Bernoulli equation at $r=a(t)$ and $r=b(t)$ and subtract, thereby eliminating its time-dependent constant. For $A=a\dot{a}$ and $\varphi =A\log r+C(t)$,where each partial derivative is taken at fixed $r$. Since $u_r=A/r$ and $p(b,t)=p_{\infty }$,Equivalently, $\dot{A}={\dot{a}}^2+a\ddot{a}$. This is the pressure in radially symmetric annular potential flow.

Let $b_0$ be the outer radius when $a=a_0$. The area relation from part (i) givessoMoreover,The term $A^2$ in the pressure is second order, while the logarithm multiplying $\dot{A}$ may be evaluated at equilibrium. The linearization is therefore

Per unit depth, the bubble volume is its area $V=\pi a^2$. At equilibrium its pressure equals $p_{\infty }$, so $p_{0}V=p_{\infty }\pi a_0^2$. HenceContinuity of fluid pressure at the interface gives $p(a,t)=p_0(t)$. Consequently

Equating the two linearized expressions for $p(a,t)-p_{\infty }$ from parts (iii) and (iv) givesThis is the equation of simple harmonic motion. Thus the bubble performs small oscillations with angular frequencyThis is the small oscillation of a planar gas bubble in an annular liquid.