= Solution
A radially symmetric <velocity potential> has $u_r=\partial_r\phi$. The <incompressible flow> condition in two-dimensional polar coordinates is
$$
\frac1r\frac{\partial}{\partial r}(ru_r)=0,
$$
so $ru_r=A(t)$. The <kinematic boundary condition> at the bubble surface gives $u_r(a,t)=\dot a$, hence
$$
\boxed{u_r(r,t)=\frac{a\dot a}{r}},
\qquad
\boxed{\phi(r,t)=a\dot a\log r+C(t)}.
$$
At the outer interface, $\dot b=u_r(b,t)=a\dot a/b$. Therefore
$$
\frac d{dt}(b^2-a^2)=2b\dot b-2a\dot a=0,
$$
and
$$
\boxed{b(t)^2-a(t)^2=b_0^2-a_0^2}.
$$
The area $\pi(b^2-a^2)$ of the incompressible liquid is conserved. These formulas are the <radially symmetric incompressible flow in a planar annulus>.
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