= Solution
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 $\phi=A\log r+C(t)$,
$$
\partial_t\phi(a,t)-\partial_t\phi(b,t)=\dot A\log\frac ab,
$$
where each partial derivative is taken at fixed $r$. Since $u_r=A/r$ and $p(b,t)=p_\infty$,
$$
\boxed{p(a,t)-p_\infty
=\rho\left[\dot A\log\frac ba
-\frac{A^2}{2}\left(\frac1{a^2}-\frac1{b^2}\right)\right]},
\qquad A=a\dot a.
$$
Equivalently, $\dot A=\dot a^2+a\ddot a$. This is the <pressure in radially symmetric annular potential flow>.
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