= Solution
The interface is the material level set $F=r-R(\theta,t)=0$. The condition $DF/Dt=0$ gives
$$
-R_t+u-\frac vrR_\theta=0,
$$
which is the stated equation.
Let the perturbation potentials inside and outside be
$$
\phi_i=A r^k e^{ik\theta+\sigma t},
\qquad
\phi_o=B r^{-k}e^{ik\theta+\sigma t}.
$$
Linearizing the kinematic condition at $r=a$ gives
$$
(\sigma+ik\omega)\eta=kAa^{k-1},
\qquad
\sigma\eta=-kBa^{-k-1}.
$$
The linearized interior Euler equations give
$$
p_i'=-\rho[\sigma+i(k-2)\omega]\phi_i,
$$
whereas the stationary exterior gives $p_o'=-\rho\sigma\phi_o$. Since the basic interior <pressure> has $dp_i/dr=\rho\omega^2r$, <pressure> continuity on the displaced boundary is
$$
p_i'(a)+\rho\omega^2a\eta=p_o'(a).
$$
Eliminating $A$ and $B$ gives the <circular vortex-sheet mode> relation
$$
\sigma^2+i\omega(k-1)\sigma
-\frac12\omega^2k(k-1)=0,
$$
so
$$
\boxed{\displaystyle
\sigma=\frac\omega2\left[-i(k-1)\pm\sqrt{k^2-1}\right]}.
$$
The $k=1$ mode is a neutral displacement. Every $k>1$ mode has one exponentially growing branch, so the circular interface rolls up through a Kelvin--Helmholtz instability. Its pattern angular <velocity> is $\omega(k-1)/(2k)$, in the direction of the gyre but slower than the solid-body motion; the disturbances therefore propagate upstream relative to the rotating water.
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