= Solution
The linearized azimuthal <Euler momentum equation> is
$$
\partial_tv_\phi+\frac{u'_R}{R}\frac{d(R^2\Omega)}{dR}=0.
$$
For the time dependence $e^{-i\omega t}$ and $u'_R=-i\omega\xi_R$, it gives
$$
\boxed{v_\phi=-\frac{\xi_R}{R}\frac{d(R^2\Omega)}{dR}.}
$$
This expresses conservation of the displaced element's <specific angular momentum>. It applies directly to nonzero-<frequency> modes, with the zero-<frequency> limit taken in the displacement formulation.
The radial advective acceleration supplies $-2\Omega v_\phi$, while the <pressure> force perturbation is $\delta\rho\,\nabla p/\rho^2-\nabla\delta p/\rho$. Eliminate $v_\phi$ and retain the vertical equation. Under the <Cowling approximation>, $\delta\Phi=0$, so
$$
\boxed{-\omega^2\boldsymbol\xi=\mathcal F\boldsymbol\xi=\frac{\delta\rho}{\rho^2}\nabla p-\frac1\rho\nabla\delta p-\kappa^2\xi_R\mathbf e_R,\qquad \kappa^2=\frac{2\Omega}{R}\frac{d(R^2\Omega)}{dR}.}
$$
The <continuity equation> gives $\delta\rho=-\nabla\cdot(\rho\boldsymbol\xi)$. The Lagrangian adiabatic relation $\Delta p/p=\gamma\Delta\rho/\rho$, with $\Delta\rho=-\rho\nabla\cdot\boldsymbol\xi$, gives
$$
\boxed{\delta\rho=-\nabla\cdot(\rho\boldsymbol\xi),\qquad \delta p=-\boldsymbol\xi\cdot\nabla p-\gamma p\nabla\cdot\boldsymbol\xi.}
$$
Here $\kappa$ is the <radial epicyclic frequency>. These formulas define the <axisymmetric adiabatic displacement operator>.
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