= Solution
In an axisymmetric <cylindrical coordinate system>, the steady <Collisionless Boltzmann equation> is
$$
v_R\frac{\partial f}{\partial R}+v_z\frac{\partial f}{\partial z}+\left(\frac{v_\phi^2}{R}-\Phi_R\right)\frac{\partial f}{\partial v_R}-\frac{v_Rv_\phi}{R}\frac{\partial f}{\partial v_\phi}-\Phi_z\frac{\partial f}{\partial v_z}=0.
$$
Let $\nu=\int f\,d^3v$ and $\overline A=\nu^{-1}\int Af\,d^3v$. Multiply by $v_R$ and integrate over velocities. The spatial terms give $\partial_R(\nu\overline{v_R^2})+\partial_z(\nu\overline{v_Rv_z})$. <Integration by parts> in $v_R$ supplies $\nu\Phi_R-\nu\overline{v_\phi^2}/R$, and the $v_\phi$ derivative supplies $\nu\overline{v_R^2}/R$. The $v_z$ force term vanishes because $v_R$ does not depend on $v_z$. This assumes sufficient decay of the <galactic distribution function> to discard all velocity boundary terms. The radial <Jeans equation> is consequently
$$
\partial_R(\nu\overline{v_R^2})+\partial_z(\nu\overline{v_Rv_z})+\frac{\nu}{R}(\overline{v_R^2}-\overline{v_\phi^2})+\nu\Phi_R=0.
$$
Reflection symmetry about $z=0$ makes $\nu$ even and $\overline{v_Rv_z}$ odd. At the midplane the latter vanishes, and $\nu^{-1}\partial_z(\nu\overline{v_Rv_z})=\partial_z\overline{v_Rv_z}$. Multiplying by $R/\nu$ proves
$$
\boxed{\frac R\nu\partial_R(\nu\overline{v_R^2})+R\partial_z\overline{v_Rv_z}+\overline{v_R^2}-\overline{v_\phi^2}+R\Phi_R=0\quad(z=0).}
$$
All quadratic velocities here are population moments, not individual stellar velocities. In particular, symmetry does not make the <midplane tilt contribution to asymmetric drift> disappear: an odd mixed moment can have a nonzero derivative at zero.
For application to the <Milky Way>, assume negligible mean radial and vertical flows, and define the circular <speed> by $v_c^2=R\Phi_R(R,0)$. Write $\overline{v_R^2}=\sigma_R^2$ and $\overline{v_\phi^2}=\overline v_\phi^{\,2}+\sigma_\phi^2$. Rearranging gives the exact stress-support relation
$$
\boxed{v_c^2-\overline v_\phi^{\,2}=\sigma_R^2\left[-\frac{d\ln(\nu\sigma_R^2)}{d\ln R}-1+\frac{\sigma_\phi^2}{\sigma_R^2}\right]-R\partial_z\overline{v_Rv_z}\big|_0.}
$$
Thus the <stellar asymmetric drift> $v_a=v_c-\overline v_\phi$ is the right-hand side divided by $v_c+\overline v_\phi$. For $v_a\ll v_c$, this denominator is approximately $2v_c$. If $\nu\propto e^{-R/h_\nu}$ and $\sigma_R^2\propto e^{-R/h_\sigma}$ locally, the bracket becomes $R/h_\nu+R/h_\sigma-1+\sigma_\phi^2/\sigma_R^2$, making the density and dispersion-gradient contributions explicit.
A radially declining random-motion stress supplies some of the force needed to support a <galactic disk>, so its stars need less ordered azimuthal motion than a population on circular orbits. Hotter, often older, <stellar populations> therefore generally rotate more slowly than cold populations in the same <gravitational potential>; this is the observed <stellar asymmetric drift>. It is a stress-support effect, not a frictional slowing of each orbit. For a <velocity ellipsoid> aligned approximately with spherical coordinates, $\overline{v_Rv_z}\simeq(\sigma_R^2-\sigma_z^2)z/R$, and the midplane tilt term contributes $-(\sigma_R^2-\sigma_z^2)$ rather than zero.
Local measurements of mean rotation and all components of the <velocity dispersion>, combined with tracer-density and dispersion gradients, can therefore constrain $v_c$ and the radial <Newtonian gravitational field>. A hotter sample's mean rotation must not be identified directly with the circular <speed>. The tracer <number density> need not be the gravitating <mass density>. Nonaxisymmetric streaming, a nonstationary disk, poorly measured radial gradients, or a neglected tilt term can bias this inference; these are assumptions of the derivation rather than extra free corrections to the circular force.
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