In a <Keplerian accretion disk>, $u_\phi=R\Omega=(GM/R)^{1/2}$. Substitution into the steady azimuthal Navier--Stokes equation and cancellation of the common Keplerian factors gives
$$
\Sigma u_R\frac d{dR}(R^2\Omega)
=\frac1R\frac d{dR}\left(\nu\Sigma R^3\frac{d\Omega}{dR}\right).
$$
Since $R^2\Omega\propto R^{1/2}$ and $d\Omega/dR=-3\Omega/(2R)$, this reduces to the standard viscous drift formula
$$
\boxed{u_R=-\frac3{\Sigma R^{1/2}}
\frac d{dR}\left(\nu\Sigma R^{1/2}\right).}
$$
The minus sign describes inward drift when $\nu\Sigma R^{1/2}$ increases outward.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 347 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-347/2/b/solution