= Solution
Steady <mass conservation> with inward-positive <accretion rate> gives
$$
\dot m=-2\pi R\Sigma u_R=\text{constant}.
$$
Write the specific angular momentum as $l=R^2\Omega$. Multiplying the azimuthal equation by $2\pi R$ and using the mass equation shows that the sum of advected and viscous angular-momentum flux is constant:
$$
\dot m,l+2\pi\nu\Sigma R^3\frac{d\Omega}{dR}
=\dot m,l_{\rm in}.
$$
The right-hand side implements the <zero-torque inner boundary condition>. For a <Keplerian accretion disk>, $l\propto R^{1/2}$ and $R^3d\Omega/dR=-(3/2)l$. Therefore
$$
\boxed{
\nu\Sigma
=\frac{\dot m}{3\pi}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right]
},
$$
and
$$
\boxed{
u_R
=-\frac{3\nu}{2R}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right]^{-1}
}.
$$
Far outside the inner edge, $\Sigma\simeq\dot m/(3\pi\nu)$ and $u_R\simeq-3\nu/(2R)$.
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