Write $y=\nu\Sigma$. A steady inward <accretion rate> has $F_M=-\dot M$, so the integrated angular-momentum equation is
$$
\partial_r\mathcal G-r\mathcal T=\dot M\frac{dh}{dr}.
$$
For $\beta=0$, $\mathcal T=0$ and therefore
$$
\mathcal G=\dot M(h-h_{\rm in})
$$
after imposing the <zero-torque inner boundary condition> at $r_{\rm in}$. In a <Keplerian accretion disk>, $h\propto r^{1/2}$ and $\mathcal G=3\pi hy$, so
$$
\boxed{\nu\Sigma
=\frac{\dot M}{3\pi}
\left[1-\left(\frac{r_{\rm in}}r\right)^{1/2}\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / c / i / Solution

The steady disk should end at the <innermost stable circular orbit>, $r_{\rm in}=3r_S$. Inside it, matter plunges inward on an orbital timescale and cannot maintain the nearly circular shear flow assumed by the viscous-disk equations. If the plunging region communicates negligible stress back across the ISCO, the natural boundary condition is the <zero-torque inner boundary condition>
$$
\boxed{\mathcal G(r_{\rm in})=0}.
$$

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1 / b / i / Solution

The zero-torque inner boundary condition sets the viscous torque to zero where nearly circular disk flow joins a plunging or stress-free region.


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