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

The steady disk should end at the innermost stable circular orbit, $r_{\mathrm{in}}=3r_S$. 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

Take $\dot{M}>0$ for inward accretion, so the outward radial mass flux is $F=-\dot{M}$. Since $F=-(dh/dr{)}^{-1}d\mathcal{G}/dr$ and $\mathcal{G}(r_{\mathrm{in}})=0$,Using $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^{3}d\Omega /dr$ and $x=r/r_S$ givesThus $\boxed{A=3\sqrt{3}/2}$. Far from the hole this approaches the Keplerian accretion disk result $\bar{\nu }\Sigma =\dot{M}[1-(r_{\mathrm{in}}/r{)}^{1/2}]/(3\pi )$, although the pseudo-Newtonian boundary factor retains a different finite-radius shape.