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Define the surface density of a disk, outward mass flux, internal torque, and surface magnetic torque byThe sign convention makes $\mathcal{G}$ positive for outward angular momentum transport in an ordinary Keplerian accretion disk. Vertical integration of mass conservation gives

The specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal equation by $r$, use the continuity equation to put its left-hand side in conservative form, and integrate over $z$. The assumed decay removes the vertical mass and viscous fluxes, whereas the magnetic surface stress remains:Subtracting $h$ times the integrated mass equation yieldsSince $F_M=-(dh/dr{)}^{-1}({\partial }_r\mathcal{G}-r\mathcal{T})$, substitution in mass conservation gives the required one-dimensional advection-diffusion equation

Write $y=\nu \Sigma$. A steady inward accretion rate has $F_M=-\dot{M}$, so the integrated angular-momentum equation isFor $\beta =0$, $\mathcal{T}=0$ and thereforeafter imposing the zero-torque inner boundary condition at $r_{\mathrm{in}}$. In a Keplerian accretion disk, $h\propto r^{1/2}$ and $\mathcal{G}=3\pi hy$, so

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.