General steady viscous flux in a Keplerian disk ($\bar{\nu }\Sigma =A+B{r}^{-1/2},\dot{M}=3\pi A$)

Two integrations of the steady Keplerian viscous diffusion equation give this general transported quantity. $A$ fixes the mass accretion rate, while $B$ fixes the additive angular-momentum flux or inner torque. A zero-torque inner boundary condition gives $B=-A\sqrt{r_{\mathrm{in}}}$; a nonaccreting constant-torque disk has $A=0$.

For steady injection into a Keplerian disk between two zero-torque boundaries, conservation of mass gives $\dot{M}={\dot{M}}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}$. Conservation of angular momentum gives $\dot{M}{h}_0={\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$, where $h(R)=\sqrt{GMR}$. Eliminating the outward mass flux gives the displayed inward fraction. This fraction does not depend on the viscosity law: the two boundary fluxes must carry away the injected mass and angular momentum. Equivalently, continuity of the viscous torque in an accretion disk joins $\nu \Sigma \sqrt{R}={\dot{M}}_{\mathrm{in}}(\sqrt{R}-\sqrt{R_{\mathrm{in}}})/(3\pi )$ inside the source to ${\dot{M}}_{\mathrm{out}}(\sqrt{R_{\mathrm{out}}}-\sqrt{R})/(3\pi )$ outside.