Two-boundary mass split of a ring-fed disk (${\dot{M}}_{\mathrm{in}}/\dot{M}=(\sqrt{R_{\mathrm{out}}}-\sqrt{R_0})/(\sqrt{R_{\mathrm{out}}}-\sqrt{R_{\mathrm{in}}})$)

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.