Steady viscous disk supplied by a radial source ($\mathcal{F}(r)=2\pi {\int }_r^{r_2}{r}^{\prime }S(r^{\prime })d{r}^{\prime }$)

For a Keplerian disk with evolution ${\partial }_t\Sigma =(2\pi r{)}^{-1}{\partial }_r\mathcal{F}+S$ and $\mathcal{F}=6\pi r^{1/2}{\partial }_r(r^{1/2}\bar{\nu }\Sigma )$, take zero viscous torque at $r_1>0$ and zero flux outside the source region at $r_2>r_1$. The steady state gives ${\mathcal{F}}^{\prime }=-2\pi rS$, and its outer condition gives the displayed flux. Integrating once more from the zero-torque inner boundary condition yieldsThe equation determines the viscosity-density product, so a thermodynamic viscosity closure can be imposed afterward. For nonnegative integrable $S$, it is nonnegative and the inner accretion rate is exactly the total supplied mass per unit time.

A narrow surface density of a disk source approaching the outer edge from inside, with $S=S_0\delta (r-r_2)$ normalized to have full weight, gives $\dot{M}=2\pi r_2{S}_0$ and the displayed steady interior profile. This follows by replacing the inner source integral by $r_2{S}_0$ and integrating $u^{-1/2}$. The flux jumps from $\dot{M}$ just inside to zero outside the injecting edge. Thus a zero exterior flux is consistent, whereas a continuous zero interior boundary trace is not. A half-weight convention for a symmetric endpoint Dirac delta function represents a different total injected mass; it must not be silently mixed with the full-weight boundary source.