= Solution
Take $0<r_1<r_2$ and positive mean <kinematic viscosity> $\bar\nu$. The <steady state> of the supplied <viscous evolution of an accretion disk> equation satisfies $\mathcal F'=-2\pi rS(r)$. The outer no-flux condition therefore gives
$$
\mathcal F(r)=2\pi\int_r^{r_2}vS(v)\,dv.
$$
Positive $S$ produces positive inward <accretion rate> in this sign convention. Set $Y=r^{1/2}\bar\nu\Sigma$. In a <Keplerian disk>, the <viscous torque in an accretion disk> is proportional to $Y$, so the <zero-torque inner boundary condition> gives $Y(r_1)=0$. Since $Y'=\mathcal F/(6\pi r^{1/2})$, a second integration yields
$$
\boxed{\Sigma(r)=\frac{1}{3\bar\nu(r)r^{1/2}}\int_{r_1}^{r}u^{-1/2}\left[\int_u^{r_2}vS(v)\,dv\right]du.}
$$
This is the <steady viscous disk supplied by a radial source>. It does not require $\bar\nu$ to be independent of radius: the equation directly determines $\bar\nu\Sigma$. If viscosity depends on the local thermodynamic state, the displayed formula is an implicit relation completed by a separate closure. For nonnegative integrable $S$ it gives nonnegative <surface density of a disk>, and the inner <accretion rate> equals the total supplied mass per unit time, $2\pi\int_{r_1}^{r_2}rS(r)dr$. Differentiating both nested integrals recovers the flux and the differential equation.
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