= Solution
Interpret the outer <Dirac delta function> as a boundary source carrying its full mass, or as the limit of a narrow source lying just inside $r_2$. Then $\int_u^{r_2}vS(v)dv=r_2S_0$ for $r_1<u<r_2$, and
$$
\boxed{\Sigma(r)=\frac{2r_2S_0}{3\bar\nu(r)}\left(1-\sqrt{\frac{r_1}{r}}\right)=\frac{\dot M}{3\pi\bar\nu(r)}\left(1-\sqrt{\frac{r_1}{r}}\right),\qquad \dot M=2\pi r_2S_0.}
$$
Thus the interior is the usual steady <Keplerian accretion disk> with a constant inward <accretion rate>. The product $\bar\nu\Sigma$ vanishes at the stress-free inner edge, and the <surface density of a disk> does too if the viscosity has a finite nonzero boundary value; far from that edge $\bar\nu\Sigma$ approaches $\dot M/(3\pi)$. Its density need not itself be constant because the viscosity may vary.
The <outer-edge source of a viscous disk> requires care about the boundary trace. The flux just inside the injecting edge is $\mathcal F(r_2^-)=\dot M$, while the no-flux value outside the source is $\mathcal F(r_2^+)=0$; the source produces the intervening jump. A simultaneously continuous interior condition $\mathcal F(r_2^-)=0$ would contradict nonzero steady injection. A symmetrically regularized delta integrated only up to its center can instead give half its full weight; that would mean a different supplied mass, not a different interior solution for the same $\dot M$. The physically normalized one-sided limiting source removes this convention ambiguity.
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