= Solution
Assume an axisymmetric, geometrically <thin disc> in a fixed <Keplerian rotation> field, neglecting the disk's self-gravity. Write $v_R$ positive outwards and let
$$
\Omega(R)=\sqrt{\frac{GM}{R^3}},\qquad h(R)=R^2\Omega=\sqrt{GMR}.
$$
The ring supplies the local <specific angular momentum>. Its source per unit area is
$$
s(R,t)=\frac{\dot M(t)}{2\pi R_0}\delta(R-R_0),
$$
where the <Dirac delta function> is normalized so that $\int2\pi Rs\,dR=\dot M(t)$. Define the positive outward <viscous torque in an accretion disk> by
$$
\mathcal G=-2\pi R^3\nu\Sigma\frac{d\Omega}{dR}=3\pi\nu\Sigma h.
$$
The height-integrated <conservation of mass> and <conservation of angular momentum> equations are
$$
\partial_t\Sigma+\frac1R\partial_R(R\Sigma v_R)=s,
$$
$$
\partial_t(\Sigma h)+\frac1R\partial_R(R\Sigma v_Rh)
=-\frac1{2\pi R}\partial_R\mathcal G+s h_0.
$$
Subtract $h$ times the mass equation. Since $h$ is time-independent, this gives
$$
\Sigma v_R h'=-\frac1{2\pi R}\mathcal G'+s(h_0-h).
$$
The last term is zero as a distribution, because the source is supported where $h=h_0$. Since $h'=h/(2R)$, substituting the <torque> gives
$$
v_R=-\frac3{\Sigma R^{1/2}}\partial_R(\nu\Sigma R^{1/2}).
$$
Inserting this into the mass equation proves the <Keplerian disk evolution with a ring source>:
$$
\boxed{\partial_t\Sigma=\frac3R\partial_R\!\left[R^{1/2}\partial_R(\nu\Sigma R^{1/2})\right]
+\frac{\dot M(t)}{2\pi R_0}\delta(R-R_0).}
$$
The <viscosity> may be a prescribed function or a function of the local disk state; the conservation derivation is the same.
To track signs in the steady problem, define the inward <mass flux>
$$
F_M=-2\pi R\Sigma v_R=6\pi R^{1/2}\partial_R(\nu\Sigma R^{1/2}).
$$
Then $0=F_M'/(2\pi R)+s$. Away from the source $F_M$ is constant, with $F_M=\dot M_{\rm in}$ on the inner side and $F_M=-\dot M_{\rm out}$ on the outer side, where both loss rates are positive. Integrating through the source gives $F_M(R_0^+)-F_M(R_0^-)=-\dot M$. Therefore
$$
\boxed{\dot M=\dot M_{\rm in}+\dot M_{\rm out}.}
$$
The outward <angular momentum> flux is $F_J=\mathcal G-F_Mh$. The steady angular-momentum equation gives $F_J'=2\pi Rs h_0$. At the two <torque>-free boundaries its values are $F_J(R_{\rm in})=-\dot M_{\rm in}h_{\rm in}$ and $F_J(R_{\rm out})=\dot M_{\rm out}h_{\rm out}$. Integrating across the disk therefore yields
$$
\dot M h_0=\dot M_{\rm in}h_{\rm in}+\dot M_{\rm out}h_{\rm out}.
$$
Eliminate $\dot M_{\rm out}=\dot M-\dot M_{\rm in}$ and use $h\propto R^{1/2}$. The <two-boundary mass split of a ring-fed disk> is
$$
\boxed{\frac{\dot M_{\rm in}}{\dot M}
=\frac{h_{\rm out}-h_0}{h_{\rm out}-h_{\rm in}}
=\frac{\sqrt{R_{\rm out}}-\sqrt{R_0}}{\sqrt{R_{\rm out}}-\sqrt{R_{\rm in}}}.}
$$
The complementary outward fraction is $(\sqrt{R_0}-\sqrt{R_{\rm in}})/(\sqrt{R_{\rm out}}-\sqrt{R_{\rm in}})$. Both fractions lie between zero and one for a source strictly between the boundaries.
As a local check, integrating the steady <mass flux> from each zero-<torque> boundary gives
$$
\nu\Sigma\sqrt R=
\begin{cases}
\dfrac{\dot M_{\rm in}}{3\pi}(\sqrt R-\sqrt{R_{\rm in}}),&R_{\rm in}<R<R_0,\\
\dfrac{\dot M_{\rm out}}{3\pi}(\sqrt{R_{\rm out}}-\sqrt R),&R_0<R<R_{\rm out}.
\end{cases}
$$
Continuity of this quantity at $R_0$ gives the same split. The split depends on boundary and source angular momenta, rather than on the detailed <viscosity> law.
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