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Assume an axisymmetric, geometrically thin disc in a fixed Keplerian rotation field, neglecting the disk's self-gravity. Write $v_R$ positive outwards and letThe ring supplies the local specific angular momentum. Its source per unit area iswhere the Dirac delta function is normalized so that $\int 2\pi RsdR=\dot{M}(t)$. Define the positive outward viscous torque in an accretion disk byThe height-integrated conservation of mass and conservation of angular momentum equations areSubtract $h$ times the mass equation. Since $h$ is time-independent, this givesThe last term is zero as a distribution, because the source is supported where $h=h_0$. Since $h^{\prime }=h/(2R)$, substituting the torque givesInserting this into the mass equation proves the Keplerian disk evolution with a ring source: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 fluxThen $0=F_M^{\prime }/(2\pi R)+s$. Away from the source $F_M$ is constant, with $F_M={\dot{M}}_{\mathrm{in}}$ on the inner side and $F_M=-{\dot{M}}_{\mathrm{out}}$ 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

The outward angular momentum flux is $F_J=\mathcal{G}-F_{M}h$. The steady angular-momentum equation gives $F_J^{\prime }=2\pi Rs{h}_0$. At the two torque-free boundaries its values are $F_J(R_{\mathrm{in}})=-{\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}$ and $F_J(R_{\mathrm{out}})={\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$. Integrating across the disk therefore yieldsEliminate ${\dot{M}}_{\mathrm{out}}=\dot{M}-{\dot{M}}_{\mathrm{in}}$ and use $h\propto R^{1/2}$. The two-boundary mass split of a ring-fed disk isThe complementary outward fraction is $(\sqrt{R_0}-\sqrt{R_{\mathrm{in}}})/(\sqrt{R_{\mathrm{out}}}-\sqrt{R_{\mathrm{in}}})$. 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 givesContinuity 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.