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Steady mass conservation with inward-positive accretion rate givesWrite the specific angular momentum as $l=R^2\Omega$. Multiplying the azimuthal equation by $2\pi R$ and using the mass equation shows that the sum of advected and viscous angular-momentum flux is constant:The right-hand side implements the zero-torque inner boundary condition. For a Keplerian accretion disk, $l\propto R^{1/2}$ and $R^{3}d\Omega /dR=-(3/2)l$. ThereforeandFar outside the inner edge, $\Sigma \simeq \dot{m}/(3\pi \nu )$ and $u_R\simeq -3\nu /(2R)$.

For inward accretion rate $\dot{M}>0$, the steady conservation of angular momentum equation is $\mathcal{G}=\dot{M}(h-h_{\mathrm{in}})$ when the inner torque vanishes. The viscous torque in an accretion disk is $\mathcal{G}=2\pi qh\bar{\nu }\Sigma$, where $h=r^2\Omega$ and $q$ is the orbital shear parameter. ThusThe familiar Keplerian accretion disk follows when $q=3/2$ and $h\propto r^{1/2}$. The general form also applies to the non-Keplerian Paczyński-Wiita circular orbit rotation law.

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.

For a steady Keplerian accretion disk with inward accretion rate $\dot{M}$ and vanishing inner torque, the outward angular-momentum torque is $\mathcal{G}=\dot{M}(\sqrt{GMr}-\sqrt{GM{r}_{\mathrm{in}}})$. It transports energy outward at rate $\Omega \mathcal{G}$. The binding-energy acquisition rate between radii $b$ and $a$ is $B_{a,b}=GM\dot{M}(1/a-1/b)/2$. Integrating the vertical viscous dissipation and then the annulus givesSince $\Omega \mathcal{G}=GM\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$, subtraction proves the displayed energy balance. Far from the inner edge $D_{a,b}\simeq 3B_{a,b}$, whereas the boundary torque powers vanish for the complete disk and $D=B=GM\dot{M}/(2r_{\mathrm{in}})$. Local dissipation is therefore not supplied solely by local changes in orbital binding energy.

A thin disk with alpha disk viscosity $\nu \sim \alpha H^2\Omega$ has a viscous timescale much longer than its dynamical infall time in the plunging region of a black-hole accretion disk. At fixed nonzero accretion rate, rapid inflow reduces its surface density of a disk. The resulting viscous torque in an accretion disk is smaller by the displayed factor than in the slowly drifting circular disk. This estimate concerns radii outside the narrow matching layers.

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.