Keplerian viscous diffusion equation (${\Sigma }_t=(3/r){\partial }_r[r^{1/2}{\partial }_r(r^{1/2}\bar{\nu }\Sigma )]$)

Use $l=\sqrt{G{M}_{\ast }r}$ and ${\Omega }^{\prime }=-(3/2)\Omega /r$ in the vertically averaged viscous disk equations. The radial drift is $v_r=-3(\Sigma \sqrt{r}{)}^{-1}{\partial }_r(\sqrt{r}\bar{\nu }\Sigma )$. Conservation of mass gives this nonlinear diffusion equation. It assumes the central rotation law remains fixed and excludes wind/surface torque terms.

For an axisymmetric thin disc supplied at $R_0$ with the local Keplerian rotation specific angular momentum, the displayed source augments the Keplerian viscous diffusion equation. The radial velocity remains $v_R=-3(\Sigma R^{1/2}{)}^{-1}{\partial }_R(\nu \Sigma R^{1/2})$: subtracting local specific angular momentum times conservation of mass from conservation of angular momentum cancels the source term. The normalization of the Dirac delta function makes the integrated supplied mass per unit time equal to $\dot{M}(t)$.

For a steady zero-mass-flux background and constant viscous transport response exponent, this change removes the first derivative in the radial diffusion operator. If ${\Sigma }_0\propto r^{-p}$, the kinematic viscosity scales as $r^{p-1/2}$, so the transformed diffusivity scales as $x^{2p-3}$. It is constant at $p=3/2$.

For source-free Keplerian viscous diffusion equation, the variables $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$ give $g_t=(3{\nu }_0/4)g_{yy}$ when $\bar{\nu }={\nu }_{0}r$. Rescaling y gives the displayed heat equation. Initial and radial boundary conditions must also be transformed; the reduction does not specify them.

With $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$, the half-line heat equation solution $g=C{t}^{-1/2}{e}^{-x^2/(4t)}$ has $g_x(0)=0$ but $g(0)>0$. It gives $\Sigma \propto r^{-3/2}{t}^{-1/2}{e}^{-r/(3At)}$, constant disk mass and angular momentum proportional to $t^{1/2}$. Its radial velocity is $r/t$, and a central torque proportional to $t^{-1/2}$ supplies the increasing angular momentum. It is an outward-spreading, torque-driven disk, not a zero-torque accreting solution.

// Target: astrophysical-fluid-dynamics.bigb

For $\bar{\nu }=Ar$ with $A>0$, use $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$. The Keplerian viscous diffusion equation becomes $g_t=g_{xx}$. The zero-torque inner boundary condition is $g(0,t)=0$, and the positive solution $g=Cx{t}^{-3/2}{e}^{-x^2/(4t)}$ gives $\Sigma \propto r^{-1}{t}^{-3/2}{e}^{-r/(3At)}$. Its mass is proportional to $t^{-1/2}$, angular momentum is constant and inward mass flow at the origin is proportional to $t^{-3/2}$. Its radial velocity is $r/t-3A/2$, exhibiting inner accretion and outer spreading.

// Target: astrophysics.bigb

Two integrations of the steady Keplerian viscous diffusion equation give this general transported quantity. $A$ fixes the mass accretion rate, while $B$ fixes the additive angular-momentum flux or inner torque. A zero-torque inner boundary condition gives $B=-A\sqrt{r_{\mathrm{in}}}$; a nonaccreting constant-torque disk has $A=0$.

For steady injection into a Keplerian disk between two zero-torque boundaries, conservation of mass gives $\dot{M}={\dot{M}}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}$. Conservation of angular momentum gives $\dot{M}{h}_0={\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$, where $h(R)=\sqrt{GMR}$. Eliminating the outward mass flux gives the displayed inward fraction. This fraction does not depend on the viscosity law: the two boundary fluxes must carry away the injected mass and angular momentum. Equivalently, continuity of the viscous torque in an accretion disk joins $\nu \Sigma \sqrt{R}={\dot{M}}_{\mathrm{in}}(\sqrt{R}-\sqrt{R_{\mathrm{in}}})/(3\pi )$ inside the source to ${\dot{M}}_{\mathrm{out}}(\sqrt{R_{\mathrm{out}}}-\sqrt{R})/(3\pi )$ outside.