Heat-equation transform for linear-radius disk viscosity ($\bar{\nu }={\nu }_{0}r,x=2\sqrt{r/(3{\nu }_0)},g=r^{3/2}\Sigma ,g_t=g_{xx}$)

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