 Compact self-similar disk with square-root density viscosity

For $\nu ={\nu }_0(\Sigma /{\Sigma }_0{)}^{1/2}(R/R_0{)}^{9/4}$, the variables $x=\sqrt{R/R_0}$, $S=(\Sigma /{\Sigma }_0)x^3$ and $\tau =3{\nu }_{0}t/(4R_0^2)$ reduce the source-free Keplerian viscous diffusion equation to $S_{\tau }=(x{S}^{3/2}{)}_{xx}$. The displayed self-similar solution has an outer edge $R_e=36R_0\tau$ and a surface-density cusp proportional to $R^{-3/2}$ at the origin. The positive-part cutoff is essential: beyond the edge, $S^{3/2}$ is the cube of the absolute value, so continuing the squared polynomial without truncation does not solve the same equation.

The disk mass is $8\pi {\Sigma }_0{R}_0^2{\tau }^{-1/2}$ and its angular momentum is $12\pi {\Sigma }_0{R}_0^2\sqrt{GM{R}_0}$, independent of time. It therefore describes accretion with outward spreading and conserved disk angular momentum, under the formal zero-torque condition at the origin. The inward mass flux changes sign at $R=9R_0\tau /4$, so the outer portion moves outward while the inner portion accretes.