 Solution

Set $S=T=0$ and choose $\boxed{\bar{\nu }(r)={\nu }_{0}r}$, with ${\nu }_0>0$ constant. Define $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$. Then $\sqrt{r}\bar{\nu }\Sigma ={\nu }_{0}g$, and ${\partial }_r=(2y{)}^{-1}{\partial }_y$. Directly transforming the Keplerian viscous diffusion equation givesThe heat-equation transform for linear-radius disk viscosity is completed by $x=2y/\sqrt{3{\nu }_0}$:Equivalently one can use $g=\sqrt{r}\bar{\nu }\Sigma$, which differs only by a constant. This is the viscosity law making the transformed diffusivity constant under this square-root-radius reduction. Initial and boundary conditions transform too; the change of variables does not supply unspecified disc-edge data.

The physical diffusion arises from angular momentum transport. Differential rotation creates a viscous shear stress, which transfers angular momentum between neighboring annuli. Gas that loses angular momentum moves inward, while gas that gains it can move outward. Combining conservation of angular momentum with mass conservation turns the stress gradient into a radial mass flux and its divergence into a second radial derivative. Positive kinematic viscosity smooths the appropriately weighted profile; matter need not all move in one direction, and an accreting disc can simultaneously spread outward. Viscosity dissipates orbital energy but redistributes angular momentum internally except for boundary losses.