= Solution
Set $S=T=0$ and choose \b[$\boxed{\bar\nu(r)=\nu_0r}$], with $\nu_0>0$ constant. Define $y=\sqrt r$ and $g=r^{3/2}\Sigma$. Then $\sqrt r\bar\nu\Sigma=\nu_0g$, and $\partial_r=(2y)^{-1}\partial_y$. Directly transforming the <Keplerian viscous diffusion equation> gives
$$
\Sigma_t=\frac{3\nu_0}{4r^{3/2}}g_{yy},\qquad g_t=\frac{3\nu_0}{4}g_{yy}.
$$
The <heat-equation transform for linear-radius disk viscosity> is completed by $x=2y/\sqrt{3\nu_0}$:
$$
\boxed{x=2\sqrt{\frac r{3\nu_0}},\qquad g=r^{3/2}\Sigma,\qquad g_t=g_{xx}.}
$$
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.
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