= Solution
Write $\Sigma=\Sigma_0x^{-5/2}g$ because the prescribed <kinematic viscosity> is $\bar\nu=\bar\nu_0x^{5/2}$. Substituting $r=r_0x$ and $\partial_t=(3\bar\nu_0/4r_0^2)\partial_\tau$ into the <viscous evolution of an accretion disk> gives
$$
x^{-5/2}g_\tau=\frac4x\partial_x\left[x^{1/2}\partial_x(x^{1/2}g)\right].
$$
Hence the dimensionless <diffusion equation> is
$$
\boxed{g_\tau=4x^{5/2}g_{xx}+6x^{3/2}g_x.}
$$
For the useful inverse-radius coordinate $y=x^{-1/2}$, this becomes simply $g_\tau=y g_{yy}$. The positive factor in $\tau=1+3\bar\nu_0t/(4r_0^2)$ makes increasing $\tau$ correspond to increasing physical time.
There is a distinction between the named torque variable and the physical <viscous torque in an accretion disk>. For <Keplerian rotation>, the outward torque magnitude is
$$
\mathcal T=3\pi\bar\nu\Sigma\sqrt{GM_\star r}=3\pi\bar\nu_0\Sigma_0\sqrt{GM_\star r_0}\,x^{1/2}g.
$$
Here $M_\star$ denotes the central <mass>. The <zero-torque inner boundary condition> therefore requires $x^{1/2}g\to0$ as $x\to0$. Requiring $g\to0$ is stronger. The nontrivial finite-mass profile used below satisfies both; this distinction matters when counting integration constants.
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