= Solution
Let $g=G(\xi)$ with $\xi=\tau^{-1}x^{-1/2}$. The derivatives of this <similarity solution> are
$$
g_\tau=-\frac\xi\tau G',\quad g_x=-\frac\xi{2x}G',\quad g_{xx}=\frac{\xi^2G''+3\xi G'}{4x^2}.
$$
The first-derivative terms cancel in the dimensionless <diffusion equation>, leaving $G''+G'=0$ for $\xi>0$. Thus
$$
G=A+B e^{-\xi}.
$$
At the inner boundary $x\to0$, $\xi\to\infty$. Interpreting the named variable as vanishing there sets $A=0$. Even if only the true <viscous torque in an accretion disk> $x^{1/2}g$ is required to vanish, a nonzero $A$ would give infinite total <mass> near $r=0$; finite <mass> again selects $A=0$. With nonnegative <surface density>, choose $B=C>0$; its value fixes the initial mass normalization. Therefore
$$
\boxed{g=C e^{-\xi},\qquad \frac\Sigma{\Sigma_0}=C x^{-5/2}\exp\left(-\frac1{\tau\sqrt x}\right).}
$$
The <surface density> vanishes exponentially at the origin and has an $x^{-5/2}$ tail at infinity. Its peak occurs at $x_{\max}=1/(25\tau^2)$, with $\Sigma_{\max}/\Sigma_0=C(5\tau)^5e^{-5}$. The peak moves inward and grows as time increases.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-321-surface-density.png]
{title=Surface-density profiles of the printed similarity solution at two times, showing an inward-moving growing peak and the outer power-law tail}
The plot uses $C=1$ and shows the actual <surface density>, with logarithmic radius to resolve both peaks. The later profile is larger at every fixed radius. Its total <mass> is finite; its total <angular momentum> is not, as the next calculation establishes.
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