= Solution
Assume <axisymmetry> in a <thin disk> orbiting a fixed central <mass> $M$. To leading order in thickness, the <Keplerian rotation> is independent of height, $\Omega=(GM/r^3)^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial <pressure> support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface <torques>. The <Cauchy stress tensor> can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the <conservation of angular momentum> equation.
Define the <surface density> and its mass-weighted radial <velocity> by
$$
\Sigma=\int\rho\,dz,\qquad\Sigma\bar u_r=\int\rho u_r\,dz,
$$
and set $W_{r\phi}=\int T_{r\phi}\,dz$. Integrating <mass conservation> vertically gives
$$
\partial_t\Sigma+\frac1r\partial_r(r\Sigma\bar u_r)=0.
$$
The azimuthal equation in <cylindrical coordinates>, multiplied by $r$ and combined with <mass conservation>, is the conservative equation for <specific angular momentum> $l=r^2\Omega$. The cylindrical <stress tensor> <divergence> contributes $r^{-1}\partial_r(r^2T_{r\phi})+\partial_z(rT_{\phi z})$. With vanishing surface <torque>, vertical integration therefore gives
$$
\partial_t(\Sigma l)+\frac1r\partial_r(r\Sigma\bar u_r l)
=\frac1r\partial_r(r^2W_{r\phi}).
$$
Because $l(r)$ is fixed in time, subtracting $l$ times the mass equation yields
$$
\Sigma\bar u_r\frac{dl}{dr}=\frac1r\partial_r(r^2W_{r\phi}).
$$
Define the <effective viscosity defined by disk stress> by
$$
\boxed{\bar\nu\Sigma r\frac{d\Omega}{dr}=W_{r\phi},\qquad
\bar\nu=-\frac{2}{3\Sigma\Omega}\int T_{r\phi}\,dz\quad\text{for Keplerian rotation}.}
$$
This sign convention treats $T_{r\phi}$ as the <shear stress> appearing with $+\nabla\cdot\mathbf T$. For the <Newtonian fluid stress tensor> it reduces to the <density-weighted viscosity of a disk>, $\bar\nu=\Sigma^{-1}\int\rho\nu dz$, rather than an unweighted height average.
Using $l=\sqrt{GMr}$ and $W_{r\phi}=-(3/2)\bar\nu\Sigma\Omega$ in the <conservation of angular momentum> equation gives
$$
\bar u_r=-\frac3{\Sigma\sqrt r}\partial_r(\sqrt r\bar\nu\Sigma).
$$
Substitution into <mass conservation> proves the <Keplerian viscous diffusion equation>:
$$
\boxed{\partial_t\Sigma=\frac3r\partial_r\left[\sqrt r\,\partial_r(\sqrt r\bar\nu\Sigma)\right].}
$$
The outward <viscous torque in an accretion disk> is $\mathcal G=-2\pi r^2W_{r\phi}=3\pi\sqrt{GM}\sqrt r\bar\nu\Sigma$.
For $\bar\nu=Ar$ with $A>0$, make the <heat-equation transform for linear-radius disk viscosity>
$$
\boxed{x=2\sqrt{\frac r{3A}},\qquad g=\sqrt r\bar\nu\Sigma=Ar^{3/2}\Sigma.}
$$
Writing $y=\sqrt r$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The <torque> is $3\pi\sqrt{GM}g$, so the <zero-torque inner boundary condition> at the origin becomes $g(0,t)=0$ on the half-line $x>0$.
A positive <similarity solution> obeying this <boundary condition> is
$$
\boxed{g=Cxt^{-3/2}e^{-x^2/(4t)},\qquad
\Sigma=\frac{2C}{A\sqrt{3A}}\frac{t^{-3/2}}r e^{-r/(3At)},\qquad C>0.}
$$
For example, substitution of $g=t^{-1}F(x/\sqrt t)$ reduces the <heat equation> to $F''+(\eta/2)F'+F=0$, and $F=C\eta e^{-\eta^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.
To determine the global integrals, transform the <mass> and <angular momentum>:
$$
M_d=2\pi\int_0^\infty\Sigma r\,dr=2\pi\sqrt{\frac3A}\int_0^\infty g\,dx,
\qquad
J_d=2\pi\sqrt{GM}\int_0^\infty\Sigma r^{3/2}dr=3\pi\sqrt{GM}\int_0^\infty xg\,dx.
$$
The <Gaussian integrals> give
$$
\boxed{M_d=4\pi C\sqrt{3/A}\,t^{-1/2},\qquad
J_d=6\pi^{3/2}C\sqrt{GM}=\text{constant}.}
$$
Thus the <accreting similarity disk with viscosity proportional to radius> loses <mass> as $t^{-1/2}$ while conserving <angular momentum>; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-\dot M_d=2\pi C\sqrt{3/A}\,t^{-3/2}$. Its radial <velocity> $\bar u_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner <density> singularity is part of the idealized extension to $r=0$.
For the other profile, direct differentiation gives
$$
\frac{g_t}{g}=-\frac1{2t}+\frac{x^2}{4t^2}=\frac{g_{xx}}g,
\qquad g=t^{-1/2}e^{-x^2/(4t)}.
$$
It also satisfies the <heat equation>, but has $g(0,t)=t^{-1/2}\ne0$ and $g_x(0,t)=0$. It therefore has a nonzero inner <torque> and zero mass flow at the origin, not the previous <zero-torque inner boundary condition>. Here
$$
\Sigma=\frac{t^{-1/2}}{Ar^{3/2}}e^{-r/(3At)},\qquad
\bar u_r=\frac rt,
$$
$$
\boxed{M_d=2\pi\sqrt{3\pi/A}=\text{constant},\qquad
J_d=6\pi\sqrt{GM}\,t^{1/2},\qquad
\mathcal G(0,t)=\dot J_d=3\pi\sqrt{GM}\,t^{-1/2}.}
$$
An arbitrary positive amplitude multiplies these quantities. This <constant-mass disk with viscosity proportional to radius> spreads outward under a central <torque> that supplies <angular momentum>, with no ongoing central mass supply. It has the reflecting <Neumann boundary condition> for the <heat equation> and can be regarded as the spreading of an initially central mass distribution driven by an applied <torque>. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.
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