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Balance the time derivative, cubic dispersion and quadratic transport using the cylindrical KdV equation similarity ansatz $q(x,t)=t^{-2/3}F(\xi )$, $\xi =x{t}^{-1/3}$ on $t>0$. For $t<0$ the same real cube-root convention can be used separately. Substitution and removal of the common factor $t^{-5/3}$ giveEvery term is a total derivative: $F{F}^{\prime }=(F^2/2{)}^{\prime }$ and $\xi F^{\prime }+F=(\xi F{)}^{\prime }$. Integration therefore gives the requested second-order equationwhere $C$ is a constant determined by any additional boundary data. No condition in the question forces $C=0$.

This part uses source-free viscous evolution of an accretion disk. Put $R=R_0{x}^2$ and $\Sigma ={\Sigma }_0{x}^{-3}S$. The prescribed kinematic viscosity then becomesso $\nu \Sigma R^{1/2}={\nu }_0{\Sigma }_0{R}_0^{1/2}x{S}^{3/2}$. Since ${\partial }_R=(2R_{0}x{)}^{-1}{\partial }_x$, the Keplerian viscous diffusion equation givesCanceling the prefactor and setting $\tau =3{\nu }_{0}t/(4R_0^2)$ proves

Use the similarity ansatz $S={\tau }^{-1}f(\xi )$ with $\xi =x/\sqrt{\tau }$. Derivatives at fixed $x$ giveOn the branch $1-k\xi \ge 0$, substitute $f=(1-k\xi {)}^2$. The two expressions reduce toEquality of the constant coefficients requires $\boxed{k=1/6}$, and then the other coefficients agree as well.

The physical compact self-similar disk with square-root density viscosity uses this branch up to its outer edge and vacuum beyond it:Here $(u{)}_{+}=\max (u,0)$ is the positive part. The cutoff matters: beyond $\xi =6$, the untruncated squared expression would have $f^{3/2}=|1-\xi /6{|}^3$, and its right-hand side would no longer be the polynomial used above. At the moving edge, $S$ vanishes quadratically and $x{S}^{3/2}$ vanishes cubically. The latter and its first derivative match the vacuum values continuously, so joining to zero creates no distributional mass-flux source. The joined profile satisfies the equation, including in the weak sense at the edge.

Returning to the surface density gives the explicit resultwith $\Sigma =0$ for $R\ge 36R_0\tau$. Thus the outer edge is $\boxed{R_e=36R_0\tau }$ and moves linearly in physical time, with speed $27{\nu }_0/R_0$. The density decreases monotonically with radius, has an integrable $R^{-3/2}$ central cusp, and drops quadratically to zero at the outer edge. The formal origin boundary has zero viscous torque in an accretion disk, because $\nu \Sigma \sqrt{R}\to 0$ there, even though the surface density diverges.

To characterize the spreading, let $a=6\sqrt{\tau }$ and integrate in $x$. The disk mass and angular momentum areThe mass decreases while the angular momentum remains constant: matter accretes at the origin and the remaining disk spreads outward. More explicitly, with $u=x/(6\sqrt{\tau })$, the inward mass flux isIt is inward for $u<1/4$ and outward for $u>1/4$. The dividing radius is $9R_0\tau /4$, and the origin flux $3\pi {\nu }_0{\Sigma }_0{\tau }^{-3/2}$ equals $-d{M}_d/dt$. The outer edge carries no flux beyond itself. This family is a self-similar accreting disk with finite mass at each positive time, rather than a regular finite-mass initial profile at $\tau =0$.