Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 2

The viscous time isWith zero stellar torque, total angular momentum is conserved and $J\sim M(G{M}_{\star }R{)}^{1/2}$, so $M\propto R^{-1/2}$. HenceandAccreting material loses specific angular momentum; conservation requires a diminishing fraction of the disk to spread outward and carry that angular momentum.

If no mass is accreted, $M$ is constant and $\Sigma \sim M/R^2$. The viscous-time estimate becomes $t\propto R^{2-m+2n}$, while $J\sim M(G{M}_{\star }R{)}^{1/2}$. ThusThe central torque supplies the angular momentum that allows the fixed disk mass to spread.

Set $r=r_0{x}^2$, $\Sigma ={\Sigma }_0{x}^{-3}S$, and $\bar{\nu }={\bar{\nu }}_0{x}^{2m-3n}{S}^n$. Then ${\partial }_r=(2r_{0}x{)}^{-1}{\partial }_x$. Substitution into the viscous evolution of an accretion disk equation, with $\tau =3{\bar{\nu }}_{0}t/(4r_0^2)$, cancels the common dimensional factors and gives

For $m=4,n=2$, the equation is the one-dimensional porous medium equation $S_{\tau }=(S^3{)}_{xx}$. Use a similarity solution $S={\tau }^{-1/4}f(\xi )$ and $\xi =x{\tau }^{-1/4}$. ThenThe no-mass-flux condition at $x=0$ sets the integration constant to zero:With $f(0)=1$,ConsequentlyIts edge is $R=12r_0{\tau }^{1/2}\propto t^{1/2}$. Moreover,is time-independent. This agrees with part (b), since $2-m+2n=2$.