Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 321 / 1 / a

Write $\Sigma ={\Sigma }_0{x}^{-5/2}g$ because the prescribed kinematic viscosity is $\bar{\nu }={\bar{\nu }}_0{x}^{5/2}$. Substituting $r=r_{0}x$ and ${\partial }_t=(3{\bar{\nu }}_0/4r_0^2){\partial }_{\tau }$ into the viscous evolution of an accretion disk givesHence the dimensionless diffusion equation isFor the useful inverse-radius coordinate $y=x^{-1/2}$, this becomes simply $g_{\tau }=y{g}_{yy}$. The positive factor in $\tau =1+3{\bar{\nu }}_{0}t/(4r_0^2)$ 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 isHere $M_{\star }$ denotes the central mass. The zero-torque inner boundary condition therefore requires $x^{1/2}g\to 0$ as $x\to 0$. Requiring $g\to 0$ is stronger. The nontrivial finite-mass profile used below satisfies both; this distinction matters when counting integration constants.

Let $g=G(\xi )$ with $\xi ={\tau }^{-1}{x}^{-1/2}$. The derivatives of this similarity solution areThe first-derivative terms cancel in the dimensionless diffusion equation, leaving $G^{\prime \prime }+G^{\prime }=0$ for $\xi >0$. ThusAt the inner boundary $x\to 0$, $\xi \to \infty$. 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. ThereforeThe 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 ${\Sigma }_{\max }/{\Sigma }_0=C(5\tau {)}^5{e}^{-5}$. The peak moves inward and grows as time increases.

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.

For the finite-mass similarity solution above, the total disk mass isSubstitute $\xi =1/(\tau \sqrt{x})$, so $x={\tau }^{-2}{\xi }^{-2}$ and $x^{-3/2}|dx|=2\tau d\xi$. This givesThe requested linear increase of mass is therefore correct. It is supplied by the infinite-radius boundary, rather than by a source-free isolated finite disk. The inward mass accretion rate iswhich tends to zero at the origin and to $3\pi C{\bar{\nu }}_0{\Sigma }_0$ at infinity.

For Keplerian rotation, the specific angular momentum is $\ell =\sqrt{G{M}_{\star }r}$. The total disk angular momentum would beThis integral diverges at $\xi =0$, corresponding to infinite radius. There is no finite constant total angular momentum for the nonzero printed similarity solution. The seemingly time-independent last expression is an infinite quantity; it cannot prove a finite conservation law.

The problem remains visible under a cutoff. With $r\le r_{0}R$, write $J_R=J_{\ast }{E}_1(1/(\tau \sqrt{R}))$, where $J_{\ast }=4\pi C{\Sigma }_0\sqrt{G{M}_{\star }}{r}_0^{5/2}$ and $E_1$ is the exponential integral. ThenEven the finite change after subtracting a common logarithmic divergence is not zero. Direct integration of the disk diffusion equation gives the boundary identityfor this profile the outer expression has a nonzero limit, giving the same finite change rate. An inner zero torque alone does not impose zero outer angular-momentum flux.

More generally, the printed family $A+B{e}^{-\xi }$ can have finite angular momentum at both ends only if $A=0$ and $A+B=0$, hence it is trivial. Thus the request for a nonzero finite constant $J_d$ is incompatible with the stated viscosity exponent, similarity ansatz and infinite radial domain. A physical finite-domain or altered similarity problem would need different boundary data or assumptions; silently assigning a finite value to the divergent integral does not repair it.