Set $r=r_0x^2$, $\Sigma=\Sigma_0x^{-3}S$, and $\bar\nu=\bar\nu_0x^{2m-3n}S^n$. Then $\partial_r=(2r_0x)^{-1}\partial_x$. Substitution into the <viscous evolution of an accretion disk> equation, with $\tau=3\bar\nu_0t/(4r_0^2)$, cancels the common dimensional factors and gives
$$
\boxed{\partial_\tau S=\partial_x^2\left(x^{2m-3n-2}S^{n+1}\right)}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-321/2/c/i/solution