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 $\xi=x\tau^{-1/4}$. Then
$$
-\frac14(f+\xi f')=(f^3)''.
$$
The no-mass-flux condition at $x=0$ sets the integration constant to zero:
$$
(f^3)'=-\frac14\xi f.
$$
With $f(0)=1$,
$$
\boxed{f(\xi)=\left(1-\frac{\xi^2}{12}\right)_+^{1/2}}.
$$
Consequently
$$
\boxed{\Sigma(r,t)=\Sigma_0\left(\frac r{r_0}\right)^{-3/2}\tau^{-1/4}
\left[1-\frac{r/r_0}{12\tau^{1/2}}\right]_+^{1/2}}.
$$
Its edge is $R=12r_0\tau^{1/2}\propto t^{1/2}$. Moreover,
$$
M=4\pi\Sigma_0r_0^2\int_0^\infty S\,dx
=4\pi\Sigma_0r_0^2\int_0^{\sqrt{12}}f(\xi)d\xi
$$
is time-independent. This agrees with part (b), since $2-m+2n=2$.

Solution

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

 Solution

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