Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 6B (Mathematical Biology)

For nonnegative insect density assume $N_0>0$. Substitution of $N=e^{\alpha t}R$ gives $R_t=D{e}^{2\alpha t}(R^2{R}_x{)}_x$, so chooseThen ${\tau }^{\prime }=D{e}^{2\alpha t}>0$ and the porous medium equation is $R_{\tau }=(R^2{R}_x{)}_x$. Its total mass is conserved when the flux vanishes at infinity. For the mass-preserving similarity solution, direct differentiation givesOn the positive support, integrating once with zero boundary flux gives $F^2{F}^{\prime }=-\xi F/4$ and hence $F^2=({\xi }_0^2-{\xi }^2)/4$. The semicircle mass integral is $\pi {\xi }_0^2/4=N_0$, confirming the profile and its symmetric compact support.

The front is at $|x|={\xi }_0{\tau }^{1/4}$. When $\alpha <0$, $\tau (t)$ increases towards $D/(2|\alpha |)$, so the supremum of the distance reached isIt is approached as $t\to \infty$, rather than attained at a finite time. If $N_0=0$, the nonnegative solution is identically zero and the corresponding distance is zero.