= Solution
For nonnegative insect density assume $N_0>0$. Substitution of $N=e^{\alpha t}R$ gives $R_t=D e^{2\alpha t}(R^2R_x)_x$, so choose
$$
\boxed{\tau(t)=\begin{cases}\dfrac{D}{2\alpha}(e^{2\alpha t}-1)&\alpha\ne0,\\Dt&\alpha=0.\end{cases}}
$$
Then $\tau'=De^{2\alpha t}>0$ and the <porous medium equation> is $R_\tau=(R^2R_x)_x$. Its total mass is conserved when the flux vanishes at infinity. For the mass-preserving <similarity solution>, direct differentiation gives
$$
R_\tau=-\frac14\tau^{-5/4}(F+\xi F'),\qquad
(R^2R_x)_x=\tau^{-5/4}(F^2F')',
\qquad \int_{\mathbb R}F\,d\xi=N_0.
$$
On the positive support, integrating once with zero boundary flux gives $F^2F'=-\xi F/4$ 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 is
$$
\boxed{x_{\max}=\sqrt{\frac{4N_0}{\pi}}\left(\frac{D}{2|\alpha|}\right)^{1/4}}.
$$
It 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.
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