 Solution

The travelling-wave reduction of a reaction-diffusion system of Fisher's equation isMultiplying by $U^{\prime }$ and integrating gives $c{\int }_{\mathbb{R}}(U^{\prime }{)}^{2}dz=1/6$, so the decreasing invasion front has $c>0$. The linearization at $U=0$ has spatial eigenvaluesFor $0<c<2$ they are complex: approaching zero spirals in the $(U,U^{\prime })$ plane and changes the sign of $U$, which is inadmissible for a density. Thus $c\ge 2$ is necessary.

It is also sufficient. Write $V=U^{\prime }$ and choose $\ell =(c-\sqrt{c^2-4})/2>0$. The triangleis forward invariant for $U^{\prime }=V$, $V^{\prime }=-cV-U(1-U)$. On $V=0$, $V^{\prime }\le 0$. On $V=-\ell U$,The boundary $U=1$ points left, and the boundary at $U=0$ consists only of the origin. The unstable branch from the saddle $(1,0)$ enters this triangle, where $U$ decreases and remains nonnegative. Its limiting state must have $V=0$ and $U(1-U)=0$; since it has left $U=1$, it tends to $(0,0)$. Therefore a monotone positive front exists for every $c\ge 2$, including the double-root endpoint:This is the Fisher-KPP minimum wave speed, obtained with an explicit invariant-region existence argument.