= Solution
The <travelling-wave reduction of a reaction-diffusion system> of <Fisher's equation> is
$$
U''+cU'+U(1-U)=0,\qquad(U,U')(-\infty)=(1,0),\quad(U,U')(\infty)=(0,0).
$$
Multiplying by $U'$ and integrating gives $c\int_{\mathbb R}(U')^2dz=1/6$, so the decreasing invasion front has $c>0$. The <linearization> at $U=0$ has spatial <eigenvalues>
$$
\lambda_\pm=\frac{-c\pm\sqrt{c^2-4}}2.
$$
For $0<c<2$ they are complex: approaching zero spirals in the $(U,U')$ plane and changes the sign of $U$, which is inadmissible for a density. Thus $c\geq2$ is necessary.
It is also sufficient. Write $V=U'$ and choose $\ell=(c-\sqrt{c^2-4})/2>0$. The triangle
$$
0\leq U\leq1,\qquad-\ell U\leq V\leq0
$$
is forward invariant for $U'=V$, $V'=-cV-U(1-U)$. On $V=0$, $V'\leq0$. On $V=-\ell U$,
$$
(V+\ell U)'=[c\ell-\ell^2-1]U+U^2=U^2\geq0.
$$
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 $c\geq2$, including the double-root endpoint:
$$
\boxed{c_{\min}=2.}
$$
This is the <Fisher-KPP minimum wave speed>, obtained with an explicit invariant-region existence argument.
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