= Solution
The top edge points down by (a), the sloping edge points upward in the inward normal sense by (b), and on $a=\eta$ the velocity $a'=b\leq0$ points left. At $a=0$ the region contains only the equilibrium $(0,0)$. Thus the closed triangle is forward invariant.
The given unstable <saddle equilibrium> branch enters its interior from $(\eta,0)$ and stays in this compact region. Along it $a'=b\leq0$, so $a$ decreases and has a limit. An interior limiting orbit with $b\ne0$ would keep changing $a$, while an interior point with $b=0$, $0<a<\eta$ has $b'<0$ and is not an equilibrium. Hence the only possible forward limit is $(0,0)$. Ordinary-differential-equation uniqueness prevents reaching and leaving an equilibrium in finite profile coordinate. The whole orbit therefore joins $(\eta,0)$ at $y=-\infty$ to $(0,0)$ at $y=+\infty$ with $a\geq0$ and $a'=b\leq0$.
Consequently $\boxed{\text{a nonnegative monotone front exists for every }c>2\eta}$. This proves the requested strict-speed result without claiming an unproved conclusion at the repeated-root threshold. It is a <trapping triangle for a saturating monostable invasion front>.
Back to article page
