Past exam of the topics in applied mathematics course of the University of Oxford / 2007 / A / 3 / ii

For $0<a<\eta$, $f(a)=a({\eta }^2-a^2)/(1+a^2)>0$. On the horizontal edge $b=0$,The vector field points downward into the proposed trapping region.

On $b=-\eta a$ with $a>0$,Thus $\boxed{db/da<-\eta }$. Since $a^{\prime }=b<0$, comparing slopes alone can reverse one's intuition about crossing direction. More directly, for $H=b+\eta a$ the inward test isso the sloping edge also points into $H\ge 0$.

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 $a^{\prime }=b\le 0$ 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 $a^{\prime }=b\le 0$, so $a$ decreases and has a limit. An interior limiting orbit with $b\ne 0$ would keep changing $a$, while an interior point with $b=0$, $0<a<\eta$ has $b^{\prime }<0$ 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 $a\ge 0$ and $a^{\prime }=b\le 0$.

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.