\\
  -d\D \un{v}(x,y)\leq(f'(0)-\d)\un{v} & \text{for } (x,y)\in\left(-\frac{\pi}{2\o},\frac{\pi}{2\o}\right)\times\left(0,L\right)  \\ 
  d\un{v}_y(x,L)=\m \un{u}(x)-\nu \un{v}(x,L) & \text{for } x\in \left(-\frac{\pi}{2\o},\frac{\pi}{2\o}\right) \\
  \un{v}(x,0)=0 & \text{for } x\in \left(-\frac{\pi}{2\o},\frac{\pi}{2\o}\right), \end{array}\right.$$ for $0<\d<f'(0)$ and therefore, by the second relation of , there exists $\e_0$ such that, for all $0<\e<\e_0$, $\e(\un{u},\un{v})$ is a strict generalized subsolution of to which Proposition \[prcp2\] can be applied. Observe that, with choice , the last two equations of are satisfied and the first two inequalities reduce to $$D\o^2\leq \frac{-\mu d\b\cos(\b L)}{d\b\cos(\b L)+\nu\sin(\b L)}, \qquad
d\o^2+d\b^2\leq f'(0)-\d.$$ Now, thanks to , it is possible to fix $\d$ in such a way that $$d\left(\frac{\pi}{2L}\right)^2<f'(0)-\d$$ and take $\frac{\pi}{2L}<\b<\frac{\pi}{L}$ in a neighborhood of $\frac{\pi}{2L}$ (we denote $\b\sim\frac{\pi}{2L}$), so that $$m:=\min\left\{\frac{f'(0)-\d}{d}-\b^2,\frac{-\mu d\b\cos(\b L)}{D\left(d\b\cos(\b L)+\nu\sin(\b L)\right)}\right\}>0.$$ As a consequence, if $\o^2\leq m$, $(\un{u},\un{v})$ satisfies . Moreover, reducing $\e$ if necessary, we can assume that $\e(\un{u},\un{v})<(u(x,1),v(x,y,1))$, because, thanks to Proposition \[prcp1\] and the Hopf lemma, we have that $$u(x,1)>0, \quad v_y(x,0,1)>0 \quad \text{ and }  \quad v(x,y,1)>0 \quad \text{ for all } (x,y)\in\O,$$ and, in addition, $v(x,L,1)>0$ for every $x\in{{\mathbb{R}}}$, since if there was $x_0$ such that $v(x_0,L,1)=0$, then from the third equation in we would have $$dv_y(x_0,L,1)=\mu u(x_0,1)>0,$$ which is impossible, since, again by Proposition \[prcp1\], $v(x,y,1)\geq 0$ 