hold, Problem may exhibit more than one solution. Consider indeed $f(s)=s(-6s^3+9s^2-4s+1)$, which satisfies but not . With this choice, the function $M$ defined in , which is ${\mathcal}{C}^1[0,1)$, satisfies $$M'(\r)=\sqrt{\frac{15\, d}{2}}\int_0^1h(\r,\xi)\,d\xi$$ where $$\label{34}
h(\r,\xi):=\frac{216\r^2(1-\xi^5)-270\r(1-\xi^4)+80(1-\xi^3)}{\left(-72\r^3(1-\xi^5)+135\r^2(1-\xi^4)-80\r(1-\xi^3)+30(1-\xi^2)\right)^{3/2}}$$ and, as a consequence, $M'(0)>0$, since $h(0,\xi)>0$ for every $\xi\in(0,1)$. On the other hand, for $\r=1/2$ the numerator in reduces to $$h_1(\xi)=-54\xi^5+135\xi^4-80\xi^3-1,$$ which satisfies $h_1(0)=-1$, is decreasing for $\xi\in(0,2/3)$ and increasing for $\xi>2/3$. Since $h_1(1)=0$, this implies that $h(1/2,\xi)<0$ for all $\xi\in(0,1)$ and, therefore, $M'(1/2)<0$. Recalling that $M(\r)\to+\infty$ as $\r\ua 1$, the previous analysis entails that $L$ can be chosen in such a way that Problem possesses at least $3$ solutions.

The last result we need in order to prove Theorem \[thltb\] is the following

\[pr34\] Assume and let $(u,v)$ be the solution of starting with a nonnegative, not equal to $(0,0)$, bounded initial datum. Then, there exists a positive bounded solution $V_2$ of such that $$U_2\leq \liminf_{t\to+\infty}u(x,t), \qquad V_2\leq \liminf_{t\to+\infty}v(x,y,t)$$ locally uniformly in $(x,y)\in\overline{\O}$, where $$U_2=\frac{\nu}{\mu}V_2(L).$$

Consider $$\label{35}
(\un{u},\un{v}):=\left\{\begin{array}{ll}
\cos(\o x)\left(1,\frac{\mu\sin(\b y)}{d\b\cos(\b L)+\nu\sin(\b L)}\right) &\text{if $(x,y)\in\left(-\frac{\pi}{2\o},\frac{\pi}{2\o}\right)\times\left[0,L\right]$} \\
(0,0) &\text{otherwise}. 
\end{array}\right.$$ We now show that $\b$ and $\o$ can be chosen so that $(\un{u},\un{v})$ satisfy $$\label{36}
  \left\{ \begin{array}{ll}
  -D\un{u}_{xx}(x)\leq\nu \un{v}(x,L)-\m \un{u}(x) & \text{for } x\in \left(-\frac{\pi}{2\o},\frac{\pi}{2\o}\right) 