\g=\frac{\m}{d\b(e^{\b L}+e^{-\b L})+\n (e^{\b L}-e^{-\b L})}>0. \end{array}\right.$$ Using the expression of $\phi(L)$ given from the third equation of , the first one reduces to $$-D\a^2+c\a+\frac{\m d\b(e^{\b L}+e^{-\b L})}{d\b(e^{\b L}+e^{-\b L})+\n (e^{\b L}-e^{-\b L})}=0,$$ whose solutions are $$\label{tipo2}
\tilde{\a}_D^{\pm}(c,\b)=\frac{1}{2D}\left(c\pm \sqrt{c^2+\tilde{\chi}(\b)}\right),$$ where we have set $$\tilde{\chi}(\b)=\frac{4\m D}{1+\frac{\n\tanh(\b L)}{d\b}}.$$ This function is positive and therefore $\tilde{\a}_D^{+}(c,\b)$ is defined for every $\b\in{{\mathbb{R}}}$. It can be easily seen that it is even and satisfies the following monotonicity conditions $$\label{417}
\p_{\b}\tilde{\a}_D^+(c,\b)>0, \quad \p_{c}\tilde{\a}_D^+(c,\b)>0, \quad \p_{D}\tilde{\a}_D^+(c,\b)<0$$ for every positive $c,\b$. Moreover we have $\tilde{\a}_D^+(c,0)=\a_D^+(c,0)$, for every $c>0$. Naturally, similar properties hold for $\tilde{\a}_D^-$, taking into account that it is symmetric to $\tilde{\a}_D^+$ with respect to $\a=\frac{c}{2D}$. Finally, observe that, for every $\b\geq 0$, $$\lim_{c\ua+\infty}\tilde{\a}_D^+(c,\b)=+\infty$$ and, therefore, the regions of the first quadrant in the $(\b,\a)-$plane delimited by $$\label{418}
\tilde{\S}_D(c):=\left\{(\b,\tilde{a}_D^-(c,\b)):\b>0\right\}\cup\left\{(\b,\tilde{a}_D^+(c,\b)):\b>0\right\}.$$ and containing $\left(\frac{c}{2D},1\right)$ invade monotonically the first quadrant of the plane $(\b,\a)$.

![The curves $\tilde{\a}_D^+$ and $\tilde{\a}_d^{\pm}$.[]{data-label="fig24"}](fig24.pdf)

On the other hand, the second equation of describes, for $c>{c_{\operatorname{KPP}}}$, as we are assuming, a circle in the $(\b,\a)-$plane, with center at $\left(0,\frac{c}{2d}\right)$ and radius $$r(c)=\frac{\sqrt{c^2-{c_{\operatorname{KPP}}}^2}}{2d}$$ (see Figure \[fig24\]). Precisely, the part of the graph of the circle which lies in the first quadrant is given by $$\tilde{\S}_d(c):=\left\{(\b,\tilde{\a}_d^{\pm}(c,\b)):\b\in[0,r(c)]\right\},$$ where $$\label{419}
