value of $\b$ for which $$\label{44}
d\b\cos(\b L)+\n \sin(\b L)=0.$$

By substituting the expression of $\g$ given by the third equation of into the first one, we get $$-D\a^2+c\a+\frac{\m}{1+\frac{\n\tan(\b L)}{d\b}}=0,$$ whose solutions are $$\label{eqdef}
\a_D^{\pm}(c,\b)=\frac{1}{2D}\left(c\pm \sqrt{c^2+\chi(\b)}\right),$$ where we have set $$\chi(\b)=\frac{4\m D}{1+\frac{\n\tan(\b L)}{d\b}}=\frac{4\m dD\b\cos(\b L)}{d\b\cos(\b L)+\n \sin(\b L)}.$$ It is easy to see that $\chi(\b)$ is continuous, even, decreases for $\b\in(0,\ov{\b})$ and satisfies $$\lim_{\b\da 0} \chi(\b)=\frac{4\m D}{1+\frac{\n L}{d}}, \qquad \chi\left(\frac{\pi}{2L}\right)=0, \qquad \lim_{\b\ua\ov{\b}} \chi(\b)=-\infty,$$ where $\ov{\b}$ is the one defined in . Hence, for every $c>0$, there exists a unique $\tilde{\b}(c)\in(\frac{\pi}{2L},\ov{\b})$, for which $$\label{45}
c^2=-\chi(\tilde{\b}(c)).$$ Moreover it satisfies $$\lim_{c\da 0} \tilde{\b}(c)=\frac{\pi}{2L}, \qquad \lim_{c\ua \infty} \tilde{\b}(c)=\ov{\b}.$$

As a consequence of these properties, for fixed $c>0$, $\a_D^{+}(c,\b)$ is a regular even function, which decreases in $(0,\tilde{\b}(c))$ and satisfies $$\begin{gathered}
\a_D^+(c,0)\!=\!\frac{1}{2D}\left(c\!+\!\sqrt{c^2+\frac{4\m D}{1+\frac{\n L}{d}}}\right), \quad \a_D^{+}\left(c,\frac{\pi}{2L}\right)\!=\!\frac{c}{D}, \quad \a_D^{+}(c,\tilde{\b}(c))\!=\!\frac{c}{2D}, \label{eqstar} \\ 
\p_\b\a_D^+(c,0)=0, \qquad \lim_{\b\ua\tilde{\b}(c)} \p_\b\a_D^+(c,\b)=-\infty. \notag\end{gathered}$$ In addition it is easy to verify that $$\label{46}
\p_c\a_D^+(c,\b)>0.$$ Since $\a_D^-$ is symmetric to $\a_D^+$ with respect to the line $\a=\frac{c}{2D}$, analogous properties can be established for $\a_D^-$. In particular observe that $$\a_D^{-}\left(c,\frac{\pi}{2L}\right)=0$$ and that, for fixed $\b\in(\frac{\pi}{2L},\ov{\b})$, we have $$\begin{aligned}
\lim_{c\ua\infty} \a_D^{-}(c,\b)=0,\end{aligned}$$ $\a_D^{-}$ decreasing monotonically in $c$. As a consequence, the bounded region of the first quadrant in 