clear from the above discussion that, for $c\sim 0$, they are disjoint, since so are their domain of definition. On the contrary, for sufficiently large $c$ such an intersection exists, since $$\a_d^-(c,0)<\a_D^+(c,0), \quad \a_d^-\left(c,\sqrt{\frac{f'(0)}{d}}\right)=0, \quad \a_D^-\left(c,\frac{\pi}{2L}\right)=0.$$

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   ![The tangency between $\S_D(c)$ and $\S_d(c)$, for $D\leq D_{\operatorname{KPP}}$.[]{data-label="fig23"}](fig231.pdf "fig:")   ![The tangency between $\S_D(c)$ and $\S_d(c)$, for $D\leq D_{\operatorname{KPP}}$.[]{data-label="fig23"}](fig232.pdf "fig:")
   ![The tangency between $\S_D(c)$ and $\S_d(c)$, for $D\leq D_{\operatorname{KPP}}$.[]{data-label="fig23"}](fig233.pdf "fig:")   ![The tangency between $\S_D(c)$ and $\S_d(c)$, for $D\leq D_{\operatorname{KPP}}$.[]{data-label="fig23"}](fig234.pdf "fig:")
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Actually, the first value of $c$, denoted by $\un{c}=\un{c}(D)$, for which an intersection can exist is the one for which the domains of $\a_D^{\pm}$ and $\a_d^{\pm}$ intersect in one point, i.e. the one for which $$\left\{ \begin{array}{l}
  c^2=-\chi(\b) \\
  c^2=\eta(\b) \end{array}\right.$$ admits a solution. Indeed, as shown in Figure \[fig23\](A), the curves $\S_D(\un{c})$ and $\S_d(\un{c})$ are tangent if $D=d$, their common tangent being vertical. Therefore in this case $c^*=\un{c}$. If $D<d$, there exists $c^*$ satisfying $\un{c}<c^*<{c_{\operatorname{KPP}}}$ and such that $\a_D^-$ and $\a_d^+$ are tangent, as described in Figure \[fig23\](B). If $D>d$ the situation is more complex, because we have to take into account the change in the nature of $\a_d^-(c,\b)$ as $c$ crosses ${c_{\operatorname{KPP}}}$. From , it follows that $$\lim_{D\ua+\infty}\tilde{\b}(c_{\operatorname{KPP}},D)=\frac{\pi}{2L}$$ 