\qquad \p_c\a_d^-(c,\b)<0.$$ Moreover, if $0<c<{c_{\operatorname{KPP}}}$, we have that $$\a_d^+(c,\hat{\b}(c))=\frac{c}{2d} \qquad \lim_{\b\da\hat{\b}(c)}\p_\b\a_d^{+}(c,\b)=+\infty,$$ while, for $c={c_{\operatorname{KPP}}}$, the hyperbolas degenerate into the straight lines with equations $$\label{ipdeg}
\a_d^{\pm}({c_{\operatorname{KPP}}},\b)=\pm\b+\frac{{c_{\operatorname{KPP}}}}{2d}.$$ Finally, for $c>{c_{\operatorname{KPP}}}$, we have that $$\a_d^{\pm}(c,0)=\frac{c\pm\sqrt{c^2-{c_{\operatorname{KPP}}}^2}}{2d}, \qquad \p_\b\a_d^{\pm}(c,0)=0$$ and, for fixed $\b$, $$\lim_{c\ua\infty} \a_d^{-}(c,\b)=0.$$ As a consequence of all the aforementioned properties, if we set $\hat{\b}(c)=0$ for $c\geq {c_{\operatorname{KPP}}}$, which is consistent with the previous notation, and define $$\S_d(c):=\left\{(\b,\a_d^-(\b)):\b\in\left[\hat{\b}(c),\sqrt{\frac{f'(0)}{d}}\right]\right\}\cup\left\{(\b,\a_d^+(\b)):\b\geq\hat{\b}(c)\right\},$$ we have that the region of the first quadrant in the $(\b,\a)-$plane bounded by $\S_d(c)$ and containing the point $\left(\sqrt{\frac{f'(0)}{d}},\frac{c}{2d}\right)$ invades monotonically, as $c\ua\infty$, the first quadrant in the $(\b,\a)-$plane. All these features have been represented in Figure \[fig22\].

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   ![The hyperbola defined by the second equation of , for $0<c<{c_{\operatorname{KPP}}}$ (A), $c={c_{\operatorname{KPP}}}$ (B) and $c>{c_{\operatorname{KPP}}}$ (C).[]{data-label="fig22"}](fig221bis.pdf "fig:")   ![The hyperbola defined by the second equation of , for $0<c<{c_{\operatorname{KPP}}}$ (A), $c={c_{\operatorname{KPP}}}$ (B) and $c>{c_{\operatorname{KPP}}}$ (C).[]{data-label="fig22"}](fig222bis.pdf "fig:")   ![The hyperbola defined by the second equation of , for $0<c<{c_{\operatorname{KPP}}}$ (A), $c={c_{\operatorname{KPP}}}$ (B) and $c>{c_{\operatorname{KPP}}}$ (C).[]{data-label="fig22"}](fig223bis.pdf "fig:")
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Another candidate to construct supersolutions to for $c>{c_{\operatorname{KPP}}}$ are the functions $$\label{412}
(u(x,t),v(x,y,t))=e^{\a(x+ct)}(1,\phi(y))$$ with $$\phi(y):=\g(e^{\b y}-e^{-\b y}),$$ where $\a,\b,\g$ are positive constants. By plugging into we are driven to $$\label{413}
  \left\{ \begin{array}{l}
  -D\a^2+c\a=\n\phi(L)-\m \\
  -d\a^2-d\b^2+c\a=f'(0) \\
  