= Cubic FitzHugh-Nagumo nullcline criterion
{c}
{title2=$k<(1-\alpha+\alpha^2)/3$}
For $\dot v=v(\alpha-v)(v-1)-w+I$, $\dot w=\beta v-\gamma w$, set $k=\beta/\gamma$. Intersections of the <nullclines> solve $I=h(v)=v^3-(1+\alpha)v^2+(\alpha+k)v$. Three simple <equilibria> are possible precisely when $k<M=(1-\alpha+\alpha^2)/3$ and $h(u_+)<I<h(u_-)$, where $u_\pm=(1+\alpha\pm\sqrt{1-\alpha+\alpha^2-3k})/3$. To prove this, observe that $h'=k-M+3(v-(1+\alpha)/3)^2$ has two real zeros exactly when $k<M$; monotonicity on the three resulting intervals counts the intersections. At either endpoint there are a double and a simple root, and at $k=M$, $I=h((1+\alpha)/3)$ there is one triple root.
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