Cubic FitzHugh-Nagumo nullcline criterion ($k<(1-\alpha +{\alpha }^2)/3$)

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 $h^{\prime }=k-M+3(v-(1+\alpha )/3{)}^2$ 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.