and the refractory phase (-): $$\begin{aligned}
\label{eq:T_fire0}
  \int_{\pm 2}^x dx_1 \frac{1-x_1^2}{x_1+a}=(a^2-1) \ln\frac{a\pm 2}{a+x}-a(\pm 2-x)+2
-\frac{x^2}{2} = t.\end{aligned}$$ Integrating from $B$ to $C$ gives the firing time $$\begin{aligned}
\label{eq:T_fire}
T_f=\int_2^1 dx_1 \frac{1-x_1^2}{x_1+a}=(a^2-1) \ln\frac{a+2}{a+1}-a+\frac{3}{2}.\end{aligned}$$ For $\epsilon=0.01$, $a=1.05$ the analytical solution is in good agreement with the numerical solution in Fig. \[fig:single\_FHN\](b), including the firing time $T_f=0.482$ (analytical approximation: $0.491$).

For a rough estimate, in the following we shall approximate the spike by a rectangular pulse $$\label{eq:spikeactivator}
  x_1(t) \approx \left\{ \begin{array}{ll}
  2& \mbox{if }\,\, t<T_f\\
  -a& \mbox{if }\,\, t\ge T_f
\end{array} \right.$$

If the first subsystem is in the rest state, and a spike of the second subsystem arrives at $t=0$ (after the propagation delay $\tau$), we can approximate the initial dynamic response by linearizing $x_1, y_1$ around the fixed point $(x_1^*, y_1^*)$ and approximating the feedback by a constant impulse during the firing time $T_f$. The fast dynamic response along the $x_1$ direction is then given by $$\begin{aligned}
\label{eq:fast}
   \epsilon \delta\dot{x}_1 &=&  \xi \delta x_1+ 2C\end{aligned}$$ with $\xi<0$. This inhomogeneous linear differential equation can be solved with the initial condition $x_1(0)=-a$: $$\begin{aligned}
\label{eq:fast1}
   x_1(t) &=&  -a + \frac{2C}{|\xi|} (1-e^{-\frac{|\xi|}{\epsilon}t})\end{aligned}$$ Note that this equation is not valid for large $t$ since (i) the 