## 2 Synchronization of Two Coupled FHN Neurons without Time Delay  
A single FHN neuron [38] is modeled in this paper by the following equation:  
\[ \begin{array}{l}\dot{u}_{1}=-u_{1}(u_{1}-1)(u_{1}-a)-u_{2},\\ \dot{u}_{2}=b(u_{1}-\gamma u_{2}).\end{array}  \tag{1}\]  
The FHN system describes the generation of the nerve impulse in a squid as well as its propagation along the giant axon. Here \(u_{1}\) represents the potential difference at time \(t\) across the membrane of the axon and \(u_{2}\) is a recovery current caused by all ion flows. According to experiments [1], for the space clamp segment of the axon it is required that \(0<a<0.5\), \(b\), \(\gamma>0\) in Equation (1). Furthermore, from biophysical considerations, it is rational to let \(1/\gamma-(1-a+a^{2})/3>0\), \((1-a+a^{2})/3-b\gamma>0\). We consider the coupled FHN neural system as:  
\[ \begin{array}{l}\dot{u}_{1}=-u_{1}(u_{1}-1)(u_{1}-a)-u_{2}+c(u_{1}-u_{3}), \\ \dot{u}_{2}=b(u_{1}-\gamma u_{2}),\\ \dot{u}_{3}=-u_{3}(u_{3}-1)(u_{3}-a)-u_{4}+c(u_{3}-u_{1}),\\ \dot{u}_{4}=b(u_{3}-\gamma u_{4}),\end{array}  \tag{2}\]  
where \(c\) is the coupling strength. It is said that the two FHN neurons are synchronized if:  
\[\lim_{t\rightarrow\infty}(u_{1}-u_{3})=0\ \,\ \ \ \lim_{t\rightarrow\infty}(u_{2}-u_{4})=0\ \.\]  
For the first and third equations in Equation (2), taking derivative with respect to time \(t\) on both sides yields:  
\[ \begin{array}{l}\ddot{u}_{1}=[-3u_{1}^{2}+2(a+1)u_{1}-a+c]\dot{u}_{1}-\dot{ u}_{2}-\dot{c}\dot{u}_{3},\\ \ddot{u}_{3}=[-3u_{3}^{2}+2(a+1)u_{3}-a+c]\dot{u}_{3}-\dot{u}_{4}-\dot{c}\dot{ u}_{1}.\end{array} \]  
Substituting the second and forth equations in Equation (2) into the above two equations, respectively, we are able to eliminate the variables \(u_{2}\) and \(u_{4}\) in Equation (2), which become:  
\[ \begin{array}{l}\ddot{u}_{1}+(3u_{1}^{2}+\alpha_{1}u_{1}+\alpha_{2})\dot{u} _{1}+\alpha_{3}u_{1}^{3}+\alpha_{4}u_{1}^{2}+\alpha_{5}u_{1}+\alpha_{6}u_{3}+ \dot{c}\dot{u}_{3}=0,\\ \ddot{u}_{3}+(3u_{3}^{2}+\alpha_{1}u_{1}+\alpha_{2})\dot{u}_{3}+\alpha_{3}u_{3} ^{3}+\alpha_{4}u_{3}^{2}+\alpha_{5}u_{3}+\alpha_{6}u_{1}+\dot{c}\dot{u}_{1}=0, \end{array}  \tag{3}\]  
where \(\alpha_{1}=-2(a+1)\), \(\alpha_{2}=b\gamma+a-c\), \(\alpha_{3}=b\gamma\), \(\alpha_{4}=-(a+1)b\gamma\), \(\alpha_{5}=b[(a-c)\gamma+1]\), \(\alpha_{6}=bc\gamma\). Therefore, we only need to consider the condition \(\lim_{t\rightarrow\infty}(u_{1}-u_{3})=0\) in system (3). By letting:  
\[e_{1}=\frac{u_{1}-u_{3}}{2},e_{2}=\frac{u_{1}+u_{3}}{2},\]  
where  
\[ \begin{array}{l}A_{1}=\frac{\sqrt{5\alpha_{3}a_{1}^{2}}}{4}>0,\,A_{2}=\frac{ \sqrt{5\alpha_{3}a_{2}^{2}+8(\alpha_{5}+a_{6})a_{2}}}{4}>0,\\ 0<A_{3}=\frac{\alpha_{3}a_{2}^{2}}{2(5\alpha_{3}a_{2}^{2}+8(\alpha_{5}+a_{6})) }<\frac{1}{2},\,\,0<A_{4}=\frac{2(\alpha_{3}a_{2}^{2}+2(\alpha_{5}+a_{6}))}{5 \alpha_{3}a_{2}^{2}+8(\alpha_{5}+a_{6})}<\frac{1}{2}.\end{array} \]  
From Equations (4) and (5), we have:  
\[ \begin{array}{l}\frac{dE_{1}}{dt}=-f_{1}(e,e)\dot{e}_{1}=B_{1}+B_{2}\cos( \theta_{1})+B_{3}\cos(2\theta_{1})+B_{4}\cos(\theta_{2})\\ \quad+B_{5}\cos(2\theta_{2})+B_{6}\cos(\theta_{1}-\theta_{2})+B_{7}\cos(2 \theta_{1}-2\theta_{2})+H(\theta_{1},\theta_{2}),\end{array}  \tag{7}\]  
where  
\[ \begin{array}{l}B_{1}=(-\frac{651a_{2}^{2}}{400}-\frac{423a_{1}^{2}}{400}+ \frac{217c}{200}-\frac{217a_{2}}{200})A_{1}^{2},\\ B_{2}=\frac{\delta}{5}(3\alpha_{3}a_{2}^{2}-2A_{22}a_{1}-2a_{6}+2\alpha_{5})a_ {1}A_{1},\\ B_{3}=(-\frac{19c}{25}+\frac{19a_{2}}{25}-\frac{199a_{2}^{2}}{400}+\frac{57a_ {2}^{2}}{50})A_{1}^{2},\\ B_{4}=-\frac{217}{200}A_{1}^{2}a_{1}a_{2},\,\,B_{5}=-\frac{651}{200}A_{1}^{2} a_{2}^{2},\\ B_{6}=\frac{\delta}{5}(2\alpha_{4}-6A_{2}+3A_{2}A_{4})a_{2}a_{1}A_{1},\,\,B_{7}= \frac{57}{100}A_{1}^{2}a_{2}^{2},\end{array} \]  
and \(H(\theta_{1},\theta_{2})\) represents the rest of the periodic functions of \(\theta_{1},2\). Obviously, \(e_{1}\to 0\) holds when \(E_{1}\to 0\). To make \(E_{1}\to 0\), a non-periodic term in \(dE_{1}/dt\), it should be less than zero, which means \(B_{1}<0\) holds.