## 2 GS between Two FHN Neurons
### Necessary Conditions for GS between Two FHN Neurons  
Based on the auxiliary system approach [11], the auxiliary system corresponding to System (3) is given by  
\[\dot{u}_{5} =-u_{5}^{3}/3+cu_{5}^{2}+du_{5}-u_{6}+k(u_{5}-H(u_{1})),\] \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]  
Then Condition (4) will lead to  
\[\lim_{t\to\infty}||u_{3}-u_{5}||=0. \tag{6}\]  
By defining  
\[e_{1}=\frac{u_{3}-u_{5}}{2},\quad e_{2}=\frac{u_{3}+u_{5}}{2}, \tag{7}\]  
Systems (3) and (5) become \[\dot{u}_{1} =-u_{1}^{3}/3+cu_{1}^{2}+du_{1}-u_{2},\] \[\dot{u}_{2} =\gamma(u_{1}-bu_{2}),\] \[\dot{u}_{4} =\gamma(e_{1}+e_{2}-bu_{4}),\] \[\dot{u}_{6} =\gamma(e_{2}-e_{1}-bu_{6}),\] \[\dot{e}_{1} =-e_{1}^{3}/3+(-e_{2}^{2}+2 ce_{2}+d+k)e_{1}-\frac{u_{4}-u_{6}}{2},\] \[\dot{e}_{2} =-e_{2}^{3}/3+c(e_{1}^{2}+e_{2}^{2})+(d+k-e_{1}^{2})e_{2}-\frac{u _{4}+u_{6}}{2}-kH(u_{1}). \tag{8}\]  
Condition (6) is equivalent to  
\[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]0  
GS in System (3) appears if Condition (9) is satisfied. Consider the Laplace transform defined as follows:  
\[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]1 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]2 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]3 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]4  
By taking the Laplace transform on both sides of equations in Equation (8), one has  
\[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]5 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]6 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]7 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]8 \[\dot{u}_{6} =\gamma(u_{5}-bu_{6}). \tag{5}\]9 \[\lim_{t\to\infty}||u_{3}-u_{5}||=0. \tag{6}\]0  
where \(u_{i0}\), \(i=1,2,4,6\), and \(e_{j0}\), \(j=1,2\), are given initial values of System (8), and \(\hat{W}_{k}\), \(k=1,2,3\) are the Laplace transform of the nonlinear parts in the first, fifth, and last equations in Equation (8), respectively  
\[\lim_{t\to\infty}||u_{3}-u_{5}||=0. \tag{6}\]1 \[\lim_{t\to\infty}||u_{3}-u_{5}||=0. \tag{6}\]2 \[\lim_{t\to\infty}||u_{3}-u_{5}||=0. \tag{6}\]3  
Substituting the third and fourth equations into the fifth equation in Equation (11) yields  
\[\lim_{t\to\infty}||u_{3}-u_{5}||=0. \tag{6}\]4where \(\beta_{1}=b\gamma-d-k\), \(\beta_{2}=[1-(k+d)b]\gamma\), \(\beta_{3}=b\gamma\), \(\beta_{4}=(u_{60}-u_{40})/2\). To analyze whether \(||e_{1}||\to 0\) holds when \(t\to\infty\), taking the inverse Laplace transform on both sides of Equation (12), produces  
\[ \begin{split} e_{1}&=e_{10}\Psi_{1}(t)+\beta_{4} \Psi_{2}(t)+\int_{0}^{t}\Psi_{1}(t-\xi)W_{2}d\xi,\\ &=e_{10}\Psi_{1}(t)+\beta_{4}\Psi_{2}(t)+\int_{0}^{t}\Psi_{1}(t- \xi)[-e_{1}(\xi)^{2}/3-e_{2}(\xi)^{2}+2 ce_{2}(\xi)]e_{1}(\xi)d\xi,\end{split}  \tag{13}\]  
where \(\Psi_{1,2}(t)\) represent the inverse Laplace transform defined in Equation (10):  
\[ \begin{split}\Psi_{1}(t)&=L^{-1}[\frac{s+\beta_{3} }{s^{2}+\beta_{1}s+\beta_{2}}]=\left\{\begin{array}{ll}e^{-\beta_{1}t/2}[ \cosh(\phi_{1}t)+\frac{2\beta_{3}-\beta_{1}}{2\beta_{1}}\sinh(\phi_{1}t)],& \beta_{1}\
eq 0\\ e^{-\beta_{1}t/2}[1+2(2\beta_{3}-\beta_{1})t],&\beta_{1}=0\\ \Psi_{2}(t)&=L^{-1}[\frac{1}{s^{2}+\beta_{1}s+\beta_{2}}]= \left\{\begin{array}{ll}\frac{\sinh(\phi_{1}t)}{\phi_{1}}e^{-\beta_{1}t/2},& \beta_{1}\
eq 0\\ te^{-\beta_{1}t/2},&\beta_{1}=0\end{array}\right.\end{split}  \tag{14}\]  
in which \(\phi_{1}=\sqrt{\beta_{1}^{2}-4\beta_{2}}/2\). Obviously, from Equation (13), the necessary condition for \(||e_{1}||\to 0\) is \(\Psi_{1,2}\to 0\) when \(t\to\infty\). Then, \(-\beta_{1}/2\pm Re\{\phi_{1}\}<0\), in which \(Re\{\phi_{1}\}\) represents the real part of \(\phi_{1}\), should be satisfied from Equation (14). Actually, this means that the two roots of equation \(s^{2}+\beta_{1}s+\beta_{2}=0\) have negative real parts.