## 3 Detection of GFPS Condition Based on the Laplace Transform  
By letting  
\[e_{1}=\frac{y-z}{2},\qquad e_{2}=\frac{\hat{\phi}-\bar{\phi}}{2},\qquad e_{3}= \frac{\hat{\psi}-\bar{\psi}}{2}\]  
\[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]  
systems (1), (3) and (5) become  
\[\dot{e}_{1}=k{e}_{1}+\frac{J(x)}{2}[f_{1}(x,e_{2}+e_{5})-f_{1}(x, e_{5}-e_{2})]+\frac{1}{2}[f_{2}(e_{1}+e_{4},\psi)-f_{2}(e_{4}-e_{1},\psi)] \tag{8}\] \[\qquad+\frac{1}{2}[f_{2}(e_{4}-e_{1},e_{6}-e_{3})-f_{2}(e_{4}+e_{1 },e_{6}+e_{3})],\] \[\dot{e}_{2}=\frac{1}{2}[g_{1}(e_{2}+e_{5}-\phi,e_{1}+e_{4}-h(x)x) -g_{1}(e_{5}-e_{2}-\phi,e_{4}-e_{1}-h(x)x)],\] \[\dot{e}_{3}=\frac{1}{2}[g_{2}(e_{3}+e_{6}-\psi,e_{1}+e_{4}-h(x)x) -g_{2}(e_{6}-e_{3}-\psi,e_{4}-e_{1}-h(x)x)],\] \[\dot{e}_{4}=k{e}_{4}+\frac{J}{2}[f_{1}(x,e_{2}+e_{5})+f_{1}(x,e_{ 5}-e_{2})]+\frac{1}{2}[f_{2}(e_{1}+e_{4},\psi)+f_{2}(e_{4}-e_{1},\psi)]\] \[\qquad-\frac{1}{2}[f_{2}(e_{4}-e_{1},e_{6}-e_{3})+f_{2}(e_{4}+e_{1 },e_{6}+e_{3})],\] \[\dot{e}_{5}=\frac{1}{2}[g_{1}(e_{2}+e_{5}-\phi,e_{1}+e_{4}-h(x)x) +g_{1}(e_{5}-e_{2}-\phi,e_{4}-e_{1}-h(x)x)],\] \[\dot{e}_{6}=\frac{1}{2}[g_{2}(e_{3}+e_{6}-\psi,e_{1}+e_{4}-h(x)x) +g_{2}(e_{6}-e_{3}-\psi,e_{4}-e_{1}-h(x)x)],\] \[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]0  
GFPS condition (7) can be written as  
\[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]1  
For sufficiently small \(e_{1,2,3}\) near \((0,0,0)\), the right-hand side of the first three equations in system (8) can be expanded as  
\[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]2 \[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]3 \[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]4  
where  
\[e_{4}=\frac{y+z}{2},\qquad e_{5}=\frac{\hat{\phi}+\bar{\phi}}{2},\qquad e_{6}= \frac{\hat{\psi}+\bar{\psi}}{2}\]5 \[F_{2}(e_{1},e_{2})=\sum_{n=1}^{\infty}\frac{1}{(2n+1)! }[ \frac{\partial^{(2n+1)}g_{1}(0,0)}{\color[rgb]{0,0,0}\definecolor[named]{ pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0} \pgfsys@color@gray@fill{0}2(^{2n+1})}e_{1}^{(2n+1)}+\frac{\partial^{(2n+1)}g_{1} (0,0)}{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2(\hat{\phi}-\phi)^{(2n+1)} }{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2(\hat{\phi}-\phi)^{(2n+1)} }{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2(\hat{\phi}-\phi)^{(2n+1)} }{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2(\hat{\phi}-\phi)^{(2n+1)} }{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2(\hat{\phi}-\phi)^{(2n+1)} }{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2(\hat{\phi}-\psi)^{(2n+1)} }{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0} \pgfsys@color@gray@stroke{0}\pgfsysConsider the Laplace transform, defined as follows:  
\[ \begin{split} E_{j}(s)&=L[e_{j}]=\int_{0}^{+\infty}e_{ j}(t)e^{-st}dt,\\ e_{j}(t)&=L^{-1}[E_{j}]=\frac{1}{2\pi i}\int_{\sigma -i\omega}^{\sigma+i\omega}E_{j}(s)e^{st}ds,\qquad j=1,2,3.\end{split}  \tag{11}\]  
Both \(e_{j}\) and \(E_{j}\) are vectors with the same dimensions for \(j=1,2,3\). Taking the Laplace transform of \(e_{1,2,3}\) on both sides of system (10) yields  
\[(S-M)[E_{1},E_{2},E_{3}]^{T}=[e_{10},e_{20},e_{30}]^{T}+[\hat{F}_{1},\hat{F}_{2 },\hat{F}_{3}]^{T}, \tag{12}\]  
where  
\[S= \begin{bmatrix}sI_{n}&0&0\\ 0&sI_{m}&0\\ 0&0&sI_{l}\end{bmatrix} ,\qquad M=