eigenvalues and eigenvectors of $\mathcal{M}'$. Let $\mathcal{A}$ denote the matrix whose columns are the eigenvalues, and $\mathcal{S}$ denote the matrix whose columns are the eigenvectors, and then we have $$\begin{aligned}
\mathcal{A}=\begin{pmatrix} -3\\ -3\gamma\\ 0\\ 0
\end{pmatrix},\quad
\mathcal{S}=\begin{pmatrix}
1& 0& -\frac{\sqrt{6}}{6}& 0\\
0& 1& 0 & \frac{1}{2}\\
0 &  0&0& 1\\
0 & 0& 1 & 0
\end{pmatrix}.\end{aligned}$$ With the help of $\mathcal{S}$, we can transform $\mathcal{M}'$ into a block diagonal matrix $$\begin{aligned}
\mathcal{S}^{-1} \mathcal{M}' \mathcal{S}=\begin{pmatrix}
-3 & 0&0&0\\
0& -3\gamma& 0& 0\\
0&0&0&0\\
0&0&0&0
\end{pmatrix}=\begin{pmatrix}
\mathcal{A}_1 &0\\
0& \mathcal{A}_2
\end{pmatrix},\end{aligned}$$ where all eigenvalues of $\mathcal{A}_1$ have negative real parts, and all eigenvalues of $\mathcal{A}_2$ have zero real parts.

Now we change the variables to be $$\begin{aligned}
    \begin{pmatrix}
    X\\Y\\Z\\ \bar{\lambda}
  \end{pmatrix}=
  \mathcal{S}^{-1}\begin{pmatrix}
    x\\ \bar{y}\\ {z}\\ \lambda
  \end{pmatrix}=
  \begin{pmatrix}
    x+\frac{\sqrt{6}}{6}\lambda\\ \bar{y}-\frac12z\\ \lambda \\z
  \end{pmatrix}.\end{aligned}$$ Then, we can rewrite the autonomous system in the form of the new variables: $$\begin{aligned}
\begin{pmatrix}\label{A9}
  \frac{dX}{dN}\\ \frac{dY}{dN}\\ \frac{dZ}{dN}\\ \frac{d\bar{\lambda}}{dN}
\end{pmatrix}=
\begin{pmatrix}
  -3 & 0&0&\\
0&-3\gamma&0&0\\
  0&0&0&0\\
  0&0&0&0
\end{pmatrix}
\begin{pmatrix}
  X\\Y\\Z\\ \bar{\lambda}
\end{pmatrix}+
\begin{pmatrix}
  G_1\\G_2\\G_3\\G_4
\end{pmatrix},\nonumber\\\end{aligned}$$ where $G_i=G_i(X,Y,Z,\bar{\lambda}),(i=1,2,3,4)$ are functions of $X,Y,Z$, and $\bar{\lambda}$. It is easy to get $G_i$ by substituting the transformations $x=X-\frac{\sqrt{6}}{6}Z,
\bar{y}=Y+\frac12\bar{\lambda}, z=\bar{\lambda}, \lambda=Z$ into the R.H.S. of Eqs. (\[X’\])-(\[L’\]).

According to the center 