I have the following nonlinear system:
\begin{align}
x_1'=& x_1-x_2-x_1^3\\
x_2'=& x_1+x_2-x_2^3
\end{align}
I wish to classify the equilibrium points and determine their stability using the linearization. First, I found the equilibrium points as follows:
\begin{align}
\frac{\partial f_1}{\partial x_1} =& 1-3x_1^2 \\
\frac{\partial f_2}{\partial x_2} =& 1-3x_2^2 \\
\frac{\partial f_1}{\partial x_2} =& -1 \\
\frac{\partial f_2}{\partial x_1} =& 1
\end{align}
Therefore we have equilibria at $(1, \pm \frac{1}{\sqrt{3}})$ and $(\pm \frac{1}{\sqrt{3}}, -1)$. Thus for the jacobian:
\begin{equation}
J = \begin{pmatrix}\frac{\partial f_1}{\partial x_1}&\frac{\partial f_1}{\partial x_2}\\
\frac{\partial f_2}{\partial x_1}&\frac{\partial f_1}{\partial x_2}\end{pmatrix}
\end{equation}
And the jacobian at all equilibria points is:
\begin{equation}
A = \begin{pmatrix}0&-1\\1&0\end{pmatrix}
\end{equation}
Thus the eigenvalues are $\pm i$ which is purely imaginary and implies that we have a center point. However, this is a borderline case and can not be approximated by a
linearization. How do I then determine the stability of these equilibrium points then?
EDIT I am having trouble solving for the equilibrium points and I also made a mistake in the equation for $x_1'$. It should be a cubic term and not a quadratic term.