I am trying to determine the stability of the equilibrium points of the following system for all $a\in\mathbb{R}$:
\begin{align}
\dot{x}&=ax-y-xz\\
\dot{y}&=x\\
\dot{z}&=-z+x^2+y^2
\end{align}
Attempt:
I only found one equilibrium (0,0,0).
To determine the stability I first found the Jacobian matrix to be, which was:
\begin{pmatrix}
a-z&-1&-x\\
1 & 0 & 0\\
2x & 2y & -1\\
\end{pmatrix}
evaluated at the origrin the matrix becomes:
\begin{pmatrix}
a&-1&0\\
1 & 0 & 0\\
0 & 0 & -1\\
\end{pmatrix}
Then I found the eigenvalues to be: $\lambda_1=-1$, $\lambda_2=\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}$ and $\lambda_3=\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}$.
I know some theory in 2D about when we have a stable/unstable node, focus, saddle, center etc. I'm just a bit confused about what happens in 3D?
I have looked, at different values of a, and written my thoughts in parenthesis:
For $a<-2$ we have 3 negative real eigenvalues. (stable node?)
For $-2<a<0$ we have 1 negative real eigenvalue and a pair of complex conjugate eigenvalues with negative real part.(focus? stable? unstable?)
For $a=0$ we have one negative real eigenvalue and two pure imaginary eigenvalues. (No clue? in 2D we have a center, if we have pure imaginary eigenvalues, but here, we also have 1 negative real eigenvalue.)
For $0<a<2$ we have 1 negative real eigenvalue and a pair of complex conjugate eigenvalues with positive real part.(focus? stable? unstable?)
For $a>2$ we have 1 negative and 2 positive real eigenvalues (unstable saddle?).
Can anyone help me with this? It would be very much appreciated! :D