linear stability analysis performed on the Rossler System yields the Jacobian
$J=\begin{pmatrix} 0& -1&-1\\ 1 &a&0\\ z_0& 0&x_0-c \end{pmatrix}$
$z_0$, $x_0$ are the corresponding components of the fixed point. The Rossler System has 2 fixed points,
$(x_0, y_0, z_0)_1={\displaystyle \left(\frac{c - \sqrt{c^2-4ab}}{2},\frac{-c + \sqrt{c^2-4ab}}{2a},\frac{c - \sqrt{c^2-4ab}}{2a}\right)},$ and
$(x_0,y_0,z_0)_2={\displaystyle \left(\frac{c + \sqrt{c^2-4ab}}{2},\frac{-c - \sqrt{c^2-4ab}}{2a},\frac{c + \sqrt{c^2-4ab}}{2a}\right)}.$
I want to obtain for both of these fixed points the range of the parameters $a$, $b$, $c$ for which the fixed point of the Rossler system is unstable.
Or an easier problem: I want to know for which parameter ranges the Jacobian $J$ has an eigenvalue with real-part greater than $0$.
I thought of applying Cardano's Formula to get an analytical expression of the eigenvalues and somehow estimate the real-part, but this looks like ugly work. Gershgorin's circle won't help for a big parameter range, too. My only idea is to use the continuity of the eigenvalues with respect to the parameters and let the computer calculate the eigenvalues for many parameter combinations, and then - the dirty way - generalize it to ranges of the parameters.
How would you approach this problem?