Classify equilibrium points and find bifurcation points of a non-linear dynamic system Context: The question refers to computational physics of non linear systems with Mathematica.
Excercise:
Given the system $\{f_1: \dot{x} = a x + y + x^3, f_2: \dot{y} = x - y \}$:


*

*Find the equilibrium points and classify them.

*What kind of bifurcation is there and for what value of $a=a^*$ does it happen ?

*Draw the phase portrait for one $a<a^*$ and for one $a>a^*$.


Solution:
I calculate the equilibrium points by solving the system of equations $\{\dot{x} = 0, \dot{y} = 0\}$. I get three solutions: $(0,0), (-\sqrt{-1-a},-\sqrt{-1-a}), (\sqrt{-1-a}, \sqrt{-1-a})$.
My lecture notes mention that in order to classify an equilibrium point, I first need to find out the topology near the equilibrium points. This is done by calculating the eigenvalues of the following matrix: 
\begin{equation}
A=
\begin{pmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y}\\\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{pmatrix}_{(x_0^*, y_0^*)}
\end{equation}
As can be seen from the following screenshot (L1 are the eigenvalues of matrix A calculated at the 1st equilibrium point, at (0,0)), we can see that:


*

*for $a <-1$, it is $\lambda_1 < 0, \lambda_2 < 0$, therefore it is stable node.

*for $a>-1$, it is $\lambda_1 < 0 < \lambda_2$, therefore it is saddle.


My 1st question is: What does it happen for $a = -1$, where the eigenvalues become (-2, 0) ? My closest guess is to use the Hartman-Grobman theorem and say that the equilibrium point is 'linearly stable'. Is there anything I am missing here ?
My 2nd questions is: Regarding the bifurcation point,  by looking at the equilibrium points of the system, we deduce that at value $a=-1$ they change (the stable node becomes saddle and vice versa). Does that suffice to say that there is a bifurcation point at $a=-1$ ? Also for $a=-1$ the three equilibrium points collide to just one. How does that this fact fit with everything else ?

Update:
These are the phase portraits for $a<a^*, a>a^*$ with $a^*=-1$ being the bifurcation point:

 A: As a check, for your linear stability matrix I get:
$$A = {\left. {\left( {\matrix{
   {3{x^2} + a} & 1  \cr 
   1 & { - 1}  \cr 
 } } \right)} \right|_{(x* = 0,y* = 0)}} = \left( {\matrix{
   a & 1  \cr 
   1 & { - 1}  \cr 
 } } \right)$$
which is presumably what you have as well.  By the way, did you switch notation from, $a \to \alpha $?  
So the determinant, $\Delta $, and trace, $\tau $, are given by $-(a + 1)$ and, $a - 1$, respectively.
As you are aware, but just so we are on the same page, the classification of a given fixed point is determined by the values of $\Delta$ and $\tau$, for a given value of $a$.  As you mention, with $a<-1$, $\Delta>0$, so the fixed point is a stable node because, ${\tau ^2} - 4\Delta> 0$ with $\tau<0$, and for $a>-1$, you have a saddle node since the $\Delta<0$.
For $a=-1$ (assuming that by $\alpha$ you meant $a$) we have $\Delta  = 0$, which signals that you are at a bifurcation point, meaning that the topological structure of the phase diagram is transitioning at this value of $a$.  In your case the transition is from a stable node to a saddle node.  
You can convince yourself of this visually in Mathematica by plotting something like:
PlotVectorField[{a x + y + x^3, x - y}, {x, -L , L}, {y, -L, L}, 
 Axes -> True]

For suitably chosen values of the parameters, $a$ and $L$. 
There are conventions on how to classify the different types of bifurcations that can occur, e.g., saddle-node, transcritical, pitchfork, critical, Hopf, transcritical, etc.  See e.g.,  Strogatz for more discussion on the classifications, and in particular, to discover the type of bifurcation point you have in your system.
Hope this helps.
