Context: The question refers to computational physics of non linear systems with Mathematica.

Excercise: Given the system $\{f_1: \dot{x} = a x + y + x^3, f_2: \dot{y} = x - y \}$:

  1. Find the equilibrium points and classify them.
  2. What kind of bifurcation is there and for what value of $a=a^*$ does it happen ?
  3. 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:

  1. for $a <-1$, it is $\lambda_1 < 0, \lambda_2 < 0$, therefore it is stable node.
  2. 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 ?

Eigenvalues of matrix for equilibrium point (0,0)


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

enter image description here

  • 1
    $\begingroup$ Nice update, your plots look cool. Also, try making some animations as a function of a. $\endgroup$
    – Bruce Dean
    Commented Dec 28, 2013 at 18:03

1 Answer 1


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.

  • $\begingroup$ Thanks @roybatty! By the way, let me fix the a versus \alpha amgibuity! $\endgroup$
    – stathisk
    Commented Dec 28, 2013 at 17:26
  • $\begingroup$ Just saw your 2nd question, so yes, you were on the right track! $\endgroup$
    – Bruce Dean
    Commented Dec 28, 2013 at 17:30
  • $\begingroup$ Just one more clarification please, before I accept your answer: Isn't the equilibrium point $(0,0)$ for $a<-1$ a stable node (sink) rather than a center node? As depicted in the phase portraits as well ? Thanks! $\endgroup$
    – stathisk
    Commented Dec 28, 2013 at 17:39
  • $\begingroup$ Yes your right, if $\Delta>0$ then the eigenvalues are either real with the same sign (nodes) or complex conjugates (spirals or centers). Nodes satisfy, ${\tau ^2} - 4\Delta > 0$, and in your case it is stable since $\tau < 0$. I just edited my answer to correct that. $\endgroup$
    – Bruce Dean
    Commented Dec 28, 2013 at 17:57

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