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.

  • $\begingroup$ Can you check coordinates of your equilibria? Vector field must be zero at them. It seems that your set of equilibria doesn't satisfy $x_1 - x_2 - x_1^2 = 0$, $x_1 + x_2 - x^3_2 = 0$. $\endgroup$
    – Evgeny
    Aug 1, 2013 at 9:36
  • $\begingroup$ Yes, I see that mistake now... I guess I don't see how to find the equilibrium points. You can see from the jacobian that the partial derivates are very similar, but how can I use that to find the equilibrium points? I see that the origin is an equilibrium but can't find the others. $\endgroup$ Aug 1, 2013 at 9:47
  • $\begingroup$ Your system is unstable. The real equilibrium points are (-0.7274,-1.2565) & (0,0), one is saddle and other is unstable focus, respectively. $\endgroup$
    – kaka
    Aug 1, 2013 at 10:20
  • $\begingroup$ How did you find the other equilibrium point? $\endgroup$ Aug 1, 2013 at 10:25
  • $\begingroup$ other equilibrium pt?? you mean how did I solve for it? $\endgroup$
    – kaka
    Aug 1, 2013 at 10:37

1 Answer 1


To find the equilibrium points, we want to find the points where $x'_1$ and $x'_2$ are zero.

So, we want to solve:

$\tag 1 x_1-x_2-x_1^3 = 0$
$\tag 2 x_1+x_2-x_2^3=0$

Using $(2)$, we have $x_1 = x_2^3 - x_2$ and substituting this into equation $(1)$ yields:

$$x_2^3 - x_2 - x^2 -(x_2^3 - x_2)^3 = -x_2 (x_2^8-3 x_2^6+3 x_2^4-2 x_2^2+2) = 0$$

  • One solution is $x_2=0$.
  • The other solutions for the eighth-order polynomial are all imaginary. I used this Wolfram Alpha solution for the roots, but you should be able to use your favorite tool for root finding.

If we substitute these back into $x_1 = x_2^3 - x_2$, we get the corresponding point as $0$. So, our critical point is $(0,0)$.

Now, you can use these two points with the Jacobian (assuming they are well behaved, do you know when you can use linearization and when you cannot) and determine how these equilibria behave.

Hint: Look at these two points in the phase portrait and your analysis should show this.

enter image description here

  • $\begingroup$ I made a type with the $x_1'$ equation. You should get a polynomial of degree 8 I believe. $\endgroup$ Aug 1, 2013 at 18:30
  • $\begingroup$ Sorry, I will do better in the future and thank you very much for updating your solution. It matches what I have derived. Except my plot is not as good... $\endgroup$ Aug 1, 2013 at 18:50
  • $\begingroup$ So cute system. Can you tell, how did you find that all other solutions are imaginary? $\endgroup$
    – Evgeny
    Aug 1, 2013 at 18:52
  • 1
    $\begingroup$ @Evgeny: Thanks. See the link to WA that I added. Regards $\endgroup$
    – Amzoti
    Aug 1, 2013 at 18:55
  • 1
    $\begingroup$ Ah, ok, I thought that was done with some trick from polynoms theory. Thanks, nevertheless. $\endgroup$
    – Evgeny
    Aug 1, 2013 at 18:57

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