An exercise from the book I am reading is: "Construct a non-linear system that has four critical points:two saddle points, one stable focus, and one unstable focus."

I have tried many systems. I found one quickly but I was lucky even if I had a few clues thanks my previous trials.

I wonder if there is any way to find such systems using a not completely "gropingly way".

Edit: with only two equations in the system.

The system I have is:



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    $\begingroup$ Is the dimension of system = 2 ? $\endgroup$ Aug 2, 2013 at 17:13
  • $\begingroup$ Yes the dimension is 2 as it was in the chapter preceding this exercise, I should have written that also ^^ Thanks. $\endgroup$ Aug 2, 2013 at 22:44
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    $\begingroup$ @Ouistiti: Was the answer helpful? What does your system look like? Where is this problem from? $\endgroup$
    – Amzoti
    Aug 4, 2013 at 1:29
  • $\begingroup$ @Amzoti I will post my system above. The answer was very helpful, I learned more from your answer than I expected. Thank you :) The problem is from "Dynamical Systems with Applications using Mathematica®" written by Stephen Lynch. $\endgroup$ Aug 5, 2013 at 12:22
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    $\begingroup$ @Ouistiti: It makes me very happy to hear that! If you write the four sets of eigenvalues for the Jacobian matrix at each critical point, It may be possible to set up an optimization problem to solve for the six parameters. Look forward to seeing your system and this was a fun problem. Regards $\endgroup$
    – Amzoti
    Aug 5, 2013 at 12:25

1 Answer 1


We need to choose a form for our system to give us exactly four critical points. This selection is not unique, hence, we can get more than one solution.

Lets choose a system with parameters that give us some degree of freedom. An example of a system that gives us four critical points and some degree of freedom is (there are other choices, this is not unique):

$$\tag 1 x' = a x + b x^2 + c x y = x(a + bx + cy)\\ y' = d y + e y^2 + f x y = y(d+ey + fx)$$

First thing we need to do is to find the critical points (recall, we need exactly four of them).

  • $x = 0 \rightarrow y = 0$ or $y = -\dfrac{d}{e}$
  • $y = 0 \rightarrow x = 0$ or $x = -\dfrac{a}{b}$
  • $x \ne 0, y \ne 0 \rightarrow x = \dfrac{cd -ae}{be - cf}, y = \dfrac{af -bd}{be - cf},~~\text{with}~~ c \ne 0, be \ne cf$

Thus, our four critical points are (note the constraints above):

$$(0,0), ~\left(0, -\dfrac{d}{e}\right), ~\left(-\dfrac{a}{b}, 0\right), \left(\dfrac{cd -ae}{be - cf}, \dfrac{af -bd}{be - cf}\right)$$

The Jacobian matrix of $(1)$ is given by:

$$\tag 2 \displaystyle J(x, y) = \begin{bmatrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{bmatrix} = \begin{bmatrix}a + 2bx + cy & cx \\ fy & d + 2 e y + fx \end{bmatrix}$$

Now, we have freedom when choosing the parameters with the goal to get two saddle points, one stable focus, and one unstable focus.

There is not a clean way to do this, but our problem now is to choose the parameters $a, b, c, d, e, f$ to make that happen by evaluating the eigenvalues of the Jacobian at each critical point, and choosing the parameters to give us the desired behavior.

I am going to crank these one at a time and keep my fingers crossed for the last critical point.

Point (0,0)

We have:

$$\displaystyle J(0, 0) = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$$

Lets choose $a = -1, d = 1$ and this is our first saddle.

Point (0,-d/e)

We have:

$$\displaystyle J(0, -d/e) = \begin{bmatrix}a - (cd)/e & 0 \\ -(fd)/e & d -(2d)/e \end{bmatrix}$$

Lets choose $e = 1, c = -3$ and this is our second saddle.

Point (-a/b, 0)

We have:

$$\displaystyle J(-a/b, 0) = \begin{bmatrix} - a - (ac)/b & 0 \\ 0 & d -(af)/b \end{bmatrix}$$

Lets choose $b = 1, f = 1$ and this is our unstable focus.

Point $\left(\dfrac{cd -ae}{be - cf}, \dfrac{af -bd}{be - cf}\right)$

Well, this is where we need luck with the parameters we chose, but I think it might be possible to cast this problem as an optimization problem, so you might want to play around with that. This would give you ranges for these six parameters, so no guesswork is needed.

Anyway, from our parameter choices above, we get $x = y = -1/2$.

$$\displaystyle J(-1/2, -1,2) = \begin{bmatrix} -1/2 & 3/2 \\ -1/2 & -1 \end{bmatrix}$$

This, thankfully, gives us our stable focus.

Thus, we have the system:

$$\tag 1 x' = - x + x^2 -3 x y \\ y' = y + y^2 + x y$$

Lets draw the phase portrait and validate this analysis.

enter image description here

We have what we need with two saddles, an unstable and a stable focus at the four critical points.

I am curious how you did it (even if it is guessing) and what your system looks like and it would help to post your solution in your question.


Here is the phase portrait of the system you wrote and it is also a good example.

enter image description here

  • $\begingroup$ Great job. For my knowledge how did you come up with the second order system in (1)? By intuition ? $\endgroup$
    – kaka
    Aug 3, 2013 at 6:14
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    $\begingroup$ @kaka: Yes, by intuition and a bit of handwaving. I wish I could say I had some wonderful approach to that. I realized that we needed something that could give us a point at (0,0) and then realized we needed a parabola that gave points at (0, w), and (f, 0) and one more at some (h, j). Thinking along those lines lead me to that system. Then, we needed freedom, so I added all of the parameters. I still think the approach can be improved using optimization for ranges on the parameters, but I'll let the OP work that. Thanks for the kind words. Regards $\endgroup$
    – Amzoti
    Aug 3, 2013 at 6:19
  • $\begingroup$ @Amzoti: Wowwww. It is complete as always. Full of points, and finally a nice illustrating graph. $\endgroup$
    – Mikasa
    Aug 3, 2013 at 14:56
  • $\begingroup$ This is a nice problem. Go for it! ;-) $\endgroup$
    – amWhy
    Aug 4, 2013 at 1:14

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