Which properties does a critical point have if the Jacobian determinant is $0$ and thus no characteristic equation exists?

Question

I have the system

\begin{equation} \begin{array} fx'=y(1-x^2-y^2)\\ y'=x-y \end{array} \end{equation}

which has several CPs. 5 of them have either of the variables at $0$. This yields a zero-valued determinant for the Jacobian for all these 5 CPs.

Since there is no characteristic equation from the Jacobian at these points, what can be said of these points?

Thanks

Answer 1

We have

$$\begin{align} y(1 - x^2 - y^2) &= 0 \\ x - y &= 0 \end{align}$$

From the second equation

$$y = x$$

Substituting that into the first equation

$$x(1 - 2 x^2) = 0 \implies x = 0, \pm \dfrac{1}{\sqrt{2}}$$

Since we know $y = x$, we have three $x$ values and end up with

$$(x, y) = (0,0), \left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right), \left(-\dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}\right)$$

You can now test these three critical points in the Jacobian.

We can look at a phase portrait and determine their behavior

Answer 2

For the matrix of the Jacobian with inserted values of (x,y) at critical point:

\begin{pmatrix} a & b\\ c & d \end{pmatrix}

If $\det A=0$, then all one can use to classify the CPs is the $a+b$ from the matrix above. If $a+b>0$, it is called a degenerated unstable node. If $a+b<0$, then it is called a degenerated stable node.