When classifying critical points for systems of differential equations, we use the Jacobian matrix. However the test is inconclusive if the Jacobian determinant is equal to four times the trace squared ($\Delta=4\tau^2$). Why is this?
An example to (maybe) help clarify what I mean by Jacobian matrix:
For a system of differential equations
$x'=x(2-x-y)$ and $y'=(1-y)(2+x)$
there are critical points at (-2, 1), (0, 1), and (1, 1).
The Jacobian is $$J_{\vec{X}'}=\begin{bmatrix} {\partial{x'}\over\partial{x}} & {\partial{x'}\over\partial{y}} \\ {\partial{y'}\over\partial{x}} & {\partial{y'}\over\partial{y}} \\ \end{bmatrix}=\begin{bmatrix} 2-2x-y & -x \\ 1-y & -2-x \\ \end{bmatrix}$$
Then you test the critical points by putting them into the matrix and comparing the determinant and the trace.
Using the critical point (0, 1) as an example: $$J_{\vec{X}'}(0, 1)=\begin{bmatrix} 2-2(0)-(1) & -(0) \\ 1-(1) & -2-(0) \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -2 \\ \end{bmatrix}$$ Then $$\det{J_{\vec{X}'}}(0, 1)=-2$$ and $$\tau_{J_{\vec{X}'}(0, 1)}=-1$$ Becase $\det{J_{\vec{X}'}}(0, 1)<0$, the critical point (0, 1) is a saddle point.
This is what (I think) we have been doing in class. If the determinant $\Delta$ is equal to $4\tau^2$, though, the test fails. I was wondering why.