What is the classification of a fixed point if the trace and determinant are equal?

Question

In the text book I am reading Nonlinear Dynamics and Chaos with applications to Physics, Biology, Chemistry, and Engineering by Steven H. Strogatz. He uses a diagram similar to this one:

to show how to classify a fixed point of a system using the determinants and trace of the matrix. But I was wondering what would happen if you are on the parabola thats given ? Prime example would be if $\tau$ is the trace and $\Delta$ is the determinant then what would be the classification of the fixed point if:

$$ \tau^2 = 4\Delta $$

This equation is given in the textbook as: $\tau^2 - 4\Delta = 0$

Answer 1

When the trace and determinant are on the indicated parabola for the system $\vec{x}'=A \vec{x}$, we are dealing with a repeated real eigenvalue $\lambda$ for $A$. This will often introduce solutions of the form $t e^{\lambda t}$, due to potentially being non-diagonalizable. Computing the matrix exponential gives $e^{tA}= (I + t(A-\lambda I)) e^{t \lambda I}$, and all solutions will be of the form $\vec{x}(t) = e^{tA} \vec{c}$.

Such an equilibrium is referred to as a "defective nodal source / sink," although it is perhaps not the most common name.

The mathlet linear phase portraits gives a nice visualization of solutions as we move around the different possibilities. Here's a screenshot showing the desired behavior: