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

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: $\tau^2 - 4\Delta = 0$

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 $\vec{x}(t) = e^{tA} \vec{c}$.
• Two moments. 1) repeated eigenvalues don't always lead to terms like $te^{\lambda t}$ — all depends on geometric multiplicity of eigenvalue of matrix $A$. What you've described always happens for system that is obtained from second order equation $\ddot{x} + a \dot{x} + b x = 0$. 2) Going from node to focus isn't a bifurcation: both are sinks or source and topologically the behaviour is the same. Apr 7, 2017 at 10:18