I have a soft question regarding the Jacobian of vector fields and isolated equilibria, and what they imply about local behavior of nearby integral curves near.


Let $V:U \subset\_{open} \mathbb{R}^n \to \mathbb{R}^n$ be a smooth vector field. Let $x^\* \in U$ be an isolated equilibrium of $V$. That is, $V(x^\*)=0$ and there is a neighborhood of $x^\*$ with no other equilibria. It is well known if the Jacobian matrix, $DV(x^\*)$, has eigenvalues with negative real part (i.e. Hurwitz stable), then $x^\*$ is asymptotically stable.


In my research, I am dealing with a vector field with a single equilibrium, $x^\*$, where $DV(x^\*)$ has 1 simple zero eigenvalue and the remaining eigenvalues have strictly negative real part. I've noted numerically that $x^\*$ is asymptotically stable despite $DV(x^\*)$ not being Hurwitz stable (In fact, I've noticed it is globally asymptotically stable, but that is outside the scope of my question).


I have long term goal to show $x^\*$ is asymptotically stable. For now, I need to learn what the Jacobian implies about the local behavior of an equilibrium point.



What does the Jacobian of a vector field at an equilibrium tell you about local behavior of integral curves when the Jacobian is not a stable?
The Jacobian alone doesn't have the information you need. For example, consider the two vector fields
$$f(x, y) = (-x^2, -y)$$
and
$$g(x, y) = (-x^3, -y)$$.


They have an isolated equilibrium at the origin, with the same Jacobian: $\begin{bmatrix}
 0 & 0 \\
 0 & -1
\end{bmatrix}$.


But the equilibrium in $f$ isn't stable, while in $g$ it is.


With the Jacobian you describe, you'd need some other information, like the existence of a Lyapunov function, to prove stability.


