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?

Question

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.

Answer 1

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.