Consider the following autonomous system of differential equations:
$$\frac{\mathrm d\mathbf x}{\mathrm dt} = \mathbf v(\mathbf x)$$
where $\mathbf x, \mathbf v \in \mathbb R^n$. Assume that $\mathbf v(\mathbf x)$ is Lipschitz continuous. Suppose that $\mathbf x^*$ is a fixed point, i.e., $\mathbf v(\mathbf x^*) = 0$. I need to characterize the stability of $\mathbf x^*$. However, $\mathbf v(\mathbf x)$ may not be differentiable. Therefore straightforward method of finding the eigenvalues of the Jacobian does not apply. I'm trying to find alternative characterizations.
My intuition tells me that the following statement is true, but I am not sure.
If there exists an $\epsilon > 0$ such that for all $\mathbf x \ne \mathbf x^*$ in an $\epsilon$-neighborhood of $\mathbf x^*$ the vector $\mathbf v(\mathbf x)$ has a negative projection on the displacement $\mathbf x - \mathbf x^*$ (i.e., $\mathbf v(\mathbf x) \cdot (\mathbf x - \mathbf x^*)<0 $), then $\mathbf x^*$ is attracting.
Maybe I am missing some additional necessary conditions... If this statement is true (or something close), that's a start... Can it be proved/disproved? Is this condition necessary and/or sufficient for an attractive node?
Note: In a separate question I am searching for references on the topic of the stability of ODEs when the right-hand side is only Lipschitz continuous (but may not be differentiable). See https://mathoverflow.net/q/261439/16615.
Notice that if you consider the function $V(\mathbf x)=\|\mathbf x-\mathbf x^*\|^2$ taking the $2$-norm, for any solution $\mathbf x=\mathbf x(t)$ you have $$ \frac d{dt} V(\mathbf x)=2(\mathbf x-\mathbf x^*)\cdot \mathbf v(\mathbf x) $$ For example, if $(\mathbf x-\mathbf x^*)\cdot \mathbf v(\mathbf x)\le0$, then the solution $\mathbf x$ remains in a neighborhood of $\mathbf x^*$ and so it is stable. So your intuition is quite right.
On the other hand, for the same reason, it is difficult to find places where this type of problem is discussed at length since it is really a matter of either giving a condition such as yours or to use the specific form of each problem.