Linear stability analysis of a dynamical systems based only on symmetric part of Jacobian

Question

Consider an autonomous dynamical system:

$$\frac{dx}{dt} = f(x)$$

where $x\in\mathbb R^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$.

Suppose 0 is a stationary point, $f(0)=0$. We Taylor expand to first-order, and obtain in the vicinity of the orign:

$$\frac{dx}{dt} = f'(0)\cdot x$$

where $f'(x)$ is the Jacobian matrix.

Typically linear stability analysis requires computing the eigenvalues of $f'(0)$, which need not be a symmetric matrix.

However I reason that if $x^Tf'(0)x<0$ for all non-zero vectors $x$, then velocity field always points into the origin in its neighborhood and therefore the origin is a stable fixed-point. Conversely, if there exists a direction for which $x^Tf'(0)x>0$, then the system can "escape" on this direction and the origin is not stable.

According to this reasoning, we therefore only need to determine the negative-definitedness of the symmetric part of $f'(0)$ (because the skey-symmetric contribution vanishes in the quadratic form) to assess the stability of the origin.

But this criterion is different from the classical one of having to analyze the real parts of the spectrum of $f'(0)$ (which includes a skew-symmetric part).

Am I missing something?

Answer 1

Counter example: $f^\prime(0)=\begin{pmatrix}0 & 1\\-\frac{1}{3}&-\frac{1}{3}\end{pmatrix}$ and $x=\begin{pmatrix}1\\1\end{pmatrix}$.

Your intuition is understandable, but it is ignoring that the "escape" is temporary because the state would get out of the "bad area" since $x$ is not an eigenvector.

Answer 2

This is connected to the Lyapunov analysis of differential equations. The linear dynamical $\dot x=Ax$ is asymptotically stable if and only if there exists a symmetric positive definite matrix $P$ such that $x^T(A^TP+PA)x$ is negative for all $x\ne0$. This is equivalent to saying that $A^TP+PA$ is negative definite. In such a case, $V(x)$ is called a Lyapunov function for the considered system.

Now, your condition that $A=f'(0)$ is such that $x^Tf'(0)x$ is negative for all $x\ne0$ implies that $x^Tf'(0)x+x^Tf'(0)^Tx$ is negative for all $x\ne0$, which is equivalent to saying that $f'(0)+f(0)^T$ is negative definite. So, your condition implies that $V(x)=x^Tx$ is a Lyapunov function of the system. The reverse is not true in general.

Your stability condition is, however, sufficient only because there are systems which are asymptotically stable but for which $x^Tf'(0)x$ is not negative for some $x\ne0$.

For instance, we can pick $$A=\begin{bmatrix}-1 & 5\\0 & -2\end{bmatrix}$$ and we have that $x^TAx=3$ for $x=(0,1)$. Yet, the system is obviously stable.