Proposition

Suppose $U\colon\Omega\to\mathbb R$ is a non-constant harmonic function, i.e. $U\in\mathcal C^\omega$, i.e. analytic, and $\Delta U=0$, where $\Omega\subseteq\mathbb R^n$ is a region. Then every equilibrium of the system $\ddot x=-\nabla U$ is unstable in the sense of Lyapunov.

(In other words, suppose $(x_0,y_0)$ is an equilibrium of $\dot x=y,\dot y=-\nabla U$, then $(x_0,y_0)$ is unstable.)

Discussion

In electrostatic, the preceding statement is known as Earnshaw's theorem. However, the proofs I saw aren't completely rigorous. They rely on the maximum principle of harmonic functions, and the following statement:

If $\nabla U(x_0)=0$ and $x_0$ isn't a strict local minimum, then the equilibrium $x=x_0,\dot x=0$ is unstable.

However, in V.I.Arnold's Mathematical Methods of Classical Mechanics, it's said that

It seems likely that in an analytic system with $n$ degrees of freedom, an equilibrium position which is not a minimum point (of the potential energy) is unstable; but this has never been proved for $n>2$.

In addition, $\Delta U=0$ implies that the Hessian $H$ of $U$ at $x_0$ satisfies $\operatorname{tr}H=0$. If $H$ has a negative eigenvalue (otherwise $H=0$), then by spectrum theorem of symmetric matrix $H$ and the theory of linearization, system is unstable at $x_0$. However, there's no evident that $H\neq0$.

Any idea? Thanks!