Does the Hessian matrix of energy function of a gradient system have to be positive semidefinite when the system has one globally stable point?

Question

Given a gradient dynamical system $$\frac{d\theta_i}{dt}=f_i(\theta_1,\cdots,\theta_n),\forall i\in\{1,\cdots,n\},$$ where $$\frac{\partial G}{\partial \theta_i}=f_i(\theta_1,\cdots,\theta_n),$$

where $G$ is the energy function.

Assume now that there exists a globally asymptotic stable equilibrium point. Does the Hessian matrix of the $G$ have to be positive semidefinite ($G$ is convex function)?

Remark

I basically curious that:

On the one hand, if we know there exists a globally asymptotic stable equilibrium point, whether we would have some properties on $G$.

On the other hand, what conditions on $G$ could deduce the globally asymptotic stability of an equilibrium point.

Answer 1

Let $U$ being an open set in $\mathbb{R}^n$ and $G:U\to \mathbb{R}$ a two times continuously differentiable function.

If $x(t)$ is a solution to the differential equation $$\begin{cases} \displaystyle \frac{dx}{dt}&=&\nabla G(x)\\x(0)&=&x_0\end{cases}\tag{1}$$ then $$\frac{dG(x(t))}{dt}=\nabla G(x(t))\cdot \nabla G(x(t))=\|\nabla G(x(t))\|^2\geq 0,\tag{2}$$ $x(t)$ follows the steepest ascent direction, and $G(x(t))$ is non decreasing.

If we assume that there exists a globally asymptotic stable equilibrium point $x^*$ to $(1)$. Then $x(t)\to x^*$, to any $x_0$, as $t\to +\infty$, and $$\max_xG(x)=G(x^*).\tag{3}$$

This, in particular, means that Hessian matrix $HG(x^*)$ has no positive eigenvalues. As so, it is not positive semidefinite.

You can find some related discussions searching for "\(\dot{x}=\nabla G(x)\)" on SearchOnMath, for instance.

Note:

1. What happens if you choose $G(x)=e^{-\|x\|^2}$?.

2. Now, if we assume that there is a unique global maximum $x^*$ for $(3)$. Then this point is an equilibrium point for $(1)$, and $x(t)\to x^*$, as $t\to +\infty$, for any $x_0$ close enough to $x^ *$ . If there is some $x_0\in U$ and some $r>0$, such that $\|x(t)-x^*\|\geq r$, as $t\to +\infty$, then $( 2)-(3)$ implies that $G(x(t))\to M$, for some $M<G(x^*)$. Therefore, if $x(t)$ is bounded, then $M$ is a local maximum, and $x(t)$ approximates some non-empty set $Y^*\subset{R}^n$ containing equilibrium points of $(1)$.

3. Therefore, if we assume that there is a unique global maximum $x^*$ for $(3)$ and there is no other local maximum, then any bounded solution $x(t)$ to $(1)$, is such that $x(t)\to x^*$, as $t\to +\infty$, when $x_0$ is not a local minimum or saddle point.