Question about Cetaev Theorem

Question

Cetaev Theorem states that:

"Considering the ode $X'=F(X)$, with an equilibrium $x_{0}$.

If there is a function $V$: $U_{0} \rightarrow $ IR and a region $\omega$ in $U_{0}$ that contains $x_{0}$, such that:

(1) $V$ is positively defined;

(2) $V'$ is also positively defined;

(3) $V=0$ in the border of $\omega$.

Then $x_{0}$ is an unstable point."

My question is, for example, if I define $V(x,y)=x^{2}+y^{2}$, V and V' were positively defined on $\omega=\mathbb{R}^{2}$, then this has no boundary but satisfies (1) and (2), so does it still work to conclude that $x_{0}$ is an unstable point?

Thanks in advance.

Answer 1

What you have given is not a correct formulation of Chetaev theorem. However, the question is clear and the answer is yes. The statement in question can be proven in a manner similar to the proofs of Lyapunov's theorems.

Consider any non-zero solution $(x(t),y(t))$, $x(t_0)=x_0$, $y(t_0)=y_0$. Since for any non-zero point $(x,y)$ we have $\dot V(x,y)>0$, the function $V(x(t),y(t))$ monotonically increases along any trajectory (except for the zero solution). This means that the solution does not tend to zero. But then there exists the number $$ m= \inf\limits_{t\ge t_0}\, \dot V(x(t),y(t)),\quad m>0 $$ ($\dot V$ is positive definite, so $\dot V=0$ only at $(x,y)=0$).

For any solution $$ \forall t\ge t_0\quad V(x(t),y(t))=V(x_0,y_0)+\int_{t_0}^t \dot V(x(t),y(t))\, dt\ge V(x_0,y_0)+m(t-t_0). $$ Hence, $V(x(t),y(t))$ grows unlimited; this means that the solution eventually leaves any predetermined neighborhood of the origin, therefore, the system is unstable.

Answer 2

In Khalil, Hassan K. "Nonlinear Systems Third Edition." the Chetaev theorem is stated as follows:

Let $x=0$ be an equilibrium point of $\dot{x}=f(x)$. Let $V: D \to \mathbb{R}$ be a continuously differentiable function on a domain $D \subset \mathbb{R}^n$ such that $V(0)=0$ and $V(x_0)>0$ for some $x_0$ with arbitrarily small $\|x_0\|$. Choose $r > 0$ such that the ball $B_r = \left\{x \in \mathbb{R}^n\,|\,\|x\| \leq r\right\}$ is contained in $D$ and let $U = \left\{x \in B_r\,|\,V(x) > 0\right\}$ and suppose that $\dot{V}(x) > 0$ in $U$. Then, $x=0$ is unstable.

Your equilibrium point $x_0$ can always be translated to the origin, so for ease of notation I will keep assuming that it is at the origin. Translating your question to the context of this definition of the theorem one gets that $U=B_r\setminus\{0\}$. For this it can be noted that $r$ does not have to be chosen as large as possible, because instability of an equilibrium point is only a local property. It would actually be preferred to choose a relatively small $r$ to also avoid including other potential equilibria inside $B_r$. Therefore, $U$ will always be a bounded in size. So if $\dot{V}(x)>0$ in $U$ then you can still conclude that that equilibrium is unstable.

To better understand Chetaev's theorem it helps to know what each part actually implies. Namely, on the interior of $U$ both $V(x)>0$ and $\dot{V}(x)>0$. So any trajectory that enters $U$ will cause $V(x)$ to increase and will have to leave $U$ at some point and can not stay indefinitely in $U$, since $U$ is a bounded set and $V(x)$ is continuously differentiable (so doesn't have singularities). The boundary of $U$, along which trajectories can enter or leave, are the outside boundary of $B_r$ and $V(x)=0$ (which at least always includes the origin/equilibrium point). Since it holds that $V(x)>0$ and $\dot{V}(x)>0$ in $U$, thus any trajectory can't leave $U$ at boundaries obtained from $V(x)=0$ (because in order to reach any $V(x)=0$ from $V(x)>0$ requires $\dot{V}(x)<0$) and thus only trajectories leave the outside boundary of $B_r$. So in layman's terms Chetaev's theorem shows that certain trajectories close to an equilibrium point all move away from that equilibrium point.