0

sadly I wasn't able to find a 'control systems' community, so I'm posting here. I'm working through the book of H. Khalil - Nonlinear Systems and I'm stuck at the following part:

He states the following:

$$\Omega_C = \{x \in \mathbb{R}^n | V(x) \le c \} $$

For $\Omega_C$ to be in the interior of a ball $B_r$, $c$ must satisfy $c < \inf_{||x||\ge r} V(x)$. If $$l = \lim_{r\to \infty} \inf_{||x|| \ge r} V(x) < \infty$$ then $\Omega_c$ will be bounded if $c<l$.

Can anyone explain this to me in simpler words. Khalil also gives the following example:

$$l = \lim_{r\to\infty} \min_{||x||=r} \left( \frac{x_{1}^2}{1+x_{1}^2} + x_2^2 \right) = \lim_{|x_1|\to\infty} \frac{x_1^2}{1+x_1^2} = 1 $$

where I also don't understand how he evaluates this.

Another big question for me is the claim, that $V(x)$ needs to be radially unbounded, hence,

$$V(x) \to \infty ~ \text{as} ~ ||x|| \to \infty $$

that the equilibrium point is globally asymptotically stable.

I'm confused regarding this last statement because at the top I want that $V(x)$ fits into a 'ball' to be compact. But if $V(x)\to\infty$ it's not compact anymore.

I would be very grateful for your explanations and help.

  • A possible approach for the evaluation of the limit: introduce polar coordinates and minimise explicitly -- one finds that the minimum is at $\theta\in2\pi\mathbb Z$. – AccidentalFourierTransform Aug 11 '19 at 17:16
  • That's many questions at once... But at least some of them should be answered here: https://math.stackexchange.com/questions/2472608/system-with-a-lyapunov-function-over-mathbbrn-but-not-globally-asymptotica – Hans Lundmark Aug 12 '19 at 04:51
  • Thanks. One of the biggest questions from above is, why it is necessary that $V(x)$ needs to be radially unbounded that the level set is compact. I mean if the function goes to infinity, it's not compact anymore. So what do I get wrong? – Steradiant Aug 12 '19 at 11:19
  • Sorry for replying late, but since you didn't ping me I didn't notice your comment. It doesn't make sense to say that the function $V$ is compact. Compactness is a property of sets (or topological spaces), in this case the sets $\Omega_C$. If $V$ tends to infinity, then every $\Omega_C$ must be compact, since whatever $C$ you take, $V(x)$ will be greater than $C$ if you're sufficiently far away from the origin. If some $\Omega_C$ is not compact, then there are points arbitrarily far out where $V(x) \le C$, so then $V$ doesn't tend to infinity. – Hans Lundmark Jun 12 '21 at 13:52

0 Answers0