0

Consider the dynamical equation $\dot{x} = f(x)$, where $f:\mathbb{R}_+ \rightarrow \mathbb{R}_+$.

  1. Is $V(x) = x^{1+\alpha}$, where $\alpha \in (0,1)$ a Lyapunov function?
  2. Since $x(t) \ge 0$ for all $t$, would a function $V(y) \not\in \mathbb{R}_+$ for $y \not\in \mathbb{R}_+$ still be a Lyapunov function?
SEJ
  • 37
  • 7
  • what have you tried so far? – Student Nov 28 '20 at 12:34
  • One example: $V(x) = x^{1.5}$. This $V$ is continuous on $(0,\infty)$ and its derivative is as well. They are both undefined (well, not real) for $x < 0$. Is such a $V$ still a Lyapunov function? – SEJ Nov 28 '20 at 12:53

0 Answers0