can someone give me a proof of http://nptel.ac.in/courses/101108047/module13/Lecture%2031.pdf page 15?
Suppose all conditions for asymptotic stability are satisfied. In addition to it, suppose $\exists$ constants $k_1,k_2,k_3,p:$ $$ k_1 \|x\|^p \le V(x) \le k_2 \|X\|^p$$ $$ \dot V(x) \le -k_3 \|X\|^p$$ Then the orgin X=0 is "exponentially stable". Moreover, if these conditions hold globally, then the origin X=0 is "globally exponentially stable".
I got a version of the time-dependend version where p=2 and the third condition $$\| V_x \| \le k_4|x|$$ is added. I ve got a proof of the second version, can somone explain me what is the difference, and which role does play p to swipe out the third condition? Thank you.
Greetings.
From the assumptions $$\dot{V}\leq -k_3\|X\|^p\leq -\frac{k_3}{k_2}V$$ Multiplying by $e^{(k_3/k_2)t} $ we have $$\frac{d}{dt}\big[V(X(t))e^{(k_3/k_2)t}\big]\leq 0$$ Integrating the above inequality over $[t_0,t]$ we obtain $$V(X(t))\leq V(X(t_0))e^{-(k_3/k_2)(t-t_0)}$$ and $$\|X(t)\|\leq \frac{V^{1/p}(X(t_0))}{k_1^{1/p}}e^{-(k_3/k_2p)(t-t_0)}$$ that completes the proof.