Let us consider the $1$D system: $ x'=-\sin x $ .We want to prove that it is locally exponentially stable around $0$ using the Lyapunov function: $V(x)= \frac{1}{2} x(t)^2.$
Attempt:
Step 1:
i) $ V $ is differentiable, $ V(0)=0 $ and $ V(x)>0$ for $x\neq 0. $
ii) $ \dot V(x) = -x\sin x \approx -x(t)^2 < 0 $ for $x\neq 0$ if $x(t)$ is sufficiently small.
With this, we conclude that $x=0$ is asymptotically stable if $x(t)$ is sufficiently small.
To see the exponential stability:
i) $ V $ is differentiable
ii) $ \frac{1}{2} \|x\|^2 \le V(x) \le \frac{1}{2} \|x\|^2 $
iii) $ \dot V(x) \le - \frac{1}{2} \|x\|^2$ if $x(t)$ is sufficiently small.
Step 2:
Now, we have to check that if $\|x(0) \| \le \epsilon $ (local stability)$ , \ \|x(t) \| \le \epsilon \ \ (\|x(t) \|$ is still small) . For this purpose, I am told to fix $\epsilon$ such that (*1) $\|\sin x\| \le \frac{1}{2}\|x\|$ $ \ $ if $ \ $ $\|x\| \le 2\epsilon.$
Let us prove first that $\|x(t) \| \le 2\epsilon$ $ \ $ if $ \ $ $\|x(0) \| \le \epsilon.$ By contradiction, let us suppose there exists $t_1$ such that $\|x(t_1)\| = 2\epsilon$. The function $\|x(t) \|$ $ \ $ is decreasing in the interval from $0$ to $t_1$ which contradicts the fact that $\|x(0) \| \le \epsilon$ and $\|x(t_1)\| = 2\epsilon$. Hence, $ \dot V(x) \le - \frac{1}{2} \|x\|^2 = -V(x)$ and then $V(x) \le V(x(0))e^{-t}.$ (*2)
Questions:
1)Is Step 1 well posed?
2)How can we conclude from (*1) that $\|x(t) \|$ is decreasing?
3)How can we finally conclude that $\|x(t) \| \le \epsilon$? Why do we need (*2)?