Check if $\frac{x'^{2}}{2}-\sin(x)$ is a lyapunov function for $x''=\cos(x)-0.1x'$ near $(x,x')=(\frac{\pi}{2},0)$
I rewrote this into a system: $x'=y$, $y'=\cos(x)-0.1y$, $F(x,y)=\frac{y^2}{2}-\sin(x)$,
And want to show that this is lyapunov function near $(x,y)=(\frac{\pi}{2},0)$
And then $F'(x,y)=yy'-\cos(x)x'=y\cos(x)-0.1y^2-y\cos(x)=-0.1y^2$
which is strictly decreasing so the sign doesn't change and $(\frac{\pi}{2},0)$ is a critical point of my system.
So do I know this is a lyapunov function?
This is my definition of Lyapunov function:
I have this lyapunov stability theorem, Assume $U$ is an open neighbourhood of $x^*, F:U\to \mathbb{R}^n$ and $F(x^*)=0$. Further assume that there is a $C^1$-function $V:U\to\mathbb{R}$ satisfying $V(x^*)=0$, $V(x)>0$ for $x\neq x^*$. Then, V is called a lyapunov function for $x'=F(x)$ and:
a. If $V'(x)\leq 0$ for all $x\in U$, then $x^*$ is L-stable b. If $V'(x)<0$ for all $x\in U\{x^*\}$, then $x^*$ is asymptotically stable. c. If $V'(x)>0$ for all $x\in U\{x^*\}$ then $x^*$ is unstable.
Let's transform the ODE into an SDE:
$y_1=x,y_2=x'\Rightarrow y_1'=y_2,y_2'=x''=\cos(x)-0.1x'=\cos(y_1)-0.1y_2\Rightarrow\begin{cases}y_1'=y_2\\y_2'=\cos(y_1)-0.1y_2\end{cases}$
Then the function would be $V(y_1,y_2)=\dfrac{y_2^2}{2}-\sin(y_1)$. We need points near $(\frac{\pi}{2},0)$, so we could consider points $(y_1,y_2)=(\frac{\pi}{2}+r,0)$ with $|r|>0$ as small as we want. For those points $V(y_1,y_2)=-\sin(\frac{\pi}{2}+r)<0$ for $0<r<\frac{\pi}{2}$, so $V$ can't be a lyapunov function.