Consider the pendulum system with time-varying friction: $$\dot{x_1}=x_2$$ $$\dot{x_2}=-\sin{x_1}-g(t)x_2$$ where, $g(t)$ is a $C^1$ function satisfying $$0<a<\alpha\leq g(t) \leq \beta<\infty$$ $$\dot{g(t)}\leq \gamma<2,\ \ \forall t\in \mathbb{R}^{+}$$ Now, considering the Lyapuniv function candidate: $$V(t,x)=\frac{(x_2+a\sin{x_1})^2}{2}+(1+ag(t)-a^2)(1-\cos{x_1})$$ It can be shown that $V$ is positive definite and decrescent for $|x_1|<\frac{\pi}{2}$.
Show that $$\dot{V}\leq -(\alpha-a)x_2^2-a(2-\gamma)(1-\cos{x_1})+O(||x||^3)$$ where, $O(||x||^3)$ is a term bounded by $k||x||^3$ in some neighborhod of the origin.
My solution: $$\dot{V}=-(g(t)-a\cos {x_1})x_2^2-a(2-\dot{g(t)})(1-\cos {x_1})-a(ax_2\sin{x_1}+\cos{x_1}-1)(1-\cos {x_1})$$ Therefore: $$\dot{V}\leq -(\alpha-a)x_2^2-a(2-\gamma )(1-\cos {x_1})+h(x)$$ $$h(x)=-a(ax_2\sin{x_1}+\cos{x_1}-1)(1-\cos {x_1})$$ Now, how we can show $h(x)=O(||x||^3)$? Moreover, how to prove that $\dot{V}$ is negative definite?
In order to show that $h(x)=O(\| x \|^3)$ use the Taylor expansion around the origin; $$h(x)=-\frac{a}{2} x_1^3(ax_2-\frac{a}{6}x_2x_1^3-\frac{x_1^2}{2}+h.o.t)$$ Hence, in a bounded neighborhood of the origin one gets: $$|h(x)|\leq k|x_1|^3\leq k \|x\|^3$$ for some $k\in \mathbb{R}^{+}$.
To show that $\dot{V}$ is negative definite, first we prove that the following function $$W(x)=(\alpha -a)x_2^2+a(2-\gamma )(1-\cos{x_1})-h(x)$$ is positive definite. Then, considering the fact that $$\dot{V}(t,0)=0,\ \forall t\geq 0$$ $$\dot{V}(t,x)\leq -W(x)$$ the negative definiteness of $\dot{V}$ can be inferred. The positive definiteness of $W$ follows from the fact that $x=0$ is indeed, a strict local minimizer since: $$W(0)=0$$ $$\nabla W(0)=0$$ $$\nabla^2 W(x)\ \text{is continuous and}\ \nabla^2 W(0)>0$$ This means that $W(x)$ is postive definite and $\dot{V}$ is negative definite indicating local asymptotic stability of the origin.