## 2 Stability of the Equilibrium Point
### Stability Analysis by Lyapunov-LaSalle Energy-Based Approach
\right]\) goes to infinity and therefore, the [PERSON] global invariant set theorem conditions are satisfied, which indicates that \(\dot{V}(\underline{x})=0\). If we just stop here, we cannot prove the equilibrium point is globally asymptotically stable, we need to further look at the fact that when \(\dot{V}(\underline{x})=0\), this indicates that \(\dot{\theta}=0\), which further indicates that \(\ddot{\theta}=-g\sin\theta\). In order to prove that the equilibrium point \(\theta=\left[ \begin{array}{c}\theta\\ \dot{\theta}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\) is globally asymptotically stable, we need to have the small-angle assumption [68], i.e., \(\sin\theta\approx\theta\) when \(\theta\) is very small, in this way, \(\ddot{\theta}=-g\cdot\theta\), which means the angular acceleration is not zero, except when we reach the equilibrium point at \(\theta=0\). Therefore, the pendulum always has an acceleration, except when it reaches to the equilibrium point and stops there [69]. Hence, the equilibrium point \(\theta=\left[ \begin{array}{c}\theta\\ \dot{\theta}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\) is globally asymptotically stable. As one can see from the above analysis because we made the small-angle assumption, we proved that equilibrium point is globally asymptotically stable. If, however, we do not make this assumption, i.e., \(\sin\theta\
eq\theta\), then we cannot prove that the equilibrium point is globally asymptotically stable by using the LaSalle global invariant set theorem. The reason is that when \(\ddot{\theta}=-g\cdot\sin\theta=0\), this indicates \(\theta=k\pi\) instead of zero. This is the flaw in the [PERSON] global invariant set theorem. In the subsequent section, we will show that by, using the graphical phase plane method, we do not have to make this small-angle assumption in order to prove that the equilibrium point is globally asymptotically stable. In the meantime, the graphical phase plane method is also able to demonstrate how the pendulum behaves when there is a very large angle [61], which cannot be demonstrated by using the analytical Lyapunov-LaSalle approach. The [PERSON] local invariant set theorem states that: \"for a chosen Lyapunov candidate function \(V(\underline{x})\), if \(\dot{V}(\underline{x})\leq 0\) always, and there exist a value \(h\) where \(V(\underline{x})\) is in a bounded region defined by \(V(\underline{x})<h\), then all trajectories tend to the largest invariant set \(\left\{\underline{x},\dot{V}(\underline{x})=0\right\}\) and stay within this set for all future times\". Typically, the [PERSON] local invariant set theorem is used to prove whether a limit cycle is stable or not. Consider the following two different examples. **Case 1**: consider the following dynamical system, and the dynamical equation is given as follows:  
\[\left\{ \begin{array}{l}\dot{x}_{1}=x_{2}+x_{1}({x_{1}}^{2}+{x_{2}}^{2}-1)\\ \dot{x}_{2}=-x_{1}+x_{2}({x_{1}}^{2}+{x_{2}}^{2}-1)\end{array} \right. \tag{2}\]  
where the state is \(\underline{x}=\left[ \begin{array}{c}x_{1}\\ x_{2}\end{array} \right]\), and we want to find the equilibrium state (equilibrium point and/or limit cycle if there exists one) of this dynamical system. First of all, we can choose the Lyapunov candidate function of \(V(\underline{x})=\left({x_{1}}^{2}+{x_{2}}^{2}-1\right)^{2}\), and by looking at the time derivative of the \(V(\underline{x})\), we can determine how \(V(\underline{x})\) changes as time goes on. \[ \begin{array}{ll}\dot{V}(\underline{x})&=2({x_{1}}^{2}+{x_{2}}^{2}-1)(2{x_{1}}{ \dot{x}_{1}}+2{x_{2}}{\dot{x}_{2}})\\ &=2({x_{1}}^{2}+{x_{2}}^{2}-1)(2{x_{1}}({x_{2}}+{x_{1}}({x_{1}}^{2}+{x_{2}}^{2}- 1))+2{x_{2}}(-{x_{1}}+{x_{2}}({x_{1}}^{2}+{x_{2}}^{2}-1)))\\ &=4({x_{1}}^{2}+{x_{2}}^{2})({x_{1}}^{2}+{x_{2}}^{2}-1)^{2}>0\end{array}  \tag{3}\]  
As \(\dot{V}(\underline{x})>0\), it means \(V(\underline{x})\) increases as time goes on.