## 2 Stability of the Equilibrium Point  
### Stability Analysis by Lyapunov-LaSalle Energy-Based Approach  
Before getting into demonstrating how to use the Lyapunov theorem to prove the stability of an equilibrium point of a two-dimensional nonlinear dynamical system, it is necessary to reiterate the [PERSON] theorem here [64; 65]. For the [PERSON] local theorem, it states as follows: \"for a chosen Lyapunov candidate function \(V(\underline{x})\), if \(V(\underline{x})\) is positive definite and \(\dot{V}(\underline{x})\) is negative semi definite, the equilibrium point being analyzed is stable. Furthermore, if the \(\dot{V}(\underline{x})\) is negative definite, the equilibrium point is asymptotically stable\". It is noted that the underscore represents a vector. For the [PERSON] global theorem, it states as follows: \"for a chosen Lyapunov candidate function \(V(\underline{x})\), if \(V(\underline{x})\) is positive definite, and \(\dot{V}(\underline{x})\) is negative definite, and \(V(\underline{x})\) tends to infinity as the state tends to infinity, the equilibrium point being analyzed is globally asymptotically stable\". Consider the pendulum in Figure 1, and the dynamical equation of the pendulum is written below:  
\[ml^{2}\ddot{\theta}+b\dot{\theta}+mgl\sin\theta=0 \tag{1}\]  
In order to show the stability of the equilibrium point \(\theta=\left[ \begin{array}{c}\theta\\ \dot{\theta}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\), a Lyapunov function is chosen as below: \(V(\underline{x})=\frac{1}{2}ml^{2}\dot{\theta}^{2}+mg(l-l\sin\theta)\). It can be shown that the Lyapunov function is positive definite as \(V(\underline{x})>0\) except \(V(\underline{0})=0\). Additionally, \(\dot{V}(\underline{x})=-b\dot{\theta}^{2}\) is negative semi-definite because it does not contain any angular position \(\theta\) information. Based on the [PERSON] local theorem, it is concluded that the equilibrium point is stable. It is noted that here we can only use the local version, not the global version of the Lyapunov theorem, as \(\dot{V}(\underline{x})=-b\dot{\theta}^{2}\) is not negative definite. The fact of the matter is 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 actually globally asymptotically stable, which can be determined just by observing how the pendulum behaves as time goes on [66; 67]. As we can imagine, the pendulum will eventually come back to the equilibrium point, that is to say, the equilibrium point is at least asymptotically stable. However, by using the Lyapunov local theorem, we cannot prove this, which is why the LaSalle theorem comes in. The LaSalle global invariant set theorem states that: \"for a chosen Lyapunov candidate function \(V(\underline{x})\), if \(\dot{V}(\underline{x})\leq 0\) always, and \(V(\underline{x})\) tends to infinity as the state \(\underline{x}\) goes to infinity, 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\". Figure 1: Pendulum. Now let us use the [PERSON] global invariant set theorem to prove that the equilibrium point \(\theta=\left[ \begin{array}{c}\theta\\ \dot{\theta}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\) of the previous pendulum case is globally asymptotically stable. As same before, we choose the Lyapunov function as \(V(\underline{x})=\frac{1}{2}ml^{2}\dot{\theta}^{2}+mg(l-l\sin\theta)\), and after calculation of the time derivative of \(V(\underline{x})\), we know that \(\dot{V}(\underline{x})=-b\dot{\theta}^{2}\), which is always negative or zero, i.e., \(\dot{V}(\underline{x})\leq 0\). It is also true that the function \(V(\underline{x})\) will go to infinity when the state \(\theta=\left[ \begin{array}{c}\theta\\ \dot{\theta}\end{array}