## 2 Stability of the Equilibrium Point
### Stability Analysis by Graphical Phase Plane Approach  
Another way to demonstrate the stability of an equilibrium point and a limit cycle is to use the graphical phase plane approach. Here, the graphical phase plane approach will be used to address the same pendulum case as we discussed in the previous section, and the advantages of using such a graphical approach will be shown.  
As shown before, we know that the dynamical equation for the pendulum in the previous section is given as \(ml^{2}\ddot{\theta}+b\dot{\theta}+mgl\sin\theta=0\). To generalize the dynamical equation and for the ease of analysis, let us replace \(\theta\) with \(x\) and assume that the length of the pendulum is 1, so the equation of motion is reduced to \(m\ddot{x}+b\dot{x}+mg\sin x=0\). Let \(\dot{x}=y\), so \(\dot{y}=\ddot{x}=-\frac{b}{m}\dot{x}-\frac{\dot{g}}{m}\sin x\). The equivalent model for the dynamical equation can be written as below:  
\[\left\{\begin{array}{l}\dot{x}=y\\ \dot{y}=-\frac{b}{m}y-g\sin x\end{array}\right. \tag{6}\]  
Now the state or the input is \(\left[\begin{array}{l}x\\ y\end{array}\right]\), and the output is \(\left[\begin{array}{l}\dot{x}\\ \dot{y}\end{array}\right]\). We will first draw the vector field for the case when \(\dot{x}=0\), and then plot the case when \(x=0\), and finally we will combine the above two cases to get the entire vector field. So, when \(\dot{x}=0\), \(\dot{y}=-g\sin x\), the vector field is plotted as follows, as shown in Figure 4.  
We then draw the vector field for the case when \(x=0\). When \(x=0\), \(\dot{y}=-\frac{b}{m}\dot{x}\). Let us first draw the case when the damping coefficient \(b\) is zero, as shown in Figure 5, then we can easily plot the case when \(b\) is not zero.  
Figure 4: Vector field when \(\dot{x}=0\).  
Figure 3: Plot of the Lyapunov candidate function of case 2.  
If we combine the above two plots together, we have the following vector field, as shown in Figure 6, and immediately we can sense the direction of the whole vector field.  
The above vector field is for the case when the damping coefficient \(b\) is zero, as we can see that the vectors form an infinite number of closed orbits, and each of those closed orbits represents how the pendulum behaves as time goes on. Additionally, it can be seen that the trajectories will not move back to the origin but also not move away from the origin, which indicates that the equilibrium point \(\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\) is stable but not asymptotically stable. Now when the damping coefficient \(b\) is not zero, in this case, all the arrows on the \(y\)-axis that are above the origin will point downwards and all the arrows on the \(y\)-axis that are below the origin will point upwards, as shown in Figure 7, and the slope of the downwards and upwards arrows depends on the damping coefficient \(b\). The larger the value of \(b\), the larger the slope will be, and vice versa. In this case, as one can see that when the damping coefficient \(b\) is not zero, the trajectories will tend towards the origin as time goes on, which indicates that the equilibrium point \(\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\) is asymptotically stable. On the contrary, if the damping coefficient \(b\) is not just nonzero, but also negative, in this case the trajectories will move away from the origin, which means the equilibrium point \(\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\) will become unstable.