## 2 Stability of the Equilibrium Point
### Stability Analysis by Linearizing around an Equilibrium Point  
The third approach to determine the stability of an equilibrium point is to linearize around the equilibrium point to see how its neighbours behave as time goes on [70; 71]. Here, we will use the previous pendulum case as an example, and we will show the case with damping and the case without damping. This method needs to be used with care as only the following cases can the linearization give qualitatively correct dynamics near the equilibrium point: a saddle, node, and spiral or sometimes referring as focus [61], whereas if the equilibrium point is a centre, star, or non-isolated point, then the linearized result cannot tell you if the equilibrium point is truly stable or not, which is the potential problem  
Figure 5: Vector field when \(x=0\). Figure 6: Entire vector field when damping coefficient \(b\) is zero. of using the linearizing-around-an equilibrium-point method to determine the stability of an equilibrium point. With damping present, we know that the dynamical equation is \(ml^{2}\ddot{x}+b\dot{x}+mgl\sin x=0\). In order to linearize around the equilibrium point \(\underset{-}{x}=\left[ \begin{array}{c}x\\ \dot{x}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\) using the Taylor formula, we will first need to convert the above equation into the following equivalent and assume the length \(l=1\) for the ease of calculation:  
\[\left\{ \begin{array}{c}\dot{x}=y\\ \dot{y}=-\frac{b}{m}y-g\sin x\end{array} \right. \tag{7}\]  
Considering a small deviation from the equilibrium point \(\underset{-}{x}=\left[ \begin{array}{c}x\\ \dot{x}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\) as follows:  
\[u(t)=x(t)-0 \tag{8}\]  
\[v(t)=y(t)-0 \tag{9}\]  
In order to look at how the small deviation changes as time goes on, we need to look at the time derivative of \(u(t)\) and \(v(t)\) as follows:  
\[\dot{u}(t)=\dot{x}(t)=y \tag{10}\]  
\[\dot{v}(t)=\dot{y}(t)=-\frac{b}{m}y-g\sin x \tag{11}\]  
By using the [PERSON] expansion formula, we can linearize the above equations, which leads to the following:  
\[\dot{u}=f(0,0)+u\frac{\partial\dot{x}}{\partial x}|_{(0,0)}+v\frac{\partial \dot{x}}{\partial y}|_{(0,0)} \tag{12}\]  
\[\dot{v}=f(0,0)+u\frac{\partial\dot{y}}{\partial x}|_{(0,0)}+v\frac{\partial \dot{y}}{\partial y}|_{(0,0)} \tag{13}\]  
We noticed that the higher-order terms are neglected here as they are extremely small compared to the first few terms. By further converting the above equations into a matrix format, we have the following:  
\[\left[ \begin{array}{c}\dot{u}\\ \dot{v}\end{array} \right]= \begin{bmatrix}\frac{\partial\dot{x}}{\partial x}& \frac{\partial\dot{x}}{\partial y}\\ \frac{\partial\dot{y}}{\partial x}&\frac{\partial\dot{y}}{\partial y}\end{bmatrix}  |_{(0,0)}\left[ \begin{array}{c}u\\ v\end{array} \right]=J\left[ \begin{array}{c}u\\ v\end{array} \right] \tag{14}\]  
where \(J\) is the Jacobian matrix. As we can see, whether the equilibrium point is stable or not depends on the Jacobian matrix, and particularly depends on the eigenvalues of the Jacobian matrix. If all eigenvalues are negative, that indicates the small deviation will  
Figure 7: Entire vector field when damping coefficient \(b\) is not zero. decrease as time goes on, which further indicates that the equilibrium point is stable. To further generalize the whole scenarios including, when eigenvalues are not negative, we can use the trace and determinant of the Jacobian matrix to determine if the equilibrium point is stable or not. By looking at the Jacobian matrix that evaluated at the equilibrium point, i.e., at \(x=\left[ \begin{array}{c}x\\ \dot{x}\end{array} \right]=\left[ \begin{array}{c}0\\ 0\end{array} \right]\), the Jacobian matrix is \(J=\left[ \begin{array}{cc}0&1\\ -g\cos x&-\frac{b}{m}\end{array} \right]_{|(0,0)}=\left[ \begin{array}{cc}0&1 \\ -g&-\frac{b}{m}\end{array} \right]\). As we can see that when \(b=0\), i.e., no damping, the Jacobian matrix is \(J=\left[ \begin{array}{cc}0&1\\ -g&0\end{array}