## 3 Stability of Limit Cycle
### Stability of Limit Cycle by Invariant Set Theorem  
Sometimes the equilibrium state of a dynamical system only has an equilibrium point, and sometimes the equilibrium state of a dynamical system contains both the equilibrium point and the limit cycle [72]. In the previous Section 2.1, we touched upon the limit cycle's  
Figure 8: Centre point. stability when we analyzed the stability of an equilibrium point. We looked at two different cases, i.e., stable limit cycle and unstable limit cycle. Here we will use a different case, i.e., a semi-stable limit cycle, to demonstrate how to use the invariant set theorem to determine the stability of the limit cycle. Consider the following dynamical system:  
\[\left\{ \begin{array}{l}\dot{x}_{1}=x_{2}-x_{1}({x_{1}}^{2}+{x_{2}}^{2}-1)^{2} \\ \dot{x}_{2}=-x_{1}-x_{2}({x_{1}}^{2}+{x_{2}}^{2}-1)^{2}\end{array} \right. \tag{15}\]  
where the state is \(\underline{x}=\left[ \begin{array}{c}x_{1}\\ x_{2}\end{array} \right]\). In order to see the entire equilibrium state of this system, we can use the global invariant set theorem. First of all, we can choose the Lyapunov function as follows:  
\[V(\underline{x})=\left({x_{1}}^{2}+{x_{2}}^{2}-1\right)^{2} \tag{16}\]  
and  
\[ \begin{array}{ll}\dot{V}(\underline{x})&=2({x_{1}}^{2}+{x_{2}}^{2}-1)(2x_{1 }\dot{x}_{1}+2x_{2}\dot{x}_{2})\\ &=-4({x_{1}}^{2}+{x_{2}}^{2})({x_{1}}^{2}+{x_{2}}^{2}-1)^{3}\end{array}  \tag{17}\]  
If \({x_{1}}^{2}+{x_{2}}^{2}>1\), geometrically, it means that, if we are outside of the cycle \({x_{1}}^{2}+{x_{2}}^{2}=1\), then \(\dot{V}(\underline{x})<0\). Based on the global invariant set theorem, we can conclude that \(\dot{V}(\underline{x})=0\) as time goes on, which indicates that \({x_{1}}^{2}+{x_{2}}^{2}=1\) or \(x_{1}=0\), \(x_{2}=0\). This means that, starting from outside of the cycle \({x_{1}}^{2}+{x_{2}}^{2}=1\), all trajectories tend towards the cycle or the origin. If, however, \({x_{1}}^{2}+{x_{2}}^{2}<1\), geometrically, it means that, if we are inside of the cycle \({x_{1}}^{2}+{x_{2}}^{2}=1\), then \(\dot{V}(\underline{x})>0\), which indicates that \(V(\underline{x})\) increases as time goes on. If we plot the Lyapunov function as a function of the state, as shown in Figure 9, we can see that inside of the cycle, the trajectories tend to the origin if there is a small perturbation. Conversely, outside of the cycle, all trajectories tend to the limit cycle, as shown in Figure 9c. The valley in Figure 9a is the limit cycle, and it can be seen from the top view as indicated in the Figure 9b and therefore, the limit cycle is semi-stable or half stable. Again, from Figure 9c, we can see that the blue arrows are approaching the equilibrium point from two sides, and so the origin is stable; whereas one set of blue arrows are approaching to the limit cycle and the other set of the blue arrows are getting away from the limit cycle, so the limit cycle is semi-stable. Figure 9: Plot of the Lyapunov candidate function that shows the limit cycle. Now, if we choose the below Lyapunov candidate function, as shown in Equation (18), it will not be able to capture the right dynamics of the system and it will not be able to demonstrate the existence of the limit cycle. \[V(\underline{x})={x_{1}}^{2}+{x_{2}}^{2} \tag{18}\]  
And  
\[ \begin{split}\dot{V}(\underline{x})&=2x_{1}\dot{x} _{1}+2x_{2}\dot{x}_{2}\\ &=-2({x_{1}}^{2}+{x_{2}}^{2})({x_{1}}^{2}+{x_{2}}^{2}-1)^{2} \end{split}  \tag{19}\]  
The \(\dot{V}(\underline{x})\) is always negative. Based on the [PERSON] global invariant set theorem, we can conclude that \(\dot{V}(\underline{x})=0\) as time goes on, which indicates that \({x_{1}}^{2}+{x_{2}}^{2}=1\), \(x_{1}=0\), \(x_{2}=0\) as time goes on. However, when we plot the Lyapunov candidate function \(V(\underline{x})\), there is no existence of the limit cycle, as shown in the Figure 10. The dashed line is the limit cycle. The plot shows the 3D view, the side view and the top view. The trajectories are supposed to converge to the limit cycle from outside of the cycle \({x_{1}}^{2}+{x_{2}}^{2}=1\).