# Using Interval Analysis to Compute the Invariant Set of a Nonlinear Closed-Loop Control System
## 4 Computation of the Invariant Set of the Closed-Loop Control System  
Once the controller \(u(\mathbf{x})\) in Equation (12) is found, it should be shown that the system that converges to the sliding surface \(s(\mathbf{x})=0\) that is characterized by (7) stays on the surface for all times. This is the definition of an invariant set due to La Salle. If a trajectory enters an invariant set, it will stay inside this set forever [33; 40]. To find this set, the switching function \(s(\mathbf{x})\) is used as a Lyapunov-like function \(\tilde{V}(\mathbf{x})=s(\mathbf{x})\), which does not necessarily comply with all requirements of a Lyapunov function, as its derivative may also be zero for states \(\mathbf{x}\), where \(V=\frac{1}{2}s^{2}(\mathbf{x})\
eq 0\). Moreover, the function \(\tilde{V}(\mathbf{x})\) may exhibit a change of sign. Under this preliminary consideration of the characteristics of invariant sets, states are not part of such sets if they are related to an inflection point in the form of a saddle point of the function \(V(\mathbf{x})\) (see Figure 1). In fact, it suffices if any finite time derivative of \(V(\mathbf{x})\) is not equal to zero for a point not belonging to the invariant set. This can also be confirmed when thinking of the Barbalat lemma, which states that, if a function has a finite limit for \(t\to\infty\), its uniformly continuous derivative will tend to zero [40]. This can also be applied for the first derivative being the function and the second derivative being its first derivative, and so on:  
\[ \begin{split}\lim_{t\to\infty}V(\mathbf{x})=c& \implies\lim_{t\to\infty}\tilde{V}(\mathbf{x})=0\implies\\ \lim_{t\to\infty}\tilde{V}(\mathbf{x})=0&\implies \lim_{t\to\infty}\tilde{V}(\mathbf{x})=0\dots,\\ &\text{where $c$ is a constant.}\end{split}  \tag{22}\]  
Using the fact that the desired sliding surface \(s(\mathbf{x})\) itself and the first three derivatives of \(s(\mathbf{x})\) have to be zero as a necessary condition if \(V(\mathbf{x})=0\) holds constantly, the invariant limit set of the system can be found. This statement can be verified by means of the following relations:  
\[V(\mathbf{x})=\frac{1}{2}s^{2}(\mathbf{x})=0\ \implies\ s(\mathbf{x})=0 \tag{23}\] \[\dot{V}(\mathbf{x})=s(\mathbf{x})\cdot\dot{s}(\mathbf{x})=0\ \implies\ s(\mathbf{x})=0 \tag{24}\]  
satisfied due to (23)  
\[\ddot{V}(\mathbf{x})=s(\mathbf{x})\cdot\ddot{s}(\mathbf{x})+\dot{s}^{2}(\mathbf{x})=0\ \implies\ s(\mathbf{x})=0 \tag{25}\]  
satisfied due to (23) \(\wedge\)\(\delta(\mathbf{x})=0\)  
\[\ddot{V}(\mathbf{x})=s(\mathbf{x})\cdot\ddot{s}(\mathbf{x})+3s(\mathbf{x})\cdot\ddot{s}(\mathbf{x} )=0\ \implies\ s(\mathbf{x})=0\ \wedge\ \dot{s}(\mathbf{x})=0. \tag{26}\]  
satisfied due to (23) and (25). Requiring furthermore that \(\ddot{s}(\mathbf{x})=0\) and \(\dddot{s}(\mathbf{x})=0\) hold, all further higher order derivatives of \(V(\mathbf{x})\) up to order seven are directly proven to be equal to zero. Moreover, accounting for the fact that \(s(\mathbf{x})=-s(\mathbf{x})\cdot p(\mathbf{x})\) holds by construction (see Equation (7)), the relations (23)-(25) above turn into:  
\[V(\mathbf{x})=\frac{1}{2}s^{2}(\mathbf{x})=0\ \implies\ s(\mathbf{x})=0 \tag{27}\] \[\dot{V}(\mathbf{x})=-s^{2}(\mathbf{x})\cdot p(\mathbf{x})=0\ \implies\ s(\mathbf{x})=0 \tag{28}\]  
satisfied due to (27)  
\[\ddot{V}(\mathbf{x})=2s^{2}(\mathbf{x})\cdot p^{2}(\mathbf{x})-s^{2}(\mathbf{x})\cdot\dot{p}( \mathbf{x})=0\ \implies\ s(\mathbf{x})=0, \tag{29}\]  
satisfied due to (27), where \(\ddot{V}(\mathbf{x})\) in (29) can be rewritten into:  
\[\ddot{V}(\mathbf{x})=s^{2}(\mathbf{x})\cdot\left(2p^{2}(\mathbf{x})-\dot{p}(\mathbf{x})\right). \tag{30}\]  
Hence, the equivalent control designed in this paper guarantees that all temporal derivatives of \(V(\mathbf{x})\) are perfectly equal to zero as soon as \(s(\mathbf{x})=0\) has been reached.