## 3 Stabilization Problem
### Main Result  
The error dynamic is given by  
\[ \begin{array}{ll}\frac{de(t)}{dt}&=\frac{dx(t)}{dt}&=A_{0}x(t)+B_{0}(x(t)) \lambda(t)\\ &=A_{0}e(t)+\left[B_{0}(x(t))\lambda(t)+A_{0}x_{e}\right]\\ &=A_{0}e(t)+\left[B_{0}(x(t))\lambda(t)-B_{0}(x_{e})\lambda_{e}\right]\end{array}  \tag{24}\]  
Because \(A_{0}\) is Hurwitz, given a positive definite symmetric matrix \(Q\in\mathbb{R}^{n\times n}\), there always exists a positive definite symmetric matrix \(P\in\mathbb{R}^{n\times n}\) satisfying the following Lyapunov inequality  
\[A_{0}^{T}P+PA_{0}+2Q<0 \tag{25}\]  
To solve the stabilization problem, we introduce the following important assumption. **Assumption A2**. : _Given \(x_{e}\in X_{eq}\) and \(x(t)\in\mathbb{R}^{n}\), \(x(t)\
eq x_{e}\), there exist \(\lambda(t)\in\Lambda_{R}\), \(\Lambda_{R}\) defined in (14) and a positive definite symmetric matrix \(Q_{1}\), \(0<Q_{1}\leq Q\) such that_  
\[e(t)^{T}P[B_{0}(x(t))\lambda(t)-B_{0}(x_{e})\lambda_{e}]<-e(t)^{T}P\Lambda_{0} e(t)-e(t)^{T}Q_{1}e(t) \tag{26}\]  
Assumption A2 ensures that there exists a control \(\lambda:\mathbb{R}\to\Lambda_{R}\) which stabilizes the equilibrium point \(x_{e}\) for (13). This assumption is always satisfied for a large class of converters, as shown in the following lemma. **Lemma 1**. : _We have the following properties:_  
1. _For a given positive definite symmetric matrix_ \(Q\)_, suppose that there exists a positive definite symmetric matrix_ \(S_{\lambda_{e}}\in\mathbb{R}^{n\times n}\)_such that_ \[\sum_{i=1}^{M}\lambda_{e_{i}}(A_{0}+A_{i})^{T}S_{\lambda_{e}}+S_{\lambda_{e}} \sum_{i=1}^{M}\lambda_{e_{i}}(A_{0}+A_{i})+2Q<0,\ \ \sum_{i=1}^{M}\lambda_{e_{i}}=1\] (27) _Then, Assumption A2 is satisfied with_ \(P=S_{\lambda_{e}}\)_._
2. _For a given positive definite symmetric matrix_ \(Q\)_, suppose that there exists a positive definite symmetric matrix_ \(S\in\mathbb{R}^{n\times n}\)_such that_ \[(A_{0}+A_{i})^{T}S+S(A_{0}+A_{i})+2Q<0,\ \ \ i=1,\ldots,M\] (28) _Then, Assumption A2 is satisfied with_ \(P=S\)_._  
Proof. : Suppose (27) is satisfied by a matrix \(S_{\lambda_{e}}\). Then,  
\[e^{T}S_{\lambda_{e}}[B_{0}(x)\lambda-B_{0}(x_{e})\lambda_{e}] =e^{T}S_{\lambda_{e}}[B_{0}(x)(\lambda-\lambda_{e})+(B_{0}(x)-B_ {0}(x_{e}))\lambda_{e}]\] \[=e^{T}S_{\lambda_{e}}[B_{0}(x)(\lambda-\lambda_{e})+\left[ \begin{array} []{cc}A_{1}e&\cdots&A_{M}e\end{array} \right]\lambda_{e}]\] \[=e^{T}S_{\lambda_{e}}B_{0}(x)(\lambda-\lambda_{e})+e^{T}S_{ \lambda_{e}}\sum_{i=1}^{M}\lambda_{e_{i}}(A_{0}+A_{i})e-e^{T}S_{\lambda_{e}}A_ {0}e\]  
For \(\lambda=\lambda_{e}\), we have  
\[e^{T}S_{\lambda_{e}}A_{0}e+e^{T}S_{\lambda_{e}}[B_{0}(x)\lambda_{e}-B_{0}(x_{ e})\lambda_{e}])=e^{T}S_{\lambda_{e}}\sum_{i=1}^{M}\lambda_{e_{i}}(A_{0}+A_{i})e<-e^ {T}Qe \tag{29}\]  
from (27). By continuity arguments, there always exists a neighborhood of \(\lambda_{e}\) such that  
\[e^{T}S_{\lambda_{e}}A_{0}e+e^{T}S_{\lambda_{e}}[B_{0}(x)\lambda-B_{0}(x_{e}) \lambda_{e}])=e^{T}S_{\lambda_{e}}\sum_{i=1}^{M}\lambda_{e_{i}}(A_{0}+A_{i})e +e^{T}S_{\lambda_{e}}B_{0}(x)(\lambda-\lambda_{e})<-e^{T}Qe \tag{30}\]  
then Assumption A2 is satisfied and \((i)\) follows. The proof of \((ii)\) is obtained remarking that, if \(S\) satisfies (28), it satisfies (27) for all \(\lambda_{e}\in\Lambda_{R}\), meaning that condition (28) is stronger than condition (27). We remark that, in Lemma 1, Assumption A2 is satisfied for a matrix \(S_{\lambda_{e}}\) or \(S\) satisfying, respectively, inequalities (27) or (28). Such non-unique matrices can be found using an LMI solver. An important fact is that, in the context of Lemma 1, the existence of \(\lambda\) satisfying Assumption A2 is independent of the selected matrices \(Q\) and \(S_{\lambda_{e}}\) (or \(S\)), only the satisfaction of (27) or (28) being needed. However, it is possible to select them to guarantee some optimality properties, as shown below. It is also worth remarking that, if there exists a matrix \(S\) satisfying (28), it also satisfies (25). This can be deduced by remarking that \(A_{1}=0\).