## 3 Stabilization Problem
### Integral Action
To extend the stabilization problem taking into account an integral action, introduce the following assumption that, in that context, is very similar to Assumption A2. First, introduce the following matrix  
\[\mathcal{Q}=\left[ \begin{array}{cc}Q&0\\ 0&0\end{array} \right] \tag{50}\]  
**Assumption A3**. : _Given \(X_{e}\in\mathcal{X}_{e}\) and \(X(t)\in\mathbb{R}^{n+p},\ X(t)\
eq X_{e}\), there exist \(\lambda(t)\in\Lambda_{R}\) and a positive definite symmetric matrix \(\mathcal{Q}_{1}\), \(0<\mathcal{Q}_{1}\leq Q\) such that_  
\[e_{I}{}^{\mathrm{T}}(t)\mathcal{P}[\mathcal{B}_{0}(X(t))\lambda-\mathcal{B}_{0 }(X_{e})\lambda_{e}]\leq-e_{I}{}^{\mathrm{T}}(t)\mathcal{A}_{0}^{\mathrm{T}} \mathcal{P}e_{I}(t)-e_{I}{}^{\mathrm{T}}(t)\mathcal{Q}e_{I}(t) \tag{51}\]We can state the following lemma involved in the proof of the main result of this paragraph. **Lemma 3**. : _Consider \(X_{e}\in\mathcal{X}_{e}\) and matrix \(\mathcal{P}\) obtained from optimization problem (49). Suppose that Assumption A3 is satisfied. Then,_  
_(i)_ _For all_ \(0<\delta\leq\delta_{\max}\)_, the matrix_ \(\mathcal{P}=\mathcal{P}^{T}>0\) _verifies_  
\[\mathcal{A}_{0}^{T}\mathcal{P}+\mathcal{P}\mathcal{A}_{0}+2\mathcal{Q}\leq 0 \tag{52}\]  
_(ii)_ _We have_  
\[\min_{\lambda\in\Lambda_{S}}e_{I}^{T}(t)\mathcal{P}[\mathcal{A}_{0}e_{I}(t)+ \mathcal{B}_{0}(X(t))\lambda+\mathcal{A}_{0}X_{e}+\mathcal{H}X_{e}+\mathcal{H} _{e}]\leq-e^{T}(t)\mathcal{Q}_{1}e(t) \tag{53}\]  
Proof. : \((i)\) is obtained by direct substitution and \((ii)\) by arguments similar to those for Lemma 2. An important remark is that, if conditions of Lemma 2 are satisfied, they are sufficient to guarantee that Assumption A3 is also satisfied; then, we can conclude that Assumption A3 is satisfied for a large class of power converters. This fact results from the structure of Assumption A3 being very similar to the one of Assumption A1. The main theorem providing a solution to the stabilization problem with integral action can now be introduced. **Theorem 2**. : _Assumptions A1 and A3 are satisfied. Then, for all \(0<\delta\leq\delta_{\max}\), the control_  
\[\lambda(t)=\arg\biggl{\{}\min_{d\in\Lambda_{S}}e_{I}^{T}(t)\mathcal{P}[ \mathcal{A}_{0}e_{I}(t)+\mathcal{B}_{0}(X(t))d+\mathcal{A}_{0}X_{e}+\mathcal{ H}X_{e}+\mathcal{H}_{e}]\biggr{\}} \tag{54}\]  
_is such that the states of system (44) are bounded and we have_  
\[\lim_{t\to\infty}x(t)=x_{e}\]  
_Consequently, \(\lim_{t\to\infty}y(t)=y_{ref}\)._  
Proof. : Consider the Lyapunov function \(V(e_{I}(t))=\frac{1}{2}e_{I}^{T}(t)\mathcal{P}e_{I}(t)\). Then  
\[\frac{dV(e_{I}(t))}{dt} =e_{I}^{T}(t)\mathcal{P}\frac{de_{I}(t)}{dt}=e_{I}(t)^{T}\mathcal{ P}(\mathcal{A}_{0}e_{I}(t)+\mathcal{B}_{0}(X(t))\lambda(t)+\mathcal{A}_{0}X_{e}+ \mathcal{H}X_{e}+\mathcal{H}_{e})r\] (55a) \[=\min_{d\in\Lambda_{S}}\Bigl{[}e_{I}^{T}(t)\mathcal{P}(Ae_{I}(t)+ \mathcal{B}_{0}(X(t))d+\mathcal{A}_{0}X_{e}+\mathcal{H}X_{e}+\mathcal{H}_{e} \Bigr{]}\] (55b) \[\leq-e_{I}^{T}\mathcal{Q}_{1}e_{I}\] by Lemma 3. (55c)  
Define the sets  
\[\Omega_{V}=\bigl{\{}e_{I}\in\mathbb{R}^{n+p}:\dot{V}(e_{I})=0\bigr{\}},\ \ \ \Omega_{\mathcal{Q}_{1}}=\Bigl{\{}e_{I}\in\mathbb{R}^{n+p}:e_{I}^{T}\mathcal{Q}_{ 1}e_{I}=0\Bigr{\}} \tag{56}\]  
A basis of \(\Omega_{\mathcal{Q}_{1}}\)is given by \(\bigl{[} \begin{array}{cc}0&I\end{array} \bigr{]}^{T}\). If, for \(e_{I}\in\mathbb{R}^{n+p}\), we have \(\dot{V}(e_{I})=0\), then \(e_{I}^{T}\mathcal{Q}_{1}e_{I}=0\). To show it, we proceed by contradiction. Suppose that, for \(e_{I}\in\mathbb{R}^{n+p}\) such that \(\dot{V}(e_{I})=0\), we have \(e_{I}^{T}\mathcal{Q}_{1}e_{I}\
eq 0\). Then, by (55c), \(\dot{V}(e_{I})<0\) which is a contradiction. Then, \(\dot{V}(e_{I})=0\) implies \(e_{I}^{T}\mathcal{Q}_{1}e_{I}=0\) and \(\Omega_{V}\subseteq\Omega_{\mathcal{Q}_{1}}\). The largest invariant set contained in \(\Omega_{V}\) is also included in \(\Omega_{\mathcal{Q}_{1}}\). By the Lasalle principle [31], we have  
\[\lim_{t\to\infty}e_{I}(t)=\left[ \begin{array}{c}0\\ x_{I_{e}}\end{array}