## 5 Formulation of the LMI Stability Criterion
### Design of the LMI Condition
#### 5.2.3 LMI Approach with Fixed Parameter \(\epsilon\) and Unknown Parameter \(\delta\)  
This section addresses the parameter \(\epsilon\) as fixed _a priori_ and the parameter \(\delta\) as unknown, then to be optimized and hence maximized. Notice that the parameter \(\delta\) exists only in the fourth and last element of the diagonal of the matrix inequality (77), which is not, in the present case, an LMI due to the presence of the terms \(\frac{1}{\delta^{2}}\mathcal{M}_{3}\) and \(\frac{1}{\delta^{2}}\mathcal{M}_{3}\). Thus, the objective is to convert the matrix inequality (77) into an LMI. This idea is not simple to achieve because of the presence of the previous two nonlinear terms. Thus, to design the third LMI condition, the first step is to transform the matrix \(\mathcal{M}_{3}\) to a scalar by setting: \(\mathcal{M}_{3}=m_{3}\mathcal{I}\). As a result, the fourth diagonal element of the matrix inequality (77) becomes as follows:  
\[\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}-\left(\sigma \Big{(}\frac{1}{\delta^{2}}-1\Big{)}^{2}+\frac{1}{\delta^{2}}\right)\mathcal{ M}_{3}=\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}-\frac{m_{3}}{ \delta^{2}}\mathcal{I}-\sigma m_{3}\Big{(}\frac{1}{\delta^{2}}-1\Big{)}^{2} \mathcal{I}<0 \tag{84}\]  
The idea is to express the quantity \(\frac{m_{3}}{\delta^{2}}\). To achieve this, relation (84) can be formulated as follows:  
\[\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}-\left(\sigma \Big{(}\frac{1}{\delta^{2}}-1\Big{)}^{2}+\frac{1}{\delta^{2}}\right)\mathcal{ M}_{3}=\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}-\frac{m_{3}}{ \delta^{2}}\mathcal{I}-\sigma\frac{1}{m_{3}}\Big{(}\frac{m_{3}}{\delta^{2}}-m _{3}\Big{)}^{2}\mathcal{I} \tag{85}\]  
The second step is to make a variable change by setting \(\lambda=\frac{m_{3}}{\delta^{2}}\). Thus, expression (85) becomes the following:  
\[\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}-\left(\sigma \Big{(}\frac{1}{\delta^{2}}-1\Big{)}^{2}+\frac{1}{\delta^{2}}\right)\mathcal{ M}_{3}=\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}- \lambda\mathcal{I}-\sigma m_{3}^{-1}(\lambda-m_{3})^{2}\mathcal{I} \tag{86}\]  
It is clear that the term \(\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{\mathrm{T}}\) is linear with respect to the variable \(\gamma\) since \(\epsilon\) was assumed to be constant in this part. Moreover, the term \(-\lambda\mathcal{I}\) is linear with respect to the variable \(\lambda\). Additionally, notice that the term \(-\sigma m_{3}^{-1}(\lambda-m_{3})^{2}\mathcal{I}\) is quadratic. Thus, since \(\sigma<0\) and \(m_{3}>0\), this previous quadratic term can be linearized utilizing the Schur complement to then obtain the subsequent subpart of the matrix inequality:  
\[\left[ \begin{array}{cc}\frac{\gamma}{\epsilon^{2}}\mathcal{B}\mathcal{B}^{ \mathrm{T}}-\lambda\mathcal{I}&\lambda-m_{3}\\ \star&\frac{m_{3}}{\sigma}\end{array} \right]<0 \tag{87}\]  
Recall that the objective is to maximize the parameter \(\delta\). Since \(\lambda=\frac{m_{3}}{\delta^{2}}\), we have \(\delta=\sqrt{\frac{m_{3}}{\lambda}}\). Accordingly, to maximize \(\delta\), we need to maximize \(m_{3}\) and minimize \(\lambda\). To transform the maximization problem of the parameter \(\delta\) into a minimization problem, we should write \(\kappa=\delta^{-1}=\sqrt{\frac{\lambda}{m_{3}}}\). Then, to maximize \(\delta\), we should minimize \(\kappa\) and consequently minimize the two variables \(\lambda\) and \(m_{3}^{-1}\), hence minimizing the quantity \(\lambda+m_{3}^{-1}\). Therefore, the maximization problem of \(\bar{\delta}\) can be recast by the following condition:  
\[\lambda+m_{3}^{-1}<\theta_{2} \tag{88}\]  
where \(\theta_{2}>0\), and that should be minimized. Hence, to maximize \(\bar{\delta}\), we should minimize \(\theta_{2}\). Then, we need to transform inequality (88) into an LMI.