LMIs in Control/Stability Analysis/Kharitonov-Bernstein-Haddad

Kharitonov-Bernstein-Haddad
Consider the set of matrices A = { $$ \begin{align} A = \begin{bmatrix} 0_{(n-1)\times 1} & 1_{(n-1)\times(n-1)}\\ -a_{0} & -a_{n-1}\\ \end{bmatrix} \qquad |\qquad \underline{a}_{j}\le \underline{a}_{j}\le \overline{a}_{j}, \qquad j=1,2,3,4,...,n-1 \end{align} $$ }

Every matrix in the set A is Hurwitz if and only if there exist $$P_{i}\in \text{S}^{n} $$, i = 1, 2, 3, 4, where Pi > 0, i = 1, 2, 3, 4, such that

\begin{align} \qquad P_{i}A_{i} + A_{i}^{T}P_{i}< 0      \qquad    i=1,2,3,4, \end{align} $$ where

$$ \begin{align} \qquad A_{i} = \begin{bmatrix} [0_{(n-1)\times 1} & 1_{(n-1)\times(n-1)}] \\ \qquad   a_{i} \\

\end{bmatrix} \end{align} ,\qquad   i=1,2,3,4, $$

$$ \begin{align} \qquad a_{1} =- \begin{bmatrix} \underline{a}_{0} & \underline{a}_{1} &\overline{a}_{2} &\overline{a}_{2} &... & \underline{a}_{n-4} & \underline{a}_{n-3} & \overline{a}_{n-2}& \overline{a}_{n-1} \end{bmatrix}, \end{align} $$

$$ \begin{align} \qquad a_{2} =- \begin{bmatrix} \underline{a}_{0} & \overline{a}_{1} &\overline{a}_{2} &\underline{a}_{2} &... & \underline{a}_{n-4} & \overline{a}_{n-3} & \overline{a}_{n-2}& \underline{a}_{n-1} \end{bmatrix}, \end{align} $$

$$ \begin{align} \qquad a_{3} =- \begin{bmatrix} \overline{a}_{0} & \underline{a}_{1} &\underline{a}_{2} &\overline{a}_{2} &... & \overline{a}_{n-4} & \underline{a}_{n-3} & \underline{a}_{n-2}& \overline{a}_{n-1} \end{bmatrix}, \end{align} $$

$$ \begin{align} \qquad a_{4} =- \begin{bmatrix} \overline{a}_{0} & \overline{a}_{1} &\underline{a}_{2} &\underline{a}_{2} &... & \overline{a}_{n-4} & \overline{a}_{n-3} & \underline{a}_{n-2}& \underline{a}_{n-1} \end{bmatrix}, \end{align} $$

Equivalently, every matrix in the set A is Hurwitz if and only if there exist $$Q_{i}\in \text{S}^{n} $$, i = 1, 2, 3, 4, where Qi > 0, i = 1, 2, 3, 4, such that

\begin{align} \qquad A_{i}Q_{i} + Q_{i}A_{i}^{T}< 0      \qquad    i=1,2,3,4, \end{align} $$