Riccati inequalities solution via LMI

Question

$${A^T}P + PA + 2{C^T}C + \frac{1}{{{\gamma ^2}}}P{G_1}G_1^TP + P{G_2}G_2^TP - PB{B^T}P < 0$$

I want to know how to calculate the positive definite $P$ through LMI. Thank you very much.

Answer 1

For this you can use Schur complement. This allows you the write two smaller inequalities, one of which has a quadratic term, into one LMI. Namely

$$ M = \begin{bmatrix} \mathcal{A} & \mathcal{B} \\ \mathcal{B}^\top & \mathcal{C} \end{bmatrix} \succ 0 \tag{1} $$

is equivalent to $\mathcal{C} \succ 0$ and $\mathcal{A} - \mathcal{B}\,\mathcal{C}^{-1}\mathcal{B}^\top \succ 0$. Now we only have to find what to choose for $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$. Since $\mathcal{B}$ is the only term that appears quadratically, therefore it should at least include $P$. For know I will assume it is of the following form

$$ \mathcal{B} = P + X \tag{2} $$

with $X$ for now an unknown constant and it is assumed that $P = P^\top$. We want that the second equivalent inequality to be your inequality, which can be rewritten as

$$ A^\top P + P\,A + 2\,C^\top C + P\left(\frac{1}{\gamma^2}\,G_1\,G_1^\top + G_2\,G_2^\top - B\,B^\top\right) P \prec 0. \tag{3} $$

So we want to solve the following

$$ A^\top P + P\,A + 2\,C^\top C + P\left(\frac{1}{\gamma^2}\,G_1\,G_1^\top + G_2\,G_2^\top - B\,B^\top\right) P = \mathcal{B}\,\mathcal{C}^{-1}\mathcal{B}^\top - \mathcal{A}. \tag{4} $$

Substituting in the expression for $\mathcal{B}$ from $(2)$ into the right hand side of $(4)$ gives

$$ (P + X)\,\mathcal{C}^{-1}(P + X)^\top - \mathcal{A} = P\,\mathcal{C}^{-1}P + P\,\mathcal{C}^{-1}X^\top + X\,\mathcal{C}^{-1}P + X\,\mathcal{C}^{-1}X^\top - \mathcal{A}. \tag{5} $$

By looking at the quadratic terms in $P$ from $(4)$, while using $(5)$ for its right hand side, it follows that

$$ \mathcal{C} = \left(\frac{1}{\gamma^2}\,G_1\,G_1^\top + G_2\,G_2^\top - B\,B^\top\right)^{-1} \tag{6} $$

which implies that the matrix within the brackets has to be positive definite, since in order for $(1)$ to hold it must be true that $\mathcal{C} \succ 0$ and the inverse of a positive definite matrix is also positive definite.

By now looking at the linear terms in $P$ from $(4)$, while using $(5)$ for its right hand side, it follows that

$$ X = A^\top \mathcal{C}. \tag{7} $$

Solving for $\mathcal{A}$ in $(4)$ after substituting in $(6)$ gives

$$ \mathcal{A} = X\,\mathcal{C}^{-1}X^\top - 2\,C^\top C = A^\top \mathcal{C}\,A - 2\,C^\top C. $$

So as long as the matrix within the brackets in $(6)$ is positive definite then $(3)$ is equivalent to the following LMI

$$ \begin{bmatrix} A^\top \mathcal{C}\,A - 2\,C^\top C & P + A^\top \mathcal{C} \\ P + \mathcal{C}\,A & \mathcal{C} \end{bmatrix} \succ 0. \tag{8} $$