Finite horizon Riccati solved by LMI

Question

Considering the system $x_{k+1}=Ax_k+Bu_k$ with quadratic cost

$J^* = \min x_N^T S x_N + \sum_{k=0}^{N-1} x_k^T Qx_k+u^T_kRu_k$

where $Q,S\succeq 0, R\succ 0$. The optimal state feedback is found as $u_k = -(R+B^\top P_{k+1}B)^{-1}B^T P(k+1)A x_k$ where $P(k)$ is from the discrete-time riccati equation

$ P_{k} = Q+ A^T P_{k+1}A -A^T P_{k+1}B(R+B^T P_{k+1}B)^{-1}B^T P_{k+1}A $

and terminal penalty $P_N = S$. This can be solved like an LMI turning the equality into

$Q+ A^T P_{k+1}A - P_{k} -A^T P_{k+1}B(R+B^T P_{k+1}B)^{-1}B^T P_{k+1}A \succeq 0$

But how can I formally prove that the Riccati equality can be turned into an inequality? is it just Schur complement considering $P(k)\succeq 0$ and $R\succ0$?

thanks for any suggestion

Answer 1

If $A = B$ then it's trivially true that $A - B \geq 0$.

Answer 2

The inequality

$$ Q + A^\top P_{k+1} A - P_{k} - A^\top P_{k+1} B \left(R + B^\top P_{k+1} B\right)^{-1} B^\top P_{k+1} A \succeq 0 \tag{1} $$

can also be seen as an equality plus some unknown positive semi-definite term $X_k=X_k^\top\succeq0$

$$ Q + A^\top P_{k+1} A - P_{k} - A^\top P_{k+1} B \left(R + B^\top P_{k+1} B\right)^{-1} B^\top P_{k+1} A + X_k = 0. \tag{2} $$

However this can also be seen as solving the time varying cost

$$ J^* = \min_u x_N^T S\,x_N + \sum_{k=0}^{N-1} x_k^\top Q_k x_k + u^\top_k R\,u_k, \tag{3} $$

with $Q_k \succeq Q$. For example setting $Q_k = Q + I$ would have a corresponding solution for $(1)$, but this minimizes the cost from $(3)$ and not the cost when $Q_k=Q$. So solving $(1)$ can give different results then solving the discrete-time Riccati equation.