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I'm facing the following problem obtaining the solution of the Discrete Algebraic Riccati Inequality.

Notation and assumption:

$\succeq, \succ,\preceq,\prec$ refers to matrix definiteness

$R\succ0$, $P\succeq 0$ $\quad (A,B)$ controllable pair

Let's consider first the continuous-time case

$(1)\quad A^TP + P A + PBR^{-1}B^TP + Q \preceq 0$

This is a quadratic matrix inequality that can be easily recast as LMI through the Schur Complement Lemma having $R^{-1} \succ 0$

$(2)\quad\begin{bmatrix}\matrix{-(A^TP + PA +Q) & (PB) \\ (PB)^T & R}\end{bmatrix} \succeq 0$

Now, I'm trying doing the same in the discrete-time case.

Consider

$(3)\quad A^TPA - P - A^TPB\ (R+B^TPB)^{-1}\ B^TPA +Q \preceq 0$

which is a Nonlinear Matrix Inequality.

Although tempting the Schur Complement does not work here because the nonlinear term inner part $-(R+B^TPB)^{-1}\prec 0$ is negative definite.

Clearly I could use standard algorithms to find the solution of the associated Matrix Equation (replacing $\preceq$ by $=$ in Eq.(3) ). However I still wonder if the above problem Eq.(3) has a simple linear form using some 'LMI trick'.


EDIT:

Eq. (1) is incorrect as suggested by Johan. I got this formulation from a rather popular article see [1,sec 4.7]. Hence the derivation in Eq. (2) is wrong as well.

The correct one is

$(1\prime)\quad A^TP + P A - PBR^{-1}B^TP + Q \preceq 0$

obtained from the Lyapunov equation

$(4)\quad (A+BK)^T P + P(A+BK) \preceq -Q - K^T R K$

and replacing the LQ optimal controller $K=-R^{-1} B^TP$. (I'm not entirely sure if such a substitution makes really sense). However as far as I can see the quadratic matrix function $(1\prime)$ is for general $P\succ 0$ indefinite. Existence of the strict solution (equality $=$) is provided by the solution of the ARE.


[1] VanAntwerp, Jeremy G., and Richard D. Braatz. "A tutorial on linear and bilinear matrix inequalities." Journal of process control 10.4 (2000): 363-385.

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Your question is incorrect to begin with, as the continuous-time ARE is

$\quad A^TP + P A - PBR^{-1}B^TP + Q = 0$

Hence, just as bad form.

The LMI formulations of LQ in both discrete-time and continuous-time would typically be done in both the Riccati matrix and the feedback matrix.

Find minimum-trace $P$ such that

$ (A-BK)^T P (A-BK) - P + K^TRK + Q \preceq 0$

Multiply with $P^{-1}$ from left and right, define $Y = P^{-1}$ and $F = KP^{-1}$, and apply suitable Schur complements to arrive at linear form.

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  • $\begingroup$ Thank you, you are right about the continuous-time case. I got the formulation from a tutorial paper (see ref in question edit section) which turns out to be wrong. I am aware of the solution you propose, however the optimization is run over two matrix variables $Y$ and $F$. The Riccati equation results to have just one symmetric matrix variable $P$ and the controller can be recovered using the classical formula from Dynamic Programming. I still wonder if one can reduce the number of variables in the case of LMI, by using some matrix lemma e.g. Finsler's lemma, elimination lemma $\endgroup$ – nicmus May 5 '17 at 9:07
  • $\begingroup$ Does the inequality hold for finite horizon DARE? $\endgroup$ – Betelgeuse Jul 4 at 8:24
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To be more precise, the discrete-time Algebraic inequality is: $0\preceq A^TPA+Q-P-A^TPB(B^TPB+R)^{-1}B^TPA$ and $P\succeq 0$. Then by using Schur complement, we have $$ \begin{bmatrix} R+B^TPB &B^TPA &0\\ A^TPB &A^TPA+Q-P &0\\ 0 &0 &P \end{bmatrix}\succeq 0 $$.

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