Converse linear quadratic optimal control. It is well known that for a linear time invariant system dotx=Ax+Bu(1)

Rihanna Bentley

Rihanna Bentley

Answered question

2022-11-02

Converse linear quadratic optimal control
It is well known that for a linear time invariant system
(1) x ˙ = A x + B u
with (A,B) controllable, there exists a static state feedback u = K x such that the cost function
(2) J = 0 x T Q x + u T R u d t
is minimized, assuming Q 0 (positive semi-definite) and R > 0 (positive definite). The gain K is the solution of the algebraic Riccati equation:
0 = A T P + P A P B R 1 B T P + Q K = R 1 B T P P = P T 0
known as linear quadratic regulator (LQR). However, I wonder whether the converse also holds?
That is, given a stabilizing K s (such that A B K s is Hurwitz), do there exist matrices Q 0 and R > 0 such that u = K s x minimizes (2) given (1)? Or put differently:
Question: Is every stabilizing linear state feedback optimal in some sense?

Answer & Explanation

brulotfao

brulotfao

Beginner2022-11-03Added 16 answers

Step 1
Given a system
x ˙ = A x + B u , x ( t 0 ) = x 0
z = [ Q 1 / 2 0 0 R 1 / 2 ] [ x u ] ,
with (A,B) is stabilizable, (Q,A) is detectable and R > 0 (positive definite). The linear quadratic regulator problem is given by minimizing
0 z T z d t .
Step 2
From Boyd et al 1994: The inverse problem of optimal control is the following. Given a matrix K, determine if there exists Q 0 and R > 0, such that (Q,A) is detectable and u = K x is the optimal control for the corresponding LQR problem. Equivalently, we seek R > 0 and Q 0 such that there exists P nonnegative and P1 positive-definite satisfying
( A + B K ) T P + P ( A + B K ) + K T R K + Q = 0 , B T P + R K = 0
and A T P 1 + P 1 A < Q. This is an LMIP in P, P 1 , R and Q. (The condition involving P 1 is equivalent to (Q,A) being detectable.)

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