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Given $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times 2}$, I am interested in solving the following problem:

\begin{array}{ll} \underset{R \in \mathbb{R}^{2\times 2}}{\text{minimize}} & \mathrm{tr} \left( R^{-1}B^T X B \right)\\ \text{subject to} & X=A^TXA-A^TXB(R+B^TXB)^{-1}B^TXA.\end{array}

Here $X$ is unique stabilizing solution to DARE, thus X is positive definite. It is required that $R>0$ (i.e. positive definite) and $B^TXB$ to be full rank.

EDIT: For a fixed $A,B,R$, we can get unique $X$ by solving DARE, for example by using matlab "idare" or "dare" command. However, here $R$ is not fixed, it is a variable, thus for each $R$, there is corresponding $X$.

My attempt: I wanted to start with simpler case when we put additional constraints on $R$. Assume that $R$ is diagonal and positive definite. WLOG we can assume that $R=\mathrm{diag}\{r_1,r_2\}$, such that $r_1+r_2=1$ and $1>r_i>0$ for $i=1,2.$

A=[3 0 0 0; 0 2 1 0; 0 0 2 0; 0 0 0 2];
B=rand(4,2);
Q=zeros(4,4);    
r1=linspace(0.01, 0.99);
for i=1:100  
    R=[r1(i) 0; 0 1-r1(i)];
    [X,~,~] = idare(A,B,Q,R);
    T(i)=trace(inv(R)*B'*X*B);
end 
plot(T)

enter image description here

It looks like that as we increase $r_1$ from $0$ to $1$, then $\mathrm{tr} \left( R^{-1}B^T X B \right)$ is continuous, moreover, it is convex. However, I am unable to prove it.

Lee
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  • What exactly is $X$ ? – P. Quinton Jul 17 '22 at 08:13
  • @P.Quinton X is the unique solution to DARE (discrete-time algebraic Riccati equation), thus it is symmetric and positive definite – Lee Jul 17 '22 at 09:38
  • Do we have a rank constraint about $B^T X B$ ? I was thinking aobut trying something like this on $(R+B^T X B)^{-1}$ : https://math.stackexchange.com/questions/17776/inverse-of-the-sum-of-matrices – P. Quinton Jul 17 '22 at 13:21
  • @P.Quinton assume that $B^TXB$ is full rank. we can also assume that both $A$ and $B$ are full rank – Lee Jul 17 '22 at 13:57
  • @user1551 done. – Lee Jul 20 '22 at 04:01
  • $X=0$ is a solution... if you say it is unique then it must be it... – Exodd Jul 22 '22 at 13:27
  • @Exodd For a fixed $A,B,R$, we can get unique $X$ by solving DARE, for example by using matlab "idare" or "dare" command. However, here $R$ is not fixed, it is a variable, thus for each $R$, there is corresponding $X$. – Lee Jul 23 '22 at 01:29
  • For every R, X=0 is always a solution... – Exodd Jul 23 '22 at 08:26
  • @Exodd not any solution, but stabilizing solution. I know that I need to make my question self contained, but sometimes it is difficult to include all information. The link of DARE is provided where it is mentioned that usually stabilizing solution is required, which is my case. And stabilizing means that we need to make sure that there exist a controller $F$, that $A+BF$ have all eigenvalues inside unit circle for discrete time system. And it turns out that there is only one such solution $X$ to DARE that is stabilizing. – Lee Jul 24 '22 at 08:43

1 Answers1

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Given the matrix inner product $(:)$ $$\eqalign{ \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\qiq{\quad\implies\quad} A:B &= \trace{A^TB} \\ }$$ The objective function can be written as $$\eqalign{ \phi &= \trace{R^{-1}B^T X B} \\ &= R^{-1}:B^TXB \\ }$$ For typing convenience, define the matrix variable $$\eqalign{ &M = (R+B^TXB)^{-1} \;=\; M^T \\ &dM = -M\LR{dR+B^TdX\,B}M \\ &\c{dm} = -\LR{M\otimes M}\c{dr} - \LR{MB^T\otimes MB^T}\c{dx} \\ }$$ Use the DARE constraint to relate the differentials of $M,R,$ and $X,\,$ and vectorize them. $$\eqalign{ &X = A^TXA-A^TXB\c{M}B^TXA \\ &dX = A^TXB\,\c{dM}\,B^TXA + A^T\c{dX}\,A - A^T\c{dX}\,BMB^TXA - A^TXBMB^T\c{dX}\,A \\ &\c{dx} = \LR{A^TXB\otimes A^TXB}\c{dm} + \LR{A^T\otimes A^T}\c{dx} \\ &\qquad\qquad\qquad - \LR{A^TXBMB^T\otimes A^T}\c{dx} - \LR{A^T\otimes A^TXBMB^T}\c{dx} \\ &\LR{A^TXB\otimes A^TXB}\LR{M\otimes M}\c{dr} \\&\qquad\qquad\qquad = \LR{A^T\otimes A^T-I\otimes I}\c{dx} \\ &\qquad\qquad\qquad - \LR{A^TXB\otimes A^TXB}\LR{MB^T\otimes MB^T}\c{dx} \\&\qquad\qquad\qquad - \LR{A^TXBMB^T\otimes A^T}\c{dx} \\&\qquad\qquad\qquad - \LR{A^T\otimes A^TXBMB^T}\c{dx} \\ &P^T\c{dr} = Q^T\c{dx} \\ }$$ Calculate the gradient of the objective function $$\eqalign{ \phi &= R^{-1}:B^TXB \\ \c{d\phi} &= R^{-1}:B^T\c{dX}\,B - B^TXB:R^{-1}\c{dR}\,R^{-1} \\ &= BR^{-1}B^T:\c{dX} - R^{-1}B^TXBR^{-1}:\c{dR} \\ &= E:\c{dX} - F:\c{dR} \\ &= e:\c{dx} - f:\c{dr} \\ &= \LR{PQ^{+}e - f}:\c{dr} \\ \grad{\phi}{r} &= {PQ^{+}e - f} \\ }$$ Solving the zero gradient condition for an optimal $R$ value seems hopeless, but a numerical solution via gradient descent is feasible given this closed-form expression for the gradient.

$$\eqalign{ }$$

greg
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