Numerical algebraic Riccati equations and stabilizing branches — Examples

Substitution gives $2X-X^2+1=0$, or $(X-1{)}^2=2$. The two real roots are $X_{+}=1+\sqrt{2}$ and $X_{-}=1-\sqrt{2}$.

The corresponding closed-loop scalars are $1-X_{+}=-\sqrt{2}<0$ and $1-X_{-}=\sqrt{2}>0$. Thus $X_{+}$ is the only stabilizing branch; it is positive, whereas the other root is negative and unstable.

Here $G=0$ and $R+B^{\ast }XB=1$. The DARE reduces to $X=1$, so its solution is $X=1$ and $A_c=0$, which is Schur stable.

The pencil matrices are $N=\left[\begin{array}{cc}0& 0\\ -1& 1\end{array}\right]$ and $M=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$. Therefore $\det (\beta N-\alpha M)=\det \left[\begin{array}{cc}-\alpha & 0\\ -\beta & \beta \end{array}\right]=-\alpha \beta \mathrm{.}$ This nonzero homogeneous polynomial has roots $[0:1]$ (the finite eigenvalue $0$) and $[1:0]$ (an infinite eigenvalue). No inverse of $A$ is used.

First take $A=C=R=1$ and $B=0$. Then $Q=1$; every feedback leaves $A-BK=1$, so the pair is not stabilizable. The CARE is $2X+1=0$, whose only root is $X=-1\mathrm{/}2$; it is not positive semidefinite and its closed loop remains $1$.

Next take $A=B=R=1$ and $C=0$, so $Q=0$. The pair is stabilizable, since $K=2$ gives $A-BK=-1$, but it is not detectable because $A-LC=1$ for every $L$. The CARE is $2X-X^2=0$, with positive-semidefinite roots $X=0$ and $X=2$; the first has unstable closed loop $1$ and the second stable closed loop $-1$. Since $Q=0$, zero control has zero cost for every initial state, so the value is zero rather than the stabilizing root $X=2$. Thus without detectability uniqueness among PSD roots and the stabilizing value conclusion in [F3] fail.

The scalar CARE is $2\epsilon X-X^2+{\epsilon }^2=0$, with positive root $X_{\ast }=(1+\sqrt{2})\epsilon$ and stable closed loop $A_c(X_{\ast })=-\sqrt{2}\epsilon$. At the candidate, ${\widehat{A}}_c=-0.1\epsilon <0$, the exact residual is $r=\mathrm{\mid }{F}_c(\widehat{X})\mathrm{\mid }=1.99{\epsilon }^2=1.99\cdot {10}^{-16}$, and $g=1$.

The scalar Lyapunov derivative is $\mathcal{L}(E)=2{\widehat{A}}_{c}E=-0.2\epsilon E$, so its inverse norm is $\kappa =1\mathrm{/}(0.2\epsilon )=5\cdot {10}^8$. For every $\rho >0$, $r+{\rho }^2\ge 2\sqrt{r}\rho$, so $\kappa (r+{\rho }^2)\ge 2\kappa \sqrt{r}\rho >\rho$ because $2\kappa \sqrt{r}>14$. Thus no radius satisfies even the self-map condition, despite the absolute residual being below $2\cdot {10}^{-16}$ and the exact stabilizing root existing. The residual alone is inconclusive because the inverse Lyapunov factor is large near the imaginary axis.

At $K_0=2$, the closed loop is $-1$ and the exact Lyapunov equation gives $-2X_0+5=0$, hence $X_0=5\mathrm{/}2$ and $K_1=5\mathrm{/}2$. The next closed loop is $-3\mathrm{/}2$, and its policy equation gives $-3X_1+1+25\mathrm{/}4=0$, hence $X_1=29\mathrm{/}12<5\mathrm{/}2$.

Consider instead an unchecked approximate policy value $\tilde{P}=0$, which proposes $\tilde{K}=0$. A full step gives closed loop $A-B\tilde{K}=1$, which is unstable. More generally, a line-search trial between $K_0=2$ and $0$ has $K\in [0,2]$. If $K\le 1$ its closed loop $1-K$ is not Hurwitz. If $1<K<2$, its exact cost matrix is $P(K)=(1+K^2)\mathrm{/}(2(K-1))$, and $P(K)-\frac{5}{2}=\frac{(K-2)(K-3)}{2(K-1)}>0.$ Thus every nonzero trial either loses stability or increases cost; the certified safeguard rejects it.

The inputs are positive semidefinite, and $I+G_0{H}_0=\left[\begin{array}{cc}2& 1\\ 0& 1\end{array}\right],W_0=\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ 0& 1\end{array}\right]\mathrm{.}$ Using [F1] gives $A_1=W_0,G_1=\left[\begin{array}{cc}\frac{3}{2}& 0\\ 0& 0\end{array}\right],H_1=\left[\begin{array}{cc}\frac{3}{2}& \frac{3}{2}\\ \frac{3}{2}& \frac{3}{2}\end{array}\right]\mathrm{.}$ Both updated symmetric blocks are positive semidefinite; $G_1$ is diagonal PSD and $H_1=\frac{3}{2}(1,1{)}^T(1,1)$.

To illustrate the diagnostic, perturb only one stored off-diagonal entry by $ϵ>0$ and report ${\tilde{H}}_1=\left[\begin{array}{cc}\frac{3}{2}& \frac{3}{2}+ϵ\\ \frac{3}{2}& \frac{3}{2}\end{array}\right]$. Its symmetry defect is $\mathrm{\parallel }{\tilde{H}}_1-{\tilde{H}}_1^{\ast }{\mathrm{\parallel }}_F=\sqrt{2}ϵ\mathrm{.}$ Even its symmetrization has determinant $\frac{9}{4}-(\frac{3}{2}+\frac{ϵ}{2}{)}^2<0$, so this perturbed stored block is not positive semidefinite. It must be reported and rechecked rather than treated as covered by the exact-arithmetic PSD invariant.