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A scalar CARE has two real roots but only one stabilizing branch

Example

Take the scalar continuous data $A = B = C = R = 1$ . Then $Q = 1$ and $G = 1$ .

Facts & Assumptions

Given: The scalar data above and the CARE convention of Standard CARE and DARE data, equations, and conventions.

[F1]

Standard CARE and DARE data, equations, and conventions: the CARE residual is $A^{*} X + X A - X G X + Q$ , and a CARE root is stabilizing when $A - G X$ is Hurwitz.

Proof

1.1F1givenalgebra

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

2.1F1step 1.1algebra∎

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.

Depends on

Standard CARE and DARE data, equations, and conventions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel Liberzon, Calculus of Variations and Optimal Control Theory: A Concise Introduction, Chapter 6 (standard reference, not scraped)