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Exact Newton-Kleinman decreases the cost while an unchecked inexact step can lose stability

Example

Take scalar CARE data A=B=C=R=1 and the stabilizing starting gain K0=2.

Facts & Assumptions

[F1]

Finite-budget exact or safeguarded inexact Newton-Kleinman policy iteration: exact mode solves (A−BKj)∗Xj+Xj(A−BKj)=−(Q+Kj∗RKj) and then sets Kj+1=R−1B∗Xj.

[F2]

Finite-budget exact or safeguarded inexact Newton-Kleinman policy iteration: inexact line-search trials are Kj+θℓ(K~j−Kj) for the finite list θℓ=2−ℓ.

[F3]

Verified Lyapunov residual, descent, and stability checks make an inexact Newton-Kleinman step safe: an accepted inexact trial must have a certified Hurwitz closed loop and an exact cost matrix no greater than the current one.

Proof

1.1F1givenalgebra

At K0=2, the closed loop is −1 and the exact Lyapunov equation gives −2X0+5=0, hence X0=5/2 and K1=5/2. The next closed loop is −3/2, and its policy equation gives −3X1+1+25/4=0, hence X1=29/12<5/2.

2.1F2F3step 1.1algebra∎

Consider instead an unchecked approximate policy value P~=0, which proposes K~=0. A full step gives closed loop A−BK~=1, which is unstable. More generally, a line-search trial between K0=2 and 0 has K∈[0,2]. If K≤1 its closed loop 1−K is not Hurwitz. If 1<K<2, its exact cost matrix is P(K)=(1+K2)/(2(K−1)), and P(K)−52=(K−2)(K−3)2(K−1)>0. Thus every nonzero trial either loses stability or increases cost; the certified safeguard rejects it.

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