Page 81 - Textos de Matemática Vol. 36

3.4.
Main result 71
To prove property 3. for i + 1, we use again (3.8) and obtain
−Ki+1(A − SKi+1) − (D − Ki+1S)Ki+1
= −Ki+1(A − SKi − S(Ki+1 − Ki)) − (D − KiS − (Ki+1 − Ki)S)Ki+1
= KiSKi + Q + Ki+1S(Ki+1 − Ki) + (Ki+1 − Ki)SKi+1
= (Ki+1 − Ki)S(Ki+1 − Ki) + Q + Ki+1SKi+1.
With this identity and R(P ) > 0 we get the inequality
(P − Ki+1)(A − SKi+1) + (D − Ki+1S)(P − Ki+1)
< (Ki+1 − Ki)S(Ki+1 − Ki) + Ki+1SKi+1 + P SP − P SKi+1 − Ki+1SP
= (Ki+1 −Ki)S(Ki+1 −Ki)+(P −Ki+1)S(P −Ki+1)
≥0 ≥0 ≤0
and deduce that
and as Λi+1 is a Z-matrix, Theorem 3.1.1 shows that it is an M-matrix.
Λi+1⃗(Ki+1 − P ) < 0
Finally, we prove property 1. for i + 1. To this end a similar calculation
and reasoning as above yields the inequality
(Ki+2 − Ki+1)(A − SKi+1) + (D − Ki+1S)(Ki+2 − Ki+1)
= (Ki+1 − Ki) S(Ki+1 − Ki)
≥0 ≤0
and Ki+2 ≥ Ki+1.
As the Newton sequence is monotonically increasing and bounded by P , the limit exists and is denoted by K, i.e. limi→∞ Ki = K ≥ 0. By construction solves K the non symmetric Riccati equation (3.3). Moreover any K′ ≥ 0 fulfilling the Riccati equation must fullfill K ≤ K′. Otherwise we see by subtracting the Riccati equations, that the inequality
(K − K′)A + D(K − K′) = K(−S)K + K′(−S)K′
= (K − K′)(−S)(K − K′) + K(−S)K′ + K′(−S)K ≥ −(K − K′)S(K − K′)
holds. This implies as Λ0 is an M-matrix that K ≤ K′, hence K is the smallest solution under all solutions ≥ 0.
Corollary 3.4.2. Under the assumptions of Theorem 3.4.1 and the additional requisite, that −A + SP and −D + PS are M-matrices, the Newton sequence converges to a left-right stabilizing solution K of the Nash Riccati equation.