Page 71 - Textos de Matemática Vol. 36

2.5. Global existence results 61 The following result holds for Hermitian (or real symmetric) solutions of
(2.67).
Theorem 2.5.10. Assume that in equation (2.67)
g(K) = −g1(K) + Kg2(K) + g3(K)K, (2.69)
where g1∗(K∗) = g1(K) and g2∗(K∗) = g3(K). If for some positive α and for all x ∈ Cn,
x∗(g1(t, K) + Q(t))x ≥ α|x|2 (2.70) for t ∈ (−∞, tf ] and K ≥ 0, then any solution of (2.67) with positive definite
terminal data Kf > 0 stays positive definite for t ≤ tf .
Proof. Let K(t) be a Hermitian solution of the equation which is piecewise continuously differentiable on a finite interval [t0,tf]. This will mean that K(t) is continuous on [t0,tf] and there exists a subdivision of [t0,tf] into a finite number of intervals such that on each interval the function K(t) is continuously differentiable, possessing one-sided finite derivatives at the end-points of these intervals. Let λ(t) denote the minimal eigenvalue of K(t,tf) = K(t) with x(t) being a corresponding unit eigenvector. Then (suppressing the variable t and involving (2.69)) with Theorem 2.5.1, we conclude
dλ = x∗K˙ x = −2x∗KAx − x∗Qx + x∗KSKx + x∗g(K)x dt
= −2λx∗Ax + λ2x∗Sx − x∗(g1(K) + Q)x + 2λx∗(g2(K))x. From (2.70) and (2.71) we now obtain
dλ ≤ −λ(2x∗Ax − 2x∗g2(K)x − λx∗Sx) − α|x|2 dt
(2.71)
a.e. in (−∞, tf ], whenever K ≥ 0.
It is now evident that for sufficiently small λ(t) > 0 we have dλ < 0. From
dt
this monotonicity property and the continuity of λ(t) we conclude λ(t) > 0 for all t < tf , whenever λf = λ(tf ) > 0, where λf denotes the minimal eigenvalue of the terminal matrix Kf > 0. Hence, the solution of (2.67) remains positive for t ∈ (−∞,tf].