Page 92 - Textos de Matemática Vol. 36

82 Appendix A.
Proof. Obviously X(t0) = X0, therefore it follows from the properties of the transition matrices, stated in Theorem A.1.1, by differentiation that the function X, defined by (A.16), solves (A.15). Notice that the solution of (A.15) is unique.

t
X(t)=ΦA(t,t0)X0Φ∗A(t,t0)+ ΦA(t,τ)R(τ)Φ∗A(t,τ)dτ, t∈R. (A.17)
t0 Furthermore, if A is constant, then
t0
for t ≥ t0.
Theorem A.1.7. Let A be stable . Then the (unique) solution of the continuous-
time algebraic Lyapunov equation CALE
Corollary A.1.6. The unique solution of the Lyapunov differential equation X˙ = A(t)X + XA∗(t) + R(t), X(t0) = X0
is defined by
has the representation
AX + XA∗ + R = 0 ˜∞∗
X=
∗t∗
X(t) = e(t−t0)AX0e(t−t0)A +
From (A.17) we infer that X0 ≥0 and R(t)≥0 for t≥t0 imply that X(t)≥0
0
e(t−τ)A R(τ) e(t−τ)A dτ, t ∈ R. (A.18)
eτAReτA dτ;
thus RP0 implies X˜ P0 for P ∈ {<,≤,=,>,≥}. Proof. The initial value problem
X˙=AX+XA∗+R, X(0)=0, has, according to (A.18), the solution
t∗t∗ X(t) = e(t−τ)A R e(t−τ)A dτ = esA R esA ds.
00
σ(A) ⊂ C< yields lim X˙ (t) = lim etA R etA∗ = 0, therefore it follows from the t→∞ t→∞
general theory of differential equations (see [KnKa74], p. 135) that
lim X(t) = t→∞
∞∗˜
esA R esA ds = X is the unique solution of the CALE.
0
The following result was already obtained by Lyapunov in 1897.