]> 7.2.3 The step response of an abstract system

7.2.3 The step response of an abstract system

In this section we define the step response of the system

(t) =𝒜x(t) + du(t) x(0) =0 y(t) =c#x(t) ,c#x = h𝒜x = 𝒜x,h H (7.4)

i.e., the response of the system for the Heaviside function u(t) = 1l(t) under null initial conditions.

Theorem 7.2.3. Suppose that the semigroup {S(t)}t0 generated by 𝒜 is EXS. Assume also that either c# is an admissible observation functional or d is an admissible factor control vector, and the compatibility assumption

d D(c#) (7.5)

holds.

If c# is an admissible observation functional then for almost all t 0 the step response has the form

y(t) = (P¯d)(t) c#d (7.6)

where P¯ L(H,L2(0,)) denotes the closure of the observability map P,

P : H D(𝒜) xh𝒜S()x L2(0,) .

If d is an admissible factor control vector then for almost all t 0 the step response can be represented as

y(t) = (Q¯h)(t) c#d (7.7)

where Q¯ L(H,L2(0,)) denotes the closure of the observability map Q,

Q : H D(𝒜) xd𝒜S()x L2(0,) .

Proof. Since [69, p. 5], [18, Theorem 2.1.10, p. 21]

0tS(τ)x 0dτ D(𝒜),𝒜0tS(τ)x 0dτ = S(t)x0 x0t 0x0 H

then by (7.3)

x(t) = 𝒜0tS(τ)ddτ = S(t)d d

satisfies (7.2) for almost all t 0 and all w D(𝒜), i.e., in the weak sense,

d dtx(t),wH = x(t) + d,𝒜w H𝒜1(t) = d dt 𝒜1x(t) = x(t) + d .

Thus the Laplace transform x̂ of x is

x̂(s) = (sI 𝒜)1d 1 sd,s Π+¯ 0 .

The first term belongs to D(𝒜) while the second one is in D(c#) due to (7.5). Now

ŷ(s) = h𝒜(sI 𝒜)1d 1 sc#d = d𝒜(sI 𝒜)1h 1 sc#d = ŷ(s¯)¯s Π+

If h𝒜 is an admissible observation functional with respect to {S(t)}t0 then (7.6) holds, while if d is an admissible factor control vector then, due to duality Theorem 7.2.2, d𝒜 is an admissible observation functional with respect to {S(t)} t0 and thus the representation (7.7) holds. □

Lemma 7.2.1. Let all assumptions of Theorem 7.2.3 hold and additionally the operator 𝒜 has a Riesz basis of eigenvectors {xn}n, corresponding to the set of its eigenvalues {λn}n, with the biorthogonal system {vn}n being the system of eigenvectors of 𝒜 corresponding to its eigenvalues {λn¯}n. Then the step response has the spectral representation

y(t) = n=1λ n eλnt 1 d,v nHxn,hH (7.8)

Proof. The result follows from Theorem 7.2.3 and the proof of Lemma 2.6.1. □