]> 2.4 The Lyapunov operator equation

2.4 The Lyapunov operator equation

Let us consider an abstract observation system on a general Hilbert space H with scalar product ,H
(t) =𝒜x(t),t 0 x(0) =x0 y =𝒞x (2.19)

with the operator 𝒜 : (D(𝒜) H)H generating a linear C0–semigroup {S(t)}t0 on H and the operator 𝒞L(D𝒜,Y) where D𝒜 denotes the domain D(𝒜) equipped with the norm induced by the scalar product x,v𝒜 := x,vH + 𝒜x,𝒜vH. Y is another Hilbert space with the scalar product ,Y. Observe also that if 𝒜1 L(H) then the norm induced by x,v𝒜 is equivalent with the norm induced by the scalar product 𝒜x,𝒜vH. In this case without loss of generality one can assume that

𝒞 = 𝒟𝒜 (2.20)

where 𝒟L(Y,H).

Definition 2.4.1. The observation operator 𝒞L(D𝒜,Y) is called admissible if there exists γ > 0 such that

0𝒞S(t)x 0 Y2dt γ x 0 H2x 0 D(𝒜) (2.21)

i.e., the observability map

P : (D(𝒜) H) x0𝒞S()x0 L2(0,; Y)

is bounded.

In this section, we prove a few results concerning the existence, uniqueness, and regularity of solution of the Lyapunov operator equation.

Theorem 2.4.1. 𝒞 is admissible iff there exists = L(H), 0, and satisfies the Lyapunov operator equation

𝒜x1,x2H + x1,𝒜x2H = 𝒞x1,𝒞x2Yx1,x2 D(𝒜) (2.22)

Proof. Notice that, for 𝒞L(D𝒜,Y), the function [0,) t𝒞S(t)x0 Y is continuous for every fixed x0 D(𝒜).

Necessity. Putting x1 = x2 = S(t)x0, for x0 D(𝒜) and t 0, in (2.22), we get

d dtS(t)x0,S(t)x0H = 𝒜S(t)x0,S(t)x0H + S(t)x0,𝒜S(t)x0H = 𝒞S(t)x0 Y2 ,

and hence

0t 𝒞S(τ)x 0 Y2dτ = x 0,x0H S(t)x0,S(t)x0H x0,x0H x0 H2.

The integrand is nonnegative and continuous, thus the function

t0t 𝒞S(τ)x 0 Y2dτ

is C1, monotone increasing, and bounded from above. This implies the convergence of the improper integral, and (2.21) holds with γ = .

Sufficiency. By (2.21), the operator P : HL2(0,; Y) defined as

(Px0)(t) = 𝒞S(t)x0,x0 D(P) = D(𝒜) (2.23)

is linear, densely defined, and bounded. Thus P is closable, and its closure P¯ is unique continuous extension of P to the whole space H [86, Theorem 4.5, p. 58, Theorem 5.2, p. 84, Theorem 5.3, p. 90]. Now, the bilinear form

x1,x2H = P¯x1,P¯x2L2(0,;Y) (2.24)

uniquely determines an operator L(H), with = and 0. Actually = P¯P¯ = PP¯. By virtue of (2.23), (2.24), and the semigroup axiom (i), for x1,x2 D(𝒜) we have

S(t)x1,S(t)x2H =t𝒞S(τ)x 2,𝒞S(τ)x2Ydτ,t 0 .

Hence, taking the right derivative at t = 0, we get (2.22). □

In [20], this theorem is proved in the case where 𝒞L(H,Y). If 𝒜 has an inverse 𝒜1 L(H), then (2.22) is equivalent to the following equation in L(H):

(𝒜)1 + 𝒜 = DD (2.25)

where D = 𝒞𝒜1 L(H,Y) by (2.20).

Theorem 2.4.2. Let 𝒞L(D𝒜,Y) be admissible. If the semigroup {S(t)}t0 is AS then (2.22) has the unique solution.

Proof. Let 1, 2 L(H), with k = k 0 (k = 1, 2), be two solutions of (2.22). Then, taking f, g D(𝒜), and proceeding as in the proof of Theorem 2.4.1 (the "necessity" part), we obtain

d dt S(t)f, (1 2)S(t)gH = 0,t 0 .

Hence, by (2.12), and noting that D(𝒜)¯ = H, we get f, (1 2)gH = 0 for all f,g H, and the application of the Hahn–Banach theorem completes the proof. □

Theorem 2.4.1 and 2.4.2, in somewhat different versions, appeared in [31] and [34, Theorem 3, p. 322, Theorem 4, p. 323].

Exercise 2.4.1. Let H = L2(0,) with scalar product 0x 1(θ)x2(θ)dθ. Assume that in (2.19) 𝒜 denotes the generator of the left–shift semigroup on H, i.e.,

𝒜x = x,D(𝒜) = W1,2[0,) .

Prove that the observation functional

𝒞x = x(0),D(𝒞) D(A)

can be represented on D(𝒜) as 𝒞x = e(),x 𝒜. Indeed,

e(),x 𝒜 = e(),x L2(0,) e(),x L2(0,) = x(0) = 𝒞x .

Hence

𝒞x e() 𝒜x𝒜 = x𝒜

and 𝒞L(D𝒜, ). It follows from (2.15) that for x0 D(𝒜)

(Px0)(t) = 𝒞S(t)x0 = x0(t + θ) θ=1 = x0(t) .

Hence (2.21) holds with equality sign and with γ = 1. We have proved that 𝒞 is admissible. Verify that = I is a unique solution of (2.22).

Exercise 2.4.2. The operator 𝒜,

𝒜x = x x,D(𝒜) = W1,2[0,)

generates on H = L2(0,) the semigroup,

S(t)x0 (θ) = etx 0(t + θ),t,θ 0,x0 H ,

which is clearly EXS. Show that the observation functional

𝒞x = 𝒜x,hH,h L2(0,),h(θ) = eθ θ4 (2.26)

is not admissible. Indeed, observe that for x D(𝒜) we have

(Px0)(t) = 𝒞S(t)x0 = 𝒜S(t)x0,hH =0etx 0(t + θ) etx 0(t + θ) eθ θ4 dθ .

It is sufficient to show that P is unbounded. If we take the sequence {xn}n D(𝒜), xn(θ) = 2nenθ, xn H = 1 then

Pxn (t) =0etn2nen(θ+t) et2nen(θ+t) eθ θ4 dθ =

= et(n+1)(n + 1)2n0e(n+1)θ θ4 dθ = et(n+1)2nn + 14Γ 3 4 .

Thus

Pxn L2(0,)2 = n n + 1 Γ 3 4 2

and P maps the bounded sequence {xn}n into an unbounded one, which means that P is unbounded. Repeating the arguments above one can prove that the observation functional (2.26) with h replaced by h(θ) = θνeθ for ν (1 2, 0) is also not admissible.

Exercise 2.4.3. The following result is proved in [44, Theorem 4.1].

Theorem 2.4.3. Assume that 𝒜 generated an EXS semigroup {S(t)}t0 on H and 𝒞L(D𝒜,Y), 𝒞x = D𝒜x for x D(𝒜) and some D L(Y,H), is an addmissible observation operator. Suppose that L and R = L denote the generators of the semigroups of left– and right–shifts on L2(0,), viz.

Lf = f,D(L) = W1,2(0,) Rf = f,D(R) = {f W1,2(0,) : f(0) = 0} .

Let V L(H,L2(0,)),

(V x)(t) = hS(t)xV f =0S(t)hf(t)dt .

Then we have

R(V ) D(L),P¯ = LV P D(R) = V R P = V 𝒜R(V ) D(𝒜),P = 𝒜V .

Write down the explicit forms of P¯, P D(R), P and = PP¯. The last representation holds due to (2.24). Derive a formula expressing the Laplace transform P¯x̂ of P¯x, x H.

The next result [34, Theorem 5, p. 324] relates to the existence of a special type of solution to (2.22).

Definition 2.4.2. Assume that H1 and H2 are Hilbert spaces with scalar products ,H 1, ,H 2, respectively. An operator 𝒯 L(H1,H2) is called a Hilbert–Schmidt operator (HS) if there exists an orthonormal basis {ei}iJ in H1 such that

iJ 𝒯 ei H 22 < .

Theorem 2.4.4. Suppose that 𝒞L(D𝒜, ) is such that the estimate

𝒞S(t)x0 k(t) x0 Hx0 D(𝒜) and for almost all t 0 (2.27)

holds for some k L2(0,). Then (2.22) has a solution L(H), with = and 0, and is a HS operator.

Proof. Since (2.27) implies (2.21), it is sufficient to show that a solution , defined by (2.23), (2.24), is an HS operator [86, Theorem 6.12, p. 140, Theorem 6.10, p. 137]). Hence there exists an orthonormal basis {ej}iJ in H for which

iJ P¯ei L2(0,)2 = iJei,eH = iJ 1 2 ei H2 < ,

Thus 1 2 is an HS operator, which implies that is an HS operator. □

The assertion of this theorem remains true for a finite–rank operator 𝒞L(D𝒜,Y) satisfying (2.27). The assumption dim Y < is essential. The implication (2.21) (2.27) does not hold generally.

Remark 2.4.1. Assume that 𝒜 is the infinitesimal generator of an EXS linear analytic semigroup, and that 𝒞 is a linear functional bounded on Hα, where Hα (0 α 1) denotes the Banach space equal to D(𝒜α), the domain of a fractional power of 𝒜, equipped with the topology induced by the norm xHα = 𝒜αxH. Then (2.27) holds for α < 1 2. Indeed, there exists δ > 0 such that

𝒞S(t)x cα S(t)xHα = cα 𝒜αS(t)x H c¯αtαeδt x H

[69, p. 74]. For α < 1 2 the function k : ttαeδt belongs to L2(0,) [22, 860.19].

The last theorem of this section enables us to express the solution to the Lyapunov operator equation in the Riesz basis of eigenvectors of the semigroup generator.

Theorem 2.4.5. Suppose that H and Y are Hilbert spaces with inner products ,H and ,Y respectively. Let 𝒜 : (D(𝒜) H)H be a linear operator which is similar to a discrete normal operator and generating EXS linear C0–semigroup {S(t)}t0 on H. Also, let 𝒞L(D𝒜,Y) be admissible. Then the unique solution of the Lyapunov operator equation (2.22) has the spectral representation

x1,x2H = iJ Res λ=λi 𝒞(λ¯I 𝒜)1x 1,𝒞(λI 𝒜)1x 2 Yx1,x2 H (2.28)

Proof. It follows from Theorems 2.4.1 and 2.4.2 that (2.22) has a nique solution L(H) with = nd 0. As in Corollary 2.3.1, denote by {φi}ij the Riesz basis built from eigenvectors of 𝒜, nd by {ψi}ij its biorthogonal system. Putting x2 = φi for i J in (2.22), we obtain

(𝒜 + λ¯iI)x1,φiH = 𝒞x1,𝒞φiYx1 D(𝒜)i J (2.29)

By EXS, λ¯i ρ(𝒜), and 2.29) is equivalent to

f,φiH = 𝒞(λ¯iI + 𝒜)1f,𝒞φ iYf Hi J (2.30)

Taking into account the fact that {φi}iJ is a Riesz basis nd L(H), we get from (2.30) that

x1,x2H = x1, iJx2,ψiHφi H = iJx1,φiHx2,ψiH¯ = = iJ 𝒞(λ¯iI 𝒜)1x 1,𝒞(x2,ψiHφi) Y = = iJ 𝒞(λ¯iI 𝒜)1x 1,𝒞 Res λ=λi(λI 𝒜)1x 2 Yx1,x2 H .

In the next four sections, we give a complete solution of Problem 2.2.1.