]> 6.5 Example

6.5 Example

In this section we consider the following thermoelastic system (see [64]),
utt uzz + γθz =0,0 z π,t 0 θt + γuzt kθzz =0,0 z π,t 0 u(0,t) = u(π,t) =0, t 0 θ(0,t) = θ(π,t) =0, t 0 (6.65)

with appropriate initial conditions. Here u denotes the displacement, θ is the temperature deviation from the reference temperature and γ, k are positive constants depending on the material properties.

Substituting v = ut we can formally reduce the first two equations of (6.65) to the form

ut =v vt =uzz γθz θt = γvz + kθzz .

This suggests taking the Hilbert space

H =H01(0,π) ×L2(0,π) ×L2(0,π) , H01(0,π) = u H1(0,π) : u(0) = 0,u(π) = 0 ,

where H1(0,π) denotes the standard Sobolev space of the first order, as the state space with the scalar product

u1 v1 θ1 , u2 v2 θ2 H = u1,u2H 01+v1,v2L2+θ1,θ2L2 =

=0πu 1(z)u 2(z)dz +0πv 1(z)v2(z)dz +0πθ 1(z)θ2(z)dz .

In this space, the system equations can be rewritten in the abstract form

d dt u v θ = 𝒜u v θ

with

𝒜u v θ = v u γθ γv + kθ , D(𝒜) = H2(0,π) H 01(0,π) ×H 01(0,π) ×H2(0,π) H 01(0,π) (6.66)

where H2(0,π) denotes the standard Sobolev space of the second order.

Lemma 6.5.1. The operator 𝒜 has a compact resolvent.

Proof. It is enough to prove that 𝒜1 is compact. The form of 𝒜1 can be determined by solving the system of equations

v =U u γθ =V γv + kθ =Θ (6.67)

with boundary conditions

u(0) = u(π) = 0,v(0) = v(π) = 0,θ(0) = θ(π) = 0 (6.68)

Here U H01(0,π) and V , Θ L2(0,π). The solution of (6.67), (6.68) is

u(x) = γ k0x1 2(x s)2Θ(s)ds + γ2 k 0x(x s)U(s)ds 1 2x2 γ kπ0π(π s)Θ(s)ds + γ2 kπ0πU(s)ds +0x(x s)V (s)ds+ + x γ 2k0π(π s)Θ(s)ds + γ2 2k0πU(s)ds 1 π0π(π s)V (s)ds γ kπ0π1 2(π s)2Θ(s)ds γ2 kπ0π(π s)U(s)ds (6.69)
v = U (6.70)
θ(x) = x 1 kπ0π(π s)Θ(s)ds γ kπ0πU(s)ds + + 1 k0x(x s)Θ(s)ds + γ k0xU(s)ds (6.71)

Furthermore, u H01(0,π) H2(0,π), v H01(0,π) and θ H01(0,π) H2(0,π).

The operator

H01(0,π) Uγ2π 4k ff,UH 01 H01(0,π) ,

where f(x) = x 1 πx2, f H01(0,π) is of rank one, see [86, Theorem 6.1, p. 129], and therefore it is compact. The value of this operator at U is the function

[0,π] xγ2π 4k x 1 πx20π 1 2s π U(s)ds = γ2 2k x 1 πx20πU(s)ds .

Define the operator

Ru = u,D(R) = {u H 01(0,π) : u H3(0,π),u H 01(0,π)} .

It is self–adjoint with the inverse

(R1U)(x) =0x(x s)U(s)ds x π0π(π s)U(s)ds .

The operator R has a system of eigenvectors, corresponding to eigenvalues λn = n2, n , which forms an orthonormal basis in H01(0,π),

en(x) = 2 π sin nx n ,n .

Since

n=1R1e n H012 = n=1 1 λn 2 = n=1 1 n2 = π2 6 < ,

the operator R1 is by definition [86, p. 136] a HS–operator belonging to L(H01(0,π)). The system

en(x) = 2 π sin nx,n .

constitutes an orthonormal basis in L2(0,π). It is a system of eigenvectors of the extension of R onto L2(0,π), which will be denoted also by R,

Ru = u,D(R) = H 01(0,π) H2(0,π) .

The operator Φ L(L2(0,π),H 01(0,π)),

(ΦV )(x) =0x(x s)V (s)ds x π0π(π s)V (s)ds

is a HS–operator and therefore it is compact. Indeed,

n=1Φe n H012 = n=1e n 11 π0πe n(s)dsL22

2 n=1e n L22 + 2 n=111 π0π(π s)e n(s)dsL22 =

= 2 n=1e n L22 + 21 π n=10πe n(x)dx2 ,

where 1 denotes the constant function 1(x) = 1, 0 x 1. Hence, applying the Schwarz inequality we get

n=1Φe n H012 4 n=1e n L22 = 4 n=1 2 n2π0π sin 2nxdx = 4 n=1 1 n2 = 2π2 3 <

and Φ is a HS–operator directly by the definition, see [86, p. 136]. Next, consider the composition of the bounded integral operator

L2(0,π) Θ0()Θ(s)ds L2(0,π)

with Φ,

[0,π] x0x1 2(x s)2Θ(s)ds x π0π1 2(π s)2Θ(s)ds .

By the result of [86, Theorem 6.4, p. 131] it is a HS–operator from L(L2(0,π),H 01(0,π)) and we have proved that all operators defined by the right–hand side of (6.69) are compact.

The next step is to prove that the injection operator from H01(0,π) into L2(0,π), corresponding to the right–hand side of (6.70), is a HS–operator. Indeed,

n=1e n L22 = n=1 2 n2π0π sin 2nxdx = n=1 1 n2 = π2 6 <

and we recall once more the definition of a HS–operator given in [86, p. 136] to conclude that the injection operator is compact.

It remains to prove that the right–hand side of (6.71) is expressed also by compact operators. Observe that (6.71) can be written shortly as

θ = 1 kΦΘ + γ kΦU .

We already know that Φ is a HS–operator and therefore it suffices to prove that the composition operator Φ d dx is a HS–operator from L(H01(0,π),L2(0,φ)). In virtue of [86, Theorem 6.4, p. 131] this requires proving that the first order differentiation operator from H01(0,π) to L2(0,φ) is bounded. However, this easily follows from the definition, together with the fact that the norm of such an operator equals 1.

Now, all operators defining the right–hand sides of (6.69), (6.70), (6.71) are compact which completes the proof of compactness of 𝒜1. □

Lemma 6.5.2. The operator (6.66) generates a semigroup {S(t)}t0 of contractions on H.

Proof. For u v θ D(𝒜) we have

𝒜u v θ , u v θ H = v u γθ γv + kθ , u v θ H =

=0πv(z)u(z)dz +0πu(z)v(z)dz 0πγθ(z)v(z)dz 0πγv(z)θ(z)dz+

+0πkθ(z)θ(z)dz = 0πk θ(z) 2dz = k θ L22 0 .

Hence 𝒜 is dissipative. From the definition of an adjoint operator [86, p. 68] we find

𝒜u v θ = v u + γθ γv + kθ ,D(𝒜) = D(𝒜)

and it is not difficult to see that

𝒲𝒜𝒲 = 𝒜,𝒲u v θ = u v θ = 𝒲u v θ = 𝒲1 u v θ .

This implies that 𝒜 is also dissipative and thus, by [69, Corollary 4.4, p. 15], 𝒜 generates a C0–semigroup of contractions on H. □

Lemma 6.5.3. The semigroup {S(t)}t0 is AS,

limtS(t) u v θ = 0 0 0 u v θ H (6.72)

Proof. By the result of [63, Lemma 3.1, p. 497] we have

S(t) u v θ D(𝒜)Hiast u v θ D(𝒜) ,

where

Hi := u v θ H : S(t) u v θ H = u v θ H

is the subspace on which S(t) acts isometrically. It follows from the proof of Lemma 6.5.2 that

D(𝒜)Hi = u v θ D(𝒜) : θ(t) = 0t .

It is an invariant subspace, on which the two first components of S(t) u v θ coincide with the unitary semigroup generated on H01(0,π) ×L2(0,π) by the skew–adjoint operator

𝒦u v = v u ,D(𝒦) = [H01(0,π)H 2(0,π)]×H01(0,π) .

Hence

u(t,z) =φ(t + z) φ(t z),t ,0 z π v(t,z) =ut(t,z), t ,0 z π

where φ H2() is a 2π–periodic function to ensure

u(t,π) = φ(t + π) φ(t π) = 0t .

But θ satisfies the equation

θt(t,z) = γvz(t,z) = γ zt φ(t + z) φ(t z)

and therefore, integrating both sides from 0 to t, we get

θ(t,z) θ(0,z) = γ z φ(t + z) φ(t z) φ(z) + φ(z) .

Since θz(t,z) = 0t we have

2 z2 φ(t + z) φ(t z) φ(z) + φ(z) = 0 .

This means that

φ(t + z) φ(t z) φ(z) + φ(z) = α(t)z + β(t) .

Substituting z = 0 one obtains β(t) = 0t , while for z = π one gets α(t)π = 0. Hence

φ(t + z) φ(t z) = φ(z) φ(z) .

However this means that the function zφ(z) φ(z) is t–periodic for any t , which is possible only if φ = 0. This yields u(t) = v(t) = θ(t) = 0 and thus

D(𝒜)Hi = 0 0 0 .

We have proved that

S(t) u v θ s 0 0 0 ast u v θ D(𝒜) .

Since {S(t)}t0 is a semigroup of contractions and D(𝒜) is dense in H, we can apply the results of [86, Problem 4.28, p. 81] to conclude that (6.72) holds. □

Lemma 6.5.3 has an important consequence when considering the lq problem (6.6) for the system (6.1) with A = 𝒜 given by (6.66) and with a compact operator . If is a compact operator and the solution of lq problem (6.6) exists then 𝒢 is compact too and, by the result of [81], {S(t)}t0 is EXS. But if {S(t)}t0 is EXS then (𝒜,) is stabilizable, (𝒜,𝒞) is detectable and the lq problem has a unique solution. Thus if is compact, the exponential stability of the semigroup {S(t)}t0 is necessary and sufficient for solvability of the lq problem. In this context the result of Lin and Zheng [64] is extremely important. They have proved, using the Prüss–Huang–Weiss theorem [71], [50], [87], that {S(t)}t0 is EXS. The proof is very technical.

Remark 6.5.1. A more fine structure of 𝒜 can be exhibited,

𝒜 = 𝒩+𝒫,𝒩 u v θ = v u kθ ,𝒫u v θ = 0 γθ γv ,D(𝒜) = D(𝒩) = D(𝒫)

where 𝒩 is a normal discrete dissipative operator and 𝒫 is a skew–adjoint operator.

Since σ(𝒩), the spectrum of 𝒩, is equally spaced on j we cannot apply any perturbation result to conclude that 𝒜 is similar to a normal operator, which is supposed to be true. If it would be the case then the proof of EXS of the semigroup {S(t)}t0 were considerably simplified. For the time being we do not know whether there exists any simpler proof that {S(t)}t0 is EXS.