]> 6.5 Example

### 6.5 Example

In this section we consider the following thermoelastic system (see ),
 $\left\{\begin{array}{ccccc}\hfill {u}_{tt}-{u}_{zz}+\gamma {\theta }_{z}& \hfill =\hfill & 0,\hfill & \hfill 0\le z\le \pi ,\hfill & t\ge 0\hfill \\ \hfill {\theta }_{t}+\gamma {u}_{zt}-k{\theta }_{zz}& \hfill =\hfill & 0,\hfill & \hfill 0\le z\le \pi ,\hfill & t\ge 0\hfill \\ \hfill u\left(0,t\right)=u\left(\pi ,t\right)& \hfill =\hfill & 0,\hfill & \hfill \hfill & t\ge 0\hfill \\ \hfill \theta \left(0,t\right)=\theta \left(\pi ,t\right)& \hfill =\hfill & 0,\hfill & \hfill \hfill & t\ge 0\hfill \end{array}\right\}$ (6.65)

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

Substituting $v={u}_{t}$ we can formally reduce the ﬁrst two equations of (6.65) to the form

$\left\{\begin{array}{ccc}\hfill {u}_{t}& \hfill =\hfill & v\hfill \\ \hfill {v}_{t}& \hfill =\hfill & {u}_{zz}-\gamma {\theta }_{z}\hfill \\ \hfill {\theta }_{t}& \hfill =\hfill & -\gamma {v}_{z}+k{\theta }_{zz}\hfill \end{array}\right\}.$

This suggests taking the Hilbert space

$\begin{array}{ccc}\hfill \text{H}& \hfill =\hfill & {\text{H}}_{0}^{1}\left(0,\pi \right)×{\text{L}}^{2}\left(0,\pi \right)×{\text{L}}^{2}\left(0,\pi \right),\hfill \\ \hfill {\text{H}}_{0}^{1}\left(0,\pi \right)& \hfill =\hfill & \left\{u\in {\text{H}}^{1}\left(0,\pi \right):\phantom{\rule{0ex}{0ex}}u\left(0\right)=0,\phantom{\rule{0ex}{0ex}}u\left(\pi \right)=0\right\},\hfill \end{array}$

where ${\text{H}}^{1}\left(0,\pi \right)$ denotes the standard Sobolev space of the ﬁrst order, as the state space with the scalar product

${〈\left[\begin{array}{c}\hfill {u}_{1}\hfill \\ \hfill {v}_{1}\hfill \\ \hfill {\theta }_{1}\hfill \end{array}\right],\left[\begin{array}{c}\hfill {u}_{2}\hfill \\ \hfill {v}_{2}\hfill \\ \hfill {\theta }_{2}\hfill \end{array}\right]〉}_{\text{H}}={〈{u}_{1},{u}_{2}〉}_{{\text{H}}_{0}^{1}}+{〈{v}_{1},{v}_{2}〉}_{{\text{L}}^{2}}+{〈{\theta }_{1},{\theta }_{2}〉}_{{\text{L}}^{2}}=$

$={\int }_{0}^{\pi }{u}_{1}^{\prime }\left(z\right){u}_{2}^{\prime }\left(z\right)dz+{\int }_{0}^{\pi }{v}_{1}\left(z\right){v}_{2}\left(z\right)dz+{\int }_{0}^{\pi }{\theta }_{1}\left(z\right){\theta }_{2}\left(z\right)dz.$

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

$\frac{d}{dt}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\mathcal{𝒜}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]$

with

 $\left\{\begin{array}{c}\mathcal{𝒜}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\left[\begin{array}{c}\hfill v\hfill \\ \hfill {u}^{\prime \prime }-\gamma {\theta }^{\prime }\hfill \\ \hfill -\gamma {v}^{\prime }+k{\theta }^{\prime \prime }\hfill \end{array}\right],\hfill \\ D\left(\mathcal{𝒜}\right)=\left[{\text{H}}^{2}\left(0,\pi \right)\cap {\text{H}}_{0}^{1}\left(0,\pi \right)\right]×{\text{H}}_{0}^{1}\left(0,\pi \right)×\left[{\text{H}}^{2}\left(0,\pi \right)\cap {\text{H}}_{0}^{1}\left(0,\pi \right)\right]\hfill \end{array}\right\}$ (6.66)

where ${\text{H}}^{2}\left(0,\pi \right)$ denotes the standard Sobolev space of the second order.

Lemma 6.5.1. The operator $\mathcal{𝒜}$ has a compact resolvent.

Proof. It is enough to prove that ${\mathcal{𝒜}}^{-1}$ is compact. The form of ${\mathcal{𝒜}}^{-1}$ can be determined by solving the system of equations

 $\left\{\begin{array}{ccc}\hfill v& \hfill =\hfill & U\hfill \\ \hfill {u}^{\prime \prime }-\gamma {\theta }^{\prime }& \hfill =\hfill & V\hfill \\ \hfill -\gamma {v}^{\prime }+k{\theta }^{\prime \prime }& \hfill =\hfill & \Theta \hfill \end{array}\right\}$ (6.67)

with boundary conditions

 $u\left(0\right)=u\left(\pi \right)=0,\phantom{\rule{2em}{0ex}}v\left(0\right)=v\left(\pi \right)=0,\phantom{\rule{1em}{0ex}}\theta \left(0\right)=\theta \left(\pi \right)=0$ (6.68)

Here $U\in {\text{H}}_{0}^{1}\left(0,\pi \right)$ and $V$, $\Theta \in {\text{L}}^{2}\left(0,\pi \right)$. The solution of (6.67), (6.68) is

 $\begin{array}{c}u\left(x\right)=\frac{\gamma }{k}{\int }_{0}^{x}\frac{1}{2}{\left(x-s\right)}^{2}\Theta \left(s\right)ds+\frac{{\gamma }^{2}}{k}{\int }_{0}^{x}\left(x-s\right)U\left(s\right)ds-\hfill \\ -\frac{1}{2}{x}^{2}\left[\frac{\gamma }{k\pi }{\int }_{0}^{\pi }\left(\pi -s\right)\Theta \left(s\right)ds+\frac{{\gamma }^{2}}{k\pi }{\int }_{0}^{\pi }U\left(s\right)ds\right]+{\int }_{0}^{x}\left(x-s\right)V\left(s\right)ds+\hfill \\ +x\left[\frac{\gamma }{2k}{\int }_{0}^{\pi }\left(\pi -s\right)\Theta \left(s\right)ds+\frac{{\gamma }^{2}}{2k}{\int }_{0}^{\pi }U\left(s\right)ds-\frac{1}{\pi }{\int }_{0}^{\pi }\left(\pi -s\right)V\left(s\right)ds-\right\hfill \\ -\frac{\gamma }{k\pi }{\int }_{0}^{\pi }\frac{1}{2}{\left(\pi -s\right)}^{2}\Theta \left(s\right)ds-\frac{{\gamma }^{2}}{k\pi }{\int }_{0}^{\pi }\left(\pi -s\right)U\left(s\right)ds]\hfill \end{array}$ (6.69)
 $v=U$ (6.70)
 $\begin{array}{c}\hfill \theta \left(x\right)=x\left[-\frac{1}{k\pi }{\int }_{0}^{\pi }\left(\pi -s\right)\Theta \left(s\right)ds-\frac{\gamma }{k\pi }{\int }_{0}^{\pi }U\left(s\right)ds\right]+\hfill \\ \hfill +\frac{1}{k}{\int }_{0}^{x}\left(x-s\right)\Theta \left(s\right)ds+\frac{\gamma }{k}{\int }_{0}^{x}U\left(s\right)ds\hfill \end{array}$ (6.71)

Furthermore, $u\in {\text{H}}_{0}^{1}\left(0,\pi \right)\cap {\text{H}}^{2}\left(0,\pi \right)$, $v\in {\text{H}}_{0}^{1}\left(0,\pi \right)$ and $\theta \in {\text{H}}_{0}^{1}\left(0,\pi \right)\cap {\text{H}}^{2}\left(0,\pi \right)$.

The operator

${\text{H}}_{0}^{1}\left(0,\pi \right)\ni U↦\frac{{\gamma }^{2}\pi }{4k}f{〈f,U〉}_{{\text{H}}_{0}^{1}}\in {\text{H}}_{0}^{1}\left(0,\pi \right),$

where $f\left(x\right)=x-\frac{1}{\pi }{x}^{2}$, $f\in {\text{H}}_{0}^{1}\left(0,\pi \right)$ 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

$\left[0,\pi \right]\ni x↦\frac{{\gamma }^{2}\pi }{4k}\left(x-\frac{1}{\pi }{x}^{2}\right){\int }_{0}^{\pi }\left(1-\frac{2s}{\pi }\right){U}^{\prime }\left(s\right)ds=\frac{{\gamma }^{2}}{2k}\left(x-\frac{1}{\pi }{x}^{2}\right){\int }_{0}^{\pi }U\left(s\right)ds.$

Deﬁne the operator

$Ru={u}^{\prime \prime },\phantom{\rule{2em}{0ex}}D\left(R\right)=\left\{u\in {\text{H}}_{0}^{1}\left(0,\pi \right):\phantom{\rule{0ex}{0ex}}u\in {\text{H}}^{3}\left(0,\pi \right),{u}^{\prime \prime }\in {\text{H}}_{0}^{1}\left(0,\pi \right)\right\}.$

It is self–adjoint with the inverse

$\left({R}^{-1}U\right)\left(x\right)={\int }_{0}^{x}\left(x-s\right)U\left(s\right)ds-\frac{x}{\pi }{\int }_{0}^{\pi }\left(\pi -s\right)U\left(s\right)ds.$

The operator $R$ has a system of eigenvectors, corresponding to eigenvalues ${\lambda }_{n}=-{n}^{2}$, $n\in \mathbb{ℕ}$, which forms an orthonormal basis in ${\text{H}}_{0}^{1}\left(0,\pi \right)$,

${e}_{n}\left(x\right)=\sqrt{\frac{2}{\pi }}\frac{sinnx}{n},\phantom{\rule{2em}{0ex}}n\in \mathbb{ℕ}.$

Since

$\sum _{n=1}^{\infty }{∥{R}^{-1}{e}_{n}∥}_{{\text{H}}_{0}^{1}}^{2}=\sum _{n=1}^{\infty }\frac{1}{{\left|{\lambda }_{n}\right|}^{2}}=\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}=\frac{{\pi }^{2}}{6}<\infty ,$

the operator ${R}^{-1}$ is by deﬁnition [86, p. 136] a HS–operator belonging to $L\left({\text{H}}_{0}^{1}\left(0,\pi \right)\right)$. The system

${e}_{n}^{\prime }\left(x\right)=\sqrt{\frac{2}{\pi }}sinnx,\phantom{\rule{2em}{0ex}}n\in \mathbb{ℕ}.$

constitutes an orthonormal basis in ${\text{L}}^{2}\left(0,\pi \right)$. It is a system of eigenvectors of the extension of $R$ onto ${\text{L}}^{2}\left(0,\pi \right)$, which will be denoted also by $R$,

$Ru={u}^{\prime \prime },\phantom{\rule{2em}{0ex}}D\left(R\right)={\text{H}}_{0}^{1}\left(0,\pi \right)\cap {\text{H}}^{2}\left(0,\pi \right).$

The operator $\Phi \in L\left({\text{L}}^{2}\left(0,\pi \right),{\text{H}}_{0}^{1}\left(0,\pi \right)\right)$,

$\left(\Phi V\right)\left(x\right)={\int }_{0}^{x}\left(x-s\right)V\left(s\right)ds-\frac{x}{\pi }{\int }_{0}^{\pi }\left(\pi -s\right)V\left(s\right)ds$

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

$\sum _{n=1}^{\infty }{∥\Phi {e}_{n}^{\prime }∥}_{{\text{H}}_{0}^{1}}^{2}=\sum _{n=1}^{\infty }{∥{e}_{n}-1\frac{1}{\pi }{\int }_{0}^{\pi }{e}_{n}\left(s\right)ds∥}_{{\text{L}}^{2}}^{2}\le$

$\le 2\sum _{n=1}^{\infty }{∥{e}_{n}∥}_{{\text{L}}^{2}}^{2}+2\sum _{n=1}^{\infty }{∥1\frac{1}{\pi }{\int }_{0}^{\pi }\left(\pi -s\right){e}_{n}\left(s\right)ds∥}_{{\text{L}}^{2}}^{2}=$

$=2\sum _{n=1}^{\infty }{∥{e}_{n}∥}_{{\text{L}}^{2}}^{2}+2\frac{1}{\pi }\sum _{n=1}^{\infty }{\left|{\int }_{0}^{\pi }{e}_{n}\left(x\right)dx\right|}^{2},$

where $1$ denotes the constant function $1\left(x\right)=1$, $0\le x\le 1$. Hence, applying the Schwarz inequality we get

$\sum _{n=1}^{\infty }{∥\Phi {e}_{n}^{\prime }∥}_{{\text{H}}_{0}^{1}}^{2}\le 4\sum _{n=1}^{\infty }{∥{e}_{n}∥}_{{\text{L}}^{2}}^{2}=4\sum _{n=1}^{\infty }\frac{2}{{n}^{2}\pi }{\int }_{0}^{\pi }{sin}^{2}nxdx=4\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}=\frac{2{\pi }^{2}}{3}<\infty$

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

${\text{L}}^{2}\left(0,\pi \right)\ni \Theta ↦{\int }_{0}^{\left(\cdot \right)}\Theta \left(s\right)ds\in {\text{L}}^{2}\left(0,\pi \right)$

with $\Phi$,

$\left[0,\pi \right]\ni x↦{\int }_{0}^{x}\frac{1}{2}{\left(x-s\right)}^{2}\Theta \left(s\right)ds-\frac{x}{\pi }{\int }_{0}^{\pi }\frac{1}{2}{\left(\pi -s\right)}^{2}\Theta \left(s\right)ds.$

By the result of [86, Theorem 6.4, p. 131] it is a HS–operator from $L\left({\text{L}}^{2}\left(0,\pi \right),{\text{H}}_{0}^{1}\left(0,\pi \right)\right)$ 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 ${\text{H}}_{0}^{1}\left(0,\pi \right)$ into ${\text{L}}^{2}\left(0,\pi \right)$, corresponding to the right–hand side of (6.70), is a HS–operator. Indeed,

$\sum _{n=1}^{\infty }{∥{e}_{n}∥}_{{\text{L}}^{2}}^{2}=\sum _{n=1}^{\infty }\frac{2}{{n}^{2}\pi }{\int }_{0}^{\pi }{sin}^{2}nxdx=\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}=\frac{{\pi }^{2}}{6}<\infty$

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

$\theta =\frac{1}{k}\Phi \Theta +\frac{\gamma }{k}\Phi {U}^{\prime }.$

We already know that $\Phi$ is a HS–operator and therefore it suffices to prove that the composition operator $\Phi \frac{d}{dx}$ is a HS–operator from $L\left({\text{H}}_{0}^{1}\left(0,\pi \right),{\text{L}}^{2}\left(0,\phi \right)\right)$. In virtue of [86, Theorem 6.4, p. 131] this requires proving that the ﬁrst order differentiation operator from ${\text{H}}_{0}^{1}\left(0,\pi \right)$ to ${\text{L}}^{2}\left(0,\phi \right)$ 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 ${\mathcal{𝒜}}^{-1}$. □

Lemma 6.5.2. The operator (6.66) generates a semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ of contractions on $\text{H}$.

Proof. For $\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\in D\left(\mathcal{𝒜}\right)$ we have

${〈\mathcal{𝒜}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right],\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]〉}_{\text{H}}={〈\left[\begin{array}{c}\hfill v\hfill \\ \hfill {u}^{\prime \prime }-\gamma {\theta }^{\prime }\hfill \\ \hfill -\gamma {v}^{\prime }+k{\theta }^{\prime \prime }\hfill \end{array}\right],\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]〉}_{\text{H}}=$

$={\int }_{0}^{\pi }{v}^{\prime }\left(z\right){u}^{\prime }\left(z\right)dz+{\int }_{0}^{\pi }{u}^{\prime \prime }\left(z\right)v\left(z\right)dz-{\int }_{0}^{\pi }\gamma {\theta }^{\prime }\left(z\right)v\left(z\right)dz-{\int }_{0}^{\pi }\gamma {v}^{\prime }\left(z\right)\theta \left(z\right)dz+$

$+{\int }_{0}^{\pi }k{\theta }^{\prime \prime }\left(z\right)\theta \left(z\right)dz=-{\int }_{0}^{\pi }k{\left[{\theta }^{\prime }\left(z\right)\right]}^{2}dz=-k{∥{\theta }^{\prime }∥}_{{\text{L}}^{2}}^{2}\le 0.$

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

${\mathcal{𝒜}}^{\ast }\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\left[\begin{array}{c}\hfill -v\hfill \\ \hfill -{u}^{\prime \prime }+\gamma {\theta }^{\prime }\hfill \\ \hfill \gamma {v}^{\prime }+k{\theta }^{\prime \prime }\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}D\left({\mathcal{𝒜}}^{\ast }\right)=D\left(\mathcal{𝒜}\right)$

and it is not difficult to see that

$\mathcal{𝒲}\mathcal{𝒜}\mathcal{𝒲}={\mathcal{𝒜}}^{\ast },\phantom{\rule{2em}{0ex}}\mathcal{𝒲}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\left[\begin{array}{c}\hfill u\\ \hfill -v\\ \hfill \theta \end{array}\right]={\mathcal{𝒲}}^{\ast }\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]={\mathcal{𝒲}}^{-1}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right].$

This implies that ${\mathcal{𝒜}}^{\ast }$ is also dissipative and thus, by [69, Corollary 4.4, p. 15], $\mathcal{𝒜}$ generates a ${\text{C}}_{0}$–semigroup of contractions on $\text{H}$. □

Lemma 6.5.3. The semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is AS,

 ${lim}_{t\to \infty }S\left(t\right)\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\forall \left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\in \text{H}$ (6.72)

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

$S\left(t\right)\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\to D\left(\mathcal{𝒜}\right)\cap {\text{H}}_{i}\phantom{\rule{1em}{0ex}}\text{as}\phantom{\rule{1em}{0ex}}t\to \infty \phantom{\rule{2em}{0ex}}\forall \left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\in D\left(\mathcal{𝒜}\right),$

where

${\text{H}}_{i}:=\left\{\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\in \text{H}:\phantom{\rule{0ex}{0ex}}{∥S\left(t\right)\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]∥}_{\text{H}}={∥\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]∥}_{\text{H}}\right\}$

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

$D\left(\mathcal{𝒜}\right)\cap {\text{H}}_{i}=\left\{\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\in D\left(\mathcal{𝒜}\right):\phantom{\rule{0ex}{0ex}}{\theta }^{\prime }\left(t\right)=0\phantom{\rule{2em}{0ex}}\forall t\in \mathbb{ℝ}\right\}.$

It is an invariant subspace, on which the two ﬁrst components of $S\left(t\right)\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]$coincide with the unitary semigroup generated on ${\text{H}}_{0}^{1}\left(0,\pi \right)×{\text{L}}^{2}\left(0,\pi \right)$ by the skew–adjoint operator

$\mathcal{𝒦}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \end{array}\right]=\left[\begin{array}{c}\hfill v\hfill \\ \hfill {u}^{\prime \prime }\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}D\left(\mathcal{𝒦}\right)=\left[{\text{H}}_{0}^{1}\left(0,\pi \right)\cap {\text{H}}_{2}\left(0,\pi \right)\right]×{\text{H}}_{0}^{1}\left(0,\pi \right).$

Hence

$\begin{array}{ccccc}\hfill u\left(t,z\right)& \hfill =\hfill & \phi \left(t+z\right)-\phi \left(t-z\right),\hfill & t\in \mathbb{ℝ},\hfill & 0\le z\le \pi \hfill \\ \hfill v\left(t,z\right)& \hfill =\hfill & {u}_{t}\left(t,z\right),\hfill & t\in \mathbb{ℝ},\hfill & 0\le z\le \pi \hfill \end{array}$

where $\phi \in {\text{H}}^{2}\left(\mathbb{ℝ}\right)$ is a $2\pi$–periodic function to ensure

$u\left(t,\pi \right)=\phi \left(t+\pi \right)-\phi \left(t-\pi \right)=0\phantom{\rule{2em}{0ex}}\forall t\in \mathbb{ℝ}.$

But $\theta$ satisﬁes the equation

${\theta }_{t}\left(t,z\right)=-\gamma {v}_{z}\left(t,z\right)=-\gamma \frac{\partial }{\partial z\partial t}\left[\phi \left(t+z\right)-\phi \left(t-z\right)\right]$

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

$\theta \left(t,z\right)-\theta \left(0,z\right)=-\gamma \frac{\partial }{\partial z}\left[\phi \left(t+z\right)-\phi \left(t-z\right)-\phi \left(z\right)+\phi \left(-z\right)\right].$

Since ${\theta }_{z}\left(t,z\right)=0\phantom{\rule{1em}{0ex}}\forall t\in \mathbb{ℝ}$ we have

$\frac{{\partial }^{2}}{\partial {z}^{2}}\left[\phi \left(t+z\right)-\phi \left(t-z\right)-\phi \left(z\right)+\phi \left(-z\right)\right]=0.$

This means that

$\phi \left(t+z\right)-\phi \left(t-z\right)-\phi \left(z\right)+\phi \left(-z\right)=\alpha \left(t\right)z+\beta \left(t\right).$

Substituting $z=0$ one obtains $\beta \left(t\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\forall t\in \mathbb{ℝ}$, while for $z=\pi$ one gets $\alpha \left(t\right)\pi =0$. Hence

$\phi \left(t+z\right)-\phi \left(t-z\right)=\phi \left(z\right)-\phi \left(-z\right).$

However this means that the function $z↦\phi \left(z\right)-\phi \left(-z\right)$ is $t$–periodic for any $t\in \mathbb{ℝ}$, which is possible only if $\phi =0$. This yields $u\left(t\right)=v\left(t\right)=\theta \left(t\right)=0$ and thus

$D\left(\mathcal{𝒜}\right)\cap {\text{H}}_{i}=\left\{\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\right\}.$

We have proved that

$S\left(t\right)\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\stackrel{s}{\to }\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\phantom{\rule{1em}{0ex}}\text{as}\phantom{\rule{1em}{0ex}}t\to \infty \phantom{\rule{2em}{0ex}}\forall \left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]\in D\left(\mathcal{𝒜}\right).$

Since ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is a semigroup of contractions and $D\left(\mathcal{𝒜}\right)$ is dense in $\text{H}$, 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 $A=\mathcal{𝒜}$ given by (6.66) and with a compact operator $\mathcal{ℬ}$. If $\mathcal{ℬ}$ is a compact operator and the solution of lq problem (6.6) exists then $-\mathcal{ℬ}\mathcal{𝒢}$ is compact too and, by the result of , ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is EXS. But if ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is EXS then $\left(\mathcal{𝒜},\mathcal{ℬ}\right)$ is stabilizable, $\left(\mathcal{𝒜},\mathcal{𝒞}\right)$ is detectable and the lq problem has a unique solution. Thus if $\mathcal{ℬ}$ is compact, the exponential stability of the semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is necessary and sufficient for solvability of the lq problem. In this context the result of Lin and Zheng  is extremely important. They have proved, using the Prüss–Huang–Weiss theorem , , , that ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is EXS. The proof is very technical.

Remark 6.5.1. A more ﬁne structure of $\mathcal{𝒜}$ can be exhibited,

$\mathcal{𝒜}=\mathcal{𝒩}+\mathcal{𝒫},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathcal{𝒩}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\left[\begin{array}{c}\hfill v\hfill \\ \hfill {u}^{\prime \prime }\hfill \\ \hfill k{\theta }^{\prime \prime }\hfill \end{array}\right],\mathcal{𝒫}\left[\begin{array}{c}\hfill u\hfill \\ \hfill v\hfill \\ \hfill \theta \hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -\gamma {\theta }^{\prime }\hfill \\ \hfill -\gamma {v}^{\prime }\hfill \end{array}\right],\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}D\left(\mathcal{𝒜}\right)=D\left(\mathcal{𝒩}\right)=D\left(\mathcal{𝒫}\right)$

where $\mathcal{𝒩}$ is a normal discrete dissipative operator and $\mathcal{𝒫}$ is a skew–adjoint operator.

Since $\sigma \left(\mathcal{𝒩}\right)$, the spectrum of $\mathcal{𝒩}$, is equally spaced on $j\mathbb{ℝ}$ we cannot apply any perturbation result to conclude that $\mathcal{𝒜}$ 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 ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ were considerably simpliﬁed. For the time being we do not know whether there exists any simpler proof that ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ is EXS.