]> 8.3 Abstract dynamical model

### 8.3 Abstract dynamical model

Let
$x\left(t\right)\left(\theta \right)=z\left(t+\theta \right),\phantom{\rule{2em}{0ex}}t\ge 0,\phantom{\rule{1em}{0ex}}\theta \in \left[-r,0\right].$

Then

$\frac{\partial }{\partial t}x\left(t\right)\left(\theta \right)=\frac{\partial }{\partial t}z\left(t+\theta \right)=\frac{\partial }{\partial \theta }z\left(t+\theta \right)=\frac{\partial }{\partial \theta }x\left(t\right)\left(\theta \right).$

In the Hilbert space $\text{H}={\text{L}}^{2}\left(-r,0\right)\oplus {\text{L}}^{2}\left(-r,0\right)$ with standard scalar product, the system–dynamics equations (8.5) can be written in the abstract form proposed in Section 7.6:

 $\left\{\begin{array}{ccc}\hfill ẋ\left(t\right)& \hfill =\hfill & \sigma x\left(t\right)\hfill \\ \hfill \tau x\left(t\right)& \hfill =\hfill & u\left(t\right)\hfill \\ \hfill y\left(t\right)& \hfill =\hfill & {c}^{#}x\left(t\right)\hfill \end{array}\right\}$ (8.6)

where

$\sigma x={x}^{\prime },\phantom{\rule{2em}{0ex}}D\left(\sigma \right)=\left\{x=\left[\begin{array}{c}{x}_{1}\hfill \\ {x}_{2}\hfill \end{array}\right]\in {\text{W}}^{1,2}\left(-r,0\right)\oplus {\text{W}}^{1,2}\left(-r,0\right):\phantom{\rule{0ex}{0ex}}{x}_{1}\left(0\right)={x}_{2}\left(-r\right)\right\},$

$\tau x={x}_{2}\left(0\right)+b{x}_{1}\left(-r\right),\phantom{\rule{2em}{0ex}}D\left(\tau \right)\supset \text{C}\left[-r,0\right]\oplus \text{C}\left[-r,0\right]\supset D\left(\sigma \right),$

${c}^{#}x={c}_{0}^{\ast }x\left(-r\right)=a{x}_{2}\left(-r\right),\phantom{\rule{2em}{0ex}}D\left({c}^{#}\right)\supset \text{C}\left[-r,0\right]\oplus \text{C}\left[-r,0\right],$

and $\sigma$ is a closed linear operator. Notice that the ﬁrst equation of the system in (8.5) was encountered into the deﬁnition of $\sigma$ while the second one deﬁnes $\tau$.

Now we seek the factor control vector $d\in D\left(\sigma \right)$ such that $\sigma d=0$ and $\tau d=1$. Elementary calculations show that

$d=\frac{1}{1+b}\left[\begin{array}{c}1\hfill \\ 1\hfill \end{array}\right]$

where $1$ denotes the constant function taking the value $1$ on $\left[-r,0\right]$. The knowledge of the factor control vector allows us to derive, as in Section 7.6, a ﬁnal form of an abstract model of the system:

 $\left\{\begin{array}{ccc}\hfill ẋ\left(t\right)& \hfill =\hfill & \mathcal{𝒜}\left[x\left(t\right)-u\left(t\right)d\right]\hfill \\ \hfill y\left(t\right)& \hfill =\hfill & {c}^{#}x\left(t\right)\hfill \end{array}\right\}$ (8.7)

where $\mathcal{𝒜}={\sigma |}_{ker\phantom{\rule{0ex}{0ex}}\tau }$, i.e.

 $\mathcal{𝒜}x={x}^{\prime },\phantom{\rule{2em}{0ex}}D\left(\mathcal{𝒜}\right)=\left\{x\in {\text{W}}^{1,2}\left(-r,0\right)\oplus {\text{W}}^{1,2}\left(-r,0\right):\phantom{\rule{0ex}{0ex}}x\left(0\right)=Ax\left(-r\right)\right\}$ (8.8)

From the theory of neutral delay systems, $\mathcal{𝒜}$ generates a ${\text{C}}_{0}$–semigroup on $\text{H}$ (or even a ${\text{C}}_{0}$–group if $detA\ne 0$). This semigroup is EXS iff $\left|\lambda \left(A\right)\right|<1$ [30, p. 148 - 154] or equivalently $\left|b\right|<1$ . Since $\rho >1$ and $\kappa \in \left[-1,1\right]$, these inequalities are always satisﬁed. Observe that the condition $\kappa =1$ corresponds to ${R}_{1}=\infty$ (unloaded line), while $\kappa =0$ means that ${R}_{1}=Z$ (the line loaded by the wave impedance), and $\kappa =-1$ means ${R}_{1}=0$ (shunted line). To get the transfer function of (8.7), we rewrite the ﬁrst equation of the system (8.7) in an equivalent form

${\mathcal{𝒜}}^{-1}ẋ\left(t\right)=\frac{d}{dt}\left[{\mathcal{𝒜}}^{-1}x\left(t\right)\right]=x\left(t\right)-u\left(t\right)d.$

Next, assuming null initial conditions and applying the Laplace transform we get

$\begin{array}{ccc}\hfill \stackrel{̂}{x}\left(s\right)& \hfill =\hfill & {\left(s{\mathcal{𝒜}}^{-1}-I\right)}^{-1}dû\left(s\right)={\left[\left(sI-\mathcal{𝒜}\right){\mathcal{𝒜}}^{-1}\right]}^{-1}û\left(s\right)=\hfill \\ \hfill & \hfill =\hfill & \mathcal{𝒜}{\left(sI-\mathcal{𝒜}\right)}^{-1}dû\left(s\right)=\left[d-{\left(sI-\mathcal{𝒜}\right)}^{-1}d\right]û\left(s\right).\hfill \end{array}$

Hence,

 $ĝ\left(\lambda \right)={c}^{#}\mathcal{𝒜}{\left(\mathcal{𝒜}-\lambda I\right)}^{-1}d={c}^{#}d-\lambda {c}^{#}{\left(\lambda I-\mathcal{𝒜}\right)}^{-1}d$ (8.9)

because $d\in D\left({c}^{#}\right)$, i.e., the compatibility assumption (7.5) holds. Now, by (8.8),

 $\begin{array}{c}\mathcal{𝒜}{\left(\mathcal{𝒜}-\lambda I\right)}^{-1}d=\hfill \\ =\frac{d}{d\theta }\left\{{e}^{\lambda \theta }\left[I-{\left(I-{e}^{-\lambda r}A\right)}^{-1}\right]{\stackrel{̂}{d}}_{F}\left(\lambda \right)-{\int }_{\theta }^{0}{e}^{\lambda \left(\theta -\tau \right)}d\phantom{\rule{0ex}{0ex}}d\tau \right\}=\hfill \\ =\frac{{e}^{\lambda \theta }}{1+b}\left[\left({e}^{-\lambda r}-1\right)A{\left(I-{e}^{-\lambda r}A\right)}^{-1}+I\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]=\hfill \\ =\frac{{e}^{\lambda \theta }}{1+b}\left(I-A\right){\left(I-{e}^{-\lambda r}A\right)}^{-1}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]=\frac{{e}^{\lambda \theta }}{1+b{e}^{-2\lambda r}}\left[\begin{array}{c}\hfill {e}^{-\lambda r}\hfill \\ \hfill 1\hfill \end{array}\right]\phantom{\rule{1em}{0ex}}\forall \lambda \notin \sigma \left(\mathcal{𝒜}\right)\hfill \end{array}$ (8.10)

where ${\stackrel{̂}{d}}_{F}\left(\lambda \right)={\int }_{-r}^{0}{e}^{-\lambda \theta }d\phantom{\rule{0ex}{0ex}}d\theta$ is the ﬁnite Laplace transform of $d$. Substituting (8.10) into (8.9), we obtain

 $ĝ\left(\lambda \right)=\frac{a{e}^{-\lambda r}}{1+b{e}^{-2\lambda r}}$ (8.11)

Exercise 8.3.1. Prove that $ĝ\in {\text{H}}^{\infty }\left({\Pi }^{+}\right)$ but $1∕ĝ\notin {\text{H}}^{\infty }\left({\Pi }^{+}\right)$. Hint: $\left|a{e}^{-sr}\right|=\left|a\right|{e}^{-rRes}$ and $1-\left|b\right|{e}^{-2rRes}\le \left|1+b{e}^{-2sr}\right|\le 1+\left|b\right|{e}^{-2rRes}$ for all $s\in \mathbb{ℂ}$, provided that $\left|b\right|<1$.