]> 6.4.2 Example 2

#### 6.4.2 Example 2

The Mach number control system

 $\left\{\begin{array}{ccc}{ż}_{1}\left(t\right)=-a{z}_{1}\left(t\right)+ka{z}_{2}\left(t-r\right),\hfill & \hfill t\ge 0\phantom{\rule{0ex}{0ex}}& \hfill \\ {ż}_{2}\left(t\right)={z}_{3}\left(t\right),\hfill & \hfill t\ge 0\phantom{\rule{0ex}{0ex}}& \hfill \\ {ż}_{3}\left(t\right)=-{\omega }^{2}{z}_{2}\left(t\right)-2\zeta \omega {z}_{3}\left(t\right)+{\omega }^{2}u\left(t\right),\hfill & \hfill t\ge 0\phantom{\rule{0ex}{0ex}}& \hfill \\ {z}_{1}\left(0\right)={z}_{1}^{0}\hfill & \hfill & \hfill \\ {z}_{2}\left(0\right)={z}_{2}^{0}\hfill & \hfill & \hfill \\ {z}_{3}\left(0\right)={z}_{3}^{0}\hfill & \hfill & \hfill \\ {z}_{2}\left(\theta \right)={\phi }_{2}\left(\theta \right),\hfill & \hfill -r\le \theta \le 0,& {\phi }_{2}\in {\text{L}}^{2}\left(-r,0\right)\hfill \end{array}\right\}$ (6.50)

where ${a}^{-1}=1.964$, $\omega =0.6$, $\zeta =0.8$, $r=0.33$, $k=-0.0117$, has been investigated in several papers – see [89, Section 4, pp. 17-20 and Section 6, p. 7] and the references therein. The problem is to ﬁnd an optimal stabilizing lq controller minimizing the performance index

 $J={\int }_{0}^{\infty }\left[10{0}^{2}{z}_{1}^{2}\left(t\right)+{u}^{2}\left(t\right)\right]dt$ (6.51)

Introduction of the new variables

 ${\xi }_{1}\left(t\right)={z}_{1}\left(t+r\right),\phantom{\rule{2em}{0ex}}{\xi }_{2}\left(t\right)={z}_{2}\left(t\right),\phantom{\rule{2em}{0ex}}{\xi }_{3}\left(t\right)={z}_{3}\left(t\right)$ (6.52)

reduces our lq problem to a ﬁnite–dimensional one

 $\left\{\begin{array}{ccc}\hfill {\stackrel{̇}{\xi }}_{1}\left(t\right)& \hfill =\hfill & -a{\xi }_{1}\left(t\right)+ka{\xi }_{2}\left(t\right)\hfill \\ \hfill {\stackrel{̇}{\xi }}_{2}\left(t\right)& \hfill =\hfill & {\xi }_{3}\left(t\right)\hfill \\ \hfill {\stackrel{̇}{\xi }}_{3}\left(t\right)& \hfill =\hfill & -{\omega }^{2}{\xi }_{2}\left(t\right)-2\zeta \omega {\xi }_{3}\left(t\right)+{\omega }^{2}u\left(t\right)\hfill \\ \hfill {\xi }_{1}\left(0\right)& \hfill =\hfill & {\xi }_{1}^{0}={z}_{1}\left(r\right)\hfill \\ \hfill {\xi }_{2}\left(0\right)& \hfill =\hfill & {\xi }_{2}^{0}\hfill \\ \hfill {\xi }_{3}\left(0\right)& \hfill =\hfill & {\xi }_{3}^{0}\hfill \end{array}\right\}$ (6.53)
 $J={\int }_{0}^{r}10{0}^{2}{z}_{1}^{2}\left(t\right)dt+{\int }_{0}^{\infty }\left[10{0}^{2}{\xi }_{1}^{2}\left(t\right)+{u}^{2}\left(t\right)\right]dt$ (6.54)

which can be solved both by classical methods and the spectral factorization. Here the restriction of ${z}_{1}$ to an interval $\left[0,r\right]$ is the solution of the nonhomogeneous differential equation

$\left\{\begin{array}{cccc}\hfill {ż}_{1}\left(t\right)& \hfill =\hfill & -a{z}_{1}\left(t\right)+ka{\phi }_{2}\left(t-r\right),\hfill & 0\le t\le r\hfill \\ \hfill {z}_{1}\left(0\right)& \hfill =\hfill & {z}_{1}^{0}\hfill & \hfill \end{array}\right\}.$

In a Hilbert space $\text{H}={\mathbb{ℝ}}^{3}\oplus {\text{L}}^{2}\left(-1,0;{\mathbb{ℝ}}^{3}\right)$, (6.50) can be written as an abstract equation (6.1), $\mathcal{𝒜}$ deﬁned by (6.42) and with

$A=\left[\begin{array}{ccc}\hfill -a\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -{\omega }^{2}\hfill & \hfill -2\zeta \omega \hfill \end{array}\right],\phantom{\rule{2em}{0ex}}B=\left[\begin{array}{ccc}\hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}ka\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}n=3,$

$x=\left[\begin{array}{c}\hfill {z}_{1}\hfill \\ \hfill {z}_{2}\hfill \\ \hfill {z}_{3}\hfill \\ \hfill {\psi }_{1}\hfill \\ \hfill {\psi }_{2}\hfill \\ \hfill {\psi }_{3}\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}b=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill {\omega }^{2}\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}c=\left[\begin{array}{c}\hfill 100\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right].$

Now, $det\left[sI-A-{e}^{-sr}B\right]=\left({s}^{2}+2\zeta \omega s+{\omega }^{2}\right)\left(s+a\right)$ and the semigroup generated by $\mathcal{𝒜}$ is EXS. The pair $\left(\mathcal{𝒜},b\right)$ is stabilizable, $\left(\mathcal{𝒜},c\right)$ is detectable and the lq problem has a unique solution.

The rank condition (6.44) is not satisﬁed and thus $\left(\mathcal{𝒜},b\right)$ is not approximately controllable. As we have mentioned at the end of Section 6.3, the full form of a controller can be found with the aid of a Lyapunov operator equation, however our intent is to ﬁnd the controller without invoking the use of such an equation. It turns out that despite the lack of approximate controllability, this is still possible. To do this, we evaluate

 ${\left(sI-\mathcal{𝒜}\right)}^{-1}b=\frac{{\omega }^{2}}{{s}^{2}+2\zeta \omega s+{\omega }^{2}}\left[\begin{array}{c}\hfill \frac{ka}{s+a}{e}^{-sr}\hfill \\ \hfill 1\hfill \\ \hfill s\hfill \\ \hfill \frac{ka}{s+a}{e}^{s\left(\theta -r\right)}\hfill \\ \hfill {e}^{s\theta }\hfill \\ \hfill s{e}^{s\theta }\hfill \end{array}\right]$ (6.55)
 ${c}^{\ast }{\left(sI-\mathcal{𝒜}\right)}^{-1}b=\frac{100{\omega }^{2}ka{e}^{-sr}}{\left({s}^{2}+2\zeta \omega s+{\omega }^{2}\right)\left(s+a\right)}$ (6.56)

Taking $f=0\in \text{H}$, we ﬁnd via (6.24), (6.25)

 $\phi \left(s\right)=\frac{{s}^{3}+\alpha {s}^{2}+\beta s+\gamma }{\left({s}^{2}+2\zeta \omega s+{\omega }^{2}\right)\left(s+a\right)}$ (6.57)

where $\alpha ,\beta ,\gamma$ are positive solutions, $\alpha \beta >\gamma$ satisfying the equations

 $\left\{\begin{array}{ccc}\hfill {a}^{2}-2{\omega }^{2}+4{\zeta }^{2}{\omega }^{2}& \hfill =\hfill & {\alpha }^{2}-2\beta \hfill \\ \hfill 2{\omega }^{2}{a}^{2}-{\omega }^{4}-4{a}^{2}{\zeta }^{2}{\omega }^{2}& \hfill =\hfill & 2\alpha \gamma -{\beta }^{2}\hfill \\ \hfill 10{0}^{2}{a}^{2}{k}^{2}{\omega }^{4}+{a}^{2}{\omega }^{4}& \hfill =\hfill & {\gamma }^{2}\hfill \end{array}\right\}$ (6.58)

Thus the values found in [89, Section 4, pp. 17 - 20] are conﬁrmed to be: $\alpha =1.5939$, $\beta =1.0398$, $\gamma =0.2821$. By Riesz representation theorem the controller is assumed to be a vector

$g=\left[\begin{array}{c}{\alpha }_{1}\hfill \\ {\alpha }_{2}\hfill \\ {\alpha }_{3}\hfill \\ {\beta }_{1}\left(\cdot \right)\hfill \\ {\beta }_{2}\left(\cdot \right)\hfill \\ {\beta }_{3}\left(\cdot \right)\hfill \end{array}\right]\in \text{H}.$

Multiplying both sides of the identity (6.23) on the characteristic polynomial, we convert (6.23) into an identity for entire functions

 $\begin{array}{c}\hfill ka{\omega }^{2}{e}^{-sr}{\alpha }_{1}+{\omega }^{2}\left(s+a\right){\alpha }_{2}+{\omega }^{2}s\left(s+a\right){\alpha }_{3}+{\int }_{-r}^{0}ka{\omega }^{2}{e}^{s\left(\theta -r\right)}{\beta }_{1}\left(\theta \right)d\theta \phantom{\rule{0ex}{0ex}}+\hfill \\ \hfill +{\int }_{-r}^{0}{\omega }^{2}\left(s+a\right){e}^{s\theta }{\beta }_{2}\left(\theta \right)d\theta +{\int }_{-r}^{0}{\omega }^{2}s\left(s+a\right){e}^{s\theta }{\beta }_{3}\left(\theta \right)d\theta =\hfill \\ \hfill ={s}^{2}\left(\alpha -a-2\zeta \omega \right)+s\left(\beta -{\omega }^{2}-2a\zeta \omega \right)+\left(\gamma -a{\omega }^{2}\right)\hfill \end{array}$ (6.59)

Since the system is not approximately controllable one can try to apply Theorem 6.3.1 to ﬁnd the optimal feedback. This requires veriﬁcation whether (6.30) is satisﬁed. By (6.21) and the Paley–Wiener theorem (support arguments) we get

 $\begin{array}{c}\hfill R\left({\mathcal{𝒜}}^{\ast },c\right)=\text{NO}{\left({c}^{\ast },\mathcal{𝒜}\right)}^{\perp }=\left\{x=\left[\begin{array}{c}\hfill z\hfill \\ \hfill \psi \hfill \end{array}\right]\in {\mathbb{𝕄}}^{2}:\phantom{\rule{3.26212pt}{0ex}}ka{e}^{-sr}\left(s+2\zeta \omega \right){z}_{2}+ka{e}^{-sr}{z}_{3}+\right\\hfill \\ \hfill {+\left({s}^{2}+2\zeta \omega s+{\omega }^{2}\right)\left[{z}_{1}+ka{\int }_{-r}^{0}{e}^{-s\left(\theta +r\right)}{\psi }_{2}\left(\theta \right)d\theta \right]=0\phantom{\rule{2em}{0ex}}\forall s\in \mathbb{ℂ}}}^{\perp }=\hfill \\ \hfill ={\left\{x\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}x=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill {\psi }_{1}\hfill \\ \hfill 0\hfill \\ \hfill {\psi }_{3}\hfill \end{array}\right]\right\}}^{\perp }=\left\{x\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}z=\left[\begin{array}{c}\hfill {z}_{1}\hfill \\ \hfill {z}_{2}\hfill \\ \hfill {z}_{3}\hfill \\ \hfill 0\hfill \\ \hfill {\psi }_{2}\hfill \\ \hfill 0\hfill \end{array}\right]\right\}\hfill \end{array}$ (6.60)

Similarly, by (6.22)

$\text{NO}\left({b}^{\ast },{\mathcal{𝒜}}^{\ast }\right)=\left\{x=\left[\begin{array}{c}\hfill z\hfill \\ \hfill \psi \hfill \end{array}\right]\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}ka{e}^{-sr}{z}_{1}+\left(s+a\right){z}_{2}+s\left(s+a\right){z}_{3}+\right\$

$+{\int }_{-r}^{0}ka{e}^{s\left(\theta -r\right)}{\psi }_{1}\left(\theta \right)d\theta +{\int }_{-r}^{0}\left(s+a\right){e}^{s\theta }{\psi }_{2}\left(\theta \right)d\theta +$

$+{\int }_{-r}^{0}s\left(s+a\right){e}^{s\theta }{\psi }_{3}\left(\theta \right)d\theta =0\phantom{\rule{2em}{0ex}}\forall s\in \mathbb{ℂ}}.$

Integrating–by–parts the second integral and applying the Paley–Wiener theorem one obtains the nontrivial closed subspace

 $\text{NO}\left({b}^{\ast },{\mathcal{𝒜}}^{\ast }\right)=\left\{x=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill {z}_{2}\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill {\psi }_{2}\hfill \\ \hfill {\psi }_{3}\hfill \end{array}\right]\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}{\psi }_{3}\left(\theta \right)={\int }_{-r}^{\theta }{\psi }_{2}\left(\tau \right)d\tau ,\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{z}_{2}=-{\int }_{-r}^{0}{\psi }_{2}\left(\theta \right)d\theta \right\}$ (6.61)

and thus its orthogonal complement

$\text{R}\left(\mathcal{𝒜},b\right)=\left\{x\in {\mathbb{𝕄}}^{2}:\phantom{\rule{0ex}{0ex}}x=\left[\begin{array}{c}\hfill {z}_{1}\hfill \\ \hfill {z}_{2}\hfill \\ \hfill {z}_{3}\hfill \\ \hfill {\psi }_{1}\hfill \\ \hfill {\psi }_{2}\hfill \\ \hfill {\psi }_{3}\hfill \end{array}\right],\phantom{\rule{0ex}{0ex}}{\psi }_{2}={z}_{2}+{\int }_{0}^{\theta }{\psi }_{3}\left(\tau \right)d\tau \right\}$

is not equal to ${\mathbb{𝕄}}^{2}$. It follows from (6.60) and (6.61) that (6.30) holds contrary to (6.29) which is not satisﬁed.

As we already know, the left–hand side of (6.59) vanishes for $g\in \text{NO}\left({b}^{\ast },{\mathcal{𝒜}}^{\ast }\right)$. The system reduced to $\text{R}\left({\mathcal{𝒜}}^{\ast },c\right)$ arises by zeroing the ﬁrst and third function components and hence ${\beta }_{1}=0$, ${\beta }_{3}=0$ in (6.59). Now, to eliminate integrals and exponential terms from the left–hand side of (6.59) and in order to get polynomial identity, the following equations should hold

 $\left\{\begin{array}{c}\hfill {\beta }_{2}^{\prime }\left(\theta \right)=a{\beta }_{2}\left(\theta \right),\phantom{\rule{2em}{0ex}}{\beta }_{2}\left(-r\right)=ak{\alpha }_{1},\hfill \\ \hfill {\omega }^{2}\left[\begin{array}{ccc}\hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}a\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}a\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}\hfill \end{array}\right]\left[\begin{array}{c}\hfill {\alpha }_{2}\hfill \\ \hfill {\alpha }_{3}\hfill \\ \hfill {\beta }_{2}\left(0\right)\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \alpha -a2\zeta \omega \hfill \\ \hfill \beta -{\omega }^{2}-2\zeta \omega a\hfill \\ \hfill \gamma -a{\omega }^{2}\hfill \end{array}\right]\hfill \end{array}\right\}$ (6.62)

(6.62) has a unique solution, determining the optimal lq controller

 $\begin{array}{cccc}{\alpha }_{1}={e}^{-ar}\frac{1}{ka}{\beta }_{2}\left(0\right),\hfill & \hfill \phantom{\rule{2em}{0ex}}{\beta }_{2}\left(0\right)& \hfill =& \frac{\alpha {a}^{2}-{a}^{3}-a\beta +\gamma }{{\omega }^{2}}\hfill \\ {\alpha }_{2}=\frac{-a\alpha +{a}^{2}+\beta -{\omega }^{2}}{{\omega }^{2}},\hfill & \hfill \phantom{\rule{2em}{0ex}}{\alpha }_{3}& \hfill =& \frac{\alpha -a-2\zeta \omega }{{\omega }^{2}}\hfill \\ {\beta }_{1}={\beta }_{3}=0,\hfill & \hfill \phantom{\rule{2em}{0ex}}{\beta }_{2}\left(\theta \right)& \hfill =& {e}^{a\theta }{\beta }_{2}\left(0\right)\hfill \end{array}$ (6.63)

(6.63) conﬁrms the results obtained by Kappel and Salamon – see [89, Section 4, pp. 17 - 20] with the aid of the Riccati operator equation.

Observe also, that eigenvectors of ${\mathcal{𝒜}}^{\ast }$ corresponding to eigenvalues ${\overline{\lambda }}_{1}$, ${\overline{\lambda }}_{2}$, $-a$, where ${\lambda }_{1}$, ${\lambda }_{2}$ are the roots of trinomial ${\lambda }^{2}+2\zeta \omega \lambda +{\omega }^{2}$ are, respectively

${f}_{i}=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill {\omega }_{2}\hfill \\ \hfill -{\overline{\lambda }}_{i}\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\phantom{\rule{0ex}{0ex}}\left(i=1,2\right),\phantom{\rule{2em}{0ex}}{f}_{3}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill \frac{ak\left(2\zeta \omega -a\right)}{{a}^{2}-2\zeta \omega a+{\omega }^{2}}{e}^{ar}\hfill \\ \hfill \frac{ak}{{a}^{2}-2\zeta \omega a+{\omega }^{2}}{e}^{ar}\hfill \\ \hfill 0\hfill \\ \hfill ak{e}^{a\left(\theta +r\right)}\hfill \\ \hfill 0\hfill \end{array}\right]$

while eigenvectors of $\mathcal{𝒜}$ corresponding to ${\lambda }_{1}$, ${\lambda }_{2}$, $-a$ are, respectively

${e}_{i}=\left[\begin{array}{c}\hfill \frac{ak}{{\lambda }_{i}+a}{e}^{-{\lambda }_{i}r}\hfill \\ \hfill 1\hfill \\ \hfill {\lambda }_{i}\hfill \\ \hfill \frac{ak}{{\lambda }_{i}+a}{e}^{{\lambda }_{i}\left(\theta -r\right)}\hfill \\ \hfill {e}^{{\lambda }_{i}\theta }\hfill \\ \hfill {\lambda }_{i}{e}^{{\lambda }_{i}\theta }\hfill \end{array}\right]\phantom{\rule{0ex}{0ex}}\left(i=1,2\right),\phantom{\rule{2em}{0ex}}{e}_{3}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill {e}^{-a\theta }\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right].$

The next result shows that the optimal controller can be expressed as a linear combination of eigenvectors of ${\mathcal{𝒜}}^{\ast }$.

Lemma 6.4.1. We have:

$\begin{array}{c}span\left\{{f}_{1},{f}_{2},{f}_{3}\right\}⊈R\left({\mathcal{𝒜}}^{\ast },c\right),\hfill \\ span\left\{{e}_{1},{e}_{2},{e}_{3}\right\}\cap R\left({\mathcal{𝒜}}^{\ast },c\right)=\left\{0\right\},\hfill \\ span\left\{{e}_{1},{e}_{2},{e}_{3}\right\}\subset R\left(\mathcal{𝒜},b\right),\hfill \end{array}$

and there exist constants ${c}_{1}$, ${c}_{2}$, and ${c}_{3}$ such that:

 ${g}_{\infty }=\sum _{i=1}^{3}{c}_{i}{f}_{i}$ (6.64)

Proof. All above properties of eigensystems of $\mathcal{𝒜}$ and ${\mathcal{𝒜}}^{\ast }$ are a consequence of the fact that their ${\mathbb{ℝ}}^{3}$ components are linearly independent, i.e. the original system as well as its dual system are ${\mathbb{ℝ}}^{3}$complete. Thus, only a proof of (6.64) will be given. We have to prove that there exist constants ${c}_{1}$, ${c}_{2}$, ${c}_{3}$ such that

$\left[\begin{array}{c}\hfill {c}_{3}\hfill \\ \hfill {\omega }^{2}\left({c}_{1}+{c}_{2}\right)+{c}_{3}\frac{ak\left(2\zeta \omega -a\right)}{{a}^{2}-2\zeta \omega a+{\omega }^{2}}{e}^{ar}\hfill \\ \hfill -{\overline{\lambda }}_{1}{c}_{1}-{\overline{\lambda }}_{2}{c}_{2}+{c}_{3}\frac{ak}{{a}^{2}-2\zeta \omega a+{\omega }^{2}}{e}^{ar}\hfill \\ \hfill 0\hfill \\ \hfill {c}_{3}ak{e}^{a\left(\theta +r\right)}\hfill \\ \hfill 0\hfill \end{array}\right]={g}_{\infty }=\left[\begin{array}{c}\hfill {\alpha }_{1}\hfill \\ \hfill {\alpha }_{2}\hfill \\ \hfill {\alpha }_{3}\hfill \\ \hfill 0\hfill \\ \hfill {\beta }_{2}\left(0\right){e}^{a\theta }\hfill \\ \hfill 0\hfill \end{array}\right].$

We take ${c}_{3}={\alpha }_{1}=\frac{1}{ak}{\beta }_{2}\left(0\right){e}^{-ar}$. Then ${c}_{3}ak{e}^{a\left(\theta +r\right)}={\beta }_{2}\left(0\right){e}^{a\theta }$ and the remaining constants ${c}_{1}$, ${c}_{2}$ can be uniquely (as ${\lambda }_{1}\ne {\lambda }_{2}$ or equivalently ${\mathbb{ℝ}}^{3}$ components of ${f}_{1}$, ${f}_{2}$, ${f}_{3}$ are linearly independent) determined from the linear system of equations

$\left[\begin{array}{cc}\hfill {\omega }^{2}& \hfill {\omega }^{2}\\ \hfill -{\overline{\lambda }}_{1}& \hfill -{\overline{\lambda }}_{2}\end{array}\right]\left[\begin{array}{c}\hfill {c}_{1}\\ \hfill {c}_{2}\end{array}\right]=\left[\begin{array}{c}\hfill {\alpha }_{2}-\frac{2\zeta \omega -a}{{a}^{2}-2\zeta \omega a+{\omega }^{2}}{\beta }_{2}\left(0\right)\hfill \\ \hfill {\alpha }_{3}-\frac{1}{{a}^{2}-2\zeta \omega a+{\omega }^{2}}{\beta }_{2}\left(0\right)\hfill \end{array}\right].$

Remark 6.4.1. Korytowski () discovered by generalization of formulae (6.33) and (6.52), a class of lq problems for degenerated time–delay systems which admit a reduction to a ﬁnite–dimensional lq problem. Both our examples are representatives of that class of retarded systems.