]> 8.7 Formal derivation of the optimal cost

### 8.7 Formal derivation of the optimal cost

It is known, see (6.11), (6.14), (6.15) that the Lyapunov operator equations for the closed–loop and open–loop systems are respectively
 $〈{\mathcal{𝒜}}_{c}x,\mathcal{ℋ}x〉+〈x,\mathcal{ℋ}{\mathcal{𝒜}}_{c}x〉=-{\left|{c}^{#}x\right|}^{2}-{\left|{g}^{#}x\right|}^{2}\phantom{\rule{2em}{0ex}}\forall x\in D\left({\mathcal{𝒜}}_{c}\right)$ (8.27)
 $〈\mathcal{𝒜}x,\mathcal{ℋ}x〉+〈x,\mathcal{ℋ}\mathcal{𝒜}x〉=-{\left|{c}^{#}x\right|}^{2}+{\left|{g}^{#}x\right|}^{2}\phantom{\rule{2em}{0ex}}\forall x\in D\left(\mathcal{𝒜}\right)$ (8.28)

In (8.27) the closed–loop operator ${\mathcal{𝒜}}_{c}$ is given by (8.20), while in (8.28) the open–loop operator $\mathcal{𝒜}$ is deﬁned as in (8.8). In both formulae (8.27) and (8.28) the functional ${g}^{#}$ is determined by (8.18). Solutions to (8.27) and (8.28) will be sought in the form

$\left(\mathcal{ℋ}x\right)\left(\theta \right)=Hx\left(\theta \right),\phantom{\rule{1em}{0ex}}H={H}^{\ast }=\left[\begin{array}{cc}\hfill {h}_{1}\hfill & \hfill \phantom{\rule{0ex}{0ex}}{h}_{2}\hfill \\ \hfill {h}_{2}\hfill & \hfill \phantom{\rule{0ex}{0ex}}{h}_{3}\hfill \end{array}\right]\in L\left({\mathbb{ℝ}}^{2}\right),$

i.e. $\mathcal{ℋ}$ is the bounded linear operator of multiplication by a constant matrix $H$. Introducing all data into (8.27) and taking into account that $x\in D\left({\mathcal{𝒜}}_{c}\right)$, we obtain the linear system of equations characterizing ${h}_{1}$, ${h}_{2}$, ${h}_{3}$:

 $\left\{\begin{array}{ccc}\hfill {h}_{2}& \hfill =\hfill & 0\hfill \\ \hfill {h}_{1}-{h}_{3}& \hfill =\hfill & -{a}^{2}\hfill \\ \hfill -\frac{{q}^{2}}{{p}^{2}}{\left({p}^{2}-1\right)}^{2}& \hfill =\hfill & {h}_{3}\frac{{q}^{2}}{{p}^{2}}-{h}_{1}\hfill \end{array}\right\}$ (8.29)

Hence the solution of (8.27) takes the form

 $\left(\mathcal{ℋ}x\right)\left(\theta \right)=Hx\left(\theta \right),\phantom{\rule{2em}{0ex}}H=\frac{{q}^{2}}{{p}^{2}-{q}^{2}}\left[\begin{array}{cc}\hfill {a}^{2}+{\left({p}^{2}-1\right)}^{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{{p}^{2}}{{q}^{2}}{a}^{2}+{\left({p}^{2}-1\right)}^{2}\hfill \end{array}\right]$ (8.30)

Clearly, $\mathcal{ℋ}$ is a coercive self–adjoint bounded operator since $\left|q∕p\right|<1$ is the necessary and sufficient condition for EXS of the closed–loop system. By EXS, $\mathcal{ℋ}$ is also a unique solution of (8.27).

Similar considerations for (8.28) with $x\in D\left({\mathcal{𝒜}}_{c}\right)$ replaced by $x\in D\left(\mathcal{𝒜}\right)$ lead to the linear system

 $\left\{\begin{array}{ccc}\hfill {h}_{2}& \hfill =\hfill & 0\hfill \\ \hfill {h}_{1}-{h}_{3}& \hfill =\hfill & -{a}^{2}\hfill \\ \hfill {q}^{2}{\left({p}^{2}-1\right)}^{2}& \hfill =\hfill & {h}_{3}{b}^{2}-{h}_{1}\hfill \end{array}\right\}$ (8.31)

instead of (8.29). Hence, a unique solution to (8.28) turns out to be

 $\left(\mathcal{ℋ}x\right)\left(\theta \right)=Hx\left(\theta \right),\phantom{\rule{2em}{0ex}}H=\frac{{q}^{2}}{1-{b}^{2}}\left[\begin{array}{cc}\hfill {p}^{2}{a}^{2}-{\left({p}^{2}-1\right)}^{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{{a}^{2}}{{q}^{2}}-{\left({p}^{2}-1\right)}^{2}\hfill \end{array}\right]$ (8.32)

Using the identities

$pq=b,\phantom{\rule{2em}{0ex}}{a}^{2}={p}^{2}+{q}^{2}-1-{p}^{2}{q}^{2},$

which follow from (8.12), one can reduce both (8.30) and (8.32) to the form

$\left(\mathcal{ℋ}x\right)\left(\theta \right)=Hx\left(\theta \right),\phantom{\rule{2em}{0ex}}H=\left({p}^{2}-1\right)\left[\begin{array}{cc}\hfill {q}^{2}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\hfill \\ \hfill 0\hfill & \hfill \phantom{\rule{0ex}{0ex}}1\hfill \end{array}\right].$

Observe that the solution of (8.27) and (8.28) is not a HS operator, contrary to the standard lq problem for a SISO system – see Section 2.4.

Another interesting feature is that, again contrary to the standard lq problem for a SISO system, the inclusion $\mathcal{ℋ}D\left(\mathcal{𝒜}\right)\subset D\left({\mathcal{𝒜}}^{\ast }\right)$ does not hold for our special system. Indeed, by (8.8),

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

Moreover,

$\mathcal{ℋ}D\left(\mathcal{𝒜}\right)=\left\{v\in {\text{W}}^{1,2}\left(-r,0\right)\oplus {\text{W}}^{1,2}\left(-r,0\right):\phantom{\rule{0ex}{0ex}}H{A}^{-1}{H}^{-1}v\left(0\right)=v\left(-r\right)\right\}.$

If the inclusion $\mathcal{ℋ}D\left(\mathcal{𝒜}\right)\subset D\left({\mathcal{𝒜}}^{\ast }\right)$ were true, then ${A}^{T}HA=H$ would hold. But, since $\left|\lambda \left(A\right)\right|<1$, this is impossible in view of the discrete Lyapunov matrix equation theory. Thus

 $\mathcal{ℋ}D\left(\mathcal{𝒜}\right)⊈D\left({\mathcal{𝒜}}^{\ast }\right)$ (8.33)