]> 3.3.1 Example 1: Nuclear reactor temperature control II

#### 3.3.1 Example 1: Nuclear reactor temperature control II

The performance index (3.4) has a form (3.16) with

$P=c{c}^{T},\phantom{\rule{2em}{0ex}}{c}^{T}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}Q=R=0\in L\left({\mathbb{ℝ}}^{2}\right).$

Treating $a$ and $r$ as ﬁxed constants, we seek for a pair $\left(b,d\right)$ belonging to the domain of stability which minimizes the performance index. The stability domain of the characteristic quasipolynomial $\left({s}^{2}-as\right)+\left(-bs-d\right){e}^{-sr}$, in the plane $0bd$ is an open bounded set with boundary ${\Gamma }_{1}\cup {\Gamma }_{2}$ where the curve ${\Gamma }_{1}$ is given parametrically

$\left\{\begin{array}{ccccc}\hfill b\left(\omega \right)\hfill & \hfill =& -acos\omega r\hfill & -\hfill & \omega sin\omega r\hfill \\ \hfill d\left(\omega \right)\hfill & \hfill =& -{\omega }^{2}cos\omega r\hfill & +\hfill & a\omega sin\omega r\hfill \end{array}\right\},\phantom{\rule{2em}{0ex}}\omega \in \left[0,{\omega }_{0}\right].$

Here ${\omega }_{0}$ denotes the smallest positive root of the equation $\omega =atan\omega r$ and ${\Gamma }_{2}$ is an interval of $0b$ axis with ends coinciding with the ends of ${\Gamma }_{1}$.

Taking into account the nature of initial conditions we conclude that

$J\left({x}_{0}\right)\left(b,d\right)={〈{x}_{0},\mathcal{ℋ}{x}_{0}〉}_{\text{H}}={v}_{0}^{T}\alpha {v}_{0}={\left({z}_{2}^{0}\right)}^{2}{\alpha }_{22},\phantom{\rule{2em}{0ex}}\alpha ={\alpha }^{T}=\left[\begin{array}{cc}\hfill {\alpha }_{11}\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}{\alpha }_{12}\hfill \\ \hfill {\alpha }_{12}\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}{\alpha }_{22}\hfill \end{array}\right].$

But, since $C=R=0\in L\left({\mathbb{ℝ}}^{2}\right)$ to ﬁnd ${\alpha }_{22}$ from (3.33) we ought to put $\gamma =0$ in its right hand side. To determine the system (3.33) we ﬁrst calculate the polynomial (3.29),

$\left[{\lambda }^{4}+{\lambda }^{2}\left({b}^{2}-{a}^{2}\right)-{d}^{2}\right]{\lambda }^{2}\left({\lambda }^{2}-{a}^{2}\right).$

It has eight roots but we write down only four of them because others may differ of signs,

$\begin{array}{c}{\lambda }_{1}=\sqrt{\frac{1}{2}\left[{a}^{2}-{b}^{2}+\sqrt{{\left({b}^{2}-{a}^{2}\right)}^{2}+4{d}^{2}}\right]},\hfill \\ {\lambda }_{2}=jK,\phantom{\rule{2em}{0ex}}K=\sqrt{\frac{1}{2}\left[{b}^{2}-{a}^{2}+\sqrt{{\left({b}^{2}-{a}^{2}\right)}^{2}+4{d}^{2}}\right]},\hfill \\ {\lambda }_{3}=0,\phantom{\rule{2em}{0ex}}{\lambda }_{4}=a.\hfill \end{array}$

From (3.31) we determine the corresponding eigenvectors of (3.28)

${L}_{i}=\left[\begin{array}{cc}\hfill d\hfill & \hfill b\hfill \\ \hfill \frac{{\lambda }_{i}^{2}\left({\lambda }_{i}+a\right)}{d-b{\lambda }_{i}}\hfill & \hfill \frac{b{\lambda }_{i}^{2}\left({\lambda }_{i}+a\right)}{d\left(d-b{\lambda }_{i}\right)}\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}{M}_{i}=\left[\begin{array}{cc}\hfill \frac{-d{\lambda }_{i}\left({\lambda }_{i}+a\right)}{d-b{\lambda }_{i}}\hfill & \hfill {\lambda }_{i}\hfill \\ \hfill \frac{-b{\lambda }_{i}\left({\lambda }_{i}+a\right)}{d-b{\lambda }_{i}}\hfill & \hfill \frac{b{\lambda }_{i}}{d}\hfill \end{array}\right]$

for $i=1,2$ and

${L}_{3}=\left[\begin{array}{cc}\hfill d\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}{M}_{3}=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\\ \hfill a\hfill & \hfill -1\end{array}\right],\phantom{\rule{2em}{0ex}}{L}_{4}=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill a\hfill & \hfill 1\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}{M}_{4}=\left[\begin{array}{cc}\hfill -d& \hfill 0\hfill \\ \hfill -b& \hfill 0\hfill \end{array}\right].$

The system (3.33) takes the form

$\left[\begin{array}{cccccccc}\hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill {s}_{1}& {s}_{8}\hfill & \hfill 2d\hfill & \hfill -2d{e}^{-ar}\hfill \\ \hfill 2& \hfill 2a& \hfill 0& \hfill 0& \hfill {s}_{2}& {s}_{9}\hfill & \hfill a+b\hfill & \hfill a-b{e}^{-ar}\hfill \\ \hfill 2& \hfill 0& \hfill 2a& \hfill 0& \hfill {s}_{2}& {s}_{9}\hfill & \hfill a+b\hfill & \hfill a-b{e}^{-ar}\hfill \\ \hfill 0& \hfill 2& \hfill 2& \hfill 4a& \hfill {s}_{3}& {s}_{10}\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill 0& \hfill -2d& \hfill 0& \hfill 0& \hfill {s}_{4}& {s}_{11}\hfill & \hfill d\hfill & \hfill -d\hfill \\ \hfill 0& \hfill -2b& \hfill 0& \hfill 0& \hfill {s}_{5}& {s}_{12}\hfill & \hfill a+b\hfill & \hfill -b\hfill \\ \hfill 0& \hfill 0& \hfill 0& \hfill -2d& \hfill {s}_{6}& {s}_{13}\hfill & \hfill 0\hfill & \hfill a{e}^{-ar}\hfill \\ \hfill 0& \hfill 0& \hfill 0& \hfill -2b& \hfill {s}_{7}& {s}_{14}\hfill & \hfill -1\hfill & \hfill {e}^{-ar}\hfill \end{array}\right]\left[\begin{array}{c}\hfill {\alpha }_{11}\hfill \\ \hfill {\alpha }_{12}\hfill \\ \hfill {\alpha }_{12}\hfill \\ \hfill {\alpha }_{22}\hfill \\ \hfill 2{\kappa }_{1}\hfill \\ \hfill 2{\kappa }_{2}\hfill \\ \hfill 2{\kappa }_{3}\hfill \\ \hfill 2{\kappa }_{4}\hfill \end{array}\right]=-\left[\begin{array}{c}\hfill 2\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\\ \hfill 0\phantom{\rule{0ex}{0ex}}\end{array}\right]$

with

${s}_{1}=2d-2{e}^{-{\lambda }_{1}r}\frac{d{\lambda }_{1}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}},\phantom{\rule{2em}{0ex}}{s}_{8}=2d-2{e}^{-{\lambda }_{2}r}\frac{d{\lambda }_{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}},$

${s}_{2}=b+\frac{{\lambda }_{1}^{2}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}}+{e}^{-{\lambda }_{1}r}\left[{\lambda }_{1}-\frac{b{\lambda }_{1}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}}\right],$

${s}_{9}=b+\frac{{\lambda }_{2}^{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}}+{e}^{-{\lambda }_{2}r}\left[{\lambda }_{2}-\frac{b{\lambda }_{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}}\right],$

${s}_{3}=\frac{2b{\lambda }_{1}^{2}\left({\lambda }_{1}+a\right)}{d\left(d-b{\lambda }_{1}\right)}+2\frac{b{\lambda }_{1}}{d}{e}^{-{\lambda }_{1}r},\phantom{\rule{2em}{0ex}}{s}_{10}=\frac{2b{\lambda }_{2}^{2}\left({\lambda }_{2}+a\right)}{d\left(d-b{\lambda }_{2}\right)}+2\frac{b{\lambda }_{2}}{d}{e}^{-{\lambda }_{2}r},$

${s}_{4}={e}^{-{\lambda }_{1}r}d-\frac{d{\lambda }_{1}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}},\phantom{\rule{2em}{0ex}}{s}_{11}={e}^{-{\lambda }_{2}r}d-\frac{d{\lambda }_{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}},$

${s}_{5}=b{e}^{-{\lambda }_{1}r}-\frac{b{\lambda }_{1}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}},\phantom{\rule{2em}{0ex}}{s}_{12}=b{e}^{-{\lambda }_{2}r}-\frac{d{\lambda }_{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}},$

${s}_{6}={e}^{-{\lambda }_{1}r}\frac{{\lambda }_{1}^{2}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}}+{\lambda }_{1},\phantom{\rule{2em}{0ex}}{s}_{13}={e}^{-{\lambda }_{2}r}\frac{{\lambda }_{2}^{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}}+{\lambda }_{2},$

${s}_{7}=\frac{b{\lambda }_{1}^{2}\left({\lambda }_{1}+a\right)}{d\left(d-b{\lambda }_{1}\right)}{e}^{-{\lambda }_{1}r}+\frac{b{\lambda }_{1}}{d},\phantom{\rule{2em}{0ex}}{s}_{14}=\frac{b{\lambda }_{2}^{2}\left({\lambda }_{2}+a\right)}{d\left(d-b{\lambda }_{2}\right)}{e}^{-{\lambda }_{2}r}+\frac{b{\lambda }_{2}}{d}.$

The third equation is redundant. Adding both sides of the ﬁfth equation multiplied by $b$ and the sixth equation multiplied by $-d$ one obtains $-ad{\kappa }_{3}=0$, whence ${\kappa }_{3}=0$. Next adding both sides of the seventh equation multiplied by $b$ and the eighth equation multiplied by $-d$ one obtains $d{\kappa }_{3}+\left(ab-d\right){e}^{-ar}{\kappa }_{4}=0$, whence ${\kappa }_{4}=0$, provided that $ab-d\ne 0$. Since the common part of the straight line $ab=d$ and the stability domain is of planar measure zero we can neglect the possible degeneration getting a preliminary reduction of the linear system to be solved,

$\left[\begin{array}{ccccc}\hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}\frac{1}{2}{s}_{1}\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}\frac{1}{2}{s}_{8}\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}2\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}2a\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}{s}_{2}\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}{s}_{9}\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}2\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}2a\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}\frac{1}{2}{s}_{3}\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}\frac{1}{2}{s}_{10}\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}-2\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}\frac{1}{d}{s}_{4}\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}\frac{1}{d}{s}_{11}\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}-2d\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}{s}_{6}\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}{s}_{13}\phantom{\rule{0ex}{0ex}}\end{array}\right]\left[\begin{array}{c}\hfill \phantom{\rule{0ex}{0ex}}{\alpha }_{11}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}{\alpha }_{12}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}{\alpha }_{22}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}2{\kappa }_{1}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}2{\kappa }_{2}\phantom{\rule{0ex}{0ex}}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \phantom{\rule{0ex}{0ex}}-1\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\end{array}\right].$

The ﬁnal reduced form arises by elimination of ${\alpha }_{11}$ and ${\alpha }_{12}$,

$\begin{array}{c}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1-{e}^{-{\lambda }_{1}r}\frac{{\lambda }_{1}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}}\hfill & \hfill 1-{e}^{-{\lambda }_{2}r}\frac{{\lambda }_{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}}\hfill \\ \hfill 2ad\hfill & \hfill {e}^{-{\lambda }_{1}r}\left(d+b{\lambda }_{1}\right)-{\lambda }_{1}\left({\lambda }_{1}+a\right)\hfill & \hfill {e}^{-{\lambda }_{2}r}\left(d+b{\lambda }_{2}\right)-{\lambda }_{2}\left({\lambda }_{2}+a\right)\hfill \\ \hfill -2d\hfill & \hfill {e}^{-{\lambda }_{1}r}\frac{{\lambda }_{1}^{2}\left({\lambda }_{1}+a\right)}{d-b{\lambda }_{1}}+{\lambda }_{1}\hfill & \hfill {e}^{-{\lambda }_{2}r}\frac{{\lambda }_{2}^{2}\left({\lambda }_{2}+a\right)}{d-b{\lambda }_{2}}+{\lambda }_{2}\hfill \end{array}\right]\left[\begin{array}{c}\hfill \phantom{\rule{0ex}{0ex}}{\alpha }_{22}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}2{\kappa }_{1}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}2{\kappa }_{2}\phantom{\rule{0ex}{0ex}}\hfill \end{array}\right]=\hfill \\ =\left[\begin{array}{c}\hfill \phantom{\rule{0ex}{0ex}}-\frac{1}{d}\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\\ \hfill \phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}\end{array}\right].\hfill \end{array}$

This yields

 $\begin{array}{cc}\hfill \frac{1}{{\left({z}_{0}^{2}\right)}^{2}}J\left({x}_{0}\right)\left(a,b\right)=& {\alpha }_{22}=\frac{{G}_{3}{F}_{3}-{F}_{2}{G}_{5}}{2{d}^{2}\left(a{F}_{3}{G}_{1}-{F}_{1}{G}_{3}+{G}_{1}{F}_{2}-a{G}_{5}{F}_{1}\right)}=\hfill \\ \hfill =& \frac{{G}_{4}{F}_{3}-{F}_{2}{G}_{6}}{2{d}^{2}\left(a{F}_{3}{G}_{2}-{F}_{1}{G}_{4}+{G}_{2}{F}_{2}-a{G}_{6}{F}_{1}\right)}\hfill \end{array}$ (3.34)

with

$\begin{array}{c}{F}_{1}=d-b{\lambda }_{1}-{e}^{-{\lambda }_{1}r}{\lambda }_{1}\left({\lambda }_{1}+a\right)\hfill \\ {F}_{2}={e}^{-{\lambda }_{1}r}\left({d}^{2}-{b}^{2}{\lambda }_{1}^{2}\right)-{\lambda }_{1}\left({\lambda }_{1}+a\right)\left(d-b{\lambda }_{1}\right)\hfill \\ {F}_{3}={e}^{-{\lambda }_{1}r}{\lambda }_{1}^{2}\left({\lambda }_{1}+a\right)+{\lambda }_{1}\left(d-b{\lambda }_{1}\right)\hfill \\ {G}_{1}=d+{K}^{2}cosKr-aKsinKr\hfill \\ {G}_{2}=-bK-{K}^{2}sinKr-aKcosKr\hfill \\ {G}_{3}=\left(d-ab\right){K}^{2}+\left({d}^{2}+{b}^{2}{K}^{2}\right)cosKr\hfill \\ {G}_{4}=-adK-b{K}^{3}-\left({d}^{2}+{b}^{2}{K}^{2}\right)sinKr\hfill \\ {G}_{5}=b{K}^{2}-a{K}^{2}cosKr-{K}^{3}sinKr\hfill \\ {G}_{6}=Kd-{K}^{3}cosKr+a{K}^{2}sinKr.\hfill \end{array}$

The results of minimization of the performance index (3.34) for $a=-5$, $r=0.5$ are presented at Figure 3.4 and they agree with calculations obtained in [36, Formula (44), p. 1049] with the use of the frequency–domain method. The minimal value of ${\left({z}_{2}^{0}\right)}^{-2}J\left({x}_{0}\right)$ is $0.01947$ and it is achieved at $b=-3.78$, $d=-6.897$. Figure 3.4: The level curves of the performance index ${\left({z}_{2}^{0}\right)}^{-2}J\left({x}_{0}\right)$ as a function of $b$ and $d$