Admissibility of retarded diagonal systems with one-dimensional input space

Abstract

We investigate infinite-time admissibility of a control operator B in a Hilbert space state-delayed dynamical system setting of the form ${\dot{z}}(t)=Az(t)+A_1 z(t-\tau )+Bu(t)$, where A generates a diagonal $C_0$-semigroup, $A_1\in {\mathcal {L}}(X)$ is also diagonal and $u\in L^2(0,\infty ;{\mathbb {C}})$. Our approach is based on the Laplace embedding between $L^2$ and the Hardy space $H^2({\mathbb {C}}_+)$. The results are expressed in terms of the eigenvalues of A and $A_1$ and the sequence representing the control operator.

Appendix

1.1 Proof of Lemma 19

We rewrite J as

$$\begin{aligned} J= & {} \frac{1}{2\pi }\int _{-\infty }^{\infty }\frac{d\omega }{\vert i\omega -\uplambda -\gamma \textrm{e}^{-i\omega \tau }\vert ^2} \nonumber \\= & {} \frac{1}{2\pi }\int _{-\infty }^{\infty }\frac{d\omega }{(i\omega -\uplambda -\gamma \textrm{e}^{-i\omega \tau })(-i\omega -\overline{\uplambda }-\overline{\gamma }\textrm{e}^{i\omega \tau })}\nonumber \\= & {} \frac{1}{2\pi i}\int _{-i\infty }^{i\infty }\frac{ds}{(s-\uplambda -\gamma \textrm{e}^{-s\tau })(-s-\overline{\uplambda }-\overline{\gamma }\textrm{e}^{s\tau })}\nonumber \\= & {} \frac{1}{2\pi i}\int _{-i\infty }^{i\infty }E_1(s)E_2(s)\,ds, \end{aligned}$$

(96)

where

$$\begin{aligned} E_1(s):=\frac{1}{s-\uplambda -\gamma \textrm{e}^{-s\tau }}, \qquad E_2(s):=\frac{1}{-s-\overline{\uplambda }-\overline{\gamma }\textrm{e}^{s\tau }}. \end{aligned}$$

(97)

Note that writing explicitly $E_1$ and $E_2$ as functions of s and parameters $\uplambda$, $\gamma$ and $\tau$ we have

$$\begin{aligned} E_1(s,\uplambda ,\gamma ,\tau )=E_2(-s,\overline{\uplambda },\overline{\gamma },\tau ). \end{aligned}$$

Let ${\mathcal {E}}_1$ be the set of poles of $E_1$ and ${\mathcal {E}}_2$ be the set of poles of $E_2$. As, by assumption, $\gamma \textrm{e}^{-ib\tau }\in \Lambda _{\tau ,a}$ Proposition 18 states that ${\mathcal {E}}_1\subset {\mathbb {C}}_-$ and ${\mathcal {E}}_2\subset {\mathbb {C}}_+$. Thus, we have that $E_1$ is analytic in ${\mathbb {C}}\setminus {\mathbb {C}}_-$ while $E_2$ is analytic in ${\mathbb {C}}\setminus {\mathbb {C}}_+$.

Let $s_n\in {\mathcal {E}}_1$, i.e. $s_n-\uplambda -\gamma \textrm{e}^{-s_n\tau }=0$. Rearranging gives

$$\begin{aligned} \frac{\gamma }{s_n-\uplambda }=\textrm{e}^{s_n\tau }. \end{aligned}$$

Substituting above to $E_2$ gives

$$\begin{aligned} E_2(s_n)=-\frac{s_n-\uplambda }{(s_n+\overline{\uplambda })(s_n-\uplambda )+|\gamma |^2}. \end{aligned}$$

(98)

and this value is finite as $s_n\not \in {\mathcal {E}}_2$. Rearranging (96) to account for (98) gives

$$\begin{aligned} J=\frac{1}{2\pi i}\int _{-i\infty }^{i\infty }\Bigg (E_1(s)\bigg (E_2(s)+\frac{s-\uplambda }{(s+\overline{\uplambda })(s-\uplambda )+|\gamma |^2}\bigg )-E_1(s)\frac{s-\uplambda }{(s+\overline{\uplambda })(s-\uplambda )+|\gamma |^2}\Bigg )\,ds.\nonumber \\ \end{aligned}$$

(99)

The above integrand has no poles at the roots $\{z_1,z_2\}$ of

$$\begin{aligned} (s+\overline{\uplambda })(s-\uplambda )+|\gamma |^2=0. \end{aligned}$$

(100)

However, as the two parts of the integrand in (99) will be treated separately, we need to consider poles introduced by $z_1$ and $z_2$ with regard to the contour of integration. Rewrite also (100) as

$$\begin{aligned} (s+\overline{\uplambda })(s-\uplambda )+|\gamma |^2=(s-z_1)(s-z_2)=0. \end{aligned}$$

(101)

From this point onwards we analyse three cases given by the right side of (63). Assume first that $|\gamma |<|a|$. Then

$$\begin{aligned} z_1=-\sqrt{a^2-|\gamma |^2}+ib,\quad z_2=\sqrt{a^2-|\gamma |^2}+ib. \end{aligned}$$

(102)

Fig. 2a shows integration contours $\Gamma _1(r)=\Gamma _I(r)+\Gamma _L(r)$ and $\Gamma _2(r)=\Gamma _I(r)+\Gamma _R(r)$ for $r\in (0,\infty )$ used for calculation of J. In particular $\Gamma _I$ runs along the imaginary axis, $\Gamma _L$ is a left semicircle and $\Gamma _R$ is a right semicircle.

Fig. 2

Contours of integration in (99): part a is used for the case $|\gamma |<|a|$, part b is used when $|\gamma |\ge |a|$. Both parts are drawn for a sufficiently large r so that $\Gamma _I(r)=C$, $\Gamma _L(r)=C_L$ and $\Gamma _R(r)=C_R$ and they enclose particular values of $z_1$ and $z_2$ in (102) and (110), respectively. The location of infinitesimally small semicircles around $z_1$ and $z_2$ in part b is to be modified depending on the location of $z_1$ and $z_2$ on the imaginary axis

Due to the above argument for a sufficiently large r we get

$$\begin{aligned} J= & {} \frac{1}{2\pi i}\lim _{r\rightarrow \infty }\int _{\Gamma _I(r)} \Bigg (E_1(s)\bigg (E_2(s)+\frac{s-\uplambda }{(s+\overline{\uplambda })(s-\uplambda ) +|\gamma |^2}\bigg )-E_1(s)\frac{s-\uplambda }{(s+\overline{\uplambda })(s-\uplambda ) +|\gamma |^2}\Bigg )\,ds \nonumber \\= & {} \frac{1}{2\pi i}\int _{C+C_L}E_1(s)\bigg (E_2(s)+\frac{s-\uplambda }{(s-z_1)(s-z_2)} \bigg )\,ds \nonumber \\{} & {} \quad -\frac{1}{2\pi i}\int _{C+C_R}E_1(s)\frac{s-\uplambda }{(s-z_1)(s-z_2)}\,ds. \end{aligned}$$

(103)

In calculation of the above we used the fact both integrals round the semicircles at infinity are zero as the integrands are, at most, of order $s^{-2}$ and for every fixed $\varphi , \uplambda , \gamma , \tau$,

$$\begin{aligned} \lim _{r\rightarrow \infty }\frac{1}{r\textrm{e}^{i\varphi }-\uplambda -\gamma \textrm{e}^{-r\textrm{e}^{i\varphi }\tau }}=0. \end{aligned}$$

(104)

Define separate parts of (103) as

$$\begin{aligned} J_L:=\frac{1}{2\pi i}\int _{C+C_L}E_1(s)\bigg (E_2(s)+\frac{s-\uplambda }{(s-z_1)(s-z_2)}\bigg )\,ds \end{aligned}$$

(105)

and

$$\begin{aligned} J_R:=-\frac{1}{2\pi i}\int _{C+C_R}E_1(s)\frac{s-\uplambda }{(s-z_1)(s-z_2)}\,ds \end{aligned}$$

(106)

and consider them separately.

To calculate $J_L$ note that from (98), it follows that for every $s_n\in {\mathcal {E}}_1$ the value

$$\begin{aligned} E_2(s_n)=-\frac{s_n-\uplambda }{(s_n-z_1)(s_n-z_2)} \end{aligned}$$

is finite and that implies that $\{z_1,z_2\}\cap {\mathcal {E}}_1=\emptyset$. Thus, the only pole of the integrand in (105) encircled by the $C+C_L$ contour is at $z_1$. Denoting this integrand by f the residue formula gives

$$\begin{aligned} \hbox {Res}_{z_1}f(s)=\lim _{s\rightarrow z_1}(s-z_1)f(s)=E_1(z_1)\frac{z_1-\uplambda }{z_1-z_2}. \end{aligned}$$

As the $C+C_L$ contour is counter-clockwise, we obtain

$$\begin{aligned} J_L=E_1(z_1)\frac{z_1-\uplambda }{z_1-z_2}. \end{aligned}$$

(107)

To calculate $J_R$ note that the only pole encircled by the $C+C_R$ contour is at $z_2$. Denoting the integrand of (106) by g the residue formula gives

$$\begin{aligned} \hbox {Res}_{z_2}g(s)=\lim _{s\rightarrow z_2}(s-z_2)g(s)=E_1(z_2)\frac{z_2-\uplambda }{z_2-z_1}. \end{aligned}$$

As the $C+C_R$ contour is clockwise, we obtain

$$\begin{aligned} J_R=E_1(z_2)\frac{z_2-\uplambda }{z_2-z_1}. \end{aligned}$$

(108)

Thus, we obtain

$$\begin{aligned} J=J_L+J_R=\frac{1}{z_2-z_1}\Big ((\uplambda -z_1)E_1(z_1)+(z_2-\uplambda )E_1(z_2)\Big ), \end{aligned}$$

(109)

where $z_1,z_2$ are given by (102). We substitute these values for $z_1$ and $z_2$ and perform tedious calculations to obtain

$$\begin{aligned} J&=\frac{1}{2\sqrt{a^2-|\gamma |^2}}\\&\quad \times \frac{\gamma \textrm{e}^{-ib\tau }\Big (\textrm{e}^{\sqrt{a^2-|\gamma |^2}\tau }(a-\sqrt{a^2-|\gamma |^2})+\textrm{e}^{- \sqrt{a^2-|\gamma |^2}\tau }(-a-\sqrt{a^2-|\gamma |^2})\Big )}{\gamma \textrm{e}^{-ib\tau }\Big (\overline{\gamma }\textrm{e}^{ib\tau }-\textrm{e}^{-\sqrt{a^2-|\gamma |^2}\tau }(-\sqrt{a^2-|\gamma |^2}-a)-\textrm{e}^{\sqrt{a^2-|\gamma |^2}\tau }(\sqrt{a^2-|\gamma |^2}-a)+\gamma \textrm{e}^{-ib\tau }\Big )}\\&=\frac{1}{2\sqrt{a^2-|\gamma |^2}}\\&\quad \frac{\textrm{e}^{\sqrt{a^2-|\gamma |^2}\tau }\big (a-\sqrt{a^2-|\gamma |^2}\big ) +\textrm{e}^{-\sqrt{a^2-|\gamma |^2}\tau }\big (-a-\sqrt{a^2-|\gamma |^2}\big )}{2\textrm{Re}(\gamma \textrm{e}^{-ib\tau }) +\textrm{e}^{\sqrt{a^2-|\gamma |^2}\tau }\big (a-\sqrt{a^2-|\gamma |^2}\big ) -\textrm{e}^{-\sqrt{a^2-|\gamma |^2}\tau }\big (-a-\sqrt{a^2-|\gamma |^2}\big )}. \end{aligned}$$

Assume now that $|\gamma |>|a|$. The roots $\{z_1,z_2\}$ of (101) are

$$\begin{aligned} z_1=i\sqrt{a^2-|\gamma |^2}+ib,\quad z_2=-i\sqrt{a^2-|\gamma |^2}+ib. \end{aligned}$$

(110)

To calculate J in (99), we now use the contour shown in Fig. 2b. We again define $J_L$ and $J_R$ as in (105) and (106), respectively, but with this new contour.

As $\gamma \textrm{e}^{-ib\tau }\in \Lambda _{\tau ,a}$ by Proposition 18 no pole of $E_1$ lies on the imaginary axis. Hence, no pole of the integrand in (105) is encircled by the $C+C_L$ contour and this gives

$$\begin{aligned} J_L=0. \end{aligned}$$

(111)

For $J_R$ the only poles of the integrand of (106) encircled by the $C+C_R$ contour are $z_1$ and $z_2$. Denoting this integrand by g the residue formula gives

$$\begin{aligned} \hbox {Res}_{z_1}g(s)=E_1(z_1)\frac{z_1-\uplambda }{z_1-z_2},\quad \hbox {Res}_{z_2}g(s)=E_1(z_2)\frac{z_2-\uplambda }{z_2-z_1}. \end{aligned}$$

As the $C+C_R$ contour is clockwise, we obtain

$$\begin{aligned} J_R=E_1(z_1)\frac{z_1-\uplambda }{z_1-z_2}+E_1(z_2)\frac{z_2-\uplambda }{z_2-z_1}. \end{aligned}$$

(112)

Thus, we obtain

$$\begin{aligned} J=J_L+J_R=\frac{1}{z_2-z_1}\Big ((\uplambda -z_1)E_1(z_1)+(z_2-\uplambda )E_1(z_2)\Big ), \end{aligned}$$

(113)

where $z_1,z_2$ are given by (110). Substituting these values, again after tedious calculations, we obtain

$$\begin{aligned} J=\,&\frac{1}{2i\sqrt{|\gamma |^2-a^2}}\\&\times \frac{a\Big (\textrm{e}^{i\sqrt{|\gamma |^2-a^2}\tau }-\textrm{e}^{-i\sqrt{|\gamma |^2-a^2}\tau }\Big ) -i\sqrt{|\gamma |^2-a^2}\Big (\textrm{e}^{i\sqrt{|\gamma |^2-a^2}\tau }+\textrm{e}^{-i\sqrt{|\gamma |^2-a^2}\tau }\Big )}{2\textrm{Re}(\gamma \textrm{e}^{-ib\tau })+a\Big (\textrm{e}^{i\sqrt{|\gamma |^2-a^2}\tau }+\textrm{e}^{-i\sqrt{|\gamma |^2-a^2}\tau }\Big ) -i\sqrt{|\gamma |^2-a^2}\Big (\textrm{e}^{i\sqrt{|\gamma |^2-a^2}\tau }-\textrm{e}^{-i\sqrt{|\gamma |^2-a^2}\tau }\Big )}\\ =\,&\frac{1}{2\sqrt{|\gamma |^2-a^2}}\\&\times \frac{a\sin \big (\sqrt{|\gamma |^2-a^2}\tau \big )-\sqrt{|\gamma |^2-a^2}\cos \big (\sqrt{|\gamma |^2-a^2}\tau \big )}{\textrm{Re}(\gamma \textrm{e}^{-ib\tau })+a\cos \big (\sqrt{|\gamma |^2-a^2}\tau \big )+\sqrt{|\gamma |^2-a^2}\sin \big (\sqrt{|\gamma |^2-a^2}\tau \big )}. \end{aligned}$$

For the last case assume that $|\gamma |=|a|>0$, as the assumption $\gamma \textrm{e}^{-b\tau }\in \Lambda _{\tau ,0}$ excludes the case $|a|=|\gamma |=0$ because $0\not \in \Lambda _{\tau ,0}$. Instead of $\{z_1,z_2\}$ we now have a single double root $z_0$ of (101) given by

$$\begin{aligned} z_0=ib. \end{aligned}$$

(114)

As $z_0$ lies on the imaginary axis we use the contour shown in Fig. 2b tailored to the case $z_1=z_2=z_0$. Define $J_L$ and $J_R$ as in (105) and (106), respectively, but with the contour tailored for $z_0$. For the same reasons as in (111) we have

$$\begin{aligned} J_L=0. \end{aligned}$$

(115)

For $J_R$ the only pole of the integrand of (106) encircled by the $C+C_R$ contour is $z_0$. Denoting this integrand by g the residue formula for a double root gives

$$\begin{aligned} \hbox {Res}_{z_0}g(s)=\lim _{s\rightarrow z_0}\frac{d}{ds}\left( (s-z_0)^2g(s)\right) =\frac{(a\tau -1)\gamma \textrm{e}^{-ib\tau }}{(a+\gamma \textrm{e}^{-ib\tau })^2}. \end{aligned}$$

With the current assumption we have that $a^2=\gamma \overline{\gamma }$. By this and the fact that the $C+C_R$ contour is clockwise, we obtain

$$\begin{aligned} J_R=\frac{1}{2}\frac{a\tau -1}{\textrm{Re}(\gamma \textrm{e}^{-ib\tau })+a}. \end{aligned}$$

(116)

As $J=J_L+J_R$ this finishes the proof. $\square$