# PARTIAL NULL-CONTROLLABILITY OF EVOLUTION EQUATIONS BY ONE-DIMENSIONAL
CONTROL

B. Shklyar  Department of Exact Sciences, Holon Academic Institute of
Technology, Holon 58102, Israel<EMAIL_ADDRESS>
###### Abstract.

The problem of partial null controllability for linear autonomous evolution
equations, which are controlled by a one-dimensional control, is under
consideration. The partial null-controllability conditions for coupled
abstract evolution systems have been obtained using the moment problem
approach.

###### Contents

  1. 1 Introduction and problem statement
  2. 2 Coupled evolution systems
  3. 3 The case $\sigma_{YZ}=\varnothing$
    1. 3.1 Eigenvectors and eigenvalues
      1. 3.1.1 Eigenvectors of operator $A$
      2. 3.1.2 Eigenvectors of adjoint operator $A^{\ast}$
  4. 4 Partial controllability conditions
    1. 4.0.1 The case $b\in Y$
    2. 4.0.2 One sufficient condition for strong minimality
    3. 4.1 Example 1
  5. 5 The case $\sigma_{YZ}\neq\varnothing$
    1. 5.1 Eigenvectors of the operators $A$ and $A^{\ast}$
  6. 6  Partial controllability conditions
    1. 6.1  Example 2
  7. 7 Conclusion

Keywords: Partial null-controllability of evolution equations, coupled
systems, eigenvalue problems, moment method, strongly minimal sequences.

AMS(MOS) Subject Classification: 93B05, 93B60

##  1\. Introduction and problem statement

Partial controllability is the ability to exert control or influence over only
a subset of components or entities within a distributed system, while leaving
others unaffected. This can be useful in scenarios where certain components
need to be modified or managed independently, while others remain unchanged.

For example, in a distributed database system, partial controllability may
involve the ability to reach data in specific partitions or shards, while
leaving other untouched. In a robotic system, partial controllability may
involve the ability to control the motion of a subset of robots while allowing
others to operate autonomously based on their local sensors and algorithms.

The controllability conditions for first-order differential control systems
are well-studied. However, high-order differential equations, partial
hyperbolic equations, delay equations, and many others can be written as a
first-order differential equation only by enlarging the dimension of the state
space. In many cases, it is necessary to modify only components of the
original state space, without affecting the rest of state space. This is why
the concept of partial controllability is important.

Let $X,Y,U$ be separable Hilbert spaces. Consider the evolution equation

$\dot{x}(t)=Ax(t)+bu\left(t\right),$ (e1) $x(0)=x_{0},$ (e2) $y(t)=Px(t),t\geq
0,$ (e3)

where $x(t)\in X$ is a current state, $x_{0}\in X$ is an initial state; the
operator $A$ generates a strongly continuous $C_{0}$-semigroup $S(t)$ of
bounded operators [14], $b\in X,P:X\rightarrowtail Y$ is a linear bounded
operator The input function $u\left(t\right)$ is assumed to be locally
$L^{2}\,$-integrated.

Following [2] we assume, that the observation operator
$P:X\mathbb{\rightarrowtail}Y$ is an orthogonal projection operator,

Denote by $x\left(t,x_{0},u\left(\cdot\right)\right)$ the mild solution of
equation (e1) with initial condition (e2).

######  Definition 1.

Equation (e1)–(e3) is said to be completely null-controllable on $[0,t_{1}]$,
if for each $x_{0}\in X$ there is a control $u(t),u(\cdot)\in L([0,t_{1}],U)$,
such that the corresponding solution
$x\left(t,x_{0},u\left(\cdot\right)\right)$ satisfies the condition

$x\left(t_{1},x_{0},u\left(\cdot\right)\right)=0.$

######  Definition 2.

Equation (e1)–(e3) is said to be partially null-controllable on $[0,t_{1}]$,
if for each $x_{0}\in X$ there is a control $u(t),u(\cdot)\in L([0,t_{1}],U)$,
such that the corresponding solution
$x\left(t,x_{0},u\left(\cdot\right)\right)$ satisfies the condition

$Px\left(t_{1},x_{0},u\left(\cdot\right)\right)=0.$ (e5.1)

Further for the sake of brevity we use $x\left(t\right)\ $instead of
$x\left(t,x_{0},u\left(\cdot\right)\right).$

The partial controllability has been well investigated for finite-dimensional
systems (see, for example, [12] and references therein). However, as far as we
know there are very few investigations on the partial controllability of
infinite-dimensional equations (see [8], [2] and reference therein).

In recent years, there has been an increasing interest in the study of the
controllability of coupled PDEs. Partial controllability of such systems is
one of important applications to control of these systems. In [2] partial
controllability conditions have been obtained for coupled parabolic systems
(see also references therein). Controllability conditions of [5] (see also
references therein) may be considered as partially null controllability
conditions for coupled hyperbolic (namely waved) equations, where only initial
condition conditions $x_{0}\in Y$ are considered.

The moment method is a very powerful tool for investigation of (complete)
controllability for linear evolution systems (see [4], [1], [5] and references
therein).

The main goal of this paper is to obtain results related to partial
controllability issues for linear evolution equations using the moment method.
The paper aims to provide insights into controlling such systems and may
discuss aspects related to hyperbolic and parabolic systems. The moment method
is a mathematical technique employed to analyze and establish controllability
in these systems.

We assume that $A$ has the properties:

(I) the operator $A$ has a purely point spectrum $\sigma_{A}$ which is either
finite or has no finite limit points, and each $\lambda\in\sigma_{A}$ has a
finite multiplicity;

(II) the family of generalized eigenvectors of the operator $A$ forms a Riesz
basic of $X.$

It is well-known, that $x\left(t\right)$ is defined by the formula 111The
integral is understood in the Bochner sense.:

$x\left(t\right)=S(t)x_{0}+\int_{0}^{t}S(t-\tau)bu(\tau)d\tau.$ (eqsol)

##  2\. Coupled evolution systems

Let $X=Y\times Z,$ where $Y$ and $Z$ are separable Hilbert spaces222We will
not distinguish between spaces $X$ and $X\times\\{0\\},Y$ and $\\{0\\}\times
Y$.. Consider the linear evolution system

$\dot{y}\left(t\right)=A_{Y}y\left(t\right)+A_{YZ}z\left(t\right)+b_{Y}u\left(t\right),$
(CS1 ) $\dot{z}\left(t\right)\
=~{}~{}~{}~{}~{}~{}~{}~{}~{}A_{Z}z\left(t\right)~{}+b_{Z}u\left(t\right).$
(CS2) $P=\left(\begin{array}[]{cc}I_{Y}&0\\\ 0&0\end{array}\right),$ (P)

where the operators $A_{Y}$ and $A_{Z}$ generate strongly continuous
semigroups $S_{Y}\left(t\right)~{}$and $S_{Z}\left(t\right)$ in $X$ and Y
correspondingly, and $S_{YZ}\left(t\right):Z\rightarrowtail Y$ is a bounded
strongly continuous operator.

This system is a partial case of system (e1)–(e3), where
$x=\left(\begin{array}[]{c}y\\\ z\end{array}\right),y\in Y,z\in
Z,~{}A=\left(\begin{array}[]{cc}A_{Y},&A_{YZ}\\\ 0&A_{Z}\end{array}\right),\
A^{\ast}=\left(\begin{array}[]{cc}A_{Y}^{\ast},&0\\\
A_{YZ}^{\ast}&A_{Z}^{\ast}\end{array}\right),$
$b=\left(\begin{array}[]{c}b_{Y}\\\ b_{Z}\end{array}\right).$

We assume, that eigenvalues of the operators $A_{Y}$ and $A_{Z}$ are simple,
and the operator

$A=\left(\begin{array}[]{cc}A_{Y},&A_{YZ}\\\ 0&A_{Z}\end{array}\right)$
satisfies the conditions (I)–(II).

The linear bounded operator $A_{YZ}$ will be called a couple operator.

We have

$S\left(t\right)=\left(\begin{array}[]{cc}S_{Y}\left(t\right),&S_{YZ}\left(t\right)\\\
0,&S_{Z}\left(t\right)\end{array}\right),$

where $S_{Y}\left(t\right)~{}$and $S_{Z}\left(t\right)$ are strongly
continuous semigroups, generated by the operators $A_{Y}$ and $A_{Z}$
correspondingly, and $S_{YZ}\left(t\right):Z\rightarrowtail Y$ is a bounded
strongly continuous operator.

Denote by $\sigma_{Y}$ the spectrum of the operator $A_{Y},$ and by
$\sigma_{Z}$ the spectrum of the operator $A_{Z}.$

Denote also:
$I_{Y}:=\left\\{i\in\mathbb{N}:\lambda_{i}\in\sigma_{Y}\right\\},I_{Z}:=\left\\{i\in\mathbb{N}:\lambda_{i}\notin\sigma_{Y}\right\\}.$
Obviously, $\sigma_{Y}\cup\sigma_{Z}=\mathbb{\sigma}_{A},~{}I_{Y}\cup
I_{Z}=\mathbb{N}.$

Denote $\sigma_{YZ}=\sigma_{Y}\cap\sigma_{Z}.$

##  3\. The case $\sigma_{YZ}=\varnothing$

###  3.1. Eigenvectors and eigenvalues

In the beginning consider the case $\sigma_{YZ}=\varnothing$.
Obviously,$~{}I_{Y}\cap I_{Z}=\varnothing$ also.

####  3.1.1. Eigenvectors of operator $A$

Let $\varphi_{i}=\left(\begin{array}[]{c}\varphi_{Yi}\\\
\varphi_{Yi}\end{array}\right),i\in\mathbb{N}$ be the eigenvector of the
operator $A,$ corresponding to its eigenvalue $\lambda_{i},i\in\mathbb{N}.$ We
have

$A\varphi_{i}=\left(\begin{array}[]{c}A_{Y}\varphi_{Yi}+A_{YZ}\varphi_{Zi}\\\
A_{Z}\varphi_{Zi}\end{array}\right)=\lambda_{i}\left(\begin{array}[]{c}\varphi_{Yi}\\\
\varphi_{Yi}\end{array}\right),i\in\mathbb{N}.$

Hence,

$\displaystyle A_{Y}\varphi_{Yi}+A_{YZ}\varphi_{Zi}$ $\displaystyle=$
$\displaystyle\lambda_{i}\varphi_{Yi},i\in\mathbb{N},$ $\displaystyle
A_{Z}\varphi_{Zi}$ $\displaystyle=$
$\displaystyle\lambda_{i}\psi_{Zi},i\in\mathbb{N}.$

1) Let $j\in I_{Y}\iff\lambda_{j}\in\sigma_{Y},~{}$ and let $\varphi_{Yj}$ be
a corresponding eigenvector of the operator $A_{Y}.$ In this case the vector
$\varphi_{j}=\left(\begin{array}[]{c}\varphi_{Yj}\\\ 0\end{array}\right)$ is
an eigenvector of the operator $A.$

2) Let $k\in I_{Z}\iff\lambda_{k}\in\sigma_{Z},$ and let $\varphi_{Zk}\neq
0~{}$be a corresponding eigenvector of the operator $A_{Z}.$ In this case the
vector $\varphi_{k}=\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right),~{}$where the vector $\varphi_{Yk}~{}$ is a
solution of the operator equation

$A_{Y}\varphi_{Yk}+A_{YZ}\varphi_{Zk}=\lambda_{k}\varphi_{Yk},\varphi_{Zk}\neq
0,k\in I_{\mathbb{Z}},$ (AphiYk)

is an eigenvector of the operator $A.$

######  Remark 1.

Since $\lambda_{k}\ $is a regular value of the operator $A_{Y},$ the operator
equation (AphiYk) is always solvable. Its solution is:

$\varphi_{Yk}=\left(A_{Y}-\lambda_{k}I_{Y}\right)^{-1}A_{YZ}\varphi_{Zk},k\in
I_{\mathbb{Z}}.$ (phiYk)

####  3.1.2. Eigenvectors of adjoint operator $A^{\ast}$

Let $\psi_{i}=\left(\begin{array}[]{c}\psi_{Yi}\\\
\psi_{Zi}\end{array}\right),i\in\mathbb{N}$ be the eigenvector of the operator
$A^{\ast},$ corresponding to its eigenvalue
$\bar{\lambda}_{i},i\in\mathbb{N}.$ We have

$A^{\ast}\psi_{i}=\left(\begin{array}[]{c}A_{Y}^{\ast}\psi_{Yi}\\\
A_{YZ}^{\ast}\psi_{Yi}+A_{Z}^{\ast}\psi_{Zi}\end{array}\right)=\bar{\lambda}_{i}\left(\begin{array}[]{c}\psi_{Yi}\\\
\psi_{Zi}\end{array}\right),i\in\mathbb{N}.$

Hence,

$\displaystyle A_{Y}^{\ast}\psi_{Yi}$ $\displaystyle=$
$\displaystyle\bar{\lambda}_{i}\psi_{Yi},i\in\mathbb{N},$ $\displaystyle
A_{YZ}^{\ast}\psi_{Yi}+A_{Z}^{\ast}\psi_{Zi}$ $\displaystyle=$
$\displaystyle\bar{\lambda}_{i}\psi_{Zi},i\in\mathbb{N}.$

1) Let $k\in I_{Z}\iff\lambda_{k}\in\sigma_{Z},$ and let $\psi_{Zk}$ be a
corresponding eigenvector of the operator $A_{Z}^{\ast}.$ In this case the
vector $\psi_{k}=\left(\begin{array}[]{c}0\\\ \psi_{Zk}\end{array}\right)$ is
an eigenvector of the operator $A^{\ast}.$

2) Let $j\in I_{Y}\iff\lambda_{j}\in\sigma_{Y},$ and let $\psi_{Zj}$ be a
corresponding eigenvector of the operator $A_{Y}^{\ast}.$ In this case the
vector $\psi_{j}=\left(\begin{array}[]{c}\psi_{Yj}\\\
\psi_{Zj}\end{array}\right),$ where the vector $\psi_{Zj}~{}$ is a solution of
the equation

$A_{YZ}^{\ast}\psi_{Yj}+A_{Z}^{\ast}\psi_{Zj}=\bar{\lambda}_{j}\psi_{Zj},j\in
I_{Y},$ (A*psiZj)

is an eigenvector of the operator $A^{\ast}.$

######  Remark 2.

Since $\bar{\lambda}_{j}\ $is a regular value of the operator $A_{Z}^{\ast},$
the operator equation (A*psiZj) is always solvable. Its solution is:

$\psi_{Zj}=\left(A_{Z}^{\ast}-\bar{\lambda}_{k}I_{Z}\right)^{-1}A_{YZ}^{\ast}\psi_{Yj},0,k\in
I_{\mathbb{Z}}.$ (psiZj)

Denote: $b_{i}=\left(b,\psi_{i}\right),~{}i\in\mathbb{N}.$

We have

$b_{i}=\left(\begin{array}[]{c}b_{Yi}\\\
b_{Zi}\end{array}\right),i\in\mathbb{N}.$

where

$b_{Yi}=\left(b_{Y},\psi_{Yi}\right),~{}b_{Zi}=\left(b_{Z},\psi_{Zi}\right),i\in\mathbb{N}.$
(bi)

If $k\in I_{Z},$ then $\psi_{Yk}=0,$ so $b_{k}=\left(\begin{array}[]{c}0\\\
b_{Zk}\end{array}\right).$

The eigenvectors $\varphi_{i},\psi_{i},i\in\mathbb{N}$ of operators $A~{}$and
$A^{\ast}$ are chosen such that following orthogonality conditions hold:

$\left(\varphi_{m},\psi_{n}\right)=\delta_{mn},~{}m,n\in\mathbb{N}.$

Hence

$\left(\varphi_{Yj},\psi_{Yl}\right)=\delta_{jl},j,l\in I_{Y},$ (ortY)
$\left(\varphi_{Zk},\psi_{Zm}\right)=\delta_{km},k,m\in I_{Z},$ (ortZ)
$\left(\varphi_{Yk},\psi_{Yj}\right)=-\left(\varphi_{Zk},\psi_{Zj}\right),k\in
I_{Z},j\in I_{Y}.$ (ortYZ)

Denote:

$\Phi_{Y}:=\left\\{\varphi_{j}=\left(\begin{array}[]{c}\varphi_{Yj}\\\
0\end{array}\right),j\in
I_{Y}\right\\},\Phi_{Z}:=\left\\{\varphi_{k}=\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right),k\in I_{Z}\right\\},$

$\Psi_{Y}:=\left\\{\psi_{j}=\left(\begin{array}[]{c}\psi_{Yj}\\\
\psi_{Zj}\end{array}\right),j\in
I_{Y}\right\\},\Psi_{Z}:=\left\\{\psi_{k}=\left(\begin{array}[]{c}0\\\
\psi_{Zk}\end{array}\right),k\in I_{Z}\right\\}$

######  Lemma 1.

The sequences $\left\\{\varphi_{Yj},j\in I_{Y}\right\\}$ and
$\left\\{\psi_{Yj},j\in I_{Y}\right\\}$ are Riesz basics of the space $Y.$

The sequences $\left\\{\varphi_{Zk},k\in I_{Z}\right\\}$ and
$\left\\{\psi_{Zk},k\in I_{Z}\right\\}$ are Riesz basics of the space $Z.$

###### Proof.

This follows from results of [19] Theorem 2.1 (see also [7]).

##  4\. Partial controllability conditions

Below all definitions and notations of strongly minimal sequences belong to
[20].

Let

$\left\\{x_{1},x_{2},...,x_{n},...\right\\}$ (s)

be a minimal sequence of elements of a separable Hilbert space $H.$

The infinite-dimensional matrix

$\left\\{\varphi_{ij}=(x_{i},x_{j}),i,j\in\mathbb{N}\right\\}$ (GM)

is called a Gram matrix of family (s).

Let

$G_{n}=\left\\{\varphi_{ij}=(x_{i},x_{j}),i,j=1,2,...,n\right\\}$ (GMn)

be $n\times n$ cutout of (GM), and let
$0\leq\lambda_{1}^{n}\leq\lambda_{2}^{n}\leq...\leq\lambda_{n}^{n}$ be
eigenvalues of (GMn)333Matrix (GMn) is a positive Hermitian matrix for any
$n\in\mathbb{N}.$. It is well-known, that $\lambda_{1}^{n}\geq 0$ is not
increasing, $\lambda_{n}^{n}$ is not decreasing, so there exists
$\lim\limits_{n\rightarrow\infty}\lambda_{1}^{n}\geq 0.$

######  Definition 3.

If $\lim\limits_{n\rightarrow\infty}\lambda_{1}^{n}>0,$ the sequence (s) is
called strongly minimal sequence444The definition of strongly minimal
sequences is equivalent to the definition of Riesz-Fisher sequences [10],
[21]. In our opinion,the strong minimality concept better explains the
relationship of such sequences with minimal systems..

Denote:

$b_{j}\left(t\right)=b_{j}-e^{\lambda_{j}t}e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}-t\right)b_{Z,},\psi_{Zj}\right),j\in
I_{Y},~{}t\in\left[0,t_{1}\right].$

######  Theorem 1.

Let the series

$\sum\limits_{j\in\mathbb{I}_{Y}}e^{-2\lambda_{j}t_{1}}\left|\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)\right|^{2}$
(x_z(t1))

converges for any $x_{Z}^{0}\in Z$.

Equation (CS1) is partially null controllable on $[0,t_{1}]$ if and only if
the family of exponents

$\left\\{e^{-\lambda_{j}t}b_{j}\left(t\right),j\in
I_{Y},t\in\left[0,t_{1}\right]\right\\}$ (-exp)

is strongly minimal.

###### Proof.

Since the family of eigenvectors
$\left\\{\varphi_{i},i\in\mathbb{N}\right\\}~{}$of the operator $A~{}$is a
Riesz basic of $X$ and the family of eigenvectors
$\left\\{\psi_{i},i\in\mathbb{N}\right\\}~{}$of the operator $A^{\ast}~{}$is a
Riesz basic of $X~{}$as well , for any $x^{0}\in X$ we have

$x^{0}=\sum\limits_{i\in\mathbb{N}}\left(x^{0},\psi_{i}\right)\varphi_{i}=\sum\limits_{j\in
I_{Y}}\left(x^{0},\psi_{j}\right)\left(\begin{array}[]{c}\varphi_{Yj}\\\
0\end{array}\right)+\sum\limits_{k\in\mathbb{I}_{Z}}\left(x^{0},\psi_{k}\right)\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right),$ (x0)

$x_{Y}^{0}=\sum\limits_{j\in\mathbb{I}_{Y}}\left(x^{0},\psi_{j}\right)\varphi_{Yj}+\sum\limits_{k\in\mathbb{I}_{Z}}\left(x^{0},\psi_{k}\right)\varphi_{Yk},$
(xY0)

$\left(x_{Y}^{0},\psi_{Yj}\right)=\left(x^{0},\psi_{j}\right)+\left(\sum\limits_{k\in\mathbb{I}_{Z}}\left(x^{0},\psi_{k}\right)\varphi_{Yk},\psi_{Yj}\right)=$
(xY0psiY)
$=\left(x_{Y}^{0},\psi_{Yj}\right)+\left(x_{Z}^{0},\psi_{Zj}\right)+\left(\sum\limits_{k\in\mathbb{I}_{Z}}\left(x^{0},\psi_{k}\right)\varphi_{Yk},\psi_{Yj}\right).$

After substituting $S\left(t_{1}\right)x^{0}$ instead of $x^{0}$ in (x0), we
obtain

$x^{0}\left(t_{1}\right):=S\left(t_{1}\right)x^{0}=\sum\limits_{i\in\mathbb{N}}\left(S\left(t_{1}\right)x^{0},\psi_{i}\right)\varphi_{i}=$

$=\sum\limits_{j\in
I_{Y}}\left(x^{0},\psi_{j}\right)e^{\lambda_{j}t_{1}}\left(\begin{array}[]{c}\varphi_{Yj}\\\
0\end{array}\right)+\sum\limits_{k\in
I_{Z}}\left(x^{0},\psi_{k}\right)e^{\lambda_{k}t_{1}}\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right),$ (S(t1)x0)

Using $b$ instead of $x^{0}$ and $t_{1}-\tau$ instead of $t_{1}$ in (S(t1)x0)
we obtain after integration, that

$\int_{0}^{t_{1}}S\left(t_{1}-\tau\right)bu\left(\tau\right)d\tau=$
$=\int_{0}^{t_{1}}\left[\sum\limits_{j\in\mathbb{I}_{Y}}b_{j}e^{\lambda_{j}\
(t_{1}-\tau)\ }\left(\begin{array}[]{c}\varphi_{Yj}\\\
0\end{array}\right)+\sum\limits_{k\in
I_{Z}}b_{k}e^{\lambda_{k}\left(t_{1}-\tau\right)}\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right)\right]u\left(\tau\right)d\tau.$ (S(t1)bu(t))

Substituting (S(t1)x0) and (S(t1)bu(t)) into (eqsol), we obtain

$x\left(t_{1}\right)=\sum\limits_{j\in
I_{Y}}\left(\left(x^{0},\psi_{j}\right)e^{\lambda_{j}t_{1}}+\left(\int_{0}^{t_{1}}b_{j}e^{\lambda_{j}\left(\
t_{1}-\tau\right)\
}u\left(\tau\right)d\tau\right)\right)\left(\begin{array}[]{c}\varphi_{Yj}\\\
0\end{array}\right)+$ (x(t1)) $+\sum\limits_{k\in
I_{Z}}\left(\left(x^{0},\psi_{k}\right)e^{\lambda_{k}t_{1}}+\left(\int_{0}^{t_{1}}\left(b,\psi_{k}\right)e^{\lambda_{k}\left(t_{1}-\tau\right)}u\left(\tau\right)d\tau\right)\right)\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right),$

$x_{Y}\left(t_{1}\right)=\sum\limits_{j\in
I_{Y}}\left(x^{0},\psi_{j}\right)e^{\lambda_{j}t_{1}}\varphi_{Yj}+\sum\limits_{k\in
I_{Z}}\left(x^{0},\psi_{k}\right)e^{\lambda_{k}t_{1}}\varphi_{Yk}+$ (xY(t1))
$+\int_{0}^{t_{1}}\sum\limits_{j\in I_{Y}}\left(b_{j}e^{\lambda_{j}\left(\
t_{1}-\tau\right)\ }\varphi_{Yj}+\sum\limits_{k\in
I_{Z}}b_{k}e^{\lambda_{k}\left(t_{1}-\tau\right)}\varphi_{Yk}\right)u\left(\tau\right)d\tau.$

Taking into account, that

$\left(x^{0},\psi_{k}\right)=\left(x_{Z}^{0},\psi_{Zk}\right)~{}~{}\mathrm{and~{}}\left(~{}\varphi_{Yk},\psi_{Yj}\right)+\left(\varphi_{Zk},\psi_{Zj}\right)~{}=0,\forall
j\in I_{Y},k\in I_{Z},$

we obtain

$\left(x_{Y}\left(t_{1}\right),\psi_{Yj}\right)=$
$=e^{\lambda_{j}t_{1}}\left(x^{0},\psi_{j}\right)+\left(\sum\limits_{k\in
I_{\mathbb{Z}}}e^{\lambda_{k}t_{1}}\left(x^{0},\psi_{k}\right)\varphi_{Yk},\psi_{Yj}\right)+$
$+\left(\int_{0}^{t_{1}}e^{\lambda_{j}\left(t_{1}-\tau\right)\
}b_{j}+\sum\limits_{k\in
I_{\mathbb{Z}}}\left(e^{\lambda_{k}\left(t_{1}-\tau\right)}\left(b_{Z},\psi_{Zk}\right)\varphi_{Yk},\psi_{Yj}\right)\right)u\left(\tau\right)d\tau=.$
$=e^{\lambda_{j}t_{1}}\left(x^{0},\psi_{j}\right)-\sum\limits_{k\in
I_{\mathbb{Z}}}\left(e^{\lambda_{k}t_{1}}\left(x_{Z}^{0},\psi_{Zk}\right)\varphi_{Zk},\psi_{Zj}\right)+$
$+\int_{0}^{t_{1}}\left(e^{\lambda_{j}\left(t_{1}-\tau\right)\
}b_{j}-e^{-\lambda_{j}t_{1}}\sum\limits_{k\in
I_{\mathbb{Z}}}\left(e^{\lambda_{k}\left(t_{1}-\tau\right)}\left(b_{Z},\psi_{Zk}\right)\varphi_{Yk},\psi_{Yj}\right)\right)u\left(\tau\right)d\tau,~{}j\in
I_{Y}~{}.$

Since the sequence $\left\\{\varphi_{Zk},k\in I_{Z}\right\\}$ is a Riesz basic
of the space $Z$ (see Lemma 1), we have

$S_{Z}\left(t\right)z=\sum\limits_{k\in
I_{\mathbb{Z}}}e^{\lambda_{k}t}\left(z,\psi_{Zk}\right)\varphi_{Zk},\forall
t>0,\forall z\in Z~{}.$ (Sz(t)Xz)

Hence

$\left(x_{Y}\left(t_{1}\right),\psi_{Yj}\right)=e^{\lambda_{j}t_{1}}\left[\left(x^{0},\psi_{j}\right)-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)\right]+$
(xYpsiYj) $+\left(\int_{0}^{t_{1}}e^{\lambda_{j}\ t_{1}\
}\left[e^{-\lambda_{j}\tau\
}b_{j}-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\psi_{Zj}\right)\right]\right)u\left(\tau\right)d\tau,~{}j\in
I_{Y}~{}$

.

Thus, the partial null-controllability we are proving is equivalent to the
existence of a solution $u\left(\cdot\right)\in L_{2}\left[0,t_{1}\right]$ to
the moment problem

$-\left(x^{0},\psi_{j}\right)+e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zj}\right)=$
$=\int_{0}^{t_{1}}\left[e^{-\lambda_{j}\tau\
}b_{j}-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\psi_{Zj}\right)\right]u\left(t\right)dt,$
( LMP )

$\forall j\in I_{Y},~{}~{}\forall x^{0}\in X.$

By convergence of series (x_z(t1)) we obtain

$\sum\limits_{j\in
I_{Y}}\left|\left(x^{0},\psi_{j}\right)-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zj}\right)\right|^{2}<+\infty,\forall
x^{0}\in X.$

Hence for any $\forall x^{0}\in X$ moment problem (LMP) admits a solution
$u\left(\cdot\right)\in L_{2}\left[0,t_{1}\right]$ if and only if the family
(-exp) is strongly minimal.

This proves the theorem.

######  Remark 3.

Of course series (x_z(t1)) converges, if $I_{Y}$ is finite. If $I_{Y}$ is
infinite, then the convergence of series (x_z(t1)) essentially depends on the
structure of operator $A_{YZ}$.

Example. Let
$X=L_{2}\left[0,\pi\right]\times\mathbb{R},~{}Y=L_{2}\left[0,\pi\right],~{}Z=\mathbb{R},A_{Y}=\Delta~{}$with
boundary conditions $y\left(0\right)=y\left(\pi\right)=0,\
A_{Z}=w\in\mathbb{R},~{}0\leq w<1,~{}A_{YZ}=a\in Y;$

$I_{Y}=\mathbb{N}\backslash\\{1\\}=2,3,...,~{}I_{Z}=\\{1\\};~{}\lambda_{1}=w,\varphi_{Z1}=\psi_{Z1}=1,$

$\lambda_{l}=-\left(l-1\right)^{2},~{}\varphi_{Yl}=\psi_{Yl}=\sqrt{\frac{2}{\pi}}\sin\left(l-1\right)x,~{}l\in\mathbb{N}\backslash\\{1\\}~{};$

$A_{YZ}^{\ast}\psi_{Yl}+w\psi_{Zl}=\lambda_{l}\psi_{Zl}\iff$

$\iff\psi_{Zl}=\frac{A_{YZ}^{\ast}\psi_{Yl}}{\lambda_{l}-w}=-\frac{\left(a,\psi_{Yl}\right)}{\left(l-1\right)^{2}+w}=-\frac{\sqrt{\frac{2}{\pi}}\int_{0}^{\pi}a\left(x\right)\sin\left(\left(l-1\right)x\right)dx}{\left(l-1\right)^{2}+w},l\in\mathbb{N}\backslash\\{1\\},$

Obviously,$~{}S_{Z}\left(t_{1}\right)x_{Z,}^{0}=e^{wt_{1}}x_{Z}^{0}.$ Using
$j=l-1.l=2,3...,$we obtain

$\psi_{Zj+1}=-\sqrt{\frac{2}{\pi}}\frac{\int_{0}^{\pi}a\left(x\right)\sin
jxdx}{j^{2}+w},j\in\mathbb{N}.$

Hence, for any $l\in I_{Y}~{}~{}$we have

$(e^{-\lambda_{l}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zl}\right)=e^{\left(l-1\right)^{2}t_{1}}e^{wt_{1}}x_{0}^{2}\psi_{Zl}=$

$=-\sqrt{\frac{2}{\pi}}x_{0}^{2}e^{wt_{1}}e^{j^{2}t_{1}}\frac{\int_{0}^{\pi}a\left(x\right)\sin
jxdx}{j^{2}+w},j\in\mathbb{N}.$

Obvious, $\sum\limits_{j=1}^{\infty}\frac{1}{j^{2}}<+\infty,$ hence by Riesz-
Fisher Theorem there exists $a\left(x\right)\in L_{2}\left[0,\pi\right],$ such
that $\int_{0}^{\pi}a\left(x\right)\sin
jxdx=\frac{1}{j},~{}~{}~{}~{}j\in\mathbb{N}.$

Also, $\sum\limits_{j=1}^{~{}\infty}e^{-2j^{2}t_{1}}<+\infty,$ hence by Riesz-
Fisher Theorem there exists $a\left(x\right)\in L_{2}\left[0,\pi\right],$ such
that $\int_{0}^{\pi}a\left(x\right)\sin jxdx=e^{-j^{2}t_{1}},j\in\mathbb{N}.$

If $\int_{0}^{\pi}a\left(x\right)\sin jxdx=\frac{1}{j},$ then

$\sum\limits_{j\in
I_{Y}}\left|e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zj}\right)\right|^{2}=\sqrt{\frac{2}{\pi}}e^{2wt_{1}}\left|x_{Z}^{0}\right|^{2}\sum\limits_{j=1}^{\infty}e^{2j^{2}t_{1}}\frac{1}{j^{2}\left(j^{2}+w\right)^{2}}=+\infty.$

If $\int_{0}^{\pi}a\left(x\right)\sin jxdx=e^{-2j^{2}t_{1}},$then

$\displaystyle\sum\limits_{j\in
I_{Y}}\left|e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zj}\right)\right|^{2}$
$\displaystyle=$
$\displaystyle\sqrt{\frac{2}{\pi}}\sum\limits_{j=1}^{\infty}e^{2j^{2}t_{1}}\sqrt{\frac{2}{\pi}}e^{2wt_{1}}\left|x_{Z,}^{0}\right|^{2}\frac{e^{-2j^{2}t_{1}}}{\left(j^{2}+w\right)^{2}}=$
$\displaystyle=$
$\displaystyle\sqrt{\frac{2}{\pi}}e^{2wt_{1}}\left|x_{Z,}^{0}\right|^{2}\sum\limits_{j=1}^{\infty}\frac{1}{\left(j^{2}+w\right)^{2}}<+\infty.$

####  4.0.1. The case $b\in Y$

In this chase problem moment (LMP) is simplified.

Let $b_{z}=0,$ i.e. $b=b_{Y}\in Y.$

Controllability of such partial parabolic control systems have important
applications to many fields such as chemistry, physics or biology (see also
Example 3 below).

If $b\in Y,$ then (CS1)–(P) is not completely null-controllable on
$\left[0,t_{1}\right],$ $\forall t_{1}>0,$ because solutions of equation (CS2)
are independent on control. However it may be partially null controllable.

In this chase problem moment (LMP) is simplified:

$-\left(x^{0},\psi_{j}\right)+e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zj}\right)=$
$=\int_{0}^{t_{1}}b_{j}e^{-\lambda_{j}\tau\ }u\left(t\right)dt,j\in I_{Y}.$ (
LMP0)

######  Corollary 1.

If the series (x_z(t1)) converges for any $x_{Z}^{0}\in Z,$ then system
(CS1)–(P) is partially null controllable on $[0,t_{1}]$ if and only if the
family of exponent

$\left\\{e^{-\lambda_{j}t}b_{j},j\in I_{Y},t\in\left[0,t_{1}\right]\right\\}$
(-exp0)

is strongly minimal.

######  Corollary 2.

If the series (x_z(t1) converges for any $x_{Z}^{0}\in Z,$ then system
(CS1)–(P) is partially null controllable on $[0,t_{1}]$ if and only if
equation

$\dot{y}\left(t\right)=A_{Y}y\left(t\right)+b_{Y}u\left(t\right)$

is null (completely) controllable on $[0,t_{1}].$

###### Proof.

Since the series (x_z(t1)) converges for any $x_{Z}^{0}\in Z,$ and the family
$\psi_{j},~{}j\in Y$ forms a Riesz basic of $Y$, then for $x_{Y}^{0}\in
Y~{}$and$~{}x_{Z}^{0}\in Z$ there exists $y^{0}\in Y,$ such that

$-\left(x^{0},\psi_{j}\right)+e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z,}^{0},\psi_{Zj}\right)=-\left(y^{0},\psi_{j}\right),j\in
Y.$

The corollary follows from the above formula and (LMP0).

####  4.0.2. One sufficient condition for strong minimality

The direct method for checking the strong minimality of any sequence (s) is
the computing of $\lim\limits_{n\rightarrow\infty}\lambda_{1}^{n}>0$ by
numeric methods. It may be rather difficult. However if $b_{Y}\in Y,$ then one
can find more simpler strong minimality conditions for sequence (-exp) in some
partial cases (see, for example, results of [1],[21], [22],[11] and references
therein) .

Consider a strongly decreasing sequence of negative numbers
$\\{\lambda_{j},j\in I_{Y}\\}$, with finite upper density [18].

Let the family

$\left\\{\exp(\lambda_{j}t),j\in I_{Y},t\in\left[0,t\right]\right\\}$
(exp(lambda_t))

be minimal, and let

$\left\\{q_{j}\left(t\right),j\in I_{Y},t\in\left[0,t_{1}\right]\right\\}$
(exp_biort)

be biorthogonal family for (exp(lambda_t)).

######  Theorem 2.

If $b_{j}\neq 0,\forall j\in
I_{Y},\lim_{j\rightarrow\infty}\frac{2\ln\left|b_{j}\right|}{\lambda_{j}}$ is
finite,
$t_{1}>\lim_{j\rightarrow\infty}\frac{2\ln\left|b_{j}\right|}{\lambda_{j}}$,
and for any $\varepsilon>0$ there exists a positive constant $K_{\varepsilon}$
such that

$\left|\left|q_{j}\left(\cdot\right)\right|\right|<K_{\varepsilon}e^{-\lambda_{j}\varepsilon},j\in
I_{Y},$ (bio_est)

then the family

$\left\\{b_{j}\exp(-\lambda_{j}t),j\in
I_{Y},t\in\left[0,t_{1}\right]\right\\}$ (-expb)

is strongly minimal.

###### Proof.

Obviously, the family

$\left\\{\frac{q_{j}\left(t_{1}-t\right)}{b_{j}}e^{\lambda_{j}t_{1}},j\in
I_{Y},t\in\left[0,t_{1}\right]\right\\}$

is biorthogonal to the family (-expb).

Let $I_{Y}^{f}$ be any finite subsequence of $I_{Y}.$ We have

$\left\|\sum_{j\in
I_{Y}^{f}}\beta_{j}\frac{q_{j}\left(t_{1}-t\right)e^{\lambda_{j}t_{1}}}{b_{j}}\right\|^{2}\leq\left(\sum_{j\in
I_{Y}^{f}}\left|\beta_{j}\right|\frac{1}{\left|b_{j}\right|}\left\|q_{j}\left(\cdot\right)\right\|e^{\lambda_{j}t_{1}}\right)^{2}\leq$
$\leq\left(\sum_{j\in
I_{Y}^{f}}\left|\beta_{j}\right|^{2}\right)\left(\sum_{j\in
I_{Y}^{f}}\frac{1}{\left|b_{j}\right|^{2}}\left\|q_{j}\left(\cdot\right)\right\|^{2}e^{2\lambda_{j}t_{1}}\right)\leq$
$\leq\left(\sum_{j\in
I_{Y}^{f}}\left|\beta_{j}\right|^{2}\right)\left(\sum_{j\in
I_{Y}^{f}}\frac{1}{\left|b_{j}\right|^{2}}K_{\varepsilon}^{2}e^{-2\lambda_{j}\varepsilon}e^{2\lambda_{j}t_{1}}\right)=$
$=K_{\varepsilon}^{2}\left(\sum_{j\in
I_{Y}^{f}}\left|\beta_{j}\right|^{2}\right)\left(\sum_{j\in
I_{Y}^{f}}\frac{1}{\left|b_{j}\right|^{2}}e^{2\lambda_{j}\left(t_{1}-\varepsilon\right)}\right)<K_{\varepsilon}^{2}\left(\sum_{j\in
I_{Y}^{f}}\left|\beta_{j}\right|^{2}\right)\left(\sum_{j\in
I_{Y}^{f}}\frac{1}{\left|b_{j}\right|^{2}}e^{\lambda_{j}t_{1}}\right)$

for $\varepsilon<\frac{t_{1}}{2}.$

The series $\sum_{j\in
I_{Y}}\frac{1}{\left|b_{j}\right|^{2}}e^{\lambda_{j}t_{1}}$ is the Dirichlet
series with abscissa of the convergence [18]

$\sigma_{c}=\lim_{j\rightarrow\infty}\frac{2\ln\left|b_{j}\right|}{\lambda_{j}}.$

Hence

$\left\|\sum_{j\in
I_{Y}^{f}}\beta_{j}\frac{q_{j}\left(t_{1}-t\right)}{b_{j}}e^{\lambda_{j}t_{1}}\right\|^{2}\leq
M^{2}\sum_{j\in
I_{Y}^{f}}\left|\beta_{j}\right|^{2},~{}\forall\beta_{j}\in\mathbb{R},j\in
I_{Y}^{f}$ (BS)

for
any$~{}t_{1}>\lim_{j\rightarrow\infty}\frac{2\ln\left|b_{j}\right|}{\lambda_{j}},$where
$M^{2}=K_{\varepsilon}^{2}\sum_{j\in
I_{Y}}\frac{1}{b_{j}^{2}}e^{\lambda_{j}t_{1}}.$ It follows from (BS), that the
family
$\left\\{\frac{q_{j}\left(t_{1}-t\right)}{b_{j}}e^{\lambda_{j}t_{1}},j\in
I_{Y},t\in\left[0,t_{1}\right]\right\\}$ is a Bessel sequence. Hence [10],
[21], family (-expb) is a Riesz-Fisher sequence, i.e. this family is strongly
minimal.

This proves the theorem.

The next theorem follows from previous Theorem 2 and results of [18], [17].

######  Theorem 3.

. Let
$I_{Y}=\\{j_{1},j_{2},...,j_{n},...\\},~{}j_{1}<j_{2}<...<j_{n}<...,~{}\lambda_{j_{n}}<0,\left|\lambda_{j_{n+1}}\right|>\left|\lambda_{j_{n}}\right|,n\in\mathbb{N}.~{}$If:

  1. (1)

there exists $\rho>0$ such that$~{}\left|\lambda_{j_{1}}\right|>\rho,$
$\left|\lambda_{j_{m}}-\lambda_{j_{n}}\right|>\left|m-n\right|\rho,$

  2. (2)

$\sum_{n\in I_{Y}}\frac{1}{\left|\lambda_{n}\right|}<+\infty,$

then the family (-exp) is strongly minimal on $\left[0,t_{1}\right]$ for any

$t_{1}>\lim\limits_{j\rightarrow\infty,j\in
I_{Y}}\frac{2\ln\left|b_{j}\right|}{\lambda_{j}}.$

###### Proof.

The proof is derived from the fact that conditions specified in the theorem
are adequate to establish the validity of (bio_est).

######  Corollary 3.

If conditions of Theorem 3 hold and series (x_z(t1)) converges for any
$x_{Z}^{0}\in Z$, then system (CS1)–(P) is partially null controllable on
$[0,t_{1}]$ for any

$t_{1}>\lim\limits_{j\rightarrow\infty,j\in
I_{Y}}\frac{2\ln\left|b_{j}\right|}{\lambda_{j}}.$

###  4.1. Example 1

Consider the system of parabolic equations in the form (e1):

$\left\\{\begin{array}[]{c}y_{t}^{\prime}=\triangle y+y+z+xu\left(t\right),\\\
z_{t}^{\prime}=z_{xx}^{\prime\prime},\end{array}\right.\left(t,x\right)\in[0,+\infty)\times\left[0,\pi\right],$
(Ex1)

$\displaystyle y\left(0,x\right)$ $\displaystyle=$ $\displaystyle
y_{0}\left(x\right),z\left(0,x\right)=z_{0}\left(x\right),x\in
L_{2}\left[0,\pi\right],$ $\displaystyle y\left(t,0\right)$ $\displaystyle=$
$\displaystyle y\left(t,\pi\right)=0,$ $\displaystyle z\left(t,0\right)$
$\displaystyle=$ $\displaystyle z\left(t,\pi\right)=0,$

where $y_{0}(x),z_{0}(x)\in L_{2}\left[0,\pi\right],$
$b\left(x\right)=\left(\begin{array}[]{c}x\\\
0\end{array}\right),u\left(t\right)\in L_{2}\left[0,t_{1}\right].$

Let $P=\left(\begin{array}[]{cc}1&0\\\ 0&0\end{array}\right)$. In this case

$Y=\left\\{\left(\begin{array}[]{c}y\left(x\right)\\\
0\end{array}\right),y\left(\cdot\right)\in
L_{2}\left[0,\pi\right]\right\\},Z=\left\\{\left(\begin{array}[]{c}0\\\
z\left(x\right)\end{array}\right),z\left(\cdot\right)\in
L_{2}\left[0,\pi\right]\right\\},$

$A=\left(\begin{array}[]{cc}\triangle+1&1\\\
0&\vartriangle\end{array}\right),b=\left(\begin{array}[]{c}x\\\
0\end{array}\right),$ $A_{Y}=\vartriangle+1,A_{Z}=\vartriangle$

with domains [6, page 214, Example 4.6.6, page 241, Example 4.10.1]

$D\left(A_{Y}\right)=D\left(A_{Z}\right)=\left\\{f,f^{\prime}~{}\mathrm{a.c.:}f^{\prime},f^{\prime\prime}\in
L_{2}\left[0,\pi\right]\mathrm{~{}~{}and~{}}f\left(0\right)=f\left(\pi\right)=0\right\\}.$

After computations and enumeration we obtain:

1) $I_{Y}=\left\\{1,3,5,...\right\\},~{}I_{z}=\left\\{2,4,6,...\right\\}$

2) the eigenvalues $\lambda_{j}~{}$and eigenvectors
$\varphi_{j},j\in\mathbb{N}~{}$of the operator $A$ are:

$\lambda_{2m-1}=1-m^{2},\varphi_{2m-1}\left(x\right)=\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}\sin
mx\\\ 0\end{array}\right),j=2m-1\in I_{Y},m\in\mathbb{N},$ (phiEY)

$\lambda_{2m}=-m^{2},\varphi_{2m}\left(x\right)=\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}-\sin
mx\\\ ~{}~{}~{}\sin mx\end{array}\right),j=2m\in I_{Z},m\in\mathbb{N},$ (
phiEZ)

3) the eigenvalues $\lambda_{i}$ and eigenvectors $\psi_{j}$ of the operator
$A^{\ast}$ are:

$\lambda_{2m-1}=1-m^{2},\psi_{2m-1}\left(x\right)=\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}\sin
mx\\\ \sin mx\end{array}\right),j=2m-1\in I_{Y},~{}m\in\mathbb{N},$ (psiEY)
$\lambda_{2m}=-m^{2},\psi_{2m}\left(x\right)=\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}0\\\
\sin mx\end{array}\right),j=2m\in I_{Z},~{}m\in\mathbb{N}.$ (psiEZ)

Substituting (psiEY) and (psiEZ) to the formula
$b_{j}=(b,\psi_{j}),~{}j\in\mathbb{N}$, we obtain:

$b_{j}=\left\\{\begin{tabular}[]{ll}$\int_{0}^{\pi}x\sin
mxdx=$&$\frac{\left(-1\right)^{m-1}}{m}\pi,j=2m-1\in I_{Y},m\in\mathbb{N},$\\\
$~{}~{}0,~{}$&$~{}~{}j=2m\in I_{Z},m\in\mathbb{N}.$\end{tabular}\right.$ (bj)

Using the characterization of Riesz bases via biorthogonal sequences developed
in [19] (see Theorem 2.1), one can prove, that

######  Lemma 2.

The sequence

$\left\\{\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}\sin mx\\\
0\end{array}\right),\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}-\sin mx\\\
~{}~{}~{}\sin mx\end{array}\right),~{}m\in\mathbb{N}\right\\}$ (B1)

and sequence

$\left\\{\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}\sin mx\\\ ~{}~{}~{}\sin
mx\end{array}\right),\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}0\\\
~{}~{}~{}\sin mx\end{array}\right),~{}m\in\mathbb{N}\right\\}$ (B1*)

are a Riesz basic of $L_{2}\left(\left[0,\pi\right],\mathbb{R}^{2}\right).~{}$

###### Proof.

We will prove, that 1) sequence (B1) is complete, and 2) sequences (B1) and
(B1*) are Bessel sequences.

1) If for $f\left(x\right)=\left(\begin{array}[]{c}f_{Y}\left(x\right)\\\
f_{Z}\left(x\right)\end{array}\right),$ where
$~{}f_{Y}\left(x\right),f_{Z}\left(x\right)\in L_{2}\left[0,\pi\right],$
equalities $\left(\varphi_{2m-1},f\right)=0,~{}\left(\varphi_{2m},f\right)=0$
hold for any $m\in\mathbb{N}$, i.e.

$\displaystyle\left(\sin mx,f_{Y}\right)$ $\displaystyle=$ $\displaystyle 0,$
$\displaystyle-\left(\sin mx,f_{Y}\right)+\left(\sin mx,f_{Z}\right)$
$\displaystyle=$ $\displaystyle 0,$ $\displaystyle\forall m$
$\displaystyle\in$ $\displaystyle\mathbb{N},$

then from the completeness of the sequence $\\{\sin mx,m\in\mathbb{N}\\}$ in
$L_{2}\left[0,\pi\right],$ it follows, $f_{Y}=f_{Z}=0.$

Hence, sequence (B1) is complete.

2) For any $f\left(x\right)=\left(\begin{array}[]{c}f_{Y}\left(x\right)\\\
f_{Z}\left(x\right)\end{array}\right),$
where$~{}f_{Y}\left(x\right),f_{Z}\left(x\right)\in L_{2}\left[0,\pi\right],$
we have

$\sum_{m=1}^{\infty}\left|\left(\varphi_{2m-1},f\right)\right|^{2}+\sum_{m=1}^{\infty}\left|\left(\varphi_{2m},f\right)\right|^{2}\leq$

$\leq 3\sum_{m=1}^{\infty}\left|\left(\sqrt{\frac{2}{\pi}}\sin
mx,f_{Y}\right)\right|^{2}+2\sum_{m=1}^{\infty}\left|\left(\sqrt{\frac{2}{\pi}}\sin
mx,f_{Z},\right)\right|^{2}<+\infty.$

Hence, sequence (B1) is a Bessel sequence.

By the same way, for any
$f\left(x\right)=\left(\begin{array}[]{c}f_{Y}\left(x\right)\\\
f_{Z}\left(x\right)\end{array}\right),$
where$~{}f_{Y}\left(x\right),f_{Z}\left(x\right)\in L_{2}\left[0,\pi\right],$
we have

$\sum_{m=1}^{\infty}\left|\left(\psi_{2m-1},f\right)\right|^{2}+\sum_{m=1}^{\infty}\left|\left(\psi_{2m},f\right)\right|^{2}\leq$

$\leq 2\sum_{m=1}^{\infty}\left|\left(\sqrt{\frac{2}{\pi}}\sin
mx,f_{Y}\right)\right|^{2}+3\sum_{m=1}^{\infty}\left|\left(\sqrt{\frac{2}{\pi}}\sin
mx,f_{Z}\right)\right|^{2}<+\infty.$

Hence the sequence (B1*) is a Bessel sequence also.

This proves the lemma.

Lemma 2 shows that equation (Ex1) satisfies all the conditions of equation
(CS1).

For $j=2m-1,m\in\mathbb{N},$ we have
$e^{-2\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)=$

$=e^{-\lambda_{2m-1}t_{1}}\left(\sum_{n=1}^{\infty}\left(x_{Z\left(2n\right)},\psi_{Z\left(2n\right)}\right)e^{\lambda_{2n}t_{1}}\varphi_{Z\left(2n\right)},\psi_{Z\left(2m-1\right)}\right)=$

$=e^{\left(m^{2}-1\right)t_{1}}\left(\mathop{\displaystyle\int}_{0}^{\pi}\left(\sum_{n=1}^{\infty}\sqrt{\frac{2}{\pi}}\left(\mathop{\displaystyle\int}_{0}^{\pi}z^{0}\left(\xi\right)\sin
n\xi d\xi\right)e^{-n^{2}t_{1}}\sqrt{\frac{2}{\pi}}\sin
nx\right)\sqrt{\frac{2}{\pi}}\sin mx\right)dx=$

$=e^{\left(m^{2}-1\right)t_{1}}e^{-m^{2}t_{1}}\frac{2}{\pi}\left(\mathop{\displaystyle\int}_{0}^{\pi}z^{0}\left(x\right)\sin
mxdx\right)dx=\sqrt{\frac{2}{\pi}}e^{-t_{1}}\left(\mathop{\displaystyle\int}_{0}^{\pi}z^{0}\left(x\right)\sin
mdx\right),$

so

$\sum_{\begin{subarray}{c}j=1,\\\
j=1,3,5,...\end{subarray}}^{\infty}\left|e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)\right|^{2}=$

$=\frac{2}{\pi}e^{-2t_{1}}\sum_{m=1}^{\infty}\left|\left(\int_{0}^{\pi}z^{0}\left(x\right)\sin
mxdx\right)dx\right|^{2}<+\infty.$

Also we have

$e^{-\lambda_{j}t_{1}}b_{j}=e^{\left(m^{2}-1\right)t_{1}}\frac{\left(-1\right)^{m-1}}{m}\pi=\pi
e^{-1}\frac{\left(-1\right)^{m-1}}{m}e^{m^{2}t_{1}},j=2m-1\in
I_{Y},m\in\mathbb{N}.$

When the family

$\left\\{\frac{\left(-1\right)^{m-1}}{m}e^{m^{2}t_{1}},m=1,2,...,.\right\\}$

is strongly minimal?

We have
$\lim\limits_{m\rightarrow\infty}\ln\left(\frac{1}{m}\right)\frac{1}{-m^{2}}=\allowbreak\lim\limits_{m\rightarrow\infty}\frac{1}{\left(\ln
m\right)m^{2}}$ $=0.$

In accordance with Theorem 3 system (Ex1) is partially null-controllable on
$[0,t_{1}]$ for any $t_{1}>0.\ $

However there is no $t_{1}>0,$ such that system (Ex1) is completely null-
controllable on $[0,t_{1}]$ , because the second equation of this system
doesn’t depend on a control.

##  5\. The case $\sigma_{YZ}\neq\varnothing$

###  5.1. Eigenvectors of the operators $A$ and $A^{\ast}$

If $j\in I_{Y}\backslash I_{YZ}$ and $k\in I_{Z}\backslash I_{YZ},~{}$we are
in conditions of the previous section, so eigenvectors of the operator $A$ and
$A^{\ast}$ are obtained in the previous section andthey satisfy satisfy
orthogonality conditions (ortYZ).

If $m\in$ $I_{YZ},$ the eigenvalue $\lambda_{m}$ of the operator

$A=\left(\begin{array}[]{cc}A_{Y}&A_{YZ}\\\ 0&A_{Z}\end{array}\right)$

may have two linear independent eigenvectors.

The first one is $\phi_{1m}$ $=\left(\begin{array}[]{c}\phi_{1Ym}\\\
0\end{array}\right),~{}$ where $\phi_{1Ym}\neq 0\ $is an eigenvector of the
operator $A_{Y}.$

The second one is:
$\phi$${}_{2m}=\left(\begin{array}[]{c}\mathrm{\phi}_{2Ym}\\\
\phi_{2Zm}\end{array}\right),~{}$where $\phi_{2Zm}$ is an eigenvector of the
operator $A_{Z},~{}$ provided that there exists a solution of the operator
equation

$A_{Y}\mathrm{\phi}_{2Ym}+A_{YZ}\phi_{2Zm}=\lambda_{m}\mathrm{\phi}_{2Ym}$
(AYZ)

for $m\in$ $I_{YZ}.$ In this case the vector $\mathrm{\phi}_{2Ym}$ is a
solution of above equation.

Since $\lambda_{m}\ $is an eigenvalue of the operator $A_{Y},$ the operator
equation (AYZ) may be not solvable. In this case the operator $A$ has an
unique eigenvector $\phi_{1m}$ $=\left(\begin{array}[]{c}\phi_{1Ym}\\\
0\end{array}\right)~{}$

The same considerations can be performed for the operator

$A^{\ast}=\left(\begin{array}[]{cc}A_{Y}^{\ast},&0\\\
A_{YZ}^{\ast}&A_{Z}^{\ast}\end{array}\right),$

i.e .for $m\in$ $I_{YZ}$ the eigenvalue $\bar{\lambda}_{m}$ may have two
linear independent eigenvectors.

The first one is $\psi$${}_{2m}\mathrm{=}\left(\begin{array}[]{c}\mathrm{0}\\\
\psi_{2Zm}\end{array}\right),~{}$ where $\phi_{1Ym}\neq 0\ $is an eigenvector
of the operator $A_{Z}^{\ast}.$

The second one is:
$\psi$${}_{1m}\mathrm{=}\left(\begin{array}[]{c}\psi_{1Ym}\\\
\mathrm{\psi}_{1Zm}\end{array}\right),~{}$where $\psi_{1Ym}$ is an eigenvector
of the operator $A_{Y}^{\ast},~{}$ provided that there exists a solution the
operator equation

$A_{YZ}^{\ast}\psi_{1Ym}+A_{Z}^{\ast}\mathrm{\psi}_{1Zm}=\bar{\lambda}_{m}\mathrm{\psi}_{1Zm}.$
(AYZ*)

Since $\bar{\lambda}_{m}\ $is an eigenvalue of the operator $A_{Z}^{\ast},$
the operator equation (AYZ*) may be not solvable. In this case the operator
$A$ has an unique eigenvector $\psi_{2m}=\left(\begin{array}[]{c}0\\\
\psi_{2Zm}\end{array}\right).$

If equation (AYZ) does not admit a solution, then the vector $\phi_{1m}$ in n
unique eigenvector of the operator $A$ for the eigenvalue $\lambda_{m.}.$

In this case consider the equation

$\left(A_{Y}-\lambda_{m}I\right)\mathrm{\phi}_{2Ym}=c_{m}\phi_{1Ym}-A_{YZ}\phi_{2Zm}.$
(AYZG)

######  Theorem 4.

The operator equation

$A\mathrm{\phi}_{2m}-\lambda_{m}\mathrm{\phi}_{2m}=c_{m}\phi_{1m},$ (gev)

has a solution $\phi$${}_{2m}=\left(\begin{array}[]{c}\mathrm{\phi}_{2Ym}\\\
\phi_{2Zm}\end{array}\right)$ if and only if

$c_{m}=\left(A_{YZ}\phi_{2Zm},\psi_{1Ym}\right).$ (c)

###### Proof.

Equation (gev) is equivalent to equation (AYZG).

Equation (AYZG) has a solution if and only if [16]

$\left(c\phi_{1Ym}-A_{YZ}\phi_{2Zm},\psi_{1Ym}\right)=0.$

We have

$\left(c_{m}\phi_{1Ym}-A_{YZ}\phi_{2Zm},\psi_{1Ym}\right)=$ $\displaystyle=$
$\displaystyle
c_{m}\left(\phi_{1Ym},\psi_{1Ym}\right)-\left(A_{YZ}\phi_{2Zm},\psi_{1Ym}\right)=$
$\displaystyle=$ $\displaystyle
c_{m}-\left(A_{YZ}\phi_{2Zm},\psi_{1Ym}\right),$

i.e. the last condition is equivalent to the condition (c).

This proves the theorem.

If $c_{m}=0,$ then $\phi$${}_{2m}\ $is a solution of the equation

$A\mathrm{\phi}_{2m}-\lambda_{m}\mathrm{\phi}_{2m}=0,$

so $\phi$2m is the second eigenvector of the operator $A$ of the same simple
eigenvalue $\lambda_{m}$ having the multiplicity $2.$

If $c_{m}\neq 0$, then the $\phi$2m is a generalized eigenvector of the
operator $A$, corresponding to the eigenvalue $\lambda_{m}$, having the
multiplicity $2$.

If equation (AYZ*) does not admit a solution, then the vector $\psi$2m in an
unique eigenvector of the operator $A$ for the eigenvalue $\lambda_{m.}.$

In this case consider the equation

$\left(A_{Z}^{\ast}-\bar{\lambda}_{m}I\right)\mathrm{\psi}_{1Zm}=c_{m}\psi_{2Zm}-A_{YZ}^{\ast}\psi_{1Ym},$
(AYZG*)

######  Theorem 5.

The operator equation

$A^{\ast}\mathrm{\psi}_{1m}-\lambda_{m}\mathrm{\psi}_{1m}=c_{m}\psi_{2m},$
(gev*)

has a solution $\psi$${}_{1m}=\left(\begin{array}[]{c}\psi_{1Ym}\\\
\mathrm{\psi}_{1Zm}\end{array}\right),~{}$where $\mathrm{\phi}_{2Zm}\neq 0$ is
an eigenvector of the operator$A_{Z},\ $corresponding to its eigenvalue
$\lambda_{m},$ if and only if

$c_{m}=\left(\phi_{2Zm},A_{YZ}^{\ast}\psi_{1Ym}\right).$ (c*)

###### Proof.

The operator equation (gev*) is equivalent to operator equation (AYZG*).
Equation (AYZG*) has a solution if and only if [16]

$\left(\phi_{2Zm},c_{m}\psi_{2Zm}-A_{YZ}^{\ast}\psi_{1Ym}\right)=0.$

We have
$\left(\phi_{2Zm},c_{m}\psi_{2Zm}-A_{YZ}^{\ast}\psi_{1Ym}\right)=c_{m}\left(\phi_{2Zm},\psi_{2Zm}\right)-\left(\phi_{2Zm},A_{YZ}^{\ast}\psi_{1Ym}\right).$
i.e. the last condition is equivalent to the condition (c*).

This proves the theorem.

Conditions (c) and (c*) are equivalent.

If $c_{m}=0,$ then we obtain equation

$A^{\ast}\mathrm{\psi}_{1m}-\lambda_{m}\mathrm{\psi}_{1m}=0$ (ev*)

so the vector $\psi$${}_{1m}\mathrm{=}\left(\begin{array}[]{c}\psi_{1Ym}\\\
\mathrm{\psi}_{1Zm}\end{array}\right)$ is a second eigenvector of the operator
$A^{\ast},$ corresponding the same simple eigenvalue $\bar{\lambda}_{m.}$of
the multiplicity $2$.

If $c_{m}\neq 0,$ then $\psi$1m is a generalized eigenvector of the operator
$A^{\ast},$ corresponding the eigenvalue $\bar{\lambda}_{m.}$ of the
multiplicity $2$.

######  Lemma 3.

The sequence $\left\\{c_{m},m\in I_{YZ}\right\\}$ is bounded.

###### Proof.

We have

$\left|c_{m}\right|=\left|\left(A_{YZ}\phi_{2Zm},\psi_{1Ym}\right)\right|\leq\left\|A_{YZ}\right\|\left\|\phi_{2Zm}\right\|\left\|\psi_{1Ym}\right\|,m\in
I_{YZ}.$

The sequences $\left\\{\phi_{2Zm},m\in I_{YZ}\right\\}$ and
$\left\\{\psi_{1Ym},m\in I_{YZ}\right\\}$ are bounded, because in accordance
with Lemma 1 the first one is a Riesz basic of $Z,$ and the second one is a
Riesz basic of $Y.$

This proves the Lemma.

The eigenvectors $\phi_{1m},~{}\psi_{2m}~{}$ and the generalized eigenvectors
$\phi$${}_{2m},$ $\psi$1m of the operators $A$ and $A^{\ast}$correspondingly
are chosen such that the following orthogonality conditions hold:

$\left(\begin{array}[]{c}\begin{array}[]{cc}\left(\phi_{1l},\mathrm{\psi}_{1j}\right)=\delta_{lj}&\left(\phi_{1l},\psi_{2j}\right)=0,\end{array}\\\
\begin{array}[]{cc}\left(\mathrm{\phi}_{2l},\mathrm{\psi}_{1j}\right)=0&\left(\mathrm{\phi}_{2l},\psi_{2j}\right)=\delta_{lj},\end{array}\end{array}\right),l,j\in
I_{YZ}.$ (Y)

Hence,
$\left(\phi_{2l},\mathrm{\psi}_{1j}\right)=\left(\mathrm{\phi}_{2Yl},\psi_{1Yj}\right)+\left(\mathbf{\phi}_{2Zl},\mathrm{\psi}_{1Zj}\right)=0,~{}l,j\in
I_{YZ}\iff$

$\left(\mathrm{\phi}_{2Yl},\mathbf{\psi}_{1Yj}\right)=-\left(\phi_{2Zl},\mathrm{\psi}_{1Zj}\right),~{}l,j\in
I_{YZ}.$ (YZ)

Lemma 1 holds true for the case $\sigma_{YZ}\neq\varnothing$ as well.

##  6\.  Partial controllability conditions

Denote:

$b_{1j}=\left\\{\begin{tabular}[]{ll}$\left(b,\psi_{j}\right),$&$j\in
I_{Y}\backslash I_{YZ},$\\\ $\left(b,\mathrm{\psi}_{1j}\right),$&$j\in
I_{YZ},$\end{tabular}\right.,b_{2j}=\left(b,\mathrm{\psi}_{2j}\right),j\in
I_{YZ}.$

$b_{j}\left(\tau\right)=\left\\{\begin{array}[]{c}\begin{tabular}[]{ll}$b_{1j}-e^{-\lambda_{j}\left(t_{1}-\tau\right)}\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\psi_{Zj}\right),$&$j\in
I_{Y}\backslash I_{YZ},$\\\
$b_{1j}+b_{2j}c_{j}\left(t_{1}-\tau\right)-$&\end{tabular}\\\
-e^{-\lambda_{j}\left(t_{1}-\tau\right)}\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\mathrm{\psi}_{1Zj}\right).j\in
I_{YZ}.\end{array}\right.$

######  Theorem 6.

Let the series

$\sum\limits_{j\in
I_{Y}}e^{-2\lambda_{j}t_{1}}\left|\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)\right|^{2}$
(xg_z(t1))

converges for any $x_{Z}^{0}\in Z.$

Equation (CS1)–(P) is partially null controllable on $[0,t_{1}]$ if and only
if the family of exponent

$\left\\{e^{-\lambda_{j}t}b_{j}\left(t\right),j\in
I_{Y},t\in\left[0,t_{1}\right]\right\\}$ (-expg)

is strongly minimal.

###### Proof.

Since the family of eigenvectors
$\left\\{\varphi_{i},i\in\mathbb{N}\right\\}~{}$of the operator $A$ is a Riesz
basic of $X$ and the family of eigenvectors
$\left\\{\psi_{i},i\in\mathbb{N}\right\\}~{}$of the operator $A^{\ast}$ is a
Riesz basic of $X$ as well , for any
$x^{0}=\left(\begin{array}[]{c}x_{Y}^{0}\\\ x_{z}^{0}\end{array}\right)\in
X=Y\times Z$ we have

$x^{0}(t_{1}):=S\left(t_{1}\right)x^{0}=\sum\limits_{l\in I_{Y}\backslash
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\psi_{l}\right)\left(\begin{array}[]{c}\varphi_{Yl}\\\
0\end{array}\right)+$

$+\sum\limits_{l\in
I_{YZ}}e^{\lambda_{l}t_{1}}\left[\left(x^{0},\psi_{1l}\right)\left(\begin{array}[]{c}\phi_{1Yl}\\\
0\end{array}\right)+\left(x^{0},\psi_{2l}\right)\left(c_{j}t_{1}\left(\begin{array}[]{c}\phi_{1Yl}\\\
0\end{array}\right)+\left(\begin{array}[]{c}\phi_{2Yl}\\\
\phi_{2Zl}\end{array}\right)\right)\right]+$

$+\sum\limits_{k\in
I_{Z}|I_{YZ}}\left(x^{0},\psi_{k}\right)e^{\lambda_{k}t_{1}}\left(\begin{array}[]{c}\varphi_{Yk}\\\
\varphi_{Zk}\end{array}\right),$

$x_{Y}^{0}\left(t_{1}\right):=PS(t_{1})x^{0}=\sum\limits_{l\in I_{Y}\backslash
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\psi_{l}\right)\varphi_{Yl}+$

$+\sum\limits_{l\in
I_{YZ,}}e^{\lambda_{l}t_{1}}\left[\left(x^{0},\psi_{1l}\right)\phi_{1Yl}+\left(x^{0},\psi_{2l}\right)\left(c_{j}t_{1}\phi_{1Yl}+\phi_{2Yl}\right)\right]+$

$+\sum\limits_{k\in
I_{Z}|I_{YZ}}e^{\lambda_{k}t_{1}}\left(x^{0},\psi_{k}\right)\varphi_{Yk}.$

Using (ortY)–(ortYZ), (Y)–(YZ), we obtain the following:

If $j\in I_{Y}\backslash I_{YZ},$ then

$\left(x_{Y}^{0}\left(t_{1}\right),\psi_{Yj}\right)=\sum\limits_{l\in
I_{Y}\backslash
I_{YZ}}\left(x^{0},\psi_{j}\right)e^{\lambda_{lk}t_{1}}\left(\varphi_{Yl},\psi_{Yj}\right)+$

$+\sum\limits_{l\in
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\psi_{2l}\right)\left[c_{m}t_{1}\left(\phi_{1Yl},\psi_{Yj}\right)+\left(\phi_{2Yl},\psi_{Yj}\right)\right]+$

$+\sum\limits_{l\in
I_{YZ},}e^{\lambda_{l}t_{1}}\left(x^{0},\psi_{1l}\right)\left(\phi_{1Yl},\psi_{Yj}\right)+$

$+\sum\limits_{k\in
I_{Z}|I_{YZ}}\left(x^{0},\psi_{k}\right)e^{\lambda_{k}t_{1}}\left(\varphi_{Yk},\psi_{Yj}\right)=$

$=e^{\lambda_{j}t_{1}}\left(x^{0},\psi_{Yj}\right)+$

$+\sum\limits_{l\in
I_{YZ}}\left(x_{Z}^{0},\psi_{2Zl}\right)e^{\lambda_{l}t_{1}}\left(\phi_{2Yl},\psi_{Yj}\right)+$

$+\sum\limits_{k\in
I_{Z}|I_{YZ}}\left(x^{0},\psi_{Zk}\right)e^{\lambda_{k}t_{1}}\left(\varphi_{Yk},\psi_{Yj}\right)=$

$=e^{\lambda_{j}t_{1}}\left(x^{0},\psi_{j}\right)-$

$-\sum\limits_{I_{Z}|I_{YZ}}\left(\ x^{0},\psi_{k}\right)\
e^{\lambda_{k}t_{1}}(\varphi_{Zk},\psi_{Zj})-\sum\limits_{k\in I_{Z}}\left(\
x^{0},\psi_{k}\right)e^{\lambda_{k}t_{1}}\left(\varphi_{2Zk},\psi_{Zj}\right),$
and since the sequence $\left\\{\varphi_{2Zk,}k\in I_{Z}\right\\}$ is a Riesz
basic of $Z,$ we obtain, that

$\sum\limits_{k\in I_{Z}}\left(\ x_{Z}^{0},\psi_{Zk}\right)\
e^{\lambda_{k}t_{1}}\varphi_{Zk}=S_{Z}\left(t_{1}\right)x_{Z}^{0},$ so

$\left(x_{Y}\left(t_{1}\right),\psi_{Yj}\right)=e^{\lambda_{j}t_{1}}\left(x^{0},\psi_{j}\right)-\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right).$

If $j\in I_{YZ},$ then

$\left(PS\left(t_{1}\right)x^{0},\mathrm{\psi}_{1Yj}\right)=$

$=\sum\limits_{l\in
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\mathrm{\psi}_{1l}\right)\left(\mathrm{\phi}_{1Yl},\mathrm{\psi}_{1Yj}\right)+$

$+\sum\limits_{l\in
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\mathrm{\psi}_{2l}\right)\left(c_{j}t_{1}\left(\mathrm{\phi}_{1Yl},\mathrm{\psi}_{1Yj}\right)+\left(\mathrm{\phi}_{2Yl},\mathrm{\psi}_{1Yj}\right)\right)+$

$+\sum\limits_{k\in
I_{Z}|I_{YZ}}e^{\lambda_{k}t_{1}}\left(x^{0},\psi_{k}\right)\left(\varphi_{Yk},\mathrm{\psi}_{1Yj}\right)=$

$=e^{\lambda_{j}t_{1}}\left[\left(x^{0},\mathrm{\psi}_{1j}\right)+c_{j}t_{1}\left(x^{0},\mathrm{\psi}_{2l}\right)\right]+\sum\limits_{l\in
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\mathrm{\psi}_{2l}\right)\left(\mathrm{\phi}_{2Yl},\mathrm{\psi}_{1Yj}\right)$

$+\sum\limits_{k\in
I_{Z}|I_{YZ}}e^{\lambda_{k}t_{1}}\left(x^{0},\psi_{k}\right)\left(\varphi_{Yk},\mathrm{\psi}_{1Yj}\right)=e^{\lambda_{j}t_{1}}\left[\left(x^{0},\mathrm{\psi}_{1j}\right)+c_{j}t_{1}\left(x^{0},\mathrm{\psi}_{2l}\right)\right]-$

$-\sum\limits_{l\in
I_{YZ}}e^{\lambda_{l}t_{1}}\left(x^{0},\mathrm{\psi}_{2l}\right)\left(\mathrm{\phi}_{2Zl},\mathrm{\psi}_{1Zj}\right)-\sum\limits_{k\in
I_{Z}|I_{YZ}}e^{\lambda_{k}t_{1}}\left(x^{0},\psi_{k}\right)\left(\varphi_{Zk},\mathrm{\psi}_{1Zj}\right),$
and again, since the sequence $\left\\{\varphi_{2Zk,}k\in I_{Z}\right\\}$ is a
Riesz basic of $Z,$ we obtain, that

$\left(x_{Y}\left(t_{1}\right),\psi_{Yj}\right)=e^{\lambda
jt_{1}}\left[\left(x^{0},\mathrm{\psi}_{1j}\right)+\left(x^{0},\mathrm{\psi}_{2j}\right)c_{j}t_{1}\right]-\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{1Zj}\right).$

Using $b,t_{1}-\tau$ instead of $x^{0},t_{1}$ in above computations, we obtain
:

if $j\in I_{Y}\backslash I_{YZ},$ then

$\left(PS\left(t_{1}-\tau\right)b,\psi_{j}\right)=e^{\lambda_{j}\left(t_{1}-\tau\right)}b_{1j}-\left(S_{Z}\left(t_{1}\right)b_{Z},\psi_{Zj}\right),$

if$~{}j\in I_{YZ},$ then

$\left(PS\left(t_{1}-\tau\right)b,\mathrm{\psi}_{1j}\right)=$

$=\left[b_{1j}+b_{2j}c_{j}\left(t_{1}-\tau\right)\right]e^{\lambda_{j}\left(t_{1}-\tau\right)}-\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\mathrm{\psi}_{1Zj}\right).$

After integration in $\left[0,t_{1}\right]$ we obtain:

if $j\in I_{Y}\backslash I_{YZ},$ then

$\left(x_{Y}(t_{1}),\psi_{j}\right)=e^{\lambda_{j}t_{1}}\left(x^{0},\psi_{j}\right)-\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)+$

$+\int_{0}^{t_{1}}\left[b_{1j}e^{\lambda_{j}\left(t_{1}-\tau\right)}-\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\psi_{Zj}\right)\right]u\left(\tau\right)d\tau;$

if $j\in I_{YZ},$ then

$\left(x_{Y}(t_{1}),\psi_{j}\right)=\left[\left(x^{0},\mathrm{\psi}_{1j}\right)+\left(x^{0},\mathrm{\psi}_{2j}\right)c_{j}t_{1}\right]e^{\lambda_{j}t_{1}}-\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{2Zj}\right)+$

$+\int_{0}^{t_{1}}\left(\left[b_{1j}+b_{2j}c_{j}\left(t_{1}-\tau\right)\right]e^{\lambda_{j}\left(t_{1}-\tau\right)}-\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\mathrm{\psi}_{2Zj}\right)\right)u\left(\tau\right)d\tau.$

Hence, equation (CS1)–(P) is partial controllable on $[0,t_{1}]$ if and only
if the moment problem

$-\left(x^{0},\psi_{j}\right)+e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)=$
(LPRY)
$=\int_{t_{0}}^{t_{1}}\left[b_{1j}e^{-\lambda_{j}\tau}-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\psi_{Zj}\right)\right]u\left(\tau\right)d\tau,\\\
j\in I_{Y}\backslash I_{YZ},$

$-\left(x^{0},\mathrm{\psi}_{1j}\right)-\left(x^{0},\mathrm{\psi}_{2j}\right)c_{j}t_{1}+e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{2Zj}\right)=$

$=\int_{0}^{t_{1}}\left[b_{1j}+b_{2j}c_{j}\left(t_{1}-\tau\right)\right]e^{-\lambda_{j}\tau}u\left(\tau\right)d\tau-$
(LPRYZ)
$-\int_{0}^{t_{1}}e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}-\tau\right)b_{Z},\mathrm{\psi}_{1Zj}\right)u\left(\tau\right)d\tau,\\\
j\in I_{YZ}.$

has a solution $u(\cdot)\in L_{2}[0,t_{1}]$.

Since the series (xg_z(t1)) converges for any $x_{z}^{0}\in Z,$ the sequence

$\left\\{\left(x^{0},\psi_{j}\right)-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right),j\in
I_{Y}\backslash I_{YZ}\right\\}\in l^{2},$

and, in accordance with Lemma 3, the sequence $\left\\{c_{j,}j\in
I_{YZ}\right\\}$ is bounded, so

$\left\\{\left(x^{0},\mathrm{\psi}_{1j}\right)+\left(x^{0},\mathrm{\psi}_{2j}\right)c_{j}t_{1}-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{2Zm}\right),j\in
I_{YZ}\right\\}\in l^{2}$ as well.

Hence [9], [22] above problem moment has a solution if and only if family of
exponential (-expg) is strongly minimal.

This proves the theorem.

If $b\in Y,~{}$i.e. $b=\left(\begin{array}[]{c}b_{Y}\\\ 0\end{array}\right),$
then moment problem (LPRY)–(LPRYZ) may be simplified:

$\left(x^{0},\psi_{j}\right)-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right)=$
(LPY) $=\int_{0}^{t_{1}}b_{1j}e^{-\lambda_{j}\tau}u\left(\tau\right)d\tau,j\in
I_{Y}\backslash I_{YZ}~{},$
$\left(x^{0},\mathrm{\psi}_{1j}\right)+\left(x^{0},\mathrm{\psi}_{2j}\right)c_{j}t_{1}-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{1Zj}\right)=$
(LPYZ)
$=\int_{0}^{t_{1}}b_{1j}e^{-\lambda_{j}\tau}u\left(\tau\right)d\tau,j\in
I_{YZ},$

or

$d_{j}=\int_{0}^{t_{1}}b_{1j}e^{-\lambda_{j}\tau}u\left(\tau\right)d\tau,$
(LPYJ)

where

$d_{j}=\left\\{\begin{tabular}[]{l}$\left(x^{0},\mathrm{\psi}_{1j}\right)-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zj}\right),~{}j\in
I_{Y}\backslash I_{YZ}$\\\
$\left(x^{0},\mathrm{\psi}_{1j}\right)+\left(x^{0},\mathrm{\psi}_{2j}\right)c_{j}t_{1}-e^{-\lambda_{j}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{1Zj}\right),~{}j\in
I_{YZ}.$\end{tabular}\right.$

Since series (xg_z(t1)) converges for any $x_{Z}^{0}\in Z,$ we have
$\left\\{d_{j}\in I_{Y}\right\\}\in l^{2}.$

Hence the following corollaries are valid.

######  Corollary 4.

If series (xg_z(t1)) converges for any $x_{Z}^{0}\in Z,$ then system
(CS1)–(P)) is partially null controllable on $[0,t_{1}]$ if and only if the
family of exponent

$\left\\{e^{-\lambda_{j}t}b_{1j},j\in I_{Y},t\in\left[0,t_{1}\right]\right\\}$
(-expg0)

is strongly minimal.

######  Corollary 5.

If the series (xg_z(t1) converges for any $x_{Z}^{0}\in Z,$ then system
(CS1)–(P) is partially null controllable on $[0,t_{1}]$ if and only if
equation

$\dot{y}\left(t\right)=A_{Y}y\left(t\right)+b_{Y}u\left(t\right)$

is null (completely) controllable on $[0,t_{1}].$

Theorems 2–3 and Corollaries 1–3 hold true for $\sigma_{YZ}\neq\varnothing$ as
well.

###  6.1.  Example 2

The authors of [1, system (12), page 4] have proved the partial
controllability conditions for the two-dimensional system

$\left\\{\begin{array}[]{c}y_{t}^{\prime}=\triangle y+\alpha z+1_{\omega}u,\\\
z_{t}^{\prime}=\triangle z\end{array}\right.~{}\left(t,x\right)\in
Q_{T}=\left[0,T\right]\times Q,$ (AMKho)
$\begin{array}[]{c}y\left(0,x\right)=y_{0}\left(x\right),z\left(0,x\right)=z_{0}\left(x\right),x\in
Q,\\\
y\left(t,x\right)=z\left(t,x\right)=0,~{}t\in\left[0,T\right],x\in\partial
Q,\end{array}$

where $Q$ is a bounded domain in $\mathbb{R}$ with a $C^{2}$-boundary
$\partial Q,~{}\omega$ is a non-empty open subset of $Q$ and $T>0$,

$y_{0}\left(\cdot\right),z_{0}\left(\cdot\right)\in
L_{2}(Q),u\left(\cdot,\cdot\right)\in
L_{2}\left(Q_{T}\right),\alpha\left(\cdot,\cdot\right)\in
L_{\infty}\left(Q_{T},\mathcal{L}(\mathbb{R}\right)).$

As it is noted in [1], controllability of such partial parabolic control
systems are investigated in many fields such as chemistry, physics or biology.

However, even if $u(t,x)\in\mathbb{R}$ for all $t\in[0,T]$ and $x\in Q$, the
control $u(\cdot,\cdot)\in L(Q_{T})$ cannot be considered as a single control
force666it is infinite-dimensional control. The given paper focuses on
achieving controllability using a single control force, so we will define
$u(t,x)$ in the form $\cdot)\in L(Q)$ is predetermined and $u(t),u(\cdot)\in
L([0,T],\mathbb{R})$ represents a single control force.

The system being studied in this section is

$\left\\{\begin{array}[]{c}y_{t}^{\prime}=\triangle y+\alpha z+\hat{b}u,\\\
z_{t}^{\prime}=\triangle z,\end{array}\right.~{}$ (exmp3) $\left(t,x\right)\in
Q_{T}=\left[0,T\right]\times\left[0,\pi\right],~{}\alpha\in\mathbb{R}\backslash\\{0\\},\hat{b}\left(\cdot\right)\in
L_{2}\left(\left[0,\pi\right],\mathbb{R}\right),u\left(\cdot\right)\in
L_{2}\left[0,T\right],$
$y\left(0,x\right)=y_{0}\left(x\right),z\left(0,x\right)=z_{0}\left(x\right),x\in[0,\pi],$
(inc3) $y\left(t,0\right)=z\left(t,0\right)=0,~{}t\in\left[0,T\right],$ (buc3)
${y_{0}}\left(\cdot\right),{z_{0}}\left(\cdot\right)\in
L_{2}\left(\left[0,\pi\right]\right).$

One can write system (exmp3)–(buc3) in form (CS1)–(P), where

$A=\left(\begin{array}[]{cc}\triangle&\alpha\\\
0&\triangle\end{array}\right),b=\left(\begin{array}[]{c}\hat{b}\\\
0\end{array}\right),~{}P=\left(\begin{array}[]{cc}1&0\\\
0&0\end{array}\right).$

Here

$Y=\left\\{\left(\begin{array}[]{c}y\left(x\right)\\\
0\end{array}\right),y\left(\cdot\right)\in
L_{2}\left[0,\pi\right]\right\\},Z=\left\\{\left(\begin{array}[]{c}0\\\
z\left(x\right)\end{array}\right),z\left(\cdot\right)\in
L_{2}\left[0,\pi\right]\right\\},$

$A_{Y}=A_{Z}=\triangle$

with domains [6, page 214, Example 4.6.6, page 241, Example 4.10.1]

$D\left(A_{Y}\right)=D\left(A_{Z}\right)=\left\\{f,f^{\prime}~{}\mathrm{a.c.:}f^{\prime},f^{\prime\prime}\in
L_{2}\left[0,\pi\right]\mathrm{~{}~{}and~{}}f\left(0\right)=f\left(\pi\right)=0\right\\}$
,
$b_{Y}=\hat{b},b_{Z}=0.$

Let $\lambda_{i}~{}$and $\varphi_{i},i\in\mathbb{N}~{}$ be eigenvalues and
eigenvectors of the operator $\Delta$ with boundary conditions (inc3).
Obviously, eigenvalues $\lambda_{n}=-n^{2},n=1,2,...,$ of $\triangle$ are
eigenvalues of $A$, and to every eigenvalues $\lambda_{n}$ correspond the
eigenvectors $\phi_{1n}=\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}\sin nx\\\
0\end{array}\right),n=1,2,...$ of the operator $A$, and the eigenvectors
$\psi_{2n}=\sqrt{\frac{2}{\pi}}\left(\begin{array}[]{c}0\\\ \sin
nx\end{array}\right)$ of the operator $A^{\ast}.$ In this example both
operator equations (AYZG) and (AYZG*) are reduced to the equation

$\displaystyle y^{\prime\prime}+\alpha\sqrt{\frac{2}{\pi}}\sin nx$
$\displaystyle=$ $\displaystyle-n^{2}y+c_{n}\sqrt{\frac{2}{\pi}}\sin nx,$
$\displaystyle y\left(0\right)$ $\displaystyle=$ $\displaystyle
y\left(\pi\right)=0.$

where according to (c) we obtain

$c_{n}=\alpha\int_{0}^{\pi}\frac{2}{\pi}\sin^{2}nxdx=\alpha\neq
0,n\in\mathbb{N}.$

Hence equations $y^{\prime\prime}+\alpha\sqrt{\frac{2}{\pi}}\sin
nx=-n^{2}y+c_{n}\sqrt{\frac{2}{\pi}}\sin nx,~{}n\in\mathbb{N},$ are equivalent
to equations

$y^{\prime\prime}=-n^{2}y~{},n\in\mathbb{N},y\left(0\right)=\left(\pi\right)=0,$
(geAA*)

so the vector

$\phi_{2n}=\left(\begin{array}[]{c}\phi_{2Yn}\left(x\right)\\\
\sqrt{\frac{2}{\pi}}\sin nx\end{array}\right),x\in\left[0,\pi\right]$

is a generalized eigenvector of $A,$ and the vector

$\psi_{1n}=\left(\begin{array}[]{c}\sqrt{\frac{2}{\pi}}\sin nx\\\
\psi_{1Zn}\left(x\right)\end{array}\right),x\in\left[0,\pi\right]$

is a generalized eigenvector of the operator $A^{\ast},$ where
$\phi_{2Yn}\left(x\right)$ and $\psi_{1Zn}\left(x\right)$ are solutions of
equation (geAA*). For any $n\in\mathbb{N}$ equation (geAA*) has the infinite
number of solutions. Let’s choose the solutions
$\phi_{2Yn}=-\sqrt{\frac{2}{\pi}}\sin nx~{}$and
$\psi_{1Zn}\left(x\right)=\sqrt{\frac{2}{\pi}}\sin nx,$ so the vector

$\phi_{2n}=\left(\begin{array}[]{c}-\sqrt{\frac{2}{\pi}}\sin nx\\\
\sqrt{\frac{2}{\pi}}\sin nx\end{array}\right),x\in\left[0,\pi\right]$

is a generalized eigenvector of the operator $A,$ and the vector

$\psi_{1n}=\left(\begin{array}[]{c}\sqrt{\frac{2}{\pi}}\sin nx\\\
\sqrt{\frac{2}{\pi}}\sin nx\end{array}\right),x\in\left[0,\pi\right]$

is a generalized eigenvector of the operator $A^{\ast}$, corresponding to the
eigenvalue $\lambda_{n}=-n^{2},n\in\mathbb{N}.$

One can prove ( see [2] and Example 1), that the sequences

$\left\\{\phi_{1n},\phi_{2n},n\in\mathbb{N}\right\\},$ (B3)

and

$\left\\{\psi_{1n},\psi_{2n},n\in\mathbb{N}\right\\},$ (B3*)

are Riesz basic in $X=L_{2}\left(\left[0,\pi\right],\mathbb{R}^{2}\right).$
Hence system (exmp3)–(buc3) satisfies to conditions of system (CS1)–(P)777See
the bottom of Pages 2 and 3., so all theorems an corollaries proven in
previous sections, are applicable to Example 2.

Denote
$b_{n}=\left(\hat{b},\sqrt{\frac{2}{\pi}}\mathrm{\sin}nx\right)=\sqrt{\frac{2}{\pi}}\mathop{\displaystyle\int}_{0}^{\pi}\hat{b}\left(x\right)\sin
nxdx,n\in\mathbb{N}.$

From (LPY)–(LPYZ) it follows, that888Here $I_{Y}\backslash I_{YZ}$ is an empty
set.

$\left(x^{0},\mathrm{\psi}_{1n}\right)+\left(x^{0},\mathrm{\psi}_{2n}\right)\alpha
t_{1}-e^{n^{2}t_{1}}\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{1Zn}\right)=\int_{0}^{t_{1}}b_{n}e^{n^{2}\tau}u\left(\tau\right)d\tau,n\in\mathbb{N},$
(LMPYZ)

where $\mathrm{\psi}_{1Zn}=\sqrt{\frac{2}{\pi}}\sin nx,n\in\mathbb{N}$.

Now we will check the convergence of series (xg_z(t1)).

We have

$S_{Z}\left(t_{1}\right)x_{Z}^{0}=\sum_{m=1}^{\infty}x^{0}_{Zm}e^{-m^{2}t_{1}}\sqrt{\frac{2}{\pi}}\sin
mxdx,0\leq x\leq\pi,$ where

$x_{Zm}^{0}=\int_{0}^{\pi}x_{Z}^{0}\left(\theta\right)\sqrt{\frac{2}{\pi}}\sin
m\theta d\theta$,

$(S_{Z}\left(t_{1}\right)x_{Z}^{0},\mathrm{\psi}_{1Zn})=\sum_{m=1}^{\infty}e^{-m^{2}t_{1}}x_{Zm}^{0}\mathop{\displaystyle\int}_{0}^{\pi}\left(\sqrt{\frac{2}{\pi}}\sin
mx\sqrt{\frac{2}{\pi}}\sin nx\right)dx=x_{Zn}^{0}e^{-n^{2}t_{1}},$

$\sum_{n=1}^{\infty}e^{2n^{2}t_{1}}\left|\left(S_{Z}\left(t_{1}\right)x_{Z}^{0},\psi_{Zn}\right)\right|^{2}=\sum_{n=1}^{\infty}e^{2n^{2}t_{1}}e^{-2n^{2}t_{1}}\left|x_{Zn}^{0}\right|^{2}=\left\|x_{Z}^{0}\right\|^{2}<+\infty.\forall
t_{1}>0.$

Hence from Corollary 4 it follows, that system (exmp3)–(buc3) is partially
null controllable on $\left[0,t_{1}\right]$ if and only if the family

$\left\\{e^{n^{2}t}b_{n},n\in\mathbb{N},t\in\left[0,t_{1}\right]\right\\}$
(-expn2)

is strongly minimal.

The eigenvalues $\lambda_{n}=-n^{2},n\in\mathbb{N}$ satisfy he sufficient
conditions for strong minimality of family (-expn2) , obtained in Theorem 3
with $\rho=1$. In accordance with Theorem 3 and Corollary 3 system
(exmp3)–(buc3) is partially null controllable on $\left[0,t_{1}\right]$ for
any $t_{1}>\lim_{n\rightarrow\infty}\frac{2\ln\left|b_{n}\right|}{-n^{2}}$.

##  7\. Conclusion

The partial exact null-controllability conditions for abstract coupled control
equations (e1)–(e3) governed by a single distributed control $u(t)$ have been
obtained using the moment problem approach. These conditions strengthen the
partial controllability conditions for one of coupled parabolic systems
obtained in Ammar Khodja et al. (2016). This is because, as demonstrated in
Example 2, even when $u(t,x)\in\mathbb{R}$ for all $t\in[0,T]$ and $x\in Q$,
the control $u(\cdot,\cdot)\in L(Q)$ cannot be treated as a single control
force in the conventional sense999it is infinite-dimensional control.

The controllability results to date have focused on control problems for
coupled PDEs, mostly of the same structure. The majority of results have been
obtained for coupled parabolic equations, and a few have been done for the
controllability of coupled wave equations. The abstract approach developed
here allows us to investigate various controllability problems, including
partial controllability, for coupled systems containing equations of different
structures.

All examples are simple; they have been chosen only for illustrative purposes.

## References

  * [1] F. Ammar-Khodja, Controllability of parabolic systems: the moment method, Evolution Equations,Long Time Behavior and Control, Cambridge, London, Mathematical Society, 2018, 1–30.
  * [2] F. Ammar-Khodja, F. Chouly and M. Duprez, Partial null controllability of parabolic linear systems, J. Control and Related Fields, AIMS, 6 (2) (2017), 185–216.
  * [3] F. Ammar-Khodja, A. Benabdallah, M. Gonzalez-Burgos and L. de Teresa, New phenomena for the null controllability of parabolic systems: Minimal time and geometrical dependence, J. Math. Anal. and Appl., HAL Id: hal-01165713, https: //hal. archives-ouvertes.fr/ hal-01165713.
  * [4] S. Avdonin and S. Ivanov, Families of exponentials: the methods of moments in controllability problems for distributed parameter systems, Cambridge University Press, 1995.
  * [5] S. Avdonin, A. Rivero and L. Teresa, Exact boundary controllability of coupled hyperbolic systems, Int. J. Appl. Math. Comput. Sci., 80 (2007), 1–7.
  * [6] A. Balakrishnan, Applied functional analysis, Springer-Verlag, New York Heilderberg Berlin, 1976.
  * [7] N. Bari, Biorthogonal systems and bases in Hilbert space, Uch. zap. MGU, 4, 148 (1951), 69–106.
  * [8] A. Bashirov, N. Mahmudov, N. Semi and H. Etikan, Partial controllability concept, Int. J. on Control, 80 (2007), 1–7.
  * [9] R. Boaz, A general moment problem, Amer. J. Math., 63 (1941), 361–370.
  * [10] P. Casazza, 0. Christensen, S. Li and A. Lindner, Riesz-Fischer Sequences and Lower Frame Bounds, Z. Anal. Anwendungen, 21 (2002), 305–314.
  * [11] H. Fattorini and D. Russel, Uniform bounds on biorthogonal functions for real exponentials with an application to the control theory of parabolic equations, Quart. Appl. Math., 32 (1974), 45–69.
  * [12] R. Gabasov and F. Kirillova, Qualitative theory of optimal processes, Nauka, Moscow, 1971.
  * [13] I. Gohberg and M. Krein, Introduction to the theory of linear nonselfadjoint operators, 18 , American Mathematical Society, Providence, R.I., 1969.
  * [14] E. Hille and R. Phillips, Functional analysis and semigroups, Providence, 1977.
  * [15] S. Krein, Linear Differential Equations in a Banach Spaces, Trans of Mathematical Monographies, 29, American Mathematical Society, Providence, R.I., 1971.
  * [16] S. Krein, Linear Equations in a Banach Spaces, Nauka, Moscow, 1971 (in Russian).
  * [17] A. Leontiev, Serieses of exponentials, Nauka, Moscow, 1976\.
  * [18] S. Mandelbrojt, Series de Dirichlet, Principes et Methodes, Cauthier-Villars, Paris, 1969.
  * [19] D. Stoeva, On a Characterization of Riesz Bases via Biorthogonal Sequences, Journal of Fourier Analysis and Applications, 26:67 (2020), 1–5.
  * [20] A. Taldykin, Vector functions and equations (with applications to the control theory). Leningrad University Publisher, 1977 (in Russian).
  * [21] R. Young, An introduction to non-harmonic Fourier serieses, Academic Press, 1980.
  * [22] R. Young, On a class of Riesz-Fisher sequences, Proceedings of AMS, 126 (1998), 1139—1142.