# A mutual control problem for semilinear systems via fixed point approach

Radu Precup  R. Precup, Faculty of Mathematics and Computer Science and
Institute of Advanced Studies in Science and Technology, Babeş-Bolyai
University, 400084 Cluj-Napoca, Romania & Tiberiu Popoviciu Institute of
Numerical Analysis, Romanian Academy, P.O. Box 68-1, 400110 Cluj-Napoca,
Romania<EMAIL_ADDRESS>and  Andrei Stan  A. Stan, Department of
Mathematics, Babeş-Bolyai University, 400084 Cluj-Napoca, Romania & Tiberiu
Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O. Box 68-1,
400110 Cluj-Napoca, Romania<EMAIL_ADDRESS>
###### Abstract.

In this paper, we introduce and discuss the concept of a mutual control
problem. Our analysis relies on a vector fixed-point approach based on the
fixed-point theorems of Perov, Schauder, Leray-Schauder, and Avramescu.
Additionally, for a related semi-observability problem, we employ a novel
technique utilizing Bielecki equivalent norms.

Key words and phrases: control; fixed point; nonlinear operator; differential
system

Mathematics Subject Classification (2010): 34B15, 34K35, 47H10

##  1\. Introduction and preliminaries

The study of systems of abstract or concrete equations has been the subject of
research for a long time, especially from the perspective of the existence and
uniqueness of solutions. In the present paper, our objective goes beyond
merely establishing the existence of solutions; we strive to identify
solutions whose components maintain a level of control over each other.

Let $D_{1},D_{2},C$ be sets with $C\subset D_{1}\times D_{2}$ and let
$E:D_{1}\times D_{2}\rightarrow Z$ be a mapping, where $Z$ is a linear space.
Consider the equation $E(x,\lambda)=0_{Z}$ and the problem of finding
$\lambda\in D_{2}$ such that the equation has a solution $x\in D_{1}$ with
$(x,\lambda)\in C$. We say that $x$ is the state variable, $\lambda$ is the
control variable, and $C$ is the controllability domain. One way to solve the
problem (see [12]) is to use the controllability condition to express
$\lambda$ as a function of $x$, $\lambda=N(x)$, and then find a solution to
the equation $E(x,N(x))=0_{Z}$. Alternatively, we can express $x$ as
$x=N(\lambda)$ and then solve the equation $E(N(\lambda),\lambda)=0_{Z}$. In
the first case, we solve the problem

$\left\\{\begin{array}[]{l}\lambda=N\left(x\right)\\\\[5.0pt]
E\left(x,N\left(x\right)\right)=0_{Z},\end{array}\right.$

while in the second case, the problem

$\left\\{\begin{array}[]{l}x=N\left(\lambda\right)\\\\[5.0pt]
E\left(N\left(\lambda\right),\lambda\right)=0_{Z}.\end{array}\right.$

For a system of two equations, we view each variable as a control over the
other, and we aim to find solutions that achieve equilibrium under certain
controllability conditions. We call this a mutual control problem.
Specifically, we consider four sets $D_{1},D_{2},C_{1},C_{2}$ with
$C_{1}\subset D_{1}\times D_{2}$ and $C_{2}\subset D_{2}\times D_{1}$,
$E_{1},E_{2}:D_{1}\times D_{2}\rightarrow Z$ two mappings and the problem

$\begin{cases}E_{1}(x,y)=0_{Z}\\\\[5.0pt] E_{2}(x,y)=0_{Z}\\\\[5.0pt] (x,y)\in
C_{1},\text{ }(y,x)\in C_{2}.\end{cases}$

The interpretation of this problem is as follows: the first equation describes
the state $x$ controlled by $y$, and the second equation describes the state
$y$ controlled by $x$. The restriction of $(x,y)$ to $C_{1}$ and $(y,x)$ to
$C_{2}$ represents the controllability conditions. One way to solve such a
problem, is to incorporate the controllability conditions into the equations
and give the problem a fixed point formulation

$\left(x,y\right)\in\left(N_{1}(x,y),\ N_{2}(x,y)\right),$

where $N_{1},N_{2}$ are set-valued mappings $N_{1}:D_{1}\times
D_{2}\rightarrow D_{1},$ $N_{2}:D_{1}\times D_{2}\rightarrow D_{2}.$ A
solution $\left(x,y\right)$ of this fixed point equation is said to be a
solution of the mutual control problem, while the problem is said to be
mutually controllable if such a solution exists.

In some cases, we can do even more, namely to express $x$ and $y$ as state
variables in terms of the controls $y$ and $x$, respectively. In such
situations, one has $x\in N_{1}(y)\ $ and $y\in N_{2}(x)$, where

$\displaystyle N_{1}:D_{2}\rightarrow D_{1},\ N_{1}(y):=\left\\{x:\text{
}E_{1}(x,y)=0_{{}_{Z}}\text{ and }(x,y)\in C_{1}\right\\},$ $\displaystyle
N_{2}:D_{1}\rightarrow D_{2},\ N_{2}(x):=\left\\{y:\text{
}E_{2}(x,y)=0_{{}_{Z}}\text{ and }(y,x)\in C_{2}\right\\}.$

Note that $N_{1}$ indicates how the state $x$ is controlled by $y$, and
$N_{2}$ shows how the state $y$ is controlled by $x$. The equilibrium is
achieved if there exists $(x,y)\in D_{1}\times D_{2}$ such that

$(x,y)\in\left(N_{1}(y),N_{2}(x)\right).$

It is interesting to note the similarity between the mutual control problem
and the Nash equilibrium problem. In case of the last one, we have two
functionals $J_{1}\left(x,y\right)$ and$\ J_{2}\left(x,y\right).$ Minimizing
$J_{1}\left(.,y\right)$ on $D_{1}$ for each $y\in D_{2}$ yields to the set of
minimum points denoted $s_{1}\left(y\right),$ while minimizing
$J_{2}\left(x,.\right)$ on $D_{2}\ $for each $x\in D_{1}$ yields to the set of
minimum points denoted $s_{2}\left(x\right).$ A point $\left(x,y\right)$ is a
Nash equilibrium with respect the two functionals if

$\left(x,y\right)\in\left(s_{1}\left(y\right),\ s_{2}\left(x\right)\right).$

In situations when the two functionals are differentiable, a Nash equilibrium
$\left(x,y\right)$ solves the system $J_{11}\left(x,y\right)=0,\
J_{22}\left(x,y\right)=0,$ where $J_{11},J_{22}$ are the derivatives of
$J_{1},J_{2}$ in the first and second variable, respectively. These equations
replace the ones associated to $E_{1},E_{2},$ while the controllability
conditions are such that $x,y$ minimize $J_{1},J_{2}$ in the first and second
variable, respectively, that is

$\displaystyle C_{1}$ $\displaystyle=$
$\displaystyle\left\\{\left(x,y\right)\in D_{1}\times D_{2}:\ x\text{
minimizes }J_{1}\left(.,y\right)\right\\},$ $\displaystyle C_{2}$
$\displaystyle=$ $\displaystyle\left\\{\left(y,x\right)\in D_{2}\times D_{1}:\
y\text{ minimizes }J_{2}\left(x,.\right)\right\\}.$

Therefore, in the differential case, the Nash equilibrium problem appears as a
particular case of the mutual control problem as specified above. For further
details on such results, we refer the reader to [1, 8, 10, 11, 14, 15, 16, 17,
18].

Compared to classical control theory, where the problems involve explicit
control variables (see, e.g., [3, 20]), mutual control problems feature
controls that are implicit by the very form of the coupling terms. In this
way, the problem of mutual control appears to be more general.

In this paper, we discuss the solvability of a mutual control problem for a
system of semilinear first-order differential equations, namely

(1.1) $\left\\{\begin{array}[]{l}x^{\prime}+Ax=f\left(x,y\right)\\\\[5.0pt]
y^{\prime}+By=g\left(x,y\right),\text{ on
}\left[0,T\right].\end{array}\right.$

Here, $T>0$; $x,y\in\mathbb{R}^{n};$ $A,B\in M_{n\times
n}\left(\mathbb{R}\right)$ and
$f,g:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ are
continuous mappings. Each of the two states, $x$ and $y$, controls the other
with the aim of achieving the equilibrium specified by the controllability
condition

(1.2) $x(T)=ky(T),$

for some given $k>0$. Thus,

$\displaystyle C_{1}$ $\displaystyle=$
$\displaystyle\left\\{(x,y)\,:\,x(T)=ky(T)\right\\},$ $\displaystyle C_{2}$
$\displaystyle=$
$\displaystyle\left\\{(y,x)\,:\,y(T)=\frac{1}{k}x(T)\right\\}.$

Such a control condition is of interest in dynamics of populations when it
expresses the requirement that at a certain moment $T$ the ratio between two
populations, for example prey and predators, should be the desired $k.$
Similarly, for the control of epidemics, it expresses the requirement that at
some time one reach a certain ratio between the infected population and that
susceptible to infection. Analogous interpretations can be given in the case
of some chemical or medical reaction models.

Our results target two aspects:

a):

System controllability, that is, the existence of a solution for any given
initial states $x\left(0\right)$ and $y\left(0\right),$ and

b):

A related semi-observability problem, that is, finding a solution of the
mutual problem when only one of the initial states is given, consequently
leading to the determination of the other initial state.

Our analysis relies on the basic fixed point theorems of Perov, Schauder,
Leray-Schauder, and Avramescu. We present both the advantages and limitations
of each theorem. Using a vector approach based on matrices instead of
constants allows us to obtain results independent of the norm of the spaces we
are working in.

###  1.1. Bielecki-type norms

For each number $\theta\geq 0$, on the space $C([0,T];\mathbb{R}^{n})$, we
define the Bielecki norm

$\left|\left|u\right|\right|_{\theta}=\max_{t\in\left[0,T\right]\ }e^{-\theta
t}\left|u\left(t\right)\right|.$

Note that if $\theta=0$, one has $\|\cdot\|_{0}=\|\cdot\|,$ where $\|\cdot\|$
is the uniform norm. We mention that, with a suitable choice of $\theta$, the
Bielecki-type norms allow us to relax the restrictions on constants required
by the Lipschitz or growth conditions when using fixed point theorems.

###  1.2. Matrices convergent to zero

A square matrix $M\in M_{n\times n}\left(\mathbb{R}_{+}\right)$ is said to be
convergent to zero if its power $M^{k}$ tends to the zero matrix as
$k\rightarrow\infty$. The next lemma provides equivalent conditions for a
square matrix to be convergent to zero (see, e.g., [2, 9, 13]).

######  Lemma 1.1.

Let $A\in M_{n\times n}\left(\mathbb{R}_{+}\right)$ be a square matrix. The
following statements are equivalent:

(a):

The matrix $A$ is convergent to zero.

(b):

The spectral radius of $A$ is less than $1$, i.e., $\rho(A)<1$.

(c):

The matrix $I-A,$ where $I$ is the unit matrix of the same size, is invertible
and its inverse has nonnegative entries, i.e., $\left(I-A\right)^{-1}\in
M_{n\times n}\left(\mathbb{R}_{+}\right).$

(d):

In case $n=2$, a matrix $A=[a_{ij}]_{1\leq i,j\leq 2}\in M_{2\times
2}\left(\mathbb{R}_{+}\right)$ is convergent to zero if and only if

$a_{11},\ a_{22}<1\ \ \text{and\ \
}a_{11}+a_{22}<1+a_{11}a_{22}-a_{12}a_{21}.$

In what follows, the vectors in $\mathbb{R}^{n}$ are treated as columns. By
$\left|x\right|$, we mean the Euclidean norm of the vector $x$; $\left\langle
x,y\right\rangle$ represents the inner product in $\mathbb{R}^{n}$; and
$\left\|x\right\|$ stands for the uniform norm in
$C\left([0,T];\mathbb{R}^{n}\right)$. For a square matrix $A$, we denote
$S_{A}(t)=e^{-tA}$ ($t\in\mathbb{R}$). Additionally, let $C_{A}$ be a upper
bound of the norms of the linear operators $e^{-tA}$ on $[-T,T]$, i.e.,

$\max_{t\in[-T,T]}\left|e^{-tA}\right|\leq C_{A}.$

We conclude this introduction by recalling two less known fixed point theorems
which are applied in order to establish our results.

######  Theorem 1.2 (Perov).

Let $\left(X_{i},d_{i}\right),$ $i=1,2$ be complete metric spaces and
$N_{i}:X_{1}\times X_{2}\rightarrow X_{i}$ be two mappings for which there
exists a square matrix $M$ of size two with nonnegative entries and the
spectral radius $\rho\left(M\right)<1$ such that the following vector
inequality

$\left(\begin{array}[]{c}d_{1}\left(N_{1}\left(x,y\right),N_{1}\left(u,v\right)\right)\\\\[5.0pt]
d_{2}\left(N_{2}\left(x,y\right),N_{2}\left(u,v\right)\right)\end{array}\right)\leq
M\left(\begin{array}[]{c}d_{1}\left(x,y\right)\\\\[5.0pt]
d_{2}\left(u,v\right)\end{array}\right)$

holds for all $\left(x,y\right),\left(u,v\right)\in X_{1}\times X_{2}.$ Then,
there exists a unique point $\left(x,y\right)\in X_{1}\times X_{2}$ with
$(x,y)=\left(N_{1}\left(x,y\right),N_{2}\left(x,y\right)\right).$

######  Theorem 1.3 (Avramescu).

Let $(D_{1},d)$ be a complete metric space,$\ D_{2}$ a closed convex subset of
a normed space$\ Y$, and $N_{i}:D_{1}\times D_{2}\rightarrow D_{i}$, $i=1,2$
be continuous mappings. Assume that the following conditions are satisfied:

  1. (i)

There is a constant $L\in[0,1)$ such that

$d(N_{1}(x,y),N_{1}(\overline{x},y))\leq Ld(x,\overline{x})$

for all $x,\overline{x}\in D_{1}$;

  2. (ii)

$N_{2}(D_{1}\times D_{2})$ is a relatively compact subset of $Y$ ;

Then, there exists $(x,y)\in D_{1}\times D_{2}$ such that

$N_{1}(x,y)=x,\quad N_{2}(x,y)=y.$

Some reference works in fixed point theory are the books [4] and [5].

##  2\. Controllability of the mutual control problem

###  2.1. Fixed point formulation of the mutual control problem

The system (1.1) is equivalent (see, e.g., [6, 19]) to

(2.1)
$\left\\{\begin{array}[]{l}x\left(t\right)=S_{A}\left(t\right)x(0)+\int_{0}^{t}S_{A}\left(t-s\right)f\left(x(s),y(s)\right)ds\\\\[5.0pt]
y\left(t\right)=S_{B}\left(t\right)y(0)+\int_{0}^{t}S_{B}\left(t-s\right)g\left(x(s),y(s)\right)ds,\end{array}\right.$

where

$S_{A}\left(t\right)=e^{-tA},\ \ \ S_{B}\left(t\right)=e^{-tB}\ \
\left(t\in\mathbb{R}\right),$

are the groups of linear operators generated by $A$ and $B$, respectively.
Using the controllability condition (1.2), one has

(2.2) $\displaystyle S_{A}\left(T\right)x(0)$
$\displaystyle+\int_{0}^{T}S_{A}\left(T-s\right)f\left(x(s),y(s)\right)ds$
$\displaystyle=kS_{B}\left(T\right)y(0)+k\int_{0}^{T}S_{B}\left(T-s\right)g\left(x(s),y(s)\right)ds.$

Then, expressing $S_{A}\left(t\right)x(0)$ and $S_{A}\left(t\right)y(0)$ from
(2.2) and replacing them into (2.1), we obtain

(2.3)
$\begin{cases}x\left(t\right)=kS_{A}(t-T)S_{B}\left(T\right)y\left(0\right)+k\int_{0}^{T}S_{A}(t-T)S_{B}\left(T-s\right)g\left(x(s),y(s)\right)ds\\\\[5.0pt]
\qquad\quad-\int_{t}^{T}S_{A}\left(t-s\right)f\left(x(s),y(s)\right)ds\\\\[5.0pt]
y\left(t\right)=\frac{1}{k}S_{B}(t-T)S_{A}\left(T\right)x\left(0\right)+\frac{1}{k}\int_{0}^{T}S_{B}(t-T)S_{A}\left(T-s\right)f\left(x(s),y(s)\right)ds\\\\[5.0pt]
\qquad\quad-\int_{t}^{T}S_{B}\left(t-s\right)g\left(x(s),y(s)\right)ds,\end{cases}$

which is a fixed point equation in
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ for the operator
$N=(N_{1},N_{2})$, where $N_{1},N_{2}$ are defined by the right sides in
(2.3). Thus, any solution of the mutual control problem
$\left(\ref{se}\right)$-$\left(\ref{cc0}\right)$ is a fixed point of the
operator $N.$

The converse statement also holds as the following lemma shows.

######  Lemma 2.1.

A pair $\left(x,y\right)\in C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$
is a solution of the problem _( 1.1)-(1.2)_ if and only if it solves _( 2.3)_.

###### Proof.

The necessity has already been explained. To prove the sufficiency, assume
that $(x,y)$ is any solution of (2.3). Letting $t=0$ in (2.3) and multiplying
the two equalities by $S_{A}\left(T\right)$ and $kS_{B}\left(T\right)$,
respectively, and then subtracting, we deduce (2.2). Next, using (2.2) in
(2.3), we obtain (2.1), which, based on the equivalence of (2.1) with (1.1),
concludes our proof. ∎

Throughout this section, the initial states $x(0)$ and $y(0)$ are arbitrarily
given and denoted as $\alpha$ and $\beta$, respectively. Accordingly, in the
expressions of the operators $N_{1}$ and $N_{2}$, $x\left(0\right)$ is fixed
as $\alpha$ while $y\left(0\right)$ is fixed as $\beta$.

###  2.2. Existence and uniqueness via Perov fixed point theorem

For our first existence result, we assume that $f$ and $g$ satisfy global
Lipschitz conditions.

######  Theorem 2.2.

Assume that there are constants $a,b,c,d\geq 0$ such that

$\displaystyle\left|f\left(x,y\right)-f\left(\overline{x},\overline{y}\right)\right|$
$\displaystyle\leq$ $\displaystyle
a\left|x-\overline{x}\right|+b\left|y-\overline{y}\right|,$
$\displaystyle\left|g\left(x,y\right)-g\left(\overline{x},\overline{y}\right)\right|$
$\displaystyle\leq$ $\displaystyle
c\left|x-\overline{x}\right|+d\left|y-\overline{y}\right|,$

for all $x,\overline{x},y,\overline{y}\in\mathbb{R}^{n}.$ If the matrix

(2.4)
$M=T\left[\begin{array}[]{ll}C_{A}\left(a+kcC_{B}\right)&C_{A}\left(b+kdC_{B}\right)\\\\[5.0pt]
\frac{1}{k}C_{B}\left(aC_{A}+kc\right)&\frac{1}{k}\left(bC_{A}+kd\right)\end{array}\right]$

is convergent to zero, then for every $\alpha,\beta\in\mathbb{R}^{n},$ the
mutual control problem _( 1.1)-(1.2)_ has in
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ a unique solution
$\left(x,y\right)$ with $x\left(0\right)=\alpha$ and $y\left(0\right)=\beta.$

###### Proof.

We apply Perov’s fixed point theorem to the operator $(N_{1},N_{2})$ on the
spaces $X_{1}=X_{2}=C([0,T];\mathbb{R}^{n})$, both of them equipped with the
uniform norm. For any $x,\overline{x},y,\overline{y}\in
C\left(\left[0,T\right];\mathbb{R}^{n}\right),$ one has

$\displaystyle\left|N_{1}\left(x,y\right)\left(t\right)-N_{1}\left(\overline{x},\overline{y}\right)\left(t\right)\right|$
$\displaystyle\leq$ $\displaystyle
TC_{A}\left(a+kcC_{B}\right)\left\|x-\overline{x}\right\|+TC_{A}\left(b+kdC_{B}\right)\left\|y-\overline{y}\right\|,$
$\displaystyle\left|N_{2}\left(x,y\right)\left(t\right)-N_{2}\left(\overline{x},\overline{y}\right)\left(t\right)\right|$
$\displaystyle\leq$ $\displaystyle
T\frac{1}{k}C_{B}\left(aC_{A}+kc\right)\left\|x-\overline{x}\right\|+T\frac{1}{k}\left(bC_{A}+kd\right)\left\|y-\overline{y}\right\|.$

Writing the above relations in the vector form, we obtain

$\left[\begin{array}[]{c}\left\|N_{1}\left(x,y\right)-N_{1}\left(\overline{x},\overline{y}\right)\right\|\\\\[5.0pt]
\left\|N_{2}\left(x,y\right)-N_{2}\left(\overline{x},\overline{y}\right)\right\|\end{array}\right]\leq
M\left[\begin{array}[]{c}\left\|x-\overline{x}\right\|\\\\[5.0pt]
\left\|y-\overline{y}\right\|\end{array}\right].$

Consequently, $N=(N_{1},N_{2})$ is a Perov contraction on
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ and the result follows
from the Perov’s fixed point theorem. ∎

###  2.3.  Existence and localization via Schauder’s fixed point theorem

Here, we use linear growth conditions on $f$ and $g$ instead of the Lipschitz
ones.

######  Theorem 2.3.

Assume that $f$ and $g$ satisfy the following linear growth conditions

(2.5) $\displaystyle\left|f(x,y)\right|$ $\displaystyle\leq
a\left|x\right|+b\left|y\right|+\gamma,$ $\displaystyle\left|g(x,y)\right|$
$\displaystyle\leq c\left|x\right|+d\left|y\right|+\delta,$

for all $x,y\in\mathbb{R}^{n}$ and some nonnegative constants
$a,b,c,d,\gamma,\delta$. If the matrix $M$ given in (2.4) converges to zero,
then for every $\alpha,\beta\in\mathbb{R}^{n}$, the mutual control problem _(
1.1)-(1.2)_ has in $C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ at
least one solution $(x,y)$ with $x(0)=\alpha$ and $y(0)=\beta$. Moreover, the
following bounds hold:

$\left\|x\right\|\leq R_{1}\quad\text{and}\quad\left\|y\right\|\leq R_{2},$

where $R_{1}$ and $R_{2}$ are chosen appropriately below.

###### Proof.

We apply the Schauder’s fixed-point theorem to the operator $(N_{1},N_{2})$ on
the bounded, closed, and convex set $B_{R_{1}}\times B_{R_{2}}$, where
$B_{R_{i}}$ ($i=1,2$) are the closed balls centered at the origin of the space
$C([0,T];\mathbb{R}^{n})$ with radius $R_{i}$. Here, $R_{1}$ and $R_{2}$ are
positive real numbers chosen conveniently later. To this aim, we need to show
that the operator $(N_{1},N_{2})$ is completely continuous on
$C([0,T];\mathbb{R}^{n})$, and that it maps the set $B_{R_{1}}\times
B_{R_{2}}$ into itself, i.e.,

(2.6) $\left\|N_{1}(x,y)\right\|\leq R_{1},\ \left\|N_{2}(x,y)\right\|\leq
R_{2}\text{ \ whenever\ }\|x\|\leq R_{1},\ \,\|y\|\leq R_{2}.$

Using standard arguments, the complete continuity is immediate (see, e.g.,
[13]).

Let $x,y\in$$C([0,T];\mathbb{R}^{n})$. Then,

$\displaystyle\left\|N_{1}\left(x,y\right)\right\|$ $\displaystyle\leq$
$\displaystyle
TC_{A}\left(a+kcC_{B}\right)\left\|x\right\|+TC_{A}\left(b+kdC_{B}\right)\left\|y\right\|+\eta_{1},$
$\displaystyle\left\|N_{2}\left(y,y\right)\right\|$ $\displaystyle\leq$
$\displaystyle
T\frac{1}{k}C_{B}\left(aC_{A}+kc\right)\left\|x\right\|+T\frac{1}{k}\left(bC_{A}+kd\right)\left\|y\right\|+\eta_{2},$

where
$\eta_{1}=C_{A}\left(kC_{B}\left|\beta\right|+kTC_{B}\delta+T\gamma\right)$
and
$\eta_{2}=C_{B}\left(\frac{1}{k}C_{A}\left|\alpha\right|+\frac{1}{k}TC_{A}\gamma+T\delta\right).$
Thus

$\left[\begin{array}[]{c}\left\|N_{1}\left(x,y\right)\right\|\\\\[5.0pt]
\left\|N_{2}\left(y,y\right)\right\|\end{array}\right]\leq
M\left[\begin{array}[]{c}\left\|x\right\|\\\\[5.0pt]
\left\|y\right\|\end{array}\right]+\left[\begin{array}[]{c}\eta_{1}\\\\[5.0pt]
\eta_{2}\end{array}\right].$

Note that the invariance condition (2.6) holds if $R_{1},R_{2}$ satisfy

$M\left[\begin{array}[]{c}R_{1}\\\\[5.0pt]
R_{2}\end{array}\right]+\left[\begin{array}[]{c}\eta_{1}\\\\[5.0pt]
\eta_{2}\end{array}\right]\leq\left[\begin{array}[]{c}R_{1}\\\\[5.0pt]
R_{2}\end{array}\right],$

or equivalently

(2.7) $\left[\begin{array}[]{c}\eta_{1}\\\\[5.0pt]
\eta_{2}\end{array}\right]\leq\left(I-M\right)\left[\begin{array}[]{c}R_{1}\\\\[5.0pt]
R_{2}\end{array}\right].$

Since the matrix $M$ is convergent to zero, from Lemma 1.1, we have that
$\left(I-M\right)^{-1}$ has nonnegative entries. Thus, we can multiply (2.7)
with $\left(I-M\right)^{-1}$ without changing the sign of inequality. If we
choose $R_{1},R_{2}$ large enough such that

$(I-M)^{-1}\begin{bmatrix}\eta_{1}\\\\[5.0pt]
\eta_{2}\end{bmatrix}\leq\begin{bmatrix}R_{1}\\\ R_{2}\end{bmatrix},$

the invariance condition is satisfied. Consequently, Schauder’s fixed point
theorem in $B_{R_{1}}\times B_{R_{2}}$ yields the result. ∎

###  2.4.  Existence and localization via Leray-Schauder fixed point theorem

The next result relies on the Leray-Schauder fixed point theorem, which allows
us to consider more general nonlinearities.

######  Theorem 2.4.

Assume the matrices $A,B$ are positive semi-definite, i.e.,

$\langle x,Ax\rangle\geq 0\ \ \text{and\ \ }\langle x,Bx\rangle\geq 0,\ \
\text{for every \ }x\in\mathbb{R}^{n}.$

In addition, assume that $f,g$ are continuous,

$f=f_{0}+f_{1}\ \ \ \text{and\ \ \ }g=g_{0}+g_{1},$

where $f_{1},g_{1}$ satisfy the linear growth conditions _( 2.5)_, while
$f_{0}$ and $g_{0}$ have the sign property

$\langle x,f_{0}\left(x,y\right)\rangle\leq 0,\ \ \ \ \langle
y,g_{0}\left(x,y\right)\rangle\leq 0,$

for all $x,y\in\mathbb{R}^{n}.$ If the matrix $M$ given in (2.4) is convergent
to zero, then for every $\alpha,\beta\in\mathbb{R}^{n},$ the mutual control
problem _( 1.1)-(1.2)_ has in
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ at least one solution
$\left(x,y\right)$ with $x\left(0\right)=\alpha$ and $y\left(0\right)=\beta.$

###### Proof.

In order to apply the Leray-Schauder fixed point theorem to the operator
$\left(N_{1},N_{2}\right)$, we first establish the existence of constants
$\theta,R>0$ such that whenever

$\left\|x\right\|_{\theta}\leq R,\quad\left\|y\right\|_{\theta}\leq R,$

and one of these inequalities becomes an equality, it follows that

$\left(x,y\right)\neq\lambda\left(N_{1}\left(x,y\right),N_{2}\left(x,y\right)\right),\text{
for all }\lambda\in\left(0,1\right).$

Assume there exists $\lambda\in\left(0,1\right)$ such that

$x=N_{1}(x,y)\quad\text{and}\quad y=N_{2}(x,y).$

Differentiating both equations with respect to $t\in[0,T]$ gives

$\frac{1}{\lambda}\left(x^{\prime}+Ax\right)=f\left(x,y\right)\quad\text{and}\quad\frac{1}{\lambda}\left(y^{\prime}+By\right)=g\left(x,y\right).$

Next, multiplying by $x$ and $y$ respectively, and integrating from $0$ to $t$
yields

$\left|x\left(t\right)\right|^{2}\leq\left|\alpha\right|^{2}+2\lambda\int_{0}^{t}\langle
x,f_{0}\left(x,y\right)\rangle
ds,\quad\left|y\left(t\right)\right|^{2}\leq\left|\beta\right|^{2}+2\lambda\int_{0}^{t}\langle
y,g_{0}\left(x,y\right)\rangle ds.$

Assume $\left\|x\right\|_{\theta}=R$ and $\left\|y\right\|_{\theta}\leq R.$
Then, from the first relation we obtain

$\displaystyle\left|x\left(t\right)\right|^{2}$ $\displaystyle\leq$
$\displaystyle\left|\alpha\right|^{2}+2\lambda\int_{0}^{t}\left(a\left|x\right|^{2}+b\left|x\right|\left|y\right|+\gamma\left|x\right|\right)ds$
$\displaystyle=$
$\displaystyle\left|\alpha\right|^{2}+2\lambda\int_{0}^{t}\left(a\left|x\right|^{2}+b\left|x\right|\left|y\right|+\gamma\left|x\right|\right)e^{-2\theta
s}e^{2\theta s}ds$ $\displaystyle\leq$
$\displaystyle\left|\alpha\right|^{2}+2\lambda\left(a\left\|x\right\|_{\theta}^{2}+b\left\|x\right\|_{\theta}\left\|y\right\|_{\theta}+\gamma\left\|x\right\|_{\theta}\right)\int_{0}^{t}e^{2\theta
s}ds$ $\displaystyle\leq$
$\displaystyle\left|\alpha\right|^{2}+2\lambda\left(a\left\|x\right\|_{\theta}^{2}+b\left\|x\right\|_{\theta}\left\|y\right\|_{\theta}+\gamma\left\|x\right\|_{\theta}\right)\frac{1}{2\theta}e^{2\theta
t}.$

Dividing by $e^{2\theta t}$ and taking the maximum over $t$, we obtain

$R^{2}<\left|\alpha\right|^{2}+\frac{1}{\theta}\left(a+b\right)R^{2}+\gamma
R.$

Thus, if $\theta>a+b$ and $R$ is large enough, the above relation is not
satisfied. Similarly, if $\left\|y\right\|_{\theta}=R$ and
$\left\|x\right\|_{\theta}\leq R,$ we derive the contradiction

$R^{2}<\left|\beta\right|^{2}+\frac{1}{\theta}\left(c+d\right)R^{2}+\delta R,$

providing that $\theta>c+d$ and $R$ is large. Since the operator
$\left(N_{1},N_{2}\right)$ is also completely continuous, the theorem of
Leray-Schauder applies and gives the conclusion. ∎

######  Remark 1.

The solution $\left(x,y\right)$ guaranteed by the previous result is
exponentially localized,

$\left|x\left(t\right)\right|\leq Re^{\theta t},\ \ \
\left|y\left(t\right)\right|\leq Re^{\theta t},\text{ for all
}t\in\left[0,T\right].$

######  Example 1.

For instance, functions $f_{0}$ and $g_{0}$ may have the form

$f_{0}\left(x,y\right)=-\left|x\right|^{2p}h_{1}\left(x,y\right)x,\ \ \
g_{0}\left(x,y\right)=-\left|y\right|^{2q}h_{2}\left(x,y\right)y,$

where $h_{1}\left(x,y\right)\geq 0$ and $h_{2}\left(x,y\right)\geq 0$ for all
$x,y\in\mathbb{R}^{n}.$

###  2.5. Existence via Avramescu’s fixed point theorem

Assuming that $f$ and $g$ are Lipschitz continuous in one of the variables
$x,y$, say in $x$, and have linear growth in both variables, we obtain the
following existence result.

######  Theorem 2.5.

Assume that there are constants $a,c\geq 0$ such that

$\left|f\left(x,y\right)-f\left(\overline{x},y\right)\right|\leq
a\left|x-\overline{x}\right|,$
$\left|g\left(x,y\right)-g\left(\overline{x},y\right)\right|\leq
c\left|x-\overline{x}\right|,$

and constants $b,d,\gamma,\delta\geq 0$ with

$\left|f\left(0,y\right)\right|\leq
b\left|y\right|+\gamma,\quad\left|g\left(0,y\right)\right|\leq
d\left|y\right|+\delta,$

for all $x,\overline{x},y\in\mathbb{R}^{n}$. If the matrix $M$ given in _(
2.4)_ is convergent to zero, then for every $\alpha,\beta\in\mathbb{R}^{n}$,
the mutual control problem _( 1.1)-(1.2)_ has in
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ at least one solution
$\left(x,y\right)$ with $x\left(0\right)=\alpha$ and $y\left(0\right)=\beta$.

###### Proof.

Clearly,

$\left|f\left(x,y\right)\right|\leq a\left|x\right|+b\left|y\right|+\gamma,$
$\left|g\left(x,y\right)\right|\leq c\left|x\right|+d\left|y\right|+\delta.$

Then, as in the proof of Theorem 2.3, we find $R_{1},R_{2}>0$ such that
$\left\|N_{1}\left(x,y\right)\right\|\leq R_{1}$ and
$\left\|N_{2}\left(x,y\right)\right\|\leq R_{2}$ for all $\left(x,y\right)$
with $\left\|x\right\|\leq R_{1}$ and $\left\|y\right\|\leq R_{2}$. Moreover,
$N_{2}\left(B_{R_{1}}\times B_{R_{2}}\right)$ is relatively compact in
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)$.

It remains to show that $N_{1}\left(\cdot,y\right)$ is a contraction on
$B_{R_{1}}$ with the contraction coefficient independent of $y\in B_{R_{2}}$.
For any $x,\overline{x},y\in C\left(\left[0,T\right];\mathbb{R}^{n}\right)$,
one has

$\left|N_{1}\left(x,y\right)\left(t\right)-N_{1}\left(\overline{x},y\right)\left(t\right)\right|\leq
TC_{A}\left(a+kcC_{B}\right)\left\|x-\overline{x}\right\|,$

hence

$\left\|N_{1}\left(x,y\right)-N_{1}\left(\overline{x},y\right)\right\|\leq
TC_{A}\left(a+kcC_{B}\right)\left\|x-\overline{x}\right\|,$

where $TC_{A}\left(a+kcC_{B}\right)<1$ since the matrix $M$ is convergent to
zero. Thus, Avramescu’s theorem applies and gives the result. ∎

######  Remark 2.

Under the conditions of the previous theorem, if $\left(x,y\right),\
\left(\overline{x},y\right)$ are two solutions of the mutual control problem
$\left(\ref{se}\right)$-$\left(\ref{cc0}\right)$ with
$x\left(0\right)=\overline{x}\left(0\right)=\alpha$ and
$y\left(0\right)=\overline{y}\left(0\right)=\beta$, then necessarily
$x=\overline{x}.$

##  3\. A semi-observability problem

The mutual control problem discussed in the previous section was performed
under the assumption that initial conditions for both unknowns $x$ and $y$ are
given, i.e., $x(0)=\alpha$ and $y(0)=\beta$ are known. Now, if we assume only
the initial condition for $y$ to be known, i.e., $y(0)=\beta$, we face a semi-
observability problem. In this case, if the initial state $\beta$ of $y$ is
known and the observability relation (1.2) holds, then we aim to determine the
initial state $x\left(0\right)$ of $x$. For more details on observability
problems we send to [20].

###  3.1. On a type of matrices convergent to zero

In the subsequent analysis based on the fixed point arguments, it is
advantageous to use the matrix

$M(\theta)=\begin{bmatrix}a_{11}&a_{12}\frac{e^{\theta
T}-1}{\theta}\\\\[3.0pt] a_{21}&a_{22}\frac{1-e^{-\theta
T}}{\theta}\end{bmatrix},$

where $\theta\geq 0,$ and we need to find $\theta$ such that $M(\theta)$ is
convergent to zero. Here, $a_{ij}\,(i,j=1,2)$ are nonnegative numbers with $\
a_{11}<1$ and $a_{22}<\frac{1}{T}.$ Notice that the last inequality guarantees
$a_{22}\frac{1-e^{-\theta T}}{\theta}<1$ since $\frac{1-e^{-\theta
T}}{\theta}\leq T$ for every $\theta\geq 0.$

From Lemma 1.1, the matrix $M(\theta)$ is convergent to zero if and only if
$h(\theta)<0$, where

$\displaystyle h(\theta)$
$\displaystyle=\operatorname{tr}(M(\theta))-1-\operatorname{det}(M(\theta))$
$\displaystyle=a_{11}+a_{22}\frac{1-e^{-\theta
T}}{\theta}-1-a_{11}a_{22}\frac{1-e^{-\theta
T}}{\theta}+a_{12}a_{21}\frac{e^{\theta T}-1}{\theta}\ \ \left(\theta\geq
0\right).$

Denoting $\tau=a_{22}(1-a_{11})-a_{12}a_{21},$ one has

$\displaystyle h^{\prime}(\theta)=\frac{1}{\theta^{2}}\left(a_{22}e^{-\theta
T}(1-a_{11})(1+\theta T)+a_{12}a_{21}(\theta T-1)e^{\theta T}-\tau\right),$
$\displaystyle
h(0)=-\left(1-a_{11}\right)\left(1-a_{22}T\right)+a_{12}a_{21}T,$
$\displaystyle
h^{\prime}\left(0\right)=-a_{22}(1-a_{11})\left(\frac{T^{2}}{2}+1\right)<0,\text{
and }$
$\displaystyle\lim_{\theta\rightarrow\infty}h\left(\theta\right)=\left\\{\begin{array}[]{l}a_{11}-1\
\ \text{if }a_{12}a_{21}=0\\\\[5.0pt] +\infty\ \ \text{if
}a_{12}a_{21}>0.\end{array}\right.$

In the following lemma, we discuss conditions to guarantee the existence of a
$\theta$ such that $M(\theta)$ is convergent to zero.

######  Lemma 3.1.

Assume $0\leq a_{11}<1$ and $0<a_{22}<\frac{1}{T}$.

  * (i)

If $h(0)<0$, then $M(0)$ converges to zero.

  * (ii)

If $h(0)\geq 0$, then there exists $\theta_{1}>0$ with
$h^{\prime}(\theta_{1})=0$ and

    * (a)

if $h(\theta_{1})<0$, then the matrix $M(\theta)$ converges to zero for every
$\theta$ between the zeroes of $h$ and does not converge to zero otherwise;

    * (b)

if $h(\theta_{1})\geq 0$, then there are no $\theta$ such that $M(\theta)$
converges to zero.

###### Proof.

(i) is obvious. The next assertions are based on the convexity of the function
$h.$ To prove this, we show its second derivative is positive everywhere on
$(0,\infty)$. Indeed, one has

$h^{\prime\prime}(\theta)=\frac{1}{\theta^{3}}\left(2\left(a_{22}-a_{22}a_{11}-a_{12}a_{21}\right)+a_{12}a_{21}\varphi(\theta)-a_{22}\left(1-a_{11}\right)\varphi(-\theta)\right),$

where

$\varphi(\theta)=e^{\theta T}\left(\left(\theta T-1\right)^{2}+1\right).$

Since

$\varphi(\theta)\geq 2\,\text{ \ and }\ \varphi(-\theta)\leq 2,\text{ for all
$\theta\geq 0,$ }$

we find

$\theta^{3}h^{\prime\prime}(\theta)\geq
2\left(a_{22}-a_{22}a_{11}-a_{12}a_{21}\right)+2a_{12}a_{21}-2a_{22}(1-a_{11})=0,$

which gives our conclusion.

(ii) Assume that $h(0)\geq 0$. Since $h^{\prime}(0)<0$ and
$\lim_{\theta\rightarrow\infty}h(\theta)=+\infty$, the function $h$ has a
minimum at some $\theta_{1}\in(0,+\infty)$, whence $h^{\prime}(\theta_{1})=0$.
Clearly, if $h(\theta_{1})<0$, then $h$ has two positive zeros and is negative
between them and nonnegative otherwise. If $h(\theta_{1})\geq 0$, then
$h(\theta)\geq 0$ for all $\theta\geq 0$, and thus there are no $\theta$ such
that $M(\theta)$ converges to zero. ∎

######  Remark 3.

In case $a_{22}=0,$ $M\left(\theta\right)$ is convergent to zero if and only
if

$a_{11}<1\text{ \ \ and\ \ \ }a_{12}a_{21}\frac{e^{\theta
T}-1}{\theta}<1-a_{11}.$

Clearly, solving the equation $h^{\prime}(\theta)=0$ analytically is
challenging. Thus, instead of $M(\theta)$, we may consider an approximate
variant, larger than $M\left(\theta\right),$ namely

$\widetilde{M}(\theta)=\begin{bmatrix}a_{11}&a_{12}\frac{e^{\theta
T}-1}{\theta}\\\\[5.0pt] a_{21}&a_{22}\frac{1}{\theta}\end{bmatrix}.$

Since $M\left(\theta\right)\leq\widetilde{M}(\theta)$ componentwise, if
$\widetilde{M}(\theta)$ is convergent to zero, then $M\left(\theta\right)$ is
so too. Following similar reasoning as above, the matrix
$\widetilde{M}(\theta)$ converges to zero if and only if $\
\widetilde{h}(\theta)<0$, where

$\widetilde{h}(\theta)=a_{11}+\frac{a_{22}}{\theta}-1-\frac{a_{11}a_{22}}{\theta}+a_{12}a_{21}\frac{e^{\theta
T}-1}{\theta}\ \ \left(\theta>0\right).$

Note that $\widetilde{h}$ cannot be extended at zero by continuity and in
general is not convex. However, under suitable conditions on $a_{ij},$ a
similar result to the previous lemma holds true. In the following,
$W:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ represents the Lambert function
restricted to $\mathbb{R}_{+}$, i.e., the inverse of the function $ze^{z}\
\left(z\in\mathbb{R}_{+}\right).$ For details we send to [7].

######  Lemma 3.2.

Assume that $a_{11}<1$.

(i):

If

(3.1) $\tau:=a_{22}(1-a_{11})-a_{12}a_{21}>0$

then the function $\tilde{h}$ is strictly convex on $(0,\infty)$.

(ii):

If

$a_{12}a_{21}Te^{\theta_{1}T}<1-a_{11},$

where
$\theta_{1}=\frac{1}{T}\left(W\left(\frac{\tau}{e\,a_{12}a_{21}}\right)+1\right)$,
then $\widetilde{h}(\theta_{1})<0$ and there is a vicinity
$V=\left(\sigma_{1},\sigma_{2}\right)$ of $\theta_{1}$ with
$0<\sigma_{1}<\sigma_{2}<+\infty$ such that matrix $\widetilde{M}(\theta)$ is
convergent to zero for every $\theta\in V$ and is not so for $\theta\notin V.$

###### Proof.

(i) Simple computations yield

$\widetilde{h}^{\prime}(\theta)=\frac{1}{\theta^{2}}\left(a_{12}a_{21}e^{\theta
T}\left(\theta T-1\right)-\tau\right).$

Differentiating again $h^{\prime}(\theta)$ with respect to
$\theta\in(0,\infty)$, and using (3.1), we deduce

$\displaystyle\tilde{h}^{\prime\prime}(\theta)$
$\displaystyle=\frac{1}{\theta^{3}}\left(a_{12}a_{21}e^{\theta
T}\theta^{2}T^{2}-2\left(-\tau+a_{12}a_{21}e^{\theta T}\theta
T-a_{12}a_{21}e^{\theta T}\right)\right)$
$\displaystyle=\frac{a_{12}a_{21}}{\theta^{3}}e^{\theta
T}\left(\theta^{2}T^{2}-2\theta
T+2\right)+\frac{2\tau}{\theta^{3}}\left(-a_{11}a_{22}-a_{12}a_{21}+a_{22}\right)$
$\displaystyle=\frac{a_{12}a_{21}}{\theta^{3}}e^{\theta T}\left(1+(\theta
T-1)^{2}\right)+\frac{2\tau}{\theta^{3}}>0.$

(ii) Note that from $a_{22}(1-a_{11})>0$, we find $\lim_{\theta\rightarrow
0}\widetilde{h}(\theta)=+\infty$, while from $a_{12}a_{21}>0$, we have
$\lim_{\theta\rightarrow\infty}\widetilde{h}(\theta)=+\infty$. Therefore,
$\widetilde{h}$ has a minimum $\theta_{1}\in(0,\infty)$ . Since
$\widetilde{h}$ is convex, we have $\widetilde{h}^{\prime}(\theta_{1})=0$,
which leads to

$a_{12}a_{21}e^{\theta_{1}T}(\theta_{1}T-1)=\tau.$

Letting $z:=\theta_{1}T-1$, we obtain

$ze^{z}=\frac{\tau}{e\,a_{12}a_{21}},$

thus $z=W\left(\frac{\tau}{e\,a_{12}a_{21}}\right)$. Consequently,

$\theta_{1}=\frac{1}{T}\left(W\left(\frac{\tau}{e\,a_{12}a_{21}}\right)+1\right).$

Next, evaluating $\widetilde{h}$ at $\theta_{1}$, we find

$\widetilde{h}(\theta_{1})=a_{11}-1+a_{12}a_{21}Te^{\theta_{1}T}.$

Therefore, under our assumption, $\widetilde{h}(\theta_{1})<0$. The conclusion
follows from the convexity of $\widetilde{h}$ and its limits as $\theta$
approaches 0 and infinity. ∎

######  Remark 4.

The advantage of using $\widetilde{M}(\theta)$ instead of $M(\theta)$ is that
we have an analytical expression of the point where the minimum of $\tilde{h}$
is attained. However, in applications, due to numerical computer power, it may
be more advantageous to find an approximate solution of the equation
$h^{\prime}(\theta)=0$, and evaluate $h$ at that point.

Since $\widetilde{M}(\theta)$ is an approximation of $M(\theta)$, there are
cases where $\widetilde{M}\left(\theta\right)$ does not converge to zero for
any $\theta>0$, however, there exists a $\theta_{0}>0$ such that
$M(\theta_{0})$ does. The example below illustrates this situation.

######  Example 2.

Let

$a_{11}=0.3,\,a_{12}=0.62,\,a_{21}=0.45,\,a_{22}=0.63,\text{ and }T=0.98.$

Simple computations show that $\tau=0.162>0$ and $\theta_{1}\approx 1.1931$ ,
but conditions (ii) is not satisfied, since

$0.88\approx a_{12}a_{21}Te^{\theta_{1}T}>1-a_{11}=0.7.$

On the other hand, solving numerically $h^{\prime}(\theta)=0$ gives an
approximate solution $\theta_{0}\approx 0.3505$, and hence
$h(\theta_{0})\approx-0.00798<0,$ i.e., $M\left(\theta_{0}\right)$ is
convergent to zero. Additionally, note that $h(0)=0.0056>0$, i.e., $M(0)$ is
not convergent to zero.

###  3.2. Fixed point formulation of the semi-observability problem

Under the initial condition $y\left(0\right)=\beta,$ system (1.1) is

(3.2)
$\begin{cases}x\left(t\right)=S_{A}\left(t\right)x(0)+\int_{0}^{t}S_{A}\left(t-s\right)f\left(x(s),y(s)\right)ds\\\\[5.0pt]
y\left(t\right)=S_{B}\left(t\right)\beta+\int_{0}^{t}S_{B}\left(t-s\right)g\left(x(s),y(s)\right)ds,\end{cases}$

while the final condition (1.2) becomes

(3.3) $\displaystyle x(0)$
$\displaystyle=kS_{A}(-T)S_{B}\left(T\right)\beta+k\int_{0}^{T}S_{A}(-T)S_{B}\left(T-s\right)g\left(x(s),y(s)\right)ds$
$\displaystyle\quad-\int_{0}^{T}S_{A}\left(-s\right)f\left(x\left(s\right),y\left(s\right)\right)ds.$

Thus, substituting relation (3.3) in (3.2), for every $t\in[0,T]$, we have

(3.4) $\begin{cases}x\left(t\right)=N_{1}(x,y)(t)\\\
y\left(t\right)=N_{2}(x,y)(t),\end{cases}$

where $N_{1},N_{2}\colon C([0,T];\mathbb{R}^{n})^{2}\rightarrow
C([0,T];\mathbb{R}^{n})$,

$\displaystyle N_{1}(x,y)(t)$
$\displaystyle=kS_{A}(t-T)S_{B}\left(T\right)\beta+k\int_{0}^{T}S_{A}(t-T)S_{B}\left(T-s\right)g\left(x(s),y(s)\right)ds$
$\displaystyle\quad-\int_{t}^{T}S_{A}\left(t-s\right)f\left(x(s),y(s)\right)ds,$

and

$N_{2}(x,y)(t)=S_{B}\left(t\right)\beta+\int_{0}^{t}S_{B}\left(t-s\right)g\left(x(s),y(s)\right)ds.$

Note that the system (3.4) is a fixed point equation in
$C([0,T];\mathbb{R}^{n})^{2}$ for the operator $(N_{1},N_{2})$.

The next result establishes the equivalence between the fixed point of the
operator $(N_{1},N_{2})$ and the solution of our semi-observability problem.

######  Lemma 3.3.

Any fixed point of the operator $(N_{1},N_{2})$ is a solution of the semi-
observability problem, i.e., it satisfies both (3.2) and (3.3).

###### Proof.

Assume $(x,y)$ satisfies (3.4). Letting $t=0$ in the first relation, we derive
the controllability condition (3.3). Next, applying the operator $S_{A}(t)$ in
(3.3), we deduce

$\displaystyle S_{A}(t)x(0)+\int_{0}^{T}S_{A}(t-s)f(x(s),y(s))ds\,ds$
$\displaystyle=kS_{A}(t-T)S_{B}(T)\beta+k\int_{0}^{T}S_{A}(t-T)S_{B}(T-s)g(x(s),y(s))\,ds,$

which, when used in the definition of $N_{1}$, gives

$x(t)=S_{A}(t)x(0)+\int_{0}^{t}S_{A}(t-s)f(x(s),y(s))\,ds,$

i.e., the first relation in (3.2) holds. The second one follows immediately
from the definition of $N_{2}$. ∎

###  3.3.  Existence and uniqueness via Perov’s fixed point theorem

To determine a fixed point for the operator $(N_{1},N_{2})$, we use Perov’s
fixed point theorem. As we are dealing with an observability problem, it is
appropriate to determine a unique solution since from the final observation
$x(T)=ky\left(T\right)$, we expect a unique initial state $x(0)$ that produces
the observed results.

######  Theorem 3.4.

Assume that there are constants $a,b,c,d\geq 0$ such that

$\displaystyle\left|f\left(x,y\right)-f\left(\overline{x},\overline{y}\right)\right|$
$\displaystyle\leq$ $\displaystyle
a\left|x-\overline{x}\right|+b\left|y-\overline{y}\right|,$
$\displaystyle\left|g\left(x,y\right)-g\left(\overline{x},\overline{y}\right)\right|$
$\displaystyle\leq$ $\displaystyle
c\left|x-\overline{x}\right|+d\left|y-\overline{y}\right|,$

for all $x,\overline{x},y,\overline{y}\in\mathbb{R}^{n}.$ If there exists
$\theta\geq 0$ such that the matrix

$M(\theta)=\left[\begin{array}[]{ll}TC_{A}\left(a+kcC_{B}\right)&C_{A}\left(b+kdC_{B}\right)\frac{e^{\theta
T}-1}{\theta}\\\\[5.0pt] TC_{B}&C_{B}\frac{1-e^{-\theta
T}}{\theta}\end{array}\right]$

is convergent to zero, then for every $\beta\in\mathbb{R}^{n},$ the problem _(
1.1)-(1.2)_ has in $C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ a
unique solution $\left(x,y\right)$ with $y\left(0\right)=\beta,$ which
uniquely gives the initial state $x\left(0\right).$

###### Proof.

We apply Perov’s fixed point theorem to the operator $(N_{1},N_{2})$ on the
spaces $X_{1}=X_{2}=C([0,T];\mathbb{R}^{n})$. The first space, $X_{1}$, is
equipped with the uniform norm, while the second space, $X_{2}$, is equipped
with a Bielecki norm $\|\cdot\|_{\theta}$, where $\theta$ is given in the
hypothesis.

For any $x,\overline{x},y,\overline{y}\in
C\left(\left[0,T\right];\mathbb{R}^{n}\right),$ one has

$\displaystyle\left|N_{1}\left(x,y\right)\left(t\right)-N_{1}\left(\overline{x},\overline{y}\right)\left(t\right)\right|$
$\displaystyle\leq TC_{A}\left(a+kcC_{B}\right)\left\|x-\overline{x}\right\|$
$\displaystyle\quad+C_{A}\left(b+kdC_{B}\right)\int_{0}^{T}e^{\theta
s}e^{-\theta s}\left|y(s)-\overline{y}(s)\right|ds,$ $\displaystyle\leq
a_{11}\left\|x-\overline{x}\right\|+a_{12}\frac{e^{\theta
T}-1}{\theta}\left\|y-\overline{y}\right\|,$

where

$a_{11}=TC_{A}\left(a+kcC_{B}\right)\text{ and
}a_{12}=C_{A}\left(b+kdC_{B}\right).$

Therefore, taking the supremmum over $t\in[0,T]$ yields

$\left\|N_{1}(x,y)-N_{1}(\overline{x},\overline{y})\right\|\leq
a_{11}\|x-\overline{x}\|+a_{12}\frac{e^{\theta
T}-1}{\theta}\|y-\overline{y}\|_{\theta}.$

For the second operator $N_{2}$, for every $t\in[0,T]$, we estimate

$\left|N_{2}\left(x,y\right)\left(t\right)-N_{2}\left(\overline{x},\overline{y}\right)\left(t\right)\right|\leq
TC_{B}\|x-\overline{x}\|+C_{B}\frac{e^{t\theta}-1}{\theta}\|y-\overline{y}\|_{\theta}.$

Multiplying both sides by $e^{-\theta t}$ and taking the supremum over
$t\in[0,T]$, we obtain

$\left\|N_{2}\left(x,y\right)-N_{2}\left(\overline{x},\overline{y}\right)\right\|_{\theta}\leq
a_{21}\|x-\overline{x}\|+a_{22}\frac{1-e^{-\theta
T}}{\theta}\|y-\overline{y}\|_{\theta},$

where

$a_{22}=C_{B}\frac{1-e^{-\theta T}}{\theta}\text{ and }a_{21}=TC_{B}.$

Writing the above relations in the vector form, one has

$\displaystyle\begin{bmatrix}\left\|N_{1}\left(x,y\right)-N_{1}\left(\overline{x},\overline{y}\right)\right\|\\\\[5.0pt]
\
\left\|N_{2}\left(x,y\right)-N_{2}\left(\overline{x},\overline{y}\right)\right\|_{\theta}\end{bmatrix}\leq
M(\theta)\begin{bmatrix}\left\|x-\overline{x}\right\|\\\\[5.0pt] \
\left\|y-\overline{y}\right\|_{\theta}\end{bmatrix}.$

Consequently, since the matrix $M(\theta)$ is convergent to zero, the operator
$(N_{1},N_{2})$ is a Perov contraction on
$C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$. Thus, in view of Lemma
3.3, there exists a unique solution $(x,y)\in
C\left(\left[0,T\right];\mathbb{R}^{n}\right)^{2}$ with $y(0)=\beta$ for the
problem $\left(\ref{se}\right)$-$\left(\ref{cc0}\right).$ ∎

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