\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 63, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2004/63\hfil Existence of $\psi$-bounded solutions]
{Existence of $\Psi$-bounded solutions for a system of differential equations}

\author[A. Diamandescu\hfil EJDE-2004/63\hfilneg]
{Aurel Diamandescu}

\address{Department of Applied Mathematics, 
University of Craiova, 13 ``Al. I. Cuza''
st., 200.585 Craiova, Rom\^{a}nia}
\email{adiamandescu@central.ucv.ro}

\date{}
\thanks{Submitted March 5, 2004. Published April 23, 2004.}
\subjclass[2000]{34D05, 34C11}
\keywords{$\Psi$-bounded, Lebesgue $\Psi$-integrable function}

\begin{abstract}
 In this article, we present a necessary and sufficient condition for
 the existence of solutions to the linear nonhomogeneous system 
 $x'=A(t)x + f(t)$. Under the condition stated, for every Lebesgue 
 $\Psi$-integrable function $f$ there is at least one $\Psi$-bounded 
 solution on the interval $(0,+\infty)$.
\end{abstract}

\maketitle
\numberwithin{equation}{section} 
\newtheorem{theorem}{Theorem}[section]

\section{Introduction}

We give a necessary and sufficient condition for 
the nonhomogeneous system 
\begin{equation}
x' = A(t)x + f(t)  \label{1}
\end{equation}
to have at least one $\Psi$-bounded solution for every Lebesgue 
$\Psi$-integrable function $f$, on the interval 
$\mathbb{R}_{+} = [0,+\infty )$. Here $\Psi$ is a continuous matrix 
function, instead of a scalar function, which allows a mixed 
asymptotic behavior of the components of the solution.

The problem of $\Psi$-boundedness of the solutions for systems of ordinary
differential equations has been studied by many authors; see for example
Akinyele \cite{a1}, Constantin \cite{c1}, Avramescu \cite{a2}, Hallam \cite
{h1}, and Morchalo \cite{m1}. In these papers, the function $\Psi $ is a
scalar continuous function: Increasing, differentiable, and bounded in \cite
{a1}; nondecreasing with $\Psi (t) \geq 1$ on $\mathbb{R}_{+}$ in \cite{c1}).

Let $\mathbb{R}^d$ be the Euclidean $d$-space. Elements in this space are
denoted by $x = (x_1, x_2,\dots x_d)^T$ and their norm by $\|x\| =\max \{
|x_1|,|x_2|,\dots |x_d|\}$. For $d\times d$ real matrices, we define the
norm $|A| = \sup_{\|x\| \leq 1}\|Ax\|$.

Let $\Psi _{i}:\mathbb{R}_{+}\to (0,\infty )$, $i=1,2,\dots d$, be
continuous functions, and let 
\begin{equation*}
\Psi =\mathop{\rm diag} [\Psi _1,\Psi _2,\dots \Psi _d].
\end{equation*}
Then the matrix $\Psi (t)$ is invertible for each $t\geq 0$. 
\smallskip

\noindent \textbf{Definition.} % 1.1.
A function $\varphi :\mathbb{R}_{+}\to R^{d}$ is said to be 
$\Psi $-bounded on $\mathbb{R}_{+}$ if $\Psi (t)\varphi (t)$ is bounded 
on $\mathbb{R}_{+}$. \smallskip

\noindent \textbf{Definition.} % 1.2.
A function $\varphi : \mathbb{R}_+ \to R^d$ is said to be Lebesgue $\Psi$%
-integrable on $\mathbb{R}_+$ if $\varphi (t)$ is measurable and $\Psi
(t)\varphi(t)$ is Lebesgue integrable on $\mathbb{R}_+$. \smallskip

By a solution of \eqref{1}, we mean an absolutely continuous function
satisfying the system for almost all $t \geq 0$.

Let $A$ be a continuous $d\times d$ real matrix and the associated linear
differential system be 
\begin{equation}
y'= A(t)y.  \label{2}
\end{equation}
Also let $Y$ be the fundamental matrix of \eqref{2} with $Y(0) = I_d$, the
identity $d\times d$ matrix.

Let $X_1$ denote the subspace of $\mathbb{R}^d$ consisting of all vectors
which are values of $\Psi $-bounded solutions of \eqref{2} at $t = 0$. Let $%
X_2$ be an arbitrary closed subspace of $\mathbb{R}^d$, supplementary to $%
X_1 $. Let $P_1$, $P_2$ denote the corresponding projections of $\mathbb{R}^d
$ onto $X_1$, X$_2$.

\section{The Main Results}

In this section, we give the main results of this Note.

\begin{theorem} \label{thm2.1}
If $A$ is a continuous $d\times d$ real
matrix, then \eqref{1} has at least one $\Psi $-bounded
solution on $\mathbb{R}_+$ for every Lebesgue $\Psi$-integrable
function $f$ on $\mathbb{R}_+$ if and only if there is a positive constant
$K$ such that
\begin{equation} \label{3}
\begin{gathered}
|\Psi (t)Y(t)P_1 Y^{-1}(s)\Psi ^{-1}(s)|\leq  K,\quad\text{for } 0 \leq  s \leq  t,  \\
|\Psi (t)Y(t)P_2 Y^{-1}(s)\Psi ^{-1}(s)|\leq  K,\quad\text{for }0 \leq  t \leq  s.
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof}
 First, we prove the ``only if'' part.
We define the sets:
\\
$C_{\Psi} = \{ x : \mathbb{R}_+ \to \mathbb{R}^d : x$
is $\Psi $-bounded and continuous on $\mathbb{R}_+ \}$,
\\
$B = \{ x : \mathbb{R}_+ \to  \mathbb{R}^d : x
\mbox{ is Lebesgue $\Psi$-integrable on }\mathbb{R}_+ \}$,
\\
$D = \{ x : \mathbb{R}_+ \to \mathbb{R}^d : x$
is absolutely continuous on all intervals $J \subset \mathbb{R}_+$,
 $\Psi $-bounded on $\mathbb{R}_+$, $x(0)$ in $X_2$, $x'(t) - A(t)x(t)$ in
 $B \}$.

It is well-known that $C_\Psi$ is a real Banach space with the norm
\[
\| x\| _{C_\Psi }  = \sup_{t\geq 0} \| \Psi (t)x(t)\| .
\]
Also, it is well-known that $B$ is a real Banach space with the norm
\[
\| x\| _B=\int_0^\infty \| \Psi(t)x(t)\| dt.
\]
The set $D$ is obviously a real linear space and
\begin{equation*}
\| x\| _D=\underset{t\geq 0}{\sup }\| \Psi(t)x(t)\|
+\|  x' - A(t)x \| _{B}
\end{equation*}
is a norm on $D$.

Now, we show that ($D, \|\cdot\|_D$) is a Banach space. Let
$(x_n)_n$ be a fundamental sequence in $D$. Then,
$(x_n)_n$ is a fundamental sequence in $C_\Psi$. Therefore, there exists a
continuous and bounded function $x : \mathbb{R}_+ \to \mathbb{R}^d$
such that
\begin{equation*}
\lim_{n\to \infty } \Psi (t)x_n (t) = x(t), \quad\mbox{uniformly on }
\mathbb{R}_+.
\end{equation*}
Denote $\bar{x}(t) = \Psi ^{-1}(t)x(t) \in C_\Psi$. From
\begin{equation*}
\|x_n(t) - \bar{x}(t)\| \leq |\Psi ^{-1}(t)|\| \Psi (t)x_n(t) - x(t) \| ,
\end{equation*}
it follows that
$\lim_{n\to \infty } x_n(t) = \bar{x}(t)$, uniformly
on every compact of $\mathbb{R}_{+}$.
Thus, $\bar{x}(0) \in  X_2$.

On the other hand, ($f_n(t)$), where $f_n(t) = \Psi (t)(x_n'(t) - A(t)x_n(t))$,
is a fundamental sequence in $L$, the Banach space of all vector functions which
are Lebesgue integrable on $\mathbb{R}_+$ with the norm
\begin{equation*}
\| f\| =\int_0^{\infty }\| \Psi (t)f(t)\| dt.
\end{equation*}
Thus, there is a function $f $ in  $L$ such that
\begin{equation*}
\lim_{n\to \infty }\int_0^{\infty }\|f_n(t) - f(t)\| dt = 0.
\end{equation*}
Putting $\bar{f}(t) = \Psi ^{-1}(t)f(t)$, it follows that $\bar{f}(t) \in  B$

For a fixed, but arbitrary,  $t \geq  0$, we have
\begin{align*}
\bar{x}(t) - \bar{x}(0)
&= \lim_{n\to \infty } (x_n(t) - x_n(0)) \\
&= \lim_{n\to \infty} \int_0^t x_n'(s)ds \\
&= \lim_{n\to \infty } \int_0^t [ (x_n'(s) - A(s)x_n(s) ) +A(s)x_n(s) ]ds \\
&= \lim_{n\to\infty}\int_0^t \{\Psi ^{-1}(s)[ f_n(s) - f(s) ] +
\bar{f}(s) + A(s)x_n(s) \}ds \\
&=\int_0^t [ \bar{f}(s) + A(s)\bar{x}(s) ]ds.
\end{align*}
It follows that $\bar{x}'(t) - A(t)\bar{x}(t) = \bar{f}(t) \in  B$ and
$\bar{x}(t)$ is absolutely continuous on all intervals
$J \subset  \mathbb{R}_+$.
Thus, $\bar{x}(t) \in  D$.
From
$\lim_{n\to \infty } \Psi (t)x_n (t) =
\Psi (t)\bar{x}(t)$, uniformly on $\mathbb{R}_{+}$ and
\begin{equation*}
\lim_{n\to\infty}\int_0^{\infty }\| \Psi (t)[( x_n'(t) - A(t)x_n (t) ) - (
\bar{x}'(t) - A(t)\bar{x}(t) )]\|  dt = 0 ,
\end{equation*}
it follows that $\lim_{n\to\infty}\| x_n-\bar{x} \| _D= 0$.
Thus, $(D,\| \cdot \| _D)$ is a Banach space.

Now, we define
\begin{equation*}
T: D\to  B, \quad Tx = x' - A(t)x.
\end{equation*}
Clearly, $T$ is linear and bounded, with $\| T \| \leq  1$.
Let $Tx = 0$. Then, $x'= A(t)x, x \in  D$. This shows that $x$ is
a $\Psi$-bounded solution of \eqref{2}. Then, $x(0) \in  X_1 \cap  X_2  = \{0\}$.
Thus, $x = 0$, such that the operator $T$ is one-to-one.

Now, let $f \in B$ and let $x(t)$ be the $\Psi $-bounded solution of the
system \eqref{1}. Let $z(t)$ be the solution of the Cauchy problem
\begin{equation*}
z'= A(t)z + f(t), \quad z(0) = P_2x(0).
\end{equation*}
Then, $x(t) - z(t)$ is a solution of \eqref{2} with $P_2(x(0) - z(0)) = 0$,
i.e. $x(0) - z(0)\in X_1$. It follows that $x(t) - z(t)$ is $\Psi$-bounded
on $R_{+}$. Thus, $z(t)$ is $\Psi$-bounded on $\mathbb{R}_{+}$.
It follows that $z(t) \in D$ and $Tz = f$. Consequently, the operator
$T$ is onto.

 From a fundamental result of Banach: ``If $T$ is a bounded one-to-one
linear operator from Banach space onto another, then the inverse operator
$T^{-1}$ is also bounded, we have that there is a positive constant
$K = \| T^{-1}\| - 1$ such that, for $f\in  B$ and for the
solution $x \in  D$ of \eqref{1},
\begin{equation*}
\sup_{t\geq 0}\| \Psi(t)x(t)\| \leq  K\int_0^{\infty } \| \Psi (t)f(t)\| .
\end{equation*}
For $s \geq  0$, $\delta > 0$, $\xi \in \mathbb{R}^d$,
we consider the function $f : \mathbb{R}_+ \to \mathbb{R}^d$,
\begin{equation*}
f(t) =\begin{cases}
\Psi ^{-1}(t)\xi ,&\text{for } s \leq  t \leq  s + \delta \\
0, &\text{elsewhere.}
\end{cases}
\end{equation*}
Then, $f \in  B$ and $\|f \| _B = \delta \| \xi \| $.
The corresponding solution $x \in  D$ is
\begin{equation*}
x(t) = \int_s^{s + \delta } G(t,u)du,
\end{equation*}
where
\begin{equation*}
G(t,u) = \begin{cases}
Y(t)P_1 Y^{-1}(u), &\text{for } 0 \leq  u \leq  t \\
-Y(t)P_2Y^{-1}(u), &\text{for } 0 \leq  t \leq  u.
\end{cases}
\end{equation*}
Clearly, $G$ is continuous except on the line $t = u$, where it has a jump
discontinuity. Therefore,
\begin{equation*}
\| \Psi(t)x(t)\| =\| \int_s^{s + \delta }\Psi
(t)G(t,u)\Psi ^{-1}(u)\xi du \| \leq K\delta \| \xi \| .
\end{equation*}
It follows that
\begin{equation*}
\| \Psi (t)G(t,s) \Psi ^{-1}(s) \xi \|\leq  K\| \xi \| .
\end{equation*}
Hence,
\begin{equation*}
|\Psi (t)G(t,s)\Psi ^{-1}(s)|\leq  K,
\end{equation*}
which is equivalent with \eqref{3}. By continuity, \eqref{3} remains true
also in the case $t = s$.

Now, we prove the ``if'' part.
We consider the function
\begin{equation*}
x(t) = \int_0^t Y(t)P_1Y^{-1}(s)f(s)ds - \int_t^{\infty }
Y(t)P_2Y^{-1}(s)f(s)ds, t \geq  0,
\end{equation*}
where $f$ is a Lebesgue $\Psi$-integrable function on $\mathbb{R}_+$
It is easy to see that $x(t)$ is a $\Psi$-bounded solution on
$\mathbb{R}_+$ of \eqref{1}.
The proof is now complete.
\end{proof}

\noindent\textbf{Remark.}\; %2.1
By taking $\Psi (t) = I_d$ in Theorem \ref{thm2.1}, the conclusion in \cite[Theorem
2, Chapter V]{c2} follows.

\begin{theorem} \label{thm2.2}
Suppose that: \begin{enumerate}
\item The fundamental matrix $Y(t)$ of \eqref{2} satisfies the conditions:
\begin{itemize}
\item[(a)] $\lim_{t\to \infty }\Psi (t)Y(t)P_1 = 0$;

\item[(b)] $|\Psi (t)Y(t)P_1Y^{-1}(s)\Psi ^{-1}(s)\leq K$, for
$0 \leq  s \leq  t$,\\
$|\Psi (t)Y(t)P_2Y^{-1}(s)\Psi ^{-1}(s)| \leq K$, for
$0 \leq  t \leq  s$,
\end{itemize}
where $K$ is a positive constant and $P_1$ and $P_2$ are as in the Introduction

\item The function $f : \mathbb{R}_{+} \to \mathbb{R}^d$ is Lebesgue
$\Psi$-integrable on $\mathbb{R}_{+}$.
\end{enumerate}
Then, every $\Psi$-bounded solution $x(t)$ of \eqref{1} is such
that
\begin{equation*}
\lim_{t\to \infty }\| \Psi(t)x(t)\| =0.
\end{equation*}
\end{theorem}

\begin{proof}
Let $x(t)$ be a $\Psi $-bounded solution of \eqref{1}. There is a
positive constant $M$ such that $\|\Psi (t)x(t) \| \leq M$, for all
$t \geq  0$.
We consider the function
\begin{equation*}
y(t) = x(t) - Y(t)P_1x(0) -\int_0^t Y(t)P_1 Y^{-1}(s)f(s)ds
+\int_t^{\infty }Y(t)P_2Y^{-1}(s)f(s)ds
\end{equation*}
for all t $\geq $ 0.

 From the hypotheses, it follows that the function $y(t)$ is a $\Psi$-bounded
solution of \eqref{2}. Then, $y(0) \in  X_1$.
On the other hand, $P_1y(0) = 0$. Therefore, $y(0) = P_2 y(0) \in  X_2$.
 Thus, $y(0) = 0$ and then $y(t) = 0$ for $t\geq 0$.

Thus, for $t \geq  0$ we have
\begin{equation*}
x(t) = Y(t)P_1 x(0) + \int_0^t Y(t)P_1
Y^{-1}(s)f(s)ds -\int_t^{\infty } Y(t)P_2 Y^{-1}(s)f(s)ds.
\end{equation*}
Now, for a given $\varepsilon >0$, there exists $t_1\geq 0$ such that
\begin{equation*}
\int_t^{\infty}\| \Psi (s)f(s)\| ds
< \frac{\varepsilon }{2K},\quad \text{for }t \geq  t_1.
\end{equation*}
Moreover, there exists $t_2>t_1$ such that, for $t \geq  t_2$,
\begin{equation*}
|\Psi (t)Y(t)P_1|\leq \frac{\varepsilon }{2}
\Big[ \| x(0)\| +\int_0^{t_1} \| Y^{-1}(s)f(s)\| ds\Big]^{-1}.
\end{equation*}
Then, for $t \geq t_2$ we have
\begin{align*}
\| \Psi(t)x(t)\| &\leq |\Psi (t)Y(t)P_1|\| x(0)\|
+\int_0^{t_1}|\Psi (t)Y(t)P_1|\| Y^{-1}(s)f(s)\| ds\\
&\quad + \int_{t_1}^{t}|\Psi (t)Y(t)P_1 Y
^{-1}(s)\Psi ^{-1}(s)|\| \Psi (s)f(s)\| ds\\
&\quad +\int_{t}^{\infty }|\Psi (t)Y(t)P_2 Y^{-1}(s)
\Psi ^{-1}(s)|\| \Psi (s)f(s)\| ds \\
&\leq |\Psi (t)Y(t)P_1|\Big[ \| x(0)\| +\int_0^{t_1}\| Y^{-1}(s)f(s)\|
ds\Big] \\
&\quad + K\int_{t_1}^{\infty }\| \Psi (s)f(s)\| ds <\varepsilon .
\end{align*}
This shows that $\lim{t\to \infty }\|\Psi (t)x(t) \| =0$.
The proof is now complete.
\end{proof}

\noindent\textbf{Remark.}% 2.2.
Theorem \ref{thm2.2} generalizes a result in Constantin \cite{c1}.

Note that Theorem \ref{thm2.2} is no longer true if we require that the
function $f$ be $\Psi $-bounded on $\mathbb{R}_{+}$, instead of condition
(2) of the Theorem. Even if the function $f$ is such that 
\begin{equation*}
\lim {t\to \infty }\Vert \Psi (t)f(t)\Vert =0,
\end{equation*}
Theorem \ref{thm2.2} does not apply. This is shown by the next example.

\noindent\textbf{Example.} % 2.1.
Consider the linear system \eqref{2} with $A(t) = O_2$. Then $Y(t) = I_2$ is
a fundamental matrix for \eqref{2}. Consider 
\begin{equation*}
\Psi(t) = 
\begin{pmatrix}
\frac{1}{t + 1} & 0 \\ 
0 & t + 1
\end{pmatrix}
\end{equation*}
We have $\Psi (t)Y(t) = \Psi (t)$, such that 
\begin{equation*}
P_1= 
\begin{pmatrix}
1 & 0 \\ 
0 & 0
\end{pmatrix}
, \quad p_2= 
\begin{pmatrix}
0 & 0 \\ 
0 & 1
\end{pmatrix}
\end{equation*}
It follows that the first hypothesis of the Theorem is satisfied with $K=1$.
When we take $f(t) = (\sqrt{t + 1},( t + 1 )^{-2} )^T$, then $\lim_{t\to
\infty }\|\Psi (t)f(t)\| = 0$. On the other hand, the solutions of the
system \eqref{1} are 
\begin{equation*}
x(t) = 
\begin{pmatrix}
\frac{2}{3} ( t + 1 )^{3/2} + c_1 \\ 
-\frac{1}{t + 1} + c_2
\end{pmatrix}
\end{equation*}
It follows that the solutions of the system \eqref{1} are $\Psi$-unbounded
on $\mathbb{R}_{+}$. \smallskip

\noindent\textbf{Remark.}\; % 2.3.\ }
When in the above example we consider 
\begin{equation*}
f(t) = \big( ( t + 1 )^{-1} , (t+1)^{-3}\big)^T ,
\end{equation*}
then we have 
\begin{equation*}
\int_0^{\infty }\| \Psi (t)f(t)\| dt = 1.
\end{equation*}
On the other hand, the solutions of the system \eqref{1} are 
\begin{equation*}
x(t) = 
\begin{pmatrix}
\ln ( t + 1 ) + c_1 \\ 
-\frac{1}{2} ( t + 1 )^{-2} + c_2
\end{pmatrix}
\end{equation*}
It is easy to see that these solutions are $\Psi $-bounded on $\mathbb{R}_+$
if and only if $c_2 = 0$. In this case, $\lim_{t\to \infty }\| \Psi
(t)x(t)\| = 0$.

Note that the asymptotic properties of the components of the solutions are
not the same. This is obtained by using a matrix $\Psi$ rather than a scalar.

\subsection*{Acknowledgment}

The author would like to thank the anonymous referee for his/her valuable
comments and suggestions.

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\end{thebibliography}

\end{document}