\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 40, pp. 1--12.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/40\hfil On the $\psi$-dichotomy]
{On the $\psi$-dichotomy for homogeneous
linear differential equations}
\author[P. N. Boi\hfil EJDE-2006/40\hfilneg]
{Pham Ngoc Boi}

\address{Pham Ngoc Boi \newline
Department of Mathematics, Vinh University, Vinh City,
Vietnam}
\email{pnboi\_vn@yahoo.com}


\date{}
\thanks{Submitted December 18, 2005. Published March 26, 2006.}
\subjclass[2000]{34A12, 34C11, 34D05}
\keywords{$\psi$-bounded; $\psi$-integrable;
 $\psi$-integrally bounded; $\psi$-exponential}

\begin{abstract}
 In this article we present some conditions for the
 $\psi$-dichotomy of the homogeneous linear differential equation
 $x'=A(t)x$. Under our condition every $\psi$-integrally bounded
 function $f$ the nonhomogeneous linear differential equation
 $x'=A(t)x +f(t)$ has at least one $\psi$-bounded solution on
 $(0,+\infty)$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}


\section{Introduction}

The problem of solutions being $\psi$-bounded and $\psi$-stable
for systems of ordinary differential equations has been studied by
many authors; see for example Akinyele \cite{a1}, Avramescu \cite{a2},
Constantin \cite{c1}. In particular, Diamandescu \cite{d2,d3}
 presented some
necessary and sufficient conditions for existence of a
$\psi$-bounded solution to the linear nonhomogeneous system
$x'=A(t)x+f(t)$.

Denote by $\mathbb{R}^d$ the $d$-dimensional Euclidean space.
Elements in this space are denoted by $x=(x_1,x_2,\dots , x_d)^T$
and their norm by
$\Vert x\Vert =\max \{ \vert x_1\vert, \vert x_2\vert,\dots ,
\vert x_d\vert\}$. For real $d\times d$  matrices, we define norm
$\vert A\vert = \sup_{\Vert x\Vert \leqslant 1}\Vert Ax\Vert$.
 Let $\mathbb{R}_+ =[0, +\infty)$ and $\psi_i :\mathbb{R}_+ \to (0,\infty)$, 
 $i=1,2,\dots,d$ be continuous functions. Set
$$
\psi =\mathop{\rm diag}[\psi_1, \psi_2,\dots ,\psi_d].
$$
\begin{definition}[\cite{d2}] \label{def1.1}  \rm
 A function $f:\mathbb{R}_+ \to \mathbb{R}^d$ is said to be
\begin{itemize}
\item $\psi$-bounded on $\mathbb{R}_+$ if $\psi(t) f(t)$ is
bounded on $\mathbb{R}_+$.
\item $\psi$-integrable on $\mathbb{R}_+$ if $f(t)$ is measurable
and $\psi(t) f(t)$ is Lebesgue integrable on $\mathbb{R}_+$.
\end{itemize}
\end{definition}

 In $\mathbb{R}^d$, consider  the following equations
\begin{gather}
        x'=A(t)x+f(t)\label{e1.1} \\
x'=A(t)x \label{e1.2}
\end{gather}
where $A(t)$ is continuous matrix on $\mathbb{R}_+$.

By solution of \eqref{e1.1}, \eqref{e1.2}, we mean an absolutely continuous
function satisfying the system  for all $t\in \mathbb{R}_+$. Let
$Y(t)$ be fundamental matrix of  \eqref{e1.2} with
$Y(0)=I_d$, the identity $d\times d$ matrix.
By $X_1$ denote the subspace of $\mathbb{R}^d$ consisting of the initial
values of all $\psi$-bounded solutions of equation \eqref{e1.2} and
let $X_2$ be the closed subspace of $\mathbb{R}^d$, supplementary to $X_1$.
Also let $P_1, P_2$ denote the corresponding projections of $\mathbb{R}^d$
on to $X_1, X_2$.

\begin{definition} \label{def1.2} \rm
 The equation \eqref{e1.2} is said to has a $\psi$-exponential dichotomy
 if there exist  positive constants $K, L, \alpha, \beta$ such that
\begin{gather}
\vert \psi(t) Y(t) P_1 Y^{-1}(s)\psi^{-1}(s) \vert \leqslant
 Ke^{-\alpha(t-s)} \quad for \quad 0\leqslant   s\leqslant   t, \label{e1.3}
\\
\vert \psi(t) Y(t) P_2 Y^{-1}(s)\psi^{-1}(s)\vert \leqslant
Ke^{\beta(t-s)} \quad for \quad 0\leqslant   t\leqslant   s. \label{e1.4}
\end{gather}
The equation \eqref{e1.2} is said to be has a $\psi$-ordinary
dichotomy if \eqref{e1.3}, \eqref{e1.4} hold with $\alpha =\beta =0$.

We say that  \eqref{e1.2} has $\psi$-bounded grow if for some fixed $h>0$
there exists a constant $C\geqslant 1$ such that every solution $x(t)$
of \eqref{e1.2} is satisfied
\begin{equation}
\|\psi(t)x(t)\|\leqslant C\|\psi(s)x(s)\|\mbox{ for } 0\leqslant s\leqslant t\leqslant s+h.
\label{e1.5}
\end{equation}
\end{definition}

\begin{remark} \label{rmk1.1} \rm
For $\psi_i=1$, $i=1,2,\dots ,d$,
we obtain the notion exponential and ordinary dichotomy
 \cite{c2,d1}.
\end{remark}

Diamandescu proved the following results.

\begin{theorem}[\cite{d2}] \label{thm1.1}
The equation \eqref{e1.1} has at least one $\psi$-bounded solution
on $\mathbb{R}_+$ for every  $\psi$-integrable function $f$ on
$\mathbb{R}_+$ if and only if  \eqref{e1.2} has a
$\psi$-ordinary dichotomy.
\end{theorem}

\begin{theorem}[\cite{d4}] \label{thm1.2}
Let
\begin{gather*}
\vert \psi(t) A(t) \psi^{-1}(t) \vert \leqslant   M \quad
\mbox{for  all }  t\geqslant 0,\\
\vert \psi(t)  \psi^{-1}(s) \vert \leqslant   L \quad
\mbox{for  } 0\leqslant   s\leqslant   t.
\end{gather*}
Then \eqref{e1.1} has at least one $\psi$-bounded solution
on $\mathbb{R}_+$ for every $\psi$-bounded function $f$ on
$\mathbb{R}_+$  if and only if \eqref{e1.2} has
$\psi$-exponential dichotomy.
\end{theorem}

In this paper we prove some condition of the $\psi$-dichotomy for a
homogeneous linear differential equations  and we concerted that
with the preceding results.
 Finally, it is noted that the concept of $\psi$-dichotomy for
linear differential equations remain valid in Banach spaces.
In this case we need a few changes for the definition of $\psi$.
It seems to us that the majority of the results of this paper remain
true for Banach spaces.

\section{Preliminaries}

\begin{lemma} \label{lem2.1}
The equation \eqref{e1.2} has a $\psi$-exponential dichotomy if there
exist positive constants $K'$, $L'$, $T$, $\alpha$, $\beta$ such that
\begin{gather}
|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)|\leqslant  K'e^{-\alpha(t-s)},
\quad\mbox{for   } T\leqslant  s\leqslant  t \label{e2.1}\\
|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)|\leqslant  L'e^{\beta (t-s)},
\quad\mbox{for   } T\leqslant  s\leqslant  t.\label{e2.2}
\end{gather}
\end{lemma}

\begin{proof}
We will show that \eqref{e1.3} holds. Using a lemma of Coppel \cite{c2},
$$
|Y^{-1}(s)|\leqslant  (2^d-1)\frac{|Y(s)|^{d-1}}{|det Y(s)|}.
$$
On the other hand $Y(s)$ is continuous, we deduce
$|Y^{-1}(s)|\leqslant  N_1<+\infty$ for $0\leqslant  s \leqslant  T$.
It follows from the continuity of $\psi (t)$, $\psi^{-1}(t)$, $Y(t)$,
that $|\psi (t)|$, $|\psi^{-1}(t)|, |Y(t)| $are bounded on $[0,T]$.
Thus $|\psi (t) Y(t) P_1Y^{-1}(s)\psi^{-1}(s)|\leqslant  N<+\infty$
for $0\leqslant  s\leqslant  T$, $0\leqslant  t\leqslant  T$.
If $0\leqslant  s\leqslant  T\leqslant  t$, then
\begin{align*}
&|\psi (t) Y(t) P_1Y^{-1}(s)\psi^{-1}(s)|\\
&\leqslant  |\psi (t) Y(t) P_1Y^{-1}(T)\psi^{-1}(T)||\psi (T) Y(T) Y^{-1}(s)
\psi^{-1}(s)|\\
&\leqslant  N|\psi (t) Y(t) P_1Y^{-1}(T)\psi^{-1}(T)|\\
&\leqslant  NK'e^{-\alpha(t-T)}
\leqslant  NK'e^{\alpha T}e^{-\alpha(t-s)}.
\end{align*}
If $0\leqslant  s\leqslant  t\leqslant  T$, then
\begin{align*}
&|\psi (t) Y(t)P_1 Y^{-1}(s)\psi^{-1}(s)|\\
&\leqslant  |\psi (t) Y(t) Y^{-1}(T)\psi^{-1}(T)|
  |\psi (T) Y(T) P_1Y^{-1}(T)\psi^{-1}(T)| \\
&\quad  |\psi (T) Y(T) Y^{-1}(s)\psi^{-1}(s)|\\
&\leqslant  N^2K'\leqslant  N^2 K'e^{\alpha T}e^{-\alpha(t-s)}.
\end{align*}
Thus the inequality \eqref{e1.3} holds for
$K=\max\{K', NK'e^{\alpha T}, N^2 K'e^{\alpha T}\}$.
Similarly, inequality \eqref{e1.4} holds for
$L=\max\{L', NL'e^{\alpha T}, N^2 L'e^{\alpha T}\}$.
\end{proof}

\begin{lemma} \label{lem2.2}
Equation \eqref{e1.2} has a $\psi$-exponential dichotomy if only
if following statements are satisfied
\begin{gather}
\|\psi(t)Y(t)P_1\xi\|\leqslant  K'e^{-\alpha (t-s)}\|\psi(s)Y(s)P_1\xi\|,
\quad\mbox{for  all } \xi\in{\mathbb{R}}^d \mbox{  and     }
t\geqslant s\geqslant 0\label{e2.3}
\\
\|\psi(t)Y(t)P_2\xi\|\leqslant  L'e^{\beta (t-s)}\|\psi(s)Y(s)P_2\xi\|,
\quad\mbox{for  all } \xi\in{\mathbb{R}}^d \mbox{  and     }s
\geqslant t\geqslant 0\label{e2.4}
\\
|\psi(t)Y(t)P_1Y^{-1}(t)\psi^{-1}(t)|\leqslant  M \quad\mbox{for   }
 t\geqslant 0\label{e2.5}
\end{gather}
where $K'$, $L'$, $M$ are positive constants.
\end{lemma}

\begin{proof}
If \eqref{e1.2} has a $\psi$-exponential dichotomy then for
any vector $y\in \mathbb{R}^d$, we get
$$
\|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)y\|\leqslant
 Ke^{-\alpha(t-s)}\|y\|\mbox {   for    } 0\leqslant  s \leqslant  t.
$$
Choose $y=\psi(s)Y(s)P_1\xi$, we obtain \eqref{e2.3}.
The proof of \eqref{e2.2} is similar. Inequality \eqref{e2.5}
evidently holds.
Conversely, if inequality \eqref{e2.3}, \eqref{e2.4}, \eqref{e2.5}
are true. For any vector $y\in\mathbb{R}^d$, putting
$\xi=Y^{-1}(s)\psi^{-1}(s)y$ we get
\begin{align*}
\|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)y\|&\leqslant
K'e^{-\alpha(t-s)}\|\psi(s)Y(s)P_1Y^{-1}(s)\psi^{-1}(s)y\|\\
&\leqslant  MK'e^{-\alpha(t-s)}\|y\| \quad \mbox{for }
 t\geqslant s\geqslant 0.
\end{align*}
Thus, we have \eqref{e1.3}. The proof of \eqref{e1.4} is similar.
\end{proof}

\begin{remark} \label{rmk2.1}
By Lemma \ref{lem2.1} and in the same way as in the proof of Lemma
 \ref{lem2.2},
we can show that \eqref{e1.2} has $\psi$-exponential dichotomy
if there exists positive constant $Q$ such that
\begin{gather}
\|\psi(t)Y(t)P_1\xi\|\leqslant  K'e^{-\alpha (t-s)}\|\psi(s)Y(s)P_1\xi\|,
\quad\mbox{for  all } \xi\in{\mathbb{R}}^d \mbox{  and     }t\geqslant 
s\geqslant Q,\label{e2.6}
\\
\|\psi(t)Y(t)P_2\xi\|\leqslant  L'e^{\beta (t-s)}\|\psi(s)Y(s)P_2\xi\|,
\quad\mbox{for  all } \xi\in{\mathbb{R}}^d \mbox{  and     }s\geqslant 
t\geqslant Q,\label{e2.7}
\\
|\psi(t)Y(t)P_1Y^{-1}(t)\psi^{-1}(t)|\leqslant  M \quad\mbox{for   } 
t\geqslant Q.\label{e2.8}
\end{gather}
\end{remark}

\begin{lemma} \label{lem2.3}
Equation \eqref{e1.2} has $\psi$-bounded grow if and only if
there exist positive constants $K$, $\gamma$ such that
\begin{equation}
|\psi(t)Y(t)Y^{-1}(s)\psi^{-1}(s)|\leqslant  Ke^{\gamma(t-s)},
\quad\mbox{for  } t\geqslant s\geqslant 0.\label{e2.9}
\end{equation}
\end{lemma}

\begin{proof}
Suppose that \eqref{e1.2} has a $\psi$-bounded grow.
For arbitrary vector $\xi\in\mathbb{R}^d$, we consider the
solution $x(t)$ of  \eqref{e1.2}, with $x(0)= Y^{-1}(s)\psi^{-1}(s)\xi$.
Setting $n=[\frac{t-s}{h}]$, we get
\begin{align*}
\|\psi(t)x(t)\|&=\|\psi(nh+s)x(nh+s)\|\\
&\leqslant  C\|\psi(nh+s-h)x(nh+s-h)\|\\
&\leqslant\dots
\leqslant  C^n\|\psi(s)x(s)\|\\
&\leqslant  C^{\frac{t-s}{h}}\|\psi(s)x(s)\|\mbox{    for     } 
0\leqslant  s\leqslant  t.
\end{align*}
Set $K=C$, $\gamma=h^{-1}\ln C$, we obtain
$$
\|\psi(t)x(t)\|\leqslant  Ke^{\gamma(t-s)}\|\psi(s)x(s)\|.
$$
Therefore, $\|\psi(t)Y(t)Y^{-1}(t)\psi^{-1}(s)\xi\|\leqslant
Ke^{\gamma(t-s)}|\xi\|$.
It follows \eqref{e2.9}.

Conversely, if \eqref{e2.9} is true, then we can take
$C= Ke^{\gamma h}$. Thus \eqref{e1.5} is satisfied.
\end{proof}

\begin{remark} \label{rmk2.2} \rm
The preceding proof shows that the condition of $\psi$-bounded grow
 of  \eqref{e1.2} is independent of the choice of $h$.
\end{remark}

\section{The main results}
\begin{theorem} \label{thm3.1}
If \eqref{e1.2} has a $\psi$-exponential dichotomy, then for
any $0<\theta<1$ there exists constants $T>0$ such that every
solution $x(t)$ of \eqref{e1.2} satisfies
\begin{equation}
\|\psi(t)x(t)\|\leqslant\theta\sup_{\|s-t\|\leqslant  T}\|\psi(s)x(s)\|
\quad \mbox{for all     } t\geqslant  T.\label{e3.1}
\end{equation}
\end{theorem}

\begin{proof}
Set $x_1(t)=Y(t)P_1Y^{-1}(t)x(t)$, $x_2(t)=Y(t)P_2Y^{-1}(t)x(t)$.
Suppose that
$$
\|\psi(s)x_2(s)\|\geqslant\|\psi(s)x_1(s)\|.
$$
It follows from \eqref{e2.3} that
$$
\|\psi(s)x_1(s)\|\leqslant  K'e^{-\alpha(t-s)}\|\psi(s)x_1(s)\|
\leqslant  K'e^{-\alpha(t-s)}\|\psi(s)x_2(s)\|\quad
 \mbox{for    } 0\leqslant  s\leqslant  t.
$$
Applying \eqref{e2.4} for $\xi=Y^{-1}(s)x_2(s)$,
\begin{align*}
\|\psi(t)x_2(t)\|&=\|\psi(t)Y(t)P_2Y^{-1}(s)x_2(s)\|\\
&\geqslant  {L'}^{-1}e^{\beta(t-s)}\|\psi(s)Y(s)P_2Y^{-1}(s)x_2(s)\|
\quad \mbox{for     } 0\leqslant  s\leqslant  t.
\end{align*}
Note that $x_2(t)=Y(t)P_2Y^{-1}(t)x_2(t)$. Thus
$$
\|\psi(t)x_2(t)\|\geqslant {L'}^{-1}e^{\beta(t-s)}\|\psi(s)x_2(s)\|
\quad\mbox{for     }0\leqslant  s \leqslant  t.
$$
Therefore,
$$
\|\psi(t)x(t)\|\geqslant\frac{1}{2}[{L'}^{-1}e^{\beta(t-s)}-K'
e^{-\alpha(t-s)}]\|\psi(s)x(s)\|\quad
\mbox{for     } 0\leqslant  s\leqslant  t.
$$
Similarly, if $\|\psi(s)x_1(s)\|\geqslant\|\psi(s)x_2(s)\|$, then
$$
\|\psi(t)x(t)\|\geqslant\frac{1}{2}[{K'}^{-1}e^{\alpha(t-s)}
-L'e^{-\beta(t-s)}]\|\psi(s)x(s)\|\quad \mbox{for     }
0\leqslant  t\leqslant  s.
$$
For any $0<\theta<1$ we can choose $T>0$  large so that
$$
{L'}^{-1}e^{\beta T}-K'e^{-\alpha T}\geqslant 2\theta^{-1}
\quad\mbox{and}\quad
{K'}^{-1}e^{\alpha T}-L'e^{-\beta T}\geqslant 2\theta^{-1}.
$$
Thus for $t\geqslant T$,
$$
\|\psi(t)x(t)\|\leqslant\max\{\theta\|\psi(t+T)x(t+T)\|,
\theta\|\psi(t-T)x(t-T)\|\}.$$
Then \eqref{e3.1} is satisfied.
\end{proof}

\begin{definition} \label{def55} \rm
The function $f:\mathbb{R}_+\to \mathbb{R}^d$ is said to be
$\psi$-integrally bounded  if it is measurable and Lebesgue integrals
$\int_{t}^{t+1}\Vert \psi(u) f(u)\Vert du$ are uniformly bounded for
any $t\in \mathbb{R}_+$.
\end{definition}

\begin{theorem} \label{thm3.2}
Equation \eqref{e1.1} has at least one $\psi$-bounded solution on
${\mathbb{ R}}_+$ for every $\psi$-integrally bounded function $f$
if and only if  \eqref{e1.2} has a $\psi$-exponential
dichotomy.
\end{theorem}

\begin{proof}
First we prove the ``if" part. Suppose that \eqref{e1.2} has a
$\psi$-exponential dichotomy. Consider the function
\begin{align*}
\tilde{x}(t)&=\int_0^t\psi(t)Y(t)P_1 Y^{-1}(s)f(s)ds
 -\int_t^{\infty}\psi(t)Y(t) P_2 Y^{-1}(s)f(s)ds\\
&=\int_0^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
\end{align*}
for $t\geqslant 0$. The function $\tilde{x}(t)$ is bounded.
In fact, suppose that
$$
\int_{t}^{t+1}\Vert \psi(s) f(s)\Vert ds \leqslant   c
\quad \mbox{for } t\geqslant 0.
 $$
Then
\begin{gather*}
\int_{0}^{t}e^{-\alpha(t-s)}\Vert \psi(s) f(s)\Vert ds
\leqslant   c(1-e^{-\alpha})^{-1},\\
\int_{0}^{\infty}e^{\beta(t-s)}\Vert \psi(s) f(s)\Vert ds\leqslant
  c(1-e^{-\beta})^{-1},
\end{gather*}
 by using a Lemma in  Massera  and Schaffer. 
Set
$$
x(t)=\psi^{-1}(t)\tilde{x}(t)=\int_0^t Y(t)P_1Y^{-1}(s)f(s)ds
-\int_t^{\infty}Y(t) P_2Y^{-1}(s)f(s)ds.
$$
Then $x(t)$ is the $\psi$-bounded and continuous function on
$\mathbb{R}_+$.
\begin{align*}
x'(t)
&=A(t)\big[\int_{0}^{t}Y(t)P_1Y^{-1}(s)f(s)ds
  -\int_{t}^{\infty}Y(t)P_2Y^{-1}(s)f(s)ds\big]\\
&\quad +Y(t)P_1Y^{-1}(t)f(t)+Y(t)P_2Y^{-1}(t)f(t)\\
&=A(t)x(t)+f(t).
\end{align*}
It follows that $x(t)$ is a solution of  \eqref{e1.1}.

Now, we prove the ``only part". We define the set
$$
C_{\psi}=\{ x:\mathbb{R}_+\to \mathbb{R}^d; x  \mbox{ is
$\psi$-bounded and continuous on } \mathbb{R}_+\}.
$$
 It is well-known that $C_{\psi}$ is real Banach space with the norm
$$
\Vert x\Vert_{C_{\psi}}= \sup_{t\geqslant  0}\Vert \psi(t)x(t)\Vert.
$$
First we show that \eqref{e1.1} has a unique $\psi$-bounded
solution $x(t)$ with $x(0)\in X_2$ for each $f\in C_{\psi}$.
Further, there exists a positive constant $r$ independent of
$f$ such that
\begin{equation}
\Vert {x}\Vert_{c_{\psi}}\leqslant   r\Vert {f}\Vert_{c_{\psi}}.\label{e3.2}
\end{equation}
We prove the existence. Suppose $f\in C_{\psi}$.
By hypothesis, there exists a $\psi$-bounded solution $x(t)$
of  \eqref{e1.1}. We denote by $y(t)$ the  solution
of the Cauchy problem
$$
y'=A(t)y;\qquad y(0)=-P_1{x}(0).
$$
This solution $y(t)$ is $\psi$-bounded by definition of the subset $X_1$.
But then $z=x+y$ is a $\psi$-bounded solution of  \eqref{e1.1}
for which
$$
P_1z(0)=P_1x(0)-P_1^2x(0)=0.
$$ Thus $z(0)\in X_2$. Hence $z(t)$ is a $\psi$-bounded solution
of \eqref{e1.1} with $z(0)\in X_2$.

 We prove the uniqueness. Let $x(t)$ and $y(t)$ be the $\psi$-bounded
  solutions of equation \eqref{e1.1} with $x(0)\in X_2, y(0)\in X_2$.
Hence $x-y$ is a $\psi$-bounded of   \eqref{e1.2} and
$x(0)-y(0)\in X_2$. But $x(0)-y(0)\in X_1$. we obtain $x(0)=y(0)$,
hence $x=y$.

 We prove the inequality \eqref{e3.2} Consider the map
$T:c_{\psi}\to c_{\psi}$ which is defined $Tf=x$, where $x$ is the
$\psi$-bounded solution of  \eqref{e1.1} with
$x(0)\in X_2$. We will show that $T$ is continuous. Suppose that
$x_n=Tf_n$, $f_n\to f$ and $x_n \to x$. For any fixed $t$, we have
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}\Vert \int_0^t[f_n(s)-f(s)]ds\Vert
&\leqslant
 \lim_{n\to\infty} \int_0^t\vert\psi^{-1}(s)\vert
 \Vert \psi(s) f_n(s)-\psi(s)f(s)\Vert ds \\
&\leqslant   \lim_{n\to\infty}\Vert
f_n-f\Vert_{C_{\psi}}\int_0^t\vert\psi^{-1}(s)\vert ds=0.
\end{aligned}\label{e3.3}
\end{equation}
On the other hand
\begin{equation}
\begin{aligned}
&\lim_{n\to\infty}\Vert\int_0^tA(s)[x_n(s)-x(s)]ds\Vert \\
&\leqslant   \lim_{n\to\infty}\int_0^t\vert A(s)\psi^{-1}(s)\vert \Vert
 \psi(s)x_n(s)-\psi(s)x(s)\Vert ds\\
&\leqslant   \lim_{n\to\infty}\Vert
x_n-x\Vert_{C_{\psi}}\int_0^t\vert A(s)\psi^{-1}(s)\vert ds=0.
\end{aligned} \label{e3.4}
\end{equation}
From \eqref{e3.3} and \eqref{e3.4} we obtain
\begin{align*}
x(t)-x(0)&=\lim_{n\to\infty}(x_n(t)-x_n(0))\\
&=\lim_{n\to\infty}\int_0^t[A(s)x_n(s)+x'_n(t)-A(s)x_n(s)] ds\\
&=\lim_{n\to\infty}\int_0^t[A(s)x_n(s)+f_n(s)]ds
 \int_0^t[A(s)x(s)+f(s)]ds.
\end{align*}
Thus $x(t)$ is a solution of  \eqref{e1.1}. Since $x(t)$ is
$\psi$-bounded and
$$
x(0)=\lim_{n\to\infty}x_n(0) \in X_2
$$
we have $x=Tf$. It follows from the Closed Graph Theorem that the
linear map $T$ is continuous. Hence \eqref{e3.2} is proved.
Now, put
  \begin{equation*}
G(t,s)=\begin{cases}
Y(t)P_1Y^{-1}(s) &\text{for} 0\leqslant   s\leqslant   t \\
-Y(t)P_2Y^{-1}(s)& \text{for} 0\leqslant   t\leqslant   s.
\end{cases}
  \end{equation*}
If $\tilde{f}\in C_{\psi}$, $\tilde{f}(t)=0$ for $t>t_1>0$, then
\begin{equation}
\tilde{x}(t)=\int_0^{t_1}G(t,s) \tilde{f}(s)ds \label{e3.5}
\end{equation}
is a solution of \eqref{e1.1}. Moreover $\tilde{x}\in C_{\psi}$, since
$$
\psi(t)\tilde{x}(t) =\int_0^{t_1}\psi(t)
Y(t)P_1Y^{-1}(s)\psi^{-1}(s)\psi(s)\tilde{f}(s) ds\quad \mbox{for }
t\geqslant t_1.
$$
On the other hand,
 $\tilde{x}(0)=-P_2\int_0^{t_1}Y^{-1}(s)\tilde{f}(s)ds \in X_2$.
Thus
\begin{equation}
\Vert \tilde{x}\Vert_{c_{\psi}}\leqslant
r\Vert \tilde{f}\Vert_{c_{\psi}}.\label{e3.6}
\end{equation}
Let $x$ is an nontrivial solution of \eqref{e1.2} and let
$ \alpha(t)$ be any continuous real-valued function such that
$0\leqslant \alpha(t)\leqslant 1$ for all $t\geqslant 0$,
$\alpha(t)=0$ for $t\geqslant t_2$, $\alpha(t)=1$ for
$0\leqslant t_0\leqslant t\leqslant t_1\leqslant t_2$. Set
  \begin{equation*}
\tilde{f}(t)= \alpha(t)x(t)\| \psi(t) x(t)\|^{-1}.
  \end{equation*}
Then $\tilde {f}\in C_{\psi}$. From \eqref{e3.5} and \eqref{e3.6}, we have
\begin{equation}
\Vert \int_{t_0}^{t_1}\psi(t)G(t,s)x(s)\Vert
\psi(s)x(s)\Vert^{-1}ds\Vert_{c_{\psi}} =r \quad \mbox{for }
t_1\geqslant t_0\geqslant 0.\label{e3.7}
\end{equation}
By continuity, \eqref{e3.7} remains true also in the case $t= s$.
 Choose $x(0)=P_1\xi,  \xi\in \mathbb{R}^d$. By the arbitrary of $t_1$,
from \eqref{e3.7} we get
$$
\Vert \psi(t)Y(t)P_1\xi\Vert\int_{t_0}^{t}\Vert
\psi(u)Y(u)P_1\xi\Vert^{-1}du\leqslant   r \quad \mbox{for }
 t\geqslant t_0\geqslant 0.
$$
Choose $x(0)=P_2\xi,  \xi\in \mathbb{R}^d$. By the arbitrary of
$t_0$, from \eqref{e3.7} we get
$$
\Vert \psi(t)Y(t)P_2\xi\Vert\int_{t}^{t_1}\Vert
\psi(u)Y(u)P_2\xi\Vert^{-1}du\leqslant   r \quad \mbox{for }  0\leqslant
 t\leqslant   t_1.
$$
Next, putting $x_1(t)=Y(t)P_1Y^{-1}(s)x(s) =Y(t)P_1\xi$, we have
\begin{equation}
\Vert \psi(t)x_1(t)\Vert \int_{t_0}^t\Vert
\psi(u)x_1(u)\Vert^{-1}du \leqslant   r \quad \mbox{for }
t\geqslant t_0\geqslant 0.\label{e3.8}
\end{equation}
Also putting $x_2(t)=Y(t)P_2Y^{-1}(s)x(s) =Y(t)P_2\xi$, we get
\begin{equation}
\Vert \psi(t)x_2(t)\Vert \int_{t}^{t_1}\Vert
\psi(u)x_2(u)\Vert^{-1}du \leqslant   r \quad \mbox{for }
t_1\geqslant t\geqslant 0.\label{e3.9}
\end{equation}
It follows  by integration that
\begin{gather}
\int_{t_0}^s\Vert \psi(u)x_1(u)\Vert^{-1}du\leqslant
e^{-r^{-1}(t-s)}\int_{t_0}^t\Vert \psi(u)x_1(u)\Vert^{-1}du \quad
\mbox{for } t\geqslant s\geqslant t_0. \label{e3.10}
\\
\int_s^{t_1}\Vert \psi(u)x_2(u)\Vert^{-1}du\leqslant
e^{r^{-1}(s-t)}\int_{t}^{t_1}\Vert \psi(u)x_2(u)\Vert^{-1}du \quad
\mbox{for } t_1\geqslant s\geqslant t. \label{e3.11}
\end{gather}
Because a  $\psi$-integrable function is  $\psi$-locally integrable,
 by Theorem \ref{thm1.1} there exists a positive constant $K$ such that
\begin{gather}
\Vert\psi(t)x_1(t)\Vert
\leqslant K\Vert \psi(s)x(s)\Vert \quad \mbox{for } 0\leqslant
s\leqslant   t,\label{e3.12}
\\
\Vert\psi(t)x_2(t)\Vert\leqslant    K\Vert \psi(s)x(s)\Vert
 \quad \mbox{for } 0\leqslant   t\leqslant   s.
\label{e3.12'}
\end{gather}
Thus
$$
rK^{-1}\Vert \psi(s)x(s)\Vert^{-1}\leqslant
\int_s^{r+s}\Vert\psi(u)x_1(u)\Vert^{-1} du\quad
\mbox{for } s\geqslant 0.
$$
Using \eqref{e3.10}, replacing $t_0$ by $s$, $s$ by $s+r$ we deduce
\begin{align*}
\int_{s}^{r+s}\Vert\psi(u)x_1(u)\Vert^{-1}du
&\leqslant   e^{-r^{-1}(t-r-s)}\int_s^t\Vert\psi(u)x_1(u)\Vert^{-1}du\\
&\leqslant   ee^{-r^{-1}(t-s)}\int_s^t\Vert\psi(u)x_1(u)\Vert^{-1}du
\quad \mbox{for } t\geqslant s+r.
\end{align*}
Hence
$$r\Big( \int_t^s\Vert\psi(u)x_1(u)\Vert^{-1}du\Big)^{-1}\leqslant
 eK\Vert\psi(s)x(s)\Vert e^{-r^{-1}(t-s)}\quad \mbox{for } t\geqslant s+r.
$$
From \eqref{e3.8}, replacing $t_0$ by $s$, $s$ by $s+r$, we get
$$
\Vert \psi(t)x_1(t)\Vert\leqslant   eK\Vert\psi(s)x(s)\Vert e^{-r^{-1}(t-s)}
\mbox{ for }t\geqslant s+r.
$$
It is easy to see that the inequality holds also for
$s\leqslant   t\leqslant   s+r$. Since
$x_1(t)=Y(t)P_1 Y^{-1}(s)x(s)$, it follows that
$$
\Vert \psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)\Vert\leqslant
K'e^{-\alpha(t-s)} \quad \mbox{for } t\geqslant s\geqslant 0
$$
where $K'=eK$, $\alpha=r^{-1}$. By the same way, using
\eqref{e3.9}, \eqref{e3.11}, \eqref{e3.12'}, we get
$$
\Vert \psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)\Vert \leqslant
 K'e^{\alpha(s-t)}\quad \mbox{for } s\geqslant t\geqslant 0.
$$
The proof is complete.
 \end{proof}

Now, we are going to show some conditions for  \eqref{e1.2} has
a $\psi$-exponential dichotomy in the case it has $\psi$-bounded grow.

\begin{theorem} \label{thm3.3}
Suppose that  \eqref{e1.2} has $\psi$-bounded grow.
Equation \eqref{e1.2} has a $\psi$-exponential dichotomy if there
exists constants $T>0$, $0<\theta<1$ such that every solution of
 \eqref{e1.2} satisfies \eqref{e3.1}.
\end{theorem}

\begin{proof}
By Remark \ref{rmk2.1}, we shall show that
\eqref{e2.6}, \eqref{e2.7}, \eqref{e2.8}
 are satisfied for some $Q>0$. We may consider $x(t)$ is nontrivial
solution of  \eqref{e1.2}. The first we prove that every solution $x(t)$
of  \eqref{e1.2} with $x(0)\in X_1$ satisfies
$$
\|\psi(t)x(t)\|\leqslant  Ke^{-\alpha(t-s)}\|\psi(s)x(s)\| \quad
\mbox{for    } 0\leqslant  s\leqslant  t.
$$
By Remark \ref{rmk2.2} we can choose $h=T$, so that
\begin{equation}
\|\psi(t)x(t)\|\leqslant  C\|\psi(s)x(s)\|\quad \mbox{for     } 0\leqslant
 s\leqslant  t\leqslant  s+T.\label{e3.13}
\end{equation}
Hence
$\|\psi(t)x(t)\|\leqslant\theta\sup_{u\geqslant s}\|\psi(u)x(u)\|$
 for  $s\geqslant  0$, $t\geqslant  s+T$.
Therefore,
$$
\sup_{u\geqslant s}\|\psi(u)x(u)\|>\|\psi(t)x(t)\|
$$
for $t\geqslant  s+T$. It follow that
\begin{equation}
\sup_{u\geqslant s}\|\psi(u)x(u)\|=\sup_{s\leqslant\tau
\leqslant  s+T}\|\psi(\tau)x(\tau)\|.\label{e3.14}
\end{equation}
Hence \eqref{e3.13} and \eqref{e3.14} yield
$\|\psi(t)x(t)\|\leqslant  C\|\psi(s)x(s)\|$ for
$0\leqslant  s\leqslant  t$.
Set $n=[\frac{t-s}{T}]$ then
\begin{align*}
&\|\psi(t)x(t)\|\\
&\leqslant  \theta \sup_{\|u-t\|\leqslant  T}\|\psi(u)x(u)\|\\
&\leqslant\theta\sup_{\|u-t\|\leqslant  T}\{\theta\sup_{\|u-v\|
\leqslant  T}\|\psi(v)x(v)\|\}\leqslant\theta^2\sup_{\|v-t\|\leqslant  2T}
\|\psi(v)x(v)\|\\
&\leqslant\theta^n\sup_{\|v-t\|\leqslant  nT}\|\psi(v)x(v)\|\
leqslant\theta^nC\|\psi(s)x(s)\|\leqslant\theta^{-1}C\theta^{\frac{t-s}{T}}\|\psi(s)x(s)\|.
\end{align*}
Put $K=\theta^{-1}C>1$, $\alpha=-T^{-1}ln\theta>0$, we get
$$
\|\psi(t)x(t)\|\leqslant  Ke^{-\alpha(t-s)}\|\psi(s)x(s)\|\quad
\mbox{for    } 0\leqslant  s\leqslant  t.
$$
Now, for each $\xi\in\mathbb{R}^d$, consider the solution $x(t)$
of the equation (1) with $x(0)=P_1\xi$. Apply this inequality we
deduce \eqref{e2.6} for any $Q\geqslant 0$.

Now, suppose that $x(t)$ is any solution $x(t)$ of  \eqref{e1.2} 
with $x(0)\in X_2$.\\
May be consider $\|\psi(0)x(0)\|=1$. We can define sequence 
$t_n\to +\infty$ by
$$
\|\psi(t_n)x(t_n)\|=\theta^{-n}C, \quad
 \|\psi(t)x(t)\|<\theta^{-n}C\quad \mbox{for    } 0
\leqslant  t\leqslant  t_n.
$$
Since $\|\psi(t)x(t)\|\leqslant  C$ for $0\leqslant  t\leqslant  T$
and $\|\psi(t_1)x(t_1)\|=C\theta^{-1}>C$ we get $T<t_1$.
Consequently,
$$T<t_1<t_2<\dots<t_n<\dots.$$
From
$$
\|\psi(t_n)x(t_n)\|\leqslant\theta\sup_{0\leqslant  u\leqslant
t_n+T}\|\psi(u)x(u)\|
$$
and
$$
\|\psi(u)x(u)\|\leqslant\theta^{-1}\|\psi(t_n)x(t_n)\|\quad
\mbox{for     } 0\leqslant  u\leqslant  t_n
$$
we get $t_{n+1}<t_n+T$.
Suppose that $0\leqslant  s\leqslant  t$ and
$t_m\leqslant  t\leqslant  t_{m+1}$,
$t_n\leqslant  s\leqslant  t_{n+1}$ $(1\leqslant  m\leqslant  n)$. Then
\begin{align*}
\|\psi(t)x(t)\|&<\theta^{-m-1}C&=\theta^{n-m}\|\psi(t_{n+1})x(t_{n+1})\|\\
&\leqslant  C\theta^{-1}\theta^{n-m+1}\|\psi(s)x(s)\|\\
&\leqslant  C\theta^{-1}\theta^{\frac{s-t}{T}}\|\psi(s)x(s)\|.
\end{align*}
Thus $\|\psi(t)x(t)\|\leqslant  Ke^{-\alpha(s-t)}\|\psi(s)x(s)\|$
for $t_1\leqslant  t\leqslant  s$.

For any unit vector $\xi\in X_2$, let $x(t,\xi)$ be the solution
of \eqref{e1.2} with $\psi(0)x(0)=\xi$. Then $x(t,\xi)$ is unbounded,
and hence there is a value $t=t_1(\xi)$ such that
$$
\|\psi(t_1)x(t_1)\|=\theta^{-1}C.
$$
We will show that the values $t_1(\xi)$ are bounded.
In fact, otherwise there exists a sequence of unit vector
$\xi_k\in X_2$ such that $t_1^k=t_1(\xi_k)\to+\infty$ as $k\to +\infty$.
 By the compactness of the unit sphere in $X_2$ we may suppose that
$\xi_k\to\xi$ as $k\to +\infty$, where $\xi$ is a unit vector.
Then $x(t,\xi_k)\to x(t,\xi)$ for every $t\geqslant  0$.
Since $\|\psi(t)x(t,\xi_k)\|<\theta^{-1}C$ for
$0\leqslant  t\geqslant  t_1^k$ and $t_1^k\to +\infty$ we get
$$
\|\psi(t)x(t,\xi)\|\leqslant\theta^{-1} C\quad \mbox{ for all }
 t\geqslant  0
$$
which is a contradiction because $\xi\in X_2$.
Thus there exists $Q>0$ such that $t_1(\zeta)$ for all unit vector
$\zeta$ and every solution $x(t)$ of equation \eqref{e1.2} with
$x(0)\in X_2$ satisfies
$$
\|\psi(t)x(t)\|\leqslant  K e^{-\alpha(s-t)}\|\psi(s)x(s)\|\quad
\mbox{for      } Q\leqslant  t\leqslant  s.
$$
Thus $|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)|\leqslant  Le^{\beta(t-s)}$,
for $Q\leqslant  t\leqslant  s$.
Thus \eqref{e2.7} is proved.
Note that \eqref{e2.8} is proved in \cite[Theorem 2.1, estimate (12)]{d4}.
So the proof is cimplete.
\end{proof}

From Theorem \ref{thm3.1} and Theorem \ref{thm3.3}, we have the following result.

\begin{corollary} \label{coro1}
Suppose that  \eqref{e1.2} has $\psi$-bounded grow.
Then equation \eqref{e1.2} has a $\psi$-exponential dichotomy
if and only if there exists constants $T>0$, $0<\theta<1$ such that
every solution of  \eqref{e1.2} is satisfied \eqref{e3.1}.
\end{corollary}

\begin{theorem} \label{thm3.4}
Suppose that  \eqref{e1.2} has $\psi$-bounded grow. Then
 \eqref{e1.1} has at least one $\psi$-bounded solution on
$\mathbb{R}_{+}$ for every $\psi$-bounded function f on
$\mathbb{R}_{+}$ if and only if  \eqref{e1.2} has
$\psi$-exponential dichotomy.
\end{theorem}

\begin{proof}
Diamandescu presented this Theorem. In the proof \cite[Theorem 1.2]{d4},
 the author proved that $|\psi(t)A(t)\psi^{-1}(t)|\leqslant  M$
for all $t\geqslant 0$ and $|\psi (t)\psi^{-1}(s)|\leqslant  L$
for $t\geqslant  s\geqslant  0$ deduce \eqref{e2.9}.
Throughout the proof, he only used condition \eqref{e2.9}.
By lemma \ref{lem2.3}, condition \eqref{e2.9} is satisfied if and only if
 \eqref{e1.2} has $\psi$-bounded grow.
The proof is complete
\end{proof}

Now, consider the perturbed equation
\begin{equation}
{x'}(t)=[A(t)+B(t)]x(t)\label{e3.15}
\end{equation}
where $B(t)$ is a $d\times d$ continuous matrix function on
 $\mathbb{R}_{+}$. We have the following result.

\begin{theorem} \label{thm3.5}
(a) Suppose that  \eqref{e1.2} has a $\psi$-exponential dichotomy.
If $\delta=\sup_{t\geqslant0}|\psi(t)B(t)\psi^{-1}(t)|$ is
sufficiently small, then  \eqref{e3.15} has a $\psi$-exponential dichotomy.
\\
(b) Suppose that  \eqref{e1.2} has a $\psi$-exponential dichotomy
or $\psi$-ordinary dichotomy. If
 $\int_0^\infty|\psi(t)B(t)\psi^{-1}(t)|dt<\infty$, then  \eqref{e3.15}
has a $\psi$-ordinary dichotomy.
\end{theorem}

\begin{proof}
(a) By Theorem \ref{thm3.2} it suffices to show that the equation
\begin{equation}
{x'}(t)=[A(t)+B(t)]x(t)+f(t)\label{e3.16}
\end{equation}
has at least a $\psi$-bounded solution for every $\psi$-integrally
bounded $f$ function. Denote $Y(t)$, $P_1$, $P_2$ as in the proof of
the Theorem \ref{thm3.2}.

Consider the map $T: {C_{\psi}}\to {C_{\psi}}$ which is defined by
\begin{align*}
Tz(t)&=\int_0^tY(t)P_1Y^{-1}(s)[B(s)z(s)+f(s)]ds\\
&\quad -\int_t^\infty Y(t)P_2Y^{-1}(s)[B(s)z(s)+f(s)]ds.
\end{align*}
It is easy verified that $Tz\in C_{\psi}$. More ever if
$z_1, z_2\in C_\psi$ then
\begin{align*}
&\|Tz_1-Tz_2\|\\
&\leqslant\int_0^t|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)|
|\psi(s)B(s)\psi^{-1}(s)|\|\psi(s)z_1(s)-\psi(s)z_2(s)\|ds\\
&\quad +\int_t^\infty|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)||\psi(s)B(s)
\psi^{-1}(s)|\|\psi(s)z_1(s)-\psi(s)z_2(s)\|ds\\
&\leqslant  K\delta\|z_1-z_2\|_{C_{\psi}}\int_0^t e^{-\alpha(t-s)}ds
+L\delta\|z_1-z_2\|_{C_{\psi}}\int_t^\infty e^{\beta(t-s)}ds\\
&\leqslant\delta(K\alpha^{-1}+L\beta^{-1})\|z_1-z_2\|_{C_{\psi}}.
\end{align*}
Hence, by the contraction principle, if $\delta(K\alpha^{-1}+L\beta^{-1})<1$, 
then the mapping $T$ has a unique fixed point. Denoting this fixed point 
by $z$, we have
$$
z(t)=\int_0^tY(t)P_1Y^{-1}(s)[B(s)z(s)+f(s)]ds \\
- \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)z(s)+f(s)]ds.
$$
It follows that $z(t)$ is a solution of  \eqref{e3.16}.

\noindent (b) We can assume that  \eqref{e1.2} has a
$\psi$-ordinary dichotomy.
By Theorem \ref{thm1.1} it suffices to show that  \eqref{e3.16} has at
least a $\psi$ - bounded solution for every $\psi$-integrable $f$.
From $\int_0^\infty|\psi(t)B(t)\psi^{-1}(t)|dt<\infty$, it follows that
$$
k=K\int_T^\infty|\psi(t)B(t)\psi^{-1}(t)|dt<1
$$
for a sufficiently large and positive $T$. Let $C_{T,\psi}$ be the
Banach space of all $\psi$ - bounded and continuous functions
$z(t)$ on $[T,\infty)$ equipped with the norm
$$
\|z\|_{C_{T,\psi}}=\sup_{t\geqslant T}\|\psi(t)z(t)\|.
$$
Consider the map $T:  C_{T,\psi}\to C_{T,\psi}$ which is defined by
\begin{align*}
&Tz(t)\\
&=\int_T^tY(t)P_1Y^{-1}(s)[B(s)z(s)+f(s)]ds
-\int_t^\infty Y(t)P_2Y^{-1}(s)[B(s)z(s)+f(s)]ds.
\end{align*}
It is easy to check that $Tz\in C_{T,\psi}$. Moreover if 
$z_1, z_2\in C_{T,\psi}$ then
\begin{align*}
\|Tz_1-Tz_2\|_{C_{T,\psi}}
&\leqslant K\int_T^\infty|\psi(s)B(s)\psi^{-1}(s)|\|\psi(s)z_1(s)
-\psi(s)z_2(s)\| ds\\
&\leqslant  k\|z_1-z_2\|_{C_{T,\psi}}.
\end{align*}
It follows from the contraction principle that the equation
$Tz=z$
has a unique solution ${\tilde{z}}\in C_{T,\psi}$. Denote by $y$
the solution of  \eqref{e3.15}, which is extension of
$\tilde{z}$ on $\mathbb{R}_{+}$. Clearly $y$ is a $\psi$ - bounded
solution of  \eqref{e3.15}. The proof is complete.
\end{proof}

We remark that \eqref{e1.2} has a $\psi$-ordinary dichotomy with
 $P_1=I_d$ if and only if it is $\psi$-uniformly stable.
Theorem \ref{thm3.5} follows \cite[Theorem 3.4]{d3}.

\subsection*{Acknowledgment}
 The author wants to thank the anonymous referee for his critical review
of the original manuscript and for the changes suggested.
 The author also wants  thank Professor Diamandescu,
at University of Craiova, for his suggestions.

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\end{document}