\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 52, pp. 1--10.\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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/52\hfil Existence of $\psi$-bounded solutions] {Existence of $\psi$-bounded solutions for nonhomogeneous linear differential equations} \author[P. N. Boi\hfil EJDE-2007/52\hfilneg] {Pham Ngoc Boi} \address{Pham Ngoc Boi \newline Department of Mathematics, Vinh University, Vinh City, Vietnam} \email{pnboi\_vn@yahoo.com} \thanks{Submitted January 26, 2007. Published April 5, 2007.} \subjclass[2000]{34A12, 34C11, 34D05} \keywords{$\psi$-bounded solution; $\psi$-integrable; $\psi$-integrally bounded; \hfill\break\indent $\psi$-exponential dichotomy} \begin{abstract} In this article we present a necessary and sufficient condition for the existence of $\psi$-bounded solution on $\mathbb{R}$ of the nonhomogeneous linear differential equation $x'=A(t)x+f(t)$. We associate that with the condition of the concept $\psi$-dichotomy on $\mathbb{R}$ of the homogeneous linear differential equation $x'=A(t)x$. \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 existence of $\psi$-bounded and $\psi$-stable solutions on $\mathbb{R}_+$ for systems of ordinary differential equations has been studied by many authors; see for example Akinyele \cite{a1}, Avramescu \cite{a2}, Constantin \cite{c1}, Diamandescu \cite{d1,d2,d3}. 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 $\|x\|=\max \{ |x_1|, |x_2|,\dots , | x_d|\}$. For real $d\times d$ matrices, we define norm $| A|= \sup_{\|x\|\leqslant 1}\|Ax\Vert$. Let $\mathbb{R}_+ =[0, \infty)$, $\mathbb{R}_-=(- \infty,0]$, $J=\mathbb{R}_-,\mathbb{R}_+$ or $\mathbb{R}$ and $\psi_i :J \to (0,\infty)$, $i=1,2,\dots,d$ be continuous functions. Set $$\psi =\mathop{\rm diag}[\psi_1, \psi_2,\dots ,\psi_d].$$ \begin{definition} \label{def1.1} \rm A function $f:J \to \mathbb{R}^d$ is said to be \begin{itemize} \item $\psi$-bounded on $J$ if $\psi(t) f(t)$ is bounded on $J$. \item $\psi$-integrable on $J$ if $f(t)$ is measurable and $\psi(t) f(t)$ is Lebesgue integrable on $J$. \item $\psi$-integrally bounded on $J$ if $f(t)$ is measurable and the Lebesgue integrals $\int_t^{t+1}\|\psi(u) f(u)\|du$ are uniformly bounded for any $t, t+1\in J$. \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 $J$, $f(t)$ is a continuous function on $J$. Let $Y(t)$ be fundamental matrix of \eqref{e1.2} with $Y(0)=I_d$, the identity $d\times d$ matrix. The $d\times d$ matrices $P_1, P_2$ is said to be the pair of the supplementary projections if $P_1^2=P_1, P_2^2=P_2, P_1+P_2=I_d$. \begin{definition} \label{def1.2} \rm The equation \eqref{e1.2} is said to have a $\psi$-exponential dichotomy on $J$ if there exist positive constants $K, L, \alpha, \beta$ and a pair of the supplementary projections $P_1, P_2$ such that \begin{gather} |\psi(t) Y(t) P_1 Y^{-1}(s)\psi^{-1}(s) |\leqslant Ke^{-\alpha(t-s)} \quad\text{for } s\leqslant t, s,t\in J, \label{e1.3} \\ |\psi(t) Y(t) P_2 Y^{-1}(s)\psi^{-1}(s)|\leqslant Le^{\beta(t-s)} \quad\text{for } t\leqslant s, s,t \in J. \label{e1.4} \end{gather} The equation \eqref{e1.2} is said to have a $\psi$-ordinary dichotomy on $J$ if \eqref{e1.3}, \eqref{e1.4} hold with $\alpha =\beta =0$. We say that \eqref{e1.2} has a $\psi$-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 $$\|\psi(t)x(t)\|\leqslant C\|\psi(s)x(s)\|\mbox{ for } s\leqslant t\leqslant s+h, s,t\in J. \label{e1.5}$$ \end{definition} \begin{remark} \label{rmk1.3} \rm It is easy to see that if \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ and on $\mathbb{R}_-$ with a pair of the supplementary projections $P_1, P_2$ then \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$ with the pair of the supplementary projections $P_1, P_2$. \end{remark} \begin{theorem}[\cite{b1,d1,d3}] \label{thm1.4} (a) 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 on $\mathbb{R}_+$. \noindent (b) The equation \eqref{e1.1} has at least one $\psi$-bounded solution on $\mathbb{R}_+$ for every $\psi$- integrally bounded function $f$ on $\mathbb{R}_+$ if and only if \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$. \noindent (c) Suppose that \eqref{e1.2} has a $\psi$-bounded grow on $\mathbb{R}_+$. 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 a $\psi$-exponential dichotomy on $\mathbb{R}_+$. \end{theorem} \begin{theorem}[\cite{d3}] \label{thm1.5} Suppose that \eqref{e1.1} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ and, $P_1\ne 0, P_2\ne 0$. If $lim_{t\to\infty}\|\psi(t)f(t)\| =0$ then every $\psi$-bounded solution $x(t)$ of \eqref{e1.1} is such that $lim_{t\to\infty}\|\psi(t)x(t)\| =0$. \end{theorem} \section{Preliminaries} \begin{lemma} \label{lem2.1} (a) Let \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with a pair of the supplementary projections $P_1, P_2$. If $Q_1, Q_2$ is a pair of the supplementary projections such that $ImP_1= ImQ_1$, then \eqref{e1.2} also has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with the pair of the supplementary projections $Q_1, Q_2$. \noindent (b) Let \eqref{e1.2} have a $\psi$-exponential dichotomy on $\mathbb{R}_-$ with a pair of the supplementary projections $P_1, P_2$. If $Q_1, Q_2$ is a pair of supplementary projections such that $ImP_2= ImQ_2$, then \eqref{e1.2} also has a $\psi$-exponential dichotomy on $\mathbb{R}_-$ with the pair of the supplementary projections $Q_1, Q_2$. \end{lemma} \begin{proof} First, we prove in the case of $J =\mathbb{R}_+$. Note that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with the pair of the supplementary projections $P_1, P_2$ 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)\xi\| \quad\mbox{for all } \xi \in \mathbb{R}^d \mbox{ and } t \geqslant s \geqslant 0 , \label{e2.1}\\ \|\psi(t)Y(t)P_2\xi\|\leqslant L'e^{\beta(t-s)}\|\psi(s)Y(s)\xi\| \quad \mbox{for all } \xi \in \mathbb{R}^d \mbox{ and } s \geqslant t \geqslant 0. \label{e2.2} \end{gather} In fact, if \eqref{e1.3} and \eqref{e1.4} are true, we have for any vector $y\in \mathbb{R}^d$ \begin{gather*} \|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)y\|\leqslant Ke^{-\alpha(t-s)}\|y\| \quad \mbox{for } t \geqslant s \geqslant 0 ,\\ \|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)y\|\leqslant Le^{\beta(t-s)}\|y\| \quad \mbox{for } s\geqslant t \geqslant 0. \end{gather*} Choose $y=\psi(s)Y(s)\xi$, we obtain \eqref{e2.1}, \eqref{e2.2}. Conversely, suppose that inequalities \eqref{e2.1}, \eqref{e2.2} are true. For any vector $y\in \mathbb{R}^d$, putting $\xi=Y^{-1}(s)\psi^{-1}(s)y$ we get \eqref{e1.3}, \eqref{e1.4}. Now prove the lemma. It follows from $Ker P_2= ImP_1= ImQ_1=KerQ_2$ that $P_2Q_1=0$. Hence $P_1Q_1= P_1Q_1+P_2Q_1=Q_1$. Similarly $Q_1P_1=P_1$. Then \begin{gather} P_1-Q_1=P^2_1- P_1Q_1= P_1(P_2-Q_2) \label{e2.3},\\ P_1-Q_1= - Q_1P_2 =P_1P_2 - Q_1P_2 = (P_1-Q_1)P_2. \label{e2.4} \end{gather} For each $u\in \mathbb{R}^d$, put $\xi=(P_1-Q_1)u$. The relation \eqref{e2.3} implies that $\xi\in ImP_1$, then $P_1\xi=\xi$. Result from \eqref{e2.1}, for $s = 0$ that $$\label{e2.5} \|\psi(t)Y(t)[P_1-Q_1]u\|\leqslant K'e^{-\alpha t}\|\psi(0)[P_1-Q_1]u\|, t\geqslant 0.$$ By \eqref{e2.4} we conclude \label{e2.6} \begin{aligned} K'e^{-\alpha t}\|\psi(0)[P_1-Q_1]u\| &=K'e^{-\alpha t}\|\psi(0)[P_1-Q_1]P_2u\|\\ &\leqslant K'|\psi(0)||P_1-Q_1| e^{-\alpha t}\|P_2u\|,\quad t\geqslant 0. \end{aligned} Applying \eqref{e2.2}, for $t=0$, we get \begin{aligned} \|P_2u\|&=\|\psi^{-1}(0)\psi(0)P_2u\|\\ &\leqslant |\psi^{-1}(0)|\|\psi(0)P_2u\|\\ &\leqslant L'e^{-\beta s}|\psi^{-1}(0)|\|\psi(s)Y(s)u\|, \quad \mbox{for} s \geqslant 0. \end{aligned} \label{e2.7} The relations \eqref{e2.5}--\eqref{e2.7} imply \label{e2.8} \begin{aligned} \|\psi(t)Y(t)[P_1-Q_1]u\|&\leqslant K'L'|\psi(0)||\psi^{-1}(0)| |P_1-Q_1|e^{-\alpha t}e^{-\beta t}\|\psi(s)Y(s)u\| \\ & \leqslant K_1e^{\beta(t-s)}\|\psi(s)Y(s)u\|, \quad \mbox{ for } t, s \geqslant 0 . \end{aligned} On the other hand, by \eqref{e2.2} we get $$\label{e2.9} \|\psi(t)Y(t)P_2u\|\leqslant L'e^{\beta(t-s)}\|\psi(sY(s))u\| ,\quad \mbox{ for } 0 \leqslant t\leqslant s.$$ It follows from $Q_2=P_2+P_1-Q_1,$ \eqref{e2.8} and \eqref{e2.9} that \label{e2.10} \begin{aligned} \|\psi(t)Y(t)Q_2u\|&\leqslant\|\psi(t)Y(t)P_2u\|+\|\psi(t)Y(t)[P_1-Q_1]u\|\\ &\leqslant (L'+K_1)e^{\beta(t-s)}\|\psi(s)Y(s)u\|\\ & \leqslant L_2e^{\beta(t-s)}\|\psi(s)Y(s)u\| ,\quad\mbox{for } 0 \leqslant t \leqslant s. \end{aligned} Similarly, for $u \in\mathbb{ R}^d$, we have $$\label{e2.11} \|\psi(t)Y(t)Q_1u\|\leqslant K_2e^{-\alpha(t-s)}\|\psi(s)Y(s)u\| ,\quad \mbox{for } 0 \leqslant s \leqslant t.$$ Then from this inequality, \eqref{e2.10} and the preceding note it follows that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with the pair of the supplementary projections $Q_1, Q_2$. In the case of $J =\mathbb{ R}_ -$, the proof is similar. \end{proof} \begin{remark} \label{rmk2.2} \rm (a) Suppose that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with a pair of supplementary projections $P_1, P_2$. The set $P_1\mathbb{R}^d$ is the subspace of $\mathbb{R}^d$ consisting of the values $x(0)$ of all $\psi$-bounded solutions $x(t)$ on $\mathbb{R}_+$ of \eqref{e1.2}. In fact, denote by $X_1$ this subspace, if $v \in P_1 \mathbb{R}^d$ then $v \in X_1$ by virtue of (2.1). Conversely if $u \in X_1$, we have to show that $P_2u = 0$. Suppose otherwise that $P_2u\ne 0$, by (2.1), (2.2) we have $\|\psi(t)Y(t)P_1u\|$ is bounded and the limit of $\|\psi(t)Y(t)P_2u\|$ is $\infty$, as $t$ tend to $\infty$. Denote $y$ the solution of \eqref{e1.2}, $y(0) = u$. The relation $\psi(t)y(t) -\psi(t)Y(t)P_1u = \psi(t)Y(t)P_2u$ follows that $y$ is non$\psi$-bounded on $\mathbb{R}_ +$, which is a contradiction. \noindent(b) Similarly if \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_-$ with a pair of supplementary projections $P_1, P_2$ then the set $P_2\mathbb{R}^d$ is the subspace of $\mathbb{R}^d$ consisting of the values $x(0)$ of all $\psi$-bounded solutions $x(t)$ on $\mathbb{R}_-$ of \eqref{e1.2}. (c) Suppose that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$, then \eqref{e1.2} has no nontrivial $\psi$-bounded solution on $\mathbb{R}$. In fact if $x(t)$ is the $\psi$-bounded solution of \eqref{e1.2} on $\mathbb{R}$ then it is $\psi$-bounded on $\mathbb{R}_+$ and on $\mathbb{R}_-$. Because equation \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$, and on $\mathbb{R}_-$ with a pair of supplementary projections $P_1, P_2$, by preceding notice we have $P_2x(0) = 0$ and $P_1x(0) = 0$. Hence $x(0) = 0$, then $x(t)$ is the trivial solution of \eqref{e1.2}. \end{remark} \begin{lemma}[\cite{m1}] \label{lem2.3} Let $h(t)$ be a non-negative, locally integrable such that $$\int_t^{t+1}h(s)ds\leqslant c, \quad \mbox{for all } t \in\mathbb{ R}$$ If $\theta > 0$ then, for all $t \in\mathbb{ R}$, \begin{gather} \int_t^{\infty} e^{-\theta (s-t)}h(s)ds\leqslant c[1-e^{-\theta}]^{-1}, \label{e2.12}\\ \int_{-\infty}^t e^{-\theta (t-s)}h(s)ds\leqslant c[1-e^{-\theta}]^{-1}. \label{e2.13} \end{gather} \end{lemma} \begin{proof} We prove \eqref{e2.12}, the proof of \eqref{e2.13} is similar. \begin{align*} \int_{t+m}^{t+m+1} e^{-\theta(s-t)}h(s)ds &\leqslant \int_{t+m}^{t+m+1} e^{-\theta(t+m)}e^{\theta t}h(s)ds \\ &=\int_{t+m}^{t+m+1} e^{-\theta m}h(s)ds\leqslant c e^{-\theta m} \end{align*} implies that $$\int_t^\infty e^{-\theta(s-t)}h(s)ds =\sum_{m=0}^\infty \int_{t+m}^{t+m+1}e^{-\theta(s-t)}h(s)ds\leqslant c\sum_{m=0}^\infty e^{-\theta m} = c[1-e^{-\theta}]^{-1}$$ \end{proof} \begin{lemma} \label{lem2.4} Equation \eqref{e1.1} has at least one $\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}$ if and only if the following three conditions are satisfied: \begin{enumerate} \item Equation \eqref{e1.1} has at least one solution on $\mathbb{R}$, $\psi$-bounded on $\mathbb{R}_+$ for every $\psi$- integrally bounded function $f$ on $\mathbb{R}_+$ \item Equation \eqref{e1.1} has at least one solution on $\mathbb{R}$, $\psi$-bounded on $\mathbb{R}_-$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}_-$. \item Every solution of \eqref{e1.2} is the sum of two solution of \eqref{e1.2}, one of that is $\psi$-bounded on $\mathbb{R}_+$, another is $\psi-$ bounded on $\mathbb{R}_-$. \end{enumerate} \end{lemma} \begin{proof} Suppose the three conditions are satisfied we have to prove that \eqref{e1.1} has at least one $\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}$. Every $\psi$-integrally bounded function $f$ on $\mathbb{R}$ is $\psi$-integrally bounded function $f$ on $\mathbb{R}_+$ and on $\mathbb{R}_-$. Then for each $\psi$-integrally bounded function $f$ on $\mathbb{R}$ exists the solution $y_1$ and $y_2$ of \eqref{e1.1}, which is defined on $\mathbb{R}$ and corresponding $\psi$-bounded on $\mathbb{R}_+$ and on $\mathbb{R}_-$. Denote by $x(t)$ the solution of \eqref{e1.2} such that $x(0) = y_2(0)-y_1(0)$. By 3, we get $x(t) = x_1(t)+x_2(t)$, here $x_1, x_2$ are two solutions of \eqref{e1.2}, that are corresponding $\psi$-bounded solution on $\mathbb{R}_+$ and $\mathbb{R}_-$. Set $z_1 = y_1+x_1, z_2=y_2-x_2$. Hence $z_1$ and $z_2$ are the solutions of \eqref{e1.1} corresponding $\psi$-bounded solution on $\mathbb{R}_+$ and on $\mathbb{R}_-$. Further, $z_2(0)=y_2(0)-x_2(0)=y_1(0)+x_1(0)=z_1(0)$, then $z_1=z_2$. Consequently $z_1$ is a $\psi$-bounded solution on $\mathbb{R}$ of \eqref{e1.1}. Conversely, now if \eqref{e1.1} has at least one $\psi$-bounded solution on $\mathbb{R }$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}$ we have to prove three condition are satisfied. The conditions 1, 2 are satisfied since every $\psi$-integrally bounded function $f$ on $\mathbb{R}_+$ , or $\mathbb{R}_-$ is the restriction of a $\psi$- integrally bounded function $f$ on $\mathbb{R}$. We prove that the condition 3 is satisfied. Set $h(t)=\begin{cases} 0& \mbox{for } | t|\geqslant 1 \\ 1 &\mbox{for } t=0\\ \mbox{linear} &\text{for } t\in[-1,0],t\in[0,1] \end{cases}$ Fix a solution $x(t)$ of \eqref{e1.2} . Then $h(t)x(t)$ is a $\psi$-integrally bounded function on $\mathbb{R}$. Set $y(t) =x(t)\int_0^th(s)ds$ , we have $$y'(t) = A(t)x(t)\int_0^th(s)ds +h(t)x(t) = A(t)y(t)+h(t)x(t).$$ By hypothesis, the equation $$y'(t)=A(t)y(t)+h(t)x(t)$$ has a solution $\widetilde y(t)$, which is $\psi$-bounded on $\mathbb{R}$. Set $x_1(t) = \widetilde y(t)-y(t)+\frac{1}{2}x(t)$ and $x_2(t)= \widetilde y(t)+y(t)+\frac{1}{2}x(t)$. It follows from $\int_{-1}^0h(t)dt =\int_0^1h(t)dt=\frac{1}{2}$ that $x_1(t) = \widetilde y(t)$ for $t \geqslant 1$; $x_2(t) =\widetilde y(t)$ for $t\leqslant -1$. Then $x_1, x_2$ are the corresponding $\psi$-bounded solutions on $\mathbb{R}_+$, $\mathbb{R}_ -$ of \eqref{e1.2}. Consequently the solution $x(t)$ of \eqref{e1.2} is the sum of two solutions $x_1(t)$ and $x_2(t)$ of \eqref{e1.2}, those solutions satisfy the condition 3. The lemma is proved. \end{proof} \section{Main results} \begin{theorem} \label{thm3.1} Equation \eqref{e1.1} has at least one $\psi$-bounded solution on $\mathbb{R}_-$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}_-$ if and only if \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_-$. \end{theorem} \begin{proof} This Theorem can be shown as in \cite[Theorem 3.3]{b1}. We give the main steps of the proof as follows. In the proof of if part'': Suppose that $\int_{t-1}^t\|\psi(s)f(s)\|ds\leqslant c$ for $t\leqslant 0$. By using Lemma \ref{lem2.3} we get \begin{align*} \|\int_{-\infty}^t\psi(t)Y(t)P_1Y^{-1}(s)ds\| &\leqslant \int_{-\infty}^t|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)|\| \psi(s)f(s)\|ds \\ &\leqslant\int_{-\infty}^te^{-\alpha(t-s)}\|\psi(s)f(s)\|ds \leqslant c(1-e^{-\alpha})^{-1} \end{align*} and \begin{align*} \|\int_t^0\psi(t)Y(t)P_2Y^{-1}(s)f(s)ds\| &\leqslant\int_t^0 e^{-\beta(s-t)}\|\psi(s)f(s)\|ds \\ &\leqslant\int_t^\infty e^{-\beta(s-t)}\|\psi(s)f(s)\|ds \leqslant c(1-e^{-\beta})^{-1}. \end{align*} It follows that the function $$\widetilde x(t)=\int_{-\infty}^t\psi(t)Y(t)P_1Y^{-1}(s)f(s)ds - \int_t^0\psi(t) Y(t)P_2Y^{-1}(s)f(s)ds$$ is bounded on $\mathbb{R}_-$. Hence the function \begin{align*} x(t)&=\psi^{-1}(t)\widetilde x(t) \\ &=\int_{-\infty}^t\psi(t)Y(t)P_1Y^{-1}(s)f(s)ds - \int_t^0\psi(t) Y(t)P_2Y^{-1}(s)f(s)ds \end{align*} is $\psi$-bounded on $\mathbb{R}_-$. On the other hand \begin{align*} x'(t)&=A(t)(\int_{-\infty}^tY(t)P_1Y^{-1}(s)f(s)ds -\int_t^0Y(t)P_2Y^{-1}(s)f(s)ds)\\ &\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 implies that $x(t)$ is a solution of \eqref{e1.1}. In the proof of only if part'': The set $$\widetilde C_\psi=\{ x:\mathbb{R}_-\to\mathbb{R}^d : x$$ is $\psi$-bounded and continuous on $\mathbb{R}_-\}$. It is a Banach space with the norm $\|x\|_{\widetilde C_\psi}=sup_{t\leqslant 0}\|\psi(t)x(t)\|$. The first step: we show that \eqref{e1.1} has a unique $\psi$-bounded solution $x(t)$ with $x(0)\in\widetilde X_1 = P_1\mathbb{R}^d$ for each $f\in \widetilde C_\psi$ and $\|x\|_{\widetilde C_\psi}\leqslant r\|f\|_{\widetilde C_\psi}$, here $r$ is a positive constant independent of $f$. The next steps of the proof are similar to the proof of \cite[Theorem 3.3]{b1}, with the corresponding replacement (for example replace $t\geqslant t_0\geqslant 0$ by $0\geqslant t_0 \geqslant t$, $P_1$ by $-P_2, P_2$ by $-P_1$, $\infty$ by $-\infty$, $- \infty$ by $\infty$, \dots ). \end{proof} \begin{theorem} \label{thm3.2} The equation \eqref{e1.1} has a unique $\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}$ if and only if \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$. \end{theorem} \begin{proof} First, we prove the if'' part. By Lemma \ref{lem2.3} and in the same way as in the proof of Theorem \ref{thm3.1}, the function $$x(t)=\int_{-\infty}^t Y(t)P_1Y^{-1}(s)f(s)ds -\int_t^\infty Y(t)P_2Y^{-1}(s)f(s)ds$$ is $\psi$-bounded and continuous on $\mathbb{R}$. Moreover, \begin{align*} x'(t)&=A(t)(\int_{-\infty}^t Y(t)P_1Y^{-1}(s)f(s)ds -\int_t^\infty Y(t)P_2Y^{-1}(s)f(s)ds)\\ &\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}. The uniqueness of the solution $x(t)$ result from \eqref{e1.2} having no nontrivial $\psi$-bounded solution on $\mathbb{R}$ (Remark \ref{rmk2.2}). Suppose that $y$ is a $\psi$-bounded solution of \eqref{e1.1} then $x - y$ is a $\psi$-bounded solution of \eqref{e1.2} on $\mathbb{R}$. We conclude $x = y$ since $x - y$ is the trivial solution of \eqref{e1.2}. We prove the only if ''part. Suppose that \eqref{e1.1} has unique $\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally bounded function $f$ on $\mathbb{R}$, we have to prove that \eqref{e1.1} has a $\psi$-exponential dichotomy on $\mathbb{R}$. By Lemma \ref{lem2.4}, Theorem \ref{thm1.4} and Theorem \ref{thm3.1} we get \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with a pair of the supplementary projections $P_1, P_2$ and has a $\psi$-exponential dichotomy on $\mathbb{R}_ -$. with a pair of the supplementary projections $Q_1, Q_2$. Remark \ref{rmk2.2} follows that $P_1\mathbb{R}^d$ is the subspace of $\mathbb{R}^d$ consisting of the values $x(0)$ of all $\psi$-bounded solutions $x(t)$ on $\mathbb{R}_+$ of \eqref{e1.2} and $Q_2\mathbb{R}^d$ is the subspace of $\mathbb{R}^d$ consisting of the values $x(0)$ of all $\psi$-bounded solutions $x(t)$ on $\mathbb{R}_ -$ of \eqref{e1.2}. We are going to prove that $$\label{e3.1} \mathbb{R}^d =P_1\mathbb{R}^d \oplus Q_2\mathbb{ R}^d .$$ For each $u \in\mathbb{R}^d$, denote by $x=x(t)$ the solution of \eqref{e1.2}, $x(0) =u$. By Lemma \ref{lem2.4} we get $x =x_1 + x_2$, where $x_1, x_2$ are the solutions of \eqref{e1.2} corresponding $\psi$-bounded on $\mathbb{R}_+$,$\mathbb{ R}_ -$. It follows from Remark \ref{rmk2.2} that $x_1(0) \in P_1\mathbb{R}^d$ and $x_2(0) \in Q_2\mathbb{R}^d$. It follows from $u = x_1(0) + x_2(0)$, that $$\label{e3.2} \mathbb{R}^d =P_1\mathbb{R}^d + Q_2\mathbb{ R}^d .$$ By hypothesis \eqref{e1.1} with $f = 0$ has unique $\psi$-bounded solution on $\mathbb{R}$ i.e. \eqref{e1.2} have no nontrivial $\psi$-bounded solution on $\mathbb{R}$. For any $v \in P_1\mathbb{R}^d\cap Q_2\mathbb{R}^d$, denote by $x(t)$ the solution of \eqref{e1.2} such that $x(0) = v$. Then $x(t)$ is the $\psi$-bounded solution of \eqref{e1.2}, it implies that $x(t)$ is the trivial solution. Hence $v = 0$. Consequently $$\label{e3.3} P_1\mathbb{R}^d\cap Q_2\mathbb{R}^d =0.$$ The relations \eqref{e3.2} and \eqref{e3.3} imply \eqref{e3.1}. Now, we prove the existence of a pair supplementary projections, for which \eqref{e1.1} has a $\psi$-exponential dichotomy on $\mathbb{R}$. Choose the projection $P$ of $\mathbb{R}^d$ such that $ImP = P_1\mathbb{R}^d$ , $\ker P = Q_2\mathbb{R}^d$. By Lemma 2.1, \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$, and have a $\psi$-exponential dichotomy on $\mathbb{R}_ -$ with the pair of the supplementary projections $P, I_d-P$. From Remark \ref{rmk1.3} it follows that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$ with the pair of the supplementary projections $P, I_d-P$. The proof is complete. \end{proof} \begin{theorem} \label{thm3.3} Suppose that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$. If $$\label{e3.4} \lim_{t\to\pm\infty}\int_t^{t+1}\|\psi(s)f(s)\|ds=0$$ then the $\psi$-bounded solution of \eqref{e1.1} is such that $$\label{e3.5} \lim_{t\to\pm\infty}\|\psi(t)x(t)\|=0.$$ \end{theorem} \begin{proof} By Theorem \ref{thm3.2}, the unique solution of \eqref{e1.1} is $$x(t) = \int_{-\infty}^tY(t)P_1Y^{-1}(s)f(s)ds -\int_t^\infty Y(t)P_2Y^{-1}(s)f(s)ds.$$ \label{e3.6} \begin{aligned} \|\psi(t)x(t)\| &\leqslant \int_{-\infty}^t\|\psi(t)Y(t)P_1Y^{-1}(s)f(s)\|ds +\int_t^\infty \|\psi(t)Y(t)P_2Y^{-1}(s)f(s)\|ds \\ & \leqslant K\int_{-\infty}^t e^{-\alpha(t-s)}\|\psi(s)f(s)\|ds + L\int_t^\infty e^{-\beta(s-t)}\|\psi(s)f(s)\|ds \\ & \leqslant K_1\{\int_{-\infty}^t e^{-\alpha(t-s)}\|\psi(s)f(s)\|ds +\int_t^\infty e^{-\beta(s-t)}\|\psi(s)f(s)\|ds\}, \end{aligned} where $K1 = \max\{K,L\}$. Denote by $\gamma = \min\{\alpha,\beta\}$. Under the hypothesis \eqref{e3.4}, for a given $\varepsilon > 0$, there exists $T > 0$ such that $\int_t^{t+1}\|\psi(s)f(s)\|ds < \frac{\varepsilon}{2K_1}(1-e^{-\gamma}) \quad \mbox{for } |t| > T.$ Then from Lemma \ref{lem2.3} and inequality \eqref{e3.6} it follow that \begin{align*} \|\psi(t)x(t)\|&\leqslant K_1\frac{\varepsilon}{2K_1}(1-e^{-\gamma}) [(1-e^{-\alpha})^{-1}+(1-e^{-\beta})^{-1}]\\ &\leqslant K_1\frac{\varepsilon}{2K_1}(1-e^{-\gamma})2(1-e^{-\gamma})^{-1} =\varepsilon\quad \mbox{for all } |t| > T , \end{align*} this implies \eqref{e3.5}. The proof is complete. \end{proof} \begin{corollary} \label{coro3.4} Suppose that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$. If $$\label{e3.7} \lim_{t\to\pm\infty}\|\psi(t)f(t)\|=0$$ then the $\psi$-bounded solution of \eqref{e1.1} is such that $$\label{e3.8} \lim_{t\to\pm\infty}\|\psi(t)x(t)\|=0.$$ \end{corollary} \begin{proof} It is easy to see that \eqref{e3.7} implies \eqref{e3.4} \end{proof} Now, we consider the perturbed equation $$\label{e3.9} {x'}(t)=[A(t)+B(t)]x(t)$$ where $B(t)$ is a $d\times d$ continuous matrix function on $\mathbb{R}$. We have the following result. \begin{theorem} \label{thm3.5} Suppose that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$. If $\delta=\sup_{t\in\mathbb{R}}\int_t^{t+1}|\psi(s)B(s)\psi^{-1}(s)|ds$ is sufficiently small, then \eqref{e3.9} has a $\psi$-exponential dichotomy on $\mathbb{R}$. \end{theorem} \begin{proof} By Theorem \ref{thm3.2} it suffices to show that the equation $${x'}(t)=[A(t)+B(t)]x(t)+f(t) \label{e3.10}$$ has a unique $\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally bounded $f$ function on $\mathbb{R}$. Denote by $G_\psi$ the set $$G_\psi=\{x:\mathbb{R}\to \mathbb{R}^d : x \mbox{ is \psi-bounded and continuous on }\mathbb{R}\}.$$ It is well-known that $G_\psi$ is a real Banach space with the norm $$\|x\|_{G_\psi}=\sup_{t\in R}\|\psi(t)x(t)\|.$$ Consider the mapping $T: {G_{\psi}}\to {G_{\psi}}$ which is defined by \begin{align*} Tz(t)&=\int_{-\infty}^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 $Tz\in G_{\psi}$. More ever if $z_1, z_2\in G_\psi$ then \begin{align*} &\|Tz_1-Tz_2\|_{G_\psi}\\ &\leqslant\int_{-\infty}^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 \end{align*} By Lemma \ref{lem2.3}, we have \begin{align*} \|Tz_1-Tz_2\|_{G_\psi} &\leqslant K\|z_1-z_2\|_{G_\psi}\int_{-\infty}^t e^{-\alpha(t-s)} |\psi(s)B(s)\psi^{-1}(s)|ds \\ &\quad + L\|z_1-z_2\|_{G_\psi}\int_t^{\infty} e^{\beta(t-s)}| \psi(s)B(s)\psi^{-1}(s)|ds \\ &\leqslant \delta[K(1-e^{-\alpha})^{-1}+L(1-e^{-\beta})^{-1}] \|z_1-z_2\|_{G_\psi} \end{align*} Hence, by the contraction principle, if $\delta[K(1-e^{-\alpha})^{-1}+L(1-e^{-\beta})^{-1}] < 1$, then the mapping $T$ has a unique fixed point. Denoting this fixed point by $z$, we have \begin{align*} z(t) &=\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)z(s)+f(s)]ds \\ &\quad - \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)z(s)+f(s)]ds. \end{align*} It follows that $z(t)$ is a solution on $\mathbb{R}$ of \eqref{e3.10}. Now, we prove the uniqueness of this solution. Suppose that $x(t)$ is a arbitrary $\psi$-bounded solution on $\mathbb{R}$ of \eqref{e3.10}. Consider the function \begin{align*} y(t) &=x(t)-\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)x(s)+f(s)]ds \\ &\quad + \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)x(s)+f(s)]ds. \end{align*} It is easy to see that $y(t)$ is a $\psi$-bounded solution on $\mathbb{R}$ of \eqref{e1.2}. Then from Theorem \ref{thm3.2} follows that $y(t)$ is the trivial solution. Then \begin{align*} x(t) &=\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)x(s)+f(s)]ds\\ &\quad - \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)x(s)+f(s)]ds. \end{align*} Hence $x(t)$ is the fixed point of mapping $T$. From the uniqueness of this point, it follows that $x=z$. The proof is complete. \end{proof} \begin{corollary} \label{coro3.6} Suppose that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$. If $\delta=\sup_{t\in\mathbb{R}}|\psi(t)B(t)\psi^{-1}(t)|$ is sufficiently small, then \eqref{e3.9} has a $\psi$-exponential dichotomy on $\mathbb{R}$. \end{corollary} \begin{thebibliography}{0} \bibitem{a1} O. Akinyele; \emph{On partial stability and boundedness of degree $k$}, Atti. Acad. Naz. Lincei Rend. Cl. Sei. Fis. Mat. Natur., (8), 65 (1978), 259-264. \bibitem{a2} C. Avramescu; \emph{Asupra comport$\tilde{a}$rii asimptotice a solutiilor unor ecuatii funcionable}, Analele Universit\~atii din Timisoara, Seria Stiinte Matamatice-Fizice, Vol. VI, 1968, 41-55. \bibitem{b1} P. N. Boi; \emph{On the $\psi$- dichotomy for homogeneous linear differential equations.} Electron. J. of Differential Equations, vol. 2006 (2006), No. 40, 1-12. \bibitem{c1} A. 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