\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 128, pp. 1--11.\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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/128\hfil Existence of $\psi$-bounded solutions] {A note on the existence of $\Psi$-bounded solutions for a system of differential equations on $\mathbb{R}$} \author[A. Diamandescu\hfil EJDE-2008/128\hfilneg] {Aurel Diamandescu} % in alphabetical order \address{Aurel Diamandescu \newline University of Craiova, Department of Applied Mathematics, 13, Al. I. Cuza'' st., 200585, Craiova, Romania} \email{adiamandescu@central.ucv.ro} \thanks{Submitted April 10, 2008. Published September 18, 2008.} \subjclass[2000]{34D05, 34C11} \keywords{$\Psi$-bounded solution; $\Psi$-boundedenss; boundedness} \begin{abstract} We prove a necessary and sufficient condition for the existence of $\Psi$-bounded solutions of a linear nonhomogeneous system of ordinary differential equations on $\mathbb{R}$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \section{Introduction} The aim of this paper is to give a necessary and sufficient condition so that the nonhomogeneous system of ordinary differential equations $$x' = A(t)x + f(t) \label{e1}$$ has at least one $\Psi$-bounded solution on $\mathbb{R}$ for every continuous and $\Psi$-bounded function f on $\mathbb{R}$. Here, $\Psi$ is a continuous matrix function on $\mathbb{R}$. The introduction of the matrix function $\Psi$ permits to obtain a mixed asymptotic behavior of the components of the solutions. The problem of boundedness of the solutions for the system \eqref{e1} was studied in \cite{c2} The problem of $\Psi$-boundedness of the solutions for systems of ordinary differential equations has been studied in many papers, as e.q. \cite{a1,c1,h1,m1}. The fact that in \cite{a1} the function $\Psi$ is a scalar continuous function and increasing, differentiable and such that $\Psi (t) \geq 1$ on $\mathbb{R}_{+}$ and $\lim_{t\to \infty }\Psi (t) = b \in \mathbb{R}_{+}$ does not enable a deeper analysis of the asymptotic properties of the solutions of a differential equation than the notions of stability or boundedness. In \cite{c1}, the function $\Psi$ is a scalar continuous function, nondecreasing and such that $\Psi$(t) $\geq$ 1 on $\mathbb{R}_{+}$. In \cite{h1,m1}, $\Psi$ is a scalar continuous function. In \cite{d1,d2,d3}, the author proposes a novel concept, $\Psi$-boundedness of solutions, $\Psi$ being a continuous matrix function, which is interesting and useful in some practical cases and presents the existence conditions for such solutions on $\mathbb{R}_{+}$. In \cite{b1}, the author associates this problem with the concept of $\Psi$-dichotomy on $\mathbb{R}$ of the system $x' = A(t)x$. Also, in \cite{h2}, the authors define $\Psi$-boundedness of solutions for difference equations via $\Psi$-bounded sequences and establish a necessary and sufficient condition for existence of $\Psi$-bounded solutions for a nonhomogeneous linear difference equation. Let $\mathbb{R}^d$ be the Euclidean $d$-space. For $x = (x_1, x_2, x_{3}, \dots, x_d)^T \in \mathbb{R}^d$, let $\|x\| = \max \{| x_1|$, $|x_2|, | x_{3}| ,\dots,|x_d| \}$ be the norm of $x$. For a $d\times d$ real matrix $A = (a_{ij}$), we define the norm $|A|$ by $|A| =\sup_{\| x\| \leq 1} \| Ax\|$. It is well-known that $|A| = \max_{1\leq i\leq d}\{\sum_{j=1}^d|a_{ij}| \}$. Let $\Psi _i: \mathbb{R}\to (0,\infty )$, $i =1,2,\dots, d$, be continuous functions and \begin{equation*} \Psi =\mathop{\rm diag}[\Psi _1,\Psi _2,\dots\Psi _d]. \end{equation*} \noindent\textbf{Definition.} A function $\varphi : \mathbb{R} \to \mathbb{R}^d$ is said to be $\Psi$-bounded on $\mathbb{R}$ if $\Psi \varphi$ is bounded on $\mathbb{R}$. By a solution of \eqref{e1}, we mean a continuously differentiable function $x :\mathbb{R}\to \mathbb{R}^d$ satisfying the system for all $t\in \mathbb{R}$. Let $A$ be a continuous $d\times d$ real matrix and the associated linear differential system $$y' = A(t)y. \label{e2}$$ Let $Y$ be the fundamental matrix of \eqref{e2} for which $Y(0) = I_d$ (identity $d\times d$ matrix). Let the vector space $\mathbb{R}^d$ be represented as a direct sum of three subspaces $X_{-}$, $X_0$, $X_{+}$ such that a solution $y(t)$ of \eqref{e2} is $\Psi$-bounded on $\mathbb{R}$ if and only if $y(0) \in X_0$ and $\Psi$-bounded on $\mathbb{R}_{+} = [0,\infty )$ if and only if $y(0) \in X_{-} \oplus X_0$. Also, let $P_{-}$, $P_0$, $P_{+}$ denote the corresponding projection of $\mathbb{R}^d$ onto $X_{-}$, $X_0$, $X_{+}$ respectively. \section*{Main result} We are now in position to prove our main result. \begin{theorem} \label{thm1} If $A$is a continuous $d\times d$ real matrix on $\mathbb{R}$, then, the system \eqref{e1} has at least one $\Psi$-bounded solution on $\mathbb{R}$ for every continuous and $\Psi$-bounded function $f : \mathbb{R}\to \mathbb{R}^d$ if and only if there exists a positive constant $K$ such that \label{e3} \begin{aligned} &\int_{-\infty }^t| \Psi (t)Y(t)P_{-}Y^{-1} (s)\Psi ^{-1}(s)| ds \\ &+ \int_t^0| \Psi (t)Y(t)(P_0 + P_{+})Y^{-1}(s)\Psi ^{-1}(s)| ds\\ &+ \int_0^{\infty }| \Psi (t)Y(t)P_{+}Y^{-1}(s)\Psi ^{-1}(s)| ds \leq K, \quad {for } t \geq 0, \end{aligned} and \begin{equation*} %\label{e3b} \begin{aligned} &\int_{-\infty }^{0}| \Psi (t)Y(t)P_{-}Y^{-1}(s)\Psi ^{-1}(s)| ds \\ &+ \int_0^t| \Psi(t)Y(t)(P_0 + P_{-})Y^{-1}(s)\Psi ^{-1}(s)| ds \\ &\int_t^{\infty }| \Psi (t)Y(t)P_{+}Y^{-1}(s)\Psi ^{-1}(s)| ds \leq K, \quad {for }t \geq 0. \end{aligned} \end{equation*} \end{theorem} \begin{proof} First, we prove the only if'' part. Suppose that the system \eqref{e1} has at least one $\Psi$-bounded solution on $\mathbb{R}$ for every continuous and $\Psi$-bounded function $f :\mathbb{R} \to \mathbb{R}^d$ on $\mathbb{R}$. We shall denote by $B$ the Banach space of all $\Psi$-bounded and continuous functions $x : \mathbb{R}\to \mathbb{R}^d$ with the norm $\|$x$\|_B = \sup_{t\in \mathbb{R}}\| \Psi (t)x(t)\|$. Let $D$ denote the set of all $\Psi$-bounded and continuously differentiable functions $x : \mathbb{R} \to \mathbb{R}^d$ such that $x(0) \in X_{-}\oplus X_{+}$ and $x'-Ax \in B$. Evidently, $D$ is a vector space. We define a norm in $D$ by setting $\| x\| _D = \| x\| _B +\| x' - Ax\| _B$. \noindent \textbf{Step 1.} $(D, \| \cdot \| _D)$ is a Banach space. Let $(x_n)_{n\in \mathbb{N}}$ be a fundamental sequence of elements of $D$. Then, $(x_n)_{n\in\mathbb{N}}$ is a fundamental sequence in $B$. Therefore, there exists a continuous and $\Psi$-bounded function $x :\mathbb{R}\to \mathbb{R}^d$ such that $\lim_{n\to \infty } \Psi (t)x_{n}(t) = \Psi (t)x(t)$, uniformly on $\mathbb{R}$. From the inequality \begin{equation*} \| x_n(t)- x(t)\| \leq | \Psi ^{-1}(t)| \| \Psi (t)x_n (t)- \Psi (t)x(t)\| , \end{equation*} it follows that $\lim_{n\to \infty } x_{n} (t) = x(t)$, uniformly on every compact of $\mathbb{R}$. Thus, $x(0) \in X_{-}\oplus X_{+}$. Similarly, $(x_n'- Ax_n)_{n\in \mathbb{N}}$ is a fundamental sequence in $B$. Therefore, there exists a continuous and $\Psi$-bounded function $f : \mathbb{R}\to \mathbb{R}^d$ such that \begin{equation*} \lim_{n\to \infty } \Psi (t)( x_{n}'(t)- A(t)x_n(t)) = \Psi (t)f(t), \quad\text{uniformly on }\mathbb{R}. \end{equation*} Similarly, \begin{equation*} \lim_{n\to \infty } ( x_{n}'(t)- A(t)x_n(t)) = f(t), \quad\text{uniformly on every compact subset of }\mathbb{R}. \end{equation*} For any fixed t $\in \mathbb{R}$, we have \begin{align*} x(t) - 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 \\ &=\int_0^t(f(s) + A(s)x(s)) ds. \end{align*} Hence, the function $x$ is continuously differentiable on $\mathbb{R}$ and \begin{equation*} x'(t) = A(t)x(t) + f(t), \quad t \in \mathbb{R}. \end{equation*} Thus, $x \in D$. On the other hand, from \begin{align*} \lim_{n\to \infty } \Psi (t)x_{n}(t) = \Psi (t)x(t),\quad\text{uniformly on }\mathbb{R}, \\ \lim_{n\to \infty } \Psi (t)(x_n'(t)-{A(t)x}_n(t)) = \Psi (t)( x'(t)-A(t)x(t)),\quad\text{uniformly on }\mathbb{R}, \end{align*} it follows that $\lim_{n\to \infty }\| x_n - x\| _D = 0$. This proves that $(D, \| \cdot \| _D)$ is a Banach space. \noindent\textbf{Step 2.} There exists a positive constant $K_0$ such that, for every $f \in B$ and for corresponding solution $x \in D$ of \eqref{e1}, we have \begin{equation*} \sup_{t\in \mathbb{R}} \| \Psi (t)x(t)\| \leq K_0 \sup_{t\in \mathbb{R}} \| \Psi (t)f(t)\| , \end{equation*} or $$\sup_{t\in \mathbb{R}} \max_{1\leq i\leq d} | \Psi _i (t)x_i(t)| \leq K_0 \sup_{t\in \mathbb{R}} \max_{1\leq i\leq d} |\Psi _i(t)f_i(t)| . \label{e4}$$ For this, define the mapping $T : D \to B$, $Tx = x' - Ax$. This mapping is obviously linear and bounded, with $\|T\|\leq 1$. Let $Tx = 0$. Then, $x'= Ax$, $x \in D$. This shows that $x$ is a $\Psi$-bounded solution on $\mathbb{R}$ of \eqref{e2}. Then, $x(0) \in X_0 \cap ( X_{-}\oplus X_{+}) = \{0\}$. Thus, $x = 0$, such that the mapping $T$ is one-to-one''. Finally, the mapping $T$ is onto''. In fact, for any $f \in B$, let $x$ be the $\Psi$-bounded solution on $\mathbb{R}$ of the system \eqref{e1} which exists by assumption. Let $z$ be the solution of the Cauchy problem $x' = A(t)x + f(t), \quad z(0) = (P_{-} + P_{+})x(0).$ Then, $u = x - z$ is a solution of \eqref{e2} with $u(0) = x(0) - (P_{-} + P_{+})x(0) = P_0x(0)$. From the Definition of $X_0$, it follows that $u$ is $\Psi$-bounded on $\mathbb{R}$. Thus, z belongs to D and Tz = f. Consequently, the mapping $T$ is onto''. From a fundamental result of $S. Banach$: If T is a bounded one-to-one linear operator of one Banach space onto another, then the inverse operator $T^{-1}$ is also bounded''. We have $\|$T$^{-1}f\| _D \leq \| T^{-1}\| \| f\| _B$, for all $f \in B$. For a given $f \in B$, let $x = T^{-1}f$ be the corresponding solution $x \in D$ of \eqref{e1}. We have $\| x\| _D = \| x \| _B + \| x' - Ax\| _B = \| x\| _B + \| f\| _B \leq \| T^{-1}\| \| f\| _B$. It follows that $\| x\| _B\leq K_0 \| f\| _B$, where $K_0 = \| T^{-1}\| -1$, which is equivalent with \eqref{e4}. \noindent\textbf{Step 3.} The end of the proof. Let $T_1 < 0 < T_2$ be fixed points but arbitrarily and let $f : \mathbb{R} \to \mathbb{R}^{d}$ be a continuous and $\Psi$-bounded function which vanishes on $(-\infty , T_1]\cup [T_2, +\infty )$. It is easy to see that the function $x : \mathbb{R} \to \mathbb{R}^d$ defined by $x(t) = \begin{cases} -\int_{ T_1}^{0} Y(t)P_0Y^{-1}(s)f(s)ds -\int_{ T_1}^{ T_2} Y(t)P_{+}Y^{-1}(s)f(s)ds, & t < T_1 \\ \int_{ T_1}^tY(t)P_{-}Y^{-1}(s)f(s)ds + \int_0^t Y(t)P_0Y^{-1}(s)f(s)ds \\ -\int_t^{ T_2}Y(t)P_{+}Y^{-1}(s)f(s)ds, & T_1 \leq t \leq T_2 \\ \int_{ T_1}^{ T_2}Y(t)P_{-}Y^{-1} (s)f(s)ds + \int_0^{ T_2}Y(t)P_0Y ^{-1} (s)f(s)ds, & t > T_2 \end{cases}$ is the solution in $D$ of the system \eqref{e1}. Putting $G(t,s) = \begin{cases} Y(t)P_{-}Y^{-1}(s), & t > 0, s \leq { 0} \\ {Y(t)(P}_0{ + P}_{-}{)Y}^{-1}{(s),} & {t> 0, s }>{ 0, s < t} \\ -Y(t)P_{+}Y^{-1}{(s),} & {t > 0, s }> { 0, s }\geq { t} \\ Y(t)P_{-}Y^{-1}{(s),} & {t }\leq { 0, s < t} \\ -{Y(t)(P}_0{ + P}_{+}{)Y}^{-1}{(s),} & {t }\leq { 0, s }\geq { t, s }<{ 0} \\ -Y(t)P_{+}Y^{-1}{(s),} & {t }\leq { 0, s }\geq { t, s }\geq { 0} \end{cases}$ we have that $x(t) = \int_{ T_1}^{ T_2} G(t,s)f(s)ds$, $t \in \mathbb{R}$. Indeed, \noindent$\bullet$ for $t > T_2$, we have \begin{align*} \int_{T_1}^{T_2} G(t,s)f(s)ds &= \int_{T_1}^0 Y(t)P_{-}Y^{-1}(s)f(s)ds + \int_0^{T_2} Y(t)(P_0 + P_{-})Y ^{-1}(s)f(s)ds \\ &=\int_{T_1}^{T_2} Y(t)P_{-}Y^{-1}(s)f(s)ds +\int_0^{T_2} Y(t)P_0Y^{-1}(s)f(s)ds = x(t), \end{align*} \noindent$\bullet$ for $t \in (0,T_2]$, we have \begin{align*} \int_{T_1}^{T_2}G(t,s)f(s)ds &= \int_{T_1}^0 Y(t)P_{-}Y^{-1}(s)f(s)ds + \int_0^tY(t)(P_0 + P_{-})Y^{-1} (s)f(s)ds\\ &\quad -\int_t^{T_2}Y(t)P_{+}Y^{-1}(s)f(s)ds \\ &=\int_{T_1}^tY(t)P_{-}Y^{-1} (s)f(s)ds + \int_0^t Y(t)P_0Y^{-1}(s)f(s)ds\\ &\quad -\int_t^{T_2}Y(t)P_{+}Y^{-1}(s)f(s)ds = x(t), \end{align*} \noindent$\bullet$ for $t \in [T_1,0]$, we have \begin{align*} \int_{T_1}^{ T_2}G(t,s)f(s)ds &= \int_{T_1}^tY(t)P_{-}Y^{-1}(s)f(s)ds -\int_t^0Y(t)(P_0 + P_{+} )Y^{-1}(s)f(s)ds \\ &\quad- \int_0^{ T_2} Y(t)P_{+}Y^{-1}(s)f(s)ds \\ &= \int_{T_1}^tY(t)P_{-}Y^{-1} (s)f(s)ds + \int_0^tY(t)P_0Y^{-1}(s)f(s)ds \\ &- \int_t^{T_2}Y(t)P_{+}Y^{-1}(s)f(s)ds = x(t), \end{align*} \noindent$\bullet$ for $t < T_1$, we have \begin{align*} &\int_{T_1}^{T_2}G(t,s)f(s)ds\\ &= - \int_{T_1}^0Y(t)(P_0 + P_{+})Y^{-1} (s)f(s)ds - \int_0^{T_2}Y(t)P_{+}Y^{-1}(s)f(s)ds \\ &= -\int_{T_1}^{0}Y(t)P_0Y^{-1}(s)f(s)ds -\int_{T_1}^{T_2}Y(t)P_{+}Y^{-1}(s)f(s)ds = x(t). \end{align*} Now, putting $\Psi (t)G(t,s)\Psi ^{-1}(s) =( {G}_{{ij}} {(t,s)})$, inequality \eqref{e4} becomes $\big| \int_{T_1}^{T_2} \sum_{k=1}^d {G}_{ik}{(t,s) \Psi _k (s)f_k (s)}\,ds\big| \leq K_0\sup_{t\in \mathbb{R}} \max_{1\leq i\leq d} | \Psi _i(t)f_i(t)|,\quad t \in \mathbb{R},$ $i = 1,2,\dots, d$, for every $f = (f_1,f_2,\dots,f_d) : \mathbb{R} \to \mathbb{R}^d$, continuous and $\Psi$-bounded, which vanishes on ($-\infty , T_1] \cup [T_2, +\infty$). For a fixed $i$ and $t$, we consider the function $f$ such that \begin{equation*} {f}_k (s) = \begin{cases} \Psi _k ^{-1}(s)\mathop{\rm sgn} G_{{ik}}{(t,s),} & T_1 \leq { s }\leq { T}_2 \\ 0, & \text{elsewhere} \end{cases} \end{equation*} The function $\Psi _k (s)f_k (s)$ is pointwise limit of a sequence of continuous functions having the same supremum 1. The above inequality continues to hold for the functions of this sequence. By the dominated convergence Theorem, we get \begin{equation*} \int_{ T_1}^{ T_2}\sum_{{k=1}}^{{ d}}| {G}_{{ik}}{(t,s)}| {ds }\leq K_0 {, t }\in \mathbb{R},\quad i = 1,2,\dots, d. \end{equation*} Since $| \Psi (t)G(t,s)\Psi ^{-1}(s)| \leq \sum_{{i, k=1}}^d| G_{{ik}}(t,s)|$, it follows that \begin{equation*} \int_{ T_1}^{ T_2}| \Psi {(t)G(t,s)}\Psi ^{-1}(s)| {ds }\leq { d} {K}_0. \end{equation*} This holds for any $T_1 < 0$ and $T_2 > 0$. Hence, $| \Psi (t)G(t,s)\Psi ^{-1}(s)|$ is integrable over $\mathbb{R}$ and \begin{equation*} \int_{-\infty }^{\infty }| \Psi {(t)G(t,s)}\Psi ^{-1}{ (s)}| {ds }\leq { d} {K}_0,\quad{for all }t \in \mathbb{R}. \end{equation*} By the Definition of $\Psi (t)G(t,s)\Psi ^{-1}(s)$, this is equivalent to \eqref{e3}, with $K = d K_0$. Now, we prove the if'' part. Suppose that the fundamental matrix $Y$ of \eqref{e2} satisfies the conditions \eqref{e3} for some $K > 0$. For a continuous and $\Psi$-bounded function $f : \mathbb{R}\to \mathbb{R}^d$, we consider the function $u : \mathbb{R} \to \mathbb{R}^d$, defined by \begin{aligned} u(t) &= \int_{-\infty }^tY(t)P_{-}Y ^{-1}(s)f(s)ds \\ &\quad + \int_0^tY(t)P_0Y ^{-1}{(s)f(s)ds }- \int_t^{\infty }Y(t)P_{+}Y^{-1}{(s)f(s)ds.} \end{aligned} \label{e5} \noindent\textbf{Step 4.} The function $u$ is well-defined on $\mathbb{R}$. For v $\geq t$, we have \begin{align*} &\int_t^{{v}}\| Y(t)P_{+}Y^{-1}(s)f(s)\| ds\\ &=\int_t^{{v}}\| \Psi ^{-1}(t)\Psi (t)Y(t)P_{+}Y^{-1}(s) \Psi ^{-1}(s)\Psi (s)f(s)\| ds \\ &\leq | \Psi ^{-1}(t)| \int_t^{{v}}| \Psi (t)Y(t)P_{+}Y^{-1}(s) \Psi ^{-1}(s)| \|\Psi (s)f(s)\| ds \\ &\leq | \Psi ^{-1}(t)| \sup_{{s}\in \mathbb{R}} \| \Psi (s)f(s)\| \int_{{t}}^{{v}}| \Psi (t)Y(t)P_{+}Y^{-1}(s) \Psi ^{-1}(s)| ds. \end{align*} This shows that the integral $\int_t^{\infty }Y(t)P_{+}Y^{-1}(s)f(s)ds$ is absolutely convergent. Similarly, the integral $\int_{-\infty }^tY(t)P_{-}Y^{-1}(s)f(s)ds$ is absolutely convergent. Thus, the function $u$ is continuously differentiable on $\mathbb{R}$. \noindent\textbf{Step 5.} The function $u$ is a solution of the equation \eqref{e1}. For $t \in \mathbb{R}$, we have \begin{align*} u'(t) &= \int_{-\infty }^t A(t)Y(t)P_{-}Y^{-1}(s)f(s)ds + Y(t)P_{-}Y^{-1}(t)f(t)\\ &\quad +\int_0^tA(t)Y(t)P_0Y^{-1} (s)f(s)ds + Y(t)P_0Y^{-1}(t)f(t) \\ &\quad - \int_t^{\infty }A(t)Y(t)P_{+}Y ^{-1} (s)f(s)ds + Y(t)P_{+}Y^{-1}(t)f(t) \\ &= A(t)u(t) + Y(t)(P_{-} + P_0 + P_{+})Y^{-1} (t)f(t) \\ &= A(t)u(t) + f(t), \end{align*} which shows that $u$ is a solution of \eqref{e1} on $\mathbb{R}$. \noindent\textbf{Step 6.} The solution $u$ is $\Psi$-bounded on $\mathbb{R}$. For $t \geq 0$, we have \begin{align*} \Psi (t)u(t) &= \int_{-\infty }^t\Psi (t)Y(t)P_{-}Y^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\\ &\quad + \int_0^t\Psi (t)Y(t)P_0Y^{-1} (s)\Psi ^{-1}(s)\Psi (s)f(s)ds \\ &\quad - \int_t^{\infty }\Psi (t)Y(t)P_{+}Y ^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds \\ &= \int_{-\infty }^0\Psi (t)Y(t)P_{-}Y^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\\ &\quad + \int_0^t\Psi (t)Y(t)( P_0 + P_{-})Y^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\\ &\quad -\int_t^{\infty }\Psi (t)Y(t)P_{+}Y ^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\,. \end{align*} Then \begin{equation*} \| \Psi {(t)u(t)}\| \leq { K }\sup_{t\in \mathbb{R}} \| \Psi {(t)f(t)}\| . \end{equation*} For t $<$ 0, we have \begin{align*} \Psi (t)u(t)& = \int_{-\infty }^t\Psi (t)Y(t)P_{-}Y^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\\ &\quad + \int_0^t\Psi (t)Y(t)P_0Y^{-1} (s)\Psi ^{-1}(s)\Psi (s)f(s)ds\\ &\quad - \int_t^{\infty }\Psi (t)Y(t)P_{+}Y ^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds \\ &=\int_{-\infty }^t\Psi (t)Y(t)P_{-}Y ^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\\ &\quad -\int_t^0\Psi (t)Y(t)( P_0 + P_{+})Y^{-1}(s)\Psi ^{-1}(s) \Psi (s)f(s)ds \\ &\quad -\int_0^{\infty }\Psi (t)Y(t)P_{+}Y ^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds\,. \end{align*} Then \begin{equation*} \| \Psi {(t)u(t)}\| \leq { K } \sup_{t\in \mathbb{R}} \| \Psi {(t)f(t)}\| . \end{equation*} Hence, \begin{equation*} \sup_{t\in \mathbb{R}} \| \Psi {(t)u(t)} \| \leq { K } \sup_{t\in \mathbb{R}} \| \Psi {(t)f(t)}\| , \end{equation*} which shows that $u$ is a $\Psi$-bounded solution on $\mathbb{R}$ of \eqref{e1}. The proof is now complete. \end{proof} As a particular case, we have the following result. \begin{theorem} \label{thm2} If the homogeneous equation \eqref{e2} has no nontrivial $\Psi$-bounded solution on $\mathbb{R}$, then, the equation \eqref{e1} has a unique $\Psi$-bounded solution on $\mathbb{R}$ for every continuous and $\Psi$-bounded function $f : \mathbb{R} \to \mathbb{R}^d$ if and only if there exists a positive constant $K$ such that for $t \in \mathbb{R}$, $$\int_{-\infty }^t| \Psi (t)Y(t)P_{-}Y^{-1}(s) \Psi ^{-1}(s)| ds +\int_t^{\infty }| \Psi (t)Y(t)P_{+}Y^{-1}(s)\Psi ^{-1}(s)| ds \leq K \label{e6}$$ \end{theorem} \begin{proof} Indeed, in this case, $P_0 = 0$. Now, the Proof goes in the same way as before. We prove finally a theorem in which we will see that the asymptotic behavior of the solutions of \eqref{e1} is determined completely by the asymptotic behavior of $f$ as $t \to \pm \infty$. \end{proof} \begin{theorem} \label{thm3} Suppose that: \begin{enumerate} \item The fundamental matrix $Y(t)$ of \eqref{e2} satisfies:\\ (a) conditions \eqref{e3} for some $K > 0$;\\ (b) the condition $\lim_{t\to \pm \infty }| \Psi (t)Y(t)P_0| = 0$; \item the continuous and $\Psi$-bounded function $f : \mathbb{R}\to \mathbb{R}^d$ is such that $$\lim_{t\to \pm\infty } \| \Psi (t)f(t)\| = 0.$$ \end{enumerate} Then, every $\Psi$-bounded solution $x$ of \eqref{e1} satisfies \begin{equation*} \lim_{t\to \pm \infty } \| \Psi{(t)x(t)}\| { = 0.} \end{equation*} \end{theorem} \begin{proof} By Theorem \ref{thm1}, for every continuous and $\Psi$-bounded function $f : \mathbb{R} \to \mathbb{R}^d$, the equation \eqref{e1} has at least one $\Psi$-bounded solution. Let $x$ be a $\Psi$-bounded solution of \eqref{e1}. Let $u$ be defined by \eqref{e5}. This function is a $\Psi$-bounded solution of \eqref{e1}. Now, let the function $y(t) = x(t) - Y(t)P_0x(0) - u(t)$, $t \in \mathbb{R}$. Obviously, $y$ is a $\Psi$-bounded solution on $\mathbb{R}$ of \eqref{e2}. Thus, $y(0) \in X_0$. On the other hand, \begin{align*} y(0) &= x(0) - Y(0)P_0x(0) - u(0) \\ &= (I - P_0)x(0) - P_{-}\int_{-\infty }^{{0}}Y^{-1}(s)f(s)ds + P_{+}\int_0^{\infty }Y^{-1}(s)f(s)ds \\ &=P_{-}(x(0) - \int_{-\infty }^0Y^{-1}(s)f(s)ds) \\ &\quad + P_{+}(x(0) + \int_0^{\infty }Y^{-1}(s)f(s)ds) \in X_{-} \oplus X_{+}. \end{align*} Therefore, $y(0) \in X_0\cap (X_{-}\oplus X_{+}) = \{0\}$ and then, $y = 0$. It follows that \begin{equation*} x(t) = Y(t)P_0x(0) + u(t), \quad t \in \mathbb{R}. \end{equation*} We prove that $\lim_{t\to \pm \infty }\|\Psi (t)u(t)\| = 0$. For a given $\varepsilon > 0$, there exists $t_1> 0$ such that $\| \Psi (t)f(t)\| < \frac{\varepsilon }{{3K}}$, for all $t \geq t_1$. For $t > 0$, write \begin{align*} \Psi (t)u(t) &= \int_{-\infty }^0\Psi (t)Y(t)P_{-}Y^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds \\ &\quad + \int_0^t\Psi (t)Y(t)( P_0 + P_{-})Y^{-1}(s) \Psi ^{-1} (s) \Psi (s)f(s)ds \\ &\quad - \int_t^{\infty }\Psi (t)Y(t)P_{+}Y ^{-1}(s)\Psi ^{-1}(s)\Psi (s)f(s)ds. \end{align*} From the hypothesis (1)(a), it follows that $\int_0^t| \Psi (t)Y(t)( P_0 + P_{-})Y^{-1}(s)\Psi ^{-1}(s)| ds \leq K, t \geq 0.$ From the \cite[Lemma 1]{d4}, it follows that $\lim_{t\to +\infty } | \Psi (t)Y(t)( P_0 + P_{-})| = 0.$ From this and from hypothesis (1)(b), it follows that $\lim_{t\to +\infty } | \Psi (t)Y(t)P_{-}| = 0$. Thus, there exists $t_2 \geq t_1$ such that, for all $t \geq t_2$, \begin{gather*} | \Psi (t)Y(t)P_{-}| < \frac{\varepsilon }{3\big( 1+\int_{-\infty }^0\| {P }_{-}Y^{-1}{(s)f(s)}\| {ds}\big) }, \\ | \Psi (t)Y(t)(P_0 + P_{-})| < \frac{\varepsilon }{3\big(1+\int_0^{t_1}\| Y^{-1}{(s)f(s)}\| {ds}\big) }. \end{gather*} Then, for $t \geq t_2$, we have \begin{align*} \| \Psi (t)u(t)\| &\leq | \Psi (t)Y(t)P_{-}| \int_{-\infty }^0\| P _{-} Y^{-1}(s)f(s)\| ds \\ &\quad + | \Psi (t)Y(t)(P_0 + P_{-})| \int_0^{t_1}\|Y^{-1}(s)f(s)\| ds \\ &\quad + \int_{t_1}^t| \Psi (t)Y(t)(P_0 + P_{-}) Y^{-1}(s)\Psi ^{-1}(s)| \| \Psi (s)f(s)\| ds \\ &\quad + \int_t^{\infty }| \Psi (t)Y(t)P_{+} Y^{-1}(s)\Psi ^{-1}(s)| \| \Psi (s)f(s)\| ds\\ &< \frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3{K}}\int_{t _1}^t| \Psi (t)Y(t)(P_0 + P_{-})Y^{-1}(s)\Psi ^{-1}(s)| ds \\ &\quad+ \frac{\varepsilon }{3{K}}\int_{t }^{\infty }| \Psi (t)Y(t)P_{+}Y^{-1}(s)\Psi ^{-1}(s)| \| \Psi (s)f(s)\| ds \\ &\leq \frac{2\varepsilon }{3}+\frac{ \varepsilon }{3{K}}K = \varepsilon . \end{align*} This shows that $\lim_{t\to +\infty }\|\Psi (t)u(t)\| = 0$. Now, from hypothesis (1)(b) it follows that $\lim_{t\to +\infty } \| \Psi (t)Y(t)P_0x(0)\| = 0$ and then, $\lim_{t\to +\infty }\|\Psi (t)x(t)\| = 0$. Similarly, $\lim_{t\to -\infty }\| \Psi (t)x(t)\| = 0$. The proof is now complete. \end{proof} \begin{corollary} \label{coro1} Suppose that: \begin{enumerate} \item The homogeneous equation \eqref{e2} has no nontrivial $\Psi$-bounded solution on $\mathbb{R}$; \item the fundamental matrix $Y$ of \eqref{e2} satisfies the condition \eqref{e6} for some $K > 0$; \item the continuous and $\Psi$-bounded function $f : \mathbb{R}\to \mathbb{R}^d$ is such that \begin{equation*} \lim_{t\to \pm \infty }\| \Psi {(t)f(t)}\| { = 0.} \end{equation*} \end{enumerate} Then, the equation \eqref{e1} has a unique solution $x$ on $\mathbb{R}$ such that \begin{equation*} \lim_{t\to \pm \infty } \| \Psi {(t)x(t)}\| = 0. \end{equation*} \end{corollary} The above result follows from the Theorems \ref{thm2} and \ref{thm3}. Furthermore, this unique solution of \eqref{e1} is $u(t) = \int_{-\infty }^tY(t)P_{-} Y^{-1}(s)f(s)ds - \int_t^{\infty }Y(t)P_{+}Y ^{-1} (s)f(s)ds.$ \begin{remark} \label{rmk1} \rm If we do not have $\lim_{t\to \pm \infty } \| \Psi (t)f(t)\| = 0$, then the solution $x$ may be such that $\lim_{t\to \pm \infty }\| \Psi (t)x(t)\| \neq 0$. This is shown by the next example: Consider the linear system \eqref{e1} with \begin{equation*} {A(t) = }\begin{pmatrix} 2 & 0 \\ 0 & -3 \end{pmatrix}, \quad f(t) = \begin{pmatrix} {e}^{{3t}} \\ {e}^{-{4t}} \end{pmatrix} \end{equation*} A fundamental matrix for the homogeneous system \eqref{e2} is \begin{equation*} {Y(t) = }\begin{pmatrix} {e}^{{2t}} & 0 \\ 0 & {e}^{-{3t}} \end{pmatrix} \end{equation*} Consider \begin{equation*} \Psi {(t) = }\begin{pmatrix} {e}^{-{3t}} & 0 \\ 0 & {e}^{{4t}} \end{pmatrix} . \end{equation*} Then, we have $\| \Psi (t)f(t)\| = 1$ for all $t \in \mathbb{R}$. The first condition of Theorem \ref{thm3} is satisfied with $K = 2$ and $P_{-} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad P_0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \quad P_{+} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\,.$ The solutions of the system \eqref{e1} are ${x(t) = }\begin{pmatrix} {c}_1{e}^{{2t}}{ + e}^{{3t}} \\ {c}_2{e}^{-{3t}} -{ e}^{-{4t}} \end{pmatrix}$ with $c_1, c_2 \in \mathbb{R}$ and $t\in \mathbb{R}$. There exists a unique $\Psi$-bounded solution on $\mathbb{R}$, \begin{equation*} {x(t) = }\begin{pmatrix} {e}^{{3t}} \\ -{e}^{-{4t}} \end{pmatrix} , \end{equation*} but $\lim_{t\to \pm \infty }\|\Psi (t)x(t)\| = 1$. \end{remark} \begin{thebibliography}{00} \bibitem{a1} Akinyele, O.; \emph{On partial stability and boundedness of degree k,} Atti. Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur., (8), 65(1978), 259 - 264. \bibitem{b1} Boi, P. 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