\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 80, 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} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/80\hfil Hyers-Ulam stability] {Laplace transform and generalized Hyers-Ulam stability of linear differential equations} \author[Q. H. Alqifiary, S.-M. Jung \hfil EJDE-2014/80\hfilneg] {Qusuay H. Alqifiary, Soon-Mo Jung} % in alphabetical order \address{Qusuay H. Alqifiary \newline Department of Mathematics, University of Belgrade, Belgrade, Serbia. \newline University of Al-Qadisiyah, Al-Diwaniya, Iraq} \email{qhaq2010@gmail.com} \address{Soon-Mo Jung \newline Mathematics Section, College of Science and Technology, Hongik University, 339--701 Sejong, Korea} \email{smjung@hongik.ac.kr} \thanks{Submitted March 5, 2014. Published March 21, 2014.} \subjclass[2000]{44A10, 39B82, 34A40, 26D10} \keywords{Laplace transform method; differential equations; \hfill\break\indent generalized Hyers-Ulam stability} \begin{abstract} By applying the Laplace transform method, we prove that the linear differential equation $$ y^{(n)}(t)+\sum_{k=0}^{n-1}{\alpha_k y^{(k)}(t)}=f(t) $$ has the generalized Hyers-Ulam stability, where $\alpha_k$ is a scalar, $y$ and $f$ are $n$ times continuously differentiable and of exponential order. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} In 1940, Ulam \cite{Ka} posed a problem concerning the stability of functional equations: ``Give conditions in order for a linear function near an approximately linear function to exist.'' A year later, Hyers \cite{r3} gave an answer to the problem of Ulam for additive functions defined on Banach spaces: Let $X_1$ and $X_2$ be real Banach spaces and $\varepsilon > 0$. Then for every function $f : X_1 \to X_2$ satisfying $$ \| f(x+y) - f(x) - f(y) \| \leq \varepsilon \quad (x, y \in X_1), $$ there exists a unique additive function $A : X_1 \to X_2$ with the property $$ \| f(x) - A(x) \| \leq \varepsilon \quad (x \in X_1). $$ After Hyers's result, many mathematicians have extended Ulam's problem to other functional equations and generalized Hyers's result in various directions (see \cite{Cz,HyIsRa,Jud,Kc}). A generalization of Ulam's problem was recently proposed by replacing functional equations with differential equations: The differential equation $\varphi(f,y,y', \dots,y^{(n)}) = 0$ has Hyers-Ulam stability if for a given $\varepsilon > 0$ and a function $y$ such that $| \varphi(f, y, y', \dots, y^{(n)}) |$ $\leq \varepsilon$, there exists a solution $y_a$ of the differential equation such that $| y(t) - y_a(t) | \leq K(\varepsilon)$ and $\lim_{\varepsilon \to 0} K(\varepsilon) = 0$. If the preceding statement is also true when we replace $\varepsilon$ and $K(\varepsilon)$ by $\varphi(t)$ and $\Phi(t)$, where $\varphi,\, \Phi$ are appropriate functions not depending on $y$ and $y_a$ explicitly, then we say that the corresponding differential equation has the generalized Hyers-Ulam stability (or Hyers-Ulam-Rassias stability). Ob\a{l}oza seems to be the first author who has investigated the Hyers-Ulam stability of linear differential equations (see \cite{ob1,ob2}). Thereafter, Alsina and Ger published their paper \cite{AlGe}, which handles the Hyers-Ulam stability of the linear differential equation $y'(t) = y(t)$: If a differentiable function $y(t)$ is a solution of the inequality $| y'(t) - y(t) | \leq \varepsilon$ for any $t \in (a, \infty)$, then there exists a constant $c$ such that $| y(t) - ce^t | \leq 3\varepsilon$ for all $t \in (a, \infty)$. Those previous results were extended to the Hyers-Ulam stability of linear differential equations of first order and higher order with constant coefficients in \cite{MMT1,TMM,TTMM} and in \cite{MMT2}, respectively. Furthermore, Jung has also proved the Hyers-Ulam stability of linear differential equations (see \cite{r19,r21,r20}). Rus investigated the Hyers-Ulam stability of differential and integral equations using the Gronwall lemma and the technique of weakly Picard operators (see \cite{Ru1,Ru2}). Recently, the Hyers-Ulam stability problems of linear differential equations of first order and second order with constant coefficients were studied by using the method of integral factors (see \cite{LiSh,WZS}). The results given in \cite{r21,LiSh,MMT1} have been generalized by Cimpean and Popa \cite{CiPo} and by Popa and Ra\c{s}a \cite{popa1,popa2} for the linear differential equations of $n$th order with constant coefficients. Recently, Rezaei, Jung and Rassias have proved the Hyers-Ulam stability of linear differential equations by using the Laplace transform method (see \cite{Ke}). In this paper, by using the Laplace transform method, we prove that the linear differential equation of the $n$th order $$ y^{(n)}(t) + \sum_{k=0}^{n-1} {\alpha_k y^{(k)}(t)} = f(t) $$ has the generalized Hyers-Ulam stability, where $\alpha_k$ is a scalar, $y$ and $f$ are $n$ times continuously differentiable and of exponential order, respectively. \section{Preliminaries} Throughout this paper, $\mathbb{F}$ will denote either the real field $\mathbb{R}$ or the complex field $\mathbb{C}$. A function $f : (0, \infty) \to \mathbb{F}$ is said to be of exponential order if there are constants $A, B \in \mathbb{R}$ such that $$ | f(t) | \leq Ae^{tB} $$ for all $t > 0$. For each function $f : (0, \infty) \to \mathbb{F}$ of exponential order, we define the Laplace transform of $f$ by $$ F(s) = \int_{0}^{\infty} f(t) e^{-st}dt. $$ There exists a unique number $-\infty \leq \sigma < \infty$ such that this integral converges if $\Re(s) > \sigma$ and diverges if $\Re(s) < \sigma$, where $\Re(s)$ denotes the real part of the (complex) number $s$. The number $\sigma$ is called the abscissa of convergence and denoted by $\sigma_{f}$. It is well known that $| F(s) | \to 0$ as $\Re(s) \to \infty$. Furthermore, $f$ is analytic on the open right half plane $\{ s \in \mathbb{C}: \Re(s)> \sigma \}$ and we have $$ \frac{d}{ds} F(s) = -\int_{0}^{\infty} te^{-st} f(t) dt \quad (\Re(s)> \sigma). $$ The Laplace transform of $f$ is sometimes denoted by $\mathcal{L}(f)$. It is well known that $\mathcal{L}$ is linear and one-to-one. Conversely, let $f(t)$ be a continuous function whose Laplace transform $F(s)$ has the abscissa of convergence $\sigma_{f}$, then the formula for the inverse Laplace transforms yields $$ f(t) = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{\alpha-iT}^{\alpha+iT} F(s) e^{st} ds = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{(\alpha+iy)t} F(\alpha+iy) dy $$ for any real constant $\alpha > \sigma_{f}$, where the first integral is taken along the vertical line $\Re(s) = \alpha$ and converges as an improper Riemann integral and the second integral is used as an alternative notation for the first integral (see \cite{Da}). Hence, we have \begin{gather*} \mathcal{L}(f)(s) = \int_{0}^{\infty} f(t) e^{-st} dt \quad (\Re(s) > \sigma_{f}) \\ \mathcal{L}^{-1}(F)(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{(\alpha+iy)t} F(\alpha+iy) dy \quad (\alpha > \sigma_{f}). \end{gather*} The convolution of two integrable functions $f, g : (0,\infty) \to \mathbb{F}$ is defined by $$ (f*g)(t) := \int_0^t {f(t - x) g(x) dx}. $$ Then $\mathcal{L}(f*g) = \mathcal{L}(f) \mathcal{L}(g)$. \begin{lemma}[\cite{Ke}] \label{lem1} Let $P(s) = \sum_{k=0}^n {\alpha_k s^k}$ and $Q(s) = \sum_{k=0}^m {\beta_k s^k}$, where $m, n$ are nonnegative integers with $m < n$ and $\alpha_k, \beta_k$ are scalars. Then there exists an infinitely differentiable function $g : (0,\infty) \to \mathbb{F}$ such that $$ \mathcal{L}(g) = \frac{Q(s)}{P(s)} \quad ( \Re(s) > \sigma_{{}_P} ) $$ and $$ g^{(i)}(0) = \begin{cases} 0 & \text{for } i \in \{ 0, 1, \dots, n-m-2 \},\\ \beta_m / \alpha_n & \text{for } i = n-m-1 \end{cases} $$ where $\sigma_{{}_P} = \max \{ \Re(s) : P(s) = 0 \}$. \end{lemma} \begin{lemma}[\cite{Ke}] \label{lem2} Given an integer $n > 1$, let $f : (0,\infty) \to \mathbb{F}$ be a continuous function and let $P(s)$ be a complex polynomial of degree $n$. Then there exists an $n$ times continuously differentiable function $h : (0,\infty) \to \mathbb{F}$ such that $$ \mathcal{L}(h) = \frac{\mathcal{L}(f)}{P(s)} \quad (\Re(s) > \max\{ \sigma_{{}_P}, \sigma_{f} \} ), $$ where $\sigma_{{}_P} = \max\{ \Re(s) : P(s) = 0 \}$ and $\sigma_{f}$ is the abscissa of convergence for $f$. In particular, it holds that $h^{(i)}(0) = 0$ for every $i \in \{ 0, 1, \dots, n-1 \}$. \end{lemma} \section{Main Results} Let $\mathbb{F}$ denote either $\mathbb{R}$ or $\mathbb{C}$. In the following theorem, using the Laplace transform method, we investigate the generalized Hyers-Ulam stability of the linear differential equation of first order \begin{equation} \label{eq:1} y'(t) + \alpha y(t) = f(t). \end{equation} \begin{theorem}\label{thm:3.1} Let $\alpha$ be a constant in $\mathbb{F}$ and let $\varphi : (0,\infty) \to (0,\infty)$ be an integrable function. If a continuously differentiable function $y : (0,\infty) \to \mathbb{F}$ satisfies the inequality \begin{equation} \label{eq:2} | y'(t) + \alpha y(t) - f(t) | \leq \varphi(t) \end{equation} for all $t > 0$, then there exists a solution $y_\alpha : (0,\infty) \to \mathbb{F}$ of the differential equation $\eqref{eq:1}$ such that $$ | y(t) - y_\alpha(t) | \leq e^{-\Re(\alpha)t} \int_0^t e^{\Re(\alpha)x} \varphi(x) dx $$ for any $t > 0$. \end{theorem} \begin{proof} If we define a function $z : (0,\infty) \to \mathbb{F}$ by $z(t) := y'(t) + \alpha y(t) - f(t)$ for each $t > 0$, then \begin{equation} \mathcal{L}(y) - \frac{y(0) + \mathcal{L}(f)}{s + \alpha} = \frac{\mathcal{L}(z)}{s + \alpha}. \label{eq:20131113-1} \end{equation} If we set $y_\alpha(t) := y(0) e^{-\alpha t} + (E_{-\alpha}*f)(t)$, where $E_{-\alpha}(t) = e^{-\alpha t}$, then $y_\alpha(0) = y(0)$ and \begin{equation} \mathcal{L} (y_\alpha) = \frac{y(0) + \mathcal{L}(f)}{s + \alpha} = \frac{{y_\alpha(0) + \mathcal{L}(f)}}{s + \alpha}. \label{eq:20131113-2} \end{equation} Hence, we get $$ \mathcal{L}\big( y'_\alpha(t) + \alpha y_\alpha(t) \big) = s\mathcal{L}(y_\alpha) - y_\alpha(0) + \alpha \mathcal{L}(y_\alpha) = \mathcal{L}(f). $$ Since $\mathcal{L}$ is a one-to-one operator, it holds that $$ y'_\alpha(t) + \alpha y_\alpha(t) = f(t). $$ Thus, $y_\alpha$ is a solution of \eqref{eq:1}. Moreover, by \eqref{eq:20131113-1} and \eqref{eq:20131113-2}, we obtain $\mathcal{L}(y) - \mathcal{L}(y_\alpha) = \mathcal{L}(E_{-\alpha}*z)$. Therefore, we have \begin{equation} y(t) - y_\alpha(t) = (E_{-\alpha}*z)(t). \label{eq:20131113-3} \end{equation} In view of \eqref{eq:2}, it holds that \begin{equation} | z(t) | \leq \varphi(t) \label{eq:20131113-4} \end{equation} for all $t > 0$, and it follows from the definition of convolution, \eqref{eq:20131113-3}, and \eqref{eq:20131113-4} that \begin{align*} | y(t) - y_\alpha(t) | & = | (E_{-\alpha}*z)(t) | \\ & = \big| \int_0^t E_{-\alpha}(t-x) z(x) dx \big| \\ & \leq \int_0^t \big| e^{-\alpha(t-x)} \big| \varphi(x) dx \\ & \leq e^{-\Re(\alpha)t} \int_0^t e^{\Re(\alpha)x} \varphi(x) dx \end{align*} for all $t > 0$. (We remark that $\int_0^t e^{\Re(\alpha)x} \varphi(x) dx$ exists for each $t > 0$ provided $\varphi$ is an integrable function.) \end{proof} \begin{corollary}\label{cor:3.1} Let $\alpha$ be a constant in $\mathbb{F}$ and let $\varphi : (0,\infty) \to (0,\infty)$ be an integrable function such that \begin{equation} \int_0^t e^{\Re(\alpha)(x-t)} \varphi(x) dx \leq K \varphi(t) \label{eq:20131121-1} \end{equation} for all $t > 0$ and for some positive real constant $K$. If a continuously differentiable function $y : (0,\infty) \to \mathbb{F}$ satisfies the inequality $\eqref{eq:2}$ for all $t > 0$, then there exists a solution $y_\alpha : (0,\infty) \to \mathbb{F}$ of the differential equation $\eqref{eq:1}$ such that $$ | y(t) - y_\alpha(t) | \leq K \varphi(t) $$ for any $t > 0$. \end{corollary} In the following remark, we show that there exists an integrable function $\varphi : (0,\infty) \to (0,\infty)$ satisfying the condition \eqref{eq:20131121-1}. \begin{remark}\rm Let $\alpha$ be a constant in $\mathbb{F}$ with $\Re(\alpha) > -1$. If we define $\varphi(t) = Ae^t$ for all $t > 0$ and for some $A > 0$, then we have \begin{align*} \int_0^t e^{\Re(\alpha)(x-t)} \varphi(x) dx &= \int_0^t e^{\Re(\alpha)(x-t)} A e^x dx \\ &= \frac{1}{1 + \Re(\alpha)} \left( Ae^t - Ae^{-\Re(\alpha)t} \right) \\ &\leq \frac{1}{1 + \Re(\alpha)} \varphi(t) \end{align*} for each $t > 0$. \end{remark} Now, we apply the Laplace transform method to the proof of the generalized Hyers-Ulam stability of the linear differential equation of second order \begin{equation} \label{eq:4} y''(t) + \beta y'(t) + \alpha y(t) = f(t). \end{equation} \begin{theorem}\label{thm:3.2} Let $\alpha$ and $\beta$ be constants in $\mathbb{F}$ such that there exist $a, b \in \mathbb{F}$ with $a + b = -\beta$, $ab = \alpha$, and $a \neq b$. Assume that $\varphi : (0,\infty) \to (0,\infty)$ is an integrable function. If a twice continuously differentiable function $y : (0,\infty) \to \mathbb{F}$ satisfies the inequality \begin{equation} \label{eq:22} | y''(t) + \beta y'(t) + \alpha y(t) - f(t) | \leq \varphi(t) \end{equation} for all $t > 0$, then there exists a solution $y_c : (0,\infty) \to \mathbb{F}$ of the differential equation $\eqref{eq:4}$ such that $$ | y(t) - y_c(t) | \leq \frac{e^{\Re(a)t}}{| a - b |} \int_0^t e^{-\Re(a)x} \varphi(x) dx + \frac{e^{\Re(b)t}}{| a - b |} \int_0^t e^{-\Re(b)x} \varphi(x) dx $$ for all $t > 0$. \end{theorem} \begin{proof} If we define a function $z : (0,\infty) \to \mathbb{F}$ by $z(t) := y''(t) + \beta y'(t) + \alpha y(t) - f(t)$ for each $t > 0$, then we have \begin{equation} \label{eq:5} \mathcal{L}(z) = \big( s^2 + \beta s + \alpha \big) \mathcal{L}(y) - [ sy(0) + \beta y(0) + y'(0) ] - \mathcal{L}(f). \end{equation} In view of \eqref{eq:5}, a function $y_0 : (0,\infty) \to \mathbb{F}$ is a solution of \eqref{eq:4} if and only if \begin{equation} \label{eq:6} \big( s^2 + \beta s + \alpha \big) \mathcal{L}(y_0) - s y_0(0) - [ \beta y_0(0) + y'_0(0) ] = \mathcal{L}(f). \end{equation} Now, since $s^2 + \beta s + \alpha = (s-a)(s-b)$, \eqref{eq:5} implies that \begin{equation} \label{eq:7} \mathcal{L}(y) - \frac{sy(0) + [\beta y(0) + y'(0)] + \mathcal{L}(f)} {(s - a)(s - b)} = \frac{\mathcal{L}(z)}{(s - a)(s - b)}. \end{equation} If we set \begin{equation} y_c(t) := y(0) \frac{ae^{at} - be^{bt}}{a - b} + [\beta y(0) + y'(0)] E_{a,b}(t) + (E_{a,b}*f)(t), \label{eq:20131119-1} \end{equation} where $E_{a,b}(t) := \frac{e^{at} - e^{bt}}{a-b}$, then $y_c(0) = y(0)$. Moreover, since \begin{align*} &y'_c(t) = y(0) \frac{a^2 e^{at} - b^2 e^{bt}}{a-b} + [\beta y(0) + y'(0)] \frac{ae^{at} - be^{bt}}{a-b} + \frac{d}{dt}(E_{a,b}*f)(t), \\ &(E_{a,b}*f)(t) = \frac{e^{at}}{a-b} \int_0^t e^{-ax} f(x) dx - \frac{e^{bt}}{a-b} \int_0^t e^{-bx} f(x) dx, \end{align*} we have \begin{align*} y'_c(0) & = y(0) \frac{a^2 - b^2}{a-b} + [\beta y(0) + y'(0)] \frac{a-b}{a-b} \\ & = (a+b) y(0) + \beta y(0) + y'(0) \\ % & = -\beta y(0) + \beta y(0) + y'(0) \\ & = y'(0). \end{align*} It follows from \eqref{eq:20131119-1} that \begin{equation} \label{eq:8} \mathcal{L}(y_c) = \frac{sy_c(0) + [ \beta y_c(0) + y'_c(0) ] + \mathcal{L}(f)} {(s - a)(s - b)}. \end{equation} Now, \eqref{eq:6} and \eqref{eq:8} imply that $y_c$ is a solution of \eqref{eq:4}. Applying \eqref{eq:7} and \eqref{eq:8} and considering the facts that $y_c(0) = y(0)$, $y'_c(0) = y'(0)$, and $\mathcal{L}(E_{a,b}*z) = \frac{\mathcal{L}(z)}{(s - a)(s - b)}$, we obtain $\mathcal{L}(y) - \mathcal{L}(y_c) = \mathcal{L}(E_{a,b}*z)$ or equivalently, $y(t) - y_c(t) = (E_{a,b}*z)(t)$. In view of \eqref{eq:22}, it holds that $| z(t) | \leq \varphi(t)$, and it follows from the definition of the convolution that \begin{align*} | y(t) - y_{c}(t) | & = | (E_{a,b}*z)(t) | \\ & \leq \frac{e^{\Re(a)t}}{| a - b |} \int_0^t {e^{-\Re(a)x} \varphi(x) dx} + \frac{e^{\Re(b)t}}{| a - b |} \int_0^t {e^{-\Re(b)x} \varphi(x) dx} \end{align*} for any $t > 0$. We remark that $\int_0^t e^{-\Re(a)x} \varphi(x) dx$ and $\int_0^t e^{-\Re(b)x} \varphi(x) dx$ exist for any $t > 0$ provided $\varphi$ is an integrable function. \end{proof} \begin{corollary}\label{cor:3.2} Let $\alpha$ and $\beta$ be constants in $\mathbb{F}$ such that there exist $a, b \in \mathbb{F}$ with $a + b = -\beta$, $ab = \alpha$, and $a \neq b$. Assume that $\varphi : (0,\infty) \to (0,\infty)$ is an integrable function for which there exists a positive real constant $K$ with \begin{equation} \int_0^t \left( e^{\Re(a)(t-x)} + e^{\Re(b)(t-x)} \right) \varphi(x) dx \leq K \varphi(t) \label{eq:20131121-2} \end{equation} for all $t > 0$. If a twice continuously differentiable function $y : (0,\infty) \to \mathbb{F}$ satisfies the inequality $\eqref{eq:22}$ for all $t > 0$, then there exists a solution $y_c : (0,\infty) \to \mathbb{F}$ of the differential equation $\eqref{eq:4}$ such that $$ | y(t) - y_c(t) | \leq \frac{K}{| a - b |} \varphi(t) $$ for all $t > 0$. \end{corollary} We now show that there exists an integrable function $\varphi : (0,\infty) \to (0,\infty)$ which satisfies the condition \eqref{eq:20131121-2}. \begin{remark} \rm Let $\alpha$ and $\beta$ be constants in $\mathbb{F}$ such that there exist $a, b \in \mathbb{F}$ with $a + b = -\beta$, $ab = \alpha$, $\Re(a) < 1$, $\Re(b) < 1$, and $a \neq b$. If we define $\varphi(t) = Ae^t$ for all $t > 0$ and for some $A > 0$, then we get \begin{align*} &\int_0^t \Big( e^{\Re(a)(t-x)} + e^{\Re(b)(t-x)} \Big) \varphi(x) dx\\ &= \int_0^t \Big( e^{\Re(a)(t-x)} + e^{\Re(b)(t-x)} \Big) Ae^x dx \\ &= \frac{A}{1 - \Re(a)} \Big( e^t - e^{\Re(a)t} \Big) + \frac{A}{1 - \Re(b)} \Big( e^t - e^{\Re(b)t} \Big) \\ &\leq \Big( \frac{1}{1 - \Re(a)} + \frac{1}{1 - \Re(b)} \Big) \varphi(t) \end{align*} for all $t > 0$. \end{remark} Similarly, we apply the Laplace transform method to investigate the generalized Hyers-Ulam stability of the linear differential equation of $n$th order \begin{equation} \label{eq:10} y^{(n)}(t) + \sum_{k=0}^{n-1} \alpha_k y^{(k)}(t) = f(t) \end{equation} \begin{theorem}\label{thm:3.3} Let $\alpha_0, \alpha_1, \dots, \alpha_{n}$ be scalars in $\mathbb{F}$ with $\alpha_n = 1$, where $n$ is an integer larger than $1$. Assume that $\varphi : (0,\infty) \to (0,\infty)$ is an integrable function of exponential order. If an $n$ times continuously differentiable function $y : (0,\infty) \to \mathbb{F}$ satisfies the inequality \begin{equation} \label{eq:11} \big| y^{(n)}(t) + \sum_{k=0}^{n-1} \alpha_k y^{(k)}(t) - f(t) \big| \leq \varphi(t) \end{equation} for all $t > 0$, then there exist real constants $M > 0$ and $\sigma_g$ and a solution $y_c : (0,\infty) \to \mathbb{F}$ of the differential equation $\eqref{eq:10}$ such that $$ | y(t) - y_c(t) | \leq M \int_0^t e^{\alpha (t-x)} \varphi(x) dx $$ for all $t > 0$ and $\alpha > \sigma_g$. \end{theorem} \begin{proof} Applying integration by parts repeatedly, we derive $$ \mathcal{L}\big( y^{(k)} \big) = s^k \mathcal{L}(y) - \sum_{j=1}^k s^{k-j} y^{(j-1)}(0) $$ for any integer $k > 0$. Using this formula, we can prove that a function $y_0 : (0,\infty) \to \mathbb{F}$ is a solution of \eqref{eq:10} if and only if \begin{equation} \label{eq:12} \begin{aligned} \mathcal{L}(f) & = \sum_{k=0}^n \alpha_k s^k \mathcal{L}(y_0) - \sum_{k=1}^n \alpha_k \sum_{j=1}^k s^{k-j} y_0^{(j-1)}(0) \\ & = \sum_{k=0}^n \alpha_k s^k \mathcal{L}(y_0) - \sum_{j=1}^n \sum_{k=j}^n \alpha_k s^{k-j} y_0^{(j-1)}(0) \\ & = P_{n,0}(s) \mathcal{L}(y_0) - \sum_{j=1}^n P_{n,j}(s) y_0^{(j-1)}(0), \end{aligned} \end{equation} where $P_{n,j}(s) := \sum_{k=j}^n \alpha_k s^{k-j}$ for $j \in \{ 0, 1, \dots, n \}$. Let us define a function $z : (0,\infty) \to \mathbb{F}$ by \begin{equation} z(t) := y^{(n)}(t) + \sum_{k=0}^{n-1} \alpha_k y^{(k)}(t) - f(t) \label{eq:20131118-1} \end{equation} for all $t > 0$. Then, similarly as in \eqref{eq:12}, we obtain $$ \mathcal{L}(z) = P_{n,0}(s) \mathcal{L}(y) - \sum_{j=1}^n P_{n,j}(s) y^{(j-1)}(0) - \mathcal{L}(f). $$ Hence, we get \begin{equation} \label{eq:13} \mathcal{L}(y) - \frac{1}{P_{n,0}(s)} \Big( \sum_{j=1}^n P_{n,j}(s) y^{(j-1)}(0) + \mathcal{L}(f) \Big) = \frac{\mathcal{L}(z)}{P_{n,0}(s)}. \end{equation} Let $\sigma_{f}$ be the abscissa of convergence for $f$, let $s_1, s_2, \dots, s_n$ be the roots of the polynomial $P_{n,0}(s)$, and let $\sigma_{{}_P} = \max\{ \Re(s_k) : k \in \{ 1, 2, \dots, n \} \}$. For any $s$ with $\Re(s) > \max\{ \sigma_{f}, \sigma_{{}_P} \}$, we set \begin{equation} \label{eq:14} G(s) := \frac{1}{P_{n,0}(s)} \Big( \sum_{j=1}^n P_{n,j}(s) y^{(j-1)}(0) + \mathcal{L}(f) \Big). \end{equation} By Lemma \ref{lem2}, there exists an $n$ times continuously differentiable function $f_0$ such that \begin{equation} \label{eq:15} \mathcal{L}(f_0) = \frac{\mathcal{L}(f)}{P_{n,0}(s)} \end{equation} for all $s$ with $\Re(s) > \max\{ \sigma_{f}, \sigma_{{}_P} \}$ and \begin{equation} f_0^{(i)}(0) = 0 \label{eq:20131116-1} \end{equation} for any $i \in \{ 0, 1, \dots, n-1 \}$. For $j \in \{ 1, 2, \dots, n \}$, we note that \begin{equation} \label{eq:16} \frac{P_{n,j}(s)}{P_{n,0}(s)} = \frac{1}{s^j} - \frac{\sum_{k=0}^{j-1} \alpha_k s^k}{s^j P_{n,0}(s)} \end{equation} for every $s$ with $\Re(s) > \max\{ 0, \sigma_{{}_P} \}$. Applying Lemma \ref{lem1} for the case of $Q(s) = \sum_{k=0}^{j-1} \alpha_k s^k$ and $P(s) = s^j P_{n,0}(s)$, we can find an infinitely differentiable function $g_j$ such that \begin{equation} \label{eq:17} \mathcal{L}(g_j) = \frac{\sum_{k=0}^{j-1} \alpha_k s^k} {s^j P_{n,0}(s)} \end{equation} and $g_j^{(k)}(0) = 0$ for $k \in \{ 0, 1, \dots, n-1 \}$. Let \begin{equation} \label{eq:18} f_j (t) := \frac{t^{j-1}}{(j-1)!} - g_j (t) \end{equation} for $j \in \{ 1, 2, \dots, n \}$. Then we have \begin{equation} f_j^{(i)}(0) = \begin{cases} 0 & \text{for } i \in \{ 0, 1, \dots, j-2, j, j+1, \dots, n-1 \}, \\ 1 & \text{for } i = j-1. \end{cases} \label{eq:20131116-2} \end{equation} If we define $$ y_c (t) := \sum_{j=1}^n y^{(j-1)}(0) f_j(t) + f_0(t), $$ then the conditions \eqref{eq:20131116-1} and \eqref{eq:20131116-2} imply that \begin{equation} y_c^{(i)}(0) = y^{(i)}(0) \label{eq:20131116-3} \end{equation} for every $i \in \{ 0, 1, \dots, n-1 \}$. Moreover, it follows from \eqref{eq:14}--\eqref{eq:20131116-3} that \begin{equation} \label{eq:19} \begin{aligned} \mathcal{L}(y_c) & = \sum_{j=1}^n y^{(j-1)}(0) \mathcal{L}(f_j) + \mathcal{L}(f_0) \\ & = \sum_{j=1}^n y^{(j-1)}(0) \Big( \frac{1}{s^j} - \mathcal{L}(g_j) \Big) + \frac{\mathcal{L}(f)}{P_{n,0}(s)} \\ & = \frac{1}{P_{n,0}(s)} \Big( \sum_{j=1}^n P_{n,j}(s) y^{(j-1)}(0) + \mathcal{L}(f) \Big) \end{aligned} \end{equation} for each $s$ with $\Re(s) > \max\{ 0, \sigma_{f}, \sigma_{{}_P} \}$. Now, \eqref{eq:12} implies that $y_c$ is a solution of \eqref{eq:10}. Moreover, by \eqref{eq:13} and \eqref{eq:19}, we have \begin{equation} \label{eq:20} \mathcal{L}(y) - \mathcal{L}(y_c) = \frac{\mathcal{L}(z)}{P_{n,0}(s)}. \end{equation} Applying Lemma \ref{lem1} for the case of $Q(s) = 1$ and $P(s) = P_{n,0}(s)$, we find an infinitely differentiable function $g : (0,\infty) \to \mathbb{F}$ such that \begin{equation} \label{eq:21} \mathcal{L}(g) = \frac{1}{P_{n,0}(s)} \end{equation} which implies that $$ g(t) = \mathcal{L}^{-1} \bigg( \frac{1}{P_{n,0}(s)} \bigg) = \frac{1}{2\pi} \int_{-\infty }^\infty e^{(\alpha + iy)t} \frac{1}{P_{n,0} (\alpha + iy)} dy $$ for any real constant $\alpha > \sigma_g$. Moreover, it holds that \begin{equation} \label{eq:23} \begin{aligned} | g(t-x) | & \leq \frac{1}{2\pi} \int_{-\infty}^\infty \big| e^{(\alpha + iy)(t-x)} \big| \frac{1}{| P_{n,0}(\alpha + iy) |} dy \\ & \leq \frac{1}{2\pi} \int_{-\infty}^\infty e^{\alpha(t-x)} \frac{1}{| P_{n,0}(\alpha + iy) |} dy \\ & \leq \frac{1}{2\pi} e^{\alpha(t-x)} \int_{-\infty}^\infty \frac{1}{| P_{n,0}(\alpha + iy) |} dy \\ & \leq M e^{\alpha(t-x)} \end{aligned} \end{equation} for all $\alpha > \sigma_g$, where $$ M = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{| P_{n,0}(\alpha + iy) |} dy < \infty, $$ because $n$ is an integer larger than $1$. By \eqref{eq:11} and \eqref{eq:20131118-1}, it also holds that $| z(t) | \leq \varphi(t)$ for all $t > 0$. In view of \eqref{eq:20}, \eqref{eq:21}, and \eqref{eq:23}, we obtain $$ \mathcal{L}(y) - \mathcal{L}(y_c) = \mathcal{L}(g) \mathcal{L}(z) = \mathcal{L}(g*z). $$ Consequently, we have $y(t) - y_c(t) = (g*z)(t)$ for any $t > 0$. Hence, it follows from \eqref{eq:11}, \eqref{eq:20131118-1}, and \eqref{eq:23} that $$ | y(t) - y_c(t) | = | (g*z)(t) | \leq \int_0^t | g(t-x) | | z(x) | dx \leq M \int_0^t e^{\alpha(t-x)} \varphi(x) dx $$ for all $t > 0$ and for any real constant $\alpha > \sigma_g$, which completes the proof. \end{proof} \begin{corollary}\label{cor:3.3} Let $\alpha_0, \alpha_1, \dots, \alpha_{n}$ be scalars in $\mathbb{F}$ with $\alpha_n = 1$, where $n$ is an integer larger than $1$. Assume that there exist real constants $\alpha$ and $K > 0$ such that a function $\varphi : (0,\infty) \to (0,\infty)$ satisfies $$ \int_0^t e^{\alpha (t-x)} \varphi(x) dx \leq K \varphi(t) $$ for all $t > 0$. Moreover, assume that the constant $\sigma_g$ given in Theorem \ref{thm:3.3} is less than $\alpha$. If an $n$ times continuously differentiable function $y : (0,\infty) \to \mathbb{F}$ satisfies the inequality $\eqref{eq:11}$ for all $t > 0$, then there exist a real constants $M > 0$ and a solution $y_c : (0,\infty) \to \mathbb{F}$ of the differential equation $\eqref{eq:10}$ such that $$ | y(t) - y_c(t) | \leq K M \varphi(t) $$ for all $t > 0$. \end{corollary} \begin{remark} \rm Assume that $\alpha < 1$. If we define $\varphi(t) = Ae^t$ for all $t > 0$ and for some $A > 0$, then we get \begin{align*} \int_0^t e^{\alpha (t-x)} \varphi(x) dx = \int_0^t e^{\alpha (t-x)} A e^x dx = \frac{A}{1-\alpha} \big( e^t - e^{\alpha t} \big) \leq \frac{1}{1-\alpha} \varphi(t) \end{align*} for all $t > 0$. \end{remark} \subsection*{Acknowledgements} This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No. 2013R1A1A2005557). \begin{thebibliography}{99} \bibitem{AlGe} C. Alsina, R. 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