\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 139, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/139\hfil Laplace transform]
{Laplace transform of fractional order differential equations}

\author[S. Liang, R. Wu, L. Chen \hfil EJDE-2015/139\hfilneg]
{Song Liang, Ranchao Wu, Liping Chen}

\address{Song Liang \newline
School of Mathematics,  Anhui University,
 Hefei 230601, China}
\email{songliangeq@163.com}

\address{Ranchao Wu \newline
School of Mathematics,  Anhui University,
 Hefei 230601, China}
\email{rcwu@ahu.edu.cn}

\address{Liping Chen \newline
 School of Electrical Engineering and Automation,
 Hefei University of Technology, Hefei 230009, China}
\email{lip\_chenhut@126.com}

\thanks{Submitted October 10, 2014. Published May 20, 2015.}
\subjclass[2010]{26A33, 34A08, 34K37, 44A10}
\keywords{Fractional-order differential equation; 
Laplace transform; \hfill\break\indent exponential order}

\begin{abstract}
 In this article, we show that Laplace transform can be applied to
 fractional system. To this end, solutions of linear  fractional-order equations
 are first derived by a direct method, without using Laplace transform.
 Then the solutions of  fractional-order differential equations are estimated
 by employing  Gronwall and H\"older inequalities. They are showed be
 to of exponential order, which are necessary to apply the Laplace transform.
 Based on the estimates of solutions, the  fractional-order and the integer-order
 derivatives of solutions are all estimated to be exponential order.
 As a result, the Laplace transform is proved to be valid  in  fractional equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

Fractional calculus is generally  believed to have stemmed from a
question raised in the year 1695 by L'Hopital and Leibniz.
It is the generalization of integer-order calculus
to arbitrary order one. Frequently, it is called fractional-order calculus,
including fractional-order derivatives and fractional-order integrals.
Reviewing its history of three centuries,  we could  find that fractional
calculus were mainly interesting to mathematicians for a long time,
due to its lack of application background.  However, in the previous decades
more and more researchers have paid their attentions to fractional calculus,
since they found that the fractional-order derivatives and fractional-order
 integrals were more suitable for the description
of the  phenomena in the real world, such as viscoelastic systems,
dielectric polarization, electromagnetic waves,
heat conduction, robotics, biological systems, finance and so on;
 see, for example,
\cite{Ahmed,Butzer,Hilfer,Kilbas,Laskin,Ozalp,Podlubny,Samko}.

Owing to great efforts of researchers, there have been  rapid developments
on the theory of fractional calculus and its applications, including
 well-posedness, stability, bifurcation and chaos in fractional differential
equations and their control. Several  useful tools for solving fractional-order
equations have been discovered, of which Laplace transform is frequently applied.
 Furthermore, it is showed to be most efficient and helpful in  analysis
and applications of fractional-order systems, from which some results could be
derived immediately. For instance, in \cite{Li1,Li2}, the authors investigated
stability of fractional-order nonlinear dynamical systems using Laplace
transform method and Lyapunov direct method, with the introduction of
Mittag-Leffler stability and generalized Mittag-Leffler stability concepts.
In \cite{Deng}, Deng  et al studied the stability of $n$-dimensional linear
fractional differential equation with time delays by  Laplace transform method.
In \cite{Sabatier}, Jocelyn Sabatier et al obtained the stability conditions in the
form of  linear matrix inequality (LMI) for fractional-order systems
by using Laplace transform. The Laplace transform was also used
in \cite{Gao,Hamamci,MaltiA,MaltiM,Mesbahi,Ye}.


Although it is often used in analyzing fractional-order systems,
the validity  of Laplace transform to fractional systems is seldom touched
upon when it is applied to fractional systems.
In this paper, its  validity to fractional systems will be justified.
It is showed that Laplace transform could be
applied to fractional systems under certain conditions. To this end,
solutions of linear  fractional-order equations are first derived by
direct method, without using the Laplace transform. The obtained results
match those obtained by the Laplace transform
very well. The method provides an alternative way of solution, different
from the Laplace transform. Then solutions of
fractional-order differential equations are estimated. They are showed to
be of exponential order, which  is necessary
to apply the Laplace transform. Finally, the Laplace transform is proved
to be feasible in  fractional equations.


The article is organized as follows. In Section 2, some preliminaries
about fractional calculus  are presented. In Section 3,  solutions of
linear fractional-order equations are expressed
by the direct method, without using  Laplace transform.
Section 4 is devoted to the estimates of solutions of
fractional-order equations. The Laplace transform is proved to be valid in
fractional-order equations in Section 5.
Finally,  some conclusions are drawn in Section 6.

\section{Preliminaries}

In fractional calculus, the traditional integer-order  integrals and derivatives
of  functions are generalized  to fractional-order  ones, which  are commonly
defined by Laplace convolution operation as follows.

\begin{definition}[{\cite[p. 92]{Kilbas}}]  \rm
Caputo fractional derivative with order $\alpha$ for a function $x(t)$ is
defined as
$$
{}^{C}D^{\alpha}_{t_0}x(t)=\frac{1}{\Gamma(m-\alpha)}
\int^{t}_{t_0}(t-\tau)^{m-\alpha-1}x^{(m)}(\tau)d\tau,
$$
where $0\leq m-1\leq\alpha<m$, $m\in Z_{+}$, and $t=t_0$ is
the initial time and $\Gamma(\cdot)$ is the Gamma function.
\end{definition}

\begin{definition}[{\cite[p. 69]{Kilbas}}] \rm
Riemann-Liouville  fractional integral of order $\alpha>0$ for a function
$x: \mathbb{R}^{+} \mathbb{R} $
is defined as
$$
I_{t_0}^{\alpha}f(t)=\frac{1}{F(\alpha)}
\int^{t}_{t_0}(t-\tau)^{\alpha-1}f(\tau)d\tau,
$$
where $t=t_0$ is the initial time and  $\Gamma(.)$ is the Gamma function.
\end{definition}

\begin{definition}[{\cite[p. 70]{Kilbas}}] \rm
Riemann-Liouville fractional derivative with order $\alpha$ for a function
$x: \mathbb{R}^{+} \mathbb{R} $ is defined as
$$
{}^{RL}D^{\alpha}_{t_0}x(t)=\frac{1}{\Gamma(m-\alpha)}\frac{d^{m}}{dt^{m}}
\int^{t}_{t_0}(t-\tau)^{m-\alpha-1}x(\tau) d\tau,
$$
where $0\leq m-1\leq\alpha<m$, $m\in Z_{+}$, and $t=t_0$ is the initial
time and  $\Gamma(.)$ is the Gamma function.
\end{definition}

\begin{definition} [{\cite[pp. 16-17]{Podlubny}}] \rm
The Mittag-Leffler  function is defined as
$$
E_{\alpha}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(ka+1)},
$$
where $\alpha>0$, $z\in C$. The two-parameter Mittag-Leffler function is defined as
$$
E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(ka+\beta)},
$$
where $\alpha>0$, $\beta>0$, $z\in C$.
\end{definition}

There are some properties between fractional-order derivatives and fractional-order
integrals, which are expressed as follows.

\begin{lemma}[{\cite[pp. 75-76, 96]{Kilbas}}]\label{lem2.5} \rm
 Let $\alpha >0$, $n=[\alpha]+1$ and $f_{n-\alpha}(t)=(I^{n-\alpha}_{a}f)(t)$.
Then fractional integrals and fractional derivatives have the following properties.
% 3.1
(1)  If $f(t)\in L^1(a,b)$ and $f_{n-\alpha}(t)\in AC^n[a,b]$, then 
$$
(I^{\alpha}_{a}{^{RL}}D^{\alpha}_{a}f)(t)
=f(t)-\sum^{n}_{j=1}\frac{f^{(n-j)}_{n-\alpha}(a)}
{\Gamma(a-j+1)}(t-a)^{\alpha-j},
$$
holds almost everywhere in $[a,b]$.

(2) If $f(t)\in AC^n[a,b]$ or $f(t)\in C^n[a,b]$, then
$$
(I^{\alpha}_{a}{^{C}}D^{\alpha}_{a}f)(t)
=f(t)-\sum^{n-1}_{k=0}\frac{f^{(k)}(a)}{k!}(t-a)^{k}.
$$
\end{lemma}

Note that the Laplace transform is a useful tool
for analyzing and solving ordinary
and partial differential equations.
The definition of Laplace transform and some applications to integer-order 
systems are recalled from \cite{Schiff}. They will be useful for later analysis.

\begin{definition}[{\cite[pp. 1-2]{Schiff}}] \rm
The Laplace transform of $f$ is defined as
$$
F(s)=\mathcal{L}(f(t))(s)=\int^{\infty}_{0}e^{-st}f(t)dt
=\lim_{\tau\to\infty} \int^{\tau}_{0}e^{-st}f(t)dt,
$$
whenever the limit exists (as a finite number).
\end{definition}

\begin{definition}[{\cite[p. 10]{Schiff}}] \rm
A function $f$ is piecewise continuous on the interval
$[0,\infty)$ if (i) $\lim_{t\to0^{+}} f(t)=f(0^{+})$ exists and
(ii) $f$ is continuous on every finite interval $(0, b)$ except possibly
at a finite number of points $\tau_1,\tau_2,\dots,\tau_n$ in $(0, b)$
at which $f$ has a jump discontinuity.
\end{definition}

\begin{definition}[{\cite[p. 12]{Schiff}}] \rm
A function $f$ is of exponential order $\gamma$ if there
exist constants $M>0$ and $\gamma$ such that for some $t_0>0$ such
that $|f(t)|\leq Me^{\gamma t}$ for $t\geq t_0$.
\end{definition}

Some existence results of Laplace transform for functions and their
derivatives are listed as follows.

\begin{theorem}[{\cite[p. 13]{Schiff}}]
If $f$ is piecewise continuous on $[0,\infty)$ and of exponential
order $\gamma$, then the Laplace transform $\mathcal{L}(f(t))$ exists for
$\operatorname{Re}(s) > \gamma$ and converges absolutely.
\end{theorem}

\begin{theorem}[{\cite[p. 56]{Schiff}}] \label{thm2.10}
If we assume that $f'$ is continuous $[0,\infty)$ and also of exponential
order, then it follows that the same is true of $f$.
\end{theorem}

\begin{theorem} [{\cite[p. 57]{Schiff}}] \label{thm2.11}
Suppose that $f(t), f'(t),\dots,f^{(n-1)}(t)$  are continuous
on $(0,\infty)$ and of exponential order, while $f^{(n)}(t)$ is piecewise continuous
on $[0,\infty)$. Then
$$
\mathcal{L}(f^{(n)}(t))(s)=s^n\mathcal{L}(f(t))-s^{n-1}f(0+)-s^{n-2}f'(0+)
-\dots-f^{(n-1)}(0+).
$$
\end{theorem}

Although the Laplace operator can be applied to many functions,
there are  some functions, to which it could not
be applied, see for example \cite[p.6]{Schiff}.
The following inequalities  will also be helpful for later analysis.

\begin{lemma}[\cite{Ye}]
Suppose $\beta >0$, $a(t)$ is a nonnegative function locally integrable
on $0\leq t<T$ (some $T\leq +\infty$) and $g(t)$ is a nonnegative,
nondecreasing continuous function defined on $0\leq t < T$,
$g(t)\leq M$ (constant), and suppose $u(t)$ is nonnegative and
locally integrable on $0\leq t <T$ with
$$
u(t)\leq a(t)+g(t)\int^{t}_{0}(t-s)^{\beta-1}u(s)\,ds
$$
on this interval. Then
$$
u(t)\leq a(t)+\int^{t}_{0}\Big[\sum^{\infty}_{n=1}\frac{(g(t)
\Gamma(\beta))^n}{\Gamma(n\beta)}
(t-s)^{n\beta-1}a(s)\Big]ds,0\leq t<T.
$$
\end{lemma}

\begin{lemma}[Cauchy inequality \cite{Kuczma}] \label{lem2.13}
 Let $n\in N$, and let $x_1, x_2, \dots, x_n$ be nonnegative real numbers.
Then for $\vartheta$,
$$
\Big(\sum^{n}_{i=1}x_i\Big)^\vartheta
\leq n^{\vartheta-1}\sum^{n}_{i=1}x_i^\vartheta.
$$
\end{lemma}

\begin{lemma}[Gronwall integral inequality \cite{Cordone}] \label{lem2.14}
If
$$
x(t)\leq h(t)+\int^{t}_{t_0}k(s)x(s)\,ds,\quad t\in[t_0,T),
$$
where all the functions involved are continuous on $[t_0,T)$,
 $T\leq+\infty$, and $k(s)\geq 0$, then $x(t)$
satisfies
$$
x(t)\leq h(t)+\int^{t}_{t_0}h(s)k(s)e^{\int^{t}_{s}k(u)du}\,ds, \quad t\in[t_0,T).
$$
If, in addition, $h(t)$ is nondecreasing, then
$$
x(t)\leq h(t)e^{\int^{t}_{t_0}k(s)\,ds}, t\in[t_0,T).
$$
\end{lemma}

\section{Solutions of linear fractional-order equations by a direct method}

Consider the one-dimensional linear fractional-order equation
\begin{equation}\label{3.1}
D^{\alpha}_0x(t)=\lambda x(t),
\end{equation}
where $D$ denotes $^{RL}D$ or$^{C}D$,  $l-1<\alpha\leq l$,
$l\in N$, $\lambda\in R$.

Take Laplace transform on both sides of \eqref{3.1}, then the solutions
of \eqref{3.1} could  be figured out,
see \cite[pp.284, 313]{Kilbas}. The solutions are presented as follows.
\smallskip

\noindent (a) When $D$ denotes $^{RL}D$, the solution is represented as
\begin{equation}
x(t)=\sum^{l}_{j=1}d_jx_j(t),
\end{equation}
where $d_j=({^{RL}D}^{\alpha-j}x)(0+)=x^{(l-j)}_{l-\alpha}(0+)$,
$x_j(t)=t^{\alpha-j}E_{\alpha,\alpha+1-j}
(\lambda t^\alpha)$, and $j=1,2,\dots,l$.
\smallskip

\noindent (b) When $D$ denotes $^{C}D$, the solution is represented as
\begin{equation}
x(t)=\sum^{l-1}_{j=0}b_j\widetilde{x}_j(t),
\end{equation}
where $b_j=x^{(j)}(0)$, $\widetilde{x}_j(t)=t^{j}E_{\alpha,j+1}(\lambda t^\alpha)$,
 and $j=1,2,\dots,l-1$.

 Now we employ the direct method to derive the solutions of \eqref{3.1}.
The whole process will be formulated  after the following theorem is introduced.

\begin{theorem} \label{thm3.1}
Suppose that $\alpha >0$, $u(t)$ and $a(t)$  are locally integrable on
$0\leq t<T$(some $T\leq +\infty$), and
$|a(t)|\leq M(constant)$. Suppose $x(t)$  is locally integrable on
$0\leq t<T$  with
$$
x(t)=u(t)+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-\tau)^{\alpha-1}
a(\tau)x(\tau)d\tau
$$
on this interval. Then
$$
x(t)=u(t)+\int^{t}_{0}\Big[\sum^{\infty}_{n=1}
\frac{1}{\Gamma(n\alpha)}(t-\tau)^{n\alpha -1}a^{n}(\tau)u(\tau)\Big]d\tau.
$$
\end{theorem}

 \begin{proof}
Let $B\phi(t)=\frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-\tau)^{\alpha -1}
a(\tau)\phi(\tau)d\tau$, $t\geq 0$, where
 $\phi$ is the locally integrable function. Then
 $x(t)=u(t)+Bx(t)$
 implies
 $$
x(t)=\sum^{n-1}_{k=0}B^{k}u(t)+B^{n}x(t).
$$
 Let us prove by mathematical induction that
 \begin{equation}\label{3.4}
B^{n}x(t)=\frac{1}{\Gamma (n\alpha)}\int^{t}_{0}(t-\tau)^{n\alpha -1}
a^{n}(\tau)x(\tau)d\tau,
\end{equation}
and $B^{n}x(t)\to 0$ as $n\to +\infty$ for each t in $0\leq t<T$.

We know that the relation \eqref{3.4} is true for $n=1$.
Assume that it is true for  $n=k$. If $n=k+1$,
 then the induction hypothesis implies
 \begin{align*}
 B^{k+1}x(t)&=B(B^{k}x(t))\\
 &= \frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-s)^{\alpha-1}a(s)
\Big[\frac{1}{\Gamma(k\alpha)}
 \int^{s}_{0}(s-\tau)^{k\alpha -1}a^{k}(\tau)x(\tau)d\tau\Big]ds.
 \end{align*}
By interchanging the order of integration, we have
$$
B^{k+1}x(t)=\frac{1}{\Gamma(\alpha) \Gamma (k\alpha)}
\int^{t}_{0}[\int^{t}_{\tau}(t-s)^{\alpha -1}
(s-\tau)^{k\alpha -1}ds]a^{k+1}(\tau)x(\tau)d\tau,
$$
where the integral
\begin{align*}
\int^{t}_{\tau}(t-s)^{\alpha -1}(s-\tau)^{k\alpha -1}ds
&= (t-\tau)^{k\alpha+\alpha-1}\int^{1}_{0}(1-z)^{\alpha-1}
z^{k\alpha-1}dz\\
&= (t-\tau)^{(k+1)\alpha-1}B(k\alpha,\alpha)\\
&= \frac{\Gamma(\alpha)\Gamma(k\alpha)}{\Gamma((k+1)\alpha)}(t-\tau)^{(k+1)\alpha-1}
\end{align*}
is evaluated with the help of the substitution
$s=\tau+z(t-\tau)$ and the definition of the beta function.
The relation \eqref{3.4} is proved.

Since
\begin{align*}
B^{n}x(t)
&= \frac{1}{\Gamma (n\alpha)}\int^{t}_{0}(t-\tau)^{n\alpha -1}a(\tau)x(\tau)d\tau\\
&\leq  \frac{1}{\Gamma (n\alpha)}\int^{t}_{0}|(t-\tau)^{n\alpha -1}
 a^{n}(\tau)x(\tau)|d\tau\\
&\leq \frac{1}{\Gamma (n\alpha)}\int^{t}_{0}(t-\tau)^{n\alpha -1}|
 a^{n}(\tau)||x(\tau)|d\tau\\
&\leq \frac{M^{n}}{\Gamma (n\alpha)}\int^{t}_{0}(t-\tau)^{n\alpha -1}|
 x(\tau)|d\tau\\
&\leq \frac{M^{n}T^{n\alpha -1}}{\Gamma (n\alpha)}\int^{t}_{0}|x(\tau)|d\tau\to 0,
\end{align*}
as $n\to +\infty$ for $t\in [0,T)$. Then the proof is complete.
\end{proof}

The way to prove the theorem can also be found in \cite{Ye} and \cite{Henry}.
The theorem provides a direct
method to solve the linear fractional-order equation \eqref{3.1}.
\smallskip

(A) When $D$ denotes $^{RL}D$, take the operator $I_{0}^{\alpha}$ on both
sides of \eqref{3.1}, then from Lemma \ref{lem2.5} we have
\[
x(t)=\sum^{l}_{j=1}\frac{x^{(l-j)}_{l-\alpha}(0+)}{\Gamma(\alpha-j+1)}t^{\alpha-j}
+\frac{1}{\Gamma (\alpha)}
\int^{t}_{0}(t-\tau)^{\alpha-1}\lambda x(\tau)d\tau.
\]
Let
\[
\sum^{l}_{j=1}\frac{x^{(l-j)}_{l-\alpha}(0+)}{\Gamma(\alpha-j+1)}t^{\alpha-j}
=\sum^{l}_{j=1} \frac{d_j}
{\Gamma(\alpha-j+1)}t^{\alpha-j}=u(t),
\]
 from Theorem \ref{thm3.1}, one obtains
\begin{equation}
x(t)=u(t)+\int^{t}_{0}\Big[\sum^{\infty}_{n=1}\frac{1}{\Gamma(n\alpha)}
(t-\tau)^{n\alpha -1}\lambda^n u(\tau)\Big]d\tau.
\end{equation}
Assume that
\begin{align*}
A_j
&= \frac{d_j}{\Gamma(\alpha-j+1)}t^{\alpha-j}
 +\int^{t}_{0}\Big[\sum^{\infty}_{n=1}\frac{1}{\Gamma(n\alpha)}
(t-\tau)^{n\alpha -1}\lambda^n \frac{d_j}{\Gamma(\alpha-j+1)}
 \tau^{\alpha-j}\Big]d\tau \\
&= \frac{d_j}{\Gamma(\alpha-j+1)}t^{\alpha-j} \\
&\quad +\sum^{\infty}_{n=1}\Big[\frac{d_j\lambda^nt^{(n+1)\alpha-j}}
{\Gamma(n\alpha)\Gamma(\alpha-j+1)}
\int^{1}_{0}(1-\frac{\tau}{t})^{n\alpha-1}
\big(\frac{\tau}{t}\big)^{\alpha-j}d\big(\frac{\tau}{t}\big)\Big] \\
&= \frac{d_j}{\Gamma(\alpha-j+1)}t^{\alpha-j}+\sum^{\infty}_{n=1}
\big[\frac{d_j\lambda^nt^{(n+1)\alpha-j}}
{\Gamma(n\alpha)\Gamma(\alpha-j+1)}
B(n\alpha,\alpha-j+1)\big] \\
&= \frac{d_j}{\Gamma(\alpha-j+1)}t^{\alpha-j}+\sum^{\infty}_{n=1}
\big[\frac{d_j\lambda^nt^{(n+1)\alpha-j}}
{\Gamma(n\alpha+\alpha-j+1)}\big] \\
&= d_jt^{\alpha-j}E_{\alpha,\alpha+1-j}(\lambda t^\alpha)
= d_jx_j(t).
\end{align*}
Then
\begin{equation}
x(t)=\sum_{j=1}^{l}A_j=\sum^{l}_{j=1}d_jx_j(t).
\end{equation}
It means that the solution of linear fractional-order differential
equations with Rieman-Liouville derivative could be
solved  by the direct method above.
\smallskip

(B) When $D$ denotes $^{C}D$, take the operator $I_{0}^{\alpha}$ on both
sides of \eqref{3.1}, then from Lemma  \ref{lem2.5} we have
\begin{align*}
x(t)=\sum^{l-1}_{j=0}\frac{x^{(j)}(0)}{j!}t^{j}
+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-\tau)^{\alpha-1}
\lambda x(\tau)d\tau.
\end{align*}
Let
\[
\sum^{l-1}_{j=0}\frac{x^{(j)}(0)}{j!}t^{j}=u(t),
\]
 from Theorem \ref{thm3.1}  one obtains
\begin{equation}
x(t)=u(t)+\int^{t}_{0}\Big[\sum^{\infty}_{n=1}\frac{1}{\Gamma(n\alpha)}
(t-\tau)^{n\alpha -1}\lambda^n u(\tau)\Big]d\tau.
\end{equation}
Assume that
\begin{align*}
B_j
&= \frac{x^{(j)}(t)}{j!}t^{j}+\int^{t}_{0}
\Big[\sum^{\infty}_{n=1}\frac{1}{\Gamma(n\alpha)}(t-\tau)^{n\alpha -1}
\lambda^n \frac{x^{(j)}(0)}{j!}t^{j}\Big]d\tau \\
&= \frac{x^{(j)}(t)}{j!}t^{j}+\sum^{\infty}_{n=1}
\Big[\frac{x^{(j)}(0)\lambda^nt^{n\alpha+j}}
{\Gamma(n\alpha)\Gamma(j+1)}
\int^{1}_{0}(1-\frac{\tau}{t})^{n\alpha-1}
\big(\frac{\tau}{t}\big)^{j}d\big(\frac{\tau}{t}\big)\Big] \\
&= \frac{x^{(j)}(t)}{j!}t^{j}+\sum^{\infty}_{n=1}
\big[\frac{x^{(j)}(0)\lambda^n t^{n\alpha+j}}{\Gamma(n\alpha)
\Gamma(j+1)} B(n\alpha,j+1)\big] \\
&= \frac{x^{(j)}(0)}{j!}t^{j}+\sum^{\infty}_{n=1}
\big[\frac{x^{(j)}(0)\lambda^nt^{n\alpha+j}}
{\Gamma(n\alpha+j+1)}\big] \\
&= x^{(j)}(0)t^{j}E_{\alpha,j+1}(\lambda t^\alpha)
= b_j\widetilde{x}_j(t).
\end{align*}
Then
\begin{equation}
x(t)=\sum_{j=0}^{l-1}B_j=\sum^{l-1}_{j=0}b_j\widetilde{x}_j(t).
\end{equation}
That is,  the solution of  linear fractional-order differential equations
with  Caputo derivative
could also  be solved by the direct method.

\section{Estimates of solutions to fractional-order differential equations}

Consider the nonlinear fractional-order differential equation
\begin{equation}\label{4.1}
D^{\alpha}_0x(t)=Ax(t)+f(x)+d(t),
\end{equation}
where $D$ denotes ${^{RL}D}$ or ${^CD}$, $l-1<\alpha\leq l$, $l\in N$,
 $\lambda\in R$, $x\in \mathbb{R}^n$, $f(x)$ is the
nonlinear part  and  continuous in $x\in \mathbb{R}^n$, $f(0)=0$,
$d(t)$ means the input of the equation. To obtain
the main results, make the following assumptions.
\begin{itemize}
\item[(i)] $f(x)$ satisfies the Lipschitz condition, that is, there
exists a constant $L>0$ such that $\|f(x) \|\leq L\|x\|$;

\item[(ii)] $d(t)$ is bounded, that is, there exists a  constant
$M>0$ such that $\|d(t)\|\leq M$.
\end{itemize}
Then we have the following result.

\begin{theorem}\label{thm4.1}
When $t>1$, $D$ denotes $^CD$ or $^{RL}D$ and  \eqref{4.1} satisfies assumptions
{\rm (i)} and {\rm (ii)}, then the
 solution of \eqref{4.1} satisfies
\begin{equation}
\|x(t)\|\leq \widetilde{M}_1e^{p_1t},
\end{equation}
where
\begin{gather*}
p_1=\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}(va-v+1)}
{hv^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}+\alpha+1,  \\
h=1+\frac{1}{\alpha},\quad  v=1+\alpha,\quad
\widetilde{M}_1=\frac{1}{2}(l+1)\overline{M}, \\
\begin{aligned}
\overline{M}
=\max\Big\{&\frac{\|x_{l-1}^{(l-\alpha)}(0)\|}{\Gamma(\alpha-1+1)},
 \frac{\|x_{l-2}^{((l-\alpha))}(0)\|}
{\Gamma(\alpha-2+1)},\dots,
\frac{\|x_{0}^{(l-\alpha)}(0)\|}{\Gamma(\alpha-l+1)},\\
&\frac{M}{\Gamma(\alpha+1)}\frac{\|x(0)\|}{0!},
\frac{\|x'(0)\|}{1!},
\dots,\frac{\|x^{(l-1)}(0)\|}{(l-1)!}\Big\}.
\end{aligned}
\end{gather*}
\end{theorem}

\begin{proof}
Applying  the operator $I_{0}^{\alpha}$ on both sides of \eqref{4.1}, we have
\begin{equation}\label{4.3}
x(t)=u(t)+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-\tau)^{\alpha-1}(Ax(\tau)+f(x(\tau))+d(\tau))d\tau,
\end{equation}
where
\[
u(t)=\begin{cases}
\sum^{l}_{j=1}\frac{x^{(l-j)}_{l-\alpha}(0)}{\Gamma(a-j+1)}t^{a-j}, & D={^{RL}D},\\
\sum^{l-1}_{j=0}\frac{x^{(j)}(0)}{j!}t^{j},& D={^{C}D}.
\end{cases}
\]
Taking norms of both sides of \eqref{4.3}, one obtains
\begin{equation}
\|x(t)\|\leq\|u(t)\|+\frac{1}{\Gamma (\alpha)}
\int^{t}_{0}(t-\tau)^{\alpha-1}(\|A\|\|x(\tau)\|+\|f(x(\tau))\|
+\|d(\tau)\|)d\tau,
\end{equation}
where
\[
\|u(t)\|\leq
\begin{cases}
\sum^{l}_{j=1}\frac{\|x^{(l-j)}_{l-\alpha}(0)\|}{\Gamma(a\alpha-j+1)}t^{\alpha-j},
& D={^{RL}D},\\
\sum^{l-1}_{j=0}\frac{\|x^{(j)}(0)\|}{j!}t^{j},& D={^{C}D}.
\end{cases}
\]

From assumptions (i) and (ii),  one  has
\begin{equation}\label{4.5}
\begin{aligned}
\|x(t)\|
&\leq \|u(t)\|+\frac{1}{\Gamma (\alpha)}
\int^{t}_{0}(t-\tau)^{\alpha-1}[(\|A\|+L)\|x(\tau)\|+M]d\tau \\
&= \|u(t)\|+\frac{Mt^{\alpha}}{\Gamma(\alpha+1)}
+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-\tau)^{\alpha-1}
 (\|A\|+L)\|x(\tau)\|d\tau.
\end{aligned}
\end{equation}
Let $t>1$, $\hat{M}=(l+1)\overline{M}$ and
\begin{align*}
\overline{M}
=\max\Big\{&\frac{\|x_{l-1}^{(l-\alpha)}(0)\|}{\Gamma(\alpha-1+1)},
\frac{\|x_{l-2}^{((l-\alpha))}(0)\|}
{\Gamma(\alpha-2+1)},\dots,\frac{\|x_{0}^{(l-\alpha)}(0)\|}{\Gamma(\alpha-l+1)} ,\\
& \frac{M}{\Gamma(\alpha+1)}\frac{\|x(0)\|}{0!},
\frac{\|x'(0)\|}{1!},
\dots,\frac{\|x^{(l-1)}(0)\|}{(l-1)!}\Big\}.
\end{align*}
Then we have
\begin{equation}\label{4.6}
\begin{aligned}
\|x(t)\|
&\leq  \hat{M}t^{\alpha}+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}
(t-\tau)^{\alpha-1}(\|A\|+L)\|x(\tau) \|d\tau \\
&=  \hat{M}t^{\alpha}+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}
 (t-\tau)^{\alpha-1}e^{\tau-t}e^{t-\tau}(\|A\|+L)
\|x(\tau)\|d\tau.
\end{aligned}
\end{equation}

Let $h=1+\frac{1}{\alpha}$, $v=1+\alpha$, from \eqref{4.6} and
H\"older inequality, we have
\begin{equation}\label{4.7}
\begin{aligned}
&\|x(t)\|\\
&\leq \hat{M}t^{\alpha}+\frac{\|A\|+L}{\Gamma (\alpha)}
\Big[\int^{t}_{0}\left((t-\tau)^{\alpha-1}
e^{\tau-t}\right)^vd\tau\Big]^{1/v}
\Big[\int^{t}_{0}\left(e^{t-\tau}\|x(\tau)
\|\right)^hd\tau\Big]^{1/h} \\
&= \hat{M}t^{\alpha}+\frac{\|A\|+L}{\Gamma (\alpha)}
\Big[\int^{t}_{0}(t-\tau)^{v\alpha-v}e^{v\tau-vt}
d\tau\Big]^{1/v}
\Big[\int^{t}_{0}e^{ht-h\tau}\|x(\tau)\|^hd\tau\Big]^{1/h}.
\end{aligned}
\end{equation}
Note that
\begin{equation}\label{4.8}
\begin{aligned}
\int^{t}_{0}(t-\tau)^{v\alpha-v}e^{-(vt-v\tau)}d\tau
&= \int^{t}_{0}s^{v\alpha-v}e^{-sv}ds \\
&= \frac{1}{v}\int^{tv}_{0}u^{v\alpha-v}e^{-u}du \\
&\leq \frac{1}{v^{v\alpha-v+1}}\int^{+\infty}_{0}u^{v\alpha-v}e^{-u}du \\
&= \frac{1}{v^{v\alpha-v+1}}\Gamma(va-v+1),
\end{aligned}
\end{equation}
where $s=t-\tau$,$u=sv$. Submitting \eqref{4.8} into \eqref{4.7}, one has
\begin{equation}\label{4.9}
\|x(t)\|\leq\hat{M}t^{\alpha}+\frac{(\|A\|+L)\Gamma^{1/v}(va-v+1)}
{v^{\alpha-1+\frac{1}{v}}\Gamma (\alpha)}
\Big[\int^{t}_{0}e^{ht-h\tau}\|x(\tau)\|^hd\tau\Big]^{1/h}.
\end{equation}
From Lemma \ref{lem2.13} and  \eqref{4.9} it follows that
\begin{equation}\label{4.10}
\begin{aligned}
&\|x(t)\|^h\\
&\leq2^{h-1}\hat{M}^h t^{h\alpha}+2^{h-1}
\frac{(\|A\|+L)^h\Gamma^{\frac{h}{v}}(va-v+1)e^{ht}}
{v^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}
\int^{t}_{0}e^{-h\tau}\|x(\tau)\|^hd\tau,
\end{aligned}
\end{equation}
then we can obtain
\begin{equation}\label{4.11}
\begin{aligned}
&\|x(t)\|^he^{-ht}\\
&\leq 2^{h-1}\hat{M}^h t^{h\alpha}e^{-ht}+\frac{2^{h-1}
(\|A\|+L)^h\Gamma^{\frac{h}{v}}
(va-v+1)}{v^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}
\int^{t}_{0}e^{-h\tau}\|x(\tau)\|^hd\tau \\
&\leq 2^{h-1}\hat{M}^h t^{h\alpha}
 +\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}(va-v+1)}{v^{\alpha h-h
+\frac{h}{v}}\Gamma^h (\alpha)}
\int^{t}_{0}e^{-h\tau}\|x(\tau)\|^hd\tau.
\end{aligned}
\end{equation}
From Lemma \ref{lem2.14}, we have
\begin{equation}\label{4.12}
\|x(t)\|^he^{-ht}\leq  2^{h-1}\hat{M^h}t^{h\alpha}e^{\tilde{K}t}
\leq  2^{h-1}\hat{M^h}e^{(\tilde{K}+h\alpha)t},
\end{equation}
where
\[
\tilde{K}=\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}(va-v+1)}
{v^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}.
\]
 Then one obtains
\begin{equation}\label{4.13}
\|x(t)\|\leq \frac{1}{2}\hat{M}e^{(\frac{\tilde{K}}{h}+\alpha+1)t}.
\end{equation}
Let $\widetilde{M}_1=\frac{1}{2}\hat{M}$,
$p_1={\frac{\tilde{K}}{h}}+\alpha+1$, then $||x(t)||\leq\widetilde{M}_1e^{p_1t}$.
 The proof is complete.
\end{proof}


\begin{theorem}\label{thm4.2}
(1) When $0<t\leq1$, $D$ denotes $^{C}D$ and the equation \eqref{4.1}
satisfy assumptions {\rm (i)} and {\rm (ii)},
then the solution of \eqref{4.1} satisfies
\begin{equation}
\|x(t)\|\leq \widetilde{M}_2e^{p_2t},
\end{equation}
where  $\widetilde{M}_2=\frac{1}{2}(l+1)\overline{M}$,
\[
p_2=\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}(va-v+1)}
{hv^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}+1.
\]
(2) When $0<t\leq1$, $D$ denotes $^{RL}D$ and  \eqref{4.1} satisfies
assumptions {\rm (i)} and {\rm (ii)}
for any $b>0$,  then the solution of \eqref{4.1} satisfies
\begin{equation}
\|x(t)\|\leq \widetilde{M}_3e^{p_3t}\quad (t>b>0),
\end{equation}
where $\widetilde{M}_3=\frac{1}{2}(l+1)\overline{M}b^{\alpha-l}$,
$p_3=\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}
(va-v+1)}{hv^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}+1$.
The expressions $\overline{M}$, $h$ and $v$ are  the same as in
 Theorem \ref{thm4.1}.
\end{theorem}

\begin{proof}
(1) When $0<t\leq1$, $D$ denotes $^{C}D$, we can write \eqref{4.5} as
\begin{equation}
\begin{aligned}
\|x(t)\|
&\leq  \hat{M}+\frac{1}{\Gamma (\alpha)}
 \int^{t}_{0}(t-\tau)^{\alpha-1}(\|A\|+L)\|x(\tau)\|d\tau \\
&=  \hat{M}+\frac{1}{\Gamma (\alpha)}\int^{t}_{0}(t-\tau)^{\alpha-1}e^{\tau-t}
e^{t-\tau}(\|A\|+L)\|x(\tau)\|d\tau.
\end{aligned}
\end{equation}
Following the same process as in Theorem \ref{thm4.1}, we have
\begin{equation}
\|x(t)\|\leq \widetilde{M}_2e^{p_2t},
\end{equation}
where
$\widetilde{M}_2=\frac{1}{2}(l+1)\overline{M}$,
\[
p_2=\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}(va-v+1)}
{hv^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}+1.
\]
(2) When $0<t\leq1$, $D$ denotes $^{RL}D$, and $t>b$, we can write \eqref{4.5} as
\begin{equation}
\begin{aligned}
\|x(t)\|
&\leq  \hat{M}b^{\alpha-l}+\frac{1}{\Gamma (\alpha)}
 \int^{t}_{0}(t-\tau)^{\alpha-1}(\|A\|+L)\|x(\tau)\|
d\tau \\
&=  \hat{M}b^{\alpha-l}+\frac{1}{\Gamma (\alpha)}
 \int^{t}_{0}(t-\tau)^{\alpha-1}e^{\tau-t}e^{t-\tau}(\|A\|+L)\|
x(\tau)\|d\tau.
\end{aligned}
\end{equation}
Following the same process as in Theorem \ref{thm4.1}, we have
\begin{equation}
\|x(t)\|\leq \widetilde{M}_3e^{p_3t},
\end{equation}
where $\widetilde{M}_3=\frac{1}{2}(l+1)\overline{M}b^{\alpha-l}$,
\[
p_3=\frac{2^{h-1}(\|A\|+L)^h\Gamma^{\frac{h}{v}}
(va-v+1)}{hv^{\alpha h-h+\frac{h}{v}}\Gamma^h (\alpha)}+1.
\]
 The proof is complete.
\end{proof}

The methods to prove Theorems \ref{thm4.1} and \ref{thm4.2} are similar to
those  in \cite{Wu} and \cite{Chenl}.
From Theorems \ref{thm4.1} and \ref{thm4.2}, we can have the following theorem.

\begin{theorem} \label{thm4.3}
(1) When $D$ denotes $^CD$, and   \eqref{4.1} satisfies {\rm (i)} and {\rm (ii)},
 there exist constants
 $\widetilde{M}>0$ and $p>0$ such that
\begin{equation}
\|x(t)\|\leq \widetilde{M}e^{pt},
\end{equation}
for all $t>0$.

(2) When $D$ denotes $^{RL}D$, and  \eqref{4.1} satisfies {\rm (i)} and {\rm (ii)},
there exist constants
$\widetilde{M}>0$ and $p>0$ for any $b>0$ such that
\begin{equation}
\|x(t)\|\leq \widetilde{M}e^{pt},
\end{equation}
for all $t>b>0$.
\end{theorem}

\section{Validity of Laplace transform for fractional-order equations}


Consider the  one-dimensional fractional-order differential equation
\begin{equation}\label{5.1}
D^{\alpha}_0x(t)=ax(t)+f_1(x(t))+d_1(t),
\end{equation}
where $D$ denotes ${^{RL}D}$ or ${^CD}$, $l-1<\alpha\leq l$, $l\in N$,
 $\lambda\in R$,  $x\in R$, $f_1(x)$ is the
nonlinear part and continuous in  $x\in R$,  $d_1(t)$ is the input of the equation.
 And $f_1$ and $d_1$ also satisfy
the assumptions (i) and (ii). Before the validity of Laplace transform method
is justified, some lemmas and theorems are needed.

\begin{lemma}[{\cite[p. 84]{Kilbas}}] \label{lem5.1}
Let $\operatorname{Re}(\alpha)>0$ and $f\in L^1(0,b)$ for any $b>0$.
Also let the estimate
$$
|f(t)|\leq Ae^{p_0t}\quad(t>b>0)
$$
hold for some constants $A>0$ and $p_0>0$. Then the relation
$\mathcal{L}(I_0^{\alpha}f(t))=s^{-\alpha}\mathcal{L}(f(t))$
is valid for $Re\{s\}>p_0$.
\end{lemma}

\begin{theorem}\label{thm5.2}
If $\alpha>0$, $n=[\alpha]+1$, and $x(t), I^{n-\alpha}_0x(t)$,
$\frac{d}{dt}I^{n-\alpha}_0x(t),\dots,\frac{d^{n-1}}{dt^{n-1}}\\I^{n-\alpha}_0x(t)$
are continuous in $(0,\infty)$ and of exponential order, while
$^{RL}D^{\alpha}_0x(t)$ is piecewise continuous on $[0,\infty)$. Then
$$
\mathcal{L}(^{RL}D^{\alpha}_0x(t))=s^\alpha \mathcal{L}(x(t))
-\sum^{n-1}_{k=0}s^{n-k-1}\frac{d^{(k-1)}}{dt^{(k-1)}}I^{n-\alpha}_0x(0+).
$$
\end{theorem}

\begin{proof}
Since $^{RL}D^{\alpha}_0x(t)=\frac{d^{(n)}}{dt^{(n)}}I^{n-\alpha}_0x(t)$,
let $f(t)=I^{n-\alpha}_0x(t)$, then
$^{RL}D^{\alpha}_0x(t)=\frac{d^{(n)}}{dt^{(n)}}f(t)$.
Under the assumptions, from  theorem \ref{thm2.11} we have
\begin{align*}
\mathcal{L}(^{RL}D^{\alpha}_0x(t))
&= \mathcal{L}(f^{(n)}(t))=s^n\mathcal{L}(f(t))-\sum^{n-1}_{k=0}
                                        s^{n-k-1}f^{(k)}(0+) \\
&= s^\alpha \mathcal{L}(x(t))-\sum^{n-1}_{k=0}s^{n-k-1}
\frac{d^{(k-1)}}{dt^{(k-1)}}I^{n-\alpha}_0x(0+).
\end{align*}
This completes the proof.
\end{proof}

\begin{theorem}
When $D$ denotes ${^{RL}D}$,  the Laplace transform can be taken on both sides of
\eqref{5.1}, if  assumptions
{\rm (i)} and {\rm (ii)} are satisfied and $x(t)$ is continuous.
\end{theorem}

\begin{proof}
From \eqref{5.1} and Theorem \ref{thm4.3}, there exist constants
$M_1>0$ and $P_1>0$ such that
\begin{align*}
|^{RL}D^{\alpha}_0x(t)|
&\leq  |a||x(t)|+|f_1(x(t))|+|d_1(t)|\\
&\leq (|a|+L)x(t)+M\\
&\leq M_1e^{p_1t}.
\end{align*}
This implies  $^{RL}D^{\alpha}_0x(t)=\frac{d^{n}}{dt^{n}}I^{n-\alpha}_0x(t)$
begin of exponential order.  Then from
Theorems \ref{thm2.10} and \ref{thm4.3} and Lemma \ref{lem5.1},
we have that $x(t)$, $I^{n-\alpha}_0x(t)$, and
$\frac{d}{dt}I^{n-\alpha}_0x(t)$, \dots, $\frac{d^{n-1}}{dt^{n-1}}I^{n-\alpha}_0x(t)$
are of exponential order.
From theorem \ref{thm5.2}, the Laplace transform can be taken on both sides of
 \eqref{5.1}.
\end{proof}

\begin{theorem}\label{thm5.4}
If $\alpha>0$, $n=[a]+1$, and $x(t), x'(t), x''(t), x^{(n-1)}(t)$ are
continuous on $[0,+\infty)$ and of exponential
order, while $^{C}D^{\alpha}_0x(t)$ is piecewise continuous
on $[0,\infty)$. Then
$$
\mathcal{L}(^{C}D^{\alpha}_0x(t))
=s^\alpha \mathcal{L}(x(t))-\sum^{n-1}_{k=0}s^{\alpha-k-1}x^{(k)}(0).
$$
\end{theorem}

\begin{proof}
Since
\begin{align*}
^{C}D^{\alpha}_0x(t)
&= \frac{1}{\Gamma(n-\alpha)}\int^{t}_0(t-\tau)^{m-\alpha-1}x^{(n)}(\tau)d\tau\\
&= \frac{1}{\Gamma(n-\alpha)}t^{m-\alpha-1}\ast x^{(n)}(t).
\end{align*}
Under the assumptions in  Theorem \ref{thm2.11}, one has
\begin{align*}
\mathcal{L}(^{C}D^{\alpha}_0x(t))
&= \frac{1}{\Gamma(n-\alpha)}\mathcal{L}(t^{n-\alpha-1})\cdot\mathcal{L}
(x^{(n)}(t))\\
&= s^\alpha \mathcal{L}(x(t))-\sum^{n-1}_{k=0}s^{\alpha-k-1}x^{(k)}(0).
\end{align*}
\end{proof}

\begin{lemma}\label{lem5.5}
If $^{C}D^{\alpha}_{0}x(t)$ is of exponential order and $n=[\alpha]+1$,
then $x^{(j)}(t)(j=1,\dots,n-1)$ is also of exponential order.
\end{lemma}

\begin{proof}
Since $^{C}D^{\alpha}_{0}x(t)$ is of exponential order, then there exist
constants $M_2>0$ and $P_2>0$ such that
\begin{align*}
|^{C}D^{\alpha}_0x(t)|=|^{C}D^{\alpha-j}_0x^{(j)}(t)|\leq M_2e^{p_2t};
\end{align*}
that is,
\begin{align*}
- M_2e^{p_2t}\leq{^{C}D^{\alpha-j}_0x^{(j)}(t)}\leq M_2e^{p_2t}.
\end{align*}
Then there exist functions  $M_1(t)\leq 0, M_2(t)\leq 0$ such that
\begin{align*}
- M_2e^{p_2t}+M_1(t)={^{C}D^{\alpha-j}_0x^{(j)}(t)}=M_2e^{p_2t}-M_2(t).
\end{align*}
This is equivalent to
\begin{align*} 
& \sum_{i=0}^{n-i-1} \frac{x^{(i+j)}(0)}{i!} t^i 
- I^{\alpha-j}_0(M_2e^{p_2t})+I^{\alpha-j}_0M_1(t)\\
&=x^{(j)}(t)
= \sum_{i=0}^{n-i-1} \frac{x^{(i+j)}(0)}{i!} t^i 
+I^{\alpha-j}_0(M_2e^{p_2t}) -I^{\alpha-j}_0M_2(t).
\end{align*}
In view of $I^{\alpha-j}_0M_1(t)\geq 0$ and $I^{\alpha-j}_0M_2(t)\geq 0$, we have
\begin{align*}
\sum_{i=0}^{n-i-1} \frac{x^{(i+j)}(0)}{i!} t^i 
- I^{\alpha-j}_0(M_2e^{p_2t})\leq x^{(j)}(t)
\leq \sum_{i=0}^{n-i-1} \frac{x^{(i+j)}(0)}{i!} t^i 
+I^{\alpha-j}_0(M_2e^{p_2t}).
\end{align*}
Note that
\begin{align*}
I^{\alpha-j}_0e^{p_2t}
&= \frac{1}{\Gamma(\alpha-j)}\int^{t}_{0}(t-\tau)^{\alpha-j-1}e^{p_2\tau}d\tau\\
&= \frac{1}{\Gamma(\alpha-j)}e^{p_2t}\int^{t}_{0}(t-\tau)^{\alpha-j-1}e^{p_2(\tau-t)}d\tau\\
&= \frac{1}{\Gamma(\alpha-j)}e^{p_2t}\int^{t}_0s^{\alpha-j-1}e^{-sp_2}ds\\
&= \frac{1}{\Gamma(\alpha-j)}e^{p_2t}\frac{1}{p_2^\alpha}\int^{p_2t}_0u^{\alpha-j-1}e^{-u}du\\
&\leq \frac{1}{\Gamma(\alpha-j)}p_2^{-\alpha+j}e^{p_2t}\int^{+\infty}_0u^{\alpha-j-1}e^{-u}du\\
&\leq p_2^{-\alpha+j}e^{p_2t},
\end{align*}
where $s=t-\tau$, $u=p_2s$. It is not difficult to get that  there exist
$M_3>0$ and $p_3>0$ such that
$$
|x^{(j)}(t)|\leq M_3e^{p_3t},
$$
where $j=0,\dots,n-1$.
\end{proof}

\begin{theorem} \label{thm5.6}
When $D$ denotes $^{C}D_0^\alpha$, the Laplace transform can be taken
on both sides of \eqref{5.1}, if
assumptions {\rm (i)} and {\rm (ii)} are satisfied and $x(t)$ is continuous.
\end{theorem}

\begin{proof}
From \eqref{5.1} and Theorem \ref{thm4.3}, then there exist constants
 $M_4>0$ and $P_4>0$ such that
\begin{align*}
|^{C}D^{\alpha}_0x(t)|&\leq  |a||x(t)|+|f_1(x(t))|+|d_1(t)|\\
&\leq (|a|+L)x(t)+M\\
&\leq M_4e^{p_4t}.
\end{align*}
It means that $^{C}D^{\alpha}_0x(t)$  is of exponential order.
Then from Lemma \ref{lem5.5} and Theorem \ref{thm4.3}  we have
$x^{(j)}(t) (j=0,\dots,n-1)$ is of exponential order.
From Theorem \ref{thm5.4}, the Laplace transform can be taken
 on both sides of \eqref{5.1}.
\end{proof}

\subsection*{Conclusions}
By Gronwall and  H\"older inequalities, solutions of fractional-order equations
are showed to be of exponential order. Based on that, the fractional-order
and integer-order derivatives are all estimated to be of exponential order.
 Consequently, the Laplace transform is proved to be valid  for fractional-order
equations under general conditions.
  So the validity of Laplace transform of fractional-order equations is justified.

\subsection*{Acknowledgments}
This work was supported by the National Science Foundation of China (No. 61403115),
by the Specialized Research Fund
for the Doctoral Program of Higher Education of China
(No.20093401120001), by the Natural Science Foundation
of Anhui Province (No. 11040606M12, 1508085QF120),
and by the 211 project of
Anhui University (No.KJJQ1102, KJTD002B).



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