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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 145, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/145\hfil Fractional evolution equations]
{Integral solutions of fractional evolution equations with nondense domain}

\author[H. B. Gu, Y. Zhou, B. Ahmad, A. Alsaedi \hfil EJDE-2017/145\hfilneg]
{Haibo Gu, Yong Zhou, Bashir Ahmad, Ahmed Alsaedi}

\address{Haibo Gu \newline
School of Mathematics Sciences,
Xinjiang Normal University,
Urumqi, Xinjiang 830054, China.\newline
Faculty of Mathematics and Computational Science,
Xiangtan University, Hunan 411105, China}
\email{hbgu\_math@163.com}

\address{Yong Zhou (corresponding author) \newline
Faculty of Mathematics and Computational Science,
 Xiangtan University, Hunan 411105, China.\newline
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group,
Faculty of Science, King Abdulaziz University,
Jeddah 21589, Saudi Arabia}
 \email{yzhou@xtu.edu.cn}

\address{Bashir Ahmad \newline
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group,
Faculty of Science, King Abdulaziz University,
Jeddah 21589, Saudi Arabia}
\email{bashirahmad\_qau@yahoo.com}

\address{Ahmed Alsaedi \newline
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group,
 Faculty of Science, King Abdulaziz University,
Jeddah 21589, Saudi Arabia}
\email{aalsaedi@hotmail.com}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted March 11, 2017. Published June 19, 2017.}
\subjclass[2010]{26A33, 34K37, 37L05, 47J35}
\keywords{Fractional evolution equation; Caputo derivative;
\hfill\break\indent integral solution; nondense domain}

\begin{abstract}
 In this article, we study the existence of integral solutions for two
 classes of fractional order evolution equations with nondensely defined
 linear operators. First, we consider the nonhomogeneous fractional
 order evolution equation and obtain its integral solution by Laplace
 transform and probability density function. Subsequently, based on
 the form of integral solution for nonhomogeneous fractional order
 evolution equation, we investigate the existence of integral solution
 for nonlinear fractional order evolution equation by noncompact measure method.
\end{abstract}

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

\section{Introduction}

We consider the nonhomogeneous fractional order evolution equation
\begin{equation}
\begin{gathered}
^C\! D_{0+}^{q}u(t)=Au(t)+f(t), \quad t\in(0, b],\\
u(0)=u_0
\end{gathered} \label{e1.1}
\end{equation}
and the nonlinear fractional order evolution equation
\begin{equation}
\begin{gathered}
^C\! D_{0+}^{q}u(t)=Au(t)+g(t,u(t)), \quad t\in (0, b],\\
u(0)=u_0,
\end{gathered} \label{e1.2}
\end{equation}
where $^C\! D_{0+}^{q}$ is the Caputo fractional derivative of order $0< q< 1$,
the state $u(\cdot)$ takes values in a Banach
space $X$ with norm $|\cdot|$, $A:D(A)\subseteq X\to X$ is a nondensely
closed linear operator on $X$, $f$ and $g$ are given functions satisfying
appropriate conditions.

For the integer order evolution equation:
 $$\left\{
\begin{gathered}
u'(t)=Au(t)+f(t,u(t)), \quad t\in (0, b],\\
u(0)=u_0,
\end{gathered}\right.\label{e1.3}
$$
in case $A$ is a Hille-Yosida operator and is densely defined
(i.e., $\overline{D(A)}=X$), the problem has been extensively studied
(see \cite{A. Pazy}). When $A$ is a Hille-Yosida operator
but its domain is nondensely defined, there have many results
(see \cite{N. Abada,X.L. Fu, 1 H.R. Thieme, 2 H.R. Thieme}
and the references therein). It is noted that Da Prato and Sinestrari
are the first to work on equations with nondense domains, see \cite{Prato}.

On the other hand, fractional order differential equations have recently
been applied in various areas of engineering, physics and
bio-engineering, and other applied sciences. For some
fundamental results in the theory of fractional calculus and
fractional differential equations, we refer the reader to the
monographs by Samko et al.\ \cite{sam}, Kilbas et al.\
 \cite{KST}, Diethelm \cite{kd-1} and Zhou \cite{Z1}, and the papers
 \cite{als,kir,z1,z2,z3,z4,z5,z6} and the references cited therein.

For nonlinear fractional evolution equation \eqref{e1.2} with initial data
 or nonlocal condition, when $A$ is densely defined,
there have been many results on the existence of mild solutions
(see \cite{M. Li, J.G. Peng, R.N. Wang, Y. Zhou 1}).
In \cite{Y. Zhou 1}, by using similar methods due to
El-Borai \cite{el-1, el-2}, Zhou and Jiao proposed a suitable concept
on mild solution by applying probability density function and Laplace transform,
which is widely used now.
When $A$ is not densely defined, there have been some investigations
(see \cite{G. M. Mophou, Z.F. Zhang1}). However, in \cite{G. M. Mophou},
there was an error in transforming integral solution into an available form.
 Zhang et al.\ \cite{Z.F. Zhang1} presented a formula for integral solution
by using the similar method described in \cite{Y. Zhou 1}, but the equivalency
 of integral equations was not proved.

Motivated by the above discussion, in this paper, we will firstly give the
integral solution for nonhomogeneous fractional evolution equation \eqref{e1.1}
by Laplace transform and probability density function, and subsequently
investigate the existence of integral solution for nonlinear fractional
order evolution equation \eqref{e1.2} by Ascoli-Arzela theorem and the
measure of noncompactness. In what follows we do not require the
$C_0-$semigroup (will be given later) to be compact.

The rest of this paper is organized as follows. In Section 2,
notation and preliminaries are given. The integral solution of nonhomogeneous
fractional evolution equation \eqref{e1.1} is given in Section 3.
In Section 4, the existence of integral solution for nonlinear fractional
order evolution equation \eqref{e1.2} is studied. The paper concludes
with a problem proposed for further research.

\section{Preliminaries}

In this section, we recall some concepts on fractional calculus and present
some lemmas and assumptions which are useful in the sequel.

Let $p>0$, $n=\lceil p\rceil$ (the least integer greater than or equal to $p$)
and $u\in L^1([0,b], X)$.
 The Riemann-Liouville fractional integral is defined by
$$
I_{0+}^p u(t)=g_p(t)*u(t)=\int_0^tg_p(t-s)u(s)ds, ~t>0,
$$
where $*$ denotes convolution and $g_p(t)=t^{p-1}/\Gamma(p)$.
In case $p=0$, we set $g_0(t)=\delta(t)$, the Dirac measure concentrated at
the origin.
For $u\in C([0,b], X)$, the Riemann-Liouville fractional derivative is defined by
$$
^L\!D_{0+}^pu(t)=\frac{d^n}{dt^n}(g_{n-p}(t)*u(t))
$$
and the Caputo fractional derivative can be defined by
$$
^C\!D_{0+}^pu(t)=g_{n-p}(t)*\frac{d^nu(t)}{dt^n}
$$
for all $t>0$. For more details, see \cite{KST}.

Next, we introduce the Hausdorff measure of noncompactness
$\beta(\cdot)$ defined on each bounded subset $\Omega$
of Banach space $X$ by
$$
\beta(\Omega)=\inf\{\epsilon>0, \Omega \text{ has a finite }
 \epsilon\text{-net in } X\}.
$$
Some basic properties of $\beta(\cdot)$ are listed in the following Lemmas.

\begin{lemma}[\cite{Banas}] \label{lem2.1}
The noncompact measure $\beta(\cdot)$ satisfies:
\begin{itemize}
\item[(i)] for all bounded subsets $B_1, B_2$ of $X$, $B_1\subseteq B_2$
implies $\beta(B_1)\leq \beta(B_2)$;

\item[(ii)] $\beta(\{x\}\cup B)=\beta( B)$ for every $x\in X$ and every
nonempty subset $B\subseteq X$;

\item[(iii)] $\beta(B)=0$ if and only if $B$ is relatively compact in $X$;

\item[(iv)] $\beta(B_{1}+B_{2})\leq \beta(B_{1})+\beta(B_{2})$, where
$B_{1}+B_{2}=\{x+y: x\in B_{1},~y\in B_{2}\}$;

\item[(v)] $\beta(B_{1}\cup B_{2})\leq
\max\{\beta(B_{1}),\beta(B_{2})\}$;

\item[(vi)] $\beta(\lambda B)\leq |\lambda| \beta(B)$ for any
$\lambda\in \mathbb{R}$.
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{Monch}] \label{lem2.2}
 Let $J=[0,b]$ and $\{u_n\}^{\infty}_{n=1}$ be
a sequence of Bochner integrable functions from $J$ into $X$ with
$|u_n(t)|\leq\tilde{m}(t)$ for almost all $t\in J$ and every $n\geq1$,
where $\tilde{m}\in L(J,\mathbb{R}^{+})$. Then the function
$\psi(t)=\beta(\{u_n(t)\}^{\infty}_{n=1})$ belongs to $L(J,\mathbb{R}^{+})$
and satisfies
$$
\beta\Big(\Big\{\int^{t}_{0}u_n(s)ds:
n\geq1\Big\}\Big)\leq2\int^{t}_{0}\psi(s)ds.
$$
\end{lemma}

Let $X_0=\overline{D(A)}$ and $A_0$ be the part of $A$ in
$\overline{D(A)}$ defined by
\[
D(A_0)=\{x\in D(A): Ax\in \overline{D(A)}\},\quad
A_0x=Ax.
\]


\begin{proposition}[\cite{A. Pazy}] \label{prop2.1}
 The part $A_0$ of $A$ generates a strongly continuous semigroup(that is,
 $C_0-$semigroup) $\{Q(t)\}_{t\geq 0}$ on $X_0$.
\end{proposition}

In the forthcoming analysis, we need the following hypothesis:
\begin{itemize}
\item[(H1)] The linear operator $A:D(A)\subset X\to X$ satisfies the
Hille-Yosida condition, that is, there exist two constant
$\omega\in \mathbb{R}$ and $\overline{M}$
such that $(\omega, +\infty)\subseteq \rho(A)$ and
$$
\|(\lambda I-A)^{-k}\|_{\mathcal{L}(X)}
\leq \frac{\overline{M}}{(\lambda-\omega)^k}, \quad \text{for all }
\lambda>\omega,\; k\geq 1.
$$

\item[(H2)] $Q(t)$ is continuous in the uniform operator topology for $t>0$,
and $\{Q(t)\}_{t\geq 0}$ is uniformly bounded, that is, there exists $M>1$
such that $\sup_{t\in[0,+\infty)} |Q(t)|<M$.
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Integral solution to nonhomogeneous Cauchy problem}

Here we derive the integral solution for nonhomogeneous fractional
order evolution equation \eqref{e1.1} with the aid of
Laplace transform and probability density function.
For Cauchy problem \eqref{e1.1}, it is assumed that $u_0\in X_0$ and
$f:J\to X$ is continuous.

\begin{definition} \label{def3.1} \rm
 A function $u(t)$ is said to be an integral solution of \eqref{e1.1} if
\begin{itemize}
\item[(i)] $u:J\to X$ is continuous;

\item[(ii)] $ I_{0+}^q u(t)\in D(A)$ for $t\in J$ and

\item[(iii)]
\begin{equation}
u(t)=u_0+AI_{0+}^q u(t)+I_{0+}^q f(t), \quad t\in J. \label{e3.1}
\end{equation}
\end{itemize}
\end{definition}

\begin{remark} \label{rmk3.1} \rm
 If $u(t)$ is an integral solution of \eqref{e1.1}, then $u(t)\in X_0$ for
 $t\in J$. In fact, by $I_{0+}^q u(t)\in D(A)$, we have
 $I_{0+}^1 u(t)=I_{0+}^{1-q}I_{0+}^q u(t)\in D(A)$ for $t\in J$.
Then $u(t)=\lim_{h\to 0^+}\frac{1}{h}\int_t^{t+h}u(s)ds\in X_0$ for $t\in J$.
\end{remark}

\begin{definition}[\cite{mai}] \label{def3.2}\rm
 The Wright function $M_q(\theta)$ defined by
$$
M_q(\theta)=\sum_{n=1}^\infty \frac{(-\theta)^{n-1}}{(n-1)!\Gamma(1-q n)}$$
is such that
$$
\int^{\infty}_{0}\theta^{\delta} M_{q}(\theta)d\theta=\frac{\Gamma
(1+\delta)}{\Gamma (1+q \delta)},\quad\text{for }\delta\geq 0.
$$
\end{definition}

Consider the auxiliary problem
\begin{equation}
\begin{gathered}
^C\! D_{0+}^{q}u(t)=A_0u(t)+f(t), \quad t\in (0, b],\\
u(0)=u_0.
\end{gathered} \label{e3.2}
\end{equation}
By Definition \ref{def3.1}, the integral solution of \eqref{e3.2} can be written as
\begin{equation}
u(t)=u_0+A_0I_{0+}^q u(t)+I_{0+}^q f(t)\label{e3.3}
\end{equation}
for $u_0\in X_0$ and $t\in J$. The following Lemma gives an equivalent
form of \eqref{e3.3} by means of Laplace transform.

\begin{lemma} \label{lem3.1}
 If $f$ take values in $X_0$, then the integral equation \eqref{e3.3} can
be expressed as
\begin{equation}
u(t)=\Big(I_{0+}^{1-q}K_q(t)\Big)u_0+\int_0^tK_{q}(t-s)f(s)ds,\quad t\in J,\label{e3.4}
\end{equation}
 where
$$
K_{q}(t)=t^{q-1}P_{q}(t),\quad
P_{q}(t)=\int_0^\infty q\theta M_q (\theta)Q(t^q\theta)d\theta.
$$
\end{lemma}

\begin{proof}
 Let $\lambda>0$. Applying the Laplace transform
\[
\chi(\lambda)=\int_0^{\infty}e^{-\lambda s}u(s)ds \quad
\text{and}\quad
\omega(\lambda)=\int_0^{\infty}e^{-\lambda s}f(s)ds
\]
 to \eqref{e3.3}, we obtain
\begin{equation}
\begin{aligned}
\chi(\lambda)
&=\lambda^{-1}u_0+\frac{1}{{\lambda}^{q}}A_0\chi(\lambda)
+\frac{1}{{\lambda}^{q}} \omega(\lambda)\\
 &=\lambda^{q-1}(\lambda^{q}I-A_0)^{-1}u_0+(\lambda^{q}I-A_0)^{-1}\omega(\lambda)\\
 &=\lambda^{q-1}\int_0^\infty e^{-\lambda^{q}s}Q(s)u_0 ds+\int_0^\infty
 e^{-\lambda^{q}s}Q(s)\omega(\lambda)ds,
\end{aligned} \label{e3.5}
\end{equation}
provided that the integrals in \eqref{e3.5} exist, where $I$ is the
identity operator defined on $X$.

The Laplace transform of
$$
\psi_{q}(\theta)=\frac{q}{\theta^{q+1}}M_{q}(\theta^{-q}),
$$
 is
\begin{equation}
\int_0^{\infty}e^{-\lambda
\theta}\psi_{q}(\theta)d\theta=e^{-\lambda^{q}},
\label{e3.6}
\end{equation}
where $q\in (0,1)$. Using \eqref{e3.6}, we have
\begin{equation}
\begin{aligned}
\int_0^\infty e^{-\lambda^{q}s}Q(s)u_0ds
&=\int_0^\infty q t^{q-1}e^{-(\lambda t)^q}Q(t^q)u_0 dt\\
&=\int_0^\infty\int_0^\infty q\psi_q(\theta)e^{-(\lambda t\theta)}
 Q(t^q)t^{q-1}u_0d\theta dt \\
&=\int_0^\infty\int_0^\infty q\psi_q(\theta)e^{-\lambda t}
 Q\Big(\frac{t^q}{\theta^q}\Big)\frac{t^{q-1}}{\theta^q}u_0d\theta dt \\
&=\int_0^\infty e^{-\lambda t}\Big[ q\int_0^{\infty}\psi_{q}(\theta)
 Q\Big(\frac{t^{q}}{\theta^{q}}\Big)
 \frac{t^{q-1}}{\theta^{q}}u_0d\theta \Big]dt\\
&= \int_0^\infty e^{-\lambda t}t^{q-1}P_q(t)u_0dt
\end{aligned}\label{e3.7}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\int_0^\infty
 e^{-\lambda^{q}s}Q(s)\omega(\lambda)ds\\
&=\int_0^\infty\int_0^\infty q
 t^{q-1}e^{-(\lambda t)^q}Q(t^q)e^{-\lambda s}f(s)dsdt\\
&=\int_0^\infty\int_0^\infty\int_0^\infty
 q\psi_q(\theta)e^{-(\lambda t\theta)}Q(t^q)e^{-\lambda s}t^{q-1}f(s)
 d\theta ds dt \\
&=\int_0^\infty\int_0^\infty\int_0^\infty
 q\psi_q(\theta)e^{-\lambda (t+s)}Q\Big(\frac{t^q}{\theta^q}\Big)\frac{t^{q-1}}{\theta^q}f(s)
 d\theta ds dt \\
&=\int_0^\infty
 e^{-\lambda t}\Big[ q\int_0^{t}\int_0^{\infty}\psi_{q}(\theta)Q\Big(\frac{(t-s)^{q}}{\theta^{q}}\Big)
 \frac{(t-s)^{q-1}}{\theta^{q}}f(s)d\theta ds
 \Big]dt\\
&=\int_0^\infty
 e^{-\lambda t}\Big[ \int_0^{t}(t-s)^{q-1}P_q(t-s)f(s)ds
 \Big]dt.
 \end{aligned}\label{e3.8}
\end{equation}
Since the Laplace inverse transform of $\lambda^{q-1}$ is
$$
\mathfrak{L}^{-1}(\lambda^{q-1})=\frac{t^{-q}}{\Gamma(1-q)}=g_{1-q}(t),
$$
therefore, by \eqref{e3.5}, \eqref{e3.7} and \eqref{e3.8}, for $t\in J$,
we obtain
\begin{equation}
\begin{aligned}
u(t)&=\Big(\mathfrak{L}^{-1}(\lambda^{q-1})*K_q(t)\Big)u_0+\int_0^t K_q(t-s)f(s)ds\\
 &=\Big(I_{0+}^{1-q}K_q(t)\Big)u_0+\int_0^t K_q(t-s)f(s)ds.\\
\end{aligned}\label{e3.9}
\end{equation}
This completes the proof.
\end{proof}

\begin{remark} \label{rmk3.2} \rm
 Let $S_{q}(t)=I_{0+}^{1-q}K_{q}(t)$. By the uniqueness of Laplace inverse
transform, it is obvious that operators $S_q(t)$ and $P_q(t)$
(obtained here) are the same as the ones given in \cite{Y. Zhou 1}.
In addition, we also obtain the relationship between $S_q(t)$ and $K_q(t)$;
 that is, $S_{q}(t)=I_{0+}^{1-q}K_{q}(t)$ for $t\geq 0$. So, we can say that
$\{K_q(t)\}_{t\geq 0}$ is generated by $A_0$.
\end{remark}

\begin{proposition}[\cite{Z1}] \label{prop3.1} \rm
 With assumption {\rm (H2)}, $P_q(t)$ is continuous in the uniform operator
topology for $t> 0$.
\end{proposition}

\begin{proposition}[\cite{Y. Zhou 1}] \label{prop3.2}
 With assumption {\rm (H2)}, for any fixed $t>0$,
$\{K_q(t)\}_{t>0}$ and $\{S_{q}(t)\}_{t>0}$ are linear operators, and for any
$x\in X_0$,
$$
|K_q(t)x|\leq\frac{Mt^{q-1}}{\Gamma(q)}|x|\quad \text{and}\quad
|S_{q}(t)x|\leq M|x|.
$$
\end{proposition}

\begin{proposition}[\cite{Y. Zhou 1}] \label{prop3.3}
 With assumption {\rm (H2)}, $\{K_q(t)\}_{t>0}$ and $\{S_{q}(t)\}_{t>0}$
are strongly continuous, that is, for any
$x\in X_0$ and $0<t'<t''\leq b$,
$$
|K_q(t')x-K_q(t'')x|\to 0\quad \text{and}\quad |S_{q}(t')x-S_{q}(t'')x|\to 0,
\quad \text{as } t''\to t'.
$$
\end{proposition}

If we assume that $f$ takes values in $X_0$, then \eqref{e3.4} can be written as
\begin{equation}
u(t)=S_{q}(t)u_0+\int_0^tK_{q}(t-s)
\lim_{\lambda\to +\infty}B_{\lambda}f(s)ds \label{e3.10}
\end{equation}
or
\begin{equation}
u(t)=S_{q}(t)u_0+\lim_{\lambda\to +\infty}
 \int_0^tK_{q}(t-s)B_{\lambda}f(s)ds,\label{e3.11}
\end{equation}
where $B_{\lambda}=\lambda(\lambda I-A)^{-1}$, since
$\lim_{\lambda\to +\infty}B_{\lambda}x=x$ for $x\in X_0$.
When $f$ takes values in $X$, but not in $X_0$, then the limit in
 \eqref{e3.11} exists (as we will prove). But the limit in \eqref{e3.10}
 will no longer exist.

\begin{lemma} \label{lem3.2}
 Any solution of integral equation \eqref{e3.1} with values in $X_0$ is
 represented by \eqref{e3.11}.
\end{lemma}

\begin{proof} Let
$$
u_{\lambda}(t)=B_{\lambda}u(t),\quad f_{\lambda}(t)=B_{\lambda}f(t),\quad
u_{\lambda}=B_{\lambda}u_0.
$$
By applying $B_{\lambda}$ to \eqref{e3.1}, we have
$$
u_{\lambda}(t)=u_{\lambda}+A_0I_{0+}^qu_{\lambda}(t)+I_{0+}^qf_{\lambda}(t).
$$
Hence, by Lemma \ref{lem3.1}, we obtain
$$
u_{\lambda}(t)=S_{q}(t)u_{\lambda}+\int_0^tK_{q}(t-s)f_{\lambda}(s)ds.
$$
As $u(t), u_0\in X_0$, we have
$$
u_{\lambda}(t)\to u(t), u_{\lambda}\to u_0,\quad
S_{q}(t)u_{\lambda}\to S_{q}(t)u_0, \quad \text{as }\lambda\to +\infty.
$$
Thus \eqref{e3.11} holds. This completes the proof.
\end{proof}

Let us define
\begin{equation}
\Phi_q(t)x=\lim_{\lambda\to +\infty}\int_0^tK_{q}(t-s)B_{\lambda}x\,ds
=\lim_{\lambda\to +\infty}\int_0^tK_{q}(s)B_{\lambda}x\,ds, \label{e3.12}
\end{equation}
for $x\in X$ and $t\geq 0$.

\begin{proposition} \label{prop3.4}
 For $x\in X$ and $t\geq 0$, the limit in \eqref{e3.12} exists and defines
a bounded linear operator $\Phi_q (t)$.
\end{proposition}

\begin{proof} Let
$$
\Phi_q^0(t)x=\int_0^tK_{q}(t-s)x\,ds=\int_0^tK_{q}(s)x\,ds,
$$
for $x_0\in X_0$ and $t\geq 0$. Then, the definition
$$
\Phi_q(t)=(\lambda I-A)\Phi^0(t)(\lambda I-A)^{-1},
$$
for $\lambda >\omega$, extends $\Phi_q^0(t)$ from $X_0$ to $X$.
This definition is independent of $\lambda$ because of the resolvent identity.
 As $\Phi_q(t)$ maps $X$ into $X_0$, we have
 $$
 \Phi_q(t)x=\lim_{\lambda\to +\infty}B_{\lambda}\Phi_q(t)x
=\lim_{\lambda\to +\infty}\Phi_q^0(t)B_{\lambda}x.
 $$
This completes the proof.
\end{proof}

\begin{proposition} \label{prop3.5}
 For $x\in X_0$ and $t\geq 0$, $^C\!D_{0+}^q \Phi_q^0(t)x=S_q(t)x$ and
$S_q(t)x=A\Phi_q^0(t)x+x$.
\end{proposition}

The proof of the above proposition follows directly from the definitions of
$S_q(t)$ and $\Phi_q^0(t)$ for $t\geq 0$.

\begin{lemma} \label{lem3.3}
 (i) For $x\in X$ and $t\geq 0$, $I_{0+}^q \Phi_q(t)\in D(A)$ and
\begin{equation}
 \Phi_q(t)x=A\big(I_{0+}^q \Phi_q(t)x\big)+\frac{t^q}{\Gamma(1+q)}x.\label{e3.13}
\end{equation}
(ii) For $x\in D(A)$,
\begin{equation}
 \Phi_q(t)Ax+x=S_q(t)x.\label{e3.14}
\end{equation}
\end{lemma}

\begin{proof}
(i) For $x\in X$ and $t\geq 0$, let
$$
V(t)=\lambda I_{0+}^q \Phi_q^0(t)(\lambda I-A)^{-1}x+\frac{t^q}
{\Gamma(1+q)}(\lambda I-A)^{-1}x-\Phi_q^0(t)(\lambda I-A)^{-1}x.
$$
Clearly $V(0)=0$. By Proposition \ref{prop3.5}, we have
\begin{align*}
&^C\!D_{0+}^q V(t) \\
&=\lambda \Phi_q^0(t)(\lambda I-A)^{-1}x+(\lambda I-A)^{-1}x
 -{^C\!D_{0+}^q}\Phi_q^0(t)(\lambda I-A)^{-1}x\\
&=\lambda \Phi_q^0(t)(\lambda I-A)^{-1}x+(\lambda I-A)^{-1}x
 -S_q(t)(\lambda I-A)^{-1}x\\
&=\lambda \Phi_q^0(t)(\lambda I-A)^{-1}x+(\lambda I-A)^{-1}x
-A\Phi_q^0(t)(\lambda I-A)^{-1}x-(\lambda I-A)^{-1}x\\
&=\lambda \Phi_q^0(t)(\lambda I-A)^{-1}x-A\Phi_q^0(t)(\lambda I-A)^{-1}x\\
&=(\lambda I-A)\Phi_q^0(t)(\lambda I-A)^{-1}x\\
&=\Phi_q(t)x.
\end{align*}
Then
$$
V(t)=I_{0+}^q\Phi_q(t)x+V(0)=I_{0+}^q\Phi_q(t)x
$$
and
$$
(\lambda I-A)V(t)=(\lambda I-A)I_{0+}^q\Phi_q(t)x
=\lambda I_{0+}^q\Phi_q(t)x+\frac{t^q}{\Gamma(1+q)}x-\Phi_q(t)x.
$$
Thus
$$
 \Phi_q(t)x=A\big(I_{0+}^q \Phi_q(t)x\big)+\frac{t^q}{\Gamma(1+q)}x.
$$

(ii) For $x\in D(A)$, it follows by Proposition \ref{prop3.5} that
\begin{align*}
\Phi_q(t)Ax
&=\lim_{\lambda\to +\infty}\int_0^tK_q(s)B_{\lambda}Ax\,ds
=\lim_{\lambda\to +\infty}A_0\int_0^tK_q(s)B_{\lambda}x\,ds\\
&=A_0\Phi_q^0(t)x=S_q(t)x-x.
\end{align*}
This completes the proof.
\end{proof}

\begin{theorem} \label{thm3.1}
 $u(t)$ is an integral solution of \eqref{e1.1} if and only if
\begin{equation}
u(t)=S_{q}(t)u_0+\lim_{\lambda\to +\infty}
\int_0^tK_{q}(t-s)B_{\lambda}f(s)ds \label{e3.15}
\end{equation}
for $t\in J$ and $u_0\in X_0$.
\end{theorem}

\begin{proof}
In view of Lemma \ref{lem3.2}, we only need to show that \eqref{e3.15} is the
integral solution of \eqref{e1.1}. Indeed it is sufficient to prove the
theorem for $u_0=0$, because it can easily be proved for the special case $f=0$.
 We complete the proof in two steps.
\smallskip

\noindent\textbf{Step I.}
 Assume that $f$ is continuously differentiable, then for $t\in J$, we have
\begin{align*}
u_{\lambda}(t)
&=\int_0^t K_q(s)B_{\lambda}f(s)ds\\
&=\int_0^t K_q(s)B_{\lambda}\Big(f(0)+\int_0^s f'(r)dr\Big)ds\\
&=\int_0^t K_q(s)B_{\lambda}f(0)ds+\int_0^t K_q(s)B_{\lambda}
 \Big(\int_0^s f'(r)dr\Big)ds\\
&=\Phi_q^0(t)B_{\lambda}f(0)+\int_0^t \Phi_q^0(t-r)B_{\lambda} f'(r)dr.
\end{align*}
By Lemma \ref{lem3.3}, for $t\in J$, we obtain
\begin{align*}
u(t)
&=\lim_{\lambda\to +\infty}u_{\lambda}(t)\\
&= \Phi_q(t)f(0)+\int_0^t \Phi_q(t-r) f'(r)dr\\
&= A\big(I_{0+}^q\Phi_q(t)f(0)\big)+\frac{t^q}{\Gamma(1+q)}f(0)\\
&\quad +\int_0^t\Big[A\big(I_{0+}^q\Phi_q(t-r)\big)+\frac{(t-r)^q}{\Gamma(1+q)}\Big]f'(r)dr\\
&= A\Big[I_{0+}^q\Phi_q(t)f(0)+\int_0^tI_{0+}^q\Phi_q(t-r)f'(r)dr\Big]\\
&\quad +\frac{t^q}{\Gamma(1+q)}f(0)+\frac{1}{\Gamma(1+q)}\int_0^t(t-r)^qf'(r)dr\\
&= A\Big[I_{0+}^q\Phi_q(t)f(0)+I_{0+}^q\Big(\int_0^t\Phi_q(t-r)f'(r)dr\Big)\Big]\\
&\quad +\frac{t^q}{\Gamma(1+q)}f(0)+\frac{1}{\Gamma(1+q)}\int_0^t(t-r)^qf'(r)dr\\
&= A\big(I_{0+}^q u(t)\big)+I_{0+}^q f(t).
\end{align*}
\smallskip

\noindent\textbf{Step II.}
 We approximate $f$ by continuously differentiable functions $f_n$ such that
$$
\sup_{t\in J}|f(t)-f_n(t)|\to 0,\quad \text{as } n\to \infty.
$$
Letting
$$
u_n(t)=\lim_{\lambda\to \infty}\int_0^t K_q(s)B_{\lambda}f_n(s)ds,
$$
we have
$$
u_n(t)=A\big(I_{0+}^q u_n(t)\big)+I_{0+}^q f_n(t).\label{e3.16}
$$
Then
\begin{align*}
 |u_n(t)-u_m(t)|
&=\big|\lim_{\lambda\to \infty}\int_0^t K_q(s)B_{\lambda}\big[f_n(s)
 -f_m(s)\big]ds\big|\\
&\leq \frac{M\overline{M}}{\Gamma(q)}\int_0^t(t-s)^{q-1}|f_n(s)-f_m(s)|ds\\
&\leq \frac{M\overline{M}b^q}{\Gamma(q)}\|f_n-f_m\|,
\end{align*}
which implies that $\{u_n\}$ is a Cauchy sequence and its limit,
denoted by $u(t)$, exists. Taking limit on both sides of \eqref{e3.16},
 we obtain
$$
u(t)=A\big(I_{0+}^q u(t)\big)+I_{0+}^q f(t), \quad \text{for }  t\in J.
$$
Therefore, \eqref{e3.15} is the integral solution of \eqref{e1.1}.
This completes the proof.
\end{proof}

\begin{remark} \label{rmk3.3} \rm
(i) Integrating the last term in \eqref{e3.15} and using Proposition \ref{prop3.3},
the integral solution \eqref{e3.15} can be expressed as
$$
u(t)=S_q(t)u_0+\frac{d}{dt}\int_0^t\Phi_q(t-s)f(s)ds.
$$

(ii) $(\lambda^q I-A)^{-1}x=\lambda\int_0^{\infty}e^{-\lambda t}\Phi_q(t)x\,ds$
for $x\in X$ and $\lambda^q>\omega$.
In fact, by taking Laplace transform of \eqref{e3.13}, we obtain
\begin{align*}
\mathfrak{L}[\Phi_q(t)x]
&=A\mathfrak{L}[I_{0+}^q\Phi_q(t)x]+\mathfrak{L}[\frac{t^q}{\Gamma(1+q)}x]\\
&=\lambda^{-q}A\mathfrak{L}[\Phi_q(t)x]+\lambda^{-q-1}x\\
&=\lambda^{-1}(\lambda^q I-A)^{-1}x.
\end{align*}

(iii) We can say that $A$ generates the operator $\{\Phi_q(t)\}_{t\geq 0}$.
 When $q=1$, $\{\Phi_q(t)\}_{t\geq 0}$ degenerates into
 $\{S(t)\}_{t\geq 0}$, which is integrated semigroup generated by $A$
in \cite{1 H.R. Thieme}.
\end{remark}

\section{Integral solution to a nonlinear Cauchy problem}

In this section, we study the existence of integral solution of nonlinear
fractional evolution equation \eqref{e1.2}.
We need the following assumptions:
\begin{itemize}
\item[(H3)] for each $t\in J$, the function $g(t,\cdot): X\to X$
is continuous and for each $x\in X$, the function
$g(\cdot,x): J\to X$ is strongly measurable;

\item[(H4)] there exists a function $m\in L(J, \mathbb{R}^+)$ such
that
\begin{gather*}
{I_{0+}^{q}}m(t)\in C(J,\mathbb{R}^+), \quad\lim_{t\to 0+}{I_{0+}^{q}}m(t)=0,\\
|g(t,x)|\leq m(t)\quad\text{for all $x\in X$ and almost all }t\in J;
\end{gather*}

\item[(H5)] there exists a constant $l>0$ such that for any bounded $D\subseteq X$,
$$
\beta(g(t, D))\leq l \beta(D),\quad\text{for a.e. }t\in J.
$$
\end{itemize}

 By Theorem \ref{thm3.1}, it is easy to see that the integral solution of
\eqref{e1.2} is equal to the solution of
\begin{equation}
 u(t)=S_q(t)u_0+\frac{d}{dt}\int_0^t\Phi_q(t-s)g(s,u(s))ds\label{e4.1}
\end{equation}
or
\begin{equation}
 u(t)=S_q(t)u_0+\lim_{\lambda\to +\infty}
\int_0^tK_q(t-s)B_{\lambda}g(s,u(s))ds.\label{e4.2}
\end{equation}

For  $u\in C(J,X_0)$, define an operator
$$
(\mathscr{T}u)(t)=(\mathscr{T}_1u)(t)+(\mathscr{T}_2u)(t),
$$
where
$$
(\mathscr{T}_1u)(t)=S_q(t)u_0\quad\text{and}\quad
(\mathscr{T}_2u)(t)=\lim_{\lambda\to +\infty}\int_0^tK_q(t-s)B_{\lambda}g(s,u(s))ds,
$$
for all $t\in J$. Let $B_r(J)=\{u\in C(J, X_0): \|u\|\leq r\}$.

\begin{lemma} \label{lem4.1}
Suppose that  conditions {\rm (H1)--(H4)} hold. Then
$\{\mathscr{T}u: u\in B_r(J)\}$ is equicontinuous.
\end{lemma}

\begin{proof}
By Proposition \ref{prop3.3}, $S_{q}(t)u_0$ is uniformly continuous on $J$.
 Consequently, $\{\mathscr{T}_1u:~u\in B_r(J)\}$ is equicontinuous.

For  $u\in B_r(J)$, taking $t_1=0$, $0<t_2\leq b$, we obtain
\begin{align*}
|(\mathscr{T}_2u)(t_{2})-(\mathscr{T}_2u)(0)
&= \big|\lim_{\lambda\to +\infty}\int_0^{t_2} K_q(t-s)B_{\lambda}g(s,u(s))ds\big|\\
&\leq \frac{M\overline{M}}{\Gamma (q)}\int_{0}^{t_2}(t_2-s)^{q-1}
m(s)ds\to 0,\quad\text{as }t_2\to 0.
\end{align*}
For $0<t_1<t_2\leq b$, we have
\begin{align*}
&|(\mathscr{T}_2u)(t_{2})-(\mathscr{T}_2u)(t_{1})|\\
&\leq \Big|\lim_{\lambda\to +\infty}\int_{t_1}^{t_2}(t_2-s)^{q-1}
 P_{q}(t_2-s)B_{\lambda}g(s,u(s))ds\Big|\\
&\quad+\Big|\lim_{\lambda\to +\infty}\int_{0}^{t_1} (t_2-s)^{q-1}
 P_{q}(t_2-s)B_{\lambda}g(s,u(s))ds\\
&-\lim_{\lambda\to +\infty}\int_{0}^{t_1} (t_1-s)^{q-1}P_{q}(t_2-s)
 B_{\lambda}g(s,u(s))ds\Big|\\
&+\Big|\lim_{\lambda\to +\infty}\int_{0}^{t_1} (t_1-s)^{q-1}P_{q}(t_2-s)
 B_{\lambda}g(s,u(s))ds\\
&- \lim_{\lambda\to +\infty}\int_{0}^{t_1} (t_1-s)^{q-1}P_{q}(t_1-s)
 B_{\lambda}g(s,u(s))ds\Big|\\
&\leq \frac{M\overline{M}}{\Gamma(q)}\Big|\int_{t_1}^{t_2} (t_2-s)^{q-1}m(s)ds\Big|\\
&\quad +\frac{M\overline{M}}{\Gamma(q)}\int_{0}^{t_1}
 [ (t_1-s)^{q-1}- (t_2-s)^{q-1}]m(s)ds\\
&\quad +\Big|\lim_{\lambda\to +\infty}\int_{0}^{t_1} (t_1-s)^{q-1}[P_{q}(t_2-s)
 -P_{q}(t_1-s)]B_{\lambda}g(s,u(s))ds\Big|\\
&\leq I_1+I_2+I_3,
\end{align*}
where
\begin{gather*}
 I_1=\frac{M\overline{M}}{\Gamma(q)}\Big|\int_{0}^{t_2} (t_2-s)^{q-1}m(s)ds
 -\int_{0}^{t_1} (t_1-s)^{q-1}m(s)ds\Big|,\\
 I_2=\frac{2M\overline{M}}{\Gamma(q)}\int_{0}^{t_1}
 \Big[ (t_1-s)^{q-1}- (t_2-s)^{q-1}\Big]m(s)ds,\\
 I_3=\Big|\lim_{\lambda\to +\infty}\int_{0}^{t_1} (t_1-s)^{q-1}[P_{q}(t_2-s)
 -P_{q}(t_1-s)]B_{\lambda}g(s,u(s))ds\Big|.
\end{gather*}
By condition (H4), one can deduce that $\lim _{t_2\to t_1}I_1=0$. Noting that
$$
\big[ (t_1-s)^{q-1}- (t_2-s)^{q-1}\big] m(s)\leq  (t_1-s)^{q-1}m(s),
$$
and $\int_{0}^{t_1} (t_1-s)^{q-1}m(s)ds$ exists, it follows by Lebesgue
 dominated convergence theorem that
\[
\int_{0}^{t_1}[ ({t_1}-s)^{q-1}- ({t_2}-s)^{q-1}]m(s)
ds\to 0,\quad\text{as } t_2\to t_1,
\]
which implies that $\lim _{t_2\to t_1}I_2=0$.

For $\varepsilon>0$  small enough, by (H4), we have
\begin{align*}
I_3&\leq \overline{M}\int_{0}^{t_1-\varepsilon} (t_1-s)^{q-1}
 |P_q(t_2-s)-P_q(t_1-s)||g(s,u(s))|ds\\
&\quad +\overline{M}\int_{t_1-\varepsilon}^{t_1} (t_1-s)^{q-1}|P_q(t_2-s)
 -P_q(t_1-s)||g(s,u(s))|ds\\
 &\leq \overline{M} \int^{t_1}_0(t_1-s)^{q-1}m(s)ds
 \sup_{s\in [0,t_1-\varepsilon]}|P_q(t_2-s)-P_q(t_1-s)|\\
 &\quad +\frac{2M\overline{M}}{\Gamma(q)} \int_{t_1-\varepsilon}^{t_1}
 (t_1-s)^{q-1}m(s)ds\\
 &\leq I_{31}+I_{32}+I_{33},
 \end{align*}
where
\begin{gather*}
 I_{31}=\frac{r\Gamma(q)}{M\overline{M}}\sup_{s\in [0,t_1-\varepsilon]}
 |P_q(t_2-s)-P_q(t_1-s)|,\\
 I_{32}=\frac{2M\overline{M}}{\Gamma(q)}
\Big|\int_{0}^{t_1} (t_1-s)^{q-1}m(s)ds-\int_{0}^{t_1-\varepsilon}
 (t_1-\varepsilon-s)^{q-1}m(s)ds\Big|,\\
 I_{33}=\frac{2M\overline{M}}{\Gamma(q)}
\int_{0}^{t_1-\varepsilon}[(t_1-\varepsilon-s)^{q-1}- (t_1-s)^{q-1}]m(s)ds.
\end{gather*}
By Proposition \ref{prop3.1}, it follows that $I_{31}\to 0$ as
$t_2\to t_1$. Applying the arguments similar to the ones employed in
 proving that $I_1,I_2$ tend to
zero, we obtain $I_{32}\to0$ and $I_{33}\to0$ as
$\varepsilon\to0$. Thus, $I_3$ tends to zero independently
of $u\in B_r(J)$ as $t_2\to t_1,~\varepsilon\to 0$. Therefore,
$|(\mathscr{T}_2u)(t_2)-(\mathscr{T}_2u)(t_1)|\to 0$
independently of $u\in B_r(J)$ as $t_2\to t_1$, which implies that
$\{\mathscr{T}_2u:~u\in B_r(J)\}$ is equicontinuous.
Therefore, $\{\mathscr{T}u:~u\in B(J)\}$ is equicontinuous. The proof is complete.
\end{proof}

\begin{lemma} \label{lem4.2}
Assume that {\rm (H1)--(H4)} hold. Then $\mathscr{T}$ maps
$B_r(J)$ into $B_r(J)$, and is continuous in $B_r(J)$.
\end{lemma}

\begin{proof}
\textbf{Claim I.} $\mathscr{T}$ maps $B_r(J)$ into $B_r(J)$.
Obviously, by (H4), there exists a constant $r>0$ such that
$$
 M\Big(|u_0|+\sup_{t\in J}\Big\{\frac{\overline{M}}{\Gamma(q)}
\int_{0}^{t}(t-s)^{q-1}m(s)ds\Big\}\Big)\leq r.
$$
For any $u\in B_r(J)$, by Proposition \ref{prop3.2}, we have
\begin{align*}
|(\mathscr{T}u)(t)|
&\leq \left|S_{q}(t)u_0\right|+\big|\lim_{\lambda\to +\infty}\int_0^{t}
K_{q}(t-s)B_{\lambda}g(s,u(s))ds\big|\\
&\leq M|u_0|+\frac{M\overline{M}}{\Gamma
(q)}\int_{0}^{t}(t-s)^{q-1}|g(s,u(s))|ds\\
 &\leq M\Big(|u_0|+\sup_{t\in J}\Big\{\frac{\overline{M}}{\Gamma(q)}
\int_{0}^{t}(t-s)^{q-1}m(s)ds\Big\}\Big)
\leq r.
\end{align*}
Hence $\|\mathscr{T}u\| \leq r$ for any $u\in B_r(J)$.
\smallskip

\noindent\textbf{Claim II.} $\mathscr{T}$ is continuous in $B_r(J)$.
For any $u_m,u\in B_r(J)$, $m=1,2,\dots$, with $\lim_{m\to \infty}u_{m}=u$,
 by (H3), we have
$$
g(t,u_m(t))\to g(t,u(t)) \quad\text{as }m\to\infty,
$$
for $t\in J$. On the one hand, using (H4), for each $t\in J$, we obtain
$$
(t-s)^{q-1}|g(s, u_m(s))-g(s, u(s))|\leq 2(t-s)^{q-1}m(s), \quad
\text{a.e. in }[0, t).
$$
As the function $s\to 2(t-s)^{q-1}m(s)$ is integrable for $s\in [0,t)$
and $t\in J$, by Lebesgue dominated convergence theorem, we obtain
$$
\int^t_0(t-s)^{q-1}|g(s, u_m(s))-g(s, u(s))|ds\to 0 \quad\text{as }m\to \infty.
$$
For $t\in J$, we obtain
\begin{align*}
&|(\mathscr{T}u_m)(t)-(\mathscr{T}u)(t)|\\
&\leq \Big|\lim_{\lambda\to +\infty}\int_0^{t}K_{q}(t-s)
 B_{\lambda}(g(s,u_m(s))-g(s,u(s)))ds \Big|\\
&\leq \frac{M \overline{M}}{\Gamma(q)}\int_0^t (t-s)^{q-1}
 |g(s,u_m(s))-g(s,u(s))|ds
 \to 0\quad \text{as } m\to \infty.
\end{align*}
Therefore, $\mathscr{T}u_m\to \mathscr{T}u$ pointwise on $J$ as
 $m\to \infty$. Hence it follows by Lemma \ref{lem4.1} that
$\mathscr{T}u_m\to \mathscr{T}u$ uniformly on $J$ as $m\to \infty$
and so $\mathscr{T}$ is continuous. The proof is complete.
\end{proof}

\begin{theorem} \label{thm4.3}
 Assume that {\rm (H1)--(H5)} hold. Then the Cauchy problem \eqref{e1.2}
has at least one integral solution in $B_r(J)$.
\end{theorem}

\begin{proof}
Let $y_0(t)=S_{q}(t)u_0$ for all $t\in J$ and $y_{m+1}=\mathscr{T}y_m$,
$m=0, 1, 2, \cdots$. Consider the set $\mathscr{H}=\{y_m: m=0,1,2,\cdots\}$,
and show that it is relatively compact.

By Lemmas \ref{lem4.1} and \ref{lem4.2}, $\mathscr{H}$ is uniformly bounded
and euqicontinuous on $J$. Next, for any $t\in J$, we just need to show that
$\mathscr{H}(t)=\{y_m(t), m=0,1,2,\cdots\}$ is relatively compact in $X_0$.

By the assumption (H5) together with Lemmas \ref{lem2.1} and \ref{lem2.2},
for any $t\in J$, we have
$$
\beta\Big(\mathscr{H}(t)\Big)=\beta\Big(\{y_m(t)\}_{m=0}^\infty\Big)
=\beta\Big(\{y_0(t)\}\cup\{y_m(t)\}_{m=1}^\infty\Big)
=\beta\Big(\{y_m(t)\}_{m=1}^\infty\Big)
$$
and
\begin{align*}
 \beta\Big(\{y_m(t)\}_{m=1}^\infty\Big)
&= \beta\Big(\{(\mathscr{T}y_m)(t)\}_{m=0}^\infty\Big)\\
&=  \beta\Big(\Big\{S_{q}(t)u_0+\lim_{\lambda\to +\infty}\int_0^{t}
K_{q}(t-s)B_{\lambda}g(s,y_m(s))ds\Big\}_{m=0}^\infty\Big)\\
&=  \beta\Big(\Big\{\lim_{\lambda\to +\infty}\int_0^{t}
K_{q}(t-s)B_{\lambda}g(s,y_m(s))ds\Big\}_{m=0}^\infty\Big)\\
&\leq \frac{2M\overline{M}}{\Gamma(q)}\int_0^{t}
(t-s)^{1-q}\beta\Big(g(s,\{y_m(s)\}_{m=0}^\infty)\Big)ds\\
&\leq \frac{2M\overline{M}l}{\Gamma(q)}\int_0^{t}
(t-s)^{1-q}\beta\Big(\{y_m(s)\}_{m=0}^\infty\Big)ds.
\end{align*}
Thus
$$
\beta(\mathscr{H}(t))\leq \frac{2M\overline{M}l}{\Gamma(q)}\int_0^{t}
(t-s)^{1-q}\beta(\mathscr{H}(s))ds.
$$
Therefore, by generalized Grownwall's inequality \cite{H.P. Ye}, we
 infer that $\beta(\mathscr{H}(t))=0$. In consequence, $\mathscr{H}(t)$
is relatively compact. Hence, it follows from Ascoli-Arzela theorem that
 $\mathscr{H}$ is relatively compact. Therefore, there exists a convergent
subsequence of $\{y_m\}_{m=0}^\infty$. For the sake of clarity, let
$\lim_{m\to \infty}y_{m}=y^*\in B_r(J)$. Thus, by continuity of the operator
 $\mathscr{T}$, we have
$$
y^*=\lim_{m\to \infty}y_m=\lim_{m\to \infty}\mathscr{T}y_{m-1}
=\mathscr{T}\big(\lim_{m\to \infty}y_{m-1}\big)
=\mathscr{T}y^*,
$$
which implies the Cauchy problem \eqref{e1.2} has least an integral solution.
\end{proof}

\section{An example}

 As an application of our results we consider the  fractional time
partial differential equation
\begin{equation}
\begin{gathered}
\frac{\partial^q}{\partial t^q}z(t,x)
=\frac{\partial^2}{\partial x^2}z(,x)+G(t,z(t,x)), \quad x\in [0,\pi],\;
t\in (0, b],\; 0<q<1,\\
z((t,0)=z(t,\pi)=0, \quad t\in (0, b],\\
z(0,x)=z_0, \quad x\in [0,\pi],
\end{gathered}\label{e5.1}
\end{equation}
where $G: [0, b]\times \mathbb{R} \to \mathbb{R}$ is a given function. Let
\begin{gather*}
u(t)(x) = z(t, x),\quad  t\in [0,b],\; x\in [0,\pi],\\
g(t,u)(x)=G(t,u(x)),\quad t\in [0,b],\; x\in [0,\pi].
\end{gather*}
We choose $X=C([0,\pi],\mathbb{R})$ endowed with the uniform topology and
consider the operator $A: D(A)\subset X\to X$ defined by:
$$
D(A)=\{u\in C^2([0,\pi], \mathbb{R}): u(0)=u(\pi)=0\}, \quad Au=u''.
$$
It is well known that the operator $A$ satisfies the Hille-Yosida condition
with $(0,+\infty)\subset \rho(A)$, $\|(\lambda I-A)^{-1}\|\leq \frac{1}{\lambda}$
for $\lambda >0$, and
$$
\overline{D(A)}=\{u\in X: u(0)=u(\pi)=0\}\neq X.
$$
We can show that problem \eqref{e1.2} is an abstract formulation of
problem \eqref{e5.1}. Under suitable conditions, Theorem \ref{thm4.3}
implies that problem \eqref{e5.1} has a unique solution $z$ on
$[0, b]\times [0, \pi]$.


\subsection*{Concluding remarks}

 In this article, we have obtained the integral solution for nonhomogeneous
Cauchy problem \eqref{e1.1} and established the relationship between
$\{S_q(t)\}_{t\geq 0}$ and $\{K_q(t)\}_{t\geq 0}$. Also sufficient conditions
ensuring the existence of integral solutions to nonlinear Cauchy
problem \eqref{e1.2}, involving a linear closed operator $A$ of Hille-Yosida
type with not densely defined domain, are presented.

For further research, we propose the following open problem:
How to establish the existence of an integral solution to fractional
evolution equation \eqref{e1.2} when linear closed operator
 $A$ is not a Hille-Yosida type and its domain is not densely defined?

\subsection*{Acknowledgements}
 The authors would like to
express their thanks to the editor and anonymous referees for their
suggestions and comments that improved the quality of the paper.
The second author is supported by National Natural Science Foundation
of China (11671339).


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