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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 249, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/249\hfil
 Riemann-Liouville fractional cosine functions]
{Riemann-Liouville fractional cosine functions}

\author[Z.-D. Mei, J.-G. Peng \hfil EJDE-2016/249\hfilneg]
{Zhan-Dong Mei, Ji-Gen Peng}

\address{Zhan-Dong Mei (corresponding author) \newline
School of Mathematics and Statistics,
Xi'an Jiaotong University, Xi'an 710049, China}
\email{zhdmei@mail.xjtu.edu.cn}

\address{Ji-Gen Peng \newline
School of Mathematics and Statistics,
 Xi'an Jiaotong University, Xi'an 710049, China}
\email{jgpeng@mail.xjtu.edu.cn}

\thanks{Submitted June 7, 2015. Published September 16, 2016.}
\subjclass[2010]{33E12}
\keywords{Riemann-Liouville fractional cosine function; fractional resolvent; 
\hfill\break\indent fractional differential equation}

\begin{abstract}
 In this article, we present the  notion of Riemann-Liouville fractional cosine
 function. we prove that a Riemann-Liouville $\alpha$-order
 fractional cosine function is equivalent to the Riemann-Liouville $\alpha$-order
 fractional resolvent introduced in \cite{Mei2015}.
\end{abstract}

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

\section{Introduction}

Let $X$ be a Banach space, and $A: D(A)\subset X\to X$,
$B: D(B)\subset X\to X$ be  closed linear operators.
It is well-known that $C_0$-semigroups are important
tools to study the abstract Cauchy problem of first order
\begin{equation}\label{firstorder}
\begin{gathered}
 \frac{du(t)}{dt}=Au(t), \quad t>0\\
 u(0)=x,
 \end{gathered}
\end{equation}
and that the cosine function essentially characterizes the abstract Cauchy
problem of second order
\begin{equation}\label{secondorder}
\begin{gathered}
 \frac{d^2u(t)}{dt^2}=Bu(t), \quad t>0\\
 u(0)=x, u'(0)=0.
 \end{gathered}
\end{equation}
Here a $C_0$-semigroup is a family $\{T(t)\}_{t\geq 0}$ of strongly
continuous and bounded linear operators defined on $X$ satisfying $T(0)=I$
and $T(t+s)=T(t)T(s)$, $t,s\geq 0$; a cosine function is a family
$\{S(t)\}_{t\geq 0}$ of strongly continuous and bounded linear operators
defined on $X$ satisfying $S(0)=I$ and $2S(t)S(s)=S(t)+S(s)$, $t\geq s\geq 0$.

Concretely, system \eqref{firstorder} is well-posed if and only if $A$
generates a $C_0$-semigroup $\{T(t)\}_{t\geq 0}$, namely,
$Ax=\lim_{t\to 0^+}t^{-1}(T(t)x-x)$ with domain
$D(A)=\{x\in D(A):\lim_{t\to 0^+}t^{-1}(T(t)x-x)$ exists $\}$;
system \eqref{secondorder} is well-posed if and only if $B$ generates
a cosine function $\{S(t)\}_{t\geq 0}$, namely,
$Bx=2\lim_{t\to 0^+}t^{-2}(S(t)x-x)$ with domain
$D(B)=\{x\in D(B):\lim_{t\to 0^+}t^{-2}(S(t)x-x)$ exists $\}$.
Therefore, pure algebraic methods can be used to study
abstract Cauchy problems of first and second orders.
For details, we refer to \cite{Engel2000,Goldstein1985}.

However, equations of integer order such as \eqref{firstorder}
and \eqref{secondorder} cannot exactly describe the behavior of many physical
systems; fractional differential equations maybe more suitable for describing
anomalous diffusion on fractals (physical objects of fractional dimension,
like some amorphous semiconductors or strongly porous materials;
see \cite{Anh2001,Metzler2001} and the references therein), fractional
random walk \cite{Germano2009,Scalas2006}, etc. Fractional derivatives appear
in the theory of fractional differential equations; they describe the property of
memory and heredity of materials, and it is the major advantage of
fractional derivatives compared with integer order derivatives.
Let $\alpha>0$ and $m=[\alpha]$, The smallest integer larger than
or equal to $\alpha$. There are mainly two types of $\alpha$-order
fractional differential equations, which are most used in the real problems.

(1) Caputo fractional abstract Cauchy problem
\begin{equation}\label{caputo}
 \begin{gathered}
 ^CD_t^\alpha u(t)=Au(t), \quad t>0,\\
 u(0)=x, u^{(k)}(0)=0,\quad k=1,2, \dots, m-1.
 \end{gathered}
\end{equation}
where $^CD_t^\alpha$ is the Caputo fractional differential operator
defined as follows:
\begin{align*}
 ^CD_t^\alpha
 u(t)=\frac{1}{\Gamma(m-\alpha)}\int_0^t(t-\sigma)^{-\alpha}u^{(m)}(\sigma)\,d\sigma;
\end{align*}

 (2) Riemann-Liouville fractional abstract Cauchy
problem
\begin{equation}\label{R-L}
  \begin{gathered}
 D_t^\alpha u(t)=Au(t), \quad \\
 (g_{2-\alpha}*u)(0)=\lim_{s\to 0^+}\int_0^s
 \frac{(s-\sigma)^{m-1-\alpha}}{\Gamma(2-\alpha)}u(\sigma)\,d\sigma= x,  \\
 (g_{2-\alpha}*u)^{(k)}(0)=\lim_{s\to 0^+}\int_0^s\frac{d^k}{dt^k}
\frac{(s-\sigma)^{m-1-\alpha}}{\Gamma(m-\alpha)}u(\sigma)\,d\sigma
=0, \\
 k=1,2, \dots, m-1.
 \end{gathered}
\end{equation}
where  the Riemann-Liouville fractional differential operator is
\begin{align*}
 D_t^\alpha
 u(t)=\frac{1}{\Gamma(m-\alpha)}\frac{d}{dt}
\int_0^t(t-\sigma)^{m-1-\alpha}u(\sigma)\,d\sigma.
\end{align*}

Obviously, \eqref{firstorder} is just the limit state of equations
\eqref{caputo} and \eqref{R-L} as $\alpha\to 1$, and \eqref{secondorder}
is just the limit state of equations \eqref{caputo} and \eqref{R-L} as
 $\alpha\to 2$. Initial conditions for the Caputo fractional derivatives
are expressed in terms of initials of integer order
derivatives \cite{Eidelman2004,Mei2014,Peng2012}.
For some real materials, initial conditions should be expressed in terms of
Riemann-Liouville fractional derivatives, and it is possible to
obtain initial values for such initial conditions by appropriate
measurements \cite{Heymans2006, Hilfer2000}.

To study Caputo fractional abstract Cauchy problem \eqref{caputo},
Bajlekova \cite{Bazhlekova2001} introduced the important notion
of solution operator for equations \eqref{caputo} as follows.

\begin{definition}\label{solutiona} \rm
A family $\{T(t)\}_{t\geq 0}$ of bounded linear operators of $X$ is
called a solution operator for \eqref{caputo} if the following three
conditions are satisfied:
\begin{itemize}
\item[(a)] $T(t)$ is strongly continuous for $t\geq 0$ and $T(0)=I$,

\item[(b)] $T(t)D(A)\subset D(A)$ and $AT(t)x=T(t)Ax$ for all $x\in D(A)$
and $t\geq 0$,

\item[(c)] for any $x\in D(A)$, it holds
\begin{align*}
 T(t)x=x+J_t^\alpha T(t)Ax,\quad  t\geq 0,
\end{align*}
where
\[
 J_t^\alpha
f(t)=\frac{1}{\Gamma(\alpha)}\int_0^t(t-\sigma)^{\alpha-1}f(t)dt.
\]
\end{itemize}
\end{definition}

Chen and Li \cite{Chen2010}  developed  a notion of $\alpha$-resolvent
operator function, which was proved to be a new characteristic of solution operator.
 Hence, Caputo fractional abstract Cauchy problem can be studied by pure
algebraic methods.
The definition of $\alpha$-resolvent operator function is as follows.

\begin{definition}\label{lichen} \rm
Let $\{S(t)\}_{t\geq 0}$ be a family of bounded linear operators on
$X$. Then $\{S(t)\}_{t\geq 0}$ is called to be an $\alpha$-resolvent
operator function, if the following assumptions are satisfied:
\begin{itemize}
\item[(1)] $S(t)$ is strongly continuous and $S(0)=I.$

\item[(2)] $S(s)S(t)=S(t)S(s)$ for all $t,s\geq 0$.

\item[(3)] $S(s)J_t^\alpha S(t)-J_s^\alpha S(s)S(t)=J_t^\alpha
S(t)-J_s^\alpha S(s)$ for all $t,s\geq 0$.
\end{itemize}
\end{definition}

Li and Peng \cite{Li2012} proposed the following notion of fractional
resolvent to study Riemann-Liouville $\alpha$-order fractional abstract
Cauchy problem \eqref{R-L} with $\alpha\in (0,1)$.

\begin{definition}[\cite{Li2012}]\label{def1} \rm
 Let $0<\alpha< 1$. A family $\{T(t)\}_{t>0}$ of bounded linear operators
on Banach space $X$ is called an  $\alpha$-order fractional resolvent
 if it satisfies the following assumptions:
\begin{itemize}
\item[(1)] for any $x\in X$, $\ T(\cdot)x\in C((0,\infty),X)$, and
\begin{equation}\label{clear}
\lim_{t\to 0+}\Gamma(\alpha)t^{1-\alpha}T(t)x=x \quad 
\text{for all }  x\in X;
\end{equation}

\item[(2)] $T(s)T(t)=T(t)T(s) \quad \text{for all } t,s>0$;

\item[(3)] for all $t,s>0$, it holds
\begin{equation}\label{fi}
\ T(t) J_{s}^{\alpha}T(s)-
J_{t}^{\alpha}T(t)T(s)=\frac{t^{\alpha-1}}{\Gamma(\alpha)}J_{s}^{\alpha}T(s)
-\frac{s^{\alpha-1}}{\Gamma(\alpha)}J_{t}^{\alpha}T(t).
\end{equation}
\end{itemize}
\end{definition}

In \cite{Mei2015}, we studied  the Riemann-Liouville $\alpha$-order fractional 
Cauchy problem \eqref{R-L} with order $\alpha\in (1,2)$. There 
 Riemann-Liouville $\alpha$-order fractional resolvent defined as follows.

\begin{definition}\label{def2} \rm
 A family $\{T(t)\}_{t>0}$ of bounded linear operators is called Riemann-Liouviille
$\alpha$-order fractional resolvent if it satisfies the following
assumptions:
\begin{itemize}
\item[(a)] For any $x\in X$, $T_{\alpha}(\cdot)x\in C((0,\infty),X)$, and
\begin{equation}\label{clear2}
\lim_{t\to 0^+}\Gamma(\alpha-1)t^{2-\alpha}T(t)x=x \quad
\text{for all } \ x\in X;
\end{equation}

\item[(b)] $T(s)T_{\alpha}(t)=T(t)T_{\alpha}(s) \quad \text{for all } t,s>0$;

\item[(c)] for all $t,s>0$, it holds
\begin{equation}\label{fi2}
 T(s) J_{t}^{\alpha}T(t)- J_{s}^{\alpha}T(s)T(t)
=\frac{s^{\alpha-2}}{\Gamma(\alpha-1)}J_{t}^{\alpha}T(t)
-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}J_{s}^{\alpha}T(s).
\end{equation}
\end{itemize}
\end{definition}

The linear operator $A$ defined by
\begin{gather*}
Ax=\lim_{t\to 0^+}\frac{t^{1-\alpha}T(t)x-\frac{1}{\Gamma(\alpha)}x}{t^{2\alpha}}, \\
\text{for } x \in D(A)=\big\{x\in X: \lim_{t\to 0^+}\frac{t^{1-\alpha}T(t)x
-\frac{1}{\Gamma(\alpha)}x}{t^{2\alpha}} \text{ exists}\big\}\,.
\end{gather*}
Operator $A$ generates a Riemann-Liouville $\alpha$-order fractional 
resolvent $\{T(t)\}_{t>0}$ in Definition \ref{def1}.

Also, we proved that $\{T(t)\}_{t>0}$ is a Riemann-Liouville $\alpha$-order 
fractional resolvent if and only if it is a solution operator defined as follows.

\begin{definition}\label{solutionb}
A family $\{T(t)\}_{t>0}$ of bounded linear operators of $X$ is
called a solution operator for \eqref{R-L} if the following three
conditions are satisfied:
\begin{itemize}
\item[(a)] $T(t)$ is strongly continuous for $t> 0$ and 
$\lim_{t\to 0^+}\Gamma(\alpha-1)t^{2-\alpha}T(t)x=x$, $x\in X$,

\item[(b)] $T(t)D(A)\subset D(A)$ and $AT(t)x=T(t)Ax$ for all $x\in D(A)$
and $t> 0$,

\item[(c)] for any $x\in D(A)$, it holds
\[
 T(t)x=\frac{t^{1-\alpha}}{\Gamma(2-\alpha)}x+J_t^\alpha T(t)Ax,\quad t> 0.
\]
\end{itemize}
\end{definition}

However, the above functional equations for fractional differential equations
 are not expressed in terms of the sum of time variables: $s+t$. 
This is very important in concrete applications of the functional equation, 
just like $C_0$-semigroups, cosine functions.
 Motivated by this, Peng and Li \cite{Peng2012} established the characterization 
of $\alpha$-order fractional semigroup with $\alpha\in (0,1)$:
\begin{align*}
&\int_0^{t+s}\frac{T(\tau)}{(t+s-\tau)^\alpha}\,d\tau-
 \int_0^t\frac{T(\tau)}{(t+s-\tau)^\alpha}\,d\tau-
 \int_0^s\frac{T(\tau)}{(t+s-\tau)^\alpha}\,d\tau\\
&=\alpha\int_0^t\int_0^s\frac{T(r_1)T(r_2)}{(t+s-r_1-r_2)^{1+\alpha}}dr_1dr_2,
\quad t,s\geq 0,
\end{align*}
where the integrals are in the sense of strong operator topology.
Concretely, they proved that $\alpha$-order
fractional semigroup is closely related to the solution operator of
Caputo fractional abstract Cauchy problem \eqref{caputo}.

Mei, Peng and Zhang \cite{Mei2013} developed the notion of 
Riemann-Liouville fractional semigroup as follows.

\begin{definition} \rm
A family $\{T(t)\}_{t>0}$ of bounded linear operators is called
 Riemann-Liouville $\alpha$-order fractional semigroup on Banach
space $X$, if the following conditions are satisfied:
\begin{itemize}
\item[(i)] for any $x\in X$, $t\mapsto T(t)x$ is continuous over
$(0,\infty)$ and
\begin{equation}\label{clear3}
\lim_{t\to 0+}\Gamma(\alpha)t^{1-\alpha}T(t)x=x;
\end{equation}

\item[(ii)] for all $t,s> 0$, it holds
\begin{equation}\label{cosin}
\Gamma(1-\alpha)T(t+s)=\alpha\int_0^t\int_0^s
\frac{T(r_1)T(r_2)}{(t+s-r_1-r_2)^{1+\alpha}}dr_1dr_2,
\end{equation}
where the integrals are in the sense of strong operator topology.
\end{itemize}
\end{definition}
It is proved in \cite{Mei2013} that $A$ generates a Rimann-Liouville 
fractional semigroup if and only if it generates a fractional 
resolvent developed in \cite{Li2010}.

To study Caputo fractional Cauchy problem of order $\alpha\in (1,2)$, 
we recently studied in \cite{Mei2014} the notion of fractional cosine 
function as follows.

\begin{definition} \rm
A family $\{T(t)\}_{t\geq 0}$ of bounded and strongly continuous operators 
is called an $\alpha$-fractional cosine function if $T(0)=I$ and it holds
\begin{equation}\label{cosin3}
\begin{aligned}
&\int_0^{t+s}\int_0^\sigma\frac{T(\tau)}{(t+s-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma-\int_0^t\int_0^\sigma\frac{T(\tau)}{(t+s-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma\\
&-\int_0^s\int_0^\sigma\frac{T(\tau)}{(t+s-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma\\
&=\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
&\quad -\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma, \quad t,s\geq 0,
\end{aligned}
\end{equation}
where the integrals are in the sense of strong operator topology.
\end{definition}

There, we proved that $A$ generates a fractional cosine function 
$\{T(t)\}_{t\geq 0}$ if and only if it generates an $\alpha$-resolvent 
operator function; that is, the following equalities hold:
\begin{equation*}
T(s)J_t^\alpha T(t)-J_s^\alpha T(s)T(t)=J_t^\alpha
T(t)-J_s^\alpha T(s), \quad t,s\geq 0.
\end{equation*}

As stated above, functional equations involving $t$, $s$ and $t+s$ have been 
discussed for Caputo fractional differential equations \eqref{caputo}
 with $\alpha\in (0,1)$ and $\alpha\in (1,2)$, Riemann-Liouville 
fractional equation \eqref{R-L} with $\alpha\in (0,1)$. 
To close the gap, we will discuss the residual case, that is, functional 
equations involving $t$, $s$ and $t+s$ for Riemann-Liouville fractional 
equation \eqref{R-L} with $\alpha\in (1,2)$. To this end, we first 
consider the special case that $T(\cdot)$ is exponentially
bounded (hence it is Laplace transformable).
Take laplace transform on both sides of \eqref{fi} with respect to $s$ 
and $t$ to obtain
\begin{equation}\label{lap}
 (\lambda^{-\alpha}-\mu^{-\alpha})\hat{T}(\mu)\hat{T}(\lambda)
 =\lambda^{1-\alpha}\mu^{1-\alpha}(\lambda^{-1}\hat{T}(\lambda)
 -\mu^{-1}\hat{T}(\mu)).
\end{equation}
It follows from \cite[(3.8)]{Mei2014} that the Laplace transform of the 
right-hand side of \eqref{cosin} satisfies
\begin{equation}\label{cosin1}
\begin{aligned}
 &\int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}\Big(\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
  &-\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma\Big) \,ds\,dt\\
&=\frac{\Gamma(2-\alpha)(\lambda^\alpha-\mu^\alpha)}{\lambda\mu(\lambda-\mu)}
\hat{T}(\mu)\hat{T}(\lambda).
\end{aligned}
\end{equation}
The combination of \eqref{lap} and \eqref{cosin1} implies
\begin{align*}
 &\int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}\Big(\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
 &-\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma\Big) \,ds\,dt\\
&=\frac{\Gamma(2-\alpha)(\lambda^{-1}\hat{T}(\lambda)
-\mu^{-1}\hat{T}(\mu))}{\mu-\lambda}.
\end{align*}
Let $m(t)=\int_0^t T(\sigma)\,d\sigma$, by similar proof of
\cite[(4.2)]{Keyantuo2009}, it holds
\begin{align*}
 \int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
m(t+s)\,ds\,dt
=\frac{\hat{m}(\mu)-\hat{m}(\lambda)}{\lambda-\mu}
=\frac{\lambda^{-1}\hat{T}(\lambda)
-\mu^{-1}\hat{T}(\mu)}{\mu-\lambda}.
\end{align*}
By the Laplace transform, it follows that
\begin{equation}\label{cosin2}
\begin{aligned}
 &\Gamma(2-\alpha)\int_0^{t+s}T(\sigma)\,d\sigma\\
&=\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
&\quad -\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma,
\end{aligned}
\end{equation}

In the following two sections, we show that \eqref{cosin2} also holds 
without the assumption that $\{T(t)\}_{t>0}$ is exponentially bounded 
and it essentiality describes a Riemann-Liouville fractional resolvent.

\section{Riemann-Liouville fractional cosine function}

Equality \eqref{fi} is an important functional equation for the solution 
of \eqref{R-L} with $\alpha\in(1,2)$. However, as  stated in the introduction, 
\eqref{fi} does not write the functional equation in terms of the sum of 
time variables: $s+t$. This is very important in concrete applications 
of the algebraic functional equation. Therefore, it is very valuable 
to study functional equation \eqref{cosin2}, which appears in the 
following definitions.

\begin{definition}\label{DEF} \rm
A family $\{T(t)\}_{t> 0}$ of bounded linear operators is called
Riemann-Liouville $\alpha$-order fractional cosine function on a
Banach space $X$, if the following conditions are satisfied:
\begin{itemize}
\item[(i)] $T(t)$ is strongly continuous, that is, for any $x\in X$, the
mapping $t\mapsto T(t)x$ is continuous over $(0,\infty)$;

\item[(ii)] it holds 
\begin{equation}\label{clear4}
\lim_{t\to
0+}t^{2-\alpha} T(t)x
=\frac{x}{\Gamma(\alpha-1)} \quad \text{for all } \ x\in X;
\end{equation}

\item[(iii)] for all $t,s> 0$, it holds
\begin{equation}\label{sin}
\begin{aligned}
 & \Gamma(2-\alpha)\int_0^{t+s}T(\sigma)\,d\sigma\\
&=\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
&\quad -\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma,
\end{aligned}
\end{equation}
where the integrals are in the sense of strong operator topology.
\end{itemize}
\end{definition}

\begin{lemma}\label{commutative}
Let $\{T(t)\}_{t> 0}$ be a Riemann-Liouville $\alpha$-order fractional cosine 
on Banach space $X$. Then $\{T(t)\}_{t> 0}$ is commutative, i.e. $T(t)T(s)=T(s)T(t)$
 for all $t,s > 0$.
\end{lemma}

\begin{proof}
Observe that the left-hand side of \eqref{sin} is symmetric with respect 
to $t$ and $s$. Hence we can obtain the  equality
\begin{align*}
&\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau \,d\sigma
+\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(s-\tau)^{\alpha-1}}\,d\tau\,d\sigma\\
& -\int_0^t\int_0^s\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma\\
&= \int_0^s\int_0^t\frac{T(\sigma)T(\tau)}{(s-\sigma)^{\alpha-1}}\,d\tau\,d\sigma
 +\int_0^s\int_0^t\frac{T(\sigma)T(\tau)}{(t-\tau)^{\alpha-1}}\,d\tau
\,d\sigma \\
&\quad -\int_0^s\int_0^t\frac{T(\sigma)T(\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma,\quad  t,s> 0.
\end{align*}
The commutative property is proved as in \cite[Proposition 3.4]{Mei2013}.
\end{proof}

\begin{definition}\label{sss} \rm
Let $\{T(t)\}_{t>0}$ be a Riemann-Liouville $\alpha$-order fractional cosine 
function on Banach space $X$. Denote by $D(A)$ the set of all $x\in X$ such 
that the limit
\begin{align*}
 \lim_{t\to 0^+}\Gamma(\alpha+1)t^{-\alpha}J_t^{2-\alpha}\Big(T(t)x-
 \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x\Big)
\end{align*}
exists. Then the operator $A:D(A)\to X$ defined by
\begin{align*}
 Ax=\lim_{t\to 0^+}\Gamma(\alpha+1)t^{-\alpha}J_t^{2-\alpha}\Big(T(t)x-
 \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x\Big)
\end{align*}
is called the generator of $\{T(t)\}_{t> 0}.$
\end{definition}

\begin{proposition}\label{pr2}
Assume $\{T(t)\}_{t>0}$ is a Riemann-Liouville $\alpha$-order fractional 
cosine function on Banach space $X$. Suppose that $A$ is the generator
 of $\{T(t)\}_{t>0}$. Then
\begin{itemize}
\item[(a)] For any $x\in X$ and $t>0$, it holds $J_t^\alpha T(t)x\in D(A)$
 and
 \begin{equation}\label{aabbcc}
 T(t)x=\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x+AJ_t^\alpha T(t)x;
 \end{equation}

\item[(b)] $T(t)D(A)\subset D(A)$ and $T(t)Ax=AT(t)x$, for all $x\in  D(A)$.

\item[(c)] For all $x\in D(A)$, we have
\begin{align*}
 T(t)x=\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x+J_t^\alpha T(t)Ax;
 \end{align*}

\item[(d)] $A$ is equivalently defined by
\begin{equation}\label{ssf}
 Ax=\Gamma(2\alpha-1)\lim_{t\to
0^+}\frac{T(t)x-\frac{t^{\alpha-2}}
{\Gamma(\alpha-1)}x}{t^{2\alpha-2}}
\end{equation}
and $D(A)$ is just consists of those $x\in X$ such that the above
limit exists.

\item[(e)] $A$ is closed and densely defined.

\item[(f)] $A$ admits at most one Riemann-Liouville $\alpha$-order fractional 
cosine function.
\end{itemize}
\end{proposition}

\begin{proof}
(a) Let $x\in X$ and $b>0$ be fixed. Denote by $g_{b}(\cdot)$ the
truncation of $T(\cdot)$ at $b$; that is,
\[
 g_{b}(\sigma)= \begin{cases}
 T(\sigma), &\text{if } 0< \sigma\leq b\\
 0, & \text{if } \sigma>b.
 \end{cases}
\]
Define the function
$H_b(r,s)$ for $r,s> 0$ by
\begin{equation}\label{Htrs}
 H_b(r,s)=\Big(g_b(r)-\frac{r^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)J_s^\alpha
g_b(s)x.
\end{equation}
Obviously, for $0< r\leq t$,
\begin{equation}\label{HET}
 H_t(r,t)=\Big(T(r)-\frac{r^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)J_t^\alpha T(t)x.
\end{equation}
Take Laplace transform with respect to $r$ and $s$ successively for
both sides of \eqref{Htrs} to obtain
\begin{equation}\label{htl}
 \hat{H}_b(\mu,\lambda)
=\lambda^{-\alpha}\hat{g}_b(\mu)\hat{g}_b(\lambda)x
 -\lambda^{-\alpha}\mu^{1-\alpha}\hat{g}_b(\lambda)x.
\end{equation}

Denote by $L(t,s)$ and $R(t,s)$ the left and right sides of 
\eqref{sin}, respectively. Moreover, denote by $R_b(t,s)$, and $L_b(t,s)$ 
the quantities resulted by replacing
$T(t)$ with $g_b(t)$ in $R(t,s)$, $L(t,s)$,
respectively.

It follows from \cite[(3.7)]{Mei2014} that the Laplace transform of 
$R_b(t,s)$ with respect to $t$ and $s$ is given by
\begin{equation}\label{R}
\hat{R}_b(\mu,\lambda)
=\frac{\Gamma(2-\alpha)(\lambda^\alpha-\mu^\alpha)}{\lambda\mu(\lambda-\mu)}
\hat{g}_b(\mu)\hat{g}_b(\lambda).
\end{equation}

For all $t> 0$, the Laplace transform of $\hat{L}_b(t,s)$ with respect to
 $s$ and $t$ can be obtained as 
\begin{equation}\label{L}
 \hat{L}_b(\mu,\lambda)
 =\Gamma(2-\alpha)\frac{\lambda^{-1}\hat{g_b}(\lambda)
-\mu^{-1}\hat{g_b}(\mu)}{\mu-\lambda}.
\end{equation}
Combine \eqref{htl}, \eqref{R} and \eqref{L} to derive
\begin{align*}
 \hat{H}_b(\mu,\lambda)
&=\mu^{-\alpha}\hat{g}_b(\mu)\hat{g}_b(\lambda)x
 -\mu^{-\alpha}\lambda^{1-\alpha}\hat{g}_b(\mu)x\\
&\quad +\frac{\lambda^{1-\alpha}\mu^{1-\alpha}(\lambda-\mu)}{\Gamma(2-\alpha)}(\hat{L}_b(\mu,\lambda)-\hat{R}_b(\mu,\lambda))x.
\end{align*}
Take inverse Laplace transform to obtain
\begin{align*}
 H_b(r,s)&= \Big(g_b(s)-\frac{s^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)J_r^\alpha g_b(r)x\\
 &+\frac{[(D_s^{2-\alpha})J_r^{\alpha-1}-(D_r^{2-\alpha})J_s^{\alpha-1}]\cdot[L_b(r,s)-R_b(r,s)]x}{\Gamma(2-\alpha)}.
\end{align*}
Here the Laplace transform formula
\[
\widehat{D_b^\beta f}(\lambda)
=\lambda^\beta  \hat{f}(\lambda)-\lim_{t\to 0^+}J_t^{\alpha-1}f(t), 
\quad 0<\beta<1,\; f\in C([0,\infty),X)
\]
is used.

 From the definition of $g_b$, it follows that
$L_b(r,s)=R_b(r,s)$ for $0< s,r\leq b$. Then we have 
\begin{align*}
 H_b(r,s)=\Big(T(s)-\frac{s^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)J_r^\alpha T(r)x, 
\quad \forall \text{ }0< r,s \leq b.
\end{align*}
This implies 
\begin{equation}\label{HTT}
 H_t(r,t)=\Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)J_r^\alpha T(r)x,
\quad \forall  0< r\leq t.
\end{equation}
Combining \eqref{HET} with \eqref{HTT}, we obtain
\begin{align*}
 &\lim_{r\to 0^+}\Gamma(\alpha+1)r^{-\alpha}J_r^{2-\alpha}
\Big(T(r)-\frac{r^{\alpha-2}}{\Gamma(\alpha-1)}\Big)J_t^\alpha
 T(t)x\\
&=\lim_{r\to
 0^+}\Gamma(\alpha+1)r^{-\alpha}
\Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)J_r^2
 T(r)x\\
&=\Gamma(\alpha+1)\Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)
\lim_{r\to 0^+}r^{-\alpha}\int_0^r(r-\sigma)
 T(\sigma)x\,d\sigma\\
&=\Gamma(\alpha+1)\Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)
\\
&\quad\times \lim_{r\to
 0^+}\int_0^1(1-\sigma)\sigma^{\alpha-2}
 (r\sigma)^{2-\alpha}T(r\sigma)x\,d\sigma.
\end{align*}
By the dominated convergence theorem and (b) of Definition \ref{DEF}, 
it follows that
\begin{align*}
&\lim_{r\to 0^+}\Gamma(\alpha+1)r^{-\alpha}J_r^{2-\alpha}
\Big(T(r)-\frac{r^{\alpha-2}}{\Gamma(\alpha-1)}\Big)J_t^\alpha T(t)x\\
&= \Gamma(\alpha+1)\Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)
\int_0^1(1-\sigma)\sigma^{\alpha-2}
 \lim_{r\to  0^+}(r\sigma)^{2-\alpha}T(r\sigma)x\,d\sigma\\
&= \frac{\Gamma(\alpha+1)}{\Gamma(\alpha-1)}
 \Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)
\int_0^1(1-\sigma)\sigma^{\alpha-2}\,d\sigma x\\
&= \frac{\Gamma(\alpha+1)}{\Gamma(\alpha-1)}
 \Big(T(t)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}I\Big)
\frac{\Gamma(\alpha-1)\Gamma(2)}{\Gamma(\alpha+1)} x\\
&= T(t)x-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x.
\end{align*}
This implies that $J_t^\alpha T(t)x\in D(A)$ and 
$$
AJ_t^\alpha  T(t)x=T(t)x-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x.
$$

Conditions (b) and (c) are directly obtained by Lemma \ref{commutative} and
(a).

(d) Denote by $D$ the set of those $x\in X$ such that the limit
$$
\lim_{t\to 0^+}\frac{T(t)x-
\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x}{t^{2\alpha-2}}
$$
exists. Let $x\in D(A)$. Then, by (b), we have 
\begin{align*}
 &\Gamma(2\alpha-1)\lim_{t\to
0^+}\frac{T(t)x-
\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x}{t^{2\alpha-2}}\\
 &= \Gamma(2\alpha-1)\lim_{t\to
 0^+}\frac{J_t^\alpha T(t)Ax}{t^{2\alpha-2}}\\
 &= \frac{\Gamma(2\alpha-1)}{\Gamma(\alpha)}\lim_{t\to
 0^+}\frac{1}{t^{2\alpha-2}}\int_0^t(t-\sigma)^{\alpha-1}T(\sigma)Ax\,d\sigma\\
 &= \frac{\Gamma(2\alpha-1)}{\Gamma(\alpha)}
 \lim_{t\to
 0^+}\int_0^1(1-\sigma)^{\alpha-1}\sigma^{\alpha-2}
 (t\sigma)^{2-\alpha}T(t\sigma)Ax\,d\sigma.
\end{align*}
The  dominated convergence theorem and (b) of Definition \ref{DEF} indicate that
\begin{align*}
 &\Gamma(2\alpha-1)\lim_{t\to
0^+}\frac{T(t)x-
\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x}{t^{2\alpha-2}}\\
 &= \frac{\Gamma(2\alpha-1)}{\Gamma(\alpha)}
 \int_0^1(1-\sigma)^{\alpha-1}\sigma^{\alpha-2}
 \lim_{t\to
 0^+}(t\sigma)^{2-\alpha}T(t\sigma)Ax\,d\sigma\\
 &= \frac{\Gamma(2\alpha-1)}{\Gamma(\alpha-1)\Gamma(\alpha)}
 \int_0^1(1-\sigma)^{\alpha-1}\sigma^{\alpha-2}Ax\,d\sigma\\
 &= \frac{\Gamma(2\alpha-1)}{\Gamma(\alpha-1)\Gamma(\alpha)}
\frac{\Gamma(\alpha-1)\Gamma(\alpha)}{\Gamma(2\alpha-1)}Ax
 = Ax.
\end{align*}
This implies that $x\in D$ and then $D(A)\subset D$. Now we prove
the converse inclusion.
Let $x\in D$, that is, the limit
$$
\lim_{t\to 0^+}\frac{T(t)x-
\frac{t^{\alpha-2}}{\Gamma(\alpha-1))}x}{t^{2\alpha-2}}.
$$
exists. By the dominated convergence theorem, it follows that
\begin{align*}
 &\lim_{t\to 0^+}\Gamma(\alpha+1)t^{-\alpha}J_t^{2-\alpha}\Big(T(t)x-
 \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}x\Big)\\
 &= \lim_{t\to
 0^+}\frac{\Gamma(\alpha+1)}{\Gamma(2-\alpha)}
 \int_0^1(1-\sigma)^{1-\alpha}\sigma^{2\alpha-2}
 \frac{T(t\sigma)x-\frac{(t\sigma)^{\alpha-2}}{\Gamma(\alpha-1)}x}
{(t\sigma)^{2\alpha-2}}\,d\sigma\\
 &= \frac{\Gamma(\alpha+1)}{\Gamma(2-\alpha)}
 \int_0^1(1-\sigma)^{1-\alpha}\sigma^{2\alpha-2}
 \lim_{t\to
 0^+}\frac{T(t\sigma)x-\frac{(t\sigma)^{\alpha-2}}{\Gamma(\alpha-1)}x}
{(t\sigma)^{2\alpha-2}}\,d\sigma\\
 &= \frac{\Gamma(\alpha+1)}{\Gamma(2-\alpha)}
 \frac{\Gamma(2-\alpha)\Gamma(2\alpha-1)}{\Gamma(\alpha+1)} \lim_{t\to
 0^+}\frac{T(t)x-\frac{(t)^{\alpha-2}}{\Gamma(\alpha-1)}x}
{t^{2\alpha-2}}.
\end{align*}
Hence, $x\in D(A)$ and
\begin{equation}\label{absc}
 Ax=\Gamma(2\alpha-1)\lim_{t\to
0^+}\frac{T(t)x-
\frac{t^{\alpha-2}}{\Gamma(\alpha-1))}x}{t^{2\alpha-2}}.
\end{equation}

(e) The properties that $A$ is closed and densely defined are followed 
directly from the combination of (d) and \cite{Li2012}.

(f) Assume that both $\{T(t)\}_{t>0}$ and $\{S(t)\}_{t> 0}$
are Riemann-Liouville $\alpha$-order fractional resolvent generated
by $A$. Then, by (c), for all $x\in D(A)$, we have
\begin{align*}
 \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}*T(t)x
&= (S(t)-J_t^\alpha AS(t))*T(t)x\\
&= S(t)*T(t)x-(J_t^\alpha AS(t))*T(t)x\\
&= S(t)*(T(t)x-J_t^\alpha AT(t)x)\\
&=  \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}*S(t)x.
 \end{align*}
By Titchmarsh's Theorem, for any $t> 0$, $T(t)=S(t)$ on $D(A)$. 
The result is obtained by the density of $A$.
\end{proof}

\begin{corollary}\label{zheng}
Assume that $A$ generates a Rimann-Liouville $\alpha$-order fractional 
cosine function on Banach space $X$. Then
$\{T(t)\}_{t> 0}$ is a Riemann-Liouville $\alpha$-order fractional resolvent.
\end{corollary}

\begin{proof}
In (a) of Theorem \ref{pr2}, replacing $x$ with $J_s^\alpha T(s)x$,
and using Lemma \ref{commutative}, we obtain 
\begin{align*}
 T(t)J_s^\alpha T(s)x
&= \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}J_s^\alpha T(s)x
+AJ_t^\alpha T(t)J_s^\alpha T(s)x\\
&= \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}J_s^\alpha T(s)x
+AJ_s^\alpha T(s)J_t^\alpha T(t)x\\
&= \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}J_s^\alpha T(s)x
+\Big(T(s)-\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}\Big)J_t^\alpha T(t)x,
\end{align*}
which is just \eqref{fi}. The proof is  complete.
\end{proof}

\section{Equivalence of Riemann-Liouville fractional resolvent}

In this section, we prove that equality \eqref{fi} essentially describes 
a Rimann-Liouville $\alpha$-order fractional cosine function.

\begin{theorem}\label{relation}
Suppose that $\{T(t)\}_{t> 0}$ is a Riemann-Liouville $\alpha$-order 
fractional resolvent on Banach space $X$.
Then, the family is a Riemann-Liouville $\alpha$-order fractional 
cosine function.
\end{theorem}

\begin{proof}
Denote by $L(t,s)$ and $R(t,s)$ the left and right sides of equality
\eqref{sin}, respectively. Obviously,  we need to prove
that $L(t,s)=R(t,s)$ for all $t,s> 0$. For brevity, we introduce the
following notation. Let
\begin{gather*}
 H(t,s)= T(t)J_s^\alpha T(s)-J_t^\alpha T(t)T(s),\\
K(t,s)= \frac{t^{\alpha-2}}{\Gamma(\alpha-1)}J_{s}^{\alpha}T(s)
-\frac{s^{\alpha-2}}{\Gamma(\alpha-1)}J_{t}^{\alpha}T(t), t,s> 0.
\end{gather*}
Moreover, for sufficiently large $b>0$ denote by $g_b(t)$ the
truncation of $T(t)$ at $b$, and by $R_b(t,s)$, $L_b(t,s)$,
$H_b(t,s)$ and $K_b(t,s)$ the quantities resulted by replacing
$T(t)$ with $g_b(t)$ in $R(t,s)$, $L(t,s)$, $H(t,s)$ and $K(t,s)$,
respectively.

We set
\begin{align*}
 P_b(t,s)&= \int_0^t\int_0^s\frac{H_b(\sigma,\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{H_b(\sigma,\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
  &\quad -\int_0^t\int_0^s\frac{H_b(\sigma,\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma
\end{align*}
and
\begin{equation}\label{limi}
\begin{aligned}
 Q_b(t,s)&= \int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma+\int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma\\
  &\quad -\int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma.
\end{aligned}
\end{equation}
Observe that the equality \eqref{fi} implies $H(t,s)=K(t,s)$ for any $t,s> 0$. 
Thus, for all $t,s>0$,
\begin{equation}\label{li}
 \lim_{b\to \infty}P_b(t,s)=\lim_{b\to \infty}Q_b(t,s).
\end{equation}
By \cite[(3.13)]{Mei2014}, it follows that
\begin{equation}\label{P}
 P_b(t,s)=(J_s^\alpha-J_t^\alpha)R_b(t,s), \quad\forall\ t,s> 0.
\end{equation}

We now compute Laplace transform of the first term of $Q_b(t,s)$
with respect to $s$ and $t$ as follows,
\begin{align*}
 &\int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma \,ds\,dt\\
 &= \int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}
J_{\tau}^{\alpha}g_b(\tau)
-\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}J_{\sigma}^{\alpha}g_b(\sigma)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma \,ds\,dt\\
 &= \int_0^\infty e^{-\mu t}\int_0^t\int_0^\infty e^{-\lambda s}
 \int_0^s\frac{\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}J_{\tau}^{\alpha}g_b(\tau)
-\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}J_{\sigma}^{\alpha}g_b(\sigma)}
{(t-\sigma)^{\alpha-1}}\,d\tau\, ds \,d\sigma dt\\
 &= \int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}J_{\tau}^{\alpha}g_b(\tau)
-\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}J_{\sigma}^{\alpha}g_b(\sigma)}{(t-\sigma)^{\alpha-1}}\,d\tau
 \,d\sigma \,ds\,dt\\
 &= \int_0^\infty e^{-\mu t}\int_0^t\frac{\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}
}{(t-\sigma)^{\alpha-1}}\int_0^\infty e^{-\lambda s}
 \int_0^sJ_{\tau}^{\alpha}g_b(\tau)\,d\tau
 ds \,d\sigma dt\\
 &-\int_0^\infty e^{-\mu t}\int_0^t\frac{J_{\sigma}^{\alpha}g_b(\sigma)
}{(t-\sigma)^{\alpha-1}}\int_0^\infty e^{-\lambda s}
 \int_0^s\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}\,d\tau
 ds \,d\sigma dt\\
 &= \Gamma(2-\alpha)\mu^{-1}\lambda^{-\alpha-1}\hat{g}_b(\lambda)
 -\Gamma(2-\alpha)\mu^{-2}\lambda^{-\alpha}\hat{g}_b(\mu),
\end{align*}
The Laplace transform of the second term of $Q_b(t,s)$
with respect to $s$ and $t$ is computed as follows
\begin{align*}
 &\int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma \,ds\,dt\\
 &= \int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{\frac{\sigma^{\alpha-2}}
{\Gamma(\alpha-1)}J_{\tau}^{\alpha}g_b(\tau)
-\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}J_{\sigma}^{\alpha}
 g_b(\sigma)}{(s-\tau)^{\alpha-1}}\,d\tau
 \,d\sigma \,ds\,dt\\
 &= \int_0^\infty e^{-\mu t}\int_0^t\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1))}
\int_0^\infty e^{-\lambda s}
 \int_0^s\frac{J_{\tau}^{\alpha}g_b(\tau)
}{(s-\tau)^{\alpha-1}}\,d\tau\, ds\,d\sigma dt\\
 &\quad -\int_0^\infty e^{-\mu t}\int_0^tJ_{\sigma}^{\alpha}g_b(\sigma)
\int_0^\infty e^{-\lambda s} \int_0^s\frac{\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}
}{(s-\tau)^{\alpha-1}}\,d\tau\, ds\,d\sigma dt\\
 &= \Gamma(2-\alpha)\mu^{-\alpha}\lambda^{-2}\hat{g}_b(\lambda)-
 \Gamma(2-\alpha)\lambda^{-1}\mu^{-\alpha-1}\hat{g}_b(\mu).
\end{align*}
We compute the Laplace transform of the third term of $Q_b(t,s)$ with respect 
to $s$ and $t$ as follows.
\begin{align*}
 &-\int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma \,ds\,dt\\
&= -\int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
 \int_0^t\int_0^s\frac{\frac{\sigma^{\alpha-2}}
{\Gamma(\alpha-1)}J_{\tau}^{\alpha}g_b(\tau)
-\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}J_{\sigma}^{\alpha}g_b(\sigma)}
{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau\,d\sigma \,ds\,dt\\
&= -\int_0^\infty e^{-\mu t}\int_0^t\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}\int_0^\infty e^{-\lambda s}
 \int_0^s\frac{J_{\tau}^{\alpha}g_b(\tau)
}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau\, ds\,d\sigma dt\\
 &\quad +\int_0^\infty e^{-\mu t}\int_0^tJ_{\sigma}^{\alpha}g_b(\sigma)
\int_0^\infty e^{-\lambda s} \int_0^s
\frac{\frac{\tau^{\alpha-2}}{\Gamma(\alpha-1)}}
{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau\, ds\,d\sigma dt\\
 &= -\int_0^\infty e^{-\mu t}\int_0^t\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}
\int_0^\infty e^{-\lambda s}\frac{1}{(t+s-\sigma)^{\alpha-1}}
 \, ds\,d\sigma dt\lambda^{-\alpha}g_b(\lambda)\\
 &\quad +\lambda^{1-\alpha}\int_0^\infty 
e^{-\mu t}\int_0^tJ_{\sigma}^{\alpha}g_b(\sigma)
\int_0^\infty e^{-\lambda s} \frac{1}{(t+s-\sigma)^{\alpha-1}}\, ds\,d\sigma dt\\
 &= -\int_0^\infty e^{-\mu t}\int_0^t\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}
 e^{\lambda(t-\sigma)}\\
&\quad\times  \Big(\int_0^\infty e^{-\lambda r}r^{1-\alpha}
 dr-\int_0^{t-\sigma} e^{-\lambda r}r^{1-\alpha}
 dr\Big)\,d\sigma dt\lambda^{-\alpha}g_b(\lambda)\\
 &\quad +\lambda^{1-\alpha}\int_0^\infty e^{-\mu t}
 \int_0^tJ_{\sigma}^{\alpha}g_b(\sigma)e^{\lambda(t-\sigma)}  \\
 &\quad\times \Big(\int_0^\infty e^{-\lambda r} r^{1-\alpha}
 dr-\int_0^{t-\sigma} e^{-\lambda r} r^{1-\alpha}\, dr\Big)\,d\sigma dt\\
 &= -\Gamma(2-\alpha)\lambda^{\alpha-2}
 \int_0^\infty e^{-\mu t}\int_0^t\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}
 e^{\lambda(t-\sigma)}\,d\sigma dt\lambda^{-\alpha}g_b(\lambda)\\
 &\quad +\int_0^\infty e^{-\mu t}\int_0^t\frac{\sigma^{\alpha-2}}{\Gamma(\alpha-1)}
 \int_0^{t-\sigma} e^{\lambda(t-\sigma-r)}r^{1-\alpha}\, dr\,d\sigma dt
\lambda^{-\alpha}g_b(\lambda)\\
 &\quad +\Gamma(2-\alpha)\lambda^{\alpha-2}\lambda^{1-\alpha}
 \int_0^\infty e^{-\mu t}\int_0^tJ_{\sigma}^{\alpha}g_b(\sigma)e^{\lambda(t-\sigma)}
 \,d\sigma dt\\
 &\quad -\lambda^{1-\alpha}\int_0^\infty e^{-\mu t}
 \int_0^tJ_{\sigma}^{\alpha}g_b(\sigma)
 \int_0^{t-\sigma} e^{\lambda(t-\sigma-r)} r^{1-\alpha}
 dr\,d\sigma dt\\
 &= -\Gamma(2-\alpha)\lambda^{\alpha-2}\lambda^{-\alpha}
 \frac{\mu^{1-\alpha}}{\mu-\lambda}g_b(\lambda)
 +\Gamma(2-\alpha)\frac{\mu^{-1}}{\mu-\lambda}\lambda^{-\alpha}g_b(\lambda)\\
 &\quad +\Gamma(2-\alpha)\lambda^{\alpha-2}\lambda^{1-\alpha}
 \frac{\mu^{-\alpha}}{\mu-\lambda}\hat{g}_b(\mu)
 -\Gamma(2-\alpha)\lambda^{1-\alpha}\mu^{\alpha-2}
 \frac{\mu^{-\alpha}}{\mu-\lambda}\hat{g}_b(\mu).
\end{align*}
Using \eqref{L}, we  obtain 
\begin{equation}\label{dd}
\begin{aligned}
&\hat{Q}_b(\mu,\lambda)\\
&= \int_0^\infty e^{-\mu t}\int_0^\infty e^{-\lambda s}
\Big(\int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(t-\sigma)^{\alpha-1}}\,d\tau
\,d\sigma+\int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(s-\tau)^{\alpha-1}}\,d\tau
\,d\sigma\\
 &\quad -\int_0^t\int_0^s\frac{K_b(\sigma,\tau)}{(t+s-\sigma-\tau)^{\alpha-1}}\,d\tau
\,d\sigma)\Big)\,ds\,dt\\
&= (\lambda^{-\alpha}-\mu^{-\alpha})\hat{L_b}(\mu,\lambda).
\end{aligned}
\end{equation}
Taking inverse Laplace transform on both sides of \eqref{dd}, we derive
\begin{equation}\label{Q}
 Q_b(t,s)=(J_s^\alpha-J_t^\alpha)L_b(t,s),\quad \forall  t,  s> 0.
\end{equation}
Form \eqref{P} and \eqref{Q}, we have 
\begin{align*}
 (J_s^\alpha-J_t^\alpha)L(t,s)=(J_s^\alpha-J_t^\alpha)R(t,s),\quad \forall  t, s>0.
\end{align*}
Therefore, $L(t,s)=R(t,s)$. This completes the proof.
\end{proof}

Combining Corollary \ref{zheng} and Theorem \ref{relation},
we can obtain the equivalent of Riemann-Liouville $\alpha$-order fractional 
resolvents and Riemann-Liouville $\alpha$-order fractional cosine functions.

\subsection*{Acknowledgments}
This work was supported by the Natural Science Foundation
of China (grant nos. 11301412 and 11131006),
Research Fund for the Doctoral Program of Higher Education
of China (grant no. 20130201120053),
Shaanxi Province Natural Science Foundation of China
(grant no. 2014JQ1017), Project funded by China Postdoctoral
Science Foundation (grant nos. 2014M550482 and 2015T81011),
the Fundamental Research Funds for the Central Universities
(grant no. 2012jdhz52).

\begin{thebibliography}{00}

\bibitem{Anh2001} V. V. Anh, N. N. Leonenko;
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