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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 122, pp. 1--16.\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-2017/122\hfil Fourier truncation method]
{Fourier truncation method for an inverse source problem for space-time
 fractional diffusion equation}

\author[N. H. Tuan, L. D. Long \hfil EJDE-2017/122\hfilneg]
{Nguyen Huy Tuan, Le Dinh Long}

\address{Nguyen Huy Tuan (corresponding author) \newline
Applied Analysis Research Group,
Faculty of Mathematics and Statistics,
Ton Duc Thang University,
Ho Chi Minh City, Viet Nam}
\email{nguyenhuytuan@tdt.edu.vn}

\address{Le Dinh Long \newline
Institute of Computational Science and Technology,
Ho Chi Minh City, Viet Nam}
\email{long04011990@gmail.com}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted March 5, 2017. Published May 4, 2017.}
\subjclass[2010]{35B53, 35R11}
\keywords{Fractional diffusion equation; Cauchy problem;
\hfil\break\indent   Ill-posed problem; convergence estimates}

\begin{abstract}
 In this article, we study an inverse problem to determine an unknown
 source term in a space time fractional diffusion equation, whereby
 the data are obtained at a certain time. In general, this problem is
 ill-posed in the sense of Hadamard, so the Fourier truncation  method
 is proposed to solve the problem. In the theoretical results,  we
 propose a priori and a posteriori parameter choice rules and analyze them.
\end{abstract}

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

\section{Introduction}

In this work, we consider  the  inverse problem of finding the source function $f$
in the  problem
\begin{equation}  \label{1}
\begin{gathered}
\frac{\partial^{\beta}}{\partial t^{\beta}} u(x,t)
= -r^{\beta}(-\Delta)^{\frac{\alpha}{2}}u(x,t) + h(t)f(x),  \quad
  (x,t) \in \Omega_{T},\\
u(-1,t) = u(1,t) = 0,  \quad 0 < t < T, \\
u(x,0) =  0,  \quad x \in \Omega, \\
u(x,T) = g(x),  \quad x \in \Omega,  \\
\end{gathered}
\end{equation}
where $\Omega_{T} = (-1,1) \times (0,T) $, $r > 0$ is a parameter,
$h \in C[0,T]$ is a given function, $\beta \in (0,1)$,
$\alpha \in (1,2)$ are fractional order of the time and the space derivatives,
respectively, and $T > 0$ is a final time.
The function  $u = u(x, t )$ denotes a concentration of contaminant at a position
$x$ and time $t.$ The symbol
$\frac{\partial^{\beta} u}{\partial t^{\beta}}$ is the  Caputo fractional
derivative of order $\beta$ {for differentiable function u}; it writes
\begin{align*}
\frac{\partial^{\beta}}{\partial t^{\beta}} u(t)
=\frac{1}{\Gamma(1-\alpha)}\int_0^t\frac{u'(s)}{(t-s)^{\beta}}ds,
\end{align*}
and $\Gamma(.)$ denotes the standard Gamma function.
Note that if the fractional order $\beta$ tends to unity,
the fractional derivative $\frac{\partial^{\beta}}{\partial t^{\beta}} u$
converges to the first-order derivative $\frac{du}{dt}$ \cite{Ki}, and thus
the problem reproduces the diffusion model. See, e.g., \cite{Ki,Po} for the
definition and properties of Caputo's derivative.

It is known that the inverse source problem mentioned above is ill-posed in
 general, i.e., a solution does not always exist, and in the
case of existence of a solution, {it} does not depend continuously on
the given data. In fact, from a small noise of a physical measurement, for
example $(h,g)$ is noised by observation data $(h^\varepsilon, g^\varepsilon )$ with order
of $\varepsilon>0$
\begin{equation}
\|h^\varepsilon-h\|_{C([0,T])}+ \|g^\varepsilon-g\|_{L^2(-1,1)} \le \varepsilon
\end{equation}
where we denote $\|\theta\|_{C([0,T])}= {\sup}_{0\le t \le T} |\theta(t)|$
for any $\theta \in C([0,T]$.
It is well-known that if  $\varepsilon$ is small,  the sought solution $f$  may have
a large error. An example for illustrating this is given in Theorem \ref{thm2.12}.
Hence some regularization method are required for stable computation of a
sought solution.

The inverse source problem attracted  many authors and its physical background
can be found in \cite{Tuan}.  Wei et al  \cite{Wei, Wei1, Wei2} studied an
inverse source problem in a spatial fractional diffusion equation by
quasi-boundary value and truncation methods.  Recently,  Kirane et al \cite{Ki1,Ki}
studied conditional well-posedness to determine a space dependent source in
one-dimensional and two-dimensional time-fractional diffusion equations.
 Rundell et al \cite{Jin,rundell} considered an inverse problem for a one-dimensional
time-fractional diffusion problem.  However, the inverse source problem for both
 time and space fractional  is limited.
Recently,  Tatar et al \cite{Salih} considered Problem \eqref{1} with a general
source $h(t,x) f(x)$. They show that the inverse source problem is well-posed
in the sense of Hadamard except for a finite set of $r>0$.
However  the source function is also  unstable  in $L^2 $ norm 
(See Theorem \ref{thm2.13} below).  The topic in this paper is to   finding  approximate solution. Hence, our purpose is different and not contradict
with the results in \cite{Salih}.
Motivated by above reasons, in this study, we apply the Fourier  regularization
method to establish a regularized solution.  We estimate a convergence rate under
an a priori bound assumption of the {sought} solution and a priori parameter
choice rule. Because the a priori bound is difficult to obtain in practical
application, so we also estimate a convergence rate under the a posteriori
parameter choice rule which is independent on the a priori bound.

This article is organized as follows. In Section 2, we give a formula of
the source function $f$ and establish some lemmas and theorems which are
useful to the  next results. Moreover, the ill-posedness of the inverse
 source problem is also given in this section.
In Section 2, we propose Fourier  regularization method and give two
convergence estimates under an a priori assumption for the exact solution
and two regularization parameter choice rules.

\section{Inverse source  problem }
\subsection{Formula of the source function}
First, we introduce a few properties of the eigenvalues of the operator
$(-\Delta)^{\alpha/2}$, see \cite{18,Po}.

\begin{theorem}[Eigenvalues of the fractional Laplacian operator] 
\label{thm2.1}
 1. Each eigenvalues of $(-\Delta)^{\alpha/2}$ is real.
The family of eigenvalues  $\{ \overline{\lambda_k}\}_{k=1}^\infty$ satisfy
 $0 \le \overline{\lambda_1} \le \overline{\lambda_2} \le \overline{\lambda_3}
\le \dots$, and $\overline{\lambda_k} \to \infty $ as $k \to \infty$.

 2. We take $\{ \overline{\lambda_k}, \phi_k \}$ the eigenvalues and
corresponding eigenvectors of the fractional Laplacian operator in $\Omega$
with Dirichlet boundary conditions on $ \partial \Omega$:
 \begin{equation}
 \begin{gathered}
 -\Delta \phi_k(x) = \overline{\lambda_k} \phi_k(x), \quad x \in \Omega,  \\
 \phi_k(x)=0 ,\quad \text{on }\partial \Omega,\quad \\
 \end{gathered} \label{e1}
 \end{equation}
for $k=1,2,\dots $.
\end{theorem}

Then we define the operator $(-\Delta)^{\frac{\alpha}{2}}$ by
\[
(-\Delta)^{\alpha/2}u := \sum_{k=0}^{\infty}c_k\phi_k(x)
= - \sum_{k=0}^{\infty} c_k \overline{\lambda}_k^{\alpha/2}\phi_k(x),
\]
which maps $H_0^{\alpha}(\Omega)$ into $L^2(\Omega)$.
Let $0 \neq \gamma < \infty$. By $H^\gamma(\Omega)$ we denote the space
of all functions $g \in L^2(\Omega)$ with the property
\begin{equation} \label{condition3}
\sum_{k=1}^\infty (1 + \overline{\lambda_k} )^{2\gamma}|g_k|^2 < \infty,
\end{equation}
where
$
g_k=\int_{\Omega}g(x)\phi_k(x)dx.$ Then we also define \\$\|g\|_{H^\gamma(\Omega)}= \sqrt{\sum_{k=1}^\infty (1 + \overline{\lambda_k})^{2\gamma}|g_k|^2 } $.  If $\gamma=0$ then $H^\gamma (\Omega)$ is $L^2(\Omega).$  \\
Now we use the separation of variables to yield the solution of $\eqref{1}$.  Suppose that the solution of $\eqref{1}$  is defined by Fourier series
\begin{equation}
u(x,t) = \sum_{k=1}^{\infty}u_k(t) \phi_k(x),\quad
\text{with } u_k(t) = \langle u(.,t),\phi_k(x)\rangle .
\end{equation}
Then the eigenfunction expansions can be defined by the Fourier method.
That is, we multiply both sides of $\eqref{1}$ by $\phi_k(x)$ and
integrate the equation with respect to $x$.
 Using the Green formular and $\phi_k|_{\partial \Omega} = 0$, we obtain an
 uncouple system of initial value problem for the fractional differential
equations for the unknown Fourier coefficient $u_k(t)$
\begin{equation} \label{4}
\begin{gathered}
\frac{\partial^{\beta}}{\partial t^{\beta}} u_k(t)
= -r^{\beta}(-\Delta)^{\frac{\alpha}{2}}u_k(t) + h(t)f_k(x),  \quad (x,t)
\in \Omega  \times (0,T),\\
u_k(0) = \langle u(x,0),\varphi_k(x)\rangle .
\end{gathered}
\end{equation}
From the result in  \cite{Salih}, the formula of solution corresponding to
the initial value problem for \eqref{4} is  obtained as follows, from $u(x,0) = 0$.
\begin{equation}
u(x,t) = \sum_{k=1}^{\infty} \Big( \int_0^t \tau^{\beta - 1}E_{\beta,\beta}
\Big(- (\frac{k\pi}{2})^{\alpha} r^{\beta} \tau^{\beta} \Big)
\langle  f(x) h(t - \tau), \phi_k(x) \rangle d\tau \Big) \phi_k(x) .
\end{equation}
By a change variable in the integral, we can rewrite
\begin{equation} \label{u}
u(x,t) = \sum_{k=1}^{\infty} \Big( \int_0^t (t-\tau)^{\beta - 1}
E_{\beta,\beta}\Big(- (\frac{k\pi}{2})^{\alpha} r^{\beta} (t - \tau)^{\beta} \Big)
 h(\tau)d\tau \Big)\langle  f(x), \phi_k(x) \rangle  \phi_k(x).
\end{equation}
Letting  $t = T$ in the latter equality, we obtain
\begin{equation} \label{so5}
f(x) = \sum_{k=1}^{\infty}\frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} (T-\tau)^{\beta - 1}E_{\beta,\beta}
\big(- (\frac{k\pi}{2})^{\alpha} r^{\beta} (T - \tau)^{\beta} \big) h(\tau) d\tau  }  .
\end{equation}
to abbreviate  notation,  we set
 \[
\Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
= (T-\tau)^{\beta - 1}E_{\beta,\beta}\Big(- (\frac{k\pi}{2})^{\alpha} r^{\beta}
(T - \tau)^{\beta} \Big),
 \]
then  the source function $f$ is rewritten as
\begin{equation}\label{nghiemchinhxaccuoicung}
f(x) = \sum_{k=1}^{\infty}\frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  }  .
\end{equation}

\begin{remark} \label{rmk2.1} \rm
 Applying \cite[Theorem 2.1]{Salih}, we obtain the existence and uniqueness
of problem \eqref{1} such that $u \in L^2(0,T; H^\alpha (\Omega))$.
 The  regularity estimate for $u$ as in \eqref{u} is mentioned in \cite{Salih}
and so, we omit it here.
\end{remark}

\subsection{Preliminary results}
Now, we  consider  the following definition and lemmas which are useful for our
main results.

\begin{definition}[\cite{Po}] \label{def2.2} \rm
 The Mittag-Leffler function is
\[
 E_{\alpha,\beta}(z) = \sum_{k=0}^{\infty} \frac{z^{k}}{ \Gamma(\alpha k + \beta) },
\quad z \in \mathbb{C}
\]
 where $\alpha > 0$ and $\beta \in \mathbb{R}$ are arbitrary {constants}.
\end{definition}

\begin{lemma}[\cite{Sa}] \label{lem2.3}
 For $\lambda > 0$ and $0<\beta<1$,  we have
 \begin{equation}  \label{bodeso1}
 \frac{d}{dt} E_{\beta,1}(-\lambda t^{\beta})
= - \lambda t^{\beta-1} E_{\beta,\beta}(-\lambda t^{\beta}),\quad t > 0 .
 \end{equation}
\end{lemma}

 \begin{lemma}[\cite{Po}] \label{lem2.4}
  For $\alpha > 0$ and $\beta \in \mathbb{R}$, we have
\[
  E_{\alpha,\beta}(z) = z E_{\alpha,\alpha+\beta}(z) + \frac{1}{\Gamma(\beta)},\quad
 z \in \mathbb{C} .
\]
 \end{lemma}

 \begin{lemma}[\cite{Sa}] \label{lem2.5}
  The following equality holds for $\lambda > 0$, $\alpha > 0$ and
$m \in \mathbb{N}$
  \begin{equation}
  \frac{d^{m}}{dt^{m}}E_{\alpha,1}(-\lambda t^{\alpha})
= -\lambda t^{\alpha-m} E_{\alpha, \alpha - m + 1}
(-\lambda t^{\alpha}),\quad t > 0 .
  \end{equation}
 \end{lemma}

\begin{lemma}[\cite{Salih}] \label{lem2.6}
 If $\alpha \le 2$, $\beta$ is arbitrary real number, $\mu$ is such that
 $ \frac{\pi \alpha}{2} < \mu < \min\{\pi \alpha, \pi \} $,
$ \mu \le |arg(z)| \le \pi $ then there exists two constants
 $A_0>0$ and $A_1>0$ such that
 \begin{equation}
 \frac{A_0}{1 + |z|} \le | E_{\alpha,\beta}(z) | \le \frac{A_{1}}{1 + |z|} .
 \end{equation}
\end{lemma}

\begin{lemma}[\cite{Sa}] \label{lem2.7}
 Let $E_{\beta,\beta} (-\eta) \ge 0$, $ 0 < \beta < 1 $, we have
 \begin{equation} \label{bodeso3}
\begin{aligned}
 \int_0^{M} \Big| t^{\beta - 1} E_{\beta,\beta}
(-\overline{\lambda_k} t^{\beta})\Big|dt
&= \int_0^{M} t^{\beta - 1}E_{\beta,\beta}(- \overline{\lambda_k}t^{\beta}) dt  \\
 & = - \frac{1}{ \overline{\lambda_k} } \int_0^{M} \frac{d}{dt}
E_{\beta,1}(-\overline{\lambda_k} t^{\beta} ) dt  \\
 &= \frac{1}{\overline{\lambda_k} }
\Big( 1 - E_{\beta,1}(-\overline{\lambda_k} M^{\beta} ) \Big) .
\end{aligned}
 \end{equation}
\end{lemma}

\begin{lemma}[\cite{Salih}] \label{bode36}
 For any $\lambda_k^{\alpha} = (\frac{k\pi}{2})^{\alpha}$ satisfying
$\lambda_k^{\alpha} \ge \lambda_{1}^{\alpha}$ there exists positive constant
$C$ depending on $\{\beta, T, \frac{\pi}{2}\}$ such that
\begin{equation}
 \frac{C}{r^{\beta} T^{\beta} \lambda_k^{\alpha} }
\le E_{\beta,\beta+1}( -\lambda_k^{\alpha} r^{\beta} T^{\beta})
 \le \frac{1}{r^{\beta} T^{\beta}\lambda_k^{\alpha}} .
 \end{equation}
\end{lemma}

\begin{lemma} \label{bodeso8}
 Let $h: [0,T] \to \mathbb{R}^+$ be {a}  continuous function such that
$|\mathcal{I}(h)| = \inf_{t \in [0,T]} |h(t)| >0 $.
Set $\|h\|_{C[0,T]}=\sup_{t \in [0,T]} |h(t)|   $. Then we have
 \begin{equation}
 \frac{ |\mathcal{I}(h)| (1 - E_{\beta,1}(- \lambda_{1}^{\alpha} r^{\beta}
T^{\beta}) ) }{\lambda_k^{\alpha} r^{\beta}}
\le \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) {h(\tau)}d\tau
\le  \frac{\|h\|_{C[0,T]}}{ \lambda_k^{\alpha} r^{\beta} } .
 \end{equation}
\end{lemma}

A proof of the above lemma can be found in \cite{Salih}.


\begin{theorem} \label{thm2.10}
Let $g \in H^\alpha (\Omega)$. Then the inverse source problem has the solution
$f \in L^2(\Omega)$.
\end{theorem}

\begin{proof}
 The solution $f$ exists if and only if the series in the right-hand side of
\eqref{nghiemchinhxaccuoicung} converges. Hence, we show  this point.
 Indeed, using Lemma \ref{bodeso8} and noting that $g \in H^\alpha (\Omega) $, we obtain
 \begin{equation}
\begin{aligned}
 \sum_{k=1}^{\infty} \Big|\frac{\langle g(x),\phi_k(x)\rangle  }{ \int_0^{T}
\Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  } \Big|^2
&\ge  \sum_{k=1}^{\infty} \frac{\lambda_k^{2\alpha} r^{2\beta}
\langle g(x),\phi_k(x)\rangle ^2 } { \|h\|_{C[0,T]}^2  } \\
 &= \frac{r^{2\beta} }{\|h\|_{C[0,T]}^2  } \|g\|^2_{H^\alpha(\Omega)},
\end{aligned}
 \end{equation}
 and
 \begin{equation}
\begin{aligned}
 \sum_{k=1}^{\infty} \Big|\frac{\langle g(x),\phi_k(x)\rangle  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  }  \Big|^2
&\le  \sum_{k=1}^{\infty} \frac{\lambda_k^{2\alpha} r^{2\beta}
 \langle g(x),\phi_k(x)\rangle ^2 } { |\mathcal{I}(h)|^2
 (1 - E_{\beta,1}(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) )^2  } \\
 &= \frac{ r^{2\beta}  } { |\mathcal{I}(h)|^2 (1 - E_{\beta,1}
(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) )^2  } \|g\|^2_{H^\alpha(\Omega)} .
\end{aligned}
 \end{equation}
 From two latter inequality, we conclude that the series
$\sum_{k=1}^{\infty}\frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  } $
is convergent. The proof is complete.
\end{proof}

\begin{theorem} \label{thm2.11}
 Let $R: [0,T] \to \mathbb{R}$ be as in Lemma \ref{bodeso8}, then
the solution {($u(x,t), f(x)$)}  of Problem (1) is unique.
\end{theorem}

\begin{proof}
 Let $f_1$ and $f_2$ be the source functions  corresponding to the final values
$g_{1}$ and $g_{2}$ respectively.
 Suppose that $g_{1} = g_{2}$ then we prove that $f_1=f_2$.
In fact, it is well-known that
$E_{\beta,\beta}(-( \frac{k\pi}{2})^{\alpha} r^{\beta} (t-\tau)^{\beta} ) \ge 0$
for $\tau \le t$. Since $ \|h\|_{C[0,T]} \ge |\mathcal{I}(h)| > 0$ for
 $t \in [0,T] $, we have
 \begin{equation}  \label{dinhli1}
\begin{aligned}
&\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau \\
&\ge |\mathcal{I}(h)| \int_0^{T}(T-\tau)^{\beta-1}E_{\beta,\beta}
\Big( -( \frac{k\pi}{2})^{\alpha} r^{\beta} (T - \tau)^{\beta} \Big)d\tau  \\
&= |\mathcal{I}(h)|T^{\beta}E_{\beta,\beta + 1}
 \Big(-( \frac{k\pi}{2})^{\alpha} (rT)^{\beta} \Big) >0 .
\end{aligned}
 \end{equation}
We have the estimate
 \begin{equation}
 f_1(x)-f_2(x) = \sum_{k=1}^{\infty}
\frac{\langle g_1(x)-g_2(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha}, \tau, r) h(\tau)d\tau}=0 .
 \end{equation}
 The proof is complete.
\end{proof}

\begin{theorem} \label{thm2.12}
 The  inverse source problem of finding the function  $f$ is ill-posed in
the  Hadamard sense in the $L^2 $ norm.
\end{theorem}

\begin{proof}
Let us Define a linear operator $\mathcal{P}:L^2(\Omega) \to L^2(\Omega) $
as follows
 \begin{equation}  \label{sudung1}
\begin{aligned}
 \mathcal{P}f(x)
&= \int_\Omega p(x,\omega) f(\omega) d\omega \\
&= \sum_{k=1}^{\infty}\Big[\int_0^{T} \Phi_{\beta}
(\lambda_k^{\alpha},\tau,r) {h(\tau)} d\tau \Big] \langle f(x),\phi_k(x)\rangle
 \phi_k(x)
\end{aligned}
 \end{equation}
where
$$
p(x,\omega) = \sum_{k = 1}^{\infty} \Big[\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},
\tau,r)  h(\tau) d\tau \Big]  \phi_k(x) \phi_k(\omega) .
$$
Because $p(x,\omega) = p(\omega,x)$ we know that $\mathcal{K}$ is self-adjoint
operator. Next, we  prove its compactness. Defining the finite rank operators
$\mathcal{K}_{N}$ as follows
 \begin{equation}  \label{sudung2}
 \mathcal{P}_{N}f(x) = \sum_{k=1}^{N}  \Big[\int_0^{T} \Phi_{\beta}
(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau \Big]  \langle f(x),\phi_k(x)\rangle
 \phi_k(x) .
 \end{equation}
Then, from \eqref{sudung1} and \eqref{sudung2}, we have
 \begin{equation}
\begin{aligned}
 \|\mathcal{P}_{N}f - \mathcal{P}f \|_{L^2(\Omega)}^2
&= \sum_{k=N+1}^{\infty} \Big[\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
 h(\tau) d\tau \Big]^2 |\langle f(x),\phi_k(x)\rangle |^2 \\
 &\le \sum_{k = N+1}^{\infty} \frac{ \|h\|^2_{C[0,T]} }
 { \lambda_k^{2\alpha} } |\langle f(x),\phi_k(x)\rangle |^2 \\
 &\le \frac{ \|h\|^2_{C[0,T]} } {\lambda_{N}^{2\alpha}}
\sum_{k = N+1}^{\infty} |\langle f(x),\phi_k(x)\rangle |^2 .
\end{aligned}
 \end{equation}
 This implies
 \begin{equation}
 \|\mathcal{P}_{N}f - \mathcal{P}f \|_{L^2(\Omega)}
\le \Big(\frac{ \|h\|^2_{C[0,T]} }{\lambda_{N}^{2\alpha} }
\|f\|^2_{L^2(\Omega)}  \Big)^{1/2}
= \frac{  \|h\|_{C[0,T]} }{\lambda_{N}^{\alpha}}  \|f\|_{L^2(\Omega)} .
 \end{equation}
 Therefore, $\|\mathcal{P}_{N} - \mathcal{P}\| \to 0$ in the sense of operator
norm in $L(L^2(\Omega);L^2(\Omega))$ as $N \to \infty$.
Also, $\mathcal{P}$ is a compact operator.
 Next, the singular values for the linear self-adjoint compact operator
$\mathcal{P}$ are
 \begin{equation}  \label{nhan21}
\begin{aligned}
 \psi_k &= \int_0^{T} (T - \tau)^{\beta-1} E_{\beta,\beta}
\Big( -(\frac{k\pi}{2})^{\alpha} r^{\beta} (T-\tau)^{\beta} \Big) h(\tau) d\tau \\
&=\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau,
\end{aligned}
 \end{equation}
and corresponding eigenvectors is $\phi_k$ which is known as an orthonormal
basis in $L^2(\Omega)$. From $\eqref{sudung1}$, the inverse source problem
we introduced above can be formulated as an operator equation.
 \begin{equation}
 \mathcal{P}f(x) = g(x)
 \end{equation}
and by  Kirsch \cite{Kirsch}, we conclude that  it is ill-posed.
 To illustrate an ill-posed problem, we present an example.
 Let us choose  the input final data
$g^{m}(x) = \frac{\phi_m(x) }{\sqrt{r^{2\beta} \lambda_{m}^{\alpha}}} $.
Following \eqref{bode36}, we know
 $\lambda_{m}^{\alpha} = (\frac{m \pi}{2})^{\alpha}$.
By \eqref{nghiemchinhxaccuoicung},
 the  source term corresponding to $g^{m}$ is
 \begin{equation}
\begin{aligned}
 f^{m}(x)
&= \sum_{k=1}^{\infty}\frac{\langle g^{m}(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau)d\tau  }  \\
 &= \sum_{k=1}^{\infty}
\frac{\langle \frac{\phi_m(x) }{\sqrt{r^{2\beta}(\frac{m\pi}{2})^{\alpha}}},\phi_k(x)
\rangle\phi_k(x)}{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
  h(\tau) d\tau} \\
 &= \frac{\phi_{m}(x) }{ \sqrt{r^{2\beta} (\frac{m\pi}{2})^{\alpha}}
\int_0^{T} \Phi_{\beta}(\lambda_{m}^{\alpha},\tau,r)  h(\tau) d\tau}.
\end{aligned}
 \end{equation}
Let us choose another input final data $g=0$.  By \eqref{so5},
 the  source term corresponding to $g$ is $f=0$.
 An error in $L^2$ norm between two input final data is
 \begin{equation}
 \|g^{m}-g\|_{L^2(\Omega)}= \|\frac{\phi_m(x) }{|r^{\beta}
|\sqrt{( \frac{m\pi}{2})^{\alpha} }}\|_{L^2(\Omega)}
= \frac{1}{|r^{\beta}|\sqrt{(\frac{m\pi}{2})^{\alpha}} },
 \end{equation}
 where $\alpha \in (1,2)$. Therefore
 \begin{equation}
 \lim_{m \to +\infty} \|g^m-g\|_{L^2(\Omega)}= \lim_{m \to +\infty}
\frac{1}{|r^{\beta}| \sqrt{( \frac{m\pi}{2})^{\alpha} } }=0 . \label{t2}
 \end{equation}
 And an error in $L^2(-1,1)$ norm between two corresponding source term is
 \begin{equation}
\begin{aligned}
 \|f^m-f\|_{L^2(\Omega)}
&= \| \frac{\phi_m(x) }{ |r^{\beta}| \sqrt{( \frac{m \pi}{2})^{\alpha} }
  \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)   h(\tau) d\tau   }
\|_{L^2(\Omega)} \\
 &= \frac{1}{|r^{\beta}| \sqrt{( \frac{m\pi}{2})^{\alpha}  }
\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau) d\tau  } ,
\end{aligned} \label{t1}
 \end{equation}
which we note that $\beta \in (0,1)$ and $r$ is positive number.
From  \eqref{t1} and using the inequality as in Lemma \ref{bodeso8},
 we obtain
 \begin{equation}
 \|f^m-f\|_{L^2(\Omega)} \ge \frac{\sqrt{ ( \frac{m\pi}{2} )^{\alpha} } }
{  \|h\|_{C[0,T]} } .
 \end{equation}
 This leads to
 \begin{equation}
 \lim_{m \to +\infty} \|f^m-f\|_{L^2(\Omega)}
> \lim_{m \to +\infty} \frac{\sqrt{( \frac{m\pi}{2})^{\alpha} } }
 { \|h\|_{C[0,T]} }=+\infty . \label{t3}
 \end{equation}
 Combining \eqref{t2} with \eqref{t3}, we conclude that  the inverse
source problem is ill-posed.
\end{proof}

\begin{theorem}[A conditional stability estimate] \label{thm2.13}
 Assume that there exists $\gamma>0$ such that
$\|f\|_{H^{\alpha \gamma}(\Omega)} \le  M$ for $M>0$. Then
 \begin{equation} \label{ondinhcuanghiem}
 \|f\|_{L^2(\Omega)} \le \mathcal{K_{\alpha,\beta}}(h,r,T)
 M^{\frac{1}{\gamma + 1}} \|g\|^{\frac{\gamma}{\gamma + 1}}_{L^2(\Omega)} .
 \end{equation}
 where
 \begin{equation}
 \mathcal{K_{\alpha,\beta}}(h,r,T)
=  \frac{(r^{\beta})^{\frac{ \gamma}{\gamma + 1}}}
{|{\mathcal{I}(h)}|^{\frac{\gamma}{\gamma+1}} \
 big( 1- E_{\alpha,1}( - \lambda_{1}^{\alpha} r^{\beta} T^{\beta})
\big)^{ \frac{\gamma}{\gamma+1}}} .
 \end{equation}
\end{theorem}

\begin{proof}
 According \eqref{nghiemchinhxaccuoicung}, by  H\"older's inequality, we have
 \begin{equation} \label{tinhdeu}
\begin{aligned}
 \|f\|^2_{L^2(\Omega)}
&=\sum_{k=1}^{\infty} \Big| \frac{ \langle g(x),\phi_k(x)\rangle \phi_k(x) }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau) d\tau  }  \Big|^2 \\
&\le \sum_{k=1}^{\infty}  \frac{ |\langle g(x),\phi_k(x)\rangle
 |^{\frac{2}{\gamma + 1}} |\langle g(x),\phi_k(x)\rangle
 |^{\frac{2 \gamma}{\gamma + 1}} }{\Big| \int_0^{T} \Phi_{\beta}
 (\lambda_k^{\alpha},\tau,r)   h(\tau) d\tau \Big|^2 }    \\
&\le \Big( \sum_{k=1}^{\infty} \frac{|\langle g(x),\phi_k(x)\rangle |^2}
{\big|\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau
 \big|^{2(\gamma+1)}} \Big)^{\frac{1}{\gamma+1}} \\
&\quad \times \Big(\sum_{k=1}^{\infty} |\langle g(x),\phi_k(x)\rangle |^2
 \Big)^{\frac{\gamma}{\gamma + 1}}  \\
&\le \Big( \sum_{k=1}^{\infty} \frac{ |\langle f(x),\phi_k(x)\rangle |^2 }
{\big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau)
 d\tau \big|^{2\gamma}} \Big)^{\frac{1}{\gamma + 1}}
\|g\|^{\frac{2\gamma}{\gamma+1}}_{L^2(\Omega)} .
\end{aligned}
 \end{equation}
Here we have used the fact that
\[
\langle g(x),\phi_k(x)\rangle =\langle f(x),\phi_k(x)\rangle
\Big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau \Big|^2.
\]
 Using Lemma \ref{bodeso8}, we have
 \begin{equation}
 \Big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau) d\tau
\Big|^{2\gamma}
\ge \frac{ |{\mathcal{I}(h)}|^{2\gamma} }{ \lambda_k^{2\alpha \gamma}
 r^{2 \beta \gamma} }  \big( 1 - E_{\beta,1}
(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) \big)^{2\gamma}
 \end{equation}
 and this inequality leads to
 \begin{equation}
\begin{aligned}
&\sum_{k=1}^{\infty}   \frac{  |\langle f(x),\phi_k(x)\rangle |^2}
{\big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
 h(\tau)d\tau  \big|^{2\gamma} } \\
&\le  \sum_{k=1}^{\infty}\frac{r^{2 \beta \gamma} \lambda_k^{2 \alpha \gamma } |
\langle f(x),\phi_k(x)\rangle |^2}{ |{\mathcal{I}(h)}|^{2\gamma}
\big( 1- E_{\beta,1} (- \lambda_{1}^{\alpha} r^{\beta} T^{\beta} ) \big)^{2\gamma}   }
  \\
&\le \frac{r^{2 \beta \gamma}  }{|{\mathcal{I}(h)}|^{2\gamma}
 \big( 1- E_{\beta,1} (- \lambda_{1}^{\alpha} r^{\beta} T^{\beta} ) \big)^{2\gamma}}
\sum_{k=1}^{\infty} \lambda_k^{2 \alpha \gamma } |\langle f(x),\phi_k(x)\rangle|^2  \\
&=  \frac{ r^{2 \beta \gamma } \|f\|^2_{H^{\alpha \gamma }(\Omega)}   }
{ |{\mathcal{I}(h)}|^{2\gamma} \big( 1- E_{\beta,1}
(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) \big)^{2\gamma}  } .
\end{aligned} \label{t8}
 \end{equation}
 Combining \eqref{tinhdeu} with \eqref{t8}, we obtain
 \begin{equation}
\begin{aligned}
 \|f\|_{L^2(\Omega)}^2
&\le  \frac{  r^{\frac{2 \beta \gamma}{\gamma + 1}}
 \|f\|^{\frac{2}{\gamma+1}}_{H^{\alpha \gamma}(\Omega)}   }
{ |{\mathcal{I}(h)}|^{\frac{2\gamma}{\gamma+1}}
\big( 1- E_{\beta,1} ( - \lambda_{1}^{\alpha} r^{\beta} T^{\beta})
 \big)^{ \frac{2\gamma}{\gamma+1}  }  }
 \|g\|_{L^2(\Omega)}^{\frac{2\gamma}{\gamma+1}}  \\
&\le {|\mathcal{K_{\alpha,\beta}}(h,r,T)|^2} M^{\frac{2}{\gamma + 1}}
\|g\|_{L^2(\Omega)}^{\frac{2\gamma}{\gamma+1}} .
\end{aligned}
 \end{equation}
\end{proof}


\section{Fourier truncation regularization and error estimate}

In this section, we eliminate all the components of large $k$ from the solution
and define the truncation regularized solution as follows:
\begin{equation} \label{re}
f^{\epsilon, N}(x) = \sum_{k = 1}^{N}
\frac{\langle g^{\epsilon}(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\epsilon}(\tau) d\tau  }
\end{equation}
where the positive integer $N$ plays the role of regularization parameter.
Next, we consider an a {\it a-priori} and an a {\it a-posteriori} choice
to find the regularization parameter. Under each choice of the regularization
 parameter, the error estimates between the exact solution $f$ given
by \eqref{so5} and the regularized approximation solution $f^{\epsilon, N}$ given
by \eqref{re} can be obtained.

\subsection{An a priori parameter choice}

Afterwards, we will give an error estimation for
$\|f(x)-f^{\epsilon,N}(x)\|_{L^2(\Omega)}$ and show convergence rate
under a suitable choice for the regularization parameter.

\begin{theorem} \label{thm3.1}
 Let $f^{\epsilon, N}$ be the regularized solution for problem $\eqref{1}$
 with noisy data $g^{\epsilon}$ and $f(x)$ be the exact solution for problem
\eqref{1}. Let us choose parameter regularization $N=[\mu]$, where $[\mu]$
denotes the largest integer less than or equal to $\mu$.
Then we have the following:
\begin{itemize}
\item  If  $0 < \gamma \le  1$ then choose
$\mu = \frac{2}{\pi} \Big(\frac{M}{\epsilon} \Big)^{\frac{1}{\alpha(\gamma + 1)}} $,
 we have the  estimate
 \begin{equation}
 \|f(x) - f^{\epsilon, N}(x)\|_{L^2(\Omega)}
\le \varepsilon^{\frac{\gamma}{\gamma + 1}}M^{\frac{1}{\gamma + 1}}
\mathcal{D_{\alpha,\beta}}(f,h,h^{\varepsilon},r,T)  .
 \end{equation}

\item If $ \gamma  > 1$, choose
$\mu = \frac{2}{\pi} \Big( \frac{M}{\epsilon} \Big)^{\frac{1}{2\alpha}} $,
we obtain the error estimate
 \begin{equation}
 \|f(x) - f^{\epsilon, N}(x)\|_{L^2(\Omega)}
\le \varepsilon^{\frac{1}{2}}M^{\frac{1}{2}} \mathcal{D_{\alpha,\beta}}(f,h,h^{\varepsilon},r,T) ,
 \end{equation}
 where
 \begin{equation}
 \mathcal{D_{\alpha,\beta}}(f,h,h^{\varepsilon},r,T)
= \Big[ 1 + \max\Big\{ \frac{ r^{\beta} }{|{\mathcal{I}(h)}|
\big( 1 - E_{\beta,1}( - \lambda_{1}^{\alpha} r^{\beta}T^{\beta}) \big) },
 \frac{ \|f\|_{L^2(\Omega)}}{|{\mathcal{I}(h^{\varepsilon})}|} \Big\} \Big] .
 \end{equation}
\end{itemize}
\end{theorem}

\begin{remark} \label{rmki3.2} \rm
 If the function $h$ depends on $x$ and $t$, i.e.\ $h=h(t,x)$ then we can not
represent $f$ as the Fourier series as \eqref{nghiemchinhxaccuoicung}.
Hence, we can not use some  usual  regularization methods.
The regularized problem is open and difficult when $h$ depends on $x$ and $t$.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
 Using  \eqref{so5} and \eqref{re} and the triangle inequality, we have
 \begin{equation} \label{quantrong}
\begin{aligned}
 f(x) - f^{\epsilon,N}(x)
&=  \sum_{k=1}^{\infty}
 \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)}
 {\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau)d\tau}
 - \sum_{k = 1}^{N} \frac{\langle g^{\epsilon}(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h^{\epsilon}(\tau) d\tau  }  \\
 &= \sum_{k=1}^{\infty} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T}\Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau) d\tau  } - \sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)  }{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau) d\tau  }  \\
 &+ \sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  }
- \sum_{k = 1}^{N} \frac{\langle g^{\epsilon}(x),\phi_k(x)\Big\rangle
\phi_k(x)  }{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
h^{\epsilon}(\tau) d\tau  } .
\end{aligned}
 \end{equation}
 Hence
 \begin{equation}
\begin{aligned}
&f(x) - f^{\epsilon,N}(x) \\
&= \underbrace{ \sum_{k=N+1}^{\infty} \frac{\langle g(x),\phi_k(x)\rangle
 \phi_k(x)}{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
 h(\tau) d\tau  } }_{ \mathcal{Q}_1 } + \underbrace{ \sum_{k=1}^{N}
 \frac{\langle g(x) - g^{\epsilon}(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\epsilon}(\tau) d\tau  }
 }_{ \mathcal{Q}_2 }  \\
&\quad + \underbrace {\sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)
\rangle \phi_k(x) }{ \int_0^{T}\Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
 h(\tau) d\tau} \times \sum_{k=1}^{N}
\frac{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
( h^{\epsilon}(\tau) - h(\tau) ) d\tau  }
{  \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
 h^{\epsilon}(\tau) d\tau } }_{ \mathcal{Q}_3} .
\end{aligned}
 \end{equation}
 First, we have the following estimate
 \begin{equation} \label{danhgiaQ1}
\begin{aligned}
 \|\mathcal{Q}_{1}\|^2_{L^2(\Omega)}
 &= \sum_{k=N+1}^{\infty}   \frac{|\langle g(x),\phi_k(x)\rangle |^2 }
{ \big|\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h(\tau) d\tau  \big|^2  } \\
 &= |\langle f(x),\phi_k(x)\rangle |^2 \\
 &\le \sum_{k=N+1}^{\infty} ( 1 + \lambda_k)^{-2\alpha \gamma}
( 1 + \lambda_k)^{2 \alpha \gamma} |\langle f(x),\phi_k(x)\rangle |^2  \\
 &\le (1 + \lambda_{N} )^{-2\alpha \gamma}M^2.
\end{aligned}
 \end{equation}
 Hence, we obtain
 \begin{equation}
 \|\mathcal{Q}_{1}\|_{L^2(\Omega)} \le (1 + \lambda_{N} )^{-\alpha \gamma}M.
 \end{equation}
 Second, the term  $\|\mathcal{Q}_{2}\|_{L^2(\Omega)}$ is bounded by
 \begin{equation} \label{danhgiaQ2}
\begin{aligned}
 \|\mathcal{Q}_{2}\|^2_{L^2(\Omega)}
&\le \sum_{k=1}^{N}    \frac{|<g(x) - g^{\epsilon}(x),\phi_k(x)>|^2}
 {\big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
 h^{\epsilon}(\tau) d\tau \big|^2 } \\
&\le \sum_{k=1}^{N}   \frac{ \lambda_k^{2\alpha} r^{2\beta}|<g{(x)} - g^{\varepsilon}{(x)},
 \phi_k{(x)}>|^2 }{ |{\mathcal{I}(h^{\varepsilon})}|^2
 \big( 1 - E_{\beta,1}( -\lambda_{1}^{\alpha} r^{\beta} T^{\beta} ) \big)^2 }    \\
&\le \sup_{ 1 \le k \le N} \frac{\lambda_k^{2\alpha} r^{2\beta}}
 { |{\mathcal{I}(h^{\varepsilon})}|^2 \big( 1 - E_{\beta,1}( - \lambda_{1}^{\alpha}
 r^{\beta}T^{\beta}) \big)^2 } \\
&\quad \times \sum_{k=1}^{N} |\langle g(x) - g^{\epsilon}(x),\phi_k(x)\rangle |^2   \\
 &\le \frac{\lambda_{N}^{2\alpha} r^{2\beta}}{ |{\mathcal{I}(h^{\varepsilon})}|^2
 \big( 1 - E_{\beta,1}( - \lambda_{1}^2 r^{\beta} T^{\beta}) \big)^2 }
 \|g^\varepsilon-g\|^2_{L^2(\Omega)}  \\
&\le  \frac{\lambda_{N}^{2\alpha} r^{2\beta}}
 { |{\mathcal{I}(h^{\varepsilon})}|^2 \big( 1 - E_{\beta,1}
 ( - \lambda_{1}^2 r^{\beta} T^{\beta}) \big)^2 } \epsilon^2.
\end{aligned}
 \end{equation}
 Hence
 \begin{equation}
 \|\mathcal{Q}_{2}\|_{L^2(\Omega)}
\le \frac{\lambda_{N}^{\alpha} r^{\beta}}{ |{\mathcal{I}(h^{\varepsilon})}|
 \big( 1 - E_{\beta,1}(-\lambda_{1}^2 r^{\beta} T^{\beta}) \big) } \epsilon.
 \end{equation}
Finally,  the  term $\|\mathcal{Q}_{3}\|_{L^2(\Omega)}$ can be estimated as
follows
\begin{align}
&\|\mathcal{Q}_{3}\|^2_{L^2(\Omega)} \nonumber \\
&\le \Big[\sum_{k=1}^{N} \Big| \frac{ \langle g(x),\phi_k(x)\rangle \phi_k(x)  }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)   h(\tau) d\tau  }
 \Big|^2 \Big] 
\Big[ \sum_{k=1}^{N} \Big| \frac{  \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
  ( h^{\epsilon}(\tau) - h(\tau) ) d\tau  }
 {  \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)  h^{\epsilon}(\tau) d\tau }
   \Big|^2 \Big]  \nonumber\\
&\le \Big[\sum_{k=1}^{N}
 \frac{\big|\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) ( h^{\varepsilon}(\tau)
 - h(\tau) ) d\tau \big|^2 }{\big| \int_0^{T}
 \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}(\tau)d\tau  \big|^2 } \Big]
 \Big[ \sum_{k=1}^{N}  \frac{|\langle g(x),\phi_k(x)\rangle |^2}
{ \big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau)d\tau \big|^2 }
 \Big]  \nonumber \\
 &\le   \frac{ \|h^{\varepsilon} - h\|^2_{C[0,T]}  }{|{\mathcal{I}(h^{\varepsilon})}|^2  }
 \sum_{k=1}^{\infty}  \frac{|\langle g(x),\phi_k(x)\rangle |^2}
{ \big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau)d\tau \big|^2 }
 \label{danhgiaQ3}  \\
 &= \frac{\|h^{\varepsilon} - h\|^2_{C[0,T]}}{|{\mathcal{I}(h^{\varepsilon})}|^2}
 \|f\|^2_{L^2(\Omega)} \nonumber\\
 &\le \frac{\varepsilon^2 \|f\|^2_{L^2(\Omega)} }{ |{\mathcal{I}(h^{\varepsilon})}|^2 } . \nonumber
\end{align}
  Hence
 \begin{equation}
 \|\mathcal{Q}_{3}\|_{L^2(\Omega)} 
\le \frac{\varepsilon \|f\|_{L^2(\Omega)} }{|{\mathcal{I}(h^{\varepsilon})}| } . \label{Q33}
 \end{equation}
 Combining \eqref{danhgiaQ1}, \eqref{danhgiaQ2} and \eqref{danhgiaQ3}, it yields
 \begin{equation}
\begin{aligned}
& \|f(x) - f^{\epsilon,N}(x)\|_{L^2(\Omega)}  \\
&\le (1 + \lambda_{N})^{-\alpha \gamma}M + \varepsilon \frac{ \|f\|_{L^2(\Omega)}}
{|{\mathcal{I}(h^{\varepsilon})}|}
+ \varepsilon \frac{\lambda_{N}^{\alpha} r^{\beta}}{|{\mathcal{I}(h^{\varepsilon})}|
 \big( 1 - E_{\beta,1}(-\lambda_{1}^2 r^{\beta} T^{\beta}) \big) } .
\end{aligned}
 \end{equation}
 This and the fact that  $N \le \mu \le N+1$ give
 \begin{align*}
 &\|f(x) - f^{\epsilon,N}(x)\|_{L^2(\Omega)} \\
&\le \Big( \frac{\mu \pi}{2} \Big)^{-\gamma \alpha} M
 + \varepsilon \Big( \frac{\mu \pi}{2} \Big)^{\alpha}
\max\Big\{ \frac{ r^{\beta} }{|{\mathcal{I}(h^{\varepsilon})}|
 \big( 1 - E_{\beta,1}(-\lambda_{1}^{\alpha} r^{\beta} T^{\beta}) \big) },
 \frac{ \|f\|_{L^2(\Omega)}}{|{\mathcal{I}(h^{\varepsilon})}|} \Big\}  \\
&\le \varepsilon^{\frac{\gamma}{\gamma + 1}}M^{\frac{1}{\gamma + 1}}
\Big[ 1 + \max~\Big\{ \frac{ r^{\beta} }{|{\mathcal{I}(h^{\varepsilon})}|
\big( 1 - E_{\beta,1}( - \lambda_{1}^{\alpha} r^{\beta}T^{\beta} ) \big) },
\frac{ \|f\|_{L^2(\Omega)}}{|{\mathcal{I}(h^{\varepsilon})}|} \Big\} \Big] .
\end{align*}
\end{proof}

\subsection{An a posteriori parameter choice}

In this subsection, we consider  an a posteriori regularization parameter
choice by the discrepancy principle. Define
\begin{equation}
F_{N}g^{\epsilon} = \sum_{k=1}^{N}\langle g(x),\phi_k(x)\rangle  \phi_k(x) .
\end{equation}
By the discrepancy principle, we  take $K = K(\varepsilon, g^{\varepsilon})$ as the solution of
\begin{equation} \label{canthiet}
\|(I - F_{N})g^{\varepsilon}\|_{L^2(\Omega)} \le m \epsilon
\le \|(I - F_{N-1})g^{\varepsilon}\|_{L^2(\Omega)},\quad m > 1.
\end{equation}

 \begin{lemma} \label{lem3.3}
  We have
  \begin{equation}
  N \le \frac{2}{\pi} \Big(\frac{\|h\|_{C[0,T]} M}{r^{\beta}  (m - 1)\epsilon }
 \Big)^{\frac{1}{\alpha(\gamma + 1)}} .
  \end{equation}
 \end{lemma}

\begin{proof}
 From $\|g^\varepsilon-g\|_{L^2(\Omega)} \le \varepsilon$ and \eqref{canthiet}, we have
 \begin{equation} \label{uuuu}
\begin{aligned}
 \|F_{N-1}g - g\|_{L^2(\Omega)}
&= \| (F_{N-1}-I)g^\varepsilon- (I-F_{N-1}) (g-g^\varepsilon) \|_{L^2(\Omega)} \\
&\ge \| (F_{N-1}-I)g^\varepsilon \|_{L^2(\Omega)}- \|  (I-F_{N-1})
 (g-g^\varepsilon) \|_{L^2(\Omega)}  \\
 & \ge (m-1) \varepsilon .
\end{aligned}
 \end{equation}
 On the  other hand, for $k \ge N$, we obtain
 \begin{equation}
\begin{aligned}
 \big| \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r)
h(\tau)d\tau \big|
&\le \|h\|_{C[0,T]} \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) d\tau \\
&= \|h\|_{C[0,T]}
 \frac{\big( 1 - E_{\beta,1}(- \lambda_k^{\alpha} r^{\beta} T^{\beta}) \big)}
 {\lambda_k^{\alpha} r^{\beta}} \\
&\le \frac{ \|h\|_{C[0,T]}}{ \lambda_N^\alpha  r^\beta } .
\end{aligned}
 \end{equation}
This implies
 \begin{equation} \label{uuuuu}
\begin{aligned}
&\|F_{N-1}g - g\|^2_{L^2(\Omega)} \\
&= \sum_{k=N}^{\infty}  |\langle g(x),\phi_k(x)\rangle |^2  \\
&= \sum_{k=N}^{\infty} \Big| \int_0^{T} \Phi_{\beta}
 (\lambda_k^{\alpha},\tau,r) h(\tau)d\tau \langle f(x),\phi_k(x)\rangle  \Big|^2 \\
&\le \frac{ \|h\|^2_{C[0,T]}}{ \lambda_N^{2\alpha}  r^{2\beta} }
 \sum_{k=N}^{\infty}  |\langle f(x),\phi_k(x)\rangle |^2  \\
&\le \frac{ \|h\|^2_{C[0,T]}}{ \lambda_N^{2\alpha}  r^{2\beta} }
\sum_{k=N}^{\infty}   ( 1 + \lambda_k )^{-2\alpha \gamma}
 ( 1 + \lambda_k )^{2\alpha \gamma}  |\langle f(x),\phi_k(x)\rangle |^2  \\
&\le \frac{ \|h\|^2_{C[0,T]}}{ \lambda_N^{2\alpha}  r^{2\beta} \lambda_N^{2\alpha \gamma}  }
 \sum_{k=N}^{\infty}   ( 1 + \lambda_k )^{2 \alpha \gamma} |
 \langle f(x),\phi_k(x)\rangle |^2  \\
& \le \frac{ \|h\|^2_{C[0,T]}}{ \lambda_N^{2\alpha}  r^{2\beta} \lambda_N^{2\alpha \gamma}  }
  \|f\|^2_{H^{\alpha \gamma (\Omega)}} \\
&\le  \|h\|^2_{C[0,T]} \frac{M^2}{r^{2\beta}}
\frac{1}{\lambda_{N}^{2\alpha(\gamma + 1)}} .
\end{aligned}
 \end{equation}
 Hence,
 \begin{equation}\label{ketqua112}
 \|F_{N-1}g - g\|_{L^2(\Omega)} \le
\frac{M}{r^{\beta}} \frac{\|h\|_{C[0,T]}}{\lambda_{N}^{\alpha(\gamma + 1)}} .
 \end{equation}
 From \eqref{uuuu} and \eqref{ketqua112}, we have
 \begin{equation} \label{tau1}
 (m - 1)\epsilon \le \frac{M}{r^{\beta}}
\frac{\|h\|_{C[0,T]}}{\lambda_{N}^{\alpha(\gamma + 1)}} .
 \end{equation}
 It follows from  $\lambda_k^{\alpha} = (\frac{k \pi}{2})^{\alpha}$ and
\eqref{tau1} that
 \begin{equation} \label{thaytheN}
 N \le \frac{2}{\pi} \Big(\frac{\|h\|_{C[0,T]} M}{r^{\beta}
(m - 1)\epsilon } \Big)^{\frac{1}{\alpha(\gamma + 1)}} .
 \end{equation}
\end{proof}

Next we present an error estimate for the approximate solution of
 problem \eqref{1}.

\begin{theorem} \label{thm3.4}
 Let $f^{\epsilon, N}$ and $f$ be as in  Theorem \ref{thm3.1}.
Then we have
 \begin{equation}
\begin{aligned}
&\| f(x) - f^{\varepsilon,N}(x) \|_{L^2(\Omega)} \\
&\le  \varepsilon^{\frac{\gamma}{\gamma + 1}} M^{\frac{1}{\gamma + 1}}
\Big[\mathcal{L_{\beta}}(f,h^{\varepsilon},h,r,m,T)
+  \mathcal{K_{\alpha,\beta}}(h,r,T) (m+1)^{\frac{\gamma}{\gamma + 1}} \Big] .
\end{aligned}
 \end{equation}
 where
 \begin{gather*}
\begin{aligned}
&\mathcal{L_{\beta}}(f,h^{\varepsilon},h,r,m,T)\\
&= \Big( \frac{\|h\|_{C[0,T]}}{r^{\beta} (m - 1){|\mathcal{I}(h^{\varepsilon})|}^{\gamma+1} }
\Big)^{\frac{1}{\gamma + 1}}
\max\Big\{ \|f\|_{L^2(\Omega)}, \frac{r^{\beta}}
{\big( 1 - E_{\beta,1}(-\lambda_{1}^{\alpha} r^{\beta}T^{\beta}) \big)} \Big\},
\end{aligned}  \\
 \mathcal{K_{\alpha,\beta}}(h,r,T)
= \frac{  (r^{\beta})^{\frac{\gamma}{\gamma + 1}} }{|{\mathcal{I}(h)}
|^{\frac{\gamma}{\gamma + 1}} \big( 1- E_{\alpha,1}
( - \lambda_{1}^{\alpha} r^{\beta} T^{\beta} ) \big)^{ \frac{\gamma}{\gamma+1} } }.
 \end{gather*}
\end{theorem}

\begin{proof}
 Using the triangle inequality,
 \begin{equation} \label{tamgiac}
 \| f(x) - f^{\varepsilon,N}(x) \|_{L^2(\Omega)}
\le \| f(x) - f^{N}(x) \|_{L^2(\Omega)}  + \| f(x)
- f^{\epsilon, N}(x) \|_{L^2(\Omega)} .
 \end{equation}
 We split the proof into three steps.
\smallskip

\noindent\textbf{Step 1:}
Estimate $\| f(\cdot) - f^{N}(\cdot) \|_{L^2(\Omega)}$.
 \begin{equation}
\begin{aligned}
 \| f(x) - f^{N}(x) \|_{H^{\gamma}(\Omega)}
&\le \| \sum_{k = N + 1}^{\infty} \langle f(x),\phi_k(x)\rangle \phi_k(x) \|  \\
&= \Big(\sum_{k = N + 1}^{\infty} \big( 1 + \lambda_k^{\alpha}
 \big)^{2\gamma} |\langle f(x),\phi_k(x)\rangle |^2 \Big)^{1/2}
 \le M .
\end{aligned}
 \end{equation}
By triangle inequality and $\eqref{canthiet}$,
 \begin{equation}
\begin{aligned}
 \| Af(x) - Af^{N}(x) \|_{L^2(\Omega)} &\le \| (I - F_{N})g \| \\
 &\le \| (I - F_{N})g^{\epsilon} + ( I - F_{N} )( g - g^{\epsilon} ) \|  \\
 &\le \| ( I - F_{N} )g^{\epsilon} \| + \| ( I - F_{N} )( g - g^{\varepsilon} ) \| \\
 &\le (m + 1) \varepsilon .
\end{aligned}
 \end{equation}

Therefore, by the conditional stability \eqref{ondinhcuanghiem}, we have
\begin{equation} \label{so1}
\| f(\cdot) - f^{N}(\cdot) \|_{L^2(\Omega)} \le \mathcal{K_{\alpha,\beta}}(h,r,T)
((m+1)\varepsilon)^{\frac{\gamma}{\gamma + 1}} .
\end{equation}
Next, we obtain
\begin{equation} \label{2thangs1s2}
\begin{aligned}
&f^{N}(x) - f^{\epsilon, N}(x) \\
&=  \sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)}
 {\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  }
 -\sum_{k=1}^{N} \frac{\langle g^{\varepsilon}(x),\phi_k(x)\rangle \phi_k(x)}
{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}(\tau) d\tau}   \\
&\le  \sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)}
 {\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau  }
- \sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)}
{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}(\tau) d\tau}   \\
&\quad +  \sum_{k=1}^{N} \frac{\langle g(x) - g^{\varepsilon}(x),\phi_k(x)\rangle
 \phi_k(x)}{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}(\tau) d\tau}   \\
&\le \underbrace {\sum_{k=1}^{N} \frac{ \int_0^{T} \Phi_{\beta}
 (\lambda_k^{\alpha},\tau,r) (h(\tau) - h^{\varepsilon}(\tau)) d\tau }
{ \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}(\tau) d\tau }
 \sum_{k=1}^{N} \frac{\langle g(x),\phi_k(x)\rangle \phi_k(x)}
 { \int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h(\tau) d\tau } }
 _{:=\mathcal{Q}_3}  \\
&\quad +  \underbrace { \sum_{k=1}^{N} 
\frac{\langle g(x) - g^{\varepsilon}(x),\phi_k(x)\rangle
\phi_k(x)}{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}
(\tau) d\tau}}_{:=\mathcal{Q}_4}  .
\end{aligned}
\end{equation}
Using \eqref{Q33}, we obtain
\begin{equation}
\| \mathcal{Q}_3 \|_{L^2(\Omega)}
\le \varepsilon \frac{\|f\|_{L^2(\Omega)} }{|{\mathcal{I}(h^{\varepsilon})}|} .
\end{equation}
We now estimate the norm of $\mathcal{Q}_4$. Using Lemma \ref{lem2.7}, we have
\begin{equation}
\begin{aligned}
\|\mathcal{Q}_4 \|^2_{L^2(\Omega)} 
&= \sum_{k=1}^{N}  \Big| \frac{\langle g(x) - g^{\varepsilon}(x),\phi_k(x)\rangle }
{\int_0^{T} \Phi_{\beta}(\lambda_k^{\alpha},\tau,r) h^{\varepsilon}(\tau) d\tau} \Big|^2 \\ 
&\le \sum_{k=1}^{N}  \frac{ \langle g(x) - g^{\varepsilon}(x),\phi_k(x)\rangle ^2 }
{ \frac{ |\mathcal{I}(h_\varepsilon)|^2 
\big(1 - E_{\beta,1}(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) \big)^2 }
{\lambda_k^{2\alpha} r^{2\beta}}   }  \\
& \le \frac{ \lambda_N^{2\alpha}  r^{2\beta} }{  |\mathcal{I}(h_\varepsilon)|^2 
\big(1 - E_{\beta,1}(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) \big)^2 } 
 \sum_{k=1}^{N}\langle g(x) - g^{\varepsilon}(x),\phi_k(x)\rangle ^2 \\
&\le \frac{ \varepsilon^2 \lambda_N^{2\alpha}  r^{2\beta} }{  |\mathcal{I}(h_\varepsilon)|^2      
 \big(1 - E_{\beta,1}(- \lambda_{1}^{\alpha} r^{\beta} T^{\beta}) \big)^2 }.
\end{aligned}
\end{equation}
Hence
\begin{equation} \label{s2}
\|\mathcal{Q}_4 \|_{L^2(\Omega)} \le \Big(\frac{N\pi}{2}\Big)^{\alpha} 
\frac{\epsilon}{|{\mathcal{I}(h^{\varepsilon})}|} 
\frac{ r^{\beta} }{\big(1 - E_{\beta,1}( -\lambda_{1}^{\alpha} r^{\beta} T^{\beta} )
\big) } .
\end{equation}
From  above observations, we deduce that
\begin{equation} \label{so2}
\begin{aligned}
&\| f^{N}(x) - f^{\epsilon, N}(x) \|_{L^2(\Omega)}  \\
&\le \Big(\frac{N\pi}{2}\Big)^{\alpha} \frac{\varepsilon}{|{\mathcal{I}(h^{\varepsilon})}|} 
\max \Big\{ \|f\|_{L^2(\Omega)}, 
 \frac{r^{\beta}}{\big( 1 - E_{\beta,1}(-\lambda_{1}^{\alpha} r^{\beta} T^{\beta} ) 
\big)} \Big\} .
\end{aligned}
\end{equation}
Substituting \eqref{thaytheN} in \eqref{so2}, we obtain
\begin{equation}
\| f^{N}(x) - f^{\epsilon, N}(x) \|_{L^2(\Omega)}  
\le \varepsilon^{\frac{\gamma}{\gamma + 1}} M^{\frac{1}{\gamma + 1}} 
\mathcal{L_{\beta}}(f,h^{\varepsilon},h,r,m,T) .
\end{equation}
Combining \eqref{so1} with \eqref{so2}, we obtain the final estimate as follows:
\begin{equation}
\begin{aligned}
&\| f(x) - f^{\epsilon, N}(x) \|_{L^2(\Omega)}  \\
&\le \varepsilon^{\frac{\gamma}{\gamma + 1}} M^{\frac{1}{\gamma + 1}} 
\big[ \mathcal{L_{\beta}}(f,h^{\varepsilon},h,r,m,T) +  \mathcal{K_{\alpha,\beta}}(h,r,T)
 (m+1)^{\frac{\gamma}{\gamma + 1}} \big] .
\end{aligned}
\end{equation}
hereby
\begin{gather*}
\begin{aligned}
&\mathcal{L_{\beta}}(f,h^{\varepsilon},h,r,m,T) \\
&= \Big( \frac{\|h\|_{C[0,T]}}{r^{\beta} (m - 1){|\mathcal{I}(h^{\varepsilon})|}^{\gamma+1} } 
\Big)^{\frac{1}{\gamma + 1}} 
\max\Big\{ \|f\|_{L^2(\Omega)}, \frac{r^{\beta}}
{\big( 1 - E_{\beta,1}(-\lambda_{1}^{\alpha} r^{\beta} T^{\beta} ) \big)} \Big\}, 
\end{aligned} \\
\mathcal{K_{\alpha,\beta}}(h,r,T) 
= \frac{  (r^{\beta})^{\frac{\gamma}{\gamma + 1}} }
{|{\mathcal{I}(h)|}^{\frac{\gamma}{\gamma + 1}} 
\big( 1- E_{\alpha,1} ( - \lambda_{1}^{\alpha} r^{\beta} T^{\beta} )
 \big)^{ \frac{\gamma}{\gamma+1}  }  } .
\end{gather*}
This completes the proof.
\end{proof}

\subsection*{Acknowledgments}
This work is  supported  by Institute for Computational Science and Technology 
Ho Chi Minh City under project named
``Determination source function for fractional diffusion equations and 
applications''.  The authors also desire to thank the handling editor and  the
anonymous referees for their helpful comments on this paper.

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\end{document}