\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 258, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/258\hfil Uniqueness results for an inverse problem]
{A uniqueness result for an inverse problem in a space-time fractional 
diffusion equation}

\author[S. Tatar, S. Ulusoy \hfil EJDE-2013/258\hfilneg]
{Salih Tatar, S\"{u}leyman Ulusoy}  % in alphabetical order

\address{Salih Tatar\newline
Department of Mathematics, Faculty of Education, 
Zirve University \newline
Sahinbey, Gaziantep,  27270, Turkey}
\email{salih.tatar@zirve.edu.tr}
\urladdr{http://person.zirve.edu.tr/statar/}


\address{S\"{u}leyman Ulusoy \newline
Department of Mathematics, Faculty of Education,
Zirve University,  \newline
Sahinbey, Gaziantep,
27270, Turkey}
\email{suleyman.ulusoy@zirve.edu.tr}
\urladdr{http://person.zirve.edu.tr/ulusoy/}

\thanks{Submitted May 8, 2013. Published November 22, 2013.}
\subjclass[2000]{45K05, 35R30, 65M32}
\keywords{Fractional derivative; fractional Laplacian; weak solution;
\hfill\break\indent inverse problem; Mittag-Leffler function; Cauchy problem}

\begin{abstract}
 Fractional (nonlocal) diffusion equations replace the integer-order
 derivatives in space and time by fractional-order derivatives.
 This article considers a nonlocal inverse problem and shows that
 the  exponents of the fractional time and space derivatives
 are determined uniquely by the data $u(t, 0)= g(t)$, $0 < t < T$.
 The uniqueness result is a theoretical background for determining
 experimentally the order of many anomalous diffusion phenomena,
 which are important in physics and in environmental engineering.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction} \label{sec:intro}


The classical diffusion equation $\partial_t u = \Delta u$ is used to 
describe a cloud of spreading particles at the macroscopic level. 
The point source solution is a Gaussian probability density that 
predicts the relative particle concentration. For microscopic picture, 
Brownian motion is employed, which describes the path of individual particles. 
 The space-time fractional diffusion equation 
 $\partial_t^{\beta} u = \Delta^{\alpha/2} u $ with $0 < \beta < 1$ 
 and $0 < \alpha < 2$ is used to model anomalous diffusion \cite{MBSB}. 
Here, the fractional derivative in time is used to describe particle 
sticking and trapping phenomena and the fractional space derivative 
is used to model long particle jumps. These two effects combined 
together produce a concentration profile with a sharper peak, and heavier tails. 
The fractional-time derivative considered here is the Caputo fractional 
derivative of order $0 < \beta < 1$ and is defined as
\begin{equation}\label{def:Caputo}
\frac{\partial^{\beta} f(t) }{\partial t^{\beta}} 
:= \frac{1}{\Gamma(1 - \beta)} \int_0^t \frac{\partial f(r)}{\partial r} 
\, \frac{dr}{(t - r)^{\beta}},
\end{equation}
where $\Gamma$ is the Gamma function. This is intended to properly handle 
initial values \cite{Cap, CMN, 10}, since its Laplace transform  
$s^{\beta} \tilde{f}(s) - s^{\beta -1}f(0)$ incorporates the initial 
value in the same way the first derivative does. Here,  $\tilde{f}(s)$ 
is the usual Laplace transform. It is well-known that the Caputo derivative 
has a continuous spectrum \cite{CMN, 11},  with eigenfunctions given in 
terms of the Mittag-Leffler function
$$ 
E_{\beta}(z)  := \sum_{k = 0}^{\infty} \frac{z^k}{\Gamma(1 + \beta k)}.
$$
In fact, it is easy to see that, $f(t) = E_{\beta}(-\lambda t^{\beta})$ 
solves the eigenvalue equation
$$ 
\frac{\partial^{\beta}f(t)}{\partial t^{\beta}} = - \lambda f(t),
$$
for any $\lambda > 0$. This is easily verified by differentiating
term-by-term and using the fact that $t^{p}$ has Caputo derivative 
$t^{p - \beta} \frac{\Gamma(p + 1)}{\Gamma(p + 1 - \beta)}$ for $p > 0$ 
and $0 < \beta \leq 1$. 
For slow diffusion we take $0 < \beta < 1$, which is  related to parameter 
specification of the
large-time behaviour of the waiting-time distribution function,
 see \cite{11} and references  therein.


For $0 < \alpha < 2$, $\Delta^{\alpha/2} f$ denotes the fractional Laplacian,
defined for
$$ 
f \in \operatorname{Dom}(\Delta^{\alpha/2})
 = \big\{   f \in L^2(\mathbb{R}^d; dx) : \int_{\mathbb{R}^d}  |\xi|^{\alpha} 
|\widehat{f}(\xi)|^2 \, d\xi < \infty \big\}  
$$
as the function with Fourier transform
\begin{equation}\label{def-fralapFT}
\widehat{\Delta^{\alpha/2} f} (\xi) = - |\xi|^{\alpha} \widehat{f}(\xi),
\end{equation}
where $\widehat{f}(\xi)$ denotes the Fourier transform of $f(x)$. For sufficiently regular
functions (for example, $C^2$ functions with bounded second derivatives), the fractional Laplacian can be defined pointwise by

\begin{equation}\label{fr-lap-pointw}
\Delta^{\alpha/2} f(x) = \int_{\mathbb{R}^d \setminus \{0\}} 
\left( f(x+y) - f(x) - \nabla f(x) \cdot y 1_{\{|y| \leq 1 \}}  \right)
 \frac{c_{d, \alpha}}{|y|^{d+\alpha}} \, dy,
\end{equation}
where $c_{d, \alpha} > 0$ is given by
$$ 
c_{d, \alpha} \int_{y \in \mathbb{R}^d} \frac{1- \cos(y_1)}{|y|^{d + \alpha}} \, dy = 1.
$$
Similar formulations have been observed before, see \cite{CMN} and references 
therein. We note that the fractional Laplacian used here is usually called 
the Riesz fractional derivative (see \cite{RFD} for details) in the fractional 
calculus community. Following \cite{CMN}, we observe that if $f$ is bounded 
and continuous on $\mathbb{R}^d$ and $f$ is $C^2$ in an open set $D$, then
$\Delta^{\alpha/2} f$ exists
pointwise and is continuous in $D$. Moreover,  if $f$ is $C^1$-function on 
$[0, \infty)$ satisfying $|f'(t)| \leq C t^{\gamma - 1}$  for some $\gamma > 0$, 
then by \eqref{def:Caputo}, the Caputo derivative
$\frac{\partial^{\beta}f(t)}{\partial t^{\beta}}$ of $f$ exists for all $t > 0$ 
and the derivative is continuous in $t >0$. The reader is referred to 
Kilbas et al.~\cite{Kil} and Podlubny \cite{11}
for  properties of the Caputo derivative.

Important references documenting the recent interest on inverse problems with
fractional derivatives are \cite{CNYY, JR, Mis2, Mis1, Mis5, Mis6, Mis3, Mis4, ZX}.
A substantial difference between our study and those in the latter references 
is that we consider fractional derivatives both in the time and space variable 
whereas those study only consider fractional derivatives in the time variable. 
On the other hand, our work shares with those studies the use of eigenfunction 
expansion of weak solutions to the initial/boundary value problem.

The main purpose of this article is to establish the  determination of  
the unique exponents $\beta$ and $\alpha$ in the fractional time and 
space derivatives by means of the observation data $u(t, 0)=g(t)$, $0 < t < T$. 
This uniqueness result may lead to the
identification of anomalous diffusions.
For the sake of simplicity and for the technical reasons on eigenvalue bounds, 
throughout this paper we consider $d =1$ and
 $\alpha \in (1/2,2)$.


This article is organized as follows: In the next section we provide a 
review of main properties of the direct problem and introduce 
the inverse problem. Section \ref{sec:main} includes both the
statement and the proof of the main result of this paper.

\section{Analysis of the direct problem and formulation of the inverse problem}
\label{sec:analy direk-inverse}

First we consider the direct problem
\begin{equation} \label{eqn01}
 \begin{gathered}
\frac{\partial^\beta}{\partial t^\beta}u(t, x) = \Delta^{\alpha/2}u(t, x),
\quad -1< x < 1, ~0< t <T,\\
u(t, -1) = u(t, 1) = 0, \quad 0 < t < T,\\
u(0, x) = f(x), \quad -1 < x < 1.
\end{gathered}
\end{equation}
Here $T > 0$ is a final time and $f$ is a given function.

\begin{definition}[\cite{CMN}]\label{def:weak-sol} \rm
A function $u(t, x)$ is said to be a weak solution of  \eqref{eqn01}
 if the following conditions hold:
\begin{equation} \label{weaksol-cond}
\begin{gathered}
u(t, \cdot) \in W_0^{\alpha/2,2}(D) \quad \text{for each } t>0,\\
\lim_{t \downarrow 0} u(x,t)=f(x)\quad \text{a.e.},\\
\frac{\partial^\beta}{\partial t^\beta}u(t, x)= \Delta^{\alpha/2} u(t, x)
\quad \text{in the distrubitional sense}.
\end{gathered}
\end{equation}
\end{definition}

Here $W_0^{\alpha/2,2}(D)$ is the $\sqrt {\epsilon_1}$-completion of the
space $C_c^\infty(D)$ of smooth functions with compact support in $D$ 
where $\epsilon_1(u,u)=\epsilon(u,u)+\int_{\mathbb{R}}u(x)^2dx$ and 
$\epsilon(u,v)=\varepsilon^D(u,v)$ for $u, v \in W_0^{\alpha/2,2}(D)$.
 We note that such a definition can be given for any bounded domain
 $D$ in $\mathbb{R}^d$, replacing $(-1, 1)$. The last condition of 
\eqref{weaksol-cond} is equivalent to the following 
(which comes from multiplying the equation by the test function 
$\phi(x)\psi(t)$, integration by parts and symmetry, 
for details see \cite{CMN}): 
for each $\psi \in C_c^1(0,\infty)$ and $\phi \in C_c^2(D)$,
\[
\int_{\mathbb{R}} \bigg(\int_0^{\infty} u(t, x) 
\frac{\partial^\beta \psi(t) }{\partial t^\beta} \, dt \bigg) \phi(x) \, dx 
= \int_0^{\infty}\varepsilon^D(u(t, \cdot), \phi) \psi(t) \, dt,
\]
where $D = (-1,1)$, $C_c^1(0,\infty)$ is the space of compactly supported 
$C^1$ (the class of continuously differentiable functions) functions,  
$C_c^2(D)$ is the space of compactly supported $C^2$ 
(the class of twice continuously differentiable functions) functions,
\[
\varepsilon^D(u,v)=\frac{c_\alpha}{2} 
\iint_{\mathbb{R} \times \mathbb{R}} \frac{\big(u(x)-u(y)\big)\big(v(x)-v(y)\big)}
{\vert x-y \vert^{1+\alpha}}\, dx \,dy,
\]
for $u,v \in \mathcal{F}$, $c_\alpha>0$ is a constant, $\varepsilon^D(u,v)$ 
comes from variational formulation and symmetry (see \cite{CMN} for details),
 $$
\mathcal{F} := W^{\frac{\alpha}{2},2}(\mathbb{R}) 
:= \Big\{    u \in L^2(\mathbb{R}; dx): \iint_{\mathbb{R} \times \mathbb{R}} \frac{(u(x)-u(y))^2} 
{\vert x-y\vert^{1+\alpha}} \, dx \,dy<\infty  \Big\}. 
$$


Following \cite{CMN}, we obtain a useful formula for the weak 
solution of \eqref{eqn01}:
\begin{equation}\label{use-form-ws}
\begin{split}
u(t, x) &= \int_0^{\infty} \mathbb{E}_x[f(X_s); s < \tau_D] f_t(s) \, ds \\
&=\int_0^{\infty} \Big( \sum_{n=1}^{\infty}e^{-s\lambda_n} \langle f,\psi_n\rangle 
 \psi_n(x) \bigg) f_t(s) \, ds \\
&=\sum_{n=1}^{\infty} E_\beta(-\lambda_n t^\beta) \langle f,\psi_n\rangle 
 \psi_n(x),
\end{split}
\end{equation}
where  the last equality comes from a conditioning argument
 (see \cite{CMN} for more details), $\mathbb{E}_x$ is the expected value 
with respect to $x$,  $\{ \lambda_n \}_{n \geq 1}$ is a sequence of 
positive numbers $0 < \lambda_1 \leq \lambda_2 \leq \dots$,
$\{\psi_n\}_{n \geq 1}$ is an orthonormal basis for $L^2(D; dx)$ and for any 
$f \in L^2(D; dx)$ has the representation
$$ 
f(x) = \sum_{n = 1}^{\infty} \langle f,\psi_n\rangle  \psi_n(x).
$$
The following lemma indicates an important property of the Mittag-Leffler 
function, see \cite{Kil, 11} for more details.

\begin{lemma} \label{lem2.2}
For each $\alpha<2$ and $\pi\alpha/2<\mu < \min \{\pi, \pi\alpha \}$ 
there exists a constant $C_0>0$ 
such that
\begin{equation}\label{(5)}
\vert  E_\beta(z)\vert \leq \frac{C_0}{1+\vert z \vert}, \quad
\mu \leq\vert arg(z) \vert \leq \pi.
\end{equation}
\end{lemma}

We recall the following result from \cite{K} about the asymptotic 
behavior of the eigenvalues of the fractional Laplacian in an interval,
 which is used for further estimates and proofs. We note that in the 
higher dimensional case, the estimates are not so explicit, though we believe 
that a similar uniqueness result holds.

\begin{theorem}\label{ev-asympexp}
The eigenvalues of the spectral problem for the one-dimensional 
fractional Laplace operator, i.e. $(-\Delta)^{\alpha/2} u(x) = \lambda u(x)$, 
in the interval $D \subset \mathbb{R}$ satisfy the asymptotic equality
\begin{equation}\label{asmpt-beha-frLap}
\lambda_n = \Big(\frac{n \pi}{2}-\frac{(2-\alpha)\pi}{8} \Big) ^\alpha 
+ O\big(\frac{1}{n}\big).
\end{equation}
\end{theorem}

Next, we show that the series on the right-hand side of \eqref{use-form-ws} 
is  uniformly convergent in $x \in [-1, 1]$ and $t\in (0,T]$. 
For this purpose, we use the following inequalities for the eigenvalues 
and corresponding eigenvectors of the one-dimensional fractional Laplacian:
\begin{equation} \label{bounds}
\begin{gathered}
C_1 n^\alpha \leq \lambda_n \leq C_2 n^\alpha, \quad n\geq 1,\\
\vert\langle f,\psi_n\rangle \vert \leq \sqrt{M} \lambda_n^{-k},\\
\vert \psi_n(x) \vert \leq C_3\lambda_n^{1/2\alpha},
\end{gathered}
\end{equation}
where $C_i $, $i = 1, 2, 3$ are  positive constants, and
\[
 M := \sum_{n=1}^\infty \lambda_n ^{2k} \langle f,\psi_n\rangle ^2<\infty,
\]
for some $k$ that satisfies the following inequality
(see \cite{BG,CMN} for details):
\begin{equation}\label{(fork)}
 k>-1+\frac{7}{2\alpha}.
\end{equation}
Then by \eqref{(5)} and \eqref{bounds} we have
\begin{equation}\label{inek-1}
\begin{split}
 \sum_{n=1}^\infty \max_{0 \leq x \leq l}
\vert E_\beta(-\lambda_nt^\beta)\langle f,\psi_n\rangle \psi_n(x) \vert 
& \leq\sqrt{M}C_0C_3\sum_{n=1}^\infty\frac{1}{1+\vert\lambda_n  t^\beta\vert}
 \lambda_n^{-k}\lambda_n^{1/2{\alpha} }  \\
& \leq C_\star\sum_{n=1}^\infty \frac{1}{1+\vert\lambda_n  t^\beta\vert}
n^{\alpha(\frac{1} {2\alpha}-k)}<\infty,
\end{split}
\end{equation}
where  $C_\star=\sqrt{M}C_0C_2^2C_3$ is a positive constant.
 Now using \eqref{asmpt-beha-frLap}, we see that the series on the last
line of \eqref{inek-1} is uniformly convergent if $k > -1 + \frac{3}{2\alpha}$,
which is already guaranteed by the condition \eqref{(fork)}.

 The inverse problem  consists of determining the unknown orders $\beta$ 
and $\alpha$ of the  time and space derivatives in the space-time fractional 
diffusion problem \eqref{eqn01} from the measured output data
(also called additional condition)
\begin{equation}\label{(8)}
u(t, 0)=g(t),\quad 0 < t < T.
\end{equation}
For technical reasons in the proof of determining the exponents $\beta$ 
and $\alpha$ uniquely,  we will need  a specific class of the initial 
functions $f(x)$ satisfying
\begin{equation}\label{(specificclass)}
\langle f(x), \psi_n(x)\rangle  > 0 \quad  
\text{(or $\langle f(x), \psi_n(x)\rangle < 0$)}
\quad \text{for all $ n \geq 1$},
\end{equation}
where $\{\psi_n(x)\}_{n \geq 1}$ is the orthonormal basis for 
$L^2(D; dx)$ considered above. In  this article, we assume 
that $g(t) \not\equiv 0$. Next section is devoted to the statement 
and the proof of the uniqueness result for the inverse problem.

\section{Statement and the proof of the main result}\label{sec:main}

The main result of the paper, whose proof is also included in this section,
reads as follows.

\begin{theorem}\label{thm:main}
Let $u$ be the weak solution of  \eqref{eqn01} and let $v$ be the weak 
solution of the following equation with the same initial and boundary 
conditions:
\begin{equation}\label{(othereq)}
 \frac{\partial^\gamma}{\partial t^\gamma}u(t, x) 
= \Delta^{\eta/2}u(t, x), \quad -1< x < 1,\; 0< t <T.
\end{equation}
If  $u(t,0) = v(t,0)$, $0 < t < T$ and \eqref{(specificclass)} holds,
 then $\beta=\gamma$ and $\alpha=\eta$.
\end{theorem}

\begin{proof} Using the explicit formula \eqref{use-form-ws}, 
the weak solutions $u(t,x)$ and $v(t,x)$ can be written as follows:
\begin{gather}\label{weak-1}
u(t,x)=\sum_{n=1}^\infty E_\beta(-\lambda_nt^\beta)\langle f,\psi_n\rangle 
\psi_n(x), \\
\label{weak-2}
v(t,x)=\sum_{n=1}^\infty E_\gamma(-\mu_nt^\gamma)\langle f,\varphi_n\rangle 
\varphi_n(x).
\end{gather}

Here $\psi_n$ and $\varphi_n$ are the eigenfunctions corresponding to 
$\lambda_n$ and $\mu_n$ which  are the eigenvalues of the equations 
$(-\Delta)^{\alpha/2} u(x)=\lambda u(x)$ and 
$ (-\Delta)^{\eta/2} v(x)=\lambda v(x)$, respectively satisfying
 $\psi_n(0)=1$ and $\varphi_n(0)=1$. Consequently, assuming that 
$u(t, 0)=v(t, 0)$ we have
\begin{equation}\label{100}
\sum_{n=1}^\infty E_\beta(-\lambda_nt^\beta)\langle f,\psi_n\rangle 
= \sum_{n=1}^\infty E_\gamma(-\mu_nt^\gamma)\langle f,\varphi_n\rangle .
\end{equation}
Now we derive an asymptotic equality for the left-hand side of \eqref{100}
 using the following well known asymptotic property of the Mittag-Leffler 
function \cite{Kil, 11}:
\begin{eqnarray}\label{prop-mit-lef}
E_\beta(-t)=\frac{1}{t\Gamma(1-\beta)}+O(\vert t \vert^{-2}).
\end{eqnarray}
By \eqref{asmpt-beha-frLap} and \eqref{prop-mit-lef}, there exists a 
constant $C_4>0$ such that
\begin{equation}\label{13}
\Big \vert E_\beta(-\lambda_n t^\beta)-\frac{1}{\Gamma(1-\beta)}
\frac{1}{\lambda_n  t^\beta}\Big\vert \leq \frac{C_4}{t^{2\beta}}.
\end{equation}
Now we use the asymptotic behaviour of \eqref{13} on the left-hand side 
of \eqref{100}. For this purpose, we add and subtract the term 
$\frac{1}{\Gamma(1-\beta)\lambda_n  t^\beta}$ to right hand side 
of \eqref{weak-1}. Then, we obtain
\begin{equation}\label{15}
\begin{split}
&\sum_{n=1}^\infty E_\beta(-\lambda_nt^\beta)\langle f,\psi_n\rangle \\
& = \sum_{n=1}^\infty\langle f,\psi_n\rangle 
\Big[\frac{1}{\Gamma(1-\beta)\lambda_n  t^\beta} 
+\big \{ E_\beta(-\lambda_nt^\beta)
-\frac{1}{\Gamma(1-\beta)\lambda_n  t^\beta}\big\}\Big].
\end{split}
\end{equation}
Finally, we get the following asymptotic equality for the left hand 
side of \eqref{100} by using \eqref{13} in \eqref{15}
\begin{equation}\label{16}
\sum_{n=1}^\infty E_\beta(-\lambda_nt^\beta)\langle f,\psi_n\rangle 
 = \sum_{n=1}^\infty\langle f,\psi_n\rangle 
\frac{1}{\Gamma(1-\beta)}\frac{1}{\lambda_n  t^\beta}
+ O\big(\vert \frac{1}{t^{2\beta} }\vert\big).
\end{equation}
Arguing similarly for 
$\sum_{n=1}^\infty E_\gamma(-\mu_nt^\gamma)\langle f,\varphi_n\rangle $, 
we have
\begin{equation}\label{17}
\sum_{n=1}^\infty E_\gamma(-\mu_nt^\gamma)\langle f,\varphi_n\rangle \\
 =\sum_{n=1}^\infty\langle f,\varphi_n\rangle \frac{1}{\Gamma(1-\gamma)}
\frac{1}{\mu_n  t^\gamma} + O\big(\vert \frac{1}{t^{2\gamma} } \vert\big).
\end{equation}
Therefore, from \eqref{100}, \eqref{16} and \eqref{17}, we have,
 as $t \to \infty$,
\begin{equation}\label{18}
\begin{split}
&\sum_{n=1}^\infty\langle f,\psi_n\rangle \frac{1}{\Gamma(1-\beta)}
\frac{1}{\lambda_n  t^\beta}+O\big(\vert \frac{1}{t^{2\beta} }\vert\big)\\
&=\sum_{n=1}^\infty\langle f,\varphi_n\rangle 
\frac{1}{\Gamma(1-\gamma)}\frac{1}{\mu_n  t^\gamma}
+O\big(\vert \frac{1}{t^{2\gamma} }\vert\big).
\end{split}
\end{equation}
To complete the proof, for the moment, we suppose that  $\beta>\gamma$. 
Then multiplying  \eqref{18} by $t^\gamma$ yields that
\begin{equation} \label{result-1}
\begin{split}
&-\frac{t^\gamma}{t^\beta}\sum_{n=1}^\infty\langle f,\psi_n\rangle
\frac{1}{\Gamma(1-\beta)}\frac{1}{\lambda_n }
+ O\big(\vert \frac{t^\gamma}{t^{2\beta} } \vert\big)\\
&+\sum_{n=1}^\infty\langle f,\varphi_n\rangle
 \frac{1}{\Gamma(1-\gamma)}\frac{1}{\mu_n}
 +O\big(\vert \frac{1}{t^{\gamma} } \vert\big)=0.
\end{split}
\end{equation}
Letting $t \to \infty$ in  \eqref{result-1}, we deduce that
\begin{eqnarray}\label{19}
\sum_{n=1}^\infty\langle f,\varphi_n\rangle \frac{1}{\Gamma(1-\gamma)}
\frac{1}{\mu_n }=0.
\end{eqnarray}
By \eqref{(specificclass)}, this is a contradiction since the left-hand
side of \eqref{19} can not be zero. Similarly, the assumption
$\gamma > \beta$   leads to a  contradiction.
Therefore, we conclude that $\gamma = \beta$, and this completes the
 first part of  the proof.


Now, we prove the second part of the theorem, 
i.e. $\alpha = \eta$. For this purpose we show
 $\lambda_n=\mu_n$, $n=1,2,3,\dots$. By \eqref{100} and $\gamma = \beta$ we have
\begin{equation}\label{byfipart}
\sum_{n=1}^\infty E_\beta(-\lambda_nt^\beta)\langle f,\psi_n\rangle 
= \sum_{n=1}^\infty E_\beta(-\mu_nt^\beta)\langle f,\varphi_n\rangle .
\end{equation}
We take the Laplace transform of $E_\beta(-\lambda_nt^\beta)$ as follows:
\begin{equation}\label{Laplace_1}
\int_0^\infty e^{-zt} E_\beta(-\lambda_n t^\beta)\, dt
=\frac{z^{\beta-1}}{z^\beta+\lambda_n}, \quad \operatorname{Re} z>0.
\end{equation}
Moreover, if we take the Laplace transform of the Mittag-Leffler 
function term by term, we obtain
\begin{equation}\label{Laplace_2}
\int_0^\infty e^{-zt} E_\beta(-\lambda_n t^\beta)\, dt
=\frac{z^{\beta-1}}{z^\beta+\lambda_n},\quad 
\operatorname{Re} z>\lambda_n^{1/\beta}.
\end{equation}
Then by \eqref{(5)}, we conclude that 
$\sup_{t\geq 0}\vert  E_\beta(-\lambda_n t^\beta)\vert < \infty$, 
and this implies that   $\int_0^\infty e^{-zt} E_\beta(-\lambda_n t^\beta)\, dt$ 
is analytic in $z$ for $\operatorname{Re}z>0$. Thus the analytic 
continuation yields \eqref{Laplace_1} for $\operatorname{Re} z>0$. 
By using \eqref{(5)}, \eqref{asmpt-beha-frLap}, \eqref{bounds} and Lebesgue's 
convergence theorem, we get that $e^{-t\operatorname{Re} z}t^{-\beta}$ 
is integrable for $t\in (0,\infty)$ with fixed $z$ satisfying
 $\operatorname{Re}z>0$ and
\begin{align*}
 \Big\vert  e^{-t\operatorname{Re}z}  
 \sum_{n=1}^{\infty} \langle f,\psi_n\rangle   
 E_\beta(-\lambda_n t^\beta)  \Big\vert 
& \leq C_0  e^{-t\operatorname{Re}z} 
\Big(\sum_{n=2}^{\infty} \langle f,\psi_n\rangle \frac{1}
{\vert \lambda_n \vert}\frac{1}{t^\beta} \Big)\\
&\leq \frac{C_0' }{t^\beta} e^{-t\operatorname{Re}z}
 \sum_{n=1}^{\infty}n^{-\alpha(k+1)},
\end{align*}
where $C_0'=\sqrt{M}C_0C_2$. Here we note that the series 
$\sum_{n=1}^{\infty}n^{-\alpha(k+1)}$ is convergent by \eqref{(fork)}. 
Then, for $\operatorname{Re}z>0$ we obtain
\begin{equation}\label{obtained1}
\int_0^\infty e^{-zt} \sum_{n=1}^{\infty} \langle f,\psi_n\rangle 
 E_\beta(-\lambda_n t^\beta)\, dt
=\sum_{n=1}^{\infty} \langle f,\psi_n\rangle 
\frac{z^{\beta-1}}{z^\beta+\lambda_n}.
\end{equation}
Similarly,
\begin{equation}\label{obtained2}
\int_0^\infty e^{-zt} \sum_{n=1}^{\infty} \langle f,\varphi_n\rangle  
E_\beta(-\mu_n t^\beta)\, dt
= \sum_{n=1}^{\infty} \langle f,\varphi_n\rangle  
\frac{z^{\beta-1}}{z^\beta+\mu_n}.
\end{equation}
Then, from \eqref{byfipart}, \eqref{obtained1} and \eqref{obtained2} 
we deduce that
\begin{equation}\label{finalywehave}
\sum_{n=1}^{\infty} \frac{\langle f,\psi_n\rangle }{z^\beta+\lambda_n}
=\sum_{n=1}^{\infty}  \frac{\langle f,\varphi_n\rangle }{z^\beta+\mu_n}, 
\quad \operatorname{Re}z>0,
\end{equation}
or equivalently,
\begin{equation}\label{finalywehave1}
\sum_{n=1}^{\infty} \frac{\langle f,\psi_n\rangle }{\rho+\lambda_n}
=\sum_{n=1}^{\infty}  \frac{\langle f,\varphi_n\rangle }{\rho+\mu_n}, \quad
\operatorname{Re}\rho>0.
\end{equation}
Since we can continue analytically both sides of \eqref{finalywehave1} 
in $\rho$, this equality holds for 
$\rho \in \mathbb{C} \setminus\big( \{ -\lambda_n\}_{n \geq 1}  
\cup \{ -\mu_n\}_{n \geq 1} \big)$. 
Now we prove that $\lambda_1=\mu_1$. For this purpose, assume that 
 $\lambda_1\neq \mu_1$. Without loss of generality, we can suppose 
$\lambda_1<\mu_1$. Then we can find a suitable disk that contains
 $-\lambda_1$ but does not contain  
$\{ -\lambda_n\}_{n \geq 2}  \cup \{ -\mu_n\}_{n \geq 1}$. 
If we integrate \eqref{finalywehave1} in this disk, the Cauchy's integral 
formula yields
\begin{equation*}
2\pi i \langle f,\psi_1\rangle =0.
\end{equation*}
By \eqref{(specificclass)}, this is a contradiction since $\langle f, \psi_1\rangle $ 
can not be zero. This means $\lambda_1=\mu_1$.  By repeating the same argument, 
we obtain $\lambda_2 = \mu_2$. Continuing inductively we finally deduce that
\begin{eqnarray}\label{finalfinal}
\lambda_n=\mu_n,\quad  n=1,2,3,\dots.
\end{eqnarray}
This means that the following equality holds for $n=1,2,3,\dots$,
\begin{equation}\label{reachresult}
\big(\frac{n \pi}{2}-\frac{(2-\alpha)\pi}{8} \bigg ) ^\alpha 
+ O\big(\frac{1}{n}\big)=\big(\frac{n \pi}{2}-\frac{(2-\eta)\pi}{8} \big) ^\eta 
+ O\big(\frac{1}{n}\big).
\end{equation}
To conclude $\alpha=\eta$, we prove that the function  
$H(\alpha)=\big(\frac{n \pi}{2}-\frac{(2-\alpha)\pi}{8} \big) ^\alpha$,
$n =1, 2, 3,\dots$ is a monotone increasing function of $\alpha$. 
For this purpose, we need to find the  derivative of the function
 $H(\alpha)$. By using the logarithmic differentiation we obtain
\begin{equation}\label{firstderH}
H'(\alpha)=H(\alpha) \Big\{\ln \Big(\frac{n \pi}{2}-\frac{(2-\alpha)\pi}{8}\Big)
+\alpha \frac{ \pi/8}{\big(\frac{n \pi}{2}-\frac{(2-\alpha)\pi}{8}\big)} \Big\}.
\end{equation}
Since the function $H(\alpha)$ and the second term in the bracket are positive 
for  $n=1,2,3,\dots$ and $\alpha \in (1/2,2)$, we estimate 
the first term in the bracket. We know that the function $\ln(x)$ 
is positive for $x>1$. This means we solve the following inequality 
with respect to $n$:
\begin{equation}\label{inequality}
\frac{n\pi}{2}-\frac{(2-\alpha)\pi}{8}\geq 1.
\end{equation}
Solving \eqref{inequality} yields 
\[
n\geq\frac{2}{\pi}+\frac{2-\alpha}{4}>\frac{2}{\pi}+\frac{2-1/2}{4}\approx 1.01.
\]
Then we deduce that the function $H(\alpha)$ is monotone increasing 
function for $n=2,3,4,\dots$. In addition, for $n=1$ we need to solve 
the  inequality
\begin{equation}\label{inequalityadditional}
\ln \Big(\frac{\pi}{2}-\frac{(2-\alpha)\pi}{8} \Big)
+\alpha \frac{ \pi/8}{\big(\frac{\pi}{2}-\frac{(2-\alpha)\pi}{8}\big)} \geq 0.
\end{equation}
Solving \eqref{inequalityadditional} yields $\alpha \geq 0.27$ which is 
already guaranteed by the condition $\alpha \in (1/2,2)$. 
This completes the proof.
\end{proof}

\subsection*{Acknowledgments} 
The authors want thank the anonymous referees for their valuable suggestions 
to improve the presentation of the original manuscript. 
The second author wants to thank E. Nane for the interesting discussions.
This research was partially supported by the Scientific
and Technological Research Council of Turkey (TUBITAK), also
by the  Zirve University Research Fund.

\begin{thebibliography}{00}

\bibitem{BG} 
R. M.~ Blumenthal, R. K.~ Getoor.
\newblock  Asymptotic distribution of the eigenvalues for a class of Markov 
operators.
\newblock {\em  Pacific J. Math.}, 9:399--408, 1959.

\bibitem{Cap}
M. ~ Caputo.
\newblock Linear models of diffuson whose Q is almost frequency independent,
 part II.
\newblock {\em Geophys. J. R. Astron. Soc.}, 13:529--539, 1967.

\bibitem{CMN}
Q. Z. ~ Chen, M. M. ~ Meerschaert and E.~ Nane.
\newblock Space-time fractional diffusion on bounded domains.
\newblock {\em Journal of Math. Analysis and its Appl.}, 393:479--488, 2012.

\bibitem{CNYY}
J. ~ Chen, J. ~ Nakagawa, M. ~Yamamoto and T.~ Yamazaki.
\newblock  Uniqueness in an inverse problem for a one-dimensional 
fractional diffusion equation.
\newblock {\em Inverse Problems}, 25:115--131, 2009.

\bibitem{10}
S. D. ~ Eidelman, S. D.~ Ivasyshen, A. N. ~ Kochubei.
\newblock Analytic Methods in the Theory of Differential and 
Pseudo-Differential Equations of Parabolic Type.
\newblock {\em  Birkh\"{a}user, Basel}, 2004.


\bibitem{JR}
B. ~ Jin,  and W.~ Rundell.
\newblock An inverse problem for a one-dimensional time-fractional diffusion 
problem.
\newblock {\em Inverse Problems}, 28:075010, 2012.

\bibitem{Kil}
A. A. ~Kilbas, H. M. ~Srivastava and J. J.~ Trujillo.
\newblock Theory and Applications of Fractional Differential Equations.
\newblock {\em Elsevier, Amsterdam}, 2006.

\bibitem{K}
M. ~ Kwasnicki.
\newblock Eigenvalues of the fractional Laplace operator
in the interval.
\newblock {\em Journal of Functional Analysis}, 262:2379--2402, 2012.

\bibitem{Mis2}
J.J ~ Liu and M. ~ Yamamato.
\newblock A backward problem for the time-fractional diffusion equation.
\newblock {\em Applicable Analysis}, 89:1769-1788, 2010.

\bibitem{Mis1}
F. ~ Mainardi, Y. ~ Luchko and G. ~ Pagnini.
\newblock The fundamental solution of the space-time fractional diffusion equation.
\newblock {\em Fractional Calculus and Applied Analysis}, 4:153-192, 2001.

\bibitem{MBSB}
M. M. ~ Meerschaert, D. A. ~Benson, H.-P. ~ Scheffler and B.~ Baeumer.
\newblock Stochastic solution of space-time fractional diffusion equations.
\newblock {\em Phys. Rev. E}, 65, 2002.

\bibitem{11}
I. ~Podlubny.
\newblock Fractional Differential Equations.
\newblock {\em Academic Press, San Diego}, 1999.

\bibitem{Mis5}
K. ~ Sakamoto and M. ~ Yamamato
\newblock Inverse source problem with a final overdetermination for a fractional 
diffusion equation.
\newblock {\em Mathematical Controls and Related Fields}, 4:509-518, 2011.
\bibitem{Mis6}
K. ~ Sakamoto and M. ~ Yamamato
\newblock Initial value/boundary value problems for fractional diffusion-wave 
equations and applications to some inverse problems
\newblock {\em Journal of Mathematical Analysis and Applications}, 
382:426-447, 2011.

\bibitem{RFD}
S.G. ~ Samko, A.A. ~ Kilbas and O.I. ~ Marichev.
\newblock Fractional Integrals and Derivatives: Theory and Applications.
\newblock {\em Taylor$\&$Francis}, 1983.

\bibitem{Mis3}
X ~ Xu, J. ~ Cheng and M. ~ Yamamato
\newblock Carleman esimate for a fractional diffusion equation with half order 
and application.
\newblock {\em Applicable Analysis}, 90:1355-1371, 2011.

\bibitem{Mis4}
M. ~ Yamamato and Y. ~ Zhang
\newblock Conditional stability in determining a zeroth-order coefficient in 
a half-order fractional diffusion equation by a Carleman estimate.
\newblock {\em Inverse Problems}, 28:105010, 2012.

\bibitem{ZX}
Y. ~ Zhang,  and X.~ Xu.
\newblock Inverse source problem for a fractional diffusion equation.
\newblock {\em Inverse Problems}, 27:035010, 2011.

\end{thebibliography}

\end{document}
