1 Paola Loreti, 1 Daniela Sforza, 2,3,4 M. Yamamoto

#  Simultaneous determination of initial value and source term for time-
fractional wave-diffusion equations

###### Abstract.

We consider initial boundary value problems for time fractional diffusion-wave
equations:

$d_{t}^{\alpha}u=-Au+\mu(t)f(x)$

in a bounded domain where $\mu(t)f(x)$ describes a source and
$\alpha\in(0,1)\cup(1,2)$, and $-A$ is a symmetric ellitpic operator with
repect to the spatial variable $x$. We assume that $\mu(t)=0$ for $t>T$:some
time and choose $T_{2}>T_{1}>T$. We prove the uniqueness in simultaneously
determining $f$ in $\Omega$, $\mu$ in $(0,T)$, and initial values of $u$ by
data $u|_{\omega\times(T_{1},T_{2})}$, provided that the order $\alpha$ does
not belong to a countably infinite set in $(0,1)\cup(1,2)$ which is
characterized by $\mu$. The proof is based on the asymptotic behavior of the
Mittag-Leffler functions.
Key words: time-fractional diffusion-wave equation, inverse source problem,
initial value, uniqueness
Mathematics Subject Classification: 35R30, 35R11, 35R25

1 Dipartimento di Scienze di Base e Applicate per l’Ingegneria, Sapienza
Università di Roma, Via Antonio Scarpa 16, 00161 Roma, Italy e-mail:
<EMAIL_ADDRESS>
1 Dipartimento di Scienze di Base e Applicate per l’Ingegneria, Sapienza
Università di Roma, Via Antonio Scarpa 16, 00161 Roma, Italy e-mail:
<EMAIL_ADDRESS>
2 Graduate School of Mathematical Sciences, The University of Tokyo, Komaba,
Meguro, Tokyo 153-8914, Japan
3 Honorary Member of Academy of Romanian Scientists, Ilfov, nr. 3, Bucuresti,
Romania
4 Correspondence member of Accademia Peloritana dei Pericolanti,
Palazzo Università, Piazza S. Pugliatti 1 98122 Messina Italy
e-mail<EMAIL_ADDRESS>
##  1\. Introduction and Main Results

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded domain with smooth boundary
$\partial\Omega$ and let $\nu=(\nu_{1}(x),...,\nu_{d}(x))$ be the unit outward
normal vector to $\partial\Omega$ at $x\in\partial\Omega$. For
$\alpha\in(0,1)\cup(1,2)$, let $d_{t}^{\alpha}$ denote the Caputo derivative:

$d_{t}^{\alpha}v(t)=\left\\{\begin{array}[]{rl}\frac{1}{\Gamma(1-\alpha)}\int^{t}_{0}(t-s)^{-\alpha}\frac{dv}{ds}(s)ds,&0<\alpha<1,\\\
\frac{1}{\Gamma(2-\alpha)}\int^{t}_{0}(t-s)^{1-\alpha}\frac{d^{2}v}{ds^{2}}(s)ds,&1<\alpha<2\end{array}\right.$

for suitable $v$, where $\Gamma(\cdot)$ denotes the gamma function.

We consider

$None$ $\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}u=-Au+\mu(t)f(x),\quad
x\in\Omega,t>0,\\\ &u|_{\partial\Omega\times(0,\infty)}=0,\\\
&u(\cdot,0)=a\quad\mbox{in $\Omega$ $\quad$ if
$0<\alpha<1$}\end{array}\right.$

and

$None$ $\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}u=-Au+\mu(t)f(x),\quad
x\in\Omega,t>0,\\\ &u|_{\partial\Omega\times(0,\infty)}=0,\\\
&u(\cdot,0)=a,\quad\partial_{t}u(\cdot,0)=b\quad\mbox{in $\Omega$ $\quad$ if
$1<\alpha<2$}.\end{array}\right.$

Here

$None$
$-Au(x)=\sum_{i,j=1}^{d}\partial_{i}(a_{ij}(x)\partial_{j}u)+c(x)u,\quad c\leq
0\quad\mbox{on $\overline{\Omega}$},$

and $a,b\in L^{2}(\Omega)$. We assume

$None$ $\mu\in L^{\infty}(0,T),\quad\mu\not\equiv 0\quad\mbox{in
$(0,\infty)$},\quad\mu(t)=0\quad\mbox{for $t\geq T$}.$

Then it is known that there exists a unique solution to each of (1.1) and
(1.2) in suitable classes. More precisely, we show
Lemma 1.
In (1.1) and (1.2) we assume $\mu\in L^{\infty}(0,\infty)$ and $f\in
L^{2}(\Omega)$.
(i): $0<\alpha<1$: In (1.1), we assume that $a\in L^{2}(\Omega)$. To (1.1),
there exists a unique solution $u\in C([0,T];L^{2}(\Omega))\cap
C((0,T];H^{2}(\Omega)\cap H^{1}_{0}(\Omega))$ to (1.1).
(ii): $1<\alpha<2$: In (1.2) we assume $a,b\in L^{2}(\Omega)$. To (1.2), there
exists a unique solution $u\in C([0,T];L^{2}(\Omega))\cap
C^{1}([0,T];\mathcal{D}(A^{-\frac{1}{\alpha}}))\cap C((0,T];H^{2}(\Omega)\cap
H^{1}_{0}(\Omega))$.

Here $A^{-\frac{1}{\alpha}}$ denotes the fractional power of the operator $A$
(e.g., Pazy [23]) defined by (1.3) with the domain
$\mathcal{D}(A)=H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$. The space
$\mathcal{D}(A^{-\frac{1}{\alpha}})$ is wider than $L^{2}(\Omega)$ and the
topology is weaker than the one in $L^{2}(\Omega)$. The proof of the lemma is
done by the eigenfunction expansions (2.1) stated below and as for the
details, see Sakamoto and Yamamoto [25] for example. We remark that the
framework for coverning more general forms of equations with non-symmetric $A$
is available and see Kubica, Ryszewska and Yamamoto [17], Yamamoto [26] for
instance.

Let $\omega\subset\Omega$ be an arbitrarily chosen subdomain and let
$T<T_{1}<T_{2}$ be arbitrary. We consider
Inverse problem: Determine $f(x)$ and $\mu(t)$ of a source term and initial
values $a$ for $0<\alpha<1$ and $a,b$ for $1<\alpha<2$ simultaneously by data
$u|_{\omega\times(T_{1},T_{2})}$.

For the statement of the main result, we introduce an index
$\ell_{0}\in\mathbb{N}\cup\\{0\\}$. Here and henceforth let
$\mathbb{N}:=\\{1,2,3,...\\}$.

We show
Lemma 2.
The space Span $\\{t^{\ell};\,\ell\in\mathbb{N}\cup\\{0\\}\\}$ spanned by all
the monomials is dense in $L^{2}(0,T)$.

The proof is direct from the Weierstrass polynomial approximation theorem and
for completeness we prove the lemma at the end of Secton 2.

By $\mu\not\equiv 0$ in $(0,T)$, Lemma 2 yields that

$\left\\{m\in\mathbb{N}\cup\\{0\\};\,\int^{T}_{0}(-s)^{m}\mu(s)ds\neq
0\right\\}\neq\emptyset.$

Indeed, if not, then $\mu\in L^{\infty}(0,T)\subset L^{2}(0,T)$ is orthogonal
to the dense set in $L^{2}(0,T)$, and so $\mu=0$ in $(0,T)$, which is
impossible by $\mu\not\equiv 0$ in $(0,T)$.

Therefore, we can define

$\ell_{0}:=\min\left\\{m\in\mathbb{N}\cup\\{0\\};\,\int^{T}_{0}(-s)^{m}\mu(s)ds\neq
0\right\\}.$

In particular, we see $\ell_{0}=0$ if $\int^{T}_{0}\mu(s)ds\neq 0$, and if
$\ell_{0}\in\mathbb{N}$, then

$None$ $\int^{T}_{0}(-s)^{\ell_{0}}\mu(s)ds\neq
0,\quad\int^{T}_{0}(-s)^{m}\mu(s)ds=0\quad\mbox{for $0\leq m\leq\ell_{0}-1$}.$

If $\mu\geq 0$ in $(0,T)$ satisfies (1.4), then $\int^{T}_{0}\mu(s)ds\neq 0$,
and so $\ell_{0}=0$.

Now we are ready to state the main result.
Theorem 1.
(i): $0<\alpha<1$. We assume

$None$ $\alpha\not\in\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}.$

Then $u|_{\omega\times(T_{1},T_{2})}=0$ implies $a=f=0$ in $\Omega$.
(ii): $1<\alpha<2$. We assume

$None$
$\alpha\not\in\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}\cup\left\\{\frac{\ell_{0}+2}{n}\right\\}_{n\in\mathbb{N}}.$

Then $u|_{\omega\times(T_{1},T_{2})}=0$ implies $a=b=f=0$ in $\Omega$.

In (1.6), (1,7) and also (1.11), (1.12) stated below, the right-hand sides
mean sequences, that is, $\ell_{0}+1$ and $n$ are not assumed to be
irreducible for example in (1.6).

For example, we see

$\displaystyle\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}=\left\\{\frac{1}{n}\right\\}_{n\in\mathbb{N}}\quad\mbox{if
$\ell_{0}=0$},$
$\displaystyle\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}=\left\\{\frac{1}{n}\right\\}_{n\in\mathbb{N}}\cup\left\\{\frac{2}{2n-1}\right\\}_{n\in\mathbb{N}}\quad\mbox{if
$\ell_{0}=1$}.$

We do not know whether we can remove the conditions (1.6) and (1.7) on
$\alpha$ in Theorem 1, and also (1.10) and (1.11) in Theorem 2 stated below.
We give more remark in 1. of Section 5.

Now we can extend the uniqueness in Theorem 1 to the inverse problem of
determining both $f(x)$ and $\mu(t)$ as well as initial values.

For the statement of Theorem 2, in addition to (1.1) and (1.2), we consider

$None$
$\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}\widetilde{u}=-A\widetilde{u}+\widetilde{\mu}(t)\widetilde{f}(x),\quad
x\in\Omega,t>0,\\\ &\widetilde{u}|_{\partial\Omega\times(0,\infty)}=0,\\\
&\widetilde{u}(\cdot,0)=\widetilde{a},\quad\mbox{in $\Omega$ $\quad$ if
$0<\alpha<1$}\end{array}\right.$

and

$None$
$\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}\widetilde{u}=-A\widetilde{u}+\widetilde{\mu}(t)\widetilde{f}(x),\quad
x\in\Omega,t>0,\\\ &\widetilde{u}|_{\partial\Omega\times(0,\infty)}=0,\\\
&\widetilde{u}(\cdot,0)=\widetilde{a},\quad\partial_{t}\widetilde{u}(\cdot,0)=\widetilde{b}\quad\mbox{in
$\Omega$ $\quad$ if $1<\alpha<2$}.\end{array}\right.$

We assume that $a,\,\widetilde{a},\,b,\,\widetilde{b},\,f,\,\widetilde{f}\in
L^{2}(\Omega)$ and $\mu,\widetilde{\mu}$ satisfy (1.4). Moreover let
$f\not\equiv 0$ in $\Omega$ or $\widetilde{f}\not\equiv 0$ in $\Omega$. We
define $\ell_{1}\in\mathbb{N}\cup\\{0\\}$ as the minimal number satisfying

$\ell_{1}:=\min\left\\{m\in\mathbb{N}\cup\\{0\\};\,\int^{T}_{0}\mu(s)(-s)^{m}ds\neq
0\quad\mbox{or}\quad\int^{T}_{0}\widetilde{\mu}(s)(-s)^{m}ds\neq 0\right\\}.$

More precisely, we define $\ell_{1}=0$ if $\int^{T}_{0}\mu(s)ds\neq 0$ or
$\int^{T}_{0}\widetilde{\mu}(s)(-s)^{m}ds\neq 0$, and $\ell_{1}\geq 0$ if and
only if

$None$
$\left\\{\begin{array}[]{rl}&\int^{T}_{0}\mu(s)(-s)^{m}ds=\int^{T}_{0}\widetilde{\mu}(s)(-s)^{m}ds=0\quad\mbox{for
$0\leq m\leq\ell_{1}-1$},\\\ &\int^{T}_{0}\mu(s)(-s)^{\ell_{1}}ds\neq
0\quad\mbox{or}\quad\int^{T}_{0}\widetilde{\mu}(s)(-s)^{\ell_{1}}ds\neq
0.\end{array}\right.$

By $\mu\not\equiv 0$ and $\widetilde{\mu}\not\equiv 0$, Lemma 2 guarantees the
existence of such $\ell_{1}\not\in\mathbb{N}\cup\\{0\\}$.
Theorem 2.
(i): $0<\alpha<1$. Let $u,\widetilde{u}$ satisfy (1.1) and (1.8) respectively.
We assume

$None$ $\alpha\not\in\left\\{\frac{\ell_{1}+1}{n}\right\\}_{n\in\mathbb{N}}.$

If $u=\widetilde{u}$ in $\omega\times(T_{1},T_{2})$, then $a=\widetilde{a}$ in
$\Omega$ and there exists constants $c_{1},c_{2}$ satisfying
$|c_{1}|+|c_{2}|>0$ such that

$c_{1}f(x)=c_{2}\widetilde{f}(x),\quad x\in\Omega\quad\mbox{and}\quad
c_{2}\mu(t)=c_{1}\widetilde{\mu}(t),\quad 0<t<T.$

(ii): $1<\alpha<2$. Let $u,\widetilde{u}$ satisfy (1.2) and (1.9)
respectively. We assume

$None$
$\alpha\not\in\left\\{\frac{\ell_{1}+1}{n}\right\\}_{n\in\mathbb{N}}\cup\left\\{\frac{\ell_{1}+2}{n}\right\\}_{n\in\mathbb{N}}$

If $u=\widetilde{u}$ in $\omega\times(T_{1},T_{2})$, then $a=\widetilde{a}$
and $b=\widetilde{b}$ in $\Omega$ and there exists constants $c_{1},c_{2}$
satisfying $|c_{1}|+|c_{2}|>0$ such that

$c_{1}f(x)=c_{2}\widetilde{f}(x),\quad x\in\Omega\quad\mbox{and}\quad
c_{2}\mu(t)=c_{1}\widetilde{\mu}(t),\quad 0<t<T.$

In (1.11) and (1.12), we recall that $\frac{\ell_{1}+1}{n}$ and
$\frac{\ell_{1}+2}{n}$ are not assumed to be irreducible.

We remark that $\mu(t)f(x)=c\mu(t)\frac{1}{c}f(x)$ for $x\in\Omega$ and $t>0$
with any constant $c\neq 0$, we can at best expect the uniqueness up to a
multiplicative constant, as is stated in Theorem 2.

Theorem 1 follows directly from Theorem 2 by setting
$\widetilde{\mu}\equiv\mu$ in $(0,T)$, and the proof is based on the same idea
for Theorem 1. For more readability, we separately state and prove both
theorems.

As for inverse problems of determining source terms of fractional differential
equations in the case where extra data are taken along a time interval from
$t=0$, there are already many works. Here not aiming at any comprehensive
list, we can refer to Jiang, Li, Liu and Yamamoto [8], Jin, Kian and Zhou
[10], Kian [11], Kian, Soccorsi, Xue and Yamamoto [13], Kian and Yamamoto
[14], Li, Liu and Yamamoto [18], Liu, Li and Yamamoto [19], Liu, Rundell and
Yamamoto [20], Liu and Zhang [21] and the references of these articles. For
determination of initial values, we here only refer to Jiang, Li, Pauron and
Yamamoto [9], Theorem 4.2 in Sakamoto and Yamamoto [25].

We emphasize that unlike the above articles, we exclusively take data not from
$t=0$ but over time interval $0<T_{1}<t<T_{2}$. It is more realistic than data
starting at $t=0$ that we can start observations after $T>0$, that is, after
the incident finishes at $t=T$, which is described by $\mu(t)=0$ for $t\geq
T$. For such a formulation of inverse source problems of determining $f(x)$
with known initial values, see Cheng, Lu and Yamamoto [2] for parabolic
equations and hyperbolic equations. As for inverse problems of determining a
source $f(x)$ with such data for time-fractional differential equations, see
Jannno and Kian [7], Kian, Liu and Yamamoto [12], Kinash and Janno [16],
Yamamoto [27].

All the above referred works discuss two types for inverse source problems:

  * •

determination of source terms such as $f(x)$ with given initial values $a$
and/or $b$.

  * •

determination of initial values with given source terms.

To the best knowledge of the authors, there are no works on the simultaneous
determination of $f(x)$ and $\mu(t)$ of a source and initial values by single
observation data $u|_{\omega\times(T_{1},T_{2})}$ after the incident finishes.
Thus Theorems 2 is the first uniqueness result for determination $f(x)$ and
$\mu(x)$ as well as initial values.

The time-fractional diffusion-wave equations have remarkably different
properties from the cases $\alpha=1$ and $\alpha=2$, and one distinguished
character is less smoothing or averaging effect for solutions, which implies
that original profiles of initial values can be preserved more compared with
the case $\alpha=1$, but less compared with $\alpha=2$. Such a difference
brings drastic differences especially for inverse problems for time-fractional
diffusion-wave equations, and such an example is well-posedness of the
backward problems in time, which is a strong contrast to $\alpha=1$, and we
can refer only to Floridia, Li, Yamamoto [4], Floridia and Yamamoto [5] as a
few recent works.

Moreover by such less smoothing effect, we can expect the uniqueness in
simultaneously detemining initial values and source terms, because the effect
from initial values and the one from the source can be separately reflected in
the observation data. This can not be correct for the cases $\alpha=1$ and
even $\alpha=2$, as the following examples show for cases of ordinary
differential equations.
Example.
Let $\lambda\neq 0$ be an arbitrarily chosen real constant. Let
$a,f\in\mathbb{R}$ and

$\mu(t)=\chi_{(0,T)}(t)=\left\\{\begin{array}[]{rl}&1,\quad 0<t<T,\\\ &0,\quad
t\geq T.\end{array}\right.$

(i) Case $\alpha=1$: We consider

$\frac{du}{dt}(t)=-\lambda u(t)+\chi_{(0,T)}(t)f,\quad u(0)=a.$

Then

$u(t)=\left\\{\begin{array}[]{rl}&e^{-\lambda t}a+e^{-\lambda
t}\frac{e^{\lambda T}-1}{\lambda}f,\quad t>T,\\\ &e^{-\lambda
t}a+\frac{1-e^{-\lambda t}}{\lambda}f,\quad 0<t<T.\end{array}\right.$

Therefore $u(t)=0$ for $T_{1}<t<T_{2}$ if and only if $a\lambda+(e^{\lambda
T}-1)f=0$, which does not imply the uniqueness for two constants $a,f$.
(ii) Case $\alpha=2$: We consider

$\frac{d^{2}u}{dt^{2}}(t)=-\lambda u(t)+\chi_{(0,T)}(t)f,\quad
u(0)=a,\quad\frac{du}{dt}(0)=b$

with $\lambda=r^{2}>0$. Then

$\displaystyle u(t)=a\cos rt+b\frac{\sin rt}{r}+f\int^{T}_{0}\frac{\sin
r(t-s)}{r}ds=a\cos rt+b\frac{\sin rt}{r}+\frac{\cos r(T-t)-\cos rt}{r^{2}}f$
$\displaystyle=$ $\displaystyle\left(a+\frac{\cos rT-1}{r^{2}}f\right)\cos
rt+\left(\frac{b}{r}+\frac{\sin rT}{r^{2}}f\right)\sin rt,\quad t>T.$

Therefore, $u(t)=0$ for $T_{1}<t<T_{2}$ if and only if

$a+\frac{\cos rT-1}{r^{2}}f=\frac{b}{r}+\frac{\sin rT}{r^{2}}f=0.$

In general we cannot conclude that $a=b=f=0$.

This article is composed of five sections. Section 2 provides some lemmata
which are used for the proofs of Theorems 1 and 2, and Sections 3 and 4 are
devoted to the proofs of Theorems 1 and 2 respectively. In Section 5, we give
concluding remarks.

##  2\. Preliminaries

It is known that the elliptic operator $A$ with the domain
$\mathcal{D}(A)=H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ has a countably infinite
number of eigenvalues, and by $c\leq 0$ in $\Omega$ in (1.3), we can number as

$0<\lambda_{1}<\lambda_{2}<\cdots\longrightarrow\infty.$

Let $P_{n}$ be the eigenprojection for the eigenvalue $\lambda_{n}$ of $A$:
$AP_{n}v=\lambda_{n}P_{n}v$ for each $n\in\mathbb{N}$. We write

$\|f\|:=\|f\|_{L^{2}(\Omega)},\quad(f,g):=\int_{\Omega}f(x)g(x)dx.$

Henceforth by $E_{\beta,\gamma}(z)$ with $\beta>0$ and $\gamma>0$, we denote
the Mittag-Leffler function (e.g., Podlubny [24]):

$E_{\beta,\gamma}(z):=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(\beta
k+\gamma)},\quad z\in\mathbb{C}.$

The power series can readily verify that $E_{\beta,\gamma}(z)$ is an entire
function in $z\in\mathbb{C}$.

First we can show
Lemma 3 (eigenfunction expansions).
Let $u$ be the solutions to (1.1) and (1.2) in Lemma 1. Then

$None$
$u(x,t)=\left\\{\begin{array}[]{rl}&\sum_{n=1}^{\infty}E_{\alpha,1}(-\lambda_{n}t^{\alpha})P_{n}a\\\
+&\sum_{n=1}^{\infty}\left(\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(t-s)^{\alpha})\mu(s)ds\right)P_{n}f,\quad
t>T\quad\mbox{if $0<\alpha<1$},\\\
&\sum_{n=1}^{\infty}E_{\alpha,1}(-\lambda_{n}t^{\alpha})P_{n}a+\sum_{n=1}^{\infty}tE_{\alpha,2}(-\lambda_{n}t^{\alpha})P_{n}b\\\
+&\sum_{n=1}^{\infty}\left(\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(t-s)^{\alpha})\mu(s)ds\right)P_{n}f,\quad
t>T\quad\mbox{if $1<\alpha<2$}.\end{array}\right.$

The series are convergent in the corresponding spaces indicated in Lemma 1.

The proof can be found for example, in [25].

Next
Lemma 4.
The mapping $t\longrightarrow u(\cdot,t)$ is analytic from $(T,\infty)$ to
$L^{2}(\Omega)$.
Proof. The proof is done by the eigenfunction expansion (2.1).

We further show
Lemma 5.
Let $N\in\mathbb{N}$ be fixed. We assume that $p_{1},...,p_{N}$ are mutually
distinct positive constants and $a_{1},...,a_{N}\in\mathbb{R}$. If

$\sum_{k=1}^{N}\frac{a_{k}}{t^{p_{k}}}+O\left(\frac{1}{t^{r}}\right)=0\quad\mbox{as
$t\to\infty$},$

then $a_{k}=0$ as long as $p_{k}<r$ with $k=1,2,...,N$.
Proof of Lemma 5. Without loss of generality, we can assume that
$p_{1}<p_{2}<\cdots<p_{N}$. We have

$a_{1}+\sum_{k=2}^{N}a_{k}\frac{1}{t^{p_{k}-p_{1}}}=O\left(\frac{1}{t^{r-p_{1}}}\right)$

as $t\to\infty$. Therefore, $a_{1}=0$. Next

$\frac{a_{2}}{t^{p_{2}}}+\sum_{k=3}^{N}a_{k}\frac{1}{t^{p_{k}}}=O\left(\frac{1}{t^{r}}\right),$

that is,

$a_{2}+\sum_{k=3}^{N}a_{k}\frac{1}{t^{p_{k}-p_{2}}}=O\left(\frac{1}{t^{r-p_{2}}}\right)$

as $t\to\infty$. Therefore, $a_{2}=0$. We can continue this argument as long
as $p_{k}<r$ for $k=1,2,...,N$. Thus we completes the proof of Lemma 5.
$\blacksquare$

Furthermore, we show
Lemma 6.
Let

$\sum_{n=1}^{\infty}|a_{n}|^{2}<\infty$

hold. We assume that there exists
$\\{\ell_{k}\\}_{k\in\mathbb{N}}\subset\mathbb{N}$ satisfying
$\lim_{k\to\infty}\ell_{k}=\infty$. If

$\sum_{n=1}^{\infty}\frac{a_{n}}{\lambda_{n}^{\ell_{k}}}=0\quad\mbox{for all
$k\in\mathbb{N}$},$

then $a_{n}=0$ for all $n\in\mathbb{N}$.
Proof of Lemma 6. We recall that $0<\lambda_{1}<\lambda_{2}<\cdots$ are the
eigenvalues of $A$. Setting
$d_{n}:=\mbox{dim}\,\\{w\in\mathcal{D}(A);\,(A-\lambda_{n})w=0\\}$, we create
a sequence $\\{\mu_{m}\\}_{m\in\mathbb{N}}$ by repeating $\lambda_{n}$
$d_{n}$-times, in other words, we number $\lambda_{n}$, $n\in\mathbb{N}$
according to the multiplicities $d_{n}$. Then it is known (e.g., Courant and
Hilbert [3]) that $\mu_{m}\sim c_{0}m^{\frac{2}{d}}$ with some constant
$c_{0}>0$ for all large $m\in N$. By the definition of $\mu_{m}$, we have
$\lambda_{n}\geq\mu_{n}$ for each $n\in\mathbb{N}$, and so

$\lambda_{n}\geq c_{0}n^{\frac{2}{d}}\quad\mbox{for all large
$n\in\mathbb{N}$}.$

Hence, since $\lim_{k\to\infty}\ell_{k}=\infty$, we see that there exists
$k_{0}\in\mathbb{N}$ such that $\frac{4\ell_{k}}{d}>1$ for all $k\geq k_{0}$.
Consequently, by $\sum_{n=1}^{\infty}|a_{n}|^{2}<\infty$, we obtain

$\sum_{n=1}^{\infty}\left|\frac{a_{n}}{\lambda_{n}^{\ell_{k}}}\right|\leq\left(\sum_{n=1}^{\infty}|a_{n}|^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty}\left|\frac{1}{\lambda_{n}^{2\ell_{k}}}\right|\right)^{\frac{1}{2}}\leq
C_{1}\left(\sum_{n=1}^{\infty}|a_{n}|^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty}\frac{1}{n^{\frac{4\ell_{k}}{d}}}\right)^{\frac{1}{2}}<\infty$

for all $k\geq k_{0}$.

Since

$\frac{a_{1}}{\lambda_{1}^{\ell_{k}}}+\sum_{n=2}^{\infty}\frac{a_{n}}{\lambda_{n}^{\ell_{k}}}=0,\quad\mbox{that
is},\quad
a_{1}+\sum_{n=2}^{\infty}a_{n}\left(\frac{\lambda_{1}}{\lambda_{n}}\right)^{\ell_{k}}=0,$

we have

$\displaystyle|a_{1}|=\left|-\sum_{n=2}^{\infty}a_{n}\left(\frac{\lambda_{1}}{\lambda_{n}}\right)^{\ell_{k}}\right|=\left|\sum_{n=2}^{\infty}\frac{a_{n}}{\lambda_{n}^{\ell_{k_{0}}}}\lambda_{1}^{\ell_{k_{0}}}\left(\frac{\lambda_{1}}{\lambda_{n}}\right)^{\ell_{k}-\ell_{k_{0}}}\right|$
$\displaystyle\leq$ $\displaystyle
C\sum_{n=2}^{\infty}\left|\frac{a_{n}}{\lambda_{n}^{\ell_{k_{0}}}}\right|\left(\frac{\lambda_{1}}{\lambda_{n}}\right)^{\ell_{k}-\ell_{k_{0}}}\leq
C_{1}\left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{\ell_{k}-\ell_{k_{0}}}$

for all $k\geq k_{0}$. By $0<\lambda_{1}<\lambda_{2}<....$, we see that
$\left|\frac{\lambda_{1}}{\lambda_{2}}\right|<1$. Letting $k\to\infty$, we see
that $\ell_{k}-\ell_{k_{0}}\to\infty$, and so $a_{1}=0$. Therefore,

$\sum_{n=2}^{\infty}\frac{a_{n}}{\lambda_{n}^{\ell_{k}}}=0,\quad
k\in\mathbb{N}.$

Repeating the above argument, we have $a_{2}=a_{3}=\cdots=0$. Thus the proof
of Lemma 6 is complete. $\blacksquare$

We conclude this section with
Proof of Lemma 2. Since it is proved by the mollifier (e.g., Adams [1]) that
$C[0,T]$ is dense in $L^{2}(0,T)$, for any $u\in L^{2}(0,T)$ and
$\varepsilon>0$, we can find $u_{\varepsilon}\in C[0,T]$ such that
$\|u-u_{\varepsilon}\|_{L^{2}(0,T)}<\varepsilon$. By the Weierstrass
polynomial approximation theorem, for any $\varepsilon>0$, we can find
$p_{\varepsilon}\in\,\mbox{Span}\,\\{s^{\ell};\,\ell\in\mathbb{N}\cup\\{0\\}\\}$
such that $\|p_{\varepsilon}-u_{\varepsilon}\|_{C[0,T]}<\varepsilon$.
Therefore,

$\displaystyle\|u-p_{\varepsilon}\|_{L^{2}(0,T)}=\|u-u_{\varepsilon}+u_{\varepsilon}-p_{\varepsilon}\|_{L^{2}(0,T)}\leq\|u-u_{\varepsilon}\|_{L^{2}(0,T)}+\|u_{\varepsilon}-p_{\varepsilon}\|_{L^{2}(0,T)}$
$\displaystyle\leq$
$\displaystyle\varepsilon+\sqrt{T}\|u_{\varepsilon}-p_{\varepsilon}\|_{C[0,T]}\leq(1+\sqrt{T})\varepsilon.$

Thus the proof of Lemma 2 is complete. $\blacksquare$

##  3\. Proof of Theorem 1

The proof relies on the asymptotic expansions of the Mittag-Leffler functions.

We define a set $\\{m(k)\\}_{k\in\mathbb{N}}\subset\mathbb{N}$ by
$\\{k\in\mathbb{N};\,\alpha k\not\in\mathbb{N}\\}$. We see that
$\\{m(k)\\}_{k\in\mathbb{N}}$ is an infinite set.
Indeed, if it is a finite set, then there exists some $N_{0}\in\mathbb{N}$
such that $n\alpha\in\mathbb{N}$ for $n\geq N_{0}$. Therefore
$N_{0}\alpha,\,(N_{0}+1)\alpha\in\mathbb{N}$, that is $\alpha=(N_{0}+1)\alpha-
N_{0}\alpha\in\mathbb{N}$, which is impossible by $0<\alpha<1$ and
$1<\alpha<2$. $\blacksquare$

More precisely, we can number

$m(k)\alpha\not\in\mathbb{N},\quad m(1)<m(2)<\cdots\longrightarrow\infty.$

In particular, we note that $m(1)=1$ and, if $\alpha\not\in\mathbb{Q}$, then
$m(k)=k$ for all $k\in\mathbb{N}$, and that

$j\in\mathbb{N}$ satisfies $\alpha j\not\in\mathbb{N}$ if and only if $j=m(k)$
with some $k\in\mathbb{N}$.

First Step: asymptotic expansions the solutions.
Let $\lambda>0$. The asymptotic behavior of the Mittag-Leffler functions
(e.g., [24]) yields

$E_{\alpha,1}(-\lambda t^{\alpha})=\sum_{k\in\\{1,...,M\\},\alpha
k\not\in\mathbb{N}}\frac{(-1)^{k+1}}{\Gamma(1-\alpha
k)}\frac{1}{\lambda^{k}}\frac{1}{t^{\alpha
k}}+O\left(\frac{1}{t^{\alpha(M+1)}}\right),$
$t^{\alpha-1}E_{\alpha,\alpha}(-\lambda
t^{\alpha})=\sum_{k\in\\{2,...,M\\},\alpha(k-1)\not\in\mathbb{N}}\frac{(-1)^{k+1}}{\Gamma((1-k)\alpha)}\frac{1}{\lambda^{k}}\frac{1}{t^{\alpha(k-1)+1}}+O\left(\frac{1}{t^{\alpha
M+1}}\right)$

and

$tE_{\alpha,2}(-\lambda t^{\alpha})=\sum_{k\in\\{1,...,M\\},\alpha
k\not\in\mathbb{N}\setminus\\{1\\}}\frac{(-1)^{k+1}}{\Gamma(2-\alpha
k)}\frac{1}{\lambda^{k}}\frac{1}{t^{\alpha
k-1}}+O\left(\frac{1}{t^{\alpha(M+1)-1}}\right)$

for all large $t>0$ and all $M\in\mathbb{N}$.

Since $\frac{1}{\Gamma(-k)}=0$ for $k\in\mathbb{N}\cup\\{0\\}$, we can
represent the series as follows in terms of $m(k)$. First

$E_{\alpha,1}(-\lambda
t^{\alpha})=\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(1-\alpha
m(k))}\frac{1}{\lambda^{m(k)}}\frac{1}{t^{\alpha m(k)}}+O\left(t^{-\alpha
m(N+1)}\right)$

as $t\to\infty$.

Next, for $j\in\mathbb{N}$, noting that $\alpha(j-1)\not\in\mathbb{N}$ if and
only if $j\in\\{m(k)+1\\}_{k\in\mathbb{N}}$, we represent

$t^{\alpha-1}E_{\alpha,\alpha}(-\lambda
t^{\alpha})=\sum_{j=1}^{N}\frac{(-1)^{m(j)}}{\Gamma(-\alpha
m(j))}\frac{1}{\lambda^{m(j)+1}}\frac{1}{t^{\alpha m(j)+1}}+O\left(t^{-\alpha
m(N+1)-1}\right)$

as $t\to\infty$.

Finally, for $1<\alpha<2$, we see that $\alpha
k\not\in\mathbb{N}\setminus\\{1\\}$ if and only if $\alpha
k\not\in\mathbb{N}$.
Indeed, it is trivial that $\alpha k\not\in\mathbb{N}$ implies $\alpha
k\not\in\mathbb{N}\setminus\\{1\\}$. Let $\alpha
k\not\in\mathbb{N}\setminus\\{1\\}$. Then $\alpha k\not\in\mathbb{N}$ or
$\alpha k=1$ with $k\in\mathbb{N}$. Then $\alpha k=1$ implies
$\alpha=\frac{1}{k}\leq 1$, which contradicts that $\alpha>1$. Hence, $\alpha
k\not\in\mathbb{N}\setminus\\{1\\}$ implies $\alpha k\not\in\mathbb{N}$.
$\blacksquare$
Consequently,

$t^{\alpha-1}E_{\alpha,\alpha}(-\lambda
t^{\alpha})=\sum_{j=1}^{N}\frac{(-1)^{m(j)+1}}{\Gamma(2-\alpha
m(j))}\frac{1}{\lambda^{m(j)}}\frac{1}{t^{\alpha m(j)-1}}+O\left(t^{-\alpha
m(N+1)+1}\right)$

as $t\to\infty$ for each $N\in\mathbb{N}$.

To sum up,

$None$
$\left\\{\begin{array}[]{rl}&E_{\alpha,1}(-\lambda_{n}t^{\alpha})=\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(1-\alpha
m(k))}\frac{1}{\lambda_{n}^{m(k)}}\frac{1}{t^{\alpha m(k)}}+O\left(t^{-\alpha
m(N+1)}\right),\cr\\\
&tE_{\alpha,2}(-\lambda_{n}t^{\alpha})=\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(2-\alpha
m(k))}\frac{1}{\lambda_{n}^{m(k)}}\frac{1}{t^{\alpha
m(k)-1}}+O\left(t^{-\alpha m(N+1)+1}\right)\quad\mbox{for $1<\alpha<2$},\cr\\\
&t^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}t^{\alpha})=\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\frac{1}{\lambda_{n}^{m(k)+1}}\frac{1}{t^{\alpha
m(k)+1}}+O\left(t^{-\alpha m(N+1)-1}\right)\end{array}\right.$

for all large $t>0$ and each $n,N\in\mathbb{N}$.
Consequently,

$\displaystyle\sum_{n=1}^{\infty}(P_{n}a)(x)E_{\alpha,1}(-\lambda_{n}t^{\alpha})$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}(P_{n}a)\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(1-\alpha
m(k))}\frac{1}{\lambda_{n}^{m(k)}}\frac{1}{t^{\alpha
m(k)}}+\sum_{n=1}^{\infty}(P_{n}a)O\left(t^{-\alpha m(N+1)}\right)$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(1-\alpha
m(k))}\left(\sum_{n=1}^{\infty}\frac{P_{n}a}{\lambda_{n}^{m(k)}}\right)\frac{1}{t^{\alpha
m(k)}}+O\left(t^{-\alpha m(N+1)}\right),$
$\displaystyle\sum_{n=1}^{\infty}(P_{n}b)(x)tE_{\alpha,2}(-\lambda_{n}t^{\alpha})$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}(P_{n}b)\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(2-\alpha
m(k))}\frac{1}{\lambda_{n}^{m(k)}}\frac{1}{t^{\alpha
m(k)-1}}+\sum_{n=1}^{\infty}(P_{n}b)O\left(t^{-\alpha m(N+1)+1}\right)$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{N}\frac{(-1)^{m(k)+1}}{\Gamma(2-\alpha
m(k))}\left(\sum_{n=1}^{\infty}\frac{P_{n}b}{\lambda_{n}^{m(k)}}\right)\frac{1}{t^{\alpha
m(k)-1}}+O\left(t^{-\alpha m(N+1)+1}\right),$
$\displaystyle\sum_{n=1}^{\infty}\left(\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(t-s)^{\alpha})\mu(s)ds\right)P_{n}f$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\left(\int^{T}_{0}\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}(t-s)^{-\alpha
m(k)-1}\mu(s)ds\right)\frac{1}{\lambda_{n}^{m(k)+1}}(P_{n}f)$ $\displaystyle+$
$\displaystyle\sum_{n=1}^{\infty}(P_{n}f)\left(\int^{T}_{0}O((t-s)^{-\alpha
m(N+1)-1})\mu(s)ds\right)$ $None$
$=\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\left(\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\mu(s)ds\right)\sum_{n=1}^{\infty}\frac{P_{n}f}{\lambda_{n}^{m(k)+1}}+O(t^{-\alpha
m(N+1)-1})$

for all large $t>0$.

For the last equality, we used $(t-s)^{-\alpha m(N+1)-1}\sim t^{-\alpha
m(N+1)-1}$ for $t>2T$ and $0<s<T$, and $\int^{T}_{0}O((t-s)^{-\alpha
m(N+1)-1})\mu(s)ds=O(t^{-\alpha m(N+1)-1})$ as $t\to\infty$.

Henceforth we assume $t>2T$ and we set

$\mu_{m}:=\int^{T}_{0}(-s)^{m}\mu(s)ds,\quad m\in\mathbb{N}\cup\\{0\\}.$

Calculations of $\int^{T}_{0}(t-s)^{-\sigma}\mu(s)ds$ with $\sigma=\alpha
m(k)+1$.
Let $t>2T$ and $\sigma>0$. Then the binomial expansion implies

$None$ $(t-s)^{-\sigma}=\sum_{\ell=0}^{L}\left(\begin{array}[]{c}-\sigma\\\
\ell\\\ \end{array}\right)t^{-\sigma-\ell}(-s)^{\ell}+R_{L}(t,s),\quad
0<s<T,\,t>2T,$

where we write

$\left(\begin{array}[]{c}-\sigma\\\ 0\\\
\end{array}\right):=1,\quad\left(\begin{array}[]{c}-\sigma\\\ \ell\\\
\end{array}\right):=\frac{(-\sigma)(-\sigma-1)\cdots(-\sigma-\ell+1)}{\ell!}\quad\mbox{for
$\ell\in\mathbb{N}$},$

and the function $R_{L}(t,s)$ satisfies

$|R_{L}(t,s)|\leq C(L,\sigma)t^{-\sigma-L-1}T^{L+1}.$

Indeed for fixed $0<\rho<1$, noting that
$(t-s)^{-\sigma}=t^{-\sigma}\left(1-\left(\frac{s}{t}\right)\right)^{-\sigma}$,
we can derive (3.2) from the Taylor series:

$(1-\eta)^{-\sigma}=\sum_{\ell=0}^{L}\left(\begin{array}[]{c}-\sigma\\\
\ell\\\ \end{array}\right)(-\eta)^{\ell}+R_{L}(\eta),\quad|\eta|\leq\rho,$

where

$|R_{L}(\eta)|\leq C(L,\sigma)|\eta|^{L+1}\quad\mbox{for $|\eta|\leq\rho$}.$

Thus (3.3) is seen. $\blacksquare$

We apply (3.3) to obtain

$None$
$\left\\{\begin{array}[]{rl}&\int^{T}_{0}(t-s)^{-\sigma}\mu(s)ds=\left(\begin{array}[]{c}-\sigma\\\
\ell_{0}\\\
\end{array}\right)t^{-\sigma-\ell_{0}}\mu_{\ell_{0}}+O(t^{-\sigma-\ell_{0}-1}),\quad
t>2T,\\\ &\mbox{where $\ell_{0}=0$ or $\ell_{0}\in\mathbb{N}$ satisfying
$\int^{T}_{0}(-s)^{m}\mu(s)ds=0$ for $0\leq
m\leq\ell_{0}-1$}.\end{array}\right.$

We note that $\left(\begin{array}[]{c}-\sigma\\\ \ell_{0}\\\
\end{array}\right)=\left(\begin{array}[]{c}-\alpha m(k)-1\\\ \ell_{0}\\\
\end{array}\right)\neq 0$ for each $k\in\mathbb{N}$, because $\alpha
m(k)\not\in\mathbb{N}$ by the definition implies $-\alpha
m(k)-1\not\in\mathbb{N}$.
Verification of (3.4). By (3.3) with $L=\ell_{0}$, since
$\mu_{0}=\cdots=\mu_{\ell_{0}-1}=0$, we have

$\displaystyle\int^{T}_{0}\mu(s)(t-s)^{-\sigma}ds$ $\displaystyle=$
$\displaystyle\sum_{\ell=0}^{\ell_{0}}\left(\begin{array}[]{c}-\sigma\\\
\ell\\\
\end{array}\right)t^{-\sigma-\ell}\int^{T}_{0}(-s)^{\ell}\mu(s)ds+\int^{T}_{0}R_{\ell_{0}}(t,s)\mu(s)ds$
$\displaystyle=$ $\displaystyle\left(\begin{array}[]{c}-\sigma\\\ \ell_{0}\\\
\end{array}\right)t^{-\sigma-\ell_{0}}\mu_{\ell_{0}}+\int^{T}_{0}R_{\ell_{0}}(t,s)\mu(s)ds.$

Moreover,

$\left|\int^{T}_{0}R_{\ell_{0}}(t,s)\mu(s)ds\right|\leq
C\|\mu\|_{L^{\infty}(0,T)}t^{-\sigma-\ell_{0}-1}T^{\ell_{0}+1}.$

Thus the verification of (3.4) is complete. $\blacksquare$.

We set

$\left\\{\begin{array}[]{rl}&Q_{k}(x):=\frac{(-1)^{m(k)+1}}{\Gamma(1-\alpha
m(k))}\sum_{n=1}^{\infty}\frac{P_{n}a(x)}{\lambda_{n}^{m(k)}},\cr\\\
&R_{k}(x):=\frac{(-1)^{m(k)+1}}{\Gamma(2-\alpha
m(k))}\sum_{n=1}^{\infty}\frac{P_{n}b(x)}{\lambda_{n}^{m(k)}},\cr\\\
&S_{k}(x):=\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\left(\sum_{n=1}^{\infty}\frac{(P_{n}f)(x)}{\lambda_{n}^{m(k)+1}}\right)\left(\begin{array}[]{c}-\alpha
m(k)-1\\\ \ell_{0}\\\ \end{array}\right).\end{array}\right.$

Now we assume

$u(x,t)=0,\quad x\in\omega,\,T_{1}<t<T_{2}.$

Then Lemma 4 implies that $u(x,t)=0$ for $x\in\omega$ and all $t>T$.

We note that $\mu_{\ell_{0}}\neq 0$ and

$t^{-\alpha m(N+1)-1}\leq Ct^{-\alpha m(N+1)},\quad t^{-\alpha m(N+1)-1}\leq
Ct^{-\alpha m(N+1)+1}$

for all $t>2T$, and in terms of $\mu_{\ell_{0}}\left(\begin{array}[]{c}-\alpha
m(k)-1\\\ \ell_{0}\\\ \end{array}\right)\neq 0$, we obtain

$\displaystyle\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\mu(s)ds=\left(\begin{array}[]{c}-\alpha m(k)-1\\\ \ell_{0}\\\
\end{array}\right)t^{-\alpha m(k)-\ell_{0}-1}\mu_{\ell_{0}}+O(t^{-\alpha
m(k)-\ell_{0}-2})$ $\displaystyle=$ $\displaystyle t^{-\alpha
m(k)-\ell_{0}-1}\mu_{\ell_{0}}\left(\begin{array}[]{c}-\alpha m(k)-1\\\
\ell_{0}\\\ \end{array}\right)\left(1+O\left(\frac{1}{t}\right)\right).$

Substituting the above asymptotic expansions into (3.2) and (2.1), we obtain:
Case: $0<\alpha<1$.

$u(x,t)=\sum_{k=1}^{N}Q_{k}t^{-\alpha
m(k)}+\sum_{k=1}^{N}S_{k}\mu_{\ell_{0}}t^{-\alpha m(k)-\ell_{0}-1}$ $None$
$+O(t^{-\alpha m(1)-\ell_{0}-2})+O(t^{-\alpha m(N+1)})=0,\quad
x\in\omega,\,t>2T$

for $N\in\mathbb{N}$.
Case: $1<\alpha<2$.

$u(x,t)=\sum_{k=1}^{N}Q_{k}t^{-\alpha m(k)}+\sum_{k=1}^{N}R_{k}t^{-\alpha
m(k)+1}$ $None$ $+\sum_{k=1}^{N}S_{k}\mu_{\ell_{0}}t^{-\alpha
m(k)-\ell_{0}-1}+O(t^{-\alpha m(1)-\ell_{0}-2})+O(t^{-\alpha
m(N+1)+1})=0,\quad x\in\omega,\,t>2T$

for $N\in\mathbb{N}$.
Second Step: Completion of the proof for $0<\alpha<1$.
Since $\alpha\not\in\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}$,
we see

$None$ $\\{\alpha m(k)\\}_{k\in\mathbb{N}}\cap\\{\alpha
m(k)+\ell_{0}+1\\}_{k\in\mathbb{N}}=\emptyset.$

Indeed, if $\alpha m(k)=\alpha m(k^{\prime})+\ell_{0}+1$ with some
$k,k^{\prime}\in\mathbb{N}$, we have $\alpha(m(k)-m(k^{\prime}))=\ell_{0}+1$,
that is, $\alpha=\frac{\ell_{0}+1}{m(k)-m(k^{\prime})}$, which is impossible
by (1.6):
$\alpha\not\in\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}$. This is
the verification of (3.7). $\blacksquare$

By (3.7), we see that $\alpha m(j),\alpha(k)+\ell_{0}+1$ for $1\leq j,k\leq N$
are mutually distinct. We choose $N_{1}\in\mathbb{N}$ large, so that

$\alpha m(1),\alpha m(1)+\ell_{0}+1<\alpha m(N_{1}+1).$

Then

$\alpha m(1),\alpha m(1)+\ell_{0}+1<\min\\{\alpha m(N_{1}+1),\,\alpha
m(1)+\ell_{0}+2\\}=:r,$

and (3.5) implies

$None$ $\sum_{k=1}^{N_{1}}Q_{k}t^{-\alpha
m(k)}+\sum_{k=1}^{N_{1}}\mu_{\ell_{0}}S_{k}t^{-\alpha
m(k)-\ell_{0}-1}=O(t^{-r})\quad\mbox{in $\omega$}$

as $t\to\infty$. Hence we can apply Lemma 5 to (3.8), so that $Q_{1}=S_{1}=0$
in $\omega$.

Therefore, in view of (3.5), we obtain

$None$ $\sum_{k=2}^{N}Q_{k}t^{-\alpha
m(k)}+\sum_{k=2}^{N}S_{k}\mu_{\ell_{0}}t^{-\alpha
m(k)-\ell_{0}-1}+O(t^{-\alpha m(2)-\ell_{0}-2})=O(t^{-\alpha
m(N+1)})\quad\mbox{in $\omega$}$

as $t\to\infty$, where $N\in\mathbb{N}$ is arbitrarily given. We can repeat
the previous argument to (3.9), and so we can obtain $Q_{2}=S_{2}=0$ in
$\omega$. Thus, continuing this argument, we can complete the proof of
$Q_{1}=Q_{2}=\cdots=Q_{N}=S_{1}=S_{2}=\cdots=S_{N}=0$ in $\omega$ for each
$N\in\mathbb{N}$.

Hence,

$\sum_{n=1}^{\infty}\frac{P_{n}a}{\lambda_{n}^{m(k)}}=\sum_{n=1}^{\infty}\frac{P_{n}f}{\lambda_{n}^{m(k)+1}}=0\quad\mbox{in
$\omega$ for all $k\in\mathbb{N}$}.$

For large $k\in\mathbb{N}$, similarly to the proof of Lemma 6, by using
$\lambda_{n}\geq c_{0}n^{\frac{2}{d}}$ for large $n\in\mathbb{N}$, we see that
the series are convergent in $L^{2}(\omega)$. Therefore, for any $\varphi\in
L^{2}(\omega)$, we see

$\sum_{n=1}^{\infty}\frac{(P_{n}a,\varphi)_{L^{2}(\omega)}}{\lambda_{n}^{m(k)}}=\sum_{n=1}^{\infty}\frac{(P_{n}f,\varphi)_{L^{2}(\omega)}}{\lambda_{n}^{m(k)+1}}=0$

for all $k\in\mathbb{N}$. Hence, Lemma 6 yields

$(P_{n}a,\varphi)_{L^{2}(\omega)}=(P_{n}f,\varphi)_{L^{2}(\omega)}=0\quad\mbox{in
$\omega$ for all $n\in\mathbb{N}$ and all $\varphi\in L^{2}(\omega)$.}$

This means that $P_{n}a=P_{n}f=0$ in $\omega$ for all $n\in\mathbb{N}$. Then
we can prove that $P_{n}a=P_{n}f=0$ in $\Omega$ for all $n\in\mathbb{N}$.
Indeed, the definition of $P_{n}$ implies $(A-\lambda_{n})P_{n}a=0$ in
$\Omega$. By $P_{n}a=0$ in $\omega$, the unique continuation (e.g., Isakov
[6]) for the elliptic operator $A-\lambda_{n}$ yields that $P_{n}a=0$ in
$\Omega$. Similarly we can obtain $P_{n}f=0$ in $\Omega$ for all
$n\in\mathbb{N}$. Since $a=\sum_{n=1}^{\infty}P_{n}a$ and
$f=\sum_{n=1}^{\infty}P_{n}f$ in $L^{2}(\Omega)$, we reach $a=f=0$ in
$\Omega$. Thus the proof of Theorem 1 is complete for the case $0<\alpha<1$.
$\blacksquare$

Third Step: Completion of the proof for $1<\alpha<2$.
Since
$\alpha\not\in\left\\{\frac{\ell_{0}+1}{n}\right\\}_{n\in\mathbb{N}}\cup\left\\{\frac{\ell_{0}+2}{n}\right\\}_{n\in\mathbb{N}}$
and $\alpha>1$, we can verify

$None$ $\left\\{\begin{array}[]{rl}&\\{\alpha
m(k)\\}_{k\in\mathbb{N}}\cup\\{\alpha m(k)-1\\}_{k\in\mathbb{N}}=\emptyset,\\\
&\\{\alpha m(k)\\}_{k\in\mathbb{N}}\cup\\{\alpha
m(k)+\ell_{0}+1\\}_{k\in\mathbb{N}}=\emptyset,\\\ &\\{\alpha
m(k)-1\\}_{k\in\mathbb{N}}\cup\\{\alpha
m(k)+\ell_{0}+1\\}_{k\in\mathbb{N}}=\emptyset.\end{array}\right.$

By (3.10), we see that

$\alpha m(i),\,\alpha m(j)-1,\,\alpha m(k)+\ell_{0}+1,\quad
i,j,k\in\\{1,2,...,N\\}$

are mutually distinct. We choose $N_{1}\in\mathbb{N}$ such that

$None$ $\alpha m(1),\,\alpha m(1)+\ell_{0}+1<\alpha m(N_{1}+1)-1.$

Then $\alpha m(1)-1<\alpha m(N_{1}+1)-1$.

With this $N_{1}$, we can write (3.6) as

$\sum_{k=1}^{N_{1}}Q_{k}t^{-\alpha m(k)}+\sum_{k=1}^{N_{1}}R_{k}t^{-\alpha
m(k)+1}+\sum_{k=1}^{N_{1}}S_{k}t^{-\alpha m(k)-\ell_{0}-1}$ $None$
$+O(t^{-\alpha m(1)-\ell_{0}-2})=O(t^{-\alpha m(N_{1}+1)+1})$

in $\omega$ as $t\to\infty$.

In view of (3.11), we apply Lemma 5 to (3.12), and so we can obtain

$None$ $Q_{1}=R_{1}=S_{1}=0\quad\mbox{in $\omega$}.$

Hence by (3.13), for all $N\in\mathbb{N}$ we obtain

$\sum_{k=s}^{N}Q_{k}t^{-\alpha m(k)}+\sum_{k=2}^{N}R_{k}t^{-\alpha
m(k)+1}+\sum_{k=2}^{N}S_{k}\mu_{\ell_{0}}t^{-\alpha m(k)-\ell_{0}-1}$ $None$
$+O(t^{-\alpha m(2)-\ell_{0}-2})=O(t^{-\alpha m(N+1)+1})$

in $\omega$ as $t\to\infty$. We apply the same argument as (3.13) with
$N=N_{1}$ to (3.14), and so we obtain $Q_{2}=R_{2}=S_{2}=0$ in $\omega$.

Continuing this argument, we reach $Q_{k}=R_{k}=S_{k}=0$ in $\omega$ for each
$k\in\mathbb{N}$. Similarly to the case $0<\alpha<1$, we can verify that
$Q_{k}=R_{k}=S_{k}=0$ in $\omega$ for all $k\in\mathbb{N}$ imply
$P_{n}a=P_{n}b=P_{n}f=0$ in $\Omega$ for all $n\in\mathbb{N}$. Therefore
$a=b=f=0$ in $\Omega$. Thus the proof of Theorem 1 is complete also for
$1<\alpha<2$. $\blacksquare$

##  4\. Proof of Theorem 2

The proof is the repeat of the same arguments as in the proof of Theorem 1
which is by equating the coefficients of the asymptotic expansions of the
solutions.

For $\ell\in\mathbb{N}\cup\\{0\\}$, $k\in\mathbb{N}$ and $x\in\Omega$ and
$a,b,f\in L^{2}(\Omega)$, we set

$\left\\{\begin{array}[]{rl}&Q_{k}(a)(x):=\frac{(-1)^{m(k)+1}}{\Gamma(1-\alpha
m(k))}\sum_{n=1}^{\infty}\frac{P_{n}a(x)}{\lambda_{n}^{m(k)}},\cr\\\
&R_{k}(b)(x):=\frac{(-1)^{m(k)+1}}{\Gamma(2-\alpha
m(k))}\sum_{n=1}^{\infty}\frac{P_{n}b(x)}{\lambda_{n}^{m(k)}},\cr\\\
&S_{k,\ell}(f)(x):=\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\left(\begin{array}[]{c}-\alpha m(k)-1\\\ \ell\\\
\end{array}\right)\left(\sum_{n=1}^{\infty}\frac{(P_{n}f)(x)}{\lambda_{n}^{m(k)+1}}\right)\end{array}\right.$

and

$\mu_{\ell}:=\int^{T}_{0}(-s)^{\ell}\mu(s)ds,\quad\widetilde{\mu}_{\ell}:=\int^{T}_{0}(-s)^{\ell}\widetilde{\mu}(s)ds\quad\mbox{for
all $\ell\in\mathbb{N}\cup\\{0\\}$}.$

Henceforth, without loss of generality, in (1.10), we assume that
$\mu_{\ell_{1}}=\int^{T}_{0}(-s)^{\ell_{1}}\mu(s)ds\neq 0$.
Proof in the case $0<\alpha<1$.
Similarly to (3.5), by means of $u(x,t)=\widetilde{u}(x,t)$ for $x\in\omega$
and $T_{1}<t<T_{2}$ and Lemma 4, we have

$\displaystyle\sum_{k=1}^{N}Q_{k}(a)t^{-\alpha
m(k)}+\sum_{k=1}^{N}\mu_{\ell_{1}}S_{k,\ell_{1}}(f)t^{-\alpha
m(k)-\ell_{1}-1}=$ $\displaystyle\sum_{k=1}^{N}Q_{k}(\widetilde{a})t^{-\alpha
m(k)}+\sum_{k=1}^{N}\widetilde{\mu}_{\ell_{1}}S_{k,\ell_{1}}(\widetilde{f})t^{-\alpha
m(k)-\ell_{1}-1}$ $\displaystyle+$ $\displaystyle O(t^{-\alpha
m(1)-\ell_{1}-2})+O(t^{-\alpha m(N+1)})$

in $\omega$ as $t\to\infty$. Here we applied also

$\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\mu(s)ds=\mu_{\ell_{1}}\left(\begin{array}[]{c}-\alpha m(k)-1\\\
\ell_{1}\\\ \end{array}\right)t^{-\alpha m(k)-1-\ell_{1}}+O(t^{-\alpha
m(k)-2-\ell_{1}})$

and

$\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\widetilde{\mu}(s)ds=\widetilde{\mu}_{\ell_{1}}\left(\begin{array}[]{c}-\alpha
m(k)-1\\\ \ell_{1}\\\ \end{array}\right)t^{-\alpha
m(k)-1-\ell_{1}}+O(t^{-\alpha m(k)-2-\ell_{1}}),$

which are proved in the same way as for (3.3) by using
$\int^{T}_{0}\mu(s)(-s)^{m}ds=\int^{T}_{0}\widetilde{\mu}(s)(-s)^{m}ds=0$ for
$0\leq m\leq\ell_{1}-1$ with $\ell_{1}\in\mathbb{N}$. We note

$None$ $\mu_{\ell_{1}}\neq 0,$

but $\widetilde{\mu}_{\ell_{1}}=0$ may occur. Since the definition of
$S_{k,\ell}(\cdot)$ yields

$\mu_{\ell_{1}}S_{k,\ell_{1}}(f)-\widetilde{\mu}_{\ell_{1}}S_{k,\ell_{1}}(\widetilde{f})=S_{k,\ell_{1}}(\mu_{\ell_{1}}f-\widetilde{\mu}_{\ell_{1}}\widetilde{f})\quad\mbox{in
$\omega$},$

we obtain

$\sum_{k=1}^{N}Q_{k}(a-\widetilde{a})t^{-\alpha
m(k)}+\sum_{k=1}^{N}S_{k,\ell_{1}}(\mu_{\ell_{1}}f-\widetilde{\mu}_{\ell_{1}}\widetilde{f})t^{-\alpha
m(k)-\ell_{1}-1}$ $None$ $=O(t^{-\alpha m(1)-\ell_{1}-2})+O(t^{-\alpha
m(N+1)})\quad\mbox{in $\omega$}$

as $t\to\infty$.

Thanks to (1.11), we can argue in the same way as Second Step of the proof of
Theorem 1 in Section 3, by (4.1) we can verify

$None$ $a=\widetilde{a}\quad\mbox{and}\quad
f=\frac{\widetilde{\mu}_{\ell_{1}}}{\mu_{\ell_{1}}}\widetilde{f}\quad\mbox{in
$\Omega$.}$

Since $u$ and $\widetilde{u}$ satisfies (1.1) and (1.8), setting
$y:=u-\widetilde{u}$ in $\Omega\times(0,\infty)$, in view of (4.3) we obtain

$\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}y=-Ay+\rho(t)\widetilde{f}(x),\quad
x\in\Omega,\,t>0,\\\ &y|_{\partial\Omega\times(0,\infty)}=0,\\\
&y(\cdot,0)=0\quad\mbox{in $\Omega$},\end{array}\right.$

where

$\rho(t):=\frac{\widetilde{\mu}_{\ell_{1}}}{\mu_{\ell_{1}}}\mu(t)-\widetilde{\mu}(t),\quad
t>0.$

Moreover by Lemma 4, we have

$y(x,t)=0,\quad x\in\omega,\,t>T.$

Therefore, in view of (2.1), we apply (3.1) to have

$\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\left(\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\rho(s)ds\right)\sum_{n=1}^{\infty}\frac{P_{n}\widetilde{f}}{\lambda_{n}^{m(k)+1}}$
$None$ $=O(t^{-\alpha m(N+1)-1})\quad\mbox{in $\omega$}$

as $t\to\infty$.

Assume that $\rho\not\equiv 0$ in $(0,T)$, that is,

$None$
$\widetilde{\mu}_{\ell_{1}}\mu(t)\not\equiv\mu_{\ell_{1}}\widetilde{\mu}(t)\quad\mbox{in
$(0,T)$}.$

Then the Weierstrass theorem yields the existence of
$\ell_{2}\in\mathbb{N}\cup\\{0\\}$ such that

$\rho_{\ell_{2}}:=\int^{T}_{0}(-s)^{\ell_{2}}\rho(s)ds\neq
0,\quad\int^{T}_{0}(-s)^{m}\rho(s)ds=0\quad\mbox{for $0\leq m\leq\ell_{2}-1$
if $\ell_{2}\in\mathbb{N}$}.$

Therefore, the same agument as in (3.3) implies

$\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\rho(s)ds=\rho_{\ell_{2}}\left(\begin{array}[]{c}-\alpha m(k)-1\\\
\ell_{2}\\\ \end{array}\right)t^{-\alpha
m(k)-1-\ell_{2}}\left(1+O\left(\frac{1}{t}\right)\right)$

as $t\to\infty$.

Substituting this into (4.4), we obtain

$\displaystyle\sum_{k=1}^{N}\rho_{\ell_{2}}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\left(\begin{array}[]{c}-\alpha m(k)-1\\\ \ell_{2}\\\
\end{array}\right)t^{-\alpha
m(k)-1-\ell_{2}}\left(\sum_{n=1}^{\infty}\frac{P_{n}\widetilde{f}}{\lambda_{n}^{m(k)+1}}\right)$
$\displaystyle=$ $\displaystyle O(t^{-\alpha m(1)-\ell_{2}-2})+O(t^{-\alpha
m(N+1)-1})$

in $\omega$ as $t\to\infty$, that is,

$\sum_{k=1}^{N}\rho_{\ell_{2}}S_{k,\ell_{2}}(\widetilde{f})t^{-\alpha
m(k)-1-\ell_{2}}=O(t^{-\alpha m(1)-\ell_{2}-2})+O(t^{-\alpha
m(N+1)-1})\quad\mbox{in $\omega$}$

as $t\to\infty$.

Arguing in the same way as in Second Step of the proof of Theorem 1, we can
see $\rho_{\ell_{2}}S_{k,\ell_{2}}(\widetilde{f})=0$ in $\omega$ for each
$k\in\mathbb{N}$. The assumption of the theorem implies $\rho_{\ell_{2}}\neq
0$. Hence, $S_{k,\ell_{2}}(\widetilde{f})=0$ in $\omega$ for each
$k\in\mathbb{N}$, that is,
$\sum_{n=1}^{\infty}\frac{P_{n}\widetilde{f}}{\lambda_{n}^{m(k)+1}}=0$ in
$\omega$ for each $k\in\mathbb{N}$. Therefore, in view of Lemma 6, applying
the same argument at the end of Second Step of the proof of Theorem 1 in
Section 3, we reach $\widetilde{f}=0$ in $\Omega$. Then (4.3) implies $f=0$ in
$\Omega$, which is a contradiction for $f\not\equiv 0$ in $\Omega$ or
$\widetilde{f}\not\equiv 0$ in $\Omega$. Hence assumption (4.5) is impossible,
and so $\widetilde{\mu}_{\ell_{1}}\mu(t)=\mu_{\ell_{1}}\widetilde{\mu}(t)$ for
$0<t<T$. Thus the proof of Theorem 2 (i) is complete. $\blacksquare$
Proof in the case $1<\alpha<2$.
We assume that $\mu_{\ell_{1}}:=\int^{T}_{0}(-s)^{\ell_{1}}\mu(s)ds\neq 0$.
Arguing the same way for (4.2), we obtain

$\displaystyle\sum_{k=1}^{N}Q_{k}(a-\widetilde{a})t^{-\alpha
m(k)}+\sum_{k-1}^{N}R_{k}(b-\widetilde{b})t^{-\alpha m(k)+1}$ $\displaystyle+$
$\displaystyle\sum_{k=1}^{N}S_{k,\ell_{1}}(\mu_{\ell_{1}}f-\widetilde{\mu}_{\ell_{1}}\widetilde{f})t^{-\alpha
m(k)-\ell_{1}-1}=O(t^{-\alpha m(1)-\ell_{1}-2})+O(t^{-\alpha m(N+1)+1})$

in $\omega$ as $t\to\infty$.

Thus in the same way as in Third Step of the proof of Theoerm 1 in Section 3,
we can obtain $a=\widetilde{a}$, $b=\widetilde{b}$ in $\Omega$ and
$\mu_{\ell_{1}}f=\widetilde{\mu}_{\ell_{1}}\widetilde{f}$ in $\Omega$.

Now the same arguments as in the case $0<\alpha<1$ after (4.3) bring
$\widetilde{\mu}_{\ell_{1}}\mu(t)=\mu_{\ell_{1}}\widetilde{\mu}(t)$ for
$0<t<T$. Thus the proof of Theorem 2 (ii) is complete. $\blacksquare$

##  5\. Concluding remarks

1. We have proved the uniqueness in simultaneously determining factors $f(x)$ and $\mu(t)$ of a source term and initial values $a$ and/or $b$, except for some values of orders $\alpha$. The proof is by equating the terms $t^{-\alpha m(k)}$, etc. with the same orders of the asymptotic expansions of the solutions. Those terms are generated by initial data and a source term: $\int^{T}_{0}(t-s)^{-\alpha m(k)-1}\mu(s)ds$ with $k\in\mathbb{N}$ (see (3.5) and (3.6)). Without conditions (1.6), (1.7), (1.11) and (1.12) on $\alpha$, infinitely many terms are overlapped and our argument cannot conclude the uniqueness. So far, in general, we do not know the uniqueness without such conditions on $\alpha$. However in a case of time-fractional ordinary differential equations, although the separation of the terms is not perfect, we can prove the uniqueness corresponding to Theorem 2, thanks to that an unknown initial value is scalar. More precisely,
Proposition 1.
Let $\alpha\in(0,1)\cup(1,2)$ be arbitrarily given. For
$\lambda\in\mathbb{R}$, we consider

$None$ $\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}u(t)=-\lambda
u(t)+\mu(t),\\\ &u(0)=a,\quad\mbox{if $0<\alpha<1$},\\\
&u(0)=a,\quad\frac{du}{dt}(0)=b,\quad\mbox{if
$1<\alpha<2$}.\end{array}\right.$

We assume that $\mu(t)$ satisfies (1.4), and $T<T_{1}<T_{2}$. Then
(i) Case $0<\alpha<1$: if $u(t)=0$ for $T_{1}<t<T_{2}$, then $a=0$ and
$\mu(t)=0$ for $0<t<T$.
(ii) Case $1<\alpha<2$: if $u(t)=0$ for $T_{1}<t<T_{2}$, then $a=b=0$ and
$\mu(t)=0$ for $0<t<T$.
Proof of Proposition 1.
It is known (e.g., Kilbas, Srivastava and Trujillo [15], p.141) that

$None$ $u(t)=\left\\{\begin{array}[]{rl}&aE_{\alpha,1}(-\lambda
t^{\alpha})+\int^{t}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda(t-s)^{\alpha})\mu(s)ds,\quad
0<\alpha<1,\\\ &aE_{\alpha,1}(-\lambda t^{\alpha})+btE_{\alpha,2}(-\lambda
t^{\alpha})\\\
+&\int^{t}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda(t-s)^{\alpha})\mu(s)ds,\quad
1<\alpha<2,\quad t>0.\end{array}\right.$

In view of (3.1) with $N=1$, we have

$None$ $\left\\{\begin{array}[]{rl}&E_{\alpha,1}(-\lambda
t^{\alpha})=\frac{(-1)^{m(1)+1}}{\Gamma(1-\alpha
m(1))}\frac{1}{\lambda^{m(1)}}\frac{1}{t^{\alpha m(1)}}+O(t^{-\alpha
m(2)}),\\\ &tE_{\alpha,2}(-\lambda
t^{\alpha})=\frac{(-1)^{m(1)+1}}{\Gamma(2-\alpha
m(1))}\frac{1}{\lambda^{m(1)}}\frac{1}{t^{\alpha m(1)-1}}+O(t^{-\alpha
m(2)+1}),\\\ &t^{\alpha-1}E_{\alpha,\alpha}(-\lambda
t^{\alpha})=\frac{(-1)^{m(1)}}{\Gamma(-\alpha
m(1))}\frac{1}{\lambda^{m(1)+1}}\frac{1}{t^{\alpha m(1)+1}}+O(t^{-\alpha
m(2)-1})\end{array}\right.$

as $t\to\infty$.

Since $u(t)=0$ for $T_{1}<t<T_{2}$, similarly to Lemma 4, we obtain

$u(t)=0,\quad t>T.$

Case: $0<\alpha<1$.
By (5.2) and $\mu(t)=0$ for $t>T$, we see

$None$ $aE_{\alpha,1}(-\lambda
t^{\alpha})+\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda(t-s)^{\alpha})\mu(s)ds=0,\quad
t>2T.$

Therefore, by (5.3), we obtain

$\displaystyle\frac{a(-1)^{m(1)+1}}{\Gamma(1-\alpha
m(1))}\frac{1}{\lambda^{m(1)}}\frac{1}{t^{\alpha m(1)}}+O(t^{-\alpha m(2)})$
$\displaystyle+$ $\displaystyle\frac{(-1)^{m(1)}}{\Gamma(-\alpha
m(1))}\frac{1}{\lambda^{m(1)+1}}\int^{T}_{0}(t-s)^{-\alpha
m(1)-1}\mu(s)ds+\int^{T}_{0}O((t-s)^{-\alpha m(2)-1})\mu(s)ds,$

and so the multiplication by $t^{\alpha m(1)}$ yields

$\frac{a(-1)^{m(1)+1}}{\Gamma(1-\alpha
m(1))}\frac{1}{\lambda^{m(1)}}+\frac{(-1)^{m(1)}}{\Gamma(-\alpha
m(1))}\frac{1}{\lambda^{m(1)+1}}\int^{T}_{0}\frac{t^{\alpha
m(1)}}{(t-s)^{\alpha m(1)+1}}\mu(s)ds$ $None$
$+O(t^{-\alpha(m(2)-m(1))})+\int^{T}_{0}O(t^{\alpha m(1)}(t-s)^{-\alpha
m(2)-1})ds=0\quad\mbox{as $t\to\infty$}.$

Here, for $0<s<T<2T<t$, we have $\frac{1}{1-\frac{s}{t}}\leq 2$ and
$\frac{1}{t-s}\leq\frac{1}{T}$, so that

$\frac{t^{\alpha m(1)}}{(t-s)^{\alpha
m(1)+1}}=\left(\frac{1}{1-\frac{s}{t}}\right)^{\alpha m(1)}\frac{1}{t-s}\leq
2^{\alpha m(1)}\frac{1}{T}\quad\mbox{for all $s<T<2T<t$}.$

Since $\lim_{t\to\infty}\frac{t^{\alpha m(1)}}{(t-s)^{\alpha m(1)+1}}=0$, the
Lebesgue convergence theorem implies

$None$ $\lim_{t\to\infty}\int^{T}_{0}\frac{t^{\alpha m(1)}}{(t-s)^{\alpha
m(1)+1}}\mu(s)ds=0.$

Therefore, letting $t\to\infty$ in (5.5), we see

$\frac{a(-1)^{m(1)+1}}{\Gamma(1-\alpha m(1))}\frac{1}{\lambda^{m(1)}}=0,$

which implies $a=0$. Hence (5.4) yields

$None$
$\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda(t-s)^{\alpha})\mu(s)ds=0,\quad
t>2T.$

Therefore, the following Lemma 7 yields that $\mu=0$ in $(0,T)$, and so
completes the proof of Proposition 1 for the case $0<\alpha<1$. Thus it
suffices to prove
Lemma 7.
Let $\mu\in L^{\infty}(0,T)$. Then (5.7) implies $\mu=0$ in $(0,T)$.
Proof of Lemma 7. Substituting the third asymptotic expansion in (3.1) into
(5.7), we have

$\displaystyle
0=\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda(t-s)^{\alpha})\mu(s)ds$
$\displaystyle=$ $\displaystyle\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))\lambda^{m(k)+1}}\left(\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\mu(s)ds\right)+O(t^{-\alpha m(N+1)-1})$

as $t\to\infty$ for all $N\in\mathbb{N}$.

Contrarily assume that $\mu\not\equiv 0$ in $(0,T)$. Then Lemma 2 implies the
existence of $\ell_{0}\in\mathbb{N}\cup\\{0\\}$ such that

$\mu_{\ell_{0}}:=\int^{T}_{0}(-s)^{\ell_{0}}\mu(s)ds,\quad\int^{T}_{0}(-s)^{m}\mu(s)ds\quad\mbox{for
$0\leq m\leq\ell_{0}-1$ if $\ell_{0}\in\mathbb{N}$.}$

Applying (3.4), we obtain

$\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))\lambda^{m(k)+1}}t^{-\alpha
m(k)-\ell_{0}-1}\mu_{\ell_{0}}\left(\begin{array}[]{c}-\alpha m(k)-1\\\
\ell_{0}\\\
\end{array}\right)\left(1+O\left(\frac{1}{t}\right)\right)+O(t^{-\alpha
m(N+1)-1})=0$

as $t\to\infty$. Lemma 5 yields

$\mu_{\ell_{0}}\frac{(-1)^{m(1)}}{\Gamma(-\alpha
m(1))\lambda^{m(1)+1}}\left(\begin{array}[]{c}-\alpha m(1)-1\\\ \ell_{0}\\\
\end{array}\right)=0.$

Since

$\frac{(-1)^{m(1)}}{\Gamma(-\alpha
m(1))\lambda^{m(1)+1}}\left(\begin{array}[]{c}-\alpha m(1)-1\\\ \ell_{0}\\\
\end{array}\right)\neq 0$

by $\alpha m(1)\not\in\mathbb{N}\cup\\{0\\}$, we obtain $\mu_{\ell_{0}}=0$.
This is a contradiction by the definition of $\mu_{\ell_{0}}$. Therefore
$\int^{T}_{0}(-s)^{m}\mu(s)ds=0$ for all $m\in\mathbb{N}\cup\\{0\\}$. Lemma 2
yields that $\mu=0$ in $(0,T)$. Thus the proof of Lemma 7 is complete.
$\blacksquare$
Case: $1<\alpha<2$.
By (5.2), we have

$None$ $aE_{\alpha,1}(-\lambda t^{\alpha})+btE_{\alpha,2}(-\lambda
t^{\alpha})\\\
+\int^{T}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda(t-s)^{\alpha})\mu(s)ds=0,\quad
t>2T.$

Substituting (5.3) into (5.8), we obtain

$\frac{a(-1)^{m(1)+1}}{\Gamma(1-\alpha
m(1))}\frac{1}{\lambda^{m(1)}}\frac{1}{t^{\alpha
m(1)}}+\frac{b(-1)^{m(1)+1}}{\Gamma(2-\alpha
m(1))}\frac{1}{\lambda^{m(1)}}\frac{1}{t^{\alpha m(1)-1}}$ $None$
$+\frac{(-1)^{m(1)}}{\Gamma(-\alpha
m(1))}\frac{1}{\lambda^{m(1)+1}}\int^{T}_{0}(t-s)^{-\alpha
m(1)-1}\mu(s)ds+O(t^{-\alpha m(2)+1})=0\quad\mbox{as $t\to\infty$}.$

Similarly to (5.6), we have

$\lim_{t\to\infty}\int^{T}_{0}\frac{t^{\alpha m(1)-1}}{(t-s)^{\alpha
m(1)+1}}\mu(s)ds=0.$

Consequently, multiplying by $t^{\alpha m(1)-1}$ and letting $t\to\infty$, we
see

$b\frac{(-1)^{m(1)+1}}{\Gamma(2-\alpha m(1))}\frac{1}{\lambda^{m(1)}}=0,$

that is, $b=0$. Hence, multiplying by $t^{\alpha m(1)}$ and using (5.6) in
(5.9), we obtain $a=0$. Therefore (5.8) implies (5.7). Lemma 7 yields
$\mu(t)=0$ for $0<t<T$. Thus the proof of Proposition 1 is complete.
$\blacksquare$

2. We take interior data $u|_{\omega\times(T_{1},T_{2})}$ as observation data. We can discuss to obtain the uniqueness with other typs of data such as boundary data $\partial_{\nu}u|_{\gamma\times(T_{1},T_{2})}$, where $\gamma\subset\partial\Omega$ is an arbitrarily chosen subboundary.

3. We should extend the uniqueness to more general elliptic operator, especially, non-symmetric $A$. As for the determination of initial values for $\alpha\in(1,2)$, see Loreti, Sforza and Yamamoto [22]. The extension of the uniqueness in the current article to non-symmetric $A$ should be a future work.

4. Our argument is applicable to other type of inverse problem. For example, for $\alpha\in(0,1)\cup(1,2)$, we consider

$\left\\{\begin{array}[]{rl}&d_{t}^{\alpha}y=-Ay+\mu(t)f(x),\quad
x\in\Omega,\,t>0,\\\ &y|_{\partial\Omega\times(0,\infty)}=0,\\\
&y(\cdot,0)=0\quad\mbox{in $\Omega$ $\quad$ if $0<\alpha<1$},\\\
&y(\cdot,0)=\partial_{t}y(\cdot,0)=0\quad\mbox{in $\Omega$ $\quad$ if
$1<\alpha<2$}.\end{array}\right.$

We assume that $\mu$ satisfies (1.4). Then we dicuss
Inverse source problem. We choose $x_{0}\in\Omega$ and $T_{1},T_{2}>0$ such
that $T<T_{1}<T_{2}$. Then determine $\mu(t)$ for $0<t<T$ by data $y(x_{0},t)$
for $T_{1}<t<T_{2}$.

For simplicity, let $f\in C^{\infty}_{0}(\Omega)$. We can prove
Proposition 2.
We assume that $f(x_{0})\neq 0$. Then $y(x_{0},t)=0$ for $T_{1}<t<T_{2}$
implies $\mu(t)=0$ for $0<t<T$.

The uniqueness for $\alpha\in(0,1)\cup(1,2)$ is different from the cases
$\alpha=1$ and $=2$. In particular, in the case where $\alpha=1$ and the
spatial dimension is one, we have no uniqueness ([2]).
Proof of Proposition 2. The proof is based on (4.4). Assume that
$\mu\not\equiv 0$ in $(0,T)$. Then Lemma 2 implies that there exists
$\ell_{0}\in\mathbb{N}\cup\\{0\\}$ satisfying (1.5), and in particular, we
have $\mu_{\ell_{0}}:=\int^{T}_{0}(-s)^{\ell_{0}}\mu(s)ds\neq 0$. By $f\in
C^{\infty}_{0}(\Omega)$, we can verify that $\sum_{n=1}^{\infty}P_{n}f$
converges in $C(\overline{\Omega})$. Therefore, similarly to (4.4), we can
obtain

$\displaystyle y(x_{0},t)=\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\left(\int^{T}_{0}(t-s)^{-\alpha
m(k)-1}\mu(s)ds\right)\sum_{n=1}^{\infty}\frac{(P_{n}f)(x_{0})}{\lambda_{n}^{m(k)+1}}$
$\displaystyle=$ $\displaystyle O(t^{-\alpha m(N+1)-1})$

for all $t>2T$ and $N\in\mathbb{N}$.

Applying (3.4), we reach

$\displaystyle\sum_{k=1}^{N}\frac{(-1)^{m(k)}}{\Gamma(-\alpha
m(k))}\mu_{\ell_{0}}\left(\begin{array}[]{c}-\alpha m(k)-1\\\ \ell_{0}\\\
\end{array}\right)t^{-\alpha
m(k)-\ell_{0}-1}\left(1+O\left(\frac{1}{t}\right)\right)\sum_{n=1}^{\infty}\frac{(P_{n}f)(x_{0})}{\lambda_{n}^{m(k)+1}}$
$\displaystyle=$ $\displaystyle O(t^{-\alpha m(N+1)-1})\quad\mbox{for all
$t>2T$ and all $N\in\mathbb{N}$}.$

In view of Lemma 5, we can prove

$\sum_{n=1}^{\infty}\frac{(P_{n}f)(x_{0})}{\lambda_{n}^{m(k)+1}}=0\quad\mbox{for
all $k\in\mathbb{N}$}.$

Hence Lemma 6 yields $(P_{n}f)(x_{0})=0$ for all $n\in\mathbb{N}$. In terms of
$f\in C^{\infty}_{0}(\Omega)$, we see that
$f(x_{0})=\sum_{n=1}^{\infty}(P_{n}f)(x_{0})=0$, which contradicts the
assumption $f(x_{0})\neq 0$. Therefore, $\mu\equiv 0$ in $(0,T)$. Thus the
proof of Proposition 2 is complete. $\blacksquare$

Acknowledgements.

The third author is supported by Grant-in-Aid for Scientific Research (A)
20H00117 and Grant-in-Aid for Challenging Research (Pioneering) 21K18142,
Japan Society for the Promotion of Science (JSPS).

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