#  Uniqueness for inverse source problems for fractional diffusion-wave
equations by data during not acting time

1,2,3 M. Yamamoto

###### Abstract.

We consider fractional diffusion-wave equations with source term which is
represented in a form of a product of a temporal function and a spatial
function. We prove the uniqueness for inveres source problem of determining
spatially varying factor by decay of data as the time tends to $\infty$,
provided that the source does not work during the observations. Our main
result asserts the uniqueness if data decay more rapidly than
$\left(\frac{1}{t^{p}}\right)$ with any $p\in\mathbb{N}$ as $t\to\infty$. Date
taken not from the initial time are realistic but the uniqueness was not known
in general. The proof is based on the analyticity and the asymptotic behavior
of a function generated by the solution.
Key words. fractional diffusion-wave equation, inverse source problem,
uniqueness
AMS subject classifications. 35R30, 35R11

1 Graduate School of Mathematical Sciences, The University of Tokyo, Komaba,
Meguro, Tokyo 153-8914, Japan
2 Honorary Member of Academy of Romanian Scientists, Ilfov, nr. 3, Bucuresti,
Romania
3 Correspondence member of Accademia Peloritana dei Pericolanti,
Palazzo Università, Piazza S. Pugliatti 1 98122 Messina Italy
e-mail<EMAIL_ADDRESS>
##  1\. Introduction

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded domain with smooth boundary
$\partial\Omega$, $\nu(x)=(\nu_{1}(x),...,\nu_{d}(x))$ be the unit outward
normal vector to $\partial\Omega$ at $x$, and let

$0<\alpha<2,\quad\alpha\neq 1.$

Throughout this article, we set

$None$
$(-Av)(x)=\sum_{i,j=1}^{d}\partial_{i}(a_{ij}(x)\partial_{j}v(x))+c(x)v(x),\quad
x\in\Omega,$

where $a_{ij}=a_{ji}\in C^{1}(\overline{\Omega})$, $1\leq i,j\leq d$ and $c\in
C(\overline{\Omega})$ and $c\leq 0$ in $\Omega$. Moreover we assume that there
exists a constant $\kappa>0$ such that

$\sum_{i,j=1}^{d}a_{ij}(x)\zeta_{i}\zeta_{j}\geq\kappa\sum_{i=1}^{d}\zeta_{i}^{2}\quad\mbox{for
all $x\in\overline{\Omega}$ and $\zeta_{1},...,\zeta_{d}\in\mathbb{R}$}.$

We set

$\partial_{\nu_{A}}v(x):=\sum_{i,j=1}^{d}a_{ij}(x)(\partial_{j}v)\nu_{i}.$

We consider an initial boundary value problem for a fractional diffusion-wave
equation whose source term is described by $\mu(t)f(x)$, where $f$ is a
spatial distribution of the source and $\mu$ is a temporally changing factor.
We can formally describe an initial boundary value problem:

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

Here we can define the pointwise classical Caputo derivative:

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

We postpone a rigorous formulation of the initial boundary value problem, and
we first describe our motivation for the inverse source problem in this
article.

The first equation in (1.2) describes anomalous diffusion in heterogeneous
media such as soil. In the heterogenous media, it is often observed that the
classical diffusion equation corresponding to $\alpha=1$ cannot simulate
diffusion phenomenon well and real density profiles are deviated from the ones
given by the equation with $\alpha=1$. Therefore the time-fractional
diffusion-wave equations in (1.2) with $\alpha\in(0,1)\cup(1,2)$ have been
proposed and studied (e.g., Metzler and Klafter [11]). The right-hand side
$\mu(t)f(x)$ of the fractional equation in (1.2) models a source term causing
diffusion of some substances, for example, harmful contaminants such as
cesium-137 after destruction of plants or storages. The source of the
diffusion can be often modelled by a product of a temporal function $\mu(t)$
and a spatial density profile $f(x)$. In a case of radioactive isotope, we can
choose $\mu(t)=e^{-\mu_{0}t}$ with some decay constant $\mu_{0}>0$.

Usually we cannot a priori know the source term, for example, when some
dangerous substances start to diffuse after some incident in an inaccesible
place. Then a serious direct problem is concerned with accurate prediction of
contaminants, and for more realistic simulations, we have to estimate or
identify $f(x)$ and/or $\mu(t)$ in the source by observation data.

Thus, in this article, we are concerned with an inverse source problem of
determining $f(x)$, $x\in\Omega$ or $\mu(t)$ in some time interval by
asymptotic behavior of data $u(\cdot,t)|_{\omega}$ and
$\partial_{\nu_{A}}u(\cdot,t)|_{\gamma}$ as $t\to\infty$ of the solution
$u(x,t)$ to (1.2). Here $\omega\subset\Omega$ is a non-empty subdomain and
$\gamma\subset\partial\Omega$ is a non-empty subboundary.

Our main interest is the uniqueness in the inverse source problem: Which
behavior of data (1.3) or (1.4) as $t\to\infty$, can uniquely determine $f(x)$
for $x\in\Omega$ or $\mu(t)$ over some time interval?

For the cases where observations start at $t=0$, there are many works
available on the stability as well as the uniqueness. Here, owing to the
numerous growing references, we are restricted to a very small number of
works: Kian, Liu and Yamamoto [6], Kian, Soccorsi, Xue and Yamamoto [7], and a
survey Liu, Li and Yamamoto [10]. The readers can consult the references in
the above articles. For the cases of $\alpha=1$ and $=2$, we can refer to
Yamamoto [15], [16], [17], [18].

By taking practical situations into consideration, it is not realistic that we
can start observations at time $t=0$. In the case where the diffusion was
caused by some drastic changes such as explosion, only after the incident, one
can start to take observation data, and it is more acceptable to observe data
after time passes sufficiently long. This is our primary interest for data,
and in this article, as is formulated by (1.7) and (1.8), we adopt the
asymptotic behavior of solution limited to a subdomain or a subboundary at
$t\to\infty$. For inverse source problems, there are very few theoretical
works by data measured after positive time. See Cheng, Lu and Yamamoto [3] for
$\alpha=1,2$. For $\alpha\in(0,1)\cup(1,2)$, we can refer to Theorem 2.2 in
Kian, Liu and Yamamoto [6], which proved the uniqueness with data
$u|_{\omega\times(t_{0},T_{0})}$ if $\mu\not\equiv 0$ but we have to extra
assume that $f=0$ in $\omega$.

To the best knowledge of the author, without such an extra condition on $f$,
there are no published results for the uniqueness of the inverse source
problem.

For the statement of the main result, we need to formulate an initial boundary
value problem and the class of solutions where we do not require the
differentiability for the sake of our choice of natural regularity: $\mu f\in
L^{2}(0,T;L^{2}(\Omega))$. First, we extend the domain of the classical Caputo
derivative.

For it, we define the Riemann-Liouville fractional integral operator
$J^{\alpha}$ for $\alpha>0$ by

$J^{\alpha}w(t):=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}w(s)ds,\quad
w\in L^{2}(0,T).$

We set

${{}_{0}C^{1}[0,T]}:=\\{w\in C^{1}[0,T];\,w(0)=0\\}.$

Henceforth $\overline{B}^{X}$ denotes the closure of a set $B$ in the space
$X$.

Recalling the Sobolev-Slobodecki space and $\|\cdot\|_{H^{\alpha}(0,T)}$ for
$0<\alpha<1$ (e.g., Adams [1]), we define

$H_{\alpha}(0,T)=\overline{{{}_{0}C^{1}[0,T]}}^{H^{\alpha}(0,T)}.$

It is known (e.g., Gorenflo, Luchko and Yamamoto [4], Kubica, Ryszewska and
Yamamoto [9]) that $J^{\alpha}:L^{2}(0,T)\longrightarrow H_{\alpha}(0,T)$ is
isomorphism for $0<\alpha<1$. Then we define

$\partial_{t}^{\alpha}:=(J^{\alpha})^{-1},\quad\mathcal{D}(\partial_{t}^{\alpha})=H_{\alpha}(0,T).$

Next we define $\partial_{t}^{\alpha}$ for $1<\alpha<2$. Let $\alpha:=1+\beta$
with $0<\beta\leq 1$ and

$H_{1}(0,T):=\\{v\in H^{1}(0,T);\,v(0)=0\\}$ $H_{1+\beta}(0,T):=\left\\{v\in
H_{1}(0,T);\,\frac{dv}{dt}\in H_{\beta}(0,T)\right\\}$

with

$\|v\|_{H_{1+\beta}(0,T)}:=\|v\|_{H^{1}(0,T)}+\left\|\frac{dv}{dt}\right\|_{H_{\beta}(0,T)}.$

By [4] or [9], we can prove that $J^{1+\beta}:L^{2}(0,T)\longrightarrow
H_{1+\beta}(0,T)$ is isomorphism, and we define

$\partial_{t}^{\alpha}=\partial_{t}^{1+\beta}=(J^{\alpha})^{-1},\quad\mathcal{D}(\partial_{t}^{\alpha})=H_{\alpha}(0,T).$

We can directly see that $\partial_{t}^{\alpha}v=d_{t}^{\alpha}v$ for $v\in
C^{2}[0,T]$ satisfying $v(0)=\frac{dv}{dt}(0)=0$. Throughout this article,
based on this extended fractional derivative $\partial_{t}^{\alpha}$ with
$\alpha\in(0,1)\cup(1,2)$, we consider the initial boundary value problem

$None$
$\left\\{\begin{array}[]{rl}&\partial_{t}^{\alpha}u(x,t)=-Au(x,t)+\mu(t)f(x),\quad
x\in\Omega,\,0<t<T,\\\ &u|_{\partial\Omega\times(0,T)}=0,\\\ &u(x,\cdot)\in
H_{\alpha}(0,T)\quad\mbox{for almost all $x\in\Omega$.}\end{array}\right.$

If $\frac{1}{2}<\alpha$ and $u(x,\cdot)\in H_{\alpha}(0,T)$, then Sobolev
embedding $u(x,\cdot)\in C[0,T]$, and $u(x,\cdot)$ can be approximated by
functions in ${{}_{0}C^{1}[0,T]}$ and so $u(x,\cdot)\in H_{\alpha}(0,T)$ is
interpreted as $u(x,t)\longrightarrow 0$ as $t\to 0$ for almost all
$x\in\Omega$. In the case $\frac{3}{2}<\alpha$, we can similarly interpret the
zero initial condition for $u(\cdot,0)$ and $\partial_{t}u(\cdot,0)$ within
the framework $H_{\alpha}(0,T)$. Thus the third condition in (1.5) means the
generalized zero initial condition in a unified manner for each
$\alpha\in(0,1)\cup(1,2)$. Then we can prove that for $f\in H^{2}(\Omega)\cap
H^{1}_{0}(\Omega)$ and $\mu\in L^{2}(0,T)$, there exists a unique solution

$None$ $u\in L^{2}(0,T;H^{2}(\Omega)\cap H^{1}_{0}(\Omega))\cap
H_{\alpha}(0,T;L^{2}(\Omega))$

to (1.3).

We can relax the regularity $H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ of $f$ for
(1.4), but we do not pursue here. The proof of (1.4) is found in [9] for
$0<\alpha<1$ and in Theorem 2.2 (ii) in [14] for $1<\alpha<2$, for example.
Indeed for $0<\alpha<1$, the work [9] establishes the unique existence of $u$
satisfying (1.4) under regulairty $f\in L^{2}(\Omega)$. Furthermore we can
weaken the condition $\mu\in L^{2}(0,T)$, but we do not discuss details.

Throughout this article, we always consider the solution class (1.4) for
(1.3).

Let $t_{0}>0$ be arbitrarily fixed. We assume

$None$ $\mu\in L_{loc}^{2}(0,\infty),\quad\mu(t)=0\quad\mbox{for $t>t_{0}$}.$

Let $\|\cdot\|$ and $(\cdot,\cdot)$ denote the norm and the scalar product in
$L^{2}(\Omega)$. For an elliptic operator $A$ defined by (1.1), we define the
domain $\mathcal{D}(A)$ by $\mathcal{D}(A)=H^{2}(\Omega)\cap
H^{1}_{0}(\Omega)$. Then the spectrum $\sigma(A)$ of $A$ is composed of
positive eigenvalues with finite multiplicities. The set
$\sigma(A)=\\{\lambda_{n}\\}_{n\in\mathbb{N}}$ is numbered as

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

Let $\\{\varphi_{nj}\\}_{1\leq j\leq d_{n}}$ be an orthonormal basis of Ker
$(\lambda_{n}-A)$. We further set

$P_{n}a=\sum_{j=1}^{d_{n}}(a,\varphi_{nj})\varphi_{nj},\quad
n\in\mathbb{N},\,a\in L^{2}(\Omega).$

Then $a=\sum_{n=1}^{\infty}P_{n}a$ in $L^{2}(\Omega)$ for all $a\in
L^{2}(\Omega)$. Moreover we can define a fractional power $A^{\sigma}$ with
$\sigma>0$ (e.g., Pazy [12]) and we note $\|v\|_{H^{2\sigma}(\Omega)}\leq
C\|A^{\sigma}v\|$ for each $v\in\mathcal{D}(A^{\sigma})$.

Let two numbers $\sigma_{1}$ and $\sigma_{2}$ satisfy

$None$
$\sigma_{1}>\max\left\\{1,\,\frac{d}{2}-2\right\\},\quad\sigma_{2}>\max\left\\{1,\,\frac{d}{2}-\frac{1}{4}\right\\}.$

Now we are ready to state the main result.
Theorem 1.
Let $\alpha\in(0,1)\cup(1,2)$ and let (1.5) hold.
(i) Let $f\in\mathcal{D}(A^{\sigma_{1}})$. We assume that for any
$p\in\mathbb{N}$, there exists a constant $C(p)>0$ such that

$None$ $\|u(\cdot,t)\|_{L^{2}(\omega)}\leq\frac{C(p)}{t^{p}}\quad\mbox{as
$t\to\infty$}.$

Then $\mu=0$ in $(0,t_{0})$ or $f=0$ in $\Omega$.
(ii) Let $f\in\mathcal{D}(A^{\sigma_{2}})$. We assume that for any
$p\in\mathbb{N}$, there exists a constant $C(p)>0$ such that

$None$
$\|\partial_{\nu_{A}}u(\cdot,t)\|_{L^{2}(\gamma)}\leq\frac{C(p)}{t^{p}}\quad\mbox{as
$t\to\infty$}.$

Then $\mu=0$ in $(0,t_{0})$ or $f=0$ in $\Omega$.

In general, as is seen by the arguments in Section 2 (especially by e.g.,
(2.7)), we can find a constant $\beta>0$ such that

$\|u(\cdot,t)\|_{L^{2}(\omega)}=O\left(\frac{1}{t^{\beta}}\right),\quad\|\partial_{\nu_{A}}u(\cdot,t)\|_{L^{2}(\omega)}=O\left(\frac{1}{t^{\beta}}\right)$

as $t\to\infty$. Thus the assumptions (1.7) and (1.8) especially require more
rapid decay, in order to conclude that $u=0$ in $\Omega\times(0,\infty)$.

By means of the analyticity of $u(\cdot,t)$ for $t>t_{0}$, from Theorem 1 we
can derive
Corollary 1.
We arbitrarily choose $T_{0}>t_{0}$. Let $\alpha\in(0,1)\cup(1,2)$ and let
(1.5) hold.
(i) Let $f\in\mathcal{D}(A^{\sigma_{1}})$. Then $u=0$ in
$\omega\times(t_{0},T_{0})$ implies $\mu=0$ in $(0,t_{0})$ or $f=0$ in
$\Omega$.
(ii) Let $f\in\mathcal{D}(A^{\sigma_{2}})$. Then $\partial_{\nu_{A}}u=0$ on
$\gamma\times(t_{0},T_{0})$ implies $\mu=0$ in $(0,t_{0})$ or $f=0$ in
$\Omega$.

This article is composed of five sections. In Section 2, we construct
$t$-functions related to the solution $u$ and prove Proposition 1 which is a
key for the proof of Theorem 1. In Section 3, we complete the proof of Theorem
1 and in Section 4 we provide the proof of Corollary 1. Finally Section 5
gives concluding remarks.

##  2\. Key proposition

For $\alpha,\beta>0$, we define the Mittag-Leffler functions by

$E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(\alpha k+\beta)},$

which is an entire function in $z\in\mathbb{C}$.

We prove
Lemma 1.
Let $T>0$ be arbitrary. For $\mu\in L^{2}(0,T)$, the unique solution $u$ in
the class (1.6) to (1.5) is given by

$None$
$u(x,t)=\sum_{n=1}^{\infty}\left(\int^{t}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(t-s)^{\alpha})\mu(s)ds\right)(P_{n}f)(x)$

in $H_{\alpha}(0,T;L^{2}(\Omega))\cap L^{2}(0,T;H^{2}(\Omega)\cap
H^{1}_{0}(\Omega))$.

The proof relies on the classical Fourier method by the separation of
variables, or is found for example in Sakamoto and Yamamoto [14].

We choose $T_{1}>0$ such that

$t_{0}<T_{1}$

and set

$None$
$\psi_{n}(t):=\int^{t_{0}}_{0}(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(t-s)^{\alpha})\mu(s)ds,\quad
t\geq T_{1},\quad n\in\mathbb{N}.$

Henceforth $C>0$, $C_{k}>0$ denote generic constants which are independent of
$n\in\mathbb{N}$, $f$ and choices of $t$ in intervals under consideration.
When we emphasize the dependence on other quantities $K$, we describe $C(K)$,
$C_{1}(K)$, etc., but we omit the dependence if there is no fear of confusion.

Then
Lemma 2.
We see $\psi_{n}\in C[T_{1},\,\infty)$ and there exists a constant $C>0$ such
that

$None$ $|\psi_{n}(t)|\leq\frac{C}{\lambda_{n}}\|\mu\|_{L^{1}(0,t_{0})}$

for all $n\in\mathbb{N}$ and $t\geq T_{1}$.
Proof.
Since $t\geq T_{1}>t_{0}$, we see that $0<T_{1}-t_{0}<t-s$ for all $0<s\leq
t_{0}$, the function
$(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(t-s)^{\alpha})$ is continous
in $t\geq T_{1}$ and $0\leq s\leq t_{0}$, so that $\psi_{n}\in
C[T_{1},\,\infty)$. Morever, by

$|E_{\alpha,\alpha}(-\eta)|\leq\frac{C}{1+\eta},\quad\eta>0,\quad 0<\alpha<2$

(e.g., Theorem 1.6 (p.35) in Podlubny [13]), using $|t-s|\geq T_{1}-t_{0}>0$,
we see

$|\psi_{n}(t)|\leq\int^{t_{0}}_{0}(t-s)^{\alpha-1}\frac{C|\mu(s)|}{1+\lambda_{n}|t-s|^{\alpha}}ds.$

Here we note that $\lambda_{n}>0$ for all $n\in\mathbb{N}$. Thus the proof of
Lemma 2 is complete. $\blacksquare$

Now, we show the key proposition.
Proposition 1.
Let $\mu\not\equiv 0$ in $(0,t_{0})$. We assume that $a_{n}\in\mathbb{R}$,
$n\in\mathbb{N}$ satisfy

$None$ $\sum_{n=1}^{\infty}\left|\frac{a_{n}}{\lambda_{n}}\right|<\infty.$

If for each $p\in\mathbb{N}$, there exist constants $T_{2}(p)>0$ and $C(p)>0$
such that

$None$
$\left|\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)\right|\leq\frac{C(p)}{t^{p}},\quad
t>T_{2}(p),$

then $a_{n}=0$ for all $n\in\mathbb{N}$.

By Lemma 3 and (2.4), we notice that the series
$\sum_{n=1}^{\infty}a_{n}\psi_{n}$ is uniformly convergent in any compact set
in $(T_{1},\,\infty)$.
Proof of Proposition 1.
We first prove
Lemma 3.
Let (2.4) 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 3.
We have

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

Hence

$|a_{1}|=\left|-\sum_{n=2}^{\infty}a_{n}\left(\frac{\lambda_{1}}{\lambda_{n}}\right)^{\ell_{k}}\right|=\left|\sum_{n=2}^{\infty}\left(a_{n}\frac{\lambda_{1}}{\lambda_{n}}\right)\left(\frac{\lambda_{1}}{\lambda_{n}}\right)^{\ell_{k}-1}\right|\leq|\lambda_{1}|\left(\sum_{n=2}^{\infty}\left|\frac{a_{n}}{\lambda_{n}}\right|\right)\left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{\ell_{k}-1}.$

By $0<\lambda_{1}<\lambda_{2}<....$, we see that
$\left|\frac{\lambda_{1}}{\lambda_{2}}\right|<1$. Letting $k\to\infty$, we see
that $\ell_{k}\to\infty$, 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 3 is complete. $\blacksquare$

We note the asymptotic expansion of the Mittag-Leffler function:

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

as $\eta\to\infty$ for each $N\in\mathbb{N}\setminus\\{1\\}$ (e.g., Theorem
1.4 (pp.33-34) in [13]). We arrange a set
$\\{\ell\in\mathbb{N}\setminus\\{1\\};\,\alpha\ell\not\in\mathbb{N}\\}$ in an
increasing manner:

$\ell_{1}<\ell_{2}<\cdots.$

Here $\\{\ell\in\mathbb{N}\setminus\\{1\\};\,\alpha\ell\not\in\mathbb{N}\\}$
is an infinite set, and so $\lim_{k\to\infty}\ell_{k}=\infty$. Indeed, if it
is not a finite set, then we see by
$\mathbb{N}\setminus\\{1\\}=\\{\ell\in\mathbb{N}\setminus\\{1\\};\,\alpha\ell\not\in\mathbb{N}\\}\cap\\{\ell\in\mathbb{N}\setminus\\{1\\};\,\alpha\ell\in\mathbb{N}\\}$
that there exists some $N_{0}\in\mathbb{N}$ such that

$\\{\ell\in\mathbb{N}\setminus\\{1\\};\,\alpha\ell\in\mathbb{N}\\}\supset\\{N_{0},N_{0}+1,N_{0}+2,\cdots\\},$

so that $\alpha(N_{0}+1),\,\alpha N_{0}\in\mathbb{N}$. Therefore,
$\alpha=\alpha(N_{0}+1)-\alpha N_{0}\in\mathbb{N}$, which is impossible by
$\alpha\in(0,1)\cup(1,2)$. $\blacksquare$

In particular, for $\alpha\not\in\mathbb{Q}$, we note that $\ell_{k}=k$ for
$k=2,3,4,...$.

Hence, for each $K\in\mathbb{N}$, we have

$\eta^{\alpha-1}E_{\alpha,\alpha}(-\lambda\eta^{\alpha})=\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\frac{1}{\lambda^{\ell_{k}}\eta^{\alpha\ell_{k}-\alpha+1}}+O\left(\frac{1}{\lambda^{\ell_{K+1}}\eta^{\alpha\ell_{K+1}-\alpha+1}}\right)$

as $\lambda\eta^{\alpha}\to\infty$. Therefore,

$\eta^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}\eta^{\alpha})=\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\frac{1}{\lambda_{n}^{\ell_{k}}\eta^{\alpha\ell_{k}-\alpha+1}}+O\left(\frac{1}{\lambda_{n}^{\ell_{K+1}}\eta^{\alpha\ell_{K+1}-\alpha+1}}\right)$

as $\lambda_{n}\eta^{\alpha}\to\infty$ for each $K\in\mathbb{N}$ and
$n\in\mathbb{N}$. Using $\lambda_{n}\geq\lambda_{1}>0$ for $n\in\mathbb{N}$,
we see: for each $K\in\mathbb{N}$, there exists some constant
$\eta_{0}=\eta_{0}(K)>0$ and $C=C(K)>0$ such that if
$\lambda_{1}\eta^{\alpha}\geq\eta_{0}(K)$, then

$None$
$\left|\eta^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}\eta^{\alpha})-\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\frac{1}{\lambda_{n}^{\ell_{k}}\eta^{\alpha\ell_{k}-\alpha+1}}\right|\leq\frac{C(K)}{\lambda_{n}^{\ell_{K+1}}\eta^{\alpha\ell_{K+1}-\alpha+1}}$

for all $n\in\mathbb{N}$.

Here we note that the constant $C(K)>0$ is independent of $n\in\mathbb{N}$.

We fix $K\in\mathbb{N}$ arbitrarily. Let $t\geq T_{1}$. Since
$\lambda_{1}(t-s)^{\alpha}\geq\lambda_{1}(t-t_{0})^{\alpha}$ for $0\leq s\leq
t_{0}$, if $\lambda_{1}(t-t_{0})^{\alpha}\geq\eta_{0}$, then
$\lambda_{1}(t-s)^{\alpha}\geq\eta_{0}$ for $0\leq s\leq t_{0}$. Then, we
substitute (2.6) into (2.2), and we obtain

$\left|\psi_{n}(t)-\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\frac{1}{\lambda_{n}^{\ell_{k}}}\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{\alpha\ell_{k}-\alpha+1}}ds\right|$
$\leq\int^{t_{0}}_{0}\frac{C_{1}}{\lambda_{n}^{\ell_{K+1}}}(t-s)^{\alpha\ell_{K+1}-\alpha+1}|\mu(s)|ds$

for $\lambda_{1}(t-t_{0})^{\alpha}\geq\eta_{0}$ and $t>T_{1}$.

We set

$T_{3}:=\max\left\\{\left(\frac{\eta_{0}}{\lambda_{1}}\right)^{\frac{1}{\alpha}}+t_{0},\,T_{1}\right\\}.$

Multiplying by $a_{n}$ and taking $\sum_{n=1}^{\infty}$, by the triangle
inequality, we obtain

$\displaystyle\left|\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)-\sum_{n=1}^{\infty}a_{n}\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\frac{1}{\lambda_{n}^{\ell_{k}}}\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{\alpha\ell_{k}-\alpha+1}}ds\right|$
$\displaystyle\leq$ $\displaystyle
C_{1}\sum_{n=1}^{\infty}\frac{|a_{n}|}{\lambda_{n}^{\ell_{K+1}}}\int^{t_{0}}_{0}\frac{|\mu(s)|}{(t-s)^{\alpha\ell_{K+1}-\alpha+1}}ds$

for $t>T_{3}$.

We write

$A_{k}:=\sum_{n=1}^{\infty}\frac{a_{n}}{\lambda_{n}^{\ell_{k}}},\quad
k\in\mathbb{N},\quad T^{*}:=\max\\{T_{1},\,T_{2}(p),\,T_{3}\\}.$

Henceforth we omit the dependence of the constants $C>0$, $C_{2}>0$, etc. on
$K,\ell_{k}$, etc.

Then, since $\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)=0$ for $t>T_{1}$ and

$\sum_{n=1}^{\infty}\frac{|a_{n}|}{\lambda_{n}^{\ell_{K+1}}}\leq
C\sum_{n=1}^{\infty}\left|\frac{a_{n}}{\lambda_{n}}\right|<\infty,$

and

$\int^{t_{0}}_{0}\frac{|\mu(s)|}{(t-s)^{\alpha\ell_{K+1}-\alpha+1}}ds\leq\frac{\|\mu\|_{L^{1}(0,t_{0})}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}\leq\frac{C_{2}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}},$

we obtain

$None$
$\left|\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)-\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\left(\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{\alpha\ell_{k}-\alpha+1}}ds\right)A_{k}\right|\leq\frac{C_{3}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}$

for all $t>T^{*}$. Thus, in view of (2.5), we reach

$None$
$\left|\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}\left(\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{\alpha\ell_{k}-\alpha+1}}ds\right)A_{k}\right|\leq\frac{C_{3}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}+\frac{C(p)}{t^{p}}$

for all $t>T^{*}$.

On the other hand, we consider the binomial asymptotic expansion. We fix
$0<r<1$ arbitrarily. Then, for each $M\in\mathbb{N}$ and
$\sigma\in\mathbb{R}$, we know

$None$ $(1-\eta)^{-\sigma}=\sum_{m=0}^{M}\left(\begin{array}[]{c}-\sigma\\\
m\\\ \end{array}\right)(-\eta)^{m}+R_{\sigma,M}(\eta),\quad|\eta|\leq r,$

where we set

$\left(\begin{array}[]{c}-\sigma\\\ m\\\
\end{array}\right):=\frac{(-\sigma)(-\sigma-1)\cdots(-\sigma-m+1)}{m!},$

and the function $R_{\sigma,M}$ in $\eta$ satisfies

$|R_{\sigma,M}(\eta)|\leq C(M,\sigma)|\eta|^{M+1}\quad\mbox{for $|\eta|\leq
r$}$

with a constant $C(M,\sigma)>0$. Whenever we fix the ranges of
$\sigma\in\mathbb{R}$ and $M\in\mathbb{N}$ among finite sets, we can write
$C(M,\sigma)$ simply by $C_{4}$.

Henceforth the variables $s$ and $t$ satisfy

$None$ $0\leq s\leq t_{0},\quad t>T^{*}.$

Then, by $t>T_{1}$, we have

$\frac{s}{t}\leq\frac{t_{0}}{T_{1}}.$

Since $t_{0}<T_{1}$, setting $r:=\frac{t_{0}}{T_{1}}$, we see that $r<1$, that
is,

$None$ $\frac{s}{t}\leq\frac{t_{0}}{T_{1}}=r<1.$

In terms of (2.11), applying (2.9), we obtain

$\frac{1}{(t-s)^{\sigma}}=\frac{1}{t^{\sigma}}\frac{1}{\left(1-\frac{s}{t}\right)^{\sigma}}=t^{-\sigma}\left(1-\frac{s}{t}\right)^{-\sigma}$
$None$ $=\sum_{m=0}^{M}\left(\begin{array}[]{c}-\sigma\\\ m\\\
\end{array}\right)t^{-\sigma-m}(-s)^{m}+R_{M}(t,s),\quad 0\leq s\leq
t_{0},\,t>T^{*},$

where

$|R_{M}(t,s)|\leq C_{4}t^{-\sigma-M-1}t_{0}^{M+1}.$

We further set

$None$ $\mu_{m}:=\int^{t_{0}}_{0}(-s)^{m}\mu(s)ds,\quad
m\in\mathbb{N}\cup\\{0\\}.$

Therefore, in view of (2.12), we see

$\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{\sigma}}ds=\sum_{m=0}^{M}\left(\begin{array}[]{c}-\sigma\\\
m\\\
\end{array}\right)\frac{1}{t^{\sigma+m}}\int^{t_{0}}_{0}(-s)^{m}\mu(s)ds+\int^{t_{0}}_{0}\mu(s)R_{M}(t,s)ds$
$None$ $=\sum_{m=0}^{M}\left(\begin{array}[]{c}-\sigma\\\ m\\\
\end{array}\right)\frac{\mu_{m}}{t^{\sigma+m}}+\int^{t_{0}}_{0}\mu(s)R_{M}(t,s)ds.$

Consequently

$\left|\int^{t_{0}}_{0}\mu(s)R_{M}(t,s)ds\right|\leq\frac{C_{4}t_{0}^{M+1}}{t^{\sigma+M+1}}\|\mu\|_{L^{1}(0,t_{0})}\leq\frac{C_{5}}{t^{\sigma+M+1}},\quad
t>T^{*}.$

Hence, in terms of (2.14), by (2.8), we have

$\biggl{|}\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}A_{k}\biggl{[}\sum_{m=0}^{M}\left(\begin{array}[]{c}-\alpha\ell_{k}+\alpha-1\\\
m\\\
\end{array}\right)\frac{\mu_{m}}{t^{\alpha\ell_{k}-\alpha+1+m}}+O\left(\frac{1}{t^{\alpha\ell_{k}-\alpha+M+2}}\right)\biggr{]}\biggr{|}$
$None$
$\leq\frac{C_{6}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}+\frac{C(p)}{t^{p}}\quad\mbox{for
all $p,K,M\in\mathbb{N}$ and $T>T^{*}$}.$

We can choose $m_{1}\in\mathbb{N}\cup\\{0\\}$ such that
$\mu_{m_{1}}:=\int^{t_{0}}_{0}(-s)^{m_{1}}\mu(s)ds\neq 0$ and $\mu_{j}=0$ for
$j=0,1,2,...,m_{1}-1$ if $m_{1}\geq 1$. Indeed, if such
$m_{1}\in\mathbb{N}\cup\\{0\\}$ does not exist, then $\mu_{m}=0$ for all
$m\in\mathbb{N}\cup\\{0\\}$. Therefore Weierstrass polynomial approximation
theorem (e.g., Yosida [22]), implies $\mu(s)=0$ for almost all
$s\in(0,t_{0})$, which contradicts the assumption $\mu\not\equiv 0$ in
$(0,t_{0})$.

By (2.15), choosing $M=m_{1}$, we have

$\biggl{|}\sum_{k=1}^{K}\frac{(-1)^{\ell_{k}+1}}{\Gamma(\alpha-\alpha\ell_{k})}A_{k}\left(\begin{array}[]{c}-\alpha\ell_{k}+\alpha-1\\\
m_{1}\\\
\end{array}\right)\frac{\mu_{m_{1}}}{t^{\alpha\ell_{k}-\alpha+1+m_{1}}}$
$None$
$+O\left(\frac{1}{t^{\alpha\ell_{k}-\alpha+m_{1}+2}}\right)\biggr{]}\biggr{|}\leq\frac{C_{6}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}+\frac{C(p)}{t^{p}},\quad
t>T^{*}.$

In (2.16), we set

$b_{k}:=\left(\begin{array}[]{c}-\alpha\ell_{k}+\alpha-1\\\ m_{1}\\\
\end{array}\right),\quad k\in\mathbb{N}.$

We remark that $b_{k}\neq 0$ because $\alpha\ell_{k}-\alpha\not\in\mathbb{N}$
by the choice of $\ell_{k}$.

Moreover in (2.16), since $K,p\in\mathbb{N}$ can be arbitrary, for already
fixed $m_{1}\in\mathbb{N}$, we can choose sufficiently large
$K,p\in\mathbb{N}$ such that

$p,\,\alpha_{\ell_{K+1}}-\alpha+1>\alpha\ell_{1}-\alpha+1+m_{1}.$

Then, we can find constants $C_{7}=C_{7}(p,K)>0$ and $C_{8}=C_{8}(p,K)$ such
that

$\displaystyle\biggl{|}$
$\displaystyle\,\frac{(-1)^{\ell_{1}+1}}{\Gamma(\alpha-\alpha\ell_{1})}A_{1}\frac{1}{t^{\alpha\ell_{1}-\alpha+1+m_{1}}}\left(b_{1}\mu_{m_{1}}+\frac{C_{7}}{t}\right)$
$\displaystyle+$
$\displaystyle\,\frac{(-1)^{\ell_{2}+1}}{\Gamma(\alpha-\alpha\ell_{2})}A_{2}\frac{1}{t^{\alpha\ell_{2}-\alpha+1+m_{1}}}\left(b_{2}\mu_{m_{1}}+\frac{C_{7}}{t}\right)$
$\displaystyle+$ $\displaystyle\,\cdots\cdots$ $\displaystyle+$
$\displaystyle\,\frac{(-1)^{\ell_{K}+1}}{\Gamma(\alpha-\alpha\ell_{K})}A_{K}\frac{1}{t^{\alpha\ell_{K}-\alpha+1+m_{1}}}\left(b_{K}\mu_{m_{1}}+\frac{C_{7}}{t}\right)\biggr{|}$
$None$ $\leq
C_{8}\left(\frac{1}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}+\frac{1}{t^{p}}\right)\quad\mbox{for
each $t>T^{*}$}.$

The multiplication by $t^{\alpha\ell_{1}-\alpha+1+m_{1}}$, yields

$\displaystyle\biggl{|}$
$\displaystyle\frac{(-1)^{\ell_{1}+1}}{\Gamma(\alpha-\alpha\ell_{1})}A_{1}\left(b_{1}\mu_{m_{1}}+\frac{C_{7}}{t}\right)+\,\frac{(-1)^{\ell_{2}+1}}{\Gamma(\alpha-\alpha\ell_{2})}A_{2}\frac{t^{\alpha\ell_{1}-\alpha+1+m_{1}}}{t^{\alpha\ell_{2}-\alpha+1+m_{1}}}\left(b_{2}\mu_{m_{1}}+\frac{C_{7}}{t}\right)$
$\displaystyle+$
$\displaystyle\,\cdots\cdots+\,\frac{(-1)^{\ell_{K}+1}}{\Gamma(\alpha-\alpha\ell_{K})}A_{K}\frac{t^{\alpha\ell_{1}-\alpha+1+m_{1}}}{t^{\alpha\ell_{K}-\alpha+1+m_{1}}}\left(b_{K}\mu_{m_{1}}+\frac{C_{7}}{t}\right)\biggr{|}$
$\displaystyle\leq$ $\displaystyle
C_{8}\left(\frac{t^{\alpha\ell_{1}-\alpha+1+m_{1}}}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}+\frac{t^{\alpha\ell_{1}-\alpha+1+m_{1}}}{t^{p}}\right)\quad\mbox{for
$t>T^{*}$}.$

In view of $\alpha\ell_{1}-\alpha+1+m_{1}<\alpha\ell_{K+1}-\alpha+1$,
$\alpha\ell_{1}-\alpha+1+m_{1}<p$ and $\ell_{1}<\ell_{2}<\cdots<\ell_{K}$,
letting $t\to\infty$, we obtain

$\frac{(-1)^{\ell_{1}+1}}{\Gamma(\alpha-\alpha\ell_{1})}A_{1}b_{1}\mu_{m_{1}}=0.$

By $\mu_{m_{1}}\neq 0$ and $b_{1}\neq 0$, we reach $A_{1}=0$.

Substituting $A_{1}=0$ into (2.17), we have

$\displaystyle\biggl{|}$
$\displaystyle\frac{(-1)^{\ell_{2}+1}}{\Gamma(\alpha-\alpha\ell_{2})}\frac{A_{2}}{t^{\alpha\ell_{2}-\alpha+1+m_{1}}}\left(b_{2}\mu_{m_{1}}+\frac{C_{7}}{t}\right)$
$\displaystyle+$ $\displaystyle\,\cdots\cdots$ $\displaystyle+$
$\displaystyle\,\frac{(-1)^{\ell_{K}+1}}{\Gamma(\alpha-\alpha\ell_{K})}\frac{A_{K}}{t^{\alpha\ell_{K}-\alpha+1+m_{1}}}\left(b_{K}\mu_{m_{1}}+\frac{C_{7}}{t}\right)\biggr{|}$
$\displaystyle\leq$
$\displaystyle\,C_{8}\left(\frac{1}{(t-t_{0})^{\alpha\ell_{K+1}-\alpha+1}}+\frac{1}{t^{p}}\right)\quad\mbox{for
$t>T^{*}$}.$

Again we choose larger $p$ and $K\in\mathbb{N}$ such that

$p,\,\alpha\ell_{K+1}-\alpha+1>\alpha\ell_{2}-\alpha+1+m_{1}.$

Multipying by $t^{\alpha\ell_{2}-\alpha+1+m_{1}}$ and letting $t\to\infty$, we
obtain

$\frac{(-1)^{\ell_{2}+1}}{\Gamma(\alpha-\alpha\ell_{2})}A_{2}b_{2}\mu_{m_{1}}=0.$

By $\mu_{m_{1}}\neq 0$ and $b_{2}\neq 0$, we reach $A_{2}=0$.

Choosing $p,K\in\mathbb{N}$ at each step and continuing this argument, we
finally prove that $A_{k}=0$ for all $k\in\mathbb{N}$, that is,

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

Thus Lemma 3 completes the proof of Proposition 1. $\blacksquare$

##  3\. Proof of Theorem 1

Assume that for each $p\in\mathbb{N}$, we can find a constant $C(p)>0$ such
that (1.7) holds. We arbitrarily choose $v\in C^{\infty}_{0}(\Omega)$
satisfying supp $v\subset\omega$, and set

$a_{n}:=(P_{n}f,\,v)_{L^{2}(\omega)},\quad n\in\mathbb{N}.$

Then by $\mu=0$ in $(t_{0},T_{0})$, by (2.1) and the assumption (1.7), there
exists $T_{2}=T_{2}(p)>0$ such that

$\left\|\sum_{n=1}^{\infty}\psi_{n}(t)P_{n}f\right\|_{L^{2}(\omega)}\leq\frac{C(p)}{t^{p}},\quad
t\geq T_{2}.$

Since the series is convergent in $L^{2}(0,T;L^{2}(\omega))$, taking the
scalar products with $v$ in $L^{2}(\omega)$, we see

$None$
$\left|\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)\right|\leq\frac{C(p)\|v\|_{L^{2}(\omega)}}{t^{p}},\quad
t\geq T_{2}.$

Next we prove

$None$ $\sum_{n=1}^{\infty}\left|\frac{a_{n}}{\lambda_{n}}\right|<\infty$

for $a_{n}=(P_{n}f,\,v)_{L^{2}(\omega)}$, $n\in\mathbb{N}$.
Proof of (3.2).
By $f\in\mathcal{D}(A^{\sigma_{1}})$, we can see $g:=A^{\sigma_{1}}f\in
L^{2}(\Omega)$. Then

$(f,\,\varphi_{nj})=(A^{-\sigma_{1}}g,\,\varphi_{nj})=(g,\,A^{-\sigma_{1}}\varphi_{nj})=\lambda_{n}^{-\sigma_{1}}(g,\,\varphi_{nj}).$

Therefore,

$P_{n}f=\sum_{j=1}^{d_{n}}(f,\,\varphi_{nj})\varphi_{nj}=\lambda_{n}^{-\sigma_{1}}\sum_{j=1}^{d_{n}}(g,\,\varphi_{nj})\varphi_{nj}=\lambda_{n}^{-\sigma_{1}}P_{n}g.$

Since $v\in C^{\infty}_{0}(\Omega)\subset\mathcal{D}(A)$, we have
$\widetilde{v}:=Av\in L^{2}(\Omega)$, and so

$a_{n}=(P_{n}f,v)_{L^{2}(\omega)}=(P_{n}f,v)_{L^{2}(\Omega)}=(\lambda_{n}^{-\sigma_{1}}P_{n}g,\,A^{-1}\widetilde{v})\\\
$ $None$
$=(\lambda_{n}^{-\sigma_{1}}A^{-1}P_{n}g,\,\widetilde{v})=\lambda_{n}^{-\sigma_{1}-2}(P_{n}g,\,\widetilde{v}).$

Therefore,

$\sum_{n=1}^{\infty}\left|\frac{a_{n}}{\lambda_{n}}\right|=\sum_{n=1}^{\infty}\lambda_{n}^{-\sigma_{1}-2}|(P_{n}g,\widetilde{v})|\leq\|g\|\|\widetilde{v}\|\sum_{n=1}^{\infty}\lambda_{n}^{-\sigma_{1}-2}.$

On the other hand, by $\mu_{n}$, $n\in\mathbb{N}$, we renumber the eigenvalues
$\lambda_{n}$ of $A$ repeatedly according to the multiplicities:

$\mu_{k}=\lambda_{1}\,\,\mbox{for $1\leq k\leq
d_{1}$},\quad\mu_{k}=\lambda_{2}\,\,\mbox{for $d_{1}+1\leq k\leq
d_{1}+d_{2}$},\cdots.$

Then $\mu_{n}\leq\lambda_{n}$ for $n\in\mathbb{N}$.

We know that there exists a constant $c_{1}>0$ such that

$\mu_{n}=c_{1}n^{\frac{2}{d}}+o(1)\quad\mbox{as $n\to\infty$}$

(e.g., Agmon [2], Theorem 15.1). Here we recall that $d$ is the spatial
dimensions. Therefore, we can find a constant $c_{2}>0$ such that

$None$ $\lambda_{n}\geq c_{2}n^{\frac{2}{d}}\quad\mbox{as $n\to\infty$}.$

Hence, (1.6) yields $\frac{2}{d}(\sigma_{1}+2)>1$, so that

$\sum_{n=1}^{\infty}\lambda_{n}^{-\sigma_{1}-2}\leq
C_{3}\sum_{n=1}^{\infty}n^{-\frac{2}{d}(\sigma_{1}+2)}<\infty,$

which means (3.2). $\blacksquare$

Thus we can apply Proposition 1 in terms of (3.1) and (3.2) to obtain
$a_{n}=(P_{n}f,\,v)_{L^{2}(\omega)}=0$ for all $n\in\mathbb{N}$ and each $v\in
C^{\infty}_{0}(\omega)$.

Since $v\in C^{\infty}_{0}(\omega)$ is arbitrary, we obtain

$None$ $P_{n}f=0\quad\mbox{in $\omega$ for all $n\in\mathbb{N}$}.$

For each $n\in\mathbb{N}$, we have $(A-\lambda_{n})P_{n}f=0$ in $\Omega$. In
view of (3.5), we can apply the unique continuation for the elliptic operator
$A-\lambda_{n}$ (e.g., Isakov [5]), and we can conclude that $P_{n}f=0$ in
$\Omega$ for each $n\in\mathbb{N}$. Hence $f=\sum_{n=1}^{\infty}P_{n}f=0$ in
$L^{2}(\Omega)$. Thus the proof of Theorem 1 (i) is complete.

Finally in the case of data $\partial_{\nu_{A}}u|_{\gamma\times(t_{0},T_{)}}$,
as test functions we choose $v\in C^{\infty}(\gamma)$ and set
$a_{n}=(\partial_{\nu_{A}}P_{n}f,\,v)_{L^{2}(\gamma)}$ for $n\in\mathbb{N}$.
Then for any $\varepsilon>0$, we have

$\|\partial_{\nu_{A}}P_{n}f\|_{L^{2}(\gamma)}\leq
C\|P_{n}f\|_{H^{\frac{3}{2}+2\varepsilon}(\Omega)}\leq
C\|A^{\frac{3}{4}+\varepsilon}P_{n}f\|\leq
C\lambda_{n}^{\frac{3}{4}+\varepsilon}\|P_{n}f\|$

by the trace theorem.

By $f\in\mathcal{D}(A^{\sigma_{2}})$ where $\sigma_{2}>0$ is given by (1.6),
we have $A^{\sigma_{2}}f=:g\in L^{2}(\Omega)$, and so

$(f,\,\varphi_{nj})=(A^{-\sigma_{2}}g,\,\varphi_{nj})=(g,\,A^{-\sigma_{2}}\varphi_{nj})=\lambda_{n}^{-\sigma_{2}}(g,\,\varphi_{nj}).$

Therefore, we can verify

$P_{n}f=\sum_{j=1}^{d_{n}}(f,\,\varphi_{nj})\varphi_{nj}=\lambda_{n}^{-\sigma_{2}}\sum_{j=1}^{d_{n}}(g,\,\varphi_{nj})\varphi_{nj}=\lambda_{n}^{-\sigma_{2}}P_{n}g.$

Therefore,

$|a_{n}|\leq\|\partial_{\nu_{A}}P_{n}f\|_{L^{2}(\gamma)}\|v\|_{L^{2}(\gamma)}\leq
C_{4}\lambda_{n}^{\frac{3}{4}+\varepsilon}\|P_{n}f\|\leq
C_{4}\lambda_{n}^{\frac{3}{4}+\varepsilon-\sigma_{2}}\|P_{n}g\|,$

and so (3.4) yields

$None$ $\sum_{n=1}^{\infty}\left|\frac{a_{n}}{\lambda_{n}}\right|\leq
C_{4}\sum_{n=1}^{\infty}\lambda_{n}^{-\frac{1}{4}+\varepsilon-\sigma_{2}}\|P_{n}g\|\leq
C_{4}\sum_{n=1}^{\infty}n^{-(\frac{1}{4}-\varepsilon+\sigma_{2})\frac{2}{d}}\|g\|.$

In terms of (1.6), for sufficiently small $\varepsilon>0$ we see that
$\left(\frac{1}{4}-\varepsilon+\sigma_{2}\right)\frac{2}{d}>1$ and (3.2) is
verified to hold in the case of
$a_{n}=(\partial_{\nu_{A}}P_{n}f,\,v)_{L^{2}(\gamma)}$.

Consequently we can argue similarly to the case of
$u|_{\omega\times(t_{0},T_{0})}$ to complete the proof of Theorem 1 (ii).
$\blacksquare$

##  4\. Proof of Corollary 1.

First Step: holomorphicity of time factors.
We consider the holomorphicity of related functions to (2.1).

For $\theta\in\left(0,\frac{\pi}{2}\right)$ and $z_{0}\in\mathbb{C}$, we
define a sector $S_{\theta}(z_{0})\subset\mathbb{C}$ by

$S_{\theta}(z_{0}):=\\{z\in\mathbb{C};\,z\neq
z_{0}.\quad|\mbox{arg}\,(z-z_{0})|<\theta\\}.$

Let $t_{0}<T_{1}<T_{0}$. Then
Lemma 4.
Let $h=h(z)$ be holomorphic in $S_{\theta}(T_{1}-t_{0})$ and let $\mu\in
L^{1}(0,t_{0})$.Then

$H(z):=\int^{t_{0}}_{0}h(z-s)\mu(s)ds$

is holomorphic in $z\in S_{\theta}(T_{1})$.
Proof of Lemma 4.
For $z\in S_{\theta}(T_{1})$, we choose $r>0$ and $0<\theta_{1}<\theta$ such
that $z=T_{1}+re^{\sqrt{-1}\theta_{1}}$. Therefore,
$z-s=(T_{1}-s)+re^{\sqrt{-1}\theta_{1}}\in S_{\theta_{1}}(T_{1}-s)\subset
S_{\theta_{1}}(T_{1}-t_{0})$ by $T_{1}-t_{0}\leq T_{1}-s$ for all
$s\in[0,t_{0}]$. Hence, $z-s\in S_{\theta_{1}}(T_{1}-t_{0})$ if $z\in
S_{\theta}(T_{1})$ and $s\in[0,t_{0}]$.

Consequently, by the Lebesgue convergence theorem and $\mu\in L^{1}(0,T)$, we
have

$\frac{\partial H}{\partial z}(z)=\int^{t_{0}}_{0}\frac{\partial h}{\partial
z}(z-s)\mu(s)ds,\quad z\in S_{\theta}(T_{1}).$

Thus the holomorphicity follows. $\blacksquare$

We see that $g(z):=z^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}z^{\alpha})$ is
holomorphic in $z\in S_{\theta}(T_{1}-t_{0})$ by $T_{1}-t_{0}>0$. We can
choose a small constant $\theta_{0}>0$ such that

$None$
$|E_{\alpha,\alpha}(-\lambda_{n}(z-s)^{\alpha})|\leq\frac{C}{1+\lambda_{n}|z-s|^{\alpha}}$

for $|\mbox{arg}\,(z-s)|\leq\theta_{0}$.
Proof of (4.1).
We choose constants $\mu_{1},\mu_{2}$ satisfying

$\left\\{\begin{array}[]{rl}&\frac{\pi}{2}\alpha<\mu_{1}<\pi\alpha\quad\mbox{if
$0<\alpha<1$},\\\ &\frac{\pi}{2}\alpha<\mu_{2}<\pi\quad\mbox{if
$1<\alpha<2$}.\end{array}\right.$

Setting $\eta:=z-s$, we apply Theorem 1.6 (p.35) in [13], so that we know that
(4.1) holds if

$\mu_{1}\leq|\mbox{arg}\,(-\lambda_{n}\eta^{\alpha})|\leq\pi\quad\mbox{if
$0<\alpha<1$}$

and

$\mu_{2}\leq|\mbox{arg}\,(-\lambda_{n}\eta^{\alpha})|\leq\pi\quad\mbox{if
$1<\alpha<2$}.$

Henceforth, for $0\leq\theta<\pi$, we use that
$\theta\leq|\mbox{arg}\,(-\xi)|\leq\pi$ is equivalent to
$|\mbox{arg}\,\xi|\leq\pi-\theta$.

Let $0<\alpha<1$. Then, $|\mbox{arg}\,\eta^{\alpha}|\leq\pi-\mu_{1}$ implies
$\mu_{1}\leq|\mbox{arg}\,(-\eta^{\alpha})|\leq\pi$. Setting
$\theta_{0}:=\pi-\mu_{1}$, we see that if $|\mbox{arg}\,\eta|\leq\theta_{0}$,
then

$|\mbox{arg}\,\eta^{\alpha}|=\alpha|\mbox{arg}\,\eta|\leq\alpha\theta_{0}=\alpha(\pi-\mu_{1})<\pi-\mu_{1},$

so that (4.1) holds for $|\mbox{arg}\,(z-s)|\leq\theta_{0}$ in the case of
$0<\alpha<1$.

Next let $1<\alpha<2$. We choose $\mu_{2}=\pi-\varepsilon_{0}$ with
sufficiently small constant $\varepsilon_{0}>0$. Let
$|\mbox{arg}\,\eta|\leq\varepsilon_{0}$. Then
$|\mbox{arg}\,\eta^{\alpha}|\leq\alpha\varepsilon_{0}$, and so

$\pi-\alpha\varepsilon_{0}\leq|\mbox{arg}\,(-\eta^{\alpha})|\leq\pi.$

For $\mu_{2}\in\left(\frac{\pi}{2}\alpha,\,\pi\right)$, we further choose
small $\varepsilon_{0}>0$ such that $\mu_{2}<\pi-\alpha\varepsilon_{0}$. Then
we obtain $\mu_{2}<|\mbox{arg}\,(-\eta^{\alpha})|\leq\pi$, that is, (4.1)
holds for $|\mbox{arg}\,(z-s)|\leq\varepsilon_{0}$ in the case of
$1<\alpha<2$.

Now, by choosing

$\theta_{0}=\left\\{\begin{array}[]{rl}\pi-\mu_{1},&0<\alpha<1,\\\
\varepsilon_{0},&1<\alpha<2,\end{array}\right.$

the estimate (4.1) holds. $\blacksquare$

Therefore, since $z\in S_{\theta_{0}}(T_{1})$ implies that
$|\mbox{arg}\,(z-s)|\leq\theta_{0}$ for any $s\in[0,t_{0}]$ by $s\leq
t_{0}<T_{1}$, we have

$None$
$|E_{\alpha,\alpha}(-\lambda_{n}(z-s)^{\alpha})|\leq\frac{C}{1+\lambda_{n}|z-s|^{\alpha}}\quad\mbox{for
all $z\in S_{\theta_{0}}(T_{1})$, $s\in[0,t_{0}]$ and all $n\in\mathbb{N}$.}$

Moreover, for arbitrarily fixed $s\in[0,t_{0}]$, the function in $z$

$g(z):=z^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}z^{\alpha})$

is holomorphic in $S_{\theta_{0}}(T_{1}-t_{0})$, because $T_{1}-t_{0}>0$.

Second Step.
We can assume that $\mu\not\equiv 0$ in $(0,t_{0})$. Indeed if $\mu\equiv 0$
in $(0,t_{0})$, the the proof is already complete.

From Lemma 4 it follows that $\psi_{n}(t)$, $n\in\mathbb{N}$ defined by (2.2)
for $t\geq T_{1}$, can be extended to a holomorphic function $\psi_{n}(z)$ for
$z\in S_{\theta_{0}}(T_{1})$.

Assume that $u(x,t)=0$ for $x\in\omega$ and $t_{0}<t<T_{0}$. We will prove

$None$ $\sum_{n=1}^{\infty}\sup_{z\in
S_{\theta_{0}}(T_{1})}|\psi_{n}(z)||a_{n}|<\infty.$

Here we set $a_{n}=(P_{n}f,v)_{L^{2}(\omega)}$, $n\in\mathbb{N}$ for $v\in
C^{\infty}_{0}(\omega)$ and
$a_{n}=(\partial_{\nu_{A}}P_{n}f,\,v)_{L^{2}(\gamma)}$, $n\in\mathbb{N}$ for
$v\in C^{\infty}(\gamma)$.
Proof of (4.3).
Since $|z-s|\geq T_{1}-t_{0}>0$ for each $z\in S_{\theta_{0}}(T_{1})$ and
$0\leq s\leq t_{0}$, we estimate $|\psi_{n}(z)|$. In view of (4.2), similarly
to Lemma 2, we have

$\displaystyle|\psi_{n}(z)|=\left|\int^{t_{0}}_{0}(z-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda_{n}(z-s)^{\alpha})\mu(s)ds\right|$
$\displaystyle\leq$ $\displaystyle
C\int^{t_{0}}_{0}\frac{|z-s|^{\alpha-1}}{1+\lambda_{n}|z-s|^{\alpha}}|\mu(s)|ds\leq\frac{C}{\lambda_{n}}\|\mu\|_{L^{1}(0,t_{0})},\quad
z\in S_{\theta_{0}}(T_{1}),\,n\in\mathbb{N}.$

Hence, applying (3.2) and (3.6) respectively for
$a_{n}=(P_{n}f,v)_{L^{2}(\omega)}$ and
$a_{n}=(\partial_{\nu_{A}}P_{n}f,\,v)_{L^{2}(\gamma)}$, we obtain

$\sum_{n=1}^{\infty}\sup_{z\in
S_{\theta_{0}}(T_{1})}|\psi_{n}(z)||a_{n}|\leq\sum_{n=1}^{\infty}\frac{C}{\lambda_{n}}\|\mu\|_{L^{1}(0,t_{0})}|a_{n}|<\infty.$

Thus the proof of (4.3) is complete. $\blacksquare$

By means of (4.3) and the holomorphicity of $\psi_{n}(z)$, we see that
$\sum_{n=1}^{\infty}a_{n}\psi_{n}(z)$ is holomorphic in any compact set in
$S_{\theta_{0}}(T_{1})$, and in particular,
$\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)$ is real analytic in $t>T_{1}$.
Consequently if $u=0$ in $\omega\times(t_{0},T_{0})$ or
$\partial_{\nu_{A}}u=0$ on $\gamma\times(t_{0},T_{0})$, then

$\sum_{n=1}^{\infty}a_{n}\psi_{n}(t)=0,\quad t>T_{1}.$

Since $v$ is arbitrarily chosen, we see that $u=0$ in
$\omega\times(t_{0},\infty)$ or $\partial_{\nu_{A}}u=0$ on
$\gamma\times(t_{0},\infty)$. Thus the assumptions in Theorem 1 are satisfied,
so that $f=0$ in $\Omega$ follows, and the proof of Corollary 1 is complete.
$\blacksquare$

##  5\. Concluding remarks

5.1. The proof of the main result relies on Proposition 1 which is concerned
with the infinite series of $\psi_{n}(t)$, $n\in\mathbb{N}$ defined by (2.2).
The proof of the proposition is based on the asymptotic expansion of
$\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{\sigma}}ds$ with $\sigma=\alpha
k-\alpha+1$. Here in the case of $0<\alpha<1$, we extract its essence and
prove as
Proposition 2.
(i) We assume that there exist $a\in\mathbb{R}$ and $t_{0}>0$ and $v-a\in
H_{\alpha}(0,T)$ satisfies

$None$ $\partial_{t}^{\alpha}(v-a)(t)=0,\quad t>t_{0}$

and for each $p\in\mathbb{N}$ we can find a constant $C(p)>0$ such that

$None$ $|v(t)|\leq\frac{C(p)}{t^{p}}\quad\mbox{as $t\to\infty$}.$

Then $a=0$ and $v(t)=0$ for $t>0$.
(ii) In particular, if $\partial_{t}^{\alpha}(v-a)(t)=v(t)=0$ for $t>t_{0}$
with some $a\in\mathbb{R}$ and $t_{0}>0$, then $v(t)=0$ for $t>0$.

Proposition 2 (ii) corresponds to Theorem 1 in Kinash and Janno [8], which
treats more general time-fractional derivative. We can relax (5.1) as the
vanishing in some finite interval $t_{0}<t<T_{0}$ but we do not discuss.
Proof of Proposition 2 (i).
Part (ii) is a direct consequence of part (i), and so it suffices to prove
part (i). Setting $\mu(t):=\partial_{t}^{\alpha}(v-a)$ for $t>0$, we have
$\mu\in L^{2}(0,T)$ by $v-a\in H_{\alpha}(0,T)$ and

$v(t)=a+\frac{1}{\Gamma(\alpha)}\int^{t}_{0}(t-s)^{\alpha-1}\mu(s)ds,\quad
t>0$

(e.g., [9], [21]). By (5.1), we obtain

$None$
$v(t)=a+\frac{1}{\Gamma(\alpha)}\int^{t_{0}}_{0}(t-s)^{\alpha-1}\mu(s)ds,\quad
t>t_{0}.$

Let $T_{1}>t_{0}$ be arbitrarily chosen. Applying (2.14), for any
$M\in\mathbb{N}$ we have

$\int^{t_{0}}_{0}\frac{\mu(s)}{(t-s)^{1-\alpha}}ds=\sum_{m=0}^{M}\left(\begin{array}[]{c}1-\alpha\\\
m\\\
\end{array}\right)\frac{\mu_{m}}{t^{1-\alpha+m}}+O\left(\frac{1}{t^{1-\alpha+M+1}}\right),\quad
t>T_{1}.$

Here we set $\mu_{m}:=\int^{t_{0}}_{0}(-s)^{m}\mu(s)ds$ for
$m\in\mathbb{N}\cup\\{0\\}$. Substituting this into (5.3), we reach

$v(t)=a+\sum_{m=0}^{M}\frac{1}{\Gamma(\alpha)}\left(\begin{array}[]{c}1-\alpha\\\
m\\\
\end{array}\right)\frac{\mu_{m}}{t^{1-\alpha+m}}+O\left(\frac{1}{t^{1-\alpha+M+1}}\right),\quad
t>T_{1}.$

We number the set $\\{m\in\mathbb{N};\,\mu_{m}\neq 0\\}$ as

$m_{1}<m_{2}<\cdots.$

Henceforth we assume that such a set of $\mu_{k}$ is an infinite set. The
proof is same for the case of the finite set.

Applying (5.2) for all $p\in\mathbb{N}$, we obtain

$None$
$\left|a+\sum_{j=1}^{m_{N}}\frac{1}{\Gamma(\alpha)}\left(\begin{array}[]{c}1-\alpha\\\
m\\\
\end{array}\right)\frac{\mu_{m_{j}}}{t^{1-\alpha+m_{j}}}+O\left(\frac{1}{t^{1-\alpha+m_{N}+1}}\right)\right|\leq\frac{C(p)}{t^{p}}$

as $t\to\infty$. Letting $t\to\infty$ in (5.4), we obtain $a=0$.

For arbitrarily fixed $N\in\mathbb{N}$, we can choose $p>1-\alpha+m_{N}+1$.
Multiplying (5.4) with $t^{1-\alpha+m_{1}}$ and letting $t\to\infty$, we
obtain

$\frac{1}{\Gamma(\alpha)}\left(\begin{array}[]{c}1-\alpha\\\ m_{1}\\\
\end{array}\right)\mu_{m_{1}}=0,$

that is, $\mu_{m_{1}}=0$. Continuing this argument, we reach
$\mu_{m_{1}}=\mu_{m_{2}}=\cdots=\mu_{m_{N}}=0$. Since $N\in\mathbb{N}$ is
arbitrary, we see that $\mu_{m_{k}}=0$ for $k\in\mathbb{N}$. The Weierstrass
polynomial approximation theorem implies $\mu=0$ in $(0,t_{0})$. Thus $a=0$
and $v=0$ in $(0,t_{0})$ by (5.3). $\blacksquare$

5.2. We sum up the uniqueness results for the inverse source problem of
determining a spatially varying function $f(x)$ in (1.5):

  * •

Data over time interval $0<t<T_{0}$:
This is a conventional formulation and there are satisfactory researches on
the uniqueness and also the conditional stability.

  * •

Data over time interval $0<t_{0}<t<T_{0}$:
The current work proved the uniqueness if $\mu\not\equiv 0$ in $(0,t_{0})$ and
$\mu=0$ in $(t_{0},\,T_{0})$.
The uniqueness remains open under other condition $\mu\not\equiv 0$ in
$(t_{0},\,T_{0})$, in particular, $\\{t\in(t_{0},\,T_{0});\,\mu(t)=0\\}$ has
no interior points.

5.3. In this article, for keeping conciseness, we are restricted to establish
the uniqueness of $f(x)$ or $\mu(t)$ by fixing the order
$\alpha\in(0,1)\cup(1,2)$. Indeed our data concerning
$u|_{\omega\times(t_{0},T_{0})}$ or
$\partial_{\nu_{A}}u|_{\gamma\times(t_{0},T_{0})}$ can contain enough
information for determining also an order $\alpha\in(0,2)$ including
$\alpha=1$. The method in Yamamoto [19], [20] is applicable, but we postpone
the details to a forthcoming work.

The order $\alpha\in(0,2)$ is an important parameter characterizing the
anomaly of the diffusion, which is a quantity describing the heterogeneity of
media under consideration. Usually also $\alpha$ cannot be determined a priori
and we should estimate or identify $\alpha$ by available observation data.
Thus the inverse problems of determining an order $\alpha$ are significant.

5.4. In this article, we assume that initial values are zero. On the other
hand, our data under assumption $\mu(t)=0$ for $t>t_{0}$ may determine also an
initial value uniquely and in a forthcoming paper we will discuss.

5.5. Our main result asserts that if

$None$ $\mu\not\equiv 0\quad\mbox{in $(0,\,t_{0})$},$

then the uniqueness in determining $f$ in $\Omega$ follows for
$\alpha\in(0,1)\cup(1,2)$. As is seen from Theorem 4 in [3], for the case of
$\alpha=1$, the condition (5.5) is not at all sufficient for the uniqueness
for the corresponding inverse source problem, and we can prove the uniqueness
under much stronger condition

$None$ $\int^{t_{0}}_{0}e^{\lambda_{n}s}\mu(s)ds\neq 0\quad\mbox{for all
$n\in\mathbb{N}$}.$

This difference in the conditions for the uniqueness, can be interpreted by a
kind of anomaly of the fractional diffusion as that the case $\alpha\neq 1$
keeps more information of the state before the observation time
$(t_{0},\,T_{0})$, while in the case of $\alpha=1$, we need more data holding
information of every eigenmode corresponding to the eigenvalue $\lambda_{n}$,
$n\in\mathbb{N}$.

Acknowledgments. The author was supported by Grant-in-Aid for Scientific
Research Grant-in-Aid (A) 20H00117 of Japan Society for the Promotion of
Science.

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