#  Uniqueness for inverse problem of determining fractional orders for time-
fractional advection-diffusion equations

1,2,3 M. Yamamoto

###### Abstract.

We consider initial boundary value problems of time-fractional advection-
diffusion equations with the zero Dirichlet boundary value
$\partial_{t}^{\alpha}u(x,t)=-Au(x,t)$, where
$-A=\sum_{i,j=1}^{d}\partial_{i}(a_{ij}(x)\partial_{j})+\sum_{j=1}^{d}b_{j}(x)\partial_{j}+c(x)$.
We establish the uniqueness for an inverse problem of determining an order
$\alpha$ of fractional derivatives by data $u(x_{0},t)$ for $0<t<T$ at one
point $x_{0}$ in a spatial domain $\Omega$. The uniqueness holds even under
assumption that $\Omega$ and $A$ are unknown, provided that the initial value
does not change signs and is not identically zero. The proof is based on the
eigenfunction expansions of finitely dimensional approximating solutions, a
decay estimate and the asymptotic expansions of the Mittag-Leffler functions
for large time.
Key words. fractional advection-diffusion equation, uniqueness, fractional
order
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 Peoples’ Friendship University of Russia (RUDN University) 6 Miklukho-
Maklaya St, Moscow, 117198, Russian Federation e-mail<EMAIL_ADDRESS>
##  1\. Introduction

Throughout this article, we assume that the spatial dimensions $d=1,2,3$. We
can similarly argue for higher dimensions $d\geq 4$, but we need more
regularity for initial values in (1.1) and (1.5) later described. Let
$\Omega\subset\mathbb{R}^{d}$ be a bounded domain with smooth boundary
$\partial\Omega$ and let

$0<\alpha,\beta<1.$

By $\partial_{t}^{\alpha}$ we denote the Caputo derivative:

$\partial_{t}^{\alpha}g(t)=\frac{1}{\Gamma(1-\alpha)}\int^{t}_{0}(t-s)^{-\alpha}\frac{d}{ds}g(s)ds$

for $\alpha\in(0,1)$ (e.g., Podlubny [25]).

Throughout this article, we set

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

where $a_{ij}=a_{ji}\in C^{1}(\overline{\Omega})$, $b_{j}\in C^{1}(\Omega)$,
$1\leq i,j\leq d$, $c\in C(\overline{\Omega})$ are all real-valued. Moreover
we assume that $c(x)\leq 0$ for $x\in\overline{\Omega}$, and there exists a
constant $\sigma=\sigma(a_{ij})>0$ such that

$None$
$\sum_{i,j=1}^{d}a_{ij}(x)\zeta_{i}\zeta_{j}\geq\sigma(a_{ij})\sum_{i=1}^{d}\zeta_{i}^{2}\quad\mbox{for
all $x\in\overline{\Omega}$ and $\zeta_{1},...,\zeta_{d}\in\mathbb{R}$}.$

We consider an initial boundary value problem for a time-fractional advection-
diffusion equation:

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

For $\alpha\in(0,1)$, the first equation in (1.3) is called a fractional
advection-diffusion equation with the first-order term
$\sum_{j=1}^{d}b_{j}\partial_{j}u$, and is a macroscopic model for anomalous
diffusion in hetrogeneous media. Fractional diffusion equations are studied
related also to diffusions in fractals and we refer for example to Mainardi
[22], Metzler, Glöckle and Nonnenmacher [23], Metzler and Klafter [24], Roman
and Alemany [27].

Let $x_{0}\in\Omega$ and $0<t<T$ be arbitrarily chosen. The main subject of
this article is
Inverse problem of determining the order $\alpha$.
Determine $\alpha$ by data $u(x_{0},t)$ for $0<t<T$ for (1.3).

Several properties such as asymptotic behavior as $t\to\infty$ of solution $u$
to (1.1) depend on the fractional order $\alpha$ of the derivative. It is
known that $\alpha$ is an essential physical parameter characterizing the
anomaly of diffusion. Thus our inverse problem is important not only from the
theoretical viewpoint but also for modelling actual anomalous advection-
diffusion of substances such as contaminants by a relevant fractional
diffusion equation.

As for inverse problems of determining orders and other parameters, we refer
to Alimov and Ashurov [3], Ashurov and Umarov [4], Cheng, Nakagawa, Yamamoto
and Yamazaki [5], Hatano, Nakagawa, Wang and Yamamoto [9], Janno [10], Janno
and Kinash [11], Jin and Kian [12], Krasnoschok, Pereverzyev, Siryk and
Vasylyeva [15], Li, Zhang, Jia and Yamamoto [17], Li and Yamamoto [20], Tatar,
Tinaztepe and Ulusoy [29], Tatar and Ulusoy [30], Yamamoto [32], [33], Yu,
Jiang and Qi [34], for example. See Li, Liu and Yamamoto [19] as a survey.

In [15], the authors proved that for smooth function $\varphi(t)$ on $[0,T]$,
the order $\alpha$ can be uniquely determined only by
$\partial_{t}^{\alpha}\varphi$ and $\varphi$ near $t=0$ only, whether or not
$\varphi(t)$ is related to a solution to a fractional equation equation, and
applied such uniqueness to other inverse problem of determining an order as
well as a solution to semilinear subdiffusion equations. In [15], the
continuity of $u(x_{0},t)$ and $\partial_{t}^{\alpha}u(x_{0},t)$ at $t=0$ is
essential and so initial value $a$ must be smoother.

We can prove the uniqueness by specified data of solution of even if the
coefficients of $A$ are unknown, and we refer to [5] as early work, and see
[12], [33] as related articles.

However, these works do not consider advection terms, that is, assume that
$b_{1}=b_{2}=\cdots=b_{d}=0$ in $\Omega$ to study the uniqueness in
determining the order $\alpha$. For such a symmetric $A$, relying on a well-
known eigenfunction expansion of solution $u$ to (1.3), we can establish the
uniqueness for the inverse problem directly. However, for non-symmetric $A$,
such a method does not work, and to the best knowledge of the author, there
are no works on the uniqueness in determining the order for non-symmetric $A$
given by (1.1).

On the other hand, since the advection term
$\sum_{j=1}^{d}b_{j}(x)\partial_{j}u$ is of lower-order and does not
drastically change the structure of the equation, for non-symmetric $A$, we
can naturally expect a similar uniqueness result to [4], [12], [32], [33]. The
main purpose of this article is to prove that such a conjecture is correct for
$A$ given by (1.1). Moreover, suggested by [5], [12] and [33], the order
$\alpha$ can be uniquely determined independently of the operator $A$ and the
domain $\Omega$.

For the formulation of our result, we introduce operators and domains. Let
$L^{2}(\Omega)$, $H^{2}(\Omega)$, $H^{1}_{0}(\Omega)$ denote usual Lebesgue
space and Sobolev spaces (e.g., Adams [1]). By $\|\cdot\|_{L^{2}(\Omega)}$ and
$(\cdot,\cdot)_{L^{2}(\Omega)}$, we denote the norm and the scalar product in
$L^{2}(\Omega)$ respectively.

We recall the Poincaré inequality: there exists a constant $C(\Omega)>0$,
depending on a bounded domain $\Omega$ with smooth boundary $\partial\Omega$
such that

$None$ $C(\Omega)\int_{\Omega}|w|^{2}dx\leq\int_{\Omega}|\nabla
w|^{2}dx\quad\mbox{for $w\in H^{1}_{0}(\Omega)$}.$

For a bounded domain $\Omega\subset\mathbb{R}^{d}$ with smooth boundary
$\partial\Omega$, we define $-A$ by (1.1) and assume (1.2), and for another
bounded domain $\widetilde{\Omega}$ with smooth boundary
$\partial\widetilde{\Omega}$, we additionally set

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

where $\widetilde{a_{ij}}=\widetilde{a_{ji}}\in C^{1}(\overline{\Omega})$,
$\widetilde{b_{j}}\in C^{1}(\Omega)$,$1\leq i,j\leq d$, $\widetilde{c}\in
C(\overline{\Omega})$ are all real-valued. Moreover we assume that
$\widetilde{c}(x)\leq 0$ for $x\in\overline{\Omega}$, and there exists a
constant $\sigma=\sigma(\widetilde{a_{ij}})>0$ such that

$None$
$\sum_{i,j=1}^{d}\widetilde{a_{ij}}(x)\zeta_{i}\zeta_{j}\geq\sigma(\widetilde{a_{ij}})\sum_{i=1}^{d}\zeta_{i}^{2}\quad\mbox{for
all $x\in\overline{\Omega}$ and $\zeta_{1},...,\zeta_{d}\in\mathbb{R}$}.$

We consider an initial boundary value problem for a time-fractional advection-
diffusion equation:

$None$
$\left\\{\begin{array}[]{rl}&\partial_{t}^{\beta}\widetilde{u}(x,t)=-\widetilde{A}\widetilde{u}(x,t),\quad
x\in\widetilde{\Omega},\,0<t<T,\\\
&\widetilde{u}|_{\partial\widetilde{\Omega}\times(0,T)}=0,\\\
&\widetilde{u}(x,0)=\widetilde{a}(x),\quad
x\in\widetilde{\Omega}.\end{array}\right.$

We define an operator $A$ in $L^{2}(\Omega)$ attaching the zero Dirichlet
boundary condition, that is, the domain $\mathcal{D}(A)$ of $A$ is
$H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$, and the same for $\widetilde{A}$:
$\mathcal{D}(\widetilde{A}):=H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$. Then it is
known that the spectrum $\sigma(A)$ of $A$ consists entirely of eigenvalues
with finite multiplicities (e.g,, Agmon [2]). With some numbering, we can set

$\sigma(A)=\\{\lambda_{n}\\}_{n\in\mathbb{N}}\subset\mathbb{C},\quad\lim_{n\to\infty}\mbox{Re}\,\lambda_{n}=\infty.$

We note $\lambda_{n}\not\in\mathbb{R}$ in general. Similarly we set
$\sigma(\widetilde{A})=\\{\widetilde{\lambda_{n}}\\}_{n\in\mathbb{N}}$.

We further asssume

$None$
$\left\\{\begin{array}[]{rl}&\inf_{x\in\Omega}\left(\frac{1}{2}\sum_{j=1}^{d}\partial_{j}b_{j}(x)-c(x)\right)+C(\Omega)\sigma(a_{ij})>0,\quad
c(x)\leq 0,\quad x\in\overline{\Omega},\\\
&\inf_{x\in\widetilde{\Omega}}\left(\frac{1}{2}\sum_{j=1}^{d}\partial_{j}\widetilde{b_{j}}(x)-\widetilde{c}(x)\right)+C(\widetilde{\Omega})\sigma(\widetilde{a_{ij}})>0,\quad\widetilde{c}(x)\leq
0,\quad x\in\overline{\widetilde{\Omega}}.\end{array}\right.$

Condition (1.8) yields

$None$
$\\{\mbox{Re}\,\lambda_{n};\,\lambda_{n}\in\sigma(A)\\}>0,\quad\\{\mbox{Re}\,\widetilde{\lambda_{n}};\,\widetilde{\lambda_{n}}\in\sigma(\widetilde{A})\\}>0.$

For completeness, we verify (1.9) in Appendix.

In this article, for simplicity, we do not discuss the case where there exist
eigenvalues $\lambda_{n}$ such that Re $\lambda_{n}<0$.

Now we are reay to state the main result in this article.
Theorem.
Let $0<\alpha,\beta<1$ and $\Omega\cap\widetilde{\Omega}\neq\emptyset$, and
$x_{0}\in\Omega\cap\widetilde{\Omega}$, $T>0$ be arbitrarily chosen. We assume
that each of initial values $a\in\mathcal{D}(A)$ and
$\widetilde{a}\in\mathcal{D}(\widetilde{A})$ does not change signs in $\Omega$
and $\widetilde{\Omega}$ respectively, and

$None$ $\mbox{either $a\not\equiv 0$ in $\Omega$ or $\widetilde{a}\not\equiv
0$ in $\widetilde{\Omega}$}.$

If

$u(x_{0},t)=\widetilde{u}(x_{0},t),\quad 0<t<T,$

then $\alpha=\beta$.

By Lemma 1 in Section 2, we see that $u(x_{0},\cdot)\in C(0,\infty)$ and so
$u(x_{0},t)$ makes sense for $t>0$.
Remark.
From the proof in Section 3, we can see the following:
Let

$None$ each of $a$ and $\widetilde{a}$ does not change signs in $\Omega$ and
$\widetilde{\Omega}$ respectively.

If $u(x_{0},t)=\widetilde{u}(x_{0},t)$ for $0<t<T$ and $\alpha\neq\beta$, then
$u(x,t)=0$ in $\Omega\times(0,T)$ and $\widetilde{u}(x,t)=0$ in
$\widetilde{\Omega}\times(0,T)$.

In other words, under assumption (1.11), we can conclude that
$u(x_{0},t)=\widetilde{u}(x_{0},t)$ for $0<t<T$ implies either

$\alpha=\beta,$

or

$u=0\quad\mbox{in
$\Omega\times(0,T)$}\quad\mbox{and}\quad\widetilde{u}=0\quad\mbox{in
$\widetilde{\Omega}\times(0,T)$}.$

The proof relies on the asymptotic behavior of $u$ and $\widetilde{u}$ as
$t\to\infty$, which is an idea similar to Sakamoto and Yamamoto [28], Yamamoto
[33], but by the non-symmetry of $A$ and $\widetilde{A}$, we cannot make use
of the eigenfunction expansions of tne solutions $u$ and $\widetilde{u}$
themselves. Alternatively we derive asymptotic expansions of eigenprojections
of $u$ and $\widetilde{u}$ in view of the completeness in $L^{2}(\Omega)$ of
generalized eigenfunctions for each of $A$ and $\widetilde{A}$.

The article is composed of three sections and one appendix. In Section 2, we
prepare fundamental qualitative properties of the solutions and a
representation formula of eigenprojections of the solution. In Section 3, on
the basis of the results in Section 2, we complete the proof of the main
result.

##  2\. Preliminaries

2.1. Qualitative properties of the solution $u$ to (1.3).

We recall that the spatial dimensions $d$ is either $1$, $2$ or $3$.
Lemma 1.
Let $a\in L^{2}(\Omega)$. Then $u\in C(\overline{\Omega}\times(0,\infty))$ for
any $t>0$.
Proof.
Similarly to Gorenflo, Luchko and Yamamoto [8], we can prove $u\in
C((0,\infty);H^{2}(\Omega))$, Since $d\leq 3$, the Sobolev embedding yields
$H^{2}(\Omega)\subset C(\overline{\Omega})$, and so we see the lemma.
$\blacksquare$
Lemma 2.
Let $a\in L^{2}(\Omega)$ and let $x_{0}\in\Omega$ be fixed. Then $u(x_{0},t)$
is analytic in $t>0$.
Proof.
We can repeatt the proof of e.g., Theorem 2.2 in Li, Imanuvilov and Yamamoto
[18], and we omit the details.

Next we show decay estimates of the solution $u$.
Proposition 1.
(i) There exists a constant $C>0$ such that

$\|u(\cdot,t)\|_{L^{2}(\Omega)}\leq\frac{C}{t^{\alpha}}\|a\|_{L^{2}(\Omega)},\quad
t>0.$

(ii) Let $d=1,2,3$ and let $x_{0}\in\Omega$. Then there exists a constant
$C>0$ such that

$|u(x_{0},t)|\leq\frac{C}{t^{\alpha}}\|Aa\|_{L^{2}(\Omega)},\quad t>0$

for each $a\in\mathcal{D}(A)$.

The part (i) of the proposition is well-known for the case of symmetric $A$
and can be proved directly by the eigenfunction expansions (e.g., [28]).
Moreover, Vergara and Zacher [31] proved the same decay estimate for symmetric
$A$ with time dependent coefficients which does not admit eigenfunction
expansions, and so the proof requires more technicality. See also Chapter 5 in
the book Kubica, Ryszewska and Yamamoto [16]. For non-symmetric $A$, to the
best knowledge of the author, there are no published works and so Proposition
1 (i) can be an independent interest.

In Appendix we provide a sketch of the proof of Proposition 1.

2.2. Spectral decomposition and representation of solution.

It is sufficient to argue for the operator $A$ in $\Omega$, because for
$\widetilde{A}$ in $\widetilde{\Omega}$, we can do similarly to obtain the
same results.

We recall that $A$ is defined by (1.1) with $\mathcal{D}(A)=H^{2}(\Omega)\cap
H^{1}_{0}(\Omega)$ and satisfies (1.2), and that the spectrum $\sigma(A)$
consists only of isolated eigenvalues $\lambda_{n}\in\mathbb{C}$ with
$n\in\mathbb{N}$ and that $\infty$ is a unique accumulation point (e.g., Agmon
[2]). For each $n\in\mathbb{N}$, we take a circle $\gamma_{n}$ centered at
$\lambda_{n}$ with sufficiently small radius such that $\gamma_{n}$ does not
enclose $\lambda_{m}$ with any $m\neq n$. We define

$None$ $P_{n}:=\frac{1}{2\pi\sqrt{-1}}\int_{\gamma_{n}}(z-A)^{-1}dz$

and then we have

$None$ $P_{n}P_{m}=\left\\{\begin{array}[]{rl}0,&n\neq m,\\\
P_{n},&n=m\end{array}\right.$

(e.g., Kato [13]). We call $P_{n}$ the eigenprojection for $\lambda_{n}$ and
$\varphi\in P_{n}L^{2}(\Omega)$, $\neq 0$ a generalized eigenfunction.

We see that dim $P_{n}L^{2}(\Omega)<\infty$, and we set

$None$ $i_{n}:=\mbox{dim}\,P_{n}L^{2}(\Omega)<\infty.$

Then

$None$ $(A-\lambda_{n})^{i_{n}}P_{n}=0$

(e.g., [13]). We note that if $A$ is symmetric, then
$\lambda_{n}\in\mathbb{R}$ $P_{n}L^{2}(\Omega)=\mbox{Ker}\,(\lambda_{n}-A)$
and $i_{n}=\mbox{dim}\,\mbox{Ker}\,(\lambda_{n}-A)$ for all $n\in\mathbb{N}$.

More we set

$None$ $D_{n}:=(\lambda_{n}-A)P_{n},\quad n\in\mathbb{N}.$

By (2.1) and (2.2), we prove $P_{n}\varphi=\varphi$ for $\varphi\in
P_{n}L^{2}(\Omega)$, and

$None$ $P_{n}L^{2}(\Omega)\subset\mathcal{D}(A),\quad
AP_{n}L^{2}(\Omega)\subset P_{n}L^{2}(\Omega),\quad
D_{n}P_{n}L^{2}(\Omega)\subset P_{n}L^{2}(\Omega),\quad D_{n}^{i_{n}}=0.$

Moreover since $0\not\in\sigma(A)$, we see that $A^{-1}$ exists and

$None$
$A^{-1}\varphi=\sum_{k=0}^{i_{n}-1}\frac{D_{n}^{k}}{\lambda_{n}^{k+1}}\varphi\in
P_{n}L^{2}(\Omega)\quad\mbox{for all $\varphi\in P_{n}L^{2}(\Omega)$}.$

Indeed (2.5) yields

$\displaystyle
A\left(\sum_{k=0}^{i_{n}-1}\frac{D_{n}^{k}}{\lambda_{n}^{k+1}}\varphi\right)=(\lambda_{n}-D_{n})\left(\sum_{k=0}^{i_{n}-1}\frac{D_{n}^{k}}{\lambda_{n}^{k+1}}\varphi\right)$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{i_{n}-1}\frac{D_{n}^{k}}{\lambda_{n}^{k}}\varphi-\sum_{k=0}^{i_{n}-1}\frac{D_{n}^{k+1}}{\lambda_{n}^{k+1}}\varphi=\sum_{k=0}^{i_{n}-1}\frac{D_{n}^{k}}{\lambda_{n}^{k}}\varphi-\sum_{k=1}^{i_{n}}\frac{D_{n}^{k}}{\lambda_{n}^{k}}\varphi=\varphi-\frac{D_{n}^{i_{n}}}{\lambda_{n}^{i_{n}}}\varphi=\varphi,$

which proves (2.7).

We state the completeness of the generalized eigenfunctions of $A$.
Lemma 3.
(i) For any $a\in L^{2}(\Omega)$, there exists a sequence $a_{N}$,
$N\in\mathbb{N}$ such that

$a_{N}\in\sum_{k=1}^{N}P_{k}L^{2}(\Omega),\quad\lim_{N\to\infty}\|a_{N}-a\|_{L^{2}(\Omega)}=0.$

(ii) For any $a\in\mathcal{D}(A)$, there exists a sequence $a_{N}$,
$N\in\mathbb{N}$ such that

$a_{N}\in\sum_{k=1}^{N}P_{k}L^{2}(\Omega)\subset\mathcal{D}(A),\quad\lim_{N\to\infty}\|Aa_{N}-Aa\|_{L^{2}(\Omega)}=0.$

In particular,

$None$ $\lim_{N\to\infty}\|a_{N}-a\|_{C(\overline{\Omega})}=0.$

Proof.
Since the linear subspace spanned by all the generalized eigenfunctions of $A$
is dense in $L^{2}(\Omega)$ (e.g., [2]), we see part (i). Next let
$a\in\mathcal{D}(A)$. Then $Aa\in L^{2}(\Omega)$. By (i) we can choose a
sequence $b_{N}$, $N\in\mathbb{N}$ such that

$b_{N}\in\sum_{k=1}^{N}P_{k}L^{2}(\Omega),\quad\lim_{N\to\infty}\|b_{N}-Aa\|_{L^{2}(\Omega)}=0.$

Since $0\not\in\sigma(A)$, we see that $A^{-1}:L^{2}(\Omega)\longrightarrow
L^{2}(\Omega)$ exists and is bounded, and

$\lim_{N\to\infty}\|A(A^{-1}b_{N})-Aa\|_{L^{2}(\Omega)}=0.$

By (2.7), we see that
$a_{N}:=A^{-1}b_{N}\in\sum_{k=1}^{N}P_{k}L^{2}(\Omega)\subset\mathcal{D}(A)$
and

$\lim_{N\to\infty}\|Aa_{N}-Aa\|_{L^{2}(\Omega)}=0.$

Therefore, in view of $d\leq 3$, the Sobolev embedding $\mathcal{D}(A)\subset
H^{2}(\Omega)\subset C(\overline{\Omega})$ implies (2.8). Thus the proof of
Lemma 3 is complete. $\blacksquare$

Henceforth, we set

$\left(\begin{array}[]{c}k\\\ j\\\
\end{array}\right):=\frac{k!}{j!\,(k-j)!},\quad 0!:=1.$

Now we establish a representation formula of solution to (1.3) with $a\in
P_{n}L^{2}(\Omega)$.
Lemma 4.
Let $n\in\mathbb{N}$ be arbitrarily fixed. Then the solution $u_{n}$ to (1.3)
with $a\in P_{n}L^{2}(\Omega)$ is given by

$None$ $u_{n}(x,t)=\sum_{k=0}^{\infty}\frac{t^{\alpha k}}{\Gamma(\alpha
k+1)}(-\lambda_{n}+D_{n})^{k}a(x)=\sum_{k=0}^{\infty}\frac{t^{\alpha
k}}{\Gamma(\alpha k+1)}\sum_{j=0}^{k}\left(\begin{array}[]{c}k\\\ j\\\
\end{array}\right)(-\lambda_{n})^{k-j}D_{n}^{j}a(x),$

where the series is convergent in $C(\overline{\Omega}\times[\delta,\infty))$
with any $\delta>0$.
Proof.
We can interpret (2.9) as

$E_{\alpha,1}(-(AP_{n})t^{\alpha})a=E_{\alpha,1}((-\lambda_{n}+D_{n})t^{\alpha})a=\sum_{k=0}^{\infty}\frac{t^{\alpha
k}}{\Gamma(\alpha k+1)}(-\lambda_{n}+D_{n})^{k}a,\quad t>0,$

and we can verify as follows.

We set $A_{n}:=AP_{n}$. Then $A_{n}:P_{n}L^{2}(\Omega)\longrightarrow
P_{n}L^{2}(\Omega)$. Since dim $P_{n}L^{2}(\Omega)<\infty$, we see that
$A_{n}$ is a bounded operator and

$\|A_{n}\|_{B(P_{n}L^{2}(\Omega))}=:\rho<\infty,$

where $\|A_{n}\|_{B(P_{n}L^{2}(\Omega))}$ denotes the operator norm of
$A_{n}:P_{n}L^{2}(\Omega)\longrightarrow P_{n}L^{2}(\Omega)$. The Sobolev
embedding yields

$\|a\|_{C(\overline{\Omega})}\leq
C\|Aa\|_{L^{2}(\Omega)}=C\|A_{n}a\|_{L^{2}(\Omega)}\leq
C\rho\|a\|_{L^{2}(\Omega)}$

for all $a\in P_{n}L^{2}(\Omega)$. Moreover

$\|A_{n}^{k}a\|_{C(\overline{\Omega})}\leq
C\|A_{n}^{k+1}a\|_{L^{2}(\Omega)}\leq
C\|A_{n}\|_{B(P_{n}L^{2}(\Omega))}^{k+1}\|a\|_{L^{2}(\Omega)}\leq
C\rho^{k+1}\|a\|_{L^{2}(\Omega)}.$

Since

$\partial_{t}^{\alpha}\left(\frac{t^{\alpha k}}{\Gamma(\alpha
k+1)}\right)=\frac{t^{(k-1)\alpha}}{\Gamma(1+(k-1)\alpha)},\quad
k\in\mathbb{N},$

we have

$\displaystyle\sum_{k=0}^{\infty}\left|\partial_{t}^{\alpha}\left(\frac{t^{\alpha
k}}{\Gamma(\alpha
k+1)}\right)\right|\|(-A_{n})^{k}a\|_{C(\overline{\Omega})}\leq\sum_{k=1}^{\infty}\frac{t^{\alpha(k-1)}}{\Gamma(1+(k-1)\alpha)}C\rho^{k+1}\|a\|_{L^{2}(\Omega)}$
$\displaystyle=$ $\displaystyle
C\|a\|_{L^{2}(\Omega)}\rho^{2}\sum_{k=0}^{\infty}\frac{(\rho
t^{\alpha})^{k}}{\Gamma(\alpha k+1)}<\infty,$

noting that

$E_{\alpha,1}(\rho t^{\alpha})=\sum_{k=0}^{\infty}\frac{(\rho
t^{\alpha})^{k}}{\Gamma(\alpha k+1)}$

and $E_{\alpha,1}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(\alpha k+1)}$ is
an entire function in $z\in\mathbb{C}$. Therefore, we can justify

$\displaystyle\partial_{t}^{\alpha}u_{n}(x,t)=\sum_{k=0}^{\infty}\partial_{t}^{\alpha}\left(\frac{t^{\alpha
k}}{\Gamma(\alpha
k+1)}\right)(-A_{n})^{k}a=\sum_{k=1}^{\infty}\frac{t^{\alpha(k-1)}}{\Gamma(1+(k-1)\alpha)}(-A_{n})^{k}a$
$\displaystyle=$ $\displaystyle-A_{n}\sum_{k=0}^{\infty}\frac{t^{\alpha
k}}{\Gamma(\alpha k+1)}(-A_{n})^{k}a=-A_{n}u_{n}(x,t)$

for $x\in\overline{\Omega}$ and $t>0$. We can directly verify that
$u_{n}(x,0)=a(x)$ and $u_{n}(\cdot,t)\in\mathcal{D}(A)\subset
H^{1}_{0}(\Omega)$, so that $u_{n}$ given by (2.9) satisfies (1.3) with $a\in
P_{n}L^{2}(\Omega)$. The proof of Lemma 4 is complete. $\blacksquare$

Next we calculate the right-hand side of (2.9). To this end, we prove
Lemma 5.
For $j\in\mathbb{N}$ and $1\leq\ell\leq j$, we have

$None$
$\left(\frac{d}{dt^{\alpha}}\right)^{j}=\sum_{\ell=1}^{j}\theta_{j\ell}t^{\ell-j\alpha}\left(\frac{d}{dt}\right)^{\ell},$

where $\theta_{j\ell}\in\mathbb{R}$ are defined by

$None$
$\theta_{11}=\frac{1}{\alpha},\quad\theta_{j+1,\ell}=\left\\{\begin{array}[]{rl}&\frac{\theta_{j1}(1-\alpha
j)}{\alpha},\quad\ell=1,\\\ &\frac{\theta_{j\ell}(\ell-\alpha
j)}{\alpha}+\frac{C_{j,\ell-1}}{\alpha},\quad 2\leq\ell\leq j,\\\
&\frac{\theta_{jj}}{\alpha},\quad\ell=j+1\end{array}\right.$

for $j\geq 1$.
Here in the case of $j=1$ in (2.11), we neglect the possibility $2\leq\ell\leq
j$, that is,

$\theta_{2,\ell}=\left\\{\begin{array}[]{rl}&\frac{\theta_{11}(1-\alpha)}{\alpha},\quad\ell=1,\\\
&\frac{\theta_{11}}{\alpha},\quad\ell=2.\end{array}\right.$

Proof.
For $j=1$, we have
$\frac{d}{dt^{\alpha}}=\frac{dt}{dt^{\alpha}}\frac{d}{dt}=\frac{1}{\alpha}t^{1-\alpha}\frac{d}{dt}$,
and (2.10) holds with $\theta_{11}=\frac{1}{\alpha}$. Let (2.10) hold for $j$.
Then

$\displaystyle\left(\frac{d}{dt^{\alpha}}\right)^{j+1}=\frac{d}{dt^{\alpha}}\left(\sum_{\ell=1}^{j}\theta_{j\ell}t^{\ell-j\alpha}\left(\frac{d}{dt}\right)^{\ell}\right)=\frac{1}{\alpha}t^{1-\alpha}\frac{d}{dt}\left(\sum_{\ell=1}^{j}\theta_{j\ell}t^{\ell-j\alpha}\left(\frac{d}{dt}\right)^{\ell}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\alpha}t^{1-\alpha}\left(\sum_{\ell=1}^{j}\theta_{j\ell}(\ell-\alpha
j)t^{\ell-j\alpha-1}\left(\frac{d}{dt}\right)^{\ell}+\sum_{\ell=1}^{j}\theta_{j\ell}t^{\ell-j\alpha}\left(\frac{d}{dt}\right)^{\ell+1}\right)$
$None$ $=\sum_{\ell=1}^{j}\frac{\theta_{j\ell}(\ell-\alpha
j)}{\alpha}t^{\ell-(j+1)\alpha}\left(\frac{d}{dt}\right)^{\ell}+\sum_{\ell=2}^{j+1}\frac{\theta_{j\ell-1}}{\alpha}t^{\ell-(j+1)\alpha}\left(\frac{d}{dt}\right)^{\ell},$

which proves (2.10) for $j+1$, and (2.10) is seen for $j\in\mathbb{N}$ by the
induction. The recurrence formula (2.11) follows from (2.12). Thus the proof
of Lemma 5 is complete. $\blacksquare$

Now we can show a represenatio formula f $u_{n}(x,t)$.
Proposition 2.

$None$
$u_{n}(\cdot,t)=\sum_{j=0}^{\infty}\frac{(-1)^{j}}{\lambda_{n}^{j}\,j!}\left(\sum_{\ell=1}^{j}\theta_{j\ell}E_{\alpha,1-\ell}(-\lambda_{n}t^{\alpha})\right)D_{n}^{j}P_{n}a\quad\mbox{in
$C(\overline{\Omega})$ for $t\geq 0$.}$

Proof.
By (2.9) and $D_{n}^{i_{n}}=0$, the absolute convergence of the series in $j$
and $k$ allows us to exchange the order of the summation, and we have

$None$ $u_{n}(x,t)=\sum_{k=0}^{\infty}\frac{t^{k\alpha}}{\Gamma(\alpha
k+1)}\sum_{j=0}^{k}\left(\begin{array}[]{c}k\\\ j\\\
\end{array}\right)(-\lambda_{n})^{k-j}D_{n}^{j}a$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\frac{D_{n}^{j}a}{j!}\sum_{k=j}^{\infty}\frac{t^{k\alpha}}{\Gamma(\alpha
k+1)}k(k-1)\,\cdots\,(k-j+1)(-\lambda_{n})^{k-j}$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\frac{D_{n}^{j}a}{j!}t^{\alpha
j}\sum_{k=j}^{\infty}\frac{k(k-1)\,\cdots\,(k-j+1)}{\Gamma(\alpha
k+1)}(-\lambda_{n}t^{\alpha})^{k-j}=\sum_{j=0}^{\infty}\frac{D_{n}^{j}a}{j!}t^{\alpha
j}(-1)^{j}\frac{d^{j}}{dz^{j}}E_{\alpha,1}(-z)|_{z=\lambda_{n}t^{\alpha}}.$

Now we calculate $\frac{d^{j}}{dz^{j}}E_{\alpha,1}(-z)$. Since the series
$E_{\alpha,1}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(\alpha k+1)}$ is
convergent uniformly in any compact set of $z\in\mathbb{C}$, the termwise
differentiation yields

$None$
$\frac{d^{\ell}}{dt^{\ell}}E_{\alpha,1}(-t^{\alpha})=t^{-\ell}E_{\alpha,1-\ell}(-t^{\alpha}),\quad\ell\in\mathbb{N}$

by

$\Gamma(\alpha k+1)=\alpha k(\alpha k-1)\cdots(\alpha k-\ell+1)\Gamma(\alpha
k-\ell+1).$

Here we note that since $|\Gamma(1-\ell)|=\infty$ for $\ell\in\mathbb{N}$, we
can have

$E_{\alpha,1-\ell}(-t^{\alpha})=\sum_{k=0}^{\infty}\frac{(-t^{\alpha})^{k}}{\Gamma(\alpha
k+1-\ell)}=\sum_{k=1}^{\infty}\frac{(-t^{\alpha})^{k}}{\Gamma(\alpha
k+1-\ell)}.$

Consequently, in view of (2.10), we have

$\displaystyle\left(\frac{d}{dt^{\alpha}}\right)^{j}E_{\alpha,1}(-t^{\alpha})=\sum_{\ell=1}^{j}\theta_{j\ell}t^{\ell-\alpha
j}\frac{d^{\ell}}{dt^{\ell}}E_{\alpha,1}(-t^{\alpha})$ $\displaystyle=$
$\displaystyle\sum_{\ell=1}^{j}\theta_{j\ell}t^{\ell-\alpha
j}t^{-\ell}E_{\alpha,1-\ell}(-t^{\alpha})=t^{-\alpha
j}\sum_{\ell=1}^{j}\theta_{j\ell}E_{\alpha,1-\ell}(-t^{\alpha}).$

Setting $t^{\alpha}=z$, we obtain

$\left(\frac{d}{dz}\right)^{j}E_{\alpha,1}(-z)=z^{-j}\sum_{\ell=1}^{j}\theta_{j\ell}E_{\alpha,1-\ell}(-z)$

if Re $z>0$. Therefore,

$\left(\frac{d}{dz}\right)^{j}E_{\alpha,1}(-z)|_{z=\lambda_{n}t^{\alpha}}=(\lambda_{n}t^{\alpha})^{-j}\sum_{\ell=1}^{j}\theta_{j\ell}E_{\alpha,1-\ell}(-\lambda_{n}t^{\alpha}).$

Substituting this into (2.14), we obtain (2.13). The proof of Proposition 2 is
complete. $\blacksquare$

Finally we prove a key proposition for the proof of Theorem. Henceforth, we
fix $t_{0}>0$ arbitrarily, and we omit the dependency of the constants on
$t_{0}>0$ and $a$.
Proposition 3.
There exists a constant $C(n)>0$ such that

$u_{n}(x,t)=\frac{A^{-1}(P_{n}a)(x)}{\Gamma(1-\alpha)t^{\alpha}}+R_{n}(x,t),\quad
t\geq t_{0},\,x\in\overline{\Omega},$

where

$\|R_{n}(\cdot,t)\|_{C(\overline{\Omega})}\leq\frac{C(n)}{t^{2\alpha}},\quad
t\geq t_{0}.$

Proof.
For $\ell\in\mathbb{N}$, we have asymptotics

$None$
$E_{\alpha,1-\ell}(-\lambda_{n}t^{\alpha})=\frac{1}{\Gamma(1-\ell-\alpha)}\frac{1}{\lambda_{n}t^{\alpha}}+R_{\alpha,\ell}(-\lambda_{n}t^{\alpha}),\quad
t\geq t_{0},$

where there exists a constant $C_{\alpha,\ell}>0$ such that

$None$
$|R_{\alpha,\ell}(-\eta)|\leq\frac{C_{\alpha,\ell}}{\eta^{2}},\quad\eta>0.$

Henceforth constants $C_{\alpha,\ell}$, $C$, $C(n)>0$, etc. depend also on the
initial value $a$ and the order $\alpha$, but we omit the dependency otherwise
we need to specify it. As for (2.16), we refer to Section 4.5 of Chapter 4 in
Gorenflo, Kilbas, Mainardi and Rogosin [7], formula (1.8.28) (p.43) in Kilbas,
Srivastava and Trujillo [14] or Theorem 1.4 (pp.33-34) in Podlubny [25], and
see also Popov and Sedletskii [26] as recent research article.

We substitite (2.16) into (2.13):

$u_{n}(x,t)=\sum_{j=0}^{\infty}\frac{1}{\lambda_{n}^{j}}\frac{(-1)^{j}}{j!}\left(\sum_{\ell=1}^{j}\left(\frac{\theta_{j\ell}}{\Gamma(1-\ell-\alpha)}\frac{1}{\lambda_{n}t^{\alpha}}+\theta_{j\ell}R_{\alpha,\ell}(-\lambda_{n}t^{\alpha})\right)\right)D_{n}^{j}P_{n}a$
$None$
$=:\sum_{j=0}^{\infty}\frac{1}{\lambda_{n}^{j+1}}\left(\frac{(-1)^{j}}{j!}\left(\sum_{\ell=1}^{j}\frac{\theta_{j\ell}}{\Gamma(1-\ell-\alpha)}\right)\right)\frac{1}{t^{\alpha}}D_{n}^{j}P_{n}a+S_{n}(t^{\alpha})(x).$

We note that by $D_{n}^{i_{n}}=0$ the series are finite. Here, by (2.16), we
can find a constant $C_{1}(n)>0$ such that

$None$
$\|S_{n}(t^{\alpha})(\cdot)\|_{C(\overline{\Omega})}\leq\left(\sum_{j=0}^{\infty}\left|\frac{(-1)^{j}}{\lambda_{n}^{j}j!}\right|\sum_{\ell=1}^{j}\frac{|\theta_{j\ell}|C_{\alpha,\ell}}{\lambda_{n}^{2}t^{2\alpha}}\right)\|D_{n}^{j}P_{n}a\|_{C(\overline{\Omega})}\leq\frac{C_{1}(n)}{t^{2\alpha}}$

for all $t\geq t_{0}$.

We calculate

$\Phi_{j}:=\frac{(-1)^{j}}{j!}\sum_{\ell=1}^{j}\frac{\theta_{j\ell}}{\Gamma(1-\ell-\alpha)},\quad
j\in\mathbb{N}.$

By (2.11) and $\Gamma(1-\alpha)=-\alpha\Gamma(-\alpha)$, we see

$None$
$\Phi_{1}=\frac{(-1)^{1}}{1!}\frac{\theta_{11}}{\Gamma(-\alpha)}=\frac{-1}{\alpha\Gamma(-\alpha)}=\frac{1}{\Gamma(1-\alpha)}.$

Now we will verify $\Phi_{j}=\Phi_{1}$ for each $j\in\mathbb{N}$. By (2.11) we
obtain

$\displaystyle\Phi_{j+1}=\frac{(-1)^{j+1}}{(j+1)!}\sum_{\ell=1}^{j+1}\frac{\theta_{j+1,\ell}}{\Gamma(1-\ell-\alpha)}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j+1}}{(j+1)!}\biggl{\\{}\frac{\theta_{j1}(1-\alpha
j)}{\alpha}\frac{1}{\Gamma(-\alpha)}+\sum_{\ell=2}^{j}\frac{1}{\Gamma(1-\ell-\alpha)}\left(\frac{\theta_{j\ell}(\ell-\alpha
j)}{\alpha}+\frac{\theta_{j,\ell-1}}{\alpha}\right)$ $\displaystyle+$
$\displaystyle\frac{\theta_{jj}}{\alpha}\frac{1}{\Gamma(-j-\alpha)}\biggr{\\}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j+1}}{(j+1)!}\biggl{\\{}\frac{\theta_{j1}(1-\alpha
j)}{\alpha\Gamma(-\alpha)}+\sum_{\ell=2}^{j}\frac{\theta_{j\ell}(\ell-\alpha
j)}{\alpha\Gamma(1-\ell-\alpha)}$ $\displaystyle+$
$\displaystyle\sum_{\ell=1}^{j-1}\frac{\theta_{j\ell}}{\alpha\Gamma(-\ell-\alpha)}+\frac{\theta_{jj}}{\alpha}\frac{1}{\Gamma(-j-\alpha)}\biggr{\\}}.$

Here we used

$\sum_{\ell=2}^{j}\frac{\theta_{j,\ell-1}}{\alpha\Gamma(1-\ell-\alpha)}=\sum_{\ell=1}^{j-1}\frac{\theta_{j\ell}}{\alpha\Gamma(-\ell-\alpha)}.$

Using $\Gamma(1-\eta)=-\eta\Gamma(-\eta)$ for
$\eta\not\in\mathbb{N}\cup\\{0\\}$, we obtain
$(-\ell-\alpha)\Gamma(-\ell-\alpha)=\Gamma(1-\ell-\alpha)$ for
$\ell=1,2,....,j$, and so

$\frac{1}{\alpha\Gamma(-\ell-\alpha)}=-\frac{\ell+\alpha}{\alpha\Gamma(1-\ell-\alpha)},$

which means

$\frac{\ell-\alpha
j}{\alpha\Gamma(1-\ell-\alpha)}+\frac{1}{\alpha\Gamma(-\ell-\alpha)}=\frac{1}{\alpha\Gamma(1-\ell-\alpha)}(\ell-\alpha
j-\ell-\alpha)=\frac{-(j+1)}{\Gamma(1-\ell-\alpha)}.$

Therefore,

$\displaystyle\Phi_{j+1}=\frac{(-1)^{j+1}}{(j+1)!}\biggl{\\{}\left(\frac{\theta_{j1}(1-\alpha
j)}{\alpha\Gamma(-\alpha)}+\frac{\theta_{j1}}{\alpha\Gamma(-1-\alpha)}\right)$
$\displaystyle+$
$\displaystyle\sum_{\ell=2}^{j-1}\left(\frac{\theta_{j\ell}(\ell-\alpha
j)}{\alpha\Gamma(1-\ell-\alpha)}+\frac{\theta_{j\ell}}{\alpha\Gamma(-\ell-\alpha)}\right)+\left(\frac{\theta_{jj}(j-\alpha
j)}{\alpha\Gamma(1-j-\alpha)}+\frac{\theta_{jj}}{\alpha\Gamma(-j-\alpha)}\right)\biggr{\\}}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j+1}}{(j+1)!}\sum_{\ell=1}^{j}\left(\frac{\ell-\alpha
j}{\alpha\Gamma(1-\ell-\alpha)}+\frac{1}{\alpha\Gamma(-\ell-\alpha)}\right)\theta_{j\ell}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{j+1}}{(j+1)!}(-1)(j+1)\sum_{\ell=1}^{j}\frac{\theta_{j\ell}}{\Gamma(1-\ell-\alpha)}=\frac{(-1)^{j}}{j!}\sum_{\ell=1}^{j}\frac{\theta_{j\ell}}{\Gamma(1-\ell-\alpha)}=\Phi_{j}$

for each $j\in\mathbb{N}$. Thus (2.20) yields
$\Phi_{j}=\Phi_{1}=\frac{1}{\Gamma(1-\alpha)}$ for all $j\in\mathbb{N}$.

Hence, applying also (2.7), we obtain

$\displaystyle\sum_{j=0}^{\infty}\frac{1}{\lambda_{n}^{j+1}}\left(\frac{(-1)^{j}}{j!}\sum_{\ell=1}^{j}\frac{\theta_{j\ell}}{\Gamma(1-\ell-\alpha)}\right)\frac{1}{t^{\alpha}}D_{n}^{j}P_{n}a=\sum_{j=0}^{\infty}\frac{1}{\lambda_{n}^{j+1}}\Phi_{j}\frac{1}{t^{\alpha}}D_{n}^{j}P_{n}a$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(1-\alpha)}\left(\sum_{j=0}^{i_{n}-1}\frac{1}{\lambda_{n}^{j+1}}D_{n}^{j}P_{n}a\right)\frac{1}{t^{\alpha}}=\frac{1}{\Gamma(1-\alpha)}A^{-1}(P_{n}a)(x)\frac{1}{t^{\alpha}}.$

Combining (2.19), we complete the proof of Proposition 3. $\blacksquare$

##  3\. Completion of Proof of Theorem

First Step.

We recall that $u$ and $\widetilde{u}$ are the solutions to (1.3) and (1.7)
with the initial values $a$ and $\widetilde{a}$ respectively. For
$a\in\mathcal{D}(A)$, by Lemma 3, for each $n\in\mathbb{N}$, we can find
$a_{N}\in\sum_{k=1}^{N}P_{k}L^{2}(\Omega)$ such that

$None$ $\lim_{N\to\infty}\rho_{N}=0,$

where $\rho_{N}:=\|A(a-a_{N})\|_{L^{2}(\Omega)}$. Let $v_{N}$ and $w_{N}$
satisfy

$None$
$\left\\{\begin{array}[]{rl}&\partial_{t}^{\alpha}v_{N}=-Av_{N}\quad\mbox{in
$\Omega\times(0,\infty)$},\\\ &v_{N}|_{\partial\Omega\times(0,\infty)}=0,\\\
&v_{N}(x,0)=a_{N}(x),\quad x\in\Omega\end{array}\right.$

and

$None$
$\left\\{\begin{array}[]{rl}&\partial_{t}^{\alpha}w_{N}=-Aw_{N}\quad\mbox{in
$\Omega\times(0,\infty)$},\\\ &w_{N}|_{\partial\Omega\times(0,\infty)}=0,\\\
&w_{N}(x,0)=a(x)-a_{N}(x),\quad x\in\Omega.\end{array}\right.$

Then $u=v_{N}+w_{N}$ in $\Omega\times(0,\infty)$.

First applying Proposition 1 to (3.3) and setting $W_{N}(t)=w_{N}(x_{0},t)$,
we obtain

$None$ $|W_{N}(t)|\leq\frac{C}{t^{\alpha}}\rho_{N},\quad t\geq t_{0}.$

Since $v_{N}=\sum_{n=1}^{N}u_{n}$ by the uniqueness of solution to (3.2), the
application of Proposition 3 to $u_{n}$ yields

$\displaystyle
v_{N}(x_{0},t)=\sum_{n=1}^{N}u_{n}(x_{0},t)=\frac{1}{\Gamma(1-\alpha)t^{\alpha}}\sum_{n=1}^{N}(A^{-1}(P_{n}a))(x_{0})+\sum_{n=1}^{N}R_{n}(x_{0},t)$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(1-\alpha)t^{\alpha}}(A^{-1}a_{N})(x_{0})+\sum_{n=1}^{N}R_{n}(x_{0},t),\quad
t\geq t_{0}.$

Therefore, we can find a constant $C_{0}=C_{0}(N)>0$ and a function $S_{N}(t)$
such that

$None$
$v_{N}(x_{0},t)=\frac{1}{\Gamma(1-\alpha)t^{\alpha}}(A^{-1}a_{N})(x_{0})+S_{N}(t),\quad
t\geq t_{0},$

where

$None$ $|S_{N}(t)|\leq\frac{C_{0}(N)}{t^{2\alpha}},\quad t\geq t_{0}.$

Consequently,

$None$
$u(x_{0},t)=\frac{1}{\Gamma(1-\alpha)t^{\alpha}}(A^{-1}a_{N})(x_{0})+S_{N}(t)+W_{N}(t),\quad
t\geq t_{0}.$

We can similarly argue for $\widetilde{A}$ to see that there exist
$\widetilde{a_{N}}\in\mathcal{D}(\widetilde{A})$ and functions
$\widetilde{S_{N}}$ and $\widetilde{W_{N}}$ for $N\in\mathbb{N}$ such that

$\widetilde{\rho_{N}}:=\|\widetilde{A}(\widetilde{a}-\widetilde{a_{N}})\|_{L^{2}(\widetilde{\Omega})}\longrightarrow
0\quad\mbox{as $N\to\infty$},$ $None$
$\widetilde{u}(x_{0},t)=\frac{1}{\Gamma(1-\beta)t^{\beta}}(\widetilde{A}^{-1}\widetilde{a_{N}})(x_{0})+\widetilde{S_{N}}(t)+\widetilde{W_{N}}(t),\quad
t\geq t_{0}$

and

$None$
$|\widetilde{S_{N}}(t)|\leq\frac{C_{0}(N)}{t^{2\beta}},\quad|\widetilde{W_{N}}(t)|\leq\frac{C}{t^{\beta}}\widetilde{\rho_{N}},\quad
t\geq t_{0}.$

Second Step.

By Lemma 2, from $u(x_{0},t)=\widetilde{u}(x_{0},t)$ for $0<t<T$, we derive

$u(x_{0},t)=\widetilde{u}(x_{0},t)\quad\mbox{for all $t>0$.}$

Therefore, (3.7) and (3.8) yield

$\displaystyle\frac{1}{\Gamma(1-\alpha)t^{\alpha}}(A^{-1}a_{N})(x_{0})+S_{N}(t)+W_{N}(t)$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(1-\beta)t^{\beta}}(\widetilde{A}^{-1}\widetilde{a_{N}})(x_{0})+\widetilde{S_{N}}(t)+\widetilde{W_{N}}(t),\quad
t\geq t_{0}.$

Assume that $\alpha<\beta$. Then we multiply by $t^{\alpha}$, we have

$\frac{1}{\Gamma(1-\alpha)}(A^{-1}a_{N})(x_{0})+t^{\alpha}S_{N}(t)+t^{\alpha}W_{N}(t)$
$None$
$=\frac{1}{\Gamma(1-\beta)t^{\beta-\alpha}}(\widetilde{A}^{-1}\widetilde{a_{N}})(x_{0})+t^{\alpha}\widetilde{S_{N}}(t)+t^{\alpha}\widetilde{W_{N}}(t),\quad
t\geq t_{0}$

for all $N\in\mathbb{N}$ and all $t\geq t_{0}$.

By (3.4), (3.6) and (3.9), for each fixed $N\in\mathbb{N}$, we have

$None$
$\left\\{\begin{array}[]{rl}&\lim_{t\to\infty}|t^{\alpha}S_{N}(t)|\leq\lim_{t\to\infty}\frac{C_{0}(N)}{t^{\alpha}}=0,\\\
&\lim_{t\to\infty}|t^{\alpha}\widetilde{S_{N}}(t)|\leq\lim_{t\to\infty}\frac{C_{0}(N)}{t^{2\beta-\alpha}}=0\end{array}\right.$

by $\beta>\alpha$. Moreover, whenever we fix $N\in\mathbb{N}$ arbitrarily, by
(3.4) and (3.9) we see

$None$ $|t^{\alpha}W_{N}(t)|\leq
C\rho_{N},\quad|t^{\alpha}\widetilde{W_{N}}(t)|\leq\frac{C\widetilde{\rho_{N}}}{t^{\beta-\alpha}}\longrightarrow
0\quad\mbox{as $t\to\infty$}$

and

$None$
$\lim_{t\to\infty}\frac{1}{\Gamma(1-\beta)}\frac{1}{t^{\beta-\alpha}}(\widetilde{A}^{-1}\widetilde{a_{N}})(x_{0})=0$

by $\beta>\alpha$.

Therefore, applying (3.11) - (3.13) in (3.10) and letting $t\to\infty$, we
obtain

$None$ $\left|\frac{1}{\Gamma(1-\alpha)}(A^{-1}a_{N})(x_{0})\right|\leq
C\rho_{N}.$

Since $a_{N}\longrightarrow a$ in $L^{2}(\Omega)$ as $N\to\infty$, we see that
$A^{-1}a_{N}\longrightarrow A^{-1}a$ in $\mathcal{D}(A)$, and the Sobolev
embedding implies

$\lim_{N\to\infty}\|A^{-1}a_{N}-A^{-1}a\|_{C(\overline{\Omega})}=0,$

that is,

$\lim_{N\to\infty}A^{-1}a_{N}(x_{0})=A^{-1}a(x_{0}).$

Hence, letting $N\to\infty$ in (3.14), we see

$A^{-1}a(x_{0})=0.$

We set $f:=A^{-1}a$ in $\Omega$. Then $Af=a$ in $\Omega$. By (1.10), without
loss of generality, we can assume that $a\leq 0$ on $\overline{\Omega}$. In
view of $c\leq 0$ from (1.8), by noting that $f|_{\partial\Omega}=0$, the weak
maximum principle (e.g., Theorem 3.1 (p.32) in Gilbarg and Trudinger [6])
yields $f(x)\leq 0$ for $x\in\overline{\Omega}$.

Since $f\leq 0$ on $\overline{\Omega}$ and $f(x_{0})=A^{-1}a(x_{0})=0$, we see
that $f$ attains the maximum $0$ at an interior point $x_{0}\in\Omega$. In
view of $(-A)f\geq 0$ in $\Omega$, we can apply the strong maximum principle
(e.g., Theorem 3.5 (p.35) in [6]) to conclude that $f(x)$ is a constant
function. Since $f\in\mathcal{D}(A)$, we see $f|_{\partial\Omega}=0$, so that
$f=0$ in $\Omega$. Then $u=0$ in $\Omega\times(0,\infty)$. In terms of (3.8)
and $\widetilde{u}(x_{0},t)=u(x_{0},t)=0$ for $t>0$, we have

$\frac{1}{\Gamma(1-\beta)t^{\beta}}(\widetilde{A}^{-1}\widetilde{a_{N}})(x_{0})+\widetilde{S_{N}}(t)+\widetilde{W_{N}}(t)=0,\quad
t\geq t_{0}.$

Multiplying with $t^{\beta}$ and letting $t\to\infty$, we similarly obtain

$\left|\frac{1}{\Gamma(1-\beta)}(\widetilde{A}^{-1}\widetilde{a_{N}})(x_{0})\right|\leq
C\widetilde{\rho_{N}}\quad\mbox{for all $N\in\mathbb{N}$.}$

Letting $N\to\infty$, we see $\widetilde{A}^{-1}\widetilde{a}(x_{0})=0$. By
the same way as for $a$, the weak and the strong maximum principles similarly
yield $\widetilde{a}=0$ in $\Omega$. This means that $a=0$ in $\Omega$ and
$\widetilde{a}=0$ in $\widetilde{\Omega}$, which is a contradiction for
(1.10). Therefore, $\alpha<\beta$ is impossible. Similarly we can prove that
$\alpha>\beta$ is impossible. Thus the proof of Theorem is complete.
$\blacksquare$

##  4\. Appendix

4.1. Proof of (1.9).

It sufficient to prove only for $A$. Let $\lambda\in\mathbb{C}$ be an
eigenvalue of $A$: $A\varphi=\lambda\varphi$ and $\varphi\not\equiv 0$ in
$\Omega$. That is,

$None$
$\left\\{\begin{array}[]{rl}&-\sum_{i,j=1}^{d}\partial_{i}(a_{ij}(x)\partial_{j}\varphi(x))-\sum_{j=1}^{d}b_{j}(x)\partial_{j}\varphi-c(x)\varphi=\lambda\varphi\quad\mbox{in
$\Omega$},\\\ &\varphi|_{\partial\Omega}=0.\end{array}\right.$

Henceforth let $\overline{\lambda}$ denote the complex conjugate of
$\lambda\in\mathbb{C}$. Since $a_{ij}$, $b_{j}$ and $c$ are real-valued, we
have

$None$
$-\sum_{i,j=1}^{d}\partial_{i}(a_{ij}(x)\partial_{j}\overline{\varphi}(x))-\sum_{j=1}^{d}b_{j}(x)\partial_{j}\overline{\varphi}-c(x)\overline{\varphi}=\overline{\lambda}\overline{\varphi}\quad\mbox{in
$\Omega$}.$

Multiplying (4.1) and (4.2) with $\overline{\varphi}$ and $\varphi$
respectively, integrating by parts in $\Omega$ and adding, we obtain

$\displaystyle\int_{\Omega}\sum_{i,j=1}^{d}a_{ij}((\partial_{i}\varphi)\overline{\partial_{j}\varphi}+(\overline{\partial_{i}\varphi})\partial_{j}\varphi)dx-\int_{\Omega}b_{j}((\partial_{j}\varphi)\overline{\varphi}+(\overline{\partial_{j}\varphi})\varphi)dx$
$\displaystyle-$
$\displaystyle\int_{\Omega}2c|\varphi|^{2}dx=(\lambda+\overline{\lambda})\int_{\Omega}|\varphi|^{2}dx.$

On the other hand,

$((\partial_{i}\varphi)\overline{\partial_{j}\varphi}+\overline{\partial_{i}\varphi}\partial_{j}\varphi)=2(\mbox{Re}\,(\partial_{i}\varphi)\mbox{Re}\,(\partial_{j}\varphi)+\mbox{Im}\,(\partial_{i}\varphi)\mbox{Im}\,(\partial_{j}\varphi))$

and so (1.2) yields

$\displaystyle\sum_{i,j=1}^{d}a_{ij}((\partial_{i}\varphi)\overline{\partial_{j}\varphi}+\overline{\partial_{i}\varphi}\partial_{j}\varphi)$
$\displaystyle=$ $\displaystyle
2\left(\sum_{i,j=1}^{d}a_{ij}\mbox{Re}\,(\partial_{i}\varphi)\mbox{Re}\,(\partial_{j}\varphi)+\sum_{i,j=1}^{d}a_{ij}\mbox{Im}\,(\partial_{i}\varphi)\mbox{Im}\,(\partial_{j}\varphi)\right)$
$\displaystyle\geq$ $\displaystyle
2\sigma(a_{ij})\sum_{i,j=1}^{d}(|\mbox{Re}\,(\partial_{i}\varphi)|^{2}+|\mbox{Im}\,(\partial_{i}\varphi)|^{2})=2\sigma|\nabla\varphi|^{2}.$

Moreover,

$\displaystyle-\int_{\Omega}\sum_{j=1}^{d}b_{j}((\partial_{j}\varphi)\overline{\varphi}+\overline{(\partial_{j}\varphi)}\varphi)dx$
$\displaystyle=$
$\displaystyle-\int_{\Omega}\sum_{j=1}^{d}b_{j}\partial_{j}(|\varphi|^{2})dx=\int_{\Omega}\left(\sum_{j=1}^{d}\partial_{j}b_{j}\right)|\varphi|^{2}dx.$

Hence, using also (1.4), we have

$\displaystyle 2\sigma
C(\Omega)\int_{\Omega}|\varphi|^{2}dx+\int_{\Omega}\left(\left(\sum_{j=1}^{d}\partial_{j}b_{j}\right)-2c(x)\right)|\varphi|^{2}dx$
$\displaystyle\leq$ $\displaystyle
2\sigma\int_{\Omega}|\nabla\varphi|^{2}dx+\int_{\Omega}\left(\sum_{j=1}^{d}\partial_{j}b_{j}\right)|\varphi|^{2}dx-2\int_{\Omega}c(x)|\varphi(x)|^{2}dx$
$\displaystyle\leq$
$\displaystyle(\lambda+\overline{\lambda})\int_{\Omega}|\varphi|^{2}dx=2(\mbox{Re}\,\lambda)\int_{\Omega}|\varphi|^{2}dx.$

In view of (1.8), we see

$2\sigma C(\Omega)+\sum_{j=1}^{d}\partial_{j}b_{j}(x)-2c(x)>0\quad\mbox{for
all $x\in\overline{\Omega}$}.$

Thus Re $\lambda>0$ follows and we complete the proof of (1.9). $\blacksquare$

4.2. Proof of Proposition 1.

For the case of symmetric $A$, in terms of the eigenfunction expansions by the
Mittag-Leffler functions, we can directly prove the proposition (e.g.,
Sakamoto and Yamamoto [28]). However, for non-symmetric $A$, we have to take
other way. We make use of the argument by Vergara and Zacher [31] which is the
first work proving Proposition 1 (i) for a case where the eigenfunction
expansion does not work, that is, $A$ is symmetric but the coefficients are
time dependent. For the proof of Proposition 1, relying on Chapter 5 in [16]
which modifies [31], we here give a sketch of the proof of Proposition 1. We
can mostly repeat the argument in [16] and so we describe the main
differences.

For the proof, we have to return to the construction of the solution $u$ to
(1.3) by the Galerkin method. More precisely, let
$\\{\psi_{k}\\}_{k\in\mathbb{N}}$ be an orthonormal basis in $L^{2}(\Omega)$
which consists of all the eigenfunctions of $-\Delta$ with the zero Dirichlet
boundary condition. We construct approximating solutions to $u$ in the form of

$None$ $u_{N}(x,t):=\sum_{k=1}^{N}p_{k}^{N}(t)\psi_{k}(x),\quad
N\in\mathbb{N}$

such that

$None$ $u_{N}(x,0)=\sum_{k=1}^{N}(u,\psi_{k})_{L^{2}(\Omega)}\psi_{k}(x)$

and

$(\partial_{t}^{\alpha}u_{N}+Au_{N},\,\psi_{\ell})_{L^{2}(\Omega)}=0,\quad
1\leq\ell\leq N,\quad t>0,$

which is rewritten as

$\int_{\Omega}\partial_{t}^{\alpha}u_{N}(x,t)\psi_{\ell}(x)dx+\int_{\Omega}\sum_{i,j=1}^{d}a_{ij}(x)(\partial_{j}u_{N})(x,t)\partial_{i}\psi_{\ell}(x)dx$
$None$
$-\int_{\Omega}\sum_{j=1}^{d}b_{j}(\partial_{j}u_{N})\psi_{\ell}(x)dx-\int_{\Omega}c(x)u_{N}(x,t)\psi_{\ell}(x)dx=0.$

Equation (4.4) is equivalent to a system of time-fractional ordinary
differential equations in $\\{p_{k}^{N}(t)\\}_{1\leq k\leq N}$. Such
$p_{k}^{N}$ exist uniquely and $u_{N}$ converges to the solution $u$ weakly in
$L^{2}(0,T;H^{2}(\Omega))$. We can refer to e.g., Chapter 4 in [16], and we
omit the details.

Multiplying (4.5) with $p_{\ell}^{N}(t)$ and adding over $\ell=1,...,N$, and
integrating by parts the third term on the left-hand side, we obtain

$\displaystyle\int_{\Omega}(\partial_{t}^{\alpha}u_{N}(x,t))u_{N}(x,t)dx+\int_{\Omega}\sum_{i,j=1}^{d}a_{ij}(x)(\partial_{j}u_{N})(x,t)\partial_{i}u_{N}(x,t)dx$
$\displaystyle+$
$\displaystyle\frac{1}{2}\int_{\Omega}\left(\sum_{j=1}^{d}\partial_{j}b_{j}\right)|u_{N}|^{2}dx-\int_{\Omega}c(x)|u_{N}(x,t)|^{2}dx=0,\quad
t>0.$

By (1.2) and (1.4), we have

$\displaystyle\int_{\Omega}(\partial_{t}^{\alpha}u_{N}(x,t))u_{N}(x,t)dx+C(\Omega)\sigma(a_{ij})\int_{\Omega}|u_{N}(x,t)|^{2}dx$
$\displaystyle+$
$\displaystyle\frac{1}{2}\int_{\Omega}\left(\sum_{j=1}^{d}\partial_{j}b_{j}\right)|u_{N}|^{2}dx-\int_{\Omega}c(x)|u_{N}(x,t)|^{2}dx\leq
0,\quad t>0.$

Applying (1.8), we can choose a constant $\mu_{0}>0$ such that

$\int_{\Omega}(\partial_{t}^{\alpha}u_{N}(x,t))u_{N}(x,t)dx+\mu_{0}\|u_{N}(\cdot,t)\|^{2}_{L^{2}(\Omega)}\leq
0,\quad t>0\quad\mbox{for all $N\in\mathbb{N}$.}$

Hence Lemma 5.1 in [16] yields

$None$
$\|u_{N}(\cdot,t)\|_{L^{2}(\Omega)}\partial_{t}^{\alpha}\|u_{N}(\cdot,t)\|_{L^{2}(\Omega)}\leq-\mu_{0}\|u_{N}(\cdot,t)\|^{2}_{L^{2}(\Omega)}\leq
0,\quad t>0.$

Then, we can formally argue as follows. By setting
$d_{N}(t):=\|u_{N}(\cdot,t)\|_{L^{2}(\Omega)}$ and assuming that $d_{N}(t)\neq
0$ for all $t>0$, inequality (4.6) yields

$None$
$\left\\{\begin{array}[]{rl}&\partial_{t}^{\alpha}d_{N}(t)\leq-\mu_{0}d_{N}(t),\quad
t>0,\\\
&d_{N}(0)=\left\|\sum_{k=1}^{N}(a,\psi_{k})_{L^{2}(\Omega)}\psi_{k}\right\|_{L^{2}(\Omega)}.\end{array}\right.$

Noting that

$\partial_{t}^{\alpha}E_{\alpha,1}(-\mu_{0}t^{\alpha})=-\mu_{0}E_{\alpha,1}(-\mu_{0}t^{\alpha}),\quad
E_{\alpha,1}(0)=1,$

we apply the comparison principle (e.g., Luchko and Yamamoto [21]) to obtain

$d_{N}(t)\leq d_{N}(0)E_{\alpha,1}(-\mu_{0}t^{\alpha}),\quad t>0.$

Therefore, for all $N\in\mathbb{N}$, we have

$d_{N}(t)\leq\|a\|_{L^{2}(\Omega)}E_{\alpha,1}(-\mu_{0}t^{\alpha})\leq\frac{C\|a\|_{L^{(}\Omega)}}{t^{\alpha}},\quad
t>0,$

where we use

$|E_{\alpha,1}(-\mu_{0}t^{\alpha})|\leq\frac{C}{1+\mu_{0}t^{\alpha}},\quad
t>0$

(e.g., Theorem 1.6 (p.35) in [25]). This is an outline and for example, but we
need to justify in deriving (4.7) by dividing with
$\|u_{N}(\cdot,t)\|_{L^{2}(\Omega)}$, because it may vanish at some $t$. Such
details are still the same as the proof of Theorem 5.1 of Chapter 5 in [16],
and so we omit.

Finally we have to prove part (ii). Let $d=1,2,3$. By $a\in\mathcal{D}(A)$, we
consider

$\left\\{\begin{array}[]{rl}&\partial_{t}^{\alpha}U=-AU(x,t),\quad
x\in\Omega,\,t>0,\\\ &U|_{\partial\Omega\times(0,\infty)}=0,\\\
&U(x,0)=-Aa(x),\quad x\in\Omega.\end{array}\right.$

There exists a unique solution $U\in C([0,\infty);L^{2}(\Omega))$ and applying
part (i) to $U$, we have

$\|U(\cdot,t)\|_{L^{2}(\Omega)}\leq\frac{C}{t^{\alpha}}\|Aa\|_{L^{2}(\Omega)},\quad
t>0.$

Moreover the uniqueness of the solution yields $U=\partial_{t}^{\alpha}u$ in
$\Omega\times(0,\infty)$. Therefore $U=-Au$ in $\Omega\times(0,\infty)$, and
so

$\|Au(\cdot,t)\|_{L^{2}(\Omega)}\leq\frac{C}{t^{\alpha}}\|Aa\|_{L^{2}(\Omega)},\quad
t>0.$

Since $d=1,2,3$, the Sobolev embedding implies

$\|u(\cdot,t)\|_{C(\overline{\Omega})}\leq\frac{C}{t^{\alpha}}\|Aa\|_{L^{2}(\Omega)},\quad
t>0.$

Thus the proof of Proposition 1 (ii) is complete. $\blacksquare$

## Acknowledgment

The author was supported by Grant-in-Aid for Scientific Research (S) 15H05740
and Grant-in-Aid (A) 20H00117 of Japan Society for the Promotion of Science
and by The National Natural Science Foundation of China (no. 11771270,
91730303). This paper has been supported by the RUDN University Strategic
Academic Leadership Program.

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