# Determination of source terms in diffusion and wave equations by
observations after incidents: uniqueness and stability

Jin Cheng111School of Mathematical Sciences, Fudan University, Shanghai,
China<EMAIL_ADDRESS><EMAIL_ADDRESS>Shuai Lu11footnotemark: 1,
Masahiro Yamamoto 222Department of Mathematical Sciences, The University of
Tokyo, Komaba, Meguro, Tokyo 153, Japan<EMAIL_ADDRESS>333Honorary
Member of Academy of Romanian Scientists, Ilfov, nr. 3, Bucuresti, Romania
444Correspondence member of Accademia Peloritana dei Pericolanti  555Peoples’
Friendship University of Russia (RUDN University) 6 Miklukho-Maklaya St,
Moscow, 117198, Russian Federation

###### Abstract

We consider a diffusion and a wave equations:

$\partial_{t}^{k}u(x,t)=\Delta u(x,t)+\mu(t)f(x),\quad x\in\Omega,\,t>0,\quad
k=1,2$

with the zero initial and boundary conditions, where
$\Omega\subset\mathbb{R}^{d}$ is a bounded domain. We establish uniqueness
and/or stability results for inverse problems of

  * •

determining $\mu(t)$, $0<t<T$ with given $f(x)$.

  * •

determining $f(x)$, $x\in\Omega$ with given $\mu(t)$

by data of $u$: $u(x_{0},\cdot)$ with fixed point $x_{0}\in\Omega$ or Neumann
data on subboundary over time interval. In our inverse problems, data are
taken over time interval $T_{1}<t<T_{1}$, by assuming that $T<T_{1}<T_{2}$ and
$\mu(t)=0$ for $t\geq T$, which means that the source stops to be active after
the time $T$ and the observations are started only after $T$. This assumption
is practical by such a posteriori data after incidents, although inverse
problems had been well studied in the case of $T=0$. We establish the non-
uniqueness, the uniqueness and conditional stability for a diffusion and a
wave equations. The proofs are based on eigenfunction expansions of the
solutions $u(x,t)$, and we rely on various knowledge of the generalized
Weierstrass theorem on polynomial approximation, almost periodic functions,
Carleman estimate, non-harmonic Fourier series.
AMS subject classifications. 35R30, 35R25, 35K20, 35L20

##  1 Introduction

In this article, we consider initial-boundary value problems for a diffusion
and a wave equations:

$\left\\{\begin{array}[]{rl}&\partial_{t}u(x,t)=\Delta u(x,t)+\mu(t)f(x),\quad
x\in\Omega,\,t>0,\\\ &u(x,0)=0,\quad x\in\Omega,\\\ &u(x,t)=0,\quad
x\in\partial\Omega,\,t>0.\end{array}\right.$ $None$
$\left\\{\begin{array}[]{rl}&\partial_{t}^{2}u(x,t)=\Delta
u(x,t)+\mu(t)f(x),\quad x\in\Omega,\,t>0,\\\
&u(x,0)=\partial_{t}u(x,0)=0,\quad x\in\Omega,\\\ &u(x,t)=0,\quad
x\in\partial\Omega,\,t>0.\end{array}\right.$ $None$

Here and henceforth $\Omega\subset\mathbb{R}^{d}$ is a bounded domain with
smooth boundary $\partial\Omega$ and we set
$x=(x_{1},...,x_{d})\in\mathbb{R}^{d}$, $\partial_{j}=\frac{\partial}{\partial
x_{j}}$, and $\Delta=\sum_{j=1}^{d}\partial_{j}^{2}$. Let $\nu=\nu(x)$ be the
unit outward normal vector to $\partial\Omega$ and let $\partial_{\nu}u=\nabla
u\cdot\nu$. We mainly consider the zero Dirichlet boundary condition and can
treat the Neumann boundary condition similarly but we omit the details.
Moreover we can consider the inverse problems for (1.1) and (1.2) where
$\Delta$ is replaced by a suitable elliptic operator with time independent
coefficients, but for simplicity, we mainly argue for $\Delta$.

The source is assumed to be represented in the form of $\mu(t)f(x)$ where
$\mu(t)$ and $f(x)$ describe changes in the time $t$ and the spacial variable
$x$ respectively. Such a form of separation of variables is frequently used in
modelling diffusion and wave phenomena.

The unique existence of solutions to (1.1) and (1.2) are standard results
(e.g., Evans [9], Lions and Magenes [17], Pazy [18]), but we need more
regularity of solutions. We sum up these results as Lemmas 1 and 2. We
arbitrarily fix $T_{0}>0$.
Lemma 1.
Let $f\in C^{\infty}_{0}(\Omega)$ and $\mu\in H^{1}(0,T_{0})$.
(i) To (1.1), there exists a unique solution $u\in
C([0,T_{0}];H^{2}(\Omega)\cap H^{1}_{0}(\Omega))\cap
C^{1}([0,T_{0}];L^{2}(\Omega))$ and we can choose a constant $C>0$, dependent
on $f$, such that

$\|u\|_{C(\Omega\times[0,T_{0}])}\leq C\|\mu\|_{L^{2}(0,T_{0})}.$ $None$

(ii) To (1.2), there exists a unique solution $u\in
C([0,T_{0}];H^{2}(\Omega))\cap C^{1}([0,T_{0}];H^{1}_{0}(\Omega))\cap
C^{2}([0,T_{0}];L^{2}(\Omega))\cap C(\overline{\Omega}\times[0,T_{0}])$ such
that $\partial_{\nu}u\in H^{1}(0,T_{0};L^{2}(\partial\Omega))$ and (1.3)
holds.
Lemma 2.
Let $f\in L^{2}(\Omega)$ and $\mu\in C^{1}[0,T_{0}]$.
(i) To (1.1), there exists a unique solution $u\in
H^{1}(0,T_{0};L^{2}(\Omega))\cap L^{2}(0,T_{0};H^{1}_{0}(\Omega))$.
(ii) To (1.2), there exists a unique solution $u\in
H^{1}(0,T_{0};L^{2}(\Omega))\cap L^{2}(0,T_{0};H^{1}_{0}(\Omega))$ such that
$\partial_{\nu}u\in H^{1}(0,T_{0};L^{2}(\partial\Omega))$.

Here we do not aim at the best possible regularity, and for completeness the
proofs of the lemmata are given in Appendix II.

Throughout the article, we assume that $\mu\in L^{2}_{loc}(\mathbb{R})$
satisfies

$\mu(t)=0\quad\mbox{if $t>T$}$

with some constant $T>0$. This means that the a diffusion source for (1.1) and
an external force for (1.2) continue to be activated only before the moment
$T>0$.

In our inverse problems, the measurements can be started after the time $T>0$,
and we are required to determine $\mu(t)$, $0<t<T$ or $f(x)$, $x\in\Omega$ of
the source term. For example, in the case where the explosion of some
equipments such as nuclear power plant, causes diffusion of contaminants or
dangerous substances, any measurements starting at $t=0$ are not realistic.
Inverse source problems of determining $\mu$ or $f$ are well studied if the
measurements of data are started at $t=0$, but to the best knowledge of the
authors, there are no publications on mathematical analysis by data starting
after the time $T>0$.

The main purpose of this article is to establish the uniqueness and the
stability for inverse source problems for (1.1) and (1.2) by data from the
time when the source stopped to be active.

Now we formulate several kinds of inverse source problems and state our main
results.
§1.1. Determination of starting time of decay of source.

In this subsection, in particular, for $t_{0}>0$, we consider

$\left\\{\begin{array}[]{rl}&\partial_{t}u(x,t)=\Delta
u(x,t)+\theta(t-t_{0})f(x),\quad x\in\Omega,\,t>0,\\\ &u(x,0)=0,\quad
x\in\Omega,\\\ &\partial_{\nu}u(x,t)=0,\quad
x\in\partial\Omega,\,t>0,\end{array}\right.$ $None$

where $\theta\in L^{2}_{loc}(\mathbb{R})$ is assumed to be known and monotone
decreasing and satisfy

$\theta(t)=\left\\{\begin{array}[]{rl}1,&t\leq 0,\\\ 0,&t\geq
a\end{array}\right.$ $None$

with arbitrarily fixed constant $a>0$. Only in this subsection, we consider
the zero Neumann boundary condition. The proof can be modified for the case of
the zero Dirichlet boundary condition.

We note that the diffusion source $\theta(t-t_{0})f(x)$ does not act for
$t\geq a+t_{0}$.

Then we consider
Inverse Problem I.
Let $f=f(x)$ be known. Let $T>0$ be sufficiently large and $x_{0}\in\Omega$ be
arbitrarily chosen. Determine a starting time $t_{0}>0$ of the source by
$u(x_{0},T)$.

For known $f$, we assume

$f\in C^{\infty}_{0}(\Omega),\quad f\geq 0,\,\not\equiv 0\quad\mbox{on
$\overline{\Omega}$}.$ $None$

By $u_{t_{0}}=u_{t_{0}}(x,t)$ we note the solution to (1.4), assuming that
$\theta,f$ are fixed. We are ready to state the uniqueness and the stability
as our main result for Inverse Problem I.
Theorem 1.
We arbitrarily fix constants $t_{*},t^{*},a>0$ with $t_{*}<t^{*}<T^{*}-a$.
Then we a priori assume that $t_{0},t_{1}$ are limited to an interval
$(t_{*},\,t^{*})$:

$t_{*}<t_{0},t_{1}<t^{*}.$

Then there exists a constant $C=C(t_{*},t^{*},a,x_{0},T^{*})>0$ such that

$|t_{1}-t_{0}|\leq C|u_{t_{0}}(x_{0},T^{*})-u_{t_{1}}(x_{0},T^{*})|.$

This theorem asserts the stability in determining a starting time of decay by
one-shot data $u(x_{0},T^{*})$, provided that the starting time is assumed to
be in an a priori fixed interval $(t_{*},t^{*})$.

Now we return to (1.1) and (1.2). By $u_{\mu}=u_{\mu}(x,t)$ and
$u_{f}=u_{f}(x,t)$ we denote the solutions to (1.1) or (1.2) in the cases
where we discuss the determination of $\mu$ and $f$ with fixed $f$ and $\mu$,
respectively. The existence and the regularity of $u_{\mu}$ and $u_{f}$ are
guaranteed by Lemmata 1 and 2.

Henceforth we will consider the following settings. First we assume that

$\mu(t)=0\quad\mbox{for $t\geq T$},\quad T<T_{1}<T_{2}.$

The exact description of the conditions of $\mu$ and $f$ are different
according to several formulations of our inverse problems, and later is
provided.

We understand that $(T_{1},T_{2})$ is an observation time interval. Let
$\gamma,\Gamma\subset\partial\Omega$ be subboundaries. Now for convenience, we
list our main results for the inverse source problems for (1.1) and (1.2).
Determination of $\mu(t)$
Diffusion equation

  * •

Theorem 2: data $u(x_{0},t)$, $T_{1}<t<T_{2}$

  * •

Proposition 2: data $\partial_{\nu}u$ on $\gamma\times(T_{1},T_{2})$

Wave equation

  * •

Theorem 3 (the one-dimensional case), Proposition 1: data $u(x_{0},t)$,
$T_{1}<t<T_{2}$

  * •

Proposition 3: data $\partial_{\nu}u$ on $\Gamma\times(T_{1},T_{2})$

Determination of $f(x)$
Diffusion equation

  * •

The uniqueness is impossible in general dimensions by data $u(x_{0},t)$,
$T_{1}<t<T_{2}$.

  * •

Theorem 4: data $\partial_{\nu}u$ on $\gamma\times(T_{1},T_{2})$

Wave equation

  * •

The uniqueness is impossible in general dimensions by data $u(x_{0},t)$,
$T_{1}<t<T_{2}$.

  * •

Theorem 5: data $\partial_{\nu}u$ on $\Gamma\times(T_{1},T_{2})$

In general dimensions $d$, the uniqueness does not hold with data $u(x_{0},t)$
for $T_{1}<t<T_{2}$, because unknown $f$ depends on $d$-variables, but the
data depend only on one variable $t$.

Here we do not discuss inverse problems with final data $u(\cdot,T)$, and for
the heat equations, we refer to Cheng and Liu [6], Choulli and Yamamoto [7]
and the references therein.

§1.2. Determination of $\mu(t)$ of the source term by pointwise data

In this subsection, we assume:

$\mu\in H^{1}_{loc}(0,\infty),\quad\not\equiv 0,\quad\mu(t)=0\quad\mbox{for
$t\geq T$}$ $None$

and

$f\in C^{\infty}_{0}(\Omega),\quad f\not\equiv 0.$ $None$

Here we set

$H^{1}_{loc}(0,\infty):=\\{\mu;\,\mu|_{(0,T_{0})}\in
H^{1}(0,T_{0})\quad\mbox{with any $T_{0}>0$}\\}.$

We can relax the regularity of $f$ but we assume (1.8) for simplicity.

Moreover, let $T_{1},T_{2}$ be given such that

$T<T_{1}<T_{2}$

and let $x_{0}\in\Omega$ be arbitrarily chosen.

We consider

Inverse Problem II.
In (1.1) and (1.2), we are given $f$. Determine $\mu(t)$, $0<t<T$, by
$u(x_{0},t)$ for $T_{1}<t<T_{2}$.

By Lemma 1, we know that $u\in C(\overline{\Omega}\times[T_{1},T_{2}])$, and
so our observation data $u(x_{0},t)$, $T_{1}<t<T_{2}$, are well-defined.

For the statement of our main results for Inverse Problem II, we introduce
notations. Let $\lambda_{j}$, $j\in\mathbb{N}$ be all the distinct eigenvalues
of the operator $A=-\Delta$ with the domain $\mathcal{D}(A)=H^{2}(\Omega)\cap
H^{1}_{0}(\Omega)$. We know that $\lambda_{j}>0$ for all $j\in\mathbb{N}$. By
$d_{j}$ we denote the multiplicity of $\lambda_{j}$, $j\in\mathbb{N}$, and
$\\{\varphi_{jk}\\}_{1\leq k\leq d_{j}}$ is an orthonormal basis of Ker
$(\lambda_{j}-A):=\\{v\in\mathcal{D}(A);\,Av=\lambda_{j}v\\}$, and we define
the orthogonal projection $P_{j}$ from $L^{2}(\Omega)$ to
Ker $(\lambda_{j}-A)$ by

$P_{j}f=\sum_{k=1}^{d_{j}}(f,\varphi_{jk})\varphi_{jk}.$ $None$

We set

$\Lambda=\Lambda(x_{0}):=\\{j\in\mathbb{N};\,(P_{j}f)(x_{0})=0\\}.$ $None$

We are ready to state our main result for Inverse Problem II for the diffusion
equation (1.1).
Theorem 2.
Assume that (1.8) holds and $\mu_{1},\mu_{2}$ satisfy (1.7). Let
$x_{0}\in\Omega$ be arbitrarily chosen. Then

$u_{\mu_{1}}(x_{0},t)=u_{\mu_{2}}(x_{0},t)\quad\mbox{for $T_{1}<t<T_{2}$}$
$None$

yields $\mu_{1}(t)=\mu_{2}(t)$ for $0<t<T$ if and only if

$\sum_{j\in\mathbb{N}\setminus\Lambda(x_{0})}\frac{1}{\lambda_{j}}=\infty.$
$None$

Corollary.
The uniqueness holds for Inverse Problem II only if the spatial dimensions
$d\geq 2$. The uniqueness always fails for $d=1$.

In the case of $d=1$, our data $u_{\mu}(x_{0},t)$, $t>T$ starting after the
incident (i.e., $\mu(t)=0$ for $t>T$), cannot give the uniqueness. Moreover,
as is seen by the proof in Section 3, even if we will take the perfect
observation data $u_{\mu}(x,t)$ with all $x\in\Omega$ and $t>T$, we can
determine at best

$\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds,\quad n\in\mathbb{N},$

which is the same information determined by currently adopted pointwise data
$u_{\mu}(x_{0},t)$, $t<T$. In other words, even the perfect observation data
cannot augment any information of poinwise data $u(x_{0},t)$, $t>T$ with fixed
$x_{0}\in\Omega$.
Example of (1.12).
Let $d=2$ and $\Omega=(0,\ell_{1})\times(0,\ell_{2})$ with
$\ell_{1},\ell_{2}>0$. Then we can directly verify that

$\\{\lambda_{j}\\}_{j\in\mathbb{N}}=\left\\{\left(\frac{m_{1}^{2}}{\ell_{1}^{2}}+\frac{m_{2}^{2}}{\ell_{2}^{2}}\right)\pi^{2};\,m_{1},m_{2}\in\mathbb{N}\right\\}.$
$None$

For simplicity, we consider only the case where
$\frac{\ell_{1}^{2}}{\ell_{2}^{2}}$ is an irrational number. Then
$\frac{m_{1}^{2}}{\ell_{1}^{2}}+\frac{m_{2}^{2}}{\ell_{2}^{2}}=\frac{n_{1}^{2}}{\ell_{1}^{2}}+\frac{n_{2}^{2}}{\ell_{2}^{2}}$
with $m_{1},m_{2},n_{1},n_{2}\in\mathbb{N}$ imply $m_{1}=n_{1}$ and
$m_{2}=n_{2}$, which means that the multiplicity of $\lambda_{j}$,
$j\in\mathbb{N}$ is one. We re-number
$\left(\frac{m_{1}^{2}}{\ell_{1}^{2}}+\frac{m_{2}^{2}}{\ell_{2}^{2}}\right)\pi^{2}$
with $m_{1},m_{2}\in\mathbb{N}$ as $0<\lambda_{1}<\lambda_{2}<\cdots$. Then
for
$\lambda_{j}=\left(\frac{m_{1}^{2}}{\ell_{1}^{2}}+\frac{m_{2}^{2}}{\ell_{2}^{2}}\right)\pi^{2}$,
we choose

$\varphi_{j}(x)=\frac{2}{\sqrt{\ell_{1}\ell_{2}}}\sin\frac{m_{1}\pi
x_{1}}{\ell_{1}}\sin\frac{m_{2}\pi x_{2}}{\ell_{2}}.$

We choose a monitoring point
$x_{0}=(x_{0}^{1},x_{0}^{2})\in(0,\ell_{1})\times(0,\ell_{2})$ such that

$\frac{x_{0}^{1}}{\ell_{1}}$ and $\frac{x_{0}^{2}}{\ell_{2}}$ are irrational
numbers. $None$

Then
$m_{1}\frac{x_{0}^{1}}{\ell_{1}},m_{2}\frac{x_{0}^{2}}{\ell_{2}}\not\in\mathbb{N}$
for all $m_{1},m_{2}\in\mathbb{N}$, and so
$\varphi_{j}(x_{0}^{1},x_{0}^{2})\neq 0$. Thus, $(P_{j}f)(x_{0})=0$ if and
only if $\int_{\Omega}\varphi_{j}(x)f(x)dx=0$, and so we see that
$\int_{\Omega}\varphi_{k}(x)f(x)dx\neq 0$ except for a finite number of
$k\in\mathbb{N}$, then we can conclude that $\Lambda(x_{0})$ is a finite set.

Under (1.14), we can verify that (1.12) is satisfied. Indeed, by Agmon [1] or
Courant and Hibert [8] for example, we know that $\lambda_{j}=\rho_{0}j+o(1)$
as $j\to\infty$, where $\rho_{0}>0$ is a constant. Therefore,
$\sum_{j=1}^{\infty}\frac{1}{\lambda_{j}}=\infty$. Thus if
$\frac{\ell_{1}^{2}}{\ell_{2}^{2}}$ is an irrational number and
$\int_{\Omega}\varphi_{k}(x)f(x)dx\neq 0$ except for a finite number of
$k\in\mathbb{N}$, then (1.12) is satisfied.

In the case where the measurement starts at $t=0$, that is, $T_{1}=0$, the
inverse source problem is even stable (e.g., Cannon and Esteva [5], Saitoh,
Tuan and Yamamoto [19]). However, for $T_{1}>0$, to the best knowledge of the
authors, there have been no publications, although such formulated inverse
problems are practical.

We can expect to establish conditional stability which holds under suitable a
priori boundedness condition on $\mu(t)$, and we conjecture that the rate is
of weak type such as logarithmic rate.

Moreover, in the case where $0<T_{1}<T_{2}<T$, that is, the observation starts
after the activities of the source but before the stop of the activities, the
uniqueness seems impossible, but here we do not discuss the details.

Next we consider the wave equation (1.2). For the uniqueness, we can prove
positive results only for the one-dimensional case $d=1$.
Theorem 3 (uniqueness for time dependent factor of wave source).
We assume (1.8) and (1.7) for $\mu_{1},\mu_{2}$. Let $d=1$ and
$\Omega=(0,\ell)$. Then $u_{\mu_{1}}(x_{0},t)=u_{\mu_{2}}(x_{0},t)$ for
$T_{1}<t<T_{2}$ yields $\mu_{1}(t)=\mu_{2}(t)$ for $0\leq t\leq T$ if and only

$\left\\{\begin{array}[]{rl}&\frac{x_{0}}{\ell}\in\mathbb{R}\setminus\mathbb{Q},\quad
T\leq 2\ell,\quad T_{2}-T_{1}\geq 2\ell,\\\
&\int^{\ell}_{0}f(x)\sin\frac{n\pi}{\ell}xdx\neq 0,\quad
n\in\mathbb{N}.\end{array}\right.$ $None$

The assumption $T_{2}-T_{1}\geq 2\ell$ in (1.15) means that we have to take
measurements longer than $2\ell$, while the width of support of $\mu(t)$
should not be long, that is, $\leq 2\ell$. We do not know the uniqueness for
general dimensions $d\geq 2$.

For the case of $d\geq 2$, we can prove the uniqueness with different
measurements of pointwise data $u(x_{0},t)$. We recall that
$\Lambda(x_{0})\subset\mathbb{N}$ is defined by (1.10).

We assume that $\mathbb{N}\setminus\Lambda(x_{0})$ is an infinite subset of
$\mathbb{N}$ and

$\lim_{k\to\infty,k\in\mathbb{N}\setminus\Lambda(x_{0})}\frac{k}{\sqrt{\lambda_{k}}}>0.$
$None$

We remark that also the existence of the limit is assumed in (1.16).
Proposition 1.
Let $d\geq 2$ and $f\in C^{\infty}_{0}(\Omega)$, and

$T<\pi\lim_{k\to\infty,k\in\mathbb{N}\setminus\Lambda(x_{0})}\frac{k}{\sqrt{\lambda_{k}}}.$
$None$

Then

$\limsup_{t\to\infty}|u_{\mu_{1}}(x_{0},t)-u_{\mu_{2}}(x_{0},t)|=0$ $None$

yields $\mu_{1}(t)=\mu_{2}(t)$ for $0\leq t\leq T$.

If
$\lim_{k\to\infty,k\in\mathbb{N}\setminus\Lambda(x_{0})}\frac{k}{\sqrt{\lambda_{k}}}=\infty$,
then we interpret that (1.17) holds true for any $T>0$.

The observation in (1.18) is concerned with the asymptotics of
$u_{\mu}(x_{0},t)$ as $t\to\infty$ and (1.18) requires that the data at
$x_{0}$ are asymptotically equal as $t\to\infty$.

We conclude this subsection with
Remark on condition (1.16).
We consider one sufficien condition for (1.16) in terms of the multiplicities
of $\lambda_{k}$. We assume
$\Lambda(x_{0}):=\\{j\in\mathbb{N};\,(P_{j}f)(x_{0})=0\\}=\emptyset$. We
recall that by $\lambda_{1}<\lambda_{2}<\cdots$ we number the set of all the
eigenvalues, not taking the multiplicities into consideration. On the other
hand, by $\sigma_{k}$, $k\in\mathbb{N}$, we number all the eigenvalues
according to their multiplicities:
$0<\sigma_{1}\leq\sigma_{2}\leq\sigma_{3}\leq\cdots\longrightarrow\infty$. In
other words, the value $\lambda_{j}$ appears $d_{j}$-times among the sequence
$\sigma_{1},\sigma_{2},\sigma_{3},\cdots$, where $d_{j}:=\mbox{dim
Ker}\,(\lambda_{j}-A)$. More precisely, setting

$m_{k}:=\sum_{i=1}^{k}d_{i},\quad k\in\mathbb{N},$

we see

$\sigma_{j}=\left\\{\begin{array}[]{rl}\lambda_{1},&1\leq j\leq m_{1},\\\
\lambda_{2},&m_{1}+1\leq j\leq m_{2},\\\ \cdots&\cdots,\\\
\lambda_{k},&m_{k-1}+1\leq j\leq m_{k}.\end{array}\right.$

By the definition, we note

$\sigma_{j}\leq\lambda_{j},\quad j\in\mathbb{N}.$

As general information of $\sigma_{j}$ the following is classical:

$\sigma_{j}=\rho_{0}j^{\frac{2}{d}}+o(d^{\frac{2}{d}})\quad\mbox{as
$j\to\infty$}$ $None$

(e.g., Theorems 14.6 and 15.1 in Agmon [1], Chapter 6 in Courant and Hilbert
[8]). Here $\rho_{0}>0$ is a constant determined by $d$ and $|\Omega|$.

By (1.19) we have

$\rho_{0}m_{k-1}^{\frac{2}{d}}(1+o(1))\leq\lambda_{k}\leq\rho_{0}m_{k}^{\frac{2}{d}}(1+o(1))$

as $k\to\infty$. We here note that if $j\to\infty$, then $k\to\infty$.
Therefore

$\lim_{k\to\infty}\frac{k}{m_{k}^{\frac{1}{d}}}>0$ $None$

yields (1.16). Indeed

$\frac{k}{\sqrt{\lambda_{k}}}\geq\frac{k}{m_{k}^{\frac{1}{d}}}\frac{1}{\sqrt{\rho_{0}}(1+o(1))}$

as $k\to\infty$.

In the following two examples, we consider (1.20).
Example 1: Let $d=2$ and $\Omega=(0,\ell_{1})\times(0,\ell_{2})$.
First we assume $\frac{\ell_{1}^{2}}{\ell_{2}^{2}}\not\in\mathbb{Q}$. By
(1.13), all the eigenvalues are simple, that is, $d_{i}=1$ for
$i\in\mathbb{N}$. Therefore $m_{k}=k$ for $k\in\mathbb{N}$, that is,

$\lim_{k\to\infty}\frac{k}{\sqrt{m_{k}}}=\infty.$

Consequently, we need not any assumption for $T$. Also in general domain
$\Omega$, if all the eigenvalues are simple except for a finite number, then
(1.16) is true in terms of (1.19), and (1.17) is satisfied for any $T>0$.

Next we assume $\frac{\ell_{1}^{2}}{\ell_{2}^{2}}\in\mathbb{Q}$. Then all the
eigenvalues are not necessarily simple, and we do not know suitable estimates
of $m(k)$ as $k\to\infty$, but we have

$\limsup_{i\to\infty}d_{i}=\infty,$

whose proof is found in Yamamoto [25] for example. We do not know how rapidly
$d_{i}$ tends to $\infty$. If they goes to $\infty$ very rapidly, then
$\inf\lim_{k\to\infty}\frac{k}{\sqrt{m_{k}}}=0$ may happen, which breaks
(1.20), and we do not know whether (1.16) holds.
Example 2: Let $d=2$ and $\Omega=\\{x\in\mathbb{R}^{2};\,|x|<1\\}$.
It is known that $d_{i}=\left\\{\begin{array}[]{rl}1,\quad\mbox{if $i=1$},\\\
2,\quad\mbox{if $i\geq 2$}.\end{array}\right.$. Therefore $m_{k}=2k-1$ for
$k\in\mathbb{N}$ and (1.16) holds. We see that (1.17) holds for any $T>0$.

§1.3. Determination of $\mu(t)$ of the source term by boundary data.

In this subsection, for the inverse problems, we adopt data $\partial_{\nu}u$
on a lateral subboundary.

We set

$\widetilde{\Lambda}:=\\{j\in\mathbb{N};\,P_{j}f=0\quad\mbox{in $\Omega$}\\}.$
$None$

For the diffusion equation, we can show
Proposition 2.
Let $\gamma\subset\partial\Omega$ be an arbitrarily chosen subboundary. We
assume (1.7) and (1.8). Then
$\partial_{\nu}u_{\mu_{1}}=\partial_{\nu}u_{\mu_{2}}$ on
$\gamma\times(T_{1},T_{2})$ yields $\mu_{1}(t)=\mu_{2}(t)$ for $0<t<T$ if and
only if

$\sum_{j\in\mathbb{N}\setminus\widetilde{\Lambda}}\frac{1}{\lambda_{j}}=\infty.$
$None$

The inverse source problem with boundary data is overdetermining because an
unknown function depends only on one variable $t$. The condition (1.22) for
the uniqueness is weaker than (1.12) for the uniqueness by the pointwise data
$u(x_{0},t)$, $T_{1}<t<T_{2}$.

In the one-dimensional case, Theorem 3 asserts the uniqueness for a wave
equation by the pointwise data $u(x_{0},t)$ for $T_{1}<t<T_{2}$, but we do not
know the corresponding uniqueness for general dimensions. On the other hand,
by boundary observations, we can conclude the uniqueness for general
dimensions $d\geq 2$, as the following proposition shows.
Proposition 3.
Let

$\lim_{k\to\infty,k\in\mathbb{N}\setminus\widetilde{\Lambda}}\frac{k}{\sqrt{\lambda_{k}}}>0.$
$None$

We assume that there exists $x_{0}\in\mathbb{R}^{d}$ such that

$\Gamma\supset\\{x\in\partial\Omega;\,((x-x_{0})\cdot\nu(x))\geq 0\\},\quad
T_{2}-T_{1}>2\max_{x\in\overline{\Omega}}|x-x_{0}|$ $None$

and

$T<\pi\lim_{k\to\infty,k\in\mathbb{N}\setminus\widetilde{\Lambda}}\frac{k}{\sqrt{\lambda_{k}}}.$
$None$

Moreover, let $\mu_{1},\mu_{2}\in C^{1}[0,T]$ satisfy
$\mu_{1}(t)=\mu_{2}(t)=0$ for $t>T$. Then
$\partial_{\nu}u_{\mu_{1}}(x,t)=\partial_{\nu}u_{\mu_{2}}(x,t)=0$ for
$x\in\Gamma$ and $T_{1}<t<T_{2}$ implies $\mu_{1}(t)=\mu_{2}(t)$ for $0<t<T$.

The condition (1.25) can be characterized similarly to (1.20). The observation
time length $T_{2}-T_{1}$ should be longer, while the duration time $T$ of the
source needs to be bounded for the uniqueness for our inverse source problem.
§1.4. Determination of $f(x)$ of source terms by boundary data.

In this subsection, we assume that $\Gamma,\gamma\subset\partial\Omega$ are
subboundaries, and $T<T_{1}<T_{2}$,

$\mu\in C^{1}[0,\infty),\quad\mu(t)=0\quad\mbox{if $t\geq T$}$ $None$

and

$f\in L^{2}(\Omega).$ $None$

We consider
Inverse Problem III.
Determine $f=f(x)$, $x\in\Omega$ by $\partial_{\nu}u$ on some subboundary over
a time interval $(T_{1},T_{2})$.

In the case of $T_{1}=0$, there are many researches on the uniqueness and the
stability. Here we limit ourselves to a few works: for the diffusion equation
(1.1), we refer to Xu and Yamamoto [21], Yamamoto [22], while we can consult
Yamamoto [23], [24] for the wave equation (1.2).

First we consider the determination of $f(x)$ for (1.1). Henceforth for
arbitrarily chosen constants $M>0$ and $\ell\in\mathbb{N}$, we define an
admissible set of unknowns $f$ by

$\mathcal{F}_{M,\ell}:=\\{f\in\mathcal{D}(A^{\ell});\,\|A^{\ell}f\|_{L^{2}(\Omega)}\leq
M\\}.$ $None$

Theorem 4 (determination of $f(x)$ in the diffusion equation).
Let $\gamma$ be an arbitrarily chosen subboundary of $\partial\Omega$.
(i) We assume

$\int^{T}_{0}e^{\lambda_{k}t}\mu(t)dt\neq 0,\quad k\in\mathbb{N}.$ $None$

Then $\partial_{\nu}u=0$ on $\gamma\times(T_{1},T_{2})$ yields $f=0$ in
$\Omega$.
(ii) We further assume

$\mu\geq 0,\,\not\equiv 0\quad\mbox{on $[0,T]$}.$ $None$

Then for $0<\theta<\ell$, we can find a constant $C=C(M,\theta)>0$ such that

$\|f\|_{L^{2}(\Omega)}\leq
C\left(\frac{1}{\log\frac{1}{\|\partial_{\nu}u\|_{L^{2}(\gamma\times(T_{1},T_{2}))}}}\right)^{\theta}$

as $\|\partial_{\nu}u\|_{L^{2}(\gamma\times(T_{1},T_{2}))}\longrightarrow 0$
for each $f\in\mathcal{F}_{M,\ell}$.

In the case of $T_{1}=0$, we have the same estimate without assumption (1.29)
(e.g., Yamamoto [22]).
Remark. We need the assumption (1.29). Indeed, let $\varphi$ be an
eigenfunction for $\lambda_{1}$ of $-\Delta$ with the zero Dirichlet boundary
condition and let (1.29) fail, for example, let
$\int^{T}_{0}e^{\lambda_{1}s}\mu(s)ds=0$. Then, we readily see that

$u(x,t):=\left(\int^{t}_{0}e^{-\lambda_{1}(t-s)}\mu(s)ds\right)\varphi_{1}(x),\quad
x\in\Omega,\,t>0,$

satisfies

$\left\\{\begin{array}[]{rl}&\partial_{t}u=\Delta
u(x,t)+\mu(t)\varphi_{1}(x),\quad x\in\Omega,\,t>0,\\\ &u(x,t)=0,\quad
x\in\partial\Omega,\,t>0,\\\ &u(x,0)=0,\quad x\in\Omega,\end{array}\right.$

and

$u(x,t)=e^{-\lambda_{1}t}\left(\int^{T}_{0}e^{\lambda_{1}s}\mu(s)ds\right)\varphi_{1}(x)=0,\quad
x\in\Omega,\,t>T,$

because $\mu(s)=0$ for $s\geq T$ and $\int^{T}_{0}e^{\lambda_{1}s}\mu(s)ds=0$.

Therefore even $u=0$ on $\overline{\Omega}\times(T_{1},T_{2})$ holds, and in
particular, $\partial_{\nu}u|_{\partial\Omega\times(T_{1},T_{2})}=0$. However,
$f=\varphi_{1}\neq 0$, that is, the uniqueness fails without (1.29).

Next we show
Theorem 5 (determination of $f(x)$ in the wave equation).
We assume (1.24) and

$\mu(0)\neq 0$

and

$\left|\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds\right|+\left|\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}(T-s)ds\right|\neq
0\quad\mbox{for all $n\in\mathbb{N}$}.$ $None$

Then there exists a constant $C>0$ such that

$C^{-1}\|\partial_{\nu}u\|_{H^{1}(T_{1},T_{2};L^{2}(\partial\Omega))}\leq\|f\|_{L^{2}(\Omega)}\leq
C\|\partial_{\nu}u\|_{H^{1}(T_{1},T_{2};L^{2}(\Gamma))}$ $None$

for each $f\in L^{2}(\Omega)$.

In the case of $T_{1}=0$, we refer to the existing works: for example,
Yamamoto [23], [24] which proves (1.32) only by (1.24) and $\mu(0)\neq 0$, not
assuming (1.31).
Example of (1.31).
We verify that $\mu(s)=T-s$ satisfies (1.31). Indeed we directly have

$\int^{T}_{0}(T-s)\sin(T-s)\sqrt{\lambda_{n}}ds=\frac{-T\sqrt{\lambda_{n}}\cos
T\sqrt{\lambda_{n}}+\sin T\sqrt{\lambda_{n}}}{\lambda_{n}}$

and

$\int^{T}_{0}(T-s)\cos(T-s)\sqrt{\lambda_{n}}ds=\frac{T\sqrt{\lambda_{n}}\sin
T\sqrt{\lambda_{n}}+\cos T\sqrt{\lambda_{n}}-1}{\lambda_{n}}.$

Assume that both are zero for some $n_{0}\in\mathbb{N}$. Setting
$\xi=T\sqrt{\lambda_{n_{0}}}\neq 0$, we see that

$\xi\cos\xi-\sin\xi=0,\quad\cos\xi+\xi\sin\xi=1.$

Hence $\cos\xi=\frac{1}{1+\xi^{2}}$ and $\sin\xi=\frac{\xi}{1+\xi^{2}}$, By
$\cos^{2}\xi+\sin^{2}\xi=1$, we obtain $\frac{1}{1+\xi^{2}}=1$, which yields
$\xi=0$, which is a contradiction by $\xi=\sqrt{\lambda_{n_{0}}}T\neq 0$. Thus
$\mu(s)=T-s$ satisfies (1.31).

This article is composed of 9 sections. In Sections 2 and 3, we prove Theorems
1 and 2 respectively. In Section 4, we prove Theorem 3 and Proposition 1.
Sections 5 and 6 are devoted to the proofs of Theorems 4 and 5. In Section 7,
we prove Propositions 2 and 3. For convenience, Section 8 provides proofs of
standard uniqueness results for the case of $T_{1}=0$ in determining $\mu(t)$.
Section 9 gives the proofs of Lemmata 1 and 2.

##  2 Proof of Theorem 1

Without loss of generality, we can assume that $0<t_{0}<t_{1}$. For $t\geq 0$,
we set

$g(t):=\theta(t-t_{1})-\theta(t-t_{0})=0\quad\mbox{if $t\leq t_{0}$ or
$t>t_{1}+a$.}$ $None$

Setting $y=u_{t_{1}}-u_{t_{0}}$, we have

$\left\\{\begin{array}[]{rl}&\partial_{t}y(x,t)=\Delta y(x,t)+g(t)f(x),\quad
x\in\Omega,\,t>0,\\\ &\partial_{\nu}y(x,t)=0,\quad
x\in\partial\Omega,\,t>0,\\\ &y(x,0)=0,\quad x\in\Omega.\end{array}\right.$
$None$

Let $G=G(x,y,s)$ be the Green function for $\partial_{t}-\Delta$ in $\Omega$
with the zero Neumann boundary condition. Then

$G(x,y,s)>0,\quad x,y\in\Omega,\,0<s<T$

(e.g., Theorem 10.1 (i) in Itô [14]). Setting $D:=\mbox{supp}\,f$, by (1.6) we
see that $D\subset\Omega$. Note that $D$ is a closed set and $T^{*}>t^{*}+a$
by the assumption. Since $G(x_{0},y,s)$ is a continuous function in $y\in D$
and $T^{*}-t^{*}-a\leq s\leq T^{*}-t_{*}$, we can choose a constant
$C_{0}=C_{0}(D,t_{*},t^{*}+a)>0$ such that

$G(x_{0},y,s)\geq C_{0}>0,\quad y\in D,\,T^{*}-t^{*}-a\leq s\leq T^{*}-t.$
$None$

Moreover, we have

$y(x,t)=\int^{t}_{0}\int_{\Omega}G(x,y,t-s)f(y)g(s)dyds,\quad
x\in\Omega,\,t>0$ $None$

(e.g., [14]). In general, (2.4) holds for $g\in C[0,T^{*}]$, and we here do
not assume that $g$ is continuous. However, (1.3) in Lemma 1 asserts

$\|y\|_{C(\overline{\Omega}\times[0,T^{*}])}\leq C\|g\|_{L^{2}(0,T^{*})}$
$None$

in view of $f\in C^{\infty}_{0}(\Omega)$. Therefore by taking an approximating
sequence $g_{n}\in C^{\infty}_{0}(0,T^{*})$ to $g$ in $L^{2}(0,T^{*})$, we can
verify that (2.4) holds for $g\in L^{2}(0,T^{*})$.

By $0<t_{*}<t^{*}+a<T^{*}$, (1.6), and (2.3), we obtain

$\displaystyle
y(x_{0},T^{*})=\int^{T^{*}}_{0}\int_{\Omega}G(x_{0},y,T^{*}-s)f(y)g(s)dyds$
$\displaystyle\geq$
$\displaystyle\int^{t^{*}+a}_{t_{*}}\int_{D}G(x_{0},y,T^{*}-s)f(y)g(s)dyds$
$\displaystyle\geq$ $\displaystyle
C_{0}\left(\int^{t^{*}+a}_{t_{*}}g(s)ds\right)\int_{D}f(y)dy\geq
C_{0}C_{1}\int^{t^{*}+a}_{t_{*}}g(s)ds.$

Here, by (1.6), we used $C_{1}:=\int_{D}f(y)dy>0$. On the other hand,

$\displaystyle\int^{t^{*}+a}_{t_{*}}g(s)ds=\int^{t^{*}+a}_{t_{*}}\theta(s-t_{1})ds-\int^{t^{*}+a}_{t_{*}}\theta(s-t_{0})ds$
$\displaystyle=$
$\displaystyle\int^{a+t^{*}-t_{1}}_{t_{*}-t_{1}}\theta(\eta)d\eta-\int^{a+t^{*}-t_{0}}_{t_{*}-t_{0}}\theta(\eta)d\eta$
$\displaystyle=$
$\displaystyle\left(\int^{0}_{t_{*}-t_{1}}+\int^{a}_{0}+\int^{a+t^{*}-t_{1}}_{a}\right)\theta(\eta)d\eta-\left(\int^{0}_{t_{*}-t_{0}}+\int^{a}_{0}+\int^{a+t^{*}-t_{0}}_{a}\right)\theta(\eta)d\eta=\int^{t_{*}-t_{0}}_{t_{*}-t_{1}}\theta(\eta)d\eta.$

Here we used $\theta(\eta)=0$ for $\eta\geq a$, and noted that
$a<a+t^{*}-t_{1}$ and $a<a+t^{*}-t_{0}$. By $t_{*}-t_{0},t_{*}-t_{1}<0$, we
obtain
$\int^{t^{*}+a}_{t_{*}}g(s)ds=\int^{t_{*}-t_{0}}_{t_{*}-t_{1}}d\eta=t_{1}-t_{0}$.

Noting that $y=u_{t_{1}}-u_{t_{0}}$, we complete the proof of Theorem 1.

##  3 Proof of Theorem 2

Let $\mu=\mu_{1}-\mu_{2}$ and $u=u_{\mu_{1}}-u_{\mu_{2}}$. By the
eigenfunction expansion, we have

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

(e.g., [14]). In terms of $f\in C^{\infty}_{0}(\Omega)$ and Lemma 1 in Section
1, we can verify that the series in (3.1) is convergent in
$C(\overline{\Omega}\times[0,T_{2}])$.

Let $t\in[T_{1},T_{2}]$. Since $\mu(s)=0$ for $s>T$, we have

$u(x_{0},t)=\sum_{n=1}^{\infty}e^{-\lambda_{n}t}\left(\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds\right)P_{n}f(x_{0})$
$=\sum_{n\in\mathbb{N}\setminus\Lambda(x_{0})}e^{-\lambda_{n}t}\left(\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds\right)(P_{n}f)(x_{0})=0,\quad
t>T.$ $None$

Here we recall that the set $\Lambda(x_{0})\subset\mathbb{N}$ is defined by
(1.10).

We know
Lemma 3.
Let $\sum_{n=1}^{\infty}|a_{n}|e^{-\lambda_{n}T_{1}}<\infty$. If

$\sum_{n=1}^{\infty}a_{n}e^{-\lambda_{n}t}=0,\quad T_{1}<t<T_{2},$

then $a_{n}=0$ for all $n\in\mathbb{N}$.
Proof of Lemma 3.
For completeness, we provide the proof although the lemma is well-known. We
set $b_{n}:=a_{n}e^{-\lambda_{n}T_{1}}$. By

$\sum_{n=1}^{\infty}|a_{n}|e^{-\lambda_{n}T_{1}}=\sum_{n=1}^{\infty}|b_{n}|<\infty,$

we see that the series $\sum_{n=1}^{\infty}b_{n}e^{-\lambda_{n}(z-T_{1})}$
converges uniformly and absolutely in any compact domain in
$\\{z\in\mathbb{C};\,\mbox{Re}\,z>T_{1}\\}$. Therefore
$h(z):=\sum_{n=1}^{\infty}b_{n}e^{-\lambda_{n}(z-T_{1})}$ is analytic if Re
$z>T_{1}$. Hence $\sum_{n=1}^{\infty}b_{n}e^{-\lambda_{n}(t-T_{1})}=0$ for all
$t>T_{1}$ by the analyticity. Since $\lambda_{1}<\lambda_{2}<\cdots$, we have

$b_{1}+\sum_{n=2}^{\infty}b_{n}e^{-(\lambda_{n}-\lambda_{1})(t-T_{1})}=0\quad\mbox{for
$t>T_{1}$}.$

Since

$\left|\sum_{n=2}^{\infty}b_{n}e^{-(\lambda_{n}-\lambda_{1})(t-T_{1})}\right|\leq
e^{-(\lambda_{2}-\lambda_{1})(t-T_{1})}\sum_{n=2}^{\infty}|b_{n}|$

and

$\sum_{n=2}^{\infty}|b_{n}|=\sum_{n=2}^{\infty}|a_{n}|e^{-\lambda_{n}T_{1}}<\infty,$

letting $t\to\infty$, we see that $b_{1}=0$, that is, $a_{1}=0$. Continuing
this argument, we reach $a_{n}=0$ for $n\in\mathbb{N}$. Thus the proof of
Lemma 3 is complete. $\blacksquare$

We return to the proof of Theorem 2. Setting

$a_{n}=\left(\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds\right)P_{n}f(x_{0}),\quad
n\in\mathbb{N},$

by (3.2) we obtain

$\sum_{n\in\mathbb{N}\setminus\Lambda(x_{0})}a_{n}e^{-\lambda_{n}t}=0,\quad
T_{1}<t<T_{2}$

and

$\displaystyle\sum_{n\in\mathbb{N}\setminus\Lambda(x_{0})}|a_{n}e^{-\lambda_{n}t}|=\sum_{n=1}^{\infty}\left|\left(\int^{T}_{0}e^{-\lambda_{n}(t-s)}\mu(s)ds\right)(P_{n}f)(x_{0})\right|$
$\displaystyle\leq$
$\displaystyle\sum_{n=1}^{\infty}\left(\int^{T}_{0}|\mu(s)|ds\right)|(P_{n}f)(x_{0})|,\quad
t\geq T_{1}>T.$

We recall that $A=-\Delta$ with $\mathcal{D}(A)=H^{2}(\Omega)\cap
H^{1}_{0}(\Omega)$. By (1.8) we see that $f\in\mathcal{D}(A^{m})$ with each
$m\in\mathbb{N}$. Since $P_{j}f\in\mathcal{D}(A^{m})$ and
$AP_{j}f=\lambda_{j}P_{j}f$, choosing even number $\beta>\frac{d}{2}$, by
Sobolev embedding, we obtain

$|P_{j}f(x_{0})|\leq\|P_{j}f\|_{C(\overline{\Omega})}\leq
C\|P_{j}f\|_{H^{\beta}(\Omega)}\leq
C\|A^{\frac{\beta}{2}}P_{j}f\|_{L^{2}(\Omega)}=C\lambda_{j}^{\frac{\beta}{2}}\|P_{j}f\|_{L^{2}(\Omega)}$

and

$\|P_{j}f\|^{2}_{L^{2}(\Omega)}=\sum_{k=1}^{d_{j}}|(P_{j}f,\varphi_{jk})|^{2}=\sum_{k=1}^{d_{j}}|(A^{m}P_{j}f,\,A^{-m}\varphi_{jk})|^{2}=\lambda_{j}^{-2m}\sum_{k=1}^{d_{j}}|(A^{m}P_{j}f,\varphi_{jk})|^{2}.$

Therefore,

$\displaystyle|P_{j}f(x_{0})|^{2}\leq
C\lambda_{j}^{-2m+\beta}\sum_{k=1}^{d_{j}}|(A^{m}P_{j}f,\varphi_{jk})|^{2}$
$\displaystyle\leq$ $\displaystyle
C\|A^{m}P_{j}f\|^{2}_{L^{2}(\Omega)}\lambda_{j}^{-2m+\beta}\leq
C\|A^{m}f\|^{2}_{L^{2}(\Omega)}\lambda_{j}^{-2m+\beta},\quad j\in\mathbb{N}.$

Moreover, we choose

$\beta_{1}>\frac{d}{4}\quad\mbox{and}\quad
m>\frac{1}{2}\left(\beta+2\beta_{1}+\frac{d}{2}\right),$

and we re-number the eigenvalues $\sigma_{j}$ according to the multiplicities:
$0<\sigma_{1}\leq\sigma_{2}\leq\sigma_{3}\leq\cdots\longrightarrow$ and we
note that $\lambda_{j}\geq\sigma_{j}$ by the re-numbering. Then
$\lambda_{j}\geq\sigma_{j}\sim\rho_{0}j^{\frac{2}{d}}$ as $j\to\infty$ (e.g.,
[1], [8]) and

$\displaystyle\sum^{\infty}_{j=1}|P_{j}f(x_{0})|=\sum_{j=1}^{\infty}\lambda_{j}^{-\beta_{1}}\lambda_{j}^{\beta_{1}}|P_{j}f(x_{0})|\leq\left(\sum_{j=1}^{\infty}\lambda_{j}^{-2\beta_{1}}\right)^{\frac{1}{2}}\left(\sum_{j=1}^{\infty}\lambda_{j}^{2\beta_{1}}|P_{j}f(x_{0})|^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
C\|A^{m}f\|_{L^{2}(\Omega)}\left(\sum_{j=1}^{\infty}\lambda_{j}^{-2\beta_{1}}\right)^{\frac{1}{2}}\left(\sum_{j=1}^{\infty}\lambda_{j}^{-2m+\beta+2\beta_{1}}\right)^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
C\|A^{m}f\|_{L^{2}(\Omega)}\left(\sum_{j=1}^{\infty}j^{\frac{-4\beta_{1}}{d}}\right)^{\frac{1}{2}}\left(\sum_{j=1}^{\infty}j^{\frac{2}{d}(-2m+\beta+2\beta_{1})}\right)^{\frac{1}{2}}.$

Here $m>\frac{1}{2}\left(\beta+2\beta_{1}+\frac{d}{2}\right)$ and
$\beta_{1}>\frac{d}{4}$ yield that $\frac{2}{d}(-2m+\beta+2\beta_{1})<-1$ and
$-\frac{4\beta_{1}}{d}<-1$, so that we see that the right-hand side is
convergent, that is,

$\sum_{j=1}^{\infty}|P_{j}f(x_{0})|<\infty.$

Hence, we can apply Lemma 3 to (3.2) for $t\geq T_{1}$, and we obtain that
$u_{\mu_{1}}(x_{0},t)=u_{\mu_{2}}(x_{0},t)$ for $T_{1}<t<T_{2}$ if and only if

$\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds=0,\quad
n\in\mathbb{N}\setminus\Lambda(x_{0}).$ $None$

We set $\eta=e^{s}$ and $h(\eta)=\mu(\log\eta)$, and we see that
$1<\eta<e^{T}$ for $0<s<T$. Then (3.3) is equivalent to

$\int^{e^{T}}_{1}\eta^{\lambda_{n}}\frac{h(\eta)}{\eta}d\eta=0,\quad
n\in\mathbb{N}\setminus\Lambda(x_{0}).$ $None$

By the Müntz theorem (e.g., E.7 on p.184 of Borwein and Erdélyi [3]) as
directly applicable version, we see that (3.4) yields $\frac{h(\eta)}{\eta}=0$
for $1<\eta<e^{T}$ if and only if (1.12) holds. Thus the proof of Theorem 2 is
complete.

In the case of $d=1$, we see that $\lambda_{j}=\sigma_{j}$ because all the
eigenvalues are simple. Therefore Corollary follows directly since
$\lambda_{j}\sim\rho_{0}j^{2}$ as $j\to\infty$, which implies
$\sum_{j=1}^{\infty}\frac{1}{\lambda_{j}}<\infty$ if $d=1$.

##  4 Proofs of Theorem 3 and Proposition 1

For general dimensions $d$, we have

$u_{\mu}(x,t)=\sum_{n=1}^{\infty}\left(\int^{t}_{0}\frac{\sin\sqrt{\lambda_{n}}(t-s)}{\sqrt{\lambda_{n}}}\mu(s)ds\right)P_{n}f(x),\quad
x\in\Omega,\,t>0$ $None$

(e.g., Komornik [16]). Setting $\mu:=\mu_{1}-\mu_{2}$ and
$u:=u_{\mu_{1}}-u_{\mu_{2}}$, since $\mu(t)=0$ for $t>T$ and $u\in
C(\overline{\Omega}\times[0,\infty))$, in terms of (1.8) and Lemma 1, we have

$\displaystyle
u_{\mu}(x_{0},t)=\sum_{n=1}^{\infty}\left(\int^{T}_{0}\frac{\sin\sqrt{\lambda_{n}}(t-s)}{\sqrt{\lambda_{n}}}\mu(s)ds\right)P_{n}f(x_{0})$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\frac{\sin\sqrt{\lambda_{n}}t}{\sqrt{\lambda_{n}}}\left(\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds\right)P_{n}f(x_{0})$
$\displaystyle-$
$\displaystyle\sum_{n=1}^{\infty}\frac{\cos\sqrt{\lambda_{n}}t}{\sqrt{\lambda_{n}}}\left(\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds\right)P_{n}f(x_{0}),\quad
t>T.$

Therefore,

$u_{\mu}(x_{0},t)=\sum_{n=1}^{\infty}(a_{n}(P_{n}f)(x_{0})\sin\sqrt{\lambda_{n}}t+b_{n}(P_{n}f)(x_{0})\cos\sqrt{\lambda_{n}}t),\quad
t>T,$ $None$

where we set

$a_{n}=\frac{1}{\sqrt{\lambda_{n}}}\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds,\quad
b_{n}=\frac{-1}{\sqrt{\lambda_{n}}}\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds,\quad
n\in N.$

The series is convergent in $C(\overline{\Omega}\times[0,\infty)])$ by (1.8)
and Lemma 1 (i).
Proof of Theorem 3.

In the case of $\Omega=(0,\ell)$, we know $d_{n}=1$,
$\lambda_{n}=\frac{n^{2}\pi^{2}}{\ell^{2}}$ and
$\varphi_{n}(x)=\sqrt{\frac{2}{\ell}}\sin\frac{n\pi}{\ell}x$,
$(P_{n}f)(x_{0})=(f,\varphi_{n})\varphi_{n}(x_{0})$. Therefore,

$u_{\mu}(x_{0},t)=\sum_{n=1}^{\infty}\left(a_{n}\sin\frac{n\pi
t}{\ell}+b_{n}\cos\frac{n\pi
s}{\ell}ds\right)(f,\varphi_{n})\varphi_{n}(x_{0})\quad\mbox{for
$T_{1}<t<T_{2}$}$ $None$

for $T_{1}<t<T_{2}$.

Since $T_{2}-T_{1}\geq 2\ell$ by (1.15), we have (4.3) for
$T_{1}<t<T_{1}+2\ell$. Since (4.3) is convergent in $C[0,\infty)$, we see that
(4.3) is convergent in $L^{2}(T_{1},T_{1}+2\ell)$. Hence, taking the scalar
products in $L^{2}(T_{1},T_{1}+2\ell)$ with $\sin\frac{m\pi t}{\ell}$ and
$\cos\frac{m\pi t}{\ell}$, we obtain
$a_{m}(f,\varphi_{m})\varphi_{m}(x_{0})=b_{m}(f,\varphi_{m})\varphi_{m}(x_{0})=0$
for $m\in\mathbb{N}$. By (1.15), we see that $(f,\varphi_{m})\neq 0$ and
$\varphi_{m}(x_{0})=\sqrt{\frac{2}{\ell}}\sin\frac{m\pi}{\ell}x_{0}\neq 0$ for
all $m\in\mathbb{N}$. Consequently, $a_{m}=b_{m}=0$, that is,

$\int^{T}_{0}\mu(s)\cos\frac{m\pi s}{\ell}ds=\int^{T}_{0}\mu(s)\sin\frac{m\pi
s}{\ell}ds=0,\quad m\in\mathbb{N}.$

We recall the assumption $T\leq 2\ell$. If $T<2\ell$, then we set
$\widetilde{\mu}(t)=\left\\{\begin{array}[]{rl}\mu(s),&0<s<T,\\\ 0,&T\leq
s<2\ell.\end{array}\right.$ Then

$\int^{2\ell}_{0}\widetilde{\mu}(s)\cos\frac{m\pi
s}{\ell}ds=\int^{2\ell}_{0}\widetilde{\mu}(s)\sin\frac{m\pi s}{\ell}ds=0,\quad
m\in\mathbb{N}.$

Therefore, $\widetilde{\mu}(s)=c_{1}$: constant for $0<s<2\ell$. In the case
of $T=2\ell$, we have $c_{1}=\mu(T)=0$ by $\mu(t)=0$ for $t\geq T$. In the
case of $T<2\ell$, since $\widetilde{\mu}=0$ in $[T,2\ell]$, we see $c_{1}=0$.
Hence $\mu(s)=0$ for $0<s<T$, and the proof of Theorem 3 is complete.
Proof of Proposition 1.
Since $f\in C^{\infty}_{0}(\Omega)$, we see (4.2), and the series
$u_{\mu}(x_{0},t)$ is convergent in $L^{\infty}(\mathbb{R})$, and so $\gamma$
is an almost periodic function (e.g., Böttcher and Silbermann [4]). By a
property of almost periodic functions (e.g., property (d) in p.493 in [4])
yields

$\|u_{\mu}(x_{0},\cdot)\|_{L^{\infty}(\mathbb{R})}=\limsup_{t\to\infty}|u_{\mu}(x_{0},t)|.$

By (1.18), we see that $\|u_{\mu}(x_{0},\cdot)\|_{L^{\infty}(\mathbb{R})}=0$,
that is,

$\sum_{n=1}^{\infty}a_{n}(P_{n}f)(x_{0})\sin\sqrt{\lambda_{n}}t+\sum_{n=1}^{\infty}b_{n}(P_{n}f)(x_{0})\cos\sqrt{\lambda_{n}}t=0,\quad
t\in\mathbb{R}.$

Substituting $-t$ and adding and subtracting, we obtain

$\sum_{n=1}^{\infty}a_{n}(P_{n}f)(x_{0})\sin\sqrt{\lambda_{n}}t=\sum_{n=1}^{\infty}b_{n}(P_{n}f)(x_{0})\cos\sqrt{\lambda_{n}}t=0,\quad
t\in\mathbb{R}.$ $None$

Now we can prove
Lemma 4.
Let $0<p_{1}<p_{2}<\cdots\,\longrightarrow\infty$. Let
$\alpha_{n}\in\mathbb{R}$ satisfy

$\sum_{n=1}^{\infty}|\alpha_{n}|<\infty.$ $None$

Then

$\sum_{n=1}^{\infty}\alpha_{n}\sin p_{n}t=0,\quad
t>0\quad\mbox{or}\quad\sum_{n=1}^{\infty}\alpha_{n}\cos p_{n}t=0,\quad t>0$

yields $\alpha_{n}=0$ for each $n\in\mathbb{N}$.

For completeness the proof of the lemma is provided at the end of this
section.

Now we complete the proof of Proposition 1. Applying Lemma 4 to (4.4), we
obtain

$a_{n}(P_{n}f)(x_{0})=b_{n}(P_{n}f)(x_{0})=0,\quad n\in\mathbb{N}.$

By the definition (1.10) of $\Lambda(x_{0})$, we reach $a_{n}=b_{n}=0$ for
$n\in\mathbb{N}\setminus\Lambda(x_{0})$, that is,

$\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds=\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds=0,\quad
n\in\mathbb{N}\setminus\Lambda(x_{0}).$

With the even extension of $\mu$ to $(-T,0)$ for example, we see

$\int^{T}_{-T}\mu(s)e^{s\sqrt{-1}\sqrt{\lambda_{n}}}ds=0\quad\mbox{for
$n\in\mathbb{N}\setminus\Lambda(x_{0})$}.$ $None$

We can prove the following lemma, whose proof is given at the end of this
section.
Lemma 5.
Under assumptions (1.16) and (1.17), the system

$\\{e^{s\sqrt{-1}\sqrt{\lambda_{n}}}\\}_{n\in\mathbb{N}\setminus\Lambda(x_{0})}$

is complete in $C[-T,T]$. In particular, (4.6) implies $\mu(s)=0$ for $0<t<T$.

Here we say that
$\\{e^{s\sqrt{-1}\sqrt{\lambda_{n}}}\\}_{n\in\mathbb{N}\setminus\Lambda(x_{0})}$
is complete in $C[-T,T]$ if $g\in C[-T,T]$ and

$\int^{T}_{-T}g(s)e^{s\sqrt{-1}\sqrt{\lambda_{n}}}ds=0\quad\mbox{for all
$n\in\mathbb{N}\setminus\Lambda(x_{0})$},$

imply $g(s)=0$ for $0\leq s\leq T$.

By applying Lemma 5 to (4.6), the proof of Proposition 1 is finished.

We conclude this section with
Proof of Lemma 4. It suffices to prove in the case
$\sum_{n=1}^{\infty}\alpha_{n}\sin p_{n}t=0$ for $t>0$. By (4.5), the series
$\sum_{n=1}^{\infty}\alpha_{n}\sin
p_{n}t=\frac{1}{2}\sum_{n=1}^{\infty}\alpha_{n}(e^{\sqrt{-1}p_{n}t}-e^{-\sqrt{-1}p_{n}t})$
is convergent in $L^{\infty}(\mathbb{R})$. Therefore, we can take the Laplace
transform term-wisely:

$0=\sum_{n=1}^{\infty}\alpha_{n}\int^{\infty}_{0}(e^{\sqrt{-1}p_{n}t}-e^{-\sqrt{-1}p_{n}t})e^{-\xi
t}dt=\sum_{n=1}^{\infty}\frac{2\sqrt{-1}\alpha_{n}p_{n}}{\xi^{2}+p_{n}^{2}}$

for Re $\xi>0$. Setting $\eta=\xi^{2}$, we have

$\sum_{n=1}^{\infty}\frac{\alpha_{n}p_{n}}{\eta+p_{n}^{2}}=0,\quad\eta\geq 0.$

For $\eta>0$ and $k\in\mathbb{N}$, since $|\eta+p_{n}^{2}|^{k}\geq p_{n}^{2k}$
and so

$\left|\frac{\alpha_{n}p_{n}}{(\eta+p_{n}^{2})^{k}}\right|\leq\frac{|\alpha_{n}|}{p_{n}^{2k-1}}\leq
C|\alpha_{n}|,\quad n\in\mathbb{N}.$

In view of (4.5), the series
$\sum_{n=1}^{\infty}\frac{\alpha_{n}p_{n}}{(\eta+p_{n}^{2})^{k}}$ is uniformly
convergent in $\eta\in[0,L]$ with arbitrary $L>0$ for each $k\in\mathbb{N}$.
Therefore, we can term-wisely differentiate in $\eta$:

$\sum_{n=1}^{\infty}\frac{\alpha_{n}p_{n}}{(\eta+p_{n}^{2})^{k}}=0,\quad\eta>0.$

Substituting $\eta=1$, we can write

$\frac{\alpha_{1}p_{1}}{(1+p_{1}^{2})^{k}}+\sum_{k=2}^{\infty}\frac{\alpha_{n}p_{n}}{(1+p_{n}^{2})^{k}}=0,$

that is,

$\frac{\alpha_{1}p_{1}}{1+p_{1}^{2}}=-\sum_{k=2}^{\infty}\alpha_{n}\frac{p_{n}}{1+p_{n}^{2}}\left(\frac{1+p_{1}^{2}}{1+p_{n}^{2}}\right)^{k-1}.$

Since $0<p_{1}<p_{2}<\cdots$, we have

$\frac{1+p_{1}^{2}}{1+p_{n}^{2}}\leq\frac{1+p_{1}^{2}}{1+p_{2}^{2}}=:r<1$

for each $n\geq 2$ and $\frac{p_{n}}{1+p_{n}^{2}}\leq 1$. Hence we see that

$\left|\frac{\alpha_{1}p_{1}}{1+p_{1}^{2}}\right|\leq\sum_{n=2}^{\infty}|\alpha_{n}|r^{k-1}\leq
Cr^{k-1}\quad\longrightarrow\quad 0$

as $k\to\infty$ in view of (4.5). Therefore, $\alpha_{1}=0$ by $p_{1}\neq 0$.
Continuing this argument, we can successively obtain $\alpha_{n}=0$ for
$n\in\mathbb{N}$, so that the proof of Lemma 4 is complete. $\blacksquare$
Proof of Lemma 5.
Considering the zero extension, we readily see that
$\\{e^{s\sqrt{-1}\sqrt{\lambda_{n}}}\\}_{n\in\mathbb{N}\setminus\Lambda(x_{0})}$
is complete in $C[-T,T]$, then so is it in $C[-\widetilde{T},\,\widetilde{T}]$
with any $0<\widetilde{T}<T$.

We set

$R:=\sup\\{L;\,\\{e^{s\sqrt{-1}\sqrt{\lambda_{n}}}\\}_{n\in\mathbb{N}\setminus\Lambda(x_{0})}\quad\mbox{is
complete in $C[-L,L]$}\\}$

and

$D:=\lim_{k\to\infty,\in\mathbb{N}\setminus\Lambda(x_{0})}\frac{k}{\sqrt{\lambda_{k}}}.$

Since $\sqrt{\lambda_{k}}>0$ for $k\in\mathbb{N}\setminus\Lambda(x_{0})$ and
is increasing in $k$, we can apply Theorem 13 (p.116) in [27], and we can
obtain

$\pi D\leq R.$

Therefore, if $T<\pi D$, then we have $T<R$. Hence,
$\\{e^{s\sqrt{-1}\sqrt{\lambda_{n}}}\\}_{n\in\mathbb{N}\setminus\Lambda(x_{0})}$
is complete in $C[-T,T]$ if $T<\pi D$. We finished the proof of Lemma 5.
$\blacksquare$

##  5 Proof of Theorem 4

First Step:
We have

$\left\\{\begin{array}[]{rl}&\partial_{t}u(x,t)=\Delta u(x,t),\quad
x\in\Omega,\,t>T,\\\ &u(x,t)=0,\quad
x\in\partial\Omega,\,t>T.\end{array}\right.$ $None$

We can prove

$\|u(\cdot,\widetilde{t})\|_{L^{2}(\Omega)}\leq
C\|\partial_{\nu}u\|_{L^{2}(\gamma\times(T_{1},T_{2}))}\quad\mbox{for
$T_{1}<\widetilde{t}<T_{2}$}.$ $None$

Proof of (5.2).
First we show a Carleman estimate. For it, let $d\in C^{2}(\overline{\Omega})$
satisfy $d>0$ in $\Omega$, $|\nabla d|>0$ on $\overline{\Omega}$ and
$\partial_{\nu}d\leq 0$ on $\partial\Omega\setminus\Gamma$. For the proof of
such $d$, see e.g., Imanuvilov [10].

We set

$\beta(x,t)=\frac{e^{\lambda
d(x)}}{(t-T_{1})(T_{2}-t)},\quad\alpha(x,t)=\frac{e^{\lambda
d(x)}-e^{2\lambda\|d\|_{C(\overline{\Omega})}}}{(t-T_{1})(T_{2}-t)},\quad
x\in\Omega,\,T_{1}<t<T_{2},$

where we choose $\lambda>0$ sufficiently large. Then
Lemma 6 (Carleman estimate).
There exist constants $C>0$ and $s_{0}>0$ such that

$\int_{\Omega\times(T_{1},T_{2})}\left(\frac{1}{s\beta}|\partial_{t}u|^{2}+s^{3}\lambda^{4}\beta^{3}|u|^{2}\right)e^{2s\alpha}dxdt\leq
C\int_{\gamma\times(T_{1},T_{2})}|\partial_{\nu}u|^{2}dSdt$

for all $s>s_{0}$ and all $u$ satisfying (5.1) and $\partial_{t}u,\Delta u\in
L^{2}(\Omega\times(T_{1},T_{2}))$.

For the proof, we refer to Imanuvilov [10], and also Imanuvilov and Yamamoto
[11], where the parabolic equation is considered in the time interval $(0,T)$,
and we can translate by change $t\longmapsto\frac{T}{T_{2}-T_{1}}(t-T_{1})$.

We choose and fix $t_{1},t_{2}$ such that
$T_{1}<t_{1}<\frac{T_{1}+T_{2}}{2}<t_{2}<T_{2}$. Since $\alpha(x,t)<0$ for
$x\in\Omega$ and $T_{1}<t<T_{2}$, there exist constants $C_{1},C_{2},C_{3}>0$
such that

$\alpha(x,t)\geq-C_{1},\quad
C_{2}\leq\beta(x,t)<C_{3},\quad(x,t)\in\overline{\Omega}\times[t_{1},t_{2}].$

Therefore, for all large $s>0$, we obtain

$\int_{\Omega\times(t_{1},t_{2})}(|\partial_{t}u|^{2}+|u|^{2})dxdt\leq
C\int_{\gamma\times(T_{1},T_{2})}|\partial_{\nu}u|^{2}dSdt.$

Here the constant $C>0$ depends on $t_{1},t_{2}$ and fixed large constants
$s>0$. Applying the Sobolev embedding $H^{1}(t_{1},t_{2};L^{2}(\Omega))\subset
C([t_{1},t_{2}];L^{2}(\Omega))$, we obtain

$\sup_{t_{1}\leq t\leq t_{2}}\|u(\cdot,t)\|^{2}_{L^{2}(\Omega)}\leq
C\|\partial_{\nu}u\|^{2}_{L^{2}(\gamma\times(t_{1},t_{2}))}.$

Since $t_{1},t_{2}$ are atbitrarily provided that $T_{1}<t_{1}<t_{2}<T_{2}$,
the proof of (5.2) is complete.

Henceforth we renumber the set $\\{\lambda_{k}\\}_{k\in\mathbb{N}}$ of all the
eigenvalues of $-\Delta$ with the zero Dirichlet boundary condition, according
to the multiplicities: $0<\sigma_{1}\leq\sigma_{2}\leq\sigma_{3}\leq\cdots$.
We can choose an eigenfunction $\varphi_{k}$ corresponding to $\sigma_{k}$
such that $\\{\varphi_{k}\\}_{k\in\mathbb{N}}$ forms an orthonormal basis in
$L^{2}(\Omega)$.

By the eigenfunction expansion, we can represent the solution $u$ to (1.1) for
$0\leq t\leq T$ as

$u(x,t)=\sum_{j=1}^{\infty}e^{-\sigma_{j}t}\left(\int^{t}_{0}e^{\sigma_{j}s}\mu(s)ds\right)(f,\varphi_{j})\varphi_{j}(x),\quad
x\in\Omega,\,0\leq t\leq T.$ $None$

Since $\mu(s)=0$ for $s\geq T$, we obtain

$u(x,t)=\sum_{j=1}^{\infty}e^{-\sigma_{j}t}\left(\int^{T}_{0}e^{\sigma_{j}s}\mu(s)ds\right)(f,\varphi_{j})\varphi_{j}(x),\quad
x\in\Omega,\,T<t\leq T_{2}.$ $None$

Second Step: Proof of Theorem 4 (i).
By (5.2) and $\partial_{\nu}u=0$ on $\gamma\times(T_{1},T_{2})$, we have
$u(x,t)=0$ for $x\in\Omega$ and $T_{1}<t<T_{2}$. Therefore, (5.4) yields

$\sum_{j=1}^{\infty}e^{-\sigma_{j}\widetilde{t}}\left(\int^{T}_{0}e^{\sigma_{j}s}\mu(s)ds\right)(f,\varphi_{j})\varphi_{j}(x)=0,\quad
x\in\Omega,\,T_{1}\leq\widetilde{t}\leq T_{2},$

which implies

$\left(\int^{T}_{0}e^{\sigma_{j}s}\mu(s)ds\right)(f,\varphi_{j})=0\quad\mbox{for
all $j\in\mathbb{N}$}.$

By the assumption (1.29), we reach $(f,\varphi_{j})=0$ for all
$j\in\mathbb{N}$. Thus $f=0$ in $\Omega$.
Third Step: Proof of Theorem 4 (ii).
We fix $\widetilde{t}\in(T_{1},T_{2})$ arbitrarily. Combining (5.2) with (5.4)
with $t=\widetilde{t}$, we obtain

$\|u(\cdot,\widetilde{t})\|_{L^{2}(\Omega)}^{2}=\left\|\sum_{j=1}^{\infty}p_{j}(f,\varphi_{j})\varphi_{j}\right\|^{2}_{L^{2}(\Omega)}\leq
C\|\partial_{\nu}u\|^{2}_{L^{2}(T_{1},T_{2};L^{2}(\gamma))}.$ $None$

Here we set

$p_{j}=e^{-\sigma_{j}\widetilde{t}}\left(\int^{T}_{0}e^{\sigma_{j}s}\mu(s)ds\right),\quad
j\in\mathbb{N}.$

Since $f\in\mathcal{D}(A^{\ell})$ with given $\ell\in\mathbb{N}$, we have

$|(f,\varphi_{j})|=|(A^{\ell}f,A^{-\ell}\varphi_{j})|=\left|\frac{1}{\sigma_{j}^{\ell}}(A^{\ell}f,\varphi_{j})\right|.$
$None$

Therefore, for each $N\in\mathbb{N}$, we obtain

$\displaystyle\sum_{j=1}^{\infty}|(f,\varphi_{j})|^{2}=\left(\sum_{j=1}^{N}+\sum_{j=N+1}^{\infty}\right)|(f,\varphi_{j})|^{2}=\sum_{j=1}^{N}|(f,\varphi_{j})|^{2}+\sum_{j=N+1}^{\infty}\frac{1}{\sigma_{j}^{2\ell}}|(A^{\ell}f,\varphi_{j})|^{2}$
$\displaystyle\leq$
$\displaystyle\sum_{j=1}^{N}|(f,\varphi_{j})|^{2}+\frac{1}{\sigma_{N}^{2\ell}}\sum_{j=N+1}^{\infty}|(A^{\ell}f,\varphi_{j})|^{2}$
$\leq\sum_{j=1}^{N}|(f,\varphi_{j})|^{2}+\frac{1}{\sigma_{N}^{2\ell}}\|A^{\ell}f\|^{2}_{L^{2}(\Omega)}.$
$None$

Next we estimate $p_{j}$. By (1.30), we arbitrarily fix constants
$0<t_{1}<t_{2}<T$ and $\delta>0$ such that $\mu(s)\geq\delta$ for $t_{1}\leq
s\leq t_{2}$. Therefore,

$\int^{T}_{0}e^{\sigma_{j}s}\mu(s)ds\geq\delta\int^{t_{2}}_{t_{1}}e^{\sigma_{j}s}ds=\frac{e^{\sigma_{j}t_{2}}-e^{\sigma_{j}t_{1}}}{\sigma_{j}}\delta,$

so that

$\frac{1}{|p_{j}|}=\frac{e^{\sigma_{j}\widetilde{t}}\sigma_{j}}{\delta}\frac{1}{e^{\sigma_{j}t_{2}}(1-e^{-\sigma_{j}(t_{2}-t_{1})})}=\frac{e^{\sigma_{j}(\widetilde{t}-t_{2})}\sigma_{j}}{\delta}\frac{1}{1-e^{-\sigma_{1}(t_{2}-t_{1})}}\leq
C_{1}\sigma_{j}e^{C_{2}\sigma_{j}}$ $None$

for $j\in\mathbb{N}$. By $e^{-\sigma_{j}(t_{2}-t_{1})}\leq
e^{-\sigma_{1}(t_{2}-t_{1})}<1$ for all $j\in\mathbb{N}$, the constants
$C_{1}>0$ and $C_{2}>0$ are independent of $j\in\mathbb{N}$. On the other
hand, substituting $t=\widetilde{t}$ in (5.4), we take the scalar product of
(5.4) with $\varphi_{j}$, and we obtain
$(u(\cdot,\widetilde{t}),\varphi_{j})=p_{j}(f,\varphi_{j})$. Hence, (5.8)
implies

$|(f,\varphi_{j})|=\frac{1}{|p_{j}|}|(u(\cdot,\widetilde{t}),\varphi_{j})|\leq
C_{1}\sigma_{j}e^{C_{2}\sigma_{j}}|(u(\cdot,\widetilde{t}),\varphi_{j})|,\quad
j\in\mathbb{N}.$

Substituting this into (5.7), we see

$\displaystyle\|f\|^{2}_{L^{2}(\Omega)}\leq
C_{1}^{2}\sum_{j=1}^{N}\sigma_{j}^{2}e^{2C_{2}\sigma_{j}}|(u(\cdot,\widetilde{t}),\varphi_{j})|^{2}+\frac{1}{\sigma_{N}^{2\ell}}\|A^{\ell}f\|^{2}_{L^{2}(\Omega)}$
$\displaystyle\leq$ $\displaystyle
C_{1}^{2}\sigma_{N}^{2}e^{2C_{2}\sigma_{N}}\|(u(\cdot,\widetilde{t})\|^{2}_{L^{2}(\Omega)}+\frac{1}{\sigma_{N}^{2\ell}}\|A^{\ell}f\|^{2}_{L^{2}(\Omega)}$

with arbitrary $N\in\mathbb{N}$.

In terms of (5.2), since we are considering the case of
$\|\partial_{\nu}u\|_{L^{2}(\gamma\times(T_{1},T_{2}))}\longrightarrow 0$, we
can assume that $\|u(\cdot,\widetilde{t})\|_{L^{2}(\Omega)}<1$. Since
$\sigma_{N}\sim\rho_{0}N^{\frac{2}{d}}$ as $N\to\infty$ (e.g., [1], [8]) and
$N^{\frac{2}{d}}\leq\exp(N^{\frac{2}{d}})$ for sufficiently large
$N\in\mathbb{N}$, we obtain

$\|f\|_{L^{2}(\Omega)}\leq
C_{1}\exp\left(C_{3}N^{\frac{2}{d}}\right)\eta+\frac{C_{1}}{N^{\theta_{0}}}M\quad\mbox{for
all $N\in\mathbb{N}$},$ $None$

where we choose a small constant $\eta_{0}\in(0,1)$ and set

$\eta:=\|u(\cdot,\widetilde{t})\|_{L^{2}(\Omega)}\leq 1-\eta_{0},\quad
M:=\|A^{\ell}f\|_{L^{2}(\Omega)},\quad\theta_{0}:=\frac{2\ell}{d}.$

Now, for given $\eta>0$, we make the right-hand side of (5.9) smaller by
choosing $N\in\mathbb{N}$. By $\eta<1$, we see $\log\frac{1}{\eta}>1$. Let
$\theta\in(0,\,\ell)$ be arbitrarily given. We can choose
$N=N(\eta)\in\mathbb{N}$ such that

$\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\theta_{0}}}\leq
N<\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\theta_{0}}}+1.$

Then

$\left(\log\frac{1}{\eta}\right)^{\theta}\leq N^{\theta_{0}},$

that is,

$\frac{C_{1}}{N^{\theta_{0}}}\leq\frac{C_{1}}{\left(\log\frac{1}{\eta}\right)^{\theta}}.$
$None$

Moreover, we know

$N^{\frac{2}{d}}\leq\left(\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\theta_{0}}}+1\right)^{\frac{2}{d}}\leq
C_{4}\left(\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\theta_{0}}\frac{2}{d}}+1\right),$

that is,

$e^{C_{3}N^{\frac{2}{d}}}\eta^{\frac{1}{2}}\leq\eta^{\frac{1}{2}}\exp\left(C_{3}C_{4}\left(\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\ell}}+1\right)\right)\leq
C_{5}\eta^{\frac{1}{2}}\exp\left(C_{5}\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\ell}}\right)$

for $\eta\leq 1-\eta_{0}$. Setting $t=\log\frac{1}{\eta}>0$, we can see

$\eta^{\frac{1}{2}}\exp\left(C_{5}\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\ell}}\right)=e^{-\frac{1}{2}t}\exp(C_{5}t^{\frac{\theta}{\ell}}),$

and so $\frac{\theta}{\ell}<1$ yields

$\sup_{\eta<1}\left|\eta^{\frac{1}{2}}\exp\left(C_{5}\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\ell}}\right)\right|<\infty.$

Hence,

$\exp(C_{3}N^{\frac{2}{d}})\eta\leq
C_{5}\eta^{\frac{1}{2}}\sup_{\eta<1}\left|\eta^{\frac{1}{2}}\exp\left(C_{5}\left(\log\frac{1}{\eta}\right)^{\frac{\theta}{\ell}}\right)\right|\leq
C_{6}\eta^{\frac{1}{2}},$

and (5.9) with this choice of $N$ and (5.10) yield

$\|f\|_{L^{2}(\Omega)}\leq
C_{7}\eta^{\frac{1}{2}}+\frac{C_{1}}{\left(\log\frac{1}{\eta}\right)^{\theta}}M.$

Since we can find a constant $C_{8}>0$ such that
$\eta^{\frac{1}{2}}\leq\frac{C_{8}}{\left(\log\frac{1}{\eta}\right)^{\theta}}$
for $\eta<1$, we obtain

$\|f\|_{L^{2}(\Omega)}\leq\frac{C_{9}}{\left(\log\frac{1}{\eta}\right)^{\theta}}M.$

By using (5.5), we complete the proof of Theorem 4 (ii).

##  6 Proof of Theorem 5

Since $\mu(t)=0$ for $t>T$, we have

$\left\\{\begin{array}[]{rl}&\partial_{t}^{2}u(x,t)=\Delta u(x,t),\quad
x\in\Omega,\,t>T,\\\ &u(x,t)=0,\quad
x\in\partial\Omega,\,t>T.\end{array}\right.$ $None$

Setting $v=\partial_{t}u$ and

$\widetilde{\mu}(t)=\left\\{\begin{array}[]{rl}\mu^{\prime}(t),&0<t\leq T,\\\
0,&t>T,\end{array}\right.$

we have

$\left\\{\begin{array}[]{rl}&\partial_{t}^{2}v(x,t)=\Delta
v(x,t)+\widetilde{\mu}(t)f(x),\quad x\in\Omega,\,t>0,\\\ &v(x,t)=0,\quad
x\in\partial\Omega,\,t>0,\\\ &v(x,0)=\partial_{t}v(x,0)=0,\quad
x\in\Omega.\end{array}\right.$ $None$

By Theorem 8.2 (p.275) in [17], we note that $v\in
C([0,\infty);H^{1}_{0}(\Omega))$ and $\partial_{t}v\in
C([0,\infty);L^{2}(\Omega))$. In particular,

$\left\\{\begin{array}[]{rl}&\partial_{t}^{2}v(x,t)=\Delta v(x,t),\quad
x\in\Omega,\,t>T,\\\ &v(x,t)=0,\quad x\in\partial\Omega,\,t>0,\\\
&v(x,T)=\partial_{t}u(x,T),\quad\partial_{t}v(x,0)=\Delta u(x,T),\quad
x\in\Omega.\end{array}\right.$ $None$

In view of (1.24), we can apply the observability inequality (e.g., Komornik
[16]) to (6.3), so that we obtain

$\displaystyle\|\partial_{t}u(\cdot,T)\|_{H^{1}(\Omega)}+\|\Delta
u(\cdot,T)\|_{L^{2}(\Omega)}=\|v(\cdot,T)\|_{H^{1}(\Omega)}+\|\partial_{t}v(\cdot,T)\|_{L^{2}(\Omega)}$
$\displaystyle\leq$ $\displaystyle
C\|\partial_{\nu}u\|_{H^{1}(T_{1},T_{2};L^{2}(\Gamma))}.$

Since $u(\cdot,T)=0$ on $\partial\Omega$, the elliptic regularity implies
$\|u(\cdot,T)\|_{H^{2}(\Omega)}\leq C\|\Delta u(\cdot,T)\|_{L^{2}(\Omega)}$,
we have

$\|u(\cdot,T)\|_{H^{2}(\Omega)}+\|\partial_{t}u(\cdot,T)\|_{H^{1}_{0}(\Omega)}\leq
C\|\partial_{\nu}u\|_{H^{1}(T_{1},T_{2};L^{2}(\Gamma))}.$ $None$

Similarly to (4.1), we can obtain

$u(x,t)=\sum_{n=1}^{\infty}\left(\int^{t}_{0}\frac{\sin\sqrt{\lambda_{n}}(t-s)}{\sqrt{\lambda_{n}}}\mu(s)ds\right)P_{n}f(x)$

in $H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ for $0<t<T$. Therefore,

$\partial_{t}u(x,t)=\sum_{n=1}^{\infty}\left(\int^{t}_{0}\cos\sqrt{\lambda_{n}}(t-s)\mu(s)ds\right)P_{n}f(x)$

in $H^{1}_{0}(\Omega)$ for $0<t<T$. Hence,

$\left\\{\begin{array}[]{rl}&u(x,T)=\sum_{n=1}^{\infty}\left(\int^{T}_{0}\frac{\sin\sqrt{\lambda_{n}}(T-s)}{\sqrt{\lambda_{n}}}\mu(s)ds\right)P_{n}f(x),\\\
&\partial_{t}u(x,T)=\sum_{n=1}^{\infty}\left(\int^{T}_{0}\cos\sqrt{\lambda_{n}}(T-s)\mu(s)ds\right)P_{n}f(x)\quad\mbox{in
$L^{2}(\Omega)$}.\end{array}\right.$ $None$

On the other hand, by $\mu(T)=0$, the integration by parts yields

$\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds=-\frac{\mu(0)\cos
T\sqrt{\lambda_{n}}}{\sqrt{\lambda_{n}}}-\frac{1}{\sqrt{\lambda_{n}}}\int^{T}_{0}\mu^{\prime}(s)\cos\sqrt{\lambda_{n}}(T-s)ds.$

The Riemann-Lebesgue theorem implies

$\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds=-\frac{\mu(0)\cos
T\sqrt{\lambda_{n}}}{\sqrt{\lambda_{n}}}+o\left(\frac{1}{\sqrt{\lambda_{n}}}\right)\quad\mbox{as
$n\to\infty$}.$ $None$

Similarly we can verify

$\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}(T-s)ds=\frac{\mu(0)\sin
T\sqrt{\lambda_{n}}}{\sqrt{\lambda_{n}}}+o\left(\frac{1}{\sqrt{\lambda_{n}}}\right)\quad\mbox{as
$n\to\infty$}.$ $None$

Moreover, by (6.5), we have

$\displaystyle\|u(\cdot,T)\|^{2}_{H^{2}(\Omega)}+\|\partial_{t}u(\cdot,T)\|^{2}_{H^{1}_{0}(\Omega)}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\lambda_{n}\left(\left|\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds\right|^{2}+\left|\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}(T-s)ds\right|^{2}\right)\|P_{n}f\|^{2}_{L^{2}(\Omega)}.$

By (6.6) and (6.7), we see

$\displaystyle J_{n}$
$\displaystyle:=\lambda_{n}\left(\left|\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds\right|^{2}+\left|\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}(T-s)ds\right|^{2}\right)$
$\displaystyle=$
$\displaystyle\mu(0)^{2}(\cos^{2}T\sqrt{\lambda_{n}}+\sin^{2}T\sqrt{\lambda_{n}})+o(1)\geq\mu(0)^{2}-o(1)$

as $n\longrightarrow\infty$. We choose large $N\in\mathbb{N}$ such that
$J_{n}\geq\frac{\mu(0)^{2}}{2}$ for $n\geq N$. By (1.31) we obtain

$\mu_{0}:=\min\left\\{\min_{1\leq n\leq
N}J_{n},\,\frac{\mu(0)^{2}}{2}\right\\}>0,$

and so

$\lambda_{n}\left(\left|\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds\right|^{2}+\left|\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}(T-s)ds\right|^{2}\right)\geq\mu_{0},\quad
n\in\mathbb{N},$

which yields

$\|u(\cdot,T)\|^{2}_{H^{2}(\Omega)}+\|\partial_{t}u(\cdot,T)\|^{2}_{H^{1}_{0}(\Omega)}\geq\mu_{0}\sum_{n=1}^{\infty}\sum_{k=1}^{d_{n}}|(f,\varphi_{nk})|^{2}=\mu_{0}\|f\|^{2}_{L^{2}(\Omega)}.$

The combination with (6.4) completes the proof of the second inequality in
(1.32).

Finally we prove the first inequality in (1.32). Applying the direct
inequality (e.g., Komornik [16]) to (6.1), for any $T_{0}>T$, we have

$\|\partial_{\nu}u\|_{H^{1}(T,T_{0};L^{2}(\partial\Omega))}\leq\|\partial_{\nu}v\|_{L^{2}(T,T_{0};L^{2}(\partial\Omega))}\leq
C(T_{0})(\|v(\cdot,T)\|_{H^{1}(\Omega)}+\|\partial_{t}v(\cdot,T)\|_{L^{2}(\Omega)}).$

Here $C(T_{0})>0$ depends on $T_{0}>T$. Hence,

$\|\partial_{\nu}u\|_{H^{1}(T_{1},T_{2};L^{2}(\partial\Omega))}\leq
C(\|v(\cdot,T)\|_{H^{1}(\Omega)}+\|\partial_{t}v(\cdot,T)\|_{L^{2}(\Omega)}).$
$None$

By $\mu^{\prime}\in L^{2}(0,T)$, we apply the usual energy estimate (or
Theorem 8.2 (p.275) in [17] for example) to (6.2) for $0<t<T$, we have

$\|v(\cdot,T)\|_{H^{1}(\Omega)}+\|\partial_{t}v(\cdot,T)\|_{L^{2}(\Omega)}\leq
C\|\mu^{\prime}f\|_{L^{2}(0,T;L^{2}(\Omega))},$

with which (6.8) completes the proof of the first inequality in (1.32). Thus
the proof of Theorem 5 is complete.

##  7 Proofs of Propositions 2 and 3

###  7.1 Proof of Proposition 2

We set $u:=u_{\mu_{1}}-u_{\mu_{2}}$ and $\mu:=\mu_{1}-\mu_{2}$. Then we have
(5.1) for $t>T$. Therefore, in view of (5.2), we obtain
$u\left(\cdot,\frac{T_{1}+T_{2}}{2}\right)=0$ in $\Omega$. The uniqueness for
a heat equation (5.1) backward in time (e.g., Imanuvilov and Yamamoto [12],
Isakov [13]) yields $u(\cdot,T)=0$ in $\Omega$. Here, by $u\in
C([0,T_{2}];L^{2}(\Omega))$, we see that

$\lim_{\varepsilon\to
0,\varepsilon>0}u(\cdot,T+\varepsilon)=\lim_{\varepsilon\to
0,\varepsilon>0}u(\cdot,T-\varepsilon),$

and (3.1) yields

$\sum_{n=1}^{\infty}e^{-\lambda_{n}T}\left(\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds\right)(P_{n}f)(x)=0,\quad
x\in\Omega.$

Therefore, by definition (1.21) of $\widetilde{\Lambda}$, we obtain

$\int^{T}_{0}e^{\lambda_{n}s}\mu(s)ds=0,\quad
n\in\mathbb{N}\setminus\widetilde{\Lambda}.$

Similarly to the argument after (3.3), we apply the Müntz theorem ([3]) to
obtain $\mu(t)=0$ for $0<t<T$ if and only if (1.22) holds. Thus the proof of
Proposition 2 is complete.

###  7.2 Proof of Proposition 3

We set $u:=u_{\mu_{1}}-u_{\mu_{2}}$ and $\mu:=\mu_{1}-\mu_{2}$. Then we have
(6.1). First, since $\partial_{\nu}u=0$ on $\Gamma\times(T_{1},T_{2})$, by
(1.24) the observability inequality (e.g., [16]) to (6.1) yields
$u(\cdot,T_{1})=\partial_{t}u(\cdot,T_{1})=0$ in $\Omega$. Therefore, in (6.1)
we see that $u(\cdot,T)=\partial_{t}u(\cdot,T)=0$ in $\Omega$. Here we note
that $u\in C^{1}([0,T_{2}];L^{2}(\Omega))$. Hence, (6.4) implies

$\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}(T-s)ds=0,\quad\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}(T-s)\mu(s)ds=0,\quad
n\in\mathbb{N}\setminus\widetilde{\Lambda},$

that is,

$\left\\{\begin{array}[]{rl}&\sin
T\sqrt{\lambda_{n}}\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds-\cos
T\sqrt{\lambda_{n}}\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds=0,\\\ &\cos
T\sqrt{\lambda_{n}}\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds+\sin
T\sqrt{\lambda_{n}}\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds=0,\quad
n\in\mathbb{N}\setminus\widetilde{\Lambda}.\end{array}\right.$

Considering the system as linear equations in
$\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds$ and
$\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds$, since the determinant of the
coefficient matrix is non-zero:
$\sin^{2}T\sqrt{\lambda_{n}}+\cos^{2}T\sqrt{\lambda_{n}}=1\neq 0$, we see

$\int^{T}_{0}\mu(s)\cos\sqrt{\lambda_{n}}sds=\int^{T}_{0}\mu(s)\sin\sqrt{\lambda_{n}}sds=0,\quad
n\in\mathbb{N}\setminus\widetilde{\Lambda}.$ $None$

In view of (1.23) and (1.25), we can argue similarly to the proof of
Proposition 1, and by Lemma 5, we can reach $\mu(s)=0$, $0<s<T$. Thus the
proof of Proposition 3 is complete.

##  8 Appendix I. Uniqueness in determining $\mu(t)$ by data over time
interval $(0,T)$

Our main interest is the inverse source problems by data over the time
interval $(T_{1},T_{2})$ where $T<T_{1}<T_{2}$, because the inverse problems
with data over $(0,T)$ have been well studied. However, uniqueness results by
data over $(0,T)$ seem a kind of folklore, and it is not easy to find relevant
articles. Thus for completeness, we here show the uniqueness limited to the
case of the determination of $\mu(t)$.

Our argument can work for the case where $\Delta$ is replaced by a general
uniformly elliptic operator with some regularity condition on the
coefficients, but we are restricted to (1.1) and (1.2) with the zero Dirichlet
boundary condition. For simplicity we assume that $f\in
C^{\infty}_{0}(\Omega)$.
Case 1: one-dimensional case and pointwise data $u(x_{0},t)$, $0<t<T$ with
$x_{0}\in\Omega:=(0,\ell)$.

We recall

$\lambda_{n}=\frac{n^{2}\pi^{2}}{\ell^{2}},\quad\varphi_{n}(x)=\frac{\sqrt{2}}{\sqrt{\ell}}\sin\frac{n\pi}{\ell}x,\quad
n\in\mathbb{N},$

and $(f,g):=\int^{\ell}_{0}f(x)g(x)dx$ for $f,g\in L^{2}(0,\ell)$. By $u_{k}$,
$k=1,2$, we denote the solution to (1.1) and (1.2) respectively. that is,
$u_{1}$ and $u_{2}$ are the solutions for the diffusion and the wave equations
respectively.
Proposition A.
We assume

$\mbox{there exists $n_{1}\in\mathbb{N}$ such that
$(f,\varphi_{n_{1}})\varphi_{n_{1}}(x_{0})\neq 0$}.$ $None$

(i): Case of diffusion equation. If $u_{1}(x_{0},t)=0$ for $0<t<T$, then
$\mu(t)=0$ for $0<t<T$.
(ii): Case of wave equation. If $u_{2}(x_{0},t)=0$ for $0<t<T$, then
$\mu(t)=0$ for $0<t<T-\ell$.

We can compare Proposition A (i) and (ii) with Theorems 2 and 3 respectively,
which naturally require more conditions. In (ii), if $T\leq\ell$, then we can
not conclude the uniqueness for any time interval $(0,\varepsilon)$ with
arbitrary $\varepsilon>0$, and in view of the propagation speed $1$, it is
natural that $T>\ell$ is needed for the meaningful uniqueness.
Remark.
If either

$f(x_{0})\neq 0,$

or

$\frac{x_{0}}{\ell}\not\in\mathbb{Q},\quad f\not\equiv 0\quad\mbox{in
$(0,\ell)$},$

then (8.1) holds. Indeed if (8.1) does not hold, then
$(f,\varphi_{n})\varphi_{n}(x_{0})=0$ for all $n\in\mathbb{N}$, which implies
$f(x_{0})=\sum_{n=1}^{\infty}(f,\varphi_{n})\varphi_{n}(x_{0})=0$. Hence, in
the first case, (8.1) holds. Furthermore, in the second case we note that
$\frac{x_{0}}{\ell}\not\in\mathbb{Q}$ implies that $\varphi_{n}(x_{0})\neq 0$
for all $n\in\mathbb{N}$.
Case 2: general dimensional case and data $\partial_{\nu}u$ on lateral
subboundary of $\partial\Omega\times(0,T)$.

For arbitrarily chosen $x_{0}\in\mathbb{R}^{d}$, we recall that $\nu=\nu(x)$
denotes the unit outward normal vector to $\partial\Omega$ at $x$, and

$\Gamma(x_{0}):=\\{x\in\partial\Omega;\,(x-x_{0})\cdot\nu(x)\geq 0\\},\quad
R(x_{0}):=\max_{x\in\overline{\Omega}}|x-x_{0}|.$

Proposition B.
We assume that $f\not\equiv 0$ in $\Omega$.
(i): Case of diffusion equation. Let $\gamma\subset\partial\Omega$ be an
arbitrarily fixed subboundary. If
$\partial_{\nu}u_{1}|_{\gamma\times(0,T)}=0$, then $\mu(t)=0$ for $0<t<T$.
(ii): Case of wave equation. If
$\partial_{\nu}u_{2}|_{\Gamma(x_{0})\times(0,T)}=0$, then $\mu(t)=0$ for
$0<t<T-R(x_{0})$.

Similarly to Proposition A, in the case of the wave equation, owing to the
finite propagation speed $1$, the time interval for the uniqueness of $\mu(t)$
is reduced by $R(x_{0})$ from the observation time length $T$. The conclusion
of (ii) makes sense only if $T>R(x_{0})$. Proposition B (i) and (ii) should be
compared with Propositions 2 and 3 respectively where data are taken over
$(T_{1},T_{2})$ with $T_{1}>0$. Also in (ii), for the uniqueness we can take
arbitrary subboundary with relevantly long time interval as long as the wave
equation has analytic coefficients. For the uniqueness we can apply Fritz
John’s global Holmgren theorem (e.g., John [15]), but we omit the details.

For the proofs of the propositions, we reduce the problems to initial boundary
value problems for equations with the zero right-hand sides. More precisely,
for $k=1,2$, let $v_{k}=v_{k}(x,t)$ satisfy

$\left\\{\begin{array}[]{rl}&\partial_{t}^{k}v_{k}(x,t)=\Delta
v_{k}(x,t),\quad x\in\Omega,\,t>0,\\\ &v_{k}(x,t)=0,\quad
x\in\partial\Omega,\,t>0,\\\
&\left\\{\begin{array}[]{rl}&v_{k}(x,0)=f(x),\quad x\in\Omega,\quad\mbox{if
$k=1$},\\\ &v_{k}(x,0)=0,\quad\partial_{t}v_{k}(x,0)=f(x),\quad
x\in\Omega,\quad\mbox{if $k=2$}.\\\ \end{array}\right.\end{array}\right.$
$None$

Then, as is easily verified, we see

$u_{k}(x,t)=\int^{t}_{0}\mu(t-s)v_{k}(x,s)ds,\quad x\in\Omega,\,t>0$ $None$

for $k=1,2$.
Proof of Proposition A.
For $k=1,2$, let $u_{k}(x_{0},t)=0$ for $0<t<T$. Then (8.3) yields

$\int^{t}_{0}\mu(t-s)v_{k}(x_{0},s)ds=0,\quad 0<t<T.$

The Titchmarsh theorem on the convolution (Titchmarsh [20]) yields that there
exists $t_{*}\in(0,T)$ such that

$\left\\{\begin{array}[]{rl}&\mu(s)=0,\quad 0<s<t_{*},\\\
&v_{k}(x_{0},s)=0,\quad 0<s<T-t_{*}.\end{array}\right.$ $None$

We assume that $t_{*}<T$. Otherwise $\mu(s)=0$ for $0<s<T$ have been already
proved.
Proof of (i).
We know

$v_{1}(x_{0},t)=\sum_{n=1}^{\infty}e^{-\lambda_{n}t}(f,\varphi_{n})\varphi_{n}(x_{0}),\quad
t>0.$

Then

$\sum_{n=1}^{\infty}e^{-\lambda_{n}t}(f,\varphi_{n})\varphi_{n}(x_{0})=0,\quad
0<t<T-t_{*}.$ $None$

We can apply Lemma 3 in Section 3 to obtain
$(f,\varphi_{n})\varphi_{n}(x_{0})=0$ for $n\in\mathbb{N}$. This contradicts
assumption (8.1). Hence $t_{*}<T$ is impossible and so $t_{*}=T$, that is,
$\mu(t)=0$ for $0<t<T$. Thus the proof of Proposition A (i) is complete.
Proof of (ii).
In the case of the wave equation, we have

$v_{2}(x,t)=\sum_{n=1}^{\infty}\frac{\sin\sqrt{\lambda_{n}}t}{\sqrt{\lambda_{n}}}(f,\varphi_{n})\varphi_{n}(x_{0})=0$
$None$

for $0<t<T-t_{*}$ (e.g., Komornik [16]). If $T-t_{*}\geq\ell$, then (8.6)
holds for $0<t<\ell$. Recalling that
$\varphi_{n}(x)=\frac{\sqrt{2}}{\sqrt{\ell}}\sin\frac{n\pi}{\ell}t$ with
$n\in\mathbb{N}$, we can obtain

$(f,\varphi_{n})\varphi_{n}(x_{0})=0\quad\mbox{for all $n\in\mathbb{N}$},$

which contradicts assumption (8.1). Hence, $T-t_{*}<\ell$, that is,
$t_{*}>T-\ell$. Thus (8.4) yields $\mu(t)=0$ for $0<t<T-\ell$, and the proof
of Proposition A (ii) is complete.
Proof of Proposition B.
Similarly to the proof of Proposition A, we obtain

$\left\\{\begin{array}[]{rl}&\mu(s)=0,\quad 0<s<t_{*},\\\
&\partial_{\nu}v_{k}=0\quad\mbox{on $\gamma\times(0,T-t_{*})$ if $k=1$},\\\
&\partial_{\nu}v_{k}=0\quad\mbox{on $\Gamma(x_{0})\times(0,T-t_{*})$ if
$k=2$}.\end{array}\right.$ $None$

Proof of (i).
Assume that $t_{*}<T$. Then $\partial_{\nu}v_{1}=0$ on
$\gamma\times(0,T-t_{*})$. With this, we apply the classical unique
continuation for the heat equation (e.g., Isakov [13], Yamamoto [26]), and so
$v_{1}=0$ in $\Omega\times(0,T)$, which yields $f=0$ in $\Omega$. This
contradicts the assumption $f\not\equiv 0$ in $\Omega$. Therefore, $t_{*}<T$
is impossible, and we reach $t_{*}=T$. The proof of (i) is complete.
Proof of (ii).
Assume that $t_{*}<T$. Then $\partial_{\nu}v_{2}=0$ on
$\Gamma(x_{0})\times(0,T-t_{*})$. If $T-t_{*}>R(x_{0})$, then the
observability inequality (e.g., [16]) or the unique continuation by Carleman
estimate (e.g., [2]) yields $v_{2}=0$ in $\Omega\times(0,T)$. That is, $f=0$
in $\Omega$. By the assumption $f\not\equiv 0$ in $\Omega$, it turns out that
$T-t_{*}>R(x_{0})$ is impossible. Hence, $T-t_{*}\leq R(x_{0})$, that is,
$t_{*}\geq T-R(x_{0})$. Consequently (8.7) implies $\mu(t)=0$ for
$0<t<T-R(x_{0})$. Thus the proof of Proposition B is complete.

##  9 Appendix II. Proofs of Lemmas 1 and 2

We number all the eigenvalues of $A=-\Delta$ with
$\mathcal{D}(A)=H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ with the multiplicities.
Let $\\{\varphi_{j}\\}_{j\in\mathbb{N}}$ be eigenfunctions forming an
orthonormal basis in $L^{2}(\Omega)$: $A\varphi_{j}=\sigma_{j}\varphi_{j}$.
Let $T_{0}>0$ be arbitrary.
Proof of Lemma 1.
(i) We refer to the example in Evans [9, Thm. 5 in Ch. 7] to know that there
exists a unique solution $u\in L^{2}(0,T_{0};H^{2}(\Omega))\cap
H^{1}(0,T_{0};L^{2}(\Omega))$ to (1.1). Moreover, similarly to (5.3), we have

$u(x,t)=\sum_{j=1}^{\infty}e^{-\sigma_{j}t}\left(\int^{t}_{0}e^{\sigma_{j}s}\mu(s)ds\right)(f,\varphi_{j})\varphi_{j}(x),\quad
x\in\Omega,\,t>0,$ $None$

where the series is convergent in $L^{2}(0,T_{0};H^{2}(\Omega))\cap
H^{1}(0,T_{0};L^{2}(\Omega))$. Since $f\in
C^{\infty}_{0}(\Omega)\subset\mathcal{D}(A^{\ell})$ with any
$\ell\in\mathbb{N}$ and $\sigma_{j}\sim\rho_{o}j^{\frac{2}{d}}$ as
$j\to\infty$ (e.g., [1], [8]), we see (5.6) and

$|(f,\varphi_{j})|\leq\frac{C}{j^{\frac{2\ell}{d}}}\|A^{\ell}f\|_{L^{2}(\Omega)},\quad\ell\in\mathbb{N}.$
$None$

Moreover, with $\ell\in\mathbb{N}$ satisfying $\ell_{0}>\frac{d}{4}$, the
Sobolev embedding yields

$\|\varphi_{j}\|_{C(\overline{\Omega})}\leq
C\|\varphi_{j}\|_{H^{2\ell_{0}}(\Omega)}\leq
C\|A^{\ell_{0}}\varphi_{j}\|_{L^{2}(\Omega)}=C\sigma_{j}^{\ell_{0}}\leq
Cj^{\frac{2\ell_{0}}{d}}.$

Therefore, for sufficiently large $\ell\in\mathbb{N}$, we see that the series
in (9.1) is convergent in $L^{\infty}(0,T_{0};H^{2}(\Omega))$, and so $u\in
C([0,T_{0}];H^{2}(\Omega))$. Thus (1.3) is verified.

Since $\partial_{t}u=-Au+\mu(t)f$ in $\Omega\times(0,T)$, in terms of (9.2),
we can similarly prove that $\partial_{t}u\in C([0,T_{0}];L^{2}(\Omega))$.
Thus the proof of Lemma 1 (i) is complete.
(ii) Setting $v=\partial_{t}u$, we have

$\left\\{\begin{array}[]{rl}&\partial_{t}^{2}v=\Delta
v+\mu^{\prime}(t)f(x),\quad x\in\Omega,\,0<t<T_{0},\\\
&v(x,0)=0,\quad\partial_{t}v(x,0)=\mu(0)f(x),\quad x\in\Omega,\\\
&v(x,t)=0,\quad x\in\partial\Omega,\,0<t<T_{0}.\end{array}\right.$

Since $\mu^{\prime}\in L^{2}(0,T_{0})$ and $f\in L^{2}(\Omega)$, we see that
$v\in C([0,T_{0}];H^{1}_{0}(\Omega))\cap C^{1}([0,T_{0}];L^{2}(\Omega))$
(e.g., [17], Theorem 8.2 (p.275)). Therefore,

$u\in C^{1}([0,T_{0}];H^{1}_{0}(\Omega))\cap C^{2}([0,T_{0}];L^{2}(\Omega)).$

Moreover, $\Delta u(\cdot,t)=\partial_{t}^{2}u(\cdot,t)-\mu(t)f$ in $\Omega$
and $u(\cdot,t)=0$ on $\partial\Omega$ for each $t\in[0,T_{0}]$, and the
elliptic regularity yields $u\in C([0,T_{0}];H^{2}(\Omega))$. Since

$u(x,t)=\sum_{j=1}^{\infty}\left(\int^{t}_{0}\frac{\sin(t-s)\sqrt{\lambda_{j}}}{\sqrt{\lambda_{j}}}\mu(s)ds\right)(f,\varphi_{j})\varphi_{j}(x),\quad
x\in\Omega,\,t>0$

(e.g., [16]), in terms of (9.2) we can see that the series is convergent on
$\overline{\Omega}\times[0,T_{0}]$, so that $u\in
C(\overline{\Omega}\times[0,T_{0}])$ follows.

Finally $\partial_{\nu}v\in L^{2}(0,T_{0};L^{2}(\partial\Omega))$ is seen by
e.g., [16], and so $\partial_{\nu}u\in H^{1}(0,T_{0};L^{2}(\partial\Omega))$.
Thus the proof of Lemma 1 is complete. $\blacksquare$
Proof of Lemma 2.
The proof is similarly done to Lemma 1 for example by means of Sections 7.1
and 7.2 of Chapter 7 in Evans [9], Theorem 8.2 (p.275) in Lions and Magenes
[17], Komornik [16] and Pazy [18], and we do not repeat the details.

## Acknowledgments

The first author was supported by NSFC (no. 11971104, 11971121). The second
author was supported by NSFC (no.11925104) and Program of Shanghai
Academic/Technology Research Leader (19XD1420500). The third author was
supported by 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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