# Increasing stability for the inverse source problems in electrodynamics

Suliang Si School of Mathematics and Statistics, Shandong University of
Technology, Shandong, 255049, China<EMAIL_ADDRESS>
###### Abstract

We are concerned with increasing stability in the inverse source problems for
the time-dependent Maxwell equations in ${\mathbb{R}}^{3}$, where the source
term is compactly supported in both time and spatial variables. By using the
Fourier transform, sharp bounds of the analytic continuation and the Huygens’
principle, increasing stability estimates of the $L^{2}$-norm of the source
function are obtained. The main goal of this paper is to understand increasing
stability for the Maxwell equations in the time domain.

##  1 Introduction

###  1.1 statement of the problem

Consider the time-dependent Maxwell equations in a homogeneous medium

$\mu\partial_{t}\textbf{H}(\textbf{x},t)+\nabla\times\textbf{E}(\textbf{x},t)=0,\quad\varepsilon\partial_{t}\textbf{E}(\textbf{x},t)-\nabla\times\textbf{H}(\textbf{x},t)=-\sigma\textbf{E}+\tilde{\textbf{F}}(\textbf{x},t),\
\textbf{x}\in{\mathbb{R}}^{3},\ t>0,$ (1.1)

where E and H are the electric and magnetic fields, respectively, the source
function $\tilde{\textbf{F}}$ is known as the electric current density,
$\varepsilon$ and $\mu$ are the dielectric permittivity and the magnetic
permeability, respectively, and $\sigma$ is the electric conductivity and is
assumed to be zero.

Let $n=\frac{\varepsilon}{\mu}$. Eliminating the magnetic field H from (1.1),
we obtain the Maxwell system for the electric field E:

$n\partial_{t}^{2}\textbf{E}(\textbf{x},t)+\nabla\times(\nabla\times\textbf{E}(\textbf{x},t))=\frac{1}{\mu}\partial_{t}\tilde{\textbf{F}}(\textbf{x},t)=:\textbf{F}(\textbf{x},t),\
\textbf{x}\in{\mathbb{R}}^{3},t>0,$ (1.2)

which is supplemented by the homogeneous initial conditions

$\textbf{E}(\textbf{x},0)=\partial_{t}\textbf{E}(\textbf{x},0)=0,\
\textbf{x}\in{\mathbb{R}}^{3}.$ (1.3)

In this work, we consider the following inverse source problems:

IP1: Assume $\textbf{F}(\textbf{x},t)=\textbf{f}(\textbf{x})g(t)$, where
$\textbf{f}\in{\mathbb{R}}^{3},\ g\in{\mathbb{R}}$ have compact supports such
that supp f $\subset\\{\textbf{x}\in\mathbb{R}^{3}|\ |\textbf{x}|<R\\}$ and
supp $g$ $\subset(0,T_{0})$ where $R>0,\ T_{0}>0$ are constants. Then we
concern the inverse problem of recovering the source term
$\textbf{f}(\textbf{x})$ from the Dirichlet boundary data measured on
$\partial B_{R}\times(0,T_{0})$.

IP2: Assume $n\in(0,b^{2})$ for $b>1$. The source function
$\textbf{F}(\textbf{x},t)$ has compact supports. The inverse problem is to
determine source term $\textbf{F}(\textbf{x},t)$ from the measurement
$\textbf{E}(\textbf{x},t)$, $\textbf{x}\in\partial B_{R}$, $t\in(0,T_{0})$.

IP3: Assume $\textbf{F}(\textbf{x},t)=\textbf{f}(x_{1},x_{2},t)g(x_{3})$,
where the function $\textbf{f}\in{\mathbb{R}}^{3}$ and $g\in{\mathbb{R}}$ have
compact supports. The inverse problem is to determine source term f from the
boundary observation data $\textbf{E}(\textbf{x},t)$, $\textbf{x}\in\partial
B_{R}$, $t\in(0,T_{0})$.

###  1.2 Motivations

Motivated by significant applications, the inverse source problems, as an
important research subject in inverse scattering theory, have continuously
attracted much attention by many researchers [1, 3, 22, 25, 34]. Consequently,
a great deal of mathematical and numerical results are available, especially
for the acoustic waves or the Helmholtz equations. In general, it is known
that there is no uniqueness for the inverse source problem at a fixed
frequency due to the existence of non-radiating sources [20]. Therefore,
additional information is required for the source in order to obtain a unique
solution, such as to seek the minimum energy solution [29]. From the
computational point of view, a more challenging issue is the lack of
stability. A small variation of the data might lead to a huge error in the
reconstruction. Recently, it has been realized that the use of multi-frequency
data is an effective approach to overcome the difficulties of non-uniqueness
and instability which are encountered at a single frequency. The first
increasing stability result in [4] was obtained in a more particular case by
quite different (spatial Fourier analysis) methods. A topical review can be
found in [10] on the inverse source problems as well as other inverse
scattering problems by using multiple frequencies to overcome the ill-
posedness and gain increased stability.

For electromagnetic waves, Ammari et al. [11] showed uniqueness and stability,
and presented an inversion scheme to reconstruct dipole sources based on a
low-frequency asymptotic analysis of the time-harmonic Maxwell equations. In
[2], Albanese and Monk discussed uniqueness and non-uniqueness of the inverse
source problems for Maxwell’s equations. A monograph can be found in [31] on
general inverse problems for Maxwell’s equations. In [5], Bao et al. develop
new techniques and establish an increasing stability theory in the inverse
source scattering problems for electromagnetic waves. Due to the inverse
source problem of electromagnetic waves, there is a phenomenon of increasing
stability. A natural question is whether there is also some increasing
stability in the inverse source problem of time-dependent Maxwell equations.
Thus the main goal of this paper is to understand increasing stability for the
Maxwell equations in the time domain.

###  1.3 Known results

Inverse sourse problems have many significant applications in scientific and
engineering areas. For instance, detection of submarines and non-destructive
measurement of industrial objects can be regarded as recovery of acoustic
sources from boundary measurements of the pressure. Other application include
biomedical imaging optical tomography [1, 22], and geophysics. For a
mathematical overview of various inverse source problems, we can see that
uniqueness and stability are discussed in [22]. For inverse source problems in
time domain, it is solved as hyperbolic systems by using Carleman estimate
[26] and unique continuation; we refer to [12, 25, 34, 35] for an incomplete
list. For time-harmonic inverse source problems, it is well-known that there
is no uniqueness for the inverse source problem with a single frequency due to
the existence of non-radiating sources [3, 20]. Therefore, the use of multiple
frequencies data is an effective way to overcome non-uniqueness and has
received a lot of attention in recent years. [16] show the uniqueness and
numerical results for Helmholtz equation with multi-frequency data. And in
[4], Bao et al. firstly get increasing stability for Helmholtz equation by
direct spatial Fourier analysis methods. Then in [13], a different method
involving a temporal Fourier transform, sharp bounds of the analytic
continuation to higher wave numbers were used to derive increasing stability
bounds for the three dimensional Helmholtz equation. Also, [13] firstly
combined the Helmholtz equation and associated hyperbolic equations to get the
stability results. Later in [27] and [5], increasing stability were extended
to Helmholtz equation and Maxwell’s equation in three dimension.

Motivated by those works, we are interested in the increasing stability for
the time-dependent Maxwell equations. Since by Fourier transform, inverse
source problem of the time-dependent Maxwell equations can be reduced to that
of the associated Electromagnetic with multi-frequency data, we derive our
increasing stability estimate by using sharp bounds of analytic continuation
given in [13]. Instead of Cauchy data on measure boundary used in the works
mentioned above, we use only Dirichlet data on the lateral boundary. For the
uniqueness, we just prove it in time domain without Carleman estimate.

The rest of this paper is organized as follows. In Section 2 and 3, we state
our main results and well-posedness of the direct problem. Sections 3 is
devoted to the increasing stability of inverse problem 1. In Section 4, we
prove general source terms in homogeneous medium. Section 5, we establish the
increasing stability of general source term. Finally, the increasing stability
of time-dependent source term can be established by using the boundary
Dirichlet data.

##  2 Main results

Let $B_{R}=B(0,R)=\\{\textbf{x}\in{\mathbb{R}}^{3}|\ |\textbf{x}|<R\\}$ for
$R>0$. We always assume that the source function $\textbf{F}(\textbf{x},t)$ is
required to be real-valued, which implies that
$\widehat{\textbf{F}}(-\xi,-\omega)=\overline{\textbf{F}}(\xi,\omega)$ for all
$(\xi,\omega)\in{\mathbb{R}}^{3}\times{\mathbb{R}}$. In addition, we also
assume that $\nabla\cdot\textbf{{F}}(\textbf{x},t)=0$. It follows from that
the source term $\textbf{F}(\textbf{x},t)$ can be decomposed into a sum of
radiation and non-radiating parts. The non-radiating part cannot be determined
and gives rise to the non-uniqueness issue. By the divergence-free condition
of $\textbf{F}(\textbf{x},t)$, we eliminate non-radiating sources in order to
ensure the uniqueness of the inverse problem. First we show increasing
stability result for the time-dependent inverse problem (IP1). Let
$\textbf{F}(\textbf{x},t)=\textbf{f}(\textbf{x})g(t)$, where
$\textbf{f}\in{\mathbb{R}}^{3}$ is compactly supported in $B_{R}$ and
$g\in{\mathbb{R}}$ is supported in $(0,T_{0})$ for some $T_{0}>0$. It is
supposed that $\textbf{f}(\textbf{x})\in H^{1}({\mathbb{R}}^{3})^{3}$ and
$g(t)\in H^{4}(0,\infty)$. The one-dimensional Fourier transform of $g$ with
respect to the time variable $t$ is defined as

$\widehat{g}(\omega):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}g(t)e^{-i\omega
t}dt=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}g(t)e^{-i\omega t}dt.$

We suppose there exist a number $b>1$ and a constant $\delta>0$ such that

$\displaystyle|\widehat{g}(\omega)|\geq\delta>0\quad\mbox{for
all}\quad\omega\in(0,b).$ (2.4)

Physically, the parameter $b$ in (2.4) is associated with the bandwidth of the
temporal signal $g(t)$. The condition (2.4) covers a large class of functions.
For example, if $g(t)=e^{-(t-1)^{2}/\eta}\chi(t)$ with some $\eta>0$ and
$\chi(t)\in C_{0}^{\infty}({\mathbb{R}})$ such that $\chi(t)=0$ for
$t\notin[0,T_{0}]$, the one can always find the parameters $b>0$ and
$\delta>0$ such that (2.4) holds true. Since the source function is real
valued, (2.4) holds true for all $\omega\in(-b,b)$. We remark that the
interval $(0,b)$ in (2.4) can be replaced by $(\omega_{0}-b,\omega_{0}+b)$ for
some $\omega_{0}\geq 0$. In this paper we take $\omega_{0}=0$ for simplicity.
The condition (2.8) is also similar.

Define a boundary operator

$T(\textbf{E}\times\nu)=(\nabla\times\textbf{E})\times\nu\quad\mbox{on}\
\partial B_{R}\times(0,T_{0}),$ (2.5)

where $\nu$ is the unit normal vector on $\partial B_{R}$.

In the following theorem, we establish increasing stability estimate of the
$L^{2}$-norm of f about $b$.

######  Theorem 2.1.

Let the condition (2.4) hold and let $T>2R+T_{0}$. Assume that $g(t)$ is given
and $||\textbf{f}||_{H^{1}({\mathbb{R}}^{3})^{3}}\leq M$ where $M>1$ is a
constant. If $\nabla\cdot\textbf{f}(x)=0$, then there exists a constant $C>0$
depending on $\delta$, $T_{0}$, $n$ and $R$ such that

$||\textbf{f}||_{L^{2}({\mathbb{R}}^{3})^{3}}^{2}\leq
C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}),$
(2.6)

where $\epsilon=\Big{(}\int_{0}^{T}\int_{\partial
B_{R}}\big{(}|T(\textbf{E}\times\nu)|^{2}+|\textbf{E}\times\nu|^{2}\big{)}ds(\textbf{x})dt\Big{)}^{\frac{1}{2}}$.

######  Remark 2.2.

There are two parts in the stability estimates: the first parts are the data
discrepancy, while this second parts comes from the high frequency tail of the
function. The coefficient appearing in the Lipschitz part of $(\ref{es0})$ is
polynomial type. However, since $b$ is fixed in practice, these coefficients
are constants and do not pose any problem. It is clear to conclude that the
ill-posedness of the inverse time-dependent source problem decreases as the
parameter $b$ increases. This is in consistent with the increasing stability
results of [5] in the frequency domain: the ill-posedness decreases when the
width of the wavenumber interval $(0,b)$ increases.

Next, we state the stability estimate for the second inverse problem (IP2).
Let $n\in(0,b^{2})$, $b>1$. In the following theorem, we establish the
increasing stability estimate of $L^{2}$-norm of F about $b$. Let
$\textbf{E}=\textbf{E}(\textbf{x},t;n)$ satisfy (1.2) and (1.3).

######  Theorem 2.3.

Let $\textbf{F}(\textbf{x},t)\in
H^{4}([0,\infty);H^{1}({\mathbb{R}}^{3}))^{3}$ be such that supp
$\textbf{F}(\textbf{x},t)$ $\subset B_{R}\times(0,T_{0})$ and let
$T>2Rn+T_{0}$. Assume that there exists $M>1$ and
$\|\textbf{F}\|_{H^{4}([0,\infty);H^{1}({\mathbb{R}}^{3}))^{3}}\leq M$. Then
there exists a constant $C>0$ depending on $T_{0}$ and $R$ such that

$\displaystyle\|\textbf{F}\|^{2}_{L^{2}({\mathbb{R}}^{4})}\leq
C\Big{(}b^{11}\,\epsilon^{2}+\frac{M^{2}}{b|\ln\epsilon|^{\frac{2}{5}}}\Big{)}.$
(2.7)

where $\epsilon=\sup\limits_{n\in(0,b^{2})}\Big{(}\int_{0}^{T}\int_{\partial
B_{R}}\big{(}|T(\textbf{E}\times\nu)|^{2}+|\textbf{E}\times\nu|^{2}\big{)}ds(\textbf{x})dt\Big{)}^{\frac{1}{2}}$.

Finally, we present the increasing stability for the third inverse problem
(IP3). Assume that $\textbf{F}(\textbf{x},t)$ takes the form

$\textbf{F}(\textbf{x},t)=\textbf{f}(\tilde{x},t)g(x_{3}),\quad\tilde{x}=(x_{1},x_{2})\in{\mathbb{R}}^{2},\quad
t\in(0,+\infty),$ (2.8)

where $\textbf{f}(\tilde{x},t)$ has compact supports such that supp f
$\subset$
$\widetilde{B}_{R_{0}}\times(0,T_{0}):=\\{(\tilde{x},t)\in{\mathbb{R}}^{2}\times{\mathbb{R}}|\
|\tilde{x}|<R_{0},\ 0<t<T_{0}\\}$. Suppose that $g$ is given and supported in
$(-R_{0},R_{0})$ for some $0<R_{0}<R/\sqrt{2}$.

The Fourier transform of $g(x_{3})$ of space variable $x_{3}$ is given by

$\widehat{g}(\xi_{3})=\frac{1}{\sqrt{2\pi}}\int_{{\mathbb{R}}^{3}}g(x_{3})e^{-i\xi_{3}\cdot
x_{3}}dx_{3}.$ (2.9)

We suppose

$\displaystyle|\widehat{g}(\xi_{3})|\geq\delta>0\quad\mbox{for
all}\quad\xi_{3}\in(-b,b),$ (2.10)

where $b>1$.

######  Theorem 2.4.

Let $\textbf{f}(\tilde{x},t)\in H^{4}([0,\infty);H^{1}({\mathbb{R}}^{2}))^{3}$
be such that
supp$\textbf{f}(\tilde{x},t)\subset\widetilde{B}_{R_{0}}\times(0,T_{0})$ and
let $T>2R+T_{0}$. Assume that there exists $M>1$ such that
$\|\textbf{f}\|_{H^{4}([0,\infty);H^{1}({\mathbb{R}}^{2}))^{3}}\leq M$. Then
there exist constants $\alpha\in(0,1)$ and $C>0$ depending on $n$, $\delta$,
$M$, $T_{0}$ and $R$ such that

$\displaystyle\|\textbf{f}\|^{2}_{L^{2}({\mathbb{R}}^{3})^{3}}\leq
C\Big{(}b^{7}\,\epsilon^{2}+b^{5}\,\,e^{2b(1-\alpha)}\epsilon^{2\alpha}+\frac{M^{2}}{b^{\frac{4}{3}}|\alpha\ln\epsilon|^{\frac{1}{2}}}\Big{)}.$
(2.11)

where $\epsilon=\Big{(}\int_{0}^{T}\int_{\partial
B_{R}}\big{(}|T(\textbf{E}\times\nu)|^{2}+|\textbf{E}\times\nu|^{2}\big{)}ds(\textbf{x})dt\Big{)}^{\frac{1}{2}}$.

##  3 Preliminaries and Proof of Theorem 2.1

In this section. The electric current density is assumed to take the form

$\textbf{F}(\textbf{x},t)=\textbf{f}(\textbf{x})g(t)$

where $g$ is given. Let $\textbf{E}^{inc}$ and $\textbf{H}^{inc}$ be the
electric and magnetic plane waves. Explicitly, we have

$\textbf{E}^{inc}=\textbf{p}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega
t)},\quad\textbf{H}^{inc}=\textbf{q}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega
t)}$

where $\textbf{d}\in{\mathbb{S}}^{2}$ is a unit vector and p, q are two unit
polarization vectors satisfying $\textbf{p}\cdot\textbf{d}=0$,
$\textbf{q}=\textbf{d}\times\textbf{p}$.

If $\kappa_{p}^{2}-n\omega^{2}=0$, it is easy to verify that
$\textbf{E}^{inc}$ and $\textbf{H}^{inc}$ satisfy the homogeneous Maxwell
equations in ${\mathbb{R}}^{3}$:

$n\partial_{t}^{2}\textbf{E}(\textbf{x},t)+\nabla\times(\nabla\times\textbf{E}(\textbf{x},t))=0$

and

$n\partial_{t}^{2}\textbf{H}(\textbf{x},t)+\nabla\times(\nabla\times\textbf{H}(\textbf{x},t))=0.$

Multiplying the both sides of (1.2) by $\textbf{E}^{inc}$, we obtain

$\int_{0}^{T}\int_{B_{R}}(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\
\mathrm{d}\textbf{x}\mathrm{d}t=\int_{0}^{T}\int_{B_{R}}\textbf{f}(\textbf{x})g(t)\textbf{E}^{inc}(\textbf{x},t)\
\mathrm{d}\textbf{x}\mathrm{d}t.$ (3.12)

Integrating by parts, one deduces from the left hand side of (3.12) that

$\displaystyle\int_{0}^{T}\int_{B_{R}}$
$\displaystyle(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\
\mathrm{d}\textbf{x}\mathrm{d}t$ $\displaystyle=$
$\displaystyle\int_{B_{R}}n(\partial_{t}\textbf{E}(\textbf{x},t)\textbf{E}^{inc}(\textbf{x},t)-\textbf{E}(\textbf{x},t)\partial_{t}\textbf{E}^{inc}(\textbf{x},t))\big{|}_{0}^{T}\;\mathrm{d}\textbf{x}$
$\displaystyle\quad+\int_{0}^{T}\int_{\partial
B_{R}}\big{(}\nu\times(\nabla\times\textbf{E})\cdot\textbf{E}^{inc}-\nu\times(\nabla\times\textbf{E}^{inc})\cdot\textbf{E}\big{)}ds(\textbf{x})dt.$
$\displaystyle=$ $\displaystyle-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt.$

Note that, in the last step we have used the fact that
$\textbf{E}(\textbf{x},t)=0$ when $|\textbf{x}|<R$ and $t>T_{0}+2R$, which
follows straightforwardly from Huygens’ principle (see e.g., [9, Lemma 2.1]).
This implies $\textbf{E}(\textbf{x},T)=\partial_{t}\textbf{E}(\textbf{x},T)=0$
for $\textbf{x}\in B_{R}$ and $T>T_{0}+2R$. Hence, the integral over $B_{R}$
on the left hand side of the previous identity vanishes.

$\int_{B_{R}}\textbf{p}e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\textbf{f}(\textbf{x})d\textbf{x}\int_{0}^{T}g(t)e^{-i\omega
t}dt=-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(x)dt$
(3.13)

By the definition of $\textbf{E}^{inc}$, one can check that

$\nabla\times\textbf{E}^{inc}=-i\kappa_{p}\textbf{d}\times\textbf{p}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega
t)}$

which gives

$|\nabla\times\textbf{E}^{inc}|=\sqrt{n}\omega.$

Using the assumption about $g$ and the fact that f is supported in $B_{R}$, we
derive from (3.13) together with the previous two relations that

$\displaystyle|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})\widehat{g}(\omega)|$
$\displaystyle=$
$\displaystyle\Big{|}(2\pi)^{-2}\int_{0}^{T}\int_{\mathbb{R}^{3}}\textbf{p}\cdot\textbf{f}(\textbf{x})g(t)e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}-i\omega
t}\ \mathrm{d}\textbf{x}\mathrm{d}t\Big{|}$ $\displaystyle=$
$\displaystyle\Big{|}(2\pi)^{-2}\int_{0}^{T}\int_{B_{R}}\textbf{p}\cdot\textbf{f}(\textbf{x})g(t)\textbf{E}^{inc}(\textbf{x},t)\
\mathrm{d}\textbf{x}\mathrm{d}t\Big{|}$ $\displaystyle\leq$ $\displaystyle
C(1+\omega)\epsilon,$

where $\epsilon=\Big{(}\int_{0}^{T}\int_{\partial
B_{R}}\big{(}|T(\textbf{E}\times\nu)|^{2}+|\textbf{E}\times\nu|^{2}\big{)}ds(\textbf{x})dt\Big{)}^{\frac{1}{2}}$
represents the measurement data on $\partial B_{R}\times(0,T)$. In view of the
assumption (2.4), one obtains for $\omega\in(0,b)$, $b>1$ and
$\kappa_{p}^{2}=n\omega^{2}$ that

$\begin{split}|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|\leq
C\frac{\omega\epsilon}{|\hat{g}(\omega)|}&\leq
C\,\delta^{-1}(1+\omega)\epsilon\\\ &\leq Cb\epsilon,\end{split}$ (3.14)

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$.

Repeating similar steps, we get

$\displaystyle|\textbf{q}\cdot\hat{\textbf{f}}(\kappa_{p}\textbf{d})|\leq
C\,b\,\epsilon\quad\mbox{for all}\ |\kappa_{p}|<\sqrt{n}b.$ (3.15)

On the other hand, Since $\nabla\cdot\textbf{f}=0$, we obtain that

$-i\kappa_{p}\textbf{d}\cdot\widehat{\textbf{f}}(-\kappa_{p}\textbf{d})=-i\kappa_{p}\int_{B_{R}}\textbf{d}e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\cdot\textbf{f}(\textbf{x})d\textbf{x}=\int_{B_{R}}\nabla
e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\cdot\textbf{f}(\textbf{x})d\textbf{x}=\int_{B_{R}}e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\nabla\cdot\textbf{f}(\textbf{x})d\textbf{x}=0,$
(3.16)

which means $\textbf{d}\cdot\widehat{\textbf{f}}(-\kappa_{p}\textbf{d})=0$.
Using the Pythagorean theorem yields

$|\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}=|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}+|\textbf{q}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}+|\textbf{d}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}=|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}+|\textbf{q}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}.$
(3.17)

Let $\xi_{p}=\kappa_{p}\textbf{d}$, we obtain from the Parseval theorem that

$\|\textbf{f}\|_{L^{2}({\mathbb{R}}^{3})^{3}}^{2}=\|\widehat{\textbf{f}}\|_{L^{2}({\mathbb{R}}^{3})^{3}}^{2}=\int_{{\mathbb{R}}^{3}}|\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}=\int_{{\mathbb{R}}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}+\int_{{\mathbb{R}}^{3}}|\textbf{q}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}.$
(3.18)

Denote

$I(s)=\int_{|\xi_{p}|\leq\sqrt{n}s}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}.$

Since the integrand $I(s)$ is an entire analytic function of $\xi$, the
integral $I(s)$ with respect to $\xi_{p}$ can be taken over by any path
joining points $0$ and $\sqrt{n}s$ in complex plane. Thus $I(s)$ is an entire
analytic function of $s=s_{1}+is_{2}$ $(s_{1},s_{2}\in\mathbb{R})$ and the
following estimate holds.

######  Lemma 3.1.

Let $\textbf{f}(\textbf{x})\in L^{2}(\mathbb{R}^{3})^{3}$, $\mathrm{supp}\
\textbf{f}\subset B_{R}$. Then

$|I(s)|\leq(\frac{4\pi}{3})^{2}\,R^{3}\,\sqrt{n}^{3}|s|^{3}e^{2R\sqrt{n}|s_{2}|}||\textbf{f}||^{2}_{L^{2}(\mathbb{R}^{3})^{3}},\qquad
s=s_{1}+is_{2}\in{\mathbb{C}}.$ (3.19)

###### Proof.

Set $l=\sqrt{n}sl^{\prime}$ for $l^{\prime}\in(0,1)$. Then it is easy to get

$\displaystyle I(s)$
$\displaystyle\leq\int_{|\xi_{p}|\leq\sqrt{n}s}|\widehat{\textbf{f}}|^{2}(\xi_{p})d\xi_{p}$
$\displaystyle=\int_{0}^{\sqrt{n}s}\int_{\mathbb{S}^{2}}|\widehat{\textbf{f}}(l\hat{\theta})|^{2}l^{2}\mathrm{d}\hat{\theta}\mathrm{d}l$
$\displaystyle=\int_{0}^{1}\int_{\mathbb{S}^{2}}|\widehat{\textbf{f}}(c_{n}sl^{\prime}\hat{\theta})|^{2}(\sqrt{n}s)^{3}l^{\prime
2}\mathrm{d}\hat{\theta}\mathrm{d}l^{\prime}.$

Noting the elementary inequality
$|e^{-i\sqrt{n}sl^{\prime}\hat{\theta}\cdot\textbf{x}}|\leq
e^{\sqrt{n}|s_{2}|R}$ for all $x\in B_{R}$ and
$\hat{\theta}\in\mathbb{S}^{2}$, we have

$\displaystyle|I(s)|$
$\displaystyle\leq\left|\int_{0}^{1}\int_{\mathbb{S}^{2}}\left|\int_{B_{R}}\textbf{f}(\textbf{x})e^{-i\sqrt{n}sl^{\prime}\hat{\theta}\cdot\textbf{x}}\mathrm{d}\textbf{x}\right|^{2}(c_{n}s)^{3}l^{\prime
2}\mathrm{d}\hat{\theta}\mathrm{d}l^{\prime}\right|$
$\displaystyle\leq\frac{4\pi}{3}R^{3}\int_{0}^{1}\int_{\mathbb{S}^{2}}\sqrt{n}^{3}|s|^{3}\left(\int_{B_{R}}|\textbf{f}(\textbf{x})|^{2}|e^{2\sqrt{n}R|s_{2}|}|\,\mathrm{d}\textbf{x}\right)\mathrm{d}\hat{\theta}\mathrm{d}l^{\prime}$
$\displaystyle\leq(\frac{4\pi}{3})^{2}\,R^{3}\,\sqrt{n}^{3}\,|s|^{3}e^{2R\sqrt{n}|s_{2}|}||\textbf{f}||^{2}_{L^{2}(\mathbb{R}^{3})^{3}}.$

This completes the Lemma 3.1. ∎

The following Lemma is essential to show the relation between $I(s)$ for
$s\in(b,\infty)$ with $I(b).$ Its proof can be found in [13].

######  Lemma 3.2.

Let $J(z)$ be an analytic function in
$S=\\{z=x+iy\in\mathbb{C}:-\frac{\pi}{4}<\arg z<\frac{\pi}{4}\\}$ and
continuous in $\overline{S}$ satisfying

$\begin{cases}|J(z)|\leq\epsilon,\ &z\in(0,L],\\\ |J(z)|\leq V,\ &z\in S,\\\
|J(0)|=0.\end{cases}$

Then there exists a function $\mu(z)$ satisfying

$\begin{cases}\mu(z)\geq\frac{1}{2},\ \ &z\in(L,2^{\frac{1}{4}}L),\\\
\mu(z)\geq\frac{1}{\pi}((\frac{z}{L})^{4}-1)^{-\frac{1}{2}},\ \
&z\in(2^{\frac{1}{4}}L,\infty)\end{cases}$

such that

$|J(z)|\leq V\epsilon^{\mu(z)},\ \ \forall z\in(L,\infty).$

Let the sector $S\subset{\mathbb{C}}$ be given as in Lemma 3.2. Now, it
follows from Lemma 3.1 that

$|I(s)e^{-(2R\sqrt{n}+1)s}|\leq CM^{2}\quad\mbox{for all}\quad s\in S,$

where $C>0$ depends on $n$ and $R$. Recalling from a priori estimate (3.14),
we obtain

$I(s)=\int_{|\xi_{p}|\leq\sqrt{n}s}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}\leq
C\,s^{3}\,b^{2}\,\epsilon^{2},\quad s\in(0,b]$

where $C>0$ depends on $n$, $T_{0}$ and $R$. Hence

$|I(s)e^{-(2R\sqrt{n}+1)s}|\leq Cb^{2}\,\epsilon^{2},\ \ s\in(0,b].$

Then applying Lemma 3.2 with $L=b$ to the function
$J(s):=\frac{1}{b^{2}}I(s)e^{-(2R\sqrt{n}+1)s}$ , we know that there exists a
function $\mu(s)$ satisfying

$\begin{cases}\mu(s)\geq\frac{1}{2},\ \ &s\in(b,2^{\frac{1}{4}}b),\\\
\mu(s)\geq\frac{1}{\pi}((\frac{s}{b})^{4}-1)^{-\frac{1}{2}},\ \
&s\in(2^{\frac{1}{4}}b,\infty)\end{cases}$

such that

$|\frac{1}{b^{2}}I(s)e^{-(2R\sqrt{n}+1)s}|\leq CM^{2}\epsilon^{2\mu}\ \
\quad\mbox{for all}\quad s\in(b,\infty),$

where $C>0$ depends on $n$, $T_{0}$ and $R$.

Now we show the proof of Theorem 2.1. We assume that $\epsilon<e^{-1}$,
otherwise the estimate is obvious. Let

$s=\begin{cases}\frac{1}{((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}},\qquad&\mbox{if}\quad
2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}<|\ln\epsilon|^{\frac{1}{4}},\\\
b,\qquad&\mbox{if}\quad|\ln\epsilon|^{\frac{1}{4}}\leq
2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}.\end{cases}$

Case (i):
$2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}<|\ln\epsilon|^{\frac{1}{4}}$.
Then we have

$\displaystyle|I(s)|$ $\displaystyle\leq$ $\displaystyle
CM^{2}b^{2}\epsilon^{2\mu}e^{(2R\sqrt{n}+1)s}$ $\displaystyle=$ $\displaystyle
CM^{2}b^{2}e^{(2R\sqrt{n}+1)s-2\mu(s)|\ln\epsilon|}$ $\displaystyle\leq$
$\displaystyle
CM^{2}b^{2}e^{\frac{(2R\sqrt{n}+3)}{((2R+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}}-\frac{2|\ln\epsilon|}{\pi}(\frac{b}{s})^{2}}$
$\displaystyle=$ $\displaystyle
CM^{2}b^{2}e^{-2\big{(}\frac{(2R\sqrt{n}+3)^{2}}{\pi}\big{)}^{\frac{1}{3}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}(1-\frac{1}{2}|\ln\epsilon|^{-\frac{1}{4}})},$

where $C>0$ depends on $n$, $T_{0}$ and $R$. Noting that
$|\ln\epsilon|^{-\frac{1}{4}}<1$ and
$\big{(}\frac{(2R\sqrt{n}+3)^{2}}{\pi}\big{)}^{\frac{1}{3}}>1$, we obtain

$|I(s)|\leq CM^{2}b^{2}e^{-b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}}.$

Using the inequality $e^{-t}\leq\frac{6!}{t^{6}}$ for $t>0$, we get

$|I(s)|\leq C\frac{M^{2}}{b^{2}|\ln\epsilon|^{3}}.$ (3.20)

Since $b^{2}|\ln\epsilon|^{3}\geq b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}$
when $b>1$ and $|\ln\epsilon|>1$. Hence

$\begin{split}I(s)+\int_{|\xi_{p}|>\sqrt{n}s}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}\,\mathrm{d}\xi_{p}&\leq
I(s)+\frac{M^{2}}{n^{2}s^{2}}\\\ &\leq
C(\frac{M^{2}}{b^{4}|\ln\epsilon|^{3}}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}})\\\
&\leq\frac{CM^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}},\end{split}$
(3.21)

where $C>0$ depends on $n$, $T_{0}$ and $R$.

Case (ii): $|\ln\epsilon|^{\frac{1}{4}}\leq
2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}$. In this case
we have $s=b$ by the choice of $s$ and $|I(b)|\leq C\,b^{5}\epsilon^{2}$.
Hence,

$\begin{split}I(b)+\int_{|\xi_{p}|>\sqrt{n}b}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}\,\mathrm{d}\xi_{p}\leq
C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{2}})\\\ \leq
C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}),\end{split}$
(3.22)

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$. Combining (3.21) and
(3.22), we finally obtain

$\int_{{\mathbb{R}}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}\leq
C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}).$

Repeating similar steps by multiplying $\textbf{H}^{inc}$ on both sides of the
equation (1.2), we get

$\int_{{\mathbb{R}}^{3}}|\textbf{q}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}\leq
C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}),$

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$. Combining the above
estimates and (3.18), we complete the proof.

##  4 Proof of Theorem 2.3

In this section. Let $n\in(0,b^{2})$ for $b>1$. Through this section, $C>0$
denotes a generic constant which is independent of $n$, may vary from line to
line. Consider the equation

$\begin{cases}n\partial_{t}^{2}\textbf{E}(\textbf{x},t)+\nabla\times(\nabla\times\textbf{E}(\textbf{x},t))=\textbf{F}(\textbf{x},t),\
&(\textbf{x},t)\in{\mathbb{R}}^{3}\times(0,\infty),\\\
\textbf{E}(\textbf{x},0)=0,\quad\partial_{t}\textbf{E}(\textbf{x},0)=0,\
&\textbf{x}\in{\mathbb{R}}^{3}.\end{cases}$ (4.23)

Assume that $\textbf{E}(\textbf{x},t):=\textbf{E}(\textbf{x},t;n)$.

Multiplying the both sides of (4.23) by $\textbf{E}^{inc}$ and using the
integration by parts over $B_{R}$, we obtain

$\int_{0}^{T}\int_{B_{R}}(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\
\mathrm{d}\textbf{x}\mathrm{d}t=\int_{0}^{T}\int_{B_{R}}\textbf{F}(\textbf{x},t)\textbf{E}^{inc}(\textbf{x},t)\
\mathrm{d}\textbf{x}\mathrm{d}t.$ (4.24)

Integrating by parts, one deduces from the left hand side of (4.24) that

$\displaystyle\int_{0}^{T}\int_{B_{R}}$
$\displaystyle(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\
\mathrm{d}\textbf{x}\mathrm{d}t$ $\displaystyle=$
$\displaystyle\int_{B_{R}}n(\partial_{t}\textbf{E}(\textbf{x},t)\textbf{E}^{inc}(\textbf{x},t)-\textbf{E}(\textbf{x},t)\partial_{t}\textbf{E}^{inc}(\textbf{x},t))\big{|}_{0}^{T}\;\mathrm{d}\textbf{x}$
$\displaystyle\quad+\int_{0}^{T}\int_{\partial
B_{R}}\big{(}\nu\times(\nabla\times\textbf{E})\cdot\textbf{E}^{inc}-\nu\times(\nabla\times\textbf{E}^{inc})\cdot\textbf{E}\big{)}ds(\textbf{x})dt.$
$\displaystyle=$ $\displaystyle-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt.$

Note that, in the last step we have used the fact that
$\textbf{E}(\textbf{x},t)=0$ when $|\textbf{x}|<R$ and $t>T_{0}+2R$, which
follows straightforwardly from Huygens’ principle (see e.g., [9, Lemma 2.1]).
This implies $\textbf{E}(\textbf{x},T)=\partial_{t}\textbf{E}(\textbf{x},T)=0$
for $\textbf{x}\in B_{R}$ and $T>T_{0}+2R$. Hence, the integral over $B_{R}$
on the left hand side of the previous identity vanishes. Thus

$\int_{B_{R}}\int_{0}^{T}\textbf{p}e^{-i(\xi_{p}\cdot\textbf{x}+\omega
t)}\cdot\textbf{F}(\textbf{x},t)d\textbf{x}dt=-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt.$
(4.25)

Define

$\widehat{\textbf{F}}(\xi,\omega)=(2\pi)^{-2}\int_{{\mathbb{R}}^{4}}\textbf{F}(\textbf{x},t)e^{-i(\xi\cdot\textbf{x}+\omega
t)}d\textbf{x}dt,$

with $(\xi,\omega)=(\xi_{1},\xi_{2},\xi_{3},\omega)\in{\mathbb{R}}^{4}$. Since
supp $\textbf{F}(\textbf{x},t)$ $\subset B_{R}\times(0,T_{0})$, we have

$\begin{split}(2\pi)^{2}\textbf{p}\cdot\widehat{\textbf{F}}(\xi_{p},\omega)&=\int_{0}^{T}\int_{B_{R}}\textbf{p}\cdot\textbf{F}(\textbf{x},t)e^{-i(\xi_{p}\cdot\textbf{x}+\omega
t)}d\textbf{x}dt\\\ &=-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt.\end{split}$
(4.26)

Consider the set (see Figure 1)

$E_{s}=\\{(\xi,\omega)\in{\mathbb{R}}^{4}|\ |\xi|^{2}-n\omega^{2}=0,\
|\omega|<s,\ n\in(0,s^{2})\\}.$

Figure 1: $E_{s}$ is the shadow area, $\zeta^{2}=|\xi|^{2}$.

Combining $(\ref{pF})$, we have

$|\textbf{p}\cdot\widehat{\textbf{F}}(\xi_{p},\omega)|\leq
Cb^{2}\epsilon,\quad(\xi_{p},\omega)\in{E_{b}}$ (4.27)

where $\epsilon=\sup\limits_{n\in(0,b^{2})}\Big{(}\int_{0}^{T}\int_{\partial
B_{R}}\big{(}|T(\textbf{E}\times\nu)|^{2}+|\textbf{E}\times\nu|^{2}\big{)}ds(\textbf{x})dt\Big{)}^{\frac{1}{2}}$.
Denote

$I(s):=\int_{E_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}(\xi_{p},\omega)|^{2}d\xi_{p}\,d\omega.$
(4.28)

It is easy to verify from (4.28) that

$\displaystyle|I(s)|\leq C\,|E_{s}|\,b^{4}\,\epsilon^{2},\quad s\in[0,b]$
(4.29)

and

$I(s)\leq\int_{E_{s}}|\widehat{\textbf{F}}(\xi_{p},\omega)|^{2}d\xi_{p}\,d\omega.$
(4.30)

Using the polar coordinates, we deduce that

$\int_{E_{s}\cap\\{(\xi,\omega)|\ \omega\geq
0\\}}|\widehat{\textbf{F}}(\xi,\omega)|^{2}d\xi
d\omega=\int_{0}^{s}\big{(}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{s\omega}|\widehat{\textbf{F}}(r\hat{\xi},\omega)|^{2}r^{2}\sin\varphi
drd\theta d\varphi\big{)}d\omega.$

Let $r=s\omega\hat{r}$ and $\omega=s\hat{\omega}$ for
$\hat{r},\hat{\omega}\in(0,1)$. A simple calculation yields

$\begin{split}\int_{E_{s}\cap\\{(\xi,\omega)|\ \omega\geq
0\\}}&|\widehat{\textbf{F}}(\xi,\omega)|^{2}d\xi
d\omega=\int_{0}^{s}\big{(}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}|\widehat{\textbf{F}}(s\omega\hat{r}\hat{\xi},\omega)|^{2}(s\omega)^{3}\hat{r}^{2}\sin\varphi
drd\theta d\varphi\big{)}d\omega\\\
&=\int_{0}^{1}\big{(}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}|\widehat{\textbf{F}}(s^{2}\omega\hat{r}\hat{\xi},\omega)|^{2}s(s^{2}\hat{\omega})^{3}\hat{r}^{2}\sin\varphi
drd\theta d\varphi\big{)}d\hat{\omega}\\\
&=\int_{0}^{1}\big{(}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}|\int_{{\mathbb{R}}^{4}}\widehat{\textbf{F}}(x,t)e^{-i(s^{2}\hat{\omega}\hat{r}\hat{\xi}\cdot
x+s\hat{\omega}t)}dxdt|^{2}s(s^{2}\hat{\omega})^{3}\hat{r}^{2}\sin\varphi
drd\theta d\varphi\big{)}d\hat{\omega}.\end{split}$

This integrals are analytic functions of $s=s_{1}+is_{2}$,
$s_{1},s_{2}\in{\mathbb{R}}$. Noting that
$e^{-i(s^{2}\hat{\omega}\hat{r}\hat{\xi}\cdot x+s\hat{\omega}t)}\leq
e^{2R|s_{1}s_{2}|+T|s_{2}|}$. We have for all $s\in S$ that

$\big{|}\int_{E_{s}\cap\\{(\xi,\omega)|\omega\geq
0\\}}|\widehat{\textbf{F}}(\xi,\omega)|^{2}d\xi d\omega\big{|}\leq
CM^{2}|s|^{6}e^{2R|s_{1}s_{2}|+T|s_{2}|}.$

Hence we also obtain

$|I(s)|\leq C\,M^{2}|s|^{6}e^{2R|s_{1}s_{2}|+T|s_{2}|},\quad s\in S.$

Let $\Delta=\max\\{2R,T\\}$, it follows for all $s\in S$ that

$|e^{-(\Delta+1)s-(\Delta+1)s^{2}}I(s)|\leq CM^{2}.$

Recalling from (4.29) a prior estimate, we obtain

$|e^{-(\Delta+1)s-(\Delta+1)s^{2}}I(s)|\leq
C\,b^{4}\,\epsilon^{2}\quad\mbox{for all}\quad s\in[0,b].$

Then applying Lemma 3.2 with $L=b$ to the function
$\frac{1}{b^{4}}e^{-(\Delta+1)s-(\Delta+1)s^{2}}I(s)$, we know that

$|I(s)|\leq CM^{2}\,b^{4}\,e^{2(\Delta+1)s^{2}}\epsilon^{2\mu},\quad\forall
s\in(b,\infty).$

Consider the set

$E^{1}_{s}=\\{(\xi,\omega)\in{\mathbb{R}}^{4}|\ \ |\omega|\geq s,\ |\xi|\leq
s|\omega|\\},$

we have by using the Parseval’s identity that

$\begin{split}I_{1}(s):=\int_{E^{1}_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega\leq\frac{1}{s^{2}}\int_{{\mathbb{R}}^{3}}|\widehat{\nabla(\textbf{p}\cdot\textbf{F})}|^{2}d\xi_{p}\,d\omega\leq\,C\,\frac{M^{2}}{s^{2}}.\end{split}$

Denote

$I_{2}(s):=\int_{E^{2}_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega,$

where

$E^{2}_{s}=\\{(\xi,\omega)\in{\mathbb{R}}^{4}|\ \ |\xi|\geq s|\omega|\\}.$

It is easy to obtain

$I_{2}(s)\leq\int_{E^{2}_{s}}|\widehat{\textbf{F}}|^{2}d\xi\,dw,$

Similarly, by using the polar coordinates. Let $r=s\hat{r}$ and
$\omega=s^{2}\hat{r}\hat{\omega}$ for $\hat{r},\hat{\omega}\in(0,1)$. We get

$\begin{split}&\int_{E^{2}_{s}\cap\\{(\xi,\omega)|\ \omega\geq
0\\}}|\textbf{p}\cdot\widehat{\textbf{F}}(\xi,\omega)|^{2}d\xi d\omega\\\
&=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\big{(}\int_{0}^{\frac{r}{s}}|\textbf{p}\cdot\widehat{\textbf{F}}(r\hat{\xi},\omega)|^{2}d\omega\big{)}r^{2}\sin\varphi
dr\,d\theta\,d\varphi\\\
&=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\big{(}\int_{0}^{1}|\textbf{p}\cdot\widehat{\textbf{F}}(r\hat{\xi},\frac{1}{s}r\hat{\omega})|^{2}\frac{1}{s}rd\hat{\omega}\big{)}r^{2}\sin\varphi
dr\,d\theta\,d\varphi\\\
&=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\big{(}\int_{0}^{1}|\textbf{p}\cdot\widehat{\textbf{F}}(r\hat{\xi},\frac{1}{s}r\hat{\omega})|^{2}\frac{1}{s}rd\hat{\omega}\big{)}r^{2}\sin\varphi
dr\,d\theta\,d\varphi\\\
&+\int_{0}^{2\pi}\int_{0}^{\pi}\int_{1}^{\infty}\big{(}\int_{0}^{1}|\textbf{p}\cdot\widehat{\textbf{F}}(r\hat{\xi},\frac{1}{s}r\hat{\omega})|^{2}\frac{1}{s}rd\hat{\omega}\big{)}r^{2}\sin\varphi
dr\,d\theta\,d\varphi.\end{split}$

Since

$\begin{split}\int_{0}^{1}|\textbf{p}\cdot\widehat{\textbf{F}}(r\hat{\xi},\frac{1}{\sqrt{n}}r\hat{\omega})|^{2}d\hat{\omega}=\frac{1}{r^{6}}\int_{0}^{1}|\int_{{\mathbb{R}}^{4}}\textbf{p}\cdot\Delta\textbf{F}(x,t)e^{-i(r\hat{\xi}\cdot
x+\frac{1}{\sqrt{n}}r\hat{\omega}t)}dxdt|^{2}d\hat{\omega}\leq\frac{M^{2}}{r^{3}},\end{split}$

which gives

$\big{|}\int_{E^{2}_{s}\cap\\{(\xi,\omega)|\ \omega\geq
0\\}}|\textbf{p}\cdot\widehat{\textbf{F}}(\xi,\omega)|^{2}d\xi
d\omega\big{|}\leq C\,\frac{M^{2}}{s}.$

Consequently

$I_{2}(s)\leq C\,\frac{M^{2}}{s}.$

Now we show the proof of Theorem 2.3. We assume that $\epsilon<e^{-1}$,
otherwise the estimate is obvious. Let

$\displaystyle
s=\begin{cases}\frac{1}{(2(\Delta+2)\pi)^{\frac{1}{4}}}b^{\frac{1}{2}}|\ln\epsilon|^{\frac{1}{5}}&\textrm{if}\
|\ln\epsilon|^{\frac{1}{5}}>2^{\frac{1}{4}}b^{\frac{1}{2}}(2(\Delta+2)\pi)^{\frac{1}{4}},\\\
b&\textrm{if}\ |\ln\epsilon|^{\frac{1}{5}}\leq
2^{\frac{1}{4}}b^{\frac{1}{2}}(2(\Delta+2)\pi)^{\frac{1}{3}}.\end{cases}$
(4.31)

Case (i):
$|\ln\epsilon|^{\frac{1}{5}}>2^{\frac{1}{4}}b^{\frac{1}{2}}(2(\Delta+2)\pi)^{\frac{1}{4}}$.
One can check that

$s>2^{\frac{1}{4}}b.$

Thus, using Lemma 3.2, we obtain

$\displaystyle|I(s)|$ $\displaystyle\leq$ $\displaystyle
C\,M^{2}\,b^{4}\,e^{2(\Delta+2)s^{2}}\epsilon^{2\mu(s)}$ (4.32)
$\displaystyle\leq$ $\displaystyle
C\,M^{2}\,b^{4}\,e^{-2\mu(s)|\ln\epsilon|+2(\Delta+2)s^{2}}$ (4.33)
$\displaystyle\leq$ $\displaystyle
C\,M^{2}\,b^{4}\,e^{-\big{(}\frac{-2|\ln\epsilon|}{\pi}(\frac{b}{s})^{2}+2(\Delta+2)s^{2}\big{)}}$
(4.34) $\displaystyle\leq$ $\displaystyle
C\,M^{2}\,b^{4}\,e^{-2(\frac{2(\Delta+2)}{\pi})^{\frac{1}{2}}b|\ln\epsilon|^{\frac{3}{5}}(1-\frac{1}{2}|\ln\epsilon|^{-\frac{1}{5}})}.$
(4.35)

Noting that $\frac{1}{2}|\ln\epsilon|^{-\alpha}<\frac{1}{2}$ and
$(\frac{2(\Delta+2)}{\pi})^{\frac{1}{2}}>1$, we have

$|I(s)|\leq C\,M^{2}\,b^{4}\,e^{-b|\ln\epsilon|^{\frac{3}{5}}}.$

Using the elementary inequality

$e^{-t}\leq\frac{5!}{t^{5}},\quad t>0,$

we get

$|I(s)|\leq C\frac{b^{4}M^{2}}{b^{5}|\ln\epsilon|^{3}}\leq
C\frac{M^{2}}{b|\ln\epsilon|^{3}}.$

Case (ii): $|\ln\epsilon|^{\frac{1}{5}}\leq
2^{\frac{1}{4}}b^{\frac{1}{2}}(2(\Delta+2)\pi)^{\frac{1}{4}}$. In this case we
have for $s=b$ that

$|I(s)|=|I(b)|\leq b^{11}\,\epsilon^{2}.$

Combining estimates of $I_{1}(s)$ and $I_{2}(s)$ , we obtain

$\displaystyle\int_{E_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{1}_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{2}_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega$
$\displaystyle=$ $\displaystyle I(s)+I_{1}(s)+I_{2}(s)$ (4.36)
$\displaystyle\leq$ $\displaystyle
C\Big{(}b^{11}\epsilon^{2}+\frac{M^{2}}{b|\ln\epsilon|^{3}}+\frac{M^{2}}{b^{\frac{1}{2}}|\ln\epsilon|^{\frac{1}{5}}}\Big{)}$
(4.37) $\displaystyle\leq$ $\displaystyle
C\Big{(}b^{11}\epsilon^{2}+\frac{M^{2}}{b^{\frac{1}{2}}|\ln\epsilon|^{\frac{1}{5}}}\Big{)}$
(4.38)

Since
$\big{(}b|\ln\epsilon|^{3}\big{)}>\big{(}b|\ln\epsilon|^{\frac{1}{5}}\big{)}$
when $b>1$ and $|\ln\epsilon|>1$.

Repeating similar steps, we get

$\int_{E_{s}}|\textbf{q}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{1}_{s}}|\textbf{q}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{2}_{s}}|\textbf{q}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega\leq
C\Big{(}b^{11}\epsilon^{2}+\frac{M^{2}}{b^{\frac{1}{2}}|\ln\epsilon|^{\frac{1}{5}}}\Big{)}.$
(4.39)

On the other hand, since $\nabla\cdot\textbf{F}=0$, we have that

$-i\kappa_{p}\textbf{d}\cdot\widehat{\textbf{F}}(\kappa_{p}\textbf{d},\omega)=\int_{{\mathbb{R}}^{4}}\nabla
e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega
t)}\cdot\textbf{F}(\textbf{x},t)d\textbf{x}dt=\int_{0}^{T}\int_{B_{R}}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega
t)}\nabla\cdot\textbf{F}(\textbf{x},t)d\textbf{x}dt=0$

Using the Pythagorean theorem yields

$|\widehat{\textbf{F}}(\xi_{p},\omega)|^{2}=|\textbf{p}\cdot\widehat{\textbf{F}}(\xi_{p},\omega)|^{2}+|\textbf{q}\cdot\widehat{\textbf{F}}(\xi_{p},\omega)|^{2}+|\textbf{d}\cdot\widehat{\textbf{F}}(\xi_{p},\omega)|^{2}.$

According to ${\mathbb{R}}^{4}=E_{s}\cup E^{1}_{s}\cup E^{2}_{s}$, we get

$\begin{split}||\textbf{F}||^{2}_{L^{2}({\mathbb{R}}^{4})^{3}}=||\widehat{\textbf{F}}||^{2}_{L^{2}({\mathbb{R}}^{4})^{3}}&=\int_{E_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{1}_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{2}_{s}}|\textbf{p}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega\\\
&+\int_{E_{s}}|\textbf{q}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{1}_{s}}|\textbf{q}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega+\int_{E^{2}_{s}}|\textbf{q}\cdot\widehat{\textbf{F}}|^{2}d\xi_{p}\,d\omega.\end{split}$
(4.40)

Combining the above estimates (4.36)-(4.40), we obtain the stability estimate
(2.7). This completes the proof.

##  5 Proof of Theorem 2.4

In this section, we assume that F takes the form

$\textbf{F}(x_{1},x_{2},x_{3},t)=\textbf{f}(\tilde{x},t)g(x_{3}),\quad\tilde{x}=(x_{1},x_{2})\in{\mathbb{R}}^{2},\
x_{3}\in{\mathbb{R}},\quad t\in(0,T_{0}).$ (5.41)

Then the equation $(\ref{eq1})$ becomes

$n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})=\textbf{f}(\tilde{x},t)\,g(x_{3}).$
(5.42)

Assume that $g$ is known, we establish an increasing stability estimate about
f from the Dirichlet data $\\{\textbf{E}(\textbf{x},t)|\ \textbf{x}\in\partial
B_{R},\ t\in(0,T_{0})\\}$.

Multiplying the both sides of (5.42) by $\textbf{E}^{inc}$, we obtain

$\int_{0}^{T}\int_{B_{R}}(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\
\mathrm{d}\textbf{x}\mathrm{d}t=\int_{0}^{T}\int_{B_{R}}\textbf{f}(\tilde{x},t)\,g(x_{3})\textbf{E}^{inc}(\textbf{x},t)\
\mathrm{d}\textbf{x}\mathrm{d}t.$ (5.43)

Integrating by parts, one deduces from the left hand side of (5.43) that

$\displaystyle\int_{0}^{T}\int_{B_{R}}$
$\displaystyle(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\
\mathrm{d}\textbf{x}\mathrm{d}t$ $\displaystyle=$
$\displaystyle\int_{B_{R}}n(\partial_{t}\textbf{E}(\textbf{x},t)\textbf{E}^{inc}(\textbf{x},t)-\textbf{E}(\textbf{x},t)\partial_{t}\textbf{E}^{inc}(\textbf{x},t))\big{|}_{0}^{T}\;\mathrm{d}\textbf{x}$
$\displaystyle\quad+\int_{0}^{T}\int_{\partial
B_{R}}\big{(}\nu\times(\nabla\times\textbf{E})\cdot\textbf{E}^{inc}-\nu\times(\nabla\times\textbf{E}^{inc})\cdot
E\big{)}ds(\textbf{x})dt.$ $\displaystyle=$
$\displaystyle-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt.$

Note that, in the last step we have used the fact that
$\textbf{E}(\textbf{x},t)=0$ when $|\textbf{x}|<R$ and $t>T_{0}+2R$, which
follows straightforwardly from Huygens’ principle (see e.g., [18, Lemma 2.1]).
This implies $\textbf{E}(\textbf{x},T)=\partial_{t}\textbf{E}(\textbf{x},T)=0$
for $\textbf{x}\in B_{R}$ and $T>T_{0}+2R$. Hence, the integral over $B_{R}$
on the left hand side of the previous identity vanishes. Thus (5.43) becomes

$\begin{split}\int_{B_{R}}\textbf{p}e^{-i(\kappa_{p}d_{1}x_{1}+\kappa_{p}d_{2}x_{2}+\omega
t)}\cdot&\textbf{f}(x_{1},x_{2},t)d\tilde{x}\int_{0}^{T}g(x_{3})e^{-i\xi_{3}x_{3}}dt\\\
&=-\int_{0}^{T}\int_{\partial
B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt\end{split}$
(5.44)

Let $\tilde{\xi}^{2}=\xi_{1}^{2}+\xi_{2}^{2}$. Define the set (see Figure 2)

$E_{s}=\\{(\xi_{1},\xi_{2},\omega)\in{\mathbb{R}}^{3}|\
n\omega^{2}-|\xi_{1}|^{2}-|\xi_{2}|^{2}=|\xi_{3}|^{2},\
|\omega|\leq\frac{s}{\sqrt{n}},\ |\xi_{3}|\leq s\\}.$

Figure 2: $E_{s}$ is the shadow area,
$\tilde{\xi}^{2}=\xi_{1}^{2}+\xi_{2}^{2}$.

Let
$\xi_{p}=\xi=\kappa_{p}\textbf{d}=(\kappa_{p}d_{1},\kappa_{p}d_{2},\kappa_{p}d_{3})$.
Combining (5.44) and $|\widehat{g}(\xi_{3})|\geq\delta>0$ for
$\xi_{3}\in(-b,b)$, we get that for
$(\kappa_{p}d_{1},\kappa_{p}d_{2},\omega)\in E_{b}$

$\begin{split}|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}d_{1},\kappa_{p}d_{2},\omega)|\leq
C\frac{(1+\kappa_{p})\epsilon}{|\hat{g}(\xi_{3})|}&\leq
C\delta^{-1}(1+\kappa_{p})\epsilon\\\ &\leq\,C\,b\epsilon.\end{split}$ (5.45)

Given

$I(s)=\int_{E_{s}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega,$

it is easy to verify that

$I(s)\leq\int_{E_{s}}|\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega.$
(5.46)

Using the polar coordinates $\xi_{1}=r\sin\theta$, $\xi_{2}=r\cos\theta$ for
$0\leq r\leq s$, $0\leq\theta\leq 2\pi$, we obtain that

$\begin{split}\int_{E_{s}\cap\\{(\xi_{1},\xi_{2},\omega)|\ \omega\geq
0\\}}|\widehat{\textbf{f}}|^{2}d\xi_{1}d\xi_{2}d\omega=\int_{0}^{s}\int_{0}^{2\pi}\int_{\frac{1}{\sqrt{n}}r}^{\frac{1}{\sqrt{n}}s}|\widehat{\textbf{f}}|^{2}r\,d\omega\,dr\,d\theta.\end{split}$

Let $r=s\hat{r}$, $r\hat{\xi}=(\xi_{1},\xi_{2})$ and $\hat{\omega}\in(0,1)$. A
simple calculation yields

$\begin{split}&\int_{E_{s}\cap\\{(\xi_{1},\xi_{2},\omega)|\ \omega\geq
0\\}}|\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega\\\
&=\int_{0}^{s}\int_{0}^{2\pi}\int_{\frac{1}{\sqrt{n}}r}^{\frac{1}{\sqrt{n}}s}|\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\omega
rdrd\theta\\\
&=\int_{0}^{1}\int_{0}^{2\pi}\big{(}\int_{\frac{1}{\sqrt{n}}s\hat{r}}^{\frac{1}{\sqrt{n}}s}|\widehat{\textbf{f}}(r\hat{\xi},\omega)|^{2}d\omega\big{)}s^{2}\hat{r}d\hat{r}d\theta\\\
&=\int_{0}^{1}\int_{0}^{2\pi}\big{(}\int_{0}^{\frac{1}{\sqrt{n}}s}|\widehat{\textbf{f}}(s\hat{r}\hat{\xi},\omega)|^{2}d\omega-\int_{0}^{\frac{1}{\sqrt{n}}s\hat{r}}|\widehat{\textbf{f}}(s\hat{r}\hat{\xi},\omega)|^{2}d\omega\big{)}s^{2}\hat{r}d\hat{r}d\theta\\\
&=\int_{0}^{1}\int_{0}^{2\pi}\big{(}\int_{0}^{1}|\textbf{f}_{1}(s\hat{r}\hat{\xi},\frac{1}{\sqrt{n}}s\hat{\omega})|^{2}\frac{1}{\sqrt{n}}s^{3}\hat{r}d\hat{\omega}-\int_{0}^{1}|\textbf{f}_{2}(s\hat{r}\hat{\xi},\frac{1}{\sqrt{n}}s\hat{r}\hat{\omega})|^{2}\frac{1}{\sqrt{n}}s^{3}\hat{r}^{2}d\hat{\omega}\big{)}d\hat{r}d\theta\end{split}$

where

$\textbf{f}_{1}(s\hat{r}\hat{\xi},\frac{1}{\sqrt{n}}s\hat{\omega})=\int_{{\mathbb{R}}^{3}}\textbf{f}(x_{1},x_{2},t)e^{-i(s\hat{r}\hat{\xi}\cdot\tilde{x}+\frac{1}{\sqrt{n}}s\hat{\omega}t)}d\tilde{x}dt$

and

$\textbf{f}_{2}(s\hat{r}\hat{\xi},\frac{1}{\sqrt{n}}s\hat{r}\hat{\omega})=\int_{{\mathbb{R}}^{3}}\textbf{p}\cdot\textbf{f}(x_{1},x_{2},t)e^{-i(s\hat{r}\hat{\xi}\cdot\tilde{x}+\frac{1}{\sqrt{n}}s\hat{r}\hat{\omega}t)}d\tilde{x}dt.$

This integral is an analytic function of $s=s_{1}+is_{2}\in{\mathbb{C}}$,
$s_{1},s_{2}\in{\mathbb{R}}$. Noting that
$|-i(s\hat{r}\hat{\xi}\cdot\tilde{x}+\frac{1}{\sqrt{n}}s\hat{\omega}t|\leq(R+\frac{1}{\sqrt{n}}T)|s_{2}|$
and
$|-i(s\hat{r}\hat{\xi}\cdot\tilde{x}+\frac{1}{\sqrt{n}}s\hat{r}\hat{\omega}t)|\leq(R+\frac{1}{\sqrt{n}}T)|s_{2}|$,
we deduce

$\begin{split}|I(s)|\leq CM^{2}|s|^{3}e^{((R+cT)+1)|s_{2}|},\ s\in
S\end{split}$

where $c:=\frac{1}{\sqrt{n}}$ and $C>0$ depends on $n$, $T_{0}$ and $R$.

From estimate (5.45), we have

$\begin{split}|I(s)|\leq C\,|E_{s}|\,b^{2}\,\epsilon^{2}\quad\mbox{for
all}\quad s\in[0,b],\end{split}$ (5.47)

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$. Then applying Lemma
3.2, we know for all $s>b$ that

$\begin{split}|I(s)|\leq
CM^{2}\,b^{2}\,e^{((R+cT)+1)s}\epsilon^{2\mu(s)},\end{split}$

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$.

Define
$E_{s}^{1}=\\{(\xi_{1},\xi_{2},\omega)\in{\mathbb{R}}^{3}|\,|\omega|>\frac{1}{\sqrt{n}}s,\
\xi_{1}^{2}+\xi_{2}^{2}\leq n\omega^{2}\\}$. The Parseval’s identity yields

$I_{1}(s)=\int_{E^{1}_{s}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega\leq\frac{C}{s^{2}}\int_{E^{1}_{s}}|\widehat{\nabla(\textbf{p}\cdot\textbf{f})}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega\leq\,C\,\frac{M^{2}}{s^{2}},$

where $C>0$ depends on $n$. Consider the set

$E^{2}_{s}=\\{(\xi_{1},\xi_{2},\omega)\in{\mathbb{R}}^{3}|\,\omega^{2}+\xi_{1}^{2}+\xi_{2}^{2}\leq\frac{1}{n}s^{2},\
\xi_{1}^{2}+\xi_{2}^{2}\geq n\omega^{2}\\}.$

Now we estimate

$I_{2}(s)=\int_{E^{2}_{s}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega.$

The following Lemma 5.1 is essential to estimate $I_{2}(s)$. For $r>0$, denote
$B(0,r)=\\{x\in{\mathbb{R}}^{d}:|x|<r\\}$. Below we state a stability estimate
for analytic continuation problems, which can be seen in [8, 33].

######  Lemma 5.1.

Let $\mathcal{O}$ be a non empty open set of the unit ball
$B(0,1)\subset\mathbb{R}^{d}$, $d\geq 2$, and let $G$ be an analytic function
in $B(0,2),$ that satisfy

$\|\partial^{\gamma}G\|_{L^{\infty}(B(0,2))}\leq
M_{0}\,|\gamma|!\;{\eta^{-|\gamma|}},\,\,\,\,\forall\,\gamma\in(\mathbb{N}\cup\\{0\\})^{d},$

for some $M_{0}>0$ and $\eta>0$. Then, we have

$\|G\|_{L^{\infty}(B(0,1))}\leq
N\,M_{0}^{1-\mu}\;\|G\|_{L^{\infty}(\mathcal{O})}^{\mu},$

where $\mu\in(0,1)$ depends on $d$, $\eta$ and $|\mathcal{O}|$ and
$N=N(\eta)>0$.

Define
$F_{s}(\xi_{1},\xi_{2},\omega)=\textbf{p}\cdot\widehat{\textbf{f}}(s\xi_{1},s\xi_{2},s\omega)$
for any $(\xi_{1},\xi_{2},\omega)\in\mathbb{R}^{3}$ and $s>0$. Since f is
compactly supported, one can see that the function $F_{s}$ is analytic and it
satisfies for $\gamma\in({\mathbb{N}}\cup\\{0\\})^{3}$ that

$\displaystyle|\partial^{\gamma}F_{s}(\xi_{1},\xi_{2},\omega)|=|\partial^{\gamma}F(s\xi_{1},s\xi_{2},s\omega)|=\Big{|}\partial^{\gamma}\int_{\mathbb{R}^{3}}\textbf{p}\cdot\textbf{f}(x_{1},x_{2},t)e^{-is(\xi_{1}\cdot
x_{1}+\xi_{2}\cdot x_{2}+\omega t)}\,dx_{1}dx_{2}dt\Big{|}$
$\displaystyle=\Big{|}\sum_{\alpha+\beta=\gamma\atop\alpha,\beta\in{\mathbb{N}}\cup
0}\int_{\mathbb{R}^{3}}(-i)^{|\gamma|}s^{|\gamma|}\tilde{x}^{\alpha}t^{\beta}\textbf{p}\cdot\textbf{f}(x_{1},x_{2},t)e^{-ib(\xi_{1}\cdot
x_{1}+\xi_{2}\cdot x_{2}+\omega t)}\,dx_{1}dx_{2}dt\Big{|}.$

Using $s^{|\gamma|}<|\gamma|!\,e^{s}$ and $\eta=\max\\{R,T\\}^{-1}$, we obtain

$\displaystyle|\partial^{\gamma}F_{s}(\xi_{1},\xi_{2},\omega)|\leq\|\textbf{f}\|_{L^{2}(B_{R}\times(0,T))}\,\eta^{-|\gamma|}\,s^{|\gamma|}\leq
C\|\textbf{f}\|_{L^{2}(B_{R}\times(0,T))}\,\eta^{-|\gamma|}\,|\gamma|!\,e^{s}\leq
CM\,\eta^{-|\gamma|}\,|\gamma|!\,e^{s},$

where $C>0$ depends on $R$ and $T$. Applying Lemma 5.1 to the set
$\mathcal{O}$ defined as $\mathcal{O}:=E_{s}^{2}$, we find a constant
$\alpha\in(0,1)$ such that

$\|F_{s}\|_{L^{\infty}(B(0,1))}\leq
C\,M\,e^{s(1-\alpha)}\|F_{s}\|^{\alpha}_{L^{\infty}(\mathcal{O})}.$

Using the fact that
$\textbf{p}\cdot\widehat{\textbf{f}}(\xi)=F_{s}(s^{-1}\xi)$, one gets the
following estimate

$\displaystyle\|\textbf{p}\cdot\widehat{\textbf{f}}\|_{L^{\infty}(B(0,s))}$
$\displaystyle=$ $\displaystyle\|F_{s}\|_{L^{\infty}(B(0,1))}\leq
CMe^{s(1-\alpha)}\|F_{s}\|_{L^{\infty}(\mathcal{O})}^{\alpha}$ (5.48)
$\displaystyle\leq$ $\displaystyle
C\,M\,e^{s(1-\alpha)}\|\textbf{p}\cdot\widehat{\textbf{f}}\|_{L^{\infty}(E_{s}^{2})}^{\alpha}\leq
C\,M\,e^{s(1-\alpha)}b^{\alpha}\epsilon^{\alpha},$

where $C>0$ depends on $R$, $T$, $\delta$ and $n$. Combining (5.45), we know
for $s\in[0,b]$ that

$|I_{2}(s)|\leq
CM^{2}|E^{2}_{s}|\,e^{2s(1-\alpha)}\,b^{2\alpha}\,\epsilon^{2\alpha}$

and similarly

$\begin{split}|I_{2}(s)|\leq
CM^{2}|s|^{3}\,e^{\big{(}(R+cT)+1\big{)}|s_{2}|}\quad\mbox{for all}\quad s\in
S.\end{split}$

Thus applying Lemma 3.2, we have

$\begin{split}|e^{-((R+cT)+3)s}I_{2}(s)|\leq
CM^{2}b^{2\alpha}(\epsilon^{\alpha})^{2\mu}\quad\mbox{for all}\quad
s>b,\end{split}$ (5.49)

where $C>0$ depends on $R$, $T$, $\delta$ and $n$.

Consider the set

$E^{3}_{s}=\\{(\xi_{1},\xi_{2},\omega)\in{\mathbb{R}}^{3}|\,\
\omega^{2}+|\xi_{1}|^{2}+|\xi_{2}|^{2}>\frac{1}{n}s^{2},\
\xi_{1}^{2}+\xi_{2}^{2}\geq n\omega^{2}\\}.$

The Parseval’s identity implies

$I_{3}(s)=\int_{E^{3}_{s}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{1},\xi_{2},\omega)|^{2}d\xi_{1}d\xi_{2}d\omega\leq\frac{CM^{2}}{s^{2}},$
(5.50)

where $C>0$ depends on $n$.

Now we proof Theorem 2.4. We assume that $\epsilon<e^{-1}$, otherwise the
estimate is obvious. Let

$\displaystyle
s=\begin{cases}\frac{1}{(((R+cT)+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}}&\textrm{if}\
|\ln\epsilon|^{\frac{1}{4}}>2^{\frac{1}{4}}b^{\frac{1}{3}}(((R+cT)+3)\pi)^{\frac{1}{3}},\\\
b&\textrm{if}\
|\ln\epsilon|^{\frac{1}{4}}>2^{\frac{1}{4}}b^{\frac{1}{3}}(((R+cT)+3)\pi)^{\frac{1}{3}}.\end{cases}$
(5.51)

Case (i):
$|\ln\epsilon|^{\frac{1}{4}}>2^{\frac{1}{4}}b^{\frac{1}{3}}(((R+cT)+3)\pi)^{\frac{1}{3}}$.
One can check that

$s>2^{\frac{1}{4}}b.$

Thus, using Lemma 3.2 we obtain

$\displaystyle|I(s)|$ $\displaystyle\leq$ $\displaystyle
CM^{2}\,b^{2}e^{((R+cT)+3)s^{2}}\epsilon^{2\mu(s)}$ (5.52) $\displaystyle\leq$
$\displaystyle CM^{2}\,b^{2}\,e^{-2\mu(s)|\ln\epsilon|+((R+cT)+3)s^{2}}$
(5.53) $\displaystyle\leq$ $\displaystyle
CM^{2}\,b^{2}\,e^{-\frac{-2|\ln\epsilon|}{\pi}(\frac{b}{s})^{2}+\frac{((R+cT)+3)}{(((R+cT)+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}}}$
(5.54) $\displaystyle\leq$ $\displaystyle
CM^{2}\,b^{2}\,e^{-2\big{(}\frac{((R+cT)+3)^{2}}{\pi}\big{)}^{\frac{1}{3}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}\big{(}1-\frac{1}{2}|\ln\epsilon|^{-\frac{1}{4}}\big{)}},$
(5.55)

where $C>0$ depends on $R$, $T$, $\delta$ and $n$. Noting that
$\frac{1}{2}|\ln\epsilon|^{-\frac{1}{4}}<\frac{1}{2}$ and
$\big{(}\frac{(2(R+2T)+3)^{2}}{\pi}\big{)}^{\frac{1}{3}}>1$, we have

$|I(s)|\leq CM^{2}\,b^{2}\,e^{-b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}}.$

Using the elementary inequality

$e^{-t}\leq\frac{6!}{t^{6}},\quad t>0,$

we get

$|I(s)|\leq\frac{C\,b^{2}\,M^{2}}{\big{(}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}\big{)}^{6}}\leq\frac{CM^{2}}{b^{2}|\ln\epsilon|^{3}},$
(5.56)

where $C>0$ depends on $R$, $T$, $\delta$ and $n$.

Case (ii):
$|\ln\epsilon|^{\frac{1}{4}}>2^{\frac{1}{4}}b^{\frac{1}{3}}(((R+cT)+3)\pi)^{\frac{1}{3}}$.
In this case we have $s=b$, and from (5.47),

$|I(s)|=|I(b)|\leq|E_{b}|b^{2}\epsilon^{2}.$

Combining the estimates of $I(s)$ and $I_{1}(s)$, we obtain

$\displaystyle\int_{E_{s}}|\textbf{p}\cdot\widehat{\textbf{f}}(\tilde{\xi},\omega)|^{2}\,d\tilde{\xi}\,d\omega$
$\displaystyle+$
$\displaystyle\int_{E_{s}^{1}}|\textbf{p}\cdot\widehat{\textbf{f}}(\tilde{\xi},\omega)|^{2}\,d\tilde{\xi}\,d\omega=I(s)+I_{1}(s)$
(5.57) $\displaystyle\leq$ $\displaystyle
C\Big{(}b^{5}\,b^{2}\epsilon^{2}++\frac{M^{2}}{b^{2}|\ln\epsilon|^{3}}+\frac{M^{2}}{\big{(}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}}\big{)}^{2}}\Big{)},$
(5.58)

where $C>0$ depends on $R$, $T$, $\delta$ and $n$.

Now we estimate
$I_{2}(s)+I_{3}(s)=\int_{E_{s}^{2}}|\textbf{p}\cdot\widehat{\textbf{f}}(\tilde{\xi},\omega)|^{2}\,d\tilde{\xi}\,d\omega+\int_{E_{s}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}(\tilde{\xi},\omega)|^{2}\,d\tilde{\xi}\,d\omega$.
Let

$\displaystyle
s=\begin{cases}\frac{1}{(((R+cT)+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\alpha\ln\epsilon|^{\frac{1}{4}}&\textrm{if}\
|\alpha\ln\epsilon|^{\frac{1}{4}}>2^{\frac{1}{4}}b^{\frac{1}{3}}(((R+cT)+3)\pi)^{\frac{1}{3}},\\\
b&\textrm{if}\
|\alpha\ln\epsilon|^{\frac{1}{4}}>2^{\frac{1}{4}}b^{\frac{1}{3}}(((R+cT)+3)\pi)^{\frac{1}{3}}.\end{cases}$

Repeating similar steps, we find out

$\displaystyle\int_{E_{s}^{2}}|\textbf{p}\cdot\widehat{\textbf{f}}(\tilde{\xi},\omega)|^{2}\,d\tilde{\xi}\,d\omega+\int_{E_{s}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}(\tilde{\xi},\omega)|^{2}\,d\tilde{\xi}\,d\omega=I_{2}(s)+I_{3}(s)$
$\displaystyle\leq
C\Big{(}b^{5}\,e^{2b(1-\alpha)}\epsilon^{2\alpha}+\frac{M^{2}}{b^{2}|\alpha\ln\epsilon|^{3}}+\frac{M^{2}}{\big{(}b^{\frac{2}{3}}|\alpha\ln\epsilon|^{\frac{1}{4}}\big{)}^{2}}\Big{)},$
(5.59)

where $C>0$ depends on $R$, $T$, $\delta$, $M$ and $n$.

Since $b^{2}|\ln\epsilon|^{3}>b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}$ and
$b^{2}|\alpha\ln\epsilon|^{3}>b^{\frac{4}{3}}|\alpha\ln\epsilon|^{\frac{1}{2}}$
when $b>1$ and $|\ln\epsilon|>1$. Applying
$b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}>b^{\frac{4}{3}}|\alpha\ln\epsilon|^{\frac{1}{2}}$,
we obtain

$\begin{split}\int_{{\mathbb{R}}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}|^{2}d\tilde{\xi}d\omega&=\int_{E_{s}}|\textbf{p}\cdot\widehat{\textbf{f}}|^{2}\,d\tilde{\xi}\,d\omega+\int_{E_{s}^{1}}|\textbf{p}\cdot\widehat{\textbf{f}}|^{2}\,d\tilde{\xi}\,d\omega+\int_{E_{s}^{2}}|\textbf{p}\cdot\widehat{\textbf{f}}|^{2}\,d\tilde{\xi}\,d\omega+\int_{E_{s}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}|^{2}\,d\tilde{\xi}\,d\omega\\\
&\leq
C\Big{(}b^{7}\epsilon^{2}+b^{5}e^{2b(1-\alpha)}\,\epsilon^{2\alpha}+\frac{M^{2}}{b^{2}|\ln\epsilon|^{3}}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}+\frac{M^{2}}{b^{2}|\alpha\ln\epsilon|^{3}}+\frac{M^{2}}{b^{\frac{4}{3}}|\alpha\ln\epsilon|^{\frac{1}{2}}}\Big{)}\\\
&\leq
C(b^{7}\epsilon^{2}+b^{5}e^{2b(1-\alpha)}\,\epsilon^{2\alpha}+\frac{M^{2}}{b^{\frac{4}{3}}|\alpha\ln\epsilon|^{\frac{1}{2}}}),\end{split}$
(5.60)

where $C>0$ depends on $R$, $T$, $\delta$, $M$ and $n$.

Repeating similar steps, we get that

$\int_{{\mathbb{R}}^{3}}|\textbf{q}\cdot\widehat{\textbf{f}}|^{2}d\tilde{\xi}d\omega\leq
C(b^{7}\epsilon^{2}+b^{5}e^{2b(1-\alpha)}\,\epsilon^{2\alpha}+\frac{M^{2}}{b^{\frac{4}{3}}|\alpha\ln\epsilon|^{\frac{1}{2}}}),$
(5.61)

where $C>0$ depends on $R$, $T$, $\delta$, $M$ and $n$. Using the Pythagorean
theorem yields

$\begin{split}||\textbf{f}||^{2}_{L^{2}({\mathbb{R}}^{3})^{3}}&=||\widehat{\textbf{f}}||^{2}_{L^{2}({\mathbb{R}}^{3})^{3}}\\\
&=\int_{{\mathbb{R}}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}|^{2}d\tilde{\xi}d\omega+\int_{{\mathbb{R}}^{3}}|\textbf{q}\cdot\widehat{\textbf{f}}|^{2}d\tilde{\xi}d\omega.\end{split}$
(5.62)

Combining (5.60)-(5.61) and (5.62), we obtain the stability estimate (2.11).
This completes the proof.

##  6 Acknowledgment

The work of S. Si is supported by the Natural Science Foundation of Shandong
Province, China(No. ZR202111240173). The author would like to thank Guanghui
Hu for helpful discussions.

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