# Inverse problem for a time-dependent Convection-diffusion equation in
admissible geometries

Rohit Kumar Mishra† and Manmohan Vashisth∗ † Mathematics Discipline, Indian
Institute of Technology, Gandhinagar, Gujarat - 382355, India.
E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>∗Department of
Mathematics, Indian Institute of Technology, Jammu, J $\&$ K - 181221, India.
E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We consider a partial data inverse problem for time-dependent convection-
diffusion equation on an admissible manifold. We prove that the time-dependent
convection term can be recovered uniquely modulo a gauge invariance while the
time-dependent potential can be recovered fully. There has been several works
on inverse problems related to the steady state Convection-diffusion operator
in Euclidean as well as in Riemannian geometry however inverse problems
related to time-dependent Convection-diffusion equation on manifold is not
studied in the prior works which is the main objective of this paper. In fact
to the best of our information, the problem considered here is the first work
related to a partial data inverse problem for recovering both first and zeroth
order time-dependent perturbartions of evolution equations in Riemannian
geometry.

Keywords: Inverse problems, time-dependent coefficients, Convection-diffusion
equation, partial boundary data, admissible manifold, Carleman estimates,
geometric optics solutions.

Mathematics subject classification 2010: 35R30, 35K20, 58J35, 58J65.

##  1\. Introduction and statement of the main result

For $n\geq 2$, let $(M,g)$ be a smooth $n$-dimensional Riemannian manifold
with smooth boundary $\partial M$. For $T>0$, let $M_{T}:=(0,T)\times M$ and
denote $\partial M_{T}:=(0,T)\times\partial M,$ the lateral boundary of
$M_{T}$. We also denote by ${T}M$ and ${T}^{*}M$ the tangent and cotangent
bundle of $M$. For $A\in C^{\infty}\left(M_{T};T^{*}M\right)$ given by
$\displaystyle A(t,x):=\sum_{j=1}^{n}A_{j}(t,x)dx^{j}$ in local coordinates
and $q\in L^{\infty}(M_{T})$, we consider the initial boundary value problem
(IBVP) given by

$\displaystyle\begin{aligned}
\begin{cases}\left[\partial_{t}-\large\sum\limits_{j,k=1}^{n}\frac{1}{\sqrt{\lvert
g\rvert}}\Big{(}\partial_{x_{j}}+iA_{j}\Big{)}\left(\sqrt{\lvert
g\rvert}g^{jk}\left(\partial_{x_{k}}+iA_{k}\right)\right)+q\right]u(t,x)=0,\
(t,x)\in M_{T}\\\ u(0,x)=0,\ x\in M\\\ u(t,x)=f(t,x),\ (t,x)\in\partial
M_{T}\end{cases}\end{aligned}$ (1.1)

where $g^{-1}:=\left(\left(g^{ij}\right)\right)_{1\leq i,j\leq n}$ denote the
inverse of $g:=\left(\left(g_{ij}\right)\right)_{1\leq i,j\leq n}$, $\lvert
g\rvert=\det(g)$ and the Dirichlet data $f$ is assumed to be non-zero.
Throughout this article, we denote by $\mathcal{L}_{A,q}$ the following
operator

$\mathcal{L}_{A,q}:=\partial_{t}-\large\sum\limits_{j,k=1}^{n}\frac{1}{\sqrt{\lvert
g\rvert}}\Big{(}\partial_{x_{j}}+iA_{j}(t,x)\Big{)}\left(\sqrt{\lvert
g\rvert}g^{jk}\left(\partial_{x_{k}}+iA_{k}(t,x)\right)\right)+q(t,x).$ (1.2)

The well-posedness (see [24, 15, 42]) of (1.1) allows us to define the
Dirichlet to Neumann (DN) map $\displaystyle\Lambda_{A,q}$ by

$\displaystyle\Lambda_{A,q}(f):=\Big{[}\partial_{\nu}u(t,x)+i\nu(x)\cdot
A(t,x)u(t,x)\Big{]}\Big{|}_{\partial M_{T}},\ \mbox{for}\ f\in H^{1}(\partial
M_{T})$ (1.3)

where $\nu(x)$ is the outward unit normal to $\partial M$ at $x\in\partial M$.
In this article, we are interested in determining the time-dependent
coefficients $A$ and $q$ from the knowledge of DN map $\Lambda_{A,q}$ given on
a proper subset of $\partial M_{T}$ for the case when $(M,g)$ is an admissible
manifold where by an admissible manifold, we mean the following.

######  Definition 1.1.

(Admissible manifold [11, 28]) We say that a compact Riemannian manifold
$(M,g)$ of dimension $n\geq 2$ with boundary $\partial M$, is admissible if
$M$ is orientable and $(M,g)$ is submanifold of
$\mathbb{R}\times(\mbox{int}(M_{0}),g^{\prime})$ where $(M_{0},g^{\prime})$ is
a compact, simply connected Riemannian manifold with boundary $\partial M_{0}$
which is strictly convex in the sense of the second fundamental form and
$M_{0}$ has no conjugate points.

In order to state the main result of this article, we first need to specify
the subset of $\partial M$ where the measurements are given. Now if we write
$x\in M$, as $x:=(x_{1},x^{\prime})\in\mathbb{R}\times M_{0}$ and
$\varphi(x):=x_{1}$ then $\partial M$ can be decomposed into the two parts
given by

$\partial{M}_{+}:=\\{x\in\partial M:\ \partial_{\nu}\varphi(x)\geq 0\\}\
\mbox{and}\ \partial{M}_{-}:=\\{x\in\partial M:\ \partial_{\nu}\varphi(x)\leq
0\\}$

where $\nu(x)$ stands for outward unit normal to $\partial M$ at $x\in\partial
M$ and $\partial_{\nu}\varphi$ denote the normal derivative of $\varphi$ with
respect to the metric $g$. In this paper, we will be assuming that our
boundary measurements are given on slightly bigger than half of $\partial M$.
To specify this portion of $\partial M$, we take $\epsilon>0$ small enough and
define $\displaystyle\partial M_{\pm,\epsilon/2}$ by

$\displaystyle\begin{aligned}
\partial{M}_{\pm,\epsilon/2}:=\left\\{x\in\partial M:\
\pm\partial_{\nu}\varphi(x)>\frac{\epsilon}{2}\right\\}\ \end{aligned}$ (1.4)

as the small enough open neighborhood of $\partial M_{\pm}$. We denote the
corresponding lateral part of $\partial M_{T}$ by
$\Sigma_{+}:=(0,T)\times\partial{M}_{+},$
$\Sigma_{+,\epsilon/2}:=(0,T)\times\partial{M}_{+,\epsilon/2}$,
$\Sigma_{-}:=(0,T)\times\partial{M}_{-}$ and
$\Sigma_{-,\epsilon/2}:=(0,T)\times\partial{M}_{-,\epsilon/2}$. Now using
these notations, we define the partial DN map by

$\displaystyle\Lambda_{A,q}(f)|_{\Sigma_{+,\epsilon/2}}:=\Big{[}\partial_{\nu}u(t,x)+i\nu(x)\cdot
A(t,x)u(t,x)\Big{]}\Big{|}_{\Sigma_{+,\epsilon/2}},\ \mbox{for}\ f\in
L^{2}(0,T;H^{1/2}(\partial M)).$ (1.5)

With this preparation, we are ready to state the main result of this paper as
follows.

######  Theorem 1.2.

Let $(M,g)$ be a simply connected admissible manifold. Let $\displaystyle
A^{(i)}\in C^{\infty}(M_{T},\mathcal{T}^{*}(M))$ for $i=1,2$ given by
$\displaystyle A^{(i)}(t,x)=\sum\limits_{j=1}^{n}A^{(i)}_{j}(t,x)dx^{j},$ in
local coordinates on $(M,g)$ and $q_{i}\in L^{\infty}(M_{T})$ for $i=1,2$.
Suppose $u_{i}$ for $i=1,2,$ is solution to (1.1) when $(A,q)=(A^{(i)},q_{i})$
for $i=1,2$ and $\Lambda_{A^{(i)},q_{i}}$ are DN maps given by (1.5)
corresponding to $u_{i}$ for $i=1,2$. Now for $\epsilon>0$ small enough if

$\displaystyle\Lambda_{A^{(1)},q_{1}}(f)|_{\Sigma_{+,\epsilon/2}}=\Lambda_{A^{(2)},q_{2}}(f)|_{\Sigma_{+,\epsilon/2}},\
\mbox{for all}\ f\in L^{2}(0,T;H^{1/2}(\partial M)$

then there exists a smooth function $\Psi\in C^{\infty}(M_{T})$ such that

$\displaystyle A^{(2)}(t,x)-A^{(1)}(t,x)=\nabla_{x}\Psi(t,x)\ \mbox{and}\
q_{1}(t,x)=q_{2}(t,x),\ \mbox{for}\ (t,x)\in M_{T}$

provided $A^{(2)}(t,x)=A^{(1)}(t,x)$, for $(t,x)\in\partial M_{T}.$

######  Remark 1.3.

We observe that because of gauge invariance (see [28, 44] and references
therein for more details about gauge invariance), it is impossible to prove
$A^{(1)}=A^{(2)}$ in $M_{T}$ from the given hypothesis of Theorem 1.2. As far
as the uniqueness issue is concerned, gauge invariance guarantees that the
result obtained in Theorem 1.2 is optimal. Of course with some extra
information on vector fields $A^{(i)}$ for $i=1,2$, like
$\nabla_{x}\cdot\left(A^{(2)}-A^{(1)}\right)=0$ in $M_{T}$, one can prove that
$A^{(1)}=A^{(2)}$ in $M_{T}$. Such divergence-free conditions on vector fields
have been considered in prior works (see for example [42, 44]).

The problem considered in this article can be put into the category of
Calderón type inverse problems for parabolic partial differential equations
(pdes) which was initially proposed by Calderón in [14] for elliptic pdes and
studied by Nachman [37] in two dimensions and by Sylvester-Uhlmann in [47] in
dimension three and higher. Analogous problems for parabolic and hyperbolic
pdes have been studied in [2, 29, 38, 41]. Choulli-Kian in [21] derived a
stability estimate for recovering the time-dependent coefficient, which is a
product of functions depending only on time and only space variables, from the
boundary measurements. We also refer to [20] where an abstract inverse problem
for parabolic pde is studied. All these works are concerned with the recovery
of zeroth order perturbation of elliptic, parabolic, and hyperbolic pdes from
full boundary data. In [22] the recovery of general time-dependent zeroth
order perturbation of heat operator from partial boundary measurements is
considered. Inverse problems of recovering the coefficients appearing in the
steady state Convection-diffusion from full and partial boundary measurements
in Euclidean geometry have been studied in [12, 17, 19, 23, 27, 39, 40, 45].
Recovery of first-order perturbation of a parabolic pde from final and single
measurement has been studied in [16] and [18] respectively. In [7], stable
recovery of time-dependent coefficients appearing in a Convection-diffusion
from full boundary data has been studied. Choulli-Kian in [22] proved a
stability estimate for recovering a time-dependent potential from partial
boundary data and motivated by their work authors of [42] proved the unique
recovery of time-dependent coefficients appearing in a Convection-diffusion
equation from partial boundary data. In [42], a uniqueness result is proved
with a smallness assumption on the convection term, which is later on removed
in a recent work of [44] where stability estimates for recovering the time-
dependent coefficients of Convection-diffusion equation from partial boundary
data are derived. We also refer to [7] and [8] where stability estimates for
Convection-diffusion equation from full and partial boundary data is studied,
respectively. Recently in [15, 26] inverse problems related to nonlinear
Convection-diffusion equation is studied. In all the above-mentioned works,
the inverse problems of recovering coefficients appearing in parabolic pdes
from full and partial boundary measurements in Euclidean geometry are
considered. The inverse problems related to steady state Convection-diffusion
equation in Riemannian geometry is considered in prior works (see for example
[11, 28, 30, 36] ) however the recovery of time-dependent coefficients
appearing in parabolic pdes in Riemannian geometry has not been considered in
prior works, and this is the main objective of this paper. To the best of our
knowledge, this is the first work that considers the partial data inverse
problem for recovering both first and zeroth order time-dependent
perturbations of evolution equations in Riemannian geometry. Next, we mention
works on inverse problems related to hyperbolic and dynamical Schrödinger
equations which are closely related to the study of this work. Inspired by
[13] and [9] authors of [31, 35] studied the unique recovery of time-dependent
coefficients appearing in a hyperbolic pde from partial boundary measurements.
We also refer to [3, 4, 5, 6, 10, 25, 33, 34, 43] for inverse problems related
to hyperbolic pdes and dynamical Schrödinger equation in Euclidean as well as
in Riemannian geometry, where time-dependent coefficients are recovered from
full boundary measurements.

The rest of the article is organized as follows. In section 2 we derive the
boundary, and interior Carleman estimates which we will use in section 3 to
construct the exponentially growing as well as decaying solutions. The main
Theorem 1.2 of the article will be proved in section 4. Finally, we conclude
the article with Acknowledgements.

##  2\. Boundary and interior Carleman Estimates

The present section is devoted to prove a boundary and interior Carleman
estimates. The boundary Carleman estimate will be required to estimate the
boundary terms in the integral identity where there is no measurement and the
interior Carleman estimate will be required to construct the geometric optics
solutions (GO) for $\mathcal{L}_{A,q}$ and its formal $L^{2}-$adjoint. For a
compact Riemannian manifold $(M,g)$ with boundary denoted by $\partial M$, we
denote by $dV_{g}$ the volume form on $(M,g)$ and by $dS_{g}$ the induced
volume form on $\partial M$. Then the $L^{2}$-norm of a function $u$ on $M$
and $f$ on $\partial M$ are given by

$\lVert u\rVert_{L^{2}(M)}:=\left(\int_{M}\lvert u(x)\rvert^{2}\
dV_{g}\right)^{1/2}\ \mbox{and}\ \lVert f\rVert_{L^{2}(\partial
M)}:=\left(\int_{\partial M}\lvert f(x)\rvert^{2}\ dS_{g}\right)^{1/2},\
\mbox{respectively}.$

We denote by $L^{2}(M)$ as the space of all functions $u$ defined on $M$ for
which $\lVert u\rVert_{L^{2}(M)}<\infty$ and $L^{2}(\partial M)$ as the space
of all functions $f$ defined on $\partial M$ for which $\lVert
f\rVert_{L^{2}(\partial M)}<\infty.$ Then
$\left(L^{2}(M),\lVert\cdot\rVert_{L^{2}(M)}\right)$ and $\left(L^{2}(\partial
M),\lVert\cdot\rVert_{L^{2}(\partial M)}\right)$ are Hilbert spaces with
respect to the inner-products defined by

$\displaystyle\langle f,g\rangle_{L^{2}(M)}:=\int_{M}f(x)\overline{g(x)}\
dV_{g}$

and

$\langle f,g\rangle_{L^{2}(\partial M)}:=\int_{\partial M}f(x)\overline{g(x)}\
dS_{g}$

respectively.

######  Theorem 2.1.

For $\displaystyle\beta\in\left(\frac{1}{\sqrt{3}},1\right)$, let
$\varphi(t,x):=\lambda^{2}\beta^{2}t+\lambda x_{1}$,
$A\in\left(L^{\infty}(M_{T})\right)^{n}$ and $q\in L^{\infty}(M_{T})$. Then
there exists a constant $C>0$ depending only on $M,T$, $A$ and $q$ such that

$\displaystyle\lambda^{2}\|e^{-\varphi}u\|_{L^{2}(M_{T})}^{2}+\|e^{-\varphi(T,\cdot)}\nabla_{g}u(T,\cdot)\|_{L^{2}(M)}^{2}+\left\|e^{-\varphi}\nabla_{g}u\right\|_{L^{2}(M_{T})}^{2}+\lambda\int_{\Sigma_{+}}\lvert
e^{-\varphi}\partial_{\nu}u(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt$ $\displaystyle\leq
C\left(\|e^{-\varphi}\mathcal{L}_{A,q}u\|_{L^{2}(M_{T})}^{2}+\lambda^{2}C\|e^{-\varphi(T,\cdot)}u(T,\cdot)\|_{L^{2}(M)}^{2}+\lambda\int_{\Sigma_{-}}\lvert
e^{-\varphi}\partial_{\nu}u(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt\right)$ (2.1)

hold for $\lambda$ large enough and for all $u\in C^{2}(\overline{M_{T}})$
satisfying the following

$\displaystyle\begin{aligned} u(0,x)=0,\ \mbox{for}\ x\in M\ \ \mbox{and}\
u(t,x)=0,\ \mbox{for}\ (t,x)\in\partial M_{T}.\end{aligned}$

###### Proof.

To prove the above $H^{1}-L^{2}$ Carleman estimate, for $s>0$, we define by
$\varphi_{s}$ the following convexified weight function.

$\displaystyle\varphi_{s}(t,x):=\varphi(t,x)-\frac{s(x_{1}+2\ell)^{2}}{2}.$
(2.2)

A direct computation gives

$\displaystyle\partial_{t}\varphi_{s}$
$\displaystyle=\lambda^{2}\beta^{2},\qquad\partial_{x_{1}}\varphi_{s}=\lambda-s(x_{1}+2\ell)$
$\displaystyle\partial_{x_{1}}^{2}\varphi_{s}$
$\displaystyle=-s,\qquad|\partial_{x_{1}}\varphi_{s}|^{2}=\left[\lambda-s(x_{1}+2\ell)\right]^{2}.$

Before we proceed further, let us define
$\tilde{q}(t,x):=q(t,x)+i\delta_{g}A(t,x)-|A(t,x)|_{g}^{2}$, where

$\delta_{g}A=\frac{1}{\sqrt{\lvert
g\rvert}}\sum\limits_{j,k=1}^{n}\partial_{j}\left(g^{jk}\sqrt{|g|}A_{k}\right).$

Now let us consider the following conjugated operator

$\displaystyle P_{s}v(t,x)$
$\displaystyle:=e^{-\varphi_{s}(t,x)}\mathcal{L}_{A,q}e^{\varphi_{s}(t,x)}v(t,x)$
$\displaystyle\
=e^{-\varphi_{s}(t,x)}\left(\partial_{t}-\partial^{2}_{x_{1}}-\triangle_{g_{0}}+2iA_{j}g^{jk}\partial_{x_{k}}+i\delta_{g}A-|A|^{2}_{g}+q\right)e^{\varphi_{s}(t,x)}v(t,x)$
$\displaystyle\
=\left\\{\partial_{t}v+(\partial_{t}\varphi_{s})v\right\\}-\left\\{\partial^{2}_{x_{1}}v+2(\partial_{x_{1}}\varphi_{s})(\partial_{x_{1}}v)+(\partial_{x_{1}}\varphi_{s})^{2}v+(\partial_{x_{1}}^{2}\varphi_{s})v\right\\}-\Delta_{g_{0}}v$
$\displaystyle\qquad+2i\langle A,\nabla
v\rangle+2i\partial_{x_{1}}\varphi_{s}g^{1k}A_{k}v+(q-|A|_{g}^{2}+i\delta_{g}A)v$
$\displaystyle\
=\partial_{t}v+\lambda^{2}\beta^{2}v-\left(\partial_{x_{1}}^{2}v+2(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v+(\lambda-s(x_{1}+2\ell))^{2}v-sv\right)-\Delta_{g_{0}}v$
$\displaystyle\qquad+2i\langle A,\nabla
v\rangle+2i(\lambda-s(x_{1}+2\ell)g^{1k}A_{k}v+\tilde{q}v$
$\displaystyle:=P_{1}v(t,x)+P_{2}v(t,x)+P_{3}v(t,x)$

where

$\displaystyle\begin{aligned}
P_{1}v(t,x)&:=\left(\partial_{t}v-2(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v+4sv\right)(t,x),\\\
P_{2}v(t,x)&:=\left(-\partial^{2}_{x_{1}}v-\Delta_{g^{\prime}}v-\lambda^{2}(1-\beta^{2})v+2\lambda
s(x_{1}+2\ell)v-s^{2}(x_{1}+2\ell)^{2}v-3sv\right)(t,x)\\\
&:=-\partial^{2}_{x_{1}}v(t,x)-\Delta_{g^{\prime}}v(t,x)+\mathcal{K}(x_{1})v(t,x),\quad\mbox{
and }\\\ P_{3}v(t,x)&:=2i\langle
A(t,x),\nabla_{g}v(t,x)\rangle+2i(\lambda-s(x_{1}+2\ell)\ell)g^{1k}A_{k}(t,x)v(t,x)+\tilde{q}(t,x)v(t,x).\end{aligned}$

In the definition of $P_{2}$ above, we used
$\mathcal{K}(x_{1}):=-\lambda^{2}(1-\beta^{2})+2\lambda
s(x_{1}+2\ell)-s^{2}(x_{1}+2\ell)^{2}.$ Now define $I_{s}$ by

$\displaystyle\begin{aligned} I_{s}&:=\int_{M_{T}}\lvert
P_{s}v(t,x)\rvert^{2}dV_{g}(x)dt=\int_{M_{T}}\lvert
P_{1}v(t,x)+P_{2}v(t,x)+P_{3}v(t,x)\rvert^{2}dV_{g}dt\\\
&\geq\frac{1}{2}\int_{M_{T}}\left(P_{1}v(t,x)+P_{2}v(t,x)\right)^{2}dV_{g}dt-\int_{M_{T}}\lvert
P_{3}v(t,x)\rvert^{2}dV_{g}dt\\\
&\geq\int_{M_{T}}P_{1}v(t,x)P_{2}v(t,x)dV_{g}dt-\int_{M_{T}}|P_{3}v(t,x)|^{2}dV_{g}dt.\end{aligned}$

This gives us

$\displaystyle\begin{aligned}
I_{s}\geq\underbrace{\int_{M_{T}}P_{1}v(t,x)P_{2}v(t,x)dV_{g}dt}_{I_{s,1}}-\underbrace{\int_{M_{T}}|P_{3}v(t,x)|^{2}dV_{g}dt}_{I_{s,2}}.\end{aligned}$
(2.3)

Our aim is to estimate the right-hand side of (2.3). We start with the first
terms in the above inequality and therefore consider

$\displaystyle\begin{aligned}
P_{1}v(t,x)P_{2}v(t,x)&=-\partial_{t}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)+\mathcal{K}(x_{1})v\partial_{t}v(t,x)-4sv\left(\partial_{x_{1}}^{2}v(t,x)+\Delta_{g^{\prime}}v\right)(t,x)\\\
&\quad+4s\mathcal{K}(x_{1})\lvert
v(t,x)\rvert^{2}+2(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\\\
&\qquad-2\mathcal{K}(x_{1})(\lambda-s(x_{1}+2\ell))v(t,x)\partial_{x_{1}}v(t,x).\end{aligned}$

Next, consider

$\displaystyle\begin{aligned} I_{s,1}&=\int_{M_{T}}P_{1}v(t,x)P_{2}v(t,x)\
dV_{g}dt=-\int_{M_{T}}\partial_{t}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt\\\ &\quad+4s\int_{M_{T}}\mathcal{K}(x_{1})\lvert v(t,x)\rvert^{2}\
dV_{g}dt+\frac{1}{2}\int_{M_{T}}\mathcal{K}(x_{1})\partial_{t}\lvert
v(t,x)\rvert^{2}\ dV_{g}dt\\\
&\quad-4s\int_{M_{T}}v(t,x)\left(\partial_{x_{1}}^{2}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt\\\ &\ \ \ \
+2\int_{M_{T}}(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt\\\ &\ \ \ \
-\int_{M_{T}}\mathcal{K}(x_{1})(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}\lvert
v(t,x)\rvert^{2}\ dV_{g}dt\\\ &\ \
:=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}\end{aligned}$

where

$\displaystyle\begin{aligned}
&I_{1}:=-\int_{M_{T}}\partial_{t}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt;\ \ I_{2}:=4s\int_{M_{T}}\mathcal{K}(x_{1})\lvert v(t,x)\rvert^{2}\
dV_{g}dt\\\
&I_{3}:=\frac{1}{2}\int_{M_{T}}\mathcal{K}(x_{1})\partial_{t}\lvert
v(t,x)\rvert^{2}\ dV_{g}dt;\ \
I_{4}:=-4s\int_{M_{T}}v(t,x)\left(\partial_{x_{1}}^{2}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt\\\
&I_{5}:=2\int_{M_{T}}(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt\\\
&I_{6}:=-\int_{M_{T}}\mathcal{K}(x_{1})(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}\lvert
v(t,x)\rvert^{2}\ dV_{g}dt.\end{aligned}$

In order to estimate the first term on the right-hand side of (2.3), we need
to estimate $I_{j}$ for each $1\leq j\leq 6$. To estimate these $I_{j}$, we
use integration by parts repeatedly along with initial and boundary conditions
on $v$.

$\displaystyle I_{1}$
$\displaystyle=-\int_{M_{T}}\partial_{t}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt=\frac{1}{2}\int_{M}\left|\nabla_{g}v(T,x)\right|_{g}^{2}dV_{g}.$
(2.4) $\displaystyle\begin{aligned}
I_{2}&=4s\int_{M_{T}}\mathcal{K}(x_{1})\lvert v(t,x)\rvert^{2}\ dV_{g}dt\\\
&=-4s\int_{M_{T}}\left(\lambda^{2}(1-\beta^{2})-2\lambda
s(x_{1}+2\ell)+s^{2}(x_{1}+2\ell)^{2}\right)\lvert v(t,x)\rvert^{2}\
dV_{g}dt.\end{aligned}$ (2.5) $\displaystyle\begin{aligned}
I_{3}&=\frac{1}{2}\int_{M_{T}}\mathcal{K}(x_{1})\partial_{t}\lvert
v(t,x)\rvert^{2}\
dV_{g}dt=\frac{1}{2}\int_{M}\mathcal{K}(x_{1})|v(T,x)|^{2}dV_{g}\\\
&=\frac{1}{2}\int_{M}\left(-\lambda^{2}(1-\beta^{2})+2\lambda
s(x_{1}+2\ell)-s^{2}(x_{1}+2\ell)^{2}\right)\lvert v(T,x)\rvert^{2}\
dV_{g}.\end{aligned}$

Recall $\ell\leq(x_{1}+2\ell)\leq 3\ell$, therefore choosing $\lambda$ large
enough, we obtain

$\displaystyle\begin{aligned}
I_{3}&\geq-C\lambda^{2}\|v(T,\cdot)\|_{L^{2}(M)}^{2},\qquad\mbox{ for constant
}C.\end{aligned}$ (2.6)

Next, consider

$\displaystyle I_{4}$
$\displaystyle=-4s\int_{M_{T}}v(t,x)\left(\partial_{x_{1}}^{2}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt=4s\int_{M_{T}}\left|\nabla_{g}v(t,x)\right|_{g}^{2}dV_{g}dt.$ (2.7)

The next integral in the line is

$\displaystyle\begin{aligned}
I_{5}&=2\int_{M_{T}}(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v(t,x)\left(\partial^{2}_{x_{1}}v+\Delta_{g^{\prime}}v\right)(t,x)\
dV_{g}dt\\\
&=2\int_{M_{T}}(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v(t,x)\Delta_{g}v(t,x)\
dV_{g}dt.\end{aligned}$

Using the integration by parts, we have

$\displaystyle\begin{aligned}
I_{5}&=-2\int_{M_{T}}\partial_{x_{1}}v(t,x)\Big{\langle}\nabla_{g}v,\nabla_{g}\left(\lambda-s(x_{1}+2\ell)\right)\Big{\rangle}_{g}\
dV_{g}dt\\\
&\qquad-2\int_{M_{T}}(\lambda-s(x_{1}+2\ell))\Big{\langle}\nabla_{g}v,\partial_{x_{1}}\nabla_{g}v\Big{\rangle}_{g}\
dV_{g}dt\\\ &\qquad\quad+2\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v(t,x)\partial_{\nu}v(t,x)\
dS_{g}dt\end{aligned}$

where $\nu(x)$ is outward unit normal vector to $\partial M$ at $x\in\partial
M$, $\partial_{\nu}v(t,x)$ stands for the normal derivative w.r.t. $x$ of $v$
at $(t,x)\in(0,T)\times\partial M$ and $dS_{g}$ denote the surface measure on
$\partial M$. Again using the integration by parts, we have that

$\displaystyle\begin{aligned}
I_{5}&=2s\int_{M_{T}}\lvert\partial_{x_{1}}v(t,x)\rvert^{2}\
dV_{g}dt-\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\langle\nu,e_{1}\rangle_{g}\lvert\nabla_{g}v(t,x)\rvert_{g}^{2}\
dS_{g}dt\\\ &\qquad-s\int_{M_{T}}\lvert\nabla_{g}v(t,x)\rvert_{g}^{2}\
dV_{g}dt+2\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}v(t,x)\partial_{\nu}v(t,x)\
dS_{g}dt.\end{aligned}$

Now $v|_{(0,T)\times\partial M}=0,$ implies that
$\displaystyle\nabla_{g}v|_{\partial M_{T}}=\left(\partial_{\nu}v\right)\nu$
and $\displaystyle\partial_{x_{1}}v|_{\partial
M_{T}}=\langle\nabla_{g}v,e_{1}\rangle_{g}|_{\partial
M_{T}}=\partial_{\nu}v\langle\nu,e_{1}\rangle_{g}$. Using these, we get

$\displaystyle
I_{5}=s\int_{M_{T}}\left(\lvert\partial_{x_{1}}v(t,x)\rvert^{2}-\lvert\nabla_{g^{\prime}}v(t,x)\rvert_{g^{\prime}}^{2}\right)dV_{g}dt+\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\lvert\partial_{\nu}v(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt.$

Combining $I_{5}$ and $I_{4}$, we have

$\displaystyle\begin{aligned}
I_{4}+I_{5}&=5s\int_{M_{T}}\lvert\partial_{x_{1}}v(t,x)\rvert^{2}dV_{g}dt-3s\int_{M_{T}}\lvert\nabla_{g^{\prime}}v(t,x)\rvert_{g^{\prime}}^{2}dV_{g}dt\\\
&\qquad+\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\lvert\partial_{\nu}v(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt\\\ &\geq
3s\left\|\nabla_{g}v\right\|_{L^{2}(M_{T})}^{2}+\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\lvert\partial_{\nu}v(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt.\end{aligned}$ (2.8)

Next, we consider last term of $I_{s,1}$

$\displaystyle\begin{aligned}
I_{6}&=-\int_{M_{T}}\mathcal{K}(x_{1})(\lambda-s(x_{1}+2\ell))\partial_{x_{1}}\lvert
v(t,x)\rvert^{2}\ dV_{g}dt\\\
&=\int_{M_{T}}\left(\lambda-s(x_{1}+2\ell)\right)\lvert
v(t,x)\rvert^{2}\partial_{x_{1}}\mathcal{K}(x_{1})\
dV_{g}dt-s\int_{M_{T}}\mathcal{K}(x_{1})\lvert v(t,x)\rvert^{2}\ dV_{g}dt\\\
&=2s\int_{M_{T}}\left(\lambda-s(x_{1}+2\ell)\right)^{2}\lvert
v(t,x)\rvert^{2}\ dV_{g}dt\\\ &\ \ \
+s\int_{M_{T}}\left(\lambda^{2}(1-\beta^{2})-2\lambda
s(x_{1}+2\ell)+s^{2}(x_{1}+2\ell)^{2}\right)\lvert v(t,x)\rvert^{2}\
dV_{g}dt.\end{aligned}$

After simplifying, we get

$\displaystyle\begin{aligned}
I_{6}&=s\lambda^{2}(3-\beta^{2})\int_{M_{T}}\lvert v(t,x)\rvert^{2}_{g}\
dV_{g}dt-6\lambda s\int_{M_{T}}(x_{1}+2\ell)\lvert v(t,x)\rvert_{g}^{2}\
dV_{g}dt\\\ &\ \ \ \ \ +3s^{3}\int_{M_{T}}(x_{1}+2\ell)^{2}\lvert
v(t,x)\rvert_{g}^{2}\ dV_{g}dt\end{aligned}$ (2.9)

Next, we estimate $I_{s,2}$ in the following way:

$\displaystyle I_{s,2}$ $\displaystyle=\int_{M_{T}}|P_{3}v(t,x)|^{2}dV_{g}dt$
$\displaystyle=\int_{M_{T}}\left|2i\langle
A(t,x),\nabla_{g}v(t,x)\rangle+2i(\lambda-s(x_{1}+2\ell)\ell)g^{1k}A_{k}(t,x)v(t,x)+\tilde{q}(t,x)v(t,x)\right|^{2}dV_{g}dt$
$\displaystyle\leq
8\|A\|^{2}_{L^{\infty}(M_{T})}\|\nabla_{g}v\|^{2}_{L^{2}(M_{T})}+8\lambda^{2}\|A\|^{2}_{L^{\infty}(M_{T})}\|v\|^{2}_{L^{2}(M_{T})}+2\|\tilde{q}\|^{2}_{L^{\infty}(M_{T})}\|v\|_{L^{2}(M_{T})}^{2}$
(2.10)

Combining $I_{2}$, $I_{4}$, $I_{5}$, $I_{6}$, and $I_{s,2}$ in the following
way:

$\displaystyle I_{2}+I_{4}+I_{5}+I_{6}-I_{s,2}$
$\displaystyle=\left\\{\lambda^{2}\left(s\left(3\beta^{2}-1\right)-8\|A\|^{2}_{L^{\infty}(M_{T})}-2\|\tilde{q}\|^{2}_{L^{\infty}(M_{T})}\right)+l.\
o.\ t.\mbox{ of }\lambda\right\\}\|v\|_{L^{2}(M_{T})}^{2}$
$\displaystyle\quad+\left(3s-8\|A\|^{2}_{L^{\infty}(M_{T})}\right)\|\nabla_{g}v\|^{2}_{L^{2}(M_{T})}$
$\displaystyle\qquad+\int_{\partial
M_{T}}(\lambda-s(x_{1}+2\ell))\lvert\partial_{\nu}v\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt.$

Since $s$ and $\lambda$ both are large parameters therefore by choosing $s$
and $\lambda$ large enough and using the fact $\beta\in(1/\sqrt{3},1)$, we can
combine all estimates obtained above to get the following estimates on
$P_{s}$:

$\displaystyle\begin{aligned} \left\|P_{s}v\right\|_{L^{2}(M_{T})}^{2}&\geq
C\left(s\lambda^{2}\|v\|_{L^{2}(M_{T})}^{2}+\|\nabla_{g}v(T,\cdot)\|_{L^{2}(M)}^{2}-\lambda^{2}\|v(T,\cdot)\|_{L^{2}(M)}^{2}+s\left\|\nabla_{g}v\right\|_{L^{2}(M_{T})}^{2}\right.\\\
&\qquad\left.+\lambda\int_{\partial
M_{T}}\lvert\partial_{\nu}v(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt\right)\end{aligned}$ (2.11)

where constant $C$ depends on $A$, $q$, $T$ and $M$.

This provides the estimate for the operator
$P_{s}=e^{-\varphi_{s}(t,x)}\mathcal{L}_{A,q}e^{\varphi_{s}(t,x)}$. To obtain
the required Carleman estimate, we put $v(t,x)=e^{-\varphi_{s}(t,x)}u(t,x)$

$\displaystyle\|e^{-\varphi_{s}}\mathcal{L}_{A,q}u\|_{L^{2}(M_{T})}^{2}+\lambda^{2}C\|e^{-\varphi_{s}(T,\cdot)}u(T,\cdot)\|_{L^{2}(M)}^{2}+\lambda\int_{\Sigma_{-}}\lvert
e^{-\varphi_{s}}\partial_{\nu}u(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt$ $\displaystyle\quad\geq
s\lambda^{2}\|e^{-\varphi_{s}}u\|_{L^{2}(M_{T})}^{2}+\|e^{-\varphi_{s}(T,\cdot)}\nabla_{g}u(T,\cdot)\|_{L^{2}(M)}^{2}+s\left\|e^{-\varphi_{s}}\nabla_{g}u\right\|_{L^{2}(M_{T})}^{2}$
$\displaystyle\qquad+\lambda\int_{\Sigma_{+}}\lvert
e^{-\varphi_{s}}\partial_{\nu}u(t,x)\rvert_{g}^{2}\langle\nu,e_{1}\rangle_{g}\
dS_{g}dt.$

∎

Our next aim of this section is to derive the interior Carleman estimates in a
Sobolev space of negative order for $\mathcal{L}_{A,q}$ and its formal the
formal $L^{2}$-adjoint $\mathcal{L}_{A,q}^{*}$. Before going to state and
prove the interior Carleman estimates, we first give some definitions and
notations for semi-classical Sobolev spaces of arbitrary order. This will help
us to represent the Carleman estimates in a nice form. Let us begin by
assuming that $(M,g)$ is embedded in a compact Riemannian manifold $(N,g)$
without boundary and denote by $N_{T}:=(0,T)\times N.$ Following [28], we
denote by $J^{s}$ where $s\in\mathbb{R}$, is the semi-classical
pseudodifferential operator of order $s$ on $(N,g)$ and it is defined by
$\displaystyle J^{s}:=\left(\lambda^{2}-\Delta_{g}\right)^{s/2}$. Using this
we define the semi-classical Sobolev space $H^{s}_{\lambda}(N)$ for
$s\in\mathbb{R}$, as the completion of $C^{\infty}(N)$ with respect to the
following norm

$\lVert u\rVert_{H^{s}_{\lambda}(N)}:=\lVert J^{s}u\rVert_{L^{2}(N)}.$

Since $(N,g)$ is a Riemannian manifold without boundary therefore the dual of
$H^{s}_{\lambda}(N)$, for any $s\in\mathbb{R}$ can be identified with
$H^{-s}_{\lambda}(N)$. Also note that for $s=1$, we have that

$\lVert u\rVert^{2}_{H^{1}_{\lambda}(N)}:=\lambda^{2}\lVert
u\rVert^{2}_{L^{2}(N)}+\lVert\nabla_{g}u\rVert^{2}_{L^{2}(N)}.$

Now following [24] the time-dependent Sobolev sapace
$L^{2}(0,T;H^{s}_{\lambda}(N))$ is defined as the set of all strongly
measurable functions $u:[0,T]\rightarrow H^{s}_{\lambda}(N)$ such that

$\displaystyle\lVert
u\rVert_{L^{2}(0,T;H^{s}_{\lambda}(N)}:=\left(\int_{0}^{T}\lVert
u(t,\cdot)\rVert_{H^{s}_{\lambda}(N)}^{2}\ dt\right)^{1/2}<\infty.$ (2.12)

Then $L^{2}(0,T;H^{s}_{\lambda}(N))$ is a Banach space with respect to the
norm $\lVert\cdot\rVert_{L^{2}(0,T;H^{s}_{\lambda}(N))}$ defined by (2.12) and
the dual of $L^{2}(0,T;H^{s}_{\lambda}(N))$ can be identified with
$L^{2}(0,T;H^{-s}_{\lambda}(N))$. If we take $v\in C_{c}^{\infty}(M_{T})$ in
(2.11) then we have the following estimate

$\displaystyle\begin{aligned} \lVert
v\rVert^{2}_{L^{2}(0,T;H^{1}_{\lambda}(N))}\leq
C\lVert\mathcal{L}_{\varphi}v\rVert^{2}_{L^{2}(0,T;L^{2}(N))}\end{aligned}$
(2.13)

where $\varphi$ is same as in Theorem 2.1 and
$\displaystyle\mathcal{L}_{\varphi}:=e^{-\varphi}\mathcal{L}_{A,q}e^{\varphi}$.
Now if we denote by
$\mathcal{L}_{\varphi}^{*}:=e^{\varphi}\mathcal{L}_{A,q}^{*}e^{-\varphi}$
where $\mathcal{L}_{A,q}^{*}$ stands for a formal $L^{2}-$adjoint of
$\mathcal{L}_{A,q}$ then using the arguments similar to the one used in
deriving (2.11), the following estimate

$\displaystyle\lVert u\rVert^{2}_{L^{2}(0,T;H^{1}_{\lambda}(N))}\leq
C\lVert\mathcal{L}^{*}_{\varphi}u\rVert^{2}_{L^{2}(0,T;L^{2}(N))}$ (2.14)

holds for all $u\in C_{c}^{\infty}(M_{T})$ where $\varphi$ is same as in
Theorem 2.1 and constant $C>0$ is independent of $\lambda$ and $u$.

In order to construct the suitable solutions to $\mathcal{L}^{*}_{A,q}u=0$ and
$\mathcal{L}_{A,q}v=0$, we need to shift the index by $-1$ for spacial
variable in (2.13) and (2.14) respectively, which we will do in the following
lemma.

######  Lemma 2.2.

Let
$\displaystyle\mathcal{L}_{\varphi}^{*}:=e^{\varphi}\mathcal{L}_{A,q}^{*}e^{-\varphi},$
and
$\displaystyle\mathcal{L}_{\varphi}:=e^{-\varphi}\mathcal{L}_{A,q}e^{\varphi},$
where $\mathcal{L}_{A,q}^{*}$ denote the formal $L^{2}$-adjoint of
$\mathcal{L}_{A,q}$ and $\varphi$, $A$ and $q$ be as in Theorem 2.1. Then
there exists a constant $C>0$ independent of $\lambda$ and $v$ such that

$\displaystyle\lVert v\rVert_{L^{2}([0,T];L^{2}(N))}\leq
C\lVert\mathcal{L}_{\varphi}^{*}v\rVert_{L^{2}(0,T;H^{-1}_{\lambda}(N))}$
(2.15)

holds for all $\lambda$ large enough and for all $v\in
C^{1}([0,T];C_{c}^{\infty}(M))$ satisfying $v(T,x)=0,$ for $x\in M$ and

$\displaystyle\lVert v\rVert_{L^{2}([0,T];L^{2}(N))}\leq
C\lVert\mathcal{L}_{\varphi}v\rVert_{L^{2}(0,T;H^{-1}_{\lambda}(N))}$ (2.16)

holds for all $\lambda$ large enough and for all $v\in
C^{1}([0,T];C_{c}^{\infty}(M))$ satisfying $v(0,x)=0,$ for $x\in M$.

###### Proof.

The proof of lemma can derived by combining the arguments from [28] and [42].
∎

The above estimates together with the Hahn-Banach theorem and the Riesz
representation theorem gives the following solvability result, proof of which
follows from [42, 44].

######  Lemma 2.3.

Let $\varphi$, $A$ and $q$ be as before and $\lambda>0$ be large enough. Then
for $F\in L^{2}(M_{T})$ there exists a solution $u\in
H^{1}(0,T;H^{-1}(M))\times L^{2}(0,T;H^{1}(M))$ of

$\displaystyle\begin{aligned}
\begin{cases}\mathcal{L}_{\varphi}w(t,x)=F(t,x),\ (t,x)\in M_{T}\\\ w(0,x)=0,\
x\in M\end{cases}\end{aligned}$

satisfying the following estimate

$\displaystyle\lVert u\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M)}\leq C\lVert
F\rVert_{L^{2}(M_{T})}$ (2.17)

for some constant $C>0$ independent of $\lambda$ and there exists a solution
$v\in H^{1}(0,T;H^{-1}(M))\times L^{2}(0,T;H^{1}(M))$ of

$\displaystyle\begin{aligned}
\begin{cases}\mathcal{L}_{\varphi}^{*}w(t,x)=F(t,x),\ (t,x)\in M_{T}\\\
w(T,x)=0,\ x\in M\end{cases}\end{aligned}$

satisfying the following estimate

$\displaystyle\lVert v\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M))}\leq C\lVert
F\rVert_{L^{2}(M_{T})}$ (2.18)

for some constant $C>0$ independent of $\lambda$ and $v$.

##  3\. Construction of geometric optics solutions

In this section, we aim to construct the exponential growing and decaying
solutions to the Convection-diffusion operator
$\displaystyle\mathcal{L}_{A,q}$ and its $L^{2}$-adjoint
$\displaystyle\mathcal{L}^{*}_{A,q}$, respectively. Construction of these
solutions will be proved with the help of the interior Carleman estimates in a
negative order Sobolev spaces stated in Lemma 2.2.

###  3.1. Construction of exponentially growing solutions

In this subsection, we will construct the exponential growing solutions to
$\mathcal{L}_{A,q}u(t,x)=0,$ in $M_{T}$ which takes the following form

$\displaystyle
u(t,x)=e^{\left(\varphi+i\psi\right)(t,x)}\Big{(}T_{g}(t,x)+R_{g,\lambda}(t,x)\Big{)}$
(3.1)

where $\varphi$ is same as in Theorem 2.1 and $\psi$, $T_{g}$ will be
constructed using the WKB construction in such a way that the correction term
$R_{g,\lambda}$ satisfies the following

$\displaystyle\begin{aligned}
e^{-\left(\varphi+i\psi\right)}\mathcal{L}_{A,q}\left(e^{\left(\varphi+i\psi\right)}R_{g,\lambda}(t,x)\right)&=F_{\lambda}(t,x),\
(t,x)\in M_{T}\\\ \end{aligned}$

for $F_{\lambda}\in L^{2}(M_{T})$ such that $\lVert
F_{\lambda}\rVert_{L^{2}(M_{T})}\leq C,$ for some constant $C>0$ independent
of $\lambda$ and $R_{g,\lambda}$ satisfies $\displaystyle\lVert
R_{g,\lambda}\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M))}\leq C\lVert
F_{\lambda}\rVert_{L^{2}(M_{T})},$ for some constant $C>0,$ not depending on
$\lambda$. More precisely, we prove the following theorem.

######  Theorem 3.1.

Let $M_{T}$, $\mathcal{L}_{A,q}$ and $\varphi$ be as before. Suppose
$(D,g_{0})$ be a simple manifold which is extension of $(M_{0},g_{0})$ in the
sense that $M_{0}\subset D$ and $y_{0}\in D$ is such that $(x_{1},y_{0})\notin
M$ for all $x_{1}$. Now if $(r,\theta)$ denote the polar normal coordinates on
$(D,g_{0})$, $(x_{1},r,\theta)$ denote the points in $M$ and $A_{1}$ and
$A_{r}$ are components of $A$ in $x_{1}$ and $r$ coordinates respectively,
then for $\lambda$ large enough the following equation

$\displaystyle\begin{aligned} \begin{cases}\mathcal{L}_{A,q}v(t,x)=0,\ \
(t,x)\in M_{T},\\\ v(0,x)=0,\ \ x\in M\end{cases}\end{aligned}$

has a solution taking the following form

$\displaystyle\begin{aligned}
u(t,x)=e^{\varphi+i\psi}\Big{(}T_{g}(t,x_{1},r,\theta)+R_{g,\lambda}(t,x_{1},r,\theta)\Big{)}\end{aligned}$
(3.2)

where

$\displaystyle\begin{aligned} \psi=\lambda(\sqrt{1-\beta^{2}})r,\ \mbox{and}\
T_{g}(t,x_{1},r,\theta)=\phi(t)e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}e^{-\mu
r}e^{i\Phi_{1}(t,x_{1},r,\theta)}b(r,\theta)^{-1/4}h(\theta)\end{aligned}$

here $\phi\in C_{c}^{\infty}(0,T)$, $\Phi_{1}$ is solution to

$\partial_{1}\Phi_{1}+i(\sqrt{1-\beta^{2}})\partial_{r}\Phi_{1}+\left(A_{1}+i(\sqrt{1-\beta^{2}})A_{r}\right)=0$

and $R_{g,\lambda}$ satisfies the following

$\displaystyle\begin{aligned}
\begin{cases}\mathcal{L}_{\varphi}\left(e^{i\psi}R_{g,\lambda}\right)(t,x)=-e^{-i\psi}\mathcal{L}_{A,q}T_{g}(t,x),\
\ \ (t,x)\in M_{T}\\\ R_{g,\lambda}(0,x)=0,\ \ \ x\in
M\end{cases}\end{aligned}$

and $\displaystyle\lVert
R_{g,\lambda}\rVert_{L^{2}\left(0,T;H^{1}_{\lambda}(M)\right)}\leq C$ for some
constant $C>0$ independent of $\lambda$.

###### Proof.

Following [28], if we denote $\rho:=\varphi+i\psi,$ then simple calculations
show that the conjugated operator
$\mathcal{L}_{\rho}:=e^{-\rho}\mathcal{L}_{A,q}e^{\rho}$ will have the
following expression

$\displaystyle\begin{aligned}
\mathcal{L}_{\rho}=\mathcal{L}_{A,q}+\left(\partial_{t}\rho-\Delta_{g}\rho-g^{jk}\partial_{j}\rho\partial_{k}\rho\right)-2\left(g^{jk}\partial_{j}\rho\partial_{k}+ig^{jk}\partial_{j}\rho
A_{k}\right).\end{aligned}$

Using $\rho=\varphi+i\psi,$ and $\varphi=\lambda^{2}\beta^{2}t+\lambda x_{1}$,
we get

$\displaystyle\begin{aligned}
\mathcal{L}_{\rho}=\mathcal{L}_{A,q}+\left(\lambda^{2}\beta^{2}-\lambda^{2}+g^{jk}\partial_{j}\psi\partial_{k}\psi\right)-\left(2\lambda\partial_{1}+2ig^{jk}\partial_{j}\psi\partial_{k}+2i\lambda
A_{1}-2g^{jk}\partial_{j}\psi A_{k}+i\Delta_{g}\psi\right).\end{aligned}$
(3.3)

Now $u$ given by (3.1) solves $\mathcal{L}_{A,q}v=0$ if and only if it
satisfies $\mathcal{L}_{\rho}u=0$. This will give us

$\displaystyle\begin{aligned}
\mathcal{L}_{\rho}R_{g,\lambda}(t,x)&=-\mathcal{L}_{A,q}T_{g}(t,x)-\left(\lambda^{2}\beta^{2}-\lambda^{2}+g^{jk}\partial_{j}\psi\partial_{k}\psi\right)T_{g}(t,x)\\\
&\ \ \ \ \ \ \
+\left(2\lambda\partial_{1}+2ig^{jk}\partial_{j}\psi\partial_{k}+2i\lambda
A_{1}-2g^{jk}\partial_{j}\psi A_{k}+i\Delta_{g}\psi\right)T_{g}(t,x),\ \
(t,x)\in M_{T}.\end{aligned}$ (3.4)

In order to have $\lVert
R_{g,\lambda}\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M)}\leq C$, we choose $\psi$
and $T_{g}$ such that

$\displaystyle
g^{jk}\partial_{j}\psi\partial_{k}\psi=\lambda^{2}(1-\beta^{2})$ (3.5)

and

$\displaystyle\left(2\lambda\partial_{1}+2ig^{jk}\partial_{j}\psi\partial_{k}+2i\lambda
A_{1}-2g^{jk}\partial_{j}\psi A_{k}+i\Delta_{g}\psi\right)T_{g}(t,x)=0,\
(t,x)\in M_{T}.$ (3.6)

To solve equations (3.5) and (3.6) for $\psi$ and $T_{g}$, we use the polar
normal coordinates $(r,\theta)$ on $(D,g_{0})$ centered at $y_{0}\in D$ as
mentioned in statement of theorem. We consider the polar normal coordinates on
$D$ which are denoted by $(r,\theta)$ and given by
$x_{0}=\exp_{y_{0}}(r\theta),$ where $r>0$ and $\theta\in S_{y_{0}}(D):\\{v\in
T_{y_{0}}D:\ \lvert v\rvert_{g}=1\\}$, here $T_{y_{0}}D$ denote the tangent
space to $D$ at $y_{0}\in D$. Then using the Gauss lemma (see Lemma $15$ in
Chapter $9$ of [46]) there exists a smooth positive definite matrix
$P(r,\theta)$ with $\det(P):=b(r,\theta)$ such that the metric $g_{0}$ in the
polar normal coordinates $(r,\theta)$, takes the following form

$\displaystyle\begin{aligned} g_{0}(r,\theta)=\begin{bmatrix}1&0\\\
0&P(r,\theta)\end{bmatrix}.\end{aligned}$ (3.7)

Now since the points in $M$ are denoted by $(x_{1},r,\theta)$ where
$(r,\theta)$ are polar normal coordinates in $(D,g_{0})$, therefore after
using the previous form of $g_{0}$, the metric $g$ has the following form

$\displaystyle\begin{aligned} g(x_{1},r,\theta)=\begin{bmatrix}1&0&0\\\
0&1&0\\\ 0&0&P(r,\theta)\end{bmatrix}.\end{aligned}$ (3.8)

Using (3.8), we see that

$\displaystyle\psi(x)=(\lambda\sqrt{1-\beta^{2}})\text{dist}_{g}(y_{0},x)=(\lambda\sqrt{1-\beta^{2}})r,$
(3.9)

solves equation (3.5) on $M$. Using this choice of $\psi$ and form of $g$
given by (3.8) in equation (3.6), we have

$\displaystyle\begin{aligned}
\left(\partial_{1}+i(\sqrt{1-\beta^{2}})\partial_{r}+iA_{1}-(\sqrt{1-\beta^{2}})A_{r}+i(\sqrt{1-\beta^{2}})\frac{\partial_{r}b(r,\theta)}{4b(r,\theta)}\right)T_{g}(t,x_{1},r,\theta)=0.\end{aligned}$

Now one can check that solution of the above equation can be given by

$\displaystyle
T_{g}(t,x_{1},r,\theta)=\phi(t)e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}e^{-\mu
r}e^{i\Phi_{1}(t,x_{1},r,\theta)}b(r,\theta)^{-1/4}h(\theta)$ (3.10)

where $\phi\in C_{c}^{\infty}(0,T)$, $\mu\in\mathbb{R}$, $h\in
C^{\infty}(S_{y_{0}}(D))$ are arbitrary but fixed and
$\Phi_{1}(t,x_{1},r,\theta)$ satisfies the following

$\displaystyle\left(\partial_{1}\Phi_{1}+i\left(\sqrt{1-\beta^{2}}\right)\partial_{r}\Phi_{1}\right)+\left(A_{1}+i(\sqrt{1-\beta^{2}})A_{r}\right)=0.$
(3.11)

Now using (3.9) and (3.10) in (3.4), we get

$\displaystyle\begin{aligned}
\begin{cases}\mathcal{L}_{\rho}R_{g,\lambda}(t,x)=-\mathcal{L}_{A,q}T_{g}(t,x),\
\ (t,x)\in M_{T}\\\ R_{g,\lambda}(0,x)=0,\ \ x\in M.\end{cases}\end{aligned}$

But
$\mathcal{L}_{\rho}R_{g,\lambda}=e^{-i\psi}\mathcal{L}_{\varphi}\left(e^{i\psi}R_{g,\lambda}\right)$
therefore if we denote $\widetilde{R}_{g,\lambda}=e^{i\psi}R_{g,\lambda}$ then
$\widetilde{R}_{g,\lambda}$ satisfies the following equation

$\displaystyle\begin{cases}\mathcal{L}_{\varphi}\widetilde{R}_{g,\lambda}(t,x)=-e^{i\psi}\mathcal{L}_{A,q}T_{g}(t,x),\
\ (t,x)\in M_{T}\\\ \widetilde{R}_{g,\lambda}(0,x)=0,\ \ x\in M\end{cases}$
(3.12)

Now using the expressions for $\psi$ and $T_{g}$ from (3.9) and (3.10)
respectively and assumptions on $A$ and $q$, we have that
$\displaystyle-e^{i\psi}\mathcal{L}_{A,q}T_{g}\in L^{2}(M_{T})$ and
$\displaystyle\lVert e^{i\psi}\mathcal{L}_{A,q}T_{g}\rVert_{L^{2}(M_{T})}\leq
C$, for some constant $C>0$ independent of $\lambda$. Hence using Lemma 2.3
together with above estimate for right hand side of (3.12), we conclude that
there exists $\widetilde{R}_{g,\lambda}\in H^{1}(0,T;H^{-1}(M))\times
L^{2}(0,T;H^{1}(M))$ solving (3.12) and it satisfies the following estimate
$\displaystyle\lVert\widetilde{R}_{g,\lambda}\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M))}\leq
C$, for some constant $C>0,$ independent of $\lambda$. Hence, finally we
conclude that $R_{g,\lambda}$ solves the required equation and satisfies the
desired estimate. This completes the proof Theorem. ∎

###  3.2. Construction of exponentially decaying solutions

Aim of this subsection, is to construct the exponential decaying solutions to
$\displaystyle\mathcal{L}^{*}_{A,q}u:=\mathcal{L}_{-A,\overline{q}}u=0,$ in
$M_{T}$ taking the following form

$\displaystyle
u(t,x)=e^{-\left(\varphi+i\psi\right)(t,x)}\Big{(}T_{d}(t,x)+R_{d,\lambda}(t,x)\Big{)}$
(3.13)

where $\varphi$ is same as in Theorem 2.1 and $\psi$, $T_{d}$ will be
constructed using the WKB construction in such a way that the correction term
$R_{d,\lambda}$ satisfies the following

$\displaystyle\begin{aligned}
\begin{cases}e^{\left(\varphi+i\psi\right)}\mathcal{L}_{A,q}\left(e^{-\left(\phi+i\psi\right)}R_{d,\lambda}(t,x)\right)=F_{\lambda}(t,x),\
(t,x)\in M_{T}\\\ R_{d,\lambda}(0,x)=0,\ x\in M\end{cases}\end{aligned}$

for some $F_{\lambda}\in L^{2}(M_{T})$ such that $\lVert
F_{\lambda}\rVert_{L^{2}(M_{T})}\leq C,$ for some constant $C>0$ independent
of $\lambda$ and $R_{d,\lambda}$ satisfies $\displaystyle\lVert
R_{d,\lambda}\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M))}\leq C\lVert
F_{\lambda}\rVert_{L^{2}(M_{T})},$ for some constant $C>0$ not depending on
$\lambda$. To construct these solutions, we first start with the construction
of $\psi$ and $T_{d}$ following the arguments used in Theorem 3.1. Denote by
$\rho:=\varphi+i\psi,$ then one can check that the conjugated operator

$\mathcal{L}_{\rho}^{*}:=e^{\rho}\mathcal{L}_{A,{q}}^{*}e^{-\rho}=e^{\rho}\mathcal{L}_{-A,\overline{q}}e^{-\rho}$

is given by

$\displaystyle\begin{aligned}
\mathcal{L}_{\rho}^{*}=\mathcal{L}_{-A,\overline{q}}+\left(\partial_{t}\rho-g^{jk}\partial_{j}\rho\partial_{k}\rho\right)+\left(2g^{jk}\partial_{j}\rho\partial_{k}-2ig^{jk}\partial_{j}\rho
A_{k}+\Delta_{g}\rho\right).\end{aligned}$

Now using $\rho=\varphi+i\psi$ and $\varphi=\lambda^{2}\beta^{2}t+x_{1}$, we
have

$\displaystyle\mathcal{L}_{\rho}^{*}=\mathcal{L}_{-A,\overline{q}}+\left(\lambda^{2}\beta^{2}-\lambda^{2}+g^{jk}\partial_{j}\psi\partial_{k}\psi\right)+\left(2\lambda\partial_{1}+2ig^{jk}\partial_{j}\psi\partial_{k}-2i\lambda
A_{1}+2g^{jk}\partial_{j}\psi A_{k}+i\Delta_{g}\psi\right).$

Now $u$ given by (3.13) solves $\mathcal{L}_{A,q}^{*}v=0$ in $M_{T}$ if and
only if $\mathcal{L}^{*}_{\rho}u=0$ in $M_{T}.$ Using this, we see that
$R_{d,\lambda}$ satisfies the following equation

$\displaystyle\begin{aligned}
\mathcal{L}_{\rho}^{*}R_{d,\lambda}(t,x)&=-\mathcal{L}_{-A,\overline{q}}T_{d}(t,x)-\left(\lambda^{2}\beta^{2}-\lambda^{2}-g^{jk}\partial_{j}\psi\partial_{k}\psi\right)T_{d}(t,x)\\\
&\ \ \ \ \
-\left(2\lambda\partial_{1}+2ig^{jk}\partial_{j}\psi\partial_{k}-2i\lambda
A_{1}+2g^{jk}\partial_{j}\psi
A_{k}+i\Delta_{g}\psi\right)T_{d}(t,x).\end{aligned}$ (3.14)

Now to have the estimate $\displaystyle\lVert
R_{d,\lambda}\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M)}\leq C,$ for some constant
$C>0$ independent of $\lambda$, we choose $\psi$ and $T_{d}$ satisfying the
following equations

$\displaystyle\lambda^{2}\beta^{2}-\lambda^{2}+g^{jk}\partial_{j}\psi\partial_{k}\psi=0$
(3.15)

and

$\displaystyle\left(2\lambda\partial_{1}+2ig^{jk}\partial_{j}\psi\partial_{k}-2i\lambda
A_{1}+2g^{jk}\partial_{j}\psi A_{k}+i\Delta_{g}\psi\right)T_{d}(t,x)=0,\
(t,x)\in M_{T}$ (3.16)

respectively. To solve equations (3.15) and (3.16) for $\psi$ and $T_{d}$, we
use the polar normal coordinates $(r,\theta)$ on $(D,g_{0})$ centered at
$y_{0}\in D$ as used in proof of Theorem 3.1. For a fixed $y_{0}\in D$, we
consider the polar normal coordinates on $D$ which are denoted by $(r,\theta)$
and given by $x_{0}=\exp_{y_{0}}(r\theta),$ where $r>0$ and $\theta\in
S_{y_{0}}(D):\\{v\in T_{y_{0}}D:\ \lvert v\rvert_{g}=1\\}$, here $T_{y_{0}}D$
denote the tangent space to $D$ at $y_{0}\in D$. Then using the Gauss lemma
(see Lemma $15$ in Chapter $9$ of [46]) there exists a smooth positive
definite matrix $P(r,\theta)$ with $\det P(r,\theta)=b(r,\theta)$ such that
the metric $g_{0}$ in the polar normal coordinates $(r,\theta)$, takes form
given by (3.7). Now since the points in $M$ are denoted by $(x_{1},r,\theta)$
where $(r,\theta)$ are polar normal coordinates in $(D,g_{0})$, therefore
after using the form of $g_{0}$ given by (3.7), the metric $g$ takes the form
given by equation (3.8) and using this, we observe that

$\displaystyle\psi(x)=\left(\lambda\sqrt{1-\beta^{2}}\right)\text{dist}_{g}(y_{0},x)=\left(\lambda\sqrt{1-\beta^{2}}\right)r,$
(3.17)

solves equation (3.15) and

$\displaystyle
T_{d}(t,x_{1},r,\theta)=\phi(t)e^{i\Phi_{2}(t,x_{1},r,\theta)}b(r,\theta)^{-1/4}h(\theta)$
(3.18)

solves equation (3.16) where $\phi\in C_{c}^{\infty}(0,T)$, $h\in
C^{\infty}(S_{y_{0}}(D))$ are arbitrary but fixed and
$\Phi_{2}(t,x_{1},r,\theta)$ satisfies the following

$\displaystyle\left(\partial_{1}\Phi_{2}+i\left(\sqrt{1-\beta^{2}}\right)\partial_{r}\Phi_{2}\right)-\left(A_{1}+i\left(\sqrt{1-\beta^{2}}\right)A_{r}\right)=0,$
(3.19)

$A_{1}$ and $A_{r}$ are components of $A$ in $x_{1}$ and $r$ coordinates
respectively. Now if we use (3.17) and (3.18) in equation (3.14) and repeating
the arguments used in showing the estimate for $R_{g,\lambda}$ in Theorem 3.1,
then we get that there exists $R_{d,\lambda}\in H^{1}(0,T;H^{-1}(M))\times
L^{2}(0,T;H^{1}(M))$ solving

$\displaystyle\begin{aligned}
\begin{cases}\mathcal{L}^{*}_{\varphi}\left(e^{i\psi}R_{d,\lambda}\right)(t,x)=-e^{-i\psi}\mathcal{L}_{A,q}T_{d}(t,x),\
\ \ (t,x)\in M_{T}\\\ R_{d,\lambda}(0,x)=0,\ \ \ x\in
M\end{cases}\end{aligned}$ (3.20)

and $R_{d,\lambda}$ satisfies the following estimate

$\displaystyle\lVert R_{d,\lambda}\rVert_{L^{2}(0,T;H^{1}_{\lambda}(M)}\leq C$
(3.21)

for some constant $C>0$ independent of $\lambda$. Combining all these, we end
up with proving the following theorem.

######  Theorem 3.2.

Let $M_{T}$, $\mathcal{L}_{A,q}$ and $\varphi$ be as before. Suppose
$(D,g_{0})$ be a simple manifold which is extension of $(M_{0},g_{0})$ in the
sense that $M_{0}\subset D$ and $y_{0}\in D$ is such that $(x_{1},y_{0})\notin
M$ for all $x_{1}$. Now if $(r,\theta)$ denote the polar normal coordinates on
$(D,g_{0})$, $(x_{1},r,\theta)$ denote the points in $M$ and $A_{1}$ and
$A_{r}$ are components of $A$ in $x_{1}$ and $r$ coordinates respectively,
then for $\lambda$ large enough the following equation

$\displaystyle\begin{aligned} \begin{cases}\mathcal{L}^{*}_{A,q}v(t,x)=0,\ \
(t,x)\in M_{T},\\\ v(0,x)=0,\ \ x\in M\end{cases}\end{aligned}$

has a solution taking the following form

$\displaystyle\begin{aligned}
v(t,x)=e^{-\left(\varphi+i\psi\right)(t,x)}\Big{(}T_{d}(t,x_{1},r,\theta)+R_{d,\lambda}(t,x_{1},r,\theta)\Big{)}\end{aligned}$
(3.22)

where $\psi,T_{d}$ are given by (3.17), (3.18) and $R_{d,\lambda}$ satisfies
(3.20) and (3.21).

##  4\. Integral identity and proof of Main Theorem

We use this section to prove an integral identity which we will need to prove
our main result. The integral identity will be proved by using the geometric
optics solutions constructed in Section 3. Let us start by recalling

$\displaystyle\mathcal{L}_{A,q}=\partial_{t}-\frac{1}{\sqrt{|g|}}\left(\partial_{j}+iA_{j}\right)\left(g^{jk}\sqrt{|g|}(\partial_{k}+iA_{k})\right)+q.$

For $j=1,2$, let $A^{(j)}$ and $q_{j}$ be as in Theorem 1.2. Further assume
$u_{j}$ for $j=1,2$ are solutions of corresponding initial boundary value
problem, that is,

$\displaystyle\left\\{\begin{array}[]{rlc}\mathcal{L}_{A^{(j)},q_{j}}u_{j}(t,x)&=0,&(t,x)\in
M_{T}\\\ u_{j}(0,x)&=0,&x\in M\\\ u_{j}(t,x)&=f(t,x)&(t,x)\in\partial
M_{T}.\end{array}\right.$ (4.4)

Then $u=u_{1}-u_{2}$ satisfies the following equation:

$\displaystyle\left\\{\begin{array}[]{rlc}\mathcal{L}_{A^{(1)},q_{1}}u(t,x)&=\mathcal{Q}u_{2}(t,x),&(t,x)\in
M_{T}\\\ u_{j}(0,x)&=0,&x\in M\\\ u_{j}(t,x)&=0,&(t,x)\in\partial
M_{T},\end{array}\right.$ (4.8)

where
$\mathcal{Q}u_{2}(t,x)=\left(\left|A^{(1)}\right|_{g}^{2}-\left|A^{(1)}\right|_{g}^{2}\right)u_{2}+2i\left\langle
A^{(2)}-A^{(1)},\nabla_{g}u_{2}\right\rangle+i\delta_{g}\left(A^{(2)}-A^{(1)}\right)u_{2}+(q_{2}-q_{1})u_{2}$
and
$\delta_{g}A=\frac{1}{|\sqrt{g}|}\sum_{j,k=1}^{n}\partial_{j}\left(g^{jk}\sqrt{|g|}A_{k}\right)$.
To simplify the notations, let us define
$\tilde{q}_{i}=|A^{(i)}|^{2}_{g}-i\delta_{g}A^{(i)}-q_{i}$ then

$\mathcal{Q}u_{2}(t,x)=2i\left\langle
A^{(2)}-A^{(1)},\nabla_{g}u_{2}\right\rangle+\left(\tilde{q}_{1}-\tilde{q}_{2}\right)u_{2}.$

Let $v(t,x)$ be the solution of adjoint equation for the operator
$\mathcal{L}_{A^{(1)},q_{1}}$, that is,

$\mathcal{L}^{*}_{A^{(1)},q_{1}}v(t,x)=0,\qquad\mbox{ for }(t,x)\in M_{T}$

where

$\displaystyle\mathcal{L}^{*}_{A,q}=-\partial_{t}-\large\sum\limits_{j,k=1}^{n}\frac{1}{\sqrt{\lvert
g\rvert}}\Big{(}\partial_{x_{j}}-iA_{j}(t,x)\Big{)}\left(\sqrt{\lvert
g\rvert}g^{jk}\left(\partial_{x_{k}}-iA_{k}(t,x)\right)\right)+\overline{q}(t,x).$

By a direct application of integration by parts, we get the following identity
for $u,v\in H^{1}(M_{T})$:

$\displaystyle\left\langle\mathcal{L}_{A,q}u,v\right\rangle_{L^{2}(M_{T})}-\left\langle
u,\mathcal{L}_{A,q}^{*}v\right\rangle_{L^{2}(M_{T})}$
$\displaystyle=\int_{M}\left\\{u(T,\cdot)\overline{v}(T,\cdot)-u(0,\cdot)\overline{v}(0,\cdot)\right\\}dV_{g}$
$\displaystyle\quad+i\int_{\Sigma}u\left(\overline{\nu_{j}g^{jk}(\partial_{k}+iA_{k})v}\right)dtdS_{g}$
$\displaystyle\qquad-i\int_{\Sigma}\left(\nu_{j}g^{jk}(\partial_{k}+iA_{k})u\right)\overline{v}dtdS_{g}.$
(4.9)

Multiplying equation (4.8) with $v(t,x)$ and integrate it over $M_{T}$:

$\displaystyle\int_{M_{T}}\mathcal{L}_{A^{(1)},q_{1}}u(t,x)v(t,x)dtdV_{g}$
$\displaystyle=2\int_{M_{T}}i\left\langle
A^{(2)}-A^{(1)},\nabla_{g}u_{2}\right\rangle
vdtdV_{g}+\int_{M_{T}}\left(\tilde{q}_{1}-\tilde{q}_{2}\right)u_{2}vdtdV_{g}.$

Next, we use the equation (4) together with $u|_{\Sigma}=0$, $u|_{t=0}=0$,
$\partial_{\nu}u|_{G}=0$ and $\mathcal{L}^{*}_{A^{(1)},q_{1}}$ with $v(T,x)=0$
to obtain the following:

$\displaystyle 2\int_{M_{T}}i\left\langle
A^{(2)}-A^{(1)},\nabla_{g}u_{2}\right\rangle
vdtdV_{g}+\int_{M_{T}}\left(\tilde{q}_{1}-\tilde{q}_{2}\right)u_{2}vdtdV_{g}=-i\int_{\Sigma\setminus
G}\left(\nu_{j}g^{jk}(\partial_{k}+iA_{k})u\right)vdtdS$ (4.10)

Next, we state a lemma which gives an estimate on the left-hand side of the
equation. The proof of this lemma follows directly from the results already
proved in [42, Lemma 5.1], see also [35, Lemma 5.1].

######  Lemma 4.1.

Let $u_{2}$ and $v$ be as above. Then

$\displaystyle\left|\int_{\Sigma\setminus\Sigma_{-,\epsilon/2}}\left(\nu_{j}g^{jk}(\partial_{k}+iA_{k})u\right)vdtdS_{g}\right|\leq\lambda^{1/2}.$
(4.11)

This lemma will be used in the equation (4.10). Recall, $u_{2}$ solves
equation (4.4) and is of the form
$e^{\varphi+i\psi}\Big{(}T_{g}(t,x)+R_{g,\lambda}(t,x)\Big{)}$. Using this, we
have

$\left\langle
A^{(2)}-A^{(1)},\nabla_{g}u_{2}\right\rangle=e^{\varphi+i\psi}\left\langle
A^{(2)}-A^{(1)},\nabla_{g}\left(\varphi+i\psi\right)\Big{(}T_{g}(t,x)+R_{g,\lambda}(t,x)\Big{)}+\nabla_{g}\Big{(}T_{g}(t,x)+R_{g,\lambda}(t,x)\Big{)}\right\rangle.$

Next, we use the fact that $v$ is of the form
$e^{-(\varphi+i\psi)}\Big{(}T_{d}(t,x)+R_{d,\lambda}(t,x)\Big{)}$ to obtain

$\left\langle A^{(2)}-A^{(1)},\nabla_{g}u_{2}\right\rangle v=\left\langle
A^{(2)}-A^{(1)},\nabla_{g}\left(\varphi+i\psi\right)\right\rangle
T_{g}(t,x)T_{d}(t,x)+\mathcal{Z}(t,x)$

where

$\displaystyle\mathcal{Z}$ $\displaystyle=\left\langle
A^{(2)}-A^{(1)},\nabla_{g}\left(\varphi+i\psi\right)\right\rangle\big{(}T_{g}R_{d,\lambda}+T_{d}R_{g,\lambda}+R_{g,\lambda}R_{d,\lambda}\big{)}+\left\langle
A^{(2)}-A^{(1)},\nabla_{g}\left(T_{g}+R_{g,\lambda}\right)\right\rangle\left(T_{d}+R_{d,\lambda}\right).$

With the estimates and bounds for terms involved in $\mathcal{Z}(t,x)$, one
can see that

$\displaystyle\lim_{\lambda\rightarrow\infty}\frac{1}{\lambda}\left|\int_{M_{T}}\left(\mathcal{Z}(t,x)+\left(\tilde{q}_{1}-\tilde{q}_{2}\right)u_{2}v\right)dtdV_{g}\right|=0.$
(4.12)

Using relations (4.11), (4.12) in equation (4.10) to get

$\displaystyle\int_{M_{T}}\left\langle
A^{(2)}-A^{(1)},\nabla_{g}\left(\varphi+i\psi\right)\right\rangle
T_{g}(t,x)T_{d}(t,x)dtdV_{g}=0.$

Let $(x_{1},r,\theta)$ be the polar coordinate and $A_{1}$ and $A_{r}$ be
components of $A$ in $x_{1}$ and $r$ coordinates respectively as in Theorem
3.1. With this in mind, we use the explicit expressions of $T_{g}$ (from
equation (3.10)) and $T_{d}$ (from equation (3.18)) to reduce the above
equation as follows

$\displaystyle\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{S_{y}M_{0}}\int_{0}^{\tau_{+}(y_{0},\theta)}\left\langle\tilde{A},\nabla_{g}\left(\varphi+i\psi\right)\right\rangle\left(\phi(t)\right)^{2}e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}e^{-\mu
r}e^{i\Phi(t,x_{1},r,\theta)}\left(h(\theta)\right)^{2}dx_{1}drd\theta dt=0,$

where $\tilde{A}=A^{(2)}-A^{(1)}$ and
$\Phi(t,x_{1},r,\theta)=\left(\Phi_{1}+\Phi_{2}\right)(t,x_{1},r,\theta)$ with
$\Phi_{1}$ and $\Phi_{2}$ satisfying equations (3.11) and (3.19) respectively.
We also used the fact $dV_{g}=b(r,\theta)^{1/2}dx_{1}drd\theta$.

As this relation is true for all cutoff functions $\phi\in
C^{\infty}_{c}(0,T)$ , we get

$\displaystyle\int_{M}\left(\tilde{A}_{1}+i\sqrt{1-\beta^{2}}\tilde{A}_{r}\right)e^{i\mu\left(t+(\sqrt{1-\beta^{2}})x_{1}\right)}e^{-\mu
r}e^{i\Phi(t,x_{1},r,\theta)}\left(h(\theta)\right)^{2}dx_{1}drd\theta=0,\quad
t\in(0,T)$

From here, we follow [28, Section 6] step by step to obtain $\Psi\in
C^{\infty}(M_{T})$ such that
$\tilde{A}(t,x)=A^{(2)}(t,x)-A^{(1)}(t,x)=\nabla_{x}\Psi(t,x)$ for $(t,x)\in
M_{T}$.

To prove the uniqueness of the potential part $q$, we start by taking
$A^{(1)}=A^{(2)}$, which can be achieved by a simple gauge transformation.
Using this in the equation (4.10) and taking limit $\lambda\rightarrow\infty$
we get

$\int_{M_{T}}\left(q_{1}-q_{2}\right)uvdtdV_{g}=0$

where $\mathcal{L}_{A^{(1)},q_{1}}u=0$ and
$\mathcal{L}^{*}_{A^{(1)},q_{2}}v=0$. Again, we use the explicit expressions
of $u$ and $v$ (equations (3.2) and (3.22) ) and simplify it to get

$\displaystyle\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{S_{y}M_{0}}\int_{0}^{\tau_{+}(y,\theta)}\left(q_{1}-q_{2}\right)\left(\phi(t)\right)^{2}e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}e^{-\mu
r}e^{i\Phi(t,x_{1},r,\theta)}\left(h(\theta)\right)^{2}dtdx_{1}drd\theta=0.$

By varying $\phi\in C^{\infty}_{c}(0,T)$, $h\in C^{\infty}(S_{y_{0}}(D))$ and
taking $\Phi=0$ (since $A^{(1)}=A^{(2)}$, if $\Phi_{1}$ solves (3.11) then we
can choose $\Phi_{2}=-\Phi_{1}$ which will solve (3.19)), we obtain

$\displaystyle\int_{\mathbb{R}}\int_{0}^{\tau_{+}(y_{0},\theta)}\left(q_{1}-q_{2}\right)e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}e^{-\mu
r}dx_{1}dr$ $\displaystyle=0,\quad\mbox{ for all }\theta\in S_{y_{0}}(D)\mbox{
and }\ t\in(0,T)$ $\displaystyle\int_{0}^{\tau_{+}(y_{0},\theta)}e^{-\mu
r}\left\\{\int_{\mathbb{R}}\left(q_{1}-q_{2}\right)e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}dx_{1}\right\\}dr$
$\displaystyle=0$
$\displaystyle\int_{\mathbb{R}}\left(q_{1}-q_{2}\right)e^{i\mu\left(\sqrt{1-\beta^{2}}\right)x_{1}}dx_{1}$
$\displaystyle=0,\mbox{ using infectivity of exponential ray transform}$
$\displaystyle\left(q_{1}-q_{2}\right)$ $\displaystyle\equiv 0,$

which completes the proof.

## Acknowledgments

MV is supported by Start-up Research Grant SRG/2021/001432 from the Science
and Engineering Research Board, Government of India.

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