# AN INVERSE PROBLEM FOR THE MINIMAL SURFACE EQUATION IN THE PRESENCE OF A
RIEMANNIAN METRIC

Janne Nurminen  Department of Mathematics and Statistics, University of
Jyväskylä<EMAIL_ADDRESS>
###### Abstract.

In this work we study an inverse problem for the minimal surface equation on a
Riemannian manifold $(\mathbb{R}^{n},g)$ where the metric is of the form
$g(x)=c(x)(\hat{g}\oplus e)$. Here $\hat{g}$ is a simple Riemannian metric on
$\mathbb{R}^{n-1}$, $e$ is the Euclidean metric on $\mathbb{R}$ and $c$ a
smooth positive function. We show that if we know the associated Dirichlet-to-
Neumann maps corresponding to metrics $g$ and $\tilde{c}g$, then the Taylor
series of the conformal factor $\tilde{c}$ at $x_{n}=0$ is equal to a positive
constant. We also show a partial data result when $n=3$.

Keywords. Inverse problem, higher order linearization, quasilinear elliptic
equation, minimal surface equation

###### Contents

  1. 1 Introduction
  2. 2 Deriving the minimal surface equation
  3. 3 First and second order linearizations
  4. 4 Proofs of Theorems 1.3 and 1.4

##  1\. Introduction

This article focuses on an inverse problem for the minimal surface equation
(MSE), which is a quasilinear elliptic PDE. In particular we consider MSE on a
manifold $(\mathbb{R}^{n},g),n\geq 3$, where

(1.1)
$g(x^{\prime},x_{n})=c(x^{\prime},x_{n})\begin{pmatrix}\hat{g}(x^{\prime})&0\\\
0&1\end{pmatrix}.$

Here $(x^{\prime},x_{n})\in\mathbb{R}^{n-1}\times\mathbb{R}$, $\hat{g}$ is a
Riemannian metric on $\mathbb{R}^{n-1}$ and $c\in C^{\infty}(\mathbb{R}^{n}),$
$c(x)>0$ for all $x\in\mathbb{R}^{n}$. Inverse problems for the MSE were
considered in [Nur23] where $\hat{g}$ was Euclidean and in [CLLO22] where
$c=1$. Now we consider metrics of the form (1.1) which includes both of these
cases. Metrics with local coordinates of the form (1.1) were introduced for
example in [DKSU09], [DKLS16] where the Calderón problem was studied.

Let $\Omega\subset\mathbb{R}^{n-1}$ be a bounded domain. In order to define a
minimal surface, we introduce notation for Riemannian Hessian and norm of the
gradient

$\displaystyle\nabla_{g}^{2}u=\left(\partial^{2}_{x_{i}x_{j}}u-\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}u\right)_{i,j=1}^{n},\quad|\nabla_{g}u|_{g}^{2}=g^{ij}\partial_{x_{i}}u\partial_{x_{j}}u.$

We use the Einstein summation convention where there is no chance to
misinterpret. Also here $\Gamma^{m}_{\hphantom{m}ij}$ are the standard
Christoffel symbols. Now define the Laplace-Beltrami operator to be the trace
of the Hessian

$\Delta_{g}u=\operatorname{Tr}(\nabla_{g}^{2}u)=g^{ij}\left(\partial^{2}_{x_{i}x_{j}}u-\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}u\right).$

Then for the metric (1.1) the MSE has the form

(1.2)
$\displaystyle\operatorname{div}_{g}\left(\frac{\nabla_{g}u}{\left(1+|\nabla_{g}u|_{g}^{2}\right)^{\frac{1}{2}}}\right)$
$\displaystyle+\frac{\Delta_{g}u(1-c^{-1})+\frac{1}{2c}(\nabla_{g}u)^{j}\partial_{x_{j}}c(1+|\nabla_{g}u|_{g}^{2}-c^{-2})+\frac{n-1}{2c^{3}}\partial_{x_{n}}c(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})}{\left(1+|\nabla_{g}u|_{g}^{2}\right)^{\frac{3}{2}}}=0.$

For the derivation of this equation, see Section 2. When $c=1$, one gets the
MSE from [CLLO22]. The graph of a function
$u\colon\Omega\subset\mathbb{R}^{n-1}\to\mathbb{R}$ is called a minimal
surface if it satisfies (1.2) for all $x^{\prime}\in\Omega$.

We will look at the boundary value problem

(1.3)
$\left\\{\begin{array}[]{ll}F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u)=0&\text{in}\,\,\Omega\\\
u=f,&\text{on}\,\,\partial\Omega\end{array}\right.$

where $F\colon\mathbb{R}^{n^{2}}\to\mathbb{R},$

$\displaystyle F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u):=$
$\displaystyle\left(-\Delta_{\hat{g}}u+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})$
$\displaystyle+\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$

(this is an equivalent formulation of (1.2), see Section 2). One can show that
(1.3) is well-posed for small enough boundary data $f$ for example by
following the arguments in [LLLS21]. In particular one can show that there
exist $C,\delta>0$ and $s>3$, $s\notin\mathbb{N}$, such that for all

$f\in U_{\delta}:=\\{h\in
C^{s}(\partial\Omega):||h||_{C^{s}(\partial\Omega)}<\delta\\}$

there is a unique small solution $u_{f}$ in $\\{v\in
C^{s}(\bar{\Omega}):||v||_{C^{s}(\bar{\Omega})}<C\delta\\}$. In addition to
full data we will be considering partial data and for this let
$\Gamma\subset\partial\Omega$ be a nonempty open subset of the boundary. We
can now define a (partial) Dirichlet-to-Neumann map (DN map)
$\Lambda_{g}^{\Gamma}\colon C_{c}^{s}(\Gamma)\to C^{s-1}(\Gamma),$

(1.4) $\Lambda_{g}^{\Gamma}f\mapsto\partial_{\nu}u_{f}|_{\Gamma}$

where $u_{f}$ is the unique small solution corresponding to the boundary value
$f$, $\partial_{\nu}u=\hat{g}^{ij}\partial_{x_{i}}u\nu_{j}$ and $\nu$ is the
unit normal of $\partial\Omega$. When $\Gamma=\partial\Omega$, we will denote
$\Lambda_{g}=\Lambda_{g}^{\partial\Omega}$.

The inverse problem we study in this work is the following: Given the
knowledge of the partial DN map for two metrics in the same conformal class,
does it hold that the metrics are the same? We note that when trying to
recover the metric from boundary measurements one expects an obstruction to
uniqueness by a boundary fixing diffeomorphism. When the metrics are in the
same conformal class such an obstruction is not present, since the only
diffeomorphism that leaves a conformal class invariant is the identity. In
this article we will give some partial answers to this inverse problem.

Before stating the results, we will introduce the class of admissible metrics
[DKSU09].

######  Definition 1.1.

A Riemannian metric $g$ is called admissible if on $\mathbb{R}^{n}$ it is of
the form (1.1) and $(\Omega,\hat{g})$ is a simple manifold.

######  Definition 1.2.

A compact manifold $(\hat{M},\hat{g})$ with boundary is called simple if for
any $p\in\hat{M}$ the exponential map $\exp_{p}$ with its maximal domain of
definition in $T_{p}\hat{M}$ is a diffeomorphism onto $\hat{M}$, and if
$\partial\hat{M}$ is strictly convex in the sense that its second fundamental
form is positive definite.

The first result is for the full data case.

######  Theorem 1.3.

Let $(\mathbb{R}^{n},g)$, $n\geq 3$, be a Riemannian manifold where $g$ is as
in (1.1) with
$\partial_{x_{n}}c(x^{\prime},0)=\partial_{x_{n}}^{2}c(x^{\prime},0)=0$ and
let $\Omega\subset\mathbb{R}^{n-1}$ be a bounded domain with $C^{\infty}$
boundary. When $n=3$ assume that $\Omega$ is simply connected and when $n>3$
assume that $\hat{g}$ is admissible. Let $\tilde{c}\in
C^{\infty}(\mathbb{R}^{n})$ be such that
$\partial_{x_{n}}\tilde{c}(x^{\prime},0)=\partial_{x_{n}}^{2}\tilde{c}(x^{\prime},0)=0$
and let $\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)|_{\partial\Omega}=0$ for
$k\geq 3$. Assume that $\Lambda_{g}(f)=\Lambda_{\tilde{c}g}(f)$ for all $f\in
U_{\delta}$, where $\delta>0$ is sufficiently small. Then

$\tilde{c}(x^{\prime},0)=\lambda,\quad\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)=0$

for some $\lambda>0$ and for all $k>2$.

Thus, if $\tilde{c}$ is real analytic with respect to $x_{n}$, then
$\tilde{c}(x)=\lambda$ for all $x\in\Omega\times\mathbb{R}$.

For the partial data we will only consider the case $n=3$. For $n>3$ we would
be able to get some results, but the partial data results for the magnetic
Schrödinger equation (see [SY23]) are more restrictive and hence we do not
record these results here.

######  Theorem 1.4.

Let $(\mathbb{R}^{3},g)$, $\tilde{c}$ and $\Omega$ be as in Theorem 1.3 and
let $\Gamma\subset\partial\Omega$, $\Gamma\neq\emptyset$, be open. Assume that
$\Lambda_{g}^{\Gamma}(f)=\Lambda_{\tilde{c}g}^{\Gamma}(f)$ for all $f\in
U_{\delta}$, where $\delta>0$ is sufficiently small. Then

$\tilde{c}(x^{\prime},0)=\lambda,\quad\partial_{x_{3}}^{k}\tilde{c}(x^{\prime},0)=0$

for some $\lambda>0$ and for all $k>2$.

Thus, if $\tilde{c}$ is real analytic with respect to $x_{3}$, then
$\tilde{c}(x)=\lambda$ for all $x\in\Omega\times\mathbb{R}$.

These results extend the results from [Nur23].

We make some comments about the assumptions made in Theorems 1.3 and 1.4.
First of all, the assumption
$\partial_{x_{n}}c(x^{\prime},0)=\partial_{x_{n}}\tilde{c}(x^{\prime},0)=0$ is
needed for the well-posedness of the boundary value problem (1.3) for small
data, that is we want it to be well-posed for both the metric $g$ and the
metric $\tilde{c}g$. The assumption
$\partial_{x_{n}}^{2}c(x^{\prime},0)=\partial_{x_{n}}^{2}\tilde{c}(x^{\prime},0)=0$
is needed in order for the method used in the proof to work and it is not yet
known if this could be removed. Lastly, there is a small gauge invariance in
(1.2) and also in the DN map. To be more precise, if one replaces the
conformal factor $c$ in (1.1) by $\mu c$, $\mu\neq 0$, then both the equation
(1.2) and the DN map remain the same. Thus we cannot have $\lambda=1$ in
Theorem 1.3.

There are also some assumptions made when $n=3$. The assumption that $\Omega$
is simply connected is needed to use the Poincaré Lemma. This is something
that one might be able relax with some amount of work by looking at the so
called gauge isomorphism mentioned in [GT11] that relates two connection
1-forms. The other assumption that
$\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)|_{\partial\Omega}=0$ for $k\geq
3$ could possibly be removed by doing a boundary determination result.

The proof will use the method of higher order linearization. We will begin by
looking at the first linearization of (1.3) and the DN map for that. The first
linearization of the DN map corresponds to an advection diffusion type
equation. From this it is possible to determine the advection term ([GT11] for
$n=3$, [KU18] for $n>3$) and this will then imply that
$\tilde{c}(x^{\prime},0)=\lambda$ for some $\lambda\in(0,\infty)$. Now fix
$x_{0}\in\Omega$ and construct a solution $v^{(0)}$ to the adjoint of the
linearized equation so that $v^{(0)}(x_{0})\neq 0$. Then using the higher
order linearizations one can obtain an integral identity

$\int_{\Omega}\tilde{c}(x^{\prime},0)^{-1}\partial_{x_{n}}^{N+2}\tilde{c}(x^{\prime},0)v^{(0)}\prod_{k=1}^{N+1}v^{l_{k}}\,dx^{\prime}=0,$

where $N+1$ is the order of linearization and $v^{l_{k}}$ are solutions to the
linearized equation. Then choosing $v^{l_{k}}\equiv 1$ for $k>2$ and using
that a product of these solutions form a complete set in $L^{1}$ (see Lemma
4.1) we can conclude that

$\tilde{c}(x^{\prime},0)^{-1}\partial_{x_{n}}^{N+2}\tilde{c}(x^{\prime},0)v^{(0)}=0.$

From this, using $v^{(0)}(x_{0})\neq 0$ and that $x_{0}$ was arbitrary, we get
$\partial_{x_{n}}^{N+2}\tilde{c}(x^{\prime},0)=0$. Then one proceeds by
induction.

For the partial data, the proof is very similar and has only a few
modifications. One of the more significant changes is that instead of using
the results of [GT11] to determine the advection term, we use the results for
partial data in [Tzo17].

As mentioned above, the main technique in this work is the method of higher
order linearization. This method, which uses the nonlinearity of the partial
differential equation as a tool, was first introduced in [KLU18] in the case
of a nonlinear wave equation and was further developed in [LLLS21], [FO20] for
nonlinear elliptic equations. In these works the equation $\Delta u+a(x,u)=0$
was considered and further results for example in partial data and obstacle
problems were obtained in [KU20], [LLLS21a]. Also there has been multiple
works when the nonlinearity is of power type (e.g. [LLST22], [ST22] and
[Nur23a]).

At the same time there have been multiple articles concerning inverse problems
for other nonlinear elliptic PDEs. For example different cases of nonlinear
conductivity equations have been considered in [CFKKU21], [KKU20] and in the
latter they also consider partial data. For the nonlinear magnetic Schrödinger
equation the full data case has been treated on conformally transversally
anisotropic manifolds in [KU20a] and in the Euclidean setting the partial data
case is treated in [LZ20].

Linearization has been used before these works mentioned above. It was used in
the parabolic case in [Isa93] where it was shown that the first linearization
of the DN map is the DN map for a linear equation. Other nonlinear elliptic
cases have been treated, for example in [IS94], [SU97].

The present work focuses on an inverse problem for the minimal surface
equation and for this equation there have been some works previously in the
literature. The Euclidean case has been treated in [MU20], in the sense that
they deal with a quasilinear conductivity depending on a function $u$ and its
gradient $\nabla u$. After that there have been two works on the minimal
surface equation in a Riemannian manifold setting, one by the author of this
article [Nur23] and another one in [CLLO22]. These were simultaneously and
independently done. The work of this article somehow brings these two articles
together in the following sense: In [Nur23] the metric was conformally
Euclidean and thus in this article the setting is more general. In [CLLO22]
the authors consider a two dimensional Riemannian manifold $(\Sigma,g)$ and
look at a minimal surface $Y\subset\Sigma\times\mathbb{R}$ given as a graph of
a function. Thus if $\Sigma\subset\mathbb{R}^{2}$ and in this article we would
have $c\equiv 1$, the settings in these two articles would be the same.

There is also the work [ABN20] that is related to minimal surfaces and inverse
problems. In that article the authors consider a generalization of the
boundary rigidity problem in the sense that instead of measuring distances of
boundary points they use measurements related to areas of minimal surfaces.

The rest of this article is organized as follows. In Section 2 we derive the
minimal surface equation in the setting mentioned above. Section 3 is
dedicated to calculating the first and second order linearizations of (1.3)
and addressing shortly the well-posedness of (1.3). Finally in Section 4 we
prove the main results and two Lemmas to help in the proof.

Acknowledgements. The author was supported by the Finnish Centre of Excellence
in Inverse Modelling and Imaging (Academy of Finland grant 284715). The author
would like to thank Mikko Salo for helpful discussions on the minimal surface
equation and everything related to inverse problems.

##  2\. Deriving the minimal surface equation

In this section we derive the equation (1.2). This is done similarly as in
[Nur23, Section 3]. Let $(M,g),$ $M=\mathbb{R}^{n}$, $n\geq 3$, be a
Riemannian manifold with the metric

(2.1)
$g(x^{\prime},x_{n})=c(x^{\prime},x_{n})\begin{pmatrix}\hat{g}(x^{\prime})&0\\\
0&1\end{pmatrix}$

where

$(x^{\prime},x_{n})\in\mathbb{R}^{n-1}\times\mathbb{R},\;c\in
C^{\infty}(\mathbb{R}^{n}),\;c(x)>0\quad\text{for all}\quad
x\in\mathbb{R}^{n}$

and $\hat{g}$ is a Riemannian metric on $\mathbb{R}^{n-1}$. We use the
standard notations $g_{ij}$ for the matrix $g$ and $g^{ij}$ for the inverse
$g^{-1}$. These assumptions are valid for the rest of the article, unless
otherwise stated.

Let $u\colon\Omega\subset\mathbb{R}^{n-1}\to\mathbb{R},$ $u\in
C^{2}(\bar{\Omega})$, and consider the graph of the function $u$

$\text{Graph}_{u}=\\{(x^{\prime},u(x^{\prime}))\colon
x^{\prime}\in\Omega\\}\subset M.$

This graph is a minimal surface if and only if its mean curvature $H$ is equal
to zero at all points on the graph. By defining

$f\colon\Omega\times\mathbb{R}\to\mathbb{R},\quad
f(x^{\prime},x_{n})=x_{n}-u(x^{\prime}),$

the graph of $u$ is the surface

$\Sigma:=\\{(x^{\prime},x_{n})\in\Omega\times\mathbb{R}:f(x^{\prime},x_{n})=0\\}.$

The mean curvature of $\Sigma$ at $x\in\Sigma$ is the sum of principal
curvatures. We omit the normalizing factor $\frac{1}{n-1}$ when calculating
the mean curvature and we use the Einstein summation convention when it does
not cause confusion. In order to calculate the principal curvatures, we
introduce the Riemannian gradient and Hessian of a function $f\colon
M\to\mathbb{R}$:

$\displaystyle\nabla_{g}f=g^{ij}\partial_{x_{i}}f\partial_{x_{j}},\quad\nabla_{g}^{2}f=\left(\partial^{2}_{x_{i}x_{j}}f-\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}f\right)_{i,j=1}^{n},$

where $g^{ij}$ is the inverse of $g_{ij}$ and
$\Gamma^{m}_{\hphantom{m}ij}=\frac{1}{2}g^{ml}(\partial_{x_{i}}g_{jl}+\partial_{x_{j}}g_{il}-\partial_{x_{l}}g_{ij})$
is the Christoffel symbol related to the metric $g$. Define also the Laplace-
Beltrami operator, which is the trace of the Hessian (this is one way of
defining it), and the norm of the gradient:

$\Delta_{g}f=\operatorname{Tr}(\nabla_{g}^{2}f)=g^{ij}\left(\partial^{2}_{x_{i}x_{j}}f-\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}f\right),\quad|\nabla_{g}f|_{g}^{2}=g^{ij}\partial_{x_{i}}f\partial_{x_{j}}f.$

Now the principal curvatures of $\Sigma$ at $x\in\Sigma$ are the eigenvalues
of $\nabla_{g}^{2}f(x)$ restricted to the tangent space $T_{x}\Sigma$ at $x$.
Since $\frac{\nabla_{g}f(x)}{|\nabla_{g}f(x)|_{g}}$ is a normal to $\Sigma$ at
the point $x$, we have $T_{x}\Sigma=\\{\nabla_{g}f(x)\\}^{\perp}$, or in other
words, the tangent space $T_{x}\Sigma$ is the orthogonal complement of the
vector $\nabla_{g}f(x)$.

Let $\\{E_{1},\ldots,E_{n-1}\\}$ be an $g$-orthonormal basis of $T_{x}\Sigma$.
Then
$\left\\{E_{1},\ldots,E_{n-1},\frac{\nabla_{g}f(x)}{|\nabla_{g}f(x)|_{g}}\right\\}$
is an orthonormal basis of $\mathbb{R}^{n}$. Now the mean curvature of
$\Sigma$ at $x\in\Sigma$ is the trace of
$\nabla_{g}^{2}f(x)|_{\\{\nabla_{g}f(x)\\}^{\perp}}$:

$\displaystyle H(x)$
$\displaystyle=\sum_{i=1}^{n-1}\langle\nabla_{g}^{2}f(x)E_{i},E_{i}\rangle$
$\displaystyle=\sum_{i=1}^{n-1}\langle\nabla_{g}^{2}f(x)E_{i},E_{i}\rangle+\left(\nabla_{g}^{2}f(x)\right)\left(\frac{\nabla_{g}f(x)}{|\nabla_{g}f(x)|_{g}},\frac{\nabla_{g}f(x)}{|\nabla_{g}f(x)|_{g}}\right)$
$\displaystyle-\left(\nabla_{g}^{2}f(x)\right)\left(\frac{\nabla_{g}f(x)}{|\nabla_{g}f(x)|_{g}},\frac{\nabla_{g}f(x)}{|\nabla_{g}f(x)|_{g}}\right)$
$\displaystyle=\operatorname{Tr}(\nabla_{g}^{2}f(x))-|\nabla_{g}f(x)|_{g}^{-2}\left(\nabla_{g}^{2}f(x)\right)\left(\nabla_{g}f(x),\nabla_{g}f(x)\right)$
$\displaystyle=\Delta_{g}f(x)-|\nabla_{g}f(x)|_{g}^{-2}\left(\nabla_{g}^{2}f(x)\right)\left(\nabla_{g}f(x),\nabla_{g}f(x)\right).$

Thus $\text{Graph}_{u}$ is a minimal surface if and only if

(2.2)
$|\nabla_{g}f(x)|_{g}^{2}\Delta_{g}f(x)-\left(\nabla_{g}^{2}f(x)\right)\left(\nabla_{g}f(x),\nabla_{g}f(x)\right)=0\quad\text{for
all}\enspace x\in\text{Graph}_{u}.$

Next we will calculate the minimal surface equation more explicitly using the
metric (2.1). Now

$g^{-1}=c^{-1}\begin{pmatrix}\hat{g}^{-1}&0\\\ 0&1\end{pmatrix}$

is the inverse matrix of (2.1). Let us calculate the first term of (2.2):

(2.3) $\displaystyle c^{2}|\nabla_{g}f(x)|_{g}^{2}\Delta_{g}f(x)$
$\displaystyle=c^{2}g^{ij}(\delta_{in}-\partial_{x_{i}}u)(\delta_{jn}-\partial_{x_{j}}u)g^{kl}\left(-\partial^{2}_{x_{k}x_{l}}u-\Gamma^{m}_{\hphantom{m}kl}(\delta_{mn}-\partial_{x_{m}}u)\right)$
$\displaystyle=c^{2}g^{ij}(\delta_{in}\delta_{jn}-\delta_{in}\partial_{x_{j}}u-\delta_{jn}\partial_{x_{i}}u+\partial_{x_{i}}u\partial_{x_{j}}u)g^{kl}(-\partial^{2}_{x_{k}x_{l}}u-\Gamma^{n}_{\hphantom{n}kl}+\Gamma^{m}_{\hphantom{m}kl}\partial_{x_{m}}u)$
$\displaystyle=c\left(1+|\nabla_{\hat{g}}u|_{\hat{g}}^{2}\right)\left(-\Delta_{g}u-g^{kl}\Gamma^{n}_{\hphantom{n}kl}\right).$

The second term of (2.2) becomes

(2.4) $\displaystyle
c^{2}\left(\nabla_{g}^{2}f(x)\right)\left(\nabla_{g}f(x),\nabla_{g}f(x)\right)$
$\displaystyle=c^{2}(\partial_{x_{i}x_{j}}^{2}f-\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}f)g^{ai}\partial_{x_{a}}fg^{bj}\partial_{x_{b}}f$
$\displaystyle=c^{2}(-\partial_{x_{i}x_{j}}^{2}u-\Gamma^{n}_{\hphantom{n}ij}+\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}u)g^{ai}g^{bj}(\delta_{an}\delta_{bn}-\delta_{an}\partial_{x_{b}}u-\delta_{bn}\partial_{x_{a}}u+\partial_{x_{a}}u\partial_{x_{b}}u)$
$\displaystyle=c^{2}(-\partial_{x_{i}x_{j}}^{2}u-\Gamma^{n}_{\hphantom{n}ij}+\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}u)(g^{ni}g^{nj}-g^{ni}g^{bj}\partial_{x_{b}}u-g^{ai}g^{nj}\partial_{x_{a}}u+g^{ai}g^{bj}\partial_{x_{a}}u\partial_{x_{b}}u)$
$\displaystyle=\Gamma^{m}_{\hphantom{m}nn}\partial_{x_{m}}u-\Gamma^{n}_{\hphantom{n}nn}-2(\Gamma^{m}_{\hphantom{m}nj}\partial_{x_{m}}u-\Gamma^{n}_{\hphantom{n}nj})cg^{bj}\partial_{x_{b}}u$
$\displaystyle+c^{2}(-\partial_{x_{i}x_{j}}^{2}u-\Gamma^{n}_{\hphantom{n}ij}+\Gamma^{m}_{\hphantom{m}ij}\partial_{x_{m}}u)g^{ai}g^{bj}\partial_{x_{a}}u\partial_{x_{b}}u$
$\displaystyle=\Gamma^{m}_{\hphantom{m}nn}\partial_{x_{m}}u-\Gamma^{n}_{\hphantom{n}nn}-2(\Gamma^{m}_{\hphantom{m}nj}\partial_{x_{m}}u-\Gamma^{n}_{\hphantom{n}nj})\left(\nabla_{\hat{g}}u\right)^{j}$
$\displaystyle-\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)-\Gamma^{n}_{\hphantom{n}ij}\left(\nabla_{\hat{g}}u\right)^{i}\left(\nabla_{\hat{g}}u\right)^{j}.$

In order to simplify further we will calculate the Christoffel symbols more
explicitly:

$\displaystyle 2\Gamma^{m}_{\hphantom{m}ij}$
$\displaystyle=g^{mr}(\partial_{x_{j}}g_{ir}+\partial_{x_{i}}g_{jr}-\partial_{x_{r}}g_{ij}),$
$\displaystyle 2\Gamma^{m}_{\hphantom{m}nn}$
$\displaystyle=g^{mr}(2\partial_{x_{n}}g_{nr}-\partial_{x_{r}}g_{nn})=2g^{mn}\partial_{x_{n}}g_{nn}-g^{mr}\partial_{x_{r}}g_{nn}$
$\displaystyle=2g^{mn}\partial_{x_{n}}c-g^{mr}\partial_{x_{r}}c,$
$\displaystyle 2\Gamma^{n}_{\hphantom{n}ij}$
$\displaystyle=g^{nr}(\partial_{x_{j}}g_{ir}+\partial_{x_{i}}g_{jr}-\partial_{x_{r}}g_{ij})=g^{nn}(\partial_{x_{j}}g_{in}+\partial_{x_{i}}g_{jn}-\partial_{x_{n}}g_{ij})$
$\displaystyle=c^{-1}(\partial_{x_{j}}g_{in}+\partial_{x_{i}}g_{jn}-\partial_{x_{n}}g_{ij}),$
$\displaystyle 2\Gamma^{m}_{\hphantom{m}nj}$
$\displaystyle=g^{mr}(\partial_{x_{j}}g_{nr}+\partial_{x_{n}}g_{jr}-\partial_{x_{r}}g_{nj}),$
$\displaystyle 2\Gamma^{n}_{\hphantom{n}nj}$
$\displaystyle=g^{nr}(\partial_{x_{j}}g_{nr}+\partial_{x_{n}}g_{jr}-\partial_{x_{r}}g_{nj})=g^{nn}(\partial_{x_{j}}g_{nn}+\partial_{x_{n}}g_{jn}-\partial_{x_{n}}g_{nj})$
$\displaystyle=c^{-1}\partial_{x_{j}}c.$

Inserting these into (2.3) we have

$\displaystyle c^{2}|\nabla_{g}f(x)|_{g}^{2}\Delta_{g}f(x)$
$\displaystyle=c\left(1+|\nabla_{\hat{g}}u|_{\hat{g}}^{2}\right)(-\Delta_{g}u-\frac{1}{2}g^{kl}c^{-1}(\partial_{x_{l}}g_{kn}+\partial_{x_{k}}g_{ln}-\partial_{x_{n}}g_{kl}))$
$\displaystyle=\left(1+|\nabla_{\hat{g}}u|_{\hat{g}}^{2}\right)(-c\Delta_{g}u-\frac{1}{2}(g^{nl}\partial_{x_{l}}c+g^{kn}\partial_{x_{k}}c-g^{kl}\partial_{x_{n}}g_{kl}))$
$\displaystyle=\left(1+|\nabla_{\hat{g}}u|_{\hat{g}}^{2}\right)(-c\Delta_{g}u-\frac{1}{2}c^{-1}(2-n)\partial_{x_{n}}c).$

For calculating the linearizations in Section 3 it is useful to transform
$\Delta_{g}u$ into $\Delta_{\hat{g}}u$:

$\displaystyle\Delta_{g}u$
$\displaystyle=g^{ij}\left(\partial_{x_{i}x_{j}}^{2}u-\frac{1}{2}g^{mr}(\partial_{x_{j}}g_{ir}+\partial_{x_{i}}g_{jr}-\partial_{x_{r}}g_{ij})\partial_{x_{m}}u\right)$
$\displaystyle=c^{-1}\hat{g}^{ij}\partial_{x_{i}x_{j}}^{2}u-\frac{1}{2c}\sum_{m,r=1}^{n-1}\hat{g}^{mr}g^{ij}(\partial_{x_{j}}g_{ir}+\partial_{x_{i}}g_{jr}-\partial_{x_{r}}g_{ij})\partial_{x_{m}}u$
$\displaystyle=c^{-1}\hat{g}^{ij}\partial_{x_{i}x_{j}}^{2}u-\frac{1}{2c}\sum_{m,r=1}^{n-1}\hat{g}^{mr}\left(\hat{g}^{ij}((\partial_{x_{j}}\hat{g}_{ir}+\partial_{x_{i}}\hat{g}_{jr}-\partial_{x_{r}}\hat{g}_{ij}))+\frac{2-n}{c}\partial_{x_{r}}c\right)\partial_{x_{m}}u$
$\displaystyle=c^{-1}\Delta_{\hat{g}}u-\frac{2-n}{2c^{2}}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u.$

Thus

$\displaystyle c^{2}|\nabla_{g}f(x)|_{g}^{2}\Delta_{g}f(x)$
$\displaystyle=\left(1+|\nabla_{\hat{g}}u|_{\hat{g}}^{2}\right)\left(-\Delta_{\hat{g}}u+\frac{2-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u-\frac{1}{2}c^{-1}(2-n)\partial_{x_{n}}c\right).$

Now the second part (2.4) becomes

$\displaystyle
c^{2}\left(\nabla_{g}^{2}f(x)\right)\left(\nabla_{g}f(x),\nabla_{g}f(x)\right)$
$\displaystyle=-\frac{1}{2}g^{mr}\partial_{x_{r}}c\partial_{x_{m}}u-\frac{1}{2}g^{nn}\partial_{x_{n}}c$
$\displaystyle-(g^{mr}(\partial_{x_{j}}g_{nr}+\partial_{x_{n}}g_{jr}-\partial_{x_{r}}g_{nj})\partial_{x_{m}}u-c^{-1}\partial_{x_{j}}c)(\nabla_{\hat{g}}u)^{j}$
$\displaystyle-\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)-\frac{1}{2}c^{-1}(\partial_{x_{j}}g_{in}+\partial_{x_{i}}g_{jn}-\partial_{x_{n}}g_{ij})(\nabla_{\hat{g}}u)^{i}(\nabla_{\hat{g}}u)^{j}$
$\displaystyle=-\frac{1}{2}c^{-1}\sum_{m,r=1}^{n-1}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u-\frac{1}{2}c^{-1}\partial_{x_{n}}c$
$\displaystyle-\sum_{m,r=1}^{n-1}c^{-1}\hat{g}^{mr}\hat{g}_{jr}\partial_{x_{n}}c\partial_{x_{m}}u(\nabla_{\hat{g}}u)^{j}+c^{-1}\partial_{x_{j}}c(\nabla_{\hat{g}}u)^{j}$
$\displaystyle-\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)+\frac{1}{2}c^{-1}\partial_{x_{n}}c\hat{g}_{ij}(\nabla_{\hat{g}}u)^{i}(\nabla_{\hat{g}}u)^{j}$
$\displaystyle=\frac{1}{2}c^{-1}\sum_{m,r=1}^{n-1}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u-\frac{1}{2}c^{-1}\partial_{x_{n}}c(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})-\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u).$

As before we would like to modify
$\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$ to have
$\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$:

$\displaystyle\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$
$\displaystyle=(\partial_{x_{i}x_{j}}^{2}u-\frac{1}{2}g^{mr}(\partial_{x_{i}}g_{rj}+\partial_{x_{j}}g_{ri}-\partial_{x_{r}}g_{ji})\partial_{x_{m}}u)\hat{g}^{ia}\partial_{x_{a}}u\hat{g}^{jb}\partial_{x_{b}}u.$

Since $\partial_{x_{n}}u=0$, all the indices run from $1$ to $n-1$. Hence

$\displaystyle\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$
$\displaystyle=(\partial_{x_{i}x_{j}}^{2}u-\frac{1}{2c}\hat{g}^{mr}(\partial_{x_{i}}c\hat{g}_{rj}+c\partial_{x_{i}}\hat{g}_{rj}+\partial_{x_{j}}c\hat{g}_{ri}+c\partial_{x_{j}}\hat{g}_{ri}$
$\displaystyle-\partial_{x_{r}}c\hat{g}_{ij}-c\partial_{x_{r}}\hat{g}_{ij})\partial_{x_{m}}u)\hat{g}^{ia}\partial_{x_{a}}u\hat{g}^{jb}\partial_{x_{b}}u$
$\displaystyle=(\partial_{x_{i}x_{j}}^{2}u-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u-\frac{1}{2c}(\delta^{m}_{j}\partial_{x_{i}}c+\delta^{m}_{i}\partial_{x_{j}}c-\hat{g}^{mr}\hat{g}_{ij}\partial_{x_{r}}c)\partial_{x_{m}}u)\hat{g}^{ia}\partial_{x_{a}}u\hat{g}^{jb}\partial_{x_{b}}u$
$\displaystyle=\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)-\frac{1}{2c}(\hat{g}^{ia}\partial_{x_{i}}c\partial_{x_{a}}u\hat{g}^{mb}\partial_{x_{m}}u\partial_{x_{b}}u+\hat{g}^{jb}\partial_{x_{j}}c\partial_{x_{b}}u\hat{g}^{mb}\partial_{x_{m}}u\partial_{x_{a}}u$
$\displaystyle-\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u\hat{g}^{jb}\partial_{x_{j}}u\partial_{x_{b}}u)$
$\displaystyle=\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)-\frac{1}{2c}\hat{g}^{ia}\partial_{x_{i}}c\partial_{x_{a}}u|\nabla_{\hat{g}}u|_{\hat{g}}^{2},$

where
$2\hat{\Gamma}^{m}_{\hphantom{m}ij}=\hat{g}^{mr}(\partial_{x_{j}}\hat{g}_{ir}+\partial_{x_{i}}\hat{g}_{jr}-\partial_{x_{r}}\hat{g}_{ij})$.

Putting these together gives

$\displaystyle 0$
$\displaystyle=c^{2}|\nabla_{g}f(x)|_{g}^{2}\Delta_{g}f(x)-c^{2}\left(\nabla_{g}^{2}f(x)\right)\left(\nabla_{g}f(x),\nabla_{g}f(x)\right)$
$\displaystyle=\left(-\Delta_{\hat{g}}u+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})$
$\displaystyle+\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u).$

Thus $\text{Graph}_{u}$ is a minimal surface if and only if the function $u$
satisfies the following minimal surface equation

(2.5)
$\displaystyle\left(-\Delta_{\hat{g}}u+\frac{1-n}{2c(x^{\prime},u(x^{\prime}))}\hat{g}^{mr}(x^{\prime})\partial_{x_{r}}c(x^{\prime},u(x^{\prime}))\partial_{x_{m}}u+\frac{n-1}{2c(x^{\prime},u(x^{\prime}))}\partial_{x_{n}}c(x^{\prime},u(x^{\prime}))\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})$
$\displaystyle+\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)=0$

for all $x^{\prime}\in\Omega$.

We will modify this a bit in order to see the more familiar Euclidean version
of the equation directly. For this, we start with the divergence of a vector
field defined in local coordinates as

$\operatorname{div}_{g}a=\sum_{i=1}^{n}\left(\partial_{x_{i}}a^{i}+\sum_{j=1}^{n}a^{j}\Gamma^{i}_{\hphantom{i}ij}\right).$

Let us denote $\eta:=\left(1+|\nabla_{g}u|_{g}^{2}\right)^{\frac{1}{2}}$ and
expand the following for a CTA metric and a function
$u\colon\Omega\to\mathbb{R}$ as before:

$\displaystyle\operatorname{div}_{g}\left(\frac{\nabla_{g}u}{\eta}\right)$
$\displaystyle=\sum_{i=1}^{n}\left(\partial_{x_{i}}\left(\frac{(\nabla_{g}u)^{i}}{\eta}\right)+\sum_{j=1}^{n}\frac{(\nabla_{g}u)^{j}}{\eta}\Gamma^{i}_{\hphantom{i}ij}\right)$
$\displaystyle=\sum_{i=1}^{n-1}\frac{\partial_{x_{i}}g^{ik}\partial_{x_{k}}u+g^{ik}\partial_{x_{i}x_{k}}^{2}u}{\eta}$
$\displaystyle-\frac{1}{2}\frac{(\nabla_{g}u)^{i}}{\eta^{3}}\left(\partial_{x_{i}}g^{ab}\partial_{x_{a}}u\partial_{x_{b}}u+g^{ab}\left(\partial_{x_{i}x_{a}}^{2}u\partial_{x_{b}}u+\partial_{x_{a}}u\partial_{x_{i}x_{b}}^{2}u\right)\right)$
$\displaystyle+\sum_{j=1}^{n-1}\frac{(\nabla_{g}u)^{j}}{\eta}\left(\Gamma^{i}_{\hphantom{i}ij}+\Gamma^{n}_{\hphantom{n}nj}\right).$

Now we will use that
$\partial_{x_{l}}g^{ik}=-(\Gamma^{i}_{\hphantom{i}ml}g^{mk}+\Gamma^{k}_{\hphantom{k}ml}g^{im})$
to have

$\displaystyle\operatorname{div}_{g}\left(\frac{\nabla_{g}u}{\eta}\right)$
$\displaystyle=\sum_{i=1}^{n-1}\frac{-(\Gamma^{i}_{\hphantom{i}il}g^{lk}+\Gamma^{k}_{\hphantom{k}il}g^{il})\partial_{x_{k}}u+g^{ik}\partial_{x_{i}x_{k}}^{2}u}{\eta}$
$\displaystyle-\frac{1}{2}\frac{(\nabla_{g}u)^{i}}{\eta^{3}}\left(-(\Gamma^{a}_{\hphantom{a}il}g^{lb}+\Gamma^{b}_{\hphantom{b}il}g^{al})\partial_{x_{a}}u\partial_{x_{b}}u+g^{ab}\left(\partial_{x_{i}x_{a}}^{2}u\partial_{x_{b}}u+\partial_{x_{a}}u\partial_{x_{i}x_{b}}^{2}u\right)\right)$
$\displaystyle+\sum_{j=1}^{n-1}\frac{(\nabla_{g}u)^{j}}{\eta}\left(\Gamma^{i}_{\hphantom{i}ij}+\Gamma^{n}_{\hphantom{n}nj}\right)$
$\displaystyle=\frac{\Delta_{g}u}{\eta}-\sum_{i,l=1}^{n-1}\frac{(\nabla_{g}u)^{l}}{\eta}\Gamma^{i}_{\hphantom{i}il}$
$\displaystyle-\frac{1}{2}\frac{(\nabla_{g}u)^{i}}{\eta^{3}}\left(-(\nabla_{g}u)^{l}\Gamma^{a}_{\hphantom{a}il}\partial_{x_{a}}u-(\nabla_{g}u)^{l}\Gamma^{b}_{\hphantom{b}il}\partial_{x_{b}}u+(\nabla_{g}u)^{a}\partial_{x_{i}x_{a}}^{2}u+(\nabla_{g}u)^{b}\partial_{x_{i}x_{b}}^{2}u\right)$
$\displaystyle+\sum_{j=1}^{n-1}\frac{(\nabla_{g}u)^{j}}{\eta}\left(\Gamma^{i}_{\hphantom{i}ij}+\Gamma^{n}_{\hphantom{n}nj}\right).$

After renaming some of the indices, using the definition of the Riemannian
Hessian of a function and
$\Gamma^{n}_{\hphantom{n}nj}=\frac{1}{2}c^{-1}\partial_{x_{j}}c$ we get

(2.6)
$\operatorname{div}_{g}\left(\frac{\nabla_{g}u}{\eta}\right)=\frac{\Delta_{g}u}{\eta}-\frac{\nabla_{g}^{2}u(\nabla_{g}u,\nabla_{g}u)}{\eta^{3}}+\sum_{j=1}^{n-1}\frac{(\nabla_{g}u)^{j}}{2c\eta}\partial_{x_{j}}c.$

Let us now go back to (2.5) and modify it to have (2.6) visible. Now

$\displaystyle\left(-\Delta_{\hat{g}}u+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})$
$\displaystyle+\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$
$\displaystyle=\left(-\Delta_{\hat{g}}u+\frac{2-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})$
$\displaystyle-\frac{1}{2}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})+\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u)$
$\displaystyle=\left(-c\Delta_{g}u+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})-\frac{1}{2}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\nabla_{g}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u).$

Then, we will multiply both sides of (2.5) by $c^{-2}$ to get

$\displaystyle 0$
$\displaystyle=\left(-\Delta_{g}u+\frac{n-1}{2c^{2}}\partial_{x_{n}}c\right)(c^{-1}+|\nabla_{g}u|^{2}_{g})-\frac{1}{2}c^{-3}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\nabla_{g}^{2}u(\nabla_{g}u,\nabla_{g}u)$
$\displaystyle=-\Delta_{g}u|\nabla_{g}u|^{2}_{g}-c^{-1}\Delta_{g}u+\Delta_{g}u-\Delta_{g}u+\frac{(\nabla_{g}u)^{j}}{2c}\partial_{x_{j}}c\eta^{2}-\frac{(\nabla_{g}u)^{j}}{2c}\partial_{x_{j}}c\eta^{2}$
$\displaystyle+\frac{n-1}{2c^{3}}\partial_{x_{n}}c(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})-\frac{1}{2}c^{-3}(\nabla_{g}u)^{r}\partial_{x_{r}}c+\nabla_{g}^{2}u(\nabla_{g}u,\nabla_{g}u)$
$\displaystyle=-\Delta_{g}u\eta^{2}+\nabla_{g}^{2}u(\nabla_{g}u,\nabla_{g}u)-\frac{(\nabla_{g}u)^{j}}{2c}\partial_{x_{j}}c\eta^{2}+\Delta_{g}u(1-c^{-1})$
$\displaystyle+\frac{n-1}{2c^{3}}\partial_{x_{n}}c(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})+\frac{1}{2c}(\nabla_{g}u)^{j}\partial_{x_{j}}c(\eta^{2}-c^{-2}).$

Dividing both sides by
$\eta^{3}=\left(1+|\nabla_{g}u|_{g}^{2}\right)^{\frac{3}{2}}$ gives

$\displaystyle\operatorname{div}_{g}\left(\frac{\nabla_{g}u}{\left(1+|\nabla_{g}u|_{g}^{2}\right)^{\frac{1}{2}}}\right)$
$\displaystyle+\frac{\Delta_{g}u(1-c^{-1})+\frac{1}{2c}(\nabla_{g}u)^{j}\partial_{x_{j}}c(1+|\nabla_{g}u|_{g}^{2}-c^{-2})+\frac{n-1}{2c^{3}}\partial_{x_{n}}c(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})}{\left(1+|\nabla_{g}u|_{g}^{2}\right)^{\frac{3}{2}}}=0.$

When $g$ is the Euclidean metric, we will have the familiar Euclidean minimal
surface equation. Moreover, when $c=1$ we get the minimal surface equation in
[CLLO22]:

$\operatorname{div}_{\hat{g}}\left(\frac{\nabla_{\hat{g}}u}{\left(1+|\nabla_{\hat{g}}u|_{\hat{g}}^{2}\right)^{\frac{1}{2}}}\right)=0.$

##  3\. First and second order linearizations

Let $g$ be as in (2.1) and define $F\colon\mathbb{R}^{n^{2}}\to\mathbb{R},$

(3.1) $\displaystyle F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u):=$
$\displaystyle\left(-\Delta_{\hat{g}}u+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u|^{2}_{\hat{g}})$
(3.2)
$\displaystyle+\nabla_{\hat{g}}^{2}u(\nabla_{\hat{g}}u,\nabla_{\hat{g}}u).$

We want to use the well-posedness result from [Nur23] to say that the boundary
value problem

(3.3)
$\left\\{\begin{array}[]{ll}F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u)=0&\text{in}\,\,\Omega\\\
u=f,&\text{on}\,\,\partial\Omega\end{array}\right.$

has a unique small solution. For this we need that $F(x^{\prime},0,0,0)=0$
which guarantees that $u\equiv 0$ is a solution to (3.3) with $f=0$. From
(3.1) we can see that this happens if and only if
$\partial_{x_{n}}c(x^{\prime},0)=0$ for all $x^{\prime}\in\Omega$.

To use the above mentioned well-posedness, we will need to calculate the first
linearization of the equation $F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u)=0$.
Let us do a formal calculation to get the first linearization. Let
$\varepsilon=(\varepsilon_{1},\ldots,\varepsilon_{m})$ where
$\varepsilon_{k}\in\mathbb{R}$ small and assume that
$u_{\varepsilon}:=u(x,\varepsilon)$ depends smoothly on $\varepsilon$ and
solves
$F(x^{\prime},u_{\varepsilon},\nabla_{g}u_{\varepsilon},\nabla_{g}^{2}u_{\varepsilon})=0$.
We will differentiate this in three parts with respect to $\varepsilon_{l}$
and evaluate at $\varepsilon=0$ (We could have calculated the first
linearization as in [Nur23], but it would be a longer calculation). Denote by
$u_{\varepsilon}^{(l)}:=\partial_{\varepsilon_{l}}u_{\varepsilon}$. Just to
make the calculations a bit nicer we will start with

(3.4)
$\displaystyle\partial_{\varepsilon_{l}}\left(\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}\right)$
$\displaystyle=-\frac{1-n}{2c^{2}}\partial_{x_{n}}cu_{\varepsilon}^{(l)}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}+\frac{1-n}{2c}\hat{g}^{mr}(\partial_{x_{r}x_{n}}^{2}cu_{\varepsilon}^{(l)}\partial_{x_{m}}u_{\varepsilon}+\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}^{(l)})$
$\displaystyle:=Q_{l}.$

Now the first term of (3.1) becomes

(3.5)
$\displaystyle\partial_{\varepsilon_{l}}\left(\left(-\Delta_{\hat{g}}u_{\varepsilon}+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u_{\varepsilon}|^{2}_{\hat{g}})\right)$
$\displaystyle=\left(-\Delta_{\hat{g}}u_{\varepsilon}^{(l)}-\frac{n-1}{2c^{-2}}(\partial_{x_{n}}c)^{2}u_{\varepsilon}^{(l)}+\frac{n-1}{2c}\partial_{x_{n}}^{2}cu_{\varepsilon}^{(l)}+Q_{l}\right)(1+|\nabla_{\hat{g}}u_{\varepsilon}|^{2}_{\hat{g}})$
$\displaystyle+\left(-\Delta_{\hat{g}}u_{\varepsilon}+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}+\frac{n-1}{2c}\partial_{x_{n}}c\right)\hat{g}^{ij}(\partial_{x_{i}}u_{\varepsilon}^{(l)}\partial_{x_{j}}u_{\varepsilon}+\partial_{x_{i}}u_{\varepsilon}\partial_{x_{j}}u_{\varepsilon}^{(l)}).$

The second term is

(3.6)
$\displaystyle\partial_{\varepsilon_{l}}\left(\nabla_{\hat{g}}^{2}u_{\varepsilon}(\nabla_{\hat{g}}u_{\varepsilon},\nabla_{\hat{g}}u_{\varepsilon})\right)$
$\displaystyle=\partial_{\varepsilon_{l}}\left((\partial_{x_{i}x_{j}}^{2}u_{\varepsilon}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon})\hat{g}^{ia}\partial_{x_{a}}u_{\varepsilon}\hat{g}^{jb}\partial_{x_{b}}u_{\varepsilon}\right)$
$\displaystyle=(\partial_{x_{i}x_{j}}^{2}u_{\varepsilon}^{(l)}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon}^{(l)})\hat{g}^{ia}\partial_{x_{a}}u_{\varepsilon}\hat{g}^{jb}\partial_{x_{b}}u_{\varepsilon}$
$\displaystyle+(\partial_{x_{i}x_{j}}^{2}u_{\varepsilon}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon})\hat{g}^{ia}\hat{g}^{jb}(\partial_{x_{a}}u_{\varepsilon}^{(l)}\partial_{x_{b}}u_{\varepsilon}+\partial_{x_{a}}u_{\varepsilon}\partial_{x_{b}}u_{\varepsilon}^{(l)})$
$\displaystyle:=P_{1}+P_{2}.$

where $P_{1}$ and $P_{2}$ are for future reference. When evaluating at
$\varepsilon=0$ we have $u_{0}=0$, also $\nabla_{g}u_{0},\nabla_{g}^{2}u_{0}$
vanish, and thus when combining the above, denoting
$v^{l}:=u_{\varepsilon}^{(l)}|_{\varepsilon=0}$,

$\displaystyle\partial_{\varepsilon_{l}}F(x^{\prime},u_{\varepsilon},\nabla_{g}u_{\varepsilon},\nabla_{g}^{2}u_{\varepsilon})|_{\varepsilon=0}$
$\displaystyle=-\Delta_{\hat{g}}v^{l}+\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{2}c(x^{\prime},0)v^{l}+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}v^{l}=0.$

Now this first linearization satisfies the assumptions of [Nur23, Proposition
2.1] and thus we have the existence and uniqueness of small solutions to
(3.3).

Next we will calculate the second linearization for the equation
$F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u)=0$. Denote
$u_{\varepsilon}^{(kl)}:=\partial_{\varepsilon_{k}\varepsilon_{l}}^{2}u_{\varepsilon}$.
We will differentiate (3.4)-(3.6) with respect to $\varepsilon_{k}$, $k\neq
l$, starting with (3.4):

$\displaystyle\partial_{\varepsilon_{k}}\partial_{\varepsilon_{l}}\left(\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}\right)$
$\displaystyle=\frac{1-n}{c^{3}}(\partial_{x_{n}}c)^{2}u_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}\hat{g}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}-\frac{1-n}{2c^{2}}\partial_{x_{n}}^{2}cu_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}\hat{g}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}$
$\displaystyle-\frac{1-n}{2c^{2}}\partial_{x_{n}}cu_{\varepsilon}^{(kl)}\hat{g}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}$
$\displaystyle-\frac{1-n}{2c^{2}}\partial_{x_{n}}cu_{\varepsilon}^{(l)}\hat{g}^{mr}(\partial_{x_{r}x_{n}}^{2}cu_{\varepsilon}^{(k)}\partial_{x_{m}}u_{\varepsilon}+\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}^{(k)})$
$\displaystyle-\frac{1-n}{2c^{2}}\partial_{x_{n}}c\hat{g}^{mr}(\partial_{x_{r}x_{n}}^{2}cu_{\varepsilon}^{(l)}\partial_{x_{m}}u_{\varepsilon}+\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}^{(l)})$
$\displaystyle+\frac{1-n}{2c}\hat{g}^{mr}\big{(}\partial_{x_{r}}\partial_{x_{n}}^{2}cu_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}\partial_{x_{m}}u_{\varepsilon}+\partial_{x_{r}x_{n}}^{2}c(u_{\varepsilon}^{(kl)}\partial_{x_{m}}u_{\varepsilon}+u_{\varepsilon}^{(l)}\partial_{x_{m}}u_{\varepsilon}^{k})$
$\displaystyle+\partial_{x_{r}x_{n}}cu_{\varepsilon}^{(k)}\partial_{x_{m}}u_{\varepsilon}^{(l)}+\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}^{(kl)}\big{)}.$

Then we differentiate (3.5) to have

$\displaystyle\partial_{\varepsilon_{k}}\partial_{\varepsilon_{l}}\left(\left(-\Delta_{\hat{g}}u_{\varepsilon}+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}+\frac{n-1}{2c}\partial_{x_{n}}c\right)(1+|\nabla_{\hat{g}}u_{\varepsilon}|^{2}_{\hat{g}})\right)$
$\displaystyle=\bigg{(}-\Delta_{\hat{g}}u_{\varepsilon}^{(kl)}+\frac{n-1}{c^{-3}}(\partial_{x_{n}}c)^{3}u_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}-\frac{n-1}{2c^{-2}}\left(2\partial_{x_{n}}c\partial_{x_{n}}^{2}cu_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}+(\partial_{x_{n}}c)^{2}u_{\varepsilon}^{(kl)}\right)$
$\displaystyle-\frac{n-1}{2c^{-2}}\partial_{x_{n}}c\partial_{x_{n}}^{2}cu_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}+\frac{n-1}{2c}\left(\partial_{x_{n}}^{3}cu_{\varepsilon}^{(k)}u_{\varepsilon}^{(l)}+\partial_{x_{n}}^{2}cu_{\varepsilon}^{(kl)}\right)+\partial_{\varepsilon_{k}}Q_{l}\bigg{)}(1+|\nabla_{\hat{g}}u_{\varepsilon}|_{\hat{g}}^{2})$
$\displaystyle+\left(-\Delta_{\hat{g}}u_{\varepsilon}^{(l)}-\frac{n-1}{2c^{-2}}(\partial_{x_{n}}c)^{2}u_{\varepsilon}^{(l)}+\frac{n-1}{2c}\partial_{x_{n}}^{2}cu_{\varepsilon}^{(l)}+Q_{l}\right)\hat{g}^{ij}(\partial_{x_{i}}u_{\varepsilon}^{(k)}\partial_{x_{j}}u_{\varepsilon}+\partial_{x_{i}}u_{\varepsilon}\partial_{x_{j}}u_{\varepsilon}^{(k)})$
$\displaystyle+\left(-\Delta_{\hat{g}}u_{\varepsilon}^{(k)}-\frac{n-1}{2c^{-2}}(\partial_{x_{n}}c)^{2}u_{\varepsilon}^{(k)}+\frac{n-1}{2c}\partial_{x_{n}}^{2}cu_{\varepsilon}^{(k)}+Q_{k}\right)\hat{g}^{ij}(\partial_{x_{i}}u_{\varepsilon}^{(l)}\partial_{x_{j}}u_{\varepsilon}+\partial_{x_{i}}u_{\varepsilon}\partial_{x_{j}}u_{\varepsilon}^{(l)})$
$\displaystyle+\left(-\Delta_{\hat{g}}u_{\varepsilon}+\frac{1-n}{2c}\hat{g}^{mr}\partial_{x_{r}}c\partial_{x_{m}}u_{\varepsilon}+\frac{n-1}{2c}\partial_{x_{n}}c\right)\hat{g}^{ij}(\partial_{x_{i}}u_{\varepsilon}^{(kl)}\partial_{x_{j}}u_{\varepsilon}+\partial_{x_{i}}u_{\varepsilon}^{(l)}\partial_{x_{j}}u_{\varepsilon}^{(k)}$
$\displaystyle+\partial_{x_{i}}u_{\varepsilon}^{(k)}\partial_{x_{j}}u_{\varepsilon}^{(l)}+\partial_{x_{i}}u_{\varepsilon}\partial_{x_{j}}u_{\varepsilon}^{(kl)}).$

Now we calculate the derivative of (3.6) in two parts, starting with $P_{1}$:

$\displaystyle\partial_{\varepsilon_{k}}P_{1}$
$\displaystyle=(\partial_{x_{i}x_{j}}^{2}u_{\varepsilon}^{(kl)}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon}^{(kl)})\hat{g}^{ia}\partial_{x_{a}}u_{\varepsilon}\hat{g}^{ja}\partial_{x_{b}}u_{\varepsilon}$
$\displaystyle+(\partial_{x_{i}x_{j}}^{2}u_{\varepsilon}^{(l)}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon}^{(l)})\hat{g}^{ia}\hat{g}^{jb}(\partial_{x_{a}}u_{\varepsilon}^{(k)}\partial_{x_{b}}u_{\varepsilon}+\partial_{x_{a}}u_{\varepsilon}\partial_{x_{b}}u_{\varepsilon}^{(k)}).$

For $P_{2}$ we get

$\displaystyle\partial_{\varepsilon_{k}}P_{2}$
$\displaystyle=(\partial_{x_{i}x_{j}}^{2}u_{\varepsilon}^{(k)}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon}^{(k)})\hat{g}^{ia}\hat{g}^{jb}(\partial_{x_{a}}u_{\varepsilon}^{(l)}\partial_{x_{b}}u_{\varepsilon}+\partial_{x_{a}}u_{\varepsilon}\partial_{x_{b}}u_{\varepsilon}^{(l)})$
$\displaystyle+(\partial_{x_{i}x_{j}}u_{\varepsilon}-\hat{\Gamma}^{m}_{\hphantom{m}ij}\partial_{x_{m}}u_{\varepsilon})\hat{g}^{ia}\hat{g}^{jb}(\partial_{x_{a}}u_{\varepsilon}^{(kl)}\partial_{x_{b}}u_{\varepsilon}+\partial_{x_{a}}u_{\varepsilon}^{(l)}\partial_{x_{b}}u_{\varepsilon}^{(k)}$
$\displaystyle+\partial_{x_{a}}u_{\varepsilon}^{(k)}\partial_{x_{b}}u_{\varepsilon}^{(l)}+\partial_{x_{a}}u_{\varepsilon}\partial_{x_{b}}u_{\varepsilon}^{(kl)}).$

Evaluating the above second derivatives at $\varepsilon=0$, denoting
$w^{kl}:=u_{\varepsilon}^{(kl)}|_{\varepsilon=0}$, gives the second
linearization of the equation $F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u)=0$:

$\displaystyle\partial_{\varepsilon_{k}\varepsilon_{l}}^{2}F(x^{\prime},u_{\varepsilon},\nabla_{g}u_{\varepsilon},\nabla_{g}^{2}u_{\varepsilon})|_{\varepsilon=0}$
$\displaystyle=-\Delta_{\hat{g}}w^{kl}+\frac{n-1}{2c(x^{\prime},0)}\left(\partial_{x_{n}}^{3}c(x^{\prime},0)v^{k}v^{l}+\partial_{x_{n}}^{2}c(x^{\prime},0)w^{kl}\right)$
$\displaystyle+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}w^{kl}=0.$

Here we used that $u_{0}=0$, $\nabla_{g}u_{0},\nabla_{g}^{2}u_{0}$ vanish,
$\partial_{x_{n}}c(x^{\prime},0)=0$ and that
$\partial_{x_{k}x_{n}}c(x^{\prime},0)=0$ for $k=1,\ldots,n-1$.

##  4\. Proofs of Theorems 1.3 and 1.4

Before going to the proof of Theorem 1.3, we state two lemmas that will be
used also in the proof of Theorem 1.4. The first lemma says that the products
of two solutions to an advection diffusion equation form a complete set in
$L^{1}$. For this we note that equations of the form $\Delta_{g}u+Xu=0$, for a
smooth real valued vector field $X$, can be written in the form of a magnetic
Schrödinger equation. In local coordinates we have

$L_{g,A,q}u=-|g|^{-\frac{1}{2}}\left(\partial_{x_{j}}+iA_{j}\right)\left(|g|^{\frac{1}{2}}g^{jk}\left(\partial_{x_{k}}+iA_{k}\right)u\right)+qu=0$

for $A=\frac{iX}{2}$ and
$q=\frac{1}{4}g(X,X)-\frac{1}{2}\operatorname{div}_{g}(X)$.

######  Lemma 4.1.

Let $(\Omega,g)$, $\Omega\subset\mathbb{R}^{n}$, be a smooth Riemannian
manifold with boundary. We have two cases:

  1. (1)

When $n=2$ let $\Gamma\subset\partial\Omega$ be a nonempty open set, let $f\in
C^{\infty}(\bar{\Omega})$ be such that $f|_{\Gamma}=0$. Assume that

(4.1) $\int_{\Omega}fu_{1}u_{2}\,dV_{g}=0$

for all $u_{j}$ solving $L_{g,A,q}u_{j}=0$ in $\Omega$ with
$u_{j}|_{\partial\Omega\setminus\Gamma}=0$. Then $f\equiv 0$ in $\Omega$.

  2. (2)

When $n>2$ let $g$ be admissible, $f\in C^{\infty}(\bar{\Omega})$ be such that
$f|_{\partial\Omega}=0$. Assume that (4.1) holds for all $u_{j}$ solving
$L_{g,A,q}u_{j}=0$ in $\Omega$. Then $f\equiv 0$ in $\Omega$.

  * Proof.

$n=2$: The proof is written in [Tzo17, Section 7.3] where the author considers
identifying a zeroth order term.

$n>2$: This can be read from the proof of [DKSU09, Theorem 1.7], more
precisely from the part where they prove that the two potentials $q_{1}$ and
$q_{2}$ agree in $M$ (using the notation of the referred article). ∎

The second lemma states that if we know the partial DN maps corresponding to
$\Delta_{g}u+X_{j}u=0$ for $j=1,2$ and $X_{1}=X_{2}$ on
$\Gamma\subset\partial\Omega$, $\Gamma\neq\emptyset$, then $X_{1}=X_{2}$ in
$\Omega$. The DN map in question is defined as $\Lambda_{X}^{\Gamma}\colon
C^{k,\alpha}(\Gamma)\to C^{k-1,\alpha}(\Gamma),$

$\Lambda_{X}^{\Gamma}f=\partial_{\nu}u|_{\Gamma}.$

######  Lemma 4.2.

Let $(\mathbb{R}^{2},g)$ be a Riemannian manifold,
$\Omega\subset\mathbb{R}^{2}$ be simply connected with smooth boundary and
$\Gamma\subset\partial\Omega$ be an open nonempty set. Let $X_{j}$ be smooth
vector fields for $j=1,2$ so that $X_{1}|_{\Gamma}=X_{2}|_{\Gamma}$. If
$\Lambda_{X_{1}}^{\Gamma}f=\Lambda_{X_{2}}^{\Gamma}f$ for all $f\in
C^{k,\alpha}(\Gamma)$, then $X_{1}=X_{2}$ in $\Omega$.

  * Proof.

By [Tzo17, Theorem 1.1] we have $iA_{1}-iA_{2}=\theta^{-1}d\theta$ and
$q_{1}=q_{2}$ in $\Omega$, where

$A_{j}=\frac{iX_{j}}{2},\quad
q_{j}=\frac{1}{4}g(X_{j},X_{j})-\frac{1}{2}\operatorname{div}_{g}(X_{j})$

and $\theta$ is a nonvanishing function with $\theta|_{\Gamma}=1$. Notice that
$d(\theta^{-1}d\theta)=d(\theta^{-1})\wedge d\theta+\theta^{-1}\wedge
d^{2}\theta=0$ which implies that $dX_{1}=dX_{2}$ and by the Poincaré Lemma we
have a function $\varphi\in C^{\infty}(\bar{\Omega})$ such that

$X_{1}-X_{2}=\nabla_{g}\varphi.$

Using that $X_{1}|_{\Gamma}=X_{2}|_{\Gamma}$ gives
$0=(X_{1}-X_{2})\cdot\nu|_{\Gamma}=\partial_{\nu}\varphi|_{\Gamma}$ and
$\varphi|_{\Gamma}=0$ (actually we have that $\varphi|_{\partial\Omega}$ is a
constant, but we can subtract this constant to obtain
$\varphi|_{\partial\Omega}=0$). Let us see what kind of an equation $\varphi$
satisfies. Now $X_{2}=X_{1}-\nabla_{g}\varphi$ and plugging this into
$q_{1}=q_{2}$ gives

$\displaystyle\frac{1}{4}g(X_{1},X_{1})-\frac{1}{2}\operatorname{div}_{g}(X_{1})$
$\displaystyle=\frac{1}{4}g^{kl}\left(X_{1,k}-(\nabla_{g}\varphi)_{k}\right)\left(X_{1,l}-(\nabla_{g}\varphi)_{l}\right)-\frac{1}{2}\operatorname{div}_{g}(X_{1})+\frac{1}{2}\operatorname{div}_{g}(\nabla_{g}\varphi)$
$\displaystyle=\frac{1}{4}g(X_{1},X_{1})-\frac{1}{2}g(X_{1},\nabla_{g}\varphi)+\frac{1}{4}g(\nabla_{g}\varphi,\nabla_{g}\varphi)-\frac{1}{2}\operatorname{div}_{g}(X_{1})+\frac{1}{2}\Delta_{g}\varphi.$

Thus $\varphi$ solves

(4.2)
$\left\\{\begin{array}[]{ll}\Delta_{g}\varphi-g(X_{1},\nabla_{g}\varphi)+\frac{1}{2}|\nabla_{g}\varphi|_{g}^{2}=0&\text{in}\,\,\Omega\\\
\varphi=0&\text{on}\,\,\Gamma\\\
\partial_{\nu}\varphi=0&\text{on}\,\,\Gamma.\end{array}\right.$

From this we get that

$\displaystyle|\Delta_{g}\varphi|=|g(X_{1},\nabla_{g}\varphi)-\frac{1}{2}|\nabla_{g}\varphi|_{g}^{2}|\leq|g(X_{1},\nabla_{g}\varphi)|+|\frac{1}{2}|\nabla_{g}\varphi|_{g}^{2}|\leq
C|\nabla_{g}\varphi|_{g}\leq C(|\nabla_{g}\varphi|_{g}+|\varphi|)$

since $|\nabla_{g}\varphi|_{g}\leq M$ for some $M>0$. Then by the unique
continuation principle for local Cauchy data $\varphi\equiv 0$ in $\Omega$
(see [KSU11, Theorem B.$1$]) and hence $X_{1}=X_{2}$. ∎

  * Proof Theorem 1.3.

The assumptions of [Nur23, Proposition 2.1] are satisfied and thus (3.3) has a
unique small solution $u_{f}\in C^{s}(\bar{\Omega})$. Let
$\varepsilon=(\varepsilon_{1},\ldots,\varepsilon_{N})$ where
$\varepsilon_{k}\in\mathbb{R}$ small enough so that (3.3) has a unique small
solution for $f_{\varepsilon}:=\sum_{k=1}^{N}f_{k}$, where $f_{k}\in
C^{s}(\partial\Omega)$ with $\lVert
f_{k}\rVert_{C^{s}(\partial\Omega)}<\delta$, $\delta>0$. Denote by
$u:=u(x,\varepsilon)$ and $\tilde{u}:=\tilde{u}(x,\varepsilon)$ the unique
small solutions to (3.3) with the mterics $g$ and $\tilde{c}g$ respectively.
Then by [Nur23, Proposition 2.1] the solutions $u$ and $\tilde{u}$ depend
smoothly on $\varepsilon$. Hence we can calculate the first and second order
linearizations of (3.3) and the linearizations of the DN map (1.4).

In section 3 we already calculated the first linearization of
$F(x^{\prime},u,\nabla_{g}u,\nabla_{g}^{2}u)=0$ and since we also assume that
$\partial_{x_{n}}^{2}c(x^{\prime},0)=0$ we have that the first linearization
of (3.3) is

(4.3)
$\left\\{\begin{array}[]{ll}-\Delta_{\hat{g}}v^{l}+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}v^{l}=0&\text{in}\,\,\Omega\\\
v^{l}=f_{l},&\text{on}\,\,\partial\Omega,\end{array}\right.$

where $v^{l}=\partial_{\varepsilon_{l}}u|_{\varepsilon=0}$. Also, the first
linearization of the DN map is $(D\Lambda_{g})_{0}\colon
C^{s}(\partial\Omega)\to
C^{s-1}(\partial\Omega),f\mapsto\partial_{\nu}v^{l}|_{\partial\Omega}$. Now
(4.3) can be written in the form of an advection diffusion equation
$-(\Delta_{\hat{g}}-X)v^{l}=0$ for the real vector field $X$ such that

$Xh=\frac{1-n}{2c(x^{\prime},0)}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}h.$

For the metric $\tilde{c}g$ the first linearization of
$F(x^{\prime},\tilde{u},\nabla_{\tilde{c}g}\tilde{u},\nabla_{\tilde{c}g}^{2}\tilde{u})=0$
is

(4.4)
$\left\\{\begin{array}[]{ll}-\Delta_{\hat{g}}\tilde{v}^{l}+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}\tilde{v}^{l}\phantom{\bigg{|}}\\\
+\frac{1-n}{2\tilde{c}(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}\tilde{c}(x^{\prime},0)\partial_{x_{j}}\tilde{v}^{l}=0&\text{in}\,\,\Omega\\\
\tilde{v}^{l}=f_{l},&\text{on}\,\,\partial\Omega,\phantom{\Big{|}}\end{array}\right.$

where $\tilde{v}^{l}=\partial_{\varepsilon_{l}}\tilde{u}|_{\varepsilon=0}$.
Similarly this can be written as an advection diffusion equation for the real
vector field $\tilde{X}$ such that
$\tilde{X}h=\frac{1-n}{2}\hat{g}^{ij}\left(\frac{\partial_{x_{i}}c(x^{\prime},0)}{c(x^{\prime},0)}+\frac{\partial_{x_{i}}\tilde{c}(x^{\prime},0)}{\tilde{c}(x^{\prime},0)}\right)\partial_{x_{j}}h$.
Since we know that $\Lambda_{g}(f)=\Lambda_{\tilde{c}g}(f)$ for all $f\in
U_{\delta}$, where $\delta>0$ is sufficiently small, we can apply
$\partial_{\varepsilon_{l}}|_{\varepsilon=0}$ to this, which implies

$(D\Lambda_{g})_{0}=(D\Lambda_{cg})_{0}.$

When $n=3$ we use lemma 4.2 together with a boundary determination [GT11,
Proposition 4.1.] and when $n>3$ we use [KU18, Theorem 1.4] (we assume that
the metric $\hat{g}$ is admissible) to deduce that $X=\tilde{X}$ and hence for
all $i=1,\ldots,n-1$

$\frac{\partial_{x_{i}}\tilde{c}(x^{\prime},0)}{\tilde{c}(x^{\prime},0)}=0,\
\text{for}\ x^{\prime}\in\Omega.$

This then implies two things. Firstly,
$\tilde{c}(x^{\prime},0)=\lambda\in(0,\infty)$. Secondly, since solutions to
an advection diffusion equation are unique, we have that
$v^{l}=\tilde{v}^{l}$.

Now the second order linearization for (3.3) is

(4.5)
$\left\\{\begin{array}[]{ll}-\Delta_{\hat{g}}w^{kl}+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}w^{kl}+\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{3}c(x^{\prime},0)v^{k}v^{l}=0&\text{in}\,\,\Omega\\\
w^{kl}=0,&\text{on}\,\,\partial\Omega.\phantom{\Big{|}}\end{array}\right.$

For the metric $\tilde{c}g$ the second linearization is

(4.6)
$\left\\{\begin{array}[]{ll}-\Delta_{\hat{g}}\tilde{w}^{kl}+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}\tilde{w}^{kl}+\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{3}c(x^{\prime},0)v^{k}v^{l}\phantom{\bigg{|}}\\\
+\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{k}v^{l}=0&\text{in}\,\,\Omega\\\
\tilde{w}^{kl}=0,&\text{on}\,\,\partial\Omega.\phantom{\Big{|}}\end{array}\right.$

Here $w^{kl}=\partial_{\varepsilon_{k}\varepsilon_{l}}^{2}u|_{\varepsilon=0}$
and
$\tilde{w}^{kl}\partial_{\varepsilon_{k}\varepsilon_{l}}^{2}\tilde{u}|_{\varepsilon=0}$.
Fix $x_{0}^{\prime}\in\Omega$. Let now $v^{(0)}$ be a solution to

(4.7) $\Delta_{\hat{g}}v^{(0)}+Xv^{(0)}+qv^{(0)}=0\text{ in}\,\,\Omega,$

where
$q:=\frac{1-n}{2c(x^{\prime},0)}\left(-\frac{1}{c(x^{\prime},0)}|\nabla_{\hat{g}}c(x^{\prime},0)|^{2}_{\hat{g}}+\partial_{x_{j}}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)+\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\Gamma^{k}_{\hphantom{k}jk}\right)$,
so that $v^{(0)}(x_{0}^{\prime})\neq 0$. The existence of such a solution can
be shown using Runge approximation (see for example [LLS20, Proposition A.2]).
More precisely, there exists a solution $\tilde{v}^{(0)}$ of (4.7) in a small
neighborhood $U\subset\Omega$ of $x_{0}^{\prime}$ such that
$\tilde{v}^{(0)}(x_{0}^{\prime})\neq 0$ [BJS64, Theorem II5.4.1]. By Runge
approximation there exists a solution $v^{(0)}$ with the desired properties.
Now subtracting equations (4.5), (4.6) and integrating against $v^{(0)}$ gives

(4.8) $\displaystyle 0$
$\displaystyle=\int_{\Omega}\left(-\Delta_{\hat{g}}(w^{kl}-\tilde{w}^{kl})+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}(w^{kl}-\tilde{w}^{kl})\right)v^{(0)}$
$\displaystyle+\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{k}v^{l}v^{(0)}\,dV_{\hat{g}}$
$\displaystyle=\int_{\Omega}-(w^{kl}-\tilde{w}^{kl})\Delta_{\hat{g}}v^{(0)}+\frac{1-n}{2c(x^{\prime},0)}\sum_{i,j=1}^{n-1}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}(w^{kl}-\tilde{w}^{kl})v^{(0)}$
$\displaystyle+\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{k}v^{l}v^{(0)}\,dV_{\hat{g}}+\int_{\partial\Omega}(w^{kl}-\tilde{w}^{kl})\partial_{\nu}v^{(0)}-v^{(0)}\partial_{\nu}(w^{kl}-\tilde{w}^{kl})\,dS.$

Since the volume form is $dV_{\hat{g}}=|\hat{g}|^{\frac{1}{2}}\,dx$, where
$|\hat{g}|$ is the determinant of $\hat{g}$, we use integration by parts for
the second term in the last equality to have

$\displaystyle+\int_{\Omega}\frac{1-n}{2c(x^{\prime},0)}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}(w^{kl}-\tilde{w}^{kl})v^{(0)}|\hat{g}|^{\frac{1}{2}}\,dx$
$\displaystyle=-\int_{\Omega}\bigg{(}\frac{1-n}{2c(x^{\prime},0)}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)\partial_{x_{j}}v^{(0)}|\hat{g}|^{\frac{1}{2}}$
$\displaystyle+\partial_{x_{j}}\left(\frac{1-n}{2c(x^{\prime},0)}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)|\hat{g}|^{\frac{1}{2}}\right)v^{(0)}\bigg{)}(w^{kl}-\tilde{w}^{kl})\,dx$
$\displaystyle+\int_{\partial\Omega}(w^{kl}-\tilde{w}^{kl})\frac{1-n}{2c(x^{\prime},0)}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)|\hat{g}|^{\frac{1}{2}}v^{(0)}\nu_{j}\,dS$
$\displaystyle=-\int_{\Omega}(w^{kl}-\tilde{w}^{kl})(Xv^{(0)}+qv^{(0)})\,dV_{\hat{g}}.$

Here we used that $w^{kl}=\tilde{w}^{kl}=0$ on $\partial\Omega$. Combining
this with (4.8) gives

$\displaystyle\int_{\Omega}\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{k}v^{l}v^{(0)}\,dV_{\hat{g}}$
$\displaystyle=\int_{\Omega}(w^{kl}-\tilde{w}^{kl})(\Delta_{\hat{g}}v^{(0)}+Xv^{(0)}+qv^{(0)})\,dV_{\hat{g}}$
$\displaystyle-\int_{\partial\Omega}(w^{kl}-\tilde{w}^{kl})\partial_{\nu}v^{(0)}-v^{(0)}\partial_{\nu}(w^{kl}-\tilde{w}^{kl})\,dS$
$\displaystyle=0.$

In the last equality we used $w^{kl}=\tilde{w}^{kl}=0$ on $\partial\Omega$ and
that applying $\partial_{\varepsilon_{k}\varepsilon_{l}}^{2}|_{\varepsilon=0}$
to
$\partial_{\nu}u|_{\partial\Omega}=\partial_{\nu}\tilde{u}|_{\partial\Omega}$
implies

$\partial_{\nu}w^{kl}|_{\partial\Omega}=\partial_{\nu}\tilde{w}^{kl}|_{\partial\Omega}.$

Then by Lemma 4.1 we have

$\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{(0)}=0\text{
for }x^{\prime}\in\Omega.$

In particular for $x^{\prime}=x_{0}^{\prime}$ this implies
$\partial_{x_{n}}^{3}\tilde{c}(x_{0}^{\prime},0)=0$ but since $x_{0}^{\prime}$
was arbitrary we get

$\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)=0\text{ for
}x^{\prime}\in\Omega.$

Now the boundary value problems (4.5) and (4.6) are the same and thus by
uniqueness of solutions $w^{kl}=\tilde{w}^{kl}$ in $\Omega$.

The rest of the proof is done very similarly as in [Nur23] but we record the
proof here for completeness. Next we use induction to show
$\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)=0$ for all $k\geq 3$. By the
above this already holds for $k=3.$ Our assumption now is

$\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)=0,\quad
x^{\prime}\in\Omega,\enspace\text{for all}\quad
k=3,\ldots,N\in\mathbb{N},N>3.$

Let us do a subinduction to prove

$\partial_{l_{1}\dots l_{k}}^{k}u(x^{\prime},0)=\partial_{l_{1}\dots
l_{k}}^{k}\tilde{u}(x^{\prime},0),\quad x^{\prime}\in\,\,\Omega,$

for all $k=1,\ldots,N$, where $\partial_{l_{1}\dots
l_{k}}^{k}=\frac{\partial^{k}}{\partial_{\varepsilon_{l_{1}}}\dots\partial_{\varepsilon_{l_{k}}}}$.
Above we have shown this for $k=1,2$. Assume that it holds for $k\leq K<N$.
Then the linearization of order $K+1$ for the metric $g$ is, when evaluated at
$\varepsilon_{1}=\dots=\varepsilon_{K+1}=0$,

(4.9) $\displaystyle-\Delta_{\hat{g}}\partial_{l_{1}\dots
l_{K+1}}^{K+1}u(x^{\prime},0)-X\partial_{l_{1}\dots
l_{K+1}}^{K+1}u(x^{\prime},0)+R_{K}(u,g(x^{\prime},0),0)$
$\displaystyle+\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{K+2}c(x^{\prime},0)\left(\prod_{k=1}^{K+1}v^{(l_{k})}\right)=0,$

$x^{\prime}\in\,\,\Omega$. Here $R$ is a polynomial of components of $g$, the
derivatives of $g$ and $\partial_{l_{1}\dots l_{k}}^{k}u_{1}(x^{\prime},0)$.
Also the linearization of order $K+1$ for the metric $\tilde{c}g$ is

(4.10) $\displaystyle-\Delta_{\hat{g}}\partial_{l_{1}\dots
l_{K+1}}^{K+1}\tilde{u}(x^{\prime},0)-X\partial_{l_{1}\dots
l_{K+1}}^{K+1}\tilde{u}(x^{\prime},0)+R_{K}(\tilde{u},g(x^{\prime},0),0)$
$\displaystyle+\left(\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{K+2}c(x^{\prime},0)+\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{K+2}\tilde{c}(x^{\prime},0)\right)\left(\prod_{k=1}^{K+1}v^{(l_{k})}\right)=0.$

Here $R_{K}$ could also have terms with
$\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)$ and the components of
$\nabla_{x^{\prime}}\left(\partial_{x_{n}}^{k}\tilde{c}(x^{\prime},0)\right)$
but terms containing these are zero by the induction assumption. Now an
integration by parts argument similar to the case of the second linearization
and together with Lemma 4.1 (choosing $v^{3}=\ldots=v^{K+1}=1$) gives
$\partial_{x_{n}}^{K+2}\tilde{c}(x^{\prime},0)=0$.

Subtracting the equations (4.9) and (4.11) we get

$\left\\{\begin{array}[]{ll}-\Delta_{\hat{g}}\left(\partial_{l_{1}\dots
l_{K+1}}^{K+1}u(x^{\prime},0)-\partial_{l_{1}\dots
l_{K+1}}^{K+1}\tilde{u}(x^{\prime},0)\right)\\\ -X\left(\partial_{l_{1}\dots
l_{K+1}}^{K+1}u(x^{\prime},0)-\partial_{l_{1}\dots
l_{K+1}}^{K+1}\tilde{u}(x^{\prime},0)\right)=0,&\text{in}\,\,\Omega\\\
\partial_{l_{1}\dots l_{K+1}}^{K+1}u(x^{\prime},0)-\partial_{l_{1}\dots
l_{K+1}}^{K+1}\tilde{u}(x^{\prime},0)=0,&\text{on}\,\,\partial\Omega.\phantom{\Big{|}}\end{array}\right.$

This is true, since by induction assumptions for all $x^{\prime}\in\Omega$ the
other terms agree for $k\leq K$. Again, by the uniqueness of solutions,
$\partial_{l_{1}\dots l_{K+1}}^{K+1}u(x^{\prime},0)=\partial_{l_{1}\dots
l_{K+1}}^{K+1}\tilde{u}(x^{\prime},0)$, $x^{\prime}\in\Omega$, which ends the
subinduction.

Returning to the original induction, the linearization of order $N+1$ at
$\varepsilon_{1}=\dots=\varepsilon_{N+1}=0$ for the metric $g$ is

$\displaystyle-\Delta_{\hat{g}}\partial_{l_{1}\dots
l_{N+1}}^{N+1}u(x^{\prime},0)-X\partial_{l_{1}\dots
l_{N+1}}^{N+1}u(x^{\prime},0)+R_{N}(u,g(x^{\prime},0),0)$
$\displaystyle+\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{N+2}c(x^{\prime},0)\left(\prod_{k=1}^{N+1}v^{(l_{k})}\right)=0,$

$x^{\prime}\in\,\,\Omega$ and for the metric $\tilde{c}g$ we have

(4.11) $\displaystyle-\Delta_{\hat{g}}\partial_{l_{1}\dots
l_{N+1}}^{N+1}\tilde{u}(x^{\prime},0)-X\partial_{l_{1}\dots
l_{N+1}}^{N+1}\tilde{u}(x^{\prime},0)+R_{N}(\tilde{u},g(x^{\prime},0),0)$
$\displaystyle+\left(\frac{n-1}{2c(x^{\prime},0)}\partial_{x_{n}}^{N+2}c(x^{\prime},0)+\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{N+2}\tilde{c}(x^{\prime},0)\right)\left(\prod_{k=1}^{N+1}v^{(l_{k})}\right)=0.$

Now the subinduction implies that
$R_{N}(u,g(x^{\prime},0),0)=R_{N}(\tilde{u},g(x^{\prime},0),0)$. Thus by
subtracting, integrating against $v^{(0)}$ (the solution of (4.7)), using
integration by parts and that $\partial_{\nu}\partial_{l_{1}\dots
l_{N+1}}^{N+1}u(x^{\prime},0)|_{\partial\Omega}=\partial_{\nu}\partial_{l_{1}\dots
l_{N+1}}^{N+1}\tilde{u}(x^{\prime},0)|_{\partial\Omega}$ we get

$\int_{\Omega}\tilde{c}(x^{\prime},0)^{-1}\partial_{x_{n}}^{N+2}\tilde{c}(x^{\prime},0)v^{(0)}\prod_{k=1}^{N+1}v^{l_{k}}\,dx^{\prime}=0.$

Choosing all but two of the functions $v^{l_{k}}$ to be equal to $1$, then by
the completeness of such solutions (Lemma 4.1) implies that

$\partial_{x_{n}}^{N+2}\tilde{c}(x_{0}^{\prime},0)=0.$

Since $x_{0}^{\prime}$ was arbitrary, we have
$\partial_{x_{n}}^{N+2}\tilde{c}(x^{\prime},0)=0$ for all
$x^{\prime}\in\Omega$. ∎

We will next prove Theorem 1.4. The proof will be very similar to the one
above and we will only point out the differences.

  * Proof of Theorem 1.4.

Let $\varepsilon$ be, as in the previous proof, small enough so that (3.3) has
a unique small solution for $f_{\varepsilon}:=\sum_{k=1}^{N}f_{k}$, where
$f_{k}\in C^{s}(\partial\Omega)$ with $\operatorname{spt}(f)\subset\Gamma$ and
$\lVert f_{k}\rVert_{C^{s}(\partial\Omega)}<\delta$, $\delta>0$.

The first linearizations are (4.3), (4.4) and for the corresponding partial DN
maps we have

(4.12) $(D\Lambda_{g}^{\Gamma})_{0}=(D\Lambda_{cg}^{\Gamma})_{0}.$

Now using Lemma 4.2 together with a boundary determination [Tzo17, Proposition
$4.1$] we get from the first linearization that for all $i=1,\ldots,n-1$

$\frac{\partial_{x_{i}}\tilde{c}(x^{\prime},0)}{\tilde{c}(x^{\prime},0)}=0,\
\text{for}\ x^{\prime}\in\Omega.$

This then implies two things. Firstly,
$\tilde{c}(x^{\prime},0)=\lambda\in(0,\infty)$. Secondly, since solutions to
an advection diffusion equation are unique, we have that
$v^{l}=\tilde{v}^{l}$.

Let us move to the second order linearization and let
$x_{0}^{\prime}\in\Omega$. We will subtract equations (4.5), (4.6) and
integrate against the function $v^{(0)}$ that solves

(4.13)
$\left\\{\begin{array}[]{ll}\Delta_{\hat{g}}v^{(0)}-Xv^{(0)}-qv^{(0)}=0&\text{in}\,\,\Omega\\\
v^{(0)}=g,&\text{on}\,\,\Gamma\\\
v^{(0)}=0,&\text{on}\,\,\partial\Omega\setminus\Gamma.\phantom{\Big{|}}\end{array}\right.$

for $g\in C^{\infty}_{c}(\Gamma)$, $g\geq 0$, $g\not\equiv 0$ and that
$v^{(0)}(x_{0}^{\prime})\neq 0$. This function can be constructed exactly as
before. We now do the integration, where the difference will be on how to deal
with the boundary terms:

$\displaystyle\int_{\Omega}\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{k}v^{l}v^{(0)}\,dV_{\hat{g}}$
$\displaystyle=\int_{\Omega}-(w^{kl}-\tilde{w}^{kl})(-\Delta_{\hat{g}}v^{(0)}+Xv^{(0)}+qv^{(0)})\,dV_{\hat{g}}$
$\displaystyle-\int_{\partial\Omega}(w^{kl}-\tilde{w}^{kl})\partial_{\nu}v^{(0)}-v^{(0)}\partial_{\nu}(w^{kl}-\tilde{w}^{kl})\,dS$
$\displaystyle+\int_{\partial\Omega}(w^{kl}-\tilde{w}^{kl})\frac{1-n}{2c(x^{\prime},0)}\hat{g}^{ij}\partial_{x_{i}}c(x^{\prime},0)|\hat{g}|^{\frac{1}{2}}v^{(0)}\nu_{j}\,dS.$

Since $v^{(0)}$ solves (4.13) and
$(w^{kl}-\tilde{w}^{kl})|_{\partial\Omega}=0$ the first and last integral
vanish and also the first term in the second integral vanishes. For the
remaining part we divide the integral in two parts

$\int_{\partial\Omega}v^{(0)}\partial_{\nu}(w^{kl}-\tilde{w}^{kl})\,dS=\int_{\Gamma}v^{(0)}\partial_{\nu}(w^{kl}-\tilde{w}^{kl})\,dS+\int_{\partial\Omega\setminus\Gamma}v^{(0)}\partial_{\nu}(w^{kl}-\tilde{w}^{kl})\,dS.$

This is zero since $v^{(0)}|_{\partial\Omega\setminus\Gamma}=0$ and from
(4.12) we have that $\partial_{\nu}(w^{kl}-\tilde{w}^{kl})|_{\Gamma}=0$. Hence

$\int_{\Omega}\frac{n-1}{2\tilde{c}(x^{\prime},0)}\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)v^{k}v^{l}v^{(0)}\,dV_{\hat{g}}=0$

for all $v_{k}$, $v^{l}$ solving (4.3). Continuing as in the proof of Theorem
1.3 we get

$\partial_{x_{n}}^{3}\tilde{c}(x^{\prime},0)=0\text{ for
}x^{\prime}\in\Omega.$

and this then implies that $w^{kl}=\tilde{w}^{kl}$ in $\Omega$.

Proceeding with induction as before and taking further care about the boundary
terms in the integrals will conclude the proof. ∎

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