# Inverse problems for third–order nonlinear perturbations of biharmonic
operators

Sombuddha Bhattacharyya  S. Bhattacharyya, Department of Mathematics
Indian Institute of Science Education and Research
Bhopal, India (IISERB<EMAIL_ADDRESS>,  Katya Krupchyk  K. Krupchyk,
Department of Mathematics
University of California, Irvine
CA 92697-3875, USA<EMAIL_ADDRESS>,  Suman Kumar Sahoo  S. K. Sahoo,
Seminar for Applied Mathematics
Department of Mathematics
ETH Zürich, Switzerland<EMAIL_ADDRESS>and  Gunther Uhlmann  G. Uhlmann,
Department of Mathematics
University of Washington
Seattle, WA 98195-4350
USA
and Institute for Advanced Study of the Hong Kong University of Science and
Technology<EMAIL_ADDRESS>
###### Abstract.

We study inverse boundary problems for third–order nonlinear tensorial
perturbations of biharmonic operators on a bounded domain in $\mathbb{R}^{n}$,
where $n\geq 3$. By imposing appropriate assumptions on the nonlinearity, we
demonstrate that the Dirichlet–to–Neumann map, known on the boundary of the
domain, uniquely determines the genuinely nonlinear tensorial third-order
perturbations of the biharmonic operator. The proof relies on the inversion of
certain generalized momentum ray transforms on symmetric tensor fields.
Notably, the corresponding inverse boundary problem for linear tensorial
third-order perturbations of the biharmonic operator remains an open question.

##  1\. Introduction and statement of results

Let $\Omega\subset\mathbb{R}^{n}$, $n\geq 3$, be a bounded open set with
$C^{\infty}$ boundary. Consider the semilinear partial differential operator,

$\displaystyle
L_{A^{(1)},A^{(2)},A^{(3)},q}u:=(-\Delta)^{2}u+\sum_{i_{1},i_{2},i_{3}=1}^{n}A^{(3)}_{i_{1}i_{2}i_{3}}(x,u)D^{3}_{i_{1}i_{2}i_{3}}u+\sum_{i_{1},i_{2}=1}^{n}A^{(2)}_{i_{1}i_{2}}(x,u)D^{2}_{i_{1}i_{2}}u$
(1.1) $\displaystyle+\sum_{i_{1}=1}^{n}A^{(1)}_{i_{1}}(x,u)D_{i_{1}}u+q(x,u),$

where $u\in C^{\infty}(\overline{\Omega})$ and $D^{j}_{i_{1}\dots
i_{j}}=\frac{1}{i^{j}}\frac{\partial^{j}}{\partial_{x_{i_{1}}}\dots\partial_{x_{i_{j}}}}$,
$i_{1},\dots,i_{j}\in\\{1,2,\dots,n\\}$, $j=1,2,3$. We let
$S^{j}:=S^{j}(\mathbb{R}^{n})$ stand for the space of symmetric $j$–tensors on
$\mathbb{R}^{n}$. Let $0<\alpha<1$, and let
$C^{0,\alpha}(\overline{\Omega};S^{j})$ stand for the space of Hölder
continuous symmetric $j$–tensor fields, $j=0,1,2,3$. We assume that
$A^{(j)}:\overline{\Omega}\times\mathbb{C}\to S^{j}$, $j=1,2,3$, and
$q:\overline{\Omega}\times\mathbb{C}\to\mathbb{C}$ satisfy the following
conditions,

  * (i)

the map $\mathbb{C}\ni z\mapsto A^{(j)}(\cdot,z)$ is holomorphic with values
in $C^{0,\alpha}(\overline{\Omega};S^{j})$ for some $0<\alpha<1$,

  * (ii)

the map $\mathbb{C}\ni z\mapsto q(\cdot,z)$ is holomorphic with values in
$C^{0,\alpha}(\overline{\Omega})$,

  * (iii)

$A^{(j)}(x,0)=0$, $q(x,0)=0$, and $\partial_{z}q(x,0)=0$ for all
$x\in\overline{\Omega}$, $j=1,2,3$.

It follows from (i), (ii), (iii) that $A^{(j)}$, $j=1,2,3$, and $q$ can be
expanded into power series

$A^{(j)}(x,z)=\sum_{k=1}^{\infty}A^{(j),k}(x)\frac{z^{k}}{k!},\quad
A^{(j),k}(x):=\partial_{z}^{k}A^{(j)}(x,0)\in
C^{0,\alpha}(\overline{\Omega};S^{j}),$ (1.2)

converging in the $C^{0,\alpha}(\overline{\Omega};S^{j})$ topology, and

$q(x,z)=\sum_{k=2}^{\infty}q^{k}(x)\frac{z^{k}}{k!},\quad
q^{k}(x):=\partial_{z}^{k}q(x,0)\in C^{0,\alpha}(\overline{\Omega}),$ (1.3)

converging in the $C^{0,\alpha}(\overline{\Omega})$ topology.

Consider the Dirichlet problem for the operator
$L_{A^{(1)},A^{(2)},A^{(3)},q}$,

$\begin{cases}L_{A^{(1)},A^{(2)},A^{(3)},q}u=0\quad\text{in}\quad\Omega,\\\
u|_{\partial\Omega}=f,\\\ \partial_{\nu}u|_{\partial\Omega}=g,\end{cases}$
(1.4)

where $\nu$ is the unit outer normal to the boundary $\partial\Omega$. It is
shown in Theorem A.1 below that under the assumptions (i), (ii), (iii), there
exist $\delta>0$ and $C>0$ such that when $(f,g)\in
B_{\delta}(\partial\Omega):=\\{(f,g)\in C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega):\|f\|_{C^{4,\alpha}(\partial\Omega)}+\|g\|_{C^{3,\alpha}(\partial\Omega)}<\delta\\}$,
the problem (1.4) has a unique solution $u=u_{f,g}\in
C^{4,\alpha}(\overline{\Omega})$ satisfying
$\|u\|_{C^{4,\alpha}(\overline{\Omega})}<C\delta$. Associated to the problem
(1.4), we define the Dirichlet–to–Neumann map

$\Lambda_{A^{(1)},A^{(2)},A^{(3)},q}(f,g)=(\partial^{2}_{\nu}u|_{\partial\Omega},\partial^{3}_{\nu}u|_{\partial\Omega}),$
(1.5)

where $(f,g)\in B_{\delta}(\partial\Omega)$ and $u=u_{f,g}$. The inverse
problem that we are interested in is whether the knowledge of the
Dirichlet–to–Neumann map $\Lambda_{A^{(1)},A^{(2)},A^{(3)},q}$ determines the
nonlinear tensorial potentials $A^{(1)}$, $A^{(2)}$, $A^{(3)}$ and $q$ in the
operator $L_{A^{(1)},A^{(2)},A^{(3)},q}$ given by (1.1), uniquely.

Our main result is as follows.

######  Theorem 1.1.

Let $\Omega\subset\mathbb{R}^{n}$, $n\geq 3$, be a bounded open set with
$C^{\infty}$ boundary. Let
$A^{(j)},\widetilde{A}^{(j)}:\overline{\Omega}\times\mathbb{C}\to S^{j}$,
$j=1,2,3$, and
$q,\widetilde{q}:\overline{\Omega}\times\mathbb{C}\to\mathbb{C}$ satisfy the
conditions (i), (ii), (iii). Assume furthermore that

$\partial_{z}^{k}A^{(j)}(\cdot,0)|_{\partial\Omega}=\partial_{z}^{k}\widetilde{A}^{(j)}(\cdot,0)|_{\partial\Omega},\quad
k=1,2,\dots,\quad j=1,2,3.$ (1.6)

If
$\Lambda_{A^{(1)},A^{(2)},A^{(3)},q}=\Lambda_{\widetilde{A}^{(1)},\widetilde{A}^{(2)},\widetilde{A}^{(3)},\widetilde{q}}$
then $A^{(j)}=\widetilde{A}^{(j)}$ in $\overline{\Omega}\times\mathbb{C}$,
$j=1,2,3$, and $q=\widetilde{q}$ in $\overline{\Omega}\times\mathbb{C}$.

######  Remark 1.2.

To the best of our knowledge, the inverse boundary problem of recovering
linear tensorial third-order perturbations of the biharmonic operator remains
an open question. Furthermore, the uniqueness of such recovery is not possible
due to the gauge invariance of the problem, as explained in [13, 51]. We refer
to [51] for the study of the linearized inverse problems for tensorial third-
order perturbations of the biharmonic operator, where third-order tensorial
perturbations are determined up to a natural gauge transformation. Theorem 1.1
can be regarded as the first result establishing the unique recovery of third-
order nonlinear tensorial perturbations of biharmonic operators.

######  Remark 1.3.

The assumption (1.6) in Theorem 1.1 can be removed by performing boundary
determination, for instance as in [37, Appendix C].

Perturbed biharmonic operators naturally arise in various areas of physics and
geometry, such as the study of the Kirchhoff plate equation in elasticity
theory and the investigation of the Paneitz-Branson operator in conformal
geometry, see [21, 53]. The inverse boundary problems concerning the recovery
of zeroth-order linear perturbations of biharmonic operators have been studied
in [24, 26], see also [34]. Furthermore, the investigations of inverse
boundary problems for determining first-order linear perturbations of
biharmonic operators, as well as more general polyharmonic operators, can be
found in [31, 32].

Further advancements have been made in the field of inverse boundary problems
for linear perturbations of biharmonic and polyharmonic operators in [6, 7, 8,
9, 12, 13, 14, 16, 17, 22, 47, 48, 58, 59, 15, 3, 4, 5], see also [11] for the
two-dimensional case. However, it should be noted that when considering linear
perturbations of biharmonic operators specifically, all of these results are
limited to determining up to second-order perturbations. The reason for this
limitation lies in the construction of complex geometric optics solutions for
the perturbed biharmonic operators, which are essential for solving the
corresponding inverse problems. This construction relies on Carleman estimates
with limiting Carleman weights. To obtain Carleman estimates for the perturbed
semiclassical biharmonic operator, an iterative procedure is employed,
involving the iteration of Carleman estimates with limiting Carleman weights
for the semiclassical Laplacian and perturbing them with lower-order terms.
However, it should be noted that the Carleman estimates with limiting Carleman
weights for the semiclassical Laplacian suffer from a semiclassical loss,
which is manifested by the presence of the factor ’$h$’ in the estimates, as
illustrated in equation (3.2) below. Consequently, the resulting Carleman
estimates for the semiclassical biharmonic operator, as depicted in equation
(3.3) below, lack the necessary strength to accommodate third-order linear
perturbations of the biharmonic operator.

Now the work [40] shows that nonlinearity might help when solving inverse
problems for hyperbolic PDE. Furthermore, similar phenomena have been observed
and utilized in the context of nonlinear elliptic equations, as demonstrated
in [20, 42]. We refer to [28, 35, 36, 37, 38, 41, 49, 43, 46, 45, 52] for
extensive recent work on inverse boundary problems for nonlinear elliptic PDE.
Notably, these works highlight that the presence of nonlinearity allows for
solving inverse problems for nonlinear PDE, even when the corresponding
inverse problems for linear equations remain unsolved. Our present paper aims
to further illustrate and explore this intriguing phenomenon.

Let us now proceed to discuss the main ideas behind the proof of Theorem 1.1.
By employing the technique of higher-order linearizations of the
Dirichlet–to–Neumann map, initially introduced in [40] (see also [20, 42]),
along with previous works on the second linearization [56, 57], the proof of
Theorem 1.1 can be reduced to establishing the following density result.

######  Theorem 1.4.

Let $\Omega\subset\mathbb{R}^{n}$, $n\geq 3$, be a bounded open set with
$C^{\infty}$ boundary. Let $A^{(j)}\in C(\overline{\Omega};S^{j})$ be such
that

$A^{(j)}|_{\partial\Omega}=0,$ (1.7)

$j=1,2,3$, and $q\in L^{\infty}(\Omega)$. If

$\displaystyle\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\int_{\Omega}A^{(j)}_{i_{1}\dots
i_{j}}(x)\big{(}v^{(1)}D^{j}_{i_{1}\dots
i_{j}}v^{(2)}+v^{(2)}D^{j}_{i_{1}\dots i_{j}}v^{(1)}\big{)}v^{(0)}dx$ (1.8)
$\displaystyle+\int_{\Omega}q(x)v^{(1)}v^{(2)}v^{(0)}dx=0,$

for all $v^{(l)}\in C^{4,\alpha}(\overline{\Omega})$ solving
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=0,1,2$, then $A^{(j)}=0$ and $q=0$ in
$\Omega$, $j=1,2,3$.

######  Remark 1.5.

The assumption (1.7) in Theorem 1.4 can be removed by performing boundary
determination, for instance as in [37, Appendix C]. This assumption ensures
that when we extend $A^{(j)}$ to $\mathbb{R}^{n}\setminus\overline{\Omega}$ by
setting it to zero outside of $\overline{\Omega}$ and denote this extension by
the same letters, we satisfy $A^{(j)}\in C_{0}(\mathbb{R}^{n};S^{j})$ for
$j=1,2,3$. The injectivity results for generalized momentum ray transforms, as
presented in Lemma 2.6 and Lemma 2.8 below, apply precisely to this class of
tensor fields. Our proof of Theorem 1.4 relies on these injectivity results.

To establish Theorem 1.4, our first step involves constructing complex
geometrical optics (CGO) solutions for the biharmonic equations, which belong
to the function space $C^{4,\alpha}(\overline{\Omega})$. Since the integral
identity (1.8) involves products of three biharmonic functions and their
derivatives up to the third order, it is advantageous to construct CGO
solutions whose remainders with all derivatives up to the third order are
small in $L^{\infty}$-norm. Additionally, we adopt specific amplitude choices
as presented in [14, Theorem 1.1] for our constructed solutions.

By testing the integral identity (1.8) with the constructed solutions, we can
effectively recover certain generalized momentum ray transforms on tensor
fields as well as the sum of tensor fields of different ranks. The concept of
momentum ray transforms was initially introduced in [54], with further
investigations conducted in [55, 29, 30, 14, 51, 25]. These transforms have
proven instrumental in the study of inverse boundary problems for polyharmonic
operators, as initially demonstrated in [14] and also explored in [51, 15]. To
conclude the proof of Theorem 1.4, we leverage the injectivity results given
in Lemma 2.6 and Lemma 2.8 below for the obtained momentum ray transforms.

######  Remark 1.6.

Upon setting $v^{(2)}=1$ in (1.8), we obtain that

$\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\int_{\Omega}A^{(j)}_{i_{1}\dots
i_{j}}(x)(D^{j}_{i_{1}\dots
i_{j}}v^{(1)})v^{(0)}dx+\int_{\Omega}q(x)v^{(1)}v^{(0)}dx=0,$ (1.9)

for all $v^{(l)}\in C^{4,\alpha}(\overline{\Omega})$ solving
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=0,1$. The integral identity (1.9)
corresponds to the same identity considered in [51, Theorem 2.3]. In the
aforementioned reference, the validity of (1.9) is required for biharmonic
functions $v^{(l)}\in H^{4}(\Omega)$, where $l=1,2$, and it is shown in [51,
Theorem 2.3] that the symmetric tensor of order three, denoted as $A^{(3)}$,
can be recovered from this integral identity, up to a natural gauge.
Therefore, it becomes apparent that the unique determination of a symmetric
tensor of order three $A^{(3)}$ from (1.8) necessitates the utilization of the
freedom provided by employing three biharmonic functions in (1.8).

######  Remark 1.7.

In the case where $A^{(3)}=0$, it appears that to uniquely recover symmetric
tensors $A^{(2)}$, $A^{(1)}$, and a function $q$ from (1.8), it is sufficient
to work with two biharmonic functions. Indeed, assuming $A^{(3)}=0$ and
setting $v^{(2)}=1$ in (1.8), we obtain the following identity,

$\sum_{j=1}^{2}\sum_{i_{1},i_{j}=1}^{n}\int_{\Omega}A^{(j)}_{i_{1}\dots
i_{j}}(x)(D^{j}_{i_{1}i_{j}}v^{(1)})v^{(0)}dx+\int_{\Omega}q(x)v^{(1)}v^{(0)}dx=0,$
(1.10)

valid for all $v^{(l)}\in C^{4,\alpha}(\overline{\Omega})$ solving
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=0,1$. The identity (1.10) coincides
with the identity considered in [51, Theorem 2.1], which holds for biharmonic
functions $v^{(l)}\in H^{4}(\Omega)$. Consequently, following the proof of
[51, Theorem 2.1], we can conclude that $A^{(j)}=0$ for $j=1,2$, and $q=0$ in
$\Omega$.

######  Remark 1.8.

It is important to note that when establishing Theorem 1.4, the recovery of
$A^{(3)}$ is not independent of that of $A^{(2)}$. Initially, we can
demonstrate that $A^{(3)}=i_{\delta}W^{(1)}$, where $W^{(1)}\in
C_{0}(\mathbb{R}^{n};S^{1})$. However, proving that $W^{(1)}=0$ requires the
inversion of the generalized momentum ray transform, which involves both
$W^{(1)}$ and the trace-free part of the tensor field $A^{(2)}$, as shown in
(4.26) below.

Finally, it is worth mentioning that one could also explore the inverse
boundary problem for the operator $L_{A^{(1)},A^{(2)},A^{(3)},q}$ with Navier
boundary conditions instead of the Dirichlet conditions addressed in this
paper. However, in the case of measurements taken on the entire boundary, the
explicit description of the Laplacian in boundary normal coordinates, as
outlined in [44], reveals that considering Navier boundary conditions will
yield the same boundary data for the inverse problem.

The paper is organized as follows. In Section 2 we collect some facts on
symmetric tensor fields and their momentum and generalized momentum ray
transforms, including the injectivity results, which are required for proving
Theorem 1.4. For the completeness and convenience of the reader, we provide
the proof of the necessary injectivity results for the generalized momentum
ray transforms in Appendix B. The construction of complex geometric optics
solutions to the biharmonic equations and the selection of appropriate
amplitudes required for the proof of Theorem 1.4 are discussed in Section 3.
The proof of Theorem 1.4 is provided in Section 4. Section 5 is dedicated to
the proof of Theorem 1.1. Lastly, in Appendix A, the well-posedness of the
Dirichlet problem for third-order nonlinear perturbations of biharmonic
operators is demonstrated in the case of small boundary data.

##  2\. Momentum ray transforms

In this section we present some facts on symmetric tensor fields and their
momentum and generalized momentum ray transforms, following [55, 14, 51],
which are required for proving Theorem 1.4.

###  2.1. Symmetric tensor fields

Let $n\geq 2$, and let $m\in\mathbb{N}\cup\\{0\\}$. Let
$T^{m}:=T^{m}(\mathbb{R}^{n})$ be the complex vector space of functions
$\underbrace{\mathbb{R}^{n}\times\dots\times\mathbb{R}^{n}}_{\text{m
times}}\to\mathbb{C}$ which are $\mathbb{R}$–linear in each of its argument,
and let $S^{m}:=S^{m}(\mathbb{R}^{n})$ be its subspace which consists of
functions symmetric in all arguments. The elements of $T^{m}$ are called
tensors of degree $m$ while the elements of $S^{m}$ are said to be symmetric
tensors of degree $m$. In particular, $S^{0}=\mathbb{C}$ and
$S^{1}=\mathbb{C}^{n}$. Let $\sigma:T^{m}\to S^{m}$ be the canonical
projection (symmetrization) defined by

$\sigma
u(x_{1},\dots,x_{m})=\frac{1}{m!}\sum_{\pi\in\Pi_{m}}u(x_{\pi(1)},\dots,x_{\pi(m)}),$
(2.1)

where the summation is taken over the group $\Pi_{m}$ of all permutations of
the set $\\{1,\dots,m\\}$.

Let $e_{1},\dots,e_{n}$ be a basis of $\mathbb{R}^{n}$. Then the numbers
$u_{i_{1}\dots i_{m}}=u(e_{i_{1}},\dots,e_{i_{m}})$ are said to be the
covariant coordinates or components of $u\in T^{m}$. A tensor $u\in T^{m}$
belongs to $S^{m}$ if and only of its covariant components $u_{i_{1}\dots
i_{m}}$ are symmetric with respect to all indices. Together with covariant
coordinates $u_{i_{1}\dots i_{m}}$ of the tensor $u$, one can define the
contravariant coordinates $u^{i_{1}\dots i_{m}}$, see [55, page 23]. In the
case of the Euclidean metric tensor $g_{ij}=\delta_{ij}$, we have
$u^{i_{1}\dots i_{m}}=u_{i_{1}\dots i_{m}}$.

The scalar product on $T^{m}$ is defined by

$\langle u,v\rangle=\sum_{i_{1},\dots,i_{m}=1}^{n}u^{i_{1}\dots
i_{m}}\bar{v}_{i_{1}\dots i_{m}},\quad u,v\in T^{m}.$ (2.2)

For $u\in T^{k}$ and $v\in T^{m}$, the tensor product $u\otimes v\in T^{k+m}$
is given by

$u\otimes
v(x_{1},\dots,x_{k+m})=u(x_{1},\dots,x_{k})v(x_{k+1},\dots,x_{k+m}).$

Together with the full symmetrization (2.1), given indices
$i_{1},\dots,i_{p}$, the operator of partial symmetrization of a tensor $u\in
T^{m}$ with respect to $i_{1},\dots,i_{p}$, $p\leq m$, is defined by

$\sigma(i_{1},\dots,i_{p})u_{i_{1},\dots,i_{m}}=\frac{1}{p!}\sum_{\pi\in\Pi_{p}}u_{i_{\pi(1)}\dots
i_{\pi(p)}i_{p+1}\dots i_{m}}.$ (2.3)

We define the operator $i_{\delta}$ of symmetric multiplication by the
Euclidean metric tensor $\delta$ as follows,

$i_{\delta}:S^{m}\to S^{m+2},\quad(i_{\delta}u)_{i_{1}\dots
i_{m+2}}=\sigma(i_{1},\dots,i_{m+2})(u_{i_{1}\dots
i_{m}}\delta_{i_{m+1}i_{m+2}}).$ (2.4)

We also define the operator $j_{\delta}$, dual to $i_{\delta}$, as follows,

$j_{\delta}:S^{m}\to S^{m-2},\quad(j_{\delta}u)_{i_{1}\dots
i_{m-2}}=\sum_{i_{m-1},i_{m}=1}^{n}u_{i_{1}\dots
i_{m}}\delta^{i_{m-1}i_{m}}=\sum_{k=1}^{n}u_{i_{1}\dots i_{m-2}kk}.$ (2.5)

We also set $j_{\delta}f=0$ for $f\in S^{0}$ and $f\in S^{1}$. If for $f\in
S^{m}$, $j_{\delta}f=0$, we say that $f$ is a trace free tensor.

For $m\geq 0$, the following commutation formula holds on $S^{m}$,

$j_{\delta}i_{\delta}=\frac{2(n+2m)}{(m+1)(m+2)}E+\frac{m(m-1)}{(m+1)(m+2)}i_{\delta}j_{\delta},$
(2.6)

where $E$ is the identity operator on $S^{m}$, see [18, Lemma 2.2].

Let $\Omega\subset\mathbb{R}^{n}$ be open. We denote by
$C^{\infty}_{0}(\Omega;S^{m})$ the space of smooth compactly supported
symmetric $m$–tensor fields on $\Omega$, and by
$\mathcal{E}^{\prime}(\Omega;S^{m})$ the space of compactly supported
$m$–tensor field-distributions. We let $L^{p}(\Omega;S^{m})$, $1\leq
p<\infty$, be the space of $L^{p}$ integrable $m$–tensor fields on $\Omega$,
and let $L^{\infty}(\Omega;S^{m})$ be the spaces of bounded almost everywhere
$m$–tensor fields on $\Omega$. We define the scalar product on
$L^{2}(\Omega;S^{m})$ by

$(u,v)_{L^{2}(\Omega;S^{m})}=\int_{\mathbb{R}^{n}}\langle u(x),v(x)\rangle
dx,$

where $\langle\cdot,\cdot\rangle$ is the scalar product on $S^{m}$ given by
(2.2).

###  2.2. Momentum ray transforms and their injectivity

Let

$T\mathbb{S}^{n-1}=\\{(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}:|\xi|=1,\
x\cdot\xi=0\\}$

be the tangent bundle of the unit sphere $\mathbb{S}^{n-1}$, and let
$k\in\mathbb{N}\cup\\{0\\}$. Following [54], see also [55, 29, 30, 14, 51],
for $f^{(m)}\in C_{0}(\mathbb{R}^{n};S^{m})$ we define the $k$th momentum ray
transform $\widetilde{I}^{m,k}$ by

$\widetilde{I}^{m,k}f^{(m)}(x,\xi)=\sum_{i_{1},\dots,i_{m}=1}^{n}\int_{\mathbb{R}}t^{k}f^{(m)}_{i_{1}\dots
i_{m}}(x+t\xi)\xi_{i_{1}}\dots\xi_{i_{m}}dt,$ (2.7)

where $(x,\xi)\in T\mathbb{S}^{n-1}$. One can easily see that

$\widetilde{I}^{m,k}:C_{0}(\mathbb{R}^{n};S^{m})\to
C_{0}(T\mathbb{S}^{n-1}),\quad\widetilde{I}^{m,k}:C^{\infty}_{0}(\mathbb{R}^{n};S^{m})\to
C^{\infty}_{0}(T\mathbb{S}^{n-1}),$ (2.8)

where $C_{0}(T\mathbb{S}^{n-1})$ ($C^{\infty}_{0}(T\mathbb{S}^{n-1})$) is the
spaces of continuous (smooth) functions on $T\mathbb{S}^{n-1}$ that have
compact support in the first variable.

When $k=0$, $\widetilde{I}^{m,0}$ is the classical ray transform. When $m=0$,
the transform $\widetilde{I}^{0,0}$ is the main mathematical tool of Computer
Tomography. When $m=1$, the operator $\widetilde{I}^{1,0}$ is known as the
Doppler transform and is the key mathematical tool of Doppler Tomography. When
$k>0$ and $m=0$, the operator $\widetilde{I}^{0,k}$ appears in the study of
the inversion of the cone transform and conical Radon transform whereas the
latter transform has an application in Compton cameras, see [39].

We observe that (2.7) also makes sense when
$(x,\xi)\in\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})$ and
$f^{(m)}\in C_{0}(\mathbb{R}^{n};S^{m})$. We thus set

$I^{m,k}f^{(m)}(x,\xi)=\sum_{i_{1},\dots,i_{m}=1}^{n}\int_{\mathbb{R}}t^{k}f^{(m)}_{i_{1}\dots
i_{m}}(x+t\xi)\xi_{i_{1}}\dots\xi_{i_{m}}dt,$ (2.9)

where $(x,\xi)\in\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})$. A
direct computation shows that the function $I^{m,k}f^{(m)}(x,\xi)$ has the
following homogeneity property in the second variable,

$(I^{m,k}f^{(m)})(x,\tau\xi)=\frac{\tau^{m-k}}{|\tau|}(I^{m,k}f^{(m)})(x,\xi),\quad\tau\in\mathbb{R}\setminus\\{0\\},$
(2.10)

and the following property in the first variable,

$(I^{m,k}f^{(m)})(x+\tau\xi,\xi)=\sum_{l=0}^{k}\begin{pmatrix}k\\\
l\end{pmatrix}(-\tau)^{k-l}(I^{m,l}f^{(m)})(x,\xi),\quad\tau\in\mathbb{R},$
(2.11)

see [29, formulas (2.3), (2.4)]. It follows from (2.10) and (2.11) that

$\displaystyle(I^{m,k}f^{(m)})(x,\xi)=$
$\displaystyle|\xi|^{m-2k-1}\sum_{l=0}^{k}(-1)^{k-l}\begin{pmatrix}k\\\
l\end{pmatrix}|\xi|^{l}(\xi\cdot x)^{k-l}$ (2.12)
$\displaystyle\times(I^{m,l}f^{(m)})\bigg{(}x-\frac{\xi\cdot
x}{|\xi|^{2}}\xi,\frac{\xi}{|\xi|}\bigg{)},$

see [29, formula (2.6)]. Note that here $(x-\frac{\xi\cdot
x}{|\xi|^{2}}\xi,\frac{\xi}{|\xi|})\in T\mathbb{S}^{n-1}$. Therefore, using
(2.8), we see from (2.12) that

$\displaystyle I^{m,k}:C_{0}(\mathbb{R}^{n};S^{m})\to
C(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})),$ (2.13)
$\displaystyle I^{m,k}:C^{\infty}_{0}(\mathbb{R}^{n};S^{m})\to
C^{\infty}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})).$

We can also consider the $k$th momentum ray transform, defined by (2.7), for
$(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$ and $f^{(m)}\in
C_{0}(\mathbb{R}^{n};S^{m})$. This transform will be denoted as $J^{m,k}$, i.e

$J^{m,k}f^{(m)}(x,\xi)=\sum_{i_{1},\dots,i_{m}=1}^{n}\int_{\mathbb{R}}t^{k}f^{(m)}_{i_{1}\dots
i_{m}}(x+t\xi)\xi_{i_{1}}\dots\xi_{i_{m}}dt,$ (2.14)

where $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. We have
$J^{m,k}f^{(m)}=I^{m,k}f^{(m)}|_{\mathbb{R}^{n}\times\mathbb{S}^{n-1}}$, and
therefore,

$J^{m,k}:C_{0}(\mathbb{R}^{n};S^{m})\to
C(\mathbb{R}^{n}\times\mathbb{S}^{n-1}),\
J^{m,k}:C^{\infty}_{0}(\mathbb{R}^{n};S^{m})\to
C^{\infty}(\mathbb{R}^{n}\times\mathbb{S}^{n-1}).$ (2.15)

When $f^{(m)}\in C_{0}(\mathbb{R}^{n};S^{m})$, (2.10) implies that

$I^{m,k}f^{(m)}(x,\xi)=|\xi|^{m-k-1}J^{m,k}f^{(m)}\bigg{(}x,\frac{\xi}{|\xi|}\bigg{)},\quad
x\in\mathbb{R}^{n},\quad\xi\in\mathbb{R}^{n}\setminus\\{0\\}.$ (2.16)

Hence, the operators $I^{m,k}$ and $J^{m,k}$, defined on
$C_{0}(\mathbb{R}^{n};S^{m})$, provide equivalent information.

In what follows, we shall only consider the momentum ray transforms $I^{m,k}$
and $J^{m,k}$. We shall next extend the definition (2.9) of $I^{m,k}$ to
$\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})$. To that end, we let
$\mathcal{D}^{\prime}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\}))$
be the spaces of distributions on
$\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})$. The $k$th momentum ray
transform $I^{m,k}$ on $\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})$ is defined
as follows,

$\displaystyle
I^{m,k}:\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})\to\mathcal{D}^{\prime}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})),\quad\langle
I^{m,k}f^{(m)},\psi\rangle=\langle f^{(m)},(I^{m,k})^{*}\psi\rangle,$ (2.17)

where $\psi\in
C^{\infty}_{0}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\}))$ and
$(I^{m,k})^{*}$ is the adjoint of $I^{m,k}$ given by

$\displaystyle(I^{m,k})^{*}:C^{\infty}_{0}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\}))\to
C^{\infty}(\mathbb{R}^{n};S^{m}),$ (2.18)
$\displaystyle((I^{m,k})^{*}\psi)_{i_{1}\dots
i_{m}}(x)=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}}t^{k}\xi_{i_{1}}\dots\xi_{i_{m}}\psi(x-t\xi,\xi)dtd\xi,\quad
1\leq i_{1},\dots,i_{m}\leq n,$

see [14, Definition 4.2]. One can easily see that when $f^{(m)}\in
C_{0}(\mathbb{R}^{n};S^{m})$, the $k$th momentum ray transform
$I^{m,k}f^{(m)}$ defined by (2.9) satisfies (2.17).

Let $C_{0}(\mathbb{R}^{n};\mathcal{S}^{m}):=C_{0}(\mathbb{R}^{n};S^{0})\oplus
C_{0}(\mathbb{R}^{n};S^{1})\dots\oplus C_{0}(\mathbb{R}^{n};S^{m})$. Then any
$F\in C_{0}(\mathbb{R}^{n};\mathcal{S}^{m})$ can be written uniquely as

$F=\sum_{l=0}^{m}f^{(l)}=f^{(0)}+\sum_{i_{1}=1}^{n}f^{(1)}_{i_{1}}dx^{i_{1}}+\dots+\sum_{i_{1},\dots,i_{m}=1}^{n}f^{(m)}_{i_{1}\dots
i_{m}}dx^{i_{1}}\dots dx^{i_{m}}.$ (2.19)

Let $k\in\mathbb{N}\cup\\{0\\}$. For $F\in
C_{0}(\mathbb{R}^{n};\mathcal{S}^{m})$, the $k$th momentum ray transform
$I^{m,k}$ is defined by

$I^{m,k}F:=\sum_{l=0}^{m}I^{l,k}f^{(l)},$

where $I^{l,k}f^{(l)}$ is given by (2.9) with $m=l$. Let
$\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m}):=\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{0})\oplus\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{1})\dots\oplus\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})$.
The $k$th momentum ray transform $I^{m,k}$ can be extended to
$\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m})$ as follows,

$\displaystyle
I^{m,k}:\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m})\to\mathcal{D}^{\prime}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})),\quad\langle
I^{m,k}F,\psi\rangle=\sum_{l=0}^{m}\langle f^{(l)},(I^{l,k})^{*}\psi\rangle,$

where $\psi\in
C^{\infty}_{0}(\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\}))$ and
$(I^{l,k})^{*}$ is given by (2.18) with $m=l$, see [14, Definition 4.4].

Next, we state the injectivity results for the momentum ray transforms. Let us
first recall that when $m\geq 1$, the ray transform $I^{m,0}$ always has a
nontrivial kernel. Indeed, any symmetric tensor field can be decomposed
uniquely into its solenoidal and potential parts, see [55, Theorem 2.6.2], and
the potential part lies in the kernel.

######  Theorem 2.1.

[14, Theorem 4.5] Let
$F\in\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m})$. If $I^{m,k}F=0$
for all $k=0,1,\dots,m$, then $F=0$.

######  Corollary 2.2.

Let $f^{(m)}\in\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})$. If
$I^{m,k}f^{(m)}=0$ for all $k=0,1,\dots,m$, then $f^{(m)}=0$.

The following stronger injectivity result holds.

######  Theorem 2.3.

[14, Theorem 4.9] Let
$F\in\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m})$. If $I^{m,m}F=0$
then $F=0$.

We extend the operator $J^{m,k}$ to
$\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})$ as follows,

$\displaystyle
J^{m,k}:\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})\to\mathcal{D}^{\prime}(\mathbb{R}^{n}\times\mathbb{S}^{n-1}),\quad\langle
J^{m,k}f^{(m)},\psi\rangle=\langle f^{(m)},(J^{m,k})^{*}\psi\rangle.$ (2.20)

Here $\psi\in C^{\infty}_{0}(\mathbb{R}^{n}\times\mathbb{S}^{n-1})$ and
$(J^{m,k})^{*}$ is the adjoint of $J^{m,k}$ given by

$\displaystyle(J^{m,k})^{*}:C^{\infty}_{0}(\mathbb{R}^{n}\times\mathbb{S}^{n-1})\to
C^{\infty}(\mathbb{R}^{n};S^{m}),$ (2.21)
$\displaystyle((J^{m,k})^{*}\psi)_{i_{1}\dots
i_{m}}(x)=\int_{\mathbb{S}^{n-1}}\int_{\mathbb{R}}t^{k}\xi_{i_{1}}\dots\xi_{i_{m}}\psi(x-t\xi,\xi)dtdS(\xi),\
1\leq i_{1},\dots,i_{m}\leq n,$

where $dS$ is the Euclidean surface element on the sphere $\mathbb{S}^{n-1}$,
see [55, Section 2.5]. One can easily see that when $f^{(m)}\in
C_{0}(\mathbb{R}^{n};S^{m})$, the $k$th momentum ray transform
$J^{m,k}f^{(m)}$ defined by (2.14) satisfies (2.20).

Let $F\in C_{0}(\mathbb{R}^{n};\mathcal{S}^{m})$ be given by (2.19). The $k$th
momentum ray transform $J^{m,k}$ is defined by

$J^{m,k}F:=\sum_{l=0}^{m}J^{l,k}f^{(l)},$

where $J^{l,k}f^{(l)}$ is given by (2.14) with $m=l$. Furthermore, $J^{m,k}$
can be extended to $\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m})$ as
follows,

$\displaystyle
J^{m,k}:\mathcal{E}^{\prime}(\mathbb{R}^{n};\mathcal{S}^{m})\to\mathcal{D}^{\prime}(\mathbb{R}^{n}\times\mathbb{S}^{n-1}),\quad\langle
J^{m,k}F,\psi\rangle=\sum_{l=0}^{m}\langle f^{(l)},(J^{l,k})^{*}\psi\rangle,$

where $\psi\in C^{\infty}_{0}(\mathbb{R}^{n}\times\mathbb{S}^{n-1})$ and
$(J^{l,k})^{*}$ is given by (2.21) with $m=l$.

In what follows, we shall need some properties of the $k$th momentum ray
transform, which are obtained by a direct computation. When $f^{(m)}\in
C_{0}(\mathbb{R}^{n};S^{m})$, we have

$J^{m,k}f^{(m)}(x,-\xi)=(-1)^{m-k}J^{m,k}f^{(m)}(x,\xi),$ (2.22)

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. When $f^{(0)}\in
C_{0}(\mathbb{R}^{n};S^{0})$ and $f^{(1)}\in C_{0}(\mathbb{R}^{n};S^{1})$, we
also have

$J^{0,k}f^{(0)}(x,\xi)=J^{2,k}(i_{\delta}f^{(0)})(x,\xi),\quad
J^{1,k}f^{(1)}(x,\xi)=J^{3,k}(i_{\delta}f^{(1)})(x,\xi),$ (2.23)

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. Note that in (2.23)
it is crucial that $\xi\in\mathbb{S}^{n-1}$. Let $\xi\in\mathbb{S}^{n-1}$ and
$f^{(m)}\in\mathcal{E}^{\prime}(\mathbb{R}^{n};S^{m})$. Then we have

$(\xi\cdot\partial_{x})J^{m,k}f^{(m)}=-kJ^{m,k-1}f^{(m)}\quad\text{in}\quad\mathcal{D}^{\prime}(\mathbb{R}^{n}\times\mathbb{S}^{n-1}),$
(2.24)

see [14, Theorem 4.8].

We shall need the following result established in [14, Theorem 4.18]. We state
it only in the special cases of $m$–tensor fields with $m=0,1,2,3$ needed
here. While the statement given in [14, Theorem 4.18] is for smooth with
compact support tensor fields, the proof works in the case of continuous with
compact support tensor fields as well, and we shall present it in Appendix B.1
in the special cases needed for completeness, and convenience of the reader.

######  Theorem 2.4.

[14, Theorem 4.18] Let $m=0,1,2,3$ and let $F=\sum_{l=0}^{m}f^{(l)}\in
C_{0}(\mathbb{R}^{n};\mathcal{S}^{m})$. We have $J^{m,m}F(x,\xi)=0$, for all
$(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$ if and only if

  * (i)

$f^{(2)}=-i_{\delta}f^{(0)}$ and $f^{(3)}=-i_{\delta}f^{(1)}$, when $m=3$,

  * (ii)

$f^{(2)}=-i_{\delta}f^{(0)}$ and $f^{(1)}=0$, when $m=2$,

  * (iii)

$f^{(1)}=0$ and $f^{(0)}=0$, when $m=1$,

  * (iv)

$f^{(0)}=0$, when $m=0$.

######  Remark 2.5.

It follows from Theorem 2.3 and Theorem 2.4 that while the kernel of $I^{m,m}$
on $C_{0}(\mathbb{R}^{n};\mathcal{S}^{m})$ is trivial, the kernel of $J^{m,m}$
on $C_{0}(\mathbb{R}^{n};\mathcal{S}^{m})$ is nontrivial when $m\geq 2$.

To prove Theorem 1.4, we need to utilize a result related to the injectivity
of the generalized momentum ray transform, as defined in (2.25) below. This
result has been established in [14, Lemma 3.7], see also [51, Lemma 3.1]. We
will only state it in the specific cases where $m=1,2,3$ – these cases are
essential for this paper. The proof of this result can be found in Appendix
B.2 for completeness and convenience.

######  Lemma 2.6.

[14, Lemma 3.7] Let $n\geq 3$, $e_{1}=(1,0,\dots,0)\in\mathbb{R}^{n}$, and let
$f^{(m)}\in C_{0}(\mathbb{R}^{n-1};S^{m}(\mathbb{R}^{n}))$ with some
$m=1,2,3$. Assume that

$\sum_{i_{1},\dots,i_{m}=1}^{n}\int_{\mathbb{R}}t^{m}f^{(m)}_{i_{1}\dots
i_{m}}(y_{0}^{\prime}+t\eta^{\prime})(e_{1}+i\eta)_{i_{1}}\dots(e_{1}+i\eta)_{i_{m}}dt=0,$
(2.25)

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$ and all
$\eta=(\eta_{1},\eta^{\prime})\in\mathbb{R}^{n}$ such that $e_{1}\cdot\eta=0$
and $|\eta|=1$. Then

$f^{(m)}=\begin{cases}i_{\delta}v^{(m-2)},&m=2,3,\\\ 0,&m=1,\end{cases}$

where

$v^{(1)}=\frac{3}{n+2}j_{\delta}f^{(3)}\in
C_{0}(\mathbb{R}^{n-1};S^{1}(\mathbb{R}^{n})),\quad
v^{(0)}=\frac{1}{n}j_{\delta}f^{(2)}\in
C_{0}(\mathbb{R}^{n-1};S^{0}(\mathbb{R}^{n})).$

We can derive the following immediate consequence of Lemma 2.6.

######  Corollary 2.7.

For $m=2$ or $3$, assuming that all the conditions of Lemma 2.6 are met, and
if, additionally, $j_{\delta}f^{(m)}=0$, it follows that $f^{(m)}=0$.

In the proof of Theorem 1.4, we will also require the following result, which
addresses the injectivity of the generalized momentum ray transforms applied
to a sum of tensor fields of order one and two, see [51, Lemma 6.7] for a
similar result. The proof of this result will be presented in Appendix B.3

######  Lemma 2.8.

Let $n\geq 3$, $e_{1}=(1,0,\dots,0)\in\mathbb{R}^{n}$, $f^{(2)}\in
C_{0}(\mathbb{R}^{n-1},S^{2}(\mathbb{R}^{n}))$, and $f^{(1)}\in
C_{0}(\mathbb{R}^{n-1},S^{1}(\mathbb{R}^{n}))$. Assume that

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}}t^{2+k}$ $\displaystyle
f^{(2)}_{i_{1}i_{2}}(y_{0}^{\prime}+t\eta^{\prime})(e_{1}+i\eta)_{i_{1}}(e_{1}+i\eta)_{i_{2}}dt$
(2.26)
$\displaystyle+\sum_{i_{1}=1}^{n}\int_{\mathbb{R}}t^{1+k}f^{(1)}_{i_{1}}(y_{0}^{\prime}+t\eta^{\prime})(e_{1}+i\eta)_{i_{1}}dt=0,$

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$, all
$\eta=(\eta_{1},\eta^{\prime})\in\mathbb{R}^{n}$ such that $e_{1}\cdot\eta=0$
and $|\eta|=1$, and $k=0,1$. Then

$f^{(2)}=i_{\delta}v^{(0)},\quad f^{(1)}=0,$

where $v^{(0)}=\frac{1}{n}j_{\delta}f^{(2)}\in
C_{0}(\mathbb{R}^{n-1},S^{0}(\mathbb{R}^{n}))$.

We have the following immediate consequence of Lemma 2.8.

######  Corollary 2.9.

Assuming that all the conditions of Lemma 2.8 are met, and additionally, if
$j_{\delta}f^{(2)}=0$, then we have that both $f^{(2)}=0$ and $f^{(1)}=0$.

##  3\. Construction of biharmonic functions

Let $\Omega\subset\mathbb{R}^{n}$, $n\geq 3$, be a bounded open set with
$C^{\infty}$ boundary. In this section, we aim to construct complex geometric
optics (CGO) solutions to the biharmonic equation

$\Delta^{2}u=0\quad\text{in}\quad\Omega,$ (3.1)

which are necessary for proving Theorem 1.4. Although the construction follows
a standard approach as outlined in [31, 32, 14], we present it here to address
the need for solutions with enhanced regularity. Specifically, the constructed
CGO solutions should belong to $C^{4,\alpha}(\overline{\Omega})$ rather than
the commonly used $H^{4}(\Omega)$, see the integral identity (1.8) as well as
(5.5) and (5.14) below. Additionally, since the integral identity (1.8) in
Theorem 1.4 involves products of three biharmonic functions and their
derivatives up to the third order, it is advantageous to estimate the norms of
the derivatives of the remainders up to the third order in the
$L^{\infty}$-norm. Therefore, we need to construct CGO solutions that exhibit
small remainders measured in the $C^{3}(\overline{\Omega})$-norm instead of
the commonly used $H^{4}_{\text{scl}}(\Omega)$-norm in existing constructions.
Consequently, we proceed with the construction of CGO solutions to (3.1) that
possess $C^{5}(\overline{\Omega})$ regularity, accompanied by small remainders
in $C^{5}(\overline{\Omega})$.

Our starting point is the following Carleman estimate for the semiclassical
Laplacian, which is established in [27].

######  Proposition 3.1.

Let $\alpha\in\mathbb{R}^{n}\setminus\\{0\\}$ and let $\varphi(x)=\alpha\cdot
x$. Then given any $s\in\mathbb{R}$, we have for all $h>0$ small enough and
all $u\in C^{\infty}_{0}(\Omega)$,

$h\|u\|_{H^{s}_{\emph{\text{scl}}}(\mathbb{R}^{n})}\leq
C\|e^{\frac{\varphi}{h}}(-h^{2}\Delta)e^{-\frac{\varphi}{h}}u\|_{H^{s}_{\emph{{scl}}}(\mathbb{R}^{n})},\quad
C>0.$ (3.2)

Here $H^{s}(\mathbb{R}^{n})$, $s\in\mathbb{R}$, is the standard Sobolev space,
equipped with the semiclassical norm
$\|u\|_{H^{s}_{\text{scl}}(\mathbb{R}^{n})}=\|\langle
hD\rangle^{s}u\|_{L^{2}(\mathbb{R}^{n})}$, where
$\langle\xi\rangle=(1+|\xi|^{2})^{1/2}$.

Iterating the Carleman estimate (3.2) twice, we obtain the following Carleman
estimate for the biharmonic operator,

$h^{2}\|u\|_{H^{s}_{\text{scl}}(\mathbb{R}^{n})}\leq
C\|e^{\frac{\varphi}{h}}(-h^{2}\Delta)^{2}e^{-\frac{\varphi}{h}}u\|_{H^{s}_{\textrm{scl}}(\mathbb{R}^{n})},\quad
C>0,$ (3.3)

for all $u\in C^{\infty}_{0}(\Omega)$ and $h>0$ small enough.

Using a standard argument, we convert the Carleman estimate (3.3) into the
following solvability result, see [33, Proposition 2.3].

######  Proposition 3.2.

Let $\alpha\in\mathbb{R}^{n}\setminus\\{0\\}$, let $\varphi(x)=\alpha\cdot x$,
and let $s\in\mathbb{R}$. If $h>0$ is small enough, then for any $v\in
H^{s}(\Omega)$, there is a solution $u\in H^{s}(\Omega)$ of the equation

$e^{\frac{\varphi}{h}}(-h^{2}\Delta)^{2}e^{-\frac{\varphi}{h}}u=v\quad\text{in}\quad\Omega,$

which satisfies the bound

$\|u\|_{H^{s}_{\emph{\text{scl}}}(\Omega)}\leq\frac{C}{h^{2}}\|v\|_{H^{s}_{\emph{\text{scl}}}(\Omega)}.$

Here

$H^{s}(\Omega)=\\{V|_{\Omega}:V\in H^{s}(\mathbb{R}^{n})\\},\quad
s\in\mathbb{R},$

equipped with the norm

$\|v\|_{H^{s}_{\text{scl}}(\Omega)}=\inf_{V\in
H^{s}(\mathbb{R}^{n}),v=V|_{\Omega}}\|V\|_{H^{s}_{\text{scl}}(\mathbb{R}^{n})}.$

Let $l\in\mathbb{N}$. We shall construct CGO solutions to the equation (3.1)
in the form

$u(x;h)=e^{\frac{x\cdot\zeta}{h}}(a(x;h)+r(x;h)),$ (3.4)

where $\zeta\in\mathbb{C}^{n}$ such that $\zeta\cdot\zeta=0$,
$|\text{Re}\,\zeta|=|\text{Im}\,\zeta|\sim 1$, the amplitude $a\in
C^{\infty}(\overline{\Omega})$, and the remainder satisfies
$\|r\|_{H^{l}_{\text{scl}}(\Omega)}\to 0$ as $h\to 0$.

Now $u$ given by (3.4) is a solution to (3.1) provided that

$e^{-\frac{x\cdot\zeta}{h}}(-h^{2}\Delta)^{2}e^{\frac{x\cdot\zeta}{h}}r=-(-h^{2}\Delta-2\zeta\cdot
h\nabla)^{2}a\quad\text{in}\quad\Omega.$ (3.5)

Letting

$T=\zeta\cdot\nabla,$

(3.5) can be rewritten as follows,

$\displaystyle
e^{-\frac{x\cdot\text{Re}\,\zeta}{h}}(-h^{2}\Delta)^{2}e^{\frac{x\cdot\zeta}{h}}r$
$\displaystyle=-e^{\frac{ix\cdot\text{Im}\,\zeta}{h}}(h^{2}\Delta+2hT)^{2}a$
(3.6)
$\displaystyle=-e^{\frac{ix\cdot\text{Im}\,\zeta}{h}}(h^{4}\Delta^{2}+4h^{3}T\Delta+4h^{2}T^{2})a\quad\text{in}\quad\Omega.$

We shall look for the amplitude in the form,

$a(x;h)=a_{0}(x)+ha_{1}(x)+\dots+h^{k}a_{k}(x),$

for some $k\in\mathbb{N}$, where $a_{0}\in C^{\infty}(\mathbb{R}^{n})$
satisfies the first transport equation,

$T^{2}a_{0}=0\quad\text{in}\quad\mathbb{R}^{n},$ (3.7)

and $a_{j}\in C^{\infty}(\overline{\Omega})$ solves the $j$th transport
equation,

$T^{2}a_{j}=-T\Delta
a_{j-1}-\frac{1}{4}\Delta^{2}a_{j-2}\quad\text{in}\quad\Omega,$ (3.8)

$j=1,2,\dots,k$. Here we set $a_{-1}=0$.

In the proof of Theorem 1.4 we shall need CGO solutions (3.1) of the form
(3.4) with $\zeta=\mu(e_{1}+i\eta_{2})$, where $0\neq\mu\in\mathbb{R}$. Here
$e_{1}=(1,0,\dots,0)\in\mathbb{R}^{n}$ and $\eta_{2}\in\mathbb{R}^{n}$ is such
that $e_{1}\cdot\eta_{2}=0$ and $|\eta_{2}|=1$. In this case, the first
transport equation (3.7) becomes

$((e_{1}+i\eta_{2})\cdot\nabla)^{2}a_{0}=0\quad\text{in}\quad\mathbb{R}^{n}.$
(3.9)

We shall proceed to solve (3.9). To that end, we complete $e_{1}$ and
$\eta_{2}$ to an orthonormal basis in $\mathbb{R}^{n}$ and denote it by
$\\{e_{1},\eta_{2},\eta_{3},\dots,\eta_{n}\\}$. We denote by
$y=(y_{1},y_{2},\dots,y_{n})$, $y_{1}=x_{1}$, the coordinates with respect to
this basis. In these coordinates (3.9) becomes

$(\partial_{y_{1}}+i\partial_{y_{2}})^{2}a_{0}=0.$ (3.10)

Letting $z=y_{1}+iy_{2}$ be a complex variable,
$\partial_{\bar{z}}=\frac{1}{2}(\partial_{y_{1}}+i\partial_{y_{2}})$, and
making a change of coordinates $y\mapsto(z,y^{\prime\prime})$,
$y^{\prime\prime}=(y_{3},\dots,y_{n})$, the equation (3.10) reduces to

$\partial_{\bar{z}}^{2}a_{0}=0.$ (3.11)

Functions $a_{0}(z)$ that satisfy (3.11) are called polyanalytic of order two,
see [10]. One can easily see that any solution of (3.11) in $\mathbb{C}$ has
the form,

$a_{0}(z)=\bar{z}f_{1}(z)+f_{2}(z),$

where $f_{1}$ and $f_{2}$ are holomorphic in $\mathbb{C}$, see [10, Page 203].
Thus, in particular, the function

$\big{(}(2i)^{-1}(z-\bar{z})-c\big{)}^{k}f(z)g(y^{\prime\prime}),$ (3.12)

satisfies (3.11). Here $k=0,1$, $c\in\mathbb{R}$, $f$ is holomorphic in
$\mathbb{C}$, and $g\in C^{\infty}(\mathbb{R}^{n-2})$. In view of (3.12), in
the proof of Theorem 1.4 we shall work with solutions $a_{0}\in
C^{\infty}(\mathbb{R}^{n})$ to (3.10) in the form,

$a_{0}(y_{1},y_{2},y^{\prime\prime})=(y_{2}-c)^{k}f(y_{1}+iy_{2})g(y^{\prime\prime}).$
(3.13)

The particular form (3.13) of solutions $a_{0}$ of the first transport
equation (3.10) will be helpful to get the generalized momentum ray transforms
of unknown perturbations of $\Delta^{2}$, see [14, 51] for the same choice of
the amplitude.

We shall now proceed to solve the $j$th transport equation (3.8), assuming
that we have already found $a_{0},a_{1},\dots,a_{j-1}\in
C^{\infty}(\overline{\Omega})$ for $j=1,\dots,k$. Making the change of
coordinates $x\mapsto(z,y^{\prime\prime})$, (3.8) becomes

$\partial_{\bar{z}}^{2}a_{j}=f\quad\text{in}\quad\Omega,$ (3.14)

where $f\in C^{\infty}(\overline{\Omega})$ is given. As in [31], in order to
solve (3.14) we first find $v\in C^{\infty}(\overline{\Omega})$ which
satisfies

$\partial_{\bar{z}}v=f\quad\text{in}\quad\Omega,$ (3.15)

and then solve

$\partial_{\bar{z}}a_{j}=v\quad\text{in}\quad\Omega.$ (3.16)

Note that (3.15) can be solved by applying the Cauchy transform, i.e.

$v(z,y^{\prime\prime})=\frac{1}{\pi}\int_{\mathbb{C}}\frac{\chi(z-\zeta,y^{\prime\prime})f(z-\zeta,y^{\prime\prime})}{\zeta}d\text{Re}\,\zeta
d\text{Im}\,\zeta,$

where $\chi\in C^{\infty}_{0}(\mathbb{R}^{n})$ is such that $\chi=1$ near
$\overline{\Omega}$. The equation (3.16) can be solved in a similar way and
therefore, (3.14) and hence, (3.8) are solvable globally near
$\overline{\Omega}$ with a solution $a_{j}\in C^{\infty}(\overline{\Omega})$.

Having chosen the amplitudes $a_{0},a_{1},\dots,a_{k}\in
C^{\infty}(\overline{\Omega})$, we get from (3.6) that

$e^{-\frac{\mu x_{1}}{h}}(-h^{2}\Delta)^{2}e^{\frac{\mu
x_{1}}{h}}(e^{\frac{i\mu x\cdot\eta_{2}}{h}}r)=-e^{\frac{i\mu
x\cdot\eta_{2}}{h}}h^{k+3}(h\Delta^{2}a_{k}+4T\Delta
a_{k})\quad\text{in}\quad\Omega.$ (3.17)

Let $l\in\mathbb{N}$. Then Proposition 3.2 implies that for all $h>0$ small
enough, there is $r\in H^{l}(\Omega)$ satisfying (3.17) such that

$\|r\|_{H^{l}_{\text{scl}}(\Omega)}=\mathcal{O}(h^{k+1}).$

Here we used that
$\|r\|_{H^{l}_{\text{scl}}(\Omega)}^{2}\sim\sum_{|\alpha|\leq
l}\|(hD)^{\alpha}r\|^{2}_{L^{2}(\Omega)}$. Summing up, we have the following
result.

######  Proposition 3.3.

Let $\mu\in\mathbb{R}\setminus\\{0\\}$,
$e_{1}=(1,0,\dots,0)\in\mathbb{R}^{n}$, and $\eta_{2}\in\mathbb{R}^{n}$ be
such that $e_{1}\cdot\eta_{2}=0$ and $|\eta_{2}|=1$. Let $k,l\in\mathbb{N}$.
Then for all $h>0$ small enough, there is $u\in H^{l}(\Omega)$ satisfying
$\Delta^{2}u=0$ in $\Omega$ of the form,

$u(x;h)=e^{\frac{\mu
x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}(x)+ha_{1}(x)+\dots+h^{k}a_{k}(x)+r(x;h)).$

Here $a_{0}\in C^{\infty}(\mathbb{R}^{n})$ satisfies the first transport
equation (3.9), and $a_{j}\in C^{\infty}(\overline{\Omega})$ satisfies the
$j$th transport equation (3.8), $j=1,\dots,k$, and $r\in H^{l}(\Omega)$ such
that $\|r\|_{H^{l}_{\text{scl}}(\Omega)}=\mathcal{O}(h^{k+1})$, as $h\to 0$.

Now

$\|r\|_{H^{l}(\Omega)}\leq
h^{-l}\|r\|_{H^{l}_{\text{scl}}(\Omega)}=\mathcal{O}(h^{k+1-l}).$

Taking $l>n/2+5$ and using the Sobolev embedding $H^{l}(\Omega)\subset
C^{5}(\overline{\Omega})$, see [2, Theorem 5.1.4], we see that $r\in
C^{5}(\overline{\Omega})$ with

$\|r\|_{C^{5}(\overline{\Omega})}=\mathcal{O}(h^{k+1-l}),$ (3.18)

as $h\to 0$. Here

$\|r\|_{C^{5}(\overline{\Omega})}=\sum_{|\alpha|\leq
5}\sup_{\overline{\Omega}}|D^{\alpha}r|.$

Letting $k=l$, in view of (3.18), we get the following consequence of
Proposition 3.3.

######  Corollary 3.4.

Let $\mu\in\mathbb{R}\setminus\\{0\\}$,
$e_{1}=(1,0,\dots,0)\in\mathbb{R}^{n}$, and $\eta_{2}\in\mathbb{R}^{n}$ be
such that $e_{1}\cdot\eta_{2}=0$ and $|\eta_{2}|=1$. Then for all $h>0$ small
enough, there is $u\in C^{5}(\overline{\Omega})$ satisfying $\Delta^{2}u=0$ in
$\Omega$ of the form,

$u(x;h)=e^{\frac{\mu x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}(x)+r(x;h)).$ (3.19)

Here $a_{0}\in C^{\infty}(\mathbb{R}^{n})$ satisfies the first transport
equation (3.9), and $r\in C^{5}(\overline{\Omega})$ such that
$\|r\|_{C^{5}(\overline{\Omega})}=\mathcal{O}(h)$, as $h\to 0$.

######  Remark 3.5.

In the proof of Theorem 1.4, to obtain the generalized momentum ray transforms
of unknown tensorial perturbations of $\Delta^{2}$, we shall work with $a_{0}$
in (3.19) given by

$a_{0}(y_{1},y_{2},y^{\prime\prime})=(y_{2}-c)^{k}f(z)g(y^{\prime\prime}),$

where $k=0,1$, $c\in\mathbb{R}$, $f$ is holomorphic in $\mathbb{C}$, and $g\in
C^{\infty}(\mathbb{R}^{n-2})$, see (3.13). Here $y=(y_{1},y_{2},\dots,y_{n})$,
$y_{1}=x_{1}$, are the coordinates with respect to the orthonormal basis
$\\{e_{1},\eta_{2},\eta_{3},\dots,\eta_{n}\\}$, and $z=y_{1}+iy_{2}$. We refer
to [14, 51] where a similar choice of the amplitude was made when solving
inverse problems.

In the proof of Theorem 1.4, to recover the Fourier transform of $q$, we will
require the following CGO solutions to biharmonic equations, constructed in
the same way as before.

######  Proposition 3.6.

Let $\alpha,\beta\in\mathbb{R}^{n}$ be such that $\alpha\cdot\beta=0$ and
$|\alpha|=|\beta|=1$. Then for all $h>0$ small enough, there is $u\in
C^{5}(\overline{\Omega})$ satisfying $\Delta^{2}u=0$ in $\Omega$ of the form,

$u(x;h)=e^{\frac{x\cdot(\alpha+i\beta)}{h}}(a_{0}(x)+r(x;h)),$ (3.20)

where $a_{0}\in C^{\infty}(\mathbb{R}^{n})$ satisfies
$((\alpha+i\beta)\cdot\nabla)^{2}a_{0}=0$ in $\mathbb{R}^{n}$. Here
$\|r\|_{C^{5}(\overline{\Omega})}=\mathcal{O}(h)$, as $h\to 0$.

##  4\. Proof of Theorem 1.4

We shall start to examine the integral identity (1.8) by employing biharmonic
functions constructed in Corollary 3.4. To that end, following [14, Theorem
1.1], we carefully select appropriate amplitudes as discussed in Remark 3.5.
This approach allows us to recover specific generalized momentum ray
transforms on tensor fields and the sum of tensor fields of different ranks.
Leveraging the injectivity results of Lemma 2.6 and Lemma 2.8 for the obtained
generalized momentum ray transforms, we shall show that $A^{(j)}=0$ in
$\Omega$, $j=1,2,3$. To demonstrate that $q=0$, we shall utilize the
biharmonic functions from Proposition 3.6 to recover the Fourier transform of
$q$.

In doing so, we first extend $A^{(j)}$ by zero to
$\mathbb{R}^{n}\setminus\overline{\Omega}$ and denote the extension again by
$A^{(j)}$, $j=1,2,3$. We see that $A^{(j)}\in C_{0}(\mathbb{R}^{n};S^{j})$ as
$A^{(j)}|_{\partial\Omega}=0$, $j=1,2,3$. Let
$e_{1}=(1,0,\dots,0)\in\mathbb{R}^{n}$ and let $\eta_{2}\in\mathbb{R}^{n}$ be
such that $e_{1}\cdot\eta_{2}=0$ and $|\eta_{2}|=1$. Thus,
$\eta_{2}=(0,\eta_{2}^{\prime})$, $\eta_{2}^{\prime}\in\mathbb{R}^{n-1}$,
$|\eta_{2}^{\prime}|=1$. By Corollary 3.4, there are
$v^{(0)},v^{(1)},v^{(2)}\in C^{5}(\overline{\Omega})$ satisfying
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$ of the form

$\displaystyle v^{(0)}$
$\displaystyle=e^{\frac{-2x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}^{(0)}(x)+r^{(0)}(x;h)),$
(4.1) $\displaystyle v^{(1)}$
$\displaystyle=e^{\frac{x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}^{(1)}(x)+r^{(1)}(x;h)),\quad
v^{(2)}=v^{(1)}.$

Here $a_{0}^{(l)}\in C^{\infty}(\mathbb{R}^{n})$ satisfies (3.9), and

$\|r^{(l)}\|_{C^{5}(\overline{\Omega})}=\mathcal{O}(h),$ (4.2)

as $h\to 0$, $l=0,1$.

Substituting $v^{(0)},v^{(1)},v^{(2)}$ given by (4.1) in the integral identity
(1.8), multiplying by $h^{3}$, letting $h\to 0$, and using (4.2), we obtain
that

$\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\Omega}A^{(3)}_{i_{1}i_{2}i_{3}}(x)(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}(a_{0}^{(1)}(x))^{2}a_{0}^{(0)}(x)dx=0.$
(4.3)

Here $(e_{1}+i\eta_{2})_{i_{j}}$ is the $i_{j}$th component of the vector
$e_{1}+i\eta_{2}$ in the standard basis in $\mathbb{R}^{n}$. Now we have the
following decomposition

$A^{(3)}=B^{(3)}+i_{\delta}W^{(1)},$ (4.4)

where $W^{(1)}\in C_{0}(\mathbb{R}^{n};S^{1})$, $B^{(3)}\in
C_{0}(\mathbb{R}^{n};S^{3})$ is such that

$j_{\delta}B^{(3)}=0,$ (4.5)

see [18, Lemma 2.3] as well as the beginning of the proof of Lemma 2.6 in
Appendix B.2. Since $(e_{1}+i\eta_{2})\cdot(e_{1}+i\eta_{2})=0$, we have

$\sum_{i_{1},i_{2},i_{3}=1}^{n}(i_{\delta}W^{(1)})_{i_{1}i_{2}i_{3}}(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}=0.$
(4.6)

Substituting (4.4) into (4.3) and using (4.6), we get

$\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}^{n}}B^{(3)}_{i_{1}i_{2}i_{3}}(x)(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}(a_{0}^{(1)}(x))^{2}a_{0}^{(0)}(x)dx=0.$
(4.7)

We shall proceed to show that $B^{(3)}=0$. In doing so, we shall complete
$e_{1}$ and $\eta_{2}$ to an orthonormal basis
$\\{e_{1},\eta_{2},\eta_{3},\dots,\eta_{n}\\}$ in $\mathbb{R}^{n}$, and write
$y=(x_{1},y_{2},y^{\prime\prime})$, $y^{\prime\prime}=(y_{3},\dots,y_{n})$,
for the coordinates with respect to this basis. According to Remark 3.5, we
choose

$a_{0}^{(1)}(x_{1},y_{2},y^{\prime\prime})=y_{2}-c,\quad
a_{0}^{(0)}(x_{1},t,y^{\prime\prime})=(y_{2}-c)e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime}),$
(4.8)

where $c\in\mathbb{R}$, $\lambda\in\mathbb{R}$, and $g\in
C^{\infty}(\mathbb{R}^{n-2})$. Writing out the integral in (4.7) in
$y$–coordinates, and substituting (4.8), we get

$\displaystyle\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}^{n}}B^{(3)}_{i_{1}i_{2}i_{3}}(x_{1},y_{2},y^{\prime\prime})(e_{1}+i\eta_{2})_{i_{1}}$
$\displaystyle(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}$ (4.9)
$\displaystyle(y_{2}-c)^{3}e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime})dx_{1}dy_{2}dy^{\prime\prime}=0.$

We get from (4.9) that

$\displaystyle\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}^{n-1}}\widehat{B^{(3)}_{i_{1}i_{2}i_{3}}}(\lambda,y_{2},y^{\prime\prime})(e_{1}+i\eta_{2})_{i_{1}}$
$\displaystyle(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}$ (4.10)
$\displaystyle(y_{2}-c)^{3}e^{\lambda
y_{2}}g(y^{\prime\prime})dy_{2}dy^{\prime\prime}=0,$

for all $c\in\mathbb{R}$, $\lambda\in\mathbb{R}$, and $g\in
C^{\infty}(\mathbb{R}^{n-2})$. Here

$\widehat{B^{(3)}}(\lambda,y^{\prime})=\int_{\mathbb{R}}e^{-i\lambda
x_{1}}B^{(3)}(x_{1},y^{\prime})dx_{1}$

is the Fourier transform of $B^{(3)}$ with respect to $x_{1}$, and
$y^{\prime}=(y_{2},\dots,y_{n})\in\mathbb{R}^{n-1}$. We observe that the
function $x_{1}\mapsto B^{(3)}(x_{1},y^{\prime})$ is of class
$C_{0}(\mathbb{R};S^{3}(\mathbb{R}^{n}))$ for any
$y^{\prime}\in\mathbb{R}^{n-1}$. Furthermore, it can be easily verified that
the function $y^{\prime}\mapsto\widehat{B^{(3)}}(\lambda,y^{\prime})$ is of
class $C_{0}(\mathbb{R}^{n-1};S^{3}(\mathbb{R}^{n}))$ for any
$\lambda\in\mathbb{R}$. Based on these observations, one can easily check
that, for any $\lambda\in\mathbb{R}$ and $c\in\mathbb{R}$, the integral with
respect to $y_{2}$ in (4.10) is a continuous function of
$y^{\prime\prime}\in\mathbb{R}^{n-2}$. Now since $g\in
C^{\infty}(\mathbb{R}^{n-2})$ is arbitrary, we conclude from (4.10) that

$\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}}\widehat{B^{(3)}_{i_{1}i_{2}i_{3}}}(\lambda,y_{2},y^{\prime\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}(y_{2}-c)^{3}e^{\lambda
y_{2}}dy_{2}=0,$ (4.11)

for all $y^{\prime\prime}\in\mathbb{R}^{n-1}$, all $c\in\mathbb{R}$, and all
$\lambda\in\mathbb{R}$.

Making the change of variables $t=y_{2}-c$ in the integral in (4.11), we
obtain that

$\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}}t^{3}\widehat{B^{(3)}_{i_{1}i_{2}i_{3}}}(\lambda,y_{0}^{\prime}+t\eta^{\prime}_{2})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}e^{\lambda
t}dt=0,$ (4.12)

for all $y_{0}^{\prime}=(c,y^{\prime\prime})\in\mathbb{R}^{n-1}$, and all
$\lambda\in\mathbb{R}$. Now, since the function $x_{1}\mapsto
B^{(3)}(x_{1},y^{\prime})$ is of class
$C_{0}(\mathbb{R};S^{3}(\mathbb{R}^{n}))$ for any
$y^{\prime}\in\mathbb{R}^{n-1}$, we have that the function
$\lambda\mapsto\widehat{B^{(3)}}(\lambda,y^{\prime})$ is of class
$C^{\infty}(\mathbb{R};S^{3}(\mathbb{R}^{n}))$ for any
$y^{\prime}\in\mathbb{R}^{n-1}$.

Letting $\lambda=0$ in (4.12), we get

$\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}}t^{3}\widehat{B^{(3)}_{i_{1}i_{2}i_{3}}}(0,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}dt=0,$
(4.13)

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$, and all
$\eta_{2}\in\mathbb{R}^{n}$ such that $e_{1}\cdot\eta_{2}=0$ and
$|\eta_{2}|=1$. Here $\widehat{B^{(3)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{3}(\mathbb{R}^{n}))$ and
$j_{\delta}(\widehat{B^{(3)}}(0,\cdot))=0$ in view of (4.5) and the fact that
$j_{\delta}$ commutes with the Fourier transform. An application of Corollary
2.7 to (4.13) shows that

$\widehat{B^{(3)}}(0,\cdot)=0.$ (4.14)

Differentiation (4.12) with respect to $\lambda$, letting $\lambda=0$, and
using (4.14), we obtain that

$\displaystyle\sum_{i_{1},i_{2},i_{3}=1}^{n}$
$\displaystyle\int_{\mathbb{R}}t^{3}\partial_{\lambda}\widehat{B^{(3)}_{i_{1}i_{2}i_{3}}}(0,y_{0}^{\prime}+t\eta^{\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(e_{1}+i\eta_{2})_{i_{3}}dt=0,$
(4.15)

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$, and all
$\eta_{2}\in\mathbb{R}^{n}$ such that $e_{1}\cdot\eta_{2}=0$ and
$|\eta_{2}|=1$. Here $\partial_{\lambda}\widehat{B^{(3)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{3}(\mathbb{R}^{n}))$ and
$j_{\delta}(\partial_{\lambda}\widehat{B^{(3)}}(0,\cdot))=0$ in view of (4.5).

Applying Corollary 2.7 to (4.15), we conclude that
$\partial_{\lambda}\widehat{B^{(3)}}(0,\cdot)=0$. Continuing in the same
fashion, we show that $\partial_{\lambda}^{k}\widehat{B^{(3)}}(0,\cdot)=0$ for
all $k=0,1,2,\dots$.

Since $B^{(3)}(x_{1},\cdot)$ is compactly supported in $x_{1}$, the function
$\lambda\mapsto\widehat{B^{(3)}}(\lambda,\cdot)$ is analytic. Therefore, we
can deduce that $\widehat{B^{(3)}}(\lambda,\cdot)=0$ for all
$\lambda\in\mathbb{R}$. Consequently, $B^{(3)}=0$, and it follows from (4.4)
that

$A^{(3)}=i_{\delta}W^{(1)}.$ (4.16)

Let us now return to the integral identity (1.8) with $A^{(3)}$ given by
(4.16) and substitute $v^{(0)},v^{(1)},v^{(2)}$ given by (4.1) in (1.8). By
multiplying (1.8) by $h^{2}$, letting $h\to 0$, and utilizing (4.6), we obtain
that

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\Omega}A^{(2)}_{i_{1}i_{2}}(x)(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(a_{0}^{(1)}(x))^{2}a_{0}^{(0)}(x)dx+\frac{3}{i}\times$
(4.17)
$\displaystyle\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\Omega}(i_{\delta}W^{(1)})_{i_{1}i_{2}i_{3}}(x)(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(\partial_{i_{3}}a_{0}^{(1)}(x))a_{0}^{(1)}(x)a_{0}^{(0)}(x)dx=0.$

Using (2.4), (2.3), and the fact that
$(e_{1}+i\eta_{2})\cdot(e_{1}+i\eta_{2})=0$, we get from (4.17) that

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}^{n}}A^{(2)}_{i_{1}i_{2}}(x)(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(a_{0}^{(1)}(x))^{2}a_{0}^{(0)}(x)dx$
(4.18)
$\displaystyle+\frac{2}{i}\sum_{i_{1}=1}^{n}\int_{\mathbb{R}^{n}}W^{(1)}_{i_{1}}(x)(e_{1}+i\eta_{2})_{i_{1}}((e_{1}+i\eta_{2})\cdot\nabla
a_{0}^{(1)}(x))a_{0}^{(1)}(x)a_{0}^{(0)}(x)dx=0.$

Now we express $A^{(2)}$ as

$A^{(2)}=B^{(2)}+i_{\delta}W^{(0)},$ (4.19)

where $W^{(0)}\in C_{0}(\mathbb{R}^{n};S^{0})$, $B^{(2)}\in
C_{0}(\mathbb{R}^{n};S^{2})$ is such that

$j_{\delta}B^{(2)}=0,$ (4.20)

see [18, Lemma 2.3] as well as the proof of Lemma 2.6 in Appendix B.2 in the
case $m=2$. Thanks to the fact that
$(e_{1}+i\eta_{2})\cdot(e_{1}+i\eta_{2})=0$, we get

$\sum_{i_{1},i_{2}=1}^{n}(i_{\delta}W^{(0)})_{i_{1}i_{2}}(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}=0.$
(4.21)

By substituting (4.19) into (4.18) and using (4.21), we obtain

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}^{n}}B^{(2)}_{i_{1}i_{2}}(x)(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}(a_{0}^{(1)}(x))^{2}a_{0}^{(0)}(x)dx$
(4.22)
$\displaystyle+\frac{2}{i}\sum_{i_{1}=1}^{n}\int_{\mathbb{R}^{n}}W^{(1)}_{i_{1}}(x)(e_{1}+i\eta_{2})_{i_{1}}((e_{1}+i\eta_{2})\cdot\nabla
a_{0}^{(1)}(x))a_{0}^{(1)}(x)a_{0}^{(0)}(x)dx=0.$

As before, we shall work in the coordinates
$y=(x_{1},y_{2},y^{\prime\prime})$, $y^{\prime\prime}=(y_{3},\dots,y_{n})$,
with respect to the orthonormal basis
$\\{e_{1},\eta_{2},\eta_{3},\dots,\eta_{n}\\}$ in $\mathbb{R}^{n}$. According
to Remark 3.5, we choose

$a_{0}^{(1)}(x_{1},y_{2},y^{\prime\prime})=y_{2}-c,\quad
a_{0}^{(0)}(x_{1},t,y^{\prime\prime})=(y_{2}-c)^{l}e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime}),$
(4.23)

where $c\in\mathbb{R}$, $\lambda\in\mathbb{R}$, $g\in
C^{\infty}(\mathbb{R}^{n-2})$, and $l=0,1$.

Writing out the integral in (4.22) in $y$–coordinates, substituting (4.23),
and using that $(e_{1}+i\eta_{2})\cdot\nabla a_{0}^{(1)}=i$, we obtain that

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}^{n}}B^{(2)}_{i_{1}i_{2}}(x_{1},y_{2},y^{\prime\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}$
(4.24)
$\displaystyle\quad\quad\quad\quad\quad(y_{2}-c)^{2+l}e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime})dx_{1}dy_{2}dy^{\prime\prime}$
$\displaystyle+2\sum_{i_{1}=1}^{n}\int_{\mathbb{R}^{n}}W^{(1)}_{i_{1}}(x_{1},y_{2},y^{\prime\prime})(e_{1}+i\eta_{2})_{i_{1}}(y_{2}-c)^{1+l}e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime})dx_{1}dy_{2}dy^{\prime\prime}=0.$

Since $B^{(2)}\in C_{0}(\mathbb{R}^{n};S^{2})$, $W^{(1)}\in
C_{0}(\mathbb{R}^{n};S^{1})$, and considering that $g\in
C^{\infty}(\mathbb{R}^{n-2})$ is arbitrary, we can argue as above and deduce
from (4.24) that

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}}t^{2+l}\widehat{B^{(2)}_{i_{1}i_{2}}}(\lambda,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}e^{\lambda
t}dt$ (4.25)
$\displaystyle+2\sum_{i_{1}=1}^{n}\int_{\mathbb{R}}t^{1+l}\widehat{W^{(1)}_{i_{1}}}(\lambda,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}e^{\lambda
t}dt=0,$

for all $y_{0}^{\prime}=(c,y^{\prime\prime})\in\mathbb{R}^{n-1}$, all
$\lambda\in\mathbb{R}$, and all $l=0,1$. Here we have made the change of
variables $t=y_{2}-c$.

Letting $\lambda=0$, (4.25) becomes

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}}t^{2+l}\widehat{B^{(2)}_{i_{1}i_{2}}}(0,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}dt$
(4.26)
$\displaystyle+2\sum_{i_{1}=1}^{n}\int_{\mathbb{R}}t^{1+l}\widehat{W^{(1)}_{i_{1}}}(0,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}dt=0,$

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$, $l=0,1$, and all
$\eta_{2}\in\mathbb{R}^{n}$ such that $\eta_{2}\cdot e_{1}=0$ and
$|\eta_{2}|=1$. Here $\widehat{B^{(2)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{2}(\mathbb{R}^{n}))$ is such that
$j_{\delta}(\widehat{B^{(2)}}(0,\cdot))=0$ thanks to (4.20), and
$\widehat{W^{(1)}}(0,\cdot)\in C_{0}(\mathbb{R}^{n-1};S^{1}(\mathbb{R}^{n}))$.

Applying Corollary 2.9 to (4.26), we conclude that

$\widehat{B^{(2)}}(0,\cdot)=0,\quad\widehat{W^{(1)}}(0,\cdot)=0.$ (4.27)

Differentiating (4.25) with respect to $\lambda$, letting $\lambda=0$, and
using (4.27), we get

$\displaystyle\sum_{i_{1},i_{2}=1}^{n}\int_{\mathbb{R}}t^{2+l}\partial_{\lambda}\widehat{B^{(2)}_{i_{1}i_{2}}}(0,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}(e_{1}+i\eta_{2})_{i_{2}}dt$
(4.28)
$\displaystyle+2\sum_{i_{1}=1}^{n}\int_{\mathbb{R}}t^{1+l}\partial_{\lambda}\widehat{W^{(1)}_{i_{1}}}(0,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}dt=0,$

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$, $l=0,1$, and all
$\eta_{2}\in\mathbb{R}^{n}$ such that $\eta_{2}\cdot e_{1}=0$ and
$|\eta_{2}|=1$. Here $\partial_{\lambda}\widehat{B^{(2)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{2}(\mathbb{R}^{n}))$ is such that
$j_{\delta}(\partial_{\lambda}\widehat{B^{(2)}}(0,\cdot))=0$ thanks to (4.20),
and $\partial_{\lambda}\widehat{W^{(1)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{1}(\mathbb{R}^{n}))$. An application of Corollary
2.9 to (4.28) shows that $\partial_{\lambda}\widehat{B^{(2)}}(0,\cdot)=0$, and
$\partial_{\lambda}\widehat{W^{(1)}}(0,\cdot)=0$. Continuing in the same
fashion, we see that $\partial_{\lambda}^{k}\widehat{B^{(2)}}(0,\cdot)=0$ and
$\partial_{\lambda}^{k}\widehat{W^{(1)}}(0,\cdot)=0$, for every
$k=0,1,2,\dots$.

As $B^{(2)}(x_{1},\cdot)$ and $W^{(1)}(x_{1},\cdot)$ are compactly supported
in $x_{1}$, the functions $\lambda\mapsto\widehat{B^{(2)}}(\lambda,\cdot)$ and
$\lambda\mapsto\widehat{W^{(1)}}(\lambda,\cdot)$ are analytic. Thus, we get
$\widehat{B^{(2)}}(\lambda,\cdot)=0$ and $\widehat{W^{(1)}}(\lambda,\cdot)=0$
for all $\lambda\in\mathbb{R}$. Hence, $B^{(2)}=0$ and $W^{(1)}=0$. Therefore,
it follows from (4.16) that $A^{(3)}=0$, completing the recovery of $A^{(3)}$.
We also conclude from (4.19) that

$A^{(2)}=i_{\delta}W^{(0)}.$ (4.29)

Let us now return to the integral identity (1.8) with $A^{(3)}=0$ and
$A^{(2)}$ from (4.29). Now it would be convenient to work with three different
biharmonic functions $v^{(0)},v^{(1)},v^{(2)}\in C^{5}(\overline{\Omega})$
given by

$\displaystyle v^{(0)}$
$\displaystyle=e^{\frac{-2x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}^{(0)}(x)+r^{(0)}(x;h)),$
(4.30) $\displaystyle v^{(1)}$
$\displaystyle=e^{\frac{x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}^{(1)}(x)+r^{(1)}(x;h)),\quad
v^{(2)}=e^{\frac{x\cdot(e_{1}+i\eta_{2})}{h}}(a_{0}^{(2)}(x)+r^{(2)}(x;h)),$

where $a_{0}^{(l)}\in C^{\infty}(\mathbb{R}^{n})$ satisfying (3.9), and

$\|r^{(l)}\|_{C^{5}(\overline{\Omega})}=\mathcal{O}(h),$ (4.31)

as $h\to 0$, $l=0,1,2$.

We will now substitute $v^{(0)},v^{(1)},v^{(2)}$ given by (4.30) into the
integral identity (1.8) with $A^{(3)}=0$ and $A^{(2)}$ from (4.29). By
utilizing (4.21) and (4.31), multiplying (1.8) by $h$ and letting $h$ tend to
$0$, we obtain that

$\displaystyle\sum_{i_{1}=1}^{n}\int_{\Omega}A^{(1)}_{i_{1}}(x)(e_{1}+i\eta_{2})_{i_{1}}a_{0}^{(1)}(x)a_{0}^{(2)}(x)a_{0}^{(0)}(x)dx$
(4.32)
$\displaystyle+\frac{1}{i}\sum_{i_{1},i_{2}=1}^{n}\int_{\Omega}(i_{\delta}W^{(0)})_{i_{1}i_{2}}(x)(e_{1}+i\eta_{2})_{i_{1}}\partial_{i_{2}}(a_{0}^{(1)}(x)a_{0}^{(2)}(x))a_{0}^{(0)}(x)dx=0.$

By using (2.4) and (2.3), we can deduce from (4.32) that

$\displaystyle\sum_{i_{1}=1}^{n}\int_{\mathbb{R}^{n}}A^{(1)}_{i_{1}}(x)(e_{1}+i\eta_{2})_{i_{1}}a_{0}^{(1)}(x)a_{0}^{(2)}(x)a_{0}^{(0)}(x)dx$
(4.33)
$\displaystyle+\frac{1}{i}\int_{\mathbb{R}^{n}}W^{(0)}(x)a_{0}^{(0)}(x)(e_{1}+i\eta_{2})\cdot\nabla(a_{0}^{(1)}(x)a_{0}^{(2)}(x))dx=0.$

As before, we shall consider an orthonormal basis
$\\{e_{1},\eta_{2},\eta_{3},\dots,\eta_{n}\\}$ in $\mathbb{R}^{n}$, and denote
the coordinates in this basis by $y=(x_{1},y_{2},y^{\prime\prime})$,
$y^{\prime\prime}=(y_{3},\dots,y_{n})$. According to Remark 3.5, we choice

$a_{0}^{(1)}(x_{1},y_{2},y^{\prime\prime})=a_{0}^{(2)}(x_{1},y_{2},y^{\prime\prime})=1,\quad
a_{0}^{(0)}(x_{1},t,y^{\prime\prime})=(y_{2}-c)e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime}),$
(4.34)

where $c\in\mathbb{R}$, $\lambda\in\mathbb{R}$, and $g\in
C^{\infty}(\mathbb{R}^{n-2})$.

Writing out the integral in (4.33) in $y$–coordinates, and substituting
(4.34), we get

$\sum_{i_{1}=1}^{n}\int_{\mathbb{R}^{n}}A^{(1)}_{i_{1}}(x_{1},y_{1},y^{\prime\prime})(e_{1}+i\eta_{2})_{i_{1}}(y_{2}-c)e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime})dx_{1}dy_{2}dy^{\prime\prime}=0.$
(4.35)

By considering that $A^{(1)}\in C_{0}(\mathbb{R}^{n};S^{1})$ and the fact that
$g\in C^{\infty}(\mathbb{R}^{n-2})$ is an arbitrary function, and arguing as
before, we can deduce from (4.35) that

$\sum_{i_{1}=1}^{n}\int_{\mathbb{R}}t\widehat{A^{(1)}_{i_{1}}}(\lambda,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}e^{\lambda
t}dt=0,$ (4.36)

for all $y_{0}^{\prime}=(c,y^{\prime\prime})\in\mathbb{R}^{n-1}$ and all
$\lambda\in\mathbb{R}$. Letting $\lambda=0$, (4.36) becomes

$\sum_{i_{1}=1}^{n}\int_{\mathbb{R}}t\widehat{A^{(1)}_{i_{1}}}(0,y_{0}^{\prime}+t\eta_{2}^{\prime})(e_{1}+i\eta_{2})_{i_{1}}dt=0,$
(4.37)

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$, and all
$\eta_{2}\in\mathbb{R}^{n}$ such that $\eta_{2}\cdot e_{1}=0$ and
$|\eta_{2}|=1$. Here $\widehat{A^{(1)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{1}(\mathbb{R}^{n}))$ and therefore, an application
of Lemma 2.6 shows that $\widehat{A^{(1)}}(0,\cdot)=0$. Proceeding as before
by differentiating (4.36) with respect to $\lambda$, letting $\lambda=0$,
applying Lemma 2.6, and inverting the Fourier transform, we conclude that
$A^{(1)}=0$.

Inserting $A^{(1)}=0$ in (4.33), we get

$\int_{\mathbb{R}^{n}}W^{(0)}(x)a_{0}^{(0)}(x)(e_{1}+i\eta_{2})\cdot\nabla(a_{0}^{(1)}(x)a_{0}^{(2)}(x))dx=0.$
(4.38)

Writing out the integral in (4.38) in $y$–coordinates, and choosing

$a_{0}^{(1)}(x_{1},y_{2},y^{\prime\prime})=y_{2},\
a_{0}^{(2)}(x_{1},y_{2},y^{\prime\prime})=1,\
a_{0}^{(0)}(x_{1},t,y^{\prime\prime})=e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime}),$

where $\lambda\in\mathbb{R}$ and $g\in C^{\infty}(\mathbb{R}^{n-2})$, we
obtain from (4.38) that

$\int_{\mathbb{R}^{n}}W^{(0)}(x_{1},y_{2},y^{\prime\prime})e^{-i\lambda(x_{1}+iy_{2})}g(y^{\prime\prime})dx_{1}dy_{2}dy^{\prime\prime}=0.$
(4.39)

Since $W^{(0)}\in C_{0}(\mathbb{R}^{n};S^{0})$, and $g\in
C^{\infty}(\mathbb{R}^{n-2})$ is arbitrary, proceeding as previously
discussed, we can deduce from (4.39) that

$\int_{\mathbb{R}}\widehat{W^{(0)}}(\lambda,y_{0}^{\prime}+t\eta_{2}^{\prime})e^{\lambda
t}dt=0.$ (4.40)

This holds for all $y_{0}^{\prime}=(c,y^{\prime\prime})\in\mathbb{R}^{n-1}$,
all $\eta_{2}^{\prime}\in\mathbb{R}^{n-1}$ such that $|\eta_{2}^{\prime}|=1$,
and all $\lambda\in\mathbb{R}$. We have
$\partial_{\lambda}^{k}\widehat{W^{(0)}}(0,\cdot)\in
C_{0}(\mathbb{R}^{n-1};S^{0})$, $k=0,1,\dots$.

When setting $\lambda$ to zero in (4.40), we obtain
$(J^{0,0}\widehat{W^{(0)}}(0,\cdot))(y_{0}^{\prime},\eta_{2}^{\prime})=0$ for
all $(y_{0},\eta_{2}^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$. As a
consequence of Theorem 2.4, we can conclude that
$\widehat{W^{(0)}}(0,\cdot)=0$. Following the same strategy as before, which
involves differentiating (4.40) with respect to $\lambda$, setting $\lambda$
to zero, utilizing Theorem 2.4, and inverting the Fourier transform, we
conclude from (4.40) that $W^{(0)}=0$. Thus, it follows from (4.29) that
$A^{(2)}=0$.

Let us return to integral identity (1.8) while setting $A^{(1)}=0$,
$A^{(2)}=0$, and $A^{(3)}=0$, and taking $v^{(0)}=1$. We have

$\int_{\Omega}qv^{(1)}v^{(2)}\,dx=0,$ (4.41)

for all $v^{(1)},v^{(2)}\in C^{4,\alpha}(\overline{\Omega})$ solving
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=1,2$. To conclude that $q=0$, we
follow a standard argument, see [26, 32]. Let $\xi\in\mathbb{R}^{n}$ be
arbitrary, and let $\alpha,\beta\in\mathbb{R}^{n}$ be such that
$|\alpha|=|\beta|=1$ and $\xi\cdot\alpha=\xi\cdot\beta=\alpha\cdot\beta=0$.
Utilizing Proposition 3.6, we take

$v^{(1)}=e^{\frac{x\cdot(\alpha+i\beta)}{h}}(e^{-ix\cdot\xi}+r^{(1)}(x;h)),\quad
v^{(2)}=e^{\frac{-x\cdot(\alpha+i\beta)}{h}}(1+r^{(2)}(x;h)),$ (4.42)

where $\|r^{(l)}\|_{C^{5}(\overline{\Omega})}=\mathcal{O}(h)$ as $h\to 0$ for
$l=1,2$. Substituting $v^{(1)},v^{(2)}$ given by (4.42) into the integral
identity (4.41) and letting $h\to 0$, we obtain

$\int_{\Omega}qe^{-ix\cdot\xi}\,dx=0,$

for all $\xi\in\mathbb{R}^{n}$. Thus, $q=0$ in $\Omega$. This completes the
proof of Theorem 1.4.

##  5\. Proof of Theorem 1.1

In this section, it will be convenient for us to write $A^{(1),(j)}:=A^{(j)}$,
$q^{(1)}:=q$, and $A^{(2),(j)}:=\widetilde{A}^{(j)}$,
$q^{(2)}:=\widetilde{q}$, $j=1,2,3$. Let
$\varepsilon=(\varepsilon_{1},\dots,\varepsilon_{m})\in\mathbb{C}^{m}$, $m\geq
2$, and consider the Dirichlet problem (1.4) for
$L_{A^{(r),(1)},A^{(r),(2)},A^{(r),(3)},q^{(r)}}$ with

$f=\sum_{k=1}^{m}\varepsilon_{k}f_{k},\quad
g=\sum_{k=1}^{m}\varepsilon_{k}g_{k},\quad f_{k}\in
C^{4,\alpha}(\partial\Omega),\quad g_{k}\in C^{3,\alpha}(\partial\Omega),\quad
k=1,\dots,m,$

where $r=1,2$. By Theorem A.1, for all $|\varepsilon|$ sufficiently small, the
problem (1.4) has a unique small solution $u_{r}(\cdot,\varepsilon)\in
C^{4,\alpha}(\overline{\Omega})$, which depends holomorphically on
$\varepsilon\in\text{neigh}(0,\mathbb{C}^{m})$. We shall use an induction
argument on $m\geq 2$ to prove that the equality
$\Lambda_{A^{(1),(1)},A^{(1),(2)},A^{(1),(3)},q^{(1)}}=\Lambda_{A^{(2),(1)},A^{(2),(2)},A^{(2),(3)},q^{(2)}}$
implies that

$A^{(1),(j),{m-1}}=A^{(2),(j),{m-1}},\ j=1,2,3,\quad q^{(1),m}=q^{(2),m}\
\text{in}\ \overline{\Omega}\times\mathbb{C},\quad m=2,3,\dots,$

where $A^{(r),(j),{m-1}}$ are given in (1.2), and $q^{(r),m}$ are given in
(1.3).

First, let $m=2$, and let us start by performing a second-order linearization
of the Dirichlet–to–Neumann map. Let $u_{r}=u_{r}(x,\varepsilon)$ be the
unique small solution to the Dirichlet problem

$\begin{cases}(-\Delta)^{2}u_{r}+\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\big{(}\sum_{k=1}^{\infty}A^{(r),(j),k}_{i_{1}\dots
i_{j}}(x)\frac{u_{r}^{k}}{k!}\big{)}D^{j}_{i_{1}\dots i_{j}}u_{r}\\\
\quad\quad\quad\quad\quad\quad\quad+\sum_{k=2}^{\infty}q^{(r),k}(x)\frac{u_{r}^{k}}{k!}=0\quad\text{in}\quad\Omega,\\\
u_{r}|_{\partial\Omega}=\varepsilon_{1}f_{1}+\varepsilon_{2}f_{2},\\\
\partial_{\nu}u_{r}|_{\partial\Omega}=\varepsilon_{1}g_{1}+\varepsilon_{2}g_{2},\end{cases}$
(5.1)

for $r=1,2$. Applying $\partial_{\varepsilon_{l}}|_{\varepsilon=0}$ to (5.1),
$l=1,2$, and using that $u_{r}(x,0)=0$, we get

$\begin{cases}(-\Delta)^{2}v_{r}^{(l)}=0\quad\text{in}\quad\Omega,\\\
v_{r}^{(l)}|_{\partial\Omega}=f_{l},\\\
\partial_{\nu}v_{r}^{(l)}|_{\partial\Omega}=g_{l},\end{cases}$ (5.2)

where $v_{r}^{(l)}=\partial_{\varepsilon_{l}}u_{r}|_{\varepsilon=0}$, $r=1,2$.
By the uniqueness for the Dirichlet problem (5.2), we see that
$v^{(l)}:=v_{1}^{(l)}=v_{2}^{(l)}\in C^{4,\alpha}(\overline{\Omega})$,
$l=1,2$, see [21, Theorem 2.19], [19, Section 6.6].

Applying
$\partial_{\varepsilon_{1}}\partial_{\varepsilon_{2}}|_{\varepsilon=0}$ to
(5.1), we get the second order linearization,

$\begin{cases}(-\Delta)^{2}w_{r}+\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}A^{(r),(j),1}_{i_{1}\dots
i_{j}}(x)\big{(}v^{(1)}D^{j}_{i_{1}\dots
i_{j}}v^{(2)}+v^{(2)}D^{j}_{i_{1}\dots i_{j}}v^{(1)}\big{)}\\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+q^{(r),2}v^{(1)}v^{(2)}=0\quad\text{in}\quad\Omega,\\\
w_{r}|_{\partial\Omega}=0,\\\
\partial_{\nu}w_{r}|_{\partial\Omega}=0,\end{cases}$ (5.3)

where
$w_{r}=\partial_{\varepsilon_{1}}\partial_{\varepsilon_{2}}u_{r}|_{\varepsilon=0}$,
$r=1,2$. Now the equality
$\Lambda_{A^{(1),(1)},A^{(1),(2)},A^{(1),(3)},q^{(1)}}(\varepsilon_{1}f_{1}+\varepsilon_{2}f_{2},\varepsilon_{1}g_{1}+\varepsilon_{2}g_{2})=\Lambda_{A^{(2),(1)},A^{(2),(2)},A^{(2),(3)},q^{(2)}}(\varepsilon_{1}f_{1}+\varepsilon_{2}f_{2},\varepsilon_{1}g_{1}+\varepsilon_{2}g_{2})$,
for all small $\varepsilon$ and all $f_{k}\in C^{4,\alpha}(\partial\Omega)$,
$g_{k}\in C^{3,\alpha}(\partial\Omega)$, $k=1,2$, implies that
$\partial_{\nu}^{2}u_{1}|_{\partial\Omega}=\partial_{\nu}^{2}u_{2}|_{\partial\Omega}$
and
$\partial_{\nu}^{3}u_{1}|_{\partial\Omega}=\partial_{\nu}^{3}u_{2}|_{\partial\Omega}$.
Hence, applying
$\partial_{\varepsilon_{1}}\partial_{\varepsilon_{2}}|_{\varepsilon=0}$, we
conclude that
$\partial_{\nu}^{2}w_{1}|_{\partial\Omega}=\partial_{\nu}^{2}w_{2}|_{\partial\Omega}$
and
$\partial_{\nu}^{3}w_{1}|_{\partial\Omega}=\partial_{\nu}^{3}w_{2}|_{\partial\Omega}$.

It follows from (5.3) that

$\displaystyle(-\Delta)^{2}($ $\displaystyle w_{1}-w_{2})$ (5.4)
$\displaystyle=\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}(A^{(2),(j),1}_{i_{1}\dots
i_{j}}(x)-A^{(1),(j),1}_{i_{1}\dots i_{j}}(x))\big{(}v^{(1)}D^{j}_{i_{1}\dots
i_{j}}v^{(2)}+v^{(2)}D^{j}_{i_{1}\dots i_{j}}v^{(1)}\big{)}$
$\displaystyle+(q^{(2),2}-q^{(1),2})v^{(1)}v^{(2)}=0\quad\text{in}\quad\Omega.$

We also have $(w_{1}-w_{2})|_{\partial\Omega}=0$,
$\partial_{\nu}(w_{1}-w_{2})|_{\partial\Omega}=0$,
$\partial_{\nu}^{2}(w_{1}-w_{2})|_{\partial\Omega}=0$, and
$\partial_{\nu}^{3}(w_{1}-w_{2})|_{\partial\Omega}=0$. Multiplying (5.4) by
$v^{(0)}\in C^{4,\alpha}(\overline{\Omega})$ such that
$(-\Delta)^{2}v^{(0)}=0$, integrating over $\Omega$, and using Green’s
formula, see [1, Theorem 10.1], we obtain that

$\displaystyle\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\int_{\Omega}(A^{(2),(j),1}_{i_{1}\dots
i_{j}}(x)-A^{(1),(j),1}_{i_{1}\dots i_{j}}(x))\big{(}v^{(1)}D^{j}_{i_{1}\dots
i_{j}}v^{(2)}+v^{(2)}D^{j}_{i_{1}\dots i_{j}}v^{(1)}\big{)}v^{(0)}dx$ (5.5)
$\displaystyle+\int_{\Omega}(q^{(2),2}-q^{(1),2})v^{(1)}v^{(2)}v^{(0)}dx=0.$

Note that (5.5) holds for all $v^{(l)}\in C^{4,\alpha}(\overline{\Omega})$
such that $(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=1,2,3$. In view of (1.6),
an application of Theorem 1.4 allows us to conclude that

$A^{(1),(j),1}=A^{(2),(j),1},\quad j=1,2,3,\quad q^{(1),2}=q^{(2),2}.$

Let $m\geq 3$ and let us assume that

$A^{(1),(j),k}=A^{(2),(j),k},\quad j=1,2,3,\quad q^{(1),k+1}=q^{(2),k+1},\quad
k=1,\dots,m-2.$ (5.6)

To prove that

$A^{(1),(j),m-1}=A^{(2),(j),m-1},\quad j=1,2,3,\quad q^{(1),m}=q^{(2),m},$

we shall carry out the $m$th order linearization of the Dirichlet–to–Neumann
map. In doing so, we let $u_{r}=u_{r}(x,\varepsilon)$ be the unique small
solution to

$\begin{cases}(-\Delta)^{2}u_{r}+\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\big{(}\sum_{k=1}^{\infty}A^{(r),(j),k}_{i_{1}\dots
i_{j}}(x)\frac{u_{r}^{k}}{k!}\big{)}D^{j}_{i_{1}\dots i_{j}}u_{r}\\\
\quad\quad\quad\quad\quad\quad\quad+\sum_{k=2}^{\infty}q^{(r),k}(x)\frac{u_{r}^{k}}{k!}=0\quad\text{in}\quad\Omega,\\\
u_{r}|_{\partial\Omega}=\varepsilon_{1}f_{1}+\dots+\varepsilon_{m}f_{m},\\\
\partial_{\nu}u_{r}|_{\partial\Omega}=\varepsilon_{1}g_{1}+\dots+\varepsilon_{m}g_{m},\end{cases}$
(5.7)

for $r=1,2$. We would like to apply
$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}|_{\varepsilon=0}$
to (5.7). To that end, we first note that

$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}\bigg{(}\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\big{(}\sum_{k=m}^{\infty}A^{(r),(j),k}_{i_{1}\dots
i_{j}}(x)\frac{u_{r}^{k}}{k!}\big{)}D^{j}_{i_{1}\dots
i_{j}}u_{r}+\sum_{k=m+1}^{\infty}q^{(r),k}(x)\frac{u_{r}^{k}}{k!}\bigg{)}$

is a sum of terms, each of them containing positive powers of $u_{r}$ which
vanish when $\varepsilon=0$. The only term in the expression for
$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}\big{(}q^{(r),m}(x)\frac{u_{r}^{m}}{m!}\big{)}$
which does not contain a positive power of $u_{r}$ is given by

$q^{(r),m}(x)\partial_{\varepsilon_{1}}u_{r}\dots\partial_{\varepsilon_{m}}u_{r}.$
(5.8)

The only term in
$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}\big{(}\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}A^{(r),(j),m-1}_{i_{1}\dots
i_{j}}(x)\frac{u_{r}^{m-1}}{(m-1)!}D^{j}_{i_{1}\dots i_{j}}u_{r}\big{)}$ which
does not contain a power of $u_{r}$ is

$\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}A^{(r),(j),m-1}_{i_{1}\dots
i_{j}}(x)\bigg{(}\sum_{k=1}^{m}\prod_{l=1,l\neq
k}^{m}\partial_{\varepsilon_{l}}u_{r}D^{j}_{i_{1}\dots
i_{j}}\partial_{\varepsilon_{k}}u_{r}\bigg{)}.$ (5.9)

Finally, the expression

$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}\bigg{(}\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\big{(}\sum_{k=1}^{m-2}A^{(r),(j),k}_{i_{1}\dots
i_{j}}(x)\frac{u_{r}^{k}}{k!}\big{)}D^{j}_{i_{1}\dots
i_{j}}u_{r}+\sum_{k=2}^{m-1}q^{(r),k}(x)\frac{u_{r}^{k}}{k!}\bigg{)}\bigg{|}_{\varepsilon=0}$
(5.10)

is independent of $r=1,2$. Indeed, this follows from (5.6), the fact that
(5.10) contains only the derivatives of $u_{r}$ of the form
$\partial^{s}_{\varepsilon_{l_{1}},\dots,\varepsilon_{l_{s}}}u_{r}|_{\varepsilon=0}$
with $s=1,\dots,m-1$,
$\varepsilon_{l_{1}},\dots,\varepsilon_{l_{s}}\in\\{\varepsilon_{1},\dots,\varepsilon_{m}\\}$,
and the fact that

$\partial^{s}_{\varepsilon_{l_{1}},\dots,\varepsilon_{l_{s}}}u_{1}|_{\varepsilon=0}=\partial^{s}_{\varepsilon_{l_{1}},\dots,\varepsilon_{l_{s}}}u_{2}|_{\varepsilon=0}.$

The latter can be established by induction on $s$, applying the operator
$\partial^{s}_{\varepsilon_{l_{1}},\dots,\varepsilon_{l_{s}}}|_{\varepsilon=0}$
to (5.7) and using (5.6) together with the unique solvability of the Dirichlet
problem for the operator $(-\Delta)^{2}$.

Thus, applying
$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}|_{\varepsilon=0}$
to (5.7), we get

$\begin{cases}(-\Delta)^{2}w_{r}+\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}A^{(r),(j),m-1}_{i_{1}\dots
i_{j}}(x)\bigg{(}\sum_{k=1}^{m}\prod_{l=1,l\neq k}^{m}v^{(l)}D^{j}_{i_{1}\dots
i_{j}}v^{(k)}\bigg{)}\\\
\quad\quad\quad\quad\quad\quad\quad+q^{(r),m}(x)v^{(1)}\dots
v^{(m)}=H_{m}\quad\text{in}\quad\Omega,\\\ w_{r}|_{\partial\Omega}=0,\\\
\partial_{\nu}w_{r}|_{\partial\Omega}=0,\end{cases}$ (5.11)

for $r=1,2$, see (5.8), (5.9). Here
$w_{r}=\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}u_{r}|_{\varepsilon=0}$
and

$\displaystyle H_{m}(x):=-$
$\displaystyle\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}$
$\displaystyle\bigg{(}\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\big{(}\sum_{k=1}^{m-2}A^{(r),(j),k}_{i_{1}\dots
i_{j}}(x)\frac{u_{r}^{k}}{k!}\big{)}D^{j}_{i_{1}\dots
i_{j}}u_{r}+\sum_{k=2}^{m-1}q^{(r),k}(x)\frac{u_{r}^{k}}{k!}\bigg{)}\bigg{|}_{\varepsilon=0}.$

Letting $w=w_{1}-w_{2}$, we obtain from (5.11) that

$\begin{cases}(-\Delta)^{2}(w_{1}-w_{2})=\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}(A^{(2),(j),m-1}_{i_{1}\dots
i_{j}}(x)-A^{(1),(j),m-1}_{i_{1}\dots i_{j}}(x))\\\
\quad\quad\quad\quad\quad\quad\quad\bigg{(}\sum_{k=1}^{m}\prod_{l=1,l\neq
k}^{m}v^{(l)}D^{j}_{i_{1}\dots i_{j}}v^{(k)}\bigg{)}\\\
\quad\quad\quad\quad\quad\quad\quad+(q^{(2),m}(x)-q^{(1),m}(x))v^{(1)}\dots
v^{(m)}\quad\text{in}\quad\Omega,\\\ w|_{\partial\Omega}=0,\\\
\partial_{\nu}w|_{\partial\Omega}=0.\end{cases}$ (5.12)

Applying
$\partial_{\varepsilon_{1}}\dots\partial_{\varepsilon_{m}}|_{\varepsilon=0}$
to
$\Lambda_{A^{(1),(1)},A^{(1),(2)},A^{(1),(3)},q^{(1)}}(\varepsilon_{1}f_{1}+\dots+\varepsilon_{m}f_{m},\varepsilon_{1}g_{1}+\dots+\varepsilon_{m}g_{m})=\Lambda_{A^{(2),(1)},A^{(2),(2)},A^{(2),(3)},q^{(2)}}(\varepsilon_{1}f_{1}+\dots+\varepsilon_{m}f_{m},\varepsilon_{1}g_{1}+\dots+\varepsilon_{m}g_{m})$,
we get $\partial_{\nu}^{2}w|_{\partial\Omega}=0$ and $\partial_{\nu}^{3}w=0$.
Let $v^{(0)}\in C^{4,\alpha}(\overline{\Omega})$ be such that
$(-\Delta)^{2}v^{(0)}=0$ in $\Omega$. Multiplying (5.12) by $v^{(0)}$ and
using Green’s formula, see [1, Theorem 10.1], we obtain that

$\displaystyle\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\int_{\Omega}(A^{(2),(j),m-1}_{i_{1}\dots
i_{j}}(x)-A^{(1),(j),m-1}_{i_{1}\dots
i_{j}}(x))\bigg{(}\sum_{k=1}^{m}\prod_{l=1,l\neq
k}^{m}v^{(l)}D^{j}_{i_{1}\dots i_{j}}v^{(k)}\bigg{)}v^{(0)}dx$ (5.13)
$\displaystyle+\int_{\Omega}(q^{(2),m}(x)-q^{(1),m}(x))v^{(1)}\dots
v^{(m)}v^{(0)}dx=0,$

for all $v^{(l)}\in C^{4,\alpha}(\overline{\Omega})$ solving
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=0,1,\dots,m$.

Letting $v^{(3)}=\dots=v^{(m)}=1$ in (5.13), we get

$\displaystyle\sum_{j=1}^{3}\sum_{i_{1},\dots,i_{j}=1}^{n}\int_{\Omega}(A^{(2),(j),m-1}_{i_{1}\dots
i_{j}}(x)-$ $\displaystyle A^{(1),(j),m-1}_{i_{1}\dots
i_{j}}(x))\big{(}v^{(1)}D^{j}_{i_{1}\dots
i_{j}}v^{(2)}+v^{(2)}D^{j}_{i_{1}\dots i_{j}}v^{(1)}\big{)}$ (5.14)
$\displaystyle
v^{(0)}dx+\int_{\Omega}(q^{(2),m}(x)-q^{(1),m}(x))v^{(1)}v^{(2)}v^{(0)}dx=0,$

for all $v^{(l)}\in C^{4,\alpha}(\overline{\Omega})$ solving
$(-\Delta)^{2}v^{(l)}=0$ in $\Omega$, $l=0,1,2$. In view of (1.6), applying
Theorem 1.4, we conclude that

$A^{(1),(j),m-1}=A^{(2),(j),m-1},\quad j=1,2,3,\quad q^{(1),m}=q^{(2),m}.$

This completes the proof of Theorem 1.1.

##  Appendix A Well-posedness of the Dirichlet problem for a third-order
nonlinear perturbation of a biharmonic operator

In this appendix, we shall show the well-posedness of the Dirichlet problem
for a third-order nonlinear perturbation of a biharmonic operator. The
argument is standard, see [35], [42], and is presented here for completeness
and convenience of the reader.

Let $\Omega\subset\mathbb{R}^{n}$, $n\geq 3$, be a bounded open set with
$C^{\infty}$ boundary. Let $k\in\mathbb{N}\cup\\{0\\}$, $0<\alpha<1$, and let
$C^{k,\alpha}(\overline{\Omega})$ be the standard Hölder space, see [23]. Note
that $C^{k,\alpha}(\overline{\Omega})$ is an algebra under pointwise
multiplication, with

$\|uv\|_{C^{k,\alpha}(\overline{\Omega})}\leq
C\big{(}\|u\|_{C^{k,\alpha}(\overline{\Omega})}\|v\|_{L^{\infty}(\Omega)}+\|u\|_{L^{\infty}(\Omega)}\|v\|_{C^{k,\alpha}(\overline{\Omega})}\big{)},\quad
u,v\in C^{k,\alpha}(\overline{\Omega}),$ (A.1)

see [23, Theorem A.7].

Consider the Dirichlet problem for the operator
$L_{A^{(1)},A^{(2)},A^{(3)},q}$, given by (1.1),

$\begin{cases}L_{A^{(1)},A^{(2)},A^{(3)},q}u=0\quad\text{in}\quad\Omega,\\\
u|_{\partial\Omega}=f,\\\ \partial_{\nu}u|_{\partial\Omega}=g,\end{cases}$
(A.2)

where $\nu$ is the unit outer normal to the boundary. Here
$A^{(j)}:\overline{\Omega}\times\mathbb{C}\to S^{j}$, $j=1,2,3$, and
$q:\overline{\Omega}\times\mathbb{C}\to\mathbb{C}$ satisfy the conditions:

  * (i)

the map $\mathbb{C}\ni z\mapsto A^{(j)}(\cdot,z)$ is holomorphic with values
in $C^{0,\alpha}(\overline{\Omega};S^{j})$ for some $0<\alpha<1$, the space of
Hölder continuous symmetric $j$–tensor fields,

  * (ii)

the map $\mathbb{C}\ni z\mapsto q(\cdot,z)$ is holomorphic with values in
$C^{0,\alpha}(\overline{\Omega})$,

  * (iii)

$A^{(j)}(x,0)=0$, $q(x,0)=0$, and $\partial_{z}q(x,0)=0$ for all
$x\in\overline{\Omega}$, $j=1,2,3$.

It follows from (i), (ii), (iii) that $A^{(j)}$, $j=1,2,3$, and $q$ can be
expanded into power series

$A^{(j)}(x,z)=\sum_{k=1}^{\infty}A^{(j),k}(x)\frac{z^{k}}{k!},\quad
A^{(j),k}(x):=\partial_{z}^{k}A^{(j)}(x,0)\in
C^{0,\alpha}(\overline{\Omega};S^{j}),$ (A.3)

converging in the $C^{0,\alpha}(\overline{\Omega};S^{j})$ topology, and

$q(x,z)=\sum_{k=2}^{\infty}q^{k}(x)\frac{z^{k}}{k!},\quad
q^{k}(x):=\partial_{z}^{k}q(x,0)\in C^{0,\alpha}(\overline{\Omega}),$ (A.4)

converging in the $C^{0,\alpha}(\overline{\Omega})$ topology.

In what follows, when $X,Y$ are Banach spaces, we equip $X\times Y$ with the
norm $\|(x,y)\|_{X\times Y}=\|x\|_{X}+\|y\|_{Y}$. Then $X\times Y$ becomes a
Banach space. We have the following result.

######  Theorem A.1.

Assume that $A^{(j)}:\overline{\Omega}\times\mathbb{C}\to S^{j}$, $j=1,2,3$,
and $q:\overline{\Omega}\times\mathbb{C}\to\mathbb{C}$ satisfy the conditions
(i), (ii), and (iii). Then there exist $\delta>0$ and $C>0$ such that for any
$(f,g)\in B_{\delta}(\partial\Omega):=\\{(f,g)\in
C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega):\|f\|_{C^{4,\alpha}(\partial\Omega)}+\|g\|_{C^{3,\alpha}(\partial\Omega)}<\delta\\}$,
the problem (A.2) has a solution $u=u_{f,g}\in
C^{4,\alpha}(\overline{\Omega})$ satisfying

$\|u\|_{C^{4,\alpha}(\overline{\Omega})}\leq
C(\|f\|_{C^{4,\alpha}(\partial\Omega)}+\|g\|_{C^{3,\alpha}(\partial\Omega)}).$

The solution is unique within the class $\\{u\in
C^{4,\alpha}(\overline{\Omega}):\|u\|_{C^{4,\alpha}(\overline{\Omega})}<C\delta\\}$,
and it depends holomorphically on $(f,g)\in B_{\delta}(\partial\Omega)$.
Furthermore, the map

$B_{\delta}(\partial\Omega)\to C^{2,\alpha}(\partial\Omega)\times
C^{1,\alpha}(\partial\Omega),\quad(f,g)\mapsto(\partial_{\nu}^{2}u|_{\partial\Omega},\partial_{\nu}^{3}u|_{\partial\Omega})$

is holomorphic.

###### Proof.

We shall follow [35, Appendix A], see also [42], and rely on the implicit
function theorem for holomorphic maps between complex Banach spaces, see [50,
p. 144]. In doing so, we consider the map

$\displaystyle F:C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega)\times C^{4,\alpha}(\overline{\Omega})\to
C^{0,\alpha}(\overline{\Omega})\times C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega),$ (A.5) $\displaystyle
F(f,g,u)=\big{(}L_{A^{(1)},A^{(2)},A^{(3)},q}u,u|_{\partial\Omega}-f,\partial_{\nu}u|_{\partial\Omega}-g\big{)}.$

Let us check that $F$ enjoys the mapping properties (A.5). To that end let
$u\in C^{4,\alpha}(\overline{\Omega})$ and first note that $(-\Delta)^{2}u\in
C^{0,\alpha}(\overline{\Omega})$. Let $j=1,2,3$ and let us now check that
$A^{(j)}(\cdot,u(\cdot))\in C^{0,\alpha}(\overline{\Omega};S^{j})$. As in
[35], using Cauchy’s estimates, we see that the coefficients $A^{(j),k}$ in
(A.3) satisfy

$\|A^{(j),k}\|_{C^{0,\alpha}(\overline{\Omega};S^{j})}\leq\frac{k!}{R^{k}}\sup_{|z|=R}\|A^{(j)}(\cdot,z)\|_{C^{0,\alpha}(\overline{\Omega};S^{j})},$
(A.6)

where $R>0$ and $k=1,2,\dots$. It follows from (A.6) and (A.1) that

$\bigg{\|}A^{(j),k}\frac{u^{k}}{k!}\bigg{\|}_{C^{0,\alpha}(\overline{\Omega};S^{j})}\leq\frac{C^{k}}{R^{k}}\|u\|^{k}_{C^{0,\alpha}(\overline{\Omega})}\sup_{|z|=R}\|A^{(j)}(\cdot,z)\|_{C^{0,\alpha}(\overline{\Omega};S^{j})}.$
(A.7)

Choosing $R=2C\|u\|_{C^{0,\alpha}(\overline{\Omega})}$, we see from (A.7) that
the series $\sum_{k=1}^{\infty}A^{(j),k}(x)\frac{u^{k}}{k!}$ converges in
$C^{0,\alpha}(\overline{\Omega};S^{j})$, and hence,
$A^{(j)}(\cdot,u(\cdot))\in C^{0,\alpha}(\overline{\Omega};S^{j})$, $j=1,2,3$.
Similarly, we see that $q(\cdot,u(\cdot))\in C^{0,\alpha}(\overline{\Omega})$.
Thus, we conclude from (1.1) that $L_{A^{(1)},A^{(2)},A^{(3)},q}u\in
C^{0,\alpha}(\overline{\Omega})$.

Let us next show that $F$, given by (A.5), is holomorphic. First, $F$ is
locally bounded as it is continuous in $(f,g,u)$. Therefore, it suffices to
check that $F$ is weakly holomorphic, see [50, p. 133]. In doing so, letting
$(f_{1},g_{1},u_{1}),(f_{2},g_{2},u_{2})\in C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega)\times C^{4,\alpha}(\overline{\Omega})$, we have
to check that the map

$\lambda\mapsto F((f_{1},g_{1},u_{1})+\lambda(f_{2},g_{2},u_{2}))$

is holomorphic in $\mathbb{C}$ with values in
$C^{0,\alpha}(\overline{\Omega})\times C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega)$. To that end, it suffices to show that the map
$\lambda\mapsto A^{(j)}(\cdot,u_{1}(\cdot)+\lambda u_{2}(\cdot))$ is
holomorphic in $\mathbb{C}$ with values in
$C^{0,\alpha}(\overline{\Omega};S^{j})$, $j=1,2,3$. The latter follows from
the fact that the series
$\sum_{k=1}^{\infty}\frac{A^{(j),k}}{k!}(u_{1}+\lambda u_{2})^{k}$ converges
in $C^{0,\alpha}(\overline{\Omega};S^{j})$, locally uniformly in
$\lambda\in\mathbb{C}$, see (A.7).

Now we have $F(0,0,0)=0$ and the partial differential

$\partial_{u}F(0,0,0):C^{4,\alpha}(\overline{\Omega})\to
C^{0,\alpha}(\overline{\Omega})\times C^{4,\alpha}(\partial\Omega)\times
C^{3,\alpha}(\partial\Omega)$ (A.8)

is given by

$\partial_{u}F(0,0,0)v=((-\Delta)^{2}v,v|_{\partial\Omega},\partial_{\nu}v|_{\partial\Omega}).$

Thus, it follows from [21, Theorem 2.19], see also [19, Section 6.6] , that
the map (A.8) is a linear isomorphism.

An application of the implicit function theorem, see [50, p. 144], gives that
there exists $\delta>0$ and a unique holomorphic map
$S:B_{\delta}(\partial\Omega)\to C^{4,\alpha}(\overline{\Omega})$ such that
$S(0,0)=0$ and $F(f,g,S(f,g))=0$ for all $(f,g)\in
B_{\delta}(\partial\Omega)$. Setting $u=S(f,g)$ and noting that $S$ is
Lipschitz continuous with $S(0,0)=0$, we get

$\|u\|_{C^{4,\alpha}(\overline{\Omega})}\leq
C(\|f\|_{C^{4,\alpha}(\partial\Omega)}+\|g\|_{C^{3,\alpha}(\partial\Omega)}).$

∎

##  Appendix B Proofs of injectivity results for generalized momentum ray
transforms

The purpose of this appendix is to provide proofs of key injectivity results
for generalized momentum ray transforms required to establish Theorem 1.4. We
will specifically present the proofs for Theorem 2.4, Lemma 2.6, and Lemma
2.8. These results are established in [14, Theorem 4.18], [14, Lemma 3.7], and
[51, Lemma 6.7], considering tensor fields of arbitrary order
$m\in\mathbb{N}$. In this appendix, we shall present the proofs given in those
works, making them more direct when tailored to the case of $m$–tensor fields
where $m=0,1,2,3$, as needed in this paper. We will also give special
consideration to continuous tensor fields with compact support. This appendix
is included for the completeness and convenience of the reader only.

###  B.1. Proof of Theorem 2.4

We shall follow the proof of [14, Theorem 4.18]. Let first $m=3$ and let us
assume that $J^{3,3}F(x,\xi)=0$, for all
$(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. Then $J^{3,3}F(x,-\xi)=0$,
for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. Using that
$J^{3,3}F(x,\xi)\pm J^{3,3}F(x,-\xi)=0$ and (2.22), we get

$J^{1,3}f^{(1)}(x,\xi)+J^{3,3}f^{(3)}(x,\xi)=0,\quad
J^{0,3}f^{(0)}(x,\xi)+J^{2,3}f^{(2)}(x,\xi)=0,$ (B.1)

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. Using (2.23), we
obtain from (B.1) that

$J^{3,3}(i_{\delta}f^{(1)}+f^{(3)})(x,\xi)=0,$ (B.2)
$J^{2,3}(i_{\delta}f^{(0)}+f^{(2)})(x,\xi)=0,$ (B.3)

for all $(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. Using (2.16), we get
from (B.2) that $I^{3,3}(i_{\delta}f^{(1)}+f^{(3)})(x,\xi)=0$, for all
$(x,\xi)\in\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})$, and
therefore, in $\mathcal{D}^{\prime}(\mathbb{R}^{n}\times\mathbb{S}^{n-1})$.
Hence, Theorem 2.3 implies that $i_{\delta}f^{(1)}+f^{(3)}=0$.

Applying the operator $\xi\cdot\partial_{x}$ to (B.3) in the dense of
distributions, and using (2.24), we obtain that
$J^{2,2}(i_{\delta}f^{(0)}+f^{(2)})=0$ in
$\mathcal{D}^{\prime}(\mathbb{R}^{n}\times\mathbb{S}^{n-1})$. As
$J^{2,2}(i_{\delta}f^{(0)}+f^{(2)})\in
C(\mathbb{R}^{n}\times\mathbb{S}^{n-1})$, we have
$J^{2,2}(i_{\delta}f^{(0)}+f^{(2)})(x,\xi)=0$ for all
$(x,\xi)\in\mathbb{R}^{n}\times\mathbb{S}^{n-1}$. In view of (2.16), we get
$I^{2,2}(i_{\delta}f^{(0)}+f^{(2)})(x,\xi)=0$ for all
$(x,\xi)\in\mathbb{R}^{n}\times(\mathbb{R}^{n}\setminus\\{0\\})$. Hence,
Theorem 2.3 implies that $i_{\delta}f^{(0)}+f^{(2)}=0$. This establishes the
implication in (i) in one direction. The proof of the converse follows by a
direct computation. The proof of (ii)-(iv) is similar. This completes the
proof of Theorem 2.4.

###  B.2. Proof of Lemma 2.6

We shall follow the proof of [14, Lemma 3.7]. First, assume that $m=3$. We
have the following decomposition,

$f^{(3)}=g^{(3)}+i_{\delta}v^{(1)},$ (B.4)

where $v^{(1)}\in C_{0}(\mathbb{R}^{n-1};S^{1}(\mathbb{R}^{n}))$ and
$g^{(3)}\in C_{0}(\mathbb{R}^{n-1};S^{3}(\mathbb{R}^{n}))$ is such that

$j_{\delta}g^{(3)}=0,$ (B.5)

see [18, Lemma 2.3]. Indeed, applying $j_{\delta}$ to (B.4) and using (B.5)
and (2.6), we get $v^{(1)}=\frac{3}{n+2}j_{\delta}f^{(3)}\in
C_{0}(\mathbb{R}^{n-1};S^{1}(\mathbb{R}^{n}))$, and therefore,
$g^{(3)}=f^{(3)}-\frac{3}{n+2}i_{\delta}j_{\delta}f^{(3)}\in
C_{0}(\mathbb{R}^{n-1};S^{3}(\mathbb{R}^{n}))$. The condition (B.5) follows.

A direct computation using (2.4), (2.3), and the fact that
$(e_{1}+i\eta)\cdot(e_{1}+i\eta)=0$ gives that $i_{\delta}v^{(1)}$ is in the
kernel of (2.25). Thus, substituting (B.4) in (2.25), we obtain that

$\sum_{i_{1},i_{2},i_{3}=1}^{n}\int_{\mathbb{R}}t^{3}g^{(3)}_{i_{1}i_{2}i_{3}}(y_{0}^{\prime}+t\eta^{\prime})(e_{1}+i\eta)_{i_{1}}(e_{1}+i\eta)_{i_{2}}(e_{1}+i\eta)_{i_{3}}dt=0,$
(B.6)

for all $y_{0}^{\prime}\in\mathbb{R}^{n-1}$ and all $\eta\in\mathbb{R}^{n}$
such that $e_{1}\cdot\eta=0$ and $|\eta|=1$.

We shall proceed to show that $g^{(3)}=0$. First, a direct computation using
that $\eta_{1}=0$ gives

$\displaystyle\sum_{i_{1},i_{2},i_{3}=1}^{n}$ $\displaystyle
g^{(3)}_{i_{1}i_{2}i_{3}}(e_{1}+i\eta)_{i_{1}}(e_{1}+i\eta)_{i_{2}}(e_{1}+i\eta)_{i_{3}}=g^{(3)}_{111}+3i\sum_{i_{1}=2}^{n}g^{(3)}_{i_{1}11}\eta_{i_{1}}$
(B.7)
$\displaystyle+3i^{2}\sum_{i_{1},i_{2}=2}^{n}g^{(3)}_{i_{1}i_{2}1}\eta_{i_{1}}\eta_{i_{2}}+i^{3}\sum_{i_{1},i_{2},i_{3}=2}^{n}g^{(3)}_{i_{1}i_{2}i_{3}}\eta_{i_{1}}\eta_{i_{2}}\eta_{i_{3}}=g^{(3)}_{111}+3i\sum_{i_{1}=1}^{n-1}g^{(3)}_{i_{1}+111}\eta^{\prime}_{i_{1}}$
$\displaystyle+3i^{2}\sum_{i_{1},i_{2}=1}^{n-1}g^{(3)}_{i_{1}+1i_{2}+11}\eta^{\prime}_{i_{1}}\eta^{\prime}_{i_{2}}+i^{3}\sum_{i_{1},i_{2},i_{3}=1}^{n-1}g^{(3)}_{i_{1}+1i_{2}+1i_{3}+1}\eta^{\prime}_{i_{1}}\eta^{\prime}_{i_{2}}\eta^{\prime}_{i_{3}}.$

Substituting (B.7) in (B.6), we get that

$\displaystyle\int_{\mathbb{R}}t^{3}\bigg{(}g^{(3)}_{111}(y_{0}^{\prime}+t\eta^{\prime})$
$\displaystyle+3i\sum_{i_{1}=1}^{n-1}g^{(3)}_{i_{1}+111}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}$
(B.8)
$\displaystyle+3i^{2}\sum_{i_{1},i_{2}=1}^{n-1}g^{(3)}_{i_{1}+1i_{2}+11}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}\eta^{\prime}_{i_{2}}$
$\displaystyle+i^{3}\sum_{i_{1},i_{2},i_{3}=1}^{n-1}g^{(3)}_{i_{1}+1i_{2}+1i_{3}+1}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}\eta^{\prime}_{i_{2}}\eta^{\prime}_{i_{3}}\bigg{)}dt=0,$

for all
$(y_{0}^{\prime},\eta^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$. An
application of Theorem 2.4, in the case of tensor fields in
$C_{0}(\mathbb{R}^{n-1};\mathcal{S}^{3}(\mathbb{R}^{n-1}))$, to (B.8), and a
subsequent use of (2.4), (2.3) give that for all
$i_{1},i_{2},i_{3}=1,\dots,n-1$,

$3i^{2}g^{(3)}_{i_{1}+1i_{2}+11}=-(i_{\delta}g^{(3)}_{111})_{i_{1}i_{2}}=-\frac{1}{2}g^{(3)}_{111}(\delta_{i_{1}i_{2}}+\delta_{i_{2}i_{1}}),$
(B.9)

and

$i^{3}g^{(3)}_{i_{1}+1i_{2}+1i_{3}+1}=-(i_{\delta}\widetilde{g}^{(1)})_{i_{1}i_{2}i_{3}},\quad(\widetilde{g}^{(1)})_{i_{1}}=3ig^{(3)}_{i_{1}+111}.$
(B.10)

Now (B.9) implies that

$g_{j_{1}j_{2}1}^{(3)}=0,\quad j_{1}\neq j_{2},\quad j_{1},j_{2}=2,\dots,n,$
(B.11)

and

$3g^{(3)}_{j_{1}j_{1}1}=g^{(3)}_{111},\quad j_{1}=2,\dots,n.$ (B.12)

It follows from (B.5) that for $j_{1}=1,\dots,n$,

$(j_{\delta}g^{(3)})_{j_{1}}=\sum_{l=1}^{n}g^{(3)}_{j_{1}ll}=g^{(3)}_{j_{1}11}+\sum_{l=2}^{n}g^{(3)}_{j_{1}ll}=0.$
(B.13)

Now (B.13) with $j_{1}=1$ and (B.12) imply that

$g^{(3)}_{j_{1}j_{1}1}=0,\quad j_{1}=1,2,\dots,n.$ (B.14)

It follows from (B.10) that

$g^{(3)}_{j_{1}j_{2}j_{3}}=0,\quad j_{1}\neq j_{2}\neq j_{3},\quad
j_{1},j_{2},j_{3}=2,\dots,n,$ (B.15)
$g^{(3)}_{j_{1}j_{1}j_{1}}=3g^{(3)}_{j_{1}11},\quad j_{1}=2,\dots,n,$ (B.16)

and

$g_{j_{1}j_{1}j_{3}}^{(3)}=g^{(3)}_{j_{3}11},\quad j_{1}\neq j_{3},\quad
j_{1},j_{3}=2,\dots,n.$ (B.17)

Combining (B.13) with $j_{1}=j_{3}$ and (B.17), we get

$\displaystyle g^{(3)}_{j_{3}11}=0,\quad j_{3}=2,\dots,n,$ (B.18)
$\displaystyle g^{(3)}_{j_{1}j_{1}j_{3}}=0,\quad j_{1}\neq j_{3},\quad
j_{1},j_{3}=2,\dots,n.$

Thus, in view of (B.18), (B.16) gives that

$g^{(3)}_{j_{1}j_{1}j_{1}}=0,\quad j_{1}=2,\dots,n.$ (B.19)

We conclude from (B.11), (B.14), (B.15), (B.18), and (B.19) that $g^{(3)}=0$.
This completes the proof in the case $m=3$.

Assume now that $m=2$. Similar to the case $m=3$, we have the following
decomposition $f^{(2)}=g^{(2)}+i_{\delta}v^{(0)}$, where
$v^{(0)}=\frac{1}{n}j_{\delta}f^{(2)}\in
C_{0}(\mathbb{R}^{n-1},S^{0}(\mathbb{R}^{n}))$ and
$g^{(2)}=f^{(2)}-\frac{1}{n}i_{\delta}j_{\delta}f^{(2)}\in
C_{0}(\mathbb{R}^{n-1},S^{2}(\mathbb{R}^{n}))$ is such that

$j_{\delta}g^{(2)}=0.$ (B.20)

We reduce the proof of the lemma to the following result: if

$\displaystyle\int_{\mathbb{R}}t^{2}\bigg{(}g^{(2)}_{11}(y_{0}^{\prime}+t\eta^{\prime})$
$\displaystyle+2i\sum_{i_{1}=1}^{n-1}g^{(2)}_{i_{1}+11}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}$
(B.21)
$\displaystyle+i^{2}\sum_{i_{1},i_{2}=1}^{n-1}g^{(2)}_{i_{1}+1i_{2}+1}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}\eta^{\prime}_{i_{2}}\bigg{)}dt=0,$

for all
$(y_{0}^{\prime},\eta^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$,
show that $g^{(2)}=0$. An application of Theorem 2.4, in the case of
$C_{0}(\mathbb{R}^{n-1},\mathcal{S}^{2}(\mathbb{R}^{n-1}))$, to (B.21), and a
subsequent use of (2.4), (2.3) give that

$g^{(2)}_{j_{1}1}=0,\quad j_{1}=2,\dots,n,$ (B.22)
$g^{(2)}_{j_{1}j_{2}}=0,\quad j_{1}\neq j_{2},\quad j_{1},j_{2}=2,\dots,n,$
(B.23)

and

$g^{(2)}_{j_{1}j_{1}}=g^{(2)}_{11},\quad j_{1}=2,\dots,n.$ (B.24)

Now (B.20) implies that

$j_{\delta}g^{(2)}=g^{(2)}_{11}+\sum_{k=2}^{n}g^{(2)}_{kk}=0,$

and therefore, (B.24) gives that

$g^{(2)}_{j_{1}j_{1}}=0,\quad j_{1}=1,\dots,n.$ (B.25)

We conclude from (B.22), (B.23), and (B.25) that $g^{(2)}=0$. This completes
the proof of Lemma in the case $m=2$.

Let $m=1$. Then (2.25) can be written as

$\int_{\mathbb{R}}t\big{(}f^{(1)}_{1}(y_{0}^{\prime}+t\eta^{\prime})+\sum_{i_{1}=1}^{n-1}if^{(1)}_{i_{1}+1}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}\big{)}dt=0,$

for all
$(y_{0}^{\prime},\eta^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$. An
application of Theorem 2.4 gives $f^{(1)}=0$. The proof of Lemma 2.6 is
complete.

###  B.3. Proof of Lemma 2.8

First, we write

$f^{(2)}=g^{(2)}+i_{\delta}v^{(0)},$ (B.26)

where

$\displaystyle v^{(0)}=\frac{1}{n}j_{\delta}f^{(2)}\in
C_{0}(\mathbb{R}^{n-1},S^{0}(\mathbb{R}^{n})),\quad
g^{(2)}=f^{(2)}-\frac{1}{n}i_{\delta}j_{\delta}f^{(2)}\in
C_{0}(\mathbb{R}^{n-1},S^{2}(\mathbb{R}^{n})),$

so that $j_{\delta}g^{(2)}=0$, see [18, Lemma 2.3]. Substituting (B.26) in
(2.26), and using that $\eta_{1}=0$, we get

$\displaystyle\int_{\mathbb{R}}t^{2+k}\bigg{(}g^{(2)}_{11}(y_{0}^{\prime}+t\eta^{\prime})+$
$\displaystyle
2i\sum_{i_{1}=1}^{n-1}g^{(2)}_{i_{1}+11}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}$
(B.27)
$\displaystyle+i^{2}\sum_{i_{1},i_{2}=1}^{n-1}g^{(2)}_{i_{1}+1i_{2}+1}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}\eta^{\prime}_{i_{2}}\bigg{)}dt$
$\displaystyle+\int_{\mathbb{R}}t^{1+k}\bigg{(}f^{(1)}_{1}(y_{0}^{\prime}+t\eta^{\prime})+i\sum_{i_{1}=1}^{n-1}f^{(1)}_{i_{1}+1}(y_{0}^{\prime}+t\eta^{\prime})\eta^{\prime}_{i_{1}}\bigg{)}dt=0,$

for all $\eta^{\prime}\in\mathbb{R}^{n-1}$, $|\eta^{\prime}|=1$, all
$y_{0}^{\prime}\in\mathbb{R}^{n-1}$, and $k=0,1$. Now (B.27) can be written as
follows,

$J^{2,2+k}G(y_{0}^{\prime},\eta^{\prime})+J^{1,k+1}F(y_{0}^{\prime},\eta^{\prime})=0,$
(B.28)

for all
$(y_{0}^{\prime},\eta^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$.
Here $G=\widetilde{g}^{(0)}+\widetilde{g}^{(1)}+\widetilde{g}^{(2)}\in
C_{0}(\mathbb{R}^{n-1},\mathcal{S}^{2}(\mathbb{R}^{n-1}))$, and
$F=\widetilde{f}^{(0)}+\widetilde{f}^{(1)}\in
C_{0}(\mathbb{R}^{n-1},\mathcal{S}^{1}(\mathbb{R}^{n-1}))$, where

$\displaystyle\widetilde{g}^{(0)}=g^{(2)}_{11},\quad\widetilde{g}^{(1)}_{i_{1}}=2ig^{(2)}_{i_{1}+11},\quad\widetilde{g}^{(2)}_{i_{1}i_{2}}=i^{2}g^{(2)}_{i_{1}+1i_{2}+1},$
$\displaystyle\widetilde{f}^{(0)}=f^{(1)}_{1},\quad\widetilde{f}^{(1)}_{i_{1}}=if^{(1)}_{i_{1}+1},\quad
1\leq i_{1},i_{2}\leq n-1.$

Applying $\eta^{\prime}\cdot\partial_{y_{0}^{\prime}}$ to (B.28) in the sense
of distributions, and using (2.24), we get

$(k+2)J^{2,1+k}G+(k+1)J^{1,k}F=0\quad\text{in}\quad\mathcal{D}^{\prime}(\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}).$
(B.29)

Now since the left hand side of (B.29) belongs to
$C(\mathbb{R}^{n-1}\times\mathbb{S}^{n-2})$, (B.29) implies that

$(k+2)J^{2,1+k}G(y_{0}^{\prime},\eta^{\prime})+(k+1)J^{1,k}F(y_{0}^{\prime},\eta^{\prime})=0,$
(B.30)

for all
$(y_{0}^{\prime},\eta^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$.
Combining (B.28) with $k=0$ and (B.30) with $k=1$, we obtain that

$J^{2,2}G(y_{0}^{\prime},\eta^{\prime})=0,\quad
J^{1,1}F(y_{0}^{\prime},\eta^{\prime})=0,$

for all
$(y_{0}^{\prime},\eta^{\prime})\in\mathbb{R}^{n-1}\times\mathbb{S}^{n-2}$.
Thus, Theorem 2.4 implies that $f^{(1)}=0$. Now the equality
$J^{2,2}G(y_{0}^{\prime},\eta^{\prime})=0$ is precisely the equality (B.21) in
the proof of Lemma 2.6. Hence, in view of the fact that $j_{\delta}g^{(2)}=0$,
we conclude that $g^{(2)}=0$. The proof of Lemma 2.8 is complete.

## Acknowledgements

The research of S.B. is partially supported by the grant: SRG/2022/00129 from
the Science and Engineering Research Board, India. The research of K.K. is
partially supported by the National Science Foundation (DMS 2109199). The
research of S. K. S is partly supported by the Academy of Finland (Centre of
Excellence in Inverse Modelling and Imaging, grant 284715) and by the European
Research Council under Horizon 2020 (ERC CoG 770924). The research of G.U. is
partially supported by NSF, a Robert R. and Elaine F. Phelps Endowed
Professorship at the University of Washington, a Si-Yuan Professorship at IAS,
HKUST.

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