# Uniqueness for the anisotropic fractional conductivity equation

Giovanni Covi  Institut fur Angewandte Mathematik, Ruprecht-Karls-Universität
Heidelberg, Im Neuenheimer Feld 205, 69120 Heidelberg, Germany
<EMAIL_ADDRESS>
###### Abstract.

In this paper we study an inverse problem for fractional anisotropic
conductivity. Our nonlocal operator is based on the well-developed theory of
nonlocal vector calculus, and differs substantially from other generalizations
of the classical anisotropic conductivity operator obtained spectrally. We
show that the anisotropic conductivity matrix can be recovered uniquely from
fractional Dirichlet-to-Neumann data up to a natural gauge. Our analysis makes
use of techniques recently developed for the study of the isotropic fractional
elasticity equation, and generalizes them to the case of non-separable,
anisotropic conductivities. The motivation for our study stems from its
relation to the classical anisotropic Calderón problem, which at the time of
writing is one of the main open problems in the field.

###### Key words and phrases:

Fractional Laplacian, fractional gradient, Calderón problem, conductivity
equation

###### 2020 Mathematics Subject Classification:

Primary 35R30; secondary 26A33, 42B37, 46F12

##  1\. Introduction

Let $s\in(0,1)$. We study the anisotropic fractional conductivity operator

$\mathbf{C}^{s}_{A}u:=(\nabla\cdot)^{s}(A(x,y)\cdot\nabla^{s}u),$

where $A\in L^{\infty}({\mathbb{R}}^{2n},{\mathbb{R}}^{n\times n})$ is an
$n\times n$ matrix function of the variables $x,y\in{\mathbb{R}}^{n}$. The
operators $(\nabla\cdot)^{s},\nabla^{s}$, which we will define in detail in
the upcoming Section 2, are the fractional divergence and gradient. They were
firstly introduced in the nonlocal vector calculus presented in [DGLZ12], and
their properties were further studied in [Cov20a, Cov20b, CRZ22], among
others. They represent fractional counterparts to the classical divergence and
gradient operators. As a matter of fact, they enjoy the expected properties
$(\nabla^{s})^{*}=(\nabla\cdot)^{s}$ and
$(\nabla\cdot)^{s}\nabla^{s}=(-\Delta)^{s}$. The operator $\mathbf{C}^{s}_{A}$
will be rigorously defined in Section 3.

Our inverse problem consists in recovering the values of $A$ in $\Omega$ from
measurements performed in the exterior
$\Omega_{e}:={\mathbb{R}}^{n}\setminus\overline{\Omega}$. These measurement
will be given in the form of a Dirichlet-to-Neumann (DN) map $\Lambda_{A}$,
associating each exterior datum $f$ of the direct problem

$\begin{array}[]{rll}\mathbf{C}^{s}_{A}u&=0&\quad\mbox{ in }\Omega\\\
u&=f&\quad\mbox{ in }\Omega_{e}\end{array}$

to the corresponding nonlocal Neumann datum
$\mathbf{C}^{s}_{A}u_{f}|_{\Omega_{e}}$. Here $u_{f}\in
H^{s}({\mathbb{R}}^{n})$ is the unique weak solution of the above problem
associated to $f$. We will show that there exists in fact only one such
$u_{f}$ in Section 4, and thus $\Lambda_{A}$ is well-defined. In order to
represent the inaccessibility to the measurements of certain regions of space,
which is a very common physical situation, we assume that $\Lambda_{A}f$ is
known only on $W_{2}$ and only for $f\in C^{\infty}_{c}(W_{1})$, where
$W_{1},W_{2}\subset\Omega_{e}$ are non-empty, open and disjoint subsets of the
exterior.

The inverse problem we are interested in can be formulated as follows:

Q1: _Does $\Lambda_{A_{1}}f|_{W_{2}}=\Lambda_{A_{2}}f|_{W_{2}}$ for all $f\in
C^{\infty}_{c}(W_{1})$ imply $A_{1}=A_{2}$?_

However, by Remark 3.1 it becomes soon evident that the answer to question Q1
can not be positive, since the anisotropic fractional conductivity operator
$\mathbf{C}^{s}_{A}$ possesses a natural gauge related to the properties of
symmetry of the matrix $A$. Therefore, we shall rather ask:

Q2: _Does $\Lambda_{A_{1}}f|_{W_{2}}=\Lambda_{A_{2}}f|_{W_{2}}$ for all $f\in
C^{\infty}_{c}(W_{1})$ imply $A_{1,s}=A_{2,s}$?_

Here $A_{s}$ is a symmetrized version of the matrix $A$, in a sense that will
be made clear in the same Remark 3.1.

The study of the fractional Calderón problem dates back to the seminal article
[GSU20], which considered the recovery of a bounded potential for the
fractional Schrödinger equation in a bounded domain from measurements
performed in the exterior in the form of a nonlocal DN map. This result was
then generalized in many different directions by the works [BGU21, GLX17,
GRSU20, Cov22, CGFR22, CMRU22, Li20b, Li20a, Li21, KRS21, RS19], to cite a
few. These contributions address the problem of uniqueness in the presence of
both local and nonlocal perturbations, the problem of stability, and that of
reconstruction. The inverse problem we study represents a generalization of
the inverse problem for the _isotropic_ fractional conductivity equation,
which was first introduced in [Cov20b] and later studied quite extensively in
the works [Cov20a, CRZ22, CRTZ22, LRZ22, RZ22], among others.

Our study is also related to the inverse problem for the anisotropic
_classical_ conductivity equation. This is one of the main open problems in
the field; it is only well understood at the time of writing in dimension
$n=2$ (see [LU01]), and only partial results are available in dimension $n\geq
3$ (see [LTU03, LU89, DSFKSU07, DSFKLS16]).

The very recent article [FGKU21] also bears a relation to ours, in that the
authors of this work also consider the problem of uniqueness for a fractional
anisotropic conductivity operator. This is however obtained spectrally, and
thus generalizes the classical anisotropic conductivity operator in a
different way than done here. Therefore, the used techniques and obtained
results differ substantially.

In order to answer question Q2, we attempt to adopt the technique from
[GSU20]. We first prove the Alessandrini identity, that is, an integral
identity which relates the difference of the DN maps
$\Lambda_{A_{1}},\Lambda_{A_{2}}$ to the differences of the matrices
$A_{1},A_{2}$ using special solutions to the direct problem. The left-hand
side of the Alessandrini identity is then equated to $0$ by the assumption on
the DN maps. By testing the resulting identity with particular solutions, we
are able to deduce the desired result. In order to produce such solutions we
prove a Runge approximation property, using a technique involving the Hahn-
Banach theorem and the unique continuation property (UCP) for the fractional
Laplacian.

However, as observed in [CdHS22, Remark 7.3], this technique will not lead to
the wanted result by itself. This is due to the fact that the operator
appearing in the right-hand side of the Alessandrini identity is nonlocal,
which causes problems with the supports of both solutions and data. This issue
may be avoided by means of the so called _fractional Liouville reduction_ ,
which transforms a conductivity-type problem in a Schrödinger-type one. This
technique, whose name and idea come from the method used for the classical
conductivity equation, was introduced in [Cov20a] for the _isotropic_
fractional conductivity equation and later generalized in [CdHS22] for
studying the inverse problem for fractional linear elasticity. Once the
reduction is performed (see Section 4), the direct problem takes the form

(1) $\begin{split}\mathbf{(}-\Delta)^{s-1}Dw+w\cdot Q=0&\quad\mbox{ in
}\Omega\\\ w=g&\quad\mbox{ in }\Omega_{e}\end{split},$

where the new solution $w$ and the new exterior datum $g$ are respectively
computed from $u$ and $f$. Here the transformed potential $Q$ contains all the
information about the matrix $A$, and $D$ is a fixed second order differential
operator. As it will be shown in Section 6, the technique based on the Runge
approximation property explained above works for the transformed problem. This
is due to the fact that the relative Alessandrini identity has a local
operator on the right-hand side, namely the matrix multiplication of
$Q_{1}-Q_{2}$ against two special solutions of the transformed problem.

There is however one more obstacle which needs to be addressed. In both
[Cov20a] and [CdHS22] the coefficient part of the considered unreduced
operators, respectively the matrix $\gamma^{1/2}(x)\gamma^{1/2}(y)$Id and the
tensor $C^{1/2}(x):C^{1/2}(y)$, were by definition separable functions of the
variables $x$ and $y$. This detail is of great importance at the time of
proving the fractional Liouville reduction. Since our matrix $A(x,y)$ is not
in general a separable function of the variables $x$ and $y$, we are prevented
from applying the fractional Liouville reduction directly. However, in Section
3 we shall prove the existence of a decomposition of the kind

(2) $A_{s}(x,y)=\sum_{k}\Phi_{k}(x)\odot\Phi_{k}(y),$

where $\\{\Phi_{k}\\}_{k\in{\mathbb{N}}}$ is a suited sequence of matrices
depending on one variable, and $\odot$ indicates the Hadamard product (see
Section 2 for the definition of this and other matrix operations). Once this
formula is achieved, it will be possible to show that a fractional Liouville
reduction holds for each element of the above sum. As a consequence, the
functions $w,Q$ and $g$ in the reduced problem will rather be sequences.

In order to obtain a suitable Alessandrini identity for the transformed
problem, we need to additionally assume that the anisotropic coefficients are
in gauge in the following sense:

######  Definition 1.1.

Let $\Omega\subset{\mathbb{R}}^{n}$ be open and bounded. We say that
$A_{1},A_{2}\in L^{\infty}({\mathbb{R}}^{2n},{\mathbb{R}}^{n\times n})$ are in
gauge if there exists $\rho\in C^{\infty}_{c}(\Omega)$ such that
$A_{2,s}(x,y)=(1+\rho(x))(1+\rho(y))A_{1,s}(x,y)$ for all
$x,y\in{\mathbb{R}}^{n}$. In this case we write $A_{1}\sim A_{2}$.

Using the techniques described above, we have obtained the following
uniqueness result:

######  Theorem 1.2.

Let $\Omega,W_{1},W_{2}\subset\mathbb{R}^{n}$ be bounded open sets such that
$W_{1},W_{2}\subset\Omega_{e}$, and assume $s\in(0,1)$. Let $A_{1},A_{2}\in
L^{\infty}({\mathbb{R}}^{2n},{\mathbb{R}}^{n\times n})$ be two anisotropic
coefficients satisfying assumptions (A1)–(A3) and (A4). Then $A_{1}\sim A_{2}$
and

$\Lambda_{A_{1}}f|_{W_{2}}=\Lambda_{A_{2}}f|_{W_{2}}\qquad\mbox{ for all }f\in
C^{\infty}_{c}(W_{1})$

hold if and only if $A_{1,s}=A_{2,s}$.

Assumptions (A1)–(A3) describe the Hölder regularity of $A$, its behaviour in
${\mathbb{R}}^{2n}\setminus\Omega^{2}$, and a uniform positivity condition
which is required for the well-posedness of the direct problem (see Section
3.1). Assumption (A4) is only needed for the proof of the Runge approximation
property, and it is an isotropicity requirement for the exterior values of the
coefficients (see Lemma 6.1).

###  1.1. Organization of the rest of the article

The article is organized as follows. The first section is the Introduction,
which presents the main problem, its connection to the literature, the
statements of the main results and the techniques involved in their proofs.
The second section is dedicated to the necessary Preliminaries and
definitions, in particular about tensors, fractional operators, Hölder and
Sobolev spaces. Section 3 discusses our assumptions for the anisotropy matrix,
introduces the anisotropic fractional conductivity operator, and proves our
fundamental reduction result. The fourth section treats the problem of well-
posedness both for our original problem and for the transformed one,
presenting the fractional Liouville reduction related to our discussion. The
DN map and Alessandrini identity are both presented in Section 5, while
Section 6 contains the Runge approximation property and the proof of the main
theorem. Finally, the last section 7 gives the relation between the classical
and fractional problems of anisotropic conductivity in the limit case
$s\rightarrow 1$.

##  2\. Preliminaries

###  2.1. Operations on tensors

In this section we introduce a number of operations between tensors and
sequences of tensors. Define ${\mathbb{N}}$ as the set of positive integers,
${\mathbb{N}}_{0}:={\mathbb{N}}\cup\\{0\\}$ as the set of non-negative
integers, and ${\mathbb{N}}_{\infty}:={\mathbb{N}}\cup\\{\infty\\}$ as the set
of _extended_ positive integers. In all the definitions below, we assume that
$m,n\in\mathbb{N}$, that
$a\in\mathbb{N}_{\infty}^{m},b\in\mathbb{N}_{\infty}^{n}$ are two (extended)
multi-indices, and that $A,B$ are two (extended) tensors respectively given by
$A_{\alpha_{1},...,\alpha_{m}}$ and $B_{\beta_{1},...,\beta_{n}}$, with
$\alpha_{i}\in\\{1,...,a_{i}\\}$ for $i=1,...,m$ and
$\beta_{j}\in\\{1,...,b_{j}\\}$ for $j=1,...,n$.

######  Definition 2.1 (Tensor product).

The tensor $A\otimes B$ of elements

$(A\otimes
B)_{\alpha_{1},...,\alpha_{m},\beta_{1},...,\beta_{n}}:=A_{\alpha_{1},...,\alpha_{m}}B_{\beta_{1},...,\beta_{n}}$

is the _tensor product_ of $A$ and $B$. If $B$ is a scalar, in which case
$n=b_{1}=1$, we simply write $AB$ for $A\otimes B$.

######  Definition 2.2 (Hadamard product).

Assume $m=n$ and $a=b$. The tensor $A\odot B$ of elements

$(A\odot
B)_{\alpha_{1},...,\alpha_{m}}:=A_{\alpha_{1},...,\alpha_{m}}B_{\alpha_{1},...,\alpha_{m}}$

is the _Hadamard product_ of $A$ and $B$.

######  Definition 2.3 (Tensor contraction of order $k$).

Let $k\in\mathbb{N}$ verify $k\leq\min\\{m,n\\}$, and let
$a_{m+\ell-k}=b_{\ell}$ for all $\ell\in\\{1,...,k\\}$. The tensor
$A\cdot_{k}B$ of elements

$(A\cdot_{k}B)_{\alpha_{1},...,\alpha_{m-k},\beta_{k+1},...,\beta_{n}}:=\sum_{\beta_{1}=1}^{b_{1}}...\sum_{\beta_{k}=1}^{b_{k}}A_{\alpha_{1},...,\alpha_{m-k},\beta_{1},...,\beta_{k}}B_{\beta_{1},...,\beta_{n}}$

is the _$k$ -th contraction_ of $A$ and $B$. For simplicity we use $\cdot$ for
$\cdot_{1}$, $:$ for $\cdot_{2}$, and $\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}$ for $\cdot_{3}$.

A sequence $\\{A_{k}\\}_{k\in{\mathbb{N}}}$ of (possibly extended) tensors can
itself be seen as an extended tensor $A_{k,\alpha_{1},...\alpha_{n}}$ with one
more index $k$ with range in ${\mathbb{N}}$. With this identification in mind,
all the above definitions can be extended to sequences of tensors. They can
also be naturally extended to functions and function spaces (see [CdHS22] for
more properties relative to these tensor operations).

######  Remark 2.4.

It is customary to interpret the symbols $\nabla u,\nabla\cdot u$ for the
gradient and divergence of a function respectively as a tensor and a scalar
product (i.e. a contraction), where $\nabla$ formally represents a vector of
partial derivatives. Building on this, we use the symbol
$\nabla^{2}u:=(\nabla\otimes\nabla)u$ for the Hessian matrix of $u$.
Similarly, with clear meaning of the symbols, $\nabla^{2}\odot$ is the
differential operator mapping an $n\times n$ matrix $M$ to $\nabla^{2}\odot
M$, with $(\nabla^{2}\odot M)_{ij}:=\partial_{i}\partial_{j}M_{ij}$.

Observe that here and everywhere else we are _not_ assuming the Einstein
summation convention on repeated indices.

###  2.2. Fractional Sobolev spaces

Let
$\hat{u}(\xi)=\mathcal{F}u(\xi):=\int_{\mathbb{R}^{n}}e^{-ix\cdot\xi}u(x)\,dx$
indicate the Fourier transform. If $r$ is any real number, we indicate the
usual $L^{2}$-based Bessel potential spaces by $H^{r}({\mathbb{R}}^{n})$, with
norm

$\|u\|_{H^{r}}:=\|(1+|\xi|^{2})^{r/2}\hat{u}(\xi)\|_{L^{2}({\mathbb{R}}^{n})}.$

Let $\Omega,F\subset{\mathbb{R}}^{n}$ respectively be an open and a closed
set. Following [McL00] and [GSU20], we define the fractional Sobolev spaces

$\displaystyle H^{r}(\Omega)$ $\displaystyle:=\\{u|_{\Omega},u\in
H^{r}(\mathbb{R}^{n})\\},$ $\displaystyle\widetilde{H}^{r}(\Omega)$
$\displaystyle:=\,\mbox{closure of }C^{\infty}_{c}(\Omega)\mbox{ in
}H^{r}(\mathbb{R}^{n}),$ $\displaystyle H^{r}_{F}(\mathbb{R}^{n})$
$\displaystyle:=\\{u\in H^{r}(\mathbb{R}^{n}):\mbox{ supp}(u)\subset F\\}.$

As usual, the quotient norm $\|u\|_{H^{r}(\Omega)}:=\inf\\{\|w\|_{H^{r}}:w\in
H^{r}(\mathbb{R}^{n}),w|_{\Omega}=u\\}$ is used for $H^{r}(\Omega)$. The
spaces defined above are related to each other in the following ways:

$\widetilde{H}^{r}(\Omega)\subseteq
H^{r}(\Omega)\,,\quad\widetilde{H}^{r}(\Omega)\subseteq
H^{r}_{\overline{\Omega}}\,,\quad H^{r}_{F}\subseteq H^{r}\,,$
$(\widetilde{H}^{r}(\Omega))^{*}=H^{-r}(\Omega)\,,\quad(H^{r}(\Omega))^{*}=\widetilde{H}^{-r}(\Omega).$

Moreover, if $\partial\Omega$ is Lipschitz, then
$\widetilde{H}^{r}(\Omega)=H^{r}_{\overline{\Omega}}$ for all $r\in\mathbb{R}$
(see [CMRU22, McL00]).

###  2.3. Fractional operators

Let $\mathcal{S}$ be the set of Schwartz functions, and assume
$s\in\mathbb{R}^{+}\setminus\mathbb{Z}$. The _fractional Laplacian_ is the
continuous map from $\mathcal{S}$ to $L^{\infty}$ defined by

$(-\Delta)^{s}u:=\mathcal{F}^{-1}(|\xi|^{2s}\hat{u}(\xi))$

for $u\in\mathcal{S}$. One can uniquely extend this definition to have
$(-\Delta)^{s}:H^{r}(\mathbb{R}^{n})\rightarrow H^{r-2s}(\mathbb{R}^{n})$
continuously for all $r\in\mathbb{R}$. It is also possible to have extensions
to Sobolev spaces of negative exponent and to $L^{p}$-based Sobolev spaces
(see e.g. [CMR21, GSU20]). Alternatively, one can define the fractional
Laplacian as the singular integral

$(-\Delta)^{s}u(x):=C_{n,s}\,PV\int_{\mathbb{R}^{n}}\frac{u(x)-u(y)}{|x-y|^{n+2s}}dy,$

where $C_{n,s}:=\frac{4^{s}\Gamma(n/2+s)}{\pi^{n/2}|\Gamma(-s)|}$. Many other
equivalent definitions are also possible (see [Kwa17]). The fractional
Laplacian verifies the UCP (unique continuation property): If $u\in
H^{r}(\mathbb{R}^{n})$ is such that $u=(-\Delta)^{s}u=0$ in a non-empty open
set $V$, then $u\equiv 0$ in $\mathbb{R}^{n}$. The proof of this fact can be
found in [GSU20] for $s\in(0,1)$ and in [CMR21] for higher exponents. The
latter paper also proves the following fractional Poincaré inequality: If
$s\geq t\geq 0$ and $u\in H_{K}^{s}({\mathbb{R}}^{n})$ for a compact set
$K\subset{\mathbb{R}}^{n}$, then there exists a constant
$\widetilde{c}=\widetilde{c}(n,K,s)>0$ such that

$\left\lVert(-\Delta)^{t/2}u\right\rVert_{L^{2}({\mathbb{R}}^{n})}\leq\widetilde{c}\left\lVert(-\Delta)^{s/2}u\right\rVert_{L^{2}({\mathbb{R}}^{n})}.$

We also need the fractional gradient and divergence operators $\nabla^{s}$ and
$(\nabla\cdot)^{s}$. These were first introduced in the nonlocal vector
calculus developed in [DGLZ12, DGLZ13], and were later studied in their
present form in [Cov20a, Cov20b, CMR21, CRZ22, CdHS22] and more. Let $u\in
C^{\infty}_{c}(\mathbb{R}^{n})$ and $s\in(0,1)$. The fractional gradient of
$u$ is the two-points function given by

$\nabla^{s}u(x,y):=(u(y)-u(x))\otimes\zeta(x,y),\quad\mbox{ where
}\zeta(x,y):=\frac{C_{n,s}^{1/2}}{\sqrt{2}}\frac{x-y}{|x-y|^{n/2+s+1}}.$

This definition can be uniquely extended to act as
$H^{s}(\mathbb{R}^{n})\rightarrow L^{2}(\mathbb{R}^{2n})$. The adjoint of
$\nabla^{s}$, which is indicated by $(\nabla\cdot)^{s}$ and acts as
$L^{2}(\mathbb{R}^{2n})\rightarrow H^{-s}(\mathbb{R}^{n})$, is by definition
the fractional divergence. One can observe that the property

$(\nabla\cdot)^{s}\nabla^{s}u=(-\Delta)^{s}u$

holds in $H^{s}(\mathbb{R}^{n})$. Because of their mapping properties and of
the particular relation they have among themselves and with the Laplacian, the
fractional gradient and divergence can be though of as nonlocal counterparts
of the classical gradient and divergence operators.

###  2.4. Hölder spaces

In this Section we introduce the Hölder spaces and their main properties,
following [Tay96] (see also [CdHS22]). If $r\in[0,1)$ and
$k\in{\mathbb{N}}_{0}$, we define

(3) $\displaystyle
C^{r}(\mathbb{R}^{n}):=\\{u:{\mathbb{R}}^{n}\rightarrow{\mathbb{C}}\;:\;u\mbox{
bounded and }|u(x)-u(y)|\leq C|x-y|^{r}\mbox{ for all
}x,y\in\mathbb{R}^{n}\\},$ (4) $\displaystyle
C^{k}(\mathbb{R}^{n}):=\\{u:{\mathbb{R}}^{n}\rightarrow{\mathbb{C}}\;:\;D^{\beta}u\mbox{
bounded continuous for all }|\beta|\leq k\\},$ (5) $\displaystyle
C^{r+k}(\mathbb{R}^{n}):=\\{u\in C^{k}(\mathbb{R}^{n})\;:\;D^{\beta}u\in
C^{r}(\mathbb{R}^{n})\mbox{ for all }|\beta|=k\\}.$

The Hölder norm of $u\in C^{r+k}({\mathbb{R}}^{n})$ is defined as

$\|u\|_{C^{r+k}({\mathbb{R}}^{n})}:=\max_{|\beta|\leq
k}\|D^{\beta}u\|_{L^{\infty}({\mathbb{R}}^{n})}+\max_{|\beta|=k}\sup_{x\neq
y}\frac{|D^{\beta}u(x)-D^{\beta}u(y)|}{|x-y|^{r}}.$

Moreover, the following properties are known to hold:

  1. (H1)

Continuous embedding property:

If $r,r^{\prime}\in{\mathbb{R}}^{+}$ are such that $r\leq r^{\prime}$, then
$C^{r^{\prime}}(\mathbb{R}^{n})\hookrightarrow C^{r}(\mathbb{R}^{n})$.

  2. (H2)

Closure under composition with smooth functions:

If $r\in{\mathbb{R}}^{+}$, $u\in C^{r}({\mathbb{R}}^{n})$ and $F\in
C^{\infty}({\mathbb{R}})$, then $F(u)\in C^{r}({\mathbb{R}}^{n})$ as well.

  3. (H3)

Algebra property:

If $r\in{\mathbb{R}}^{+}\setminus\mathbb{Z}^{+}$ and $u,v\in
C^{r}({\mathbb{R}}^{n})$, then $uv\in C^{r}({\mathbb{R}}^{n})$ as well.

  4. (H4)

Mapping property for pseudodifferential operators:

If $r\in{\mathbb{R}}^{+}\setminus\mathbb{Z}^{+}$ and $m\in{\mathbb{R}}$ verify
$r-m\in{\mathbb{R}}^{+}\setminus\mathbb{Z}^{+}$, and $\Psi\in OPS^{m}_{1,0}$
is a $\Psi$DO with symbol in the Hörmander class $S^{m}_{1,0}$, then $\Psi$
maps $C^{r}({\mathbb{R}}^{n})$ continuously to $C^{r-m}({\mathbb{R}}^{n})$.

  5. (H5)

Mapping property for $(-\Delta)^{s-1}\partial_{i}\partial_{j}$:

If $r\in{\mathbb{R}}^{+}\setminus\mathbb{Z}^{+}$ and $s\in(0,1)$ verify
$r-2s\in{\mathbb{R}}^{+}\setminus\mathbb{Z}^{+}$, then
$(-\Delta)^{s-1}\partial_{i}\partial_{j}$ maps $C^{r}({\mathbb{R}}^{n})$
continuously to $C^{r-2s}({\mathbb{R}}^{n})$.

We will also make use of the following Lemma:

######  Lemma 2.5.

Let $s\in(0,1)$ and $\varepsilon>0$. If $h\in
C^{s+\varepsilon}({\mathbb{R}}^{n})$ and $u\in H^{s}({\mathbb{R}}^{n})$, then
$hu\in H^{s}({\mathbb{R}}^{n})$ as well, and there exists a constant
$C=C(n,s,\varepsilon)>0$ such that

$\|hu\|_{H^{s}({\mathbb{R}}^{n})}\leq
C\|h\|_{C^{s+\varepsilon}({\mathbb{R}}^{n})}\|u\|_{H^{s}({\mathbb{R}}^{n})}.$

###### Proof.

Let $u,h$ be as in the statement. Then we know that

$\|hu\|^{2}_{H^{s}({\mathbb{R}}^{n})}\lesssim\|hu\|^{2}_{L^{2}({\mathbb{R}}^{n})}+[hu]^{2}_{H^{s}({\mathbb{R}}^{n})},$

(see e.g. [DNPV12]) where the Gagliardo seminorm is given by

$[hu]^{2}_{H^{s}({\mathbb{R}}^{n})}:=\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\frac{|h(x)u(x)-h(y)u(y)|^{2}}{|x-y|^{n+2s}}dxdy.$

Since $h\in L^{\infty}({\mathbb{R}}^{n})$, the first term on the right hand
side is immediately estimated as

$\|hu\|_{L^{2}({\mathbb{R}}^{n})}\leq\|h\|_{L^{\infty}({\mathbb{R}}^{n})}\|u\|_{L^{2}({\mathbb{R}}^{n})}\leq\|h\|_{L^{\infty}({\mathbb{R}}^{n})}\|u\|_{H^{s}({\mathbb{R}}^{n})}.$

For the seminorm we compute

$\displaystyle[hu]^{2}_{H^{s}({\mathbb{R}}^{n})}$
$\displaystyle\lesssim\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\frac{|h(x)|^{2}|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}dxdy+\int_{{\mathbb{R}}^{n}}\int_{{\mathbb{R}}^{n}}\frac{|h(x)-h(y)|^{2}|u(y)|^{2}}{|x-y|^{n+2s}}dxdy=:I_{1}+I_{2},$

and then observe that

$\displaystyle I_{1}$
$\displaystyle\leq\|h\|_{L^{\infty}({\mathbb{R}}^{n})}^{2}[u]^{2}_{H^{s}({\mathbb{R}}^{n})}\leq\|h\|_{L^{\infty}({\mathbb{R}}^{n})}^{2}\|u\|^{2}_{H^{s}({\mathbb{R}}^{n})},$

so that we only need to estimate $I_{2}$. With a change of variables we get

$\displaystyle I_{2}$
$\displaystyle=\int_{{\mathbb{R}}^{n}}|u(y)|^{2}\int_{{\mathbb{R}}^{n}}\frac{|h(y+z)-h(y)|^{2}}{|z|^{n+2s}}dzdy,$

and then we split the second integral as follows, where $B$ is the unitary
ball centered at the origin:

$\displaystyle\int_{{\mathbb{R}}^{n}}\frac{|h(y+z)-h(y)|^{2}}{|z|^{n+2s}}dz$
$\displaystyle=\int_{B}\frac{|h(y+z)-h(y)|^{2}}{|z|^{n+2s}}dz+\int_{{\mathbb{R}}^{n}\setminus
B}\frac{|h(y+z)-h(y)|^{2}}{|z|^{n+2s}}dz$
$\displaystyle\lesssim\|h\|_{C^{s+\varepsilon}({\mathbb{R}}^{n})}^{2}\int_{B}\frac{|z|^{2s+2\varepsilon}}{|z|^{n+2s}}dz+\|h\|^{2}_{L^{\infty}({\mathbb{R}}^{n})}\int_{{\mathbb{R}}^{n}\setminus
B}\frac{1}{|z|^{n+2s}}dz.$

Observe that the remaining integrals converge to constant values depending on
$n,s,\varepsilon$. Thus

$I_{2}\lesssim\|h\|_{C^{s+\varepsilon}({\mathbb{R}}^{n})}^{2}\int_{{\mathbb{R}}^{n}}|u(y)|^{2}dy=\|h\|_{C^{s+\varepsilon}({\mathbb{R}}^{n})}^{2}\|u\|^{2}_{L^{2}({\mathbb{R}}^{n})}\leq\|h\|_{C^{s+\varepsilon}({\mathbb{R}}^{n})}^{2}\|u\|^{2}_{H^{s}({\mathbb{R}}^{n})}.$

∎

###  2.5. Spaces of tensors

If $A=A_{\alpha_{1},...,\alpha_{m}}$ with $\alpha_{i}\in\\{1,...,a_{i}\\}$ for
$i=1,...,m$ and $a\in\mathbb{N}_{\infty}^{m}$ for $m\in{\mathbb{N}}$ is any
(extended) tensor whose elements belong to $C^{r}$, $r\in{\mathbb{R}}$, we
define its $C^{r}$ norm as

$\|A\|_{C^{r}}:=\sum_{\alpha_{1}=1}^{a_{1}}...\sum_{\alpha_{m}=1}^{a_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r}}.$

Similarly, if $B=B_{\beta_{1},...,\beta_{n}}$ with
$\beta_{j}\in\\{1,...,b_{j}\\}$ for $j=1,...,n$ and
$b\in\mathbb{N}_{\infty}^{n}$ for $n\in{\mathbb{N}}$ is any (extended) tensor
whose elements belong to $H^{r}$, we define its $H^{r}$ norm as

$\|B\|_{H^{r}}:=\sum_{\beta_{1}=1}^{b_{1}}...\sum_{\beta_{n}=1}^{b_{n}}\|B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}.$

The result of Lemma 2.5 extends to products of tensors: if $A\in
C^{r+\varepsilon}$ and $B\in H^{r}$, then $A\otimes B,A\odot B$ and
$A\cdot_{k}B$ all belong to $H^{r}$ whenever the operations make sense, since

(6) $\begin{split}\|A\otimes
B\|_{H^{r}}&=\sum_{\alpha_{1},...,\alpha_{m},\beta_{1},...,\beta_{n}}\|A_{\alpha_{1},...,\alpha_{m}}B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}\\\
&\lesssim\sum_{\alpha_{1},...,\alpha_{m},\beta_{1},...,\beta_{n}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r+\varepsilon}}\|B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}\\\
&=\sum_{\alpha_{1},...,\alpha_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r+\varepsilon}}\sum_{\beta_{1},...,\beta_{n}}\|B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}=\|A\|_{C^{r+\varepsilon}}\|B\|_{H^{r}},\end{split}$
(7) $\begin{split}\|A\odot
B\|_{H^{r}}&=\sum_{\alpha_{1},...,\alpha_{m}}\|A_{\alpha_{1},...,\alpha_{m}}B_{\alpha_{1},...,\alpha_{m}}\|_{H^{r}}\\\
&\lesssim\sum_{\alpha_{1},...,\alpha_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r+\varepsilon}}\|B_{\alpha_{1},...,\alpha_{m}}\|_{H^{r}}\\\
&\leq\sum_{\alpha_{1},...,\alpha_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r+\varepsilon}}\sum_{\beta_{1},...,\beta_{n}}\|B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}=\|A\|_{C^{r+\varepsilon}}\|B\|_{H^{r}},\end{split}$

and

(8)
$\begin{split}\|A\cdot_{k}B\|_{H^{r}}&=\sum_{\alpha_{1},...,\alpha_{m-k},\beta_{k+1},...,\beta_{n}}\|\sum_{\beta_{1},...,\beta_{k}}A_{\alpha_{1},...,\alpha_{m-k},\beta_{1},...,\beta_{k}}B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}\\\
&\leq\sum_{\alpha_{1},...,\alpha_{m-k},\beta_{1},...,\beta_{n}}\left\|A_{\alpha_{1},...,\alpha_{m-k},\beta_{1},...,\beta_{k}}B_{\beta_{1},...,\beta_{n}}\right\|_{H^{r}}\\\
&\lesssim\sum_{\alpha_{1},...,\alpha_{m-k},\beta_{1},...,\beta_{n}}\left\|A_{\alpha_{1},...,\alpha_{m-k},\beta_{1},...,\beta_{k}}\|_{C^{r+\varepsilon}}\|B_{\beta_{1},...,\beta_{n}}\right\|_{H^{r}}\\\
&\leq\sum_{\alpha_{1},...,\alpha_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r+\varepsilon}}\sum_{\beta_{1},...,\beta_{n}}\|B_{\beta_{1},...,\beta_{n}}\|_{H^{r}}=\|A\|_{C^{r+\varepsilon}}\|B\|_{H^{r}}.\end{split}$

Moreover, if $A\neq 0$ we have $A/|A|^{2}\in
C^{r+\varepsilon}({\mathbb{R}}^{n})$ by (H2), since the function
$F(x)=x/|x|^{2}$ is smooth away from the origin. Therefore by the above
computation

(9) $\begin{split}\|B\|_{H^{r}}=\left\|\frac{A}{|A|^{2}}\cdot_{m}(A\otimes
B)\right\|_{H^{r}}&\lesssim\left\|\frac{A}{|A|^{2}}\right\|_{C^{r+\varepsilon}}\left\|A\otimes
B\right\|_{H^{r}}.\end{split}$

It now easily follows that if $A\in C^{r+\varepsilon}$, then $A\otimes H^{r}$
is a Hilbert subspace of $H^{r}$.

######  Remark 2.6.

We might as well have chosen to define

(10)
$\|A\|^{2}_{C^{r}}:=\sum_{\alpha_{1}=1}^{a_{1}}...\sum_{\alpha_{m}=1}^{a_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|^{2}_{C^{r}},$

and similarly for the $H^{r}$ norm of a tensor. If the tensors are not
extended, i.e. if $a\in\mathbb{N}^{m}$ rather than $\mathbb{N}_{\infty}^{m}$,
the two definitions of course coincide in light of the Cauchy-Schwarz
inequality for series, since

$\sum_{\alpha_{1}=1}^{a_{1}}...\sum_{\alpha_{m}=1}^{a_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r}}\leq\left(\sum_{\alpha_{1}=1}^{a_{1}}...\sum_{\alpha_{m}=1}^{a_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|^{2}_{C^{r}}\right)^{1/2}\prod_{j=1}^{m}a_{j}^{1/2}$

and

$\sum_{\alpha_{1}=1}^{a_{1}}...\sum_{\alpha_{m}=1}^{a_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|^{2}_{C^{r}}\leq\left(\sum_{\alpha_{1}=1}^{a_{1}}...\sum_{\alpha_{m}=1}^{a_{m}}\|A_{\alpha_{1},...,\alpha_{m}}\|_{C^{r}}\right)^{2}.$

However, since the first inequality does not hold for extended tensors, the
two definitions are not equivalent for $a\in\mathbb{N}_{\infty}^{m}$. In
particular, definition (10) would not allow us to get the desired estimate for
$\|A\cdot_{k}B\|_{H^{r}}$.

##  3\. The anisotropic fractional conductivity equation

###  3.1. Assumptions

Any square matrix $A$ depending on two variables $x,y$ is linked to two
different concepts of symmetry. One can define the _matrix-wise symmetric
part_ $A_{ms}$ of $A$ as

$A_{ms}(x,y):=\frac{A(x,y)+A^{T}(x,y)}{2},$

where $T$ indicates transposition, and the _variable-wise symmetric part_
$A_{vs}$ of $A$ as

$A_{vs}(x,y):=\frac{A(x,y)+A(y,x)}{2}.$

One sees that $A_{ms}$ is a symmetric matrix in the sense that
$A_{ms}=A_{ms}^{T}$, while $A_{vs}$ is a symmetric function of the variables
$x,y$ in the sense that $A_{vs}(x,y)=A_{vs}(y,x)$. We also define the
_(matrix-wise and variable-wise) antisymmetric parts_ of $A$ as

$A_{ma}:=A-A_{ms},\qquad A_{va}:=A-A_{vs}.$

Thus we can write

$A=A_{vs}+A_{va}=(A_{vs})_{ms}+(A_{vs})_{ma}+A_{va}=A_{s}+A_{a},$

where $A_{s}:=(A_{vs})_{ms}$ is symmetric both matrix-wise and variable-wise,
and $A_{a}:=(A_{vs})_{ma}+A_{va}$.

Let now $s\in(0,1)$ and $\Omega\subset{\mathbb{R}}^{n}$ be bounded and open.
Moreover, let $A\in L^{\infty}({\mathbb{R}}^{2n},{\mathbb{R}}^{n\times n})$
verify the following assumptions:

  1. (A1)

There exist two families of matrix-wise symmetric functions
$\\{\beta_{k}\\}_{k\in{\mathbb{N}}}\subset
L^{\infty}({\mathbb{R}}^{n},{\mathbb{R}}^{n\times n})$, each constant in
$\Omega_{e}$, and $\\{\phi_{k}\\}_{k\in{\mathbb{N}}}\subset
L^{2}(\Omega,{\mathbb{C}}^{n\times n})$, each compactly supported in $\Omega$,
such that for almost every $(x,y)\in{\mathbb{R}}^{2n}$

(11)
$A_{s}(x,y)=\tilde{a}(x,y)+\sum_{k\in{\mathbb{N}}}\phi_{k}(x)\odot\phi_{k}(y),$

where

$\tilde{a}(x,y):=\sum_{k\in{\mathbb{N}}}\beta_{k}(x)\odot\beta_{k}(y)$

coincides with $A_{s}(x,y)$ whenever $(x,y)\not\in\Omega^{2}$. We also define
a third family of matrix-wise symmetric functions
$\\{\Phi_{k}\\}_{k\in{\mathbb{N}}}$ given by

$\Phi_{2k}:=\beta_{k},\qquad\Phi_{2k+1}:=\phi_{k}\qquad\qquad\mbox{for all
}k\in{\mathbb{N}},$

so that we can write

(12) $A_{s}(x,y)=\sum_{k\in{\mathbb{N}}}\Phi_{k}(x)\odot\Phi_{k}(y).$

  2. (A2)

$A_{s}$ is uniformly positive definite, i.e. there exists a constant $\nu>0$
such that

$A_{s}(x,y):(\xi\otimes\xi)\geq\nu|\xi|^{2},\qquad\mbox{ for all
}\xi\in{\mathbb{R}}^{n}\mbox{ and }x,y\in{\mathbb{R}}^{n}.$

  3. (A3)

There exists $\varepsilon>0$ such that $\Phi\in
C^{2s+\varepsilon}({\mathbb{R}}^{n})$.

In the next section we will use these assumptions in order to construct the
anisotropic fractional conductivity operator having $A$ as coefficient matrix.
The rest of this section is dedicated to comments about the assumptions.

  * •

Assumption (A3) implies that the functions $\beta_{k}$ and $\phi_{k}$ belong
to $C^{2s+\varepsilon}({\mathbb{R}}^{n})$ for all $k\in{\mathbb{N}}$. It also
gives $A_{s}\in C^{2s+\varepsilon}({\mathbb{R}}^{2n})$, since

$\displaystyle\|A_{s}\|_{C^{2s+\varepsilon}({\mathbb{R}}^{2n})}$
$\displaystyle\leq\sum_{k\in{\mathbb{N}}}\|\Phi_{k}(x)\odot\Phi_{k}(y)\|_{C^{2s+\varepsilon}({\mathbb{R}}^{2n})}$
$\displaystyle=\sum_{k\in{\mathbb{N}}}\sum_{i,j=1}^{n}\|\Phi_{k,ij}(x)\Phi_{k,ij}(y)\|_{C^{2s+\varepsilon}({\mathbb{R}}^{2n})}$
$\displaystyle\lesssim\sum_{k\in{\mathbb{N}}}\sum_{i,j=1}^{n}\|\Phi_{k,ij}\|_{C^{2s+\varepsilon}({\mathbb{R}}^{2n})}^{2}$
$\displaystyle\lesssim\sum_{k\in{\mathbb{N}}}\sum_{i,j=1}^{n}\|\Phi_{k,ij}\|_{C^{2s+\varepsilon}({\mathbb{R}}^{n})}^{2}\leq\|\Phi\|_{C^{2s+\varepsilon}({\mathbb{R}}^{n})}^{2},$

where at the third step we used property (H3), and at the following one just
the definition of Hölder norm. Similarly, $\tilde{a}\in
C^{2s+\varepsilon}({\mathbb{R}}^{2n})$.

  * •

The particular structure of the exterior values $\tilde{a}$ of $A_{s}$ emerges
naturally as a generalization of the exterior condition for the isotropic
fractional conductivity equation studied in [Cov20a]. In that case it holds

$A(x,y)=A_{s}(x,y)=\gamma^{1/2}(x)\gamma^{1/2}(y)Id=(\gamma^{1/2}(x)Id)\odot(\gamma^{1/2}(y)Id),$

and thus it is sufficient to take $\tilde{a}=A$.

  * •

The existence of a sequence $\\{\phi_{k}\\}_{k\in{\mathbb{N}}}$ as in (A1)
such that (11) holds can be proved in the assumption that the scalar function
$(A_{s}-\tilde{a})_{ij}$ is compactly supported in $\Omega^{2}$. Fix
$i,j\in\\{1,...,n\\}$, and let $\psi:=(A_{s}-\tilde{a})_{ij}$. In this case we
define the Hilbert-Schmidt integral operator $\Psi:L^{2}(\Omega)\rightarrow
L^{2}(\Omega)$ by

$\Psi u(x):=\int_{\Omega}\psi(x,y)u(y)dy.$

Since $\psi$ is real-valued, symmetric in $x,y$, and compactly supported, the
operator $\Psi$ is compact and self-adjoint. By the spectral theorem there
exists an orthonormal basis $\\{\phi_{i}\\}_{i\in{\mathbb{N}}}$ of
$L^{2}(\Omega)$ composed of eigenfunctions of $\Psi$ with real eigenvalues
$\\{\lambda_{i}\\}_{i\in{\mathbb{N}}}$ such that
$|\lambda_{1}|\geq|\lambda_{2}|\geq...\geq 0$. One immediately sees that the
new double sequence $\\{\phi_{i}(x)\phi_{j}(y)\\}_{i,j\in{\mathbb{N}}}$ is an
orthonormal basis of $L^{2}(\Omega^{2})$. By writing $\psi$ in this basis, we
see that for all $x,y\in\Omega$

$\displaystyle\psi(x,y)$
$\displaystyle=\sum_{i,j}\left(\int_{\Omega\times\Omega}\psi(w,z)\phi_{i}(w)\phi_{j}(z)dwdz\right)\phi_{i}(x)\phi_{j}(y)$
$\displaystyle=\sum_{i,j}\left(\int_{\Omega}\phi_{i}(w)\Psi\phi_{j}(w)dw\right)\phi_{i}(x)\phi_{j}(y)$
$\displaystyle=\sum_{i,j}\lambda_{j}\left(\int_{\Omega}\phi_{i}(w)\phi_{j}(w)dw\right)\phi_{i}(x)\phi_{j}(y)$
$\displaystyle=\sum_{i}\lambda_{i}\phi_{i}(x)\phi_{i}(y),$

since we have the equalities

$\Psi\phi_{j}=\lambda_{j}\phi_{j},\quad\mbox{for all
}j\in{\mathbb{N}},\qquad\mbox{and}\qquad\int_{\Omega}\phi_{i}(w)\phi_{j}(w)dw=\delta_{ij}.$

If $\phi^{\prime}_{i}$ is the extension by $0$ of $\sqrt{\lambda_{i}}\phi_{i}$
from $\Omega$ to ${\mathbb{R}}^{n}$, then the decomposition

$\psi(x,y)=\sum_{i}\phi^{\prime}_{i}(x)\phi^{\prime}_{i}(y)$

holds for all $x,y\in{\mathbb{R}}^{n}$, and formula (11) follows. Thus we see
that the decomposition given in equation (11) is actually an assumption just
on the exterior values of $A_{s}$.

If in particular $\psi$ is a positive semidefinite kernel in $\Omega^{2}$,
i.e.

$\int_{\Omega}\int_{\Omega}\psi(x,y)g(x)g(y)dxdy\geq 0,\qquad\mbox{for all
}g\in L^{2}(\Omega),$

then by Mercer’s theorem [Jö82, Section 8.7] all the eigenvalues $\lambda_{i}$
are non-negative,

$\mbox{tr}\Psi:=\int_{\Omega}\psi(x,x)dx=\sum_{i}\lambda_{i}<\infty,$

and the functions $\phi^{\prime}_{i}$ are real valued.

  * •

If $A_{1}\sim A_{2}$ as in Definition 1.1, then the decomposition (12) of
$A_{2,s}$ can be taken such that $\Phi_{2,k}(x)=(1+\rho(x))\Phi_{1,k}(x)$ for
all $x\in{\mathbb{R}}^{n}$ and $k\in{\mathbb{N}}$. In particular,
$\tilde{a}_{1}=\tilde{a}_{2}$ on $(\Omega_{e})^{2}$.

###  3.2. The operator

For this subsection we do not need any of the assumptions (A1)-(A3). The fact
that $A\in L^{\infty}$, the mapping properties of the fractional gradient
$\nabla^{s}$ and the Cauchy-Schwartz inequality are enough to imply that
$A\cdot\nabla^{s}u\in L^{2}({\mathbb{R}}^{2n})$ for all $u\in
H^{s}({\mathbb{R}}^{n})$. With this in mind, we can define the _anisotropic
fractional conductivity operator_
$\mathbf{C}^{s}_{A}:H^{s}({\mathbb{R}}^{n})\rightarrow
H^{-s}({\mathbb{R}}^{n})$ as

$\mathbf{C}^{s}_{A}u:=(\nabla\cdot)^{s}(A(x,y)\cdot\nabla^{s}u)$

for $u\in H^{s}({\mathbb{R}}^{n})$. If there exists
$\sigma:{\mathbb{R}}^{2n}\rightarrow{\mathbb{R}}$ such that
$A(x,y)=\sigma(x,y)Id$, the fractional conductivity operator is said to be
_isotropic_ , and

$\mathbf{C}^{s}_{\sigma Id}u=(\nabla\cdot)^{s}(\sigma(x,y)\nabla^{s}u).$

If the two-points function $\sigma$ admits the decomposition
$\sigma(x,y)=\gamma^{1/2}(x)\gamma^{1/2}(y)$ for some function
$\gamma:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}$, then we recover the usual
fractional conductivity operator $\mathbf{C}^{s}_{\gamma}$ studied in
[Cov20a]. Finally, in the case $\gamma(x)\equiv 1$ our operator reduces to the
fractional Laplacian $(-\Delta)^{s}$.

######  Remark 3.1.

For any $u,v\in H^{s}$ we see that

$\displaystyle\langle\mathbf{C}^{s}_{A}u,v\rangle$ $\displaystyle=\langle
A,\nabla^{s}v\otimes\nabla^{s}u\rangle$ $\displaystyle=\langle
A_{s},\nabla^{s}v\otimes\nabla^{s}u\rangle+\langle(A_{vs})_{ma},\nabla^{s}v\otimes\nabla^{s}u\rangle+\langle
A_{va},\nabla^{s}v\otimes\nabla^{s}u\rangle.$

Observe that the matrix $\nabla^{s}v\otimes\nabla^{s}u$ is symmetric both
matrix- and variable-wise. This makes the last two terms on the right hand
side vanish, and we are left with

(13) $\langle\mathbf{C}^{s}_{A}u,v\rangle=\langle
A_{s},\nabla^{s}v\otimes\nabla^{s}u\rangle=\langle\mathbf{C}^{s}_{A_{s}}u,v\rangle.$

This means that the operator $\mathbf{C}^{s}_{A}$ actually contains no
information about $A_{a}$, which is therefore unrecoverable. Thus the
anisotropic fractional conductivity equation possesses a natural gauge. This
is reminiscent of the situation emerging for the fractional magnetic
Schrödinger equation studied in [Cov20b].

######  Lemma 3.2.

The operator $\mathbf{C}^{s}_{A}$ is self-adjoint.

###### Proof.

By Remark 3.1 we have

$\langle\mathbf{C}^{s}_{A}u,v\rangle=\langle\mathbf{C}^{s}_{A_{s}}u,v\rangle=\langle
u,\mathbf{C}^{s}_{A_{s}}v\rangle=\langle u,\mathbf{C}^{s}_{A}v\rangle,$

since the symmetries of $A_{s}$ ensure that $C^{s}_{A_{s}}$ is itself self-
adjoint. ∎

###  3.3. The reduction lemma

The next result is a reduction lemma, by means of which we can write
$\mathbf{C}^{s}_{A}$ as a combination of fractional Laplacian operators acting
on sequences. This result is related to the fractional Liouville reduction
studied in [Cov20a] for the isotropic, separable case, and it also resembles
the reduction lemma presented in [CdHS22] for the fractional elasticity
equation.

######  Lemma 3.3.

In weak sense it holds that

$\mathbf{C}^{s}_{A}u=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\left\\{(-\Delta)^{s-1}D(\Phi u)+(\Phi u)\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q\right\\},$

where $D:=-\frac{\left(\Delta Id+2s\nabla^{2}\right)\odot}{n+2s}$ is a second
order operator acting between $n\times n$ matrices and
$Q:=-\frac{\Phi\otimes(-\Delta)^{s-1}D\Phi}{|\Phi|^{2}}$ is the transformed
potential.

###### Proof.

Let $u,v\in H^{s}$, and compute

$\displaystyle\langle\mathbf{C}^{s}_{A}u,v\rangle$ $\displaystyle=\langle
A_{s}\cdot\nabla^{s}u,\nabla^{s}v\rangle$
$\displaystyle=\langle\sum_{k\in{\mathbb{N}}}(\Phi_{k}(x)\odot\Phi_{k}(y))\cdot\nabla^{s}u,\nabla^{s}v\rangle$
$\displaystyle=\sum_{k\in{\mathbb{N}}}\langle(\Phi_{k}(x)\odot\Phi_{k}(y))\cdot[(u(y)-u(x))\otimes\zeta],(v(y)-v(x))\otimes\zeta\rangle$
$\displaystyle=\sum_{k\in{\mathbb{N}}}\langle(u(y)-u(x))(\Phi_{k}(x)\odot\Phi_{k}(y)):(\zeta\otimes\zeta),v(y)-v(x)\rangle.$

Observe that the order of summation and integration can be exchanged, given
that for all $k\in{\mathbb{N}}$ and almost all $x,y\in{\mathbb{R}}^{n}$

$\displaystyle|(\Phi_{k}(x)\odot\Phi_{k}(y)):(\nabla^{s}v\otimes\nabla^{s}u)|$
$\displaystyle\leq|\Phi_{k}(x)\odot\Phi_{k}(y)|\,|\nabla^{s}v\otimes\nabla^{s}u|$
$\displaystyle\leq|\Phi_{k}(x)|\,|\Phi_{k}(y)|\,|\nabla^{s}u|\,|\nabla^{s}v|$
$\displaystyle\leq\|\Phi_{k}\|_{L^{\infty}({\mathbb{R}}^{n})}^{2}|\nabla^{s}u|\,|\nabla^{s}v|$
$\displaystyle\leq\|\Phi_{k}\|_{C^{2s+\varepsilon}({\mathbb{R}}^{n})}^{2}|\nabla^{s}u|\,|\nabla^{s}v|$
$\displaystyle\leq\|\Phi\|_{C^{2s+\varepsilon}({\mathbb{R}}^{n})}^{2}|\nabla^{s}u|\,|\nabla^{s}v|=:g(x,y),$

and $g\in L^{1}({\mathbb{R}}^{2n})$. For constants $\alpha=(n+2s)^{-1}$ and
$\beta=(n+2s)^{-1}(n+2s-2)^{-1}$ we have (see e.g. [CdHS22])

$\frac{(x-y)\otimes(x-y)}{|x-y|^{n+2s+2}}=\alpha
Id|x-y|^{-(n+2s)}-\beta\nabla_{y}\nabla_{x}(|x-y|^{-(n+2s-2)}),$

and thus

$\zeta\otimes\zeta=\frac{C_{n,s}}{2}\left(\alpha
Id|x-y|^{-(n+2s)}-\beta\nabla_{y}\nabla_{x}(|x-y|^{-(n+2s-2)})\right).$

It follows that we can write
$\langle\mathbf{C}^{s}_{A}u,v\rangle=\sum_{k}\left(\alpha
I_{1,k}-\frac{C_{n,s}}{2}\beta I_{2,k}\right)$, with

$\displaystyle I_{1,k}$ $\displaystyle:=\frac{C_{n,s}}{2}\langle
Id:(\Phi_{k}(x)\odot\Phi_{k}(y))(u(y)-u(x))|x-y|^{-(n+2s)},v(y)-v(x)\rangle,$

and

$I_{2,k}:=\langle(u(y)-u(x))(\Phi_{k}(x)\odot\Phi_{k}(y)):\nabla_{y}\nabla_{x}(|x-y|^{-(n+2s-2)}),v(y)-v(x)\rangle.$

For $I_{1,k}$ we compute

$\displaystyle I_{1,k}$
$\displaystyle=\sum_{i}\langle\Phi_{k,i,i}(x)\Phi_{k,i,i}(y)\nabla^{s}u,\nabla^{s}v\rangle$
$\displaystyle=\sum_{i}\langle\Phi_{k,i,i}(-\Delta)^{s}(\Phi_{k,i,i}u)-\Phi_{k,i,i}u(-\Delta)^{s}\Phi_{k,i,i},v\rangle$
$\displaystyle=Id:\langle\Phi_{k}\odot(-\Delta)^{s}(\Phi_{k}u)-(\Phi_{k}u)\odot(-\Delta)^{s}\Phi_{k},v\rangle,$

where at the second line we recognized the scalar fractional conductivity
operator studied in [Cov20a] with conductivity $\Phi_{k,i,i}^{2}$, and thus
applied the usual fractional Liouville reduction. We want to compute the
second term $I_{2,k}$ integrating by parts. We let $w_{k}:=\Phi_{k}u$ and
compute

$\displaystyle I_{2,k}$
$\displaystyle=\int_{{\mathbb{R}}^{2n}}(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))(v(y)-v(x)):\nabla_{y}\nabla_{x}(|x-y|^{-(n+2s-2)})dydx$
$\displaystyle=-\int_{{\mathbb{R}}^{2n}}(\nabla_{y}\cdot)\left\\{(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))(v(y)-v(x))\right\\}\cdot\nabla_{x}(|x-y|^{-(n+2s-2)})dydx$
$\displaystyle=-\int_{{\mathbb{R}}^{2n}}\left\\{(v(y)-v(x))\nabla_{y}\cdot[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]+\right.$
$\displaystyle\qquad\qquad\qquad+\left.\nabla v(y)\cdot[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]\right\\}\cdot\nabla_{x}(|x-y|^{-(n+2s-2)})dydx$
$\displaystyle=\int_{{\mathbb{R}}^{2n}}(\nabla_{x}\cdot)\left\\{(v(y)-v(x))(\nabla_{y}\cdot)[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]+\right.$
$\displaystyle\qquad\qquad\qquad+\left.\nabla v(y)\cdot[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]\right\\}|x-y|^{-(n+2s-2)}dydx$
$\displaystyle=\int_{{\mathbb{R}}^{2n}}\left\\{(v(y)-v(x))(\nabla_{x}\cdot)(\nabla_{y}\cdot)[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]+\right.$
$\displaystyle\qquad\qquad\qquad-\nabla
v(x)\cdot(\nabla_{y}\cdot)[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]+$
$\displaystyle\qquad\qquad\qquad+\left.\nabla
v(y)\cdot(\nabla_{x}\cdot)[(\Phi_{k}(x)\odot
w_{k}(y)-w_{k}(x)\odot\Phi_{k}(y))]\right\\}|x-y|^{-(n+2s-2)}dydx.$

Now observe that

$\\{(\nabla_{y}\cdot)(A(x)\odot
B(y))\\}_{j}=\sum_{i}\partial_{y_{i}}(A_{ij}(x)B_{ij}(y))=\sum_{i}A_{ij}(x)\partial_{y_{i}}B_{ij}(y)$

and similarly

$(\nabla_{x}\cdot)(\nabla_{y}\cdot)(A(x)\odot
B(y))=\sum_{j}\partial_{x_{j}}\\{(\nabla_{y}\cdot)(A(x)\odot
B(y))\\}_{j}=\sum_{i,j}\partial_{x_{j}}A_{ij}(x)\partial_{y_{i}}B_{ij}(y).$

This implies

$\displaystyle I_{2,k}$
$\displaystyle=\sum_{i,j}\int_{{\mathbb{R}}^{2n}}v(y)\\{\partial_{j}\Phi_{k,ij}(x)\partial_{i}w_{k,ij}(y)-\partial_{j}w_{k,ij}(x)\partial_{i}\Phi_{k,ij}(y)\\}|x-y|^{-(n+2s-2)}dydx$
$\displaystyle\quad+\sum_{i,j}\int_{{\mathbb{R}}^{2n}}\partial_{j}v(y)\\{\partial_{i}\Phi_{k,ij}(x)w_{k,ij}(y)-\partial_{i}w_{k,ij}(x)\Phi_{k,ij}(y)\\}|x-y|^{-(n+2s-2)}dydx$
$\displaystyle\quad-\sum_{i,j}\int_{{\mathbb{R}}^{2n}}v(x)\\{\partial_{j}\Phi_{k,ij}(x)\partial_{i}w_{k,ij}(y)-\partial_{j}w_{k,ij}(x)\partial_{i}\Phi_{k,ij}(y)\\}|x-y|^{-(n+2s-2)}dydx$
$\displaystyle\quad-\sum_{i,j}\int_{{\mathbb{R}}^{2n}}\partial_{j}v(x)\\{\Phi_{k,ij}(x)\partial_{i}w_{k,ij}(y)-w_{k,ij}(x)\partial_{i}\Phi_{k,ij}(y)\\}|x-y|^{-(n+2s-2)}dydx.$

Now we exchange $x$ and $y$ in the last two integrals, observing that the
$n+2s-2<n$ exponent ensures the well-definiteness of all the integrals
involved. At this step we also use that $\Phi_{k}$ and $w_{k}$ are symmetric
matrices for all values of $k$. We get

$\displaystyle\frac{I_{2,k}}{2}$
$\displaystyle=\sum_{i,j}\int_{{\mathbb{R}}^{2n}}v(y)\\{\partial_{j}\Phi_{k,ij}(x)\partial_{i}w_{k,ij}(y)-\partial_{j}w_{k,ij}(x)\partial_{i}\Phi_{k,ij}(y)\\}|x-y|^{-(n+2s-2)}dydx$
$\displaystyle\quad+\sum_{i,j}\int_{{\mathbb{R}}^{2n}}\partial_{j}v(y)\\{\partial_{i}\Phi_{k,ij}(x)w_{k,ij}(y)-\partial_{i}w_{k,ij}(x)\Phi_{k,ij}(y)\\}|x-y|^{-(n+2s-2)}dydx.$

Let us rewrite the last formula equivalently as

$\displaystyle\frac{C_{n,s-1}I_{2,k}}{2}$
$\displaystyle=\sum_{i,j}\int_{{\mathbb{R}}^{n}}v\\{R\partial_{j}\Phi_{k,ij}\,\partial_{i}w_{k,ij}-R\partial_{j}w_{k,ij}\,\partial_{i}\Phi_{k,ij}\\}dy$
$\displaystyle\quad+\sum_{i,j}\int_{{\mathbb{R}}^{n}}\partial_{i}v\\{R\partial_{j}\Phi_{k,ij}\,w_{k,ij}-R\partial_{j}w_{k,ij}\,\Phi_{k,ij}\\}dy,$

where for the sake of readability we let $R:=(-\Delta)^{s-1}$. Since $R$
commutes with the derivatives, one last integration by parts in the last
integral gives

$\displaystyle\int_{{\mathbb{R}}^{n}}\partial_{i}v\\{R\partial_{j}\Phi_{k,ij}\,w_{k,ij}-R\partial_{j}w_{k,ij}\,\Phi_{k,ij}\\}dy$
$\displaystyle=-\int_{{\mathbb{R}}^{n}}v\partial_{i}\\{R\partial_{j}\Phi_{k,ij}\,w_{k,ij}-R\partial_{j}w_{k,ij}\,\Phi_{k,ij}\\}dy$
$\displaystyle=-\int_{{\mathbb{R}}^{n}}v\\{w_{k,ij}\,R\partial_{i}\partial_{j}\Phi_{k,ij}-\Phi_{k,ij}\,R\partial_{i}\partial_{j}w_{k,ij}\\}dy$
$\displaystyle\quad-\int_{{\mathbb{R}}^{n}}v\\{\partial_{i}w_{k,ij}\,R\partial_{j}\Phi_{k,ij}-\partial_{i}\Phi_{k,ij}\,R\partial_{j}w_{k,ij}\\}dy,$

and so eventually

$\displaystyle\frac{C_{n,s-1}I_{2,k}}{2}$
$\displaystyle=\sum_{i,j}\int_{{\mathbb{R}}^{n}}v\\{\Phi_{k,ij}\,R\partial_{i}\partial_{j}w_{k,ij}-w_{k,ij}\,R\partial_{i}\partial_{j}\Phi_{k,ij}\\}dy$
$\displaystyle=\langle\Phi_{k}:(-\Delta)^{s-1}(\nabla^{2}\odot
w_{k})-w_{k}:(-\Delta)^{s-1}(\nabla^{2}\odot\Phi_{k}),v\rangle.$

Putting together the computations for $I_{1,k}$ and $I_{2,k}$ we get the
following equality

$\displaystyle\mathbf{\langle}\mathbf{C}^{s}_{A}u,v\rangle$
$\displaystyle=\sum_{k\in{\mathbb{N}}}\left(\alpha
I_{1,k}-\frac{C_{n,s}}{2}\beta I_{2,k}\right)$
$\displaystyle=\alpha\sum_{k\in{\mathbb{N}}}Id:\langle\Phi_{k}\odot(-\Delta)^{s}w_{k}-w_{k}\odot(-\Delta)^{s}\Phi_{k},v\rangle$
$\displaystyle\quad+\frac{C_{n,s}\beta}{C_{n,s-1}}\sum_{k\in{\mathbb{N}}}\langle
w_{k}:(-\Delta)^{s-1}(\nabla^{2}\odot\Phi_{k})-\Phi_{k}:(-\Delta)^{s-1}(\nabla^{2}\odot
w_{k}),v\rangle$
$\displaystyle=\sum_{k\in{\mathbb{N}}}\langle\Phi_{k}:(-\Delta)^{s-1}Dw_{k}-w_{k}:(-\Delta)^{s-1}D\Phi_{k},v\rangle$
$\displaystyle=\lim_{K\rightarrow\infty}\langle\sum_{k=1}^{K}(\Phi_{k}:(-\Delta)^{s-1}Dw_{k}-w_{k}:(-\Delta)^{s-1}D\Phi_{k}),v\rangle=:\lim_{K\rightarrow\infty}F_{K}(v),$

where the differential operator with constant coefficients $D$ is

$D:=-\left(\alpha\Delta
Id+\beta\frac{C_{n,s}}{C_{n,s-1}}\nabla^{2}\right)\odot=-\frac{\left(\Delta
Id+2s\nabla^{2}\right)\odot}{n+2s}.$

By Lemma 2.5 and property (H1), $\Phi_{k}$ is a multiplier on
$H^{s}({\mathbb{R}}^{n})$ (and thus also in $H^{-s}({\mathbb{R}}^{n})$, see
[MS09, CMRU22]) for every $k\in{\mathbb{N}}$. Moreover,
$(-\Delta)^{s-1}D\Phi_{k}\in C^{\varepsilon}({\mathbb{R}}^{n})$ by property
(H5). This ensures that the operators $F_{K}$ belong to
$H^{-s}({\mathbb{R}}^{n})$ for all $K\in{\mathbb{N}}$. They converge to
$F:=\Phi\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(-\Delta)^{s-1}D(\Phi u)-\Phi u\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}(-\Delta)^{s-1}D\Phi$ in
the norm of $H^{-s}$ as $K\rightarrow\infty$, since

$\displaystyle\|F_{K}-F\|_{H^{-s}}$
$\displaystyle\leq\|\sum_{k=1}^{K}(\Phi_{k}:(-\Delta)^{s-1}D(\Phi_{k}u))-\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}(-\Delta)^{s-1}D(\Phi
u)\|_{H^{-s}}$
$\displaystyle\quad+\|\sum_{k=1}^{K}(\Phi_{k}u:(-\Delta)^{s-1}D\Phi_{k})-\Phi
u\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(-\Delta)^{s-1}D\Phi\|_{H^{-s}}$
$\displaystyle=\|\sum_{k>K}\Phi_{k}:(-\Delta)^{s-1}D(\Phi_{k}u)\|_{H^{-s}}+\|\sum_{k>K}\Phi_{k}u:(-\Delta)^{s-1}D\Phi_{k}\|_{H^{-s}}$
$\displaystyle\leq\sum_{k>K}\left(\|\Phi_{k}:(-\Delta)^{s-1}D(\Phi_{k}u)\|_{H^{-s}}+\|\Phi_{k}u:(-\Delta)^{s-1}D\Phi_{k}\|_{H^{-s}}\right)$
$\displaystyle\lesssim\sum_{k>K}\left(\|\Phi_{k}\|_{C^{2s+\varepsilon}}\|(-\Delta)^{s-1}D(\Phi_{k}u)\|_{H^{-s}}+\|\Phi_{k}u\|_{H^{s}}\|(-\Delta)^{s-1}D\Phi_{k}\|_{C^{\varepsilon}}\right)$
$\displaystyle\lesssim\sum_{k>K}\|\Phi_{k}\|_{C^{2s+\varepsilon}}\|\Phi_{k}u\|_{H^{s}}$
$\displaystyle\lesssim\|u\|_{H^{s}}\sum_{k>K}\|\Phi_{k}\|^{2}_{C^{2s+\varepsilon}}\leq\|u\|_{H^{s}}\left(\sum_{k>K}\|\Phi_{k}\|_{C^{2s+\varepsilon}}\right)^{2},$

which converges to $0$ as $K\rightarrow\infty$ by assumption (A3). Given that

$|F_{K}(v)-F(v)|\leq\|v\|_{H^{s}}\sup_{\|\xi\|_{H^{s}}=1}|F_{K}(\xi)-F(\xi)|=\|v\|_{H^{s}}\|F_{K}-F\|_{H^{-s}}\rightarrow
0,$

we deduce that $\mathbf{C}^{s}_{A}u=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}(-\Delta)^{s-1}D(\Phi
u)-\Phi u\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(-\Delta)^{s-1}D\Phi$ holds in weak sense. Equivalently, if the transformed
potential $Q$ is defined as
$Q:=-\frac{\Phi\otimes(-\Delta)^{s-1}D\Phi}{|\Phi|^{2}}$, one can write

$\displaystyle\mathbf{C}^{s}_{A}u$ $\displaystyle=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\left\\{(-\Delta)^{s-1}D(\Phi u)-u(-\Delta)^{s-1}D\Phi\right\\}$
$\displaystyle=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\left\\{(-\Delta)^{s-1}D(\Phi u)-\frac{(\Phi u)\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\Phi}{|\Phi|^{2}}(-\Delta)^{s-1}D\Phi\right\\}$
$\displaystyle=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\left\\{(-\Delta)^{s-1}D(\Phi u)+(\Phi u)\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q\right\\}.$

Observe that all the terms on the right hand side make sense in
$H^{-s}({\mathbb{R}}^{n})$. In fact, since $\Phi\in
C^{2s+\varepsilon}({\mathbb{R}}^{n})$ and $|\Phi|>0$ by assumptions (A3) and
(A2), following the reasoning of Section 2.5 we have that $\Phi/|\Phi|^{2}\in
C^{2s+\varepsilon}({\mathbb{R}}^{n})$. As observed, $(-\Delta)^{s-1}D\Phi\in
C^{\varepsilon}({\mathbb{R}}^{n})$ by property (H5). Since by (H1) we have
$C^{2s+\varepsilon}({\mathbb{R}}^{n})\subset
C^{\varepsilon}({\mathbb{R}}^{n})$, a new application of (H3) ensures $Q\in
C^{\varepsilon}({\mathbb{R}}^{n})\subset L^{\infty}({\mathbb{R}}^{n})$. Given
that $\Phi$ is a multiplier on $H^{s}({\mathbb{R}}^{n})$ and
$H^{-s}({\mathbb{R}}^{n})$ by Lemma 2.5 and property (H1), we get
$(-\Delta)^{s-1}D(\Phi u)+(\Phi u)\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q\in
H^{-s}({\mathbb{R}}^{n})$, and eventually $\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\left\\{(-\Delta)^{s-1}D(\Phi u)+(\Phi u)\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q\right\\}\in
H^{-s}({\mathbb{R}}^{n})$. ∎

###  3.4. Weakly anisotropic and isotropic matrices

In this section we look at $\mathbf{C}^{s}_{A}$ for matrices $A$ of special
forms. In some of these cases, the structure of the operator simplifies
noticeably, and we recover familiar operators related to the fractional
Laplacian $(-\Delta)^{s}$.

Since $A_{s}(x,y)$ is positive definite and symmetric for all
$x,y\in{\mathbb{R}}^{n}$, there exist a diagonal matrix $L(x,y)$ with positive
entries and an orthonormal matrix $U(x,y)$ such that

$A_{s}(x,y)=U^{T}(x,y)\,L(x,y)\,U(x,y)$

for all $x,y\in{\mathbb{R}}^{n}$. The columns of $U$ (that is, the
eigenvectors of $A_{s}$ of unit norm) represent the principal directions of
$A_{s}$ at $(x,y)$, while the entries of $L$ (i.e. the eigenvalues of $A_{s}$)
are the corresponding scalar conductivities in the given directions. The
anisotropy of $A_{s}$ is due to two factors: the principal directions of
$A_{s}$ change according to position, and the $n$ directional conductivities
differ among themselves even at any given point. As a result, in general the
electric field and the current density are not parallel to each other.

However, there exist materials showing a weaker form of anisotropy, in the
sense that the principal directions do not change with respect to position.
This is the case of many crystalline materials (such as graphite, high
temperature superconductors, and some metals), which are characterized by
atomic structures repeating periodically in fixed directions. In this case,
the matrices of the family $\\{A_{s}(x,y):x,y\in{\mathbb{R}}^{n}\\}$ are
_simultaneously diagonalizable_ , which means that there exists a unique
orthonormal matrix $U$ such that the formula

$A_{s}(x,y)=U^{T}\,L(x,y)\,U$

holds for all $x,y\in{\mathbb{R}}^{n}$. By letting $x^{\prime}:=U\cdot x$,
$u^{\prime}:=u\circ U^{T}$ and
$L^{\prime}:=L\circ\left(\begin{array}[]{cc}U^{T}&0\\\
0&U^{T}\end{array}\right)$ for all $x\in{\mathbb{R}}^{n}$, we see that

$U\cdot\nabla^{s}v(x,y)=\frac{C_{n,s}^{1/2}}{\sqrt{2}}\frac{v(y)-v(x)}{|x-y|^{n/2+s+1}}U\cdot(x-y)=\frac{C_{n,s}^{1/2}}{\sqrt{2}}\frac{v^{\prime}(y^{\prime})-v^{\prime}(x^{\prime})}{|x^{\prime}-y^{\prime}|^{n/2+s+1}}(x^{\prime}-y^{\prime})=\nabla^{s}v^{\prime}(x^{\prime},y^{\prime}),$

and thus the fractional conductivity operator becomes

$\displaystyle\langle\mathbf{C}^{s}_{A}u,v\rangle$ $\displaystyle=\langle
A_{s}(x,y)\cdot\nabla^{s}u,\nabla^{s}v\rangle$ $\displaystyle=\langle
L(x,y)\cdot(U\cdot\nabla^{s}u),U\cdot\nabla^{s}v\rangle$
$\displaystyle=\langle
L^{\prime}(x^{\prime},y^{\prime})\cdot\nabla^{s}u^{\prime},\nabla^{s}v^{\prime}\rangle$
$\displaystyle=\langle\mathbf{C}^{s}_{L^{\prime}}u^{\prime},v^{\prime}\rangle$

Thus we see that it is interesting to consider $\mathbf{C}^{s}_{A}$ for a
diagonal matrix $A$. We shall also assume that the exterior value matrices
$\beta_{k}$ are diagonal for all $k\in{\mathbb{N}}$, so that the matrices
$\phi_{k}$ can themselves be taken to be diagonal for all $k\in{\mathbb{N}}$,
and $\Phi$ is a sequence of diagonal matrices. Given that the usual matrix
product and the Hadamard product coincide for diagonal matrices, in this case
we can write

$A_{s}(x,y)=\sum_{k\in{\mathbb{N}}}\Phi_{k}(x)\cdot\Phi_{k}(y).$

Assume now that the operator is isotropic, that is $A_{s}(x,y)=\sigma(x,y)$Id
for a symmetric scalar function $\sigma$, and $\beta_{k}=b_{k}$Id for all
$k\in{\mathbb{N}}$. In this case, the functions $\phi_{k}^{(i,j)}$ can be
taken to vanish whenever $i\neq j$, and there exists a function $\phi_{k}$
such that $\phi_{k}^{(i,i)}=\phi_{k}$ for all $i\in\\{1,...,n\\}$. In other
words, $\Phi_{k}=\phi_{k}$Id, and so we have

$\displaystyle Id:D(u\Phi_{k})$ $\displaystyle=-Id:\frac{\left(\Delta
Id+2s\nabla^{2}\right)}{n+2s}\odot(u\phi_{k}Id)=-\Delta(u\phi_{k}).$

This implies that

$\displaystyle\mathbf{C}^{s}_{A}u$
$\displaystyle=\sum_{k}\phi_{k}Id:(-\Delta)^{s-1}D(u\Phi_{k})-u\phi_{k}Id:(-\Delta)^{s-1}D\Phi_{k}$
$\displaystyle=\sum_{k}\phi_{k}\left((-\Delta)^{s}(u\phi_{k})-u(-\Delta)^{s}\phi_{k}\right)$
$\displaystyle=\phi\cdot((-\Delta)^{s}(u\phi)+(u\phi)\cdot Q),$

where $Q:=-\frac{\phi\otimes(-\Delta)^{s}\phi}{|\phi|^{2}}$. We see that in
this case the equation retains its form, but it is greatly simplified. If
moreover $\sigma$ is a separable function of $x,y$, then $\mathbf{C}^{s}_{A}$
is the usual fractional conductivity operator from [Cov20a], and Lemma 3.3
gives the relative fractional Liouville reduction.

##  4\. Well-posedness of the direct problems and reduction

We consider the direct problem

(14) $\begin{split}\mathbf{C}^{s}_{A}u=F&\quad\mbox{ in }\Omega,\\\
u=f&\quad\mbox{ in }\Omega_{e},\end{split}$

for $F\in H^{-s}(\Omega)$ and $f\in H^{s}({\mathbb{R}}^{n})$. Define the
bilinear form

$B^{s}_{A}(u,v):=\langle A(x,y)\cdot\nabla^{s}u,\nabla^{s}v\rangle=\langle
A_{s}(x,y)\cdot\nabla^{s}u,\nabla^{s}v\rangle$

for $u,v\in C^{\infty}_{c}({\mathbb{R}}^{n})$. The boundedness estimate

(15)
$|B^{s}_{A}(u,v)|\leq\|A\cdot\nabla^{s}u\|_{L^{2}({\mathbb{R}}^{2n})}\|\nabla^{s}v\|_{L^{2}({\mathbb{R}}^{2n})}\lesssim\|u\|_{H^{s}}\|v\|_{H^{s}}$

holds since $A\in L^{\infty}$. Thus $B^{s}_{A}$ can be extended by density to
act as a bounded linear operator on $H^{s}({\mathbb{R}}^{n})\times
H^{s}({\mathbb{R}}^{n})$. We also define the potential energy $U^{s}_{A}$ for
all $u\in H^{s}({\mathbb{R}}^{n})$ as

$U^{s}_{A}(u):=B^{s}_{A}(u,u).$

Since by assumption (A2) $A_{s}$ is uniformly positive definite, it holds that

$U^{s}_{A}(u)=\langle
A_{s}\cdot\nabla^{s}u,\nabla^{s}u\rangle\geq\nu\|\nabla^{s}u\|_{L^{2}({\mathbb{R}}^{2n})}^{2}\approx\|(-\Delta)^{s/2}u\|_{L^{2}({\mathbb{R}}^{n})}^{2},$

so that if $u\in H^{s}_{K}({\mathbb{R}}^{n})$ for some compact set $K$, by the
fractional Poincaré inequality (see e.g. [GSU20] and [CdHS22]) we obtain the
coercivity estimate

(16) $B^{s}_{A}(u,u)=U^{s}_{A}(u)\gtrsim\|u\|_{H^{s}({\mathbb{R}}^{n})}^{2}.$

We say that $u\in H^{s}(\mathbb{R}^{n})$ is a weak solution to the
inhomogeneous problem (14) if and only if $B^{s}_{A}(u,v)=F(v)$ holds for all
$v\in\widetilde{H}^{s}(\Omega)$, and $u-f\in\widetilde{H}^{s}(\Omega)$. The
next Proposition gives the well-posedness for the direct problem:

######  Proposition 4.1.

Let $s\in(0,1)$ and assume $\Omega\subset\mathbb{R}^{n}$ is a bounded open
set. For any $f\in H^{s}(\mathbb{R}^{n})$ and $F\in H^{-s}(\Omega)$ there
exists a unique $u\in H^{s}(\mathbb{R}^{n})$ such that
$u-f\in\widetilde{H}^{s}(\Omega)$ and

$B^{s}_{A}(u,v)=F(v)\quad\mbox{for all}\quad v\in\widetilde{H}^{s}(\Omega).$

Moreover, the following estimate holds:

$\|u\|_{H^{s}(\mathbb{R}^{n})}\leq
C\left(\|f\|_{H^{s}(\mathbb{R}^{n})}+\|F\|_{H^{-s}(\Omega)}\right).$

###### Proof.

The proof of this statement follows by the Lax-Milgram theorem, given the
boundedness and coercivity estimates (15), (16). It goes along the same lines
as the corresponding proposition in [CdHS22], see also [GSU20], and it is by
now well understood. We thus omit it here. ∎

Next, we study the well-posedness of the transformed direct problem

(17) $\begin{split}\mathbf{(}-\Delta)^{s-1}Dw+w\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q=G&\quad\mbox{ in
}\Omega,\\\ w=\Phi f&\quad\mbox{ in }\Omega_{e},\end{split}$

for $G\in(\Phi\widetilde{H}^{s}(\Omega))^{*}$ and $f\in
H^{s}({\mathbb{R}}^{n})$, where
$Q:=-\frac{\Phi\otimes(-\Delta)^{s-1}D\Phi}{|\Phi|^{2}}$. As for the original
direct problem, we define the bilinear form

$B^{s}_{Q}(w,v):=\langle(-\Delta)^{s-1}Dw,v\rangle+\langle w\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q,v\rangle$

for $w,v\in\Phi H^{s}({\mathbb{R}}^{n})$. We say that $w\in\Phi
H^{s}(\mathbb{R}^{n})$ is a weak solution to the inhomogeneous problem (17) if
and only if $B^{s}_{Q}(w,v)=G(v)$ holds for all
$v\in\Phi\widetilde{H}^{s}(\Omega)$, and $w-\Phi
f\in\Phi\widetilde{H}^{s}(\Omega)$. By virtue of the reduction Lemma 3.3, we
can compute

(18) $\begin{split}B^{s}_{Q}(\Phi u,\Phi v)&=\langle(-\Delta)^{s-1}D(\Phi
u),\Phi v\rangle+\langle(\Phi u)\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q,\Phi v\rangle\\\
&=\langle\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\\{(-\Delta)^{s-1}D(\Phi
u)+(\Phi u)\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}Q\\},v\rangle\\\
&=\langle\mathbf{C}^{s}_{A}u,v\rangle=B^{s}_{A}(u,v)\end{split}$

for all $u,v\in H^{s}({\mathbb{R}}^{n})$, and thus $B^{s}_{Q}$ is immediately
bounded and coercive in the space $\Phi H^{s}({\mathbb{R}}^{n})\times\Phi
H^{s}({\mathbb{R}}^{n})$ by formulas (15), (16), and Lemma 2.5. Therefore, we
obtain the following well-posedness result:

######  Proposition 4.2.

Let $s\in(0,1)$ and assume $\Omega\subset\mathbb{R}^{n}$ is a bounded open
set. For any $f\in H^{s}(\mathbb{R}^{n})$ and
$G\in(\Phi\widetilde{H}^{s}(\Omega))^{*}$ there exists a unique $w\in\Phi
H^{s}(\mathbb{R}^{n})$ such that $w-\Phi f\in\Phi\widetilde{H}^{s}(\Omega)$
and

$B^{s}_{Q}(w,v)=G(v)\quad\mbox{for all}\quad
v\in\Phi\widetilde{H}^{s}(\Omega).$

The next proposition follows immediately from Lemma 3.3.

######  Proposition 4.3 (Fractional Liouville reduction).

Let $s\in(0,1)$ and assume $\Omega\subset\mathbb{R}^{n}$ is a bounded open
set. If $u\in H^{s}(\mathbb{R}^{n})$ solves the original problem

(19) $\begin{array}[]{rll}\mathbf{C}^{s}_{A}u&=F&\quad\mbox{ in }\Omega\\\
u&=f&\quad\mbox{ in }\Omega_{e}\end{array}$

in weak sense in $\widetilde{H}^{s}(\Omega)$ for some $f\in
H^{s}(\mathbb{R}^{n})$ and $F\in H^{-s}(\Omega)$, then $w:=\Phi u$ solves the
transformed problem

(20) $\begin{array}[]{rll}\mathbf{(}-\Delta)^{s-1}Dw+w\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q&=G&\quad\mbox{ in
}\Omega\\\ w&=\Phi f&\quad\mbox{ in }\Omega_{e}\end{array}$

in weak sense in $\Phi\widetilde{H}^{s}(\Omega)$, where $G:=\frac{\Phi
F}{|\Phi|^{2}}$. Conversely, if $w\in\Phi H^{s}(\mathbb{R}^{n})$ solves (20)
in weak sense in $\Phi\widetilde{H}^{s}(\Omega)$ for some
$G\in(\Phi\widetilde{H}^{s}(\Omega))^{*}$, then
$u:=\frac{\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}w}{|\Phi|^{2}}$ solves
(19) in weak sense in $\widetilde{H}^{s}(\Omega)$, where
$F=\Phi\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}G$.

###### Proof.

Observe that $u\in H^{s}({\mathbb{R}}^{n})$ solves (19) in weak sense if and
only if $u-f\in\widetilde{H}^{s}(\Omega)$, and also for all
$v\in\widetilde{H}^{s}(\Omega)$ we have $B^{s}_{A}(u,v)=F(v)$. Then of course
$w:=\Phi u$ belongs to $\Phi H^{s}({\mathbb{R}}^{n})$, it verifies $w-\Phi
f\in\Phi\widetilde{H}^{s}(\Omega)$, and also by (18)

$B^{s}_{Q}(w,\Phi v)=B^{s}_{A}(u,v)=\langle F,v\rangle=\langle G,\Phi
v\rangle$

for $G:=\frac{\Phi F}{|\Phi|^{2}}\in(\Phi\widetilde{H}^{s}(\Omega))^{*}$.
Conversely, if $w\in\Phi H^{s}(\mathbb{R}^{n})$ solves (20) in weak sense in
$\Phi\widetilde{H}^{s}(\Omega)$ for some
$G\in(\Phi\widetilde{H}^{s}(\Omega))^{*}$, then $w-\Phi
f\in\Phi\widetilde{H}^{s}(\Omega)$, and also $B^{s}_{Q}(w,\Phi v)=\langle
G,\Phi v\rangle$ for all $v\in\widetilde{H}^{s}(\Omega)$. Then
$u:=\frac{\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}w}{|\Phi|^{2}}\in
H^{s}({\mathbb{R}}^{n})$, $u-f=\frac{\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}(w-\Phi
f)}{|\Phi|^{2}}\in\widetilde{H}^{s}(\Omega)$, and again by (18)

$B^{s}_{A}(u,v)=B^{s}_{Q}(w,\Phi v)=\langle G,\Phi v\rangle=\langle
F,v\rangle$

for $F:=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}G\in H^{-s}(\Omega)$. ∎

##  5\. The DN map and the Alessandrini identity

Following [GSU20], define the abstract trace space
$X:=H^{s}({\mathbb{R}}^{n})/\widetilde{H}^{s}(\Omega)$. Observe that
$X=H^{s}(\Omega_{e})$ holds for all Lipschitz $\Omega$. Since problem (14) is
well-posed, we can define the Poisson operator $P_{A}:X\rightarrow
H^{s}({\mathbb{R}}^{n})$ associating to each exterior value $f\in X$ the
unique solution $u_{f}\in H^{s}({\mathbb{R}}^{n})$ to (14) with $F=0$. It is
then possible to define a continuous, self-adjoint, linear map
$\Lambda_{A}:X\rightarrow X^{*}$ by

$\langle\Lambda_{A}[f],[g]\rangle:=B^{s}_{A}(P_{A}f,g),$

where $f,g\in H^{s}(\mathbb{R}^{n})$. This can be proved making use of the
well-posedness of the direct problem and the properties of the bilinear form
(see e.g. [GSU20], [CMRU22], [CdHS22] for the standard proof). In particular,
the self-adjointness follows from the symmetry of the bilinear form
$B^{s}_{A}$, which is due to formula (13). Equation (18) now ensures the
symmetry of the bilinear form $B^{s}_{Q}$ in $\Phi H^{s}({\mathbb{R}}^{n})$
(which may be non-trivial at first sight).
The DN map $\Lambda_{A}$ is related to the transformed potential $Q$ by the
following integral identity:

######  Proposition 5.1 (Alessandrini identity).

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded open set and $s\in(0,1)$. Let
$A_{1},A_{2}$ be two anisotropic conductivity matrices satisfying assumptions
(A1)–(A3) and $A_{1}\sim A_{2}$. For $j=1,2$, let $\Phi_{j},Q_{j}$ be the
sequence and transformed potential corresponding to $A_{j}$. Then the
following integral identity holds for all $f_{1},f_{2}\in
C^{\infty}_{c}(\Omega_{e})$

$\langle(\Lambda_{A_{1}}-\Lambda_{A_{2}})[f_{1}],[f_{2}]\rangle=\langle
w_{1}\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{2}-Q_{1}),w_{2}\rangle,$

where $w_{1}:=\Phi_{1}P_{A_{1}}f_{1}$ and $w_{2}:=\Phi_{2}P_{A_{2}}f_{2}$.

###### Proof.

By definition of the DN maps and formula (18) we have

$\displaystyle\langle(\Lambda_{A_{1}}-\Lambda_{A_{2}})[f_{1}],[f_{2}]\rangle$
$\displaystyle=\langle\Lambda_{A_{1}}[f_{1}],[f_{2}]\rangle-\langle\Lambda_{A_{2}}[f_{1}],[f_{2}]\rangle$
$\displaystyle=\langle\Lambda_{A_{1}}[f_{1}],[f_{2}]\rangle-\langle\Lambda_{A_{2}}[f_{2}],[f_{1}]\rangle$
$\displaystyle=B^{s}_{A_{1}}(P_{A_{1}}f_{1},f_{2})-B^{s}_{A_{2}}(P_{A_{2}}f_{2},f_{1})$
$\displaystyle=B^{s}_{Q_{1}}(\Phi_{1}P_{A_{1}}f_{1},\Phi_{1}f_{2})-B^{s}_{Q_{2}}(\Phi_{2}P_{A_{2}}f_{2},\Phi_{2}f_{1})$

Observe that $P_{A_{2}}f_{2}-f_{2}\in\widetilde{H}^{s}(\Omega)$ by the
definition of the Poisson operator. Thus
$\Phi_{2}(P_{A_{2}}f_{2}-f_{2})\in\Phi_{2}\widetilde{H}^{s}(\Omega)$, and by
the assumption $A_{1}\sim A_{2}$ we can also deduce
$\Phi_{2}(P_{A_{2}}f_{2}-f_{2})\in\Phi_{1}\widetilde{H}^{s}(\Omega)$. This
implies
$B^{s}_{Q_{1}}(\Phi_{1}P_{A_{1}}f_{1},\Phi_{2}(P_{A_{2}}f_{2}-f_{2}))=0$.
Since moreover $A_{1}\sim A_{2}$ implies $\Phi_{1}=\Phi_{2}$ in $\Omega_{e}$,
we eventually deduce

$B^{s}_{Q_{1}}(\Phi_{1}P_{A_{1}}f_{1},\Phi_{1}f_{2})=B^{s}_{Q_{1}}(\Phi_{1}P_{A_{1}}f_{1},\Phi_{2}f_{2})=B^{s}_{Q_{1}}(\Phi_{1}P_{A_{1}}f_{1},\Phi_{2}P_{A_{2}}f_{2}),$

and similarly for $B^{s}_{Q_{2}}$. Therefore

$\displaystyle\langle(\Lambda_{A_{1}}-\Lambda_{A_{2}})[f_{1}],[f_{2}]\rangle$
$\displaystyle=B^{s}_{Q_{1}}(\Phi_{1}P_{A_{1}}f_{1},\Phi_{2}P_{A_{2}}f_{2})-B^{s}_{Q_{2}}(\Phi_{2}P_{A_{2}}f_{2},\Phi_{1}P_{A_{1}}f_{1})$
$\displaystyle=\langle w_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2}),w_{2}\rangle,$

where we used the symmetry of $B^{s}_{Q_{2}}$ and set
$w_{1}:=\Phi_{1}P_{A_{1}}f_{1}$ and $w_{2}:=\Phi_{2}P_{A_{2}}f_{2}$. ∎

##  6\. Proof of the main theorem

We now move forward to proving a Runge-type approximation result. The proof is
based on the UCP for the fractional Laplacian and a standard technique (see
[GSU20, CMRU22]).

######  Lemma 6.1 (Runge approximation property).

Let $\Omega,W\subset\mathbb{R}^{n}$ be bounded open sets such that
$W\subset\Omega_{e}$. Assume that the exterior coefficients are isotropic,
i.e.

  1. (A4)

there exist scalar functions $b_{k}$ such that $\beta_{k}=b_{k}Id$ for all
$k\in{\mathbb{N}}$.

The set

$\mathcal{R}:=\left\\{P_{A}f-f:f\in C^{\infty}_{c}(W)\right\\}$

is dense in $L^{2}(\Omega)$.

###### Proof.

Assume that $F\in L^{2}(\Omega)$ is such that $\langle F,v\rangle=0$ for all
$v\in\mathcal{R}$, and let $u\in\widetilde{H}^{s}(\Omega)$ be the unique
solution of (19) with vanishing exterior value. Moreover, let
$w\in\Phi\widetilde{H}^{s}(\Omega)$ correspond to $u$ by the fractional
Liouville reduction. Then for all $f\in C^{\infty}_{c}(W)$

$\displaystyle 0$ $\displaystyle=\langle
F,P_{A}f-f\rangle=\langle\mathbf{C}^{s}_{A}u,P_{A}f-f\rangle=B^{s}_{A}(u,P_{A}f-f)=-B^{s}_{A}(u,f)=-B^{s}_{Q}(\Phi
u,\Phi f).$

Since the supports of $u$ and $f$ have empty intersection, we are left with

$0=\langle(-\Delta)^{s-1}Dw,\Phi f\rangle=\langle\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(-\Delta)^{s-1}Dw,f\rangle,$

which by the arbitrariety of $f\in C^{\infty}_{c}(W)$ implies the vanishing of
$\Phi\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(-\Delta)^{s-1}Dw$ in $\Omega_{e}$. Recall that by assumption (A1) in
$\Omega_{e}$ we have $\Phi_{2k+1}(x)=0$ and $\Phi_{2k}(x)=\beta_{k}(x)$ for
all $k\in{\mathbb{N}}$. Thus in $\Omega_{e}$ it holds

$\displaystyle 0=\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(-\Delta)^{s-1}Dw=\sum_{k}\Phi_{k}:(-\Delta)^{s-1}D(\Phi_{k}u)=\sum_{k}\beta_{k}:(-\Delta)^{s-1}D(\beta_{k}u).$

Let $\tilde{\beta}_{k}$ be the constant exterior value of $\beta_{k}$, and
define

$\gamma(x):=\sum_{k}\tilde{\beta}_{k}\odot\beta_{k}(x)=\sum_{k}\beta_{k}(y)\odot\beta_{k}(x)=\tilde{a}(x,y)$

for all $x\in{\mathbb{R}}^{n}$ and $y\in\Omega_{e}$. We have that in
$\Omega_{e}$

$\displaystyle 0$
$\displaystyle=(-\Delta)^{s-1}\left(\sum_{k}\tilde{\beta}_{k}:D(\beta_{k}u)\right)$
$\displaystyle=-\frac{1}{n+2s}(-\Delta)^{s-1}\left(\sum_{k}\tilde{\beta}_{k}:(\Delta
Id+2s\nabla^{2})\odot(\beta_{k}u)\right)$
$\displaystyle=-\frac{1}{n+2s}(-\Delta)^{s-1}(\Delta
Id+2s\nabla^{2}):\left(\sum_{k}\tilde{\beta}_{k}\odot\beta_{k}u\right)$
$\displaystyle=-\frac{(-\Delta)^{s-1}(\Delta Id+2s\nabla^{2}):\left(\gamma
u\right)}{n+2s},$

and thus $(-\Delta)^{s-1}(\Delta Id+2s\nabla^{2}):(\gamma u)=0$ in
$\Omega_{e}$. Since $(-\Delta)$ and $(\Delta Id+2s\nabla^{2}):$ are local
operators, and also $u=0$ in $\Omega_{e}$, we deduce

$(-\Delta)^{s}[(\Delta Id+2s\nabla^{2}):(\gamma u)]=0,\qquad(\Delta
Id+2s\nabla^{2}):(\gamma u)=0$

in $\Omega_{e}$. Therefore $(\Delta Id+2s\nabla^{2}):(\gamma u)=0$ in all
${\mathbb{R}}^{n}$ by the UCP for the fractional Laplacian.

Since $\gamma(x)$ is a symmetric matrix for all $x\in{\mathbb{R}}^{n}$, we can
compute

$\displaystyle\nabla^{2}:(\gamma u)$
$\displaystyle=\sum_{i,j}\partial_{i}\partial_{j}(\gamma_{ij}u)$
$\displaystyle=\sum_{i,j}\partial_{i}\partial_{j}\gamma_{ij}u+\partial_{i}\gamma_{ij}\partial_{j}u+\partial_{j}\gamma_{ji}\partial_{i}u+\gamma_{ij}\partial_{i}\partial_{j}u$
$\displaystyle=\gamma:\nabla^{2}u+2(\nabla\cdot\gamma)\cdot\nabla
u+(\nabla^{2}:\gamma)u$

and

$\displaystyle\Delta Id:(\gamma u)$ $\displaystyle=\Delta(u\mbox{ tr}\gamma)$
$\displaystyle=\mbox{ tr}\gamma\Delta u+2(\nabla\mbox{tr}\gamma)\cdot\nabla
u+(\Delta\mbox{tr}\gamma)u$
$\displaystyle=(\mbox{tr}\gamma\,Id):\nabla^{2}u+2(\nabla\mbox{tr}\gamma)\cdot\nabla
u+(\Delta\mbox{tr}\gamma)u,$

which gives

$\displaystyle 0$
$\displaystyle=(\mbox{tr}\gamma\,Id+2s\gamma):\nabla^{2}u+(2\nabla\mbox{tr}\gamma+4s\nabla\cdot\gamma)\cdot\nabla
u+(\Delta\mbox{tr}\gamma+2s\nabla^{2}:\gamma)u$
$\displaystyle=\Gamma:\nabla^{2}u+(2\nabla\cdot\Gamma)\cdot\nabla
u+(\nabla^{2}:\Gamma)u$ $\displaystyle=\nabla^{2}:(\Gamma u):=Lu$

in all ${\mathbb{R}}^{n}$, with the definition
$\Gamma:=\mbox{tr}\gamma\,Id+2s\gamma$.

Only at this point we use the assumption that the exterior coefficients are
isotropic, i.e. that there exist scalar functions $b_{k}$ such that
$\beta_{k}=b_{k}Id$ for all $k\in{\mathbb{N}}$. This gives

$\gamma(x)=Id\sum_{k}\tilde{b}_{k}b_{k}(x)\qquad\mbox{ and
}\qquad\Gamma(x):=(n+2s)\,Id\sum_{k}\tilde{b}_{k}b_{k}(x),$

and therefore

$0=Lu(x)=\nabla^{2}:(\Gamma(x)u(x))=(n+2s)\Delta(u(x)\sum_{k}\tilde{b}_{k}b_{k}(x)),$

for all $x\in{\mathbb{R}}^{n}$. The function $u\sum_{k}\tilde{b}_{k}b_{k}$ is
thus seen to be harmonic, and since it vanishes on $\Omega_{e}$ it must be
equal to $0$ in all of ${\mathbb{R}}^{n}$. However, given that by assumption
(A2) we have $\sum_{k}\tilde{b}_{k}b_{k}(x)>0$ for all $x\in{\mathbb{R}}^{n}$,
the function $u$ must itself vanish everywhere.

We have proved that any $F\in L^{2}(\Omega)$ such that $\langle F,v\rangle=0$
for all $v\in\mathcal{R}$ vanishes identically. This gives the wanted result
by the Hahn-Banach theorem.

∎

######  Remark 6.2.

The above result can be generalized to a larger class of exterior
coefficients. In particular, any exterior coefficients $\beta_{k}$ such that
the weak maximum principle holds for the operator $L$ would suffice. Let us
make for the sake of this discussion the slightly stronger regularity
assumption that $\Phi\in C^{2+\varepsilon}({\mathbb{R}}^{n})$. Then $\Gamma\in
C^{2+\varepsilon}({\mathbb{R}}^{n})$, and all the coefficients of the operator
$L$ belong to $C^{\varepsilon}({\mathbb{R}}^{n})$. Moreover, $L$ is uniformly
elliptic, since for all $\xi,x\in{\mathbb{R}}^{n}$ and $y\in\Omega_{e}$

$\displaystyle\Gamma(x):(\xi\otimes\xi)$
$\displaystyle=(\mbox{tr}\gamma(x)\,Id+2s\gamma(x)):(\xi\otimes\xi)$
$\displaystyle=\mbox{tr
}\tilde{a}(x,y)|\xi|^{2}+2s\tilde{a}(x,y):(\xi\otimes\xi)$
$\displaystyle\geq(n\lambda_{min}(x,y)+2s\nu)|\xi|^{2}$
$\displaystyle\geq(n+2s)\nu|\xi|^{2}.$

Here we used assumptions (A1), (A2) and observed that
$\nu\in(0,\lambda_{min}(x,y)]$ for all $x,y\in{\mathbb{R}}^{n}$, where
$\lambda_{min}(x,y)$ is the minimal eigenvalue of $\tilde{a}(x,y)$. Due to the
Hölder continuity of the coefficients of $L$, $u\in
C^{2+\varepsilon}({\mathbb{R}}^{n})$ is actually a classical solution. If now
the weak maximum principle holds for the operator $L$, we may consider
equation $Lu=0$ in a large enough ball $B$ containing $\Omega$, and the result
would follow from the fact that $u$ vanishes on $\partial B$.

We are now ready to give the proof of our main result.

###### Proof of Theorem 1.2.

Assume without loss of generality that $W_{1}$ and $W_{2}$ are disjoint, as
this condition can always be achieved by restriction of the data. Let
$f_{j}\in C^{\infty}_{c}(W_{j})$ and $u_{j}:=P_{A_{j}}f_{j}$ for $j=1,2$.
Using the Alessandrini identity and the assumption on the DN maps, we can
write

(21)
$\begin{split}0&=\langle(\Lambda_{A_{1}}-\Lambda_{A_{2}})[f_{1}],[f_{2}]\rangle\\\
&=\langle\Phi_{1}u_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2}),\Phi_{2}u_{2}\rangle\\\ &=\langle\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\Phi_{2},u_{1}u_{2}\rangle.\end{split}$

We now apply the Runge approximation property. For all $g_{1},g_{2}\in
C^{\infty}_{c}(\Omega)$ we can find two sequences of exterior values
$\\{f_{j,i}\\}_{i}\subset C^{\infty}_{c}(W_{j})$, $j=1,2$, such that

$u_{j,i}:=f_{j,i}+g_{j}+r_{j,i},\qquad\mbox{with}\quad\|r_{j,i}\|_{L^{2}(\Omega)}\leq
1/i.$

Substituting these solutions into the previous equation gives

$\displaystyle 0$ $\displaystyle=\langle\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\Phi_{2},u_{1,i}u_{2,i}\rangle$
$\displaystyle=\langle\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\Phi_{2},(f_{1,i}+g_{1}+r_{1,i})(f_{2,i}+g_{2}+r_{2,i})\rangle$
$\displaystyle=\langle\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\Phi_{2},(g_{1}+r_{1,i})(g_{2}+r_{2,i})\rangle$

where we used the support assumptions and the locality of the operators
involved. Since $\Phi_{1},\Phi_{2}\in C^{2s+\varepsilon}({\mathbb{R}}^{n})$
and $Q_{1},Q_{2}\in L^{\infty}({\mathbb{R}}^{n})$ (as showed in the proof of
Lemma 3.3), we have

$\displaystyle|\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\Phi_{2}|$
$\displaystyle=\left|\sum_{k,i,j,k^{\prime},i^{\prime},j^{\prime}}\Phi_{1,k,ij}(Q_{1}-Q_{2})_{k,ij,k^{\prime},i^{\prime}j^{\prime}}\Phi_{2,k^{\prime},i^{\prime}j^{\prime}}\right|$
$\displaystyle\leq\left(\sum_{k,i,j,k^{\prime},i^{\prime},j^{\prime}}|(Q_{1}-Q_{2})_{k,ij,k^{\prime},i^{\prime}j^{\prime}}|^{2}\right)^{1/2}\left(\sum_{k,i,j,k^{\prime},i^{\prime},j^{\prime}}|\Phi_{1,k,ij}|^{2}|\Phi_{2,k^{\prime},i^{\prime}j^{\prime}}|^{2}\right)^{1/2}$
$\displaystyle\leq\|Q_{1}-Q_{2}\|_{L^{\infty}}\|\Phi_{1}\|_{C^{2s+\varepsilon}}\|\Phi_{2}\|_{C^{2s+\varepsilon}}<\infty.$

Thus $\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\Phi_{2}\in
L^{\infty}({\mathbb{R}}^{n})$, and the error terms vanish in the limit by the
Cauchy-Schwartz inequality. Therefore,

$\displaystyle 0$ $\displaystyle=\langle\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\Phi_{2},g_{1}g_{2}\rangle,$

and the arbitrariety of $g_{1},g_{2}\in C^{\infty}_{c}(\Omega)$ ensures
$\Phi_{1}\mathinner{\vbox{ \fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\Phi_{2}=0$ in $\Omega$.

Since $A_{1}\sim A_{2}$, there exists a function $\rho\in
C^{\infty}_{c}(\Omega)$ such that $\Phi_{2}=(\rho+1)\Phi_{1}$. Thus

$\displaystyle Q_{1}-Q_{2}$
$\displaystyle=\frac{\Phi_{2}\otimes(-\Delta)^{s-1}D\Phi_{2}}{|\Phi_{2}|^{2}}-\frac{\Phi_{1}\otimes(-\Delta)^{s-1}D\Phi_{1}}{|\Phi_{1}|^{2}}$
$\displaystyle=\frac{\Phi_{1}\otimes(-\Delta)^{s-1}D((\rho+1)\Phi_{1})}{(\rho+1)|\Phi_{1}|^{2}}-\frac{\Phi_{1}\otimes(-\Delta)^{s-1}D\Phi_{1}}{|\Phi_{1}|^{2}}$
$\displaystyle=\Phi_{1}\otimes\frac{(-\Delta)^{s-1}D(\rho\Phi_{1})-\rho(-\Delta)^{s-1}D\Phi_{1}}{(\rho+1)|\Phi_{1}|^{2}},$

and so in $\Omega$ it holds that

$\displaystyle 0$ $\displaystyle=\Phi_{1}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}(Q_{1}-Q_{2})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\Phi_{2}$
$\displaystyle=\left\\{(-\Delta)^{s-1}D(\rho\Phi_{1})-\rho(-\Delta)^{s-1}D\Phi_{1}\right\\}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\Phi_{1}$
$\displaystyle=((-\Delta)^{s-1}D(\rho\Phi_{1})+(\rho\Phi_{1})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}Q_{1})\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\Phi_{1}.$

This means that $\rho\Phi_{1}$ weakly solves the transformed problem (17) in
$\Phi_{1}\widetilde{H}^{s}(\Omega)$ with vanishing inhomogeneity and exterior
value. Since this problem is well-posed, it must be $\rho\Phi_{1}\equiv 0$,
which entails $\Phi_{1}\equiv\Phi_{2}$. By formula (12), we eventually obtain

$A_{1,s}(x,y)=\sum_{k}\Phi_{1,k}(x)\odot\Phi_{1,k}(y)=\sum_{k}\Phi_{2,k}(x)\odot\Phi_{2,k}(y)=A_{2,s}(x,y).$

∎

##  7\. The limit case $s\rightarrow 1$

In this section we consider the limit case $s\rightarrow 1$ for the fractional
anisotropic conductivity operator. The following Proposition shows that, as
expected, under slightly stronger conditions one recovers the classical
anisotropic conductivity operator:

######  Proposition 7.1.

Let $K\subset{\mathbb{R}}^{n}$ compact, and assume that $A\in
L^{\infty}({\mathbb{R}}^{2n},{\mathbb{R}}^{n\times n})$ verifies assumptions
(A1), (A2) and

  1. (A3)enumi

there exists $\varepsilon>0$ such that $\Phi\in
C^{2+\varepsilon}({\mathbb{R}}^{n})$.

Then $A^{\prime}:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}^{n\times n}$ defined
for all $x\in{\mathbb{R}}^{n}$ by

$A^{\prime}(x):=\frac{Id\emph{ tr}A_{s}(x,x)+2A_{s}(x,x)}{n+2}$

belongs to $C^{2+\varepsilon}({\mathbb{R}}^{n},{\mathbb{R}}^{n\times n})$, is
constant in $\Omega_{e}$, symmetric, and uniformly positive definite with the
same constant $\nu$ as $A_{s}$. Moreover, the following limit holds for all
$u\in H^{2}_{K}$:

$\displaystyle\lim_{s\rightarrow
1}\mathbf{C}^{s}_{A}u=-\nabla\cdot(A^{\prime}\cdot\nabla u).$

###### Proof.

Since assumption (A3) is substituted by (A3)enumi, $\Phi$ is a multiplier of
$H^{2}({\mathbb{R}}^{n})$ into itself. The proof of this fact follows the one
of Lemma 2.5. Therefore, $\Phi u\in H^{2}_{K}$ and $D(\Phi u)\in
L^{2}({\mathbb{R}}^{n})$ with support in $K$. Moreover, the facts that
$\Phi_{2k+1}$ has compact support in $\Omega$ and that $\beta_{k}$ is constant
in $\Omega_{e}$ for all $k\in{\mathbb{N}}$ imply that $D\Phi\in
L^{2}({\mathbb{R}}^{n})$ with compact support.

If $v\in L^{2}({\mathbb{R}}^{n})$ with compact support, then
$\lim_{s\rightarrow 1^{-}}(-\Delta)^{s-1}v=v$ almost everywhere. This can be
proved directly by Fourier methods first for
$v\in\mathcal{S}({\mathbb{R}}^{n})$, the set of Schwartz functions, then for
general $v$ by density. In particular, $(-\Delta)^{s-1}D(\Phi u)\rightarrow
D(\Phi u)$ and $(-\Delta)^{s-1}D\Phi\rightarrow D\Phi$ as $s\rightarrow 1^{-}$
almost everywhere in ${\mathbb{R}}^{n}$. Therefore, by the reduction Lemma 3.3
we compute

$\displaystyle\lim_{s\rightarrow 1}\mathbf{C}^{s}_{A}u$
$\displaystyle=\lim_{s\rightarrow 1}\Phi\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.}
}}\left\\{(-\Delta)^{s-1}D(\Phi u)-u(-\Delta)^{s-1}D\Phi\right\\}$
$\displaystyle=-\frac{\Phi}{n+2}\mathinner{\vbox{
\fontcharht:-\fontcharht.\hbox{.}\hbox{.}\hbox{.} }}\left\\{\left(\Delta
Id+2\nabla^{2}\right)\odot(\Phi u)-u\left(\Delta
Id+2\nabla^{2}\right)\odot\Phi\right\\}$
$\displaystyle=-\sum_{k}\frac{\Phi_{k}}{n+2}:\left\\{\left(\Delta
Id+2\nabla^{2}\right)\odot(\Phi_{k}u)-u\left(\Delta
Id+2\nabla^{2}\right)\odot\Phi_{k}\right\\}$
$\displaystyle=-\sum_{k}\frac{\Phi_{k}}{n+2}:\left\\{Id\odot\left(\Delta(\Phi_{k}u)-u\Delta\Phi_{k}\right)\right\\}-\sum_{k}\frac{2\Phi_{k}}{n+2}:\left\\{\nabla^{2}\odot(\Phi_{k}u)-u\nabla^{2}\odot\Phi_{k}\right\\}$
$\displaystyle=:S_{1}+S_{2}.$

For the first term $S_{1}$ we write

$\displaystyle-(n+2)S_{1}$
$\displaystyle=\sum_{k}\Phi_{k}:\left\\{Id\odot\left(\Phi_{k}\Delta u+2\nabla
u\cdot\nabla\Phi_{k}\right)\right\\}$
$\displaystyle=Id:\sum_{k}\left\\{\left(\Phi_{k}\Delta u+2\nabla
u\cdot\nabla\Phi_{k}\right)\odot\Phi_{k}\right\\}$
$\displaystyle=Id:\sum_{k}\left\\{\left(\Phi_{k}\odot\Phi_{k}\right)\Delta
u+\nabla u\cdot\nabla\left(\Phi_{k}\odot\Phi_{k}\right)\right\\}$
$\displaystyle=Id:\left\\{A_{s}(x,x)\Delta u+\nabla u\cdot\nabla
A_{s}(x,x)\right\\}$ $\displaystyle=\mbox{tr}A_{s}(x,x)\Delta u+\nabla
u\cdot\nabla\mbox{tr}A_{s}(x,x)$
$\displaystyle=\nabla\cdot(\mbox{tr}A_{s}(x,x)\nabla u).$

Similarly, for the second term $S_{2}$ we see that

$\displaystyle-(n+2)S_{2}$
$\displaystyle=2\sum_{k}\sum_{i,j}\Phi_{k,ij}\left\\{\nabla^{2}\odot(\Phi_{k}u)-u\nabla^{2}\odot\Phi_{k}\right\\}_{ij}$
$\displaystyle=2\sum_{k}\sum_{i,j}\Phi_{k,ij}\left\\{\partial_{i}\partial_{j}(\Phi_{k,ij}u)-u\partial_{i}\partial_{j}\Phi_{k,ij}\right\\}$
$\displaystyle=2\sum_{k}\sum_{i,j}\Phi_{k,ij}^{2}\partial_{i}\partial_{j}u+\Phi_{k,ij}\partial_{i}\Phi_{k,ij}\partial_{j}u+\Phi_{k,ij}\partial_{j}\Phi_{k,ij}\partial_{i}u$
$\displaystyle=\sum_{i,j}2(\sum_{k}\Phi_{k,ij}^{2})\partial_{i}\partial_{j}u+\partial_{i}(\sum_{k}\Phi_{k,ij}^{2})\partial_{j}u+\partial_{j}(\sum_{k}\Phi_{k,ij}^{2})\partial_{i}u.$

However,

$\sum_{k}\Phi_{k,ij}^{2}=\\{\sum_{k}\Phi_{k}\odot\Phi_{k}\\}_{ij}=A_{s}(x,x)_{ij}=A_{s}(x,x)_{ji},$

and thus

$\displaystyle-(n+2)S_{2}$
$\displaystyle=2\sum_{i,j}A_{s}(x,x)_{ij}\partial_{i}\partial_{j}u+\partial_{j}u\,\partial_{i}A_{s}(x,x)_{ij}$
$\displaystyle=2\sum_{i,j}\partial_{i}(A_{s}(x,x)_{ij}\partial_{j}u)$
$\displaystyle=2\nabla\cdot(A_{s}(x,x)\cdot\nabla u).$

This gives the expected formula for $A^{\prime}$. Since $A_{s}\in
C^{2+\varepsilon}({\mathbb{R}}^{2n})$, the function $x\mapsto A_{s}(x,x)$ is
itself in $C^{2+\varepsilon}({\mathbb{R}}^{n})$, and we have $A^{\prime}\in
C^{2+\varepsilon}({\mathbb{R}}^{n})$. The facts that $A^{\prime}$ is symmetric
and that it is constant in $\Omega_{e}$ follow from the related properties of
$A_{s}$. The new matrix $A^{\prime}$ is also uniformly positive definite,
since for all $x,\xi\in{\mathbb{R}}^{n}$ it holds

$\displaystyle A^{\prime}(x):(\xi\otimes\xi)$ $\displaystyle=\frac{Id\mbox{
tr}A_{s}(x,x)+2A_{s}(x,x)}{n+2}:(\xi\otimes\xi)$ $\displaystyle=\frac{\mbox{
tr}A_{s}(x,x)|\xi|^{2}+2A_{s}(x,x):(\xi\otimes\xi)}{n+2}$
$\displaystyle\geq\frac{n\nu|\xi|^{2}+2\nu|\xi|^{2}}{n+2}=\nu|\xi|^{2}.$

The inequality in the above computation is due to (A2) and the consideration
that $\nu\in(0,\lambda_{min}(x,y)]$ for all $x,y\in{\mathbb{R}}^{n}$, where
$\lambda_{min}(x,y)$ is the minimal eigenvalue of $A_{s}(x,y)$.

∎

######  Remark 7.2.

When the matrix $A_{s}$ is isotropic, i.e. of the form $\sigma(x,y)$Id for a
scalar function $\sigma$, then the limit matrix $A^{\prime}$ is

$A^{\prime}(x):=\frac{Id\emph{
tr}A_{s}(x,x)+2A_{s}(x,x)}{n+2}=\frac{Id(n\sigma(x,x))+2\sigma(x,x)Id}{n+2}=\sigma(x,x)Id=A_{s}(x,x),$

and thus it is itself isotropic. In this case the above result can be
inverted, since every isotropic, positive definite matrix
$A^{\prime}(x)=\gamma(x)$Id can be realized as the limit of a separable,
isotropic matrix $A_{s}(x,y)=\gamma^{1/2}(x)\gamma^{1/2}(y)$Id. There exist
however many different, non-separable matrices $A_{s}$ giving rise to the same
limit matrix $A^{\prime}$.

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