Qualitative properties of positive solutions

for mixed integro-differential equations

Patricio Felmer and Ying Wang

Departamento de Ingeniería Matemática and Centro de Modelamiento Matemático,
UMR2071 CNRS-UChile, Universidad de Chile

<EMAIL_ADDRESS>and<EMAIL_ADDRESS>)

###### Abstract

This paper is concerned with the qualitative properties of the solutions of
mixed integro-differential equation

$\left\\{\begin{array}[]{lll}(-\Delta)_{x}^{\alpha}u+(-\Delta)_{y}u+u=f(u)\quad\
\ {\rm in}\ \ \mathbb{R}^{N}\times\mathbb{R}^{M},\\\\[5.69054pt] u>0\ \
{\rm{in}}\ \mathbb{R}^{N}\times\mathbb{R}^{M},\ \
\quad\lim_{|(x,y)|\to+\infty}u(x,y)=0,\end{array}\right.$ (0.1)

with $N\geq 1$, $M\geq 1$ and $\alpha\in(0,1)$. We study decay and symmetry
properties of the solutions to this equation. Difficulties arise due to the
mixed character of the integro-differential operators. Here, a crucial role is
played by a version of the Hopf’s Lemma we prove in our setting. In studying
the decay, we construct appropriate super and sub solutions and we use the
moving planes method to prove the symmetry properties.

Key words: Integro-differential equation, Hopf’s Lemma, Decay, Symmetry.
MSC2010: 35R11, 35B06, 35B40, 35B50.

##  1 Introduction

The study of qualitative properties of positive solutions to semi-linear
elliptic equations in $\mathbb{R}^{N}$ has been the concern of numerous
authors along the last several decades. The asymptotic behavior of the
solution at infinity, the actual rate of decay and symmetry properties have
been the most studied qualitative properties for these equations. It was the
seminal work by Gidas, Ni and Nirenberg [19] that settled these two main
qualitative properties for the semi-linear elliptic equation

$\left\\{\begin{array}[]{lll}-\Delta u+u=f(u)\quad{\rm
in}\quad\mathbb{R}^{M},\\\\[5.69054pt] u>0\ \ {\rm{in}}\ \ \mathbb{R}^{M},\ \
\quad\lim_{|y|\to+\infty}u(y)=0,\end{array}\right.$ (1.1)

when the non-linearity is merely Lipschitz continuous, super-linear at the
zero, in the sense that

$f(s)=O(s^{p})\quad{\rm as}\ s\to 0,$ (1.2)

for some $p>1$, and $M\geq 3$. Gidas, Ni and Nirenberg proved that the
solutions of (1.1) are radially symmetric and they satisfy the precise decay
estimate

$\lim_{|y|\to+\infty}u(y)e^{|y|}|y|^{\frac{M-1}{2}}=c,$ (1.3)

for certain constant $c>0$. After this work, many authors extended the results
in various directions, generalizing the non-linearity, the elliptic operator
or the hypotheses on the solutions. Out of the very many contributions in this
direction we mention here only a few: Berestycki and Lions [5], Berestycki and
Nirenberg [6], Brock [7], Busca and Felmer [8], Cortázar, Elgueta and Felmer
[13], Da Lio and Sirakov [14], Dolbeault and Felmer [16], Gui [20], Kwong
[21], Li and Ni [24] and Pacella and Ramaswamy [26].

Recently, much attention has been given to the study of elliptic equations of
fractional order. In this direction, Felmer, Quaas and Tan in [17] studied the
problem

$\left\\{\begin{array}[]{lll}(-\Delta)^{\alpha}u+u=f(u)\quad{\rm
in}\quad\mathbb{R}^{N},\\\\[5.69054pt] u>0\ \ {\rm{in}}\ \ \mathbb{R}^{N},\ \
\quad\lim_{|x|\to+\infty}u(x)=0.\end{array}\right.$ (1.4)

They proved existence and regularity of positive solutions, and also decay and
symmetry results. Precisely, it was proved that the solutions $u$ of (1.4)
satisfy

$\frac{c^{-1}}{|x|^{N+2\alpha}}\leq
u(x)\leq\frac{c}{|x|^{N+2\alpha}},\qquad|x|\geq 1,$ (1.5)

for some $c>1$, when $f$ is superlinear at $0$ in the sense that

$\lim_{s\to 0}\frac{f(s)}{s}=0.$

The radial symmetry of the solutions of (1.4) is derived by using the moving
planes method in integral form developed in [11, 25], assuming further that
$f\in C^{1}(\mathbb{R}),$ it is increasing and there exists $\tau>0$ such that

$\lim_{s\to 0}\frac{f^{\prime}(s)}{s^{\tau}}=0.$ (1.6)

This symmetry result was generalized by the authors in [18], using an
appropriate truncation argument together with the moving planes method with
ideas developed in [23]. We refer to some other papers with more discussions
on qualitative properties of solutions to fractional elliptic problems as
Cabré and Sire [9], Caffarelli and Silvestre [10], Chen, Li and Ou [11],
Barles, Chasseigne, Ciomaga and Imbert [12], Dipierro Palatucci, Valdinoci
[15], Li [25], Quaas and Xia [28], Ros-Oton and Serra [29] and Sire and
Valdinoci [32].

Both operators, the laplacian and the fractional laplacian, are particular
cases of a general class of elliptic operators connected to backward
stochastic differential equations associated to Brownian and Levy-Itô
processes, see for example Barles, Buckdahn and Pardoux [1], Benth, Karlsen
and Reikvam [4] and Pham [27]. Recently, Barles, Chasseigne, Ciomaga and
Imbert in [2, 3] and Ciomaga in [12] considered the existence and regularity
of solutions for equations involving mixed integro-differential operators
belonging to the general class of backward stochastic differential equations
mentioned above. A particular case of elliptic integro-differential operator
of mixed type is the one considering the laplacian in some of the variables
and the fractional laplacian in the others, modeling diffusion sensible to the
direction. In view of (1.1) and (1.4) we may write similarly

$\left\\{\begin{array}[]{lll}(-\Delta)_{x}^{\alpha}u+(-\Delta)_{y}u+u=f(u),\ \
\ \ (x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M},\\\\[5.69054pt] u>0\ \
{\rm{in}}\ \mathbb{R}^{N}\times\mathbb{R}^{M},\ \
\quad\lim_{|(x,y)|\to+\infty}u(x,y)=0,\end{array}\right.$ (1.7)

where $N\geq 1$, $M\geq 1$. The operator $(-\Delta)_{y}$ denotes the usual
laplacian with respect to $y$, while $(-\Delta)_{x}^{\alpha}$ denotes the
fractional laplacian of exponent $\alpha\in(0,1)$ with respect to $x$, i.e.

$(-\Delta)_{x}^{\alpha}u(x,y)=\int_{\mathbb{R}^{N}}\frac{u(x,y)-u(z,y)}{|x-z|^{N+2\alpha}}dz,$
(1.8)

for all $(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$. Here the integral is
understood in the principal value sense.

In view of the known results on decay and symmetry for solutions of equations
(1.1) and (1.4) just described above, it is interesting to ask if these
results still hold for solutions of the equation of mixed type (1.7), where
the elliptic operator represents diffusion depending on the direction in
space. Regarding the asymptotic decay of solution at infinity, the question is
interesting since a proper mix of the two variables should be obtained for the
decay estimates. The natural way to estimate the decay is through the
construction of super and sub solutions involving the fundamental solution of
the elliptic operator, which in this case is singular in
$\mathbb{R}^{N}\times\\{0\\}$. Moreover, the solution of (1.7) cannot be
radially symmetric, so this property cannot be used to estimate the decay. On
the other hand, regarding radial symmetry, we may still have symmetry in $x$
and $y$, but the moving planes method would require an adequate version of the
Hopf’s Lemma, that we prove here.

Our first theorem concerns the decay of solutions for (1.7) with general
nonlinearity and it states as follows.

###### Theorem 1.1

Let $\alpha\in(0,1)$, $N,M\in\mathbb{N}$, $N\geq 1$ and $M\geq 1$ and let us
assume that the function $f:(0,+\infty)\to\mathbb{R}$ is continuous and it
satisfies

$-\infty<B:=\liminf_{v\to 0^{+}}\frac{f(v)}{v}\leq A:=\limsup_{v\to
0^{+}}\frac{f(v)}{v}<1.$ (1.9)

Let $u$ be a positive classical solution of problem (1.7), then for any
$\epsilon>0$ small, there exists $C_{\epsilon}>1$ such that for any
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$,

$C_{\epsilon}^{-1}(1+|x|)^{-N-2\alpha}e^{-\theta_{2}|y|}\leq u(x,y)\leq
C_{\epsilon}(1+|x|)^{-N-2\alpha}e^{-\theta_{1}|y|},$ (1.10)

where

$\theta_{1}=\sqrt{1-A}-\epsilon\quad\mbox{
and}\quad\theta_{2}=\sqrt{1-B}+\epsilon.$ (1.11)

When we compare estimate (1.10) with (1.3) for $N=0$, we first observe that in
ours an exponential decay is obtained, but with a constant $C_{\epsilon}$
depending on $\epsilon$, which is a parameter controlling the rate of
exponential decay. This is more clear when $A=B=0$. On the other hand we are
making much more general assumptions on $f$ and, in particular, we are not
making any assumption on the radial symmetry of the solution, which is crucial
in proving (1.3). We do not know of a decay estimate better than

$C_{\epsilon}^{-1}e^{-\theta_{2}|y|}\leq u(y)\leq
C_{\epsilon}e^{-\theta_{1}|y|},\qquad y\in\mathbb{R}^{M},$ (1.12)

for solutions of (1.1) under assumption (1.9) for $f$, and where radial
symmetry of the solutions is not available, like in a case where $f$ may
depend on $y$. On the other hand, when $M=0$, we recover (1.5) from (1.10).
For the proof of the decay estimate (1.10) we construct suitable super and sub
solutions and we use comparison principle with a version of Hopf’s lemma.

When we assume further hypothesis we can get sharper estimates for the decay
of the solutions of equation (1.7). Precisely, we have the following result:

###### Theorem 1.2

Assume that $\alpha\in(0,1)$, $N\geq 1$, $M\geq 5$ and the non-linearity
$f:(0,+\infty)\to\mathbb{R}$ is non-negative and it satisfies (1.2). Let $u$
be a positive classical solution of (1.7), then there exists a constant $c>1$
such that for all $(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$,

$\frac{1}{c}{\rho}(x,y)\leq u(x,y)\leq c{\rho}(x,y)(1+|y|)^{\frac{1}{2}},$
(1.13)

where the function $\rho$ is defined as

${\rho}(x,y)=\min\\{\frac{1}{(1+|x|)^{N+2\alpha}}\,,\,e^{-|y|}|y|^{-\frac{N}{2\alpha}-\frac{M}{2}}\,,\,\frac{e^{-|y|}|y|^{1-\frac{M}{2}}}{(1+|x|)^{N+2\alpha}}\\}.$
(1.14)

We notice that this theorem gives the expected exponential decay for positive
solutions, as suggested by (1.3), assuming the dimension of the space
satisfies $M\geq 5$. Moreover, it gives the expected polynomial correction for
the lower bound with a gap in the power for the upper bound. This theorem is
proved under the assumption (1.2) on the non-linearity, constructing super and
sub solutions devised upon the fundamental solution of
$(-\Delta)_{x}^{\alpha}+(-\Delta)_{y}+id$. In our argument, a crucial role is
played by the estimate already obtained in Theorem 1.1. Since the fundamental
solution of $(-\Delta)_{x}^{\alpha}+(-\Delta)_{y}+id$ has
$\mathbb{R}^{N}\times\\{0\\}$ as singular set, we cannot use the method in
[19] in order to derive our estimate. Moreover, some other arguments in [19]
cannot be used either because the solutions of (1.7) are not radial, since the
differential operator is not radially invariant and there are no solutions
depending only on one of the $x$ or $y$ variables, as can be seen from (1.13),

Even though solutions of (1.7) are not radially symmetric, we can prove
partial symmetry in each of the variables $x$ and $y$ and this is the content
of our third theorem.

###### Theorem 1.3

Assume that $\alpha\in(0,1)$, $N\geq 1$, $M\geq 1$ and the function
$f:(0,+\infty)\to\mathbb{R}$ is locally Lipschitz and it satisfies (1.9).
Moreover, we assume that $f$ also satisfies

  * $(F)\ $

there exist $u_{0}>0$, $\gamma>\frac{N}{N+M}\cdot\frac{2\alpha}{N+2\alpha}$
and $\bar{c}>0$ such that

$\frac{f(v)-f(u)}{v-u}\leq\bar{c}v^{\gamma}\ \ \ \ {\rm for\ all}\quad
0<u<v<u_{0}.$ (1.15)

Then, every positive classical solution $u$ of equation (1.7) satisfies

$u(x,y)=u(r,s)$

and $u(r,s)$ is strictly decreasing in $r$ and $s$, where $r=|x|$ and $s=|y|$.

When $N=0$, we see that assumption $(F)$ implies $\gamma>0$ and (1.15)
coincides with the assumption considered in [23]. When $M=0$, assumption $(F)$
implies that $\gamma>\frac{2\alpha}{N+2\alpha}$ and it coincides with the
assumption considered in [18], when the solutions is assumed to decay as a
power $N+2\alpha$ at infinity. We remark that the operator
$(-\Delta)_{x}^{\alpha}+(-\Delta)_{y}$ is a combination of two operators with
different differential orders in $x-$variable and $y-$variable, and this
produced a combined polynomial-exponential decay and does not allow for radial
symmetry, but only partial symmetry as stated in Theorem 1.3.

The proof of Theorem 1.3 is based on the moving planes method as developed in
[18, 23]. In these arguments, the strong maximum principle plays a crucial
role and it is available for the laplacian and for the fractional laplacian.
However, in the case of our mixed integro-differential operator some
difficulties arise and we overcome them with a version of the Hopf’s Lemma.

The rest of the paper is organized as follows. In Section §2, we introduce a
version of the Hopf’s Lemma and a strong maximum principle. In Section §3, we
prove the decay of solutions as in Theorem 1.1 and Theorem 1.2 by constructing
suitable super and sub solutions. Section §4 is devoted to prove symmetry
results presented in Theorem 1.3.

##  2 Preliminaries

This section is devoted to study the Strong Maximum Principle for mixed
integro-differential operators as in equation (1.7). To this end, we prove
first a suitable form of the Hopf’s Lemma.

However, before to go to this, we recall some basic properties of the Sobolev
embeddings. If we denote the Sobolev spaces

$H(\mathbb{R}^{N+M})=\\{w\in
L^{2}(\mathbb{R}^{N+M})|\int_{\mathbb{R}^{M}}\int_{\mathbb{R}^{N}}(|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2}+1)|\hat{w}(\xi_{1},\xi_{2})|^{2}d\xi_{1}d\xi_{2}<\infty\\}$

and

$H^{\alpha}(\mathbb{R}^{N+M})=\\{w\in L^{2}(\mathbb{R}^{N+M})\ |\
\int_{\mathbb{R}^{N+M}}(|\xi|^{2\alpha}+1)|\hat{w}(\xi)|^{2}d\xi<\infty\\},$

with norms

$\|w\|_{H}=(\int_{\mathbb{R}^{M}}\int_{\mathbb{R}^{N}}(|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2}+1)|\hat{w}(\xi_{1},\xi_{2})|^{2}d\xi_{1}d\xi_{2})^{\frac{1}{2}}$

and

$\|w\|_{H^{\alpha}}=(\int_{\mathbb{R}^{N+M}}(|\xi|^{2\alpha}+1)|\hat{w}(\xi)|^{2}d\xi)^{\frac{1}{2}},$

respectively, then it is not difficult to see that the following proposition
holds.

###### Proposition 2.1

For $\alpha\in(0,1)$, we have that

$H(\mathbb{R}^{N+M})\subset H^{\alpha}(\mathbb{R}^{N+M})\subset
L^{p}(\mathbb{R}^{N+M}),$

where the first inclusion is continuous and the second inclusion is continuous
if $1\leq p\leq\frac{2(N+M)}{N+M-2\alpha}$. Moreover,

$H(\mathbb{R}^{N+M})\subset L^{p}_{loc}(\mathbb{R}^{N+M})$

is compact if $1\leq p<\frac{2(N+M)}{N+M-2\alpha}$.

We devote the rest of this section to prove the Strong Maximum Principle in
our context and to this end, we start with versions of the Maximum Principle
and the Hopf’s Lemma. In what follows, given $\Omega$ an open subset in
$\mathbb{R}^{N}\times\mathbb{R}^{M}$, we define its closed cylindrical
extension in the direction $x$ as

$\displaystyle\tilde{\Omega}=\\{(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}:\
\exists\ x^{\prime}\in\mathbb{R}^{N}\ {\rm{s.t.}}\
(x^{\prime},y)\in\bar{\Omega}\\}.$

Given a function $h$ defined in an appropriate domain, we consider the mixed
integro-differential operator

$\mathcal{L}w(x,y)=(-\Delta)^{\alpha}_{x}w(x,y)+(-\Delta)_{y}w(x,y)+h(x,y)w(x,y).$

###### Lemma 2.1

Assume that $\Omega$ is an open domain of $\mathbb{R}^{N}\times\mathbb{R}^{M}$
and the function $h:\Omega\to\mathbb{R}$ satisfies $h\geq 0$ in $\Omega$. If
the function $w\in C(\bar{\Omega})\cap L^{\infty}(\tilde{\Omega})$ satisfies

$\left\\{\begin{array}[]{ll}\mathcal{L}w\geq 0\quad{\rm in}\ \ \Omega,\quad
w\geq 0\quad{\rm in}\ \ \tilde{\Omega}\setminus\Omega,\\\\[5.69054pt]
\liminf_{(x,y)\in\Omega,|(x,y)|\to\infty}w(x,y)\geq 0\end{array}\right.$ (2.1)

then $w\geq 0$ in $\tilde{\Omega}.$

Proof. If not, we may assume that there exists some $(x_{0},y_{0})\in\Omega$
such that

$w(x_{0},y_{0})=\min_{(x,y)\in\tilde{\Omega}}w(x,y)<0.$

Then

$\displaystyle(-\Delta)_{x}^{\alpha}w(x_{0},y_{0})=\int_{\mathbb{R}^{N}}\frac{w(x_{0},y_{0})-w(z,y_{0})}{|x_{0}-z|^{N+2\alpha}}dz<0$

and

$(-\Delta)_{y}w(x_{0},y_{0})\leq 0$

and then, since $h$ is non-negative we have $\mathcal{L}w(x_{0},y_{0})<0,$
which contradicts (2.1), completing the proof. $\Box$

It what follows we prove a version of the Hopf’s Lemma and for this purpose we
need to give some conditions to the boundary of the domain where the function
is defined. We say that the domain
$\Omega\subset\mathbb{R}^{N}\times\mathbb{R}^{M}$ satisfies _interior cylinder
condition at $(x_{0},y_{0})\in\partial\Omega$_ if there exist $r>0$ and
$\tilde{y}\in\mathbb{R}^{M}$ such that $O_{r}=B_{r}^{N}(x_{0})\times
B_{r}^{M}(\tilde{y})$ satisfies

$O_{r}\subset\Omega\quad{\rm{and}}\quad(x_{0},y_{0})\in\partial O_{r},$ (2.2)

where $B_{r}^{N}(x_{0})=\\{x\in\mathbb{R}^{N}:|x-x_{0}|<r\\}$ and
$B_{r}^{M}(\tilde{y})=\\{y\in\mathbb{R}^{M}:|y-\tilde{y}|<r\\}$ and, obviously
$|\tilde{y}-y_{0}|=r$. We define also

$D=\\{(x,y)\in O_{r}:|x-x_{0}|<\frac{r}{2},\ |y-\tilde{y}|>\frac{r}{2}\\}.$
(2.3)

###### Lemma 2.2

[Hopf’s Lemma] Let $\Omega$ be an open set satisfying interior cylinder
condition at $(x_{0},y_{0})\in\partial\Omega$. Assume that $h\in
L^{\infty}(D)$ and $w\in C(\bar{\Omega})\cap\L^{\infty}(\tilde{\Omega})$
satisfies

$\mathcal{L}w\geq 0\quad{\rm in}\ \Omega$

and

$0=w(x_{0},y_{0})<w(x,y),\quad\forall(x,y)\in\Omega.$

Further assume that for $r>0$ be given in (2.3) and for any $(x,y)\in D$ we
have

$\int_{\mathbb{R}^{N}\setminus
B_{r}^{N}(x_{0})}\frac{w(z,y)}{|x-z|^{N+2\alpha}}dz\geq 0.$ (2.4)

Then

$\limsup_{s\to 0^{+}}\frac{w(x_{0},y_{0})-w(x_{0},y_{0}+s\tilde{y})}{s}<0,$
(2.5)

moreover, if the limit exists, then

$\frac{\partial w}{\partial n}(x_{0},y_{0})<0,$ (2.6)

where $n$ is the unit exterior normal vector of $\Omega$ at the point
$(x_{0},y_{0})$.

Proof. Let us define

$\varphi_{M}(y)=e^{-\beta|y-\tilde{y}|^{2}}-e^{-\beta r^{2}},\quad
y\in\bar{B}_{r}^{M}(\tilde{y}),$ (2.7)

where $\beta>0$ will be chosen later. By direct computation, we have that

$-\Delta\varphi_{M}(y)=(2M\beta-4\beta^{2}|y-\tilde{y}|^{2})e^{-\beta|y-\tilde{y}|^{2}}.$
(2.8)

Next we consider the function

$v(x,y)=\varphi_{N}(x)\varphi_{M}(y),\quad(x,y)\in\tilde{O}_{r},$

where $\varphi_{N}$ is the first eigenfunction of Dirichlet problem

$\left\\{\begin{array}[]{lll}(-\Delta)^{\alpha}\varphi_{N}(x)=\lambda_{1}\varphi_{N}(x),&x\in
B_{r/2}^{N}(x_{0}),\\\\[5.69054pt]
\varphi_{N}(x)=0,&x\in\mathbb{R}^{N}\setminus
B_{r/2}^{N}(x_{0}),\end{array}\right.$ (2.9)

where $\varphi_{N}$ is positive and bounded in $B_{r/2}^{N}(x_{0})$ and the
first eigenvalue $\lambda_{1},$ is positive, see Propositions 9 and 4 in [30]
and [31], respectively.

For $(x,y)\in D$, by (2.8) and (2.9), we obtain that

$\displaystyle\mathcal{L}v(x,y)$ $\displaystyle=$
$\displaystyle\varphi_{M}(y)(-\Delta)^{\alpha}\varphi_{N}(x)+\varphi_{N}(x)(-\Delta\varphi_{M}(y))+h(x,y)\varphi_{N}(x)\varphi_{M}(y)$
$\displaystyle=$
$\displaystyle\varphi_{N}(x)[\lambda_{1}\varphi_{M}(y)+(2M\beta-4\beta^{2}|y-\tilde{y}|^{2})e^{-\beta|y-\tilde{y}|^{2}}+h(x,y)\varphi_{M}(y)]$
$\displaystyle\leq$
$\displaystyle\varphi_{N}(x)e^{-\beta|y-\tilde{y}|^{2}}(\lambda_{1}+2M\beta-\beta^{2}r^{2}+\|h\|_{L^{\infty}(D)}),$

where the last inequality holds by the fact that
$0\leq\varphi_{M}(y)<e^{-\beta|y-\tilde{y}|^{2}}$ and $|y-\tilde{y}|>r/2$ in
$D$. Let us choose $\beta>0$ big enough such that

$\mathcal{L}v\leq 0\quad{\rm in}\ D.$ (2.10)

On the other hand, since $\varphi_{N}(x)=0$ for $|x-x_{0}|\geq r/2$ and
$\varphi_{M}(y)=0$ for $|y-\tilde{y}|=r$, it is obvious that $v=0$ in
$A_{1}\cup A_{2}$ where $A_{1}=\\{(x,y)\in\tilde{D}:|x-x_{0}|\geq r/2\\}$ and
$A_{2}=\\{(x,y)\in\bar{D}:|y-\tilde{y}|=r\\}$. If we define the set
$A_{3}:=\\{(x,y)\in\bar{D}:|y-\tilde{y}|=r/2\\}$, we see that
$\tilde{D}\setminus D=A_{1}\cup A_{2}\cup A_{3}$. We also observe that $v$ is
a bounded function in $\tilde{O}_{r}$.

Next we prove (2.5) assuming $h\geq 0$. Defining

$W(x,y)=\left\\{\begin{array}[]{lll}w(x,y),&(x,y)\in\bar{O}_{r},\\\\[5.69054pt]
0,&(x,y)\in\tilde{O}_{r}\setminus\bar{O}_{r}\end{array}\right.$ (2.11)

and using (2.4), we have that for any $(x,y)\in D$,

$\displaystyle\mathcal{L}W(x,y)=\mathcal{L}w(x,y)+\int_{\mathbb{R}^{N}\setminus
B_{r}^{N}(x_{0})}\frac{w(z,y)}{|x-z|^{N+2\alpha}}dz\geq 0.$

Combining with (2.10), we have that, for every $\epsilon>0$

$\mathcal{L}(W-\epsilon v)\geq 0\quad{\rm{in}}\ D.$ (2.12)

Since $v$ is bounded in $\tilde{O}_{r}$, the set $A_{3}$ is a compact subset
of $O_{r}$ and $w>0$ in $O_{r}$, then there exists $\epsilon>0$ small such
that

$W=w\geq\epsilon v\quad{\rm{in}}\ A_{3}.$

Since $v=0$ in $A_{1}\cup A_{2}$, $w\geq 0$ in $\bar{O}_{r}$ and (2.11), we
have $W\geq 0=\epsilon v$ in $A_{1}\cup A_{2}$. Consequently,

$W-\epsilon v\geq 0\quad{\rm{in}}\ \ \tilde{D}\setminus D.$

Then we can use Lemma 2.1, recalling that $h\geq 0$ to obtain that

$W-\epsilon v\geq 0\quad{\rm{in}}\ D.$

In view of the definition of $W$, since $D\subset\bar{O}_{r}$, we find that
$w-\epsilon v\geq 0$ in $D$ and noticing that
$w(x_{0},y_{0})=v(x_{0},y_{0})=0$ we obtain that

$\frac{w(x_{0},y_{0})-w(x_{0},y_{0}+s\tilde{y})}{s}\leq\epsilon\frac{v(x_{0},y_{0})-v(x_{0},y_{0}+s\tilde{y})}{s},$

for all $s\in(0,r/2)$. Thus, we have

$\displaystyle\limsup_{s\to
0^{+}}\frac{w(x_{0},y_{0})-w(x_{0},y_{0}+s\tilde{y})}{s}$ $\displaystyle\leq$
$\displaystyle\epsilon\lim_{s\to
0^{+}}\frac{v(x_{0},y_{0})-v(x_{0},y_{0}+s\tilde{y})}{s}$ $\displaystyle=$
$\displaystyle\epsilon\varphi_{N}(x_{0})\lim_{s\to
0^{+}}\frac{\varphi_{M}(y_{0})-\varphi_{M}(y_{0}+s\tilde{y})}{s}$
$\displaystyle=$ $\displaystyle-2\epsilon\beta r^{2}e^{-\beta
r^{2}}\varphi_{N}(x_{0})<0,$

completing the proof of (2.5).

The case for general $h$ can be done simply by replacing $h$ by $h^{+}$. In
fact, since $w>0$ in $\Omega$, we have

$(-\Delta)^{\alpha}_{x}w(x,y)+(-\Delta)_{y}w(x,y)+h^{+}(x,y)w(x,y)\geq
0,\quad(x,y)\in\Omega$

and similarly we obtain that

$(-\Delta)^{\alpha}_{x}v(x,y)+(-\Delta)_{y}v(x,y)+h^{+}(x,y)v(x,y)\leq
0,\quad(x,y)\in D,$

so we may proceed as before to get (2.5) and the proof is complete. $\Box$

In order to state the Strong Maximum Principle to be used in our moving planes
procedure, it is convenient to consider property $(P)$:

  * $(P)$

We say that a function $w:\tilde{\Omega}\to\mathbb{R}$ satisfies property
$(P)$ if whenever $(x_{0},y_{0})\in\Omega$ such that

$0=w(x_{0},y_{0})=\inf_{(x,y)\in\Omega}w(x,y),$

then

$w(x,y_{0})\equiv 0,\quad\forall x\in\mathbb{R}^{N}.$

The following lemma is in preparation of the strong maximum principle.

###### Lemma 2.3

Let $\Omega$ be an open set in $\mathbb{R}^{N}\times\mathbb{R}^{M}$ and $w$
have property $(P)$. We denote

$\Omega_{0}=\\{(x,y)\in\Omega:w(x,y)=\inf_{\Omega}w=0\\}.$ (2.13)

If $\O\not=\Omega_{0}\subsetneqq\Omega$, then $\Omega\setminus\Omega_{0}$
satisfies interior cylinder condition at any point
$(x_{0},y_{0})\in\partial\Omega_{0}\cap\Omega$.

Proof. Since $\O\not=\Omega_{0}\subsetneqq\Omega$, we have that
$\O\not=\partial\Omega_{0}\cap\Omega\subset\partial(\Omega\setminus\Omega_{0})$.
For any $(x_{0},y_{0})\in\partial\Omega_{0}\cap\Omega$, let us denote
$r=\frac{1}{4}dist((x_{0},y_{0}),\partial\Omega)$ and let
$\tilde{y}\in\mathbb{R}^{M}$ such that
$(x_{0},\tilde{y})\in\Omega\setminus\Omega_{0}$ and $|\tilde{y}-y_{0}|=r$.
Since $w$ has property $(P)$, then $w=0$ in $\tilde{\Omega}_{0}$, where
$\tilde{\Omega}_{0}$ is the extension of $\Omega_{0}$ in $x$-direction and as
$\Omega\setminus\Omega_{0}$ is open, we have that $B_{r}^{N}(x_{0})\times
B_{r}^{M}(\tilde{y})\subset\Omega\setminus\Omega_{0}$. Therefore,
$\Omega\setminus\Omega_{0}$ satisfies interior cylinder condition at
$(x_{0},y_{0})\in\partial\Omega_{0}\cap\Omega$. $\Box$

###### Theorem 2.1

[Strong Maximum Principle] Let $\Omega$ be an open set of
$\mathbb{R}^{N}\times\mathbb{R}^{M}$, the function $h\in
L_{loc}^{\infty}(\Omega)$ and $w\in C(\bar{\Omega})\cap
L^{\infty}(\tilde{\Omega})$ has the property $(P)$ satisfying

$\mathcal{L}w\geq 0\quad{\rm{in}}\ \Omega\quad{\rm{and}}\quad w\geq
0\quad{\rm{in}}\ \Omega.$ (2.14)

Assume that $\Omega_{0}\not=\O$ defined by (2.13) and there exists some
$(x_{0},y_{0})\in\partial\Omega_{0}\cap\Omega$ such that (2.4) holds in
corresponding $D$.

Then $w$ must be 0 in $\tilde{\Omega}$.

Proof. Assume that $\Omega_{0}\not=\Omega.$ By Lemma 2.3,
$\Omega\setminus\Omega_{0}$ satisfies interior cylinder condition at
$(x_{0},y_{0})\in\partial\Omega_{0}\cap\Omega$ and then $w(x_{0},y_{0})=0$ by
$w\in C(\bar{\Omega})$ and the definition of $\Omega_{0}$. Furthermore, we
observe that $\bar{D}$ is compact in $\Omega$ and then $h\in
L^{\infty}(\bar{D})$. Using Lemma 2.2, we obtain (2.5), which is impossible by
the fact of $w(x_{0},y_{0})=\inf_{\Omega}w=0$. Therefore, $\Omega_{0}=\Omega$,
i.e. $w\equiv 0$ in $\Omega$. Since $w$ has property $(P)$, then $w\equiv 0$
in $\tilde{\Omega}$. $\Box$

##  3 Decay estimate

###  3.1 Proof of Theorem 1.1

In this subsection, we prove Theorem 1.1 on decay estimates for positive
classical solutions of equation (1.7). The main work is to construct
appropriate super and sub solutions and then the decay estimate is derived by
Lemma 2.1.

Before proving Theorem 1.1, we introduce some computations gathered in the
next proposition. For $\alpha\in(0,1)$ and $\mu>0$, we define the function
$\psi_{\mu}:\mathbb{R}^{N}\to\mathbb{R}$ as follows

$\psi_{\mu}(x)=\left\\{\begin{array}[]{lll}\mu^{-N-2\alpha},\quad\ \
|x|<\mu,\\\\[5.69054pt] |x|^{-N-2\alpha},\quad\ |x|\geq\mu.\end{array}\right.$
(3.1)

###### Proposition 3.1

For any $\mu>0$, there exists $R_{0}>3\mu$ and $c>0$, independent of $\mu$,
such that

$-c\mu^{-2\alpha}\psi_{\mu}(x)\leq(-\Delta)^{\alpha}\psi_{\mu}(x)\leq-c^{-1}\mu^{-2\alpha}\psi_{\mu}(x),\quad
x\in B_{R_{0}}^{c}.$ (3.2)

Proof. We consider along the proof that $\mu>0$ and $x\in\mathbb{R}^{N}$
satisfies $|x|>3\mu$. We define

$\displaystyle
A(\mu,x,z)=\frac{\psi_{\mu}(x+z)+\psi_{\mu}(x-z)-2\psi_{\mu}(x)}{|z|^{N+2\alpha}},\quad\quad
z\in\mathbb{R}^{N}$

and we observe that

$\displaystyle(-\Delta)^{\alpha}\psi_{\mu}(x)=-\frac{1}{2}\int_{\mathbb{R}^{N}}A(\mu,x,z)dz.$
(3.3)

Now we compute the integral above by decomposing the domain in various pieces.
First we consider the integral over $B_{\frac{|x|}{3}}(0)$. We observe that
$|x\pm z|\geq\mu$ for all $z\in B_{\frac{|x|}{3}}(0)$, then by (3.1) we obtain

$\displaystyle|\int_{B_{\frac{|x|}{3}}(0)}A(\mu,x,z)dz|$ $\displaystyle=$
$\displaystyle|\int_{B_{\frac{|x|}{3}}(0)}\frac{|x+z|^{-N-2\alpha}+|x-z|^{-N-2\alpha}-2|x|^{-N-2\alpha}}{|z|^{N+2\alpha}}dz|$
(3.4) $\displaystyle=$
$\displaystyle|x|^{-N-4\alpha}|\int_{B_{\frac{1}{3}}(0)}\frac{|z+e_{x}|^{-N-2\alpha}+|z-e_{x}|^{-N-2\alpha}-2}{|z|^{N+2\alpha}}dz|$
$\displaystyle\leq$ $\displaystyle
c_{1}|x|^{-N-4\alpha}\int_{B_{\frac{1}{3}}(0)}\frac{|z|^{2}}{|z|^{N+2\alpha}}dz\leq
c_{2}|x|^{-N-4\alpha},$

where $e_{x}=\frac{x}{|x|}$ and $c_{1},c_{2}>0$ are independent of $\mu$. Next
we consider the integral over $B_{\frac{|x|}{3}}(x)\setminus B_{\mu}(x)$. We
observe that for all $z\in B_{\frac{|x|}{3}}(x)\setminus B_{\mu}(x)$ we have
$|x+z|\geq|x-z|\geq\mu$ and then we obtain

$\displaystyle\int_{B_{\frac{|x|}{3}}(x)\setminus B_{\mu}(x)}A(\mu,x,z)dz$
$\displaystyle=$ $\displaystyle\int_{B_{\frac{|x|}{3}}(x)\setminus
B_{\mu}(x)}\frac{|x+z|^{-N-2\alpha}+|x-z|^{-N-2\alpha}-2|x|^{-N-2\alpha}}{|z|^{N+2\alpha}}dz$
$\displaystyle=$
$\displaystyle|x|^{-N-4\alpha}\int_{B_{\frac{1}{3}}(e_{x})\setminus
B_{\frac{\mu}{|x|}}(e_{x})}\frac{|z+e_{x}|^{-N-2\alpha}+|z-e_{x}|^{-N-2\alpha}-2}{|z|^{N+2\alpha}}dz$
$\displaystyle\leq$ $\displaystyle
c_{3}|x|^{-N-4\alpha}\int_{B_{\frac{1}{3}}(e_{x})\setminus
B_{\frac{\mu}{|x|}}(e_{x})}{|z-e_{x}|^{-N-2\alpha}}dz\leq
c_{4}\mu^{-2\alpha}|x|^{-N-2\alpha},$

where the first inequality holds since $|z+e_{x}|\geq|z-e_{x}|$ for $z\in
B_{\frac{1}{3}}(e_{x})\setminus B_{\frac{\mu}{|x|}}(e_{x})$ and
$|z|\geq\frac{2}{3}$ for $z\in B_{\frac{1}{3}}(e_{x})$. For the inequality on
the other side, we obtain

$\displaystyle\int_{B_{\frac{|x|}{3}}(x)\setminus B_{\mu}(x)}A(\mu,x,z)dz$
$\displaystyle=$
$\displaystyle|x|^{-N-4\alpha}\int_{B_{\frac{1}{3}}(e_{x})\setminus
B_{\frac{\mu}{|x|}}(e_{x})}\frac{|z+e_{x}|^{-N-2\alpha}+|z-e_{x}|^{-N-2\alpha}-2}{|z|^{N+2\alpha}}dz$
$\displaystyle\geq$
$\displaystyle|x|^{-N-4\alpha}(\int_{B_{\frac{1}{3}}(e_{x})\setminus
B_{\frac{\mu}{|x|}}(e_{x})}\frac{|z-e_{x}|^{-N-2\alpha}}{|z|^{N+2\alpha}}dz-\int_{B_{\frac{1}{3}}(e_{x})}\frac{2}{|z|^{N+2\alpha}}dz)$
$\displaystyle\geq$ $\displaystyle
c_{5}|x|^{-N-4\alpha}\int_{B_{\frac{1}{3}}(e_{x})\setminus
B_{\frac{\mu}{|x|}}(e_{x})}|z-e_{x}|^{-N-2\alpha}dz-c_{6}|x|^{-N-4\alpha}$
$\displaystyle\geq$ $\displaystyle
c_{7}\mu^{-2\alpha}|x|^{-N-2\alpha}-c_{8}|x|^{-N-4\alpha},$

where the second inequality holds by $|z|\leq\frac{4}{3}$ for $z\in
B_{\frac{1}{3}}(e_{x})$. Consequently,

$c_{7}\mu^{-2\alpha}|x|^{-N-2\alpha}-c_{8}|x|^{-N-4\alpha}\leq\int_{B_{\frac{|x|}{3}}(x)\setminus
B_{\mu}(x)}A(\mu,x,z)dz\leq c_{4}\mu^{-2\alpha}|x|^{-N-2\alpha},$ (3.5)

where the constants $c_{4},c_{7},c_{8}>0$ are independent of $\mu$. The
estimate for the integral over $B_{\frac{|x|}{3}}(-x)\setminus B_{\mu}(-x)$ is
similar.

Next we consider the integral over $B_{\mu}(x)$. We observe that, for $z\in
B_{\mu}(x)$ we have since $|x+z|>\mu>|x-z|$ and
$|z|\geq|x|-\mu\geq\frac{2|x|}{3}$, thus

$\displaystyle\int_{B_{\mu}(x)}A(\mu,x,z)dz=\int_{B_{\mu}(x)}\frac{|x+z|^{-N-2\alpha}+\mu^{-N-2\alpha}-2|x|^{-N-2\alpha}}{|z|^{N+2\alpha}}dz$
$\displaystyle\leq
2\int_{B_{\mu}(x)}\frac{\mu^{-N-2\alpha}}{|z|^{N+2\alpha}}dz\leq
c_{9}\mu^{-2\alpha}(|x|-\mu)^{-N-2\alpha}\leq
c_{10}\mu^{-2\alpha}|x|^{-N-2\alpha}$

and, for the other inequality

$\displaystyle\int_{B_{\mu}(x)}A(\mu,x,z)dz$ $\displaystyle\geq$
$\displaystyle\int_{B_{\mu}(x)}\frac{-2|x|^{-N-2\alpha}}{|z|^{N+2\alpha}}dz$
$\displaystyle\geq$ $\displaystyle-
c_{11}\mu^{N}|x|^{-N-2\alpha}(|x|-\mu)^{-N-2\alpha}\geq-
c_{12}|x|^{-N-4\alpha},$

where $c_{9},c_{10},c_{11}$ and $c_{12}$ are positive constant independent of
$\mu$. Therefore,

$\displaystyle-c_{12}|x|^{-N-4\alpha}\leq\int_{B_{\mu}(x)}A(\mu,x,z)dz\leq
c_{10}\mu^{-2\alpha}|x|^{-N-2\alpha}.$ (3.6)

The integral over $B_{\mu}(-x)$ is exactly the same. Finally, we consider the
complementary integral over
$D(x)=\mathbb{R}^{N}\setminus(B_{\frac{|x|}{3}}(0)\cup{B_{\frac{|x|}{3}}(x)}\cup{B_{\frac{|x|}{3}}(-x)})$.
For $|x|>3\mu$ and $z\in D(x)$, we have that $|x\pm z|\geq\frac{|x|}{3}$, thus

$\displaystyle|\int_{D(x)}A(\mu,x,z)dz|$ $\displaystyle\leq$
$\displaystyle\int_{D(x)}\frac{|x+z|^{-N-2\alpha}+|x-z|^{-N-2\alpha}+2|x|^{-N-2\alpha}}{|z|^{N+2\alpha}}dz$
(3.7) $\displaystyle\leq$ $\displaystyle
c_{13}|x|^{-N-2\alpha}\int_{\mathbb{R}^{N}\setminus
B_{\frac{|x|}{3}}(0)}\frac{1}{|z|^{N+2\alpha}}dz$ $\displaystyle\leq$
$\displaystyle c_{14}|x|^{-N-4\alpha},$

where $c_{13}>0$ and $c_{14}>0$ are independent of $\mu$. Therefore, by
(3.4)-(3.7), there exist $c_{15},c_{16}>1$ independent of $\mu$ such that

$\displaystyle
c_{15}^{-1}\mu^{-2\alpha}|x|^{-N-2\alpha}-c_{15}|x|^{-N-4\alpha}\leq\int_{\mathbb{R}^{N}}A(\mu,x,z)dz$
$\displaystyle\leq
c_{16}\mu^{-2\alpha}|x|^{-N-2\alpha}+c_{16}|x|^{-N-4\alpha}\leq
c_{15}\mu^{-2\alpha}|x|^{-N-2\alpha},$

where we used that $|x|>3\mu$. Choosing $R_{0}>3\mu$ such that
$c_{15}^{-1}\mu^{-2\alpha}-c_{15}|x|^{-2\alpha}\geq\frac{1}{2}c_{15}^{-1}\mu^{-2\alpha}$
for $|x|\geq R_{0}$, together with (3.3), we obtain (3.2). $\Box$

In what follows we provide a proof of our first theorem on the decay of the
positive solutions of our equation.

Proof of Theorem 1.1. By definition of $A$ and $B$ in (1.9), for any
$\epsilon>0$, there exits $\delta_{\epsilon}>0$ such that

$(B-\epsilon^{2})t\leq f(t)\leq(A+\epsilon^{2})t,\quad\ \forall\
t\in(0,\delta_{\epsilon}).$ (3.8)

Since $u$ is a positive solution of (1.7) vanishing at infinity, there exists
$R_{\epsilon}>0$ such that $0<u(x,y)<\delta_{\epsilon}$ for any $(x,y)\in
B^{c}_{R_{\epsilon}}$. Therefore,

$(-\Delta)_{x}^{\alpha}u+(-\Delta)_{y}u+(1-A-\epsilon^{2})u\leq
0\quad{\rm{in}}\ B^{c}_{R_{\epsilon}}$ (3.9)

and

$(-\Delta)_{x}^{\alpha}u+(-\Delta)_{y}u+(1-B+\epsilon^{2})u\geq
0\quad{\rm{in}}\ B^{c}_{R_{\epsilon}}.$ (3.10)

Next we define the function $\phi_{\nu}:\mathbb{R}^{M}\to\mathbb{R}$ as
$\phi_{\nu}(y)=e^{-\nu|y|},$ where $\nu>0$ and we find that for
$y\in\mathbb{R}^{M}\setminus\\{0\\}$,

$-\Delta\phi_{\nu}(y)=\nu\left(\frac{M-1}{|y|}-\nu\right)\phi_{\nu}(y).$
(3.11)

Step 1._There exists $C(\epsilon)>1$ such that_

$u(x,y)\leq
C(\epsilon)e^{-\theta_{1}|y|},\quad\quad(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$
(3.12)

To prove (3.12) we let $U_{1}(x,y)=\phi_{\theta_{1}}(y)$, for
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$ and then, by (3.11), we have

$\displaystyle(-\Delta)_{x}^{\alpha}U_{1}+(-\Delta)_{y}U_{1}+(1-A-\epsilon^{2})U_{1}$
$\displaystyle=\left[\theta_{1}\left(\frac{M-1}{|y|}-\theta_{1}\right)+1-A-\epsilon^{2}\right]U_{1}\geq
0,$ (3.13)

if $\epsilon\leq\sqrt{1-A}$. By definition of $U_{1}$ and $\phi_{\theta_{1}}$
we have that $U_{1}=1$ in $\mathbb{R}^{N}\times\\{0\\}$ and $U_{1}\geq
e^{-\theta_{1}R_{\epsilon}}$ in $\bar{B}_{R_{\epsilon}}$ and, since $u$ is
bounded, there exists $\rho_{1}>0$ depending on $\epsilon$, such that

$W_{1}=\rho_{1}U_{1}-u\geq 0\quad{\rm{in}}\ \
\bar{B}_{R_{\epsilon}}\cup(\mathbb{R}^{N}\times\\{0\\}).$

Combining (3.9) with (3.13), we obtain

$(-\Delta)_{x}^{\alpha}W_{1}+(-\Delta)_{y}W_{1}+(1-A-\epsilon^{2})W_{1}\geq
0\quad{\rm{in}}\ \
\bar{B}_{R_{\epsilon}}^{c}\cap(\mathbb{R}^{N}\times\\{0\\})^{c}.$

By Lemma 2.1, this implies that $W_{1}\geq 0$ in
$\mathbb{R}^{N}\times\mathbb{R}^{M}$ and then

$u(x,y)\leq\rho_{1}U_{1}(x,y)=\rho_{1}\phi_{\theta_{1}}(y)=\rho_{1}e^{-\theta_{1}|y|},\quad(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$
(3.14)

Step 2. _There exists $C(\epsilon)>1$ such that_

$u(x,y)\leq
C(\epsilon)|x|^{-N-2\alpha},\quad\quad(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$
(3.15)

Let $c$ and $R_{0}$ be as in Proposition 3.1
$\mu=({c}/({2\epsilon\sqrt{(1-A)}-2\epsilon^{2}}))^{\frac{1}{2\alpha}}$ and
consider the function $U_{2}(x,y)=\psi_{\mu}(x),$ for
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$ Then, by (3.2), we have for all
$(x,y)\in(B^{N}_{R_{0}}(0))^{c}\times\mathbb{R}^{M}$ that

$\displaystyle(-\Delta)_{x}^{\alpha}U_{2}+(-\Delta)_{y}U_{2}+(1-A-\epsilon^{2})U_{2}$
$\displaystyle\geq(-c\mu^{-2\alpha}+1-A-\epsilon^{2})U_{2}\geq 0$ (3.16)

for $0<\epsilon<\sqrt{1-A}$. Let us denote $W_{2}=\rho_{2}U_{2}-u$, where
$\rho_{2}>0$ is such that

$W_{2}\geq\rho_{2}(R_{0}+R_{\epsilon})^{-N-2\alpha}-u\geq 0\quad{\rm{in}}\
\bar{B}_{R_{\epsilon}}\cup(\overline{B_{R_{0}}^{N}(0)}\times\mathbb{R}^{M}).$

Combining (3.9) with (3.16), we obtain that

$\\!(-\Delta)_{x}^{\alpha}W_{2}+(-\Delta)_{y}W_{2}+(1-A-\epsilon^{2})W_{2}\geq
0\quad{\rm{in}}\ \
\bar{B}_{R_{\epsilon}}^{c}\cap(\overline{B_{R_{0}}^{N}(0)}\times\mathbb{R}^{M})^{c}.$

By Lemma 2.1, we have that $W_{2}=\rho_{2}U_{2}-u\geq 0$ in
$R^{N}\times\mathbb{R}^{M}$ and then, for all
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$,

$u(x,y)\leq\rho_{2}U_{2}(x,y)=\rho_{2}\psi_{\mu}(x)\leq\rho_{2}|x|^{-N-2\alpha}.$

Step 3. _There exists $C(\epsilon)>1$ such that_

$u(x,y)\leq C(\epsilon)|x|^{-N-2\alpha}e^{-\theta_{1}|y|},\quad\ \
(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$ (3.17)

Let us consider the function $V(x,y)=\psi_{\mu}(x)\phi_{\theta_{1}}(y),$ for
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M},$ with $\mu$ as defined above.
From (3.2) and (3.11), we have that

$\displaystyle(-\Delta)_{x}^{\alpha}V+(-\Delta)_{y}V+(1-A-\epsilon^{2})V$
$\displaystyle\geq\left[-c\mu^{-2\alpha}+\theta_{1}\left(\frac{M-1}{|y|}-\theta_{1}\right)+1-A-\epsilon^{2}\right]V\geq
0,$ (3.18)

for $(x,y)\in(B^{N}_{R_{0}}(0))^{c}\times(\mathbb{R}^{M}\setminus\\{0\\})$ and
assuming that $0<\epsilon<\sqrt{1-A}$. Since $u$, $V$ are bounded in
$\bar{B}_{R_{\epsilon}}$ and $V$ is positive, there is $\bar{\rho}_{1}>0$
large such that

$\bar{\rho}_{1}V-u\geq 0\quad{\rm{in}}\ \ \bar{B}_{R_{\epsilon}}.$

By (3.12) and (3.14), we may choose $\bar{\rho}_{2}>0$ such that

$\displaystyle\bar{\rho}_{2}V-u$ $\displaystyle\geq$
$\displaystyle\bar{\rho}_{2}R_{0}^{-N-2\alpha}\phi_{\theta_{1}}(y)-u\geq
0\quad{\rm{in}}\ \
\overline{B^{N}_{R_{0}}(0)}\times\mathbb{R}^{M}\quad\mbox{and}$
$\displaystyle\bar{\rho}_{2}V-u$ $\displaystyle\geq$
$\displaystyle\bar{\rho}_{2}\psi_{\mu}(x)-u\geq 0\quad{\rm{in}}\ \
\mathbb{R}^{N}\times\\{0\\}.$

Taking $\bar{\rho}=\max\\{\bar{\rho}_{1},\bar{\rho}_{2}\\}$, defining
$W=\bar{\rho}V-u$ and combining (3.9) with (3.18), we have that

$W\geq 0\ \ {\rm{in}}\ \
\bar{B}_{R_{\epsilon}}\cup(\overline{B^{N}_{R_{0}}(0)}\times\mathbb{R}^{M})\cup(\mathbb{R}^{N}\times\\{0\\})\quad\mbox{and}$
$(-\Delta)_{x}^{\alpha}W+(-\Delta)_{y}W+(1-A-\epsilon^{2})W\geq 0\ \
{\rm{in}}\ \
\bar{B}_{R_{\epsilon}}^{c}\cap((B^{N}_{R_{0}}(0))^{c}\times(\mathbb{R}^{M}\setminus\\{0\\})).$

Then, by Lemma 2.1, we have that $\bar{\rho}V-u\geq 0$ in
$\mathbb{R}^{N}\times\mathbb{R}^{M}.$ Thus, there exists $C(\epsilon)>1$ such
that

$u(x,y)\leq C(\epsilon)\psi_{\mu}(x)\phi_{\theta_{1}}(y)\leq
C(\epsilon)|x|^{-N-2\alpha}e^{-\theta_{1}|y|},\qquad(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$

Step 4. _There exists $C_{1}(\epsilon)>0$ and $R>0$ such that_

$u(x,y)\geq
C_{1}(\epsilon)e^{-\theta_{2}|y|},\quad\quad(x,y)\in\overline{B_{R}^{N}(0)}\times\mathbb{R}^{M}.$
(3.19)

Let $R_{0}$ be as in Proposition 3.1 and let $R>R_{0}$ such that
$\lambda_{1}<\epsilon^{2}$, where $\lambda_{1}$ is the first eigenvalue of the
fractional Dirichlet problem (2.9) with $x_{0}=0$ and $r=4R$. Let
$\varphi_{N}$ be the first eigenfunction of (2.9) and define
$V_{1}(x,y)=\varphi_{N}(x)\phi_{\theta_{2}}(y)$ for
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$ From (2.9) and (3.11), for
$(x,y)\in B_{2R}^{N}(0)\times(B_{R_{1}}^{M}(0))^{c}$ with
$R_{1}=\frac{M-1}{\epsilon}$, we have

$\displaystyle(-\Delta)_{x}^{\alpha}V_{1}+(-\Delta)_{y}V_{1}+(1-B+\epsilon^{2})V_{1}$
$\displaystyle=\left[\lambda_{1}+{\theta_{2}}\left(\frac{M-1}{|y|}-{\theta_{2}}\right)+1-B+\epsilon^{2}\right]V_{1}$
$\displaystyle\leq[\epsilon^{2}+{\theta_{2}}(\epsilon-{\theta_{2}})+1-B+\epsilon^{2}]V_{1}\leq
0,$ (3.20)

if $\epsilon<\sqrt{1-B}$. Let us define $w_{1}=u-r_{1}V_{1}$, where $r_{1}>0$
is such that

$w_{1}\geq 0\quad{\rm{in}}\
\overline{B}_{R_{\epsilon}}\cup(\overline{B_{2R}^{N}(0)\times
B_{R_{1}}^{M}(0)})$

and observe that $w_{1}\geq 0$ in $(B_{2R}^{N}(0))^{c}\times\mathbb{R}^{M}$
since $V_{1}=0$. Combining (3.10) with (3.20), we obtain that

$(-\Delta)_{x}^{\alpha}w_{1}+(-\Delta)_{y}w_{1}+(1-B+\epsilon^{2})w_{1}\geq
0\quad{\rm{in}}\ \ (B_{2R}^{N}(0)\times(B_{R_{1}}^{M}(0))^{c})\cap
B_{R_{\epsilon}}^{c}$

and then, by Lemma 2.1, we have that

$w_{1}=u-r_{1}V_{1}\geq 0\quad{\rm{in}}\ \
\mathbb{R}^{N}\times\mathbb{R}^{M}.$

Since $\varphi_{N}$ is classical solution of (2.9) with $r=4R$ and $x_{0}=0$
then $\varphi_{N}(x)$ is positive in
$\overline{B_{R}^{N}(0)}\subset\mathbb{R}^{N}$, we can finally choose
$C_{1}(\epsilon)>0$ such that

$u(x,y)\geq r_{1}\varphi_{N}(x)\phi_{\theta_{2}}(y)\geq
C_{1}(\epsilon)e^{-\theta_{2}|y|},\ \
\forall(x,y)\in\overline{B_{R}^{N}(0)}\times\mathbb{R}^{M}.$ (3.21)

Step 5._There exists $C_{1}(\epsilon)>0$ such that, for $R$ and $R_{1}$ as in
Step 4,_

$u(x,y)\geq
C_{1}|x|^{-N-2\alpha},\quad\quad(x,y)\in(B_{R}^{N}(0))^{c}\times\overline{B_{R_{1}}^{M}(0)}.$
(3.22)

To prove this, we define $V_{2}(x,y)=\psi_{\mu}(x)\eta_{M}(y)$ for
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M},$ where $\eta_{M}$ is the solution
of

$\left\\{\begin{array}[]{lll}-\Delta\eta_{M}(y)=\bar{\lambda}_{1}\eta_{M}(y),&y\in
B_{R_{2}}^{M}(0),\\\\[5.69054pt]
\eta_{M}(y)=0,&y\in(B_{R_{2}}^{M}(0))^{c},\end{array}\right.$ (3.23)

with $R_{2}>R_{1}$ such that $\bar{\lambda}_{1}<\epsilon^{2}$. Here
$\mu=[c(1-B+2\epsilon^{2})]^{\frac{-1}{2\alpha}}$ with $c$ as in Proposition
3.1 and $\psi_{\mu}$ defined in (3.1). By (3.2) and (3.23), for
$(x,y)\in((B_{R}^{N}(0))^{c}\times\mathbb{R}^{M})\cap(\mathbb{R}^{N}\times
B_{R_{2}}^{M}(0))$, we have that

$\displaystyle(-\Delta)_{x}^{\alpha}V_{2}+(-\Delta)_{y}V_{2}+(1-B+\epsilon^{2})V_{2}$
$\displaystyle\leq(-c^{-1}\mu^{-2\alpha}+\bar{\lambda}_{1}+1-B+\epsilon^{2})V_{2}=0.$
(3.24)

Let $w_{2}=u-r_{2}V_{2}$, with $r_{2}>0$ such that

$w_{2}\geq 0\quad{\rm{in}}\
\bar{B}_{R_{\epsilon}}\cup\overline{(B_{R}^{N}(0)}\times\mathbb{R}^{M})\cup(\mathbb{R}^{N}\times(B_{R_{2}}^{M}(0))^{c}).$

Combining (3.10) with (3.24), we obtain that

$(-\Delta)_{x}^{\alpha}w_{2}+(-\Delta)_{y}w_{2}+(1-B+\epsilon^{2})w_{2}\geq 0$

in
$B_{R_{\epsilon}}^{c}\cap((B_{R}^{N}(0))^{c}\times\mathbb{R}^{M})\cap(\mathbb{R}^{N}\times
B_{R_{2}}^{M}(0)).$ By Lemma 2.1, we have then

$w_{2}=u-r_{2}V_{2}\geq 0\quad{\rm{in}}\ \
\mathbb{R}^{N}\times\mathbb{R}^{M}.$

Since $\eta_{M}$ is positive in
$\overline{B_{R_{1}}^{M}(0)}\subset{B_{R_{2}}^{M}(0)}$, there exists
$C_{1}(\epsilon)>0$ such that for any
$(x,y)\in(B_{R}^{N}(0))^{c}\times\overline{B_{R_{1}}^{M}(0)}$, we have that

$u(x,y)\geq r_{2}\psi_{\mu}(x)\eta_{M}(y)\geq
C_{1}(\epsilon)|x|^{-N-2\alpha}.$

Step 6. _There exist $C_{1}(\epsilon)>0$ such that, for $R$ as in Step 4,_

$u(x,y)\geq
C_{1}(\epsilon)|x|^{-N-2\alpha}e^{-\theta_{2}|y|},\quad\quad(x,y)\in(B_{R}^{N}(0))^{c}\times\mathbb{R}^{M}.$
(3.25)

To prove this we let $\tilde{V}(x,y)=\psi_{\mu}(x)\phi_{\theta_{2}}(y),$ for
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$ with $\mu$ as defined above.
Using (3.2) and (3.11), for
$(x,y)\in(B^{N}_{R}(0))^{c}\times(B^{M}_{R_{1}}(0))^{c}$ with
$R_{1}=\frac{M-1}{\epsilon}$, we have that

$\displaystyle(-\Delta)_{x}^{\alpha}\tilde{V}+(-\Delta)_{y}\tilde{V}+(1-B+\epsilon^{2})\tilde{V}$
$\displaystyle\leq\left[-c^{-1}\mu^{-2\alpha}+{\theta_{2}}\left(\frac{M-1}{|y|}-{\theta_{2}}\right)+1-B+\epsilon^{2}\right]\tilde{V}$
$\displaystyle\leq[{\theta_{2}}(\epsilon-{\theta_{2}})+1-B+\epsilon^{2}]\tilde{V}\leq
0,$ (3.26)

if $0<\epsilon<\sqrt{1-B}$. Since $u$ is positive and $V$ is bounded in
$\overline{B}_{R_{\epsilon}}$, we can choose $\tilde{r}_{1}>0$ such that

$u-\tilde{r}_{1}V\geq 0\quad{\rm{in}}\ \ \overline{B}_{R_{\epsilon}}.$

Since $\psi_{\mu}$ is bounded in $\overline{B^{N}_{R}(0)}$, using (3.21),
there exists $\tilde{r}_{2}>0$ such that

$u-\tilde{r}_{2}V\geq u-\tilde{r}_{2}c_{1}e^{-\theta_{2}|y|}\geq
0\quad{\rm{in}}\ \ \overline{B^{N}_{R}(0)}\times\mathbb{R}^{M},$

and by (3.22), there exists $\tilde{r}_{3}>0$ such that

$u-\tilde{r}_{3}V\geq u-\tilde{r}_{3}|x|^{-N-2\alpha}\geq 0\quad{\rm{in}}\ \
(B_{R}^{N}(0))^{c}\times\overline{B_{R_{1}}^{M}(0)}.$

Taking $\tilde{r}=\min\\{\tilde{r}_{1},\tilde{r}_{2},\tilde{r}_{3}\\}$ and
combining (3.10) with (3.26), we obtain that

$\displaystyle w=u-\tilde{r}V\geq 0\quad{\rm{in}}\ \
\bar{B}_{R_{\epsilon}}\cup(\overline{B^{N}_{R}(0)}\times\mathbb{R}^{M})\cup((B_{R}^{N}(0))^{c}\times\overline{B_{R_{1}}^{M}(0)})\
\ \mbox{and}$
$\displaystyle(-\Delta)_{x}^{\alpha}w+(-\Delta)_{y}w+(1-B+\epsilon)w\geq
0\quad{\rm{in}}\ \
\bar{B}_{R_{\epsilon}}^{c}\cap((B^{N}_{R}(0))^{c}\times(B^{M}_{R_{1}}(0))^{c}).$

Thus Lemma 2.1, we have that $w\geq 0$ in $\mathbb{R}^{N}\times\mathbb{R}^{M}$
and then (3.25) holds.

Finally, Step 1 $-$ Step 6 completes the proof. $\Box$

###  3.2 Proof of Theorem 1.2

This subsection is devoted to prove Theorem 1.2. Our proof is based on the
fundamental solution of the mixed integro-differential operator. We first
study the fundamental solution $\mathcal{K}$ for

$(-\Delta)_{x}^{\alpha}u+(-\Delta)_{y}u+u=0\quad{\rm
in}\quad\mathbb{R}^{N}\times(\mathbb{R}^{M}\setminus\\{0\\}),$

which can be characterized by

$\mathcal{K}(x,y)=\int_{0}^{\infty}e^{-t}\mathcal{H}(x,y,t)dt,$ (3.27)

where

$\mathcal{H}(x,y,t)=\int_{\mathbb{R}^{M}}\int_{\mathbb{R}^{N}}e^{-2\pi
i(x,y)\cdot(\xi_{1},\xi_{2})-t(|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2})}d\xi_{1}d\xi_{2}.$
(3.28)

In fact, for $\phi\in\mathcal{S}$, we have that

$\begin{array}[]{lll}\langle\mathcal{K},\phi\rangle=\int_{\mathbb{R}^{N+M}}\int_{0}^{\infty}\int_{\mathbb{R}^{N+M}}e^{-2\pi
i(x,y)\cdot(\xi_{1},\xi_{2})-t(|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2}+1)}\phi(x,y)d\xi_{1}d\xi_{2}dtdxdy\\\\[5.69054pt]
\phantom{-}=\int_{\mathbb{R}^{N+M}}\left[\int_{0}^{\infty}e^{-t(|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2}+1)}dt\int_{\mathbb{R}^{N+M}}e^{-2\pi
i(x,y)\cdot(\xi_{1},\xi_{2})}\phi(x,y)dxdy\right]d\xi_{1}d\xi_{2}\\\\[5.69054pt]
\phantom{-}=\int_{\mathbb{R}^{N+M}}\left[\frac{1}{|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2}+1}\int_{\mathbb{R}^{N+M}}e^{-2\pi
i(x,y)\cdot(\xi_{1},\xi_{2})}\phi(x,y)dxdy\right]d\xi_{1}d\xi_{2}\\\\[5.69054pt]
\phantom{-}=\left\langle\frac{1}{|\xi_{1}|^{2\alpha}+|\xi_{2}|^{2}+1},\mathcal{F}\phi\right\rangle.\end{array}$

Next we want to find some properties of $\mathcal{H}$. To this end, we
consider

$\mathcal{H}_{\alpha}(x,t)=\int_{\mathbb{R}^{N}}e^{-2\pi
ix\cdot\xi_{1}-t|\xi_{1}|^{2\alpha}}d\xi_{1}\quad{\rm
and}\quad\mathcal{H}_{1}(y,t)=\int_{\mathbb{R}^{M}}e^{-2\pi
iy\cdot\xi_{2}-t|\xi_{2}|^{2}}d\xi_{2}.$

It is well known that the function $\mathcal{H}_{\alpha}$ has the following
properties:

$\mathcal{H}_{\alpha}(x,t)=t^{-\frac{N}{2\alpha}}\mathcal{H}_{\alpha}(t^{-\frac{1}{2\alpha}}x,1)\quad{\rm
and}\quad\lim_{|x|\to\infty}|x|^{N+2\alpha}\mathcal{H}_{\alpha}(x,1)=C,$

where $C>0$, which imply that there exists $c_{1}>0$ and $c_{2}>$ such that

$c_{1}\min\\{t^{-\frac{N}{2\alpha}},t|x|^{-N-2\alpha}\\}\leq\mathcal{H}_{\alpha}(x,t)\leq
c_{2}\min\\{t^{-\frac{N}{2\alpha}},t|x|^{-N-2\alpha}\\},$ (3.29)

see [22, 17]. By the definition of $\mathcal{H}$, we have that

$\mathcal{H}(x,y,t)=\mathcal{H}_{\alpha}(x,t)\mathcal{H}_{1}(y,t).$ (3.30)

Since we have

$\mathcal{H}_{1}(y,t)=(4\pi t)^{-\frac{M}{2}}e^{-\frac{|y|^{2}}{4t}},$ (3.31)

see [22], together with (3.27)-(3.30), for $|y|>2$,

$\displaystyle\mathcal{K}(x,y)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-t}\mathcal{H}_{\alpha}(x,t)\mathcal{H}_{1}(y,t)dt$
$\displaystyle\geq$ $\displaystyle
c_{1}\int_{0}^{\infty}e^{-t}\min\\{t^{-\frac{N}{2\alpha}},t|x|^{-N-2\alpha}\\}(4\pi
t)^{-\frac{M}{2}}e^{-\frac{|y|^{2}}{4t}}dt$ $\displaystyle\geq$ $\displaystyle
c_{1}\int_{\frac{|y|}{2}}^{\frac{|y|}{2}+1}e^{-t}\min\\{t^{-\frac{N}{2\alpha}},t|x|^{-N-2\alpha}\\}(4\pi
t)^{-\frac{M}{2}}e^{-\frac{|y|^{2}}{4t}}dt$ $\displaystyle\geq$ $\displaystyle
c_{3}\min\\{e^{-|y|}|y|^{-\frac{N}{2\alpha}-\frac{M}{2}},|x|^{-N-2\alpha}e^{-|y|}|y|^{1-\frac{M}{2}}\\},$

for some $c_{3}>0$. On the other hand, since for $n\geq 3$ we have

$\int_{0}^{\infty}e^{-t}(4\pi t)^{-\frac{n}{2}}e^{-\frac{|y|^{2}}{4t}}dt\leq
c_{4}e^{-|y|}|y|^{2-n}(1+|y|)^{\frac{n-3}{2}}$

with $c_{4}>0$ (see [22]), for $M\geq 5$ we have that

$\displaystyle\mathcal{K}(x,y)=\int_{0}^{\infty}e^{-t}\mathcal{H}_{\alpha}(x,t)\mathcal{H}_{1}(y,t)dt$
$\displaystyle\leq$ $\displaystyle
c_{2}\int_{0}^{\infty}e^{-t}\min\\{t^{-\frac{N}{2\alpha}},t|x|^{-N-2\alpha}\\}(4\pi
t)^{-\frac{M}{2}}e^{-\frac{|y|^{2}}{4t}}dt$ $\displaystyle\leq$ $\displaystyle
c_{5}\min\\{\int_{0}^{\infty}e^{-t}(4\pi
t)^{-\frac{N}{2\alpha}-\frac{M}{2}}e^{-\frac{|y|^{2}}{4t}}dt,|x|^{-N-2\alpha}\int_{0}^{\infty}e^{-t}(4\pi
t)^{1-\frac{M}{2}}e^{-\frac{|y|^{2}}{4t}}dt\\}$ $\displaystyle\leq$
$\displaystyle
c_{6}\min\\{e^{-|y|}|y|^{2-\frac{N}{\alpha}-M}(1+|y|)^{\frac{N}{2\alpha}+\frac{M}{2}-\frac{3}{2}},|x|^{-N-2\alpha}e^{-|y|}|y|^{4-M}(1+|y|)^{\frac{M-5}{2}}\\}$

Therefore, for $N\geq 1$ and $M\geq 5$, there exist $c_{8}>c_{7}>0$ such that

$c_{7}\rho(x,y)\leq\mathcal{K}(x,y)\leq
c_{8}\rho(x,y)|y|^{\frac{1}{2}},\quad(x,y)\in\mathbb{R}^{N}\times(B_{2}^{M}(0))^{c},$
(3.32)

where $\rho(x,y)$ is defined in (1.14). In what follows, we construct super
and sub-solutions to obtain the decay estimate given in Theorem 1.2.

Proof of Theorem 1.2. By the estimate in Theorem 1.1, we observe that, for
constants $c_{10}>c_{9}>0$ such that

$c_{9}(1+|x|)^{-N-2\alpha}\leq u(x,y)\leq
c_{10}(1+|x|)^{-N-2\alpha},\quad(x,y)\in\mathbb{R}^{N}\times{B_{2}^{M}(0)},$

so we only need to prove (1.13) holds for
$(x,y)\in\mathbb{R}^{N}\times{(B_{2}^{M}(0))^{c}}$.

Step 1: Lower bound. Let
$\tilde{u}=\mathcal{K}*\chi_{B_{1}^{N}(0)\times{B_{1}^{M}(0)}}$, where
$\chi_{B_{1}^{N}(0)\times B_{1}^{M}(0)}$ is the characteristic function of
$B_{1}^{N}(0)\times B_{1}^{M}(0)$. By (3.32), we have that

$\tilde{u}(x,y)\geq
c_{11}\min\\{e^{-|y|}|y|^{-\frac{N}{2\alpha}-\frac{M}{2}},(1+|x|)^{-N-2\alpha}e^{-|y|}|y|^{1-\frac{M}{2}}\\},$
(3.33)

for all $(x,y)\in\mathbb{R}^{N}\times{(B_{2}^{M}(0))^{c}}$, where $c_{11}>0$.
By definition of $\tilde{u}$, we have

$\displaystyle(-\Delta)_{x}^{\alpha}\tilde{u}+(-\Delta)_{y}\tilde{u}+\tilde{u}=0\quad{\rm
in}\ \
\mathbb{R}^{N}\times(\mathbb{R}^{M}\setminus\\{0\\})\setminus(B_{1}^{N}(0)\times
B_{1}^{M}(0))$

and, by (3.32) and Theorem 1.1, there exists $c_{12}>0$ such that $u\geq
c_{11}\tilde{u}$ in $\mathbb{R}^{N}\times\\{y\in\mathbb{R}^{M}:|y|=2\\}$.
Since $f$ is nonnegative, we use the Comparison Principle to obtain that, for
any $(x,y)\in\mathbb{R}^{N}\times(B_{2}^{M}(0))^{c}$

$\displaystyle u(x,y)\geq c_{11}\tilde{u}(x,y)\geq
c_{12}\min\\{e^{-|y|}|y|^{-\frac{N}{2\alpha}-\frac{M}{2}},(1+|x|)^{-N-2\alpha}e^{-|y|}|y|^{1-\frac{M}{2}}\\}.$

Step 2: Upper bound. For $y\in\mathbb{R}^{M}$ with $|y|\geq 2$, there exists
$1\leq i\leq M$ such that $|y_{i}|>1$, we may assume that $y_{1}>1$. Let
$\bar{u}(x,y)=\mathcal{K}(x,y)(1-|y_{1}|^{-1})$, then by direct computation

$\displaystyle(-\Delta)_{y}\bar{u}$ $\displaystyle=$
$\displaystyle(1-|y_{1}|^{-1})(-\Delta)_{y}\mathcal{K}-2y_{1}^{-2}\partial_{y_{1}}\mathcal{K}+2\mathcal{K}y_{1}^{-3}$
$\displaystyle\geq$
$\displaystyle(-\Delta)_{y}\mathcal{K}(1-|y_{1}|^{-1})+2\mathcal{K}y_{1}^{-3},$

where the last inequality holds since $y_{1}>0$ and
$\partial_{y_{1}}\mathcal{K}<0.$ Therefore, by (3.32), we have that for
$(x,y)\in\mathbb{R}^{N}\times(B_{2}^{M}(0))^{c}$,

$\displaystyle(-\Delta)_{x}^{\alpha}\bar{u}(x,y)+(-\Delta)_{y}\bar{u}(x,y)+\bar{u}(x,y)$
(3.34) $\displaystyle\geq$
$\displaystyle[(-\Delta)_{x}^{\alpha}\mathcal{K}+(-\Delta)_{y}\mathcal{K}+\mathcal{K}](1-|y_{1}|^{-1})+2\mathcal{K}(x,y)y_{1}^{-3}\geq
2\mathcal{K}(x,y)|y|^{-3}$ $\displaystyle\geq$ $\displaystyle
2c_{8}\min\\{e^{-|y|}|y|^{-\frac{N}{2\alpha}-\frac{M}{2}-3},|x|^{-N-2\alpha}e^{-|y|}|y|^{-\frac{M}{2}-2}\\}.$

Since $f(u)=O(u^{p})$ near $u=0$ for some $p>1$, by Theorem 1.1 with
$\epsilon=\frac{p-1}{4p}$, we have that

$(-\Delta)_{x}^{\alpha}u+(-\Delta)_{y}u+u)=f(u)\leq
c_{13}(1+|x|)^{-(N+2\alpha)p}e^{-\frac{3p+1}{4}|y|},$

where $c_{13}>0$. We notice that $\frac{3p+1}{4}>1$. By definition of
$\bar{u}$, (3.32) and Theorem 1.1 with $\epsilon=\frac{p-1}{4p}$, there exists
$c_{14}>0$ such that $u\leq c_{14}\bar{u}$ in
$\mathbb{R}^{N}\times\\{y\in\mathbb{R}^{M}:|y|=2\\}$. By Comparison Principle,
we have that

$\displaystyle u(x,y)$ $\displaystyle\leq$ $\displaystyle
c_{14}\bar{u}(x,y)\leq c_{14}\mathcal{K}(x,y)$ $\displaystyle\leq$
$\displaystyle
c_{15}\min\\{e^{-|y|}|y|^{\frac{1}{2}-\frac{N}{2\alpha}-\frac{M}{2}},(1+|x|)^{-N-2\alpha}e^{-|y|}|y|^{\frac{3}{2}-\frac{M}{2}}\\}$

for all $(x,y)\in\mathbb{R}^{N}\times(B_{2}^{M}(0))^{c}$ and some $c_{15}>0$.
This complete the proof.

$\Box$

##  4 Symmetry results

In this section, we prove Theorem 1.3 by moving planes method. Let $u$ be a
classical positive solution of (1.7) and consider first the $y$-direction. Let

$\Sigma^{y_{1}}_{\lambda}=\\{(x,y_{1},y^{\prime})\in\mathbb{R}^{N}\times\mathbb{R}\times\mathbb{R}^{M-1}\
|\ y_{1}>\lambda\\},$
$T^{y_{1}}_{\lambda}=\\{(x,y_{1},y^{\prime})\in\mathbb{R}^{N}\times\mathbb{R}\times\mathbb{R}^{M-1}\
|\ y_{1}=\lambda\\}$

and $u_{\lambda}(x,y_{1},y^{\prime})=u(x,2\lambda-y_{1},y^{\prime})$ for
$\lambda\in\mathbb{R}$. We introduce a preliminary inequality which plays a
crucial role in the procedure of moving planes.

###### Lemma 4.1

Under the assumptions of Theorem 1.3, for any $\lambda\in\mathbb{R}$, there
exists $c_{1}>0$, independent of $\lambda$, such that

$\displaystyle
c_{1}(\int_{\Sigma^{y_{1}}_{\lambda}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}$
$\displaystyle\leq$
$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}({u_{\lambda}}-u)+(-\Delta)_{y}({u_{\lambda}}-u)+({u_{\lambda}}-u)](u_{\lambda}-u)^{+}dxdy<\infty.$

Proof. First we show that the integrals are finite. We observe that
$u_{\lambda}$ satisfies the same equation (1.7) as $u$ in
$\Sigma^{y_{1}}_{\lambda}$. Taking $(u_{\lambda}-u)^{+}$ as test function in
the equations for $u$ and $u_{\lambda}$, subtracting and integrating in
$\Sigma^{y_{1}}_{\lambda}$, we find

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}({u_{\lambda}}-u)+(-\Delta)_{y}({u_{\lambda}}-u)+({u_{\lambda}}-u)](u_{\lambda}-u)^{+}dxdy$
$\displaystyle=\int_{\Sigma^{y_{1}}_{\lambda}}(f(u_{\lambda})-f(u))(u_{\lambda}-u)^{+}dxdy.$
(4.1)

Now we only need to prove that

$\int_{\Sigma^{y_{1}}_{\lambda}}(f(u_{\lambda})-f(u))(u_{\lambda}-u)^{+}dxdy<+\infty.$
(4.2)

In fact, for any given $\lambda\in\mathbb{R}$, using (1.10), we choose $R>1$
such that

$0<u_{\lambda}(x,y)\leq
C_{\epsilon}(1+|x|)^{-N-2\alpha}e^{-\theta_{1}|y_{\lambda}|}<s_{0},\ \ \
\forall(x,y)\in B^{c}_{R},$

where $y_{\lambda}=(2\lambda-y_{1},y^{\prime})$ for
$y=(y_{1},y^{\prime})\in\mathbb{R}^{M}$ and $s_{0}$ is from $(F)$.

If $u_{\lambda}(x,y)>u(x,y)$ for some $(x,y)\in{\Sigma^{y_{1}}_{\lambda}}\cap
B^{c}_{R}$, we have $0<u(x,y)<u_{\lambda}(x,y)<s_{0}$. Using (1.15) with
$v=u_{\lambda}(x,y)$, then

$\frac{f(u_{\lambda}(x,y))-f(u(x,y))}{u_{\lambda}(x,y)-u(x,y)}\leq\bar{c}u^{\gamma}_{\lambda}(x,y),$

then

$\displaystyle(f(u_{\lambda}(x,y))-f(u(x,y)))^{+}(u_{\lambda}(x,y)-u(x,y))^{+}\leq\bar{c}u^{\gamma+2}_{\lambda}(x,y).$

The inequality above is obvious if $u_{\lambda}(x,y)\leq u(x,y)$ for some
$(x,y)\in{\Sigma^{y_{1}}_{\lambda}}\cap B^{c}_{R}$. Then

$\displaystyle(f(u_{\lambda})-f(u))^{+}(u_{\lambda}-u)^{+}\leq\bar{c}u^{\gamma+2}_{\lambda}\
\ \ {\rm in}\ \ {\Sigma^{y_{1}}_{\lambda}}\cap B^{c}_{R}.$

Therefore,

$\displaystyle\int_{{\Sigma^{y_{1}}_{\lambda}}\cap
B^{c}_{R}}(f(u_{\lambda})-f(u))^{+}(u_{\lambda}-u)^{+}dxdy$
$\displaystyle\leq$ $\displaystyle\bar{c}\int_{{\Sigma^{y_{1}}_{\lambda}}\cap
B^{c}_{R}}u^{\gamma+2}_{\lambda}(x,y)dxdy$ $\displaystyle\leq$
$\displaystyle\bar{c}C_{\epsilon}\int_{\Sigma^{y_{1}}_{\lambda}}(1+|x|)^{-(N+2\alpha)(\gamma+2)}e^{-(\gamma+2)\theta_{1}|y_{\lambda}|}dxdy$
$\displaystyle\leq$
$\displaystyle\bar{c}C_{\epsilon}\int_{\mathbb{R}^{N}}(1+|x|)^{-(N+2\alpha)(\gamma+2)}dx\int_{\mathbb{R}^{M}}e^{-(\gamma+2)\theta_{1}|y|}dy<+\infty,$

where the last inequality holds by $\gamma>\frac{2\alpha
N}{(N+M)(N+2\alpha)}$. Since $u$ and $u_{\lambda}$ are bounded and $f$ is
locally Lipschitz, we have

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}\cap
B_{R}}(f(u_{\lambda})-f(u))^{+}(u_{\lambda}-u)^{+}dxdy<+\infty.$

Therefore, (4.2) holds. Together with (4.1), we have the second inequality in
the result.

Next we show that the first inequality holds in Lemma 4.1. Let us denote

$w(x,y)=\left\\{\begin{array}[]{lll}(u_{\lambda}-u)^{+}(x,y),&(x,y)\in\Sigma^{y_{1}}_{\lambda},\\\\[5.69054pt]
(u_{\lambda}-u)^{-}(x,y),&(x,y)\in(\Sigma^{y_{1}}_{\lambda})^{c}\end{array}\right.$
(4.3)

and

${\rm{supp}}(w)=\overline{\\{(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}\ |\
w(x,y)\not=0\\}},$

where $(u_{\lambda}-u)^{+}(x,y)=\max\\{(u_{\lambda}-u)(x,y),\ 0\\}$,
$(u_{\lambda}-u)^{-}(x,y)=\min\\{(u_{\lambda}-u)(x,y),\ 0\\}$. We observe that
$w(x,y_{1},y^{\prime})=-w(x,2\lambda-y_{1},y^{\prime})$ for
$(x,y_{1},y^{\prime})\in\mathbb{R}^{N}\times\mathbb{R}\times\mathbb{R}^{M-1}$
and

$w=u_{\lambda}-u\ \quad{\rm in}\ \ {\rm{supp}}(w).$ (4.4)

It is obvious that for $(x,y)\in\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)$,
$\\{z\in\mathbb{R}^{N}|\ (z,y)\in(\Sigma^{y_{1}}_{\lambda})^{c}\\}=\O$ and

$\displaystyle\mathbb{R}^{N}$ $\displaystyle=$
$\displaystyle\\{z\in\mathbb{R}^{N}|\
(z,y)\in{\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)}\\}\cup$
$\displaystyle\\{z\in\mathbb{R}^{N}|\
(z,y)\in{\Sigma^{y_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}}\\}\cup\\{z\in\mathbb{R}^{N}|\
(z,y)\in(\Sigma^{y_{1}}_{\lambda})^{c}\\}.$

Combining with (4.4), then for
$(x,y)\in\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)$,

$\displaystyle(-\Delta)_{x}^{\alpha}w(x,y)-(-\Delta)_{x}^{\alpha}(u_{\lambda}-u)(x,y)=\int_{\mathbb{R}^{N}}\frac{(u_{\lambda}-u)(z,y)-w(z,y)}{|x-z|^{N+2\alpha}}dz$
$\displaystyle=\int_{\\{z\in\mathbb{R}^{N}:(z,y)\in{\Sigma^{y_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}}\\}}\frac{(u_{\lambda}-u)(z,y)}{|x-z|^{N+2\alpha}}dz\leq
0,$ (4.5)

where the last inequality holds by $u_{\lambda}-u\leq 0$ in
$\Sigma^{y_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}$. On one hand, from (4.5)
and $w=(u_{\lambda}-u)^{+}>0$ in $\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)$,
we have that

$\int_{\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)}(-\Delta)_{x}^{{\alpha}}w\,wdxdy\leq\int_{\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)}(-\Delta)_{x}^{{\alpha}}(u_{\lambda}-u)(u_{\lambda}-u)^{+}dxdy.$
(4.6)

On the other hand, we know that $w(x,y)=(u_{\lambda}-u)(x,y)$ and
$(-\Delta)_{y}w(x,y)=(-\Delta)_{y}(u_{\lambda}-u)(x,y)$ for
$(x,y)\in\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)$. Together with (4.6),
then

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)}[(-\Delta)_{x}^{\alpha}w+(-\Delta)_{y}w+w]\,wdxdy$
$\displaystyle\leq$
$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}\cap{\rm{supp}}(w)}[(-\Delta)_{x}^{\alpha}(u_{\lambda}-u)+(-\Delta)_{y}(u_{\lambda}-u)+(u_{\lambda}-u)](u_{\lambda}-u)^{+}dxdy$

and then by the fact of $w=(u_{\lambda}-u)^{+}=0$ in
$\Sigma^{y_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}$, we have that

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}w+(-\Delta)_{y}w+w]\,wdxdy$
(4.7) $\displaystyle\leq$
$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}(u_{\lambda}-u)+(-\Delta)_{y}(u_{\lambda}-u)+(u_{\lambda}-u)](u_{\lambda}-u)^{+}.$

By the definition of $w$, we have that

$\int_{\mathbb{R}^{N+M}}|w|^{2}dxdy=2\int_{\Sigma^{y_{1}}_{\lambda}}|w|^{2}dxdy,$
$\int_{\mathbb{R}^{N+M}}|w|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy=2\int_{\Sigma^{y_{1}}_{\lambda}}|w|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy,$
$\int_{\mathbb{R}^{N+M}}(-\Delta)_{y}w\,wdxdy=2\int_{\Sigma^{y_{1}}_{\lambda}}(-\Delta)_{y}w\,wdxdy,$
$\int_{\mathbb{R}^{N+M}}(-\Delta)_{x}^{\alpha}w\,wdxdy=2\int_{\Sigma^{y_{1}}_{\lambda}}(-\Delta)_{x}^{\alpha}w\,wdxdy,$

then, together with Proposition 2.1, we obtain that

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}w+(-\Delta)_{y}w+w]\,wdxdy$
(4.8) $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{N+M}}[(-\Delta)_{x}^{\alpha}w+(-\Delta)_{y}w+w]\,wdxdy$
$\displaystyle\geq$ $\displaystyle
c_{3}(\int_{\mathbb{R}^{N+M}}|w|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{{N+M-2\alpha}}{N+M}}$
$\displaystyle=$ $\displaystyle
c_{3}(2\int_{\Sigma^{y_{1}}_{\lambda}}|w|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{{N+M-2\alpha}}{N+M}},$

for some $c_{3}>0$. Combining (4.7) with (4.8), by $w=(u_{\lambda}-u)^{+}$ in
$\Sigma^{y_{1}}_{\lambda}$, we get the first inequality in Lemma 4.1. The
proof is complete. $\Box$

###### Lemma 4.2

Under the assumptions of Theorem 1.3, for any $\lambda\in\mathbb{R}$, there
exists $c_{4}>0$ independent of $\lambda$ such that

$\displaystyle
c_{4}(\int_{\Sigma^{x_{1}}_{\lambda}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}$
$\displaystyle\leq$
$\displaystyle\int_{\Sigma^{x_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}({u_{\lambda}}-u)+(-\Delta)_{y}({u_{\lambda}}-u)+({u_{\lambda}}-u)](u_{\lambda}-u)^{+}dxdy<\infty,$

where
$\Sigma^{x_{1}}_{\lambda}=\\{(x_{1},x^{\prime},y)\in\mathbb{R}\times\mathbb{R}^{N-1}\times\mathbb{R}^{M}\
|\ x_{1}>\lambda\\}.$

Proof. The proof proceeds similarly to the proof of Lemma 4.1, the only
difference is to show (4.5) with
$(x,y)\in\Sigma^{x_{1}}_{\lambda}\cap{\rm{supp}}(w)$. It is obvious that

$\displaystyle\mathbb{R}^{N}$ $\displaystyle=$
$\displaystyle\\{z\in\mathbb{R}^{N}|\
(z,y)\in{\Sigma^{x_{1}}_{\lambda}\cap{\rm{supp}}(w)}\\}\cup$
$\displaystyle\\{z\in\mathbb{R}^{N}|\
(z,y)\in{\Sigma^{x_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}}\\}\cup$
$\displaystyle\\{z\in\mathbb{R}^{N}|\
(z,y)\in(\Sigma^{x_{1}}_{\lambda})^{c}\cap({\rm{supp}}(w))^{c}\\}\cup$
$\displaystyle\\{z\in\mathbb{R}^{N}|\
(z,y)\in(\Sigma^{x_{1}}_{\lambda})^{c}\cap{\rm{supp}}(w)\\}$

and $w=u_{\lambda}-u$ in ${\rm{supp}}(w)$, then for
$(x,y)\in\Sigma^{x_{1}}_{\lambda}\cap{\rm{supp}}(w)$,

$\displaystyle(-\Delta)_{x}^{\alpha}w(x,y)-(-\Delta)_{x}^{\alpha}(u_{\lambda}-u)(x,y)=\int_{\mathbb{R}^{N}}\frac{(u_{\lambda}-u)(z,y)-w(z,y)}{|x-z|^{N+2\alpha}}dz$
$\displaystyle=$ $\displaystyle\int_{\\{z\in\mathbb{R}^{N}|\
(z,y)\in{\Sigma^{x_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}}\\}}(\frac{1}{|x-z|^{N+2\alpha}}-\frac{1}{|x-z_{\lambda}|^{N+2\alpha}})(u_{\lambda}-u)(z,y)dz$
$\displaystyle\leq$ $\displaystyle 0,$

where $z_{\lambda}=(2\lambda-z_{1},z^{\prime})$ for
$z=(z_{1},z^{\prime})\in\mathbb{R}^{N}$ and the last inequality holds by
$u_{\lambda}-u\leq 0$ in $\Sigma^{x_{1}}_{\lambda}\cap({\rm{supp}}(w))^{c}$.
$\Box$

###### Theorem 4.1

Under the assumptions of Theorem 1.3, for $x\in\mathbb{R}^{N}$, we have

$u(x,y)=u(x,|y|)$

and $u$ is strictly decreasing in $y$-direction.

Proof. We divide the proof into three steps.

Step 1: $\lambda_{0}:=\sup\\{\lambda\ |\ u_{\lambda}\leq u\ {\rm in}\
\Sigma^{y_{1}}_{\lambda}\\}$ is finite. Since $u$ decays at infinity, we
observe that the set $\\{\lambda\ |\ u_{\lambda}\leq u\ {\rm in}\
\Sigma^{y_{1}}_{\lambda}\\}$ is nonempty. Using $(u_{\lambda}-u)^{+}$ as a
test function in the equation for $u$ and $u_{\lambda}$, by (1.15) and Hölder
inequality, for $\lambda$ big (negative), we find that

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}({u_{\lambda}}-u)+(-\Delta)_{y}({u_{\lambda}}-u)+({u_{\lambda}}-u)](u_{\lambda}-u)^{+}dxdy$
$\displaystyle=\int_{\Sigma^{y_{1}}_{\lambda}}(f(u_{\lambda})-f(u))(u_{\lambda}-u)^{+}dxdy$
$\displaystyle=\int_{\Sigma^{y_{1}}_{\lambda}}\frac{f(u_{\lambda})-f(u)}{u_{\lambda}-u}[(u_{\lambda}-u)^{+}]^{2}dxdy\leq\bar{c}\int_{\Sigma^{y_{1}}_{\lambda}}{u^{\gamma}_{\lambda}}[(u_{\lambda}-u)^{+}]^{2}dxdy$
$\displaystyle\leq
c_{5}\int_{\Sigma^{y_{1}}_{\lambda}}(1+|x|)^{-\gamma(N+2\alpha)}e^{-\gamma\theta_{1}|y_{\lambda}|}[(u_{\lambda}-u)^{+}]^{2}dxdy$
$\displaystyle\leq
c_{5}(\int_{\Sigma^{y_{1}}_{\lambda}}(1+|x|)^{-a}e^{-b|y_{\lambda}|}dxdy)^{\frac{2\alpha}{N+M}}(\int_{\Sigma^{y_{1}}_{\lambda}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}},$

where $a=\frac{\gamma(N+2\alpha)(N+M)}{2\alpha}$ and
$b=\frac{\theta_{1}\gamma(N+M)}{2\alpha}$. Since $\gamma>\frac{2\alpha
N}{(N+2\alpha)(N+M)}$, we have that $a>N$. Then we can choose $R>0$ such that
for all $\lambda<-R$,

$c_{5}(\int_{\Sigma_{\lambda}^{y_{1}}}(1+|x|)^{-a}e^{-b|y_{\lambda}|}dxdy)^{\frac{2\alpha}{N+M}}\leq\frac{1}{4}.$

By Lemma 4.1, we obtain that

$\displaystyle\int_{\Sigma^{y_{1}}_{\lambda}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy=0,\
\ \quad\forall\ \lambda<-R.$

Thus $u_{\lambda}\leq u$ in $\Sigma^{y_{1}}_{\lambda}$ for all $\lambda<-R$
and then conclude that $\lambda_{0}\geq-R.$ On the other hand, since $u$
decays at infinity, then there exist $\lambda_{1}\in\mathbb{R}$ and
$(x,y)\in\Sigma^{y_{1}}_{\lambda}$ such that $u(x,y)<u_{\lambda_{1}}(x,y)$.
Hence $\lambda_{0}$ is finite.

Step 2: $u\equiv u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{y_{1}}$. Assuming
the contrary, we have that $u\not\equiv u_{\lambda_{0}}$ and $u\geq
u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{y_{1}}$, in this case the following
claim holds.

Claim 1. _If $u\not\equiv u_{\lambda_{0}}$ and $u\geq u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{y_{1}}$, then $u>u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{y_{1}}$._

Let us assume, for the moment, that Claim 1 is true, then for any given
$\lambda\in({\lambda_{0}},{\lambda_{0}}+\epsilon)$, where $\epsilon>0$ is
chosen later. Let $P=(0,\cdots,\lambda,\cdots,0)\in T_{\lambda}^{y_{1}}$ and
$B(P,R)$ be the ball centered at $P$ and with radius $R>1$ to be chosen later.
Define $B_{1}=\Sigma_{\lambda}^{y_{1}}\cap B(P,R)$ and let us consider
$(u_{\lambda}-u)^{+}$ test function in the equation for $u$ and $u_{\lambda}$
in $\Sigma_{\lambda}^{y_{1}}$, then from Lemma 4.1 we obtain

$\displaystyle(\int_{\Sigma^{y_{1}}_{\lambda}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}$
(4.9) $\displaystyle\leq$ $\displaystyle
c_{6}\int_{\Sigma^{y_{1}}_{\lambda}}[(-\Delta)_{x}^{\alpha}({u_{\lambda}}-u)+(-\Delta)_{y}({u_{\lambda}}-u)+({u_{\lambda}}-u)](u_{\lambda}-u)^{+}dxdy$
$\displaystyle=$ $\displaystyle
c_{6}\int_{\Sigma_{\lambda}^{y_{1}}}(f(u_{\lambda})-f(u))(u_{\lambda}-u)^{+}dxdy.$

We estimate the integral on the right. Proceeding as in Step 1, we can choose
$R>1$ big enough such that

$c_{7}(\int_{\Sigma_{\lambda}^{y_{1}}\setminus{B_{1}}}(1+|x|)^{-a}e^{-b|y_{\lambda}|}dxdy)^{\frac{2\alpha}{N+M}}\leq\frac{1}{4}$

for some $c_{7}>0$, where $a=\frac{\gamma(N+2\alpha)(N+M)}{2\alpha}$ and
$b=\frac{\theta_{1}\gamma(N+M)}{2\alpha}$. Then

$\displaystyle\int_{\Sigma_{\lambda}^{y_{1}}\setminus{B_{1}}}(f(u_{\lambda})-f(u))(u_{\lambda}-u)^{+}dxdy\leq\bar{c}\int_{\Sigma_{\lambda}^{y_{1}}\setminus{B_{1}}}u^{\gamma}_{\lambda}|(u_{\lambda}-u)^{+}|^{2}dxdy$
$\displaystyle\leq
c_{7}(\int_{\Sigma_{\lambda}^{y_{1}}\setminus{B_{1}}}(1+|x|)^{-a}e^{-b|y_{\lambda}|}dxdy)^{\frac{2\alpha}{N+M}}(\int_{\Sigma^{y_{1}}_{\lambda}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}$
$\displaystyle\leq\frac{1}{4}(\int_{\Sigma_{\lambda}^{y_{1}}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}.$
(4.10)

Now using Claim 1, we choose $\epsilon>0$ such that
$c_{8}|B_{1}\cap{{\rm{supp}}{(u_{\lambda}-u)^{+}}}|^{\frac{2\alpha}{N+M}}<1/4,$
for some $c_{8}>0$. Since $f$ is locally Lipschitz, using Hölder inequality,
we have

$\displaystyle\int_{B_{1}}(f(u_{\lambda})-f(u))(u_{\lambda}-u)^{+}dxdy\leq
c_{46}\int_{B_{1}}|(u_{\lambda}-u)^{+}|^{2}\chi_{{\rm{supp}}{(u_{\lambda}-u)^{+}}}dxdy$
$\displaystyle=c_{8}|B_{1}\cap{{\rm{supp}}{(u_{\lambda}-u)^{+}}}|^{\frac{2\alpha}{N+M}}(\int_{B_{1}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}$
$\displaystyle\leq\frac{1}{4}(\int_{B_{1}}|(u_{\lambda}-u)^{+}|^{\frac{2(N+M)}{N+M-2\alpha}}dxdy)^{\frac{N+M-2\alpha}{N+M}}.$
(4.11)

From (4.9), (4.10) and (4.11), it follows that $(u_{\lambda}-u)^{+}=0$ in
$\Sigma_{\lambda}^{y_{1}}$. Then $u_{\lambda}\leq u$ in
$\Sigma_{\lambda}^{y_{1}}$ for
$\lambda\in({\lambda_{0}},{\lambda_{0}}+\epsilon)$, which contradicts the
definition of $\lambda_{0}$. As a consequence, we have $u\equiv
u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{y_{1}}$.

In order to complete Step 2, we only need to prove Claim 1.

Proof of Claim 1. By contradiction, if there exists
$(\bar{x},\bar{y})\in\Sigma_{\lambda_{0}}^{y_{1}}$ such that
$u(\bar{x},\bar{y})=u_{\lambda_{0}}(\bar{x},\bar{y}),$ then

$\displaystyle(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})+(-\Delta)_{y}(u-u_{\lambda_{0}})(\bar{x},\bar{y})+(u-u_{\lambda_{0}})(\bar{x},\bar{y})$
$\displaystyle=f(u(\bar{x},\bar{y}))-f(u_{\lambda_{0}}(\bar{x},\bar{y}))=0.$

Since
$(u-u_{\lambda_{0}})(\bar{x},\bar{y})=\min_{\Sigma_{\lambda_{0}}^{y_{1}}}(u-u_{\lambda_{0}})=0$,
we have $(-\Delta)_{y}(u-u_{\lambda_{0}})(\bar{x},\bar{y})\leq 0$, then

$(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})\geq 0.$ (4.12)

The other side, we observe that $\\{z\in\mathbb{R}^{N}|\
(z,\bar{y})\in(\Sigma_{\lambda_{0}}^{y_{1}})^{c}\\}=\O$ when
$(\bar{x},\bar{y})\in\Sigma_{\lambda_{0}}^{y_{1}}$. By
$u(\bar{x},\bar{y})=u_{\lambda_{0}}(\bar{x},\bar{y})$ and then

$\displaystyle(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})=-\int_{\mathbb{R}^{N}}\frac{(u-u_{\lambda_{0}})(z,\bar{y})}{|\bar{x}-z|^{N+2\alpha}}dz$
$\displaystyle=-\int_{\\{z\in\mathbb{R}^{N}|\
(z,\bar{y})\in\Sigma_{\lambda_{0}}^{y_{1}}\\}}\frac{(u-u_{\lambda_{0}})(z,\bar{y})}{|\bar{x}-z|^{N+2\alpha}}dz\leq
0,$ (4.13)

where the last inequality holds by $u\geq u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{y_{1}}$.

Combining (4.12) with (4.13), we obtain that
$(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})=0$ and then from
(4.13), we have that

$u(z,\bar{y})=u_{\lambda_{0}}(z,\bar{y}),\quad\forall z\in\mathbb{R}^{N},$
(4.14)

this means that $u-u_{\lambda_{0}}$ has property $(P)$ and by $u\neq
u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{y_{1}}$ we have

$(\bar{x},\bar{y})\in(\Sigma_{\lambda_{0}}^{y_{1}})_{0}:=\\{(x,y)\in\Sigma_{\lambda_{0}}^{y_{1}}\
|\
(u-u_{\lambda_{0}})(x,y)=\inf_{\Sigma_{\lambda_{0}}^{y_{1}}}(u-u_{\lambda_{0}})=0\\}\subsetneqq\Sigma_{\lambda_{0}}^{y_{1}}.$

Moreover, by Proposition 2.3 with $\Omega=\Sigma_{\lambda_{0}}^{y_{1}}$, we
observe that
$\Sigma_{\lambda_{0}}^{y_{1}}\setminus(\Sigma_{\lambda_{0}}^{y_{1}})_{0}$
satisfies interior cylinder condition at point
$(x_{0},y_{0})\in\partial(\Sigma_{\lambda_{0}}^{y_{1}})_{0}\cap\Sigma_{\lambda_{0}}^{y_{1}}$.
Then there exist $r>0$ small and $\tilde{y}\in\mathbb{R}^{M}$ such that

$O_{r}:=B_{r}^{N}(x_{0})\times
B_{r}^{M}(\tilde{y})\subset\Sigma_{\lambda_{0}}^{y_{1}}\setminus(\Sigma_{\lambda_{0}}^{y_{1}})_{0}\quad{\rm{and}}\quad(x_{0},y_{0})\in\partial
O_{r}.$

Let $D$ be defined by (2.3). Since $u\geq u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{y_{1}}$, then for any $(x,y)\in D$, we have

$\displaystyle\int_{\mathbb{R}^{N}\setminus
B_{r}^{N}(x_{0})}\frac{(u-u_{\lambda_{0}})(z,y)}{|x-z|^{N+2\alpha}}dz\geq 0.$

Finally, it is obvious that

$(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})+(-\Delta)_{y}(u-u_{\lambda_{0}})+h(u-u_{\lambda_{0}})=0\quad{\rm{in}}\
\Sigma_{\lambda_{0}}^{y_{1}},$

where $h=1-\frac{f(u)-f(u_{\lambda_{0}})}{u-u_{\lambda_{0}}}\in
L_{loc}^{\infty}(\Sigma_{\lambda_{0}}^{y_{1}})$. Then we use Theorem 2.1 to
obtain

$u\equiv u_{\lambda_{0}}\quad{\rm{in}}\ \tilde{\Sigma}_{\lambda_{0}}^{y_{1}},$

which contradicts the condition of $u\neq u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{y_{1}}$, then we obtain the results in Claim 1.

Step 3. By translation, we may say that $\lambda_{0}=0.$ Repeating the
argument from the other side, we find that $u$ is symmetric about
$y_{1}$-axis. Using the same argument in any $y$-direction, we conclude that

$u(x,y)=u(x,|y|),\quad(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}.$

Finally, we prove that $u(x,|y|)$ is strictly decreasing in $|y|>0$. Indeed,
for any given $y_{1}<\widetilde{y}_{1}<0$ and letting
$\lambda=\frac{y_{1}+\widetilde{y}_{1}}{2}$. Then, as proved above we have

$u>u_{\lambda}\quad\ {\rm{in}}\quad\Sigma_{\lambda}^{y_{1}}.$

For any given $x\in\mathbb{R}^{N}$, we observe that
$(x,\widetilde{y}_{1},0,\cdots,0)\in\Sigma_{\lambda}^{y_{1}}$, then

$\displaystyle
u(x,\widetilde{y}_{1},0,\cdots,0)>u_{\lambda}(x,\widetilde{y}_{1},0,\cdots,0)=u(x,y_{1},0,\cdots,0).$

Using the result of $u(x,y)=u(x,|y|)$ for all
$(x,y)\in\mathbb{R}^{N}\times\mathbb{R}^{M}$ and
$|\widetilde{y}_{1}|<|y_{1}|$, we conclude monotonicity of $u$ respect to $y$.
This completes the proof. $\Box$

Next we study the symmetry result in $x$-direction.

###### Theorem 4.2

Under the assumptions of Theorem 1.3, for $y\in\mathbb{R}^{M}$, we have

$u(x,y)=u(|x|,y)$

and $u$ is strictly decreasing in $x$-direction.

Proof. The proof of this theorem goes like the one for Theorem 4.1. The only
place where there is a difference is in the following property: _if
$u\not\equiv u_{\lambda_{0}}$ and $u\geq u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{x_{1}}$, then $u>u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{x_{1}}$._ By contradiction, if there exists
$(\bar{x},\bar{y})\in\Sigma_{\lambda_{0}}^{x_{1}}$ such that
$u(\bar{x},\bar{y})=u_{\lambda_{0}}(\bar{x},\bar{y})$, then

$\displaystyle(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})+(-\Delta)_{y}(u-u_{\lambda_{0}})(\bar{x},\bar{y})+(u-u_{\lambda_{0}})(\bar{x},\bar{y})$
$\displaystyle=f(u(\bar{x},\bar{y}))-f(u_{\lambda_{0}}(\bar{x},\bar{y}))=0.$

Since $u\geq u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{x_{1}}$, we have
$(u-u_{\lambda_{0}})(\bar{x},\bar{y})=\min_{\Sigma_{\lambda_{0}}^{x_{1}}}(u-u_{\lambda_{0}})=0$
and $(-\Delta)_{y}(u-u_{\lambda_{0}})(\bar{x},\bar{y})\leq 0$ and then

$(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})\geq 0.$

The other side, by direct computation, we have that

$\displaystyle(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})(\bar{x},\bar{y})=\int_{\mathbb{R}^{N}}\frac{(u_{\lambda_{0}}-u)(z,\bar{y})}{|\bar{x}-z|^{N+2\alpha}}dz$
$\displaystyle=\int_{\\{z\in\mathbb{R}^{N}|\
(z,\bar{y})\in\Sigma_{\lambda_{0}}^{x_{1}}\\}}(\frac{1}{|\bar{x}-z|^{N+2\alpha}}-\frac{1}{|\bar{x}-z_{\lambda_{0}}|^{N+2\alpha}})(u_{\lambda_{0}}-u)(z,\bar{y})dz\leq
0,$

where $z_{\lambda_{0}}=(2{\lambda_{0}}-z_{1},z^{\prime})$ for
$z=(z_{1},z^{\prime})\in\mathbb{R}^{N}$ and the last inequality holds by
$u\geq u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{x_{1}}$. Therefore,

$u(z,\bar{y})=u_{\lambda_{0}}(z,\bar{y}),\quad\forall z\in\mathbb{R}^{N},$
(4.15)

this means that $u-u_{\lambda_{0}}$ has property $(P)$ and by $u\neq
u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{x_{1}}$ we have

$(\bar{x},\bar{y})\in(\Sigma_{\lambda_{0}}^{x_{1}})_{0}:=\\{(x,y)\in\Sigma_{\lambda_{0}}^{x_{1}}\
|\
(u-u_{\lambda_{0}})(x,y)=\inf_{\Sigma_{\lambda_{0}}^{x_{1}}}(u-u_{\lambda_{0}})=0\\}\subsetneqq\Sigma_{\lambda_{0}}^{x_{1}}.$

Moreover, by Proposition 2.3, we observe that
$\Sigma_{\lambda_{0}}^{x_{1}}\setminus(\Sigma_{\lambda_{0}}^{x_{1}})_{0}$
satisfies interior cylinder condition at point
$(x_{0},y_{0})\in\partial(\Sigma_{\lambda_{0}}^{x_{1}})_{0}\cap\Sigma_{\lambda_{0}}^{x_{1}}$.
Then there exist $r_{1}>0$ and $\tilde{y}\in\mathbb{R}^{M}$ such that for all
$r\in(0,r_{1}]$,

$O_{r}:=B_{r}^{N}(x_{0})\times
B_{r}^{M}(\tilde{y})\subset\Sigma_{\lambda_{0}}^{x_{1}}\setminus(\Sigma_{\lambda_{0}}^{x_{1}})_{0}\quad{\rm{and}}\quad(x_{0},y_{0})\in\partial
O_{r}.$

Next we show that there exists some $r\in(0,r_{1}]$ such that for any
$(x,y)\in D$,

$\int_{\mathbb{R}^{N}\setminus
B_{r}^{N}(x_{0})}\frac{(u-u_{\lambda_{0}})(z,y)}{|x-z|^{N+2\alpha}}dz\geq 0,$
(4.16)

where $D$ is defined by (2.3). Indeed, since $u\not\equiv u_{\lambda_{0}}$ and
$u\geq u_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}^{x_{1}}$, then for $(x,y)\in
D\subset\Sigma_{\lambda_{0}}^{x_{1}}$, we have that

$\int_{\mathbb{R}^{N}}\frac{(u-u_{\lambda_{0}})(z,y)}{|x-z|^{N+2\alpha}}dz>0.$

Let us define

$r(x,y)=\sup\\{r\in(0,r_{1}]:\int_{\mathbb{R}^{N}\setminus
B_{r}^{N}(x_{0})}\frac{(u-u_{\lambda_{0}})(z,y)}{|x-z|^{N+2\alpha}}dz\geq
0\\}.$ (4.17)

Let $r_{m}=\inf_{(x,y)\in D}r(x,y)$, it is obvious that $r_{m}\in[0,r_{1}]$.
Now we prove that $r_{m}>0$. By contradiction, if $r_{m}=0$, then there exist
a sequence $(x_{n},y_{n})\in D$ and $(\tilde{x},\tilde{y})\in\bar{D}$ such
that $(x_{n},y_{n})\to(\tilde{x},\tilde{y})$ and $r(x_{n},y_{n})\to 0,$ as
$n\to+\infty.$ Since $r(x,y)$ is continuous, then $r(\tilde{x},\tilde{y})=0.$
If
$(\tilde{x},\tilde{y})\in\bar{D}\setminus(\Sigma_{\lambda_{0}}^{x_{1}})_{0}$,
i.e. $u(\tilde{x},\tilde{y})>u_{\lambda_{0}}(\tilde{x},\tilde{y})$, we have

$\displaystyle\int_{\mathbb{R}^{N}}\frac{(u-u_{\lambda_{0}})(z,\tilde{y})}{|\tilde{x}-z|^{N+2\alpha}}dz$
$\displaystyle=$ $\displaystyle\int_{\\{z\in\mathbb{R}^{N}|\
(z,\tilde{y})\in\Sigma_{\lambda_{0}}^{x_{1}}\\}}(u-u_{\lambda_{0}})(z,\tilde{y})(\frac{1}{|\tilde{x}-z|^{N+2\alpha}}-\frac{1}{|\tilde{x}-z_{\lambda_{0}}|^{N+2\alpha}})dz>0.$

By the continuity of the integration and (4.17), we obtain that
$r(\tilde{x},\tilde{y})>0$, which is impossible.

Then $(\tilde{x},\tilde{y})\in\bar{D}\cap(\Sigma_{\lambda_{0}}^{x_{1}})_{0}$,
i.e. $u(\tilde{x},\tilde{y})=u_{\lambda_{0}}(\tilde{x},\tilde{y})$. Since the
function $u-u_{\lambda_{0}}$ has property $(P)$, then for any $\tilde{r}>0$,

$\int_{\mathbb{R}^{N}\setminus
B_{\tilde{r}}^{N}(x_{0})}\frac{(u-u_{\lambda_{0}})(z,\tilde{y})}{|\tilde{x}-z|^{N+2\alpha}}dz=0.$

Combining with (4.17), we obtain that $r(\tilde{x},\tilde{y})=r_{1}>0$, which
contradicts $r(\tilde{x},\tilde{y})=0$. As a consequence, we have that
$0<r_{m}\leq r_{1}$. Taking $r=r_{m}$, then (4.16) holds for any $(x,y)\in D$.
Finally, it is obvious that

$(-\Delta)_{x}^{\alpha}(u-u_{\lambda_{0}})+(-\Delta)_{y}(u-u_{\lambda_{0}})+h(u-u_{\lambda_{0}})=0\quad{\rm{in}}\
\Sigma_{\lambda_{0}}^{x_{1}},$

where $h=1-\frac{f(u)-f(u_{\lambda_{0}})}{u-u_{\lambda_{0}}}\in
L_{loc}^{\infty}(\Sigma_{\lambda_{0}}^{x_{1}})$. Then we use Theorem 2.1 to
obtain that

$u\equiv u_{\lambda_{0}}\quad{\rm{in}}\ \tilde{\Sigma}_{\lambda_{0}}^{x_{1}},$

which contradicts the condition of $u\neq u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{x_{1}}$. Then $u>u_{\lambda_{0}}$ in
$\Sigma_{\lambda_{0}}^{x_{1}}$, to complete the proof. $\Box$

Acknowledgements: P.F. was partially supported by Fondecyt Grant # 1110291 and
BASAL-CMM projects. Y.W. was partially supported by Becas CMM.

## References

  * [1] G. Barles, R. Buckdahn and E. Pardoux, Backward stochastic differential equations and integral partial differential equations, Stochastics Stochastics Rep., 60, 57-83 (1997).
  * [2] G. Barles, E. Chasseigne, A. Ciomaga and C. Imbert, Lipschitz regularity of solutions for mixed integro-differential equations, J. Differential Equations, 252, 6012-6060 (2012).
  * [3] G. Barles, E. Chasseigne, A. Ciomaga and C. Imbert, Large time behavior of periodic viscosity solutions for uniformly elliptic integro-differential equations, Calc. Var. Partial Differential Equations, 50, 283-304 (2014).
  * [4] F. Benth, K. Karlsen and K. Reikvam, Optimal portfolio selection with consumption and nonlinear integro-differential equations with gradient constraint: a viscosity solution approach, Finance Stoch., 5, 275-303 (2001).
  * [5] H. Berestycki and P.L. Lions, Non linear scalar field equations I: Existence of a ground state, Arch. Rational Mech. Anal., 82, 313-345 (1983).
  * [6] H. Berestycki and L. Nirenberg, On the method of moving planes and the sliding method, Bol. Soc. Brasileira Mat., 22(1), 1991.
  * [7] F. Brock, Continuous Steiner-Symmetrization, Mathematische Nachrichten, 172(1), 25-48 (1995).
  * [8] J. Busca and P. Felmer, Qualitative properties of some bounded positive solutions to scalar field equations, Calc. Var. Partial Differential Equations, 13, 191-211 (2001).
  * [9] X. Cabré and Y. Sire, Nonlinear equations for fractional laplacians II: existence, uniqueness and qualitative properties of solutions, Trans. Amer. Math. Soc. to appear, arXiv:1111.0796.
  * [10] L. Caffarelli and L. Silvestre, An extension problem related to the fractional laplacian, Comm. Partial Differential Equations, 32, 1245-1260 (2007).
  * [11] W. Chen, C. Li and B. Ou, Classification of solutions for an integral equation, Comm. Pure Appl. Math., 59, 330-343 (2006).
  * [12] A. Ciomaga, On the strong maximum principle for second order nonlinear parabolic integro-differential equations, Advances in Differential Equations, 17, 635-671 (2012).
  * [13] C. Cortázar, M. Elgueta and P. Felmer, On a semilinear elliptic problem in $\mathbb{R}^{N}$ with a non-lipschitzian non-linearity, Advances in Differential Equations, 1, 199-218 (1996).
  * [14] F. Da Lio, B. Sirakov, Symmetry results for viscosity solutions of fully nonlinear uniformly elliptic equations, J. Eur. Math. Soc., 9, 317-330 (2007).
  * [15] S. Dipierro, G. Palatucci and E. Valdinoci, Existence and symmetry results for a schrödinger type problem involving the fractional laplacian, Le Matematiche (Catania), 68, (2013) no 1.
  * [16] J. Dolbeault and P. Felmer, Symmetry and monotonicity properties for positive solution of semi-linear elliptic PDE’s, Comm. Partial Differential Equations, 25, 1153-1169 (2000).
  * [17] P. Felmer, A. Quaas and J. Tan, Positive solutions of non-linear Schrödinger equation with the fractional laplacian, Proceedings of the Royal Society of Edinburgh: Section A Mathematics, 142, 1237-1262 (2012).
  * [18] P. Felmer and Y. Wang, Radial symmetry of positive solutions to equations involving the fractional laplacian, Comm. Contem. Math., 16, No. 01 (2013).
  * [19] B. Gidas, W.M. Ni and L. Nirenberg, Symmetry of positive solutions of non-linear elliptic equations in $\mathbb{R}^{N}$, Math. Anal. Appl., Part A, Advances in Math. Suppl. Studied, 7A, 369-403 (1981).
  * [20] C. Gui, Symmetry of the blow up set of a porous medium type equation, Comm. Pure Appl. Math. XLVIII, 471-500 (1995).
  * [21] M.K. Kwong, Uniqueness of positive solutions of $\Delta u-u+u^{p}=0$ in $\mathbb{R}^{N}$, Arch. Rational Mech. Anal., 105, 243-266 (1989).
  * [22] N.S. Landkof, Foundations of Modern Potential Theory, Springer-Verlag, Berlin, 1972.
  * [23] C.M. Li, Monotonicity and symmetry of solutions of fully non-linear elliptic equations on unbounded domains, Comm. Partial Differential Equations, 16, 585-615 (1991).
  * [24] Y. Li and W.M. Ni, Radial symmetry of positive solutions of non-linear elliptic equations in $\mathbb{R}^{N}$, Comm. Partial Differential Equations, 18, 1043-1054 (1993).
  * [25] Y.Y. Li, Remark on some conformally invariant integral equations: the method fo moving spheres, J. Eur. Math. Soc., 6, 153-180 (2004).
  * [26] F. Pacella and M. Ramaswamy, Symmetry of solutions of elliptic equations via maximum principles, Handbook of Differential Equations (M. Chipot, ed.), Elsevier, 269-312 (2012).
  * [27] H. Pham, Optimal stopping of controlled jump diffusion processes: a viscosity solution approach, J. Math. Systems Estim. Control, 8, 1-27 (1998).
  * [28] A. Quaas and A. Xia, Liouville type theorems for nonlinear elliptic equations and systems involving fractional laplacian in the half space, Calc. Var. Partial Differential Equations, DOI 10.1007/s00526-014-0727-8.
  * [29] X. Ros-Oton and J. Serra, The Dirichlet problem for the fractional laplacian: regularity up to the boundary, J. Math. Pures Appl., 101(3), 275-302 (2014).
  * [30] R. Servadei and E. Valdinoci, Variational methods for non-local operators of elliptic type, Discrete Contin. Dyn. Syst., 33(5), 2105-2137 (2013).
  * [31] R. Servadei and E. Valdinoci, A Brezis-Nirenberg result for non-local critical equations in low dimension, Comm. Pure Appl. Anal., 12, 2445-2464 (2013).
  * [32] Y. Sire and E. Valdinoci, Fractional laplacian phase transitions and boundary reactions: a geometric inequality and a symmetry result, J. Funct. Anal., 256, 1842-1864 (2009).