# Matching Cells

###### Abstract.

A (complete) matching of the cells of a triangulated manifold can be thought
as a combinatorial or discrete version of a nonsingular vector field. We give
several methods for constructing such matchings.

###### 1991 Mathematics Subject Classification:

M.S.C. 2010: 05C70, 05E45, 37C10, 37F20, 57Q15

Gaël Meigniez

On a triangulated manifold (all triangulations being understood smooth), a
“complete matching” (for short we just say “matching”) is a partition of the
set of the cells into pairs such that in each pair, one of the two cells is a
hyperface of the other. Such objects are regarded as a combinatorial
equivalent to nonsingular vector fields — a viewpoint inspired by Forman’s
works [4], see also [5]. The present note intends to provide some methods for
the construction of matchings, either allowing oneself to subdivide the
triangulation, or not. We feel that the methods are more important than the
existence results themselves. A first approach is algebraic, playing with
Hall’s “marriage theorem” and cellular homology; a second one is geometric: a
matching is deduced from an ambiant nonsingular vector field transverse to the
cells, or from a round handle decomposition of the manifold.

In a first time, manifolds are not mandatory, nor simplices. Consider
generally a polyhedral cellular complex $X$ (the cells are convex polyhedra,
finiteness is understood everywhere) and a subcomplex $Y\subset X$. Write
$\Sigma(X,Y)$ the set of the cells of $X$ not lying in $Y$. Call two cells
incident to each other if one is a hyperface of the other.

######  definition 1.

A _matching_ on $X$ _relative to $Y$,_ or a matching _on the pair $(X,Y)$,_ is
a partition of $\Sigma(X,Y)$ into incident pairs.

As usual, for $Y=\emptyset$ we write $\Sigma(X)$ instead of
$\Sigma(X,\emptyset)$ and we speak of “a matching on $X$”.

The cases of the complexes of dimension $1$ and of the triangulations of
surfaces will easily follow from a few general remarks.

######  remark 2 (Euler characteristic).

If $(X,Y)$ is matchable, then the relative Euler characteristic
$\chi(X,Y)=\chi(X)-\chi(Y)$ vanishes.

######  remark 3 (Collapse).

Every collapse of a polyhedral complex $X$ onto a subcomplex $Y$ gives a
matching on $X$ rel. $Y$.

Indeed, a collapse is nothing but filtration of $X$ by subcomplexes $X_{n}$,
$0\leq n\leq N$, such that $X_{0}=Y$ and $X_{N}=X$ and $\Sigma(X_{n},X_{n-1})$
consists, for each $1\leq n\leq N$, of exactly two incident cells. (A collapse
of $X$ onto $Y$ amounts more precisely to a matching of $X$ rel. $Y$ _without
cyclic orbit,_ where an orbit is defined as a finite sequence $\sigma_{0}$,
$\sigma_{1}$, $\dots\in\Sigma(X,Y)$, alternately of dimensions $d$ and $d+1$
for some $d\geq 0$, such that $\sigma_{k-1}$ and $\sigma_{k}$ are incident for
every $k\geq 1$, and mates for $k$ odd, and not mates for $k$ even.)

######  remark 4 (Top-dimensional cycle).

Every cellulation of the circle admits two matchings.

More generally, let $X$ be a polyhedral cellulation of a manifold; let $\ell$
be a simple loop in the $1$-skeleton of the dual cellulation; let $Y\subset X$
be the union of the cells of $X$ disjoint from $\ell$. Then, $X$ admits two
matchings rel. $Y$.

######  example 5 (Graphs).

_Every connected graph whose Euler characteristic vanishes is matchable._

Indeed, such a graph collapses onto a circle.

######  example 6 (Surfaces).

_Let $M$ be a compact, connected $2$-manifold such that $\chi(M)=0$. Then,
every polyhedral cellulation $X$ of $M$ is matchable absolutely, and
relatively to $\partial M$._

###### Proof.

First case: $M$ is the annulus or the Möbius strip. Then, the $1$-skeleton of
the cellulation dual to $X$ contains an essential simple loop $\ell$ such that
$M\setminus\ell$ collapses onto $\partial M$. Consider the union $Y\subset X$
of the cells of $X$ disjoint from $\ell$. The pair $(X,Y)$ is matchable
(Remark 4), the pair $(Y,\partial M)$ is matchable (Remark 3), and $X|\partial
M$ is matchable (Remark 4).

Second case: $M$ is the $2$-torus or the Klein bottle. Then, the $1$-skeleton
of the cellulation dual to $X$ contains an essential simple loop $\ell$ such
that $M\setminus\ell$ is an annulus. Consider the union $Y\subset X$ of the
cells of $X$ disjoint from $\ell$. The pair $(X,Y)$ is matchable (Remark 4)
and the annulus $Y$ is matchable. ∎

Next, recall Hall’s so-called “marriage theorem”. Let
$\Sigma:=\Sigma_{0}\sqcup\Sigma_{1}$ be a finite, ${\bf Z}/2{\bf Z}$-graded
set and let $I$ be a symmetric relation in $\Sigma$, of degree $1$. For every
subset $A\subset\Sigma$, denote by $|A|$ its cardinality, and denote by
$I(A)\subset\Sigma$ the subset of the elements $I$-related to at least one
element of $A$. A _matching_ on $\Sigma$ w.r.t. $I$ is a partition of $\Sigma$
into $I$-related pairs.

######  theorem 7 (Hall [6]).

The following properties are equivalent:

  1. (1)

The relation $I$ is matchable;

  2. (2)

One has $|A|\leq|I(A)|$ for every $A\subset\Sigma$ ;

  3. (3)

$|\Sigma_{1}|=|\Sigma_{0}|$ and one has $|A_{0}|\leq|I(A_{0})|$ for every
$A_{0}\subset\Sigma_{0}$.

Also recall that the Ford-Fulkerson algorithm [3][2] computes a matching, if
any, in time $O(|\Sigma|^{2}|I|)$, thus giving some (moderate) effectiveness
to the existence results below.

Coming back to a pair of polyhedral complexes $(X,Y)$, we write
$\Sigma_{0}(X,Y)$ (resp. $\Sigma_{1}(X,Y)$) for the set of the cells of $X$ of
even (resp. odd) dimension not lying in $Y$. Some counterexamples of
unmatchable complexes will follow from the _trivial_ sense of Hall’s
criterium.

######  example 8.

_A connected simplicial $2$-complex whose Euler characteristic vanishes,
unmatchable as well as its subdivisions._

Let $S$ (resp. $T$) be a triangulated $2$-sphere (resp. circle); let
$X:=S*T*T$ be the bouquet, at some common vertex $v$, of $S$ with two $T$’s.
Then, $\chi(X)=0$, but $X$ does not admit any matching. Indeed, for
$A:=\Sigma_{0}(S,v)$, one has $I(A)=\Sigma_{1}(S)$, thus $|I(A)|=|A|-1$. The
same holds for any subdivision of $X$.

######  example 9.

_An unmatchable triangulated closed $4$-manifold whose Euler characteristic
vanishes._

Let $n$ be even and at least $4$. Let $M_{0}$ be a closed $(n-1)$-manifold
which is the boundary of a compact $n$-manifold. Let $X_{0}$ be a
triangulation of $M_{0}$. It is easy to make two compact $n$-manifolds
$M_{i}$, $i=1,2$, bounded by $M_{0}$, and whose Euler characteristics verify :

$\chi(M_{2})=-\chi(M_{1})>|\Sigma_{0}(X_{0})|$

Extend $X_{0}$ to some triangulation $X_{1}$ of $M_{1}$ and to some
triangulation $X_{2}$ of $M_{2}$. Let $M$ (resp. $X$) be the union of $M_{1}$
(resp. $X_{1}$) with $M_{2}$ (resp. $X_{2}$) over $M_{0}$ (resp. $X_{0}$).
Then $\chi(M)=0$, but its triangulation $X$ does not admit any matching.
Indeed, the set $A:=\Sigma_{0}(X_{2},X_{0})$ has $I(A)=\Sigma_{1}(X_{2})$;
hence:

$|I(A)|=|\Sigma_{0}(X_{2})|-\chi(X_{2})=|A|+|\Sigma_{0}(X_{0})|-\chi(X_{2})<|A|$

On the other hand, in dimension $3$, E. Gallais has proved that every closed
$3$-manifold admits a matchable triangulation [5].

######  question 10.

_Is _every_ triangulation of every closed $3$-manifold matchable? Is every
triangulation of every closed odd-dimensional manifold matchable?_

######  lemma 11 (Acyclic pair).

If $H_{*}(X,Y)=0$, then the pair $(X,Y)$ is matchable.

Rational coefficients are understood everywhere; one could as well use ${\bf
Z}/2{\bf Z}$.

###### Proof.

This is an application of Hall’s criterium. For $n\geq 0$, consider as usual
the set $\Sigma^{n}$ of the $n$-dimensional cells of $X$ not lying in $Y$; the
chain vector space $C_{n}$ of basis $\Sigma^{n}$; the differential
$\partial_{n}:C_{n}\to C_{n-1}$; and its kernel $Z_{n}$. Consider the union
$X_{n}$ of $Y$ with the $(n-1)$-skeleton of $X$ and with some $n$-cells which
span a linear subspace complementary to $Z_{n}$ in $C_{n}$. The sequence
$(X_{n})$ is a filtration of the pair $(X,Y)$ by subcomplexes, and
$H_{*}(X_{n},X_{n-1})=0$.

One is thus reduced to the case where moreover,
$\Sigma(X,Y)=\Sigma^{n}\cup\Sigma^{n-1}$ for some $n\geq 1$. Note that
necessarily, $|\Sigma^{n}|=|\Sigma^{n-1}|$. For every $A\subset\Sigma(X,Y)$,
let $\langle A\rangle\subset C_{*}$ denote the spanned linear subspace; recall
that $I(A)\subset\Sigma(X,Y)$ is the set of the cells incident to at least one
cell belonging to $A$; hence $\partial\langle A\rangle\subset\langle
I(A)\rangle$. If $A\subset\Sigma^{n}$, since $\partial_{n}$ is linear and one-
to-one:

$|A|=\dim(\langle A\rangle)=\dim(\partial\langle A\rangle)\leq|I(A)|$

By the equivalence of (1) with (3) in the marriage theorem, the pair $(X,Y)$
is matchable. ∎

######  corollary 12 (Subdivision).

Let $(X,Y)$ be a pair of polyhedral complexes. Assume that $(X,Y)$ is
matchable.

Then, every polyhedral subdivision $(X^{\prime},Y^{\prime})$ of $(X,Y)$ is
also matchable.

###### Proof.

Consider a matching on $(X,Y)$. For each matched pair
$\sigma,\tau\in\Sigma(X,Y)$ with $\tau\subset\sigma$, consider the union
$\hat{\partial}\sigma:=\partial\sigma\setminus Int(\tau)$ of the other
hyperfaces of $\sigma$. The restriction
$(X^{\prime}|\sigma,X^{\prime}|\hat{\partial}\sigma)$ is a pair of polyhedral
complexes, matchable by Lemma 11. Clearly, the collection of all these partial
matchings constitutes a global matching for the pair of complexes
$(X^{\prime},Y^{\prime})$. ∎

######  corollary 13 (Rational homology sphere).

Let $M$ be a rational homology sphere of odd dimension $n$.

Then, every polyhedral cellulation $X$ of $M$ is matchable.

###### Proof.

One can assume that $n\geq 3$. Fix a $(n-1)$-cell $\sigma$ of $X$ and a
hyperface $\tau\subset\sigma$. Consider in $X$ the union $Y$ of $\tau$ with
the cells of $X$ not containing $\tau$. First, the pair $(X,Y)$ is matchable
(Remark 4). Second, $H_{*}(Y,\partial\sigma)=0$, hence the pair
$(Y,\partial\sigma)$ is matchable (Lemma 11). Third, the polyhedral complex
$\partial\sigma$, being homeomorphic to the $(n-2)$-sphere, is matchable by
induction on $n$. ∎

######  corollary 14 (Betti number $1$).

Let $M$ be a closed $3$-manifold whose first Betti number is $1$.

Then, every polyhedral cellulation $X$ of $M$ is matchable.

###### Proof.

The 1-skeleton of $X$ and the $1$-skeleton of the dual cellulation contain
respectively two homologous essential simple loops $\ell$, $\ell^{*}$.
Consider the union $Y\subset X$ of the cells of $X$ disjoint from $\ell^{*}$.
First, the pair $(X,Y)$ is matchable (Remark 4). Second, $H_{*}(Y,\ell)=0$,
hence the pair $(Y,\ell)$ is matchable (Lemma 11). Third, the circle $\ell$ is
matchable (Remark 4). ∎

Now, consider a triangulation $X$ of a compact manifold $M$ of dimension
$n\geq 1$ with smooth boundary $\partial M$ (maybe empty). If a nonsingular
vector field $\nabla$ is transverse to every $(n-1)$-simplex of $X$, we say
for short that $\nabla$ is _transverse to $X$._ Note that in particular,
$\nabla$ is then transverse to $\partial M$; thus $\partial M$ splits as the
disjoint union of $\partial_{s}(M,\nabla)$, where $\nabla$ enters $M$, with
$\partial_{u}(M,\nabla)$, where $\nabla$ exits $M$.

######  theorem 15 (Transverse nonsingular vector field).

If the nonsingular vector field $\nabla$ is transverse to the triangulation
$X$, then $X$ is matchable rel. $\partial_{u}(M,\nabla)$.

###### Proof.

Because of the transversality, for every simplex $\sigma\in\Sigma(X)$ of
dimension less than $n$ and not contained in ${\partial_{u}(M,\nabla)}$ (resp.
${\partial_{s}(M,\nabla)}$), there is a unique downstream (resp. _upstream_)
$n$-simplex $d(\sigma)$ (resp. $u(\sigma)$) $\in\Sigma^{n}(X)$ containing
$\sigma$ and such that the vector field $\nabla$ enters $d(\sigma)$ (resp.
exits $u(\sigma)$) at every point of $Int(\sigma)$. For
$\sigma\in\Sigma^{n}(X)$, we put $d(\sigma):=u(\sigma):=\sigma$.

Consider any $n$-simplex $\delta\in\Sigma^{n}(X)$ and any face
$\sigma\subset\delta$ (the case $\sigma=\delta$ is included.) We call $\sigma$
_stable_ (resp. _unstable_) with respect to $\delta$ if
$\sigma\not\subset{\partial_{u}(M,\nabla)}$ (resp.
$\sigma\not\subset{\partial_{s}(M,\nabla)}$) and if $d(\sigma)=\delta$ (resp.
$u(\sigma)=\delta$). Note that

  * •

Every _hyperface_ of $\delta$ is either stable or unstable;

  * •

$\delta$ has at least one stable hyperface (for degree reasons);

  * •

$\sigma$ is stable if and only if every hyperface of $\delta$ containing
$\sigma$ is stable.

Next, for each $\delta\in\Sigma^{n}(X)$ we pick arbitrarily a base vertex
$v(\delta)$ _in the intersection $\partial_{-}\delta$ of the unstable
hyperfaces of $\delta$_ (here of course, it is mandatory that $\delta$ is a
simplex rather than a general convex polytope.) To this choice there
corresponds canonically a matching, as follows. For every simplex
$\sigma\in\Sigma(X,\partial_{u}(M,\nabla))$ we define its mate $\bar{\sigma}$
by:

  1. (1)

If $v(d(\sigma))\in\sigma$ then $\bar{\sigma}$ is the hyperface of $\sigma$
opposed to $v(d(\sigma))$;

  2. (2)

If $v(d(\sigma))\notin\sigma$ then $\bar{\sigma}$ is the join of $\sigma$ with
$v(d(\sigma))$.

These rules do define a matching on the pair $(X,\partial_{u}(M,\nabla))$: the
point here is that $\bar{\sigma}$ is also a stable face of $d(\sigma)$.
Indeed, if not, then $\bar{\sigma}$ would be contained in some unstable
hyperface $\eta$ of $d(\sigma)$; but in both cases (1) and (2) above, this
would imply that $\sigma$ itself would be contained in $\eta$, a
contradiction. In other words, $d(\bar{\sigma})=d(\sigma)$: the map
$\sigma\mapsto\bar{\sigma}$ induces locally, for each $n$-simplex $\delta$, an
involution in the set of the stable faces of $\delta$; and thus globally a
matching on $\Sigma(X,\partial_{u}(M,\nabla))$.

Note — It can be suggestive, for $n=2$ and $n=3$ and for each $0\leq i\leq
n-1$, to figure out in ${\bf R}^{n}$, endowed with the parallel vector field
$\nabla:=-\partial/\partial x_{n}$, a linear $n$-simplex $\delta$ in general
position with respect to $\nabla$ and such that $\dim(\partial_{-}\delta)=i$;
to list the stable faces and the unstable faces; to choose a base vertex
$v\in\partial_{-}\delta$; and to compute the corresponding matching between
the stable faces. ∎

In particular, the Hall cardinality conditions also constitute some
_combinatorial_ necessary conditions for a triangulation to admit a transverse
nonsingular vector field. For example, in Example 9, not only $X$ does not
admit any transverse nonsingular vector field (which is obvious since such a
field would be transverse to $M_{0}$, in contradiction with
$\chi(M_{1},M_{0})\neq 0$), but this holds also for every triangulation of $M$
combinatorially isomorphic with $X$ (e.g. every jiggling of $X$).

######  corollary 16 (Matchable subdivision).

Let $M$ be a compact connected manifold with smooth boundary, and let
${\partial_{0}M}$ be a union of connected components of $\partial M$ such that
$\chi(M,{\partial_{0}M})=0$.

Then, every triangulation $X$ of $M$ admits a subdivision matchable rel.
${\partial_{0}M}$.

Either of the two sets ${\partial_{0}M}$ and $\partial_{1}M:=\partial
M\setminus{\partial_{0}M}$ may be empty, or both.

###### Proof.

Since $\chi(M,{\partial_{0}M})=0$, there is on $M$ a nonsingular vector field
$\nabla$ transverse to $\partial M$, which exits $M$ through
${\partial_{0}M}$, and which enters $M$ through ${\partial_{1}M}$. Then, by W.
Thurston’s famous Jiggling lemma [7], one has on $M$ a triangulation
$X^{\prime}$ which is combinatorially isomorphic to some iterated crystalline
subdivision of $X$, and which is transverse to $\nabla$, hence matchable by
Theorem 15. ∎

Finally, we give an alternative construction for Corollary 16 which works in
every dimension, but $3$. Note that, by Corollary 12 and the Hauptvermutung
for smooth triangulations, it is enough to construct _one_ triangulation of
$M$ matchable relatively to $\partial_{0}M$.

Since $n\geq 4$ and $\chi(M,{\partial_{0}M})=0$, the pair
$(M,{\partial_{0}M})$ admits a _round handle decomposition_ [1]. One chooses a
triangulation of $M$ for which each handle is a subcomplex. Hence, one is
reduced to the case of a round handle

$M={\bf S}^{1}\times{\bf D}^{i}\times{\bf D}^{n-i-1}$ ${\partial_{0}M}={\bf
S}^{1}\times{\bf S}^{i-1}\times{\bf D}^{n-i-1}$

for some $0\leq i\leq n-1$ (we agree that ${\bf S}^{-1}=\emptyset$). Since
${\bf D}^{i}\times{\bf D}^{n-i-1}$ retracts by deformation onto the union of
${\bf D}^{i}\times 0$ with ${\bf S}^{i-1}\times{\bf D}^{n-i-1}$, by Lemma 11
the proof is reduced to the case where $M={\bf S}^{1}\times{\bf D}^{i}$ and
${\partial_{0}M}={\bf S}^{1}\times{\bf S}^{i-1}$. In that case, let $X$ be any
triangulation of $M$. The $1$-skeleton of the dual subdivision contains a
simple loop $\ell$ homologous to the core ${\bf S}^{1}\times 0$. Consider the
union $Y\subset X$ of the cells of $X$ disjoint from $\ell$. On the one hand,
the pair $(X,Y)$ is matchable (Remark 4). On the other hand, since
$H_{*}(Y,{\partial_{0}M})=0$, the pair $(Y,{\partial_{0}M})$ is matchable
(Lemma 11).

## References

  * [1] D. Asimov, Round Handles and Non-Singular Morse-Smale Flows, Annals of Mathematics Second Series, vol. 102, No. 1 (1975), 41-54.
  * [2] J. Edmonds, R.M. Karp, Theoretical improvements in algorithmic efficiency for network flow problems, Journal of the Association for Computing Machinery 19 (2)(1972), 248–264.
  * [3] L.R. Ford, D.R. Fulkerson, Maximal flow through a network, Canadian Journal of Mathematics 8 (1956), 399–404.
  * [4] R. Forman, Combinatorial vector fields and dynamical systems, R. Math Z. (1998) 228–629.
  * [5] E. Gallais, Combinatorial realization of the Thom-Smale complex via discrete Morse theory,  Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5) vol. IX (2010), 229-252.
  * [6] P. Hall, On Representatives of Subsets, J. London Math. Soc., 10 (1) (1935), 26–30.
  * [7] W. Thurston, The theory of foliations of codimension greater than one, Comment. Math. Helv. 49 (1974), 214–231.

Université de Bretagne Sud

Laboratoire de Mathématiques de Bretagne Atlantique

LMBA UBS (UMR CNRS 6205)

B.P. 573

F-56019 VANNES CEDEX, France

<EMAIL_ADDRESS>


On Schrödinger equations involving the regional fractional Laplacian

in a ball with the zero boundary condition

Huyuan<EMAIL_ADDRESS>Huihuan Peng222<EMAIL_ADDRESS>and
Yanqing Sun333<EMAIL_ADDRESS>
Department of Mathematics, Jiangxi Normal University, Nanchang,

Jiangxi 330022, PR China

###### Abstract

Our purpose in this article to show the existence of positive classical
solutions of

$(-\Delta)_{B_{1}}^{s}u+u=h_{1}u^{p}+\epsilon h_{2}\quad{\rm in}\
\,B_{1},\qquad u=0\quad{\rm on}\ \partial B_{1}$

for $\epsilon>0$ small enough, where $(-\Delta)_{B_{1}}^{s}$ is the regional
fractional Laplacian, $p>1$, $h_{i}$ with $i=1,2$ are Hölder continuous and
satisfy some additional conditions. Our existence is based on the solution of

$(-\Delta)_{B_{1}}^{s}u+u=1\quad{\rm in}\ \,B_{1},\qquad u=0\quad{\rm on}\
\partial B_{1}.$

While this existence is very different from the case that

$(-\Delta)_{B_{1}}^{s}u=1\quad{\rm in}\ \,B_{1},\qquad u=0\quad{\rm on}\
\partial B_{1}$

which has no positive solution for $s\in(0,\frac{1}{2}]$.

Key Words: Schrödinger equations; Regional Fractional Laplacian; the method of
moving planes.

MSC2010: 35J10, 35R11, 35A01.

##  1 Introduction

Let $s\in(0,1)$, $\Omega$ be a $C^{2}$ domain in $\mathbb{R}^{N}$ with $N\geq
2$, the regional fractional Laplacian be defined in $\Omega$ in the principle
sense

$(-\Delta)^{s}_{\Omega}u(x)=c_{N,s}{\rm
p.v.}\int_{\Omega}\frac{u(x)-u(y)}{|x-y|^{N+2s}}dy:=c_{N,s}\lim_{\varepsilon\to
0^{+}}\int_{\Omega\setminus
B_{\varepsilon}(x)}\frac{u(x)-u(z)}{|z-x|^{N+2s}}\,dz,$

where $B_{r}(x)$ is the ball with the radius $r$ and the center at $x$,
particularly, $B_{r}=B_{r}(0)$, $c_{N,s}>0$ is the normalized constant of the
fractional Laplacian $(-\Delta)^{s}_{\mathbb{R}^{N}}$ (simply we use the
notation $(-\Delta)^{s}$), see [16]. For simplicity, let’s consider
$c_{N,s}=1$. Our interest in this article is to study the qualitative
properties of solutions to equations

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}u(x)+u=h(x,u)&{\rm in}\
\,B_{1},\\\\[5.69054pt] \qquad\qquad\qquad\,\ u=0&{\rm on}\ \partial
B_{1},\end{array}\right.$ (1.1)

where $h:B_{1}\times\mathbb{R}\to\mathbb{R}$ is a measurable function.

In the last few years, the study of nonlocal problems are now widely used in
physics, operation research, queuing theory, mathematical finance, risk
estimation, and others (see [5]). The regional fractional laplacian is a
typical operator as the generator of the censored stable process. In fact,
from the view of the Probability, a symmetric $2s$-stable process in
$\mathbb{R}^{N}$ killed upon leaving the set $\Omega$ is called a symmetric
$2s$-stable process in $\Omega$. Bogdan, Burdzy and Chen [3] (also see Guan
and Ma [8, 20]) extended this type of processes to obtain a version of the
strong Markov process, named censored symmetric stable process. A censored
stable process in an open set $\Omega\subset\mathbb{R}^{N}$ is obtained from
the symmetric stable process by suppressing its jumps from $\Omega$ to
$\mathbb{R}^{N}\backslash\Omega$. It should be pointed out that the censored
stable process has some interesting properties, which imply some differences
between $(\frac{1}{2},\,1)$ and $(0,\,\frac{1}{2}]$ (see [3, Theorem 1]):

  * for $s\in(\frac{1}{2},\,1)$, the censored symmetric $2s$-stable process in $\Omega$ has a finite lifetime and will approach $\partial\Omega$;

  * for $s\in(0,\,\frac{1}{2}]$, the censored symmetric $2s$-stable process in $\Omega$ is conservative and will never approach $\partial\Omega$.

From the purely analysis, there are some new phenomenas for elliptic problems
with the regional fractional laplacian. Let ${\bf H}^{s}_{0}(B_{1})$ be the
closure of $C^{\infty}_{c}(B_{1})$ under the norm that

$\|u\|_{s,\Omega}=\sqrt{\int_{\Omega\times\Omega}\frac{(u(x)-u(y))^{2}}{|x-y|^{N+2s}}dxdy},$

and [21] proves that

  * for $s\in(\frac{1}{2},\,1)$, Hilbert space ${\bf H}^{s}_{0}(\Omega)$ has the zero trace;

  * for $s\in(0,\,\frac{1}{2}]$, Hilbert space ${\bf H}^{s}_{0}(\Omega)$ has no the zero trace.

Let $H^{s}_{0}(\Omega)$ be the closure of $C^{\infty}_{c}(B_{1})$ under the
norm that

$\|u\|_{s,\Omega}=\sqrt{\int_{\Omega\times\Omega}\frac{(u(x)-u(y))^{2}}{|x-y|^{N+2s}}dxdy+\int_{\Omega}u^{2}dx}.$

It is could be shown that for $s\in(0,\,\frac{1}{2}]$, Hilbert space
$H^{s}_{0}(\Omega)$ has no the zero trace, therefore it is much more difficult
to obtain the solution of (1.1) with zero boundary condition in the weak
sense.

When $s\in(\frac{1}{2},\,1)$, [7, 8] give estimates on the heat kernel and
Green kernel related to the regional fractional Laplacian; [19] builds a
version of integration by formula for regional fractional Laplacian and [9]
extends this formula to solve regional fractional problem with kinds of
inhomogenous terms. Via building the formula of integral by part and related
embedding results, [9] obtains the existence of solutions to

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{\Omega}u=f&{\rm in}\
\,\Omega,\\\\[4.2679pt] \phantom{(-\Delta)^{\alpha}}\displaystyle u=0&{\rm
on}\ \partial\Omega\end{array}\right.$ (1.2)

for $s\in(\frac{1}{2},1)$ when $\Omega$ is a bounded regular domain. More
study of regional fractional laplacian with $s\in(\frac{1}{2},1)$ could see
[10] boundary blowing-up solutions, [17] boundary regularity. More related
study could see [1] for other type regional fractional order laplacians and
[14, 2] on related topics.

For $s\in(0,\frac{1}{2}]$, the structure of solutions of elliptic equations
are very different. [11] shows the nonexistence of solution of Poisson problem

$(-\Delta)^{s}_{\Omega}u=1\quad{\rm in}\ \,\Omega$

and nonexistence of positive solutions for Lane-Emden type equation

$(-\Delta)^{s}_{\Omega}u=u^{p}\quad{\rm in}\ \,\Omega.$ (1.3)

Replace the zero boundary restriction by $\int_{\Omega}fdx=0$, [24] shows the
existence of weak solution for the Poisson problem

$(-\Delta)^{s}_{\Omega}u=f\quad{\rm in}\ \,\Omega.$

In this article we deal the Schrödinger equation

$h(x,t)=h_{1}(x)t^{p}+\epsilon h_{2}(x)$

where $h_{i}$ with $i=1,2$ and $\epsilon>0$. Our main result states as
following.

###### Theorem 1.1

Let $p>1$,

$h(x,t)=h_{1}(x)t^{p}+\epsilon h_{2}(x)$

where $h_{i}\in C^{\theta}(\bar{B}_{1})$ with $i=1,2$, $\theta\in(0,1)$ be
nonnegative, nontrivial.

Then there exists $\epsilon_{0}>0$ depending on
$\|h_{i}\|_{L^{\infty}(B_{1})}$ $i=1,2$ such that (1.1) has a minimal positive
solution $u_{\epsilon}$ for $\epsilon\in(0,\epsilon_{0})$.

Moreover, if we assume that

  * $(\mathcal{H}_{1})$

$h_{1},\ h_{2}:B_{1}\setminus\\{0\\}\to\mathbb{R}_{+}:=[0,+\infty)$ is
continuous, radially symmetric and decreasing,

then $u_{\epsilon}$ is radially symmetric and decreasing with respect to
$|x|$.

For $s\in(\frac{1}{2},1)$, the existence could be obtained obtain in
$H^{s}_{0}(B_{1})$ by the variational method, since the the space
$H^{s}_{0}(B_{1})$ has the boundary trace in [3]. While it is fails for
$s\in(0,\frac{1}{2}]$. We emphasize from [11] that for $s\in(0,\frac{1}{2}]$
the Lane-Emden equation

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u=u^{p}&{\rm
in}\ \,B_{1},\\\\[4.2679pt] \phantom{(-\Delta)^{\alpha}_{B_{1}}}\displaystyle
u=0&{\rm on}\ \partial B_{1}\end{array}\right.$ (1.4)

has no positive solutions.

Our method of the existence is based on the proposition:

###### Proposition 1.1

The Poisson problem

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}u+u=1&{\rm in}\
\,B_{1},\\\\[5.69054pt] \qquad\qquad\quad u=0&{\rm in}\ \,\partial
B_{1}\end{array}\right.$

has a unique classical solution $u_{1}\in C_{0}(B_{1})$. Moreover, for any
$k\in(0,s)$,

$\lim_{|x|\to 1^{-}}u_{1}(x)(1-|x|)^{-k}=+\infty.$

In fact, we can show that $u_{1}$ could be approximated by solutions
$u_{r_{0}}$, as $r_{0}\to 1^{-}$, of

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}u(x)+u=1&{\rm in}\
\,B_{r_{0}},\\\\[5.69054pt] \qquad\qquad\qquad\,\ u=0&{\rm in}\
\,\overline{B_{1}}\setminus B_{r_{0}}.\end{array}\right.$

By the radial symmetry and the decreasing monotonicity we can show $u_{1}$
verifies the zero boundary condition. It is could be shown that for
$s\in(\frac{1}{2},1)$, $u_{1}$ decays with the rate $(1-|x|)^{2s-1}$.

###### Remark 1.1

$(i)$ For $s\in(0,\frac{1}{2}]$, a challenging open problem is to get the
precise decaying behavior;

$(ii)$ For $s\in(0,\frac{1}{2}]$ and given a general regular, bounded domain
$\Omega$, the challenging open problem is to determine if

$(-\Delta)_{\Omega}^{s}u(x)+u=1\quad\ {\rm in}\ \,\Omega$

has a solution decaying at the boundary $\partial\Omega$.

As we have already mentioned, the properties of the radial symmetry and the
decreasing monotonicity play an important role in our derivation of boundary
decay of $u_{1}$. To this end, the positive solutions of

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}u+u=h(x,u)&{\rm in}\
\,B_{r_{0}},\\\\[5.69054pt] \qquad\qquad\quad\ u=0&{\rm in}\ \
\overline{B_{1}}\setminus B_{r_{0}}\end{array}\right.$ (1.5)

with $r_{0}\in(0,1]$ is derived via the method of moving planes. Precisely, we
have that

###### Theorem 1.2

Let

$h(x,t)=h_{1}(x)h_{3}(t)+h_{2}(x),$

where $h_{1},\ h_{2}$ verify $(\mathcal{H}_{1})$ and $h_{3}$ does

  * $(\mathcal{H}_{2})$

$h_{3}:\mathbb{R}_{+}\to\mathbb{R}_{+}$ is nondecreasing, locally Lipschtiz
continuous in $[0,+\infty)$.

Then any nonnegative solution $u\in C_{0}(B_{1})$ of (1.5) is radially
symmetric and strictly decreasing in $r=|x|$ for $r\in(0,1)$.

It is worth noting that we can apply the method of moving planes of the
fractional Laplacian defined in $\mathbb{R}^{N}$ via transform problem (1.5
into the equation

$\left\\{\begin{array}[]{lll}(-\Delta)^{s}u+u=h(x,u)+\varphi_{B_{1}}u&{\rm
in}\ \,B_{r_{0}},\\\\[5.69054pt] \qquad\qquad\qquad\,\ u=0&{\rm in}\ \
\mathbb{R}^{N}\setminus B_{r_{0}},\end{array}\right.$

since $\varphi_{B_{1}}$ is not decreasing with respect to $|x|$, where
$\varphi_{B_{1}}(x)=\int_{\mathbb{R}^{N}\setminus
B_{1}}\frac{dy}{|x-y|^{N+2s}}$, which is called as killing measure. The
properties of $\varphi_{B_{1}}$ could see Section 2.1.

Note that the essential for the method of moving planes and we obtain the
maximum principle is maximum principle for small domains, which we obtain only
from the nonlocal property of the regional fractional Laplacian.

The rest of this paper is organized as follows. In Section 2, we recall the
connection of the regional fractional Laplacian and the fractional Laplacian,
the properties of viscosity solution and regularities estimates. In Section 3,
we prove Theorem 1.2 by the method of moving planes. Section 4 and Section 5
are devoted to the Poisson problem and Schrödinger equations respectively.
Finally, we annex properties of killing measure $\varphi_{\Omega}$.

##  2 Preliminary

###  2.1 Connections with the fractional Laplacian

Note that by the zero extension of the function in
$\mathbb{R}^{N}\setminus\Omega$, we can build the connection between the
regional fractional laplacian and the fractional Laplacian.

Given $u\in C_{0}(\Omega)$, we denote

$\tilde{u}(x)=\left\\{\begin{array}[]{ll}u(x),&x\in\Omega,\\\\[5.69054pt]
0,&x\notin\Omega,\end{array}\right.$ (2.1)

Then for $x\in\Omega$

$\displaystyle(-\Delta)^{s}\tilde{u}(x)$ $\displaystyle=$ $\displaystyle{\rm
p.v.}\int_{\mathbb{R}^{N}}\frac{\tilde{u}(x)-\tilde{u}(y)}{|x-y|^{N+2s}}dy$
$\displaystyle=$ $\displaystyle{\rm
p.v.}\int_{\Omega}\frac{u(x)-u(y)}{|x-y|^{N+2s}}dy+\int_{\Omega^{c}}\frac{u(x)}{|x-y|^{N+2s}}dy$
$\displaystyle=$
$\displaystyle(-\Delta)^{s}_{\Omega}u(x)+u(x)\varphi_{{}_{\Omega}}(x),$

where $\Omega^{c}=\mathbb{R}^{N}\setminus\Omega$ and

$\varphi_{{}_{\Omega}}(x)=\int_{\Omega^{c}}\frac{dy}{|x-y|^{N+2s}}.$ (2.2)

Related Hardy potentials could be seen in [4, 15].

###### Proposition 2.1

Let $\varphi_{{}_{\Omega}}$ be defined in (2.2).

$(i)$ If $\Omega$ is a $C^{2}$ domain, then $\varphi_{{}_{\Omega}}$ is locally
Lipschitz continuous.

$(ii)$ If $\Omega=B_{1}$, then $\varphi_{B_{1}}$ is radially symmetric,
decreasing and there exists $c_{5}>0$ such that

$\lim_{x\to\partial B_{1}}\varphi_{B_{1}}(x)(1-|x|)^{2s}=c_{5}.$ (2.3)

Proof.  For $x_{1},x_{2}\in\Omega$ and any
$z\in\mathbb{R}^{N}\setminus\Omega$, we have that

$|z-x_{1}|\geq\rho(x_{1})+\rho(z),\qquad|z-x_{2}|\geq\rho(x_{2})+\rho(z)$

and

$||z-x_{1}|^{N+2s}-|z-x_{2}|^{N+2s}|\leq
c_{2}|x_{1}-x_{2}|(|z-x_{1}|^{N+2s-1}+|z-x_{2}|^{N+2s-1}),$

where $\rho(x)={\rm dist}(x,\partial\Omega)$, $c_{2}>0$ is independent of
$x_{1}$ and $x_{2}$. Then

$\displaystyle|\varphi_{{}_{\Omega}}(x_{1})-\varphi_{{}_{\Omega}}(x_{2})|\leq\int_{\Omega^{c}}\frac{||z-x_{2}|^{N+2s}-|z-x_{1}|^{N+2s}|}{|z-x_{1}|^{N+2s}|z-x_{2}|^{N+2s}}dz$
$\displaystyle\leq
c_{2}|x_{1}-x_{2}|\left[\int_{\Omega^{c}}\frac{dz}{|z-x_{1}||z-x_{2}|^{N+2s}}+\int_{\Omega^{c}}\frac{dz}{|z-x_{1}|^{N+2s}|z-x_{2}|}\right].$

By direct computation, we have that

$\displaystyle\int_{\Omega^{c}}\frac{1}{|z-x_{1}||z-x_{2}|^{N+2s}}dz$
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{N}\setminus{B_{\rho(x_{1})}(x_{1})}}\frac{1}{|z-x_{1}|^{N+2s+1}}dz$
$\displaystyle+\int_{\mathbb{R}^{N}\setminus{B_{\rho(x_{2})}(x_{2})}}\frac{1}{|z-x_{2}|^{N+2s+1}}dz$
$\displaystyle\leq$ $\displaystyle
c_{3}[\rho(x_{1})^{-1-2s}+\rho(x_{2})^{-1-2s}]$

and similar to obtain that

$\int_{\Omega^{c}}\frac{1}{|z-x_{1}|^{N+2s}|z-x_{2}|}dz\leq
c_{4}[\rho(x_{1})^{-1-2s}+\rho(x_{2})^{-1-2s}],$

where $c_{4}>0$ is independent of $x_{1},\,x_{2}$. Then

$|\varphi_{{}_{\Omega}}(x_{1})-\varphi_{{}_{\Omega}}(x_{2})|\leq
c_{2}(c_{3}+c_{4})[\rho(x_{1})^{-1-2s}+\rho(x_{2})^{-1-2s}]|x_{1}-x_{2}|,$

that is, $\varphi_{{}_{\Omega}}$ is $C^{0,1}$ locally in $\Omega$.

$(ii)$ Firstly, we prove that $\varphi_{B_{1}}(x)=\varphi_{B_{1}}(z)$ if
$|x|=|z|$. Let ${\bf A}$ be a matrix with $|{\bf A}|=1$ and $z={\bf A}x$, then

$\displaystyle\varphi_{B_{1}}(z)=\varphi_{B_{1}}({\bf A}x)$ $\displaystyle=$
$\displaystyle\int_{B_{1}^{c}}\frac{dy}{|{\bf A}x-y|^{N+2s}}$ $\displaystyle=$
$\displaystyle\int_{B_{1}^{c}}\frac{d\tilde{y}}{|x-\tilde{y}|^{N+2s}}=\varphi_{B_{1}}(x)$

where $\tilde{y}={\bf A}^{-1}y$.

Next we prove that the monotonicity. From the radial symmetry of $\varphi$, we
let

$\varphi(r)=\varphi_{B_{1}}(x),\ \ r=|x|\in(0,1).$

Now we fix $x_{1}=t_{1}e_{1},$ $x_{2}=t_{2}e_{1},$ where
$e_{1}=(1,0,\cdots,0)\in\mathbb{R}^{N},$ $0<t_{1}<t_{2}<1.$

$\displaystyle\varphi(t_{1})-\varphi(t_{2})$ $\displaystyle=$
$\displaystyle\int_{B_{1}^{c}}(\frac{1}{|t_{1}e_{1}-y|^{N+2s}}-\frac{1}{|t_{2}e_{1}-y|^{N+2s}})dy$
$\displaystyle=$
$\displaystyle\int_{\mathcal{A}_{1}\cup\mathcal{A}_{2}}(\frac{1}{|t_{1}e_{1}-y|^{N+2s}}-\frac{1}{|t_{2}e_{1}-y|^{N+2s}})dy$
$\displaystyle+\int_{\mathcal{A}_{0}}(\frac{1}{|t_{1}e_{1}-y|^{N+2s}}-\frac{1}{|t_{2}e_{1}-y|^{N+2s}})dy,$

where $\mathcal{A}_{0}=B_{1}\big{(}(t_{1}+t_{2})e_{1}\big{)}\setminus B_{1},$

$\mathcal{A}_{1}=\left\\{(x^{1},x^{\prime})\in\big{(}-\infty,\frac{t_{1}+t_{2}}{2}\big{)}\times\mathbb{R}^{N-1}\setminus
B_{1}\right\\}$

and

$\mathcal{A}_{2}=\left\\{(x^{1},x^{\prime})\in\big{(}\frac{t_{1}+t_{2}}{2},+\infty\big{)}\times\mathbb{R}^{N-1}\setminus
B_{1}\big{(}(t_{1}+t_{2})e_{1}\big{)}\right\\}.$

By symmetric of $\mathcal{A}_{1}\cup\mathcal{A}_{2}$

$\int_{\mathcal{A}_{1}\cup\mathcal{A}_{2}}(\frac{1}{|t_{1}e_{1}-y|^{N+2s}}-\frac{1}{|t_{2}e_{1}-y|^{N+2s}})dy=0.$

Since $|t_{1}e_{1}-y|>|t_{2}e_{1}-y|$ for any $y\in\mathcal{A}_{0}$, then we
have

$\varphi(t_{1})-\varphi(t_{2})=\int_{\mathcal{A}_{0}}(\frac{1}{|t_{1}e_{1}-y|^{N+2s}}-\frac{1}{|t_{2}e_{1}-y|^{N+2s}})dy<0.$

Finally,

$\displaystyle\varphi_{B_{1}}(x)=\int_{B_{1}^{c}}\frac{dy}{|x-y|^{N+2s}}$
$\displaystyle=$ $\displaystyle\int_{B_{1}^{c}(x)}\frac{dz}{|z|^{N+2s}}$
$\displaystyle=$
$\displaystyle\int_{B_{\frac{1}{1-|x|}}^{c}(\frac{x}{1-|x|})}\frac{(1-|x|)^{N}d\widetilde{z}}{(1-|x|)^{N+2s}|\widetilde{z}|^{N+2s}}$
$\displaystyle=$
$\displaystyle\frac{1}{(1-|x|)^{2s}}\int_{B_{\frac{1}{1-|x|}}^{c}(\frac{x}{1-|x|})}\frac{d\widetilde{z}}{|\widetilde{z}|^{N+2s}}.$

Since
$\bigcap{B_{\frac{1}{1-|x|}}^{c}(\frac{x}{1-|x|})}=(-\infty,-1)\times\mathbb{R}^{N-1}$
and

$\displaystyle\int_{(-\infty,-1)\times\mathbb{R}^{N-1}}\frac{d\widetilde{z}}{|\widetilde{z}|^{N+2s}}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{-1}{d\widetilde{z}_{1}}\int_{R^{N-1}}\frac{d\widetilde{z}^{\prime}}{(|\widetilde{z}_{1}|^{2}+|\widetilde{z}^{\prime}|^{2})^{\frac{N+2s}{2}}}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{-1}{d\widetilde{z}_{1}}\int_{R^{N-1}}\frac{{\widetilde{z}_{1}}^{N-1}dt^{\prime}}{|\widetilde{z}_{1}|^{N+2s}{(1+|t^{\prime}|^{2})^{\frac{N+2s}{2}}}}$
$\displaystyle=$
$\displaystyle\int_{R^{N-1}}\frac{dt^{\prime}}{{(1+|t^{\prime}|^{2})^{\frac{N+2s}{2}}}}\int_{-\infty}^{-1}\frac{d\widetilde{z}_{1}}{{\widetilde{z}_{1}}^{2s+1}}$
$\displaystyle=$ $\displaystyle c_{5}.$

Then (2.3) is proved.

###### Remark 2.1

From Proposition 2.1, the method of moving planes of fractional problem fails
for $\Omega=B_{1}$.

###  2.2  Viscosity solution

We start with the definition of viscosity solution, inspired by the definition
of viscosity sense for nonlocal problem in [6].

###### Definition 2.1

We say that a function $u\in C(\Omega)$ is a viscosity super solution (sub-
solution) of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{\Omega}u=f&{\rm in}\
\,\Omega,\\\\[4.2679pt] \phantom{(-\Delta)^{\alpha}}\displaystyle u=0&{\rm
on}\ \partial\Omega,\end{array}\right.$ (2.4)

if $u\geq 0$ (resp. $u\leq 0$) on $\partial\Omega$ and for every point
$x_{0}\in\Omega$ and some neighborhood $V$ of $x_{0}$ with
$\bar{V}\subset\Omega$ and for any $\varphi\in C^{2}(\bar{V})$ such that
$u(x_{0})=\varphi(x_{0})$ and $x_{0}$ is the minimum (resp. maximum) point of
$u-\varphi$ in $V$, let

$\displaystyle\tilde{u}=\left\\{\begin{array}[]{lll}\varphi&\rm{in}&V,\\\\[2.84526pt]
u&\rm{in}&\Omega\setminus V,\end{array}\right.$

we have that

$(-\Delta)^{s}_{\Omega}\tilde{u}(x_{0})\geq f(x_{0})\quad(resp.\
(-\Delta)^{s}_{\Omega}\tilde{u}(x_{0})\leq f(x_{0})).$

We say that $u$ is a viscosity solution of (2.4) if it is a viscosity super-
solution and also a viscosity sub-solution of (2.4).

###### Theorem 2.1

Assume that the functions $f:\Omega\to\mathbb{R}$,
$h:\partial\Omega\to\mathbb{R}$ are continuous. Let $u$ and $v$ be a viscosity
super-solution and sub-solution of (2.4) respectively. Then

$v\leq u\quad{\rm in}\quad\Omega.$ (2.6)

Proof. Let us define $w=u-v$, then

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{\Omega}w\geq 0&{\rm
in}\ \,\Omega,\\\\[4.2679pt] \phantom{(-\Delta)^{s}}\displaystyle w\geq 0&{\rm
on}\ \partial\Omega.\end{array}\right.$ (2.7)

If (2.6) fails, then there exists $x_{0}\in\Omega$ such that

$w(x_{0})=u(x_{0})-v(x_{0})=\min_{x\in\Omega}w(x)<0,$

then in the viscosity sense,

$(-\Delta)^{s}_{\Omega}w(x_{0})\geq 0.$ (2.8)

Since $w$ is a viscosity super solution, $x_{0}$ is the minimum point in
$\Omega$ and $w\geq 0$ on $\partial\Omega$, then we can take a small
neighborhood $V_{0}$ of $x_{0}$ such that $\tilde{w}=w(x_{0})$ in $V_{0}$.
From (2.8), we have that

$(-\Delta)^{s}_{\Omega}\tilde{w}(x_{0})\geq 0.$

But the definition of regional fractional Laplacian implies that

$(-\Delta)^{s}_{\Omega}\tilde{w}(x_{0})=\int_{\Omega\setminus
V_{0}}\frac{w(x_{0})-w(y)}{|x_{0}-y|^{N+2\alpha}}dy<0,$

which is impossible.

###### Remark 2.2

Let $u$ be a continuous function in $\Omega$ and $x_{0}$ is a minimum point of
$u$, then $(-\Delta)^{s}_{\Omega}u(x_{0})>0$ in the viscosity sense.

We recall the stability property for viscosity solutions in our setting.

###### Theorem 2.2

[10, Theorem 2.2] Assume that the function $g:\Omega\to\mathbb{R}$ is
continuous. Let $u_{n}$, $(n\in\mathbb{N})$ be a sequence of functions in
$C(\Omega)$, uniformly bounded in $L^{1}(\Omega)$, $g_{n}$ and $g$ be
continuous in $\Omega$ such that

$(-\Delta)^{s}_{\Omega}u_{n}\geq g_{n}\ ({\rm resp.}\
(-\Delta)^{s}_{\Omega}u_{n}\leq g_{n})$ in $\Omega$ in viscosity sense,
$u_{n}\geq g_{n}$ ( resp. $u_{n}\geq g_{n})$ on $\partial\Omega$.

$u_{n}\to u$ locally uniformly in $\Omega$,

$u_{n}\to u$ in $L^{1}(\Omega)$,

$h_{n}\to h$ locally uniformly in $\Omega$.
Then $(-\Delta)^{s}_{\Omega}u\geq g\ ({\rm resp.}\ (-\Delta)^{s}_{\Omega}u\leq
g)$ in $\Omega$ in the viscosity sense.

Next we have an interior regularity result. For simplicity, we denote by
$C^{t}$ the space $C^{t_{0},t-t_{0}}$ for $t\in(t_{0},t_{0}+1)$, $t_{0}$ is a
positive integer.

###### Proposition 2.2

Assume that $s\in(0,1)$, $g\in C_{\rm loc}^{\theta}(\Omega)$ with $\theta>0$,
$w\in C_{\rm loc}^{2s+\epsilon}(\mathcal{O})\cap L^{1}(\Omega)$ with
$\epsilon>0$ and $2s+\epsilon$ not being an integer, is a solution of

$(-\Delta)^{s}_{\Omega}w=g\quad{\rm in}\ \ \mathcal{O}.$ (2.9)

Let $\mathcal{O}_{1},\mathcal{O}_{2}$ be open $C^{2}$ sets such that

$\bar{\mathcal{O}_{1}}\subset\mathcal{O}_{2}\subset\bar{\mathcal{O}_{2}}\subset\mathcal{O}\subset\Omega.$

Then
$(i)$ for any $\gamma\in(0,2s)$ not an integer, there exists $c_{6}>0$ such
that

$\|w\|_{C^{\gamma}(\mathcal{O}_{1})}\leq
c_{6}\left[\|w\|_{L^{\infty}(\mathcal{O}_{2})}+\|w\|_{L^{1}(\Omega)}+\|g\|_{L^{\infty}(\mathcal{O}_{2})}\right];$
(2.10)

$(ii)$ for any $\epsilon^{\prime}\in(0,\min\\{\theta,\epsilon\\})$,
$2s+\epsilon^{\prime}$ not an integer, there exists $c_{7}>0$ such that

$\|w\|_{C^{2s+\epsilon^{\prime}}(\mathcal{O}_{1})}\leq
c_{7}\left[\|w\|_{L^{\infty}(\mathcal{O}_{2})}+\|w\|_{L^{1}(\Omega)}+\|g\|_{c^{\theta}(\mathcal{O}_{2})}\right].$
(2.11)

Proof. The proof is similar to the one of [10, Proposition 2.1] for
$s\in(\frac{1}{2},1)$. For the reader’s convenience, we give the details. Let
$\tilde{w}=w$ in $\Omega$, $\tilde{w}=0$ in
$\mathbb{R}^{N}\setminus\bar{\Omega}$, we have that

$\displaystyle(-\Delta)^{s}\tilde{w}(x)=(-\Delta)^{s}_{\Omega}w(x)+w(x)\varphi_{{}_{\Omega}}(x),\quad\forall
x\in\mathcal{O},$

where $\varphi_{{}_{\Omega}}$ is defined as (2.2). Note that
$\varphi_{{}_{\Omega}}\in C^{0,1}_{\rm loc}(\Omega)$. Combining with (2.9), we
have that

$\displaystyle(-\Delta)^{s}\tilde{w}(x)=g(x)+w(x)\varphi_{{}_{\Omega}}(x),\quad\forall
x\in\mathcal{O}.$

By [13, Lemma 3.1], for any $\gamma\in(0,2s)$, we have that

$\displaystyle\|w\|_{C^{\gamma}(\mathcal{O}_{1})}$ $\displaystyle\leq$
$\displaystyle
c_{8}\left[\|w\|_{L^{\infty}(\mathcal{O}_{2})}+\|w\|_{L^{1}(\Omega)}+\|g+w\varphi_{{}_{\Omega}}\|_{L^{\infty}(\mathcal{O}_{2})}\right]$
$\displaystyle\leq$ $\displaystyle
c_{9}\left[\|w\|_{L^{\infty}(\mathcal{O}_{2})}+\|w\|_{L^{1}(\Omega)}+\|g\|_{L^{\infty}(\mathcal{O}_{2})}\right]$

and by [22, Lemma 2.10], for any
$\epsilon^{\prime}\in(0,\min\\{\theta,\epsilon\\})$, we have that

$\displaystyle\|w\|_{C^{2s+\epsilon^{\prime}}(\mathcal{O}_{1})}$
$\displaystyle\leq$ $\displaystyle
c_{10}\left[\|w\|_{C^{\epsilon^{\prime}}(\mathcal{O}_{2})}+\|g+w\phi\|_{C^{\epsilon^{\prime}}(\mathcal{O}_{2})}\right]$
$\displaystyle\leq$ $\displaystyle
c_{11}\left[\|w\|_{L^{\infty}(\mathcal{O}_{2})}+\|w\|_{L^{1}(\Omega)}+\|g\|_{C^{\epsilon^{\prime}}(\mathcal{O}_{2})}\right],$

where $c_{8},\cdots,c_{11}>0$. This ends the proof. $\Box$

##  3 Radial symmetry and decreasing monotonicity

###  3.1  Maximum Principle for small domain

The basic tools for the moving planes in a ball is the Maximum Principle for
small domain:

###### Proposition 3.1

Let $\mathcal{O}$ be an open set in ${B_{1}}$ such that $|\mathcal{O}|\leq
2^{-N}|B_{1}|$. Suppose that $\phi:\mathcal{O}\to\mathbb{R}$ is in
$L^{\infty}(\mathcal{O})$ satisfying

$\|\phi\|_{L^{\infty}(\mathcal{O})}<+\infty.$ (3.1)

Let $w\in L^{1}(B_{1})\cap C(\bar{\mathcal{O}})$ is a solution of

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}w\geq\phi w&{\rm in}\
\mathcal{O},\\\\[5.69054pt] \qquad\quad\,w\geq 0&{\rm in}\
{B_{1}}\setminus\mathcal{O},\end{array}\right.$ (3.2)

then there exists $\delta>0$ such that for $|\mathcal{O}|<\delta$, $w$ has to
be non-negative in $\mathcal{O}$.

In order to prove Proposition 3.1, we need the following estimate.

###### Lemma 3.1

Assume that $\mathcal{O}\subset B_{1}$ such that $|\mathcal{O}|\leq
2^{-N}|B_{1}|$, and $g:\mathcal{O}\to\mathbb{R}$ is in
$L^{\infty}(\mathcal{O})$.

Let $w\in L^{1}(B_{1})\cap C(\bar{\mathcal{O}})$ be a solution of

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}w\geq g&{\rm in}\
\mathcal{O},\\\\[5.69054pt] \qquad\quad\ w\geq 0&{\rm in}\ \
B_{1}\setminus\mathcal{O}.\end{array}\right.$ (3.3)

Then there exists $c_{12}>0$ such that

$-\inf_{\mathcal{O}}w\leq
c_{12}\|g\|_{L^{\infty}(\mathcal{O})}|\mathcal{O}|^{\frac{2s}{N}}.$ (3.4)

Proof. The result is obvious if $\inf_{\mathcal{O}}w\geq 0$. Now we assume
that $\inf_{\mathcal{O}}w<0$, then there exists $x_{0}\in\mathcal{O}$ such
that

$w(x_{0})=\inf_{x\in\mathcal{O}}w(x)<0.$

Combining with (3.3), we have that

$-\|g\|_{L^{\infty}(\mathcal{O})}\leq
g(x_{0})\leq(-\Delta)_{B_{1}}^{s}w(x_{0}).$ (3.5)

By the definition of $(-\Delta)_{B_{1}}^{s}$, we have that

$\displaystyle(-\Delta)_{B_{1}}^{s}w(x_{0})$ $\displaystyle=$
$\displaystyle{\rm
p.v.}\int_{B_{1}}{\frac{w(x_{0})-w(y)}{{|{x_{0}}-y|}^{N+2s}}}dy$
$\displaystyle=$ $\displaystyle{\rm
p.v.}\int_{\mathcal{O}}{\frac{w(x_{0})-w(y)}{{|{x_{0}}-y|}^{N+2s}}}dy+\int_{B_{1}\setminus\mathcal{O}}{\frac{w(x_{0})-w(y)}{{|{x_{0}}-y|}^{N+2s}}}dy$
$\displaystyle\leq$
$\displaystyle\int_{B_{1}\setminus\mathcal{O}}{\frac{w(x_{0})}{{|{x_{0}}-y|}^{N+2s}}}dy.$

Let

$r=(|\omega_{{}_{N}}|^{-1}|\mathcal{O}|)^{\frac{1}{N}}\leq\frac{1}{2}$

by the fact that $|\mathcal{O}|\leq 2^{-N}|B_{1}|$, then we see that
$|\mathcal{O}|=|B_{r}(x_{0})|$.

Thanks to the decreasing monotonicity of the kernel $\frac{1}{r^{N+2s}}$, we
obtain that

$\displaystyle\int_{B_{1}\setminus{B_{r}(x_{0})}}{\frac{1}{|x_{0}-y|^{N+2s}}dy}\leq\int_{B_{1}\setminus{\mathcal{O}}}{\frac{1}{|x_{0}-y|^{N+2s}}dy}.$

If $x_{0}=0$, then we get that

$\displaystyle(-\Delta)_{B_{1}}^{s}w(0)$ $\displaystyle\leq$ $\displaystyle
w(0)\int_{B_{1}\setminus{\mathcal{O}}}{\frac{1}{|y|^{N+2s}}dy}$
$\displaystyle\leq$ $\displaystyle
w(0)\int_{B_{1}\setminus{B_{r}}}{\frac{1}{|y|^{N+2s}}dy}$ $\displaystyle\leq$
$\displaystyle c_{12}^{-1}|\mathcal{O}|^{-\frac{2s}{N}}w(0),$

since $w(0)<0$ and

$\displaystyle
c_{12}^{-1}|\mathcal{O}|^{-\frac{2s}{N}}\leq\int_{B_{1}\setminus{B_{r}}}{\frac{1}{|y|^{N+2s}}dy}.$

If $x_{0}\not=0$. We let the vertical plane with respect to $x_{0}$

$\mathcal{P}(x_{0})=\big{\\{}z\in\mathbb{R}^{N}:\,z\cdot x_{0}=0\big{\\}}$

and denote the cone

$\mathcal{C}(x_{0})=\big{\\{}t(z-\frac{x_{0}}{|x_{0}|}):t\in(-\frac{1}{2},0),\
z\in\mathcal{P}(x_{0})\cap B_{\frac{1}{2}}\big{\\}}.$

Then $\mathcal{C}(x_{0})\subset B_{1}(-x_{0})\setminus B_{r}$.

Thus, we get that

$\displaystyle(-\Delta)_{B_{1}}^{s}w(x_{0})$ $\displaystyle\leq$
$\displaystyle
w(x_{0})\int_{B_{1}\setminus{\mathcal{O}}}{\frac{1}{|x_{0}-y|^{N+2s}}dy}$
$\displaystyle\leq$ $\displaystyle
w(x_{0})\int_{B_{1}\setminus{B_{r}(x_{0})}}{\frac{1}{|x_{0}-y|^{N+2s}}dy}$
$\displaystyle=$ $\displaystyle
w(x_{0})\int_{B_{1}(-x_{0})\setminus{B_{r}}}{\frac{1}{|y|^{N+2s}}dy}$
$\displaystyle\leq$ $\displaystyle
w(x_{0})\int_{\mathcal{C}(x_{0})}{\frac{1}{|y|^{N+2s}}dy}$ $\displaystyle\leq$
$\displaystyle c_{12}^{-1}|\mathcal{O}|^{-\frac{2s}{N}}w(x_{0}),$

since $w(x_{0})<0$ and

$\displaystyle
c_{12}^{-1}|\mathcal{O}|^{-\frac{2s}{N}}\leq\int_{\mathcal{C}(x_{0})}{\frac{1}{|x_{0}-y|^{N+2s}}dy}.$

By (3.5), we have that

$-\|g\|_{L^{\infty}(\Omega)}\leq(-\Delta)_{B_{1}}^{s}w(x_{0})\leq
c_{12}^{-1}w(x_{0})|\mathcal{O}|^{-\frac{2s}{N}}.$

Therefore,

$w(x_{0})\geq-c_{12}\|g\|_{L^{\infty}(\Omega)}|\mathcal{O}|^{\frac{2s}{N}},$

or

$-\inf_{\Omega}w\leq
c_{12}\|g\|_{L^{\infty}(\mathcal{O})}|\mathcal{O}|^{\frac{2s}{N}}.$

We complete the proof. $\Box$

Proof of Proposition 3.1. Let us define $\mathcal{O}^{-}=\\{x\in\mathcal{O}\
|\ w(x)<0\\}$, then we observe that

$\left\\{\begin{array}[]{lll}(-\Delta)_{B_{1}}^{s}w(x)\geq\phi(x)w(x),&x\in\mathcal{O}^{-},\\\\[5.69054pt]
\qquad\quad\ w(x)\geq
0,&x\in{B_{1}}\setminus{\mathcal{O}^{-}}.\end{array}\right.$ (3.6)

Using Lemma 3.1 with $g=\phi w$, we have that

$\displaystyle\|w\|_{L^{\infty}(\mathcal{O}^{-})}=-\inf_{\mathcal{O}^{-}}w\leq
c_{12}\|\phi\|_{L^{\infty}(\mathcal{O})}\|w\|_{L^{\infty}(\mathcal{O}^{-})}|\mathcal{O}|^{\frac{2s}{N}}.$

Then there exists $\delta>0$ such that for $|\mathcal{O}|\leq\delta$, we have
that

$c_{12}\|\phi\|_{L^{\infty}(\mathcal{O})}|\mathcal{O}|^{\frac{2s}{N}}\leq
c_{12}\|\phi\|_{L^{\infty}(\mathcal{O})}\delta^{\frac{2s}{N}}<1,$

then $\|w\|_{L^{\infty}(\mathcal{O}^{-})}=0$, that is, $\mathcal{O}^{-}$ is
empty, which ends the proof. $\Box$

###  3.2 Moving planes

Proof of Theorem 1.2.  Given $\lambda\in(0,r_{0})$, let us define

$\Sigma_{\lambda}=\\{x=(x^{1},x^{\prime})\in B_{1}\ |\ x^{1}>\lambda\\},\
T_{\lambda}=\\{x=(x^{1},x^{\prime})\in B_{1}\ |\ x^{1}=\lambda\\},$
$\Sigma=\Sigma_{\lambda}\cup{(\Sigma_{\lambda})^{\lambda}},\
w_{\lambda}(x)=u_{\lambda}(x)-u(x),$
$u_{\lambda}(x)=\left\\{\begin{array}[]{lll}u(x_{\lambda}),&x\in\Sigma,\\\
u(x),&x\in B_{1}\setminus\Sigma,\end{array}\right.$

where $x_{\lambda}=(2\lambda-x^{1},x^{\prime})$ for $x=(x^{1},x^{\prime})\in
B_{1}.$ For any subset $A$ of $B_{1}$, we write
$A_{\lambda}=\\{x_{\lambda}:\,x\in A\\}$.

Step 1: we first prove that $w_{\lambda}\geq 0$ in $\Sigma_{\lambda}$ if
$\lambda\in(0,r_{0})$ is close to $r_{0}$.  Indeed, let

$\Sigma_{\lambda}^{-}=\\{x\in\Sigma_{\lambda}\ |\ w_{\lambda}(x)<0\\}$

and

$w_{\lambda}^{-}(x)=\left\\{\begin{array}[]{lll}w_{\lambda}(x),&x\in\Sigma_{\lambda}^{-},\\\\[5.69054pt]
0,&x\in B_{1}\setminus\Sigma_{\lambda}^{-},\end{array}\right.\qquad
w_{\lambda}^{+}(x)=\left\\{\begin{array}[]{lll}0,&x\in\Sigma_{\lambda}^{-},\\\\[5.69054pt]
w_{\lambda}(x),&x\in B_{1}\setminus\Sigma_{\lambda}^{-}.\end{array}\right.$

By the linearity of the regional fractional laplacian, we have that for all
$0<\lambda<1$,

$(-\Delta)_{B_{1}}^{s}w_{\lambda}^{+}(x)\leq 0,\ \ \ \ \forall\
x\in\Sigma_{\lambda}^{-}.$

In fact, for $x\in\Sigma_{\lambda}^{-},$ $w_{\lambda}^{+}(x)=0$ and

$\displaystyle(-\Delta)_{B_{1}}^{s}w_{\lambda}^{+}(x)$ $\displaystyle=$
$\displaystyle-\int_{{B_{1}}\backslash\Sigma_{\lambda}^{-}}\frac{w_{\lambda}(y)}{|x-y|^{N+2s}}dy$
$\displaystyle=$
$\displaystyle-\int_{{B_{1}}\backslash{\Sigma}}{\frac{{w_{\lambda}}(y)}{|x-y|^{N+2s}}}dy-\int_{({\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-})\cup({\Sigma_{\lambda}}\backslash\Sigma_{\lambda}^{-})^{\lambda}}{\frac{{w_{\lambda}}(y)}{|x-y|^{N+2s}}}dy$
$\displaystyle-\int_{(\Sigma_{\lambda}^{-})^{\lambda}}{\frac{w_{\lambda}(y)}{|x-y|^{N+2s}}}dy$
$\displaystyle=:$ $\displaystyle-I_{1}-I_{2}-I_{3}.$

Note that

${I_{1}}=\int_{{B_{1}}\setminus{\Sigma}}({u_{\lambda}}(y)-u(y)){\frac{1}{|x-y|^{N+2s}}}dy=0.$

Since ${w_{\lambda}}(y^{\lambda})=-{w_{\lambda}}(y)$ for any $y\in B_{1}$,
then

$\displaystyle I_{2}$ $\displaystyle=$
$\displaystyle\int_{({\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-})\cup({\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-})^{\lambda}}{w_{\lambda}}(y){\frac{1}{|x-y|^{N+2s}}}dy$
$\displaystyle=$
$\displaystyle\int_{{\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-}}{w_{\lambda}}(y){\frac{1}{|x-y|^{N+2s}}}dy+\int_{{\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-}}{w_{\lambda}}(y^{\lambda}){\frac{1}{|x-y^{\lambda}|^{N+2s}}}dy$
$\displaystyle=$
$\displaystyle\int_{{\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-}}{w_{\lambda}}(y)({\frac{1}{|x-y|^{N+2s}}}-{\frac{1}{|x-{y}^{\lambda}|^{N+2s}}})dy.$

For $x\in\Sigma_{\lambda}^{-}$ and
$y\in{\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-},$ we have
$x-y=(x^{1}-y^{1},x^{{}^{\prime}}-y^{{}^{\prime}})$,
$x-y^{\lambda}=(x^{1}+y^{1}-2{\lambda},x^{{}^{\prime}}-y^{{}^{\prime}})$,
$|x^{1}+y^{1}-2{\lambda}|>|x^{1}-y^{1}|$, then

$\frac{1}{|x-y|^{N+2s}}\geq\frac{1}{|x-{y}^{\lambda}|^{N+2s}}.$

Combing with $w_{\lambda}\geq 0$ in
${\Sigma_{\lambda}}\setminus\Sigma_{\lambda}^{-}$, we have

${I_{2}}\geq 0.$

Since $w_{\lambda}(y)<0$ for $y\in\Sigma_{\lambda}^{-}$ and
${w_{\lambda}}(y^{\lambda})=-{w_{\lambda}}(y)$ for any $y\in B_{1}$, we have
that

$\displaystyle
I_{3}=\int_{(\Sigma_{\lambda}^{-})^{\lambda}}{w_{\lambda}}(y){\frac{1}{|x-y|^{N+2s}}}dy$
$\displaystyle=$
$\displaystyle\int_{\Sigma_{\lambda}^{-}}{\frac{w_{\lambda}(y^{\lambda})}{|x-y^{\lambda}|^{N+2s}}}dy$
$\displaystyle=$
$\displaystyle-\int_{\Sigma_{\lambda}^{-}}{\frac{{w_{\lambda}}(y)}{|x-y^{\lambda}|^{N+2s}}}dy\geq
0.$

Hence, we obtain that for $\lambda\in(0,1)$

$(-\Delta)_{B_{1}}^{s}w_{\lambda}^{+}(x)\leq 0,\quad\forall
x\in\Sigma_{\lambda}^{-}$

i.e.

$(-\Delta)_{B_{1}}^{s}[w_{\lambda}-w_{\lambda}^{-}](x)\leq 0.$

Then for $x\in\Sigma_{\lambda}^{-}$

$(-\Delta)_{B_{1}}^{s}w_{\lambda}(x)\leq(-\Delta)_{B_{1}}^{s}[w_{\lambda}^{-}](x)$

and

$\displaystyle(-\Delta)_{B_{1}}^{s}w_{\lambda}^{-}(x)\geq(-\Delta)_{B_{1}}^{s}w_{\lambda}(x)$
$\displaystyle=$
$\displaystyle(-\Delta)_{B_{1}}^{s}u_{\lambda}(x)-(-\Delta)_{B_{1}}^{s}u(x)$
$\displaystyle=$ $\displaystyle h(x_{\lambda},u_{\lambda})-h(x,u)$
$\displaystyle=$
$\displaystyle\big{(}h_{1}(x_{\lambda})-h_{1}(x)\big{)}h_{3}(u_{\lambda})+h_{1}(x)\big{(}h_{3}(u_{\lambda})-h_{3}(u)\big{)}+h_{2}(x_{\lambda})-h_{2}(x)$
$\displaystyle\geq$ $\displaystyle
h_{1}(x)\psi(x)\big{(}u_{\lambda}(x)-u(x)\big{)},$

where the last inequality holds by the assumption $(\mathcal{H}_{1})$ and
$(\mathcal{H}_{2})$,

$\psi(x)=\frac{h_{3}(u_{\lambda})-h_{3}(u)}{u_{\lambda}(x)-u(x)},$

which is bounded for $x\in\Sigma_{\lambda}^{-}$.

Choosing $\lambda\in(0,r_{0})$ close enough to $r_{0}$, then
$|\Sigma_{\lambda}^{-}|$ is small enough, by $w_{\lambda}^{-}=0$ in
$(\Sigma_{\lambda}^{-})^{c}$ it follows by Proposition 3.1 that
$w_{\lambda}=w_{\lambda}^{-}\geq 0$ in $\Sigma_{\lambda}^{-}$. Then
$\Sigma_{\lambda}^{-}$ is empty, that is, $w_{\lambda}\geq 0$ in
$\Sigma_{\lambda}$.

Step 2: we claim that for $0<\lambda<1$, if $w_{\lambda}\geq 0$ and
$w_{\lambda}\not\equiv 0$ in $\Sigma_{\lambda}$, then $w_{\lambda}>0$ in
$\Sigma_{\lambda}$.

If this is not true, then there exists $x_{0}\in\Sigma_{\lambda}$ such that
$w_{\lambda}(x_{0})=0$ and then $u_{\lambda}(x_{0})=u(x_{0})$ and

$\displaystyle(-\Delta)_{B_{1}}^{s}w_{\lambda}(x_{0})$ $\displaystyle=$
$\displaystyle(-\Delta)_{B_{1}}^{s}u_{\lambda}(x_{0})-(-\Delta)_{B_{1}}^{s}u(x_{0})$
(3.7) $\displaystyle=$ $\displaystyle
h((x_{0})_{\lambda},u_{\lambda}(x_{0}))-h(x_{0},u(x_{0}))$ $\displaystyle\geq$
$\displaystyle\big{(}h_{1}((x_{0})_{\lambda}-h_{1}(x_{0})\big{)}h_{3}(u_{\lambda}(x_{0}))$
$\displaystyle\geq$ $\displaystyle 0.$

However, since $x_{0}$ is the minimal of $w_{\lambda}$ and by the definition
of the regional fractional laplacian

$\displaystyle(-\Delta)_{B_{1}}^{s}[w_{\lambda}](x_{0})$ $\displaystyle=$
$\displaystyle-\int_{B_{1}}{\frac{{w_{\lambda}}(y)}{|{x_{0}}-y|^{N+2s}}}dy$
$\displaystyle=$
$\displaystyle-\int_{\Sigma_{\lambda}^{-}}{\frac{w_{\lambda}^{-}(y)}{|{x_{0}}-y|^{N+2s}}}dy-\int_{{B_{1}}\setminus\Sigma_{\lambda}^{-}}{\frac{w_{\lambda}^{+}(y)}{|{x_{0}}-y|^{N+2s}}}dy$
$\displaystyle\leq$
$\displaystyle-\int_{{B_{1}}\setminus\Sigma_{\lambda}^{-}}{\frac{w_{\lambda}^{+}(y)}{|{x_{0}}-y|^{N+2s}}}dy$
$\displaystyle<$ $\displaystyle 0.$

By the fact that $w_{\lambda}^{+}\geq 0$ and $w_{\lambda}^{+}\not=0$ for
$\Sigma_{\lambda}^{-}$. So we obtain a contradiction from (3.7). Thus,
$w_{\lambda}>0$ in $\Sigma_{\lambda}$ if ${\lambda}\in(0,1)$ is close to 1.

Step 3: we prove $\lambda_{0}=0$, where

$\lambda_{0}=\inf\\{\lambda\in(0,r_{0})\ |\ w_{\lambda}>0\ \ \rm{in}\ \
\Sigma_{\lambda}\\}.$

If it is not true, i.e. $\lambda_{0}>0$, by the definition of $\lambda_{0}$,
we have that $w_{\lambda_{0}}\geq 0$ in $\Sigma_{\lambda_{0}}$ and
$w_{\lambda_{0}}\not\equiv 0$ in $\Sigma_{\lambda_{0}}$. By Step 2, we have
$w_{\lambda_{0}}>0$ in $\Sigma_{\lambda_{0}}$.

Claim 1. If $w_{\lambda}>0$ in $\Sigma_{\lambda}$ for $\lambda\in(0,1)$, then
there exists $\epsilon\in(0,\lambda)$ such that $w_{\lambda_{\epsilon}}>0$ in
$\Sigma_{\lambda_{\epsilon}}$, where $\lambda_{\epsilon}=\lambda-\epsilon$.
Now we assume that Claim 1 is true, then we get some
$\epsilon\in(0,\lambda_{0})$ such that $w_{\lambda_{0}-\epsilon}>0$ in
$\Sigma_{\lambda_{0}-\epsilon}$, which implies that

$\lambda_{0}-\epsilon\geq\lambda_{0},$

it is impossible. Then we obtain $\lambda_{0}=0.$

Now we only need to prove Claim 1 to complete Step 3.

Proof of Claim 1. Let $D_{\mu}=\\{x\in\Sigma_{\lambda}\ |\
dist(x,\partial\Sigma_{\lambda})\geq\mu\\}$ for $\mu>0$ small. Since
$w_{\lambda}>0$ in $\Sigma_{\lambda}$ and $D_{\mu}$ is compact, then there
exists $\mu_{0}>0$ such that $w_{\lambda}\geq\mu_{0}$ in $D_{\mu}$. By
continuity of $w_{\lambda}(x)$, for $\epsilon>0$ small enough, we denote
$\lambda_{\epsilon}=\lambda-\epsilon,$ then

$w_{\lambda_{\epsilon}}(x)\geq 0\ \ \rm{in}\ \ D_{\mu}.$

As a consequence,

$\Sigma_{\lambda_{\epsilon}}^{-}\subset\Sigma_{\lambda_{\epsilon}}\setminus
D_{\mu}$

and $|\Sigma_{\lambda_{\epsilon}}^{-}|$ small if $\epsilon$ and $\mu$ small.
By Step 1, $(-\Delta)_{B_{1}}^{s}w_{\lambda_{\epsilon}}^{-}(x)\leq 0$ in
$x\in\Sigma_{\lambda_{\epsilon}}^{-}$, Since $w_{\lambda_{\epsilon}}^{+}=0$ in
$(\Sigma_{\lambda_{\epsilon}}^{-})^{c}$ with
$|\Sigma_{\lambda_{\epsilon}}^{-}|$ small for $\epsilon$ and $\mu$ small,
$\varphi(x)=\frac{{u_{\lambda_{\epsilon}}^{p}}(x)-u^{p}(x)}{u_{\lambda_{\epsilon}}(x)-u(x)},$
similar with Step 1, then we have $w_{\lambda_{\epsilon}}\geq 0$ in
$\Sigma_{\lambda_{\epsilon}}$. And since $\lambda_{\epsilon}>0$,
$w_{\lambda_{\epsilon}}\not\equiv 0$ in $\Sigma_{\lambda_{\epsilon}}$, we have
$w_{\lambda_{\epsilon}}>0$ in $\Sigma_{\lambda_{\epsilon}}$. Claim 1 is
proved.

We conclude from the fact that $\lambda_{0}=0$ that

$u((-x^{1},x^{\prime}))\geq u((x^{1},x^{\prime}))\ \ \rm{for}\ \ x^{1}\geq 0.$

Using the same way, do moving plane from left side to 0, we have

$u((-x^{1},x^{\prime}))\leq u((x^{1},x^{\prime}))\ \ \rm{for}\ \ x^{1}\geq 0.$

So

$u((-x^{1},x^{\prime}))=u((x^{1},x^{\prime}))\ \ \rm{for}\ \ x^{1}\geq 0.$

Step 4: we prove $u(x)$ is strictly decreasing in the $x_{1}$ direction for
$x=(x_{1},x^{\prime})\in B_{r_{0}},$ $x_{1}>0$.  If not, there exists
$(x^{1},x^{\prime}),(\tilde{x}^{1},x^{\prime})\in\Omega$,
$0<x^{1}<\tilde{x}^{1}$ such that

$u((x^{1},x^{\prime}))\leq u((\tilde{x}^{1},x^{\prime})).$ (3.8)

Let $\lambda=\frac{x^{1}+\tilde{x}^{1}}{2}$ and by arguments above, we have

$w_{\lambda}(x)>0\ \ \rm{for}\ \ x\in\Sigma_{\lambda}.$

Since $(\tilde{x}^{1},x^{\prime})\in\Sigma_{\lambda}$, then

$\displaystyle 0<w_{\lambda}((\tilde{x}^{1},x^{\prime}))$ $\displaystyle=$
$\displaystyle
u_{\lambda}((\tilde{x}^{1},x^{\prime}))-u((\tilde{x}^{1},x^{\prime}))$
$\displaystyle=$ $\displaystyle
u((x^{1},x^{\prime}))-u((\tilde{x}^{1},x^{\prime})),$

i.e.

$u((x^{1},x^{\prime}))>u((\widetilde{x}^{1},x^{\prime}))$

which is impossible with (3.8). Hence, $u(x)$ is strictly decreasing in the
$x_{1}$ direction for $x=(x^{1},x^{\prime})\in\Omega,$ $x^{1}>0$. $\Box$

##  4 Poisson problems

We first consider the Poisson problem

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=f&{\rm in}\
\,B_{1},\\\\[4.2679pt] \phantom{(-\Delta)^{\alpha}+u\,}\displaystyle u=0&{\rm
on}\ \partial B_{1}.\end{array}\right.$ (4.1)

Let $H^{s}_{0}(B_{1})$ be the closure of $C^{\infty}_{c}(B_{1})$ under the
norm that

$\|u\|_{s}=\sqrt{\int_{B_{1}\times
B_{1}}\frac{(u(x)-u(y))^{2}}{|x-y|^{N+2s}}dxdy+\int_{B_{1}}u^{2}dx},$

which is a Hilbert space with the inner product

$\langle u,v\rangle_{s}=\int_{B_{1}\times
B_{1}}\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}}dxdy+\int_{B_{1}}uvdx.$

The existence result states as following.

###### Theorem 4.1

Let $f\in C^{\theta}(\bar{B}_{1})$ with $\theta\in(0,1)$, then (4.1) has a
unique classical solution. Moreover, we have that $u_{f}\in H^{1}_{0}(B_{1})$
and

$\langle u_{f},\xi\rangle_{s}=\int_{B_{1}}f\xi dx,\quad\xi\in
H^{s}_{0}(B_{1}).$ (4.2)

Before the proof of Theorem 4.1, we need the following existence results:

###### Lemma 4.1

Let $r_{0}\in(0,1)$, $f:B_{1}\to[0,+\infty)$ be Hölder continuous, then

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=f&{\rm in}\
\,B_{r_{0}},\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+u\,}\displaystyle
u=0&{\rm in}\ \overline{B}_{1}\setminus B_{r_{0}}\end{array}\right.$ (4.3)

has a unique positive solution $u_{r_{0},f}$.

Moreover, $(i)$if $f$ is radially symmetric function decreasing with respect
to $|x|$, then $u_{r_{0},f}$ is radially symmetric and decreasing with respect
to $|x|$;
$(ii)$ $r_{0}\to u_{r_{0},f}$ is non-decreasing, i.e.

$u_{r_{1},f}\leq u_{r_{2},f}\quad{\rm if}\ \ 0<r_{1}<r_{2}<1.$

Proof.  Let $H^{s}_{0}(B_{r_{0}})$ be the closure of
$C^{\infty}_{0}(B_{r_{0}})$, with zero value in $\mathbb{R}^{N}\setminus
B_{r_{0}}$, under the norm that

$\|u\|_{s,r_{0}}=\sqrt{\int_{B_{1}\times
B_{1}}\frac{(u(x)-u(y))^{2}}{|x-y|^{N+2s}}dxdy+\int_{B_{1}}u^{2}dx},$

which is a Hilbert space with the inner product

$\langle u,v\rangle_{s,r_{0}}=\int_{B_{1}\times
B_{1}}\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}}dxdy+\int_{B_{1}}uvdx.$

Note that $H^{s}_{0}(B_{r_{0}})\subset H^{s}_{0}(B_{1})$ and from [16,
Corollary 7.2], the embedding $H^{s}_{0}(B_{1})\hookrightarrow L^{q}(B_{1})$
is compact for $q\in(2,\frac{2N}{N-2s})$. Since $f$ is Hölder continuous, it
follows by the standard argument of variational methods to find the critical
point of

$\mathcal{J}_{s,r_{0}}(u)=\frac{1}{2}\|u\|_{s,1}^{2}-\int_{B_{r_{0}}}fudx,\quad
u\in H^{s}_{0}(B_{r_{0}}).$

Then (4.3) has a weak solution $u_{r_{0},f}\in H^{s}_{0}(B_{r_{0}})$ in the
sense that

$\langle u_{r_{0},f},\xi\rangle_{s,r_{0}}=\int_{B_{r_{0}}}f\xi dx,\quad\xi\in
H^{s}_{0}(B_{r_{0}}).$ (4.4)

and taking $\xi=u_{r_{0},f}$, the Hölder inequality and fractional Sobolev
embedding in [16] implies that

$\displaystyle\|u_{r_{0},f}\|_{s,r_{0}}^{2}$ $\displaystyle=$
$\displaystyle\int_{B_{r_{0}}}fu_{r_{0},f}dx$ $\displaystyle\leq$
$\displaystyle\|u_{r_{0},f}\|_{L^{2^{*}_{s}}(B_{r_{0}})}\|f\|_{L^{p^{*}}(B_{r_{0}})}\leq
c_{13}\|u_{r_{0},f}\|_{s,r_{0}}\|f\|_{L^{p^{*}}(B_{1})},$

that is,

$\|u_{r_{0},f}\|_{s,r_{0}}\leq c_{13}\|f\|_{L^{p^{*}}(B_{1})},$ (4.5)

where $2^{*}_{s}=\frac{2N}{N-2s}$, $p^{*}=\frac{2N}{N+2s}$ and $c_{13}>0$ is
independent of $r_{0}$.

Let $\tilde{H}^{s}_{0}(B_{r_{0}})$ be the closure of
$C^{\infty}_{0}(B_{r_{0}})$, with zero value in $\mathbb{R}^{N}\setminus
B_{r_{0}}$, under the norm that

$\|\\!|u|\\!\|_{s,r_{0}}=\sqrt{\int_{\mathbb{R}^{N}\times\mathbb{R}^{N}}\frac{(u(x)-u(y))^{2}}{|x-y|^{N+2s}}dxdy+\int_{B_{r_{0}}}u^{2}dx}$

with the inner product

$\prec
u,v\succ_{s,r_{0}}=\int_{\mathbb{R}^{N}\times\mathbb{R}^{N}}\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}}dxdy+\int_{B_{r_{0}}}uvdx.$

Direct computation shows that

$\|\\!|u|\\!\|_{s,r_{0}}^{2}=\|u\|_{s,r_{0}}^{2}+2\int_{B_{1}}\varphi_{B_{1}}(x)u^{2}(x)dx,$

where $\varphi_{B_{1}}$ is defined in (2.2) with $\Omega=B_{1}$, i.e.

$\varphi_{B_{1}}(x)=\int_{B_{1}^{c}}\frac{dy}{|x-y|^{N+2s}}.$

which by Proposition 2.1 that $\varphi_{B_{1}}$ is Lipschitz in
$\overline{B_{r_{0}}}$. Thus $u_{r_{0},f}$ verifies that

$\prec u_{r_{0},f},\xi\succ_{s,r_{0}}=\int_{B_{r_{0}}}f\xi
dx+2\int_{B_{1}}\varphi_{B_{1}}(x)u_{r_{0},f}\xi dx,\quad\xi\in
H^{s}_{0}(B_{r_{0}}).$

This means, $u_{r_{0},f}$ is a weak solution of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}u+u=f+\varphi_{B_{1}}u&{\rm
in}\ \,B_{r_{0}},\\\\[4.2679pt] \phantom{(-\Delta)^{\alpha}+u\,}\displaystyle
u=0&{\rm in}\ \,B_{r_{0}}^{c}.\end{array}\right.$ (4.6)

Note that the solution could be expressed by

$u_{r_{0},f}=\Phi_{r_{0}}\ast F$

where $F=f+(\varphi_{B_{1}}-1)u_{r_{0},f}$ and $\Phi_{r_{0}}$ is the Green
kernel of $B_{r_{0}}$ under the zero condition $\Phi_{r_{0}}=0$ in
$\mathbb{R}^{N}\setminus B_{r_{0}}$. Since $(\varphi_{B_{1}}-1)$ is uniformly
bounded in $B_{r_{0}}$ and $u_{r_{0},f}\in L^{2^{*}_{s}}$, then $F\in
L^{2^{*}_{s}}$ and if follows by Lemma B.1 that $u_{r_{0},f}\in
L^{\infty}(B_{r_{0}})$ if $2^{*}_{s}>\frac{N}{2s}$ and we are done; or
$u_{r_{0},f}\in L^{q_{1}}(B_{r_{0}})$ if
$q_{1}=\frac{Nq_{0}}{2sq_{0}-N}>q_{0}$ if $q_{0}:=2^{*}_{s}>\frac{N}{2s}$, in
this case, $F\in L^{q_{1}}$. Repeat the procedure, we can find a $i_{0}\geq 1$
such that $u_{r_{0},f}\in L^{\infty}(B_{r_{0}})$ and

$\|u_{r_{0},f}\|_{L^{\infty}}\leq c_{14}\|f\|_{L^{\infty}(B_{r_{0}})}.$

where $c_{14}>0$ depends on $r_{0}$. From (4.5), we have that $u_{r_{0},f}$
has a uniform bound in $H^{s}_{0}(B_{r_{0}})$ from Proposition 2.2, we obtain
that for any $\mathcal{O}\subset\bar{\mathcal{O}}\subset B_{r_{0}}$

$\displaystyle\|u_{r_{0},f}\|_{C^{\gamma}(\mathcal{O})}$ $\displaystyle\leq$
$\displaystyle
c_{15}\left[\|u_{r_{0},f}\|_{L^{\infty}(B_{r_{0}})}+\|u_{r_{0},f}\|_{L^{1}(B_{r_{0}})}+\|f\|_{L^{\infty}(B_{r_{0}})}\right]$
$\displaystyle\leq$ $\displaystyle c_{16}\|f\|_{L^{\infty}(B_{1})}$

and then by (2.11), we have that

$\displaystyle\|u_{r_{0},f}\|_{C^{2s+\epsilon^{\prime}}(\mathcal{O})}$
$\displaystyle\leq$ $\displaystyle
c_{17}\left[\|u_{r_{0},f}\|_{L^{\infty}(B_{r_{0}})}+\|u_{r_{0},f}\|_{L^{1}(B_{r_{0}})}+\|f\|_{C^{\theta}(B_{r_{0}})}\right]$
$\displaystyle\leq$ $\displaystyle c_{18}\|f\|_{C^{\theta}(B_{r_{0}})}.$

It follows by boundary regularity in [22] that $u_{r_{0},f}$ is a classical
solution of (4.6), so is (4.3). Then for $\xi\in
C^{2s+\theta}_{loc}(B_{r_{0}})\cap C^{s}_{0}(B_{r_{0}})$

$\displaystyle\int_{B_{r_{0}}}f\xi dx$ $\displaystyle=$
$\displaystyle\int_{B_{r_{0}}}\big{(}\xi(-\Delta)^{s}_{B_{1}}u_{r_{0},f}+u_{r_{0},f}\xi\big{)}dx$
$\displaystyle=$
$\displaystyle\int_{B_{r_{0}}}\big{(}\xi(-\Delta)^{s}u_{r_{0},f}+u_{r_{0},f}\xi-\varphi_{B_{1}}(x)u_{r_{0},f}\xi\big{)}dx$
$\displaystyle=$
$\displaystyle\int_{B_{r_{0}}}\big{(}u_{r_{0},f}(-\Delta)^{s}\xi+u_{r_{0},f}\xi-\varphi_{B_{1}}(x)u_{r_{0},f}\xi\big{)}dx$
$\displaystyle=$
$\displaystyle\int_{B_{r_{0}}}\big{(}u_{r_{0},f}(-\Delta)^{s}_{B_{1}}\xi+u_{r_{0},f}\xi\big{)}dx,$

that is,

$\int_{B_{r_{0}}}u_{r_{0},f}\big{(}(-\Delta)^{s}_{B_{1}}\xi+\xi\big{)}dx=\int_{B_{r_{0}}}f\xi
dx.$ (4.7)

By the method of moving planes in Theorem 1.2 with $h_{1}=0$ and $h_{2}=f$,
the solution $u_{r_{0},f}$ is radially symmetric and decreasing to $|x|$.

For $0<r_{1}<r_{2}<1$, we see that the solution $u_{r_{2},f}$ is a super
solution of (4.3) with $r_{0}=r_{1}$, then maximum principle shows that
$u_{r_{2},f}\geq u_{r_{1},f}$ in $B_{1}$. $\Box$

###### Corollary 4.1

Let $f\in L^{q}(B_{1})$ with $q>\frac{2N}{N+2s}$, then there exists a unique
weak solution $u_{f}\in H^{s}_{0}(B_{1})$ in the sense that

$\langle u_{r_{0},f},\xi\rangle_{s,1}=\int_{B_{1}}f\xi dx,\quad\forall\xi\in
H^{s}_{0}(B_{1}).$

Proof.  The weak solution is the critical point of

$\mathcal{J}_{s}(u)=\frac{1}{2}\|u\|_{s,1}^{2}-\int_{B_{1}}fudx,\quad u\in
H^{s}_{0}(B_{1}),$

whose critical point is unique. $\Box$

Proof of Theorem 4.1.  Case 1: $f\in C^{\theta}(\overline{B_{1}})$ is a
radially symmetric function decreasing with respect to $|x|$.  It follows by
Lemma 4.1, problem (4.3) has a unique classical solution $u_{r_{0},f}$, which
is positive, radially symmetric and decreasing with respect to $|x|$ and
$r_{0}:\to u_{r_{0},f}$ is increasing. From (4.5)

$\|u_{r_{0},f}\|_{s,r_{0}}\leq C\|f\|_{L^{p^{*}}(B_{1})},$

where $p^{*}=\frac{2N}{N+2s}$ and $C>0$ is independent of $r_{0}$. Note that
$r_{0}\to u_{r_{0},f}$ is increasing, then passing to the limit as $r_{0}\to
1^{-}$, there exists $u_{f}$ such that $u_{r_{0},f}\to u_{f}$ as $r_{0}\to
1^{-}$ weakly in $H^{s}_{0}(B_{1})$, strong in $L^{p}(B_{1})$ and a.e. in
$B_{1}$ for $p\in[2,2^{*}_{s})$. From (4.4), we have that

$\langle u_{f},\xi\rangle_{s}=\int_{B_{1}}f\xi dx,\quad\forall\xi\in
H^{s}_{0}(B_{1})$ (4.8)

and for $\xi\in C^{2s+\theta}_{loc}(B_{1})\cap C^{s}_{c}(B_{1})$

$\int_{B_{1}}u_{f}\big{(}(-\Delta)^{s}_{B_{1}}\xi+\xi\big{)}dx=\int_{B_{1}}f\xi
dx.$ (4.9)

Moreover, $u_{f}$ inherits the positivity, the symmetry property and
decreasing monotonicity of $u_{r_{0},f}$, then $u_{f}$ is locally bounded in
$B_{1}\setminus\\{0\\}$.

Now we show $u_{f}\in L^{\infty}(B_{1})$.  Let $t>1$ be a parameter and

$w_{t}=(u_{f}-t)_{+}\quad{\rm in}\ \,B_{1},$

which is $H^{s}_{0}(B_{1})$. For $\sigma\in(0,\frac{1}{2}]$ we have that

$\displaystyle\int_{B_{1}}fw_{t}dx$ $\displaystyle=$ $\displaystyle\langle
u_{f},w_{t}\rangle_{s}$ $\displaystyle=$ $\displaystyle\int\int_{\\{x,y\in
B_{1}:|x-y|<\sigma\\}}\frac{\big{(}u_{f}(x)-u_{f}(y)\big{)}\big{(}w_{t}(x)-w_{t}(y)\big{)}}{|x-y|^{N+2s}}dxdy$
$\displaystyle+\int\int_{\\{x,y\in
B_{1}:|x-y|\geq\sigma\\}}\frac{\big{(}u_{f}(x)-u_{f}(y)\big{)}\big{(}w_{t}(x)-w_{t}(y)\big{)}}{|x-y|^{N+2s}}dxdy$
$\displaystyle\geq$ $\displaystyle\int\int_{\\{x,y\in
B_{1}:|x-y|<\sigma\\}}\frac{\big{(}w_{t}(x)-w_{t}(y)\big{)}^{2}}{|x-y|^{N+2s}}dxdy$
$\displaystyle\,+2c_{1}\Big{(}\sigma^{-2s}\int_{B_{1}}u_{f}w_{t}dx-\int_{B_{1}}(\kappa_{\sigma}\ast
u_{f})w_{t}dx\Big{)}$ $\displaystyle:=$
$\displaystyle\mathcal{F}_{\sigma}+2c_{1}(\mathcal{E}_{1,\sigma}-\mathcal{E}_{2,\sigma})\quad{\rm
resp.},$

where $c_{1}=\frac{\omega_{N}}{2s}$,

$\kappa_{\sigma}=\chi_{\mathbb{R}^{N}\setminus B_{\sigma}}|\cdot|^{-N-2s}$

and we used the fact that for $x,y\in B_{1}$,

$\displaystyle\big{(}u_{f}(x)-u_{f}(y)\big{)}\big{(}w_{t}(x)-w_{t}(y)\big{)}$
$\displaystyle=$
$\displaystyle\big{(}(u_{f}(x)-t)-(u_{f}(y)-t)\big{)}\big{(}w_{t}(x)-w_{t}(y)\big{)}$
$\displaystyle=$
$\displaystyle(u_{f}(x)-t)w_{t}(x)+(u_{f}(y)-t)w_{t}(y)-(u_{f}(x)-t)w_{t}(y)-(u_{f}(y)-t)w_{t}(x)$
$\displaystyle=$ $\displaystyle
w_{t}^{2}(x)+w_{t}^{2}(y)-2w_{t}(x)w_{t}(y)+(u_{f}(x)-t)_{-}w_{t}(x)+(u_{f}(y)-t)_{-}w_{t}(y)$
$\displaystyle\geq$ $\displaystyle\big{(}w_{t}(x)-w_{t}(y)\big{)}^{2}$

with $t_{-}=\max\\{-t,0\\}$.

Direct computations show that

$\displaystyle\|\kappa_{\sigma}\ast u_{f}\|_{L^{\infty}(B_{1})}$
$\displaystyle\leq$
$\displaystyle\|u_{f}\|_{L^{2^{*}_{s}}(B_{1})}\big{(}\int_{\mathbb{R}^{N}\setminus
B_{\sigma}}|y|^{-(N+2s)p^{*}}ds\big{)}^{\frac{1}{p^{*}}}$ $\displaystyle\leq$
$\displaystyle c_{19}\|f\|_{L^{p^{*}}(B_{1})}\sigma^{-\frac{N+2s}{2}}$
$\displaystyle\leq$ $\displaystyle
c_{20}\|f\|_{L^{\infty}(B_{1})}\sigma^{-\frac{N+2s}{2}},$

which implies that

$\mathcal{E}_{2,\sigma}\leq
c_{20}\|f\|_{L^{\infty}(B_{1})}\sigma^{-\frac{N+2s}{2}}\int_{B_{1}}w_{t}dx.$

Moreover,

$\int_{B_{1}}fw_{t}dx\leq\|f\|_{L^{\infty}(B_{1})}\int_{B_{1}}w_{t}dx,\qquad\mathcal{E}_{1,\sigma}\geq
c_{21}\sigma^{-2s}t\int_{B_{1}}w_{t}dx.$

As a consequence, if $t\geq t_{0}$ for some $t_{0}>0$ large enough, we obtain
that

$\mathcal{F}_{\sigma}\leq\big{(}\|f\|_{L^{\infty}(B_{1})}(c\sigma^{-\frac{N+2s}{2}}+1)-c_{21}\sigma^{-2s}t\big{)}\int_{B_{1}}w_{t}dx\leq
0,$

where

$\|f\|_{L^{\infty}(B_{1})}(c\sigma^{-\frac{N+2s}{2}}+1)-c_{21}\sigma^{-2s}t<0\quad{\rm
if}\ \,t\ {\rm is\ large}.$

So we have that $w_{t}=0$ a.e. in $\Omega$, which means $u_{f}\leq t_{0}$. Now
the $L^{\infty}$ bound of $u_{f}$ is proved.

Note that for $\mathcal{O}\subset\bar{\mathcal{O}}\subset B_{1}$, then for
$r_{0}$ close to 1 such that $\mathcal{O}\subset\bar{\mathcal{O}}\subset
B_{r_{0}}$,

$\displaystyle\|u_{r_{0},f}\|_{C^{\gamma}(\mathcal{O})}$ $\displaystyle\leq$
$\displaystyle
c_{22}\left[\|u_{f}\|_{L^{\infty}(B_{1})}+\|u_{r_{0},f}\|_{L^{1}(B_{r_{0}})}+\|f\|_{L^{\infty}(B_{r_{0}})}\right]$
$\displaystyle\leq$ $\displaystyle
c_{23}\big{(}\|f\|_{L^{\infty}(B_{1})}+\|u_{f}\|_{L^{\infty}(B_{r_{0}}\setminus
B_{\frac{\epsilon}{2}})}\big{)}$

and by (2.11), we have that

$\displaystyle\|u_{r_{0},f}\|_{C^{2s+\epsilon^{\prime}}(\mathcal{O})}$
$\displaystyle\leq$ $\displaystyle
c_{24}\left[\|u_{f}\|_{L^{\infty}(B_{1})}+\|w\|_{L^{1}(B_{r_{0}})}+\|f\|_{C^{\theta}(B_{r_{0}})}\right]$
$\displaystyle\leq$ $\displaystyle c_{25}\|f\|_{C^{\theta}(B_{r_{0}})}.$

Then $u_{r_{0},f}\to u_{f}$ as $r_{0}\to 1^{-}$ locally in
$C^{2s+\epsilon^{\prime\prime}}(B_{1})$. By the stability results, Theorem
2.2, we obtain that $u_{f}\in C^{2s+\epsilon}(B_{1})$ and it verifies that

$(-\Delta)^{s}_{B_{1}}u_{f}+u_{f}=f\quad{\rm in}\ \,B_{1}.$

Until now, we don’t show boundary condition of $u_{f}$.

Particular case that $f=1$. Let

$t_{0}=\lim_{|x|\to 1^{-}}u_{f=1}(x)\geq 0.$

Let

$\tilde{u}_{1}(x)=u_{f=1}-t_{0},$

which is a solution of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=1+t_{0}&{\rm
in}\ \,B_{1},\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}_{B_{1}}+u}\displaystyle u=0&{\rm on}\ \partial
B_{1}.\end{array}\right.$

Let

$u_{1}(x)=\frac{1}{1+t_{0}}\tilde{u}_{1}(x)\quad{\rm for}\ \,x\in B_{1},$

then it is a solution of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=1&{\rm in}\
\,B_{1},\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}_{B_{1}}+u}\displaystyle
u=0&{\rm on}\ \partial B_{1}.\end{array}\right.$ (4.10)

Now we use $\|f\|_{L^{\infty}}u_{1}$ as an upper bound of $u_{r_{0},f}$, which
implies that $u_{f}$ decays at the boundary.

Case 2: $f\in C^{\theta}(\overline{B_{1}})$.  Note that
$\|f\|_{L^{\infty}}u_{1}$ is the upper bound for (4.3), the limit of
$u_{r_{0},f}$ is controlled by $\|f\|_{L^{\infty}}u_{1}$. We obtain that
$u_{r_{0},f}\to u_{f}$ in $C^{2s+\epsilon^{\prime}}_{loc}(B_{1})\cap
C_{0}(B_{1})$ and $H^{1}_{0}(B_{1})$, then $u_{f}$ is a classical solution of
(4.1). Moreover, passing to the limit of (4.4), we obtain (4.2). It is worth
noting that $\tilde{u}_{1}=u_{f=1}=u_{1}$ in $B_{1}$, which is the limit of
$u_{r_{0},1}$ as $r_{0}\to 1^{-}$. This means $t_{0}=0$. $\Box$

From the proof of Theorem 4.1, we have the following formula.

###### Corollary 4.2

Let $f_{i}\in C^{\theta}(\bar{B}_{1})$ with $i=1,2$ and $u_{i}$ be the
solutions of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=f_{i}&{\rm
in}\ \,B_{1},\\\\[4.2679pt] \phantom{(-\Delta)^{\alpha}+u\,}\displaystyle
u=0&{\rm on}\ \partial B_{1}.\end{array}\right.$

respectively. Then

$\int_{B_{1}}u_{1}\big{(}(-\Delta)^{s}_{B_{1}}u_{2}+u_{2}\big{)}dx=\int_{B_{1}}f_{1}u_{2}dx\quad\
{\rm and}\quad\
\int_{B_{1}}u_{2}\big{(}(-\Delta)^{s}_{B_{1}}u_{1}+u_{1}\big{)}dx=\int_{B_{1}}f_{1}u_{2}dx.$

###### Lemma 4.2

Let $s\in(0,\frac{1}{2}]$, $f\geq 0$ be nontrivial and $u_{f}$ be the unique
solution of (4.1), then for any $k\in(0,s)$,

$\lim_{|x|\to 1^{-}}u_{f}(x)(1-|x|)^{-k}=+\infty.$

or for any $k\in(0,2s)$, there exists $a_{k}>0$ such that

$u_{f}(x)\geq a_{k}(1-|x|)^{k}\quad{\rm in}\ \ B_{1}.$

Proof.  For $\tau\in(0,1)$, let

$V_{\tau}(x)=(1-|x|)^{\tau}\quad{\rm for}\ \,x\in B_{1}.$

Similar proof of [11, Proposition 2.1], we can prove that form any
$\tau\in(0,1)$, there exist $\delta_{1}\in(0,\frac{1}{4})$ and $C_{\tau}>1$
such that

$\frac{1}{C_{\tau}}(1-|x|)^{\tau-2s}\leq-(-\Delta)^{s}_{B_{1}}V_{\tau}(x)\leq
C_{\tau}(1-|x|)^{\tau-2s}\ \ \mbox{for all}\,\,x\in B_{1}\setminus
B_{1-\delta_{1}}.$ (4.11)

We address the proof in the appendix.

Since $f\geq 0$ is nontrivial, for above $\delta_{1}>0$ $u_{f}\geq C>0$ n
$\bar{B}_{1-\delta_{1}}$ and there is $c_{26}>0$ such that $c_{26}V_{\tau}\leq
u_{f}$ in $\bar{B}_{1-\delta_{1}}$, then it follows by the comparison
principle that

$u_{f}(x)\geq c_{26}V_{\tau}(x)\ \ \mbox{for all}\,\,x\in\bar{B}_{1}\setminus
B_{1-\delta_{1}}.$

The proof is complete. $\Box$

Proof of Proposition 1.1.  It follows by Theorem 4.1 with $f\equiv 1$.

By Lemma 4.2 we obtain that for any $k\in(0,s)$,

$\lim_{|x|\to 1^{-}}u_{1}(x)(1-|x|)^{-k}=+\infty.$

We complete the proof. $\Box$

##  5 The Schrödinger equation

Under the assumption of Theorem 1.1, the Schrödinger equation (1.1) could be
written as

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=h_{1}u^{p}+\epsilon
h_{2}&{\rm in}\ \,B_{1},\\\\[5.69054pt]
\phantom{(-\Delta)^{\alpha}_{B_{1}}+u}\displaystyle u=0&{\rm on}\ \partial
B_{1},\end{array}\right.$ (5.1)

where $p>1$ and $\epsilon>0$.

Proof of Theorem 1.1.  Recall that $u_{h_{2}}$ is the unique solution of
(4.3), we first define the iterating sequence

$v_{0}:=\epsilon u_{h_{2}}>0,$

when

$f=\epsilon h_{2},$

and by Theorem 4.1, $v_{n}$ is the unique solution of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=h_{1}v_{n-1}^{p}+\epsilon
h_{2}&{\rm in}\ \,B_{1},\\\\[5.69054pt]
\phantom{(-\Delta)^{s}_{B_{1}}+u}\displaystyle u=0&{\rm on}\ \partial
B_{1}.\end{array}\right.$ (5.2)

By the comparison principle, we have that $v_{1}>v_{0}$ and assuming that

$v_{n-1}\geq v_{n-2}\quad{\rm in}\quad B_{1},$

then

$\displaystyle(-\Delta)^{s}_{B_{1}}(v_{n}-v_{n-1})+(v_{n}-v_{n-1})=h_{1}(v_{n-1}^{p}-v_{n-2}^{p})\geq
0\quad{\rm in}\ \,B_{1}$

and $v_{n}-v_{n-1}=0\quad{\rm on}\ \,\partial B_{1},$ which imply that
$v_{n}\geq v_{n-1}$.

Thus the sequence $\\{v_{n}\\}_{n}$ is increasing with respect to $n$.

We next build an upper bound for the sequence $\\{v_{n}\\}_{n}$. For $t>0$,
denote

$w_{t}=tu_{1},$

then

$\displaystyle(-\Delta)^{s}_{B_{1}}w_{t}+w_{t}-h_{1}w_{t}^{p}$
$\displaystyle=$ $\displaystyle t-t^{p}h_{1}u_{1}^{p}$ $\displaystyle\geq$
$\displaystyle
t\Big{(}1-t^{p-1}\|h_{1}^{\frac{1}{p}}u_{1}\|_{L^{\infty}(B_{1})}^{p}\Big{)}$

and letting

$L(t)=t-t^{p}\|h_{1}^{\frac{1}{p}}u_{1}\|_{L^{\infty}(B_{1})}^{p},$

$L(\cdot)$ has maximum
$\frac{p-1}{p}\big{(}p\|h_{1}^{\frac{1}{p}}u_{1}\|_{L^{\infty}(B_{1})}^{p}\big{)}^{-\frac{1}{p-1}}$
in at
$t_{p}=(p\|h_{1}^{\frac{1}{p}}u_{1}\|_{L^{\infty}(B_{1})})^{-\frac{1}{p-1}}$.

In order to find the upper solution, we take $t=t_{p}$ and if

$\epsilon\|h_{2}\|_{L^{\infty}(B_{1})}\leq\frac{p-1}{p}\big{(}p\|u_{1}\|_{L^{\infty}(B_{1})}^{p}\big{)}^{-\frac{1}{p-1}}.$
(5.3)

then

$(-\Delta)^{s}_{B_{1}}w_{t_{p}}+w_{t_{p}}\geq h_{1}w_{t_{p}}^{p}+\epsilon
h_{2}.$ (5.4)

Note that (5.3) holds if

$\epsilon\leq\epsilon_{p}:=\frac{p-1}{p}\big{(}p\|h_{1}^{\frac{1}{p}}u_{1}\|_{L^{\infty}(B_{1})}^{p}\big{)}^{-\frac{1}{p-1}}\|h_{2}\|_{L^{\infty}(B_{1})}^{-1}$

Obviously, we have that $w_{t_{p}}\geq v_{0}$. Inductively, we obtain

$v_{n}\leq w_{t_{p}}$ (5.5)

for all $n\in\mathbb{N}$. Therefore, the sequence $\\{v_{n}\\}_{n}$ converges.
Let $u_{\epsilon}:=\lim_{n\to\infty}v_{n}$. By the regularity results,
$u_{\epsilon}$ is a solution of (5.1).

We claim that $u_{\epsilon}$ is the minimal solution of (1.1), that is, for
any nonnegative solution $u$ of (1.3), we always have $u_{\epsilon}\leq u$.
Indeed, there holds

$(-\Delta)^{s}_{B_{1}}u+u=h_{1}u^{p}+\epsilon
h_{2}\geq(-\Delta)^{s}_{B_{1}}v_{0}+v_{0}\quad{\rm in}\ \,B_{1},\qquad
u_{\epsilon}=u\quad{\rm on}\ \,\partial B_{1}$

then $u\geq v_{0}$,

$(-\Delta)^{s}_{B_{1}}u+u=h_{1}u^{p}+\epsilon h_{2}\geq
h_{1}v_{0}^{p}+\epsilon h_{2}=(-\Delta)^{s}_{B_{1}}v_{1}+v_{1}\quad{\rm in}\
\,B_{1},\qquad u_{\epsilon}=u\quad{\rm on}\ \,\partial B_{1}$

which implies that $u\geq v_{1}$ in $B_{1}$. We may show inductively that

$u\geq v_{n}$

for all $n\in\mathbb{N}$. The claim follows.

Similarly, if problem (5.1) has a nonnegative solution $u$ for
$\epsilon_{1}>0$, then (5.1) admits a minimal solution $u_{\epsilon}$ for all
$\epsilon\in(0,\epsilon_{1}]$. As a result, the mapping $\epsilon\mapsto
u_{\epsilon}$ is increasing. So we may define

$\epsilon^{*}=\sup\big{\\{}\epsilon>0:\ (\ref{eq 4.1-S})\ {\rm has\ minimal\
solution\ for\ }\epsilon\big{\\}}$

and we have that

$\epsilon^{*}\geq\epsilon_{p}.$

Finally, we prove that $\epsilon^{*}<+\infty$. Assume that (5.1) has a
positive solution for $\epsilon>0$. Our above proof shows that (5.1) has a
minimal solution $u_{\epsilon}$. Recall that $u_{h_{1}}$ be the solution of

$\left\\{\begin{array}[]{lll}\displaystyle(-\Delta)^{s}_{B_{1}}u+u=h_{1}&{\rm
in}\ \,B_{1},\\\\[5.69054pt] \phantom{(-\Delta)^{\alpha}+u\,}\displaystyle
u=0&{\rm on}\ \partial B_{1}.\end{array}\right.$

Use $u_{h_{1}}$ as test function, we have that

$\displaystyle\int_{B_{1}}u_{\epsilon}^{p}h_{1}u_{h_{1}}dx+\epsilon\int_{B_{1}}h_{2}u_{h_{1}}dx$
$\displaystyle=$
$\displaystyle\int_{B_{1}}[(-\Delta)^{s}_{B_{1}}u_{\epsilon}+u_{\epsilon}]u_{h_{1}}dx$
$\displaystyle=$
$\displaystyle\int_{B_{1}}u_{\epsilon}[(-\Delta)^{s}_{B_{1}}u_{h_{1}}+u_{h_{1}}]dx$
$\displaystyle=$ $\displaystyle\int_{B_{1}}u_{\epsilon}h_{1}dx$
$\displaystyle\leq$
$\displaystyle\Big{(}\int_{B_{1}}u_{\epsilon}^{p}h_{1}u_{h_{1}}\,dx\Big{)}^{\frac{1}{p}}\Big{(}\int_{B_{1}}h_{1}(u_{h_{1}})^{-\frac{1}{p-1}}\,dx\Big{)}^{1-\frac{1}{p}}$
$\displaystyle\leq$ $\displaystyle
c_{0}\Big{(}\int_{B_{1}}u_{\epsilon}^{p}h_{1}u_{h_{1}}\,dx\Big{)}^{\frac{1}{p}},$

where from Lemma 4.2 with $k\in(0,\frac{1}{p-1})$

$\displaystyle c_{0}$ $\displaystyle:=$
$\displaystyle\big{(}\int_{B_{1}}h_{1}(u_{h_{1}})^{-\frac{1}{p-1}}\,dx\big{)}^{1-\frac{1}{p}}$
$\displaystyle\leq$
$\displaystyle\|h_{1}\|_{L^{\infty}(B_{1})}^{1-\frac{1}{p}}\big{(}a_{k}\int_{B_{1}}(1-|x|)^{-\frac{1}{p-1}k}\,dx\big{)}^{1-\frac{1}{p}}<+\infty.$

Thus, we have that

$\int_{B_{1}}u_{\epsilon}^{p}h_{1}u_{h_{1}}\,dx\leq c_{0}^{\frac{p}{p-1}}$

and we have that

$\epsilon\leq\frac{c_{0}^{\frac{p}{p-1}}}{\int_{B_{1}}h_{2}u_{h_{1}}dx}=\frac{\int_{B_{1}}(h_{1}u_{h_{1}})^{-\frac{1}{p-1}}\,dx}{\int_{B_{1}}h_{2}u_{h_{1}}dx},$
(5.6)

which means

$\epsilon^{*}\leq\frac{\int_{B_{1}}(h_{1}u_{h_{1}})^{-\frac{1}{p-1}}\,dx}{\int_{B_{1}}h_{2}u_{h_{1}}dx}<+\infty.$

Finally, the properties of the radial symmetry and decreasing monotonicity
follows by Theorem 1.2 with $h_{3}(t)=t^{p}$. The proof ends. $\Box$

##  Appendix A Appendix: Proof of (4.11)

Proof of (4.11).  By compactness we prove that the corresponding inequality
holds in a neighborhood of any point $\bar{x}\in\partial B_{1}(e_{1})$ with
$e_{1}:=(1,0,\cdots,0)\in\mathbb{R}^{N}$, and then without loss of generality
we set $\bar{x}=0$ and we note that $-e_{1}$ is the outer normal vector $0$.

For a given $0<\eta\leq\delta$, we define

$Q_{\eta}=\\{y=(y_{1},y^{\prime})\in\mathbb{R}\times\mathbb{R}^{N-1},\,0<y_{1}<\eta,\,|y^{\prime}|<\eta\\}$

and

$\tilde{Q}_{\eta}=\\{y=(y_{1},y^{\prime})\in\mathbb{R}\times\mathbb{R}^{N-1},\,|y_{1}|<\eta,\,|y^{\prime}|<\eta\\}.$

There is a $C^{2}$ function $\varphi:\mathbb{R}^{N-1}\to\mathbb{R}$ such that
$(z_{1},z^{\prime})\in B_{1}(e_{1})\cap\tilde{Q}_{2\eta}$ if and only if
$z_{1}=\varphi(z^{\prime})=1-\sqrt{1-|z^{\prime}|^{2}}$ for
$|z^{\prime}|<2\eta$ for $\eta>0$ small enough. In fact, we take $\eta>0$
small enough, then for any $z\in A_{t}:=\\{x\in B_{1}(e_{1}):\,\rho(x)=t\\}$
with $t\in(0,\eta]$, there exists a unique point $z_{0}\in\partial
B_{1}(e_{1})$ such that $|z-z_{0}|=t$, $z-z_{0}$ is a normal vector at $z_{0}$
and $A_{t}$ is $C^{2}$, where $\rho(x)=1-|x|$.

Let $\Phi:\mathbb{R}^{N}\to\mathbb{R}^{N}$ be a $C^{2}$ diffeomorphism such
that

$\Phi(z)=(z^{\prime},z_{1}+\varphi(z^{\prime}))\quad{\rm for}\
\,z=(z_{1},\,z^{\prime})\in Q_{\eta}.$

Note that

$\Phi(te_{1})=te_{1}\ \ {\rm for}\
\,t\in(0,\eta),\quad\quad\Phi(0,y^{\prime})\in\partial
B_{1}(e_{1})\cap\tilde{Q}_{\eta}\quad{\rm for}\ \,|y^{\prime}|<\eta$

and

$\tilde{Q}_{\frac{\eta}{2}}\cap
B_{1}(e_{1})\subset\Phi(Q_{\eta})\subset\tilde{Q}_{2\eta}\cap B_{1}(e_{1}).$

Moreover, $\rho(\Phi(y))=y_{1}+O(|y^{\prime}|^{2})$ for $y\in Q_{\eta}$ and
there exists $c_{6}>0$ such that

$\Phi(0)=0,\ \ |D\Phi(0)-E_{1}|\leq c_{6}|z^{\prime}|\quad{\rm
and}\quad|\Phi(y)-y|\leq c_{6}|y^{\prime}|^{2}\quad{\rm for}\
y=(y_{1},y^{\prime})\in Q_{\eta},$

where $E_{1}=(d_{ij})_{N\times N}$ is the unit matrix, i.e. $d_{ii}=1$ and
$d_{ij}=0$ if $i\not=j$.

Let $x_{1}\in(0,\eta/4)$, $x=(x_{1},0)$ be a generic point in $A_{\eta/4}$. We
observe that $|x-\bar{x}|=\rho(x)=x_{1}$. By definition we have

$\displaystyle(-\Delta)^{s}_{B_{1}(e_{1})}V_{\tau}(x)=c_{N,s}{\rm
p.v.}\int_{\Phi(Q_{\eta})}\frac{x_{1}^{\tau}-\rho^{\tau}(z)}{|x-z|^{N+2s}}dz+c_{N,s}\int_{B_{1}(e_{1})\setminus\Phi(Q_{\eta})}\frac{V_{\tau}(x)-V_{\tau}(z)}{|x-z|^{N+2s}}dz$
(A.1)

and we see that

$|\int_{B_{1}(e_{1})\setminus\Phi(Q_{\eta})}\frac{V_{\tau}(x)-V_{\tau}(z)}{|x-z|^{N+2s}}dz|\leq
c_{7}(x_{1}^{\tau}+1),$ (A.2)

where the constant $c_{7}$ is independent of $x$. Thus we only need to study
the asymptotic behavior of the first integral, that from now on, we denote by

$\mathbb{E}(x_{1})={\rm
p.v.}\int_{\Phi(Q_{\eta})}\frac{x_{1}^{\tau}-\rho^{\tau}(z)}{|x-z|^{N+2s}}dz.$

Let $z=\Phi(y)$, then we have that

$\mathbb{E}(x_{1})={\rm
p.v.}\int_{Q_{\eta}}\frac{x_{1}^{\tau}-y_{1}^{\tau}}{|x-\Phi(y)|^{N+2s}}|D\Phi(y)|dy+\int_{Q_{\eta}}\frac{O(|y^{\prime}|^{2})}{|x-\Phi(y)|^{N+2s}}|D\Phi(y)|dy.$

Observe that

$\displaystyle|D\Phi(y)|=1+O(|y^{\prime}|),$ (A.3)
$\displaystyle|x-\Phi(y)|^{2}=|x-y+y-\Phi(y)|^{2}=|x_{1}-y_{1}|^{2}+|y^{\prime}|^{2}+O(|y^{\prime}|^{4})$
(A.4)

and

$\displaystyle\Big{|}\int_{Q_{\eta}}\frac{O(|y^{\prime}|^{2})}{|x-\Phi(y)|^{N+2s}}|D\Phi(y)|dy\Big{|}\leq
c_{8},$ (A.5)

where we use the fact that $|y-\Phi(y)|=O(|y^{\prime}|^{2})$. Then we have
that

$\displaystyle\tilde{\mathbb{E}}(x_{1})$ $\displaystyle:=$ $\displaystyle{\rm
p.v.}\int_{Q_{\eta}}\frac{x_{1}^{\tau}-y_{1}^{\tau}}{\left(|x_{1}-y_{1}|^{2}+|y^{\prime}|^{2}+O(|y^{\prime}|^{4})\right)^{\frac{N+2s}{2}}}(1+O(|y^{\prime}|))dy$
$\displaystyle=$ $\displaystyle{\rm
p.v.}\int_{0}^{\eta}(x_{1}^{\tau}-y_{1}^{\tau})\int_{B^{N-1}_{\eta}(0)}\frac{(1+O(|y^{\prime}|))}{\left(|x_{1}-y_{1}|^{2}+|y^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dy^{\prime}dy_{1}$
$\displaystyle=$ $\displaystyle{\rm
p.v.}\int_{0}^{\eta}\frac{x_{1}^{\tau}-y_{1}^{\tau}}{|x_{1}-y_{1}|^{1+2s}}\int_{B^{N-1}_{\frac{\eta}{|x_{1}-y_{1}|}(0)}}\frac{(1+O(|x_{1}-y_{1}||z^{\prime}|))}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}dy_{1}$
$\displaystyle=$ $\displaystyle{\rm
p.v.}\int_{0}^{\eta}\frac{x_{1}^{\tau}-y_{1}^{\tau}}{|x_{1}-y_{1}|^{1+2s}}F_{s}(|x_{1}-y_{1}|)dy_{1},$

where

$\displaystyle F_{s}(|x_{1}-y_{1}|)$ $\displaystyle=$
$\displaystyle\int_{B^{N-1}_{\frac{\eta}{|x_{1}-y_{1}|}(0)}}\frac{(1+O(|x_{1}-y_{1}||z^{\prime}|))}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}$
$\displaystyle=$
$\displaystyle\int_{B^{N-1}_{\frac{\eta}{|x_{1}-y_{1}|}(0)}}\frac{1}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}+O(|x_{1}-y_{1}|)\int_{B^{N-1}_{\frac{\eta}{|x_{1}-y_{1}|}(0)}}\frac{|z^{\prime}|}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N-1}}\frac{1}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}+O(|x_{1}-y_{1}|^{1+2s})+\int_{\mathbb{R}^{N-1}}\frac{|z^{\prime}|}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}O(|x_{1}-y_{1}|)$
$\displaystyle=$ $\displaystyle d_{s}+O(|x_{1}-y_{1}|),$ (A.7)

with
$d_{s}=\int_{\mathbb{R}^{N-1}}\frac{1}{\left(1+|z^{\prime}|^{2})\right)^{\frac{N+2s}{2}}}dz^{\prime}>0$
and $B^{N-1}_{r}(0)$ being the ball with center at the origin and radius $r$
in $\mathbb{R}^{N-1}$. Thus we have that

$\displaystyle\mathbb{E}(x_{1})$ $\displaystyle=$ $\displaystyle d_{s}\,{\rm
p.v.}\int_{0}^{\eta}\frac{x_{1}^{\tau}-y_{1}^{\tau}}{|x_{1}-y_{1}|^{1+2s}}dy_{1}+O\Big{(}{\rm
p.v.}\int_{0}^{\eta}\frac{x_{1}^{\tau}-y_{1}^{\tau}}{|x_{1}-y_{1}|^{2s}}dy_{1}\Big{)}+O(1)$
(A.8) $\displaystyle=$ $\displaystyle d_{s}\left(x_{1}^{\tau-2\alpha}{\rm
p.v.}\int_{0}^{\frac{\eta}{x_{1}}}\frac{1-t^{\tau}}{|1-t|^{1+2s}}dt+O(x_{1}^{\tau-2s+1}{\rm
p.v.}\int_{0}^{\frac{\eta}{x_{1}}}\frac{1-t^{\tau}}{|1-t|^{2s}}dt)\right)+O(1)$
$\displaystyle=$ $\displaystyle
d_{\alpha}\Big{[}x_{1}^{\tau-2s}\left(c_{s}(\tau)-(\frac{\eta}{x_{1}})^{\tau-2s}\right)+O(x_{1}^{\tau-2s+1}\left(c_{s}(\tau)-(\frac{\eta}{x_{1}})^{\tau-2s+1}\right)\Big{]}+O(1)$
$\displaystyle=$ $\displaystyle
d_{s}c_{s}(\tau)x_{1}^{\tau-2s}+O(x_{1}^{\tau-2s+1})+O(1).$

Therefore, we conclude from (A.2) and (A.8) that

$\displaystyle(-\Delta)^{s}_{B_{1}(e_{1})}V_{\tau}(x)=d_{s}c_{s}(\tau)x_{1}^{\tau-2s}+O(x_{1}^{\tau})+O(x_{1}^{\tau-2s+1})+O(1)\quad{\rm
as}\ \ {x_{1}\to 0^{+}}.$

From [11, Lemma 2.1], $c_{s}(\tau)<0$ for $\tau\in(0,2s)$ and then (4.11) is
proved. $\Box$

##  Appendix B  Potential estimates

For $r_{0}>0$, let $\Phi_{r_{0}}$ be the Green kernel of $(-\Delta)^{s}$ in
$B_{r_{0}}$ with the zero Dirichlet boundary condition in
$\mathbb{R}^{2N}\setminus(\Omega\times\Omega)$, then

$\Phi_{r_{0}}(x,y)\leq c_{28}|x-y|^{2s-N}$ (B.9)

for some $c_{28}>0$ independent of $r_{0}$.

###### Lemma B.1

Assume that $s\in(0,1)$ and integer $N\geq 2$.
$(i)$ if

$\frac{1}{q}<\frac{2s}{N},$

then there exists some $C>0$ such that

$\|\Phi_{r_{0}}\ast h\|_{\infty}\leq c_{29}\|h\|_{q};$ (B.10)

$(ii)$ if

$\frac{1}{q}\leq\frac{1}{r}+\frac{2s}{N}\quad\rm{and}\quad q>1,$

then there exists some $c_{30}>0$ such that

$\|\Phi_{r_{0}}\ast h\|_{r}\leq c_{30}\|h\|_{q}.$ (B.11)

$(iii)$ if

$1<\frac{1}{r}+\frac{2s}{N},$

then there exists some $c_{31}>0$ such that

$\|\Phi_{r_{0}}\ast h\|_{r}\leq c_{31}\|h\|_{1}.$ (B.12)

Proof. Proof of (B.10). By the Hölder inequality and (B.9), we infer that for
any $x\in\Omega$,

$\displaystyle\|\Phi_{r_{0}}\ast h\|_{\infty}$ $\displaystyle\leq$
$\displaystyle\Big{\|}\Big{(}\int_{B_{r_{0}}}\Phi_{r_{0}}^{q^{\prime}}dy\Big{)}^{\frac{1}{q^{\prime}}}\Big{(}\int_{B_{r_{0}}(x)}|h(y)|^{q}dy\Big{)}^{\frac{1}{q}}\Big{\|}_{L^{\infty}(\Omega)}$
$\displaystyle\leq$ $\displaystyle
c_{32}\|h\|_{q}\Big{(}\int_{B_{r_{0}}(x)}\frac{1}{|x-y|^{(N-2s)q^{\prime}}}dy\Big{)}^{\frac{1}{q}}$
$\displaystyle\leq$ $\displaystyle c_{33}\|h\|_{q},$

where $q^{\prime}=\frac{q}{q-1}$, thanks to the assumption
$\frac{1}{q}<\frac{2s}{N}$, $(N-2s)q^{\prime}<N$ and
$\int_{B_{r_{0}}(x)}\frac{1}{|x-y|^{(N-2s)q^{\prime}}}dy<+\infty.$

Proofs of (B.11) and (B.12) with $r\leq q$. By using the integral Minkowski’s
inequality we have that

$\displaystyle\|(\Phi_{r_{0}}\ast h\|_{r}$ $\displaystyle=$
$\displaystyle\|h\ast\Phi_{r_{0}}\|_{r}$ $\displaystyle\leq$ $\displaystyle
c_{34}\Big{[}\int_{\mathbb{R}^{N}}\Big{(}\int_{B_{r_{0}}}\frac{|h|(x-y)\chi_{{}_{B_{2}(0)}}(y)}{|y|^{N-2s}}dx\Big{)}^{r}dy\Big{]}^{\frac{1}{r}}$
$\displaystyle\leq$ $\displaystyle
c_{35}\Big{[}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|h(x-y)|^{r}\chi_{{}_{B_{2}(0)}}(y)}{|y|^{(N-2s)r}}dxdy\Big{]}^{\frac{1}{r}}$
$\displaystyle\leq$ $\displaystyle
c_{35}\Big{[}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}|h(x-y)|^{r}dx\frac{\chi_{{}_{B_{2}(0)}}(y)}{|y|^{N-2s}}dy\Big{]}^{\frac{1}{r}}$
$\displaystyle\leq$ $\displaystyle c_{36}\|h\|_{L^{r}(\mathbb{R}^{N})}.$

Part 3. To prove (B.11) and (B.12) with $r>q\geq 1$ and
$\frac{1}{q}\leq\frac{1}{r}+\frac{2s}{N}$. We claim that if $r>s$ and
$\frac{1}{r^{*}}=\frac{1}{q}-\frac{2s}{N}$, the mapping
$h\to(\Phi_{r_{0}}\eta_{0})\ast h$ is of weak-type $(q,r^{*})$, in the sense
that

$|\\{x\in\mathbb{R}^{N}:|(\Phi_{r_{0}}\eta_{0})\ast
h(x)|>t\\}|\leq\Big{(}A_{q,r^{*}}\frac{\|h\|_{L^{q}(\Omega)}}{t}\Big{)}^{r^{*}},\quad
h\in L^{q}(B_{r_{0}}),\quad\rm{all}\ \ t>0,$ (B.13)

where constant $A_{q,r^{*}}>0$.

Denote for $\nu>0$,

$G_{1}=\Phi_{r_{0}}\eta_{0}\chi_{{}_{B_{\nu}}}\quad{\rm and}\quad
G_{2}=\Phi_{r_{0}}\eta_{0}\chi_{{}_{B_{\nu}^{c}}}.$

Then we have that

$|\\{x\in B_{r_{0}}:|(\Phi_{r_{0}}\eta_{0})\ast
h(x)|>2t\\}|\leq|\\{x\in\mathbb{R}^{N}:|G_{1}\ast
h(x)|>t\\}|+|\\{x\in\mathbb{R}^{N}:|G_{2}\ast h(x)|>t\\}|.$

By Part 2 and the integral Minkowski’s inequality, we have

$\displaystyle|\\{x\in\mathbb{R}^{N}:|G_{1}\ast h(x)|>t\\}|$
$\displaystyle\leq$ $\displaystyle\frac{\|G_{1}\ast
h\|^{s}_{s}}{t^{s}}=\frac{\|h\ast G_{1}\|^{s}_{s}}{t^{s}}$ $\displaystyle\leq$
$\displaystyle\frac{[\int_{\mathbb{R}^{N}}(\int_{\mathbb{R}^{N}}|h(x-y)|^{s}dx)^{\frac{1}{s}}|y|^{2\alpha-N}\chi_{{}_{B_{\nu}}}(y)dy]^{s}}{t^{s}}$
$\displaystyle\leq$
$\displaystyle\frac{\|h\|^{s}_{s}}{t^{s}}\int_{B_{\nu}}|y|^{2\alpha-N}dy$
$\displaystyle=$ $\displaystyle
c_{37}\nu^{2\alpha}\frac{\|h\|^{s}_{s}}{t^{s}}.$

On the other hand, direct computation shows that

$\displaystyle\|G_{2}\ast h\|_{\infty}$ $\displaystyle\leq$ $\displaystyle
c_{38}\Big{\|}\int_{\mathbb{R}^{N}}\chi_{{}_{B_{\nu}^{c}}}(x-y)(\Phi_{r_{0}}\eta_{0})|h(y)|dy\Big{\|}_{\infty}$
$\displaystyle\leq$
$\displaystyle\Big{(}\int_{\mathbb{R}^{N}}|h(y)|^{q}dy\Big{)}^{\frac{1}{q}}\Big{\|}\Big{(}\int_{B_{2}(x)}\chi_{{}_{B_{\nu}^{c}}}(\Phi_{r_{0}}\eta_{0})^{q^{\prime}}dy\Big{)}^{\frac{1}{q^{\prime}}}\Big{\|}_{\infty}$
$\displaystyle\leq$
$\displaystyle\|h\|_{q}\|\Phi_{r_{0}}\eta_{0}\chi_{{}_{B_{\nu}^{c}}}\|_{q^{\prime}},$

where $q^{\prime}=\frac{q}{q-1}$ if $q>1$, if not, $q^{\prime}=\infty$.

Since

$\|\Phi_{r_{0}}\eta_{0}\chi_{{}_{B_{\nu}^{c}}}\|_{L^{q^{\prime}}(\mathbb{R}^{N})}=\Big{[}\int_{B_{2}\setminus
B_{\nu}}|x|^{(2s-N)q^{\prime}}dx\Big{]}^{\frac{1}{q^{\prime}}}=c_{39}\nu^{2s-\frac{N}{q}},$

by choosing $\nu=(\frac{t}{c_{39}\|h\|_{q}})^{\frac{1}{2s-\frac{N}{q}}}$, then

$\|G_{2}\ast h\|_{\infty}\leq t,$

that means

$\big{|}\\{x\in\mathbb{R}^{N}:|G_{2}\ast h(x)|>t\\}\big{|}=0.$

With this $\nu$, we have that

$\big{|}\\{x\in\mathbb{R}^{N}:|(\Phi_{r_{0}}\eta_{0})\ast h|>2t\\}\big{|}\leq
c_{37}\frac{\|h\|^{q}_{q}\nu^{2sq}}{t^{q}}\leq
c_{39}\Big{(}\frac{\|h\|_{q}}{t}\Big{)}^{r^{*}}.$

The argument of $(ii)$ and $(iii)$ with $r>s$ follows by the Marcinkiewicz
Interpolation Theorem. $\Box$

Acknowledgements: This work is supported by NNSF of China, No: 12071189, by
the Jiangxi Provincial Natural Science Foundation, No: 20212ACB211005 and
20202ACBL201001.

## References

  * [1] Alves, C. Torres, C.: Existence and multiplicity of solutions for a non-linear Schrödinger equation with non-local regional diffusion. J. Math. Phys. 58(11), 111507, 19 pp. (2017)
  * [2] Bhattacharyya, S., Ghosh,T. , Uhlmann, G.: Inverse problems for the fractional-Laplacian with lower order non-local perturbations. Trans. Amer. Math. Soc. 374(5), 3053–3075 (2021).
  * [3] Bogdan, K., Burdzy, K., Chen Z. : Censored stable processes. Probability theory and related fields, 127, 89–152 (2003)
  * [4] Bogdan, K., Dyda, B.: The best constant in a fractional Hardy inequality. Math. Nachr., 284, 629–638 (2011)
  * [5] Caffarelli, L., Salsa, S., Silvestre, L.: Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian. Invent. Math. 171(2), 425–461 (2008)
  * [6] Caffarelli, L., Silvestre, L.: Regularity theory for fully nonlinear integro-differential equations. Comm. Pure Appl. Math., 62, 597–638 (2009)
  * [7] Chen, Z., Kim, P.: Green function estimate for censored stable processes. Probability theory and related fields, 124, 595–610 (2002)
  * [8] Chen, Z., Kim, P., Song, R.: Two-sided heat kernel estimates for censored stable-like processes. Probability theory and related fields, 146, 361–399 (2010)
  * [9] Chen, H.: The Dirichlet problem involving regional fractional Laplacian. J. Math. Phys., 59(7), 071504 (2018)
  * [10] Chen, H., Hajaiej H.: Boundary blow-up solutions of elliptic equations involving regional fractional Laplacian. Comm. Math. Sci., 17(4), 989–1004 (2019)
  * [11] Chen, H., Wei, Y.: Non-existence of Poisson problem involving regional fractional Laplacian with order in $(0,\frac{1}{2}]$. ArXiv: 2007.05775 (2020).
  * [12] Chen, H., Felmer, P., Quaas, A. Large solutions to elliptic equations involving fractional Laplacian. Ann. Inst. H. Poincaré C-AN 32(6), 1199–1228 (2015)
  * [13] Chen, H., Véron, L.: Weakly and strongly singular solutions of semilinear fractional elliptic equations. Asym. Anal., 88, 165-184 (2014)
  * [14] Djitte, S., Fall, M., Weth, T.: A fractional Hadamard formula and applications. Calc. Var. PDE 60(6), No. 231, 31 pp. (2021)
  * [15] Dyda, B.: A fractional order Hardy inequality. Illinois J. Math., 48, 575–588 (2004)
  * [16] Di Nezza, E., Palatucci, G., Valdinoci, E.: Hitchhiker’s guide to the fractional Sobolev spaces. Bull. Sci. Math., 136 (5), 521–573 (2012)
  * [17] M. Fall, Regional fractional Laplacians: boundary regularity. J. Diff. Eq. 320, (2022) 598–658.
  * [18] Gressman, P.: Fractional Poincaré and logarithmic Sobolev inequalities for measure spaces. J. Funct. Anal. 265(6) 867–889 (2003)
  * [19] Guan, Q.: Integration by parts formula for regional fractional laplacian. Comm. Math. Phys., 266, 289–329 (2006)
  * [20] Guan, Q., Ma Z.: Reflected symmetric A-stable processes and regional fractional Laplacian. Probability theory and related fields, 134, 649–694 (2006)
  * [21] Mikhailov, S.: Traces, extensions and co-normal derivatives for elliptic systems on Lipschitz domains. J. Math. Anal. Appl., 378, 324–342 (2011)
  * [22] Ros-Oton X., Serra, J.: The Dirichlet problem for the fractional laplacian: regularity up to the boundary. J. Math. Pures Appl., 101, 275–302 (2014)
  * [23] Sire, Y., Valdinoci, E.: Fractional Laplacian phase transitions and boundary reactions: a geometric inequality and a symmetry result. J. Funct. Anal. 256(6),  1842–1864 (2009)
  * [24] Temgoua, R. : On the $s$-derivative of weak solutions of the Poisson problem for the regional fractional Laplacian, arXiv:2112.09547.