\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 74, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/74\hfil Fu\v{c}ik spectrum]
{First curve of Fu\v{c}ik spectrum for the $p$-fractional Laplacian operator
with nonlocal normal boundary conditions}

\author[D. Goel, S. Goyal, K. Sreenadh \hfil EJDE-2018/74\hfilneg]
{Divya Goel, Sarika Goyal, Konijeti Sreenadh}

\address{Divya Goel \newline
Department of Mathematics,
Indian Institute of Technology Delhi,
Hauz Khas, New Delhi-110016, India}
\email{divyagoel2511@gmail.com}

\address{Sarika Goyal \newline
Department of Mathematics,
Bennett University, Greater Noida,
Uttar Pradesh - 201310, India}
\email{sarika1.iitd@gmail.com}

\address{Konijeti Sreenadh \newline
Department of Mathematics,
Indian Institute of Technology Delhi,
Hauz Khaz, New Delhi-110016, India}
\email{sreenadh@maths.iitd.ac.in}

\dedicatory{Communicated by Vicentiu Radulescu}

\thanks{Submitted November 22, 2017. Published March 17, 2018.}
\subjclass[2010]{35A15, 35J92, 35J60}
\keywords{Nonlocal operator; Fu\v{c}ik spectrum; Steklov problem;
 Non-resonance}

\begin{abstract}
 In this article, we study the Fu\v{c}ik spectrum of the $p$-fractional
 Laplace operator with nonlocal normal derivative conditions which
 is defined as the set of all $(a,b)\in\mathbb{R}^2$ such that
 \begin{gather*}
 \Lambda_{n,p}(1-\alpha)(-\Delta)_{p}^{\alpha} u
 + |u|^{p-2}u = \frac{\chi_{\Omega_\epsilon}}{\epsilon} (a (u^{+})^{p-1}
 - b (u^{-})^{p-1}) \quad  \text{in }\Omega,  \\
 \mathcal{N}_{\alpha,p} u = 0  \quad  \text{in }\mathbb{R}^n \setminus
 \overline{\Omega},
 \end{gather*}
 has a non-trivial solution $u$, where $\Omega$ is a bounded domain in
 $\mathbb{R}^n$ with Lipschitz boundary, $p \geq 2$, $n>p \alpha$,
 $\epsilon, \alpha \in(0,1)$ and
 $\Omega_{\epsilon}:=\{x \in \Omega:  d(x,\partial \Omega)\leq \epsilon \}$.
 We show existence of the first non-trivial curve $\mathcal{C}$ of the
 Fu\v{c}ik  spectrum which is used to obtain the  variational 
 characterization of a second eigenvalue of the problem defined above.
 We also discuss some  properties of  this curve $\mathcal{C}$, 
 e.g.\ Lipschitz continuous, strictly decreasing
 and asymptotic behavior and non-resonance with respect to the Fu\v{c}ik spectrum.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

The Fu\v{c}ik spectrum of $p$-fractional Laplacian with nonlocal normal
derivative is defined as the set $\Sigma_p$ of all $(a,b)\in\mathbb{R}^2$ such that
\begin{equation}\label{eq1}
\begin{gathered}
 \Lambda_{n,p}(1-\alpha)(-\Delta)_{p}^{\alpha} u  + |u|^{p-2}u
 = \frac{\chi_{\Omega_\epsilon}}{\epsilon} (a (u^{+})^{p-1}
 - b (u^{-})^{p-1}) \quad\text{in }\Omega,\\
 \mathcal{N}_{\alpha,p} u = 0 \quad  \text{in }\mathbb{R}^n
 \setminus \overline{\Omega},
\end{gathered}
\end{equation}
has a non-trivial solution $u$, where $\Omega$ is a bounded domain in
$\mathbb{R}^n$ with Lipschitz boundary, $p \geq 2$, $\alpha, \epsilon \in (0,1)$
and $\Omega{_\epsilon}:=\{x \in \Omega:  d(x,\partial \Omega)\leq \epsilon \}$.
The $(-\Delta)^{\alpha}_p$ is the $p$-fractional Laplacian operator defined as
\begin{equation*}
(-\Delta)_{p}^{\alpha} u(x):= 2 \operatorname{p.v.}
\int_{\mathbb{R}^n}\frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{n+p\alpha}} dy
\quad \text{for all } x\in\mathbb{R}^n,
\end{equation*}
and $\mathcal{N}_{\alpha,p}$ is the associated nonlocal derivative defined
in \cite{va} as
\begin{equation*}
\mathcal{N}_{\alpha,p}u(x):= 2  \int_{\Omega} \frac{|u(x)-u(y)|^{p-2}(u(x)
-u(y))}{|x-y|^{n+p\alpha}} dy \quad \text{for all }
x\in\mathbb{R}^n \setminus \overline{\Omega} .
\end{equation*}
  Bourgain, Brezis and Mironescu \cite{bre}
 proved that for any smooth bounded domain  $\Omega \subset\mathbb{R}^n$,
$ u \in W^{1,p}(\Omega)$, there exist a constant $\Lambda_{n,p}$ such that
\begin{equation*}
\lim_{\alpha \to 1 ^-} \Lambda_{n,p} (1-\alpha)\int _{\Omega \times \Omega }
\frac{|u(x)-u(y)|^{p}}{|x-y|^{n+p\alpha}}\,dx  dy= \int_{\Omega}|\nabla u|^p\,dx.
\end{equation*}
The constant $\Lambda_{n,p}$ can be explicitly computed and is given by
\begin{equation*}
\Lambda_{n,p}= \frac{p\Gamma(\frac{n+p}{2})}{2\pi^{\frac{n-1}{2}}
\Gamma(\frac{p+1}{2}) }.
\end{equation*}
For $a=b=\lambda $, the Fu\v{c}ik spectrum in \eqref{eq1} becomes the usual
spectrum that satisfies
\begin{equation}\label{eq2}
\begin{gathered}
\Lambda_{n,p}(1-\alpha)(-\Delta)_{p}^{\alpha} u  + |u|^{p-2}u
 = \frac{ \lambda}{\epsilon} \chi_{\Omega_\epsilon} |u|^{p-2}u  \quad \text{in }
\Omega,\\
 \mathcal{N}_{\alpha,p} u = 0 \quad \text{in }\mathbb{R}^n \setminus
\overline{\Omega}.
\end{gathered}
\end{equation}
In \cite{pe}, authors proved that there exists a sequence of eigenvalues
$\lambda_{k,\epsilon}(\Omega_{\epsilon})$ of \eqref{eq2}  such that
$\lambda_{k,\epsilon}(\Omega_{\epsilon})\to \infty$ as $k\to \infty$.
Moreover, $0<\lambda_{1,\epsilon}(\Omega_{\epsilon})<\lambda_{2,\epsilon}(\Omega_{\epsilon})
\leq\dots \leq\lambda_{k,\epsilon}(\Omega_{\epsilon})\leq\dots$,
and the first eigenvalue $\lambda_{1,\epsilon}(\Omega_{\epsilon})$ of \eqref{eq2}
is simple, isolated and can be characterized as follows
\begin{align*}
&\lambda_{1,\epsilon}(\Omega_{\epsilon}) \\
&= \inf_{u\in \mathcal{W}^{\alpha,p}}\Big\{  \Lambda_{n,p}(1-\alpha)
\int_{Q}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
+ \int_{\Omega} |u|^p\,dx :  \int_{\Omega_{\epsilon}}|u|^p\,dx= \epsilon \Big\}.
\end{align*}
 The  Fu\v{c}ik spectrum was introduced by Fu\v{c}ik (1976) who studied
the problem in one dimension with periodic boundary
conditions.  In higher dimensions, the non-trivial first curve in the
Fu\v{c}ik spectrum of Laplacian with Dirichlet boundary for bounded
domain has been studied in \cite{fg}. Later
in \cite{cfg} Cuesta, de Figueiredo and Gossez studied this problem
for $p$-Laplacian operator with Dirichlet boundary condition.

The Fu\v{c}ik spectrum in the case of Laplacian, $p$-Laplacian operator
with Dirichlet, Neumann and Robin boundary condition has been studied by
many authors, for instance \cite{al, cg, ro, ros, pe,kpe}.
Goyal and Sreenadh \cite{sa} extended the results of \cite{cfg} to
nonlocal linear operators which include fractional Laplacian.
The existence of  Fu\v{c}ik eigenvalues for $p$-fractional Laplacian operator
with Dirichlet boundary conditions has been studied by many authors,
for instance refer \cite{dan, inf}. Also, in \cite{hardy}, Goyal discussed
the Fu\v{c}ik spectrum of of $p$-fractional Hardy Sobolev-Operator
with weight function. A non-resonance
problem with respect to Fu\v{c}ik spectrum is also discussed in many
papers \cite{cfg,kpr,mP}. We also refer to the related papers
 \cite{ana1, ana2, ana3, ana4}.

 The inspiring point of our work is \cite{sa, hardy}, where the existence
of a nontrivial curve is studied only for $p=2$ but the nature of the curve
is left open for $p\ne 2$. In the present work, we extend the results
obtained in \cite{sa} to the nonlinear case of $p$-fractional operator
for any $p \geq 2$ and also show that this curve is the first curve.
 We also showed the variational characterization of the second eigenvalue
of the operator associated  with \eqref{eq1}. There is a substantial
difference while handling the nonlinear nature of the operator.
This difference is reflected while constructing the paths below a
mountain-pass level (see the proof of Theorem \ref{thm1.1}).  To the best of our
knowledge, no work has been done on the Fu\v{c}ik spectrum for nonlocal
operators with nonlocal normal derivative. We would like to remark that
the main result obtained in this paper is new even for the following
 $p$-fractional Laplacian equation with Dirichlet boundary condition:
\[
(-\Delta)_{p}^{\alpha} u + |u|^{p-2} u = a (u^+)^{p-1}-b (u^-)^{p-1}\quad \text{in }
\Omega, \quad u=0  \text{ on } \mathbb{R}^n\backslash \overline{\Omega}.
\]
 With this introduction, we state our main result.

\begin{theorem}\label{thm1.1}
Let $s\geq 0$ then the point $(s+c(s),c(s))$ is the first nontrivial
point of $\Sigma_p$ in the intersection between $\Sigma_p$ and the line
$(s,0)+t(1,1)$ of \eqref{eq1}.
\end{theorem}

 This article is organized as follows: In section 2 we give some
 preliminaries. In section 3 we construct a
first nontrivial curve in $\Sigma_p$, described as $(s+c(s),c(s))$.
In section 4 we prove that the lines
$\lambda_{1,\epsilon}(\Omega_{\epsilon})\times\mathbb{R}$ and
$\mathbb{R}\times\lambda_{1,\epsilon}(\Omega_{\epsilon})$ are isolated in $\Sigma_p$,
the curve that we obtained
in section 3 is the first nontrivial curve and give the variational
characterization of second eigenvalue of \eqref{eq1}.  In section 5 we
prove some properties of the first curve and non resonance problem.

\section{Preliminaries}

 In this section we assemble some requisite material.
 By \cite{va} we know the nonlocal analogue of divergence theorem which
 states that for any bounded functions $u$ and $v\in C^2$, it holds that
\begin{equation*}
\int_{\Omega}(-\Delta)_{p}^{\alpha} u(x)\,dx
= -\int _{\Omega ^c}\mathcal{N}_{\alpha,p}u(x)\,dx.
\end{equation*}
More generally, we have following integration by parts  formula
\begin{equation*}
\mathcal{H}_{\alpha,p}(u,v)
=\int_{\Omega}v(x)(-\Delta)_{p}^{\alpha} u(x)\,dx
+\int _{\Omega ^c} v(x) \mathcal{N}_{\alpha,p}u(x)\,dx,
\end{equation*}
where $\mathcal{H}_{\alpha,p}(u,v)$ is defined as
\begin{equation*}
\mathcal{H}_{\alpha,p}(u,v)
:= \int _Q \frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))(v(x)-v(y))}{|x-y|^{n+p\alpha}} dy,
\quad Q:=\mathbb{R}^{2n} \setminus (\Omega^c)^2.
\end{equation*}
Now, given a measurable function $ u:\mathbb{R}^n\to \mathbb{R} $, we set
\begin{equation} \label{feq12}
\|u\|_{\alpha,p}:= (\|u\|^p_{L^p(\Omega)}+[u]^p_{\alpha,p})^{1/p},\quad
\text{where } [u]_{\alpha,p}:=(\mathcal{H}_{\alpha,p}(u,u))^{1/p}.
\end{equation}
 Then $\|\cdot\|_{\alpha,p}$ defines a norm on the space
\begin{equation*}
\mathcal{W}^{\alpha,p}:= \{ u:\mathbb{R}^n\to \mathbb{R}
\text{ measurable }:\|u\|_{\alpha,p} < \infty \}.
\end{equation*}
Clearly $\mathcal{W}^{\alpha,p}\subset W^{\alpha,p}(\Omega)$,  where
$ W^{\alpha,p}(\Omega)$ denotes the usual fractional Sobolev space endowed
with the norm
\[
\|u\|_{W^{\alpha,p}}=\|u\|_{L^p}
+ \Big(\int_{\Omega\times\Omega} \frac{(u(x)-u(y))^{p}}{|x-y|^{n+p \alpha }}\,dx\,dy
\Big)^{1/p}.
\]
To study the fractional Sobolev space in detail see \cite{ms}.

\begin{definition} \label{def2.1} \rm
A function $u \in \mathcal{W}^{\alpha,p}$ is a weak solution of \eqref{eq1},
if for every $v\in \mathcal{W}^{\alpha,p}$, $u$ satisfies
\[
 \Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u,v)
 + \int_{\Omega}|u|^{p-2}uv-\frac{a}{\epsilon}
\int_{\Omega_{\epsilon}} (u^{+})^{ p-1}v+\frac{b}{\epsilon}
\int_{\Omega_{\epsilon}} (u^{-})^{ p-1}v=0.
\]
\end{definition}

 Now, we define the functional $J$ associated to problem \eqref{eq1} as
$J : \mathcal{W}^{\alpha,p} \to \mathbb{R}$ such that
\begin{align*}
J(u)&=  \Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
 + \int_{\Omega}|u|^p\,dx \\
&\quad - \frac{a}{\epsilon} \int_{\Omega_{\epsilon}} (u^{+})^p\,dx
+\frac{b}{\epsilon} \int_{\Omega_{\epsilon}} (u^{-})^p\,dx.
\end{align*}
Then $J$ is Fr\'echet differentiable in $ \mathcal{W}^{\alpha,p}$
and for all $v\in \mathcal{W}^{\alpha,p}$.
\[
 \langle J^\prime (u),v\rangle
=\Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u,v)
+ \int_{\Omega}|u|^{p-2}uv-\frac{a}{\epsilon}
\int_{\Omega_{\epsilon}} (u^{+})^{ p-1}v+\frac{b}{\epsilon}
\int_{\Omega_{\epsilon}} (u^{-})^{ p-1}v.
\]
For the sake of completeness, we describe the Steklov problem
\begin{equation}\label{stek}
\begin{gathered}
(-\Delta)_{p} u  + |u|^{p-2}u = 0 \quad  \text{in }\Omega,\\
 |\nabla u|^{p-2}\frac{\partial u}{\partial \nu}=\lambda |u|^{p-2}u   \quad
\text{on } \partial \Omega,
\end{gathered}
\end{equation}
 where $\Omega$ is a bounded domain and $p>1$. By \cite{pe}, \eqref{eq1}
is related to \eqref{stek} in the sense that if $\Omega $ be a bounded smooth
domain in $\mathbb{R}^n$ with Lipschitz boundary and $p\in(1,\infty)$. For a fixed
$u\in W^{1,p}(\Omega)\setminus W^{1,p}_0(\Omega)$, we have
\begin{equation*}
\lim_{\epsilon \to 0^+}\frac{1}{\epsilon}\int_{\Omega_{\epsilon}}|u|^pdx
=\int_{\partial \Omega}|u|^pdS \quad \text{and}\quad
 \lim_{\alpha \to 1^-}\Lambda_{n,p}(1-\alpha)[Eu]^p_{\alpha,p}= \|\nabla u\|^p_{L^p(\Omega)},
\end{equation*}
where $E$ is a bounded linear extension operator from $W^{1,p}(\Omega)$ to
$W^{1,p}_0(B_R)$ such that $Eu=u$ in $\Omega$ and $\Omega$ is relatively
compact in $B_R$, the ball of radius $R$ in $\mathbb{R}^n$. This leads to
the following Lemma in \cite{ros}.

\begin{lemma} \label{f4}
 Let $\Omega $ be a smooth domain in $\mathbb{R}^n$ with Lipschitz boundary and
$p \in (1,\infty)$. For a fixed $u\in W^{1,p}(\Omega)\setminus W^{1,p}_0(\Omega)$, it holds
\begin{equation*}
\lim_{\alpha \to 1^-}\frac{\Lambda_{n,p}(1-\alpha)[Eu]^p_{\alpha,p}
+\|Eu\|^p_{L^p(\Omega)}}{\frac{1}{1-\alpha}\|Eu\|^p_{L^p(\Omega_{1-\alpha})}}
= \frac{\|\nabla u\|^p_{L^p(\Omega)}+\|u\|^p_{L^p(\Omega)}}{\|u\|^p_{L^p(\partial \Omega)}}.
\end{equation*}
\end{lemma}

Taking $\epsilon=1-\alpha$, by Lemma \ref{f4} the eigenvalue
$\lambda_{1,1-\alpha}(\Omega_{1-\alpha}) \to \lambda_1$ as $\alpha \to 1^-$, where
$\lambda_1$ is the first eigenvalue of the operator associated with \eqref{stek}.
Similarly, we obtain  that as $ \alpha \to 1^-$ the Fu\v{c}ik Spectrum of
the operator associated with \eqref{eq1} tends to Fu\v{c}ik Spectrum of the
Steklov problem.

 We shall throughout use the function space $\mathcal{W}^{\alpha,p}$ with the norm
$\|\cdot\|$ and we use the standard $L^{p}(\Omega)$ space whose norms
are denoted by $\|u\|_{L^p(\Omega)}$. Also, we denote
$\lambda_{n,\epsilon}(\Omega_{\epsilon})$ by $\lambda_{n,\epsilon}$.
Here $\phi_{1,\epsilon}$ is the eigenfunction  corresponding to $\lambda_{1,\epsilon}$.


\section{The Fu\v{c}ik spectrum $\Sigma_p$}
In this section, we study  existence of the first nontrivial curve in
the Fu\v{c}ik spectrum $\Sigma_p$ of \eqref{eq1}. We find that the points
in $\Sigma_p$ are associated with the critical value of some restricted
functional. For  this, for fixed $s\in\mathbb{R}$ and  $s\geq 0$, we consider the
functional $J_{s}: \mathcal{W}^{\alpha,p} \to \mathbb{R} $ defined by
\begin{align*}
J_{s}(u)= \Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
+ \int_{\Omega}|u|^p\,dx - \frac{s}{\epsilon} \int_{\Omega_{\epsilon}} (u^{+})^p\,dx.
\end{align*}
Then $J_{s}\in C^{1}(\mathcal{W}^{\alpha,p},\mathbb{R})$ and for any
$\phi\in \mathcal{W}^{\alpha,p}$
\begin{equation*}
\langle J_{s}'(u),\phi \rangle
=  p\text{ }\Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u,\phi)
 + p \int_{\Omega}|u|^{p-2}u \phi\,dx - \frac{ps}{\epsilon}
\int_{\Omega_{\epsilon}} (u^{+})^{ p-1}\phi\,dx.
\end{equation*}
 Also $\tilde{J_{s}}:= J_{s}|_{\mathcal{S}}$ is $C^1(\mathcal{W}^{\alpha,p},\mathbb{R})$,
 where $\mathcal{S}$ is defined as
\[
\mathcal{S} :=\big\{u\in \mathcal{W}^{\alpha,p}:
 I(u):= \frac{1}{\epsilon}\int_{\Omega_{\epsilon}}|u|^p=1\big\}.
\]
 We first note that $u\in\mathcal{S}$ is a critical point of $\tilde{J_{s}}$
if and only if there exists $t\in\mathbb{R}$ such that
\begin{equation}\label{feq2}
 \Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u,v)
-\frac{s}{\epsilon}\int_{\Omega_{\epsilon}} (u^{+})^{p-1} v\,dx
= \frac{t}{\epsilon}\int_{\Omega_{\epsilon}} |u|^{p-2} u v\,dx,
\end{equation}
for all $v\in \mathcal{W}^{\alpha,p}$. Hence $u\in \mathcal{S}$ is a
nontrivial weak solution of the problem
\begin{gather*}
\Lambda_{n,p}(1-\alpha)(-\Delta)_{p}^{\alpha} + |u|^{p-2}u
= \frac{\chi_{\Omega_{\epsilon}}}{\epsilon}\left((s+t)(u^{+})^{p-1} - {t}
(u^{-})^{p-1}\right)\quad \text{in }\Omega, \\
 \mathcal{N}_{\alpha,p} u = 0 \quad\text{in }\mathbb{R}^n \setminus
\overline{\Omega},
\end{gather*}
which exactly means $(s+t,t)\in \Sigma_p$. Substituting $v=u$ in \eqref{feq2},
we obtain $t= \tilde{J_{s}}(u)$. Thus we obtain the following Lemma
which links the critical point of $\tilde{J_{s}}$ and the spectrum $\Sigma_p$.

\begin{lemma} \label{lem3.1}
For $s\geq 0$, $(s+t,t)\in\mathbb{R}^2$ belongs to the spectrum $\Sigma_p$
if and only if there exists a critical point $u\in \mathcal{S}$ of
$\tilde{J_{s}}$ such that $t= \tilde{J_{s}}(u)$, a critical value.
\end{lemma}

\begin{proposition} \label{prop3.2}
The first eigenfunction $\phi_{1,\epsilon}$ is a global minimum for
$\tilde{J_{s}}$ with $\tilde{J_{s}}(\phi_{1,\epsilon})=\lambda_{1,\epsilon}-s$.
The corresponding point in $\Sigma_p$ is $(\lambda_{1,\epsilon},\lambda_{1,\epsilon}-s)$
which lies on the
vertical line through $(\lambda_{1,\epsilon},\lambda_{1,\epsilon})$.
\end{proposition}

\begin{proof}
We have
\begin{align*}
\tilde{J_{s}}(u)
=&  \Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
  + \int_{\Omega}|u|^{p}dx-\frac{s}{\epsilon} \int_{\Omega_{\epsilon}} (u^{+})^p\,dx\\
\geq & \frac{\lambda_{1,\epsilon}}{\epsilon}  \int_{\Omega_{\epsilon}} |u|^p\,dx
  -\frac{s}{\epsilon} \int_{\Omega_{\epsilon}} (u^{+})^p\,dx
\geq \lambda_{1,\epsilon} -s.
\end{align*}
Thus $\tilde{J_{s}}$ is bounded below by $\lambda_{1,\epsilon}-s$.
Moreover,
\[
\tilde{J_{s}}(\phi_{1,\epsilon})= \lambda_{1,\epsilon}
- \frac{s}{\epsilon} \int_{\Omega_{\epsilon}} (\phi_{1,\epsilon}^{+})^p\,dx
= \lambda_{1,\epsilon}-s.
\]
Thus $\phi_{1,\epsilon}$ is a global minimum of $\tilde{J_{s}}$ with
$\tilde{J_{s}}(\phi_{1,\epsilon})=\lambda_{1,\epsilon}-s$.
\end{proof}

\begin{proposition} \label{prop3}
The negative eigenfunction $-\phi_{1,\epsilon}$ is a strict local minimum for
$\tilde{J_{s}}$ with $\tilde{J_{s}}(-\phi_{1,\epsilon})=\lambda_{1,\epsilon}$. The
corresponding point in $\Sigma_p$ is $(\lambda_{1,\epsilon}+s,\lambda_{1,\epsilon})$,
which lies on the horizontal line through $(\lambda_{1,\epsilon},\lambda_{1,\epsilon})$.
\end{proposition}

\begin{proof}
Suppose by contradiction that there exists a sequence $u_k\in\mathcal{S}$,
$u_k\ne -\phi_{1,\epsilon}$ with $\tilde{J_{s}}(u_k) \leq \lambda_{1,\epsilon}$,
$u_k\to -\phi_{1,\epsilon}$ in $\mathcal{W}^{\alpha,p}$.
We claim that $u_k$ changes sign for sufficiently large $k$.
Since $u_k\to -\phi_{1,\epsilon}$, $u_k$ must be $<0$
for sufficiently large $k$. If $u_k \leq 0$ for a.e $x\in \Omega$, then
\begin{align*}
\tilde{J_{s}}(u_k)= \Lambda_{n,p}(1-\alpha)
\int_{Q}\frac{|u_k(x)-u_k(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
+\int_{\Omega}|u_k|^{p}\,dx >\lambda_{1,\epsilon},
\end{align*}
since $u_k \not\equiv \pm\phi_{1,\epsilon}$ and we obtain contradiction as
$\tilde{J_{s}}(u_k) \leq \lambda_{1,\epsilon}$. Therefore the claim is proved.

Now, define
$w_k := \frac{\epsilon^{1/p} u_{k}^{+}}{\|u_{k}^{+}\|_{L^p(\Omega_{\epsilon})}}$
and
\[
r_k := \Lambda_{n,p}(1-\alpha)\int_{Q}\frac{|w_{k}(x)-w_{k}(y)|^p}{|x-y|^{n+p\alpha}}
\,dx\,dy + \int_{\Omega}|w_k|^{p}\,dx.
\]
 We claim that $r_k \to \infty$ as $k\to \infty$. Assume by contradiction that
$r_k$ is bounded. Then there exists a
subsequence (still denoted by $\{w_k\}$) of $\{w_k\}$ and
$w\in \mathcal{W}^{\alpha,p}$ such that $w_k \rightharpoonup w$ weakly in
$\mathcal{W}^{\alpha,p}$ and $w_k\to w$ strongly in $L^{p}(\Omega)$.
It implies $w_k\to w$ strongly in $L^{p}(\Omega_{\epsilon})$.
Therefore $\frac{1}{\epsilon} \int_{\Omega_{\epsilon}} w^p\,dx =1$, $w\geq 0$ a.e.
 in $\Omega_{\epsilon}$ and so for some $\eta>0$,
$\delta =|\{x\in \Omega_{\epsilon}: w(x)\geq \eta\}|>0$.
Since, $u_k\to -\phi_{1,\epsilon}$ in $\mathcal{W}^{\alpha,p}$ and hence in
$L^{p}(\Omega)$. Therefore, for each $\eta>0$,
$|\{x\in \Omega_{\epsilon} : u_k(x)\geq \eta\}|\to 0$ as $k\to \infty$
and $|\{x\in \Omega_{\epsilon} : w_k(x)\geq \eta\}|\to 0$ as $k\to \infty$,
 which is a contradiction to $\eta>0$. Hence, $r_k \to \infty$.
Clearly, one can have
\begin{align*}
&|u_k(x)-u_k(y)|^p \\
&=(|u_k(x)-u_k(y)|^2)^{p/2}=[((u_{k}^{+}(x)-u_{k}^{+}(y))
 -(u_{k}^{-}(x)-u_{k}^{-}(y)))^2]^{p/2} \\
&=[(u_{k}^{+}(x)-u_{k}^{+}(y))^2+ (u_{k}^{-}(x)-u_{k}^{-}(y))^2
 - 2(u_{k}^{+}(x)-u_{k}^{+}(y))(u_{k}^{-}(x)-u_{k}^{-}(y))]^{p/2} \\
&= [(u_{k}^{+}(x)-u_{k}^{+}(y))^2+ (u_{k}^{-}(x)-u_{k}^{-}(y))^2
 + 2u_{k}^{+}(x) u_{k}^{-}(y)+ 2 u_{k}^{-}(x) u_{k}^{+}(y)]^{p/2} \\
&\geq  |u_{k}^{+}(x)-u_{k}^{+}(y)|^p +|u_{k}^{-}(x)-u_{k}^{-}(y)|^p.
\end{align*}
Using the above inequality, we have
\begin{equation} \label{fe12}
\begin{aligned}
\tilde{J_{s}}(u_k)
&= \Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u_k(x)-u_k(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
 + \int_{\Omega}|u_k|^{p}
 - \frac{s}{\epsilon} \int_{\Omega_{\epsilon}} (u^{+}_k)^p\,dx \\
&\geq  \left[ \Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u_{k}^{+}(x)
 -u_{k}^{+}(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy + \int_{\Omega}|u_k^+|^{p}\right]  \\
&\quad +\Big[\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u_{k}^{-}(x)
 -u_{k}^{-}(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
 + \int_{\Omega}|u_k^-|^{p} \\
&\quad - \frac{s}{\epsilon} \int_{\Omega_{\epsilon}}
  (u^{+}_{k})^p\,dx\Big] \\
&\geq  \frac{(r_k-s)}{\epsilon} \int_{\Omega_{\epsilon}} (u_{k}^{+})^p\,dx
 + \frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{\epsilon}} (u_{k}^{-})^p\,dx.
\end{aligned}
\end{equation}
On the other hand, since $u_k\in \mathcal{S}$, we obtain
\begin{equation}\label{fe13}
\tilde{J_{s}}(u_k)\leq \lambda_{1,\epsilon}
=   \frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{}\epsilon} (u_{k}^{+})^p\,dx+
\frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{}\epsilon} (u_{k}^{-})^p\,dx.
\end{equation}
 From \eqref{fe12} and \eqref{fe13}, we have
\[
\frac{(r_k -s-\lambda_{1,\epsilon})}{\epsilon} \int_{\Omega_{\epsilon}} (u_{k}^{+})^p\,dx
\leq 0,
\]
 and this implies
$r_k- s\leq \lambda_{1,\epsilon}$,
 which contradicts  that $r_k\to +\infty$. Therefore,
$-\phi_{1,\epsilon}$ is the strict local minimum.
\end{proof}

\begin{proposition}[\cite{AR}] \label{fpr1}
 Let $Y$ be a Banach space, $g,f \in C^{1}(Y,\mathbb{R})$,
$M=\{u\in Y : g(u)=1\}$ and $u_0$, $u_1\in M$. Let $\epsilon>0$ such that
$\|u_1-u_0\|>\epsilon$ and
\[
\inf\{f(u): u\in M \text{ and }\|u-u_0\|_{Y}=\epsilon\}
>\max\{f(u_0),f(u_1)\}.
\]
 Assume that $f$ satisfies the (PS) condition on $M$ and that
\[
\Gamma =\{\gamma \in C([-1,1], M): \gamma(-1)=u_0 \text{ and } \gamma(1)=u_1\}
\]
is non empty. Then $ c=\inf_{\gamma \in \Gamma}\max_{u\in\gamma[-1,1]}
f(u)$ is a critical value of $f|_M$.
\end{proposition}

 We  now find the third critical point via mountain pass Theorem as stated above.
A norm of derivative of the restriction $\tilde{J_{s}}$ of $J_{s}$
at $u\in \mathcal{S}$ is defined as
\[
\|\tilde{J}_{s}'(u)\|_{*}=\min\{\|\tilde{J}_{s}'(u)- t I'(u)\|:  t\in\mathbb{R}\}.
\]

\begin{lemma}  \label{fle21}
$J_{s}$ satisfies the (PS) condition on $\mathcal{S}$.
\end{lemma}

\begin{proof}
Let $J_{s}(u_k)$ and $t_k\in\mathbb{R}$ be a sequences such that  for some $K>0$,
\begin{gather}\label{feq3}
|J_{s}(u_k)|\leq K, \\
\label{feq4}
\begin{aligned}
&\Big|\Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u_k, v)
 +\int_{\Omega}|u_k|^{p-2}u_k v
 - \frac{s}{\epsilon}\int_{\Omega_{\epsilon}} (u^{+}_{k})^{p} v\,dx \\
&- \frac{t_k}{\epsilon} \int_{\Omega_{\epsilon}}
|u_{k}|^{p-2} u_{k}v\,dx \Big| \\
&\leq \eta_{k}\|v\|
\end{aligned}
\end{gather}
 for all $v\in \mathcal{W}^{\alpha,p}$, $\eta_k\to 0$. From \eqref{feq3},
using fractional Sobolev embedding, we obtain $\{u_k\}$ is bounded in
$\mathcal{W}^{\alpha,p}$ which implies there is a subsequence denoted
by $u_k$ and $u_0\in \mathcal{W}^{\alpha,p}$ such that
$u_k\rightharpoonup u_0$ weakly in $\mathcal{W}^{\alpha,p}$, and $u_{k}\to u_0$
strongly in $L^{p}(\Omega)$ for all $1\leq p< p_{\alpha}^*$. Substituting $v=u_k$
in \eqref{feq4}, we obtain
\begin{align*}
|t_k| &\leq \Lambda_{n,p}(1-\alpha)\int_{Q}
 \frac{|u_k(x)-u_k(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
 + \int_{\Omega}|u_k|^{p} + \frac{s}{\epsilon}\int_{\Omega_{\epsilon}} (u^{+}_{k})^p\,dx
 + \eta_{k}\|u_k\| \\
&\leq C.
\end{align*}
 Hence, $t_k$ is a bounded sequence so has a convergent subsequence
say $t_k$ that converges to $t$.  Next, we claim that $u_k\to u_0$
strongly in $\mathcal{W}^{\alpha,p}$. Since $u_k\rightharpoonup u_0$
weakly in $\mathcal{W}^{\alpha,p}$, we obtain
\begin{equation} \label{feq7}
\begin{aligned}
&\int_{Q}\frac{|u_0(x)-u_0(y)|^{p-2} (u_0(x)-u_0(y))(u_k(x)-u_k(y))}
 {|x-y|^{n+p\alpha}}\,dx\,dy \\
&\to \int_{Q} \frac{|u_0(x)-u_0(y)|^{p}}{|x-y|^{n+p\alpha}}\,dx\,dy
\quad \text{as }k\to \infty.
\end{aligned}
\end{equation}
Also $\langle \tilde{J_{s}'}(u_k),(u_k-u_0)\rangle= o(\eta_k)$. This implies
\begin{align*}
&\Big|\Lambda_{n,p}(1-\alpha) \int_{Q}
\frac{1}{|x-y|^{n+p\alpha}}
\Big(|u_k(x)-u_k(y)|^{p-2} \\
&\times (u_k(x)-u_k(y))((u_k- u_0)(x)-(u_k-u_0)(y))\Big) \,dx\,dy \Big| \\
& \leq o(\eta_k)+\|u_{k}\|_{L^p(\Omega)}^{p-1}\|u_k-u_0\|_{L^p(\Omega)}
 + s\|u_{k}^{+}\|_{L^p(\Omega_{\epsilon})}^{p-1} \|u_k-u_0\|_{L^p(\Omega_{\epsilon})} \\
&\quad +|t_k|\|u_k\|_{L^p(\Omega_{\epsilon})}^{p-1}\|u_k-u_0\|_{L^p(\Omega_{\epsilon})}
\to  0
\end{align*}
as $k\to \infty$.
Thus,
\begin{equation} \label{feq8}
\begin{aligned}
&\int_{Q}\frac{|u_k(x)-u_k(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy\\
&- \int_{Q}\frac{|u_k(x)-u_k(y)|^{p-2}(u_k(x)-u_k(y))(u_0(x)-u_0(y))}
 {|x-y|^{n+p\alpha}}\,dx\,dy \to 0,
\end{aligned}
\end{equation}
as $k\to \infty$. As we know that $|a-b|^p \leq 2^p(|a|^{p-2}a-|b|^{p-2}b)(a-b)$
for all $a, b \in \mathbb{R}$.
Therefore, from \eqref{feq7} and \eqref{feq8} we obtain
\[
\int_{Q}\frac{|(u_k-u_0)(x)-(u_k-u_0)(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy\to 0 \quad
\text{as } k \to \infty
\]
Hence,  $u_k$ converges strongly to  $u_0$ in $\mathcal{W}^{\alpha,p}$.
\end{proof}

\begin{lemma} \label{fle22}
Let $\eta_0>0$  be such that
\begin{align}\label{feq13}
\tilde{J_{s}}(u)>\tilde{J_{s}}(-\phi_{1,\epsilon})
\end{align}
for all $u\in B(-\phi_{1,\epsilon}, \eta_0)\cap \mathcal{S}$ with
$u \not\equiv -\phi_{1,\epsilon}$, where the ball is
taken in $\mathcal{W}^{\alpha,p}$. Then for any $0<\eta<\eta_0$,
\begin{align}\label{feq14}
\inf\{\tilde{J_{s}}(u): u\in \mathcal{S}\quad\text{and}\quad
\|u-(-\phi_{1,\epsilon})\|=\eta\}> \tilde{J_{s}}(-\phi_{1,\epsilon}).
\end{align}
\end{lemma}

\begin{proof}
If possible, let  infimum in \eqref{feq14} is equal to
$\tilde{J_{s}}(-\phi_{1,\epsilon})=\lambda_{1,\epsilon}$ for some
$\eta$ with $0<\eta<\eta_0$. It implies
there exists a sequence $u_k\in \mathcal{S}$ with
$\|u_k-(-\phi_{1,\epsilon})\|=\eta$ such that
\begin{align}\label{feq15}
\tilde{J_{s}}(u_k)\leq \lambda_{1,\epsilon} + \frac{1}{2k^2}.
\end{align}
Consider the set $V =\{u\in\mathcal{S}: \eta-\delta\leq\|u-(-\phi_{1,\epsilon})\|
\leq \eta+\delta\}$, where $\delta$ is chosen such
that $\eta-\delta>0$ and $\eta+\delta<\eta_0$. From \eqref{feq14} and given hypotheses,
it follows that $\inf\{\tilde{J_{s}}(u): u\in V\}=\lambda_{1,\epsilon}$.
Now for each $k$, we apply Ekeland's variational principle to the functional
$\tilde{J_{s}}$ on $V$ to get the
existence of $v_k\in V$ such that
\begin{gather}
\tilde{J_{s}}(v_k)\leq \tilde{J_{s}}(u_k),\; \|v_k-u_k\|
  \leq \frac{1}{k}, \label{feq018}\\
\tilde{J_{s}}(v_k)\leq \tilde{J_{s}}(u) +
\frac{1}{k}\|u-v_k\|,\;{\text{for all}}\; u\in V. \label{feq020}
\end{gather}
 We claim that $v_k$ is a Palais-Smale sequence for $\tilde{J_{s}}$ on
$\mathcal{S}$. That is, there exists $M>0$ such that $|\tilde{J_{s}}(v_k)|<M$  and
$\|\tilde{J}_{s}'(v_k)\|_{*}\to 0 \text{ as } k \to \infty$.
Once this is proved then by Lemma \ref{fle21}, there exists a subsequence
denoted by $v_k$ of $v_k$ such that $v_k\to v$ strongly in
$\mathcal{W}^{\alpha,p}$. Clearly, $v\in \mathcal{S}$ and satisfies
 $\|v-(-\phi_{1,\epsilon})\|\leq\eta+\delta<\eta_0$ and
$\tilde{J_{s}}(v)= \lambda_{1,\epsilon}$ which contradicts \eqref{feq13}.

 Now, the boundedness of  $\tilde{J_{s}}(v_k)$  follows from \eqref{feq15}
and \eqref{feq018}. So, we only need to prove that
$\|\tilde{J}_{s}'(v_k)\|_{*}\to 0$. Let $k>\frac{1}{\delta}$ and take
$w\in \mathcal{W}^{\alpha,p}$ tangent to $\mathcal{S}$ at $v_k$.
That is, $ \frac{1}{\epsilon} \int_{\Omega_{\epsilon}} |v_k|^{p-2} v_k w\,dx =0$.
Then by taking
$ u_t:= \frac{\epsilon^{1/p}(v_k+tw)}{\|v_k+tw\|_{L^p(\Omega_{\epsilon})}}$ for
$t\in\mathbb{R}$, we obtain
\begin{align*}
\lim_{t\to 0}\|u_t-(-\phi_{1,\epsilon})\|
&= \|v_k-(-\phi_{1,\epsilon})\|
\leq \|v_k-u_k\|+\|u_k-(-\phi_{1,\epsilon})\| \\
&\leq \frac{1}{k}+\eta<\delta+\eta,
\end{align*}
and
\begin{align*}
\lim_{t\to 0}\|u_t-(-\phi_{1,\epsilon})\|
&= \|v_k-(-\phi_{1,\epsilon})\|
 \geq\|u_k-(-\phi_{1,\epsilon})\|-\|v_k -u_k\| \\
&\geq \eta-\frac{1}{k}>\eta-\delta.
\end{align*}
 Hence, for $t$ small enough $u_t\in  V$ and replacing $u$ by $u_t$
in \eqref{feq020}, we obtain
\[
\tilde{J_{s}}(v_k) \leq \tilde{J_{s}}(u_t)+\frac{1}{k}\|u_t-v_k\|.
\]
Let $r(t): = \epsilon^{1/p}\|v_k + tw\|_{L^p(\Omega_{\epsilon})}$, then
\begin{align*}
&\frac{J_{s}(v_k)-J_{s}(v_k+tw)}{t} \\
&\leq \frac{J_{s}(u_t)+\frac{1}{k}\|u_t-v_k\|-J_{s}(v_k+tw)}{t}\\
&=\frac{1}{k\; t\; r(t)}\|v_k(1-r(t)+tw)\|+\frac{1}{t}
\Big(\frac{1}{r(t)^p}-1\Big)J(v_k+tw).
\end{align*}
Now since
\[
\frac{d}{dt}r(t)^p |_{t=0}= \frac{p}{\epsilon}
\int_{\Omega_{\epsilon}} |v_k|^{p-2} v_k w=0,
\]
we obtain  $\frac{r(t)^p-1}{t}\to 0\;\text{as}\; t\to 0$, and then
 $\frac{1-r(t)}{t}\to 0\; \text{as}\; t\to 0$. Therefore, we obtain
\begin{align}\label{feq16}
|\langle{J_{s}'}(v_k),w\rangle|\leq \frac{1}{k}\|w\|.
\end{align}
 Since $w$ is arbitrary in $\mathcal{W}^{\alpha,p}$, we choose $a_k$
such that
$\frac{1}{\epsilon} \int_{\Omega_{\epsilon}} |v_k|^{p-2} v_k(w-a_k v_k)\,dx =0$.
Replacing $w$ by $w-a_k v_k$ in \eqref{feq16}, we obtain
\[
\big|\langle{J_{s}'}(v_k),w\rangle - a_k \langle{J_{s}'}(v_k), v_k\rangle\big|
\leq \frac{1}{k}\|w-a_k v_k\|.
\]
 Since $\|a_k v_k\|\leq C\|w\|$, we obtain
 $\big|\langle{J_{s}'}(v_k),w \rangle - t_k \int_{\Omega}|v_k|^{p-2}v_k w\,dx \big|
\leq \frac{C}{k}\|w\|$,
where $t_k=\langle{J_{s}'}(v_k), v_k \rangle$. Hence,
$\|\tilde{J_{s}'}(v_k)\|_*\to 0$ as $k\to \infty$, as we required.
\end{proof}

\begin{proposition} \label{fga}
Let $\mathcal{W}^{\alpha,p}$ be a Banach Space. Let $\eta>0$ such that \\
$\|\phi_{1,\epsilon}-(-\phi_{1,\epsilon})\|>\eta$ and
\[
\inf\{\tilde{J_s}(u): u\in \mathcal{S} \text{ and }
\|u-(-\phi_{1,\epsilon})\|=\eta\}>\max\{\tilde{J_s}(-\phi_{1,\epsilon}),
\tilde{J_s}(\phi_{1,\epsilon})\}.
\]
Then
$\Gamma =\{\gamma \in C([-1,1], \mathcal{S}): \gamma(-1)
=- \phi_{1,\epsilon} \text{ and } \gamma(1)=\phi_{1,\epsilon}\}$
is non empty and
\begin{align}\label{feq18}
c(s)=\inf_{\gamma \in \Gamma}\max_{u\in\gamma[-1,1]} J_{s}(u)
\end{align}
is a critical value of $\tilde{J_{s}}$. Moreover $c(s)>\lambda_{1,\epsilon}$.
\end{proposition}

\begin{proof}
We prove that $\Gamma$ is non-empty. To end this, we take
$\phi\in \mathcal{W}^{\alpha,p}$
such that $\phi\not\in\mathbb{R}\phi_{1,\epsilon}$ and consider the path
$t\phi_{1,\epsilon}+(1-|t|)\phi$ then
\[
 w=  \frac{\epsilon^{1/p} (t\phi_{1,\epsilon}+(1-|t|)\phi)}{\| t\phi_{1,\epsilon}
+(1-|t|)\phi\|_{L^p(\Omega_{\epsilon})}}.
\]
 Moreover the (PS)
condition and the geometric assumption are satisfied by the Lemmas
\ref{fle21} and \ref{fle22}. Then by Proposition \ref{fpr1}, $c(s)$ is
a critical value of $\tilde{J_s}$. Using the definition of $c(s)$ we
have $c(s)>\max\{\tilde{J_s}(-\phi_{1,\epsilon}),
\tilde{J_s}(\phi_{1,\epsilon})\}=\lambda_{1,\epsilon}$.
\end{proof}
 Thus we have proved the following result.

\begin{theorem}\label{fth1}
For each $s\geq 0$, the point $(s+c(s),c(s))$, where $c(s)>\lambda_{1,\epsilon}$ is
defined by the minimax formula \eqref{feq18}, then the point
$(s+c(s),c(s))$ belongs to $\Sigma_p$.
\end{theorem}

 It is a trivial fact that $\Sigma_p$ is symmetric with respect to
diagonal. The whole curve, that we obtain using Theorem \ref{fth1}
is denoted by
\[
\mathcal{C}:= \{(s+c(s),c(s)),(c(s),s+c(s)): s\geq 0\}.
\]

\section{First nontrivial curve}

We start this section by establishing that the lines
$\mathbb{R}\times\{\lambda_{1,\epsilon}\}$ and
$\{\lambda_{1,\epsilon}\}\times\mathbb{R}$ are isolated in $\Sigma_p$. Then
we state some topological properties of the functional $\tilde{J_{s}}$
and some Lemmas. Finally, we prove that the curve $\mathcal{C}$
constructed in the previous section is the first non trivial curve in
the spectrum $\Sigma_p$. As a consequence of this, we also obtain a
variational characterization of the second eigenvalue $\lambda_{2,\epsilon}$.

\begin{proposition} \label{prop4.1}
The lines $\mathbb{R}\times \{\lambda_{1,\epsilon}\}$ and
$\{\lambda_{1,\epsilon}\}\times\mathbb{R}$ are isolated in $\Sigma_p$.
In other words, there exists no sequence $(a_k,b_k)\in \Sigma_p$ with
$a_k>\lambda_{1,\epsilon}$ and $b_k > \lambda_{1,\epsilon}$ such that
$(a_k, b_k)\to (a,b)$ with $a=\lambda_{1,\epsilon}$ or $b=\lambda_{1,\epsilon}$.
\end{proposition}

\begin{proof}
Suppose by contradiction that there exists a sequence
$(a_k, b_k)\in\Sigma_p$ with
$a_k$, $b_k>\lambda_{1,\epsilon}$ and $(a_k, b_k)\to (a, b)$ with $a$ or
$b=\lambda_{1,\epsilon}$. Let $u_k\in \mathcal{W}^{\alpha,p}$ be a solution of
\begin{equation}\label{feq17}
\begin{gathered}
\Lambda_{n,p}(1-\alpha)(-\Delta)_{p}^{\alpha} u_k  + |u_k|^{p-2}u_k
= \frac{\chi_{\Omega_\epsilon}}{\epsilon} (a_k(u_k^{+})^{p-1}
- b_k(u_k^{-})^{p-1})\quad \text{in }\Omega,\\
 \mathcal{N}_{\alpha,p} u_k =0 \quad\text{in }\mathbb{R}^n \setminus
 \overline{\Omega},
\end{gathered}
\end{equation}
 with $\frac{1}{\epsilon}\int_{\Omega_{\epsilon}} |u_k|^pdx=1$. Multiplying by
$u_k$ in \eqref{feq17} and integrate, we have
 \begin{align*}
&\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u_{k}(x)-u_{k}(y)|^p }{ |x-y|^{n+p\alpha}}\,dx\,dy
 + \int_{\Omega}|u_k|^{p}dx \\
&= \frac{a_k}{\epsilon} \int_{\Omega_{\epsilon}}(u_k^+)^p\,dx
- \frac{b_k}{\epsilon}\int_{\Omega_{\epsilon}}(u_k^-)^p\,dx\leq a_k.
\end{align*}
Thus $\{u_k\}$ is a bounded sequence in $\mathcal{W}^{\alpha,p}$.
Therefore up to a subsequence $u_k \rightharpoonup u$ weakly in
$\mathcal{W}^{\alpha,p}$ and $u_k\to u$ strongly in $L^p(\Omega_{\epsilon})$.
Then taking limit $k\to \infty$ in the weak formulation of  \eqref{feq17}, we obtain
\begin{equation}\label{feq017}
\begin{gathered}
\Lambda_{n,p}(1-\alpha)(-\Delta)_{p}^{\alpha} u  + |u|^{p-2}u
= \frac{\chi_{\Omega_\epsilon}}{\epsilon} (\lambda_{1,\epsilon}
(u^{+})^{p-1} - b (u^{-})^{p-1}) \quad \text{in } \Omega,\\
 \mathcal{N}_{\alpha,p} u = 0 \quad\text{in }\mathbb{R}^n \setminus
\overline{\Omega}.
\end{gathered}
\end{equation}
 Taking $u^+$ as test function  in \eqref{feq017}  we obtain
\begin{equation}\label{f02}
\Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u,u^+)
 +\int_{\Omega} (u^{+})^{p}dx
= \frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{\epsilon}}(u^+)^p\,dx.
\end{equation}
Observe that
\begin{equation} \label{03}
((u(x)-u(y))(u^{+}(x)-u^{+}(y))= 2 u^{-}(x)u^{+}(y)+(u^{+}(x)-u^{+}(y))^2,
\end{equation}
and
\begin{equation}\label{033}
\begin{aligned}
|u(x)-u(y)|^{p-2} 
&= (|u(x)-u(y)|^2)^{\frac{p-2}2} \\
&=(|u^{+}(x)-u^{+}(y)|^2+ |u^{-}(x)-u^{-}(y)|^2  +2 u^{+}(x)u^{-}(y) \\
&\quad  + 2u^{+}(y)u^{-}(x))^{\frac{p-2}2} \\
&\geq |u^{+}(x)-u^{+}(y)|^{p-2}.
\end{aligned}
\end{equation}
 Using \eqref{03} and \eqref{033} in \eqref{f02} and the definition
of $\lambda_{1,\epsilon}$, we obtain
\begin{align*}
\frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{\epsilon}} (u^+)^p\,dx
&\leq \Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u^{+}(x)-u^{+}(y)|^p}{ |x-y|^{n+p\alpha}}
 \,dx\,dy  +\int_{\Omega} (u^{+})^{p}dx \\
&\leq \frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{\epsilon}} (u^+)^p\,dx.
\end{align*}
Thus
\[
\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u^{+}(x)-u^{+}(y)|^p}{ |x-y|^{n+p\alpha}}\,dx\,dy
 +\int_{\Omega} (u^{+})^{p}dx
 =\frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{\epsilon}} (u^+)^p\,dx,
\]
so either $u^+\equiv 0$ or $u=\phi_{1,\epsilon}$.
If $u^+\equiv 0$ then $u\leq 0$ and \eqref{feq017} implies that
 $u$ is an eigenfunction with $u\leq 0$ so that $u=-\phi_{1,\epsilon}$.
So, in any case $u_k$ converges to either $\phi_{1,\epsilon}$ or
$-\phi_{1,\epsilon}$ in $L^{p}(\Omega_{\epsilon})$. Thus
\begin{equation} \label{feq019}
\text{either } |\{x\in \Omega_{\epsilon} : u_k(x)<0\}|\to 0 \text{ or }
|\{x\in\Omega_{\epsilon} :u_k(x)> 0\}|\to 0
\end{equation}
as $k\to \infty$.
 On the other hand, taking $u_k^+$ as test function in \eqref{feq17}, we obtain
\begin{align}\label{feq016}
\Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u_k,u_{k}^{+})
+\int_{\Omega} |u_{k}|^{p-2}u_{k}u_{k}^+
 = \frac{a_k}{\epsilon}\int_{\Omega_{\epsilon}}(u_{k}^+)^p.
\end{align}
 Using H\"{o}lders inequality, fractional Sobolev embeddings and \eqref{feq016},
we obtain
\begin{align*}
&\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u^{+}_k(x)-u^{+}_k(y)|^p}{|x-y|^{n+p\alpha}}
 \,dx\,dy +\int_{\Omega} (u_{k}^{+})^{p}dx\\
&\leq\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|u_{k}(x)-u_{k}(y)|^{p-2}
 (u_{k}(x)-u_{k}(y))(u^{+}_k(x)-u^{+}_k(y))}{|x-y|^{n+p\alpha}}\,dx\,dy \\
&\quad +\int_{\Omega} |u_{k}|^{p-2}u_{k}u_{k}^+dx \\
&=\Lambda_{n,p}(1-\alpha) \mathcal{H}_{\alpha,p}(u_k, u_{k}^{+})
 +\int_\Omega |u_k|^{p-2} u_k u_{k}^{+}dx\\
&= \frac{a_k}{\epsilon}\int_{\Omega_{\epsilon}}(u_{k}^+)^p\,dx\\
&\leq \frac{a_k}{\epsilon} C |\{x\in \Omega_{\epsilon} :
 u_k(x)>0\}|^{1-\frac{p}{q}}\|u_k^+\|^{p}
\end{align*}
 with a constant $C>0$, $p<q\leq p^*=\frac{np}{n-p\alpha}$. Then we have
\[
|\{x\in \Omega : u_k(x)>0\}|^{1-\frac{p}{q}}
\geq \epsilon a_k ^{-1}C^{-1}\min\{\Lambda_{n,p}(1-\alpha),1\}.
\]
Similarly, one can show that
\[
|\{x\in \Omega : u_k(x)<0\}|^{1-\frac{p}{q}}
\geq \epsilon b_k ^{-1}C^{-1}\min\{\Lambda_{n,p}(1-\alpha),1\}.
\]
Since $(a_k,b_k)$ does not belong to the trivial lines
$\lambda_{1,\epsilon}\times\mathbb{R}$ and $\mathbb{R}\times \lambda_{1,\epsilon}$
 of $\Sigma_p$,  by  \eqref{feq17} we conclude that $u_k$ changes sign.
Hence, from the above inequalities, we obtain a
contradiction with \eqref{feq019}.
Therefore, the trivial lines $\lambda_{1,\epsilon}\times\mathbb{R}$ and
$\mathbb{R}\times \lambda_{1,\epsilon}$ are isolated in $\Sigma_p$.
\end{proof}

\begin{lemma}[\cite{cfg}] \label{fle1}
Let $\mathcal{S}= \{u\in \mathcal{W}^{\alpha,p} :
\frac{1}{\epsilon}\int_{\Omega_{\epsilon}}|u|^p\,dx =1\}$ then
\begin{enumerate}
\item $\mathcal{S}$ is locally arcwise connected.
\item Any open connected subset $\mathcal{O}$ of $\mathcal{S}$ is arcwise connected.
\item If $\mathcal{O}^{'}$ is any connected component of an open set
$\mathcal{O}\subset \mathcal{S}$, then $\partial \mathcal{O}'\cap \mathcal{O}=\emptyset$.
\end{enumerate}
\end{lemma}

\begin{lemma} \label{fle01}
Let $\mathcal{O}=\{u\in \mathcal{S} : \tilde{J_s}(u)<r\}$, then any connected
component of $\mathcal{O}$ contains a critical point of $\tilde{J_s}$.
\end{lemma}

\begin{proof}
Let $\mathcal{O}_1$ be any connected component of $\mathcal{O}$, let
$d=\inf\{\tilde{J_{s}}(u): u\in  \overline{\mathcal{O}}_1\}$, where
$\overline{\mathcal{O}}_1$ denotes the closure of $\mathcal{O}_1$ in
 $\mathcal{W}^{\alpha,p}$. We show that there exists
$u_0\in \mathcal{W}^{\alpha,p}$ such that
$\tilde{J_{s}}(u_0)=d$. For this let $u_k\in \mathcal{O}_1$ be a minimizing
sequence such that $\tilde{J_{s}}(u_k)\leq d+\frac{1}{2k^2}$.
For each $k$, by applying Ekeland's Variational principle, we obtain
a sequence $v_k\in \overline{\mathcal{O}}_1$ such that
\[
\tilde{J_{s}}(v_k)\leq \tilde{J_{s}}(u_k),\;
 \|v_k-u_k\| \leq \frac{1}{k},\quad
\tilde{J_{s}}(v_k)\leq \tilde{J_{s}}(v) + \frac{1}{k}\|v-v_k\|\quad\forall
  v\in \overline{\mathcal{O}}_1.
\]
For $k$ large enough, we have
\[
\tilde{J_{s}}(v_k)\leq \tilde{J_{s}}(u_k)\leq d+ \frac{1}{2k^2}<r,
\]
then $v_k\in \mathcal{O}$. By Lemma \ref{fle1}, we obtain
$v_k\not\in \partial\mathcal{O}_1$ so $v_k\in \mathcal{O}_1$. On the other hand,
for $t$ small enough and $w$ such that
$\frac{1}{\epsilon}\int_{\Omega_{\epsilon}}  |v_k|^{p-2}v_k w\,dx=0$, we have
\[
 u_t:=\frac{\epsilon^{1/p}(v_k+tw)}{\|v_k+tw\|_{L^p{(\Omega_{\epsilon})}}}
\in \overline{\mathcal{O}}_1.
\]
Then $\tilde{J_{s}}(v_k)\leq \tilde{J_{s}}(u_t) + \frac{1}{k}\|u_t-v_k\|$.
 Following the same calculation as in Lemma \ref{fle22}, we have that
$v_k$ is a Palais-Smale sequence for
$\tilde{J_{s}}$ on $\mathcal{S}$ i.e $\tilde{J_{s}}(v_k)$ is bounded and
$\|\tilde{J_{s}}(v_k)\|_{*}\to 0$. Again by Lemma \ref{fle21}, up to
a subsequence $v_k\to u_0$ strongly in $\mathcal{W}^{\alpha,p}$ and hence
$\tilde{J_{s}}(u_0)=d<r$ and moreover $u_0\in \mathcal{O}$. By part $3$ of
Lemma \ref{fle1}, $u_0\not\in \partial \mathcal{O}_1$ so $u_0\in \mathcal{O}_1$.
Hence $u_0$ is a critical point of $\tilde{J_{s}}$, which completes
the proof.
\end{proof}

Before proving the main Theorem \ref{thm1.1}, we state some Lemmas and the
details of the proof can be found in \cite{second} and \cite{pal}.

\begin{lemma}[{\cite[Lemma B.1]{second}}]  \label{f2}
Let $1 \leq p \leq \infty$ and $U,V \in\mathbb{R}$ such that $U.V \leq 0$.
Define the following function
\[
g(t)=|U -tV|^p+|U-V|^{p-2}(U-V)V|t|^p,\; t \in\mathbb{R}.
\]
Then we have
\[
g(t)\leq g(1)=|U-V|^{p-2}(U-V)U,\; t \in\mathbb{R}.
\]
\end{lemma}

\begin{lemma}[{\cite[Lemma 4.1]{pal}}] \label{f3}
Let $\alpha\in (0,1)$ and $p>1$. For any non-negative functions $u$,
$v \in \mathcal{W}^{\alpha,p}$, consider the function
$\sigma_t:= \left[(1-t)v^p(x)+ tu^p(x)\right]^{1/p}$ for all $t\in[0, 1]$. Then
\begin{equation*}
[\sigma_t]_{\alpha,p} \leq (1-t)[v]_{\alpha,p}
 + t[u]_{\alpha,p},\quad\text{for all } t \in [0,1],
\end{equation*}
where $ [u]_{\alpha,p}$ is defined in \eqref{feq12}.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 Assume by contradiction that there exists $\mu$ such that
$\lambda_{1,\epsilon}<\mu<c(s)$ and $(s+\mu,\mu)\in \Sigma_p$.
Using the fact that $\{\lambda_{1,\epsilon}\}\times\mathbb{R}$ and
$\mathbb{R}\times \{\lambda_{1,\epsilon}\}$ are isolated in
$\Sigma_p$ and $\Sigma_p$ is closed we can choose such a point with
$\mu$ minimum. In other words, $\tilde{J_{s}}$ has a critical value $\mu$
with $\lambda_{1,\epsilon}<\mu<c(s)$, but there is no critical value in
$(\lambda_{1,\epsilon},\mu)$. If we construct a path connecting from
$\phi_{1,\epsilon}$ to $-\phi_{1,\epsilon}$ such that
$\tilde{J_s}\leq \mu$, then we obtain a
contradiction with the definition of $c(s)$, which wiil complete the proof.

 Let $u\in \mathcal{S}$ be a critical point of $\tilde{J_s}$ at level
$\mu$. Then $u$ satisfies
\begin{align}
&\Lambda_{n,p}(1-\alpha) \int_{Q} \frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))(v(x)-v(y))}
 { |x-y|^{n+p\alpha}}\,dx\,dy 
+\int_{\Omega}|u|^{p-2}uv\,dx  \nonumber \\
& = \frac{(s+\mu)}{\epsilon}\int_{\Omega_{\epsilon}}(u^{+})^{p-1} v\,dx
-\frac{\mu}{\epsilon} \int_{\Omega_{\epsilon}}(u^{-})^{p-1}v\,dx \label{feq20}
\end{align}
 for all $v\in \mathcal{W}^{\alpha,p}$. Substituting $v=u^{+}$ in \eqref{feq20},
we have
\begin{align}
&\Lambda_{n,p}(1-\alpha) \int_{Q} \frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))
 (u^+(x)-u^+(y))}{ |x-y|^{n+p\alpha}}\,dx\,dy
+\int_{\Omega}(u^+)^{p}dx \nonumber \\
& = \frac{(s+\mu)}{\epsilon}\int_{\Omega_{\epsilon}}(u^{+})^p\,dx. \label{feq21}
\end{align}
Since, $|u^+(x)-u^{+}(y)|^p \leq|u(x)-u(y)|^{p-2}(u(x)-u(y))(u^+(x)-u^+(y)$,
we obtain
\[
\Lambda_{n,p}(1-\alpha)\int_{Q} \frac{|u^+(x)-u^{+}(y)|^p}{ |x-y|^{n+p\alpha}}\,dx\,dy
+\int_{\Omega}(u^+)^pdx-\frac{s}{\epsilon}\int_{\Omega_{\epsilon}}(u^{+})^pdx
\leq \mu.
\]
Again substituting  $v=u^-$ in \eqref{feq20}, we have
\begin{align}
&\Lambda_{n,p}(1-\alpha) \int_{Q} \frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))
 (u^-(x)-u^-(y))}{ |x-y|^{n+p\alpha}}\,dx\,dy 
-\int_{\Omega}(u^-)^{p}dx  \nonumber\\
&= -\frac{\mu}{\epsilon}\int_{\Omega_{\epsilon}} (u^{-})^p\,dx. \label{feq22}
\end{align}
Therefore,
\begin{align*}
&\Lambda_{n,p}(1-\alpha) \int_{Q} \frac{|u(x)-u(y)|^{p-2}((u^-(x)-u^-(y))^2
 +2u^+(x)u^-(y))}{ |x-y|^{n+p\alpha}}\,dx\,dy \\
&+\int_{\Omega}(u^-)^{p}dx \\
&= \frac{\mu}{\epsilon}\int_{\Omega_{\epsilon}}(u^{-})^p\,dx
\end{align*}
Since $|u^-(x)-u^-(y)|^{p}\leq |u(x)-u(y)|^{p-2}[(u^-(x)
-u^-(y))^2+2u^+(x)u^-(y)]$, ti follows that
\[
\Lambda_{n,p}(1-\alpha)\int_{Q} \frac{|u^-(x)-u^{-}(y)|^p}{ |x-y|^{n+p\alpha}}
\,dx\,dy+\int_{\Omega}|u^-|^pdx\leq \mu.
\]
 Therefore, from all above relations, one can easily verify that
\[
\tilde{J_{s}}(u)=\mu,\; \tilde{J_{s}}
 \Big(\frac{\epsilon^\frac{1}{p} u^+}{ \|u^+\|_{L^p(\Omega_{\epsilon})}}\Big)
\leq \mu,\tilde{J_{s}}\Big(\frac{\epsilon^\frac{1}{p}u^-}{\|u^-\|_{L^p(\Omega_{\epsilon})}}
 \Big)
 \leq \mu -s,\tilde{J_{s}}
 \Big( \frac{-\epsilon^\frac{1}{p}u^-}{\|u^-\|_{L^p(\Omega_{\epsilon})}}\Big)\leq \mu .
\]
Since, $u$ changes sign (see Proposition \ref{prop3}), the following paths are
 well-defined on $\mathcal{S}$:
\begin{gather*}
u_{1}(t)=\frac{u^+- (1-t)u^-}{\epsilon^\frac{-1}{p} \|u^+
 - (1-t)u^-\|_{L^p(\Omega_{\epsilon})}},\\
 u_2(t)  =\frac{[(1-t)(u^{+})^p+ t(u^{-})^p]^{1/p}}{\epsilon^\frac{-1}{p} \|(1-t)(u^{+})^p
 + t(u^{-})^p\|_{L^p(\Omega_{\epsilon})}},\\
u_3(t)=\frac{(1-t)u^{+}-u^{-}}{\epsilon^\frac{-1}{p}
 \|(1-t)u^{+}-u^{-}\|_{L^p(\Omega_{\epsilon})}}.
\end{gather*}
Then, using the above calculations and Lemma \ref{f2} for $U=u^+(x)-u^+(y)$ and
 $V=u^-(x)-u^-(y)$, one can easily obtain that for all $t\in[0,1]$,
\begin{align*}
\tilde{J_{s}}(u_1(t))
&\leq \frac{\Lambda_{n,p}(1-\alpha)
\int_{Q}\frac{|U-V|^{p-2}(U-V)U}{|x-y|^{n+p\alpha}}
 +\int_{\Omega}(u^+)^p-\frac{s}
 {\epsilon}\int_{\Omega_{\epsilon}}(u^+)^p }{\epsilon^{-1}
 \|u^+- (1-t)u^-\|_{L^p(\Omega_{\epsilon})}^p}\\
&\quad +\frac{|1-t|^p\big[-\Lambda_{n,p}(1-\alpha)
 \int_{Q}\frac{|U-V|^{p-2}(U-V)V}{|x-y|^{n+p\alpha}}+\int_{\Omega}(u^-)^p\big]}
 {\epsilon^{-1}\|u^+- (1-t)u^-\|_{L^p(\Omega_{\epsilon})}^p}
=\mu,
\end{align*}
 by using \eqref{feq21} and \eqref{feq22}.
Now using Lemma \ref{f3} we have
\begin{align*}
\tilde{J_{s}}(u_2(t))
&\leq \frac{(1-t)\big[\Lambda_{n,p}(1-\alpha)
 \int_{Q}\frac{|u^+(x)-u^+(y)|^p}{|x-y|^{n+p\alpha}}
 +\int_{\Omega}(u^+)^p -\frac{s}{\epsilon}
 \int_{\Omega_{\epsilon}}(u^+)^p\big] }{\epsilon^{-1} \|(1-t)(u^{+})^p
 + t(u^{-})^p\|^p_{L^p(\Omega_{\epsilon})}}\\
& \quad +\frac{t \big[\Lambda_{n,p}(1-\alpha)
 \int_{Q}\frac{|u^-(x)-u^-(y)|^p}{|x-y|^{n+p\alpha}}
 +\int_{\Omega}(u^-)^p- \frac{s}{\epsilon} \int_{\Omega_{\epsilon}}(u^-)^p\big]}{\epsilon^{-1}
  \|(1-t)(u^{+})^p+ t(u^{-})^p\|^p_{L^p(\Omega_{\epsilon})}}\\
& \leq \mu - \frac{s t \int_{\Omega_{\epsilon}}(u^-)^p}{\epsilon^{-1} \|(1-t)(u^{+})^p
 + t(u^{-})^p\|^p_{L^p(\Omega_{\epsilon})}}
 \leq \mu.
\end{align*}
Again, by Lemma \ref{f2}, for $U=u^-(y)-u^-(x)$ and $V=u^+(y)-u^+(x)$, we obtain
\begin{align*}
&\tilde{J_{s}}(u_3(t)) \\
&\leq \frac{\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|U-V|^{p-2}(U-V)U}
 {|x-y|^{n+p\alpha}}+\int_{\Omega}(u^-)^p  }
 {\epsilon^{-1} \|(1-t)u^{+}-u^{-}\|_{L^p(\Omega_{\epsilon})}^p}\\
&\quad + \frac{ |1-t|^p \big[ -\Lambda_{n,p}(1-\alpha)
\int_{Q}\frac{|U-V|^{p-2}(U-V)V}{|x-y|^{n+p\alpha}}
 + \int_{\Omega}(u^+)^p-\frac{s}{\epsilon}\int_{\Omega_{\epsilon}}(u^+)^p \big]}
{\epsilon^{-1}\|(1-t)u^{+}-u^{-}\|_{L^p(\Omega_{\epsilon})}^p}\\
&=\mu,\quad \text{by using \eqref{feq21} and \eqref{feq22}}.
\end{align*}
 Let $\mathcal{O} = \{v\in\mathcal{S}: \tilde{J_s}(v)<\mu-s\}$. Then clearly
$\phi_{1,\epsilon}\in \mathcal{O}$, while $-\phi_{1,\epsilon}\in \mathcal{O}$ if
$\mu- s>\lambda_{1,\epsilon}$.
Moreover $\phi_{1,\epsilon}$ and $-\phi_{1,\epsilon}$ are the only possible
critical points of $\tilde{J_s}$ in $\mathcal{O}$ because of the choice of $\mu$.
 We note that
\[
\tilde{J_s}\Big(\frac{\epsilon^{1/p} u^-}{\|u^-\|_{L^p(\Omega_{\epsilon})}}\Big)\leq \mu-s,
\]
$\epsilon^{1/p} u^-/\|u^-\|_{L^p(\Omega_{\epsilon})}$ does not change sign and vanishes on a
set of positive measure, it is not a critical point of
$\tilde{J_s}$. Therefore, there exists a $C^1$ path
 $\eta:[-\delta,\delta]\to \mathcal{S}$ with
 $\eta(0)= \epsilon^{1/p}u^-/\|u^-\|_{L^p(\Omega_{\epsilon})}$ and
$\frac{d}{dt}\tilde{J_s}(\eta(t))|_{t=0}\ne 0$. Using this path we
can move from $\epsilon^{1/p} u^-/\|u^-\|_{L^p(\Omega_{\epsilon})}$
to a point $v$ with $\tilde{J_s}(v)<\mu-s$. Taking a connected component
of $\mathcal{O}$ containing $v$ and applying Lemma \ref{fle01} we have that either
$\phi_{1,\epsilon}$ or $-\phi_{1,\epsilon}$ is in this component.
Let us assume that it is $\phi_{1,\epsilon}$. So we continue by a path
$u_{4}(t)$ from $\epsilon^{1/p} u^-/\|u^-\|_{L^p(\Omega_{\epsilon})}$
to $\phi_{1,\epsilon}$ which is at level less than $\mu$.
Then the path $-u_{4}(t)$ connects
$-\epsilon^{1/p} u^-/\|u^-\|_{L^p(\Omega_{\epsilon})}$ to $-\phi_{1,\epsilon}$.
We observe that 
\[
|\tilde{J_s}(u)- \tilde{J_s}(-u)|\leq s.
\]
 Then it follows that
\[
\tilde{J_s}(-u_4(t))\leq \tilde{J_s}(u_4(t))+s\leq \mu-s+s
= \mu \quad \text{for all } t.
\]
Connecting $u_1(t)$, $u_2(t)$ and $u_4(t)$, we obtain a path from $u$
to $\phi_{1,\epsilon}$ and joining $u_3(t)$ and $-u_4(t)$ we obtain a path from
$u$ to $-\phi_{1,\epsilon}$. These yields a path $\gamma(t)$ on $\mathcal{S}$ joining
from $\phi_{1,\epsilon}$ to $-\phi_{1,\epsilon}$ such that
$\tilde{J_s}(\gamma(t))\leq \mu$ for all $t$, which concludes the proof.
\end{proof}

As a consequence of Theorem \ref{thm1.1}, we give a variational characterization
of the second value of \eqref{eq2}.

\begin{corollary}
The second eigenvalue $\lambda_2$ of \eqref{eq2} has the variational
characterization given by
\[
  \lambda_2 :=  \inf_{\gamma \in \Gamma}\sup_{u\in \gamma}
\Big(\Lambda_{n,p}(1-\alpha)\int_{Q}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
+\int_{\Omega}|u|^p\,dx\Big),
\]
where $\Gamma$ is as in Proposition \ref{fga}.
\end{corollary}

\begin{proof}
Taking $s=0$ in Theorem \ref{thm1.1} and using \eqref{feq18} we have  
$c(0)=\lambda_2$.
\end{proof}


\section{Properties of the curve $\mathcal{C}$}

In this section, we prove that the curve $\mathcal{C}$ is Lipschitz
continuous, has a certain asymptotic behavior and is strictly
decreasing.
 For $A \subset \Omega_{\epsilon}$, define the eigenvalue problem
\begin{equation}\label{feq51}
\begin{gathered}
 \Lambda_{n,p}(1-\alpha)(-\Delta)^{\alpha}_p u +|u|^{p-2}u
= \frac{\chi_{A}}{\epsilon}(\lambda |u|^{p-2}u) \quad \text{in } \Omega, \\
 \mathcal{N}_{\alpha,p} u = 0  \quad  \text{in }
\mathbb{R}^n \setminus \overline{\Omega},
\end{gathered}
\end{equation}
Let $\lambda_{1,\epsilon}(A)$ denotes the first eigenvalue of \eqref{feq51}, then
\begin{align*}
\lambda_{1,\epsilon}(A) &= \inf_{u\in \mathcal{W}^{\alpha,p}}
\Big\{ \Lambda_{n,p}(1-\alpha)\int_{Q}\frac{|u(x)-u(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy \\
&\quad + \int_{\Omega} |u|^p\,dx : \int_{A}|u|^p\,dx= \epsilon \Big\}.
\end{align*}

\begin{lemma} \label{f111}
Let $A$, $B$ be two bounded open sets in $\Omega_{\epsilon}$, with
$A\subsetneq B$ and $B$ is connected then
$\lambda_{1,\epsilon}(A)>\lambda_{1,\epsilon}(B)$.
\end{lemma}

\begin{proof}
Clearly from the definition of $\lambda_{1,\epsilon}$, we have
$\lambda_{1,\epsilon}(A)\geq\lambda_{1,\epsilon}(B)$. Let if possible equality
holds and let $\phi_{1,\epsilon}$ be a non-negative normalized eigenfunction
associated to $\lambda_{1,\epsilon}(A)$ such that $\phi_{1,\epsilon}$
is equal to zero outside
$A$. Therefore, from the definition of $\lambda_{1,\epsilon}(A)$, we have
\begin{align*}
&\Lambda_{n,p}(1-\alpha)\int_{Q} \frac{|\phi_{1,\epsilon}(x)
 -\phi_{1,\epsilon}(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
 +\int_{\Omega}|\phi_{1,\epsilon}|^pdx \\
&= \frac{\lambda_1(A)}{\epsilon}\int_{A} \phi_{1,\epsilon}^p\,dx
  = \frac{\lambda_1(B)}{\epsilon}\int_{B} \phi_{1,\epsilon}^p\,dx.
\end{align*}
It implies $\phi_{1,\epsilon}$ is an eigenfunction associated to
$\lambda_{1,\epsilon}(B)$. But this is impossible since $B$ is connected and
$\phi_{1,\epsilon}$ vanishes on
$B\setminus A \ne\emptyset$.
\end{proof}

\begin{proposition} \label{prop5.2}
The curve $s\to (s+c(s), c(s))$, $s\in\mathbb{R}^+$ is Lipschitz
continuous and strictly decreasing $($in the sense that $s_1<s_2$
implies $s_1+c(s_1)<s_2+c(s_2)$ and	$c(s_1)>c(s_2))$.
\end{proposition}

\begin{proof}
Let $s_1<s_2$ then $\tilde{J}_{s_1}(u)>\tilde{J}_{s_2}(u)$ for all
 $u\in \mathcal{S}$. So we have $c(s_1)>c(s_2)$. Now for every $\eta>0$
 there exists $\gamma\in\Gamma$ such that
$\max_{u\in \gamma[-1,1]}\tilde{J}_{s_2}(u)\leq c(s_2)+\eta$,
and so
\[
0\leq c(s_1)- c(s_2)\leq \max_{u\in \gamma[-1,1]}\tilde{J}_{s_1}(u)
- \max_{u\in \gamma[-1,1]}\tilde{J}_{s_2}(u)+\eta.
\]
Let $u_0\in\gamma[-1,1]$ such that
$\max_{u\in \gamma[-1,1]}\tilde{J}_{s_1}(u)=\tilde{J}_{s_1}(u_0)$.
 Then
\[
0\leq c(s_1)- c(s_2)\leq \tilde{J}_{s_1}(u_0)-\tilde{J}_{s_2}(u_0)+\eta
\leq s_2-s_1+\eta,
\]
as $\eta>0$ is arbitrary so the curve $\mathcal{C}$ is Lipschitz continuous
with constant $\leq 1$.

 Next,  to prove that the curve is decreasing, it suffices to argue
for $s>0$. Let $0<s_1<s_2$ then it implies $c(s_1)> c(s_2)$.
 On the other hand, since $(s_1+c(s_1), c(s_1))$, $(s_2+c(s_2), c(s_2))\in\Sigma_p$,
 Theorem \ref{thm1.1} implies that $s_1+c(s_1)< s_2+c(s_2)$, which completes
the proof.
\end{proof}

 As $c(s)$ is decreasing and positive so the limit of $c(s)$ exists as
$s\to \infty$.

\begin{theorem} \label{thm5.3}
If $n\geq p \alpha$, then the limit of $c(s)$ as $s\to \infty$ is	
$\lambda_{1,\epsilon}$.
\end{theorem}

\begin{proof}
For $n\geq  p \alpha $, we can choose a function $\phi\in \mathcal{W}^{\alpha,p} $
such	that there does not exist $r\in\mathbb{R}$ such that
 $\phi(x)\leq r \phi_{1,\epsilon}(x)$ a.e. in $\Omega_{\epsilon}$
(it suffices to take $\phi\in \mathcal{W}^{\alpha,p}$ such that it is
unbounded from above in a neighborhood of some point $0\ne x\in\Omega_{\epsilon}$).
Suppose that the result is not true then there exists $\delta>0$ such that
$\max_{u\in \gamma[-1,1]}\tilde{J}_{s}(u)
\geq \lambda_{1,\epsilon} +\delta$ for all $\gamma\in \Gamma$ and all $s\geq 0$.
Consider a path	$\gamma\in \Gamma$ by
\[
\gamma(t)= \frac{\epsilon^{1/p}(t\phi_{1,\epsilon} +(1-|t|)\phi)}{\|t\phi_{1,\epsilon}
+(1-|t|)\phi\|_{L^{p}(\Omega_{\epsilon})}}\quad \text{for all }t\in[-1,1].
\]
Now, for every $s>0$, let $t_s\in [-1,1]$ satisfy
$\max_{t\in[-1,1]}\tilde{J_s}(\gamma(t))= \tilde{J_s}(\gamma(t_s))$.
Let $v_{t_s}= {t_s\phi_{1,\epsilon} +(1-|t_s|)\phi}$. Then we have
\begin{equation} \label{feq4.1}
\tilde{J_s}(v_{t_s})\geq \frac{(\lambda_{1,\epsilon}+\delta)}{\epsilon}
\int_{\Omega_{\epsilon}} |v_{t_s}|^p.
\end{equation}
Letting $s\to \infty$, we can assume a subsequence $t_s\to \tilde t\in[-1,1]$.
Then $v_{t_s}$ is bounded in $\mathcal{W}^{\alpha,p}$. So, from last inequality
we obtain $\int_{\Omega_{\epsilon}}(v_{t_s}^{+})^p\,dx \to 0$ as $s\to \infty$,
 which forces
\[
\int_{\Omega_{\epsilon}}((\tilde{t}\phi_{1,\epsilon} +(1-|\tilde{t}|)\phi)^+)^p\,dx=0.
\]
Hence, $\tilde{t}\phi_{1,\epsilon} +(1-|\tilde{t}|)\phi \leq 0$.
By the choice of $\phi$, $\tilde{t}$ must be equal to $-1$.
Passing to the limit in \eqref{feq4.1}, we obtain
\begin{align*}
\frac{\lambda_{1,\epsilon}}{\epsilon}\int_{\Omega_{\epsilon}}\phi_{1,\epsilon}^{p}dx
&=\Lambda_{n,p}(1-\alpha) \int_{Q}\frac{|\phi_{1,\epsilon}(x)
 -\phi_{1,\epsilon}(y)|^p}{|x-y|^{n+p\alpha}}\,dx\,dy
 +\int_{\Omega}|\phi_{1,\epsilon}|^p\,dx   \\
& \geq \frac{(\lambda_{1,\epsilon}+\delta)}{\epsilon}\int_{\Omega_{\epsilon}} |v_{t_s}|^p.
\end{align*}
We arrive at a contradiction that $\delta \leq 0$. Hence
$c(s)\to \lambda_{1,\epsilon}$ as $s\to \infty$.
\end{proof}

\section{Non resonance between $(\lambda_1,\lambda_1)$ and $\mathcal{C}$}

 In this section, we study the non-resonance problem with respect
to the Fu\v{c}ik spectrum for $p=2$ case.

\begin{lemma} \label{fle31}
Let $(a,b)\in \mathcal{C}$, and let $m(x)$, $b(x)\in L^{\infty}(\Omega)$ satisfying
\begin{equation} \label{feq31}
\lambda_{1,\epsilon}\leq m(x)\leq a,\quad \lambda_{1,\epsilon}\leq b(x)\leq b.
\end{equation}
 Assume that
\begin{equation} \label{feq32}
\lambda_{1,\epsilon}<m(x)\text{ and $\lambda_{1,\epsilon}<b(x)$
 on subsets of positive measure of } \Omega_{\epsilon}.
\end{equation}
Then any non-trivial solution $u$ of
\begin{equation}\label{feq33}
\begin{gathered}
\Lambda_{n,2}(1-\alpha)(-\Delta)^{\alpha} u +u
 = \frac{\chi_{\Omega_{\epsilon}}}{\epsilon}(m(x)u^{+} - b(x) u^{-}) \quad \text{in }
\Omega, \\
 \mathcal{N}_{\alpha,2} u = 0 \quad  \text{in }
\mathbb{R}^n \setminus \overline{\Omega},
\end{gathered}
\end{equation}
changes sign in $\Omega_{\epsilon}$ and
\[
m(x)=a \text{ a.e. on }\{x\in \Omega_{\epsilon} : u(x)>0\}\, \quad
b(x)=b \text{ a.e. on }\{x\in \Omega_{\epsilon} : u(x)<0\}.
\]
\end{lemma}

\begin{proof}
Let $u$ be a nontrivial solution of \eqref{feq33}. Replacing $u$ by	$-u$
if necessary, we can assume that the point $(a,b)$ in $\mathcal{C}$
is such that $a\geq b$. We first claim that $u$ changes sign in
 $\Omega_{\epsilon}$. Suppose by contradiction that this is not true,
first consider the case $u\geq 0$, (case $u\leq 0$ can be proved similarly).
 Then $u$ solves
\[
\Lambda_{n,2}(1-\alpha)(-\Delta)^{\alpha}u +u
= \frac{\chi_{\Omega_{\epsilon}}}{\epsilon}m(x) u^{+} \quad \text{ in }\Omega, \quad
\mathcal{N}_{\alpha,2} u = 0 \quad  \text{in }
\mathbb{R}^n \setminus \overline{\Omega}.
\]
This means that $u $ is an eigenfunction of the problem with weight $m(x)$
 corresponding to the eigenvalue equal to one. From the definition of the first eigenvalue of the problem with weight $m(x)\geq \lambda_{1,\epsilon}$, we have
\begin{equation}\label{feq34}
\begin{aligned}
&\lambda_{1,\epsilon}(m(x)) \\
&=  \inf_{0 \not \equiv u\in \mathcal{W}^{\alpha,2} }
\Big\{ \frac{\Lambda_{n,2}(1-\alpha)\int_{Q}
 \frac{|u(x)-u(y)|^2}{|x-y|^{n+2\alpha}}\,dx\,dy
 +\int_{\Omega}|u|^2(x)\,dx}
 {\frac{1}{\epsilon}\int_{\Omega_{\epsilon}}m(x)  |u|^2\,dx}:  \Big\}=1.
\end{aligned}
\end{equation}
 From \eqref{fle31}, \eqref{feq31} and \eqref{feq34}, we have	
\begin{align*}
1=&\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} {|\phi_{1,\epsilon}(x)
 -\phi_{1,\epsilon}(y)|^2}{|x-y|^{-(n+2\alpha)}}\,dx\,dy
 +\int_{\Omega}|\phi_{1,\epsilon}|^2(x)\,dx}{\lambda_{1,\epsilon}}\\
>&\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} {|\phi_{1,\epsilon}(x)
 -\phi_{1,\epsilon}(y)|^2}{|x-y|^{-(n+2\alpha)}}\,dx\,dy
 +\int_{\Omega}|\phi_{1,\epsilon}|^2(x)\,dx}
 { \frac{1}{\epsilon}\int_{\Omega_{\epsilon}} m(x)|\phi_{1,\epsilon}|^2\,dx}\\
&\geq 1,
\end{align*}
which is a contradiction. Hence, $u$ changes sign on $\Omega_{\epsilon}$.

 Let suppose by contradiction that either
\begin{equation}\label{feq35}
|\{x\in \Omega_{\epsilon} : m(x)<a \text{ and } u(x)>0\}|>0
\end{equation}
or
\begin{equation} \label{feq36}
|\{x\in \Omega_{\epsilon} : b(x)<b \text{ and } u(x)<0\}|>0.
\end{equation}
 Suppose that \eqref{feq35} holds (a similar argument will hold for \eqref{feq36}).
Put $a-b= s\geq 0$. Then $b= c(s)$, where $c(s)$
is given by \eqref{feq18}. We show that there exists a path $\gamma\in\Gamma$ such that
\begin{align}\label{feq37}
\max_{u\in \gamma[-1,1]}\tilde{J}_{s}(u)<b,
\end{align}
which gives a contradiction with the definition of $c(s)$, prove the last
part of the Lemma.

 To construct $\gamma$ we show that there exists of a function
$v\in \mathcal{W}^{\alpha,2}$ such that it changes sign and satisfies
\begin{equation} \label{feq38}
\begin{gathered}
\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} {|v^{+}(x)-v^{+}(y)|^2}{|x-y|^{-(n+2\alpha)}}
\,dx\,dy + \int_{\Omega}(v^{+})^2\,dx}{\frac{1}{\epsilon}
\int_{\Omega_{\epsilon}} (v^{+})^2\,dx}
<a , \\
\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} {|v^{-}(x)-v^{-}(y)|^2}{|x-y|^{-(n+2\alpha)}}
\,dx\,dy +\int_{\Omega}(v^{-})^2\,dx}{\frac{1}{\epsilon}
\int_{\Omega_{\epsilon}} (v^{-})^2\,dx} <b.
\end{gathered}
\end{equation}
Let  $\mathcal{O}_1$ be a component of $\{x\in \Omega_{\epsilon} : u(x)>0\}$ such that
 $|\{x\in\mathcal{O}_1:m(x)<a\}| >0$ and $\mathcal{O}_2$ be a component of
$\{x\in \Omega_{\epsilon} : u(x)<0\}$ such that
$|\{x\in\mathcal{O}_2:b(x)<b\}| >0$.
 Define the eigenvalue problem
\begin{equation}\label{feq42}
\begin{gathered}
\Lambda_{n,2}(1-\alpha)(-\Delta)^{\alpha} u +u
= \frac{\chi_{\mathcal{O}_{i}}}{\epsilon}(\lambda u) \quad \text{in }\Omega, \\
 \mathcal{N}_{\alpha,2} u = 0  \quad  \text{in }
 \mathbb{R}^n \setminus \overline{\Omega},\quad i=1,2.
\end{gathered}
\end{equation}
Let $\lambda_{1,\epsilon}(\mathcal{O}_i)$ denote the first eigenvalue of
 \eqref{feq42}. Next, we claim that
\begin{align}\label{feq39}
\lambda_{1,\epsilon}(\mathcal{O}_1)<a \quad \text{and} \quad
\lambda_{1,\epsilon}(\mathcal{O}_2)< b,
\end{align}
where $\lambda_{1,\epsilon}(\mathcal{O}_i)$ denotes the first eigenvalue of
$\Lambda_{n,2}(1-\alpha)(-\Delta)^{\alpha} u+u$  on $\mathcal{W}^{\alpha,2}$ and
\begin{align*}
\lambda_{1,\epsilon}(\mathcal{O}_1)
&=\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} {|u(x)-u(y)|^2}{|x-y|^{-(n+2\alpha)}}\,dx\,dy
 + \int_{\Omega}|u|^2\,dx}{\frac{1}{\epsilon}\int_{\mathcal{O}_1} |u|^2\,dx}\\
&<a\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} {|u(x)-u(y)|^2}{|x-y|^{-(n+2\alpha)}}\,dx\,dy
 + \int_{\Omega}|u|^2\,dx}{\frac{1}{\epsilon}\int_{\mathcal{O}_1} m(x)  |u|^2\,dx}
=a,
\end{align*}
since $|x\in \mathcal{O}_1: m(x)< a|>0$. This implies
$\lambda_{1,\epsilon}(\mathcal{O}_1)<a$. The other inequality can be proved similarly.
 Now with some modification on the sets $\mathcal{O}_1$ and $\mathcal{O}_2$,
we construct the sets $\tilde{\mathcal{O}}_1$ and $\tilde{\mathcal{O}}_2$
such that $\tilde{\mathcal{O}}_1\cap \tilde{\mathcal{O}}_2=\emptyset$ and
$\lambda_{1,\epsilon}(\tilde{\mathcal{O}}_1)<a$ and
$\lambda_{1,\epsilon}(\tilde{\mathcal{O}}_2)<b$. For $\nu\geq 0$,
${\mathcal{O}}_1(\nu)=\{x\in {\mathcal{O}}_1 :
 \operatorname{dist}(x, (\Omega_{\epsilon})^c )>\nu\}$.
By Lemma \ref{f111}, we have
$\lambda_{1,\epsilon}(\mathcal{O}_1(\nu))\geq \lambda_{1,\epsilon}(\mathcal{O}_1))$
and moreover
$\lambda_{1,\epsilon}(\mathcal{O}_1(\nu))\to \lambda_{1,\epsilon}(\mathcal{O}_1))$ as
$\nu\to0$. Then there exists $\nu_0>0$ such that
\begin{equation}\label{feq40}
\lambda_{1,\epsilon}(\mathcal{O}_1(\nu))<a\quad \text{for all }0\leq \nu\leq \nu_0.
\end{equation}
Let $x_0\in \partial \mathcal{O}_2\cap \Omega_{\epsilon}$
(not empty as $\mathcal{O}_1\cap\mathcal{O}_2=\emptyset)$, choose
 $0<\nu<\min\{\nu_0, \operatorname{dist}(x_0,\Omega_{\epsilon}^c)\}$ and
$\tilde{\mathcal{O}}_1=\mathcal{O}_1(\nu)$ and
$\tilde{\mathcal{O}}_2=\mathcal{O}_2 \cup B(x_0, \frac{\nu}{2})$.
Then $\tilde{\mathcal{O}}_1\cap \tilde{\mathcal{O}}_2=\emptyset$ and by
\eqref{feq40}, $\lambda_{1,\epsilon}(\tilde{\mathcal{O}}_1)<a$.
Since $\tilde{\mathcal{O}}_2$ is connected,  by \eqref{feq39} and Lemma \ref{f111},
 we obtain $\lambda_1(\tilde{\mathcal{O}}_2)<b$. Now, we define $v=v_1-v_2$,
where $v_i$ are the eigenfunctions associated to
$\lambda_{i,\epsilon}(\tilde{\mathcal{O}}_i)$. Then $v$ satisfies \eqref{feq38}.

 Thus there exist $v\in \mathcal{W}^{\alpha,2}$ which changes sign, satisfies
condition \eqref{feq38}. Moreover we have
\begin{gather*}
\begin{aligned}
\tilde{J_s}\Big(\frac{\epsilon^{\frac{1}{2}} v}{\|v\|_{L^2(\Omega_{\epsilon})}}\Big)
&= \frac{\Lambda_{n,2}(1-\alpha)\int_{Q}\frac{|v^{+}(x)-v^{+}(y)|^2}{|x-y|^{n+2\alpha}}
 \,dx\,dy+ \int_{\Omega}(v^{+})^2}{\frac{1}{\epsilon}\|v\|^2_{L^2(\Omega_{\epsilon})}}
 -s\frac{\int_{\Omega_{\epsilon}}(v^{+})^2\,dx}{\|v\|^2_{L^2(\Omega_{\epsilon})}}\\
&\quad +\frac{\Lambda_{n,2}(1-\alpha)\int_{Q}\frac{|v^{-}(x)-v^{-}(y)|^2}
 {|x-y|^{n+2\alpha}}\,dx\,dy
 + \int_{\Omega}(v^{-})^2\,dx}{\frac{1}{\epsilon}\|v\|^2_{L^2(\Omega_{\epsilon})}}\\
&\quad +4\frac{\Lambda_{n,2}(1-\alpha)\int_{Q} \frac{v^{+}(x)v^{-}(y)}
 {|x-y|^{n+2\alpha}}\,dx\,dy}{\frac{1}{\epsilon}\|v\|^2_{L^2(\Omega_{\epsilon})}}\\
& < (a-s)\frac{\int_{\Omega_{\epsilon}}(v^{+})^2\,dx}{\|v\|^2_{L^2(\Omega_{\epsilon})}}
 +b\frac{\int_{\Omega_{\epsilon}}(v^{-})^2\,dx}{\|v\|^2_{L^2(\Omega_{\epsilon})}}=b.
\end{aligned} \\
\tilde{J_s}\Big(\frac{\epsilon^{\frac{1}{2}} v^+}{\|v^+\|_{L^2(\Omega_{\epsilon})}}\Big)
 < a-s =b,\quad
\tilde{J_s}\Big(\frac{\epsilon^{\frac{1}{2}} v^-}{\|v^-\|_{L^2(\Omega_{\epsilon})}}\Big)< b-s.
\end{gather*}
Using Lemma \ref{fle01}, we have that there exists a critical point in the
connected component of the set
$\mathcal{O}=\{u\in\mathcal{S}:\tilde{J_{s}}(u)<b-s\}$. As the point
$(a,b)\in \mathcal{C}$, the only possible critical point is $\phi_{1,\epsilon}$,
then we can construct a path from $\phi_{1,\epsilon}$ to $-\phi_{1,\epsilon}$
exactly in the same manner as in Theorem \ref{thm1.1} only by taking $v$
in place of $u$. Thus we have construct a path satisfying \eqref{feq37},
and hence the result follows.
\end{proof}

\begin{corollary}\label{fle32}
Let $(a,b)\in \mathcal{C}$ and let $m(x)$, $b(x)\in L^{\infty}(\Omega)$
satisfying $\lambda_{1,\epsilon}\leq m(x)\leq a$ a.e.,
 $\lambda_{1,\epsilon}\leq b(x)\leq b$ a.e.
Assume that $\lambda_{1,\epsilon}<m(x)$ and $\lambda_{1,\epsilon}<b(x)$ on
subsets of positive measure on $\Omega_{\epsilon}$. If either $m(x)<a$ a.e.
 in $\Omega_{\epsilon}$ or $b(x)<b$ a.e. in $\Omega_{\epsilon}$.
Then \eqref{feq33} has only the trivial solution.
\end{corollary}

\begin{proof}
By Lemma \ref{fle31}, any non-trival solution of \eqref{feq33}
 changes sign and $m(x)=a$ a.e. on $\{x\in\Omega_{\epsilon}:u(x)>0\}$ or $b(x)=b$
a.e. on $\{x\in \Omega_{\epsilon}: u(x)<0\}$. So, by our hypotheses, \eqref{feq33}
 has only trivial solution.
\end{proof}

 Now, we study the non-resonance between $(\lambda_1,\lambda_1)$ and $\mathcal{C}$,
\begin{equation} \label{041}
\begin{gathered}
\Lambda_{n,2}(1-\alpha)(-\Delta)^{\alpha} u+u
=  \frac{\chi_{\Omega_{\epsilon}}f(x,u)}{\epsilon} \quad \text{in }\Omega,\\
 \mathcal{N}_{\alpha,2}u=0\quad \text{in } \mathbb{R}^n \setminus \overline{\Omega},
\end{gathered}
\end{equation}
where $f(x,u)/u$ lies asymptotically between $(\lambda_{1,\epsilon},\lambda_{1,\epsilon})$
and $(a,b)\in \mathcal{C}$. 

Let $f:\Omega\times\mathbb{R}\to \mathbb{R}$ be a
function satisfying $L^{\infty}(\Omega)$ Caratheodory conditions.
 Given a point $(a,b)\in \mathcal{C}$, we assume that
\begin{align}\label{41}
\gamma_{\pm}(x)\leq \liminf_{s\to \pm\infty}\frac{f(x,s)}{s}
\leq \limsup_{s\to \pm \infty} \frac{f(x,s)}{s}\leq \Gamma_{\pm}(x)
\end{align}
holds uniformly with respect to $x$, where
$\gamma_{\pm}(x)$ and $\Gamma_{\pm}(x)$ are $L^{\infty}(\Omega)$ functions which satisfy
\begin{equation} \label{42}
\begin{gathered}
\lambda_{1,\epsilon}\leq \gamma_{+}(x)\leq \Gamma_{+}(x)\leq a\quad
 \text{a.e. in }\Omega_{\epsilon}\\
\lambda_{1,\epsilon}\leq \gamma_{-}(x)\leq \Gamma_{-}(x)\leq b\quad
\text{a.e. in }\Omega_{\epsilon}.
\end{gathered}
\end{equation}
The function $F(x,s)=\int_{0}^{s}f(x,t) dt$, we also satisfies
\begin{equation} \label{43}
 \delta_{\pm}(x)\leq \liminf_{s\to \pm\infty}\frac{2F(x,s)}{|s|^2}
\leq \limsup_{s\to \pm \infty} \frac{2F(x,s)}{|s|^2}\leq \Delta_{\pm}(x)
 \end{equation}
 uniformly with respect to $x$, where $\delta_{\pm}(x)$ and
$\Delta_{\pm}(x)$ are $L^{\infty}(\Omega)$ functions which satisfy
\begin{equation}\label{44}
\begin{gathered}
 \lambda_{1,\epsilon}\leq \delta_{+}(x)\leq \Delta_{+}(x)\leq a
 \text{ a.e. in }\Omega_{\epsilon},\quad
\lambda_{1,\epsilon}\leq \delta_{-}(x)\leq \Delta_{-}(x)\leq b\text{ a.e. in }\Omega_{\epsilon},\\
 \delta_{+}(x)>\lambda_{1,\epsilon} \text{ and }
\delta_{-}(x)>\lambda_{1,\epsilon} \text{ on subsets of positive measure,}\\
 \text{either } \Delta_{+}(x)< a \text{ a.e. in }
\Omega_{\epsilon}\text{ or } \Delta_{-}(x)< b \text{ a.e. in }\Omega_{\epsilon}.
\end{gathered}
\end{equation}

\begin{theorem}\label{th51}
Let \eqref{41}, \eqref{42}, \eqref{43} and \eqref{44} hold and
$(a,b)\in\mathcal{C}$. Then  \eqref{041} admits at least one
solution $u$ in $\mathcal{W}^{\alpha,2}$.
\end{theorem}

Define the energy functional $\Psi:\mathcal{W}^{\alpha,2}\to \mathbb{R}$ as
\[
\Psi(u)=\frac{\Lambda_{n,2}(1-\alpha)}{2}\int_{Q}
\frac{|u(x)-u(y)|^2}{|x-y|^{n+2\alpha}}\,dx\,dy
+ \frac{1}{2}\int_{\Omega}|u|^2\,dx
- \frac{1}{\epsilon}\int_{\Omega_{\epsilon}} F(x,u)\,dx
\]
Then $\Psi$ is a $C^{1}$ functional on $\mathcal{W}^{\alpha,2}$ and for all
$v\in \mathcal{W}^{\alpha,2}$,
\begin{align*}
\langle\Psi'(u),v\rangle
&=\Lambda_{n,2}(1-\alpha) \int_{Q}
\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{n+2\alpha}}\,dx\,dy
+\int_{\Omega}u v\,dx \\
&-\frac{1}{\epsilon}\int_{\Omega_{\epsilon}}  {f(x,u)v}\,dx
\end{align*}
and critical points of $\Psi$ are exactly the weak solutions of
\eqref{041}.

Next, we state some Lemmas, whose proofs can be found in
\cite[Lemma 5.2 and 5.3]{sa}.

\begin{lemma} \label{le51}
 $\Psi$ satisfies the (PS) condition on $\mathcal{W}^{\alpha,2}$.
 \end{lemma}

\begin{lemma}  \label{le52}
There exists $R>0$ such that
\[
\max\{\Psi(R\phi_{1,\epsilon}),\Psi(-R\phi_{1,\epsilon})\}
< \max_{u\in \gamma[-1,1]}\Psi(u),
\]
for any $\gamma\in \Gamma_1:=\{\gamma\in C([-1,1],\mathcal{S}) : \gamma(\pm 1)= \pm
R\phi_{1,\epsilon}\}$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{th51}]
 Lemmas \ref{le51} and \ref{le52} complete the proof.
\end{proof}


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 \end{document}