\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 88, pp. 1--17.\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-2018/88\hfil Existence and multiplicity of solutions]
{Existence and multiplicity of solutions to semilinear elliptic equation
 with nonlinear term of superlinear and subcritical growth}

\author[X.-F. Ke, C.-L. Tang \hfil EJDE-2018/88\hfilneg]
{Xiao-Feng Ke, Chun-Lei Tang}

\address{Xiao-Feng Ke \newline
School of Mathematics and Statistics,
Southwest University,
Chongqing 400715, China}
\email{kexf@swu.edu.cn}

\address{Chun-Lei Tang (corresponding author)\newline
School of Mathematics and Statistics,
Southwest University,
Chongqing 400715, China}
\email{tangcl@swu.edu.cn}

\dedicatory{Communicated by Paul Rabinowitz}

\thanks{Submitted February 4, 2018. Published April 10, 2018.}
\subjclass[2010]{35J20, 35J61, 35D30}
\keywords{Semilinear elliptic equation; new superlinear conditionl; 
\hfill\break\indent general subcritical condition}

\begin{abstract}
 This article concerns the existence and multiplicity of solutions to
 the superlinear elliptic problems. We introduce a new superlinear
 condition which is proved to be weaker than the Ambrosetti-Rabinowitz
 condition, the nonquadratic condition, the monotonicity conditions.
 As an application, positive solution and infinitely many solutions to
 semilinear elliptic equation with general subcritical growth are obtained,
 which generalize some known results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction and statement of main results}

 Consider the semilinear elliptic equation Dirichlet problem
\begin{equation}\label{sep}
\begin{gathered}
-\triangle u+a(x)u=f(x,u)\quad \text{in } \Omega,\\
 u=0 \quad \text{on } \partial\Omega,
 \end{gathered}
\end{equation}
where $\triangle$ is the Laplacian operator, $\Omega$ is a bounded domain
in $\mathbb{R}^N$$(N\geq3)$ with smooth boundary
$\partial\Omega$, and $a\in L^{\frac{N}{2}}(\Omega)$.
The inner product and induced norm in $H^{1}_{0}(\Omega)$ are 
respectively given by
$$
\langle u,v\rangle:=\int_{\Omega}(\nabla u,\nabla v)dx,\quad
\|u\|:=\Big(\int_{\Omega}|\nabla u|^2dx\Big)^{1/2},\quad \forall
 u, v\in H^{1}_{0}(\Omega),
$$
where $(\cdot,\cdot)$ is the Euclidean inner product. The operator
$-\triangle+a:H^{1}_{0}(\Omega)\cap H^2(\Omega)\to L^2{(\Omega)}$
possesses a unbounded eigenvalues sequence
$$
\lambda_1<\lambda_2\leq\dots\leq\lambda_n\to
+\infty\ \text{as}\ n\to \infty,
$$
where $\lambda_1$ is simple and characterized by
$$
\lambda_1={\inf_{u\in H^{1}_{0}(\Omega),
 u\neq0}}\frac{\int_{\Omega}|\nabla u|^2+a(x)u^2dx}{\int_{\Omega}u^2dx},
$$
the infimum is achieved by a positive function $\phi_1$ which is exactly
a eigenfunction corresponding to $\lambda_1$, and $u$ is a eigenfunction
corresponding to $\lambda_1$ if and only if 
$u\in H_{0}^{1}(\Omega)\setminus\{0\}$
is such that $\int_{\Omega}|\nabla u|^2dx+\int_{\Omega}a(x)u^2dx
=\lambda_1\int_{\Omega}u^2dx$. Besides this, it is well known that the
embedding mapping $H^{1}_{0}(\Omega)\hookrightarrow L^r(\Omega)$ is
continuous for $r\in[1,2^*]$ and is compact for $r\in[1,2^*)$, where
$2^*:=\frac{2N}{N-2}$. We denote by $|\cdot|_{r}$ the norm in
$L^{r}(\Omega)$ and $S_{r}$ the best constant to the corresponding embedding
mapping, that is, $S_{r}|u|_{r}\leq\|u\|$, for all $u\in H^{1}_{0}(\Omega)$.

In the celebrated paper \cite{Ambrosetti1973+Rabinowitz}, Ambrosetti and
 Rabinowitz established the famous mountain pass theorem and applied it
to obtain nontrivial solution and multiple solutions to problem \eqref{sep}
by assuming
\begin{itemize}
\item[(A1)] $f$ is H\"older continuous in $\overline{\Omega}\times\mathbb{R}$
 and $f(x,0)=0$,

\item[(A2)] there exist positive constants $a_1, a_2$ and $q\in(2,2^*)$ 
such that
 $$
|f(x,s)|\leq a_1+a_2|s|^{q-1}
$$
 for $s\in\mathbb{R}$ and $x\in\Omega$,

\item[(A3)] $\lim_{s\to 0} f(x,s)/s=0$ uniformly in $x\in\overline{\Omega}$,

\item[(A4)] $\lim_{|s|\to \infty} f(x,s)/s=+\infty$ uniformly in
 $x\in\overline{\Omega}$,
\item[(A5)] there exist constants $s'_0>0$ and $\theta>2$ such that
 $$
\theta F(x,s)\leq sf(x,s)
$$
 for $|s|\geq s'_0$ and $x\in\overline{\Omega}$, where
$F(x,s):=\int_{0}^{s}f(x,t)dt$,
\item[(A6)] $f$ is odd in $s$,
\end{itemize}
where (A4) shows that $f$ is essentially superlinear at $\infty$.
Moreover, (A4) together with (A5) leads to
\begin{itemize}
\item[(A7)]  there exist constants $s'_1>0$ and $\theta>2$ such that
 $$
0<\theta F(x,s)\leq sf(x,s)
$$
 for $|s|\geq s'_1$ and $x\in\overline{\Omega}$,
\end{itemize}
which is hereafter called the Ambrosetti-Rabinowitz  condition  and
plays a key role in ensuring that the Euler-Lagrange functional associated
to problem \eqref{sep} admits a mountain pass geometry and the
Palais-Smale sequences are bounded. Integrating,
from the continuity of $f$ one deduces that
\begin{equation}
F(x,s)\geq \xi|s|^{\theta} \label{e1.2}
\end{equation}
for $|s|\geq s'_1$ and $x\in\Omega$,
where
\[
\xi:=\big(\frac{1}{s'_1}\big)^{\theta}
\min\Big\{{\min_{x\in\overline{\Omega}}F(x,s'_1),
  \min_{x\in\overline{\Omega}}F(x,-s'_1)}\Big\}>0.
\]
Here we note two things. Firstly, in order to obtain \eqref{e1.2}, one should
add the assumption $\operatorname{ess\,inf}_{x\in\Omega}F(x,\pm s'_1)>0$
if (A7) is satisfied only on $\Omega$ rather than $\overline{\Omega}$ or
$f(\cdot,\pm s'_{1}):\overline{\Omega}\to \mathbb{R}$ is discontinuous,
see \cite{Mugnai2012} for more details. Secondly, $\eqref{e1.2}$
eliminates many interesting superlinear functions, such as
$F(x,s)=s^2\ln(1+|s|)$. For this reason, this technique has been subsequently
improved in order to include more superlinear functions and extended to
deal with more complicated variational problems by a large number of researchers,
see \cite{Costa1994+Magalhaes,Ding2004+Luan,Lan2014+Tang,Li2010+Yang,
Liu2004+Wang,Miyagaki2008+Souto,Pan2016+Tang,Qin2013+Tang,Schechter2004+Zou,
Szulkin2009+Weth, Tang2014+Wu,Tang2014,Wang1991,Willem2003+Zou,
Ye2014+Tang,Zhang2012+Liu,Zou2001} and references therein.

In \cite{Costa1994+Magalhaes}, Costa and Magalh\~{a}es replaced (A7) by
\begin{itemize}
\item[(A8)] (i) there exist constants $q\in(2, 2^*)$ and
$a_3>0$ such that
\[
\limsup_{|s|\to \infty} \frac{F(x,s)}{|s|^q}\leq a_3\text{ uniformly in a.e. }
x\in\Omega,
\]
(ii) there exist constants $\delta>0$ and $\mu>\frac{N(q-2)}{2}$ such that
\[
\liminf_{|s|\to \infty}\frac{sf(x,s)-2F(x,s)}{|s|^\mu} \geq \delta
\]
uniformly in a.e. $x\in\Omega$,
 \end{itemize}
then a nontrivial solution was obtained provided
$$
\limsup_{s\to 0}\frac{2F(x,s)}{s^2} <\lambda_1<{\liminf_{|s|\to \infty}
\frac{2F(x,s)}{s^2}}
\quad \text{uniformly in a.e. } x\in\Omega.
$$
Under these assumptions, they can deal with both superlinear situation
and sublinear situation.

In \cite{Ding2004+Luan}, Ding and Luan investigated a class of
 Schr\"odinger equations with the nonlinear term satisfying
\begin{itemize}
\item[(A9)]
 (i)  $\lim_{|s|\to \infty}\frac{F(x,s)}{s^2}=+\infty$ uniformly in $x\in\Omega$,\\
 (ii) $H(x,s):=sf(x,s)-2F(x,s)>0$ for $s\neq0$,\\
 (iii) there exist positive constants $s'_2,a_4$ and $\sigma>N/2$ such that
 $\big(\frac{f(x,s)}{s}\big)^\sigma \leq a_4 H(x,s)$ for $|s|\geq s'_2$ and
 $x\in\Omega$,
 \end{itemize}
where (A9)(iii) can be deduced from (A7) and a subcritical growth condition,
see \cite[Lemma 2.2]{Ding2007+Szulkin}.

In \cite{Willem2003+Zou}, Willem and Zou studied a class of superlinear
Schr\"odinger equation by assuming
\begin{itemize}
\item[(A10)]
 (i) there exist positive constants $a_5, a_6$ and $\nu\in(2, 2^*)$ such that
 $a_5|s|^{\nu}\leq f(x,s)s\leq a_6|s|^{\nu}$ for $s\in\mathbb{R}$ and
 $x\in\Omega$,\\
 (ii) $sf(x,s)-2F(x,s)>0$ for $s\neq0$ and $x\in\Omega$,\\
 (iii) there exist constants $\delta>0$ and
 $\mu>\frac{2^{*}\nu(\nu-2)}{2^{*}\nu-2^{*}-\nu}$ such that
\[
\liminf_{|s|\to \infty}\frac{sf(x,s)-2F(x,s)}{|s|^\mu}
 \geq \delta\text{ uniformly in }x\in\Omega.
\]
 \end{itemize}

In \cite{Miyagaki2008+Souto}, Miyagaki and Souto studied a eigenvalue problem
under $\rm(S_2)(i)$ and
\begin{itemize}
\item[(A11)]  there exists constant $s'_3>0$ such that
$$\frac{f(x,s)}{s} \text{ is increasing for } s\geq s'_3\text{ and decreasing for }
  s\leq-s'_3,\quad\forall x\in\Omega,
$$
\end{itemize}
which implies
\begin{itemize}
\item[(A12)] there exists constant $s'_4>0$ such that
$$
H(x,s)\text{ is increasing for } s\geq s'_4 \text{ and decreasing for }
 s\leq-s'_4,\quad \forall x\in\Omega.
$$
\end{itemize}
It is remarkable that (A12) also implies that (A11) when $f(x,s)$
is differentiable with respect to $s$ (see \cite{Li2010+Yang}).
Furthermore, (A12) can be generalized in two directions.
One is the following generalized monotonic condition
\begin{itemize}
\item[(A13)] there exists a constant $D\geq1$ such that
$$
H(x,t)\leq DH(x,s)\quad \text{for }s'_4<t<s \text{ or }
 s<t<-s'_4,\; \forall x\in\Omega,
$$
\end{itemize}
which was first introduced in \cite{Jeanjean1999}.
The other is the following \lq\lq quasi-monotonic" condition
\begin{itemize}
\item[(A14)] there exists a nonnegative function $W_1\in L^1(\Omega)$ such that
$$
H(x,t)\leq H(x,s)+W_1(x)\quad \text{for } 0<t<s \text{ or } s<t<0,\; \forall
 x\in\Omega.
$$
\end{itemize}
A weaker condition than (A13) and (A14) is
\begin{itemize}
\item[(A15)] there exist a constant $D\geq1$ and a nonnegative function
$W_1\in L^1(\Omega)$ such that
$$
H(x,t)\leq DH(x,s)+W_1(x)\quad \text{for } 0<t<s \text{ or } s<t<0,\;
 \forall\ x\in\Omega.
$$
\end{itemize}
Using (A15) instead of (A11), Lan and Tang in \cite{Lan2014+Tang}
generalized the result in \cite{Miyagaki2008+Souto}.

In addition, Schechter and Zou in \cite{Schechter2004+Zou} established
the existence of nontrivial solution for problem \eqref{sep} provided
\begin{itemize}
 \item $H(x,s)$ is convex in $s$, $\forall\ x\in\Omega$, or there are constants
 $a_{7}>0$, $\theta>2$ and $s'_5$ such that
 $$
 \theta F(x,s)-sf(x,s)\leq a_{7}(s^2+1)
 $$
 for $|s|\geq s'_5$.
\end{itemize}
As remarked in \cite{Miyagaki2008+Souto}, the convexity of $H$
in the above assumption is stronger than (A11), while the second
alternative is equivalent to (A7).

Under (A7), Wang in \cite{Wang1991} proved that problem \eqref{sep}
had at least three nontrivial solutions via the mountain pass theorem
and Morse theory. By assuming (A12) holds, Liu and Wang
in \cite{Liu2004+Wang} also obtained at least three nontrivial solutions
via the Nehari manifold method, and infinitely many solutions via
the Ljusternik-Schnirelmann theory. Recently, Tang in \cite{Tang2014}
investigated a superlinear Schr\"odinger equation with the nonlinear term satisfying
\begin{itemize}
 \item  there exists $\theta_0\in(0, 1)$ such that $sf(x,s)\geq0$ and
 $$
\frac{1-\theta^2}{2}sf(x,s)\geq\int_{\theta s}^{s}f(x,t)dt=F(x,s)-F(x,\theta s),\quad
\forall\ \theta\in[0, \theta_0]
$$
 for $s\in\mathbb{R}$ and a.e. $x\in\Omega$.
\end{itemize}
Tang and Wu in \cite{Tang2014+Wu} also introduced a new superquadratic condition
to guarantee the existence of nontrivial solution to a second order Hamiltonian
systems.

In this paper, we assume that $f:\overline{\Omega}\times\mathbb{R}\to \mathbb{R}$
is a Carath\'{e}odory function, and satisfies
\begin{itemize}
\item[(A16)] for every $M>0$, there exists a constant $L_{M}>0$ such that
\[
 |f(x,s)|\leq L_{M}
\]

 for $|s|\leq M$ and a.e. $x\in\Omega$,
\item[(A17)] ${\lim_{|s|\to \infty}\frac{f(x,s)}{|s|^{2^*-2}s}=0}$
 uniformly in a.e. $x\in\Omega$,

\item[(A18)] there exist a function $m\in L^{\frac{N}{2}}(\Omega)$
and a subset $\Omega'\subset\Omega$ with $|\Omega'|>0$ such that
$$
\limsup_{s\to 0}\frac{2F(x,s)}{s^2}\leq m(x)\leq\lambda_1
$$
 uniformly in a.e. $x\in\Omega$, and $m<\lambda_1$ in $\Omega'$, where
$F(x,s)=\int_{0}^{s}f(x,t)dt$ and $|\cdot|$ is the Lebesgue measure,

\item[(A19)] $\lim_{|s|\to \infty}\frac{F(x,s)}{s^2}=+\infty$ uniformly in a.e.
$x\in\Omega$,

\item[(A20)] there exist constants $s_0>0$, $\alpha>0$,
 $\sigma>\frac{N}{2}$ and a nonnegative function
$W\in L^1(\Omega)$ such that
\[
 \Big(\frac{F(x,s)}{s^2}\Big)^\sigma\leq \alpha H(x,s)+W(x)
\]
 for $|s|\geq s_0$ and a.e. $x\in\Omega$, where $H(x,s)=sf(x,s)-2F(x,s)$.
\end{itemize}

\begin{remark} \label{rmk1.1} \rm
 Obviously, (A16) holds when $f:\overline{\Omega}\times\mathbb{R}\to \mathbb{R}$
is continuous. (A17) is essentially weaker than
\begin{itemize}
\item[(A21)] there exist positive constants $a_{8}, a_{9}$ and
$q\in(2,2^*)$ such that
 $$
|f(x,s)|\leq a_{8}+a_{9}|s|^{q-1}
$$
 for $s\in\mathbb{R}$ and a.e. $x\in\Omega$.
\end{itemize}
which is equivalent to (A8)(i) when (A16) holds. Besides,
if $\lambda_{1}>0$, (A18) is obviously weaker than
\begin{itemize}
\item [(A22)] ${\lim_{s\to 0}\frac{2F(x,s)}{s^2}}=0$
 uniformly in a.e. $x\in\Omega$.
\end{itemize}
\end{remark}

\begin{remark} \label{rmk1.2} \rm
There exist functions which satisfy our conditions and do not satisfy 
(A7)--(A10), and (A15). For example, when
$a(x)\equiv0$ and $N=4$, let $\Omega_{0}\subset\Omega$ be such that
$|\Omega_{0}|>0$ and $|\Omega\setminus\Omega_{0}|>0$, $\chi_{\Omega_{0}}$
denotes the characteristic function of $\Omega_{0}$,
set $h:[1,+\infty)\to \mathbb{R}$ as follows
$$
 h(s)=\begin{cases}
 n^{3}(\frac{1}{n^2}-|s-n|)+\frac{1}{s}, &\text{if }
  |s-n|\leq\frac{1}{n^2},\; n=2,3,4,\dots,\\
 \frac{1}{s}, &\text{otherwise},
 \end{cases}
$$
and
$$
 f(x,s)=\begin{cases}
 \frac{2s\int_{1}^{s}h(t)dt+s^2h(s)}{(\ln s+1)^{1/2}}
-\frac{s\int_{1}^{s}h(t)dt}{2(\ln s+1)^{3/2}}
 +3\chi_{\Omega_{0}}(x)s^2,
& s\geq1,\; x\in\Omega,\\
 2(s-\frac{1}{2})(1+3\chi_{\Omega_{0}}(x)),
&s\in(\frac{1}{2},1),\; x\in\Omega,\\
 0, & s\leq\frac{1}{2},\; x\in\Omega.
 \end{cases}
$$
By simple calculation, we have $\frac{2N}{N-2}=4$, $\frac{N}{2}=2$,
$$
h(n)=n+\frac{1}{n},\quad h(n+\frac{1}{n^2})
=\frac{1}{n+\frac{1}{n^2}},\quad n=2,3,4,\dots,
$$
$$
F(x,s)=\begin{cases}
 \frac{s^2\int_{1}^{s}h(t)dt}{(\ln s+1)^{1/2}}
 +\chi_{\Omega_{0}}(x)s^{3}+\frac{1}{4}(1-\chi_{\Omega_{0}}(x)),
&s\geq1,\ x\in\Omega,\\
 (s-\frac{1}{2})^2(1+3\chi_{\Omega_{0}}(x)),
&s\in(\frac{1}{2},1),\; x\in\Omega,\\
 0, & s\leq\frac{1}{2},\; x\in\Omega,
 \end{cases}
$$
and
$$
sf(x,s)-2F(x,s)=\frac{s^{3}h(s)}{(\ln s+1)^{1/2}}
-\frac{s^2\int_{1}^{s}h(t)dt}{2(\ln s+1)^{3/2}}
+\chi_{\Omega_{0}}(x)s^{3}-\frac{1}{2}(1-\chi_{\Omega_{0}}(x)),
$$
for $s\geq1$, $x\in\Omega$.
Besides this, for $s\geq1$,
$$
\int_{1}^{s}h(t)dt=\int_{1}^{s}\big(h(t)-\frac{1}{t}\big)dt
 +\int_{1}^{s}\frac{1}{t}dt
=\int_{1}^{s}\big(h(t)-\frac{1}{t}\big)dt+\ln s,
$$
then for $3\leq n\leq s\leq n+1$, one has
\begin{gather*}
\int_{1}^{s}\Big(h(t)-\frac{1}{t}\Big)dt
\leq\sum_{k=2}^{n+1}\int^{k+\frac{1}{k^2}}_{k-\frac{1}{k^2}}k^{3}
\Big(\frac{1}{k^2}-|s-k|\Big)ds
=\sum_{k=2}^{n+1}\frac{1}{k},\\
\int_{1}^{s}\Big(h(t)-\frac{1}{t}\Big)dt
\geq\sum_{k=2}^{n-1}\int^{k+\frac{1}{k^2}}_{k-\frac{1}{k^2}}k^{3}
\Big(\frac{1}{k^2}-|s-k|\Big)ds
=\sum_{k=2}^{n-1}\frac{1}{k}.
\end{gather*}
From the above two inequalities and
${\lim_{n\to \infty}\frac{{\sum_{k=1}^{n}}\frac{1}{k}}{\ln n}}=1$
it follows that
$$
{\lim_{s\to +\infty}\frac{\int_{1}^{s}\left(h(t)-\frac{1}{t}\right)dt}{\ln s}}=1,
$$
which leads to
$$
{\lim_{s\to +\infty}\frac{\int_{1}^{s}h(t)dt}{\ln s}}=2.
$$
Thus, it is easy to verify that assumptions (A16)--(A19) hold.
Furthermore, (A20) holds for arbitrary $\sigma\in(2,3)$. However,
 we can draw the following conclusions.

(i) Condition (A7) is not satisfied. Indeed, for $\theta>2$,
$x\in\Omega\setminus\Omega_{0}$ and $s_n:=n+\frac{1}{n^2}$,
$$
s_nf(x,s_n)-\theta F(x,s_n)
\leq-\frac{(\theta-2)s_n^2\int_{1}^{s_n}h(t)dt}{(\ln s_n+1)^{1/2}}
+\frac{s_n^{3}h(s_n)}{(\ln s_n+1)^{1/2}}\to -\infty
$$
as $n\to +\infty$.

(ii) Conditions (A8) and (A10) are not satisfied. Indeed, it needs
$q\geq3$ to ensure (A8)(i) holds. But for $x\in\Omega\setminus\Omega_{0}$
and arbitrary $\mu>\frac{4(q-2)}{2}\geq2$, one has
\begin{align*}
\liminf_{|s|\to \infty}\frac{sf(x,s)-2F(x,s)}{|s|^\mu}
&\leq \lim_{n\to \infty}\frac{s_nf(x,s_n)-2F(x,s_n)}{(s_n)^\mu} \\
&\leq \lim_{n\to \infty}\frac{s_n^{3}h(s_n)}{(\ln s_n+1)^{1/2}(s_n)^\mu}=0,
\end{align*}
which is in contradiction with (A8)(ii). Similarly, there are not constants
$\nu\in(2,4)$ and $\mu>\frac{4\nu(\nu-2)}{4\nu-4-\nu}$ such that (A10) holds.

(iii) Condition (A9) is not satisfied. In fact, for
 $x\in\Omega\setminus\Omega_{0}$, we have
$$
\frac{f(x,n)}{n}\geq \frac{n^2}{(\ln n+1)^{1/2}},\quad
 nf(x,n)-2F(x,n)\leq \frac{2n^{4}}{(\ln n+1)^{1/2}}
$$
for $n$ large, so there does not exist constant $\sigma>2$ such that (A9) holds.

(iv) Condition (A15) is not satisfied. In fact, for
$x\in\Omega\setminus\Omega_{0}$ and $s'_n:=n-\frac{1}{n^2}$,
it is not difficult to prove that
\begin{gather*}
nf(x,n)-2F(x,n)-(s_nf(x,s_n)-2F(x,s_n))\to +\infty, \\
nf(x,n)-2F(x,n)-(s'_nf(x,s'_n)-2F(x,s'_n))\to +\infty
\end{gather*}
as $n\to \infty$. Then there do not exist constant $D\geq1$ and nonnegative
function $W_{1}\in L^{1}(\Omega)$ such that (A15) holds.
\end{remark}

Our main results are the following theorems.

\begin{theorem}\label{thm1.1}
 Assume that {\rm (A16)--(A20)} hold, and  that
$a\in L^{\infty}(\Omega)$ and $sf(x,s)\geq0$ for $s\in\mathbb{R}$
and a.e. $x\in\Omega$. Then problem \eqref{sep} has at least a positive
solution and a negative solution.
\end{theorem}

\begin{remark} \label{rmk1.3} \rm
In the next section, we will prove that (A20) indeed weaker than 
(A7)--(A10), (A15) under the assumptions (A21) and (A19).
In addition, if $f$ satisfies (A15)--(A17), (A19), and (A22),,
so does the term $\lambda f$ for $\lambda>0$.
 Therefore, Theorem \ref{thm1.1} generalizes \cite[Theorem 1.1]{Miyagaki2008+Souto}
and complements \cite[Theorem 2.1]{Liu2004+Wang},
\cite[Theorem 1.2]{Lan2014+Tang}.
 It is necessary to point out here that the integrability requirement
 $a\in L^{\infty}(\Omega)$ and sign condition $sf(x,s)\geq0$
for $s\in\mathbb{R}$ and a.e. $x\in\Omega$ are only used to obtain
a positive solution, especially in order to guarantee the validity
of the strong Maximum principle in \cite{Vazquez1984}.
In fact, $a\in L^{\frac{N}{2}}(\Omega)$ is enough to ensure the existence
of nontrivial solution.
\end{remark}

\begin{theorem}\label{thm1.2}
 Assume that {\rm (A6), (A16), (A17), (A19), (A20)} hold,
then problem \eqref{sep} has infinitely many solutions.
\end{theorem}

\begin{remark} \label{rmk1.4} \rm
Theorem \ref{thm1.2} unifies and generalizes \cite[Theorem 3.7]{Willem1996},
\cite[Theorem 3.2]{Szulkin2009+Weth},
\cite[Theorem 1.3]{Ye2014+Tang}, \cite[Theorem 1.1]{Zhang2012+Liu},
\cite[Theorems 1.2 and 1.3]{Qin2013+Tang}.
Besides this, Theorem \ref{thm1.2} complements \cite[Theorem 2.3]{Liu2004+Wang},
\cite[Theorem 3.1]{Zou2001}, \cite[Theorem 1.4]{Pan2016+Tang}.
\end{remark}

\begin{remark} \label{rmk1.5}\rm
 A condition similar to (A20) was introduced in \cite{Qin2013+Tang}.
 However, compared with the description in \cite{Qin2013+Tang},
firstly, our description is more general and simple.
Secondly, we point out the relations between (A20) and several
famous superlinear conditions for the first time.
Thirdly, we can deal with the superlinear problems with nonlinear
term satisfying the general subcritical condition (A17).
\end{remark}

\section{Preliminaries}

Let $E:=H_{0}^{1}(\Omega)$, the Euler-Lagrange functional associated to
problem \eqref{sep} is
$$
\Phi(u)=\frac{1}{2}\int_{\Omega}|\nabla u|^2dx
+\frac{1}{2}\int_{\Omega}a(x)u^2dx-\int_{\Omega}F(x,u)dx,\quad u\in E.
$$
By (A16) and (A17), it is standard to verify that $\Phi\in C^1(E,\mathbb{R})$ and
$$
\langle\Phi'(u),v\rangle=\int_{\Omega}(\nabla u,\nabla v)dx
 +\int_{\Omega}a(x)uv\,dx-\int_{\Omega}f(x,u)v\,dx,
$$
for all $u,v\in E$.
Moreover, the weak solutions of problem \eqref{sep} are exactly the critical
points of $\Phi$ in $E$. In order to obtain positive solution and
negative solution, we let $\widetilde{f}(x,s):=f(x,s)-m(x)s$ and
truncate $\widetilde{f}$ above or below $s=0$, i.e., let
$$
\widetilde{f}_{+}(x,s):=\begin{cases}
 \widetilde{f}(x,s), & s\geq0, \\
 0, & s<0,
 \end{cases} \quad
\widetilde{f}_{-}(x,s):=\begin{cases}
 \widetilde{f}(x,s), & s\leq0, \\
 0, & s>0,
 \end{cases}
$$
and $\widetilde{F}_{+}(x,s)=\int^{s}_{0}\widetilde{f}_{+}(x,t)dt$,
$\widetilde{F}_{-}(x,s)=\int^{s}_{0}\widetilde{f}_{-}(x,t)dt$.
Under (A16) and (A17), the functionals $\widetilde{\Phi}_{+}$ and
$\widetilde{\Phi}_{-}$ defined as follows
\begin{gather*}
\widetilde{\Phi}_{+}(u)
 =\frac{1}{2}\int_{\Omega}|\nabla u|^2dx
 +\frac{1}{2}\int_{\Omega}a(x)u^2dx-\frac{1}{2}\int_{\Omega}m(x)u^2dx
 -\int_{\Omega}\widetilde{F}_{+}(x,u)dx,\\
\widetilde{\Phi}_{-}(u)
=\frac{1}{2}\int_{\Omega}|\nabla u|^2dx+\frac{1}{2}\int_{\Omega}a(x)u^2dx
 -\frac{1}{2}\int_{\Omega}m(x)u^2dx-\int_{\Omega}\widetilde{F}_{-}(x,u)dx,
\end{gather*}
 belong to $C^{1}(E,\mathbb{R})$ and
\begin{gather*}
\langle\widetilde{\Phi}'_{+}(u),v\rangle
 =\int_{\Omega}(\nabla u,\nabla v)dx+\int_{\Omega}a(x)uv\,dx
 -\int_{\Omega}m(x)uv\,dx-\int_{\Omega}\widetilde{f}_{+}(x,u)v\,dx,\\
\langle\widetilde{\Phi}'_{-}(u),v\rangle
 =\int_{\Omega}(\nabla u,\nabla v)dx+\int_{\Omega}a(x)uv\,dx
 -\int_{\Omega}m(x)uv\,dx-\int_{\Omega}\widetilde{f}_{-}(x,u)v\,dx,
\end{gather*}
for all $u,v\in E$.
The following lemmas show that our superlinear situation, i.e., (A19) and (A20),
indeed includes all the superlinear situations implied by 
(A7)--(A10), (A15).

\begin{lemma} \label{lem2.1}
Under assumption {\rm (A21)},  assumption (A7) implies {\rm (A19), (A20)}.
\end{lemma}

\begin{proof}
 Clearly, (A19) naturally holds because of \eqref{e1.2}. Additionally,
from $q\in(2, 2^{*})$ we derive $\frac{N}{2}<\frac{q}{q-2}$.
Taking arbitrarily $\sigma\in(\frac{N}{2}, \frac{q}{q-2})$,
one has $q<\frac{2\sigma}{\sigma-1}$. Then (A21) leads to
$$
\lim_{|s|\to \infty}\frac{F(x,s)}{|s|^{\frac{2\sigma}{\sigma-1}}}=0
\quad \text{uniformly in a.e. } x\in\Omega.
$$
From this and (A19) it follows that there exists a constant $s_{1}>s'_{1}$
such that
$$
0<\frac{F(x,s)}{|s|^{\frac{2\sigma}{\sigma-1}}}\leq(\theta-2)^{\frac{1}{\sigma-1}}
$$
for $|s|\geq s_{1}$ and a.e. $x\in\Omega$, from this and (A7) we obtain that
$$
\Big(\frac{F(x,s)}{s^2}\Big)^{\sigma}\leq(\theta-2)F(x,s)\leq sf(x,s)-2F(x,s)
$$
for $|s|\geq s_{1}$ and a.e. $x\in\Omega$.
\end{proof}



\begin{lemma} \label{lem2.2}
Under assumption {\rm (A19)}, assumption {\rm (A8)} implies {\rm (A20)}.
\end{lemma}

\begin{proof}
From (A8) and (A19) it follows that there exists constant $s_{2}>1$ such that
$$
0<\frac{F(x,s)}{|s|^q}\leq a_3+1\quad \text{and}\quad  H(x,s)\geq \delta|s|^\mu
$$
for $|s|\geq s_2$ and a.e. $x\in\Omega$. From $\mu>\frac{N(q-2)}{2}$
we deduce that $\frac{N}{2}<\frac{\mu}{q-2}$. Taking arbitrarily
 $\sigma\in(\frac{N}{2},\frac{\mu}{q-2})$, one has $\sigma(q-2)<\mu$.
Then
$$
\Big(\frac{F(x,s)}{s^2}\Big)^\sigma
=\Big(\frac{F(x,s)}{|s|^q}\Big)^\sigma |s|^{\sigma(q-2)}
\leq (a_3+1)^{\sigma}|s|^\mu
\leq\frac{(a_3+1)^\sigma}{\delta}H(x,s)
$$
for $|s|\geq s_{2}$ and a.e. $x\in\Omega$.
\end{proof}

\begin{lemma} \label{lem2.3}
Condition {\rm (A9)} implies {\rm (A19)} and {\rm (A20)}.
\end{lemma}

\begin{proof} (A9)(i) implies (A19).
From this and $\rm(A9)(ii)$ it follows that there exists a constant
$s_3>s'_{2}$ such that
$$
sf(x,s)\geq 2F(x,s)>0
$$
for $|s|>s_3$ and a.e. $x\in\Omega$, which together with (A9)(iii) leads to
$$
\Big(\frac{F(x,s)}{s^2}\Big)^{\sigma}
\leq\frac{1}{2^{\sigma}}\Big(\frac{f(x,s)}{s}\Big)^{\sigma}
\leq \frac{a_4}{2^{\sigma}} H(x,s)
$$
for $|s|\geq s_3$ and a.e. $x\in\Omega$.
\end{proof}

\begin{lemma} \label{lem2.4}
Condition {\rm (A10)} implies {\rm (A19)} and {\rm (A20)}.
\end{lemma}

\begin{proof}
(A10)(i) implies (A19). Moreover, from this and $\rm(A10)(iii)$ we deduce that
there exists a constant $s_4>1$ such that
$$
\frac{F(x,s)}{s^2}\geq 1\quad \text{and}\quad
f(x,s)s-2F(x,s)\geq \delta|s|^{\mu}
$$
for $|s|\geq s_4$ and a.e. $x\in\Omega$. Additionally, from $\nu\in(2, 2^{*})$
it follows that
$$
\nu-2>0\quad \text{and} \quad \frac{2^{*}\nu}{2^{*}\nu-2^{*}-\nu}>\frac{N}{2}.
$$
Taking arbitrarily $\sigma\in(\frac{N}{2},\frac{2^{*}\nu}{2^{*}\nu-2^{*}-\nu})$,
one derives from $\mu>\frac{2^{*}\nu(\nu-2)}{2^{*}\nu-2^{*}-\nu}$ and from
(A10)(ii) that
\begin{align*}
f(x,s)s-2F(x,s)
&\geq \delta|s|^{\mu}>\delta\big(|s|^{\nu-2}
\big)^{\frac{2^{*}\nu}{2^{*}\nu-2^{*}-\nu}} \\
&\geq\frac{\delta}{a_6^{\sigma}}\Big(\frac{sf(x,s)}{s^2}
 \Big)^{\sigma}
\geq \frac{2\delta}{a_6^{\sigma}}\Big(\frac{F(x,s)}{s^2}\Big)^{\sigma}
\end{align*}
for $|s|\geq s_4$ and a.e. $x\in\Omega$.
\end{proof}


\begin{lemma} \label{lem2.5}
Under assumptions{(A19)} and {\rm (A21)}, condition {\rm (A15)}  implies {\rm (A20)}.
\end{lemma}

\begin{proof}
From (A21) and (A19) it follows that there exists constant $s_5>0$ such that
\begin{equation}
\frac{F(x,s)}{s^2}>0\quad \text{and}\quad
\frac{F(x,s)}{|s|^q}\leq \frac{a_{9}}{q}+1\label{e2.1}
\end{equation}
for $|s|\geq s_5$ and $x\in\Omega$. The fact $q\in\left(2,\frac{2N}{N-2}\right)$
yields $\frac{q}{q-2}>\frac{N}{2}$.
Then taking arbitrarily $\sigma\in\left(\frac{N}{2},\frac{q}{q-2}\right)$,
we have $2>(\sigma-1)(q-2)>0$.
Let $\varsigma:=2-(\sigma-1)(q-2)>0$, one gets
\begin{align*}
\frac{d}{ds}\Big[\Big(\frac{F(x,s)}{s^2}\Big)^\sigma\Big]
&= \sigma\Big(\frac{F(x,s)}{s^2}\Big)^{\sigma-1}\frac{sf(x,s)-2F(x,s)}{s^3}\\
&= \sigma\Big(\frac{F(x,s)}{|s|^q}\Big)^{\sigma-1}\frac{H(x,s)}{|s|^{\varsigma}s}
\end{align*}
for a.e. $x\in\Omega$. From this, \eqref{e2.1} and (A15) it follows that
\begin{align*}
\Big(\frac{F(x,s)}{s^2}\Big)^\sigma
 -\Big(\frac{F(x,s_5)}{s_5^2}\Big)^\sigma 
&= \int_{s_5}^{s}\frac{d}{dt}\Big[\Big(\frac{F(x,t)}{t^2}\Big)^\sigma\Big]dt\\
&= \int_{s_5}^{s}\sigma\Big(\frac{F(x,t)}{|t|^q}\Big)^{\sigma-1}
 \frac{H(x,t)}{t^{\varsigma+1}}dt\\
&\leq \sigma \Big(\frac{a_{9}}{q}+1\Big)^{\sigma-1}
 \big(D H(x,s)+W_1(x)\big)\int_{s_5}^{s}\frac{1}{t^{\varsigma+1}}dt\\
&\leq \sigma \Big(\frac{a_{9}}{q}+1\Big)^{\sigma-1}
 \big(D H(x,s)+W_1(x)\big)\frac{s_5^{-\varsigma}}{\varsigma}
\end{align*}
for $s\geq s_5$ and a.e. $x\in\Omega$, where in the last inequality
we use the fact that $DH(x,s)+W_1(x)\geq0$ for $s\neq0$ and
a.e. $x\in\Omega$ which can be deduced from condition (A15)
and $H(x,0)=0$ a.e. $x\in\Omega$. Then we have
$$
\Big(\frac{F(x,s)}{s^2}\Big)^\sigma\leq\alpha H(x,s)+W(x),
$$
for $s\geq s_5$ and a.e. $x\in\Omega$, where
$\alpha=\frac{\sigma D s_5^{-\varsigma}(a_{9}+q)^{\sigma-1}}{q^{\sigma-1}\varsigma}$,
$W(x)=\frac{\sigma s_5^{-\varsigma}(a_{9}+q)^{\sigma-1}}{q^{\sigma-1}\varsigma}
 W_1(x)+\left(\frac{F(x,s_5)}{s_5^2}\right)^\sigma.$
In a similar way, it is easy to verify that the above inequality holds for
$s\leq-s_5$ and a.e. $x\in\Omega$.
\end{proof}

 \section{Proof of main results}

To prove Theorems \ref{thm1.1} and \ref{thm1.2}, we recall two abstract critical 
point theorems,  i.e., the mountain pass theorem and the symmetric mountain pass 
theorem under the $\rm(C)$ condition, the readers can refer to
\cite{Bartolo1983+Benci+Fortunato} and \cite{Rabinowitz1986}.

\begin{theorem} \label{thmA}
Let $(X,\|\cdot\|_{X})$ be a Banach space, suppose that
$\varphi\in C^{1}(X, \mathbb{R})$ satisfies $\varphi(0)=0$ and
\begin{itemize}
 \item[(i)] there exist positive constants $R_0$ and $\alpha_{0}$ such that
$$
\varphi(u)\geq\alpha_{0}\quad \text{for all } u\in X\ \text{with}\ \|u\|_{X}=R_0,
$$

 \item[(ii)] there exists $e\in X$ with $\|e\|_{X}>R_0$ such that $\varphi(e)<0$,

 \item[(iii)] $\varphi$ satisfies the $\rm(C)$ condition, that is, for
$c\in \mathbb{R}$, every sequence $\{u_n\}\subset X$ such that
$$
\varphi(u_n)\to c,\quad \|\varphi'(u_n)\|(1+\|u_n\|)\to 0
$$
has a convergent subsequence.
\end{itemize}
Then $c:={\inf_{\gamma\in \Gamma}\sup_{s\in[0,1]}\varphi(\gamma(s))}$
is a critical value of $\varphi$, where
$$
\Gamma:=\{\gamma\in C([0,1], X); \gamma(0)=0, \gamma(1)=e\}.
$$
\end{theorem}

\begin{theorem} \label{thmB}
Let $(X,\|\cdot\|_{X})$ be an infinite dimensional Banach space, and let
$\varphi\in C^{1}(X,\ \mathbb{R})$ be even. Suppose that $\varphi$
satisfies $\varphi(0)=0$ and
\begin{itemize}
 \item[(i)] there exist a closed subspace $X^1$ of $X$ with
$\operatorname{codim} X^1<+\infty$ and positive constants $R_1,\alpha_1$ such that
 $$
\varphi(u)\geq\alpha_{1}\quad \text{for } u\in X^1 \text{ with } \|u\|_{X}=R_1,
$$

 \item[(ii)] for every finite dimensional subspace $X^2$ of $X$, there exists
positive constant $R_2$ such that
 $$
\varphi(u)\leq0\quad \text{for $u\in X^2$  with } \|u\|_{X}=R_2,,
$$

 \item[(iii)] $\varphi$ satisfies the {\rm (C)} condition in Theorem \ref{thmA}.
\end{itemize}
Then $\varphi$ possesses an unbounded sequence of critical values.
\end{theorem}

In addition, we  need the following lemmas.


\begin{lemma}[{\cite[Lemma 2.13]{Willem1996}}] \label{lem3.1}
 Assume that $N\geq3$ and $\vartheta\in L^{\frac{N}{2}}(\Omega)$,
then the functional
$$
\psi(u):=\int_{\Omega}\vartheta(x)u^2dx,\ u\in H^{1}_{0}(\Omega)
$$
is weakly continuous.
\end{lemma}

\begin{lemma} \label{lem3.2}
 Assume that $m\in L^{\frac{N}{2}}(\Omega)$, and there exists a subset
$\Omega'\subset\Omega$ with $|\Omega'|>0$ such that
$$
m\leq\lambda_1 \text{ in } \Omega\quad\text{and}\quad
m<\lambda_1 \text{ in } \Omega',
$$
then
$$
d:=\inf_{u\in H^{1}_{0}(\Omega),\ u\neq0}
\frac{\int_{\Omega}|\nabla u|^2dx+\int_{\Omega}a(x)u^2dx
-\int_{\Omega}m(x)u^2dx}{\int_{\Omega}|\nabla u|^2dx}>0.
$$
\end{lemma}

\begin{proof}
From the characteristic of $\lambda_1$ and the assumption $m\leq\lambda_1$
in $\Omega$ it follows that $d\geq0$. The reminder is to prove that $d\neq0$.
Let
\begin{gather*}
J(u):=\int_{\Omega}a(x)u^2dx,\ u\in H^{1}_{0}(\Omega), \\
K(u):=\int_{\Omega}m(x)u^2dx,\ u\in H^{1}_{0}(\Omega), \\\
L(u):=\|u\|^2+J(u)-K(u),\ u\in H^{1}_{0}(\Omega).
\end{gather*}
We argue by contradiction. If $d=0$, there exists a sequence
$\{u_n\}\subset H^{1}_{0}(\Omega)$ such that
$$
\|u_n\|=1\quad \text{and}\quad \lim_{n\to \infty}L(u_n)=0.
$$
By the boundedness of $\{u_n\}$, up to subsequence we may assume that
$u_n\rightharpoonup u$ in $H^{1}_{0}(\Omega)$. From this, the weak continuity
of $J, K$, and the weak lower continuity of $L$ it follows that
\begin{equation}
\lim_{n\to \infty}J(u_n)=J(u),\quad \lim_{n\to \infty}K(u_n)=K(u)\label{e3.1}
\end{equation}
and
$$
0\leq L(u)\leq{\liminf_{n\to \infty}L(u_n)}={\lim_{n\to \infty}L(u_n)}=0.
$$
Then we have
\begin{equation}
L(u)=\|u\|^2+J(u)-K(u)={\lim_{n\to \infty}L(u_n)}=0,\label{e3.2}
\end{equation}
which implies
$$
\|u\|^2+J(u)=K(u)\leq\lambda_1\int_{\Omega}u^2dx,
$$
this together with the characteristic of $\lambda_1$ leads to
\begin{equation}
\|u\|^2+J(u)=\lambda_1\int_{\Omega}u^2dx.\label{e3.3}
\end{equation}
If $u=0$, from \eqref{e3.1} and \eqref{e3.2} it follows that
$\|u_n\|\to 0\ \text{as}\ n\to \infty,$ which is in contradiction with
$\|u_n\|=1$. So $u\neq0$, then $u$ is a eigenfunction corresponding to
$\lambda_1$, so $u=l_0\phi_1$ for some $l_0\in\mathbb{R}\setminus\{0\}$ as
$\lambda_1$ is simple. Thus, from $\phi_1>0$, $m\leq\lambda_1$ in
$\Omega$ and $m<\lambda_1$ in $\Omega'$ with $|\Omega'|>0$ it follows that
\begin{align*}
\|u\|^2+J(u)&=K(u)=\int_{\Omega}m(x)u^2dx\\
&=l_0^2\int_{\Omega}m(x)\phi_1^2dx
<l_0^2\lambda_1\int_{\Omega}\phi_1^2dx\\
&=\lambda_1\int_{\Omega}u^2dx,
\end{align*}
which is in contradiction with $\eqref{e3.3}$. Hence, $d>0$.
The proof is complete.
\end{proof}


 \begin{lemma} \label{lem3.3}
 Assume that {\rm (A16)--(A18)} hold.
Then $\widetilde{\Phi}_{+}$ satisfies {\rm(i)} of Theorem \ref{thmA}.
\end{lemma}

\begin{proof}
By (A18), for $\varepsilon\in\left(0,\ \frac{dS_{2}^2}{2}\right)$,
there exists a positive constant $M_1<1$ such that
\begin{equation}
F_{+}(x,s)=F(x,s^{+})
\leq\frac{1}{2}(m(x)+\varepsilon)(s^{+})^2\label{e3.4}
\end{equation}
for $|s|\leq M_1$ and a.e. $x\in\Omega$, where and in what follows we denote
 by $s^{+}:=\max\{s,0\}$ and $s^{-}:=\max\{-s,0\}$. For above $\varepsilon$,
 from (A16), (A17) and \eqref{e3.4} it follows that there exists a constant
$M_2>1$ such that
\begin{equation}
|f_{+}(x,s)|=|f(x,s^{+})|\leq\varepsilon(s^{+})^{2^*-1}+L_{M_{2}}\label{e3.5}
\end{equation}
and
\begin{equation}
F_{+}(x,s)\leq\frac{1}{2}(m(x)+\varepsilon)(s^{+})^2
 +\Big(\frac{L_{M_2} M_2}{M_1^{2^*}}+\frac{\varepsilon}{2^*}\Big)(s^{+})^{2^*}
\label{e3.6}
\end{equation}
for $s\in\mathbb{R}$ and a.e. $x\in\Omega$. From $\eqref{e3.6}$ and
Lemma \ref{lem3.2} we obtain
\begin{align*}
\widetilde{\Phi}_{+}(u) 
&\geq \frac{1}{2}\int_{\Omega}|\nabla u|^2dx
 +\frac{1}{2}\int_{\Omega}a(x)u^2dx-\frac{1}{2}\int_{\Omega}m(x)u^2dx \\
&\quad -\frac{1}{2}\int_{\Omega}(m(x)+\varepsilon)(u^{+})^2dx\\
&\quad -\int_{\Omega}\Big(\frac{L_{M_2} M_2}{M_1^{2^*}}
 +\frac{\varepsilon}{2^*}\Big)(u^{+})^{2^*}dx
 +\frac{1}{2}\int_{\Omega}m(x)(u^{+})^2dx\\
&\geq \frac{d}{2}\|u\|^2-\frac{\varepsilon}{2S_{2}^2}\|u\|^2
 -\Big(\frac{L_{M_2} M_2}{M_1^{2^*}}+\frac{\varepsilon}{2^*}\Big)
 \Big(\frac{1}{S_{2^*}}\Big)^{2^*}\|u\|^{2^*}\\
&=\|u\|^2\Big[\frac{d}{4}-\Big(\frac{L_{M_2} M_2}{M_1^{2^*}}
 +\frac{\varepsilon}{2^*}\Big)\Big(\frac{1}{S_{2^*}}\Big)^{2^*}\|u\|^{2^*-2}
 \Big],\quad \forall u\in E.
\end{align*}
Let
\[
C_1=\Big(\frac{L_{M_2} M_2}{M_1^{2^*}}+\frac{\varepsilon}{2^*}\Big)
 \Big(\frac{1}{S_{2^*}}\Big)^{2^*},  \quad
R_0=\Big(\frac{d}{8 C_1}\Big)^{\frac{1}{2^*-2}}, \quad
\alpha_0=\frac{d}{8} R_0^2\,.
\]
Then $\widetilde{\Phi}_{+}$ satisfies $\rm(i)$ of Theorem \ref{thmA}.
\end{proof}

\begin{lemma} \label{lem3.4}
Assume that {\rm (A16), (A19)} hold. Then $\widetilde{\Phi}_{+}$ satisfies
{\rm(ii)} of Theorem \ref{thmA}.
\end{lemma}

\begin{proof}
 From (A16) and (A19) it follows that for
$\Lambda>\frac{\|\phi_1\|^2+\int_{\Omega}a(x)\phi_1^2dx}{2|\phi_1|_{2}^2}$,
there exists a constant $M_3>0$ such that
$$
F_{+}(x,s)\geq \Lambda(s^{+})^2-L_{M_3} M_3
$$
for $s\in\mathbb{R}$ and a.e.\ $x\in\Omega$. Then for $t>0$, one obtains
$$
\widetilde{\Phi}_{+}(t\phi_1)\leq t^2\Big(\frac{1}{2}\|\phi_1\|^2
+\frac{1}{2}\int_{\Omega}a(x)\phi_1^2dx-\Lambda|\phi_1|_{2}^2\Big)
+M_3 L_{M_3}|\Omega|.
$$
Let $C_2=\frac{1}{2}\big(\|\phi_1\|^2+\int_{\Omega}a(x)\phi_1^2dx\big)
-\Lambda|\phi_1|_{2}^2<0$, $C_3=M_3 L_{M_3}|\Omega|>0$,
$t_0=\sqrt{\frac{2C_3}{-C_2}}+R_0$ and $e=t_0\phi_1$, then
$\widetilde{\Phi}_{+}$ satisfies $\rm(ii)$ of Theorem \ref{thmA}.
\end{proof}

\begin{lemma} \label{lem3.5}
Assume that {\rm (A16), (A17), (A19), (A20)} hold.
Then $\widetilde{\Phi}_{+}$ satisfies the {\rm (C)} condition in 
Theorem \ref{thmA}.
\end{lemma}

\begin{proof}
For $c\in\mathbb{R}$ and $\{u_n\}\subset E$ such that
\begin{equation}
\|\widetilde{\Phi}'_{+}(u_n)\|(1+\|u_n\|)\to 0 \text{ and }
 \widetilde{\Phi}_{+}(u_n)\to c\quad \text{as } n\to \infty,\label{e3.7}
\end{equation}
we first prove that $\{u_n\}$ is bounded. Arguing by contradiction,
if $\{u_n\}$ is unbounded, then $\|u_n\|\to +\infty$ as $n\to \infty$
after passing to a subsequence.
Set $w_n=\frac{u_n}{\|u_n\|}$, then $\|w_n\|=1$.
Hence, up to subsequence, we may assume that
$$
w_n\rightharpoonup w\quad \text{weakly in }E,
$$
which results in
\begin{equation}
 \begin{gathered}
 w_n\to w \quad \text{strongly in }
 L^{r}(\Omega)\text{ for } r\in[1,\ 2^*),\\
 w^{\pm}_n\rightharpoonup w^{\pm}\quad \text{weakly in } E,\\
 w^{\pm}_n(x)\to w^{\pm}(x)\quad \text{a.e. in } \Omega,\\
 w^{\pm}_n\to w^{\pm} \quad \text{strongly in } L^{r}(\Omega) \text{ for }
 r\in[1,\ 2^*).
 \end{gathered}\label{e3.8}
\end{equation}
From (A16) and (A19) it follows that there exists a constant
 $M_4>\max\{M_1, s_0\}$ such that
\begin{equation}
|F_{+}(x,s)|\leq L_{M_4}(s^{+})\leq L_{M_4} M_4\label{e3.9}
\end{equation}
for $|s|\leq M_4$ and a.e.\ $x\in\Omega$, and
$F_{+}(x,s)\geq (s^{+})^2$
for $|s|\geq M_4$ and a.e.\ $x\in\Omega$.
Then we have
\begin{equation}
F_{+}(x,s)\geq (s^{+})^2-M_4 L_{M_4}-M_4^2\geq-M_4 L_{M_4}-M_4^2\label{e3.10}
\end{equation}
for $s\in\mathbb{R}$ and a.e.\ $x\in\Omega$.

Now we claim that $w=0$. In fact, if $w^{+}\neq0$, that is, $|\Omega_{+}|>0$,
where $\Omega_{+}:=\{x\in\Omega:w(x)>0\}$. Then, for a.e.\
 $x\in\Omega_{+}$, one has
$u^{+}_n(x)=w^{+}_n(x)\|u_n\|\to +\infty$ as $n\to \infty$, which implies that
\begin{equation}
\lim_{n\to \infty}\frac{F(x,u^{+}_n(x))}{(u^{+}_n(x))^2}=+\infty.\label{e3.11}
\end{equation}
From \eqref{e3.7} and \eqref{e3.10} it follows that
\begin{align*}
&\frac{1}{2}\Big(1+\int_{\Omega}a(x)w_n^2dx
 -\int_{\Omega}m(x)(w^{-}_n)^2dx\Big)-\frac{c+o(1)}{\|u_n\|^2}\\
&= \int_{\Omega}\frac{F_{+}(x,u_n)}{\|u_n\|^2}dx\\
&\geq \int_{\Omega_{+}}\frac{F(x,u^{+}_n)}{(u^{+}_n)^2}(w^{+}_n)^2dx
 +\int_{\Omega\setminus\Omega_{+}}\frac{-M_4 L_{M_4}-M_4^2}{\|u_n\|^2}dx\\
&\geq \int_{\Omega_{+}}\frac{F(x,u^{+}_n)}{(u^{+}_n)^2}(w^{+}_n)^2dx
 -\frac{(M_4 L_{M_4}+M_4^2)|\Omega|}{\|u_n\|^2}.
\end{align*}
Then by Lemma \ref{lem3.1}, Fatou's lemma and \eqref{e3.11}, one obtains
\begin{align*}
&\frac{1}{2}\Big(1+\int_{\Omega}a(x)w^2dx-\int_{\Omega}m(x)(w^{-})^2dx\Big)\\
&\geq\liminf_{n\to +\infty}\Big(\int_{\Omega_{+}}\frac{F(x,u^{+}_n)}{(u^{+}_n)^2}
 |w_n|^2dx \Big)=+\infty,
\end{align*}
a contradiction. Hence $|\Omega_{+}|=0$, that is, $w^{+}=0$.

In addition, from \eqref{e3.7} and Lemma \ref{lem3.2} it follows that
\begin{align*}
d\|u^{-}_n\|^2
&\leq \int_{\Omega}|\nabla (u^{-}_n)|^2dx
 +\int_{\Omega}a(x)(u^{-}_n)^2dx
 -\int_{\Omega}m(x)(u^{-}_n)^2dx\\
&=\int_{\Omega}|\nabla (u^{-}_n)|^2dx+\int_{\Omega}a(x)(u^{-}_n)^2dx
 -\int_{\Omega}m(x)(u^{-}_n)^2dx \\
&\quad -\int_{\Omega}\widetilde{f}_{+}(x,u_n)u^{-}_ndx\\
&= \langle\widetilde{\Phi}'_{+}(u_n),u^{-}_n\rangle\to 0
\end{align*}
as $n\to \infty$, that is, $u^{-}_n\to 0$ in $E$ as $n\to \infty$.
This together with \eqref{e3.8} shows $w^{-}=0$. To sum up, we have
$w=w^{+}-w^{-}=0$, so the claim is proved.

From (A16) it follows that the term $|sf_{+}(x,s)-2F_{+}(x,s)|$ is bounded
in $[0, M_4]\times\Omega$. Set
$$
\varpi:=\min_{(x,s)\in\Omega\times[0, M_4]}|sf_{+}(x,s)-2F_{+}(x,s)|,\quad
\Omega_n:=\{x\in\Omega: u_n(x)\geq M_4\}.
$$
Then from \eqref{e3.9} and (A20) we have
\begin{align*}
&\frac{1}{2}\Big(1+\int_{\Omega}a(x)w_n^2dx-\int_{\Omega}m(x)(w^{-}_n)^2dx\Big)
 -\frac{c+o(1)}{\|u_n\|^2}\\
&= \int_{\Omega\setminus\Omega_n}\frac{F_{+}(x,u_n)}{\|u_n\|^2}dx
 +\int_{\Omega_n}\frac{F_{+}(x,u_n)}{\|u_n\|^2}dx\\
&\leq \int_{\Omega\setminus\Omega_n}\frac{L_{M_4} M_4}{\|u_n\|^2}dx
 +\Big[\int_{\Omega_n}\Big(\frac{F(x,u_n^{+})}{(u_n^{+})^2}\Big)^{\sigma}dx
 \Big]^{1/\sigma}
\Big[\int_{\Omega_n}(w_n^{+})^{\frac{2\sigma}{\sigma-1}}dx
 \big]^{\frac{\sigma-1}{\sigma}}\\
&\leq \frac{L_{M_4} M_4|\Omega|}{\|u_n\|^2}
 +\Big[\int_{\Omega_n}\alpha\Big(u_n^{+}f(x,u_n^{+})-2F(x,u_n^{+})\Big)
 +W(x)dx\Big]^{1/\sigma}|w_n^{+}|_{\frac{2\sigma}{\sigma-1}}^2\\
&\leq \frac{L_{M_4} M_4|\Omega|}{\|u_n\|^2}
 +\Big[\alpha\Big(2\widetilde{\Phi}_{+}(u_n)-\widetilde{\Phi}_{+}'(u_n)u_n\Big)
 +\alpha\varpi|\Omega|+|W|_{1}\Big]^{\frac{1}{\sigma}}
 |w_n^{+}|_{\frac{2\sigma}{\sigma-1}}^2.
\end{align*}
Since $\sigma>\frac{N}{2}$, one has $\sigma>1$ and
$\frac{2\sigma}{\sigma-1}\in(1,2^*)$. By \eqref{e3.7} and \eqref{e3.8},
letting $n\to \infty$ in the above inequality gives the contradiction
$1/2 \leq 0$.,
Hence $\{u_n\}$ is bounded, that is, $\|u_n\|\leq C_4$ for all $n$,
where $C_4$ is a positive constant independent of $n$. Hence, up to subsequence,
there exists a $u\in E$ such that
\begin{equation}
\begin{gathered}
 u_n\rightharpoonup u \quad \text{weakly in } E,\\
 u_n\to u \quad \text{strongly in } L^{r}(\Omega) \text{ for }
 r\in[1,\ 2^*),
 \end{gathered} \label{e3.12}
\end{equation}
Then by the weak lower semicontinuity of norm, we have
 $\|u\|\leq{\liminf_{n\to \infty}\|u_n\|}\leq C_4$, which implies that
 $\|u_n-u\|\leq\|u_n\|+\|u\|\leq 2C_4$.


Additionally, for $\varepsilon$ in $\eqref{e3.5}$, from \eqref{e3.12}
there exists a positive constant $N(\varepsilon)$ such that
$$
|u_n-u|_1<\varepsilon\quad \text{for } n>N(\varepsilon),
$$
from this and \eqref{e3.5} it follows that for $n>N(\varepsilon)$,
\begin{align*}
\big|\int_{\Omega}f_{+}(x,u_n)(u_n-u)dx\big|
&\leq \int_{\Omega}\left(\varepsilon(u^{+}_n)^{2^*-1}+L_{M_2}\right)|u_n-u|dx\\
&\leq \varepsilon|u_n|_{2^*}^{2^*-1}|u_n-u|_{2^*}+L_{M_2}|u_n-u|_1\\
&\leq \varepsilon 2\Big(\frac{C_4}{S_{2^*}}\Big)^{2^{*}}+\varepsilon L_{M_2},
\end{align*}
that is, $\int_{\Omega}f_{+}(x,u_n)(u_n-u)dx\to 0$ as $n\to \infty$.
From this, \eqref{e3.7}, \eqref{e3.12}, and Lemma \ref{lem3.1} it follows that
$$
\int_{\Omega}(\nabla u_n, \nabla(u_n-u))dx\to 0
$$
as $n\to \infty$. Then one has $\|u_n-u\|\to 0$ as $n\to \infty$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 By Lemmas \ref{lem3.3}, \ref{lem3.4} and \ref{lem3.5}, 
$\widetilde{\Phi}_{+}$ has a nontrivial critical
point $u$ via Theorem \ref{thmA}, that is, for any $v\in E$,
\[
\langle\widetilde{\Phi}'_{+}(u),v\rangle
=\int_{\Omega}(\nabla u,\nabla v)dx+\int_{\Omega}a(x)uv\,dx
 -\int_{\Omega}m(x)uv\,dx 
 -\int_{\Omega}\widetilde{f}_{+}(x,u)v\,dx=0.
\]
Letting $v=u^-$ in the above equation gives $\|u^-\|=0$, so $u=u^{+}\geq0$.
Then $u$ is also a critical point of $\Phi_{+}$; that is,
$$
\langle\Phi'_{+}(u),v\rangle=\int_{\Omega}(\nabla u,\nabla v)dx
 +\int_{\Omega}a(x)uv\,dx-\int_{\Omega}f_{+}(x,u)v\,dx=0,\ \forall\ v\in E.
$$
In addition, from (A16), (A17) and $a\in L^{\infty}(\Omega)$ it follows that
there exists positive constant $C_\varepsilon$ such that
$$
|-a(x)u+f(x,u)|\leq C_\varepsilon\left(1+|u|^{2^*-1}\right)
$$
for $s\in\mathbb{R}$ and a.e. $x\in\Omega$. Let
$b(x):=\frac{-a(x)u(x)+f(x,u(x))}{1+|u(x)|}$, then
$b\in L^{\frac{N}{2}}(\Omega)$ and
$$
-\triangle u=b(x)(1+|u|).
$$
\cite[Lemma B.3]{Struwe} shows $u\in L^{p}(\Omega)$ for any $p<\infty$,
 which implies that $f(x,u)\in L^{p}(\Omega)$ for any $p<\infty$.
By \cite[Lemma B.2]{Struwe}, we have $u\in H^{2,p}(\Omega)\cap H_{0}^{1}(\Omega)$
for any $p<\infty$. Therefore, $u\in C^{1,\beta}(\Omega)$ for some
$\beta\in(0,1)$ by the Sobolev embedding theorem. Moreover, from
$sf(x,s)\geq0$ it follows that
$$
\triangle u=a(x)u-f(x,u)\leq|a|_{\infty} u:=\zeta(u),
$$
where $\zeta:[0,+\infty)\to \mathbb{R}$ is continuous and nondecreasing,
and satisfies $\zeta(0)=0$, $\zeta(s)>0$ for all $s>0$, and
$\int_{0}^{1}(\zeta(s)s)^{-\frac{1}{2}}ds=+\infty$.
Then we can conclude that $u>0$ in $\Omega$ by \cite[Theorem 5]{Vazquez1984}.
In a similar way, we can obtain a negative solution for problem \eqref{sep}
by treating with $\widetilde{\Phi}_{-}$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 Without loss of generality, we assume that
$$
\lambda_{1}<\lambda_{2}\leq\lambda_3\leq\dots \leq\lambda_{k_{0}}\leq0
<\lambda_{k_{0}+1}\leq\dots\leq\lambda_{k}\dots
$$
and $e_{k}$ is eigenfunction corresponding to $\lambda_{k}$.
Set $E_{k}=\operatorname{span}\{e_1,e_2,\dots,e_k\}$ and
$E_{k}^{\perp}$ be the orthogonal complement of $E_{k}$ in $E$. Then one has
\begin{gather*}
\int_{\Omega}|\nabla u|^2dx+\int_{\Omega}a(x)u^2dx
\leq\lambda_{k}\int_{\Omega}u^2dx,\quad \forall u\in E_{k}, \\
\int_{\Omega}|\nabla u|^2dx+\int_{\Omega}a(x)u^2dx
 \geq\lambda_{k+1}\int_{\Omega}u^2dx,\quad \forall u\in E_{k}^{\perp}.
\end{gather*}
Hence in $E_{k}^{\perp}$ with $k\geq k_{0}$,
$\|u\|_{\star}:=\left\{\int_{\Omega}|\nabla u|^2dx
+\int_{\Omega}a(x)u^2dx\right\}^{1/2}$ is also a norm and is equivalent to
$\|u\|$. Hence for $k\geq k_{0}$, there exists a positive constant $C_5$ such that
$$
\|u\|_{\star}\geq \sqrt{C_5}\|u\|,\quad \forall u\in E _{k}^{\perp}.
$$
Similar to \eqref{e3.5}, from (A16) and (A17) it follows that
$$
|f(x,s)|\leq \varepsilon|s|^{2^*-1}+L_{M_{2}}
$$
for $s\in\mathbb{R}$ and a.e.\ $x\in\Omega$.
Set $\varrho_k:={\sup_{u\in E_{k}^{\perp}, \|u\|=1}}|u|_1$.
It was shown in \cite[Lemma 3.8]{Willem1996} that
 $\varrho_k\to 0$ as $k\to \infty$.
Let $X^1=E_{k}^{\perp}$ with $k\geq k_{0}$ such that
$\varrho_{k}<\frac{C_5}{8L_{M_2}}$,
\[
R_1=\Big(\frac{2^* S_{2^*}^{2^*}}{4\varepsilon} C_5\Big)^{\frac{1}{2^{\ast}-1}}>0,
\]
 for $u\in E_{k}^{\perp}$ with $\|u\|=R_1$,  we have
\begin{align*}
\Phi(u)
&\geq \frac{1}{2}\|u\|^2_{\star}-\frac{\varepsilon}{2^*}|u|_{2^*}^{2^*}-L_{M_2}|u|_1\\
&\geq \|u\|\Big(\frac{C_5}{2}\|u\|-\frac{\varepsilon}{2^* S_{2^*}^{2^*}}\|u\|^{2^*-1}
 -\varrho_{k}L_{M_2}\Big)\\
&\geq \frac{1}{8}C_5R_{1}.
\end{align*}
then $\Phi$ satisfies $\rm(i)$ of Theorem \ref{thmB} with $\alpha_1=\frac{1}{8}C_5R_{1}>0$.

For every $E_{k}$, there exists a positive constant $C_6$ such that
$$
\|u\|\leq \sqrt{C_6}|u|_{2},\quad \forall\ u\in E_{k},
$$
because all the norms on the finite dimension space $E_{k}$ are equivalent.
 From (A19), there exists a positive constant $C_7$ such that
$$
F(x,s)\geq\Big(\frac{|\lambda_{k}|}{2}+1\Big) s^2-C_7
$$
for $s\in\mathbb{R}$ and a.e.\ $x\in\Omega$. Set
$R_2=\sqrt{C_6C_7|\Omega|}$, for $u\in E_{k}$ with $\|u\|=R_{2}$,
\[
\Phi(u)
\leq\frac{\lambda_{k}}{2}|u|_{2}^2-\Big(\frac{|\lambda_{k}|}{2}+1\Big)|u|_{2}^2
 +C_7|\Omega|
\leq-\frac{1}{C_6}\|u\|^2+C_7|\Omega|\leq0,
\]
then $\Phi$ satisfies $\rm(ii)$ of Theorem \ref{thmB}.

Lastly, in a way similar to treat with $\widetilde{\Phi}_{+}$
in Lemma \ref{lem3.5}, we can prove that $\Phi$ satisfies the $\rm{(C)}$ condition.
Therefore Theorem \ref{thmB} shows that $\Phi$ has a unbounded sequence of critical values.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by National Natural Science Foundation of China
(No. 11471267, No. 11226118),by the Fundamental Research Funds for the
Central Universities(XDJK2014C161), by the Doctoral Fund of Southwest
 University (SWU111060).

The authors express their gratitude to the reviewers for the valuable comments
and helpful suggestions which lead to the improvement of the manuscript.




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\end{document}

 Xiao-Feng Ke (kexf@swu.edu.cn), Chun-Lei Tang(tangcl@swu.edu.cn).
If you have any queries, please contact me. Thank you and best regards.

Sincerely Yours,
Xiao-Feng Ke
E-mail: kexf@swu.edu.cn