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\title[Positive solutions for a class of singular elliptic equations]{Multiplicity of solutions for resonant Neumann problems with an indefinite and unbounded potential}


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\author[N.S. Papageorgiou]{Nikolaos S. Papageorgiou}
\address{National Technical University, Department of Mathematics,
				Zografou Campus, Athens 15780, Greece}
\email{\tt npapg@math.ntua.gr}

\author[V. R\u{a}dulescu]{Vicen\c{t}iu D. R\u{a}dulescu}
\address{Department of Mathematics, Faculty of Science, King Abdulaziz University,
Jeddah, Saudi Arabia \\ and Institute of Mathematics ``Simion Stoilow" of the Romanian Academy, P.O. Box 1-764,
          014700 Bucharest, Romania}
\email{\tt vicentiu.radulescu@math.cnrs.fr}

\keywords{Indefinite and unbounded potential, reduction method, resonance, unique continuation property, regularity, critical groups.\\
\phantom{aa} 2010 AMS Subject Classification:
35J20, 35J60, 58E05}

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\begin{abstract}
We examine semilinear Neumann problems driven by the Laplacian plus on unbounded and indefinite potential. The reaction is a Carath\'eodory function which exhibits linear growth near $\pm\infty$. We allow for resonance to occur with respect to a nonprincipal nonnegative eigenvalue. We prove several multiplicity results. Our approach uses critical point theory, Morse theory and the reduction method (the Lyapunov-Schmidt method).
\end{abstract}

\maketitle

%\tableofcontents

\section{Introduction}\label{sec1}

Let $\Omega\subseteq \RR^{N}$ be a bounded domain with a $C^2$-boundary $\partial\Omega$. In this paper, we study the following semilinear Neumann problem:
\begin{eqnarray}\label{eq1}
-\Delta u(z)+\beta(z)u(z)=f(z,u(z))\ \mbox{in}\ \Omega,\ \frac{\partial u}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}

Here $n(\cdot)$ denotes the outward unit normal on $\partial\Omega$. The potential function $\beta(\cdot)$ is in general unbounded and sign changing.  More  precisely, $\beta\in L^{s}(\Omega)$ with $s>N$. Also, the reaction $f(z,x)$ is a Carath\'eodory function (that is, for all $x\in\RR, z\mapsto f(z,x)$ is measurable and for a.a. $z\in\Omega, x\mapsto f(z,x)$ is continuous), which exhibits linear growth near $\pm\infty$. We allow for resonance to occur with respect to any nonnegative nonprincipal eigenvalue of $(-\Delta+\beta(\cdot),\ H^1(\Omega))$. So, we assume that asymptotically at $\pm\infty$ the quotient $\frac{f(z,x)}{x}$ is located in the spectral interval $[\hat{\lambda}_m,\hat{\lambda}_{m+1}]$ with $m\geq \max\{m_0,2\}$, where $\hat{\lambda}_{m_0}$ is the first nonnegative eigenvalue of $(-\Delta+\beta(\cdot),\ H^1(\Omega))$. Hence, if $\beta\equiv 0$, then $m_0=2$ and so $m\geq 2$. We allow resonance with respect to the left end $\hat{\lambda}_m$ and nonuniform n
 onresonance with respect to the right end $\hat{\lambda}_{m+1}$. Problems with double resonance (that is, possible resonance at both ends of the spectral interval), were studied by O'Regan, Papageorgiou \& Smyrlis \cite{25}, with $\beta\equiv 0$ (see also Hu \& Papageorgiou \cite{16} for Dirichlet problems with $\beta\neq 0$). After that, we deal with equations which are resonant at the origin and their energy functional is indefinite. Our aim is to prove multiplicity theorems for these problems. In the past analogous multiplicity results were proved for Dirichlet problems with $\beta\equiv 0$ by Bartsch \& Wang \cite{4}, Castro, Cossio \& Velez \cite{6}, Castro \& Lazer \cite{7}, Hofer \cite{15}, Liu \cite{18}, Liu \& Li \cite{20} and Thews \cite{32}. Dirichlet problems with an unbounded, indefinite potential, were studied recently by Gasinski \& Papageorgiou \cite{12}, Kyritsi \& Papageorgiou \cite{17}, Papageorgiou \& Papalini \cite{26}. Resonant Neumann problems were inv
 estigated by Filippakis \& Papageorgiou \cite{10}, Gasinski \& Papageorgiou \cite{13}, Motreanu, Motreanu \& Papageorgiou \cite{22}, Tang \cite{30} and Tang \& Wu \cite{31}. Neumann equations with unbounded, indefinite potential were studied by Papageorgiou \& R\u adulescu \cite{27} (problems with crossing nonlinearity) and Papageorgiou \& Smyrlis \cite{28} (coercive problems).

Our approach uses variational methods based on the critical point theory, coupled with the reduction technique (the Lyapunov-Schmidt method) and with
Morse theory (critical groups). In the next section, for the convenience of reader, we review the main mathematical tools which we will use in the sequel.

\section{Mathematical Background}\label{sec2}

 Let $X$ be a Banach space and $X^{\ast}$ its topological dual. By $\langle \cdot,\cdot \rangle$ we denote the duality brackets for the pair $(X^{\ast},X)$. Let $\varphi\in C^{1}(X)$. We say that $\varphi$ satisfies the ``$C$-condition", if the following holds:
\\
``Every sequence $\{u_n\}_{n\geq 1}\subseteq X$ such that  $\{\varphi(u_n)\}_{n\geq 1}\subseteq \RR$ is bounded and
$$(1+\|u_n\|)\varphi'(u_n)\rightarrow 0\ \mbox{in}\ X^{\ast}\ \mbox{as}\ n\rightarrow\infty,$$
admits a strongly convergent subsequence".

This compactness-type condition on the functional $\varphi$, is more general than the better known Palais-Smale condition. Nevertheless, the $C$-condition suffices to prove a deformation theorem and from it derive the minimax theory of certain critical values of $\varphi$. One such minimax theorem, which we will use later, is the so-called ``mountain pass theorem".
\begin{theorem}\label{th1}
 Assume that $X$ is a Banach space, $\varphi\in C^1(X)$ satisfies the $C$-condition, $u_0, u_1\in X$, with $\|u_1-u_0\|>\varrho>0$ and
\begin{eqnarray*}
&&\max\{\varphi(x_0),\varphi(x_1)\}<\inf\left[\varphi(u): \|u-u_0\|=\varrho\right]=\eta_{\varrho}.\\
&and &c=\inf_{\gamma\in\Gamma}\max_{0\leq t\leq 1}\varphi(\gamma(t))\ \mbox{with}\ \Gamma=\{\gamma\in C([0,1],X): \gamma(0)=u_0, \gamma(1)=u_1\}.
\end{eqnarray*}
Then $c\geq \eta_{\varrho}$ and $c$ is a critical value of $\varphi$.
\end{theorem}
In the analysis of problem (\ref{eq1}), in addition to the Sobolev space $H^1(\Omega)$, we will also use the Banach space $C^1(\overline{\Omega})$. This is an ordered Banach space with positive cone
$$C_{+}=\{u\in C^1(\overline{\Omega}): u(z)\geq 0\ \mbox{for\ all}\ z\in\overline{\Omega}\}.$$
This cone has a nonempty interior given by
$$\mbox{int}\, C_{+}=\{u\in C_{+}: u(z)>0\ \mbox{for\ all}\ z\in\overline{\Omega}\}.$$

For a Banach space $(X,\|\cdot\|)$, we say that it has the {\it Kadec-Klee property} if the following is true (see Gasinski \& Papageorgiou \cite[p. 911]{11}):
\begin{eqnarray*}
``u_n\overset{w}{\longrightarrow} u\ \mbox{in}\ X\ \mbox{and}\ \|u_n\|\rightarrow \|u\|\ \mbox{imply}\ u_n\rightarrow u\ \mbox{in}\ X".
\end{eqnarray*}
It is well-known that locally uniformly convex Banach spaces (in particular Hilbert spaces), have the Kadec-Klee property.

In what follows, by $\|\cdot\|$ we denote the norm of $H^1(\Omega)$. Hence
$$\|u\|=\Big[\|u\|_2^2+\|Du\|_2^2\Big]^{1/2}\ \mbox{for\ all}\ u\in H^1(\Omega).$$
(here by $\|\cdot\|_2$ we denote the norm of $L^2(\Omega)$ and $L^2(\Omega,\RR^N)$). Also, for every $x\in\RR$, we set $x^{\pm}=\max\{0,\pm x\}$. Then for every $u\in H^1(\Omega)$  we define $u^{\pm}(\cdot)=u(\cdot)^{\pm}$. We know that
\begin{eqnarray*}
u^{\pm}\in H^1(\Omega),\ u=u^{+}-u^{-}\ \mbox{and}\ |u|=u^{+}+u^{-}\ \mbox{for\ all}\ u\in H^1(\Omega).
\end{eqnarray*}
By $|\cdot|_N$ we denote the Lebesgue measure on $\RR^N$. Finally, if $h:\Omega\times\RR\rightarrow \RR$ is a measurable function (for example, a Carath\'eodory function), then we set
\begin{eqnarray*}
N_h(u)(\cdot)=h(\cdot, u(\cdot))\ \mbox{for\ all}\ u\in H^1(\Omega).
\end{eqnarray*}

The following linear eigenvalue problem, has a pivotal role in the analysis of problem (\ref{eq1}).
\begin{eqnarray}\label{eq2}
-\Delta u(z)+\beta(z)u(z)=\lambda u(z)\ \mbox{in}\ \Omega,\ \frac{\partial u}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}
This eigenvalue problem was studied by Papageorgiou \& Smyrlis \cite{28}. So, suppose that $\beta\in L^{N/2}(\Omega)$ if $N\geq 3$, $\beta\in L^r(\Omega)$ with $r>1$ if $N=2$ and $\beta\in L^1(\Omega)$ if $N=1$. Let $\tau:H^1(\Omega)\rightarrow \RR$ be the functional defined by
\begin{eqnarray*}
\tau(u)=\|Du\|^2_2+\int_{\Omega}\beta(z)u(z)^2dz\ \mbox{for\ all}\ u\in H^1(\Omega).
\end{eqnarray*}
Then the eigenvalue problem (\ref{eq2}) has a smallest eigenvalue $\hat{\lambda}_1>-\infty$ given by
\begin{eqnarray}\label{eq3}
\hat{\lambda}_1=\inf\left[\frac{\tau(u)}{\|u\|^2_2}:\ u\in H^1(\Omega),\ u\neq 0\right].
\end{eqnarray}
From (\ref{eq3}) it follows that we can find $\xi_0>\max\{-\hat{\lambda}_1,0\}$ such that
\begin{eqnarray}\label{eq4}
\tau(u)+\xi_0\|u\|^2_2\geq c_1\|u\|^2\ \mbox{for\ all}\ u\in H^1(\Omega)\ \mbox{and\ some}\ c_1>0.
\end{eqnarray}
Using (\ref{eq4}) and the spectral theorem for compact self-adjoint operators (see, for example, Gasinski \& Papageorgiou \cite[p. 297]{11}), we obtain a sequence $\{\hat{\lambda}_k\}_{k\geq 1}$ consisting of all the eigenvalues of (\ref{eq2}) such that  $\hat{\lambda}_k\rightarrow +\infty$ when $k\rightarrow\infty$. To these eigenvalues corresponds a sequence $\{\hat{u}_n\}_{n\geq 1}\subseteq H^1(\Omega)$ of eigenfunctions which form an othonormal basis of $L^2(\Omega)$ and an orthogonal basis of $H^1(\Omega)$. Moreover, if $\beta\in L^s(\Omega)$ with $s>N$, the the regularity results of Wang \cite{34}, imply that $\{\hat{u}_n\}_{n\geq 1}\subseteq C^1(\overline{\Omega})$. These eigenvalues admit variational characterizations in terms of the Rayleigh quotient $\frac{\tau(u)}{\|u\|^2_2}$ for all $u\in H^1(\Omega)\setminus\{0\}$. In what follows, by $E(\hat{\lambda}_k)$, we denote the eigenspace corresponding to the eigenvalue $\hat{\lambda}_k,\ k\geq 1$. For $\hat{\lambda}_1$ 
 the variational characterization is given by (\ref{eq3}) and the infimum is in fact a minimum which is realized on $E(\hat{\lambda}_1)$. For the higher eigenvalues, we have:
\begin{equation}\label{eq5}\begin{array}{ll}
\hat{\lambda}_i&\displaystyle=\inf\left[\frac{\tau(u)}{\|u\|^2_2}:\ u\in \overline{\oplus_{k\geq i}E(\lambda_k)},\ u\neq 0\right]\\
&\displaystyle=\sup\left[\frac{\tau(u)}{\|u\|^2_2}:\ u\in  \oplus^i_{k=1} E(\hat{\lambda}_i),\ u\neq 0\right]\ \mbox{for\ all}\ i\geq 2.
\end{array}\end{equation}
The infimum and the supremum in (\ref{eq5}) are realized on $E(\hat{\lambda}_i)$, $i\geq 2$. The first eigenvalue $\hat{\lambda}_1$ is simple (that is, dim $E(\hat{\lambda}_1)=1$) and from (\ref{eq3}) it is clear that the nontrivial elements of $E(\hat{\lambda}_1)$ do not change sign. All the other eigenvalues have nodal (sign changing) eigenfunctions. By $\hat{u}_1$ we denote the $L^2(\Omega)$-normalized (that is, $\|\hat{u}_1\|_2=1$) positive eigenfunction corresponding to $\hat{\lambda}_1$. If $\beta\in L^{s}(\Omega)$ with $s>N$, then $\hat{u_1}\in C_{+}\setminus\{0\}$ (see Wang \cite{34}). In fact, using the Harnack inequality (see Pucci \& Serrin \cite[p. 163]{29}) we have $u_1(z)>0$ for all $z\in\Omega$. Finally, if $\beta^{+}\in L^{\infty}(\Omega)$, then by the Hopf theorem (see, for example, Pucci \& Serrin \cite[p. 120]{29}), we have that
$\hat{u}_1\in \mbox{int}\, C_{+}$. When $\beta\in L^{s}(\Omega)$ with $s>\frac{N}{2}$, all the eigenspaces have the ``unique continuation property" (UCP for short). Namely, if $u\in E(\hat{\lambda}_k)$ and vanishes on a set of positive measure, then $u\equiv 0$.

A similar analysis can be conducted for a weighted version of problem (\ref{eq2}). So, let $m\in L^{\infty}(\Omega)$ with $m\geq 0$, $m\neq 0$. We consider the following weighted linear eigenvalue problem
\begin{eqnarray}\label{eq6}
-\Delta u(z)+\beta(z)u(z)=\lambda m(z)u(z)\ \mbox{in}\ \Omega,\ \frac{\partial u}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}
As was the case for problem (\ref{eq2}), problem (\ref{eq6}) above has a strictly increasing sequence $\{\hat{\lambda}_k(m)\}_{k\geq 1}$ of distinct eigenvalues such that  $\hat{\lambda}_k(m)\rightarrow+\infty$ as $k\rightarrow\infty$. The first eigenvalue $\hat{\lambda}_1(m)>-\infty$ is simple and has eigenfunctions of constant sign. All the other eigenvalues have nodal eigenfunctions. The Rayleigh quotient is $\frac{\tau(u)}{\int_{\Omega}m|u|^2dz}$ and using it we have variational characterizations similar to those in (\ref{eq5}). Again the eigenspaces $E(\hat{\lambda}_k(m))$, $k\geq 1$ have the UCP and as a consequence of this property we obtain the following result.
\begin{prop}\label{prop2}
 Assume that $m_1,\, m_2\in L^{\infty}(\Omega)_{+}\setminus\{0\}$, $m_1(z)\leq m_2(z)$ a.e. in $\Omega$, $m_1\neq m_2$. Then $\hat{\lambda}_k(m_2)<\hat{\lambda}_k(m_1)$ for all $k\geq 1$.
\end{prop}
Let $\overline{H}_k=\oplus_{i=1}^k E(\hat{\lambda}_i)$ and $\hat{H}_k=\overline{H}_k^{\perp}=\overline{\oplus_{i\geq k+1} E(\hat{\lambda}_i)}.$ As a consequence of the UCP, we have the following simple but useful inequalities.

\begin{prop}\label{prop3}
(a) If $k\geq 0$ is an integer, $\vartheta\in L^{\infty}(\Omega)$ and $\vartheta(z)\leq \hat{\lambda}_{k+1}$ a.e. in $\Omega$, $\vartheta\neq\hat{\lambda}_{k+1}$ then there exists $\xi_1>0$ such that
    $$\tau(u)-\int_{\Omega}\vartheta u^2dz\geq \xi_1\|u\|^2\ \mbox{for\ all}\ u\in\hat{H}_k.$$
(b) If $k\geq 1$ is an integer, $\eta\in L^{\infty}(\Omega)$ and $\eta(z)\geq \hat{\lambda}_{k}$ a.e. in $\Omega$, $\vartheta\neq\hat{\lambda}_{k}$ then there exists $\xi_2>0$ such that
    $$\tau(u)-\int_{\Omega}\eta u^2dz\leq -\xi_2\|u\|^2\ \mbox{for\ all}\ u\in\overline{H}_k.$$
\end{prop}
Next we recall some definitions and facts from Morse theory (critical groups) which we will need in the sequel.

So, let $X$ be a Banach space and $\varphi\in C^1(X)$, $c\in\RR $. We introduce the following sets
$$\varphi^{c}=\{x\in X: \varphi(x)\leq c\},\ K_{\varphi}=\{x\in X: \varphi'(x)=0\}\ \mbox{and}\ K_{\varphi}^c=\{x\in K_{\varphi}: \varphi(x)=c\}.$$

Let $(Y_1,Y_2)$ be a topological pair such that  $Y_2\subseteq Y_1\subseteq X$. For every integer $k\geq 0$ by $H_k(Y_1,Y_2)$ we denote the $\mbox{k}^{th}$-relative singular homology group for the pair $(Y_1,Y_2)$  with integer coefficients. Recall that for $k<0$, $H_k(Y_1,Y_2)=0$. The critical groups of $\varphi$ at an isolated critical point $u_0\in K_{\varphi}^c$, are defined by
$$C_k(\varphi,u_0)=H_k\left(\varphi^{c}\cap U, \varphi^{c}\cap U\setminus \{u_0\}\right)\ \mbox{for\ every integer}\ k\geq 0.$$
Here $U$ is a neighborhood of $u_0$ such that  $K_{\varphi}\cap\varphi^c\cap U=\{u_0\}.$  The excision property of singular homology implies that this definition of critical groups is independent of the choice of the neighborhood $U$.

Suppose that $\varphi\in C^1(X)$ satisfies the $C$-condition and $\inf \varphi(K_{\varphi})>-\infty$. Let $c<\inf \varphi(K_{\varphi})$. Then the  {\it critical groups} of $\varphi$ at infinity are defined by
$$C_k(\varphi,\infty)=H_k(X,\varphi^{c})\ \mbox{for\ all}\ k\geq 0.$$
The second deformation theorem (see, for example, Gasinski \& Papageorgiou \cite[p. 628]{11}), implies that this definition of critical groups at infinity is independent of the choice of the level $c<\inf \varphi(K_{\varphi})$.

Suppose that $K_{\varphi}$ is finite. We introduce the following quantities:
\begin{eqnarray*}
&&M(t,u)=\sum_{k\geq 0}\mbox{rank}\ C_k(\varphi,u)t^k\ \mbox{for\ all}\ t\in\RR,\ \mbox{all}\ u\in K_{\varphi},\\
&&P(t,\infty)=\sum_{k\geq 0}\mbox{rank}\ C_k(\varphi,\infty)t^k\ \mbox{for\ all}\ t\in\RR.
\end{eqnarray*}
The {\it Morse relation} says that
\begin{eqnarray}\label{eq7}
\sum\limits_{u\in K_{\varphi}} M(t,u)=P(t,\infty)+(1+t)Q(t),
\end{eqnarray}
where $Q(t)=\sum\limits_{k\geq 0}\beta_{k}t^{k}$ is a formal series in $t\in\RR$ with nonnegative integer coefficients $\beta_k$.

Suppose that $X=Y\oplus V$ with $\dim Y<\infty$. We say that $\varphi\in C^1(X)$ admits a {\it local linking} at the origin, if there exist $\varrho>0$ such that
\begin{eqnarray*}
&&\varphi(y)\leq 0\ \mbox{for\ all}\ y\in Y\ \mbox{with}\ \|y\|\leq\varrho,\\
&\mbox{and}&\varphi(v)\geq 0\ \mbox{for\ all}\ v\in V\ \mbox{with}\ \|v\|\leq\varrho.
\end{eqnarray*}
For functionals $\varphi$ with local linking at the origin, we have the following multiplicity result due to Brezis \& Nirenberg \cite{5} (see also Liu \& Li \cite{19}).
\begin{prop}\label{prop4}
  Assume that $X=Y\oplus V$ with $\dim Y<\infty$, $\varphi\in C^1(X)$, $\varphi(0)=0$, $\varphi$ is bounded below
$$\inf_{X}\varphi<0,$$
$\varphi$ satisfies the $C$-condition and admits local linking at the origin. Then $\varphi$ has at least two nontrivial critical points.
\end{prop}
Throughout this work, our hypotheses on the potential function $\beta(\cdot)$ are the following:

\medskip
$H_0$: $\beta\in L^{s}(\Omega)$ with $s>N$ and $\beta^{+}\in L^{\infty}(\Omega).$

\section{Three Nontrivial Solutions}\label{sec3}

In this section, by combining variational methods with the Lyapunov-Schmidt reduction technique and with Morse theory, we prove a ``three solutions" theorem for problem (\ref{eq1}), when resonance can occur at $\pm\infty$ with respect to any nonnegative nonprincipal eigenvalue of $(-\Delta-\beta, H^1(\Omega))$. So, in what follows, $\hat{\lambda}_{m_0}$ denotes the first nonnegative eigenvalue of this operator.

The hypotheses on the reaction term $f(z,x)$ are the following:

$H_1:\ f:\Omega\times\RR\rightarrow\RR$ is a Carath\'eodory function such that  $f(z,0)=0$ for a.a. $z\in\Omega$ and
\begin{itemize}
\item [(i)] there exist an integer $m\geq \max\{m_0,2\}$ and a function $\eta\in L^{\infty}(\Omega)_{+}$ such that
\begin{eqnarray*}
&&\eta(z)\leq \hat{\lambda}_{m+1}\ \mbox{a.e.\ in}\ \Omega,\ \eta\neq\hat{\lambda}_{m+1}\\
&&(f(z,x)-f(z,y))(x-y)\leq\eta(z)(x-y)^2\ \mbox{for\ a.a}\ z\in\Omega,\ \mbox{all}\ x,y \in\RR.
\end{eqnarray*}
\item [(ii)] $\hat{\lambda}_{m}\leq \liminf\limits_{x\rightarrow\pm\infty}\frac{f(z,x)}{x}$ uniformly for a.a. $z\in\Omega$;
\item [(iii)] if $F(z,x)=\int\limits_0^x f(z,s)ds$, then we have
$$\lim\limits_{x\rightarrow\pm\infty}[f(z,x)x-2F(z,x)]=-\infty\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega;$$
\item [(iv)] there exists a function $\vartheta\in L^{\infty}(\Omega)$ such that
\begin{eqnarray*}
&&\vartheta(z)\leq \hat{\lambda}_{1}\ \mbox{a.e.\ in}\ \Omega,\ \vartheta\neq\hat{\lambda}_{1}\\
&&\limsup\limits_{x\rightarrow 0}\frac{2F(z,x)}{x^2}\leq\vartheta(z)\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega;
\end{eqnarray*}
\item [(v)] for every $\varrho>0$, there exists $\xi_{\varrho}>0$ such that
$$f(z,x)x+\xi_{\varrho} x^2\geq 0\ \mbox{for\ a.a}\ z\in\Omega,\ \mbox{all},\ |x|\leq \varrho.$$
\end{itemize}
\begin{remark}
Hypotheses $H_1 (i), (ii)$ imply that asymptotically at $\pm\infty$, the quotient $\frac{f(z,x)}{x}$ is in the spectral interval $[\hat{\lambda}_{m}, \hat{\lambda}_{m+1}]$ with possible resonance with respect to $ \hat{\lambda}_{m}$ (see $H_1(ii)$), while at the other end we have nonuniform nonresonance (see $H_1(i)$).
\end{remark}
\begin{ex}
The following function satisfies hypotheses $H_1$ above. For the sake of simplicity, we drop the $z$-dependence:
\begin{eqnarray*}
f(x)=\left\{  \begin{array}{ll}
  \vartheta x+\xi |x|^{p-2}x&\mbox{if}\ |x|\leq 1 \\
    \lambda x+\frac{c}{x} &\mbox{if}\ 1<|x|,
    \end{array}\right.
\end{eqnarray*}
with $\vartheta<\hat{\lambda}_{1}$, $p>2$, $\xi=\lambda+c-\vartheta$, $\lambda\in [\hat{\lambda}_{m},\hat{\lambda}_{m+1})$ for some integer $m\geq \max\{m_0,2\}$, $c>0$, $2c<\lambda$.
\end{ex}

Let $\xi_0>0$ be as in (\ref{eq4}) and consider the following truncation-perturbation of the reaction $f(z,\cdot):$

$\hat{f}_{\pm}(z,x)=f(z,\pm x^{\pm})+\xi_0(\pm x^{\pm}).$

Both are Carath\'eodory functions. We set $\hat{F}_{\pm}(z,x)=\int\limits_0^x \hat{f}_{\pm}(z,s)ds$ and introduce the $C^1$- functionals $\varphi,\hat{\varphi}_{\pm}:H^1(\Omega)\rightarrow\RR$ defined by
\begin{eqnarray*}
&&\varphi(u)=\frac{1}{2}\tau(u)-\int_{\Omega} F(z,u(z))dz\\
&&\hat{\varphi}_{\pm}(u)=\frac{1}{2}\tau(u)+\frac{\xi_0}{2}\|u\|_2^2-\int_{\Omega} \hat{F}_{\pm}(z,u(z))dz\ \mbox{for\ all}\ u\in H^1(\Omega).
\end{eqnarray*}
\begin{prop}\label{prop5}
If hypotheses $H_0$ and $H_1$ hold, then the functionals $\hat{\varphi}_{\pm}$ satisfy the $C$-condition.
\end{prop}
\begin{proof}
Let $\{u_n\}_{n\geq 1}\subseteq H^1(\Omega)$ be a sequence such that
\begin{eqnarray}\label{eq8}
&&|\hat{\varphi}_{+}(u_n)|\leq M_1\ \mbox{for\ some}\ M_1>0,\ \mbox{all}\ n\geq 1.
\end{eqnarray}
\begin{eqnarray}\label{eq9}
&\mbox{and}& (1+\|u_n\|)\hat{\varphi}'_{+}(u_n)\rightarrow 0\ \mbox{in}\ H^1(\Omega)^{\ast}\ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
From (\ref{eq9}) we have
\begin{eqnarray}\label{eq10}
|\langle A(u_n),h\rangle +\int\limits_{\Omega}(\beta(z)+\xi_0)u_n h dz-\int\limits_{\Omega}\hat{f}_{+}(z,u_n)h dz|\leq \frac{\varepsilon_n \|h\|}{1+\|u_n\|}\\
\mbox{for\ all}\ h\in H^1(\Omega)\ \mbox{with}\ \varepsilon_n\rightarrow 0^{+}\nonumber.
\end{eqnarray}
Here $A\in \mathcal{L}(H^1(\Omega),H^1(\Omega)^{\ast})$ is defined by
$$\langle A(u),y\rangle=\int\limits_{\Omega}(Du,Dy)_{\RR^N}dz\ \mbox{for\ all}\ u,y\in H^1(\Omega).$$
In (\ref{eq10}) we choose $h=-u_n^{-}\in H^1(\Omega)$ and obtain
\begin{eqnarray}\label{eq11}
&&\tau(u_n^{-})+\xi_0\|u_n^{-}\|_2^2\leq \varepsilon_n\ \mbox{for\ all}\ n\geq 1,\nonumber \\
&\Rightarrow& c_1\|u_n^{-}\|^2\leq \varepsilon_n\ \mbox{for\ all}\ n\geq 1\ (\mbox{see}\ (\ref{eq4})),\nonumber \\
&\Rightarrow& u_n^{-} \rightarrow 0\ \mbox{in}\ H^1(\Omega).
\end{eqnarray}
If we use (\ref{eq11}) in (\ref{eq10}), then
\begin{eqnarray}\label{eq12}
|\langle A(u_n^{+}),h\rangle +\int\limits_{\Omega}\beta(z)u_n^{+} h dz-\int\limits_{\Omega}f(z,u_n^{+})h dz|\leq \varepsilon_n' \|h\|\\
\mbox{for\ all}\ h\in H^1(\Omega)\ \mbox{with}\ \varepsilon_n'\downarrow 0\nonumber.
\end{eqnarray}
Suppose that $\|u_n^{+}\|\rightarrow\infty$. We set $y_n=\frac{u_n^{+}}{\|u_n^{+}\|},\ n\geq 1$. Then $\|y_n\|=1$ for all $n\geq 1$ and so we may assume that
\begin{eqnarray}\label{eq13}
y_n\overset{w}{\longrightarrow} y\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y_n{\rightarrow} y\ \mbox{in}\ L^{2s'}(\Omega)\ \left(\frac{1}{s}+\frac{1}{s'}=1\right).
\end{eqnarray}
Multiplying (\ref{eq12}) with $\frac{1}{\|u_n^{+}\|}$, we obtain
\begin{eqnarray}\label{eq14}
|\langle A(y_n),h\rangle +\int\limits_{\Omega}\beta(z)y_n h dz-\int\limits_{\Omega}\frac{N_f(u_n^{+})}{\|u_n^{+}\|}h dz|\leq \varepsilon_n'' \|h\|\ \mbox{with}\ \varepsilon_n''\rightarrow 0.
\end{eqnarray}
From hypothesis $H_1(i)$, we have
\begin{eqnarray}\label{eq15}
f(z,x)\leq \eta(z)x\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\geq 0.
\end{eqnarray}
On the other hand hypotheses $H_1(ii),(iv)$ imply that we can find $c_2>0$ such that
\begin{eqnarray}\label{eq16}
-c_2 x\leq f(z,x)\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\geq 0.
\end{eqnarray}
From (\ref{eq15}) and (\ref{eq16}) it follows that
\begin{eqnarray}\label{eq17}
&&-c_2 y_n(z)\leq \frac{f(z,u_n^{+}(z))}{\|u_n^{+}\|}\leq \eta(z)y_n(z)\ \mbox{a.e.\ in}\ \Omega\\
&\Rightarrow& \left\{\frac{N_f(u_n^{+})}{\|u_n^{+}\|}\right\}_{n\geq 1}\subseteq L^2(\Omega)\ \mbox{is\ bounded}\nonumber.
\end{eqnarray}
Therefore, by passing to a subsequence if necessary and using hypotheses $H_1(ii)$ and (\ref{eq17}), we have
\begin{eqnarray}\label{eq18}
\frac{N_f(u_n^{+})}{\|u_n^{+}\|}\overset{w}{\longrightarrow}\hat{\xi}y\ \mbox{in}\ L^2(\Omega),
\end{eqnarray}
with $\hat{\lambda}_m\leq \hat{\xi}(z)\leq \eta(z)$ a.e. in $\Omega$ (see Aizicovici, Papageorgiou \& Staicu \cite{1}, proof of Proposition $30$). In (\ref{eq14}) we choose $h=y_n-y\in H^1(\Omega)$ and pass to the limit as $n\rightarrow\infty$. Using (\ref{eq13}) and (\ref{eq18}), we obtain
\begin{eqnarray}\label{eq19}
&&\lim\limits_{n\rightarrow\infty}\langle A(y_n),y_n-y\rangle=0,\nonumber\\
&\Rightarrow&\|Dy_n\|_2\rightarrow \|Dy\|_2,\nonumber\\
&\Rightarrow& y_n\rightarrow y\ \mbox{in}\ H^1(\Omega),\ \mbox{hence}\ \|y\|=1,\ y\geq 0\\
&&(\mbox{see}\ (\ref{eq13})\ \mbox{and\ use\ the\ Kadec-Klee\ property})\nonumber.
\end{eqnarray}
In (\ref{eq14}) we pass to the limit as $n\rightarrow\infty$ and use (\ref{eq18}) and (\ref{eq19}). Then
\begin{eqnarray}\label{eq20}
&&\langle A(y),h\rangle +\int_{\Omega}\beta y h dz=\int_{\Omega}\hat{\xi} y h dz\ \mbox{for\ all}\ h\in H^1(\Omega),\nonumber\\
&\Rightarrow& A(y)+\beta y=\hat{\xi} y,\nonumber\\
&\Rightarrow& -\Delta y(z)+\beta(z)y(z)=\hat{\xi}(z) y(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial y}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}
From the properties of $\hat{\xi}$ and Proposition \ref{prop2}, we have
\begin{eqnarray}\label{eq21}
\hat{\lambda}_1(\hat{\xi})<\hat{\lambda}_1(\hat{\lambda}_1)=1.
\end{eqnarray}
Since $y\neq 0$ (see (\ref{eq19})), from (\ref{eq20}) and (\ref{eq21}) it follows that $y$ must be nodal, a contradiction to (\ref{eq19}). This proves that the sequence $\{u_n\}_{n\geq 1}\subseteq H^1(\Omega)$ is bounded and so, we may assume that
\begin{eqnarray}\label{eq22}
u_n\overset{w}{\longrightarrow} u\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ u_n{\rightarrow} u\ \mbox{in}\ L^{2s'}(\Omega).
\end{eqnarray}
In (\ref{eq10}), we choose $h=u_n-u\in H^1(\Omega)$, pass to the limit as $n\rightarrow\infty$ and use (\ref{eq22}).
Then we obtain
\begin{eqnarray*}
&&\lim\limits_{n\rightarrow\infty}\langle A(u_n),u_n-u\rangle=0,\\
&\Rightarrow& u_n\rightarrow u\ \mbox{in}\ H^1(\Omega)\ (\mbox{by\ the\ Kadec-Klee\ property}),\\
&\Rightarrow& \hat{\varphi}_{+}\  \mbox{satisfies\ the\ C-condition}.
\end{eqnarray*}
Similarly for the functional $\hat{\varphi}_{-}$.
\end{proof}

\begin{prop}\label{prop6}
If hypotheses $H_0$ and $H_1$ hold, then the functional $\varphi$ satisfies the $C$-condition.
\end{prop}
\begin{proof}
Let $\{u_n\}_{n\geq 1}\subseteq H^1(\Omega)$ be a sequence such that
\begin{eqnarray}\label{eq23}
&&|\varphi(u_n)|\leq M_2\ \mbox{for\ some}\ M_2>0,\ \mbox{all}\ n\geq 1.
\end{eqnarray}
\begin{eqnarray}\label{eq24}
&\mbox{and}& (1+\|u_n\|)\varphi'(u_n)\rightarrow 0\ \mbox{in}\ H^1(\Omega)^{\ast}\ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
From (\ref{eq24}) we have
\begin{eqnarray}\label{eq25}
|\langle A(u_n),h\rangle +\int_{\Omega}\beta(z)u_n h dz-\int_{\Omega}f(z,u_n)h dz|\leq \frac{\varepsilon_n \|h\|}{1+\|u_n\|}\\
\mbox{for\ all}\ h\in H^1(\Omega)\ \mbox{with}\ \varepsilon_n\rightarrow 0^{+}\nonumber.
\end{eqnarray}
\begin{claim}
$\{u_n\}_{n\geq 1}\subseteq H^1(\Omega)$ is bounded.
\end{claim}
Arguing by contradiction, suppose that $\|u_n\|\rightarrow\infty$. We set $y_n=\frac{u_n}{\|u_n\|},\ n\geq 1$. Then $\|y_n\|=1$ for all $n\geq 1$ and so we may assume that
\begin{eqnarray}\label{eq26}
y_n\overset{w}{\longrightarrow} y\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y_n{\rightarrow} y\ \mbox{in}\ L^{2s'}(\Omega).
\end{eqnarray}
From (\ref{eq25}), we have
\begin{eqnarray}\label{eq27}
|\langle A(y_n),h\rangle +\int_{\Omega}\beta(z)y_n h dz-\int_{\Omega}\frac{N_f(u_n)}{\|u_n\|}h dz|\leq \varepsilon_n\
\mbox{for\ all}\ n\geq 1.
\end{eqnarray}
As in the proof of Proposition \ref{prop5}, using hypotheses $H_1(i),(ii),(iv)$, we see that $\left\{\frac{N_f(u_n)}{\|u_n\|}\right\}_{n\geq 1}\subseteq L^2(\Omega)$ is bounded. Hence, by passing to a subsequence if necessary and using hypotheses $H_1(i),(ii)$, we can say that
\begin{eqnarray}\label{eq28}
\frac{N_f(u_n)}{\|u_n\|}\overset{w}{\longrightarrow} \hat{\xi}y\ \mbox{in}\ L^2(\Omega)\ \mbox{with}\ \hat{\lambda}_m\leq \hat{\xi}(z)\leq\eta(z)\ \mbox{a.e.\ in}\ \Omega.
\end{eqnarray}
Choosing $h=y_n-y\in H^1(\Omega)$ in (\ref{eq27}), passing to the limit as $n\rightarrow\infty$ and using (\ref{eq26}) and (\ref{eq28}), we obtain
\begin{eqnarray}\label{eq29}
&&\lim\limits_{n\rightarrow\infty}\langle A(y_n),y_n-y\rangle=0,\nonumber\\
&\Rightarrow& y_n\rightarrow y\ \mbox{in}\ H^1(\Omega)\ (\mbox{by\ the\ Kadec-Klee\ property}),\ \mbox{hence}\ \|y\|=1.
\end{eqnarray}
So, if in (\ref{eq27}) we pass to the limit as $n\rightarrow\infty$ and use (\ref{eq28}) and (\ref{eq29}) then
\begin{eqnarray}\label{eq30}
&&\langle A(y),h\rangle +\int_{\Omega}\beta(z) y h dz=\int_{\Omega}\hat{\xi} y h dz\ \mbox{for\ all}\ h\in H^1(\Omega),\nonumber\\
&\Rightarrow& A(y)+\beta y=\hat{\xi} y,\nonumber\\
&\Rightarrow& -\Delta y(z)+\beta(z)y(z)=\hat{\xi}(z) y(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial y}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}
From (\ref{eq29}) we know that $y\neq 0$. Recall that $\hat{\lambda}_m\leq \hat{\xi}(z)\leq\eta(z)$ a.e. in $\Omega$. If $\hat{\xi}\neq \hat{\lambda}_m$, then from (\ref{eq30}) it follows that $y=0$, a contradiction. If $\hat{\xi}= \hat{\lambda}_m$ then from (\ref{eq30}) we see that $y\in E(\hat{\lambda}_m)\setminus\{0\}$ and so by the UCP we have that $y(z)\neq 0$ for a.a. $z\in\Omega$ and so $|u_n(z)|\rightarrow+\infty$ for a.a. $z\in\Omega$. Then by virtue of hypothesis $H_1(iii)$, we have
\begin{eqnarray}\label{eq31}
&&f(z,u_n(z))u_n(z)-2F(z,u_n(z))\rightarrow -\infty\ \mbox{for\ a.a.}\ z\in\Omega,\nonumber\\
&\Rightarrow&\int_{\Omega}\left[f(z,u_n)u_n-2F(z,u_n)\right]dz\rightarrow -\infty\ (\mbox{by\ Fatou's\ lemma,\ see}\ H_1(iii)).
\end{eqnarray}
From (\ref{eq23}) we have
\begin{eqnarray}\label{eq32}
\tau(u_n)-\int_{\Omega}2F(z,u_n)dz\geq -2M_2\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray}
In (\ref{eq25}) we choose $h=u_n\in H^1(\Omega)$ and obtain
\begin{eqnarray}\label{eq33}
-\tau(u_n)+\int_{\Omega}f(z,u_n)u_ndz\geq -\varepsilon_n\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray}
Adding (\ref{eq32}) and (\ref{eq33}), we obtain
\begin{eqnarray}\label{eq34}
\int_{\Omega}\left[f(z,u_n)u_n-2F(z,u_n)\right]dz\geq -M_3\ \mbox{for\ some}\ M_3>0,\ \mbox{ all}\ n\geq 1.
\end{eqnarray}
Comparing (\ref{eq31}) and (\ref{eq34}), we reach a contradiction. This proves the Claim.

By virtue of the Claim, we may assume that
\begin{eqnarray}\label{eq35}
u_n\overset{w}{\longrightarrow} u\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ u_n{\rightarrow} u\ \mbox{in}\ L^{2s'}(\Omega).
\end{eqnarray}
From (\ref{eq25}) with $h=u_n-u\in H^1(\Omega)$ and using (\ref{eq35}) we have
\begin{eqnarray*}
&&\lim\limits_{n\rightarrow\infty}\langle A(u_n),u_n-u\rangle=0,\\
&\Rightarrow& u_n\rightarrow u\ \mbox{in}\ H^1(\Omega)\ (\mbox{by\ the\ Kadec-Klee\ property}),\\
&\Rightarrow& \varphi\  \mbox{satisfies\ the\ C-condition}.
\end{eqnarray*}
\end{proof}

\begin{prop}\label{prop7}
If hypotheses $H_0$ and $H_1$ hold, then $u=0$ is a minimizer for the functionals $\hat{\varphi}_{\pm}$ and $\varphi$.
\end{prop}
\begin{proof}
We do the proof for $\hat{\varphi}_{+}$, the proofs for $\hat{\varphi}_{-}$ and $\varphi$ being similar.

By virtue of hypothesis $H_1(iv)$, given $\varepsilon>0$ we can find $\delta=\delta(\varepsilon)>0$ such that
\begin{eqnarray}\label{eq36}
F(z,x)\leq \frac{1}{2}(\vartheta(z)+\varepsilon)x^2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\leq \delta.
\end{eqnarray}
Then for $u\in C^1(\overline{\Omega})$ with $\|u\|_{C^1(\overline{\Omega})}\leq\delta$, we have
\begin{eqnarray}\label{eq37}
\hat{\varphi}_{+}(u)&=&\frac{1}{2}\tau(u)+\frac{\xi_0}{2}\|u\|_2^2-\int_{\Omega}\hat{F}_{+}(z,u)dz\nonumber\\
&\geq&\frac{1}{2}\left[\tau(u)-\int_{\Omega}\vartheta u^2dz-\varepsilon\|u\|^2\right]\ (\mbox{see}\ (\ref{eq36}))\nonumber\\
&\geq&\frac{\xi_1-\varepsilon}{2}\|u\|^2\ (\mbox{see\ Proposition}\ \ref{prop3}(a)).
\end{eqnarray}
Choosing $\varepsilon\in (0,\xi_1)$, from (\ref{eq37}) it follows that $u=0$ is a local $C^1(\overline{\Omega})$-minimizer of $\hat{\varphi}_{+}$. From Motreanu \& Papageorgiou \cite{24}, it follows that $u=0$ is a local $H^1(\Omega)$-minimizer of $\hat{\varphi}_{+}$.
Similarly for the functionals $\hat{\varphi}_{-}$, $\varphi$.
\end{proof}

\begin{prop}\label{prop8}
If hypotheses $H_0$ and $H_1$ hold, then $\hat{\varphi}_{\pm}(t\hat{u}_1)\rightarrow -\infty$ as $t\rightarrow \pm\infty$.
\end{prop}
\begin{proof}
By virtue of hypotheses $H_1(i),(ii),(v)$, given $\varepsilon>0$ we can find $c_2=c_2(\varepsilon)>0$ such that
\begin{eqnarray}\label{eq38}
F(z,x)\leq \frac{1}{2}[\hat{\lambda}_m-\varepsilon]x^2-c_2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.
\end{eqnarray}
Then for $t>0$, we have
\begin{eqnarray}\label{eq39}
\hat{\varphi}_{+}(t\hat{u}_1)&=&\frac{t^2}{2}\hat{\lambda}_1-\int_{\Omega}F(z,t\hat{u}_1)dz\nonumber\\
&\leq& \frac{t^2}{2}[\hat{\lambda}_1+\varepsilon-\hat{\lambda}_m]+c_2|\Omega|_{N}\ (\mbox{see}\ (\ref{eq38})).
\end{eqnarray}
We choose $\varepsilon\in (0,\hat{\lambda}_m-\hat{\lambda}_1)$ (recall that $m\geq 2$). Then from (\ref{eq39}) it follows that
$$\hat{\varphi}_{+}(t\hat{u}_1)\rightarrow -\infty\ \mbox{as}\ t\rightarrow +\infty.$$
Similarly, we show that
$$\hat{\varphi}_{-}(t\hat{u}_1)\rightarrow -\infty\ \mbox{as}\ t\rightarrow +\infty.$$
\end{proof}
Now we are ready to produce two nontrivial solutions of constant sign for problem (\ref{eq1}).
\begin{prop}\label{prop9}
If hypotheses $H_0$ and $H_1$ hold, then problem (\ref{eq1}) admits at least two nontrivial constant sign solutions
$$u_0\in \mbox{int}\, C_{+}\ \mbox{and}\ v_0\in -\mbox{int}\, C_{+}.$$
\end{prop}
\begin{proof}
According to Proposition \ref{prop7}, $u=0$ is a local minimizer of the functional $\hat{\varphi}_{+}$. So, we can find $\varrho\in (0,1)$ small such that
\begin{eqnarray}\label{eq40}
\hat{\varphi}_{+}(0)=0<\inf\left[\hat{\varphi}_{+}(u): \|u\|=\varrho\right]=\hat{\eta}_{\varrho}^{+}
\end{eqnarray}
(see Aizicovici, Papageorgiou \& Staicu \cite{1} (proof of Proposition $29$)). Then relation (\ref{eq40}) together with Propositions \ref{prop5} and \ref{prop8}, permit the use of Theorem \ref{th1} (the mountain pass theorem). So, we can find $u_0\in H^1(\Omega)$ such that
\begin{eqnarray}\label{eq41}
\hat{\eta}_{\varrho}^{+}\leq \hat{\varphi}_{+}(u_0)\ \mbox{and}\ \hat{\varphi}'_{+}(u_0)=0.
\end{eqnarray}
From (\ref{eq40}) and (\ref{eq41}) it follows that
\begin{eqnarray}\label{eq42}
A(u_0)+(\beta+ \xi_0)u_0=N_{\hat{f}_{+}}(u_0),\ u_0\neq 0.
\end{eqnarray}
On (\ref{eq42}) we act with $-u_{0}^{-}\in H^1(\Omega)$ and using (\ref{eq4}), we obtain $u_0\geq 0$, $u_0\neq 0$. So, we have
\begin{eqnarray*}
&&A(u_0)+\beta u_0=N_f(u_0),\\
&\Rightarrow& -\Delta u_0(z)+\beta(z)u_0(z)=f(z,u_0(z))\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial u_0}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray*}
Thus, $u_0$ is a nontrivial positive solution of (\ref{eq1}). We set
\begin{eqnarray*}
k(z)=\left\{  \begin{array}{ll}
   \frac{f(z,u_0(z))}{u_0(z)}&\mbox{if}\ u_0(z)\neq 0 \\
    0 &\mbox{otherwise}.
    \end{array}\right.
\end{eqnarray*}
Then $k\in L^{\infty}(\Omega)$. We have
\begin{eqnarray}\label{eq43}
-\Delta u_0(z)=(k-\beta)(z)u_0(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial u_0}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}
Note that $k-\beta \in L^s(\Omega)$ (see $H_0$ and recall that $k\in L^{\infty}(\Omega)$). Then from Lemma $5.1$ of Wang \cite{34}, we have that $u_0\in L^{\infty}(\Omega)$. From (\ref{eq43}) it follows that $\Delta u_0\in L^s(\Omega)$ and so Lemma $5.2$ of Wang \cite{34} implies that $u_0\in W^{2,s}(\Omega)$. Since $s>N$ (see hypothesis $H_0$), by virtue of the Sobolev embedding theorem, we have
$$W^{2,s}(\Omega)\hookrightarrow C^{1+\alpha}(\overline{\Omega}) \ \mbox{with}\ \alpha=1-\frac{N}{s}>0.$$
Therefore, we infer that $u_0\in C_{+}\setminus\{0\}$. From (\ref{eq43}) we have
\begin{eqnarray*}
&&\Delta u_0(z)=(\beta-k)(z)u_0(z)\leq (\|\beta^{+}\|_{\infty}+\|k\|_{\infty})u_0(z)\ \mbox{a.e.\ in}\ \Omega,\\
&\Rightarrow& u_0\in \mbox{int}\, C_{+}\ (\mbox{see\ Vazquez \cite{33}}).
\end{eqnarray*}
In a similar fashion, working this time with $\hat{\varphi}_{-}$, we produce a second nontrivial constant sign solution $v_0\in -\mbox{int}\, C_{+}$.
\end{proof}

To produce a third nontrivial solution, we will employ the so-called Lyapunov-Schmidt reduction technique as this was formulated for elliptic equations by Amann \cite{2}, Castro \& Lazer \cite{7}, and Thews \cite{32}.

To this end, we introduce the following subspaces of $H^1(\Omega)$:
$$Y=\oplus_{i=1}^{m}E(\hat{\lambda}_i)\ \mbox{and}\ \hat{H}=Y^{\bot}=\overline{\oplus_{i\geq m+1}E(\hat{\lambda}_i)}.$$
We have the following orthogonal direct sum decomposition
$$H^1(\Omega)=Y\oplus \hat{H}.$$
The reduction method will be based on this decomposition.
\begin{prop}\label{prop10}
If hypotheses $H_0$ and $H_1$ hold, then there exists a continuous map $\gamma_0:Y\rightarrow \hat{H}$ such that
$$\varphi(y+\gamma_0(y))=\inf \left[\varphi(y+\hat{u}): \hat{u}\in \hat{H}\right]\ \mbox{for\ all}\ y\in Y.$$
\end{prop}
\begin{proof}
We fix $y\in Y$ and consider the $C^1$-functional $\varphi_{y}:H^1(\Omega)\rightarrow\RR$ defined by
$$\varphi_{y}(u)=\varphi (y+u)\ \mbox{for\ all}\ u\in H^1(\Omega).$$
Let $i:\hat{H}\rightarrow H^1(\Omega)$ be the inclusion map and set $\hat{\varphi}_{y}=\varphi_{y}\circ i:\hat{H}\rightarrow \RR$. The chain rule implies that
$$\hat{\varphi}_{y}'(\hat{u})=p_{\hat{H}^{\ast}}\varphi_{y}'(\hat{u})\ \mbox{for\ all}\ \hat{u}\in \hat{H},$$
where $p_{\hat{H}^{\ast}}$ is the orthogonal projection of $H^1(\Omega)$ onto $\hat{H}^{\ast}$.
Let $\langle\cdot,\cdot\rangle_{\hat{H}}$ denote  the duality brackets for the pair $(\hat{H}^{\ast},\hat{H})$.
For $\hat{u}_1,\hat{u}_2\in \hat{H}$ we have
\begin{eqnarray}
&&\langle \hat{\varphi}_{y}'(\hat{u}_1)-\hat{\varphi}_{y}'(\hat{u}_2),\hat{u}_1-\hat{u}_2\rangle _{\hat{H}} \nonumber \\
&=& \langle \varphi_{y}'(\hat{u}_1)-\varphi_{y}'(\hat{u}_2),\hat{u}_1-\hat{u}_2\rangle \nonumber \\
&=& \langle A(\hat{u}_1)-A(\hat{u}_2),\hat{u}_1-\hat{u}_2\rangle+\int_{\Omega}\beta(z)(\hat{u}_1-\hat{u}_2)^2 dz\nonumber \\ &-&\int_{\Omega}\left[f(z,y+\hat{u}_1)
-f(z,y+\hat{u}_2)\right](\hat{u}_1-\hat{u}_2)dz \nonumber \\
&\geq& \tau (\hat{u}_1-\hat{u}_2)-\int_{\Omega}\eta(\hat{u}_1-\hat{u}_2)^2 dz\ (\mbox{see\ hypothesis}\ H_1(i)) \nonumber
\end{eqnarray}
\begin{eqnarray}\label{eq44}
&\geq& \xi_1 \|\hat{u}_1-\hat{u}_2\|^2\ (\mbox{see\ Proposition}\ \ref{prop3}(a)\ \mbox{and\ recall\ that}\ \hat{u}_1,\hat{u}_2\in \hat{H}),
\end{eqnarray}
\begin{eqnarray}\label{eq45}
&\Rightarrow& \hat{\varphi}_{y}' \ \mbox{is\ strongly\ monotone,\ hence}\ \hat{\varphi}_{y}\ \mbox{is\ strictly\ convex}.
\end{eqnarray}
Also, note that
\begin{eqnarray}\label{eq46}
\langle \hat{\varphi}_{y}'(\hat{u}),\hat{u}\rangle_{\hat{H}}&=&\langle \hat{\varphi}_{y}'(\hat{u})-\hat{\varphi}_{y}'(0),\hat{u}\rangle_{\hat{H}}+\langle\hat{\varphi}_{y}(0),\hat{u}\rangle_{\hat{H}}\nonumber\\
&\geq& \xi_1\|\hat{u}\|^2-c_3\|\hat{u}\|\ \mbox{for\ some}\ c_3>0\ (\mbox{see}\ (\ref{eq44}))\\
\Rightarrow \hat{\varphi}_{y}'\ \mbox{is\ coercive} \nonumber.
\end{eqnarray}
Since $\hat{\varphi}_{y}'$ is strongly monotone (hence maximal monotone too, being continuous) and coercive, we have that $\hat{\varphi}_{y}'$ is surjective (see, for example, Papageorgiou \& Kyritsi \cite[p. 172]{26}). So, we can find $\hat{u}_0\in \hat{H}$ such that
\begin{eqnarray}\label{eq47}
\hat{\varphi}_{y}'(\hat{u}_0)=0.
\end{eqnarray}

In fact, relation (\ref{eq45}) implies that $\hat{u}_0\in \hat{H}$ is unique and is the unique minimizer on $\hat{H}$ of the strictly convex functional $\hat{\varphi}_{y}=\varphi_y|_{\hat{H}}$. So, we can define the map $\gamma_0:Y\rightarrow \hat{H}$ by setting $\gamma_0(y)=\hat{u}_0$. We have
\begin{eqnarray}\label{eq48}
p_{\hat{H}^{\ast}}\varphi'(y+\gamma_0(y))=0\ \mbox{and}\ \varphi(y+\gamma_0(y))=\inf[\varphi(y+\hat{u}):\hat{u}\in\hat{H}].
\end{eqnarray}
Next we examine the continuity properties of the map $y\mapsto \gamma_0(y)$. To this end, suppose that $y_n\rightarrow y$ in $Y$. For every $n\geq 1$, we have
\begin{eqnarray*}
0&=&\langle \hat{\varphi}_{y_{n}}'(\gamma_0(y_{n})),\gamma_0(y_{n})\rangle_{\hat{H}}\ (\mbox{see}\ (\ref{eq47}))\\
&\geq& \xi_1\|\gamma_0(y_{n})\|^2-c_3\|\gamma_0(y_{n})\|\ (\mbox{see}\ (\ref{eq46})),
\end{eqnarray*}
\begin{eqnarray*}
\Rightarrow \{\gamma_0(y_{n})\}_{n\geq 1}\subseteq \hat{H}\ \mbox{is\ bounded}.
\end{eqnarray*}
So, at least for a subsequence we have
\begin{eqnarray*}
\gamma_0(y_{n})\overset{w}{\longrightarrow} h\ \mbox{in}\ H^{1}(\Omega)\ \mbox{and}\ h\in\hat{H}.
\end{eqnarray*}
Using the Sobolev embedding theorem, we deduce that $\varphi$ is sequentially weakly lower semicontinuous. Hence
\begin{eqnarray}\label{eq49}
\varphi(y+h)\leq \liminf_{n\rightarrow\infty}\varphi({y_{n}}+\gamma_0(y_{n})).
\end{eqnarray}
From (\ref{eq48}), we have
\begin{eqnarray}\label{eq50}
&&\varphi({y_{n}}+\gamma_0(y_{n}))\leq \varphi({y_{n}}+\hat{u})\ \mbox{for\ all}\ \hat{u}\in\hat{H},\ \mbox{all}\ n\geq 1,\nonumber\\
&\Rightarrow&\varphi(y+h)\leq \varphi(y+\hat{u})\ \mbox{for\ all}\ \hat{u}\in\hat{H},\nonumber\\
&\Rightarrow& h=\gamma_0(y),\nonumber\\
&\Rightarrow& \gamma_0(y_{n})\overset{w}{\longrightarrow}\gamma_0(y)\ \mbox{in}\ H^{1}(\Omega).
\end{eqnarray}
Also, we have
\begin{eqnarray*}
&&p_{\hat{H}^{\ast}}\varphi'(y_{n}+\gamma_0(y_{n}))=0\ \mbox{for\ all}\ n\geq 1\ (\mbox{see}\ (\ref{eq47})),\\
&\Rightarrow& p_{\hat{H}^{\ast}}[A(y_{n}+\gamma_0(y_{n}))+\beta(y_{n}+\gamma_0(y_{n}))]=p_{\hat{H}^{\ast}}N_f (y_{n}+\gamma_0(y_{n}))\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray*}
On this equation, we act with $\gamma_0(y_{n})-\gamma_0(y)$ and obtain
\begin{eqnarray*}
&&\langle A(y_{n}+\gamma_0(y_{n})),\gamma_0(y_{n})-\gamma_0(y)\rangle+\int_{\Omega}\beta(z)(y_n+\gamma_0(y_{n}))(\gamma_0(y_{n})-\gamma_0(y))dz\\
&=&\int_{\Omega} f(z,y_n+\gamma_0(y_{n}))(\gamma_0(y_{n})-\gamma_0(y))dz\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray*}
Passing to the limit as $n\rightarrow\infty$ and using (\ref{eq50}), we obtain
\begin{eqnarray*}
&&\lim_{n\rightarrow\infty}\langle A(y_{n}+\gamma_0(y_{n})),\gamma_0(y_{n})-\gamma_0(y)\rangle=0,\\
&\Rightarrow& \|D(y_{n}+\gamma_0(y_{n}))\|_2\rightarrow \|D(y+\gamma_0(y))\|_2 ,\\
&\Rightarrow& y_{n}+\gamma_0(y_{n})\rightarrow y+\gamma_0(y)\ \mbox{in}\ H^1(\Omega)\ (\mbox{by\ the\ Kadec-Klee\ property})\\
&\Rightarrow& \gamma_0(y_{n})\rightarrow \gamma_0(y)\ \mbox{in}\ H^1(\Omega),\\
&\Rightarrow& \gamma_0\ \mbox{is\ continuous}\ (\mbox{by\ Urysohn's\ convergence\ criterion}).
\end{eqnarray*}
\end{proof}
Let $\psi(y)=\varphi(y+\gamma_0(y))$ for all $y\in Y$. The next proposition is not immediately clear, since $\gamma_0$ is only continuous.
\begin{prop}\label{prop11}
If hypotheses $H_0$ and $H_1$ hold, then $\psi\in C^1(Y)$.
\end{prop}
\begin{proof}
Let $y,v \in Y$ and $t>0$ (the analysis is similar if $t<0$). We have
\begin{eqnarray}\label{eq51}
&&\frac{\psi(y+tv)-\psi(y)}{t}\leq \frac{\varphi(y+tv+\gamma_0(y))-\varphi(y+\gamma_0(y))}{t},\nonumber\\
&\Rightarrow& \limsup_{t\rightarrow 0}\frac{\psi(y+tv)-\psi(y)}{t}\leq \langle \varphi'(y+\gamma_0(y)),v\rangle.
\end{eqnarray}
Also, we have
\begin{eqnarray}\label{eq52}
&&\frac{\psi(y+tv)-\psi(y)}{t}\geq \frac{\varphi(y+tv+\gamma_0(y+tv))-\varphi(y+\gamma_0(y+tv))}{t},\nonumber\\
&\Rightarrow& \liminf_{t\rightarrow 0}\frac{\psi(y+tv)-\psi(y)}{t}\geq \langle \varphi'(y+\gamma_0(y)),v\rangle\\
&&(\mbox{since}\ \varphi\in C^1(H^1(\Omega))\ \mbox{and}\ \gamma_0\ \mbox{is\ continuous}\ (\mbox{see\ Proposition\ \ref{prop10}}))\nonumber.
\end{eqnarray}
From (\ref{eq51}) and (\ref{eq52}) it follows that
\begin{eqnarray*}
\psi\in C^1(Y)\ \mbox{and}\ \langle \psi'(y),v\rangle_{Y}=\langle\varphi'(y+\gamma_0(y)),v\rangle\ \mbox{for\ all}\ y,v\in Y.
\end{eqnarray*}
\end{proof}

\begin{prop}\label{prop12}
If hypotheses $H_0$ and $H_1$ hold, then $\psi$ is anticoercive (that is, if $\|y\|\rightarrow\infty$, then $\psi(y)\rightarrow -\infty$).
\end{prop}
\begin{proof}
We argue indirectly. So, suppose that we can find $\{y_n\}_{n\geq 1}\subseteq Y$ such that
\begin{eqnarray*}
\|y_n\|\rightarrow\infty\ \mbox{and}\ \psi(y_n)\geq -M_4\ \mbox{for\ some}\ M_4>0,\ \mbox{all}\ n\geq 1.
\end{eqnarray*}
We have
\begin{eqnarray}\label{eq53}
-M_4\leq \psi(y_n)\leq \varphi(y_n)=\frac{1}{2}\tau(y_n)-\int_{\Omega}F(z,y_n)dz\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray}
Let $v_n=\frac{y_n}{\|y_n\|},\ n\geq 1$. Then $\|v_n\|=1$, $v_n\in Y$ for all $n\geq 1$. Recall that the space $Y$ is finite dimensional. So, we may assume that
\begin{eqnarray}\label{eq54}
v_n\rightarrow v\ \mbox{in}\ H^1(\Omega),\ v\in Y\ \mbox{and}\ \|v\|=1.
\end{eqnarray}
From (\ref{eq53}) we have
\begin{eqnarray}\label{eq55}
-\frac{M_4}{\|y_n\|^2}\leq \frac{1}{2}\tau(v_n)-\int_{\Omega}\frac{F(z,y_n)}{\|y_n\|^2}dz.
\end{eqnarray}
As before (see the proof of Proposition \ref{eq5}), we have that
$$\left\{\frac{F(\cdot,y_n(\cdot))}{\|y_n\|^2}\right\}_{n\geq 1}\subseteq L^1(\Omega)\ \mbox{is\ uniformly\ integrable}.$$
Then the Dunford-Pettis theorem and hypotheses $H_1(i),(ii)$ imply that
\begin{eqnarray}\label{eq56}
\frac{F(\cdot,y_n(\cdot))}{\|y_n\|^2}\overset{w}{\longrightarrow} \frac{1}{2}\xi^{\ast}v^2\ \mbox{in}\ L^1(\Omega),\ \mbox{with}\ \hat{\lambda}_m\leq \xi^{\ast}(z)\leq \eta(z)\ \mbox{a.e.\ in}\ \Omega.
\end{eqnarray}
If in (\ref{eq55}) we pass to the limit as $n\rightarrow \infty$ and use (\ref{eq54}) and (\ref{eq56}), then
\begin{eqnarray}\label{eq57}
0\leq \frac{1}{2}\tau(v)-\frac{1}{2}\int_{\Omega}\xi^{\ast}v^2dz.
\end{eqnarray}
If $\xi^{\ast}\neq \hat{\lambda}_m$, then since $v\in Y$ (see (\ref{eq54})), from (\ref{eq56}), (\ref{eq57}) and Proposition \ref{prop3}(b) we reach a contradiction, that is, $0\leq \frac{1}{2}\tau(v)-\frac{1}{2}\int_{\Omega}\xi^{\ast}v^2dz<0$.

So, suppose that $\xi^{\ast}\equiv \hat{\lambda}_m$ (see (\ref{eq56})). Then from (\ref{eq5}) and (\ref{eq57}) we have that $v\in E(\hat{\lambda}_m)\setminus \{0\}$ (see also (\ref{eq54})). The UCP implies that $v(z)\neq 0$ for a.a. $z\in\Omega$ and so $|y_n(z)|\rightarrow\infty$ for a.a. $z\in\Omega$.

By virtue of hypothesis $H_1(iii)$ we know that given any $\xi>0$, we can find $M_5=M_5(\xi)>0$ such that
\begin{eqnarray}\label{eq58}
f(z,x)x-2F(z,x)\leq -\xi\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\geq M_5.
\end{eqnarray}
We have
\begin{eqnarray*}
&&\frac{d}{dx}\left(\frac{F(z,x)}{x^2}\right)=\frac{f(z,x)x-2F(z,x)}{x^3}\leq -\frac{\xi}{x^3}\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\geq M_5\ (\mbox{see (\ref{eq58})}),\\
&\Rightarrow& \frac{F(z,x)}{x^2}-\frac{F(z,u)}{u^2}\leq \frac{\xi}{2}\left[\frac{1}{x^2}-\frac{1}{u^2}\right]\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\geq u\geq M_5.
\end{eqnarray*}
Passing to the limit as $x\rightarrow +\infty$ and using hypothesis $H_1(ii)$, we obtain
$$\frac{\hat{\lambda}_m}{2}u^2-F(z,u)\leq -\frac{\xi}{2}\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ u\geq M_5.$$
Since $\xi>0$ is arbitrary, we infer that
\begin{eqnarray}\label{eq59}
\lim_{u\rightarrow +\infty}\left[\frac{\hat{\lambda}_m}{2}u^2-F(z,u)\right]=-\infty\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega.
\end{eqnarray}
From (\ref{eq53}), we have
\begin{eqnarray}\label{eq60}
-M_4&\leq& \frac{1}{2}\tau(y_n)-\int_{\Omega}F(z,y_n)dz\nonumber \\
&\leq&\frac{\hat{\lambda}_m}{2}\|y_n\|_2^2-\int_{\Omega}F(z,y_n)dz\ (\mbox{since}\ y_n\in Y,\ \mbox{see}\ (\ref{eq5}))\nonumber \\
&=&\int_{\Omega}\left[\frac{\hat{\lambda}_m}{2}y_{n}^2-F(z,y_n)\right]dz.
\end{eqnarray}
From (\ref{eq59}) and Fatou's lemma (since $|y_n(z)|\rightarrow +\infty$ for a.a. $z\in\Omega$), we have
\begin{eqnarray}\label{eq61}
\int_{\Omega}\left[\frac{\hat{\lambda}_m}{2}y_{n}^2-F(z,y_n)\right]dz\rightarrow -\infty\ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
Comparing (\ref{eq60}) and (\ref{eq61}), we reach a contradiction. This proves the anticoercivity of the functional $\psi$.
\end{proof}

\begin{remark}
Since $-\psi$ is coercive, it follows that functional $\psi$ satisfies the $C$-condition.
\end{remark}
Now we are ready for the ``three solutions theorem" for problem (\ref{eq1}).
\begin{theorem}\label{th13}
If hypotheses $H_0$ and $H_1$ hold, then problem (\ref{eq1}) has at least three nontrivial solutions
$$u_0\in \mbox{int}\, C_{+},\  v_0\in -\mbox{int}\, C_{+}\ \mbox{and}\ y_0\in C^1(\overline{\Omega}).$$
\end{theorem}
\begin{proof}
From Proposition \ref{prop9} we already have two nontrivial constant sign solutions
$$u_0\in \mbox{int}\, C_{+}\ \mbox{and}\ v_0\in -\mbox{int}\, C_{+}.$$
From the proof of Proposition \ref{prop9}, we know that:

(i) $u_0$ is a critical point of mountain pass type for the functional $\hat{\varphi}_{+}$;

(ii) $v_0$ is a critical point of mountain pass type for the functional $\hat{\varphi}_{-}$.

Hence we have
\begin{eqnarray}\label{eq62}
C_1(\hat{\varphi}_{+},u_0)\neq 0\ \mbox{and}\ C_1(\hat{\varphi}_{-},v_0)\neq 0\ (\mbox{see\ Chang\ \cite{8}}).
\end{eqnarray}
Note that
\begin{eqnarray}\label{eq63}
\hat{\varphi}_{+}|_{C_{+}}=\varphi|_{C_{+}}\ \mbox{and}\ \hat{\varphi}_{-}|_{-C{+}}=\varphi|_{-{C_{+}}}.
\end{eqnarray}
From Chang \cite[p. 14]{8} (see also Bartsch \cite{3}), we have
\begin{eqnarray}\label{eq64}
C_k(\hat{\varphi}_{+},u_0)=C_k(\hat{\varphi}_{+}|_{C^1(\overline{\Omega})},u_0)\ \mbox{and}\ C_k(\varphi,u_0)=C_k(\varphi|_{C^1(\overline{\Omega})},u_0)\ \mbox{for\ all}\ k\geq 0.
\end{eqnarray}
\begin{eqnarray}\label{eq65}
C_k(\hat{\varphi}_{-},v_0)=C_k(\hat{\varphi}_{-}|_{C^1(\overline{\Omega})},v_0)\ \mbox{and}\ C_k(\varphi,v_0)=C_k(\varphi|_{C^1(\overline{\Omega})},v_0)\ \mbox{for\ all}\ k\geq 0.
\end{eqnarray}
From relations (\ref{eq62})--(\ref{eq65}) and since $u_0\in \mbox{int}\, C_{+},\ v_0\in -\mbox{int}\, C_{+}$, it follows that
\begin{eqnarray}\label{eq66}
C_1(\varphi,u_0)\neq 0\ \mbox{and}\ C_1(\varphi,v_0)\neq 0.
\end{eqnarray}
If $\overline{u}_0=p_Y(u_0)$, $\overline{v}_0=p_Y(v_0)$ ($p_Y$ being the orthogonal projection of $H^1(\Omega)$ onto $Y$), then from Liu \& Li \cite{20}, we have
\begin{eqnarray}\label{eq67}
&&C_k(\varphi,u_0)=C_k(\psi,\overline{u}_0)\ \mbox{and}\ C_k(\varphi,v_0)=C_k(\psi,\overline{v}_0)\ \mbox{for\ all}\ k\geq 0,\nonumber\\
&\Rightarrow& C_1(\psi,\overline{u}_0)\neq 0\ \mbox{and}\ C_1(\psi,\overline{v}_0)\neq 0\ (\mbox{see}\ (\ref{eq66})).
\end{eqnarray}
From Proposition \ref{prop7}, we have
\begin{eqnarray}\label{eq68}
&&C_k(\varphi,0)=\delta_{k,0}\ZZ\ \mbox{for\ all}\ k\geq 0,\nonumber\\
&\Rightarrow& C_k(\psi,0)=\delta_{k,0}\ZZ\ \mbox{for\ all}\ k\geq 0\ (\mbox{see\ \cite{20}}).
\end{eqnarray}
From Proposition \ref{prop12}, we have that $\psi$ is anticoercive. Also, it is continuous (recall that the space $Y$ is finite dimensional). Hence by the Weierstrass theorem we can find $\overline{y}_0\in Y$ such that  $\psi(\overline{y}_0)=\sup\limits_{Y}\psi$. Then we have
\begin{eqnarray}\label{eq69}
C_k(\psi,\overline{y}_0)=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0,\ \mbox{with}\ d_m=\dim Y\geq 2.
\end{eqnarray}
Comparing (\ref{eq69}) with (\ref{eq67}) and (\ref{eq68}), we infer that
\begin{eqnarray*}
&&\overline{y}_0\notin \{0,\overline{u}_0,v_0\},\\
&\Rightarrow& y_0=\overline{y}_0+\gamma_0(\overline{y}_0)\ \mbox{is\ the\ third\ distinct\ nontrivial\ solution}.
\end{eqnarray*}
As before, the regularity results of Wang \cite{34} imply that $y_0\in C^1(\overline{\Omega})$.
\end{proof}

\section{Four Nontrivial Solutions}\label{sec4}

By strengthening the regularity on $f(z,\cdot)$, we can improve Theorem \ref{th13} and produce four nontrivial solutions.

We start by computing the critical groups of the functional $\psi$ at infinity. To do this, we do not need any additional conditions on $f(z,\cdot)$. Our proof uses some ideas from Liu \cite{18}.
\begin{prop}\label{prop14}
If hypotheses $H_0$ and $H_1$ hold, then $C_k(\psi,\infty)=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0,\ \mbox{with}\ d_m=\dim Y\geq 2.$
\end{prop}
\begin{proof}
As always, we assume that $K_{\psi}$ is finite (otherwise we already have an infinity of nontrivial solutions for problem (\ref{eq1})).

Let $\hat{m}<\inf\psi(K_{\psi})$. In what follows, for every $r>0$, we introduce the set
$$C_r=\{y\in Y:\|y\|\geq r\}.$$

Recall that $\psi$ is anticoercive (see Proposition \ref{prop12}). So, we can find $\lambda<\mu<\hat{m}$ and $0<\varrho<R$ such that
$$C_R\subseteq \psi^{\lambda}\subseteq C_{\varrho}\subseteq\psi^{\mu}.$$
Then for the triples $(C_R, C_{\varrho},Y)$ and $(\psi^{\lambda},\psi^{\mu},Y)$, we consider the corresponding long exact sequences of singular homology groups


\begin{equation}\label{eq70}
\begin{tabular}{l}
$\cdots$ $\longrightarrow H_{k}(C_{\rho},C_{R})$  $\overset{i_{\ast}}{\longrightarrow} H_{k} (Y,C_{R})$ $\overset{j_{\ast}}{\longrightarrow} H_{k} (Y,C_{\rho}) $ $\overset{\partial_{\ast}}{\longrightarrow} H_{k-1} $$(C_\rho,C_{R}) $ $\longrightarrow$ $\cdots$\\
\hspace{23mm}	$\left\downarrow\rule{0cm}{0.5cm}\right.\phantom{} h_{\ast|_{C_{\rho}}}$ \hspace{12mm}  $\left\downarrow\rule{0cm}{0.5cm}\right.\phantom{} h_{\ast}$  \hspace{15mm} $\left\downarrow\rule{0cm}{0.5cm}\right.\phantom{} h_{\ast}$ \hspace{15mm} $\left\downarrow\rule{0cm}{0.5cm}\right.\phantom{} h_{\ast|_{C_{\rho}}}$\\
$\cdots$ $\longrightarrow H_{k}(\psi^{\mu},\psi^{\lambda})$ $\overset{\hat{i}_{\ast}}{\longrightarrow} H_{k}$ $(Y,\psi^{\lambda})$  $\overset{\hat j_{\ast}}\longrightarrow H_{k} (Y,\psi^{\mu}) $  $\overset{\hat{\partial}_{\ast}} {\longrightarrow} H_{k-1}$  $(\psi^{\lambda},\psi^{\mu}) $$\longrightarrow$ $\cdots$
\end{tabular}
\end{equation}

In (\ref{eq70}) all squares are commutative (see Granas \& Dugundji \cite[p. 377]{14}), $i_{\ast},j_{\ast},\hat{i}_{\ast},\hat{j}_{\ast},h_{\ast}$ are the group homomorphisms induced by the corresponding inclusion maps. Also, $\partial_{\ast}$ and $\hat{\partial_{\ast}}$ are the corresponding boundary homomorphisms. Recall that $\lambda<\mu<\hat{m}<\inf \psi(K_{\psi})$. So, by the second deformation theorem (see, for example, Gasinski \& Papageorgiou \cite[p. 628]{11}), we have that $\psi^{\lambda}$ is a strong deformation retract of $\psi^{\mu}$. Hence
\begin{eqnarray}\label{eq71}
H_k(\psi^{\mu},\psi^{\lambda})=0\ \mbox{for\ all}\ k\geq 0.
\end{eqnarray}

Let $\chi:C_{\varrho}\rightarrow C_R$ be the map defined by
\begin{eqnarray*}
\chi(u)=\left\{  \begin{array}{ll}
   R \frac{u}{\|u\|} &\mbox{if}\ \varrho\leq\|u\|\leq R \\
     u &\mbox{if}\ R<\|u\|.
    \end{array}\right.
\end{eqnarray*}
Evidently $\chi$ is continuous and $\chi|_{C_R}=\mbox{Id}|_{C_R}$. Therefore $C_R$ is a retract of $C_\varrho$. Consider the deformation $h:[0,1]\times C_{\varrho}\rightarrow Y$ defined by
$$h(t,u)=(1-t)u+tR\frac{u}{\|u\|}\ \mbox{for\ all}\ t\in [0,1],\ \mbox{all}\ u\in C_{\varrho}.$$
Then $h(0,u)=u$ and $h(1,u)=R\frac{u}{\|u\|}\in C_R$. So, we see that the set $C_{\varrho}$ is deformable onto $C_R$ over $Y$. Invoking Theorem $6.5$ of Dugundji \cite[p. 325]{9}, we infer that $C_R$ is a deformation retract of $C_{\varrho}$. Therefore from Granas \& Dugundji \cite[p. 387]{14}, we have
\begin{eqnarray}\label{eq72}
H_k(C_{\varrho},C_R)=0\ \mbox{for\ all}\ k\geq 0.
\end{eqnarray}
The exactness of the two long homology sequences in (\ref{eq70}) and the rank theorem imply that
$$0=\mbox{im}\, i_{\ast}=\ker j_{\ast}\ \mbox{and}\  \mbox{im}\, j_{\ast}=\ker \partial_{\ast}=H_k(Y,C_{\varrho})\ (\mbox{see}\ (\ref{eq72})),$$
$$0=\mbox{im}\, \hat{i}_{\ast}=\ker \hat{j}_{\ast}\ \mbox{and}\  \mbox{im}\, \hat{j}_{\ast}=\ker \hat{\partial}_{\ast}=H_k(Y,\psi^{\mu})\ (\mbox{see}\ (\ref{eq71})).$$
So, we see that $j_{\ast}$ and $\hat{j}_{\ast}$ are group isomorphisms. Then from Granas \& Dugundji \cite[p. 610]{14}, it follows that $h_{\ast}$ is an isomorphism. Hence
\begin{eqnarray}\label{eq73}
&&H_k(Y,C_{\varrho})=H_k(Y,\psi^{\mu})\ \mbox{for\ all}\ k\geq 0,\nonumber\\
&\Rightarrow& H_k(Y,C_{\varrho})=C_k(\psi,\infty)\ \mbox{for\ all}\ k\geq 0\ (\mbox{recall}\ \mu<\hat{m}).
\end{eqnarray}

Using the radial retraction and Theorem $6.5$ of Dugundji \cite[p. 325]{9}, we see that $\partial B_{\varrho}^{Y}=\{y\in Y:\|y\|=\varrho\}$ is a deformation retract of $C_{\varrho}$. Therefore
\begin{eqnarray*}
&&H_k(Y,C_{\varrho})=H_k(Y,\partial B_{\varrho})\ \mbox{for\ all}\ k\geq 0,\\
&\Rightarrow& H_k(Y,C_{\varrho})=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0,\ (\mbox{see\ Maunder \cite[p. 121]{21}},\\
&\Rightarrow& C_k(\psi,\infty)=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0,\ (\mbox{see}\ (\ref{eq73})).
\end{eqnarray*}
\end{proof}
We introduce the new stronger conditions on the reaction $f(z,x)$.

\medskip
$H_2:\ f:\Omega\times\RR\rightarrow\RR$ is a measurable function such that  for a.a. $z\in\Omega$, $f(z,0)=0$, $f(z,\cdot)\in C^1(\RR)$ and
\begin{itemize}
\item [(i)] there exist an integer $m\geq \max\{m_0,2\}$ and a function $\eta\in L^{\infty}(\Omega)_{+}$ such that
\begin{eqnarray*}
&&\eta(z)\leq \hat{\lambda}_{m+1}\ \mbox{a.e.\ in}\ \Omega,\ \eta\neq \hat{\lambda}_{m+1},\\
&&|f_x'(z,x)|\leq \eta(z)\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR;
\end{eqnarray*}
\item [(ii)] $\hat{\lambda}_{m}\leq \liminf\limits_{x\rightarrow\pm\infty}\frac{f(z,x)}{x}\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega$;
\item [(iii)] $\lim\limits_{x\rightarrow\pm\infty}[f(z,x)x-2F(z,x)]=-\infty\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega$;
\item [(iv)] there exists a function $\vartheta\in L^{\infty}(\Omega)_{+}$ such that
\begin{eqnarray*}
&&\vartheta(z)\leq \hat{\lambda}_1\ \mbox{a.e.\ in}\ \Omega,\ \vartheta\neq \hat{\lambda}_1,\\
&&\limsup\limits_{x\rightarrow 0}\frac{2F(z,x)}{x^2}\leq \vartheta(z)\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega.
\end{eqnarray*}
\end{itemize}

\begin{remark}
Note that hypothesis $H_2(i)$ and the mean value theorem imply that
$$|f(z,x)|\leq\eta(z)|x|\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.$$
Also, for every $\varrho>0$, there exists $\xi_{\varrho}>0$ such that  for a.a. $z\in\Omega$, $x\mapsto f(z,x)+\xi_{\varrho}x$ is nondecreasing on $[-\varrho,\varrho]$.
\end{remark}

\begin{ex}
The following function satisfies hypotheses $H_2$. As before, for the sake of simplicity we drop the $z$-dependence.
\begin{eqnarray*}
f(x)=\left\{  \begin{array}{ll}
 \displaystyle  \vartheta x+\frac{\xi x}{1+|x|} &\mbox{if}\ |x|\leq 1 \\
   \displaystyle  \lambda x+\frac{c}{x} &\mbox{if}\ 1<|x|,
    \end{array}\right.
\end{eqnarray*}
with $\vartheta<\hat{\lambda}_1,\ \lambda\in (\hat{\lambda}_m,\hat{\lambda}_{m+1})$ for some integer $m\geq \max\{m_0,2\},\ \xi,\ c>0,\ \vartheta+\xi<\hat{\lambda}_{m+1}$.
\end{ex}
\begin{theorem}\label{th15}
If hypotheses $H_0$ and $H_2$ hold, then problem (\ref{eq1}) admits at least four nontrivial solutions
$$u_0\in \mbox{int}\, C_{+},\ v_0\in -\mbox{int}\, C_{+}\ \mbox{and}\ y_0,\hat{y}\in C^1(\overline{\Omega}).$$
\end{theorem}
\begin{proof}
From Theorem \ref{th13} we already have three nontrivial solutions
$$u_0\in \mbox{int}\, C_{+},\ v_0\in -\mbox{int}\, C_{+}\ \mbox{and}\ y_0\in C^1(\overline{\Omega}).$$
In this case $\varphi\in C^2(H^1(\Omega))$. From the proof of Proposition \ref{prop9} and since $\hat{\varphi}_{+}|_{C_{+}}=\varphi |_{C_{+}}$, $\hat{\varphi}_{-}|_{-C_{+}}=\varphi |_{-C_{+}}$, we see that $u_0\in \mbox{int}\, C_{+}\ \mbox{and}\ v_0\in -\mbox{int}\, C_{+}$ are critical points of mountain pass type for $\varphi$ and so
\begin{eqnarray}\label{eq74}
C_k(\varphi,u_0)=C_k(\varphi,v_0)=\delta_{k,1}\ZZ\ \mbox{for\ all}\ k\geq 0\ (\mbox{see\ Bartsch\ \cite{3}}).
\end{eqnarray}
From (\ref{eq69}) we have that $C_k(\psi,\overline{y}_0)=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0\ (\mbox{recall\ that}\ \overline{y}_0=p_{Y}(y_0))$. From Liu \& Li \cite{20} we know that $C_k(\varphi,y_0)=C_k(\psi,\overline{y}_0)$. Therefore
\begin{eqnarray}\label{eq75}
C_k(\varphi,y_0)=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0.
\end{eqnarray}
Also, we have
\begin{eqnarray}\label{eq76}
C_k(\varphi,0)=\delta_{k,0}\ZZ\ \mbox{for\ all}\ k\geq 0\ (\mbox{see\ Proposition}\ \ref{prop7}),
\end{eqnarray}
\begin{eqnarray}\label{eq77}
C_k(\varphi,\infty)=\delta_{k,d_m}\ZZ\ \mbox{for\ all}\ k\geq 0\ (\mbox{see\ Proposition}\ \ref{prop14}).
\end{eqnarray}
Suppose that $K_{\varphi}=\{0,u_0,v_0,y_0\}$. Then from (\ref{eq74}), (\ref{eq75}), (\ref{eq76}), (\ref{eq77}) and the Morse relation (see (\ref{eq7})) with $t=-1$, we have
\begin{eqnarray*}
&&(-1)^0+2(-1)^1+(-1)^{d_m}=(-1)^{d_m},\\
&\Rightarrow& (-1)^1=0,\ \mbox{a\ contradiction}.
\end{eqnarray*}
So, we can find $\hat{y}\in K_{\varphi}\setminus \{0,u_0,v_0,y_0\}$. Evidently $\hat{y}$ is the fourth nontrivial solution of (\ref{eq1}) and as before, using the regularity results of Wang \cite{34}, we have that $\hat{y}\in C^1(\overline{\Omega})$.
\end{proof}

\section{Coercive Problems}\label{sec5}

In this section, we study what happens when the limit in hypothesis $H_1(iii)$ is $+\infty$. In this case, the geometry of the problem changes since, as we will see, the reduced functional $\psi$ is coercive. We can still prove a multiplicity theorem producing at least two nontrivial solutions.

The new hypotheses on the reaction $f(z,x)$ are the following:

\medskip
$H_3:\ f:\Omega\times\RR\rightarrow\RR$ is a Carath\'eodory function such that  $f(z,0)=0$ for a.a. $z\in\Omega$ and
\begin{itemize}
\item [(i)] there exist an integer $m\geq \max\{m_0,2\}$ and a function $\eta\in L^{\infty}(\Omega)_{+}$ such that
\begin{eqnarray*}
&&\eta(z)\leq \hat{\lambda}_{m+1}\ \mbox{a.e.\ in}\ \Omega,\ \eta\neq \hat{\lambda}_{m+1},\\
&&(f(z,x)-f(z,y))(x-y)\leq \eta(z)(x-y)^2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x,y\in\RR;
\end{eqnarray*}
\item [(ii)] $\hat{\lambda}_{m}\leq \liminf\limits_{x\rightarrow\pm\infty}\frac{f(z,x)}{x}\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega$;
\item [(iii)] $\lim\limits_{x\rightarrow\pm\infty}[f(z,x)x-2F(z,x)]=+\infty\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega$;
\item [(iv)] there exist an integer $1<k<m-1$, a function $\eta_0\in L^{\infty}(\Omega)$ and $\delta_0>0$ such that
\begin{eqnarray*}
&&\hat{\lambda}_k\leq \eta_0(z)\ \mbox{a.e.\ in}\ \Omega,\ \eta_0\neq \hat{\lambda}_k,\\
&&\eta_0(z)\leq \liminf\limits_{x\rightarrow 0}\frac{2F(z,x)}{x^2}\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega,\\
&&F(z,x)\leq\frac{\hat{\lambda}_{k+1}}{2}x^2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\leq\delta_0.
\end{eqnarray*}
\end{itemize}
\begin{remark}
Note that compared to hypothesis $H_1(iv)$, now hypothesis $H_3(iv)$ implies a different geometry for which $u=0$ is no longer a local minimizer of the energy functional.
\end{remark}
\begin{ex}
The following function satisfies hypotheses $H_3$. As always, we drop the $z$-dependence
\begin{eqnarray*}
f(x)=\left\{  \begin{array}{ll}
 \displaystyle  \beta x+\xi |x|^{p-2}x &\mbox{if}\ |x|\leq 1 \\
\displaystyle     \lambda x-\frac{c}{x} &\mbox{if}\ 1<|x|,
    \end{array}\right.
\end{eqnarray*}
with $ \beta\in (\hat{\lambda}_k,\hat{\lambda}_{k+1}),\ \xi,\ c>0,\ \lambda\in (\hat{\lambda}_m,\hat{\lambda}_{m+1}),\ m\geq \max\{m_0,2\},\ m>k+1,\ \beta+\xi<\hat{\lambda}_{m+1}$ and $\beta+\xi=\lambda-c$.
\end{ex}
Again we consider the following orthogonal direct sum decomposition of the space $H^1(\Omega)$:
$$H^1(\Omega)=Y\oplus\hat{H}\ \mbox{with}\ Y=\oplus_{i=1}^m E(\hat{\lambda}_i),\ \hat{H}=\overline{\oplus_{i\geq m+1} E(\hat{\lambda}_i)}.$$
As before (see Proposition \ref{prop10} and its proof), using the reduction method, we produce a continuous map $\gamma: Y\rightarrow\hat{H}$ such that
\begin{eqnarray*}
&&\psi(y)=\varphi(y+\gamma(y))=\inf[\varphi(y+\hat{u}):\hat{u}\in\hat{H}]\\
&\mbox{and}&\ \psi\in C^1(Y)\ (\mbox{see\ Proposition}\ \ref{prop11}).
\end{eqnarray*}
Reasoning as in the proof of Proposition \ref{prop8}, we obtain:
\begin{prop}\label{prop16}
If hypotheses $H_0$ and $H_3$ hold, then the reduced functional $\psi$ is coercive.
\end{prop}

Using this proposition, we obtain the following multiplicity theorem.

\begin{theorem}\label{th17}
If hypotheses $H_0$ and $H_3$ hold, then problem (\ref{eq1}) admits at least two nontrivial solutions
$$u_0,\ \hat{u}\in C^1(\overline{\Omega}).$$
\end{theorem}
\begin{proof}
From Proposition \ref{prop16}, we know that $\psi$ is coercive, hence it is bounded from below.

We consider the following orthogonal direct sum decomposition of $Y$:
$$Y=\overline{H}_k\oplus V,$$
where $\overline{H}_k=\oplus_{i=1}^k E(\hat{\lambda}_i)$ and $V=\oplus_{i=k+1}^m E(\hat{\lambda}_i)$ (recall $k<m-1$).

By virtue of hypotheses $H_3(i),(iv)$, given $\varepsilon>0$ and $r>2$, we can find $c_3=c_3(\varepsilon,r)>0$ such that
\begin{eqnarray}\label{eq78}
F(z,x)\geq\frac{1}{2}(\eta_0(z)-\varepsilon)x^2-c_3|x|^r\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.
\end{eqnarray}
Then for all $u\in \overline{H}_k$ we have
\begin{eqnarray}\label{eq79}
\psi(u)&=&\varphi(u+\gamma(u))\nonumber\\
&\leq& \varphi(u)\nonumber\\
&=& \frac{1}{2}\tau(u)-\int_{\Omega}F(x,u)dz\nonumber\\
&\leq& \frac{1}{2}\tau(u)-\frac{1}{2}\int_{\Omega}\eta_0u^2dz+\frac{\varepsilon}{2}\|u\|^2+c_4\|u\|^r\ \mbox{for\ some}\ c_4>0\ (\mbox{see}\ (\ref{eq78}))\nonumber\\
&\leq& \frac{-\hat{\xi}_2+\varepsilon}{2}\|u\|^2+c_4\|u\|^r\ (\mbox{see\ Proposition}\ \ref{prop3}(b)).
\end{eqnarray}
Choosing $\varepsilon\in (0,\hat{\xi}_2)$ and since $r>2$, from (\ref{eq79}) we see that we can find $\varrho_2\in (0,1)$ small such that
\begin{eqnarray}\label{eq80}
\psi(u)\leq 0\ \mbox{for\ all}\ u\in\overline{H}_k,\ \mbox{with}\ \|u\|\leq \varrho_2.
\end{eqnarray}
Moreover, from (\ref{eq79}) it is clear that
\begin{eqnarray}\label{eq81}
\inf_{Y}\psi<0.
\end{eqnarray}
Since $V$ is finite dimensional, all norms are equivalent and so, we can find $\delta_1>0$ such that
\begin{eqnarray}\label{eq82}
u\in V\ \mbox{with}\ \|u\|\leq \delta_1\Rightarrow |u(z)|\leq \delta_0\ \mbox{for\ all}\ z\in\overline{\Omega}.
\end{eqnarray}
Here $\delta_0>0$ is as postulated by hypothesis $H_3(iv)$. Exploiting the continuity of the map $\gamma(\cdot)$ and since $\gamma(0)=0$, we can find $\varrho_3=\varrho_3(\delta_1)>0$ such that
\begin{eqnarray}\label{eq83}
v\in V\ \mbox{with}\ \|v\|\leq \varrho_3\Rightarrow \|v+\gamma(v)\|\leq \delta_1.
\end{eqnarray}
Then from (\ref{eq82}) and (\ref{eq83}), we see that if $v\in V$ with $\|v\|\leq \varrho_3$, then
\begin{eqnarray}\label{eq84}
\psi(v)&=&\varphi(v+\gamma(v))\nonumber\\
&=& \frac{1}{2}\tau(v+\gamma(v))-\int_{\Omega}F(z,v+\gamma(v))dz\nonumber\\
&\geq& \frac{1}{2}\tau(v+\gamma(v))-\frac{\hat{\lambda}_{k+1}}{2}\|v+\gamma(v)\|_2^2\ (\mbox{see}\ (\ref{eq82}),\ (\ref{eq83})\ \mbox{and}\ H_3(iv)) \nonumber\\
&\geq& 0\ (\mbox{since}\ v+\gamma(v)\in \overline{\oplus_{i\geq k+1} E(\hat{\lambda}_i)},\ \mbox{see}\ (\ref{eq5})).
\end{eqnarray}
From (\ref{eq80}), (\ref{eq81}) and (\ref{eq84}), we see that we can apply Proposition \ref{prop4} and produce two nontrivial critical points $v_0,\hat{v}\in V$ of $\psi$. Then $u_0=v_0+\gamma(v_0)$ and $\hat{u}=\hat{v}+\gamma(\hat{v})$ are critical points of $\varphi$ (see \cite{20}), hence solutions of (\ref{eq1}). Moreover, as before the regularity theory (see Wang \cite{34}), implies that $u_0,\hat{u}\in C^1(\overline{\Omega})$.
\end{proof}

\section{Anticoercive Problems}\label{sec6}

Keeping the same geometry near origin, we see what happens when the asymptotic condition $\pm\infty$ make the reduced functional $\psi$ anticoercive. Again we can prove a multiplicity theorem producing two nontrivial solutions. As in Section \ref{sec5} our main tools is Proposition \ref{prop4} (the local linking multiplicity result).

In this section we assume that $0$ is in the spectrum of $-\Delta+\beta$ (denoted by $\sigma(-\Delta+\beta)$) and this spectrum has also negative elements. In what follows by $\hat{\lambda}_m>0$ (resp. $\hat{\lambda}_k<0$ ) we denote the first strictly positive (resp. strictly negative) eigenvalue of $-\Delta+\beta$.

The conditions on the reaction term $f(z,x)$ are the following:

\medskip
$H_4:\ f:\Omega\times\RR\rightarrow\RR$ is a Carath\'eodory function such that  $f(z,0)=0$ for a.a. $z\in\Omega$ and
\begin{itemize}
\item [(i)] there exist a set $B\subseteq\Omega$ with $|B|_N>0$ and $\xi\in L^1(\Omega)$ such that
\begin{eqnarray*}
&&F(z,x)\rightarrow +\infty\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{as}\ x\rightarrow\pm\infty,\\
&&F(z,x)\geq \xi(z)\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR;
\end{eqnarray*}
\item [(ii)] there exists a function $\eta\in L^{\infty}(\Omega)_{+}$ such that
\begin{eqnarray*}
&&\eta(z)\leq \hat{\lambda}_{m}\ \mbox{for\ a.a.}\ z\in\Omega,\ \eta\neq \hat{\lambda}_{m},\\
&&(f(z,x)-f(z,y))(x-y)\leq \eta(z)(x-y)^2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x,y\in\RR;
\end{eqnarray*}
\item [(iii)] there exist a function $\eta_0\in L^{\infty}(\Omega)$ and  $\delta_0>0$ such that
\begin{eqnarray*}
&&\eta_0(z)\leq 0\ \mbox{a.e.\ in}\ \Omega,\ \eta_0\neq 0,\\
&& \limsup\limits_{x\rightarrow 0}\frac{2F(z,x)}{x^2}\leq \eta_0(z)\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega,\\
&&\frac{\hat{\lambda}_{k}}{2}x^2\leq F(z,x)\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\leq\delta_0.
\end{eqnarray*}
\end{itemize}
\begin{remark}
We see that the condition near zero (see $H_4(iv)$), is similar to that in $H_3(iv)$. In particular, both imply that the origin can not be a local minimizer of the energy functional. So, the geometry of the problem differs from that assumed in Sections \ref{sec3} and \ref{sec4}.
\end{remark}
\begin{ex}
The following function satisfies hypotheses $H_4$. Again, we drop the $z$-dependence.
\begin{eqnarray*}
f(x)=\left\{  \begin{array}{ll}
   \eta_0 x &\mbox{if}\ |x|\leq 1 \\
     \lambda x-\beta &\mbox{if}\ 1<|x|,
    \end{array}\right.
\end{eqnarray*}
with $ \eta_0\in (\hat{\lambda}_k,0),\ \lambda\in (0,\hat{\lambda}_{m})$ and $\beta=\lambda-\eta_0$.
\end{ex}

We introduce the following subspaces of $H^1(\Omega)$:
$$\overline{H}=\oplus_{i=1}^k E(\hat{\lambda}_i),\ E= E(0),\ Y=\overline{H}\oplus E\ \mbox{and}\ \hat{H}=Y^{\perp}.$$
Reasoning as in the proof of Proposition \ref{prop10}, we obtain:
\begin{prop}\label{prop18}
If hypotheses $H_0$ and $H_4$ hold, then there exists a continuous map $\hat{\gamma}: Y\rightarrow\hat{H}$ such that
$$\varphi(y+\hat{\gamma}(y))=\inf[\varphi(y+\hat{u}): \hat{u}\in\hat{H}].$$
\end{prop}
We set $\psi(y)=\varphi(y+\hat{\gamma}(y))$ for all $y\in Y$ and as in Proposition \ref{prop11}, we show that:
\begin{prop}\label{prop19}
If hypotheses $H_0$ and $H_4$ hold, then $\psi\in C^1(Y)$.
\end{prop}
Using the chain rule, we have
$$\psi'(y)=p_{Y^{\ast}}\varphi'(y+\hat{\gamma}(y))\ \mbox{for\ all}\ y\in Y,$$
with $p_{Y^{\ast}}$ being the orthogonal projection onto $Y^{\ast}$.

Also, we introduce the functional
$$\vartheta(y)=-\psi(y)\ \mbox{for\ all}\ y\in Y.$$
\begin{prop}\label{prop20}
If hypotheses $H_0$ and $H_4$ hold, then there exists $\varrho>0$ such that
\begin{eqnarray*}
&&\vartheta(y)\leq 0\ \mbox{for\ all}\ y\in\overline{H}\ \mbox{with}\ \|y\|\leq \varrho,\\
&&\vartheta(y)\geq 0\ \mbox{for\ all}\ y\in E\ \mbox{with}\ \|y\|\leq \varrho.
\end{eqnarray*}
\end{prop}
\begin{proof}
By virtue of hypotheses $H_4(ii),(iii)$, given $\varepsilon>0$ and $r>2$, we can find $c_4=c_4(\varepsilon,r)>0$ such that
\begin{eqnarray}\label{eq85}
F(z,x)\leq \frac{\eta_0(z)+\varepsilon}{2}x^2+c_4|x|^r\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.
\end{eqnarray}
Then for all $y\in E$, we have
\begin{eqnarray}\label{eq86}
\vartheta(y)=-\psi(y)&=&-\varphi(y+\hat{\gamma}(y))\nonumber\\
&=&-\frac{1}{2}\tau(y+\hat{\gamma}(y))+\int_{\Omega}F(z,y+\hat{\gamma}(y))dz\nonumber\\
&\leq& -\frac{1}{2}\tau(y+\hat{\gamma}(y))+\frac{1}{2}\int_{\Omega}\eta_0(y+\hat{\gamma}(y))^2dz+\frac{\varepsilon}{2}\|y+\hat{\gamma}(y)\|^2+
c_5\|y+\hat{\gamma}(y)\|^r\nonumber\\
&&(\mbox{for\ some}\ c_5>0\ (\mbox{see}\ (\ref{eq85}))) \nonumber\\
&\leq& -\frac{\xi_1-\varepsilon}{2}\|y+\hat{\gamma}(y)\|^2+c_5\|y+\hat{\gamma}(y)\|^r\ (\mbox{see\ Proposition}\ \ref{prop3}(a)).
\end{eqnarray}
From hypothesis $H_4(ii)$, we have
\begin{eqnarray}\label{eq87}
F(z,x)\leq \frac{\eta(z)}{2}x^2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.
\end{eqnarray}
For every $\hat{u}\in\hat{H}$, we have
\begin{equation}\label{eq88}\begin{array}{ll}
\varphi(\hat{u})&=\displaystyle\frac{1}{2}\tau(\hat{u})-\int_{\Omega}F(z,\hat{u})dz\\
&\geq\displaystyle\frac{1}{2}\tau(\hat{u})-\frac{1}{2}\int_{\Omega}\eta\hat{u}^2dz\ (\mbox{see}\ (\ref{eq87}))\\
&\geq\displaystyle\frac{\xi_1}{2}\|\hat{u}\|^2\ (\mbox{see\ Proposition}\ \ref{prop3}(a)),\\
\Rightarrow& \inf\limits_{\hat{H}}\varphi=0,\\
\Rightarrow& \hat{\gamma}(0)=0.
\end{array}\end{equation}

Recalling that $\hat{\gamma}$ is continuous (see Proposition \ref{prop18}), from (\ref{eq86}) and (\ref{eq88}) it follows that we can find $\varrho_1\in (0,1)$ small such that
\begin{eqnarray}\label{eq89}
\vartheta(y)\leq 0\ \mbox{for\ all}\ y\in E\ \mbox{with}\ \|y\|\leq \varrho_1.
\end{eqnarray}
On the other hand, since $\overline{H}$ is finite dimensional, all norms are equivalent and so we can find $c_6>0$ such that
\begin{eqnarray}\label{eq90}
\|y\|_{\infty}\leq c_6\|y\|\ \mbox{for\ all}\ y\in\overline{H}.
\end{eqnarray}
Let $\delta_0>0$ be as postulated by hypothesis $H_4(iii)$. Then for $\varrho_2=\frac{\delta_0}{c_6}>0$ we have
$$y\in\overline{H},\ \|y\|\leq \varrho_2\Rightarrow |y(z)|\leq\delta_0\ \mbox{for\ all}\ z\in\overline{\Omega}\ (\mbox{see}\ (\ref{eq90})).$$
Then by virtue of hypotheses $H_4(iii)$ we have
\begin{eqnarray}\label{eq91}
F(z,y(z))\geq \frac{\hat{\lambda}_k}{2}y(z)^2\ \mbox{for\ a.a.}\ z\in\Omega.
\end{eqnarray}
\begin{eqnarray}\label{eq92}
\Rightarrow \vartheta(y)=-\psi(y)&\geq&-\varphi(y)\nonumber\\
&\geq&-\frac{1}{2}\tau(y)+\int_{\Omega}F(z,y)dz\nonumber\\
&\geq& -\frac{1}{2}\tau(y)+\frac{\hat{\lambda}_k}{2}\|y\|_2^2\ (\mbox{see}\ (\ref{eq91}))\nonumber\\
&\geq& 0,\ \mbox{for\ all}\ y\in \hat{H}\ \mbox{with}\ \|y\|\leq \varrho_2\ (\mbox{see}\ (\ref{eq5})).
\end{eqnarray}
Choosing $\varrho=\min\{\varrho_1,\varrho_2\}$ from (\ref{eq89}) and (\ref{eq92}), we infer the result of the proposition (that is, we have local linking for $\vartheta$).
\end{proof}

In order to use Proposition \ref{prop4}, we need to check that the functional $\vartheta$ is bounded below and satisfies the $C$-condition. To this end, we will need the following simple lemma.
\begin{lemma}\label{lem21}
For every $\varepsilon>0$, there exists $\xi_{\varepsilon}>0$ such that
$$\Big|\{z\in\Omega:|y(z)|<\xi_{\varepsilon}\|y\|\}\Big|_{N}<\varepsilon\ \mbox{for\ all}\ y\in E.$$
\end{lemma}
\begin{proof}
We argue by contradiction. So, suppose we can find $\varepsilon>0$ and $\{y_n\}_{n\geq 1}\subseteq E$ such that  $|C_n|_{N}\geq \varepsilon$ for all $n\geq 1$, where $C_n=\{z\in\Omega:|y_n(z)|<\frac{1}{n}\|y_n\|\}$. Let $v_n=\frac{y_n}{\|y_n\|},\ n\geq 1.$ Then $\|v_n\|=1$, $v_n\in E=E(0)$ for all $n\geq 1$. Since $E$ is finite dimensional, by passing to a suitable subsequence if necessary, we may assume that
\begin{eqnarray}\label{eq93}
v_n\rightarrow v\ \mbox{in}\ H^1(\Omega),\ \mbox{hence}\ \|v\|=1,\ v\in E.
\end{eqnarray}
We have
\begin{eqnarray}\label{eq94}
&&\limsup_{n\rightarrow\infty} C_n\subseteq C_0=\{z\in\Omega: y(z)=0\},\nonumber\\
&\Rightarrow& \varepsilon\leq \limsup_{n\rightarrow\infty} |C_n|_N\leq |\limsup_{n\rightarrow\infty} C_n|_N\leq |C_0|_N.
\end{eqnarray}
From (\ref{eq93}) we know that $y\neq 0$, $y\in E=E(0)$. So, by the UCP, we have $y(z)\neq 0$ a.e. in $\Omega$, which contradicts (\ref{eq94}).
\end{proof}

Using this lemma, we can establish the desired properties of $\vartheta$ in order eventually to apply Proposition \ref{prop4}.
\begin{prop}\label{prop22}
If hypotheses $H_0$ and $H_4$ hold, then $\vartheta$ is coercive.
\end{prop}
\begin{proof}
From the definition of $\vartheta$, we have
$$\vartheta=-\psi\geq -\varphi|_{Y}.$$
So, to prove the proposition, it suffices to show that $-\varphi|_{Y}$ is coercive. We argue indirectly. So, suppose that  $-\varphi|_{Y}$ is not coercive. Then we can find $M_6>0$ and $\{y_n\}_{n\geq 1}\subseteq Y$ such that
\begin{eqnarray}\label{eq95}
-M_6\leq \varphi(y_n)\ \mbox{for\ all}\ n\geq 1\ \mbox{and}\ \|y_n\|\rightarrow\infty.
\end{eqnarray}
From hypothesis $H_4(i)$ and Tang \& Wu \cite{31}, we know that given $\varepsilon>0$, we can find $C_{\varepsilon}\subseteq B$ measurable such that
\begin{eqnarray}\label{eq96}
|B\setminus C_{\varepsilon}|_N<\varepsilon\ \mbox{and}\ F(z,x)\rightarrow +\infty\ \mbox{uniformly\ for\ a.a.}\ z\in C_{\varepsilon}\ \mbox{as}\ x\rightarrow\pm\infty
\end{eqnarray}
and also there exist functions $h\in L^1(B)_{+}$ and $g\in C(\RR)$ such that
\begin{eqnarray}\label{eq97}
g\geq 0,\ g\ \mbox{is\ subadditive},
\end{eqnarray}
\begin{eqnarray}\label{eq98}
F(z,x)\geq g(x)-h(z)\ \mbox{for\ a.a.}\ z\in C_{\varepsilon},\ \mbox{all}\ x\in\RR,
\end{eqnarray}
\begin{eqnarray}\label{eq99}
g\ \mbox{is\ coercive}\ (\mbox{that is}\ g(x)\rightarrow +\infty\ \mbox{as}\ x\rightarrow\pm\infty),
\end{eqnarray}
\begin{eqnarray}\label{eq100}
|g(x)|\leq 4+|x|\ \mbox{for\ all}\ x\in\RR.
\end{eqnarray}
From (\ref{eq95}) we have
\begin{eqnarray*}
-M_6\leq \varphi(y_n)&=& \frac{1}{2}\tau(y_n)-\int_{\Omega}F(z,y_n)dz\\
&=& \frac{1}{2}\tau(y_n)-\int_{C_{\varepsilon}}F(z,y_n)dz-\int_{\Omega\setminus C_{\varepsilon}}F(z,y_n)dz \\
&\leq& \frac{1}{2}\tau(y_n)-\int_{C_{\varepsilon}}g(y_n)dz+c_7\ \mbox{for\ some}\ c_7>0\ (\mbox{see}\ (\ref{eq98})\ \mbox{and}\ H_4(i))\\
&\leq&-\frac{c_8}{2}\|\overline{y}_n\|^2+c_7,\ \mbox{for\ some}\ c_8>0\ \mbox{with}\ y_n=\overline{y}_n+y_n^0,\ \overline{y}_n\in \overline{H},\ y_n^0\in E\ \\
&&(\mbox{since}\ g\geq 0,\ \mbox{see}\ (\ref{eq97}))
\end{eqnarray*}
$\Rightarrow\{\overline{y}_n\}_{n\geq 1}\subseteq \overline{H}\subseteq H^1(\Omega)\ \mbox{is\ bounded}.$

We have
\begin{eqnarray}\label{eq101}
&&\|y_n\|\leq \|\overline{y}_n\|+\|y_n^0\|\ \mbox{for\ all}\ n\geq 1\ \mbox{and}\ \|y_n^0\|\rightarrow +\infty\ \mbox{as}\ n\rightarrow\infty,\nonumber\\
&\Rightarrow& \|y_n^0\|\rightarrow\infty\ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
We fix $\delta>0$. By virtue of Lemma \ref{lem21}, we can find $\xi_{\delta}>0$ such that
$$\Big|\{z\in\Omega:|y^0(z)|<\xi_{\delta}\|y^0\|\}\Big|_{N}<\delta\ \mbox{for\ all}\ y^0\in E.$$
Let $D_n=\{z\in\Omega:|y_n^0(z)|\geq\xi_{\delta}\|y_n^0\|\}$. Then $|\Omega\setminus D_n|_N<\delta$. Since $\overline{H}$ is finite dimensional and $\{\overline{y}_n\}_{n\geq 1}$ is bounded, we can find $c_9>0$ such that
\begin{eqnarray}\label{eq102}
\|\overline{y}_n\|_{\infty}\leq c_9\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray}
From (\ref{eq99}) we know that $g$ is coercive. So, given $\mu>0$, we can find $M_7=M_7(\mu)>0$ such that
\begin{eqnarray}\label{eq103}
g(x)\geq \mu\ \mbox{for\ all}\ |x|\geq M_7.
\end{eqnarray}
Let $\Gamma_n=\{z\in\Omega:|y_n(z)|\geq M_7\}$. Then
\begin{eqnarray}\label{eq104}
&&|y_n(z)|\geq |y_n^0(z)|-|\overline{y}_n(z)|\geq\xi_{\delta}\|y_n^0\|-c_9\ \mbox{for\ a.a.}\ z\in D_n,\nonumber\\
&&\mbox{all}\ n\geq 1\ (\mbox{see}\ (\ref{eq102})),\nonumber\\
&\Rightarrow& |y_n(z)|\geq M_7\ \mbox{for\ all}\ z\in D_n,\ \mbox{all}\ n\geq n_0\ (\mbox{see}\ (\ref{eq101}))\nonumber\\
&\Rightarrow& D_n\subseteq\Gamma_n\ \mbox{for\ all}\ n\geq n_0.
\end{eqnarray}
We have
\begin{eqnarray}\label{eq105}
\int_{C_{\varepsilon}}g(y_n)dz&=& \int_{C_{\varepsilon}\cap\Gamma_n}g(y_n)dz+\int_{C_{\varepsilon}\setminus\Gamma_n}g(y_n)dz\nonumber\\
&\geq&\mu|C_{\varepsilon}\cap\Gamma_n|_N\ (\mbox{see}\ (\ref{eq103})\ \mbox{and\ recall}\ g\geq 0\ (\mbox{see}\ (\ref{eq97}))).
\end{eqnarray}
Also,
\begin{eqnarray*}
|C_{\varepsilon}\cap D_n|_N &=& |C_{\varepsilon}|_N-|C_{\varepsilon}\setminus D_n|_N\\
&\geq&|C_{\varepsilon}|_N-|\Omega\setminus D_n|_N\\
&\geq&|B|_N-|B\setminus C_{\varepsilon}|_N-\delta\\
&\geq&|B|_N-\varepsilon-\delta>0\ \mbox{for}\ \varepsilon,\ \delta>0\ \mbox{small},
\end{eqnarray*}
\begin{eqnarray}\label{eq106}
\Rightarrow|C_{\varepsilon}\cap\Gamma_n|_N>0\ \mbox{for\ all}\ n\geq n_0\ (\mbox{see}\ (\ref{eq104})).
\end{eqnarray}
Since $\mu>0$ is arbitrary, from (\ref{eq105}) and (\ref{eq106}) it follows that
\begin{eqnarray}\label{eq107}
\lim\limits_{n\rightarrow\infty}\int_{C_{\varepsilon}}g(y_n)dz=+\infty.
\end{eqnarray}
Recall that
\begin{eqnarray*}
-M_6\leq \varphi(y_n)&\leq& \frac{1}{2}\tau(y_n)-\int_{\Omega}g(y_n)dz+c_7\\
&\leq& -\int_{\Omega}g(y_n)dz+c_7\ (\mbox{since}\ y_n\in Y,\ (\mbox{see}\ (\ref{eq5}))),
\end{eqnarray*}
which contradicts (\ref{eq107}) above. This proves the coercivity of the functional $\vartheta$.
\end{proof}

The coercivity of $\vartheta$ implies that $\vartheta$ satisfies the $C$-condition and it is bounded below. Moreover, from (\ref{eq86}) and since $2<r$, we see that $\inf\limits_{Y}\vartheta<0$. These facts and Proposition \ref{prop20}, permit the use of Proposition \ref{prop4} and so we obtain the following multiplicity theorem for problem (\ref{eq1}).
\begin{theorem}\label{th23}
If hypotheses $H_0$ and $H_4$ hold, then problem (\ref{eq1}) admits at least two nontrivial solutions
$$u_0,\ \hat{u}\in C^1(\overline{\Omega}).$$
\end{theorem}

\section{Reduction on an Infinite Dimensional Space}\label{sec7}

In this section we examine what happens when the perturbed monotonicity condition in hypothesis $H_4(i)$ is reversed. Then the reduction method is implemented on an infinite dimensional subspace of $H^1(\Omega)$ and so the situation is more delicate.

The new hypotheses on the reaction $f(z,x)$ are the following:

\medskip
$H_5:\ f:\Omega\times\RR\rightarrow\RR$ is a Carath\'eodory function such that  $f(z,0)=0$ for a.a. $z\in\Omega$ and
\begin{itemize}
\item [(i)] $|f(z,x)|\leq a(z)x\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR\ \mbox{with}\ a\in L^{\infty}(\Omega)_{+};$
\item [(ii)] there exist a function $\eta\in L^{\infty}(\Omega)$ and $M_8>0$ such that
\begin{eqnarray*}
&&\eta(z)\geq \hat{\lambda}_{k}\ \mbox{for\ a.a.}\ z\in\Omega,\ \eta\neq \hat{\lambda}_{k},\\
&&(f(z,x)-f(z,y))(x-y)\geq \eta(z)(x-y)^2\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x,y\in\RR,\\
&&F(z,x)\leq 0\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\geq M_8;
\end{eqnarray*}
\item [(iv)] $2F(z,x)-f(z,x)x\rightarrow\pm\infty$ uniformly for a.a. $z\in\Omega$ as $x\rightarrow\pm\infty$;
\item [(iv)] there exist a function $\eta_0\in L^{\infty}(\Omega)_{+}$ and $\delta_0>0$ such that
\begin{eqnarray*}
&&\eta_0(z)\leq \hat{\lambda}_m\ \mbox{a.e.\ in}\ \Omega,\ \eta_0\neq \hat{\lambda}_m,\\
&& \limsup\limits_{x\rightarrow 0}\frac{2F(z,x)}{x^2}\leq \eta_0(z)\ \mbox{uniformly\ for\ a.a.}\ z\in\Omega,\\
&&0\leq F(z,x)\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\leq\delta_0.
\end{eqnarray*}
\end{itemize}
\begin{remark}
Note that in this case the perturbed monotonicity condition $H_5(ii)$ is in the opposite direction than before.
\end{remark}
\begin{ex}
The following function satisfies hypotheses $H_5$. As before, for simplicity no $z$-dependence is assumed:
\begin{eqnarray*}
f(x)=\left\{  \begin{array}{ll}
   \eta_0 x &\mbox{if}\ |x|\leq 1 \\
     \lambda x-(\lambda-\eta_0) &\mbox{if}\ 1<|x|,
    \end{array}\right.
\end{eqnarray*}
with $\hat{\lambda}_{k}<\lambda<0<\eta_0<\hat{\lambda}_{m}$.
\end{ex}

As before (see Section \ref{sec6}), we assume that $0\in\sigma(-\Delta+\beta)$ and this spectrum has a negative part. We set
$$E=E(0),\ \hat{H}=\overline{\oplus_{i\geq m}E(\hat{\lambda}_{i})}\ \mbox{and}\ V=E\oplus\hat{H}.$$
Evidently $V$ is infinite dimensional.

Let $\varphi$ be the energy functional of the problem and $\psi=\varphi|_{V}$. Evidently $\psi\in C^1(V)$.
\begin{prop}\label{prop24}
If hypotheses $H_0$ and $H_5$ hold, then the functional $\psi$ satisfies the $C$-condition.
\end{prop}
\begin{proof}
We do the proof for the case when the limit in $H_5(iii)$ is $+\infty$. The proof is similar if the limit is $-\infty$.

So, let $\{y_n\}_{n\geq 1}\subseteq V$ be such that
\begin{eqnarray}\label{eq108}
&&|\psi(y_n)|\leq M_9\ \mbox{for\ some}\ M_9>0, \mbox{all}\ n\geq 1,
\end{eqnarray}
\begin{eqnarray}\label{eq109}
&&(1+\|y_n\|)\psi'(y_n)\rightarrow 0\ \mbox{in}\ V^{\ast}\ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
\begin{claim}
The sequence $\{y_n\}_{n\geq 1}\subseteq H^1(\Omega)$ is bounded.
\end{claim}
Arguing indirectly, we assume that $\|y_n\|\rightarrow\infty$. Let $w_n=\frac{y_n}{\|y_n\|}, n\geq 1$. Then $\|w_n\|=1$ for all $n\geq 1$ and so we may assume that
\begin{eqnarray}\label{eq110}
w_n\overset{w}{\longrightarrow} w\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ w_n{\rightarrow} w\ \mbox{in}\ L^{2s'}(\Omega).
\end{eqnarray}
From (\ref{eq109}) we have
\begin{eqnarray}\label{eq111}
|\langle A(y_n),h\rangle +\int_{\Omega}\beta(z)y_n h dz-\int_{\Omega}f(z,y_n)h dz|\leq \frac{\varepsilon_n \|h\|}{1+\|y_n\|},\\
\mbox{for\ all}\ h\in Y\ \mbox{with}\ \varepsilon_n\rightarrow 0^{+}\nonumber.
\end{eqnarray}
We multiply (\ref{eq111}) with $\frac{1}{\|y_n\|}$, and choose $h=w_n-w\in H^1(\Omega)$. Then we have
\begin{eqnarray*}
&&|\langle A(w_n),w_n-w\rangle +\int_{\Omega}\beta(z)w_n(w_n-w) dz-\int_{\Omega}\frac{f(z,y_n)}{\|y_n\|}(w_n-w) dz|\\
&&\leq \frac{\varepsilon_n}{1+\|y_n\|} \|w_n-w\|\ \mbox{for\ all}\ n\geq 1,\\
&\Rightarrow&\lim\limits_{n\rightarrow\infty}\langle A(w_n),w_n-w\rangle=0\ (\mbox{see}\ (\ref{eq110})),\\
&\Rightarrow&\|D w_n\|_2\rightarrow \|D w\|_2,\\
&\Rightarrow& w_n\rightarrow w\ \mbox{in}\ H^1(\Omega)\ (\mbox{by\ the\ Kadec-Klee\ property\ of\ Hilbert\ spaces}),\\
&\Rightarrow& \|w\|=1.
\end{eqnarray*}

Let $\Omega_0=\{z\in\Omega: w(z)\neq 0\}$. Evidently $|\Omega_0|_N>0$ and we have
$$|y_n(z)|\rightarrow+\infty\ \mbox{for\ all}\ z\in\Omega_0.$$
Then hypothesis $H_5(iii)$ implies that
$$2F(z,y_n(z))-f(z,y_n(z))y_n(z)\rightarrow+\infty\ \mbox{for\ all}\ z\in\Omega_0.$$
Using  $H_5(iii)$ and Fatou's lemma, we have
\begin{eqnarray}\label{eq112}
\int_{\Omega}\left[2F(z,y_n)-f(z,y_n)y_n\right]dz\rightarrow+\infty.
\end{eqnarray}
On the other hand from (\ref{eq108}), we have
\begin{eqnarray}\label{eq113}
-\tau(y_n)+\int_{\Omega}2F(z,y_n)dz\leq 2M_9\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray}
Also, if in (\ref{eq111}) we choose $h=y_n\in H^1(\Omega)$, then
\begin{eqnarray}\label{eq114}
\tau(y_n)-\int_{\Omega}f(z,y_n)y_n dz\leq \varepsilon_n\ \mbox{for\ all}\ n\geq 1.
\end{eqnarray}
Adding (\ref{eq113}) and (\ref{eq114}), we obtain
\begin{eqnarray}\label{eq115}
\int_{\Omega}\left[2F(z,y_n)-f(z,y_n)y_n\right]dz\leq M_{10}\ \mbox{for\ some}\ M_{10}>0,\ \mbox{all}\ n\geq 1.
\end{eqnarray}
Comparing (\ref{eq112}) and (\ref{eq115}), we reach a contradiction. This proves the Claim.

By virtue of the Claim, we may assume that
\begin{eqnarray}\label{eq116}
y_n\overset{w}{\longrightarrow} y\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y_n{\rightarrow} y\ \mbox{in}\ L^{2s'}(\Omega).
\end{eqnarray}
As before,  if in (\ref{eq111}) we choose $h=y_n-y\in H^1(\Omega)$, pass to the limit as $n\rightarrow\infty$ and use (\ref{eq116}), then via the Kadec-Klee property of Hilbert spaces, we have
\begin{eqnarray*}
&&y_n\rightarrow y\ \mbox{in}\ H^1(\Omega),\\
&\Rightarrow& \psi\ \mbox{satisfies\ the\ C-condition}.
\end{eqnarray*}
\end{proof}

We consider the following orthogonal direct sum decomposition of $H^1(\Omega)$:
$$H^{\perp}(\Omega)=\overline{H}\oplus V.$$
where $\overline{H}=\oplus_{i=1}^{k}E(\hat{\lambda}_i)$ and $V=E\oplus\hat{H}$ (recall $E=E(0),\ \hat{H}=\overline{\oplus_{i\geq m}E(\hat{\lambda}_{i})}$ ).

As we already mentioned, the reduction method will occur on the infinite dimensional space $V$.
\begin{prop}\label{prop25}
If hypotheses $H_0$ and $H_5$ hold, then there exists a continuous map $\tilde{\gamma}:V\rightarrow \overline{H}$ such that
$$\varphi(v+\tilde{\gamma}(v))=\inf \left[\varphi(v+\overline{u}): \overline{u}\in \overline{H}\right].$$
\end{prop}
\begin{proof}
The proof is similar to the proof of Proposition \ref{prop10}.

We fix $v\in V$ and consider the functional $\varphi_{v}:H^1(\Omega)\rightarrow\RR$ defined by
$$\varphi_{v}(u)=\varphi (v+u)\ \mbox{for\ all}\ u\in H^1(\Omega).$$
Let $i:\overline{H}\rightarrow H^1(\Omega)$ be the inclusion map and let $\overline{\varphi}_{v}:\overline{H}\rightarrow \RR$ be defined by
$$\overline{\varphi}_{v}=\varphi_{v}\circ i.$$
The chain rule implies that
\begin{eqnarray}\label{eq117}
\overline{\varphi}_{v}'(\overline{u})=p_{\overline{H}^{\ast}}\varphi_{v}'(\overline{u})\ \mbox{for\ all}\ \overline{u}\in \overline{H},
\end{eqnarray}
with $p_{\overline{H}^{\ast}}$ being the orthogonal projection on $\overline{H}^{\ast}$.

Let $\overline{u}_1,\overline{u}_2\in \overline{H}$. We have
\begin{eqnarray}
&&\langle \overline{\varphi}_{v}'(\overline{u}_1)-\overline{\varphi}_{v}'(\overline{u}_2),\overline{u}_1-\overline{u}_2\rangle _{\overline{H}}\nonumber \\
&=& \langle \varphi'(v+\overline{u}_1)-\varphi'(v+\overline{u}_2),\overline{u}_1-\overline{u}_2\rangle \ (\mbox{see\ (\ref{eq117})})\nonumber \\
&=& \tau(\overline{u}_1-\overline{u}_2)-\int_{\Omega}\left[f(z,v+\overline{u}_1)-f(z,v+\overline{u}_2)\right](\overline{u}_1-\overline{u}_2)dz \nonumber \\
&\leq& \tau (\overline{u}_1-\overline{u}_2)-\int_{\Omega}\eta(z)(\overline{u}_1-\overline{u}_2)^2 dz\ (\mbox{see\ hypothesis}\ H_5(ii)) \nonumber\\
&\leq&-\xi_2\|\overline{u}_1-\overline{u}_2\|^2\ (\mbox{see\ Proposition}\ \ref{prop3}(a))\nonumber,
\end{eqnarray}
\begin{eqnarray}\label{eq118}
\Rightarrow \overline{u}\longmapsto -\overline{\varphi}_{v}(\overline{u})+
\frac{\xi_2}{2}\|\overline{u}\|^2=\psi_v(\overline{u})\ \mbox{is\ convex\ on}\ \overline{H}.
\end{eqnarray}
Note that for all $\overline{u}\in\overline{H}$, we have
\begin{eqnarray}\label{eq119}
&&\langle -\overline{\varphi}_{v}'(\overline{u}),\overline{u}\rangle_{\overline{H}}\nonumber\\
&&=\langle-\overline{\varphi}_{v}'(\overline{u})+\overline{\varphi}_{v}'(0),\overline{u}\rangle_{\overline{H}}-\langle\overline{\varphi}_{v}'(0),
\overline{u}\rangle_{\overline{H}}\nonumber\\
&&\geq \xi_2\|\overline{u}\|^2-c_8\|\overline{u}\|\ \mbox{for\ some}\ c_8>0\ (\mbox{see}\ (\ref{eq118})),
\end{eqnarray}
\begin{eqnarray}\label{eq120}
\Rightarrow \overline{u}\longmapsto -\overline{\varphi}_{v}'(\overline{u})\ \mbox{is\ coercive}.
\end{eqnarray}
Then from (\ref{eq118}) and (\ref{eq120}), it follows that $\overline{u}\longmapsto -\overline{\varphi}_{v}'(\overline{u})$ is surjective (see, for example, Papageorgiou \& Kyritsi \cite[p. 172]{26}). So, we can find $\overline{u}_0\in \overline{H}$ such that
\begin{eqnarray*}
&&\overline{\varphi}_{v}'(\overline{u}_0)=0,\\
&\Rightarrow& \overline{u}_0\ \mbox{is\ the\ unique\ minimizer\ of\ the\ strictly\ concave\ functional}\ \overline{u}\longmapsto \overline{\varphi}_v(\overline{u})\ (\mbox{see}\ (\ref{eq118})).
\end{eqnarray*}
So, we can define the map $\tilde{\gamma}:V\rightarrow \overline{H}$ by
$$\tilde{\gamma}(v)=\overline{u}_0.$$
We have
\begin{eqnarray}\label{eq121}
\overline{\varphi}_v'(\tilde{\gamma}(v))=0\ \mbox{and}\ \overline{\varphi}(\tilde{\gamma}(v))=\max\left[\varphi(v+\overline{u}):\overline{u}\in\overline{H}\right].
\end{eqnarray}
\begin{claim}
$\tilde{\gamma}:V\rightarrow \overline{H}$ is continuous.
\end{claim}
Let $v_n\rightarrow v$ in $V$. From (\ref{eq121}), we have
\begin{eqnarray*}
&&\overline{\varphi}_{v_{n}}'(\tilde{\gamma}(v_n))=0\ \mbox{for\ all}\ n\geq 1,\\
&\Rightarrow&\langle \overline{\varphi}_{v_{n}}(\tilde{\gamma}(v_{n})),\tilde{\gamma}(v_{n})\rangle_{\overline{H}}=0\ \mbox{for\ all}\ n\geq 1,\\
&\Rightarrow&\langle -\overline{\varphi}_{v_{n}}(\tilde{\gamma}(v_{n}))+\overline{\varphi}_{v_{n}}'(0),\tilde{\gamma}(v_{n})\rangle_{\overline{H}}-\langle \overline{\varphi}_{v_{n}}'(0),\tilde{\gamma}(v_{n})\rangle_{\overline{H}}=0\ \mbox{for\ all}\ n\geq 1,\\
&\Rightarrow& \xi_2\|\tilde{\gamma}(v_{n})\|^2\leq c_8\|\tilde{\gamma}(v_{n})\|\ \mbox{for\ all}\ n\geq 1 \ (\mbox{see}\ (\ref{eq119})),\\
&\Rightarrow& \{\tilde{\gamma}(v_{n})\}_{n\geq 1}\subseteq \overline{H}\ \mbox{is\ bounded}.
\end{eqnarray*}
Since $\overline{H}$ is finite dimensional, by passing to a suitable subsequence if necessary, we have
\begin{eqnarray*}
&&\tilde{\gamma}(v_{n})\rightarrow \overline{u}_0\ \mbox{in}\ \overline{H}\\
&\Rightarrow&\varphi(v_n+\tilde{\gamma}(v_n))\rightarrow\varphi(v_n+\overline{u}_0).
\end{eqnarray*}
From (\ref{eq121}) we have
\begin{eqnarray*}
&&\varphi(v_{n}+\overline{u})\leq\varphi(v_{n}+\tilde{\gamma}(v_{n})) \ \mbox{for\ all}\ \overline{u}\in\overline{H},\ \mbox{all}\ n\geq 1,\\
&\Rightarrow&\varphi(v+\overline{u})\leq \varphi(v+\overline{u}_0)\ \mbox{for\ all}\ \overline{u}\in\overline{H},\\
&\Rightarrow& \overline{u}_0=\tilde{\gamma}(v),\\
&\Rightarrow& \tilde{\gamma}\ \mbox{is\ continuous}.
\end{eqnarray*}
\end{proof}
We introduce $\xi(v)=\varphi(v+\tilde{\gamma}(v))$ for all $v\in V$. Reasoning as in the proof of Proposition \ref{prop10}, we obtain.
\begin{prop}\label{prop26}
If hypotheses $H_0$ and $H_5$ hold, then $\xi\in C^1(V)$.
\end{prop}
Recall that we are assuming that
$$\sigma(-\Delta+\beta I)\cap (-\infty,0)\neq\emptyset\ \mbox{and}\ 0\in\sigma(-\Delta+\beta I).$$

We can now have our last multiplicity theorem.
\begin{theorem}\label{th27}
If hypotheses $H_0$ and $H_5$ hold, then problem (\ref{eq1}) has at least two nontrivial solutions
$$u_0,\ \hat{u}\in C^1(\overline{\Omega}).$$
\end{theorem}
\begin{proof}
Hypotheses $H_5(i),(ii)$ imply that we can find $c_9>0$ such that
\begin{eqnarray}\label{eq122}
F(z,x)\leq c_9\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.
\end{eqnarray}
Let $v\in V$. We have
\begin{eqnarray*}
\psi(v)=\varphi(v)&=&\frac{1}{2}\tau(v)-\int\limits_{\Omega}F(z,v)dz\ (\mbox{since}\ \psi=\varphi|_{V})\nonumber\\
&\geq&-\int\limits_{\Omega}F(z,v)dz\ (\mbox{since}\ v\in V=E\oplus\hat{H})\nonumber\\
&\geq& -c_9|\Omega|_N\ (\mbox{see}\ (\ref{eq122})),\nonumber
\end{eqnarray*}
\begin{eqnarray}\label{eq123}
\Rightarrow \psi\ \mbox{is\ bounded\ below}.
\end{eqnarray}
From Proposition \ref{prop24} we know that $\psi$ satisfies the $C$-condition. This fact and (\ref{eq123}) imply that $\psi$ is coercive (see Papageorgiou \& Kyristi \cite[p. 272]{26}).
\begin{claim}\label{cl1}
$\xi$ satisfies the $C$-condition.
\end{claim}
Let $\{v_n\}_{n\geq 1}\subseteq V$ such that
\begin{eqnarray}\label{eq124}
|\xi(v_n)|\leq M_{11}\ \mbox{for\ some}\ M_{11}>0,\ \mbox{all}\ n\geq 1,
\end{eqnarray}
\begin{eqnarray}\label{eq125}
\mbox{and}\ (1+\|v_n\|)\xi'(v_n)\rightarrow 0\ \mbox{in}\ V^{\ast}\ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
By definition we have
\begin{eqnarray*}
&&\xi(v)=\varphi(v+\tilde{\gamma}(v))\geq\varphi(v)=\psi(v)\ \mbox{for\ all}\ v\in V,\\
&\Rightarrow&\xi(\cdot)\ \mbox{is\ coercive}\ (\mbox{since}\ \psi\ \mbox{is\ coercive}).
\end{eqnarray*}
From this fact and (\ref{eq124}) we infer that $\{v_n\}_{n\geq 1}\subseteq V$ is bounded. So, we may assume that
\begin{eqnarray}\label{eq126}
v_n\overset{w}{\longrightarrow} v\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ v_n\rightarrow v\ \mbox{in}\ L^{2s'}(\Omega).
\end{eqnarray}
From (\ref{eq125}) we have
\begin{eqnarray}\label{eq127}
|\langle A(v_n),h\rangle +\int_{\Omega}\beta(z)v_n h dz-\int_{\Omega}f(z,v_n)h dz|\leq \frac{\varepsilon_n \|h\|}{1+\|v_n\|},\\
\mbox{for\ all}\ v\in H^1(\Omega)\ \mbox{with}\ \varepsilon_n\rightarrow 0^{+}\nonumber.
\end{eqnarray}
In (\ref{eq127}) we choose $h=v_n-v\in H^1(\Omega)$, pass to the limit as $n\rightarrow\infty$ and use (\ref{eq126}). Then
\begin{eqnarray*}
&&\lim\limits_{n\rightarrow\infty}\langle A(v_n),v_n-v\rangle=0,\\
&\Rightarrow&\|D v_n\|_2\rightarrow \|D v\|_2,\\
&\Rightarrow& v_n\rightarrow v\ \mbox{in}\ H^1(\Omega)\ (\mbox{by\ the\ Kadec-Klee\ property,\ see}\ (\ref{eq126})).
\end{eqnarray*}
This proves Claim \ref{cl1}.

\begin{claim}\label{cl2}
$\xi'(v)=p_{V^{\ast}}\varphi'(v+\tilde{\gamma}(v))$ for all $v\in V$ ($p_{V^{\ast}}$ being the orthogonal projection of $H^1(\Omega)^{\ast}$ onto $V^{\ast}$).
\end{claim}
For every $v,h\in V$, we have
\begin{eqnarray*}
\langle \xi'(v),h\rangle_{V}&=&\lim\limits_{t\rightarrow 0}\frac{\xi(v+th)-\xi(v)}{t}\\
&=&\lim\limits_{t\rightarrow 0}\frac{\varphi(v+th+\tilde{\gamma}(v+th))-\varphi(v+\tilde{\gamma}(v))}{t}\\
&\leq&\lim\limits_{t\rightarrow 0}\frac{\varphi(v+th+\tilde{\gamma}(v+th))-\varphi(v+\tilde{\gamma}(v+th))}{t}\\
&=&\langle \varphi'(v+\tilde{\gamma}(v)),i_0(h)\rangle\\
&&(\mbox{since}\ \varphi\in C^1(H^1(\Omega)),\ \mbox{with}\ i_0:V\rightarrow H^1(\Omega)\ \mbox{being\ the\ inclusion\ map})\\
&=&\langle p_{V^{\ast}}\varphi'(v+\tilde{\gamma}(v)),h\rangle\ \mbox{for\ all}\ h\in V,
\end{eqnarray*}
$\Rightarrow \xi'(v)=p_{V^{\ast}}\varphi'(v+\tilde{\gamma}(v)).$

This proves Claim \ref{cl2}.

\begin{claim}\label{cl3}
$\xi$ has a local linking with respect to $V=E\oplus\hat{H}$.
\end{claim}
Hypotheses $H_5(i),(iv)$ imply that given $\varepsilon>0$ and $r>2$, we can find $c_{10}=c_{10}(\varepsilon,r)>0$ such that
\begin{eqnarray}\label{eq128}
F(z,x)\leq \frac{1}{2}(\eta_0(z)+\varepsilon)x^2+c_{10}|x|^r\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR.
\end{eqnarray}

Let $\hat{u}\in\hat{H}$. We have
\begin{eqnarray*}
\xi(\hat{u})&=&\varphi(\hat{u}+\tilde{\gamma}(\hat{u}))\\
&\geq&\varphi(\hat{u})\\
&=&\frac{1}{2}\tau(\hat{u})-\int_{\Omega}F(z,\hat{u})dz\\
&\geq&\frac{1}{2}\left[\tau(\hat{u})-\int_{\Omega}\eta_0(z)u^2dz\right]-\frac{\varepsilon}{2}\|\hat{u}\|^2-c_{11}\|\hat{u}\|^r\ \mbox{for\ some}\ c_{11}>0\\
&\geq&\frac{\xi_1-\varepsilon}{2}\|\hat{u}\|^2-c_{11}\|\hat{u}\|^r\ (\mbox{see\ Proposition}\ \ref{prop3}(a)).
\end{eqnarray*}
Choosing $\varepsilon\in (0,\xi_1)$ and since $r>2$, we can find $\varrho_1\in (0,1)$ small such that
\begin{eqnarray}\label{eq129}
\xi(\hat{u})\geq 0\ \mbox{for\ all}\ \hat{u}\in\hat{H}\ \mbox{with}\ \|\hat{u}\|\leq \varrho_1.
\end{eqnarray}
The space $Y=\overline{H}\oplus E$ is finite dimensional and so all norms are equivalent. Therefore, we can find $\delta_1>0$ such that
\begin{eqnarray}\label{eq130}
y\in Y\ \mbox{with}\ \|y\|\leq \delta_1\Rightarrow |y(z)|\leq\delta_0\ \mbox{for\ all}\ z\in\overline{\Omega}\ (\mbox{recall}\ Y\subseteq C^1(\overline{\Omega})).
\end{eqnarray}
Here $\delta_0>0$ is as postulated by hypothesis $H_5(iv)$. Since $\tilde{\gamma}$ is continuous (see Proposition \ref{prop25}), we can find $\vartheta=\vartheta(\delta_1)>0$ such that
\begin{eqnarray}\label{eq131}
e\in E\ \mbox{with}\ \|e\|\leq \vartheta\Rightarrow \|e+\tilde{\gamma}(e)\|\leq\delta_1\ (\mbox{recall}\ \tilde{\gamma}(0)=0).
\end{eqnarray}
From (\ref{eq130}) and (\ref{eq131}) it follows that, if $e\in E$ with $\|e\|\leq \vartheta$, then
\begin{eqnarray*}
\xi(e)&=&\varphi(e+\tilde{\gamma}(e))\\
&=&\frac{1}{2}\tau(e+\tilde{\gamma}(e))-\int_{\Omega}F(z,e+\tilde{\gamma}(e))dz\\
&\leq& -\int_{\Omega}F(z,e+\tilde{\gamma}(e))dz\ (\mbox{since}\ e+\tilde{\gamma}(e)\in E\oplus\overline{H}=Y)\\
&\leq& 0\ (\mbox{see}\ (\ref{eq131})\ \mbox{and}\ H_5(iv)),
\end{eqnarray*}
\begin{eqnarray}\label{eq132}
\Rightarrow \xi(e)\leq 0\ \mbox{for\ all}\ e\in E\ \mbox{with}\ \|e\|\leq \delta_1.
\end{eqnarray}
If $\varrho=\min\{\varrho_1,\delta_1\}$, then from (\ref{eq129}) and (\ref{eq132}), we infer that $\xi$ has a local linking with respect to $V=E\oplus\hat{H}$.

This proves Claim \ref{cl3}.

Note that $\xi(0)=0$ and so $\inf\limits_{V}\xi\leq 0$. If $\inf\limits_{V}\xi= 0$ then from (\ref{eq132}) we see that for all $e\in E$ with $0<\|e\|\leq\delta_1$, we have $\xi(e)=0=\inf\limits_{V}\xi$. So, from Claim \ref{cl2} we have
\begin{eqnarray*}
&&0=\xi'(e)=p_{V^{\ast}}\varphi'(e+\tilde{\gamma}(e)),\\
&\Rightarrow& e+\tilde{\gamma}(e)=\hat{u}\ \mbox{for\ all}\ e\in E\ \mbox{with}\ 0<\|e\|\leq\delta_1\ \mbox{is\ a\ solution\ of}\ (\ref{eq1}).
\end{eqnarray*}
If $\inf\limits_{V}\xi<0$, then we can apply Proposition \ref{prop4} and produce $\hat{u}_0$ and $\tilde{u}$ two nontrivial critical points of $\xi$. Then
$$u_0=\hat{u}_0+\tilde{\gamma}(\hat{u}_0)\ \mbox{and}\ \hat{u}=\tilde{u}+\tilde{\gamma}(\tilde{u})$$
are two nontrivial solutions of problem (\ref{eq1}). Regularity theory (see Wang \cite{34}) implies $u_0$, $\hat{u}\in C^1(\overline{\Omega})$.
\end{proof}

\begin{thebibliography}{99}
\bibitem{1} {\sc S. Aizicovici, N.S. Papageorgiu, V. Staicu}, {\em Degree theory for operators of monotone type and nonlinear elliptic equations with inequality constraints}, Memoirs Amer. Math. Soc., Vol. \textbf{196}, No. \textbf{905}, November 2008.
\bibitem{2} {\sc H. Amann}, {\em Saddle points and multiple solutions of differential equations}, Math. Z. \textbf{169} (1979), 127-166.
\bibitem{3} {\sc T. Bartsch}, {\em Critical point theory on partially ordered Hilbert spaces}, J. Functional Anal. \textbf{186} (2001), 117-152.
\bibitem{4} {\sc T. Bartsch, Z.Q. Wang}, {\em On the existence of sign changing solutions for semilinear Dirichlet problems}, Topol. Meth. Nonlinear Anal. \textbf{7} (1996), 115-131.
\bibitem{5} {\sc H. Brezis, L. Nirenberg}, {\em Remarks on finding critical points}, Comm. Pure Appl. Math. \textbf{44} (1991), 939-963.
\bibitem{6} {\sc A. Castro, J. Cossio, C. Velez}, {\em Existence of seven solutions for asymptotically linear Dirichlet problems without symmetries}, Annali. Mat. Pura Appl. DOI: 10.1007/s.10.231-011-0239-5.
\bibitem{7} {\sc A. Castro, A. Lazer}, {\em Critical point theory and the number of solutions of nonlinear Dirichlet problem}, Annali. Mat. Pura Appl. \textbf{120} (1979), 113-137.
\bibitem{8} {\sc K.C. Chang}, {\em Infinite Dimensional Morse Theory and Multiple Solution Problems}, Birkh\"auser, Boston (1993).
\bibitem{9} {\sc J. Dugundji}, {\em Topology}, Allyn and Bacon Inc., Boston (1996).
\bibitem{10} {\sc M. Filippakis, N.S. Papageorgiou}, {\em Multiple nontrivial solutions for resonant Neumann problems}, Math. Nachr. \textbf{283} (2010), 1000-1014.
\bibitem{11} {\sc L. Gasinski, N.S. Papageorgiou}, {\em Nonlinear Analysis}, Chapman \& Hall/CRC, Boca Raton (2006).
\bibitem{12} {\sc L. Gasinski, N.S. Papageorgiou}, {\em Dirichlet problems with double resonance and an indefinite potential}, Nonlinear Analysis {\bf 75} (2012), 4560-4595.
\bibitem{13} {\sc L. Gasinski, N.S. Papageorgiou}, {\em Pairs of nontrivial solutions for resonant Neumann problems}, J.~Math. Anal. Appl. \textbf{398} (2013), 649-663.
\bibitem{14} {\sc A. Granas, J. Dugundji}, {\em Fixed Point Theory}, Springer-Verlag, New-York (2003).
\bibitem{15} {\sc H. Hofer}, {\em Variational and topological methods in partially ordered Hilbert spaces}, Math. Ann. \textbf{261} (1982), 493-514.
\bibitem{16} {\sc S. Hu, N.S. Papageorgiou}, {\em Double resonance for Dirichlet problems with unbounded indefinite potential and combined nonlinearities}, Comm. Pure Appl. Anal. {\bf 11} (2012), 2005-2021.
\bibitem{17} {\sc S. Kyritsi, N.S. Papageorgiou}, {\em Multiple solutions for superlinear Dirichlet problems with an indefinite potential}, Annali. Math. Pura Appl. \textbf{192} (2013), 297-315.
\bibitem{18} {\sc S. Liu}, {\em Remarks on multiple solutions for elliptic resonant problems}, J. Math. Anal. Appl. \textbf{336} (2007), 498-505.
\bibitem{19} {\sc J.P. Liu, S. Li}, {\em An existence theorem for multiple critical points and its applications}, Kexue Tongbao (Chinese) \textbf{29} (1984), 1025-1027.
\bibitem{20} {\sc S. Liu, S. Li}, {\em Critical groups at infinity, saddle point reduction and elliptic resonant problems}, Commun. Contemp. Math. \textbf{5} (2003), 761-773.
\bibitem{21} {\sc C.R.F. Maunder}, {\em Algebraic Topology}, Dover Publications Inc., Mineola, N.Y. (1996).
\bibitem{22} {\sc D. Motreanu, V. Motreanu, N.S. Papageorgiou}, {\em On resonant Neumann problems}, Math. Annalen \textbf{354} (2012), 1117-1145.
\bibitem{23} {\sc D. Motreanu, N.S. Papageorgiou}, {\em Multiple solutions for nonlinear Neumann problems driven by a nonhomogeneous differential operator}, Proc. Amer. Math. Soc. \textbf{139} (2011), 3527-3535.
\bibitem{24} {\sc D. O'Regan, N.S. Papageorgiou, G. Smyrlis}, {\em Neumann problems with double resonance}, Topol. Meth. Nonlin. Anal. \textbf{39} (2012), 151-173.
\bibitem{25} {\sc N.S. Papageorgiou, S.T. Kyritsi}, {\em Handbook of Applied Analysis}, Springer, New York (2009).
\bibitem{26} {\sc N.S. Papageorgiou, F. Papalini}, {\em Seven solutions with sign information for sublinear equations with unbounded and indefinite potential and no symmetries}, Israel J. Math., in press.
\bibitem{27} {\sc N.S. Papageorgiou, V. R\u adulescu}, {\em Semilinear Neumann problems with indefinite and unbounded potential and crossing nonlinearity}, Contemp. Math. \textbf{595} (2013), 293-315.
\bibitem{28} {\sc N.S. Papageorgiou, G. Smyrlis}, {\em On a class of parametric Neumann problem with indefinite and unbounded potential}, Forum Math. DOI: \textbf{10.1515}/forum-2012-0042.
\bibitem{29} {\sc P. Pucci, J. Serrin}, {\em The Maximum Principle}, Birkh\"auser, Basel (2007).
\bibitem{30} {\sc C.-L. Tang}, {\em Multiple solutions of Neumann problems for elliptic equations}, Nonlinear Anal. \textbf{54} (2003), 637-650.
\bibitem{31} {\sc C.-L. Tang, X.P. Wu}, {\em Existence and multiplicity for solutions of Neumann problems for elliptic equations}, J. Math. Anal. Appl. \textbf{288} (2003), 660-670.
\bibitem{32} {\sc K. Thews}, {\em Nontrivial solutions of elliptic equations at resonance}, Proc. Royal Soc. Edinburgh \textbf{85A} (1980), 119-129.
\bibitem{33} {\sc J. Vazquez}, {\em A strong maximum principle for some quasilinear elliptic equations}, Appl. Math. Optim. \textbf{12} (1984), 191-202.
\bibitem{34} {\sc X. Wang}, {\em Neumann problems of semilinear elliptic equations involving critical Sobolev exponents}, J.~Differential Equations \textbf{93} (1991), 283-310.

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