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\begin{document}

\title[Neumann problems with indefinite and unbounded potential]
{Semilinear Neumann problems with indefinite and unbounded potential and crossing nonlinearity}

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

\author[V. R\u{a}dulescu]{Vicen\c{t}iu D. R\u{a}dulescu}
\address[V. R\u{a}dulescu]{Institute of Mathematics ``Simion Stoilow" of the Romanian Academy, P.O. Box 1-764, 014700 Bucharest, Romania \& Department of Mathematics, University of Craiova,  Street A.I. Cuza No. 13, 200585 Craiova, Romania}
\email[V. R\u{a}dulescu]{vicentiu.radulescu@@imar.ro}

\date{}

\keywords{indefinite and unbounded potential, crossing nonlinearity, maximum principle, critical groups, truncation and perturbation techniques, nodal solutions.}

\thanks{{\em 2010 Mathematics Subject Classification.} Primary 35J20; Secondary 35J60,
58E05}

\begin{abstract}
We consider a semilinear Neumann problem with an indefinite and unbounded potential and an asymmetric reaction
that crosses at least the principal eigenvalue of the operator $-\Delta +\beta I$ in $ H^1(\Omega)$, $\beta$ being the potential function.
Using a combination of variational methods, with truncation and perturbation techniques and Morse theory, we prove multiplicity theorems providing precise sign information for all the solutions.
\end{abstract}


\maketitle

\section{Introduction}
Let $\Omega\subset\RR^N$ be a bounded domain with a $C^2$--boundary $\partial\Omega$. In this paper we study the following semilinear
Neumann problem
\begin{equation}
	\label{eq:1}
		\left\{
			\begin{array}{lll}
				&\di\Delta u(z)+\beta(z)u(z)=f(z,u(z))&\di\qquad \mbox{in $\Omega$} \\
				&\di \frac{\partial u}{\partial n}=0&\di\qquad\mbox{on $\partial\Omega$}.
			\end{array}
\right.
\end{equation}

Here, $\beta\in L^s(\Omega)$ with $s>N$ and, in general, it is indefinite (sign changing) and unbounded. We assume that the reaction $f(z,x)$ is a measurable function which is $C^1$ in the $x$--variable. The aim of this paper is to prove a multiplicity theorem for problem \eqref{eq:1} providing information for all the solutions, provided that the reaction $x\ri f(z,x)$ exhibits an asymmetric behavior at $+\infty$ and $-\infty$ (crossing and jumping nonlinearity).

The multiplicity of solutions for such semilinear elliptic equations was first studied by Hofer \cite{ref13}, who examined a Dirichlet problem with $\beta\equiv 0$. Assuming that $f(z,x)=f(x)$ with $f\in C^1(\RR)$, $f(0)=0$, $f'(0)\in (\lambda_i,\lambda_{i+1})$ for some $i\geq 2$ (here $\{\lambda_i\}_{i\geq 1}$ denotes the sequence of distinct eigenvalues of $(-\Delta, H^1_0(\Omega))$ and that $\limsup_{x\ri\pm\infty}f(x)/x<\lambda_1$, Hofer \cite{ref13} proved that the equation has at least four nontrivial solutions, two of which have constant sign (one positive and the other negative). Later, Bartsch \& Wang \cite{ref3} proved that from the other two solutions, one is nodal (sign-changing). In fact, Dancer \& Du \cite{ref7} and Li \& Wang \cite{ref18} established that both solutions are nodal.
In the aforementioned works it is assumed that asymptotically at $\pm\infty$ the quotient $f(z,x)/x$ stays below $\lambda_1$ and this makes the energy functional of the problem coercive. Problems with asymmetric (crossing) reaction, are usually studied using the so-called ``Fu\v{c}ik spectrum". We refer to the works C\'ac \cite{ref5}, Cuesta \& Gossez \cite{ref6}, Magalhaes \cite{ref19}, and Perera \& Schechter \cite{ref22}. However, this approach has two serious limitations. First, the use of the Fu\v cik spectrum requires that the limits $\lim_{x\ri\pm\infty}f(z,x)/x$ do exist. Second, our knowledge of the Fu\v cik spectrum is limited (see Schechter \cite{ref24}). More recently, Liu \& Sun \cite{ref17} considered the asymmetric Dirichlet problem with $\beta=0$ and without any use of the Fu\v cik spectrum. Their method of proof is based on some elaborate flow invariance arguments. In fact, Liu \& Sun \cite[p. 1071]{ref17} mention that alternatively ``Morse theory could work, but then the techniques will be more complicated".
In the present paper, working in the framework of Neumann problems (which in principle are more difficult to deal with, due to the failure of the Poincar\'e inequality) with an indefinite and unbounded potential $\beta (\cdot)$, using a combination of variational methods and Morse theory, we prove multiplicity results with precise sign information for all the solutions, under weaker conditions on the reaction $f(z,x)$ than in Liu \& Sun \cite{ref17}. Our approach is based on the critical point theory, together with suitable perturbation, truncation and comparison techniques and with the use of Morse theory (critical groups). In the next section, for the convenience of the reader, we recall the main
mathematical tools which we will use in the sequel. Also, we examine  the spectral properties of the operator $ H^1(\Omega)\ni u\longmapsto -\Delta u+\beta u$.

\section{Mathematical Background}
\label{sec:mathback}
In the study of problem \eqref{eq:1}, 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
$$\mbox{int}\,C_+=\{u\in C_+;\ u(z)> 0\ \mbox{for all}\ z\in\overline\Omega\}.$$

Let $g:\Omega\times\RR\ri\RR$ be a Carath\'eodory function (that is, for all $x\in\RR$, the mapping $z\longmapsto g(z,x)$ is measurable, and for a.a. $z\in\Omega$, the function $x\longmapsto f(z,x)$ is continuous) with subcritical growth in $x\in\RR$, namely
$$|g(z,x)|\leq\alpha (z)\, (1+|x|^{r-1})\qquad\mbox{for a.a. $z\in\Omega$ and for all $x\in\RR$}$$
with $\alpha\in L^\infty (\Omega)_+$ and $1<r<2^*$, where $2^*=2N/(N-2)$ if $N\geq 3$ and $2^*=+\infty$ if $N=1,2$.
Also, let $\beta\in L^s(\Omega)$ with $s>N$, $G(z,x)=\int_0^xg(z,t)dt$, and consider the $C^1$--functional $\Psi_0:H^1(\Omega)\ri\RR$ defined by
$$\Psi_0(u)=\frac 12\,\|Du\|^2_2+\frac 12\intom\beta u^2dz-\intom G(z,u)dz\qquad\mbox{for all}\ u\in H^1(\Omega)\,.$$


The next result was first proved by Brezis \& Nirenberg \cite{ref4} for the ``Dirichlet" space $H^1_0(\Omega)$ and was later extended to the space $W_0^{1,p}(\Omega)$ (with $1<p<\infty$) by Garcia-Azorero, Manfredi \& Peral Alonso \cite{ref11} and to the space $W^{1,p}(\Omega)$ (Neumann case) by Iannizzotto \& Papageorgiou \cite{ref14}. The proof of \cite{ref14} applies in the present setting using the regularity results of Wang \cite{ref26}.  So, we have:

\begin{proposition}
\label{prop:2.1}
Assume that $u_0\in H^1(\Omega)$ is a local $C^1(\overline\Omega)$--minimizer of $\Psi_0$, that is, there exists $\rho_0>0$ such that
$\Psi_0(u_0)\leq\Psi_0(u_0+h)$ for all $h\in C^1 (\overline\Omega)$ with $\|h\|_{C^1(\overline\Omega)}\leq\rho_0\,.$
Then $u_0\in C^{1,\gamma}(\overline\Omega)$ for some $\gamma\in (0,1)$ and $u_0$ is a local $H^1(\Omega)$--minimizer of $\Psi_0$, that is, there exists
$\rho_1>0$ such that
$\Psi_0(u_0)\leq\Psi_0(u_0+h)$ for all $h\in H^1 (\Omega)$ with $\|h\|_{H^1(\Omega)}\leq\rho_1\,.$
\end{proposition}

Here and in the sequel, we denote by $\|\,\cdot\,\|$ the norm in $H^1(\Omega)$, that is,
$$\|u\|=\left(\|u\|^2_2+\| Du\|^2_2\right)^{1/2}\qquad\mbox{for all}\ u\in H^1(\Omega).$$

Next we recall some basic definitions and facts from critical point theory. For details, we refer to the books by Gasinski \& Papageorgiou \cite{ref12} and Kristaly, R\u adulescu \& Varga \cite{ref15}.

Let $X$ be a Banach space and let $X^*$ be its topological dual. By $\langle \cdot ,\cdot\rangle$ we denote the duality brackets for the dual pair $(X^*,X)$. Let $\phi\in C^1(X)$. We say that $\phi$ satisfies the ``Cerami condition" (the ``C--condition" for short), if the following is true:\\
``Every sequence $\{x_n\}_{n\geq 1}\subset X$ such that $\{\phi(x_n)\}_{n\geq 1}\subset\RR$ is bounded and
$$\left(1+\| x_n\|_X\right)\phi'(x_n)\ri 0\ \ \mbox{in}\ X^*,$$
admits a strongly convergent subsequence."

This compactness-type condition is in general weaker than the more usual Palais-Smale condition. Nevertheless it suffices to prove a deformation theorem and to deduce the minimax theory for certain critical values of $\phi$. In particular, we have the following result, known in the literature as the ``mountain pass theorem".

\begin{theorem}
\label{th:2.2}
Assume that $\phi\in C^1(X)$, $x_0,\, x_1\in X$, $\|x_1-x_0\|_X>\rho>0$,
$$\max\|\phi (x_0),\phi(x_1)\}<\inf\{\phi(x);\ \|x-x_0\|_X=\rho\}=\eta_l$$
$$c=\inf_{\gamma\in\Gamma}\max_{0\leq t\leq 1}\phi (\gamma(t)), \ \ \mbox{where}\ \Gamma =\{\gamma\in C([0,1];X);\ \gamma(0)=x_0,\ \gamma(1)=x_1\}\,,$$
and $\phi$ satisfies the C--condition. Then $c\geq\eta_l$ and $c$ is a critical value of $\phi$.
\end{theorem}

Next, from Morse theory, we recall the definition of critical groups and the Morse relation. So, let $\phi\in C^1(X)$ and $c\in\RR$. We define the following sets:
$$\phi^c=\{x\in X;\ \phi(x)\leq c\},\qquad K_\phi =\{x\in X;\ \phi'(x)=0\},\qquad K_\phi^c=\{x\in K_\phi ;\ \phi(x)=c\}\,.$$



Let $(Y_1,Y_2)$ be a topological pair such that $Y_2\subseteq Y_1\subseteq X$ and $k\geq 0$ be an integer. By $H_k(Y_1,Y_2)$ we denote the $k$th singular homology group for the topological pair $(Y_1,Y_2)$ with integer coefficients. The critical groups of $\phi$ at an isolated $x\in K_\phi^c$ are defined by
$$C_k(\phi,x)=H_k(\phi^c\cap {\mathcal U},\phi^c\cap {\mathcal U}\setminus\{x\})\qquad\mbox{for all}\ k\geq 0,$$
where ${\mathcal U}$ is a neighborhood of $x$ such that $K_\phi\cap\phi^c\cap{\mathcal U}=\{x\}$. The excision property of singular homology implies that the above definition of critical groups is independent of the neighborhood ${\mathcal U}$ of $x$.

Suppose that $\phi\in C^1(X)$ satisfies the C--condition and $\inf\phi (K_\phi)>-\infty$. Let $c< \inf\phi (K_\phi)$.
The critical groups of $\phi$ at infinity are defined by
$$C_k(\phi,\infty)=H_k(X,\phi^c)\qquad\mbox{for all}\ k\geq 0.$$

The second deformation theorem (see for example Gasinski \& Papageorgiou \cite[p. 628]{ref12}) implies that the above definition of critical groups of $\phi$ at infinity is independent of the choice of the level $c< \inf\phi (K_\phi)$.

Suppose that $K_\phi$ is finite. We define
$$M(t,x)=\sum_{k\geq 0}\mbox{rank}\, C_k(\phi ,x)t^k\qquad\mbox{for all}\ t\in\RR\ \mbox{and all}\ x\in K_\phi$$
$$P(t,\infty)=\sum_{k\geq 0}\mbox{rank}\, C_k(\phi ,\infty )t^k\qquad\mbox{for all}\ t\in\RR\,.$$
The Morse relation says
\begin{equation}\label{eq2}\sum_{x\in K_\phi}M(t,x)=P(t,\infty)+(1+t)Q(t),\end{equation}
where $Q(t)=\sum_{k\geq 0}a_kt^k$ is a formal series with nonnegative integer coefficients $a_k$, $k\geq 0$.

Suppose that $X=H$ is a Hilbert space, $x\in H$, ${\mathcal U}$ is a neighborhood of $x$ and $\phi\in C^2({\mathcal U})$. If $x\in K_\phi$, then the Morse index of $x$ denoted by $\mu=\mu(x)$, is defined as the supremum of the dimensions of vector subspaces of $H$ in which $\phi ''(x)$  is negative definite. The nullity of $\phi$ at $x\in K_\phi$, denoted by $\nu=\nu(x)$, is defined to be the dimension of $\mbox{Ker}\, \phi ''(x)$. We say that $x\in K_\phi$ is nondegenerate if $\nu(x)=0$, that is, $\phi ''(x)$ is invertible. If $x\in K_\phi$ is nondegenerate with Morse index $\mu$, then (see Mawhin \& Willem \cite{ref20})
\begin{equation}\label{eq3}
C_k(\phi ,x)=\delta_{k,\mu}\ZZ\qquad\mbox{for all $k\geq 0$},\end{equation}
where $\delta_{k,\mu}$ denotes the Kronecker symbol.

Now we develop the spectrum of $-\Delta u+\beta u$ for $u\in H^1(\Omega)$. We follow Willem \cite{ref27}, where the Dirichlet eigenvalue problem is examined. For completeness we provide the details. So, we examine the following liner eigenvalue problem
\begin{equation}\label{ref4}
\left\{\begin{array}{lll}
& -\Delta u(z)+\beta (z)u(z)=\lambda u(z)&\qquad\mbox{in $\Omega$}\\
& \di \frac{\partial u}{\partial n}=0&\qquad\mbox{on $\partial\Omega$}.\end{array}\right.
\end{equation}

Set
$$\sigma (u)=\frac 12\, \|Du\|^2_2+\frac 12\intom \beta u^2dz,\qquad u\in H^1(\Omega).$$

To analyze the linear eigenvalue problem \eqref{ref4} it suffices to impose the following condition on the potential $\beta$. Eventually, in order to deal with problem \eqref{eq:1} we will have to strengthen this condition.

\smallskip
\noindent $H_{0}$: $\beta\in L^{N/2}(\Omega)$ if $N\geq 3$, $\beta\in L^r(\Omega)$ with $r>1$ if $N=2$ and $\beta\in L^1(\Omega)$ if $N=1$.

\begin{lemma}
	\label{lem:2.3}
		If hypothesis $H_{0}$ holds, then
		\begin{center}
		 $\widehat{\lambda}_{1}=\inf\{\sigma(u); u\in H^{1}(\Omega), \left\|u\right\|_{2}=1\}>-\infty.$
		\end{center}
	\end{lemma}
	
		\textit{Proof.} We treat the case $N\geq3$, the other two cases being similar using the Sobolev embedding theorem.
			
			We proceed by contradiction. So, suppose that the conclusion of Lemma is not true. Then we can find $\{u_{n}\}_{n\geq 1} \subset H^{1}(\Omega)$ such that $\left\|u_{n}\right\|_{2}=1$
			for all $n\geq 1$
				and $\sigma(u_{n})\rightarrow-\infty$ as $n\rightarrow\infty$. So, we can find $n_{0}\geq 1$ such that
				\begin{equation}
					\label{eq:5}
						\sigma (u_{n})\leq-1\qquad \mbox{for all}\ n\geq n_{0}.
				\end{equation}

		Suppose that $\left\|u_{n}\right\|\rightarrow\infty$ as $n\rightarrow\infty$ and let $y_{n}=\frac{u_{n}}{||u_{n}||}$. Then $||y_{n}||=1$ for all $n\geq 1$ and so we may assume that
			\begin{equation}
				\label{eq:6}
					y_{n}\rightarrow y\ \mbox{weakly in}\ \ H^1(\Omega)\ \mbox{and}\ y_{n}\rightarrow y\ \mbox{in}\ L^2(\Omega).
			\end{equation}
		Note that $\{y^{2}_{n}\}_{n\geq 1} \subset L^{\frac{N}{N-2}}(\Omega)$ is bounded (by the Sobolev embedding theorem) and so, by (\ref{eq:6}), we may assume that
			\begin{equation}
					\begin{array}{l}
						y^{2}_{n} \rightarrow y^{2}\quad \mbox{weakly in}\ L^{\frac{N}{N-2}}(\Omega).
					\end{array}
			\end{equation}
							Therefore
			\begin{equation}
				\label{eq:7}
					\begin{array}{l}
						\int_{\Omega} \beta y^{2}_{n} dz \rightarrow \int_{\Omega} \beta y^{2}dz \quad\mbox{(since $\frac{N-2}{N}+\frac{2}{N}=1$, see $H_{0}$)}.
				\end{array}
			\end{equation}
		Thus, by (\ref{eq:5}) and (\ref{eq:7}) and passing at the limit as $n\rightarrow\infty$, we obtain 		
			$\sigma(y)\leq 0.$
		
		If $y=0$, then $y_{n}\rightarrow0$ in $H^{1}(\Omega)$, a contradiction to the fact that $||y_{n}||=1$ for all $n\geq 1$. Hence $y\neq 0$. On the other hand	
			\begin{center}
				$||y_{n}||_{2}=\frac{||u_{n}||_{2}}{||u_{n}||}=\frac{1}{||u_{n}||}\rightarrow 0$ as $n\rightarrow\infty$,
			\end{center}
			which implies $||y||_{2}=0$ (see (\ref{eq:6})), hence $y=0$, a contradiction.
		This proves that $\{u_{n}\}_{n\geq 1} \subset H^{1}(\Omega)$ is bounded. So we may assume that
		\begin{equation*}
			\begin{array}{l}
				u_{n}\rightarrow u\ \mbox{weakly in}\ H^{1}(\Omega) \\
				u_{n}\rightarrow u\ \mbox{weakly in}\ L^{2^*}(\Omega)\ \ \ \mbox{(recall $2^*=\frac{2N}{N-2}$)}
			\end{array}
		\end{equation*}
			Therefore
		\begin{equation*}
			\begin{array}{l}
				\int_{\Omega} \beta u^{2}_{n} dz \rightarrow \int_{\Omega} \beta u^{2} dz .			 
			\end{array}
		\end{equation*}
		
		Taking the limit as $n\rightarrow\infty$ we obtain
		$\sigma (u)\leq \widehat{\lambda}_{1}= -\infty$,
		a contradiction. So, we conclude that ${\widehat{\lambda}}_{1}> -\infty$.
		\qed
		
		\smallskip
			By virtue of Lemma \ref{lem:2.3}, we see that we can find $\widehat{\gamma}> \max\{-\widehat{\lambda}_{1},0\}$ such that
			\begin{equation}
				\label{eq:8}
					\sigma (u)+\widehat{\gamma}||u||^{2}_{2}\geq \widehat{c}\,||u||^{2}\qquad \mbox{for all}\ u\in H^{1}(\Omega)\ \mbox{with}\ \widehat{c}>0.			
			\end{equation}
			
		Then relation (\ref{eq:8}) suggests the introduction of the following inner product on $H^{1}(\Omega)$:	
		 $$(u,y)_{*}=\int_{\Omega}(Du,Dy)_{\RR^{N}}dz+\int_{\Omega}(\beta(z)+\widehat{\gamma})uydz\qquad \mbox{for all}\ u,y\in H^{1}(\Omega).$$
		
			Given $h\in L^{2}(\Omega)$, by the Riesz representation theorem, we can find a unique $u\in H^{1}(\Omega)$ such that
			\begin{equation}
				\label{eq:9}
					(u,v)_{*}=\int hvdz\qquad \mbox{for all}\ v\in H^{1}(\Omega)		.
			\end{equation}
			
			So, we can define the continuous linear map
					$K_{0}: L^{2}(\Omega)\rightarrow H^{1}(\Omega)$ which to each $h\in L^{2}(\Omega)$			
			assigns the unique $u\in H^{1}(\Omega)$ satisfying (\ref{eq:9}). Let $i:H^{1}(\Omega)\rightarrow L^{2}(\Omega)$ be the embedding map. By virtue of the Sobolev embedding theorem, $i$ is compact and so
			$K_{0}\circ i$ is compact, self-adjoint and positive. Then by the spectral theorem for such operators (see, for example, Gasinski \& Papageorgiou \cite[p. 296]{ref12}), we can find $\{\mu_{n}\}_{n\geq 1}$ a sequence of eigenvalues of
			$K_{0}\circ i$ such that 			
			$\mu_{1}>\mu_{2}>\ldots>\mu_{n}>\ldots>0$ and $\mu_{n}\rightarrow 0$.

				We set $\widehat{\lambda_{n}}=\frac{1}{\mu_{n}}-\widehat{\gamma}$ for all $n\geq 1$. Then $\{\widehat{\lambda_{n}}\}_{n\geq 1}$ is the sequence of distinct eigenvalues of (\ref{ref4}). We have
					$-\infty <\widehat{\lambda_{1}}<\widehat{\lambda_{2}}<\ldots<\widehat{\lambda_{n}}<\ldots,$ and $\widehat{\lambda_{n}}\rightarrow+\infty$ as $n\rightarrow\infty\,.$

To these eigenvalues corresponds a sequence $\{\widehat{u_{n}}\}_{n\geq 1}\subset  H^{1}(\Omega)$ of eigenfunctions, which form an orthonormal basis of $L^{2}(\Omega)$ and an
			orthogonal basis of $H^{1}(\Omega)$. Moreover, if $\beta \in L^{s}(\Omega)$ with $s>N$, then the regularity results of Wang \cite{ref26} imply $\{\widehat{u_{n}}\}_{n\geq 1}\subset C^{1}(\overline{\Omega})$.
			These eigenvalues admit variational characterizations of Courant type using the Rayleigh quotient $\frac{\sigma(u)}{||u||^{2}_{2}}$ for all $u\in H^{1}(\Omega)\setminus\{0\}$.
			So, denoting by $E(\widehat{\lambda}_{i})$ the eigenspace corresponding to $\widehat{\lambda}_{i}$, we have
				\begin{equation}
					\label{eq:10}
						{\widehat{\lambda}}_{1}=\inf\left\{\frac{\sigma(u)}{||u||^{2}_{2}}; u\in H^{1}(\Omega), u\neq 0 \right\}.
				\end{equation}				
				\begin{eqnarray}
						\widehat{\lambda}_{k}& = & \inf \left\{ \frac{\sigma(u)}{||u||^{2}_{2}}; u\in \overline{ {\underset{\mathrm{i\geq k}}\oplus} E(\widehat{\lambda}_{i})}, u\neq 0 \right\} \label{eq:11}\\	
						& = & \sup \left\{ \frac{\sigma(u)}{||u||^{2}_{2}}; u\in \overset{k}{\underset{\mathrm{i=1}}\oplus}E(\widehat{\lambda}_{i}), u\neq 0\right\}\ \mbox{for}\ k\geq 2. \nonumber
				\end{eqnarray}
			
			
			The infimum in (\ref{eq:10}) and both the infimum and the supremum in (\ref{eq:11}) are realized on $E(\widehat{\lambda}_k)$, $k\geq 1$. The first eigenvalue $\widehat{\lambda}_{1}$ is simple
			(that is, $\mbox{dim}\,E(\widehat{\lambda}_{1})=1$) and from (\ref{eq:10}) it is clear that the nontrivial elements of $E(\widehat{\lambda}_{1})$ do not change sign. In fact $\widehat{\lambda}_{1}$ is the only
			eigenvalue with eigenfunctions of constant sign. All the other eigenvalues have nodal (sign changing) eigenfunctions. By $\widehat{u}_{1}$ we denote the $L^{2}$- normalized
			(that is, $||\widehat{u}_{1}||_{L_{2}}=1$) positive eigenfunction corresponding to $\widehat{\lambda}_{1}$. If $\beta \in L^{s}(\Omega)$ with $s>N $, then $\widehat{u}_{1}\in C_{+} \backslash \{0\} $ and in fact by the
			Harnack inequality of Pucci \& Serrin \cite[p. 163]{ref23}, we have $\widehat{u}_{1}(z)>0$ for all $z\in \Omega$. Finally, if $\beta^{+}\in L^{\infty}(\Omega)$, then the boundary point theorem of
			Pucci \& Serrin \cite[p. 120]{ref23}  implies that $\widehat{u}_{1}\in \mbox{int}\,C_{+}$. When $\beta\in L^{s}(\Omega)$ with $s>N/2$, the eigenspaces $E(\widehat{\lambda}_{k})$ have the so-called
			``Unique Continuation Property" (UCP for short). Namely, if $u\in E(\widehat{\lambda}_{k})$ and $u$ vanishes on a set of positive measure, then $u\equiv 0$ (see de Figueiredo \& Gossez \cite{ref10}).
			
				A similar analysis can be conducted for a weighted version of the eigenvalue problem (4). So, let $m\in L^{\infty}(\Omega)$, $m\geq 0$, $m\neq 0$ and consider the following linear eigenvalue problem
				\begin{equation}
						\label{eq:12}
							\left\{
										 \begin{array}{cccc}
											 -\Delta u(z)+\beta (z) u(z)   & = & \lambda m(z)u(z) & \mbox{in}\ \Omega\\
											 \di\frac{\partial u}{\partial n} & = & 0  						  & \mbox{on}\ \partial\Omega\,.
										 \end{array} \right.
				\end{equation}
					
		As for (\ref{ref4}), the eigenvalue problem (\ref{eq:12}) has a strictly increasing sequence $\{\tilde{\lambda}_{k}(m)\}_{k\geq 1}$ of eigenvalues such that $\tilde{\lambda}_{k}(m)\rightarrow+\infty$ as
$k\rightarrow\infty$. These eigenvalues admit variational characterizations in terms of Rayleigh quotient $\frac{\sigma(u)}{\int_{\Omega}mu^{2}dz}$ for all $u\in H^{1}(\Omega), u\neq 0$
(see \eqref{eq:10}, \eqref{eq:11}). The first eigenvalues $\tilde{\lambda}_{j}(m)>0$ is simple and has eigenfunctions of constant sign. These eigenspaces $E(\tilde{\lambda}_{k}(m))$ have the UCP and this leads to the following monotonicity property for the eigenvalues:
				\begin{proposition}
				\label{prop:2.4}
							If $m_{1},m_{2}\in\ L^{\infty}(\Omega)\setminus \{0\}$, $m_{1}(z)\leq\ m_{2}(z)$ a.e. in $\Omega$ and $m_{1}\neq m_{2}$,
									then $\tilde{\lambda}_{k}(m_{2})<\tilde{\lambda}_{k}(m_{1})$ for all $k\geq 1$.										 \end{proposition}
				
		Also, as a consequence of the Harnack inequality (see Pucci \& Serrin \cite[p. 163]{ref23}), we have the following useful inequality.
		
				\begin{proposition}
				\label{prop:2.5}
							If\ $\vartheta\in\ L^{s}(\Omega)$\ with\ $s>\frac{N}{2}$, $\vartheta(z)\leq \widehat{\lambda}_{1}$\ a.e.\ in\ $\Omega$\ and\ $\vartheta\neq \widehat{										 \lambda}_{1}$
				then\ there\ exists\ $c_{0}>0$\ such that
		$$\eta(u)=\sigma(u)-\int_{\Omega}\vartheta\ u^{2}dz\geq c_{0}||u||^{2}\qquad \mbox{for\ all}\ u\in H^{1}(\Omega).$$ 				
				\end{proposition}
				
		{\it Proof}. Evidently $\eta\geq 0$. Suppose that the result is not true. Then by virtue of the 2-homogeneity of the functional $\eta$, we can find $\{u_n\}_{n\geq 1}\subseteq H^{1}(\Omega)$ such that
			$||u_{n}||=1$  for all $n\geq 1$ and $\eta(u_{n})\downarrow 0$ as $n\rightarrow\infty.$
			We may assume that
					\begin{equation}
						\label{eq:13}
							u_{n}\stackrel{w}{\rightarrow}u\ \mbox{in}\ H^{1}(\Omega)\ \mbox{and}\ u_{n}\rightarrow u\ \mbox{in}\ L^{2}(\Omega).
					\end{equation}
			The functional $\sigma(\cdot)$ is sequentially weakly lower semi-continuous. Thus, for some $\xi\in\RR$,
					\begin{equation}
						\label{eq:14}
								\sigma(u)\leq \int_\Omega\vartheta u^{2}dz\quad \mbox{(see\  \eqref{eq:13})}
									\Rightarrow \sigma(u)=\widehat{\lambda}_{1}||u||^{2}_{2}\quad \mbox{(see\  \eqref{eq:10})}
									\Rightarrow u=\xi \widehat{u}_{1}\,.
					\end{equation}
			If $\xi=0$ then $u_{n}\rightarrow 0$ in $H^{1}(\Omega)$, which contradicts the fact that $||u_{n}||=1$ for all $n\geq 1$. So $\xi\neq 0$. By the Harnack inequality $\widehat{u}_1(z)>0$
		for all $z\in\Omega$ (see Pucci \& Serrin \cite[p. 163]{ref23}). Hence $|u(z)|>0$ for all $z\in\Omega$ and so from (\ref{eq:14}) we have
		$\sigma(u)<\widehat{\lambda}||u||^{2}_{2},$
		which contradicts (\ref{eq:10}).
		\qed
		
\smallskip
			For every $x\in \RR$, we set $x^{\pm}=\max\{\pm x,0\}$ and for $u\in H^{1}(\Omega)$, we define
		$u^{\pm}(\cdot)=u(\cdot)^{\pm}.$
			We know that
			$u^{\pm}\in H^{1}(\Omega)$, $|u|=u^{+}+u^{-}$, and $u=u^{+}-u^{-}.$
			
			Given a measurable function $h:\Omega \times \RR\rightarrow \RR$ (for example a Carath\'eodory function), we define
			$$N_{h}(u)(\cdot)=h(\cdot,u(\cdot))\qquad \mbox{for\ all}\ u\in H^{1}(\Omega)$$
		(the Nemytsky map corresponding to $h$). Also, $A\in$ $\mathcal{L}$$(H^{1}(\Omega),H^{1}(\Omega)^{*})$ is defined by
			$$\left\langle A(u),y \right\rangle=\int_{\Omega}(Du,Dy)_{\RR^{N}}dz\qquad \mbox{for\ all}\ u,y\in H^{1}(\Omega).$$
		Finally, by $|\cdot|_{N}$ we denote the Lebesgue measure on $\RR^{N}$.
			
		\section{Solutions of Constant Sign}
				In this section we produce solutions of constant sign for problem (1). In section 4 we have the full multiplicity theorems.
				
				The hypotheses on the data of (1), are the following:\\
					$H_{1}:\ \beta\in L^{s}(\Omega)\ with\ s>N\ and\ \beta^{+}\in L^{\infty}(\Omega).$\\
					$H_{2}:\ f:\Omega\times \RR\rightarrow \RR\ is\ measurable\ such\ that\ for\ a.a.\ z\in\Omega, f(z,0)=$ $0, f(z,\cdot)\in C^{1}(\RR)\ and$
					$\ (i)\ |f^{'}_{x}(z,x)|\leq a(z)(1+|x|^{r-2})\ for\ a.a.\ z\in\Omega, all\ x\in \RR,\ with\ a\in L^{\infty}(\Omega), 2\leq r<2^{*};$\\
					$(ii)\ there\ exist\ functions\ \theta,\widehat{\theta}\in L^{\infty}(\Omega) \ such\ that$
						\begin{eqnarray}
							&&	\theta(z)\leq \widehat{\lambda},\ a.e.\ in\ \Omega,\ \theta\neq\widehat{\lambda}_{1}\ and \nonumber \\
							&&	\widehat{\theta}(z)\leq \liminf_{x\rightarrow+\infty}\frac{f(z,x)}{x}\leq\limsup_{x\rightarrow+\infty}\frac{f(z,x)}{x}\leq\theta(z)\ uniformly\ for\ 										a.a.\ z\in\Omega;\nonumber
						\end{eqnarray}
					$(iii)\ there\ are\ functions\ \eta,\widehat{\eta}\in L^{\infty}(\Omega)\ such\ that$
						\begin{eqnarray}
							&&	\eta(z)\geq\widehat{\lambda},\ a.e.\ in\ \Omega,\ \eta\neq\widehat{\lambda}_{1}\ and \nonumber \\
							&&	 \eta(z)\leq\liminf_{x\rightarrow-\infty}\frac{f(z,x)}{x}\leq
\limsup_{x\rightarrow-\infty}\frac{f(z,x)}{x}\leq\widehat{\eta}(z)\ uniformly\ for\ a.a.\ 										 z\in\Omega;\nonumber
						\end{eqnarray}
					$(iv)\ there\ exists\ integer\ \ell\geq 2\ such\ that$
						\begin{eqnarray}
							&&	 f^{'}_{x}(z,0)\in[\widehat{\lambda}_{\ell},\widehat{\lambda}_{\ell+1}]\ a.e.\ in\ \Omega\,\ f_x'(\cdot ,0)\not=\widehat{\lambda}_{\ell},\ f_x'(\cdot ,0)\not=\widehat{\lambda}_{\ell+1} \ \mbox{and}\nonumber \\
							&&	f^{'}_{x}(z,0)=\lim_{x\rightarrow 0}\frac{f(z,x)}{x}\ uniformly\ for\ a.a.\ z\in \Omega.\nonumber
						\end{eqnarray}
							
				\textbf{Remark:} Hypotheses $H_{2}(ii),\,(iii)$ classify this nonlinearity as ``crossing" or ``jumping" since as we move from $-\infty$ to $+\infty$ the quotient $\frac{f(z,x)}{x}$ crosses at least the principal eigenvalue $\widehat{\lambda}_{1}$. This asymmetric behavior of $f(z,\cdot)$ makes it impossible to use the methods and techniques of the papers mentioned in the Introduction.
			
				Let $\varphi:H^{1}(\Omega)\rightarrow \RR$ be the energy functional for problem (1) defined by								 $$\varphi(u)=\frac{1}{2}\sigma(u)-\int_{\Omega}F(z,u)dz\quad \mbox{for\ all}\ u\in H^{1}(\Omega).$$
				
				Evidently $\varphi\in C^{2}(H^{1}(\Omega))$. Also let $\widehat{\gamma}>0$ be as in (\ref{eq:8}). We introduce the following perturbations-truncations of the reaction $\rho(z,\cdot)$:		 
				$$	\widehat{f}_{+}(z,x)=f(z,x^{+})+\widehat{\gamma}x^{+} \qquad\mbox{and}\qquad
						\widehat{f}_{-}(z,x)=f(z,-x^{-})+\widehat{\gamma}(-x^{-}).
				$$
				
				Both are Carath\'eodory functions. We set $\widehat{F}_{\pm}(z,x)=\int^{x}_{0}\widehat{f}_{\pm}(z,s)ds$ and consider the $C^{1}$- functionals $\widehat{\varphi}_{\pm}:H^{1}(\Omega)\rightarrow {\mathbb R}$ defined by			 $$\widehat{\varphi}_{\pm}(u)=\frac{1}{2}\sigma(u)+\frac{\widehat{\gamma}}{2}||u||^{2}_{2}-
\int_{\Omega}\widehat{F}_{\pm}(z,u)dz\qquad \mbox{for\ all}\ u\in H^{1}(\Omega).$$
					
				\begin{proposition}
					\label{prop:3.1}
						If\ hypotheses\ $H_{1}$\ and\ $H_{2}$\ hold,
	then\  $\varphi$\ satisfies\ the\ C-condition.
				\end{proposition}
				
				\textit{Proof.} Let $\{u_{n}\}_{n\geq 1}\subseteq H^{1}(\Omega)$ be a sequence such\ that $\{\varphi(u_{n})\}_{n\geq 1}\subseteq {\mathbb R}$ is bounded and					
					\begin{equation}
						\label{eq:15}
							(1+||u_{n}||)\varphi^{'}(u_{n})\rightarrow 0\ \mbox{in}\ H^{1}(\Omega)^{*}.
					\end{equation}
								From (\ref{eq:15}) we have
										\begin{equation}
											\label{eq:16}
											\begin{array}{l}
													|\left\langle \varphi^{'}(u_{n}),h \right\rangle|\leq \frac{\epsilon_{n}||h||}{1+||u_{n}||}\ \mbox{for\ all}\ h\in H^{1}(\Omega)\ \mbox{with}\ \epsilon_{n}\rightarrow 0 \\
													\Rightarrow |\int_\Omega u_nhdz+ \int_{\Omega}\beta u_{n}hdz-\int_{\Omega}f(z,u_{n})hdz|\leq \frac{\epsilon_{n}||h||}{1+||u_{n}||}\ \mbox{for\ all}\ n\geq 1.
											\end{array}
										\end{equation}								
								In (\ref{eq:16}) we choose $h=u^{+}_{n}\in H^{1}(\Omega)$. Then																				 \begin{equation}
											\label{eq:17}												 ||Du^{+}_{n}||^{2}_{2}+\int_{\Omega}\beta(u^{+}_{n})^{2}dz-\int_{\Omega}f(z,u^{+}_{n})u^{+}_{n}dz\leq \epsilon_{n}\qquad \mbox{for\ all}\ n\geq 1.
										\end{equation}											
								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{equation}
											\label{eq:18}
												y_{n}\stackrel{w}{\rightarrow}y\ \mbox{in}\ H^{1}(\Omega)\ \mbox{and}\ y_{n}\rightarrow y\ \mbox{in}\ L^{2s^{'}}(\Omega)\ (\frac{1}{s}+\frac{1}{s^{'}}=1).
										\end{equation}										
								From (\ref{eq:17}) we have										
										\begin{equation}
											\label{eq:19}												 \sigma(y_{n})-\int_{\Omega}\frac{f(z,u^{+}_{n})}{||u^{+}_{n}||}y_{n}dz\ \leq\ \frac{\epsilon_{n}}{||u^{+}_{n}||^{2}}\ \mbox{for\ all}\ n\geq 1.
										\end{equation}
								From hypotheses $H_{2}(i),(ii)$ we obtain
									$$\left\{\frac{N_{f}(u^{+}_{n})}{||u^{+}_{n}||}\right\}_{n\geq 1}\subseteq L^{2}(\Omega)\ \mbox{is\ bounded}.$$
								So, we may assume that
										\begin{equation}
											\label{eq:20}
												 \frac{N_{f}(u^{+}_{n})}{||u^{+}_{n}||}\stackrel{w}{\rightarrow}g\ \mbox{in}\ L^{2}(\Omega).
										\end{equation}
								Moreover, hypothesis $H_2(ii)$ implies that
										\begin{equation}
											\label{eq:21}
												g=\theta_{0}y\ \mbox{with}\	 \widehat{\theta}(z)\leq\theta_{0}(z)\leq\theta(z)\ \mbox{a.e.\ in}\ \Omega.
										\end{equation}
								So, if in (\ref{eq:19}) we pass to the limit as $n\rightarrow\infty$ and use (\ref{eq:18}), (\ref{eq:20}), (\ref{eq:21}), then
									$$	\sigma(y)-\int_{\Omega}\theta_{0}y^{2}dz\leq 0
											\Rightarrow\ C_{0}||y||^{2}\leq 0\ \mbox{(see\ Proposition\ \ref{prop:2.5})}\Rightarrow\ y=0.$$									
								Then
								$$	\sigma(y_{n})\rightarrow 0
\Rightarrow\ Dy_{n}\rightarrow 0\ \mbox{in}\ L^{2}(\Omega,\ \RR)
											\Rightarrow\ y_{n}\rightarrow 0\ \mbox{in}\ H^{1}(\Omega)\ \mbox{(see\ (\ref{eq:18}))},$$												 which contradicts the fact that $||y_{n}||=1$	for all $n\geq 1$. This proves that
										\begin{equation}
											\label{eq:22}
												\{u_n\}_{n\leq 1}\subset H^{1}(\Omega)\ \mbox{is\ bounded.}
										\end{equation}
								From (\ref{eq:16}) and (\ref{eq:22}), we have for all $n\geq 1$ and for some $M_1>0$,
										\begin{equation}
											\label{eq:23}
												|\left\langle A(-u^{-}_{n}),h \right\rangle +\int_{\Omega}\beta(-u^{-}_{n})hdz-\int_{\Omega}f(z,-u^{-}_{n})hdz|\leq M_{1}||h||\,.
										\end{equation}
								Suppose that $||u^{-}_{n}||\rightarrow\infty$. We set $v_{n}=\frac{u^{-}_{n}}{||u^{-}_{n}||},\ n\geq 1$. Then $||v_{n}||=1$ for all $n\geq 1$ and so we may assume that
										\begin{equation}
											\label{eq:24}
												v_{n}\stackrel{w}{\rightarrow}v\ \mbox{in}\ H^{1}(\Omega)\ \mbox{and}\ v_{n}\rightarrow v\ \mbox{in}\ L^{2}(\Omega).
										\end{equation}
								From (\ref{eq:23}), we have for\ all\ $n\geq 1$,
										\begin{equation}
											\label{eq:25}
											|\left\langle A(-v^{-}_{n}),h \right\rangle +\int_{\Omega}\beta(-v_{n})hdz-\int_{\Omega}\frac{f(z,-u^{-}_{n})}{||u^{-}_{n}||}hdz|\leq\frac{M_{1}||h||}{||u^{-}_{n}||}
												\,.
										\end{equation}
								Again we have that $\left\{\frac{N_{f}(-u^{-}_{n})}{||u^{-}_{n}||}\right\}_{n\geq 1}\subseteq L^{2}(\Omega)$ is bounded and by virtue of hypothesis $H_{2}(iii)$ we have
										\begin{equation}
											\label{eq:26}												 \frac{N_{f}(-u^{-}_{n})}{||u^{-}_{n}||}\stackrel{w}{\rightarrow}g_{*}=\eta_{0}v\ \mbox{in}\ L^{2}(\Omega)\ \mbox{with}\ \eta(z)\leq\eta_{0}(z)\leq\widehat{\eta}(z)\ \mbox{for\ a.a.}\ z\in\Omega.
										\end{equation}
								So, if in (\ref{eq:25}) we pass to the limit as $n\rightarrow\infty$ and use (\ref{eq:24}) and (\ref{eq:26}), then
										\begin{equation}
											\label{eq:27}
												\begin{array}{l}
													\left\langle A(-v),h \right\rangle +\int_{\Omega}\beta(-v)hdz=\int_{\Omega}\eta_{0}(-v)hdz\ \mbox{for\ all}\ h\in H^{1}(\Omega)\\
													\Rightarrow A(v)+\beta v=\eta_{0}v,\\
													\Rightarrow -\Delta v(z)+\beta(z)v(z)=\eta_{0}(z)v(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\theta v}{\theta n}=0\ \mbox{on}\ \theta\Omega.
												\end{array}	
										\end{equation}
										
								Also, if in (\ref{eq:25}) we choose $h=v_{n}-v\in H^{1}(\Omega)$, pass to the limit as $n\rightarrow\infty$ and use (\ref{eq:24}) and (\ref{eq:26}), then
										\begin{equation}
											\label{eq:28}
												\begin{array}{l}
													\lim_{n\rightarrow\infty}\left\langle A(-v_{n}), v_{n}-v \right\rangle=0\Rightarrow||Dv_{n}||_{2}\rightarrow||Dv||_{2}\\													 
													\Rightarrow v_{n}\rightarrow v\ \mbox{in}\ H^{1}(\Omega)\ \mbox{(by\ the\ Kadec-Klee\ property,\ see\ (\ref{eq:24}))}\\
													\Rightarrow ||v||=1,v\geq 0.
												\end{array}
										\end{equation}	
								By virtue of Proposition \ref{prop:2.4}, we have
										\begin{equation}
											\label{eq:29}
												 \tilde{\lambda}_{1}(\eta_{0})\leq\tilde{\lambda}_{1}(\eta)<\tilde{\lambda}_{1}(\widehat{\lambda}_{1})=1.
										\end{equation}
																From (\ref{eq:27}) and (\ref{eq:29}) it follows that $v\neq 0$ (see (\ref{eq:28})) is nodal, a contradiction.
						This means that
								$$	\{u^{-}_{n}\}_{n\geq 1}\subseteq H^{1}(\Omega)\ \mbox{is\ bounded}
							\Rightarrow\ \{u_{n}\}_{n\geq 1}\subseteq H^{1}(\Omega)\ \mbox{is\ bounded\ (see\ (\ref{eq:22}))}.$$
																Hence we may assume that	
									\begin{equation}
										\label{eq:30}
											u_{n}\stackrel{w}{\rightarrow}u\ \mbox{in}\ H^{1}(\Omega)\ \mbox{and}\ u_{n}\rightarrow u\ \mbox{in}\ L^{2}(\Omega).
									\end{equation}
								In (\ref{eq:16}) we choose $h=u_{n}-u\in H^{1}(\Omega)$, pass to the limit as $n\rightarrow\infty$ and use (\ref{eq:30}). Then
								$$	\lim_{n\rightarrow\infty}\left\langle A(u_{n}),u_{n}-u \right\rangle=0	 \Rightarrow||Du_{n}||_{2}\rightarrow||Du||_{2}	\Rightarrow u_{n}\rightarrow u\ \mbox{in}\ H^{1}(\Omega)\,.
								$$								
								This proves that $\varphi$ satisfies the C-condition. \qed
								
								
							\begin{proposition}
								\label{prop:3.2}
										If\ hypotheses\ $H_{1}$\ and\ $H_{2}$\ hold,
	then\ problem\ \eqref{eq:1}\ has\ a\ solution\ $u_{0}$ in $\mbox{int}\,C_{+}$\
											which\ is\ a\ local\ minimizer\ of\ the\ functional\ $\varphi$.
							\end{proposition}
					
	\textit{Proof.}	Hypotheses $H_{2}(i),(ii)$ imply that given $\epsilon>0$, we can find $C_{\epsilon}>0$ such\ that
		\begin{equation}
			\label{eq:31}
				F(z,x)\leq\frac{1}{2}(\theta(z)+\epsilon)x^{2}+C_{\epsilon}\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{ and for all}\ x\geq 0.
		\end{equation}
					Then for all $u\in H^{1}(\Omega)$ we have							
							\begin{eqnarray}
										\widehat{\varphi}_{+}(u)= && \frac{1}{2}\sigma(u)+\frac{\widehat{\gamma}}{2}||u||^{2}_{2}-\int_{\Omega}\widehat{F}_{+}(z,u)dz \nonumber \\
										&&	\geq\frac{1}{2}[\sigma (u)-\int_{\Omega}\theta u^{2}dz]-\frac{\epsilon}{2}||u||^{2}-C_{\epsilon}|\Omega|_{N}\ (\mbox{see}\ (\ref{eq:31})) \nonumber \\
										&&	 \geq\frac{C_{0}-\epsilon}{2}||u||^{2}-C_{\epsilon}|\Omega|_{N}\ (\mbox{see Prop.}\ \ref{prop:2.5}).\label{eq:32}
							\end{eqnarray}	
							
					Choosing $\epsilon\in(0,C_{0})$, from (\ref{eq:32}) we infer that $\widehat{\varphi}_{+}$ is coercive. Also, it is sequentially weakly lower semi-continuous.
						So, by the Weierstrass theorem, we can find $v_{0}\in H^{1}(\Omega)$ such\ that
							\begin{equation}
								\label{eq:33}
									\widehat{\varphi}_{+}(u_0)=\inf\{\widehat{\varphi}_{+}(u):u\in H^{1}(\Omega)\}.
							\end{equation}
					Hypothesis $H_{2}(iv)$ implies that we can find $\delta>0$ and $\epsilon>\widehat{\lambda}_{1}$ (recall $\ell\geq 2$, see $H_{2}(iv)$) such\ that
							\begin{equation}
								\label{eq:34}
									F(z,x)\geq\frac{\epsilon}{2}x^{2}\ \mbox{for\ a.a.}\ z\in\Omega,\ \mbox{all}\ x\in[0,\delta]\,.
							\end{equation}
					Since $\widehat{u}_{1}\in \mbox{int}\,C_{+}$ (see Section \ref{sec:mathback}), we can find $t\in(0,1)$ small such\ that $t\widehat{u}_{1}(z)\in[0,\delta]$ for all $z\in\bar{\Omega}$. Then
					\begin{eqnarray}
						\widehat{\varphi}_{+}(t\widehat{u}_{1})= && \frac{t^{2}}{2}\sigma(\widehat{u}_{1})-\int_{\Omega}F(z,t\widehat{u}_{1})\ \mbox{(recall\ the\ definition\ of\ $\widehat{f}_{+}(z,x)$)}						\nonumber \\
						&&	\leq\frac{t^{2}}{2}[\widehat{\lambda}_{1}-\epsilon]\ \mbox{(see\ (\ref{eq:34})\ and\ recall\ that\ $||\widehat{u}_{1}||_{2}=1$)}\nonumber \\
						&&	<0\ \mbox{(since\ $\epsilon>\widehat{\lambda}_{1}$)}
							\Rightarrow\widehat{\varphi}_{+}(u_{0})<0=\widehat{\varphi}_{+}\ \mbox{(see\ (\ref{eq:33}))},\nonumber
					\end{eqnarray} hence\ $u_{0}\neq 0$.
										Therefore
							\begin{equation}
								\label{eq:35}
										\widehat{\varphi}^{'}_{+}(u_{0})=0
										\Rightarrow A(u_{0})+(\beta+\widehat{\gamma})u_{0}=N_{\widehat{f}_{+}}(u_{0}).
															\end{equation}
					On (\ref{eq:35}) we act with $-u^{-}_{0}\in H^{1}(\Omega)$. Then
											 $$\sigma(u^{-}_{0})+\widehat{\gamma}||u^{-}_{0}||^{2}_{2}=0
							\Rightarrow \widehat{c}||u^{-}_{0}||^{2}\leq 0\ \mbox{(see\ (\ref{eq:8})),\ hence\ $u_{0}\geq 0$,\ $u_{0}\neq 0$.}
					$$
						
					Then relation (\ref{eq:35}) becomes
\begin{equation}\label{eq36}
		A(u_0)+\beta u_0 = N_f(u_0)
  	\Rightarrow -\Delta u_0+\beta u_0=f(z,u_0)\:\mbox{in}\:\Omega, \frac{\partial u_0}{\partial n}=0\:\mbox{on}\:\partial \Omega \,.
\end{equation}

	Hypotheses $H_2$ imply
\begin{equation}
	\label{eq:37}
		|f(z,x)|\leq\:c_1|x|\:\mbox{for\:a.a.}\:z\in\Omega,\:\mbox{all}\:x\in\RR\: \mbox{and\:some}\:c_1>0\,.
\end{equation}

	We set
\begin{center}
 $\zeta(s)=
  \left\{
   \begin{array}{cc}
    \frac{f(z,u_0(z))}{u_0(z)}		& \mbox{if}\:u_0(z)\neq0\\
		\\
		0                             & \mbox{otherwise}.
     \end{array}
   \right.$
\end{center}

	Evidently $\zeta \in L^ \infty(\Omega)$ (see (\ref{eq:37})). From (\ref{eq36}) we have
\begin{equation}
	\label{eq:38}
		-\Delta u_0(z)=(\zeta-\beta)(z)u_0(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial u_0}{\partial n}=0\ \mbox{on}\ \partial\Omega\,.
\end{equation}

	Note that $\zeta - \beta \in L^s(\Omega)$ with $s>N$ (see $H_1$). Lemma 5.1 of Wang \cite{ref26} implies that $u_0 \in L^\infty(\Omega)$. Then from (\ref{eq:38}) we have $\Delta u_0 \in L^s(\Omega)$. Invoking Lemma 5.2 of Wang \cite{ref26}, we have $u_0 \in W^{2,s}(\Omega)$. Since $s>N$, from the Sobolev embedding theorem we have $W^{2,s}(\Omega)\subset C^{1+\alpha}(\bar{\Omega})$ with $\alpha=1-\frac{N}{s}>0$. Therefore $u_0 \in C_+ \setminus \{0\}$. From (\ref{eq:38}) we have
\begin{eqnarray}
	&&	\Delta u_0(z) \leq\ (||\zeta||_\infty + ||\beta^+||_\infty)\ u_0(z)\ \mbox{a.e.\ in}\ \Omega\ \mbox{(see\ $H_{1}$)}\nonumber\\
	&&	\Rightarrow\ u_0\ \in\ \mbox{int}\, C_+\ \mbox{(see Pucci \& Serrin\ \cite[p. 120]{ref23}\ and\ V\'{a}zquez\ \cite{ref25})}. \nonumber
\end{eqnarray}

	Note that $\widehat{\varphi}_+|_{C_+} = \varphi|_{C_+}$. So, $u_0 \in \mbox{int}\,C_+$ is a local $C^{1}(\bar{\Omega})$--minimizer of $\varphi$, hence by virtue of Proposition \ref{prop:2.1} $u_o \in \mbox{int}\,C_+$ is a local $H^1(\Omega)$--minimizer of $\varphi$.
\qed

	In fact we can show that problem (1) has a smallest nontrivial positive solution.	
	
\begin{proposition}
	\label{prop:3.3}
		Assume that hypotheses\ $H_1$\ and\ $H_2$\ hold.
			Then\ problem\ \eqref{eq:1}\ has\ a\ smallest\ nontrivial\ positive\ solution\ $u_+ \in \mbox{int}\, C_+$
			(that is,\ if\ $u$\ is\ a\ nontrivial\ positive\ solution\ of\ \eqref{eq:1},\ then\ $u_+ \leq u$).	
\end{proposition}

\textit{Proof}: Let $S_+$ be the set of nontrivial positive solutions of (1). From Proposition \ref{prop:3.2} and its proof, we have
$S_+ \neq \O$ and $S_+ \subseteq\;\mbox{int}\,C_+$.

	We know that $S_+$ is downward directed (that is, if $u_1,u_2 \in S_+$, then we can find $u \in S_+$ such\ that $u\leq u_1, u\leq u_2$, see Aizicovici, Papageorgiou \& Staicu \cite[p. 703]{ref1}). So, without any loss of generality, we may assume that there exists $M_2 > 0$ such\ that $u(z)\leq M_2$ for all $z \in \bar{\Omega}$, all $u \in S_+$.

	Let $C\subseteq S_+$ be a chain (a totally ordered subset of $S_+$). From Dunford \& Schwartz \cite[p. 336]{ref8}, we know that we can find $\{u_n\}_{n\geq 1} \subseteq C$ such\ that
$\inf C=\inf_{n\geq 1} u_n\,.$

	We have for all $n\geq 1$,
\begin{equation}\label{eq:39}	A(u_n)+\beta u_n\:=\:N_f(u_n)
			\Rightarrow \{u_n\}_{n\geq 1} \subseteq H^1(\Omega)\ \mbox{is\;bounded}.
\end{equation}
So, we may assume that
\begin{equation}
	\label{eq:40}
		u_n \stackrel{w}{\rightarrow} u\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ u_n\rightarrow u\ \mbox{in}\ L^2(\Omega).
\end{equation}

	Moreover, acting on (\ref{eq:39}) with $u_n-u \in H^1(\Omega)$, passing to the limit as $n\rightarrow \infty$ and using (\ref{eq:40}) and the Kadec-Klee property of Hilbert spaces, we obtain
\begin{equation}
	\label{eq:41}
		u_n \rightarrow u\ \mbox{in}\ H^1(\Omega)\,.
\end{equation}

	Then passing to the limit as $n \rightarrow \infty$ in (\ref{eq:39}) and using (\ref{eq:40}), we have
$$
		A(u)+ \beta u\:=\:N_f(u)
		\Rightarrow\:u\!\in\!C_+\ \mbox{is\ a\ solution\ of}\ \eqref{eq:1}.
$$

	If we show that $u\neq 0$, then $u \in S_+$. Suppose that $u=0$ and let $y_n = \frac{u_n}{||u_n||}\,,\ n\geq1$. Then $||y_n||=1$ for all $n\geq 1$ and so we may assume that
\begin{equation}
	\label{eq:42}
		y_n \stackrel{w}{\rightarrow} y\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y_n\:\rightarrow\:y\ \mbox{in}\ L^2(\Omega),\;y\geq 0.
\end{equation}
	From (\ref{eq:39}) we have
\begin{equation}
	\label{eq:43}
		A(y_n)+ \beta y_n\:=\:\frac{N_f(u_n)}{||u_n||}\ \mbox{for\ all}\ n\geq 1.
\end{equation}

	Since $u_n \rightarrow 0$ in $H^1(\Omega)$ (see (\ref{eq:41})), by virtue of hypothesis $H_2(iv)$, we have
\begin{equation}
	\label{eq:44}
		\frac{N_f(u_n)}{||u_n||}\:\stackrel{w}{\rightarrow}\:\mu_0\:=\:my\ \mbox{in}\ L^2(\Omega),\ \mbox{where}\ m(\cdot)=f^{'}_{x}(\cdot,0)\:\in\:L^\infty(\Omega).
\end{equation}

	So, from (\ref{eq:42}) and (\ref{eq:43}) it follows that
\begin{eqnarray}
	&&	A(y)+\beta y\!=\!my\nonumber \\
	&&	\Rightarrow\:-\Delta y(z)+\beta(z)y(z)\:=\:m(z)y(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial y}{\partial n}=0\ \mbox{on}\ \partial\Omega. \label{eq:45}
\end{eqnarray}

	From Proposition \ref{prop:2.4} and hypothesis $H_2(iv)$, we have
\begin{equation}
	\label{eq:46}
		\tilde{\lambda}_l(m)<\tilde{\lambda}_l(\tilde{\lambda}_l)\!=\!1\ \mbox{and}\ 1\!=\!\tilde{\lambda}_{l+1}(\tilde{\lambda}_{l+1})<\tilde{\lambda}_{l+1}(m).
\end{equation}

	From (\ref{eq:45}) and (\ref{eq:46}) it follows that $y=0$. On the other hand, acting on (\ref{eq:43}) with $y_n-y \in H^1(\Omega)$ and using (\ref{eq:42}) and (\ref{eq:44}), we obtain
$$	\lim_{n\ri\infty}\left\langle A(y_n),\!y_n-y\right\rangle\:=\:0
	\Rightarrow\:y_n\!\rightarrow y\ \mbox{in}\ H^1(\Omega)\ \mbox{(as\ before)}
	\Rightarrow\:||y||\!=\!1,
$$
a\ contradiction.
	Therefore
$u \in S_+$ and\ $u\in \mbox{int}\,C$.

	Since C is an arbitrary chain, from the Kuratowski-Zara lemma it follows that $S_+$ has a minimal element $u_+ \in S_+ \subseteq \mbox{int}\, C_+$. If $u \in S_+$, then since $S_+$ is downward directed, we can find $\tilde{u} \in S_+$ such\ that $\tilde{u} \leq u_t, \tilde{u} \leq u$. The minimality of $u_+$ implies that $\tilde{u}=u_+$ and so $u_+ \leq u$. Therefore $u_+$ is the smallest nontrivial positive solution of problem (1).
	\qed

\smallskip
	Let $S\_$ be the set of nontrivial negative solutions of problem \eqref{eq:1}.
	
	In general hypotheses $H_2$ do not guarantee that $S\_\not=\emptyset$. If $S\_\neq \O$, then $S\_ \subseteq -\mbox{int}\,C_+$ and it is upward directed (that is, $v_1, v_2 \in S\_$, then we can find $v \in S\_$ such\ that $v_1 \leq v_1,\ v_2 \leq v$, see \cite{ref1}).
	
		Reasoning as in the proof of Proposition \ref{prop:3.3}, we have:
		
\begin{proposition}
	\label{prop:3.4}
	Assume that hypotheses $H_1$ and $H_2$ hold and $S\_ \neq \O$.
		Then problem\: \eqref{eq:1}\:has\:a\:biggest\:nontrivial\:negative\:solution\: $v\_\in -\mbox{int}\,C_+$
			(that is,\:if\: $v$ is a nontrivial negative solution of \eqref{eq:1},\:then\: $v \leq v\_$).
\end{proposition}

	If we strengthen the conditions on $f(z,\cdot)$ we can guarantee that $S\_ \neq \O$. These stronger conditions on the reaction $f$ are the following:
\begin{eqnarray}
	& H_3: &	f:\Omega \times \RR \rightarrow \RR\ is\ a\ measurable\ function\ such\ that\ \nonumber \\
	&			 &	for\ a.a.\ z\in\Omega,\ f(z,0)=0,\ f(z,\cdot)\ \in C^{1}(\RR),\ hypotheses\nonumber \\
	& 			&	H_3(i)\rightarrow (iv)\ are\ the\ same\ as\ the\ corresponding\ hypotheses\ H_2(i)\rightarrow (iv)\ \nonumber \\
	&			&	and (v)\ there\ exists\ \xi_x>0\ such\ that\ f(z,-\xi_x)\geq(-\xi_x)\beta(z)\ a.e.\ in\ \Omega \,. \nonumber
\end{eqnarray}

\begin{remark}
	$If\ \beta \equiv 0,\ then\ condition\ H_3(v)\ implies\ that\ for\ a.a.\ z\in \Omega,\ f(z,\cdot)\\
	\ has\ a\ zero\ in\ (-\infty, 0).$
\end{remark}

\begin{proposition}
\label{prop:3.5}
	If\ hypotheses\ $H_1$\ and\ $H_3$\ hold,\ then\ $S\_\neq \O$,\ $S\_ \subseteq -\mbox{int}\,C_+$.
\end{proposition}

\textit{Proof.} We consider the following perturbation-truncation of the reaction $f(z,\cdot)$:
\begin{equation}
\label{eq:47}
	\tau(z,x)=
	\left\{
     \begin{array}{lc}
       f(z, - \xi_x) + \widehat{\gamma}(-\xi_x) & \mbox{if}\ x<-\xi_x\\
			 f(z, x) + \widehat{\gamma} x           & \mbox{if}\ -\xi_x \leq x \leq 0\\
			 0                                  & \mbox{if}\ 0<x.
     \end{array}
    \right.
\end{equation}

Clearly $\tau(\cdot,\cdot)$ is a Carath\'eodory function. We set $T(z,x)=\int^{x}_{0}\tau(z,s)ds$ and consider the $C^1$-functional $\chi_0:H^1(\Omega)\rightarrow \RR$ defined by
$$\chi_0(u) = \frac{1}{2}\,\sigma (u)\:+\:\frac{\widehat{\gamma}}{2}\,||u||^{2}_{2}\:-\:\int_{\Omega}T(z,u)dz\qquad \mbox{for\ all}\ u\in H^1(\Omega).$$

	From (\ref{eq:47}) and (\ref{eq:8}), we see that $\chi_0$ is coercive. Also, it is sequentially weakly lower semi-continuous. Se, we can find $v_0 \in H^1(\Omega)$ such\ that
\begin{equation}
	\label{eq:48}
		\chi_0(v_0)\ =\inf\{\chi_0(v):v \in H^1(\Omega)\}.
\end{equation}

	As in the proof of Proposition \ref{prop:3.2}, for $t \in (0,1)$ small such\ that at least we have $t \widehat{u}_1(z) \in [0,\xi_x]$ for all $z \in \bar{\Omega}$, we have $\chi_0(-t\widehat{u}_1)\ <\ 0$. Therefore
$\chi_0(v_0)<0= \chi_0(0)$ (see (\ref{eq:48})), hence $v_0\neq 0$.


	From (\ref{eq:48}) we have
\begin{equation} \label{eq:49}	\chi^{'}_{0}(v_0)=0
		\Rightarrow\ A(v_0)\:+\:(\beta+\widehat{\gamma})v_0\ =\ N_{\tau}(v_0)\,.
\end{equation}

	On (\ref{eq:49}) we act with $(-\xi_x - v_0)^+ \in H^1(\Omega)$. Then
\begin{eqnarray}
	&&	\left\langle A(v_0),\ (-\xi_x -v_0)^+\right\rangle\ +\ \int_{\Omega}(\beta(z)+\widehat{\gamma})\:v_0\:(-\xi_x -v_0)^+\ dz \nonumber \\
	&&	=\ \int_{\Omega}\ \tau(z,v_0)\:(-\xi_x-v_0)^+\ dz \nonumber \\
	&&	=\ \int_{\Omega}\ [f(z_1-\xi_x)\:+\:\widehat{\gamma}-\xi_x]\ (-\xi_z-v_0)^+\ dz\ \ \mbox{(see\ (\ref{eq:47}))}\nonumber \\
	&&	\geq \ \left\langle A(-\xi_x),(-\xi_x-v_0)^+\right\rangle\ +\ \int_{\Omega}(\beta(z)+\widehat{\gamma})(-\xi_x)(-\xi_x-v_0)^+\ dz,\nonumber \\
	&&	\Rightarrow\ \left\langle A(-\xi_x-v_0),\ (-\xi_x-v_0)^+\right\rangle\ +\ \int_{\Omega}\ (\beta(z)+\widehat{\gamma})(-\xi_x-v_0)(-\xi_x-v_0)^+\ dz\ \leq\ 0, \nonumber \\	
	&&	\Rightarrow\ \sigma((-\xi_x-v_0)^+)\ +\ \widehat{\gamma}\:||(-\xi_x-v_0)^+||^{2}_{2}\ \leq\ 0,\nonumber \\
	&&	\Rightarrow\ -\xi_x\ \leq\ v_0\ \ \mbox{(see\ (\ref{eq:8}))}.\nonumber
\end{eqnarray}
	Also, acting on (\ref{eq:49}) with $v^{+}_{0} \in H^1(\Omega)$, we obtain $v_0 \leq 0, v_0 \neq 0$. Therefore	
$$v_0\ \in\ [-\xi_x,0]\ =\ \{v\in H^1(\Omega):\:-\xi_x\:\leq\:v(z)\:\leq\:0\ \mbox{a.e.\ in}\ \Omega\}.$$
	Therefore (\ref{eq:49}) yields
$	A(v_0)+\beta v_0=N_f(v_0)$, hence
	$v_0$ is a nontrivial negative solution of problem \eqref{eq:1}.
	As before, from Wang \cite{ref26} and the strong maximum principle, we deduce that $v_0 \in -\mbox{int}\,C_+.$
	\qed


\section{Nodal Solutions}
	
	In this section we present the full multiplicity theorems for problem (1) by producing nodal solutions.
	
	First we treat the case $S\_ \neq \O$.

\begin{theorem}\label{th:4.1}
	Assume that hypotheses\ $H_1$\ and\ $H_2$\ hold\ and\  $S\_\neq \O$.
		Then\ problem\ \eqref{eq:1}\ has\ at\ least\ four\ nontrivial\ solutions:
		$u_0\in\mbox{int}\,C_+$, $v_0\in -\mbox{int}\, C_+$ and $y_0,\, \widehat{y}\in\mbox{int}_{C^1(\bar{\Omega})}[v_0,u_0]$\ both\ nodal.
\end{theorem}

	\textit{Proof.} From Proposition \ref{prop:3.2} we already have a nontrivial positive solution $u_0 \in \mbox{int}\,C_+$. In fact, by virtue of Proposition \ref{prop:3.3}, we may assume that $u_0$ is the smallest nontrivial positive solution of (1) (that is, $u_0=u_+ \in \mbox{int}\,C_+$). Similarly, since by hypothesis $S\_\neq \O$, from Proposition \ref{prop:3.4} we can have a nontrivial negative solution $v_0 \in -\mbox{int}\,C_+$ which can be taken to be the biggest such solution of (1) (that is, $v_0=v\_ \in -\mbox{int}\,C_+$). We introduce the following perturbation-truncation of $f(z,\cdot)$:	
\begin{equation}
\label{eq:50}
	g(z,x) =
	\left\{
      \begin{array}{lc}
        f(z, v_0(z))+\widehat{\gamma}v_0(z)      &  \mbox{if}\ x<v_0(z)\\
				f(z, x)+\widehat{\gamma}x                &  \mbox{if}\ v_0(z)\leq x\leq u_0(z)\\
				f(z,u_0(z))+\widehat{\gamma}u_0(z)       &  \mbox{if}\ u_0(z)<x.
      \end{array}
   \right.
\end{equation}
	This is a Carath\'eodory function. We set $G(z,x) = \int^{x}_{0} g(z,s) ds$ and introduce the $C^1$-functional $\Psi: H^1(\Omega) \rightarrow \RR$ defined by	
$$\Psi(u)\ =\ \frac{1}{2}\sigma (u)\:+\:\frac{\widehat{\gamma}}{2}||u||^{2}_{2}\:-\:\int_{\Omega}\ G(z,u)dz\qquad \mbox{for\ all}\ u\in H^1(\Omega).$$

	Also, let $g_{\pm}(z,x) = g(z,\pm x^{\pm})$ and $G_{\pm}(z,x) = \int^{x}_{0} g_{\pm}(z,s)ds$. We introduce the $C^1$--functional $\Psi_{\pm}: H^1(\Omega) \rightarrow \RR$ defined by
$$\Psi_{\pm}(u)\ =\ \frac{1}{2}\sigma (u)\:+\:\frac{\widehat{\gamma}}{2}||u||^{2}_{2}\:-\:\int_{\Omega}\ G_{\pm}(z,u)dz\qquad \mbox{for\ all}\ u\in H^1(\Omega).$$
	
{\sc Claim 1:} $K_{\Psi} \subseteq[v_0,u_0], K_{\Psi_{+}} = \{0,u_0\}, K_{\Psi_{-}} =\{0,v_0\}.$

	Let $u \in K_{\Psi}$. Then	
\begin{equation}
	\label{eq:51}
		A(u)\:+\:(\beta+\widehat{\gamma})u\:=\: N_g(u).
\end{equation}

	On (\ref{eq:51}) we act with $(u-u_0)^+ \in H^1(\Omega)$. Then
\begin{eqnarray}		
	&&	\left\langle A(u),(u-u_0)^+\right\rangle\ +\ \int_{\Omega}(\beta(z)+\widehat{\gamma})\;u(u-u_0)^+\ dz \nonumber \\
	&&	=\ \int_{\Omega}\:g(z,u)(u-u_0)^+\ dz \nonumber \\
	&&	=\ \int_{\Omega}[f(z,u_0)+\widehat{\gamma}u_0](u-u_0)^+\ dz\ \ (\mbox{see}\ (\ref{eq:50})) \nonumber \\
	&&	=\ \left\langle A(u_0),(u-u_0)^+\right\rangle\ +\ \int_{\Omega}\: (\beta(z)+\widehat{\gamma})u_0(u-u_0)^+\ dz \nonumber \\
	&&	\Rightarrow\ \sigma((u-u_0)^+)+\widehat{\gamma}||(u-u_0)^+||^{2}_{2}\ =\ 0 \nonumber \\	
	&&	\Rightarrow\ \widehat{c}||(u-u_0)^+||^2\ \leq \ 0\ \ \mbox{(see (\ref{eq:8}))}\nonumber \\
	&&	\Rightarrow\ u\ \leq\ u_0.\nonumber
\end{eqnarray}

	In a similar manner acting on (\ref{eq:51}) with $(v_0-u)^+ \in H^1(\Omega)$, we obtain $v_0 \leq u$. Hence
$$	u\:\in\:[v_0,u_0]\ =\ \{y\in H^1(\Omega): v_0(z)\:\leq\:y(z)\:\leq\:u_0(z)\ \mbox{ a.e.\ in}\ \Omega\}
		\Rightarrow\ K_{\Psi}\ \subseteq\ [v_0,u_0].$$

	Similarly we show that	
$$	K_{\Psi_{+}}\:\subseteq\:[0,u_0]\ =\ \{y\in H^1(\Omega):\: 0\leq y(z)\leq u_0(z)\ \mbox{a.e.\ in}\ \Omega\} $$
and $$ K_{\Psi_-}\subseteq [v_0,0]\ =\ \{y\in H^1(\Omega):\ v_0(z)\:\leq\:y(z)\:\leq\:0\ \mbox{a.e.\ in}\ \Omega\}\,. $$
	
	Recall that $u_0$ and $v_0$ are extremal constant sign solutions. So, it follows that
$K_{\Psi_+}\ =\ \{0,u_0\}$ and\ $K_{\Psi_-}\ =\ \{0,v_0\}.$
	This proves Claim 1.
	
{\sc Claim 2:} $u_0 \in \mbox{int}\,C_+\ \mbox{and}\ v_0 \in -\mbox{int}\,C_+$ are local minimizers of $\Psi$.
	
	From (\ref{eq:8}) and (\ref{eq:50}) it follows that the functional $\Psi_+$ is coercive. Also, it is sequentially weakly lower semi-continuous. So, by the Weierstrass theorem, we can find $\tilde{u} \in H^1(\Omega)$ such\ that
\begin{equation}
	\label{eq:52}
		\Psi_+(\tilde{u})\ =\inf\{\Psi_+(u):\:u\in H^1(\Omega)\}.
\end{equation}
	
	As before (see the proof of Proposition \ref{prop:3.2}), for $t\in(0,1)$ small (at least such that $t\widehat{u}_1 \in [0,u_0]$, recall $u_0 \in \mbox{int}\,C_+$), we have
$\Psi_+(t\widehat{u}_1)<0$. Therefore
	$\Psi_+(\tilde{u})<0 = \Psi_+(0)$  (see\ (\ref{eq:52})),\ hence\ $\tilde{u}\neq 0$.


	From (\ref{eq:52}) we have $\tilde{u}\in K_{\Psi_+} = \{0,u_0\}$ (see Claim 1). So, $\tilde{u} = u_0$. Note that
$\Psi_+|_{C_+}\ =\ \Psi|_{C_+}$ and $u_0\in\mbox{int}\,C_+$.

	It follows that $u_0$ is a local $C^1(\bar{\Omega})$--minimizer of $\Psi$. From Proposition \ref{prop:2.1} we infer that $u_0$ is a local $H^1(\Omega)$--minimizer of $\Psi$. Similarly for $v_0 \in -\mbox{int}\,C_+$ using this time the functional $\Psi_-$. This proves Claim 2.
	
	Without any loss of generality, we may assume that $\Psi(v_0) \leq \Psi(u_0)$ (the analysis is similar if the opposite inequality holds). By virtue of Claim 2, we can find $\rho \in (0,1)$ such\ that
\begin{equation}
	\label{eq:53}
		\Psi(v_0)\:\leq\:\Psi(u_0)<\inf\{\Psi(u):\:||u-u_0||=\rho\} = \eta_{\rho}\,.
\end{equation}

	The functional $\Psi$ is coercive (see (\ref{eq:8}) and (\ref{eq:51})), hence it satisfies the C-condition. This fact and (\ref{eq:53}) permit the use of Theorem \ref{th:2.2} (the mountain pass theorem). So, we can find $y_0 \in H^1(\Omega)$ such\ that	
\begin{equation}
	\label{eq:54}
		y_0\:\in\:K_\Psi\:\subseteq\:[v_0,u_0]\ \mbox{(see\ Claim\:1)\ and}\ \eta_{\rho}\leq \Psi(y_0).
\end{equation}

	From (\ref{eq:53}) and (\ref{eq:54}) we have $y_0\notin \{v_0,u_0\}$.
	
	Hypothesis $H_2(iv)$ via the UCP implies that $u=0$ is a nondegenerate critical point of $\varphi \in C^2(H^1(\Omega))$. Therefore
\begin{equation}
	\label{eq:55}
		C_k(\varphi,0)\ =\ \delta_{k,d_l}\ZZ\ \mbox{for\ all}\ k\geq 0\ \mbox{with}\ d_l\:=\mbox{dim}\overset{l}{\underset{\mathrm{i=1}}\oplus} E(\widehat{\lambda_i})\geq2.
\end{equation}

	Note that
\begin{eqnarray}	
	&&	\Psi|_{[v_0,u_0]}\ =\ \varphi|_{[v_0,u_0]}\ \ \mbox{(see\ (\ref{eq:50}))\ and}\ v_0\in -\mbox{int}\,C_+,\:u_0\in \mbox{int}\,C_+,\nonumber \\
	&&	\Rightarrow\ C_k(\Psi|_{C^{1}(\bar{\Omega})},0)\ =\ C_K(\varphi|_{C^{1}(\bar{\Omega})},0)\ \mbox{for\ all}\ k\geq 0,\nonumber \\
	&&	\Rightarrow\ C_k(\Psi,0)\ =\ C_k(\varphi,0)\ \mbox{for\ all}\ k\geq 0\ \ \mbox{(see\ Palais\ \cite{ref21}\ and\ Bartsch\ \cite{ref2})},\nonumber \\
	&&	\Rightarrow\ C_k(\Psi,0)\ =\ \delta_{k,d_l}\ZZ\ \mbox{for\ all}\ k\geq0\ \ \mbox{(see\ (\ref{eq:55}))}.\label{eq:56}
\end{eqnarray}
	Recall that $y_0$ is a critical point of mountain pass type for $\Psi$. Hence
\begin{equation}
	\label{eq:57}
		C_1(\Psi,y_0)\ \neq\ 0.
\end{equation}

	Since $d_l \geq 2$, comparing (\ref{eq:56}) and (\ref{eq:57}), we infer that $y_0 \neq 0$. Since $y_0 \in [v_0,u_0]$, the extremality of the solutions $u_0, v_0$ and (\ref{eq:50}) implies that $y_0$ is a solution of (1). Moreover, the regularity results of Wang \cite{ref26} imply $y_0 \in C^{1}({\bar\Omega})$.

	We have
$$-\Delta y_0(z)\:+\:\beta(z)y_0(z)\ =\ f(z,y_0(z))\ \ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial y_0}{\partial n}\ =\ 0\ \mbox{on}\ \partial \Omega.$$
	
	Hypotheses $H(i)$ and the mean value theorem imply that if $\rho=\max\{||u_0||_{\infty}, ||v_0||_{\infty}\}$, then we can find $\xi_\rho>0$ such\ that for a.a. $z \in \Omega$ the function $x\rightarrow f(z,x)+\xi_{\rho}x$ is nondecreasing on $[-\rho,\rho]$. We have
\begin{eqnarray}
	&&	-\Delta y_0(z)\ +\ (\beta(z)\ +\ \xi_{\rho})y_0(z)
	=\ f(z,y_0(z))\ +\ \xi_{\rho}y_0(z) \nonumber \\
	&&	\leq\ f(z,u_0(z)) + \xi_{\rho}u_0(z)\ \ \mbox{(since\ $y_0\leq u_0$)} \nonumber \\
	&&	=\ -\Delta u_0(z) + (\beta(z)\ +\ \xi_{\rho})u_0(z)\ \ \mbox{a.e.\ in}\ \Omega\nonumber \\
	&&	\Rightarrow\ \Delta(u_0\:-\:y_0)(z)\ \leq\ (||\beta^+||_{\infty}\ +\ \xi_{\rho})u_0(z)\ \mbox{a.e.\ in}\ \Omega\ \ \mbox{(see $H_1$)} \\
	&&	\Rightarrow\ u_0-y_0\ \in\ \mbox{int}\, C_+\ \mbox{(see\ V\'azquez\ \cite{ref25})}. \nonumber
\end{eqnarray}
	Similarly we show that
$$	y_0-v_0\in \mbox{int}\,C_+
		\Rightarrow\ y_0\in\mbox{int}_{C^{1}(\bar{\Omega})}[v_0,u_0].
$$
	Recall that $\Psi|_{[v_0,u_0]} = \varphi|_{[v_0,u_0]}$ (see (\ref{eq:50})). Hence
\begin{eqnarray}	
	&&	C_k(\Psi|_{C^{1}(\bar{\Omega})},y_0)\ =\ C_k(\varphi|_{C^{1}(\bar{\Omega})},y_0)\ \mbox{for\ all}\ k\geq 0,\nonumber \\
	&&	\Rightarrow\ C_k(\Psi,y_0)\ =\ C_k(\varphi,y_0)\ \mbox{for\ all}\ k\geq 0\ \ \mbox{(see\ \cite{ref21},\ \cite{ref2}),\label{eq:58}} \\
	&&	\Rightarrow\ C_1(\varphi,y_0)\neq 0\ \ (\mbox{see (\ref{eq:57})}). \label{eq:59}
\end{eqnarray}

	Hypotheses $H_2$ imply that $\varphi \in C^2(H^1(\Omega))$ and
for\ all\ $u,v\in H^1(\Omega)$
$$\langle \varphi\sp{\prime\prime} (y_0)u,v\rangle = \int_{\Omega}(Du,Dv)_{{\RR}^N}dz\ +\ \int_{\Omega}\beta(z)uv\ dz\ -\ \int_{\Omega}\widehat{m}_0\:uv\ dz,$$  where ${\widehat{m}}_0(z)\ =\ f^{'}_{x}(z,y_0(z)),\ \widehat{m}_0\in L^{\infty}(\Omega)_+$ (see\ $H_2(i)$). Therefore $\varphi ''(y_0)$ is a Fredholm operator. Let $\sigma(\varphi ''(y_0))$ denote the spectrum of $\varphi ''(y_0)$ and suppose that $\sigma(\varphi ''(y_0))\subseteq [0,\infty)$. Then we have
\begin{equation}
	\label{eq:60}
		\int_{\Omega}\:\widehat{m}_0 u^2\:dz\:\leq\sigma(u)\qquad \mbox{for\ all}\ u\in H^1(\Omega).
\end{equation}

	Let $u\in \mbox{Ker}\, \varphi ''(y_0)$. Then
\begin{equation}
	\label{eq:61}
		-\Delta\:u(z)\ =\ (\widehat{m}_0-\beta)(z)u(z)\ \ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial u}{\partial n}=0\ \mbox{on}\ \partial\Omega.
\end{equation}

	We have $\widehat{m}_0-\beta \in L^s(\Omega)$ with $s>N$ (see $H_1$). If $(\widehat{m}-\beta)^+=0$, then from (\ref{eq:61}) it follows that $u=0$. If $(\widehat{m}_0-\beta)^+\neq 0$, then from (\ref{eq:60}) and de Figueiredo \cite{ref9}, we have $\mbox{Ker}\, \varphi ''(y_0)\leq 1$. So, we can apply Proposition 2.5 of Bartsch \cite{ref2} and obtain
\begin{eqnarray}
	&&	C_k(\varphi,y_0)\:=\:\delta_{k,1}\ZZ\ \mbox{for\ all}\ k\geq 0 \label{eq:62}\\
	&&	\Rightarrow\ C_k(\Psi,y_0)\ =\ \delta_{k,1}\ZZ\ \mbox{for\ all}\ k\geq 0\ \mbox{(see (\ref{eq:58}))}.\nonumber
\end{eqnarray}

	From Claim 2 we know that $v_0$ and $u_0$ are local minimizers of $\Psi$. Hence
\begin{equation}
	\label{eq:63}
		C_k(\Psi,u_0)\ =\ C_k(\Psi,v_0)\ =\ \delta_{k,0}\ZZ\quad \mbox{for\ all}\ k\geq 0.
\end{equation}

	Finally recall that $\Psi$ is coercive. Therefore
\begin{equation}
	\label{eq:64}
		C_k(\Psi,\infty)\ =\ \delta_{k,0}\ZZ\quad \mbox{for\ all}\ k\geq 0.
\end{equation}

	Suppose that $K_{\Psi} = \{0, u_0, v_0, y_0\}$. Then from (\ref{eq:56}), (\ref{eq:62}), (\ref{eq:63}), (\ref{eq:64}) and the Morse relation (see (\ref{eq2})) with $t=-1$, we have
$	(-1)^{d_l}+2(-1)^0+(-1)^1\ =\ (-1)^0$,
 a\ contradiction.

	This means that there exists $\widehat{y}\in K_{\Psi}, \widehat{y}\notin \{0,u_0,v_0,y_0\}.$ From Claim 1 and Wang \cite{ref26}, we have
	$\widehat{y}\in [v_0,u_0]\cap C^{1}(\bar{\Omega}),$
hence $\widehat{y}$ is the second nodal solution of problem \eqref{eq:1}.
As we did for $y_0$, using the strong maximum principle of V\'{a}zquez \cite{ref25}, we have
$	\widehat{y} \in \mbox{int}_{C^1(\Omega)} [ v_0 , y_0 ]$.

Next we deal with the case when $S\_ = \oslash$. In this case there is no extremal negative solution. So, we introduce the following ``unilateral" perturbation--truncation of the reaction $f(z,\cdot)$:
\begin{equation}
	\label{eq:65}
		g_*(z,x)=\left\{
			\begin{array}{ll}
				\di f(z,x)+\widehat{\gamma}x&\di\qquad \mbox{if $x\leq u_+(z)$} \\
				\di f(z,u_+(z))+\widehat{\gamma}u_+(z)&\di\qquad \mbox{if $u_+(z)<x$}.
			\end{array}
		\right.
\end{equation}
 This is a Carath\'eodory function. We set $G_*(z,x)=\int_0^x{g_*(z,s)} ds$ and consider the $C^1$-functional $\phi_*:H^1(\Omega)\ri\RR$ defined by
\begin{equation*}
	\phi_*(u) = \frac{1}{2}\sigma(u)+\frac{\widehat{\gamma}}{2} ||u||_2^2 - \int_\Omega {G_*(z,u)dz}\quad \mbox{for\ all}\ u\in H^1(\Omega).
\end{equation*}

\begin{proposition}
	\label{prop:4.2}
		Assume that hypotheses\ $H_1$\ and\ $H_2$\ hold.
				Then\ the\ functional\ $\varphi_*$\ satisfies\ the\ C--condition.
\end{proposition}

 \textit{Proof.} Let $\{u_n\}_{n\geq 1}\subseteq H^1(\Omega)$ be a sequence such that $\{\varphi_*(u_n)\}_{n\geq 1}\subseteq \RR $ is bounded and
\begin{equation}
	\label{eq:65'}
		(1+||u_n||)\varphi_*^{'}(u_n)\rightarrow 0\quad \mbox{in}\ H^1(\Omega)^*.
\end{equation}

From \eqref{eq:65} we have for all $n\geq 1$,
		\begin{eqnarray}
			&&	|\left\langle\varphi_* '(u_n),h\right\rangle| \leq \frac{\epsilon_n ||h||}{1+||u_n||}\quad \mbox{for\ all}\ h\in H^1(\Omega)\ \mbox{with}\ \epsilon_n \rightarrow 0^+ \nonumber \\
			&&	\Rightarrow |\left\langle A(u_n),h\right\rangle + \int_\Omega (\beta(z)+\widehat{\gamma})u_n\ hdz - \int_\Omega g_*(z,u_n)hdz| \leq \frac{\epsilon_n||h||}{1+||u_n||}\,.\label{eq:66}
		\end{eqnarray}

Let $h=u^{+}_{n} \in H^1(\Omega)$. We have
		\begin{eqnarray}
				&&	\sigma(u^{+}_{n})+\widehat{\gamma}||u_n^+||_2^2 \leq M_3\ \mbox{for\ some}\ M_3 > 0\ (\mbox{see\ (\ref{eq:65})}) \nonumber \\
				&&	\Rightarrow \widehat{c} ||u^{+}_{n}||^2 \leq M_3\ \mbox{for\ all}\ n\geq 1\ \mbox{(see\ (\ref{eq:81}))}\nonumber \\
				&&	\Rightarrow \{u^{+}_{n}\}_{n\geq 1} \subseteq H^1 (\Omega)\ \mbox{is\ bounded}.\label{eq:67}
		\end{eqnarray}

We assume that $||u_n^-||\rightarrow \infty$ and set $y_n = \frac{u_n^-}{||u_n^-||}$. Then $||y_n||=1$ for\ all\ $n \geq 1$. So we may assume that
\begin{equation}
	\label{eq:68}
		y_n\stackrel{w}{\rightarrow} y\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y_n \rightarrow y\ \mbox{in}\ L^{2s^{'}}(\Omega)\ (\frac{1}{s} + \frac{1}{s^{'}} = 1)
\end{equation}

From (\ref{eq:66}) and (\ref{eq:67}), we have
\begin{equation}
	\label{eq:69}
			|\left\langle A(-y_n),h\right\rangle - \int_{\Omega}(\beta(z)+\widehat{\gamma})y_n hdz - \int_{\Omega} \frac{g_*(z,-u_n^-)}{||u_n^-||}hdz|\leq M_4 \frac{||h||}{||u_n^-||}\,.
\end{equation}

Hypotheses $H_2$ (i),(ii),(iii) imply that
\begin{equation*}
	\left\{\frac{N_{g_*}(-u_n^-)}{||u_n^-||}\right\}_{n\geq 1} \subset L^2(\Omega)\ \mbox{is\ bounded}.
\end{equation*}

So, if in (\ref{eq:69}) we choose $h = y_n - y \in H^1(\Omega)$ and pass to the limit as $n\rightarrow \infty$, then
\begin{equation}
	\label{eq:70}
		\begin{array}{l}
			\lim_{n\ri\infty} \left\langle A(-y_n),y_n-y\right\rangle = 0
			\Rightarrow ||Dy_n||_2 \rightarrow ||Dy||_2 \\
			\Rightarrow y_n \rightarrow y\ \mbox{in}\ H^1(\Omega)\ \mbox{(by\ the\ Kadec - Klee\ property)}\\
			\Rightarrow||y||=1,\ y\geq 0.
		\end{array}
\end{equation}

Then using hypothesis $H_2(iii)$ and by passing to a suitable suitable subsequence if necessary we have
\begin{equation}
	\label{eq:71}
		\frac{Ng_*(-u_n^-)}{||u_n^-||} \stackrel{w}{\rightarrow} -(\eta_0+\widehat{\gamma})y\ \mbox{in}\ L^2(\Omega)\ \mbox{with}\ \eta(z)\leq\eta_0(z)\leq \widehat{\eta}(z)\ \mbox{a.e.\ in}\ \Omega\,.
\end{equation}

So, if in (\ref{eq:69}) we pass to the limit as $n\longrightarrow\infty$, and use (\ref{eq:70}), then
\begin{equation}
	\label{eq:72}
		\begin{array}{l}
			\left\langle A(y),h\right\rangle+\int_{\Omega}\beta y\ hdz = \int_r \eta_0 y\ hdz\ \mbox{for\ all}\ h\in H^1(\Omega),\\
			\Rightarrow A(y)+\beta y = \eta_0y,\\
			\Rightarrow -\Delta y(z)+\beta (z)y(z) = \eta_0(z)y(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial y}{\partial n} = 0\ \mbox{on}\ \partial \Omega.
		\end{array}
\end{equation}

From Proposition \ref{prop:2.4} we have
$\tilde{\lambda}_1(\eta_0)\leq \tilde{\lambda}_1(\eta) < \tilde{\lambda}_1(\widehat{\lambda}_1)=1$.
So, from (\ref{eq:72}) it follows that $y$ is nodal, witch contradicts (\ref{eq:70}). It follows that
\begin{equation*}
		\{u^{-}_{n}\}_{n\geq 1}\subset H^1(\Omega)\ \mbox{is\ bounded}
		\Rightarrow \{u_n\}_{n\geq 1} \subset H^1(\Omega)\ \mbox{is\ bounded\ (see\ (\ref{eq:67}))}.
\end{equation*}
Therefore, we may assume that
\begin{equation}
	\label{eq:73}
		u_n \stackrel{w}{\rightarrow} u\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ u_n \rightarrow u\ \mbox{in}\ L^{2s'}(\Omega).
\end{equation}

In (\ref{eq:66}) we choose $h=u_n-u\in H^1(\Omega)$, pass to the limit as $n\rightarrow \infty$ and use (\ref{eq:73}). Then
\begin{equation*}
		\lim_{n\ri\infty}\left\langle A(u_n),u_n-u\right\rangle=0
		\Rightarrow u_n \rightarrow u\ \mbox{in}\ H^1(\Omega)
		\Rightarrow \varphi_*\ \mbox{satisfies\ the\ C--condition.}
\end{equation*}
\qed

The next result permits the computation of the critical groups of $\varphi_*$ at infinity.

\begin{proposition}
	\label{prop:4.3}
		If\ hypotheses $H_1$ and $H_2$ hold, then $C_k(\varphi_*,\infty)=0\ for\ all\ k\geq 0$.
\end{proposition}

\textit{Proof.} Let $\lambda > \max\{\widehat{\lambda}_1, \frac{1}{|\Omega|_N}\int_{\Omega}\beta dz\}$ and consider the $C^1$--functional $\Psi_* : H^1(\Omega)\rightarrow R$ defined by
\begin{equation*}
	\Psi_*(u) = \frac{1}{2}\sigma(u)+\frac{\widehat{\gamma}}{2}||u||_2^2 - \frac{1}{2}(\lambda + \widehat{\gamma})||u^-||_2^2\quad \mbox{for\ all}\ u \in H^1(\Omega)\,.
\end{equation*}

We consider the homotopy
\begin{equation*}
	h(t,u)=h_t(u)=(1-t)\varphi_*(u)+t\Psi_*(u)\quad \mbox{for\ all}\ (t,u)\in [0,1]\times H^1(\Omega).
\end{equation*}


{\sc Claim 1.} There exist $\epsilon \in \RR$ and $\delta >0$ such\ that
\begin{equation*}
	h_t(u)\leq \epsilon \Rightarrow (1+||u||)||(h_t)^{'}(u)||_* \geq \delta ||u||^2\quad \mbox{for\ all}\ t\in [0,1].
\end{equation*}
We argue by contradiction. So, suppose that the Claim is not true. Since $h$ maps bounded sets to bounded sets, we can find $\{t_n\}_{n\geq 1}\subset [0,1]$ and $\{u_n\}_{n\geq1} \subset H^1(\Omega)$ such\ that
\begin{equation}
	\label{eq:74}
		t_n\rightarrow t, ||u_n||\rightarrow \infty, h_{t_n}(u_n)\rightarrow -\infty\ \mbox{and}\ ||(h_{t_n})^{'}(u_n)||<\frac{1}{n}||u_n||\,.
\end{equation}

From  (\ref{eq:74}) we have
\begin{equation}
	\label{eq:75}
		\begin{array}{l}
			|\left\langle A(u_n),h\right\rangle+\int_{\Omega}(\beta(z)+\widehat{\gamma})u_n hdz-(1-t_n)\int_{\Omega}g_*(z,u_n)hdz-(\lambda+\widehat{\gamma})t_n\int_{\Omega} u_n^-hdz|\\
			<\frac{1}{n}\,||u_n|| \cdot ||h||\quad \mbox{for\ all}\ n\geq 1.
		\end{array}
\end{equation}

In (\ref{eq:75}) first we choose $h=u_n^+ \in H^1(\Omega)$ then
\begin{equation}
	\label{eq:76}
		\sigma(u_n^+)+\widehat{\gamma}||u_n^+||_2^2\leq\frac{1}{n}\,||u_n||\cdot ||u_n^+||\quad \mbox{for\ all}\ n\geq 1.
\end{equation}

Let $y_n = \frac{u_n}{||u_n||}$. Then from (\ref{eq:76}) we have
\begin{equation}
	\label{eq:77}
		\begin{array}{l}
			\sigma(y_n^+)+\widehat{\gamma}||y_n^+||_2^2 \leq \frac{1}{n}\,||y_n||\cdot ||y_n^+||\\
			\Rightarrow \widehat{c}||y_n^+||^2 \leq \frac{1}{n}\,||y_n^+||\ \mbox{(see\ (\ref{eq:8})\ and\ recall\ that\ $||y_n||=1$)}\\
			\Rightarrow y_n^+ \rightarrow 0\ \mbox{in}\ H^1(\Omega).
		\end{array}
\end{equation}

From (\ref{eq:75}) and (\ref{eq:77}), we have
\begin{equation}
	\label{eq:78}
		\begin{array}{l}
			|\left\langle A(-y_n^-),h\right\rangle+\int_{\Omega}(\beta(z)+\widehat{\gamma})(-y_n^-)dz-(1-t_n)\int_{\Omega}\frac{q_*(z,-u_n^-)}{||u_n||}hdz+(\lambda + \widehat{\gamma})t_n\int_{\Omega}y_n^- hdz|\\
			\leq\epsilon_n||h||\ \mbox{with}\ \epsilon_n\ \downarrow\ 0.
		\end{array}
\end{equation}
Since $||y_n||=1$ for all $n\geq 1$, by passing to a suitable subsequence if necessary, we may assume that
\begin{equation}
	\label{eq:79}
		\begin{array}{l}
			y_n \stackrel{w}{\rightarrow} y\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y_n\rightarrow y\ \mbox{in}\ L^{2s'}(\Omega),\\
			\Rightarrow y_n^- \stackrel{w}{\rightarrow} y^-\ \mbox{in}\ H^1(\Omega)\ \mbox{and}\ y^{-}_{n} \rightarrow y^-\ \mbox{in}\ L^{2s'}(\Omega).
		\end{array}
\end{equation}

In (\ref{eq:78}) we choose $h=y_n^- - y^- \in H^1(\Omega)$ and pass to the limit as $n\rightarrow\infty$. Recalling that $\left\{\frac{Ng_*(-u_n^-)}{||u_n^-||}\right\}_{n\geq 1} \subseteq L^2(\Omega)$ is bounded and using (\ref{eq:79}), we obtain
\begin{equation}
	\label{eq:80}
			\lim_{n\ri\infty} \left\langle A(-y_n^-), y_n^- - y^-\right\rangle = 0
			\Rightarrow y_n^- \rightarrow y^-\ \mbox{in}\ H^1(\Omega).
\end{equation}

Note that $y^- \neq 0$ or otherwise from (\ref{eq:77}) and (\ref{eq:80}), we have $y_n\rightarrow 0$ in $H^1(\Omega)$ which contradicts the fact that $||y_n||=1$ for all $n\geq1$. Now, using hypothesis $H_2(iii)$, we have
\begin{equation}
	\label{eq:81}
		\frac{Ng_*(-u_n^-)}{||u_n||} \stackrel{w}{\rightarrow} \xi = -(\eta_0+\widehat{\gamma})y^-\ \mbox{in}\ L^2(\Omega),\ \mbox{with}\ \eta\leq\eta_0\leq\widehat{\eta}.
\end{equation}

So, if in (\ref{eq:78}) we pass to the limit as $n\rightarrow\infty$ and use (\ref{eq:80}) and (\ref{eq:81}), then
\begin{equation}
	\label{eq:82}
		\begin{array}{l}
			\left\langle A(y^-),h\right\rangle+\int_{\Omega}\beta y^- hdz= \int_{\Omega}\eta_0 y^- hdz\ \mbox{for\ all}\ h\in H^1(\Omega)\\
			\Rightarrow A(y^-)+\beta y^- = \eta_0y^-\\
			\Rightarrow -\Delta y^-(z)+\beta(z)y^-(z) = \eta_0(z)y^-(z)\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial y^-}{\partial n} = 0\ \mbox{on}\ \partial\Omega\,.
		\end{array}
\end{equation}

By virtue of Proposition \ref{prop:2.4}, we have
$\tilde{\lambda}_1(\eta_0)\leq\tilde{\lambda}_1(\eta)<\tilde{\lambda}_1(\widehat{\lambda}_1)=1.
$

Therefore (\ref{eq:82}) implies that $y^-$ must be nodal (recall $y^- \neq 0$), contradiction. This proves Claim 1.
%%%%%%%%

Note that
\begin{equation*}
	|\partial_t h_t(u)|=|\Psi_*(u)-\varphi_*(u)|\leq C_1||u||^2\ \mbox{for\ some}\ c_1>0\ \mbox{and}\ u\in H^1(\Omega).
\end{equation*}

Finally note that $h_0(\cdot) = \varphi_* $ satisfies the C-condition (see Proposition \ref{prop:4.2}) while $h_1(\cdot)=\Psi_*$ also satisfies the C-condition since $\lambda>\widehat{\lambda}_1$. So, we have
\begin{equation}
	\label{eq:83}
		C_k(\varphi_*, \infty) = C_k(\Psi_*, \infty)\quad \mbox{for\ all}\ k\geq 0\ \mbox{(see\ Liang \& Su \cite{ref16})}.
\end{equation}

It is easy to check that $K_{\Psi_*} \subset - C_+$ and since $\lambda > \widehat{\lambda_1}$, it follows that $K_{\Psi_*}=\{0\}$ (recall that $\widehat{\lambda}_1$ is the only eigenvalue with eigenfunctions of constant sign). Hence
\begin{equation}
	\label{eq:84}
		C_k(\Psi_*, \infty) = C_k(\Psi_*,0)\ \mbox{for\ all}\ k\geq 0.
\end{equation}



Let $\zeta \in L^\infty(\Omega), \zeta\leq 0,\ \zeta \neq\ 0$ and consider the homotopy
\begin{equation*}
	\tilde{h}_t(u) = \Psi_*(u) - t\zeta u\ \mbox{for\ all}\ t\in [0,1],\ \mbox{and}\ u\in H^1(\Omega).
\end{equation*}

{\sc Claim 2:} $(\tilde{h}_t)^{'}(u)=0$ for all $t\in [0,1]$ and for all $u\in H^1(\Omega) \setminus \{0\}$.

Again we argue indirectly. So suppose that we can find $t\in [0,1]$ and $u\in H^1(\Omega),u\neq 0$ such\ that
	\begin{equation} \label{eq:85}(\tilde{h}_t)^{'}(u) = 0 \Rightarrow A(u) + (\beta+\widehat{\gamma})u = -(\lambda + \widehat{\gamma})u^- + t\zeta. 	
	\end{equation}

On \eqref{eq:85} we act with $u^+ \in H^1(\Omega)$. Then
$\sigma(u^+)+\widehat{\gamma}||u^+||_2^2 = t\int_\Omega\zeta u^+ dz \leq 0$. Thus, by \eqref{eq:8},
		 $ \widehat{c}||u^+||^2\leq 0$,  hence $u\leq 0,\, u\neq 0. $
Then relation (\ref{eq:85}) becomes
\begin{equation}
	\label{eq:86}
		\begin{array}{l}
			A(u)+\beta u = \lambda u - t\zeta\\
			\Rightarrow -\Delta u(z) + \beta(z)u(z) = \lambda u(z) - t\zeta\ \mbox{a.e.\ in}\ \Omega,\ \frac{\partial u}{\partial n} = 0\ \mbox{on}\ \partial \Omega\\
			\Rightarrow u \in -\mbox{int}\, C_+\ \mbox{(as\ before\ using\ \cite{ref26}, \cite{ref25})}.
		\end{array}
\end{equation}

Let $v\in \mbox{int}\,C_+$ and consider the function
$$
	R(v,-u)(z)=||Dv(z)||^2-(-Du(z), D(\frac{v^2}{-u})(z))_{\RR^{N}}\,.
$$

From Picone's identity (see, for example, Gasinski \& Papageorgiou \cite[p. 785]{ref12}), we have
$$
	R(v,-u)(z)\geq 0\quad \mbox{for\ a.a.}\ z\in\Omega.
$$
Therefore
\begin{eqnarray}
	0 & \leq & \int_\Omega R(v,-u)dz
		 =  ||Dv||_2^2 - \int_\Omega (-\Delta u)\frac{v^2}{u}dz\ \mbox{(by\ Green's\ identity)} \nonumber\\
		& =  & ||Dv||_2^2 - \int_\Omega (\lambda - \beta (z))v^2 dz + t\int_\Omega \zeta v^2 dz \nonumber \\
	  & \leq & ||Dv||_2^2 - \int_\Omega (\lambda - \beta(z))v^2 dz\ \mbox{(since\ $\zeta \leq 0$)}. \nonumber
\end{eqnarray}


Choose $v\equiv 1 \in \mbox{int}\, C_+$. Then
\begin{equation*}
	0\leq -\lambda|\Omega|_N + \int_\Omega \beta(z)dz < 0\ \mbox{(recall\ the\ choice\ of\ $\lambda$)},
\end{equation*}
a contradiction. This proves Claim 2.

Then Claim 2 and the homotopy invariance of critical groups imply that for all $k\geq 0$,
$$
		C_k(\Psi_*,0)=0
		\Rightarrow C_k(\varphi_*,\infty) = 0\quad \mbox{(see\ \eqref{eq:83},\ \eqref{eq:84})}.
	$$
\qed

Now we are ready for the second multiplicity theorem for problem (1), taking care of the case $S\_ = \O $.
\begin{theorem}
	\label{th:4.4}
		Assume that \ hypotheses\ $H_1$\ and\ $H_2$\ hold\ and\ $S\_ = \O$.
Then\ problem\ (1)\ has\ at\ least\ three\ nontrivial\ solutions
			 $u_{0} \in \mbox{int}\, C_{+}$\ and\ $y_{0},\,\widehat{y}\in	C^1 (\bar{\Omega})$\ nodal solutions
			 such\ that\ $u_0 - y_0$, $u_0 - \widehat{y} \in \mbox{int}\, C_{+}$.
\end{theorem}

	\textit{Proof.} From Proposition \ref{prop:3.2} we already have a nontrivial positive solution $u_0 \in \mbox{int}\,C_+$.
	By virtue of Proposition \ref{prop:3.3} we can always assume that $u_0$ is the smallest nontrivial positive solution of (1) (that is, $u_0 = u_+\in \mbox{int}\,C_+$).
	We still consider the $C^1$-functional $\varphi_* : H^1(\Omega)\rightarrow \RR$ introduced in the beginning of this section. Let
\begin{equation*}
	g_*^+(z,x) = g_*(z,x^+)\ \mbox{(see\ (\ref{eq:65}))\ and}\ G_*^+(z,x)= \int_0^x g_*^+(z,s)ds.
\end{equation*}

We consider the $C^1$-functional $\varphi_*^+:H^1(\Omega)\rightarrow \RR$ defined by
\begin{equation*}
	\varphi_*^+(u) = \frac{1}{2}\sigma(u)+\frac{\widehat{\gamma}}{2}||u||_2^2 - \int_\Omega G_*^+(z,u)dz\quad \mbox{for\ all}\ u\in H^1(\Omega)\,.
\end{equation*}

{\sc Claim 1.}
We have
\begin{equation*}
	K_{\varphi_*} \subset (u_0] = \{ u\in H^1(\Omega): u(z)\leq u_0(z)\ \mbox{a.e.\ in}\ \Omega \}\ \mbox{and}\ K_{\varphi_*^+} = \{0,u_0 \}.
\end{equation*}

Let $u\in K_{\varphi_*}$. Then
\begin{equation}
	\label{eq:87}
		A(u)+(\beta + \widehat{\gamma})u = N_{g_*}(u).
\end{equation}

On (\ref{eq:87}) we act with $(u-u_0)^+\in H^1(\Omega)$. Then
\begin{equation*}
	\begin{array}{l}
		\left\langle A(u),(u-u_0)^+ \right\rangle + \int_\Omega (\beta(z)+\widehat{\gamma})u(u-u_0)^{+} dz
		= \int_\Omega g_*(z,u)(u-u_0)^{+}dz\\
		= \int_\Omega [f(z,u_0)+\widehat{\gamma}u_0](u-u_0)^{+}dz\ \mbox{(see\ \eqref{eq:65}\ and\ recall\ $u_0 = u_+$)}\\
		= \left\langle A(u_0),(u-u_0)^{+} \right\rangle + \int_\Omega (\beta(z)+\widehat{\gamma})u_0(u-u_0)^{+}dz\\
		\Rightarrow \left\langle A((u-u_0)^{+}),(u-u_0)^{+} \right\rangle + \int_\Omega(\beta(z)+\widehat{\gamma})[(u-u_0)^{+}]^2dz = 0\\
		\Rightarrow\widehat{c}\,||(u-u_0)^{+}||^2 \leq 0\ \mbox{(see\ \eqref{eq:8})},
	\end{array}
\end{equation*}
hence\ $u\leq u_0$.
This proves that
$K_{\varphi_*} \subseteq (u_0] = \{u\in H^1(\Omega):u(z)\leq u_0(z)\ \mbox{a.e.\ in}\ \Omega\}\,.
$
In a similar fashion, we  show that
$
	K_{\varphi_*^+} \subseteq [0,u_0] = \{u \in H^1(\Omega): 0 \leq u(z)\leq u_0(z)\ \mbox{a.e.\ in}\ \Omega\}.
$

The extremality of $u_0 = u_+ \in \mbox{int}\, C_+$ (see Proposition \ref{prop:3.3}) implies
$K_{\varphi_*^+} = \{0,u_0\}\,.
$
This proves Claim 1.

{\sc Claim 2:} $u_0 \in \mbox{int}\, C_+$ is a local minimizer of the functional $\varphi_*$.

Evidently $\varphi_*^+$ is coercive (see \eqref{eq:65}). Also, it is sequentially weakly lower semi-continuous. So, we can find $\widehat{u}_0 \in H^1(\Omega)$ such\ that
\begin{equation}
	\label{eq:88}
		\varphi_*^+(\widehat{u}_0) = \inf\{\varphi_*^+(u):u \in H^1(\Omega)\}.
\end{equation}

For $t\in (0,1)$ small and using hypothesis $H_2(iv)$, we have
$$	\varphi_*^+(t\widehat{u}_1) < 0\ \mbox{(see\ the\ proof\ of\ Proposition\ \ref{prop:3.2})}
		\Rightarrow \varphi_*^+(\widehat{u}) < 0 = \varphi_k^+(0)\ \mbox{(see\ \eqref{eq:88})},
	$$
hence\ $\widehat{u}_0 \neq 0$.
Then from \eqref{eq:88} and Claim 1, we have
$\widehat{u}_0 = u_0 \in \mbox{int}\, C_+.
$
Since $\varphi_* |_{C_+} = \varphi_*^t|_{C_{+}}$ (see (\ref{eq:65})), it follows that $u_0\in \mbox{int}\,C_+$ is a local $C^1(\bar{\Omega})$-minimizer of $\varphi_*$. Invoking Proposition \ref{prop:2.1}, we have that $u_0 \in \mbox{int}\,C_+$ in a local $H^1(\Omega)$-minimizer of $\varphi_* $. This proves Claim 2.

Claim 2 implies that we can find $\rho \in(0,1)$ small such that
\begin{equation}
	\label{eq:89}
		\varphi_*(u_0)<\inf\{\varphi_*(u):||u-u_0||=\rho\} = \eta_\rho^*,\ ||u_0||>\rho\,.
\end{equation}

Hypothesis $H_2(iii)$ implies that
\begin{equation}
	\label{eq:90}
		\varphi_*(t\widehat{u}_1) \rightarrow -\infty \ \mbox{as}\ \ t\rightarrow -\infty\,.
\end{equation}

Finally recall that $\varphi_*$ satisfies the C-condition (see Proposition \ref{prop:4.2}). This fact together with (\ref{eq:89}) and (\ref{eq:90}), implies that we can apply Theorem \ref{th:2.2} (the mountain pass theorem). So, we can find $y_0 \in H^1(\Omega)$ such\ that
\begin{equation}
	\label{eq:91}
		y_0 \in K_{\varphi_*} \subseteq (u_0]\ \mbox{and}\ \eta_\rho^* \leq \varphi_*(y_0)\,.
\end{equation}
From (\ref{eq:89}) and (\ref{eq:91}), we see that $y_0 \neq u_0$. Since $y_0$ is a critical point of $\varphi_*$ of mountain pass type, we have
\begin{equation}
	\label{eq:92}
		C_1(\varphi_*, y_0) \neq 0.
\end{equation}

On the other hand, as in the proof of Theorem \ref{th:4.1} (see (\ref{eq:56})), using hypothesis $H_2(iv)$, we have
\begin{equation}
	\label{eq:93}
		C_k(\varphi_*,0)=\delta_{k_1 d_l}\ZZ\quad \mbox{for\ all}\ k\geq 0\ \mbox{and}\ d_l\geq 2.
\end{equation}

Comparing (\ref{eq:92}) and (\ref{eq:93}), we conclude that $y_0 \neq 0$. Since $y_0 \in (u_0]$ (see Claim 1) and due to the extremality of $u_0$, we infer that $y_0$ is nodal and $y_0 \in C^1(\bar{\Omega})$ (see Wang \cite{ref26}). Moreover, we have
\begin{eqnarray}
					&& u_0 - y_0 \in \mbox{int}\,C_+\ \mbox{(that is,\ $y_0 \in \mbox{int}_{C^1(\bar{\Omega})}(u_0]$)} \nonumber \\
			\mbox{and}	&& \ C_k(\varphi_*, y_0) = \delta_{k,1}\ZZ\ \mbox{for\ all}\ k\geq 0. \label{eq:94}
\end{eqnarray}


From Claim 2 we have
\begin{equation}
	\label{eq:95}
		C_k(\varphi_*, u_0)= \delta_{k,0}\ZZ\ \mbox{for\ all}\ k\geq 0
\end{equation}
while from Proposition \ref{prop:4.3} we have
\begin{equation}
	\label{eq:96}
		C_k(\varphi_*, \infty) = 0\ \mbox{for\ all}\ k\geq 0.
\end{equation}

Suppose $K_{\varphi_*} = \{0,u_0,y_0\}$. From (\ref{eq:93}), (\ref{eq:94}), (\ref{eq:95}), (\ref{eq:96}) and the Morse relation (see \eqref{eq2}), we have
$(-1)^{d_l}+(-1)^0 + (-1)^1 = 0 $,
		 a\ contradiction.
So we can find $\widehat{y}\in K_{\varphi_*} , \widehat{y} \notin \{0, u_0,y_0\}$. We have
$\widehat{y} \in (u_0]$ (see\ Claim\ 1), hence
$\widehat{y} \in C^1 (\bar{\Omega})$ (see\ Wang\ \cite{ref26})\ is\ nodal.
Moreover, as for $y_0$, we have $u_0 - \widehat{y} \in \mbox{int}\,C_+$\ (that is,\ $\widehat{y} \in \mbox{int}_{C^1(\bar{\Omega})}(u_0])$.
\qed

\medskip
 {\bf Acknowledgements.}  V.~R\u adulescu acknowledges the support through  Grant CNCS PCE--47/2011.

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