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\title[Robin problems with indefinite and unbounded potential]{ROBIN PROBLEMS WITH INDEFINITE AND UNBOUNDED POTENTIAL, RESONANT AT $-\infty$, SUPERLINEAR AT $+\infty$}


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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.D. R\u{a}dulescu]{Vicen\c{t}iu D. R\u{a}dulescu}
\address{Department of Mathematics, Faculty of Sciences, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia \& Institute of Mathematics ``Simion Stoilow" of the Romanian Academy, P.O. Box 1-764,
          014700 Bucharest, Romania}
\email{\tt vicentiu.radulescu@imar.ro}

\keywords{Indefinite and unbounded potential, Robin boundary condition, asymmetric reaction, critical groups, multiple nontrivial solutions.\\
\phantom{aa} 2010 AMS Subject Classification: Primary 35J20; Secondary 35J60, 58E05}


\begin{abstract}
We consider a semilinear Robin problem with an indefinite and unbounded potential and a reaction which exhibits asymmetric behavior as $x\rightarrow\pm\infty$. More precisely it is sublinear near $-\infty$ with possible resonance  with respect to the principal eigenvalue of the negative Robin Laplacian and it is superlinear at $+\infty$. Resonance is also allowed at zero with respect to any nonprincipal eigenvalue. We prove two multiplicity results. In the first one, we obtain two nontrivial solutions and in the second, under stronger regularity conditions on the reaction, we produce three nontrivial solutions. Our work generalizes the recent one by Recova-Rumbos (Nonlin. Anal. 112 (2015), 181-198).
\end{abstract}

\maketitle

%\tableofcontents

\section{Introduction}
Let $\Omega\subseteq\RR^{N}$ be a bounded domain with $C^{2}$-boundary $\partial\Omega$. In this paper, we study the following semilinear Robin problem:
\label{1}
\begin{equation}
\left \{
\begin{array}{l l}
-\Delta u(z)+\xi(z)u(z)=f(z,u(z))\ & \text{in}\ \Omega,\\
\di\frac{\partial u}{\partial n}+\beta(z)u(z)=0\ & \text{on}\ \partial\Omega.\\
\end{array} \right. \label{l1}
\end{equation}

In \eqref{1} the potential function $\xi(\cdot)$ is indefinite (that is, sign changing) and unbounded. The reaction $f(z,x)$ is a Carath\'{e}odory function (that is, for all $x\in\RR$ the mapping  $z\longmapsto f(z,x)$ is measurable and $x\longmapsto f(z,x)$ is continuous for a.a. $z\in\Omega$), which exhibits asymmetric behavior as $x\rightarrow\pm\infty$. More precisely, we assume that $f(z,\cdot)$ is sublinear as $x\rightarrow-\infty$ allowing for resonance to occur with respect to the principal eigenvalue $\hat{\lambda}_{1}$ of $u\longmapsto -\Delta u+\xi(z)u$ with Robin boundary condition, while $f(z,\cdot)$ is superlinear as $x\rightarrow+\infty$ but without satisfying the usual in such cases Ambrosetti-Rabinowitz condition (AR-condition for short). Instead we employ a weaker condition which incorporates in our framework also superlinear reactions with slower growth near $+\infty$.

Our goal is to prove multiplicity theorems for problem \eqref{1}. We have two such results. In the first theorem we produce two nontrivial solutions. In the second result, by strengthening the regularity on $f(z,\cdot)$ (we assume that for a.a. $z\in\Omega$, $f(z,\cdot)\in C^{1}(\RR)$), we produce three nontrivial solutions. Our approach combines variational tools (critical point theory) with elements from Morse theory (critical groups).

The starting point for this paper has been a recent work of Recova \& Rumbos \cite{24}, who examined a similar asymmetric semilinear Dirichlet problem with no potential (that is, $\xi\equiv0$) and with a reaction belonging in $C^{1}(\overline{\Omega}\times\RR)$ and satisfying more restrictive hypotheses (see hypotheses ($L_{1})-(L_{7}$) of \cite{24}) and in particular for the superlinearity in the positive direction, they employ the classical in such cases AR-condition (see hypotheses ($L_{3}$) and ($L_{4}$) in \cite{24}). In the past, there have been some other works on such asymmetric problems. We mention the papers of Arcoya \& Villegas \cite{3}, de Figueiredo \& Ruf \cite{8}, Motreanu, Motreanu \& Papageorgiou \cite{13} and Perera \cite{23}. All these works consider equations with zero potential (that is, $\xi\equiv0$) and the unilateral superlinearity of the reaction is expressed in terms of the AR-condition. In Arcoya \& Villegas \cite{3} and de Figueiredo \& Ruf \cite{8} the reaction is jointly continuous, in Perera \cite{23} it is jointly $C^{1}$ and in Motreanu, Motreanu \& Papageorgiou \cite{13} it is assumed that $f(z,\cdot)\in C^{1}(\RR)$. Note that in de Figueiredo \& Ruf \cite{8} and Perera \cite{23} it is assumed that $N=1$ (that is, the boundary value problem is an ordinary differential equation). We mention that Arcoya \& Villegas \cite{3} and de Figueiredo \& Ruf \cite{8} prove only existence theorems, while Motreanu, Motreanu \& Papageorgiou \cite{13} and Perera \cite{23} obtain multiplicity results producing two nontrivial solutions. Motreanu, Motreanu \& Papageorgiou \cite{13} consider Dirichlet problems, while the others deal with Neumann boundary conditions. We mention that our formulation here covers also the Neumann case (when $\beta\equiv0$).

Finally we mention that semilinear problems with indefinite and unbounded potential were studied recently by Kyritsi \& Papageorgiou \cite{10}, Papageorgiou \& Papalini \cite{17} (Dirichlet problems) and Papageorgiou \& R\u{a}dulescu \cite{18}, \cite{20}, Papageorgiou \& Smyrlis \cite{22} (Neumann problems).

\section{Mathematical Background}\label{Section2}

Let $X$ be a Banach space and $X^{*}$ be its topological dual. By $\langle\cdot,\cdot\rangle$ we denote the duality brackets for the pair $(X^{*},X)$. Given $\varphi\in C^{1}(X,\RR)$, we say that $\varphi$ satisfies the ``Cerami condition" (the ``C-condition" for short), if the following is true:
\begin{quote}
        ``Every sequence $\{u_{n}\}_{n\geq1}\subseteq X$ such that $\{\varphi(u_{n})\}_{n\geq1}\subseteq\RR$ is bounded and
    $$(1+\|u_{n}\|)\varphi'({u_{n}})\rightarrow0\ \mbox{in}\ X^{*}\ \mbox{as}\ n\rightarrow\infty,$$
    admits a strongly convergent subsequence".
\end{quote}

This is a compactness condition on the functional $\varphi$. So, the necessary compactness requirement is transferred from the space $X$, which is not in general locally compact (being infinite dimensional) to the functional $\varphi$. The C-condition is more general than the more common Palais-Smale condition. Nevertheless, the C-condition suffices to prove a deformation theorem and from it derive the minimax theory of the critical values of $\varphi$. Prominent in that theory is the so-called ``mountain pass theorem", due to Ambrosetti \& Rabinowitz \cite{2}. Here we state the result in a slightly more general form (see, for example, Gasinski \& Papageorgiou \cite[p. 648]{9}).

\begin{theorem}\label{Theorem1}
Assume that $\varphi\in C^{1}(X,\RR)$ satisfies the C-condition, $u_{0}, u_{1}\in X$, $\|u_{1}-u_{0}\|>r>0$
$$\max\{\varphi(u_{0}), \varphi(u_{1})\}<\inf\left[\varphi(u): \|u-u_{0}\|=r\right]=m_{r}$$
and $c=\displaystyle\inf_{\gamma\in\Gamma}\displaystyle\max_{0\leq t\leq1}\varphi(\gamma(t))$ with $\Gamma=\{\gamma\in C([0,1],X): \gamma(0)=u_{0}, \gamma(1)=u_{1}\}$. Then $c\geq m_{r}$ and $c$ is a critical value of $\varphi$.
\end{theorem}

The function spaces which we will use in the study of problem \eqref{1} are
\begin{itemize}
  \item the Sobolev space $H^{1}(\Omega)$;
  \item the Banach space $C^{1}(\overline{\Omega})$;
  \item the ``boundary" Lebesgue spaces $L^{p}(\partial\Omega)$ ($1\leq p\leq\infty$).
\end{itemize}

By $\|\cdot\|$ we will denote the norm of the Sobolev space $H^{1}(\Omega)$ defined by
$$\|u\|=\left[\|u\|_{2}^{2}+\|Du\|_{2}^{2}\right]^{1/2}\ \mbox{for all}\ u\in H^{1}(\Omega).$$
The Banach space $C^{1}(\overline{\Omega})$ is an ordered Banach space with positive cone
$$C_{+}=\left\{u\in C^{1}(\overline{\Omega}): u(z)\geq0\ \mbox{for all}\ z\in\overline{\Omega}\right\}.$$
This cone has a nonempty interior given by
$$\textrm{int}\,C_{+}=\left\{u\in C_{+}: u(z)>0\ \mbox{for all}\ z\in\overline{\Omega}\right\}.$$

The Lebesgue spaces $L^{p}(\partial\Omega)$ ($1\leq p\leq\infty$) are defined as follows. On $\partial\Omega$ we consider the $(N-1)$-dimensional Hausdorff (surface) measure $\sigma(\cdot)$. Having this measure on $\partial\Omega$, we can define in the usual way the Lebesgue spaces $L^{p}(\partial\Omega)$ ($1\leq p\leq\infty$). From the trace theorem, we know that there exists a unique linear continuous map $\gamma_{0}:H^{1}(\Omega)\rightarrow L^{2}(\partial\Omega)$, known as the ``trace map" such that
$$\gamma_{0}(u)=u|_{\partial\Omega}\ \mbox{for all}\ u\in C^{1}(\overline{\Omega}).$$
So, the trace map gives meaning to the notion of boundary values of a Sobolev function. We know that $\gamma_{0}$ is compact into $L^{q}(\partial\Omega)$ for all $q\in\left[1,\frac{2(N-1)}{N-2}\right)$. Moreover, we have
$$\textrm{im}\,\gamma_{0}=H^{\frac{1}{2},2}(\partial\Omega)\ \mbox{and}\ \textrm{ker}\,\gamma_{0}=H^{1}_{0}(\Omega).$$
In the rest of this paper, for the sake of notational simplicity, we drop the use of the trace map $\gamma_{0}$. All the restrictions on $\partial\Omega$ of functions from $H^{1}(\Omega)$ are understood in the sense of traces.

For $x\in\RR$, we set $x^{\pm}=\max\{\pm x, 0\}$. Then given $u\in H^{1}(\Omega)$, we define
$$u^{\pm}(\cdot)=u(\cdot)^{\pm}.$$
We know that
$$u^{\pm}\in H^{1}(\Omega)\ \mbox{and}\ u=u^{+}-u^{-},\, |u|=u^{+}+u^{-}.$$

Also by $|\cdot|_{N}$ we denote the Lebesgue measure on $\RR^{N}$ and if $h:\Omega\times\RR\rightarrow\RR$ is a Carath\'{e}odory function, we set
$$N_{h}(u)(\cdot)=h(\cdot,u(\cdot))\ \mbox{for all}\ u\in H^{1}(\Omega)$$
(the Nemytski or superposition operator corresponding to $h$).

By $2^{*}$ we denote the critical Sobolev exponent defined by
\[2^{*}=\left\{ \begin{array}{l l}\di \frac{2N}{N-2}, & \quad \text{if} \ N\geq3\\
+\infty, & \quad \text{if} \ N=1,2.
\end{array} \right. \]

We impose the following conditions on the potential function $\xi(\cdot)$ and on the boundary weight $\beta(\cdot)$.

\smallskip
$H(\xi):\, \xi\in L^{s}(\Omega)\ \mbox{with}\ s>N\ \mbox{and}\ \xi^{+}\in L^{\infty}(\Omega).$

\smallskip
$H(\beta):\, \beta\in W^{1,\infty}(\partial\Omega),\, \beta\geq0.$

\smallskip
Evidently if $\beta\equiv0$, then \eqref{1} reduces to the Neumann problem.

We introduce the $C^{1}$-functional $\vartheta:H^{1}(\Omega)\rightarrow\RR$ defined by
$$\vartheta(u)=\|Du\|_{2}^{2}+\int_{\Omega}\xi(z)u^{2}\,dz+\int_{\partial\Omega}\beta(z)u^{2}\,d\sigma\ \ \mbox{for all}\ u\in H^{1}(\Omega).$$

Consider a Carath\'{e}odory function $f_{0}:\Omega\times\RR\rightarrow\RR$ satisfying
$$|f_{0}(z,x)|\leq a_{0}(z)(1+|x|^{q-1})\ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ x\in\RR,$$
with $a_{0}\in L^{\infty}(\Omega)$ and $q\in(1,2^{*})$. We set $F_{0}(z,x)=\int_{0}^{x}f_{0}(z,s)ds$ and consider the $C^{1}$-functional $\varphi_{0}:H^{1}(\Omega)\rightarrow\RR$ defined by
$$\varphi_{0}(u)=\frac{1}{2}\vartheta(u)-\int_{\Omega}F_{0}(z,u)\,dz\ \mbox{for all}\ u\in H^{1}(\Omega).$$

As in Papageorgiou \& R\u{a}dulescu \cite{19}, using the regularity result of Wang \cite{26}, we have the following property.

\begin{prop}\label{Proposition2}
 Assume that hypotheses $H(\xi)$, $H(\beta)$ hold and $u_{0}\in H^{1}(\Omega)$ is a local $C^{1}(\overline{\Omega})$-minimizer of $\varphi_{0}$, that is, there exists $\rho_{0}>0$ such that
$$\varphi_{0}(u_{0})\leq\varphi_{0}(u_{0}+h)\ \mbox{for all}\ h\in C^{1}(\overline{\Omega})\ \mbox{with}\ \|h\|_{C^{1}(\overline{\Omega})}\leq\rho_{0}.$$
Then $u_{0}\in C^{1,\alpha}(\overline{\Omega})$ for some $\alpha\in(0,1)$ and $u_{0}$ is also  a local $H^{1}(\Omega)$-minimizer of $\varphi_{0}$, that is, there exists $\rho_{1}>0$ such that
$$\varphi_{0}(u_{0})\leq\varphi_{0}(u_{0}+h)\ \mbox{for all}\ h\in H^{1}(\Omega)\ \mbox{with}\ \|h\|\leq\rho_{1}.$$
\end{prop}

We will also need the spectrum of $u\longmapsto -\Delta u+\xi(z)u$ with the Robin boundary condition. So, we consider the following linear eigenvalue problem
\label{2}
\begin{equation}
\left \{
\begin{array}{l l}
-\Delta u(z)+\xi(z)u(z)=\hat{\lambda}u(z)\ & \text{in}\ \Omega,\\
\di\frac{\partial u}{\partial n}+\beta(z)u=0\ & \text{on}\ \partial\Omega.\\
\end{array} \right. \label{l11}
\end{equation}
This eigenvalue problem for $\beta\equiv0$ (Neumann problem) was studied in Papageorgiou \& R\u{a}dulescu \cite{18} and Papageorgiou \& Smyrlis \cite{22} and for the $p$-Laplacian by Mugnai \& Papageor\-giou \cite{16}. For the Robin problem with no potential (that is, $\xi\equiv0$), we have the work of Papageorgiou \& R\u{a}dulescu \cite{19} and for the $p$-Laplacian we refer to Papageorgiou \& R\u{a}dulescu \cite{19}, \cite{21}. An analogous study can be conducted for problem \eqref{2} and leads to similar results.

So, problem \eqref{2} has a smallest eigenvalue $\hat{\lambda}_{1}>-\infty$, which is given by
\begin{eqnarray}\label{3}
\hat{\lambda}_{1}=\inf\left[\frac{\vartheta(u)}{\|u\|_{2}^{2}}: u\in H^{1}(\Omega),\; u\not\equiv0\right].
\end{eqnarray}

We can find $\hat{\xi}>\max\{-\hat{\lambda}_{1},1\}$ such that
\begin{eqnarray}\label{4}
\vartheta(u)+\hat{\xi}\|u\|_{2}^{2}\geq c_{0}\|u\|^{2}\ \ \mbox{for all}\ u\in H^{1}(\Omega)\ \mbox{and some}\ c_{0}>0.
\end{eqnarray}
Using \eqref{4} and the spectral theory of compact self-adjoint linear operators, exactly as in \cite{18} and \cite{22}, we can generate the spectrum of \eqref{2}, which consists of a sequence $\{\hat{\lambda}_{k}\}_{k\geq1}$ of distinct eigenvalues such that $\hat{\lambda}_{k}\rightarrow+\infty$ as $k\rightarrow\infty$. Also, there is a corresponding sequence $\{\hat{u}_{k}\}_{k\geq1}\subseteq C^{1}(\overline{\Omega})$ of eigenfunctions which form an orthogonal basis of $H^{1}(\Omega)$ and an orthonormal basis of $L^{2}(\Omega)$. By $E(\hat{\lambda}_{k})$\; $k\geq1$, we denote the eigenspace corresponding to the eigenvalue $\hat{\lambda}_{k}$. We have that $E(\hat{\lambda}_{k})\subseteq C^{1}(\overline{\Omega})$ and it is finite dimensional. Moreover, from de Figueiredo \& Gossez \cite{7}, we know that $E(\hat{\lambda}_{m})$ has the ``unique continuation property" (the UCP for short), that is, if $u\in E(\hat{\lambda}_{m})$ vanishes on a set of positive Lebesgue measure, then $u\equiv0$.

We have
$$H^{1}(\Omega)=\overline{\underset{k\geq1}\oplus E(\hat{\lambda}_{k})}.$$
Using these eigenspaces, we can obtain the classical variational characterizations of $\hat{\lambda}_{m}$ for $m\geq2$, namely
\begin{align}\label{5}
\hat{\lambda}_{m}&=\inf\left[\frac{\vartheta(u)}{\|u\|_{2}^{2}}: u\in\widehat{H}_{m}=\overline{\underset{k\geq m}\oplus E(\hat{\lambda}_{k})},\; u\neq0\right]\nonumber \\
&=\sup\left[\frac{\vartheta(u)}{\|u\|_{2}^{2}}: u\in \overline{H}_{m}=\oplus_{k=1}^{m}E(\hat{\lambda}_{k}),\; u\neq0\right].
\end{align}

The infimum in \eqref{3} and both the infimum and the supremum in \eqref{5}, are realized on the corresponding eigenspace $E(\hat{\lambda}_{m})$\; $m\geq1$. Using Picone's identity (see, for example, Gasinski \& Papageorgiou \cite[p. 785]{9}), we show that $\hat{\lambda}_{1}$ is simple. Moreover, from \eqref{3} it is clear that the elements of $E(\hat{\lambda}_{1})$ do not change sign. Let $\hat{u}_{1}\in H^{1}(\Omega)$ be the $L^{2}$-normalized (that is, $\|\hat{u}_{1}\|_{2}=1$) positive eigenfunction corresponding to $\hat{\lambda}_{1}$. We have $\hat{u}_{1}\in C_{+}\setminus\{0\}$. Moreover, using hypothesis $H(\xi)$ and the maximum principle, we have $\hat{u}_{1}\in\textrm{int}\, C_{+}$. As in Gasinski \& Papageorgiou \cite[p. 743]{9}, via Picone's identity, we show that every nonprincipal eigenvalue $\hat{\lambda}_{m}$\, ($m\geq2$) has nodal (sign changing) eigenfunctions.

From \eqref{3}, \eqref{5} and the UCP, we derive easily the following useful inequalities.

\begin{prop}\label{Proposition3}
(a) If $\eta\in L^{\infty}(\Omega)$ and $\eta(z)\leq\hat{\lambda}_{m}$ for a.a. $z\in\Omega$ with $m\geq1$ and the inequality is strict on a set of positive measure, then $\vartheta(u)-\int_{\Omega}\eta(z)u^{2}\, dz\geq c_{1}\|u\|^{2}$ for some $c_{1}>0$, all $u\in\widehat{H}_{m}=\overline{\underset{k\geq m}\oplus E(\hat{\lambda}_{k})}$.\\
(b) If $\eta\in L^{\infty}(\Omega)$ and $\eta(z)\geq\hat{\lambda}_{m}$ for a.a. $z\in\Omega$ with $m\geq1$ and the inequality is strict on a set of positive measure, then $\vartheta(u)-\int_{\Omega}\eta(z)u^{2}\, dz\leq-c_{2}\|u\|^{2}$ for some $c_{2}>0$, all $u\in\overline{H}_{m}=\oplus_{k=1}^{m}E(\hat{\lambda}_{k})$.
\end{prop}

We will also use the following weighted linear eigenvalue problem
\begin{eqnarray}\label{6}
-\Delta u(z)=\tilde{\lambda}m(z)u(z)\ \mbox{in}\ \Omega,\; \frac{\partial u}{\partial n}+\beta(z)u=0\ \mbox{on}\ \partial\Omega.
\end{eqnarray}
Here $m\in L^{s}(\Omega)$ with $s>N$ and is in general indefinite. As in de Figueiredo \cite{6} (see also Gasinski \& Papageorgiou \cite[Section 6.1]{9}),  we show that \eqref{6} has in general a double sequence of eigenvalues $\{(\hat{\lambda}_{\pm})_{k}\}_{k\geq1}$ such that
$$-\infty\leftarrow\hat{\lambda}_{-k}<\cdots <\hat{\lambda}_{-2}<\hat{\lambda}_{-1}\leq0\leq\hat{\lambda}_{1}<\hat{\lambda}_{2}<\cdots <\hat{\lambda}_{k}
\rightarrow\infty\ \mbox{as}\ k\rightarrow\infty.$$

If $m^{+}=0$ there are no positive eigenvalues, while if $m^{-}=0$, there are no negative eigenvalues. If $m^{+}\neq0$ (resp. $m^{-}\neq0$), then $\hat{\lambda}_{1}$ (resp. $\hat{\lambda}_{-1}$) is simple. Moreover, we have
\begin{eqnarray}\label{7}
\tilde{\lambda}_{1}=\inf\left[\|Du\|_{2}^{2}+\int_{\partial\Omega}\beta(z)u^{2}\, d\sigma:\; u\in H^{1}(\Omega),\; \int_{\Omega}m(z)u^{2}\, dz=1 \right]
\end{eqnarray}
and
\begin{eqnarray}\label{8}
 \tilde{\lambda}_{-1}=\inf\left[\|Du\|_{2}^{2}+\int_{\partial\Omega}\beta(z)u^{2}\, d\sigma:\; u\in H^{1}(\Omega),\; \int_{\Omega}m(z)u^{2}\, dz=-1 \right]
\end{eqnarray}
with both infima realized on the corresponding one dimensional eigenspace.

Next let us recall some basic definitions and facts from Morse theory (critical groups). For details we refer to Motreanu, Motreanu \& Papageorgiou \cite{14}.

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

Let $(Y_{1},Y_{2})$ be a topological pair such that $Y_{2}\subseteq Y_{1}\subseteq X$. For every integer $k\geq0$, we denote by $H_{k}(Y_{1},Y_{2})$  the $k$th-relative singular homology group for the pair $(Y_{1},Y_{2})$ with integer coefficients. Recall that for every integer $k<0$, we have $H_{k}(Y_{1},Y_{2})=0$. Let $u_{0}\in K_{\varphi}^{c}$ be an isolated critical point. The critical groups of $\varphi$ at $u_{0}$ are defined by $$C_{k}(\varphi,u_{0})=H_{k}(\varphi^{c}\cap U,\, \varphi^{c}\cap U\setminus\{u_{0}\})\ \mbox{for every integer}\ k\geq0.$$
Here $U$ is an open 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 $\varphi\in C^{1}(X,\RR)$ satisfies the C-condition and $\inf \varphi(K_{\varphi})>-\infty$. Let $c<\inf \varphi(K_{\varphi})$. The critical groups of $\varphi$ at infinity are defined by
$$C_{k}(\varphi,\infty)=H_{k}(X,\varphi^{c})\ \mbox{for every integer}\ k\geq0.$$

The second deformation theorem (see, for example, Gasinski \& Papageorgiou \cite[p. 628]{9}) implies that this definition of critical groups at infinity, is independent of the choice of the level $c<\inf \varphi(K_{\varphi})$.

Assume that $K_{\varphi}$ is finite. We introduce the following quantities
\begin{align*}
& M(t,u)=\sum_{k\geq0}\textrm{rank}\,C_{k}(\varphi,u)t^{k}\ \mbox{for all}\ t\in\RR,\ \mbox{all}\ u\in K_{\varphi},\\
& P(t,\infty)=\sum_{k\geq0}\textrm{rank}\,C_{k}(\varphi,\infty)t^{k}\ \mbox{for all}\ t\in\RR.
\end{align*}
The {\it Morse relation} says that
\begin{eqnarray}\label{9}
\sum_{u\in K_{\varphi}} M(t,u)=P(t,\infty)+(1+t)Q(t),
\end{eqnarray}
where $Q(t)=\sum_{k\geq0}\beta_{k}t^{k}$ is a formal series in $t\in\RR$ with nonnegative integer coefficients $\beta_{k}$.

Suppose that $X=Y\oplus V$ with $\textrm{dim}\,Y<\infty$ and $\varphi\in C^{1}(X,\RR)$. We say that $\varphi$ has a local linking at $u=0$, if there exists $r>0$ such that
\begin{align*}
& \varphi(y)\leq0\ \ \mbox{for all}\ y\in Y,\ \|y\|\leq r\\
& \varphi(v)>0\ \ \mbox{for all}\ v\in V,\, 0<\|v\|\leq r.
\end{align*}
In this case we have
$$C_{k}(\varphi,0)\neq0\ \ \mbox{for}\ k=\textrm{dim}\,Y<\infty.$$

If $\varphi\in C^{1}(X,\RR)$ satisfies the C-condition and for some integer $k\geq0$, we have
$$C_{k}(\varphi,0)\neq0\ \ \mbox{and}\ \ C_{k}(\varphi,\infty)=0,$$
then there exists $u\in K_{\varphi}, u\neq0$ such that
$$either\ \ [\varphi(u)<0\ \mbox{and}\ C_{k-1}(\varphi,u)\neq0]\ \ or\ \ [\varphi(u)>0\ \mbox{and}\ C_{k+1}(\varphi,u)\neq0].$$

In what follows we denote by $A\in{\mathscr L}(H^{1}(\Omega), H^{1}(\Omega)^{*})$ the continuous linear operator defined by
$$\langle A(u),h\rangle=\int_{\Omega}(Du,Dh)_{\RR^{N}}\,dz\ \mbox{for all}\ u,h\in H^{1}(\Omega).$$

Also, we say that a Banach space $X$ has the {\it Kadec-Klee property}, if the following is true:
\begin{center}
  ``If $u_{n}\overset{w}{\rightarrow}u$ in $X$ and $\|u_{n}\|\rightarrow\|u\|$, then $u_{n}\rightarrow u$ in $X$".
\end{center}
We know (see \cite{9}) that locally uniformly convex Banach spaces, in particular Hilbert spaces, have the Kadec-Klee property.

\section{Two Nontrivial Solutions}

In this section, we prove a multiplicity theorem producing two nontrivial solutions for problem \eqref{1}. For this purpose, we introduce the following hypotheses on the reaction $f(z,x)$.

\medskip
$H_{1}$: $f:\Omega\times\RR\rightarrow\RR$ is a Carath\'{e}odory function such that $f(z,0)=0$ for a.a. $z\in\Omega$ and
\begin{itemize}
\item[(i)] $|f(z,x)|\leq a(z)(1+|x|^{r-1})$ for a.a. $z\in\Omega$, all $x\in\RR$, with $a\in L^{\infty}(\Omega)_{+}$, $2<r<2^{*}$;
\item[(ii)] if $F(z,x)=\int_{0}^{x}f(z,s)\,ds$, then
$$\lim_{x\rightarrow+\infty}\frac{F(z,x)}{x^{2}}=+\infty\ \mbox{uniformly for a.a.}\ z\in\Omega,$$
and if $\varrho(z,x)=f(z,x)x-2F(z,x)$, then there exists $\tau\in L^{1}(\Omega)_{+}$ such that
\begin{eqnarray}\label{10}
\varrho(z,x)\leq\varrho(z,y)+\tau(z)\ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ 0\leq x\leq y;
\end{eqnarray}
\item[(iii)] there exists a real number $\xi_0$ such that $\xi_{0}\leq\displaystyle\liminf_{x\rightarrow-\infty}\frac{f(z,x)}{x}\leq\displaystyle\limsup_{x\rightarrow-\infty}\frac{f(z,x)}{x}\leq\hat{\lambda}_{1}$ uniformly for a.a. $z\in\Omega$ and
$$\lim_{x\rightarrow-\infty}\left[f(z,x)x-2F(z,x)\right]=+\infty\ \mbox{uniformly for a.a.}\ z\in\Omega;$$
\item[(iv)] there exist an integer $m\geq2$, $\eta\in L^{\infty}(\Omega)$ and $\delta_{0}>0$ such that
\begin{align*}
&\eta(z)\leq\hat{\lambda}_{m+1}\ \mbox{for a.a.}\ z\in\Omega,\ \eta\not\equiv\hat{\lambda}_{m+1}\\
&\limsup_{x\rightarrow0}\frac{f(z,x)}{x}\leq\eta(z)\ \mbox{uniformly for a.a.}\ z\in\Omega\\
&f(z,x)x\geq\hat{\lambda}_{m}x^{2}\ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\leq\delta_{0}.
\end{align*}
\end{itemize}
%.........................

\textbf{Remarks}: Hypothesis $H_{1}$(ii) implies that for a.a. $z\in\Omega$ the reaction $f(z,\cdot)$ is superlinear near $+\infty$, but without satisfying the usual in such cases AR-condition. We recall that the AR-condition (unilateral version) says that there exist $q>2$ and $M>0$ such that
\begin{align}
\begin{split}\label{11}
&(a)\quad 0<qF(z,x)\leq f(z,x)x\ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ x\geq M,\\
&(b)\quad \essinf_{\Omega}F(\cdot,M)>0
\end{split}
\end{align}
(see Ambrosetti \& Rabinowitz \cite{2} and Mugnai \cite{15}). Integrating \eqref{11}-(a) and using \eqref{11}-(b), we obtain the weaker condition
\begin{eqnarray}\label{12}
c_{3}x^{q}\leq F(z,x)\ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ x\geq M\ \mbox{and some}\ c_{3}>0.
\end{eqnarray}
From \eqref{12} we infer the much weaker condition
\begin{eqnarray}\label{13}
\lim_{x\rightarrow+\infty}\frac{F(z,x)}{x^{2}}=+\infty\ \mbox{uniformly for a.a.}\ z\in\Omega\ (\mbox{recall}\ q>2).
\end{eqnarray}

The AR-condition ensures that the C-condition holds for the energy functional of problem \eqref{1}. From \eqref{12} we see that the AR-condition implies that the primitive $F(z,\cdot)$ has at least $q$-polynomial growth near $+\infty$. In this way, we exclude from consideration superlinear reactions which exhibit ``slower" growth near $+\infty$. For example, consider the function
$$f(x)=x\ln{x}\quad \mbox{for all}\ x\geq M>1.$$
To be able to treat also such reactions, instead of the AR-reaction, we employ \eqref{10} and \eqref{13}. Note that \eqref{10} which is a quasimonotonicity condition, is satisfied if for a.a. $z\in\Omega$\, $\varrho(z,\cdot)$ is nondecreasing on $[M,+\infty)$ (in which case we can take $\tau\equiv0$). This property is equivalent to saying that for a.a. $z\in\Omega$, the mapping $x\longmapsto\frac{f(z,x)}{x}$ is nondecreasing on $[M,+\infty)$, see Li \& Yang \cite{11paper} and Miyagaki \& Souto \cite{12}.

Hypothesis $H_{1}$\,(iii) implies that in the negative direction, $f(z,\cdot)$ is sublinear and can be resonant with respect to the principal eigenvalue $\hat{\lambda}_{1}$. Resonance can also occur at zero with respect to a nonprincipal eigenvalue $\hat{\lambda}_{m}$ (see hypothesis $H_{1}$\,(iv)). So, we are dealing with a problem which can be resonant at both $-\infty$ and zero (double resonance).

Let $\varphi: H^{1}(\Omega)\rightarrow\RR$ be the energy functional for problem \eqref{1} defined by
$$\varphi(u)=\frac{1}{2}\vartheta(u)-\int_{\Omega}F(z,u(z))\,dz\quad \mbox{for all}\ u\in H^{1}(\Omega).$$
Evidently $\varphi\in C^{1}(H^{1}(\Omega))$.

\begin{prop}\label{Proposition4}
Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{1}$ hold. Then the energy functional $\varphi$ satisfies the C-condition.
\end{prop}

\begin{proof}
Let $\{u_{n}\}_{n\geq1}\subseteq H^{1}(\Omega)$ be a sequence such that
\begin{align}
&|\varphi(u_{n})|\leq M_{1}\quad \mbox{for some}\ M_{1}>0,\; \mbox{all}\ n\geq1,\label{14}\\
&(1+\|u_{n}\|)\varphi'(u_{n})\rightarrow0\quad \mbox{in}\ H^{1}(\Omega)^{*}\; \mbox{as}\ n\rightarrow\infty.\label{15}
\end{align}
From \eqref{15} we have for all $n\geq 1$, all $h\in H^1(\Omega)$
\begin{eqnarray}\label{16}
&&\left|\langle A(u_{n}),h\rangle+\int_{\Omega}\xi(z)u_{n}h\,dz+\int_{\partial\Omega}\beta(z)u_{n}h\,d\sigma-\int_{\Omega}f(z,u_{n})h\,dz\right|
\leq\frac{\varepsilon_{n}\|h\|}{1+\|u_{n}\|}\,,
\end{eqnarray}
with $\varepsilon\rightarrow 0^+$ as $n\rightarrow\infty$.

First we show that $\{u_{n}^{-}\}_{n\geq1}\subseteq H^{1}(\Omega)$ is bounded. Arguing by contradiction, suppose that for at least a subsequence, we have $\|u_{n}^{-}\|\rightarrow+\infty$ as $n\rightarrow\infty$. We set $y_{n}=\frac{u_{n}^{-}}{\|u_{n}^{-}\|}$ for all $n\geq1$. Then $\|y_{n}\|=1$, $y_{n}\geq0$ for all $n\geq1$. So, by passing to a suitable subsequence if necessary and using the Sobolev embedding theorem and the trace theorem, we have
\begin{eqnarray}\label{17}
y_{n}\rightarrow y\ \mbox{in}\ H^{1}(\Omega)\ \mbox{and}\ y_{n}\rightarrow y\ \mbox{in}\ L^{\frac{2s}{s-1}}(\Omega)\ \mbox{and in}\ L^{2}(\partial\Omega),\; y\geq0.
\end{eqnarray}
In \eqref{16} we choose $h=-u_{n}^{-}\in H^{1}(\Omega)$ and obtain
\begin{align}\label{18}
&\vartheta(u_{n}^{-})-\int_{\Omega}f(z,u_{n})(-u_{n}^{-})\,dz\leq\varepsilon_{n}\quad \mbox{for all}\ n\geq1,\nonumber\\
\Rightarrow&\vartheta(y_{n})-\int_{\Omega}\frac{N_{f}(-u_{n}^{-})}{\|u_{n}^{-}\|}(-y_{n})\,dz\leq\frac{\varepsilon_{n}}{\|u_{n}^{-}\|^{2}}\quad \mbox{for all}\ n\geq1.
\end{align}
We have that $u_{n}^{-}(z)\rightarrow+\infty$ for a.a. $z\in\{y>0\}$. Hypotheses $H_{1}$\,(i), (iii), (iv) imply that
\begin{align*}
&|f(z,x)|\leq c_{4}|x|\quad \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq0\; \mbox{and some}\ c_{4}>0,\\
\Rightarrow&\left\{\frac{N_{f}(-u_{n}^{-})}{\|u_{n}^{-}\|}\right\}_{n\geq1}\subseteq L^{2}(\Omega)\quad \mbox{is bounded}.
\end{align*}
Using this fact and hypothesis $H_{1}$\,(iii),  we have up to a subsequence
\begin{align}\label{19}
\frac{N_{f}(-u_{n}^{-})}{\|u_{n}^{-}\|}\overset{w}\rightarrow-ky\quad \mbox{in}\ L^{2}(\Omega),\; k(z)\leq\hat{\lambda}_{1}\quad \mbox{for a.a.}\ z\in\Omega
\end{align}
(see Aizicovici, Papageorgiou \& Staicu \cite{1}, proof of Proposition 16). So, if in \eqref{18} we pass to the limit as $n\rightarrow\infty$ and use \eqref{17} and \eqref{19}, then
\begin{eqnarray}\label{20}
\vartheta(y)\leq\int_{\Omega}k(z)y^{2}\,dz.
\end{eqnarray}
If $k\not\equiv\hat{\lambda}_{1}$, then from \eqref{20} and Proposition \ref{Proposition3}, we have
$$
c_{1}\|y\|^{2}\leq0
\Rightarrow y=0.
$$
Then from \eqref{18} we see that
\begin{align*}
&\|Dy_{n}\|_{2}\rightarrow0,\\
\Rightarrow&y_{n}\rightarrow0\quad \mbox{in}\ H^{1}(\Omega)\ (\mbox{see}\ \eqref{17}),
\end{align*}
a contradiction to the fact that $\|y_{n}\|=1$ for all $n\geq1$.

If $k(z)=\hat{\lambda}_{1}$ for a.a. $z\in\Omega$, then from \eqref{20} and \eqref{3} we have
\begin{align*}
&\vartheta(y)=\hat{\lambda}_{1}\|y\|_{2}^{2},\\
\Rightarrow&y=t\hat{u}_{1}\; \mbox{with}\ t\geq0\; (\mbox{recall}\ y\geq0,\; \mbox{see}\ \eqref{17}).
\end{align*}
If $t=0$, then $y=0$ and as above we reach a contradiction. Hence $t>0$ and we have $y\in\textrm{int}\,C_{+}$ (recall $\hat{u}_{1}\in\textrm{int}\,C_{+}$). Therefore
\begin{align}
&u_{n}^{-}(z)\rightarrow+\infty\quad \mbox{for a.a.}\ z\in\Omega,\nonumber\\
\Rightarrow&f(z,-u_{n}^{-}(z))(-u_{n}^{-}(z))-2F(z,-u_{n}^{-}(z))\rightarrow+\infty\quad \mbox{for a.a.}\ z\in\Omega\label{21}\\
&\hspace{6cm}(\mbox{see hypothesis}\ H_{1}\, \mbox{(iii)}).\nonumber
\end{align}
Using hypothesis $H_{1}$\,(iii), Fatou's lemma and \eqref{21}, we have
\begin{eqnarray}\label{22}
\int_{\Omega}\left[f(z,-u_{n}^{-})(-u_{n}^{-})-2F(z,-u_{n}^{-})\right]dz\rightarrow+\infty\quad \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
From \eqref{14} we have
\begin{eqnarray}\label{23}
\vartheta(u_{n})-\int_{\Omega}2F(z,u_{n})\,dz\leq2M_{1}\quad \mbox{for all}\ n\geq1.
\end{eqnarray}
Also, if in \eqref{16} we choose $h=u_{n}\in H^{1}(\Omega)$, then
\begin{eqnarray}\label{24}
-\vartheta(u_{n})+\int_{\Omega}f(z,u_{n})u_{n}\,dz\leq\varepsilon_{n}\quad \mbox{for all}\ n\geq1.
\end{eqnarray}
Adding \eqref{23} and \eqref{24}, we obtain for all $n\geq 1$
\begin{align}\label{25}
&\int_{\Omega}\left[f(z,u_{n})u_{n}-2F(z,u_{n})\right]dz\leq M_{2}\; \mbox{for some}\ M_{2}>0,\; \mbox{all}\ n\geq1,\nonumber\\
\Rightarrow&\int_{\Omega}\left[f(z,-u_{n}^{-})(-u_{n}^{-})-2F(z,-u_{n}^{-})\right]dz+\int_{\Omega}\left[f(z,u_{n}^{+})u_{n}^{+}
-2F(z,u_{n}^{+})\right]dz\leq M_{2}\,.
\end{align}
Note that hypothesis $H_{1}$\,(ii) implies that
\begin{eqnarray}\label{26}
-\tau(z)\leq f(z,x)x-2F(z,x)\quad \mbox{for a.a.}\ z\in\Omega,\, \mbox{all}\ x\geq0.
\end{eqnarray}
Using \eqref{26} in \eqref{25}, we obtain
\begin{eqnarray}\label{27}
\int_{\Omega}\left[f(z,-u_{n}^{-})(-u_{n}^{-})-2F(z,-u_{n}^{-})\right]dz\leq M_{3}\quad \mbox{for some}\ M_{3}>0,\; \mbox{all}\ n\geq1.
\end{eqnarray}
Comparing \eqref{22} and \eqref{27}, we reach a contradiction. This proves that
\begin{eqnarray}\label{28}
\{u_{n}^{-}\}_{n\geq1}\subseteq H^{1}(\Omega)\quad \mbox{is bounded}.
\end{eqnarray}
From \eqref{14} and \eqref{28}, we have
\begin{eqnarray}\label{29}
\vartheta(u_{n}^{+})-\int_{\Omega}2F(z,u_{n}^{+})\,dz\leq M_{4}\quad \mbox{for some}\ M_{4}>0,\ \mbox{all}\ n\geq1.
\end{eqnarray}
Also, in \eqref{16} we choose $h=u_{n}^{+}\in H^{1}(\Omega)$. Then
\begin{eqnarray}\label{30}
-\vartheta(u_{n}^{+})+\int_{\Omega}f(z,u_{n}^{+})u_{n}^{+}\,dz\leq \varepsilon_{n}\quad \mbox{for all}\ n\geq1.
\end{eqnarray}
We add \eqref{29} and \eqref{30} and obtain
\begin{align}\label{31}
&\int_{\Omega}\left[f(z,u_{n}^{+})u_{n}^{+}-2F(z,u_{n}^{+})\right]dz\leq M_{5}\quad \mbox{for some}\ M_{5}>0,\ \mbox{all}\ n\geq1,\nonumber\\
\Rightarrow&\int_{\Omega}\varrho(z,u_{n}^{+})\,dz\leq M_{5}\quad \mbox{for all}\ n\geq1.\vspace{0.5cm}
\end{align}
\textbf{Claim}: $\{u_{n}^{+}\}_{n\geq1}\subseteq H^{1}(\Omega)$ is bounded.\\
We argue indirectly. So, suppose that the Claim is not true. We may assume that
$$\|u_{n}^{+}\|\rightarrow\infty\; \mbox{as}\ n\rightarrow\infty.$$
Let $v_{n}=\frac{u_{n}^{+}}{\|u_{n}^{+}\|}$\, $n\geq1$. Then $\|v_{n}\|=1$ for all $n\geq1$ and so we may assume that
\begin{eqnarray}\label{32}
v_{n}\overset{w}\rightarrow v\; \mbox{in}\ H^{1}(\Omega)\; \mbox{and}\ v_{n}\rightarrow v\; \mbox{in}\ L^{\frac{2s}{s-1}}(\Omega)\; \mbox{and in}\ L^{2}(\partial\Omega).
\end{eqnarray}
First suppose that $v\neq0$ and let $\Omega_{0}=[v=0]$. Then
$$u_{n}^{+}(z)\rightarrow+\infty\; \mbox{for a.a.}\ z\in\Omega\setminus\Omega_{0}.$$
From hypothesis $H_{1}$\,(ii), we have
$$\frac{F(z,u_{n}^{+}(z))}{\|u_{n}^{+}\|^{2}}=\frac{F(z,u_{n}^{+}(z))}{u_{n}^{+}(z)^{2}}v_{n}(z)^{2}\rightarrow+\infty\; \mbox{for a.a.}\ z\in\Omega\setminus\Omega_{0}.$$
Then, by virtue of hypothesis $H_{1}$\,(ii) and Fatou's lemma, we have
\begin{eqnarray}\label{33}
\lim_{n\rightarrow\infty}\int_{\Omega}\frac{F(z,u_{n}^{+})}{\|u_{n}^{+}\|^{2}}\,dz=+\infty.
\end{eqnarray}
On the other hand, from \eqref{14} and \eqref{28}, we have
\begin{align}
&\int_{\Omega}F(z,u_{n}^{+})\,dz\leq M_{6}+|\vartheta(u_{n}^{+})|\quad \mbox{for all}\ n\geq1,\; \mbox{with}\ M_{6}=2M_{1},\nonumber\\
\Rightarrow&\int_{\Omega}\frac{F(z,u_{n}^{+})}{\|u_{n}^{+}\|^{2}}\,dz\leq\frac{M_{6}}{\|u_{n}^{+}\|^{2}}+|\vartheta(v_{n})|\quad \mbox{for all}\ n\geq1,\nonumber\\
\Rightarrow&\int_{\Omega}\frac{F(z,u_{n}^{+})}{\|u_{n}^{+}\|^{2}}\,dz\leq M_{7}\quad \mbox{for some}\ M_{7}>0,\; \mbox{all}\ n\geq1\label{34}\\
&\hspace{6cm} (\mbox{note that}\ \{\vartheta(v_{n})\}_{n\geq1}\subseteq\RR\; \mbox{is bounded})\nonumber.
\end{align}
Comparing \eqref{33} and \eqref{34} we reach a contradiction.

Next assume $v=0$. Let $\gamma>0$ and define
\begin{align*}
&w_{n}=(2\gamma)^{1/2}v_{n}\in H^{1}(\Omega)\quad \mbox{for all}\ n\geq1,\\
\Rightarrow\ &w_{n}\rightarrow0\quad \mbox{in}\ L^{r}(\Omega)\; (\mbox{see}\ \eqref{24}\; \mbox{and recall}\ v=0).
\end{align*}
Then from the theorem of Krasnoselskii (see, for example, Gasinski \& Papageorgiou \cite[p. 407]{9}), we have
\begin{eqnarray}\label{35}
\int_{\Omega}F(z,w_{n})\,dz\rightarrow0\quad \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
Since $\|u_{n}^{+}\|\rightarrow\infty$, we can find $n_{0}\in\NN$ such that
\begin{eqnarray}\label{36}
0<(2\gamma)^{1/2}\frac{1}{\|u_{n}^{+}\|}\leq1\quad \mbox{for all}\ n\geq n_{0}.
\end{eqnarray}
Let $t_{n}\in[0,1]$ be such that
$$\varphi(t_{n}u_{n}^{+})=\max_{0\leq t\leq1}\varphi(tu_{n}^{+}),\quad n\geq1.$$
From \eqref{36}, we have
\begin{align}
\varphi(t_{n}u_{n}^{+})&\geq\varphi(w_{n})\nonumber\\
&=2\gamma\vartheta(v_{n})-\int_{\Omega}F(z,w_{n})\,dz\quad \mbox{for all}\ n\geq1.\label{37}
\end{align}
From \eqref{4}, we have
\begin{eqnarray}\label{38}
\vartheta(u)\geq c_{0}\|u\|^{2}-\hat{\xi}\|u\|_{2}^{2}\quad \mbox{for all}\ u\in H^{1}(\Omega).
\end{eqnarray}
Using \eqref{38} in \eqref{37} and recalling that $\|v_{n}\|=1$ for all $n\geq1$, we obtain
$$\varphi(t_{n}u_{n}^{+})\geq2\gamma c_{0}-2\gamma\hat{\xi}\|v_{n}\|_{2}^{2}-\int_{\Omega}F(z,w_{n})\,dz\quad \mbox{for all}\ n\geq1.$$
Since $\|v_{n}\|_{2}\rightarrow0$ (see \eqref{32} and recall that $v=0$) and $\int_{\Omega}F(z,w_{n})\,dz\rightarrow0$ (see \eqref{35}), given $\delta>0$, we can find $n_{1}\in\NN$, $n_{1}\geq n_{0}$ such that
$$\varphi(t_{n}u_{n}^{+})\geq2\gamma c_{0}-\delta\quad \mbox{for all}\ n\geq n_{1}.$$
Recall that $\gamma>0$ and $\delta>0$ are arbitrary. So, let $\gamma\rightarrow+\infty$ and $\delta\rightarrow0^{+}$ to conclude that
\begin{eqnarray}\label{39}
\varphi(t_{n}u_{n}^{+})\rightarrow+\infty\quad \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
We have $$\varphi(0)=0\; \mbox{and}\ \varphi(u_{n}^{+})\leq M_{8}\; \mbox{for some}\ M_{8}>0,\; \mbox{all}\ n\geq1$$
\hspace{9cm}(see \eqref{14} and \eqref{28}).\\
From \eqref{39} it follows that we can find $n_{2}\in\NN$ such that $t_{n}\in(0,1)$ for all $n\geq n_{2}$. Hence
\begin{align}
&\frac{d}{dt}\varphi(tu_{n}^{+})|_{t=t_{n}}=0\quad \mbox{for all}\ n\geq n_{2},\nonumber\\
\Rightarrow&\langle\varphi'(t_{n}u_{n}^{+}),u_{n}^{+}\rangle=0\quad \mbox{for all}\ n\geq n_{2}\ (\mbox{by the chain rule}),\nonumber\\
\Rightarrow&\langle\varphi'(t_{n}u_{n}^{+}),t_{n}u_{n}^{+}\rangle=0\quad \mbox{for all}\ n\geq n_{2}\ (\mbox{recall}\ t_{n}\in(0,1)\; \mbox{for all}\ n\geq n_{2}),\nonumber\\
\Rightarrow&\vartheta(t_{n}u_{n}^{+})=\int_{\Omega}f(z,t_{n}u_{n}^{+})(t_{n}u_{n}^{+})\,dz\; \mbox{for all}\ n\geq n_{2}.\label{40}
\end{align}
Hypothesis $H_{1}$\,(ii) implies that
\begin{align}
&\int_{\Omega}\varrho(z,t_{n}u_{n}^{+})\,dz\leq\int_{\Omega}\varrho(z,u_{n}^{+})\,dz+\|\tau\|_{1}\quad \mbox{for all}\ n\geq n_{2}\label{41}\\
&\hspace{4cm}(\mbox{recall}\ t_{n}\in(0,1)\; \mbox{for all}\ n\geq n_{2}).\nonumber
\end{align}
Using \eqref{41} in \eqref{40} and recalling the definition of the function $\varrho(z,x)$, we obtain
\begin{eqnarray}\label{42}
2\varphi(t_{n}u_{n}^{+})\leq M_{9}\quad \mbox{for some}\ M_{9}>0,\; \mbox{all}\ n\geq n_{2}.
\end{eqnarray}
Comparing \eqref{39} and \eqref{42}, we reach a contradiction. So, the Claim is true.

From the Claim and \eqref{28} it follows that $\{u_{n}\}_{n\geq1}\subseteq H^{1}(\Omega)$ is bounded. So, we may assume that
\begin{eqnarray}\label{43}
u_{n}\overset{w}\rightarrow u\ \ \mbox{in}\ H^{1}(\Omega)\ \ \mbox{and}\ u_{n}\rightarrow u\ \ \mbox{in}\ L^{\frac{2s}{s-1}}\ \ \mbox{and in}\ L^{2}(\partial\Omega).
\end{eqnarray}
In \eqref{16} we choose $h=u_{n}-u\in H^{1}(\Omega)$, pass to the limit as $n\rightarrow\infty$ and use \eqref{43}. Then
\begin{align*}
&\lim_{n\rightarrow\infty}\langle A(u_{n}),u_{n}-u\rangle=0,\\
\Rightarrow&\|Du_{n}\|_{2}\rightarrow\|Du\|_{2},\\
\Rightarrow&u_{n}\rightarrow u\ \ \mbox{in}\ H^{1}(\Omega)\ (\mbox{by the Kadec-Klee property}\ (\mbox{see}\ \eqref{43}),\\
\Rightarrow&\varphi\ \ \mbox{satisfies the C-condition}.
\end{align*}
This completes the proof of Proposition \ref{Proposition4}.
\end{proof}

In addition to the energy functional $\varphi$, we will also consider its ``negative" truncation-perturbation. More precisely, let $\hat{\xi}>0$ be as in \eqref{4} and define the following Carath\'{e}odory function
\[\hat{f}_{-}(z,x)=\left\{ \begin{array}{l l} f(z,x)+\hat{\xi}x & \quad \text{if} \ x<0\\
0 & \quad \text{if} \ 0\leq x.
\end{array} \right. \]

We set $\widehat{F}_{-}(z,x)=\displaystyle\int_{0}^{x}\hat{f}_{-}(z,s)\,ds$ and consider the $C^{1}$-functional $\hat{\varphi}_{-}:H^{1}(\Omega)\rightarrow\RR$ defined by
$$\hat{\varphi}_{-}(u)=\frac{1}{2}\vartheta(u)+\frac{\hat{\xi}}{2}\|u\|_{2}^{2}-\int_{\Omega}\widehat{F}_{-}(z,u)\,dz\ \ \mbox{for all}\ u\in H^{1}(\Omega).$$
A careful reading of the proof of Proposition \ref{Proposition4}, leads to the following result.

\begin{prop}\label{Proposition5}
Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{1}$ hold. Then the functional $\hat{\varphi}_{-}$ is coercive.
\end{prop}
\begin{proof}
We argue by contradiction. So, suppose that $\hat{\varphi}_{-}$ is not coercive. Then we can find $\{u_{n}\}_{n\geq1}\subseteq H^{1}(\Omega)$ and $M_{10}>0$ such that
\begin{eqnarray}\label{44}
\|u_{n}\|\rightarrow\infty\ \ \mbox{and}\ \hat{\varphi}_{-}(u_{n})\leq M_{10}\ \ \mbox{for all}\ n\geq1.
\end{eqnarray}
From \eqref{44} we have
\begin{eqnarray}\label{45}
\frac{1}{2}\vartheta(u_{n})+\frac{\hat{\xi}}{2}\|u_{n}\|_{2}^{2}-\int_{\Omega}\widehat{F}_{-}(z,u_{n})\,dz\leq M_{10}\ \ \mbox{for all}\ n\geq1.
\end{eqnarray}
Let $y_{n}=\frac{u_{n}}{\|u_{n}\|}$\ \ $n\geq1$. Then $\|y_{n}\|=1$ for all $n\geq1$. Using the Sobolev embedding theorem and the trace theorem and by passing to a subsequence if necessary, we may assume that
\begin{eqnarray}\label{46}
y_{n}\overset{w}\rightarrow y\ \ \mbox{in}\ H^{1}(\Omega)\ \ \mbox{and}\ y_{n}\rightarrow y\ \ \mbox{in}\ L^{2}(\Omega)\ \mbox{and in}\ L^{2}(\partial\Omega).
\end{eqnarray}
From \eqref{45} we have
\begin{eqnarray}\label{47}
\frac{1}{2}\vartheta(y_{n})+\frac{\hat{\xi}}{2}\|y_{n}^{+}\|-\int_{\Omega}\frac{F(z,-u_{n}^{-})}{\|u_{n}\|^{2}}\,dz\leq\frac{M_{10}}{\|u_{n}\|^{2}}\ \ \mbox{for all}\ n\geq1.
\end{eqnarray}
Hypotheses $H_{1}$\,(i), (iii) imply that
\begin{eqnarray}\label{48}
|F(z,x)|\leq c_{5}(1+|x|^{2})\ \ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ x\leq0\ \mbox{and some}\ c_{5}>0.
\end{eqnarray}
From \eqref{46} and \eqref{48} it follows that
$$\left\{\frac{N_{F}(-u_{n}^{-})}{\|u_{n}\|^{2}}\right\}_{n\geq1}\subseteq L^{1}(\Omega)\ \ \mbox{is uniformly integrable}.$$
Then by the Dunford-Pettis theorem and at least for a subsequence, we have
\begin{eqnarray}\label{49}
\frac{N_{F}(-u_{n}^{-})}{\|u_{n}\|^{2}}\overset{w}\rightarrow \psi\ \ \mbox{in}\ L^{1}(\Omega)\ \ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
Hypothesis $H_{1}$\,(iii) implies that for some $\xi_{0}^{*}\in\RR$, we have
\begin{eqnarray}\label{50}
-\infty<\xi_{0}^{*}\leq\liminf_{x\rightarrow-\infty}\frac{F(z,x)}{|x|^{2}}
\leq\limsup_{x\rightarrow-\infty}\frac{F(z,x)}{|x|^{2}}\leq\frac{\hat{\lambda}_{1}}{2}\ \ \mbox{uniformly for a.a.}\ z\in\Omega.
\end{eqnarray}
From \eqref{50} it follows that
\begin{eqnarray}\label{51}
\psi=\frac{1}{2}g(y^{-})^{2}\ \ \mbox{with}\ \xi_{0}^{*}\leq g(z)\leq\hat{\lambda}_{1}\ \ \mbox{for a.a.}\ z\in\Omega
\end{eqnarray}
(see Aizicovici, Papageorgiou \& Staicu \cite{1}, proof of Proposition 16). So, if in \eqref{47} we pass to the limit as $n\rightarrow\infty$ and use \eqref{46}, \eqref{49} and \eqref{51}, then
\begin{align}
&\vartheta(y)+\hat{\xi}\|y^{+}\|_{2}^{2}\leq\int_{\Omega}g(z)(y^{-})^{2}\,dz,\nonumber\\
\Rightarrow&\vartheta(y^{-})+\vartheta(y^{+})+\hat{\xi}\|y^{+}\|_{2}^{2}\leq\int_{\Omega}g(z)(y^{-})^{2}\,dz,\label{52}\\
\Rightarrow&\vartheta(y^{-})\leq\int_{\Omega}g(z)(y^{-})^{2}\,dz\ \ (\mbox{see}\ \eqref{4})\label{53}.
\end{align}
If $g\not\equiv\hat{\lambda}_{1}$, then from \eqref{53} and Proposition \ref{Proposition3} we have
\begin{align}
&c_{1}\|y^{-}\|^{2}\leq0,\nonumber\\
\Rightarrow&y^{-}=0.\label{54}
\end{align}
So, from \eqref{52} and \eqref{4} we have
\begin{align*}
&c_{0}\|y^{+}\|^{2}\leq0,\\
\Rightarrow&y^{+}=0,\ \mbox{that is},\ y=0\ \ (\mbox{see}\ \eqref{54}).
\end{align*}
Then from \eqref{47} we see that
\begin{align*}
&\|Dy_{n}\|_{2}\rightarrow0,\\
\Rightarrow&y_{n}\rightarrow0\ \mbox{in}\ H^{1}(\Omega),\ \ \mbox{a contradiction since}\ \|y_{n}\|=1\ \mbox{for all}\ n\geq1.
\end{align*}
If $g(z)=\hat{\lambda}_{1}$ for a.a. $z\in\Omega$, then from \eqref{53} and \eqref{3} we have
$$y^{-}=k\hat{u}_{1}\ \ \mbox{with}\ k\geq0.$$

If $k=0$, then $y=0$ and so as above we reach a contradiction.

If $k>0$, then $y^{-}\in\textrm{int}\, C_{+}$ (since $\hat{u}_{1}\in\textrm{int}\,C_{+}$) and so $y=-y^{-}\in-\textrm{int}\,C_{+}$. Then
\begin{eqnarray}\label{55}
u_{n}(z)=-u_{n}^{-}(z)\rightarrow-\infty\ \ \mbox{for a.a.}\ z\in\Omega.
\end{eqnarray}
For a.a. $z\in\Omega$ and all $x<0$, we have
\begin{align}
\frac{d}{dx}\left(\frac{F(z,x)}{|x|^{2}}\right)&=\frac{f(z,x)|x|^{2}-2xF(z,x)}{|x|^{4}}\nonumber\\
&=\frac{f(z,x)x-2F(z,x)}{|x|^{2}x}\,.\label{56}
\end{align}
Hypothesis $H_{1}$\,(iii) implies that given any $\nu>0$, we can find $M_{11}=M_{11}(\nu)>0$ such that
\begin{eqnarray}\label{57}
f(z,x)x-2F(z,x)\geq\nu\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq-M_{11}.
\end{eqnarray}
Using \eqref{57} in \eqref{56}, we obtain
\begin{align}
&\frac{d}{dx}\left(\frac{F(z,x)}{|x|^{2}}\right)\geq\frac{\nu}{|x|^{2}\nu}\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq-M_{11},\nonumber\\
\Rightarrow&\frac{F(z,y)}{|y|^{2}}-\frac{F(z,x)}{|x|^{2}}\geq-\frac{\nu}{2}\left[\frac{1}{|y|^{2}}-\frac{1}{|x|^{2}}\right]\label{58}\\
&\hspace{3cm}\mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ y\leq x\leq-M_{11}.\nonumber
\end{align}
Letting $y\rightarrow-\infty$ and using \eqref{50}, we obtain
\begin{align*}
&\frac{1}{2}\hat{\lambda}_{1}-\frac{F(z,x)}{|x|^{2}}\geq\frac{\nu}{2}\frac{1}{|x|^{2}}\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq-M_{11},\\
\Rightarrow&\hat{\lambda}_{1}|x|^{2}-2F(z,x)\geq\nu\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq-M_{11}.
\end{align*}
Since $\nu>0$ is arbitrary, we infer that
\begin{eqnarray}\label{59}
\hat{\lambda}_{1}|x|^{2}-2F(z,x)\rightarrow+\infty\ \ \mbox{uniformly for a.a.}\ z\in\Omega\; \mbox{as}\ x\rightarrow-\infty.
\end{eqnarray}
From \eqref{45}, \eqref{4} and \eqref{3}, we have
\begin{eqnarray}\label{60}
0\leq c_{0}\|u_{n}^{+}\|^{2}\leq M_{10}-\int_{\Omega}\left[\hat{\lambda}_{1}(u_{n}^{-})^{2}-2F(z,-u_{n}^{-})\right]dz\ \ \mbox{for all}\ n\geq1.
\end{eqnarray}
From \eqref{55}, \eqref{59} and Fatou's lemma, we have
\begin{eqnarray}\label{61}
\int_{\Omega}\left[\hat{\lambda}_{1}(u_{n}^{-})^{2}-2F(z,-u_{n}^{-})\right]dz\rightarrow+\infty\ \ \mbox{as}\ n\rightarrow\infty.
\end{eqnarray}
Then \eqref{60} and \eqref{61} lead to a contradiction. This proves the coercivity of the functional $\hat{\varphi}_{-}$.
\end{proof}
%.........................
\textbf{Remark}. If $H_{+}=\{u\in H^{1}(\Omega):\, u(z)\geq0\; \mbox{for a.a.}\ z\in\Omega\}$, then clearly $\hat{\varphi}_{-}|_{-H_{+}}=\varphi|_{-H_{+}}$. So, it follows that $\varphi|_{-H_{+}}$ is coercive.

Next we compute the critical groups of $\varphi$ at infinity.

\begin{prop}\label{Proposition6}
Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{1}$ hold. Then $C_{k}(\varphi,\infty)=0$ for all $k\geq0$.
\end{prop}
\begin{proof}
Let $$\partial B_{1}^{+}=\{u\in H^{1}(\Omega):\ \|u\|=1,\, u^{+}\neq0\}.$$
We consider the function $h:[0,1]\times\partial B_{1}^{+}\rightarrow\partial B_{1}^{+}$ defined by
\begin{eqnarray*}
h(t,u)=\frac{(1-t)u+t\hat{u}_{1}}{\|(1-t)u+t\hat{u}_{1}\|}\ \ \mbox{for all}\ (t,u)\in[0,1]\times\partial B_{1}^{+}.
\end{eqnarray*}
We see that
$$h(0,\cdot)=\textrm{id}|_{\partial B_{1}^{+}}\ \ \mbox{and}\ h(1,\cdot)\equiv\frac{\hat{u}_{1}}{\|\hat{u}_{1}\|}\in\partial B_{1}^{+}.$$
Hence $\partial B_{1}^{+}$ is contractible in itself. Hypothesis $H_{1}$\,(ii) implies that given any $\tilde{\xi}>0$, we can find $M_{12}=M_{12}(\xi)>0$ such that
\begin{eqnarray}\label{62}
F(z,x)\geq\frac{\tilde{\xi}}{2}x^{2}\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\geq M_{12}.
\end{eqnarray}
Similarly hypothesis $H_{1}$\,(iii) implies that we can find $c_{6}>0$ and $M_{12}^{*}>0$ such that
\begin{eqnarray}\label{63}
F(z,x)\geq-\frac{c_{6}}{2}|x|^{2}\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq-M_{12}^{*}.
\end{eqnarray}
Finally, because of hypothesis $H_{1}$\,(i), we have
\begin{eqnarray}\label{64}
|F(z,x)|\leq c_{7}\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\in[-M_{12}^{*}, M_{12}]\ \ \mbox{and some}\ c_{7}>0.
\end{eqnarray}
Let $u\in\partial B_{1}^{+}$ and $t>0$. We have
\begin{align}
\varphi(tu)&=\frac{t^{2}}{2}\vartheta(u)-\int_{\Omega}F(z,tu)\,dz\nonumber\\
&=\frac{t^{2}}{2}\vartheta(u)-\int_{\{tu\geq M_{12}\}}F(z,tu)\,dz-\int_{\{tu\leq-M_{12}^{*}\}}F(z,tu)\,dz-\int_{I^{*}}F(z,tu)\,dz\nonumber\\
&\hspace{9cm}\mbox{with}\ I^{*}=(-M_{12}^{*},M_{12})\nonumber\\
&\leq\frac{t^{2}}{2}\vartheta(u)-\frac{t^{2}}{2}\tilde{\xi}\int_{\{tu\geq M_{12}\}}u^{2}\,dz+\frac{t^{2}}{2}c_{6}\int_{\{tu\leq-M_{12}^{*}\}}u^{2}\,dz+c_{7}|\Omega|_{N}\nonumber\\
&\hspace{9cm}(\mbox{see}\ \eqref{62}, \eqref{63}, \eqref{64})\nonumber\\
&\leq\frac{t^{2}}{2}\left[c_{8}-\tilde{\xi}\int_{\{tu\geq M_{12}\}}u^{2}\,dz\right]+c_{7}|\Omega|_{N}\label{65}\\
&\hspace{5cm}(\mbox{see hypotheses}\ H(\xi),\, H(\beta)\ \mbox{and recall}\ \|u\|=1).\nonumber
\end{align}
Because $u\in\partial B_{1}^{+}$, we can find $t^{*}>0$ and $J>0$ such that
$$\int_{\{tu\geq M_{12}\}}u^{2}\,dz\geq J\ \ \mbox{for all}\ t\geq t^{*}.$$
Recall that $\tilde{\xi}>0$ is arbitrary. So, choosing $\tilde{\xi}>0$ big, from \eqref{65} we have
$$\tilde{\xi}J-c_{8}>0\ \ (\mbox{recall}\ \|u\|=1).$$
Therefore from \eqref{65} it follows that
\begin{eqnarray}\label{66}
\varphi(tu)\rightarrow-\infty\ \ \mbox{as}\ t\rightarrow+\infty.
\end{eqnarray}
Hypothesis $H_{1}$\,(ii) implies that
\begin{eqnarray}\label{67}
0=\varrho(z,0)\leq\varrho(z,x)+\tau(z)\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\geq0.
\end{eqnarray}
Also, hypotheses $H_{1}$\,(ii),\, (iii) imply that we can find $c_{9}>0$ such that
\begin{eqnarray}\label{68}
-c_{9}\leq\varrho(z,x)\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq0.
\end{eqnarray}
Using \eqref{67} and \eqref{68} that for all $u\in H^{1}(\Omega)$ we have
\begin{align*}
\int_{\Omega}\varrho(z,u)\,dz&=\int_{\Omega}\varrho(z,u^{+})\,dz+\int_{\Omega}\varrho(z,-u^{-})\,dz\\
&\geq-\|\tau\|_{1}-c_{9}|\Omega|_{N},
\end{align*}
\begin{eqnarray}\label{69}
\Rightarrow-c_{10}+2\int_{\Omega}F(z,u)\,dz\leq\int_{\Omega}f(z,u)u\,dz\ \ \mbox{with}\ c_{10}=\|\tau\|_{1}+c_{9}|\Omega|_{N}>0.
\end{eqnarray}
Then for $u\in\partial B_{1}^{+}$ and $t\geq1$, we have
\begin{align}
\frac{d}{dt}\varphi(tu)&=\langle\varphi'(tu),u\rangle\ \ \mbox{(by the chain rule)}\nonumber\\
&=\frac{1}{t}\langle\varphi'(tu),tu\rangle\nonumber\\
&=\frac{1}{t}\left[\vartheta(tu)-\int_{\Omega}f(z,tu)(tu)\,dz\right]\nonumber\\
&\leq\frac{1}{t}\left[2\varphi(tu)+c_{10}\right]\ \ (\mbox{see}\ \eqref{69}).\label{70}
\end{align}
From \eqref{66} and \eqref{70} it follows that for $t\geq1$ big, we have
\begin{eqnarray}\label{71}
\frac{d}{dt}\varphi(tu)<0.
\end{eqnarray}
Recall that $H_{+}=\{u\in H^{1}(\Omega):\, u(z)\geq0\ \mbox{for a.a.}\ z\in\Omega\}$ and that $\varphi|_{-H_{+}}$ is coercive (see Proposition \ref{Proposition5} and the Remark following it). So, we can find $c_{11}>0$ such that
$$\varphi|_{-H_{+}}\geq-c_{11}.$$
We choose $\lambda<\min\left\{-c_{10},\,-c_{11},\, \inf_{\partial B_{1}}\varphi\right\}$, where $\partial B_{1}=\{u\in H^{1}(\Omega): \|u\|=1\}$.
From \eqref{71} we infer that there exists a unique $k(u)>1$ such that
\begin{align}
&\varphi(tu)>\lambda\ \ \mbox{for}\; t\in[0,k(u))\nonumber\\
&\varphi(tu)=\lambda\ \ \mbox{for}\; t=k(u)\label{72}\\
&\varphi(tu)<\lambda\ \ \mbox{for}\; t>k(u).\nonumber
\end{align}
Moreover, from the implicit function theorem we have that $k\in C(\partial B_{1}^{+},[1,\infty))$. The choice of $\lambda$ and \eqref{72} imply that
$$\varphi^{\lambda}\subseteq\{tu:\, u\in\partial B_{1}^{+},\, t\geq k(u)\}.$$
Let $D_{+}=\{tu:\, u\in\partial B_{1}^{+},\,t\geq1\}$. Then $\varphi^{\lambda}\subseteq D_{+}$.

Consider the deformation $h^{*}:[0,1]\times D_{+}\rightarrow D_{+}$ defined by
\[h^{*}(s,u)=\left\{ \begin{array}{l l} (1-s)tu+sk(u)u & \quad \text{if} \ t\in[1,k(u)]\\
tu & \quad \text{if} \ k(u)<t.
\end{array} \right. \]
We have
$$h^{*}(0,D_{+})\subseteq\varphi^{\lambda}\ \ \mbox{and}\ h^{*}(s,\cdot)|_{\varphi^{\lambda}}=\textrm{id}|_{\varphi^{\lambda}}\ \ \mbox{for all}\ s\in[0,1]\ (\mbox{see}\ \eqref{72}).$$
This means that $\varphi^{\lambda}$ is a strong deformation retract of $D_{+}$. Using the radial retraction, we see that $D_{+}$ and $\partial B_{1}^{+}$ are homotopy equivalent (see Dugundji \cite[Theorem 6.5, p. 325]{5}). Therefore
\begin{eqnarray}\label{73}
H_{k}(H^{1}(\Omega),\varphi^{\lambda})=H_{k}(H^{1}(\Omega), D_{+})=H_{k}(H^{1}(\Omega),\partial B_{1}^{+})\ \ \mbox{for all}\ k\geq0.
\end{eqnarray}
Recall that $\partial B_{1}^{+}$ is contractible in itself. Hence
\begin{align}
&H_{k}(H^{1}(\Omega),\partial B_{1}^{+})=0\ \ \mbox{for all}\ k\geq0,\nonumber\\
\Rightarrow&H_{k}(H^{1}(\Omega),\varphi^{\lambda})=0\ \ \mbox{for all}\ k\geq0\ \ (\mbox{see}\ \eqref{73}).\label{74}
\end{align}
Choosing $|\lambda|$ even bigger if necessary, from \eqref{74} we infer that
$$C_{k}(\varphi,\infty)=0\ \ \mbox{for all}\ k\geq0\ (\mbox{see Section}\, \ref{Section2}).$$
\end{proof}

\textbf{Remark}. We mention that Wang \cite{27} was the first to compute the critical groups at infinity for the energy functional of problems with Dirichlet boundary condition, zero potential (that is, $\xi\equiv0$) and with a $C^{1}$-reaction $x\rightarrow f(x)$ which exhibits symmetric behavior as $x\rightarrow\pm\infty$ and is superlinear satisfying the AR-condition. Note that in the context of Dirichlet problems, the Poincar\'{e} inequality simplifies the argument.

Using Proposition \ref{Proposition5} and the direct method, we can produce a negative solution for problem \eqref{1}.

\begin{prop}\label{Proposition7}
 Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{1}$ hold. Then problem \eqref{1} admits a negative solution $u_{0}\in-{\rm int}\,C_{+}$, which is a local minimizer of the energy functional $\varphi$.
\end{prop}
\begin{proof}
From Proposition \ref{Proposition5}, we know that $\hat{\varphi}_{-}$ is coercive. Moreover, using the Sobolev embedding theorem and the trace theorem, we see that $\hat{\varphi}_{-}$ is sequentially weakly lower semicontinuous. Then, by the Weierstrass theorem, we can find $u_{0}\in H^{1}(\Omega)$ such that
\begin{eqnarray}\label{75}
\hat{\varphi}_{-}(u_{0})=\inf\left[\hat{\varphi}_{-}(u):\, u\in H^{1}(\Omega)\right].
\end{eqnarray}
Let $\delta_{0}>0$ be as in hypothesis $H_{1}$\,(iv). We choose $t\in(0,1]$ small such that $t\hat{u}_{1}(z)\in(0,\delta_{0}]$ for all $z\in\overline{\Omega}$ (recall $\hat{u}_{1}\in\textrm{int}\,C_{+}$). We have
\begin{align*}
\hat{\varphi}_{-}(-t\hat{u}_{1})&=\frac{t^{2}}{2}\vartheta(\hat{u}_{1})-\int_{\Omega}F(z,-t\hat{u}_{1})\,dz\\
&\leq\frac{t^{2}}{2}\hat{\lambda}_{1}-\frac{t^{2}}{2}\hat{\lambda}_{m}\ \ (\mbox{see hypothesis}\ H_{1}\mbox{(iv) and recall}\ \|\hat{u}_{1}\|_{2}=1)\\
&=\frac{t^{2}}{2}\left[\hat{\lambda}_{1}-\hat{\lambda}_{m}\right]<0\ \ (\mbox{since}\ m\geq2),
\end{align*}
$$\Rightarrow\,\hat{\varphi}_{-}(u_{0})<0=\hat{\varphi}_{-}(0)\ \ (\mbox{see}\ \eqref{75}),\; \mbox{hence}\ u_{0}\neq0.$$
From \eqref{75} we have
$$\hat{\varphi}_{-}'(u_{0})=0,$$
\begin{align}
&\Rightarrow\,\langle A(u_{0}),h\rangle+\int_{\Omega}(\xi(z)+\hat{\xi})u_{0}h\,dz+\int_{\partial\Omega}\beta(z)u_{0}h\,d\sigma=\int_{\Omega}\hat{f}_{-}(z,u_{0})h\,dz
\label{76}\\
&\hspace{9cm}\mbox{for all}\ h\in H^{1}(\Omega).\nonumber
\end{align}
In \eqref{76} we choose $h=u_{0}^{+}\in H^{1}(\Omega)$. Then
\begin{align*}
&\vartheta(u_{0}^{+})+\hat{\xi}\|u_{0}^{+}\|_{2}^{2}=0,\\
\Rightarrow&c_{0}\|u_{0}^{+}\|^{2}\leq0\ \ (\mbox{see}\ \eqref{4}),\; \mbox{hence}\ u_{0}\leq0,\; u_{0}\neq0.
\end{align*}
So, relation \eqref{76} becomes
\begin{align*}
&\langle A(u_{0}),h\rangle+\int_{\Omega}\xi(z)u_{0}h\,dz+\int_{\partial\Omega}\beta(z)u_{0}h\,d\sigma=\int_{\Omega}f(z,u_{0})h\,dz\ \ \mbox{for all}\ h\in H^{1}(\Omega),\\
\Rightarrow&-\Delta u_{0}(z)+\xi(z)u_{0}(z)=f(z,u_{0}(z))\ \ \mbox{for a.a.}\ z\in\Omega,\; \frac{\partial u_{0}}{\partial n}+\beta(z)u_{0}=0\; \mbox{on}\ \partial\Omega\\
&\hspace{8cm}(\mbox{see Papageorgiou \& R\u{a}dulescu\ \cite{21}}).
\end{align*}
So, $u_{0}$ is a negative solution of problem \eqref{1}.
We define
\[\chi(z)=\left\{ \begin{array}{l l} \di \frac{f(z,u_{0}(z))}{u_{0}(z)} & \quad \text{if} \ u_{0}(z)\neq0\\
0 & \quad \text{otherwise}.
\end{array} \right. \]
Hypotheses $H_{1}$\,(i), (iii), (iv) imply that
\begin{align*}
&|f(z,x)|\leq c_{12}|x|\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\leq0\ \ \mbox{and some}\ c_{12}>0,\\
\Rightarrow&\left|\frac{f(z,u_{0}(z))}{u_{0}(z)}\right|\leq c_{12}\ \ \mbox{for a.a.}\ z\in\{u_{0}>0\}.
\end{align*}
Hence $\chi\in L^{\infty}(\Omega)$. We have
\begin{eqnarray}\label{77}
-\Delta u_{0}(z)=(\chi(z)-\xi(z))u_{0}(z)\ \ \mbox{for a.a.}\ z\in\Omega.
\end{eqnarray}
Note that $\chi-\xi\in L^{s}(\Omega)$ with $s>N$ (see hypothesis $H(\xi)$). Using Lemma 5.1 of Wang \cite{26}, we have $u_{0}\in L^{\infty}(\Omega)$. Then \eqref{77} implies
$$\Delta u_{0}\in L^{s}(\Omega).$$
Using Lemma 5.2 of Wang \cite{26} (the Agmon-Douglis-Nirenberg or Calderon-Zygmund estimates), we obtain that $u_{0}\in W^{2,s}(\Omega)$. Since $s>N$ (see hypothesis $H(\xi)$) from the Sobolev embedding theorem, we have $W^{2,s}(\Omega)\hookrightarrow C^{1+\alpha}(\overline{\Omega})$ with $\alpha=1-\frac{N}{s}$. Hence $u_{0}\in -C_{+}\setminus\{0\}$. From \eqref{77} we have
\begin{align*}
\Delta(-u_{0})(z)&=(\xi(z)-\chi(z))(-u_{0})(z)\\
&\leq(\|\xi^{+}\|_{\infty}+\|\chi\|_{\infty})(-u_{0})(z)\ \ \mbox{for a.a.}\ z\in\Omega,
\end{align*}
\begin{align*}
\Rightarrow&\, u_{0}\in-\textrm{int}\,C_{+}\\
&(\mbox{by the maximum principle, see, for example, Gasinski \& Papageorgiou \cite[p. 738]{9}}).
\end{align*}
Note that $\hat{\varphi}_{-}|_{-C_{+}}=\varphi|_{-C_{+}}$ and so we have that $u_{0}$ is a local $C^{1}(\overline{\Omega})$-minimizer of $\varphi$. Invoking Proposition \ref{Proposition2}, we conclude that $u_{0}$ is a local $H^{1}(\Omega)$-minimizer of $\varphi$.
\end{proof}

In order to generate a second nontrivial solution of problem \eqref{1}, we need to examine the critical groups of $\varphi$ at the origin.

\begin{prop}\label{Proposition8}
 Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{1}$ hold. Then $C_{d_{m}}(\varphi,0)\neq0$ where $d_{m}=\textrm{dim}\,\overline{H}_{m}=\textrm{dim}\,\oplus_{k=1}^{m}E(\hat{\lambda}_{k})\geq2$.
\end{prop}
\begin{proof}
We know that
\begin{eqnarray}\label{78}
H^{1}(\Omega)=\overline{H}_{m}\oplus\widehat{H}_{m+1},
\end{eqnarray}
where $\overline{H}_{m}=\oplus_{k=1}^{m}E(\hat{\lambda}_{k})$ and $\widehat{H}_{m+1}=\overline{H}_{m}^{\bot}=\overline{\underset{k\geq m+1}\oplus E(\hat{\lambda}_{k})}$. The space $\overline{H}_{m}$ is finite dimensional. So, all norms are equivalent. Therefore, we can find $\delta_{1}>0$ such that
$$u\in\overline{H}_{m},\; \|u\|\leq\delta_{1}\, \Rightarrow\, |u(z)|\leq\delta_{0}\ \mbox{for all}\ z\in\overline{\Omega}\ \ (\mbox{recall that}\ \overline{H}_{m}\subseteq C(\overline{\Omega})).$$
Using hypothesis $H_{1}$\,(iv), we have
\begin{align}
&\varphi(u)\leq\frac{1}{2}\vartheta(u)-\frac{\hat{\lambda}_{m}}{2}\|u\|_{2}^{2}\leq0\ \ \mbox{for all}\ u\in\overline{H}_{m}\ \ \mbox{with}\ \|u\|\leq\delta_{1}\label{79}\\
&\hspace{9cm}(\mbox{see \eqref{5}}).\nonumber
\end{align}
Hypotheses $H_{1}$\,(i), (iv) imply that given $\varepsilon>0$, we can find $c_{13}=c_{13}(\varepsilon)>0$ such that
\begin{eqnarray}\label{80}
F(z,x)\leq\frac{1}{2}(\eta(z)+\varepsilon)x^{2}+c_{13}|x|^{r}\ \ \mbox{for a.a.}\ z\in\Omega,\; \mbox{all}\ x\in\RR.
\end{eqnarray}
Then for all $u\in\widehat{H}_{m+1}$, we have
\begin{align*}
\varphi(u)&\geq\frac{1}{2}\vartheta(u)-\frac{1}{2}\int_{\Omega}\eta(z)u^{2}\,dz-\frac{\varepsilon}{2}\|u\|_{2}^{2}-c_{14}\|u\|^{r}\ \ \mbox{for some}\ c_{14}>0\ \ (\mbox{see \eqref{80}})\\
&\geq\frac{1}{2}(c_{1}-\varepsilon)\|u\|^{2}-c_{14}\|u\|^{r}\ \ (\mbox{see Proposition \ref{Proposition3}}).
\end{align*}
Choosing $\varepsilon\in(0,c_{1})$, we see that
$$\varphi(u)\geq c_{15}\|u\|^{2}-c_{14}\|u\|^{r}\ \ \mbox{for some}\ c_{15}>0.$$
Since $r>2$, we can find $\delta_{2}>0$ such that
\begin{eqnarray}\label{81}
\varphi(u)>0\ \ \mbox{for all}\ u\in\widehat{H}_{m+1}\ \ \mbox{with}\ 0<\|u\|\leq\delta_{2}.
\end{eqnarray}
If $\delta=\min\{\delta_{1},\delta_{2}\}$, then from \eqref{78}, \eqref{79} and \eqref{81}, we see that $\varphi$ has a local linking at the origin. Therefore $C_{d_{m}}(\varphi,0)\neq0$ with $d_{m}=\textrm{dim}\,\overline{H}_{m}\geq2$ (see \cite[p. 171]{14}).
\end{proof}

Now we are ready for our first multiplicity theorem in which we produce two nontrivial solutions.

\begin{theorem}\label{Theorem9}
Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{1}$ hold. Then problem \eqref{1} has at least two nontrivial solutions
$$u_{0}\in-\textrm{int}\,C_{+}\ \ \mbox{and}\ \hat{u}\in C^{1}(\overline{\Omega}).$$
\end{theorem}
\begin{proof}
From Proposition \ref{Proposition7} we already have a negative solution $u_{0}\in-\textrm{int}\,C_{+}$ which is a local minimizer of the energy functional $\varphi$. Hence
\begin{eqnarray}\label{82}
C_{k}(\varphi,u_{0})=\delta_{k,0}\ZZ\ \ \mbox{for all}\ k\geq0.
\end{eqnarray}
We know that
\begin{align*}
&C_{k}(\varphi,\infty)=0\ \ \mbox{for all}\ k\geq0\ \ (\mbox{see Proposition \ref{Proposition6}})\\
&C_{d_{m}}(\varphi,0)\neq0\ \ (\mbox{see Proposition \ref{Proposition8}}).
\end{align*}
So, we can find $\hat{u}\in K_{\varphi}$ such that
\begin{align}
\begin{split}\label{83}
&\varphi(\hat{u})<0=\varphi(0)\ \ \mbox{and}\ C_{d_{m}-1}(\varphi,\hat{u})\neq0\\
\mbox{or}\ &\varphi(\hat{u})>0=\varphi(0)\ \ \mbox{and}\ C_{d_{m}+1}(\varphi,\hat{u})\neq0.
\end{split}
\end{align}
Evidently $\hat{u}\neq0$. Also since $d_{m}\geq2$, we have
\begin{align*}
&1\leq d_{m}-1<d_{m}+1,\\
\Rightarrow\,&\hat{u}\neq u_{0}\ \ (\mbox{see \eqref{82} and \eqref{83}}).
\end{align*}
So, $\hat{u}$ is a second nontrivial solution of problem \eqref{1}. As before, using the regularity result of Wang \cite{26}, we have $\hat{u}\in C^{1}(\overline{\Omega})$.
\end{proof}

\textbf{Remark}: Our theorem improves Theorem 1 of Recova \& Rumbos \cite{24}, which produces only one nontrivial solution for Dirichlet problems with zero potential and an asymmetric reaction satisfying more restrictive conditions (compare hypotheses $H_{1}$ with hypotheses $(L_{1})-(L_{7})$ in \cite{24}).

\section{Three Nontrivial Solutions}

In this section, by strengthening the regularity of the reaction $f(z,\cdot)$, we can improve Theorem \ref{Theorem9} and establish the existence of at least three nontrivial solutions for problem \eqref{1}.

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

\smallskip
$H_{2}$: $f:\Omega\times\RR\rightarrow\RR$ 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)] $|f_{x}'(z,x)|\leq a(z)(1+|x|^{r-2})$ for a.a. $z\in\Omega$, all $x\in\RR$, with $a\in L^{\infty}(\Omega)$, $2<r<2^{*}$;
\item[(ii)] if $F(z,x)=\int_{0}^{x}f(z,s)\,ds$, then
$$\lim_{x\rightarrow+\infty}\frac{F(z,x)}{x^{2}}=+\infty\ \mbox{uniformly for a.a.}\ z\in\Omega$$
and if $\varrho(z,x)=f(z,x)x-2F(z,x)$, then there exists $\tau\in L^{1}(\Omega)_{+}$ such that
\begin{eqnarray*}
\varrho(z,x)\leq\varrho(z,y)+\tau(z)\ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ 0\leq x\leq y;
\end{eqnarray*}
\item[(iii)] $-\infty<\xi_{0}\leq\displaystyle\liminf_{x\rightarrow-\infty}\frac{f(z,x)}{x}\leq\displaystyle\limsup_{x\rightarrow-\infty}\frac{f(z,x)}{x}\leq\hat{\lambda}_{1}$ uniformly for a.a. $z\in\Omega$ and
$$\lim_{x\rightarrow-\infty}\left[f(z,x)x-2F(z,x)\right]=+\infty\ \mbox{uniformly for a.a.}\ z\in\Omega;$$
\item[(iv)] there exist an integer $m\geq2$ and $\delta_{0}>0$ such that
\begin{align*}
&f_{x}^{'}(z,x)\leq\hat{\lambda}_{m+1}\ \ \mbox{for a.a.}\ z\in\Omega,\; f_{x}^{'}(\cdot,0)\not\equiv\hat{\lambda}_{m+1}\\
&f(z,x)x\geq\hat{\lambda}_{m}x^{2}\ \ \mbox{for a.a.}\ z\in\Omega,\ \mbox{all}\ |x|\leq\delta_{0}.
\end{align*}
\end{itemize}

\begin{theorem}\label{Theorem10}
Assume that hypotheses $H(\xi)$, $H(\beta)$ and $H_{2}$ hold. Then problem \eqref{1} admits at least three nontrivial solutions
$$u_{0}\in-{\rm int}\,C_{+},\, \hat{u},\, \tilde{u}\in C^{1}(\overline{\Omega}).$$
\end{theorem}
\begin{proof}
From Proposition \ref{Proposition7} we know that there is a negative solution $u_{0}\in-\textrm{int}\,C_{+}$ which is a local minimizer of $\varphi$. We assume that $K_{\varphi}$ is finite (otherwise we already have an infinity of nontrivial solutions which belong in $C^{1}(\overline{\Omega})$ (as before, using the result of Wang \cite{26}) and so we are done). Then we can find $\varrho\in(0,1)$ small such that
\begin{eqnarray}\label{84}
\varphi(u_{0})<\inf\left[\varphi(u):\, \|u-u_{0}\|=\varrho\right]=m_{\varrho}
\end{eqnarray}
(see Aizicovici, Papageorgiou \& Staicu \cite{1}, proof of Proposition 29). Hypothesis $H_{2}$\,(ii) implies that
\begin{eqnarray}\label{85}
\varphi(t\hat{u}_{1})\rightarrow-\infty\ \ \mbox{as}\ t\rightarrow+\infty.
\end{eqnarray}
From Proposition \ref{Proposition4} we know that
\begin{eqnarray}\label{86}
\varphi\ \ \mbox{satisfies the C-condition}.
\end{eqnarray}
Then \eqref{84}, \eqref{85}, \eqref{86} permit the use of Theorem \ref{Theorem1} (the mountain pass theorem). So, we can find $\hat{u}\in H^{1}(\Omega)$ such that
\begin{eqnarray}\label{87}
\hat{u}\in K_{\varphi}\ \ \mbox{and}\ m_{\varrho}\leq\varphi(\hat{u}).
\end{eqnarray}
From \eqref{84} and \eqref{87} it follows that $\hat{u}\neq u_{0}$ and $\hat{u}$ is a solution of problem \eqref{1} with $\hat{u}\in C^{1}(\overline{\Omega})$ (as in the proof of Proposition \ref{Proposition7}, using the result of Wang \cite{26}). Since $\hat{u}$ is a critical point of $\varphi$ of mountain pass type, we have
\begin{eqnarray}\label{88}
C_{1}(\varphi,\hat{u})\neq0.
\end{eqnarray}
Hypotheses $H_{2}$ imply that $\varphi\in C^{2}(H^{1}(\Omega))$ and
\begin{align*}
&\langle\varphi'(u),h\rangle=\int_{\Omega}(Du,Dh)_{\RR^{N}}\,dz+\int_{\Omega}\xi(z)uh\,dz+\int_{\partial\Omega}\beta(z)uh\,d\sigma
-\int_{\Omega}f(z,u)h\,dz\\
&\hspace{10cm}\mbox{for all}\ h\in H^{1}(\Omega),\\
\Rightarrow&\,\langle\varphi''(u)y,h\rangle=\int_{\Omega}(Dy,Dh)_{\RR^{N}}\,dz+\int_{\Omega}\xi(z)yh\,dz+\int_{\partial\Omega}\beta(z)yh\,d\sigma
-\int_{\Omega}f_{x}'(z,u)yh\,dz\\
&\hspace{10cm}\mbox{for all}\ y,h\in H^{1}(\Omega).
\end{align*}
Suppose that the spectrum $\sigma(\varphi''(\hat{u}))\subseteq[0,\infty)$. Then we have
\begin{eqnarray}\label{89}
\vartheta(y)\geq\int_{\Omega}m(z)y^{2}\,dz\ \ \mbox{for all}\ y\in H^{1}(\Omega),
\end{eqnarray}
with $m(z)=f_{x}'(z,\hat{u}(z))$, $m\in L^{\infty}(\Omega)$ (see hypothesis $H_{2}$\,(i)). Suppose $u\in\textrm{ker}\,\varphi''(\hat{u})$. Then
\begin{eqnarray}\label{90}
-\Delta u(z)=(m-\xi)(z)u(z)\ \ \mbox{in}\ \Omega,\; \frac{\partial u}{\partial n}+\beta(z)u=0\ \ \mbox{on}\ \partial\Omega.
\end{eqnarray}
We have $(m-\xi)(\cdot)\in L^{s}(\Omega)$, $s>N$. If $(m-\xi)^{+}=0$, then using \eqref{90} and the Green identity, we have
\begin{align*}
&\|Du\|_{2}^{2}+\int_{\partial\Omega}\beta(z)u^{2}\,d\sigma\leq0,\\
\Rightarrow\,&u\equiv c\in\RR,\ \ \mbox{that is}\ \textrm{dim\,ker}\,\varphi''(\hat{u})=1.
\end{align*}
If $(m-\xi)^{+}\neq0$, then we know that
\begin{eqnarray}\label{91}
\qquad \tilde{\lambda}_{1}=\inf\left[\|Du\|_{2}^{2}+\int_{\partial\Omega}\beta(z)u^{2}\,d\sigma:\, u\in H^{1}(\Omega),\; \int_{\Omega}(m-\xi)(z)u^{2}\,dz=1\right]\ \ (\mbox{see \eqref{7}}).
\end{eqnarray}
From \eqref{89} and \eqref{91} we see that
\begin{eqnarray}\label{92}
\tilde{\lambda}_{1}\geq0.
\end{eqnarray}
From \eqref{90} and \eqref{92} we infer that
$$u=0\ \ \mbox{or}\ \ \tilde{\lambda}_{1}=1.$$
But we know that $\tilde{\lambda}_{1}$ is simple (see Section \ref{Section2}). So, it follows that
$$\textrm{dim\,ker}\,\varphi''(\hat{u})\leq1.$$
Then using \eqref{88} and Proposition 2.5 of Bartsch \cite{4}, we conclude that
\begin{eqnarray}\label{93}
C_{k}(\varphi,\hat{u})=\delta_{k,1}\ZZ\ \ \mbox{for all}\ k\geq0.
\end{eqnarray}
From Proposition \ref{Proposition6}, we know that
\begin{eqnarray}\label{94}
C_{k}(\varphi,\infty)=0\ \ \mbox{for all}\ k\geq0.
\end{eqnarray}
From the proof of Proposition \ref{Proposition8}, we know that $\varphi$ has a local linking at the origin. Since $\varphi\in C^{2}(H^{1}(\Omega))$, using Proposition 2.3 of Su \cite{25}, we have
\begin{eqnarray}\label{95}
C_{k}(\varphi,0)=\delta_{k,d_{m}}\ZZ\ \ \mbox{for all}\ k\geq0\ \ \mbox{with}\ d_{m}=\textrm{dim}\overline{H}_{m}\geq2.
\end{eqnarray}
Finally since $u_{0}\in-\textrm{int}\,C_{+}$ is a local minimizer of $\varphi$, we have
\begin{eqnarray}\label{96}
C_{k}(\varphi,u_{0})=\delta_{k,0}\ZZ\ \ \mbox{for all}\ k\geq0.
\end{eqnarray}
Suppose $K_{\varphi}=\{0,u_{0},\hat{u}\}$. Then from \eqref{93}, \eqref{94}, \eqref{95}, \eqref{96} and the Morse relation with $t=-1$ (see \eqref{9}), we have
\begin{align*}
&(-1)^{d_{m}}+(-1)^{0}+(-1)^{1}=0,\\
\Rightarrow\,&(-1)^{d_{m}}=0, \ \ \mbox{a contradiction}.
\end{align*}
So, there exists $\tilde{u}\in K_{\varphi}$, $\tilde{u}\not\in\{0,u_{0},\hat{u}\}$. Then $\tilde{u}$ is the third nontrivial solution of \eqref{1}. As before we have $\tilde{u}\in C^{1}(\overline{\Omega}).$
\end{proof}

\textbf{Remark}: Theorem \ref{Theorem10} extends Theorem 1.2 of Recova \& Rumbos \cite{24}, which is for Dirichlet problems with no potential (that is, $\xi\equiv0$) and under more restrictive conditions on the reaction $f(z,x)$. In particular, the authors of \cite{24} introduce also the extra hypothesis that there exists $x_{0}>0$ such that $f(z,x_{0})=0$ for all $z\in\Omega$. No such condition is used here.


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\end{thebibliography}


\end{document}
