\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 207, pp. 1--29.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/207\hfil Multiple solutions to asymmetric problems] {Multiple solutions to asymmetric semilinear elliptic problems via Morse theory } \author[L. Recova, A. Rumbos \hfil EJDE-2014/207\hfilneg] {Leandro Recova, Adolfo Rumbos} % in alphabetical order \address{Leandro L. Recova \newline Institute of Mathematical Sciences, Claremont Graduate University, Claremont, California 91711, USA} \email{leandro.recova@cgu.edu} \address{Adolfo J. Rumbos \newline Department of Mathematics, Pomona College, Claremont, California 91711, USA} \email{arumbos@pomona.edu} \thanks{Submitted February 28, 2014. Published October 7, 2014.} \subjclass[2000]{35J20} \keywords{Morse theory; critical groups, local linking} \begin{abstract} In this article we study the existence of solutions to the problem \begin{gather*} -\Delta u = g(x,u) \quad \text{in } \Omega; \\ u = 0 \quad\text{on } \partial\Omega, \end{gather*} where $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$ $(N\geq 2)$ and $g:\overline{\Omega}\times\mathbb{R}\to\mathbb{R}$ is a differentiable function with $g(x,0)=0$ for all $x\in\Omega$. By using minimax methods and Morse theory, we prove the existence of at least three nontrivial solutions for the case in which an asymmetric condition on the nonlinearity $g$ is assumed. The first two nontrivial solutions are obtained by employing a cutoff technique used by Chang et al in \cite{KC3}. For the existence of the third nontrivial solution, first we compute the critical group at infinity of the associated functional by using a technique used by Liu and Shaoping in \cite{LJWS}. The final result is obtained by using a standard argument involving the Morse relation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\N}{\mathbb{N}} \section{Introduction} The goal of this article is to study the existence and multiplicity of solutions of the boundary-value problem \begin{equation}\label{e1} \begin{gathered} -\Delta u = g(x,u) \quad \text{in } \Omega; \\ u = 0 \quad\text{on } \partial\Omega, \end{gathered} \end{equation} where $\Omega\subset\R^{n}$ is an open bounded set with smooth boundary, $\partial\Omega$, and $g$ is a differentiable function. By a solution of \eqref{e1} we mean a weak solution, i.e., a function $u\in H_0^{1}(\Omega)$ satisfying \begin{equation} \int_{\Omega}\nabla u\cdot\nabla v dx = \int_{\Omega}g(x,u)vdx, \label{weakcondd} \end{equation} for any $v\in H_0^{1}(\Omega)$, where $H_0^{1}(\Omega)$ is the Sobolev space obtained through completion of $C_0^{\infty}(\Omega)$ with respect to the metric induced by the norm $$ \|u\|=\Big(\int_{\Omega}|\nabla u|^{2}dx\Big)^{1/2},\quad \text{for all }u\in H_0^{1}(\Omega). $$ We will denote by $0<\lambda_1<\lambda_2<\lambda_3<\dots$ the distinct eigenvalues of the linear problem \begin{gather*} -\Delta u = \lambda u \quad \text{ in } \Omega; \\ u = 0 \quad \text{on } \partial\Omega. \end{gather*} The following conditions on $g$, and its primitive, $G(x,s)=\int_0^{s}g(x,\xi)d\xi$, for all $x\in\Omega$ and $s\in\R$, will be assumed throughout this article: \begin{itemize} \item [(G1)] $g:\overline{\Omega}\times\R\to\R$ is differentiable, $g(x,0)=0$, and $g'(x,0)=\lambda_{m}$ with $m>1$. \item [(G2)] There is $\lambda > 0$ with $\lambda\ne\lambda_1$, and $0\leq\alpha < 1$ such that $$ \lim_{s\to -\infty}\frac{g(x,s)-\lambda s}{|s|^{\alpha}}=0. $$ \item [(G3)] There are $\theta$ and $s_0$ with $0<\theta<1/2$ and $s_0>0$ such that $$ 0s_0$ and all $x\in\Omega$}. $$ \item [(G4)] $\lim_{s\to\infty} g(x,s)/s^{\sigma}=0$, where $\frac{1}{\theta}-1<\sigma\leq\frac{N+2}{N-2}$, if $N\geq 3$, or $1<\sigma<\infty$ if $N=2$. \item [(G5)] $\sigma\theta < \min\big\{\frac{1}{1+\alpha},\frac{N+2}{2N}\big\}$. \item [(G6)] There exists $s^{-}<0$ such that $$ 2G(x,s)-g(x,s)s \leq 0, \text{ for all }s0$ such that $g(x,t_0)=0$. Then problem \eqref{e1} has at least three nontrivial solutions. \end{theorem} The work in this article was motivated by that of De Figueiredo's in \cite{DF}. In that paper, the author was interested in studying the solvability of the problem \begin{equation}\label{e1111} \begin{gathered} -\Delta u = \lambda u + f(x,u) +t\varphi + h \quad \text{in } \Omega; \\ u = 0 \quad\text{on } \partial\Omega, \end{gathered} \end{equation} where $\varphi$ is a positive eigenfunction associated with the the first eigenvalue $\lambda_1$ of $(-\Delta, H_0^{1}(\Omega))$, $t\in\R$ and $h\in C^{\nu}(\overline{\Omega})$, $0<\nu \leq 1$, $\int_{\Omega}h\varphi dx=0$. In \cite{DF}, the author assumed the following conditions on the nonlinearity $f$ and its primitive $F$: \begin{itemize} \item [(F1)] $f:\overline{\Omega}\times\R\to\R$ is a $C^{1}$ function. \item [(F2)] There exists $0<\alpha<1$ such that $\lim_{s\to -\infty} f(x,s)|s|^{-\alpha}=0$. \item [(F3)] $\lim_{s\to -\infty}f_{s}'(x,s)=0$. \item [(F4)] There are $\theta$ and $s_0$ with $0<\theta<1/2$ and $s_0>0$ such that $0s_0$ and all $x\in\Omega$. \item [(F5)] $\lim_{s\to +\infty}f(x,s)s^{-\sigma}=0$, where $\sigma\leq(N+2)/(N-2)$ if $N\geq 3$ or $1<\sigma<\infty$ if $N=2$. \item [(F6)] $f_{s}'(x,s)\geq -\mu$ where $\mu<\lambda - \lambda_{k}$. \item [(F7)] $\sigma\theta < \min\big\{\frac{1}{1+\alpha},\frac{N+2}{N-2}\big\}$. \end{itemize} De Figueiredo proved that under the assumptions (F1)--(F7), there exists $\hat{t}>0$ such that, for all $t\geq\hat{t}$, problem \eqref{e1111} has at least two solutions. De Figueiredo used a generalized version of the mountain pass theorem (see \cite[Theorem $5.3$]{RB}) which required the Palais-Smale (PS) condition to be verified. In \cite{DF}, the author proved the (PS) condition for a general class of superlinear elliptic problems of the type \eqref{e1} under the conditions (G1)--(G5), without the assumption that $g'(x,0)=\lambda_{m}$, with $m\ne 1$. In this articld, we will study the solvability of problem \eqref{e1} under the conditions (G1)--(G6) and for the case in which $0$ is a degenerate critical point of the associated functional of \eqref{e1}. Many authors have studied problem \eqref{e1} under different assumptions on $g$ (See \cite{BLi,KC,ChangAl,DF,LJWS,PE,RB,Wang}). Rabinowitz considered a similar problem in \cite{RB} where condition (G3) was valid for all $|s|>s_0$ and $g(x,s)=o(|s|)$ for small values of $s$. First, he proved the existence of a nontrivial solution by using the mountain pass theorem. Next, by assuming that $g$ is Lipschitz continuous, Rabinowitz proved the existence of two nontrivial solutions $u^{-},u^{+}$ such that $u^{-} < 0 < u^{+}$. Wang \cite{Wang} also assumed condition (G3) for $|s|>s_0$, in addition to $g(0)=0$ and $g'(0)=0$. He proved the existence of three nontrivial solutions by using a Morse theory approach. In \cite{PE}, Perera approached this problem by assuming that condition (G3) is valid for all $|s|>s_0$, and the existence of a constant $a>0$ such that $g(0)=g(a)=0$, and $g'(0)=\lambda$. Perera proved the existence of four nontrivial solutions for the cases where $\lambda \in (\lambda_{j},\lambda_{j+1})$, $\lambda=\lambda_{j}<\lambda_{j+1}$, and $\lambda_{j}<\lambda=\lambda_{j+1}$, and $j\geq 3$. In this article, we are only assuming condition (G3) for large positive values of $s$. For large negative values of $s$ we are assuming conditions (G2) and (G6). In this sense, $g$ is said to be an asymmetric nonlinearity. We will show that problem \eqref{e1} has at least three nontrivial solutions by using variational methods and Morse Theory. Another work on asymmetric nonlinearities related to the work in this article is that of Liu and Shaoping \cite{LJWS}. In \cite{LJWS}, the authors considered the model problem \begin{equation}\label{e11111} \begin{gathered} -\Delta u = \lambda u + (u^{+})^{p} \quad \text{in } \Omega; \\ u = 0 \quad \text{on } \partial\Omega, \end{gathered} \end{equation} where $u^{+}=\max\{0,u\}$, $10$, \begin{equation}\label{linkc1} \begin{gathered} J(u)\leq 0, \quad\text{for } u\in H^{-},\|u\|\leq \delta; \\ J(u) > 0, \quad\text{for } u\in H^{+},0<\|u\|\leq \delta. \end{gathered} \end{equation} \end{definition} Assume $u$ is a critical point of $J$ such that $J''(u)$ is a Fredholm operator. The Morse index of $u$, denoted by $\mu_0(u)$, is defined as the supremum of the dimensions of the vector subspaces of $H$ on which $J''(u)$ is negative definite. The nullity of $u$, denoted by $\nu_0=\nu_0(u)$, is defined as the dimension of the kernel of $J''(u)$. We say that a functional $J$ satisfies the Palais-Smale (PS) condition if any sequence $(u_n)\subset H$ for which $J(u_n)$ is bounded and $J'(u_n)\to 0$ as $n\to\infty$ possesses a convergent subsequence. We will say that $(u_n)\subset H$ is a (PS) sequence for $J$ if $$ |J(u_n)|\leq C,\quad \text{ for all $n$ and some constant }C, $$ and $J'(u_n)\to 0$ as $n\to\infty$. Based on the notion of local linking at the origin, and assuming that $J$ satisfies the (PS) condition, the critical groups $C_{q}(J,0)$ can be calculated based on a result from Su \cite{SuJ}. \begin{proposition}[{\cite[Proposition 2.3]{SuJ}}] \label{SucgOrigin1} Assume $J$ satisfies the (PS) condition and that it has a local linking at $0$ with respect to $H=H^{-}\oplus H^{+}$, where $0$ has a Morse index $\mu_0$ and nullity $\nu_0$. Set $d=\dim H^{-}$. Then \[ C_{q}(J,0)=\begin{cases} \delta_{q,\mu_0}\Z, & \text{if }d=\mu_0;\\ \delta_{q,\mu_0+\nu_0}\Z, & \text{if }d=\mu_0+\nu_0. \end{cases} \] \end{proposition} Thus, to compute the critical groups of $J$ at the origin, we will show that the functional $J$ satisfies the (PS) condition and that $J$ satisfies a local linking condition at the origin with respect to the decomposition $H=H^{-}\oplus H^{+}$, where $H^{-}=\oplus_{j=1}^{m}\ker(-\Delta -\lambda_{j}I)$ and $H^{+}=(H^{-})^{\perp}$. This will be the content of Section $4$. Let $\mathcal{K}=\{u\in H: J'(u)=0\}$ be the set of critical points of $J$ and assume $J$ satisfies the (PS) condition; then, $\mathcal{K}$ is a finite set. Set $a<\inf J(\mathcal{K})$. The critical groups of $J$ at infinity are defined as in Bartsch and Li \cite{BLi} by \begin{equation} C_{q}(J,\infty)=H_{q}(H,J^{a}),\quad q=0,1,2,\dots. \label{eq22} \end{equation} Finally, we will need the Morse relation. Let $J:H\to\R$ be a functional that satisfies the (PS) condition. If the functional $J:H\to\R$ has a finite number of critical points, we can define the Morse--type number of the pair $(H,J^{a})$ by \begin{equation} M_{q}:=M_{q}(H,J^{a})=\sum_{u\in \mathcal{K}} \dim C_{q}(J,u),\quad q=0,1,2,\dots . \label{Su23} \end{equation} Applying the infinite dimensional Morse Theory developed in \cite{KC,MW}, we can derive the Morse relation \begin{equation} \sum_{q=0}^{\infty}M_{q}t^{q}=\sum_{q=0}^{\infty}\beta_{q}t^{q}+(1+t) \sum_{q=0}^{\infty}a_{q}t^{q}, \label{Su26} \end{equation} where $\beta_{q}=\dim C_{q}(J,\infty)$, and $a_{q}$ are non-negative numbers. The numbers $\beta_{q}$ are also called the Betti numbers of the pair $(H,J^{a})$. As a consequence of equation \eqref{Su26}, if $\beta_{q}\ne 0$ for some $q$, then $J$ must have a critical point, say $w$, with $C_{q}(J,w)\not\cong 0$. In fact, by expanding the equation \eqref{Su26}, we have that $$ M_0+M_1t+\dots +M_{q}t^{q}+\dots=(\beta_0+a_0)+(\beta_1+a_1+a_0)t+\dots +(\beta_{q}+a_{q}+a_{q-1})t^{q}+\dots $$ Observe that the term $\beta_{q}+a_{q}+a_{q-1} > 0$ since $\beta_{q}\ne 0$ and $a_{q},a_{q-1}\geq 0$. Therefore, $M_{q}\ne 0$. This implies that there is at least one critical point $w\in \mathcal{K}$ such that $C_{q}(J,u)\not\cong 0$. \section{Estimates on $G(x,s)$ and $g(x,s)$} In this section we establish some estimates for $g(x,s)$ and $G(x,s)$ that will be used throughout this work. First, from condition (G2), there exists $t_{-}<0$ such that, for $s0$ such that \begin{equation} -C_1 + \lambda s - |s|^{\alpha} \leq g(x,s) \leq C_1+\lambda s + |s|^{\alpha}, \label{littlegg2} \end{equation} for all $s\leq 0$ and $x\in\Omega$. From \eqref{littlegg2}, we have \begin{equation} |sg(x,s) -\lambda s^{2}| \leq C_1|s| + |s|^{1+\alpha},\quad\text{for }s\leq 0. \label{intermed1} \end{equation} Applying Young's Inequality, \begin{equation} ab\leq\frac{a^{p}}{p}+\frac{b^{q}}{q},\quad\text{for }a,b\geq 0, \label{young1} \end{equation} with $a=|s|$, $b=1$, $p=1+\alpha$, and $q=(1+\alpha)/\alpha$, we can rewrite \eqref{intermed1} as \begin{equation} |sg(x,s)-\lambda s^{2}| \leq \frac{C_1\alpha}{1+\alpha} + \big(1+\frac{C_1}{1+\alpha}\big)|s|^{1+\alpha}. \label{intermed2} \end{equation} Setting $C_2=\max\big(1+\frac{C_1}{1+\alpha},\frac{C_1\alpha}{1+\alpha}\big)$ in \eqref{intermed2}, we obtain \begin{equation} |g(x,s)s-\lambda s^{2}|\leq C_2+C_2|s|^{1+\alpha},\quad\text{for } s\leq 0,\text{ and }x\in\Omega. \label{eq9Dj} \end{equation} By integrating the inequality in \eqref{littlegg2} and using the definition of $G$, we obtain \begin{equation} -C_1|s|+\frac{\lambda}{2}s^{2} -\frac{1}{\alpha + 1}|s|^{\alpha + 1} \leq G(x,s) \leq C_1|s|+\frac{\lambda}{2}s^{2}+\frac{1}{\alpha + 1}|s|^{\alpha + 1}, \label{numbereq1} \end{equation} for all $s\leq 0$ and a.e $x\in\Omega$, or, \begin{equation} |G(x,s)-\frac{\lambda}{2}s^{2}|\leq C_1|s| + \frac{1}{\alpha+1}|s|^{\alpha+1}, \label{mybigG2} \end{equation} for $s\leq 0$ and $x\in\Omega$. Next, we show that \begin{equation} |g(x,s)s-2G(x,s)|\leq C_4+C_4|s|^{1+\alpha}, \label{eq8Dj} \end{equation} for some constant $C_4>0$, $s\leq 0$ and $x\in\Omega$. In fact, multiplying \eqref{littlegg2} by $s\leq 0$, we obtain \begin{equation} C_1|s| + \lambda s^{2}+|s|^{1+\alpha}\geq g(x,s)s \geq -C_1|s|+\lambda s^{2} - |s|^{1+\alpha}. \label{littlegg3} \end{equation} Similarly, from \eqref{numbereq1}, we have \begin{equation} 2C_1s-\lambda s^{2}+\frac{2}{1+\alpha}|s|^{1+\alpha}\geq -2G(x,s) \geq -2C_1|s|-\lambda s^{2}-\frac{2}{1+\alpha}|s|^{1+\alpha}. \label{mybigGG4} \end{equation} Then, adding \eqref{littlegg3} and \eqref{mybigGG4}, we obtain \[ 3C_1|s|+\big(1+\frac{2}{1+\alpha}\big)|s|^{1+\alpha} \geq g(x,s)s-2G(x,s)\geq -3C_1|s|-\big(1+\frac{2}{1+\alpha}\big)|s|^{1+\alpha}, \] so that \[ |g(x,s)s-2G(x,s)|\leq 3C_1|s|+\big(1+\frac{2}{1+\alpha}\big)|s|^{1+\alpha}, \quad\text{for }s\leq 0. \] Applying Young's Inequality \eqref{young1} with $a=|s|$, $b=1$, $p=1+\alpha$, and $q=(1+\alpha)/\alpha$, we obtain \begin{equation} |g(x,s)s-2G(x,s)|\leq \frac{3C_1}{1+\alpha} +\big(\frac{2+3C_1\alpha}{1+\alpha}+1\big)|s|^{1+\alpha}, \label{youn22} \end{equation} for $s\leq0$. Therefore, defining $C_5$ by $$ C_5=\max\Big(\frac{3C_1}{1+\alpha},\frac{2+3C_1\alpha}{1+\alpha}+1\Big), $$ we obtain \eqref{eq8Dj} from \eqref{youn22}. Combining \eqref{numbereq1} and condition (G4), we find a global estimate for $G(x,s)$ given by \begin{equation} G(x,s)\leq C_6|s|+\frac{\lambda}{2}s^{2}+\frac{1}{1+\alpha}|s|^{1+\alpha} +\frac{1}{\sigma+1}|s|^{\sigma+1}, \label{estbigG} \end{equation} for all $s\in\R$ and $x\in\Omega$, and $C_6=C_1+C_2$. Finally, from condition (G3), we can find $C_7,C_{8} > 0$ such that \begin{equation} G(x,s)\geq C_7|s|^{\mu} - C_{8}, \label{superq} \end{equation} for all $s\geq 0$, where $\mu=1/\theta> 2$. In fact, from condition (G3) we have \begin{equation} 0\leq \frac{\partial G}{\partial s}(x,s) - \frac{1}{s\theta}G(x,s), \label{ded1} \end{equation} for $s>s_0$. Multiplying \eqref{ded1} by the integrating factor $s^{-1/\theta}$ and integrating over the interval $[s_0,s]$, we obtain $$ 0\leq -\frac{1}{s_0^{1/\theta}}G(x,s_0)+\frac{1}{s^{1/\theta}}G(x,s), \quad \text{for all }s>s_0. $$ Then, setting $C_7=\frac{1}{s_0^{1/\theta}}G(x,s_0)$, we can find a constant $C_{8}>0$ such that $$ G(x,s)\geq C_7|s|^{\mu} - C_{8}, $$ for all $s>0$ and $x\in\Omega$, which is \eqref{superq}. The next lemma will be used in the proof of a local linking condition at the origin. \begin{lemma} \label{lemalinkcond} Assume that $g$ satisfies condition {\rm (G1)} and let $\varepsilon > 0$ be such that $\lambda_{m}+\varepsilon < \lambda_{m+1}$. Then, there exists $\delta_1>0$ such that \begin{equation} |G(x,s)|\leq \big(\frac{\lambda_{m}+\varepsilon}{2}\big)|s|^{2}, \label{estbigG11} \end{equation} for $|s|<\delta_1$ and $x\in\Omega$, where $m$ is as given in {\rm (G1)}. \end{lemma} \begin{proof} Since $g_{s}'(x,0)=\lambda_{m}$ for all $x\in\Omega$, there exists $\delta_1 > 0$ such that, for $|s|<\delta_1$, $$ |g(x,s)-\lambda_{m}s|\leq \varepsilon |s|,\quad\text{for all }x\in\Omega; $$ then, \begin{equation} |g(x,s)|\leq (\lambda_{m}+\varepsilon)|s|, \quad\text{for }|s|<\delta_1, \label{littleg1} \end{equation} Therefore, we can show that \[ |G(x,s)|\leq \big(\frac{\lambda_{m}+\varepsilon}{2}\big)|s|^{2}, \quad\text{for }|s|<\delta_1,\text{ and } x\in\Omega, \] which is \eqref{estbigG11}. \end{proof} \section{Palais-Smale condition} Assuming (G1)--(G5), we can show that the functional $J:H\to\R$ defined in \eqref{myfunc1} satisfies the Palais-Smale (PS) condition. The proof was done by De Figueiredo in \cite{DF} assuming that $\lambda\ne\lambda_{j}$ for all $j\in\N$. It turns out the result is true if we assume that $\lambda\ne\lambda_1$. We present the proof here for the reader's convenience. \begin{lemma}[{\cite[Lemma 1, page 291]{DF}}] \label{myPScond} If $g$ and $G$ satisfy {\rm (G1)--(G5)}, then the functional $J:H\to\R$ defined by $$ J(u)=\frac{1}{2}\int_{\Omega}|\nabla u|^{2}dx - \int_{\Omega}G(x,s)dx,\quad\text{for }u\in H, $$ satisfies the Palais-Smale condition. \end{lemma} \begin{proof} In what follows, we use the same symbol $C$ to denote all constants that come up in the estimates. Let $(u_n)$ be a (PS) sequence for $J$ in $H=H_0^{1}(\Omega)$; that is, $(u_n)$ satisfies \begin{equation} |J(u_n)|=\big|\frac{1}{2}\int_{\Omega}|\nabla u_n|^{2}dx - \int_{\Omega}G(x,u_n)dx\big|\leq C,\quad\text{for all }n, \label{eqDj10} \end{equation} and some constant $C>0$, and \begin{equation} |\langle J'(u_n),v\rangle| =\big|\int_{\Omega}\nabla u_n\cdot\nabla vdx - \int_{\Omega}g(x,u_n)vdx\big| \leq\varepsilon_n\|v\|, \quad\text{for all }n, \label{eqDj11} \end{equation} where $\varepsilon_n\to 0$ as $n\to\infty$ and $v\in H$. By virtue of the subcritical growth condition in (G4), it suffices to prove that $(\|u_n\|)$ is bounded (\cite[Proposition $2.2$, p.73]{Str}). First, notice that \begin{align*} &\int_{\Omega} [g(x,u_n)u_n-2G(x,u_n)]dx \\ & = \int_{\Omega}\big[g(x,u_n)u_n-|\nabla u_n|^{2} + |\nabla u_n|^{2}-2G(x,u_n)\big]dx \\ &\leq \big|\int_{\Omega}[|\nabla u_n|^{2}-g(x,u_n)u_ndx]| + 2|J(u_n)|. \end{align*} Thus, setting $v=u_n$ in \eqref{eqDj11} and using \eqref{eqDj10} we obtain \begin{equation} \int_{\Omega}\left[g(x,u_n)u_n-2G(x,u_n)\right]dx \leq \varepsilon_n\|u_n\| + C,\quad\text{for all }n. \label{eqDj12} \end{equation} The integral on the left side of \eqref{eqDj12} can be split in three parts, $$ \int_{\Omega}[g(x,u_n)u_n-2G(x,u_n)]dx =\Big[\int_{\Omega_n^{-}}+\int_{\Omega_n^{0}} + \int_{\Omega_n^{+}}\Big][g(x,u_n)u_n-2G(x,u_n)]dx, $$ where $\Omega_n^{-}=\{x\in\Omega:u_n\leq 0\}$, $\Omega_n^{0}=\{x\in\Omega: 0\leq u_n\leq s_0\}$, and $\Omega_n^{+}=\{x\in\Omega: u_n>s_0\}$. The first integral is estimated using \eqref{eq8Dj} as follows, \begin{equation} \int_{\Omega_n^{-}} [g(x,u_n)u_n-2G(x,u_n)]dx \leq C+C\int|u_n^{-}|^{1+\alpha}dx,\quad\text{for all }n, \label{stareq} \end{equation} where $u^{-}=\max\{0,-u\}$. The second integral taken over $\Omega_n^{0}$ is bounded uniformly with respect to $n$. The third integral can be estimated using (G3) as follows: \begin{equation} \int_{\Omega_n^{+}} [g(x,u_n)u_n-2G(x,u_n)]dx \geq \big(\frac{1}{\theta}-2\big)\int_{\Omega_n^{+}}G(x,u_n)dx, \quad\text{for all }n. \label{adstar} \end{equation} Thus, combining \eqref{adstar} with \eqref{eqDj12} and \eqref{stareq}, we obtain \begin{equation} \int_{\Omega_n^{+}}G(x,u_n)dx \leq C+\varepsilon_n\|u_n\|+C\|u_n^{-}\|_{L^{1+\alpha}}^{1+\alpha}, \quad\text{for all }n, \label{eqDj13} \end{equation} and some constant $C>0$. Set $v=u_n^{-}$ in \eqref{eqDj11}. Then, it follows that \begin{equation} \big|\int_{\Omega}|\nabla u_n^{-}|^{2}dx-\int_{u_n<0}g(x,u_n)u_ndx\big| \leq\varepsilon_n\|u_n^{-}\|,\quad\text{for all }n. \label{eqDj14} \end{equation} Next, compute \begin{align*} &\big|\int |\nabla u_n^{-}|^{2}dx - 2\int_{u_n<0}G(x,u_n)dx\big| \\ & = \big|\int |\nabla u_n^{-}|^{2}dx -\int_{u_n<0}g(x,u_n)u_ndx +\int_{u_n<0}[g(x,u_n)u_ndx- 2G(x,u_n)]dx\big|, \end{align*} and use \eqref{eq8Dj} and \eqref{eqDj14} to obtain \begin{equation} \big|\int |\nabla u_n^{-}|^{2}dx - 2\int_{u_n0}|g(x,u_n)v|dx\\ &\quad + \int_{u_n<0}|g(x,u_n)v-\lambda u_n^{-}v|dx; \end{align*} so that, by \eqref{eqDj11}, \begin{equation} \big|\int\nabla u_n^{-}\cdot\nabla vdx-\lambda\int u_n^{-}v\big| \leq \varepsilon_n\|v\|+I_1+I_2+I_3,\quad \text{for all }n, \label{eqDj18} \end{equation} where \begin{gather*} I_1 =\int|\nabla u_n^{+}\cdot\nabla v| dx,\quad I_2 = \int_{u_n\geq 0}|g(x,u_n)v|dx,\\ I_3 = \int_{u_n<0}|g(x,u_n)v-\lambda u_n^{-}v|dx. \end{gather*} We will estimate $I_1$, $I_2$, and $I_3$ separately. To estimate $I_1$, use H\"{o}lder's inequality to get \begin{equation} I_1=\int|\nabla u_n^{+}\cdot\nabla v|dx\leq \|u_n^{+}\|\|v\|. \label{integral1} \end{equation} To estimate $I_2$, apply H\"older's inequality with \begin{equation}\label{pDfn05} p=\frac{2N}{N+2} \end{equation} and $q=2N/(N-2)$ for $N\geq 3$. If $N=2$, take $1\leq p \leq 1/(\sigma\theta)$ which can be done since $(G_5)$ implies $\sigma\theta<1$. Then, \begin{align*} I_2=\int_{u_n>0}|g(x,u_n)v|dx & \leq\Big(\int |g(x,u_n)|^{p}\Big)^{1/p}\Big(\int |v|^{q}\Big)^{1/q}\\ &\leq\Big(\int |C+C|u_n^{+}|^{\sigma}|^{p}\Big)^{1/p}\|v\|_{L^{q}}, \end{align*} so that, \begin{equation} I_2\leq \left(C+C\|u_n^{+}\|_{L^{p\sigma}}^{\sigma}\right)\|v\|_{L^{q}}. \label{integral2} \end{equation} Finally, use \eqref{littlegg2} to obtain the following estimate for $I_3$: \begin{equation} I_3=\int_{u_n<0}|g(x,u_n)v-\lambda u_n^{-}vdx|dx \leq (C+C\|u_n^{-}\|^{\alpha})\|v\|. \label{integral3} \end{equation} Combining \eqref{integral1}, \eqref{integral2}, and \eqref{integral3} into \eqref{eqDj18}, we have the estimate \begin{equation} \big|\int\nabla u_n^{-}\cdot\nabla vdx-\lambda\int u_n^{-}v\big| \leq (C+\|u_n^{+}\|+C\|u_n^{+}\|_{L^{p\sigma}}^{\sigma}+C\|u_n^{-}\|^{\alpha})\|v\|, \label{Keq} \end{equation} for all $n$. Set \begin{equation} K_n=C+\|u_n^{+}\|+C\|u_n^{+}\|_{L^{p\sigma}}^{\sigma}+C\|u_n^{-}\|^{\alpha}, \quad\text{for all }n; \label{kneq} \end{equation} then \eqref{Keq} can be written as \begin{equation} \big|\int\nabla u_n^{-}\cdot\nabla vdx-\lambda\int u_n^{-}v\big| \leq K_n\|v\|,\quad\text{for all }n. \label{Keq2} \end{equation} The goal next is to show that \begin{equation} \frac{K_n}{\|u_n^{-}\|}\to 0 \quad\text{as } n\to\infty. \label{kngoal2} \end{equation} where $K_n$ is as given by \eqref{kneq}. First, from \eqref{eqDj17}, and the fact that $\alpha < 1$, it follows that \begin{equation} \frac{\|u_n^{+}\|}{\|u_n^{-}\|}\to 0,\quad \text{as }n\to\infty. \label{eqDj19} \end{equation} Secondly, we claim that \begin{equation} \frac{\|u_n^{+}\|_{L^{p\sigma}}^{\sigma}}{\|u_n^{-}\|}\to 0\quad \text{as }n\to\infty. \label{eqDj20} \end{equation} In fact, using the estimates in \eqref{superq} and \eqref{eqDj13}, we obtain the estimate \begin{equation} \int(u_n^{+})^{1/\theta}dx\leq C+\varepsilon_n\|u_n\| +C\|u_n^{-}\|_{L^{1+\alpha}}^{1+\alpha},\text{ for all }n. \label{eqDj21} \end{equation} Now, by the Sobolev inequality, we obtain from \eqref{eqDj21} that \begin{equation} \begin{aligned} \Big(\int(u_n^{+})^{1/\theta}dx\Big)^{\theta} &\leq\left( C+\varepsilon_n\|u_n\|+C\|u_n^{-}\|^{1+\alpha}\right)^{\theta} \\ &\leq C+C\varepsilon_n\|u_n\|^{\theta}+C\|u_n^{-}\|^{\theta(1+\alpha)}. \end{aligned} \label{eqDj211} \end{equation} Choose $\alpha'\in (\alpha,1)$, and divide both sides of \eqref{eqDj211} by $\|u_n^{-}\|^{\theta(1+\alpha')}$, to obtain \[ \frac{\|u_n^{+}\|_{L^{1/\theta}}}{\|u_n^{-}\|^{\theta(1+\alpha')}} \leq \frac{C}{\|u_n^{-}\|^{\theta(1+\alpha')}} +C\frac{\varepsilon_n\|u_n|\|^{\theta}}{\|u_n^{-}\|^{\theta(1+\alpha')}} +\frac{C}{\|u_n^{-}\|^{\theta(\alpha'-\alpha)}},\quad\text{for all }n, \] so that, in view of \eqref{eqDj19}, \begin{equation} \frac{\|u_n^{+}\|_{L^{1/\theta}}}{\|u_n^{-}\|^{\theta(1+\alpha')}} \to 0,\quad \text{ as }n\to\infty. \label{eqDj212} \end{equation} We claim that \begin{equation} \frac{u_n^{+}}{\|u_n^{-}\|^{1/\sigma}}\to 0 \quad \text{as $n\to\infty$ in the $L^{p\sigma}$ norm}; \label{mystar22} \end{equation} this will establish \eqref{eqDj20}. To show \eqref{mystar22}, set $p_1=1/(p\sigma\theta)$, where $p$ is as given in \eqref{pDfn05}, and note that $p_1>1$ by $(G_5)$. Next, use H\"{o}lder's inequality to obtain \begin{align*} \int_{\Omega}\Big(\frac{|u_n^{+}|}{\|u_n^{-}\|^{\frac{1}{\sigma}}}\Big)^{p\sigma}dx &\leq C\Big(\int_{\Omega}\frac{|u_n^{+}|^{1/\theta}}{\|u_n^{-}\| ^{\frac{1}{\sigma\theta}}}dx\Big)^{p\sigma\theta} \\ &\leq\frac{C}{\|u_n^{-}\|^{p}}\Big(\int_{\Omega} \frac{|u_n^{+}|^{1/\theta}}{\|u_n^{-}\|^{1+\alpha'}}\|u_n^{-}\|^{1+\alpha'}dx \Big)^{p\sigma\theta} \end{align*} so that \begin{equation} \int_{\Omega}\Big(\frac{|u_n^{+}|}{\|u_n^{-}\|^{\frac{1}{\sigma}}}\Big)^{p\sigma}dx \leq C\|u_n^{-}\|^{p\sigma\theta(1+\alpha')-p} \Big(\int_{\Omega}\frac{|u_n^{+}|^{1/\theta}}{\|u_n^{-}\|^{1+\alpha'}}dx \Big)^{p\sigma\theta}. \label{eqDj2122} \end{equation} By using \eqref{eqDj212}, we see that the term in parenthesis in \eqref{eqDj2122} approaches zero as $n\to\infty$. Next, notice that $$ \|u_n^{-}\|^{p\sigma\theta(1+\alpha')-p} \to 0,\quad \text{as }n\to\infty, $$ by condition (G5), since we are assuming $\|u_n^{-}\|\to \infty$ as $n\to\infty$. Hence, $$ \int_{\Omega}\Big(\frac{|u_n^{+}|}{\|u_n^{-}\|^{\frac{1}{\sigma}}}\Big)^{p\sigma}dx \to 0,\quad\text{as }n\to\infty, $$ which is \eqref{mystar22}. We have therefore established \eqref{eqDj20}. Use \eqref{kneq} to obtain \begin{equation} \frac{K_n}{\|u_n^{-}\|}=\frac{C}{\|u_n^{-}\|}+\frac{\|u_n^{+}\|}{\|u_n^{-}\|} +C\frac{\|u_n^{+}\|_{L^{p\sigma}}^{\sigma}}{\|u_n^{-}\|} +C\frac{1}{\|u_n^{-}\|^{1-\alpha}}, \label{laststar1} \end{equation} where $$ \frac{\|u_n^{+}\|}{\|u_n^{-}\|}\to 0\quad\text{as }n\to\infty, $$ by \eqref{eqDj19}, and $$ \frac{\|u_n^{+}\|_{L^{p\sigma}}^{\sigma}}{\|u_n^{-}\|}\to 0 \quad\text{as }n\to\infty $$ by \eqref{eqDj20}. Hence, since $\alpha < 1$ and $\|u_n^{-}\|\to\infty$ as $n\to\infty$, we obtain from \eqref{laststar1} that $$ \frac{K_n}{\|u_n^{-}\|}\to 0\quad\text{as }n\to\infty, $$ which is \eqref{kngoal2}. Next, combine \eqref{Keq2} and \eqref{kngoal2} to obtain \begin{equation} \lim_{n\to\infty}\Big(\int_{\Omega}\frac{\nabla u_n^{-}}{\|u_n^{-}\|}\cdot\nabla v dx -\lambda\int_{\Omega}\frac{u_n^{-}}{\|u_n^{-}\|}v\Big)=0,\quad \text{for all } v\in H. \label{kgoal2} \end{equation} Set $w_n=u_n^{-}/\|u_n^{-}\|$ for all $n$. Since $\|w_n\|=1$, for all $n$, passing to a subsequence if necessary, we may assume that there exists $w_0\in H$ such that $w_n\rightharpoonup w_0$ (weakly) in $H_0^{1}(\Omega)$ and $w_n\to w_0$ strongly in $L^{2}(\Omega)$. We may also assume that $w_n(x)\to w(x)$ for a.e $x\in\Omega$. It follows from \eqref{kgoal2} with $v=w_n$ that $$ \int w_0^{2}=\lambda^{-1}, $$ since we are assuming that $\lambda > 0$. It then follows from \eqref{kgoal2} that \[ \int_{\Omega}\nabla w_0\cdot\nabla v dx -\lambda\int_{\Omega}w_0vdx =0,\text{ for all }v\in H_0^{1}(\Omega); \] that is, $w_0$ is a nontrivial weak solution of the problem \begin{equation}\label{e123} \begin{gathered} -\Delta w_0 = \lambda w_0 \text{in } \Omega; \\ w_0 = 0 \quad\text{on } \partial\Omega. \end{gathered} \end{equation} By the maximum principle, $w_0<0$ in $\Omega$. Thus, $w_0$ is an eigenfunction of \eqref{e123} that does not change sign in $\Omega$. Hence, $\lambda=\lambda_1$, the first eigenvalue of \eqref{e123}, which is the case we are excluding. Therefore, we obtain a contradiction. The proof of Lemma \ref{myPScond} is now completed. \end{proof} \section{Local linking at the origin} The notion of local linking at the origin was introduced by Li and Liu in \cite{LiLiu2} and \cite{LiLiu3}. We present the definition given in Li and Willem \cite{LiWi}. \begin{definition}[{\cite[Section $0$]{LiWi}}] \label{def5.1} Let $J$ be a $C^{1}$ function defined on a Banach space $H$. We say that $J$ has a local linking near the origin if $H$ has a direct sum decomposition $H=H^{-}\oplus H^{+}$ with $\dim H^{-}<\infty$, $J(0)=0$, and, for some $\delta >0$, \begin{equation}\label{linkc1b} \begin{gathered} J(u)\leq 0, \quad \text{ for } u\in H^{-},\; \|u\|\leq \delta; \\ J(u) > 0, \quad\text{for } u\in H^{+},\;0<\|u\|\leq \delta. \end{gathered} \end{equation} \end{definition} \begin{lemma} \label{lemma52} Assume {\rm (G1)--(G5)} hold. Then, $J$ has a local linking at $0$ with respect to the decomposition $H=H^{-}\oplus H^{+}$, where $H^{-}=\oplus_{j\leq m}\ker(-\Delta-\lambda_{j}I)$, and $H^{+}=(H^{-})^{\perp}$. \end{lemma} \begin{proof} The proof is based on arguments presented in papers by Li and Willem in \cite[Theorem 4]{LiWi}, and by Li and Liu in \cite[Theorem 3.1]{LiLiu4}. First, let us show that there exists $\delta >0$ such that $J(u)\leq 0$ for $u\in H^{-}$ if $\|u\|<\delta$. In fact, by the definition of $H^{-}=\oplus_{j\leq m}\ker(-\Delta-\lambda_{j}I)$, we have \begin{equation} \int_{\Omega}|\nabla u|^{2}dx\leq\lambda_{m}\int_{\Omega}u^{2}dx, \quad\text{for }u\in H^{-}. \label{liwi1} \end{equation} Since $H^{-}$ is finite--dimensional, there exists $C > 0$ such that \begin{equation} \|u\|_{\infty}\leq C\|u\|,\quad\text{for }u\in H^{-}, \label{normf} \end{equation} where $\|u\|_{\infty}=\sup\{|u(x)|:x\in\Omega\}$. Select $u\in H^{-}$ such that $\|u\|\leq \frac{\delta_1}{C}$, so that $|u(x)|\leq\delta_1$, for a.e $x\in\Omega$, where $\delta_1$ is given in Lemma \ref{lemalinkcond}. Then, from Lemma \ref{lemalinkcond} it follows that \begin{equation} -\big(\frac{\lambda_{m}+\varepsilon}{2}\big)|u(x)|^{2} \leq G(x,u)\leq\big(\frac{\lambda_{m}+\varepsilon}{2}\big)|u(x)|^{2}, \quad\text{for }|u(x)|<\delta_1. \label{bigGlema31} \end{equation} It follows from \eqref{liwi1} that \begin{equation} J(u) \leq \int_{\Omega} [\frac{\lambda_{m}}{2}u^{2}-G(x,u(x))]dx,\quad \text{for }u\in H^{-},\|u\|\leq\frac{\delta_1}{C}. \label{lincdd} \end{equation} Hence, using the estimate in \eqref{bigGlema31}, we obtain from \eqref{lincdd} that \begin{equation} J(u)\leq -\frac{\varepsilon}{2}\int_{\Omega}u^{2}dx\leq 0,\quad\text{for } u\in H^{-}\text{ and }\|u\|\leq\frac{\delta_1}{C}. \label{lincdd2} \end{equation} Next, we need to show that $J(u)>0$ for $0<\|u\|<\delta$, for $u\in H^{+}$, where $\delta$ will be chosen shortly. Since we already have the estimate \eqref{littleg1} in Lemma \ref{lemalinkcond} for $|s|\leq\delta_1$, we need an estimate for $|s|>\delta_1$. In fact, using the estimate \eqref{estbigG}, for $\frac{|s|}{\delta_1}>1$, and the assumption that $0\leq\alpha<1$ in ($G_1$), we have that \begin{align*} G(x,s) & \leq C|s|+\frac{\lambda}{2}s^{2}+\frac{1}{1+\alpha}|s|^{1+\alpha} +\frac{1}{\sigma+1}|s|^{\sigma+1}, \\ & = C\delta_1\frac{|s|}{\delta_1}+\frac{\delta_1^{2}\lambda}{2} \big(\frac{s}{\delta_1}\big)^{2}+\frac{\delta_1^{1+\alpha}}{1+\alpha} \big(\frac{|s|}{\delta_1}\big)^{1+\alpha}+\frac{\delta_1^{\sigma+1}}{\sigma+1} \big(\frac{|s|}{\delta_1}\big)^{\sigma+1} \\ &\leq\frac{1}{\delta_1^{\sigma+1}} \Big( C\delta_1+\frac{\delta_1^{2}\lambda}{2}+ \frac{\delta_1^{1+\alpha}}{1+\alpha} + \frac{\delta_1^{\sigma+1}}{\sigma+1} \Big)|s|^{\sigma+1}, \end{align*} so that \begin{equation} G(x,s)\leq C_{\varepsilon}|s|^{\sigma+1},\quad\text{for all }|s|>\delta_1, \label{estbigG12} \end{equation} where $C_{\varepsilon}$ is given by $$ C_{\varepsilon}=\frac{1}{\delta_1^{\sigma+1}} \Big( C\delta_1+\frac{\delta_1^{2}\lambda}{2} + \frac{\delta_1^{1+\alpha}}{1+\alpha} + \frac{\delta_1^{\sigma+1}}{\sigma+1} \Big). $$ Combining \eqref{estbigG11} and \eqref{estbigG12}, it follows that \begin{equation} G(x,s)\leq \big(\frac{\lambda_{m}+\varepsilon}{2}\big)|s|^{2} +C_{\varepsilon}|s|^{\sigma+1}, \label{finalBiGG} \end{equation} for all $s\in\R$ and $x\in\Omega$. Then, for $u\in H^{+}$ and using \eqref{finalBiGG}, we have \begin{align*} J(u) & = \frac{1}{2}\|u\|^{2}-\int_{\Omega}G(x,u)dx \\ & \geq \frac{1}{2}\|u\|^{2}-\big(\frac{\lambda_{m}+\varepsilon}{2}\big) \int_{\Omega}u^{2}dx-C_{\varepsilon}\int_{\Omega}|u|^{\sigma+1}dx. \end{align*} Thus, applying the Sobolev inequality, and the fact that $\|u\|^{2}\geq\lambda_{m+1}\|u\|_{L^{2}}^{2}$ for $u\in H^{+}$, we obtain \begin{equation} J(u)\geq \frac{1}{2}\big[1-\big(\frac{\lambda_{m}+\varepsilon}{\lambda_{m+1}} \big)-\tilde{C}_{\varepsilon}\|u\|^{\sigma-1} \big]\|u\|^{2}\quad\text{for } u\in H^{+}. \label{lincddd} \end{equation} Next, choose $\rho > 0$ such that $$ \rho<\Big[\frac{1}{2\tilde{C}_{\varepsilon}} \Big(1-\big(\frac{\lambda_{m}+\varepsilon}{\lambda_{m+1}}\big)\Big)\Big] ^{\frac{1}{\sigma-1}}. $$ Then, for $u\in H^{+}$ such that $\|u\|<\delta$, where $\delta=\min\{\frac{\delta_1}{C},\rho\}$, we obtain from \eqref{lincddd} that $$ J(u)>0,\quad\text{for }u\in H^{+},0<\|u\|<\delta, $$ and the lemma is proved. \end{proof} By Lemma \ref{lemma52}, $J$ satisfies a local linking condition at the origin with respect to the decomposition $H=H^{-}\oplus H^{+}$. In this case, $0$ has Morse Index $\mu_0$ and nullity $\nu_0$ given by \begin{gather} \mu_0=\sum_{j=1}^{m-1}\dim \ker (-\Delta-\lambda_{j}I),\label{m1}\\ \nu_0=\dim\ker(-\Delta-\lambda_{m}I),\label{m2} \end{gather} respectively, where we are assuming that $m > 1$ by (G1). Therefore, using Proposition \ref{SucgOrigin1}, \eqref{m1}, \eqref{m2}, and since $\dim H^{-}=\mu_0+\nu_0=d$, we obtain \begin{equation} C_{q}(J,0)=\delta_{q,d}\Z. \label{prob1cgrouporign1} \end{equation} \section{Existence of two nontrivial solutions} In this section we prove the existence of two nontrivial solutions of problem \eqref{e1} under the assumptions (G1)--(G5) and $g(x,t_0)=0$ for all $x\in\overline{\Omega}$ and some $t_0> 0$. We will employ the cutoff technique used by Chang, Li and Liu in \cite[Theorem $B$]{KC3}. \begin{proposition} \label{secondmaintheo} Assume $g$ satisfies {\rm (G1)--(G5)} and suppose there exists $t_0> 0$ such that $g(x,t_0)=0$ for all $x\in\overline{\Omega}$. Then, problem \eqref{e1} has a nontrivial solution, $u_0$, such that \begin{equation} C_{q}(J,u_0)=\delta_{q,0}\Z. \label{cgroupminimizer} \end{equation} \end{proposition} \begin{proof} Define $\overline{g}:\overline{\Omega}\times\R\to\R$ by \[ \overline{g}(x,s)= \begin{cases} g(x,s), &\text{if }s\in[0,t_0];\\ 0, &\text{if }s\notin[0,t_0]. \end{cases} \] Define the functional $\overline{J}:H\to\R$ by \begin{equation} \overline{J}(u)=\frac{1}{2}\|u\|^{2}-\int_{\Omega}\overline{G}(x,u)dx, \quad\text{for }u\in H, \label{mynewg} \end{equation} where $\overline{G}(x,s)=\int_0^{s}\overline{g}(x,\xi)d\xi$, for $x\in\Omega$, $s\in\R$. In order to show the existence of a nontrivial solution for problem \eqref{e1}, we will first show that $\overline{J}$ has a minimizer. Let $M=\sup_{x\in\overline{\Omega},s\in[0,t_0]}|\overline{g}(x,s)|$; then, using H\"{o}lder and Poincar\'{e}'s inequalities we have \begin{align*} \overline{J}(u) &\geq \frac{1}{2}\|u\|^{2}-M\int_{\Omega}|u|dx \\ &\geq \frac{1}{2}\|u\|^{2}-M|\Omega|^{1/2}\|u\|_{L^{2}(\Omega)} \\ &\geq \frac{1}{2}\|u\|^{2}-c\|u\|,\quad\text{for all }u\in H, \end{align*} which shows that $\overline{J}$ is coercive and bounded below. Also, $\overline{J}$ is weakly lower semicontinuous. Thus, there exists a global minimizer $u_0$ of $\overline{J}$ such that $$\overline{J}(u_0)=\inf_{u\in H} \overline{J}(u).$$ (See Evans \cite[Page $488$]{EV}). The function $\overline{g}$ is locally Lipschitz continuous; thus, it follows that $u_0$ is a classical solution of the problem \begin{equation}\label{u0classic} \begin{gathered} -\Delta u = \overline{g}(x,u) \quad \text{in } \Omega; \\ u = 0 \quad\text{on } \partial\Omega. \end{gathered} \end{equation} (See Agmon \cite{Ag}). Let $\Omega_{-}=\{x\in\Omega:u_0(x)<0\}$. Then, by the definition of $\overline{g}$, $u$ solves the BVP, \begin{equation}\label{u0classic1} \begin{gathered} -\Delta u = 0 \quad \text{ in } \Omega_{-} ;\\ u = 0 \quad\text{on } \partial\Omega_{-}, \end{gathered} \end{equation} which has only the trivial solution $u\equiv 0$. It then follows that $\Omega_{-}=\emptyset$. Similarly, if we consider the set $\Omega_{t_0}=\{x\in\Omega:u_0(x)>t_0\}$, it can be shown that $\Omega_{t_0}=\emptyset$. Therefore, we have $0\leq u_0\leq t_0$ in $\Omega$. Using the strong maximum principle, we can show that \begin{gather} 00$ such that $u\in C_0^{1}(\Omega)$ and $\|u-u_0\|_{C^{1}} < \delta$ imply that $0s_0\}, \] for all $n$, where $s_0$ is given by (G3). Then, \begin{equation} C+\mu^{-1}\|v_n\| \geq \big(\frac{1}{2}-\frac{1}{\mu}\big)\|v_n\|^{2} +\int_{\Omega_{1,n}} T_ndx+ \int_{\Omega_{2,n}} T_n\,dx. \label{mypss} \end{equation} Note that we have also used \eqref{myvnn222}. By (G3), $(2^{-1}-\mu^{-1}) > 0$, and the second integral in \eqref{mypss} is nonnegative. Define $$ K_2=\max_{x\in\overline{\Omega},s\leq s_0}|\mu^{-1}\widetilde{g}_+(x,s) -\widetilde{G}_+(x,s)|. $$ Then, \begin{equation} \big|\int_{\Omega_{1,n}} T_n dx\big| \leq K_2|\Omega|\quad\text{for all }n. \nonumber \end{equation} Thus, \eqref{mypss} becomes \begin{equation} C+\frac{1}{\mu}\|v_n\|\geq \big(\frac{1}{2}-\frac{1}{\mu}\big)\|v_n\|^{2}-K_2|\Omega|, \quad\text{for }n\geq n_1. \label{mypssf} \end{equation} Therefore, it follows from \eqref{mypssf} that $(v_n)$ is bounded. Hence, $\widetilde{J}_+$ satisfies the (PS) condition. \smallskip Before we proceed with the proof, we will derive an estimate for $\widetilde{G}(x,s)$ for positive values of $s$. Apply \eqref{superq} to \eqref{mygg1}, using the estimate in \eqref{BoundOnU0}, to get that \[ \widetilde{G}(x,s) \geq C_5|s+u_0(x)|^{\mu} - C_6 - |G(x,u_0(x))|-|s||g(x,u_0(x))|, \] for $s\geq 0$ and $x\in\Omega$. Thus, there exists a constant $C_{9}>0$ such that \begin{equation}\label{RumbosEst0005} \widetilde{G}(x,s) \geq C_{9}|s|^{\mu}-C_{9}|u_0(x)|^{\mu} - C_6 - C_{10}-C_{11} |s|, \end{equation} for $s\geq 0$ and $x\in\Omega$, where we have set $$ C_{10} = \max_{x\in\overline{\Omega}, 0\leq \xi\leq t_o} |G(x,\xi)|,\quad C_{11} = \max_{x\in\overline{\Omega}, 0\leq \xi\leq t_o} |g(x,\xi)|. $$ Thus, setting $$ C_{12}=C_{9}t_{o}^{\mu}+C_6+ C_{10} $$ we obtain from \eqref{RumbosEst0005} that \begin{equation}\label{RumbosEst0010} \widetilde{G}(x,s) \geq C_{9}|s|^{\mu}-C_{11} |s|- C_{12} \end{equation} Hence, using the assumption that $\mu >2$, we deduce from \eqref{RumbosEst0010} the existence of positive constants $C_{13}$ and $C_{14}$ such that \begin{equation} \widetilde{G}(x,s)\geq C_{13}|s|^{\mu}-C_{14},\quad \text{for }s\geq 0 \text{ and } x\in\Omega. \label{superq22} \end{equation} Next, we show that \begin{equation} \lim_{t\to\infty} \widetilde{J}_+(t\varphi_1)=-\infty. \label{inflim} \end{equation} In fact, use the estimate on \eqref{superq22} to obtain \begin{align*} \widetilde{J}_+(t\varphi_1) & = \frac{t^{2}}{2}\|\varphi_1\|^{2}-\int_{\Omega} \widetilde{G}_+(x,t\varphi_1)dx \\ &\leq \frac{t^{2}}{2}\|\varphi_1\|^{2}-C_{13}\frac{t^{\mu}}{2}\int_{\Omega}|\varphi_1|^{\mu}dx +C_{14}|\Omega|, \end{align*} so that $$ \lim_{t\to\infty}\widetilde{J}_+(t\varphi_1) = -\infty, $$ since $\mu > 2$, which is \eqref{inflim}. We have already noted that, since we are assuming $u_0$ is a strict local minimizer of $J$, it follows that $0$ is a strict local minimizer of $\widetilde{J}_+$. It then follows from \eqref{inflim} and the intermediate value theorem that there exists $v_0\in H$ such that $v_0\ne 0$ and $\widetilde{J}_+(v_0)=0$. Then, by the variant of the Mountain Pass Lemma in Chang \cite{KC} (See Proposition \ref{corokc}), $\widetilde{J}_+$ has a nontrivial critical point $v_1$ of mountain--pass type. We note that $v_1$ is a solution to the boundary-value problem \begin{gather*} -\Delta v = \widetilde{g}_+ (x, v(x)),\quad\text{for } x\in\Omega;\\ v = 0, \quad \text{on } \partial\Omega. \end{gather*} It then follows from the definition of $\overline{g}_+$ in \eqref{gtildeDfn10}, elliptic regularity theory, and the maximum principle that $v_1(x)>0$ for all $x\in\Omega$. Consequently, $v_1$ solves the boundary-value problem \begin{gather*} -\Delta v = \widetilde{g} (x, v(x)),\quad\text{for } x\in\Omega;\\ v = 0, \quad\ \text{on } \partial\Omega. \end{gather*} Hence, in view of the definition of $\widetilde{g}$ in \eqref{gtildeDfn05}, the function $u_1=u_0+v_1$ is the critical point of $J$ of mountain-pass type satisfying $$ u_0(x) < u_1(x), \quad\text{for all } x\in\Omega. $$ Moreover, if the critical points of the level set $K_{c_1}$, with $c_1=J(u_1)$, are isolated, then, using \cite[Corollary $8.5$]{MW}, there exists $\widetilde{u}_1 \in K_{c_1}$ such that \begin{equation} C_{q}(J,\widetilde{u}_1)\cong\delta_{q,1}\Z. \label{morse3} \end{equation} \end{proof} \section{Critical groups $C_{q}(J,\infty)$} In this section we compute the critical groups $C_{q}(J,\infty)$, for $q=1,2,\dots$, as defined in \eqref{eq22}. We will assume that conditions (G1)--(G6) are satisfied. We will use the technique outlined by Liu and Shaoping in \cite[Proposition 3.1]{LJWS}. Let $a=\inf J(\mathcal{K})$, where $\mathcal{K}=\{u\in H: J'(u)=0\}$ is the critical set of $J$. First, we will show that any compact set $A\subset J^{-M}$ is contractible in $J^{-M}$, for some constant $M>-a$. This will imply that $\widetilde{H}_{q}(J^{-M})=0$ for all $q\in\Z$. Then, by using the exact homology sequence of the pair $(H,J^{-M})$, we will show that $\widetilde{H}_{q}(H,J^{-M})=0$ for all $q\in\Z$. First, note that, by combining the conditions (G3) and (G6), we can find a constant $K_1>0$ such that \begin{equation} 2G(x,s)-sg(x,s) \leq K_1,\quad\text{for all } s\in\R \text{ and } x\in\Omega. \label{g7cond} \end{equation} The following proposition is based on a result from Liu and Shaoping in \cite[Proposition $3.1$]{LJWS}. \begin{proposition} \label{cinftycomp} Under the conditions {\rm (G1)--(G6)}, any compact subset $A$ of the sublevel set $J^{-M}=\{u\in H: J(u)\leq -M\}$ is contractible in $J^{-M}$ for $$ M\geq\max\{K_1|\Omega|,-a\}, $$ where $K_1$ is given in \eqref{g7cond}. \end{proposition} \begin{proof} \noindent\textbf{Step 1:} Let $A$ be a compact subset of $J^{-M}$, where $M>\max\{K_1|\Omega|,-a\}$. First, we show how to deform $A$ to a subset $A_1\subset J^{-2M}$ in $J^{-M}$. Compute \begin{equation} J(tu)=\frac{t^{2}}{2}\|u\|^{2}-\int_{\Omega}G(x,tu)dx, \quad\text{for }t\in\R, \label{jtu} \end{equation} and $u\in A$. Then, using \eqref{jtu}, we can show that \begin{equation} \frac{d}{dt}[J(tu)] = \frac{1}{t} \Big[2J(tu) + \int_{\Omega}\big(2G(x,tu)-g(x,tu)tu\big)dx\Big], \label{djtu1} \end{equation} for $t>0$. Using \eqref{g7cond} and \eqref{jtu} we obtain from \eqref{djtu1} that \begin{equation} \frac{d}{dt}[J(tu)]-\frac{2}{t}J(tu)\leq \frac{K_1|\Omega|}{t} \label{djtu2} \end{equation} for all $t\geq 1$. Multiply \eqref{djtu2} by the integrating factor $1/t^{2}$ and integrate from $1$ to $t>1$ to obtain \begin{equation} J(tu)\leq t^{2}J(u) + \frac{t^{2}}{2}K_1|\Omega| - \frac{1}{2}K_1|\Omega| , \label{jtu3} \end{equation} for all $t\geq 1$. Define a map $\eta_1$ on $[0,1]\times A$ by \begin{equation} \eta_1(t,u)=(1+t)u, \quad \text{for }u\in A. \label{myflow} \end{equation} Then, $\eta_1$ is continuous. Also, $\eta_1(t,u)\in J^{-M}$ for all $t\in [0,1]$. In fact, from \eqref{jtu3}, we have \begin{equation} J(\eta_1(t,u))\leq (1+t)^{2}J(u)+\frac{K_1|\Omega|}{2}[(1+t)^{2}-1]. \label{step1star} \end{equation} Since $K_1|\Omega|<-M$, and $J(u)\leq -M$, we obtain from \eqref{step1star} that \[ J(\eta_1(t,u)) \leq (1+t)^{2}J(u)+M(1+t)^{2}-M\leq -M \] Thus, \[ J((1+t)u)\leq -M,\text{ for all }0\leq t\leq 1. \] Therefore, $\eta_1$ defines a continuous map from $[0,1]\times A$ to $J^{-M}$. Set $A_1=\eta_1(1,A)$. Then, $A_1$ is a compact set. We claim that $A_1\subset J^{-2M}$. In fact, setting $t=1$ in \eqref{jtu3}, and using the assumption that $K_1|\Omega|\leq M$, we obtain $J(2u)\leq -2M$. Therefore, $A_1\subset J^{-2M}$. Thus, $\eta_1$ defines a deformation from $A$ to $A_1$ in $J^{-M}$. In what follows, we will use the fact that, if $u\in J^{-M}$, then \begin{equation} J(tu)\leq -M,\quad\text{for } t\geq 1; \label{notestep1} \end{equation} this is a consequence of \eqref{jtu3}. The remainder of the argument follows the same steps as in Liu and Shaoping in \cite[Proposition $2.1$]{LJWS}. \smallskip \noindent\textbf{Step 2:} In this step, we show how to deform the set $A_1$ obtained in Step $1$ to a subset of smooth functions. Since the functional $J:H\to \R$ is continuous on $H$ and $A_1$ is compact, there exists $\varepsilon>0$ such that, for all $u\in A_1$, \begin{equation} \|v-u\| < \varepsilon \Rightarrow |J(v)-J(u)|<\frac{M}{2}. \label{cont11} \end{equation} On the other hand, since the set $C_0^{1}(\Omega)$ is dense in $H$, for each $u\in A_1$, there exists $u^{\varepsilon}\in C_0^{1}(\Omega)$ such that \begin{equation} \|u-u^{\varepsilon}\|<\varepsilon. \label{mystep2star} \end{equation} Note that $\{B_{\varepsilon}(u^{\varepsilon})\}_{u\in A_1}$ is an open cover for $A_1$. Thus, since $A_1$ is compact, there exist smooth functions $u_1^{\varepsilon},u_2^{\varepsilon},\dots,u_n^{\varepsilon}$ such that $$ A_1\subset\cup_{i=1}^{n}B_{\varepsilon}(u_{i}^{\varepsilon}). $$ Let $\{\beta_{i}\}_{i=1}^{n}$ be a partition of unity subordinate to the cover $\{B_{\varepsilon}(u_{i}^{\varepsilon})\}_{i=1}^{n}$, where the functions $\{\beta_{i}\}_{i=1}^{n}$ are Lipschitz continuous. Then, for any $u\in A_1$, \[ \|\sum_{i=1}^{n}\beta_{i}(u)u_{i}^{\varepsilon}-u\| =\|\sum_{i=1}^{n}\beta_{i}(u)u_{i}^{\varepsilon} -\sum_{i=1}^{n}\beta_{i}(u)u\| \leq \|u_{i}^{\varepsilon}-u \|,\quad\text{for some }j. \] Hence, \begin{equation} \|\sum_{i=1}^{n}\beta_{i}(u)u_{i}^{\varepsilon}-u \| \leq \varepsilon, \label{etastar1} \end{equation} where we used \eqref{mystep2star} and the fact $\sum_{i=1}^{n}\beta_{i}(u)=1$. Let $u^{*}(u)=\sum_{i}\beta_{i}(u)u_{i}^{\varepsilon}$, for all $u\in A_1$. Then, $u^{*}$ is continuous. Let $\eta_2$ be a map defined on $[0,1]\times A_1$ by \begin{equation} \eta_2(t,u)=(1-t)u+tu^{*}(u),\quad\text{for }t\in[0,1] \text{ and }u\in A_1. \label{flow2} \end{equation} Note that $\eta_2$ is continuous. Next, we show that $\eta_2(t,u)\in J^{-\frac{3}{2}M}$ for all $t\in [0,1]$ and $u\in A_1$. Indeed, setting $v=(1-t)u+tu^{*}(u)$ and using \eqref{etastar1} and \eqref{flow2}, we obtain \begin{equation} \|v-u\| = t\|u^{*}-u\|<\varepsilon. \label{cont22} \end{equation} Then, using \eqref{cont11} we obtain \begin{equation} |J(v)-J(u)|<\frac{M}{2}, \label{cont223} \end{equation} in view of \eqref{cont22}. Since $J(u)\leq -2M$, we obtain from \eqref{cont223} that \[ J(v) < -\frac{3M}{2}, \quad\text{ for all }t\in[0,1],\text{ and }u\in A_1. \] Define $A_2=\eta_2(A_1,1)$. We have $A_2\subset J^{-\frac{3}{2}M}\cap C_0^{1}(\Omega)$. Therefore, we have deformed the set $A_1$ into a compact subset $A_2$ of $J^{-\frac{3}{2}M}\cap C_0^{1}(\Omega)$. Note that there exists a constant $\overline{M} > 0$ such that \begin{equation} |\nabla u(x)|\leq\overline{M},\quad\text{for all }u\in A_2. \label{estmmm} \end{equation} In fact, $$ \overline{M}=\max_{1\leq i\leq n}\max_{x\in\overline{\Omega}} |\nabla u_{i}^{\varepsilon}(x)|. $$ \smallskip \noindent\textbf{Step 3:} In this step, we will deform the subset $A_2$ from Step $2$ to a subset of functions with nonzero positive part. First, note that, since $J:H\to \R$ is continuous and $A_2$ is compact, there exists $\varepsilon_1>0$ such that, for all $u\in A_3$, \begin{equation} \|v-u\| < \varepsilon_1 \Rightarrow |J(v)-J(u)|<\frac{M}{2}. \label{contstep3} \end{equation} Let $d(x)=\text{dist}(x,\partial\Omega)$, for $x\in\Omega$. By \cite[Lemma $14.16$]{DN}, there exists $\nu > 0$ such that $d$ is smooth in the set $\Gamma_{\nu}=\{x\in\overline{\Omega}:d(x)<\nu\}$. Define $\varphi_{\varepsilon}:\overline{\Omega}\to\R$ by \[ \varphi_{\varepsilon}(x) = \begin{cases} 2\overline{M}d(x), &\text{if }x\in\Gamma_{\varepsilon};\\ 2\overline{M}\varepsilon, &\text{if }x\in\overline{\Omega} \backslash\Gamma_{\varepsilon}, \end{cases} \] where $\varepsilon > 0$ is such that $\varepsilon < \nu$ and \begin{equation} \int_{\Gamma_{\varepsilon}}|\nabla d|^{2}dx < \frac{\varepsilon_1^{2}}{4\overline{M}^{2}}. \label{estond11} \end{equation} It follows from \eqref{estond11} that \begin{equation} \|\varphi_{\varepsilon}\|<\varepsilon_1. \label{estgradphi} \end{equation} Furthermore, for every $u\in A_2$, we have \begin{equation} u(x)+\varphi_{\varepsilon}(x) > 0,\quad\text{for }x\text{ near }\partial\Omega. \label{step32star} \end{equation} In fact, if $u(x)>0$ for $x$ near $\partial\Omega$, the statement in \eqref{step32star} is true. If not, there exists $x_0\in\partial\Omega$ such that $u(x)<0$ for $x\in B_{\delta_0}(x_0)\cap\Omega$ for some $\delta_0>0$. Let $\vec{n}$ be a unit normal vector to $\partial\Omega$ that points towards $\Omega$. Define $f:\R\to\R$ by $$ f(t)=u(x_0+t\vec{n}),\quad\text{for all }t\in\R. $$ By the intermediate value theorem, there exists $\xi\in (0,t)$ such that $f(t)=f'(\xi)t$, for $t>0$ in some neighborhood of $0$; then, \[ u(x_0+t\vec{n})=(\nabla u(x_0+\xi\vec{n})\cdot\vec{n})t, \quad\text{for }t>0\text{ small enough}. \] Since $u(x_0+t\vec{n})<0$, $|u(x_0+t\vec{n})|=-u(x_0+t\vec{n})$, for $t > 0$ small enough. Then, using \eqref{estmmm}, we obtain \begin{equation} -u(x_0+t\vec{n})=|\nabla u(x_0+\xi\vec{n})|t\leq \overline{M}t, \quad\text{for }t>0\text{ small enough}. \nonumber \end{equation} So that, \begin{equation} -u(x_0+t\vec{n})<2\overline{M}t,\quad\text{for }t >0 \text{ small enough}. \label{step34star} \end{equation} Observe that, for $t>0$ small enough, $d(x_0+t\vec{n})=t$. We can therefore rewrite \eqref{step34star} as $$ -u(x_0+t\vec{n})<2\overline{M}d(x_0+t\vec{n}), \quad\text{for }t>0\text{ small enough}; $$ so that $$ -u(x_0+t\vec{n})<\varphi_{\varepsilon}(x_0+t\vec{n}), \quad\text{for }t>0\text{ small enough}. $$ Therefore, $v(x_0+t\vec{n})>0$ for $t>0$ small enough. Thus, $v=u+\varphi_{\varepsilon}$ has a positive part, $v^{+}$. Define a map $\eta_3$ on $[0,1]\times A_2$ by $$ \eta_3(t,u)=u+t\varphi_{\varepsilon},\quad\text{for all }u\in A_2,t\in[0,1]. $$ Then, $\eta_3$ is continuous. We claim that $\eta_3(t,u)\in J^{-M}$, for $u\in A_2$ and $t\in [0,1]$. Indeed, for $v=u+t\varphi_{\varepsilon}$, and $0\leq t \leq 1$, using \eqref{estgradphi}, we obtain $$ \|v-u\|=t\|\varphi_{\varepsilon}\|\leq\|\varphi_{\varepsilon}\|\leq\varepsilon_1. $$ Then, it follows from \eqref{contstep3} that $$ |J(v)-J(u)|<\frac{M}{2}. $$ Since $J(u)\leq -\frac{3}{2}M$, we obtain $$ J(v)< \frac{M}{2} + J(u) < \frac{M}{2}-\frac{3M}{2}=-M. $$ Thus $\eta_3:[0,1]\times A_2\to J^{-M}$ is a continuous map. Put $A_3=\eta_3(1,A_2)$. Therefore, $A_3$ is a compact subset of the level set $J^{-M}$ whose elements have nonzero positive part. This concludes the proof of Step $3$. \smallskip \noindent\textbf{Step 4:} In this step, our goal is to deform the set $A_3$ into a set of functions $u$, for which $J(u^{+})<0$. For each element $u\in A_3$, we have \begin{equation} J(tu^{+})=\frac{t^{2}}{2}\|u^{+}\|^{2}-\int_{\Omega}G(x,tu^{+})dx. \label{funcpos} \end{equation} Noting that $A_3$ is compact, we set \begin{equation} M_1=\max_{\xi\in A_3} \|\xi^{+}\|^{2}. \label{mymax} \end{equation} Similarly, $\int_{\Omega}G(x,u^{+})dx$ attains a minimum $m_1$ in $A_3$ given by \begin{equation} m_1=\inf_{\xi^{+}\in A_3}\int_{\Omega}G(x,\xi^{+})dx. \label{mymin} \end{equation} It follows from \eqref{mymax}, \eqref{mymin} and \eqref{funcpos} that \begin{equation} J(tu^{+})\leq t^{2}[\frac{M_1}{2}- \frac{m_1}{t^{2}}],\quad\text{for }u\in A_3,\text{ and }t>0. \label{funcpos3} \end{equation} Next, choose $T_1>0$ such that \begin{equation} \beta = \frac{m_1}{T_1^{2}} - \frac{M_1}{2} > 0. \label{step44stars} \end{equation} Then, by virtue of \eqref{funcpos3} and \eqref{step44stars}, $$ J(tu^{+})\leq -\beta t^{2},\quad\text{for }u\in A_3, \text{ and }t\geq T_1. $$ Since we want $J(tu^{+})\leq -M$, we can choose $t$ such that $$ t\geq \big(\frac{M}{\beta}\big)^{1/2}. $$ Put \begin{equation} T_4=\max\big\{T_1,\big(\frac{M}{\beta}\big)^{1/2}\big\}. \label{myt4} \end{equation} Now, define a map $\eta_4$ on $[0,1]\times A_3$ by \begin{equation} \eta_4(t,u)=\left[(1-t)+tT_4\right]u, \quad\text{for }t\in[0,1],\text{ and }u\in A_3. \nonumber \end{equation} Then, $\eta_4$ is continuous, and $\eta_4(t,u)\in J^{-M}$ by \eqref{notestep1}. Thus, $\eta_4$ defines a continuous map from $[0,1]\times A_3$ to $J^{-M}$. Put $A_4=\eta_4(1,A_3)$. Then, $A_4$ is a compact set. Also, $A_4\subset J^{-M}$, and $J(u^{+})\leq -M$, for all $u\in A_4$. This concludes the proof of Step $4$. \smallskip \noindent\textbf{Step 5:} In this step, the goal is to deform the set $A_4$ to a set of functions, $u$, in $J^{-M}$ such that $J(u^{+})$ is negatively large enough. First, notice that, for $0\leq s \leq 1$, \begin{align*} J(-su^{-}) &=\frac{s^{2}}{2}\int_{\Omega}[|\nabla u^{-}|^{2}-G(x,-su^{-})]dx \\ &\leq\frac{1}{2}\max_{\xi\in A_4,0\leq s\leq 1} \big|\int_{\Omega}[|\nabla \xi^{-}|^{2}-G(x,-s\xi^{-})]dx\big|. \end{align*} Set \[ C_{11}=\frac{1}{2}\max_{\xi\in A_4,0\leq s\leq 1} \big|\int_{\Omega} [|\nabla \xi^{-}|^{2}-G(x,-s\xi^{-})]dx\big|. \] Then, \begin{equation} J(-su^{-})\leq C_{11},\quad\text{for }s\in[0,1],\text{ and }u\in A_4. \label{jsminus} \end{equation} This estimate will also be used in Step $6$. Next, using the estimate \eqref{superq}, we obtain \[ J(tu^{+}) \leq \frac{t^{2}}{2}\|u^{+}\|^{2} -\frac{C_7t^{\mu}}{2}\|u^{+}\|_{L^{\mu}}^{\mu} - C_{8}|\Omega|. \] So that $$ J(tu^{+})\to -\infty,\quad\text{as }t\to\infty, $$ since $\mu > 2$. Thus, we can choose $T_5$ large enough such that \[ J(T_5u^{+})\leq - M - C_{11},\quad \text{for all }u\in A_4. \] Define a map $\eta_5$ on $[0,1]\times A_4$ by \[ \eta_5(t,u)=\left[(1-t)+tT_5\right]u^{+}-u^{-}, \quad\text{for }u\in A_4,\text{ and } t\in[0,1]. \] Then, $\eta_5$ is continuous and $\eta_5(t,u)\in J^{-M}$ for all $t\in [0,1]$ and $u\in A_4$ by \eqref{notestep1}. Thus, $\eta_5$ defines a map from $[0,1]\times A_4$ to $J^{-M}$. Put $A_5=\eta_5(1,A_4)$. Thus, $A_5$ is a compact set and \begin{equation} A_5\subset J^{-M}\cap\{u\in A_4|J(u^{+})\leq -M-C_{11}\}, \label{24starr} \end{equation} and $\eta_5$ defines a deformation from $A_4$ to $A_5$. This concludes the proof of Step $5$. \smallskip \noindent\textbf{Step 6:} In this step, we will deform the set $A_5$ obtained in Step $5$ to a subset of nonnegative functions in $J^{-M}$. For each element $u\in A_5$, we have \[ J(u^{+}-su^{-}) = J(u^{+}) + J(-su^{-})\quad\text{for all }u\in A_5. \] So that, using \eqref{jsminus} and \eqref{24starr}, we obtain \begin{equation} J(u^{+}-su^{-})\leq - M,\quad \text{for all }s\in[0,1],\text{ and }u\in A_5. \label{6stepstar} \end{equation} Define a map $\eta_6$ on $[0,1]\times A_5$ by $$ \eta_6(t,u)=u^{+}-(1-t)u^{-}. $$ Then $\eta_6$ is continuous and $\eta_6(t,u)\in J^{-M}$ for all $t\in[0,1]$ and $u\in A_5$, by virtue of \eqref{6stepstar}. Thus, $\eta_6$ defines a map from $[0,1]\times A_5$ to $J^{-M}$. Put $A_6=\eta_6(1,A_5)$. Then, $A_6$ is a compact set and its elements are nonnegative functions. This concludes the proof of Step $6$. \smallskip \noindent\textbf{Step 7:} Define $B_1^{+}=\{u\in H|\|u\|=1\text{ and }u\geq 0\}$. We saw in Step $6$ that $A_6$ is compact and $$ A_6\subset J^{-M}\cap\{u\in H:u\geq 0\}. $$ We show that $J^{-M}\cap\{u\in H:u\geq 0\}$ and $B_1^{+}$ are homotopic. Observe that for every $u\in H\backslash\{0\}$ such that $u\geq 0$ on $\Omega$, there exists a unique $t^{*}(u) > 0$ such that \begin{equation} J(t^{*}(u)u)=-M. \label{step7eqm} \end{equation} Furthermore, the map $u\mapsto t^{*}(u)$ is continuous for $u\in \{u\in H:u\geq 0\}$. In fact, it follows from \eqref{superq} that $$ J(tu)\leq\frac{t^{2}}{2}\|u\|^{2}+ C_{8}|\Omega|-C_7t^{\mu}\int_{\Omega}|u|^{\mu}dx. $$ So that, since $\mu > 2$, we have that \begin{equation} \lim_{t\to\infty}J(tu)=-\infty, \quad\text{for }u\in H\backslash\{0\},\;u\geq 0\text{ in }\Omega. \label{step7eqm1} \end{equation} Also, $J(0)=0$. It then follows by the intermediate value theorem and \eqref{step7eqm1} that, for each $u\in H\backslash\{0\}$ with $u\geq 0$, there exists $t^{*}>0$ such that, $$ J(t^{*}u)=-M. $$ So that, using \eqref{djtu2}, $$ \frac{d}{dt}J(tu)\big|_{t=t^{*}}\leq -\frac{1}{t^{*}}M < 0. $$ Hence, by the implicit function theorem, $t^{*}$ is unique and is a continuous function of $u$ for $u\in H\backslash\{0\}$, $u\geq 0$ in $\Omega$, which proves \eqref{step7eqm}. Furthermore, $J(tu)\leq -M$ for all $t\geq t^{*}(u)$. Next, set \begin{equation} B=\{tv : v\in B_1^{+}\text{ and }t\geq t^{*}(v)\}. \label{step71eq} \end{equation} We show that \begin{equation} B=J^{-M}\cap \{u\in H:u\geq 0\text{ in }\Omega\}. \label{step7eqm12} \end{equation} To see why \eqref{step7eqm12} is true, take $u\in H$ with $u\geq 0$ in $\Omega$, and $J(u)\leq -M$; so that $$ u=\|u\|u_1,\quad\text{where }u_1=\frac{1}{\|u\|}u\in B_1,$$ and $\|u\|\geq t^{*}(u_1)$, since $J(\|u\|u_1)\leq -M$. Hence, \begin{equation} J^{-M}\cap\{u\in H:u\geq 0\text{ in }\Omega\}\subseteq B. \label{secondincl1} \end{equation} Next, let $u\in B$. Then, there exists $v\in B_1^{+}$ and $t\geq t^{*}(v)$ such that $u=tv$, where $v\in B_1^{+}$ and $t\geq t^{*}(v)$. Then, by the definition of $t^{*}(v)$ it follows that $$ J(tv)\leq -M, $$ which shows that $u\in J^{-M}$. Hence, $u\in J^{-M}\cap\{u\in H:u\geq 0\text{ in }\Omega\}$. Thus \begin{equation} B\subseteq J^{-M}\cap\{u\in H:u\geq 0,\text{ in }\Omega\}. \label{secondincl2} \end{equation} The inclusion \eqref{secondincl1} and \eqref{secondincl2} establish \eqref{step7eqm12}. Next, we show that $B$ and $B_1^{+}$ are homotopic. This will imply that $$ J^{-M}\cap\{u\in H:u\geq 0\text{ in }\Omega\}\cong B_1^{+}. $$ Define $f:B\to B_1^{+}$ as follows: For each $u\in B$, $u\in J^{-M}$ and $u\geq 0$ in $\Omega$, so that $u\ne 0$; thus, we can define $$ f(u)=\frac{1}{\|u\|}u,\quad\text{for all }u\in B. $$ Define $g:B_1^{+}\to B$ by $g(u)=t^{*}(u)u$ for all $u\in B_1^{+}$. Then, $$ f\circ g(u)=\frac{1}{t^{*}(u)\|u\|}(t^{*}(u)u)=u, $$ for all $u\in B_1^{+}$. So, $f\circ g=id_{B_1^{+}}$. On the other hand, for $u\in B$, \begin{equation} g\circ f(u)=t^{*}\big(\frac{u}{\|u\|}\big)\frac{u}{\|u\|}. \label{gcircf1} \end{equation} We claim that \begin{equation} t^{*}(u)=\frac{1}{\|u\|} t^{*}\big(\frac{u}{\|u\|}\big). \label{claimstep7} \end{equation} By the definition of $t^{*}$, we have $J(t^{*}(u)u)=-M$. Similarly, $$ J\Big(t^{*}\big(\frac{u}{\|u\|}\big)\frac{u}{\|u\|}\Big)=-M. $$ Then, by the uniqueness of $t^{*}$ we obtain \eqref{claimstep7}. Therefore, we can rewrite \eqref{gcircf1} as \begin{equation} g\circ f(u)=t^{*}(u)u. \label{gcircf2} \end{equation} Next, we build a homotopy from $g$ to $id_{B}$ by $H:[0,1]\times B\to B$ given by \begin{equation} H(s,u)=[st^{*}(u)+(1-s)]u. \label{hstep7} \end{equation} Then, $H(0,u)=u$ and $H(1,u)=t^{*}(u)u=g\circ f(u)$. Note that $$ t^{*}(u)\leq st^{*}(u)+(1-s)\leq 1,\quad\text{for all }s\in[0,1], $$ since $t^{*}(u)\leq 1$ for $u\in B$, by virtue of \eqref{step7eqm12} Hence, $J(H(s,u))\leq -M$ for all $s\in [0,1]$. It follows that $B$ and $B_1^{+}$ are homotopic. Therefore, since $$ B=J^{-M}\cap\{u\in H:u\geq 0\}, $$ we obtain that $J^{-M}\cap\{u\in H:u\geq 0\}$ and $B_1^{+}$ are homotopic. \smallskip \noindent\textbf{Step 8:} In this step, we show that $B_1^{+}$ is contractible. Let $u_0$ be any element in $B_1^{+}$, so that $\|u_0\|=1$ and $u_0\geq 0$. Define $H:[0,1]\times B_1^{+}\to B_1^{+}$ by $$ H(t,u)=\frac{tu_0+(1-t)u}{\|tu_0+(1-t)u\|},\quad t\in[0,1],u\in B_1^{+}. $$ Note that, for any $t\in[0,1]$ and $u\in B_1^{+}$, $tu_0+(1-t)u\geq 0$. Furthermore, $$ \|tu_0+(1-t)u\|\ne 0,\quad\text{for all }t\in [0,1],\text{ and }u\in B_1^{+}. $$ Otherwise there would exist $t_1\in [0,1]$ and $u_1\in B_1^{+}$ such that $$ \|t_1u_0+(1-t_1)u_1\|=0. $$ Then, $t_1u_0+(1-t_1)u_1=0$. So that $$ t_1u_0=-(1-t_1)u_1, $$ where $t_1u_0\geq 0$ and $u_1\geq 0$, so that $t_1u_0\leq 0$. Thus, $t_1=0$, so that $u_1=0$, which is impossible. Thus, $H:[0,1]\times B_1^{+}\to B_1^{+}$ defines a homotopy with $H(0,u)=u$, that is, $H(0,.)=id_{B_1^{+}}$, and $H(1,u)=u_0$ for all $u\in B_1^{+}$. \smallskip \noindent\textbf{Step 9:} By Step 6 we have that $$ A_6\subset J^{-M}\cap \{u\in H|u\geq 0\}. $$ In Step 7, we showed that $J^{-M}\cap\{u\in H|u\geq 0\}$ is homotopic to $B_1^{+}$. In Step 8, we showed that $B_1^{+}$ is contractible in $J^{-M}$. Therefore, $A_6$ is contractible in $J^{-M}$. This concludes the proof of Proposition \ref{cinftycomp}. \end{proof} The previous proposition implies that $\widetilde{H}_{q}(J^{-M})=0$, for all $q\in\Z$. Then, it follows from the exactness of the homology sequence of the pair $(H,J^{-M})$, \begin{align*} \dots\stackrel{i_{*}}{\to}\widetilde{H}_{q+1}(H) &\cong\{0\} \stackrel{j_{*}}{\to}H_{q+1}(H,J^{-M}) \stackrel{\partial_{*}}{\to}\widetilde{H}_{q}(J^{-M})\\ &\cong\{0\} \stackrel{i_{*}}{\to}\widetilde{H}_{q}(H)\cong\{0\} \stackrel{\partial_{*}}{\to}\dots, \end{align*} that $\partial_{*}$ is an isomorphism. Therefore, \begin{equation} C_{q}(J,\infty)=H_{q}(H,J^{-M})=0, \label{cginfty1} \end{equation} for all $q\in\Z$. Now we present the proof of the main result. \begin{proof}[Proof of Theorem \ref{fourthmaintheo111}] Let $u_0$ be as given in Theorem \ref{secondmaintheo} and $u_1$ as given in Theorem \eqref{thirdmaintheo}. Assume by way of contradiction that $0,u_0$, and $u_1$ are the only critical points of $J$. Then, $\mathcal{K}=\{0,u_0,u_1\}$. Using the Morse relation \eqref{Su26} with $t=-1$, we obtain \begin{equation} \sum_{q=0}^{\infty}M_{q}(-1)^{q}=\sum_{q=0}^{\infty}\beta_{q}(-1)^{q}, \label{leftside} \end{equation} where $M_{q}$ are the Morse type numbers defined in \eqref{Su23} and $\beta_{q}=\dim C_{q}(J,\infty)$ are the Betti numbers for $q=0,1,2,\dots$. First, the left side of \eqref{leftside} is given by $$ \sum_{q=0}^{\infty}M_{q}(-1)^{q} = M_0-M_1+(-1)^{d}M_{d}, $$ where \begin{equation} \begin{gathered} M_{d}=\dim C_{d}(J,0)=1,\quad M_0=\dim C_0(J,u_0)=1,\\ M_1=\dim C_1(J,u_1)=1, \end{gathered}\label{morsenumbers1} \end{equation} where we have used \eqref{prob1cgrouporign1}, \eqref{cgroupu0}, and \eqref{cgroupminimizer}, respectively. The Betti numbers are given by $\beta_{q}=\dim C_{q}(J,\infty)$, where the critical groups at infinity were computed in \eqref{cginfty1}, so that \[ C_{q}(J,\infty)=0,\quad\text{for }q=0,1,2,\dots . \] Then, $\beta_{q}=0$ for all $q=0,1,2,\dots$. Hence, substituting \eqref{morsenumbers1} in \eqref{leftside}, we obtain $$ (-1)^{d}=0, $$ which is a contradiction. Therefore, $J$ must have at least four critical points; that is, problem \eqref{e1} must have at least three nontrivial weak solutions. \end{proof} \subsection*{Acknowledgements} The authors would like to thank the anonymous referees for their careful and close reading of the original manuscript. In particular, the authors are very appreciative of the suggested corrections to several misprints in the manuscript. \begin{thebibliography}{99} \bibitem{Ag} Agmon, S.; The $L^{p}$ approach to the Dirichlet problem, \emph{Ann. Scuola Norm. Sup. Pisa}, 13, 405-448, \textbf{1959}. \bibitem{AR} Ambrosetti, Antonio; Rabinowitz, P. 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