\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 112, pp. 1--9.\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{8mm}} \begin{document} \title[\hfilneg EJDE-2014/112\hfil Pairs of sign-changing solutions] {Pairs of sign-changing solutions for sublinear elliptic equations with Neumann boundary conditions} \author[C. Li, Q. Zhang, F. Chen \hfil EJDE-2014/112\hfilneg] {Chengyue Li, Qi Zhang, Fenfen Chen} % in alphabetical order \address{Chengyue Li \newline Department of Mathematics, Minzu University of China, Beijing 100081, China} \email{cunlcy@163.com} \address{Qi Zhang \newline Department of Mathematics, Minzu University of China, Beijing 100081, China} \email{709341427@qq.com} \address{Fenfen Chen \newline Department of Mathematics, Minzu University of China, Beijing 100081, China} \email{chenfenfen359@163.com} \thanks{Submitted November 5, 2013. Published April 16, 2014.} \subjclass[2000]{58E05, 34C37, 70H05} \keywords{Elliptic equation; sublinear potential; Neumann problem; \hfill\break\indent Clark Theorem; critical point} \begin{abstract} We consider the Neumann problem for a sublinear elliptic equation in a convex bounded domain of $\mathbb{R}^{N}$. Using an variant of Clark Theorem, we obtain the existence and multiplicity of its pairs of sign-changing solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Consider the Neumann problem for a semilinear elliptic equation \begin{equation} \begin{gathered} -\Delta u(x )=f(u(x ) ),\quad x\in \Omega, \\ \frac{\partial u}{\partial n}| _{\partial \Omega}=0,\label{eNP} \end{gathered} \end{equation} where $\Omega \subset \mathbb{R}^{N}$ $(N\geqslant 1 )$ is a convex and bounded domain with the smooth boundary $\partial \Omega $ and the outward normal $n,f(u ):\mathbb{R}\to \mathbb{R}$. Let $F(u )=\int_0^{u}f(s )ds$, the primitive of $f$, and assume it satisfies \begin{equation} \limsup _{|u|\to \infty }F(u)/|u|^2\leqslant a< \infty ,\label{e1} \end{equation}\\ then we say that \eqref{eNP} is sublinear (or subquadratic). If \begin{equation} \lim_{|u|\to \infty }F(u )/| u|^2=\infty,\label{e2} \end{equation} then \eqref{eNP} is superlinear (or superquadratic). For sublinear problem \eqref{eNP}, there is a vast of literature. Under the assumptions of sign conditions \cite{G,IN1}, or monotonicity conditions \cite{M}, or periodicity conditions\cite{R1}, or Landesman-Lazer type conditions \cite{IN2,K}, it has been showed that problem \eqref{eNP} possesses at least one solution. Tang \cite{T,TW1,TW2} supposed that $F$ satisfies the hypothesis \begin{equation} \lim_{u\in X_0,\| u \|_0\to \infty }\| u \|_0^{-2\alpha }\int _{\Omega } F(u(x ) )dx\to \infty , \label{e3} \end{equation} where $X_0= \{ u\in H^{1}(\Omega ):\int _{\Omega }u(x )dx=0 \}$, $\| u \|_0=(\int _{\Omega }| \nabla u(x )| ^2dx)^{1/2}$ for $u\in X_0$, and $0< \alpha < 1$. He proved the existence and multiplicity results of problem \eqref{eNP} by minimax methods. Costa \cite{C1} assumed that $f$ satisfies \begin{itemize} \item[(F1)] $f\in \mathbb{C}^{1}(\mathbb{R},\mathbb{R} )$, strictly increasing and $f(0 )=0$, \item[(F2')] the limits $f'(\pm \infty )=\lim_{u\to \pm \infty }f'(u )$ exist, $0 0 $ and $\rho > 0 $ such that $d_{\Omega }> \frac{2p \pi }{\sqrt{M}},M>4p^2f'(\pm \infty )$, and \[ F(u )\geqslant\frac{1}{2}M| u|^2,\forall | u|\leqslant\rho; \] \item[(F4)] for $\Omega$ there are continuous functions $e_1(x ),e_2(x ),\dots ,e_{p}(x )\in X_0\setminus \{ 0\}$, which are orthogonal in $H^{1}(\Omega )$ and $L^2(\Omega )$, such that \[ \int _{\Omega }|\nabla e _j(x )|^2dx\leqslant \frac{2(j+1)\pi )}{d_{\Omega }} ]^2\int _{\Omega }|e_j(x )|^2dx,\quad \forall 1\leqslant j\leqslant p. \] \end{itemize} Then \eqref{eNP} has $p$ distinct pairs $(u(x ),-u(x ) )$ of sign-changing classical solutions, and has no positive and negative solution, provided that $d_{\Omega }\in(\frac{2p\pi }{\sqrt{M}},\frac{\pi }{\sqrt{f'(\pm \infty )}} )$. \end{theorem} \begin{remark} \label{rmk1} \rm If $f$ satisfies $\lim_{| u|\to 0}F(u )/| u|^2=\infty $, then, for all $d_{\Omega } >0,p\geqslant 1$, we can find $M> 0$ and $\rho > 0$ such that (F4) holds. \end{remark} \begin{remark} \label{rmk2} \rm Usually, in some applications, the role of $e_1(x ),e_2(x ),\dots ,e_{p}(x )$ is played by the eigenfunctions $\phi _j\in X_0$ $(j\geqslant 1 )$ of problem \eqref{e4} (see \eqref{e16} and \eqref{e17} below). \end{remark} This article is organized as follows. In Section 2, we give some Lemmas. In Section 3, we prove Theorem \ref{thm1} by using a variant of Clark Theorem as stated next. \begin{theorem}[\cite{R2,S}] \label{thm2} Let $\hat{X}$ be a Banach space, $\hat{\varphi }\in C^{1}(\hat{X},\mathbb{R} )$ be even, and $\hat{\mathcal{M}}\subset \hat{X}$ be $C^{1}-$ submanifold. Suppose that $\hat{\varphi}|_{\hat{\mathcal{M}}}$ satisfies the Palais-Smale condition; $\hat{\varphi }$ is bounded from below on $\hat{\mathcal{M}}$; there exist a closed, symmetric subset $\hat{K}\subset \hat{\mathcal{M}}$ and $\hat{p}\in \mathbb{N}$ such that $\hat{K}$ is homeomorphism to $S^{\hat{p}-1}\subset \mathbb{R}^{\hat{p}}$ by an odd map, and $sup\{ \hat{\varphi }(x ): x\in \hat{K}\}<\hat{\varphi } (0 )$. Then $\hat{\varphi}|_{\hat{\mathcal{M}}}$ possesses at least $\hat{p}$ distinct pairs $(u,-u )$ of critical point with corresponding critical values less than $\hat{\varphi } (0 )$. \end{theorem} As an example, we apply Theorem \ref{thm1} to $B_{\gamma }(0 )\subset \mathbb{R}^2$ and yield an interesting result. \begin{theorem} \label{thm3} Let $f$ satisfy {\rm (F1)--(F4)} with $\Omega =\{ (x_1,x_2 )\in \mathbb{R}^2:x_1^2+x_2^2< r^2 \}$, then, for $r\in(\frac{2p\pi }{\sqrt{M}},\frac{\pi }{2\sqrt{f'(\pm \infty )}})$, the problem \begin{equation} \begin{gathered} -\Delta u(x )=f(u(x ) ),\quad x_1^2+x_2^2< r^2 \\ \frac{\partial u}{\partial n}| _{x_1^2+x_2^2= r^2}=0, \end{gathered} \label{eNP1} \end{equation} has $p$-distinct pairs $(u(x ),-u(x) )$ of sign-changing classical solutions. \end{theorem} In the proof of Theorem \ref{thm3}, by some accurate analysis about the corresponding eigenvalue problem with zero points of Bessel functions, we find that condition (F5) is naturally satisfied. \section{Preliminaries} In this article, for simplicity, for $u\in L^2(\Omega )$, we denote by $\| u \|_{L^2}$ its ${L^2}$-norm. Clearly, problem \eqref{eNP} has the trivial solution $u(x )=0$. In order to find its nontrivial solutions, we consider the functional \begin{equation} \begin{aligned} \varphi _{\Omega }(u ) &=\frac{1}{2}\int _{\Omega }| \nabla u(x )|^2dx -\int _{\Omega }F(u(x ) )dx\\ &=\frac{1}{2}\int _{\Omega }|\nabla u(x )|^2dx -\psi _{\Omega}(u ),u\in X, \end{aligned} \label{e7} \end{equation} where $\psi _{\Omega }(u )=\int _{\Omega }F(u(x ) )dx$, $X=H^{1}(\Omega )$ is the usual Sobolev space with the inner product \begin{equation} (u,w )=\int _{\Omega }[ \dot{u}(x )\dot{w} (x )+u(x )w(x )]dx \label{e8} \end{equation} and the corresponding norm \begin{equation} \| u \|=(u,u )^{1/2}=(\int _{\Omega }[ | \nabla u(x )|^2 +| u(x )|^2]dx )^{1/2}. \label{e9} \end{equation} Under the assumptions (F1) and (F2), we know that $\varphi _{\Omega }\in \mathbb{C}^2(X,\mathbb{R} )$, $\psi _{\Omega }(u )$ is weakly continuous in $X$, and $\psi _{\Omega }'(u ):X\to X^{*}$ is completely continuous. Moreover, critical points of $\varphi _{\Omega }$ in $X$ are classical solutions of problem \eqref{eNP}. Next, we decompose the Sobolev space $X=H^{1}(\Omega )$ as \begin{equation} X=X_0\oplus X_1,\quad X_0=\{ u\in X:\int _{\Omega }u(x ) dx=0\},\quad X_1=\mathbb{R}. \label{e10} \end{equation} Let us recall the problem \eqref{e4} has eigenvalues \begin{equation} 0=\lambda _0< \lambda _1< \lambda _2\leqslant\lambda _{3}\leqslant \dots \to \infty , \label{e11} \end{equation} and the corresponding eigenfunctions \begin{equation} \phi _0 (x )\equiv 1, \phi _1 (x ), \phi _2 (x ), \phi _{3} (x ), \quad \dots . \label{e12} \end{equation} In particular, for the first positive eigenvalue $ \lambda _1$, one has the Poincare type inequality \begin{equation} \int _{\Omega }| u(x )|^2dx\leqslant \frac{1}{\lambda _1}\int _{\Omega } | \nabla u(x )|^2dx, \forall u\in X_0. \label{e13} \end{equation} Using the estimate of lower bound for $\lambda _1$, \cite{L}, \begin{equation} \lambda _1\geqslant (\frac{\pi }{d_{\Omega }} )^2, \label{e14} \end{equation} we have \begin{equation} \int _{\Omega }| u(x )|^2dx\leqslant(\frac{d_{\Omega }}{\pi } )^2 \int_{\Omega } | \nabla u(x )|^2dx,\forall u\in X_0. \label{e15} \end{equation} In addition, it is a well-known that \begin{equation} \int _{\Omega }\phi _j(x )dx=0,\quad \forall j\geqslant 1,\label{e16} \end{equation} which implies \begin{equation} \phi _j\in X_0,\quad \forall j\geqslant 1.\label{e17} \end{equation} Under assumptions of (F1) and (F2), Costa \cite{C1} proved that \begin{itemize} \item[(i)] $\mathcal{M}=\{ u\in X=H^{1}(\Omega ):\int _{\Omega }f(u(x ) )dx=0 \}\subset X$ is a $\mathbb{C}^{1}-$ manifold of codimension 1; \item[(ii)] $u\in X$ is a critical point of $\varphi _{\Omega }$ in $X$ if and only if $u\in \mathcal{M} $ and it is a critical point of $\varphi_{\Omega}|_{\mathcal{M}}$. \end{itemize} Also for $u\in X$, writing $u=\nu +c$, $\nu\in X_0$, $c\in \mathbb{R}$, he also obtained that \begin{itemize} \item[(iii)] $\int _{\Omega }F(\nu +c )dx\leqslant \int _{\Omega }F(\nu )dx$; \item[(iv)] $\| \nu _{n} \|\to \infty$ as $\| \nu _{n} + c _{n} \|\to \infty, \nu _{n} + c _{n}\in \mathcal{M}$. \end{itemize} \begin{lemma} \label{lem1} If $f$ satisfies {\rm (F1)} and {\rm (F2)}, then \begin{itemize} \item[(v)] for each $\nu \in X_0$, there exists a unique $c(\nu )\in \mathbb{R}$ such that $\nu +c(\nu )\in \mathcal{M}$; \item[(vi)] $c(-\nu )=-c(\nu )$ for all $\nu \in X_0$ if $f$ is also odd. \end{itemize} \end{lemma} \begin{proof} (v) For any fixed $ \nu \in X_0$, define \begin{equation} g_{\nu }(c )=\int _{\Omega }f(\nu +c )dx,\quad \forall c\in \mathbb{R}.\label{e18} \end{equation} If $\nu \in \mathbb{C}^{1}(\bar{\Omega } )$, we easily know that $f(\nu (x ) +c_1)> 0$ for all $x\in \bar{\Omega }$ and $c_1> \max_{\bar{\Omega}} | \nu (x )|$, while $f(\nu (x ) +c_2)< 0$ for all $ x\in \bar{\Omega}$ and $c_2<- \max_{\bar{\Omega }} | \nu (x )|$. Therefore, by the continuity of $g_{\nu }( \cdot )$, there exists $c=c(\nu )\in \mathbb{R}$ such that $\int _{\Omega }f(\nu +c(\nu ) )dx=0$. For the general case $\nu \in X_0$, one can take $\nu _{k}\in \mathbb{C}^{1}(\bar{\Omega } )\cap X_0$, $\nu _{k}\to \nu $ in $X$. There are $c(\nu _{k} )\in \mathbb{R} $ such that $\int _{\Omega }f(\nu _{k}+c(\nu _{k} ) )dx=0$. We claim that $ \{ c(\nu _{k} ) \}$ is bounded. Otherwise, $| c(\nu _{k} )|\to \infty$, then $\| \nu _{k}+c(\nu_{k} ) \|\to \infty$. Since $\nu _{k}+c(\nu _{k})\in \mathcal{M}$, by $(iv)$, we have $\| \nu_{n} \|\to \infty $, a contraction. Therefore, we may assume that $c(\nu _{k} )\to c(\nu)\in \mathbb{R}$. By (F2), there are constants $\eta >0$ such that $0\leqslant f'(u )\leqslant \eta$, for all $u\in \mathbb{R}$. Thus, we have \begin{equation} \begin{aligned} | \int _{\Omega }f(\nu +c(\nu ) )dx|&=| \int _{\Omega }f(\nu +c(\nu ) )dx-\int _{\Omega } f(\nu_{k} +c( \nu_{k} ) )dx|\\ &\leqslant \eta\int_{\Omega }[ | \nu-\nu _{k}| +| c( \nu _{k} )-c(\nu )|]dx\to 0; \end{aligned}\label{e19} \end{equation} that is, $\nu +c(\nu )\in \mathcal{M}$. The uniqueness of $c(\nu )$ can be obtained from the monotonicity of $f(u)$. (vi) for all $\nu \in X_0$ naturally $-\nu \in X_0$, by (v), there is $c(-\nu )\in \mathbb{R}$ such that \begin{equation} \int _{\Omega }f(-\nu +c(-\nu ) )dx=0.\label{e20} \end{equation} Since $f(u )$ is odd, we have \begin{align} \int _{\Omega }f(\nu -c(-\nu ) )dx=0.\label{e21} \end{align} By the uniqueness of $c(\nu )$, we obtain $c(\nu )=-c(-\nu )$, namely, $c(-\nu )=-c(\nu )$. \end{proof} \begin{lemma} \label{lem2} Suppose $f$ satisfies {\rm (F1)} and {\rm (F2)}. Then the functional $\varphi _{\Omega }(u )$ is bounded from below on $\mathcal{M}$ and satisfies the Palais-Smale condition on $\mathcal{M}$. \end{lemma} \begin{proof} By (F2), there exist $m$ and $b$, \begin{align} 00,\label{e22} \end{align} such that \begin{equation} F(s)\leqslant b+\frac{1}{2}m| s|^2,\quad \forall s\in \mathbb{R}.\label{e23} \end{equation} For $u \in \mathcal{M}$, writing $u=\nu +c\in X_0\oplus X_1$, we have \begin{equation} \begin{aligned} \varphi _{\Omega }(u ) &=\frac{1}{2}\int _{\Omega}| \nabla \nu (x )|^2dx-\int_{\Omega }F(\nu +c )dx\\ &\geqslant \frac{1}{2}\int _{\Omega }| \nabla \nu (x )|^2dx -\int _{\Omega }F(\nu )dx \\&\geqslant \frac{1}{2}\| \nabla \nu \|^2_{L^2} -\frac{1}{2}m\| \nu \|^2_{L^2}-b| \Omega|\,. \end{aligned} \label{e24} \end{equation} This inequality and \eqref{e15} implies \begin{equation} \varphi _{\Omega }(u )\geqslant \frac{1}{2}[ 1-m(\frac{d_{\Omega }} {\pi } )^2 ]\| \nabla\nu \|^2_{L^2}-b| \Omega|=\frac{1}{2}D\| \nabla\nu \|^2_{L^2}-b| \Omega|\geqslant -b| \Omega| \label{e25} \end{equation} with $D=1-m(\frac{d_{\Omega }}{\pi } )^2>0$. Thus, $\varphi _{\Omega }(u )$ is bounded from below on $\mathcal{M}$. Let $\{ u_j \}\subset \mathcal{M}$ be such that $\{ \varphi _{\Omega } (u_j )\}$ is bounded and $(\varphi _{\Omega }|_{\mathcal{M}} )'(u_j )\to 0$. Let $u_j=\nu _j +c_j \in X_0\bigoplus X_1$. Then \eqref{e25} implies $\| \nabla \nu_j \|^2_{L^2}\leqslant \frac{2}{D}(\varphi _{\Omega } (u_j ) +b| \Omega|)$, thus $\{ \nu _j \}$ is bounded in $X$. The fact $u_j=\nu _j+c_j\in \mathcal{M}$ and (iv) derives $\{ u _j \}$ is also bounded in $X$, so we may assume that, by passing to a subsequence if necessary, \begin{gather} u_j\rightharpoonup u \in X\quad \text{weakly in }X.\label{e26}\\ u_j\to u \in X\quad \text{strongly in $L^{1}(\Omega )$ and in }L^2(\Omega ).\label{e27} \end{gather} Thus, for $j\geqslant 1$, noticing \begin{equation} \int _{\Omega }f(u(x ) )dx =\int _{\Omega }[f(u(x ) ) -f(u_j(x ) ) ]dx =\int _{\Omega }f'(\zeta )(u(x )-u_j (x ))dx, \label{e28} \end{equation} with $\zeta$ between $u(x )$ and $u_j (x )$, it follows that \[ | \int _{\Omega }f(u(x ) )dx|\leqslant\int _{\Omega } | f'(\zeta )| | u(x )-u_j(x )|dx \leqslant \eta\int _{\Omega } | u(x )-u_j(x)|dx\to 0; \] consequently, $u\in \mathcal{M}$. Let us denote by $\nabla \varphi _{\Omega},\nabla J _{\Omega }:X\to X$ the gradient of $\varphi _{\Omega },J _{\Omega }$, respectively, which are defined by the Riesz-Frechet representation theorem, namely, $\nabla\varphi _{\Omega },\nabla J_{\Omega }\in X$ are unique elements such that \begin{equation} \varphi _{\Omega }'(w )h=\langle \nabla \varphi _{\Omega }(w ) ,h\rangle,J _{\Omega }'( w )h =\langle \nabla J _{\Omega }(w),h\rangle,\forall w,h\in X. \label{e29} \end{equation} Then, from the boundness of $\{ u_j \}$, we easily know that $\nabla \varphi _{\Omega }(u_j ),\nabla J _{\Omega }(u_j )$ are bounded. Moreover, \begin{equation} \begin{aligned} &[ (\varphi _{\Omega }|_{\mathcal{M}})'(u_j )-(\varphi _{\Omega }| _{\mathcal{M}})'(u )](u_j-u ) \\ &= (\varphi _{\Omega } '(u_j )-\varphi _{\Omega }'(u ))(u_j-u )-\frac{ \langle \nabla \varphi _{\Omega }(u_j ),\nabla J _{\Omega }(u_j ) \rangle}{\| \nabla J _{\Omega }(u_j ) \|^2}J _{\Omega }'(u_j )(u_j-u )\\ &\quad +\frac{\langle \nabla \varphi _{\Omega }(u ),\nabla J _{\Omega }(u ) \rangle}{\| \nabla J _{\Omega }(u ) \|^2}J _{\Omega } '(u )(u_j-u )\\ &=\| \nabla u_j-\nabla u \|_{L^2}^2-\int _{\Omega }(f(u_j )-f(u ) )(u_j-u )dx-C_jJ_{\Omega }'(u_j )(u_j-u )\\ &\quad +C_0J_{\Omega }'(u)(u_j-u ), \end{aligned} \label{e30} \end{equation} where $C_0=\frac{\langle \nabla \varphi _{\Omega }( u ),\nabla J _{\Omega }(u) \rangle}{\| \nabla J _{\Omega }(u) \|^2}$ is a constant, $C_j=\frac{\langle \nabla \varphi _{\Omega }(u_j ),\nabla J _{\Omega }(u_j ) \rangle}{\| \nabla J _{\Omega }(u_j ) \|^2}$ is bounded since $\nabla J_{\Omega }(u_j )\to \nabla J_{\Omega }(u )\neq 0$. So $\| \nabla u_j -\nabla u\|_{L^2}\to 0$. Thus, with the aid of \eqref{e27}, we conclude that $u_j\to u\in \mathcal{M}$ in $X$. \end{proof} \section{Proof of Theorem \ref{thm1}} For $p\in \mathbb{N}$, $\rho > 0$, and $e_1(x),e_2(x),\dots ,e_{p}(x )$ in (F4) and (F5), we define the subset $K\subset \mathcal{M}$ as follows \begin{equation} K=\{ \nu +c(\nu ) \in \mathcal{M}:\nu =\sum_{j=1}^{p}\mu _je_j(x ) ,\mu _j\in \mathbb{R} (1\leqslant j\leqslant p ), \sum_{j=1}^{p}\mu _j^2=\hat{\rho} ^2\}, \label{e31} \end{equation} where $\hat{\rho}=\rho /(2\sqrt{p})$. Then, by Lemma \ref{lem1}, the map \begin{equation} \nu +c(\nu )\mapsto \big(-\frac{\mu _1}{\hat{\rho} },-\frac{\mu _2}{\hat{\rho} } ,\dots ,-\frac{\mu _{p}}{\hat{\rho} } \big) \label{e32} \end{equation} is an odd homeomorphism from $K$ to $S^{p-1}\subset \mathbb{R}^{p}$. \begin{proof}[Proof of Theorem \ref{thm1}] We consider the subset $K\subset \mathcal{M}$ in (31). Without loss of generality, we may assume that $| e_j(x )| \leqslant 1$, for all $1\leqslant j\leqslant p, x\in \Omega$. Thus, for any $u(x )=\nu (x )+c(\nu )=\sum_{j=1}^{p}\mu _je_j(x )+c(\nu )\in K$, we have \begin{equation} | \nu (x )|^2\leqslant \sum_{j=1}^{p}\mu _j ^2\sum_{j=1}^{p}| e_j(x )|^2\leqslant p\hat{\rho }^2,\quad \forall x\in \Omega. \label{e33} \end{equation} From $\int _{\Omega }f(\nu(x ) +c(\nu) )dx=0$, we know that there exists $\hat{x}\in\Omega$ such that $f(\nu(\hat{x} ) +c(\nu ) )=0$. By $ (f_1 )$, we obtain $\nu(\hat{x} ) +c(\nu )=0$, namely, $c(\nu )=-\nu(\hat{x} ) $. Thus \begin{gather} | c(\nu )|=|\nu (\hat{x})|\leqslant \sqrt{p}\hat{\rho },\label{e34}\\ | u(x ) |\leqslant | \nu (x )| +|c(\nu )| \leqslant 2\sqrt{p}\hat{\rho}=\rho,\quad \forall x\in \Omega, \label{e35} \end{gather} so, combining \eqref{e35} with (F4)--(F5) shows that \begin{align*} \varphi _{\Omega }(u ) &=\frac{1}{2}\int _{\Omega }| \nabla u(x )|^2dx-\int _{\Omega }F(u(x ) )dx\\ &=\frac{1}{2} \int _{\Omega }| \nabla \nu (x )|^2dx-\int _{\Omega }F(\nu (x )+c(\nu ) )dx\\ &\leqslant \frac{1}{2}\int _{\Omega }| \nabla \nu (x )|^2dx-\frac{M}{2}\int _{\Omega }(\nu (x )+c(\nu ) )^2dx\\ &=\frac{1}{2}\sum_{j=1}^{p}\mu _j^2\| \nabla e_j (x ) \|_{L^2}^2-\frac{1}{2}M\sum_{j=1}^{p}\mu _j^2\| e_j (x )\|_{L^2}^2-Mc( \nu )\sum_{j=1}^{p}\mu _j\int _{\Omega }e_j(x )dx\\ &\quad -\frac{1}{2}Mc^2(\nu )| \Omega|\\ &\leqslant \frac{1}{2}\sum_{j=1}^{p}\mu _j^2\| \nabla e_j (x )\|_{L^ {2}}^2-\frac{1}{2}M\sum_{j=1}^{p}\mu _j^2\| e_j (x )\|_{L^2}^2\\ &\leqslant \frac{1}{2}\sum_{j=1}^{p}\mu _j^2[ (\frac{2(j+1 )\pi }{d_{\Omega }} )^2 -M ]\| e_j (x )\|_{L^2}^2< 0, \end{align*} %\label{e36} using $\int _{\Omega }e_j(x )dx=0$. Thus $sup\{ \varphi _{\Omega}(u ):u\in K \}< 0=\varphi _{\Omega } (0 ) $. Hence, by Lemma \ref{lem2} and Theorem \ref{thm2}, $\varphi _{\Omega }|_{\mathcal{M}}$ possesses at least $p$ distinct pairs $(u_j ,-u_j)$ of critical points on $\mathcal{M}$ such that $\varphi _{\Omega }(u_j)<0$ with $u_j\neq 0(1\leqslant j\leqslant p )$. Since $u_j\in \mathcal{M}\setminus \{ 0 \}$; that is, $\int _{\Omega }f(u_j(x ) )dx=0$, however, the continuous function $f(s )$ satisfies $f(s)>0$ if $s>0$, and $f(s )<0$ if $s<0$, thus, we conclude that $u_j$ must change its sign. In addition, from (ii), we also know that there is no positive and negative critical point of $\varphi _{\Omega }$. In other words, problem \eqref{eNP} possesses $p$ distinct pairs $(u_j(x ),- u_j(x))$ of sign-changing classical solutions $(1\leqslant j\leqslant p )$, and has no positive and negative solution. \end{proof} \begin{remark} \label{rmk3} \rm Costa \cite[Theorem 3.7]{C1}, by minimizing method, shows that there exists $u_0=u_0(x )\in \mathcal{M}\setminus \{ 0 \}$ such that \begin{equation} \varphi _{\Omega }(u_0 )=\underset{u\in \mathcal{M}}{inf}\varphi _ {\Omega }(u )< 0. \label{e37} \end{equation} In fact, by our previous arguments in Theorem \ref{thm1}, we know that $ u_0(x )$ is a sign-changing classical solution of \eqref{eNP}. \end{remark} As an application and illustration of Theorem \ref{thm1}, Theorem \ref{thm3} is applied to the Neumann problem: \begin{equation} \begin{gathered} -\Delta u(x )=f(u(x ) ),\quad x_1^2+x_2^2< r^2\\ \frac{\partial u}{\partial n}\mid _{x_1^2+x_2^2=r^2}=0, \end{gathered} \label{eNP2} \end{equation} where $x=(x_1,x_2 )\in \mathbb{R}^2$, $r>0$. To prove Theorem \ref{thm3}, we shall use some properties of Bessel functions. \begin{proof}[Proof of Theorem \ref{thm3}] First consider the eigenvalue problem \begin{equation} \begin{gathered} -\Delta u(x )= \lambda u(x ) ,\quad x_1^2+x_2^2< r^2\\ \frac{\partial u}{\partial n}| _{x_1^2+x_2^2=r^2}=0. \end{gathered}\label{e38} \end{equation} By \cite[Chpater 5,Section 5]{CH}, positive eigenvalues $\lambda _{k}^{j}$ of \eqref{e38} satisfy \begin{equation} J_j'\big(r\sqrt{\lambda _{k}^{j}} \big)=0,\quad j=0,1,2\dots ,\; k=1,2\dots , \label{e39} \end{equation} where $J_j(\cdot )$ is the $j$-order Bessel function, and the corresponding eigenfunctions are \[ u_{k}^{j}(x )=u_{k}^{j}(x_1, x_2)=J_j(\sqrt{\lambda _{k}^{j}} \tau )(\cos j\theta +\sin j\theta) \] with $x_1=\tau \cos\theta$, $x_2=\tau \sin\theta$, $0\leqslant \tau \leqslant r$, $0\leqslant\theta \leqslant 2\pi $. Now we choose \begin{equation} e_j(x )=u_j^{0}(x )=J_0(\sqrt{\lambda _j^{0}} \tau),\quad j=1,2,\dots p. \label{e40} \end{equation}\\ From the integral expression \begin{equation} J_0(t )=\frac{1}{2\pi }\int_{-\pi }^{\pi }\cos(t\sin\theta )d\theta , \label{e41} \end{equation} we know that $| e_j(x )|\leqslant 1$. And since $-\triangle e_j(x )=\lambda _j^{0}e_j(x )$, by Green's formula, we obtain \begin{equation} \int _{\Omega }| \nabla e_j(x )|^2dx =\lambda _j^{0}\int _{\Omega }| e_j(x )|^2dx. \label{e42} \end{equation} Since $J_0'(t )=-J_1(t )$ for all $t\in \mathbb{R}$, by \eqref{e39}, we have \begin{equation} J_1(r\sqrt{\lambda _j^{0}} )=0,\quad j=1,2,\dots . \label{e43} \end{equation} Let $a _j^{0}$ be the $j$th positive zero point of $J_0(t )$. Then according to Schafheitlin's investigation of the zero points of $J_0(t )$ \cite[Section 15.32, P.489]{W}, $a_j^{0}$ satisfies the estimate \begin{equation} (j-1 )\pi +\frac{3}{4}\pi < a_j^{0}< (j-1 )\pi+\frac{7}{8}\pi ,\quad j=1,2,\dots , \label{e44} \end{equation} thus by \eqref{e44} and the property of positive zero points of $J_1(t )$, we obtain \begin{equation} (j-1 )\pi +\frac{3}{4}\pi