\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 137, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/137\hfil Multiple solutions for ODE's] {Existence of multiple solutions for a class of second-order ordinary differential equations} \author[X. B. Shu \& Y. T. Xu\hfil EJDE-2004/137\hfilneg] {Xiao-Bao Shu, Yuan-Tong Xu} % in alphabetical order \address{Xiao-Bao Shu \hfill\break Department of Mathematics\\ Sun Yat-Sen University\\ Guangzhou, 510275, China} \email{sxb0221@163.com} \address{Yuan-Tong Xu \hfill\break Department of Mathematics\\ Sun Yat-Sen University\\ Guangzhou, 510275, China} \email{xyt@zsu.edu.cn} \date{} \thanks{Submitted October 5, 2004. Published November 25, 2004.} \thanks{Supported by grant 10471155 from NNSF of China, grant 031608 from the Foundation \hfill\break\indent of the Guangdong province Natural Science Committee, and grant 20020558092 from \hfill\break\indent Foundation for PhD Specialities of Educational Department of China.} \subjclass[2000]{34B15, 34B05, 65K10, 34B24} \keywords{Variational structure; $Z_2$ group index theory; critical points; \hfill\break\indent boundary value problems} \begin{abstract} By means of variational structure and $Z_2$ group index theory, we obtain multiple solutions for the second-order differential equation $$ \frac{d}{dt}(p(t)\frac{du}{dt})+q(t)u+f(t,u)=0 ,\quad 0 0$ is a parameter, $\alpha, \beta, \gamma, \delta \geq 0$ and $\alpha\delta + \alpha\gamma + \beta\gamma > 0, f \in C((0, 1) \times R,R), r \in C([0, 1], (0,\infty))$ and $q \in C([0, 1], [0, \infty))$. In this paper we are interested in the study of second-order ordinary differential equation \begin{equation}\label{eq1} \frac{d}{dt}(p(t)\frac{du}{dt})+q(t)u+f(t,u)=0 ,\quad 00 $ such that $$ \sup_{x \in X \cap S_{\rho}} f(x)<0, $$ then we have $i_1{(f)}\geq m$ \item[(P2)] If there exists a $ j$-dimensional subspace $\widetilde{X}$ of $E$ such that $$ \inf_{x\in\widetilde{X}^{\perp}} f(x)>-\infty, $$ we have $i_2(f)\leq j$ \end{itemize} If $m\geq j$, (P1) and (P2) hold, then $f$ has at least $2(m-j) $ distinct critical points. \end{lemma} \begin{lemma}[\cite{Rabinowitz}] \label{lem1.5} Let $f\in C^1(X,\mathbb{R})$ be an even functional which satisfies the Palais-Smale condition and $f(\theta)=0$. If \begin{itemize} \item[(F1)] There exists $\rho >0$, $\alpha > 0 $ and a finite dimensional subspace $E$ of $X$, such that $ f|_{E^\perp \cap S_\rho}\geq \alpha $ \item[(F2)] For all finite dimensional subspace $\widetilde{E}$ of $X$, there is a $r = r(\widetilde{E}) > 0$, such that $f(x)\leq 0$ for $x\in \widetilde{E} \backslash B_r$ \end{itemize} Then $f$ possesses an unbounded sequence of critical values. \end{lemma} \section{Main Results} \label{sec:spectral_basis} In this paper, we use Lemma \ref{lem1.4} and \ref{lem1.5} to study the boundary-value problems \eqref{eq1}-\eqref{eq2} and \eqref{eq1}-\eqref{eq3} \begin{theorem}\label{thm1} Let $f$, $p(t)$ and $q(t)$ satisfy the following conditions: \begin{itemize} \item[(i)] $p(t) \in C[0,1]$ and $0 < m \leq p(t) \leq M$ for $t \in [0,1]$ \item[(ii)] $f\in C([0,1]\times \mathbb{R},\mathbb{R})$ \item[(iii)] $\lim_{u\to 0}\frac{f(t,u)}{u}=\xi(t)>0$ uniformly for $t\in [0,1]$, $\lambda= \min_{0\leq t\leq 1}\xi(t)$ \item[(iv)] There exists $\alpha>0$ such that $f(t,\alpha)\leq 0$ \item[(v)] $f(t,u)$ is odd in $u$ \item[(vi)] $-\frac{\lambda}{2} - \frac{m}{2p(1)} $ \end{theorem} \begin{proof} Set $h: [0,1]\times \mathbb{R}\to \mathbb{R}$, $$ h (t,u)=\begin{cases} f(t, \alpha) &\mbox{if } u>\alpha,\\ f(t, u) &\mbox{if } |u|\leq \alpha,\\ f(t,-\alpha) &\mbox{if } u<- \alpha \end{cases} $$ Let us consider the functional defined on $H_{0}^{1}(0,1)$ by \begin{equation}\label{lin1} I(u)=\int^{1}_{0}[\frac{1}{2}p(t)|u'(t)|^2 - \frac{1}{2}q(t)|u(t)|^2 - G(t,u(t))]dt+\frac{p(1)}{2}\gamma u^2(1) , \end{equation} Where $G(t,u)=\int^{u}_{0}h(t,v)dv$. The norm $\| \cdot \|$ and inner product $(\cdot ,\cdot )$ can be defined respectively by $$ \|u\|=(\int^{1}_{0}(|u'(t)|^2+|u(t)|^2)dt)^{1/2}; \quad (u,v)=\int^{1}_{0}(u'(t)v'(t)+u(t)v(t))dt\,. $$ Thus $H^{1}_{0}(0,1)=W_{0}^{1,2}(0,1)$ will be a Hilbert space. Let $E=H_{0}^{1}(0,1)$, since $h(t,u)$ is an odd continuous map in $u$, we know that $I\in C^1(E,R)$ is even in $u$ and $I(\theta)=0$. First, we will show that the critical points of the $I(u)$ are the solutions of \eqref{eq1}-\eqref{eq2} in $C^2[0,1]$. Since \begin{equation}\label{lin2} \begin{aligned} I(u+sv)&=I(u)+s\{ \int_{0}^{1}[p(t)u'(t)v'(t) -q(t)u(t)v(t)-h(t,u+\theta(t) s v)v(t)] dt\\ &\quad +p(1)\gamma u(1)v(1)\} + \frac{s^2}{2} \{ \int_{0}^{1} (p(t)|v'(t)|^2 \\ &\quad - q(t)|v(t)|^2)dt+p(1)\gamma v^2(1)\} \quad \forall u,v \in E, 0<\theta(t) <1 \end{aligned} \end{equation} We have, for all $u,v \in E$, \begin{equation}\label{int3} (I'(u),v)=\int^{1}_{0}[p(t) u'(t)v'(t)-q(t)u(t)v(t) - h(t,u(t))v]dt+p(1) \gamma u(1)v(1) \,. \end{equation} By $I'(u)=0$, one gets \begin{equation}\label{lin4} \int^{1}_{0}[p(t) u'(t)v'(t)-q(t)u(t)v(t) - h(t,u(t))v]dt+ p(1)\gamma u(1)v(1)=0 \end{equation} for all $v\in E$. On the other hand, \begin{equation}\label{lin5} \begin{aligned} &\int_{0}^{1}p(t)u'(t)v'(t)dt +\int_{0}^{1} \frac{d}{dt} (p(t)\frac{du}{dt} ) v dt \\ &=\int_{0}^{1}p(t)u'(t)v'(t)dt +\int_{0}^{1}p(t)u''(t)v(t)dt +\int_{0}^{1}p'(t)u'(t)v(t)dt\\ &= p(1) v(1)u'(1)-p(0)u'(0)v(0)=0 \end{aligned} \end{equation} So, it is easy to see that \begin{align*} &\int_{0}^{1} v [ \frac{d}{dt}(p(t)\frac{du}{dt} )+q(t)u(t)+h(t,u(t))]dt\\ &= p(1)v(1)(u'(1) + \gamma u(1))-p(0)u'(0)v(0)=0 \end{align*} Hence we obtain $$ \frac{d}{dt} (p(t)\frac{du}{dt})+q(t)u(t)+h(t,u(t))=0 $$ Thus the critical points of $I(u)$ are the solutions of \eqref{eq1}-\eqref{eq2} in $C^2[0,1]$. For convenience, we transform \eqref{lin1} into \begin{align*} I(u)&=\int^{1}_{0}[\frac{1}{2}p(t)(|u'(t)|^2+|u(t)|^2)\\ &\quad - \frac{1}{2}(q(t)+p(t))|u(t)|^2 - G(t,u(t))]dt+\frac{p(1)}{2}\gamma u^2(1)\,. \end{align*} By condition (iv) of Theorem \ref{thm1}, we have $uh(u)\leq 0$ when $|u(t)|\geq \alpha $. So $$ \int^{1}_{0}G(t,u)=\int^{1}_{0}\int^{u}_{0}h(t,v) dv dt \leq \int^{1}_{0} \int^{\alpha}_{-\alpha}|h(t,v)|dv\,dt\,. $$ Denote by $c$ the value of $\int^{1}_{0}\int^{\alpha}_{-\alpha}|h(t,v)|dvdt $. On the other hand, $q(t)+p(t)\leq 0$, then $-\int_{0}^{1}(q(t)+p(t))|u(t)|^2 dt\geq 0$. So, we have $$ I(u)\geq \frac{m}{2}\|u\|^2- c +\frac{p(1)}{2}\gamma u^2(1) \quad \forall u\in E . $$ Next, we show that $I(u)$ has lower bound. We divide our proof into two parts \noindent (I) When $\gamma\geq 0$, it is easy to see \begin{equation}\label{int 6} I(u)\geq \frac{m}{2}\|u\|^2- c \quad \forall u\in E \end{equation} (II) When $ - \frac{m}{2p(1)} < \gamma <0 $, we divide again our proof into two parts in order to show $I(u)$ has lower bound: \noindent (a) If there exist $t_0 \in [0,1] $ such that $u(t_0)=0$, then $$ |u(1)|= |\int_{t_0}^{1}u'(s)ds|\leq \int_{0}^{1}|u'(s)|ds \leq \sqrt{2} \|u\|\,. $$ So, we get \begin{equation}\label{int 7} I(u)\geq \frac{1}{2} (m+ 2 \gamma p(1)) \|u\|^2- c \quad \forall u\in E , \end{equation} (b) If does not exist $t_1 \in [0,1] $ such that $u(t_1)=0$, then $u(t)>0$ or $u(t)< 0$, for all $t\in [0,1]$. We might as well let $u(t)>0$ for all $t\in [0,1]$. When $\max_{0\leq t\leq 1}u(t)\leq 1$,we have $u(1)\leq 1$ and $$ I(u)\geq \frac{m}{2}\|u\|^2-c+\frac{1}{2}\gamma p(1), $$ i.e., $I(u)$ has lower bound. When $\max_{0\leq t \leq 1}|u(t)|>1$, since $u \in C^2[0,1]$, there exist $t_2\in [0,1]$ such that $u(t_2)=\min_{0\leq t\leq 1}u(t)$. So $$ u(1)-u(t_2)=|\int_{t_2}^{1}u'(s)ds|\leq \int_{0}^{1}|u'(s)|ds $$ i.e. \begin{align*} u(1)&\leq u(t_2) +\int_{0}^{1}|u'(s)|ds \leq (\int_{0}^{1} u^2(t)dt)^{ \frac{1}{2} } +(\int_{0}^{1} |u'(t)|^2)^\frac{1}{2}\\ &\leq \sqrt{2}(\int_{0}^{1} u^2(t)dt + \int_{0}^{1} |u'(t)|^2 )^{1/2}=\sqrt{2}\|u\| \end{align*} As in the proof of (I), we have \begin{equation}\label{int 8} I(u)\geq \frac{1}{2} (m+ 2 \gamma p(1)) \|u\|^2- c. \quad \forall u\in E . \end{equation} By (a) and (b), it is easy to see $I(u)$ has lower bound when $ - \frac{m}{2p(1)} < \gamma <0 $. From (I) and (II), we get that $I(u)$ has lower bound for all $u\in H^{1}_{0}(0,1)$, i.e., $i_2(I)=0$. Next, we verify that $I(u)$ satisfies the Palais-Smale condition. Suppose that $\{u_n\}\subset E$ with and \begin{gather}\label{int 9} c_1\leq I(u_n) \leq c_2,\\ \label{int 10} I'(u_n)\to 0 \quad \mbox{as} \quad n \to \infty\,. \end{gather} Then \begin{equation}\label{int 11} \begin{gathered} \sup\{\int^{1}_{0} [p(t)u_n'v' - q(t)u_n v- h(t,u_n(t))v]dt+ \gamma p(1) u_n(1)v(1)\}\to 0 , \\ \mbox{as } n \to \infty, \quad \forall u,v \in E, \|v\|=1 \end{gathered} \end{equation} with $\|z_n\|=\|I'(x_n)\|$. Let us denote $\varepsilon_n=\|z_n\|$, then $\varepsilon_n\to 0 $ as $n\to \infty$. Replace $v$ by $u_n$ in above equality. By \eqref{int3} and \eqref{int 11}, we have $$ \int^{1}_{0}[ p(t)|u'_n(t)|^2-q(t)|u_n(t)|^2]dt =\int^{1}_{0} h(t,u_n)u_n(t) dt+\langle z_n,u_n\rangle . $$ The above equality is equivalent to \begin{align*} &\int^{1}_{0} p(t)[|u'_n(t)|^2+|u_n(t)|^2]dt \\ &=\int^{1}_{0}[(q(t)+p(t))|u_n(t)|^2+ h(t,u_n)u_n(t)] dt+\langle z_n,u_n\rangle \end{align*} So, there exist $\xi \in [0,1]$ such that \begin{equation} \label{int 12} p(\xi)\|u_n\|^2=\int^{1}_{0} [(q(t)+p(t))|u_n(t)|^2+ h(t,u_n)u_n(t)] dt +\langle z_n,u_n\rangle\,. \end{equation} Next, we show that $\{u_n\}$ satisfying condition \eqref{int 9} and \eqref{int 10} is bounded. We divide again our proof into two parts. \noindent (c) When $\gamma\geq 0$, by \eqref{int 6}, one gets $$ \|u_n\|^2 \leq \frac{2}{m} (I(u_n)+ c) \leq \frac{2}{m}(c_2+ c), $$ i.e., $\|u_n\|\leq \sqrt{\frac{2}{m}(c_2+ c)}$. \noindent (d) When $ - \frac{m}{2p(1)} < \gamma <0 $, by the above proof and \eqref{int 7} and \eqref{int 8}, we have $$ \|u_n\|^2 \leq \frac{2}{m+2\gamma p(1)}(I(u_n)+ c) \leq \frac{2}{1+2\gamma p(1)}(c_2+ c) $$ or $$ \|u\|^2 \leq \frac{2}{m}[c_2+c- \frac{1}{2}\gamma p(1)] $$ i.e., $$ \|u_n\|\leq \sqrt{\frac{2}{m+2\gamma p(1)}(c_2+ c)} \quad\mbox{or}\quad \|u\|\leq \sqrt{\frac{2}{m}(c_2+c- \frac{1}{2}\gamma p(1) )}. $$ By (c) and (d), it is easy to see $\{u_n\}$ is bounded in the space $H_{0}^{1}(0,1)$. Reflexivity of $H_{0}^{1}(0,1)$ implies that there exists a subsequence of $\{u_n\}$ which is weak convergent in $H_{0}^{1}(0,1)$. We still denote it by $\{u_n\}$ and suppose that $u_n\rightharpoonup u_0$ in $H_{0}^{1}(0,1)$ as $ n\to\infty $. On the one hand, by boundedness of $\{u_n\}$ and \eqref{int 12}, we have $$ p(\xi)\|u_n\|^2 -\int^{1}_{0} [(q(t)+p(t))|u_n(t)|^2+ h(t,u_n)u_n(t)] dt \to 0 \quad \mbox{as } n\to\infty $$ Note that the weak convergent of $\{u_n\}$ in $H_{0}^{1}(0,1)$ implies the uniform convergence of $\{u_n\}$ in $C([0,1],R)$ \cite[Proposition 1.2 ]{Mawhin}. Hence $$ p(\xi)\|u_n\|^2 \to \int^{1}_{0} [(q(t)+p(t))|u_0(t)|^2+ h(t,u_0)u_0(t)] dt \quad \mbox{as } n\to\infty $$ This means that $\{u_n\}$ converges in $H_{0}^{1}(0,1)$. So the P.S. condition holds. Thirdly, we show that Theorem \ref{thm1} holds by Lemma \ref{lem1.4}. Denote $ \beta_k(t)=\frac{\sqrt{2}} {k \pi }\cos k\pi t$, $k=1,2,3,\dots,n,\dots$, then \[ \int^{1}_{0}|\beta_k(t)|^2dt= \frac{1}{k^2\pi^2} ,\quad \int^{1}_{0}|\beta'_k(t)|^2dt=1 \] Definite $n$-dimensional linear space $$ E_n=\mathop{\rm span}\{\beta_1(t),\beta_2(t),\dots \beta_n(t) \}. $$ It is obvious that $E_n$ is a symmetric set. Suppose $\rho>0$, then $$ E_n \cap S_\rho =\big\{\sum_{k=0}^{n}b_k\beta_k : \sum_{k=0}^{n}b_{k}^{2}(1+ \frac{1}{k^2\pi^2}) =\rho^2\big\} $$ Let $g(t,u)=\frac{1}{\xi(t)}h(t,u)-u$, by condition (iii) of Theorem \ref{thm1}, $\lim_{u\to 0}\frac{g(u)}{u} =0,$ uniformly for $t\in [0,1]$. We choose $\varepsilon$ such that $$ 0<\varepsilon<\frac{1}{n^2}-\frac{2(M+p(1)|\gamma|)(1+\pi^2)}{\lambda}. $$ By condition (iii) of Theorem \ref{thm1}, there exist $\delta>0$ such that $|g(t,u)|\leq \varepsilon |u|$ whenever $|u|\leq\delta$. We can choose $\rho$ such that $0<\rho<\min\{\alpha,\delta \}$, and have \begin{align*} \max_{0\leq t \leq 1}u(t) &\leq \sum_{k=0}^{n} \frac{ \sqrt{2} }{k\pi } |b_k| \leq \|u\|=\|\sum_{k=0}^{n}b_k \beta_k \|\\ &=(\sum_{k=0}^{n}b_{k}^{2}(1+ \frac{1}{k^2\pi^2}))^{1/2} =\rho<\min\{\alpha,\delta\} \end{align*} when $u\in E_n \cap S_\rho$. So \begin{align*} G(t,u)&= \xi(t) \int_{0}^{u}[ v+g(t,v))]dv\\ &= \frac{1}{2}\xi(t)|u(t)|^2+\xi(t) \int_{0}^{u} g(t,v)dv\\ &\geq \frac{1}{2}\xi(t)|u(t)|^2-\xi(t)\int_{0}^{u}\varepsilon vdv\\ &= \frac{1}{2} \xi(t)(1-\varepsilon)|u(t)|^2\\ &\geq \lambda (1-\varepsilon)|u(t)|^2 \end{align*} From $q(t)+p(t)\geq -\frac{\lambda}{2}$, we get that $$ -\int_{0}^{1}(q(t)+p(t))|u(t)|^2dt\leq \frac{\lambda}{2}\sum_{k=0}^{n} \frac{b_{k}^{2}}{k^2 \pi^2} $$ So \begin{align*} I(u)&= \int^{1}_{0}[\frac{1}{2}p(t)(|u'(t)|^2+ |u(t)|^2)-\frac{1}{2} (q(t)+p(t))|u(t)|^2]dt\\ &\quad -\int^{1}_{0} G(t,u) dt+\frac{p(1)}{2}\gamma u^2(1)\\ &\leq M\int^{1}_{0}[\frac{1}{2}(|u'(t)|^2+|u(t)|^2) -\frac{1}{2}\lambda(1-\varepsilon)|u(t)|^2]dt\\ &\quad + \frac{\lambda}{4}\sum_{k=0}^{n} \frac{b_{k}^{2}}{k^2 \pi^2} +\frac{p(1)}{2}|\gamma|\|u\|_{C}^{2} \\ &\leq \frac{M}{2}\sum_{k=0}^{n}b_{k}^{2}(1+ \frac{1}{k^2\pi^2}) -\frac{1}{4} \lambda(1-2\varepsilon)(\sum_{k=0}^{n}\frac{b_{k}^{2}} {k^2 \pi^2})+ \frac{p(1)}{2}|\gamma|(\sum_{k=0}^{n}\frac{ \sqrt{2} |b_{k}|} {k \pi} )^2 \\ &\leq \frac{M+p(1)|\gamma| }{2}\sum_{k=0}^{n}b_{k}^{2}(1+ \frac{1}{k^2\pi^2}) -\frac{1}{4} \lambda(1-2\varepsilon)\sum_{k=0}^{n}\frac{b_{k}^{2}} {k^2 \pi^2}\\ &< \frac{M+p(1)|\gamma| }{2}\sum_{k=0}^{n}b_{k}^{2}(1+ \frac{1}{\pi^2}) -\frac{1}{4} \lambda(1-\varepsilon)\sum_{k=0}^{n}\frac{b_{k}^{2}} {n^2 \pi^2}\\ &\leq \frac{\lambda}{2}(\frac{M+p(1)|\gamma|}{\lambda}\frac{\pi^2+1}{\pi^2} -\frac{1}{2n^2 \pi^2}+\varepsilon) \sum_{k=0}^{n}b_{k}^{2}\\ &\leq \frac{\lambda}{2\pi^2}(\frac{(M+p(1)|\gamma|)(1+\pi^2)}{\lambda} -\frac{1}{2n^2 }+\varepsilon) \sum_{k=0}^{n}b_{k}^{2}\\ &\leq \frac{\lambda}{4\pi^2}( \frac{2(M+p(1)|\gamma|)(1+\pi^2)}{\lambda} -\frac{1}{ n^2 }+\varepsilon) \sum_{k=0}^{n}b_{k}^{2}< 0 \end{align*} By Lemma \ref{lem1.4} and the above result, we have $i_1{(I)}\geq n$ and $I$ has $2n $ distinct critical points, i.e., boundary-value problem \eqref{eq1}-\eqref{eq2} has at least $2n$ nontrivial solutions in $C^2[0,1]$. \end{proof} Next we consider the boundary-value problem \eqref{eq1}-\eqref{eq3}. Similar to Theorem \ref{thm1}, we have the following result. \begin{theorem}\label{thm2} Let $f$, $p(t)$ and $q(t)$ satisfy the following conditions: \begin{itemize} \item[(i)] $p(t) \in C[0,1]$ and $0 < m \leq p(t) \leq M$ for $t \in [0,1]$ \item[(ii)] $f\in C([0,1]\times \mathbb{R},\mathbb{R})$ \item[(iii)] $\lim_{u\to 0}\frac{f(t,u)}{u}=\xi(t)>0$ uniformly for $t\in [0,1]$, $\lambda= \min_{0\leq t\leq 1}\xi(t)$ \item[(iv)] There exists $\alpha>0$, such that $f(t,\alpha)\leq 0$ \item[(v)] $f(t,u)$ is odd in $u$ \item[(vi)] $-\frac{\lambda}{2} \frac{2M}{m}\geq 2$ and $\alpha>0$ such that $$ 0 -\frac{m\theta-2M}{2(\theta-2)p(1)}$, then \eqref{eq1}-\eqref{eq2} has infinite nontrivial solutions in $C^2[0,1]$. \end{theorem} \begin{proof} It is easy to see that for $u\in H_{0}^{1}(0,1)$, the functional \begin{equation}\label{int 13} I(u)=\int^{1}_{0}[\frac{1}{2}p(t)|u'(t)|^2 - \frac{1}{2}q(t)|u(t)|^2 G(t,u(t))]dt+\frac{p(1)}{2}\gamma u^2(1) \end{equation} is well defined. The solutions of boundary-value problems \eqref{eq1}-\eqref{eq2} are the critical points of the functional $I(u)$. Note that $I(u)$ is equivalent to \[ I(u)=\int^{1}_{0}[\frac{1}{2}p(t)(|u'(t)|^2+|u(t)|^2)- \frac{1}{2}(q(t)+p(t))|u(t)|^2 - \lambda G(t,u)]dt+\frac{p(1)}{2}\gamma u^2(1) \] We show that Theorem \ref{thm3} holds by using Lemma \ref{lem1.5}. Since $f(t,u)$ is an odd continuous map in $u$, we know that $I\in C^1(E,R)$ is even in $u$ and $I(\theta)=0$. Moreover, As in the proof of Theorem \ref{thm1}, one gets \[ (I'(u),v)=\int^{1}_{0}[p(t)u'(t)v'(t)-q(t)u(t)v(t))- f(t,u)v(t)]dt\\ + \gamma p(1)u(1)v(1) , \] for all $u,v \in E$. The above equality is equivalent to \begin{align*} \int^{1}_{0}f(t,u)v(t)dt & = \int^{1}_{0}p(t)(u'(t)v'(t)+u(t)v(t))dt\\ &\quad - \int^{1}_{0}(p(t)+q(t))u(t)v(t)dt - (I'(u),v) + p(1)\gamma u(1)v(1) . \end{align*} Next, we verify that $I(u)$ satisfies the Palais-Smale condition. Suppose that ${u_n}\subset E$ with \begin{gather}\label{int 14} c_1\leq I(u_n) \leq c_2, \\ \label{int 15} I'(u_n)\to 0 \quad\mbox{as } n \to \infty \end {gather} Now, we show that $\{u_n\}$ of satisfying \eqref{int 14} and \eqref{int 15} is bounded. Denote $E_1=\{t \in [0,1] | |u_n(t)|\geq \alpha\}$, $E_2=[0,1]\setminus E_1$. We divide our proof into two parts . \noindent (A)\quad When $ - \frac{m\theta-2M}{2(\theta-2)p(1)} <\gamma <0$, by (iv), we have \begin{align*} I(u_n) &= \int_{0}^{1}\frac{p(t)}{2}(|u_n(t)|^2+|u'_n(t)|^2)dt - \int_{0}^{1}\frac{1}{2}(q(t) +p(t))|u_n(t)|^2dt\\ &\quad-\int^{1}_{0}G(t,u_n(t))dt + \frac{p(1)}{2}\gamma u_{n}^{2}(1) \\ &\geq \frac{m}{2}\|u_n\|^2-\int_{E_1}G(t,u_n(t))dt-\int_{E_2}G(t,u_n(t))dt\\ &\quad+ \frac{p(1)}{2}\gamma u_{n}^{2}(1)- \int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt\\ &\geq \frac{m}{2}\|u_n\|^2-\int_{E_1}\frac{1}{\theta}u_n(t)f(t,u_n(t))dt-c_3\\ &\quad + \frac{p(1)}{2}\gamma u_{n}^{2}(1)- \int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt\\ & \geq \frac{m}{2}\|u_n\|^2-\int_{ 0}^{1}\frac{1}{\theta}u_n(t)f(t,u_n(t))dt-c_4\\ &\quad + \frac{p(1)}{2}\gamma u_{n}^{2}(1)- \int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt\\ & = \frac{m}{2}\|u_n\|^2-\frac{1}{\theta}(\int_{0}^{1}p(t)( |u'_n(t)|^2+|u_n(t)|^2)dt -(I'(u_n),u_n)\\ &\quad +p(1)\gamma u_{n}^{2}(1))-c_4 + \frac{p(1)}{2}\gamma u_{n}^{2}(1) - (\frac{1}{2}- \frac{1}{\theta} )\int_{0}^{1}(q(t)+p(t))|u_n(t)|^2dt\\ &\geq ( \frac{m}{2}- \frac{M}{\theta}) \|u_n\|^2 +\frac{1}{\theta}(I'(u_n),u_n)-c_4+( \frac{1}{2}-\frac{1}{\theta})p(1)\gamma u_{n}^{2}(1) \\ & \geq (\frac{m}{2} -\frac{M}{\theta})\|u_n\|^2+\frac{1}{\theta}\|I'(u_n)\| \|u_n\|-c_4+( \frac{1}{2}-\frac{1}{\theta})p(1)\gamma u_{n}^{2}(1) \end{align*} \noindent\textbf{Remarks:} (1) for the rest of this article, $c_i>0$. (2) The above equality makes use of $-(\frac{1}{2}- \frac{1}{\theta} )\int_{0}^{1}(q(t)+p(t))|u_n(t)|^2dt\geq 0$ To show that $\{u_n\}$ is bounded, we divide again our proof into two parts.\\ (I')\quad When $\max_{0\leq t \leq 1}|u(t)| \leq 1$, as in the proof of Theorem \ref{thm1}, we have $u^2(1)\leq 1$ and \begin{gather*} I(u_n)\geq (\frac{m}{2} -\frac{M}{\theta})\|u_n\|^2+\frac{1}{\theta}\|I'(u_n)\| \|u_n\|-c_4+( \frac{1}{2}-\frac{1}{\theta})\gamma \\ (\frac{m}{2} -\frac{M}{\theta})\|u_n\|^2\leq I(u_n) - \frac{1}{\theta}\|I'(u_n)\| \|u_n\|+c_4-( \frac{1}{2}-\frac{1}{\theta})\gamma \end{gather*} Using $\theta > \frac{2M}{m} $, \eqref{int 14} and \eqref{int 15}, it is not difficulty to see that $\{\|u_n\|\}$ is bounded. \noindent (II') When $\max_{0\leq t\leq 1}|u(t)|>1$, as in the proof of Theorem \ref{thm1}, we have $u^2(1)\leq 2 \|u\|^2$ and \[ [( \frac{m}{2}-\frac{M}{\theta})+( \frac{1}{2}-\frac{1}{\theta} )2p(1)\gamma] \|u_n\|^2 \leq I(u_n)-\frac{1}{\theta} \|I'(u_n)\| \|u_n\|+c_4\leq c_5\|u_n\|+c_5\,. \] Since $ \theta>\frac{2M}{m}\geq 2 $ and $\gamma >- \frac{m\theta-2M}{2(\theta-2)p(1)}$, it follows that $\{\|u_n\|\}$ is bounded. From (I') and (II'), we get that $\{\|u_n\|\}$ is bounded when $- \frac{m\theta-2M}{2(\theta-2)p(1)} <\gamma <0 $. \noindent (B)\quad when $\gamma >0$, as in the proof of (A), it is not difficult to see that \begin{align*} I(u_n) &= \int_{0}^{1}\frac{p(t)}{2}(|u_n(t)|^2+|u'_n(t)|^2dt -\int^{1}_{0}G(t,u_n(t))dt\\ &\quad + \frac{p(1)}{2} \gamma u_{n}^{2}(1) - \int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt\\ &\geq \frac{m}{2}\|u_n\|^2-\int_{ 0}^{1}\frac{1}{\theta}u_n(t)f(t,u_n(t))dt\\ &\quad + \frac{p(1)}{2}\gamma u_{n}^{2}(1)- \int_{0}^{1}\frac{1}{2}(q(t) +p(t))|u_n(t)|^2dt \\ &\geq (\frac{m}{2}-\frac{M}{\theta})\|u_n\|^2+\frac{1}{\theta}\|I'(u_n)\| \|u_n\|-c_4+(\frac{1}{2}-\frac{1}{\theta} )p(1)\gamma u_{n}^{2}(1)\\ &\geq (\frac{m}{2}-\frac{M}{\theta})\|u_n\|^2+\frac{1}{\theta}\|I'(u_n)\| \|u_n\|-c_4 \end{align*} (We remark that the above equality use $(\frac{1}{2}-\frac{1}{\theta} )p(1)\gamma u^2(1)\geq 0 $ ) and \begin{align*} (\frac{m}{2}-\frac{M}{\theta})\|u_n\|^2 \leq I(u_n)-\frac{1}{\theta} \|I'(u_n)\| \|u_n\|+c_4\leq c_5\|u_n\|+c_5. \end{align*} So, we get that $\{\|u_n\|\}$ is bounded when $\gamma >0$. From (A) and (B), we obtain that $\{u_n\}$ satisfying \eqref{int 14} and \eqref{int 15} is bounded. So the P.S. condition holds. Thirdly, we show that Theorem \ref{thm3} holds by using Lemma \ref{lem1.5}. First, we verify condition (F1) of Lemma \ref{lem1.5}. Let $\beta_j(t)=\cos jt$, $j=1,2,\dots$. Consider the $n$-dimensional subspace $$ E_n=\mbox{\rm span}\{\beta_1(t),\beta_2(t),\dots, \beta_n(t)\} $$ and let $X=V^{\perp}$. By (ii), we have $\delta>0$ such that $|f(t,u(t))|\leq T|u|$, whenever $|u|\leq\delta $. Let $\rho$ with $\rho=\delta$. For any $u\in S_\rho \cap X$, we have $\|u\|_C \leq \|u\|=\rho=\delta$. From $$ \int_{0}^{1}|u(t)|^2dt \leq \frac{1}{n^2}\int_{0}^{1}|u'(t)|^2dt $$ it is easy to see $$ \int_{0}^{1}|u(t)|^2dt \leq \frac{\rho^2}{n^2+1} $$ So, when $\gamma >0$, we have \begin{align*} I(u_n) &= \int_{0}^{1}\frac{p(t)}{2}(|u_n(t)|^2+|u'_n(t)|^2)dt-\int^{1}_{0}G(t,u_n(t))dt\\ &\quad + \frac{p(1)}{2} \gamma u_{n}^{2}(1)- \int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt \\ &\geq \frac{m}{2}\|u\|^2-\int^{1}_{0}G(t,u(t))dt\\ &\geq \frac{m}{2}\rho^2-\int^{1}_{0}(\int^{|u(t)|}_{0}T vdv)dt\\ &\geq \frac{m}{2}\rho^2-\frac{T}{2(n^2+1)}\rho^2 =\frac{1}{2}(m-\frac{T}{n^2+1})\rho^2>0\,. \end{align*} Note that in the above equality, we use $n^2>\max\{\frac{T}{m},\frac{T}{m+p(1) \gamma }\} $ and $$ -\int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt>0. $$ When $- \frac{m\theta-2M}{2(\theta-2)p(1)} <\gamma <0 $, we get \begin{align*} I(u_n) &= \int_{0}^{1}\frac{p(t)}{2}(|u_n(t)|^2+|u'_n(t)|^2)dt-\int^{1}_{0}G(t,u_n(t))dt\\ &\quad + \frac{p(1)}{2} \gamma u_{n}^{2}(1)- \int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u_n(t)|^2dt \\ &\geq \frac{m}{2}\|u\|^2-\int^{1}_{0}G(t,u(t))dt+ \frac{p(1)}{2}\gamma u_{n}^{2}(1)\\ &\geq \frac{m+p(1)\gamma}{2}\rho^2-\int^{1}_{0}(\int^{|u(t)|}_{0}T vdv)dt\\ &\geq \frac{m+p(1)\gamma}{2}\rho^2-\frac{T}{2(n^2+1)}\rho^2\\ &=\frac{1}{2}(m+p(1)\gamma -\frac{T}{n^2+1})\rho^2>0\,. \end{align*} Note that the above equality uses $n^2>\max\{\frac{T}{m},\frac{T}{m+p(1)\gamma}\} $. We sum up the conclusions above to obtain that $I(u)>0$ for all $u\in S_\rho \cap X$, i.e., condition (F1) of Lemma \ref{lem1.5} holds. Finally, we verify condition (F2) of Lemma \ref{lem1.5}. By (iv), one gets $$G(t,u(t))\geq c_7 |u|^{\theta}-c_8 .$$ For all finite dimensional subspace $E_1$ of $E$, there exist $c_9$ such that $$ \Big(\int^{1}_{0}|u(t)|^{\theta}dt\Big)^{1/\theta}\geq c_9\|u\|, \quad \forall u\in E_1. $$ On the other hand, since $p(t)\in C^1[0,1], q(t)\in C[0,1]$ and $p(t)+q(t)\leq 0$, there exist a positive number $Q$ such that $-Q=\min_{0\leq t\leq 1}p(t)+q(t)$, so $$ -\int_{0}^{1}(p(t)+q(t))|u(t)|^2dt\leq Q\int_{0}^{1}|u(t)|^2dt 1$, we have $u^2(1)\leq 2\|u\|^2$ and \begin{align*} I(u)&=\int^{1}_{0}[\frac{p(t)}{2}(|u'(t)|^2+|u(t)|^2)- G(t,u(t))]dt\\ &\quad +\frac{p(1)}{2}\gamma u^2(1)-\int_{0}^{1}\frac{1}{2}(q(t)+p(t))|u(t)|^2dt\\ &\leq \frac{M+Q+2\gamma}{2}\|u\|^2-c_7\int^{1}_{0}|u(t)|^{\theta}dt+c_8\\ &\leq \frac{M+Q+2\gamma}{2}\|u\|^2-c_7c_{9}^{\theta}\|u\|^{\theta}+c_8\\ &= (\frac{M+Q+2\gamma}{2}-c_7c_{9}^{\theta}\|u\|^{\theta-2})\|u\|^2+c_8 \end{align*} We sum up the conclusions above to obtain that $I(u)\leq 0 $, for all $u\in E_1\backslash B_R$ when $R=R(E_1)$ is adequately big, i.e., condition (F2) of Lemma \ref{lem1.5} holds. So $I$ possesses infinite critical point, i.e. the boundary-value problem \eqref{eq1}-\eqref{eq2} has infinitely many nontrivial solutions in $C^2[0,1]$. \end{proof} Using a technique similar to the one above, we can show that the following theorem. \begin{theorem}\label{thm4} Let $f$, $p(t)$ and $q(t)$ be the function satisfying the following conditions: \begin{itemize} \item[(i)] $p(t) \in C[0,1]$ and $0 < m \leq p(t) \leq M$ for $t \in [0,1]$ \item[(ii)] $f\in C([0,1]\times \mathbb{R},\mathbb{R})$ \item[(iii)] There exists $T$ such that $\limsup_{u\to 0}\frac{f(t,u)}{u}\leq T$ \item[(iv)] There exists $\theta> \frac{2M}{m}\geq 2$ and $\alpha>0$ such that $$ 0 < G(t,u)=\int^{u}_{0}f(t,v)dv\leq\frac{1}{\theta}u f(t,u),\quad \forall |u|\geq \alpha $$ \item[(v)] $f(t,u)$ is odd in $u$ \item[(vi)] $q(t) \in C[0,1]$, $q(t) + p(t) \leq 0$ for $ 0 \leq t \leq 1$. \end{itemize} Then boundary-value problem \eqref{eq1}-\eqref{eq3} has infinitely solutions in $C^2[0,1]$. \end{theorem} \section{Examples} \label{sec:axbf} \begin{example} \label{ex1}\rm For $0 \sqrt{2} \\ 0<5\leq p(t) \leq 7, \quad \theta=3 >\frac{2M}{m} =\frac{14}{5} \end{gather*} So conditions (iv) holds. On the other hand, it is easy to see that $$ p(t)+q(t)=6+\sin t + (-9+t^2)=-3+t^2+ \sin t \leq 0 \quad \forall t\in [0,1] $$ So conditions (i) and (vi) of Theorem \ref{thm3} hold. By Theorem \ref{thm3}, we obtain that \eqref{ 18} has infinitely many nontrivial solutions in $C^2[0,1]$. \end{example} \begin{example} \label{ex4} \rm Consider boundary-value problem \begin{equation} \label{ 19} \begin{gathered} \frac{d}{dt}((6+\sin t) \frac{du}{dt}) +(-9+t^2)u(t)+ t(u^3(t)+u(t))=0,\quad 0