\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 91, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/91\hfil Existence of three solutions] {Existence of three solutions for a Kirchhoff-type boundary-value problem} \author[S. Heidarkhani, G. A. Afrouzi, D. O'Regan \hfil EJDE-2011/91\hfilneg] {Shapour Heidarkhani, Ghasem Alizadeh Afrouzi, Donal O'Regan} % in alphabetical order \address{Shapour Heidarkhani \newline Department of Mathematics, Faculty of Sciences, Razi University, 67149 Kermanshah, Iran} \email{s.heidarkhani@razi.ac.ir} \address{Ghasem Alizadeh Afrouzi \newline Department of Mathematics, Faculty of Basic Sciences, University of Mazandaran, 47416-1467 Babolsar, Iran} \email{afrouzi@umz.ac.ir} \address{Donal O'Regan \newline Department of Mathematics, National University of Ireland, Galway, Ireland} \email{donal.oregan@nuigalway.ie} \thanks{Submitted April 26, 2011. Published July 6, 2011.} \subjclass[2000]{35J20, 35J25, 35J60} \keywords{Kirchhoff-type problem; multiple solutions; critical point} \begin{abstract} In this note, we establish the existence of two intervals of positive real parameters $\lambda$ for which the boundary-value problem of Kirchhoff-type \begin{gather*} -K\big(\int_{a}^b |u'(x)|^2dx\big)u''=\lambda f(x,u),\\ u(a)=u(b)=0 \end{gather*} admits three weak solutions whose norms are uniformly bounded with respect to $\lambda$ belonging to one of the two intervals. Our main tool is a three critical point theorem by Bonanno. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In the literature many results focus on the existence of multiple solutions to boundary-value problems. For example, certain chemical reactions in tubular reactors can be mathematically described by a nonlinear two-point boundary-value problem and one is interested if multiple steady-states exist. For a recent treatment of chemical reactor theory and multiple solutions see \cite[section 7]{ATT} and the references therein. Bonanno in \cite{B} established the existence of two intervals of positive real parameters $\lambda$ for which the functional $\Phi-\lambda \Psi$ has three critical points whose norms are uniformly bounded in respect to $\lambda$ belonging to one of the two intervals and he obtained multiplicity results for a two point boundary-value problem. In the present paper as an application, we shall illustrate these results for a Kirchhoff-type problem. Problems of Kirchhoff-type have been widely investigated. We refer the reader to the papers \cite{ACM,CL,HZ,M,MZ,PZ,ZP} and the references therein. Ricceri \cite{R2} established the existence of at least three weak solutions to a class of Kirchhoff-type doubly eigenvalue boundary value problem using \cite[Theorem 2]{R1}. Consider the Kirchhoff-type problem $$\label{e1} \begin{gathered} -K\big(\int_{a}^b |u'(x)|^2dx\big)u''=\lambda f(x,u),\\ u(a)=u(b)=0 \end{gathered}$$ where $K:[0,+\infty[\to \mathbb{R}$ is a continuous function, $f :[a,b]\times \mathbb{R}\to \mathbb{R}$ is a Carath\'eodory function and $\lambda>0$. In the present paper, our approach is based on a three critical points theorem proved in \cite{B}, which is recalled in the next section for the reader's convenience (Theorem \ref{thmA}). Our main result is Theorem \ref{thm1} which, under suitable assumptions, ensures the existence of two intervals $\Lambda_1$ and $\Lambda_2$ such that, for each $\lambda\in \Lambda_1\cup \Lambda_2$, the problem \eqref{e1} admits at least three classical solutions whose norms are uniformly bounded in respect to $\lambda\in \Lambda_2$. Let $X$ the the Sobolev space $H^{1}_{0}([a,b])$ with the norm $$\| u\| =\Big(\int_{a}^b (|u'(x)|^2)dx \Big)^{1/2}.$$ We say that $u$ is a weak solution to \eqref{e1} if $u\in X$ and $$K\big(\int_{a}^b |u'(x)|^2dx\big)\int_{a}^b u'(x) v'(x)dx-\lambda\int_{a}^b f(x,u(x))v(x)dx=0$$ for every $v\in X$. For other basic notations and definitions, we refer the reader to \cite{BMV,GR,KRV,Z}. \section{Results} For the reader's convenience, dirst we here recall \cite[Theorem 2.1]{B}. \begin{theorem} \label{thmA} Let $X$ be a separable and reflexive real Banach space, $\Phi:X \to \mathbb{R}$ a nonnegative continuously G\^{a}teaux differentiable and sequentially weakly lower semicontinuous functional whose G\^{a}teaux derivative admits a continuous inverse on $X^{*}$, $J:X\to \mathbb{R}$ a continuously G\^{a}teaux differentiable functional whose G\^{a}teaux derivative is compact. Assume that there exists $x_{0}\in X$ such that $\Phi(x_{0})=J(x_{0})=0$ and that $$\lim_{\|x\| \to +\infty} (\Phi(x)-\lambda J(x))=+\infty \quad \text{for all } \lambda\in[0,+\infty[.$$ Further, assume that there are $r>0$, $x_1\in X$ such that $r <\Phi(x_1)$ and $$\sup\nolimits_{x\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}J(x) < \frac{r}{r+\Phi(x_1)}J(x_1);$$ here $\overline{\Phi^{-1}(]-\infty,r[)}^{w}$ denotes the closure of $\Phi^{-1}(]-\infty,r[)$ in the weak topology (in particular note $J(x_1) \geq 0$ since $x_0 \in \overline{\Phi^{-1}(]-\infty,r[)}^{w}$ (note $J(x_0)=0$) so $\sup_{x\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}J(x) \geq 0$). Then, for each $$\lambda\in \Lambda_1= ]\frac{\Phi(x_1)}{J(x_1)-\sup_{x\in\overline{\Phi^{-1} (]-\infty,r[)}^{w}}J(x)}, \frac{r}{\sup_{x\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}J(x)}[,$$ the equation $$\label{e2} \Phi'(u)+\lambda J'(u)=0$$ has at least three solutions in $X$ and, moreover, for each $\eta>1$, there exist an open interval $$\Lambda_2\subseteq[ 0,\frac{\eta r}{r \frac{J(x_1)}{\Phi(x_1)}- \sup_{x\in\overline{\Phi^{-1}(-\infty,r[)}^{w}}J(x)}]$$ and a positive real number $\sigma$ such that, for each $\lambda\in\Lambda_2$, the equation \eqref{e3} has at least three solutions in $X$ whose norms are less than $\sigma$. \end{theorem} Let $K:[0,+\infty[\to \mathbb{R}$ be a continuous function such that there exists a positive number $m$ with $K(t)\geq m$ for all $t\geq 0$, and let $f:[a,b]\times \mathbb{R}\to \mathbb{R}$ be a Carath\'eodory function such that $\sup_{|\xi|\leq s}|f(.,\xi)|\in L^{1}(a,b)$ for all $s>0$. Corresponding to $K$ and $f$ we introduce the functions $\tilde{K}:[0,+\infty[\to \mathbb{R}$ and $F:[a,b]\times \mathbb{R}\to \mathbb{R}$, respectively as follows $$\label{e3} \tilde{K}(t)=\int_{0}^{t}K(s)ds\quad \text{for all } t\geq 0$$ and $$\label{e4} F(x,t)=\int_{0}^{t}f(x,s)ds\quad \text{for all } (x,t)\in [a,b]\times \mathbb{R}.$$ Now, we state our main result. \begin{theorem} \label{thm1} Assume that there exist positive constants $r$ and $\theta$, and a function $w\in X$ such that: \begin{itemize} \item[(i)] $\tilde{K}(\|w\|^2)>2r$, \item[(ii)] $$\int_{a}^b \sup\nolimits_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx < r\frac{\int_a^b F(x,w(x))dx}{r+ \frac{1}{2}\, \tilde{K}(\|w\|^2)},$$ \item[(iii)] $\frac{(b-a)^2}{2m}\limsup_{|t|\to +\infty}\frac{F(x,t)}{t^2}<\frac{1}{\theta}$ uniformly with respect to $x\in [a,b]$. \end{itemize} Further, assume that there exists a continuous function $h:[0,+\infty[\to \mathbb{R}$ such that $h(tK(t^2))=t$ for all $t\geq 0$. Then, for each $\lambda$ in the interval \begin{align*} \Lambda_1&=] \frac{\frac{1}{2}\tilde{K}(\|w\|^2)}{\int_a^b F(x,w(x))dx-\int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx}, \\ &\quad \frac{r}{\int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}},\sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx }[, \end{align*} problem \eqref{e1} admits at least three weak solutions in $X$ and, moreover, for each $\eta>1$, there exist an open interval $$\Lambda_2\subseteq[ 0,\frac{\eta r }{2r\frac{\int_a^b F(x,w(x))dx}{\tilde{K}(\|w\|^2)}- \int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx}]$$ and a positive real number $\sigma$ such that, for each $\lambda\in\Lambda_2$, problem \eqref{e1} admits at least three weak solutions in $X$ whose norms are less than $\sigma$. \end{theorem} Let us first give a particular consequence of Theorem \ref{thm1} for a fixed test function $w$. \begin{corollary} \label{coro1} Assume that there exist positive constants $c$, $d$, $\alpha$, $\beta$ and $\theta$ with $\beta-\alpha\frac{4mc^2}{b-a}$, \item[(ii)] $F(x,t) \geq0$ for each $(x,t)\in ([a, a+\alpha]\cup [b-\beta,b])\times [0,d]$, \item[(iii)] $\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx< \frac{2mc^2}{b-a}\frac{\int_{ a+\alpha}^{b-\beta}F(x,d)dx} {\frac{2mc^2}{b-a}+\frac{1}{2}\tilde{K}(d^2 (\frac{\alpha+\beta}{\alpha\beta}))}$. \end{itemize} Further, assume that there exists a continuous function $h:[0,+\infty[\to \mathbb{R}$ such that $h(tK(t^2))=t$ for all $t\geq 0$. Then, for each $$\lambda\in \Lambda'_1 =]\frac{\frac{1}{2}\tilde{K}(d^2 (\frac{\alpha+\beta}{\alpha\beta}))}{\int_{ a+\alpha}^{b-\beta}F(x,d)dx-\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx },\frac{\frac{2mc^2}{b-a}}{\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx }[,$$ problem \eqref{e1} admits at least three weak solutions in $X$ and, moreover, for each $\eta>1$, there exist an open interval \begin{align*} \Lambda_2 &\subseteq[ 0,\big(\frac{2\eta mc^2}{b-a}\big)\\ &\quad \div \Big(\frac{4mc^2}{b-a}\frac{\int_{a}^{ a+\alpha}F(x,\frac{d}{\alpha}(x-a))dx +\int_{a+\alpha}^{b-\beta}F(x,d)dx +\int_{b-\beta}^b F(x,\frac{d}{\beta}(b-x))dx}{\tilde{K} (d^2(\frac{\alpha+\beta}{\alpha\beta}))}\\ &\quad -\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx\Big)] \end{align*} and a positive real number $\sigma$ such that, for each $\lambda\in\Lambda_2$, problem \eqref{e1} admits at least three weak solutions in $X$ whose norms are less than $\sigma$. \end{corollary} \begin{proof} We claim that the all the assumptions of Theorem \ref{thm1} are fulfilled with \label{e5} w(x)= \begin{cases} \frac{d}{\alpha}(x-a)&\text{if } a\leq x< a+\alpha, \\ d &\text{if } a+\alpha\leq x \leq b-\beta,\\ \frac{d}{\beta}(b-x) &\text{if } b-\beta\frac{4p_1c^2}{b-a}$, \item[(ii)] $$\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx< \frac{2p_1c^2}{b-a}\frac{\int_{ a+\alpha}^{b-\beta}F(x,d)dx}{\frac{2p_1c^2}{b-a} +\frac{p_1}{2}d^2(\frac{\alpha+\beta}{\alpha\beta}) +\frac{p_2}{4}d^{4}(\frac{\alpha+\beta}{\alpha\beta})^2},$$ \item[(iii)]$\frac{(b-a)^2}{2p_1}\limsup_{|t|\to +\infty}\frac{F(x,t)}{t^2}<\frac{1}{\theta}$uniformly with respect to$x\in [a,b]$. \end{itemize} Then, for each $$\lambda\in \Lambda''_1 =]\frac{ \frac{p_1}{2}d^2(\frac{\alpha+\beta}{\alpha\beta}) +\frac{p_2}{4}d^{4}(\frac{\alpha+\beta}{\alpha\beta})^2}{\int_{ a+\alpha}^{b-\beta}F(x,d)dx-\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx}, \frac{\frac{2p_1c^2}{b-a}}{\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx }[,$$ the problem $$\label{e7} \begin{gathered} -(p_1+p_2\int_{a}^b |u'(x)|^2dx)u''=\lambda f(x,u),\\ u(a)=u(b)=0 \end{gathered}$$ admits at least three weak solutions in$X$and, moreover, for each$\eta>1, there exist an open interval \begin{align*} \Lambda_2&\subseteq [0,\big(\frac{2\eta p_1c^2}{b-a}\big)\\ &\quad \div\Big(\frac{4p_1c^2}{b-a}\frac{\int_{a}^{ a+\alpha}F(x,\frac{d}{\alpha}(x-a))dx+\int_{ a+\alpha}^{b-\beta}F(x,d)dx+\int_{b-\beta}^b F(x, \frac{d}{\beta}(b-x))dx}{p_1d^2 (\frac{\alpha+\beta}{\alpha\beta}) +\frac{p_2}{2}d^{4}(\frac{\alpha+\beta}{\alpha\beta})^2}\\ &\quad -\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx\Big)] \end{align*} and a positive real number\sigma$such that, for each$\lambda\in\Lambda_2$, problem \eqref{e7} admits at least three weak solutions in$X$whose norms are less than$\sigma$. \end{corollary} \begin{proof} For fixed$p_1, p_2>0$, set$K(t)=p_1+p_2t$for all$t\geq 0$. Bearing in mind that$m=p_1$, from (i)--(iii), we see that (i)--(iii) of Corollary \ref{coro2} hold respectively. Also we note that there exists a continuous function$h:[0,+\infty[\to \mathbb{R}$such that$h(tK(t^2))=t$for all$t\geq 0$because the function$K$is nondecreasing in$[0,+\infty[$with$K(0)>0$and$t\to tK(t^2)\ (t\geq 0)$is increasing and onto$[0,+\infty[$. Hence, Corollary \ref{coro1} yields the conclusion. \end{proof} \begin{corollary} \label{coro3} Assume that there exist positive constants$c$,$d$,$\alpha$,$\beta$and$\theta$with$\beta-\alpha\frac{4c^2}{b-a}$, \item[(ii)] $$\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx< \frac{2c^2}{b-a}\frac{\int_{ a+\alpha}^{b-\beta}F(x,d)dx}{\frac{2c^2}{b-a} +\frac{d^2}{2}(\frac{\alpha+\beta}{\alpha\beta})},$$ \item[(iii)]$\frac{(b-a)^2}{2}\limsup_{|t|\to +\infty}\frac{F(x,t)}{t^2}<\frac{1}{\theta}$uniformly with respect to$x\in [a,b]$. \end{itemize} Then, for each $$\lambda\in \Lambda'''_1=] \frac{ \frac{d^2}{2}(\frac{\alpha+\beta}{\alpha\beta})}{\int_{ a+\alpha}^{b-\beta}F(x,d)dx-\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx },\frac{\frac{2c^2}{b-a}}{\int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx }[,$$ the problem $$\label{e8} \begin{gathered} -u''=\lambda f(x,u),\\ u(a)=u(b)=0 \end{gathered}$$ admits at least three weak solutions in$X$and, moreover, for each$\eta>1, there exist an open interval \begin{align*} \Lambda_2 &\subseteq[ 0,\big(\frac{2\eta c^2}{b-a}\big)\\ &\quad\div\Big(\frac{4c^2}{b-a}\frac{\int_{a}^{ a+\alpha}F(x,\frac{d}{\alpha}(x-a))dx +\int_{ a+\alpha}^{b-\beta}F(x,d)dx+\int_{b-\beta}^b F(x, \frac{d}{\beta}(b-x))dx}{ d^2(\frac{\alpha+\beta}{\alpha\beta})}\\ &\quad - \int_{a}^b \sup_{t\in[-c,c]}F(x,t)dx\Big)] \end{align*} and a positive real number\sigma$such that, for each$\lambda\in\Lambda_2$, the problem \eqref{e8} admits at least three weak solutions in$X$whose norms are less than$\sigma$. \end{corollary} We conclude this section by giving an example to illustrate our results applying by Corollary \ref{coro2}. \begin{example} \label{examp1} \rm Consider the problem $$\label{e9} \begin{gathered} -(\frac{1}{128}+\frac{1}{64}\int_{0}^{1}|u'(x)|^2dx)u'' =\lambda (e^{-u}u^{11}(12-u)),\\ u(0)=u(1)=0 \end{gathered}$$ where$\lambda>0$. Set$p_1=\frac{1}{128},\ p_2=\frac{1}{64}$and$f(x,t)=e^{-t}t^{11}(12-t)$for all$(x,t)\in [0,1]\times \mathbb{R}$. A direct calculation yields$F(x,t)=e^{-t}t^{12}$for all$(x,t)\in [0,1]\times \mathbb{R}$. Assumptions (i) and (ii) of Corollary \ref{coro2} are satisfied by choosing, for example$d=2$,$c=1$,$[a,b]=[0,1]$and$\alpha=\beta=1/4$. Also, since$\limsup_{|t|\to +\infty}\frac{F(x,t)}{t^2}=0$, Assumption (iii) of Corollary \ref{coro2} is fulfilled. Now we can apply Corollary \ref{coro2}. Then, for each $$\lambda\in \Lambda''_1=] \frac{ 33}{2^{14}e^{-2}-8e },\frac{1}{64 e }[$$ problem \eqref{e9} admits at least three weak solutions in$H^{1}_{0}([0,1])$and, moreover, for each$\eta>1$, there exist an open interval $$\Lambda_2\subseteq[ 0,\frac{\eta }{\frac{8}{33}\left(8^{12}\int_{0}^{\frac{1}{4}}e^{-8t}t^{12}dt+2^{11}e^{-2}+ 8^{12}\int_{\frac{3}{4}}^{1}e^{-8(1-t)}(1-t)^{12}dt\right)-64e}]$$ and a positive real number$\sigma$such that, for each$\lambda\in\Lambda_2$, problem \eqref{e9} admits at least three weak solutions in$H^{1}_{0}([0,1])$whose norms are less than$\sigma$. \end{example} \section{Proof of Theorem \ref{thm1}} We begin by setting \begin{gather} \label{e10} \Phi(u)=\frac{1}{2}\tilde{K}(\|u\|^2),\\ \label{e11} J(u)=\int_{a}^b F(x,u(x))dx \end{gather} for each$u \in X$, where$\tilde{K}$and$F$are given in \eqref{e3} and \eqref{e4}, respectively. It is well known that$J$is a G\^{a}teaux differentiable functional whose G\^{a}teaux derivative at the point$u\in X$is the functional$J'(u)\in X^{*}$, given by $$J'(u)v=\int_{a}^b f(x,u(x))v(x)dx$$ for every$v\in X$, and that$J':X \to X^{*}$is a continuous and compact operator. Moreover,$\Phi$is a continuously G\^{a}teaux differentiable and sequentially weakly lower semi continuous functional whose G\^{a}teaux derivative at the point$u\in X$is the functional$\Phi'(u)\in X^{*}$, given by $$\Phi'(u)v=K(\int_{a}^b | u'(x)|^2dx)\int_{a}^b u'(x) v'(x)dx$$ for every$v\in X$. We claim that$\Phi'$admits a continuous inverse on$X$(we identity$X$with$X^{*}$). To prove this fact, arguing as in \cite{R2} we need to find a continuous operator$T:X\to X$such that$T(\Phi'(u))=u$for all$u\in X$. Let$T:X\to X$be the operator defined by $$T(v)=\begin{cases} \frac{h(\|v\|)}{\|v\|}v & \text{if } v\neq 0\\ 0 &\text{if } v= 0, \end{cases}$$ where$h$is defined in the statement of Theorem \ref{thm1}. Since,$h$is continuous and$h(0)=0$, we have that the operator$T$is continuous in$X$. For every$u\in X$, taking into account that$\inf_{t\geq0}K(t)\geq m>0$, we have since$h(t\,K(t^2))=t$for all$t\geq 0that \begin{align*} T(\Phi'(u))&=T(K(\|u\|^2)u)\\ &=\frac{h(K(\|u\|^2)\|u\|)}{K(\|u\|^2)\|u\|}K(\|u\|^2)u\\ &=\frac{\|u\|}{K(\|u\|^2)\|u\|}K(\|u\|^2)u=u, \end{align*} so our claim is true. Moreover, sincem\leq K(s)$for all$s\in [0,+\infty[$, from \eqref{e10} we have $$\label{e12} \Phi(u)\geq \frac{m}{2}\|u\|^2\quad \text{for all } u\in X.$$ Furthermore from (iii), there exist two constants$\gamma, \tau\in \mathbb{R}$with$0<\gamma<1/\theta$such that $$\frac{(b-a)^2}{2m}F(x,t)\leq \gamma t^2+\tau \quad \text{for all } x\in (a,b) \text{ and all } t\in \mathbb{R}.$$ Fix$u\in X$. Then $$\label{e13} F(x,u(x))\leq \frac{2m}{(b-a)^2}(\gamma |u(x)|^2+\tau)\quad \text{for all }x\in (a,b).$$ Fix$\lambda \in ]0,+\infty[$. Then there exists$\theta>0$with$\lambda \in ]0, \theta]. Now since $$\label{e14} \max_{x\in [a,b]}|u(x)|\leq \frac{(b-a)^{1/2}}{2}\|u\|,$$ from \eqref{e12}, \eqref{e13} and \eqref{e14}, we have \begin{align*} \Phi(u)-\lambda J(u) &= \frac{1}{2}\tilde{K}(\|u\|^2)-\lambda\int^b _{a}F(x,u(x))dx\\ &\geq \frac{m}{2}\|u\|^2- \frac{2\theta m}{(b-a)^2} \Big(\gamma \int^b _{a}|u(x)|^2+\tau(b-a)\Big)\\ &\geq \frac{m}{2}\|u\|^2-\frac{2\theta m}{(b-a)^2}\Big(\gamma \frac{(b-a)^2}{4}\|u\|^2+\tau(b-a)\Big)\\ &= \frac{m}{2}(1-\gamma\theta)\|u\|^2-\frac{2\theta \tau m}{b-a}, \end{align*} and so $$\lim_{\| u\|\rightarrow+\infty} (\Phi(u)-\lambda J(u))=+\infty.$$ Also from \eqref{e10} and (i) we have\Phi(w)>r. Using \eqref{e12} and \eqref{e14}, we obtain \begin{align*} \Phi^{-1}(]-\infty,r[) &=\big\{ u\in X; \Phi(u)< r\big\}\\ &\subseteq \big\{ u\in X; \|u\|< \sqrt{2r/m}\big\}\\ &\subseteq \big\{ u\in X; |u(x)|\leq \sqrt{r(b-a)/(2m)}, \text{ for all }\ x\in [a,b]\big\}, \end{align*} so, we have $$\sup\nolimits_{\overline{u\in\Phi^{-1}(]-\infty,r[)}^{w}}J(u) \leq \int_{a}^b \sup\nolimits_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx.$$ Therefore, from (ii), we have \begin{align*} \sup\nolimits_{\overline{u\in\Phi^{-1}(]-\infty,r[)}^{w}}J(u) &\leq \int_{a}^b \sup\nolimits_{t\in[-\sqrt{\frac{r(b-a)}{2m}},\sqrt{\frac{r(b-a)}{2m}}]}F(x,t) dx\\ &< \frac{r}{r+ \frac{1}{2}\tilde{K}(\|w\|^2)}\int_{a}^b F(x,w(x))dx \\ &= \frac{r}{r+\Phi(w)}J(w). \end{align*} Now, we can apply Theorem \ref{thmA}. Note for eachx\in [a,b], $$\frac{r}{\sup_{u\in\overline{\Phi^{-1}(]-\infty,r[)}^{w}}J(u)}\geq \frac{r}{\int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}},\sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx }$$ and \begin{align*} &\frac{\Phi(w)}{J(w)-\sup_{u\in\overline{\Phi^{-1} (]-\infty,r[)}^{w}}J(u)}\\ &\leq \frac{\frac{1}{2}\tilde{K}(\|w\|^2)}{\int_a^b F(x,w(x))dx-\int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx}. \end{align*} Note also that (ii) immediately implies \begin{align*} &\frac{\frac{1}{2}\tilde{K}(\|w\|^2)}{\int_a^b F(x,w(x))dx-\int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx}\\ &< \frac{\frac{1}{2}\tilde{K}(\|w\|^2)}{ \big( \frac{r+ \frac{1}{2}\tilde{K}(\|w\|^2)}{r} \,-\,1 \big) \int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx} \\ & = \frac{r}{\int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx}. \end{align*} Also \begin{align*} &\frac{\eta r}{r\frac{J(w)}{\Phi(w)}- \sup_{u\in\overline{\Phi^{-1}(-\infty,r[)}^{w}}J(u)}\\ &\leq\frac{\eta r }{2r\frac{\int_a^b F(x,w(x))dx} {\tilde{K}(\|w\|^2)}- \int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx}=\rho. \end{align*} Note from (ii) that \begin{align*} & 2r\frac{\int_a^b F(x,w(x))dx}{\tilde{K}(\|w\|^2)}- \int_{a}^b \sup_{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx \\ & > \Big( \frac{2r}{\tilde{K}(\|w\|^2)} - \frac{r}{ r+ \frac{1}{2}\tilde{K}(\|w\|^2)} \Big) \int_{a}^b F(x,w(x))dx \\ & \geq \Big( \frac{2r}{\tilde{K}(\|w\|^2)} - \frac{2r}{ \tilde{K}(\|w\|^2)} \Big) \int_a^b F(x,w(x))dx=0 \end{align*} since\int_a^b F(x,w(x))dx \geq 0$(note$F(x,0)=0$so $$\int_{a}^b \sup\nolimits _{t\in[-\sqrt{\frac{r(b-a)}{2m}}, \sqrt{\frac{r(b-a)}{2m}}]}F(x,t)dx \geq 0$$ and now apply (ii). 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