\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 13, 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 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/13\hfil Multiplicity of solutions] {Multiplicity of solutions for some fourth-order m-point boundary-value problems} \author[H. Li, Y. Liu\hfil EJDE-2010/13\hfilneg] {Haitao Li, Yansheng Liu} % in alphabetical order \address{Haitao Li \newline Department of Mathematics, Shandong Normal University, Jinan, 250014, China} \email{haitaoli09@gmail.com} \address{Yansheng Liu \newline Department of Mathematics, Shandong Normal University, Jinan, 250014, China} \email{yanshliu@gmail.com} \thanks{Submitted December 17, 2009. Published January 21, 2010.} \thanks{Supported by grants 209072 from Key Project of Chinese Ministry of Education, \hfill\break\indent and ZR2009AM006 from the Natural Science Foundation of Shandong Province.} \subjclass[2000]{34B16} \keywords{Fixed point index; Leray-Schauder degree; fourth order; \hfill\break\indent $m$-point boundary-value problems; sign-changing solution} \begin{abstract} Using the theory of the fixed point index in a cone and the Leray-Schauder degree, this paper investigates the existence and multiplicity of nontrivial solutions for a class of fourth order $m$-point boundary-value problems. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Consider the following fourth order $m$-point boundary-value problem \begin{equation} \begin{gathered} u^{(4)}(t) = f(u(t),-u''(t)), \quad t\in (0, 1);\\ u'(0)=0,\quad u(1)=\sum_{i=1}^{m-2}\alpha_{i}u(\eta_{i});\\ u'''(0)=0,\quad u''(1)=\sum_{i=1}^{m-2}\alpha_{i}u''(\eta_i), \end{gathered} \label{e1.1} \end{equation} where $f: \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ is a given sign-changing continuous function, $m \geq 3$, $0<\eta_1<\eta_2<\dots<\eta_{m-2}<1$ and $\alpha_i > 0$ for $i = 1, \dots, m - 2$, with \begin{equation} \sum_{i=1}^{m-2}\alpha_{i}<1.\label{e1.2} \end{equation} The multi-point boundary-value problems for ordinary differential equations arise in many areas of applied mathematics and physics. The existence of solutions of the fourth order two-point boundary-value problems and the second order $m$-point boundary-value problems have been studied intensively because of their interest to physics(see [1,2,6,7,9-11] and [5,8,13,14], resp.). However, to our best knowledge, the multiplicity of nontrivial solutions of the nonlinear multi-point boundary-value problems for fourth order differential equations has not been studied intensively. Recently in [12], Wei and Pang investigated the existence and multiplicity of nontrivial solutions for the following fourth order $m$-point boundary-value problems \begin{equation} \begin{gathered} x^{(4)}(t) = f(x(t),-x''(t)), \quad t\in (0, 1);\\ x(0)=0,\quad x(1)=\sum_{i=1}^{m-2}\alpha_{i}x(\eta_{i});\\ x''(0)=0,\quad x''(1)=\sum_{i=1}^{m-2}\alpha_{i}x''(\eta_i), \end{gathered}\label{e1.3} \end{equation} where $m \geq 3$, $0<\eta_1<\eta_2<\dots<\eta_{m-2}<1$ are constants and $\alpha_i \in (0, 1)$ for $i = 1, \dots, m - 2$ satisfies \eqref{e1.2}. $f: \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ satisfies the following conditions: \begin{itemize} \item[(S0)] The sequence of positive solutions of $$ \sin(\sqrt{s})=\sum_{i=1}^{m-2}\alpha_{i}\sin(\eta_{i}\sqrt{s}) $$ is $0<\lambda_1<\lambda_{2}<\dots<\lambda_n<\lambda_{n+1}<\dots$. \item[(S1)] $f(0,0)=0$; and for $u>0$, $v>0$, $f(u,v)\geq0$; for $u<0$, $v<0$, $f(u,v)\leq0$; for $uv>0$, $f(u,v)$ does not vanished. \item[(S2)] $f(u,v)$ has a continuous partial derivative at the point $(0,0)$, and there exists a positive integer $n_{0}$ such that $\mu_{2n_{0}}<1<\mu_{2n_{0}+1}$, where $\mu_n=\frac{\lambda_n^{2}}{a_{0}+b_{0}\lambda_n}$, $a_{0}=f_{u}'(0,0)>0$, $b_{0}=f_{v}'(0,0)>0$. \item[(S3)] There exist $a_1>0$, $b_1>0$ such that $$ \lim_{|u|+|v|\to +\infty}\frac{|f(u,v)-a_1u-b_1v|}{|u|+|v|}=0, $$ and there exists a positive integer $n_1$ such that $\gamma_{2n_1}<1<\gamma_{2n_1+1}$, where $\gamma_n=\frac{\lambda_n^{2}}{a_1+b_1\lambda_n}$. \item[(S4)] There exists a constant $T>0$ such that $|f(u,v)|0$, $v>0$, $f(u,v)\geq0$; for $u<0$, $v<0$, $f(u,v)\leq0$; for $uv>0$, $f(u,v)$ does not vanish. \item[(H2)] There exist $a_0>0$, $b_0>0$, such that $$ f(u, v)= a_0u+b_0v +o(|(u, v)|),\quad \text{as } |(u, v)|\to 0, $$ where $(u, v)\in\mathbb{R}\times\mathbb{R}$, and $|(u, v)|:=\max\{|x|,\ |y|\}$. And there exists a positive integer $n_{0}$ such that $\mu_{n_{0}}<1<\mu_{n_{0}+1}$, where $\mu_n=\frac{s_n^{2}}{a_{0}+b_{0}s_n}$. \item[(H3)] There exist $a_1>0$, $b_1>0$, such that $$ f(u, v)= a_1u+b_1v +o(|(u, v)|),\quad \text{as } |(u, v)|\to +\infty, $$ where $(u, v)\in\mathbb{R}\times\mathbb{R}$, and $|(u, v)|:=\max\{|x|,\ |y|\}$. And there exists a positive integer $n_1$ such that $\gamma_{n_1}<1<\gamma_{n_1+1}$, where $\gamma_n=\frac{s_n^{2}}{a_1+b_1s_n}$. \item[(H4)] There exists a constant $T>0$ such that $|f(u,v)|0$ such that $i(A,P\cap B(\theta,r),P)=0$ for any $00$ such that $i(A,P\cap B(\theta,R),P)=0$ for any $R>\zeta$. \end{lemma} \begin{lemma}[\cite{g2}] \label{lem2.2} Let $\theta\in\Omega$ and $A:P\cap\overline{\Omega}\to P$ be condensing. Suppose that $Ax\neq\mu x$, for all $x\in P\cap\partial\Omega$ and $\mu\geq 1$. Then $i(A,P\cap\Omega,P)=1$. \end{lemma} We first transform \eqref{e1.1} into another form. Suppose $u(t)$ is a solution of \eqref{e1.1}. Let $v(t)=-u''(t)$. Note that \begin{equation} \begin{gathered} u''(t)+v(t)=0, \quad t\in (0, 1);\\ u'(0)=0,\quad u(1)=\sum_{i=1}^{m-2}\alpha_{i}u^(\eta_i), \end{gathered} \label{e2.1} \end{equation} thus $u(t)$ can be written as \begin{equation} u(t)=Lv(t), \label{e2.2} \end{equation} where the operator $L$ is defined by $Lv(t)=\int_{0}^{1}H(t,s)v(s)ds$, for all $v\in Y$, and \begin{gather*} H(t, s)=G(t,s)+\frac{\sum\limits_{i=1}^{m-2}\alpha_{i}G(\eta_{i}, s)}{1-\sum\limits_{i=1}^{m-2}\alpha_{i}\eta_{i}}t, \\ G(t, s)= \begin{cases} 1-t,& 0\leq s\leq t\leq 1; \\ 1-s, & 0\leq t\leq s\leq 1. \end{cases} \end{gather*} Therefore, we obtain the following equivalent form of \eqref{e1.1}: \begin{equation} \begin{gathered} v''(t)+f((Lv)(t),v(t))=0, \quad t\in (0, 1);\\ v'(0)=0,\quad v(1)=\sum_{i=1}^{m-2}\alpha_{i}v^(\eta_i). \end{gathered} \label{e2.3} \end{equation} Similar to \eqref{e2.1} and \eqref{e2.2}, $v(t)$ can be written as \begin{equation} v(t)=(LF)u(t), \label{e2.4} \end{equation} where $(Fu)(t)=f(u(t)$, $-u''(t))$, $t\in (0, 1)$, for all $u\in E$. From \eqref{e2.2} and \eqref{e2.4} we obtain $u(t)=(L^{2}F)u(t)$. Define $A=L^{2}F$, then it is easy to get the following lemma. \begin{lemma} \label{lem2.3} $u(t)$ is a solution of \eqref{e1.1} if and only if $u(t)$ is a solution of the operator equation \begin{equation} u(t)=Au(t). \label{e2.5} \end{equation} \end{lemma} \begin{lemma} \label{lem2.4} Suppose (H1) holds. Then $A:P\to P $ is completely continuous. \end{lemma} \begin{proof} By the continuity of $f$, it is easy to see that $A: E\to E $ is completely continuous. Suppose $x(t)\in P$, condition (H1) implies $$ Ax(t)=(L^{2}F)x(t)\geq 0,\quad -(Ax)''(t)=(LF)x(t)\geq 0,\quad \forall t\in [0, 1]. $$ Therefore ,$Ax(t)\in P$. \end{proof} \begin{remark} \label{rmk2.1}\rm Similarly to the above, if $f$ satisfies (H1), then $A: -P\to -P $ is completely continuous. \end{remark} Set \begin{gather} Kx(t)=L^{2}x(t), \label{e2.6} \\ Qx(t)=L^{2}(-x'')(t). \label{e2.7} \end{gather} \begin{lemma} \label{lem2.5} \begin{itemize} \item[(i)] $K: C[0,1]\to E $ is a completely continuous linear operator; \item[(ii)] $F: E\to C[0,1]$ is a continuous bounded operator, and $A=KF$; \item[(iii)] $Q: E\to E $ is a completely continuous linear operator; \item[(iv)] the sequences of all eigenvalues of the operators $a_{0}K+b_{0}Q$ and $a_1K+b_1Q$ are $\{\frac{1}{\mu_n}\}$, and $\{\frac{1}{\gamma{n}}\}$, respectively, where $\mu_n$ and $\gamma_n$ are respectively defined by {\rm (H2)} and {\rm (H3)}. \end{itemize} \end{lemma} \begin{proof} Items (i)-(iii) have obvious proofs. To prove (iv), let $\mu$ be a positive eigenvalue of the linear operator $a_{0}K+b_{0}Q$, and $y\in E\setminus \{\theta\}$ be an eigenfunction corresponding to the eigenvalue $\mu$. By \eqref{e2.6} and \eqref{e2.7}, we have \begin{equation} \begin{gathered} \mu y^{(4)} = a_{0}y+b_{0}(-y'');\\ y'(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_{i}y(\eta_i);\\ y'''(0)=0,\quad y''(1)=\sum_{i=1}^{m-2}\alpha_{i}y''(\eta_i). \end{gathered} \label{e2.8} \end{equation} Define $D=\frac{d}{dt}$, $G=\mu D^{4}-a_{0}+b_{0}D^{2}$, then there exist complex constants $r_1, r_{2}$ such that $$ Gu=\mu (D^{2}+r_1)(D^{2}+r_{2})u. $$ By the properties of differential operators, if \eqref{e2.8} has a nonzero solution, then there exists $r_{s}, s\in \{1, 2\}$ such that $r_{s}=s_k, k\in N_{+}$. In this case, $\cos t\sqrt{s_k}$ is a nonzero solution of \eqref{e2.8}. On substituting this solution into \eqref{e2.8}, we have $$ \mu s_k^{2}-(a_{0}+b_{0}s_k)=0. $$ Hence, $\{\frac{a_{0}+b_{0}s_k}{s_k^{2}}=\frac{1}{\mu_k}\}$, $k=1, 2, \dots$ is the sequence of eigenvalues of the operator $a_{0}K+b_{0}Q$. Then $\mu$ is one of the values $$ \frac{1}{\mu_1}>\frac{1}{\mu_{2}}>\dots >\frac{1}{\mu_n}>\dots $$ and the eigenfunction corresponding to the eigenvalue $1/\mu_n$ is $$ y_n(t)=C\cos (t\sqrt{s_n}),\quad t\in [0,1], $$ where $C$ is a nonzero constant. Similarly, we can show that the sequence of eigenvalues of the operator $a_1K+b_1Q$ is $\{1/\mu_n\}$, $n=1, 2,\dots$. \end{proof} \begin{lemma} \label{lem2.6} Suppose (H2) and (H3) hold. Then the operator $A$ is Frechet differentiable at $\theta$ and $\infty$. Moreover, $A'(\theta)=a_{0}K+b_{0}Q$ and $A'(\infty)=a_1K+b_1Q$. \end{lemma} \begin{proof} For any $x\in E$, we have \begin{gather} \begin{aligned}{} [Ax-A\theta-(a_{0}Kx+b_{0}Qx)](t) &=L^{2}[f(x(t),-x''(t))-(\alpha_{0}x(t)-\beta_{0}x''(t))]\\ &=L^{2}Bx(t), \quad \forall t\in [0, 1] \end{aligned} \label{e2.9}\\ [Ax-A\theta-(a_{0}Kx+b_{0}Qx)]'(t) =-\int^{t}_{0}LBx(s)ds, \label{e2.10}\\ [Ax-A\theta-(a_{0}Kx+b_{0}Qx)]''(t)=-LBx(t), \label{e2.11}\\ [Ax-A\theta-(a_{0}Kx+b_{0}Qx)]'''(t)=\int^{t}_{0}Bx(s)ds, \label{e2.12} \end{gather} where $Bx(t)=f(x(t),-x''(t))-(a_{0}x(t)-b_{0}x''(t))$. For each $\varepsilon > 0$, by (H2), there exists a $\delta > 0$ such that for any $0<|u|,\ |v|<\delta$, $$ |\frac{f(u,v)-a_{0}u-b_{0}v}{\sqrt{u^{2}+v^{2}}}|<\varepsilon. $$ This means \begin{equation} |f(u,v)-(a_{0}u+b_{0}v)|<\varepsilon \sqrt{u^{2}+v^{2}},\quad \forall 0<|u|,\; |v|<\delta. \label{e2.13} \end{equation} Then, for any $x\in E$ with $\| x\|<\delta$, by \eqref{e2.9}-\eqref{e2.13}, we get \begin{equation} \|Ax-A\theta-(a_{0}Kx+b_{0}Qx)\| \leq \sqrt{2}M\varepsilon\|x\|. \label{e2.14} \end{equation} Consequently, \[ \lim_{\|x\|\to 0}\frac{\|Ax-A\theta-(a_{0}Kx+b_{0}Qx)\|}{\|x\|}=0. \] Therefore, $A$ is Frechet differentiable at $\theta$, and $A'(\theta)=a_{0}K+b_{0}Q$. For each $\varepsilon > 0$, by (H3), there exists a constant $R_1> 0$ such that $|f(u,v)-a_1u-b_1v|<\varepsilon(|u|+|v|)$, for $|u|+|v|>R_1$. Let \[ b=\max_{|u|+|v|\leq R_1}|f(u,v)-a_1u-b_1v|, \] then we have $$ |f(u,v)-a_1u-b_1v|\leq \varepsilon(|u|+|v|)+b,\quad \forall u,\ v\in\mathbb{R}. $$ By a consideration similar to \eqref{e2.14}, we get $$ \|Ax-(a_1Kx+b_1Qx)\|\leq(2\varepsilon\|x\|+b)M, \quad \forall x\in E. $$ Consequently, $\lim\limits_{\|x\|\to \infty}\frac{\|Ax-(a_1Kx+b_1Qx)\|}{\|x\|}=0$. This implies that $A$ is Frechet differentiable at $\infty$, and $A'(\infty)=a_1K+b_1Q$. \end{proof} \begin{lemma} \label{lem2.7} Suppose that {\rm (H1)--(H3)} hold. Then \begin{itemize} \item[(i)] there exists a constant $r_{0}$ such that $0T$ such that for any $R\geq R_{0}$, $i(A,P\cap B(\theta,R),P)=0$, $i(A,-P\cap B(\theta,R),-P)=0$. \end{itemize} \end{lemma} \begin{proof} We prove conclusion (i) only; conclusion (ii) can be proved in the same way. By Lemma \ref{lem2.6}, $A: P\to P$ is completely continuous and Frechet differentiable along $P$ at $\theta$, and $A'_{+}(\theta)=a_{0}K+b_{0}Q, A\theta=\theta$. By Lemma \ref{lem2.5} and (H2), $A'_{+}(\theta)$ has an eigenvalue $\frac{1}{\mu_1}=\frac{a_{0}+b_{0}s_1}{s_1^{2}}>1$, and $\frac{1}{\mu_1}>\frac{1}{\mu_{2}}>\dots >\frac{1}{\mu_{n_{0}}}>1>\frac{1}{\mu_{n_{0}+1}}>\dots>0$, so $1$ is not an eigenvalue of $A'_{+}(\theta)$ corresponding to a positive eigenfunction. The eigenfunction corresponding to $\frac{1}{\mu_1}$ is $y(t)=\cos t\sqrt{s_1}, t\in[0,1]$, where $s_1$ is the smallest positive solution of the equation $$ \cos (\sqrt{s})=\sum_{i=1}^{m-2}\alpha_{i}\cos (\eta_{i}\sqrt{s}). $$ Since \begin{gather*} \cos (\sqrt{0})-\sum_{i=1}^{m-2}\alpha_{i}\cos 0=1-\sum_{i=1}^{m-2}\alpha_{i}>0,\\ \cos (\sqrt{(\pi/2)^{2}})-\sum_{i=1}^{m-2}\alpha_{i}\cos (\eta_{i}\sqrt{(\pi/2)^{2}}) =-\sum_{i=1}^{m-2}\alpha_{i}\cos (\frac{\pi}{2}\eta_{i})<0. \end{gather*} Then by the mean-value theorem, $s_1\in(0,(\frac{\pi}{2})^{2})$. Consequently $$ y(t)=\cos (t\sqrt{s_1})\geq 0, \quad t\in[0,1]. $$ And then it follows from Lemma \ref{lem2.1} that there exists an $\tau_{0}>0$ such that $i(A,P\cap B(\theta,r),P)=0$ for any $00$ such that $i(A,-P\cap B(\theta,r),-P)=0$ for any $00$ such that $|f(x)|