\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 14, pp. 1--10.\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-2010/14\hfil Multiple sign-changing solutions] {Multiple sign-changing solutions for sub-linear impulsive three-point boundary-value problems} \author[G. Bao, X. Xu\hfil EJDE-2010/14\hfilneg] {Gui Bao, Xian Xu} % in alphabetical order \address{Department of Mathematics, Xuzhou Normal University, Xuzhou, Jiangsu, 221116, China} \email[Gui Bao]{baoguigui@163.com} \email[Xian Xu]{xuxian68@163.com} \thanks{Submitted August 26, 2009. Published January 21, 2010.} \subjclass[2000]{34B15, 34B25} \keywords{Impulsive three-point boundary-value problem; \hfill\break\indent Leray-Schauder degree; sign-changing solution; strict upper and lower solutions} \begin{abstract} In this article, we study the existence of sign-changing solutions for some second-order impulsive boundary-value problem with a sub-linear condition at infinity. To obtain the results we use the Leray-Schauder degree and the upper and lower solution method. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \section{Introduction} This article concerns the impulsive differential equation $$\begin{gathered} y''(t )+f(t,y(t),y'(t))=0 , \quad t\in J, \; t\neq t_k ,\\ \Delta y'|_{t=t_k}= \bar{I}_k (y(t_k)), \quad k = 1,2,\dots,m,\\ y(0)=0 , \quad y(1)=\alpha y(\eta) , \end{gathered}\label{e1.1}$$ where $J = [0,1]$, $f \in C[J\times \mathbb{R}^2,\mathbb{R}^1]$, $\bar{I}_k \in C[ \mathbb{R}^{1},\mathbb{R}^{1}]$, $k = 1,2,\dots,m$, $0\leq \alpha<1$, $0=t_00, \quad t\neq t_k,\\ \Delta u'|_{t=t_k}> \bar{I}_k (u(t_k)), \quad k = 1,2,\dots,m,\\ u(0)< 0 , \quad u(1)-\alpha u(\eta)< 0 . \end{gathered} \label{e2.1} A function$v\in PC^1[J,\mathbb{R}^1]\cap C^2[J',\mathbb{R}^1]$is called a strict upper solution of \eqref{e1.1}, if $$\begin{gathered} v''(t)+f(t,v(t),v'(t))<0, \quad t\neq t_k,\\ \Delta v'|_{t=t_k}< \bar{I}_k (v(t_k)), \quad k = 1,2,\dots,m,\\ v(0)> 0 , \quad v(1)-\alpha v(\eta)> 0 .\\ \end{gathered} \label{e2.2}$$ \end{definition} Let us introduce the following constants: $$\begin{gathered} \beta= \limsup_{|x|+|y|\to \infty}\max_{t\in J}\frac{|f(t,x,y)|}{|x|+|y|},\\ \bar{\beta_k}=\limsup_{|x|\to \infty}\frac{|\bar{I}_k(x)|}{|x|}, \quad k=1,2,\dots m, \\ \gamma =\frac{4}{1-\alpha\eta}(2\beta +\sum_{k=1}^{m}\bar{\beta}_k). \end{gathered}\label{e2.3}$$ To state the main results in this paper we need the following assumptions: \begin{itemize} \item[(H1)] For each$k\in \{1,2,\dots,m\}$,$\bar{I}_k(0)=0$and $$\lim_{x\to 0}\frac{\bar{I}_k(x)}{x}=d_0>0.$$ \item[(H2)]$f:[0,1]\times \mathbb{R}^2 \to \mathbb{R}^1$is continuous,$f(t, 0, 0)=0$and $$\lim_{x\to 0}\frac{f(t,x,y)}{x}=d_1<0,$$ uniformly for$t\in[0,1]$. \end{itemize} From \cite[Lemma 5.4.1]{g1}, we have the following result. \begin{lemma} \label{lem2.1}$H\subset PC^1[J,\mathbb{R}^1]$is a relatively compact set if and only if for any$x\in H$,$x(t)$and$x'(t)$are uniformly bounded on$J$and equicontinuous at any$ J_k (k=1,2,\dots,m)$, where$J_1=[0,t_1]$,$J_i=(t_{i-1},t_i]$,$ i=2,3,\dots,m+1$. \end{lemma} Now we define the operator$A: PC^1[J,\mathbb{R}^1]\to PC^1[J,\mathbb{R}^1]as follows: \begin{align*} &(Ax)(t)\\ &=\frac{t}{1-\alpha\eta}\int_0^1(1-s)f(s,x(s),x'(s))ds -\frac{\alpha t}{1-\alpha\eta}\int_0^{\eta}(\eta-s)f(s,x(s),x'(s))ds \\ &\quad - \int_0^t(t-s)f(s,x(s),x'(s))ds +\sum_{00 large enough such that $$\deg(I-A,\Omega,\theta)=1,$$ where $\Omega= \{ x \in B(\theta,R_0) : \sigma_1\prec x \}$, $\sigma_1 (t)= \sup_{t\in J}\{u_1(t),u_2(t)\}$. \end{theorem} \begin{proof} If we let $I_k=0$ in the proof of \cite[Theorem 2.1]{x1}, we can easily get this theorem by slight modification. But for the completeness of this paper we will give details of the proof of this theorem. For $0\leq \gamma<1$, we take $\beta'>\beta$, $\bar{\beta}_k'>{\bar{\beta}}_k$, $(k =1,2,\dots,m)$ with $$\gamma ':= \frac{4}{1-\alpha\eta} (2\beta'+\sum_{k=1}^{m}\bar{\beta'}_k) <1.\label{e2.4}$$ From the definition of $\beta$, there exists $N>0$, such that $$|f(t,x,y)|<\beta'(|x|+|y|), \quad \forall t \in J,|x|+|y|\geq N,$$ and so $$|f(t,x,y)|\leq \beta'(|x|+|y|)+ M,\quad \forall t \in J, \; x,y \in \mathbb{R}^1, \label{e2.5}$$ where $M=\sup_{(t,x,y)\in J\times \mathbb{R}^2,\, |x|+|y|\leq N} |f(t,x,y)|$. Similarly, we have $$|\bar{I}_k(x)|\leq \bar{\beta'}_k|x|+ \overline{M}_k , \quad \forall x\in \mathbb{R}^1 ,\label{e2.6}$$ where $\overline{M}_k$ is a positive constant. Take $$R_0 > \max\{\|u_1\|_{PC^1},\|u_2\|_{PC^1}, \frac{1}{1-\gamma'}\frac{4}{1-\alpha\eta} (M+\sum_{k=1}^{m}\overline{M}_k)\}.\label{e2.7}$$ Let $\sigma_1 (t)=\sup_{t\in J}\{u_1(t),u_2(t)\}$ for all $t\in J$. Then $\sigma_1\in PC[J,\mathbb{R}^1]$. Now we define $h_1:J \times \mathbb{R}^2 \to \mathbb{R}^1$, $\bar{J}_{k,1}:\mathbb{R}^1\to \mathbb{R}^1$, $(k=1,2,\dots,m)$ as follows: $$h_1(t,x,y)=\begin{cases} f(t,\sigma_1(t),y), & x<\sigma_1(t),\\ f(t,x,y), & x\geq \sigma_1(t), \end{cases}\label{e2.8}$$ $$\bar{J}_{k,1}(x)=\begin{cases} \bar{I}_k(\sigma_1(t_k)), & x<\sigma_1(t_k),\\ \bar{I}_k(x), & x\geq\sigma_1(t_k). \end{cases}\label{e2.9}$$ Define the nonlinear operator $A_1:PC^1[J,\mathbb{R}^1]\to PC^1[J,\mathbb{R}^1]$ as follows: \begin{align*} &(A_1x)(t)\\ &= \frac{t}{1-\alpha\eta}\int_0^1(1-s)h_1(s,x(s),x'(s))ds -\frac{\alpha t}{1-\alpha\eta}\int_0^{\eta}(\eta-s)h_1(s,x(s),x'(s))ds \\ &\quad - \int_0^t(t-s)h_1(s,x(s),x'(s))ds +\sum_{00$small enough such that$[\tau_0-\delta_0,\tau_0+\delta_0]\subset(t_{k_0-1},t_{k_0})$and$\sigma_1(t)=u_1(t)$for all$t\in [\tau_0-\delta_0,\tau_0+\delta_0]$. Then$\omega(t)=u_1(t)-x_0(t)$for all$t\in [\tau_0-\delta_0,\tau_0+\delta_0] $. Thus,$\omega\in C^2[\tau_0-\delta_0,\tau_0+\delta_0]$and$\omega(\tau_0)$is a local maximum of$\omega$in$[\tau_0-\delta_0,\tau_0+\delta_0]$. Therefore$\omega'(\tau_0)=0$,$\omega''(\tau_0)\leq 0and so \begin{align*} 0&\geq \omega''(\tau_0)=u_1''(\tau_0)-x_0''(\tau_0)\\ &=u_1''(\tau_0)+h_1(\tau_0,x_0(\tau_0),x_0'(\tau_0))\\ &= u_1''(\tau_0)+f(\tau_0,u_1(\tau_0),u_1'(\tau_0))>0, \end{align*} which is a contradiction. For case (3B), let\omega_1(t)=u_2(t)-x_0(t)$for all$t\in (t_{k_0-1},t_{k_0})$. For$t'\in (t_{k_0-1},t_{k_0}), we have \begin{align*} \omega_1(\tau_0) &= u_2(\tau_0)-x_0(\tau_0)\\ &=\sigma_1(\tau_0)-x_0(\tau_0) = \omega(\tau_0)\\ &\geq\omega(t')=\sigma_1(t')-x_0(t')\\ &\geq u_2(t')-x_0(t')=\omega_1(t'). \end{align*} Then\omega_1(\tau_0)$is a local maximum of$\omega_1$in$(t_{k_0-1},t_{k_0})$. Thus$\omega_1'(\tau_0)=0$,$\omega_1''(\tau_0)\leq 0. Therefore \begin{align*} 0&\geq \omega_1''(\tau_0) =u_2''(\tau_0)-x_0''(\tau_0)\\ &=u_2''(\tau_0)+h_1(\tau_0,x_0(\tau_0),x_0'(\tau_0))\\ &= u_2''(\tau_0)+f(\tau_0,u_2(\tau_0),u_2'(\tau_0))>0, \end{align*} which is a contradiction. (4) There existsk_0\in\{1,2,\dots,m\}$such that$\omega(t_{k_0})=\sup_{t \in J}\omega(t)\geq 0$. We take$\delta_0>0$small enough such that$\omega(t_{k_0})$is a local maximum of$\omega(t)$in$[t_{k_0}-\delta_0,t_{k_0}+\delta_0]$, then we have$\omega'(t_{k_0})\geq0$and$\omega'(t^+_{k_0})\leq 0. Thus, \begin{align*} 0&\geq \omega'(t_{k_0}^+)=u_1'(t_{k_0}^+)-x_0'(t_{k_0}^+)\\ &>[u_1'(t_{k_0})+\bar{I}_{k_0}(u_1(t_{k_0}))] -[x_0'(t_{k_0})+\bar{J}_{k_0,1}(x_0(t_{k_0}))]\\ &= u_1'(t_{k_0})-x_0'(t_{k_0})\\ &=\omega'(t_{k_0})\geq0, \end{align*} which is a contradiction. From the discussion of cases (1)-(4), we see that \eqref{e2.14} holds. Since\Omega=\{x\in B(\theta,R_0)|\sigma_1\prec x \}$, it follows that$\Omega\subset PC^1[J,\mathbb{R}^1] $is an open set. We see from \eqref{e2.12} \eqref{e2.14} and the properties of topological degree that $$\deg(I-A_1,\Omega,\theta)=1.$$ Notice that$A_1x=Ax$for all$x\in \overline{\Omega}$, and so we have $$\deg(I-A,\Omega,\theta)=1 .$$ This completes the proof. \end{proof} \begin{corollary} \label{coro2.1} Assume that$u_1$is a strict lower solution of \eqref{e1.1},$0\leq \gamma<1$, then there exists$R_0>0$large enough such that $$\deg(I-A,\Omega,\theta)=1,$$ where$\Omega= \{ x \in B(\theta,R_0) : u_1 \prec x \}$. \end{corollary} Also we have the following Theorems. \begin{theorem} \label{thm2.2} Assume that$v_1$and$v_2$are two strict upper solutions of \eqref{e1.1},$0\leq \gamma<1$, then there exists$R_0>0$large enough such that $$\deg(I-A,\Omega,\theta)=1,$$ where$\Omega$=$ \{ x \in B(\theta,R_0): x\prec\sigma_2 \}$,$\sigma_2 (t)=\inf_{t\in J}\{v_1(t),v_2(t)\}$. \end{theorem} \begin{corollary} \label{coro2.2} Assume that$v_1$is a strict upper solution of \eqref{e1.1},$0\leq \gamma<1$, then there exists$R_0>0$large enough such that $$\deg(I-A,\Omega,\theta)=1,$$ where$\Omega= \{ x \in B(\theta,R_0): x\prec v_1 \}$. \end{corollary} \begin{theorem} \label{thm2.3} Assume that$u_1$is a strict lower solution and$v_1$is a strict upper solution of \eqref{e1.1},$0\leq \gamma<1$, then there exist$R_0>0$large enough such that $$\deg(I-A,\Omega,\theta)=1,$$ where$\Omega=\{x\in B(\theta,R_0):u_1\prec x\prec v_1 \}$. \end{theorem} \section{Main Results} \begin{theorem} \label{thm3.1} Assume that {\rm (H1), (H2)} are satisfied,$0\leq \gamma<1$and \eqref{e1.1} has a strict lower solution$u_1$and a strict upper solution$v_1$, such that$u_1\prec v_1$and$u_1$,$v_1$are sign-changing on$[0,1]$. Then \eqref{e1.1} has at least four sign-changing solutions, two positive solutions and two negative solutions. \end{theorem} \begin{proof} From (H2), there exists$0<\varepsilon_00, \quad f(t,\varepsilon,0)<0, \quad \forall t\in[0,1], \; \forall\varepsilon\in(0,\varepsilon_0). $$Let u_{1,i}(t)=-1/i, v_{1,j}(t)=\frac{1}{j}, i,j=1,2,\dots. Then there exists a natural number n_0>\frac{1}{\varepsilon_0} such that$$ u_{1,i}\not\preceq v_1, \quad u_1 \not\preceq v_{1,j}, $$for each i,j\geq n_0. Since u_{1,i}(t)=-\frac{1}{i}<0, it follows that u_{1,i}(t_k)=-1/i<0, k=1,2,3,\dots ,m. By (H1) and (H2), we can easily show that \begin{gather*} u_{1,i}''(t)+f(t,u_{1,i}(t),u_{1,i}'(t))>0, \quad t\neq t_k,\\ \Delta u_{1,i}'|_{t=t_k}> \bar{I}_k (u_{1,i}(t_k)), \quad k = 1,2,\dots,m,\\ u_{1,i}(0)< 0 , \quad u_{1,i}(1)-\alpha u_{1,i}(\eta)< 0 . \end{gather*} So, u_{1,i}(t) is a strict lower solution of \eqref{e1.1}. Similarly, we know v_{1,j} is a strict upper solution of \eqref{e1.1}. Take u_{1,n_0} and v_{1,n_0}, let \begin{gather*} O_1 = \{x \in B(\theta,R_0)| u_1\prec x\}, \quad O_2 = \{x \in B(\theta,R_0)|x\prec v_1\},\\ O_3 = \{ x\in B(\theta,R_0)|u_{1,n_0}\prec x\} ,\quad O_4 = \{ x\in B(\theta,R_0)|x\prec v_{1,n_0}\},\\ \Omega_1=O_1\setminus (\overline{O_1\cap O_2}) \cup (\overline{O_1\cap O_3}),\quad \Omega_2=O_2\setminus (\overline{O_1\cap O_2})\cup (\overline{O_2\cap O_4}),\\ \Omega_3=O_3\setminus (\overline{O_1\cap O_3})\cup (\overline{O_3\cap O_4}), \quad \Omega_4=B(\theta,R_0)\setminus (\overline{O}_1\cup\overline{O}_4\cup\overline{\Omega}_2 \cup\overline{\Omega}_3). \end{gather*} From Theorems \ref{thm2.1}-\ref{thm2.3} and Corollaries \ref{coro2.1}-\ref{coro2.2}, we have \begin{gather} \deg(I-A,O_1,\theta)=1, \label{e3.1}\\ \deg(I-A,O_2,\theta)=1, \label{e3.2}\\ \deg(I-A,O_1\cap O_2,\theta)=1,\label{e3.3}\\ \deg(I-A,O_1\cap O_3,\theta)=1,\label{e3.4}\\ \deg(I-A,O_2\cap O_4,\theta)=1.\label{e3.5} \end{gather} Thus, \begin{gather} \deg(I-A,\Omega_1,\theta)=1-1-1=-1,\label{e3.6}\\ \deg(I-A,\Omega_2,\theta)=1-1-1=-1.\label{e3.7}\end{gather} So, there exist x_1\in O_1\cap O_2, x_2\in \Omega_1, x_3\in \Omega_2, which are sign-changing solutions of \eqref{e1.1}. From Corollaries \ref{coro2.1}-\ref{coro2.2} and Theorem \ref{thm2.3}, we have \begin{gather} \deg(I-A,O_3,\theta)=1,\label{e3.8}\\ \deg(I-A,O_4,\theta)=1,\label{e3.9}\\ \deg(I-A,O_3\cap O_4,\theta)=1.\label{e3.10} \end{gather} Thus, from \eqref{e3.4}, \eqref{e3.7} and \eqref{e3.9}, we have $$\deg(I-A,\Omega_3,\theta)=1-1-1=-1.\label{e3.11}$$ From the proof of \eqref{e2.12}, it is easy to get \begin{gather} \deg(I-A_,B(\theta,R_0),\theta)=1.\label{e3.12} \end{gather} Then we have from \eqref{e3.1}, \eqref{e3.7}, \eqref{e3.9}, \eqref{e3.11} and \eqref{e3.12} that$$ \deg(I-A,\Omega_4,\theta)=1-1-1-(-1)-(-1)=1. $$So, there exists a fourth sign-changing solution x_4\in \Omega_4. By \eqref{e3.4}, we can get a solution x_{5,i}\in O_1\cap O_3 for i\geq n_0. From \|x_{5,i}\|=\| Ax_{5,i}\|0$$ and $$D_4=\{(t,x,y)\in \mathbb{R}^3:d((t,x,y),\widetilde{D}_4)\leq r_0\}.$$ Define the function $\widetilde{f}$ by $$\widetilde{f}(t,x,y)=\begin{cases} 30, & (t,x,y)\in D_1,\\ -30, & (t,x,y)\in D_2,\\ 1, & (t,x,y)\in D_3,\\ \frac{1}{100}(-x+y), & (t,x,y)\in D_4. \end{cases}$$ From Dugundji's extension theorem, see \cite{g2}, there exists a continuous function $f:[0,1]\times \mathbb{R}^2\mapsto \mathbb{R}^1$ such that $f(t,x,y)=\widetilde{f}(t,x,y)$ while $(t,x,y)\in D_i$ for each $i=1,2,3,4$, and $f([0,1]\times \mathbb{R}^2)\subset \widetilde{f}([0,1]\times \mathbb{R}^2)\subset B(\theta,100)$. Consider the impulsive three-point boundary-value problem $$\begin{gathered} y''(t )+f(t,y(t),y'(t))=0 , \quad t\in J, \; t\neq t_k ,\\ \Delta y'|_{t=t_k}= \bar{I}_k (y(t_k)), \quad k = 1,2,\\ y(0)=0 , \quad y(1)=\alpha y(\eta) ,\\ \end{gathered} \label{e3.13}$$ where $t_1=\frac{1}{10}$, $t_2=\frac{2}{3},\alpha=\frac{1}{2}$, $\eta=\frac{3}{4}$ and $\bar{I}_k(x)=\frac{1}{50k}x, k=1,2$. From the definition of $u_1(t)$ and $f$ we have $$u_1''(t)+f(t,u_1(t),u_1'(t))=-\frac{9}{4}\pi^2\sin\frac{3}{2}\pi t +\widetilde{f}(t,u_1(t),u_1'(t))>-\frac{9}{4}\pi^2+30>0,$$ for all $t\in[0,1]$, \begin{gather*} \bar{I}_1(u_1(t_1))=\frac{1}{50}(\sin\frac{3}{20}\pi-\frac{1}{2})<0 =\Delta u'_1|_{t=t_1},\\ \bar{I}_2(u_1(t_2))=\frac{1}{100}(\sin\pi-\frac{1}{2})<0 =\Delta u'_1|_{t=t_2},\\ u_1(0)<0, \quad \alpha u_1(\eta)=\frac{1}{2}(\sin\frac{9}{8}\pi -\frac{1}{2})>-\frac{3}{2}=u_1(1). \end{gather*} Then $u_1(t)$ is a strict lower solution of \eqref{e3.12}. Similarly, $v_1(t)$ is a strict upper solution of \eqref{e3.12}. From $$\lim_{x\to 0}\frac{\bar{I}_k(x)}{x}=\frac{1}{50k}>0, \quad \bar{I}_k(0)=0, \quad k=1,2,$$ we see that (H1) holds. Next note $$\lim_{x\to0}\frac{f(t,x,0)}{x} =\lim_{x\to0}\frac{\widetilde{f}(t,x,0)}{x}=-\frac{1}{100}<0,f(t,0,0)=0,$$ uniformly for $t\in[0,1]$, then (H2) holds. Since \begin{gather*} \beta= \limsup_{|x|+|y|\to \infty} \max_{t\in J}\frac{|f(t,x,y)|}{|x|+|y|}=0, \\ \bar{\beta_k}=\limsup_{|x|\to \infty} \frac{|\bar{I}_k(x)|}{|x|}=\frac{1}{50k}, \quad k=1,2, \end{gather*} it follows that $$\gamma =\frac{4}{1-\alpha\eta}(2\beta +\bar{\beta} _1+\bar{\beta}_2)=\frac{24}{125}<1.$$ Now all conditions of Theorem \ref{thm3.1} hold. 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