\input amstex \documentstyle{amsppt} \loadmsbm \magnification=\magstephalf \hcorrection{1cm} \vcorrection{-6mm} \nologo \TagsOnRight \NoBlackBoxes \headline={\ifnum\pageno=1 \hfill\else% {\tenrm\ifodd\pageno\rightheadline \else \leftheadline\fi}\fi} \def\rightheadline{EJDE--2000/65\hfil Positive solutions to a second order boundary-value problem \hfil\folio} \def\leftheadline{\folio\hfil Daomin Cao \& Ruyun Ma \hfil EJDE--2000/65} \def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt % Electronic Journal of Differential Equations, Vol.~{\eightbf 2000}(2000), No.~65, pp.~1--8.\hfil\break ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \hfill\break ftp ejde.math.swt.edu (login: ftp)\bigskip} } \topmatter \title Positive solutions to a second order multi-point boundary-value problem \endtitle \thanks {\it 2000 Mathematics Subject Classifications:} 34B10.\hfil\break\indent {\it Key words:} Multi-point boundary value problem, positive solution, fixed point theorem. \hfil\break\indent \copyright 2000 Southwest Texas State University. \hfil\break\indent Submitted September 18, 2000. Published Ocotber 30, 2000. \hfil\break\indent R.Ma was supported by the Natural Science Foundation of China (grant 19801028) \endthanks \author Daomin Cao \& Ruyun Ma \endauthor \address Daomin Cao \hfill\break\indent Institute of Applied Mathematics, Academy of Mathematics and System Sciences, \hfill\break\indent Beijing 100080, People's Republic of China \endaddress \email cao\@amath6.amt.ac.cn \endemail \address Ruyun Ma \hfill\break\indent Department of Mathematics, Northwest Normal University,\hfill\break\indent Lanzhou 730070, Gansu, People's Republic of China \endaddress \email mary\@mx.amss.ac.cn \endemail \abstract We prove the existence of positive solutions to the boundary-value problem $$ \displaylines{ u''+\lambda a(t)f(u,u')=0 \cr u(0)=0,\quad u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i) \,, }$$ where $a$ is a continuous function that may change sign on $[0,1]$, $f$ is a continuous function with $f(0,0)>0$, and $\lambda$ is a samll positive constant. For finding solutions we use the Leray-Schauder fixed point theorem. \endabstract \endtopmatter \document \head 1. Introduction \endhead The study of multi-point boundary value problems for linear second order ordinary differential equations was initiated by Il'in and Moiseev [8, 9]. Motivated by the study of Il'in and Moiseev [8, 9], Gupta [4] studied certain three point boundary value problems for nonlinear ordinary differential equations. Since then, more general nonlinear multi-point boundary value problems have been studied by several authors using the Leray-Schauder Continuation Theorem, Nonlinear Alternative of Leray-Schauder, coincidence degree theory or fixed point theorem in cones. We refer the reader to [1-3, 5, 10-12] for some existence results of nonlinear multi-point boundary value problems. Recently, the second author[12] proved the existence of positive solutions for the three-point boundary value problem $$ \gather u''+b(t)g(u)=0,\quad t\in (0,1) \tag 1.1\\ u(0)=0,\quad \alpha u(\eta)=u(1)\,, \tag 1.2\\ \endgather$$ where $\eta\in (0,1)$, $0<\alpha<\frac 1{\eta}$, $b\ge 0$, and $g\ge 0$ is either superlinear or sublinear by the simple application of a fixed point theorem in cones. In this paper, we consider the nonlinear eigenvalue $m$-point boundary value problem $$ \gather u''+\lambda a(t)f(u,u')=0 \tag 1.3 \\ u(0)=0,\quad u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i) \tag 1.4 \endgather $$ where $\lambda$ is a positive parameter. \medskip We make the following assumptions: \roster \item"(A1)" $a_i\geq 0$ for $i=1,\cdots, m-3$ and $a_{m-2}>0$; $\xi_i: 0<\xi_1<\xi_2<\dots <\xi_{m-2}<1$ and $\sum^{m-2}_{i=1} a_i \xi_i<1$. \item"(A2)" $f: [0,\infty)\times R\to R$ is continuous and $f(0, 0)>0$; \item"(A3)" $a\in C[0,1]$ and there exist $r_0\in [0,1]$ and $\theta>0$ such that $a(r_0)\neq 0$, and the solution of the linear problem $$\gathered u''+ a^+(t)-(1+\theta)a^-(t)=0,\quad t\in (0,1)\\ u(0)=0,\quad u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i) \endgathered $$ is nonnegative in $[0,1]$, where $a^+$ is the positive part of $a$ and $a^-$ is the negative part of $a$. \item"(A4)" There exist a constant $k$ in $(1,\infty)$ such that $$ P(t)\geq kQ(t) \tag 1.5 $$ where $$ P(t)=\int^t_0 a^+(s)ds +\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^+(s)ds} {1-\sum^{m-2}_{i=1} a_i\xi_i } +\frac {\int^1_0 (1-s)a^+(s)ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } $$ and $$ Q(t)=\int^t_0 a^-(s)ds +\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^-(s)ds} {1-\sum^{m-2}_{i=1} a_i\xi_i } +\frac {\int^1_0 (1-s)a^-(s)ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } $$ \endroster Our main result is \proclaim{Theorem 1} Let (A1), (A2), (A3), and (A4) hold. Then there exists a positive number $\lambda^*$ such that (1.3)-(1.4) has at least one positive solution for $0<\lambda<\lambda^*$. \endproclaim The proof of this theorem is based upon the Leray-Schauder fixed point theorem and motivated by [7]. \head 2. Preliminary lemmas \endhead In the sequel we shall denote by $I$ the interval $[0,1]$ of the real line. $E$ will stand for the space of functions $u:I\to R$ such that $u(0)=0$, $u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i)$ and $u'$ is continuous on $I$. We furnish the set $E$ with the norm $|u|_E=\max\{|u|_0, |u'|_0\}=|u'|_0$, where $|u|_0=\max\{u(t)\;|\;t\in I\}$. Then $E$ is a Banach space. To prove Theorem 1, we need the following preliminary results. \proclaim {Lemma 1 [6]} Let $a_i\geq 0$ for $i=1,\cdots,m-2$, and $\sum^{m-2}_{i=1} a_i \xi_i\neq 1$, then for $y\in C(I)$, the problem $$ \gather u''+y(t)=0,\quad t\in(0,1) \tag 2.1 \\ u(0)=0,\quad u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i) \tag 2.2 \endgather $$ has a unique solution, $$ u(t)=-\int^t_0 (t-s)y(s)ds -t\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)y(s)ds} {1-\sum^{m-2}_{i=1} a_i\xi_i } +t\frac {\int^1_0 (1-s)y(s)ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } $$ \endproclaim The following two results extend Lemma 2 and Lemma 3 of [12]. \proclaim {Lemma 2} Let $a_i\geq 0$ for $i=1,\cdots,m-2$, and $\sum^{m-2}_{i=1} a_i \xi_i < 1$. If $y\in C(I)$ and $y\geq 0$, then the unique solution $u$ of the problem (2.1)-(2.2) satisfies $$u(t)\geq 0,\quad\forall t\in I$$ \endproclaim \noindent{\it Proof} From the fact that $u''(x)=-y(x)\leq 0$, we know that the graph of $u(t)$ is concave down on $I$. So, if $u(1)\geq 0$, then the concavity of $u$ together with the boundary condition $u(0)=0$ implies that $u\geq 0$ for all $t\in I$. If $u(1)<0$, then from the concavity of $u$ we know that $$\frac {u(\xi_i)}{\xi_i}\geq \frac {u(1)}{1},\quad \text{for }i=1,\cdots,m-2 \tag 2.3 $$ This implies $$ u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i)\geq \sum^{m-2}_{i=1} a_i\xi_i u(1) \tag 2.4 $$ This contradicts the fact that $\sum^{m-2}_{i=1} a_i \xi_i < 1$. \proclaim {Lemma 3} Let $a_i\geq 0$ for $i=1,\cdots,m-3$, $a_{m-2}>0$, and $\sum^{m-2}_{i=1} a_i \xi_i>1$. If $y\in C(I)$ and $y(t)\geq 0$ for $t\in I$, then (2.1)-(2.2) has no positive solution. \endproclaim \noindent{\it Proof} Assume that (2.1)-(2.2) has a positive solution $u$, then $u(\xi_i)>0$ for $i=1,\cdots, m-2$, and $$ \aligned u(1)=&\sum^{m-2}_{i=1} a_i u(\xi_i) =\sum^{m-2}_{i=1} a_i\xi_i \frac {u(\xi_i)}{\xi_i}\\ \geq& \sum^{m-2}_{i=1} a_i\xi_i \frac {u(\bar\xi)}{\bar\xi} > \frac {u(\bar\xi)}{\bar\xi} \endaligned \tag 2.5 $$ (where $\bar\xi\in \{\xi_1,\cdots,\xi_{m-2}\}$ satisfies $\frac {u(\bar\xi)}{\bar\xi} =\min\{ \frac {u(\xi_i)}{\xi_i}| i=1,\cdots, m-2\} $). \noindent This contradicts the concavity of $u$. If $u(1)=0$, then applying $a_{m-2}>0$ we know that $$u(\xi_{m-2})=0 \tag 2.6 $$ From the concavity of $u$, it is easy to see that $u(t)\leq 0$ for all $t$ in $I$. \vskip 3mm In the rest of this paper, we assume that $a_i\geq 0$ for $i=1,\cdots,m-3$, $a_{m-2}>0$, and $\sum^{m-2}_{i=1} a_i \xi_i < 1$. We also assume that $f(u,p)=f(0,p)$ for $(u,p)\in (-\infty, 0)$. \proclaim {Lemma 4} Let (A1) and (A2) hold. Then for every $0<\delta<1$, there exists a positive number $\bar\lambda$ such that, for $0<\lambda<\bar \lambda$, the problem $$\gather u''+\lambda a^+(t)f(u,u')=0 \tag 2.7 \\ u(0)=0,\;u(1)=\sum^{m-2}_{i=1} a_i u(\xi_i) \tag 2.8 \endgather $$ has a positive solution $\tilde u_\lambda$ with $|\tilde u_\lambda|_E\to 0$ and $|\tilde u_\lambda'|_0\to 0$ as $\lambda\to 0$ , and $$\tilde u_\lambda\geq\lambda\delta f(0, 0)p(t),\quad t\in I \tag 2.9 $$ where $$ p(t)= -\int^t_0 (t-s)a^+(s)ds -t\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^+(s)ds} {1-\sum^{m-2}_{i=1} a_i\xi_i } +t\frac {\int^1_0 (1-s)a^+(s)ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } $$ \endproclaim \noindent{\it Proof.} By Lemma 2, we know that $p(t)\geq 0$ for $t\in I$. From Lemma 1, (2.7)-(2.8) is equivalent to the integral equation $$ \aligned u(t)=&\lambda\Bigl [ -\int^t_0 (t-s)a^+(s)f(u(s), u'(s))ds\\ &-t\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^+(s)f(u(s), u'(s))ds} {1-\sum^{m-2}_{i=1} a_i\xi_i }\\ &+t\frac {\int^1_0 (1-s)a^+(s)f(u(s),u'(s))ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } \Bigr]\\ &\overset\text{def}\to=Au(t) \endaligned $$ where $u\in C^1(I)$. Further, we have that $$ \aligned (Au)'(t)=&\lambda\Bigl [ -\int^t_0 a^+(s)f(u(s), u'(s))ds\\ &-\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^+(s)f(u(s), u'(s))ds} {1-\sum^{m-2}_{i=1} a_i\xi_i }\\ &+\frac {\int^1_0 (1-s)a^+(s)f(u(s),u'(s))ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } \Bigr] \endaligned \tag 2.10 $$ Then $A:C^1(I)\to C^1(I)$ is completely continuous and fixed points of $A$ are solutions of (2.7)-(2.8). We shall apply the Leray-Schauder fixed point theorem to prove $A$ has a fixed point for $\lambda$ small. Let $\epsilon>0$ be such that $$f(u,y)\geq \delta f(0,0),\quad\text{for } (u,y)\in [0,\epsilon]\times [-\epsilon,\epsilon] \tag 2.11 $$ Suppose that $$\lambda<\frac {\epsilon}{2|P|_0\tilde f(\epsilon)}:=\bar\lambda \tag 2.12 $$ where $\tilde f(r)=\underset{(u,y)\in [0,r]\times [-r,r]}\to\max f(u,y)$. By (A2) we know that $$\underset{r\to 0^+}\to\lim \frac {\tilde f(r)}r=+\infty. \tag 2.13 $$ It follows that there exists $r_\lambda\in (0,\epsilon)$ such that $$\frac {\tilde f(r_\lambda)}{r_\lambda}= \frac 1{2\lambda |P|_0} \tag 2.14 $$ We note that (2.14) implies $$r_\lambda\to 0,\quad \text{as}\; \lambda\to 0 \tag 2.15 $$ Now, consider the homotopy equations $$u=\theta Au,\quad \theta\in (0,1) \tag 2.16 $$ Let $u\in C^1(I)$ and $\theta \in (0,1)$ be such that $u=\theta Au$. We claim that $|u|_E\neq r_\lambda$. In fact, $$ \aligned u'(t)=&\theta \lambda\Bigl [ -\int^t_0 a^+(s)f(u(s),u'(s))ds\\ &-\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^+(s)f(u(s),u'(s))ds} {1-\sum^{m-2}_{i=1} a_i\xi_i }\\ &+\frac {\int^1_0 (1-s)a^+(s)f(u(s), u'(s))ds}{1-\sum^{m-2}_{i=1} a_i\xi_i }\Bigr] \endaligned \tag 2.17 $$ This implies that $$|u'(t)|\leq \lambda \tilde f(|u|_E)P(t),\quad t\in[0,1] \tag 2.18 $$ hence $$|u|_E\le \lambda |P|_0\tilde f(|u|_E) \tag 2.19 $$ or $$\frac {\tilde f(|u|_E)}{|u|_E}\ge \frac 1{\lambda|P|_0} \tag 2.20 $$ which implies that $|u|_E\neq r_\lambda$. Thus by Leray-Schauder fixed point theorem, $A$ has a fixed point $\tilde u_\lambda$ with $$|\tilde u_\lambda|_E \le r_\lambda< \epsilon \tag 2.21 $$ Moreover, combining (2.21) and (2.11) and using (2.10) and Lemma 2, we have that $$\tilde u_\lambda(t)\ge \lambda \delta f(0,0)p(t), \tag 2.20 $$ for $t\in I$, $\lambda\le\bar\lambda$ . \head 3. Proof of the main reuslt \endhead \noindent{\bf Proof of Theorem 1.} Let $$ q(t)=-\int^t_0 (t-s)a^-(s)ds -t\frac {\sum^{m-2}_{i=1} a_i\int^{\xi_i}_0 (\xi_i-s)a^-(s)ds} {1-\sum^{m-2}_{i=1} a_i\xi_i } +t\frac {\int^1_0 (1-s)a^-(s)ds}{1-\sum^{m-2}_{i=1} a_i\xi_i } \tag 3.1 $$ then from Lemma 2, we know that $q(t)\geq 0$. By (A3) and (A4), there exist positive numbers $c,d\in (0,1)$ such that for $t\in I$, $$\gathered q(t)\max\{|f(u,y)|\;|\;0\le u\le c,-c\le y\le c\}\leq d p(t)f(0,0),\\ Q(t)\max\{|f(u,y)|\;|\;0\le u\le c,-c\le y\le c\}\leq d P(t)f(0,0)\,. \endgathered \tag 3.2 $$ Fix $\delta\in (d,1)$ and let $\lambda^*>0$ be such that $$|\tilde u_\lambda|_E+\lambda \delta f(0,0)|P|_0\leq c \tag 3.3 $$ for $\lambda<\lambda^*$, where $\tilde u_\lambda$ is given by Lemma 4, and $$|f(u_1,y_1)-f(u_2, y_2)|\leq f(0,0)\bigl(\frac {\delta-d}2\bigr ) \tag 3.4 $$ for $(u_1,y_1),(u_2,y_2)\in [0,c]\times [-c,c]$ with $$\max\{|u_1-u_2|,|y_1-y_2|\}\le \lambda^*\delta f(0, 0)|P|_0.$$ Let $\lambda<\lambda^*$. We look for a solution $u_\lambda$ of the form $\tilde u_\lambda+v_\lambda$. Here $v_\lambda$ solves $$\gather v''+\lambda a^+(t)(f(\tilde u_\lambda+v,\tilde u_\lambda '+v') -f(\tilde u_\lambda,\tilde u_\lambda ')) -\lambda a^-(t)f(\tilde u_\lambda+v,\tilde u_\lambda '+v')=0 \tag 3.5 \\ v(0)=0,\; v(1)=\sum^{m-2}_{i=1} a_i v(\xi_i) \tag 3.6 \endgather$$ For each $w\in C^1(I)$, let $v=T(w)$ be the solution of $$\gather v''+\lambda a^+(t)(f(\tilde u_\lambda+w,\tilde u_\lambda '+w') -f(\tilde u_\lambda,\tilde u_\lambda ')) -\lambda a^-(t)f(\tilde u_\lambda+w, \tilde u_\lambda '+w')=0 \\ v(0)=0,quad v(1)=\sum^{m-2}_{i=1} a_i v(\xi_i) \endgather $$ Then $T:C^1(I)\to C^1(I)$ is completely continuous. Let $v\in C^1(I)$ and $\theta \in (0,1)$ be such that $v=\theta Tv$. Then we have $$\gather v''+\theta\lambda a^+(t)(f(\tilde u_\lambda+v,\tilde u_\lambda '+v') -f(\tilde u_\lambda, \tilde u_\lambda')) -\theta\lambda a^-(t)(f(\tilde u_\lambda+v,\tilde u_\lambda '+v'))=0 \tag 3.7 \\ v(0)=0,\; v(1)=\sum^{m-2}_{i=1} a_i v(\xi_i) \tag 3.8 \endgather $$ We claim that $|v|_E\neq \lambda\delta f(0,0)|P|_0$. Suppose to the contrary that $|v|_E= \lambda\delta f(0,0)|P|_0$. Then by (3.3), we obtain $$\gathered |\tilde u_\lambda+v|_E\leq |\tilde u_\lambda|_E +|v|_E\le c,\\ |\tilde u_\lambda+v|_0\leq |\tilde u_\lambda|_0 +|v|_0\le c\,. \endgathered \tag 3.9 $$ These inequalities and (3.4) imply $$|f(\tilde u_\lambda+v,\tilde u_\lambda '+v') -f(\tilde u_\lambda,\tilde u_\lambda ')|_0\leq f(0,0)\bigl(\frac {\delta-d}2\bigr )\,. \tag 3.10 $$ Using (3.10)and (3.2) and applying Lemma 1 and Lemma 2, we have that $$ \aligned |v(t)| &\leq \lambda\frac {\delta-d}2 f(0,0)p(t)+\lambda \max\{|f(u,y)|\;|\;0\le u\le c,-c\le y\le c\}q(t)\\ &\leq \lambda\frac {\delta-d}2 f(0,0)p(t)+\lambda d f(0,0) p(t)\\ &= \lambda\frac {\delta+d}2 f(0,0)p(t), \quad t\in I \endaligned \tag 3.11 $$ and $$ \aligned |v'(t)| &\leq \lambda\frac {\delta-d}2 f(0,0)P(t)+\lambda \max\{|f(u,y)|\;|\;0\le u\le c,-c\le y\le c\}Q(t)\\ &\leq \lambda\frac {\delta-d}2 f(0,0)P(t)+\lambda d f(0,0) P(t)\\ &= \lambda\frac {\delta+d}2 f(0,0)P(t), \quad t\in I \endaligned \tag 3.12 $$ In particular $$|v|_E\leq \lambda\frac {\delta+d}2 f(0,0)|P|_0< \lambda\delta f(0,0)|P|_0 \tag 3.13 $$ a contradiction, and the claim is proved. Thus by Leray-Schauder fixed point theorem, $T$ has a fixed ponit $ v_\lambda$ with $$|v_\lambda|_E\le \lambda\delta f(0,0)|P|_0 \tag 3.14 $$ Finally, using (2.9) and (3.11), we obtain $$ \aligned u_\lambda&\geq \tilde u_\lambda-|v_\lambda|\\ &\geq\lambda\delta f(0,0)p(t)-\lambda\frac {\delta+d}2 f(0,0)p(t)\\ &=\lambda\frac {\delta-d}2 f(0,0)p(t),\quad t\in I \endaligned \tag 3.15 $$ i.e., $u_\lambda$ is a positive solution of (1.3)-(1.4). \head References \endhead \roster \item"[1]" W. Feng and J. R. L. Webb,\quad {\it Solvability of a three-point boundary value problems at resonance}, Nonlinear Analysis TMA {\bf 30} (1997), 3227-3238. \item"[2]" W. Feng and J. R. L. Webb, {\it Solvability of a $m$-point boundary value problems with nonlinear growth}, J. Math. Anal. Appl. {\bf 212} (1997), 467-480. \item"[3]" W. Feng, {\it On an $m$-point boundary value problem}, Nonlinear Analysis TMA {\bf 30} (1997), 5369-5374. \item"[4]" C. P. 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