\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 16, pp. 1--8.\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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/16\hfil Positive solutions for fourth-order BVPs] {Positive solutions for semipositone fourth-order \\ two-point boundary value problems} \author[D. Yang, H. Zhu, C. Bai, \hfil EJDE-2007/16\hfilneg] {Dandan Yang, Hongbo Zhu, Chuanzhi Bai} % not in alphabetical order \address{Dandan Yang \newline Department of Mathematics\\ Yanbian University \\ Yanji, Jilin 133000, China. \newline Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsu 223001, China} \email{yangdandan2600@sina.com} \address{Hongbo Zhu \newline Department of Mathematics \\ Yanbian University \\ Yanji, Jilin 133000, China. \newline Department of Mathematics\\ Huaiyin Teachers College\\ Huaian, Jiangsu 223001, China} \email{zhuhongbo8151@sina.com} \address{Chuanzhi Bai \newline Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsu 223001, China} \email{czbai8@sohu.com} \thanks{Submitted August 3, 2006. Published January 23, 2007.} \thanks{Supported by the Natural Science Foundation of Jiangsu Education Office and by Jiangsu \hfill\break\indent Planned Projects for Postdoctoral Research Funds} \subjclass[2000]{34B16} \keywords{Boundary value problem; Positive solution; semipositone; fixed point} \begin{abstract} In this paper we investigate the existence of positive solutions of the following nonlinear semipositone fourth-order two-point boundary-value problem with second derivative: \begin{gather*} u^{(4)}(t) = f(t, u(t), u''(t)), \quad 0 \leq t \leq 1, \\ u'(1) = u''(1) = u'''(1) = 0, \quad k u(0) = u'''(0), \end{gather*} where $-6 < k < 0$, $f \geq - M$, and $M$ is a positive constant. Our approach relies on the Krasnosel'skii fixed point theorem. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} Recently an increasing interest in studying the existence of positive solutions for fourth-order two-point boundary value problems is observed. Among others we refer to \cite{a1, a2, b1, g1,g2, h1, l1, p1,t1}. In this paper we consider the positive solutions of the following nonlinear semipositone fourth-order two-point boundary value problem with second derivative: \begin{equation} \begin{gathered} u^{(4)}(t)=f(t,u(t),u''(t)), \quad 0\le t\le 1,\\ u'(1)=u''(1)=u'''(1)=0, \quad ku(0)=u'''(0), \end{gathered} \label{e1.1} \end{equation} where $ -6 < k <0$, $f$ is continuous and there exists $M > 0$ such that $f \geq - M$. This implies that $f$ is not necessarily nonnegative, monotone, superlinear and sublinear. And also this assumption implies that the problem \eqref{e1.1} is semipositone . The purpose of this paper is to establish the existence of positive solutions of problem \eqref{e1.1} by using Krasnosel'skii fixed point theorem in cones. The rest of this paper is organized as follows: in section 2, we present some preliminaries and lemmas. Section 3 is devoted to proving the existence of positive solutions of problem \eqref{e1.1}. An example is considered in section 4 to illustrate our main results. \section{Preliminaries and lemmas} Let $C^2[0,1]$ be the Banach space with norm $\|u\|_0 = \max\{\|u\|, \|u''\| \}$, where $$ \|u\|= \max_{0\le t \le 1} |u(t)|, \quad u \in C[0,1]. $$ By routine calculation, we easily obtain the following Lemma. \begin{lemma}\label{lem2.1} If $k \neq 0$, then \begin{gather*} u^{(4)}(t)=h(t), \quad 0 \le t \le 1,\\ u'(1)=u''(1)=u'''(1)=0, \quad ku(0)=u'''(0), \end{gather*} has a unique solution $$ u(t)= \int_0^1 G(t,s) h(s)ds, $$ where the Green function is \[ G(t,s) = -\frac{1}{6} \begin{cases} \frac 6 k + s^{3}, & 0 \le s \le t \le 1, \\ \frac 6 k -(s-t)^{3}+s^{3}, & 0 \le t \le s \le 1. \end{cases} \] \end{lemma} \begin{remark} \label{rmk2.2} {\rm If $- 6 < k < 0$, then \begin{equation} 0 < (1+\frac k6)G(0,s)\le G(t,s) \le G(0,s)=\max _{0 \le t \le 1}G(t,s)=- \frac 1k \label{e2.1} \end{equation} in closed bounded region $D=\{(t,s):0 \le t \le 1,0 \le s\le 1\}$. } \end{remark} Let $$ p(t) := \int_0^1 G(t, s) ds = \frac 1 {24} t^4 - \frac 16 t^3+\frac 14 t^2 -\frac 16 t - \frac 1k, \quad 0\le t \le 1. $$ Since \begin{gather*} p'(t)=\frac 16 t^3-\frac 12t ^2 +\frac 12t -\frac 16 =-\frac 16 (1-t)^3 \leq 0, \quad 0 \leq t \leq 1,\\ p''(t)=\frac 12t^2 -t +\frac 12 =\frac 12(1-t)^2 \geq 0, \quad 0 \leq t \leq 1, \end{gather*} we have \begin{gather} \|p\|=\max_{0 \le t\le 1} p(t) = p(0) = - \frac 1k ,\quad \min_{0\le t \le 1}p(t)=p(1)=-\frac 1k-\frac 1 {24}, \label{e2.2} \\ \|p''\|=\max _{0 \le t\le 1} |p''(t)|=\frac 12. \label{e2.3} \end{gather} Our approach is based on the following Krasnosel'skii fixed point theorem. \begin{lemma}\label{lem2.2} Let $X$ be a Banach space, and $K \subset X$ be a cone in $X$. Assume $\Omega _1, \Omega _2$ are bounded open subsets of $K$ with $0 \in \Omega _1 \subset \overline \Omega _1 \subset \Omega_2$, and let $F:K\to K $ be a completely continuous operator such that either \begin{itemize} \item[(1)] $\|Fu\|\le \|u\|, u\in \partial \Omega_1 $, and $\|Fu \|\ge \|u\|, u\in \partial \Omega_2$, or \item[(2)] $\|Fu \|\ge \|u\|, u\in \partial \Omega_1 $, and $ \|Fu\|\le \|u\|, u\in \partial \Omega_2$. \end{itemize} Then $F$ has a fixed point in $ \overline \Omega_2 \setminus \Omega_1$. \end{lemma} To apply the Krasnosel'skii fixed point theorem, we need to construct a suitable cone. Let \begin{align*} C_0^2 [0,1]=\{&u\in C^2[0,1]: u(t)\ge 0, \; u''(t) \ge 0, \ 0 \leq t \leq 1,\\ & u'(1)=u''(1)=u'''(1)=0, \; ku(0)=u'''(0)\}. \end{align*} It is easy to check that the following set $ P$ is a cone in $C^2[0,1]$: $$ P = \big\{u\in C^2_0 [0,1]: \min _{0\le t \le 1 }u(t) \ge (1+ \frac k6)\|u\|\big\}, $$ where $- 6 < k < 0$. For convenience, let \begin{gather} \alpha (r) = \max \{ f(t,u,v) : (t, u, v) \in D_1(r)\}, \label{e2.4} \\ \beta (r) = \min\{ f(t,u,v): (t, u, v) \in D_2(r)\}, \label{e2.5} \end{gather} where $$ D_1(r) = \big\{(t,u,v) : 0 \le t \le 1, \ \frac{M}{k} \le u \le r + (\frac{1}{k} + \frac{1}{24})M, \ - \frac{M}{2} \le v \le r \big\}, $$ \begin{align*} D_2(r) = \big\{&(t,u,v) : \frac 14 \le t \le \frac 34,\; (\frac 1k + \frac{175}{6144})M \le u \le r +(\frac 1k +\frac {85}{2048})M,\;\\ &-\frac 9 {32}M \le v \le r-\frac 1{32}M \big\}\,. \end{align*} \[ C_1 = \min \Big\{\Big[\max _{0 \le t \le 1 }\int ^1_0 G(t,s)ds \Big]^{-1}, \Big[\max _{0 \le t\le1}\int _0^1 |G''(t,s)|ds\Big]^{-1}\Big\} = \min \{-k, 2\}, \] \begin{align*} C_2 &= \max \Big\{\Big[\max _{0 \le t \le 1} \int ^\frac 3 4 _ \frac 14 G(t,s)ds \Big]^{-1}, \Big[\max_{0 \le t\le 1}\int _\frac14^\frac 34 |G''(t,s)|ds\Big]^{-1}\Big\}\\ &= \max \Big\{\big(-\frac 1{2k}+ \frac{1}{6144}\big)^{-1}, \frac{32}{9}\big\}. \end{align*} Obviously, $0 < C_1 < C_2$. \section{Main results} \begin{theorem}\label{thm3.1} Let $- 6 < k < 0$. Assume that \begin{equation} f : [0, 1] \times [\frac{M}{k}, + \infty) \times [-\frac{M}{2}, + \infty) \to [- M, + \infty) \label{e3.1} \end{equation} is continuous, where $M > 0$ is a constant. Suppose there exist two positive numbers $r_1$ and $ r_2$ with $ \min \{r_1, r_2\} > \frac{-6}{6k + k^2}M$ such that \begin{equation} \alpha (r_1) \le r_1 C_1 - M, \quad \beta (r_2) \ge r_2 C_2 - M, \label{e3.2} \end{equation} where $\alpha, \beta$ are as in (\ref{e2.4}) and (\ref{e2.5}), respectively. Then problem \eqref{e1.1} has at least one positive solution. \end{theorem} \begin{proof} Let $u_0(t) = M p(t), 0\le t\le 1$. Then by \eqref{e2.1} and (\ref{e2.3}) we have \begin{equation} (-\frac 1k -\frac 1 {24})M \le u_0(t)\le -\frac{M}{k}, \quad 0\le u_0''(t) \le \frac 12M , \quad 0\le t \le 1. \label{e3.3} \end{equation} Consider the fourth-order two-point boundary-value problem \begin{equation} \begin{gathered} u^{(4)}(t) = f(t,u(t)-u_0(t),u''(t)-u''_0(t))+M, \quad 0\le t\le 1,\\ u'(1)=u''(1)=u'''(1)=0,\\ ku(0)=u'''(0), \end{gathered} \label{e3.4} \end{equation} This problem is equivalent to the integral equation $$ u(t) = \int_0^1 G(t,s)[f(s, u(s)-u_0(s), u''(s)-u''_0(s)) + M]ds. $$ For $u\in C^2_0[0,1]$, we define the operator $A$ as follows $$ (A u)(t) = \int _0^1 G(t,s)[f(s, u(s)-u_0(s), u''(s)-u''_0(s)) + M]ds, \quad 0\le t \le1. $$ Computing the second derivative of $(A u)(t)$, we obtain $$ (A u)''(t) = \int_t^1 (s - t) [f(s, u(s)-u_0(s), u''(s)-u''_0(s)) + M]ds, \quad 0\le t \le1. $$ Noticing \eqref{e3.3} and that $u \in C_0^2[0, 1]$, we have \begin{gather*} \frac Mk \le u(t)-u_0(t) < + \infty,\\ -\frac 12 M \le u''(t)-u''_0(t)< + \infty, \quad 0\le t \le 1. \end{gather*} Thus, from \eqref{e3.1} we get $$ (A u)(t) \geq 0, \quad (A u)''(t) \geq 0, \quad t \in [0, 1]. $$ By the definition of $G(t, s)$, $$ G'(1, s) = G''(1, s) = G'''(1, s) = 0, \quad {\rm and} \quad G'''(0, s) = k G(0, s) = - 1,$$ which implies that $$ (A u)'(1) = (A u)''(1) = (A u)'''(1) = 0, \quad \text{and} \quad k (Au)(0) = (A u)'''(0). $$ Hence, $ A : C_0^2[0,1] \to C_0^2[0,1]$. Moreover, for each $t \in [0, 1]$, (By \eqref{e2.1} we have \begin{align*} (A u)(t)&=\int_0^1 G(t,s)[f(s,u(s)-u_0(s),u''(s)-u''_0(s))+M]ds \\ &\ge(1+\frac k6 ) \int _0^1 G(0,s) [f(s,u(s)-u_0(s),u''(s)-u''_0(s))+M]ds\\ &\ge (1+\frac k6 ) \max _{0 \leq t \leq 1} \int_0^1 G(t,s) [f(s,u(s)-u_0(s),u''(s)-u''_0(s))+M]ds\\ &=(1+\frac k6 )\|Au\|. \end{align*} Thus, $ A : P \to P $. We can check that $A$ is completely continuous by routine method. Since $C_1 < C_2$, it is easy to check that $r_1\ne r_2 $. Without loss of generality, we assume $r_1 < r_2$. Let $$ \Omega _1 = \{u\in P : \|u\|_0 < r_1 \}, \quad \Omega_2 = \{u\in P: \|u\|_0 < r_2 \}. $$ If $u\in \partial \Omega_1$, then $\|u\|_0 = r_1$. So, $\|u\| \le r_1 $ and $\|u''\| \le r_1$. This implies $$ 0 \le u(t) \le r_1 \quad0\le u''(t)\le r_1, \quad 0 \le t \le1. $$ By (\ref{e2.2}), for $ 0 \le t \le 1$, we have $$ \frac1k M \le u(t)-u_0(t)\le r_1 + \big(\frac1k +\frac1{24}\big)M, \quad -\frac12 M \le u''(t) - u''_0(t)\le r_1. $$ By (\ref{e3.2}), $$ f(t,u(t)-u_0(t),u''(t)-u''_0(t)) \le \alpha (r_1) \le r_ 1 C_1 - M, \quad 0 \leq t \leq 1. $$ It follows that \begin{align*} \|A u\| &= \max _{0 \le t \le 1}\int _0^1 G(t,s)[f(s,u(s) - u_0(s), u''(s) - u''_0(s)) + M]ds\\ &\le r_1 C_1 \max _{0\le t \le 1}\int _0^1 G(t,s)ds\le r_1, \end{align*} \begin{align*} \|(A u)''\|& = \max _{0\le t \le 1}\int_0^1 |G''(t,s)| [f(s,u(s)-u_0(s),u''(s)-u''_0(s))+M]ds\\ &\le r_1 C_1 \max _{0\le t\le1}\int _0^1 |G''(t,s)| ds \le r_1. \end{align*} Therefore, $\|Au\|_0 \le r_1 = \|u\|_0$. If $u\in \partial \Omega_2$, then $\|u\|_0 = r_2$. So, $\|u\|\le r_2 $ and $\|u''\| \le r_2$. This implies that $$ 0\le u(t)\le r_2, \quad 0\le u''(t)\le r_2,\quad 0\le t \le1. $$ Since \begin{gather*} -\frac {85} {2048}- \frac 1k = p(\frac{3}{4}) \leq p(t) \leq p(\frac{1}{4}) = -\frac {175} {6144} - \frac 1 k, \quad \frac{1}{4} \leq t \leq \frac{3}{4},\\ \frac{1}{32} \leq p''(t) = \frac{1}{2} (1 - t)^2 \leq \frac{9}{32}, \quad \frac{1}{4} \leq t \leq \frac{3}{4}, \end{gather*} we have $$ (\frac 1k +\frac {175}{6144})M \le u(t)-u_0(t) \le r_2 + (\frac 1k + \frac {85}{2048}) M, \quad \frac{1}{4} \leq t \leq \frac{3}{4}, $$ and $$ -\frac 9 {32}M \le u''(t) -u''_0(t) \le r_2 - \frac{M}{32}, \quad \frac{1}{4} \leq t \leq \frac{3}{4}. $$ Thus, by (\ref{e3.2}) we obtain $$ f(t,u(t)-u_0(t),u''(t)-u''_0(t))\ge \beta (r_2)\ge r_2 C_2 - M, \quad \frac{1}{4} \leq t \leq \frac{3}{4}. $$ From this, \begin{align*} \|Au\| &\ge \max _{0 \le t \le 1}\int_\frac 14 ^\frac 34 G(t,s)[f(s,u(s)-u_0(s),u''(s)-u''_0(s))+M]ds\\ &\ge r_2 C_2 \max _{0\le t \le 1 }\int _\frac 14 ^\frac 34 G(t,s) ds \ge r_2, \end{align*} and \begin{align*} \|(Au)''\| &\ge \max _{0\le t \le 1}\int _\frac 14 ^\frac 34 G''(t,s) [f(s,u(s)-u_0(s),u''(s)-u''_0(s))+M]ds \\ & \ge r_2 C_2 \max _{0\le t \le1 }\int _\frac 14^\frac 34 G''(t,s) ds \ge r_2 . \end{align*} It follows that $ \|A u\|_0 \ge r_2 = \|u\|_0$. By Lemma \ref{lem2.2}, we assert that the operator $A$ has at least one fixed point $\overline u \in P$ with $ r_1 \le \|\overline u \|_0 \le r_2$. This implies that (\ref{e3.4}) has at least one solution $ \overline u\in P$ with $ r_1 \le \|\overline u \|_0 \le r_2$. Let $u_*(t)=\overline{u}(t) - u_0 (t)$, $0 \le t \le 1$. We will check that $u_*$ is a solution of the problem \eqref{e1.1}. In fact, since $A \overline u=\overline u $, we have \begin{align*} u_*(t) + u_0(t) &=\overline u(t)=(A \overline u)(t)\\ &=\int _0^1 G(t,s)[f(s,\overline u (s)-u_0(s),\overline u''(s)-u''_0(s))+M]ds \\ & =\int_0^1 G(t,s) f(s,u_*(s),u_*''(s))ds + u_0(t). \end{align*} It follows that $$ u_*(t)=\int_0^1 G (t,s) f(s,u_*(s), u_*''(s))ds, \quad 0 \le t \le 1. $$ In other words, $u_*$ is a solution of \eqref{e1.1}. Therefore, the problem \eqref{e1.1} has at least one solution $u_*$ satisfying $u_* + u_0 \in P$ and $r_1 \le \|u^*+u_0\|_0\le r_2$. Since $r_1 = \min \{r_1, r_2\} > -\frac{6}{6k+k^2}M$, we have \begin{align*} u_*(t)&=[u_*(t)+u_0(t)] - u_0(t) = [u_*(t) + u_0(t)] - Mp(t)\\ &\ge (1+\frac k6)\|u_*(t)+u_0(t)\| + \frac{M}{k}\\ &\ge (1+\frac k6) [r_1 + \frac{6}{6k+k^2}M] > 0, \quad 0 \le t \le 1, \end{align*} which implies that $u_*$ is a positive solution of \eqref{e1.1}. \end{proof} Using Theorem \ref{thm3.1}, we can prove following result. \begin{theorem}\label{thm3.2} Let $- 6 < k < 0$. Assume that \begin{equation} f : [0, 1] \times [\frac{M}{k}, + \infty) \times [- \frac{M}{2}, + \infty) \to [- M, + \infty) \label{e3.5} \end{equation} is continuous, where $M \geq 0$ is a constant. Suppose that there exist three positive numbers $r_1 < r_2 < r_3$ with $r_1 > -\frac 6 {6k+k^2}M$ such that one of the following conditions is satisfied: \begin{itemize} \item[(1)] $\alpha (r_1)\le r_1 C_1 - M$, $\beta (r_2) > r_2 C_2- M$, $\alpha (r_3)\le r_3 C_1 - M$; \item[(2)] $\beta (r_1)\ge r_1 C_2 - M$, $\alpha (r_2) < r_2 C_1 - M$, $\beta (r_3)\ge r_3 C_2 - M$. \end{itemize} Then problem \eqref{e1.1} has at least two positive solutions. \end{theorem} \section {Examples} \begin{example} \label{exa4.1} \rm Consider the boundary-value problem \begin{equation} \begin{gathered} u^{(4)}(t) = f(t,u(t),u''(t)), \quad 0\le t\le 1,\\ u'(1)=u''(1)=u'''(1)=0, \quad -2 u(0)=u'''(0), \end{gathered} \label{e4.1} \end{equation} where $f: [0, 1] \times [-1, + \infty) \times [-1, + \infty) \to [-2, + \infty) $ is defined by \[ f(t, u, v) \!=\! \begin{cases} t^2 + \sqrt{u+1} + 9 \sqrt{v+1} - 2,& (t, u, v) \in [0, 1] \times [-1, -\frac{1}{2}] \times [-1, -\frac{1}{2}],\\ t^2 + \frac{u}{4} + 9 \sqrt{v+1} + \frac{\sqrt{2}}{2} - \frac{15}{8}, & (t, u, v) \in [0, 1] \times [-\frac{1}{2}, \infty) \times [-1, -\frac{1}{2}],\\ t^2 + \sqrt{u+1} + \frac{v}{5} + \frac{9}{2}\sqrt{2} - \frac{19}{10}, & (t, u, v) \in [0, 1] \times [-1, -\frac{1}{2}] \times [-\frac{1}{2}, \infty),\\ t^2 + \frac{u}{4} + \frac{v}{5} + 5 \sqrt{2} - \frac{71}{40},& (t, u, v) \in [0, 1] \times [-\frac{1}{2}, \infty) \times [-\frac{1}{2}, \infty). \end{cases} \] Thus, $k = -2$, $M = 2$, $C_1 = 2$ and $C_2 =\frac{6144}{1537}$. For $$ D_1(r) = \big\{(t,u,v) : 0 \le t \le 1, \; -1 \le u \le r - \frac{11}{12}, \; - 1 \le v \le r \big\}, $$ $$ D_2(r) = \big\{(t,u,v) : \frac 14 \le t \le \frac 34, \; - \frac{2897}{3072} \le u \le r - \frac{939}{1024}, \; - \frac{9}{16} \le v \le r - \frac{1}{16}\big\}. $$ By simple computations, we obtain \begin{align*} \alpha (6) &= \max \{ f(t,u,v) : (t, u, v) \in D_1(6)\} \\ & = \max \big\{f(1, \frac{61}{12}, 6), \; f(1, \frac{61}{12}, - \frac{1}{2}), \; f(1, - \frac{1}{2}, 6), \; f(1, - \frac{1}{2}, - \frac{1}{2}) \big\}\\ & = f(1, \frac{61}{12}, 6) = 8.76 < 10 = 6 C_1 - M, \end{align*} and \begin{align*} &\beta (\frac{13}{8}) \\ &= \min \big\{ f(t,u,v) : (t, u, v) \in D_2(\frac{13}{8}) \big\} \\ & = \min \big\{f(\frac{1}{4}, - \frac{2897}{3072}, - \frac{9}{16}), \; f(\frac{1}{4}, - \frac{2897}{3072}, - \frac{1}{2}), \; f(\frac{1}{4}, - \frac{1}{2}, - \frac{9}{16}), \; f(\frac{1}{4}, - \frac{1}{2}, - \frac{1}{2})\big\} \\ &= f(\frac{1}{4}, - \frac{2897}{3072}, - \frac{9}{16}) = 4.76 > 4.49 = \frac{13}{8} C_2 - M. \end{align*} Take $r_1 = 6$ and $r_2 = \frac{13}{8}$. Then (\ref{e3.2}) holds. Moreover, we have $$ \min \{r_1, r_2\} = \frac{13}{8} > \frac{3}{2} = - \frac{6}{6k+k^2}M. $$ So, by Theorem \ref{thm3.1}, problem (\ref{e4.1}) has at least one positive solution. \end{example} \begin{thebibliography}{00} \bibitem{a1} A. R. Aftabizadeh; \emph{Existence and uniqueness theorems for fourth-order boundary problems}, J. Math. Anal. Appl. 116 (1986) 415-426. \bibitem{a2} R. P. Agarwal; \emph{Focal Boundary Value Problems for Differential and Difference Equations}, Kluwer Academic, Dordrecht, 1998. \bibitem{b1} Z. Bai, H. Wang; \emph{On positive solutions of some nonlinear fourth-ordr beam equations}, J. Math. Anal. Appl. 270 (2002) 357-368. \bibitem{g1} J. R. Graef, B. Yang; \emph{On a nonlinear boundary value problem for fourth order equations}, Appl. Anal. 72 (1999) 439-448. \bibitem{g2} Yanping Guo, Weigao Ge, Ying Gao; \emph{Twin positive solutions for higher order m-point boundary value problems with sign changing nonlinearities}, Appl. Anal. 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