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# Multiple positive solutions of nonlinear BVPs for differential systems involving integral conditions

Boundary Value Problems20142014:61

https://doi.org/10.1186/1687-2770-2014-61

• Accepted: 6 March 2014
• Published:

## Abstract

In this paper, we consider the following system of nonlinear third-order nonlocal boundary value problems (BVPs for short):

$\left\{\begin{array}{c}-{u}^{‴}\left(t\right)=f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ -{v}^{‴}\left(t\right)=g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ u\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}a{u}^{\prime }\left(0\right)-b{u}^{″}\left(0\right)=\alpha \left[u\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}c{u}^{\prime }\left(1\right)+d{u}^{″}\left(1\right)=\beta \left[u\right],\hfill \\ v\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}a{v}^{\prime }\left(0\right)-b{v}^{″}\left(0\right)=\alpha \left[v\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}c{v}^{\prime }\left(1\right)+d{v}^{″}\left(1\right)=\beta \left[v\right],\hfill \end{array}$

where $f,g\in C\left(\left[0,1\right]×{R}^{+}×{R}^{+},{R}^{+}\right)$, $\alpha \left[u\right]={\int }_{0}^{1}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dA\left(t\right)$ and $\beta \left[u\right]={\int }_{0}^{1}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dB\left(t\right)$ are linear functionals on $C\left[0,1\right]$ given by Riemann-Stieltjes integrals and are not necessarily positive functionals; a, b, c, d are nonnegative constants with $\rho :=ac+ad+bc>0$. By using the Guo-Krasnoselskii fixed point theorem, some sufficient conditions are obtained for the existence of at least one or two positive solutions and nonexistence of positive solutions to the above problem. Two examples are also included to illustrate the main results.

MSC:34B15.

## Keywords

• differential system
• multiple positive solution
• integral conditions

## 1 Introduction

The theory of BVPs with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics can be reduced to nonlocal problems with integral boundary conditions. Moreover, BVPs with Riemann-Stieltjes integral boundary condition (BC for short) have been considered recently as both multipoint and Riemann integral type BCs are treated in a single framework. For more comments on Stieltjes integral BC and its importance, we refer the reader to the papers by Webb and Infante [13] and their other related works.

In recent years, third-order nonlocal BVPs have received much attention from many authors; see, for example [414]. It is worth mentioning that Sun and Li [12] studied the third-order BVP with integral boundary conditions
$\left\{\begin{array}{c}{u}^{‴}\left(t\right)+f\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}0
(1.1)
Their main tool was the Guo-Krasnoselskii fixed point theorem. Recently, we [14] were concerned with the existence of a monotone positive solution for the third-order BVP
$\left\{\begin{array}{c}{u}^{‴}\left(t\right)+f\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}0
(1.2)

by applying monotone iterative techniques, where $f\in C\left(\left[0,1\right]×{R}^{+}×{R}^{+},{R}^{+}\right)$, $\alpha \left[u\right]={\int }_{0}^{1}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dA\left(t\right)$ and $\beta \left[u\right]={\int }_{0}^{1}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dB\left(t\right)$ are linear functionals on $C\left[0,1\right]$ given by Riemann-Stieltjes integrals.

Furthermore, motivated by the wide applications of systems of differential equations in biomathematics, the study of systems of BVPs has received increased interest; see [1528] and the references therein. In particular, Henderson and Luca [17] established the existence of positive solutions for the system of BVPs with multi-point boundary conditions
$\left\{\begin{array}{c}-{u}^{″}\left(t\right)+\lambda c\left(t\right)f\left(u\left(t\right),v\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}t\in \left(0,T\right),\hfill \\ -{v}^{″}\left(t\right)+\mu d\left(t\right)g\left(u\left(t\right),v\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}t\in \left(0,T\right),\hfill \\ \alpha u\left(0\right)-\beta {u}^{\prime }\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u\left(T\right)={\sum }_{i=1}^{m-2}{a}_{i}u\left({\xi }_{i}\right),\phantom{\rule{1em}{0ex}}m\ge 3,\hfill \\ \gamma v\left(0\right)-\delta {v}^{\prime }\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}v\left(T\right)={\sum }_{i=1}^{n-2}{b}_{i}v\left({\eta }_{i}\right),\phantom{\rule{1em}{0ex}}m\ge 3\hfill \end{array}$
(1.3)

by applying the fixed point index theory.

Yang [27] studied the existence of positive solutions for the system of second-order nonlocal BVPs
$\left\{\begin{array}{c}-{u}^{″}\left(t\right)=f\left(t,u,v\right),\hfill \\ -{v}^{″}\left(t\right)=g\left(t,u,v\right),\hfill \\ u\left(0\right)=v\left(0\right)=0,\hfill \\ u\left(1\right)={H}_{1}\left({\int }_{0}^{1}u\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\alpha \left(\tau \right)\right),\hfill \\ v\left(1\right)={H}_{2}\left({\int }_{0}^{1}v\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\beta \left(\tau \right)\right)\hfill \end{array}$
(1.4)

by using fixed point index theory in a cone.

Infante and Pietramala [19] studied the existence of positive solutions for a system of perturbed Hammerstein integral equations by fixed point index theory for compact maps and illustrated their theory by studying the following system of BVPs:
$\left\{\begin{array}{c}{u}^{″}\left(t\right)+{g}_{1}\left(t\right){f}_{1}\left(t,u\left(t\right),v\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}0
(1.5)

The result was quite general and covered a wide class of systems of BVPs. Here ${\beta }_{ij}\left[w\right]$ was of the form ${\beta }_{ij}\left[w\right]={\int }_{0}^{1}w\left(t\right)\phantom{\rule{0.2em}{0ex}}d{B}_{ij}\left(t\right)$ involving positive Riemann-Stieltjes measures.

Inspired greatly by the above-mentioned excellent works, in this paper, we are concerned with the following system of third-order BVPs:
$\left\{\begin{array}{c}-{u}^{‴}\left(t\right)=f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ -{v}^{‴}\left(t\right)=g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ u\left(0\right)=0,\phantom{\rule{2em}{0ex}}a{u}^{\prime }\left(0\right)-b{u}^{″}\left(0\right)=\alpha \left[u\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}c{u}^{\prime }\left(1\right)+d{u}^{″}\left(1\right)=\beta \left[u\right],\hfill \\ v\left(0\right)=0,\phantom{\rule{2em}{0ex}}a{v}^{\prime }\left(0\right)-b{v}^{″}\left(0\right)=\alpha \left[v\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}c{v}^{\prime }\left(1\right)+d{v}^{″}\left(1\right)=\beta \left[v\right],\hfill \end{array}$
(1.6)

where $f,g\in C\left(\left[0,1\right]×{R}^{+}×{R}^{+},{R}^{+}\right)$, $\alpha \left[u\right]={\int }_{0}^{1}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dA\left(t\right)$ and $\beta \left[u\right]={\int }_{0}^{1}u\left(t\right)\phantom{\rule{0.2em}{0ex}}dB\left(t\right)$ are linear functionals on $C\left[0,1\right]$ given by Riemann-Stieltjes integrals with signed measures; a, b, c, d are nonnegative constants with $\rho :=ac+ad+bc>0$. To the best of our knowledge, the study of existence of positive solutions of third-order differential systems (1.6) has not been done.

A vector $\left(u,v\right)\in {C}^{2}\left(0,1\right)×{C}^{2}\left(0,1\right)$ is said to be a positive solution of BVP (1.6) if and only if u, v satisfy BVP (1.6) and u, v are positive on $\left(0,1\right)$. The proof of our main results is based on the well-known Guo-Krasnoselskii fixed point theorem, which we present now.

Theorem 1.1 Let E be a Banach space, $K\subset E$ be a cone, and ${\mathrm{\Omega }}_{1}$ and ${\mathrm{\Omega }}_{2}$ be bounded open subsets of E with $0\in {\mathrm{\Omega }}_{1}$, ${\overline{\mathrm{\Omega }}}_{1}\subset {\mathrm{\Omega }}_{2}$. Assume that $A:K\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)\to K$ is a completely continuous operator such that either
1. (i)

$\parallel Au\parallel \le \parallel u\parallel$ for $u\in K\cap \partial {\mathrm{\Omega }}_{1}$ and $\parallel Au\parallel \ge \parallel u\parallel$ for $u\in K\cap \partial {\mathrm{\Omega }}_{2}$; or

2. (ii)

$\parallel Au\parallel \ge \parallel u\parallel$ for $u\in K\cap \partial {\mathrm{\Omega }}_{1}$ and $\parallel Au\parallel \le \parallel u\parallel$ for $u\in K\cap \partial {\mathrm{\Omega }}_{2}$.

Then A has a fixed point in $K\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)$.

## 2 Preliminary lemmas

In this section, we adopt the ideas and the method which have been widely used and which are due to Webb and Infante in [1, 2].

In our case, the existence of positive solutions of nonlocal BVP
$\left\{\begin{array}{c}{u}^{‴}\left(t\right)+f\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right)=0,\phantom{\rule{1em}{0ex}}0
(2.1)
with two nonlocal boundary terms $\alpha \left[u\right]$, $\beta \left[u\right]$ can be studied via a perturbed Hammerstein integral equation of the type
$u\left(t\right)=\gamma \left(t\right)\alpha \left[u\right]+\delta \left(t\right)\beta \left[u\right]+{\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds=:{T}_{1}u\left(t\right).$
(2.2)
Here $\gamma \left(t\right)$, $\delta \left(t\right)$ are linearly independent and given by
$\begin{array}{r}-{\gamma }^{‴}\left(t\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\gamma \left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}a{\gamma }^{\prime }\left(0\right)-b{\gamma }^{″}\left(0\right)=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}c{\gamma }^{\prime }\left(1\right)+d{\gamma }^{″}\left(1\right)=0,\\ -{\delta }^{‴}\left(t\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\delta \left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}a{\delta }^{\prime }\left(0\right)-b{\delta }^{″}\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}c{\delta }^{\prime }\left(1\right)+d{\delta }^{″}\left(1\right)=1,\end{array}$
which imply $\gamma \left(t\right)=\frac{2ct+2dt-c{t}^{2}}{2\rho }$ and $\delta \left(t\right)=\frac{a{t}^{2}+2bt}{2\rho }$, $t\in \left[0,1\right]$. Let ${\parallel \cdot \parallel }_{\mathrm{\infty }}$ be the usual supremum norm in $C\left[0,1\right]$. A direct calculation shows that for $t\in \left[\theta ,1-\theta \right]$, $0<\theta <\frac{1}{2}$,
$\begin{array}{r}\gamma \left(t\right)\ge {c}_{1}{\parallel \gamma \parallel }_{\mathrm{\infty }},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\delta \left(t\right)\ge {c}_{2}{\parallel \delta \parallel }_{\mathrm{\infty }},\\ {\gamma }^{\prime }\left(t\right)\ge {d}_{1}{\parallel {\gamma }^{\prime }\parallel }_{\mathrm{\infty }}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\delta }^{\prime }\left(t\right)\ge {d}_{2}{\parallel {\delta }^{\prime }\parallel }_{\mathrm{\infty }},\end{array}$
(2.3)
where ${c}_{1}=\frac{2c\theta +2d\theta -c{\theta }^{2}}{c+2d}$, ${c}_{2}=\frac{a{\theta }^{2}+2b\theta }{a+2b}$, ${d}_{1}=\frac{c\theta +d}{c+d}$ and ${d}_{2}=\frac{a\theta +b}{a+b}$; $G\left(t,s\right)$ is Green’s function for the corresponding problem with local terms when $\alpha \left[u\right]$ and $\beta \left[u\right]$ are identically 0, i.e.,
$G\left(t,s\right)=\left\{\begin{array}{cc}\frac{\left(a{t}^{2}+2bt\right)\left(c\left(1-s\right)+d\right)}{2\rho }-\frac{{\left(t-s\right)}^{2}}{2},\hfill & 0\le s\le t\le 1,\hfill \\ \frac{\left(a{t}^{2}+2bt\right)\left(c\left(1-s\right)+d\right)}{2\rho },\hfill & 0\le t\le s\le 1.\hfill \end{array}$

In the remainder of this paper, we always assume that

(H1) $0\le \alpha \left[\gamma \right],\beta \left[\delta \right]<1$, $\alpha \left[\delta \right],\beta \left[\gamma \right]\ge 0$ and $D:=\left(1-\alpha \left[\gamma \right]\right)\left(1-\beta \left[\delta \right]\right)-\alpha \left[\delta \right]\beta \left[\gamma \right]>0$;

(H2) A, B are functions of bounded variation, and ${\mathcal{K}}_{A}\left(s\right),{\mathcal{K}}_{B}\left(s\right)\ge 0$ for $s\in \left[0,1\right]$, where
${\mathcal{K}}_{A}\left(s\right):={\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}dA\left(t\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathcal{K}}_{B}\left(s\right):={\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}dB\left(t\right).$
As shown in Theorem 2.3 in [1], if u is a fixed point of ${T}_{1}$ in (2.2), then u is a fixed point of S, which is now given by
$\begin{array}{rcl}Su\left(t\right)& :=& \frac{\gamma \left(t\right)}{D}\left(\left(1-\beta \left[\delta \right]\right){\int }_{0}^{1}{\mathcal{K}}_{A}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\alpha \left[\delta \right]{\int }_{0}^{1}{\mathcal{K}}_{B}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ +\frac{\delta \left(t\right)}{D}\left(\beta \left[\gamma \right]{\int }_{0}^{1}{\mathcal{K}}_{A}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\left(1-\alpha \left[\gamma \right]\right){\int }_{0}^{1}{\mathcal{K}}_{B}\left(s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ =:& {\int }_{0}^{1}{G}_{S}\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\end{array}$

in our case. The kernel ${G}_{S}$ is Green’s function corresponding to BVP (2.1). By Lemma 2.1 and Lemma 2.2 in [14], we can get the following properties of Green’s function.

Lemma 2.1 Let $\rho :=ac+ad+bc>0$, ${c}_{3}=\frac{\rho {\int }_{0}^{\theta }\mathrm{\Phi }\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau }{\left(a+b\right)\left(c+d\right)}$, $0<\theta <1$. Then $G\left(t,s\right)$ satisfies

where $\mathrm{\Phi }\left(s\right)=\frac{1}{\rho }\left(b+as\right)\left(d+c\left(1-s\right)\right)$, $s\in \left[0,1\right]$.

Lemma 2.2 Let ${c}_{0}=min\left\{{c}_{1},{c}_{2},{c}_{3}\right\}$. Then ${G}_{S}\left(t,s\right)$ satisfies

where ${\mathrm{\Phi }}_{1}\left(s\right):=\frac{{\parallel \gamma \parallel }_{\mathrm{\infty }}}{D}\left(\left(1-\beta \left[\delta \right]\right){\mathcal{K}}_{A}\left(s\right)+\alpha \left[\delta \right]{\mathcal{K}}_{B}\left(s\right)\right)+\frac{{\parallel \delta \parallel }_{\mathrm{\infty }}}{D}\left(\beta \left[\gamma \right]{\mathcal{K}}_{A}\left(s\right)+\left(1-\alpha \left[\gamma \right]\right){\mathcal{K}}_{B}\left(s\right)\right)+\mathrm{\Phi }\left(s\right)$, $s\in \left[0,1\right]$.

Lemma 2.3 Let ${d}_{0}=min\left\{{d}_{1},{d}_{2},{d}_{3}\right\}$, ${d}_{3}=\frac{\rho {min}_{t\in \left[\theta ,1-\theta \right]}\mathrm{\Phi }\left(t\right)}{\left(a+b\right)\left(c+d\right)}$, $0<\theta <1$, and
$\begin{array}{rcl}{\mathrm{\Phi }}_{2}\left(s\right)& :=& \frac{{\parallel {\gamma }^{\prime }\parallel }_{\mathrm{\infty }}}{D}\left[\left(1-\beta \left[\delta \right]\right){\mathcal{K}}_{A}\left(s\right)+\alpha \left[\delta \right]{\mathcal{K}}_{B}\left(s\right)\right]\\ +\frac{{\parallel {\delta }^{\prime }\parallel }_{\mathrm{\infty }}}{D}\left[\beta \left[\gamma \right]{\mathcal{K}}_{A}\left(s\right)+\left(1-\alpha \left[\gamma \right]\right){\mathcal{K}}_{B}\left(s\right)\right]+\mathrm{\Phi }\left(s\right).\end{array}$
Then $\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}$ satisfies
Proof For $s\in \left[0,1\right]$, by the fact
$\frac{\frac{\partial G\left(t,s\right)}{\partial t}}{\mathrm{\Phi }\left(s\right)}=\left\{\begin{array}{cc}\frac{\left(b+as\right)\left(d+c\left(1-t\right)\right)}{\left(b+as\right)\left(d+c\left(1-s\right)\right)}=\frac{\left(b+at\right)\left(d+c\left(1-t\right)\right)}{\left(b+at\right)\left(d+c\left(1-s\right)\right)}\ge \frac{\rho \mathrm{\Phi }\left(t\right)}{\left(a+b\right)\left(c+d\right)},\hfill & 0\le s\le t\le 1,\hfill \\ \frac{\left(b+at\right)\left(d+c\left(1-s\right)\right)}{\left(b+as\right)\left(d+c\left(1-s\right)\right)}=\frac{\left(b+at\right)\left(d+c\left(1-t\right)\right)}{\left(b+as\right)\left(d+c\left(1-t\right)\right)}\ge \frac{\rho \mathrm{\Phi }\left(t\right)}{\left(a+b\right)\left(c+d\right)},\hfill & 0\le t\le s\le 1,\hfill \end{array}$

so $\frac{\partial G\left(t,s\right)}{\partial t}\ge \frac{\rho {min}_{t\in \left[\theta ,1-\theta \right]}\mathrm{\Phi }\left(t\right)}{\left(a+b\right)\left(c+d\right)}\mathrm{\Phi }\left(s\right)$, $t\in \left[\theta ,1-\theta \right]$, $s\in \left[0,1\right]$, which together with (2.3) shows that $\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}\ge {d}_{0}{\mathrm{\Phi }}_{2}\left(s\right)$ holds. □

Let $E=\left\{{C}^{1}\left[0,1\right]:u\left(0\right)=0\right\}$ equipped with the norm $\parallel u\parallel =max\left\{{\parallel u\parallel }_{\mathrm{\infty }},{\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}\right\}$, where ${\parallel u\parallel }_{\mathrm{\infty }}$ is the usual supremum norm in $C\left[0,1\right]$. Similar to Lemma 2.1 in [29], we can get the following lemma.

Lemma 2.4 If $u\in E$, then ${\parallel u\parallel }_{\mathrm{\infty }}\le {\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}$. And so, E is a Banach space when it is endowed with the norm $\parallel u\parallel ={\parallel {u}^{\prime }\parallel }_{\mathrm{\infty }}$.

Define
$K=\left\{u\in E:u\left(t\right)\ge 0,{u}^{\prime }\left(t\right)\ge 0,t\in \left[0,1\right],\underset{t\in \left[\theta ,1-\theta \right]}{min}{u}^{\prime }\left(t\right)\ge {d}_{0}\parallel u\parallel \right\}.$

Then it is easy to verify that K is a cone in E.

For $u\in K$, we define
$\begin{array}{rcl}\left(Tu\right)\left(t\right)& =& {\int }_{0}^{1}{G}_{S}\left(t,s\right)f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\end{array}$
It is easy to see that if x is a fixed point of T in K, then BVP (1.6) has one solution $\left(u,v\right)$, where
$\left\{\begin{array}{c}u\left(t\right)=x\left(t\right),\hfill \\ v\left(t\right)={\int }_{0}^{1}{G}_{S}\left(t,s\right)g\left(s,x\left(s\right),{x}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

Lemma 2.5 $T:K\to K$.

Proof It is obvious that $\left(Tu\right)\left(t\right)\ge 0$ and ${\left(Tu\right)}^{\prime }\left(t\right)\ge 0$. Moreover, for $t\in \left[0,1\right]$, by Lemma 2.3, we have
$\begin{array}{rcl}{\left(Tu\right)}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\end{array}$
and hence
$\begin{array}{rcl}\parallel Tu\parallel & \le & {\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
Moreover, it follows from Lemma 2.2 that for $t\in \left[\theta ,1-\theta \right]$,
$\begin{array}{rcl}{\left(Tu\right)}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}\parallel Tu\parallel .\end{array}$
Then we get
$\underset{t\in \left[\theta ,1-\theta \right]}{min}{\left(Tu\right)}^{\prime }\left(t\right)\ge {d}_{0}\parallel Tu\parallel ,$

which shows that $TK\subset K$. □

Similar to the proof of Lemma 2.4 in [28], we can get the following lemma.

Lemma 2.6 $T:K\to K$ is completely continuous.

## 3 Main results

Denote
$\begin{array}{c}{f}^{0}=\underset{x+y\to {0}^{+}}{lim sup}\underset{t\in \left[0,1\right]}{max}\frac{f\left(t,x,y\right)}{x+y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{g}^{0}=\underset{x+y\to {0}^{+}}{lim sup}\underset{t\in \left[0,1\right]}{max}\frac{g\left(t,x,y\right)}{x+y},\hfill \\ {f}_{0}=\underset{x+y\to {0}^{+}}{lim inf}\underset{t\in \left[\theta ,1-\theta \right]}{min}\frac{f\left(t,x,y\right)}{x+y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{g}_{0}=\underset{x+y\to {0}^{+}}{lim inf}\underset{t\in \left[\theta ,1-\theta \right]}{min}\frac{g\left(t,x,y\right)}{x+y},\hfill \\ {f}^{\mathrm{\infty }}=\underset{x+y\to +\mathrm{\infty }}{lim sup}\underset{t\in \left[0,1\right]}{max}\frac{f\left(t,x,y\right)}{x+y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{g}^{\mathrm{\infty }}=\underset{x+y\to +\mathrm{\infty }}{lim sup}\underset{t\in \left[0,1\right]}{max}\frac{g\left(t,x,y\right)}{x+y},\hfill \\ {f}_{\mathrm{\infty }}=\underset{x+y\to +\mathrm{\infty }}{lim inf}\underset{t\in \left[\theta ,1-\theta \right]}{min}\frac{f\left(t,x,y\right)}{x+y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{g}_{\mathrm{\infty }}=\underset{x+y\to +\mathrm{\infty }}{lim inf}\underset{t\in \left[\theta ,1-\theta \right]}{min}\frac{g\left(t,x,y\right)}{x+y},\hfill \\ {A}_{1}={\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{1}=2{\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(s\right)+{\mathrm{\Phi }}_{2}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ {A}_{2}={d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{2}={d}_{0}{\int }_{\theta }^{1-\theta }\left({c}_{0}{\mathrm{\Phi }}_{1}\left(s\right)+{d}_{0}{\mathrm{\Phi }}_{2}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

Theorem 3.1 Assume that ${A}_{1}{f}^{0}<1<{A}_{2}{f}_{\mathrm{\infty }}$ and ${B}_{1}{g}^{0}<1<{B}_{2}{g}_{\mathrm{\infty }}$. Then BVP (1.6) has at least one positive solution.

Proof In view of ${A}_{1}{f}^{0}<1$ and ${B}_{1}{g}^{0}<1$, there exists ${\epsilon }_{1}>0$ such that
${A}_{1}\left({f}^{0}+{\epsilon }_{1}\right)\le 1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{1}\left({g}^{0}+{\epsilon }_{1}\right)\le 1.$
(3.1)
By the definition of ${f}^{0}$, ${g}^{0}$, we may choose ${\sigma }_{1}>0$ so that
$\begin{array}{r}f\left(t,x,y\right)\le \left({f}^{0}+{\epsilon }_{1}\right)\left(x+y\right),\\ g\left(t,x,y\right)\le \left({g}^{0}+{\epsilon }_{1}\right)\left(x+y\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],\left(x+y\right)\in \left[0,{\sigma }_{1}\right].\end{array}$
(3.2)
Set ${\mathrm{\Omega }}_{1}=\left\{u\in E|\parallel u\parallel <{\sigma }_{1}/2\right\}$. It follows from (3.1), (3.2), Lemmas 2.2 and 2.3 that for any $u\in K\cap \partial {\mathrm{\Omega }}_{1}$, $s\in \left[0,1\right]$,
$\begin{array}{c}{\int }_{0}^{1}\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(\tau \right)+{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\left({g}^{0}+{\epsilon }_{1}\right)\left(u\left(\tau \right)+{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \hfill \\ \phantom{\rule{1em}{0ex}}\le 2\parallel u\parallel \left({g}^{0}+{\epsilon }_{1}\right){\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(\tau \right)+{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \hfill \\ \phantom{\rule{1em}{0ex}}\le {\sigma }_{1}.\hfill \end{array}$
(3.3)
Then, by (3.1), (3.2) and (3.3), we have
$\begin{array}{rcl}{\left(Tu\right)}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\left({f}^{0}+{\epsilon }_{1}\right)\left({\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ +{\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\left({f}^{0}+{\epsilon }_{1}\right)\left({\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(\tau \right)+{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\left({g}^{0}+{\epsilon }_{1}\right)\left(u\left(\tau \right)+{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\left({f}^{0}+{\epsilon }_{1}\right)\left({\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(s\right)+{\mathrm{\Phi }}_{2}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right)\left({g}^{0}+{\epsilon }_{1}\right)2\parallel u\parallel \\ \le & \parallel u\parallel ,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\end{array}$
Therefore,
$\parallel Tu\parallel \le \parallel u\parallel ,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\mathrm{\Omega }}_{1}.$
(3.4)
On the other hand, since $1<{A}_{2}{f}_{\mathrm{\infty }}$ and $1<{B}_{2}{g}_{\mathrm{\infty }}$, there exists ${\epsilon }_{2}>0$ such that
${A}_{2}\left({f}_{\mathrm{\infty }}-{\epsilon }_{2}\right)\ge 1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{2}\left({g}_{\mathrm{\infty }}-{\epsilon }_{2}\right)\ge 1.$
(3.5)
By the definition of ${f}_{\mathrm{\infty }}$, ${g}_{\mathrm{\infty }}$, we may choose ${\sigma }_{2}^{\prime }>{\sigma }_{1}$ so that
$\begin{array}{r}f\left(t,x,y\right)\ge \left({f}_{\mathrm{\infty }}-{\epsilon }_{2}\right)\left(x+y\right),\\ g\left(t,x,y\right)\ge \left({g}_{\mathrm{\infty }}-{\epsilon }_{2}\right)\left(x+y\right),\phantom{\rule{1em}{0ex}}t\in \left[\theta ,1-\theta \right],\left(x+y\right)\in \left[{\sigma }_{2}^{\prime },+\mathrm{\infty }\right).\end{array}$
(3.6)
Let ${\sigma }_{2}=max\left\{2{\sigma }_{1},{\sigma }_{2}^{\prime }/{d}_{0}\right\}$ and set ${\mathrm{\Omega }}_{2}=\left\{u\in E|\parallel u\parallel <{\sigma }_{2}\right\}$. Then $u\in K\cap \partial {\mathrm{\Omega }}_{2}$ implies that ${\sigma }_{2}^{\prime }\le {d}_{0}\parallel u\parallel \le {u}^{\prime }\left(t\right)$, $t\in \left[\theta ,1-\theta \right]$. So, for $s\in \left[\theta ,1-\theta \right]$, in view of Lemmas 2.2 and 2.3, we have
$\begin{array}{c}{\int }_{0}^{1}\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \hfill \\ \phantom{\rule{1em}{0ex}}\ge {\int }_{\theta }^{1-\theta }\left({c}_{0}{\mathrm{\Phi }}_{1}\left(\tau \right)+{d}_{0}{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\left({g}_{\mathrm{\infty }}-{\epsilon }_{2}\right)\left(u\left(\tau \right)+{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \hfill \\ \phantom{\rule{1em}{0ex}}\ge \parallel u\parallel \left({g}_{\mathrm{\infty }}-{\epsilon }_{2}\right){d}_{0}{\int }_{\theta }^{1-\theta }\left({c}_{0}{\mathrm{\Phi }}_{1}\left(\tau \right)+{d}_{0}{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \ge {\sigma }_{2}.\hfill \end{array}$
(3.7)
Then, for $t\in \left[\theta ,1-\theta \right]$, by (3.5), (3.6), (3.7), Lemmas 2.2 and 2.3, we have
$\begin{array}{rcl}{\left(Tu\right)}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\left({f}_{\mathrm{\infty }}-{\epsilon }_{2}\right)\left({\int }_{0}^{1}\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\left({f}_{\mathrm{\infty }}-{\epsilon }_{2}\right){\sigma }_{2}\\ \ge & \parallel u\parallel ,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\end{array}$
Therefore,
$\parallel Tu\parallel \ge \parallel u\parallel ,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\mathrm{\Omega }}_{2}.$
(3.8)
Therefore, it follows from the first part of Theorem 1.1 that T has a fixed point ${u}_{1}\in K\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)$. Consequently, BVP (1.6) has a positive solution $\left(u,v\right)\in K×K$, here
$\left\{\begin{array}{c}u\left(t\right)={u}_{1}\left(t\right),\hfill \\ v\left(t\right)={\int }_{0}^{1}{G}_{S}\left(t,s\right)g\left(s,{u}_{1}\left(s\right),{u}_{1}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$

□

Theorem 3.2 Assume that ${A}_{1}{f}^{\mathrm{\infty }}<1<{A}_{2}{f}_{0}$ and ${B}_{1}{g}^{\mathrm{\infty }}<1<{B}_{2}{g}_{0}$. Then BVP (1.6) has at least one positive solution.

Proof The proof is similar to Theorem 3.1 and therefore omitted. □

Theorem 3.3 Assume that ${A}_{2}{f}_{0}>1$, ${A}_{2}{f}_{\mathrm{\infty }}>1$, ${B}_{2}{g}_{0}>1$, ${B}_{2}{g}_{\mathrm{\infty }}>1$, ${B}_{2}{g}^{0}<2$ and there is a $\mu >0$ such that
$max\left\{g\left(t,x,y\right),t\in \left[0,1\right],\left(x+y\right)\in \left[0,\mu \right]\right\}<\frac{2\mu }{{B}_{1}};$
(3.9)
$max\left\{f\left(t,x,y\right),t\in \left[0,1\right],\left(x+y\right)\in \left[0,\mu \right]\right\}<\frac{\mu }{2{A}_{1}}.$
(3.10)

Then BVP (1.6) has at least two positive solutions.

Proof Firstly, in view of ${A}_{2}{f}_{0}>1$ and ${B}_{2}{g}_{0}>1$, there exists $\epsilon >0$ such that
${A}_{2}\left({f}_{0}-\epsilon \right)\ge 1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{2}\left({g}_{0}-\epsilon \right)\ge 1.$
(3.11)
By the definition of ${f}_{0}$, ${g}_{0}$, we may choose ${\stackrel{ˆ}{\sigma }}_{1}>0$ so that
$\begin{array}{r}f\left(t,x,y\right)\ge \left({f}_{0}-\epsilon \right)\left(x+y\right),\\ g\left(t,x,y\right)\ge \left({g}_{0}-\epsilon \right)\left(x+y\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],\left(x+y\right)\in \left[0,{\stackrel{ˆ}{\sigma }}_{1}\right].\end{array}$
(3.12)
Moreover, from ${B}_{2}{g}^{0}<2$, take ${\rho }_{1}$ satisfying $0<{\rho }_{1}<\frac{{B}_{2}}{{B}_{1}}\stackrel{ˆ}{{\sigma }_{1}}<\mu$ such that
$g\left(t,x,y\right)\le \frac{2{\rho }_{1}}{{B}_{2}},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left[0,1\right],x+y\in \left[0,{\rho }_{1}\right].$
Set ${\mathrm{\Omega }}_{1}=\left\{u\in E|\parallel u\parallel <{\rho }_{1}/2\right\}$. It follows from (3.11), (3.12), Lemmas 2.2 and 2.3 that for any $u\in K\cap \partial {\mathrm{\Omega }}_{1}$,
$\begin{array}{rcl}{\int }_{0}^{1}\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau & \le & \frac{2{\rho }_{1}}{{B}_{2}}{\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(\tau \right)+{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ =& \frac{{B}_{1}}{{B}_{2}}{\rho }_{1}<{\stackrel{ˆ}{\sigma }}_{1},\phantom{\rule{1em}{0ex}}s\in \left[0,1\right].\end{array}$
(3.13)
Then, for $t\in \left[\theta ,1-\theta \right]$, by (3.11), (3.12), (3.13), Lemmas 2.2 and 2.3, we have
$\begin{array}{rcl}{\left(Tu\right)}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\left({f}_{0}-\epsilon \right)\left({\int }_{0}^{1}\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\left({f}_{0}-\epsilon \right)\left({\int }_{\theta }^{1-\theta }\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\left({f}_{0}-\epsilon \right)\\ ×\left({\int }_{\theta }^{1-\theta }\left({c}_{0}{\mathrm{\Phi }}_{1}\left(\tau \right)+{d}_{0}{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\left({g}_{0}-\epsilon \right)\left(u\left(\tau \right)+{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {d}_{0}{\int }_{\theta }^{1-\theta }{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\left({f}_{0}-\epsilon \right)\left({\int }_{\theta }^{1-\theta }\left({c}_{0}{\mathrm{\Phi }}_{1}\left(\tau \right)+{d}_{0}{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\left({g}_{0}-\epsilon \right){d}_{0}\parallel u\parallel \\ \ge & \parallel u\parallel ,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\end{array}$
Thus,
$\parallel Tu\parallel \ge \parallel u\parallel ,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\mathrm{\Omega }}_{1}.$
(3.14)
Secondly, similar to the proof of (3.8), we may choose ${\sigma }_{2}>\mu$ and set ${\mathrm{\Omega }}_{2}=\left\{u\in E|\parallel u\parallel <{\sigma }_{2}\right\}$, and easily get
$\parallel Tu\parallel \ge \parallel u\parallel ,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\mathrm{\Omega }}_{2}.$
(3.15)
Let ${\mathrm{\Omega }}_{3}=\left\{u\in E|\parallel u\parallel <\mu /2\right\}$. Then, for any $u\in K\cap \partial {\mathrm{\Omega }}_{3}$, it follows by (3.9) and (3.10) that
$\begin{array}{rcl}{\int }_{0}^{1}\left({G}_{S}\left(s,\tau \right)+\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}\right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau & \le & \frac{2\mu }{{B}_{1}}{\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(\tau \right)+{\mathrm{\Phi }}_{2}\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ \le & \mu ,\phantom{\rule{1em}{0ex}}s\in \left[0,1\right]\end{array}$
(3.16)
and
$\begin{array}{rcl}{\left(Tu\right)}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ <& {\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\frac{\mu }{2{A}_{1}}=\frac{\mu }{2}=\parallel u\parallel ,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\end{array}$
Thus,
$\parallel Tu\parallel <\parallel u\parallel ,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\mathrm{\Omega }}_{3},$
(3.17)

which together with (3.13), (3.14) shows that T has at least two fixed points in ${u}_{1}\in K\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{3}\right)$ and ${u}_{2}\in K\cap \left({\overline{\mathrm{\Omega }}}_{3}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)$. □

Similarly, we can get the following theorem.

Theorem 3.4 Assume that ${A}_{1}{f}^{0}<1$, ${A}_{1}{f}^{\mathrm{\infty }}<\frac{1}{2}$, ${B}_{1}{g}^{0}<1$, ${B}_{2}{g}_{\mathrm{\infty }}>{d}_{0}$ and there exists $\eta >0$ such that
$max\left\{g\left(t,x,y\right),t\in \left[\theta ,1-\theta \right],\left(x+y\right)\in \left[{d}_{0}\eta ,+\mathrm{\infty }\right)\right\}>\frac{{d}_{0}^{2}\eta }{{B}_{2}};$
(3.18)
$max\left\{f\left(t,x,y\right),t\in \left[\theta ,1-\theta \right],\left(x+y\right)\in \left[{d}_{0}\eta ,+\mathrm{\infty }\right)\right\}>\frac{\eta }{{A}_{2}}.$
(3.19)

Then BVP (1.6) has at least two positive solutions.

Theorem 3.5 If ${A}_{1}f\left(t,x,y\right)<\left(x+y\right)$ and ${B}_{1}g\left(t,x,y\right)<\left(x+y\right)$ for $t\in \left[0,1\right]$ and $\left(x+y\right)\in \left[0,+\mathrm{\infty }\right)$, then BVP (1.6) has no monotone positive solution.

Proof Suppose on the contrary that u is a monotone positive solution of BVP (1.6). Then $u\left(t\right)\ge 0$ and ${u}^{\prime }\left(t\right)\ge 0$ for $t\in \left[0,1\right]$, and
$\begin{array}{rcl}{u}^{\prime }\left(t\right)& =& {\int }_{0}^{1}\frac{\partial {G}_{S}\left(t,s\right)}{\partial t}f\left(s,{\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau ,\\ {\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ <& \frac{1}{{A}_{1}}{\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\left({\int }_{0}^{1}{G}_{S}\left(s,\tau \right)g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ +{\int }_{0}^{1}\frac{\partial {G}_{S}\left(s,\tau \right)}{\partial s}g\left(\tau ,u\left(\tau \right),{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\phantom{\rule{0.2em}{0ex}}ds\\ <& \frac{1}{{A}_{1}}\frac{1}{{B}_{1}}{\int }_{0}^{1}{\mathrm{\Phi }}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds{\int }_{0}^{1}\left({\mathrm{\Phi }}_{1}\left(s\right)+{\mathrm{\Phi }}_{2}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds{\int }_{0}^{1}\left(u\left(\tau \right)+{u}^{\prime }\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \\ <& \parallel u\parallel ,\end{array}$

which shows that $\parallel u\parallel <\parallel u\parallel$. This is a contradiction. □

Similarly, we can prove the following theorem.

Theorem 3.6 If ${A}_{2}f\left(t,x,y\right)>\left(x+y\right)$ and ${B}_{2}g\left(t,x,y\right)>\left(x+y\right)$ for $t\in \left[\theta ,1-\theta \right]$ and $\left(x+y\right)\in \left[0,+\mathrm{\infty }\right)$, then BVP (1.6) has no monotone positive solution.

## 4 Example

In this section, we give an example to illustrate our main results.

Consider the BVP:
$\left\{\begin{array}{c}-{u}^{‴}\left(t\right)=f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ -{v}^{‴}\left(t\right)=g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ u\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{u}^{\prime }\left(0\right)=\alpha \left[u\right]={\int }_{0}^{1}\left(1-s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{u}^{\prime }\left(1\right)=\beta \left[u\right]={\int }_{0}^{1}su\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ v\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{v}^{\prime }\left(0\right)=\alpha \left[v\right]={\int }_{0}^{1}\left(1-s\right)v\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{v}^{\prime }\left(1\right)=\beta \left[v\right]={\int }_{0}^{1}sv\left(s\right)\phantom{\rule{0.2em}{0ex}}ds.\hfill \end{array}$
(4.1)
For this BCs, the corresponding $\gamma \left(t\right)=\frac{2t-{t}^{2}}{2}$ and $\delta \left(t\right)=\frac{{t}^{2}}{2}$. A simple calculation shows that
$\begin{array}{c}\alpha \left[\gamma \right]=\frac{1}{8},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\alpha \left[\delta \right]=\frac{1}{24},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\beta \left[\gamma \right]=\frac{5}{24},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\beta \left[\delta \right]=\frac{1}{8},\hfill \\ D=\left(1-\alpha \left[\gamma \right]\right)\left(1-\beta \left[\delta \right]\right)-\alpha \left[\delta \right]\beta \left[\gamma \right]=\frac{109}{144},\hfill \\ {\mathcal{K}}_{A}\left(s\right):={\int }_{0}^{1}G\left(t,s\right)\left(1-t\right)\phantom{\rule{0.2em}{0ex}}dt=\frac{s}{8}-\frac{{s}^{2}}{4}+\frac{{s}^{3}}{6}-\frac{{s}^{4}}{24},\hfill \\ {\mathcal{K}}_{B}\left(s\right):={\int }_{0}^{1}G\left(t,s\right)t\phantom{\rule{0.2em}{0ex}}dt=\frac{5s}{24}-\frac{{s}^{2}}{4}+\frac{{s}^{4}}{24},\hfill \\ {\mathrm{\Phi }}_{1}\left(s\right)=\frac{265s}{218}-\frac{145{s}^{2}}{109}+\frac{13{s}^{3}}{109}-\frac{{s}^{4}}{218},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{\mathrm{\Phi }}_{2}\left(s\right)=\frac{156s}{109}-\frac{181{s}^{2}}{109}+\frac{26{s}^{3}}{109}-\frac{{s}^{4}}{109}.\hfill \end{array}$

Let $\theta =1/4$, then ${A}_{1}=719/3\text{,}270\approx 0.2199$, ${A}_{2}=126\text{,}517/3\text{,}348\text{,}480\approx 0.0378$, ${B}_{1}=2\text{,}767/3\text{,}270\approx 0.8462$, ${B}_{2}=4\text{,}438\text{,}339/428\text{,}605\text{,}440\approx 0.0104$.

Example 4.1 Let
$\begin{array}{c}f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right)=\frac{1}{1+t}\left[\frac{v\left(t\right)+{v}^{\prime }\left(t\right)}{{e}^{v\left(t\right)+{v}^{\prime }\left(t\right)}}+\frac{1\text{,}000{\left(v\left(t\right)+{v}^{\prime }\left(t\right)\right)}^{2}}{1+v\left(t\right)+{v}^{\prime }\left(t\right)}\right],\hfill \\ g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right)=\frac{1}{10\left(1+t\right)}\left[\frac{u\left(t\right)+{u}^{\prime }\left(t\right)}{{e}^{u\left(t\right)+{u}^{\prime }\left(t\right)}}+\frac{2\text{,}000{\left(u\left(t\right)+{u}^{\prime }\left(t\right)\right)}^{2}}{1+u\left(t\right)+{u}^{\prime }\left(t\right)}\right].\hfill \end{array}$

It is easy to compute that ${f}^{0}=1$, ${f}_{\mathrm{\infty }}=\frac{4\text{,}000}{7}$, ${g}^{0}=\frac{1}{10}$, ${g}_{\mathrm{\infty }}=\frac{800}{7}$, which show that ${A}_{1}{f}^{0}<1<{A}_{2}{f}_{\mathrm{\infty }}$ and ${B}_{1}{g}^{0}<1<{B}_{2}{g}_{\mathrm{\infty }}$. So, it follows from Theorem 3.1 that BVP (4.1) has at least one positive solution.

Example 4.2 Let
$\begin{array}{c}f\left(t,v\left(t\right),{v}^{\prime }\left(t\right)\right)=\frac{1}{1+t}\left[\frac{50\left(v\left(t\right)+{v}^{\prime }\left(t\right)\right)}{{e}^{v\left(t\right)+{v}^{\prime }\left(t\right)}}+\frac{50{\left(v\left(t\right)+{v}^{\prime }\left(t\right)\right)}^{2}}{1\text{,}000+v\left(t\right)+{v}^{\prime }\left(t\right)}\right],\hfill \\ g\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right)=\left[1+{\left(t-\frac{3}{4}\right)}^{2}\right]\left[\frac{100\left(u\left(t\right)+{u}^{\prime }\left(t\right)\right)}{{e}^{200\left(u\left(t\right)+{u}^{\prime }\left(t\right)\right)}}+\frac{100{\left(u\left(t\right)+{u}^{\prime }\left(t\right)\right)}^{2}}{1\text{,}000+u\left(t\right)+{u}^{\prime }\left(t\right)}\right].\hfill \end{array}$

It is easy to compute that ${f}_{0}=\frac{200}{7}$, ${f}_{\mathrm{\infty }}=\frac{200}{7}$, ${g}_{0}=100$, ${g}_{\mathrm{\infty }}=100$ and ${g}^{0}=\frac{625}{4}$, which show that ${A}_{2}{f}_{0}>1$, ${A}_{2}{f}_{\mathrm{\infty }}>1$, ${B}_{2}{g}_{0}>1$, ${B}_{2}{g}_{\mathrm{\infty }}>1$ and ${B}_{2}{g}^{0}<2$.

Choose $\mu =14$,
$\begin{array}{c}max\left\{g\left(t,x,y\right),t\in \left[0,1\right],\left(x+y\right)\in \left[0,14\right]\right\}=\frac{50}{e}+\frac{4\text{,}900}{507}<29<\frac{2\mu }{{B}_{1}};\hfill \\ max\left\{f\left(t,x,y\right),t\in \left[0,1\right],\left(x+y\right)\in \left[0,14\right]\right\}=\frac{25}{16}\left(\frac{1}{2e}+\frac{9\text{,}800}{507}\right)<21<\frac{\mu }{2{A}_{1}}.\hfill \end{array}$

So, it follows from Theorem 3.3 that BVP (4.1) has at least two positive solutions.

## Declarations

### Acknowledgements

The work is supported by the Science and Technology Foundation of Hebei Province (Z2013016) and the Science and Technology Plan Foundation of Tangshan (12110233b) and the Scientific Research Foundation of Tangshan College (13011B). The author would like to thank the anonymous referees very much for helpful comments and suggestions which led to the improvement of presentation and quality of the work.

## Authors’ Affiliations

(1)
Department of Basic Science, Tangshan College, Tangshan, Hebei, 063000, People’s Republic of China

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