Open Access

Monotone positive solution of a fourth-order BVP with integral boundary conditions

Boundary Value Problems20152015:172

https://doi.org/10.1186/s13661-015-0441-2

Received: 29 July 2015

Accepted: 12 September 2015

Published: 22 September 2015

Abstract

In this paper, we investigate the existence of concave and monotone positive solutions for a nonlinear fourth-order differential equation with integral boundary conditions of the form \(x^{(4)}(t)=f(t,x(t),x'(t),x''(t))\), \(t\in[0,1]\), \(x(0)=x'(1)=x'''(1)=0\), \(x''(0)=\int_{0}^{1}g(s)x''(s) \, \mathrm{d}s\), where \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\). By using a fixed point theorem of cone expansion and compression of norm type, the existence and nonexistence of concave and monotone positive solutions for the above boundary value problems is obtained. Meanwhile, as applications of our results, some examples are given.

Keywords

fourth-order differential equationintegral boundary conditionmonotone positive solutionexistencenonexistence

MSC

34B1534B18

1 Introduction

This paper is the follow-up of [1]. In [1], by using a fixed point theorem for the sum of two operators due to O’Regan [2], we obtained existence of solutions for a fully nonlinear fourth-order equation with integral boundary conditions of type
$$\left \{\textstyle\begin{array}{l} x^{(4)}(t)=f(t,x(t),x'(t),x''(t),x'''(t)), \quad t\in[0,1], \\ x(0)=x'(1)=x'''(1)=0, \\ x''(0)=\int_{0}^{1}h(s, x(s),x'(s),x''(s))\, \mathrm{d}s. \end{array}\displaystyle \right . $$
In this paper, we study the existence of concave and monotone positive solutions for its simplified form
$$ x^{(4)}(t)=f\bigl(t,x(t),x'(t),x''(t) \bigr), \quad t\in[0,1] $$
(1.1)
subject to the integral boundary conditions
$$ \left \{\textstyle\begin{array}{l} x(0)=x'(1)=x'''(1)=0, \\ x''(0)=\int_{0}^{1}g(s)x''(s)\, \mathrm{d}s, \end{array}\displaystyle \right . $$
(1.2)
where \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\).

It is well known that fourth-order boundary value problems models bending equilibria of elastic beams, and have been studied extensively. Among a substantial number of works dealing with fourth-order boundary value problems, we mention [1, 331]. We notice that if \(g(\cdot)\equiv0\) in (1.2), the models are known as the one endpoint simply supported and the other one sliding clamped beam. The study of this class of problems was considered by some authors via various methods, we refer the reader to [4, 7, 10, 14, 15, 23, 26].

The aim of this paper is to establish the existence and nonexistence results of concave and monotone positive solutions for the problems (1.1), (1.2). Here, a solution \(x(t)\) of the BVP (1.1), (1.2) is said to be monotone and positive if \(x'(t)\geq0\) on \([0,1]\) and \(x(t)>0\) on \(t\in(0,1]\). Our main tool is the fixed point theorem of cone expansion and compression of norm type [32]. The paper [33] motivated our study.

2 Preliminary

In this section, we present some lemmas which are needed for our main results.

Throughout this paper, we assume that \(f:[0,1]\times [0,+\infty)^{2}\times(-\infty,0]\rightarrow[0,+\infty)\) and \(g: [0,1]\rightarrow[0,+\infty)\) are continuous, moreover, \(\mu:=\int_{0}^{1}g(s)\, \mathrm{d}s<1\).

Simple computations lead to the following lemma.

Lemma 2.1

For any \(h\in C[0,1]\), the BVP
$$ \left \{ \textstyle\begin{array}{l} x^{(4)}(t)=h(t), \quad t\in[0,1], \\ x(0)=x'(1)=x'''(1)=0, \\ x''(0)=\int_{0}^{1}g(s)x''(s)\, \mathrm{d}s, \end{array}\displaystyle \right . $$
(2.1)
has a unique solution
$$x(t)=\int_{0}^{1} \biggl[G_{1}(t,s)+\frac{2t-t^{2}}{2(1-\mu)}\int_{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d}\tau \biggr]h(s)\,\mathrm{d}s, $$
where
$$\begin{aligned}& G_{1}(t,s)=\left \{ \textstyle\begin{array}{l@{\quad}l} t(s-\frac{1}{2}s^{2})-\frac{1}{6}t^{3}, & 0\leq t\leq s \leq1, \\ s(t-\frac{1}{2}t^{2})-\frac{1}{6}s^{3}, & 0\leq s\leq t \leq1, \end{array}\displaystyle \right .\qquad \\& G_{2}(t,s)=\left \{ \textstyle\begin{array}{l@{\quad}l} t, & 0\leq t\leq s \leq1, \\ s, & 0\leq s\leq t \leq1. \end{array}\displaystyle \right . \end{aligned}$$

Lemma 2.2

Let \(G_{1}(t,s)\) be as in Lemma  2.1. Then
$$G_{1}(t,s)\geq\frac{1}{2}\biggl(t-\frac{1}{2}t^{2} \biggr)s,\quad (t,s)\in[0,1]\times[0,1]. $$

Proof

For \(0\leq t\leq s \leq1\), one has
$$\begin{aligned} G_{1}(t,s) = & t\biggl(s-\frac{1}{2}s^{2}\biggr)- \frac{1}{6}t^{3} \\ \geq& ts\biggl(1-\frac{1}{2}s\biggr)- \frac{1}{6}t^{2}s \\ \geq& \frac{1}{2}ts-\frac{1}{6}t^{2}s \\ \geq& \frac{1}{2}\biggl(t-\frac{1}{2}t^{2}\biggr)s. \end{aligned}$$
On the other hand, for \(0\leq s\leq t \leq1\), we have \(\frac{1}{6}s^{2}+\frac{1}{6}t^{2}\leq\frac{1}{3}t\), and then
$$\begin{aligned} G_{1}(t,s) = & s\biggl(t-\frac{1}{2}t^{2}\biggr)- \frac{1}{6}s^{3} \\ \geq& s\biggl(t-\frac{1}{2}t^{2}\biggr)- \biggl( \frac{1}{3}t-\frac{1}{6}t^{2}\biggr)s \\ = & \frac{2}{3}\biggl(t-\frac{1}{2}t^{2}\biggr)s \\ \geq& \frac{1}{2}\biggl(t-\frac{1}{2}t^{2}\biggr)s. \end{aligned}$$
This completes the proof of the lemma. □

Lemma 2.3

If \(h\in C[0,1]\) with \(h(t)\geq0\) on \([0,1]\), then the unique solution \(x=x(t)\) of the BVP (2.1) satisfies:
  1. (1)

    \(x(t)\geq0\) for \(t\in[0,1]\);

     
  2. (2)
    \(x'(t)\geq0\), \(x''(t)\leq0\) for \(t\in[0,1]\), and
    $$x(t)\geq\frac{1}{2}\biggl(t-\frac{1}{2}t^{2}\biggr)\bigl\Vert x''\bigr\Vert _{\infty}, \quad t\in[0,1]. $$
     

Proof

(1) From Lemma 2.2 and the fact
$$ts\leq G_{2}(t,s)\leq s,\quad \forall(t,s)\in[0,1]\times[0,1], $$
it follows that
$$ x(t)=\int_{0}^{1} \biggl[G_{1}(t,s)+ \frac{2t-t^{2}}{2(1-\mu)}\int_{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d}\tau \biggr]h(s)\,\mathrm{d}s\geq0, \quad t\in[0,1]. $$
(2.2)
(2) Note that whenever \((t,s)\in[0,1]\times[0,1]\),
$$\frac{\partial}{\partial t}G_{1}(t,s)\geq0,\qquad \frac{\partial^{2}}{\partial t^{2}}G_{1}(t,s)=-G_{2}(t,s) \leq0, $$
it follows that
$$ \begin{aligned} &x'(t)=\int_{0}^{1} \biggl[\frac{\partial}{\partial t}G_{1}(t,s)+\frac{1-t}{1-\mu }\int _{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d} \tau \biggr]h(s)\,\mathrm{d}s\geq0, \quad t\in[0,1], \\ &x''(t)=\int_{0}^{1} \biggl[ -G_{2}(t,s)-\frac{1}{1-\mu}\int_{0}^{1}G_{2}( \tau,s) g(\tau)\,\mathrm{d}\tau \biggr]h(s)\,\mathrm{d}s\leq0,\quad t\in[0,1]. \end{aligned} $$
(2.3)
On the one hand, by (2.3), we have
$$ \bigl\Vert x''\bigr\Vert _{\infty}\leq\int _{0}^{1} \biggl[s+\frac{1}{1-\mu}\int _{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d} \tau \biggr]h(s)\,\mathrm{d}s. $$
(2.4)
On the other hand, in view of (2.2) and Lemma 2.2, we have
$$\begin{aligned} x(t) = & \int_{0}^{1} \biggl[G_{1}(t,s)+ \frac{2t-t^{2}}{2(1-\mu)}\int_{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d}\tau \biggr]h(s)\,\mathrm{d}s \\ \geq& \int_{0}^{1} \biggl[\frac{1}{2} \biggl(t-\frac{1}{2}t^{2}\biggr)s+\frac {2t-t^{2}}{2(1-\mu)}\int _{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d} \tau \biggr]h(s)\,\mathrm{d}s \\ \geq& \frac{1}{2}\biggl(t-\frac{1}{2}t^{2}\biggr)\int _{0}^{1} \biggl[s+\frac{1}{1-\mu }\int _{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d} \tau \biggr]h(s)\,\mathrm{d}s, \quad t\in[0,1]. \end{aligned}$$
(2.5)
It follows from (2.4) and (2.5) that
$$x(t)\geq \frac{1}{2}\biggl(t-\frac{1}{2}t^{2}\biggr)\bigl\Vert x''\bigr\Vert _{\infty}, \quad t\in[0,1]. $$
This completes the proof of the lemma. □
Let
$$E=\bigl\{ x\in C^{2}[0,1]: x(0)=x'(1)=0\bigr\} $$
be endowed with the norm \(\|x\|=\max_{t\in[0,1]}|x''(t)|=:\|x''\| _{\infty}\). Then E is a Banach space. If we denote
$$K= \biggl\{ x\in E:x(t)\geq\frac{1}{2}\biggl(t-\frac{1}{2}t^{2} \biggr)\|x\|, x'(t)\geq0,x''(t)\leq0, t \in[0,1] \biggr\} , $$
then it is easy to see that K is a cone in E.
Now, we define an operator T on K as follows: for \(x\in K\),
$$(Tx) (t)=\int_{0}^{1} \biggl[G_{1}(t,s)+ \frac{2t-t^{2}}{2(1-\mu)}\int_{0}^{1}G_{2}(\tau,s) g(\tau)\,\mathrm{d}\tau \biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s. $$
By Lemma 2.3, we know that \(T(K)\subset K\) and if x is a fixed point of T, then x is a concave and monotone positive solution of the BVP (1.1), (1.2).

Lemma 2.4

\(T:K\rightarrow K\) is completely continuous.

Proof

First, we show that T is continuous. To do this, suppose \(x_{n}, x_{0}\in K\) and \(\|x_{n}-x_{0}\|\rightarrow0\) (\(n\rightarrow\infty\)). Then there exists \(M_{1}>0\) such that \(\|x_{0}\|, \|x_{n}\|\leq M_{1}\) for all \(n\in\mathbb{N}=\{ 1,2,\ldots\}\). Hence from the continuity of f on \([0,1]\times[0,M_{1}]^{2}\times[-M_{1},0]\), we have
$$f\bigl(t,x_{n}(t),x_{n}'(t),x_{n}''(t) \bigr)\rightarrow f\bigl(t,x_{0}(t),x_{0}'(t),x_{0}''(t) \bigr)\quad (n\rightarrow\infty) $$
uniformly on \([0,1]\). Also, since
$$\begin{aligned} \begin{aligned} 0 & \leq G_{2}(t,s)+\frac{1}{1-\mu}\int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau \\ & \leq 1+\frac{1}{1-\mu}\int_{0}^{1}g(\tau)\, \mathrm{d}\tau \\ & = \frac{1}{1-\mu},\quad (t,s)\in[0,1]\times[0,1], \end{aligned} \end{aligned}$$
we have
$$\begin{aligned} (Tx_{n})''(t) = & \int _{0}^{1}\biggl[-G_{2}(t,s)- \frac{1}{1-\mu}\int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr] f\bigl(s,x_{n}(s),x_{n}'(s),x_{n}''(s) \bigr)\,\mathrm{d}s \\ \rightarrow& \int_{0}^{1}\biggl[-G_{2}(t,s)- \frac{1}{1-\mu}\int_{0}^{1}G_{2}(\tau ,s)g(\tau)\,\mathrm{d}\tau\biggr] f\bigl(s,x_{0}(s),x_{0}'(s),x_{0}''(s) \bigr)\,\mathrm{d}s \\ = & (Tx_{0})''(t) (n\rightarrow\infty) \quad \mbox{uniformly on }[0,1], \end{aligned}$$
i.e.,
$$\bigl\Vert (Tx_{n})''-(Tx_{0})'' \bigr\Vert _{\infty}=\|Tx_{n}-Tx_{0}\|\rightarrow0 \quad (n\rightarrow\infty). $$
Therefore \(T:K\rightarrow K\) is continuous.
Next, we prove that T is relatively compact. With this aim, let \(D\subset K\) be a bounded set, then there exists a constant \(M_{2}>0\) such that \(\|x\|\leq M_{2}\) for all \(x\in D\). Suppose that \(\{y_{n}\}\subset T(D)\), there exist \(\{x_{n}\} \subset D\) such that \(Tx_{n}=y_{n}\). Let
$$M_{3}=\sup\bigl\{ f(t,x_{0},x_{1},x_{2}):(t,x_{0},x_{1},x_{2}) \in[0,1]\times[0,M_{2}]^{2}\times [-M_{2},0]\bigr\} . $$
For all \(n\in\mathbb{N}\), we have
$$\begin{aligned} \bigl\vert y_{n}''(t)\bigr\vert = & \bigl\vert (Tx_{n})''(t)\bigr\vert \\ = & \int_{0}^{1} \biggl[G_{2}(t,s)+ \frac{1}{1-\mu}\int_{0}^{1}G_{2}( \tau,s)g(\tau )\,\mathrm{d}\tau \biggr] f\bigl(s,x_{n}(s),x_{n}'(s),x_{n}''(s) \bigr)\,\mathrm{d}s \\ \leq& M_{3} \int_{0}^{1}\biggl[s+ \frac{1}{1-\mu}\int_{0}^{1}sg(\tau)\,\mathrm{d} \tau \biggr]\,\mathrm{d}s \\ = & \frac{M_{3}}{2(1-\mu)},\quad t\in[0,1] \end{aligned}$$
and
$$\begin{aligned} \bigl\vert y_{n}^{(4)}(t)\bigr\vert = & \bigl\vert (Tx_{n})^{(4)}(t)\bigr\vert \\ = & f\bigl(t,x_{n}(t),x_{n}'(t),x_{n}''(t) \bigr) \\ \leq& M_{3},\quad t\in[0,1]. \end{aligned}$$
Consequently there exists a constant \(M_{4}>0\) such that, for all \(n\in \mathbb{N}\),
$$\bigl\vert y_{n}'''(t)\bigr\vert \leq M_{4}, \quad t\in[0,1]. $$
By the Arzela-Ascoli theorem, we know that \(\{y_{n}''\}\) has a convergent subsequence in supremum norm, i.e., \(\{y_{n}\}\) has a convergent subsequence in E, which indicates that \(T(D)\subset K\) is relatively compact in E. This completes the proof of the lemma. □

The following fixed point theorem of cone expansion and compression of norm type plays a crucial role in our paper.

Lemma 2.5

([32])

Let E be a Banach space and let K be a cone in E. Assume that \(\Omega_{1}\) and \(\Omega_{2}\) are bounded open subsets of E such that \(\theta\in\Omega_{1}\subset\overline{\Omega}_{1}\subset\Omega_{2}\), and let \(T:K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\rightarrow K\) be a completely continuous operator such that either
  1. (i)

    \(\|Tx\|\leq\|x\|\) for \(x\in K\cap\partial\Omega_{1}\) and \(\|Tx\|\geq\|x\|\) for \(x\in K\cap\partial\Omega_{2}\), or

     
  2. (ii)

    \(\|Tx\|\geq\|x\|\) for \(x\in K\cap\partial\Omega_{1}\) and \(\|Tx\|\leq\|x\|\) for \(x\in K\cap\partial\Omega_{2}\).

     
Then T has a fixed point in \(K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\).

3 Main results

For convenience, firstly we introduce some notations:
$$\begin{aligned}& f^{0}={ \limsup_{x_{0}+x_{1}-x_{2}\rightarrow0^{+}}\max_{t\in [0,1]}} \frac{f(t,x_{0},x_{1},x_{2})}{x_{0}+x_{1}-x_{2}},\qquad f_{0}={ \liminf_{x_{0}+x_{1}-x_{2}\rightarrow0^{+}}\min _{t\in [0,1]}}\frac{f(t,x_{0},x_{1},x_{2})}{x_{0}+x_{1}-x_{2}}, \\& f^{\infty}={ \limsup_{x_{0}+x_{1}-x_{2}\rightarrow+\infty}\max_{t\in[0,1]}} \frac{f(t,x_{0},x_{1},x_{2})}{x_{0}+x_{1}-x_{2}},\qquad f_{\infty}={ \liminf_{x_{0}+x_{1}-x_{2}\rightarrow+\infty }\min _{t\in[0,1]}}\frac{f(t,x_{0},x_{1},x_{2})}{x_{0}+x_{1}-x_{2}}, \\& H_{1}=\frac{3}{2(1-\mu)}, \qquad H_{2}=\frac{1}{4} \int_{0}^{1}s^{2}\biggl(1- \frac {1}{2}s\biggr) \biggl[\frac{1}{2}+\frac{1}{1-\mu} \int _{0}^{1}\tau g(\tau)\,\mathrm{d}\tau \biggr]\, \mathrm{d}s. \end{aligned}$$

Theorem 3.1

If \(H_{1}f^{0}<1<H_{2}f_{\infty}\), then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.

Proof

Since \(H_{1}f^{0}<1\), there exists \(\varepsilon_{1}>0\) such that
$$ H_{1}\bigl(f^{0}+\varepsilon_{1}\bigr)< 1. $$
(3.1)
By the definition of \(f^{0}\) and the continuity of f, there exists \(\rho _{1}>0\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[0,\rho_{1}]\),
$$ f(t,x_{0},x_{1},x_{2})< \bigl(f^{0}+ \varepsilon_{1}\bigr) (x_{0}+x_{1}-x_{2}). $$
(3.2)
Let \(\Omega_{1}=\{x\in E:\|x\|<\rho/3\}\). For all \(x\in K\cap\partial \Omega_{1}\), from (3.1) and (3.2), we have
$$\begin{aligned} \bigl\vert (Tx)''(t)\bigr\vert =& \int _{0}^{1}\biggl[G_{2}(t,s)+ \frac{1}{1-\mu} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s \\ \leq& \int_{0}^{1}\biggl[s+\frac{1}{1-\mu}\int _{0}^{1}sg(\tau)\,\mathrm{d}\tau\biggr] \bigl(f^{0}+\varepsilon_{1}\bigr) \bigl(x(s)+x'(s)-x''(s) \bigr)\,\mathrm{d}s \\ \leq& H_{1}\bigl(f^{0}+\varepsilon_{1}\bigr)\|x \| \\ < & \|x\|, \quad t\in[0,1], \end{aligned}$$
which implies that
$$ \|Tx\|=\bigl\Vert (Tx)''\bigr\Vert _{\infty} < \|x\|, \quad \forall x\in K\cap\partial \Omega_{1}. $$
(3.3)
On the other hand, in view of \(H_{2}f_{\infty}>1\), there exists \(\varepsilon_{2}>0\) such that
$$ H_{2}(f_{\infty}-\varepsilon_{2})>1. $$
(3.4)
By the definition of \(f_{\infty}\), there exists \(\rho_{2}>\rho_{1}\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[\rho_{2},+\infty)\),
$$ f(t,x_{0},x_{1},x_{2})>(f_{\infty}- \varepsilon_{1}) (x_{0}+x_{1}-x_{2}). $$
(3.5)
Let \(\Omega_{2}=\{x\in E:\|x\|<\rho_{2}\}\). Then for all \(x\in K\cap \partial\Omega_{2}\), from Lemma 2.2, (3.4), and (3.5) it follows that
$$\begin{aligned} (Tx) (1) =& \int_{0}^{1}\biggl[G_{1}(1,s)+ \frac{1}{2(1-\mu)} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s \\ \geq& \int_{0}^{1}\biggl[\frac{1}{4}s+ \frac{1}{2(1-\mu)}\int_{0}^{1}\tau sg(\tau )\, \mathrm{d}\tau\biggr] (f_{\infty}-\varepsilon_{2}) \bigl(x(s)+x'(s)-x''(s)\bigr)\,\mathrm{d}s \\ \geq& \frac{1}{2}\int_{0}^{1}\biggl[ \frac{1}{2}s+\frac{s}{1-\mu}\int_{0}^{1} \tau g(\tau)\,\mathrm{d}\tau\biggr] (f_{\infty}-\varepsilon_{2}) \frac{1}{2}\biggl(s-\frac{1}{2}s^{2}\biggr)\|x\|\, \mathrm{d}s \\ = & H_{2}(f_{\infty}-\varepsilon_{2})\|x\| \\ > & \|x\|,\quad t\in[0,1], \end{aligned}$$
which implies that
$$ \|Tx\|\geq\|Tx\|_{\infty}\geq(Tx) (1)>\|x\|, \quad \forall x\in K\cap \partial\Omega_{2}. $$
(3.6)
Therefore, it follows from (3.3), (3.6), and Lemma 2.5 that the operator T has one fixed point \(x\in K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\), which is a concave and monotone positive solution of the BVP (1.1), (1.2). This completes the proof of the theorem. □

Corollary 3.1

Suppose that f is superlinear, i.e.,
$$f^{0}=0, \qquad f_{\infty}=+\infty. $$
Then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.

Theorem 3.2

If \(H_{1}f^{\infty}<1<H_{2}f_{0}\), then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.

Proof

Since \(H_{2}f_{0}>1\), there exists \(\varepsilon_{1}>0\) such that
$$ H_{2}(f_{0}-\varepsilon_{1})>1. $$
(3.7)
By the definition of \(f_{0}\), there exists \(\rho_{1}>0\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[0,\rho_{1}]\),
$$ f(t,x_{0},x_{1},x_{2})>(f_{0}- \varepsilon_{1}) (x_{0}+x_{1}-x_{2}). $$
(3.8)
Let \(\Omega_{1}=\{x\in E:\|x\|<\rho_{1}\}\). Then, for all \(x\in K\cap \partial\Omega_{1}\), from Lemma 2.2, (3.7), and (3.8) it follows that
$$\begin{aligned} (Tx) (1) =& \int_{0}^{1}\biggl[G_{1}(1,s)+ \frac{1}{2(1-\mu)} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s \\ \geq& \int_{0}^{1}\biggl[\frac{1}{4}s+ \frac{1}{2(1-\mu)}\int_{0}^{1}\tau sg(\tau )\, \mathrm{d}\tau\biggr] (f_{0}-\varepsilon_{1}) \bigl(x(s)+x'(s)-x''(s)\bigr)\,\mathrm{d}s \\ \geq& \frac{1}{2}\int_{0}^{1}\biggl( \frac{1}{2}s+\frac{s}{1-\mu}\int_{0}^{1} \tau g(\tau)\,\mathrm{d}\tau\biggr) (f_{0}-\varepsilon_{1}) \frac{1}{2}\biggl(s-\frac{1}{2}s^{2}\biggr)\|x\|\, \mathrm{d}s \\ = & H_{2}(f_{\infty}-\varepsilon_{1})\|x\| \\ > & \|x\|,\quad t\in[0,1], \end{aligned}$$
which implies that
$$ \|Tx\|>\|x\|,\quad \forall x\in K\cap\partial\Omega_{1}. $$
(3.9)
On the other hand, in view of \(H_{1}f^{\infty}<1\), there exists \(\varepsilon_{2}>0\) such that
$$ H_{2}\bigl(f^{\infty}+\varepsilon_{2}\bigr)< 1. $$
(3.10)
By the definition of \(f^{\infty}\), there exists \(\rho^{*}>3\rho_{1}\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[\rho^{*},+\infty)\),
$$f(t,x_{0},x_{1},x_{2})< \bigl(f^{\infty}+ \varepsilon_{2}\bigr) (x_{0}+x_{1}-x_{2}). $$
Let
$$\beta=\max\bigl\{ f(t,x_{0},x_{1},x_{2}):(t,x_{0},x_{1},x_{2}) \in[0,1]\times\bigl[0,\rho ^{*}\bigr]^{2}\times\bigl[-\rho^{*},0\bigr]\bigr\} . $$
Then for \((t,x_{0},x_{1},x_{2})\in[0,1]\times[0,+\infty)^{2}\times(-\infty,0]\) one has
$$ f(t,x_{0},x_{1},x_{2})< \bigl(f^{\infty}+ \varepsilon_{2}\bigr) (x_{0}+x_{1}-x_{2})+ \beta. $$
(3.11)
Now, we choose \(\rho_{2}>\frac{1}{3}\max\{\rho^{*},\frac{\beta H_{1}}{1-H_{1}(f^{\infty}+\varepsilon_{2})}\}\) and let
$$\Omega_{2}=\bigl\{ x\in E:\|x\|< \rho_{2}\bigr\} . $$
For all \(x\in K\cap\partial\Omega_{2}\), from (3.10) and (3.11) it follows that
$$\begin{aligned} \bigl\vert (Tx)''(t)\bigr\vert \leq& \biggl\Vert \int_{0}^{1}\biggl[G_{2}(t,s)+ \frac{1}{1-\mu} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s \biggr\Vert _{\infty} \\ \leq& { \max_{0\leq t \leq1}}\int_{0}^{1} \biggl[G_{2}(t,s)+\frac {1}{1-\mu} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]\,\mathrm{d}s \bigl\Vert f \bigl(s,x(s),x'(s),x''(s)\bigr)\bigr\Vert _{\infty} \\ \leq& \int_{0}^{1}\biggl(s+\frac{s}{1-\mu} \int_{0}^{1}g(\tau)\,\mathrm{d}\tau\biggr)\, \mathrm{d}s \bigl[\bigl(f^{\infty}+\varepsilon_{2}\bigr)\bigl\Vert x(s)+x'(s)-x''(s)\bigr\Vert _{\infty} +\beta\bigr] \\ \leq& H_{1}\bigl(f^{\infty}+\varepsilon_{2}\bigr)\|x \|+\frac{\beta}{3}H_{1} \\ < & \rho_{2}=\|x\|, \quad t\in[0,1], \end{aligned}$$
which implies that
$$ \|Tx\|< \|x\|, \quad \forall x\in K\cap\partial\Omega_{2}. $$
(3.12)
Therefore, it follows from (3.9), (3.12), and Lemma 2.5 that the operator T has one fixed point \(x\in K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\), which is a concave and monotone positive solution of the BVP (1.1), (1.2). This completes the proof of the theorem. □

Corollary 3.2

Suppose that f is sublinear, i.e.,
$$f_{0}=+\infty, \qquad f^{\infty}=0. $$
Then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.

Theorem 3.3

Suppose that
$$H_{1}f(t,x_{0},x_{1},x_{2})< x_{0}+x_{1}-x_{2} $$
for \((t,x_{0},x_{1},x_{2})\in[0,1]\times[0,+\infty)^{2}\times(-\infty,0]\) with \(x_{0}+x_{1}-x_{2}>0\). Then the BVP (1.1), (1.2) has no concave and monotone positive solution.

Proof

By contradiction, assume that x is a concave and monotone positive solution of the BVP (1.1), (1.2). Then
$$x(t)\geq0, \qquad x'(t)\geq0, \qquad x''(t) \leq0, \quad t\in[0,1] $$
and
$$x(t)+ x'(t)-x''(t)>0, \quad t\in(0,1]. $$
Hence
$$\begin{aligned} \bigl\vert x''(t)\bigr\vert = & \int _{0}^{1}\biggl[G_{2}(t,s)+ \frac{1}{1-\mu} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s \\ \leq& \int_{0}^{1}\biggl(s+\frac{1}{1-\mu} \int_{0}^{1}sg(\tau)\,\mathrm{d}\tau\biggr)f \bigl(s,x(s),x'(s),x''(s)\bigr)\, \mathrm{d}s \\ < & \int_{0}^{1}s\biggl(1+\frac{\mu}{1-\mu} \biggr) \frac{1}{H_{1}}\bigl(x(s)+x'(s)-x''(s) \bigr)\,\mathrm{d}s \\ \leq& \|x\|,\quad t\in[0,1], \end{aligned}$$
which implies that
$$\|x\|=\bigl\Vert x''\bigr\Vert _{\infty}< \|x \|. $$
This is a contradiction. Therefore the BVP (1.1), (1.2) has no concave and monotone positive solution. This completes the proof of the theorem. □

Theorem 3.4

Suppose that
$$H_{2}f(t,x_{0},x_{1},x_{2})>x_{0}+x_{1}-x_{2} $$
for \((t,x_{0},x_{1},x_{2})\in[0,1]\times[0,+\infty)^{2}\times(-\infty,0]\) with \(x_{0}+x_{1}-x_{2}>0\). Then the BVP (1.1), (1.2) has no concave and monotone positive solution.

Proof

Suppose on the contrary that x is a concave and monotone positive solution of the BVP (1.1), (1.2). Then from Lemma 2.2 we have
$$\begin{aligned} x(1) =& \int_{0}^{1}\biggl[G_{1}(1,s)+ \frac{1}{2(1-\mu)} \int_{0}^{1}G_{2}( \tau,s)g(\tau)\,\mathrm{d}\tau\biggr]f\bigl(s,x(s),x'(s),x''(s) \bigr)\,\mathrm{d}s \\ > & \int_{0}^{1}\biggl[\frac{1}{4}s+ \frac{1}{2(1-\mu)}\int_{0}^{1}\tau sg(\tau )\, \mathrm{d}\tau\biggr] \frac{1}{H_{2}}\bigl(x(s)+x'(s)-x''(s) \bigr)\,\mathrm{d}s \\ \geq& \frac{1}{2H_{2}}\int_{0}^{1}\biggl( \frac{1}{2}s+\frac{s}{1-\mu}\int_{0}^{1} \tau g(\tau)\,\mathrm{d}\tau\biggr) \frac{1}{2}\biggl(s- \frac{1}{2}s^{2}\biggr)\|x\|\,\mathrm{d}s \\ = & \|x\|, \end{aligned}$$
which is a contradiction. This completes the proof of the theorem. □

Finally, we give some examples to demonstrate applications of our results.

Example 3.1

Consider the fourth-order boundary value problem
$$\begin{aligned}& x^{(4)}(t)=\frac{1}{1+t} \biggl[\frac{x+x'-x''}{4e^{x+x'-x''}}+ \frac {34(x+x'-x'')^{2}}{1+x+x'-x''} \biggr],\quad t\in[0,1], \end{aligned}$$
(3.13)
$$\begin{aligned}& x(0)=x'(1)=x'''(1)=0, \qquad x''(0)= \int_{0}^{1}sx''(s) \,\mathrm{d}s. \end{aligned}$$
(3.14)
Let
$$f(t,x_{0},x_{1},x_{2})=\frac{1}{1+t} \biggl[ \frac{x_{0}+x_{1}-x_{2}}{4e^{x_{0}+x_{1}-x_{2}}} +\frac{34(x_{0}+x_{1}-x_{2})^{2}}{1+x_{0}+x_{1}-x_{2}} \biggr],\qquad g(t)=t. $$
Then \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\), and \(\mu=\int_{0}^{1}g(s)\,\mathrm{d}s=\frac{1}{2}<1\). It is easy to compute that
$$f^{0}=\frac{1}{4}, \qquad f_{\infty}=17, \qquad H_{1}=3, \qquad H_{2}=\frac{35}{576}, $$
and hence
$$H_{1}f^{0}< 1< H_{2}f_{\infty}. $$
So, it follows from Theorem 3.1 that the BVP (3.13), (3.14) has at least one concave and monotone positive solution.

Example 3.2

Consider the fourth-order boundary value problem
$$\begin{aligned}& x^{(4)}(t)=\frac{1}{1+t} \biggl[\frac{14(x+x'-x'')}{1+\ln(1+x+x'-x'')} + \frac{(x+x'-x'')^{2}}{8(1+x+x'-x'')} \biggr],\quad t\in[0,1], \end{aligned}$$
(3.15)
$$\begin{aligned}& x(0)=x'(1)=x'''(1)=0, \qquad x''(0)= 3\int_{0}^{1}s^{3}x''(s) \,\mathrm{d}s. \end{aligned}$$
(3.16)
Let
$$f(t,x_{0},x_{1},x_{2})=\frac{1}{1+t} \biggl[ \frac{14(x_{0}+x_{1}-x_{2})}{1+\ln(1+x_{0}+x_{1}-x_{2})} +\frac{(x_{0}+x_{1}-x_{2})^{2}}{8(1+x_{0}+x_{1}-x_{2})} \biggr],\qquad g(t)=3t^{3}. $$
Then \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\), and \(\mu=\int_{0}^{1}g(s)\,\mathrm{d}s=\frac{3}{4}<1\). It is easy to compute that
$$f^{\infty}=\frac{1}{8}, \qquad f_{0}=7, \qquad H_{1}=6, \qquad H_{2}=\frac{29}{192}, $$
and hence
$$H_{1}f^{\infty}< 1< H_{2}f_{0}. $$
So, it follows from Theorem 3.2 that the BVP (3.15), (3.16) has at least one concave and monotone positive solution.

Declarations

Acknowledgements

This work was supported by the National Natural Science Foundation of China (11201008).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Beihua University

References

  1. Li, H, Wang, L, Pei, M: Solvability of a fourth-order boundary value problem with integral boundary conditions. J. Appl. Math. 2013, Article ID 782363 (2013) MathSciNetGoogle Scholar
  2. O’Regan, D: Fixed-point theory for the sum of two operators. Appl. Math. Lett. 9, 1-8 (1996) MATHView ArticleGoogle Scholar
  3. Agarwal, RP, Chow, YM: Iterative methods for a fourth order boundary value problem. J. Comput. Appl. Math. 10, 203-217 (1984) MATHMathSciNetView ArticleGoogle Scholar
  4. Bai, Z: The upper and lower solution method for some fourth-order boundary value problems. Nonlinear Anal. 67, 1704-1709 (2007) MATHMathSciNetView ArticleGoogle Scholar
  5. Cabada, A, Minhós, FM: Fully nonlinear fourth-order equations with functional boundary conditions. J. Math. Anal. Appl. 340, 239-251 (2008) MATHMathSciNetView ArticleGoogle Scholar
  6. Del Pino, MA, Manasevich, RF: Existence for a fourth-order boundary value problem under a two parameter nonresonance condition. Proc. Am. Math. Soc. 112, 81-86 (1991) MATHView ArticleGoogle Scholar
  7. Du, J, Cui, M: Constructive proof of existence for a class of fourth-order nonlinear BVPs. Comput. Math. Appl. 59, 903-911 (2010) MATHMathSciNetView ArticleGoogle Scholar
  8. Ehme, J, Eloe, PW, Henderson, J: Upper and lower solution methods for fully nonlinear boundary value problems. J. Differ. Equ. 180, 51-64 (2002) MATHMathSciNetView ArticleGoogle Scholar
  9. Franco, D, O’Regan, D, Perán, J: Fourth-order problems with nonlinear boundary conditions. J. Comput. Appl. Math. 174, 315-327 (2005) MATHMathSciNetView ArticleGoogle Scholar
  10. Feng, H, Ji, D, Ge, W: Existence and uniqueness of solutions for a fourth-order boundary value problem. Nonlinear Anal. 70, 3561-3566 (2009) MATHMathSciNetView ArticleGoogle Scholar
  11. Graef, JR, Kong, L: A necessary and sufficient condition for existence of positive solutions of nonlinear boundary value problems. Nonlinear Anal. 66, 2389-2412 (2007) MATHMathSciNetView ArticleGoogle Scholar
  12. Graef, JR, Qian, CX, Yang, B: A three point boundary value problem for nonlinear fourth order differential equations. J. Math. Anal. Appl. 287, 217-233 (2003) MATHMathSciNetView ArticleGoogle Scholar
  13. Grossinho, MR, Tersian, SA: The dual variational principle and equilibria for a beam resting on a discontinuous nonlinear elastic foundation. Nonlinear Anal. 41, 417-431 (2000) MATHMathSciNetView ArticleGoogle Scholar
  14. Gupta, CP: Existence and uniqueness theorems for a bending of an elastic beam equation. Appl. Anal. 26, 289-304 (1988) MATHMathSciNetView ArticleGoogle Scholar
  15. Jankowski, T: Positive solutions for fourth-order differential equations with deviating arguments and integral boundary conditions. Nonlinear Anal. 73, 1289-1299 (2010) MATHMathSciNetView ArticleGoogle Scholar
  16. Jiang, DQ, Gao, WJ, Wan, AY: A monotone method for constructing extremal solutions to fourth-order periodic boundary value problems. Appl. Math. Comput. 132, 411-421 (2002) MATHMathSciNetView ArticleGoogle Scholar
  17. Kang, P, Wei, Z, Xu, J: Positive solutions to fourth-order singular boundary value problems with integral boundary conditions in abstract spaces. Appl. Math. Comput. 206, 245-256 (2008) MATHMathSciNetView ArticleGoogle Scholar
  18. Korman, P: Computation of displacements for nonlinear elastic beam models using monotone iterations. Int. J. Math. Math. Sci. 11, 121-128 (1988) MATHMathSciNetView ArticleGoogle Scholar
  19. Ma, TF: Positive solutions for a beam equation on a nonlinear elastic foundation. Math. Comput. Model. 39, 1195-1201 (2004) MATHView ArticleGoogle Scholar
  20. Ma, R, Xu, J: Bifurcation from interval and positive solutions of a nonlinear fourth-order boundary value problem. Nonlinear Anal. 72, 113-122 (2010) MATHMathSciNetView ArticleGoogle Scholar
  21. Minhós, F, Gyulov, T, Santos, AI: Lower and upper solutions for a fully nonlinear beam equation. Nonlinear Anal. 71, 281-292 (2009) MATHMathSciNetView ArticleGoogle Scholar
  22. O’Regan, D: Solvability of some fourth (and higher) order singular boundary value problems. J. Math. Anal. Appl. 161, 78-116 (1991) MATHMathSciNetView ArticleGoogle Scholar
  23. Pietramala, P: A note on a beam equation with nonlinear boundary conditions. Bound. Value Probl. 2011, Article ID 376782 (2011) MathSciNetView ArticleGoogle Scholar
  24. Pei, M, Chang, SK: Monotone iterative technique and symmetric positive solutions for a fourth-order boundary value problem. Math. Comput. Model. 51, 1260-1267 (2010) MATHMathSciNetView ArticleGoogle Scholar
  25. Shanthi, V, Ramanujam, N: A numerical method for boundary value problems for singularly perturbed fourth-order ordinary differential equations. Appl. Math. Comput. 129, 269-294 (2002) MATHMathSciNetView ArticleGoogle Scholar
  26. Sun, JP, Wang, XQ: Existence and iteration of monotone positive solution of BVP for an elastic beam equation. Math. Probl. Eng. 2011, Article ID 705740 (2011) Google Scholar
  27. Webb, JRL, Infante, G: Positive solutions of nonlocal boundary value problems: a unified approach. J. Lond. Math. Soc. (2) 74, 673-693 (2006) MATHMathSciNetView ArticleGoogle Scholar
  28. Webb, JRL, Infante, G, Franco, D: Positive solutions of nonlinear fourth-order boundary-value problems with local and non-local boundary conditions. Proc. R. Soc. Edinb., Sect. A 138, 427-446 (2008) MATHMathSciNetView ArticleGoogle Scholar
  29. Wei, Z: A class of fourth order singular boundary value problems. Appl. Math. Comput. 153, 865-884 (2004) MATHMathSciNetView ArticleGoogle Scholar
  30. Yao, QL: Positive solutions for eigenvalue problems of fourth-order elastic beam equations. Appl. Math. Lett. 17, 237-243 (2004) MATHMathSciNetView ArticleGoogle Scholar
  31. Zhang, X, Ge, W: Positive solutions for a class of boundary value problems with integral boundary conditions. Comput. Math. Appl. 58, 203-215 (2009) MATHMathSciNetView ArticleGoogle Scholar
  32. Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, Boston (1988) MATHGoogle Scholar
  33. Sun, JP, Li, HB: Monotone positive solution of nonlinear third-order BVP with integral boundary conditions. Bound. Value Probl. 2010, Article ID 874959 (2010) View ArticleGoogle Scholar

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