Existence of positive solutions for singular fourth-order three-point boundary value problems

Abstract

In this article, we consider the boundary value problem $u^{(4)}(t)+f(t,u(t))=0$, $0<t<1$, subject to the boundary conditions $u(0)=u^{\prime }(0)=u^{″}(0)=0$ and $u^{″}(1)-\alpha u^{″}(\eta )=\lambda$. In this setting, $0<\eta <1$ and $\alpha \in [0,\frac{1}{\eta })$ are constants and $\lambda \in [0,+\mathrm{\infty })$ is a parameter. By imposing a sufficient structure on the nonlinearity $f(t,u)$, we deduce the existence of at least one positive solution to the problem. The novelty in our setting lies in the fact that $f(t,u)$ may be singular at $t=0$ and $t=1$. Our results here are achieved by making use of the Krasnosel’skii fixed point theorem. We conclude with examples illustrating our results and the improvements that they present.

MSC:34B15, 34B25, 34B18.

1 Introduction

In this paper, we consider the following nonlinear singular fourth-order three-point boundary value problem:

$$\{\begin{array}{c}{u}^{(4)}(t)+f(t,u(t))=0,0<t<1,\hfill \\ u(0)=u^{\prime }(0)=u^{″}(0)=0,\hfill \\ u^{″}(1)-\alpha u^{″}(\eta )=\lambda ,\hfill \end{array}$$

(1.1)

where $0<\eta <1$, $\alpha \in [0,\frac{1}{\eta })$ are constants, $\lambda \in [0,+\mathrm{\infty })$ is a parameter, $f(t,u(t))$ may be singular at $t=0$ and/or $t=1$. Here, by a positive solution we mean a function $u^{\ast }(t)$ which is positive on $(0,1)$ and satisfies problem (1.1).

The theory of boundary value problems for ordinary differential equations arises in different areas of applied mathematics, physics and so on. The existence of positive solutions for boundary value problems has become an important area of investigation and received a great deal of attention in recent years (see [1–20] and the references cited therein). In [14], by making use of the fixed point theorem and degree theory, Bai and Wang proved the existence, uniqueness and multiplicity of positive solutions for the following fourth-order two-point boundary value problem:

$$\{\begin{array}{c}{u}^{(4)}(t)-\lambda f(t,u(t))=0,0<t<1,\hfill \\ u(0)=u(1)=u^{″}(0)=u^{″}(1)=0.\hfill \end{array}$$

In [8], Yao studied the following nonlinear fourth-order ordinary differential equation:

$$\{\begin{array}{c}{u}^{(4)}(t)=f(t,u(t),u^{\prime }(t)),t\in [0,1]\mathrm{\setminus }E,\hfill \\ u(0)=u^{\prime }(0)=u^{″}(1)=u^{‴}(1)=0,\hfill \end{array}$$

where $E\subset [0,1]$ is a closed set with measure zero and the nonlinear term $f(t,x,y)$ may be singular for $t\in E$. The author showed the existence of n positive solutions by constructing a suitable integral equation and applying fixed point theorems on a cone.

In [9], Sun considered the following third-order boundary value problems:

$$\{\begin{array}{c}{u}^{‴}(t)+a(t)f(u(t))=0,0<t<1,\hfill \\ u(0)=u^{\prime }(0)=0,\hfill \\ u^{\prime }(1)-\alpha u^{\prime }(\eta )=\lambda .\hfill \end{array}$$

The author obtained the existence and nonexistence of positive solutions by applying the Guo-Krasnosel’skii fixed point theorem and Schauder’s fixed point theorem.

In [10], Zhang and Wang studied the following nonlinear singular fourth-order boundary value problem:

$$\{\begin{array}{c}{u}^{(4)}(t)=f(t,u(t)),0<t<1,\hfill \\ u(0)=u(1)=u^{\prime }(0)=u^{\prime }(1)=0,\hfill \end{array}$$

where the nonlinear term $f(t,u)$ may be singular at $t=0$, $t=1$ and $u=0$. The author presented the existence of a positive solution by using the fixed point index theorem and the properties of Green’s function.

In [15], by applying the Krasnosel’skii fixed point theorem, Graef, Qian and Yang established the existence and nonexistence of positive solutions for the following fourth-order three-point boundary value problem:

$$\{\begin{array}{c}{u}^{(4)}(t)=\lambda g(t)f(u(t)),0<t<1,\hfill \\ u(0)=u^{\prime }(1)=u^{″}(0)=u^{″}(p)-u^{″}(1)=0,\hfill \end{array}$$

where $p\in (0,1)$ is a constant.

Inspired and motivated by the works mentioned above, we deal with the existence and nonexistence of positive solutions to problem (1.1) by making use of the fixed point theorem together with the properties of Green’s function. The main features of the paper are as follows. Firstly, we apply the Taylor expansion formula to prove a lemma, and then we give a comparison lemma and construct a special cone. Secondly, we present the existence of positive solutions for problem (1.1). To our best knowledge, no paper has considered problem (1.1). The arguments are based upon the fixed point theorem for the special cone.

The paper is organized as follows. In Section 2, we give some properties of Green’s function associated with problem (1.1) and construct a suitable cone and transform problem (1.1) into an integral equation. In Section 3, we discuss the existence of at least one positive solution for problem (1.1).

2 Preliminary lemmas

Let $E=C[0,1]$ be a Banach space of all continuous functions with the norm $\parallel u\parallel ={max}_{0\le t\le 1}|u(t)|$, $C^{+}[0,1]=\{u\in C[0,1]:u(t)>0,t\in [0,1]\}$.

Throughout the paper, we assume that

(H₁) $f:(0,1)\times [0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is continuous.

(H₂) There exists a continuous function $q:(0,1)\to [0,+\mathrm{\infty })$ such that

$$0<{\int }_a^{b}s(1-s)q(s)ds\le {\int }_0^{1}s(1-s)q(s)ds<+\mathrm{\infty }\text{for }[a,b]\subset (0,1).$$

(H₃) There exists a continuous function $g:[0,1]\times [0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ such that

$$f(t,u)\le q(t)g(t,u),(t,u)\in (0,1)\times [0,+\mathrm{\infty }).$$

Lemma 2.1 Suppose that $p(t)\in L^1(0,1)$ and $p(t)>0$. Then the linear boundary value problem

$$\{\begin{array}{c}{u}^{(4)}(t)+p(t)=0,\hfill \\ u(0)=u^{\prime }(0)=u^{″}(0)=0,\hfill \\ u^{″}(1)-\alpha u^{″}(\eta )=\lambda \hfill \end{array}$$

(2.1)

has a unique positive solution, which can be expressed by

$$u(t)={\int }_0^{1}G(t,s)p(s)ds+\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)p(s)ds+\frac{\lambda t^3}{6(1-\alpha \eta )},$$

where

$$G(t,s)=\{\begin{array}{cc}\frac{1}{6}{t}^3(1-s)-\frac{1}{6}{(t-s)}^3,\hfill & 0⩽s⩽t⩽1,\hfill \\ \frac{1}{6}{t}^3(1-s),\hfill & 0⩽t⩽s⩽1,\hfill \end{array}$$

and

$$\frac{\partial }{\partial t}G(t,s)=\{\begin{array}{cc}\frac{1}{2}{t}^2(1-s)-\frac{1}{2}{(t-s)}^2,\hfill & 0⩽s⩽t⩽1,\hfill \\ \frac{1}{2}{t}^2(1-s),\hfill & 0⩽t⩽s⩽1,\hfill \end{array}$$

and

$$K(t,s)=\frac{{\partial }^2}{\partial t^2}G(t,s)=\{\begin{array}{cc}s(1-t),\hfill & 0⩽s⩽t⩽1,\hfill \\ t(1-s),\hfill & 0⩽t⩽s⩽1.\hfill \end{array}$$

Proof In fact, if $u(t)$ is a solution of problem (2.1), by the Taylor expansion formula, we have

$$u(t)=a_0+a_{1}t+\frac{a_2}{2!}{t}^2+\frac{a_3}{3!}{t}^3-\frac{1}{6}{\int }_0^t{(t-s)}^{3}p(s)ds,$$

then

$$\begin{array}{c}{u}^{\prime }(t)=a_1+a_{2}t+\frac{a_3}{2}{t}^2-\frac{1}{2}{\int }_0^t{(t-s)}^{2}p(s)ds,\hfill \\ u^{″}(t)=a_2+a_{3}t-{\int }_0^t(t-s)p(s)ds,\hfill \end{array}$$

which together with the boundary condition implies $a_0=a_1=a_2=0$ and

$$a_3=\frac{1}{1-\alpha \eta }{\int }_0^1(1-s)p(s)ds+\frac{\alpha }{1-\alpha \eta }{\int }_0^{\eta }(s-\eta )p(s)ds+\frac{\lambda }{1-\alpha \eta }.$$

Therefore

$$\begin{array}{rcl}u(t)& =& -\frac{1}{6}{\int }_0^t{(t-s)}^{3}p(s)ds+\frac{t^3}{6(1-\alpha \eta )}{\int }_0^1(1-s)p(s)ds\\ +\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{\eta }(s-\eta )p(s)ds+\frac{\lambda t^3}{6(1-\alpha \eta )}\\ =& \frac{1}{6}{\int }_0^t[t^3(1-s)-{(t-s)}^3]p(s)ds+\frac{1}{6}{\int }_t^1{t}^3(1-s)p(s)ds\\ +\frac{\alpha t^3}{6(1-\alpha \eta )}[{\int }_0^{\eta }s(1-\eta )p(s)ds+{\int }_{\eta }^1\eta (1-s)p(s)ds]+\frac{\lambda t^3}{6(1-\alpha \eta )}\\ =& {\int }_0^{1}G(t,s)p(s)ds+\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)p(s)ds+\frac{\lambda t^3}{6(1-\alpha \eta )}.\end{array}$$

The proof is complete. □

Lemma 2.2 For all $(t,s)\in [0,1]\times [0,1]$, we have

$$\frac{1}{6}{t}^{3}s(1-s)\le G(t,s)\le s(1-s).$$

Proof If $0\le t\le s\le 1$, then

$$G(t,s)=\frac{1}{6}{t}^3(1-s)\le \frac{1}{6}{s}^3(1-s)\le s(1-s),$$

and

$$G(t,s)=\frac{1}{6}{t}^3(1-s)\ge \frac{1}{6}{t}^{3}s(1-s).$$

If $0\le s\le t\le 1$, then

$$\begin{array}{rcl}G(t,s)& =& \frac{1}{6}{t}^3(1-s)-\frac{1}{6}{(t-s)}^3\\ \le & \frac{1}{6}s[t^2-t^3+3t(t-s)]\\ \le & \frac{1}{6}s[t^2(1-s)+3t(1-s)]\le s(1-s),\end{array}$$

and

$$\begin{array}{rcl}G(t,s)& =& \frac{1}{6}{t}^3(1-s)-\frac{1}{6}{(t-s)}^3\\ \ge & \frac{1}{6}{t}^{3}s(1-s)+\frac{1}{6}{t}^3{(1-s)}^3-\frac{1}{6}{(t-s)}^3\\ \ge & \frac{1}{6}{t}^{3}s(1-s)+\frac{1}{6}s(1-t)[t^2{(1-s)}^2+t(1-s)(t-s)+{(t-s)}^2]\\ \ge & \frac{1}{6}{t}^{3}s(1-s).\end{array}$$

Therefore

$$\frac{1}{6}{t}^{3}s(1-s)\le G(t,s)\le s(1-s).$$

 □

Define a cone $K\subset C[0,1]$ by

$$K=\{u(t)\in C^{+}[0,1]:u(t)\ge \frac{1}{6}{t}^3\parallel u\parallel ,0\le t\le 1\},$$

then K is a positive cone in $C[0,1]$. Denote

$${\mathrm{\Omega }}_r=\{u\in K:\parallel u\parallel <r\},\partial {\mathrm{\Omega }}_r=\{u\in K:\parallel u\parallel =r\}.$$

Fix $R>r>0$. Define an operator $A:({\overline{\mathrm{\Omega }}}_R\mathrm{\setminus }{\mathrm{\Omega }}_r)\cap K\to K$ by

$$Au(t)={\int }_0^{1}G(t,s)f(s,u(s))ds+\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda t^3}{6(1-\alpha \eta )}.$$

It is well known that problem (1.1) has a positive solution $u=u(t)$ if and only if u is a fixed point of A.

Lemma 2.3 Suppose that (H₁)∼(H₃) hold. Then $A(K)\subseteq K$.

Proof From (H₂) and (H₃), we know that

$$\begin{array}{rcl}0& \le & (Au)(t)\\ \le & {\int }_0^{1}s(1-s)f(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)q(s)g(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)g(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ <& +\mathrm{\infty }.\end{array}$$

On the other hand, for any $u\in K$, we have $u(t)\ge \frac{1}{6}{t}^3\parallel u\parallel$, $t\in [0,1]$, and

$$\parallel Au\parallel \le {\int }_0^{1}s(1-s)f(s,u(s))ds+\frac{\alpha }{1-\alpha \eta }{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{1-\alpha \eta }.$$

Therefore

$$\begin{array}{rcl}(Au)(t)& \ge & \frac{1}{6}{t}^3[{\int }_0^{1}s(1-s)f(s,u(s))ds+\frac{\alpha }{1-\alpha \eta }{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{1-\alpha \eta }]\\ \ge & \frac{1}{6}{t}^3\parallel Au\parallel .\end{array}$$

The proof is complete. □

Lemma 2.4 Suppose that (H₁)∼(H₃) hold. Then $A:({\overline{\mathrm{\Omega }}}_R\mathrm{\setminus }{\mathrm{\Omega }}_r)\cap K\to K$ is completely continuous.

Proof For any $u\in ({\overline{\mathrm{\Omega }}}_R\mathrm{\setminus }{\mathrm{\Omega }}_r)\cap K$, we have $0\le \frac{1}{6}{t}^{3}r\le \frac{1}{6}{t}^3\parallel u\parallel \le u(t)\le R$.

Let

$$q_n(t)=\{\begin{array}{cc}{inf}_{t\le s\le \frac{1}{n}}q(s),\hfill & 0\le t\le \frac{1}{n},\hfill \\ q(t),\hfill & \frac{1}{n}\le t\le \frac{n-1}{n},\hfill \\ {inf}_{\frac{n-1}{n}\le s\le t}q(s),\hfill & \frac{n-1}{n}\le t\le 1.\hfill \end{array}$$

Then, from (H₂) and (H₃), we have ${lim}_{n\to \mathrm{\infty }}{\int }_0^1(q(t)-q_n(t))dt=0$ for $0\le t\le 1$ and $u\in K$. Let

$$f_n(t,u)=\{\begin{array}{cc}f(t,u),\hfill & f(t,u)\le q_n(t)g(t,u),\hfill \\ q_n(t)g(t,u),\hfill & f(t,u)>q_n(t)g(t,u).\hfill \end{array}$$

It is easy to see that $f_n$ is a continuous function on $[0,1]\times [0,+\mathrm{\infty })$ and $f_n$ is bounded on any bounded set. Define

$$(A_{n}u)(t)={\int }_0^{1}G(t,s)f_n(s,u(s))ds+\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f_n(s,u(s))ds+\frac{\lambda t^3}{6(1-\alpha \eta )}.$$

By the Arzela-Ascoli theorem, we know that $A_n:{\overline{\mathrm{\Omega }}}_R\mathrm{\setminus }{\mathrm{\Omega }}_r\to C[0,1]$ is completely continuous.

Let $M(R)=max\{g(t,u):(t,u)\in [0,1]\times [0,R]\}$. For $u\in ({\overline{\mathrm{\Omega }}}_R\mathrm{\setminus }{\mathrm{\Omega }}_r)\cap K$, we know that

$$\begin{array}{rcl}\parallel Au-A_{n}u\parallel & =& \underset{0\le t\le 1}{max}\{{\int }_0^{1}G(t,s)[f(s,u(s))-f_n(s,u(s))]ds\}\\ +\underset{0\le t\le 1}{max}\{\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)[f(s,u(s))-f_n(s,u(s))]ds\}\\ \le & \underset{0\le t\le 1}{max}\{{\int }_0^{1}G(t,s)[q(s)g(s,u(s))-q_n(s)g(s,u(s))]ds\}\\ +\underset{0\le t\le 1}{max}\{\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)[q(s)g(s,u(s))-q_n(s)g(s,u(s))]ds\}\\ \le & M(R)\underset{0\le t\le 1}{max}\{{\int }_0^{1}G(t,s)[q(s)-q_n(s)]ds\}\\ +M(R)\underset{0\le t\le 1}{max}\{\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)[q(s)-q_n(s)]ds\}\\ \to & 0\text{as }n\to \mathrm{\infty }.\end{array}$$

It shows that a completely continuous operator $A_n$ converges to an operator A uniformly on $({\overline{\mathrm{\Omega }}}_R\mathrm{\setminus }{\mathrm{\Omega }}_r)\cap K$. Hence A is continuous.

Suppose that $D\subset K$ is a bounded set, then there exists $d>0$ such that $\parallel u\parallel \le d$ for any $u\in D$. From (H₃), we know that $|f(t,u)|\le q(t)g(t,u)\le M(d)q(t)$ for $(t,u)\in (0,1)\times [0,d]$. Then we have

$$\begin{array}{rcl}\parallel Au\parallel & \le & {\int }_0^{1}s(1-s)f(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)q(s)g(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)g(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & M(d){\int }_0^{1}s(1-s)q(s)ds+M(d)\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)ds+\frac{\lambda }{6(1-\alpha \eta )}\\ <& +\mathrm{\infty }.\end{array}$$

Hence A is uniformly bounded.

On the other hand, for any $u\in D$, we know that

$$\begin{array}{rcl}|{(Au)}^{\prime }(t)|& \le & \frac{1}{2}{\int }_0^{t}s(1-s)f(s,u(s))ds+\frac{1}{2}{\int }_t^{1}s(1-s)f(s,u(s))ds\\ +\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda t^2}{2(1-\alpha \eta )}\\ \le & \frac{1}{2}{\int }_0^{t}s(1-s)q(s)g(s,u(s))ds+\frac{1}{2}{\int }_t^{1}s(1-s)q(s)g(s,u(s))ds\\ +\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)g(s,u(s))ds+\frac{\lambda t^2}{2(1-\alpha \eta )}\\ \le & \frac{1}{2}M(d){\int }_0^{t}s(1-s)q(s)ds+\frac{1}{2}M(d){\int }_t^{1}s(1-s)q(s)ds\\ +\frac{\alpha t^2}{2(1-\alpha \eta )}M(d){\int }_0^{1}K(\eta ,s)q(s)ds+\frac{\lambda t^2}{2(1-\alpha \eta )}.\end{array}$$

Let

Therefore A is equicontinuous. Consequently, A is completely continuous. □

Lemma 2.5 [21, 22]

Let E be a Banach space, and let $P\subset E$ be a cone in E. Assume that ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$ are open subsets of E with $0\in {\mathrm{\Omega }}_1$ and ${\overline{\mathrm{\Omega }}}_1\subset {\mathrm{\Omega }}_2$. Let $T:P\cap ({\overline{\mathrm{\Omega }}}_2\mathrm{\setminus }{\mathrm{\Omega }}_1)\to P$ be a completely continuous operator such that either

• (i) $\parallel Tu\parallel \le \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_1$ and $\parallel Tu\parallel \ge \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_2$, or

• (ii) $\parallel Tu\parallel \ge \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_1$ and $\parallel Tu\parallel \le \parallel u\parallel$, $u\in P\cap \partial {\mathrm{\Omega }}_2$.

Then T has a fixed point in $P\cap ({\overline{\mathrm{\Omega }}}_2\mathrm{\setminus }{\mathrm{\Omega }}_1)$.

3 Main results

Theorem 3.1 Suppose that (H₁), (H₂) and (H₃) hold. In addition, assume that the following conditions hold:

(H₄) ${lim}_{u\to 0^{+}}sup{max}_{0\le t\le 1}\frac{g(t,u)}{u}=0$;

(H₅) ${lim}_{u\to +\mathrm{\infty }}inf{min}_{a\le t\le b}\frac{f(t,u)}{u}=+\mathrm{\infty }$.

Then problem (1.1) has at least one positive solution for λ small enough, and problem (1.1) has no positive solution for λ large enough.

Proof For $\lambda >0$ small enough, let

$$N=\frac{5}{6}{[{\int }_0^{1}s(1-s)q(s)ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)ds]}^{-1}.$$

(3.1)

From (H₄), there exists a constant $R_1>0$ such that $g(t,u)\le Nu$ for $(t,u)\in [0,1]\times (0,R_1]$.

Let ${\mathrm{\Omega }}_1=\{u\in K:\parallel u\parallel <R_1\}$, $0<\lambda \le (1-\alpha \eta )R_1$. For any $u\in K\cap \partial {\mathrm{\Omega }}_1$, we get

$$\begin{array}{rcl}Au(t)& =& {\int }_0^{1}G(t,s)f(s,u(s))ds+\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda t^3}{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)f(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)q(s)g(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)g(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & N{\int }_0^{1}s(1-s)q(s)u(s)ds+\frac{N\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)u(s)ds+\frac{R_1(1-\alpha \eta )}{6(1-\alpha \eta )}\\ \le & N\parallel u\parallel {\int }_0^{1}s(1-s)q(s)ds+\frac{N\alpha \parallel u\parallel }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)ds+\frac{R_1}{6}\\ =& \frac{5}{6}{R}_1+\frac{1}{6}{R}_1=R_1=\parallel u\parallel .\end{array}$$

Therefore

$$\parallel Au\parallel \le \parallel u\parallel ,u\in K\cap \partial {\mathrm{\Omega }}_1.$$

On the other hand, let

$$M={[{\int }_a^b\frac{1}{6}{a}^{3}s(1-s)ds+\frac{\alpha a^3}{6(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)ds]}^{-1}.$$

(3.2)

From (H₅), there exists $R>0$ such that $f(t,u)\ge 48Mu$ for $(t,u)\in [a,b]\times [R,+\mathrm{\infty })$. Let $R_2>\frac{6R}{a^3}>R_1$, and let ${\mathrm{\Omega }}_2=\{u\in K:\parallel u\parallel <R_2\}$. For any $u\in K\cap \partial {\mathrm{\Omega }}_2$ and $t\in [a,b]$, we know that $u(t)\ge \frac{1}{6}{t}^3\parallel u\parallel =\frac{1}{6}{a}^3{R}_2>R$ and

$$\begin{array}{rcl}(Au)\left(\frac{1}{2}\right)& =& {\int }_0^{1}G(\frac{1}{2},s)f(s,u(s))ds\\ +\frac{1}{8}\cdot \frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{1}{8}\cdot \frac{\lambda }{6(1-\alpha \eta )}\\ \ge & \frac{1}{8}{\int }_0^1\frac{1}{6}s(1-s)f(s,u(s))ds+\frac{\alpha }{48(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds\\ \ge & \frac{1}{8}{\int }_a^b\frac{1}{6}s(1-s)f(s,u(s))ds+\frac{\alpha }{48(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)f(s,u(s))ds\\ \ge & M{\int }_a^{b}s(1-s)u(s)ds+\frac{\alpha M}{1-\alpha \eta }{\int }_a^{b}K(\eta ,s)u(s)ds\\ \ge & M{\int }_a^b\frac{1}{6}{a}^{3}s(1-s)u(s)ds+\frac{\alpha M{a}^3}{6(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)u(s)ds\\ =& M[{\int }_a^b\frac{1}{6}{a}^{3}s(1-s)ds+\frac{\alpha a^3}{6(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)ds]\parallel u\parallel \\ =& \parallel u\parallel ,\end{array}$$

which implies that

$$\parallel Au\parallel \ge \parallel u\parallel \text{for }u\in K\cap \partial {\mathrm{\Omega }}_2.$$

By Lemma 2.5 we know that problem (1.1) has at least one positive solution.

For λ large enough, we prove that problem (1.1) has no positive solution. Otherwise, there exists $0<{\lambda }_1<{\lambda }_2<{\lambda }_3<\cdots <{\lambda }_n<\cdots$ with ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=+\mathrm{\infty }$ such that problem (1.1) has a positive solution $u_n(t)$, then we get

$$\begin{array}{rcl}{u}_n\left(\frac{1}{2}\right)& =& {\int }_0^{1}G(\frac{1}{2},s)f(s,u_n(s))ds\\ +\frac{\alpha }{48(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u_n(s))ds+\frac{{\lambda }_n}{48(1-\alpha \eta )}\\ \ge & \frac{{\lambda }_n}{48(1-\alpha \eta )}\to +\mathrm{\infty }\text{as }n\to +\mathrm{\infty }.\end{array}$$

Hence $\parallel u_n\parallel \to +\mathrm{\infty }$ as $n\to +\mathrm{\infty }$.

Again from (H₅), there exist $R^{\prime }>0$ and $M>0$ such that

$$f(t,u)\ge 96Mu\text{for }(t,u)\in [a,b]\times [R^{\prime },+\mathrm{\infty }),$$

where M is defined by (3.2). Let n be large enough. Choose $R_2^{\mathrm{\prime }}>\frac{6a^3}{R^{\prime }}$ such that $\parallel u_n\parallel \ge R_2^{\mathrm{\prime }}$. Thus, we get

$$\begin{array}{rcl}\parallel u_n\parallel & \ge & u_n\left(\frac{1}{2}\right)\\ =& {\int }_0^{1}G(\frac{1}{2},s)f(s,u_n(s))ds\\ +\frac{\alpha }{48(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u_n(s))ds+\frac{{\lambda }_n}{48(1-\alpha \eta )}\\ \ge & \frac{1}{8}{\int }_0^1\frac{1}{6}s(1-s)f(s,u_n(s))ds+\frac{\alpha }{48(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u_n(s))ds\\ \ge & \frac{1}{8}{\int }_a^b\frac{1}{6}s(1-s)f(s,u_n(s))ds+\frac{\alpha }{48(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)f(s,u_n(s))ds\\ \ge & 2M{\int }_a^{b}s(1-s)u_n(s)ds+\frac{2M\alpha }{1-\alpha \eta }{\int }_a^{b}K(\eta ,s)u_n(s)ds\\ \ge & 2M[{\int }_a^b\frac{1}{6}{a}^{3}s(1-s)ds+\frac{\alpha a^3}{6(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)ds]\parallel u_n\parallel =2\parallel u_n\parallel ,\end{array}$$

which is a contradiction. The proof is complete. □

Remark 3.1 The conclusion of Theorem 3.1 also holds if $\lambda =0$.

Theorem 3.2 Suppose that (H₁), (H₂) and (H₃) hold. In addition, assume that

(H₆) ${lim}_{u\to +\mathrm{\infty }}inf{min}_{0\le t\le 1}\frac{g(t,u)}{u}=0$;

(H₇) ${lim}_{u\to 0^{+}}sup{max}_{a\le t\le b}\frac{f(t,u)}{u}=+\mathrm{\infty }$.

Then problem (1.1) has at least one positive solution for any $\lambda \in [0,+\mathrm{\infty })$.

Proof From (H₇), there exist constants $R_1>0$ and $M>0$ such that

$$f(t,u)\ge 48Mu\text{for }(t,u)\in [a,b]\times (0,R_1],$$

where M is defined by (3.2). Let ${\mathrm{\Omega }}_1=\{u\in K:\parallel u\parallel <R_1\}$. For any $u\in K\cap \partial {\mathrm{\Omega }}_1$, we know that

$$\begin{array}{rcl}(Au)\left(\frac{1}{2}\right)& =& {\int }_0^{1}G(\frac{1}{2},s)f(s,u(s))ds\\ +\frac{\alpha }{48(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{48(1-\alpha \eta )}\\ \ge & \frac{1}{8}{\int }_0^1\frac{1}{6}s(1-s)f(s,u(s))ds+\frac{\alpha }{48(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds\\ \ge & \frac{1}{8}{\int }_a^b\frac{1}{6}s(1-s)f(s,u(s))ds+\frac{\alpha }{48(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)f(s,u(s))ds\\ \ge & M{\int }_a^{b}s(1-s)u(s)ds+\frac{M\alpha }{1-\alpha \eta }{\int }_a^{b}K(\eta ,s)u(s)ds\\ \ge & M[{\int }_a^b\frac{1}{6}{a}^{3}s(1-s)ds+\frac{\alpha a^3}{6(1-\alpha \eta )}{\int }_a^{b}K(\eta ,s)ds]\parallel u\parallel =\parallel u\parallel .\end{array}$$

Therefore

$$\parallel Au\parallel \ge \parallel u\parallel \text{for }u\in K\cap \partial {\mathrm{\Omega }}_1.$$

On the other hand, from (H₆), there exists a constant $R_0>0$ such that $g(t,u)\le Nu$ for $u\ge R_0$, where N is defined by (3.1). Since $g(t,u)$ is continuous on $[0,1]\times [0,+\mathrm{\infty })$, there exists $M^{\ast }>0$ such that ${max}_{0\le t\le 1}g(t,u)\le M^{\ast }$ for $(t,u)\in [0,1]\times [0,R_0]$. Choose

$$R_2\ge max\{2R_1,\frac{M^{\ast }}{N},R_0,\frac{\lambda }{1-\alpha \eta }\}.$$

Let ${\mathrm{\Omega }}_2=\{u\in K:\parallel u\parallel <R_2\}$. For any $u\in [0,R_2]$, we get

$$f(t,u)\le q(t)g(t,u)\le \frac{1}{2}{M}^{\ast }+\frac{1}{2}N{R}_2.$$

Thus, for any $u\in K\cap \partial {\mathrm{\Omega }}_2$ and $t\in (0,1)$, we know that

$$\begin{array}{rcl}Au(t)& =& {\int }_0^{1}G(t,s)f(s,u(s))ds+\frac{\alpha t^3}{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda t^3}{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)f(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)f(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)q(s)g(s,u(s))ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)g(s,u(s))ds+\frac{\lambda }{6(1-\alpha \eta )}\\ \le & {\int }_0^{1}s(1-s)q(s)(\frac{1}{2}{M}^{\ast }+\frac{1}{2}N{R}_2)ds\\ +\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)(\frac{1}{2}{M}^{\ast }+\frac{1}{2}N{R}_2)ds+\frac{1}{6}{R}_2\\ \le & \frac{1}{2}{M}^{\ast }[{\int }_0^{1}s(1-s)q(s)ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)ds]\\ +\frac{1}{2}N{R}_2[{\int }_0^{1}s(1-s)q(s)ds+\frac{\alpha }{6(1-\alpha \eta )}{\int }_0^{1}K(\eta ,s)q(s)ds]+\frac{1}{6}{R}_2\\ \le & \frac{5M^{\ast }}{12N}+\frac{5}{12N}N{R}_2+\frac{1}{6}{R}_2\\ \le & \frac{5}{12}{R}_2+\frac{5}{12}{R}_2+\frac{1}{6}{R}_2=R_2=\parallel u\parallel .\end{array}$$

Therefore

$$\parallel Au\parallel \le \parallel u\parallel \text{for }u\in K\cap \partial {\mathrm{\Omega }}_2.$$

It follows from Lemma 2.5 that problem (1.1) has at least one positive solution. □

4 Examples

Now, we give examples to illustrate the main results in the paper.

Example 4.1

Consider the following boundary value problem:

$$\{\begin{array}{c}{u}^{(4)}(t)+\frac{1}{t^{1+{\omega }_1+{\omega }_2+{\omega }_3}{(1-t)}^{3+{\omega }_4}}(u^2+\sqrt{u}){sin}^{2}2u=0,0<t<1,\hfill \\ u(0)=u^{\prime }(0)=u^{″}(0)=0,\hfill \\ u^{″}(1)-\frac{1}{3}{u}^{″}(\frac{1}{2})=1.\hfill \end{array}$$

(4.1)

Then problem (4.1) has at least one positive solution if ${\omega }_1+{\omega }_2+{\omega }_3<1$ and ${\omega }_4<-1$.

Let

$$q(t)=\frac{1}{t^{1+{\omega }_1+{\omega }_2+{\omega }_3}{(1-t)}^{3+{\omega }_4}},g(t,u)=2(u^2+\sqrt{u}){sin}^{2}2u.$$

(4.2)

Take $[a,b]=[\frac{1}{4},\frac{3}{4}]$. Notice, for any fixed $t\in (0,1)$, that $f(t,x)\le q(t)g(t,x)$ and $0<{\int }_0^{1}s(1-s)q(s)ds<+\mathrm{\infty }$ for ${\omega }_1+{\omega }_2+{\omega }_3<1$ and ${\omega }_4<-1$.

Obviously, conditions (H₁)∼(H₃) are satisfied.

Now, for any fixed $t\in (0,1)$, (H₄) and (H₅) follow immediately from

$$\begin{array}{c}\underset{u\to 0^{+}}{lim}sup\underset{0\le t\le 1}{max}\frac{2(u^2+\sqrt{u}){sin}^{2}2u}{u}=0,\hfill \\ \underset{u\to +\mathrm{\infty }}{lim}inf\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\frac{(u^2+\sqrt{u}){sin}^{2}2u}{t^{1+{\omega }_1+{\omega }_2+{\omega }_3}{(1-t)}^{3+{\omega }_4}u}=+\mathrm{\infty }.\hfill \end{array}$$

Thus, the existence of a positive solution follows from Theorem 3.1 if ${\omega }_1+{\omega }_2+{\omega }_3<1$ and ${\omega }_4<-1$.

Example 4.2

Consider the following boundary value problem:

$$\{\begin{array}{c}{u}^{(4)}(t)+\frac{2\text{,}012|lnu(t)|+{sin}^{2}u(t)}{\sqrt{t(1-t)}}=0,0<t<1,\hfill \\ u(0)=u^{\prime }(0)=u^{″}(0)=0,\hfill \\ u^{″}(1)-\frac{1}{12}{u}^{″}(\frac{1}{6})=2\text{,}013.\hfill \end{array}$$

(4.3)

Then problem (4.3) has at least one positive solution.

Let

$$q(t)=\frac{1}{\sqrt{t(1-t)}},g(t,u)=2\text{,}013|lnu(t)|+{sin}^{2}u(t).$$

(4.4)

Take $[a,b]=[\frac{1}{4},\frac{3}{4}]$. Notice, for any fixed $t\in (0,1)$, that $f(t,x)\le q(t)g(t,x)$ and $0<{\int }_0^{1}s(1-s)q(s)ds<+\mathrm{\infty }$.

Obviously, conditions (H₁)∼(H₃) are satisfied.

Now, for any fixed $t\in (0,1)$, (H₆) and (H₇) follow immediately from

$$\begin{array}{c}\underset{u\to 0^{+}}{lim}sup\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}\frac{2\text{,}012|lnu(t)|+{sin}^{2}u(t)}{\sqrt{t(1-t)}u}=+\mathrm{\infty },\hfill \\ \underset{u\to +\mathrm{\infty }}{lim}inf\underset{0\le t\le 1}{min}\frac{2\text{,}012|lnu(t)|+{sin}^{2}u(t)}{u}=0.\hfill \end{array}$$

Thus, the existence of a positive solution follows from Theorem 3.2.