## Boundary Value Problems

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# Positive solutions for nonlocal fourth-order boundary value problems with all order derivatives

Boundary Value Problems20122012:29

https://doi.org/10.1186/1687-2770-2012-29

Accepted: 2 March 2012

Published: 2 March 2012

## Abstract

In this article, by the fixed point theorem in a cone and the nonlocal fourth-order BVP's Green function, the existence of at least one positive solution for the nonlocal fourth-order boundary value problem with all order derivatives

$\left\{\begin{array}{c}{u}^{\left(4\right)}\left(t\right)+A{u}^{″}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right),{u}^{‴}\left(t\right)\right),\phantom{\rule{1em}{0ex}}0

is considered, where f is a nonnegative continuous function, λ > 0, 0 < A < π2, p, q L[0, 1], p(s) ≥ 0, q(s) ≥ 0. The emphasis here is that f depends on all order derivatives.

### Keywords

fourth-order boundary value problem fixed point theorem Green's function positive solution

## 1 Introduction

The deformation of an elastic beam in equilibrium state, whose two ends are simply supported, can be described by a fourth-order ordinary equation boundary value problem. Owing to its significance in physics, the existence of positive solutions for the fourth-order boundary value problem has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point index theory, the Krasnosel'skii's fixed point theorem and the method of upper and lower solutions, in reference [110].

In recent years, there has been much attention on the question of positive solutions of the fourth-order differential equations with one or two parameters. By the Krasnosel'skii's fixed point theorem in cone [11], Bai [5] investigated the following fourth-order boundary value problem with one parameter
$\left\{\begin{array}{c}{u}^{\left(4\right)}\left(t\right)+\beta {u}^{″}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{″}\left(t\right)\right),\phantom{\rule{1em}{0ex}}0

where λ > 0, 0 < β < π2, f: C([0, 1] × [0, ∞) × (-∞, 0], [0, ∞)) is continuous, p, q L[0, 1], p(s) ≥ 0, q(s) ≥ 0, ${\int }_{0}^{1}p\left(s\right)ds<1$, ${\int }_{0}^{1}q\left(s\right)sin\sqrt{\beta }sds+{\int }_{0}^{1}q\left(s\right)sin\sqrt{\beta }\left(1-s\right)ds.

By the fixed point index in cone, Ma [7] proved the existence of symmetric positive solutions for the nonlocal fourth-order boundary value problem
$\left\{\begin{array}{c}{u}^{\left(4\right)}\left(t\right)=h\left(t\right)f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}0
All the above works were done under the assumption that all order derivatives u', u″, u‴ are not involved explicitly in the nonlinear term f. In this article, we are concerned with the existence of positive solutions for the nonlocal fourth-order boundary value problem
$\left\{\begin{array}{c}{u}^{\left(4\right)}\left(t\right)+A{u}^{″}\left(t\right)=\lambda f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{u}^{″}\left(t\right),{u}^{‴}\left(t\right)\right),\phantom{\rule{1em}{0ex}}0
(1.1)

Throughout, we assume

(H1) λ > 0, 0 < A < π2;

(H2) f: [0, 1] × R4R+ is continuous, p, q L[0, 1], p(s) ≥ 0, q(s) ≥ 0, ${\int }_{0}^{1}p\left(s\right)ds<1$, ${\int }_{0}^{1}q\left(s\right)sin\sqrt{A}sds+{\int }_{0}^{1}q\left(s\right)sin\sqrt{A}\left(1-s\right)ds.

We will impose all order derivatives in f and make use of two continuous convex functionals which will ensure the existence of at least one positive solution to (1.1). Bai [5] applied Krasnoselskii's fixed point theorem. Ma [8] used fixed point index in cone and Leray-Schauder degree. In this article, to show the existence of positive solutions to (1.1), we define two positive continuous convex functionals. Then, using the new fixed point theorem [12] in a cone and the nonlocal fourth-order BVP's Green function, we give some new criteria for the existence of positive solutions to (1.1).

## 2 The preliminary lemmas

Let Y = C[0, 1] be the Banach space equipped with the norm
${||u\left(t\right)||}_{0}=\underset{t\in \left[0,1\right]}{max}|u\left(t\right)|.$

Set λ1, λ2 be the roots of the polynomial P(λ) = λ2 + , namely λ1 = 0, λ2 = -A. By (H1), it is obviously that -π2< λ2< 0.

Let Q1(t, s), Q2(t, s) be, respectively the Green's functions of the following problems
$\left\{\begin{array}{c}-{u}^{″}\left(t\right)+{\lambda }_{1}u\left(t\right)=0,\phantom{\rule{1em}{0ex}}0
Then, carefully calculation yield
${Q}_{1}\left(t,s\right)={G}_{1}\left(t,s\right)+\frac{{\int }_{0}^{1}{G}_{1}\left(s,x\right)p\left(x\right)dx}{1-{\int }_{0}^{1}p\left(x\right)dx},$
${Q}_{2}\left(t,s\right)={G}_{2}\left(t,s\right)+\frac{\left[sin\sqrt{A}t+sin\sqrt{A}\left(1-t\right)\right]{\int }_{0}^{1}{G}_{2}\left(s,x\right)q\left(x\right)dx}{sin\sqrt{A}-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}xdx-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}\left(1-x\right)dx},$
${G}_{1}\left(t,s\right)=\left\{\begin{array}{c}s\left(1-t\right),\phantom{\rule{1em}{0ex}}0\le s\le t\le 1,\hfill \\ t\left(1-s\right),\phantom{\rule{1em}{0ex}}0\le t\le s\le 1,\hfill \end{array}\right\$
${G}_{2}\left(t,s\right)=\left\{\begin{array}{c}\frac{sin\sqrt{A}ssin\sqrt{A}\left(1-t\right)}{\sqrt{A}sin\sqrt{A}},\phantom{\rule{1em}{0ex}}0\le s\le t\le 1,\hfill \\ \frac{sin\sqrt{A}tsin\sqrt{A}\left(1-s\right)}{\sqrt{A}sin\sqrt{A}},\phantom{\rule{1em}{0ex}}0\le t\le s\le 1.\hfill \end{array}\right\$
Denote
$\begin{array}{c}{\omega }_{1}=\frac{1}{1-{\int }_{0}^{1}p\left(x\right)dx},\hfill \\ {\omega }_{2}\left(t\right)=\frac{sin\sqrt{A}t+sin\sqrt{A}\left(1-t\right)}{sin\sqrt{A}-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}xdx-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}\left(1-x\right)dx}.\hfill \end{array}$
Lemma 2.1. [5] Suppose that (H1) and (H2) hold. Then for any y(t) C[0, 1], the problem
$\left\{\begin{array}{c}{u}^{\left(4\right)}\left(t\right)+A{u}^{″}\left(t\right)=y\left(t\right),\phantom{\rule{1em}{0ex}}0
(2.1)
has a unique solution
$u\left(t\right)=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{Q}_{1}\left(t,s\right){Q}_{2}\left(s,\tau \right)y\left(\tau \right)d\tau ds,$
(2.2)
where
$\begin{array}{c}{Q}_{1}\left(t,s\right)={G}_{1}\left(t,s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,x\right)p\left(x\right)dx,\hfill \\ {Q}_{2}\left(s,\tau \right)={G}_{2}\left(s,\tau \right)+{\omega }_{2}\left(s\right){\int }_{0}^{1}{G}_{2}\left(\tau ,x\right)q\left(x\right)dx.\hfill \end{array}$
By (2.2), we get
${u}^{\prime }\left(t\right)=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{Q}_{2}\left(s,\tau \right)y\left(\tau \right)d\tau ds-\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}s{Q}_{2}\left(s,\tau \right)y\left(\tau \right)d\tau ds;$
(2.3)
${u}^{″}\left(t\right)=-\underset{0}{\overset{1}{\int }}{Q}_{2}\left(t,\tau \right)y\left(\tau \right)d\tau ,$
(2.4)
${u}^{‴}\left(t\right)=-\underset{0}{\overset{1}{\int }}\frac{\partial {Q}_{2}\left(t,\tau \right)}{\partial t}y\left(\tau \right)d\tau .$
(2.5)

Lemma 2.2. [5] Assume that (H1) and (H2) hold. Then one has

(i) Q i (t, s) ≥ 0, t, s [0, 1]; Q i (t, s) > 0, t, s (0, 1);

(ii) G i (t, s) ≥ b i G i (t, t)G i (s, s), t, s [0, 1];

(iii) G i (t, s) ≤ c i G i (s, s), t, s [0, 1].

where b1 = 1, ${b}_{2}=\sqrt{A}sin\sqrt{A}$; c1 = 1, ${c}_{2}=\frac{1}{sin\sqrt{A}}$.

Let
${d}_{i}=\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{b}_{i}{G}_{i}\left(t,t\right),\left(i=1,2\right);\xi =\frac{\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)}{\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right)}.$

Lemma 2.3. [5] Suppose that (H1) and (H2) hold and w2, d i , ξ i are given as above. Then

one has

(i) $\underset{0\le t\le 1}{max}{\omega }_{2}\left(t\right)={\omega }_{2}\left(\frac{1}{2}\right);$

(ii) $0<{d}_{i}<1,\phantom{\rule{1em}{0ex}}0<\xi <1.$

Lemma 2.4. If y(t) C[0, 1] and y(t) ≥ 0, then the unique solution u(t) of problem (2.1)

satisfies
$\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}u\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\ge {d}_{1}{||u||}_{0},\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\left(-{u}^{″}\left(t\right)\right)\ge \frac{{d}_{2}\xi }{{c}_{2}}{||{u}^{″}||}_{0}.$
Proof. By (2.2) and (iii) of Lemma 2.2, we get
$\begin{array}{cc}u\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\hfill & ={\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ \le {\int }_{0}^{1}{\int }_{0}^{1}\left[{c}_{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ ={\int }_{0}^{1}{\int }_{0}^{1}\left[{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ ={\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)d\tau ds.\hfill \end{array}$
So,
${||u||}_{0}\le {\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)d\tau ds.$
Using (ii) of Lemma 2.2, we have
$\begin{array}{cc}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}u\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\hfill & =\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ \ge \underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\int }_{0}^{1}{\int }_{0}^{1}\left[{b}_{1}{G}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}t\right){G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)\hfill \\ +{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ ={\int }_{0}^{1}{\int }_{0}^{1}\left[{d}_{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ \ge {d}_{1}{\int }_{0}^{1}{\int }_{0}^{1}\left[{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ ={d}_{1}{\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau ds\hfill \\ \ge {d}_{1}||u|{|}_{0}.\hfill \end{array}$
By (2: 4) and (iii) of Lemma 2.2, we get
$\begin{array}{cc}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}\phantom{\rule{0.3em}{0ex}}\left(-{u}^{″}\left(t\right)\right)\hfill & =\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\int }_{0}^{1}{Q}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau \hfill \\ \le {\int }_{0}^{1}\left[{c}_{2}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]y\left(\tau \right)d\tau \hfill \\ \le {c}_{2}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right){\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]y\left(\tau \right)d\tau .\hfill \end{array}$
So,
${||{u}^{″}||}_{0}\le {c}_{2}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right){\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]\phantom{\rule{0.3em}{0ex}}y\left(\tau \right)d\tau .$
Using (ii) of Lemma 2.2, we have
$\begin{array}{cc}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\phantom{\rule{0.3em}{0ex}}\left(-{u}^{″}\left(t\right)\right)\hfill & =\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\int }_{0}^{1}{Q}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\tau \right)y\left(\tau \right)d\tau \hfill \\ \ge \underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\int }_{0}^{1}\left[{b}_{2}{G}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}t\right){G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\omega }_{2}\left(t\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]y\left(\tau \right)d\tau \hfill \\ \ge {\int }_{0}^{1}\left[{b}_{2}{G}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}t\right){G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right){\int }_{0}^{1}{G}_{2}\phantom{\rule{2.77695pt}{0ex}}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]y\left(\tau \right)d\tau \hfill \\ ={\int }_{0}^{1}\left[{d}_{2}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]y\left(\tau \right)d\tau \hfill \\ \ge {d}_{2}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}{\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]y\left(\tau \right)d\tau \hfill \\ \ge \frac{{d}_{2}}{{c}_{2}}\frac{\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)}{\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right)}||{u}^{″}|{|}_{0}\hfill \\ \ge \frac{{d}_{2}\xi }{{c}_{2}}||{u}^{″}|{|}_{0}.\hfill \end{array}$

The proof is completed.

Let X be a Banach space and K X a cone. Suppose α, β: ×R+ are two continuous convex functionals satisfying α(λu) = |λ|α(u), β(λu) = |λ|β(u), for u X, λ R, and ||u|| ≤ M max{α(u), β(u)}, for u X and α(u) ≤ α(v) for u, v K, uv, where M > 0 is a constant.

Theorem 2.1. [12] Let r2> r1> 0, L > 0 be constants and
${\mathrm{\Omega }}_{i}=\left\{u\in X\phantom{\rule{2.77695pt}{0ex}}:\phantom{\rule{2.77695pt}{0ex}}\alpha \left(u\right)<{r}_{i},\phantom{\rule{2.77695pt}{0ex}}\beta \left(u\right)
two bounded open sets in X. Set
${D}_{i}=\left\{u\in X\phantom{\rule{2.77695pt}{0ex}}:\phantom{\rule{2.77695pt}{0ex}}\alpha \left(u\right)={r}_{i}\right\},\phantom{\rule{1em}{0ex}}i=1,2.$

Assume T: KK is a completely continuous operator satisfying

(A1) α(Tu) < r1, u D1K; α(Tu) > r2, u D2K;

(A2) β(Tu) < L, u K;

(A3) there is a p 2K) \ {0} such that α(p) ≠ 0 and α(u + λp) ≥ α(u), for all u K and λ ≥ 0.

Then T has at least one fixed point in $\left({\mathrm{\Omega }}_{2}\{\stackrel{̄}{\mathrm{\Omega }}}_{1}\right)\cap K$.

## 3 The main results

Let X = C4[0, 1] be the Banach space equipped with the norm $||u||\phantom{\rule{0.3em}{0ex}}=\underset{t\in \left[0,1\right]}{max}|u\left(t\right)|+\underset{t\in \left[0,1\right]}{max}|{u}^{\prime }\left(t\right)|+\underset{t\in \left[0,1\right]}{max}|{u}^{″}\left(t\right)|+\underset{t\in \left[0,1\right]}{max}|{u}^{‴}\left(t\right)|,$ and $K=\left\{u\in X:u\left(t\right)\ge 0,\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(t\right)\le 0,\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}u\left(t\right)\ge {d}_{1}{||u||}_{0},\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\phantom{\rule{0.3em}{0ex}}\left(-{u}^{″}\left(t\right)\right)\ge \frac{{d}_{2}\xi }{{c}_{2}}{||{u}^{″}||}_{0}\right\}$ is a cone in X.

Define two continuous convex functionals $\alpha \left(u\right)=\underset{t\in \left[0,1\right]}{max}|u\left(t\right)|+\underset{t\in \left[0,1\right]}{max}|{u}^{″}\left(t\right)|$ and $\beta \left(u\right)=\underset{t\in \left[0,1\right]}{max}|{u}^{\prime }\left(t\right)|+\underset{t\in \left[0,1\right]}{max}|{u}^{‴}\left(t\right)|$, for each u X, then ||u|| ≤ 2 max{α(u), β(u)} and α(λu) = |λ|α(u), β(λu) = |λ|β(u), for u X, λ R; α(u) ≤ α(v) for u, v K, uv.

In the following, we denote
$\begin{array}{c}B\phantom{\rule{2.77695pt}{0ex}}={\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)d\tau ds,\hfill \\ D={\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\omega }_{2}\left(\frac{1}{2}\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]d\tau ,\hfill \\ F=\frac{1}{sin\sqrt{A}}{\int }_{0}^{1}sin\sqrt{A}\tau d\tau \hfill \\ \phantom{\rule{1.5em}{0ex}}+\frac{\sqrt{A}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{2}\left(\tau ,x\right)q\left(x\right)dxd\tau }{sin\sqrt{A}-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}xdx-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}\left(1-x\right)dx},\hfill \\ {\eta }_{0}=\frac{1}{B+{c}_{2}D},\phantom{\rule{2.77695pt}{0ex}}{\eta }_{1}=\frac{1}{{\int }_{\frac{1}{4}}^{\frac{3}{4}}{Q}_{2}\left(\frac{1}{2},\tau \right)d\tau },\phantom{\rule{2.77695pt}{0ex}}{\eta }_{2}=\frac{2}{3{c}_{2}D+4F},\phantom{\rule{2.77695pt}{0ex}}\theta =min\left\{\frac{{d}_{1}}{2},\phantom{\rule{2.77695pt}{0ex}}\frac{{d}_{2}\xi }{2{c}_{2}}\right\}.\hfill \end{array}$
We will suppose that there are L > b > θb > c > 0 such that f(t, u, v, u0, v0) satisfies the following growth conditions:
$\begin{array}{c}\left({H}_{3}\right)\phantom{\rule{0.3em}{0ex}}f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)<\frac{c{\eta }_{0}}{\lambda },\phantom{\rule{0.3em}{0ex}}\text{for}\phantom{\rule{0.3em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[0,\phantom{\rule{2.77695pt}{0ex}}c\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right]×\left[-c,\phantom{\rule{2.77695pt}{0ex}}0\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right],\hfill \\ \left({H}_{4}\right)\phantom{\rule{0.3em}{0ex}}f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\ge \frac{b{\eta }_{1}}{\lambda },\text{for}\phantom{\rule{0.3em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[\frac{1}{4},\frac{3}{4}\right]×\left[\theta b,\phantom{\rule{2.77695pt}{0ex}}b\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right]×\left[-b,\phantom{\rule{2.77695pt}{0ex}}0\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right]\hfill \\ \phantom{\rule{22em}{0ex}}\cup \left[\frac{1}{4},\frac{3}{4}\right]×\left[0,\phantom{\rule{2.77695pt}{0ex}}b\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right]×\left[-b,\phantom{\rule{2.77695pt}{0ex}}-\theta b\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right],\hfill \\ \left({H}_{5}\right)\phantom{\rule{0.3em}{0ex}}f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)<\frac{L{\eta }_{2}}{\lambda },\text{for}\phantom{\rule{0.3em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[0,\phantom{\rule{2.77695pt}{0ex}}b\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right]×\left[-b,\phantom{\rule{2.77695pt}{0ex}}0\right]×\left[-L,\phantom{\rule{2.77695pt}{0ex}}L\right].\hfill \end{array}$
Let ${f}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)={f}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}{u}^{*},\phantom{\rule{2.77695pt}{0ex}}{v}^{*},\phantom{\rule{2.77695pt}{0ex}}{u}_{0}^{*},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}^{*}\right)$, where
$\begin{array}{cc}{u}^{*}\hfill & =min\left\{max\left(u,0\right),\phantom{\rule{0.3em}{0ex}}b\right\},\phantom{\rule{0.3em}{0ex}}{v}^{*}=min\left\{max\left(v,-\text{L}\right),\phantom{\rule{0.3em}{0ex}}\text{L}\right\},\hfill \\ {u}_{0}^{*}\hfill & =min\left\{max\left({u}_{0},\phantom{\rule{2.77695pt}{0ex}}-b\right),\phantom{\rule{2.77695pt}{0ex}}0\right\},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}^{*}=min\left\{max\left(v,\phantom{\rule{2.77695pt}{0ex}}-L\right),\phantom{\rule{2.77695pt}{0ex}}L\right\}.\hfill \end{array}$
We denote
$\left(Tu\right)\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{Q}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds,$
(3.1)
(3.2)
${\left(Tu\right)}^{″}\left(t\right)=-\lambda \underset{0}{\overset{1}{\int }}{Q}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),{u}^{‴}\left(\tau \right)\right)d\tau ,$
(3.3)
${\left(Tu\right)}^{‴}\left(t\right)=-\lambda \underset{0}{\overset{1}{\int }}\frac{\partial {Q}_{2}\left(t,\tau \right)}{\partial t}{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau .$
(3.4)

Lemma 3.1. Suppose that (H1) and (H2) hold. Then T: KK is completely continuous.

Proof. For u K, by (3.1), (3.3) and Lemma 2.2, it is obviously that Tu ≥ 0, (Tu)″ ≤ 0. In view of c1 = 1, c2> 1, so
$\begin{array}{cc}\hfill ||Tu|{|}_{0}& =\underset{t\in \left[0,1\right]}{max}\left|\lambda {\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\right|\hfill \\ \le \lambda {\int }_{0}^{1}{\int }_{0}^{1}\left[{c}_{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)\hfill \\ \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\hfill \\ =\lambda {\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),{u}^{‴}\left(\tau \right)\right)d\tau ds,\hfill \end{array}$
$\begin{array}{cc}\hfill ||{\left(Tu\right)}^{″}|{|}_{0}& =\underset{t\in \left[0,1\right]}{max}\left|-\lambda {\int }_{0}^{1}{Q}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \right|\hfill \\ \le \lambda {\int }_{0}^{1}\left[{c}_{2}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)\hfill \\ \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}+\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \hfill \\ \le \underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}{\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]\hfill \\ ×{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau .\hfill \end{array}$
By Lemma 2.3, (3.1) and (3.3), we have
$\begin{array}{cc}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\phantom{\rule{0.3em}{0ex}}\left(Tu\right)\left(t\right)\hfill & =\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\lambda {\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\hfill \\ \phantom{\rule{1em}{0ex}}\ge \lambda {\int }_{0}^{1}{\int }_{0}^{1}\left[{b}_{1}{G}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}t\right){G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)\hfill \\ \phantom{\rule{1em}{0ex}}+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\hfill \\ =\lambda {\int }_{0}^{1}{\int }_{0}^{1}\left[{d}_{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]\hfill \\ \phantom{\rule{1em}{0ex}}×\phantom{\rule{0.3em}{0ex}}{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\hfill \\ \ge {d}_{1}\lambda {\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\hfill \\ \ge {d}_{1}||Tu|{|}_{0},\hfill \end{array}$
$\begin{array}{cc}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\left(-{\left(Tu\right)}^{″}\left(t\right)\right)\hfill & =\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\int }_{0}^{1}{Q}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \hfill \\ \ge \underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\int }_{0}^{1}\left[{b}_{2}{G}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}t\right){G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)\hfill \\ \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}+{\omega }_{2}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \hfill \\ \ge {\int }_{0}^{1}\left[{b}_{2}{G}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}t\right){G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)\hfill \\ +\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \hfill \\ ={\int }_{0}^{1}\left[{d}_{2}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)\hfill \\ +\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \hfill \\ \ge {d}_{2}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}{\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)\hfill \\ \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}+{\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \hfill \\ \ge \frac{{d}_{2}}{{c}_{2}}\frac{\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}{\omega }_{2}\left(t\right)}{\underset{\frac{1}{4}\le t\le \frac{3}{4}}{max}{\omega }_{2}\left(t\right)}||{\left(Tu\right)}^{″}|{|}_{0}\hfill \\ \ge \frac{{d}_{2}\xi }{{c}_{2}}||{\left(Tu\right)}^{″}|{|}_{0}.\hfill \end{array}$

So we can get T(K) K: Let B K is bounded, it is clear that T(B) is bounded. Using f1, Q1(t, s), Q2(t, s) is continuous, we show that T(B) is equicontinuous. By the Arzela-Ascoli theorem, a standard proof yields T: KK is completely continuous.

Theorem 3.1. Suppose that (H1)-(H5) hold. Then BVP (1.1) has at least one positive solution u(t) satisfying
$c<\alpha \left(u\right)
Proof. Take
${\mathrm{\Omega }}_{1}=\left\{u\in X\phantom{\rule{2.77695pt}{0ex}}:\phantom{\rule{2.77695pt}{0ex}}\alpha \left(u\right)
two bounded open sets in X, and
${D}_{1}=\left\{u\in X\phantom{\rule{2.77695pt}{0ex}}:\phantom{\rule{2.77695pt}{0ex}}\alpha \left(u\right)=c\right\},\phantom{\rule{1em}{0ex}}{D}_{2}=\left\{u\in X\phantom{\rule{2.77695pt}{0ex}}:\phantom{\rule{2.77695pt}{0ex}}\alpha \left(u\right)=b\right\}.$

By Lemma 3.1, T: KK is completely continuous. Let $p=\frac{b}{2}\in \left({\mathrm{\Omega }}_{2}\cap K\right)\\left\{0\right\},\phantom{\rule{2.77695pt}{0ex}}\alpha \left(p\right)\ne 0$. It is easy to see that α(u + λp) ≥ α(u), for all u K and λ ≥ 0.

Let u D1, we have
$\begin{array}{cc}||Tu|{|}_{0}\hfill & =\underset{t\in \left[0,1\right]}{max}\left|\lambda {\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\right|\hfill \\ \le \lambda {\int }_{0}^{1}{\int }_{0}^{1}\left[{c}_{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right)+{\omega }_{1}{\int }_{0}^{1}{G}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)p\left(x\right)dx\right]{Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)d\tau ds×\frac{c{\eta }_{0}}{\lambda }\hfill \\ =c{\eta }_{0}{\int }_{0}^{1}{\int }_{0}^{1}{Q}_{1}\left(s,\phantom{\rule{2.77695pt}{0ex}}s\right){Q}_{2}\left(s,\phantom{\rule{2.77695pt}{0ex}}\tau \right)d\tau ds\hfill \\ =Bc{\eta }_{0},\hfill \end{array}$
$\begin{array}{cc}||{\left(Tu\right)}^{″}|{|}_{0}\hfill & =\underset{t\in \left[0,1\right]}{max}\left|-\lambda {\int }_{0}^{1}{Q}_{2}\left(t,\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \right|\hfill \\ <\lambda {\int }_{0}^{1}\left[{c}_{2}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\omega }_{2}\left(\frac{1}{2}\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]d\tau ×\frac{c{\eta }_{0}}{\lambda }\hfill \\ \le {c}_{2}c{\eta }_{0}{\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}\tau \right)+{\omega }_{2}\left(\frac{1}{2}\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}x\right)q\left(x\right)dx\right]d\tau \hfill \\ ={c}_{2}Dc{\eta }_{0},\hfill \end{array}$
Hence, for u D1K, α(u) = c, we get
$\alpha \left(Tu\right)=\phantom{\rule{0.3em}{0ex}}{||Tu||}_{0}+{||{\left(Tu\right)}^{″}||}_{0}
Whereas for u D2K, α(u) = b, there is $||u|{|}_{0}\phantom{\rule{0.3em}{0ex}}\ge \frac{b}{2}$ or $||{u}^{″}|{|}_{0}\phantom{\rule{0.3em}{0ex}}\ge \frac{b}{2}$, By Lemma 2.4, we get
$\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}u\left(t\right)\ge {d}_{1}{||u||}_{0}\phantom{\rule{0.3em}{0ex}}\ge \frac{{d}_{1}b}{2}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\text{or}\underset{\frac{1}{4}\le t\le \frac{3}{4}}{min}\left(-{u}^{″}\left(t\right)\right)\ge \frac{{d}_{2}\xi }{{c}_{2}}{||{u}^{″}||}_{0}\ge \frac{{d}_{2}\xi b}{2{c}_{2}}.$
Therefore, using (H4) and (3.3), we have
$\begin{array}{cc}|{\left(Tu\right)}^{″}\left(\frac{1}{2}\right)|\phantom{\rule{0.3em}{0ex}}\hfill & =\left|\lambda {\int }_{0}^{1}{Q}_{2}\left(\frac{1}{2},\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \right|\hfill \\ >\left|\lambda {\int }_{\frac{1}{4}}^{\frac{3}{4}}{Q}_{2}\left(\frac{1}{2},\phantom{\rule{2.77695pt}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \right|\hfill \\ \ge \lambda ×\frac{b{\eta }_{1}}{\lambda }{\int }_{\frac{1}{4}}^{\frac{3}{4}}{Q}_{2}\phantom{\rule{2.77695pt}{0ex}}\left(\frac{1}{2},\phantom{\rule{2.77695pt}{0ex}}\tau \right)d\tau \hfill \\ =b.\hfill \end{array}$
Hence,
$\alpha \left(Tu\right)\ge \phantom{\rule{2.77695pt}{0ex}}\left|{\left(Tu\phantom{\rule{2.77695pt}{0ex}}\right)}^{″}\left(\frac{1}{2}\right)\right|>b.$
By (3.2), (3.4), and (H5), for u K, we have
$\begin{array}{l}{‖{\left(Tu\right)}^{\prime }‖}_{0}=\underset{t\in \left[0,1\right]}{\mathrm{max}}|\lambda {\int }_{t}^{1}{\int }_{0}^{1}{Q}_{2}\left(s,\phantom{\rule{0.1em}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{0.1em}{0ex}}u\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}-\lambda {\int }_{0}^{1}{\int }_{0}^{1}s{Q}_{2}\left(s,\phantom{\rule{0.1em}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{0.1em}{0ex}}u\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds|\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le \underset{t\in \left[0,1\right]}{\mathrm{max}}|\lambda {\int }_{t}^{1}{\int }_{0}^{1}{Q}_{2}\left(s,\phantom{\rule{0.1em}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{0.1em}{0ex}}u\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds|\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.1em}{0ex}}+\underset{t\in \left[0,1\right]}{\mathrm{max}}|\lambda {\int }_{0}^{1}{\int }_{0}^{1}s{Q}_{2}\left(s,\phantom{\rule{0.1em}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{0.1em}{0ex}}u\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds|\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le \lambda |{\int }_{0}^{1}{\int }_{0}^{1}\left(1+s\right){Q}_{2}\left(s,\phantom{\rule{0.1em}{0ex}}\tau \right){f}_{1}\left(\tau ,\phantom{\rule{0.1em}{0ex}}u\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.1em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau ds|\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le \lambda ×\frac{{\eta }_{2}L}{\lambda }|{\int }_{0}^{1}{\int }_{0}^{1}\left(1+s\right)\left[{c}_{2}{G}_{2}\left(\tau ,\phantom{\rule{0.1em}{0ex}}\tau \right)+{\omega }_{2}\left(\frac{1}{2}\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{0.1em}{0ex}}x\right)q\left(x\right)dx\right]d\tau ds|\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le {\eta }_{2}L×\frac{3}{2}{c}_{2}{\int }_{0}^{1}\left[{G}_{2}\left(\tau ,\phantom{\rule{0.1em}{0ex}}\tau \right)+{\omega }_{2}\left(\frac{1}{2}\right){\int }_{0}^{1}{G}_{2}\left(\tau ,\phantom{\rule{0.1em}{0ex}}x\right)q\left(x\right)dx\right]d\tau \\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.1em}{0ex}}=\frac{3}{2}{c}_{2}D{\eta }_{2}L,\end{array}$
$\begin{array}{cc}{∥{\left(Tu\right)}^{‴}∥}_{0}\hfill & =\underset{t\in \left[0,1\right]}{max}\left|\lambda {\int }_{0}^{1}\frac{\partial {Q}_{2}\left(t,\tau \right)}{\partial t}{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{u}^{‴}\left(\tau \right)\right)d\tau \right|\hfill \\ \le 2\lambda {\int }_{0}^{1}\left[\frac{sin\sqrt{A}\tau }{sin\sqrt{A}}+\frac{\sqrt{A}{\int }_{0}^{1}{G}_{2}\left(\tau ,x\right)q\left(x\right)dx}{sin\sqrt{A}-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}xdx-{\int }_{0}^{1}q\left(x\right)sin\sqrt{A}\left(1-x\right)dx}\right]\hfill \\ \phantom{\rule{3em}{0ex}}×|{f}_{1}\left(\tau ,\phantom{\rule{2.77695pt}{0ex}}u\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(\tau \right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(\tau \right)\right)|d\tau \hfill \\ \phantom{\rule{2em}{0ex}}<\lambda 2F×\frac{{\eta }_{2}L}{\lambda }\hfill \\ \phantom{\rule{2em}{0ex}}=2F{\eta }_{2}L.\hfill \end{array}$
So,
$\beta \left(Tu\right)=∥{\left(Tu\right)}^{\prime }∥{}_{0}+{∥{\left(Tu\right)}^{‴}∥}_{0}<\frac{3}{2}{c}_{2}D{\eta }_{2}L+2F{\eta }_{2}L=\left(\frac{3}{2}{c}_{2}D+2F\right){\eta }_{2}L=L.$
Theorem 2.1 implies there is $\left({\mathrm{\Omega }}_{2}\{\stackrel{̄}{\mathrm{\Omega }}}_{1}\right)\cap K$ such that u = Tu. So, u(t) is a positive solution for BVP (1.1) satisfying
$c<\alpha \left(u\right)

Thus, Theorem 3.1 is completed.

## 4 Example

Example 4.1. Consider the following boundary value problem
$\left\{\begin{array}{c}{u}^{\left(4\right)}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}+\frac{{\pi }^{2}}{9}{u}^{″}\left(t\right)\phantom{\rule{2.77695pt}{0ex}}={\pi }^{2}f\left(t,\phantom{\rule{2.77695pt}{0ex}}u\left(t\right),\phantom{\rule{2.77695pt}{0ex}}{u}^{\prime }\left(t\right),\phantom{\rule{2.77695pt}{0ex}}{u}^{″}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}{u}^{‴}\left(t\right)\right),\phantom{\rule{1em}{0ex}}0
(4.1)
where
$f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)=\left\{\begin{array}{c}\frac{1}{20}\left(u-{u}_{0}\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|,\hfill \\ \phantom{\rule{1em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[0,2\right]×\left[-16000,16000\right]×\left[-2,0\right]×\left[-16000,16000\right],\hfill \\ \frac{1}{20}\left(2-{u}_{0}\right)\left(3-u\right)+\frac{27}{2}\left(3-{u}_{0}\right)\left(u-2\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|,\hfill \\ \phantom{\rule{1em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[2,3\right]×\left[-16000,16000\right]×\left[-2,0\right]×\left[-16000,16000\right],\hfill \\ \frac{1}{20}\left(u+2\right)\left({u}_{0}+3\right)-\frac{27}{2}\left(u+3\right)\left({u}_{0}+2\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|,\hfill \\ \phantom{\rule{1em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[0,2\right]×\left[-16000,16000\right]×\left[-3,\phantom{\rule{2.77695pt}{0ex}}-2\right]×\left[-16000,16000\right],\hfill \\ \frac{1}{5}\left(3-u\right)\left({u}_{0}+3\right)+\frac{135}{2}\left(u-2\right)\left({u}_{0}+3\right)-\frac{27}{2}\left(u+3\right)\left({u}_{0}+2\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|,\hfill \\ \phantom{\rule{1em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[2,3\right]×\left[-16000,16000\right]×\left[-3,\phantom{\rule{2.77695pt}{0ex}}-2\right]×\left[-16000,16000\right],\hfill \\ \frac{27}{2}\left(u-{u}_{0}\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|,\hfill \\ \phantom{\rule{1em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[3,40\right]×\left[-16000,16000\right]×\left[-40,0\right]×\left[-16000,16000\right],\hfill \\ \phantom{\rule{9em}{0ex}}\cup \phantom{\rule{0.3em}{0ex}}\left[0,1\right]×\left[0,40\right]×\left[-16000,16000\right]×\left[-40,\phantom{\rule{2.77695pt}{0ex}}-3\right]×\left[-16000,16000\right].\hfill \end{array}\right\$
In this problem, we know that $A=\frac{{\pi }^{2}}{9},\phantom{\rule{2.77695pt}{0ex}}\lambda ={\pi }^{2},\phantom{\rule{2.77695pt}{0ex}}p\left(t\right)=t,\phantom{\rule{2.77695pt}{0ex}}q\left(t\right)=0$, then we can get ${b}_{1}=1,{b}_{2}=\frac{\sqrt{3}\pi }{6},\phantom{\rule{2.77695pt}{0ex}}{c}_{1}=1,\phantom{\rule{2.77695pt}{0ex}}{c}_{2}=\frac{2\sqrt{3}}{3},\phantom{\rule{2.77695pt}{0ex}}{\omega }_{1}=2,\phantom{\rule{2.77695pt}{0ex}}{\omega }_{2}=\frac{2\sqrt{3}sin\frac{\pi }{3}\left(1+t\right)}{3},\phantom{\rule{2.77695pt}{0ex}}{d}_{1}=\frac{3}{16},\phantom{\rule{2.77695pt}{0ex}}{d}_{2}=\frac{\sqrt{3}-1}{4},\phantom{\rule{2.77695pt}{0ex}}\xi =\frac{\sqrt{2+\sqrt{3}}}{2}.$Further more, we obtain
$B=\frac{1944\sqrt{3}-972\pi -9{\pi }^{3}}{4{\pi }^{5}},\phantom{\rule{1em}{0ex}}D=\frac{9-\sqrt{3}\pi }{2{\pi }^{2}},\phantom{\rule{1em}{0ex}}F=\frac{\sqrt{3}}{\pi }.$

then ${\eta }_{0}=\frac{12{\pi }^{5}}{5832\sqrt{3}-2916\pi -27{\pi }^{3}+36\sqrt{3}{\pi }^{3}-12{\pi }^{4}}$, ${\eta }_{1}=\frac{{\pi }^{2}}{3\sqrt{6+3\sqrt{3}}-9}$, ${\eta }_{2}=\frac{2{\pi }^{2}}{9\sqrt{3}+4\sqrt{3}\pi -3\pi }$, $\theta =min\left\{\frac{{d}_{1}}{2},\phantom{\rule{2.77695pt}{0ex}}\frac{{d}_{2}\xi }{2{c}_{2}}\right\}=\frac{\sqrt{2+\sqrt{3}}\left(3-\sqrt{3}\right)}{32}$, θb ≈ 3.06 > 3.

If we take c = 2, b = 40, L = 16000, then we get
$\begin{array}{c}f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)=\frac{1}{20}\left(u-{u}_{0}\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|\le 0.7<\frac{c{\eta }_{0}}{\lambda }\approx 0.8,\hfill \\ \text{for}\phantom{\rule{0.3em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[0,2\right]×\left[-16000,16000\right]×\left[-2,0\right]×\left[-16000,16000\right],\hfill \\ f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)=\frac{27}{2}\left(u-{u}_{0}\right)+\frac{1}{2}|cos\left(v+{v}_{0}\right)|\ge 40>\frac{b{\eta }_{1}}{\lambda }\approx 38,\hfill \\ \text{for}\phantom{\rule{0.3em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[\frac{1}{4},\frac{3}{4}\right]×\left[\theta b,\phantom{\rule{2.77695pt}{0ex}}40\right]×\left[-16000,16000\right]×\left[-40,0\right]×\left[-16000,16000\right]\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{8em}{0ex}}\cup \phantom{\rule{0.3em}{0ex}}\left[\frac{1}{4},\phantom{\rule{2.77695pt}{0ex}}\frac{3}{4}\right]×\left[0,40\right]×\left[-16000,16000\right]×\left[-40,\phantom{\rule{2.77695pt}{0ex}}-\theta b\right]×\left[-16000,16000\right],\hfill \\ f\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\le 1080.5<\frac{L{\eta }_{2}}{\lambda }\approx 1146,\hfill \\ \text{for}\phantom{\rule{0.3em}{0ex}}\left(t,\phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}v,\phantom{\rule{2.77695pt}{0ex}}{u}_{0},\phantom{\rule{2.77695pt}{0ex}}{v}_{0}\right)\in \left[0,1\right]×\left[0,40\right]×\left[-16000,16000\right]×\left[-40,0\right]×\left[-16000,16000\right].\hfill \end{array}$
Then all the conditions of Theorem 3.1 are satisfied. Therefore, by Theorem 3.1 we know that boundary value problem (4.1) has at least one positive solution u(t) satisfying
$2<\alpha \left(u\right)<40,\phantom{\rule{0.3em}{0ex}}\beta \left(u\right)<16000.$

## Declarations

### Acknowledgements

The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2009000664, A2011208012). The research item financed by the talent training project funds of Hebei Province. The authors would like to thank the referee for helpful comments and suggestions.

## Authors’ Affiliations

(1)
College of Electrical Engineering and Information, Hebei University of Science and Technology
(2)
College of Sciences, Hebei University of Science and Technology

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