## Boundary Value Problems

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# Positive solutions for a sixth-order boundary value problem with four parameters

Boundary Value Problems20132013:184

DOI: 10.1186/1687-2770-2013-184

Accepted: 30 July 2013

Published: 14 August 2013

## Abstract

This paper investigates the existence and multiplicity of positive solutions of a sixth-order differential system with four variable parameters using a monotone iterative technique and an operator spectral theorem.

MSC:34B15, 34B18.

### Keywords

positive solutions variable parameters fixed point theorem operator spectral theorem

## 1 Introduction

It is well known that boundary value problems for ordinary differential equations can be used to describe a large number of physical, biological and chemical phenomena. In recent years, boundary value problems for sixth-order ordinary differential equations, which arise naturally, for example, in sandwich beam deflection under transverse shear have been studied extensively, see [14] and the references therein. The deformation of the equilibrium state of an elastic circular ring segment with its two ends simply supported can be described by a boundary value problem involving a sixth-order ordinary differential equation
$\begin{array}{r}{u}^{\left(6\right)}+2{u}^{\left(4\right)}+{u}^{\mathrm{\prime }\mathrm{\prime }}=f\left(t,u\right),\phantom{\rule{1em}{0ex}}0
(1)
Liu and Li [5] studied the existence and nonexistence of positive solutions of the nonlinear fourth-order beam equation
$\begin{array}{r}{u}^{\left(4\right)}\left(t\right)+\beta {u}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)-\alpha u\left(t\right)=\lambda f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}0
(2)

They showed that there exists a ${\lambda }^{\star }>0$ such that the above boundary value problem has at least two, one, and no positive solutions for $0<\lambda <{\lambda }^{\star }$, $\lambda ={\lambda }^{\star }$ and $\lambda >{\lambda }^{\star }$, respectively.

In this paper, we discuss the existence of positive solutions for the sixth-order boundary value problem
$\begin{array}{c}\begin{array}{r}-{u}^{\left(6\right)}+A\left(t\right){u}^{\left(4\right)}+B\left(t\right){u}^{\mathrm{\prime }\mathrm{\prime }}+C\left(t\right)u=\left(D\left(t\right)+u\right)\phi +\lambda f\left(t,u\right),\phantom{\rule{1em}{0ex}}0
(3)

For this, we shall assume the following conditions throughout

(H1) $f\left(t,u\right):\left[0,1\right]×\left[0,\mathrm{\infty }\right)⟶\left[0,\mathrm{\infty }\right)$ is continuous;

(H2) $a,b,c\in R$, $a={\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}>-{\pi }^{2}$, $b=-{\lambda }_{1}{\lambda }_{2}-{\lambda }_{2}{\lambda }_{3}-{\lambda }_{1}{\lambda }_{3}>0$, $c={\lambda }_{1}{\lambda }_{2}{\lambda }_{3}<0$ where ${\lambda }_{1}\ge 0\ge {\lambda }_{2}\ge -{\pi }^{2}$, $0\le {\lambda }_{3}<-{\lambda }_{2}\phantom{\rule{0.25em}{0ex}}$ and ${\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c>0$, and $A,B,C,D\in C\left[0,1\right]$ with $a={sup}_{t\in \left[0,1\right]}A\left(t\right)$, $b={inf}_{t\in \left[0,1\right]}B\left(t\right)$ and $c={sup}_{t\in \left[0,1\right]}C\left(t\right)$.

Let $K={max}_{0\le t\le 1}\left[-A\left(t\right)+B\left(t\right)-C\left(t\right)-\left(-a+b-c\right)\right]$ and $\mathrm{\Gamma }={\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c$.

Assumption (H2) involves a three-parameter nonresonance condition.

More recently Li [6] studied the existence and multiplicity of positive solutions for a sixth-order boundary value problem with three variable coefficients. The main difference between our work and [6] is that we consider boundary value problem not only with three variable coefficients, but also with two positive parameters λ and μ, and the existence of the positive solution depends on these parameters. In this paper, we shall apply the monotone iterative technique [7] to boundary value problem (3) and then obtain several new existence and multiplicity results. In the special case, in [8] by using the fixed point theorem and the operator spectral theorem, we establish a theorem on the existence of positive solutions for the sixth-order boundary value problem (3) with $\lambda =1$.

## 2 Preliminaries

Let $Y=C\left[0,1\right]$ and ${Y}_{+}=\left\{u\in Y:u\left(t\right)\ge 0,t\in \left[0,1\right]\right\}$. It is well known that Y is a Banach space equipped with the norm ${\parallel u\parallel }_{0}={sup}_{t\in \left[0,1\right]}|u\left(t\right)|$. Set $X=\left\{u\in {C}^{4}\left[0,1\right]:u\left(0\right)=u\left(1\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(0\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(1\right)=0\right\}$. For given $\chi \ge 0$ and $\nu \ge 0$, we denote the norm ${\parallel \cdot \parallel }_{\chi ,\nu }$ by
${\parallel \cdot \parallel }_{\chi ,\nu }=\underset{t\in \left[0,1\right]}{sup}\left\{|{u}^{\left(4\right)}\left(t\right)|+\chi |{u}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)|+\nu |u\left(t\right)|\right\},\phantom{\rule{1em}{0ex}}u\in X.$
We also need the space X, equipped with the norm
${\parallel u\parallel }_{2}=max\left\{{\parallel u\parallel }_{0},{\parallel {u}^{\mathrm{\prime }\mathrm{\prime }}\parallel }_{0},{\parallel {u}^{\left(4\right)}\parallel }_{0}\right\}.$

In [8], it is shown that X is complete with the norm ${\parallel \cdot \parallel }_{\chi ,\nu }$ and ${\parallel u\parallel }_{2}$, and moreover $\mathrm{\forall }u\in X$, ${\parallel u\parallel }_{0}\le {\parallel {u}^{\mathrm{\prime }\mathrm{\prime }}\parallel }_{0}\le {\parallel {u}^{\left(4\right)}\parallel }_{0}$.

For $h\in Y$, consider the linear boundary value problem
$\begin{array}{r}-{u}^{\left(6\right)}+a{u}^{\left(4\right)}+b{u}^{\mathrm{\prime }\mathrm{\prime }}+cu=h\left(t\right),\phantom{\rule{1em}{0ex}}0
(4)
where a, b, c satisfy the assumption
${\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c>0,$
(5)

and let $\mathrm{\Gamma }={\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c$. Inequality (5) follows immediately from the fact that $\mathrm{\Gamma }={\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c$ is the first eigenvalue of the problem $-{u}^{\left(6\right)}+a{u}^{\left(4\right)}+b{u}^{\mathrm{\prime }\mathrm{\prime }}+cu=\lambda u$, $u\left(0\right)=u\left(1\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(0\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(1\right)={u}^{\left(4\right)}\left(0\right)={u}^{\left(4\right)}\left(1\right)=0$, and ${\varphi }_{1}\left(t\right)=sin\pi t$ is the first eigenfunction, i.e., $\mathrm{\Gamma }>0$. Since the line ${l}_{1}=\left\{\left(a,b,c\right):{\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c=0\right\}$ is the first eigenvalue line of the three-parameter boundary value problem $\phantom{\rule{0.25em}{0ex}}-{u}^{\left(6\right)}+a{u}^{\left(4\right)}+b{u}^{\mathrm{\prime }\mathrm{\prime }}+cu=0$, $u\left(0\right)=u\left(1\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(0\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(1\right)={u}^{\left(4\right)}\left(0\right)={u}^{\left(4\right)}\left(1\right)=0$, if $\left(a,b,c\right)$ lies in ${l}_{1}$, then by the Fredholm alternative, the existence of a solution of the boundary value problem (4) cannot be guaranteed.

Let $P\left(\lambda \right)={\lambda }^{2}+\beta \lambda -\alpha$, where $\beta <2{\pi }^{2}$, $\alpha \ge 0$. It is easy to see that the equation $P\left(\lambda \right)=0$ has two real roots ${\lambda }_{1},{\lambda }_{2}=\frac{-\beta ±\sqrt{{\beta }^{2}+4\alpha }}{2}$ with ${\lambda }_{1}\ge 0\ge {\lambda }_{2}>-{\pi }^{2}$. Let ${\lambda }_{3}$ be a number such that $0\le {\lambda }_{3}<-{\lambda }_{2}$. In this case, (4) satisfies the decomposition form
$-{u}^{\left(6\right)}+a{u}^{\left(4\right)}+b{u}^{\mathrm{\prime }\mathrm{\prime }}+cu=\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{1}\right)\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{2}\right)\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{3}\right)u,\phantom{\rule{1em}{0ex}}0
(6)
Suppose that ${G}_{i}\left(t,s\right)$ ($i=1,2,3$) is the Green’s function associated with
$-{u}^{\mathrm{\prime }\mathrm{\prime }}+{\lambda }_{i}u=0,\phantom{\rule{2em}{0ex}}u\left(0\right)=u\left(1\right)=0.$
(7)

We need the following lemmas.

Lemma 1 [5, 9]

Let ${\omega }_{i}=\sqrt{|{\lambda }_{i}|}$, then ${G}_{i}\left(t,s\right)$ ($i=1,2,3$) can be expressed as
1. (i)
when ${\lambda }_{i}>0$,
${G}_{i}\left(t,s\right)=\left\{\begin{array}{ll}\frac{sinh{\omega }_{i}tsinh{\omega }_{i}\left(1-s\right)}{{\omega }_{i}sinh{\omega }_{i}},& 0\le t\le s\le 1,\\ \frac{sinh{\omega }_{i}ssinh{\omega }_{i}\left(1-t\right)}{{\omega }_{i}sinh{\omega }_{i}},& 0\le s\le t\le 1\end{array}\right\};$

2. (ii)
when ${\lambda }_{i}=0$,
${G}_{i}\left(t,s\right)=\left\{\begin{array}{ll}t\left(1-s\right),& 0\le t\le s\le 1,\\ s\left(1-t\right),& 0\le s\le t\le 1\end{array}\right\};$

3. (iii)
when $-{\pi }^{2}<{\lambda }_{i}<0$,
${G}_{i}\left(t,s\right)=\left\{\begin{array}{ll}\frac{sin{\omega }_{i}tsin{\omega }_{i}\left(1-s\right)}{{\omega }_{i}sin{\omega }_{i}},& 0\le t\le s\le 1,\\ \frac{sin{\omega }_{i}ssin{\omega }_{i}\left(1-t\right)}{{\omega }_{i}sin{\omega }_{i}},& 0\le s\le t\le 1\end{array}\right\}.$

Lemma 2 [5]

${G}_{i}\left(t,s\right)$ ($i=1,2,3$) has the following properties
1. (i)

${G}_{i}\left(t,s\right)>0$, $\mathrm{\forall }t,s\in \left(0,1\right)$;

2. (ii)

${G}_{i}\left(t,s\right)\le {C}_{i}{G}_{i}\left(s,s\right)$, $\mathrm{\forall }t,s\in \left[0,1\right]$;

3. (iii)

${G}_{i}\left(t,s\right)\ge {\delta }_{i}{G}_{i}\left(t,t\right){G}_{i}\left(s,s\right)$, $\mathrm{\forall }t,s\in \left[0,1\right]$,

where ${C}_{i}=1$, ${\delta }_{i}=\frac{{\omega }_{i}}{sinh{\omega }_{i}}$, if ${\lambda }_{i}>0$; ${C}_{i}=1$, ${\delta }_{i}=1$, if ${\lambda }_{i}=0$; ${C}_{i}=\frac{1}{sin{\omega }_{i}}$, ${\delta }_{i}={\omega }_{i}sin{\omega }_{i}$, if $-{\pi }^{2}<{\lambda }_{i}<0$.

In what follows, we let ${D}_{i}={max}_{t\in \left[0,1\right]}{\int }_{0}^{1}{G}_{i}\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds$.

Lemma 3 [10]

Let X be a Banach space, K a cone and Ω a bounded open subset of X. Let $\theta \in \mathrm{\Omega }$ and $T:K\cap \overline{\mathrm{\Omega }}\to K$ be condensing. Suppose that $Tx\ne \upsilon x$ for all $x\in K\cap \partial \mathrm{\Omega }$ and $\upsilon \ge 1$. Then $i\left(T,K\cap \mathrm{\Omega },K\right)=1$.

Lemma 4 [10]

Let X be a Banach space, let K be a cone of X. Assume that $T:{\overline{K}}_{r}\to K$ (here ${K}_{r}=\left\{x\in K\mid \parallel x\parallel , $r>0$) is a compact map such that $Tx\ne x$ for all $x\in \partial {K}_{r}$. If $\parallel x\parallel \le \parallel Tx\parallel$ for $x\in \partial {K}_{r}$, then $i\left(T,{K}_{r},K\right)=0$.

Now, since
$\begin{array}{rl}-{u}^{\left(6\right)}+a{u}^{\left(4\right)}+b{u}^{\mathrm{\prime }\mathrm{\prime }}+cu& =\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{1}\right)\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{2}\right)\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{3}\right)u\\ =\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{2}\right)\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{1}\right)\left(-\frac{{d}^{2}}{d{t}^{2}}+{\lambda }_{3}\right)u=h\left(t\right),\end{array}$
(8)
the solution of boundary value problem (4) can be expressed as
$u\left(t\right)={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)h\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(9)

Thus, for every given $h\in Y$, the boundary value problem (4) has a unique solution $u\in {C}^{6}\left[0,1\right]$, which is given by (9).

We now define a mapping $T:C\left[0,1\right]\to C\left[0,1\right]$ by
$\left(Th\right)\left(t\right)={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)h\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(10)

Throughout this article, we shall denote $Th=u$ the unique solution of the linear boundary value problem (4).

Lemma 5 [8]

$T:Y⟶\left(X,{\parallel \cdot \parallel }_{\chi ,\nu }\right)$ is linear completely continuous, where $\chi ={\lambda }_{1}+{\lambda }_{3}$, $\nu ={\lambda }_{1}{\lambda }_{3}$ and $\parallel T\parallel \le {D}_{2}$. Moreover, $\mathrm{\forall }h\in {Y}_{+}$, if $u=Th$, then $u\in X\cap {Y}_{+}$, and ${u}^{\mathrm{\prime }\mathrm{\prime }}\le 0$, ${u}^{\left(4\right)}\ge 0$.

We list the following conditions for convenience

(H3) $f\left(t,u\right)$ is nondecreasing in u for $t\in \left[0,1\right]$;

(H4) $f\left(t,0\right)>\stackrel{ˆ}{c}>0$ for all $t\in \left[0,1\right]$;

(H5) ${f}_{\mathrm{\infty }}={lim}_{u\to \mathrm{\infty }}\frac{f\left(t,u\right)}{u}=\mathrm{\infty }$ uniformly for $t\in \left[0,1\right]$;

(H6) $f\left(t,\rho u\right)\ge {\rho }^{\alpha }f\left(t,u\right)$ for $\rho \in \left(0,1\right)$ and $t\in \left[0,1\right]$, where $\alpha \in \left(0,1\right)$ is independent of ρ and u.

Suppose that $G\left(t,s\right)$ is the Green’s function of the linear boundary value problem
$-{u}^{\mathrm{\prime }\mathrm{\prime }}+\varkappa u=0,\phantom{\rule{2em}{0ex}}u\left(0\right)=u\left(1\right)=0.$
(11)
Then, the boundary value problem
$-{\phi }^{\mathrm{\prime }\mathrm{\prime }}+\varkappa \phi =\mu u,\phantom{\rule{2em}{0ex}}\phi \left(0\right)=\phi \left(1\right)=0,$
can be solved by using Green’s function, namely,
$\phi \left(t\right)=\mu {\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}0
(12)
where $\varkappa >-{\pi }^{2}$. Thus, inserting (12) into the first equation in (3), yields
$\begin{array}{r}-{u}^{\left(6\right)}+A\left(t\right){u}^{\left(4\right)}+B\left(t\right){u}^{\mathrm{\prime }\mathrm{\prime }}+C\left(t\right)u=\mu \left(D\left(t\right)+u\left(t\right)\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\lambda f\left(t,u\right),\\ u\left(0\right)=u\left(1\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(0\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(1\right)={u}^{\left(4\right)}\left(0\right)={u}^{\left(4\right)}\left(1\right)=0.\end{array}$
(13)
Let us consider the boundary value problem
$\begin{array}{r}-{u}^{\left(6\right)}+A\left(t\right){u}^{\left(4\right)}+B\left(t\right){u}^{\mathrm{\prime }\mathrm{\prime }}+C\left(t\right)u=h\left(t\right),\phantom{\rule{1em}{0ex}}0
(14)

Now, we consider the existence of a positive solution of (14). The function $u\in {C}^{6}\left(0,1\right)\cap {C}^{4}\left[0,1\right]$ is a positive solution of (14), if $u\ge 0$, $t\in \left[0,1\right]$, and $u\ne 0$.

Let us rewrite equation (13) in the following form
$\begin{array}{rl}-{u}^{\left(6\right)}+a{u}^{\left(4\right)}+b{u}^{\mathrm{\prime }\mathrm{\prime }}+cu=& -\left(A\left(t\right)-a\right){u}^{\left(4\right)}-\left(B\left(t\right)-b\right){u}^{\mathrm{\prime }\mathrm{\prime }}-\left(C\left(t\right)-c\right)u\\ +\mu \left(D\left(t\right)+u\left(t\right)\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+h\left(t\right).\end{array}$
(15)
For any $u\in X$, let
$Gu=-\left(A\left(t\right)-a\right){u}^{\left(4\right)}-\left(B\left(t\right)-b\right){u}^{\mathrm{\prime }\mathrm{\prime }}-\left(C\left(t\right)-c\right)u+\mu D\left(t\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds.$
The operator $G:X\to Y$ is linear. By Lemmas 2 and 3 in [8], $\mathrm{\forall }u\in X$, $t\in \left[0,1\right]$, we have
$\begin{array}{rl}|\left(Gu\right)\left(t\right)|& \le \left[-A\left(t\right)+B\left(t\right)-C\left(t\right)-\left(-a+b-c\right)\right]{\parallel u\parallel }_{2}+\mu C{d}_{1}{\parallel u\parallel }_{0}\\ \le \left(K+\mu C{d}_{1}\right){\parallel u\parallel }_{2}\le \left(K+\mu C{d}_{1}\right){\parallel u\parallel }_{\chi ,\nu },\end{array}$
where $C={max}_{t\in \left[0,1\right]}D\left(t\right)$, $K={max}_{t\in \left[0,1\right]}\left[-A\left(t\right)+B\left(t\right)-C\left(t\right)-\left(-a+b-c\right)\right]$, ${d}_{1}={max}_{t\in \left[0,1\right]}{\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds$, $\chi ={\lambda }_{1}+{\lambda }_{3}\ge 0$, $\nu ={\lambda }_{1}{\lambda }_{3}\ge 0$. Hence ${\parallel Gu\parallel }_{0}\le \left(K+\mu C{d}_{1}\right){\parallel u\parallel }_{\chi ,\nu }$, and so $\parallel G\parallel \le \left(K+\mu C{d}_{1}\right)$. Also $u\in {C}^{4}\left[0,1\right]\cap {C}^{6}\left(0,1\right)$ is a solution of (13) if $u\in X$ satisfies $u=T\left(Gu+{h}_{1}\right)$, where ${h}_{1}\left(t\right)=\mu u\left(t\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+h\left(t\right)$, i.e.,
$u\in X,\phantom{\rule{1em}{0ex}}\left(I-TG\right)u=T{h}_{1}.$
(16)

The operator $I-TG$ maps X into X. From $\parallel T\parallel \le {D}_{2}$ together with $\parallel G\parallel \le \left(K+\mu C{d}_{1}\right)$ and the condition ${D}_{2}\left(K+\mu C{d}_{1}\right)<1$, and applying the operator spectral theorem, we find that ${\left(I-TG\right)}^{-1}$ exists and is bounded. Let $\mu \in \left(0,\frac{1-{D}_{2}K}{{D}_{2}C{d}_{1}}\right)$, where $1-{D}_{2}K>0$, then the condition ${D}_{2}\left(K+\mu C{d}_{1}\right)<1$ is fulfilled. Let $\phantom{\rule{0.25em}{0ex}}L={D}_{2}\left(K+\mu C{d}_{1}\right)$, and let ${\mu }^{\ast \ast }=\frac{1-{D}_{2}K}{{D}_{2}C{d}_{1}}$.

Let $H={\left(I-TG\right)}^{-1}T$. Then (16) is equivalent to $u=H{h}_{1}$. By the Neumann expansion formula, H can be expressed by
$H=\left(I+TG+\cdots +{\left(TG\right)}^{n}+\cdots \right)T=T+\left(TG\right)T+\cdots +{\left(TG\right)}^{n}T+\cdots .$
(17)

The complete continuity of T with the continuity of ${\left(I-TG\right)}^{-1}$ guarantees that the operator $H:Y\to X$ is completely continuous.

Now $\mathrm{\forall }h\in {Y}_{+}$, let $u=Th$, then $u\in X\cap {Y}_{+}$, and ${u}^{\mathrm{\prime }\mathrm{\prime }}\le 0$, ${u}^{\left(4\right)}\ge 0$. Thus, we have
$\begin{array}{rl}\left(Gu\right)\left(t\right)=& -\left(A\left(t\right)-a\right){u}^{\left(4\right)}-\left(B\left(t\right)-b\right){u}^{\mathrm{\prime }\mathrm{\prime }}-\left(C\left(t\right)-c\right)u\\ +\mu D\left(t\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\ge 0,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\end{array}$
Hence
$\mathrm{\forall }h\in {Y}_{+},\phantom{\rule{1em}{0ex}}\left(GTh\right)\left(t\right)\ge 0,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],$
(18)

and so, $\left(TG\right)\left(Th\right)\left(t\right)=T\left(GTh\right)\left(t\right)\ge 0$, $t\in \left[0,1\right]$.

It is easy to see [11] that the following inequalities hold: $\mathrm{\forall }h\in {Y}_{+}$,
$\frac{1}{1-L}\left(Th\right)\left(t\right)\ge \left(Hh\right)\left(t\right)\ge \left(Th\right)\left(t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],$
(19)
and, moreover,
${\parallel \left(Hh\right)\parallel }_{0}\le \frac{1}{1-L}{\parallel \left(Th\right)\parallel }_{0}.$
(20)

Lemma 6 [8]

$H:Y⟶\left(X,{\parallel \cdot \parallel }_{\varkappa ,\nu }\right)$ is completely continuous, where $\chi ={\lambda }_{1}+{\lambda }_{3}$, $\nu ={\lambda }_{1}{\lambda }_{3}$ and $\mathrm{\forall }h\in {Y}_{+}$, $\frac{1}{1-L}\left(Th\right)\left(t\right)\ge \left(Hh\right)\left(t\right)\ge \left(Th\right)\left(t\right)$, $t\in \left[0,1\right]$, and, moreover, ${\parallel Th\parallel }_{0}\ge \left(1-L\right){\parallel Hh\parallel }_{0}$.

For any $u\in {Y}_{+}$, define $Fu=\mu u\left(t\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\lambda f\left(t,u\right)$. From (H1), we have that $F:{Y}_{+}\to {Y}_{+}$ is continuous. It is easy to see that $u\in {C}^{4}\left[0,1\right]\cap {C}^{6}\left(0,1\right)$, being a positive solution of (13), is equivalent to $u\in {Y}_{+}$, being a nonzero solution of
$u=HFu.$
(21)
Let us introduce the following notations
$\begin{array}{c}\begin{array}{rl}{T}_{\lambda ,\mu }u\left(t\right):=& TFu\left(t\right)={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)\\ ×\left(\mu u\left(\tau \right){\int }_{0}^{1}G\left(\tau ,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\lambda f\left(\tau ,u\left(\tau \right)\right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv,\end{array}\hfill \\ \begin{array}{rl}{Q}_{\lambda ,\mu }u& :=HFu=TFu+\left(TG\right)TFu+{\left(TG\right)}^{2}TFu+\cdots +{\left(TG\right)}^{n}TFu+\cdots \\ ={T}_{\lambda ,\mu }u+\left(TG\right){T}_{\lambda ,\mu }u+{\left(TG\right)}^{2}{T}_{\lambda ,\mu }u+\cdots +{\left(TG\right)}^{n}{T}_{\lambda ,\mu }u+\cdots ,\end{array}\hfill \end{array}$

i.e., ${Q}_{\lambda ,\mu }u=HFu$. Obviously, ${Q}_{\lambda ,\mu }:{Y}_{+}\to {Y}_{+}$ is completely continuous. We next show that the operator ${Q}_{\lambda ,\mu }$ has a nonzero fixed point in ${Y}_{+}$.

Let $P=\left\{u\in {Y}_{+}:u\left(t\right)\ge {\delta }_{1}\left(1-L\right){g}_{1}\left(t\right){\parallel u\left(t\right)\parallel }_{0},t\in \left[\frac{1}{4},\frac{3}{4}\right]\right\}$, where ${g}_{1}\left(t\right)=\frac{1}{{C}_{1}}{G}_{1}\left(t,t\right)$. It is easy to see that P is a cone in Y, and now, we show ${Q}_{\lambda ,\mu }\left(P\right)\subset P$.

Lemma 7 ${Q}_{\lambda ,\mu }\left(P\right)\subset P$ and ${Q}_{\lambda ,\mu }:P\to P$ is completely continuous.

Proof It is clear that ${Q}_{\lambda ,\mu }:P\to P$ is completely continuous. Now $\mathrm{\forall }u\in P$, let ${h}_{1}=Fu$, then ${h}_{1}\in {Y}_{+}$. Using Lemma 6, i.e., $\left({Q}_{\lambda ,\mu }u\right)\left(t\right)=\left(HFu\right)\left(t\right)\ge \left(TFu\right)\left(t\right)$, $t\in \left[0,1\right]$ and by Lemma 2, for all $u\in P$, we have
$\left(TFu\right)\left(t\right)\le {C}_{1}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(v,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)\left(Fu\right)\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left[0,1\right].$
Thus,
${\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(v,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)\left(Fu\right)\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv\ge \frac{1}{{C}_{1}}{\parallel TFu\parallel }_{0}.$
(22)
On the other hand, by Lemma 6 and (22), we have
$\begin{array}{rl}\left(TFu\right)\left(t\right)& \ge {\delta }_{1}{G}_{1}\left(t,t\right){\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(v,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)\left(Fu\right)\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv\\ \ge {\delta }_{1}{G}_{1}\left(t,t\right)\frac{1}{{C}_{1}}{\parallel TFu\parallel }_{0}\ge {\delta }_{1}{G}_{1}\left(t,t\right)\frac{1}{{C}_{1}}\left(1-L\right){\parallel Qu\parallel }_{0},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left[0,1\right].\end{array}$

Thus, ${Q}_{\lambda ,\mu }\left(P\right)\subset P$. □

## 3 Main results

Lemma 8 Let $f\left(t,u\right)$ be nondecreasing in u for $t\in \left[0,1\right]$ and $f\left(t,0\right)>\stackrel{ˆ}{c}>0$ for all $t\in \left[0,1\right]$, where $\stackrel{ˆ}{c}$ is a constant and $L<1$. Then there exists ${\lambda }^{\divideontimes }>0$ and ${\mu }^{\divideontimes }>0$ such that the operator ${Q}_{\lambda ,\mu }$ has a fixed point ${u}^{\star }$ at $\left({\lambda }^{\divideontimes },{\mu }^{\divideontimes }\right)$ with ${u}^{\star }\in P\mathrm{\setminus }\left\{\theta \right\}$.

Proof Set ${\stackrel{ˆ}{u}}_{1}\left(t\right)=\left({Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{\star }\right)\left(t\right)$, where
${u}_{\star }\left(t\right)=\frac{2}{1-L}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv.$
It is easy to see that ${u}_{\star }\left(t\right)\in P$. Let ${\lambda }^{\divideontimes }={M}_{fu}^{-1}$ and ${\mu }^{\divideontimes }=min\left({N}_{fu}^{-1};{\mu }^{\ast \ast }\right)$, where ${M}_{fu}={max}_{t\in \left[0,1\right]}f\left(t,{u}_{\star }\left(t\right)\right)$ and ${N}_{fu}={max}_{t\in \left[0,1\right]}{u}_{\star }\left(t\right){\int }_{0}^{1}G\left(t,s\right){u}_{\star }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$, respectively. Then ${M}_{fu}>0$ and ${N}_{fu}>0$, and from Lemma 6, we obtain
$\begin{array}{rl}{\stackrel{ˆ}{u}}_{0}\left(t\right)=& {u}_{\star }\left(t\right)=\frac{2}{1-L}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge & \frac{1}{1-L}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\\ ×\left({\lambda }^{\divideontimes }f\left(s,{u}_{\star }\left(s\right)\right)+{\mu }^{\divideontimes }{u}_{\star }\left(t\right){\int }_{0}^{1}G\left(t,s\right){u}_{\star }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ =& \frac{1}{1-L}\left({T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{\star }\right)\left(t\right)\ge \left({Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{\star }\right)\left(t\right)={\stackrel{ˆ}{u}}_{1}\left(t\right).\end{array}$
It is easy to see that
$\begin{array}{rl}{\stackrel{ˆ}{u}}_{n}\left(t\right)=& \left({Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-1}\right)\left(t\right)\\ =& \left({T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-1}+\left(TG\right){T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-1}+{\left(TG\right)}^{2}{T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-1}+\cdots \\ +{\left(TG\right)}^{n}{T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-1}+\cdots \right)\\ \ge & \left({T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-2}+\left(TG\right){T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-2}+{\left(TG\right)}^{2}{T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-2}+\cdots \\ +{\left(TG\right)}^{n}{T}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-2}+\cdots \right)={\stackrel{ˆ}{u}}_{n-1}\left(t\right).\end{array}$
Indeed, for ${h}_{1},{h}_{2}\in {Y}_{+}$, let ${h}_{1}\left(t\right)\ge {h}_{2}\left(t\right)$, then from (10), we have ${u}_{1}\left(t\right)=T{h}_{1}\ge T{h}_{2}={u}_{2}\left(t\right)$. Using equation (4) and (6), we obtain
$-{u}^{\mathrm{\prime }\mathrm{\prime }}+{\lambda }_{2}u={\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{3}\left(v,\tau \right)h\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right]$
(23)
and
${u}^{\left(4\right)}-\left({\lambda }_{2}+{\lambda }_{3}\right){u}^{\mathrm{\prime }\mathrm{\prime }}+{\lambda }_{2}{\lambda }_{3}u={\int }_{0}^{1}{G}_{1}\left(t,v\right)h\left(v\right)\phantom{\rule{0.2em}{0ex}}dv,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(24)
Then by (23), we have for $t\in \left[0,1\right]$
${u}_{1}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)-{u}_{2}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)={\lambda }_{2}\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right)-{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{3}\left(v,\tau \right)\left({h}_{1}\left(t\right)-{h}_{2}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\le 0,$
because ${\lambda }_{2}<0$, and finally, from (24), we have
$\begin{array}{rl}{u}_{1}^{\left(4\right)}\left(t\right)-{u}_{2}^{\left(4\right)}\left(t\right)=& \left({\lambda }_{2}+{\lambda }_{3}\right)\left({u}_{1}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)-{u}_{2}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)\right)-{\lambda }_{2}{\lambda }_{3}\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right)\\ +{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right)\left({h}_{1}\left(t\right)-{h}_{2}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\ge 0,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right]\end{array}$
because ${\lambda }_{2}+{\lambda }_{3}\le 0$ and ${\lambda }_{2}{\lambda }_{3}\le 0$. From the equation
$\begin{array}{r}\left(Gu\right)\left(t\right)=-\left(A\left(t\right)-a\right){u}^{\left(4\right)}-\left(B\left(t\right)-b\right){u}^{\mathrm{\prime }\mathrm{\prime }}-\left(C\left(t\right)-c\right)u+\mu D\left(t\right){\int }_{0}^{1}G\left(t,s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\ge 0,\\ \phantom{\rule{1em}{0ex}}t\in \left[0,1\right]\end{array}$
we have
$\begin{array}{r}\left(G{u}_{1}\right)\left(t\right)-\left(G{u}_{2}\right)\left(t\right)\\ \phantom{\rule{1em}{0ex}}=-\left(A\left(t\right)-a\right)\left({u}_{1}^{\left(4\right)}\left(t\right)-{u}_{2}^{\left(4\right)}\left(t\right)\right)-\left(B\left(t\right)-b\right)\left({u}_{1}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)-{u}_{2}^{\mathrm{\prime }\mathrm{\prime }}\left(t\right)\right)\\ \phantom{\rule{2em}{0ex}}-\left(C\left(t\right)-c\right)\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right)+\mu D\left(t\right){\int }_{0}^{1}G\left(t,s\right)\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}ds\ge 0,\\ \phantom{\rule{2em}{0ex}}t\in \left[0,1\right]\end{array}$
(25)
i.e., $\left(G{u}_{1}\right)\left(t\right)\ge \left(G{u}_{2}\right)\left(t\right)$ for all $t\in \left[0,1\right]$. Finally, if ${h}_{1}=F{u}_{1}$ and ${h}_{2}=F{u}_{2}$, then
$\begin{array}{rl}\left(H{h}_{1}\right)\left(t\right)& =T\left({h}_{1}\right)+\left(TG\right)\left(T{h}_{1}\right)+{\left(TG\right)}^{2}\left(T{h}_{1}\right)+\cdots +{\left(TG\right)}^{n}\left(T{h}_{1}\right)+\cdots \\ \ge \left(H{h}_{2}\right)\left(t\right)=T\left({h}_{2}\right)+\left(TG\right)\left(T{h}_{2}\right)+{\left(TG\right)}^{2}\left(T{h}_{2}\right)+\cdots +{\left(TG\right)}^{n}\left(T{h}_{2}\right)+\cdots \end{array}$
i.e.,
$\begin{array}{r}{Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{1}\\ \phantom{\rule{1em}{0ex}}=\left(HF{u}_{1}\right)\left(t\right)=T\left(F{u}_{1}\right)+\left(TG\right)\left(TF{u}_{1}\right)+{\left(TG\right)}^{2}\left(TF{u}_{1}\right)+\cdots +{\left(TG\right)}^{n}\left(TF{u}_{1}\right)+\cdots \\ \phantom{\rule{1em}{0ex}}\ge \left(HF{u}_{2}\right)\left(t\right)=T\left(F{u}_{2}\right)+\left(TG\right)\left(TF{u}_{2}\right)+{\left(TG\right)}^{2}\left(TF{u}_{2}\right)+\cdots +{\left(TG\right)}^{n}\left(TF{u}_{2}\right)+\cdots \\ \phantom{\rule{1em}{0ex}}={Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{2},\end{array}$
(26)
and from (26), it follows that for ${u}_{1},{u}_{2}\in {Y}_{+}$, if ${u}_{1}\left(t\right)\ge {u}_{2}\left(t\right)$ then, we have
${Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{1}\ge {Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{u}_{2}.$
(27)
Set ${\stackrel{ˆ}{u}}_{0}\left(t\right)={u}_{\star }\left(t\right)$ and ${\stackrel{ˆ}{u}}_{n}\left(t\right)=\left({Q}_{{\lambda }^{\star },{\mu }^{\star }}{\stackrel{ˆ}{u}}_{n-1}\right)\left(t\right)$, $n=1,2,\dots$ , $t\in \left[0,1\right]$. Then
${\stackrel{ˆ}{u}}_{0}\left(t\right)={u}_{\star }\left(t\right)\ge {\stackrel{ˆ}{u}}_{1}\left(t\right)\ge \cdots \ge {\stackrel{ˆ}{u}}_{n}\left(t\right)\ge \cdots \ge {L}_{1}{G}_{1}\left(t,t\right),$
where
${L}_{1}={\lambda }^{\divideontimes }{\delta }_{1}{\delta }_{2}{\delta }_{3}\stackrel{ˆ}{c}{C}_{23}{C}_{12}{C}_{3}.$

□

Indeed, by Lemma 6, we have
$\begin{array}{rl}{\stackrel{ˆ}{u}}_{n}\left(t\right)& ={Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{n-1}=\left(HF\right)\left({\stackrel{ˆ}{u}}_{n-1}\right)\ge \left(TF\right)\left({\stackrel{ˆ}{u}}_{n-1}\right)\\ \ge {\lambda }^{\divideontimes }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{\stackrel{ˆ}{u}}_{n-1}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\lambda }^{\divideontimes }\stackrel{ˆ}{c}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\lambda }^{\divideontimes }\stackrel{ˆ}{c}{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{C}_{3}{G}_{1}\left(t,t\right).\end{array}$

Now, $f\left(t,u\right)$ nondecreasing in u for $t\in \left[0,1\right]$, Lemma 2, and the Lebesgue convergence theorem guarantee that ${\left\{{u}_{n}\right\}}_{n=0}^{\mathrm{\infty }}={\left\{{Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}{\stackrel{ˆ}{u}}_{0}\right\}}_{n=0}^{\mathrm{\infty }}$ decreases to a fixed point ${u}^{\star }\in P\mathrm{\setminus }\left\{\theta \right\}$ of the operator ${Q}_{{\lambda }^{\divideontimes },{\mu }^{\divideontimes }}$.

Lemma 9 Suppose that (H3)-(H5) hold, and $L<1$. Set
${S}_{\lambda ,\mu }=\left\{u\in P:{Q}_{\lambda ,\mu }u=u,\left(\lambda ,\mu \right)\in A\right\},$

where $A\subset \left[a,\mathrm{\infty }\right)×\left[b,\mathrm{\infty }\right)$ for some constants $a>0$, $b>0$. Then there exists a constant ${C}_{A}$ such that ${\parallel u\parallel }_{0}<{C}_{A}$ for all $u\in {S}_{\lambda ,\mu }$.

Proof Suppose, to the contrary, that there exists a sequence ${\left\{{u}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ such that ${lim}_{n\to \mathrm{\infty }}{\parallel {u}_{n}\parallel }_{0}=+\mathrm{\infty }$, where ${u}_{n}\in P$ is a fixed point of the operator ${Q}_{\lambda ,\mu }$ at $\left({\lambda }_{n},{\mu }_{n}\right)\in A$ ($n=1,2,\dots$). Then

where $k=\frac{{\delta }_{1}}{{C}_{1}}\left(1-L\right){min}_{t\in \left[\frac{1}{4},\frac{3}{4}\right]}{G}_{1}\left(t,t\right)$.

Choose ${J}_{1}>0$, so that
${J}_{1}a{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{m}_{1}{m}_{2}k>2,$
and ${l}_{1}>0$ such that
and ${N}_{0}$, so that $\parallel {u}_{{N}_{0}}\parallel >\frac{{l}_{1}}{k}$. Now,
$\begin{array}{rl}\left({Q}_{{\lambda }_{{N}_{0}},{\mu }_{{N}_{0}}}{u}_{{N}_{0}}\right)\left(\frac{1}{2}\right)& \ge \left(TF{u}_{{N}_{0}}\right)\left(\frac{1}{2}\right)\\ \ge {\lambda }_{{N}_{0}}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(\frac{1}{2},v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{N}_{0}}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\lambda }_{{N}_{0}}{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{G}_{1}\left(\frac{1}{2},\frac{1}{2}\right){\int }_{0}^{1}{G}_{3}\left(s,s\right)f\left(s,{u}_{{N}_{0}}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\lambda }_{{N}_{0}}{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{G}_{1}\left(\frac{1}{2},\frac{1}{2}\right){\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}_{3}\left(s,s\right)f\left(s,{u}_{{N}_{0}}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge \frac{1}{2}a{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{m}_{1}{m}_{2}{J}_{1}{u}_{{N}_{0}}\left(t\right)\\ \ge \frac{1}{2}a{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{m}_{1}{m}_{2}{J}_{1}k{\parallel {u}_{{N}_{0}}\parallel }_{0}>{\parallel {u}_{{N}_{0}}\parallel }_{0},\end{array}$
and so,
${\parallel {u}_{{N}_{0}}\parallel }_{0}={\parallel {Q}_{{\lambda }_{{N}_{0}},{\mu }_{{N}_{0}}}{u}_{{N}_{0}}\parallel }_{0}\ge {\parallel \left(TF\right){u}_{{N}_{0}}\parallel }_{0}\ge \left(TF{u}_{{N}_{0}}\right)\left(\frac{1}{2}\right)>{\parallel {u}_{{N}_{0}}\parallel }_{0},$

Lemma 10 Suppose that $L<1$, (H3) and (H4) hold and that the operator ${Q}_{\lambda ,\mu }$ has a positive fixed point in P at $\stackrel{ˆ}{\lambda }>0$ and $\stackrel{ˆ}{\mu }>0$. Then for every $\left({\lambda }_{\star },{\mu }_{\star }\right)\in \left(0,\stackrel{ˆ}{\lambda }\right)×\left(0,\stackrel{ˆ}{\mu }\right)$ there exists a function ${u}_{\star }\in P\mathrm{\setminus }\left\{\theta \right\}$ such that ${Q}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{\star }={u}_{\star }$.

Proof Let $\stackrel{ˆ}{u}\left(t\right)$ be a fixed point of the operator ${Q}_{\lambda ,\mu }$ at $\left(\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }\right)$. Then
$\stackrel{ˆ}{u}\left(t\right)={Q}_{\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }}\stackrel{ˆ}{u}\left(t\right)\ge {Q}_{{\lambda }_{\star },{\mu }_{\star }}\stackrel{ˆ}{u}\left(t\right),$
where $0<{\lambda }_{\star }<\stackrel{ˆ}{\lambda }$, $0<{\mu }_{\star }<\stackrel{ˆ}{\mu }$. Hence
$\begin{array}{r}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\left(\stackrel{ˆ}{\lambda }f\left(s,\stackrel{ˆ}{u}\left(s\right)\right)+\stackrel{ˆ}{\mu }\stackrel{ˆ}{u}\left(s\right){\int }_{0}^{1}G\left(s,p\right)\stackrel{ˆ}{u}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{1em}{0ex}}\ge {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\\ \phantom{\rule{2em}{0ex}}×\left({\lambda }_{\star }f\left(s,\stackrel{ˆ}{u}\left(s\right)\right)+{\mu }_{\star }\stackrel{ˆ}{u}\left(s\right){\int }_{0}^{1}G\left(s,p\right)\stackrel{ˆ}{u}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv.\end{array}$
Set
$\begin{array}{rl}\left({T}_{{\lambda }_{\star },{\mu }_{\star }}u\right)\left(t\right)=& {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\\ ×\left({\lambda }_{\star }f\left(s,u\left(s\right)\right)+{\mu }_{\star }u\left(s\right){\int }_{0}^{1}G\left(s,p\right)u\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\end{array}$
and
$\left({Q}_{{\lambda }_{\star },{\mu }_{\star }}u\right)\left(t\right)={T}_{{\lambda }_{\star },{\mu }_{\star }}u+\left(TG\right){T}_{{\lambda }_{\star },{\mu }_{\star }}u+{\left(TG\right)}^{2}{T}_{{\lambda }_{\star },{\mu }_{\star }}u+\cdots +{\left(TG\right)}^{n}{T}_{{\lambda }_{\star },{\mu }_{\star }}u+\cdots$
${u}_{0}\left(t\right)=\stackrel{ˆ}{u}\left(t\right)$ and ${u}_{n}\left(t\right)={Q}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}$. Then
$\begin{array}{rl}{u}_{0}\left(t\right)& =\stackrel{ˆ}{u}\left(t\right)={T}_{\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }}\stackrel{ˆ}{u}+\left(TG\right){T}_{\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }}\stackrel{ˆ}{u}+{\left(TG\right)}^{2}{T}_{\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }}\stackrel{ˆ}{u}+\cdots +{\left(TG\right)}^{n}{T}_{\stackrel{ˆ}{\lambda },\stackrel{ˆ}{\mu }}\stackrel{ˆ}{u}+\cdots \\ \ge {T}_{{\lambda }_{\star },{\mu }_{\star }}\stackrel{ˆ}{u}+\left(TG\right){T}_{{\lambda }_{\star },{\mu }_{\star }}\stackrel{ˆ}{u}+{\left(TG\right)}^{2}{T}_{{\lambda }_{\star },{\mu }_{\star }}\stackrel{ˆ}{u}+\cdots +{\left(TG\right)}^{n}{T}_{{\lambda }_{\star },{\mu }_{\star }}\stackrel{ˆ}{u}+\cdots ={u}_{1}\left(t\right)\end{array}$
and
$\begin{array}{rl}{u}_{n}\left(t\right)={Q}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}=& {T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}+\left(TG\right){T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}+{\left(TG\right)}^{2}{T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}+\cdots \\ +{\left(TG\right)}^{n}{T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}+\cdots \\ \ge & {T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-2}+\left(TG\right){T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-2}+{\left(TG\right)}^{2}{T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-2}+\cdots \\ +{\left(TG\right)}^{n}{T}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-2}+\cdots ={u}_{n-1}\left(t\right)\end{array}$
because $f\left(t,u\right)$ is nondecreasing in u for $t\in \left[0,1\right]$ and ${T}_{{\lambda }_{\star },{\mu }_{\star }}u$ is also nondecreasing in u. Thus
${u}_{0}\left(t\right)\ge {u}_{1}\left(t\right)\ge \cdots \ge {u}_{n}\left(t\right)\ge {u}_{n+1}\left(t\right)\ge \cdots \ge {L}_{2}{G}_{1}\left(t,t\right),$
(28)
where
${L}_{2}={\lambda }_{\divideontimes }\stackrel{ˆ}{c}{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{C}_{3}.$

□

Indeed, by Lemma 6, we have
$\begin{array}{rl}{u}_{n}\left(t\right)& ={Q}_{{\lambda }_{\star },{\mu }_{\star }}{u}_{n-1}=\left(HF\right)\left({u}_{n-1}\right)\ge {T}_{{}_{{\lambda }_{\star },{\mu }_{\star }}}\left({u}_{n-1}\right)\\ \ge {\lambda }_{\star }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{n-1}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\lambda }_{\star }\stackrel{ˆ}{c}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\ge {\lambda }_{\star }\stackrel{ˆ}{c}{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{C}_{3}{G}_{1}\left(t,t\right).\end{array}$

Lemma 2 implies that ${\left\{{Q}_{{\lambda }^{\star }}^{n}u\right\}}_{n=1}^{\mathrm{\infty }}$ decreases to a fixed point ${u}_{\star }\in P\mathrm{\setminus }\left\{\theta \right\}$.

Lemma 11 Suppose that $L<1$, (H3)-(H5) hold. Let

Then Λ is bounded.

Proof Suppose, to the contrary, that there exists a fixed point sequence ${\left\{{u}_{n}\right\}}_{n=0}^{\mathrm{\infty }}\subset P$ of ${Q}_{\lambda ,\mu }$ at $\left({\lambda }_{n},{\mu }_{n}\right)$ such that ${lim}_{n\to \mathrm{\infty }}{\lambda }_{n}=\mathrm{\infty }$ and $0<{\mu }_{n}<{\mu }^{\ast \ast }$. Then there are two cases to be considered: (i) there exists a subsequence ${\left\{{u}_{{n}_{i}}\right\}}_{n=0}^{\mathrm{\infty }}$ such that ${lim}_{i\to \mathrm{\infty }}{\parallel {u}_{{n}_{i}}\parallel }_{0}=\mathrm{\infty }$, which is impossible by Lemma 9, so we only consider the next case: (ii) there exists a constant $H>0$ such that ${\parallel {u}_{n}\parallel }_{0}\le H$, $n=0,1,2,3,\dots$ . In view of (H3) and (H4), we can choose ${l}_{0}>0$ such that $f\left(t,0\right)>{l}_{0}H$, and further, $f\left(t,{u}_{n}\right)>{l}_{0}H$ for $t\in \left[0,1\right]$. We know that
${u}_{n}={Q}_{{\lambda }_{n},{\mu }_{n}}{u}_{n}\ge {T}_{{\lambda }_{n},{\mu }_{n}}{u}_{n}.$
Let ${v}_{n}\left(t\right)={T}_{{\lambda }_{n},{\mu }_{n}}{u}_{n}$, i.e., ${u}_{n}\left(t\right)\ge {v}_{n}\left(t\right)$. Then it follows that
$-{v}_{n}^{\left(6\right)}+a{v}_{n}^{\left(4\right)}+b{v}_{n}^{\mathrm{\prime }\mathrm{\prime }}+c{v}_{n}={\lambda }_{n}f\left(t,{u}_{n}\right)+{\mu }_{n}{u}_{n}\left(t\right){\int }_{0}^{1}G\left(t,p\right){u}_{n}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp,\phantom{\rule{1em}{0ex}}0
(29)
Multiplying (29) by $sin\pi t$ and integrating over $\left[0,1\right]$, and then using integration by parts on the left side of (29), we have
$\mathrm{\Gamma }{\int }_{0}^{1}{v}_{n}\left(t\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt={\lambda }_{n}{\int }_{0}^{1}f\left(t,{u}_{n}\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt+{\mu }_{n}{\int }_{0}^{1}{u}_{n}\left(t\right)sin\pi t{\int }_{0}^{1}G\left(t,p\right){u}_{n}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}dt.$
Next, assume that (ii) holds. Then
$\begin{array}{r}\mathrm{\Gamma }{\int }_{0}^{1}{u}_{n}\left(t\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt\ge \mathrm{\Gamma }{\int }_{0}^{1}{v}_{n}\left(t\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt\\ ={\lambda }_{n}{\int }_{0}^{1}f\left(t,{u}_{n}\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt+{\mu }_{n}{\int }_{0}^{1}{u}_{n}\left(t\right)sin\pi t{\int }_{0}^{1}G\left(t,p\right){u}_{n}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}dt\end{array}$
and
$\begin{array}{rl}\mathrm{\Gamma }H{\int }_{0}^{1}sin\pi t\phantom{\rule{0.2em}{0ex}}dt& \ge \mathrm{\Gamma }{\parallel {u}_{n}\parallel }_{0}{\int }_{0}^{1}sin\pi t\phantom{\rule{0.2em}{0ex}}dt\ge \mathrm{\Gamma }{\int }_{0}^{1}{u}_{n}\left(t\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt\ge \mathrm{\Gamma }{\int }_{0}^{1}{v}_{n}\left(t\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt\\ ={\lambda }_{n}{\int }_{0}^{1}f\left(t,{u}_{n}\right)sin\pi t\phantom{\rule{0.2em}{0ex}}dt+{\mu }_{n}{\int }_{0}^{1}{u}_{n}\left(t\right)sin\pi t{\int }_{0}^{1}G\left(t,p\right){u}_{n}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}dt\\ \ge {\lambda }_{n}{l}_{0}H{\int }_{0}^{1}sin\pi t\phantom{\rule{0.2em}{0ex}}dt\end{array}$

lead to $\mathrm{\Gamma }\ge {\lambda }_{n}{l}_{0}$, which is a contradiction. The proof is complete. □

Lemma 12 Suppose that $L<1$, (H3)-(H4) hold. Let

and let ${\stackrel{˜}{\lambda }}_{\mu }=sup{\mathrm{\Lambda }}_{\mu }$. Then ${\mathrm{\Lambda }}_{\mu }=\left(0,{\stackrel{˜}{\lambda }}_{\mu }\right]$, where Λ is defined in Lemma 11.

Proof By Lemma 10, it follows that $\left(0,\stackrel{˜}{\lambda }\right)×\left(0,\mu \right)\subset \mathrm{\Lambda }$. We only need to prove $\left({\stackrel{˜}{\lambda }}_{\mu },\mu \right)\in \mathrm{\Lambda }$. We may choose a distinct nondecreasing sequence ${\left\{{\lambda }_{n}\right\}}_{n=1}^{\mathrm{\infty }}\subset \mathrm{\Lambda }$ such that ${lim}_{n\to \mathrm{\infty }}{\lambda }_{n}={\stackrel{˜}{\lambda }}_{\mu }$. Set ${u}_{n}\in P$ as a fixed point of ${Q}_{\lambda ,\mu }$ at $\left({\lambda }_{n},\mu \right)$, $n=1,2,\dots$ , i.e., ${u}_{n}={Q}_{{\lambda }_{n},\mu }{u}_{n}$. By Lemma 9, ${\left\{{u}_{n}\right\}}_{n=1,}^{\mathrm{\infty }}$ is uniformly bounded, so it has a subsequence, denoted by ${\left\{{u}_{{n}_{k}}\right\}}_{k=1}^{\mathrm{\infty }}$, converging to $\stackrel{˜}{u}\in P$. Note that
$\begin{array}{rl}{u}_{n}& ={T}_{{\lambda }_{n},\mu }{u}_{n}+\left(TG\right){T}_{{\lambda }_{n},\mu }{u}_{n}+{\left(TG\right)}^{2}{T}_{{\lambda }_{n},\mu }{u}_{n}+\cdots +{\left(TG\right)}^{n}{T}_{{\lambda }_{n},\mu }{u}_{n}+\cdots \\ ={Q}_{{\lambda }_{n},\mu }{u}_{n}.\end{array}$
(30)
Taking the limit as $n\to \mathrm{\infty }$ on both sides of (30), and using the Lebesgue convergence theorem, we have
$\stackrel{˜}{u}={T}_{\stackrel{˜}{\lambda },\mu }\stackrel{˜}{u}+\left(TG\right){T}_{\stackrel{˜}{\lambda },\mu }{\stackrel{˜}{u}}_{n}+{\left(TG\right)}^{2}{T}_{\stackrel{˜}{\lambda },\mu }\stackrel{˜}{u}+\cdots +{\left(TG\right)}^{n}{T}_{\stackrel{˜}{\lambda },\mu }\stackrel{˜}{u}+\cdots$

which shows that ${Q}_{\lambda ,\mu }$ has a positive fixed point $\stackrel{˜}{u}$ at $\left(\stackrel{˜}{\lambda },\mu \right)$. □

Theorem 1 Suppose that (H3)-(H5) hold, and $L<1$. For fixed ${\mu }^{\star }\in \left(0,{\mu }^{\star \star }\right)$, then there exists at ${\lambda }^{\star }>0$ such that (3) has at least two, one and has no positive solutions for $0<\lambda <{\lambda }^{\star }$, $\lambda ={\lambda }^{\star }$ for $\lambda >{\lambda }^{\star }$, respectively.

Proof Suppose that (H3) and (H4) hold. Then there exists ${\lambda }^{\star }>0$ and ${\mu }^{\star }>0$ such that ${Q}_{\lambda ,\mu }$ has a fixed point ${u}_{{\lambda }^{\star },{\mu }^{\star }}\in P\mathrm{\setminus }\left\{\theta \right\}$ at $\lambda ={\lambda }^{\star }$ and $\mu ={\mu }^{\star }$. In view of Lemma 12, ${Q}_{\lambda ,\mu }$ also has a fixed point ${u}_{\underline{\lambda },\underline{\mu }}<{u}_{{\lambda }^{\star },{\mu }^{\star }}$, ${u}_{\underline{\lambda },\underline{\mu }}\in P\mathrm{\setminus }\left\{\theta \right\}$, and $0<\underline{\lambda }<{\lambda }^{\star }$, $0<\underline{\mu }<{\mu }^{\star }$, ${\mu }^{\star }\in \left(0,{\mu }^{\star \star }\right)$. For $0<\underline{\lambda }<{\lambda }^{\star }\mathrm{\setminus }$, there exists ${\delta }_{0}>0$ such that
$f\left(t,{u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)-f\left(t,{u}_{{\lambda }^{\star },{\mu }^{\star }}\right)\le f\left(t,0\right)\left(\frac{{\lambda }^{\star }}{\underline{\lambda }}-1\right)$
for $t\in \left[0,1\right]$, $0<\delta \le {\delta }_{0}$. In this case, it is easy to see that
$\begin{array}{rl}{T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star },}+\delta \right)=& \underline{\lambda }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{\lambda }^{\star },{\mu }^{\star }}\left(s\right)+\delta \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ +\underline{\mu }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\left(s\right)+\delta \right)\\ ×{\int }_{0}^{1}G\left(s,p\right)\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\left(p\right)+\delta \right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \le & {\lambda }^{\star }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ +{\mu }^{\star }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right){u}_{{\lambda }^{\star },{\mu }^{\star }}\left(s\right)\\ ×{\int }_{0}^{1}G\left(s,p\right){u}_{{\lambda }^{\star },{\mu }^{\star }}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv={T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}.\end{array}$
Indeed, we have
$\begin{array}{r}\underline{\lambda }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)+\delta \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{2em}{0ex}}-{\lambda }^{\star }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{1em}{0ex}}=\underline{\lambda }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\left\{f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)+\delta \right)-f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)\right)\right\}\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{2em}{0ex}}-\left({\lambda }^{\star }-\underline{\lambda }\right){\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{1em}{0ex}}\le \left({\lambda }^{\star }-\underline{\lambda }\right){\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,0\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{2em}{0ex}}-\left({\lambda }^{\star }-\underline{\lambda }\right){\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{1em}{0ex}}=\left({\lambda }^{\star }-\underline{\lambda }\right){\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\left\{f\left(s,0\right)-f\left(s,{u}_{{\lambda }^{\star }}\left(s\right)\right)\right\}\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\le 0.\end{array}$
Similarly, it is easy to see that
$\begin{array}{r}\underline{\mu }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\left(s\right)+\delta \right){\int }_{0}^{1}G\left(s,p\right)\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\left(p\right)+\delta \right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \phantom{\rule{1em}{0ex}}-{\mu }^{\star }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right){u}_{{\lambda }^{\star },{\mu }^{\star }}\left(s\right){\int }_{0}^{1}G\left(s,p\right){u}_{{\lambda }^{\star },{\mu }^{\star }}\left(p\right)\phantom{\rule{0.2em}{0ex}}dp\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\le 0.\end{array}$
Moreover, from (25), it follows that for ${T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\le {T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}$ we have
$G\left({T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\right)\le G\left({T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}\right).$
Finally, we have
$\left(TG\right){T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\le \left(TG\right){T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}.$
By induction, it is easy to see that
${\left(TG\right)}^{n}{T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\le {\left(TG\right)}^{n}{T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }},\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(31)
Hence, using (31), we have
$\begin{array}{rl}{Q}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)=& {T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+\left(TG\right){T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\\ +{\left(TG\right)}^{2}{T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+\cdots +{\left(TG\right)}^{n}{T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+\cdots \\ \le & {T}_{{}_{{\lambda }^{\star },{\mu }^{\star }}}{u}_{{\lambda }^{\star },{\mu }^{\star }}+\left(TG\right){T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}+{\left(TG\right)}^{2}{T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}+\cdots \\ +{\left(TG\right)}^{n}{T}_{{\lambda }^{\star },{\mu }^{\star }}{u}_{{\lambda }^{\star },{\mu }^{\star }}+\cdots \\ =& {Q}_{{\lambda }^{\star },{\mu }^{\star }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\right)\end{array}$
i.e.,
${Q}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)-{Q}_{{\lambda }^{\star },{\mu }^{\star }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\right)\le 0,$
so that
${Q}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\le {Q}_{{\lambda }^{\star },{\mu }^{\star }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}\right)={u}_{{\lambda }^{\star },{\mu }^{\star }}<{u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta .$
Set ${D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}}=\left\{u\in C\left[0,1\right]:-\delta . Then ${Q}_{\underline{\lambda },\underline{\mu }}:P\cap {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}}\to P$ is completely continuous. Furthermore, ${Q}_{\underline{\lambda },\underline{\mu }}u\ne \upsilon u$ for $\upsilon \ge 1$ and $u\in P\cap \partial {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}}$. Indeed set $u\in P\cap \partial {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}}$. Then there exists ${t}_{0}\in \left[0,1\right]$ such that $u\left({t}_{0}\right)={\parallel u\parallel }_{0}={\parallel {u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \parallel }_{0}$ and
$\begin{array}{rl}\left({Q}_{\underline{\lambda },\underline{\mu }}u\right)\left({t}_{0}\right)=& \left({T}_{\underline{\lambda },\underline{\mu }}\left(u\right)+\left(TG\right){T}_{\underline{\lambda },\underline{\mu }}\left(u\right)+{\left(TG\right)}^{2}{T}_{\underline{\lambda },\underline{\mu }}\left(u\right)+\cdots +{\left(TG\right)}^{n}{T}_{\underline{\lambda },\underline{\mu }}\left(u\right)+\cdots \right)\left({t}_{0}\right)\\ \le & \left({T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+\left(TG\right){T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+{\left(TG\right)}^{2}{T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+\cdots \\ +{\left(TG\right)}^{n}{T}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)+\cdots \right)\left({t}_{0}\right)={Q}_{\underline{\lambda },\underline{\mu }}\left({u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \right)\left({t}_{0}\right)\\ <& {u}_{{\lambda }^{\star },{\mu }^{\star }}\left({t}_{0}\right)+\delta =u\left({t}_{0}\right)\le \upsilon u\left({t}_{0}\right),\phantom{\rule{1em}{0ex}}\upsilon \ge 1.\end{array}$

By Lemma 3, $i\left({Q}_{\underline{\lambda },\underline{\mu }},P\cap \partial {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}},P\right)=1$.

Let k be such that
We know that ${lim}_{u\to \mathrm{\infty }}\frac{f\left(t,u\right)}{u}=\mathrm{\infty }$ uniformly for $t\in \left[0,1\right]$, so we may choose ${J}_{3}>0$, so that
$\underline{\lambda }{J}_{3}{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{m}_{1}{C}_{3}k>2,$
${l}_{3}>{\parallel {u}_{{\lambda }^{\star },{\mu }^{\star }}+\delta \parallel }_{0}>0$, so that
Set ${R}_{1}=\frac{{l}_{3}}{k}$ and ${P}_{{R}_{1}}=\left\{u\in P:{\parallel u\parallel }_{0}<{R}_{1}\right\}$. Then ${Q}_{\underline{\lambda },\underline{\mu }}:{\overline{P}}_{{R}_{1}}\to P$ is completely continuous. It is easy to obtain
$\begin{array}{rl}\left({Q}_{\underline{\lambda },\underline{\mu }}u\right)\left(t\right)& \ge \left({T}_{\underline{\lambda },\underline{\mu }}u\right)\left(t\right)\ge \underline{\lambda }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge \underline{\lambda }{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{G}_{1}\left(t,t\right){\int }_{0}^{1}{G}_{3}\left(s,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge \underline{\lambda }{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{G}_{1}\left(t,t\right){\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}_{3}\left(s,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge \frac{1}{2}\underline{\lambda }{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{m}_{1}{C}_{3}{J}_{3}u\left(t\right)\ge \frac{1}{2}\underline{\lambda }{\delta }_{1}{\delta }_{2}{\delta }_{3}{C}_{12}{C}_{23}{m}_{1}{C}_{3}{J}_{3}k{\parallel u\parallel }_{0}>{\parallel u\parallel }_{0}\end{array}$
for $t\in \left[0,1\right]$ and $u\in \partial {P}_{{R}_{1}}$. Now $u\left(t\right)\ge k{\parallel u\parallel }_{0}=k{R}_{1}={l}_{3}$, and so
${\parallel {Q}_{\underline{\lambda },\underline{\mu }}u\parallel }_{0}>{\parallel u\parallel }_{0}.$
In view of Lemma 4, $i\left({Q}_{\underline{\lambda },\underline{\mu }},{P}_{{R}_{1}},P\right)=0$. By the additivity of the fixed point index,
$i\left({Q}_{\underline{\lambda },\underline{\mu }},{P}_{{R}_{1}}\mathrm{\setminus }{\overline{P\cap D}}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}},P\right)=i\left({Q}_{\underline{\lambda },\underline{\mu }},{P}_{{R}_{1}},P\right)-i\left({Q}_{\underline{\lambda },\underline{\mu }},P\cap {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}},P\right)=-1.$

Thus ${Q}_{\underline{\lambda },\underline{\mu }}$ has a fixed point in $\left\{P\cap {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}}\right\}\mathrm{\setminus }\left\{\theta \right\}$ and has another fixed point in ${P}_{{R}_{1}}\mathrm{\setminus }P\cap {D}_{{u}_{{\lambda }^{\star },{\mu }^{\star }}}$ by choosing ${\lambda }^{\star }=\stackrel{˜}{\lambda }$. □

Let us introduce the notation $\mu =0$ in the equation of (13), then we have
$\begin{array}{r}-{u}^{\left(6\right)}+A\left(t\right){u}^{\left(4\right)}+B\left(t\right){u}^{\mathrm{\prime }\mathrm{\prime }}+C\left(t\right)u=\lambda f\left(t,u\right),\\ u\left(0\right)=u\left(1\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(0\right)={u}^{\mathrm{\prime }\mathrm{\prime }}\left(1\right)={u}^{\left(4\right)}\left(0\right)={u}^{\left(4\right)}\left(1\right)=0.\end{array}$
(32)

In this case, we can prove the following theorem, which is similar to Theorem 1.

Theorem 2 Suppose that (H3)-(H5) hold, and $L<1$. Then there exists at ${\lambda }^{\star }>0$ such that (32) has at least two, one and has no positive solutions for $0<\lambda <{\lambda }^{\star }$, $\lambda ={\lambda }^{\star }$ for $\lambda >{\lambda }^{\star }$, respectively.

We follow exactly the same procedure, described in detail in the proof of Theorem 1 for $\mu =0$.

Let us introduce the following notations for $\mu =0$ and $\lambda =1$
$\begin{array}{r}TFu\left(t\right)={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)f\left(\tau ,u\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv,\\ Qu:=HFu=TFu+\left(TG\right)TFu+{\left(TG\right)}^{2}TFu+\cdots +{\left(TG\right)}^{n}TFu+\cdots ,\end{array}$
(33)

i.e., $Qu={Q}_{1,0}u=HFu$.

Lemma 13 Suppose that (H3), (H4) and (H6) hold, and $L<1$. Then for any $u\in {C}^{+}\left[0,1\right]\mathrm{\setminus }\left\{\theta \right\}$, there exist real numbers ${S}_{u}\ge {s}_{u}>0$ such that

where $g\left(t\right)={\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,\tau \right){G}_{2}\left(\tau ,v\right){G}_{3}\left(v,v\right)\phantom{\rule{0.2em}{0ex}}dv\phantom{\rule{0.2em}{0ex}}d\tau$.

Proof For any $u\in {C}^{+}\left[0,1\right]\mathrm{\setminus }\left\{\theta \right\}$ from Lemma 6, we have
Note that for any $u\in {C}^{+}\left[0,1\right]\mathrm{\setminus }\left\{\theta \right\}$, there exists an interval $\left[{a}_{1},{b}_{1}\right]\subset \left(0,1\right)$ and a number $p>0$ such that $u\left(t\right)\ge p$ for $t\in \left[{a}_{1},{b}_{1}\right]$. In addition, by (H6), there exists $\phantom{\rule{0.25em}{0ex}}{s}_{0}>0$ and ${u}^{0}\in \left(0,\mathrm{\infty }\right)$ such that $f\left(t,{u}^{0}\right)\ge {s}_{0}$ for $t\in \left[{a}_{1},{b}_{1}\right]$. If $p\ge {u}^{0}$, then $f\left(t,u\right)\ge f\left(t,p\right)\ge f\left(t,{u}^{0}\right)\ge {s}_{0}$; if $p<{u}^{0}$, then $f\left(t,u\right)\ge f\left(t,p\right)\ge f\left(t,\frac{p}{{u}^{0}}p\right)\ge {\left(\frac{p}{{u}^{0}}\right)}^{\alpha }{s}_{0}$. Hence
$\begin{array}{rl}\left(Qu\right)\left(t\right)& \ge \left(TFu\right)\left(t\right)\\ ={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\delta }_{3}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,\tau \right){G}_{3}\left(s,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\delta }_{3}g\left(t\right){\int }_{{a}_{1}}^{{b}_{1}}{G}_{3}\left(s,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge \left({b}_{1}-{a}_{1}\right){\delta }_{3}g\left(t\right){m}_{G}{\left(\frac{p}{{u}^{0}}\right)}^{\alpha }={s}_{u}g\left(t\right),\end{array}$

where ${m}_{G}={min}_{s\in \left[{a}_{1},{b}_{1}\right]}{G}_{3}\left(s,s\right)$, $g\left(t\right)={\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv$, ${s}_{u}=\left({b}_{1}-{a}_{1}\right){\delta }_{3}{m}_{G}{\left(\frac{p}{{u}^{0}}\right)}^{\alpha }$. □

Theorem 3 Suppose that (H3), (H4) and (H6) hold, $L<1$ and $\lambda =1$. Then
1. (i)
(32) has a unique positive solution ${u}^{\star }\in {C}^{+}\left[0,1\right]\mathrm{\setminus }\left\{\theta \right\}$ satisfying

where $0<{m}_{u}<{M}_{u}$ are constants.
1. (ii)
For any ${u}_{0}\left(t\right)\in {C}^{+}\left[0,1\right]\mathrm{\setminus }\left\{\theta \right\}$, the sequence
$\begin{array}{rl}{u}_{n}\left(t\right)& =\left(Q{u}_{n-1}\right)\left(t\right)=\left(HF{u}_{n-1}\right)\left(t\right)\\ =TF{u}_{n-1}+\left(TG\right)TF{u}_{n-1}+{\left(TG\right)}^{2}TF{u}_{n-1}+\cdots +{\left(TG\right)}^{n}TF{u}_{n-1}+\cdots \end{array}$

($n=1,2,\dots$) converges uniformly to the unique solution ${u}^{\star }$, and the rate of convergence is determined by
$\parallel {u}_{n}\left(t\right)-{u}^{\star }\left(t\right)\parallel =O\left(1-{d}^{{\alpha }^{n}}\right),$

where $0 is a positive number.

Proof In view of (H3), (H4) and (H6), $Q:{C}^{+}\left[0,1\right]\to {C}^{+}\left[0,1\right]$ is a nondecreasing operator and satisfies $Q\left(\rho u\right)\ge {\rho }^{\alpha }Q\left(u\right)$ for $t\in \left[0,1\right]$ and $u\in {C}^{+}\left[0,1\right]$. Indeed, let ${u}_{\star }\left(t\right)\le {u}_{\star \star }\left(t\right)$, ${u}_{\star },{u}_{\star \star }\in {C}^{+}\left[0,1\right]$, since $f\left(s,u\right)$ is nondecreasing in u, then by using $f\left(s,{u}_{\star }\left(s\right)\right)\le f\left(s,{u}_{\star \star }\left(s\right)\right)$, for $t\in \left[0,1\right]$, it follows that
$\begin{array}{rl}TF{u}_{\star }\left(t\right)& ={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{\star }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \le {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{u}_{\star \star }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv=TF{u}_{\star \star }\left(t\right).\end{array}$
Moreover, from (25), it follows that for $TF{u}_{\star }\left(t\right)\le TF{u}_{\star \star }\left(t\right)$
(34)
Finally, since $f\left(s,u\right)$ is nondecreasing in u, then by using form (34), $f\left(s,G\left(TF{u}_{\star }\right)\left(t\right)\right)\le f\left(s,G\left(TF{u}_{\star \star }\right)\left(t\right)\right)$, for $t\in \left[0,1\right]$, we have
$\begin{array}{rl}\left(TG\right)TF\left({u}_{\star }\right)& ={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,G\left(TF{u}_{\star }\right)\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \le {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,G\left(TF{u}_{\star \star }\right)\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ =\left(TG\right)TF{u}_{\star \star },\end{array}$
i.e.,
$\left(TG\right)TF\left({u}_{\star }\right)\le \left(TG\right)TF{u}_{\star \star }.$
By induction, it is easy to see that
${\left(TG\right)}^{n}TF\left({u}_{\star }\right)\le {\left(TG\right)}^{n}TF{u}_{\star \star },\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(35)
Hence, using (35), we have
$\begin{array}{rl}Q\left({u}_{\star }\right)& =TF\left({u}_{\star }\right)+\left(TG\right)TF\left({u}_{\star }\right)+{\left(TG\right)}^{2}TF\left({u}_{\star }\right)+\cdots +{\left(TG\right)}^{n}TF\left({u}_{\star }\right)+\cdots \\ \le TF\left({u}_{\star \star }\right)+\left(TG\right)TF\left({u}_{\star \star }\right)+{\left(TG\right)}^{2}TF\left({u}_{\star \star }\right)+\cdots +{\left(TG\right)}^{n}TF\left({u}_{\star \star }\right)+\cdots \\ =Q\left({u}_{\star \star }\right).\end{array}$
(36)
Now, we show that $Q:{C}^{+}\left[0,1\right]\to {C}^{+}\left[0,1\right]$ satisfies $Q\left(\rho u\right)\ge {\rho }^{\alpha }Q\left(u\right)$ for $t\in \left[0,1\right]$ and $u\in {C}^{+}\left[0,1\right]$. Note that
$\begin{array}{rl}TF\left(\rho u\right)& ={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,\rho u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\rho }^{\alpha }{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ ={\rho }^{\alpha }TF\left(u\right).\end{array}$
Moreover, from (25), it follows that for $TF\left(\rho u\right)\ge {\rho }^{\alpha }TF\left(u\right)$,
Finally, we have
$\begin{array}{rl}\left(TG\right)TF\left(\rho u\right)\left(t\right)& ={\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,G\left(TF\rho u\right)\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,{\rho }^{\alpha }G\left(TF\left(u\right)\right)\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ \ge {\rho }^{{\alpha }^{2}}{\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,\tau \right){G}_{3}\left(\tau ,s\right)f\left(s,G\left(TF\left(u\right)\right)\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}dv\\ ={\rho }^{{\alpha }^{2}}\left(TG\right)TF\left(u\right)\left(t\right),\end{array}$
i.e.,
$\left(TG\right)\left(TF\rho u\right)\left(t\right)\ge {\rho }^{{\alpha }^{2}}\left(TG\right)TF\left(u\right)\left(t\right).$
By induction, it is easy to see that
${\left(TG\right)}^{n}\left(TF\rho u\right)\left(t\right)\ge {\rho }^{{\alpha }^{n+1}}\left(TG\right)TF\left(\rho u\right)\left(t\right),\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(37)
Hence, using (35) and $\rho \in \left(0,1\right)$, $\alpha \in \left(0,1\right)$, we have
$\begin{array}{rl}Q\left(\rho u\right)& =TF\left(\rho u\right)+\left(TG\right)TF\left(\rho u\right)+{\left(TG\right)}^{2}TF\left(\rho u\right)+\cdots +{\left(TG\right)}^{n}TF\left(\rho u\right)+\cdots \\ \ge {\rho }^{\alpha }TF\left(u\right)+{\rho }^{{\alpha }^{2}}\left(TG\right)TF\left(u\right)+{\rho }^{{\alpha }^{3}}{\left(TG\right)}^{2}TF\left(u\right)+\cdots +{\rho }^{{\alpha }^{n+1}}{\left(TG\right)}^{n}TF\left(u\right)+\cdots \\ \ge {\rho }^{\alpha }TF\left(u\right)+{\rho }^{\alpha }\left(TG\right)TF\left(u\right)+{\rho }^{\alpha }{\left(TG\right)}^{2}TF\left(u\right)+\cdots +{\rho }^{\alpha }{\left(TG\right)}^{n}TF\left(u\right)+\cdots \\ ={\rho }^{\alpha }\left(TF\left(u\right)+\left(TG\right)TF\left(u\right)+{\left(TG\right)}^{2}TF\left(u\right)+\cdots +{\left(TG\right)}^{n}TF\left(u\right)+\cdots \right)\\ ={\rho }^{\alpha }Q\left(u\right).\end{array}$
(38)
By Lemma 13, there exists $0<{s}_{g}\le {S}_{g}$ such that
${s}_{u}g\left(t\right)\le Qg\left(t\right)\le {S}_{u}g\left(t\right).$
Let
$s=sup\left\{{s}_{g}:{s}_{u}g\left(t\right)\le Qg\left(t\right)\right\},\phantom{\rule{2em}{0ex}}S=inf\left\{{S}_{g}:Qg\left(t\right)\le {S}_{u}g\left(t\right)\right\}.$
Pick ${m}_{s}$ and ${M}_{s}$ such that
$0<{m}_{s}
(39)
and
$max\left\{1,{S}^{\frac{1}{1-\alpha }}\right\}={M}_{s}<\mathrm{\infty }.$
(40)
Set ${u}_{0}\left(t\right)={m}_{s}g\left(t\right)$, ${v}_{0}\left(t\right)={M}_{s}g\left(t\right)$, ${u}_{n}=Q{u}_{n-1}$, and ${v}_{n}=Q{v}_{n-1}$, $n=1,2,\dots$ . From (36) and (38), we have
${m}_{s}g\left(t\right)={u}_{0}\left(t\right)\le {u}_{1}\left(t\right)\le \cdots \le {u}_{n}\left(t\right)\le \cdots \le {v}_{n}\left(t\right)\le \cdots \le {v}_{1}\left(t\right)\le {v}_{0}\left(t\right)={M}_{s}g\left(t\right).$
(41)
Indeed, from (39) ${m}_{s}<1$, and ${m}_{s}^{\alpha -1}s>1$, we have
$\begin{array}{rl}{u}_{1}\left(t\right)& =Q\left({u}_{0}\right)=Q\left({m}_{s}g\left(t\right)\right)\ge {m}_{s}^{\alpha }Q\left(g\left(t\right)\right)\ge {m}_{s}^{\alpha }sg\left(t\right)\\ ={m}_{s}^{\alpha -1}s{m}_{s}g\left(t\right)={m}_{s}^{\alpha -1}s{u}_{0}\left(t\right)\ge {u}_{0}\left(t\right),\end{array}$
and by induction
${u}_{n+1}\left(t\right)=Q\left({u}_{n}\right)\ge Q\left({u}_{n-1}\right)={u}_{n}\left(t\right).$
From (40), ${M}_{s}>1$, and ${M}_{s}^{\alpha -1}S<1$, we have
$\begin{array}{rl}{v}_{1}\left(t\right)& =Q\left({v}_{0}\right)\le {M}_{s}^{\alpha }Q\left(g\left(t\right)\right)={M}_{s}^{\alpha }Q\left(\frac{1}{{M}_{s}}{v}_{0}\right)={M}_{s}^{\alpha }Q\left(g\right)\\ \le {M}_{s}^{\alpha }Sg\le S{M}_{s}^{\alpha -1}{M}_{s}g=S{M}_{s}^{\alpha -1}{v}_{0}\left(t\right)\le {v}_{0}\left(t\right),\end{array}$
and by induction
${v}_{n+1}\left(t\right)=Q\left({v}_{n}\right)\le Q\left({v}_{n-1}\right)={v}_{n}\left(t\right).$
Let $d=\frac{{m}_{s}}{{M}_{s}}$. Then
${u}_{n}\ge {d}^{{\alpha }^{n}}{v}_{n}.$
(42)
In fact ${u}_{0}=\phantom{\rule{0.2em}{0ex}}d{v}_{0}$ is clear. Assume that (42) holds with $n=k$ (k is a positive integer), i.e., ${u}_{k}\ge {d}^{{\alpha }^{k}}{v}_{k}$. Then
${u}_{k+1}=Q\left({u}_{k}\right)\ge Q\left({d}^{{\alpha }^{k}}{v}_{k}\right)\ge {\left({d}^{{\alpha }^{k}}\right)}^{\alpha }Q\left({v}_{k}\right)={d}^{{\alpha }^{k+1}}Q\left({v}_{k}\right)={d}^{{\alpha }^{k+1}}{v}_{k+1}.$
By induction, it is easy to see that (42) holds. Furthermore, in view of (38), (41) and (42), we have
$0\le {u}_{n+z}-{u}_{n}\le {v}_{n}-{u}_{n}\le \left(1-{d}^{{\alpha }^{n}}\right){v}_{0}=\left(1-{d}^{{\alpha }^{n}}\right){M}_{s}g\left(t\right)$
and
$\parallel {u}_{n+z}-{u}_{n}\parallel \le \parallel {v}_{n}-{u}_{n}\parallel \le \left(1-{d}^{{\alpha }^{n}}\right){M}_{s}\parallel g\parallel ,$
where z is a nonnegative integer. Thus, there exists ${u}^{\star }\in {C}^{+}\left[0,1\right]$ such that
and ${u}^{\star }\left(t\right)$ is a fixed point of Q and satisfies
${m}_{g}g\left(t\right)\le {u}^{\star }\left(t\right)\le {M}_{g}g\left(t\right).$

This means that ${u}^{\star }\in {C}_{\star }^{+}\left[0,1\right]$, where .

Next we show that ${u}^{\star }$ is the unique fixed point of Q in ${C}_{\star }^{+}\left[0,1\right]$. Suppose, to the contrary, that there exists another $\overline{u}\in {C}_{\star }^{+}\left[0,1\right]$ such that $Q\overline{u}=\overline{u}$. We can suppose that
Let $\stackrel{ˆ}{\tau }=sup\left\{0<\tau <1:\tau {u}^{\star }\le \overline{u}\le {\tau }^{-1}{u}^{\star }\right\}$. Then $0<\stackrel{ˆ}{\tau }\le 1$ and $\stackrel{ˆ}{\tau }{u}^{\star }\le \overline{u}\le {\stackrel{ˆ}{\tau }}^{-1}{u}^{\star }$. We assert $\stackrel{ˆ}{\tau }=1$. Otherwise, $0<\stackrel{ˆ}{\tau }<1$, and then
$\begin{array}{c}\overline{u}=Q\overline{u}\ge Q\left(\stackrel{ˆ}{\tau }{u}^{\star }\right)\ge {\stackrel{ˆ}{\tau }}^{\alpha }Q\left({u}^{\star }\right)={\stackrel{ˆ}{\tau }}^{\alpha }{u}^{\star },\hfill \\ {u}^{\star }=Q{u}^{\star }\ge Q\left(\stackrel{ˆ}{\tau }\overline{u}\right)\ge {\stackrel{ˆ}{\tau }}^{\alpha }Q\left(\overline{u}\right)={\stackrel{ˆ}{\tau }}^{\alpha }\overline{u}.\hfill \end{array}$

This means that ${\stackrel{ˆ}{\tau }}^{\alpha }{u}^{\star }\le \overline{u}\le {\left({\stackrel{ˆ}{\tau }}^{\alpha }\right)}^{-1}{u}^{\star }$, which is a contradiction of the definition of $\stackrel{ˆ}{\tau }$, because $\stackrel{ˆ}{\tau }<{\stackrel{ˆ}{\tau }}^{\alpha }$.

Let us introduce the following notations for $\mu =0$
$\begin{array}{c}{T}_{\lambda }u\left(t\right):=TFu\left(t\right)=\lambda {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}{G}_{1}\left(t,v\right){G}_{2}\left(v,s\right){G}_{3}\left(s,\tau \right)f\left(\tau ,u\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dv,\hfill \\ \begin{array}{rl}{Q}_{\lambda }u& :=HFu=TFu+\left(TG\right)TFu+{\left(TG\right)}^{2}TFu+\cdots +{\left(TG\right)}^{n}TFu+\cdots \\ ={T}_{\lambda }u+\left(TG\right){T}_{\lambda }u+{\left(TG\right)}^{2}{T}_{\lambda }u+\cdots +{\left(TG\right)}^{n}{T}_{\lambda }u+\cdots ,\end{array}\hfill \end{array}$

i.e., ${Q}_{\lambda }u=\lambda Qu$, where Q is given by (33). □

Theorem 4 Suppose that (H3), (H4), (H6) and $L<1$ hold. Then (32) has a unique positive solution ${u}_{\lambda }\left(t\right)$ for any $0<\lambda \le 1$.

Proof Theorem 3 implies that for $\lambda =1$, the operator ${Q}_{\lambda }$ has a unique fixed point ${u}_{1}\in {C}^{+}\left[0,1\right]$, that is ${Q}_{1}{u}_{1}={u}_{1}$. Then from Lemma 10, for every ${\lambda }_{\star }\in \left(0,1\right)$, there exists a function ${u}_{\star }\in P\mathrm{\setminus }\left\{\theta \right\}$ such that ${Q}_{{\lambda }_{\star }}{u}_{\star }={u}_{\star }$.

Thus, ${u}_{\lambda }$ is a unique positive solution of (32) for every $0<\lambda \le 1$. □

## 4 Application

As an application of Theorem 1, consider the sixth-order boundary value problem
$\begin{array}{r}-{u}^{\left(6\right)}+\left(1-0.5{t}^{2}\right){u}^{\left(4\right)}+\left(4.5-0.5sin\pi t\right){u}^{\mathrm{\prime }\mathrm{\prime }}+C\left(-5+cos0.5\pi t\right)u\\ \phantom{\rule{1em}{0ex}}=\left(0.5t\left(1-t\right)+u\right)\phi +\lambda \left(1+sin\pi t+{u}^{2}\right),\phantom{\rule{1em}{0ex}}0
(43)

for a fixed ${\lambda }_{1}=2$, ${\lambda }_{2}=-2$, ${\lambda }_{3}=1$ and $\varkappa =2$. In this case, $a={\lambda }_{1}+{\lambda }_{2}+{\lambda }_{3}=1$, $b=-{\lambda }_{1}{\lambda }_{2}-{\lambda }_{2}{\lambda }_{3}-{\lambda }_{1}{\lambda }_{3}=4$, and $c={\lambda }_{1}{\lambda }_{2}{\lambda }_{3}=-4$. We have $A\left(t\right)=1-0.5{t}^{2}$, $B\left(t\right)=4.5-0.5sin\pi t$, $C\left(t\right)=-5+cos0.5\pi t$, $D\left(t\right)=0.5t\left(1-t\right)$ and $f\left(t,u\right)=1+sin\pi t+{u}^{2}$. It is easy to see that ${\pi }^{6}+a{\pi }^{4}-b{\pi }^{2}+c=1,015.3>0$, $a={sup}_{t\in \left[0,1\right]}A\left(t\right)$, $b={inf}_{t\in \left[0,1\right]}B\left(t\right)$ and $c={sup}_{t\in \left[0,1\right]}C\left(t\right)$. Note also that $K={max}_{0\le t\le 1}\left[-A\left(t\right)+B\left(t\right)-C\left(t\right)-\left(-a+b-c\right)\right]=2$, ${D}_{2}={max}_{t\in \left[0,1\right]}{\int }_{0}^{1}{G}_{2}\left(t,v\right)\phantom{\rule{0.2em}{0ex}}dv=0.15768$, $C={max}_{t\in \left[0,1\right]}D\left(t\right)=0.125$, ${d}_{1}={max}_{t\in \left[0,1\right]}{\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds=0.10336$, ${\mu }^{\ast \ast }=\frac{1-{D}_{2}K}{{D}_{2}C{d}_{1}}=336.1$ and ${D}_{2}K=0.3153<1$. Thus, if $0<\mu <336.1$, then the conditions of Theorem 1 (note $L={D}_{2}\left(K+\mu C{d}_{1}\right)<1$) are fulfilled (in particular, (H3)-(H5) are satisfied). As a result, Theorem 1 can be applied to (43).

## Authors’ Affiliations

(1)
Department of Mathematics, Texas A&M University-Kingsville
(2)
Department of Analysis, University of Miskolc
(3)
School of Mathematics, Statistics and Applied Mathematics, National University of Ireland

## References

1. Evans JD, Galactionov VA, King JR: Unstable sixth-order thin film equation: II. Global similarity patterns. Nonlinearity 2007, 20: 1843-1881. 1