\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 266, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/266\hfil Symmetric positive solutions]
{Symmetric positive solutions for $\phi$-Laplacian boundary-value
problems with integral \\ boundary conditions}

\author[W. Yang \hfil EJDE-2013/266\hfilneg]
{Wengui Yang}  % in alphabetical order

\address{Wengui Yang\newline
Ministry of Public Education, Sanmenxia Polytechnic, 
Sanmenxia 472000, China}
\email{wgyang0617@yahoo.com}

\thanks{Submitted June 17, 2013. Published November 30, 2013.}
\subjclass[2000]{34B15, 34B18}
\keywords{Boundary value problems; integral boundary conditions;
\hfill\break\indent symmetric positive solutions; $\phi$-Laplacian operator; 
fixed point theorem}

\begin{abstract}
 In this article, we study the existence, multiplicity,
 and nonexistence of symmetric positive solutions for a class of
 four-order integral boundary value problems with $\phi$-Laplacian operator.
 The arguments mainly rely on the Guo-Krasnosel'skii fixed point theorem
 of cone expansion and compression of norm type and Leggett-Williams
 fixed point theorem. Finally, some examples are presented to illustrate
 the main results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}\label{sect1}

In this article, we consider the following $\phi$-Laplacian boundary-value
problems (BVP) with integral boundary conditions
\begin{equation} \label{e1.1}
\begin{gathered}
 \big[q(t)[\phi((p(t)u'(t))')]'\big]'=w(t)f(t,u(t)),\quad t\in[a,b],\\
 \alpha u(a)-\beta p(a)u'(a)=\int_a^bg(s)u(s)ds,\\
 \alpha u(b)+\beta p(b)u'(b)=\int_a^bg(s)u(s)ds,\\
\gamma \phi((p(a)u'(a))')-\delta q(a)[\phi((p(a)u'(a))')]'
 =\int_a^bh(s)\phi((p(s)u'(s))')ds,\\
\gamma \phi((p(b)u'(b))')+\delta q(b)[\phi((p(b)u'(b))')]'
 =\int_a^bh(s)\phi((p(s)u'(s))')ds.
\end{gathered}
\end{equation}
We use the following assumptions throughout this article:
\begin{itemize}
\item[(H0)] $\phi$ is an odd, increasing homeomorphism from $\mathbb{R}$
 onto $\mathbb{R}$ and there exist two increasing homeomorphisms
 $\psi_1$ and  $\psi_2$ of $(0,\infty)$ onto $(0,\infty)$ such that
 $\psi_1(u)\phi(v)\leq\phi(uv)\leq\psi_2(u)\phi(v)$ for all $u,v>0$.
 Moreover, $\phi,\phi^{-1}\in C^1(\mathbb{R})$, where $\phi^{-1}$ denotes
 the inverse of $\phi$.
\item[(H1)] $\alpha>0$, $\gamma>0$, and $\beta\geq0$, $\delta\geq0$.
\item[(H2)] $p(t),q(t):[a,b]\to (0,\infty)$ are continuous functions,
 and $p(t),q(t)$ are symmetric on $[a,b]$.
\item[(H3)] $w\in L^1[a,b]$ is nonnegative, symmetric on $[a,b]$ and
 $w(t)\not\equiv0$ on any subinterval of $[a,b]$.
\item[(H4)] $f:[a,b]\times[0,\infty)\to [0,\infty)$ is continuous
 and $f(\cdot,u)$ is symmetric on $[a,b]$ for all $u\in[0,\infty)$.
\item[(H5)] $g,h\in C([a,b],[0,\infty))$ are symmetric functions on
 $[a,b]$, and $\mu\in[0,\alpha)$, $\nu\in[0,\gamma)$, where
 \[
 \mu=\int_a^bg(s)ds, \quad \nu=\int_a^bh(s)ds.
 \]
\end{itemize}

Boundary-value problems with integral boundary conditions have
gained considerable popularity and importance due to their application in
different areas of applied mathematics and physics and so on.
They include two, three, multi-point and nonlocal boundary-value problems
 as special cases. There have been some papers dealing with the existence
and multiplicity of solutions or positive solutions for such problems
by the use of some well-known fixed point theorems and upper and lower
solutions method. For some recent developments on the subject, see
\cite{3,2,4,6,1,9,7,5,8} and the references therein.

Recently, many researchers have extensively studied the existence,
multiplicity and nonexistence of symmetric positive solutions of
boundary-value problems by using fixed point theorem; i.e.,
fixed point theorem of cone expansion and compression of norm type,
fixed point theory in cones and Leggett-Williams fixed point theorem.
To identify a few, we refer the reader to \cite{13,16,10,11,17,14,12,15}
and the references therein. In particular, we would like to mention some
results of Feng \cite{18}, Ma \cite{19}, Xu \cite{20}, Luo and Luo \cite{21},
and Zhang et al.~\cite{22}, Zhang and Ge \cite{22A}.
Feng \cite{18} considered the following second-order nonlinear
ordinary differential equation with integral boundary conditions
\begin{gather*}
(g(t)u'(t))'+w(t)f(t,u(t))=0,\quad 0<t<1,\\
au(0)-b\lim_{t\to 0^+}g(t)u'(t)=\int_0^1h(s)u(s)ds,\\
au(1)+b\lim_{t\to 1^-}g(t)u'(t)=\int_0^1h(s)u(s)ds,
\end{gather*}
where $a,b>0$, $g\in C^1([0,1],(0,\infty))$ is symmetric on
$[0,1]$, $w\in L^p[0,1]$ for some $1\leq p\leq+\infty$, and is
symmetric on $[0,1]$, $f:[0,1]\times[0,+\infty)$ is continuous,
$f(1-t,u)=f(t,u)$ for all $(t,u)\in[0,1]\times[0,+\infty)$,
 and $h\in L^p[0,1]$ is nonnegative and symmetric on $[0,1]$.
The author investigated the existence of at least one symmetric
positive solution by applying the theory of fixed point index in cones.

Ma \cite{19} obtained the existence of at least one symmetric positive
solution for fourth-order boundary-value
problem with integral boundary conditions by using the fixed point index in cones.
\begin{equation} \label{e1.2}
\begin{gathered}
u''''(t)=w(t)f(u(t)),\quad 0<t<1,\\
u(0)=u(1)=\int_0^1p(s)u(s)ds,\quad  u''(0)=u''(1)=\int_0^1q(s)u''(s)ds,
\end{gathered}
\end{equation}
where $p,q\in L^1[0,1]$, $w:(0,1)\to [0,+\infty)$ is continuous,
symmetric on $[0,1]$ and may be singular at $t=0$ and $t=1$,
$f:[0,1]\times[0,+\infty)$ is continuous, $f(1-t,u)=f(t,u)$
for all $(t,u)\in[0,1]\times[0,+\infty)$. Xu \cite{20} studied the existence
of three positive solutions for fourth-order singular nonlocal boundary-value
problems $u''''(t)=w(t)f(t,u(t)),~ 0<t<1,$ with integral boundary
conditions \eqref{e1.2} by using Leggett-Williams fixed point theorem.

Luo and Luo \cite{21} investigated the existence, multiplicity and
nonexistence of symmetric positive solutions of the following fourth
order boundary value problem with integral boundary conditions
\begin{equation} \label{e1.3}
\begin{gathered}
\phi(u''(t))''=w(t)f(t,u(t),u'(t)),\quad 0<t<1,\\
u(0)=u(1)=\int_0^1p(s)u(s)ds,\quad
\phi(u''(0))=\phi(u''(1))=\int_0^1q(s)\phi(u''(s))ds,
\end{gathered}
\end{equation}
where $\phi$ is defined as in (H0), $p,q:[0,1]\to (0,\infty)$ are
continuous functions, and $p,q$ are symmetric on $[0,1]$,
$f:[0,1]\times[0,+\infty)$ is continuous, $f(1-t,u)=f(t,u)$ for
all $(t,u)\in[0,1]\times[0,+\infty)$. Zhang et al.~\cite{22}
 studied the existence and nonexistence of symmetric positive solutions
 for a special case of fourth order boundary value problem \eqref{e1.3}
when $\phi(t)=\phi_p(t)=|t|^{p-2}t$, $p>1$, and $f(t,u(t),u'(t))=f(t,u(t))$.

Zhang and Ge \cite{22A} considered the following second-order nonlinear
ordinary differential equation with integral boundary conditions
\begin{gather*}
(q(t)u'''(t))'=w(t)f(t,u(t)),\quad 0<t<1,\\
u(0)=u(1)=\int_0^1g(s)u(s)ds\\
au''(0)-b\lim_{t\to 0^+}g(t)u'''(t)=\int_0^1h(s)u''(s)ds,\\
au''(1)+b\lim_{t\to 1^-}g(t)u'''(t)=\int_0^1h(s)u''(s)ds,
\end{gather*}
where $a,b>0$, $q\in C^1([0,1],(0,\infty))$ is symmetric on
$[0,1]$, $w\in L^p[0,1]$ for some $1\leq p\leq+\infty$, and is
symmetric on $[0,1]$, $f:[0,1]\times[0,+\infty)$ is continuous,
$f(1-t,u)=f(t,u)$ for all $(t,u)\in[0,1]\times[0,+\infty)$, and
$g,h\in L^p[0,1]$ is nonnegative and symmetric on $[0,1]$.
The authors obtained the existence of at least one symmetric positive
solution based upon a specially
constructed cone and the fixed point theory in a cone.

Motivated greatly by the above mentioned works, we establish sufficient
conditions for the existence, multiplicity and nonexistence of symmetric
positive solutions of BVP \eqref{e1.1} by applying the Guo-Krasnosel'skii
fixed point theorem of cone expansion and compression of norm type
and Leggett-Williams fixed point theorem. Our paper improves and
generalizes the results of mentioned results to some degree.
At the end of this paper,  some examples are presented to illustrate
the main results.

\section{Preliminaries}\label{sect2}

We recall that the function $u$ is said to be concave on $[a,b]$,
if $u(\lambda t_2+(1-\lambda)t_1)\geq\lambda u(t_1)+(1-\lambda)u(t_1)$,
$t_1,t_2\in[a,b]$, $\lambda\in[0,1]$  and the function $u$
is said to be symmetric on $[a,b]$, if $u(b-t+a)=u(t)$, $t\in[a,b]$.
A function $u$ is called a symmetric positive solution of  \eqref{e1.1}
provided $u$ is symmetric and positive on $[a,b]$, and satisfies the
differential equation and the boundary value conditions in  \eqref{e1.1}.

To prove the main results in this article, we will use the following lemmas.

\begin{lemma}\label{L2.1}
Assume {\rm (H0)--(H2)} hold and $\mu\neq\alpha$.
 Then for any $v\in C[a,b]$, the BVP
\begin{equation} \label{e2.1}
\begin{gathered}
(p(t)u'(t))'=\phi^{-1}(v(t)),\quad t\in[a,b],\\
\alpha u(a)-\beta p(a)u'(a)=\int_a^bg(s)u(s)ds,\\
\alpha u(b)+\beta p(b)u'(b)=\int_a^bg(s)u(s)ds,
\end{gathered}
\end{equation}
has a unique solution $u$ and $u$ can be expressed in the form
\begin{equation}\label{e2.2}
u(t)=-\int_a^bH_1(t,s)\phi^{-1}(v(s))ds,
\end{equation}
where
\begin{gather}\label{e2.3}
H_1(t,s)=G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau,\\
\label{e2.4}
G_1(t,s)=\frac{1}{\Delta_1} \begin{cases}
 \big(\beta+\alpha\int_a^s\frac{d\tau}{p(\tau)}\big)
 \big(\beta+\alpha\int_t^b\frac{d\tau}{p(\tau)}\big), &a\leq s\leq t\leq b,\\[3pt]
 \big(\beta+\alpha\int_a^t\frac{d\tau}{p(\tau)}\big)
\big(\beta+\alpha\int_s^b\frac{d\tau}{p(\tau)}\big), &a\leq t\leq s\leq b.
\end{cases}
\end{gather}
Here
$$\Delta_1=\alpha\Big(2\beta+\alpha\int_a^b\frac{ds}{p(s)}\Big), \quad
\mu=\int_a^bg(s)ds.
$$
\end{lemma}

\begin{proof}
First suppose that $u$ is a solution of problem \eqref{e2.1}.
It is easy to see by integration of both sides of \eqref{e2.1} on $[a,t]$ that
\[
 p(t)u'(t)=p(a)u'(a)+\int_a^t\phi^{-1}(v(s))ds.
\]
Letting $A=p(a)u'(a)$, then we have
\[
u'(t)=\frac{A}{p(t)}+\frac{1}{p(t)}\int_a^t\phi^{-1}(v(s))ds.
\]
Integrating again, we can get
\[
u(t)=u(a)+A\int_a^t\frac{1}{p(s)}ds+\int_a^t\frac{1}{p(s)}
\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds.
\]
By the boundary condition, we obtain
\begin{gather*}
\alpha u(a)-\beta A=\int_a^bg(s)u(s)ds,\\
\begin{aligned}
\alpha u(a)+\Big(\alpha\int_a^b\frac{ds}{p(s)}+\beta\Big)A
&= \int_a^bg(s)u(s)ds-\beta\int_a^bg(s)\phi^{-1}(v(s))ds \\
&\quad -\alpha\int_a^b\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds.
\end{aligned}
\end{gather*}
Then
\begin{align*}
A=-\frac{1}{2\beta+\alpha\int_a^b\frac{ds}{p(s)}}
\Big(\alpha\int_a^b\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds
+\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big),
\end{align*}
and
\begin{align*}
u(a)&=-\frac{1}{\Delta_1}\Big(\alpha\int_a^b\frac{1}{p(s)}
\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds
+\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big)\\
&\quad +\frac{1}{\alpha}\int_a^bg(s)u(s)ds.
\end{align*}
Thus
\begin{align*}
u(t)&=\frac{1}{\alpha}\int_a^bg(s)u(s)ds
 +\int_a^t\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds\\
&\quad -\frac{1}{\Delta_1}\Big(\alpha\int_a^b\frac{1}{p(s)}
 \int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds
+\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big)\\
&\quad -\frac{\int_a^t\frac{1}{p(s)}ds}{2\beta+\alpha\int_a^b\frac{ds}{p(s)}}
\Big(\alpha\int_a^b\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds
 +\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big)\\
&=\frac{1}{\alpha}\int_a^bg(s)u(s)ds-\int_a^bG_1(t,s)\phi^{-1}(v(s))ds,
\end{align*}
where $G_1(t,s)$ is defined by \eqref{e2.4}.
 Multiplying the above equation with $g(t)$ and integrating it again, we obtain
\[
\int_a^bg(s)u(s)ds=-\frac{\alpha}{\alpha
 -\int_a^bg(s)ds}\int_a^bG_1(t,s)\phi^{-1}(v(s))ds.
\]
It follows that
\begin{align*}
u(t)&=-\int_a^bG_1(t,s)\phi^{-1}(v(s))ds-\frac{1}{\alpha-\mu}\int_a^bG_1(t,s)\phi^{-1}(v(s))ds\\
&=-\int_a^bH_1(t,s)\phi^{-1}(v(s))ds,
\end{align*}
where $H_1(t,s)$ is defined in \eqref{e2.3}. The proof is complete.
 \end{proof}

\begin{lemma}\label{L2.2}
Assume (H1)-(H4) hold and $\nu\neq\gamma$. Then for any $u\in C[a,b]$, the BVP
\begin{equation*}
\begin{gathered}
(q(t)v'(t))'=w(t)f(t,u(t)),\quad t\in[a,b],\\
\gamma v(a)-\delta q(a)v'(a)=\int_a^bh(s)v(s)ds,\quad  \gamma v(b)+\delta q(b)v'(b)=\int_a^bh(s)v(s)ds,
\end{gathered}
\end{equation*}
has a unique solution $v$ and $v$ can be expressed in the form
\[
v(t)=-\int_a^bH_2(t,s)w(s)f(s,u(s))ds,
\]
where
\begin{gather}\label{e2.5}
H_2(t,s)=G_2(t,s)+\frac{1}{\gamma-\nu}\int_a^bG_2(s,\tau)h(\tau)d\tau, \\
\label{e2.6}
G_2(t,s)=\frac{1}{\Delta_2}\begin{cases}
 \big(\delta+\gamma\int_a^s\frac{d\tau}{q(\tau)}\big)
 \big(\delta+\gamma\int_t^b\frac{d\tau}{q(\tau)}\big), &a\leq s\leq t\leq b,\\[3pt]
 \big(\delta+\gamma\int_a^t\frac{d\tau}{q(\tau)}\big)
 \big(\delta+\gamma\int_s^b\frac{d\tau}{q(\tau)}\big), &a\leq t\leq s\leq b.
\end{cases}
\end{gather}
Here 
$$
\Delta_2=\gamma\Big(2\delta+\gamma\int_a^b\frac{ds}{q(s)}\Big), 
\quad\nu=\int_a^bh(s)ds.
$$
\end{lemma}

The above lemma can be proved in a way similar to Lemma \ref{L2.1}.
 
\begin{lemma}\label{L2.3}
Assume {\rm (H5)} holds. Then for any $t,s\in[a,b]$, the following 
results are true.
\begin{itemize}
\item[(i)] $G_1(t,s)\geq0$, $H_1(t,s)\geq0$, $G_2(t,s)\geq0$, $H_2(t,s)\geq0$;

\item[(ii)] $\frac{\beta^2}{\Delta_1}\leq G_1(t,s)\leq G_1(s,s)
 \leq\frac{\Lambda_1}{\Delta_1}$, $\frac{\beta^2\sigma_1}{\Delta_1}
 \leq H_1(t,s)\leq H_1(s,s)\leq\frac{\sigma_1\Lambda_1}{\Delta_1}$,
  $\frac{\delta^2}{\Delta_2}\leq G_2(t,s)\leq G_2(s,s)
 \leq\frac{\Lambda_2}{\Delta_2}$, $\frac{\delta^2\sigma_2}{\Delta_2}
 \leq H_2(t,s)\leq H_2(s,s)\leq\frac{\sigma_2\Lambda_2}{\Delta_2}$ with
   \begin{gather*}
    \Lambda_1=\Big(\beta+\alpha\int_a^b\frac{d\tau}{p(\tau)}\Big)^2,\quad 
 \sigma_1=\frac{\alpha}{\alpha-\mu},\\
    \Lambda_2=\Big(\delta+\gamma\int_a^b\frac{d\tau}{q(\tau)}\Big)^2,\quad 
 \sigma_2=\frac{\gamma}{\gamma-\nu};
    \end{gather*}

\item[(iii)] $G_1(b-t+a,b-s+a)=G_1(t,s)$, $H_1(b-t+a,b-s+a)=H_1(t,s)$, 
$G_2(b-t+a,b-s+a)=G_2(t,s)$, $H_2(b-t+a,b-s+a)=H_2(t,s)$;

\item[(iv)] $\rho_1G_1(s,s)\leq H_1(t,s)\leq\sigma_1 G_1(s,s)$, 
$\rho_2G_2(s,s)\leq H_2(t,s)\leq\sigma_2 G_2(s,s)$ with
$$
\rho_1=\frac{\Delta_1}{(\alpha-\mu)\Lambda_1}\int_a^bG_1(s,s)g(s)ds,\quad  
\rho_2=\frac{\Delta_2}{(\gamma-\nu)\Lambda_2}\int_a^bG_1(s,s)h(s)ds;
$$
\end{itemize}
where $H_1(t,s)$, $G_1(t,s)$, $H_2(t,s)$ and $G_2(t,s)$ are defined 
by \eqref{e2.3}--\eqref{e2.6}, respectively.
\end{lemma}

\begin{proof}
 By simple computations, we have (i) and (ii). 
Firstly, we prove that  (iii) holds. If $a\leq t\leq s\leq b$, 
then $b-t+a\geq b-s+a$. In view of \eqref{e2.4} and the assumption (H2), we get
\begin{align*}
&G_1(b-t+a,b-s+a)\\
&=\frac{1}{\Delta_1}\Big(\beta+\alpha\int_a^{b-s+a}\frac{d\tau}{p(\tau)}\Big)
\Big(\beta+\alpha\int_{b-t+a}^b\frac{d\tau}{p(\tau)}\Big)\\
&= \frac{1}{\Delta_1}\Big(\beta+\alpha\int_b^s\frac{d(b-\tau+a)}{p(b-\tau+a)}
\Big)\Big(\beta+\alpha\int_t^a\frac{d(b-\tau+a)}{p(b-\tau+a)}\Big)\\
&= \frac{1}{\Delta_1}\Big(\beta+\alpha\int_s^b\frac{d\tau}{p(\tau)}\Big)
\Big(\beta+\alpha\int_a^t\frac{d\tau}{p(\tau)}\Big)\\
&=G_1(t,s), \quad a\leq t\leq s\leq b.
\end{align*}
Similarly, we can prove that $G_1(b-t+a,b-s+a)=G_1(t,s)$, $a\leq s\leq t\leq b$. 
So, we have $G_1(b-t+a,b-s+a)=G_1(t,s)$, for any $t,s\in[a,b]$. 
By \eqref{e2.3} and (H5), we have
\begin{align*}
&H_1(b-t+a,b-s+a)\\
&= G_1(b-t+a,b-s+a)+\frac{1}{\alpha-\mu}\int_a^bG_1(b-s+a,\tau)g(\tau)d\tau\\
&= G_1(t,s)+\frac{1}{\alpha-\mu}\int_b^aG_1(b-s+a,b-\tau+a)
 g(b-\tau+a)d(b-a\tau+a)\\
&= G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau=H_1(t,s).
\end{align*}
Similarly, we can prove that $G_2(b-t+a,b-s+a)=G_2(t,s)$,
 $H_2(b-t+a,b-s+a)=H_2(t,s)$.

Next, we prove that (iv). First, we prove that 
$\frac{\Delta_1}{\Lambda_1}G_1(t,t)G_1(s,s)\leq G_1(t,s)$
and $\frac{\Delta_2}{\Lambda_2}G_2(t,t)G_2(s,s)\leq G_2(t,s)$. 
If $a\leq t\leq s\leq b$, then
\[
\frac{G_1(t,s)}{G_1(t,t)G_1(s,s)}
=\frac{\Delta_1}{\big(\beta+\alpha\int_a^s\frac{d\tau}{p(\tau)}\big)
\big(\beta+\alpha\int_t^b\frac{d\tau}{p(\tau)}\big)}
\geq\frac{\Delta_1}{\Lambda_1}.
\]
Similarly, we can prove that 
\[
\frac{G_1(t,s)}{G_1(t,t)G_1(s,s)}\geq\frac{\Delta_1}{\Lambda_1},
\]
for  $a\leq s\leq t\leq b$. So, we have
$\frac{\Delta_1}{\Lambda_1}G_1(t,t)G_1(s,s)\leq G_1(t,s)$, for any $t,s\in[a,b]$. From \eqref{e2.3}, then we obtain
\begin{align*}
H_1(t,s)
&= G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau\\
&\leq G_1(s,s)+\frac{1}{\alpha-\mu}\int_a^bg(s)dsG_1(s,s)\\
&=\frac{\alpha}{\alpha-\mu}G_1(s,s)=\sigma_1G_1(s,s).
\end{align*}
On the other hand, we have
\begin{align*}
H_1(t,s)
&= G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau\\
&\geq\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau\\
&\geq \frac{\Delta_1}{(\alpha-\mu)\Lambda_1}
 \int_a^bG_1(\tau,\tau)g(\tau)d\tau G_1(s,s)=\rho_1G_1(s,s).
\end{align*}
Hence we get $\rho_1G_1(s,s)\leq H_1(t,s)\leq\sigma_1 G_1(s,s)$.
 Similarly, we can prove that 
$\rho_2G_2(s,s)\leq H_2(t,s)\leq\sigma_2 G_2(s,s)$.
\end{proof}

From Lemmas \ref{L2.1} and \ref{L2.2}, we have the following result.

\begin{lemma}\label{L2.4}
Assume {\rm (H0)--(H5)} hold. If $u$ is a solution of \eqref{e1.1}, then
\[
u(t)=\int_a^bH_1(t,s)\phi^{-1}\Big(\int_a^bH_2(s,\tau)w(\tau)
f(\tau,u(\tau))d\tau\Big)ds.
\]
\end{lemma}

Now, we let $\mathbb{C}=C[a,b]$, then $\mathbb{C}$ is a Banach space 
with $\|u\|=\max_{t\in[a,b]}|u(t)|$, and define a cone $\mathbb{P}$
by $\mathbb{P}=\{x\in\mathbb{C} : u(t)\geq0$, $u''(t)\leq0$, 
$u(t)$ is a symmetric and concave function on $[a,b]$ and 
$u(t)\geq\omega\|u\|\}$, where $\omega$ is defined as in 
Lemma \ref{L2.6}. Also, define, for $0<r<R$ two positive numbers, 
$\Omega_r$ and $\overline{\Omega}_{r,R}$ by 
$\Omega_r=\{u\in\mathbb{C}:\|u\|<r\}$ and 
$\overline{\Omega}_{r,R}=\{u\in\mathbb{C}:r\leq u\leq R\}$. 
Note that $\partial\Omega_r=\{u\in\mathbb{C}:\|u\|=r\}$.

\begin{lemma}[\cite{23}]\label{L2.5}
Assume {\rm (H0)} holds. Then for any $u,v\in(0,\infty)$, we have 
$\psi_2^{-1}(u)v\leq\phi^{-1}(u\phi(v))\leq\psi_1^{-1}(u)v$, where 
$\psi_1^{-1}$ and $\psi_2^{-1}$ denote the inverse of $\psi_1$ and 
 $\psi_2$, respectively.
\end{lemma}


\begin{lemma}\label{L2.6}
Assume {\rm (H0)--(H5)} hold. Then the solution $u(t)$ of \eqref{e1.1}
is positive and symmetric on $[a,b]$
and $\min_{t\in[a,b]}u(t)\geq\omega\|u\|$, where 
$\omega=\frac{\rho_1\psi_2^{-1}(\rho_2)}{\sigma_1\psi_1^{-1}(\sigma_2)}$.
\end{lemma}

\begin{proof} From {\rm (H0)--(H5)} and Lemma \ref{L2.3}, it is easy 
to prove that the solution $u(t)$ of \eqref{e1.1} is positive on $[a,b]$.
We  need to  prove only that $u(t)$ is symmetric on $[a,b]$. 
Combining (H3) and (H4) and Lemma \ref{L2.3} and \ref{L2.4}, we obtain
\begin{align*}
&u(b-t+a)\\
&= \int_a^bH_1(b-t+a,s)\phi^{-1}
 \Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\
&= \int_b^aH_1(b-t+a,b-s+a)\\
&\quad \times \phi^{-1}
\Big(\int_a^bH_2(b-s+a,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)d(b-s+a)\\
&= \int_b^aH_1(b-t+a,b-s+a)\phi^{-1}
\Big(\int_b^aH_2(b-s+a,b-\tau+a)\\
&\times w(b-\tau+a)f(b-\tau+a,u(b-\tau+a))d(b-\tau+a)\Big)d(b-s+a)\\
&= \int_b^aH_1(b-t+a,b-s+a)\phi^{-1}
 \Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)d(b-s+a)\\
&= \int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds=u(t).
\end{align*}
Therefore, $u(t)$ is symmetric on $[a,b]$. From (H0) and Lemma 
\ref{L2.3}---\ref{L2.5}, we obtain
\begin{align*}
u(t)&= \int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\
&\leq \sigma_1\int_a^b G_1(s,s)\phi^{-1}
 \Big(\sigma_2\int_a^b G_2(\tau,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\
&\leq \sigma_1\psi_1^{-1}(\sigma_2)\int_a^b G_1(s,s)ds\phi^{-1}
\Big(\int_a^b G_2(\tau,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big).
\end{align*}
So we have
\begin{equation}\label{e2.7}
\|u(t)\|\leq\sigma_1\psi_1^{-1}(\sigma_2)\int_a^b G_1(s,s)ds\phi^{-1}
\Big(\int_a^b G_2(\tau,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big).
\end{equation}
On the other hand, by Lemma \ref{L2.3}---\ref{L2.5},  we have
\begin{align}\label{e2.8}
u(t)&= \int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds \\
&\geq \rho_1\int_a^b G_1(s,s)\phi^{-1}
 \Big(\rho_2\int_a^b G_2(\tau,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds \\
&\geq \rho_1\psi_2^{-1}(\rho_2)\int_a^b G_1(s,s)ds\phi^{-1}
\Big(\int_a^b G_2(\tau,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big) \\
&\geq \frac{\rho_1\psi_2^{-1}(\rho_2)}{\sigma_1\psi_1^{-1}(\sigma_2)}\|u\|
=\omega\|u\|.
\end{align}
Combined \eqref{e2.7} with \eqref{e2.8}, we deduce inequality 
$\min_{t\in[a,b]}u(t)\geq\omega\|u\|$.
\end{proof}

We define the integral operator $T:\mathbb{C}\to \mathbb{C}$ by
\begin{equation}\label{e2.9}
(Tu)(t)=\int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds.
\end{equation}

\begin{lemma}\label{L2.7}
Assume {\rm (H0)--(H5)} hold. Then $u\in\mathbb{C}$ is a solution of 
\eqref{e1.1} if and only if $u$ is a fixed point of the operator $T$.
\end{lemma}

\begin{lemma}\label{L2.8}
Assume {\rm (H0)-(H5)} hold. Then $T:\mathbb{P}\to \mathbb{P}$ 
is a completely continuous operator.
\end{lemma}

The next lemma is the Fixed point theorem of cone expansion and compression 
of norm type,  see \cite{24}.

\begin{lemma}\label{L2.9} 
Let $P$ be a cone of real Bananch space $E$, $\Omega_1$ and $\Omega_2$ 
be two bounded open sets in $E$ such that 
$0\in\Omega_1\subset\overline{\Omega}_1\subset\Omega_2$.
 Let operator $T:P\cap(\overline{\Omega}_2\setminus\Omega_1)\to  P$ be 
completely continuous. Suppose that one of the two conditions is satisfied.
\begin{itemize}
\item[(i)] $\|Tu\|\leq\|u\|$, $\forall u\in P\cap\partial\Omega_1$,
  and $\|Tu\|\geq\|u\|$, $\forall u\in P\cap\partial\Omega_2$.
\item[(ii)] $\|Tu\|\geq\|u\|$, $\forall u\in P\cap\partial\Omega_1$,
  and $\|Tu\|\leq\|u\|$, $\forall u\in P\cap\partial\Omega_2$.
\end{itemize}
Then $T$ has at least one fixed point in 
$P\cap(\overline{\Omega}_2\setminus\Omega_1)$.
\end{lemma}

Let $Q$ be a cone in a real Banach space $E$, a functional 
$\vartheta: Q\to  Q$ is said to be increasing on $Q$ provided
$\vartheta(u)\leq\vartheta(v)$, for all $u,v\in Q$ with $u\leq v$. 
Let $\chi$ be a nonnegative continuous functional on $Q$ in a real 
Banach space $E$, we define for each $d>0$ the following set 
$Q(\chi,d)=\{u\in Q|\chi(u)<d\}$. Let $Q$ be a cone in a real Banach 
space $E$, $\kappa$ is said to be nonnegative continuous concave on
 $Q$ provided 
$\kappa(\lambda u+(1-\lambda)v)\geq\lambda \kappa(u)+(1-\lambda)\kappa(v)$, 
$u,v\in[a,b]$ with $\lambda\in[0,1]$.

Let $a,b,r>0$ be constants with $Q$ and $\kappa$ as defined above, we note
$$
Q_r=\{u\in Q:\|u\|<r\},\quad Q(\kappa,a,b)
=\{u\in Q:\kappa(u)\geq a,\; \|u\|\leq b\}.
$$
Next, we have the Leggett-Williams fixed-point theorem, see \cite{25,26}.

\begin{lemma}\label{L2.10} 
Assume $E$ be a real Banach space and $Q\subset E$ be a cone.
 Let $T:\overline{Q}_c\to  \overline{Q}_c$ be completely continuous and 
$\kappa$ be a nonnegative continuous concave functional on $Q$ such that 
$\kappa(u)\leq\|u\|$, for $u\in \overline{Q}_c$. Suppose that there 
exist $0<a<b<d\leq c$ such that
\begin{itemize}
\item[(i)] $\{u\in Q(\kappa,b,d)~|~\kappa(u)>b\} \neq\emptyset$ and 
 $\kappa(Tu)>b$, for all $u\in Q(\kappa,b,d)$.
\item[(ii)] $\|Tu\|\leq a$, for all $\|u\|\leq a$.
\item[(iii)] $\kappa(Tu)\geq b$, for all $u\in Q(\kappa,b,c)$ with $\|Tu\|>d$.
\end{itemize}
Then $T$ has at least three fixed points $u_1,u_2$ and $u_3$ satisfying
$$
\|u_1\|<a,\quad b<\kappa(u_2), \quad a<\|u_3\|,\quad \kappa(u_3)<b.
$$
\end{lemma}

\section{Main results}\label{sect3}

To state the following results, we need to introduce the symbols.
\begin{gather*}
f^0=\limsup_{u\to 0}\max_{t\in[a,b]}\frac{f(t,u)}{\phi(u)},\quad
f^\infty=\limsup_{u\to \infty}\max_{t\in[a,b]}\frac{f(t,u)}{\phi(u)},\\
f_0=\liminf_{u\to 0}\min_{t\in[a,b]}\frac{f(t,u)}{\phi(u)},\quad
f_\infty=\liminf_{u\to \infty}\min_{t\in[a,b]}\frac{f(t,u)}{\phi(u)},\\
k_1=\rho_1\int_a^b G_1(s,s)ds\psi_2^{-1}
\Big(\rho_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\Big),\\
k_2=\sigma_1\int_a^b G_1(s,s)ds\psi_1^{-1}
\Big(\sigma_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\Big).
\end{gather*}

\begin{theorem}\label{th3.1}
Assume {\rm (H0)-(H5)} hold. Furthermore, suppose one of the following 
conditions are satisfied.
\begin{itemize}
\item[(A1)]  There exist two constants $r$ and $R$ with 
$0<r\leq\frac{k_1}{k_2}R$ such that $f(t,u)\geq\phi\big(\frac{r}{k_1}\big)$
for $(t,u)\in[a,b]\times[0,r]$, and $f(t,u)\leq\phi\left(\frac{r}{k_2}\right)$ 
for $(t,u)\in[a,b]\times[0,R]$.

\item[(A2)] $f_0>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and \\
$f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$\\ 
 (particularly, $f_0=\infty$ and $f^\infty=0$).

\item[(A3)] $f^0<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and \\
$f_\infty>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ 
(particularly, $f^0=0$ and $f_\infty=\infty$).
\end{itemize}
Then the BVP \eqref{e1.1} has at least one symmetric positive solution.
\end{theorem}

\begin{proof} Let the operator $T$ be defined by \eqref{e2.9}.

(A1) For $u\in\mathbb{P}\cap\partial\Omega_r$, we have $u\in[0,r]$, 
which implies $f(t,u)\geq\phi\big(\frac{r}{k_1}\big)$.
Hence for $t\in[a,b]$, by Lemma \ref{L2.3}, we obtain
\begin{align*}
u(t)
&\geq \rho_1\int_a^b G_1(s,s)\phi^{-1}
\Big(\rho_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\phi\big(\frac{r}{k_1}\big)\Big)ds\\
&\geq \rho_1\int_a^b G_1(s,s)ds\psi_2^{-1}
\Big(\rho_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\Big)\frac{r}{k_1}=r=\|u\|,
\end{align*}
which implies that for $u\in\mathbb{P}\cap\partial\Omega_r$,
\begin{equation}\label{e3.1}
\|Tu\|\geq\|u\|.
\end{equation}
Next, for $u\in\mathbb{P}\cap\partial\Omega_R$, we have $u\in[0,R]$,
 which implies $f(t,u)\leq\phi\big(\frac{R}{k_2}\big)$.
Hence for $t\in[a,b]$, by Lemma \ref{L2.3}, we obtain
\begin{align*}
u(t)
&\leq \sigma_1\int_a^b G_1(s,s)\phi^{-1}
\Big(\sigma_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\phi\big(\frac{R}{k_2}\big)\Big)ds\\
&\leq \sigma_1\int_a^b G_1(s,s)ds\psi_1^{-1}
\Big(\sigma_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\Big)\frac{R}{k_2}=R=\|u\|,
\end{align*}
which implies that for $u\in\mathbb{P}\cap\partial\Omega_R$
\begin{equation}\label{e3.2}
\|Tu\|\leq\|u\|.
\end{equation}

(A2) At first, in view of 
\[
f_0>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},
\]
 there exists $r>0$ such that $f(t,u)\geq(f_0-\varepsilon_1)\phi(\|u\|)$, 
for $t\in[a,b]$, $\|u\|\in[0,r]$, where $\varepsilon\geq0$ satisfies 
$$
\rho_1\int_a^bG_1(s,s)ds\psi_2^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau(f_0-\varepsilon_1)\Big)\geq1.
$$
Then, for $t\in[a,b]$, $u\in\mathbb{P}\cap\partial\Omega_r$, which implies 
$\|u\|\leq r$, we have
\begin{align*}
(Tu)(t)
&\geq \int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)(f_0-\varepsilon_1)\phi(\|u\|)d\tau\Big)ds\\
&\geq \rho_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau(f_0-\varepsilon_1)\phi(\|u\|)\Big)ds\\
&\geq \rho_1\int_a^bG_1(s,s)ds\psi_2^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau(f_0-\varepsilon_1)\Big)\|u\|
\geq\|u\|,
\end{align*}
which implies that for $u\in\mathbb{P}\cap\partial\Omega_r$,
\begin{equation}\label{e3.3}
\|Tu\|\geq\|u\|.
\end{equation}

Next, turning to 
\[
f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},
\]
 there exists $\overline{R}>0$ large enough such that 
$f(t,u)\leq(f^\infty+\varepsilon_2)\phi(\|u\|)$, for $t\in[a,b]$,
$\|u\|\in(\overline{R},\infty)$, where $\varepsilon_2>0$ satisfies 
$$
\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}-f^\infty-\varepsilon_2>0.
$$
Set $M=\max_{\|u\|\leq\overline{R},t\in[a,b]}f(t,u)$. Then
 $f(t,u)\leq M+(f^\infty+\varepsilon_2)\phi(\|u\|)$. Choose
\begin{align*}
R>&\max\Big\{r,\overline{R},\phi^{-1}\Big(M\Big[\psi_1
\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)\\
&\times\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}
-f^\infty-\varepsilon_2\Big]\Big)\Big\}.
\end{align*}
Hence for $u\in\mathbb{P}\cap\partial\Omega_R$, we have
\begin{align*}
(Tu)(t)
&\leq \sigma_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)(M+(f^\infty+\varepsilon_2)
 \phi(\|u\|))d\tau\Big)ds\\
&\leq \sigma_1\int_a^bG_1(s,s)ds\phi^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau
\Big(\frac{M}{\phi(R)}+f^\infty+\varepsilon_2\Big)\phi(R)\Big)\\
&\leq \sigma_1\int_a^bG_1(s,s)ds\psi_1^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau
\Big(\frac{M}{\phi(R)}+f^\infty+\varepsilon_2\Big)\Big)R\\
&\leq  R=\|u\|,
\end{align*}
which implies that for $u\in\mathbb{P}\cap\partial\Omega_R$,
\begin{equation}\label{e3.4}
\|Tu\|\leq\|u\|.
\end{equation}

(A3) Considering 
\[
f^0<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},
\]
there exists $r>0$ such that $f(t,u)\leq\eta\phi(u)$, for any 
$u\in[0,r]$, $t\in[a,b]$, where 
$$
\eta\leq\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.
$$
Then if $\Omega_r$ is the ball in $\mathbb{C}$ centered at the origin 
with radius $r$ and if $u\in\mathbb{P}\cap\partial\Omega_r$, then we have
\begin{align*}
\|Tu\|
&= \max_{t\in[a,b]}\int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\
&\leq \sigma_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)\eta\phi(u(\tau))d\tau\Big)ds\\
&\leq \sigma_1\int_a^bG_1(s,s)ds\phi^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)\eta\phi(r)d\tau\Big)\\
&\leq r\sigma_1\int_a^bG_1(s,s)ds\psi_1^{-1}
\Big(\eta\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)\leq r=\|u\|,
\end{align*}
which implies that for $u\in\mathbb{P}\cap\partial\Omega_r$
\begin{equation}\label{e3.5}
\|Tu\|\leq\|u\|.
\end{equation}

On the other hand, we use the assumption 
$$
f_\infty>\psi_2
\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.
$$ 
Then there exists $R>0$ large enough such that $f(t,u)\geq\varrho\phi(u)$ 
for any $u\in[R,\infty)$, $t\in[a,b]$, where 
\[
\varrho\geq\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.
\]
If we define $\Omega_R=\{u\in\mathbb{C}:\|u\|<R\}$, for $t\in[a,b]$ 
and $u\in\mathbb{P}\cap\partial\Omega_R$, we get
\begin{align*}
(Tu)(t)
&\geq \rho_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)\varrho\phi(u(\tau))d\tau\Big)ds\\
&\geq \rho_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)\varrho\phi(R)d\tau\Big)ds\\
&\geq R\rho_1\int_a^bG_1(s,s)ds\psi_2^{-1}
\Big(\varrho\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)\geq r=\|u\|,
\end{align*}
which implies that for $u\in\mathbb{P}\cap\partial\Omega_R$
\begin{equation}\label{e3.6}
\|Tu\|\geq\|u\|.
\end{equation}

Applying Lemma \ref{L2.9} to \eqref{e3.1} and \eqref{e3.2}, \eqref{e3.3} 
and \eqref{e3.4} or \eqref{e3.5} and \eqref{e3.6} yields that $T$ 
has a fixed point $u\in\mathbb{P}\cap\overline{\Omega}_{r,R}$ 
with $0\leq r\leq\|u\|\leq R$. It follows from Lemma \ref{L2.9} 
that problem \eqref{e1.1} has at least one symmetric positive solution $u$.
 \end{proof}

\begin{theorem}\label{th3.2}
Assume {\rm (H0)-(H5)} hold. Furthermore, suppose one of the following 
conditions is satisfied.
\begin{itemize}
\item[(A4)] \begin{itemize}
\item[(i)]  $f_0>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
 \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and
    $f_\infty>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
  \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$  \\
  (particularly, $f_0=f_\infty=\infty$).
\item[(ii)] There exists $c>0$ satisfying 
$f(t,u)<\phi\big(\frac{c}{k_2}\big)$, $(t,b)\in[a,b]\times[0,c]$.
\end{itemize}
  \item[(A5)] \begin{itemize}
\item[(i)]  $f^0<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and 
$f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\
 (particularly, $f^0=f^\infty=0$).
\item[(ii)] There exists $c>0$ satisfying 
$f(t,u)>\phi\big(\frac{c}{k_1}\big)$, $(t,b)\in[a,b]\times[0,c]$.
\end{itemize}
\end{itemize}
Then \eqref{e1.1} has at least two symmetric positive solutions $u_1(t)$ 
and $u_2(t)$, which satisfy $0<\|u_1\|<c<\|u_2\|$.
\end{theorem}

\begin{proof} (A4) Consider (i). If 
$$
f_0>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},
$$
it follows from the proof of \eqref{e3.3} that we can choose $r$ with 
 $0<r<c$ such that
\begin{equation}\label{e3.7}
\|Tu\|\geq\|u\|, \quad\text{for }u\in\mathbb{P}\cap\partial\Omega_r.
\end{equation}
If 
\[
f_\infty>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},
\]
 then as in the proof of \eqref{e3.6}, we can choose $R$ with $c<R$ such that
\begin{equation}\label{e3.8}
\|Tu\|\geq\|u\|, \quad\text{for }u\in\mathbb{P}\cap\partial\Omega_R.
\end{equation}
Next, for $u\in\mathbb{P}\cap\partial\Omega_c$, we have $u\in[0,c]$, 
then from (ii), we obtain $f(t,u)<\phi\big(\frac{c}{k_2}\big)$. 
Thus for $t\in[a,b]$, like in the proof of \eqref{e3.2}, we get
\begin{equation}\label{e3.9}
\|Tu\|\leq\|u\|, \quad\text{for }u\in\mathbb{P}\cap\partial\Omega_c.
\end{equation}
Applying Lemma \ref{L2.9} to \eqref{e3.7} and \eqref{e3.9}, 
or \eqref{e3.2} and \eqref{e3.9} yields that $T$ has a fixed point
$u_1\in\mathbb{P}\cap\overline{\Omega}_{r,c}$, and a fixed point 
$u_2\in\mathbb{P}\cap\overline{\Omega}_{c,R}$.
It follows from Lemma \ref{e2.7} that problem \eqref{e1.1} 
has at least two symmetric positive solutions $u_1(t)$ and $u_2(t)$, 
which satisfy $0<\|u_1\|<c<\|u_2\|$.

(A5) It can be proved in a way similar to (A2) and (A3) of 
Theorem \ref{th3.1} and (A4) of Theorem \ref{th3.2}. 
The proof is complete.
\end{proof}


Now we define the nonnegative, continuous concave functional
 $\varphi: \mathbb{P}\to [0,\infty)$ by
$\varphi(u)=\min_{t\in[a,b]}u(t)$. Obviously, for every $u\in\mathbb{P}$, 
we have $\varphi(u)\leq\|u\|$.

\begin{theorem}\label{th3.3}
Assume {\rm (H0)--(H5)} hold. In addition, there exist three positive 
constants $x,y$ and $z$ with $0<x<y<\omega z$ such that
\begin{itemize}
\item[(A6)] $f(t,u)<\phi\big(\frac{z}{k_2}\big)$, for all $t\in[a,b]$,
  and $0\leq u\leq z$.
\item[(A7)] $f(t,u)\geq\phi\big(\frac{y}{k_1}\big)$, for all 
 $t\in[a,b]$, and $y\leq u\leq \frac{y}{\omega}$.
\item[(A8)] $f(t,u)\leq\phi\big(\frac{x}{k_2}\big)$, for all
$t\in[a,b]$, and $0\leq u\leq x$.
\end{itemize}
Then \eqref{e1.1} has at least three symmetric positive solutions 
$u_1(t), u_2(t)$ and $u_3(t)$ such that
$$
\|u_1\|<x,\quad y<\varphi(u_2), \quad x<\|u_3\|,\quad \varphi(u_3)<y.
$$
\end{theorem}

\begin{proof} We show that all the conditions of Lemma \ref{L2.10} are satisfied. 
We first assert that there exists a positive number $z$ such that 
$T(\overline{Q}_z)\subset\overline{Q}_z$. By (A6), we obtain
\begin{align*}
\|Tu\|
&= \max_{t\in[a,b]}(Tu)(t)\\
&=\max_{t\in[a,b]}\int_a^bH_1(t,s)\phi^{-1}
\Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\
&\leq \sigma_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)\phi\big(\frac{z}{k_2}\big)d\tau\Big)ds\\
&\leq \frac{z\sigma_1}{k_2}\int_a^bG_1(s,s)ds\psi_1^{-1}
\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)=z.
\end{align*}
Therefore, we have $T(\overline{Q}_z)\subset\overline{Q}_z$.
 Especially, if $u\in\overline{Q}_x$, then assumption (A8) yields
$T:\overline{Q}_x\to \overline{Q}_x$.

We now show that condition (i) of Lemma \ref{L2.10} is satisfied. 
Clearly, $\{u\in Q(\varphi,y,\frac{y}{\omega}
):\varphi(u)>y\}\neq\emptyset$. 
Moreover, if $u\in Q(\varphi,y,\frac{y}{\omega})$, then 
$\varphi(u)\geq y$, so $y\leq\|u\|\leq\frac{y}{\omega}$. 
By the definition of $\varphi$ and (A7), we obtain
\begin{align*}
\|Tu\|
&= \min_{t\in[a,b]}(Tu)(t)\\
&=\min_{t\in[a,b]}\int_a^bH_1(t,s)\phi^{-1}
 \Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\
&\geq \rho_1\int_a^bG_1(s,s)\phi^{-1}
 \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)\phi\big(\frac{y}{k_1}\big)d\tau\Big)ds\\
&\geq \frac{y\rho_1}{k_1}\int_a^bG_1(s,s)ds\psi_2^{-1}
 \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)=y.
\end{align*}
Therefore, condition (i) of Lemma \ref{L2.10} is satisfied.

Finally, we address condition  $(\rm iii)$ of Lemma \ref{L2.10}.
 For this we choose $u\in Q(\varphi,y,z)$ with $\|Tu\|>\frac{y}{\omega}$. 
Then from Lemma \ref{L2.6}, we deduce
$$
\varphi(Tu)=\min_{t\in[a,b]}(Tu)(t)\geq\omega\|Tu\|>y.
$$
Hence, $(\rm iii)$ of Lemma  \ref{L2.10}  holds. By Lemma \ref{L2.10}, 
then we obtain that \eqref{e1.1} has at least three symmetric positive 
solutions $u_1(t), u_2(t)$ and $u_3(t)$ such that
$$
\|u_1\|<x,\quad y<\varphi(u_2), \quad x<\|u_3\|,\quad \varphi(u_3)<y.
$$
The proof is complete. 
\end{proof}

\begin{theorem}\label{th3.4}
Assume {\rm (H0)--(H5)} hold. Furthermore, suppose one of the 
following conditions are satisfied:
\begin{itemize}
\item[(A9)]  $f(t,u)>\phi\big(\frac{\|u\|}{k_1}\big)$ 
for all $t\in[a,b], u\in[0,\infty)$.
\item[(A10)] $f(t,u)<\phi\big(\frac{\|u\|}{k_2}\big)$ 
for all $t\in[a,b], u\in[0,\infty)$.
\end{itemize}
Then  \eqref{e1.1} has no positive solution.
\end{theorem}

\begin{proof} 
Assume $u(t)$ is a positive solution of \eqref{e1.1}, we have
\begin{align*}
\|u\|&= \|Tu\|\geq(Tu)(t)\\
>&\rho_1\int_a^bG_1(s,s)\phi^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)\phi\big(\frac{\|u\|}{k_1}\big)d\tau
 \Big)ds\\
&\geq \frac{\|u\|\rho_1}{k_1}\int_a^bG_1(s,s)ds\psi_2^{-1}
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)=\|u\|.
\end{align*}
which is a contradiction. So, due to (A9), equation \eqref{e1.1} has no 
positive solution. Similarly, due to (A9), we obtain that \eqref{e1.1} 
has no positive solution. 
\end{proof}

\section{Further remarks}\label{sect4}

Consider the  fourth-order integral BVP with with $\phi$-Laplacian operator
\begin{equation} \label{e4.1}
\begin{gathered}
 \big[q(t)\phi(p(t)u''(t))\big]''=w(t)f(t,u(t)),\quad t\in[a,b],\\
\alpha u(a)-\beta u'(a)=\int_a^bg(s)u(s)ds,\quad 
  \alpha u(b)+\beta u'(b)=\int_a^bg(s)u(s)ds,\\
\gamma q(a)\phi(p(a)u''(a))-\delta [q(a)\phi(p(a)u''(a))]'
 =\int_a^bh(s)q(s)\phi(p(s)u''(s))ds,\\
\gamma q(b)\phi(p(b)u''(b))+\delta [q(b)\phi(p(b)u''(b))]'
 =\int_a^bh(s)q(s)\phi(p(s)u''(s))ds.
\end{gathered}
\end{equation}

By methods analogous to the ones abov, we have the following results.

\begin{lemma}\label{L4.1}
Assume {\rm (H0)--(H2)} hold and $\mu\neq\alpha$. 
Then for any $v\in C[a,b]$, the BVP
\begin{equation} \label{e4.2}
\begin{gathered}
 q(t)\phi(p(t)u''(t))=v(t),\quad t\in[a,b],\\
 \alpha u(a)-\beta u'(a)=\int_a^bg(s)u(s)ds,\quad  
\alpha u(b)+\beta u'(b)=\int_a^bg(s)u(s)ds,
\end{gathered}
\end{equation}
has a unique solution $u$ and $u$ can be expressed in the form
\begin{equation}\label{e4.3}
u(t)=-\int_a^bH_1^\ast(t,s)\frac{1}{p(s)}\phi^{-1}
\Big(\frac{v(s)}{q(s)}\Big)ds,
\end{equation}
where
\begin{gather}\label{e4.4}
H_1^\ast(t,s)=G_1^\ast(t,s)+\frac{1}{\alpha-\mu}
\int_a^bG_1^\ast(s,\tau)g(\tau)d\tau, \\
\label{e4.5}
G_1^\ast(t,s)=\frac{1}{\Delta_1^\ast}
\begin{cases}
 [\beta+\alpha(s-a)][\beta+\alpha(b-t)], &a\leq s\leq t\leq b,\\
 [\beta+\alpha(t-a)][\beta+\alpha(b-s)], &a\leq t\leq s\leq b.
\end{cases}
\end{gather}
Here 
$$
\Delta_1^\ast=\alpha[2\beta+\alpha(b-a)], \quad\mu=\int_a^bg(s)ds.
$$
\end{lemma}

\begin{lemma}\label{L4.2}
Assume {\rm (H1)--(H4)} hold and $\nu\neq\gamma$. Then for any 
$u\in C[a,b]$, the BVP
\begin{gather*}
 v''(t)=w(t)f(t,u(t)),\quad t\in[a,b],\\
 \gamma v(a)-\delta v'(a)=\int_a^bh(s)v(s)ds,\quad  
\gamma v(b)+\delta v'(b)=\int_a^bh(s)v(s)ds,
\end{gather*}
has a unique solution $v$ that can be expressed as
\[
v(t)=-\int_a^bH_2^\ast(t,s)w(s)f(s,u(s))ds,
\]
where
\begin{gather*}\label{e4.6}
H_2^\ast(t,s)=G_2^\ast(t,s)
+\frac{1}{\gamma-\nu}\int_a^bG_2^\ast(s,\tau)h(\tau)d\tau, \\
\label{e4.7}
G_2^\ast(t,s)=\frac{1}{\Delta_2^\ast} \begin{cases}
 [\delta+\gamma(s-a)][\delta+\gamma(b-t)], &a\leq s\leq t\leq b,\\
 [\delta+\gamma(t-a)][\delta+\gamma(b-s)], &a\leq t\leq s\leq b.
\end{cases}
\end{gather*}
Here 
$$
\Delta_2^\ast=\gamma[2\delta+\gamma(b-a)], \quad\nu=\int_a^bh(s)ds.
$$
\end{lemma}

\begin{lemma}\label{L4.3}
Assume {\rm (H5)} holds. Then  for any $t,s\in[a,b]$, 
the following results are true.
\begin{itemize}
\item[(i)] $G_1^\ast(t,s)\geq0$, $H_1^\ast(t,s)\geq0$, 
$G_2^\ast(t,s)\geq0$, $H_2^\ast(t,s)\geq0$;

\item[(ii)] $\frac{\beta^2}{\Delta_1^\ast}\leq G_1^\ast(t,s)
\leq G_1^\ast(s,s)\leq\frac{\Lambda_1^\ast}{\Delta_1^\ast}$, 
$\frac{\beta^2\sigma_1}{\Delta_1^\ast}\leq H_1^\ast(t,s)
\leq H_1^\ast(s,s)\leq\frac{\sigma_1\Lambda_1^\ast}{\Delta_1^\ast}$, 
$\frac{\delta^2}{\Delta_2^\ast}\leq G_2^\ast(t,s)
\leq G_2^\ast(s,s)\leq\frac{\Lambda_2^\ast}{\Delta_2^\ast}$, 
$\frac{\delta^2\sigma_2}{\Delta_2^\ast}\leq H_2^\ast(t,s)
\leq H_2^\ast(s,s)\leq\frac{\sigma_2\Lambda_2^\ast}{\Delta_2^\ast}$ 
with
    $$
\Lambda_1^\ast=[\beta+\alpha(b-a)]^2,\quad \sigma_1
=\frac{\alpha}{\alpha-\mu},\quad \Lambda_2^\ast=[\delta+\gamma(b-a)]^2,\quad 
\sigma_2=\frac{\gamma}{\gamma-\nu};
$$

\item[(iii)] 
$G_1^\ast(b-t+a,b-s+a)=G_1^\ast(t,s)$, 
$H_1^\ast(b-t+a,b-s+a)=H_1^\ast(t,s)$, 
$G_2^\ast(b-t+a,b-s+a)=G_2^\ast(t,s)$, $H_2^\ast(b-t+a,b-s+a)=H_2^\ast(t,s)$;

\item[(iv)] $\rho_1G_1^\ast(s,s)\leq H_1^\ast(t,s)\leq\sigma_1 G_1^\ast(s,s)$, 
$\rho_2G_2^\ast(s,s)\leq H_2^\ast(t,s)\leq\sigma_2 G_2^\ast(s,s)$ with
$$
\rho_1^\ast=\frac{\Delta_1^\ast}{(\alpha-\mu)\Lambda_1^\ast}
\int_a^bG_1^\ast(s,s)g(s)ds,\quad  \rho_2^\ast
=\frac{\Delta_2^\ast}{(\gamma-\nu)\Lambda_2^\ast}\int_a^bG_1^\ast(s,s)h(s)ds;
$$
\end{itemize}
where $H_1^\ast(t,s)$, $G_1^\ast(t,s)$, $H_2^\ast(t,s)$ and $G_2^\ast(t,s)$ 
are defined by \eqref{e4.4}--\eqref{e4.7}, respectively.
\end{lemma}

\begin{lemma}\label{L4.4}
Assume {\rm (H0)--(H5)} hold. If $u$ is a solution of \eqref{e4.1}, then
\[
u(t)=\int_a^bH_1^\ast(t,s)\phi^{-1}
\Big(\int_a^bH_2^\ast(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds.
\]
\end{lemma}

\begin{lemma}\label{L4.5}
Assume {\rm (H0)-(H5)} hold. the solution $u(t)$ of  \eqref{e4.1} 
is positive and symmetric on $[a,b]$
and $\min_{t\in[a,b]}u(t)\geq\omega^\ast\|u\|$, where
 $\omega^\ast=\frac{\rho_1^\ast\psi_2^{-1}(\rho_2^\ast)}{\sigma_1\psi_1^{-1}
(\sigma_2)}$.
\end{lemma}

Let $\mathbb{P}^\ast=\{x\in\mathbb{C} : u(t)\geq0$, $u''(t)\leq0$,
 $u(t)$ is a symmetric and concave function on $[a,b]$ and 
$u(t)\geq\omega^\ast\|u\|\}$.
We introduce the integral operator $T^\ast:\mathbb{C}\to \mathbb{C}$ by
\begin{equation}\label{e4.8}
(T^\ast u)(t)=\int_a^bH_1^\ast(t,s)\phi^{-1}
\Big(\int_a^bH_2^\ast(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds.
\end{equation}

\begin{lemma}\label{L4.6}
Assume {\rm (H0)--(H5)} hold. Then $u\in\mathbb{C}$ is a solution of 
\eqref{e1.1} if and only if $u$ is a fixed point of the operator $T^\ast$.
\end{lemma}

\begin{lemma}\label{L4.7}
Assume {\rm (H0)-(H5)} hold. Then $T^\ast:\mathbb{P}\to \mathbb{P}$ 
is a completely continuous operator.
\end{lemma}

Now we need to introduce new notation.
\begin{gather*}
k_1^\ast=\rho_1^\ast\int_a^b G_1^\ast(s,s)ds\psi_2^{-1}
\Big(\rho_2^\ast\int_a^b G_2^\ast(\tau,\tau)w(\tau)d\tau\Big),\\
k_2^\ast=\sigma_1\int_a^b G_1^\ast(s,s)ds\psi_1^{-1}
\Big(\sigma_2\int_a^b G_2^\ast(\tau,\tau)w(\tau)d\tau\Big).
\end{gather*}

\begin{theorem}\label{th4.1}
Assume {\rm (H0)--(H5)} hold. Furthermore, suppose one of the 
following conditions are satisfied.
\begin{itemize}
\item[(A1*)]  There exist two constants $r$ and $R$ with 
$0<r\leq\frac{k_1^\ast}{k_2^\ast}R$ such that 
$f(t,u)\geq\phi\big(\frac{r}{k_1^\ast}\big)$ for $(t,u)\in[a,b]\times[0,r]$,
 and $f(t,u)\leq\phi\big(\frac{r}{k_2^\ast}\big)$ for $(t,u)\in[a,b]\times[0,R]$.
\item[(A2*)] $f_0>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and\\
$f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$  \\
(particularly, $f_0=\infty$ and $f^\infty=0$).
\item[(A3*)] $f^0<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and\\
$f_\infty>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\
 (particularly, $f^0=0$ and $f_\infty=\infty$).
\end{itemize}
Then  \eqref{e4.1} has at least one symmetric positive solution.
\end{theorem}

\begin{theorem}\label{th4.2}
Assume {\rm (H0)--(H5)} hold. Furthermore, suppose one of the 
following conditions are satisfied.
\begin{itemize}
\item[(A4*)] \begin{itemize}
\item[(i)]  $f_0>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and \\
$f_\infty>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\
 (particularly, $f_0=f_\infty=\infty$).
\item[(ii)] There exists $c>0$ satisfying 
$f(t,u)<\phi\big(\frac{c}{k_2^\ast}\big)$, $(t,b)\in[a,b]\times[0,c]$.
\end{itemize}
  \item[(A5*)] \begin{itemize}
\item[(i)]  $f^0<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and\\
$f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big)
\Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$  \\
(particularly, $f^0=f^\infty=0$).
\item[(ii)] There exists $c>0$ satisfying 
$f(t,u)>\phi\Big(\frac{c}{k_1^\ast}\Big)$, $(t,b)\in[a,b]\times[0,c]$.
\end{itemize}
\end{itemize}
Then  \eqref{e4.1} has at least two symmetric positive solutions
 $u_1(t)$ and $u_2(t)$, which satisfy $0<\|u_1\|<c<\|u_2\|$.
\end{theorem}

\begin{theorem}\label{th4.3}
Assume {\rm (H0)--(H5)} hold. In addition, there exist three positive 
constants $x,y$ and $z$ with $0<x<y<\omega^\ast z$ such that
\begin{itemize}
\item[(A6*)] $f(t,u)<\phi\big(\frac{z}{k_2^\ast}\big)$, for all 
 $t\in[a,b]$, and $0\leq u\leq z$.
\item[(A7*)] $f(t,u)\geq\phi\big(\frac{y}{k_1^\ast}\big)$, for all 
 $t\in[a,b]$, and $y\leq u\leq \frac{y}{\omega^\ast}$.
\item[(A8*)] $f(t,u)\leq\phi\big(\frac{x}{k_2^\ast}\big)$, for all 
 $t\in[a,b]$, and $0\leq u\leq x$.
\end{itemize}
Then  \eqref{e4.1} has at least three symmetric positive solutions
 $u_1(t), u_2(t)$ and $u_3(t)$ such that
$$
\|u_1\|<x,\quad y<\varphi(u_2), \quad x< \|u_3\|,\quad \varphi(u_3)<y.
$$
\end{theorem}

\begin{theorem}\label{th4.4}
Assume {\rm (H0)-(H5)} hold. Furthermore, suppose one of the following 
conditions are satisfied.
\begin{itemize}
\item[(A9*)]  $f(t,u)>\phi\big(\frac{\|u\|}{k_1^\ast}\big)$ for all
  $t\in[a,b], u\in[0,\infty)$.
\item[(A10*)] $f(t,u)<\phi\big(\frac{\|u\|}{k_2^\ast}\big)$ for all
  $t\in[a,b], u\in[0,\infty)$.
\end{itemize}
Then \eqref{e4.1} has no positive solution.
\end{theorem}

\begin{remark} \label{re4.1} \rm
$\phi$ defined as in (H0) generalizes the projection 
$\varphi:\mathbb{R}\to \mathbb{R}$ which is an increasing homeomorphism 
and homomorphism with $\varphi(0)=0$. A projection 
$\varphi:\mathbb{R}\to \mathbb{R}$, which generates 
the $p$-Laplacian operator $\varphi_p(u)=|u|^{p-2}u$ for $p>1$, 
is called an increasing homeomorphism and homomorphism if the 
following conditions are satisfied:
\begin{itemize}
\item[(i)] If $x\leq y$, then $\varphi(x)\leq\varphi(y)$, 
for all $x,y\in\mathbb{R}$; 
\item[(ii)] $\varphi$ is a continuous 
bijection and its inverse mapping is also continuous; 
\item[(iii)] $\varphi(xy)=\varphi(x)\varphi(y)$, for all 
$x,y\in\mathbb{R}$.  If $\phi$ is replaced by $\varphi$, 
so all of our results also hold.
\end{itemize}
\end{remark}


\section{Examples}\label{sect5}

\begin{example} \label{ex5.1} \rm
Consider the following fourth-order integral BVP with $\phi$-Laplacian operator
\begin{equation} \label{e5.1}
   \begin{gathered}
 \Big[\frac{1}{1+\sin(\pi t)}\Big[\phi\Big(\big(\frac{u'(t)}{1+\sin(\pi t)}
\big)'\Big)\Big]'\Big]'=w(t)f(t,u(t)),\quad t\in[0,1],\\
 u(0)-u'(0)=\frac{1}{2}\int_0^1u(s)ds,\quad 
 u(1)+u'(1)=\frac{1}{2}\int_0^1u(s)ds,\\
 \phi((u'(0))')-[\phi((u'(0))')]'
=\frac{1}{2}\int_0^1\phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds,\\
 \phi((u'(1))')+[\phi((u'(1))')]'
=\frac{1}{2}\int_0^1\phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds,
\end{gathered}
\end{equation}
where $\phi(u)=|u|u$, $w(t)=1/100$ and $f(t,u)=(1+\sin(\pi t))(1+u)$ 
for $(t,u)\in[0,1]\times[0,\infty)$.

Let $\psi_1(u)=\psi_2(u)=u^2$, $u>0$. Then, by calculations we obtain 
that $\mu=\nu=1/2$, $\Delta=\Delta_1=\Delta_2=3+\frac{2}{\pi}$,
\begin{align*}
&G_1(t,s)=G_2(t,s)\\
&=\frac{1}{\Delta} \begin{cases}
 \big(1+\frac{1}{\pi}+s-\frac{1}{\pi}\cos(\pi s)\big)
\big(2+\frac{1}{\pi}-t+\frac{1}{\pi}\cos(\pi t)\big),
&0\leq s\leq t\leq 1,\\
 \big(1+\frac{1}{\pi}+t-\frac{1}{\pi}\cos(\pi t)\big)
\big(2+\frac{1}{\pi}-s+\frac{1}{\pi}\cos(\pi s)\big),
&0\leq t\leq s\leq 1,
\end{cases}
\end{align*}
\begin{gather*}
\rho_1=\rho_2
=\frac{(2+3\pi)(-24+3\pi+18\pi^2+13\pi^3)}{24\pi^2(1+\pi)^2}\approx1.59199,
\quad \sigma_1=\sigma_2=2,\\
\omega\approx0.710176,\quad
k_1=\frac{(2+3\pi)^\frac{3}{2}(-24+3\pi+18\pi^2+13\pi^3)^3}
 {17280\pi^\frac{15}{2}(1+\pi)^3}\approx1.06639,\\
k_2=\frac{1}{10\pi^\frac{3}{2}}
 \Big(\frac{13}{3}-\frac{8}{\pi^3}+\frac{1}{\pi^2}+\frac{6}{\pi}\Big)
\sqrt{\frac{1}{3}(-24+3\pi+18\pi^2+13\pi^3)}\approx1.50159.
\end{gather*}
Clearly, the conditions (H0)--(H5) hold. Next, we prove that the condition 
(A1) of Theorem \ref{th3.1} is satisfied. In fact, choosing $r=1$ and 
$R=6$, we have $r<\frac{k_1}{k_2}R$, 
$\phi\big(\frac{r}{k_1}\big)\approx0.879356$, 
$\phi\big(\frac{R}{k_2}\big)\approx15.9661$. 
For $(t,u)\in[0,1]\times[0,1]$, then 
$f(t,u)=(1+\sin(\pi t))(1+u)\geq1>0.879356
\approx\phi\big(\frac{r}{k_1}\big)$.

For $(t,u)\in[0,1]\times[0,10]$, then 
$f(t,u)=(1+\sin(\pi t))(1+u)\leq14<15.9661 \approx\phi\big(\frac{R}{k_2}\big)$.
Hence, by (A1) of Theorem \ref{th3.1}, then BVP \eqref{e5.1} has at 
least one symmetric positive solution.
\end{example}

\begin{example} \label{ex5.2} \rm
Consider the following fourth-order integral BVP with $\phi$-Laplacian operator
\begin{equation} \label{e5.2}
\begin{gathered}
 \Big[\frac{1}{1+t(1-t)}\Big[\phi\Big(\big(\frac{u'(t)}{1+t(1-t)}\big)'\Big)
\Big]'\Big]'
=w(t)f(t,u(t)),\quad t\in[0,1],\\
 u(0)-u'(0)=\int_0^1su(s)ds,\quad  u(1)+u'(1)=\int_0^1su(s)ds,\\
 \phi((u'(0))')-[\phi((u'(0))')]'=
\int_0^1s \phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds,\\
 \phi((u'(1))')+[\phi((u'(1))')]'
=\int_0^1s\phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds,
\end{gathered}
\end{equation}
where $\phi(u)=u$, $w(t)=t^2(1-t)^2$ and
 $f(t,u)=\big(\big(t-\frac{1}{2}\big)^2+1\big)\big(\frac{1}{2}+u\big)
\big(\frac{1}{4}+u^2\big)$ for $(t,u)\in[0,1]\times[0,\infty)$.

Let $\psi_1(u)=\psi_2(u)=u$, $u>0$. Then, by calculations we obtain 
that $\mu=\nu=1/2$, $\Delta=\Delta_1=\Delta_2=19/6$,
\begin{align*}
&G_1(t,s)=G_2(t,s)\\
&=\frac{1}{\Delta} \begin{cases}
 \big(1-\frac{s}{6}(2s^2-3s-6)\big) 
 \big(\frac{13}{6}+\frac{t}{6}(2t^2-3t-6)\big),
&0\leq s\leq t\leq 1,\\
 \big(1-\frac{t}{6}(2t^2-3t-6)\big)
 \big(\frac{13}{6}+\frac{s}{6}(2s^2-3s-6)\big),
&0\leq t\leq s\leq 1,
\end{cases}
\end{align*}
\begin{gather*}
\rho_1=\rho_2=\frac{1745663}{1171170}\approx1.49053, \quad
\sigma_1=\sigma_2=2,\quad 
\omega=\frac{3047339309569}{5486556675600}\approx0.555419,\\
k_1=\frac{263200328806935685377871}{711426606363060588000000}\approx0.369961,
\quad k_2=\frac{86370535759}{129667230000}\approx0.666094.
\end{gather*}
Clearly, the conditions (H0)-(H5) hold. Next, we prove that the condition 
(A4) of Theorem \ref{th3.2} is satisfied.
$$
f_0=f_\infty=\infty>2.70299\approx
\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big)
\Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.
$$
On the other hand, choosing $c=\frac{1}{2}$, for $(t,u)\in[0,1]\times[0,c]$, 
we have $f(t,u)\leq f(0,c)=0.625<0.750645\approx\phi\big(\frac{c}{k_2}\big)$.
Hence, by (A4) of Theorem \ref{th3.2},  BVP \eqref{e5.2} has at 
least two symmetric positive solutions $u_1$ and $u_2$ satisfying 
$0<\|u_1\|<\frac{1}{2}<\|u_2\|$.
\end{example}

\begin{example} \label{ex5.3} \rm
Consider the BVP \eqref{ex5.2} with
\begin{equation}\label{e5.3}
f(t,u)= \begin{cases}
 \frac{1}{t(1-t)+79u}+8u^3, &0\leq t\leq 1,\; u\leq1,\\
 \frac{u}{t(1-t)+79u}+u+7, &0\leq t\leq 1,\; 1<u,
\end{cases}
\end{equation}
and other conditions also hold. Choosing $x=1/8$, $y=2$ and $z=20$, 
then $0<x<y<\omega z$.
Now, we can verify the validity of conditions (A6)-(A8) in 
Theorem \ref{th3.3}. Indeed, by direct computations, we have
\[
f(t,u)\leq\frac{1}{79}+20+7\approx27.0127<30.0258
\approx\phi\big(\frac{z}{k_2}\big),
\]
for all $t\in[0,1]$, and $0\leq u\leq z$.
\[
 f(t,u)\geq\frac{1}{80}+2+7=9.0125>5.40598\approx\phi\big(\frac{y}{k_1}\big),
\]
 for all $t\in[0,1]$, and $y\leq u\leq \frac{y}{\omega}$.
\[
f(t,u)\leq\frac{1}{79}+\frac{8}{8^3}\approx0.0282832<0.187661
\approx\phi\big(\frac{x}{k_2}\big),
\]
for all $t\in[0,1]$, and $0\leq u\leq x$.
Thus, according to Theorem \ref{th3.3},  BVP \eqref{e5.2} with \eqref{e5.3} 
has at least three positive solutions $u_1$, $u_2$, and $u_3$ satisfying
$$
\|u_1\|<\frac{1}{8},\quad 2<\varphi(u_2), \quad  \frac{1}{8}<\|u_3\|,\quad
 \varphi(u_3)<2.
$$
\end{example}

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\end{document}