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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 336, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/336\hfil 
 Existence of positive symmetric solutions]
{Existence of positive symmetric solutions for an integral
boundary-value problem with $\phi$-Laplacian operator}

\author[Y. Ding \hfil EJDE-2016/336\hfilneg]
{Yonghong Ding}

\address{Yonghong Ding \newline
Department of Mathematics,
Tianshui Normal University,
Tianshui 741000, China}
\email{dyh198510@126.com}

\thanks{Submitted May 18, 2016. Published December 28, 2016.}
\subjclass[2010]{34B15, 34B18}
\keywords{$\phi$-Laplacian; fixed point; cone; positive symmetric solutions}

\begin{abstract}
 In this article, we show the existence of three positive symmetric
 solutions for the integral boundary-value problem with $\phi$-Laplacian
 \begin{gather*}
 (\phi(u'(t)))'+f(t,u(t),u'(t))=0,\quad t\in[0,1],\\
 u(0)=u(1)=\int_0^1u(r)g(r)\,\mathrm{d}r,
 \end{gather*}
 where $\phi$ is an odd, increasing homeomorphism from $\mathbb{R}$ onto
 $\mathbb{R}$. Our main tool is a fixed point theorem due to
 Avery and Peterson. An example shows an applications of the obtained results.
\end{abstract}

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

\section{Introduction}

The aim of this article is to show the existence of positive symmetric
solutions for the integral boundary-value problem
\begin{equation}
 \begin{gathered}
 (\phi(u'(t)))'+f(t,u(t),u'(t))=0,\quad t\in[0,1],\\
 u(0)=u(1)=\int_0^1u(r)g(r)\,\mathrm{d}r,
 \end{gathered} \label{e1.1}
\end{equation}
where $\phi,f,g$ satisfy the following assumptins:
\begin{itemize}
\item[(H1)] $\phi$ is an odd, increasing homeomorphism from
$\mathbb{R}$ onto $\mathbb{R}$, and
there exist two increasing homeomorphism $\psi_1$ and $\psi_2$
of $(0, \infty)$ onto $(0, \infty)$ such that
$$
\psi_1(u)\phi(v) \leq \phi(uv) \leq
\psi_2(u)\phi(v) \quad\forall u, v > 0.
$$
Moreover,
$\phi, \phi^{-1}\in C^{1}(\mathbb{R})$, where
$\phi^{-1}$ denotes the inverse of $\phi$.

\item[(H2)] $f:[0, 1]\times [0, +\infty)\times
(-\infty, +\infty)\to (0, +\infty)$ is continuous, and
$$
f(t,u,v)=f(1-t,u,-v),\quad
 (t,u,v)\in[0, 1]\times[0, +\infty)\times(-\infty, +\infty).
$$

\item[(H3)] $ g\in L^{1}[0, 1]$ is nonnegative, and
$0<\int_0^1g(t)\,\mathrm{d}t<1$, $g(t)=g(1-t)$, $t\in[0, 1]$.

\end{itemize}
Condition (H1) was first introduced by
Wang \cite{w1,w2}, and it includes two important cases when $\phi(u)=u$ and
$\phi(u)=|u|^{p-2}u$, $p>1$.
Many authors have studied the positive solutions for two-point and multi-point
boundary-value problems when $\phi(u)=u$ and
$\phi(u)=|u|^{p-2}u$, $p>1$, see \cite{a2,a3,f1,g1,i1,j1,j2,k1,k2,l1,o1,w3,w4}
and references therein.
In 1997, Wang \cite{w3} proved the existence of at least one positive
for the equation
\begin{equation}
(\phi_{p}(u'(t)))'+a(t)f(u(t))=0, t\in(0,\;1), \label{e1.2}
\end{equation}
where $\phi_{p}(s)=|s|^{p-2}s$, $p>1$.
The technique used there is the well-known Krasnoselskii's
fixed point theorem.
In \cite{k1}, by using the upper and lower solutions method associated
with Leray-Schauder degree theory, Kim showed the existence of three-solutions
to the boundary-value problem
\begin{equation}
\begin{gathered}
 (w(t)\phi_{p}(u'(t)))'+\lambda f(t,u(t),u'(t))=0, \quad t\in(0,\;1),\\
 u(0)=u(1)=0,
 \end{gathered} \label{e1.3}
\end{equation}
where $\phi_{p}(s)=|s|^{p-2}s$, $p>1$.

Boundary-value problem with integral boundary conditions constitute a very
interesting and important class of problems. They include two, three and
multi-point boundary value problems as special cases, hence they have been
considerably developed, and the numerous properties of their solutions have
been studied, see \cite{b1,i2,k3,y1,y2} and references therein.
The main tools in these papers are the a priori estimate method
and the fixed point theorems. However, there are few papers dealing with
the existence of positive solutions for integral boundary-value problem with
$p$-Laplacian operator, especially, when $\phi$ satisfies (H1) and $f$ depends
on both $u$ and $u'$. The purpose of this paper is to investigate the
existence of positive symmetric solutions for the
integral boundary-value problem \eqref{e1.1}.
 By means of a fixed point theorem due to Avery and Peterson, sufficient
conditions are obtained that guarantee the
existence of at least three positive symmetric solutions for \eqref{e1.1}.

By a positive symmetric solution of \eqref{e1.1}, we mean
a solution $u$ of \eqref{e1.1} satisfying $u(t)>0, t\in(0, 1)$ and
$u(t)=u(1-t)$ for $t\in[0, 1]$.

For the convenience of the reader, we provided some background material
from the theory of cones in Banach spaces.
We also state in this section the Avery-Peterson fixed point theorem.


\begin{definition} \label{def1.1} \rm
A map $\alpha$ is said to be nonnegative continuous concave functional
on a cone $K$ of a real Banach
space $E$ provided that $\alpha:K\to [0, \infty)$ is
continuous and
\[
\alpha(tx+(1-t)y)\geq t\alpha(x)+(1-t)\alpha(y)
\]
for all $x, y\in K$ and $0\leq t\leq 1$. Similarly,
we say the map $\gamma$ is a nonnegative continuous convex
functional on a cone $K$ of a real Banach space $E$ provided that
$\gamma:K\to [0, \infty)$ is continuous and
\[
\gamma(tx+(1-t)y)\leq t\gamma(x)+(1-t)\gamma(y)
\]
 for all $x, y\in K$ and $0\leq t\leq 1$.
\end{definition}

Let $\gamma$ and $\theta$ be a nonnegative continuous convex
functionals on a cone $K$, $\alpha$ be a nonnegative continuous concave
functional on $K$, and $\psi$ be a nonnegative continuous
functional on $K$. Then for positive real number $a, b, c$ and
$d$, we define the following convex sets:
\begin{gather*}
 P(\gamma, d)=\{u\in K:\gamma(u)<d\},\\
 P(\gamma, \alpha, b, d)=\{u\in K:\alpha(u)\geq b, \gamma(u)\leq d\},\\
 P(\gamma, \theta, \alpha, b, c, d)=\{u\in K: \alpha(u)\geq b,
 \theta(u)\leq c, \gamma(u)\leq d\},\\
 R(\gamma, \psi, a, d)=\{u\in K|\psi(u)\geq a, \gamma(u)\leq d\}.
\end{gather*}
The following theorem due to Avery and Peterson is fundamental in the proofs
of our main results.

\begin{theorem}[\cite{a3}] \label{thm1.2}
Let $K$ be a cone in a real Banach space $E$. Let $\gamma$ and $\theta$ be
a nonnegative continuous convex functionals on $K$, $\alpha$ be a
nonnegative continuous concave functional on $K$, and $\psi$ be a
nonnegative continuous functional on $K$ satisfying
$\psi(\lambda u)\leq \lambda \psi(u)$ for $0\leq \lambda\leq 1$, such that
for positive number $M$ and $d$,
\begin{equation}
\alpha(u)\leq \psi(u),\quad \|u\|\leq M\gamma(u) \label{e1.4}
\end{equation}
for all $u\in \overline{P(\gamma, d)}$. Suppose
$T:\overline{P(\gamma, d)}\to\overline{P(\gamma, d)}$ is
completely continuous and there exist positive number
$a, b$ and $c$ with $a<b$ such that
\begin{itemize}
\item[(1)] $\{u\in P(\gamma, \theta, \alpha, b, c,
 d)|\alpha(u)>b\}\neq\emptyset$ and
$\alpha(Tu)>b$ for $u\in P(\gamma, \theta, \alpha, b, c, d)$;

\item[(2)] $\alpha(Tu)>b$ for $u\in P(\gamma, \alpha, b, d)$ with
$\theta(Tu)>c$;

\item[(3)] $0\not\in R(\gamma, \psi, a, d)$ and
 $\psi(Tu)<a$ for $u\in R(\gamma, \psi, a, d)$ with $\psi(u)=a$.

\end{itemize}
Then $T$ has at least three fixed points $u_1, u_2, u_3\in
\overline{P(\gamma, d)}$, such that
$\gamma(u_{i})\leq d$ for $i=1, 2, 3$,
$ \alpha(u_1)>b$,
$ \psi(u_2)>a$ with $\alpha(u_2)<b$,
$ \psi(u_3)<a$.
\end{theorem}

The organization of this paper is as follows.
In Section 2, some lemmas will be established
In Section 3, the main results of problem \eqref{e1.1} will be stated
 and proved. An example is also given to show our main results.

\section{Preliminaries}

 The basic space used in this paper is a real Banach space
 $C^{1}[0, 1]$ endowed the norm $\|\cdot\|_1$ defined by
 $\|u\|_1=\max\{\|u\|_{c}, \|u'\|_{c}\}$, where
$\|u\|_{c}=\max_{0\leq t\leq 1}|u(t)|$. Let
\begin{align*}
K=\Big\{&u\in C^{1}[0, 1] : u(t)\geq 0,
 u(0)=u(1)=\int_0^1u(t)g(t)\,\mathrm{d}t,\\
& u \text{ is concave and } u(t)=u(1-t), \; t\in[0, 1]\Big\}.
\end{align*}
It is obvious that $K$ is a cone in $C^{1}[0, 1]$.

\begin{lemma}[\cite{l1}] \label{lem2.1}
 Let $u\in K, \eta\in (0, 1/2)$,
 then $u(t)\geq \eta\max_{0\leq t\leq
 1}|u(t)|$, $t\in [\eta, 1-\eta]$.
\end{lemma}

\begin{lemma} \label{lem2.2}
Let $u\in K$, then there exists a constant $M>0$ such that
$\max_{0\leq t\leq 1}|u(t)|\leq M\max_{0\leq t\leq 1}|u'(t)|$.
\end{lemma}

\begin{proof}
The Mean Value Theorem implies that there exists $\rho\in [0, 1]$, such that
 $$
u(1)=u(\rho)\int_0^1g(t)\,\mathrm{d}t.
 $$ 
Furthermore, the Mean Value  Theorem of differential implies that there exists 
$\sigma\in  [\rho, 1]$, such that
 $$
\Big(\int_0^1g(t)\,\mathrm{d}t-1\Big)u(\rho)
=u(1)-u(\rho)=(1-\rho)u'(\sigma).
$$
Therefore,
$$
u(\rho)=\frac{(1-\rho)u'(\sigma)}{1-\int_0^1g(t)\,\mathrm{d}t}. 
$$
 So we obtain
\begin{align*}
|u(t)|&\leq |u(\rho)|+|\int_\rho^t u'(s)\,\mathrm{d}s| \\
&\leq \big(\frac{1-\rho}{1-\int_0^1g(t)\,\mathrm{d}t}+1\big)\max_{0\leq t\leq
 1}|u'(t)| \\
&\leq \frac{2-\int_0^1g(t)\,\mathrm{d}t}{1-\int_0^1g(t)\,\mathrm{d}t}
\max_{0\leq t\leq  1}|u'(t)|.
\end{align*}
Denote $M=\frac{2-\int_0^1g(t)\,\mathrm{d}t}{1-\int_0^1g(t)\,\mathrm{d}t}$,
then the proof is complete.
\end{proof}

Now, by a similar argument as in \cite{d1}, we define an operator 
$T:C^{1}[0, 1]\to C^{1}[0, 1]$ by
\begin{equation}
Tx(t)=\begin{cases}
\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\big(\int_s^{1/2}
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\big)\,\mathrm{d}s\,\mathrm{d}r\\
+\int_0^t\phi^{-1}\big(\int_s^{1/2} f(\tau,x(\tau),x'(\tau))\,\mathrm{d}
\tau\big)\,\mathrm{d}s, & 0\leq t\leq \frac{1}{2},\\[4pt]
\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)
 \int_r^1\phi^{-1}\big(\int_\frac{1}{2}^s
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\big)\,\mathrm{d}s\,\mathrm{d}r\\
+\int_t^1\phi^{-1}\big(\int_\frac{1}{2}^s
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\big)\,\mathrm{d}s,
 & \frac{1}{2}\leq t\leq 1 .
\end{cases}
\label{e2.1}
\end{equation}

\begin{lemma} \label{lem2.3} 
$T:K\to K$ is completely continuous.
\end{lemma}

 \begin{proof} 
Let $u\in K$. It follows from the definition  of $T$ that
\begin{equation}
(Tu)'(t)=\begin{cases}
\phi^{-1}\big(\int_t^{1/2} f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\big)\geq 0,
& 0\leq t\leq \frac{1}{2} ,\\[4pt]
-\phi^{-1}\big(\int_\frac{1}{2} ^t
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\big)\leq 0,
& \frac{1}{2} \leq t\leq 1 .
\end{cases}  \label{e2.2}
\end{equation}
From \eqref{e2.1}, \eqref{e2.2}, we have $(Tu)'(0)\geq 0 $ and
 $ (Tu)'(t)$ is positive on $[0, 1/2]$.
 Moreover, $ (Tu)'(t)$ is monotone decreasing continuous and
$(Tu)'(\frac{1}{2} )=0$. It implies that $(Tu)(t)$ is nonnegative and concave on
$[0, 1]$. By computation, we have
$Tu(0)=Tu(1)=\int_0^1Tu(t)g(t)\,\mathrm{d}t$.
Now, we show that $Tu$ is symmetric about $t$ on $[0, 1]$.

In fact, for all $t\in [0,1/2]$, we note that 
$(1-t)\in [1/2, 1]$. So, by H(2), H(3), we have
\begin{align*}
(Tu)(1-t)
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_r^1\phi^{-1}
\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad + \int_{1-t}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&= -\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_1^0g(1-r)
 \int_{1-r}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_{1-t}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\big)\,\mathrm{d}s\\
&= \frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_{1-r}^1\phi^{-1}
\Big(\int_{\frac{1}{2}}^s f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\big)
 \,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_{1-t}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_{\frac{1}{2}}^{1-s}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_{1-t}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{1/2}
f(1-\tau,u(1-\tau),u'(1-\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_{1-t}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{1/2}
f(1-\tau,u(\tau),-u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_{1-t}^1\phi^{-1}\Big(\int_{\frac{1}{2}}^s
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)
 \int_0^r\phi^{-1}\Big(\int_s^{1/2}
f(1-\tau,u(\tau),-u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r \\
&\quad +\int_0^t\phi^{-1}\Big(\int_{\frac{1}{2}}^{1-s}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_0^t\phi^{-1}\Big(\int_{\frac{1}{2}}^{1-s}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
 \Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_0^t\phi^{-1}\Big(\int_s^{1/2}
f(1-\tau,u(1-\tau),u'(1-\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&= \frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_0^t\phi^{-1}\Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&=(Tu)(t).
\end{align*}

Using the same method, we may prove that $Tu(t)=Tu(1-t)$ for
$t\in[1/2, 1]$. Thus, $T(K)\subset K$. Next, we prove $T$ is compact.

Let $D$ be a bounded subset of $K$ and $m>0$ is a constant such that
$\int_0^1f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau< m$ for $u\in D$. 
We know from \eqref{e2.1} and \eqref{e2.2} that for any $u\in D$,
\begin{gather*} 
|Tu(t)|<\begin{cases}
\frac{\phi^{-1}(m)}{1-\int_0^1g(r)\,\mathrm{d}r}, 
 & 0\leq t\leq \frac{1}{2},\\[4pt]
\frac{\phi^{-1}(m)}{1-\int_0^1g(r)\,\mathrm{d}r},
 & \frac{1}{2}\leq t\leq 1,
\end{cases} \\
 |(Tu)'(t)|<\phi^{-1}(m), 0\leq t\leq 1.
\end{gather*}
Hence, $TD$ is uniformly bounded and equicontinuous. By Arzela-Ascoli theorem, 
$TD$ is compact on $C[0, 1]$. From \eqref{e2.2}, we know that
for all $\varepsilon>0$ there exists $\kappa>0$, such that when
$|t_1-t_2|<\kappa$, we have
$|\phi(Tu)'(t_1)-\phi(Tu)'(t_2)|<\varepsilon$.
 So $\phi(TD)'$ is compact on $C[0, 1]$, it follows that $(TD)'$ is compact on 
$C[0,1]$. Therefore, $TD$ is compact on $C^{1}[0, 1]$.

 Thus, $T:K\to K$ is completely continuous. 
\end{proof}

It is easy to verify that each fixed point of $T$ is a solution for
problem \eqref{e1.1}.

 \begin{lemma}[\cite{w1}] \label{lem2.4} 
Assume that {\rm (H1)} holds. Then for $u, v\in (0, \infty)$,
$$
\psi_2^{-1}(u)v \leq \phi^{-1}(u\phi(v))
\leq \psi_1^{-1}(u)v.
$$
\end{lemma}

\section{Existence of three positive symmetric
solutions} 

For convenience, we introduce the nonation
$$
L=\frac{\int_0^1\psi_1^{-1}(1-s)\,\mathrm{d}s}{1-\int_0^1g(s)\,\mathrm{d}s},\quad
N=\int_\eta^{1/2}\psi_2^{-1}(\frac{1}{2}-s)\,\mathrm{d}s.
$$
In this section, we impose growth conditions on $f$ which allow us
to apply Theorem \ref{thm1.2} to establish the existence of three positive symmetric 
solutions of problem \eqref{e1.1}.

Let the nonnegative continuous concave functional $\alpha$, the
nonnegative continuous convex functionals $\gamma, \theta$, and
nonnegative continuous functional $\psi$ be defined on cone $K$ by
\begin{equation}
\gamma(u)=\max_{0\leq t\leq 1}|u'(t)|, \quad
\psi(u)=\theta(u)=\max_{0\leq t\leq 1}|u(t)|, \quad
\alpha(u)=\min_{\eta\leq t\leq 1-\eta}|u(t)|. \label{e3.1}
\end{equation}
Lemmas \ref{lem2.1} and \ref{lem2.2} imply that the functionals defined above satisfy
\begin{equation}
\eta\theta(u)\leq\alpha(u)\leq\psi(u)=\theta(u), \quad
\|u\|_1=\max\{\gamma(u), \theta(u)\}\leq M\gamma(u), \label{e3.2}
\end{equation}
 for all $u\in K$.
Therefore, the condition \eqref{e1.4} of Theorem \ref{thm1.2} is satisfied.

\begin{theorem} \label{thm3.1} 
Assume that {\rm (H1)} and {\rm (H2)} hold. Let
\[
0<a<b\leq\frac{d\eta}{1+\frac{1-\int_0^1g(t)\,\mathrm{d}t}
{\int_0^1g(t)t(1-t)\,\mathrm{d}t}}.
\]
If $f$ satisfies the following conditions:
\begin{itemize}
\item[(H6)] $f(t,u,v)\leq\phi(d)$ for $(t,u,v)\in [0, 1]\times[0, Md]\times[-d, d]$;

\item[(H7)] $ f(t,u,v)>\phi(\frac{b}{\eta N})$ for $(t,u,v)\in
[\eta, 1-\eta]\times[b, \frac{b}{\eta}(1
+\frac{1-\int_0^1g(t)\,\mathrm{d}t}{\int_0^1g(t)t(1-t)\,\mathrm{d}t})]\times[-d, d]$;

\item[(H8)] $f(t,u,v)<\phi(\frac{a}{L})$ for 
$(t,u,v)\in [0, 1]\times[0, a]\times[-d, d]$.
\end{itemize}
Then problem \eqref{e1.1} has at least three positive symmetric solutions
 $u_1, u_2$, and $u_3$ satisfying
\begin{gather*}
 \max_{0\leq t\leq 1}|u_{i}'(t)|\leq d \quad\text{for }
i=1, 2, 3, \min_{\eta\leq t\leq 1-\eta}|u_1(t)|>b,\\
 \max_{0\leq t\leq 1}|u_2(t)|>a\quad\text{ with }
\min_{\eta\leq t\leq 1-\eta}|u_2(t)|<b, \;\max_{0\leq t\leq
1}|u_3(t)|<a.
\end{gather*}
\end{theorem}

\begin{proof} 
We shall show that all the conditions of Theorem \ref{thm1.2} are satisfied.
If $u\in \overline{P(\gamma, d)}$, then
$\gamma(u)=\max_{0\leq t\leq 1}|u'(t)|\leq d$. Lemma \ref{lem2.2}
implies that $\max_{0\leq t\leq 1}|u(t)|\leq Md$, so by (H6), we
have $f(t,u(t),u'(t))\leq\phi(d)$ when $0\leq t\leq 1$. Thus
\begin{align*}
\gamma(Tu)
&=\max_{0\leq t\leq 1}|(Tu)'(t)| \\
&=\max\big\{\phi^{-1}\Big(\int_t^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big),
\phi^{-1}\Big(\int_\frac{1}{2}^t
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big) \big\} \\
&\leq\phi^{-1}\Big(\int_0^1 f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big) \\
&\leq\phi^{-1}(\phi(d))=d
\end{align*}
 This proves that $T:\overline{P(\gamma, d)}\to\overline{P(\gamma, d)}$.

To check condition (1) of Theorem \ref{thm1.2}, we choose
$$
u_{0}(t)=\frac{b}{\eta}+\frac{b(1-\int_0^1g(t)\,\mathrm{d}t)}{\eta\int_0^1g(t)t(1-t)
\,\mathrm{d}t}t(1-t), 0\leq
t\leq 1. 
$$ 
Let
$$
c=\frac{b}{\eta}\Big(1+\frac{1-\int_0^1g(t)\,\mathrm{d}t}{\int_0^1g(t)t(1-t)
\,\mathrm{d}t}\Big). 
$$
Then $u_{0}(t)\in P(\gamma, \theta, \alpha, b, c, d)$ and
$\alpha(u_{0})> b$, so $\{u\in P(\gamma, \theta, \alpha, b, c,
 d)|\alpha(u)>b\}\neq\emptyset$. Therefore, for 
$u\in P(\gamma, \theta,  \alpha, b, c, d)$, we have 
$$
b\leq u(t)\leq c, |u'(t)|\leq d, \quad
\eta\leq t\leq 1-\eta. 
$$ 
Thus, assumption (H7) implies that
\begin{equation}
f(t,u(t),u'(t))>\phi(\frac{b}{\eta N}) t\in
[\eta, 1-\eta]. \label{e3.3}
\end{equation}
This inequality and  Lemmas \ref{lem2.1}, and \ref{lem2.4} imply
\begin{align*}
\alpha(Tu)&=\min_{\eta\leq t\leq 1-\eta}|(Tu)(t)|
\geq \eta\max_{0\leq t\leq 1}|(Tu)(t)|\\
&= \frac{\eta}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\eta\int_0^{1/2}\phi^{-1}\Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&> \eta\int_0^{1/2}\phi^{-1}\Big(\int_s^{1/2}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&> \eta\int_\eta^{1/2}\phi^{-1}\big(\phi(\frac{b}{\eta
N})(\frac{1}{2}-s)\big)
\,\mathrm{d}s\\
&\geq \frac{b}{N}\int_\eta^{1/2}\psi_2^{-1}(\frac{1}{2}-s)\,\mathrm{d}s
= b.
\end{align*}
 This shows that condition (1) of Theorem \ref{thm1.2} is satisfied.

Secondly, for $u\in P(\gamma, \alpha, b, d)$ with $\theta(Tu)>c$,
we have
$$
 \alpha(Tu)\geq\eta\theta(Tu)\geq\eta c>b.
$$
Thus condition (2) of Theorem \ref{thm1.2} holds.

Finally, as $\psi(0)=0<a$, there holds 
$0\not\in R(\gamma, \psi,  a, d)$. Suppose that $u\in R(\gamma, \psi, a, d)$ with
$\psi(u)=a$, then by the assumption (H8),
\begin{equation}
f(t,u(t),u'(t))<\phi(\frac{a}{L}), \quad t\in [0, 1]. \label{e3.4}
\end{equation}
This inequality and Lemma \ref{lem2.4} imply
\begin{align*}
\psi(Tu)&=\max_{0\leq t\leq 1}|(Tu)(t)|\\
&<\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^1\phi^{-1}\Big(\int_s^1
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_0^1\phi^{-1}\Big(int_s^1f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&<\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1\phi^{-1}\big((1-s)\phi(\frac{a}{L})\big)\,\mathrm{d}s\\
&\leq \frac{a}{L}\frac{\int_0^1\psi_1^{-1}(1-s)\,\mathrm{d}s}{1-\int_0^1g(s)
\,\mathrm{d}s}
= a.
\end{align*}
Hence condition (3) of Theorem \ref{thm1.2} is also satisfied.

Therefore,  \eqref{e1.1}
has at least three positive symmetric solutions $u_1, u_2$, and $u_3$
satisfying
\begin{gather*}
 \max_{0\leq t\leq 1}|u_{i}'(t)|\leq d\quad\text{for }i=1, 2, 3, \min_{\eta\leq t\leq
1-\eta}|u_1(t)|>b,\\
 \max_{0\leq t\leq 1}|u_2(t)|>a\quad\text{with }\min_{\eta\leq t\leq
1-\eta}|u_2(t)|<b,]; \max_{0\leq t\leq 1}|u_3(t)|<a.
\end{gather*}
\end{proof}


\begin{example} \label{examp1} \rm
Let $\phi(u)=u^{3}$ and $ g(t)=1/2$. 
Consider the boundary-value problem
\begin{equation}
\begin{gathered}
  (\phi(u'))'+f(t,u(t),u'(t))=0, \quad t\in[0,1],\\
 u(0)=u(1)=\frac{1}{2}\int_0^1u(t)\,\mathrm{d}t,
 \end{gathered} \label{e3.5}
\end{equation}
 where
$$
f(t,u,v)= \begin{cases}
\frac{t(1-t)}{10^{7}}+85000u^{6}+\frac{1}{10^{7}}(\frac{v}{10^{6}})^{2} ,
& u\leq 21; \\[4pt]
\frac{t(1-t)}{10^{7}}+85000\cdot
21^{6}+\frac{1}{10^{7}}(\frac{v}{10^{6}})^{2}, & u>21.
\end{cases}
$$
Let $\psi_1^{-1}(u)=\psi_2^{-1}(u)=u^{3}, u>0$. Choosing
$a=1/100$, $b=1$, $\eta=1/3$, $d=10^6$, we have
$$
L=\frac{3}{2}, \quad N=\frac{3}{4}\big(\frac{1}{6}\big)^{4/3}, \quad
\phi\big(\frac{b}{\eta N}\big)=82944, \quad
\phi\big(\frac{a}{L}\big)=\frac{1}{3375000}.
$$
So, $f(t,u,v)$ satisfies
\begin{gather*}
f(t,u,v)<\phi(d)=10^{18},\quad  0\leq t\leq 1, \; 0\leq u\leq
3\cdot 10^{6},\; -10^{6}\leq v\leq 10^{6},\\
f(t,u,v)>82944, \quad \frac{1}{3}\leq t\leq \frac{2}{3},\;
 1\leq u\leq 21 ,\;  -10^{6}\leq v\leq 10^{6}, \\
f(t,u,v)<\frac{1}{3375000},\quad  0\leq t\leq 1,\; 0\leq u\leq
\frac{1}{100},\; -10^{6}\leq v\leq 10^{6}.
\end{gather*}
Therefore, by Theorem \ref{thm3.1}, boundary-value problem \eqref{e3.5} has at least
three positive symmetric solutions $u_1, u_2, u_3$ such that
\begin{gather*}
\max_{0\leq t\leq 1}|u_{i}'(t)|\leq 10^{6}, \; i=1, 2, 3, \quad
\min_{\frac{1}{4}\leq t\leq \frac{3}{4}}|u_1(t)|>1, \\
 \max_{0\leq t\leq 1}|u_2(t)|>\frac{1}{10}\quad
\min_{\frac{1}{4}\leq t\leq \frac{3}{4}}|u_2(t)|<1, \quad
\max_{0\leq t\leq 1}|u_3(t)|<\frac{1}{10}.
\end{gather*}
\end{example}

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\end{document}