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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 70, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/70\hfil Existence of positive pseudo-symmetric solutions]
{Existence of positive pseudo-symmetric solutions for
one-dimensional $p$-Laplacian boundary-value problems}

\author[Y. Yang\hfil EJDE-2007/70\hfilneg]
{Yitao Yang}  % in alphabetical order

\address{Yitao Yang \newline
Institute of Automation, Qufu Normal University,
Qufu, Shandong 273165, China}
\email{yitaoyangqf@163.com}

\thanks{Submitted March 12, 2007. Published May 10, 2007.}
\subjclass[2000]{34B15, 34B18}
\keywords{Iterative; pseudo-symmetric positive solution;  $p$-Laplacian}

\begin{abstract}
 We prove the existence of positive pseudo-symmetric solutions
 for four-point boundary-value problems with $p$-Laplacian.
 Also we present an monotone iterative scheme for approximating
 the solution.  The interesting point here is that the nonlinear
 term $f$ involves the first-order derivative.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In this paper, we consider the  four-point boundary value
problem
\begin{gather}\label{eq1}
(\phi_p(u'))'(t)+ q(t)f(t,u(t),u'(t))=0,\quad t\in (0,1), \\
\label{eq2} u(0)-\alpha u'(\xi)=0,\quad
u(\xi)-\gamma u'(\eta)=u(1)+\gamma u'(1+\xi-\eta),
\end{gather}
where $\phi_{p}(s)=|s|^{p-2}s$, $p>1$,
$(\phi_{p})^{-1}=\phi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$,
$\alpha,\gamma \geq 0$, $\xi, \eta\in (0,1)$ are prescribed and
$\xi<\eta$.

The study of  multipoint boundary-value problems for linear
second-order ordinary differential  equations was initiated by Il'in
and Moiseer \cite{i1,i2}. Since then, the more general nonlinear
multipoint boundary-value problems have been studied by many authors
by using the Leray-Schauder continuation theorem, nonlinear
alternative of Leray-Schauder and coincidence degree theory, we
refer the reader to \cite{a1,a2,a3,h1} for some recent results.
Recently, Avery and Henderson \cite{a4} consider the existence of
three positive solutions for the  problem
\begin{gather}\label{eq3}
(\phi_p(u'))'(t)+ q(t)f(t,u(t))=0,\quad t\in (0,1), \\
\label{eq4} u(0)=0,\quad u(\eta)=u(1).
\end{gather}
The definition of pseudo-symmetric was introduced in their paper.
Based on this definition, Ma \cite{m1} studied the existence and
iteration of positive pseudo-symmetric solutions for the problem
(\ref{eq3})-(\ref{eq4}). However, to the best of our knowledge, no
work has been done for BVP (\ref{eq1})-(\ref{eq2}) using the
monotone iterative technique. The aim of this paper is to fill the
gap in the relevant literatures. We obtain not only the existence of
positive solutions for (\ref{eq1})-(\ref{eq2}), but also give an
iterative scheme for approximating the solutions. It is worth
stating that the first term of our iterative scheme is a constant
function or a simple function. Therefore, the iterative scheme is
significant and feasible. At the same time, we give a way to find
the solution which will be useful from an application viewpoint.

We consider the Banach space $E=C^{1}[0,1]$ equipped with norm
$$
\|u\|:=\max\{\|u\|_{0}, \|u'\|_{0}\},
$$
where $\|u\|_{0}=\max_{0\leq t\leq 1}|u(t)|$,
$\|u'\|_{0}=\max_{0\leq t\leq 1}|u'(t)|$. In this paper, a
positive solution $u(t)$ of BVP (\ref{eq1}), (\ref{eq2}) means a
solution $u(t)$ of (\ref{eq1}), (\ref{eq2}) satisfying
$u(t)>0$, for $0<t<1$.

We recall that a function $u$ is said to be
concave on $[0,1]$, if
$$
u(\lambda t_{2}+(1-\lambda)t_{1})\geq \lambda u(t_{2})+(1-\lambda)u(t_{1}),
\quad   t_{1}, t_{2}, \lambda \in [0,1].
$$

\begin{definition} \label{defnF11} \rm
 For $\xi\in (0,1)$ a function $u\in E$
is said to be pseudo-symmetric if $u$ is symmetric over the interval
$[\xi,1]$. That is, for $t\in [\xi,1]$ we have $u(t)=u(1+\xi-t)$.
\end{definition}

\begin{remark} \label{remF11} \rm
For $\xi\in (0,1)$, if $u\in E$
is pseudo-symmetric, we have $u'(t)=-u'(1+\xi-t)$, $t\in [\xi,1]$.
\end{remark}

Define the cone $K$  of $E$ as
$$
K=\{u\in C^{1}[0,1]: u(t)\geq 0,\; u \mbox{ is concave on}
[0,1] \mbox{ and }  u  \mbox{ is symmetric on } [\xi,1]\}.
$$
For $x,y$ in $K$ a cone of $E$, recall that
$x\leq y$ if $y-x\in K$.

 In the rest of the paper, we make the following
assumptions:
\begin{itemize}
\item[(H1)]  $q(t)\in L^{1}[0,1]$ is nonnegative and
$q(t)=q(1+\xi-t)$,  a.e. $t\in [\xi,1]$, and
$q(t)\not\equiv 0$ on any subinterval of [0,1];

\item[(H2)] $f\in C([0,1]\times [0,\infty)\times R,[0,\infty))$
and $f(t,x,y)=f(1+\xi-t,x,-y)$, $(t,x,y)\in [\xi,1]\times
[0,\infty)\times R$. Moreover, $f(t,\cdot,y)$ is nondecreasing for
$(t,y)\in [0,\frac{\xi+1}{2}]\times R$, $f(t,x,\cdot)$ is
nondecreasing for $(t,x)\in [0,\frac{\xi+1}{2}]\times[0,\infty)$.
\end{itemize}

\section{Existence Result}

\begin{lemma}[\cite{m1}] \label{lemF21}
Each  $u\in K$ satisfies the following properties:
\begin{itemize}
\item[(i)] $u(t)\geq \frac{2}{1+\xi}\|u\|_{0}\min\{t,1+\xi-t\}$,
$t\in [0,1]$;
\item[(ii)] $u(t)\geq \frac{2\xi}{1+\xi}\|u\|_{0}$,
$t\in [\xi,\frac{1+\xi}{2}]$;

\item[(iii)] $\|u\|_{0}=u(\frac{1+\xi}{2})$.
\end{itemize}
\end{lemma}

For  $x\in K$, we define a mapping $T: K\to E$ given
by
\begin{equation}\label{eq5}
(Tx)(t)= \begin{cases}
\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)\\
+\int_{0}^{t}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}\
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds,
& 0\leq t\leq \frac{1+\xi}{2},\\[4pt]
\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau) \\
+\int_{0}^{\xi}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}\
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
+\int_{t}^{1}\phi_{q}(\int_{\frac{1+\xi}{2}}^{s}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds,
&\frac{1+\xi}{2}\leq t\leq 1.
\end{cases}
\end{equation}
Obviously, $Tx\in E$, and we can prove
$Tx$ is a solution of the  boundary-value problem
\begin{gather}\label{eq6}
(\phi_p(u'))'(t)+ q(t)f(t,x(t),x'(t))=0,\quad t\in (0,1), \\
\label{eq7} u(0)-\alpha u'(\xi)=0,\quad
u(\xi)-\gamma u'(\eta)=u(1)+\gamma u'(1+\xi-\eta).
\end{gather}
Therefore, each fixed point of $T$ is a  solution of
problem (\ref{eq1})-(\ref{eq2}).

\begin{lemma}\label{lemF22}
 Suppose (H1), (H2) hold, then $T: K\to K$ is
completely continuous and nondecreasing.
\end{lemma}

\begin{proof} For $t\in [\xi,\frac{1+\xi}{2}]$, we
have $1+\xi-t\in [\frac{1+\xi}{2},1]$. Therefore,
\begin{equation}\label{eq8}
\begin{aligned}
&(Tx)(1+\xi-t)\\
&=\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)
+\int_{0}^{\xi}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
&\quad  +\int_{1+\xi-t}^{1}\phi_{q}(\int_{\frac{1+\xi}{2}}^{s}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
&=\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)
+\int_{0}^{\xi}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
&\quad +\int_{\xi}^{t}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
& =\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)
+\int_{0}^{t}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}\
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
&=(Tx)(t).
\end{aligned}
\end{equation}
So that, $Tx$ is symmetric on $[\xi,1]$ with respect to $\frac{1+\xi}{2}$.
Obviously, $(Tx)(t)\geq 0, Tx$ is concave on $[0,1]$.
Therefore, $T: K\to K$. It is easy to see that $T:
K\to K$ is completely continuous.

For $x_{1}(t),x_{2}(t)\in K$ and $x_{1}(t)\leq x_{2}(t)$, thus
$x_{2}(t)-x_{1}(t)\in K$. So that $x_{2}'(t)-x_{1}'(t)\geq 0$,
$t\in [0,\frac{1+\xi}{2}]$ and $x_{2}'(t)-x_{1}'(t)\leq 0$,
$t\in [\frac{1+\xi}{2},1]$. Assumption (H2) implies
$(Tx_{1})(t)\leq (Tx_{2})(t)$.
\end{proof}

\begin{lemma}[\cite{g1}] \label{lemF23}
Let $u_{0},v_{0}\in E,\ u_{0}<v_{0}$
and $T: [u_{0},v_{0}]\to E$ be an increasing operator such
that
$$
u_{0}\leq T u_{0},\quad T v_{0}\leq v_{0}.
$$
Suppose that one of the following conditions is satisfied:
\begin{itemize}
\item[(C1)]  $K$ is normal and $T$ is condensing;

\item[(C2)]  $K$ is regular and $T$ is semicontinuous, i.e.,
$x_{n}\to x$ strongly implies $Tx_{n}\to Tx$ weakly.
\end{itemize}
Then $T$ has a maximal fixed point $x^{*}$ and a minimal fixed point
$x_{*}$ in $[u_{0},v_{0}];$\ moreover
$$
x^{*}=\lim_{n\to \infty}v_{n},\quad
x_{*}=\lim_{n\to \infty}u_{n},
$$
where $v_{n}=Tv_{n-1}$, $u_{n}=Tu_{n-1}$
$(n=1,2,3,\dots)$, and
$$
u_{0}\leq u_{1}\leq \dots\leq u_{n}\leq \dots\leq v_{n}\leq
\dots\leq v_{1}\leq v_{0}.
$$
\end{lemma}

Denote the positive quantities
\begin{gather}\label{eq9}
A=1/\max\big\{\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)d\tau)+
\int_{0}^{\frac{1+\xi}{2}}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)d\tau)ds,\; \phi_{q}(\int_{0}^{\frac{1+\xi}{2}}
q(\tau)d\tau)\big\},\\
\label{eq10}B=1/\int_{\xi}^{\frac{1+\xi}{2}}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)d\tau)ds.
\end{gather}

\begin{theorem} \label{theF24}
Assume {\rm (H1), (H2)} hold. If there
exist two positive numbers $a,b$ with $\frac{2b}{1+\xi}<a$, such
that
\begin{equation}\label{eq11}
\sup_{t\in [0,1]}f(t,a,a)\leq \phi_{p}(aA),\quad
\inf_{t\in [\xi,\frac{1+\xi}{2}]}f(t,\frac{2\xi
b}{1+\xi},0)\geq \phi_{p}(bB).
\end{equation}
Then, (\ref{eq1})-(\ref{eq2}) has at least one positive
pseudo-symmetric solution $\upsilon^{*}\in K$ with
$$
b\leq \|\upsilon^{*}\|_{0}\leq a,\quad
0\leq \|(\upsilon^{*})'\|_{0}\leq a,\quad
\lim_{n\to\infty}T^{n}\upsilon_{0}=\upsilon^{*},
$$
where $\upsilon_{0}(t)=\frac{2b}{1+\xi}\min\{t,1+\xi-t\}$,
$t\in [0,1]$.
\end{theorem}

\begin{proof} We denote $K[b,a]=\{\omega\in K: b\leq
\|\omega\|_{0}\leq a,\; 0\leq \|\omega'\|_{0}\leq a\}$.
Next, we first prove $TK[b,a]\subset K[b,a]$.
Let $\omega \in K[b,a]$, then $0\leq \omega (t)\leq
\max_{t\in [0,1]}\omega (t)\leq\|\omega\|_{0}\leq a$,\\
$ t\in [0,1]$, $0\leq \omega' (t)\leq \max_{t\in [0,1]}|\omega
'(t)|=\|\omega'\|_{0}\leq a,\ t\in [0,\frac{1+\xi}{2}]$. By lemma
\ref{lemF21} (ii), $\min_{t\in [\xi,\frac{1+\xi}{2}]}\omega (t)\geq
\frac{2\xi}{1+\xi}\|\omega\|_{0}\geq \frac{2\xi b}{1+\xi}$,
$\min_{t\in [\xi,\frac{1+\xi}{2}]}\omega' (t)\geq \omega'
(\frac{1+\xi}{2})=0$. So, by assumption \eqref{eq11}, we have
\begin{gather}\label{eq12}
0\leq f(t,\omega(t),\omega'(t))\leq
f(t,a,a)\leq \sup_{t\in [0,1]} f(t,a,a)\leq \phi_{p}(aA),\quad
t\in [0,\frac{1+\xi}{2}], \\
\label{eq13} f(t,\omega(t),\omega'(t))\geq f(t,\frac{2\xi
b}{1+\xi},0)\geq \inf_{t\in [\xi,\frac{1+\xi}{2}]} f(t,\frac{2\xi
b}{1+\xi},0)\geq \phi_{p}(bB),\quad t\in [\xi,\frac{1+\xi}{2}].
\end{gather}
By lemma \ref{lemF22}, we know $T\omega \in K$. So,
lemma \ref{lemF21} (iii) implies
$\|T\omega\|_{0}=(T\omega)(\frac{1+\xi}{2})$. As a result,
\begin{align*}
&\|T\omega\|_{0}\\
&=(T\omega)(\frac{1+\xi}{2})\\
&=\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)
 +\int_{0}^{\frac{1+\xi}{2}}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}\
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)ds\\
& \leq
aA(\alpha\phi_{q}(\int_{\xi}^{\frac{1+\xi}{2}} q(\tau)d\tau)+
\int_{0}^{\frac{1+\xi}{2}}\phi_{q}(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)d\tau)ds)\leq a;
\end{align*}
\begin{align*}
\|(T\omega)'\|_{0}&=(T\omega)'(0)\\
&=\phi_{q}(\int_{0}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau)\\
&\leq aA\phi_{q}(\int_{0}^{\frac{1+\xi}{2}} q(\tau)d\tau)\leq a;
\end{align*}
\begin{align*}
\|T\omega\|_{0}&=(T\omega)(\frac{1+\xi}{2})\\
&\geq \int_{\xi}^{\frac{1+\xi}{2}}\phi_{q}\Big(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\\
&\geq bB\int_{\xi}^{\frac{1+\xi}{2}}\phi_{q}\Big(\int_{s}^{\frac{1+\xi}{2}}
q(\tau)d\tau\Big)ds=b.
\end{align*}
Thus, $b\leq \|T\omega\|_{0}\leq a$,
$0\leq \|(T\omega)'\|_{0}\leq a$, which implies
 $TK[b,a]\subset K[b,a]$.

Let $\upsilon_{0}(t)=\frac{2b}{1+\xi}\min\{t,1+\xi-t\}$,
$t\in [0,1]$, then $\|\upsilon_{0}\|_{0}=b$ and
$\|\upsilon_{0}^{'}\|_{0}=\frac{2b}{1+\xi}<a$, so
$\upsilon_{0}\in K[b,a]$. Let $\upsilon_{1}=T\upsilon_{0}$, then
$\upsilon_{1}\in K[b,a]$, we denote
\begin{equation}\label{eq14}
\upsilon_{n+1}=T\upsilon_{n}=T^{n+1}\upsilon_{0},\quad
(n=0,1,2,\dots).
\end{equation}
Since $TK[b,a]\subset K[b,a]$, we have $\upsilon_{n}\in K[b,a]$,
($n=0,1,2,\dots$). From $\upsilon_{1}\in K[b,a]$, thus
$$
\upsilon_{1}(t)\geq \frac{2}{1+\xi}\|\upsilon_{1}\|_{0}\min\{t,1+\xi-t\}\geq
\frac{2b}{1+\xi}\min\{t,1+\xi-t\}=\upsilon_{0}(t),\quad t\in [0,1],
$$
which implies $T\upsilon_{0}\geq\upsilon_{0}$. $K$ is normal and $T$
is completely continuous. By Lemma \ref{lemF23}, we have $T$ has a
fixed point $\upsilon^{*}\in K[b,a]$. Moreover,
$\upsilon^{*}=\lim_{n\to \infty}\upsilon_{n}$. Since
$\|\upsilon^{*}\|_{0}\geq b>0$ and $\upsilon^{*}$ is a nonnegative
concave function on $[0,1]$, we conclude that
$\upsilon^{*}(t)>0$, $t\in  (0,1)$. Therefore, $\upsilon^{*}$ is a positive
pseudo-symmetric solution of (\ref{eq1})-(\ref{eq2}).
\end{proof}

\begin{corollary}\label{corF25}
Assume {\rm (H1),(H2)} hold. If
\begin{equation}\label{eq15}
\limsup_{l\to 0}\inf_{t\in
[\xi,\frac{1+\xi}{2}]}\frac{f(t,l,0)} {\phi_{p}(l)}\geq
\phi_{p}(\frac{1+\xi}{2\xi}B),
\end{equation}
particularly, $\limsup_{l\to 0}\inf_{t\in [\xi,\frac{1+\xi}{2}]}\frac{f(t,l,0)}
{\phi_{p}(l)}=+\infty$,
\begin{equation}
\label{eq16}\liminf_{l\to +\infty}\sup_{t\in [0,1]}\frac{f(t,l,l)}
{\phi_{p}(l)}\leq \phi_{p}(A),
\end{equation}
particularly, $\liminf_{l\to +\infty}\sup_{t\in
[0,1]}\frac{f(t,l,l)} {\phi_{p}(l)}=0$. Where $A,B$ are defined as
(\ref{eq9}), (\ref{eq10}). Then there exist two positive numbers
$a,b$ with $\frac{2b}{1+\xi}<a$, such that problem (\ref{eq1}),
(\ref{eq2}) has at least one positive pseudo-symmetric solution
$\upsilon^{*}\in K$ with
$$
b\leq \|\upsilon^{*}\|_{0}\leq
a,\ 0\leq \|(\upsilon^{*})'\|_{0}\leq a,\
\lim_{n\to\infty}T^{n}\upsilon_{0}=\upsilon^{*},
$$
where $\upsilon_{0}(t)=\frac{2b}{1+\xi}\min\{t,1+\xi-t\}$,
$t\in [0,1]$.
\end{corollary}


\begin{remark}\label{remF21} \rm
Problem (\ref{eq1})-(\ref{eq2}) may have two
positive pseudo-symmetric solutions $\omega^{*},\upsilon^{*}\in K$,
if we make another iteration by choosing $\omega_{0}(t)=a$ and
$\omega_{n}=\lim_{n\to\infty}T^{n}\omega_{0}=\omega^{*}$.
However,  $\omega^{*}$ and $\upsilon^{*}$ may be the same solution.
\end{remark}

\begin{example}\label{exaF21} \rm
We consider the  problem
\begin{gather}\label{eq17}
(|u'|^{3}u')'(t)+\frac{1}{t^{\frac{1}{2}}(\frac{4}{3}-t)^{\frac{1}{2}}}
[(u'(t))^{2} +\ln((u(t))^{2}+1)]=0,\quad  t\in (0,1),\\
\label{eq18} u(0)-2u'(\frac{1}{3})=0,\quad
 u(\frac{1}{3})-3u'(\frac{1}{2})=u(1)+3u'(\frac{5}{6}).
\end{gather}
We notice that $p=5$, $\alpha=2$, $\xi=\frac{1}{3}$,
$\gamma=3$, $\eta=\frac{1}{2}$. Obviously,
$f(t,u,u')=(u'(t))^{2}+\ln((u(t))^{2}+1)$ is nondecreasing
for $(t,u')\in [0,\frac{2}{3}]\times R$,
$f(t,u,u')=(u'(t))^{2} +\ln((u(t))^{2}+1)$ is nondecreasing
for $(t,u)\in [0,\frac{2}{3}]\times[0,\infty)$,
$q(t)=\frac{1}{t^{\frac{1}{2}}(\frac{4}{3}-t)^{\frac{1}{2}}}$ is
nonnegative and pseudo-symmetric about $\frac{2}{3}$. So, conditions
(H1),(H2) are satisfied. On the other hand,
\begin{gather*}
\limsup_{l\to 0}\inf_{t\in [\xi,\frac{1+\xi}{2}]}\frac{f(t,l,0)}
{\phi_{p}(l)}=\limsup_{l\to 0}\inf_{t\in
[\frac{1}{3},\frac{2}{3}]}\frac{\ln (l^{2}+1)} {l^{4}}=\infty,\\
\liminf_{l\to +\infty}\sup_{t\in
[0,1]}\frac{f(t,l,l)} {\phi_{p}(l)}=\liminf_{l\to
+\infty}\sup_{t\in [0,1]}\frac{l^{2} +\ln(l^{2}+1)}
{l^{4}}=0.
\end{gather*}
 Therefore, from Corollary \ref{corF25}, it follows that
(\ref{eq17})-(\ref{eq18}) has at least one positive pseudo-symmetric
solution.
\end{example}

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\end{document}