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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 94, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/94\hfil Necessary and sufficient conditions]
{Necessary and sufficient conditions for the existence of periodic solution to
 singular problems with impulses}

\author[J. Sun, J. Chu \hfil EJDE-2014/94\hfilneg]
{Juntao Sun, Jifeng Chu} 

\address{Juntao Sun \newline
School of Science, Shandong University of
Technology,
Zibo, 255049 Shandong, China}
\email{sunjuntao2008@163.com}

\address{Jifeng Chu \newline
College of Science, Hohai University,
Nanjing, 210098 Jiangsu, China}
\email{jifengchu@126.com}

\thanks{Submitted October 30, 2013. Published April 10, 2014.}
\subjclass[2000]{34B15}
\keywords{Positive periodic solution; singular differential equations; 
\hfill\break\indent impulses; variational methods}

\begin{abstract}
 In this article we give a necessary sufficient conditions for the existence
 of periodic solutions to impulsive periodic solution for a singular differential
 equation.  The proof is based on the variational method.
\end{abstract}

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


\section{Introduction}

In this article we discuss the $T$-periodic solution for the
second-order singular problem with impulsive effects
\begin{equation}
\begin{gathered}
u''(t)-\frac{1}{u^{\alpha}(t)}=e(t),\quad\text{a.e. } t\in(0,T),\\
\Delta u'(t_j)=b_j,\quad  j=1,2,\dots,p-1,
\end{gathered}  \label{eSP}
\end{equation}
where $\alpha\geq 1$, $e\in L^1([0,T],\mathbb{R})$ is $T$-periodic,
$\Delta u'(t_j)=u'(t_j^+)-u'(t_j^-)$ with
$u'(t_j^\pm)=\lim_{t\to t_j^\pm} u'(t)$; $t_j$,
$j=1,2,\dots,p-1,$ are the instants where the impulses occur and
$0=t_0<t_1<t_2<\dots<t_{p-1}<t_p=T$, $t_{j+p}=t_j+T$; and $b_j$
$(j=1,2,\dots,p-1)$ are constants.

Impulsive effects occur widely in many evolution processes in which
their states are changed abruptly at certain moments of time. In the
past few decades, impulsive differential equations have been
extensively studied by many researchers
\cite{2008-Chu-p143-150,2009-Nieto-p680-690,2013-Sun-p562-569,
2012-Sun-p193-204,2010-Sun-p4575-4586,2011-Sun-p544-555,2008-Tian-p509-527}.
In particular, In 2008, Tian and Ge \cite{2008-Tian-p509-527}
studied the existence of solutions for impulsive differential
equations by using a variational method. Later, Nieto and O'Regan
\cite{2009-Nieto-p680-690} further developed the variational
framework for impulsive problems and established existence results
for a class of impulsive differential equations with Dirichlet
boundary conditions. From then on, the variational method has been a
powerful tool in the study of impulsive differential equations.
On the other hand, singular
differential equations with different kinds of boundary conditions
have also been investigated widely in the literature by using
either topological methods or variational methods; see
\cite{2002-Agarwal-p409-433,2005-Agarwal-p817-824,2008-Boucherif-p147-158,
2007-Chen-p233-244,
2012-Chu-p665-675,2009-Chu-p323-338,2007-Chu-p196-212,2010-Hakl-p111-126}
and the references therein.

In 1987, Lazer and Solimini \cite{1987-Lazer-p109-114} considered a
the second order singular problem
\begin{equation}
\label{1-1}
u''(t)-\frac{1}{u^{\alpha}(t)}=e(t),\quad t\in(0,T).
\end{equation}
By using the method of upper and lower solutions, they obtained a
famous sufficient and necessary condition on positive $T$-periodic
solution for Problem \eqref{1-1} as follows
\begin{theorem} [\cite{1987-Lazer-p109-114}]\label{th1}
Assume that $e\in L^1([0,T],\mathbb{R})$ is
$T$-periodic. Then Problem \eqref{1-1} has a positive $T$-periodic
weak solution if and only if $\int_0^{T}e(t)dt<0$.
\end{theorem}

Motivated by the above fact, in the present paper we shall consider
Problem \eqref{1-1} with impulsive effects, i.e., Problem \eqref{eSP},
and also obtain a sufficient and necessary condition on $T$-periodic
solution. It is worth emphasizing that the method used by us is a
variational method, which is different from that in Theorem
\ref{th1}. Furthermore, we also point out the dynamical differences
between singular problems and singular problems with impulses.

Our results are presented as follows.

\begin{theorem}\label{th2}
Assume that $e\in L^1([0,T],\mathbb{R})$ is
$T$-periodic. Then Problem \eqref{eSP} has a positive $T$-periodic weak
solution $u\in H^1_T$ if and only if
$\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt<0$.
\end{theorem}

\begin{remark} \rm
From Theorem \ref{th2} we can see that if $\int_0^{T}e(t)dt\geq0$,
but $\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt<0$, then
Problem \eqref{eSP} still admits a positive $T$-periodic solution. This
shows that the existence of positive $T$-periodic solution for
Problem \eqref{eSP} depends on the forced term $e$ and impulsive
functions $b_j$ together, not single one.
\end{remark}


\section{Preliminaries}

Set
\begin{align*}
H_T^1= \big\{&u: \mathbb{R}\to \mathbb{R}| u
\text{ is absolutely continuous}, \; 
u'\in L^2((0,T), \mathbb{R})\\
&  \text{and }u(t)=u(t+T) \text{ for } t\in\mathbb{R}\big\}
\end{align*}
with the inner product
\[
(u,v)=\int_0^{T}u(t)v(t)dt+\int_0^{T}u'(t) v'(t)dt,\quad\forall
u,v\in H_T^1.
\]
The corresponding norm is defined by
\[
\|u\|_{H_T^1}=\Big(\int_0^{T}|u(t)|^2dt+\int_0^{T}|u'(t)|^2dt\Big)
^{1/2},\quad\forall u\in H_T^1.
\]
Then $H_T^1$ is a Banach space (in fact it is a Hilbert space).

To study Problem \eqref{eSP}, for any $\lambda\in(0,1)$ we
consider the following modified problem
\begin{equation}
\begin{gathered}
u''(t)+f_{\lambda}(u(t))=e(t),\quad\text{a.e. } t\in(0,T),\\
\Delta u'(t_j)=b_j,\quad j=1,2,\dots,p-1,
\end{gathered}  \label{eSPl}
\end{equation}
where $f_{\lambda}:\mathbb{R}\to\mathbb{R}$ is defined by
\[
f_{\lambda}(s)= \begin{cases}
-\frac{1}{s^{\alpha}},& s\geq\lambda,\\
-\frac{1}{\lambda^{\alpha}},& s<\lambda.
\end{cases}
\]
Now we introduce the following concept of a weak solution for
Problem \eqref{eSPl}.

\begin{definition} \label{def1} \rm
We say that a function $u\in H_T^1$ is a weak
solution of Problem \eqref{eSPl} if
\[
\int_0^{T}u'(t)
v'(t)dt+\sum_{j=1}^{p-1}b_jv(t_j)-\int_0^{T}f_{\lambda}(u(t))v(t)dt
+\int_0^{T}e(t)v(t)dt=0
\]
holds for any $v\in H_T^1$.
\end{definition}

Let $F_{\lambda}\in C^1(\mathbb{R},\mathbb{R})$ be defined by
\[
F_{\lambda}(s)=\int_1^{s}f_{\lambda}(t)dt
\]
and consider the functional
$\Phi_{\lambda}:H_T^1\to\mathbb{R}$
defined by
\begin{equation} \label{2-1}
\Phi_{\lambda}(u)
:= \frac{1}{2}\int_0^{T}|u'(t)|^2dt
 +\sum_{j=1}^{p-1}b_ju(t_j)-\int_0^{T}F_{\lambda}(u(t))dt
 +\int_0^{T}e(t)u(t)dt.
\end{equation}
Clearly, $\Phi_\lambda$ is well defined on $H_T^1$, continuously
G\^ateaux differentiable functional whose
 derivative is
\[
\Phi_{\lambda}'(u)v
= \int_0^{T}u'(t)v'(t)dt+\sum_{j=1}^{p-1}b_jv(t_j)
-\int_0^{T}f_{\lambda}(u(t))v(t)dt
+\int_0^{T}e(t)v(t)dt,
\]
for any $v\in H_T^1$. Moreover, it is easy to verify that
$\Phi_{\lambda}$ is weakly lower semi-continuous. Furthermore, by
the standard discussion, the critical points of $\Phi_\lambda$ are
the weak solutions of Problem \eqref{eSPl}.

\section{Proof of Theorem \ref{th2}}\label{sec4}

\begin{proof}
First we show that if $u\in H_T^1$ is  a positive
$T$-periodic weak solution of Problem \eqref{eSP}, then
$\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt<0$.

Integrating the first equation of Problem \eqref{eSP} from $0$ to $T$,
one has
\begin{equation}
\label{3-1}
\int_0^{T}u''(t)dt-\int_0^{T}\frac{1}{u^{\alpha}(t)}dt=\int_0^{T}e(t)dt.
\end{equation}
The first term one the left-hand side satisfies
\[
\int_0^{T}u''(t)dt=\sum_{j=0}^{p-1}\int_{t_j}^{t_{j+1}}u''(t)dt,
\]
and
\[
\int_{t_j}^{t_{j+1}}u''(t)dt=u'(t^{-}_{j+1})-u'(t^{+}_j).
\]
Thus,
\begin{equation} \label{3-2}
\begin{aligned}
\int_0^{T}u''(t)dt
&= \sum_{j=0}^{p-1}(u'(t^{-}_{j+1})-u'(t^{+}_j)) \\
&= -\sum_{j=1}^{p-1}\Delta u'(t_j)+u'(T)-u'(0) \\
&= -\sum_{j=1}^{p-1}b_j.
\end{aligned}
\end{equation}
By \eqref{3-1} and \eqref{3-2} we have
\[
0>-\int_0^{T}\frac{1}{u^{\alpha}(t)}dt=\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt.
\]

Now we prove that if
$\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt<0$, then Problem
\eqref{eSP} has a positive $T$-periodic weak solution $u\in H_T^1$.
The proof is based on the mountain pass theorem, see
\cite{1989-Mawhin-p}. We divide it into four steps.

\textbf{Step 1.} Let a sequence $\{u_n\}$ in $H_T^1$ satisfy
$\Phi_{\lambda}(u_n)$ be bounded and
$\Phi_{\lambda}'(u_n)\to0$, i.e., there exist a constant
$c_1>0$ and a sequence $\{\epsilon_n\}_{n\in\mathbb{N}}\subset
\mathbb{R}^+$ with $\epsilon_n\to0$ as $n\to+\infty$
such that for all $n$,
\begin{equation} \label{3-3}
\Big|\int_0^{T}\big[\frac{1}{2}|u_n'(t)|^2-F_{\lambda}(u_n(t))+e(t)u_n(t)
\big]dt+\sum_{j=1}^{p-1}b_ju_n(t_j)\Big| \leq c_1,
\end{equation}
and for every $v\in H_T^1$,
\begin{equation} \label{3-4}
\Big|\int_0^{T}[u_n'(t)
v'(t))-f_{\lambda}(u_n(t))v(t)+e(t)v(t)]dt+\sum_{j=1}^{p-1}b_jv(t_j)\Big|
\leq\epsilon_n\|v\|_{H_T^1}.
\end{equation}
Now we show that $\{u_n\}$ is bounded in $H_T^1$. Taking
$v(t)\equiv-1$ in \eqref{3-4} one has
\[
\Big|\int_0^{T}[f_{\lambda}(u_n(t))-e(t)]dt-\sum_{j=1}^{p-1}b_j\Big|
\leq\epsilon_n\sqrt{T}\quad
\text{for all}\ n,
\]
which implies
\[
\big|\int_0^{T}f_{\lambda}(u_n(t))dt\big|
\leq\epsilon_n\sqrt{T}+\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}|b_j|:=c_2.
\]
Note that for any $t\in[0,T]$, $f_{\lambda}(u_n(t))<0$. Thus
\[
\int_0^{T}|f_{\lambda}(u_n(t))|dt
=\big|\int_0^{T}f_{\lambda}(u_n(t))dt\big|
\leq c_2.
\]
On the other hand, take, in \eqref{3-4}, $v(t)\equiv w_n(t):=u_n(t)-\bar{u}_n$,
where
$\bar{u}_n=\frac{1}{T}\int_0^{T}u_n(t)dt$, by
\cite[Proposition 1.1]{1989-Mawhin-p} we have
\begin{align*}
c_3\|w_n\|_{H_T^1}
&\geq \int_0^{T}[w_n'(t)^2-f_{\lambda}(u_n(t))w_n(t)+e(t)w_n(t)]dt
 +\sum_{j=1}^{p-1}b_jw_n(t_j)\\
&\geq \|w'_n\|^2_{L^2}-(c_2+\|e\|_{L^1})\|w_n\|_{L^{\infty}}
 -\sum_{j=1}^{p-1}|b_j|\|w_n\|_{L^{\infty}}\\
&= \|w'_n\|^2_{L^2}-(c_2+\|e\|_{L^1}
 +\sum_{j=1}^{p-1}|b_j|)\|w_n\|_{L^{\infty}}\\
&\geq \|w'_n\|^2_{L^2}-c_4\|w_n\|_{H_T^1},
\end{align*}
where $c_3$ and $c_4$ are two positive constants. Thus,
\[
\|w'_n\|^2_{L^2}\leq(c_3+c_4)\|w_n\|_{H_T^1}.
\]
Consequently, using the Wirtinger inequality, we obtain the existence
of a positive constant $c_5$ such that
\begin{equation} \label{3-5}
\|u'_n\|^2_{L^2}\leq c_5.
\end{equation}
Now, suppose that
$\|u_n\|_{H_T^1}\to+\infty$ as $n\to+\infty$.
Since \eqref{3-5} holds, we have, passing to subsequence if
necessary, that either
\begin{gather*}
M_n:=\max u_n\to+\infty\quad\text{as } n\to+\infty,\quad\text{or}\\
m_n:=\min u_n\to-\infty\quad\text{as } n\to+\infty.
\end{gather*}
(i) Assume that the first possibility occurs. In view to the fact
that $f_{\lambda}<0$, one has
\begin{align*}
&\int_0^{T}[F_{\lambda}(u_n(t))-e(t)u_n(t)]dt-\sum_{j=1}^{p-1}b_ju_n(t_j)\\
&= \int_0^{T}\Big[\Big(\int_1^{M_n}f_{\lambda}(s)ds-\int_{u_n(t)}^{M_n}
f_{\lambda}(s)ds\Big)-e(t)u_n(t)\Big]dt-M_n\sum_{j=1}^{p-1}b_j\\
&\quad -\sum_{j=1}^{p-1}b_j\left(u_n(t_j)-M_n\right)
\\
&\geq \int_0^{T}F_{\lambda}(M_n)dt-\int_0^{T}M_ne(t)dt
 -\max_{t\in[0,T]}|M_n-u_n(t)|\int_0^{T}|e(t)|dt-M_n\sum_{j=1}^{p-1}b_j\\
&\quad -\max_{t\in[0,T]}|M_n-u_n(t)|\sum_{j=1}^{p-1}|b_j|\\
&\geq TF_{\lambda}(M_n)-M_n\Big(\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}b_j\Big)
 -\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)|M_n-m_n|\\
&= TF_{\lambda}(M_n)-M_n\Big(\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}b_j\Big)
 -\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)
 \Big|\int_{\bar{t}_n}^{\hat{t}_n}u'_n(t)dt\Big|\\
&\geq TF_{\lambda}(M_n)-M_n\Big(\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}b_j\Big)
 -\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)
 \int_{\bar{t}_n}^{\hat{t}_n}|u'_n(t)|dt,
\end{align*}
where $u_n(\hat{t}_n)=M_n$ and $u_n(\bar{t}_n)=m_n$.
Thus, using the H$\ddot{\text{o}}$lder inequality, one has
\begin{equation} \label{3-6}
\begin{aligned}
&-M_n\Big(\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}b_j\Big)
 +TF_{\lambda}(M_n)\\
&\leq\int_0^{T} [F_{\lambda}(u_n(t))-e(t)u_n(t)]dt
 -\sum_{j=1}^{p-1}b_ju_n(t_j)+\sqrt{T}\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|
 \Big)\|u_n'\|_{L^2}.
\end{aligned}
\end{equation}
If $\alpha=1$, then $F_{\lambda}(M_n)=-\ln M_n$. By \eqref{3-6}
one has
\[
-M_n\Big(\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}b_j\Big)-T\ln
M_n\to+\infty\quad\text{as } n\to+\infty.
\]
If $\alpha>1$, then
$F_{\lambda}(M_n)=-\frac{1}{\alpha-1}(\frac{1}{M_n^{\alpha-1}}-1)$.
By \eqref{3-6} we obtain
\[
-M_n\Big(\int_0^{T}e(t)dt+\sum_{j=1}^{p-1}b_j\Big)
-\frac{1}{\alpha-1}(\frac{1}{M_n^{\alpha-1}}-1)\to+\infty\quad\text{as }
n\to+\infty.
\]
From \eqref{3-3} and \eqref{3-5}, we see that the right hand side of
\eqref{3-6} is bounded, which is a contradiction.

(ii) Assume the second possibility occurs; i.e.,
$m_n\to-\infty$ as $n\to+\infty$. We replace $M_n$
by $-m_n$ in the preceding arguments, and we also get a
contradiction. So $\{u_n\}$ is bounded in $H_T^1$.

Since $H_T^1$ is a reflexive Banach space, there exists a
subsequence of $\{u_n\}$, denoted again by $\{u_n\}$ for
simplicity, and $u\in H_T^1$ such that $u_n\rightharpoonup u$ in
$H_T^1$; then, by the Sobolev embedding theorem, we get
$u_n\to u$ in $C([0, T])$ and $u_n\to u$ in
$L^2([0, T])$. So
\begin{equation} \label{3-7}
\begin{gathered}
\int_0^{T}(f_{\lambda}(u_n(t))-f_{\lambda}(u(t)))(u_n(t)-u(t))dt\to0, \\
\sum_{j=1}^{p-1}b_j(u_n(t_j)-u(t_j))\to0,\\
\int_0^{T}e(t)(u_n(t)-u(t))dt\to0, \\
(\Phi_{\lambda}'(u_n)-\Phi_{\lambda}'(u))(u_n-u)\to0,\quad\text{as }
n\to\infty.
\end{gathered}
\end{equation}
By \eqref{3-6}, \eqref{3-7} and the fact that $u_n\to u$ in
$L^2([0, T])$, we have $\|u_n-u\|_{H_T^1}\to0$ as
$n\to\infty$. That is, $\{u_n\}$ strongly converges to $u$
in $H^1_T$, which means that the Palais-Smale condition holds
for $\Phi_{\lambda}$.

\textbf{Step 2.} Let
\[
\Omega=\big\{u\in H^1_T|\min_{t\in[0,T]}u(t)>1\big\},
\]
and
\[
\partial\Omega
=\{u\in H^1_T|u(t)\geq1\ \text{for all}\
t\in(0,T),\ \exists t_u\in(0,T): u(t_u)=1\}.
\]
We  show that there exists $d>0$ such that
$\inf_{u\in\partial\Omega}\Phi_{\lambda}(u)\geq-d$ whenever
$\lambda\in(0,1)$.

For any $u\in\partial\Omega$, there exists some $t_{u}\in(0,T)$ such
that $\min_{t\in[0,T]}u(t)=u(t_{u})=1$. By \eqref{2-1}, and
extending the functions by $T$-periodicity, we have
\begin{align*}
\Phi_{\lambda}(u)
&= \int_{t_{u}}^{t_{u}+T}\big[\frac{1}{2}|u'(t)|^2-F_{\lambda}(u(t))+e(t)u(t)\big]dt
+\sum_{j=1}^{p-1}b_ju(t_j)\\
&\geq \frac{1}{2}\int_{t_{u}}^{t_{u}+T}|u'(t)|^2dt+\int_{t_{u}}^{t_{u}+T}e(t)(u(t)-1)
 dt+\int_{t_{u}}^{t_{u}+T}e(t)dt\\
&\quad +\sum_{j=1}^{p-1}b_j(u(t_j)-1)+\sum_{j=1}^{p-1}b_j\\
&\geq \frac{1}{2}\|u'\|^2_{L^2}-\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)
 \max_{t\in[0,T]}(u(t)-1)-\|e\|_{L^1}+\sum_{j=1}^{p-1}b_j\\
&= \frac{1}{2}\|u'\|^2_{L^2}-\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)
 \int_{t_{u}}^{\check{t}_{u}}u'(t)dt-\|e\|_{L^1}+\sum_{j=1}^{p-1}b_j\\
&\geq \frac{1}{2}\|u'\|^2_{L^2}-\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)
\int_{t_{u}}^{t_{u}+T}|u'(t)|dt-\|e\|_{L^1}+\sum_{j=1}^{p-1}b_j,
\end{align*}
where $\check{t}_{u}\in[0,T]$ and
$\max_{t\in[0,T]}u(t)=u(\check{t}_{u})$.  Applying the
H$\ddot{\text{o}}$lder inequality, we get
\[
\Phi_{\lambda}(u)
\geq\frac{1}{2}\|u'\|^2_{L^2}-\sqrt{T}\Big(\|e\|_{L^1}+\sum_{j=1}^{p-1}|b_j|\Big)
\|u'\|_{L^2}-\|e\|_{L^1}+\sum_{j=1}^{p-1}b_j.
\]
The above inequality shows that
\[
\Phi_{\lambda}(u)\to+\infty\quad{as }\|u'\|_{L^2}\to+\infty.
\]

For any $u\in\partial\Omega$, it is easy to verify the fact that
$\|u\|_{H_T^1}\to+\infty$ is equivalent to
$\|u'\|_{L^2}\to+\infty$. Indeed, when
$\|u'\|_{L^2}\to+\infty$, it is clear that
$\|u\|_{H_T^1}\to+\infty$. When
$\|u\|_{H_T^1}\to+\infty$. Assume that $\|u'\|_{L^2}$
is bounded, then $\|u\|_{L^2}\to+\infty$. Since
$\min_{t\in[0,T]}u(t)=1$, we have
\[
u(t)-1=\int_{t_{u}}^{t}u'(s)ds\leq\int_0^{T}|u'(s)|ds\leq\sqrt{T}
\Big(\int_0^{T}|u'(t)|^2dt\Big)^{1/2}.
\]
Therefore, $u$ is bounded in $L^2(0,T)$, which is a contradiction.
Hence
\[
\Phi_{\lambda}(u)\to+\infty\quad\text{as }
\|u\|_{H_T^1}\to+\infty,\; \forall u\in\partial\Omega,
\]
which shows that $\Phi_{\lambda}$ is coercive. Thus it has a
minimizing sequence. The weak lower semi-continuity of
$\Phi_{\lambda}$ yields
\[
\inf_{u\in\partial\Omega}\Phi_{\lambda}(u)>-\infty.
\]
It follows that there exists $d>0$ such that
$\inf_{u\in\partial\Omega}\Phi_{\lambda}(u)>-d$ for all
$\lambda\in(0,1)$.

\textbf{Step 3.} We prove that there exists $\lambda_0\in(0,1)$ with
the property that for every $\lambda\in(0,\lambda_0)$, any solution
$u$ of Problem \eqref{eSPl} satisfying $\Phi_{\lambda}(u)>-d$ is
such that $\min_{u\in[0,T]}u(t)\geq\lambda_0$, and hence $u$ is a
solution of Problem \eqref{eSP}.

Assume on the contrary  that there are sequences
$\{\lambda_n\}_{n\in\mathbb{N}}$ and $\{u_n\}_{n\in\mathbb{N}}$
such that
\begin{itemize}
\item[(i)] $\lambda_n\leq \frac{1}{n}$;
\item[(ii)] $u_n$ is a solution of Problem \eqref{2-1} with $\lambda=\lambda_n$;
\item[(iii)] $\Phi_{\lambda_n}(u_n)\geq-d$;
\item[(iv)] $\min_{t\in[0,T]}u_n(t)<\frac{1}{n}$.
\end{itemize}
Since $f_{\lambda_n}<0$ and
$\int_0^{T}[f_{\lambda_n}(u_n(t))-e(t)]dt=0$, one has
\[
\|f_{\lambda_n}(u_n(\cdot))\|_{L^1}\leq c_7,\quad
\text{for some constant } c_7>0.
\]
Hence
\begin{equation}
\label{3-8} \|u'_n\|_{L^{\infty}}\leq c_8,\quad\text{for some
constant } c_8>0.
\end{equation}
From $\Phi_{\lambda_n}(u_n)\geq-d$ it follows that there must
exist two constants $l_1$ and $l_2$, with $0<l_1<l_2$ such that
\[
\max\{u_n(t);t\in[0,T]\}\subset[l_1,l_2].
\]
If not, $u_n$ would tend uniformly to $0$ or $+\infty$. In both
cases, by \eqref{3-8}, we have
\[
\Phi_{\lambda_n}(u_n)\to-\infty\quad\text{as }n\to+\infty,
\]
which contradicts $\Phi_{\lambda_n}(u_n)\geq-d$.

Let $\tau_n^1, \tau_n^2$ be such that, for $n$ large enough
$$
u_n(\tau_n^1)=\frac{1}{n}<l_1=u_n(\tau_n^2).
$$
Multiplying the differential equation in \eqref{eSPl} by $u'_n$
and integrating it on $[\tau_n^1,\tau_n^2]$, or on
$[\tau_n^2,\tau_n^1]$, we get
\begin{equation} \label{3-9}
\Psi:= \int_{\tau_n^1}^{\tau_n^2}u''_n(t)u'_n(t)dt
+\int_{\tau_n^1}^{\tau_n^2}f_{\lambda_n}(u_n(t))u'_n(t)dt
= \int_{\tau_n^1}^{\tau_n^2}e(t)u'_n(t)dt.
\end{equation}
It is easy to verify that
\[
\Psi=\Psi_1+\frac{1}{2}[u'^2_n(\tau_n^2)-u'^2_n(\tau_n^1)],
\]
where
\[
\Psi_1=\int_{\tau_n^1}^{\tau_n^2}f_{\lambda_n}(u_n(t))u'_n(t)dt.
\]
From \eqref{3-5} and \eqref{3-9} it follows that $\Psi$ is bounded,
and consequently $\Psi_1$ is bounded.

On the other hand, it is easy to see that
\[
f_{\lambda_n}(u_n(t))u'_n(t)=\frac{d}{dt}[F_{\lambda_n}(u_n(t))].
\]
Thus, we have
\[
\Psi_1=F_{\lambda_n}(l_1)-F_{\lambda_n}\big(\frac{1}{n}\big).
\]
From the fact that
$F_{\lambda_n}\left(\frac{1}{n}\right)\to+\infty$ as
$n\to+\infty$, we obtain $\Psi_1\to-\infty$, i.e.,
$\Psi_1$ is unbounded. This is a contradiction.

\textbf{Step 4.} $\Phi$ has a mountain-pass geometry for
$\lambda\leq\lambda_0$.
Fix $\lambda\in(0,\lambda_0]$, one has
\begin{equation}\label{3-10}
\begin{aligned}
F_{\lambda}(0)
&= \int_1^{0}f_\lambda(s)ds=-\int_0^1f_\lambda(s)ds \\
&= -\int_0^{\lambda}f_\lambda(s)ds-\int_{\lambda}^1f_\lambda(s)ds \\
&= \frac{1}{\lambda^{\alpha-1}}-\int_{\lambda}^1f_\lambda(s)ds,
\end{aligned}
\end{equation}
which implies that
\[
 F_{\lambda}(0)>-\int_{\lambda}^1f_\lambda(s)ds
=\int_1^{\lambda}f_\lambda(s)ds=F_{\lambda}(\lambda).
\]
Thus we have
\begin{equation} \label{3-11}
\Phi_{\lambda}(0)
= -TF_{\lambda}(0)<-TF_{\lambda}(\lambda) 
= \begin{cases}
T\ln\lambda,&\text{if } \alpha=1,\\
-\frac{T}{\alpha-1}\big(\frac{1}{\lambda^{\alpha-1}}-1\big),
&\text{if }\alpha>1.
\end{cases}
\end{equation}
We choose
$\lambda\in(0,\lambda_0]\cap(0,e^{-d})\cap
(0,[\frac{T}{T+d(\alpha-1)}]^{1/(\alpha-1)})$,
then it follows from \eqref{3-11} that $\Phi_{\lambda}(0)<-d$.

Also, we can choose a constant $R>1$ enough large such that
\[
-\Big(\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt\Big)R
-\frac{T}{\alpha-1}\big(1-\frac{1}{R^{\alpha-1}}\big)>d,
\]
and
\[
-\Big(\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt\Big)R-T\ln R>d.
\]
Thus, $R\in H_T^1$ and
\begin{align*}
\Phi_{\lambda}(R)
&= R\sum_{j=1}^{p-1}b_j-TF_{\lambda}(R) +R\int_0^{T}e(t)dt\\
&\leq \begin{cases}
\sum_{j=1}^{p-1}b_jR+T\ln R+R\int_0^{T}e(t)dt,&\text{if } \alpha=1,\\[4pt]
\sum_{j=1}^{p-1}b_jR+\frac{T}{\alpha-1}\big(1-\frac{1}{R^{\alpha-1}}\big)
+R\int_0^{T}e(t)dt,&\text{if }\alpha>1.
\end{cases}\\
&\leq \begin{cases}
\big(\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt\big)R+T\ln R,& \text{if } \alpha=1,\\[4pt]
\big(\sum_{j=1}^{p-1}b_j+\int_0^{T}e(t)dt\big)R+\frac{T}{\alpha-1}
\big(1-\frac{1}{R^{\alpha-1}}\big),&\text{if }\alpha>1.
\end{cases}\\
&<-d.
\end{align*}
Since $\Omega$ is a neighborhood of $R$, $0\not\in\Omega$ and
\[
\max\{\Phi_{\lambda}(0),
\Phi_{\lambda}(R)\}<\inf_{x\in\partial\Omega}\Phi_{\lambda}(u),
\]
Step 1 and Step 2 imply that $\Phi_{\lambda}$ has a critical point
$u_\lambda$ such that
\[
\Phi_{\lambda}(u_\lambda)
=\inf_{h\in\Gamma}\max_{s\in[0,1]}\Phi_{\lambda}(h(s))
\geq\inf_{x\in\partial\Omega}\Phi_{\lambda}(u),
\]
where
\[
\Gamma=\{h\in C([0,1],H_T^1):h(0)=0,h(1)=R\}.
\]
Since $\inf_{u\in\partial\Omega}\Phi_{\lambda}(u_\lambda)\geq-d$, it
follows from Step 3 that $u_\lambda$ is a positive solution of
Problem \eqref{eSP}. The proof of the main result is complete.
\end{proof}

\subsection*{Acknowledgments} 
Juntao Sun was supported by the National
Natural Science Foundation of China (Grant No. 11201270), Shandong
Natural Science Foundation (Grant No. ZR2012AQ010), and Young
Teacher Support Program of Shandong University of Technology. 
Jifeng Chu was supported by the National Natural Science Foundation of
China (Grant Nos. 11171090, 11271078, and 11271333), the Program for
New Century Excellent Talents in University (Grant No.
NCET-10-0325), and China Postdoctoral Science Foundation funded
project (Grant Nos. 20110491345 and 2012T50431).



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