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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 35, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2011/35\hfil Impulsive differential equations]
{Mixed two-point boundary-value problems for impulsive
differential equations}

\author[Z. Han, S. Wang \hfil EJDE-2011/35\hfilneg]
{Zhiqing Han, Suqin Wang}  % in alphabetical order

\address{Zhiqing Han \newline
School of Mathematical Sciences, 
Dalian University of Technology,
Dalian, Liaoning, 116024,  China}
\email{hanzhiq@dlut.edu.cn}

\address{Suqin Wang \newline
School of Mathematical Sciences, 
Dalian University of Technology,
 Dalian, Liaoning, 116024,  China}
\email{wangsuqinwsq@126.com}

\thanks{Submitted November 22, 2010. Published March 3, 2011.}
\subjclass[2000]{34B15, 34B25}
\keywords{Impulsive differential equations;
mixed boundary value problem; \hfill\break\indent
variational methods; resonant and non-resonant}

\begin{abstract}
 In this article, we prove the existence of solutions to
 mixed two-point boundary-value problem for impulsive
 differential equations by variational methods, in
 both  resonant and the non resonant cases.
\end{abstract}

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

\section{Introduction}

In this article, we study the existence of solutions to the
impulsive problem
\begin{equation}\label{e1.1}
\begin{gathered}
 -u''+\lambda u=f(t,u), \quad\text{a.e. }t\in (0,T),\\
 u(0)=0,\quad u'(T)=0,\\
 \Delta u'(t_j)=I_j(u(t_j)), \quad j=1,2,\dots p,
\end{gathered}
\end{equation}
 where $I_j: \mathbb{R}\to\mathbb{R}$, $j=1,2,\dots, p$,
are continuous functions and $f(t,u)$ satisfies the condition
\begin{itemize}
\item[(A)] $f(t,u)$ is measurable in $t$ for each
$u\in\mathbb{R}$, continuous in $u$ for a.e. $t\in[0,T]$, and
there exist functions $g\in C(\mathbb{R}^{+},\mathbb{R}^{+})$ and
$h\in L^{1} (0,T;\mathbb{R}^{+})$ such that $|f(t,u)|\leq g(|u|)h(t)$,
for all $u\in \mathbb{R}$ and a.e. $t\in[0,T]$.
\end{itemize}

 The theory of impulsive differential equations describes
processes which experience a sudden change of their states at
certain time. It can be successfully used for mathematical
simulation in some problems from theoretical physics, chemistry,
medicine, population dynamics, optimal control and in some other
processes and phenomena in science and technology, see
\cite{VL,SP} for  the general aspects of the theory. Some
classical tools such as upper and lower solutions, monotone
iterative technique, fixed point theory, degree theory and so on
have been widely used to such equations, we refer to
\cite{AR,CTY,CS,CN,CH,GD,VL,NJA,QL} for some references. In recent
years, the variational methods \cite{MW,RA} have been applied to
such equations and are proved to be very effective, we refer to
\cite{NJ,TYG,TYW,HZ,ZR} for some  results.

Let us recall some results related to the impulsive  problem
\eqref{e1.1}
obtained  by the variational methods. Nieto and O'Regan
(\cite{NJ}) first noticed that the equation in \eqref{e1.1}
 coupled with the Dirichlet boundary value condition and
the impulsive conditions has a variational
structure and obtained some existence  results for the problem.
The results are extended to more general nonlinearities in
\cite{ZR}. Recently, Tian and Ge \cite{TYG,TYW}  investigated
the equation with the more general (Sturm-Liouville) boundary
value conditions. That is they considered the equation with the
Sturm-Liouville boundary value conditions
$\alpha u'(0)-\beta u(0)=0$, $\gamma u'(T)+\sigma u(T)=0$
but with a restriction
$\alpha,\gamma>0$, $\beta,\sigma \geq0$. The restriction excludes
the boundary value conditions of the  problem \eqref{e1.1} investigated
in this paper.

In this paper, we  investigate the  problem \eqref{e1.1} still  by  the
variational methods. The  framework involved is different from
those for the other kinds of boundary value conditions. We obtain
some existence results both by the Ambrosetti-Rabinowitz type
condition in the non-resonant case   and the generalized
Ahmad-Lazer-Paul type  condition in the resonant case.

\section{Preliminaries}

 Denote $H=\{u(t)|u(t)$ is
absolutely continuous on $[0,T]$, $u(0)=0$,
$u'(t)\in L^2(0,T;\mathbb{R})\}$. It is easy to see that
$H_0^{1}(0,T)\subset H\subset H^{1}(0,T)$ and  $H$ is a closed
subset of $H^{1}(0,T)$. So $H$ is a Hilbert space with the usual
inner product in $H^{1}(0,T)$.

\begin{proposition} \label{prop2.1}
If  $u\in H$, then
$\|u\|_{c}\leq \sqrt{T}\|u'\|_{L^2}$, where
$$
\|u\|_{c}=\max_{t\in[0,T]}|u(t)|.
$$
\end{proposition}

\begin{proof} For  $u\in H$, we have
$$
|u(t)|=|\int_0^{t}u'(s)ds|\leq\sqrt{t}
\Big(\int_0^{t}|u'(s)|^2ds\Big)^{1/2}
\leq\sqrt{T}\|u'\|_{L^2}.
$$
\end{proof}

For the linear problem
\begin{equation} \label{LP}
\begin{gathered}
-u''(t)=\lambda u(t), \quad t\in (0,T),\\
 u(0)=0,\quad  u'(T)=0,
\end{gathered}
\end{equation}
there is a sequence of eigenvalues
$\lambda_k=(2k+1)^2\pi^2/(4T^2)$ and the corresponding
$L^2(0,T)$-normalized eigenfunctions
$\phi_k(t)=\sqrt{(2/T)} \sin((2k+1)\pi t/(2T)$, $k=0,1,2\dots$.

If the equation in the problem \eqref{LP} is coupled with the boundary
value condition $u'(0)=0$, $u(T)=0$, the eigenvalues are as before
and the corresponding $L^2(0,T)$-normalized eigenfunctions are
$\psi_k(t)=\sqrt{(2/T)}\cos((2k+1)\pi t/(2T))$, $k=0,1,2 \dots$.

According to the Sturm-Liouville theory, $\{\phi_k\}$ and
$\{\psi_k\}$ are complete bases in the Hilbert space $L^2(0,T)$.


\noindent\textbf{Claim} If $u\in H$, then
$\|u'\|^2_{L^2}\geq\lambda_0\|u\|^2_{L^2}$ (Poincar\'e inequality).

In fact, for any $u\in H$,  set
$u=\sum_{k=0}^{\infty}c_k\phi_k$,
where
\[
c_k=(u,\phi_k)=\frac{2}{T}\int_0^Tu\phi_kdt.
\]
By the Parseval identity,
$\int_0^Tu^2dt=\sum_{k=0}^{\infty}c_k^2$. Let
$u'=\sum_{k=0}^{\infty}a_k\psi_k$, where
$a_k=(u',\psi_k)=\frac{2}{T}\int_0^Tu'\psi_kdt$.
By an easy calculation,
$$
a_k=\frac{2}{T}\int_0^Tu'\psi_kdt
=-\frac{2}{T}\int_0^Tu\psi'_kdt=\frac{(2k+1)\pi}{2T}
\cdot\frac{2}{T}\int_0^Tu\phi_kdt
=\frac{(2k+1)\pi}{2T}c_k.
$$
Then we have
$$
\int_0^T|u'|^2dt=\sum_{k=0}^{\infty}a_k^2
=\sum_{k=0}^{\infty}\frac{(2k+1)^2\pi^2}{4T^2}c_k^2
\geq\frac{\pi^2}{4T^2}\sum_{k=0}^{\infty}c_k^2
=\lambda_0\int_0^T|u|^2dt.
$$
Similarly, we can prove that
$\|u'\|^2_{L^2}\geq\lambda_1\|u\|^2_{L^2}$, for all
$u\in H\ominus \operatorname{span}\{\phi_0\}$, which will
 be used later.

According to the Poincar\'e inequality, we can define an inner
product in $H$, $(u,v)=\int_0^Tu'(t)v'(t)dt$. The induced norm
$\|u\|=\sqrt{\int_0^Tu'^2dt}$  is equivalent to the old one.
Throughout the paper, we will use this norm.

\begin{proposition} \label{prop2.2}
Assume that
$\mu(t)\in L^1(0,T)$ satisfies
\begin{equation}
0\leq\mu(t)\leq\lambda_0,\label{e2.1}
\end{equation} for a.e. $t\in [0,T]$ and $\mu(t)<\lambda_0$
holds on a subset of $[0,T]$ with positive measure. Then
there exists $\delta>0$ such that for all $u\in H$, one has
$$
\int_0^T[u'^2(t)-\mu(t)u^2(t)]dt\geq\delta\int_0^Tu'^2(t)dt.
$$
\end{proposition}

\begin{proof}
We use some known arguments \cite{RMN} to prove it.
By \eqref{e2.1} and the Poincar\'e inequality, we see that for
all $u\in H$,
\begin{equation}
\int_0^T[u'^2(t)-\mu(t)u^2(t)]dt\geq\int_0^T[u'^2(t)
-\lambda_0u^2(t)]dt\geq0.\label{e2.2}
\end{equation}
Now we assume that the  proposition were false. Then we can find a
sequence $\{u_n\}$ in $H$,  $u_n\neq0$   such that
$$
\int_0^T[u_n'^2(t)-\mu(t)u_n^2(t)]dt
<\frac{1}{n}\int_0^Tu_n'^2(t)dt.
$$

Let $v_n=u_n/\|u_n\|$. We obtain
$$
\int_0^T[v_n'^2(t)-\mu(t)v_n^2(t)]dt<\frac{1}{n}.
$$
We can assume that $v_n\rightharpoonup v$ in $H$ and
$v_n\to v$ in $C[0,T]$. We obtain
\begin{equation}
1\leq\int_0^T\mu(t)v^2(t)dt.\label{e2.3}
\end{equation}
 From \eqref{e2.3}, we have
\begin{equation}
v\neq0.\label{e2.4}
\end{equation}
Since $v_n\rightharpoonup v $ in $H$, we have
$$
\|v\|^2\leq\liminf_{n\to \infty}\|v_n\|^2=1.
$$
Hence, we have
$$
\int_0^T[v'^2(t)-\mu(t)v^2(t)]dt\leq0.
$$
Therefore,
\begin{equation}
\int_0^T[v'^2(t)-\mu(t)v^2(t)]dt=0.\label{e2.5}
\end{equation}
From \eqref{e2.2}, we have
$$
\int_0^T[v'^2(t)-\lambda_0v^2(t)]dt=0,
$$
which shows that $v(t)$ is an eigenfunction corresponding to
$\lambda_0$. So we set $v(t)=C\phi_0(t)$, where $C$ is a
constant. Substituting it to \eqref{e2.5} and noticing that
$\mu(t)<\lambda_0$ holds on a subset of $[0,T]$ with positive
measure, we obtain $C=0$, which contradicts to \eqref{e2.4}. This
completes the proof.
\end{proof}

\section{Non-resonant case}

In this section, we study
the existence of solutions for the non-resonant impulsive problem
\begin{equation} \label{e3.1}
\begin{gathered}
 -u''+\lambda u=f(t,u), \quad\text{a.e. }t\in (0,T),\\
  u(0)=0,\quad  u'(T)=0, \\
  \Delta u'(t_j)=I_j(u(t_j)), \quad j=1,2,\dots p
\end{gathered}
\end{equation}
with
$\lambda>-\lambda_0$, where
$0=t_0<t_1<\dots<t_p<t_{p+1}=T$,
$\Delta u'(t_j)=u'(t_j^{+})-u'(t_j^{-})$,
$f(t,u)$ satisfies the  condition (A) and
$I_j:\mathbb{R}\to \mathbb{R}$, $j=1,2,\dots, p$, are continuous
functions. Here $u'(t_j^{+})$ (respectively $u'(t_j^{-})$)
 denotes the right limit (respectively left limit) of $u'(t)$
at $t=t_j$.

\begin{definition} \label{def3.1} \rm
A function $u$ is said to be a classical
solution of \eqref{e3.1} if $u$ satisfies:
\begin{enumerate}
\item  $u\in C[0,T]$;
\item  For all $j=0,1,2,\dots, p$,
 $u_j=u|_{(t_j,t_{j+1})}\in H^{2,1}(t_j,t_{j+1})$;
\item  $u$ satisfies the equation in  \eqref{e3.1} a.e. on $(0,T)$,
the boundary value  condition and the impulsive conditions.
\end{enumerate}
\end{definition}

Set
$F(t,u)=\int_0^{u}f(t,\xi)d\xi$ and define the functional $J$ on
$H$ by
$$
J(u)=\frac{1}{2}\int_0^Tu'^2dt
+\frac{\lambda}{2}\int_0^Tu^2dt
 +\sum_{j=1}^p\int_0^{u(t_j)}I_j(s)ds-\int_0^TF(t,u)dt.
$$
It is easily verified that $J$ is continuously differentiable on
$H$ and
$$
(J'(u),v)=\int_0^Tu'v'dt+\lambda\int_0^Tuv\,dt
+\sum_{j=1}^pI_j(u(t_j))v(t_j)-\int_0^Tf(t,u)v\,dt
$$
for all $u,v\in H$.

\begin{definition} \label{def3.2} \rm
A weak solution for problem \eqref{e3.1} is a function $u\in H$
satisfying  $(J'(u),v)=0$ for all $v\in H$.
\end{definition}

\begin{lemma} \label{lem3.3}
 If $u\in H$ is a weak
solution of \eqref{e3.1}, then it is a classical solution.
\end{lemma}

\begin{proof}
 Let $u\in H$ be a weak solution of \eqref{e3.1}. Then
\begin{equation} \label{e3.2}
0=(J'(u),v)=\int_0^Tu'v'dt+\lambda\int_0^Tuv\,dt
+\sum_{j=1}^pI_j(u(t_j))v(t_j)
-\int_0^Tf(t,u)v\,dt
\end{equation}
for all $v\in H$. For $j\in \{0,1,\dots, p\}$ and every
$v\in H_0^{1}(t_j,t_{j+1})$, set
$$
\widetilde{v}(t)=\begin{cases}
v(t), &t_j\leq t\leq t_{j+1},\\
0 ,& t\in [0,T]\setminus [t_j,t_{j+1}].
\end{cases}
$$
Then $\widetilde{v}\in H_0^{1}(0,T)\subset H$. Then
replacing  $v$ in \eqref{e3.2} by  $\widetilde{v}$, we have
$$
\int_{t_j}^{t_{j+1}}u'v'dt+\lambda\int_{t_j}^{t_{j+1}}uv\,dt
-\int_{t_j}^{t_{j+1}}f(t,u)v\,dt=0
$$
 for every $v\in H_0^{1}(t_j,t_{j+1})$. Hence by standard
results, $u\in H^{2,1}(t_j,t_{j+1})$ and
\[
-u''+\lambda u=f(t,u), \quad\text{a.e. on }(t_j,t_{j+1}).
\]
 Hence   $u$ satisfies
\begin{equation} \label{e3.3}
-u''+\lambda u=f(t,u), \quad\text{a. e. on }(0,T).
\end{equation}
 Fix a positive number $\delta$ such that $t_p+\delta<T$.
 Now, for every
$v_1\in H_0^{1}(0,t_p+\delta)\subset H_0^{1}(0,T)\subset H$.
Multiplying \eqref{e3.3} by $v_1$ and integrating it between $0$ and
$T$, we obtain
$$
-\int_0^Tu''v_1dt+\lambda\int_0^Tuv_1dt
=\int_0^Tf(t,u)v_1dt.
$$
That is,
$$
\sum_{j=1}^p\Delta u'(t_j)v_1(t_j)
+\int_0^Tu'v_1'dt+\lambda\int_0^Tuv_1dt
=\int_0^Tf(t,u)v_1dt.
$$
According to \eqref{e3.2}, we have
$$
\sum_{j=1}^p\Delta u'(t_j)v_1(t_j)
=\sum_{j=1}^pI_j(u(t_j))v_1(t_j).
$$
Hence  we obtain $\Delta u'(t_j)=I_j(u(t_j))$ for
all $j=1,2,\dots,p$.


Finally, we prove that $u$ satisfies the condition $u'(T)=0$.
For all $v_2\in H^{1}(t_p,T)$ with $v_2(t_p)=0$, we define
$$
v^{*}(t)=\begin{cases}
v_2(t), & t\in[t_p,T],\\
v_2(t_p), & t\in [0,t_p].
\end{cases}
$$
It is easy to see that $v^{*}\in H$. Hence,
$$
\int_{t_p}^Tu'v_2'dt+\lambda\int_{t_p}^Tuv_2dt
-\int_{t_p}^Tf(t,u)v_2dt=0.
$$
Set $u^{*}(t)=\int_{t}^T(-\lambda u(s)+f(s,u(s)))ds$. By the
Fubini theorem, we obtain
\begin{align*}
\int_{t_p}^Tu^{*}(t)v_2'(t)dt
&= \int_{t_p}^T\Big[\int_{t}^T(-\lambda
u(s)+f(s,u(s)))v_2'(t)ds\Big]dt\\
&= \int_{t_p}^T\Big[\int_{t_p}^{s}(-\lambda
u(s)+f(s,u(s)))v_2'(t)dt\Big]ds\\
&= \int_{t_p}^T(-\lambda u(s)+f(s,u(s)))v_2(s)dt\\
&=\int_{t_p}^Tu'(t)v_2'(t)dt.
\end{align*}
Hence, for every $v_2\in H^1(t_p,T)$ with $v_2(t_p)=0$, we
have
$$
\int_{t_p}^T(u^{*}(t)-u'(t))v_2'(t)dt=0.
$$
In particular, we can choose
$v_2(t)=\phi^{\ast}_k(t)=\sqrt{2/(T-t_p)}
 \sin(\frac{(2k+1)\pi}{2(T-t_p)}(t-t_p))$, $k=0,1,2 \dots$, then
$$
\int_{t_p}^T(u^{*}(t)-u'(t))\psi^{\ast}_k(t)dt=0,~~\forall k=0,1,2
\dots$$ where
$$
\psi^{\ast}_k(t)=\sqrt{\frac{2}{(T-t_p)}}
\frac{(2k+1)\pi}{2(T-t_p)}\cos(\frac{(2k+1)\pi}{2(T-t_p)}(t-t_p)).
$$

By noticing that $\{\psi^{\ast}_k(t)\}$ is the sequence of the
eigenfunctions of the  eigenvalue problem
\[
-u''(t)=\lambda u(t), \quad t\in (t_p,T)
\]
with the boundary value condition $u(t_p)=0$, $u'(T)=0$,
hence it is complete  in $L^2(t_p,T)$. Thus, $u^{*}(t)=u'(t)$ a.e.
on $(t_p,T)$; that is,
\[
u'(t)=\int_{t}^T(-\lambda u(s)+f(s,u(s)))ds.
\]
Hence, we  obtain $u'(T)=0$ and $u\in H^{2,1}(t_p,T)$ by
condition (A).
So, $u$ is a classical solution of \eqref{e3.1}. This completes
the proof.
\end{proof}

\begin{theorem} \label{thm3.4}
Suppose that $f(t,u)$ satisfies
condition {\rm (A)}. Moreover, the following conditions hold.
\begin{itemize}
\item[(F1)]  $F(t,u)\leq a(t)|u|^2+b(t)|u|+c(t)$, where
$a(t),b(t),c(t)\in L^{1}(0,T)$;

\item[(I1)] $|I_j(u)|\leq a_j+b_j|u|,\forall u\in
\mathbb{R}$, where $a_j,b_j\geq 0$, $j=1,2,\dots, p$, and
$\frac{1}{2}-\frac{T}{2}(\sum_{j=1}^pb_j)>0$;

\item[(I2)]
$a(t)\leq\lambda_0[\frac{1}{2}-\frac{T}{2}(\sum_{j=1}^pb_j)]
+\frac{\lambda}{2} =a^{*}$ with $a(t)< a^{*}$ holds on a subset of
$[0,T]$ with positive measure.
\end{itemize}
Then  problem \eqref{e3.1} has at least one solution.
\end{theorem}

\begin{proof} For all $u\in H$,
\begin{align*}
\sum_{j=1}^p\int_0^{u(t_j)}I_j(s)ds
&\leq \sum_{j=1}^p\int_0^{|u(t_j)|}|I_j(s)|ds
\leq\sum_{j=1}^p\int_0^{|u(t_j)|}(a_j+b_j|s|)ds\\
&\leq (\sum_{j=1}^pa_j)\|u\|_{c}+\frac{1}{2}(\sum_{j=1}^pb_j)\|u\|_{c}^2\\
&\leq (\sqrt{T}\sum_{j=1}^pa_j)\|u\|+\frac{T}{2}(\sum_{j=1}^pb_j)\|u\|^2
\end{align*}
and
\[
\int_0^TF(t,u)dt\leq\int_0^T(a(t)|u|^2+b(t)|u|+c(t))dt\leq\int_0^Ta(t)|u|^2dt+C\|u\|+C
\]
where and in the following $C$ denotes a universal  constant.
Then
\begin{align*}
J(u)
&= \frac{1}{2}\int_0^Tu'^2dt+\frac{\lambda}{2}\int_0^Tu^2dt
+\sum_{j=1}^p\int_0^{u(t_j)}I_j(s)ds-\int_0^TF(t,u)dt\\
&\geq \frac{1}{2}\int_0^Tu'^2dt+\frac{\lambda}{2}\int_0^Tu^2dt
-\sum_{j=1}^p\int_0^{|u(t_j)|}|I_j(s)|ds-\int_0^Ta(t)|u|^2dt\\
&\quad -C\|u\|-C\\
&\geq \frac{1}{2}\int_0^Tu'^2dt+\int_0^T
(\frac{\lambda}{2}-a(t))|u|^2dt-(\sqrt{T}\sum_{j=1}^pa_j)\|u\|
-\frac{T}{2}(\sum_{j=1}^pb_j)\|u\|^2\\
&\quad -C\|u\|-C\\&= [\frac{1}{2}-\frac{T}{2}(\sum_{j=1}^pb_j)]
\int_0^Tu'^2dt
-\int_0^T(a(t)-\frac{\lambda}{2})|u|^2dt-C\|u\|-C.
\end{align*}
By Proposition \ref{prop2.2} and the condition (I2), there exists
$\delta>0$ such that
$$
\Big[\frac{1}{2}-\frac{T}{2}(\sum_{j=1}^pb_j)\Big]
\int_0^Tu'^2dt -\int_0^T(a(t)-\frac{\lambda}{2})|u|^2dt
\geq\delta\int_0^Tu'^2dt,\quad \forall u\in H.
$$
Thus
$$
J(u)\geq\delta\|u\|^2-C\|u\|-C,\quad \forall u\in H.
$$
So
\[
\lim_{\|u\|\to\infty,u\in H}J(u)=+\infty.
\]
Hence every minimizing sequence is bounded. It is easily verified
by the condition (A) and a compact imbedding result that $J(u)$
is weakly lower semi-continuous. Hence by a standard result,
$J(u)$ has a minimizing point $u$, which is a classical solution
of the problem \eqref{e3.1}.
\end{proof}

Since $\lambda>-\lambda_0$, we can choose the equivalent norm
$\|u\|^2_1=\int_0^T(u'^2+\lambda u^2)dt$ in $H$. Hence
there exist positive constants $m_1$ and $m_2$ such that
$m_1\|u\|\leq\|u\|_1\leq m_2\|u\|$.

\begin{theorem} \label{thm3.5}
Suppose that $f(t,u)$
satisfies the condition {\rm (A)}. Moreover, we assume that
{\rm (I1)} and the following conditions hold:
\begin{itemize}
\item[(F2)]  $\lim_{u\to 0}\frac{f(t,u)}{u}<k_1\lambda_0$
uniformly for a.e. $t\in[0,T]$, where $k_1<m_1^2$;

\item[(F3)] There exist $\mu>2$ and $R>0$ such that for a.e.
$t\in[0,T]$ and $|u|\geq R$,
 $0<\mu F(t,u)\leq uf(t,u)$ (Ambrosetti-Rabinowitz type condition);

\item[(I3)] $\lim_{u\to0}\frac{I_j(u)}{u}\to0$ for all
$1\leq j\leq p$;

\item[(I*)]
$[(\frac{1}{2}-\frac{1}{\mu})-\frac{1}{m_1^2}(\frac{1}{2}
+\frac{1}{\mu})T(\sum_{j=1}^pb_j)]>0$.
\end{itemize}
Then  problem \eqref{e3.1} has at least one nontrivial solution.
\end{theorem}

\begin{proof}
Obviously, $J(0)=0$. By (F3) and  condition (A), there
exist nonnegative functions  $d_1(t)$, $d_2(t)\in L^1(0,T)$
such that $d_1(t)>0$ a.e. on $[0,T]$ and
$ F(t,u)\geq d_1(t)|u|^{\mu}-d_2(t)$ for a.e. $t\in[0,T]$ and
all $u \in \mathbb{R}$.

Choosing  $u\in H\backslash\{0\}$, then for $q>0$, we have
\begin{align*}
J(qu)
&\leq \frac{q^2}{2}\|u\|^2_1
+\sum_{j=1}^p\int_0^{|qu(t_j)|}|I_j(s)|ds-\int_0^TF(t,qu)dt\\
&\leq \frac{q^2}{2}\|u\|^2_1+C q\|u\|+C q^2\|u\|^2
-\int_0^T(d_1(t)|qu|^{\mu}-d_2(t))dt\\
&\leq \frac{q^2}{2}\|u\|^2_1+C q\|u\|_1+C q^2\|u\|_1^2
-|q|^{\mu} \int_0^T d_1(t)|u|^{\mu}dt+C\\
&\to -\infty
\end{align*}
as $q\to\infty$. Setting  $e=qu$, then for $q$ large, we
obtain $\|e\|_1>R$ and $J(e)\leq0$.

By (F2) and (I3), for some proper $\varepsilon>0$, there
exists $0<r<R$ such that for $|u|\leq r$, there holds
\[
|F(t,u)|\leq\frac{k_1\lambda_0}{2}|u|^2, \quad
|I_j(u)|\leq\varepsilon|u|.
\]
For $u\in H$ with $\|u\|<\frac{r}{\sqrt{T}}$, we have
$\|u\|_{c}<r$. Then $\|u\|_1\leq
m_2\|u\|<\frac{m_2r}{\sqrt{T}}$. So
\begin{align*}
J(u)
&= \frac{1}{2}\|u\|^2_1+\sum_{j=1}^p
 \int_0^{u(t_j)}I_j(s)ds
 -\int_0^TF(t,u)dt\\
&\geq \frac{1}{2}\|u\|^2_1-\frac{Tp\varepsilon}{2m_1^2}\|u\|^2_1
-\frac{k_1\lambda_0}{2}\frac{1}{\lambda_0}
 \frac{1}{m_1^2}\|u\|^2_1\\
&= (\frac{1}{2}-\frac{k_1}{2m_1^2})\|u\|^2_1
 -\frac{Tp\varepsilon}{2m_1^2}\|u\|^2_1.
\end{align*}
There exists $0<\rho<\min\{R,\frac{m_2r}{\sqrt{T}}\}$ and
$\sigma>0$ such that for  $\|u\|_1=\rho$, we have $J(u)>\sigma$.

 Let $\{u_k\}$ be a PS sequence in $H$; that is,
$$
|J({u}_k)|\leq C,\; \forall k\in \mathbb{N},\quad
|(J'(u_k),h)|\leq o(1)\|h\|_1\quad \text{for all }k\in
\mathbb{N}\text{ and }h\in H.
$$
We only need to prove that
$\{u_k\}$ is bounded. For $k$ large,
\begin{align*}
C+\frac{1}{\mu}\|u_k\|_1
&\geq J({u}_k)-\frac{1}{\mu}J'(u_k)u_k=(\frac{1}{2}
 -\frac{1}{\mu})\int_0^T(|u'_k|^2+\lambda u_k^2)dt\\
&\quad +\sum_{j=1}^p\int_0^{u_k(t_j)}I_j(s)ds
 -\frac{1}{\mu}\sum_{j=1}^pI_j(u_k(t_j))u_k(t_j)
 -\int_0^TF(t,u_k)dt\\
&\quad +\frac{1}{\mu}\int_0^Tf(t,u_k)u_kdt,
\end{align*}
where
\begin{align*}
-\int_0^TF(t,u_k)dt+\frac{1}{\mu}\int_0^Tf(t,u_k)u_k\,dt
&=-\frac{1}{\mu}\int_0^T(\mu F(t,u_k)dt-f(t,u_k)u_k)dt \\
&= -\frac{1}{\mu}\int_{|u_k|\geq R}(\mu
F(t,u_k)dt-f(t,u_k)u_k)dt\\
&\quad -\frac{1}{\mu}\int_{|u_k|\leq R}(\mu F(t,u_k)dt-f(t,u_k)u_k)dt.
\end{align*}
The above  first term  is nonnegative  by (F3), the second  is
bounded by the condition (A). Moreover,
\begin{align*}
& |\sum_{j=1}^p\int_0^{u_k(t_j)}I_j(s)ds
-\frac{1}{\mu}\sum_{j=1}^pI_j(u_k(t_j))u_k(t_j)|\\
&\leq |\sum_{j=1}^p\int_0^{u_k(t_j)}I_j(s)ds|
+\frac{1}{\mu}|\sum_{j=1}^pI_j(u_k(t_j))u_k(t_j)|
\\
&\leq (\sum_{j=1}^pa_j)\|u_k\|_{c}+\frac{1}{2}
 (\sum_{j=1}^pb_j)\|u_k\|_{c}^2+
\frac{1}{\mu}(\sum_{j=1}^pa_j)\|u_k\|_{c}+\frac{1}{\mu}
(\sum_{j=1}^pb_j)\|u_k\|_{c}^2\\
&\leq (1+\frac{1}{\mu})\sqrt{T}(\sum_{j=1}^pa_j)\|u_k\|
 +(\frac{1}{2}+\frac{1}{\mu})T(\sum_{j=1}^pb_j)\|u_k\|^2\\
&\leq \frac{1}{m_1}(1+\frac{1}{\mu})\sqrt{T}
 (\sum_{j=1}^pa_j)\|u_k\|_1+\frac{1}{m_1^2}(\frac{1}{2}
+\frac{1}{\mu})T(\sum_{j=1}^pb_j)\|u_k\|^2_1.
\end{align*}
Hence
\begin{align*}
C+\frac{1}{\mu}\|u_k\|_1
&\geq (\frac{1}{2}-\frac{1}{\mu})\|u_k\|_1^2
-\frac{1}{m_1}(1+\frac{1}{\mu})\sqrt{T}(\sum_{j=1}^pa_j)\|u_k\|_1
\\
&\quad -\frac{1}{m_1^2}(\frac{1}{2}+\frac{1}{\mu})T(\sum_{j=1}^pb_j)\|u_k\|_1^2\\
&= [(\frac{1}{2}-\frac{1}{\mu})
-\frac{1}{m_1^2}(\frac{1}{2}+\frac{1}{\mu})
T(\sum_{j=1}^pb_j)]\|u_k\|_1^2\\
&\quad -\frac{1}{m_1}(1+\frac{1}{\mu})\sqrt{T}
(\sum_{j=1}^pa_j)\|u_k\|_1.
\end{align*}
Therefore, by the condition (I*),  $\{u_k\}$ is bounded in $H$.

By  the Mountain Pass Lemma \cite{MW,RA}, $J(u)$ possesses
a critical point $u\in H$ such that  $J(u)\geq \sigma>0$; hence
$u$ is a nontrivial weak solution of \eqref{e3.1}.
\end{proof}

\section{Resonance case}

In this section, we study the existence of solutions for the
resonant impulsive problem
\begin{equation} \label{e4.1}
\begin{gathered}
         u''+\lambda_0u=f(t,u), \quad\text{a.e. }t\in (0,T),\\
        u(0)=0,\quad  u'(T)=0, \\
        \Delta u'(t_j)=I_j(u(t_j)), \quad j=1,2,\dots,p
 \end{gathered}
\end{equation}
where  $f(t,u)$ satisfies  condition (A),
 $I_j:\mathbb{R}\to \mathbb{R}$, $j=1,2,\dots, p$, are
continuous functions.

The corresponding  functional $J:H\to \mathbb{R}$ is
defined by
$$
J(u)=\frac{1}{2}\int_0^Tu'^2dt-\frac{\lambda_0}{2}
\int_0^Tu^2dt+\sum_{j=1}^p\int_0^{u(t_j)}I_j(s)ds
+\int_0^TF(t,u)dt.
$$

We will consider the case where the nonlinearity satisfies some
kind of sublinear  and  generalized  Ahmad, Lazer, Paul type
coercive conditions; e.g. see  \cite{ZH} and \cite{HZ} and the
references therein  for some applications to the periodic boundary
value problems.

Decompose $H$ as $H=\bar{H}\oplus\widetilde{H}$,  where
$\bar{H}=\operatorname{span}\{\phi_0\}$ and
$\widetilde{H}=\overline{\operatorname{span}\{\phi_1,\phi_2,\dots\}}$.

For all $u\in H$, we write as
$u=\bar{u}\oplus\widetilde{u}$,
$\bar{u}\in\bar{H}$, $\widetilde{u}\in\widetilde{H}$.
We recall the inequality,
$$
\int_0^T\widetilde{u}'^2dt\geq\lambda_1\int_0^T\widetilde{u}^2dt
$$
for all $\widetilde{u}\in\widetilde{H}$.

\begin{theorem}\label{thm4.1}
 Suppose that $f(t,u)$ satisfies  condition (A). Moreover,
the following conditions hold.
\begin{itemize}
\item[(F4)]   There exists $\alpha$ with $\frac{1}{2}\leq\alpha<1$
such that $|f(t,u)|\leq g(t)|u|^{\alpha}+h(t)$,
where $g(t),h(t)\in L^{1}(0,T)$;

\item[(F5)]  $\lim_{|\bar{u}|\to\infty}\int_0^TF(t,\bar{u})dt
/|\bar{u}|^{2\alpha}=+\infty$, $\bar{u}\in\bar{H}$;

\item[(I4)]  $|I_j(u)|\leq p_j+q_j|u|^{\gamma_j}$ where
$p_j,q_j\geq 0$, $j=1,2,\dots,p$, and
$0\leq\gamma_j\leq2\alpha-1$.
\end{itemize}
Then problem \eqref{e4.1} has at least one solution.
\end{theorem}

\begin{proof}
For all $u\in H$, $u=\bar{u}\oplus\widetilde{u}$,
$\bar{u}\in\bar{H}$, $\widetilde{u}\in\widetilde{H}$,
we have
\begin{equation} \label{e4.2}
\begin{split}
|\sum_{j=1}^p\int_0^{u(t_j)}I_j(s)ds|
&\leq \sum_{j=1}^p\int_0^{|u(t_j)|}|I_j(s)|ds
\leq\sum_{j=1}^p\int_0^{|u(t_j)|}(p_j+q_j|s|^{\gamma_j})ds\\
&\leq (\sum_{j=1}^pp_j)\|u\|_{c}+\sum_{j=1}^pq_j
 \frac{1}{\gamma_j+1}\|u\|^{\gamma_j+1}\\
&\leq C|\bar{u}|+C\|\widetilde{u}\|+C|\bar{u}|^{\gamma+1}
+C\|\widetilde{u}\|^{\gamma+1}+C,
\end{split}
\end{equation}
where $\gamma=\max\{\gamma_j, j=1,2,\dots,p\}$.
By (F4), we have
\begin{equation}
\begin{split}
\int_0^T(F(t,u)-F(t,\bar{u}))dt
&\leq \int_0^T[\int_0^{1}f(t,\bar{u}+s\widetilde{u})
 \widetilde{u}ds]dt\\
&\leq \int_0^T[\int_0^{1}(g(t)|\bar{u}+s\widetilde{u}|^{\alpha}
 +h(t))\widetilde{u}ds]dt\\
&\leq C|\bar{u}|^{\alpha}\|\widetilde{u}\|_{c}+C\|\widetilde{u}
 \|_{c}^{\alpha+1}+C\|\widetilde{u}\|_{c}\\
&\leq C|\bar{u}|^{\alpha}\|\widetilde{u}\|+C\|\widetilde{u}\|^{\alpha+1}
 +C\|\widetilde{u}\|\\
&\leq \varepsilon\|\widetilde{u}\|^2+C(\varepsilon)
 |\bar{u}|^{2\alpha}+C\|\widetilde{u}\|^{\alpha+1}
 +C\|\widetilde{u}\|.
\end{split}\label{e4.3}
\end{equation}
Then
\begin{align*}
J(u)&= \frac{1}{2}\int_0^Tu'^2dt-\frac{\lambda_0}{2}\int_0^Tu^2dt
+\sum_{j=1}^p\int_0^{u(t_j)}I_j(s)ds\\
&\quad +\int_0^T(F(t,u)-F(t,\bar{u}))dt+\int_0^TF(t,\bar{u})dt\\
&\geq \frac{1}{2}\int_0^T|\widetilde{u}'|^2dt-\frac{\lambda_0}{2}\cdot\frac{1}{\lambda_1}\int_0^T|\widetilde{u}'|^2dt-C|\bar{u}|-C\|\widetilde{u}\|
-C|\bar{u}|^{\gamma+1}-C\|\widetilde{u}\|^{\gamma+1}\\
&\quad -\varepsilon\|\widetilde{u}\|^2
-C(\varepsilon)|\bar{u}|^{2\alpha}-C\|\widetilde{u}\|^{\alpha+1}
-C\|\widetilde{u}\|+\int_0^TF(t,\bar{u})dt-C\\
&\geq \big[\frac{1}{2}(1-\frac{\lambda_0}{\lambda_1})
-\varepsilon\big]\|\widetilde{u}\|^2-C\|\widetilde{u}\|^{\alpha+1}
-C\|\widetilde{u}\|^{\gamma+1}-C\|\widetilde{u}\|
-C(\varepsilon)|\bar{u}|^{2\alpha}-C\\
&\quad +\int_0^TF(t,\bar{u})dt\\
&\geq \big[\frac{1}{2}(1-\frac{\lambda_0}{\lambda_1})
-\varepsilon\big]\|\widetilde{u}\|^2-C\|\widetilde{u}\|^{\alpha+1}
-C\|\widetilde{u}\|^{\gamma+1}-C\|\widetilde{u}\|
+|\bar{u}|^{2\alpha}[-C(\varepsilon)\\
&\quad +\frac{1}{|\bar{u}|^{2\alpha}}\int_0^TF(t,\bar{u})dt]-C,
\end{align*}
where we have used $0\leq\gamma\leq 2\alpha-1$. Fixing some
$0<\epsilon<\frac{1}{2}(1-\frac{\lambda_0}{\lambda_1})$, by
(F5), we have $J(u)\to+\infty$ as $\|u\|\to\infty$.
Noticing also the weak lower
semi-continuity of $J(u)$, we complete the proof.
\end{proof}

\begin{theorem} \label{thm4.2}
 Suppose that {\rm (F4), (I4)} and the following condition
 are satisfied:
\begin{itemize}
\item[(F6)]
$\lim_{|\bar{u}|\to\infty}\int_0^TF(t,\bar{u})dt/|\bar{u}|^{2\alpha}
=-\infty$, $\bar{u}\in\bar{H}$.
\end{itemize}
Then  problem \eqref{e4.1} has at least one solution.
\end{theorem}

\begin{proof}
We apply the saddle point theorem \cite{MW,RA}
to prove the theorem.

Step 1. For $u=\bar{u}\in \bar{H}$,
\begin{align*}
J(\bar{u})
&= \frac{1}{2}\int_0^T|\bar{u}'|^2dt
 -\frac{\lambda_0}{2}\int_0^T|\bar{u}|^2dt
 -\sum_{j=1}^p\int_0^{\bar{u}(t_j)}I_j(s)ds
 +\int_0^TF(t,\bar{u})dt\\
&\leq C|\bar{u}|+C|\bar{u}|^{\gamma+1}+\int_0^TF(t,\bar{u})dt+C\\
&\leq C|\bar{u}|^{2\alpha}+\int_0^TF(t,\bar{u})dt+C\\
&= |\bar{u}|^{2\alpha}(C+\frac{1}{|\bar{u}|^{2\alpha}}
 \int_0^TF(t,\bar{u})dt)+C.
\end{align*}
Hence, we have $J(\bar{u})\to-\infty$ as $|\bar{u}|\to\infty$.

Step 2. For $u=\widetilde{u}\in \widetilde{H}$, we have
\begin{align*}
\int_0^T(F(t,\widetilde{u})-F(t,0))dt
&= \int_0^T[\int_0^{1}f(t,s\widetilde{u})\widetilde{u}ds]dt\\
&\leq \int_0^T[\int_0^{1}(g(t)|s\widetilde{u}|^{\alpha}
 +h(t))\widetilde{u}ds]dt\\
&\leq C\|\widetilde{u}\|^{\alpha+1}+C\|\widetilde{u}\|.
\end{align*}
By some arguments in the proof of Theorem \ref{thm4.1}, we obtain
\begin{align*}
J(\widetilde{u})
&\geq \frac{1}{2}(1-\frac{\lambda_0}{\lambda_1})
 \int_0^T|\widetilde{u}'|^2dt
 +\sum_{j=1}^p\int_0^{\widetilde{u}(t_j)}I_j(s)ds
 +\int_0^TF(t,\widetilde{u})dt\\
&\geq \frac{1}{2}(1-\frac{\lambda_0}{\lambda_1})
 \int_0^T|\widetilde{u}'|^2dt-C\|\widetilde{u}\|
 -C\|\widetilde{u}\|^{\gamma+1}+\int_0^T(F(t,\widetilde{u})
 -F(t,0))dt \\
&\quad +\int_0^TF(t,0)dt\\
&\geq \frac{1}{2}(1-\frac{\lambda_0}{\lambda_1})\|\widetilde{u}\|^2
-C\|\widetilde{u}\|-C\|\widetilde{u}\|^{\gamma+1}
-C\|\widetilde{u}\|^{\alpha+1} -C\|\widetilde{u}\|-C.
\end{align*}
Since $\frac{1}{2}\leq\alpha<1$ and $0\leq\gamma\leq2\alpha-1$, we
obtain that $J(u)$ is bounded below on $\widetilde{H}$.
So there exists $R>0$ such that
$$
\sup_{u\in S_{R}}J(u)<\inf_{u\in \widetilde{H}}J(u)
$$
where $S_{R}=\{u|~\|u\|=R,~~u\in \bar{H}\}$.

Step 3. We show that $J$ satisfies the PS condition. Let
$\{u_k\}$ be a PS sequence in $H$, then there exists constant
$C$ such that
$$
|J({u}_k)|\leq C,\; \forall k\in \mathbb{N},\quad
|(J'(u_k),h)|\leq o(1)\|h\|\text{ for all }k\in \mathbb{N}, \;
h\in H.
$$ Since
\begin{align*}
&|\sum_{j=1}^pI_j(u_k(t_j))\widetilde{u}_k(t_j)|\\
&\leq \sum_{j=1}^pp_j|\widetilde{u}_k(t_j)|+\sum_{j=1}^pq_j|\bar{u}_k(t_j)|^{\gamma_j}|\widetilde{u}_k(t_j)|
 +\sum_{j=1}^pq_j|\widetilde{u}_k(t_j)|^{\gamma_j+1}\\
&\leq C\|\widetilde{u}_k\|+C|\bar{u}_k|^{\gamma}\|\widetilde{u}_k\|
+C\|\widetilde{u}_k\|^{\gamma+1}
\end{align*}
and
\begin{align*}|
\int_0^Tf(t,u_k)\widetilde{u}_kdt|
&\leq \int_0^T(g(t)|u_k|^{\alpha}+h(t))|\widetilde{u}_k|dt\\
&\leq C|\bar{u}_k|^{\alpha}\|\widetilde{u}_k\|
 +C\|\widetilde{u}_k\|^{\alpha+1}+C\|\widetilde{u}_k\|,
\end{align*}
we have, for $k$ large,
\begin{align*}
\|\widetilde{u}_k\|
&\geq |(J'(u_k),\widetilde{u}_k)|=|\int_0^Tu'_k\widetilde{u}'_kdt-\lambda_0\int_0^Tu_k\widetilde{u}_kdt
-\sum_{j=1}^pI_j(u_k(t_j))\widetilde{u}_k(t_j)\\
&\quad +\int_0^Tf(t,u_k)\widetilde{u}_kdt|\\
&\geq (1-\frac{\lambda_0}{\lambda_1})\|\widetilde{u}_k\|^2
-C\|\widetilde{u}_k\|-C|\bar{u}_k|^{\gamma}\|\widetilde{u}_k\|
-C\|\widetilde{u}_k\|^{\gamma+1}\\
&\quad -C|\bar{u}_k|^{\alpha}\|\widetilde{u}_k\|
-C\|\widetilde{u}_k\|^{\alpha+1}-C\|\widetilde{u}_k\|.
\end{align*}
Therefore, we obtain
\begin{equation}
\|\widetilde{u}_k\|\leq C|\bar{u}_k|^{\gamma}
+C|\bar{u}_k|^{\alpha}+C\leq C|\bar{u}_k|^{\alpha}+C.\label{e4.4}
\end{equation}
Hence, noticing the arguments in \eqref{e4.2} and \eqref{e4.3},
we have
\begin{align*}
J(u_k)
&= \frac{1}{2}\int_0^T|\tilde{u}_k'|^2dt
 -\frac{\lambda_0}{2}\int_0^T\tilde{u}_k^2dt
+\sum_{j=1}^p\int_0^{u_k(t_j)}I_j(s)ds \\
&\quad +\int_0^T(F(t,u_k)-F(t,\bar{u}_k))dt
 +\int_0^TF(t,\bar{u}_k)dt\\
&\leq \frac{1}{2}\|\widetilde{u}_k\|^2
+C|\bar{u}_k|+C\|\widetilde{u}_k\|
+C|\bar{u}_k|^{\gamma+1}+C\|\widetilde{u}_k\|^{\gamma+1}\\
&\quad +\varepsilon\|\widetilde{u}_k\|^2
+C(\varepsilon)|\bar{u}_k|^{2\alpha}
 +C\|\widetilde{u}_k\|^{\alpha+1}+C\|\widetilde{u}_k\|
 +\int_0^TF(t,\bar{u}_k)dt\\
&\leq (\frac{1}{2}+\varepsilon)\|\widetilde{u}_k\|^2
+C\|\widetilde{u}_k\|^{\gamma+1}+C\|\widetilde{u}_k\|^{\alpha+1}
+C\|\widetilde{u}_k\|+C|\bar{u}_k|
+C|\bar{u}_k|^{\gamma+1}\\
&\quad +C(\varepsilon)|\bar{u}_k|^{2\alpha}
 +\int_0^TF(t,\bar{u}_k)dt+C.
\end{align*}
 From \eqref{e4.4}, we obtain
\begin{gather*}
\|\widetilde{u}_k\|^2\leq (C|\bar{u}_k|^{\alpha}+C)^2
\leq C|\bar{u}_k|^{2\alpha}+C,\\
\|\widetilde{u}_k\|^{\gamma+1}\leq (C|\bar{u}_k|^{\alpha}
 +C)^{\gamma+1}\leq C|\bar{u}_k|^{2\alpha}+C,\\
\|\widetilde{u}_k\|^{\alpha+1}\leq (C|\bar{u}_k|^{\alpha}+C)^{\alpha+1}
\leq C|\bar{u}_k|^{2\alpha}+C.
\end{gather*}
So
\begin{align*}
J(u_k)&\leq C(\varepsilon)|\bar{u}_k|^{2\alpha}
 +\int_0^TF(t,\bar{u}_k)dt+C\\
&\leq |\bar{u}_k|^{2\alpha}[C(\varepsilon)
+\frac{1}{|\bar{u}_k|^{2\alpha}}\int_0^TF(t,\bar{u}_k)dt]+C.
\end{align*}
Hence, if  $\{|\bar{u}_k|\}$ has a unbounded subsequence, we
will get a contradiction by ($F_6$).  Therefore,
$\{|\bar{u}_k|\}$ is bounded and moreover $\{u_k\}$ is bounded
in $H$. By a standard argument, $\{u_k\}$ has a convergent
subsequence. We obtain that $J(u)$ satisfies the PS condition.

Then existence of a critical point for $J$ then  follows  from the
saddle point theorem. The proof  is complete.
\end{proof}

Examining carefully the proofs of the Theorems \ref{thm4.1} and
\ref{thm4.2},  we
can also get the following two theorems.

\begin{theorem} \label{thm4.3}
If the conditions of
Theorem \ref{thm4.1} are replaced by the following conditions:
\begin{itemize}
\item[(F4')]  There exists  $\alpha$ with $0\leq\alpha<1$ such that
$ |f(t,u)|\leq g(t)|u|^{\alpha}+h(t)$, where
$g(t),h(t)\in L^{1}(0,T)$;

\item[(I4')]  $|I_j(u)|\leq p_j+q_j|u|^{\gamma_j}$, where
$p_j,q_j\geq 0$, $j=1,2,\dots,p$, $0\leq\gamma_j<1$;

\item[(F5')]
$\lim_{|\bar{u}|\to\infty,\bar{u}\in\bar{H}
}(\int_0^TF(t,\bar{u})dt)/(|\bar{u}|^{\beta})=+\infty$,
where $\beta=\max\{\gamma+1,2\alpha\}$,
$\gamma=\max\{\gamma_j$, $j=1,2,\dots,p\}$.
\end{itemize}
Then  problem \eqref{e4.1} has at least one solution.
\end{theorem}

\begin{theorem} \label{thm4.4}
Suppose that the conditions {\rm (F4'), (I4')} and the following
condition are satisfied:
\begin{itemize}
\item[(F5'')]
$\lim_{|\bar{u}|\to\infty}(\int_0^TF(t,\bar{u})dt)/(|\bar{u}|^{\beta})
=-\infty$, $\bar{u}\in\bar{H}$ where $\beta=\max\{\gamma+1,2\alpha\}$.
\end{itemize}
Then  problem  \eqref{e4.1} has at least one solution.
\end{theorem}


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