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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 147, pp. 1--11.\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-2014/147\hfil Dirichlet impulsive differential equations]
{Existence of solutions to Dirichlet impulsive differential equations through
a local minimization principle}

\author[G. A. Afrouzi, S. Shokooh, A. Hadjian \hfil EJDE-2014/147\hfilneg]
{Ghasem A. Afrouzi, Saeid Shokooh, Armin Hadjian}  % in alphabetical order

\address{Ghasem A. Afrouzi \newline
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran}
\email{afrouzi@umz.ac.ir}

\address{Saeid Shokooh \newline
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran}
\email{saeid.shokooh@stu.umz.ac.ir}

\address{Armin Hadjian \newline
Department of Mathematics, Faculty of Basic Sciences, University of
Bojnord, P.O. Box 1339, Bojnord 94531, Iran}
\email{hadjian83@gmail.com}

\thanks{Submitted January 20, 2014. Published June 24, 2014.}
\subjclass[2000]{34B37, 34B15, 58E05}
\keywords{Impulsive differential equations; Dirichlet condition; 
\hfill\break\indent classical solution; variational methods}

\begin{abstract}
 A critical point theorem (local minimum result) for differentiable
 functionals is used for proving that a Dirichlet  impulsive differential
 equation admits at least one non-trivial  solution. Some particular
 cases and a concrete example are also presented.
\end{abstract}

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


\section{Introduction}

In this article, we study the existence of at least one
non-trivial classical solution to the  nonlinear Dirichlet
boundary-value problem
\begin{equation}\label{e1.1}
\begin{gathered}
-(p(t)u'(t))'+q(t)u(t)=\lambda f(t,u(t)), \quad t\in [0,T],\; t\not= t_j,\\
u(0)=u(T)=0,\\
\Delta u'(t_j)=\lambda I_j(u(t_j)),\quad j=1,2,\dots,n,
\end{gathered}
\end{equation}
where $T>0$, $p\in C^1([0,T],]0,+\infty[)$, 
$q\in L^{\infty}([0,T])$, $\lambda \in ]0,+\infty[$, 
$f:[0,T]\times \mathbb{R} \to \mathbb{R}$ is an $L^1$-Carath\'{e}odory function,
$0=t_0<t_1<t_2<\dots<t_n<t_{n+1}=T$, 
$\Delta u'(t_j)=u'(t^+_j)-u'(t^-_j)= \lim_{t \to t^+_j}u'(t) -\lim_{t \to
t^-_j}u'(t)$ and $I_j: \mathbb{R} \to \mathbb{R}$ are continuous for
every $j=1,2,\dots,n$.

The study of impulsive boundary-value problems is important due to
its various applications in which abrupt changes at certain times in
the evolution process appear. The dynamics of evolving processes is
often subjected to abrupt changes such as shocks, harvesting, and
natural disasters. Often these short-term perturbations are treated
as having acted instantaneously or in the form of impulses. Such
problems arise in physics, population dynamics, biotechnology,
pharmacokinetics, industrial robotics.

Recently, many researchers pay their attention to impulsive
differential equations by variational method and critical point
theory, and we refer the reader to
\cite{chta1,nior,such,tige,xini,jizh} and references cited therein.


Our analysis is mainly based on the critical point theorem by
Bonanno \cite{Bonanno}, contained in Theorem \ref{the2.1} below.
This theorem has been used in several works  to obtain
existence results for different kinds of problems (see, for
instance,
\cite{AfHadHei,BonaDiOreg,BonaHeiOreg,BonaMoliRad1,BonaMoliRad2,
BonaPizz1,BonaPizz2,BonaSci1,BonaSci2,Hei1,Hei2}).


\section{Preliminaries}

Our main tool is Ricceri's variational principle 
\cite[Theorem 2.5]{Ricceri1} as given in \cite[Theorem 5.1]{Bonanno} which is
below recalled (see also \cite[Proposition 2.1]{Bonanno}).

For a given non-empty set $X$, and two functionals
$\Phi,\Psi:X\to\mathbb{R}$, we define the  functions
$$
\beta(r_1,r_2):=\inf_{v\in \Phi^{-1}(]r_1,r_2[)}
\frac{\sup_{u\in
\Phi^{-1}(]r_1,r_2[)}\Psi(u)-\Psi(v)}{r_2-\Phi(v)}
$$ 
and
$$
\rho(r_1,r_2):=\sup_{v\in \Phi^{-1}(]r_1,r_2[)}
\frac{\Psi(v)-\sup_{u\in
\Phi^{-1}(]-\infty,r_1])}\Psi(u)}{\Phi(v)-r_1}
$$ 
for all $r_1,r_2\in\mathbb{R},$ with $r_1<r_2$.


\begin{theorem}[{\cite[Theorem 5.1]{Bonanno}}]\label{the2.1}
Let $X$ be a reflexive real Banach space; $\Phi:X\to\mathbb{R}$ be a
sequentially weakly lower semicontinuous, coercive and continuously
G\^ateaux differentiable function whose G\^ateaux derivative admits
a continuous inverse on $X^\ast$; $\Psi:X\to\mathbb{R}$ be a
continuously G\^ateaux differentiable function whose G\^ateaux
derivative is compact. Assume that there are $r_1,r_2\in\mathbb{R}$,
with $r_1<r_2$, such that
\begin{equation}\label{e2.1}
\beta(r_1,r_2)<\rho(r_1,r_2).
\end{equation}
Then, setting $I_{\lambda}:=\Phi-\lambda \Psi$, for each
$\lambda\in]\frac{1}{\rho(r_1,r_2)},\frac{1}{\beta(r_1,r_2)}[$
there is $u_{0,\lambda}\in\Phi^{-1}(]r_1,r_2[)$ such that
$I_{\lambda}(u_{0,\lambda})\leq I_\lambda(u)$ for all $u\in
\Phi^{-1}(]r_1,r_2[)$ and $I'_\lambda(u_{0,\lambda})=0.$
\end{theorem}

In the Sobolev space $X:=H^1_0(0,T)$, consider the inner product
$$
(u,v):=\int_0^T p(t)u'(t)v'(t)\,dt+\int_0^T q(t)u(t)v(t)\,dt \,,
$$
which induces the norm
$$
\|u\|:=\Big(\int_0^T p(t)(u'(t))^2\,dt +\int_0^T
q(t)(u(t))^2\,dt\Big)^{1/2}.
$$
Then the following Poincar\'{e}-type inequality holds:
\begin{equation}\label{e2.2}
\Big[\int_0^Tu^2(t)\,dt\Big]^{1/2}\leq \frac{T}{\pi}
\Big[\int_0^T(u')^2(t)\,dt\Big]^{1/2}.
\end{equation}
Let us introduce some notation that will be used later. Assume that
$\frac{T^2q^-}{\pi^2}>-p^-$, where
$$
p^-:=\operatorname{ess\,inf}_{t\in[0,T]}p(t)>0, \quad
q^-:=\operatorname{ess\,inf}_{t\in[0,T]}q(t).
$$
Moreover, put $\sigma_0:=\min\{T^2q^-/\pi^2,0\}$
and $\delta:=\sqrt{p^-+\sigma_0}$. Then, we have the following
useful proposition.

\begin{proposition}\label{prop2.2}
Let $u \in X$. Then
\begin{gather}\label{e2.3}
\|u'\|_{L^2([0,T])}\leq \frac{1}{\delta} \|u\|, \\
\label{e2.4}
\|u\|_{\infty }\leq \frac{\sqrt T}{2\delta}\|u\|.
\end{gather}
\end{proposition}

\begin{proof}
First we prove \eqref{e2.3}. To this end, let $q^-\geq 0$. Then,
$\sigma_0=0$ and $\delta=\sqrt{p^-}$. Therefore,
\begin{align*}
\|u'\|_{L^2([0,T])}^2
&\leq\frac{1}{p^-}\int_0^T p(t)|u'(t)|^2\,dt\\
&\leq\frac{1}{p^-}\int_0^T \left(p(t)|u'(t)|^2+q(t)|u(t)|^2\right)dt
 =\frac{1}{\delta^2}\|u\|^2.
\end{align*}
Thus, the desired inequality \eqref{e2.3} follows. On the other
hand, if $q^-<0$, we have $\sigma_0=\frac{T^2q^-}{\pi^2}$ and
$\delta=\sqrt{p^-+\frac{T^2q^-}{\pi^2}}$. Obviously,
$$
q^-\|u\|_{L^2([0,T])}^2\leq\int_0^T q(t)|u(t)|^2\,dt.
$$
Now, applying inequality \eqref{e2.2} and bearing in mind that
$q^-<0$, one has
$$
\frac{T^2q^-}{\pi^2}\|u'\|_{L^2([0,T])}^2\leq
q^-\|u\|_{L^2([0,T])}^2.
$$
By the above inequalities we have
$$
\frac{T^2q^-}{\pi^2}\|u'\|_{L^2([0,T])}^2\leq\int_0^T
q(t)|u(t)|^2\,dt.
$$
This inequality together with
$$
p^-\|u'\|_{L^2([0,T])}^2\leq\int_0^T p(t)|u'(t)|^2\,dt,
$$
imply \eqref{e2.3}.

In view of H\"{o}lder's inequality and \eqref{e2.3}, one has
$$
\|u\|_{\infty }\leq \frac{\sqrt T}{2}\|u'\|_{L^2([0,T])}\leq
\frac{\sqrt T}{2\delta}\|u\|,
$$
which completes and the proof.
\end{proof}

Put $k:=\big(\|p\|_{\infty }+\frac{T^2}{\pi^2}\|q\|_{\infty
}\big)^{1/2}$. Then, from \eqref{e2.2} we have
\begin{equation}\label{e2.5}
\|u\|\leq k\|u'\|_{L^2([0,T])}.
\end{equation}
Here and in the sequel $f:[0,T]\times\mathbb{R}\to\mathbb{R}$ is an
$L^1$-Carath\'eodory function, namely:
\begin{itemize}
\item[(a)] the mapping $t\mapsto f(t,x)$ is measurable for every $x\in\mathbb{R}$;
\item[(b)] the mapping $x\mapsto f(t,x)$ is continuous for almost every $t\in [0,T]$;
\item[(c)] for every $\rho>0$ there exists a function $l_\rho\in L^1([0,T])$
such that
$$
\sup_{|x|\leq \rho}|f(t,x)|\leq l_{\rho}(t)
$$
for almost every $t\in [0,T]$.
\end{itemize}
Corresponding to $f$ we introduce the function
$F:[0,T]\times\mathbb{R}\to\mathbb{R}$ as follows
$$
F(t,x):=\int_0^x f(t,\xi)\,d\xi,
$$
for all $(t,x)\in [0,T]\times\mathbb{R}$.

By a \textit{classical solution} of problem \eqref{e1.1}, we mean a
function
$$
u\in\left\{w\in C([0,T]): w|_{{[t_j,t_{j+1}]}}\in
H^2([t_j,t_{j+1}])\right\}
$$
that satisfies the equation in \eqref{e1.1} a.e. on 
$[0,T]\setminus \{t_1,\dots,t_n\}$, the limits $u'(t^+_j)$, $u'(t^-_j)$,
$j=1,\dots,n$, exist, satisfy the impulsive conditions 
$\Delta u'(t_j)=\lambda I_j(u(t_j))$ and the boundary condition
$u(0)=u(T)=0$.

We say that a function $u \in X$ is a \textit{weak solution} of
problem \eqref{e1.1}, if $u$ satisfies
\begin{align*}
&\int_0^T p(t)u'(t)v'(t)\,dt+\int_0^Tq(t)u(t)v(t)\,dt \\
&-\lambda\Big(\int_0^T f(t,u(t))v(t)\,dt-\sum_{j=1}^n
p(t_j)I_j(u(t_j))v(t_j)\Big)=0,
\end{align*}
for any $v\in X$.

\begin{lemma}[{\cite[Lemma 2.3]{bodihe}}]\label{lem2.3}
The function $u\in X$ is a weak solution of problem \eqref{e1.1} if
and only if $u$ is a classical solution of \eqref{e1.1}.
\end{lemma}

\begin{lemma}[{\cite[Lemma 3.1]{bodihe}}]\label{lem2.4}
Assume that
\begin{itemize}
\item [(A1)] there exist constants $\eta, \theta>0$ and $\sigma\in [0,1[$
such that
$$
|I_j(x)|\leq2\eta|x|+\theta|x|^{\sigma+1}\quad \text{for all }
 x \in \mathbb{R}, \; j=1,2,\dots,n.
$$
\end{itemize}
Then, for any $u\in X$, we have
\begin{equation} \label{e2.6}
\big|\sum_{j=1}^n p(t_j)\int_0^{u(t_j)}I_j(x)\,dx\big|\le
\sum_{j=1}^n p(t_j)\Big(\eta\|u\|_\infty^2+\frac{\theta}{\sigma+2}
\|u\|_{\infty}^{\sigma+2}\Big).
\end{equation}
\end{lemma}

Also put
\[
\tilde p:=\sum_{j=1}^n p(t_j),\quad
\mu(\tau):=\frac{\sqrt{2}k\tau}{\delta},
\quad\Gamma_c:=\frac{\eta}{c}+
\Big(\frac{\theta}{\sigma+2}\Big)c^{\sigma-1},
\]
where $\eta$, $\theta$, $\sigma$ are given by (A1) and $c,\tau$
are positive constants.
We assume throughout, and without further mention, that the
assumption (A1) holds.


\section{Main results}

For a given non-negative constant $\nu$ and a positive constant
$\tau$ with $\delta^2\nu^2\neq 2k^2\tau^2,$ put
$$
a_\tau(\nu):=\frac{\int_0^T\max_{|x|\leq\nu}F(t,x)\,dt
+\tilde{p}\nu^3\Gamma_\nu
+\tilde{p}(\mu(\tau))^3\Gamma_{\mu(\tau)}-\int_{T/4}^{3T/4}F(t,\tau)\,dt}
{\delta^2\nu^2-2k^2\tau^2}.
$$

\begin{theorem}\label{the3.1}
Assume that there exist a non-negative constant $\nu_1$ and two
positive constants $\nu_2$ and $\tau$, with
$\nu_1<\sqrt 2\tau<\delta\nu_2/k$, such that
\begin{itemize}
\item[(A2)] $F(t,\xi)\geq 0$ for all $(t,\xi)\in([0,\frac{T}{4}]
\cup[\frac{3T}{4},T])\times [0,\tau]$;
\item[(A3)] $a_\tau(\nu_2)<a_\tau(\nu_1)$.
\end{itemize}
Then, for each
$\lambda\in]\frac{2}{T\,a_\tau(\nu_1)},\frac{2}{T\,a_\tau(\nu_2)}[,$
problem \eqref{e1.1} admits at least one non-trivial classical
solution $u_0\in X$ such that
$$
\frac{2\delta \nu_1}{\sqrt T}<\|u_0\|<\frac{2\delta \nu_2}{\sqrt T}.
$$
\end{theorem}

\begin{proof}
The aim is to apply Theorem \ref{the2.1} to our problem. To this
end, we introduce the functionals $\Phi,\Psi:X\to\mathbb{R}$
by setting
\begin{gather*}
\Phi(u):=\frac{1}{2}\|u\|^2, \\
\Psi(u):=\int_0^T F(t,u(t))\,dt-\sum_{j=1}^n
p(t_j)\int_0^{u(t_j)}I_j(x)\,dx,
\end{gather*}
for every $u\in X$, and put
$$
I_{\lambda}(u):=\Phi(u)-\lambda\Psi(u)\quad \forall\ u\in X.
$$
Clearly, $\Phi$ and $\Psi$ are well defined and continuously
G\^{a}teaux differentiable functionals whose G\^{a}teaux derivatives
at the point $u\in X$ are the functionals $\Phi'(u),\Psi'(u)\in
X^\ast$, given by
\begin{gather*}
\Phi'(u)(v)=\int_0^T p(t)u'(t)v'(t)\,dt + \int_0^T q(t)u(t)v(t)\,dt,\\
\Psi'(u)(v)=\int_0^Tf(t,u(t))v(t)\,dt-\sum_{j=1}^n
p(t_j)I_j(u(t_j))v(t_j),
\end{gather*}
for every $v\in X$. Moreover, $\Phi$ is coercive and sequentially
weakly lower semicontinuous and $\Psi$ is sequentially weakly upper
semicontinuous. Also, $\Phi'$ admits a continuous inverse on
$X^\ast$ and $\Psi'$ is compact. Note that the critical points of
the functional $I_\lambda$ in $X$ are exactly the weak solutions of
problem \eqref{e1.1}. We verify condition \eqref{e2.1} of Theorem
\ref{the2.1}. To this end, we put
\begin{gather*}
r_1:=\frac{2\delta^2}{T}\nu_1^2,\quad
r_2:=\frac{2\delta^2}{T}\nu_2^2, \\
w(t):=\begin{cases}
\frac{4\tau}{T}t, & \text{if } t\in [0,{T}/{4}[,\\
\tau, & \text{if } t\in [T/4,{3T}/{4}],\\
\frac{4\tau}{T}(T-t), & \text{if } t\in ]{3T}/{4},T].
\end{cases}
\end{gather*}
It is easy to verify that $w\in X$ and, in particular, taking
\eqref{e2.3} and \eqref{e2.5} into account, one has
$$
\frac{8\delta^2\tau^2}{T}\leq\|w\|^2 \leq \frac{8k^2\tau^2}{T}.
$$
So, we have
$$
\frac{4\delta^2\tau^2}{T}\leq\Phi(w)\leq \frac{4k^2\tau^2}{T}.
$$
From the condition $\nu_1<\sqrt 2\tau<\delta\nu_2/k$, we obtain 
$r_1<\Phi(w)<r_2$. Moreover,
for all $u\in X$ such that $u\in\Phi^{-1}(]-\infty,r_2[)$, from
\eqref{e2.4}, one has $|u(t)|<\nu_2$ for all $t\in [0,T]$, from
which it follows
\begin{align*}
\sup_{u\in \Phi^{-1}(]-\infty,r_2[)}\Psi(u)
&= \sup_{u\in  \Phi^{-1}(]-\infty,r_2[)}
\Big(\int_0^T F(t,u(t))\,dt
-\sum_{j=1}^n p(t_j)\int_0^{u(t_j)}I_j(x)\,dx\Big)\\
&\leq  \int_0^T \max_{|x|\leq \nu_2}F(t,x)\,dt+\tilde p\nu_2^3
\Gamma_{\nu_2}.
\end{align*}
Arguing as before, we obtain
$$
\sup_{u\in \Phi^{-1}(]-\infty,r_1])}\Psi(u) \leq \int_0^T
\max_{|x|\leq \nu_1}F(t,x)\,dt+\tilde p\nu_1^3 \Gamma_{\nu_1}.
$$
Since $0\leq w(t)\leq \tau$ for each $t\in [0,T]$, assumption
(A2) ensures that
$$
\int_0^{T/4}F(t,w(t))\,dt+\int_{3T/4}^{T}F(t,w(t))\,dt\geq 0,
$$
and so
\begin{align*}
\Psi(w)&\geq \int_{T/4}^{3T/4}F(t,\tau)\,dt
-\sum_{j=1}^n p(t_j)\int_0^{w(t_j)}I_j(x)\,dx\\
&\geq \int_{T/4}^{3T/4}F(t,\tau)\,dt-\tilde
p\,(\mu(\tau))^3\Gamma_{\mu(\tau)}.
\end{align*}
Therefore, we have
\begin{align*}
\beta(r_1,r_2)
&\leq \frac{\sup_{u\in
\Phi^{-1}(]-\infty,r_2[)}\Psi(u)-\Psi(w)}{r_2-\Phi(w)}\\
&\leq \frac{\int_0^T\max_{|x|\leq\nu_2}F(t,x)\,dt
 +\tilde p\nu_2^3 \Gamma_{\nu_2}+\tilde
p(\mu(\tau))^3\Gamma_{\mu(\tau)}-\int_{T/4}^{3T/4}F(t,\tau)\,dt}
{\frac{2\delta^2 \nu_2^2}{T}-\frac{4k^2\tau^2}{T}}\\
&= \frac{T}{2}a_\tau(\nu_2).
\end{align*}
On the other hand, we have
\begin{align*}
\rho(r_1,r_2)
&\geq \frac{\Psi(w)-\sup_{u\in \Phi^{-1}(]-\infty,r_1])}\Psi(u)}{\Phi(w)-r_1}\\
&\geq \frac{\int_{T/4}^{3T/4}F(t,\tau)\,dt-\tilde
p (\mu(\tau))^3 \Gamma_{\mu(\tau)}-\tilde p \nu_1^3
 \Gamma_{\nu_1}-\int_0^T\max_{|x|\leq \nu_1}F(t,x)\,dt}
{\frac{4k^2\tau^2}{T}-\frac{2\delta^2 \nu_1^2}{T}}\\
&= \frac{T}{2}a_\tau(\nu_1).
\end{align*}
So, from assumption (A3), it follows that
$\beta(r_1,r_2)<\rho(r_1,r_2)$. Therefore, from Theorem
\ref{the2.1}, for each
$\lambda\in]\frac{2}{Ta_\tau(\nu_1)},\frac{2}{Ta_\tau(\nu_2)}[$,
the functional $I_\lambda$ admits at least one critical point $u_0$
such that
$r_1<\Phi(u_0)<r_2$;
that is,
$$
\frac{2\delta \nu_1}{\sqrt T}<\|u_0\|<\frac{2\delta \nu_2}{\sqrt T},
$$
and the proof is complete.
\end{proof}


Now, we point out the following consequence of Theorem \ref{the3.1}.

\begin{theorem}\label{the3.2}
Assume that there are two positive constants $\nu$ and $\tau$,
with $\sqrt 2k\tau<\delta\nu$, such that assumption {\rm (A2)} in
Theorem \ref{the3.1} holds. Furthermore, suppose that
\begin{itemize}
\item[(A4)] $\frac{\int_0^T\max_{|x|\leq\nu}F(t,x)\,dt
+\tilde
p\nu^3\Gamma_\nu}{\nu^2}<\frac{\delta^2}{2k^2}\frac{\int_{T/4}^{3T/4}
F(t,\tau)\,dt-\tilde p\,(\mu(\tau))^3\Gamma_{\mu(\tau)}}{\tau^2}$.
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{\frac{4k^2\tau^2}{T}}{
\int_{T/4}^{3T/4}F(t,\tau)\,dt-\tilde p\,{(\mu(\tau))^3}
\Gamma_{\mu(\tau)}},\frac{\frac{2\delta^2\nu^2}{T}}{
\int_0^T\max_{|x|\leq\nu}F(t,x)\,dt+\tilde p\nu^3
\Gamma_\nu}\Big[,
$$
problem \eqref{e1.1} admits at least one non-trivial classical
solution $u_0\in X$ such that $|u_0(t)|<\nu$ for all $t\in[0,T]$.
\end{theorem}

\begin{proof}
The conclusion follows from Theorem \ref{the3.1}, by taking
$\nu_1=0$ and $\nu_2=\nu$. Indeed, owing to assumption (A4), one has
\begin{align*}
a_\tau(\nu)
&< \frac{\big(1-\frac{2k^2\tau^2}{\nu^2\delta^2}\big)
\big(\int_0^T\max_{|x|\leq\nu}F(t,x)\,dt+\tilde
p\nu^3\Gamma_{\nu}\big)}{\delta^2\nu^2-2k^2\tau^2}\\
&= \frac{\int_0^T\max_{|x|\leq\nu}F(t,x)\,dt+\tilde
p\nu^3\Gamma_{\nu}}{\delta^2\nu^2}.
\end{align*}
On the other hand,
\begin{align*}
a_\tau(0)=\frac{\int_{T/4}^{3T/4}F(t,\tau)\,dt-\tilde
p(\mu(\tau))^3\Gamma_{\mu(\tau)}}{2k^2\tau^2}.
\end{align*}
Now, owing to assumption {\rm (A4)} and \eqref{e2.4}, it is
sufficient to invoke Theorem \ref{the3.1} for concluding the proof.
\end{proof}


The following result gives the existence of at least one non-trivial
classical solution in $X$ to problem \eqref{e1.1} in the autonomous
case. Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Put
$ F(x):=\int_0^x f(\xi)\,d\xi$ for all
$x\in\mathbb{R}$. We have the following result as a direct
consequence of Theorem \ref{the3.1}.


\begin{theorem}\label{the3.3}
Assume that there exist a non-negative constant $\nu_1$ and two
positive constants $\nu_2$ and $\tau$, with
$\nu_1<\sqrt 2\tau<\delta \nu_2/k$, such that
\begin{itemize}
\item[(A5)] $f(x)\geq 0$ for all $x\in [-\nu_2,\max\{\nu_2,\tau\}]$;
\item[(A6)]
\begin{align*}
&\frac{TF(\nu_2)+\tilde p{\nu_2}^3
\Gamma_{\nu_2}+\tilde p\,(\mu(\tau))^3
\Gamma_{\mu(\tau)}-\frac{T}{2}F(\tau)}{\delta^2
\nu_2^2-2k^2\tau^2}\\
&<\frac{TF(\nu_1)+\tilde p{\nu_1}^3
\Gamma_{\nu_1}+\tilde p(\mu(\tau))^3
\Gamma_{\mu(\tau)}-\frac{T}{2}F(\tau)}{\delta^2 \nu_1^2-2k^2\tau^2}.
\end{align*}
\end{itemize}
Then, for each
\begin{align*}
&\lambda\in\Big]\frac{2\delta^2\nu_2^2-4k^2\tau^2}{T\left(TF(\nu_2)+
\tilde p\,{\nu_2}^3\Gamma_{\nu_2}+\tilde p\,(\mu(\tau))^3
\Gamma_{\mu(\tau)}-\frac{T}{2}F(\tau)\right)},\\
&\frac{2\delta^2\nu_2^2-4k^2\tau^2}{T\left(TF(\nu_1)+\tilde
p\,{\nu_1}^3\Gamma_{\nu_1}+\tilde p\,(\mu(\tau))^3
\Gamma_{\mu(\tau)}-\frac{T}{2}F(\tau)\right)}\Big[,
\end{align*}
the problem
\begin{equation}\label{e3.1}
\begin{gathered}
-(p(t)u'(t))'+q(t)u(t)=\lambda f(u(t)), \quad t\in [0,T],\; t\not=t_j,\\
u(0)=u(T)=0,\\
\Delta u'(t_j)=\lambda I_j(u(t_j)),\quad j=1,2,\dots,n,
\end{gathered}
\end{equation}
admits at least one non-trivial classical solution $u_0\in X$ such
that
$$
\frac{2\delta \nu_1}{\sqrt T}<\|u_0\|<\frac{2\delta \nu_2}{\sqrt T}.
$$
\end{theorem}

\begin{proof}
Since $\delta \leq k,$ from the condition 
$\nu_1<\sqrt 2\tau<\frac{\delta \nu_2}{k}$ we obtain $\nu_1<\nu_2$.
 Therefore, assumption $\rm(A5)$ means $f(x)\geq 0$ for each
$x\in[-\nu_1,\nu_1]$ and $f(x)\geq 0$ for each $x\in[-\nu_2,\nu_2]$,
which implies
$$
\max_{x\in[-\nu_1,\nu_1]}F(x)=F(\nu_1)\quad
\text{and}\quad\max_{x\in[-\nu_2,\nu_2]}F(x)=F(\nu_2).
$$
So, from assumptions $\rm(A5)$ and $\rm(A6)$, we arrive at
assumptions (A2) and $\rm(A3)$, respectively. Hence, Theorem
\ref{the3.1} yields the conclusion.
\end{proof}


\begin{theorem}\label{the3.4}
Let $f:\mathbb{R}\to\mathbb{R}$ be a non-negative continuous
function such that
\begin{itemize}
\item[(A7)] $\lim_{\xi\to 0^+}\frac{f(\xi)}{\xi}=+\infty$.
\end{itemize}
Then, for each
$\lambda\in]0,\frac{2\delta^2}{T}\sup_{\nu>0}\frac{\nu^2}{TF(\nu)+\tilde
p\nu^3\Gamma_\nu}[$, problem \eqref{e3.1} admits at least
one non-trivial classical solution $u_0\in X$.
\end{theorem}

\begin{proof}
For fixed $\lambda$ as in the conclusion, there exists a positive
constant $\nu$ such that
\begin{equation}\label{e3.2}
\lambda(TF(\nu)+\tilde p\nu^3\Gamma_\nu)<\frac{2\delta^2\nu^2}{T}.
\end{equation}
Moreover, assumption (A7) implies that 
$\lim_{t\to 0^+}\frac{F(\xi)}{\xi^2}=+\infty$. On the other hand,
$$
\lim_{\xi\to 0^+}\frac{{(\mu(\xi))^3}
\Gamma_{\mu(\xi)}}{\xi^2} 
=\begin{cases}
\eta\big(\frac{\sqrt2\,k}{\delta}\big)^2, & \text{if } 0<\sigma<1,\\
\Gamma_1\big(\frac{\sqrt2\,k}{\delta}\big)^2, & \text{if }
\sigma=0.
\end{cases}
$$
Therefore,
$$
\lim_{\xi\to 0^+}\frac{F(\xi)-{(\mu(\xi))^3}
\Gamma_{\mu(\xi)}}{\xi^2}=+\infty.
$$
So, a positive constant $\tau$ satisfying $\sqrt 2k\tau <\delta\nu$
can be chosen such that
\begin{equation}\label{e3.3}
\lambda\Big(\frac{\frac{T}{2}F(\tau)-\tilde p\,{(\mu(\tau))^3}
\Gamma_{\mu(\tau)}}{\tau^2}\Big)>\frac{4k^2}{T}.
\end{equation}
Hence, taking \eqref{e3.2} and \eqref{e3.3} into account, Theorem
\ref{the3.1} ensures the conclusion.
\end{proof}


\begin{remark}\label{rem3.5}\rm
Taking (A7) into account, fix $\rho>0$ such that $f(\xi)>0$ for all
$\xi\in ]0,\rho[$. Then, put
$$
\lambda_\rho:=\frac{2\delta^2}{T}\sup_{\nu\in
]0,\rho[}\frac{\nu^2}{TF(\nu)+\tilde p\nu^3\Gamma_\nu}.
$$
The result of Theorem \ref{the3.4} for every $\lambda \in
]0,\lambda_\rho[$ holds with $|u_0(t)|<\rho$ for all $t\in[0,T]$,
where $u_0$ is the ensured non-trivial classical solution in $X$.
\end{remark}

\begin{example}\label{ex3.6}\rm
Let $I(u(t_1))=u(t_1)$ for some $t_1\in(0,1)$. Then
$I:\mathbb{R}\to\mathbb{R}$ is a continuous function
satisfying the sublinear growth condition (A1) with
$\eta=\theta=\frac{1}{3}$ and $\sigma=0$. Now, put
$p(t)=1$, $q(t)=\frac{-\pi^2}{2}$ for all $t\in [0,1]$ and
$f(\xi)=(1+\xi)e^\xi$ for every $\xi\in\mathbb{R}$. Clearly, one has
$\delta=\frac{1}{\sqrt 2}$. Hence, since
$$
\sup_{\nu\in]0,1[}\frac{\nu^2}{\int_0^\nu
f(\xi)\,d\xi+\nu^3\Gamma_\nu}=\sup_{\nu\in]0,1[} \frac{\nu^2}{\nu
e^\nu+\nu^3\Gamma_\nu}=\frac{2}{2e+1},
$$
from Remark \ref{rem3.5}, for every
$\lambda\in\big]0,\frac{2}{2e+1}\big[$ the problem
\begin{gather*}
-u''(t)-\frac{\pi^2}{2}u(t)=\lambda(1+u(t))e^{u(t)} , \quad 
\text{ a.e. in } [0,1],\\
u(0)=u(1)=0,\\
\Delta u'(t_1)=\lambda u(t_1),
\end{gather*}
has at least one non-trivial classical solution $u_0\in H_0^1(0,1)$
such that $|u_0(t)|<1$ for all $t\in[0,1]$.
\end{example}


Here, we point out a special situation of our main result when the
nonlinear term has separable variables. To be precise, let
$\alpha\in L^1([0,T])$ such that $\alpha(t)\geq 0$ a.e.
 $t\in [0,T]$, $\alpha\not\equiv 0$, and let
$g:\mathbb{R}\to\mathbb{R}$ be a nonnegative continuous
function. Consider the following Dirichlet boundary-value problem
\begin{equation}\label{e3.4}
\begin{gathered}
-(p(t)u'(t))'+q(t)u(t)=\lambda \alpha(t)g(u(t)), \quad t\in [0,T],\; t\not=t_j,\\
u(0)=u(T)=0,\\
\Delta u'(t_j)=\lambda I_j(u(t_j)),\quad j=1,2,\dots,n.
\end{gathered}
\end{equation}
Put $ G(x):=\int_0^x g(\xi)\,d\xi$ for all
$x\in\mathbb{R}$, and set
$\|\alpha\|_1:=\int_0^T\alpha(t)\,dt$ and
$\alpha_0:=\int_{T/4}^{{3T}/4}\alpha(t)\,dt$.

\begin{corollary}\label{cor3.7}
Let $I_j(x)\leq 0$ for all $x\in\mathbb{R}$, $j=1,\ldots,n$. Assume
that there exist a non-negative constant $\nu_1$ and two positive
constants $\nu_2$ and $\tau$, with
$\nu_1<\sqrt{2}\tau<\delta\nu_2/k$, such that
\begin{itemize}
\item[(A8)]
\begin{align*}
&\frac{G(\nu_2)\|\alpha\|_1+\tilde{p}\nu_2^3\Gamma_{\nu_2}
+\tilde{p}(\mu(\tau))^3\Gamma_{\mu(\tau)}-G(\tau)\alpha_0}
{\delta^2\nu_2^2-2k^2\tau^2}\\
&<\frac{G(\nu_1)\|\alpha\|_1+\tilde{p}\nu_1^3\Gamma_{\nu_1}
+\tilde{p}(\mu(\tau))^3\Gamma_{\mu(\tau)}-G(\tau)\alpha_0}
{\delta^2\nu_1^2-2k^2\tau^2}.
\end{align*}
\end{itemize}
Then, for each
\begin{align*}
&\lambda\in\Big]\frac{2\delta^2\nu_1^2-4k^2\tau^2}
{T\big(G(\nu_1)\|\alpha\|_1+\tilde{p}\nu_1^3\Gamma_{\nu_1}
+\tilde{p}(\mu(\tau))^3\Gamma_{\mu(\tau)}-G(\tau)\alpha_0\big)},\\
&\frac{2\delta^2\nu_2^2-4k^2\tau^2}
{T\big(G(\nu_2)\|\alpha\|_1+\tilde{p}\nu_2^3\Gamma_{\nu_2}
+\tilde{p}(\mu(\tau))^3\Gamma_{\mu(\tau)}-G(\tau)\alpha_0\big)}\Big[,
\end{align*}
problem \eqref{e3.4} admits at least one positive classical solution
$u_0\in X$, such that
$$
\frac{2\delta \nu_1}{\sqrt T}<\|u_0\|<\frac{2\delta \nu_2}{\sqrt T}.
$$
\end{corollary}

\begin{proof}
Put $f(t,\xi):=\alpha(t)g(\xi)$ for all
$(t,\xi)\in[0,T]\times\mathbb{R}$. Clearly, $F(t,x)=\alpha(t)G(x)$
for all $(t,x)\in [0,T]\times\mathbb{R}$. Therefore, taking into
account that $G$ is a non-decreasing function, Theorem \ref{the3.1}
and \cite[Lemma 3.6]{bodihe} ensure the conclusion.
\end{proof}

An immediate consequence of Corollary \ref{cor3.7} is the following.

\begin{corollary}\label{cor3.8}
Let $I_j(x)\leq 0$ for all $x\in\mathbb{R},\,j=1,\ldots,n$. Assume
that there exist two positive constants $\nu$ and $\tau$, with
$\sqrt{2}k\tau<\delta\nu$, such that
\begin{itemize}
\item[(A9)] 
$\frac{G(\nu)\|\alpha\|_1+\tilde{p}\nu^3\Gamma_\nu}{\nu^2}
<\frac{\delta^2}{2k^2}\frac{G(\tau)\alpha_0-\tilde{p}(\mu(\tau))^3
\Gamma_{\mu(\tau)}}{\tau^2}$.
\end{itemize}
Then, for each
$$
\lambda\in\Big]\frac{4k^2\tau^2}{T\big(G(\tau)\alpha_0
 -\tilde{p}(\mu(\tau))^3\Gamma_{\mu(\tau)}\big)},
\frac{2\delta^2\nu^2}{T\big(G(\nu)\|\alpha\|_1+\tilde{p}\nu^3\Gamma_\nu\big)}
\Big[,
$$
problem \eqref{e3.4} admits at least one positive classical solution
$u_0\in X$, such that $|u_0(t)|<\nu$ for all $t\in[0,T]$.
\end{corollary}

The above corollary follows directly from Theorem \ref{the3.2} and 
\cite[Lemma 3.6]{bodihe}.

Now, consider the nonlinear Dirichlet boundary-value problem
\begin{equation}\label{e3.5}
\begin{gathered}
-u''(t)+a(t)u'(t)+b(t)u(t)=\lambda h(t,u(t)), \quad t\in [0,T],\; t\neq t_j,\\
u(0)=u(T)=0,\\
\Delta u'(t_j)=\lambda I_j(u(t_j)),\quad j=1,2,\dots,n,
\end{gathered}
\end{equation}
where $h:[0,T]\times \mathbb{R} \to \mathbb{R}$ is an
$L^1$-Carath\'{e}odory function and $a,b \in L^{\infty}([0,T])$
satisfy the following conditions
$$
\operatorname{ess\,inf}_{t\in[0,T]}a(t)\geq
0,\quad\operatorname{ess\,inf}_{t\in[0,T]}\big\{b(t)e^{-A(t)}\big\}
>-\frac{\pi^2}{T^2}e^{-A(T)},
$$
where $A(t)$ be a primitive of $a(t)$.

It is easy to see that the solutions of  \eqref{e1.1} are
solutions of \eqref{e3.5} if
\[
p(t)=e^{-\int_0^t a(\xi)\,d\xi},\quad 
q(t)=b(t)e^{-\int_0^t a(\xi)\,d\xi},\quad 
f(t,u)=h(t,u)e^{-\int_0^t a(\xi)\,d\xi}.
\]
Let $H(t,x):=\int_0^x h(t,\xi)\,d\xi$. Then, by a
simple computation, we obtain
$$
F(t,x)=e^{-A(t)}H(t,x),\quad\forall(t,x)\in[0,T]\times\mathbb{R}.
$$
Set
\begin{gather*}
\tilde{k}:=\Big(1+\frac{T^2}{\pi^2}\|be^{-A}\|_\infty\Big)^{1/2},
\quad\tilde{\sigma}_0:=\min\big\{\frac{T^2}{\pi^2}
\operatorname{ess\,inf}_{t\in[0,T]}\big(b(t)e^{-A(t)}\big),0\big\},\\
\tilde{\delta}:=\sqrt{e^{-A(T)}+\tilde{\sigma}_0}.
\end{gather*}
For a given non-negative constant $\nu$ and a positive constant
$\tau$ with $\tilde{\delta}^2\nu^2\neq 2\tilde{k}^2\tau^2,$ put
$\tilde{\mu}(\tau):=\sqrt{2}\tilde{k}\tau/ \tilde{\delta}$ and
\begin{align*}
\tilde{a}_\tau(v)
&:=\Big( \int_0^T e^{-A(t)}\max_{|x|\leq \nu}H(t,x)\,dt
 +\tilde p\nu^3\Gamma_\nu+\tilde p\,({\tilde{\mu}(\tau)})^3
 \Gamma_{\tilde{\mu}(\tau)}\\
&\quad -\int_{T/4}^{3T/4}e^{-A(t)}H(t,\tau)\,dt\Big)
\big/ \big(\tilde{\delta}^2 \nu^2-2\tilde{k}^2\tau^2\big).
\end{align*}
With the above notation and by Theorem \ref{the3.1}, we obtain the
following existence property for problem \eqref{e3.5}.

\begin{theorem}\label{the3.9}
Assume that there exist a non-negative constant $\nu_1$ and two
positive constants $\nu_2$ and $\tau$, with
$\nu_1<\sqrt 2\tau<\tilde{\delta}\nu_2/\tilde{k}$, such that
\begin{itemize}
\item[(A10)] $H(t,\xi)\geq 0$ for all $(t,\xi)
 \in([0,\frac{T}{4}]\cup[\frac{3T}{4},T])\times [0,\tau]$;
\item[(A11)] $\tilde{a}_\tau(\nu_2)<\tilde{a}_\tau(\nu_1)$.
\end{itemize}
Then, for each
$\lambda\in]\frac{2}{T\,\tilde{a}_\tau(\nu_1)},
\frac{2}{T\,\tilde{a}_\tau(\nu_2)}[$,
problem \eqref{e3.5} admits at least one non-trivial classical
solution $u_0\in X$ such that
$$
\frac{2\tilde{\delta} \nu_1}{\sqrt T}<\|u_0\|<\frac{2\tilde{\delta}
\nu_2}{\sqrt T}.
$$
\end{theorem}

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\end{document}