\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 234, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/234\hfil Existence of weak solutions]
{Existence of weak solutions for  three-point
 boundary-value problems of Kirchhoff-type}

\author[G. A. Afrouzi, S. Heidarkhani, S. Moradi \hfil EJDE-2016/234\hfilneg]
{Ghasem A. Afrouzi, Shapour Heidarkhani, Shahin Moradi}

\address{Ghasem A. Afrouzi \newline
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran}
\email{afrouzi@umz.ac.ir}

\address{Shapour Heidarkhani \newline
Department of Mathematics, Faculty of Sciences,
Razi University,
67149 Kermanshah, Iran}
\email{s.heidarkhani@razi.ac.ir}

\address{Shahin Moradi \newline
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran}
\email{shahin.moradi86@yahoo.com}

\thanks{Submitted  March 5, 2016. Published August 16, 2016.}
\subjclass[2010]{35J20, 34B15}
\keywords{Weak solution; three-point boundary-value problem;
\hfill\break\indent Kirchhoff-type problem; variational method}

\begin{abstract}
 We show the existence of at least one weak solution for a three-point
 boundary-value problem of Kirchhoff-type. Our technical approach is
 based on variational methods. In addition, an
 example to illustrate our results is given.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

The purpose of this paper is to establish the existence of at least one 
weak solution for the  three-point
boundary-value problem of Kirchhoff-type
 \begin{equation}\label{e1.1}
 \begin{gathered}
 -K\Big(\int_{a}^{b}|u'(t)|^2dt\Big)u''(t)=f(t,u(t))+h(u(t)),\quad  t\in(a,b), \\
 u(a)=0,\quad u(b)=\alpha u(\eta)
 \end{gathered}
 \end{equation}
 where $K : [0, +\infty[\to\mathbb{R}$ is a continuous function such that 
there exist positive numbers $m$ and
 $M$ with $m\leq K(x)\leq M$ for all $x\geq 0$, $a,b\in\mathbb{R}$ with $a<b$,
 $f:[a,b]\times\mathbb{R}\to\mathbb{R}$ is an $L^1$-Carath\'{e}odory function,
 $h : \mathbb{R}\to\mathbb{R}$ is a Lipschitz continuous function with the
 Lipschitz constant $L > 0$, i.e.,
 $$
|h(\xi_1)-h(\xi_2)| \leq L|\xi_1-\xi_2|
$$
 for every $\xi_1,\xi_2 \in\mathbb{R}$ and $h(0) = 0,$ $\alpha\in\mathbb{R}$
 and $\eta\in(a,b)$.

Multi-point boundary-value problems of ordinary differential equations play 
an important role in applied mathematics, physics and the vibration of
cables with nonuniform weights \cite{Mo}, and as a consequence, have 
attracted a great deal of interest over the years. The study of these problems
for linear second-order ordinary differential
equations was initiated by Ii'in and Moiseev \cite{IM}. Motivated by the 
study of Ii'in and Moiseev \cite{IM},  Gupta \cite{G} studied certain 
three-point boundary-value problems for nonlinear ordinary differential
 equations.

 In the past few years, there has been much attention focused on questions of
 solutions of three-point boundary-value problems for nonlinear ordinary 
differential equations.
 For background and recent results, we refer the reader to 
\cite{A,DXG,HG,G,L,M,MW,S,X} and the references therein for details.
 For example, Xu in \cite{X} by employing the fixed point index method, obtained
 some multiplicity results for positive solutions of some singular semi-positone 
three-point boundary-value problem.
 Sun in \cite{S} by using a fixed point theorem of cone
 expansion-compression type due to Krasnosel'skii, 
established various results on the existence of
 single and multiple positive solutions for the nonlinear singular
third-order three-point boundary-value problem
\begin{gather*}
 u''(t)-\lambda a(t)F(t,u(t))=0,\quad  0<t<1, \\
 u(0)=u'(\eta)=u''(1)=0
 \end{gather*}
 with $\lambda>1$, $\eta\in[\frac{1}{2}, 1)$ where $a(t)$ is a non-negative 
continuous function defined on $(0, 1)$
 and $F : [0, 1] \times [0,\infty)\to [0,\infty)$ is continuous.
 Du et al. in \cite{DXG} based upon Leray-Schauder degree theory, 
ensured the existence of at least three
 solutions for the  problem
\begin{gather*}
 u''(t)+ f(t,u(t),u'(t))=0,\quad
 0<t<1, \\
 u(0)=0,\quad u(1)=\xi u(\eta)
\end{gather*}
 where $\xi > 0$, $0 < \eta < 1$ such that $\xi\eta < 1$ and
 $f : [0, 1]\times\mathbb{R}^2\to\mathbb{R}$ is continuous. 
Lin \cite{L} by using variational method
 and three-critical-point theorem, studied the existence of at least 
three solutions for a three-point boundary-value problem
\begin{gather*}
 u''(t)+\lambda f(t,u)=0,\quad  t\in[0,1], \\
 u(0)=0,\quad u(1)=\alpha u(\eta).
\end{gather*}
 Kirchhoff's model takes into account the changes in length of the string 
produced by transverse vibrations.
 Similar nonlocal problems also model several physical and biological 
systems where $u$ describes a process that depends on the average of itself, 
for example,  the population density.
 Problems of Kirchhoff-type have been widely investigated. We refer the reader to
 the papers \cite{ANA1,ANA2,HZ,HAO,HFK,PZ,Ricceri1} and the references therein.
For example in \cite{HAO} based on a three critical
 point theorem, the existence of an interval of positive
 real parameters $\lambda$ for which the boundary-value problem of Kirchhoff-type
\begin{gather*}
 -K\Big(\int_{a}^{b}|u'(x)|^2dx\Big)u''=\lambda f(x,u),\quad  t\in[a,b], \\
 u(a)=u(b)=0
 \end{gather*}
 where $K : [0, +\infty[\to\mathbb{R}$ is a continuous function, 
$f : [a, b]\times\mathbb{R}\to\mathbb{R}$ is a Carath\'eodory
 function and $\lambda> 0$, was discussed. Also, in \cite{HFK} by using 
variational methods and critical point theory, multiplicity results of 
nontrivial solutions for one-dimensional fourth-order Kirchhoff-type 
equations was studied. In recent years, the existence and multiplicity of
stationary higher order problems of Kirchhoff type (in
$n$-dimensional domains, $n\geq 1$) has been studied, via
variational methods like the symmetric mountain pass theorem in
\cite{new1} and via a three critical point theorem in \cite{new2}.
Furthermore, in \cite{new4,new3} some evolutionary higher order
Kirchhoff problems were studied, largely concentrate on the
qualitative properties of the solutions. In \cite{MolRad2} Molica
Bisci and R\u{a}dulescu, applying mountain pass results studied
the existence of solutions to nonlocal equations involving the
$p$-Laplacian. More precisely, they proved the existence of at
least one nontrivial weak solution, and under additional
assumptions, the existence of infinitely many weak solutions. In
\cite{MolRad1}, they also by using an abstract linking theorem for
smooth functionals established a multiplicity result on the
existence of weak solutions for a nonlocal Neumann problem driven
by a nonhomogeneous elliptic differential operator.

 Inspired by the above results, in the present paper, we study
the existence of at  least one  weak solution for \eqref{e1.1}. 
Precisely, in Theorem \ref{t2} we establish the existence of at least one weak
 solution for  \eqref{e1.1} requiring an algebraic condition on $f$. 
Example \ref{example1} illustrates Theorem \ref{t2}.
Also in Theorem \ref{t3} a parametric  version of this result is successively 
discussed in which, for small values of the parameter and
 requiring an additional asymptotical behaviour of the potential at zero, 
the existence of at least  one weak solution is established. 
We also list some consequences the main results. As a consequence of 
Theorem \ref{t3}, we obtain Theorem \ref{t4} for the
 autonomous case. Finally, we present Example \ref{example2} in which 
the hypotheses of Theorem  \ref{t4} are fulfilled.

 We refer to the recent monograph by Molica Bisci, R\u adulescu and 
Servadei \cite{MRS} for related problems concerning the variational 
analysis of solutions of some classes of nonlocal problems.
For a thorough discussion of this subject we refer the reader to
\cite{ANA3}.

 \section{Preliminary results}

 We shall prove the existence of at least one weak
 solution to the problem \eqref{e1.1} applying the following version of
 Ricceri's variational principle \cite[Theorem 2.1]{Ricceri} that we now
 recall as follows (For a refinement see also \cite{BonMol}):

 \begin{theorem}\label{t1}
Let $X$ be a reflexive real Banach
 space, let $\Phi, \Psi:X \to \mathbb{R}$ be two
 G\^ateaux differentiable functionals such that $\Phi$ is
 sequentially weakly lower semicontinuous, strongly continuous and
 coercive in $X$ and $\Psi$ is sequentially weakly upper
 semicontinuous in $X$. Let $I_\lambda$ be the functional defined
 as $I_\lambda:=\Phi-\lambda\Psi$, $\lambda\in\mathbb{R}$, and for
 every $r>\inf_X \Phi$, let $\varphi$ be the function defined as
 $$
\varphi(r):=\inf_{u\in\Phi^{-1}(-\infty,r)}
\frac{\sup_{v\in\Phi^{-1}(-\infty,r)}\Psi(v)
 -\Psi(u)}{r-\Phi(u)}.
$$ 
Then, for every $r > \inf_X \Phi$ and
 every $\lambda\in(0,\frac{1}{\varphi(r)})$, the restriction of the
 functional $I_\lambda$ to $\Phi^{-1}(-\infty,r)$ admits a global
 minimum, which is a critical point (precisely a local minimum) of
 $I_\lambda$ in $X$.
 \end{theorem}

 The above result is related to the celebrated {\it three critical
points theorem} of Pucci and Serrin \cite{PS1,PS2}.
 We refer the interested reader to the papers 
\cite{AHB,GalMol,HeiAMCM,HFACM,HeiZCAM,
MolRad,MolRe1,MolRe2,MolSer1,MolSer2} in
 which Theorem \ref{t1} has been successfully employed to the existence 
of at least one nontrivial solution for boundary-value problems.

 Here and in the sequel, we take
 $$X=W^{1,2}_1(a,b):=\{u\in W^{1,2}(a,b):\; u(a)=0,\ u(b)=\alpha u(\eta)\}.$$
 The space $X$,
 equipped with the norm
 $$
\|u\|:=\Big(\int_{a}^{b}|u'(t)|^2dt\Big)^{1/2}.
$$
 Let $f:[a,b]\times \mathbb{R}\to \mathbb{R}$ be an
 $L^1$-Carath\'{e}odory function, that means:
\begin{itemize}
\item[(a)] $t \mapsto f(t,x) $ is measurable for every $x \in \mathbb{R}$,

\item[(b)] $x \mapsto f(t,x) $ is continuous for a.e. $t \in [a,b]$,

\item[(c)] for every $\rho>0$ there exists a function 
$ l_\rho \in L^1([a,b])$ such that
 $$ 
\sup_{|x| \leq \rho}|f(t,x)| \leq l_\rho(t)
$$
 for a.e. $t\in[a,b]$.
\end{itemize}

 Corresponding to the functions $f$, $K$ and $h$, we introduce the functions 
$F : [a,b]\times\mathbb{R}\to\mathbb{R}$,
 $\tilde{K}: [0,+\infty[\to\mathbb{R}$ and $H :\mathbb{R}\to\mathbb{R}$, 
defined as follows
\begin{gather*}
F(t,x) :=\int^x_ 0 f(t,\xi)d\xi\quad
 \text{for every } (t,x)\in[a,b]\times\mathbb{R},\\
\tilde{K}(x) :=\int^x_ 0 K(\xi)d\xi\quad  \text{for every } x\geq0, \\
H(x) :=\int^x_ 0 h(\xi)d\xi\quad  \text{for every } x\in\mathbb{R}.
\end{gather*}
 We say that a function $u\in X$ is a weak solution of  \eqref{e1.1} if
 $$
K\Big(\int_{a}^{b}|u'(t)|^2dt\Big)\int_{a}^{b}u'(t)v'(t)dt-\int_{a}^{b}h(u(t))v(t)dt
-\int_{a}^{b}f(t,u(t))v(t)dt=0
$$
 holds for all $v \in X$.

 We assume throughout and without further mention, that the following
 condition holds:
\begin{itemize}
\item[(H1)] $m>\frac{L(1+|\alpha|)^2(b-a)^2}{4}$.
\end{itemize}

 \begin{theorem}[{\cite[Theorem 3.2]{L}}] \label{thm2.2}
The set $X$ is a separable and reflexive real Banach space.
 \end{theorem}

 The following lemma is needed in the proof of our main result.

 \begin{lemma}[{\cite[Lemma 2.3]{HS}}] \label{lem2.3}
 For all $u\in X$, we have 
\begin{equation}  \label{emb}
\max_{t\in[a,b]}|u(t)|\leq \frac{(1+|\alpha|)\sqrt{b-a}}{2}\|u\|.
 \end{equation}
 \end{lemma}

 \section{Main results}

Our main result reads as follows.

 \begin{theorem}\label{t2}
Assume that
 \begin{equation}\label{e4}
 \sup_{\gamma>0}\frac{{\gamma}^2}{\int_{a}^{b}\sup_{|x|\leq
 {{\gamma}}}F(t,x)dt}>\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}.
 \end{equation}
 Then,  problem \eqref{e1.1} admits at least one weak solution
 in $X$. 
 \end{theorem}

 \begin{proof}
 Our goal is to apply Theorem \ref{t1} to  \eqref{e1.1}.
 We define the functionals $\Phi,\Psi$ for
 every $u\in X$, as follows
 \begin{gather}\label{e8}
 \Phi(u)=\frac{1}{2}\tilde{K}(\|u\|^2)-\int_{a}^{b}H(u(t))dt , \\
\label{e9}
 \Psi(u)=\int_{a}^{b}F(t,u(t))dt,
 \end{gather}
and we put  $I(u)=\Phi(u)-\Psi(u)$ for every $u\in X$. Let us show that the
functionals $\Phi$ and $\Psi$ satisfy the required conditions in
Theorem \ref{t1}. It is well known that $\Psi$ is a differentiable
functional whose differential at the point $u\in X$ is
$$
 \Psi'(u)(v)=\int_{a}^{b}f(t,u(t))v(t)dt
$$
for every $v\in X$, as well as is sequentially weakly upper semicontinuous.
 Moreover, since $m\leq K(s)\leq M$ for all $s\in[0, +\infty[$, 
from \eqref{e8} we have
 \begin{equation} \label{eq3}
\frac{(4m-L(1+|\alpha|)^2(b-a)^2)}{8}
 \|u\|^{2}\leq\Phi(u)\leq\frac{(4M+L(1+|\alpha|)^2(b-a)^2)}{8}\|u\|^{2}
 \end{equation}
for all $u \in X$ and bearing  (H1) in mind, it
follows that $\lim_{\|u\|\to+\infty}\Phi(u)=+\infty$, namely
 $\Phi$ is coercive.
 Moreover, $\Phi$ is continuously differentiable whose
 differential at the point $u\in X$ is
\[
\Phi'(u)(v)
 =K\Big(\int_{a}^{b}|u'(t)|^2dt\Big)\int_{a}^{b}u'(t)v'(t)dt
-\int_{a}^{b}h(u(t))v(t)dt
\]
for every $v\in X$. Furthermore, $\Phi$ is
 sequentially weakly lower semicontinuous. Therefore, we see that the 
regularity assumptions on $\Phi$  and $\Psi$, as requested in Theorem \ref{t1},
 are verified. Note that the critical points of the functional
 $I$ are the solutions of the problem \eqref{e1.1}.
We  now look on the existence of a critical point of the functional
 $I$ in $X$. The condition \eqref{e4} ensures that there exists
 $\bar{\gamma}>0$ such that
\begin{equation} \label{e6}
 \frac{{\bar{\gamma}}^{2}}{\int_{a}^{b}\sup_{|x|\leq
 {\bar{\gamma}}}F(t,x)dt}>\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}.
\end{equation}
 Choose
 $$
r=\frac{4m-L(1+|\alpha|)^2(b-a)^2}{2(1+|\alpha|)^2(b-a)}\bar{\gamma}^2.
$$
 Bearing in mind relation \eqref{emb}, we see that
 \begin{align*}
 \Phi^{-1}(-\infty,r)
&= \{u\in X;\Phi(u)<r\}\subseteq \Big\{u\in X;
 \|u\|\leq\sqrt{\frac{2r(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}}\Big\}
\\
&\subseteq \{u\in X;  |u|\leq\bar{\gamma}\},
 \end{align*}
and it follows that
 $$
\Psi(u)\leq\sup_{u\in
 \Phi^{-1}(-\infty,r)}\int_{a}^{b}F(t,u(t))dt\leq
 \int_{a}^{b}\sup_{|x|
\leq  {\bar{\gamma}}}F(t,x)dt
$$
for every $u\in  X$ such that $\Phi(u)<r$. Then
 $$
\sup_{\Phi(u)<r}\Psi(u)\leq \int_{a}^{b}\sup_{|x|\leq
 {\bar{\gamma}}}F(t,x)dt.
$$
By simple calculations and from the definition of
 $\varphi(r)$, since $0\in \Phi^{-1}(-\infty, r)$ and
 $\Phi(0)=\Psi(0)=0$, one has
 \begin{align*}
 \varphi(r)
&=\inf_{u\in\Phi^{-1}(-\infty,r)} \frac{(\sup_{v\in
 \Phi^{-1}(-\infty,r)}\Psi(v))- \Psi(u)}{r-\Phi(u)}\\
&\leq \frac{\sup_{v\in  \Phi^{-1}(-\infty,r)}
 \Psi(v)}{r}\\
&\leq \frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}
\frac{\int_{a}^{b}\sup_{|x|\leq  {\bar{\gamma}}}F(t,x)dt}{\bar{\gamma}^2}.
 \end{align*}
At this point, we observe that
 \begin{equation} \label{e7}
 \varphi(r)\leq\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}
\frac{\int_{a}^{b}\sup_{|x|\leq
 {\bar{\gamma}}}F(t,x)dt}{\bar{\gamma}^2}.
 \end{equation}
 Consequently, from \eqref{e6} and \eqref{e7} one has $\varphi(r) < 1$.
Hence, since $1\in  (0, \frac{1}{\varphi(r)})$, applying Theorem \ref{t1}
the functional $I$ admits at least
 one critical point (local minima) $\tilde{u} \in \Phi^{-1}(-\infty, r)$.
The proof is complete.
 \end{proof}

Now we present an example in which the hypotheses of Theorem
 \ref{t2} are satisfied.

\begin{example}\label{example1} \rm
Consider the problem
 \begin{equation}\label{14}
 \begin{gathered}
 -K\Big(\int_{0}^{1}|u'(t)|^2dt\Big)u''(t)=f(u)+h(u),\quad  t\in(0,1), \\
 u(0)=0,\quad u(1)=\frac{1}{4}u(\frac{1}{2})
 \end{gathered}
 \end{equation}
where $K(x)=\frac{3}{2}+\frac{\arctan(x))}{\pi}$ for all $x\in\mathbb{R}$, 
$$
f(x)=\frac{1}{10^3}(2x+e^x)
$$ 
and  $h(x)=\sin(x)$  for every $x\in\mathbb{R}$. By the expression of $f$ 
we have 
$$
F(x)=\frac{1}{10^3}(x^2+e^x-1)
$$
 for every $x\in\mathbb{R}$. By simple calculations, we obtain $m=1$. Since
 $$
 \sup_{\gamma>0}\frac{{\gamma}^2}{\sup_{|x|\leq
 {{\gamma}}}F(x)}>\frac{50}{39},
$$ 
we observe that all assumptions of  Theorem \ref{t2} are
 fulfilled. Hence, Theorem \ref{t2} implies that  problem  \eqref{14},
 admits at least one weak solution
 in $W^{1,2}_1(0,1)$.
 \end{example}

 We note that Theorem \ref{t2} can be exploited showing the
 existence of at least one solution for the following
 parametric version of  \eqref{e1.1},
\begin{equation}\label{newe1.1}
 \begin{gathered}
 -K\Big(\int_{a}^{b}|u'(t)|^2dt\Big)u''(t)=\lambda f(t,u(t))+h(u(t)),\quad
 t\in(a,b), \\
 u(a)=0,\quad u(b)=\alpha u(\eta)
 \end{gathered}
 \end{equation}
where $\lambda$ is a positive parameter. More precisely, we have
the following existence result.

 \begin{theorem}\label{t3}
 For every $\lambda$ small enough, more precisely,
$$
\lambda\in\Big(0,\frac{4m-L(1+|\alpha|)^2(b-a)^2}{2(1+|\alpha|)^2(b-a)} 
\sup_{\gamma>0}\frac{{\gamma}^{2}}{\int_{a}^{b}\sup_{|x|\leq
 {{\gamma}}}F(t,x)dt}\Big),
$$
problem \eqref{newe1.1} admits at least one weak solution $u_\lambda\in X$.
 \end{theorem}

 \begin{proof}
Fix $\lambda$ as in the conclusion. Take $\Phi$ and $\Psi$ as given in the 
proof of  Theorem \ref{t2}, and put $I_\lambda(u)=\Phi(u)-\lambda\Psi(u)$ 
for every  $u\in X$.
 Let us pick 
$$
0<\lambda<\frac{4m-L(1+|\alpha|)^2(b-a)^2}{2(1+|\alpha|)^2(b-a)} 
\sup_{\gamma>0}\frac{{\gamma}^{2}}{\int_{a}^{b}\sup_{|x|\leq
 {{\gamma}}}F(t,x)dt}.
$$
 Hence, there exists $\bar{\gamma}>0$ such that
\[
 \lambda \frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}
< \frac{{\bar{\gamma}}^{2}}{\int_{a}^{b}\sup_{|x|\leq
 {\bar{\gamma}}}F(t,x)dt}.
\]
 Choose
 $$
r=\frac{4m-L(1+|\alpha|)^2(b-a)^2}{2(1+|\alpha|)^2(b-a)}\bar{\gamma}^2.
$$
With the same notation as in the proof of Theorem \ref{t2}, one has
 \begin{align*}
 \varphi({r})
&\leq  \frac{\sup_{v\in\Phi^{-1}(-\infty,{r})}\Psi(v)}{{r}}\\
&\leq  \frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}
 \frac{\int_{a}^{b}\sup_{|x|\leq
 {\bar{\gamma}}}F(t,x)dt}{\bar{\gamma}^2}
<\frac{1}{\lambda}.
 \end{align*}
Hence, since $\lambda\in(0, \frac{1}{\varphi(r)})$, Theorem \ref{t1}
 ensures that the functional $I_\lambda$ admits at least one
 critical point (local minima)
 $u_\lambda\in\Phi^{-1}(-\infty,r)$ and since the critical points of the functional
 $I_\lambda$ are the solutions of the problem 
\eqref{newe1.1} we have the conclusion.
 \end{proof}

 \begin{remark} \rm
 In Theorem \ref{t3} we looked for the critical points of the
 functional $I_\lambda$ naturally associated with the problem
 \eqref{newe1.1}. We note that, in general, $I_\lambda$ can be
 unbounded from the following in $ X$.
 Indeed, for example,  when
 $f(t,\xi)=1+|\xi|^{\gamma-{2}}\xi$ for $(t,\xi)\in
 [a,b]\times\mathbb{R}$ with $\gamma>2$, for any
 fixed $u \in X\backslash\{0\}$ and 
$\iota\in \mathbb{R}$, we obtain
\begin{align*}
I_\lambda(\iota u)
&=\Phi(\iota u)-\lambda  \int_{a}^bF(t,\iota u(t))dt\\
&\leq \iota^2\frac{(4M+L(1+|\alpha|)^2(b-a)^2)}{8}\|u\|^{2}
 - \lambda\iota \|u\|_{L^1} 
-\lambda\frac{\iota^\gamma}{\gamma}\|u\|_{L^\gamma}^\gamma\\
&\to -\infty
\end{align*}
as $\iota\to+\infty$. Hence, we can not use direct minimization to
 find critical points of the functional $I_\lambda$.
 \end{remark}

 \begin{remark} \rm
 For a fixed $\bar{\gamma}>0$ let
 $$
\frac{{\bar{\gamma}}^{2}}{\int_{a}^b\sup_{|x|\leq  {\bar{\gamma}}}F(t,x)dt}
>\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}.
$$ 
Then the result of Theorem  \ref{t3} holds with
 $\|u_{\lambda}\|_\infty\leq\bar{\gamma}$ where
 $u_\lambda$ is the ensured
 weak solution in $X$.
 \end{remark}

 \begin{remark} \rm
 If in Theorem \ref{t2} the function $f(t,\xi)\geq0$ for every
 $t\in[a,b]$ and $\xi\in\mathbb{R}$, then the
 condition \eqref{e4} assumes the simpler form
 \begin{align}\label{eq 8}
 \sup_{\gamma>0}\frac{{{\gamma}}^{2}}{\int_{a}^bF(t,\gamma)dt}
>\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}.
 \end{align}
Moreover, if the assumption
 $$
 \limsup_{\gamma\to+\infty}\frac{{{\gamma}}^{2}}{\int_{a}^bF(t,\gamma)dt}>\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2},
 $$
 is satisfied, then  condition \eqref{eq 8} automatically holds.
 \end{remark}

 \begin{remark} \rm
If in Theorem \ref{t3}, $f(t,0)\neq0$ for all $t\in  [a,b]$, then the 
ensured weak solution is obviously  non-trivial. 
On the other hand, the non-triviality of the weak solution
 can be achieved also in the case $f(t,0)=0$ for a.e. $t\in  [a,b]$ 
requiring the extra condition at zero, that is
 there are a non-empty open set $D\subseteq[a,b]$ and
 a set $B\subset D$  of positive Lebesgue measure such that
 \begin{gather}\label{zero1}
 \limsup_{\xi\to0^+}\frac{\operatorname{ess\,inf}_{t\in
 B}F(t,\xi)}{|\xi|^{2}}=+\infty , \\
\label{zero2}
 \liminf_{\xi\to0^+}\frac{\operatorname{ess\,inf}_{t\in
 D}F(t,\xi)}{|\xi|^{2}}>-\infty.
\end{gather}
 Indeed, let $0 <  \bar{\lambda } <\lambda ^*$ where
 $$
\lambda ^*=\frac{4m-L(1+|\alpha|)^2(b-a)^2}{2(1+|\alpha|)^2(b-a)} 
\sup_{\gamma>0}\frac{{\gamma}^{2}}{\int_{a}^{b}\sup_{|x|\leq
 {{\gamma}}}F(t,x)dt}.
$$ 
Then, there exists $\bar{\gamma} >0$ such that
\[
 \bar{\lambda }\frac{2(1+|\alpha|)^2(b-a)}{4m-L(1+|\alpha|)^2(b-a)^2}
<\frac{{\bar{\gamma}}^{2}}{\int_{a}^{b}\sup_{|x|\leq
 {\bar{\gamma}}}F(t,x)dt}.
\]
 Let $\Phi$ and $\Psi$ be as given in \eqref{e8} and \eqref{e9},
 respectively. Due to Theorem \ref{t1}, for every 
$\lambda \in(0,\bar{\lambda })$ there exists a critical point of
 $I_\lambda=\Phi-\lambda\Psi$ such that 
$u_\lambda \in \Phi^{-1}(-\infty,r_\lambda )$ where 
$r_\lambda =\frac{4m-L(1+|\alpha|)^2(b-a)^2}{2(1+|\alpha|)^2(b-a)}\bar{\gamma}^2$. 
In particular, $u_\lambda$ is a
 global minimum of the restriction of $I_\lambda $ to
 $\Phi^{-1}(-\infty,r_\lambda )$.
 We will prove that the function $u_\lambda $ cannot be trivial.
Let us show that
\begin{equation}\label{e10}
\limsup_{\|u\|\to0^+}\frac{\Psi(u)}{\Phi(u)}=+\infty.
 \end{equation}
 Owing to the assumptions \eqref{zero1} and \eqref{zero2}, we can
 consider a sequence $\{\xi_n \}\subset\mathbb{R^+}$ converging to
 zero and two constants $\sigma, \kappa$ (with $\sigma>0$) such
 that
\begin{gather*} 
\lim_{n\to +\infty}\frac{\operatorname{ess\,inf}_{t\in B}F(t,\xi_n)}
{|\xi_n|^{2}}=+\infty, \\
\operatorname{ess\,inf}_{t\in D}F(t, \xi)\geq\kappa |\xi|^{2}
\end{gather*} for every 
$\xi  \in [0, \sigma]$. We consider a set $\mathcal{G}\subset B$ 
of positive measure and a function $v\in X$ such that
\begin{itemize}
\item[(1)] $v(t)\in [0, 1]$ for every $t\in[a,b]$,
\item[(2)] $v(t)=1$ for every $t\in \mathcal{G}$,
\item[(3)] $v(t)=0$ for every $x\in[a,b]\setminus D$.
\end{itemize}
Hence, for a fixed  $M>0$ we consider a real positive number $\eta$ with
 $$ 
M<\frac{\eta \operatorname{meas}(\mathcal{G}) +\kappa\int_{D \setminus 
\mathcal{G} }|v(t)|^{2} dt}{\frac{4M+L(1+|\alpha|)^2(b-a)^2}{8}\|v\|^{2}}. 
$$
 Then, there is $n_0\in \mathbb{N}$ such that $\xi_n<\sigma$ and
 $$
\operatorname{ess\,inf}_{t\in B}F(t, \xi_n)\geq\eta|\xi_n|^{2}
$$
for every $n>n_0$. Now, for every $n>n_0$, by considering the
properties of the function $v$ (that is $0\leq\xi_nv(t)<\sigma$
for $n$ large enough), by \eqref{eq3},
 one has
\begin{align*}
 \frac{ \Psi(\xi_nv)}{\Phi(\xi_nv)}
&=\frac{\int_\mathcal{G} F(t,\xi_n)dt+\int_{D
 \setminus \mathcal{G} }F(t,\xi_nv(t))dt}{\Phi(\xi_nv)} \\
&> \frac{\eta \operatorname{meas}(\mathcal{G}) 
+\kappa\int_{D \setminus \mathcal{G} }|v(t)|^{2}dt }
{\frac{4M+L(1+|\alpha|)^2(b-a)^2}{8}\|v\|^{2}}>M.
 \end{align*}
Since $M$ can be arbitrarily large, we obtain
 $$
\lim_{n\to\infty}\frac{ \Psi(\xi_nv)}{\Phi(\xi_nv)}=+\infty,
$$
from which \eqref{e10} clearly follows. So, there exists a
sequence $\{w_n\}\subset X$ strongly converging to zero such that,
for $n$ large enough, $w_n\in \Phi^{-1}(-\infty, r)$ and
 $$
I_\lambda(w_n)=\Phi(w_n)-\lambda\Psi(w_n)<0.
$$
 Since $u_\lambda$ is a global minimum of the restriction of
$I_\lambda$ to $\Phi^{-1}(-\infty, r)$, we obtain
\begin{equation}\label{e11}
 I_\lambda(u_\lambda)<0,
\end{equation}
so that $u_\lambda$ is not trivial.
 \end{remark}

 \begin{remark}\label{rem34} \rm
 By using \eqref{e11},  without difficulty we observe that the map
 \begin{equation}\label{e12}
 (0, \lambda^*)\ni \lambda \mapsto I_\lambda(u_\lambda)
 \end{equation}
 is negative. Also, one has 
$$
\lim_{\lambda\to  0^+}\|u_\lambda\|=0.
$$ 
Indeed, taking into account the fact that $\Phi$ is coercive and for every
 $\lambda\in(0,\lambda^*)$ the solution $u_\lambda\in
 \Phi^{-1}(-\infty, r)$, one has that there exists a positive
 constant $L$ such that $\|u_\lambda\|\leq L$ for every
 $\lambda\in(0,\lambda^*)$. After that, it is easy to see that
 there exists a positive constant $N$ such that
 \begin{equation}\label{e13}
 \Big|\int_{a}^bf(t,u_\lambda(t))u_\lambda(t)dt\Big|
\leq  N\|u_\lambda\|\leq NL
 \end{equation}
for every $\lambda\in(0,\lambda^*)$. Since $u_\lambda$ is a
critical point of $I_\lambda$, we have
 $I'_\lambda(u_\lambda)(v)=0$ for every $v\in X$ and every
 $\lambda\in(0,\lambda^*)$. In particular
 $I'_\lambda(u_\lambda)(u_\lambda)=0, $ that is,
 \begin{equation}\label{e14}
 \Phi'(u_\lambda)(u_\lambda)=\lambda\int_{a}^bf(t,u_\lambda(t))u_\lambda(t)dt
 \end{equation}
for every $\lambda\in(0,\lambda^*)$. Then, since
 $$ 
0\leq (m-\frac{L(1+|\alpha|)^2(b-a)^2}{4})\|u_\lambda\|^{2} 
\leq\Phi'(u_\lambda)(u_\lambda),
$$
 by considering \eqref{e14}, it follows that
\begin{equation} \label{newe14}
 0\leq (m-\frac{L(1+|\alpha|)^2(b-a)^2}{4})\|u_\lambda\|^{2}
\leq \lambda\int_{a}^bf(t,u_\lambda(t))u_\lambda(t)dt
\end{equation}
 for any $\lambda\in(0,\lambda^*)$. Letting $\lambda\to 0^+$, by \eqref{newe14}
together with \eqref{e13} we obtain
 $$
\lim_{\lambda\to0^+}\|u_\lambda\|=0.
$$
 Then, we have obviously the desired conclusion.
 At last, we have to show that the map
$$
\lambda\mapsto  I_\lambda(u_\lambda)
$$
 is strictly decreasing in $(0, \lambda^*)$. For our goal we see that for any
$u\in X$, one has
 \begin{equation}\label{e16}
 I_\lambda(u)=\lambda\Big(\frac{\Phi(u)}{\lambda}-\Psi(u)\Big).
 \end{equation}
Now, let us fix $0<\lambda_1<\lambda_2<\lambda^*$ and let
 $u_{\lambda_ i}$ be the global minimum of the functional
 $I_{\lambda_ i}$ restricted to $\Phi(-\infty, r)$ for $i=1,2$.
 Also, set
 $$
m_{\lambda_ i}=\Big(\frac{\Phi(u_{\lambda_ i})}{\lambda_ i}-\Psi(u_{\lambda_ i})
\Big)
 =\inf_{v\in\Phi^{-1}(-\infty, r)}\Big(\frac{\Phi(v)}{\lambda_i}-\Psi(v)\Big)
$$
 for every $i=1,2$. Clearly, \eqref{e12} together with
 \eqref{e16} and the positivity of $\lambda$ imply that
 \begin{equation}\label{e17}
 m_{\lambda i}<0 \quad \text{for } i=1, 2.
 \end{equation}
 Moreover
 \begin{equation}\label{e18}
 m_{\lambda_ 2}\leq m_{\lambda_ 1},
 \end{equation}
because $0<\lambda_1<\lambda_2$. Then, by \eqref{e16}-\eqref{e18}
and again by the fact that  $0<\lambda_1<\lambda_2$, we obtain that
 $$
I_{\lambda _2} (u_{\lambda_ 2})=\lambda_2 m_{\lambda_ 2}
\leq \lambda_2 m_{\lambda_ 1}<\lambda_1 m_{\lambda_ 1}
=I_{\lambda _1 }(u_{\lambda_ 1}),
$$
so that the map $ \lambda \mapsto I_\lambda(u_\lambda)$ is strictly
decreasing in $\lambda \in (0, \lambda^*)$. The arbitrariness of
 $\lambda<\lambda^*$ shows that $ \lambda \mapsto I_\lambda(u_\lambda)$
is strictly decreasing in $(0, \lambda^*)$.
 \end{remark}

 \begin{remark}\label{rem35} \rm
 If $f$ is non-negative then the solution ensured in Theorem
 \ref{t3} is non-negative. Indeed, let $u_{\ast}$ be a non-trivial
 weak solution of the problem \eqref{newe1.1}, then $u_{\ast}$ is non-negative.
Arguing by a  contradiction, assume that the set 
$\mathcal{A}=\{t\in[a,b]; u_{\ast}(t)<0\}$ is non-empty and of positive measure. 
Put  $\bar {v}(t)=\min\{u_{\ast}(t),0\}$. Using
 this fact that $u_{\ast}$ also is a solution of \eqref{newe1.1},
so for every $\bar {v}\in X$ we have
\[
K\Big(\int_{a}^{b}|u_{\ast}'(t)|^2dt\Big)
\int_{a}^{b}u_{\ast}'(t)\bar {v}'(t)dt-\int_{a}^{b}h(u_{\ast}(t))\bar {v}(t)dt
 -\lambda\int_{a}^{b}f(t,u_{\ast}(t))\bar {v}(t)dt=0
\]
and by choosing $\bar {v}= u_{\ast}$ and since $f$ is
non-negative, we have
 \begin{align*}
 0&\leq\Big(m-\frac{L(1+|\alpha|)^2(b-a)^2}{4}\Big)
 \|u_{\ast}\|^{2}_{\mathcal{A}} \\
&\leq K\Big(\int_{\mathcal{A}}|u_{\ast}'(t)|^2dt\Big)
\int_{\mathcal{A}}|u_{\ast}'(t)|^2dt
 -\int_{\mathcal{A}}h(u_{\ast}(t))u_{\ast}(t)dt\\
&=\lambda\int_{\mathcal{A}}f(t,u_{\ast}(t))u_{\ast}(t)dt\leq0
\end{align*}
since $m>\frac{L(1+|\alpha|)^2(b-a)^2}{4}$, we have
 $\|u_{\ast}\|^{2}_{\mathcal{A}}\leq 0$ which contradicts 
 fact that $u_{\ast}$ is a non-trivial solution. Hence, $u_{\ast}$ is positive.
 \end{remark}

 \begin{remark} \rm
 We observe that Theorem \ref{t3} is a bifurcation result in the
 sense that the pair $(0,0)$ belongs to the closure of the set
 $$
\{ (u_\lambda, \lambda)\in X\times (0, +\infty): u_\lambda \
 \text{is a non-trivial weak solution of}\ \eqref{newe1.1} \}
$$
in $X\times \mathbb{R}$. Indeed, by Theorem \ref{t3}
 we have that
 $$
\|u_\lambda\|\to 0 \quad \text{as}\quad \lambda \to 0.
$$
 Hence, there exist two sequences $\{u_j\}$ in $X$ and $\{\lambda_j\}$ in 
$\mathbb{R}^+$ (here $u_j=u_{\lambda_j}$) such that
 $\lambda_j \to 0^+$ and $\|u_j\|\to 0$,
as $j\to+\infty$. Moreover, we emphasis that due to the fact that
 the map
 $$
(0, \lambda^*)\ni \lambda \mapsto I_\lambda(u_\lambda)
$$ 
is strictly decreasing, for every
 $\lambda_1,\lambda_2\in(0,\lambda^*)$, with
 $\lambda_1\neq\lambda_2$, the solutions $u_{\lambda_1}$ and
 $u_{\lambda_2}$ ensured by Theorem \ref{t3} are different.
 \end{remark}

 When $f$ does not depend on $t$, we obtain the following
 consequence of Theorem \ref{t3}.

 \begin{theorem}\label{t4}
Let $f:\mathbb{R}\to\mathbb{R}$ be a non-negative continuous
 function. Put $F(x)=\int_{0}^{x}f(\xi)d\xi$ for all
 $x\in\mathbb{R}$. Assume that
 $$
\lim_{\xi\to0^+}\frac{F(\xi)}{\xi^{2}}=+\infty.
$$
Then, for each
 $$
\lambda\in \Lambda=\Big(0,\frac{4m-L(1+|\alpha|)^2(b-a)^2}
 {2(1+|\alpha|)^2(b-a)^2}\sup_{\gamma>0}\frac{{\gamma}^{2}}{F(\gamma)}\Big),
$$ 
the problem 
\begin{gather*}
 -K\Big(\int_{a}^{b}|u'(t)|^2dt\Big)u''(t)=\lambda f(u(t))+h(u(t)),\
 t\in[a,b], \\
 u(a)=0,\; u(b)=\alpha u(\eta) 
 \end{gather*}
 admits at least one positive weak solution  $u_\lambda\in X$ such that
 $$
\lim_{\lambda\to 0^+}\|u_\lambda\|=0
$$ 
and the real function
 \begin{align*}
 \lambda\to\frac{1}{2}\tilde{K}(\|u_\lambda\|^2)-\int_{a}^{b}H(u_\lambda(t))dt-\int_{a}^bF(u_\lambda(t))dt
 \end{align*}
 is negative and strictly decreasing in $\Lambda$.
 \end{theorem}

 We conclude this paper by giving an example that illustrates Theorem
 \ref{t4}.

 \begin{example}\label{example2} \rm
We consider  the problem
 \begin{equation}\label{18}
 \begin{gathered}
 -K\Big(\int_{0}^{1}|u'(t)|^2dt\Big)u''(t)=f(u)+h(u),\quad  t\in(0,1), \\
 u(0)=0,\quad u(1)=\frac{1}{3}u(\frac{1}{2})
 \end{gathered}
 \end{equation}
 where 
$$
K(x)=  \begin{cases}
 1+x-[x], &[x] \text{ is even},\\
 1+|x-[x+1]|,&[x] \text{ is odd},
 \end{cases}
$$
where $[x]$ is the integer part of $x$, 
$$
f(x)= 2x+e^x+\frac{2x}{1+x^2} 
$$ 
and  $h(x)=1-\cos(x)$  for every $x\in\mathbb{R}$. By the expression of $f$ 
we have 
\[
F(x)=  x^2+e^x+\ln(1+x^2)-1\,.
\]
Direct  calculations give $m=1$ and 
$\sup_{\gamma>0}\frac{{\gamma}^{2}}{F(\gamma)}=1$. Then all conditions 
in Theorem \ref{t4} are  satisfied. Hence, for each
 $\lambda\in \big(0,\frac{5}  {8}\big)$,  problem \eqref{18}
 admits at least one positive weak solution
 in $u_\lambda\in W^{1,2}_1(0,1)$ such that
 $\lim_{\lambda\to 0^+}\|u_\lambda\|=0$ and the real function
\[
 \lambda\to\frac{1}{2}\tilde{K}(\|u_\lambda\|^2)
-\int_{0}^{1}(u_\lambda(t)-\sin(u_\lambda(t)))\,dt
-\int_{0}^1(u^2_\lambda(t)+e^{u_\lambda(t)}+\ln(1+u^2_\lambda (t) )-1)\,dt
\]
 is negative and strictly decreasing in $(0,5/8)$.
 \end{example}

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