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\begin{document}
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	\title[Existence of solutions for a class of Kirchhoff-type equations on $\R^+$]	% at most 50 characters including spaces
	{Existence and asymptotic behavior of solutions for a class of Kirchhoff-type equations on the real half-line} 	% at most 150 characters including spaces ()
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	\author[I.~Kettaf, S.~Khoutir, and H.~Kasri] % put here short author names for header
	{Ishak Kettaf\orcidnumber{0009-0004-8591-334X},
		Sofiane Khoutir\corrauth\orcidnumber{0000-0002-3762-2474}, and
		Hicham Kasri\orcidnumber{0009-0003-8997-9293}
		% if Second and Third authors share the same affiliation
	}		 
	
	\address{Faculty of Mathematics, Laboratory AMNEDP, University of Science and Technology Houari Boumediene, PB 32 El-Alia, Bab Ezzouar 16\,111, Algiers, Algeria
		% \affilnum2  Department of Mathematics, Vali-e-Asr University of Rafsanjan,
		%Rafsanjan, Iran.
	} 
	
	\emails{
		\email{ikettaf@usthb.dz} (I.~Kettaf),
		\email{skhoutir@usthb.dz} (S.~Khoutir),
		\email{hkasri@usthb.dz}(H.~Kasri)
	}
	
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	\begin{abstract}
		This paper is concerned with the following Kirchhoff-type equation: 
		\begin{equation*}
			-\left(a+b\Int |u'(x)|^2dx\right)u''+p(x)u=q(x)f(u)
			,\quad x\in (0,+\infty),
		\end{equation*}
		where $a>0$, $b\geq0$, are constants, $f\in C(\R)$, $p\in C(\R^+,\R^+_*)$ and $q\in L^1(0,+\infty)$. Firstly, by using Ekeland's variational principle, we show the existence of solutions to the above equation in the case where $f$ is a sublinear function. Then, we establish the existence of solutions in the case where $f$ is a superlinear function by using the mountain pass theorem. Moreover, we discuss the asymptotic behavior of the obtained solutions in both cases with respect to the parameter $b$. Some recent results are complemented and extended.
	\end{abstract}
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	\keywords{Kirchhoff-type equation; sublinear; Ekeland's variational principle; superlinear; mountain pass theorem; asymptotic behavior}
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	\section{Introduction}
	
	In this paper, we consider the following Kirchhoff-type equation: 
	
	\begin{equation}\label{equ-1}
		\tag{$\mathcal{P}_b$}
		\begin{cases}
			-\left(a+b\Int |u'(x)|^2dx\right)u''+p(x)u=q(x)f(u)
			,\quad x\in (0,+\infty),\\
			u(0)=0,
		\end{cases}
	\end{equation}
	where $a>0$ and $b\geq0$ are constants, $ f\in C(\R)$, $p\in C(\R^+,\R^+_*)$, and $q\in L^1(0,+\infty)$.
	
	The problem \eqref{equ-1} is related to the stationary analogue of the following Kirchhoff equation:
	\begin{equation*}\label{t-equ}
		\frac{\partial^2 u}{\partial t^2}-\left( \alpha+\beta \int_0^{+\infty} \left| \frac{\partial u}{\partial x}\right|^{2}dx \right) \frac{\partial^2 u}{\partial x^2}=g(x,u),
	\end{equation*}
	which was introduced by Kirchhoff \cite{kirchhoff} as a generalization of the well-known D'Alembert's wave equation
	\begin{equation}\label{equ-2}
		\rho  \frac{\partial^2 u}{\partial t^2} -\left(\frac{\rho_0}{h} +\frac{E}{2L}\int_{0}^{L}\left| \frac{\partial u}{\partial x}\right|^{2}dx\right)\frac{\partial^2 u}{\partial x^2}=g(x,u),
	\end{equation}
	for free vibrations of elastic strings. Kirchhoff's model takes into account the changes in length of the string produced by transverse vibrations.\\
	In \eqref{equ-2}, $u$ denotes the displacement, $g(x,u)$ is the external force and the other parameters have the following meaning: $L$ is the length of the string, $h$ is the area of cross section, $E$ is the Young modulus of the material, $\rho$ is the mass density, and $\rho_0$ is the initial tension. 

	We notice that problem $\eqref{equ-2}$ appears in other fields such as biological systems, where $u$ describes a process which depends on the average of the density itself (for instance, population density). For more information on the physical background of problem \eqref{equ-2}, we refer the readers to \cite{2,3,4,Witold} and the references therein. 
	
	In the recent decades, the following Kirchhoff-type equation: 
	\begin{equation*}\label{kir-eq}
		-\left( a+b\int_{\Omega} \vert \nabla u\vert^{2}dx \right)\Delta  u+V(x)u=f(x,u),\quad x\in \Omega,
	\end{equation*} 
	where $\Omega$ is a smooth bounded domain of $\R^N$ or $\Omega=\R^N$, has been extensively investigated, and many interesting results have been obtained by means of variational methods, see for instance \cite{david,bitao,Furtado,24,meng,Rado,9,21,16} and the references therein.
	
	However, there are a few papers dealing with the original case \eqref{equ-2} posed on intervals $I\subseteq \R$, see \cite{Boulaiki,shapour,khoutir}. In \cite{shapour}, the authors considered the following class of the Kirchhoff-type second-order impulsive differential problem:
	\[
	\left\{\begin{array}{l}
		K\left(\int_0^{+\infty}\left(\left|u^{\prime}(t)\right|^2+q(t)|u(t)|^2\right) d t\right)\left(-u^{\prime \prime}(t)+q(t) u(t)\right)
		=\lambda f(t, u(t)), \quad t \in[0,+\infty), ~~~t \neq t_j, \\
		\Delta\left(u^{\prime}\left(t_j\right)\right)=\lambda I_j\left(u\left(t_j\right)\right), \\
		u^{\prime}\left(0^{+}\right)=g(u(0)), \quad u^{\prime}(+\infty)=0,
	\end{array}\right.\]
	where $K:[0,+\infty)\rightarrow \mathbb{R}$ is a continuous function, $q \in L^{\infty}[0,+\infty)$, $\lambda$ is a control parameter, $I_j \in C(\mathbb{R}, \mathbb{R})$ for $1 \leqslant j \leqslant p,~0=t_0<t_1<t_2<$ $\cdots<t_p<+\infty, ~\Delta\left(u^{\prime}\left(t_j\right)\right)=u^{\prime}\left(t_j^{+}\right)-u^{\prime}\left(t_j^{-}\right)=\displaystyle\lim _{t \rightarrow t_j^{+}} u^{\prime}(t)-\lim _{t \rightarrow t_j^{-}} u^{\prime}(t), \linebreak f:[0,+\infty) \times \mathbb{R} \rightarrow \mathbb{R}$ is an $L^2$-Carathéodory function, and $g: \mathbb{R} \rightarrow \mathbb{R}$ is a Lipschitz continuous function. Using variational methods, they established the existence of at least one weak solution as well as infinitely many weak solutions for the above problem.
	
	In \cite{Boulaiki}, the authors investigated the following Kirchhoff-type second-order boundary value problem:
	\[
	\left\{\begin{array}{l}
		\left(a+\lambda \int_0^{+\infty}\left(u^{\prime}(t)^2+b u(t)^2\right) \mathrm{d} t\right)\left(-u^{\prime \prime}(t)+b u(t)\right)=f(u(t)) \text { for a.e. } t \in(0,+\infty), \\
		u(0)=u(+\infty)=0,
	\end{array}\right.\]
	where $a$ and $b$ are positive constants, $\lambda \geq 0$ is a parameter and $f \in C\left(\mathbb{R}^{+}, \mathbb{R}^{+}\right)$. By using the mountain pass lemma in combination with the Pohozaev identity, they established the existence of a positive solution for the above equation in the case where $f$ is superlinear.
	
	When $b=0$ and $a=1$ in \eqref{equ-1}, equation \eqref{equ-1} becomes the following semilinear equation:
	\begin{equation}\label{semi-lin}
		-u''+p(x)u=f(x,u).
	\end{equation}
	Some interesting studies related to \eqref{semi-lin} by variational methods can be found in \cite{Marek,Bouafia,Boulaiki2,Moussaoui,pankov,sato}. 
	
	Motivated by the above works, in the present paper, by using Ekeland's variational principle we first investigate the existence of solutions to problem \eqref{equ-1} with sublinear nonlinearity $f$. Then, we study the existence of solutions in the case where $f$ is a superlinear function by using the mountain pass theorem. Finally, we discuss the asymptotic behavior of the obtained solutions in both cases with respect to the parameter $b$. 
	
	The outline of this paper is as follows. In Section 2, we introduce the variational framework associated with problem \eqref{equ-1}. Section 3 is devoted to the study of the sublinear case and the proof of Theorems \ref{1thm} and \ref{1thm2}. In Section 4, we focus on the study of the superlinear case and establish the proof of Theorem \ref{2thm}.
	
	
	\section{Variational framework}\label{sec2}
	
Throughout this paper, we use the following notations:
\begin{itemize}
	\item $\|.\|_{r}$ denotes the usual $L^r-$norm for $r\in [1,+\infty]$;
	
	\item $X^*$ denotes the topological dual space of the Banach space $X$;
	
	%	\item $\langle .,. \rangle$ denotes action of the dual;
	
	\item $C_c(0,+\infty)$ denotes the space of continuous functions with compact support in $(0,+\infty)$;
	
	\item $\rightharpoonup$ denotes the weak convergence in $X$;
	
	%	\item $C$, $C_i$ and $c_i$ denote a positive constant, which may vary from line to line;
	
	%\item $\overline{A}$ denotes the closure of the set $A$;
	\item $B_\rho$ denotes the open ball of center $0$ and radius $\rho$;
	\item $\overline{B}_\rho$ denotes the closed ball of center $0$ and radius $\rho$;
	\item $\partial B_\rho := \{y\in X : \|y\| =\rho\}$; 
	\item $o_n(1)\rightarrow 0$ as $n\rightarrow \infty$;
\end{itemize}

We consider the Sobolev space
\begin{equation*}
	H^1_0(0,+\infty)=\left\{ u\in L^2(0,+\infty) \: : u'\in L^2(0,+\infty), u(0)=0 \right\},
\end{equation*}
and let $H$ be the subspace of $H^1_0(0,+\infty)$ defined by
\begin{equation*}
	H=\Big\{u\in H^1_0(0,+\infty) : \Int p(x)u^2 dx <+\infty\Big\}.
\end{equation*}
Obviously, $H$ is a Hilbert space with a scalar product and norm given by
\[(u,v)=\Int(a u'v' + p(x) u v)dx \text{~~and~~} \|u\|^2=\Int(a |u'|^2 + p(x) u^2)dx,\]
for all $u,v\in H$.\\
%===A comment that will be deleted========================
%\textcolor{red}{The requirement $p(x)>0$ for all $x \in (0,+\infty)$ is used here to ensure that the  bilinear form above is indeed a scalar product.}\\
%=========================================================
Under the assumption $(P)$, it is easy to show that $H$ is embedded continuously into $H^1_0(0,+\infty)$ and therefore also into $L^r(0,+\infty)$ for $r\in[2,+\infty]$. Thus, there exists $\mu_r>0$ such that 
\begin{equation}\label{embed}
	\|u\|_r\leq \mu_r\|u\|,\qquad \forall u\in H.
\end{equation}
We notice that if $u\in H$, then $\displaystyle\lim_{x\rightarrow +\infty} u(x)=0$.% \cite{Brezis}
%we get
%\begin{align*}
%||u||^2 &\geq \Int (a|u'|^2 +p_0 u^2)dx\\
%        &\geq C ||u||^2_{H^1_0 (0,+\infty)} \text{~~ where~~} C=\min\{a,p_0\}>0,~~\forall u\in H
%\end{align*}
%which means that $H$ is embedded continuously into $H^1_0(0,+\infty)$ and therefore also into $L^r(0,+\infty)$ for $r\in[2,+\infty]$. Thus, there exists $\mu_r>0$ such that 
%\begin{equation}\label{embed}
%||u||_r\leq \mu_r||u||,\qquad \forall u\in H.
%\end{equation}
%Moreover, the embedding of $H$ into $L^r(0,+\infty)$ is compact for all $r\in[2,+\infty)$. For the proof of this result, we refer the readers to \cite{Badial}. \\% page 105.


For the problem \eqref{equ-1}, the associated energy functional is defined on $H$ as follows:
\begin{equation}\label{FE}
	I_b(u)=\dfrac{1}{2}\|u\|^2+\dfrac{b}{4}\Big(\Int |u'|^2dx\Big)^2-\Int q(x) F(u) dx,
\end{equation}  
where $F(u)=\displaystyle\int_0^u f(s)ds.$\\
We have the following result.
\begin{lemma}\label{lem1}
	The functional $I_b$ is of class $C^1$ on $H$, and
	\begin{equation*}
		\langle I_b'(u),v\rangle = (u,v) + b\Big(\Int |u'|^2dx\Big)\Int u'v'dx - \Int q(x) f(u)v dx,
	\end{equation*}	
	for all $u,v\in H$.
\end{lemma}
\begin{proof}
	We consider the functional $J$ defined on $H$ by
	\[J(u)=\Int q(x)F(u)dx.\]
	By combining $(f_1)$ and $(F_1)$ (which are introduced in Section \ref{sec3} and Section \ref{sec4}) with \eqref{embed}, we can easily derive that for any $u\in H$, the inequality
	\begin{align}\label{const1}
		|f(u)|&\leq  C_{\|u\|},
	\end{align}
	where $C_{\|u\|}>0$ is the constant given by
	\begin{equation*}
		C_{\|u\|}=\begin{cases}
			\alpha_1\mu_\infty^{\theta-1}\|u\|^{\theta-1},& \text{ \qquad if $(f_1) $ holds},\\
			\alpha_2+\beta \mu_\infty^{\theta-1}\|u\|^{\theta-1},& \text{ \qquad if $(F_1) $ holds}.
		\end{cases}
	\end{equation*}
	Then, it follows that
	\begin{equation*}\label{weldef}
		|F(u)|\leq  \mu_\infty C_{\|u\|}\|u\|,\qquad \forall u\in H,
	\end{equation*}
	which implies that $J$ is well-defined on $H$ since $q\in L^1(0,+\infty)$.\\
	To prove that $I_b$ is of class $C^1$ on $H$, it is sufficient to prove this property only for $J$. For this purpose, firstly we prove that $J$ is Gâteaux differentiable, and then we show that $J_G'$ is continuous.
	
	\textbf{Claim 1.} $J$ is Gâteaux differentiable.\\
	It is obvious that for all $u,v\in H$, and almost every $x\in(0,+\infty)$
	\[\displaystyle\lim_{t \longrightarrow 0} q(x)\dfrac{F(u(x)+t v(x))-F(u(x))}{t}=q(x) f(u(x))v(x).\]
	Indeed, by the mean value theorem there exists a real number $0<\theta_t<|t|$ with $|t|\leq 1$ such that
	\begin{equation}\label{Lagr}
		q(x)\Big(F\big(u(x)+t v(x)\big)-F\big(u(x)\big)\Big) = tq(x) f\big(u(x)+\theta_t v(x)\big)v(x),
	\end{equation}
	and by continuity of $f$ we obtain the result.\\ 
	Once again, from $(f_1)$, $(F_1)$, \eqref{embed} and the inequality $|a+b|^r \leq \gamma_r(|a|^r + |b|^r)$ with  $a,b \in \R$ we check that
	\begin{equation}\label{const2}
		\left|f\big(u(x)+\theta_t v(x)\big)v(x)\right|\leq C_{\|u\|,\|v\|},
	\end{equation}
	where $C_{\|u\|,\|v\|}>0$ denotes the constant given by
	\begin{equation*}
		C_{\|u\|,\|v\|}=\mu_\infty\|v\|\begin{cases}
			\alpha_1\gamma_{\theta-1}\mu_\infty^{\theta-1}\left(\|u\|^{\theta-1}+\|v\|^{\theta-1}\right),& \text{ \qquad if $(f_1) $ holds},\\
			\alpha_2+\beta\gamma_{\theta-1}\mu_\infty^{\theta-1}\left(\|u\|^{\theta-1}+\|v\|^{\theta-1}\right),& \text{ \qquad if $(F_1) $ holds}.
		\end{cases}
	\end{equation*}
	Hence, from \eqref{Lagr} and \eqref{const2}, we conclude that
	\begin{align*}
		q(x)\Bigg|\dfrac{F\big(u(x)+t v(x)\big)-F\big(u(x)\big)}{t}\Bigg| &=q(x) \Big|f\big(u(x)+\theta_t v(x)\big)v(x)\Big|\\
		&\leq C_{\|u\|,\|v\|} q(x).
	\end{align*}
	As the function $q\in L^1(0,+\infty)$, by the Lebesgue dominated convergence theorem we have
	\[\lim_{t \longrightarrow 0}\Int q(x)\dfrac{F(u+t v)-F(u)}{t} dx = \Int q(x) f(u)v dx. \]
	Since the right-hand side, as a function of $v$, is a continuous and linear functional on $H$, it is the Gâteaux differential $J'_G$ of $J$. 
	
	\textbf{Claim 2.} $J'_G$ is continuous.\\
	We complete the proof by checking that the function $J'_G$ is continuous on $H^*$. For this purpose, let take $\{u_n\}$ in $H$ such that $u_n \longrightarrow u$ as $n\longrightarrow +\infty$. Up to a subsequence, we may assume that
	%\begin{equation*}
	\begin{enumerate}
		\item	$u_n \longrightarrow u$ \qquad in  $L^r(0,+\infty)$,~~  $\forall r\in[2,+\infty]$;
		\item	$u_n(x) \longrightarrow u(x)$ \qquad a.e in $(0,+\infty)$.
		%\item there is $w_r\in L^r(0,+\infty)$ such that  $|v_n(x)|\leq w_r(x)$ \text{ a.e in } $(0,+\infty) $ and for all $n\in\N$.    
	\end{enumerate}
	We have for all $v\in H$
	\[\Big|\langle J'_G(u_n)-J'_G(u),v \rangle\Big| \leq \Int q(x)|f(u_n)-f(u)||v| dx.\]
	By continuity of $f$, it is clear that
	\[f\big(u_n(x)\big) \longrightarrow f\big(u(x)\big),\qquad \text{a.e } x\in (0,+\infty).\]
	Since the sequence $\{u_n\}$ is convergent, it is bounded in $H$, and therefore there exists a constant $M > 0$ such that
	\begin{equation}\label{BOUND}
		\|u_n\|\leq M,\qquad \forall n\in\N.
	\end{equation}
	Thus, taking into account \eqref{const1} and \eqref{BOUND}, one has 
	\begin{align*}
		q(x)\Big|f\big(u_n(x)\big)\Big|& \leq C q(x)\in L^1(0,+\infty),
	\end{align*}
	where $C>0$ is defined as
	\begin{equation*}
		C=\begin{cases}
			\alpha_1\mu_\infty^{\theta-1}M^{\theta-1},& \text{ \qquad if $(f_1) $ holds},\\
			\alpha_2+\beta \mu_\infty^{\theta-1} M^{\theta-1},& \text{ \qquad if $(F_1) $ holds},
		\end{cases}
	\end{equation*}
	and once again by the Lebesgue dominated convergence theorem, we get
	\[\Int q(x) f(u_n) dx \longrightarrow \Int q(x) f(u) dx\text{ \qquad as $n\longrightarrow +\infty.$}\]
	Hence
	\begin{align*}
		\Big|\langle J'_G(u_n)-J'_G(u),v \rangle\Big| &\leq \Int q(x)|f(u_n)-f(u)||v| dx,\\
		&\leq \mu_\infty \|v\| \Int q(x) |f(u_n)-f(u)| dx,\qquad \forall v\in H,
	\end{align*}
	and this implies that
	\begin{equation*}
		\sup_{\|v\|\leq 1}\Big|\langle J'_G(u_n)-J'_G(u),v \rangle\Big|\leq \mu_\infty \Int q(x) |f(u_n)-f(u)| dx\longrightarrow 0,\text{~~as~~}  n\longrightarrow +\infty.
	\end{equation*}
	Consequently, we conclude that
	\[	\|J'_G(v_n)-J'_G(v)\|_{H^*} \longrightarrow 0,\quad \text{ as ~~ } n\longrightarrow +\infty, \]
	which implies the continuity of $J'_G$. The proof is completed.
\end{proof}
\begin{definition}\label{def1}
	We say that $u$ is a weak solution of problem \eqref{equ-1} if for any $v\in H$ we have 
	\begin{equation*}\label{WS}
		(u,v) + b\Big(\Int |u'|^2 dx\Big)\Int u'v' dx -\Int q(x) f(u)v dx =0.
	\end{equation*}
\end{definition}

\begin{remark}
	From Lemma \ref{lem1} and Definition \ref{def1}, we deduce that the critical points of $I_b$ correspond to the weak solutions of \eqref{equ-1}.
\end{remark}
\section{The sublinear case}\label{sec3}
In this section, we investigate problem \eqref{equ-1} in the case where the nonlinear term $f$ has sublinear growth. The main results of this section are stated  as follows.
\begin{theorem}\label{1thm}
	Assume that
	\begin{enumerate}
		\item[($P$)] $p\in C(\R^+)$ and $\displaystyle\inf_{x\in\R^+} p(x)\geq p_0>0$;	% and $\displaystyle\lim_{|x|\rightarrow +\infty} p(x)=+\infty$;
		\item[($Q$)] $q : \R^+ \rightarrow \R^+_*$ such that $q\in L^1(0,+\infty)$.
		\item[($f_1$)] There exist $\alpha_1>0$ and $\theta\in(1,2)$ such that
		\[|f(s)|\leq \alpha_1|s|^{\theta-1},\qquad \forall s\in\R;\]
		\item[($f_2$)] $\displaystyle\lim_{s\rightarrow 0} \dfrac{f(s)}{|s|}=+\infty$.
	\end{enumerate}
Then problem \eqref{equ-1} has at least one nontrivial solution.
\end{theorem} 

\begin{theorem}\label{1thm2}
	Let $(P)$, $(Q)$, $(f_1)$ and $(f_2)$ hold. Then, for any sequence $\{b_n\}\subset(0,+\infty)$ with $b_n\longrightarrow 0$ as $n\longrightarrow +\infty$, there exist a subsequence, still denoted by $\{b_n\}$, and $u_*\in H$ such that $u_n\longrightarrow u_*$ in $H$, where $u_*$ is a solution of the equation
	\begin{equation}\label{equ-22}
		\tag{$\mathcal{P}_0$}
		\begin{cases}
			-au''+p(x)u=q(x)f(u)
			,\quad x\in (0,+\infty),\\
			u(0)=0.
		\end{cases}
	\end{equation}
\end{theorem}
\begin{remark}
	Since equation \eqref{equ-1} is set on the unbounded interval $[0,\infty)$, the main difficulty is the lack of compactness of the embedding $ H^1_0(0,+\infty)\hookrightarrow L^r(0,+\infty)$ for $r\geq 2$. Indeed, when using variational methods, we need to prove that the energy functional associated to \eqref{equ-1} satisfies the Palais-Smale compactness condition, that is, any sequence $\{u_n\}\subset H^1(0,+\infty)$ satisfying \eqref{PS} has a convergent subsequence. To this end, a careful analysis of the energy functional and its derivative is given (see pp. 14-15) to prove the convergence of the $(PS)$ sequence.    
\end{remark}

\subsection{Technical lemmas}
In this subsection, we will prove some lemmas which will be used for proving theorem \ref{1thm}.
\begin{lemma}\label{1lem2}
	The functional $I_b$ is bounded from below on $\overline{B}_{\rho}$, where $\rho>0.$
\end{lemma}
\begin{proof}
	From ($f_1$), one has
	\[\dfrac{|f(s)|}{|s|}\leq \alpha_1 |s|^{\theta -2},\qquad\forall s\in\R^*.\]
	Since $1<\theta<2$, we deduce that 
	\[\displaystyle\lim_{s\rightarrow +\infty} \dfrac{f(s)}{s} = 0,\]
	and then, for all $\varepsilon>0$, we get
	%============= More details=========
	%there exists $B_\varepsilon>0$ such that 
	%\begin{equation}\label{eq1}	%|u|\geq B_\varepsilon\Longrightarrow |f(u)|\leq \varepsilon |u|. 
	%\end{equation}
	%In the other hand, it follows from $(f_1)$ that
	%\begin{equation}\label{eq2}
	%|f(u)|\leq \alpha |u|^{\theta-1},\qquad \forall |u|\leq B_\varepsilon.
	%\end{equation}
	%Thus, from \eqref{eq1} and \eqref{eq2} it follows that
	\begin{equation}\label{2eq3}
		|f(u)|\leq \varepsilon |u| + \alpha_1 |u|^{\theta-1}\text{~~and~~}
		|F(u)|\leq \dfrac{\varepsilon}{2} |u|^2 + \dfrac{\alpha_1}{\theta} |u|^{\theta},\qquad \forall u\in\R.
	\end{equation}
	Therefore, by \eqref{embed}, \eqref{FE}, \eqref{2eq3}, and the fact that $b\geq0$, for all $u\in H$ we get 
	\begin{align*}
		I_b(u) &=  \dfrac{1}{2}\|u\|^2 +\dfrac{b}{4}\Big(\Int |u'|^2dx\Big)^2-\Int q(x) F(u) dx\\
		& \geq  \dfrac{1}{2}\|u\|^2 -\Int q(x)\Big[ \dfrac{\varepsilon}{2} |u|^2 + \dfrac{\alpha_1}{\theta} |u|^{\theta}\Big] dx\\
		&=  \dfrac{1}{2}\|u\|^2 -\dfrac{\varepsilon}{2}\Int q(x) |u|^2 dx - \dfrac{\alpha_1}{\theta} \Int q(x)|u|^{\theta}dx\\
		&\geq  \dfrac{1}{2}\|u\|^2 -\dfrac{\mu_\infty^2\varepsilon}{2}\|q\|_1\|u\|^2 - \dfrac{\mu_\infty^{\theta}\alpha_1}{\theta}\|q\|_1\|u\|^{\theta}\\
		&=\dfrac{1}{2}\Big(1-\varepsilon\mu_\infty^2\|q\|_1\Big)\|u\|^2 - \dfrac{\alpha_1\mu_\infty^{\theta}}{\theta}\|q\|_1\|u\|^{\theta}.
	\end{align*}
	By choosing $0<\varepsilon \leq \dfrac{1}{2\mu^2_\infty\|q\|_1}$, we obtain
	\begin{align*}
		I_b(u) & \geq \dfrac{1}{4} \|u\|^2 - \dfrac{\alpha_1\mu_\infty^{\theta}}{\theta}\|q\|_1\|u\|^{\theta}\\
		& = \dfrac{1}{4} \Big(1 - C\|u\|^{\theta-2}\Big)\|u\|^2. %\text{~~ where ~~} C=\dfrac{4\alpha\mu_\infty^{\theta+1}}{\theta+1}||q||_1>0.
	\end{align*}
	%======== More details =============
	%Now, we consider the function $g$ defined on $(0,+\infty)$ as follows
	%\[g(\rho)=1-C\rho^{\theta -1}.\]
	%It is clear that
	%\[\displaystyle\lim_{\rho \rightarrow +\infty} g(\rho)=1\]
	%since $0<\theta<1$. \\
	%Then, for all $\varepsilon>0$, there exists $\rho^*>0$ such that 
	%\[\rho\geq\rho^*\Longrightarrow \big|g(\rho)-1\big|\leq \varepsilon.\]
	%Hence, for all $\rho\geq\rho^*$ we have
	%\[g(\rho) \geq 1-\varepsilon\]
	%and by taking $\varepsilon=\dfrac{1}{2}>0$ we obtain
	%\[g(\rho) \geq \dfrac{1}{2}>0.\]
	%Consequently, there exist $\rho:=\rho^*>0$ and $\alpha :=\dfrac{1}{4}\rho^2g(\rho)>0$ such that
	Since $1<\theta < 2$, choosing $\rho > 0$ sufficiently large, we
	conclude that there exists a constant $\alpha > 0$ such that
	\[I_b(u)\geq \alpha>0 \text{~~~whenever~~~} \|u\|=\rho, \text{~~~and then~~~}\displaystyle\inf_{u\in \partial B_{\rho}} I_b(u)>0.\]
	
	If $\|u\|\leq \rho$, then
	\[ I_b(u)\geq \dfrac{1}{4} \Big(\|u\|^2 - C\|u\|^{\theta}\Big)\geq - \dfrac{C}{4}{\rho}^{\theta}:=-C_1.\]
	%where $C_1=\dfrac{C}{4}\rho^{\theta+1}>0.$
	Thus, the functional $I_b$ is bounded from below on $\overline{B}_\rho$. \\
	Therefore, $\displaystyle\inf_{u\in \overline{B}_{\rho}} I_b(u)$ exists, and we put
	\[c_b:=\displaystyle\inf_{u\in \overline{B}_{\rho}} I_b(u).\]
\end{proof}
\begin{lemma}\label{1lem3}
	We have $c_b<0$.
\end{lemma}
\begin{proof}
	By combining $(f_1)$ and $(f_2)$, it follows that for all $M>0$
	\begin{equation}\label{eq6}
		F(u)\geq M|u|^2-\dfrac{\alpha_1}{\theta}|u|^\theta,\qquad \forall u\in \R.
	\end{equation}
	Let $R>1$ and let $\psi_R$ be the function defined on $[0,+\infty)$ by
	\begin{equation}\label{psi}	\psi_R(x)=\begin{cases}
			\dfrac{x}{R^2},&\text{ if } x\in[0,R],\\
			\dfrac{1}{R},&\text{ if } x\in[R,2R],\\
			-\dfrac{x}{R^2}+\dfrac{3}{R},&\text{ if } x\in[2R,3R],\\
			0,&\text{ if } x\in[3R,+\infty).\\
		\end{cases}
	\end{equation}
	Since $\psi_R\in C_c(0,+\infty)$, it follows that $\psi_R\in H.$ Let 
	\[\varphi_R=\dfrac{\rho}{2\|\psi_R\|}\psi_R.\]
	It is clear that $\varphi_R\in H$ and by a straightforward computation we get 
	\begin{equation}\label{Norm}
		\|\varphi_R\|=\dfrac{\rho}{2},\qquad\|\psi'\|^2_2=\dfrac{2}{R^3}\text{\qquad and \qquad }	\|\varphi'_R\|_2^2=\dfrac{\rho^2}{2\|\psi_R\|^2R^3},
	\end{equation}
	which implies that $\varphi_R\in \overline{B}_\rho$ for all $R>1$.
	
	Moreover, by \eqref{eq6}, \eqref{psi} and \eqref{Norm}, we have
	\begin{align*}
		I_b(\varphi_R)&=\dfrac{1}{2}\|\varphi_R\|^2+\dfrac{b}{4}\|\varphi'_R\|_2^4-\Int q(x)F(\varphi_R)dx\\
		&=\dfrac{\rho^2}{8}+\dfrac{b}{4}\|\varphi'_R\|_2^4-\Int q(x)F(\varphi_R)dx\\
		&\leq \dfrac{\rho^2}{8}+ \dfrac{b\rho^4}{16\|\psi_R\|^4R^6}-M\Int q(x)\varphi^2_R(x)dx+\dfrac{\alpha_1}{\theta} \Int q(x)\varphi_R^\theta(x)dx \\
		&\leq \dfrac{\rho^2}{8} +\dfrac{b\rho^4}{64  a^2} -M\Int q(x)\varphi^2_R(x)dx+\dfrac{\alpha_1\rho^\theta}{2^\theta\theta}\mu_{\infty}^{\theta}\|q\|_1,
	\end{align*}
	%	where $g(R)=1+ \dfrac{b\rho^2}{2\|\psi_R\|^4R^6}$.\vspace{3pt}\\
	%	It is clear that $\displaystyle\lim_{R\longrightarrow +\infty} g(R)=1$, then it follows that there exists $R_1>1$ such that
	%	\begin{equation*}
		%		g(R)\leq 2,\qquad \forall R\geq R_1
		%	\end{equation*}
	and then 
	\begin{equation*}
		I_b(\varphi_{R})\leq g(\rho)-M\Int q(x)\varphi^2_{R}(x)dx,
	\end{equation*}
	where $g(\rho)=\dfrac{\rho^2}{8} +\dfrac{b\rho^4}{64  a^2}+\dfrac{\alpha_1 \mu_{\infty}^{\theta}}{2^\theta\theta}\|q\|_1 \rho^\theta>0$.\\ 
	By choosing \[M>\dfrac{g(\rho)}{\int_0^{+\infty} q(x)\varphi^2_{R}(x)dx},\] one has
	\[I_b(\varphi_{R})<0.\]
	Consequently, we get
	\begin{equation*}
		c_b=\displaystyle\inf_{u\in\overline{B}_{\rho}}I_b(u)\leq I_b(\varphi_{R})<0.
	\end{equation*}
	This completes the proof.
\end{proof}

From Lemma \ref{1lem2} and Lemma \ref{1lem3}, we deduce that
\begin{equation*}\label{ineq}
	\inf_{u\in \overline{B}_{\rho}} I_b(u)< 0<\inf_{u\in \partial B_{\rho}} I_b(u).
\end{equation*}
\subsection{Proof of the main result}
In this subsection, we shall give the proof of Theorem \ref{1thm} with the use of the following variational principle.
%===A comment that will be deleted========================
%\textcolor{red}{I took directly the version cited in \cite{Bouafia}.}
%=========================================================
\begin{theorem}[Ekeland's variational principle, \cite{Ekeland}]
	%	Let $(E,d)$ be a complet metric space and let $J : E \longrightarrow \R$ a functional that is lower semi-continuous and bounded from below. Then, for each $\varepsilon>0$, there exists $u_\varepsilon\in E$ with
	%	\[J(u_\varepsilon) \leq \displaystyle\inf_E J +\varepsilon\]
	%	and whenever $w\in E$ with $w\not=u_\varepsilon$, then
	%	\[J(u_\varepsilon) < J(w) + \varepsilon d(w,u_\varepsilon).\]
	Let $M$ be a complete metric space with metric $d$ and let $J : M \longrightarrow \R$ a lower semicontinuous functional bounded from below. Then, for each $\varepsilon>0$, there exists $u_\varepsilon \in M$ such that
	\[J(u_\varepsilon) \leq \displaystyle\inf_M J +\varepsilon,\]
	and whenever $w\in M$
	\[J(u_\varepsilon) \leq J(w) + \varepsilon d(u_\varepsilon,w).\]%\qquad \text{ with } w\not=u_\varepsilon.
	
\end{theorem}
\begin{proof}[\textbf{Proof of Theorem 1}] From Lemma \ref{lem1} and Lemma \ref{1lem2}, it is clear that the functional $I_b$ is lower semicontinuous and bounded from below in the complete metric space $\overline{B}_{\rho}$. Then, by applying Ekeland's variational principle, there exists a sequence $\{u_n\}\subset \overline{B}_{\rho}$ such that
	\begin{equation*}\label{eq7}
		I_b(u_n)\leq c_b+\dfrac{1}{n}
		\text{\qquad and \qquad }
		I_b(u_n)\leq I_b(w) +\dfrac{1}{n}\|w-u_n\|,\qquad \forall w\in \overline{B}_\rho.
	\end{equation*}
	
	Note that for $n\in\N$ with
	\[\dfrac{1}{n}\in \Big(0,\inf_{ \partial B_{\rho}} I_b-\inf_{ \overline{B}_{\rho}} I_b\Big),\]
	one has
	\[I_b(u_n)\leq c_b+\dfrac{1}{n}< \inf_{ \partial B_{\rho}} I_b,\]
	which implies that $u_n\in B_{\rho}$ for $n\in\N$ large enough.\\
	On the one hand, for $w=u_n+tv$ with $t>0$ and $v\in H$, we get
	%\[	I(u_n)\leq I(u_n+tv) +\dfrac{t}{n}||v||,\]
	%or else
	\[\dfrac{I_b(u_n) - I_b(u_n+tv)}{t}\leq \dfrac{1}{n}\|v\|\]
	and then
	\begin{equation}\label{ineq2}
		-\langle I_b'(u_n),v\rangle \leq  \dfrac{1}{n}\|v\|.
	\end{equation}
	On the other hand, if we put  $w=u_n-tv$, we obtain
	\begin{equation}\label{ineq3}
		\langle I_b'(u_n),v\rangle \leq  \dfrac{1}{n}\|v\|.
	\end{equation}
	Combining \eqref{ineq2} with \eqref{ineq3}, we can assert  that
	\[|\langle I_b'(u_n),v\rangle| \leq  \dfrac{1}{n}\|v\|,\qquad\forall v\in H,\]
	and then
	\[\displaystyle\sup_{\|v\|\leq 1}|\langle I_b'(u_n),v\rangle| \leq  \dfrac{1}{n},\]
	which implies that 
	\[\|I_b'(u_n)\|_{_{H*}}\longrightarrow 0,\qquad \text{ as } n\longrightarrow +\infty.\]
	Consequently, we have proved that
	\begin{equation}\label{1conv}
		I_b(u_n)\longrightarrow c_b \text{~~~and~~~} I_b'(u_n)\longrightarrow 0 \text{ in } H^*.
	\end{equation}
	
	Since $\{u_n\}$ is bounded in $\overline{B}_{\rho}$, which is a closed set, there exist a subsequence still denoted by  $\{u_n\}$  and $u_0\in \overline{B}_{\rho}\subset H$ such that
	\begin{equation}\label{1conv2}
		\begin{array}{l}
			u_n\rightharpoonup u_0 \text{~~weakly in ~~} H,\\
			u'_n\rightharpoonup  u'_0 \text{~~weakly in ~~} L^2(0,+\infty),\\
			u_n(x)\rightarrow u_0(x)\qquad \text{ a.e in } (0,+\infty).
		\end{array}
	\end{equation}
	It follows from \eqref{1conv2} and the continuity of $f$ that 
	%\[f\big(u_n(x)\big)\longrightarrow f\big(u_0(x)\big),\qquad \text{ as ~~ } n\longrightarrow +\infty\]
	%and then
	\[q(x)\Big[f\big(u_n(x)\big)-f\big(u_0(x)\big)\Big]\big(u_n(x)-u_0(x)\big)\longrightarrow 0,\qquad \text{ as ~~ } n\longrightarrow +\infty.\]
	Moreover, by $(f_1)$, \eqref{embed} and the boundedness of $\{u_n\}$, one has
	\begin{align*}
		q(x)\Big|\big[f\big(u_n(x)\big)-f\big(u_0(x)\big)\big]\big(u_n(x)-u_0(x)\big)\Big|& \leq  q(x)\Big[\big|f\big(u_n(x)\big)\big|+\big|f\big(u_0(x)\big)\big|\Big]\big(\big|u_n(x)\big|+\big|u_0(x)\big|\big)\\
		& \leq \alpha_1^2 q(x) \big(\|u_n\|^{\theta-1}_\infty + \|u_0\|^{\theta-1}_\infty\big)\big(\|u_n\|_\infty + \|u_0\|_\infty\big)\\
		& \leq \alpha_1^2 \mu_\infty^{\theta} q(x) \big(\|u_n\|^{\theta-1} + \|u_0\|^{\theta-1}\big)\big(\|u_n\| + \|u_0\|\big)\\
		&\leq  Cq(x).
	\end{align*}
	As the function $q$ is in $L^1(0,+\infty)$, by the Lebesgue dominated convergence theorem we get
	\begin{equation}\label{1conv1}
		\displaystyle\Int q(x)\big[f(u_n)-f(u_0)\big](u_n-u_0)~dx\longrightarrow 0,\qquad \text{ as~~ } n\longrightarrow +\infty.
	\end{equation}
	On the other hand, since $I_b'(u_n)\longrightarrow 0$ in $H^*$ and by \eqref{1conv2}, one has
	\begin{equation}\label{conv3}
		\langle I_b'(u_n)-I_b'(u_0),u_n-u_0\rangle\longrightarrow 0,\qquad \text{ as~~ } n\longrightarrow +\infty.
	\end{equation} 
	%=======Additional detail===========================
	%\textcolor{red}{Here, we have
		%\[|\langle I_b'(u_n),u_n-u_0\rangle|\leq C||I'_b(u_n)||_{H^*} \longrightarrow 0,\]
		%and 
		%$$\langle I_b'(u_0),u_n-u_0\rangle \longrightarrow 0
		%$$
		%since $I'_b(u_0)\in H^*$ and $u_n\rightharpoonup u_0$. (We use the definition of the weak convergence in $H$ without identifying $H$ and $H^*$)\\
		%}
	%==============================================
	By a straightforward computation we get
	\begin{align*}
		\|u_n-u_0\|^2 =&\, \langle I_b'(u_n)-I_b'(u_0),u_n-u_0\rangle-b\|u_n'\|_2^2\Int u'_n(u'_n-u'_0)dx +b\|u_0'\|_2^2\Int u'_0(u'_n-u'_0)dx \\
		&+\Int q(x)\Big[f(u_n)-f(u_0)\Big](u_n-u_0)dx,
	\end{align*}
	and taking into account \eqref{1conv1}, \eqref{conv3}, Young's inequality and Hölder's inequality,  we obtain
	\begin{align}\label{1conv4}
		o_n(1)&=\|u_n-u_0\|^2+b\|u_n'\|_2^2\Int u'_n(u'_n-u'_0)dx -b\|u_0'\|_2^2\Int u'_0(u'_n-u'_0)dx\nonumber\\
		&=\|u_n-u_0\|^2+b\Big(\|u_n'\|_2^4-\|u_n'\|_2^2\Int u'_n u'_0 dx-\|u_0'\|_2^2\Int u'_n u'_0 dx + \|u_0'\|_2^4\Big)\nonumber\\
		&\geq \|u_n-u_0\|^2+b\Big(\|u_n'\|_2^4-\dfrac{1}{2}\|u_n'\|_2^4-\dfrac{1}{2}\|u_n'\|_2^2\|u_0'\|_2^2 -\dfrac{1}{2}\|u_0'\|_2^2\|u_n'\|_2^2 -\dfrac{1}{2}\|u_0'\|_2^4+ \|u_0'\|_2^4\Big)\nonumber\\
		&=\|u_n-u_0\|^2+b\Big(\dfrac{1}{2}\|u_n'\|_2^4-\|u_n'\|_2^2\|u_0'\|_2^2  +\dfrac{1}{2}\|u_0'\|_2^4\Big)\nonumber\\
		&=\|u_n-u_0\|^2+\dfrac{b}{2}\Big(\|u_n'\|_2^2- \|u_0'\|_2^2\Big)^2.
	\end{align}
	By passing to the limit in \eqref{1conv4} as $ n\longrightarrow +\infty$, one has 
	\begin{equation}\label{1conv6}
		u_n\longrightarrow u_0 \text{ strongly in } H \: \text{ and } \: \|u_n'\|_2\longrightarrow \|u_0'\|_2.
		%u'_n\longrightarrow u'_0 \text{ in } L^2(0,+\infty).
	\end{equation}
	From \eqref{1conv2} and \eqref{1conv6} we deduce that
	\[\displaystyle\lim_{n\longrightarrow +\infty} \Big(\Int|u_n'| ^2dx\Big)\Int u_n'v' dx = \Big(\Int|u_0'| ^2dx\Big)\Int u_0'v' dx.\]
	%=====Additional detail ===================================
	%\textcolor{red}{We have
		%\[v'\in L^2(0,+\infty)\text{\qquad and \qquad } u'_n\rightharpoonup u'_0\]
		%so 
		%\[\Int u_n'v' dx = \Int u_0'v' dx\]
		%}
	%==========================================================
	Consequently, by passing to the limit in $	\langle I_b'(u_n),v\rangle$ as $n\longrightarrow +\infty$ and by using once again the dominated convergence theorem, we obtain for all $v\in H$
	\[\displaystyle\lim_{n\longrightarrow +\infty}	\langle I_b'(u_n),v\rangle =(u_0,v) + b \Big(\Int|u_0'| ^2dx\Big)\Int u_0'v' dx -\Int 	q(x)f(u_0)v dx,\]
	and from \eqref{1conv} we deduce that
	\[(u_0,v) + b \Big(\Int|u_0'| ^2dx\Big)\Int u_0'v' dx -\Int 	q(x)f(u_0)v dx=0,\qquad\forall v\in H.\]
	Finally, from \eqref{1conv}, \eqref{1conv6} and the fact that $I\in C^1(H,\R)$, we deduce that
	\[ I_b'(u_0)=0\text{\qquad and \qquad } I_b(u_0)=c_b,\]
	which means that $u_0$ is a critical point of $I_b$ and then a weak solution of problem \eqref{equ-1}. The proof is complete.\\
	%==== Important comment =============================
	%	\textcolor{red}{It is clear that $u_0\not\equiv 0$. Indeed, suppose that  $u_0\equiv 0$. Then it follows that $I_b(u_0)=0$. Thus, $c_0=0$ which is a contradiction.}
\end{proof}
\subsection{The asymptotic behavior of solutions}
In this subsection, we investigate the asymptotic behavior of solutions obtained in Theorem \ref{1thm} with respect to the parameter $b$.
\begin{proof}[\textbf{Proof of Theorem 2}] 
	Noticing that $b=0$ is allowed in the proof of Theorem \ref{1thm}. Therefore, there exists a solution $v_*\in H$ to problem \eqref{equ-22} such that
	\[I_0(v_*)=c_0\text{\qquad and \qquad } I'_0(v_*)=0,\]
	where $c_0=\displaystyle\inf_{u\in\overline{B}_{\rho}}I_0(u)$.\\
	For any sequence $\{b_n\} \subset (0,+\infty)$ with $b_n\longrightarrow 0$ as $n\longrightarrow +\infty$, let $u_{n}:=u_{b_n}\in\overline{B}_{\rho}\subset H$ be a solution of problem \eqref{equ-1} obtained by Theorem \ref{1thm}. Then we have
	\begin{equation}\label{info1}
		I_{b_n}(u_n)=c_{b_n}<0 \text{\qquad and \qquad } I'_{b_n}(u_n)=0 \text{~~in~~ } H^*
	\end{equation}
	with $c_{b_n}=\displaystyle\inf_{u\in\overline{B}_{\rho}}I_{b_n}(u)$. Therefore
	\begin{equation}\label{info3}
		(u_n,v)+b_n \Big(\Int |u'_n|^2 dx\Big) \Int u_n'v'~dx -\Int q(x) f(u_n)v~dx =0,~~\forall v\in H.
	\end{equation}
	Since $u_n\in \overline{B}_{\rho}$, it follows that the sequence $\{u_n\} $ is bounded in $H$, and then there exists $u_* \in H$ such that up to a subsequence
	\begin{equation}\label{conv5}
		\begin{array}{l}
			u_n\rightharpoonup u_* \text{~~weakly in ~~} H,\\
			u'_n\rightharpoonup  u'_* \text{~~weakly in ~~} L^2(0,+\infty),\\
			u_n(x)\rightarrow u_*(x)\qquad \text{ a.e in } (0,+\infty).
		\end{array}
	\end{equation}
	By using \eqref{info1}, \eqref{conv5}, the dominated convergence theorem and the fact that $b_n\longrightarrow 0$, it is easy to show that
	\begin{equation*}\label{info2}
		\langle I'_{b_n}(u_n)-I'_{b_n}(u_*), u_n-u_* \rangle	\longrightarrow 0,\text{\qquad as\qquad } n\longrightarrow +\infty.
	\end{equation*}
	By similar arguments as in \eqref{1conv4}, we prove that
	\begin{equation}\label{conv7}
		\begin{array}{l}
			u_n\longrightarrow u_* \text{ strongly in } H \text{\qquad and \qquad } 	\|u_n'\|_2\longrightarrow \|u_*'\|_2. \\
			
			%u'_n\longrightarrow u'_0 \text{ in } L^2(0,+\infty).
		\end{array}
	\end{equation}
	By passing to the limit in \eqref{info3} as $n\longrightarrow \infty$, one has
	\begin{equation*}
		(u_*,v) -\Int q(x) f(u_*)v~dx =0,~~\forall v\in H,
	\end{equation*}
	which means that $u_*$ is a weak solution of the following equation:
	\begin{equation*}
		\begin{cases}
			-au''+p(x)u=q(x)f(u)
			,\quad x\in (0,+\infty),\\
			u(0)=0.
		\end{cases}
	\end{equation*}
	Next, we prove that $I_0(u_*)=c_0.$ By using once again the dominated convergence theorem, and combining \eqref{conv5} and \eqref{conv7} with the fact that $b_n\longrightarrow 0$,  it holds that
	%By \eqref{conv5}, \eqref{conv7}, dominated convergence Theorem and since $b_n\longrightarrow 0$, we have
	\begin{equation*}\label{conv8}
		c_{b_n}\longrightarrow I_0(u_*).
	\end{equation*} 
	In view of the definition of $c_0$, we assert that 
	\begin{equation}\label{ineq1}
		I_0(u_*)\geq c_0.
	\end{equation}
	Thus, it follows from \eqref{FE} and \eqref{ineq1}
	\begin{align*}
		c_0 =I_0(v_*)=I_{b_n}(v_*)-\dfrac{b_n}{4}||v_*'||^4_2 \geq c_{b_n}-\dfrac{b_n}{4}||v_*'||^4_2,
	\end{align*}
	which yields
	\[\displaystyle\lim_{n\longrightarrow +\infty} c_{b_n} \leq c_0,\]
	or else
	\begin{equation}\label{ineqq2}
		I_0(u_*)\leq c_0.
	\end{equation}
	Hence, by \eqref{ineq1} and \eqref{ineqq2} we deduce that
	\[I_0(u_*)=c_0.\]
	The proof is complete.
\end{proof}
\subsection{Example}
Let $f(x)=x^\frac{1}{5}$, $p(x)=e^{x}$ and $q(x)=\dfrac{1}{1+x^2}.$
It can be seen that $(P)$, $(Q)$, $(f_1)$ and $(f_2)$ are satisfied.

Then, by Theorem \ref{1thm}, the problem
\begin{equation*}
	\begin{cases}
		%	\tag{$\mathcal{P}$}
		-\Big(a+b\Int |u'|dx\Big)u'' +e^{x} u = \dfrac{1}{1+x^2}u^\frac{1}{5},\qquad x\in(0,+\infty);\\
		u(0)=0
	\end{cases}	
\end{equation*} 
has at least one nontrivial solution.
%==== Remark ===============================================
%\textcolor{red}{If I take the second member of \eqref{equ-1} as in \cite{Bouafia}, namely under the form $q(x)f(x,u)$, I ask the question "Can I obtain exactly the same results proved in this article without any additional condition on $f$ ?" Of course by taking account that $(f_1)$ holds for all $x\in\R^+$ and $(f_2)$ uniformly for $x\in(0,+\infty)$.}
%===========================================================
\section{The superlinear case}\label{sec4}
In this section, we analyze problem \eqref{equ-1}  in the case where the nonlinear term $f$ exhibits superlinear growth. 
The main results of this section are stated  in the following theorem.

\begin{theorem}\label{2thm}
	Assume that $(P)$, $(Q)$ hold and $f$ satisfies 
\begin{list}{}{}
	\item[$(F_1)$] $f\in C(\R)$ and  there exist $\theta>2$, $\alpha_2,\beta>0$ such that
	\[|f(s)|\leq \alpha_2  +\beta |s|^{\theta-1}\qquad \forall s\in \R;\]  
	\item[$(F_2)$]  $\displaystyle\lim _{s \rightarrow 0} \dfrac{f(s)}{s}=0;$
	\item[$(F_3)$]
	% there exists $\mu > 4$ such that
	$\displaystyle\lim _{s \rightarrow \infty} \dfrac{F(s)}{s^2}=+\infty,$
	where $F(s)=\displaystyle\int_0^s f(t)dt$; 
	\item[$(F_4)$] 	there exists $L>0$ such that 
	\[ 4 F(s) \leq f(s)s,\qquad \forall |s|\geq L.\]
\end{list}	
Then, problem \eqref{equ-1} has at least one nontrivial solution. Moreover, for every vanishing sequence $\{b_n\}$, let $u_{b_n}$ be a solution of problem \eqref{equ-1}. Then, the sequence $\{u_{b_n}\}$ converges to $u_0$ in $H$, where $u_0$ is a solution of the problem
	\begin{equation}\label{equt-2}
		\tag{$\mathcal{P}_0$}
		\begin{cases}
			-au''+p(x)u=q(x)f(u)
			,\quad x\in (0,+\infty),\\
			u(0)=0.
		\end{cases}
	\end{equation}
\end{theorem}
\begin{remark}
	Since the energy functional associated to \eqref{equ-1} involves a 4-order homogeneous term (i.e., $\left(\Int |u'|^2dx\right)^2 $), it is natural to impose the well-known Ambrosetti–Rabinowitz condition (see \cite{AR}), namely,
	\begin{equation*}\tag{AR}\text{ there exists } \:\mu>4 \: \text{ such that }\: 0<\mu F(u) \leq uf(u)\:  \text{ for all } \: u \in \mathbb{R}.\end{equation*}
	This condition has two crucial uses. The first one is to check the mountain pass geometry for the energy functional $I_b$ and the second one is to guarantee the Palais-Smale compactness condition. Our assumptions $(F_3)$ and $(F_4)$ are very relaxed compared with (AR)-condition. To see this, consider the function $F(u)=u^4\ln (1+u^2)$. It is easy to check that $F$ and its derivative $f$ satisfy $(F_3)$-$(F_4)$ but not (AR).
\end{remark}

In order to prove Theorem \ref{2thm}, we will need the following definition and theorem.
\begin{definition}\label{P.S.}
	A functional $I\in C^1(X,\R)$ satisfies the Palais-Smale condition at level $c\in\R$, denoted by $(PS)_c$ if every sequence $\{u_n\}\subset X$ satisfies
	\begin{equation}\label{PS}
		I(u_n)\longrightarrow c \qquad\text{ and }\qquad I'(u_n)\longrightarrow 0,~~ n\longrightarrow +\infty,
	\end{equation}
	possesses a strongly convergent subsequence.
\end{definition}
\begin{remark}
	If $I$ satisfies the $(PS)_c$ condition for every $c\in\R$, then we say that $I$ satisfies the $(PS)$ condition.
\end{remark}
\begin{theorem}(\cite[Theorem 1.15]{willem}, mountain pass theorem)\label{mp}
	Let $X$ be a Banach space, $I\in C^1(X,\R)$ satisfies the $(PS)$ condition, $I(0)=0$ and
	\begin{enumerate}
		\item There exist $\rho, \alpha>0$ such that $I(v)\geq \alpha$ whenever $\|v\|=\rho.$
		\item There exists $e\in X$ with $\|e\|>\rho$ such that $I(e)\leq 0$.
	\end{enumerate}
	Then, $I$ has at least a critical value $c\geq \alpha$, which is characterized by
	\begin{equation*}\label{c}
		c=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]} I\big(\gamma(t)\big),
	\end{equation*}
	where 
	\[\Gamma=\big\{\gamma\in C([0,1],X) : \gamma(0)=0, \gamma(1)=e\big\}.\] 
\end{theorem} 
\subsection{Some useful lemmas}
In this section, we introduce some technical lemmas which will be used to prove our main result.
\begin{lemma}\label{2lem2}
	Assume that $(F_1)$ and $(F_2)$ hold. Then there exist $\rho_*,\alpha_* > 0$ such that $I_b(u)\geq \alpha_*$ whenever $\|u\|=\rho_*$. 
\end{lemma}
\begin{proof}
	From $(F_1)$ and $(F_2)$, for all $\varepsilon>0$, there exists $C_{\varepsilon}>0$ such that
	\begin{equation}\label{eq1}
		\left|f(u)\right|\leq  \varepsilon |u|+ C_{\varepsilon} |u|^{\theta-1}\text{\qquad and \qquad} \left|F(u)\right|\leq  \dfrac{\varepsilon}{2} |u|^2+\dfrac{C_{\varepsilon}}{\theta} |u|^{\theta},~~~\forall u\in\R.
	\end{equation}
	Hence, from \eqref{embed}, \eqref{eq1} and Hölder's inequality we obtain
	\begin{align*}
		I_b(u)&=\dfrac{1}{2}\|u\|^2+\dfrac{b}{4}\Big(\Int |u'|^2dx\Big)^2-\Int q(x) F(u) dx\\
		&\geq \dfrac{1}{2}\|u\|^2-\Int q(x) F(u) dx\\
		&\geq \dfrac{1}{2}\|u\|^2- \Int q(x)\left( \dfrac{\varepsilon}{2} |u|^2+\dfrac{C_{\varepsilon}}{\theta} |u|^{\theta}\right)dx\\
		%	&\geq \dfrac{1}{2}\|u\|^2-\dfrac{\varepsilon}{2}\mu_\infty^2\|q\|_1\|u\|^2-\dfrac{D_\varepsilon}{\theta}\mu_\infty^\theta\|q\|_1\|u\|^\theta\\
		&\geq\dfrac{1}{2}\left(1-\varepsilon \mu_\infty^2\|q\|_1\right)\|u\|^2-\dfrac{C_{\varepsilon}}{\theta}\mu_\infty^\theta\|q\|_1\|u\|^\theta.
	\end{align*}
	By taking $0<\varepsilon\leq \dfrac{1}{2\mu^2_\infty\|q\|_1}$, one has
	\begin{align*}\label{eq2}
		I_b(u)\geq \dfrac{1}{4}\|u\|^2-\dfrac{C_{\varepsilon}}{\theta}\mu_\infty^\theta\|q\|_1\|u\|^\theta,\qquad \forall u\in H.
	\end{align*}
	Taking $\rho_*=\Big[\dfrac{\theta}{8C_{\varepsilon}\mu_\infty^\theta\|q\|_1}\Big]^{\frac{1}{\theta-2}}$, then for all $u\in H$ with $\|u\|=\rho_*$ we get 
	\begin{align*}
		I_b(u)\geq \dfrac{1}{4}\left(1-\dfrac{4C_{\varepsilon}}{\theta}\mu_\infty^\theta\|q\|_1\rho_*^{\theta-2}\right)\rho_*^2=\dfrac{1}{8}\rho_*^2:=\alpha_*>0,
	\end{align*}
	%If $0<\alpha_2<\frac{\rho_*-\frac{4\beta}{\theta}\mu_\infty^\theta\|q\|_1\rho_*^\theta}{4\mu_\infty\|q\|_1}:=\delta$ then it follows that
	%\begin{align}\label{eq3}
	%	I_b(u)\geq \dfrac{1}{4}\left(\rho_*-4\alpha_2\mu_\infty\|q\|_1-\dfrac{4\beta}{\theta}\mu_\infty^\theta\|q\|_1\rho_*^\theta\right)\rho_*:=\alpha_*>0,
	%\end{align}
	and this completes the proof.
\end{proof}
\begin{lemma}\label{2lem3}
	Assume that $(F_1)$ and $(F_3)$ hold. Then there exists a function $e\in H$ with $\|e\|> \rho_*$ such that $I_b(e)\leq 0$.
\end{lemma}
\begin{proof}
	From $(F_1)$ and $(F_3)$, it follows that for all $M>0$ there exists a constant $C_M>0$ such that
	\begin{equation}\label{2eq5}
		F(u)\geq M|u|^2-C_M|u|,\qquad \forall u\in \R.
	\end{equation}
	Let $R>1$ and we consider the function $\varphi_R$ defined on $[0,+\infty)$ by 
	\begin{equation}\label{phi}
		\varphi_R=\dfrac{2\rho_*}{\|\psi_R\|}\psi_R,
	\end{equation}
	where $\psi_R$ is introduced in \eqref{psi}.\\
	It is clear that $\varphi_R\in H$ and by a straightforward computation we get 
	\begin{equation}\label{norm}	\|\varphi_R\|=2\rho_* \text{\qquad and \qquad}\|\varphi'_R\|_2^2=\dfrac{8\rho_*{^2}}{\|\psi_R\|^2R^3},
	\end{equation}
	which implies that $\varphi_R\in H\backslash \overline{B}_{\rho_*}$ for all $R>1$.\\
	Moreover, by \eqref{psi}, \eqref{Norm}, \eqref{2eq5}, \eqref{phi} and \eqref{norm}, one has
	\begin{align*}
		I_b(\varphi_R)&=\dfrac{1}{2}\|\varphi_R\|^2+\dfrac{b}{4}\|\varphi'_R\|_2^4-\Int q(x)F(\varphi_R)dx\\
		&\leq 2\rho_{*}^2+ \dfrac{16b{\rho_*}^4}{\|\psi_R\|^4R^6}-M\Int q(x)\varphi^2_R(x)dx+C_M \Int q(x)\varphi_R(x)dx \\
		&\leq 2\rho_{*}^2 +\dfrac{4b}{a^2}{\rho_*}^4-M\Int q(x)\varphi^2_R(x)dx+2C_M \|q\|_1\mu_{\infty}\rho_*,
	\end{align*}
	%	where $h(R)= 1+ \dfrac{8b\rho_*^2}{\|\psi_R\|^4R^6}$.\vspace{3pt}\\
	%	It is clear that $\displaystyle\lim_{R\longrightarrow +\infty} h(R)=1$, then it follows that there exists $R_2>1$ such that
	%	\begin{equation*}
		%		h(R)\leq 2,\qquad \forall R\geq R_2
		%	\end{equation*}
	and then 
	\begin{equation*}
		I_b(\varphi_{R})\leq h(\rho_{*})-M\Int q(x)\varphi^2_{R}(x)dx,
	\end{equation*}
	where $h(\rho_{*})= 2\rho_{*}^2 +\dfrac{4b}{a^2}{\rho_*}^4+2C_M \|q\|_1\mu_{\infty}\rho_*>0$.\\
	By choosing $M>\displaystyle \dfrac{ h(\rho_{*}) }{\int_0^{+\infty} q(x)\varphi^2_{R}(x)dx},$ we obtain\vspace{3pt}
	\[I_b(\varphi_{R})<0.\]
	Thus, we complete the proof by taking $e = \varphi_{R}\in H\backslash\overline{B}_\rho$.
\end{proof}
\begin{lemma}\label{2lem4}
	Assume that $(F_1)$-$(F_3)$ hold. Then the functional $I_b$ satisfies the $(PS)$ condition.
\end{lemma}
\begin{proof}
	Let $\{u_n\}\subset H$ be a Palais-Smale sequence at level $c\in\R$, namely satisfying \eqref{PS}. We easily see that there exists $C_1>0$ such that
	\begin{equation}\label{PSC}
		\left|I_b(u_n)\right|\leq C_1 \text{\qquad and \qquad} \left|\langle I_b'(u_n),u_n\rangle\right|\leq C_1\|u_n\|,\qquad\forall n\in\N.
	\end{equation} 
	We divide the proof into two steps.
	
	\textbf{Step 1. } We shall prove that $\{u_n\}$ is bounded in $H$. \\
	Reasoning by contradiction, assume that the sequence $\{u_n\}$ is unbounded in $H$, that is 
	\begin{equation}\label{2contr}
		\|u_n\|\longrightarrow+\infty,\text{\qquad as\qquad} n\longrightarrow +\infty,
	\end{equation}
	and set \[\Omega_n=\big\{x\in(0,+\infty):~~ |u_n(x)|\leq L\big\}\text{\qquad and\qquad} \Omega_n'={(0,+\infty)\backslash\Omega_n}.\]
	% that is to say that
	%\[\displaystyle\lim_{n\longrightarrow +\infty} \|u_n\|=+\infty.\]
	From $(F_4)$ and \eqref{PSC}, through a direct computation we obtain
	\begin{align}\label{eq8}
		C_1\left(1+\dfrac{1}{4}\|u_n\|\right) &\geq I_b(u_n)-\dfrac{1}{4}\langle I_b'(u_n),u_n\rangle\nonumber\\
		&=\dfrac{1}{2}\|u_n\|^2+\dfrac{b}{4}\|u_n'\|_2^4-\!\Int \!\!\!\!q(x)F(u_n)dx
		-\dfrac{1}{4}\left(\|u_n\|^2+b\|u_n'\|_2^4-\Int \!\!\!\!q(x)f(u_n)u_ndx\right)  \nonumber\\
		&=\dfrac{1}{4}\|u_n\|^2+\dfrac{1}{4}\int_{\Omega_n}q(x)\left[f(u_n)u_n-4F(u_n)\right]dx+\dfrac{1}{4}\int_{\Omega'_n}q(x)\left[f(u_n)u_n-4F(u_n)\right]dx\nonumber\\
		&\geq \dfrac{1}{4}\|u_n\|^2+\dfrac{1}{4}\int_{\Omega_n}q(x)\left[f(u_n)u_n-4F(u_n)\right]dx\nonumber,
	\end{align}
	which yields
	\begin{equation}\label{contrad}
		C_1\left(\dfrac{1}{\|u_n\|^2}+\dfrac{1}{4\|u_n\|}\right)\geq \dfrac{1}{4}+\dfrac{1}{4\|u_n\|^2}\int_{\Omega_n}q(x)\left[f(u_n)u_n-4F(u_n)\right]dx.
	\end{equation}
	On the other hand, for $x\in\Omega_n$, by $(F_1)$ and \eqref{2contr}, it follows that
	\begin{align}
		q(x)\left|f(u_n)u_n-4F(u_n)\right|&\leq q(x)\left(\alpha_2 |u_n|+\beta|u|^\theta +4\alpha_2|u_n|+\dfrac{4\beta}{\theta}|u_n|^\theta\right)\nonumber\\
		&\leq \left(5\alpha_2 L +\dfrac{(\theta+4)\beta}{\theta}L^\theta\right)q(x),\nonumber
	\end{align}
	and then
	\begin{align*}
		\dfrac{1}{\|u_n\|^2}\left|\int_{\Omega_n}q(x)f(u_n)u_n-4F(u_n)dx\right|&\leq   \dfrac{1}{\|u_n\|^2}\left(5\alpha_2 L +\dfrac{(\theta+4)\beta}{\theta}L^\theta\right) \int_{\Omega_n} q(x)dx\nonumber\\
		&\leq \dfrac{1}{\|u_n\|^2}\left(5\alpha_2 L +\dfrac{(\theta+4)\beta}{\theta}L^\theta\right)\|q\|_1\longrightarrow 0,
	\end{align*}
	as $n\longrightarrow +\infty$. Hence,
	\begin{align}\label{2lim}
		\dfrac{1}{\|u_n\|^2}\int_{\Omega_n}q(x)\left[f(u_n)u_n-4F(u_n)\right]dx\longrightarrow 0.%,\qquad \text{ as } n\longrightarrow +\infty.
	\end{align}
	Taking  into account \eqref{2contr} and \eqref{2lim}, by passing to the limit in \eqref{contrad} as $n\longrightarrow +\infty$, we obtain a contradiction. Consequently, $\{u_n\}$ is bounded in $H$ and that what needs to be demonstrated.
	
	\textbf{Step 2. } We will prove that $\{u_n\}$ converges strongly in $H$.\\
	In Step 1, it can be seen that the sequence $\{u_n\}$ is bounded in $H$; then we may assume for a subsequence that
	\begin{equation}\label{conv}
		\begin{array}{l}
			u_n\rightharpoonup u_0 \text{~~weakly in ~~} H,\\
			%v_n\rightarrow v_0 \text{~~strongly in ~~} L^2(0,R),\\
			u_n(x)\rightarrow u_0(x)\qquad \text{ a.e in } (0,+\infty).
		\end{array}
	\end{equation}
	An easy computation shows that
	\begin{align}\label{strong}
		\|u_n-u_0\|^2 =&\, \langle I_b'(u_n)-I_b'(u_0),u_n-u_0\rangle-b\|u_n'\|_2^2\Int u'_n(u'_n-u'_0)dx +b\|u_0'\|_2^2\Int u'_0(u'_n-u'_0)dx\nonumber\\
		\qquad &+\Int q(x)\Big(f(u_n)-f(u_0)\Big)(u_n-u_0)dx.
	\end{align}
	By \eqref{conv} and the continuity of $f$, it clear that for almost every $x\in(0,+\infty)$ 
	\begin{equation*}
		\displaystyle\Int q(x)\Big(f\big(u_n(x)\big)-f\big(u_0(x)\big)\Big)\big(u_n(x)-u_0(x)\big)~dx\longrightarrow 0,\qquad \text{ as~~ } n\longrightarrow +\infty.
	\end{equation*}
	Moreover, from $(F_1)$, \eqref{embed} and the boundedness of $\{u_n\}$ one has
	\begin{align*}
		\left|q(x)\Big(f\big(u_n\big)-f\big(u_0\big)\Big)\big(u_n-u_0\big)\right|&\leq \left[2\alpha_2 +\beta |u_n|^{\theta-1} +\beta |u_0|^{\theta-1}\right]\left( |u_n|+|u_0|\right) q(x)\\
		&\leq \left[2\alpha_2+\beta\mu_\infty^{\theta-1} \Big(C^{\theta-1}+\|u_0\|^{\theta-1}\Big) \right]\left( \mu_\infty C+\mu_\infty \|u_0\|\right) q(x)\\
		&\leq Cq(x)\in L^1(0,+\infty).
	\end{align*}
	By the dominated convergence theorem we get
	\begin{equation}\label{conv1}
		\displaystyle\Int q(x)\Big(f(u_n)-f(u_0)\Big)(u_n-u_0)~dx\longrightarrow 0,\qquad \text{ as~~ } n\longrightarrow +\infty.
	\end{equation}
	Taking into account \eqref{PSC} and the fact that $u_n\rightharpoonup u_0$ in $H$, we get
	\begin{equation}\label{2conv2}
		\langle I'(u_n)-I'(u_0),u_n-u_0\rangle \longrightarrow 0\text{,\qquad as } n\longrightarrow +\infty.
	\end{equation}
	Combining \eqref{conv1} and \eqref{2conv2} with \eqref{strong}, by the same reasoning as in \eqref{1conv4}, we prove that
	\begin{align}\label{2conv4}
		o_n(1)&=\|u_n-u_0\|^2+b\|u_n'\|_2^2\Int u'_n(u'_n-u'_0)dx -b\|u_0'\|_2^2\Int u'_0(u'_n-u'_0)dx\nonumber\\
		&\geq \|u_n-u_0\|^2+\dfrac{b}{2}\Big(\|u_n'\|_2^2- \|u_0'\|_2^2\Big)^2,
	\end{align}
	which yields
	\begin{equation*}\label{2conv6}
		u_n\longrightarrow u_0 \text{ strongly in } H,
	\end{equation*}
	and this proves that $I_b$ satisfies the $(PS)$ condition at any level $c\in\R$. 
\end{proof}
\subsection{Proof of the main result}
In this subsection, we will give the proof of Theorem \ref{2thm} which is divided into two steps. The first step refers to the existence of solutions of \eqref{equ-1}, and the second one to the study of the asymptotic behavior of solutions by considering $b$ as a parameter.

\begin{proof}[\textbf{Proof of Theorem 4}]
	\textbf{Step 1. } We have $I_b\in C^1\big(H,\R\big)$ and $I_b(0)=0$. By Lemmas \ref{2lem2} and \ref{2lem3}, the functional $I_b$ satisfies the geometric property of the mountain pass theorem. Lemma \ref{2lem4} implies that the functional $I_b$ satisfies the $(PS)$ condition. Therefore, applying the mountain pass theorem, we deduce that there exists $v_0\in H$ such that 
	\[I_b(v_0)=c \geq \alpha_* >0 \text{ and } I_b'(v_0)=0,\]
	which means that $v_0$ is a weak solution of \eqref{equ-1}, and this completes Step 1. 
	
	\textbf{Step 2. } Let $\{b_n\}\subset (0,+\infty)$ be a sequence such that
	\begin{equation}\label{lim}
		b_n\longrightarrow 0,\text{\qquad as \quad } n\longrightarrow +\infty,
	\end{equation}
	and let $u_n:=u_{b_n} \in H$ be a solution of \eqref{equ-1}. %Then, it follows from Step 1. that
	%\begin{equation}\label{Prop}
	%I_{b_n}(u_n)=c_{b_n}\text{\qquad and \qquad} %I'_{b_n}(u_n)=0
	%\end{equation}
	%where $c_{b_n}$ is given by \eqref{c}.\\
	Then
	\begin{equation}\label{WS2}
		(u_n,v)+b_n\|u'_n\|^2_2\Int u'_nv'~dx -\Int q(x) f(u_n)v~dx =0,\qquad \forall v\in H.
	\end{equation}
	In the same way as in Step 1 in the proof of Lemma \ref{2lem4}, we can prove that $\{u_{n}\}$ is bounded in $H$, and then there exists $u_0\in H$ such that up to a subsequence 
	\begin{equation}\label{Conv}
		\begin{array}{l}
			u_n\rightharpoonup u_0 \text{~~weakly in ~~} H,\\
			u'_n\rightharpoonup u'_0 \text{~~weakly in ~~} L^2(0,+\infty),\\
			u_n(x)\rightarrow u_0(x)\qquad \text{ a.e in } (0,+\infty).
		\end{array}
	\end{equation}
	Similarly to \eqref{2conv4}, we show that
	\begin{equation}\label{strong2}
		u_n\longrightarrow u_0 \text{~~in~~} H.
	\end{equation}
	Hence, from $(F_1)$, \eqref{lim}, \eqref{Conv}, \eqref{strong2}, and by the dominated convergence theorem we get
	\begin{equation*}
		(u_n,v)\longrightarrow (u_0,v),~~ b_n\|u'_n\|^2_2\Int u'_nv'~dx \longrightarrow 0 \end{equation*}
		and 
		\begin{equation*}
		\Int q(x) f(u_n)v~dx\longrightarrow \Int q(x) f(u_0)v~dx,
	\end{equation*}
	as $n\longrightarrow +\infty$. Consequently, by passing to the limit in \eqref{WS2} as $n\longrightarrow +\infty$, we obtain
	\begin{equation*}
		(u_0,v) -\Int q(x) f(u_0)v~dx =0,\qquad \forall v\in H,
	\end{equation*}
	which means that $u_0$ is a weak solution of the problem \eqref{equt-2}, and this completes the proof.
\end{proof}
\subsection{Example}
Let $f(u)=u^3\ln(1+u^2)+\frac{u^5}{2(1+u^2)}$, $p(x)=\ln(1+x^2)+1$ and $q(x)=e^{-x}$. Is is
easy to check that $(P)$, $(Q)$ and $(F_1)$-$(F_4)$ are satisfied.

Then, by Theorem \ref{2thm}, the problem
\begin{equation*}
	\begin{cases}
		%	\tag{$\mathcal{P}$}
		-\Big(a+b\Int |u'|dx\Big)u'' +\Big(\ln(1+x^2)+1\Big) u = e^{-x}\left(u^3\ln(1+u^2)+\frac{u^5}{2(1+u^2)}\right),\qquad x\in(0,+\infty);\\
		u(0)=0.
	\end{cases}	
\end{equation*} 
has at least one nontrivial solution.

%%===============================================================%%
%% ACKNOWLEDGEMENTS                                              %%
%% Acknowledgements can be added here.                           %%
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\section*{Acknowledgements}
%%===============================================================%%
The authors would like to thank the handling editor and the anonymous referee for careful reading the manuscript and suggesting many valuable comments.

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%\bibliography{ref}
\bibliography{references}

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%% APPENDICES	                                                 %%
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%\normalsize
%\begin{appendices}
%{Some Appendix}
%Appendices should be placed at the end of the manuscript, after the references list. 
%\end{appendices}
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%		%	\bibitem{Watson1983}
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