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\title[Nontrivial solution of a nonlinear fourth-order three-point...] {Nontrivial solution of a nonlinear fourth-order three-point
boundary value problem}
%between [] goes the title that will appear on the top of every
%odd page, between {} goes the real title.
%\thanks{...} put your grant

\author[Z. BEKRI  and S. BENAICHA]{Zouaoui BEKRI  and Slimane BENAICHA}
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%even page, between {} goes the real author.

\address{Zouaoui BEKRI\\ Laboratory of fundamental and applied mathematics, \\
University of Oran $1$, Ahmed Ben Bella, \\ Es-senia, 31000 Oran,
Algeria} \email{:zouaouibekri@yahoo.fr}
\address{Slimane BENAICHA\\ Laboratory of fundamental and applied mathematics, \\
University of Oran $1$, Ahmed Ben Bella, \\ Es-senia, 31000 Oran,
Algeria}
\email{:slimanebenaicha@yahoo.fr}

\maketitle \numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\newtheorem{proof}[theorem]{Proof}

\begin{abstract}
In this paper, we study the existence of nontrivial solution for the
fourth-order three-point boundary value problem having the following
form
\begin{gather*}
u^{(4)}(t)+f(t,u(t))=0,\quad\text 0<t<1,
\\
u(0)=\alpha u(\eta),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=\beta
u(\eta),
\end{gather*}
where $\eta\in(0,1)$, $\alpha, \beta\in\mathbb{R}$, $f\in
C([0,1]\times\mathbb{R},\mathbb{R})$. We give sufficient conditions
that allow us to obtain the existence of a nontrivial solution. And
by using the Leray-Schauder nonlinear alternative we prove the
existence of at least one solution of the posed problem. As an
application, we also given some examples to illustrate the results
obtained.
\end{abstract}
%%put here Key words

\keywords Green's function, Nontrivial solution, Leary-Schauder
nonlinear alternative, Fixed point theorem, Boundary value problem.


\tableofcontents



%% put here the subject class
\2000mathclass{34B10, 34B15, 34K10}


\section{Introduction}\label{sec1}
The study of fourth-order three-point boundary value problems (BVP)
for ordinary differential equations arise in a variety of different
areas of applied mathematics and physics.

Many authors studied the existence positive solutions for nth-order
m-point boundary value problems using different methods such that
fixed point theorems in cones, nonlinear alternative of
Leray-Schauder, and Krasnoselskii's fixed point theorem, see
(\cite{d1,l1,p1,r1}) and the references therein.

In $2003$, by using the Leray-Schauder degree theory, Yuji Liu and
Weigao Ge (\cite{v1}) proved the existence positive solutions for
$(n-1, 1)$ three-point boundary value problems with coefficient that
changes sign given as follows
\begin{gather*}
u^{(n)}(t)+\lambda a(t)f(u(t))=0,\quad\text t\in(0,1),\\
u(0)=\alpha u(\eta),\quad u(1)=\beta u(\eta),\quad u^{(i)}(0)=0~for~i=1,2,...,n-2,\\
and~u^{(n-2)}(0)=\alpha u^{(n-2)}(\eta),~u^{(n-2)}(1)=\beta
u^{(n-2)}(\eta),~u^{(i)}(0)=0~for~i=1,2,..,n-3,
\end{gather*}
where $\eta\in(0,1)$, $\alpha\geq0$, $\beta\geq0$, and $a:
(0,1)\rightarrow \mathbb{R}$ may change sign and
$\mathbb{R}=(-\infty,\infty)$, $f(0)>0$, $\lambda>0$ is a parameter.

In $2005$, Paul W. Eloea and Bashir Ahmad (\cite{z1}) studied the
existence positive solutions of a nonlinear nth-order boundary value
problem with nonlocal conditions as follows
\begin{gather*}
u^{(n)}(t)+a(t)f(u(t))=0,\quad\text t\in(0,1),\\
u(0)=0,\quad u^{'}(0)=0,...,u^{(n-2)}(0)=0,\quad \alpha
u(\eta)=u(1),
\end{gather*}
where $0<\eta<1$, $0<\alpha\eta^{n-1}<1$, $f$ is either superlinear
or sublinear. The methods used is the fixed point theorem in cones
due to Krasnoselki\^{\i} and Guo.

Then in the year $2009$, Xie, Liu and Bai (\cite{n1}) used
fixed-point theory to study the existence positive solutions of a
singular nth-order three-point boundary value problem on time scales
represented in the following figure
\begin{gather*}
u^{n}(t)+a(t)f(u(t))=0,\quad\text t\in(0,1),\\
u(a)=\alpha u(\eta),\quad u^{'}(a)=0,...,u^{(n-2)}(a)=0,\quad
u(b)=\beta u(\eta),
\end{gather*}
where $a<\eta<b$, $0\leq a<1$,
$0<\beta(\eta-a)^{n-1}<(1-\alpha)(b-a)^{n-1}+\alpha(\eta-a)^{n-1}$,
$f\in C([a,b]\times[0,\infty), [0,\infty))$ and $h\in C([a,b],
[0,\infty))$ may be singular at $t=a$ and $t=b$.

In $2013$, Yan Sun and Cun Zhu (\cite{u1}), considered the singular
fourth-order three point boundary value problem
\begin{gather*}
u^{''''}(t)+f(t,u(t))=0,\quad\text 0\leq t\leq 1,\\
u(0)=u^{'}(0)=u^{''}(0)=0,\quad u^{''}(1)-\alpha
u^{''}(\eta)=\lambda,
\end{gather*}
where $\eta\in(0,1)$ and $\alpha\in[0,\frac{1}{\eta})$ are constants
and $\lambda\in[0,\infty)$ is a parameter, The authors presented the
existence of positive solutions by using the Krasnosel’skii fixed
point theorem.

For Some other results on fourth-order boundary value problem, we
refer the reader to the papers (\cite{i1,o1,k1,w1,c1}).

Motivated by the above works, the aim of this paper is to establish
some sufficient conditions for the existence of nontrivial solution
for the fourth-order three-point boundary value problem (BVP)
\begin{gather}
u^{(4)}(t)+f(t,u(t))=0,\quad\text 0<t<1. \label{e1} \\
u(0)=\alpha u(\eta),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=\beta
u(\eta), \label{e2}
\end{gather}
where $\eta\in(0,1)$, $\alpha, \beta\in\mathbb{R}$, $f\in
C([0,1]\times\mathbb{R},\mathbb{R})$, and
$\mathbb{R}=(-\infty,+\infty).$

This paper is organized as follows. In section $2$, we present two
lemmas that will be used to prove the some results. Then, in section
$3$, we present and prove our main results which consist in
existence theorems and corollarys for nontrivial solution of the BVP
$(1.1)-(1.2)$, and we establish some existence criteria of at least
one solution by using the Leray-Schauder nonlinear alternative.
Finally, in section $4$, as an application, we give some examples to
illustrate the results we obtained.

\section{Preliminaries}\label{sec2}
Let $E=C[0,1]$ with the norm $\|y\|=\sup_{t\in[0,1]}|y(t)|$ for any
$u\in E$. A solution $u(t)$ of the BVP $(1.1)-(1.2)$ is called
nontrivial solution if $u(t)\neq0$. To get our results, we need to
provide the following lemma.

\begin{lemma} \label{lem2.1} Let $y\in C([0,1])$, $\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))\neq0$, then three-point BVP
\begin{gather*}
u^{(4)}(t)+ y(t)=0,\quad  0<t<1,
\\
u(0)=\alpha u(\eta),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=\beta
u(\eta),
\end{gather*}
has a unique solution
$$u(t)=-\frac{1}{6}\int_{0}^{t}(t-s)^{3}y(s)ds+\frac{(1-\alpha)(t^{3}(1-\alpha)+\alpha\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}y(s)ds+$$
$$\frac{(1-\alpha)(t^{3}(\alpha-\beta)-\alpha)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}y(s)ds.$$
\end{lemma}
\begin{proof} Rewriting the differential equation as $u^{(4)}(t)=-y(t)$, and integrating four times from $0$ to $1$, we obtain
\begin{equation}\label{e1}
u(t)=-\frac{1}{6}\int_{0}^{t}(t-s)^{3}y(s)ds+\frac{t^{3}}{6}c+\frac{t^{2}}{2}c_{1}+tc_{2}+c_{3}.
\end{equation}
By the boundary conditions $(1.2)$, we have $u'(0)=u''(0)=0$, i.e.
$c_{1}=c_{2}=0,$

and $u(0)=\alpha u(\eta)$, implies
\begin{equation}\label{e2}
c_{3}=-\frac{\alpha}{6(1-\alpha)}\int_{0}^{\eta}(\eta-s)^{3}y(s)ds+\frac{\alpha\eta^{3}}{6(1-\alpha)}c,
\end{equation}
also $u(1)=\beta u(\eta)$, we find
\begin{equation}\label{e3}
c=\frac{1}{(1-\beta\eta^{3})}\int_{0}^{1}(1-s)^{3}y(s)ds-\frac{\beta}{(1-\beta\eta^{3})}\int_{0}^{\eta}(\eta-s)^{3}y(s)ds+
\frac{6(\beta-1)}{(1-\beta\eta^{3})}c_{3}.
\end{equation}
Compensate equation $(2.2)$ in the equation $(2.3)$, we obtain
$$c=\frac{(1-\alpha)}{((1-\alpha)+\eta^{3}(\alpha-\beta))}\int_{0}^{1}(1-s)^{3}y(s)ds+
\frac{(\alpha-\beta)}{((1-\alpha)+\eta^{3}(\alpha-\beta))}\int_{0}^{\eta}(\eta-s)^{3}y(s)ds,$$
and
$$c_{3}=\frac{\alpha\eta^{3}}{6((1-\alpha)+\eta^{3}(\alpha-\beta))}\int_{0}^{1}(1-s)^{3}y(s)ds-
\frac{\alpha(1-\alpha)}{6(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))}\int_{0}^{\eta}(\eta-s)^{3}y(s)ds.$$
Substituting $c$ and $c_{3}$ by their values in $(2.1)$, we obtain
the solution in the statement of the lemma. this completes the
proof.
\end{proof}
Define the integral operator $T: E\longrightarrow E$, by
$$Tu(t)=-\frac{1}{6}\int_{0}^{t}(t-s)^{3}f(s,u(s))ds+\frac{(1-\alpha)(t^{3}(1-\alpha)+\alpha\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}f(s,u(s))ds+$$
\begin{equation}\label{e4}
\frac{(1-\alpha)(t^{3}(\alpha-\beta)-\alpha)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}f(s,u(s))ds.
\end{equation}
By Lemma $2.1$, the BVP $(1.1)-(1.2)$ has a solution if and only if
the operator $T$ has a fixed point in $E$. So we only need to seek a
fixed point of $T$in $E$. By Ascoli-Arzela theorem, we can prove
that $T$ is a completely continuous operator. Now we cite the
Leray-Schauder nonlinear alternative.

\begin{lemma} \label{lem2.3} $([1])$. Let $E$ be a Banach space and
$\Omega$ be a bounded open subset of $E$, $0\in\Omega$.
$T:\overline{\Omega}\rightarrow E$ be a completely continuous
operator. Then, either

$(i)$ there exists $u\in \partial \Omega$ and $\lambda>1$ such that
$T(u)=\lambda u$, or

$(ii)$ there exists a fixed point $u^{\ast}\in \overline {\Omega}$
of $T$.
\end{lemma}

\section{Existence of nontrivial solution}\label{sec3}
In this section, we prove the existence of a nontrivial solution for
the BVP $(1.1)-(1.2)$. Suppose that $f\in
C([0,1]\times\mathbb{R},\mathbb{R}).$

\begin{theorem}  Suppose that $f(t,0)\neq 0$, $\zeta\neq0$, and there exist
nonnegative functions $k,h \in L^{1}[0,1]$ such that
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in[0,1]\times \mathbb{R},$$
$$(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds<1.$$
Then the BVP $(1.1)-(1.2)$ has at least one nontrivial solution
$u^{\ast}\in C[0,1].$
\end{theorem}

\begin{proof} Let
$$M=(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds,$$
$$N=(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}h(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}h(s)ds.$$
Then $M<1$. Since $f(t,0)\neq 0$, there exists an interval
$[a,b]\subset [0,1]$ such that $\min_{a\leq t\leq b}|f(t,0)|>0$. And
as $h(t)\geq |f(t,0)|$, a.e. $t\in [0,1]$, we know that $N>0$.

Let $A=N(1-M)^{-1}$ and $\Omega=\{u\in E: \|u\|<A\}$. Assume that
$u\in\partial \Omega$ and $\lambda>1$ such that $Tu=\lambda u$, then
~$\lambda A=\lambda \|u\|=\|Tu\|=\max_{0\leq t\leq 1}|(Tu)(t)|$
$$\leq \frac{1}{6}\max_{0\leq t\leq1}\int_{0}^{t}(t-s)^{3}|f(s,u(s))|ds+\max_{0\leq
t\leq1}|\frac{(1-\alpha)(t^{3}(1-\alpha)+\alpha\eta^{3})}{6\zeta}|\int_{0}^{1}(1-s)^{3}|f(s,u(s))|ds+$$
$$\max_{0\leq t\leq1}|\frac{(1-\alpha)(t^{3}(\alpha-\beta)-\alpha)}{6\zeta}|\int_{0}^{\eta}(\eta-s)^{3}|f(s,u(s))|ds$$
\newpage
$$\leq(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}|f(s,u(s))|ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}|f(s,u(s))|ds$$
$$\leq[(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}k(s)|u(s)|ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}k(s)|u(s)|ds]+$$
$$[(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}h(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}h(s)ds]$$
$$= M \|u\|+N.\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
Therefore,
$$\lambda \leq M+\frac{N}{A}=M+\frac{N}{N(1-M)^{-1}}=M+(1-M)=1.$$
This contradicts $\lambda>1$. By Lemma $2.3$, T has a fixed point
$u^{\ast}\in\overline{\Omega}$. In view of $f(t,0)\neq0$, the BVP
$(1.1)-(1.2)$ has a nontrivial solution $u^{\ast}\in E$.

This completes the proof.
\end{proof}
\begin{theorem}  Suppose that $f(t,0)\neq0,~~\zeta>0$, and there exist
nonnegative functions $k,h\in L^{1}[0,1]$ such that
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R}.$$
If one of the following conditions is fulfilled

$(1)$ There exists a constant $p>1$ such that
$$\int_{0}^{1}k(s)^{p}ds<[\frac{6\zeta(1+3q)^{1/q}}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{(1+3q)/q})}]^{p},\quad(\frac{1}{p}+\frac{1}{q}=1).$$
$(2)$ There exists a constant $\mu>-1$ such that
$$k(s)\leq \frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}s^{\mu}\}>0.$$
$(3)$ There exists a constant $\mu>-4$ such that
$$k(s)\leq \frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}(1-s)^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}(1-s)^{\mu}\}>0.$$
$(4)$ $k(s)$ satisfies
$$k(s)\leq\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}\}>0.$$
$(5)$ $f(t,x)$ satisfies
$$Q=\limsup_{|x|\rightarrow \infty}\max_{t\in[0,1]}|\frac{f(t,x)}{x}|<
\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}.$$
Then the BVP $(1.1)-(1.2)$ has at least one nontrivial solution
$u^{\ast}\in E.$
\end{theorem}

\begin{proof} Let $M$ be defined as in the proof of Theorem $3.1$.
To prove Theorem $3.3$, we only need to prove that $M<1$. Since
$\zeta>0$, we have
$$M=\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds.$$
$(1)$ Using the H\"{o}lder inequality, we have
$$M\leq[\int_{0}^{1}k(s)^{p}ds]^{1/ p}\{\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}[\int_{0}^{1}(1-s)^{3q}ds]^{1/ q}
+\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}[\int_{0}^{\eta}(\eta-s)^{3q}ds]^{1/q}\}\quad$$
$$M\leq[\int_{0}^{1}k(s)^{p}ds]^{1/ p}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}(\frac{1}{1+3q})^{1/ q}
+\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}(\frac{\eta^{1+3q}}{1+3q})^{1/q}]$$
$$<\frac{6\zeta(1+3q)^{1/q}}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{(1+3q)/q})}.\frac
{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{(1+3q)/q})}{6\zeta(1+3q)^{1/q}}$$
$$=1.\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
$(2)$ In this case, we have
$$M<\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}s^{\mu}ds+$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}s^{\mu}ds]$$
$$\leq\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{\zeta}\times$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{1}{(1+\mu)(2+\mu)(3+\mu)(4+\mu)}+\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{\zeta}\times\frac{\eta^{4+\mu}}{(1+\mu)(2+\mu)(3+\mu)(4+\mu)}]$$
$$=\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}.
\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}=1.$$

$(3)$ In this case, we have
$$M<\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3+\mu}ds+$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}(1-s)^{\mu}ds]$$
$$\leq\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3+\mu}ds+$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{1}(1-s)^{3+\mu}ds]$$
$$=\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}.\frac{1}{4+\mu}+$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}.\frac{1}{4+\mu}]$$
$$=\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}.\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}{6\zeta(4+\mu)}=1.$$

$(4)$ In this case, we have
$$M<\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}ds+$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}ds]$$
$$=\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}[\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{24\zeta}+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)\eta^{4}}{24\zeta}]=1.$$

$(5)$ Let
$\epsilon=\frac{1}{2}[\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}-Q]$,
then there exists $c>0$ such that
$$|f(t,x)|\leq[\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}-\epsilon]|x|,~~~(t,x)\in[0,1]\times\mathbb{R}\(-c,c).$$
Set $A=\max\{|f(t,x)|:(t,x)\in[0,1]\times[-c,c]\}$, then
$$|f(t,x)|\leq[\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}-\epsilon]|x|+A,~~~(t,x)\in[0,1]\times\mathbb{R}.$$
Set
$k(s)=\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}-\epsilon$,
$h(s)=A$, then $(4)$ holds.

This completes the proof.
\end{proof}

\begin{corollary} Suppose $f(t,0)\neq0$, $\zeta>0$, and there exist nonnegative
functions $k, h\in L^{1}[0,1]$ such that
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in[0,1]\times \mathbb{R}.$$
If one of following conditions is holds

$(1)$ There exists a constant $p>1$ such that
$$\int_{0}^{1}k(s)^{p}ds<[\frac{6\zeta(1+3q)^{1/q}}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}]^{p},\quad(\frac{1}{p}+\frac{1}{q}=1).$$
$(2)$ There exists a constant $\mu>-1$ such that
$$k(s)\leq \frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}s^{\mu}\}>0.$$
$(3)$ $k(s)$ satisfies
$$k(s)\leq\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}\}>0.$$
$(4)$ $f(t,x)$ satisfies
$$Q=\limsup_{|x|\rightarrow \infty}\max_{t\in[0,1]}|\frac{f(t,x)}{x}|<
\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}.$$
Then the BVP $(1.1)-(1.2)$ has at least one nontrivial solution
$u^{\ast}\in E.$
\end{corollary}

\begin{proof} In this case, we have
$$M=\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds$$
$$\quad\leq\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds$$
$$=\frac{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}{6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds.\quad\quad\quad\quad\quad$$
Proof of this corollary $3.5$ is the same method in the proof
theorem $3.3.$
\end{proof}

\begin{theorem} Suppose $f(t,0)\neq0$, $\alpha>0$, $\beta>0$,
$\zeta<0$, and there exist nonnegative functions $k, h\in
L^{1}[0,1]$ such that
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in[0,1]\times \mathbb{R}.$$
If one of the following conditions is fulfilled

$(1)$ There exists a constant $p>1$ such that
$$\int_{0}^{1}k(s)^{p}ds<[\frac{-6\zeta(1+3q)^{1/q}}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)\eta^{(1+3q)/q}}]^{p},\quad(\frac{1}{p}+\frac{1}{q}=1).$$
$(2)$ There exists a constant $\mu>-1$ such that
$$k(s)\leq \frac{-\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)\eta^{4+\mu}}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{-\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)\eta^{4+\mu}}s^{\mu}\}>0.$$
$(3)$ There exists a constant $\mu>-4$ such that
$$k(s)\leq \frac{-6\zeta(4+\mu)}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}(1-s)^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{-6\zeta(4+\mu)}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}(1-s)^{\mu}\}>0.$$
$(4)$ $k(s)$ satisfies
$$k(s)\leq\frac{-24\zeta}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)\eta^{4}},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{-24\zeta}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)\eta^{4}}\}>0.$$
$(5)$ $f(t,x)$ satisfies
$$Q=\limsup_{|x|\rightarrow \infty}\max_{t\in[0,1]}|\frac{f(t,x)}{x}|<
\frac{-24\zeta}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)\eta^{4}}.$$
Then the BVP $(1.1)-(1.2)$ has at least one nontrivial solution
$u^{\ast}\in E.$
\end{theorem}

\begin{proof} Let $M$ be given as in the proof of Theorem $3.1$.
To prove Theorem $3.4$, we only need to prove that $M<1$. Since
$\alpha>0$, $\beta>0$, and $\zeta<0$, we have
$$M=\frac{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha}{-6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+\alpha)(\beta+2\alpha)}{-6\zeta}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds.$$
Proof of this theorem $3.7$ is the same method in the proof theorem
$3.3.$
\end{proof}

\begin{corollary} Suppose $f(t,0)\neq0$, $\alpha>0$, $\beta>0$,
$\zeta<0$, and there exist nonnegative functions $k, h\in
L^{1}[0,1]$ such that
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in[0,1]\times \mathbb{R}.$$
If one of the following conditions is holds

$(1)$ There exists a constant $p>1$ such that
$$\int_{0}^{1}k(s)^{p}ds<[\frac{-6\zeta(1+3q)^{1/q}}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}}]^{p},\quad(\frac{1}{p}+\frac{1}{q}=1).$$
$(2)$ There exists a constant $\mu>-1$ such that
$$k(s)\leq \frac{-\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{-\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}s^{\mu}\}>0.$$
$(3)$ $k(s)$ satisfies
$$k(s)\leq\frac{-24\zeta}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)<\frac{-24\zeta}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}\}>0.$$
$(4)$ $f(t,x)$ satisfies
$$Q=\limsup_{|x|\rightarrow \infty}\max_{t\in[0,1]}|\frac{f(t,x)}{x}|<
\frac{-24\zeta}{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}.$$
Then the BVP $(1.1)-(1.2)$ has at least one nontrivial solution
$u^{\ast}\in E.$
\end{corollary}

\begin{proof} In this case, we have
$$M=\frac{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha}{-6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+\alpha)(\beta+2\alpha)}{-6\zeta}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds$$
$$\quad\leq\frac{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha}{-6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+\alpha)(\beta+2\alpha)}{-6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds$$
$$=\frac{\eta^{3}(2\alpha^{2}+\beta(1-\alpha))+4\alpha+(1+\alpha)(\beta+2\alpha)}{-6\zeta}\int_{0}^{1}(1-s)^{3}k(s)ds.\quad\quad\quad\quad\quad$$
The rest procedure is the same as for theorem $3.7.$
\end{proof}


\section{Examples}\label{sec4}
In order to illustrate the above results, we consider some examples.

\begin{example} Consider the following problem
\begin{equation}\label{e1}
                          \left
                          \begin{array}{ll}
                          u^{(4)}+\frac{t}{5}|u|\cos\sqrt[3]{u}+2t+1=0,\quad 0<t<1, \\
                          ~~~~\quad\\
                          u(0)=3u(1/2),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=-u(1/2).
                       \end{array}\end{equation}
Set $\eta=\frac{1}{2}$, $\alpha=3$, $\beta=-1$, and
$$f(t,x)=\frac{t}{5}|x|\cos\sqrt[3]{x}+2t+1,$$
$$k(t)=\frac{t}{2},\quad h(t)=2t+1,$$
It is easy to prove that $k, h\in L^{1}[0,1]$ are nonnegative
functions, and
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R},$$
and
$$\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))=3\neq0.$$
Moreover, we have
$$M=(\frac{1}{6}+\frac{(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3})}{6|\zeta|})\int_{0}^{1}(1-s)^{3}k(s)ds+
\frac{(1+|\alpha|)(|\beta|+2|\alpha|)}{6|\zeta|}\int_{0}^{\eta}(\eta-s)^{3}k(s)ds$$
$$M=\frac{41}{36}\int_{0}^{1}(1-s)^{3}.\frac{s}{2}ds+\frac{14}{9}\int_{0}^{1/2}(\frac{1}{2}-s)^{3}.\frac{s}{2}ds=\frac{41}{1440}+\frac{7}{11520}=0.029<1.\quad\quad\quad\quad$$
Hence, by theorem $3.1$, the BVP $(4.1)$ has at least one nontrivial
solution $u^{\ast}$ in $E.$
\end{example}

\begin{example} Consider the following problem
\begin{equation}\label{e2}
                          \left\
                          \begin{array}{ll}
                          u^{(4)}+\frac{2/3\sqrt[3]{7+t}u}{1+u^{5}}\sin u^{2}-e^{t}-3=0,\quad 0<t<1, \\
                          ~~~~\\
                          u(0)=2u(1/4),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=u(1/4).
                       \end{array}\end{equation}
Set $\eta=1/4$, $\alpha=2$, $\beta=1$, and
$$f(t,x)=\frac{2/3\sqrt[3]{7+t}x}{1+x^{5}}\sin x^{2}-e^{t}-3,$$
$$k(t)=\frac{2}{3}\sqrt[3]{7+t},\quad h(t)=e^{t}+3.$$
It is easy to prove that $k, h\in L^{1}[0,1]$ are nonnegative
functions, and
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R}.$$
and
$$\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))=\frac{63}{64}>0.$$
Let $p=3, q=\frac{3}{2}$, such that $\frac{1}{p}+\frac{1}{q}=1,$
then
$$\int_{0}^{1}k(s)^{p}ds=\int_{0}^{1}\frac{8}{27}(7+s)ds=\frac{60}{27}.$$
Moreover, we have
$$[\frac{6\zeta(1+3q)^{1/q}}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{(1+3q)/q})}]^{p}=4.221.$$
Therefore,
$$\int_{0}^{1}k(s)^{p}ds<[\frac{6\zeta(1+3q)^{1/q}}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{(1+3q)/q})}]^{p}.$$
Hence, by theorem $3.3~(1)$, the BVP $(4.2)$ has at least one
nontrivial solution $u^{\ast}$ in $E.$
\end{Example}

\begin{example} Consider the following problem
\begin{equation}\label{e3}
                          \left\
                          \begin{array}{ll}
                          u^{(4)}+\frac{u^{3}}{(9+2u^{2})\sqrt[3]{t}}e^{-\sin u^{2}}-\sqrt{t}-1=0
                          ,\quad 0<t<1, \\
                          ~~~~\\
                          u(0)=4u(1/3),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=3u(1/3).
                       \end{array}\end{equation}
Set $\eta=1/3$, $\alpha=4$, $\beta=3$, and
$$f(t,x)=\frac{x^{3}}{(9+2x^{2})\sqrt[3]{t}}e^{-\sin x^{2}}-\sqrt{t}-1,$$
$$k(t)=\frac{1}{9\sqrt[3]{t}},\quad h(t)=\sqrt{t}+1.$$
It is easy to prove that $k, h\in L^{1}[0,1]$ are nonnegative
functions, and
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R}.$$
and
$$\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))=9>0.$$
Let $\mu=-\frac{1}{3}>-1$, then
$$\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}=2.724.$$
Therefore,
$$k(s)=\frac{1}{9\sqrt[3]{s}}=\frac{1}{9}s^{-\frac{1}{3}}<2.734.s^{-\frac{1}{3}},$$
$$meas\{s\in[0,1]: k(s)<\frac{\zeta(1+\mu)(2+\mu)(3+\mu)(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4+\mu})}s^{\mu}\}>0.$$
Hence, by theorem $3.3~(2)$, the BVP $(4.3)$ has at least one
nontrivial solution $u^{\ast}$ in $E.$
\end{example}

\begin{example} Consider the following problem
\begin{equation}\label{e4}
                          \left\
                          \begin{array}{ll}
                          u^{(4)}+\frac{u}{7(3+u^{2})\sqrt[5]{(1-t)^{2}}}e^{-2t}\cos u+t^{3}-2=0
                          ,\quad 0<t<1, \\
                          ~~~~\\
                          u(0)=5u(1/2),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=4u(1/2).
                       \end{array}\end{equation}
Set $\eta=1/2$, $\alpha=5$, $\beta=4$, and
$$f(t,x)=\frac{x}{7(3+x^{2})\sqrt[5]{(1-t)^{2}}}e^{-2t}\cos x+t^{3}-2,$$
$$k(t)=\frac{1}{6\sqrt[5]{(1-t)^{2}}},\quad h(t)=t^{3}+2.$$
It is easy to prove that $k, h\in L^{1}[0,1]$ are nonnegative
functions, and
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R}.$$
and
$$\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))=62>0.$$
Let $\mu=-\frac{2}{5}>-4$, then
$$\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}=\frac{26784}{3715}.$$
Therefore,
$$k(s)=\frac{1}{6\sqrt[5]{(1-s)^{2}}}=\frac{1}{6}(1-s)^{-\frac{2}{5}}<\frac{26784}{3715}(1-s)^{-\frac{2}{5}},$$
$$meas\{s\in[0,1] : k(s)<\frac{6\zeta(4+\mu)}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|))}(1-s)^{\mu}\}>0.$$
Hence, by theorem $3.3~(3)$, the BVP $(4.4)$ has at least one
nontrivial solution $u^{\ast}$ in $E.$
\end{example}

\begin{example} Consider the following problem
\begin{equation}\label{e5}
                          \left\
                          \begin{array}{ll}
                          u^{(4)}+\frac{tu^{5}}{8(1+u^{2})}-e^{4t}-1=0
                          ,\quad 0<t<1, \\
                          ~~~~\\
                          u(0)=3u(1/5),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=-2u(1/5).
                       \end{array}\end{equation}
Set $\eta=1/5$, $\alpha=3$, $\beta=-2$, and
$$f(t,x)=\frac{tx^{5}}{8(1+x^{2})}-e^{4t}-1,$$
$$k(t)=\frac{t}{5},\quad h(t)=e^{4t}+1.$$
It is easy to prove that $k, h\in L^{1}[0,1]$ are nonnegative
functions, and
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R}.$$
and
$$\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))=\frac{98}{25}>0.$$
Moreover, we have
$$\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}=\frac{29400}{6271}.$$
Therefore,
$$k(s)=\frac{s}{5}<\frac{29400}{6271},\quad s\in[0,1],$$
$$meas\{s\in[0,1] : k(s)<\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}\}>0.$$
Hence, by theorem $3.3~(4)$, the BVP $(4.5)$ has at least one
nontrivial solution $u^{\ast}$ in $E.$
\end{example}

\begin{example} Consider the following problem
\begin{equation}\label{e6}
                          \left\
                          \begin{array}{ll}
                          u^{(4)}+\frac{7t^{2}u}{3(1+e^{t})^{2}}-t+1=0
                          ,\quad 0<t<1, \\
                          ~~~~\\
                          u(0)=4u(1/6),\quad u^{'}(0)=u^{''}(0)=0,\quad u(1)=-2u(1/6).
                       \end{array}\end{equation}
Set $\eta=1/6$, $\alpha=4$, $\beta=-2$, and
$$f(t,x)=\frac{7t^{2}x}{3(1+e^{t})^{2}}-t+1,$$
$$k(t)=7t^{2},\quad h(t)=t+1.$$
It is easy to prove that $k, h\in L^{1}[0,1]$ are nonnegative
functions, and
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in [0,1]\times\mathbb{R}.$$
and
$$\zeta=(1-\alpha)((1-\alpha)+\eta^{3}(\alpha-\beta))=\frac{321}{36}>0.$$
Moreover, we have
$$\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}=\frac{138672}{22063},$$
and
$$Q=\limsup_{|x|\rightarrow \infty}\max_{t\in[0,1]}|\frac{f(t,x)}{x}|=\limsup_{|x|\rightarrow\infty}(\frac{7}{3(1+e)^{2}}+\frac{2}{|x|})=0.17.$$
Therefore,
$$Q=\limsup_{|x|\rightarrow \infty}\max_{t\in[0,1]}|\frac{f(t,x)}{x}|<
\frac{24\zeta}{\zeta+(1+|\alpha|)((1+|\alpha|)+|\alpha|\eta^{3}+(|\beta|+2|\alpha|)\eta^{4})}.$$
Hence, by theorem $3.3~(5)$, the BVP $(4.6)$ has at least one
nontrivial solution $u^{\ast}$ in $E.$
\end{example}

\begin{remark} We can give examples similar in relation to the corollary 3.5,
theorem 3.7, and corollary 3.9.
\end{remark}


%%% to create the Acknowledgments

\ack The authors want to thank the anonymous referee for the
throughout reading of the manuscript and several suggestions that
help us improve the presentation of the paper.

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