1. Introduction
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. Various authors studied the existence of positive solutions for \(n\)th-order
\(m\)-point boundary value problems using different methods, for example,
fixed point theorems in cones, nonlinear alternative of
Leray-Schauder, and Krasnoselskii’s fixed point theorem, see
[1, 2, 3, 4, 5] and the references therein.
In 2003, by using the Leray-Schauder degree theory, Liu and
Ge [6] proved the existence of 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, Eloea and Ahmad [7] studied the
existence of positive solutions of a nonlinear \(n\)th-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.
In 2009, Bai et al. [8] used fixed-point theory to study the existence
of positive solutions of a singular nth-order three-point boundary
value problem on time scales represented as:
\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 2004, Sun [9] studied the existence of
nontrivial solution for the three-point boundary value problem:
\begin{gather*}
u^{”}(t)+f(t,u(t))=0,\quad\text 0\leq t\leq 1,
\\
u^{‘}(0)=0,\quad u(1)=\alpha u^{‘}(\eta),
\end{gather*}
where \(\eta\in(0,1)\), \(\alpha\in\mathbb{R}\), \(f\in
C([0,1]\times\mathbb{R},\mathbb{R})\). The same author in [10], studied solvability of a nonlinear second-order
three-point boundary value problem:
\begin{gather*}
u^{”}(t)+f(t,u(t))=0,\quad\text 0\leq t\leq 1,
\\
u^{‘}(0)=0,\quad u(1)=\alpha u(\eta),
\end{gather*}
where \(\eta\in(0,1)\), \(\alpha\in\mathbb{R}\), \(\alpha\neq0\), \(f\in
C([0,1]\times\mathbb{R},\mathbb{R})\).
Li and Sun [11], also used the same method to
study nontrivial solution of a nonlinear second-order three-point
boundary value problem:
\begin{gather*}
u^{”}(t)+f(t,u(t))=0,\quad\text 0\leq t\leq 1,
\\
au(0)-bu^{‘}(0)=0,\quad u(1)-\alpha u(\eta)=0,
\end{gather*}
where \(\eta\in(0,1)\), \(a,b,\alpha\in\mathbb{R}\), with
\(a^{2}+b^{2}>0.\)
Motivated by the above work, we extend the results obtained for
second-order boundary value problem to fourth-order boundary value
problem using a different method from [7]. We prove the existence of
nontrivial solution for the fourth-order three-point boundary value
problem (BVP):
\begin{equation}
u^{(5)}(t)+f(t,u(t))=0,\quad\text 0< t< 1.
\end{equation}
(1)
\begin{equation}
u(0)=0,\quad u^{‘}(0)=u^{”}(0)=u^{”’}(0)=0,\quad u(1)=\alpha
u(\eta),
\end{equation}
(2)
where \(0< \eta< 1\), \(\alpha\in\mathbb{R}\) , \(\alpha\eta^{4}\neq1\),
\(f\in C([0,1]\times\mathbb{R},\mathbb{R})\),
\(\mathbb{R}=(\mathbb{-\infty,\infty})\).
This paper is organized as follows: in Section 2, we present two
lemmas that will be helpful in proving our main results, in Section 3, we present our main results and finally,
in Section 4, we illustrated our results with examples.
2. Preliminaries
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)-(2) is called
nontrivial solution if \(u(t)\neq0\). To get our results, we need to
the following lemma.
Lemma 1.
Let \(y\in C([0,1])\), \(\alpha\eta^{4}\neq1\), then the boundary value
problem
\begin{gather*}
u^{(5)}(t)+ y(t)=0,\quad 0< t< 1,
\\
u(0)=0,\quad u^{'}(0)=u^{''}(0)=u^{'''}(0)=0,\quad u(1)=\alpha
u(\eta),
\end{gather*}
has a unique solution
$$u(t)=-\frac{1}{24}\int_{0}^{t}(t-s)^{4}y(s)ds+\frac{t^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}y(s)ds-
\frac{\alpha
t^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}y(s)ds.$$
Proof. Rewriting the differential equation as \(u^{(5)}(t)=-y(t)\), and integrating five times from \(0\) to \(1\), we obtain
\begin{equation}\label{e3}
u(t)=-\frac{1}{24}\int_{0}^{t}(t-s)^{4}y(s)ds+\frac{t^{4}}{24}c+\frac{t^{3}}{6}c_{1}+\frac{t^{2}}{2}c_{2}+tc_{3}+c_{4}.
\end{equation}
(3)
By the boundary conditions (2), we have
$$u(0)=0,\quad u'(0)=u”(0)=u^{”’}(0)=0,~~i.e.~~c_{1}=c_{2}=c_{3}=c_{4}=0,$$
and \(u(1)=\alpha u(\eta)\), we get
\begin{equation}\label{e4}
c=\frac{1}{(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}y(s)ds-\frac{\alpha}{(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}y(s)ds.
\end{equation}
(4)
Compensate Equation (3) in the Equation (4), we obtain
$$u(t)=-\frac{1}{24}\int_{0}^{t}(t-s)^{4}y(s)ds+\frac{t^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}y(s)ds-
\frac{\alpha
t^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}y(s)ds.$$
This completes the proof.
Define the integral operator \(T: E\longrightarrow E\), by
\begin{equation}\label{e5}Tu(t)=-\frac{1}{24}\int_{0}^{t}(t-s)^{4}f(s,u(s))ds+\frac{t^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}f(s,u(s))ds-
\frac{\alpha
t^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}f(s,u(s))ds.
\end{equation}
(5)
By Lemma 1, the BVP (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.
Lemma 2. [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 \(T\).
3. Existence of nontrivial solution
In this section, we prove the existence of a nontrivial solution for
the BVP (1)-(2). Suppose that \(f\in
C([0,1]\times\mathbb{R},\mathbb{R}).\)
Theorem 3. \label{thm1} Suppose that \(f(t,0)\neq 0\), \(\alpha\eta^{4}\neq1\), and there exist
nonnegative functions \(k,h \in L^{1}[0,1]\), such that
$$|f(t,x)|\leq k(t)|x|+h(t),\quad a.e.~~(t,x)\in
[0,1]\times \mathbb{R},$$
$$\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds< 1.$$
Then the BVP (1)-(2) has at least one nontrivial solution
\(u^{\ast}\in C[0,1].\)
Proof.
Let
$$M=\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds,$$
$$N=\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}h(s)ds+
\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}h(s)ds.$$
Then \(M0\), and
as \(h(t)\geq |f(t,0)|\), a.e. \(t\in [0,1]\), we have \(N>0\).
Let \(A=N(1-M)^{-1}\) and \(\Omega=\{u\in E: \|u\|1\) such that \(Tu=\lambda u\),
then
\begin{eqnarray*}
\lambda A&=&\lambda \|u\|=\|Tu\|=\max_{0\leq t\leq 1}|(Tu)(t)|\\
&\leq& \frac{1}{24}\max_{0\leq t\leq1}\int_{0}^{t}(t-s)^{4}|f(s,u(s))|ds+\max_{0\leq
t\leq1}\left|\frac{t^{4}}{24(1-\alpha\eta^{4})}\right|\int_{0}^{1}(1-s)^{4}|f(s,u(s))|ds\\
&&+\max_{0\leq t\leq1}\left|\frac{\alpha t^{4}}{24(1-\alpha\eta^{4})}\right|\int_{0}^{\eta}(\eta-s)^{4}|f(s,u(s))|ds\\
&\leq&\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}|f(s,u(s))|ds+\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}|f(s,u(s))|ds\\
&\leq&\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}k(s)|u(s)|ds+\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}k(s)|u(s)|ds\\
&&+\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}h(s)ds+\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}h(s)ds\\
&=& M \|u\|+N.
\end{eqnarray*}
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, \(T\) has a fixed point
\(u^{\ast}\in\overline{\Omega}\). In view of \(f(t,0)\neq0\), the BVP
(1)-(2) has a nontrivial solution \(u^{\ast}\in E\).
This completes the proof.
Theorem
Suppose that \(f(t,0)\neq0\), \(\alpha\eta^{4}< 1\), 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 holds:
- there exists a constant \(p>1\) such that
$$\int_{0}^{1}k(s)^{p}ds< \left[\frac{24(1-\alpha\eta^{4})(1+4q)^{1/q}}{2-\alpha\eta^{4}+|\alpha|\eta^{(1+4q)/q}}\right]^{p},
\quad\frac{1}{p}+\frac{1}{q}=1;$$
- there exists a constant
\(\mu>-1\) such that
$$k(s)\leq \frac{(1-\alpha\eta^{4})(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{2-\alpha\eta^{4}+|\alpha|\eta^{5+\mu}}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)0;$$
- there exists a constant \(\mu>-5\) such that
$$k(s)\leq \frac{24(1-\alpha\eta^{4})(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}(1-s)^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\left\{s\in[0,1] :
k(s)0;$$
- \(k(s)\) satisfies
$$k(s)\leq\frac{120(1-\alpha\eta^{4})}{2-\alpha\eta^{4}+|\alpha|\eta^{5}},\quad a.e.~~~s\in [0,1], meas\left\{s\in[0,1] :
k(s)0,$$
then the BVP (1)-(2) has at least one nontrivial solution
\(u^{\ast}\in E.\)
Proof.
Let \(M\) be defined as in the proof of Theorem 3.
To prove Theorem 4, we only need to prove that \(M< 1\). Since
\(\alpha\eta^{4}< 1\), we have
$$M=\left(\frac{1}{24}+\frac{1}{24(1-\alpha\eta^{4})}\right)\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds$$
$$=\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds.~~~~\quad$$
- Using the H\”{o}lder inequality, we have
\begin{eqnarray*}
M&\leq&\left[\int_{0}^{1}k(s)^{p}ds\right]^{1/ p}\left\{\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\left[\int_{0}^{1}(1-s)^{4q}ds\right]^{1/ q}
+\frac{|\alpha|}{24(1-\alpha\eta^{4})}\left[\int_{0}^{\eta}(\eta-s)^{4q}ds\right]^{1/q}\right\}\\
&\leq&\left[\int_{0}^{1}k(s)^{p}ds\right]^{1/ p}\left[\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}(\frac{1}{1+4q})^{1/q}
+\frac{|\alpha|}{24(1-\alpha\eta^{4})}(\frac{\eta^{1+4q}}{1+4q})^{1/q}\right]\\
&<&\frac{24(1-\alpha\eta^{4})(1+4q)^{1/q}}{2-\alpha\eta^{4}+|\alpha|\eta^{(1+4q)/q}}\times
\frac{2-\alpha\eta^{4}+|\alpha|\eta^{(1+4q)/q}}{24(1-\alpha\eta^{4})(1+4q)^{1/q}}\\
&=&1.
\end{eqnarray*}
- Here, we have
\begin{eqnarray*}M&<&\frac{(1-\alpha\eta^{4})(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{2-\alpha\eta^{4}+|\alpha|\eta^{5+\mu}}\\&&\times\left[\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}s^{\mu}ds+\frac{|\alpha|}{24(1-\alpha\eta^{4})}
\int_{0}^{\eta}(\eta-s)^{4}s^{\mu}ds\right]\\
&\leq&\frac{(1-\alpha\eta^{4})(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{2-\alpha\eta^{4}+|\alpha|\eta^{5+\mu}}\left[\frac{2-\alpha\eta^{4}}{(1-\alpha\eta^{4})}\frac{1}
{(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}\right.\\&&\left.+
\frac{|\alpha|}{(1-\alpha\eta^{4})}\frac{\eta^{5+\mu}}{(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}\right]\\&=&\frac{(1-\alpha\eta^{4})(1+\mu)
(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{2-\alpha\eta^{4}+|\alpha|\eta^{5+\mu}}.
\frac{2-\alpha\eta^{4}+|\alpha|\eta^{5+\mu}}{(1-\alpha\eta^{4})(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}\\&=&1.\end{eqnarray*}
- Here, we have
\begin{eqnarray*}M&<&\frac{24(1-\alpha\eta^{4})(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}\left[\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4+\mu}ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}(1-s)^{\mu}ds\right]\\
&\leq&\frac{24(1-\alpha\eta^{4})(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}\left[\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4+\mu}ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4+\mu}ds\right]\\
&=&\frac{24(1-\alpha\eta^{4})(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}\left[\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}.\frac{1}{5+\mu}+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}.\frac{1}{5+\mu}\right]\\
&=&\frac{24(1-\alpha\eta^{4})(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}.\frac{2-\alpha\eta^{4}+|\alpha|}{24(1-\alpha\eta^{4})(5+\mu)}=1.\end{eqnarray*}
- Here, we have
\begin{eqnarray*}M&<&\frac{120(1-\alpha\eta^{4})}{2-\alpha\eta^{4}+|\alpha|\eta^{5}}\left[\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}ds\right]\\
&=&\frac{120(1-\alpha\eta^{4})}{2-\alpha\eta^{4}+|\alpha|\eta^{5}}.\frac{2-\alpha\eta^{4}+|\alpha|\eta^{5}}{120(1-\alpha\eta^{4})}=1.\end{eqnarray*}
This completes the proof.
Theorem 5.
Suppose that \(f(t,0)\neq0\), \(\alpha\eta^{4}>1\), 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 holds:
- there exists a constant \(p>1\) such that
$$\int_{0}^{1}k(s)^{p}ds< \left[\frac{24(\alpha\eta^{4}-1)(1+4q)^{1/q}}{\alpha(\eta^{4}+\eta^{(1+4q)/q})}\right]^{p},
\quad\left(\frac{1}{p}+\frac{1}{q}=1\right);$$
- there exists a constant
\(\mu>-1\) such that
$$k(s)\leq \frac{(\alpha\eta^{4}-1)(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{\alpha(\eta^{4}+\eta^{5+\mu})}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\left\{s\in[0,1] : k(s)0;$$
- there exists a constant \(\mu>-5\) such that
$$k(s)\leq \frac{24(\alpha\eta^{4}-1)(5+\mu)}{\alpha(\eta^{4}+1)}(1-s)^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] : k(s)0;$$
- \(k(s)\) satisfies
$$k(s)\leq\frac{120(\alpha\eta^{4}-1)}{\alpha(\eta^{4}+\eta^{5})},\quad a.e.~~~s\in [0,1],$$
$$meas\left\{s\in[0,1] : k(s)0,$$
then the BVP (1)-(2) has at least one nontrivial solution
\(u^{\ast}\in E.\)
Proof. Let \(M\) be defined as in the proof of Theorem 3.
To prove Theorem 5, we only need to prove that \(M1\), we have
\begin{eqnarray*}M&=&\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds\\
&=&\frac{\alpha}{24(\alpha\eta^{4}-1)}[\eta^{4}\int_{0}^{1}(1-s)^{4}k(s)ds+\int_{0}^{\eta}(\eta-s)^{4}k(s)ds].\end{eqnarray*}
- Using the H\”{o}lder inequality, we have
\begin{eqnarray*}M&\leq&\left[\int_{0}^{1}k(s)^{p}ds\right]^{1/ p}\left\{\frac{\alpha\eta^{4}}{24\left(\alpha\eta^{4}-1\right)}\left[\int_{0}^{1}(1-s)^{4q}ds\right]^{1/q}
+\frac{\alpha}{24\left(\alpha\eta^{4}-1\right)}\left[\int_{0}^{\eta}(\eta-s)^{4q}ds\right]^{1/q}\right\}\\
&\leq&\left[\int_{0}^{1}k(s)^{p}ds\right]^{1/ p}\left[\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}(\frac{1}{1+4q})^{1/q}
+\frac{\alpha}{24(\alpha\eta^{4}-1)}(\frac{\eta^{1+4q}}{1+4q})^{1/q}\right]\\
&<&\frac{24\left(\alpha\eta^{4}-1\right)(1+4q)^{1/q}}{\alpha(\eta^{4}+\eta^{(1+4q)/q})}\times
\frac{\alpha(\eta^{4}+\eta^{(1+4q)/q})}{24(\alpha\eta^{4}-1)(1+4q)^{1/q}}=1.\end{eqnarray*}
- Here, we have
\begin{eqnarray*}M&<&\frac{(\alpha\eta^{4}-1)(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{\alpha(\eta^{4}+\eta^{5+\mu})}\left[\frac{\alpha\eta^{4}}{24
(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}s^{\mu}ds\right.\end{eqnarray*}
\begin{eqnarray*}
\\&&+\left.\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{\eta}(\eta-s)^{4}s^{\mu}ds\right]\\
&\leq&\frac{(\alpha\eta^{4}-1)(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{\alpha(\eta^{4}+\eta^{5+\mu})}\left[\frac{\alpha\eta^{4}}
{(\alpha\eta^{4}-1)}\frac{1}{(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}\right.\\&&\left.+\frac{\alpha}{(\alpha\eta^{4}-1)}\times\frac{\eta^{5+\mu}}{(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}\right]\\
&=&\frac{(\alpha\eta^{4}-1)(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{\alpha(\eta^{4}+\eta^{5+\mu})}.
\frac{\alpha(\eta^{4}+\eta^{5+\mu})}{(\alpha\eta^{4}-1)(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}\\&=&1.\end{eqnarray*}
- Here, we have
\begin{eqnarray*}M&<&\frac{24(\alpha\eta^{4}-1)(5+\mu)}{\alpha(\eta^{4}+1)}\left[\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4+\mu}ds+
\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{\eta}(\eta-s)^{4}(1-s)^{\mu}ds\right]\\
&\leq&\frac{24(\alpha\eta^{4}-1)(5+\mu)}{\alpha(\eta^{4}+1)}\left[\frac{\alpha\eta^{4}}{24(\alpha(\eta^{4}-1)}\int_{0}^{1}(1-s)^{4+\mu}ds+
\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4+\mu}ds\right]\\
&=&\frac{24(\alpha\eta^{4}-1)(5+\mu)}{\alpha(\eta^{4}+1)}\left[\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}.\frac{1}{5+\mu}+
\frac{\alpha}{24(\alpha\eta^{4}-1)}.\frac{1}{5+\mu}\right]\\
&=&\frac{24(\alpha\eta^{4}-1)(5+\mu)}{\alpha(\eta^{4}+1)}.\frac{\alpha(\eta^{4}+1)}{24(\alpha\eta^{4}-1)(5+\mu)}=1.\end{eqnarray*}
- Here, we have
\begin{eqnarray*}M&<&\frac{120(\alpha\eta^{4}-1)}{\alpha(\eta^{4}+\eta^{5})}\left[\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}ds+
\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{\eta}(\eta-s)^{4}ds\right]\\
&=&\frac{120(\alpha\eta^{4}-1)}{\alpha(\eta^{4}+\eta^{5})}.\frac{\alpha(\eta^{4}+\eta^{5})}{120(\alpha\eta^{4}-1)}=1.\end{eqnarray*}
This completes the proof.
Corollary 6.
Suppose \(f(t,0)\neq0\), \(\alpha\eta^{4}< 1\), 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 holds:
- there exists a constant \(p>1\) such that
$$\int_{0}^{1}k(s)^{p}ds< \left[\frac{24(1-\alpha\eta^{4})(1+4q)^{1/q}}{2-\alpha\eta^{4}+|\alpha|}\right]^{p};
\quad\frac{1}{p}+\frac{1}{q}=1;$$
- there exists a constant
\(\mu>-1\) such that
$$k(s)\leq \frac{(1-\alpha\eta^{4})(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\{s\in[0,1] :
k(s)0;$$
- \(k(s)\) satisfies
$$k(s)\leq\frac{120(1-\alpha\eta^{4})}{2-\alpha\eta^{4}+|\alpha|},\quad a.e.~~~s\in [0,1],$$
$$meas\left\{s\in[0,1] :
k(s)0,$$
then the BVP (1)-(2) has at least one nontrivial solution
\(u^{\ast}\in E.\)
Proof. We have
\begin{eqnarray*}M&=&\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds\\
&\leq&\frac{2-\alpha\eta^{4}}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}k(s)ds\\
&=&\frac{2-\alpha\eta^{4}+|\alpha|}{24(1-\alpha\eta^{4})}\int_{0}^{1}(1-s)^{4}k(s)ds.\end{eqnarray*}
Proof of this corollary 6 is the same method in the proof Theorem
4. The proof is complete.
Corollary 7.
Suppose that \(f(t,0)\neq0\), \(\alpha\eta^{4}>1\), 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 holds:
- there exists a constant \(p>1\) such that
$$\int_{0}^{1}k(s)^{p}ds< \left[\frac{24(\alpha\eta^{4}-1)(1+4q)^{1/q}}{\alpha(\eta^{4}+1)}\right]^{p}; \quad\frac{1}{p}+\frac{1}{q}=1;$$
- there exists a constant \(\mu>-1\) such that
$$k(s)\leq \frac{\left(\alpha\eta^{4}-1\right)(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{\alpha(\eta^{4}+1)}s^{\mu},\quad a.e.~~~s\in [0,1],$$
$$meas\left\{s\in[0,1] : k(s)0;$$
- \(k(s)\) satisfies
$$k(s)\leq\frac{120(\alpha\eta^{4}-1)}{\alpha(\eta^{4}+1)},\quad a.e.~~~s\in [0,1],$$
$$meas\left\{s\in[0,1] : k(s)0,$$
then the BVP (1)-(2) has at least one nontrivial solution
\(u^{\ast}\in E.\)
Proof. We have
\begin{eqnarray*}M&=&\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds\\
&\leq&\frac{\alpha\eta^{4}}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{\alpha}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}k(s)ds\\
&=&\frac{\alpha(\eta^{4}+1)}{24(\alpha\eta^{4}-1)}\int_{0}^{1}(1-s)^{4}k(s)ds.\end{eqnarray*}
The rest procedure is the same as for Theorem 5. This completes
the proof.
4. Examples
In order to illustrate the above results, we consider some examples.
Example 1. Consider the following problem
\begin{equation}\label{e6}
\begin{array}{ll}
u^{(5)}+\frac{t}{3}u\sin u^{2}-\sqrt{t}+2=0,\quad 0< t< 1, \\ ~~~~\quad\\
u(0)=0,\quad u^{'}(0)=u^{''}(0)=u^{'''}(0)=0,\quad u(1)=-3u(1/2).
\end{array}
\end{equation}
(6)
Set \(\eta=1/2\), \(\alpha=-3\), and
$$f(t,x)=\frac{t}{3}x\sin x^{2}-\sqrt{t}+2,$$
$$k(t)=t,\quad h(t)=\sqrt{t}+2,$$
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 $$\alpha\eta^{4}=-\frac{3}{16}\neq1.$$
Moreover, we have
$$M=\left(\frac{1}{24}+\frac{1}{24|1-\alpha\eta^{4}|}\right)\int_{0}^{1}(1-s)^{4}k(s)ds+
\frac{|\alpha|}{24|1-\alpha\eta^{4}|}\int_{0}^{\eta}(\eta-s)^{4}k(s)ds\quad\quad\quad$$
$$M=\frac{105}{1368}\int_{0}^{1}(1-s)^{4} sds+\frac{6}{57}\int_{0}^{1/2}\left(\frac{1}{2}-s\right)^{4} sds=\frac{105}{41040}+\frac{6}{109440}< 1.\quad\quad\quad\quad$$
Hence, by Theorem 3, the BVP (6) has at least one nontrivial
solution \(u^{\ast}\) in \(E.\)
Example 2. Consider the following problem
\begin{equation}\label{e7}
\begin{array}{ll}
u^{(5)}+\frac{2/3\sqrt{8+t}}{5+u^{2}}u\cos u^{3}-e^{t}+1=0,\quad 0< t< 1, ~~~~\\
u(0)=0,\quad u^{'}(0)=u^{''}(0)=u^{'''}(0)=0,\quad u(1)=-4u(1/4).
\end{array}
\end{equation}
(7)
Set \(\eta=1/4\), \(\alpha=-4\), and
$$f(t,x)=\frac{2/3\sqrt{8+t}}{5+x^{2}}x\cos x^{3}-e^{t}+1,$$
$$k(t)=\frac{2}{3}\sqrt{8+t},\quad h(t)=e^{t}+1.$$
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
$$\alpha\eta^{4}=-\frac{1}{64}< 1.$$
Let \(p=q=2\), such that \(\frac{1}{p}+\frac{1}{q}=1,\) then
$$\int_{0}^{1}k(s)^{p}ds=\int_{0}^{1}\frac{4}{9}(8+s)ds=\frac{34}{9}.$$
Moreover, we have
$$\left[\frac{24(1-\alpha\eta^{4})(1+4q)^{1/q}}{2-\alpha\eta^{4}+|\alpha|\eta^{(1+4q)/q}}\right]^{p}=1307.88.$$
Therefore,
$$\int_{0}^{1}k(s)^{p}ds< \left[\frac{24(1-\alpha\eta^{4})(1+4q)^{1/q}}{2-\alpha\eta^{4}+|\alpha|\eta^{(1+4q)/q}}\right]^{p}.$$
Hence, by Theorem 4(1), the BVP (7) has at least one
nontrivial solution \(u^{\ast}\) in \(E.\)
Example 3.
Consider the following problem
\begin{equation}\label{e8}
\begin{array}{ll}
u^{(5)}+\frac{u}{7(4+u^{2})\sqrt[3]{t}}e^{-\cos u}-2t-1=0
,\quad 0< t< 1, \\
~~~~\\
u(0)=0,\quad u^{'}(0)=u^{''}(0)=u^{'''}(0)=0,\quad u(1)=-3u(1/3).
\end{array}\end{equation}
(8)
Set \(\eta=1/3\), \(\alpha=-3\), and
$$f(t,x)=\frac{x}{7(4+x^{2})\sqrt[3]{t}}e^{-\cos x}-2t-1,$$
$$k(t)=\frac{1}{7\sqrt[3]{t}},\quad h(t)=2t+1.$$
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 $$\alpha\eta^{4}=-\frac{1}{27}-1\), then
$$\frac{(1-\alpha\eta^{4})(1+\mu)(2+\mu)(3+\mu)(4+\mu)(5+\mu)}{2-\alpha\eta^{4}+|\alpha|\eta^{5+\mu}}=16.873.$$
Therefore,
$$k(s)=\frac{1}{7\sqrt[3]{s}}=\frac{1}{7}s^{-\frac{1}{3}}< 16.873.s^{-\frac{1}{3}},$$
$$meas\left\{s\in[0,1]: k(s)0.$$
Hence, by Theorem 4(2), the BVP (8) has at least one
nontrivial solution \(u^{\ast}\) in \(E.\)
Example 4. Consider the following problem
\begin{equation}
\begin{array}{ll}
u^{(5)}+\frac{3u^{3}}{7(1+u^{2})\sqrt[3]{(1-t)^{2}}}\sin u+t^{5}-1=0
,\quad 0< t< 1, \\
~~~~\\
u(0)=0,\quad u^{'}(0)=u^{''}(0)=u^{'''}(0)=0,\quad u(1)=-2u(1/2).
\end{array}
\end{equation}
(9)
Set \(\eta=1/2\), \(\alpha=-2\), and
$$f(t,x)=\frac{3x^{3}}{7(1+x^{2})\sqrt[3]{(1-t)^{2}}}\sin x+t^{5}-1,$$
$$k(t)=\frac{3}{7\sqrt[3]{(1-t)^{2}}},\quad h(t)=t^{5}+1.$$
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 $$\alpha\eta^{4}=-\frac{1}{8}-4\), then
$$\frac{24(1-\alpha\eta^{4})(5+\mu)}{2-\alpha\eta^{4}+|\alpha|}=\frac{2808}{33}.$$
Therefore,
$$k(s)=\frac{3}{7\sqrt[3]{(1-s)^{2}}}=\frac{3}{7}(1-s)^{-\frac{2}{3}}< \frac{2808}{33}(1-s)^{-\frac{2}{3}},$$
$$meas\left\{s\in[0,1] : k(s)0.$$
Hence, by Theorem 4(3), the BVP (9) has at least one
nontrivial solution \(u^{\ast}\) in \(E.\)
Example 5. Consider the following problem
\begin{equation}\label{e10}
\begin{array}{ll}
u^{(5)}+\frac{tu}{2(3+u^{2})}+e^{t}-3=0
,\quad 0< t< 1, \\
~~~~\\
u(0)=0,\quad u^{'}(0)=u^{''}(0)=u^{'''}(0)=0,\quad u(1)=-5u(1/5).
\end{array}\end{equation}
(10)
Set \(\eta=1/5\), \(\alpha=-5\), and
$$f(t,x)=\frac{tx}{2(3+x^{2})}+e^{t}-3,$$
$$k(t)=\frac{t}{2},\quad h(t)=e^{t}+3.$$
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
$$\alpha\eta^{4}=-\frac{1}{125}< 1.$$
Moreover, we have
$$\frac{120(1-\alpha\eta^{4})}{2-\alpha\eta^{4}+|\alpha|\eta^{5}}=\frac{9450}{157}.$$
Therefore,
$$k(s)=\frac{s}{2}< \frac{9450}{157},\quad s\in[0,1],$$
$$meas\left\{s\in[0,1] : k(s)0.$$
Hence, by Theorem 4(4), the BVP (10) has at least one
nontrivial solution \(u^{\ast}\) in \(E.\)
Author Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the
final manuscript.
Competing Interests
The author(s) do not have any competing interests in the manuscript.
