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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 08, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/08\hfil A necessary and sufficient condition]
{A necessary and sufficient condition for the existence of
positive solutions to singular boundary-value problems of higher
order differential equations}

\author[C. L. Zhao, Y. Y. Yuan, Y. S. Liu  \hfil EJDE-2006/08\hfilneg]
{Chenglong  Zhao, Yanyan Yuan, Yansheng Liu} 

\address{Chenglong Zhao\hfill\break
Department of Mathematics, Shandong Normal University, Jinan,
250014, China} 
\email{jnzhchl@sohu.com}

\address{Yanyan Yuan \hfill\break
Department of Mathematics, Shandong Normal University, Jinan,
250014,  China} 
\email{yanyanyuan0311@163.com}

\address{Yansheng Liu \hfill\break
Department of Mathematics, Shandong Normal University,
Jinan, 250014, China}
\email{ysliu6668@sohu.com}


\date{}
\thanks{Submitted September 8, 2005. Published January 19, 2006.}
\thanks{Supported by grants 10571111 from the
National Science Foundation of China  and \hfill\break\indent
Y2005A07 from Natural Science Foundation, Shandong Province,
China} 
\subjclass[2000]{34B16} 
\keywords{Singular sublinear
boundary-value problem; positive solution; \hfill\break\indent
fixed point theorem; cone; higher order differential equation}

\begin{abstract}
 By constructing some special cones and using fixed point
 theorem of cone expansion and compression, this paper
 presents some necessary and sufficient conditions for
 the existence of  $C^{4n-2}$ positive solutions to a
 class of singular boundary-value problems.
 Some examples are presented to illustrate our main results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction and Preliminary}

   Singular boundary-value problems (SBVP) for ordinary differential
equations arise in the field of gas dynamics, fluid mechanics,
 theory of boundary layer, and so on. These problems are also an important
  branch in the field of differential equations
\cite{a1,a2,a3,a4,a5,l1,o1,s1,t1,w1,w2,w3,w4,y1,z1}.
In recent years, the positive solutions of singular boundary-value problems
for higher order
nonlinear differential equations have been studied extensively; see for
example \cite{a3,l1,o1,s1,w1,w2,w3,w4,y1,z1}.

For instance, in the superlinear case, Shi \cite{s1} obtained some
necessary and sufficient conditions for the existence of
$C^2[0,1]$ or $C^3[0,1]$
 positive solutions of differential equations under some conditions.
In the sublinear case,
Wei \cite{w2} gave a necessary and sufficient condition for the
existence of $C^2$ and $C^3$
 positive solutions by means of the method of lower and upper
solutions with the maximum principle for
\begin{gather*}
x^{(4)}(t)= f(t,x(t)),\quad \mbox{for all } 0<t<1,\\
x(0)=x(1)=x''(0)=x''(1)=0. \end{gather*}
 In \cite{z1}, Zhang discussed
the  boundary-value problem
\begin{equation} \label{e*}
\begin{gathered}
 x^{(4n)}(t)= f(t,x(t)),\quad \mbox{for all } 0<t<1,\\
x^{(2k)}(0)=x^{(2k)}(1)=0,\quad  k=0, 1, 2, \dots, 2n-1,
\end{gathered}
\end{equation}
 by using the method of lower and upper solutions,
where $f \in C[(0,1)\times (0,+\infty)$, $[0,+\infty)]$,
$f(t,x)\not\equiv 0$, and there exist constants $\lambda, \mu,
N, M$ with $-\infty<\lambda\leq 0\leq \mu < 1$,
 $\frac{2(\mu-\lambda)}{1+\mu}<1$, $0<N\leq 1\leq M$ such that for any
$ 0<t<1$, $x\in (0,\infty)$, satisfying
\begin{equation} \label{G}
\begin{gathered} {c}^\mu f(t,x)\leq
f(t,cx)\leq c ^\lambda f(t,x),\quad 0\leq c \leq N,\\
c^\lambda f(t,x)\leq f(t,cx)\leq c^\mu f(t,x),\quad c\geq M.
\end{gathered}
\end{equation}
The main results of \cite{z1} are the following two theorems.

\begin{theorem} \label{thmA}
Under assumption \eqref{G}, \eqref{e*} has a
$C^{4n-2}$ positive solution if and only if
\begin{gather*}
0<\int_0^1t(1-t)f(t,t(1-t))dt< +\infty,\\
\lim_{t \to {0+}}t\int_{t}^{1}(1-s)f(s,s(1-s))ds=0,\\
\lim_{t \to {1-}}t\int_{t}^{1}(1-s)f(s,s(1-s))ds=0.
\end{gather*}
\end{theorem}

\begin{theorem} \label{thmB}
Under assumption \eqref{G}, \eqref{e*} has a
$C^{4n-1}$ positive solution if and only if
$$
0<\int_0^1f(t,t(1-t))dt<+ \infty.
$$
\end{theorem}

Note that when $n=1$, Theorems \ref{thmA}  and \ref{thmB} are the
results in \cite{s1}.
Inspired by above results, this paper investigates the
 boundary-value problem
\begin{equation}
\begin{gathered}
u^{(4n)}(t)= f(t,u(t),u^{(4n-2)}(t)),\quad 0<t<1,\\
u(0)=u(1)=0,\\
R_1(u)=:au^{(2k)}(0)-bu^{(2k+1)}(0)=0,\\
R_2(u)=:cu^{(2k)}(1)+du^{(2k+1)}(1)=0,\quad
 k= 1, 2, \dots, 2n-1.
\end{gathered} \label{e1.1}
\end{equation}
where
 $a\geq 0$, $b\geq 0$, $c\geq 0$, $d\geq 0$, $\Delta =ac+ad+bc>0$,
and $f\in C[(0,1)\times(0,+\infty)\times(-\infty,0)$,
$[0,+\infty)]$ is quasi-homogeneous
with respect to the last two variables, that is, there are
constants $\lambda$, $\mu$, $\alpha$, $\beta$; $N_1$, $M_1$,
$N_2$, $M_2$ with $-\infty<\lambda\leq 0\leq \mu<\infty$,
$0\leq \alpha\leq\beta<1$, $\mu+\beta<1$; $0<N_1\leq1\leq M_1$,
 $0<N_2\leq 1 \leq M_2$ such that for any $0<t<1$, $u>0$, $v\leq0$
satisfying
\begin{equation} \label{H}
\begin{gathered}
 \bar{c}^\mu f(t,u,v)\leq f(t,\bar{c} u,v)\leq
\bar{c} ^\lambda f(t,u,v),\quad 0<\bar{c} \leq N_1,\\
 \bar{c}^\lambda f(t,u,v)\leq f(t,\bar{c}u,v)\leq
\bar{c}^\mu f(t,u,v),\quad \bar{c}\geq M_1;\\
  \bar{c}^\beta f(t,u,v)\leq f(t,u,\bar{c} v)\leq \bar{c}^\alpha
f(t,u,v),\quad 0<\bar{c} \leq N_2,\\
  \bar{c}^\alpha
f(t,u,v)\leq f(t,u,\bar{c} v)\leq \bar{c}^\beta f(t,u,v),\quad
\bar{c} \geq M_2.
\end{gathered}
\end{equation}
A typical function satisfying the above hipothesis is
$$
f(t,u,v)=\sum_{i=1}^np_i(t)u^{\alpha_i}(-v)^{\beta_i},
$$
where $p_i(t)\in C[(0,1), R^+]$,
$\lambda=\alpha_1\leq\alpha_2\leq\dots\leq\alpha_k<0<\alpha_{k+1}
\leq\dots\leq\alpha_n=\mu$,
 $0\leq\beta_i<1$, $k=1, 2, \dots, n-1$, $i=1, 2, \dots, n$.


To the best of our knowledge, there is no paper that considers
\eqref{e1.1} with general  boundary-value conditions. As a result,
the goal of present paper discusses and treats  the extension of
focal boundary- value problems to more general n-th order boundary
value problems and hence fill the gap in this area. The main
features here are as follows. Firstly, the nonlinear term $f$
include $u^{(4n-2)}$. Secondly, the  boundary- value conditions
are more extensive. Thirdly, the singularity of $f$ on $u$ is
arbitrary.

The main techniques used in this paper are some new constructed
construct cones  and  cone expansion and compression fixed point
theorems. Comparing with previous literature to study the singular
problems, neither the approximation method nor upper-lower
solution approach is applied. In this paper, we obtain some
necessary and sufficient conditions for the existence of
$C^{4n-2}$ positive solutions.


We say $u \in C^{4n-2}[0,1]\cap C^{4n}(0,1)$ is  a
$C^{4n-2}[0,1]$ positive solution of \eqref{e1.1} if
$u(t)$ satisfies \eqref{e1.1} and $u(t)>0$ for $t\in (0,1)$.

Now we state the following lemma from the
literature which will be used in section 2.

 \begin{lemma}[\cite{g1}] \label{lem1.1}
Let $K$ be a cone of real Banach
space $E$, $\Omega_1$, $\Omega_2$ be bounded open sets of $E$,
$0\in \overline{\Omega}_1\subset\Omega_2$.
 Suppose that $A: K\cap(\overline{\Omega}_2\setminus\Omega_1)\rightarrow K$
is completely continuous such that
 one of the following  two conditions is satisfied:
\begin{itemize}
\item[(i)] $\|Ax\|\leq\|x\|$ for $x\in K\cap\partial\Omega_1$;
$\|Ax\|\geq\|x\|$ for $x\in K\cap\partial\Omega_2$,

\item[(ii)] $\|Ax\|\leq\|x\|$ for  $x\in K\cap\partial\Omega_2$;
$\|Ax\|\geq\|x\|$ for  $x\in K\cap\partial\Omega_1$.
\end{itemize}
Then $A$ has a fixed point in
$K\cap(\overline{\Omega}_2\setminus\Omega_1)$.
\end{lemma}

 \section{Main Results}

\begin{theorem} \label{thm2.1} Suppose \eqref{H} holds and  $b=d=0$.
Then \eqref{e1.1} has a
$C^{4n-2}[0,1]$ positive solution if and only if
\begin{gather}
0<\int_0^1t(1-t)f(t,t(1-t),-1)dt<+\infty;\label{e2.1} \\
\lim_{t \to {0+}}t\int_{t}^{1}(1-s)f(s,s(1-s),-1)ds=0; \label{e2.2} \\
\lim_{t \to {1-}}t\int_{t}^{1}(1-s)f(s,s(1-s),-1)ds=0. \label{e2.3}
\end{gather}
\end{theorem}

\begin{theorem} \label{thm2.2}
Suppose \eqref{H} holds and $b=0, d> 0$.
 Then \eqref{e1.1} has a
$C^{4n-2}[0,1]\cap C^{4n-1}(0,1]$ positive solution if and only
if
\begin{gather}
0<\int_0^1tf(t,t(1-t),-1)dt<+\infty,\label{e2.4} \\
\lim_{t \to {0+}}t\int_{t}^{1}f(s,s(1-s),-1)ds=0. \label{e2.5}
\end{gather}
\end{theorem}

\begin{theorem} \label{thm2.3}
Suppose \eqref{H} holds and $b>0, d= 0$.
 Then \eqref{e1.1} has a
$C^{4n-2}[0,1]\cap C^{4n-1}[0,1)$ positive solution if and only
if
\begin{gather}
0<\int_0^1(1-t)f(t,t(1-t),-1)dt<+\infty,\label{e2.6} \\
\lim_{t \to {1^-}}(1-t)\int_{0}^{t}f(s,s(1-s),-1)ds=0. \label{e2.7}
\end{gather}
\end{theorem}

It is well known that
\begin{equation}
G(t, s)=
\frac{1}{\Delta}\begin{cases}
(b+as)[d+c(1-t)],& s< t;\\
(b+at)[d+c(1-s)],& t\leq s,
\end{cases}
\label{e2.8}
\end{equation}
is the Green function of homogeneous boundary-value problem
\begin{equation}
\begin{gathered}
 -u''(t)= 0\quad 0\leq t \leq 1,\\
au(0)-bu'(0)=0,\\
cu(1)+du'(1)=0.
\end{gathered}
\label{e2.9}
\end{equation}
 It is easy to see that
 \begin{equation}
\begin{gathered}
G(t,s)\geq  \frac{[c(1-s)(b+as)+ads][c(1-t)(b+at)+adt]}{\Delta^2},\\
G(t,s)\leq G(t,t),\quad G(t,s)\leq G(s,s).
\end{gathered}
\label{e2.10}
\end{equation}
Since
$$
\frac{G(t,s)}{G(\tau,s)}=\begin{cases}
 \frac{(b+as)[d+c(1-t)]}{(b+a \tau)[d+c(1-s)]},& \tau< s< t;\\
 \frac{d+c(1-t)}{d+c(1-\tau)},& s\leq t, \tau;\\
 \frac{b+at}{b+a \tau},&  t, \tau\leq s;\\
 \frac{(b+at)[d+c(1-s)]}{(b+as)[d+c(1-\tau)]},& t< s< \tau,
\end{cases}
 $$
we know that
\begin{equation}
G(t,s)\geq e(t)G(\tau,s),\label{e2.11}\end{equation} where
\begin{equation}e(t)=\frac{(b+at)[d+c(1-t)]}{(b+a)(c+d)}.
\label{e2.12}
\end{equation}
It follows from \eqref{e2.8} that some special Green function of
different
  homogeneous boundary-value problems corresponding to \eqref{e2.9}
are
\begin{equation}
G_1(t,s)=\begin{cases}
 s(1-t),\ s< t;\\
t(1-s),\ t\leq s, \end{cases}\quad (b=0,\; d=0)  \label{e2.13}
\end{equation}
\begin{equation}
G_2(t,s)=\frac{1}{c+d}\begin{cases}
s[d+c(1-t)],\ s< t;\\
t[d+c(1-s)],\ t\leq s,  \end{cases}\quad  (b=0,\ d> 0)
 \label{e2.14}
\end{equation}
\begin{equation}
 G_3(t,s)=\frac{1}{a+b} \begin{cases}
(b+as)(1-t),\ s< t;\\
(b+at)(1-s),\ t\leq s. \end{cases} \quad ( b> 0,\ d= 0)
\label{e2.15}
\end{equation}
Let $E=\{u\in C^{4n-2}[0,1]: u(0)=u(1)=0\}$,  and define the
norm $\|u\|=\max\{\|u\|_0, \|u\|_{4n-2}\}$,
for all  $u \in E$, where
$$
\|u\|_0=\sup_{0\leq t\leq1}|u(t)|,\quad
\|u\|_{4n-2}=\sup_{0\leq t\leq1}|u^{(4n-2)}(t)|,\quad \forall u\in E.
$$
Then $(E,\|\cdot\|)$ is a Banach space.
Define
\begin{equation}
\begin{aligned}
P=\big\{&u\in E :R_1(u)= R_2(u)=0,\; u(t)\geq 0,\;
u^{(4n-2)}(t)\leq e(t)u^{(4n-2)}(s)\leq 0,\\
 & u(t)\geq -kt(1-t)u^{(4n-2)}(s), \forall t, s\in [0,1]
\big\}.
\end{aligned}\label{e2.16}
\end{equation}
 where
$e(t)$ is given by \eqref{e2.12}, $R_1(u)$ and $R_2(u)$ are
defined by \eqref{e1.1}, and
\begin{equation}
\begin{aligned}
k=&\Big(2ac+5bc+5ad)(15abcd+15b^2cd+15abd^2+10b^2c^2+5abc^2\\
& +5a^2cd+a^2c^2+10a^2d^2\Big)/\big(1800(a+b)(c+d))\\
&\times \Big(\frac{5abc^2+10b^2c^2+10abcd+a^2c^2+10a^2d^2+5a^2cd}{30}
\Big)^{2n-3}\frac{1}{\Delta^{4n-4}}.
\end{aligned}\label{e2.17}
\end{equation}
 It is easy to see that $P$ is a
cone of $E$. From
\begin{equation}
\begin{aligned}
u(t)
&=\int_0^1\dots \int_0^1
G_1(t,s_{2n-1})G(s_{2n-1},s_{2n-2})\dots
G(s_2,s_1)(-u^{(4n-2)}(s_1))\\
&\quad\times ds_1 \dots ds_{2n-1}\\
&\leq\int_0^1G_1(t,s_{2n-1})ds_{2n-1}\int_0^1\frac{(b+a s_{2n-2})[d+c(1-s_{2n-2})]}
{\Delta}ds_{2n-2}\\
&\quad\times \dots \int_0^1\frac{(b+a s_1)[d+c(1-s_1)]}{\Delta}ds_1\cdot \|u\|_{4n-2}\\
&=\frac{l^{2n-2}}{2}t(1-t)\|u\|_{4n-2},
\end{aligned}\label{e2.18}
\end{equation}
where
\begin{equation}
l=\frac{ac+3ad+3bc+6bd}{6\Delta}, \label{e2.19}
 \end{equation}
for fixed $u \in P$, we have
\begin{equation}
kt(1-t)\|u\|_{4n-2}\leq u(t)\leq \frac{l^{2n-2}}{2}
t(1-t)\|u\|_{4n-2}. \label{e2.20}
\end{equation}
 Moreover, for $u \in P$, $t\in J_0=[\tau,\gamma]$, $0<\tau< \gamma< 1$,
 we get $\tau(1-\gamma)\leq t(1-t)\leq 1/4$,
$(t,s)\in J_0\times J_0$. The inequality \eqref{e2.20} together
with \eqref{e2.16} yields
\begin{equation}
k\tau (1-\gamma)\|u\|_{4n-2}\leq \|u\|_0 \leq
\frac{l^{2n-2}}{8}\|u\|_{4n-2},\label{e2.21}
\end{equation}
where $k$ and $l$ are defined by \eqref{e2.17} and \eqref{e2.19},
respectively.

Also, for  $e(t), l, k$ corresponding to different
settings of boundary-value problem \eqref{e1.1}, we have:
(1) For $b=d=0$,
\begin{equation}
e_1(t)=t(1-t),\quad l_1=\frac{1}{6},\quad
k_1=\frac{1}{30^{2n-1}}.\label{e2.22}
\end{equation}
(2) For $b=0, d> 0$,
\begin{equation}
\begin{gathered}
e_2(t)=\frac{t[d+c(1-t)]}{c+d},\quad l_2=\frac{c+3d}{6(c+d)},\\
 k_2=\frac{(2c+5d)(5a^2cd+a^2c^2+10a^2d^2)}{1800(c+d)}
 \Big(\frac{a^2c^2+10a^2d^2+5a^2cd}{30}\Big)^{2n-3}\\
\times \frac{1}{(ac+ad)^{4n-4}}.
\end{gathered} \label{e2.23}
\end{equation}
(3) For $b> 0, d=0$,
\begin{equation}
\begin{gathered}
e_3(t)=\frac{(b+at)(1-t)}{b+a},\quad  l_3=\frac{a+3b}{6(a+b)},\\
k_3=\frac{(2a+5b)(10b^2c^2+5abc^2+a^2c^2)}{1800(a+b)}
\Big(\frac{5abc^2+10b^2c^2+a^2c^2}{30}\Big)^{2n-3}\\
 \times \frac {1}{(ac+bc)^{4n-4}}.
\end{gathered}\label{e2.24}
\end{equation}
In the following,  we give the proof of Theorems \ref{thm2.1},  \ref{thm2.2},
  and  \ref{thm2.3}.

\begin{proof}[Proof of Theorem \ref{thm2.1}]
\textbf{Sufficiency.}  In this theorem,  the cone $P$ is
\begin{equation}
\begin{aligned}
P_1=\big\{&u\in E :R_1(u)= R_2(u)=0,\; u(t)\geq 0,\;
u^{(4n-2)}(t)\leq e_1(t)u^{(4n-2)}(s)\leq 0,\\
 & u(t)\geq -k_1t(1-t)u^{(4n-2)}(s), \forall t, s\in [0,1]
\big\}.
\end{aligned}\label{e2.25}
\end{equation}
where $e_1(t), k_1$ are given by \eqref{e2.23},
$R_1(u)=u^{(2k)}(0)$, $R_2(u)=u^{(2k)}(1)$, $k=1, 2, \dots,2n-1$.
By \eqref{e2.21} and \eqref{e2.22}, we get
\begin{equation}
\|u\|=\|u\|_{4n-2},\quad  \forall u \in P_{1}. \label{e2.26}
\end{equation}
Furthermore, from \eqref{e2.16},\eqref{e2.20} and \eqref{e2.26},
we have
\begin{equation}
\begin{gathered}
\frac{1}{30^{2n-1}}t(1-t)\|u\|\leq u(t)\leq  \frac{1}{2\times
6^{2n-2}}t(1-t)\|u\|,\\
 t(1-t)\|u\| \leq -u^{(4n-2)}(t)\leq \|u\|.\label{e2.27}
\end{gathered}
\end{equation}
 Define an operator
$A$  on $P_1\setminus \{0\}$ by
\begin{equation}(Au)(t)=\int_0^1
h_1(t,s)f(s,u(s),u^{(4n-2)}(s))ds,\quad \forall\ u \in
P_1\setminus \{0\}, \label{e2.28}
\end{equation}
where
$$
h_1(t,s)= \int_0^1\dots \int_0^1
G_1(t,s_{2n-1})G_1(s_{2n-1}),s_{2n-2})\dots G_1(s_1,s)ds_1 \dots
ds_{2n-1};
$$
and $ G_1(t,s)$ is defined by \eqref{e2.13}. Clearly,
$$
G_1(t,s)\leq G_1(s,s),\quad G_1(t,s)\leq G_1(t,t),\quad
G_1(t,s)\geq t(1-t)s(1-s),
$$
for all $t,\ s\in [0,1]$.
 Then
\begin{equation}
\begin{aligned}
& h_1(t,s) \\
&\leq \int_0^1t(1-t)ds_{2n-1}\int_0^1s_{2n-1}(1-s_{2n-1})ds_{2n-2}\dots \int_0^1s_2(1-s_2)s(1-s)ds_1\\
&\leq t(1-t)\int_0^1 \dots \int_0^1 s_{2n-1}(1-s_{2n-1})\dots s_2(1-s_2)s(1-s)ds_{2n-1} \dots ds_2\\
&\leq t(1-t)s(1-s)\\
&\leq s(1-s),\quad \forall\ t,s\in [0,1].
\end{aligned}\label{e2.29}
\end{equation}
 Now we claim that $Au$ is well defined on $P_1\setminus \{0\}$.
First, for $\forall u\in P_1\setminus \{0\},$\ we can see that
$\|u\|\neq 0.$\ At the same time, notice that $G_1(t,s)\leq
G_1(s,s),\ \forall\ t,\ s\in [0,1]$. This together with \eqref{e2.1}
yields that $ \int_0^1 G_1(s_1,s)f(s,u(s),u^{(4n-2)}(s))ds$ is
convergent.

 In fact, for $\forall\ u\in P_1\setminus \{0\}$, choose positive numbers
 $c_1\leq \min\{N_1,\ \frac{\|u\|}{30^{2n-1}M_1}\}$ and
 $c_2\geq \max \{M_2,\ \frac{\|u\|}{N_2}\}$.\ By (1.4) and \eqref{e2.27}, we obtain
\begin{equation}
\begin{aligned} &\int_0^1 G_1(s_1,s)f(s,u(s),u^{(4n-2)}(s))ds
\leq \int_0^1 s(1-s)f(s,u(_s),u^{(4n-2)}(s))ds\\
&\leq \int_0^1 s(1-s)f(s,c_1\frac{u(s)}{c_1s(1-s)}s(1-s),(-1)c_2
 \frac{-u^{(4n-2)}(s)}{c_2})ds\\
&\leq \int_0^1 s(1-s)c_1^\lambda (\frac{u(s)}{c_1s(1-s)})^\mu c_2^\beta 
 (\frac{-u^{(4n-2)}(s)}{c_2})^\alpha f(s,s(1-s),-1)ds\\
&\leq \int_0^1 s(1-s)c_1^{\lambda-\mu} (\frac{\|u\|}{2\times6^{2n-2}})^\mu 
 c_2^\beta (\frac{\|u\|}{c_2})^\alpha f(s,s(1-s),-1)ds\\
&\leq (\frac{1}{2\times6^{2n-2}})^\mu c_1^{\lambda-\mu}c_2^{\beta-\alpha}
 \|u\|^{\mu+\alpha}\int_0^1 s(1-s)f(s,s(1-s),-1)ds\\
&= c_3\|u\|^{\mu+\alpha}\int_0^1 s(1-s)f(s,s(1-s),-1)ds <\infty,
\end{aligned}\label{e2.30}
\end{equation}
 where \begin{equation}c_3=(\frac{1}{2\times6^{2n-2}})^\mu
c_1^{\lambda-\mu}c_2^{\beta-\alpha}. \label{e2.31}
\end{equation}
Also, by \eqref{e2.29} and the process similar to the proof of
\eqref{e2.30}, for $\forall\  u\in P_1\setminus \{0\}$, there
exist positive constants $c_1$ and $c_2$ such that
\begin{equation}
\begin{aligned}
 Au(t)&=\int_0^1 h_1(t,s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq \int_0^1s(1-s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq c_3\|u\|^{\mu+\alpha}\int_0^1 s(1-s)f(s,s(1-s),-1)ds
<\infty, \end{aligned}\label{e2.32}
\end{equation}
where $c_3$ is the same as \eqref{e2.31}.
This together with \eqref{e2.1} yields that $A$ is
well defined on $P_1\setminus \{0\}$.
Obviously, if \eqref{e2.1}-\eqref{e2.3} hold, then
  \eqref{e1.1}$(b=d=0)$ has a positive solution  $u$ if and only if $A$
has a fixed
   point  in $P_1\setminus \{0\}$. So we need to prove only  that
   $A$ has a fixed point  in $P_1\setminus \{0\}$.

 Now we show that $A:\ P_1\setminus \{0\}\to P_1$ is completely continuous.
Firstly, we show that $A(P_1\setminus\{0\})\subset P_1$. To see this,
 for all $u \in  P_1\setminus \{0\}$, notice that
\begin{align*}
(Au)^{(4n-2)}(t)&=-\int_0^1G_1(t,\tau)
f(\tau,u(\tau),u^{(4n-2)}(\tau))d\tau\\
&\leq -t(1-t)\int_0^1G_1(s,\tau)f(\tau,u(\tau),u^{(4n-2)}(\tau))d\tau\\
&=t(1-t) (Au)^{(4n-2)}(s)\leq 0,\quad \forall\ t,\ s \in [0,1],
\end{align*}
and
\begin{align*}
(Au)(t)&=\int_0^1 \dots \int_0^1
G_1(t,s_{2n-1})G_1(s_{2n-1},s_{2n-2})
\dots G_1(s_2,s_1)\\
&\quad\times (-(Au)^{(4n-2)}(s_1))ds_1ds_2\dots ds_{2n-1}\\
&\geq -t(1-t)\int_0^1 \dots \int_0^1
s_{2n-1}^2(1-s_{2n-1}))^2\dots s_2^2(1-s_2)^2\\
&\quad\times s_1^2(1-s_1)^2 (Au)^{(4n-2)}(s)ds_1 \dots ds_{2n-1}\\
&\geq -\frac{t(1-t)}{30^{2n-1}}(Au)^{(4n-2)}(s).
\end{align*}
Then we have $A(P_1\setminus\{0\})\subset P_1$.

Secondly, we show that $A$ is bounded. In fact, let $V\subset
P_1\setminus \{0\}$ be  a bounded set.  There exists a positive
constant $L$ satisfying $\|u\|\leq L$,  for all $u \in V$.
Choose $ c_1\leq\min\{N_1, \frac{L}{30^{2n-1}M_1}\}$
and $ c_2\geq \max\{M_2, \frac{L}{N_2}\}$. By
\eqref{e2.1},  \eqref{e2.30}, and \eqref{e2.31}, we get
\begin{align*}
|(Au)^{(4n-2)}(t)|&=\int_0^1G_1(t,s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq \int_0^1s(1-s)f(s,c_1\frac{u(s)}{c_1s(1-s)}s(1-s),(-1)c_2
\frac{-u^{(4n-2)}(s)}{c_2})ds\\
&\leq c_3 L^{\mu+\alpha}
\int_0^1s(1-s)f(s,s(1-s),-1)ds\\
&<+\infty,\quad \forall t\in [0,1], \quad\forall u\in
P_1\setminus \{0\}.
\end{align*}
Therefore, this together with
\eqref{e2.26} implies
\begin{equation}
\|Au\|\leq c_3 L^{\mu+\alpha} \int_0^1 s(1-s)f(s,s(1-s),-1)ds <+\infty,
\label{e2.33}
\end{equation}
where $c_3$ is defined by  \eqref{e2.31}. Namely, $AV$ is uniformly
bounded.

Thirdly, by \eqref{e2.33} and  the Ascoli-Arzela theorem, we
need to show only that $AV$ is equicontinuous on $[0,1]$. Therefore,
we need to prove only that ($Au)^{(4n-2)}(t) \to 0$ as $t\to 0^+$
and $t\to 1^-$ uniformly with respect to $u\in V$ and $AV$ are
equicontinuous on any closed  subinterval of (0,1).
In fact, notice that
\begin{align*}
& -(Au)^{(4n-2)}(t)\\
&=\int_0^1 G_1(t,s) f(s,u(s),u^{(4n-2)}(s))ds\\
&=(1-t)\int_0^t sf(s,u(s),u^{(4n-2)}(s))ds
 +t\int_t^1(1-s)f(s,u(s),u^{(4n-2)}(s))ds,
\end{align*}
Then this together with \eqref{e2.1} and \eqref{e2.2}  guarantees
$(Au)^{(4n-2)}(t)\to 0$, as $t\to 0^+$ or $t\to 1^-$,  uniformly
with respect to $u\in V$.

Now, we are in position to show that for $\forall\ a\in (0,\frac{1}{2}),\ AV$
are equicontinuous on $[a,1-a]$.
For all $t_1,\ t_2 \in [a,1-a]$, $t_1< t_2$, for all
$u \in V$, by  \eqref{e2.31}, we get
\begin{align*}
& |(Au)^{(4n-2)}(t_2)-(Au)^{(4n-2)}(t_1)|\\
& =\Big|\int_0^{t_1} (t_1-t_2)sf(s,u(s),u^{(4n-2)}(s))ds
+\int_{t_1}^{t_2} (1-t_2) sf(s,u(s),u^{(4n-2)}(s))ds\\
&\quad +\int_{t_2}^1 (t_2-t_1) (1-s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\quad -\int_{t_1}^{t_2} t_1(1-s) f(s,u(s),u^{(4n-2)}(s))ds\Big|\\
& \leq c_3 L^{\mu+\alpha}[(t_2-t_1)\int_0^{1-(t_2-t_1)}
sf(s,s(1-s),-1)ds \\
&\quad +(t_2-t_1)\int_{t_2-t_1}^1(1-s) f(s,s(1-s),-1)ds
+2\int_{t_1}^{t_2} s(1-s)f(s,s(1-s),-1)ds].
\end{align*} %\label{e2.34}
Also, as  $|t_1-t_2|\to 0$,  \eqref{e2.1}-\eqref{e2.3}
imply
\begin{gather*}
(t_2-t_1)\int_0^{1-(t_2-t_1)} sf(s,s(1-s),-1)ds\to 0,\\
(t_2-t_1)\int_{t_2-t_1}^1(1-s)f(s,s(1-s),-1)ds\to 0,\\
\int_{t_1}^{t_2} s(1-s)f(s,s(1-s),-1)ds\to 0.
\end{gather*}
This guarantees
$|(Au)^{(4n-2)}(t_2)-(Au)^{(4n-2)}(t_1)|\to 0 (|t_1-t_2|\to 0)$.

Similar to the  above proof, we can get $ (Au)(t)\to 0$, as
$t\to 0^+$ or $t\to 1^-$  uniformly with respect to $u\in V$ and
 for all $t_1, t_2 \in [a,1-a]$, $t_1< t_2$, for all $u \in V$,
 we have
 $|(Au)(t_2)-(Au)(t_1)|\to 0 (|t_1-t_2|\to 0)$.
Therefore, $AV$ is relatively compact.

Finally, it remains to show $A$ is continuous. Suppose
$u_n, u_0\in P$, and $\|u_n-u_0\|\to 0\ (n\to \infty)$. Then $\{u_n\}$
is a bounded set and
$$
\|u_n-u_0\|_0\to 0,\quad  \|u_n-u_0\|_{4n-2}\to 0\quad
(n\to \infty).
$$
Let $M=\sup\{\|u_n\|,\ n=0,\ 1,\ 2,\ \dots\}$. Then we may
still choose  positive constants $ c_1\leq\min\{N_1,
\frac{M}{30^{2n-1}M_1}\}$ and $ c_2\geq \max\{M_2,
\frac{M}{N_2}\}$. Similar to the proof of \eqref{e2.30}, we get
\begin{equation}
f(t,u_n(t),u_n^{(4n-2)}(t))\leq c_3 M^{\mu+\alpha}
f(t,t(1-t),-1),\quad t\in (0,1), \label{e2.35}
\end{equation}
 \begin{equation}
\begin{aligned}
&|(Au_n)^{(4n-2)}(t)-(Au_0)^{(4n-2)}(t)|\\
& \leq \int_0^1 s(1-s) |f(s,u_n (s),u_n^{(4n-2)}(s))
-f(s,u_0(s),{u_0}^{(4n-2)}(s))|ds,
\end{aligned}\label{e2.36}
\end{equation}
and
\begin{align*}
&|(Au_n)(t)-(Au_0)(t)|\\
&\leq \int_0^1 s(1-s) |f(s,u_n (s),u_n^{(4n-2)}(s))
 -f(s,u_0(s),{u_0}^{(4n-2)}(s))|ds.
\end{align*}
The above inequality,  \eqref{e2.1}, \eqref{e2.35}, \eqref{e2.36},
 the Lebesgue dominated convergence theorem, and Ascoli-Arzela theorem
guarantee that
$$\|Au_n-Au_0\|\to 0\quad (n\to \infty),
$$
that is, $A$ is continuous.
Summing up,  $A: P_1\setminus {0} \to P_1$ is completely
continuous.

For $0< r<1 < R$, let
$$
P_{1,r}=\big\{u\in P_1: \|u\|\leq r\big\},\quad
P_{1,R}=\big\{u\in P_1:\|u\|\leq R\big\}.
$$
Choose $r$ such that
\begin{align*}
 0<r\leq\min\big\{&(2^{-(2+(10n-5)\mu+4\beta)}
 3^\beta\int_{\frac{1}{4}}^{\frac{3}{4}}s(1-s)
 f(s,s(1-s),-1)ds)^{\frac{1}{1-(\mu+\beta)}}, \\
&2\times 6^{2n-2}N_1, N_2 \big\}.
\end{align*}
 Then for $u\in \partial P_{1,r}$,
we have
\begin{gather*}
\frac{r}{30^{2n-1}}t(1-t)\leq u(t)\leq \frac{r}{2\times
6^{2n-2}}t(1-t)\leq N_1t(1-t),\\
 \frac{3r}{16}\leq rt(1-t)\leq
-u^{(4n-2)}(t)\leq r\leq N_2,\quad\forall\ t\in [\frac{1}{4},
\frac{3}{4}].
\end{gather*}
 By the properties of $G_1(t,s)$,  \eqref{e2.1}, and
(1.4), we get
\begin{align*}
-(Au)^{(4n-2)}(t)&= \int_0^1G_1(t,s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\geq\frac{1}{4}\int_{\frac{1}{4}}^{\frac{3}{4}}s(1-s)f(s,
\frac{u(s)}{s(1-s)}s(1-s), (-1)(-u^{(4n-2)}(s)))ds\\
&\geq\frac{1}{4}\int_{\frac{1}{4}}^{\frac{3}{4}}s(1-s)
(\frac{u(s)}{s(1-s)})^\mu(-u^{(4n-2)}(s))^\beta
f(s,s(1-s),-1)ds\\
&\geq\frac{1}{4}\int_{\frac{1}{4}}^{\frac{3}{4}}s(1-s)
(\frac{r}{30^{2n-1}})^\mu(\frac{3r}{16})^\beta f(s,s(1-s),-1)ds\\
&\geq\frac{1}{2^2}\int_{\frac{1}{4}}^{\frac{3}{4}}(\frac{r}
{2^{(10n-5)}})^\mu(\frac{3r}{2^4})^\beta
s(1-s)f(s,s(1-s),-1)ds\\
&= 2^{-(2+(10n-5)\mu+4\beta)} 3^\beta
r^{\mu+\beta}\int_{\frac{1}{4}}^{\frac{3}{4}}s(1-s)f(s,s(1-s),-1)ds\\
&\geq r=\|u\|_{4n-2}=\|u\|,\quad \forall\ u\in \partial P_{1,r}.
\end{align*}
This guarantees
\begin{equation}
\|Au\|\geq \|u\|,\quad \forall u\in \partial P_{1,r}.\label{e2.37}
\end{equation}
On the other hand,  Choose $R$ such that
\begin{align*}
R\geq \max \big\{&30^{2n-1}M_1, M_2N_2, \\
&[(2\times
6^{2n-2})^{-\mu}N_2^{\alpha-\beta}\int_0^1s(1-s)f(s,s(1-s),-1)ds)]
^{\frac{1}{1-(\mu+\beta)}}\big\}.
\end{align*}
and let $ c=\frac{N_2}{R}$. Then for $u\in \partial P_{1,R}$, we have
\begin{gather*}
M_1t(1-t)\leq \frac{R}{30^{2n-1}}t(1-t)\leq u(t)
\leq \frac{R}{2\times 6^{2n-2}}t(1-t),\\
-cu^{(4n-2)}(t)\leq c\|u\|=cR=N_2.
\end{gather*}
Therefore,
\begin{align*}
-(Au)^{(4n-2)}(t)
&= \int_0^1G_1(t,s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq \int_0^1s(1-s)f(s,\frac{u(s)}{s(1-s)}s(1-s),(-1)\frac{1}{c}(-cu^{(4n-2)}(s))ds\\
&\leq \int_0^1s(1-s)(\frac{u(s)}{s(1-s)})^\mu(\frac{1}{c})^\beta(-cu^{(4n-2)}(s))
 ^\alpha f(s,s(1-s),-1)ds\\
&\leq \int_0^1s(1-s)(\frac{R}{2\times 6^{2n-2}})^\mu c^{\alpha-\beta}R^\alpha 
f(s,s(1-s),-1)ds\\
&= \int_0^1s(1-s)(\frac{R}{2\times 6^{2n-2}})^\mu(\frac{N_2}{R})^{\alpha-\beta}
R^\alpha f(s,s(1-s),-1)ds\\
&= (2\times 6^{2n-2})^{-\mu}N_2^{\alpha-\beta}R^{\mu+\beta}\int_0^1s(1-s)f(s,s(1-s),
-1)ds\\
&\leq  R=\|u\|_{4n-2}=\|u\|,\quad \forall u\in\partial
P_{1,R},
\end{align*}
This implies
 \begin{equation}\|Au\|\leq \|u\|, \quad \forall u\in
\partial P_{1,R}.\label{e2.38}
\end{equation}
By the complete continuity of $A$,  \eqref{e2.37}, and
\eqref{e2.38}, and  Lemma \ref{lem1.1}, we obtain that  $A$  has a fixed
point $u_*(t)$  in  $\overline{P_{1,R}}\setminus P_{1,r}$.
Consequently, \eqref{e1.1}
 has a $C^{4n-2}[0,1]$ positive solution $u_*(t)$ in
 $\overline{P_{1,R}}\setminus P_{1,r}$. \smallskip


\noindent\textbf{Necessity.} Let $u(t)$ be a $C^{(4n-2)}[0,1]$  positive
solution of \eqref{e1.1}. It follows from the boundary value conditions
that there exists $t_0\in (0,1)$ such that $u^{(4n-1)}(t_0)=0$.
Obviously, by virtue of $u^{(4n)}(t)\geq 0,\ t\in (0,1)$, we get
$$
u^{(4n-1)}(t)\leq 0,\quad t \in (0,t_0);\quad  u^{(4n-1)}(t)\geq 0,\quad
t\in (t_0,1).
$$
  Hence, $ u^{(4n-2)}(t)\leq 0,\ t\in [0,1]$. Similarly,
when $t\in [0,1]$, by induction we know
$$
u^{(4n-4)}(t)\geq 0,\quad u^{(4n-6)}(t)\leq 0,\quad
 \dots,\quad u^{(4)}(t)\geq 0,\quad  u''(t)\leq 0.
$$
Therefore, there exists $0<m_1 < 1 <m_2$
 such that for all $t\in[0, 1]$
\begin{equation}
m_1 t(1-t)\leq u(t)\leq m_2 t(1-t).   \label{e2.39}
\end{equation}
Choose positive numbers $ c_1\leq\min\{N_1, \frac{1}{M_1m_2}\}$ and
$ c_2\geq \max\{\frac{1}{N_2}, M_2\|u\|\}$. Then we can get
\begin{equation}
\begin{aligned} f(t,t(1-t),
-1)&=f(t,c_1\frac{t(1-t)}{c_1u(t)}u(t),\frac{1}{c_2}\frac{c_2}{-u^{(4n-2)}(t)}
 u^{(4n-2)}(t))\\
&\leq
c_1^\lambda(\frac{t(1-t)}{c_1u(t)})^\mu(\frac{1}{c_2})^\alpha(\frac{c_2}{-u^{(4n-2)}
(t)})^\beta
f(t,u(t),u^{(4n-2)}(t))\\
&\leq
c_1^\lambda(\frac{1}{c_1m_1})^\mu(\frac{1}{c_2})^\alpha(-\frac{c_2}{u^{(4n-2)}(t)})^\beta
f(t,u(t),u^{(4n-2)}(t))\\
&=
c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}(-u^{(4n-2)}(t))^{-\beta}f(t,u(t),u^{(4n-2)}(t)).
\end{aligned}\label{e2.40}
\end{equation}
 By  \eqref{e2.40}, and  $- u^{(4n-2)}(t)$ being  nondecreasing on
$(0,t_0)$, integrate the first equality of  \eqref{e1.1} from $ t_0$
to $t$ to obtain
\begin{equation}- u^ {(4n-1)}(t)=\int_t^
{t_0}f(s,u(s),u^{(4n-2)}(s))ds,\ t\in (0,t_0),\label{e2.41}
\end{equation}
and
\begin{equation}
\begin{aligned} 0 &< \int_0^{t_0}t f(t,t(1-t),-1)dt=\int_0^{t_0}dt
\int_t^{t_0}f(s,s(1-s),-1)ds\\
&\leq
\int_0^{t_0}dt\int_t^{t_0}c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
(-u^{(4n-2)}(s))^{-\beta}f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq
c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}\int_0^{t_0}(-u^{(4n-2)}(t)
 )^{-\beta}dt\int_t^{t_0} f(s,u(s),u^{(4n-2)}(s))ds\\
&= c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}\int_0^{t_0}
 \frac{-u^{(4n-1)}(t)}{(-u^{(4n-2)}(t))^ {\beta}}dt\\
&= c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}(1-\beta)^{-1}
(-u^{(4n-2)}(t_0))^{1-\beta} <\infty;
\end{aligned}\label{e2.42}
\end{equation}
Similar to  \eqref{e2.41} and \eqref{e2.42}, we can also prove that
\begin{equation}
u^ {(4n-1)}(t)=\int_{t_0}^t f(s,u(s),u^{(4n-2)}(s))ds,\ t\in
(t_0,1), \label{e2.43}
\end{equation}
\begin{equation}
\begin{aligned}
0 &<\int_{t_0}^1 (1-t) f(t,t(1-t),-1)dt=\int_{t_0}^1 dt\int_{t_0}^t
f(s,s(1-s),-1)ds\\
&\leq c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu} (1-\beta)^{-1}
(-u^{(4n-2)}(t_0))^{1-\beta} <\infty.
\end{aligned}\label{e2.44}
\end{equation}
Consequently, inequalities  \eqref{e2.42} and  \eqref{e2.44}
yield  \eqref{e2.1}.
For $t\in (0,t_0)$, by   \eqref{e2.40}, and integrating
\eqref{e2.41}, we have
\begin{equation}
\begin{aligned}
 &t\int_t^{t_0}f(s,s(1-s),-1)ds \leq \int_0^t d\xi \int_\xi^ {t_0}f(s,s(1-s),-1)ds\\
 &\leq \int_0^t d\xi \int_\xi^ {t_0}c_1^{\lambda-\mu}c_2^{\beta-\alpha}
  m_1^{-\mu}(-u^{(4n-2)}(s))^{-\beta}f(s,u(s),u^{(4n-2)}(s))ds\\
 &=c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu} (1-\beta)^{-1} (-u^{(4n-2)}(t)).
\end{aligned}\label{e2.45}
\end{equation}
Noticing $u^{(4n-2)}(0)=0$ and letting $t \to 0^+$ in
\eqref{e2.45}, we obtain
 $$
\lim_{t \to {0+}}t\int_{t}^{t_0}f(s,s(1-s),-1)ds=0.
$$
This implies  \eqref{e2.2}.
 For $t\in (t_0 ,1)$, noticing  \eqref{e2.40} and integrating
 \eqref{e2.43}, we get
 \begin{equation}
\begin{aligned} &(1- t)\int_{t_0}^t f(s,s(1-s),-1)ds\\
& \leq \int_t^1 d\xi \int_{t_0}^\xi f(s,s(1-s),-1)ds\\
 &\leq \int_t^1 d\xi \int_ {t_0}^\xi c_1^{\lambda-\mu}
 c_2^{\beta-\alpha}m_1^{-\mu}(-u^{(4n-2)}(s))^{-\beta}f(s,u(s),u^{(4n-2)}(s))ds\\
& =c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
  (1-\beta)^{-1}(-u^{(4n-2)}(t))
\end{aligned}\label{e2.46}
\end{equation}
By $u^{(4n-2)}(1)=0$,  and letting  $t \to 1^-$ in  \eqref{e2.46}, we have
 $$
 \lim_{t \to {1^-}}(1-t)\int_{t_0}^t f(s,s(1-s),-1)ds=0.
 $$
This yields  \eqref{e2.3}.
Summing up, the necessity follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.2}]
{\bf Sufficiency.} In this theorem,  the cone $P$ is
\begin{align*}
P_2=&\big\{u\in E : R_1(u)=R_2(u)=0,\; u(t)\geq 0,\;
u^{(4n-2)}(t)\leq e_2(t)u^{4n-2}(s)\leq 0,\\
&u(t)\geq -k_2 t(1-t)u^{(4n-2)}(s), \forall t, s\in [0,\ 1]\big\},
\end{align*}
where $e_2(t),k_2$ are given by  \eqref{e2.24},  and $R_1(u)=u^{(2k)}(0)$,
$R_2(u)=cu^{(2k)}(1)+du^{(2k+1)}(1)$, $k= 1, 2, \dots, 2n-1$.
By  \eqref{e2.16},  \eqref{e2.20}, \eqref{e2.21}, and
\eqref{e2.23}, we have
 \begin{gather}
\|u\|=\|u\|_{4n-2},\quad  \forall u \in P_2. \label{e2.47} \\
 k_2t(1-t)\|u\|\leq u(t)\leq  \frac{l_2^{2n-2}}{2}  t(1-t)\|u\|,
\quad e_2(t) \|u\|\leq -u^{(4n-2)}(t)\leq \|u\|. \label{e2.48}
\end{gather}
where $l_2$ is given by  \eqref{e2.24}. Clearly, for
$t\in [\frac{1}{4},\frac{3}{4}]$, we have
\begin{equation}
e_2(t)\geq \frac{4d+c}{16(c+d)}. \label{e2.49}
\end{equation}
  Suppose  \eqref{e2.4} and  \eqref{e2.5} hold. Then \eqref{e1.1}
    has a $C^{4n-2}[0,1]\cap C^{4n-1}(0,1] $
   positive solution $u$ if and only if
$u$ is a positive solution of the following integral equation
\begin{equation}
u(t)=(A u)(t)=\int_0^1 h_2(t,s)f(s,u(s),u^{(4n-2)}(s))ds,\quad
\forall u \in P_2\setminus \{0\}, \label{e2.50}
\end{equation}
where
$$
h_2(t,s)= \int_0^1\dots \int_0^1
G_1(t,s_{2n-1})G_2(s_{2n-1},s_{2n-2})\dots G_2(s_1,s)ds_1 \dots
ds_{2n-1};
$$
and $G_1(t,s)$ and $G_2(t,s)$ are given  by
\eqref{e2.13} and  \eqref{e2.14}, respectively. For rall
$u\in P_2\setminus \{0\}$, it follows from \eqref{e2.11},
\eqref{e2.14}, and \eqref{e2.24} that
\begin{align*}
(A u)^{(4n-2)}(t)
&=-\int_0^1G_2(t,\tau)
f(\tau,u(\tau),u^{(4n-2)}(\tau))d\tau\\
&\leq -\frac{t[d+c(1-t)]}{c+d}\int_0^1G_2(s,\tau)
f(\tau,u(\tau),u^{(4n-2)}(\tau))d\tau\\
&=e_2(t)(Au)^{(4n-2)}(s)\leq 0,\quad \forall\ t,\ s \in
[0,1].
\end{align*}
Obviously, this together with \eqref{e2.10} implies
\begin{align*}
&(A u)(t)\\
 &=\int_0^1 \dots \int_0^1
G_1(t,s_{2n-1})G_2(s_{2n-1},s_{2n-2})
\dots G_2(s_2,s_1)(-(Au)^{(4n-2)}(s_1)) \\
&\quad\times  ds_1ds_2\dots ds_{2n-1}\\
&\geq \int_0^1t(1-t)s_{2n-1}(1-s_{2n-1})ds_{2n-1}\\
&\quad\times \int_0^1 \frac{[ac (1-s_{2n-1})s_{2n-1}
+a d s_{2n-1}][ac (1-s_{2n-2})s_{2n-2}
+a d s_{2n- 2}]}{a^2(c+d)^2}ds_{2n-2}\\
&\quad\times \dots \int_0^1 \frac{[ac (1-s_3)s_3+ads_3][ac
(1-s_2)s_2+ads_2]}{a^2(c+d)^2}ds_2 \\
&\quad \times  \int_0^1 \frac{[ac
(1-s_2)s_2+ads_2][ac (1-s_1)s_1+ad s_1]}{a^2(c+d)^2}
\cdot (Au)^{(4n-2)}(s_1)ds_1\\
&\geq -t(1-t)\frac{2ac+5ad}{60}\cdot
(\frac{a^2c^2+10a^2d^2+5a^2cd}{30})^{2n-3}
\cdot \frac{(5a^2cd+a^2c^2+10a^2d^2)}{30a(c+d)}\\
&\quad\times \frac {1}{(ac+ad)^{4n-4}} (A u)^{(4n-2)}(s),\\
&= - k_2 t(1-t) (A u)^{(4n-2)}(s).
\end{align*}
Thus,  $A(P_2\setminus \{0\})\subset P_2$.

Similar to the proof of  Theorem \ref{thm2.1}, we can show that
$A:\quad P_2\setminus \{0\}\to P_2$ is completely continuous.
For $0< r<1 < R$, let
$$
P_{2,r}=\big\{u\in P_2:\|u\|\leq r\big\},\quad
P_{2,R}=\big\{u\in P_2:\|u\|\leq R\big\},
$$
On the one hand, choose
$$
r\leq\min\{[\frac{(c+4d)^2)}{256(c+d)^2}k_2^\mu
(\frac{(c+4d)}{16(c+d)})^\beta
\int_{\frac{1}{4}}^{\frac{3}{4}}sf(s,s(1-s),-1)ds]^{\frac{1}{1-(\mu+\beta)}},
\frac{2N_1 }{l_2 ^{2n-2} }, N_2 \}.
$$
For $u\in \partial P_{2,r}$, combining  \eqref{e2.48} and
\eqref{e2.49}, then we have
\begin{gather*}
k_2 rt(1-t)\leq u(t)\leq \frac{ l_2 ^{2n-2} r}{2}t(1-t)\leq
N_1t(1-t), \\
\frac{c+4d }{16(c+d)}r \leq e_2(t)r\leq
-u^{(4n-2)}(t)\leq r\leq N_2,\quad\forall\ t\in [\frac{1}{4},
\frac{3}{4}].
\end{gather*}
In addition, by  \eqref{e2.4}, \eqref{e2.50}, (1.4), and the
properties of $G_2(t,s)$, we get
\begin{align*}
-(Au)^{(4n-2)}(t)
&= \int_0^1G_2(t,s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\geq \int_{\frac{1}{4}}^{\frac{3}{4}} \frac{[c(1-s)s+d
s][c(1-t)t+dt]}{(c+d)^2}f(s,
\frac{u(s)}{s(1-s)}s(1-s), (-1)\\
&\quad\times (-u^{(4n-2)}(s)))ds\\
&\geq \frac{(c+4d)^2)}{256(c+d)^2}\int_{\frac{1}{4}}^{\frac{3}{4}}s
(\frac{u(s)}{s(1-s)})^\mu(-u^{(4n-2)}(s))^\beta
f(s,s(1-s),-1)ds\\
&\geq \frac{(c+4d)^2)}{256(c+d)^2}\int_{\frac{1}{4}}^{\frac{3}{4}}
(k_2 r)^\mu(\frac{(c+4d)r}{16(c+d)})^\beta sf(s,s(1-s),-1)ds\\
&= \frac{(c+4d)^2)}{256(c+d)^2}k_2^\mu
(\frac{(c+4d)}{16(c+d)})^\beta
r^{\mu+\beta}\int_{\frac{1}{4}}^{\frac{3}{4}}sf(s,s(1-s),-1)ds\\
&\geq r=\|u\|_{4n-2}=\|u\|,\quad \forall\ u\in \partial P_{2,r}.
\end{align*}
Therefore, by \eqref{e2.47}, and the above inequality, we have
\begin{equation}
\|Au\|\geq \|u\|,\quad \forall u\in \partial P_{2,r}.
\label{e2.51}
\end{equation}
On the other hand, choose
$$
 R\geq \max
\big\{[(\frac{l_2^{2n-2}}{2})^\mu N_2^{\alpha-\beta} \int_0^1s
f(s,s(1-s),-1)ds)]^{\frac{1}{1-(\mu+\beta)}}],M_2N_2,\frac{M_1}{k_2}\big\}.
$$
and Let  $ c=\frac{N_2}{R}$. Then for $u\in \partial P_{2,R}$,
 we obtain
\begin{gather*}
M_1t(1-t)\leq k_2 t(1-t)R \leq u(t)\leq \frac{R l_2^{2n-2}}{2}t(1-t),\\
 -cu^{(4n-2)}(t)\leq c\|u\|=cR=N_2.
\end{gather*}
Consequently,
\begin{align*}
-(Au)^{(4n-2)}(t)
&= \int_0^1G_2(t,s)f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq \int_0^1\frac{s[d+c(1-s)]}{c+d}f(s,\frac{u(s)}{s(1-s)}s(1-s),(-1)
\frac{1}{c}(-cu^{(4n-2)}(s))ds\\
&\leq \int_0^1s(\frac{u(s)}{s(1-s)})^\mu(\frac{1}{c})^\beta
(-cu^{(4n-2)}(s))^\alpha
 f(s,s(1-s),-1)ds\\
&\leq \int_0^1s(\frac{R l_2^{2n-2}}{2})^\mu c^{\alpha-\beta}R^\alpha
 f(s,s(1-s),-1)ds\\
&= (\frac{l_2^{2n-2}}{2})^{\mu} N_2^{\alpha-\beta}R^{\mu+\beta}
\int_0^1s f(s,s(1-s),-1)ds\\
&\leq R=|u|_{4n-2}=\|u\|,\quad \forall u\in\partial
P_{2,R},
\end{align*}
which implies
\begin{equation}
\|Au\|\leq \|u\|, \quad \forall\ u\in \partial P_{2,R}.\label{e2.52}
\end{equation}
 By   the  complete continuity of $A$,  \eqref{e2.51}, and
\eqref{e2.52},  we know that $A$ has a fixed point $u_*(t)$ in
$\overline{P_{2,R}}\setminus P_{2,r}$. Consequently,  \eqref{e1.1}
 has a $C^{4n-2}[0,1] \cap C^{4n-1}(0,1]$ positive solution
${u}_*(t)$  in $\overline{P_{2,R}} \setminus P_{2,r}$.
\smallskip

\noindent\textbf{Necessity.}
 Let $u(t)$ be a  $C^{4n-2}[0,1]\cap
C^{4n-1}(0,1]\cap C^{4n}(0,1)$ positive solution of \eqref{e1.1}.
Then we get
\begin{gather*}
u^{(4n-2)}(t)=-\int_0^1G_2(t,s)f(s,u(s),u^{(4n-2)}(s))ds\leq 0,\\
u^{(4n-1)}(1)=-\frac{c}{d}u^{(4n-2)}(1)\geq  0.
\end{gather*}
Clearly, it follows from  boundary-value conditions that there
exists $t_0\in (0,1]$ such that $u^{(4n-1)}(t_0)=0$.

 The following argument is broken into two  cases:  $t_0<1$ and  $t_0=1$.

\noindent Case (1): Suppose that $t_0 <1$. Then $ u^{(4n-1)}(1)> 0$. Since
$u^{(4n)}(t)\geq 0, t\in (0,1)$, we find
$$
u^{(4n-1)}(t)\leq 0,\quad t \in (0,t_0);\quad
  u^{(4n-1)}(t)\geq 0,\quad t\in (t_0,1),
$$
and hence
$u^{(4n-2)}(t)\leq 0,\ t\in [0,1]$. By the same way, we know
$u''(t)\leq 0$ for $t\in  [0,1]$.  Therefore, this implies
\eqref{e2.39}-\eqref{e2.42}, \eqref{e2.45}, \eqref{e2.46}, and
\begin{align*}
\int_{t_0}^1f(t,t(1-t),-1)dt
&\leq \int_{t_0}^1 c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
(-u^{(4n-2)}(t))^{-\beta} f(t,u(t),u^{(4n-2)}(t))dt\\
&\leq c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
(-u^{(4n-2)}(1))^{-\beta}u^{(4n-1)}(1)<\infty.
\end{align*}
Clearly, it follows from the above inequality and
\eqref{e2.42} that  \eqref{e2.4}  holds. Moreover, by virtue of
\eqref{e2.45}, \eqref{e2.5} is satisfied.

\noindent Case (2). If $t_0=1$. Then $ u^{(4n-1)}(1)= 0, u^{(4n-2)}(1)<
0$, and \eqref{e2.39}- \eqref{e2.41} hold. Also, by \eqref{e2.40},
we have
\begin{align*}
0 &< \int_0^1 t f(t,t(1-t),-1)dt=\int_0^1 dt\int_t^1 f(s,s(1-s),-1)ds\\
&\leq \int_0^1 dt\int_t^1
c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}(-u^{(4n-2)}(s))^{-\beta}
f(s,u(s),u^{(4n-2)}(s))ds.
\end{align*}
Notice that $-u^{(4n-2)}(s)$ is nondecreasing  in  $s$ on $(0,1)$.
Then we have
 \begin{align*}
0&<\int_0^1 t f(t,t(1-t),-1)dt\\
&\leq c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
\int_0^1(-u^{(4n-2)}(t))^{-\beta} (-u^{(4n-1)}(t))dt\\
&= c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
\frac{(-u^{(4n-2)}(1))^{1-\beta}
}{1-\beta} < \infty,\ t\in (0,1).
\end{align*}
 Namely,  \eqref{e2.4} holds.
By  \eqref{e2.41}, and integrating  \eqref{e2.40}, we obtain
\begin{equation}
\begin{aligned}
&t\int_t^1 f(s,s(1-s),-1)ds\\
&\leq \int_0^t d\xi \int_\xi^1 f(s,s(1-s),-1)ds\\
&\leq c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}\int_0^t d\xi
\int_\xi^1(-u^{(4n-2)}(s))^{-\beta}
f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq
c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}\frac{(-u^{(4n-2)}(t))^{1-\beta}
}{1-\beta}.
\end{aligned}\label{e2.53}
\end{equation}
Noting $u^{(4n-2)}(0)=0,\ \beta<1$, and letting $t\to 0^+$
in  \eqref{e2.53}, we have
 $$\lim_{t \to {0+}}t\int_{t}^1 f(s,s(1-s),-1)ds=0. $$
  This implies  \eqref{e2.5}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.3}]
{\bf Sufficiency}. In this theorem, the cone $P$ is
\begin{align*}
P_3=\big\{&u\in E :R_1(u)=R_2(u)=0,\; u(t)\geq 0,\;
u^{(4n-2)}(t)\leq e_3(t)u^{(4n-2)}(s)\leq 0,\\
&u(t)\geq -k_3 t(1-t)u^{(4n-2)}(s), \forall t, s\in [0,\ 1]\big\},
\end{align*}
where $e_3(t), k_3$ are given by  \eqref{e2.24},
$R_1(u)=au^{(2k)}(0)-bu^{(2k+1)}(0)$,
$R_2(u)=u^{(2k)}(1)$,  $k= 1, 2, \dots, 2n-1$.
According to  \eqref{e2.21} and  \eqref{e2.24}, we show
\begin{gather*}
\|u\|=\|u\|_{4n-2},\quad \forall\ u \in P_3, \\
k_3t(1-t)\|u\|\leq u(t)\leq  \frac{l_3^{2n-2}}{2}  t(1-t)\|u\|,\quad
e_3(t) \|u\|\leq -u^{(4n-2)}(t)\leq \|u\|,
\end{gather*}
where $l_3$ is defined by \eqref{e2.24}.

 Assume  \eqref{e2.6} and  \eqref{e2.7} hold. Then  \eqref{e1.1} has
a $C^{4n-2}[0,1]\cap C^{4n-1}[0,1)$ positive solution  $u$ if and only if
$u$ is a positive solution of the following integral equation
$$
u(t)=(A u)(t)=\int_0^1 h_3(t,s)f(s,u(s),u^{(4n-2)}(s))ds,\quad
\forall\ u \in P_3\setminus \{0\},
$$
where
\begin{align*}
&h_3(t,s)\\
&= \int_0^1\dots \int_0^1
G_1(t,s_{2n-1})G_3(s_{2n-1},s_{2n-2})\dots
G_3(s_2,s_1)G_3(s_1,s)ds_1 \dots ds_{2n-1};
\end{align*}
and $G_1(t,s)$, $G_3(t,s)$
are  defined by \eqref{e2.13} and \eqref{e2.15}, respectively.
The rest of the proof  is very similar to  Theorem \ref{thm2.1}  and
Theorem \ref{thm2.2}. So it is omitted. \smallskip

\noindent\textbf{\bf Necessity.}
 Let $u(t)$ be a $C^{4n-2}[0,1]\cap C^{4n-1}[0,1)\cap C^{4n}(0,1)$
positive solution of \eqref{e1.1}.
Then we claim that there is  a constant $t_0\in [0,1)$ satisfying
$$
u^{(4n-1)}(t_0)=0,\quad  u^{(4n-1)}(0)=-\frac{a}{b}u^{(4n-2)}(0)\leq 0.
$$
Similar to the proof of necessity of Theorem \ref{thm2.2},
the argument can be broken into two cases:
$t_0<0$ and $t_0=0$.

\noindent Case (1):  Assume $t_0<0$. Then $ u^{(4n-1)}(0)< 0$. This
implies \eqref{e2.39}-\eqref{e2.40}, \eqref{e2.43},
\eqref{e2.44}, \eqref{e2.46}, and
$$
\int_0^{t_0}f(t,t(1-t),-1)dt \leq
c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}(-u^{(4n-2)}(0))^{-\beta}
(-u^{(4n-1)}(0))<\infty .
$$
Therefore, the above inequality and \eqref{e2.44} guarantee
\eqref{e2.6}. Also, by \eqref{e2.46}, we can deduce \eqref{e2.7}.

\noindent Case (2). If $t_0=0$, then $u^{(4n-1)}(0)= 0$,
$u^{(4n-2)}(0)< 0$, and \eqref{e2.39}-\eqref{e2.40}, \eqref{e2.43} hold.
 Notice that $-u^{(4n-2)}(s)$ is decreasing in $s$ on (0,1).
Similar to the case (2)  of Theorem \ref{thm2.2},
  by  \eqref{e2.40}, we have
\begin{align*}
0 &< \int_0^1(1- t)f(t,t(1-t),-1)dt=\int_0^1 dt\int_0^t f(s,s(1-s),-1)ds\\
&\leq c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}
\frac{(-u^{(4n-2)}(0))^{1-\beta}}{1-\beta}
 < \infty,\ t\in (0,1).
\end{align*}
Namely, \eqref{e2.6} holds.
By \eqref{e2.43}, integrating \eqref{e2.40}, we get
\begin{equation}
\begin{aligned}
&(1-t)\int_0^t f(s,s(1-s),-1)ds\\
&\leq \int_t^1 d\xi \int_0^\xi f(s,s(1-s),-1)ds\\
&\leq c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}\int_0^t d\xi
\int_\xi^1(-u^{(4n-2)}(s))^{-\beta}
f(s,u(s),u^{(4n-2)}(s))ds\\
&\leq
c_1^{\lambda-\mu}c_2^{\beta-\alpha}m_1^{-\mu}\frac{(-u^{(4n-2)}(t))^{1-\beta}
}{1-\beta}. \end{aligned}\label{e2.54}
\end{equation}
 By  $u^{(4n-2)}(1)=0,\ $  and letting $t\to 1^-$ in \eqref{e2.54}, we
 obtain
 $$\lim_{t \to {1^-}}(1-t)\int_0^t f(s,s(1-s),-1)ds=0. $$
  This implies \eqref{e2.7}.
\end{proof}

\section{Examples}

\begin{example} \label{ex1} \rm
Consider \eqref{e1.1} with $(b=d=0)$ and
$$
f(t,u,v)=p_1(t)u^{-20}(-v)^{1/6}+p_2(t)u^{1/5}(-v)^{\frac{1}{5}},
$$
where $p_i\in C[(0,1),R^+]$ $(i=1,2)$.

 It is easy to see, by  Theorem \ref{thm2.1}, that
\eqref{e1.1} with $(b=d=0)$  has a $C^{4n-2}$ positive solution if and only
if
\begin{gather*}
0<\int_0^1[p_1(t)(t(1-t))^{-19}+p_2(t)(t(1-t))^{6/5}]dt<+\infty,\\
\lim_{t \to {0+}}t\int_{t}^{1}[p_1(s)s^{-20}(1-s)^{-19}+p_2(s)
s^{\frac{1}{5}}(1-s)^{6/5}]ds=0,\\
\lim_{t \to {1-}}t\int_{t}^{1}[p_1(s)s^{-20}(1-s)^{-19}
+p_2(s)s^{\frac{1}{5}}(1-s)^{6/5}]ds=0.
\end{gather*}
\end{example}

\begin{example} \label{ex2} \rm
Consider  \eqref{e1.1} with $(b=0,d>0)$ and
$$
f(t,u,v)=q_1(t)u^{-18}(-v)^{\frac{1}{3}}+q_2(t)u^{\frac{1}{17}}
(-v)^{\frac{1}{13}},
$$
 where $q_i\in C[(0,1),R^+](i=1,2)$.

 Obviously, by  Theorem \ref{thm2.2} , one can see
that \eqref{e1.1} with  $(b=0,d>0)$ has a $C^{4n-2}[0,1]\cap C^{4n-1}(0,1]$
positive solution if and only if
\begin{gather*}
0<\int_0^1[q_1(t)t^{-17}(1-t)^{-18}+q_2(t)t^{\frac{18}{17}}(1-t)
^{1/17}]dt<+\infty.\\
\lim_{t \to {0+}}t\int_{t}^{1}[q_1(s)(s(1-s))^{-18}+q_2(s)(s(1-s)
)^{1/17}]ds=0.
\end{gather*}
\end{example}

\begin{example} \label{ex3} \rm
Consider  \eqref{e1.1} with $(b>0,d=0)$ and
$$
f(t,u,v)=m_1(t)u^{-\frac{1}{2}}(-v)^{\frac{1}{21}}
+m_2(t)u^{\frac{1}{81}}(-v)^{\frac{18}{73}},
$$  where
$m_i\in C[(0,1),R^+](i=1,2)$.

Clearly, according to Theorem \ref{thm2.3}, \eqref{e1.1} with
$(b>0,d=0)$ has a $C^{4n-2}[0,1]\cap C^{4n-1}[0,1)$ positive
solution if and only if
\begin{gather*}
0<\int_0^1[m_1(t)t^{-\frac{1}{2}}(1-t)^{\frac{1}{2}}
+m_2(t)t^{\frac{1}{81}}(1-t)^{\frac{82}{81}}]dt<+\infty,\\
\lim_{t \to {1^-}}(1-t)\int_{0}^{t}[m_1(s)(s(1-s))^{-\frac{1}{2}}
+m_2(s)(s(1-s))^{\frac{1}{81}}]ds=0.
\end{gather*}
\end{example}

\subsection*{Acknowledgements}
The authors are grateful to the anonymous referee for his or her helpful
comments on original manuscript.


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