\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 89, pp. 1--12.\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 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/89\hfil Positive solutions of boundary-value problems]
{Positive solutions of boundary-value problems for
2m-order differential equations}

\author[Yuji Liu \& Weigao Ge\hfil EJDE--2003/89\hfilneg]
{Yuji Liu \& Weigao Ge}

\address{Yuji Liu \newline
Department of Mathematics, Beijing Institute of Technology\\
Beijing, 100081, China\newline
Department of Applied Mathematics,
Hunan Institute of Technology\\
Hunan, 414000, China}
 \email{liuyuji888@sohu.com}

\address{Weigao Ge \newline
Department of Applied Mathematics,
Beijing Institute of Technology\\
Beijing, 100081, China}


\date{}
\thanks{Submitted June 23, 2003. Published September 4, 2003.}
\thanks{Both authors were supported by by the National Natural Science
Foundation of China. \hfill\break\indent
 Y. Liu was supported by the Science
Foundation of Educational Committee of Hunan Province.}

\subjclass[2000]{34B18, 34B15, 34B27} 
\keywords{Higher-order differential equation, boundary-value problem, \hfill\break\indent
  positive solution, fixed point theorem}


\begin{abstract}
 This article concerns the existence of positive
 solutions to the differential equation
 $$
 (-1)^{m} x^{(2m)}(t)=f(t,x(t),x'(t),\dots,x^{(m)}(t)),
 \quad  0<t<\pi,
 $$
 subject to boundary condition
 $$  x^{(2i)}(0)=x^{(2i)}(\pi)=0, $$
 or to the boundary condition
 $$  x^{(2i)}(0)=x^{(2i+1)}(\pi)=0,  $$
 for $i=0,1,\dots,m-1$.  Sufficient conditions for the
 existence of at least one positive solution of each
 boundary-value problem are established.
 Motivated by references \cite{ch,h,p}, the  emphasis in
 this paper is that $f$ depends on all higher-order derivatives.
\end{abstract}

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

   \section{Introduction}

The study of the existence of positive solutions of boundary-value
problems for second-order and higher-order ordinary differential
equations has gained prominence recently and is a rapidly growing
field. This happens because of the applications of this problem, especially
fourth-order differential equations; see for example the articles
\cite{b,ch,d,gy1,gy2,ht,h,mw,m,p} and the monographs
\cite{a1,a2,aow}.

For the second-order case, the existence of positive solutions of
boundary-value problems for nonlinear differential equations has
been studied by many authors. The  differential equation
\begin{equation} \label{e1}
x''(t)+f(t,x(t))=0,\quad 0<t<1,
\end{equation}
subjected to different boundary conditions has received much attention.
Specially in seeking conditions on the nonlinearity $f$ for which there are
at least one,  at least two, or at least three positive solutions.
See for example \cite{ach,deh,ew,et,wo}.

However, there are not many publication about the existence of positive
solutions of the differential equation
\begin{equation} \label{e2}
x''(t)+f(t,x(t),x'(t))=0,\quad 0<t<1,
\end{equation}
under various boundary conditions.  This because the presence of $x'$ in the
nonlinearity $f$ causes considerable difficulties \cite{cm,h,kt,pa}.

Recently, Chyan and Henderson \cite{ch} studied the $2m$-order
differential equation
\begin{equation} \label{e3}
x^{(2m)}(t)=f(t,x(t),x''(t),\dots,x^{2(m-1)}(t)),\quad 0<t<1,
\end{equation}
with either the Lidstone boundary condition
\begin{equation} \label{e4}
x^{(2i)}(0)=x^{(2i)}(1)=0 \quad\mbox{for }i=0,1,\dots,m-1,
\end{equation}
or with the focal boundary condition
\begin{equation} \label{e5}
 x^{(2i+1)}(0)=x^{(2i)}(1)=0 \quad\mbox{for }i=0,1,\dots,m-1.
\end{equation}
They proved the existence of at least one positive solution when
$f$ is either super-linear or $f$ is sub-linear.

Similar problems were also investigated by Palamides \cite{p}
using an analysis of the corresponding field on the face-plane
and the Sperner's Lemma. The method there is different
from that in \cite{ch,h}. In the papers mentioned above, the
nonlinearity $f$ depends on $x,x'',\dots,x^{(2(m-1))}$.

In this paper, we consider the  $2m$-order differential
equation
\begin{equation} \label{e6}
(-1)^{m} x^{(2m)}(t)=f(t,x(t),x'(t),\dots,x^{(m)}(t)),\quad 0<t<\pi,
\end{equation}
with either the Lidstone boundary conditions
\begin{equation} \label{e7}
x^{(2i)}(0)=x^{(2i)}(\pi)=0\quad\mbox{for }i=0,1,\dots,m-1,
\end{equation}
 or the focal boundary conditions
\begin{equation} \label{e8}
x^{(2i)}(0)=x^{(2i+1)}(\pi)=0\quad\mbox{for }i=0,1,\dots,m-1.
\end{equation}
 We assume $f:[0,\pi]\times I_0\times I_1\times \dots\times
 I_m\to [0,+\infty)$ is continuous, where $I_0=[0,+\infty)$,
 $I_1=R$, $I_2=(-\infty,0]$, $\dots$ for BVP \eqref{e6}--\eqref{e7}, and
 $I_0=I_1=[0,+\infty)$, $I_2=I_3=(-\infty,0]$, $\dots\dots$ for
 BVP \eqref{e6} and \eqref{e8}. It is easy to check that if $x(t)$ is
 a positive solution of BVP \eqref{e6}--\eqref{e7}, then
 $$
 (-1)^mx^{(2m)}(t)\ge 0,\quad (-1)^{m-1}x^{2(m-1)}(t)\ge
 0,\quad \dots \quad x(t)\ge 0
 $$
 for $t\in [0,\pi]$ and
 $$
 (-1)^mx^{(2m)}(t)\ge 0,\quad (-1)^mx^{(2m-1)}(t)\le 0,
 \quad \dots \quad x'(t)\ge 0,\quad x(t)\ge 0
 $$
 for $t\in [0,\pi]$ if $x(t)$ is a positive solution of BVP \eqref{e6} and
 \eqref{e8}.

 The emphasis of this paper is that $f$ depends on each of the $m$
 higher-order derivatives; i.e., $f$ depends on
 $x,x',\dots,x^{(m)}$. To obtain the main results, we need the
 following notation and an abstract existence theorem, whose
 proof can be found in the text books \cite{gsl,wa}.

\noindent{\bf Definition:}\quad  Let $X$ be a real Banach space. A non-empty
closed convex set $P\subset X$ is called a  cone of $X$ if it
satisfies the following conditions:
\\
(i)\quad $x\in P$ and $\lambda \ge 0$ implies $\lambda x\in P$.
\\
(ii)\quad  $x\in P$ and $-x\in P$ implies $x=0$.
\\
Every cone $P\subset X$ induces an ordering in $X$, which is
given by $x\le y$ if and only if $y-x\in P$.

Let $X$ and $Y$ be Banach spaces, $L:\mathop{\rm dom} L\subset X\to Y$ be a
Fredholm operator of index zero, $P:X\to X$ and
$Q: Y\to Y$ be projectors such that
$$
\mathop{\rm Im}P=\mathop{\rm Ker}L,\mathop{\rm Ker}Q=\mathop{\rm Im}L,\quad
X=\mathop{\rm Ker}L\oplus \mathop{\rm Ker}P,\quad Y=\mathop{\rm Im}L
\oplus \mathop{\rm Im}Q.
$$
It follows that
$$ L|_{\mathop{\rm dom} L\cap \mathop{\rm Ker}P}: \mathop{\rm dom} L\cap \mathop{\rm Ker}
P\to \mathop{\rm Im}L
$$
is invertible, we denote the inverse of that map by $K_p$.

If $\Omega $ is an open bounded subset of $X$, $\mathop{\rm dom} L\cap
\overline{\Omega }\neq \Phi$, the map $N: X\to Y$ will be
called $L-$compact on $\overline{\Omega}$ if
$QN(\overline{\Omega})$ is bounded and $K_p(I-Q)N: \overline{\Omega }\to X$
is compact. Now, we present the fixed point theorem.


\begin{theorem}[\cite{gsl,wa}] \label{thm1}
Let $X$ and $Y$ be Banach spaces,
$K_1\subset X$ and $K\subset Y$ be cones in $X$ and $Y$,
respectively, and the operators $L$ and $N$ be defined above such
that $NX\subset K$, $L^{-1}(K)\subset K_1$ and $\mathop{\rm Ker} L=\{0\}$.
Let $\Omega_1$ and $\Omega _2$ be open bounded subsets in $X$ such
that $0\in \Omega_1\subset \overline{\Omega_1}\subset \Omega_2$.
If $N: \overline{\Omega_2}\to Y$ is L-compact on
$\Omega_2$ and there is $h\in L^{-1}(K)$ with $h\neq 0$ such that
\begin{itemize}
\item[(i)]  $Lx\neq \lambda Nx$ for $\lambda \in (0,1)$ and
$x\in \mathop{\rm dom} L\cap \partial \Omega_1\cap K_1$; $Lx-Nx\neq \lambda Lh$
for $\lambda >0$ and $x\in \mathop{\rm dom} L\cap \partial \Omega_2\cap K_1$,
or
\item[(ii)] $Lx-Nx\neq \lambda Lh$ for $\lambda >0$ and $x\in
\mathop{\rm dom} L\cap \partial \Omega_1\cap K_1$; $Lx\neq \lambda Nx$ for
$\lambda \in (0,1)$ and
$x\in \mathop{\rm dom} L\cap \partial \Omega_2\cap K_1$,
\end{itemize}
then $Lx=Nx$ has at least one solution $x\in \mathop{\rm dom} L\cap
\left(\overline{\Omega_2}/\Omega_1\right)\cap K_1$.
\end{theorem}

\section{Positive solutions of boundary-value problems}

In this section, we present the main results and then
give some examples to illustrate the main results.

\begin{theorem} \label{thm2.1}
 Suppose \begin{itemize}
\item[(A)] The following inequality holds uniformly in $t$:
$$
\limsup_{\sum_{i=0}^{m}|x_i|\to +\infty}
\frac{f(t,x_0,x_1,\dots,x_m)}{\sum_{i=0}^{m}|x_i|}<\frac{1}{m+1}\,.
$$
\item[(B)] The following inequality holds uniformly in $t$:
$$
\liminf_{\sum_{i=0}^{m}|x_i|\to 0}
\frac{f(t,x_0,x_1,\dots,x_m)}{\sum_{i=0}^{m}|x_i|}>1\,.
$$
\end{itemize}
Then BVP  \eqref{e6}--\eqref{e7} has at least one
positive solution.
\end{theorem}

\begin{proof}  Let $X=C^{m}[0,\pi]$ and $Y=C^0[0,\pi]$ be endowed
with the norms
$\|x\|=\max\{\|x_\infty,\|x'\|_\infty,\dots,\|x^{(m)}\|_\infty\}$
and $\|x\|_\infty=\max_{t\in [0,\pi]}|x(t)|$, respectively. For
$x\in Y$, denote
$$ \|x\|_1=\int_0^\pi
|x(t)|dt,\quad\quad\;\|x\|_2=\Big(\int_0^\pi |x(t)|^2dt\Big)^{1/2}.
$$
Define
$$
\mathop{\rm dom} L=\{x\in C^{2m}[0,\pi]: x^{(2i)}(0)=x^{(2i)}(\pi)=0,
\;i=0,1,\dots,m-1\}.
$$
Define the linear operator $L: \mathop{\rm dom} L\cap X\to Y$ and the
nonlinear operator $N: X\to Y$ by
\begin{gather*}
Lx(t)=(-1)^mx^{(2m)}(t) \quad\mbox{for } x\in \mathop{\rm dom} L\cap X\,,\\
Nx(t)=f(t,x(t),x'(t),\dots,x^{(m)}(t)) \quad\mbox{for }x\in X\,.
\end{gather*}
Then the differential equation \eqref{e6} can be written as $Lx=Nx$. It
is easy to see that $Ker L=\{0\}$ and $Im L=Y$. Define the
projectors $P: X\to X$ by $Px(t)=0$ for all $t\in [0,\pi]$ and
$Q: Y\to Y$ by $Qy(t)=0$ for all $t\in [0,\pi]$, respectively.
So $L$ is a Fredholm operator of index
zero, and $L^{-1}: Y\to X\cap \mathop{\rm dom} L$ can be written by
$$
L^{-1}y(t)=\int_0^\pi G_m(s,t)y(s)ds, $$
where
\begin{gather*}
G_0(s,t)=\begin{cases}
\frac{s(\pi-t)}{\pi},&0\le s\le t\le \pi\\
\frac{t(\pi-s)}{\pi},&0\le t\le s\le \pi,
\end{cases}\\
G_k(s,t)=\int_0^\pi G_0(s,u)G_{k-1}(u,t)du\quad\mbox{for }k=1,\dots,m\,.
\end{gather*}
It is easy to check that $L^{-1}$ is completely continuous,
together with that $N: X\to Y$ is continuous and bounded,
it follows that $N$ is $L$-compact. We divide the proof into two
steps.

\noindent{\bf Step 1.}\quad Prove the first part of (ii) in Theorem \ref{thm1}.
 By (B), there is $r>0$ such that if $\sum_{i=0}^{m}|x_i|\le r$, then
$$
f(t,x_0,x_1,\dots,x_m)> \sum_{i=0}^{m}|x_i|\ge x_0\,.
$$
Choose
\begin{gather*}
 \Omega_1=\{x\in X: \|x\|\le r/(m+1)\},\\
K_1=\{x\in \mathop{\rm dom} L\cap X: x(t)\ge 0\mbox{ and }(-1)^mx^{(2m)}(t)\ge 0\mbox{ for }
t\in [0,\pi]\},\\
K=\{x\in Y: x(t)\ge 0\mbox{ for }t\in [0,\pi]\}.
\end{gather*}
Then $\mathop{\rm Ker} L=\{0\}$, $NX\subset K$, $L^{-1}(K)\subset K_1$  and
$K_1\subset X$ and $K\subset Y$ are cones.

If $x\in \mathop{\rm dom} L\cap \partial \Omega_1\cap K_1$, then $\|x\|\le
r/(m+1),$ so
$$ \sum_{i=0}^m|x^{(i)}(t)|\le
\sum_{i=0}^m\|x^{(i)}\|_\infty\le (m+1)\|x\|\le r\,.
$$
It follows that
\begin{equation} \label{e9}
f(t,x(t),x'(t),\dots,x^{(m)}(t))\ge x(t)\;for\;t\in [0,\pi].
\end{equation}
Thus
$$
\sin t f(t,x(t),x'(t),\dots,x^{(m)}(t))\ge x(t)\sin t\quad
\mbox{for }t\in [0,\pi].
$$
Integrating the above inequality from 0 to $\pi$,, we obtain
\begin{align*}
\int_0^\pi \sin tf(t,x(t),x'(t),\dots,x^{(m)}(t))dt
&\ge \int_0^\pi \sin tx(t)dt\\
&=-\cos tx(t)\left|_0^\pi\right.+\int_0^\pi x'(t)\cos tdt\\
&=\sin tx'(t)\left|_0^\pi \right.-\int_0^\pi \sin t x''(t)dt\\
&=\dots\\
&=\int_0^\pi \sin t(-1)^mx^{(2m)}(t)dt.
\end{align*}
i.e.,
\begin{equation} \label{e10}
\int_0^\pi \sin tNx(t)dt\ge \int_0^\pi \sin tLx(t)dt.
\end{equation}
On the other hand, let $h(t)$ be the unique solution of the
following problem (it is easy to know, from \cite{ch}, that it has
unique solution)
\begin{gather*}
(-1)^{m} x^{(2m)}(t)=1,\quad 0<t<\pi,\\
 x^{(2i)}(0)=x^{(2i)}(\pi)=0\quad i=0,1,\dots,m-1.
\end{gather*}
Then $h\in \mathop{\rm dom} L$ and $Lh(t)=1$. We will prove that
 $$
 Lx-Nx\neq \lambda Lh
 $$
 for $\lambda >0$ and $x\in \mathop{\rm dom} L\cap \partial \Omega_1\cap K_1$.
 In fact, if there is $\lambda _1>0$ and $x_1\in \mathop{\rm dom} L\cap
 \partial \Omega_1\cap K_1$ such that
 $$
 Lx_1-Nx_1=\lambda _1Lh,
 $$
 then
 \begin{align*}
 \int_0^\pi \sin tLx_1(t)dt
 &=\int_0^\pi \sin tNx_1(t)dt+\lambda _1\int_0^\pi \sin tdt\\
 &>\int_0^\pi \sin tNx_1(t)dt,
\end{align*}
which contradicts \eqref{e10}. So the  first part of (ii) in Theorem \ref{thm1} is
satisfied.

\noindent{\bf Step 2.}\quad Prove the second part of (ii) in Theorem \ref{thm1}.
Choose $1/(m+1)>\epsilon >0$ and $M>0$ such that
\begin{equation} \label{e11}
f(t,x_0,x_1,x_2,\dots,x_{m})\le \big(\frac{1}{m+1}-\epsilon\big)
\sum_{i=0}^{m}|x_i|+M
\end{equation}
for all $t\in [0,\pi]$ and $x_i\in I_i$ for $i=0,\dots,m$. In
fact, from (A), there is $H>0$ such that
$$
f(t,x_0,x_1,x_2,\dots,x_{m})\le
\big(\frac{1}{m+1}-\epsilon\big)\sum_{i=0}^{m}|x_i|
$$
for $t\in [0,\pi]$ and $\sum_{i=0}^m|x_i|\ge H$, where $x_i\in
I_i$ for $i=0,1,\dots, m$. Let
$$
M=\max_{t\in [0,\pi],\sum_{i=0}^m|x_i|\le H}f(t,x_0,x_1,\dots,x_m),
$$
then we have \eqref{e11}. So for $x\in \mathop{\rm dom} L\cap K_1$, we have
$$
f(t,x(t),x'(t),\dots,x^{(m)}(t))\le
\big(\frac{1}{m+1}-\epsilon\big)\Big(\sum_{i=1}^{m}|x_i|+x(t)\Big)+M.
$$
In order to get $\Omega _2$, we now prove that the set
$$
S=\{x\in \mathop{\rm dom} L\cap K_1,\;Lx=\lambda Nx,\;0<\lambda <1\}
$$
is bounded. In fact, if $S$ is unbounded, then there is $\lambda
\in (0,1)$, and $x\in S$ such that $x$ satisfies
\begin{equation} \label{e12}
(-1)^{m} x^{(2m)}(t)=\lambda
f(t,x(t),x'(t),\dots,x^{(m)}(t)),\;t\in [0,\pi].
\end{equation}
Thus
\begin{align*}
(-1)^mx^{(2m)}(t)x(t)
&=\lambda x(t)f(t,x(t),x'(t),\dots,x^{(m)}(t))\\
&\le \lambda
\Big(\frac{1}{m+1}-\epsilon\Big)\Big(x^2(t)+\sum_{i=1}^{m}x(t)|x^{(i)}(t)|\Big)+
x(t)M.
\end{align*}
Integrating above inequality from 0 to $\pi$, we get
\begin{align*}
 \int_0^\pi (-1)^mx(t)x^{(2m)}(t)dt\\
\le \lambda \Big(\frac{1}{m+1}-\epsilon\Big)\int_0^\pi
\Big(x^2(t)+\sum_{i=1}^{m}x(t)|x^{(i)}(t)|\Big)dt
+M\int_0^\pi x(t)dt.
\end{align*}
Since
\begin{align*}
(-1)^m\int_0^\pi x(t)x^{(2m)}(t)dt
&=(-1)^m\int_0^\pi x(t)dx^{(2m-1)}(t)\\
&=(-1)^mx(t)x^{(2m-1)}\Big|_0^\pi+(-1)^{m-1}\int_0^\pi
x^{(2m-1)}(t)x'(t)dt\\
&=(-1)^{m-1}\int_0^\pi x'(t)dx^{(2m-2)}(t)\\
&=\dots\\
&=\int_0^\pi \left(x^{(m)}(t)\right)^2dt,
\end{align*}
we obtain
\begin{align*}
\|x^{(m)}\|_2^2
&\le \lambda \big(\frac{1}{m+1}-\epsilon\big)\Big[\int_0^\pi
x^2(t)dt+\sum_{i=1}^m\int_0^\pi
x(t)|x^{(i)}(t)|dt\Big] +M\int_0^\pi x(t)dt\\
&\le \lambda \big(\frac{1}{m+1}-\epsilon
\big)\Big(\|x\|_2^2+\sum_{i=1}^m\|x\|_2\|x^{(i)}\|_2\Big)
+\pi M\|x\|_\infty.
\end{align*}
Since
$x(t)\sim \sum_{n=1}^\infty a_n \sin nt$,
where $a_n$ is the Fourier  coefficient  of $x$ and
$$
x'(t)\sim \sum_{n=1}^\infty na_n \cos nt,
$$
by Parseval equality, $\|x\|_2\le \|x'\|_2$.
Similarly, we have
 $$
 \|x\|_2\le \|x'\|_2\le \cdots\le \|x^{(m)}\|_2.
 $$
 Again,
 \begin{align*}
 |x(t)|&=|x(t)-x(0)|=|\int_0^tx'(s)ds|\\
 &\le \int_0^t|x'(s)|ds\le \int_0^\pi |x'(s)|ds\\
 &\le \left(\int_0^\pi |x'(t)|^2dt\int_0^\pi dt\right)^{1/2}
 =\pi^{1/2}\|x'\|_2.
 \end{align*}
 Then we obtain
 $\|x\|_\infty\le \pi^{1/2}\|x'\|_2$.
Thus
$$ \|x^{(m)}\|_2^2\le \lambda
\big(\frac{1}{m+1}-\epsilon\big)(m+1)\|x^{(m)}\|_2^2+M\pi^{3/2}\|x^{(m)}\|_2.
$$
Hence
$$
\|x^{(m)}\|_2\le \frac{M\pi^{3/2}}{\epsilon (m+1)}=:c_1.
$$
Thus, we obtain
\begin{gather*}
\|x\|_\infty\le \pi ^{1/2}\|x^{(m)}\|_2\le
\frac{M\pi^{2}}{\epsilon (m+1)}=:c_2\,,\\
\|x^{(i)}\|_2\le \|x^{(m)}\|_2\le
\frac{M\pi^{3/2}}{\epsilon (m+1)}=:c_1\quad\mbox{for }i=0,1,\dots,m.
\end{gather*}
Similarly, we have
$$
\|x^{(i)}\|_\infty\le \pi^{1/2}\|x^{(i+1)}\|_2\le \frac{M\pi^2}{\epsilon
(m+1)}=c_2\quad\mbox{for }i=1,\dots,m-1.
$$
>From \eqref{e11},
\begin{align*}
|x^{(2m)}(t)|
&\le\big(\frac{1}{m+1}-\epsilon\big)\Big(x(t)+\sum_{i=1}^m|x^{(i)}(t)|\Big)+M\\
&\le\big(\frac{1}{m+1}-\epsilon\big)\Big(\|x\|_\infty+\frac{m}{2}+\frac{1}{2}
\sum_{i=1}^m|x^{(i)}(t)|^2\Big)+M\\
&\le\big(\frac{1}{m+1}-\epsilon\big)\Big(c_2+\frac{m}{2}+\frac{1}{2}
\sum_{i=1}^m|x^{(i)}(t)|^2\Big)+M.
\end{align*}
Integrating above inequality from 0 to $\pi$, we get
$$
\|x^{(2m)}\|_1\le \pi
\big(\frac{1}{m+1}-\epsilon\big)(c_2+\frac{m}{2})
+\frac{1}{2}\big(\frac{1}{m+1}-\epsilon\big)c_1^2+M\pi=:c_3.
$$
Since $x^{(2m-2)}(0)=x^{(2m-2)}(\pi)=0,$ there is $\xi \in
[0,\pi]$ such that $x^{(2m-1)}(\xi)=0$, thus
$$
|x^{(2m-1)}(t)|\le \|x^{(2m)}\|_1.
$$
So $\|x^{(2m-1)}\|_\infty\le c_3$.
Similarly, one gets
$$
\|x^{(2i-1)}\|_\infty\le c_3,\quad i=1,\dots,m-1.
$$
This implies $\|x\|\le \max\{c_3,c_2,c_1\}+1$ for all $x\in S$.

 Choose $R>\max\{\max\{c_1,c_2,c_3\}+1, r/(2m+1)\}$.
Let
$$
\Omega _2=\{x\in X: \|x\|<R\}.
$$
Then $S\subset \Omega_2$. So
$Lx\neq \lambda Nx$ for $\lambda \in (0,1)$ and
$x\in \mathop{\rm dom} L\cap\partial \Omega_2\cap K_1$.
Thus by  Theorem \ref{thm1}, $Lx=Nx$ has at least one solution $x\in \mathop{\rm dom}
L\cap \left(\overline{ \Omega_2}/\Omega_1\right)\cap K_1$. $x$ is
a solution of BVP \eqref{e6}--\eqref{e7}.

Next, we prove that $x(t)>0$ for $t\in [0,\pi]$. Since
$(-1)^{m}x^{(2m)}(t)\ge 0$ for all $t\in [0,\pi]$, together with
the boundary value conditions (1.7), we get $x(t)\ge 0$ and
$x''(t)\le 0$ for all $t\in [0,\pi]$. If there is $t_0\in (0,\pi)$
such that $x(t_0)=0$, then the concavity of $x(t)$ implies
\begin{align*}
0&=x(t_0)=x\Big(\frac{\pi-t_0}{\pi-t}t+\frac{t_0-t}{\pi-t}\pi\Big)\\
&\ge \frac{\pi-t_0}{\pi-t}x(t)+\frac{t_0-t}{\pi-t}x(\pi)\\
&= \frac{\pi-t_0}{\pi-t}x(t).
\end{align*}
This implies that $x(t)=0$ for all $t\in [0,\pi]$, which
contradicts $x\in \overline{\Omega_2}/\Omega_1$. The proof is
complete. \end{proof}

For our convenience, we introduce the following notation:
\begin{align*}
&\Delta_1=\max_{t\in [0,\pi]}\int_0^\pi G_m(s,t)ds,\\
&\Delta_2=\max\Big\{\Delta_1,\;\max_{t\in [0,\pi]}\left(\int_0^t
\frac{s}{\pi}G_{m-1}(s,t)ds+\int_t^\pi(1-\frac{s}{\pi})G_{m-1}(s,t)ds\right)\Big\},\\
&\Delta_3=\max\Big\{\Delta_2,\;\max_{t\in [0,\pi]}\int_0^\pi G_{m-1}(s,t)ds\Big\},\\
&\dots\\
&\Delta_m=\max\Big\{\Delta_{m-1},\;\max_{t\in [0,\pi]}
\Big(\int_0^t\frac{s}{\pi}G_{m/2}(s,t)ds
+\int_t^\pi(1-\frac{s}{\pi})G_{m/2}(s,t)ds\Big)\Big\}\\
&\hspace{15mm} \hbox{if m is an even integer,}\\
&\Delta_m=\max\Big\{\Delta_m,\;\max_{t\in [0,\pi]}\int_0^\pi G_{(m-1)/2}(s,t)ds\Big\},
\quad\hbox{if $m$ is an odd integer.}
\end{align*}
Clearly, we have $\Delta_m\ge \Delta_i$ for $i=1,2,\dots,m$.

\begin{theorem} \label{thm2.2}
Assume the following two conditions are satisfied:
\begin{itemize}
\item[(C)] The inequality
$ f(t,x_0,x_1,\dots,x_m)\ge x_0$ holds for all
$(x_0,x_1,\dots,x_m)$ in $R^{m+1}$ and all  $t$ in $[0,\pi]$.

\item[(D)] The following inequality holds uniformly
for $t\in [0,\pi]$:
$$
\limsup_{\sum_{i=0}^m|x_i|\to 0} \frac{f(t,x_0,x_1,\dots,x_m)}{\sum_{i=0}^m|x_i|}
<\frac{1}{(m+1)\Delta_m}\,.
$$
\end{itemize}
 Then BVP \eqref{e6}--\eqref{e7} has at least one positive solution.
\end{theorem}

\begin{proof}  We divide the proof of the theorem into two steps.

\noindent{\bf Step 1.}\quad
 To prove the first part of (i),
choose $r>0$ and $\delta \in (0,1/[(m+1)\Delta_m]$ such that
\begin{equation} \label{e13}
f(t,x_0,x_1,\dots,x_m)\le \delta \sum_{i=0}^m|x_i|
\end{equation}
for $t\in [0,\pi] $ and $(x_0,x_1,\dots,x_m)\in R^{m+1}$ with
$\sum_{i=0}^m|x_i|\le r$. Let
$$
\Omega_1=\{\;x\in \mathop{\rm dom} L\cap K_1,\;\|x\|< \frac{r}{m+1}\}.
$$
For $x\in \partial \Omega_1$, we have $\|x\|=\frac{r}{m+1}$, then
$$
\sum_{i=0}^m|x^{(i)}(t)|\le \sum_{i=0}^m\|x^{(i)}\|_\infty\le (m+1)\|x\|=r.
$$
So, we get
$$
f(t,x(t),x'(t),\dots,x^{(m)}(t))\le \delta
\sum_{i=1}^m|x^{(i)}(t)|,\quad\hbox{for }t\in [0,\pi].
$$
If $Lx=\lambda Nx$ with $\lambda \in (0,1)$ has a solution $x\in
\mathop{\rm dom} L\cap K_1\cap \partial \Omega_1$, then
$$
x(t)=\lambda L^{-1}Nx(t)=\lambda \int_0^\pi
G_m(t,s)f(s,x(s),x'(s),\dots,x^{(m)}(s))ds.
$$
Hence, we get
\begin{align*}
\|x\|_\infty
&=\lambda \max_{t\in [0,\pi]}\int_0^\pi G_m(t,s)f(s,x(s),x'(s),\dots,x^{(m)}(s))ds\\
&\le \delta\max_{t\in [0,\pi]}\int_0^\pi G_m(t,s)\sum_{i=0}^m|x^{(i)}(s)|ds\\
&\le \delta \Delta_1(m+1)\|x\|.
\end{align*}
It is easy  to check that
\begin{align*}
 \|x'\|_\infty
&=\lambda \max_{t\in  [0,\pi]}\Big[-\int_0^t\frac{s}{\pi}G_{m-1}(t,s)f(s,x(s),x'(s),
  \dots,x^{(m)}(s))ds \\
 &\quad + \int_t^\pi \big(1-\frac{s}{\pi}\big)G_{m-1}(t,s)f(s,x(s),x'(s),
  \dots,x^{(m)}(s))ds\Big]\\
 &\le \max_{t\in [0,\pi]}\Big(\int_0^t\frac{s}{\pi}G_{m-1}(t,s)f(s,x(s),x'(s),
 \dots,x^{(m)}(s))ds\\
 &\quad + \int_t^\pi\big(1-\frac{s}{\pi}\big)G_{m-1}(t,s)f(s,x(s),x'(s),
 \dots,x^{(m)}(s))ds\Big)\\
 &\le \max_{t\in [0,\pi]}\Big(\int_0^t\frac{s}{\pi}G_{m-1}(t,s)ds+
 \int_t^\pi\big(1-\frac{s}{\pi}\big)G_{m-1}(t,s)ds\Big)
 \delta (m+1)\|x\|\\
 &\le \Delta_2\delta (m+1)\|x\|.
 \end{align*}
Finally, we can get
$\|x^{(m)}\|_\infty\le \delta \Delta_m(m+1)\|x\|$.
Hence, we have
$$
\|x\|\le \delta \Delta_m(m+1)\|x\|.
$$
Thus $(m+1)\delta\Delta_m\ge 1$, which contradicts
$\delta \in (0,1/[\Delta_m(m+1)])$. The first step is complete.

\noindent{\bf Step 2.}\quad
 Choose $\Omega_2$ sufficiently large such that
$\Omega_1\subset \overline{\Omega_1}\subset \Omega_2$, by
condition (C), we have that
$$f(t,x_0,x_1,\dots,x_m)\ge x_0
$$
holds for all $t\in [0,\pi]$ all $(x_0,x_1,\dots,x_m)\in R^{m+1}$.
Hence,
$$
f(t,x(t),x'(t),\dots,x^{(m)}(t))\ge x(t)
$$
holds for all $t\in [0,\pi]$, i.e. \eqref{e9} holds.
Similar to Step 1 in Theorem \ref{thm1}, we can get a contradiction,
hence the second part of (i) in Theorem \ref{thm1} is satisfied.
It follows from (i) of Theorem \ref{thm1} that BVP \eqref{e6} and \eqref{e8}
has at least one positive solution $x(t)$.
The proof is complete.
\end{proof}

\subsection*{Remark}
Consider the  boundary-value problem
\begin{equation} \label{e14}
\begin{gathered}
 (-1)^{m}
x^{(2m)}(t)=f(t,x(t),x'(t),\dots,x^{(m)}(t)),\quad 0<t<T,\\
 x^{(2i)}(0)=x^{(2i)}(T)=0\quad\mbox{for }i=0,1,\dots,m-1,
 \end{gathered}
 \end{equation}
where $T>0$ is a constant, $f$ and $m$ are defined in
\eqref{e6}--\eqref{e7}.
Let $s=\pi t/T$, we transform BVP \eqref{e14} into a BVP similar
to BVP \eqref{e6}--\eqref{e7}. Then a similar existence result can be
obtained.

\begin{theorem} \label{thm2.3}
Suppose $(A)$ and $(B)$ of Theorem  \ref{thm2.1}
hold. Then BVP \eqref{e6} and \eqref{e8} has at least one positive solution.
\end{theorem}

\begin{proof}  Consider the boundary-value problem
\begin{gather*}
 (-1)^{m}
x^{(2m)}(t)= \begin{cases}
f(t,x(t),x'(t),\dots,x^{(m)}(t)), &\mbox{for } 0\le t\le\pi,\\
f(2\pi-t,x(2\pi-t),-x'(2\pi-t),\dots,\\
(-1)^mx^{(m)}(2\pi-t)) & \mbox{for } \pi\le t\le 2\pi,
\end{cases}\\
 x^{(2i)}(0)=x^{(2i)}(2\pi)=0\quad\mbox{for }i=0,1,\dots,m-1\,.
 \end{gather*}
This problem is exactly similar to that of Theorem \ref{thm2.1}, we can obtain at
least one positive solution $x(t)$, which is defined on
$[0,2\pi]$, of above BVP and so $x(t)(t\in [0,\pi])$ is a positive
solution of BVP \eqref{e6} and \eqref{e8}. The proof completed.
\end{proof}

\begin{theorem} \label{thm2.4}
Suppose Conditions $(C)$ and $(D)$ of Theorem \ref{thm2.2}
hold. Then BVP \eqref{e6} and \eqref{e8} has at least one positive solution.
\end{theorem}

The proof is similar to that of Theorem \ref{thm2.3} and is omitted.
Next, we present two examples to illustrate the main results.

\begin{example} \rm
Consider the  boundary-value problem
\begin{equation} \label{e15}
\begin{gathered}
x^{(4)}(t)=f(t,x(t),x'(t),x''(t)), \quad 0<t<\pi,\\
x(0)=x''(0)=x(\pi)=x''(\pi)=0,
\end{gathered}
\end{equation}
where $f$ is a nonnegative continuous function.
 From Theorem \ref{thm2.1}, if
$$
\limsup_{|x|+|y|+|z|\to \infty}\frac{f(t,x,y,z)}{|x|+|y|+|z|}<\frac{1}{3},
$$
and
$$
\liminf_{|x|+|y|+|z|\to 0}\frac{f(t,x,y,z)}{|x|+|y|+|z|}>1
$$
hold uniformly, then \eqref{e15} has at least one positive solution.
\end{example}

\begin{example} \rm
Consider the boundary-value problem
\begin{equation} \label{e16}
\begin{gathered}
x^{(6)}(t)=-\frac{2}{1+|x(t)|+|x'(t)|+x''(t)|+|x'''(t)|},\quad 0<t<\pi,\\
x(0)=x''(0)=x''''(0)=x(\pi)=x''(\pi)=x''''(\pi)=0.
\end{gathered}
\end{equation}
It is easy to check that all conditions of Theorem \ref{thm2.1} are
satisfied. So \eqref{e16} has at least one positive solution.
\end{example}

{\bf Acknowledgement.}\ \ The authors wish to express their
gratitude to the referee and the editors of Electronic Journal
of Differential Equations.

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