\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 120, 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 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/120\hfil Solvability of a (p,n-p)-problem]
{Solvability of a (p, n-p)-type multi-point boundary-value problem
      for higher-order differential equations}

\author[Yuji Liu \& Weigao Ge \hfil EJDE--2003/120\hfilneg]
{Yuji Liu \& Weigao Ge }

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

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

\date{}
\thanks{Submitted August 19, 2003. Published December 1, 2003.}
\thanks{Y. Liu is supported by the Science Foundation of Educational Committee
 of Hunan Province. \hfill\break\indent
 Both authors are supported by grant 10371006 from
 the National Natural Science  \hfill\break\indent Foundation of China}
\subjclass[2000]{34K20,  92D25} 
\keywords{Solvability, resonance, non-resonance, \hfill\break\indent 
multi-point boundary-value problem, higher order differential equation}

\begin{abstract}
 In this article, we study the differential equation
 $$
 (-1)^{n-p} x^{(n)}(t)=f(t,x(t),x'(t),\dots,x^{(n-1)}(t)),\;\;0<t<1,
 $$
 subject to the multi-point boundary conditions
 \begin{gather*}
 x^{(i)}(0)=0 \quad \mbox{for }i=0,1,\dots,p-1,\\
 x^{(i)}(1)=0 \quad \mbox{for }i=p+1,\dots,n-1,\\
 \sum_{i=1}^m\alpha_ix^{(p)}(\xi_i)=0,
 \end{gather*}
 where $1\le p\le n-1$. We establish sufficient conditions for the
 existence of at least one solution at resonance and another at
 non-resonance. The emphasis in this paper is that $f$ depends on
 all higher-order derivatives. Examples are given to illustrate the
 main results of this article.
\end{abstract}

\maketitle
\numberwithin{equation}{section}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

 \section{Introduction}

In recent years, there have been many studies concerning
the solvability of multi-point boundary-value problems for
second order differential equations at resonance case; see for example
 \cite{fw1,fw2,gnt,im,l1,l2,ly3} and the references therein.
However, there has no publication concerning the solvability of
multi-point boundary-value problems for higher order differential
equations at resonance.

 In this paper, we consider the differential equation
\begin{equation} \label{e1}
(-1)^{n-p} x^{(n)}(t)=f(t,x(t),x'(t),\dots,x^{(n-1)}(t)),\quad 0<t<1,
\end{equation}
subject to the  boundary conditions
\begin{equation} \label{e2}
\begin{gathered}
 x^{(i)}(0)=0 \quad \mbox{for }i=0,1,\dots,p-1,\\
x^{(i)}(1)=0 \quad \mbox{for }i=p+1,\dots,n-1,\\
\sum_{i=1}^m\alpha_ix^{(p)}(\xi_i)=0,
\end{gathered}
\end{equation}
where $f:[0,1]\times \mathbb{R}^n\to \mathbb{R}$ is continuous, $m\ge 2$,
$n\ge 2$ are integers, $1\le p\le n-1$ is a fixed value, $\alpha _i\in \mathbb{R}$
$(i=1,2,\dots,m)$ and $0\le \xi_1<\xi_2<\dots<\xi_m\le 1$ are fixed.

When $\sum_{i=1}^m\alpha_i\neq 0$, the linear operator
$Lx(t)=(-1)^{n-p}x^{(n)}(t)$, defined in a suitable Banach space,
is invertible. This is called the non-resonance case; otherwise,
it is called the resonance case.

If $n=3$, $m=1$, $p=1$, $f(t,x,y)\equiv g(x)$ and $0<\xi_1<1$,
the boundary-value problem \eqref{e1}--\eqref{e2} becomes
\begin{equation} \label{e3}
\begin{gathered}
 x'''(t)=g(x),\quad 0<t<1,\\
x(0)=0,\alpha_1x'(\xi_1)=0,\quad x''(1)=0,
\end{gathered}
\end{equation}
where $g$ is continuous. Anderson \cite{a} studied the existence of
multiple positive solutions of \eqref{e3} when $\alpha_1\neq 0$.

 The boundary-value problem
\begin{equation} \label{e4}
\begin{gathered}
x^{(n)}(t)=f(t,x(t)),\quad 0<t<1,\\
x^{(i)}(0)=0,\quad \mbox{for }i=0,1,\dots,p-1,\\
x^{(i)}(1)=0 \quad \mbox{for }i=p,\dots,n-1,
\end{gathered}
\end{equation}
is called the $(p,n-p)$ right focal boundary-value problem
\cite{a1,ao1, ao2, ao3,aow,eh4,hg}, and is a special case of
\eqref{e1}-\eqref{e2}. Many authors studied \eqref{e4} and its
special cases; see for example \cite{a1,eh4,hg,wa}. We remark that
in the papers mentioned above, $f$ depends only on $t$ and $x(t)$,
or on $t$ and even order derivatives $x(t),x''(t),\dots$. Since
\eqref{e1}--\eqref{e2} is a generalization of \eqref{e4}, we call
this $(p, n-p)$-type boundary-value problem.

To the best of our knowledge, \eqref{e1}--\eqref{e2} has not been
studied till now. Motivated and inspired by \cite{eh1,fw2,ht,ly2},
we establish  sufficient conditions for the existence of at least
one solution of \eqref{e1}--\eqref{e2} at resonance and another
solution at non-resonance. The emphasis in this paper is that $f$
depends on all higher-order derivatives. The method used is based
on the coincidence degree method developed by Gaines and Mawhin
\cite{gm} and on Shaeffer's theorem \cite{sh}.

This paper can be placed in the existence theory of boundary-value
problems for ordinary differential equations. The foundations and
many important results on this theory were established by
Agarwal, O'Regan and Wong, whose scientific output
is summarized in the monographs \cite{a1,aol}. It is observed
that this particular branch of differential equations has been
developed and gained prominence since the early 1980s.
In recent years, many authors have discussed the boundary-value
problems at non-resonance or resonance for second-order
differential equations \cite{a1,gm,l1,ly3}.

This paper is organized as follows. In Section 2, we establish
existence results for solutions of \eqref{e1}--\eqref{e2} at resonance.
In section 3, we show the existence of solutions of \eqref{e1}--\eqref{e2} at
non-resonance. In section 4, we give some examples to illustrate the main
results of this paper.

\section{Solvability of \eqref{e1}--\eqref{e2} at resonance}

In this section, we establish sufficient conditions for the
existence of at least one solution of \eqref{e1}--\eqref{e2} in the resonance
case, i.e. $\sum_{i=1}^m\alpha_i=0$. In this case, the operator
$Lx(t)=(-1)^{n-p}x^{(n)}(t)$ is not invertible. We assume that
$\sum_{i=1}^m\alpha_i^2\neq 0$. For convenience, we first
introduce some notation and an abstract existence theorem proved by
Gaines and Mawhin \cite{gm}.

Let $X$ and $Y$ be Banach spaces, $L:\mathop{\rm dom}L\subset X\to
Y$ be a Fredholm operator of index zero, $P:X\to X$,
$Q:Y\to Y$ be projectors such that
$$
\mathop{\rm Im}P=\ker L,\quad \ker Q=\mathop{\rm Im}L,
\quad X=\ker L\oplus \ker P,\quad Y=\mathop{\rm Im}L\oplus \mathop{\rm Im}Q.
$$
It follows that
$$
L|_{\mathop{\rm dom}L\cap \ker P}:\mathop{\rm dom}L\cap \ker P\to \mathop{\rm Im}L
$$
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.

\begin{theorem}[\cite{gm}] \label{thmGM}
Let $L$ be a Fredholm operator of index zero and let $N$ be $L$-compact on
$\Omega$. Assume that the following conditions are satisfied:
\begin{itemize}
\item[(i)] $Lx\neq \lambda Nx$ for every
$(x,\lambda )\in [(\mathop{\rm dom} L/\ker L)\cap \partial \Omega ]\times (0,1)
$
\item[(ii)]  $Nx\notin \mathop{\rm Im} L$ for every
$x\in \ker L\cap \partial \Omega $;

\item[(iii)] $\deg(\Lambda QN\big|_{\ker L},\Omega \cap\ker L, 0)\neq 0$,
where $\Lambda:Y/\mathop{\rm Im} L\to \ker L$ is an isomorphism.
\end{itemize}
Then the equation $Lx=Nx$ has at least one solution in
$\mathop{\rm dom} L\cap \overline{\Omega}$.
\end{theorem}

We use the classical Banach spaces $C^k[0,1]$. Let $X=C^{n-1}[0,1]$
and $Y=c^0[0,1]$. The space $Y$ is endowed with the norm
$\|y\|_\infty=\max_{t\in [0,1]}|y(t)|$. The space $X$ is endowed with the
norm
$\|x\|=\max\{\|x\|_\infty,\|x'\|_\infty,\dots,\|x^{(n-1)}\|_\infty\}$.
 Define the linear operator $L$ and the nonlinear operator $N$ by
\begin{gather*}
L: X\cap \mathop{\rm dom} L\to Y,\quad
Lx(t)=(-1)^{n-p}x^{(n)}(t),\\
N:X\to Y, \quad Nx(t)=f(t,x(t),x'(t),\dots,x^{(n-1)}(t)),
\end{gather*}
where
\begin{align*}
\mathop{\rm dom} L=&\big\{x\in C^n[0,1]: x^{(i)}(0)=0\mbox{ for }
i=0,1,\dots,p-1,\\
& x^{(i)}(1)=0\mbox{ for } i=p+1,\dots,n-1,\;
\sum_{i=1}^m\alpha_ix^{(p)}(\xi_i)=0\big\}.
\end{align*}


\begin{lemma}\label{lm2.1}
The following results hold.
\begin{itemize}
\item[(i)]  $\ker L=\{ct^p, \; t\in [0,1],\; c\in \mathbb{R}\}$

\item[(ii)] $\mathop{\rm Im }L
=\big\{y\in Y,\;\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
y(s)ds=0\big\}$

\item[(iii)]  $L$ is a Fredholm operator of index zero

\item[(iv)]  There are projectors $P:X\to X$ and
$Q:Y\to Y$ such that $\ker L=\mathop{\rm Im} P$ and
$\ker Q=\mathop{\rm Im} L$.
Furthermore, let $\Omega \subset X$ be an open bounded subset with
$\overline{\Omega }\cap \mathop{\rm dom} L\neq  \Phi$, then $N$ is $L$-compact
on $\overline{\Omega}$

\item[(v)]  $x(t)$ is a solution of \eqref{e1}--\eqref{e2} if and only if
$x$ is a solution of the operator equation $Lx=Nx$ in $\mathop{\rm dom} L$.
\end{itemize}
\end{lemma}


\begin{proof}
(i) Let $x\in \ker L$, then $x^{(n)}(t)=0$ and $x^{(i)}(0)=0$ for
$i=0,1,\dots,p-1$ and $x^{(i)}(1)=0$ for $i=p+1,\dots,n-1$ and
$\sum_{i=1}^m\alpha x^{(p)}(\xi_i)=0$. It is easy to get $x(t)=ct^p$,
thus $x\in \{ct^p: t\in [0,1],\;c\in \mathbb{R}\;\}$.
On the other hand, if $x(t)=ct^p$, then we find that $x\in \ker L$.
This completes the proof of (i).

\noindent(ii) For $y\in \mathop{\rm Im} L$, then there is $x\in \mathop{\rm dom} L$
such that $(-1)^{n-p}x^{(n)}(t)=y(t)$ and $x^{(i)}(0)=0$ for
$i=0,1,\dots,p-1$ and $x^{(i)}(1)=0$ for $i=p+1,\dots,n-1$ and
$\sum_{i=1}^m\alpha x^{(p)}(\xi_i)=0$. Thus
$$
x^{(p)}(t)=\int_t^1\frac{(s-t)^{n-p-1}}{(n-p-1)!}y(s)ds+A.
$$
Then
$$
x^{(p)}(\xi_i)=\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}y(s)ds+A\;for\;i=1,\dots,m.
$$
Hence
\begin{equation} \label{e5}
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}y(s)ds=0.
\end{equation}
On the other hand, if \eqref{e5} holds, we let
$$
x(t)=\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\int_s^1\frac{(u-s)^{n-p-1}}
{(n-p-1)!}y(u)du\,ds+\frac{At^p}{p!},\quad t\in[0,1].
$$
Then $x\in \mathop{\rm dom} L\cap X$ and $Lx=y$. Thus the proof of (ii) is completed.


\noindent(iii) From (i), $\dim \ker L=1$. On the other hand, we
claim that there is $k\in \{0,1,\dots,m-1\}$ such that
$$
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}s^k ds\neq 0.
$$
In fact, if for all $k\in \{0,1,\dots,m-1\}$, we have
$$
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}s^k ds=0.
$$
It is easy to see that the determinant of coefficients of above
equations is
\begin{align*}
&\Big|\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}s^kds\Big|_{m\times m}\\
&=\left|\begin{matrix}
\int_{\xi_1}^1\frac{(s-\xi_1)^{n-p-1}}{(n-p-1)!}ds&\dots &\int_{\xi_1}^1
\frac{(s-\xi_1)^{n-p-1}}{(n-p-1)!}s^{m-1}ds\\
\vdots& &\vdots\\
\int_{\xi_m}^1\frac{(s-\xi_m)^{n-p-1}}{(n-p-1)!}ds&\dots
&\int_{\xi_m}^1\frac{(s-\xi_m)^{n-p-1}}{(n-p-1)!}s^{m-1}ds
\end{matrix}\right|\\
&=\Big|\frac{(1-\xi_i)^{n-p}}{(n-p)!}-k\frac{(1-\xi_i)^{n-p+1}}{(n-p+1)!}+k(k-1)
\frac{(1-\xi_i)^{n-p+2}}{(n-p+2)!}-\dots\\
&\quad +(-1)^k k!\frac{(1-\xi_i)^{n-p+k}}{n-p+k)!}\Big|_{m\times m}\\
&=\left|\begin{matrix}
\frac{(1-\xi_1)^{n-p}}{(n-p)!}&-k\frac{(1-\xi_1)^{n-p+1}}{(n-p+1)!}&\dots&(-1)^
{m-1}(m-1)!\frac{(1-\xi_1)^{n-p+m-1}}{(n-p+m-1)!}\\
\vdots&&&\vdots\\
\frac{(1-\xi_m)^{n-p}}{(n-p)!}&-k\frac{(1-\xi_m)^{n-p+1}}{(n-p+1)!}&\dots&(-1)^
{m-1}(m-1)!\frac{(1-\xi_m)^{n-p+m-1}}{(n-p+m-1)!}
\end{matrix}\right|
\neq 0\,
\end{align*}
since it can be transformed into a Vandermon dominant and $0\le
\xi_1<\xi_2<\dots<\xi_m\le 1$. Hence, we get
$\alpha_1=\dots=\alpha_m=0$, which contradicts
$\sum_{i=1}^m\alpha_i^2\neq 0$.

Now, for $y\in Y$, let
$$
y_0=y-\Big(\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
y(s)ds\;t^k\Big)/\Big(\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}s^kds\Big).
$$
It is easy to check that $y_0\in \mathop{\rm Im} L$. Let
$\overline{R}=\{ct^k: t\in [0,1],\;c\in \mathbb{R}\}$.
Then $Y=\overline{R}+\mathop{\rm Im} L$. Again, $\overline{R}\cap \mathop{\rm Im}
L=\{0\}$, so $Y=\overline{R}\oplus \mathop{\rm Im} L$.
Hence $\dim Y/\mathop{\rm Im} L=1$. On the other hand, $\mathop{\rm Im} L$
is closed. So $L$ is a Fredholm operator of index zero.

\noindent (iv) Define the projectors $Q: Y\to Y$ and $P: X\to X$ by
\begin{gather*}
Qy(t)=t^k\sum_{i=1}^m\alpha_i \int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
y(s)ds\quad\mbox{for }y\in Y, \\
Px(t)=x^{(p)}(1)t^p\quad \mbox{for } x\in X.
\end{gather*}
It is easy to prove that $\ker L=\mathop{\rm Im} P$ and $\mathop{\rm Im} L=\ker
Q$. Then the inverse $K_p: \mathop{\rm Im} L\to \mathop{\rm dom} L\cap \ker P$ of
the map $L:  \mathop{\rm dom} L\cap \ker P\to \mathop{\rm Im} L$ can be written by
$$
K_py(t)=\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\int_s^1\frac{(u-s)^{n-p-1}}
{(n-p-1)!}y(u)du\,ds\quad\mbox{for } y\in \mathop{\rm Im} L.
$$
In fact, for $y\in \mathop{\rm Im} L$, we have
\[
(LK_p)y(t)=L\Big(\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\int_s^1\frac{(u-s)^
{n-p-1}}{(n-p-1)!}y(u)du\,ds\Big)
=y(t).
\]
On the other hand, for $x\in \ker P\cap \mathop{\rm dom} L$, it follows that
\begin{align*}
(K_pL)x(t)&=K_p((-1)^{n-p}x^{(n)}(t))\\
&=\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\int_s^1\frac{(u-s)^{n-p-1}}{(n-p-1)!}
(-1)^{n-p}x^{(n)}(u) du\, ds\\
&=\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}(-x^{(p)}(1)+x^{(p)}(s))ds\\
&=\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}x^{(p)}(s)ds\\
&=x(t).
\end{align*}
Furthermore, one has
\begin{align*}
QNx(t)&=Qf(t,x(t),x'(t),\dots,x^{(n-1)}(t))\\
&=\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,x(s),x'(s),\dots,x^{(n-1)}(s)) ds
\end{align*}
and
\begin{align*}
&K_p(I-Q)Nx(t)\\
&=K_p\Big[f(t,x(t),x'(t),\dots,x^{(n-1)}(t))\\
&\quad -\sum_{i=1}^m\alpha_i \int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds \Big]\\
&=\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\Big(\int_s^1\frac{(u-s)^{n-p-1}}{(n-p-1)!}
f(u,x(u),x'(u),\dots,x^{(n-1)}(u))du\Big)ds\\
&\quad-\sum_{i=1}^m\alpha_i
\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\\
&\quad \times
\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\Big(\int_s^1\frac{(u-s)^{n-p-1}}{(n-p-1)!}\Big)ds.
\end{align*}
Since $f$ is continuous, using the Ascoli-Arzela theorem, we can
prove that $QN(\overline{\Omega})$ is bounded and
$K_p(I-Q)N: \overline{\Omega}\to X$ is compact, thus $N$
is $L$-compact on $\overline{\Omega}$.

\noindent (v) The proof is simple and is omitted.
\end{proof}

For the next theorem, we set the following asumptions:
\begin{itemize}
\item[(A1)] There is $M>0$ such that for any $x\in \mathop{\rm dom} L/\ker
L$, if $|x^{(p)}(t)|>M$ for all $t\in (0,\frac{1}{2})$, then
$$
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}f(s,x(s),
x'(s),\dots,x^{(n-1)}(s))ds\neq 0
$$

\item[(A2)] There is a function $a\in C^0[0,1]$ and positive
numbers $a_i(i=0,1,\dots,n-1)$ and $\beta_i\in [0,1)$ $(i=0,1,\dots,n-1)$ such that
$$
|f(t,x_0,x_1,\dots,x_{n-1})|\le a(t)+\sum_{i=0}^{n-1}a_i|x_i|^{\beta_i}
$$
for $t\in [0,1]$ and
$(x_0,x_1,\dots,x_{n-1})\in \mathbb{R}^n$

\item[(A3)] There is $M^*>0$ such that for any $c\in \mathbb{R}$ then
either
$$
c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds<0\quad\forall |c|>M^*
$$
or
$$
c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds>0\quad\forall |c|>M^*.
$$
\end{itemize}

\begin{theorem} \label{thm2.1}
Under Assumptions (A1)--(A3), the boundary-value problem  \eqref{e1}--\eqref{e2}
has at least one solution.
\end{theorem}

\begin{proof}
To apply Theorem \ref{thmGM}, we define an open
bounded subset $\Omega$ of $X$ so that (i), (ii) and (iii) of
Theorem \ref{thmGM} hold. To  obtain $\Omega$, we do three steps.
The proof of this theorem is divide into four steps.

\noindent{\bf Step 1.} Let
$$
\Omega_1=\{x\in \mathop{\rm dom} L/\ker L,\;Lx=\lambda
Nx\;\hbox{for}\;\;some\;\lambda \in (0,1)\}.
$$
For $x\in \Omega_1$, $x\notin \ker L,\lambda \neq 0$ and $Nx\in \mathop{\rm Im}
L$, thus $QNx=0$. Then
$$
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),
\dots,x^{(n-1)}(s))ds=0.
$$
Hence by (A1), we know that there is $t_0\in (0,\frac{1}{2})$
such that $|x^{(p)}(t_0)|\le M$. Thus
\begin{align*}
|x^{(p)}(t)|&\le |x^{(p)}(t_0)|+\Big|\int_{t_0}^tx^{(p+1)}(s)ds\Big|\\
&\le M+\int_0^1|x^{(p+1)}(s)|ds\\
&\le M+\|x^{(p+1)}\|_\infty,
\end{align*}
i.e. $\|x^{(p)}\|_\infty\le M+\|x^{(p+1)}\|_\infty$. On the other
hand, it is easy to prove that
$$
\|x\|_\infty\le \|x'\|_\infty\le \dots\le
\|x^{(p)}\|_\infty\;and\;\|x^{(p+1)}\|_\infty\le \dots\le
\|x^{(n-1)}\|_\infty.
$$
So $\|x\|=\max\{\|x^{(p)}\|_\infty,\;\|x^{(n-1)}\|_\infty\}$.
Now, we prove that there is $t_1\in [0,1]$ such that
\begin{equation} \label{e6}
|x^{(n-1)}(t_1)|\le \frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}.
\end{equation}
In fact, if
$$ |x^{(n-1)}(t)|>
\frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}\quad\mbox{for all }t\in [0,1],
$$
then either
\begin{equation} \label{e7}
x^{(n-1)}(t)> \frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}\quad \mbox{for all
 } t\in [0,1]
\end{equation}
or
\begin{equation} \label{e8}
x^{(n-1)}(t)<- \frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}\quad\mbox{for all
} t\in [0,1],
\end{equation}
or
\begin{equation} \label{e9}
\begin{gathered}
 x^{(n-1)}(t)> \frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}\quad\mbox{for some }t\in [0,1]\\
x^{(n-1)}(t)<- \frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}\quad\mbox{for other }t\in [0,1].
\end{gathered}
\end{equation}
It is easy to see that if \eqref{e9} holds, there exists $t_1\in [0,1]$
such that $x^{(n-1)}(t_1)=(n-p-1)!M/(1-t_0)^{n-p-1}$, thus \eqref{e6}
holds, which is a contradiction.
Therefore, for all $t\in [0,1]$, we have
$$
(-1)^{n-p-1}x^{(p)}(t)>\frac{(1-t)^{n-p-1}}{(n-p-1)!}
\frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}
$$
or
$$
(-1)^{n-p-1}x^{(p)}(t)<-\frac{(1-t)^{n-p-1}}{(n-p-1)!}
\frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}.
$$
Hence
$$
|x^{(p)}(t)|>\frac{(1-t)^{n-p-1}}{(n-p-1)!}\frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}.
$$
Then we obtain
$$
|x^{(p)}(t_0)|>\frac{(1-t_0)^{n-p-1}}{(n-p-1)!}\frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}=M,
$$
which contradicts $|x^{(p)}(t_0)|\le M$. Hence there is $t_1\in
[0,1]$ such that
$$
|x^{(n-1)}(t_1)|\le \frac{(n-p-1)!M}{(1-t_0)^{n-p-1}}\le
2^{n-p-1}(n-p-1)!M.
$$
Thus we get
\begin{align*}
|x^{(n-1)}(t)|&\le |x^{(n-1)}(t_1)|+|\int_{t_1}^tx^{(n)}(s)ds|\\
&\le 2^{n-p-1}(n-p-1)!M+\int_0^1|f(s,x(s),x'(s),\dots,x^{(n-1)}(s))|ds\\
&\le 2^{n-p-1}(n-p-1)!M+\int_0^1a(s)ds+\sum_{i=0}^{n-1}a_i\int_0^1|x^{(i)}(s)|^{\beta_i}ds\\
&\le 2^{n-p-1}(n-p-1)!M+\int_0^1a(s)ds+
\sum_{i=0}^{n-1}a_i\|x^{(i)}\|^{\beta_i}_\infty\\
&\le 2^{n-p-1}(n-p-1)!M+\int_0^1a(s)ds+\Big(\sum_{i=0}^{p}a_i\Big)\|x^{(p)}\|^{\beta_i}
_\infty\\
&\quad +\Big(\sum_{i=p+1}^{n-1}a_i\Big)\|x^{(n-1)}\|_\infty^{\beta_i}.
\end{align*}
and
\begin{align*}
|x^{(p)}(t)|&\le |x^{(p)}(t_0)|+\Big|\int_{t_0}^tx^{(p+1)}(s)ds\Big|\\
&\le M+\int_0^1|x^{(p+1)}(s)|ds\\
&= M+\int_0^1\int_s^1\frac{(u-s)^{n-p-2}}{(n-p-2)!}|f(u,x(u),x'(u),\dots,x^{(n-1)}(u))|du\,ds\\
&\le M+\int_0^1\frac{s^{n-p-2}}{(n-p-2)!}|f(s,x(s),x'(s),\dots,x^{(n-1)}(s))|ds\\
&\le M+\int_0^1\frac{s^{n-p-2}}{(n-p-2)!}a(s)ds+\frac{1}{(n-p-1)!}\sum_{i=0}^{n-1}
a_i\|x^{(i)}\|_\infty^{\alpha_i}\\
&\le M+\int_0^1\frac{s^{n-p-2}}
{(n-p-2)!}a(s)ds+\frac{1}{(n-p-1)!}\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\\
&\quad + \frac{1}{(n-p-1)!}\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}.
\end{align*}
Without loss of generality, suppose that $\|x^{(n-1)}\|_\infty>1$,
then
\begin{align*}
&\|x^{(n-1)}\|_\infty \\
&\le 2^{n-p-1}(n-p-1)!M
+\int_0^1a(s)ds+\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}+
\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\\
&\le 2^{n-p-1}(n-p-1)!M
+\int_0^1a(s)ds+\sum_{i=0}^{p}a_i(M+\|x^{(p+1)}\|_\infty)^{\beta_i}\\
&\quad +\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\\
&\le 2^{n-p-1}(n-p-1)!M
+\int_0^1a(s)ds+\sum_{i=0}^{p}a_i(M+\|x^{(n-1)}\|_\infty)^{\beta_i}\\
&\quad +\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i} .
\end{align*}
It follows from $\beta_i\in [0,1)$ that there is $M_1>0$ such that
$\|x^{(n-1)}\|_\infty\le M_1$. Hence
\begin{align*}
\|x^{(p)}\|_\infty &\le
M+\int_0^1\frac{s^{n-p-2}}{(n-p-2)!}a(s)ds+\frac{1}{(n-p-1)!}\sum_{i=0}
^{p}a_i\|x^{(p)}\|_\infty)^{\beta_i}\\
&\quad +\frac{1}{(n-p-1)!}\sum_{i=p+1}^{n-1}a_iM^{\beta_i}.
\end{align*}
We see from above inequality and $\beta_i\in [0,1)$ that there is
$M_2>0$ such that $\|x^{(p)}\|_\infty\le M_2$. Hence we get
$\|x\|\le \max\{M_1,\;M_2\}=M'$.
It follows that $\Omega _1$ is bounded.

\noindent{\bf Step 2.} Let
$\Omega_2=\{x\in \ker L: Nx\in \mathop{\rm Im} L\}$.
For $x\in \Omega _2$, then $x(t)=ct^p$ for some $c\in [0,1]$. It
suffices to prove that there is $M''>0$ such that
 $|c|\le M''$. $Nx\in \mathop{\rm Im} L$ implies
 $$
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds=0.
$$
By (A3), we get
 $|c|\le M^*$. Thus $\Omega _2$ is bounded.

\noindent{\bf Step 3.} According to (A3), for any $c\in \mathbb{R}$ if
 $|c|>M^*$, then either
\begin{equation} \label{e10}
c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds<0\end{equation}
 or
\begin{equation} \label{e11}
\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds>0.
\end{equation}
If \eqref{e10} holds, let
 $$
 \Omega _3=\{x\in \ker L: -\lambda \wedge x+(1-\lambda )QNx=0,\;\lambda \in [0,1]\},
 $$
where $\wedge$ is the isomorphism given by
$\wedge (ct^p)=ct^k$ for all  $c\in \mathbb{R}$.
Now, we shall show that $\Omega _3$ is bounded. Since
 for $ct^p\in \Omega_3$, we have
 $$
 \lambda c=(1-\lambda )\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds.
$$
If $\lambda =1$, it follows from above equality that $c=0$.
Otherwise, if $|c|>M^*$, in view of \eqref{e6}, one has
$$
 \lambda c^2=(1-\lambda )c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
 ^{n-p-1}}{(n-p-1)!}
f(s,cs^p,cps^{p-1},\dots,cp!,0,\dots,0)ds<0,
$$
which contradicts $\lambda c^2\ge 0$. Thus $\Omega_3$ is bounded.

If \eqref{e11} holds, let
$$
 \Omega _3=\{x\in \ker L: \lambda \wedge x+(1-\lambda )QNx=0,\;\lambda \in
 [0,1]\}.
$$
Similarly to above argument, we can prove that $\Omega_3$ is
bounded.

Next, we  show that all conditions of Theorem \ref{thmGM}
are satisfied. Set $\Omega$ be a open bounded subset of $X$ such
that $\Omega \supset \cup_{i=1}^3\overline{\Omega_i}$. By Lemma
\ref{lm2.1}, $L$ is a Fredholm operator of index zero and $N$ is
$L$-compact on $\overline{\Omega}$. From the definition of $\Omega$,
we have the first two conditions for Theorem \ref{thmGM}:
\begin{itemize}
\item  $Lx\neq \lambda Nx$ for $x\in (\mathop{\rm dom} L/\ker L)\cap
\partial \Omega $ and $\lambda \in (0,1)$

\item $Nx\notin \mathop{\rm Im} L$ for $x\in \ker L\cap \partial \Omega$.
\end{itemize}

\noindent{\bf Step 4.} We shall prove the third condition for applying
Theorem \ref{thmGM}:
\begin{itemize}
\item $\deg(QN|_{\ker L},\;\Omega \cap \ker L, 0)\neq 0$.
\end{itemize}
Let $H(x,\lambda )=\pm\lambda \wedge x+(1-\lambda )QNx$.
According the definition of $\Omega $, we know $H(x,\lambda)\neq 0$ for
$x\in \partial \Omega \cap \ker L$, thus by homotopy property of degree,
\begin{align*}
\deg(QN|{\ker L},\Omega\cap \ker L,0)
&=\deg(H(\cdot,0),\Omega\cap \ker L,0)\\
&=\deg(H(\cdot,1),\Omega\cap \ker L,0)\\
&=\deg(\pm\wedge,\Omega\cap \ker L,0)
\neq  0.
\end{align*}
Thus by Theorem \ref{thmGM}, $Lx=Nx$ has at least one solution in $\mathop{\rm dom}
L\cap \overline{\Omega}$, which is a solution of \eqref{e1}--\eqref{e2}.
\end{proof}


For the following theorem, we need the following assumptions:
\begin{itemize}
\item[(A4)] There exists $M>0$ such that for all $x\in \mathop{\rm dom}L$
if $|x^{(p)}(t)|>M$ for all $t\in [0,1]$, then
 $$
 \sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-1-p}}{(n-1-p)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\neq 0.
 $$
\item[(A5)]  There exists $a_0\in C^0[0,1]$ and non-negative
 numbers $a_i$ such that
 $$
 |f(t,x_0,x_1,\dots,x_{n-1})|\le a_0(t)+\sum_{i=0}^{n-1}a_i|x_i|
 $$
 for all $t\in [0,1]$ and $(x_0,\dots,x_{n-1})\in \mathbb{R}^n$.
\end{itemize}

\begin{theorem} \label{thm2.2}
Under Assumptions (A3), (A4), (A5), the boundary-value problem
\eqref{e1}--\eqref{e2} has at least one solution provided that
\begin{align*}
 \sum_{i=0}^pa_i<(n-1-p)!,\quad \sum_{i=p+1}^{n-1}a_i<1,\\
 \sum_{i=p+1}^{n-1}a_i
+\frac{\Big(\sum_{i=0}^pa_i\Big)\Big(\sum_{i=p+1}^{n-1}a_i\Big)}
{(n-1-p)!-\sum_{i=0}^pa_i}<1.
\end{align*}
\end{theorem}

\begin{proof}
 The proof is similar to that of Theorem \ref{thm2.1}. We
 need to do four steps. Let $\Omega_i(i=1,2,3)$ be defined in
 the proof of Theorem \ref{thm2.1}.

\noindent{\bf Step 1.} Prove that $\Omega_1$ is bounded.
 For $x\in \Omega_1$,
$$
 \sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)^{n-1-p}}
 {(n-1-p)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds=0.
 $$
It follows from (A4) that there is $t_0\in [0,1]$ such that
$|x^{(p)}(t_0)|\le M$. On the other hand, $x\in \Omega_1$ implies
$$
x^{(n)}(t)=\lambda f(t,x(t),x'(t),\dots,x^{(n-1)}(t)),\quad t\in
(0,1).
$$
Integrating it from 0 to $t$ if $p\le n-2$, or from $t_0$ to $t$
if $p=n-1$, we get
\begin{align*}
|x^{(n-1)}(t)|&=\begin{cases}
\big|x^{(n-1)}(0)+\lambda\int_0^tf(s,x(s),\dots,x^{(n-1)}(s))ds\big|
&\mbox{for }p\le n-2,\\
\big|x^{(n-1)}(t_0)+\lambda\int_{t_0}^tf(s,x(s),\dots,x^{(n-1)}(s))ds\big|
&\mbox{for }p=n-1 \end{cases}\\
&\le \begin{cases}
\int_0^1|f(s,x(s),\dots,x^{(n-1)}(s))|ds,\\
M+\int_0^1|f(s,x(s),\dots,x^{(n-1)}(s))|ds
\end{cases}\\
&\le M+\int_0^1(a_0(s)+\sum_{i=0}^{n-1}a_i|x^{(i)}(s)|)ds\\
&\le M+\int_0^1a_0(s)ds+\sum_{i=0}^{n-1}a_i\int_0^1|x^{(i)}(s)|ds.
\end{align*}
It is easy to see that $x^{(i)}(t)|\le \|x^{(p)}\|_\infty$ for
$i=0,1,\dots,p$ and $|x^{(i)}(t)|\le \|x^{(n-1)}\|_\infty$ for all
$i=p+1,\dots,n-1$ and $t\in [0,1]$. Hence
$$
|x^{(n-1)}(t)|\le
M+\int_0^1a_0(s)ds+\Big(\sum_{i=0}^pa_i\Big)\|x^{(p)}\|_\infty
+\Big(\sum_{i=p+1}^{n-1}a_i\Big)\|x^{(n-1)}\|_\infty.
$$
Thus
$$\|x^{(n-1)}\|_\infty\le
M+\int_0^1a_0(s)ds+\Big(\sum_{i=0}^pa_i\Big)\|x^{(p)}\|_\infty
+\Big(\sum_{i=p+1}^{n-1}a_i\Big)\|x^{(n-1)}\|_\infty.
$$
On the other hand, we have
$$
x^{(p+1)}(t)=\lambda
\int_t^1\frac{(s-t)^{n-1-p}}{(n-1-p)!}f(s,x(s),\dots,x^{(n-1)}(s))ds.
$$
Integrating from $t_0$ to $t$, we get
\begin{align*}
|x^{(p)}(t)|
&=\Big|x^{(p)}(t_0)+\lambda\int_{t_0}^tf(s,x(s),\dots,x^{(n-1)}(s))ds\Big|\\
&\le M+\int_0^1\int_s^1\frac{(u-s)^{n-1-p}}{(n-1-p)!}f(u,x(u),
 \dots,x^{(n-1)}(u))du\,ds\\
&\le M+\frac{1}{(n-1-p)!}\int_0^1|f(s,x(s),\dots,x^{(n-1)}(s))|ds\\
&\le
M+\frac{1}{(n-1-p)!}\Big(\int_0^1a_0(s)ds+\sum_{i=0}^{n-1}a_i|x^{(i)}(s)|ds
\Big).
\end{align*}
Similarly, we get
$$
\|x^{(p)}\|_\infty\le
M+\frac{1}{(n-1-p)!}\Big(\int_0^1a_0(s)ds+\sum_{i=0}^pa_i\|x^{(p)}\|_\infty+
\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty\Big).
$$
Hence
\begin{gather*}
\Big(1-\sum_{i=p+1}^{n-1}a_i\Big)\|x^{(n-1)}\|_\infty
\le M+\int_0^1a_0(s)ds+\Big(\sum_{i=0}^pa_i\Big)\|x^{(p)}\|_\infty,\\
\begin{aligned}
&\Big(1-\frac{1}{(n-1-p)!}\sum_{i=0}^{p}a_i\Big)\|x^{(p)}\|_\infty\\
&\le M+\frac{1}{(n-1-p)!}\Big(\int_0^1a_0(s)ds
+\sum_{i=n-1-p}^{n-1}a_i\|x^{(n-1)}\|_\infty\Big).
\end{aligned}
\end{gather*}
Thus we get from the assumptions of the Theorem \ref{thm2.2}
\begin{align*}
\Big(1-\sum_{i=p+1}^{n-1}a_i\Big)\|x^{(n-1)}\|_\infty \le&
M+\int_0^1a_0(s)ds
+\frac{\sum_{i=0}^pa_i}{1-\frac{1}{(n-1-p)!}\sum_{i=0}^pa_i}
\Big[M\\
&+\frac{1}{(n-1-p)!}\Big(\int_0^1a_0(s)ds+\sum_{i=n-1-p}^{n-1}a_i
\|x^{(n-1)}\|_\infty\Big)\Big].
\end{align*}
i.e.,
\begin{align*}
&\Big(1-\sum_{i=p+1}^{n-1}a_i-\frac{\left(\sum_{i=0}^pa_i\right)
\big(\sum_{i=p+1}^{n-1}a_i\big)}{(n-1-p)!-\sum_{i=0}^pa_i}
\Big)\|x^{(n-1)}\|_\infty\\
&\le M+\int_0^1a_0(s)ds+\frac{(n-1-p)!\sum_{i=0}^pa_i}{(n-1-p)!
-\sum_{i=0}^pa_i}\Big[M+\frac{1}{(n-1-p)!}\int_0^1a_0(s)ds\Big].
\end{align*}
It follows from the assumptions of Theorem \ref{thm2.2} that there
is $M_1>0$ such that $\|x^{(n-1)}\|_|infty\le M_1$. Thus there is
$M_2>0$ such that $\|x^{(p)}\|_\infty\le M_2$. So $\|x\|\le
\max\{M_1,M_2\}$. Thus $\Omega_1$ is bounded.

\noindent{\bf Step2.}  Prove that $\Omega_2$ is bounded. It
similar to the Step 2 of the proof of Theorem \ref{thm2.1} and is
omitted.

\noindent{\bf Step 3.} Prove that$\Omega_3$ is bounded. It is same
to the Step 3 of the proof of Theorem \ref{thm2.1} and is omitted.

\noindent{\bf Step 4.} It is same to the Step 4 of the proof of
Theorem \ref{thm2.1} and is omitted.

Thus the proof is complete.
\end{proof}
\section{Solvability of \eqref{e1}--\eqref{e2} at non-resonance}

In this section, we obtain sufficient conditions for the
existence of at least one solution of \eqref{e1}--\eqref{e2} at non-resonance,
  i.e. when $\sum_{i=1}^n\alpha_i\neq 0$. In this case, the
operator $Lx(t)=(-1)^{n-p}x^{(n)}(t)$ is invertible. The method
employed is based on Scheaffer's theorem, see for example
\cite[Theorem 4.3.2]{sm}  or [\cite{sh}.

\begin{theorem}[\cite{sh,sm}] \label{thmSH}
Let $(X,\|\ast\|)$ be a Banach space. $T$ is a continuous mapping of $X$
into $X$ which is compact on each bounded subset of $X$. Then either
\begin{itemize}
\item[(i)] The equation $x=\lambda Tx$ has a solution for
$\lambda =1$, or

\item[(ii)] The set of all such solutions $x$, for $\lambda\in (0,1)$, is unbounded.
\end{itemize}
\end{theorem}

Combining the differential equation \eqref{e1} with the boundary
conditions \eqref{e2}, a solutions $x(t)$ satisfies
$$
x^{(p)}(1)-x^{(p)}(t)=\int_t^1\frac{(s-t)^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s)
,\dots,x^{(n-1)}(s))ds.
$$
Since $\sum_{i=1}^m\alpha_ix^{(p)}(\xi_i)=0$, we have
$$
x^{(p)}(1)=\frac{1}{\sum_{i=1}^m
\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds.
$$
Thus
\begin{align*}
x^{(p)}(t)&=
\frac{1}{\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\\
&\quad-\int_t^1\frac{(s-t)^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds.
\end{align*}
Integrating above equation, we have
\begin{align*}
x(t)&=\frac{1}{\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\frac{t^p}{p!}\\
&\quad-\int_0^t\frac{(t-s)^{p-1}}{(p-1)!}\Big(\int_s^1\frac{(u-s)
^{n-p-1}}{(n-p-1)!}f(u,x(u),x'(u),\dots,x^{(n-1)}(u))du\Big)ds\\
&=\frac{1}{\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\frac{t^p}{p!}\\
&\quad+\sum_{j=0}^{n-p}\frac{(-1)^{j}t^{j+p}}{(j+p)!}\int_0^1\frac{s^{n-p-1-j}}{(n-p-1-j)!}
f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\\
&\quad+(-1)^{n-p+1}\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s,x(s),x'(s),
\dots,x^{(n-1)}(s))ds.
\end{align*}
Define the Banach space
\begin{align*}
 X=\big\{& x\in C^{n-1}[0,1]: x^{(i)}(0)=0\mbox{ for }i=0,1,\dots,p-1\\
&\mbox{and } x^{(i)}(1)=0\mbox{ for }i=p+1,\dots,n-1\big\},
\end{align*}
whose norm is $\|x\|=\max\{\|x\|_\infty,\dots,\|x^{(n-1)}\|_\infty\}$,
where $\|x\|_\infty=\max_{t\in [0,1]}|x(t)|$. It is easy to show
that
$$
\|x\|=\max\{\|x^{(p)}\|_\infty,\;\|x^{(n-1)}\|_\infty\}.
$$
Define the nonlinear operator $T: X\to X$ as
\begin{align*}
Tx(t)&=\frac{1}{\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\frac{t^p}{p!}\\
&\quad+\sum_{j=0}^{n-p}\frac{(-1)^{j}t^{j+p}}{(j+p)!}\int_0^1\frac{s^{n-p-1-j}}{(n-p-1-j)!}
f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\\
&\quad+(-1)^{n-p+1}\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s,x(s),x'(s),
\dots,x^{(n-1)}(s))ds\,.
\end{align*}


\begin{theorem} \label{thm3.1}
Assume that the nonlinearity $f$ is bounded. Then \eqref{e1}--\eqref{e2} has
at least one solution.
\end{theorem}

\begin{proof} Let $M>0$ be such that
 $|f(t,x(t),x'(t),\dots,x^{(n-1)}(t))|\le M$ for
 $t\in [0,1]$, $(x_0,x_1,\dots,x_{n-1})\in \mathbb{R}^n$.
 For $\mu\in [0,1]$, consider the equation
 \begin{equation} \label{e12}
 x=\mu Tx.
 \end{equation}
If $x(t)$ is a solution of this equation, then:
\begin{align*}
x(t)&=\mu\Big[\frac{1}{\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i
 \int_{\xi_i}^1\frac{(s-\xi_i)
 ^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\frac{t^p}{p!}\\
&\quad +\sum_{j=0}^{n-p}\frac{(-1)^{j}t^{j+p}}{(j+p)!}\int_0^1
 \frac{s^{n-p-1-j}}{(n-p-1-j)!}
 f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\\
&\quad +(-1)^{n-p+1}\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s,x(s),x'(s),\dots,
x^{(n-1)}(s))ds\Big],
\end{align*}
\begin{align*}
 x^{(p)}(t)&=\mu\Big[\frac{1}{\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i
 \int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds\\
&\quad -\int_t^1\frac{(s-t)^{n-p-1}}{(n-p-1)!}f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds
\Big],
\end{align*}
and
\[
(-1)^{n-p-1}x^{(n-1)}(t) =\mu
\int_t^1f(s,x(s),x'(s),\dots,x^{(n-1)}(s))ds.
\]
So, we have
 \begin{gather*}
 |x^{(p)}(t)|\le \mu M\Big[\frac{1}{\sum_{i=1}^m\alpha_i}
 \sum_{i=1}^m\alpha_i\int_{\xi_i}^1\frac{(s-\xi_i)
^{n-p-1}}{(n-p-1)!}ds+\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}ds\Big],\\
|x^{(n-1)}(t)|\le \mu M.
\end{gather*}
This shows that all solutions of (12) satisfy
$\|x\|=\max\{\|x^{(p)}\|_\infty,\;\|x^{(n-1)}\|_\infty\}$ is bounded.
Taking into account that $T$ is continuous and compact on each
bounded subset of $X$ and using Schaeffer's theorem, we obtain
that $T$ has a fixed point, which is a
solution of \eqref{e1}--\eqref{e2}. \end{proof}

We remark that the hypotheses in Theorem \ref{thm3.1} are strong,
but it is convenient to apply them. Next, we give another
existence result.

\begin{theorem} \label{thm3.2}
 Assume there exist $a_i\in [0,+\infty)$ $(i=0,1,\dots,n-1)$ and
$a\in C[0,1]$ and $\beta_i\in [0,1](i=0,1,\dots,n-1)$ such that
\begin{equation} \label{e13}
|f(t,x_0,x_1,\dots,x_{n-1})|\le a(t)+a_0|x_0|^{\beta_0}+\dots+a_{n-1}|x_{n-1}|^{\beta_{n-1}}
\end{equation}
for $t\in [0,1]$ and $(x_0,x_1,\dots,x_{n-1})\in \mathbb{R}^n$ and
$\sum_{i=p+1}^{n-1}a_i<1$. Then
\eqref{e1}--\eqref{e2} has at least one solution.
\end{theorem}

\begin{proof}
For $x\in X$, we have
$$
|f(t,x(t),x'(t),\dots,x^{(n-1)}(t))|\le
a(t)+\sum_{i=0}^{n-1}a_i|x^{(i)}(t)|^{\beta_i}.
$$
If $x(t)$ is a solution of \eqref{e12}, then
\begin{align*}
|f(t,x(t),x'(t),\dots,x^{(n-1)}(t))|
&= a(t)+\sum_{i=0}^{p-1}a_it|x^{(i+1)}(\xi_i)|^{\beta_i}+a_p|x^{(p)}(t)|^{\beta_p}\\
&\quad +\sum_{i=p+1}^{n-2}a_it|x^{(i+1)}(\xi_i)|^{\beta_i}+a_{n-1}|x^{(n-1)}(t)|^{\beta_{n-1}}\\
&\le a(t)+\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}+\sum_{i=p+1}^{n-1}a_i
\|x^{(n-1)}\|_\infty^{\beta_i}.
\end{align*}
Thus
\begin{align*}
&|x^{(p)}(t)|\\
&\le \mu \Big[\frac{1}{|\sum_{i=1}^m\alpha_i|}\sum_{i=1}^m|\alpha_i|
\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}
\Big(a(s)+\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\\
&\quad +\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big)ds
+\int_0^1 \frac{s^{n-p-1}}{(n-p-1)!}
\Big(a(s)+\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\\
&\quad +\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty\Big)ds\Big]
\\
&\le \mu\Big\{\frac{1}{|\sum_{i=1}^m\alpha_i|}\sum_{i=1}^m|\alpha_i|
\Big[\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}a(s)ds
+\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}ds\\
&\quad \times \Big(\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\Big)
+\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}ds
\Big(\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big)\Big]\\
&\quad +\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}a(s)ds+
\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}a(s)ds
\Big(\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\Big)\\
&\quad+\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}a(s)ds
\Big(\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big)\Big\}
\\
&=\mu \Big\{\frac{1}{|\sum_{i=1}^m\alpha_i|}\sum_{i=1}^m|\alpha_i|
\Big[\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}a(s)ds\\
&\quad +\frac{(1-\xi_i)^{n-p}}{(n-p)!}\Big(\sum_{i=0}^{p}a_i\|x^{(p)}
\|_\infty^{\beta_i}\Big)+\frac{(1-\xi_i)^{n-p}}{(n-p)!}
\Big(\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big)\Big]\\
&\quad +\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}a(s)ds
+\frac{1}{(n-p)!}\Big(\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\Big)\\
&\quad +\frac{1}{(n-p)!}\Big(\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big)\Big\}
\\
&=\mu \Big[\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}
 \int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}a(s)ds
 +\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}a(s)ds\\
&\quad+\frac{1}{(n-p)!}\Big(\frac{\sum_{i=0}^{m}|\alpha_i|}{|\sum_{i=1}^m\alpha_i|}
(1-\xi_i)^{n-p}+1\Big)\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\\
&\quad +\frac{1}{(n-p)!}\Big(\frac{\sum_{i=0}^{m}|\alpha_i|}{|\sum_{i=1}^m
\alpha_i|}(1-\xi_i)^{n-p}+1\Big)\sum_{i=p+1}^{n-1}a_i
\|x^{(n-1)}\|_\infty^{\beta_i}\Big].
\end{align*}
and
\[
|x^{(n-1)}(t)|\le \mu\Big[\int_0^1a(s)ds+\sum_{i=0}^{p}a_i
\|x^{(p)}\|_\infty^{\beta_i}+\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big].
\]
Hence
$$
\|x^{(n-1)}\|_\infty\le
\mu\Big[\int_0^1a(s)ds+\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}
+\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}\Big].
$$
Without loss of generality, suppose $\|x^{(n-1)}\|_\infty\ge 1$,
then
$$
\|x^{(n-1)}\|_\infty\le
\int_0^1a(s)ds+\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}
+\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty.
$$
Thus
$$
\|x^{(n-1)}\|_\infty\le
\Big(1-\sum_{i=p+1}^{n-1}a_i\Big)^{-1}\Big(\int_0^1a(s)ds+
\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i} \Big).
$$
Hence
\begin{align*}
&\|x^{(p)}\|_\infty\\
&\le \mu \Big[\big|\sum_{i=1}^m\alpha_i\big|^{-1}
\sum_{i=1}^m|\alpha_i|\int_{\xi_i}^1\frac{(s-\xi_i)^{n-p-1}}{(n-p-1)!}a(s)ds
+\int_0^1\frac{s^{n-p-1}}{(n-p-1)!}a(s)ds \\
&\quad+\frac{1}{(n-p)!}\Big(\frac{\sum_{i=0}^{m}|\alpha_i|}
{|\sum_{i=1}^m\alpha_i|}(1-\xi_i)^{n-p}+1
\Big)\Big(\sum_{i=0}^{p}a_i\|x^{(p)}\|_\infty^{\beta_i}\Big)\\
&\quad+\frac{1}{(n-p)!}\Big(\frac{\sum_{i=0}^{m}|\alpha_i|}
{|\sum_{i=1}^m\alpha_i|}(1-\xi_i)^{n-p}+1
\Big)\sum_{j=p+1}^{n-1}a_j\Big(1-\sum_{i=p+1}^{n-1}a_i\Big)^{-\beta_j}\\
&\quad \times \Big(\int_0^1a(s)ds
+\sum_{i=0}^pa_i\|x^{(p)}\|_\infty^{\beta_i}\Big)^{\beta_j}\Big].
\end{align*}
Since $\beta_i\in [0,1)$, there exists $M_1>0$ sufficiently large
and independent on $\mu$ such that $\|x^{(p)}\|_\infty\le M_1$,
and
$$
\|x^{(n-1)}\|_\infty\le \int_0^1a(s)ds+
\sum_{i=1}^pa_iM_1^{\beta_i}+\sum_{i=p+1}^{n-1}a_i\|x^{(n-1)}\|_\infty^{\beta_i}.
$$
Similarly, it follows that there is $M_2>0$ sufficiently large and
independent on $\mu$ such that $\|x^{(n-1)}\|_\infty \le M_2$.
These show that
$\|x\|=\max\{\|x^{(p)}\|_\infty,\;\|x^{(n-1)}\|_\infty\;\}$ is
 bounded. On the other hand, $T$ is continuous and compact on each bounded subset of $X$.
 Therefore, by Schaeffer's theorem, we obtain the
 existence of at least one fixed point for the operator $T$, which
 is a solution of \eqref{e1}--\eqref{e2}. The proof is complete.
\end{proof}



 \section{Examples}

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

\subsection*{Example 1}  Consider the following boundary-value problem
\begin{equation} \label{e14}
\begin{gathered}
x''(t)=e(t)+\frac{1}{14}[x'(t)]^{2/3}+\frac{t}{7}\cos^2t\;\sin[x(t)]^{2/3},\\
x(0)=0,\quad x'(1)=\frac{1}{2}x'(\frac{1}{2})+\frac{1}{2}x'(0).
\end{gathered}
\end{equation}
Corresponding to \eqref{e1} and \eqref{e2}, we find
$n=2$, $\xi_1=0$, $\xi_2=\frac{1}{2}$, $\xi_3=1$, and
$\alpha_1=\frac{1}{2}$, $\alpha_2=\frac{1}{2}$, $\alpha_3=-1$.
 $f(t,x,y)=e(t)+\frac{1}{14}y^{2/3}+\frac{t}{7}\cos ^2t\;\sin x^{2/3}$.
 It is easy to see
$|f(t,x,y)|\le |e(t)+\frac{1}{7}|x|^{2/3}+\frac{1}{14}|y|^{2/3}$ with
 $\beta_0=\frac{2}{3}$ and $\beta_1=\frac{2}{3}$. So Assumption (A2)
holds.   Since
\begin{align*}
&\frac{1}{2}\int_0^1f(s,x(s),x'(s))ds+\frac{1}{2}\int_{1/2}^1f(s,x(s),x'(s))ds
-\int_1^1f(s,x(s),x'(s))ds\\
&=\frac{1}{2}\int_0^{1/2}f(s,x(s),x'(s))ds+\int_{1/2}^1f(s,x(s),x'(s))ds,
\end{align*}
it is easy to see that if
$|x'(t)|>14^{3/2}\left(\|e\|_\infty+\frac{1}{7}\right)^{3/2}$ for
all $t\in [0,\frac{1}{2}]$, and $e(t)\ge \frac{t}{7}\cos ^2t$ for
$t\in [\frac{1}{2},1]$, choosing
$M=14^{3/2}\left(\|e\|_\infty+\frac{1}{7}\right)^{3/2}$,
Assumption (A1) holds. Furthermore,
\begin{align*}
&c\Big[\frac{1}{2}\int_0^1f(s,x(s),x'(s))ds+\frac{1}{2}\int_{1/2}^1f(s,x(s),x'(s))ds
-\int_1^1f(s,x(s),x'(s))ds\Big]\\
&=\frac{1}{2}\int_0^{1/2}\left(ce(s)+\frac{1}{14}c^{5/3}+\frac{cs}{7}
\cos ^2s\sin (cs)^{2/3}\right)ds\\
&\quad+\int_{1/2}^1\left(ce(s)+\frac{1}{14}c^{5/3}+\frac{cs}{7}\cos ^2s
\sin (cs)^{2/3}\right)ds > 0
\end{align*}
for sufficiently large $|c|$.
 So (A3) of Theorem \ref{thm2.1} holds. From Theorem \ref{thm2.1},
 \eqref{e14} has at least one solution for every $e\in C^0[0,1]$ with
$e(t)\ge \frac{t}{7}\cos ^2t$ for all $t\in [\frac{1}{2},1]$.

\subsection*{Example 2} Consider the boundary-value problem
\begin{equation} \label{e15}
\begin{gathered}
x'''(t)=e(t)+\frac{1}{14}[x'(t)]^{2/3}+\frac{t}{7}\cos ^2t\;\sin [x(t)]^{2/3}
+\frac{t^2}{8}\sin ^2t\cos [x''(t)]^{4/5},\\
x(0)=0,\quad x'(1)=\frac{1}{2}x'(\frac{1}{2})+\frac{1}{2}x'(0),\;x''(0)=0.
\end{gathered}
\end{equation}
Corresponding to \eqref{e1}--\eqref{e2} we find
$n=3$, $\xi_1=0$, $\xi_2=\frac{1}{2}$, $\xi_3=1$, and
$\alpha_1=\frac{1}{2}$, $\alpha_2=\frac{1}{2}$, $\alpha_3=-1$.
 $f(t,x,y,z)=e(t)+\frac{1}{14}y^{2/3}+\frac{t}{7}\cos ^2t\;\sin x^{2/3}
 +\frac{t^2}{8}\sin ^2t\;\cos z^{4/5}$.
 It is easy to see $|f(t,x,y,z)|\le
 |e(t)+\frac{1}{7}|x|^{2/3}+\frac{1}{14}|y|^{2/3}+\frac{1}{8}|z|^{4/5}$ with
 $\beta_0=\frac{2}{3}$ and $\beta_1=\frac{2}{3}$ and $\beta_2=\frac{4}{5}$.
 So Assumption (A2) holds.
  Similarly, we can prove that (A1) and (A3) hold if
$e(t)\ge \frac{t}{7}\cos ^2t+\frac{t^2}{8}\sin ^2t$ for all
$t\in [\frac{1}{2},1]$. Hence from Theorem \ref{thm2.1}, \eqref{e15} has at least
one solution for every $e\in C^0[0,1]$ with
$e(t)\ge  \frac{t}{7}\cos ^2t+\frac{t^2}{8}\sin ^2t$ for all
$t\in  [\frac{1}{2},1]$.

\subsection*{Example 3} Consider the boundary-value problem
\begin{equation} \label{e16}
\begin{gathered}
x^{(n)}(t)=\sum_{i=0,i\neq p}^{n-1}a_i\sin x^{(i)}(t)+a_px^{(p)}(t)+e(t),\\
x^{(i)}(0)=0,\quad \mbox{for }i=0,1,\dots,p-1,p+1,\dots,n-1,\\
x^{(p)}(1)=\sum_{i=1}^m\alpha_ix^{(p)}(\xi_i),
\end{gathered}
\end{equation}
where $1\le p\le n-1$, $0<\xi_1<\dots<\xi_m<1$, $a_p>0$,
$\alpha_i\ge 0$ for all $i\neq p$ with $\sum_{i=1}^m\alpha_i=1$.
It is easy to see above problem is a special case of
\eqref{e1}--\eqref{e2}. Furthermore, $ |f(t,x_0,\dots,x_{n-1})|\le
|e(t)|+\sum_{i=1}^{n-1}a_i|x_i|$. So (A5) holds. Since
$|f(t,x_0,\dots, x_{n-1})|\ge a_p|x_p|-\sum_{i=1,i\neq
p}^{n-1}|a_i\|x_i|-\|e\|_\infty$, we find that there is $M>0$ such
that if $|x^{(p)}(t)|\ge M$ for all $t\in [0,1]$, then  (A4)
holds. As in Example 1, we find that there is $M^*>0$ such that
(A3) holds. Thus from Theorem \ref{thm2.2}, \eqref{e16} has at
least one solution provided that
\begin{gather*}
\sum_{i=0}^p|a_i|<(n-1-p)!,\quad \sum_{i=p+1}^{n-1}|a_i|<1,\\
\sum_{i=p+1}^{n-1}|a_i|+\frac{\left(\sum_{i=0}^{p}|a_i|\right)
\left(\sum_{i=p+1}^{n-1}|a_i|\right)}{(n-1-p)!-\sum_{i=0}^p|a_i|}<1.
\end{gather*}

\subsection*{Acknowledgement.} The authors are very thankful
to the editors and one of the referees for their valuable
suggestions.

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