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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 159, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/159\hfil Existence of solutions]
{Existence of solutions for nonlinear second-order two-point
 boundary-value problems}

\author[R.-J. Du \hfil EJDE-2009/159\hfilneg]
{Rui-Juan Du}

\address{Rui-Juan Du \newline
Department of Computer Science,
Gansu Political Science and Law Institute\\
Lanzhou, Gansu, 730070, China}
\email{drjlucky@163.com}

\thanks{Submitted May 25, 2009. Published December 15, 2009.}
\thanks{Supported by the Gansu Political
Science and Law Institute Research Projects}
\subjclass[2000]{34B15}
\keywords{Two-point boundary value problem;
existence; Leray-Schauder theory}

\begin{abstract}
 We consider the existence of solutions for the nonlinear
 second-order two-point ordinary differential equations
 \begin{gather*}
 u''(t)+\lambda u(t)+g(u(t))=h(t),\quad t\in[0,1] \\
 u(0)=u(1)=0, \quad\text{or} \quad
 u'(0)=u'(1)=0
 \end{gather*}
 where $g:\mathbb{R}\to \mathbb{R}$ is continuous, and
 $h\in L^1(0,1)$.
\end{abstract}

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

\section{Introduction}

We consider the existence of solutions for the seconder-order
two-point ordinary differential equation
\begin{equation}
 \ u''(t)+\lambda u(t)+g(u(t))=h(t),\quad t\in[0,1]\label{e1.1}
\end{equation}
satisfying either
\begin{equation}
u(0)=u(1)=0,\label{e1.2}
\end{equation}
or
\begin{equation}
u'(0)=u'(1)=0\label{e1.2'}
\end{equation}
where $g:\mathbb{R}\to \mathbb{R}$ is continuous, $
h\in L^1(0,1)$.
The parameter $\lambda\in \mathbb{R}$  is allowed change near
 $m^2\pi^2(m=1,2,\dots)$, the $m$-th eigenvalue of the linear
eigenvalue problem
\begin{equation}
\begin{gathered}
     u''(t)+\lambda u(t)=0,\quad t\in[0,1],\\
     u(0)=u(1)=0,
\end{gathered}  \label{e1.1lamb}
\end{equation}
    and
\begin{equation} \begin{gathered}
     u''(t)+\lambda u(t)=0,\quad t\in[0,1],\\
     u'(0)=u'(1)=0.
    \end{gathered}  \label{e1.1lamb'}
\end{equation}
The linear problem associated with
 $\eqref{e1.1lamb}, \eqref{e1.1lamb'}$ are
\begin{equation}
\begin{gathered}
     u''(t)+\lambda u(t)=h(t),\quad t\in[0,1],\\
     u(0)=u(1)=0,
    \end{gathered} \label{e1.3}
\end{equation}
and
\begin{equation} \begin{gathered}
     u''(t)+\lambda u(t)=h(t),\quad t\in[0,1],\\
     u'(0)=u'(1)=0,\quad
    \end{gathered} \label{e1.3'}
\end{equation}
and the corresponding existence results are known from the linear
theory. Namely, if $\lambda \neq m^2\pi^2(m=1,2,\dots)$, then
$\eqref{e1.1lamb}, \eqref{e1.1lamb'}$ have a unique solution for each
given $h$; While for $\lambda= m^2\pi^2(m=1,2,\dots)$ a solution
exists if, and only if, $h$ satisfies the orthogonality conditions
$$
\int_0^1h(t)\phi_i(t)dt=0 \quad(i=1,2),
$$
where $\phi_1(t)=\sin m\pi t$, $\phi_2(t)=\cos m\pi t$ are the
eigenfunctions associated with the eigenvalue $m^2\pi^2$. In this
case, there are infinity many solutions $u(t)=u_0(t)+a \sin \pi t$,
$v(t)=v_0(t)+b \cos \pi t$, $a,b \in \mathbb{R}$ with $u_0, v_0$ are
the any particular solution of \eqref{e1.1lamb}, \eqref{e1.1lamb'}.

A similar situation arises when introducing a sufficiently
nonlinearity $g$. Assuming for the moment $g$ uniformly bounded, it
is easy to see that $\lambda\neq m^2\pi^2$,
\eqref{e1.1}-\eqref{e1.2}, \eqref{e1.1}-\eqref{e1.2'} again have a
solution for each given $h$. If $\lambda= m^2\pi^2$, there are more
difficulties to hold the existence of solutions of
\eqref{e1.1}-\eqref{e1.2}, \eqref{e1.1}-\eqref{e1.2'}.
In \cite{i1}, only
when $m=1$, \eqref{e1.1}-\eqref{e1.2} is solvable if $h$ satisfies
so called the Landesman-Lazer condition
$$
\limsup_{ t\to -\infty}g(t)<\int_0^1h(t)\phi_1(t)dt
< \liminf_{ t\to +\infty}g(t)\,.
$$
In \cite{s1}, the authors assumed the nonlinearity  $f(t, u)=g(u)-h(t)$ did
not satisfy Landesman-Lazer conditions, were also proved that the
boundary value problem \eqref{e1.1}-\eqref{e1.2} has at least one
solution , but $m$ is only allowed  equal to 1.

It is not difficulty to see that when $m=2,3,\dots$, the case
became more complex, there are only a few scholars to study it. In
addition, in most of the papers about second-order
 two-point are using the same method as \cite{i1,s1}. There aren't much
 more method to solve those problems.

Inspired by the above results, in this paper, we try to establish
the existence results of  boundary value problems
\eqref{e1.1}-\eqref{e1.2}, \eqref{e1.1}-\eqref{e1.2'},
$\lambda$ is allowed to change near $m^2\pi ^2(m=1,2,\dots)$,
the nonlinearity $g$ has  weaker
conditions than \cite{s1}, and the methods are  different
from the methods in \cite{s1}.

\section{Preliminaries}

In this paper, we use the following assumptions in $g$ and $h$:
\begin{itemize}

\item[(H1)]  $g:\mathbb{R}\to\mathbb{R}$ is continuous, there is
$\alpha\in[0, 1)$, $c, d\in (0, +\infty)$ such that
\begin{equation}
|g(u)|\leq c|u|^{\alpha}+d,\quad u\in \mathbb{R};\label{e2.1}
\end{equation}

\item[(H2)] There exists  $r>0$ such that
\begin{equation}
ug(u)>0,\quad |u|>r;\label{e2.2}
\end{equation}

\item[(H2')] There exists  $r>0$ such that
\begin{equation}
ug(u)<0,\quad |u|>r;\label{e2.2'}
\end{equation}

\item[(H3)]  $h:[0, 1]\to \mathbb{R}$, $h\in L^1(0, 1)$ satisfying
\begin{equation}
\int_0^1h(t)\phi_1(t)dt=0.\label{e2.3}
\end{equation}

\item[(H3')] $h:[0, 1]\to \mathbb{R}$, $h\in L^1(0, 1)$ satisfying
\begin{equation}
\int_0^1h(t)\phi_2(t)dt=0. \label{e2.3'}
\end{equation}
\end{itemize}

\textbf{Remark.} % 2.1}
For convenience, we rewrite
$\lambda=\lambda_m+\bar{\lambda}$, where $\lambda_m$ is the
$m$-th eigenvalue of the linear eigenvalue problem
\eqref{e1.1}-\eqref{e1.2}, or of
\eqref{e1.1}-\eqref{e1.2'}. Then the ordinary differential
equation \eqref{e1.1} is equivalent to
\begin{equation}
  u''(t)+\lambda_m u(t)+\bar{\lambda}u(t)+g(u(t))=h(t).\label{e2.4}
\end{equation}
Let $X, Y$ be the linear Banach space $C^1[0,1]$, $L^1(0,1)$, whose
norms are denoted by
$$
\|u\|=\max\{\|u\|_0,\|u'\|_0\},\quad \|u\|_1=\int_0^1|u(s)|ds,
$$
where $\|u\|_0$ denotes the max norm $\|u\|_0=\max\{u(t), t\in[0,1]\}$.

Let $L_i: \mathop{\rm dom}L_i\subset X\to Y$ ($i=1,2$) be linear operators
defined for $u\in \mathop{\rm dom}L_i$ as
\begin{equation}
L_iu:=u''+\lambda_m,\label{e2.5}
\end{equation}
 where $\mathop{\rm dom}L_1=\{u\in W^{2,1}(0,1): u(0)=u(1)=0\}$,
$\mathop{\rm dom}L_2=\{u\in W^{2,1}(0,1): u'(0)=u'(1)=0\}$.

\begin{lemma} \label{lem2.1}
Let $L_i(i=1,2)$ be the linear operator as defined
in \eqref{e2.5}. Then
\begin{gather*}
\ker L_i=\{u\in X: u(t)=\rho\phi_i(t), \rho\in \mathbb{R}\}, \\
 \mathop{\rm Im}L_i=\{u\in Y: \int_0^1 u(t)\phi_i(t)dt=0\}.
\end{gather*}
\end{lemma}

 Defined the operator $P_i: X\to \ker L_i\cap X$, $Q_i: X\to
\mathop{\rm Im}Q_i\cap Y$,
\begin{gather}
(P_iu)(t)=\phi_i(t)\int_0^1u(s)\phi_i(s)ds,\label{e2.6} \\
(Q_iu)(t)=u(t)-(\int_0^1u(s)\phi_i(s)ds)\phi_i(t).\label{e2.7}
\end{gather}
It is easy to check that $\mathop{\rm Im}P_i=\ker L_i$,
$Y/\mathop{\rm Im}Q_i=\mathop{\rm Im}L_i$ ($i=1,2$), and to show
the following Lemma.

\begin{lemma} \label{lem2.2}
 Let $X_{P_i}=\ker L_i$, $X_{I-{P_i}}=\ker P_i$,
$Y_{Q_i}=\mathop{\rm Im}L_i$, $Y_{I-{P_i}}=\mathop{\rm Im}{Q_i}$.
Then
$$
 X=X_{P_i}\oplus X_{I-{P_i}}, \quad
 Y=Y_{I-{Q_i}}\oplus Y_{Q_i}.
$$
\end{lemma}

It is easy to check that the restriction of $L_i$ to $X_{I-{P_i}}$ is a
bijection from $X_{I-{P_i}}$ onto $\mathop{\rm Im}L_i$ ($i=1,2$).
We define
$K_i: \mathop{\rm Im}L_i\to X_{I-{P_i}}$ by
\begin{equation}
K_i=L_i|^{-1}_{X_{I-{P_i}}}.\label{e2.8}
\end{equation}
Define the nonlinear operator $G: X\to Y$ by
$$
(Gu)(t)=g(u(t))\quad t\in[0,1].
$$
It is easy to check that $G: X\to Y$ is completely
continuous. Obviously \eqref{e1.1}-\eqref{e1.2},
\eqref{e1.1}-\eqref{e1.2'} are equivalent to
\begin{equation}
L_iu+\bar{\lambda }u+Gu=h,\quad u\in D(L_i).\label{e2.9}
\end{equation}
Since $\ker L_i= \mathop{\rm span} \{\phi_i(t)\}$ ($i=1,2$),
we see that each $x\in X$ can be uniquely decomposed as
$$
x(t)=\rho \phi_i(t)+v(t)\quad t\in[0, 1],
$$
for some $\rho\in \mathbb{R}$ and $v\in X_{I-{P_i}}$.

For $y\in Y$, we also have the decomposition
$$
y(t)=\tau \phi_i(t)+w(t),\quad t\in[0, 1],
$$
with $\tau \in \mathbb{R}$, $w\in Y_{Q_i}$ ($i=1,2$).

\begin{lemma} \label{lem2.3}
The boundary-value problems \eqref{e2.9} are equivalent
to the system
\begin{equation}
\begin{gathered}
   L_iv(t)+\bar{\lambda} v(t)+Q_iG(\rho \phi_i(t)+v(t))=h(t),\\
   \bar{\lambda }\int_0^1( \phi_i(t))^2dt
 +\int_0^1\phi_i(t)G(\rho \phi_i(t)+v(t))dt=0.
\end{gathered} \label{e2.10}
\end{equation}
\end{lemma}


\section{Main results}

\begin{theorem} \label{thm3.1}
Assume {\rm (H1), (H2), (H3)}.  Then there
exists $\lambda_+>0$ such that \eqref{e1.1}-\eqref{e1.2} has
at least one solutions in $C^1[0,1]$ if $\lambda\in[0, \lambda_+]$.
\end{theorem}

\begin{theorem} \label{thm3.2}
Assume {\rm (H1), (H2'), (H3)}.  Then
there exists $\lambda_-<0$ such that  \eqref{e1.1}-\eqref{e1.2}
has at least one solutions in $C^1[0,1]$ if
$\lambda\in[\lambda_-,0]$.
\end{theorem}


\begin{theorem} \label{thm3.1'}
Assume {\rm (H1), (H2), (H3')}. Then
there exists $\lambda_+>0$ such that  \eqref{e1.1}-\eqref{e1.2'}
has at least one solutions in $C^1[0,1]$ if
$\lambda\in[0, \lambda_+]$.
\end{theorem}

\begin{theorem} \label{thm3.2'}
Assume {\rm (H1), (H2'), (H3')}. Then there exists $\lambda_-<0$
such that  \eqref{e1.1}-\eqref{e1.2'} has at least one solutions
in $C^1[0,1]$ if $\lambda\in[\lambda_-,0]$.
\end{theorem}

 In this article, we  prove only Theorem \ref{thm3.1};
 the other theorems can be proved by using the similarly method.

\begin{lemma} \label{lem3.1}
Assume {\rm (H1), (H2), (H3)}. Then there
exists $M>0$, such that any solution $u\in D(L_1)$ of \eqref{e2.9}
satisfies
$\|u\|< M$,
as long as
\begin{equation}
0\leq\bar{\lambda} \leq \delta
:=\frac{1}{2\|K_1J_1\|_{Y_{Q_1}\to X_{I-{P_1}}}} \label{e2.11}
\end{equation}
where $J_1: X\to Y$ is defined by
$(J_1u)(t)=u(t), t\in[0, 1]$.
\end{lemma}

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

\textbf{Step I.} Obviously
$(L_1+\bar{\lambda} J_1)|_{X_{I-{P_1}}}:X_{I-{P_1}}\to Y_{Q_1}$
is invertible for $\bar{\lambda}\leq \delta$. Moreover, by \eqref{e2.11},
\begin{align*}
\|(L_1+\bar{\lambda} J_1)|^{-1}_{X_{I-{P_1}}}\|_{Y_{Q_1}\to
X_{I -{P_1}}}
&=\|L_1|^{-1}_{X_{I-{P_1}}}(I+\bar{\lambda}
K_1J_1)^{-1}\|_{Y_{Q_1}\to X_{I-{P_1}}}\\
&=\|K_1\|_{Y_{Q_1}\to X_{I -{P_1}}}\|(I+\bar{\lambda}
K_1J_1)^{-1}\|_{X_{I-P}\to X_{I-{P_1}}}\\
&\leq 2\|K_1\|_{Y_{Q_1}\to X_{I -{P_1}}}.
\end{align*}
Let $u(t)=\rho \phi_1(t)+v(t)=\rho \sin m\pi t+v(t)$ is a solution
of \eqref{e2.9} for some $\rho\neq 0$. Then
\begin{align*}
\|v\|
&=\|(L_1+\bar{\lambda} J_1)|^{-1}_{X_{I-{P_1}}}Q_1(h-g(\rho
\sin m\pi
t+v(t)))\|\\
&\leq \|(L_1+\bar{\lambda}
J_1)|^{-1}_{X_{I-{P_1}}}\|_{Y_{Q_1}\to X_{I
-{P_1}}}\|Q_1\|_{Y\to Y_{Q_1}}\\
&\quad\times \big[\|h\|_1+c(|\rho|\|\sin m\pi
  t\|_1+\|v\|_1)^\alpha+d\big ]\\
&\leq 2\|K_1\|_{Y_{Q_1}\to X_{I
-{P_1}}}\|Q_1\|_{Y\to Y_{Q_1}}[\|h\|_1+c(|\rho|\|\sin m\pi
t\|+\|v\|)^\alpha+d]\\
&\leq 2\|K_1\|_{Y_{Q_1}\to X_{I
-{P_1}}}\|Q_1\|_{Y\to Y_{Q_1}}[\|h\|_1+c(|\rho| m\pi
+\|v\|)^\alpha+d]\\
&= 2\|K\|_{Y_{Q_1}\to X_{I -{P_1}}}\|Q_1\|_{Y\to
Y_{Q_1}}[\|h\|_1+c m\pi|\rho|^\alpha (1
+\frac{\|v\|}{m\pi|\rho|})^\alpha+d]\\
&\leq 2\|K_1\|_{Y_{Q_1}\to X_{I -P}}\|Q_1\|_{Y\to
Y_{Q_1}}[\|h\|_1+c m\pi|\rho|^\alpha (1
+\frac{\alpha\|v\|}{m\pi|\rho|})+d]\\
&= 2\|K_1\|_{Y_{Q_1}\to X_{I -{P_1}}}\|Q_1\|_{Y\to
Y_{Q_1}}\\
&\quad\times\big[\|h\|_1+c m\pi|\rho|^\alpha (1
+\frac{\alpha}{(m\pi|\rho|)^{1-\alpha}}\cdot
\frac{\|v\|}{(m\pi|\rho|)^{\alpha}} )+d\big]
\end{align*}
Hence,
\[
\frac{\|v\|}{(m\pi|\rho|)^{\alpha}}\leq
\frac{c_0}{(m\pi|\rho|)^{\alpha}}+c_1+\frac{\alpha
c_1}{(m\pi|\rho|)^{1-\alpha}}\cdot
\frac{\|v\|}{(m\pi|\rho|)^{\alpha}}.
\]
where
\begin{gather*}
c_0=2\|K_1\|_{Y_{Q_1}\to X_{I-{P_1}}}\|Q_1\|_{Y\to
Y_{Q_1}}(\|h\|_1+d), \\
c_1=2c\|K_1\|_{Y_{Q_1}\to
 X_{I-{P_1}}}\|Q_1\|_{Y\to Y_{Q_1}}.
\end{gather*}
If
$$
|\rho|\geq \frac{(2\alpha
c_1)^{-\frac{1}{1-\alpha}}}{m\pi}:=\tilde{c},
$$
then
\begin{equation}
\frac{\|v\|}{(m\pi|\rho|)^{\alpha}}\leq \frac{2c_0}{(m\pi
\tilde{c})^{\alpha}}+2c_1:=\bar{c}.\label{e3.1}
\end{equation}

\textbf{Step II.} If we assume that the conclusion of the lemma
is false, we obtain a sequence $\{\bar{\lambda}_n\}$ with $0\leq
\bar{\lambda}_n \leq \delta, \bar{\lambda}_n\to 0$ and a
sequence of corresponding solutions $\{u_n\}: u_n=\rho_n\phi_1(t) dt
+(t), \rho_n\in \mathbb{R}, v_n\in X_{I-{P_1}}, n\in N$, such that
$$
\|u_n\|\to +\infty.
$$
From \eqref{e3.1}
\begin{equation}
\|v\|\leq
\bar{c}(m\pi)^{\alpha}(|\rho|)^{\alpha}:=\hat{c}|\rho|^{\alpha}.
\label{e3.2}
\end{equation}
we conclude that $|\rho_n|\to +\infty$. We may assume that
$\rho_n\to +\infty$, the other case can be treated in the
same way. Then for all $n\in N$, we get that
$\rho_n\geq \tilde{c}$.

Now, from \eqref{e2.9} we obtain
\begin{equation}
\bar{\lambda}_n \rho_n\int_0^1( \sin m\pi t)^2dt+\int_0^1\sin
     m\pi tg(\rho_n \sin m\pi t+v_n(t))dt=0.\label{e3.3}
\end{equation}
Since $\bar{\lambda}_n\geq 0$,
$\int_0^1\bar{\lambda}_n\rho_n( \sin
m\pi t)^2dt\geq0$,  for all $n\in N$, so we have
\begin{equation}
\int_0^1\sin
     m\pi tg(\rho_n \sin m\pi t+v_n(t))dt\leq0.\label{e3.4}
\end{equation}
Let $I^+:=\{t: t\in[0, 1], \sin \pi t>0\}$,
$I^-:=\{t: t\in[0, 1], \sin \pi t<0\}$.
It is easy to see that $I^+\cap I^-\neq 0,$ and
\begin{equation}
\min\{|\sin m\pi t|t\in I^+\cap I^-\}>0.\label{e3.5}
\end{equation}
Combining \eqref{e3.5} and \eqref{e3.2},we conclude
\begin{gather}
\lim_{\rho_n\to +\infty}\min\{\rho_n\sin m\pi t+v_n(t)|t\in
I^+\}=+\infty.\label{e3.6}
\\
\lim_{\rho_n\to +\infty}\min\{\rho_n\sin m\pi t+v_n(t)|t\in
I^-\}=-\infty.\label{e3.7}
\end{gather}
Applying \eqref{e3.3},\eqref{e3.6} and \eqref{e3.7} and (H2),
we conclude
\begin{align*}
\int_0^1\sin m\pi tg(\rho_n \sin m\pi t+v_n(t))dt
&=\int_{t\in
I^+}\sin m\pi tg(\rho_n \sin m\pi t+v_n(t))dt\\
&\quad +\int_{t\in I^-}\sin m\pi tg(\rho_n \sin m\pi t+v_n(t))dt>0
\end{align*}
hold for some $n$  large enough. This contradicts
\eqref{e3.4}.
\end{proof}

Similarly, we obtain the following result.

 \begin{lemma} \label{lem3.2}
Assume {\rm (H1), (H2'), (H3)}.  Then there
exists $M'>0$, such that any solution $u\in D(L_1)$ of \eqref{e2.9}
satisfies
$$
\|u\|< M',
$$
as long as
$-\delta\leq \lambda \leq 0$,
where $\delta$  and $J_1$ as lemma \ref{lem3.1}
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
 Consider the linear operator $L: X\to Y$, defined for
$u\in \mathop{\rm dom}L$ by
$$
Lu=L_1u+\bar{\lambda}u=\lambda_m u+\bar{\lambda}u,
$$
and the family maps $T_{\mu}: X\to Y$ ($0\leq\mu\leq 1$),
$$
(T_{\mu}u)(t)=\mu(h(t)-g(u(t))),\quad t\in[0,1].
$$
where $\mathop{\rm dom}L:= \{u\in W^{2,1}(0,1): u(0)=u(1)=0\}$.
Observe that $L$ is invertible with, let $K: Y\to X$, then
$K=L^{-1}$, and
\begin{equation}
u(t)=K(G(u(t))-h(t)),\quad t\in[0,1].\label{e3.8}
\end{equation}
If
$$
 R=\{u\in X: \|u\|\leq M+1\},
$$
we can define a compact homotopy $H_{\mu}: R\to \mathop{\rm dom}L$,
$$
H_{\mu}=L^{-1}\circ(T_{\mu}u)\circ J_1.
$$
We can see that the fixed points of $H_{\mu}$ are exactly the
solution of \eqref{e1.1}-\eqref{e1.2}, and the choice of $R$
enables us to say that the homotopy $H_{\mu}$ is fixed-point
free on the boundary of $R$. since $H_0=0$, by the Leray-Schauder
theory \cite{c1}, we obtain that
$H_1$ has a fixed point and so there is a solution to
 \eqref{e1.1}-\eqref{e1.2}.
\end{proof}


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\end{document}