\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2002(2002), No. 59, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2002 Southwest Texas State University.}
\vspace{1cm}}

\begin{document}

\title[\hfilneg EJDE--2002/59\hfil
 Positive solutions and nonlinear eigenvalue problems]
{Positive solutions and nonlinear eigenvalue problems
 for retarded second order differential equations}

\author[G. L. Karakostas \& P. Ch. Tsamatos \hfil EJDE--2002/59\hfilneg]
{G. L. Karakostas \& P. Ch. Tsamatos }

\address{G. L. Karakostas \hfill\break
Department of Mathematics, University of Ioannina, 451 10
Ioannina, Greece}
\email{gkarako@cc.uoi.gr}

\address{P. Ch. Tsamatos \hfill\break
Department of Mathematics, University of Ioannina, 451 10
Ioannina, Greece}
\email{ptsamato@cc.uoi.gr}

\date{}
\thanks{Submitted March 27, 2002. Published June 21, 2002.}
\subjclass[2000]{34K10}
\keywords{Nonlocal boundary value problems, positive solutions, \hfill\break\indent
concave solutions, retarded second order differential equations }

\begin{abstract}
  We investigate the eigenvalues of a nonlocal boundary
  value problem for a second  order retarded differential equation.
  We provide information on norm estimates, uniqueness, and
  continuity of solutions.
\end{abstract}

\maketitle

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{fact}[theorem]{Fact}
\numberwithin{equation}{section}

\section{Introduction}

We study the set of positive values $\lambda$ for which
second order nonlinear differential equations with retarded arguments
admit a positive, nondecreasing, concave solution. Consider
\begin{equation}
 (p(t)x'(t))'+\lambda \sum_{j=0}^k
q_j(t)f_j(x(t),x(h_j(t)))=0,\quad \hbox{a.a.}\quad t\in [0,1]
\label{1.1}
\end{equation}
with the initial condition
\begin{equation}
x(0)=0\label{1.2}
\end{equation}
and the nonlocal boundary condition
\begin{equation}
 x'(1)=\int_{0}^{1}x'(s)dg(s),\label{1.3}
\end{equation}
where $g$ is a nondecreasing function and the
integral is meant in the  Riemann-Stieljes sense.
Boundary-value problems involving retarded and functional
differential equations were
recently studied by many authors using various methods. We
especially refer to \cite{a1,d1,j1,h1,h3,w1,w2}
and to  \cite{h2,k1,k3} which  were
the motivation for this work. Our main results in this paper refer
to the values of the positive real
parameter $\lambda$ for
which the problem (\ref{1.1})-(\ref{1.3}) has a solution. Note that the
problem of finding eigenvalues, for which a second or a higher order
differential equation with various
boundary conditions has positive solutions, has been studied by
several authors in the last decade. See for example
the papers \cite{d1,e1,h2,h3,k3} and the references therein.

Problem of this type are usually transformed into operator
equations of the form
\begin{equation}
 Ax=\lambda^{-1} x, \label{1.4}
\end{equation}
where $A$ is an appropriate completely
continuous operator. (Obviously the
form (\ref{1.4}) justifies the term ``eigenvalue problems" we use in the
title of this article.)

 Equation (\ref{1.4}) is written as $x=Tx$, where $T:=\lambda A$ and in a
great number of works the following theorem is applied.

\begin{theorem}[Krasnoselskii \cite{k4}] \label{thm1.1}
Let $\mathcal{B}$ be a Banach space and let  $\mathbb{K}$
be a cone in $\mathcal{B}$. Assume that $\Omega
_1$ and $\Omega _2 $ are open bounded subsets of $\mathcal{B}$,
with $0\in\Omega _1 \subset \overline
{\Omega _1 }\subset \Omega _2$, and let
$$ T: \mathbb{K}\cap (\overline{\Omega _2}\setminus
\Omega _1  )\to \mathbb{K}
$$
be a completely continuous operator such that, either
$$ \|Tu\|\le \|u\|,\quad u\in \mathbb{K} \cap \partial
\Omega _1 \quad\hbox{and}\quad \|Tu\|\ge \|u\|,\quad
u\in \mathbb{K} \cap \partial \Omega _2 ,
$$
or
$$ \|Tu\|\ge \|u\|,\quad u\in \mathbb{K} \cap \partial
\Omega _1 \quad \hbox{and}\quad \|Tu\|\le \|u\|,\quad
u\in \mathbb{K} \cap \partial \Omega _2 .
$$
Then $T$ has a fixed point in
$\mathbb{K}\cap (\overline{\Omega _2}\setminus \Omega _1  )$.
\end{theorem}

In this paper we are interested in the existence of positive solutions and
our approach is based on Theorem \ref{thm1.1}.

Note that, as the literature  shows, in almost all the cases where
Theorem \ref{thm1.1} applies, concavity is the most significant property
of te solutions. Indeed, the idea is to use concavity of the real valued
functions $x$ defined on the interval $[0,1]=:I$ and which constitute
the elements of a cone $\mathbb{K}$, the domain of the operator $T$.
Then two elementary facts are the major steps
in our proofs. The first fact read as follows:

\begin{fact} \label{fact1.2}
Let $x:I\to \mathbb{R}$ be a nonnegative, nondecreasing and concave
function. Then, for any
$\tau\in [0,1]$ it holds
$$x(t)\ge\tau\|x\|, \quad t\in [\tau , 1],$$
where $\|x\|$ is the sup-norm of $x$.
\end{fact}

\paragraph{Proof}
  From the concavity of $x$ we have
$$ x(t)\ge x(\tau)=x\left ( (1-\tau)0+\tau 1\right ) \ge
(1-\tau)x(0)+\tau x(1)\ge\tau x(1)=\tau
\|x\|,$$ for all $t\in [\tau, 1]$.
\qed \smallskip

The second fact is that the image $Ax$ of a point $x$ of the
cone $\mathbb{K}$ is a concave function. And in case $p(t)=1, t\in I$
this fact is obvious. (Indeed, one can show that the second
derivative is nonnegative.) In the general case an additional
assumption on
$p$ is needed. This step, which notice that, though it seems to be obvious, it
should be added to the proofs of the main theorems in \cite{k1,k2}, lies on
the following elementary lemma:

\begin{lemma} \label{lm1.3}
Let $a,b$ two real valued functions defined on $I$. If the
product $ab$ is a non-increasing function, then $b$ is also
non-increasing provided that, either
\begin{enumerate}
\item[(i)] $a, b$ are nonnegative functions and $a$ is nondecreasing, or
\item[(ii)] $a$ is nonnegative and non-increasing and $b$ is non-positive.
\end{enumerate}
\end{lemma}

\paragraph{Proof} For each $t_1,t_2\in I$ with $t_1\le t_2$, it holds
\begin{align*}
a(t_1)[b(t_2)-b(t_1)]&=a(t_1)b(t_2)-a(t_1)b(t_1)\\
&\le
a(t_1)b(t_2)-a(t_2)b(t_2)= [a(t_1)-a(t_2)]b(t_2)\le 0.
\end{align*}
Thus, in any case, we have $b(t_2)\le b(t_1)$.
\qed \smallskip

  From this lemma we get the following statement.
\begin{fact} \label{fact1.4}
If $y:I\to \mathbb{R}$ is a differentiable function with $y'\ge 0$ and
$p:I\to \mathbb{R}$ is a
positive and nondecreasing function such that $(p(t)y'(t))'\le 0$, for all $t\in I$, then $y$ is concave.
\end{fact}

\paragraph{Proof}
We apply Lemma \ref{lm1.3}(i) with $a=p$, $b=y'$ and conclude that $y'$ is
non-increasing. This implies that $y$ is concave.
\qed \smallskip

Apart of positivity and concavity properties of the solutions which
are guaranteed by applying
Theorem \ref{thm1.1} we know also monotonicity of them. Moreover
we can have some information on the estimates of their sup-norm.
Finally, some Lipschitz type
conditions may provide uniqueness results as well as continuous dependence
of the solutions under the
corresponding eigenvalues.

\section{Preliminaries and the assumptions}

In the sequel we shall denote by $\mathbb{R}$ the real
line and by $I$ the interval $[0,1]$.
Then $C(I)$ will denote the space of all continuous
functions $x:I\to \mathbb{R}$. This is a Banach space
when it is furnished with the usual supremum norm $\|\cdot \| $.

Consider equation (\ref{1.1}) associated with the conditions (1.2 ),
(1.3 ). By a solution of the problem (\ref{1.1})-(\ref{1.3})
we mean a function $x\in C(I)$, whose the first derivative $x'$ is
absolutely continuous on $I$ and which satisfies
equation $(\ref{1.1})$ for almost all $t\in I$, as well as conditions
(\ref{1.2}), (\ref{1.3}).

The basic assumptions on the functions involved are the following:
\begin{enumerate}
\item[(H1)] The function $p: I\to (0,+\infty)$ is continuous and
nondecreasing.

\item[(H2)] The functions $q_j: I\to \mathbb{R}$, $j=0,1,\dots ,k$ are
continuous and such that $q_j(t)\ge 0$,
$t\in I$, $j=0,\dots ,k$, as well as $q_0(1)>0$.

\item[(H3)] The function $g\colon I\to \mathbb{R}$ is nondecreasing
and such that
$$ \int_{0}^{1}\frac{1}{p(s)}dg(s)<\frac{1}{p(1)}.$$

\item[(H4)] The retardations $h_j:I\to I$ ($j=0,\dots ,k$) satisfy
$$ 0\le h_j (t)\le h_0 (t)\le t, \quad t\in I, \quad j=1,\dots ,k
$$
and moreover $h_0$ is a nondecreasing
function not identically zero.

\item[(H5)] The functions $f_j\colon \mathbb{R}\times\mathbb{R}\to
\mathbb{R}$, $0=1,\dots ,k$ are continuous and such that
$f_j (u,v)\ge 0$, when $u\ge 0$ and $v\ge 0$, for all $j=0,1,\dots ,k$.
Also, if for some $j_0\in\{1,2,\dots ,k\}$ there is a point
$t\in I$ such that $h_{j_0}(t)<h_0(t)$, then we assume that the
function $f_{j_0}(u,v)$ is nondecreasing with
respect to $v$ for all $u\ge 0$.
\end{enumerate}
%
The first step in our approach is to reformulate the problem
(\ref{1.1})-(\ref{1.3}) as an operator equation of the form (\ref{1.4}) for an
appropriate operator $A$, which does not depend on the parameter $\lambda$.
Note that our requirement is $\lambda>0$.
To find such an operator $A$ we integrate (\ref{1.1}) from
$t$ to $1$ and get
\begin{equation}
 x'(t)=\frac{1}{p(t)}p(1)x'(1)+\frac{\lambda}{p(t)}\int_{t}^{1}z(s)ds,
\label{2.1}
\end{equation}
where
$$ z(t):=\sum_{j=0}^k q_j(t)f_j(x(t),x(h_j (t))).
$$
Taking into account condition (\ref{1.3}) we obtain
$$
x'(1)=\int_{0}^{1}x'(s)dg(s)=p(1)x'(1)\int_{0}^{1}\frac{1}{p(s)}dg(s)+
\int_{0}^{1}\frac{\lambda}{p(s)}\int_{s}^{1}z(r)dr\,dg(s),
$$
from which it follows that
$$ p(1)x'(1)=\gamma
\lambda\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}z(r)dr dg(s),
$$
where the constant $\gamma$ is
$$ \gamma :=\Big(\frac{1}{p(1)}-\int_{0}^{1}\frac{1}{p(s)}dg(s)\Big) ^{-1}.
$$
Then, from (\ref{2.1}) and (\ref{1.2}), we derive
$$
x(t)=\lambda\gamma\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}z(r)dr\,dg(s)
\int_{0}^{t}\frac{1}{p(s)}ds
+\lambda\int_{0}^{t}\frac{1}{p(s)}\int_{s}^{1}z(r)dr\,ds.
$$
This fact shows that if $x$ solves the boundary-value problem
(\ref{1.1})-(\ref{1.3}), then it solves the operator equation
$\lambda Ax=x$, where $A$ is the operator defined by
\begin{equation} \begin{aligned}
Ax(t):=&\gamma P(t)\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}\sum_{j=0}^k
q_j(r)f_j(x(r), x(h_j (r)))dr\,dg(s)\\
&+\int_{0}^{t}\frac{1}{p(s)}\int_{s}^{1}\sum_{j=0}^k
q_j(r)f_j(x(r), x(h_j (r)))dr\,ds.
\end{aligned} \label{2.2}
\end{equation}
Here we have set
$$ P(t):=\int_{0}^{t}\frac{1}{p(s)}\,ds, \quad t\in I.
$$

\begin{lemma} \label{lm2.1}
A function $x\in C(I)$ is a solution of the boundary value problem
(\ref{1.1})-(\ref{1.3}) if and only if $x$ solves the operator equation (\ref{1.4}),
where $A$ is defined by (\ref{2.2}). Also, any nonnegative
solution of (\ref{1.4}) is an increasing and concave function.
\end{lemma}

\paragraph{Proof}
The ``only if" part was shown above. For the ``if" part assume that $x$
solves (\ref{1.4}). Then, for every $t\in I$ we have
$$
x(t)=\lambda Ax(t)=\lambda\gamma
P(t)\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}z(r)dr\,dg(s)
+\lambda\int_{0}^{t}\frac{1}{p(s)}\int_{s}^{1}z(r)dr\,ds.
$$
Therefore
$$
x'(t)=\lambda\gamma
\frac{1}{p(t)}\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}z(r)dr\,dg(s)
+\lambda\frac{1}{p(t)}\int_{t}^{1}z(r)dr\,ds\,.
$$
and
$$(p(t)x'(t))'=-\lambda z(t)=-\lambda\sum_{j=0}^k q_j(t)f_j(x(t),
x(h_j (t))).
$$
Hence, if $x=\lambda Ax$, then $x$ satisfies (\ref{1.1}) and, moreover, since
$x(0)=\lambda Ax(0)=0$, it follows that $x$ satisfies
(\ref{1.2}). Also, for every $t\in I$ we have
\begin{align*}
\int_0^1x'(t)dg(t)=&\lambda\gamma\int_0^1\frac{1}{p(t)}dg(t)\cdot\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}z(r)dr\,dg(s)\\
&+\lambda\int_0^1\frac{1}{p(t)}\int_{t}^{1}z(r)dr\,dg(t)\\
=&\lambda\big [\gamma\int_0^1\frac{1}{p(t)}dg(t)+1\big
]\int_0^1\frac{1}{p(t)}\int_{t}^{1}z(r)dr\,dg(t)\\
=&\lambda\Big [ \frac{\int_0^1\frac{1}{p(t)}dg(t)}{\frac{1}{p(1)}-
\int_{0}^{1}\frac{1}{p(t)}dg(t)}+1\Big ]
\int_0^1\frac{1}{p(t)}\int_{t}^{1}z(r)dr\,dg(t)\\
=&\frac{\lambda\gamma}{p(1)}\int_{0}^{1}\frac{1}{p(t)}\int_{t}^{1}z(r)dr\,dg(t)=x'(1).
\end{align*}
Thus $x$ satisfies (\ref{1.3}). The additional properties, which the
lemma claims that any $x\ge 0$
with $x=\lambda Ax$ has, are implied from the fact that $x'\ge 0$,
$(p(t)x'(t))'\le 0$ and
Fact \ref{fact1.4}. We keep in mind that $\lambda>0$. \qed
\smallskip

  By using the continuity of the functions $f_j, q_j$ and $p$ it is not hard
to show that $A$ is a completely continuous operator.
\smallskip

Now consider the set
$$ \mathbb{K} :=\{x\in C(I): x(0)=0,\quad x\ge 0, \quad x'\ge 0
\quad\hbox{and $x$ concave}\},
$$
which, obviously, is a cone in $C(I)$. We show that the operator $\lambda A$
maps the cone $\mathbb{K}$ into
itself. Indeed we have the following statement.

\begin{lemma} \label{lm2.2}
Consider functions $p, g, f_j, q_j, h_j$, ($j=0,1,\dots ,k$),
satisfying the assumptions  (H1)-(H5).
Then
$$\lambda A(\mathbb{K} )\subset\mathbb{K} .$$
\end{lemma}

\paragraph{Proof}
Let $x\in \mathbb{K}$ be fixed. Then we observe that $Ax(0)=0$, $Ax\ge 0$
and $(Ax)'\ge 0$. Moreover,
since, obviously,
$\big ( p(t)(Ax)'(t)\big ) ' \le 0$ for all $t\in I$, by Fact \ref{fact1.4}, we
know that the function
$y=\lambda Ax$ is concave and the proof is complete.
\qed

\section{Existence Results}

Let $x$ be a function in the cone $\mathbb{K}$. Then $x$ is nondecreasing
and nonnegative, hence
$\|x\|=x(1)$. Also, from Lemma \ref{lm2.1} we have $\lambda Ax\in \mathbb{K}$,
thus $\|Ax\|=Ax(1)$.  But
then we have
\begin{align*}
\|Ax\|=Ax (1)=&\gamma
P(1)\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}\sum_{j=0}^k q_j(r)f_j(x(r),
x(h_j (r)))dr\,dg(s)\\
&+\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}\sum_{j=0}^k q_j(r)f_j(x(r),
x(h_j (r)))drds\\
=&\int_{0}^{1}\frac{1}{p(s)}\int_{s}^{1}\sum_{j=0}^k q_j(r)f_j(x(r),
x(h_j (r)))drdk(s),
\end{align*}
where $k(s):=s+\gamma P(1)g(s)$, $s\in I$. Applying Fubini's Theorem we get
\begin{equation}
\|Ax\|=  \int_{0}^{1}\sum_{j=0}^k q_j(s)f_j(x(s),
x(h_j (s)))R(s)ds,\label{3.1}
\end{equation}
where
$$ R(s):=\int_0^s \frac{1}{p(r)}\,dk(r).$$
Next, let $0<K<S<M< +\infty$ be fixed and define the functions
\begin{gather*}
\Phi(u,v):=sup\{f_0(u', v'): 0\le u'\le u, \quad 0\le v'\le v\},
\quad u,v\in [0,K]\\
\phi(u,v):=inf\{f_0 (u', v'): u\le u'\le M, \quad v\le v'\le M\},
\quad u,v\in [S,M].
\end{gather*}
It is clear that both the functions  $\Phi$ and $\phi$ are
nondecreasing with respect to their
variables and they satisfy
\begin{gather}
f_0 (u,v)\le\Phi(u,v), \quad  u,v\in [0,K]\label{3.2} \\
f_0 (u,v)\ge\phi(u,v), \quad  u,v\in [S,M]. \label{3.3}
\end{gather}
Also we make the following assumption
\begin{enumerate}
\item[(H6)] The following quantities are finite numbers:
$$L_ j := \sup_{0< v\le u<K}\frac{f_j (u,v)}{f_0 (u,v)}, \quad
j=1,2,\dots ,k,$$
\end{enumerate}
%
Now we define the set
$$E(S,M):=\big\{ \eta\in I : S<h_0 (\eta)M\big \},
$$
which maybe empty. Also we define the following real numbers:
\begin{gather*}
\zeta:=\sup\Big\{ h_0 (\eta)\int_{\eta}^1 q_0 (s)R(s)ds: \eta\in
E(S,M)\Big\},\\
\xi:=\int_0^1\big ( q_0 +\sum_{j=1}^k L_j q_j (s)\big ) R(s)ds,\\
b(S,M):=\frac{1}{\zeta}\sup_{u\in[S,M]}\frac{u}{\phi(u,u)},\\
B(K):=\frac{1}{\xi}\frac{K}{\Phi(K,K)}.
\end{gather*}

\begin{theorem} \label{thm3.1}
Assume that $p, g, f_j ,  q_j , h_j $ ($j=0,\dots ,k$) are functions which
satisfy assumptions  (H1)-(H6). Then for every $\lambda$ such that
$b(S,M)<\lambda< B(K)$,
the boundary value problem (\ref{1.1})-(\ref{1.3}) admits
at least one  positive, nondecreasing and concave solution $x$ such
that $K<\|x\|<M$.
\end{theorem}

\paragraph{Proof}
We set $T:=\lambda A$. First we shall prove that
\begin{equation}
 \hbox{if}\quad x\in \mathbb{K}\quad \hbox{and}\quad
\|x\|=M,\quad \hbox{then}\quad
\|Tx\|> M.\label{3.4}
\end{equation}
Indeed, assume on the
contrary, that there is a $x$ in $\mathbb{K}$ such that
\begin{equation}
 \|x\|=M \quad\hbox{and} \quad \|Tx\|\le M\label{3.5}
\end{equation}
and  consider any $\eta\in E(S,M)$ fixed. Then $\eta>0$ and
$h_0 (\eta)M>S$. Also for all $s\in [\eta,1]$ we have $h_0 (s)\ge h_0
(\eta):=\tau$. From the concavity
and monotonicity  of $x$ and Fact \ref{fact1.2} we have
$$ M=\|x\|\ge x (s) \ge x (h_0 (s))\ge\tau\|x\|=h_0 (\eta)M.
$$
Now taking into account (3.1) and (3.3) from (3.5) we get
\begin{align*}
M&\ge \lambda\int_{0}^{1}q_0 (s)f_0 (x(s), x(h_0 (s)))R(s)ds\\
&\ge\lambda\int_{\eta}^{1}q_0 (s)f_0
(x(s), x(h_0 (s)))R(s)ds\\ &\ge\lambda\phi\big ( h_0 (\eta)M, h_0 (\eta)M\big )
\int_{\eta}^{1}q_0 (s)R(s)ds.
\end{align*}
So
\begin{equation}
\frac{h_0 (\eta)M}{\phi\big ( h_0 (\eta)M, h_0 (\eta)M\big ) }\ge
\lambda h_0 (\eta)\int_{\eta}^{1}q_0 (s)R(s)ds. \label{3.6}
\end{equation}
This implies that
\begin{equation}
\sup_{u\in [S,M]}\frac{u}{\phi (u,u)}\ge
\lambda \zeta,\label{3.7}
\end{equation}
which contradicts to the fact that $\lambda>b(S,M)$. Thus (3.4) holds.

Next we claim that
\begin{equation}
\hbox{if}\quad x\in \mathbb{K}\quad \hbox{and}\quad
\|x\|=K,\quad \hbox{then}\quad
\|Tx\|<K.\label{3.8}
\end{equation}
Indeed if not, then assume that for some $x\in \mathbb{K}$ with
$\|x\|=K$ we have $\|Tx\|\ge K $. The first
one implies that $0\le x(s)\le K$,
$s\in I$ and so, taking into account $(3.1)$, $(3.2)$, assumptions
(H5), (H6) and the fact that $x$
is nondecreasing we get in any case that
\begin{equation}
\begin{aligned}
K &\le
{\lambda}\int_{0}^{1}\sum_{j=0}^k q_j(s)f_j
(x(s),x(h_j (s)))R(s)ds\\
&\le \lambda\int_{0}^{1}\sum_{j=0}^k q_j(s)f_j
(x(s),x(h_0 (s)))R(s)ds\\
&\le \lambda\int_{0}^{1}\Big( q_0 (s)+\sum_{j=1}^k L_j q_j(s)\Big)f_0
(x(s),x(h_0 (s)))R(s)ds.
\end{aligned} \label{3.9}
\end{equation}
Now, from (3.2) it follows that
\begin{equation}
 K\le\lambda\Phi(K,K)\int_{0}^{1}\Big ( q_0
(s)+\sum_{j=1}^k L_j q_j(s)\Big) R(s)ds =\lambda\Phi(K,K)\xi.\label{3.10}
\end{equation}
Hence it holds
$$ \frac{K}{\Phi(K,K)}< \lambda \xi, $$
which  contradicts to the fact that $\lambda <B(K)$ and  our claim is proved.

Finally, we set $\Omega _1 :=\{x\in C(I): \|x\|<K\}$ and $\Omega _2
:=\{x\in C(I): \|x\|<M\}$. Note that $K<M$. Taking into account
that $T$ is a completely continuous operator and Lemma \ref{lm2.2},
 from Theorem \ref{thm1.1} we conclude
that there exists a solution $x$ of the boundary value problem $(\ref{1.1})-(\ref{1.3})$
such that $K\le\|x\|\le M$.  From (3.4) and (3.8) we see that
equalities $\|x\|=K$ and $\|x\|=M$ cannot
hold.\qed \smallskip

One of the main questions, on the existence problem
solved above, is whether we
may enlarge the set of eigenvalues. Partial answers to this question
are given in the
following theorems.

\begin{theorem} \label{thm3.2}
Assume that $p,g,f_j ,q_j ,h_j$, ($j=0,\dots ,k$) satisfy assumptions
(H1)-(H6) and moreover that
$$ f_0(u,v)=0\quad and \quad u\ge v\ge 0\quad  imply \quad v=0.
$$
Also let $h_0$ be a continuous function, with $h_0 (1)>0$.
If $b(S,M)\le \lambda< B(K))$,
then there is a  positive, nondecreasing and concave solution $x$ of the boundary value
problem (\ref{1.1})-(\ref{1.3}) such that $K<\|x\|<M$.
\end{theorem}

\paragraph{Proof}
As in Theorem \ref{thm3.1} we have that $x\in \mathbb{K}$ and $\|x\|=K$
imply $\|Ax\|<K$ and we will show
that $x\in \mathbb{K}$ and $\|x\|=M$ imply $\|Ax\|>M$.
To do this we proceed as in Theorem \ref{thm3.1} and obtain (3.6).

Now, if for some $\eta'\in I$ equality holds, we must have  $h_0
(\eta)>\frac{S}{M}$ and
$$ \int_0^{\eta} q_0 (s)f_0 (x(s), x(h_0 (s)))R(s)ds=0
$$
for all $\eta\in [\eta', 1]$. Since $q_0(s)\ge 0$ and $q_0(1)>0$
it follows that for all $s$ close to $1$ it holds
$$f_0(x(s), x(h_0(s)))=0
$$
and so, by our hypothesis we have $x(h_0(s))=0$ for all $s>0$ close to $1$.
This gives $h_0(s)=0$, because  of Fact \ref{fact1.2}, hence, by continuity,
$h_0(1)=0$, a contradiction.

 Therefore in (3.6) we have the strict inequality. Since both sides are
continuous functions of
$\eta$, it follows that (3.7) holds as a strict inequality,
which contradicts to $b(S,M)\le\lambda$.
Now the  result follows as in Theorem \ref{3.1}.\qed

\begin{theorem} \label{thm3.3}
 Assume that $p,g,f_j ,q_j ,h_j$, ($j=0,\dots ,k$) satisfy (H1)-(H6) and
moreover assume that there is an index $j_1\in\{1,\dots ,k\}$ such  that
$\mathop{\rm meas}\{s\in I: q_{j_1} (s)\ne 0\}>0$ and
for all $u\in (0,K]$ and
$v\in (0,u]$ it holds
\begin{equation}
 f_{j_1}(u,v)<L_{j_1}f_0 (u,v).\label{3.11}
\end{equation}
If $b(S,M)<\lambda\le B(K)$,
then there is a  positive, nondecreasing and concave solution $x$ of the
boundary-value problem (\ref{1.1})-(\ref{1.3}) such that $K<\|x\|<M$.
\end{theorem}

\paragraph{Proof}
As in Theorem \ref{thm3.1} we can show  that $x\in \mathbb{K}$ and $\|x\|=M$
imply $\|Ax\|>M$. It remains to show
that if $x\in \mathbb{K}$ and $\|x\|=K$, then $\|Ax\|<K$.
To do this we obtain $(3.10)$ and we will show that  equality  cannot hold.
Indeed if it is so, then taking into account (3.9) we
conclude that for all $j=1,\dots ,k$  it holds
$$ q_j(s)f_j(x(s), x(h_j(s)))=q_j(s)L_j f_0(x(s), x(h_j(s))),
$$
where all these quantities are nonnegative. But for $j=j_1$, (3.11) cannot
be true, thus in (3.10) we have the strict inequality, a contradiction.
\qed

\section{Uniqueness and continuous dependence Results}

Here we give results on the uniqueness and the continuous dependence of the
solutions on the eigenvalues. For this we make the following condition.
\begin{enumerate}
\item[(H7)]
For every $j=0,\dots ,k$ there exist real nonnegative constants
$\rho_j , \sigma_j$ such that
$$ |f_j(u,v)-f_j(u',v')|\le \rho_j |u-v|+ \sigma_j |u'-v'|
$$
for all $u,v,u',v'\in [0,+\infty)$.
\end{enumerate}

\begin{theorem} \label{thm4.1}
Assume that $p,g,f_j ,q_j ,h_j, \rho_j, \sigma_j $, ($j=0,\dots ,k$) satisfy
(H1)-(H7) and
\begin{equation}
c:=B(K)\sum_{j=0}^k (\rho_j +\sigma_j)\int_0^1q_j(s)R(s)ds<1.\label{4.1}
\end{equation}
Then for every $\lambda\in (b(S,M), B(K))$ there exists exactly one positive
(nondecreasing and concave) solution $x_{\lambda}$  of the boundary value
problem $(\ref{1.1})-(\ref{1.3})$. Also the function $\lambda\to x_{\lambda}$ is
uniformly continuous.
\end{theorem}

\paragraph{Proof}
It is clear that conditions (H7) and (4.1) imply that the operator
$T=\lambda A$ is a contraction. Hence, by the Contraction Principle, for every
$\lambda\in (b(S,M), B(K))$ the solution $x_{\lambda}$, say, obtained by
Theorem \ref{thm3.1} is unique.

Now, consider the solutions $x_{\lambda_1},x_{\lambda_2}$, where
$\lambda_1, \lambda_2 \in (b(S,M),B(K))$. Then, for any $t\in I$ we have
\begin{align*}
|x_{\lambda_1}(t)-x_{\lambda_2}(t)|&\le |\lambda_1 -\lambda_2
||Ax_{\lambda_1}(t)|+B(K) |Ax_{\lambda_1}(t)|-Ax_{\lambda_2}(t)|\\
&\le\frac{|\lambda_1
-\lambda_2|}{b(S,M)}\|x_{\lambda_1}\|+c\|x_{\lambda_1}-x_{\lambda_2}\|.
\end{align*}
Therefore,
$$
\|x_{\lambda_1}-x_{\lambda_2}\|\le
\frac{1}{b(S,M)(1-c)}|\lambda_1 -\lambda_2|
$$
which completes our proof. \qed

\section{Some Applications}
\textbf{(a)}\quad Consider the equation
\begin{equation}
 x''(t)+\lambda \Big [ x^{\mu}(t)x^{\nu}(h(t))+\theta
\big|sin[x(t)x(h(t))]\big|
x^{\mu-1}(t)x^{\nu-1}(h(t))\Big]=0,
\quad t\in I,\label{5.1}
\end{equation}
 associated with the conditions (\ref{1.2}) and (\ref{1.3}). The function $h$ is any
retardation, $\theta\ge  0$ and $\mu,\nu>0 $
with $\mu+\nu>1$. Also consider the constants $\xi$ and $\zeta$ as
in Theorem \ref{thm3.1}. Here we have
\begin{gather*}
f_0(u,v)=\phi(u,v)=\Phi (u,v):=u^{\mu}v^{\nu}, \\
f_1(u,v):= |sin(uv)|u^{\mu-1}v^{\nu-1}.
\end{gather*}
Take any constants
$\epsilon, \Theta$ such that $0<\epsilon\zeta <\Theta\xi <+\infty$ and consider
constants $K,S(>0)$ so that
$$ K^{\mu+\nu-1}<(\Theta\xi)^{-1}<(\epsilon\zeta)^{-1}<S^{\mu+\nu-1}.
$$
Fix any $M>S$. Then observe that $L_1=1$, as well as
$$ b(S,M)=\frac{1}{\zeta}\sup_{u\in [S,M]}\frac{u}{\Phi(u,u)}
=\frac{1}{\zeta}\sup_{u\in [S,M]}\frac{1}{u^{\mu+\nu-1}}
=\frac{1}{\zeta}\frac{1}{S^{\mu+\nu-1}}<\epsilon
$$
and
$$ B(K)=\frac{1}{\xi}\frac{1}{K^{\mu+\nu-1}}>\Theta.
$$
Since $\epsilon, \Theta$ are arbitrary, we have the following statement.

\begin{corollary} \label{coro5.1}
Assume that $g,h:I\to \mathbb{R}$ are nondecreasing functions (with $h$
not identically zero) and such
that $g(1)-g(0)<1$ and $0\le h(t)\le t$, $t\in I$.
Then for every $\lambda >0$ the boundary value problem
(5.1),(\ref{1.2}),(\ref{1.3}) admits at least one  positive, nondecreasing and
concave solution $x$ such that $K<\|x\|<M$.
\end{corollary}

\noindent\textbf{(b)} \quad Consider the retarded differential equation
\begin{equation}
 x''(t)+\lambda[x^{m+1} (t)+\rho x^{n+1}(t^2)]=0, \quad t\in
[0,1],\label{5.2}
\end{equation}
with initial condition (\ref{1.2}), i.e. $x(0)=0$ and the boundary condition
\begin{equation}
 x'(1)=\delta x(1)\label{5.3}
\end{equation}
where $m, n, \lambda, \rho, \delta$ are real positive numbers with
$\delta<1$.
To apply Theorem \ref{thm3.1} we
write this problem in the form (\ref{1.1})-(\ref{1.3}) by setting
$$ p(t):=1, \quad q_0 (t):=1, \quad g(t):=\delta t,\quad h_0
(t):=t^2,\quad t\in I,
$$
and
$$ f_0(u,v):=\phi(u,v)=\Phi(u,v)=u^{m+1}+\rho v^{n+1}, \quad
f_j(u,v):=0, \quad (j=1,\dots ,k).
$$
Then we obtain
$$
\gamma=\frac{1}{1-\delta} \quad{\text{and}} \quad
R(t)=\frac{t}{1-\delta},\quad t\in I.
$$
Choose $S:=1$, $M:=2$ and observe that $E(S,M)= (2^{-1/2}, 1]$.
Thus, we have
\begin{gather*}
\zeta=\sup\big\{ \frac{1}{2(1-\delta)}\eta^2 (1-\eta^2 ): \quad \eta\in
   (2^{-1/2}, 1] \big\}=\frac{1}{8(1-\delta)}, \\
\xi=\frac{1}{2(1-\delta)}, \hskip .2 in\sup_{u\in [1,2]}\frac{u}{\phi
(u,u)}=\frac{1}{1+\rho} \quad
 \frac{K}{\phi(K,K)}=\frac{1}{K^{m+1}+\rho K^{n+1}}.
\end{gather*}
Therefore, Theorem \ref{thm3.2} applies and the following result follows.

\begin{corollary} \label{coro5.2}
Assume that $m, n, \lambda, \rho, \delta$ are real positive numbers
with $\delta<1$.
Then for any $K>0$, with $$K^{m+1}+\rho K^{n+1}<\frac{1}{4}(1+\rho)$$ and any
$\lambda$ such that
$$ \frac{8(1-\delta)}{1+\rho}\le\lambda<\frac{2(1-\delta)}{K^{m+1}+\rho
K^{n+1}},
$$
the boundary value problem (5.2), (\ref{1.2}), (5.3)  admits
at least one  solution $x$ which is a positive, nondecreasing and
concave function such that $K<\|x\|<2$.
\end{corollary}

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\end{document}