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\markboth{\hfil Positive solutions of Fredholm integral equations
\hfil EJDE--2002/30}
{EJDE--2002/30\hfil G. L. Karakostas \& P. Ch. Tsamatos \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 30, pp. 1--17. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
 Multiple positive solutions of some Fredholm integral
 equations arisen from nonlocal boundary-value problems
 %
\thanks{{\em Mathematics Subject Classifications:} 45M20, 34B18.
\hfil\break\indent
{\em Key words:} Integral equations, multiple solutions, nonlocal
boundary-value problems.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted January 22, 2002. Published March 21, 2002.} }
\date{}
%
\author{G. L. Karakostas \& P. Ch. Tsamatos}
\maketitle

\begin{abstract}
  By applying the Krasnoselskii's fixed point theorem on a
  suitable cone, several existence results for multiple positive
  solutions  of a Fredholm  integral equation are provided.
  Applications of these results to some nonlocal boundary-value
  problems are also given.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We study the existence of multiple positive solutions of the
abstract Fredholm integral equation
\begin{equation*}
x(t)=\int_0^1 K(t,s )f(x(s ))ds, \enskip t\in [0,1],\tag{1.1}
\end{equation*}
where the kernel $K(t,s)$ satisfies a continuity
assumption in the $L^1$-sense and it is monotone and concave in $t$
for a.a. $s$. Then we apply our
results to the second order ordinary differential equation
\begin{equation*}
(p(t)x'(t))'+\mu(t)f(x(t))=0,\enskip t\in [0,1],\tag{1.2}
\end{equation*}
associated with the nonlocal boundary conditions
\begin{equation*}
x'(0)=\int_0^1x'(s)dg(s)\quad\text{and}\quad
x(1)=-\int_0^1x'(s)dh(s) \tag{1.3}
\end{equation*}
or
\begin{equation*}
x'(1)=\int_0^1x'(s)dg_1(s)\quad\text{and}\quad
x(0)=\int_0^1x'(s)dh_1(s).\tag{1.4}
\end{equation*}
Notice that these problems are our motivation in investigating the
abstract problem (1.1). Here $f:\mathbb{R}\to\mathbb{R}$,
$p:[0,1]\to (0,\infty )$,
$\mu:[0,1]\to [0,\infty )$  are continuous functions and $g,
h,g_1,h_1\colon [0,1]\to \mathbb{R}$ are nondecreasing functions.
In these boundary conditions all
integrals are meant in the sense of
Riemann-Stieljes.

Boundary-value problems with integral boundary conditions
constitute a very interesting class of  problems,
because they include as special cases two,
three, multi-point and nonlocal boundary-value problems. For such
problems and comments on their
importance, we refer to the recent papers \cite{g1,k2,k8,l2}. Especially,
the existence of positive solutions for
such problems is the subject of several recent papers
\cite{c1,k1,k2,k3,k4,k5,k6,k8,k9,m1}. Moreover, since a boundary-value
problem may usually be reduced to an integral equation, the existence
of positive solutions for a boundary-value problem is closely connected to
an analogous problem for an
integral equation. Indeed, the existence
of positive solutions for integral equations, or, generally, for
operator equations, is a problem which
appeared early in the literature. For more details we refer to the
books by Krasnoselskii \cite{k9} and Agarwal,
O'Regan and Wong \cite{a1}, as well as to the recent papers \cite{
a2,a3,l1,m2,o1}
and the references therein. For more
information about the general theory of integral equations and their
relation with boundary-value problems we refer to the book of Corduneanu
\cite{c2} on finite intervals
and the most recent book of Agarwal
and O'Regan \cite{a4} on infinite intervals. We find it convenient to
compare a little our setting with
the ideas expressed in \cite{a2}. The subject studied  in \cite{a2}
is an
integral equation of the form
\begin{equation*}
y(t)=\theta(t)+\int_0^1k(t,s)[g(y(s))+h(y(s))]ds, \enskip t\in
[0,1],\tag{1.4}
\end{equation*}
where $h,g$ are continuous
functions, $g$ is sub-multiplicative on  $(0,+\infty)$ and
these two functions and $h/g$
satisfy some monotonicity conditions. The procedure in \cite{a2} is to
succeed existence via the solutions of an
approximation of (1.4). In the present paper we do not assume any
monotonicity condition on $f$ and our
existence results are obtained by a direct application of the
Krasnoselskii's theorem. Notice that, as one can
  see, all conditions required for our integral problem (1.1) are
inherited from the corresponding properties of
the boundary-value problems (1.2)-(1.3) and (1.2)-(1.4).

And indeed the boundary-value problems (1.2)-(1.3) and (1.2)-(1.4)
(though they are equivalent in a sense
explained in the last section) are reduced to the abstract Fredholm
integral equation (1.1).
That is why we investigate when equation (1.1) admits multiple
positive solutions. The conditions established on the kernel $K$ and
the force $f$ are rather simple and such
that the obtained results are original and may give as corollaries
new results for the boundary-value problems (1.2)-(1.3) and (1.2)-(1.4). Our
paper is
motivated mainly by the papers \cite{k6,k9,l2} and, among others, our
results generalize and improve
several recent results of the authors \cite{k7} and Ma and Castaneda \cite{
m1}.

 The results here are obtained by applying the well known fixed point
theorem due to Krasnoselskii \cite{k9}, which states as follows:

\begin{theorem} \label{thm1.1}
Let $\mathcal{B}$ be a Banach space and let  $\mathbb{K}$  be a cone in
$\mathcal{B}$. Assume $\Omega _1 $,
$\Omega _2 $ are open subsets of $\mathcal{B}$, with
$0\in\Omega _1\subset \mathop{\rm cl}{\Omega _1}\subset \Omega _2$, and let
$$A\colon\enskip\mathbb{K}\cap (\mathop{\rm cl}\Omega _2\setminus  {\Omega
_1 } )\to
\mathbb{K}$$
be a completely continuous operator such that either
$$\|Au\|\le \|u\|,\enskip u\in \mathbb{K} \cap \partial \Omega
_1\quad\text{and}\quad\|Au\|\ge
\|u\|,\enskip u\in \mathbb{K} \cap
\partial \Omega _2, $$
or
$$\|Au\|\ge \|u\|,\enskip u\in \mathbb{K} \cap
\partial \Omega _1
\quad\text{and}\quad \|Au\|\le
\|u\|,\enskip u\in \mathbb{K} \cap
\partial \Omega _2 .$$
Then $A$ has a fixed point in $\mathbb{K}\cap (\mathop{\rm cl}\Omega
_2\setminus
{\Omega _1 } )$.
\end{theorem}

This paper is organized as follows: In Section 2 we show how the
boundary-value problems
(1.2)-(1.3) and (1.2)-(1.4) can be written as a Fredholm integral
equation of the form (1.1). Section 3
contains the basic lemmas concerning equation (1.1). These lemmas
imply several corollaries, which
in Section 4 lead to our main existence results. In Section 5 we
specify the results of Section 4 to the
boundary-value problems (1.2)-(1.3) and (1.2)-(1.4) and we close the
paper with a discussion.

\section{The boundary-value problems (1.2)-(1.3) and (1.2)-(1.4)
reformulated}

In what follows we shall denote by $\mathbb{R}$ the real
line, by $I$ the interval $[0,1]$ and by $C(I)$ the space of all
continuous functions
$x:I\to\mathbb{R}$.
The space $C(I)$ endowed  with the sup-norm $\|\cdot \| $ is a Banach
space. Also we denote
by $L^1(I)$ the space of all functions $x:I\to \mathbb{R}$ which are
Lebesgue
integrable on $I$ endowed with the usual norm $\|\cdot\|_{L^1}$.

  In the sequel we shall use nondecreasing functions $g,h,g_1,h_1:I\to
[0,\infty)$, with
$g(0)=h(0)=g_1(1)=h_1(1)=0$ and the continuous function $p :I\to
(0,\infty)$ such that
\begin{equation*}
\int_0^1\frac{1}{p(s)}dg(s)<\frac{1}{p(0)},\tag{2.1}
\end{equation*}
 if we have
the conditions (1.3) and
\begin{equation*}
\int_0^1\frac{1}{p(s)}dg_1(s)<\frac{1}{p(1)},\tag{2.2}
\end{equation*}
if we have the conditions (1.4).

As we pointed out in the introduction, to search for solutions
to  (1.2)-(1.3) and (1.2)-(1.4), we first reformulate them as
operator equations of the form $x=Ax$, where $A$ is an
appropriate integral operator.

 To find operator $A$ consider first the boundary-value problem
$(1.2)-(1.3),$
where for simplicity we put
$$z(t):=\mu(t)f(x(t)),\enskip t\in I. $$
Integrate $(1.2)$ from $0$ to $t$ and get
\begin{equation*}
x'(t)=q(t)p(0)x'(0)-q(t)\int_0^t z(s)ds,\tag{2.3}
\end{equation*}
where we have set
$q(t):=1/p(t)$, $t\in I$.
Taking into account the first condition in $(1.3)$ we obtain
$$x'(0)=p(0)x'(0)\int_0^{1}q(s)dg(s)-\int_0^{1}q(s)\int_0^s z(\theta
)d\theta dg(s)$$
or
$$p(0)x'(0)\left [ q(0)-\int_0^{1}q(s)dg(s)\right ]
=-\int_0^{1}q(s)\int_0^s z(\theta )d\theta
dg(s)$$ and so
$$p(0)x'(0)=-\alpha\int_0^{1}q(s)\int_0^s z(\theta )d\theta dg(s),$$
where $\alpha$ is a constant defined by
$$\alpha :=\frac{1}{q(0)-\int_0^1 q(s)dg(s)}.$$
Then, from (2.3) we get
$$x'(t)=-\alpha q(t)\int_0^1 q(s)\int_0^s z(\theta )d\theta
dg(s)-q(t)\int_0^t z(\theta)d\theta . $$
Thus for $t\in I$, we have
$$x(t)=x(0)-\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta
dg(s)\int_0^t q(r)dr-\int_0^t
q(s)\int_0^s z(\theta )d\theta ds, $$
and so
$$x(1)=x(0)-\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta
dg(s)\int_0^1 q(r)dr-\int_0^1
q(s)\int_0^s z(\theta )d\theta ds.$$
On the other hand taking into account the second condition in (1.3) we
obtain
\begin{align*}
x(1)&=- \int_0^1\Big ( -\alpha q(s)\int_0^1 q(r)\int_0^r z(\theta
)d\theta dg(r)-q(s)\int_0^s
z(\theta )d\theta\Big ) dh(s)=\\
&=\alpha\int_0^1 q(s)dh(s)\int_0^1q(r)\int_0^r z(\theta )d\theta
dg(r)+\int_0^1
q(s)\int_0^s z(\theta )d\theta dh(s).
\end{align*}
Therefore it follows that
\begin{align*}
x(0)&=\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta dg(s)\int_0^1
q(r)dr+\int_0^1
q(s)\int_0^s z(\theta )d\theta ds+\\
&+\alpha \int_0^1 q(s)dh(s)\int_0^1 q(r)\int_0^r z(\theta
)d\theta dg(r)+\int_0^1 q(s)\int_0^s z(\theta )d\theta dh(s).\\
\end{align*}
Hence the solution $x$ has the form
\begin{align*}
x(t)=&\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta dg(s)\int_0^1 q(r)dr+\\
&+\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta dg(s)\int_0^1 q(r)dh(r)+\\
&+\int_0^1 q(s)\int_0^s z(\theta )d\theta ds+\int_0^1 q(s)\int_0^s
z(\theta )d\theta dh(s)-\\
&-\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta dg(s)\int_0^t
q(r)dr- \int_0^t q(s)\int_0^s
z(\theta )d\theta ds=\\
=&\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta dg(s)\int_t^1 q(r)dr+\\
&+\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta
dg(s)\int_0^1q(r)dh(r)+\int_0^1q(s)\int_0^sz(\theta
)d\theta dh(s)+\\  &+\int_t^1 q(s)\int_0^s z(\theta )d\theta ds,
\enskip t\in I.
\end{align*}
Now for simplicity we set
$b:=\int_0^1q(s)dh(s)$
and get
\begin{align*}
x(t)&=\alpha \int_0^1 q(s)\int_0^s z(\theta )d\theta dg(s) \Big (
b+\int_t^1 q(s)ds\Big )+\\
  &\hskip .2 in +\int_0^1 q(s)\int_0^s z(\theta)d\theta dh(s)+
\int_0^1q(s)\chi_{[t,1]}(s)\int_0^s z(\theta
)d\theta ds=\\  &=\int_0^1 q(s)\int_0^s z(\theta )d\theta
d_{s}\rho(t,s), \enskip t \in I,
\end{align*}
where
$$\rho(t,s):=\alpha g(s)\Big( b+\int_t^1
q(r)dr\Big)+h(s) +s\chi_{[t,1]}(s).$$
Finally by using Fubini' s theorem we obtain
$$
x(t)=\int_0^1 z(\theta )\int_{\theta}^1q(s)d_{s}
\rho(t,s)d\theta=\int_0^1 f(x(\theta
))\Big(\mu(\theta)\int_{\theta}^1q(s)d_{s} \rho(t,s)\Big)d\theta.$$
  This is an equation of the form (1.1), where the kernel
$K(t,\theta)$ is defined by
\begin{equation*}
K(t,\theta):=\mu(\theta)\int_{\theta}^1q(s)d_s \rho(t,s).\tag{2.4}
\end{equation*}
Here we have the following:

\begin{theorem} \label{thm2.1}
A function $x\in C^1 (I)$ is a solution of the boundary-value problem
$(1.2)-(1.3)$ if and only
if $x$ is a solution of equation $(1.1)$, whose the kernel $K$ is
given by $(2.4)$.
\end{theorem}

\paragraph{Proof.}
The {\it{only if}} part is proved above. For the {\it{if}} part,
first we assume that $x:I\to \mathbb{R}$ is a
continuous function satisfying (1.1). Then $x$ is differentiable with
derivative
$$x'(t)=\int_0^1\frac{
\partial K(t,\theta)}{
\partial t}f(x(\theta))d\theta,$$
where
\begin{equation*}
\frac{
\partial K(t,\theta)}{
\partial t}=
\begin{cases}
  -(\alpha \phi(\theta)+1)\mu(\theta)q(t),& \mbox{if } \theta<t\\
-\alpha \phi(\theta)\mu(\theta)q(t),& \mbox{if } t<\theta
\end{cases}
\tag{2.5}
\end{equation*}
and
$\phi(\theta ):=\int_{\theta}^1 q(r)dg(r)$.
Thus we get
\begin{equation*}
x'(t)=q(t)\Big [ -\alpha\int_0^1 z(\theta )\phi(\theta )d\theta
-\int_0^t z(\theta
)d\theta\Big ],\tag{2.6}
\end{equation*}
where, recall that, $z(t)=\mu(t)f(x(t)),\enskip t\in I$.

Now, from (2.6) it follows that equation (1.2) is satisfied. Moreover
the first condition in (1.3) is
satisfied. Indeed, from the obvious relation
$$-\alpha q(0)+1=-\alpha\int_0^1q(s)dg(s)$$
we get
\begin{align*}
\int_0^1x'(s)dg(s)&
=-\alpha\int_0^1
z(\theta)\phi(\theta)d\theta\int_0^1q(s)dg(s)-\int_0^1q(s)\int_0^s
z(\theta)d\theta dg(s)=\\
&=-\alpha q(0)\int_0^1 z(\theta)\phi(\theta)d\theta +\int_0^1
z(\theta)\phi(\theta)d\theta -\\
&\hskip .15 in-\int_0^1q(s)\int_0^s z(\theta)d\theta dg(s)=\\
&=x'(0)+\int_0^1z(\theta)\int_{\theta}^1q(s)dg(s)d\theta-\int_0^1q(s)\int_0^s
z(\theta)d\theta dg(s)=\\
&=x'(0),
\end{align*}
since from Fubini's theorem we have
\begin{equation*}
\int_0^1z(\theta)\int_{\theta}^1q(s)dg(s)d\theta=\int_0^1q(s)\int_0^s
z(\theta)d\theta dg(s). \tag{2.7}
\end{equation*}
Finally, we can use (2.7) with $h$ in place of $g$ to show that the
second condition in (1.3) is also
satisfied.
\hfill$\Box$ \smallskip

Next following the same procedure as above it is not hard to see that
a function $x\in C^1(I)$ is a solution of
the boundary-value problem $(1.2)-(1.4)$ if and only if $x$ is a
solution of equation $(1.1)$, where now the
kernel $K$ is given by
\begin{equation*}
K(t,\theta):=\mu(\theta)\int_{0}^{\theta}q(s)d_s \rho_1(t,s),\tag{2.8}
\end{equation*}
  the measure $\rho_1$ is defined by
$$\rho_1(t,s):=\alpha_1 g_1(s)\Big( b_1+\int_0^{t}q(r)dr\Big)
+h_1(s) +s\chi_{[0,t]}(s)$$
and the constants $\alpha_1, b_1$ are given by
$$\alpha_1:=\frac{1}{q(1)-\int_0^1q(s)dg_1(s)}\quad \text{and}\quad
 b_1:=\int_{0}^{1}q(s)dh_1(s).$$

\section{On the Fredholm integral equation (1.1)}

In this section we present some auxiliary facts needed to show the
existence of solutions of the
general integral equation (1.1).
Then we will return to the specific problems (1.2)-(1.3) and (1.2)-(1.4).

To unify our results in the cases appeared later, we
consider a fixed number $\delta\in\{-1,+1\}$
and assume that the following conditions hold:
\begin{enumerate}
\item[($H_f$)]  $f:\mathbb{R}\to \mathbb{R}$ is a continuous function
with
$f(x)>0$, when $x>0$.

\item[($H_K$)]   $K(\cdot,\cdot)$ maps the square $I\times I$ into
$(0,+\infty)$ and it is such that:
\begin{enumerate}
\item[a)] For a.a. $s\in I$ the function $K(\cdot, s)$ is
concave and the function $\delta K(\cdot,s)$ is
non-increasing.

\item[b)] For all $t\in I$ we have $K(t, \cdot)\in L^1(I)$ and
the function
$t\to K(t,\cdot)$ is uniformly $L^1$-continuous.
\end{enumerate}
\end{enumerate}
Next define the cone
$$\mathbb{K}_{\delta}:=\{x\in C(I): x\ge 0, \enskip x \enskip \hbox{is
concave and} \enskip \delta x\enskip
\hbox{is nonincreasing}\}$$
as well as the operator
$$(Ax)(t):=\int_0^1K(t,s)f(x(s))ds,\enskip x\in C(I).$$

\begin{lemma} \label{lm3.1}
Under the assumptions ($H_f$) and ($H_K$) the operator $A$ is
completely continuous and
it maps $\mathbb{K}_{\delta}$ into $\mathbb{K}_{\delta}$.
\end{lemma}

  \paragraph{Proof.}
For all $x,y\in C(I)$ we have
$$\big |(Ax)(t)-(Ay)(t)\big | =\Big |
\int_0^tK(t,s)(f(x(s))-f(y(s)))ds\Big | \le
k\hskip.02 in\|f(x(\cdot))-f(y(\cdot))\|,$$ where
$$k:=\sup_{t\in I}\int_0^1K(t,s)ds \quad(<+\infty).$$
Thus $A$ is a continuous operator.

Let $B$ be a
bounded subset of $C(I)$. Then there is a certain $c>0$ such that
$|x(t)|\le c,$ for all $x\in B$
and $t\in I$. So we have
$$|(Ax)(t)|\le k\ \sup_{0\le\xi\le c}|f(\xi)|$$
for all $x\in B$ and $t\in I$.

Also, let $x\in B$. Then, for all $t_1 , t_2 \in I,$ it holds
$$|(Ax)(t_1 )-(Ax)(t_2 )|\le \|K(t_1 ,\cdot)-K(t_2
,\cdot)\|_{L^1}\sup_{0\le\xi\le c}|f(\xi)|.$$
Therefore $A$ maps bounded sets into compact sets. The previous facts
show the complete continuity of
$A$.
 Let $x\in \mathbb{K}_{\delta}$. Then $f(x(s))\ge 0$, $s\in I$
and therefore $Ax(t)\ge 0$, $t\in I$.

Also, let $E\subset I$ be a set with Lebesgue measure zero such that for all
$s\in I\setminus E$ it holds
$$\delta K(t_1 ,s)\le \delta K(t_2 ,s),\enskip \hbox{whenever}\enskip
t_2\le t_1,$$
as well as
$$K(\lambda t_1+(1-\lambda)t_2, s)\ge \lambda
K(t_1,s)+(1-\lambda)K(t_2,s),\enskip
\hbox{for all }
t_1, t_2\in I, \enskip \lambda\in [0,1].$$
%
These facts are guaranteed by ($H_K$). The first inequality implies
that $\delta Ax$ is a non-increasing
function and the second one that
$Ax$ is concave. The proof is complete. \hfill$\Box$ \smallskip

Next we let $i:=\frac{1}{2}(1-\delta)$
and  observe that, for each $x\in \mathbb{K}_{\delta}$, it holds
$$\|x\|=x(i).$$
Define the
continuous function
$$\Phi_i(\theta):=\Big|\int_i^\theta K(i,s)ds\Big|, \enskip \theta \in I.$$
and make the following assumptions:
\begin{enumerate}
\item[(H1)] There is a $v>0$ such that
$\Phi_i(1-i)\sup_{\xi\in [0,v]}f(\xi )< v$.

\item[(H2)] There are $\eta\in (0,1)$ and $u>0$ such that
$$\Phi_i(\eta)\inf_{\xi\in [\zeta_i u,u]}f(\xi)> u,$$
where $\zeta_i:=1-i+\eta(2i-1)$.
\end{enumerate}

\begin{lemma} \label{lm3.2}
For each $x\in \mathbb{K}_{\delta}$ it holds
$$\zeta_i\|x\|\le x(t), \enskip t\in [i\eta,i+(1-i)\eta].$$
\end{lemma}

\paragraph{Proof.}
Concavity of $x$ on $I$ implies that
$$\frac{x(1)-x(0)}{1}\ge\frac{x(1)-x(\eta)}{1-\eta},$$
which gives
$$x(\eta)\ge (1-\eta)x(0)+\eta x(1).$$
Since $x$ is nonnegative, we get
$$x(\eta)\ge(1-\eta)x(0) \enskip {\text{and}}\enskip x(\eta)\ge\eta
x(1),$$ which is written as
\begin{equation*}
 x(\eta)\ge \zeta_i x(i). \tag{3.1}
\end{equation*}
Now, if $\delta=+1,$ we have $i=0,$ $\zeta_i=1-\eta,$ $x$ is
non-increasing and $\|x\|=x(0).$ If $\delta=-1,$
then $i=1$, $\zeta_i=\eta,$ $x$ is nondecreasing and $\|x\|=x(1).$ It
is clear that these facts together with
(3.1) complete the proof.
\hfill$\Box$

\begin{lemma} \label{lm3.3}
If the  assumptions ($H_f$), ($H_K$) and (H1) hold, then for all
$x\in\mathbb{K}_{\delta}$,
with $\|x\|=v$, we have $\|Ax\|<\|x\|$.
\end{lemma}

\paragraph{Proof.}
If $\|x\|=v$, then $0\le x(s)\le v$, for every $s\in I$ and, by
(H1) and Lemma \ref{lm3.1}, we have
\begin{align*}
\|Ax\|&=(Ax)(i)=\int_0^1 K(i,\theta)f(x(\theta))d\theta\le\\
&\le sup_{\xi\in [0,v]}f(\xi)\int_0^1 K(i,\theta)d\theta=
\Phi_i(1-i)\ sup_{\xi\in [0,v]}f(\xi)< v=\|x\|.
\end{align*}

\begin{lemma} \label{lm3.4}
If the  assumptions ($H_f$), ($H_K$) and (H2) hold, then for all
$x\in\mathbb{K}_{\delta}$, with $\|x\|=u$, we have $\|Ax\|>\|x\|$.
\end{lemma}

\paragraph{Proof.}
Take a point $x\in\mathbb{K}_{\delta}$, with $\|x\|=u.$ Then by
Lemma \ref{lm3.2} we have
$\zeta_i\|x\|\le x(s)\le\|x\|=x(i)$, for every
$s\in [i\eta,i+(1-i)\eta].$ Therefore from Lemma \ref{lm3.1} it follows that
\begin{align*}
\|Ax\|&=(Ax)(i)=\int_0^1 K(i,\theta)f(x(\theta))d\theta\ge\\
&\ge inf_{\xi\in [\zeta_iu,u]}f(\xi)\Big|\int_i^{\eta}
K(i,\theta)d\theta\Big|= \Phi_i(\eta)inf_{\xi\in
[\zeta_iu,u]}f(\xi)>\\ &> u=\|x\|.
\end{align*}
 \quad \hfill$\Box$

Now we assume that the quantities
$$T_0 :=\lim_{u\to 0}\frac{f(u)}{u},\quad\text{and}\quad
T_{\infty} :=\lim_{u\to
\infty}\frac{f(u)}{u}$$
exist. The previous results imply the following corollaries.

\begin{corollary} \label{cor3.5}
Let the assumptions ($H_f$), ($H_K$) be satisfied. If $T_0 =0$,
there is $m_0 >0$ such that for every $m\in (0,m_0]$ and for every
$x\in\mathbb{K}_{\delta}$, with $\|x\|=m$, we have $\|Ax\|<\|x\|$.
\end{corollary}

\paragraph{Proof.}
Let $\epsilon \in(0, \frac{1}{\Phi_i(1-i)}].$  Since $T_0 =0$, there
is a point $m_0 >0$ such that for every
$y\in (0,m_0]$ we have
$f(y)< \epsilon y$. Let $m\in (0,m_0]$ be fixed. For every $y\in (0,m]$ we
have
$f(y)<\epsilon y\le \epsilon m$. Thus assumption (H1) is valid with
$v:=m$. So, Lemma \ref{lm3.3} applies.

\begin{corollary} \label{cor3.6}
Let the assumptions ($H_f$), ($H_K$) be satisfied. If
$T_{\infty} =+\infty$, then there is $M_0 >0$ such that for every
$M\ge M_0$ and for every
$x\in\mathbb{K}_{\delta}$, with $\|x\|=M$, we have $\|Ax\|>\|x\|$.
\end{corollary}

\paragraph{Proof.}
Since $T_{\infty} =+\infty$, there is a $M_1>0$
such that for every $y\ge M_1$ we have
$$f(y)>\frac{1}{\Phi_i(\eta)\zeta_i}y.$$
Set $M_0 :=\frac{1}{\zeta_i}M_1$. Then for any $M\ge M_0$
we have
$\zeta_i M\ge M_1$. So, if $y\in [\zeta_i M,M]$, then $y\ge \zeta_i
M\ge M_1$ and so
$$\Phi_i(\eta)f(y)>\frac{1}{\zeta_i}y\ge\frac{1}{\zeta_i}\zeta_i M=M.$$
Therefore assumption (H2) is valid with $u:=M$ and Lemma \ref{lm3.4}
applies.

\begin{corollary} \label{cor3.7}
Let the assumptions ($H_f$), ($H_K$) be satisfied. If $T_0=+\infty$,
then there is $n_0 >0$ such that for every $n\in (0,n_0 )$
and for every
$x\in\mathbb{K}_{\delta}$, with $\|x\|=n$, we have
$\|Ax\|>\|x\|$.
\end{corollary}

\paragraph{Proof.}
Since $T_0 =+\infty$, there is a $n_1>0$ such that for every $y\in
(0,n_1]$ it holds
$\Phi_i(\eta)f(y)>\frac{1}{\zeta_i}y$. Set $n_0 :=\zeta_i n_1$. Then
for any $n\in (0,n_0]$ and $y\in
[\zeta_i n,n]$ we have
$0<y\le n\le n_0 =\zeta_i n_1 <n_1$ and thus
$$\Phi_i (\eta)f(y)>\frac{1}{\zeta_i}y\ge\frac{1}{\zeta_i}\zeta_i n=n.$$
Therefore
  assumption (H2) is valid with $u:=n$ and Lemma \ref{lm3.4} applies.

\begin{corollary} \label{cor3.8}
  Let the assumptions ($H_f$), ($H_K$)  be satisfied. If
$T_{\infty} =0$, then there is $N_0 >0$ large as we want such that for every
$x\in\mathbb{K}_{\delta},$ with
$\|x\|=N_0$, we have
$\|Ax\|<\|x\|$.
\end{corollary}

\paragraph{Proof.}
Let $\epsilon \in(0, \frac{1}{\Phi_i(1-i)}).$ We distinguish two cases:
\\
1) Assume, first, that $f$ is bounded. Then there is a $k>0$ such
that $f(y)\le k,$ for all $y\ge
0$. We set $N_0 :=\frac{k}{\epsilon}$. Then, for all $y\le N_0,$ it
holds $$f(y)\le k=\epsilon
N_0<\frac{1}{\Phi_i(1-i)}N_0.$$
 2)
If $f$ is not bounded, then there is a $N_0$ so large as we want such that
$$
\sup_{[0,N_0]}f(y)=f(N_0)\le \epsilon N_0<\frac{1}{\Phi_i(1-i)}N_0.
$$
In any case, assumption (H1) holds with $v:=N_0$
and Lemma \ref{lm3.3} applies. \hfill$\Box$

\section{Existence results for equation (1.1)}

In this section we state and prove our main results.

\begin{theorem} \label{thm4.1}
Assume that the functions $f, K$ satisfy assumptions ($H_f$), ($H_K$)
and, moreover, one of the following
statements:
\begin{enumerate}
\item[(i)]  (H1) and (H2).

\item[(ii)] (H1) and $T_0=+\infty$.

\item[(iii)] (H1) and $T_{\infty}=+\infty$.

\item[(iv)] (H2) and $T_0=0$.

\item[(v)] (H2) and $T_{\infty}=0$.

\item[(vi)] $T_0=0\enskip and\enskip T_{\infty}=+\infty$.

\item[(vii)] $T_0=+\infty\enskip and\enskip T_{\infty}=0$.
\end{enumerate}
Then equation $(1.1)$ admits
at least one  positive solution.
\end{theorem}

\paragraph{Proof.}
The result of the theorem is obtained if we apply Theorem \ref{thm1.1} to
the completely
continuous operator $A$ on the cone $\mathbb{K}_{\delta}$ and use
Lemmas \ref{lm3.3} and \ref{lm3.4} if $(i)$ holds,
Lemma \ref{lm3.3} and Corollary \ref{cor3.7} if $(ii)$ holds,
Lemma \ref{lm3.3} and Corollary \ref{cor3.6} if $(iii)$ holds,
Lemma \ref{lm3.4} and Corollary \ref{cor3.5} if $(iv)$ holds,
Lemma \ref{lm3.4} and Corollary \ref{cor3.8} if $(v)$ holds,
Corollaries \ref{cor3.5} and \ref{cor3.6}, if $(vi)$ holds, and
Corollaries \ref{cor3.7} and \ref{cor3.8} if $(vii)$ holds.
In all cases we keep in mind Lemma \ref{lm3.1}.
\hfill$\Box$

\paragraph{Remark.} In any case the application of Theorem \ref{thm1.1}
provides information on the norm of the
fixed points, namely information on the maximum value of the
corresponding solution of the
problem. For instance, in case
$(i)$ the norm of the solution lies in the interval $(min\{u,v\},
max\{u,v\}).$ The way of getting exact
information about the bounds of the norms of the solutions is
exhibited in \cite{k1}, where the meaning of the
index of convergence is used.

\begin{theorem} \label{thm4.2}
Assume that the functions $f, k$ satisfy
the assumptions ($H_f$), ($H_K$), (H1) and (H2). Moreover, let one of the
following statements hold:
\begin{enumerate}
\item[(i)]  $u<v$ and $T_0=0$.

\item[(ii)] $u<v$ and $T_{\infty}=+\infty$.

\item[(iii)] $v<u$ and $T_0=+\infty$.

\item[(iv)] $v<u$ and $T_{\infty}=0$.
\end{enumerate}
Then equation $(1.4)$ admits
at least two  positive solutions. (In any case the remark of Theorem
\ref{thm4.1}
keeps in force.)
\end{theorem}

\paragraph{Proof.}
  As in the proof of Theorem \ref{thm4.1}, we apply (twice) Theorem
\ref{thm1.1} on the completely
continuous operator $A$ defined on the cone $\mathbb{K}_{\delta}$ and use
Lemmas \ref{lm3.3}, \ref{lm3.4} in connection with
Corollary \ref{cor3.5} if $(i)$ holds, Corollary \ref{cor3.6} if $(ii)$
holds,
Corollary \ref{cor3.7} if $(iii)$ holds and
Corollary \ref{cor3.8} if
$(iv)$ holds. Again, keep in mind Lemma \ref{lm3.1}. \hfill$\Box$

\begin{theorem} \label{thm4.3}
Assume that the functions $f, k$ satisfy the assumptions ($H_f$),
($H_K$), (H1) and (H2). Moreover, assume that one of the following
conditions holds:
\begin{enumerate}
\item[(i)] $u<v$, $T_0=0$ and $T_{\infty}=+\infty$.

\item[(ii)] $v<u$, $T_0=+\infty$ and $T_{\infty}=0$.
\end{enumerate}
Then equation $(1.4)$ admits at least three  positive solutions. (In any
case the
remark of Theorem \ref{thm4.1} keeps
in force.)
\end{theorem}

\paragraph{Proof.}
The result follows as in the previous theorems, where, now, we use
Lemmas \ref{lm3.3}, \ref{lm3.4} in
connection with Corollaries \ref{cor3.5}, \ref{cor3.6}, if $(i)$
holds and Corollaries
\ref{cor3.7}, \ref{cor3.8}, if $(ii)$ holds, q.e.d.

\section{Back to the boundary-value problems (1.2)-(1.3) and
(1.2)-(1.4)}

In this section we apply the previous results to get existence
results for the boundary-value problems (1.2)-(1.3) and (1.2)-(1.4). We need
the following lemma.

\begin{lemma} \label{lm5.1}
If the function $p$ is non-increasing, then the function
$K(\cdot, \theta):I\to \mathbb{R}$ defined by (2.4) is concave for a.a.
$\theta \in I$. Also if the function $p$ is
nondecreasing, then the function $K(\cdot, \theta):I\to \mathbb{R}$
defined by (2.8) is concave for a.a. $\theta \in I$.
\end{lemma}

\paragraph{Proof.}
First we consider the function $K$ defined by (2.4). It takes the form
\begin{align*}
K(t,\theta)&=\mu(\theta)\int_{\theta}^1 q(s)d_s\Big [ \alpha \big (
b+\int_t^1
q(r)dr\big) g(s)+h(s)+s\chi_{[t,1]}(s)\Big ]=\\
&=\mu(\theta)\Big [ \alpha b\int_{\theta}^1q(s)dg(s)+\alpha
\int_t^1q(r)dr\int_{\theta}^1q(s)dg(s)+\\
&\hskip .2 in +\int_{\theta}^1q(s)dh(s)+\int_{max\{\theta,t\}}^1q(s)ds\Big
]=\\
&=\mu(\theta)\Big [\alpha b\phi(\theta)+\alpha
\phi(\theta)\int_t^1q(s)ds+\int_{\theta}^1q(s)dh(s)+\\
&\hskip .2 in +\int_{max\{\theta,t\}}^1q(s)ds\Big ].
\end{align*}
Now fix  $t, r, \theta\in I$ and consider the quantity
\begin{align*}
U:&=K(t,\theta)-K(r, \theta)-\frac{
\partial K(r,\theta)}{
\partial r}(t-r)=\\
&=\mu(\theta)\Big (\alpha
\phi(\theta)\int_t^1q(s)ds+\int_{max\{\theta,t\}}^1q(s)ds-\\
&\hskip .2 in-\alpha
\phi(\theta)\int_r^1q(s)ds-\int_{max\{\theta,r\}}^1q(s)ds\Big )
-\frac{
\partial
K(r,\theta)}{
\partial r}(t-r).
\end{align*}

We distinguish the following six cases, where we take into account that
$\phi, q$ are nonnegative functions and $q$ is nondecreasing.\\
i) $t>r>\theta$. Then
\begin{align*}
U&=\mu(\theta)\Big (-\alpha \phi(\theta)\int_r^t
q(s)ds-\int_r^tq(s)ds+(\alpha
\phi(\theta)+1)q(r)\int_r^tds\Big )=\\
&=\mu(\theta)( \alpha \phi(\theta)+1)\int_r^t[q(r)-q(s)]ds \le 0.
\end{align*}
ii) $t>\theta>r$. Then
\begin{align*}
U&=\mu(\theta)\Big (\alpha \phi(\theta)\int_t^1 q(s)ds-\int_t^1
q(s)ds-\alpha
\phi(\theta)\int_r^1q(s)ds-\int_{\theta}^1q(s)ds\Big )=\\
&=\mu(\theta)\Big ( -\alpha \phi(\theta)\int_r^t
q(s)ds-\int_{\theta}^t q(s)ds+\alpha
\phi(\theta)q(r)\int_r^t ds\Big )\le\\
&\le\mu(\theta)\alpha \phi(\theta)\int_r^t [q(r)-q(s)]ds\le 0.
\end{align*}
iii) $\theta>t>r$. Then
\begin{align*}
U&=\mu(\theta)\Big ( -\alpha \phi(\theta)\int_r^t q(s)ds+\alpha
\phi(\theta)\int_r^tq(r)ds\Big )=\\
&=\mu(\theta)\alpha \phi(\theta)\int_r^t (q(r)-q(s))ds\le 0.
\end{align*}
iv) $t<r<\theta$. Then
\begin{align*}
U&=\mu(\theta)\Big ( \alpha \phi(\theta)\int_t^r q(s)ds+\alpha
\phi(\theta)q(r)(t-r)\Big )=\\
&=\mu(\theta)\alpha \phi(\theta)\int_t^r (q(s)-q(r))ds\le 0.
\end{align*}
v) $\theta<t<r$. Then
$$
U=-\mu(\theta)(\alpha \phi(\theta)+1)\int_t^r [q(r)-q(s)]ds\le 0.$$
vi) $t<\theta<r$. Then
\begin{align*}
U&=\mu(\theta)\Big (\alpha \phi(\theta)\int_t^r
q(s)ds+\int_{\theta}^r q(s)ds+(\alpha
\phi(\theta)+1)q(r)\int_r^t ds\Big )\le \\
&\le\mu(\theta)\Big (\alpha \phi(\theta)\int_t^r q(s)ds+\int_t^r
q(s)ds+(\alpha
\phi(\theta)+)\int_r^t q(r)ds\Big )=\\
&=\mu(\theta)(\alpha \phi(\theta)+1)\int_t^r [q(s)-q(r)]ds\le 0.
\end{align*}
Therefore in any case we have $U\le0$, which proves the result in
case of (2.4). The other case can be proved
by the same way. \hfill$\Box$\smallskip

It is not hard to see that the kernel $K$ defined by either (2.4), or
(2.8) is a continuous,
nonnegative function. Also, since obviously for $\delta=1$ we have
$\frac{ \partial K(t,\cdot)}{ \partial t}\le 0$ for a.a. $t\in I$,
if $K$ is defined by
(2.4) and for $\delta=-1$ we have
$\frac{ \partial K(t,\cdot)}{\partial t}\ge 0$ for a.a. $t\in I$, if
$K$ is defined by (2.8), it follows that the
function $t\to \delta K(t,\cdot)$ is non-increasing.
 Moreover, in
both cases the function  $t\to K(t,\cdot)$
is $L_1$-uniformly continuous. To check this fact, we observe that in
case (2.4) for $t_1<t_2$ it holds
\begin{align*}
&{}\Big |\int_0^1[K(t_1,\theta)-K(t_2,\theta)]d\theta\Big |= \\
&=\Big | \int_0^1\mu(\theta)\int_{\theta}^1 q(s)d_s
\Big [ g(s)\int_{t_1}^{t_2}q(r)dr+s\chi_{[t_1,t_2]}(s)\Big]d\theta\Big |\\
&=\int_{t_1}^{t_2}q(r)dr\int_0^1\mu(\theta)\int_{\theta}^1q(s)dg(s)d\theta+
\int_0^1\mu(\theta)\int_{[\theta,1]\cap [t_1,t_2]}q(s)dsd\theta\\
&\le \int_{t_1}^{t_2}q(r)dr\Big [ \int_0^1\mu(\theta)d\theta
\int_0^1q(s)dg(s)+\int_0^1\mu(\theta)d\theta \Big ].
\end{align*}
But $\int_{t_1}^{t_2}q(r)dr\to 0,$ when $|t_1-t_2 |\to 0$.

Similarly for the case (2.8).
 Now, taking into account
Lemma \ref{lm5.1}, we conclude that assumption ($H_K$) is satisfied.
Therefore, in case assumptions (H1) and/or (H2) hold, Theorems
4.1, 4.2 and 4.3 apply also to the
boundary-value problems (1.2)-(1.3) and (1.2)-(1.4) and the
corresponding results follow.

\section{Concluding remarks}

The multi-point boundary conditions for the problem considered in
\cite{m1} are special cases of the
conditions
$$x'(0)=-\sum_{i=1}^k b_i x'(\xi_i)+\sum_{i=1}^k c_i x'(\xi_i) $$
  and
  $$x(1)=\sum_{i=1}^{\lambda} d_i x'(\zeta_i)-\sum_{i=1}^{\lambda} r_i
x'(\zeta_i),$$
which, obviously, are the discrete version of the conditions
 \begin{equation*}
 x'(0)=-\int_0^1x(s)dB(s)+\int_0^1x'(s)dC(s)\tag{6.1}
\end{equation*}
  and
\begin{equation*}
x(1)=\int_0^1x(s)dD(s)-\int_0^1x'(s)dR(s),\tag{6.2}
\end{equation*}
respectively, with $B, C, D, R$ nondecreasing functions and
(without loss of generality) $B(0)=D(0)=0$. But it is easy to see
that in case $D(1)<1$, (6.1), (6.2) can be
written in the form  (1.3) where
$$g(s):=\frac{B(1)}{1-D(1)}\Big(R(s)+\int_0^s
D(\theta)d\theta\Big)+C(s)+\int_0^s
B(\theta)d\theta$$
and
$$h(s):=\frac{1}{1-D(1)}\left (R(s)+\int_0^s
D(\theta)d\theta\right ) .$$
Hence \cite[Theorem1]{m1} is a special case of our Theorem \ref{thm4.1}.

\paragraph{Remark.} We proved above that the boundary-value problems
(1.2)-(1.3) and (1.2)-(1.4) admit one, or
two, or three solutions according to the conditions which they
satisfy. And it was their behavior
which motivated us to distinguish the two cases for the kernel $K$ of
the integral equation (1.1). However, as
the two problems is concerned, one can observe that existence of
solutions of one of them guarantees
existence of solutions of the other. Indeed, it is easy to see that
they are equivalent and their equivalence
follows by applying the transformation $x(t)\to x(1-t)$, 
$p(t)\to p(1-t)$, $\mu(t)\to \mu(1-t)$
and $g_1(t)=-g(1-t)$, $h_1(t)=-h(1-t)$.

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\noindent\textsc{G. L. Karakostas } (e-mail: gkarako@cc.uoi.gr)\\
\textsc{P. CH. Tsamatos} (e-mail: ptsamato@cc.uoi.gr)\\[3pt]
Department of Mathematics, University of Ioannina,\\
451 10 Ioannina, Greece

\end{document}