\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 $h_{j_0}(t)0$.
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 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\|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 $K<\|x\|M$.
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)M$. It remains to show
that if $x\in \mathbb{K}$ and $\|x\|=K$, then $\|Ax\|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
$$ K^{\mu+\nu-1}<(\Theta\xi)^{-1}<(\epsilon\zeta)^{-1}~~S$. Then observe that $L_1=1$, as well as
$$ 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.
$$
Since $\epsilon, \Theta$ are arbitrary, we have the following statement.
\begin{corollary} \label{coro5.1}
Assume that $g,h:I\to \mathbb{R}$ are nondecreasing functions (with $h$
not identically zero) and such
that $g(1)-g(0)<1$ and $0\le h(t)\le t$, $t\in I$.
Then for every $\lambda >0$ the boundary value problem
(5.1),(\ref{1.2}),(\ref{1.3}) admits at least one positive, nondecreasing and
concave solution $x$ such that $K<\|x\|0$, with $$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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