\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 119, 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  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/119\hfil Semipositone $m$-point boundary-value problems]
{Semipositone $m$-point boundary-value problems}

\author[Nickolai Kosmatov\hfil EJDE-2004/119\hfilneg]
{Nickolai Kosmatov}

\address{Department of Mathematics and Statistics\\
    University of Arkansas at Little Rock\\
    Little Rock, AR 72204-1099, USA}
\email{nxkosmatov@ualr.edu}

\date{}
\thanks{Submitted April 23, 2004. Published October 10, 2004.}
\subjclass[2000]{34B10, 34B18}
\keywords{Green's function; fixed point theorem;
 positive solutions; \hfill\break\indent multi-point boundary-value problem}

\begin{abstract}
  We study the $m$-point  nonlinear boundary-value problem
  $$
  \displaylines{
  -[p(t)u'(t)]' = \lambda f(t,u(t)), \quad 0 < t < 1, \cr
  u'(0) = 0, \quad \sum_{i=1}^{m-2}\alpha_i u(\eta_i) = u(1),
  }$$
  where $0 < \eta_1 < \eta_2 < \dots  < \eta_{m-2} < 1$, $\alpha_i > 0$
  for $1 \leq i \leq m-2$ and $\sum_{i=1}^{m-2}\alpha_i < 1$, $m \geq 3$.
  We assume that   $p(t)$ is non-increasing continuously differentiable
  on $(0,1)$ and $p(t) > 0$ on $[0,1]$.
  Using a cone-theoretic approach we provide sufficient conditions
  on continuous $f(t,u)$ under which the problem admits a positive solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}

\section{Introduction}

In this note we consider the nonlinear $m$-point eigenvalue problem
\begin{gather}
-[p(t)u'(t)]' = \lambda f(t,u(t)), \quad 0 < t < 1, \label{e}\\
u'(0) = 0, \quad \sum_{i=1}^{m-2}\alpha_i u(\eta_i) = u(1), \label{bc}
\end{gather}
where  $0 < \eta_1 < \eta_2 < \dots  < \eta_{m-2} < 1$, $\alpha_i >
0$ for $1 \leq i \leq m-2$, $\sum_{i=1}^{m-2}\alpha_i < 1$. We
also assume that the function $p(t)$ is non-increasing
continuously differentiable on $(0,1)$ and $p(t) > 0$ on $[0,1]$.
The inhomogeneous term in (\ref{e}) is allowed to change its sign.
Other assumptions on $f(t,u(t))$ will be made later.

The study of multi-point boundary-value problems was initiated by
Il'in and Moiseev in \cite{im1,im2}. Many authors since then
considered nonlinear multi-point boundary-value problems (see,
e.g., \cite{da, fe, fw, ghm, ek, ma1, ma2, w1, w2} and the
references therein). In particular, Ma studied in \cite{ma2}
positive solutions to the three-point nonlinear boundary-value problem
\begin{gather*}
-u''(t) = a(t)f(u(t)), \quad 0 < t < 1, \\
u(0) = 0, \quad \alpha u(\eta) = u(1),
\end{gather*}
where $0 < \alpha$, $0 < \eta < 1$ and $\alpha\eta < 1$.
The results of \cite{ma2} were complemented in the works of Webb \cite{w2},
Kaufmann \cite{ek}, Kaufmann and Kosmatov \cite{kk}, and Kaufmann and
Raffoul \cite{kar}.

Among the studies dealing with semipositone multi-point boundary-value problems,
 we mention the papers by Cao and Ma \cite{cm} and Liu \cite{liu}.
Cao and Ma considered the
boundary-value problem
\begin{gather*}
-u''(t) = \lambda a(t)f(u(t),u'(t)), \quad 0 < t < 1, \\
u(0) = 0, \quad \sum_{i=1}^{m-2}\alpha_i u(\eta_i) = u(1).
\end{gather*}
The authors applied the Leray-Schauder fixed point theorem to obtain an interval
of eigenvalues for which at least one positive solution exists.
Liu applied a fixed point index method to obtain such an interval for
\begin{gather*}
-u''(t) = \lambda a(t)f(u(t)), \quad 0 < t < 1, \\
u'(0) = 0, \quad \alpha u(\eta) = u(1).
\end{gather*}
Our approach is based on Krasnosel'ski\u{\i}'s cone-theoretic theorem \cite{kra}
and enables us to
show the existence of a positive solution for the semipositone problem
(\ref{e}), (\ref{bc}). Other applications of Krasnosel'ski\u{\i}'s fixed point
theorem to semipositone problems can, for example, be found in \cite{ag}.

\section{Preliminaries}

We now proceed with the auxiliaries.
Consider the equation
\begin{equation}\label{eq1}
-[p(t)u'(t)]' = g(t), \quad 0 < t < 1,
\end{equation}
with the boundary conditions (\ref{bc}).

For convenience we set $\alpha = \sum_{i=1}^{m-2}\alpha_i$. Recall that $\alpha < 1$.
\begin{lemma}\label{solutn}
If $g \in C[0,1]$ and $g(t) \geq 0$ on $[0,1]$, then
\begin{equation}\label{sol1}
\begin{aligned}
u(t) &=-\int_0^t \Big( \int_s^t \frac{d \tau}{p(\tau)} \Big) g(s) \, ds
+\frac{1}{1-\alpha}\int_0^1
        \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s) \, ds \\
&\quad -\frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i
    \int_0^{\eta_i} \left ( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \right ) g(s)\, ds
\end{aligned}
\end{equation}
is the unique nonnegative solution on $[0,1]$ of the problem (\ref{eq1}),
(\ref{bc}).
\end{lemma}

\begin{proof}
Integration of (\ref{eq1}) from $0$ to $t$ with the use of the boundary condition
(\ref{bc}) at $0$ yields
\[
u'(t) = - \frac{1}{p(t)} \int_0^t g(s) \, ds \leq 0.
\]
Integrating again we get
\[
u(t) = - \int_0^t \frac{1}{p(s)}  \Big( \int_0^s g(\tau) \, d\tau \Big) \,ds + A
     = - \int_0^t \Big( \int_s^t \frac{d \tau}{p(\tau)} \Big) g(s) \, ds + A.
\]
Using the multi-point condition in (\ref{bc}) we determine $A$ and obtain
(\ref{sol1}).
Since $u'(t) \leq 0$,
\begin{align*}
u(t) &\geq u(1) \\
     &= \frac{\alpha}{1-\alpha}
\int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s) \, ds
        -\frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i
\int_0^{\eta_i} \Big( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \Big) g(s) \, ds \\
     &= \frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i
\Big[
  \int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s) \, ds
- \int_0^{\eta_i} \Big( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \Big) g(s) \, ds
\Big] \geq 0
\end{align*}
on $[0,1]$ and the proof is complete.
\end{proof}

For $g(t) = 1$ on $[0,1]$, we denote by $u_0(t)$ the unique solution (\ref{sol1}).
Then we have
\begin{align*}
C &= \max_{t\in[0,1]} u_0(t) = u_0(0) \\
&=\frac{1}{1-\alpha}\int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s)ds
 - \frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i
    \int_0^{\eta_i} \Big( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \Big) g(s)ds.
\end{align*}
The Green's function for $-[p(t)u'(t)]' = 0$ with (\ref{bc}) is given by
\begin{align*}
G(t,s) &= \frac{1}{1-\alpha} \int_s^1 \frac{d \tau}{p(\tau)} \nonumber \\
       &\quad - \begin{cases}
       \int_s^t \frac{d \tau}{p(\tau)},  & s \leq t\\
        0,                               & s > t
        \end{cases}
        - \begin{cases}
        \frac{1}{1-\alpha} \sum_{i=1}^{m-2} \alpha_i \chi_i(s)
        \int_s^{\eta_i} \frac{d \tau}{p(\tau)},
        & s \leq \eta_{m-2}\\
        0,  & s > \eta_{m-2},
        \end{cases}
\end{align*}
where
\[
\chi_i(s) = \begin{cases}
                   1, & s \leq \eta_i\\
                   0, & s > \eta_i.
            \end{cases}
\]
Note that
\begin{equation}\label{est11}
\max_{t\in[0,1]}\int_{0}^1 G(t,s) \, ds = C.
\end{equation}
The integral operator $T \colon \mathcal{B} \to \mathcal{B}$ associated with
(\ref{e}), (\ref{bc}) is defined by
\[
Tu(t) = \int_0^1 G(t,s)f(s,u(s)) \, ds
\]
A routine argument shows that $T$ is completely continuous.

\begin{definition} \rm
Let $\mathcal{B}$ be a Banach space and let $\mathcal{C} \subset \mathcal{B}$ be closed and
nonempty. Then $\mathcal{C}$ is said to be a cone if
\begin{enumerate}
    \item $\alpha u + \beta v \in \mathcal{C}$ for all $u, v \in \mathcal{C}$ and
for all $\alpha, \beta \geq 0$, and
    \item $u, -u \in \mathcal{C}$ implies $u \equiv 0$.
\end{enumerate}
\end{definition}

Our Banach space, $\mathcal{B}$, is the space $C[0,1]$ with the norm
$\|u\| = \max_{t \in [0,1]} |u(t)|$. We will show now that the unique solution
(\ref{sol1}) satisfies
\begin{equation}\label{coneconst1}
\min_{t\in[0,1]} u(t) \geq {\gamma} \| u \|,
\end{equation}
where
\[
\gamma = \max_{1 \leq i \leq m-2} \frac{\alpha_i (1-\eta_i)}{1-\alpha_i \eta_i}.
\]
To this end, note that the solution (\ref{sol1}) is concave, since $g(t) \geq 0$ and
$u'(t),p'(t) \leq 0$ on $[0,1]$. By concavity and since $u(1) > \alpha_i u(\eta_i)$
for each $1 \leq i \leq m-2$,
\begin{align*}
\|u\| &= u(0) \\
&\leq u(1) + \frac{u(1) - u(\eta_i)}{1 - \eta_i}(0 - 1) \\
&   < u(1) \frac{1 - \alpha_i \eta_i}{\alpha_i (1 - \eta_i)} \\
&   = \frac{1-\alpha_i \eta_i}{\alpha_i (1-\eta_i)} \min_{t \in [0,1]} u(t)
\end{align*}
and hence (\ref{coneconst1}) holds.

The estimate (\ref{coneconst1}) is used for defining our cone
$\mathcal{C} \subset \mathcal{B}$ by
\begin{equation}\label{cone}
{\mathcal{C}} = \{u(t) \in \mathcal{B}: u(t) \geq 0 \; \text{on}
\; [0,1], \; \min_{t \in [0,1]} u(t) \geq \gamma \| u \| \}.
\end{equation}
It turns out that our operator $T$ is cone-preserving. Fixed points of $T$ are
solutions of (\ref{e}), (\ref{bc}). The existence of a fixed point of $T$
follows from a fixed point theorem due to Krasnosel'ski\u{\i} \cite{kra},
which we now state.

\begin{theorem}\label{Kr}
Let $\mathcal{B}$ be a Banach space and
let $\mathcal{C}\subset \mathcal{B}$ be a cone in $\mathcal{B}$. Assume that $\Omega_{1}$, $\Omega_{2}$
are open with $0 \in \Omega_1$, ${\overline{\Omega}}_1\subset \Omega_2$, and let
$$T\colon {\mathcal{C}}\cap({\overline{\Omega}}_2 \setminus \Omega_1)\to \mathcal{C}$$
 be a completely continuous operator such that either
\begin{itemize}
\item[(i)]  $\| Tu \| \leq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_1$,
and        $\| Tu \| \geq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_2$, or
\item[(ii)] $\| Tu \| \geq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_1$,
and       $\| Tu \| \leq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_2$.
\end{itemize}
Then $T$ has a fixed point in
${\mathcal{C}}\cap({\overline{\Omega}}_2 \setminus \Omega_1)$.
\end{theorem}

The following assumptions will stand throughout the remainder of this note:
\begin{itemize}
\item[(A1)]
$f(t,z)$ is a continuous function on $[0,1] \times [0,\infty)$
\item[(A2)] There exists $M > 0$ such that $f(t,z) + M \geq 0$
            on $[0,1] \times [0,\infty)$
\item[(A3)] There exist continuous nonnegative nondecreasing on $[0,\infty)$
            functions $\psi_a(z)$ and $\psi_b(z)$ with
            $\psi_b(z) \leq f(t,z) + M \leq \psi_a(z)$ on $[0,1] \times [0,\infty)$.
\end{itemize}

\section{Positive solutions}

We now state our main results.

\begin{theorem}\label{pos1}
Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that
\[
\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = 0 \quad\mbox{and}\quad
\lim_{z \to \infty} \frac{\psi_b(z)}{z} = \infty.
\]
Then, for a sufficiently small $\lambda > 0$, the problem (\ref{e}), (\ref{bc}) has
a positive solution.
\end{theorem}

\begin{proof}
 Consider the equation
\begin{equation}\label{aue}
-[p(t)u'(t)]' = \lambda f_p(t,u(t)-u_{\lambda}(t)), \quad 0 < t < 1,
\end{equation}
with the boundary conditions (\ref{bc}), where
\[
     f_p(t,z) =    \begin{cases}
        f(t,z) + M, & z \geq 0\\
        f(t,0) + M, & z \leq 0
               \end{cases}
\]
and $u_{\lambda}(t)=\lambda M u_0(t)$ ($u_0(t)$ is given by (\ref{sol1}) for
$g \equiv 1$).
Our objective is to show that the problem (\ref{aue}), (\ref{bc}) has a positive
solution.

Our completely continuous and cone-preserving operator associated with
(\ref{aue}), (\ref{bc}) is defined by
\[
T_{\lambda}u(t) = \lambda \int_0^1 G(t,s)f_p(s,u(s)-u_{\lambda}(s)) \, ds
\]
Since $\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = 0$, there exists $R_1 > 0$
such that
\[
\psi_a(z) \leq \frac{1}{\lambda C} z
\]
for all $z \leq R_1$.

Define $\Omega_1 = \{ u \in \mathcal{B} :  \|u\| < R_1 \}$, then for
$u \in {\mathcal{C}} \cap \partial \Omega_1$ we have
\begin{equation}\label{nearzero}
\psi_a(u(s)) \leq \psi_a(\|u\|) \leq \frac{1}{\lambda C} R_1
\end{equation}
for all $s \in [0,1]$, since $\psi_a(z)$ is nondecreasing.
Now, if $u(s) \geq u_{\lambda}(s)$ for
$s \in [0,1]$, then
\[
f_p(s,u(s)-u_{\lambda}(s)) = f(s,u(s)-u_{\lambda}(s)) + M
                         \leq {\psi}_a(u(s)-u_{\lambda}(s))
                           \leq {\psi}_a(u(s)).
\]
If $u(s) \leq u_{\lambda}(s)$, then
\[
f_p(s,u(s)-u_{\lambda}(s)) = f(s,0) + M \leq {\psi}_a(0) \leq {\psi}_a(u(s))
\]
(we know that $u(s) \geq 0$ as an element of $\mathcal{C}$).
Combining both cases and using (\ref{nearzero})
and (\ref{est11}), we get
\begin{align*}
\|T_{\lambda}u\|
&=  \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) f_p(s,u(s)-u_{\lambda}(s))
 \, ds\\
&\leq  \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) {\psi_a}(u(s)) \, ds \\
&\leq  \lambda \max_{t \in [0,1]} \int_0^1 \! G(t, s) \, ds \,
                        \frac{1}{\lambda C} R_1
 =  R_1,
\end{align*}
that is, $\|T_{\lambda}u\| \leq \|u\|$ on ${\mathcal{C}} \cap \partial \Omega_1$.

Since $\lim_{z \to \infty} \frac{\psi_b(z)}{z} = \infty$, then also
$\lim_{z \to \infty} \frac{\psi_b(\gamma z - \lambda M C)}{z} = \infty$.
Thus, there exists $R_2 > 0$ large enough
(so that $R_2 > \frac{\lambda M C}{\gamma}$ and $R_2 > R_1$) such that
\[
\psi_b(\gamma z - \lambda M C) \geq \frac{1}{\lambda C} z
\]
for all $z \geq R_2$. In fact,
\begin{equation}\label{nearinf}
\psi_b(\gamma R_2 - \lambda M C) \geq \frac{1}{\lambda C} R_2.
\end{equation}
Define $\Omega_2 = \{ u \in \mathcal{B} :  \|u\| < R_2 \}$, then for
$u \in {\mathcal{C}} \cap \partial \Omega_2$ we have
\[
u(s)-u_{\lambda}(s) \geq \gamma \|u\| - \lambda M u_0(s)
                    \geq \gamma R_2 - \lambda M C > 0.
\]
Now, for all $s \in [0,1]$,
\[
f_p(s,u(s)-u_{\lambda}(s)) =    f(s,u(s)-u_{\lambda}(s)) + M
                           \geq {\psi}_b(u(s)-u_{\lambda}(s))
                           \geq {\psi}_b(\gamma R_2 - \lambda M C),
\]
since $\psi_b(z)$ is nondecreasing. Therefore, by (\ref{nearinf}) and (\ref{est11}),
\begin{align*}
\|T_{\lambda}u\|
&=  \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) f_p(s,u(s)-u_{\lambda}(s)) \, ds\\
&\geq  \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s)
                             {\psi}_b(\gamma R_2 - \lambda M C) \, ds \\
&\geq  \lambda \max_{t \in [0,1]} \int_0^1 \! G(t, s) \, ds \,
                        \frac{1}{\lambda C} R_2
 =  R_2,
\end{align*}
that is, $\|T_{\lambda}u\| \leq \|u\|$ on ${\mathcal{C}} \cap \partial \Omega_2$.

Since the assumptions of Theorem \ref{Kr} are satisfied, we conclude that
the problem (\ref{aue}), (\ref{bc}) has a positive
solution in ${\mathcal{C}}\cap({\overline{\Omega}}_2 \setminus \Omega_1)$,
which we denote by $u_p$.

Let $\lambda$ be small enough so that $R_1 > \frac{\lambda M C}{\gamma}$.
Now we have
$u_p(t) \geq \gamma \|u_p\| \geq \gamma R_1 > \lambda M C \geq u_{\lambda}(t)$ for all
$t \in [0,1]$. Set $u(t) = u_p(t) - u_{\lambda}(t)$, then
\begin{align*}
-[p(t)u'(t)]' &= -[p(t)u_p'(t)]' - \lambda M \\
           &= \lambda f_p(t,u_p(t) - u_{\lambda}(t)) - \lambda M \\
           &= \lambda (f(t,u_p(t) - u_{\lambda}(t)) + M) - \lambda M \\
           &= \lambda f(t,u(t)),
\end{align*}
which shows that $u(t)$ is a positive solution of (\ref{e}), (\ref{bc}). The proof
is complete.
\end{proof}


\subsection*{Example}
To illustrate our main result, we consider the inhomogeneous term
in the form of the function
\[
f(t,z) = -1 + z^2(2+\sin{(4 \pi z (1+t^3))}).
\]
The function $f(t,z)$ is continuous and, setting $M = 1$, we get $f(t,z)+ M \geq 0$
on $[0,1] \times [0,\infty)$. In addition, for $\psi_b(z)= z^2$ and $\psi_a(z)=
3z^2$, we have $\psi_b(z) \leq f(t,z)+ M \leq\psi_a(z)$ and
\[
\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = 0 \quad \text{and} \quad \lim_{z
\to \infty} \frac{\psi_b(z)}{z} = \infty.
\]
Thus, Theorem \ref{pos1} applies.

With only minor adjustments to the argument above one can prove our next theorem.

\begin{theorem}\label{pos2}
Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that
\[
\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = \infty \quad \text{and} \quad
 \lim_{z\to \infty} \frac{\psi_b(z)}{z} = 0.
\]
Then, for a sufficiently small $\lambda > 0$, the problem (\ref{e}),
(\ref{bc}) has a positive solution.
\end{theorem}

\subsection*{Remark}
If problem (\ref{e}), (\ref{bc}) has a positive solution for some $\lambda_1 > 0$,
there is also a positive solution for
each $\lambda \in (0,\lambda_1]$.

We say that a function $\psi(z)$ is sublinear if
\[
\lim_{z \to 0^+} \frac{\psi(z)}{z} = \infty \quad \text{and} \quad
\lim_{z \to \infty} \frac{\psi(z)}{z} = 0.
\]
On the other hand, if
\[
\lim_{z \to 0^+} \frac{\psi(z)}{z} = 0 \quad \text{and} \quad
 \lim_{z \to \infty} \frac{\psi(z)}{z} = \infty,
\]
then the function $\psi$ is called  superlinear.

If in the assumption (A3) we take $\psi_a(z) = \psi_b(z)$,
then the following
corollary to Theorems \ref{pos1} and \ref{pos2} becomes immediate.

\begin{corollary}\label{pos4}
Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that
$\psi_a(z) = \psi_b(z)$ is either sublinear or superlinear.
Then, for a sufficiently small $\lambda > 0$, the problem (\ref{e}),
(\ref{bc}) has a positive solution.
\end{corollary}


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