\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 70, pp. 1--12.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/70\hfil Multiple positive solutions]
{Multiple positive solutions for singular semi-positone delay
differential equation}
\author[X. Xu \hfil EJDE-2005/70\hfilneg]
{Xian Xu}

\address{Xu, Xian \hfill\break
Department of  Mathematics,  Xuzhou Normal  University\\
 Xuzhou, Jiangsu, 221116,  China}
\email{xuxian68@163.com}

\date{}
\thanks{Submitted May 7, 2005. Published June 30, 2005.}
\thanks{Supported by grant 04KJB110138 from the Natural Science Foundation  of the
\hfill\break\indent Jiangsu Education Committee, China}
\subjclass[2000]{34B15, 34B25}
\keywords{Multiple positive solutions;  fixed point index; \hfill\break\indent
  semi-positone delay differential equation}

\begin{abstract}
 In this paper, we obtain new existence results for multiple positive
 solutions of a  delay singular differential boundary-value problem.
 Our main tool is the fixed point index method.
\end{abstract}

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

\section{Introduction}

Singular differential boundary-value problems arise from many
branches of basic mathematics and applied mathematics. Many
techniques have been developed to establish  the existence of
positive solutions of various classes of singular differential
boundary-value problems; \cite{a1,a2,c1,d1,e1,j1,l1,o1,o2,x1,x2,w1}
and the references therein. In
particular, the authors of \cite{a1,e1} obtained some existence results
for positive solutions of some singular functional differential
equations. Motivated  by \cite{a1,e1}, in this paper we will consider the
following singular delay differential equation
\begin{equation}
\begin{gathered}
y''+f(t, y(t-a))=0,\quad t\in (0, 1]\backslash\{a\},\\
y(t)=\mu(t),\quad t\in [-a, 0],\\
y(1)=0,
\end{gathered} \label{e1.1}
\end{equation}
where $0<a<1$,
\begin{equation}
\mu(t)\in C[-a,0],\quad \mu(0)=0,\quad \mu(t)>0,\quad \forall t\in [-a, 0),
\label{e1.2}
\end{equation}
and the nonlinear term $f(t, y)$ satisfies
\begin{equation}
\phi_0(t)h_0(y)-p(t)\leq f(t, y)\leq\phi(t)(g(y)+h(y)),\label{e1.3}
\end{equation}
$\phi, \phi_0,  p$ are in $C((0,1], R^+)$,
$g$ is in $C((0, +\infty),R^+)$, $h_0$, $h$ are in $C(R^+, R^+)$, and
$R^+=[0,+\infty)$.

      Problem \eqref{e1.1} is a singular semi-positone boundary-value problem
because $p$ is allowed to be nonnegative on interval $(0,1)$ and
$f$ may have singularity at $t=0$ and $y=0$. Recently, there have
been many papers considered the singular semi-positone boundary-value
problems; see \cite{a2,l1,w1,x1,j1} and the references therein.  But most
of these papers paid  attention to the existence of positive
solutions of  the singular semi-positone boundary-value problems
and there were fewer papers that discuss the existence of multiple
positive solutions of the singular semi-positone problems. To
cover up this gap, in this paper we will establish some existence
results for multiple positive solutions of singular delay
differential semi-positone boundary value problem \eqref{e1.1}. It
is difficult to show the existence and multiplicity results for
positive solutions of semi-positone problems.  For our purpose, a
special cone will be needed to establish the multiplicity results
for  positive solutions of semi-positone problems.

Let $P=\{x\in C[-a,1]:x(t)\geq 0 \mbox{ for all } t\in [-a,1]\}$.
It is well known that $C[-a,1]$ is a real Banach space
with the maximum norm $\|x\|=\max_{t\in [-a,1]}|x(t)|$, and
$P$ a normal cone of $C[-a,1]$. By a positive solution of
\eqref{e1.1} we mean a function $x\in P$ satisfying \eqref{e1.1}
and $x(t)\not\equiv 0$.

Throughout this paper, we will assume that \eqref{e1.2} and \eqref{e1.3} hold.

\section {Preliminary Lemmas}

Let us list some assumptions to be used.
\begin{itemize}
\item[(H1)] $g:(0,+\infty)\to R^+$ is continuous and
decreasing, $h, \ h_0:R^+\to R^+$ are continuous
  and increasing.

\item[(H2)] For any constant  $k_0>0$,
 $$
 \int^a_0[\phi(s)g(\mu(s-a))+g(k_0s)+p(s)]ds<+\infty.
 $$

\item[(H3)] There exists $R_0\geq 2\int^1_0p(s)ds$ such that
\begin{equation}
R_0-2\sqrt{B(R_0)}>A, \label{e*}
\end{equation}
where
\begin{gather*}
 A=\int^a_0[\phi(s)(g(\mu(s-a))+h(\mu(s-a)+1))+p(s)]ds
 +  \int^1_a\phi(s)g(\frac{1}{2}R_0a(s-a))ds,
 \\
 B(R_0)=2\sup_{t\in  [a,1]}[\phi(t)+p(t)]\int^{R_0}_0[g(\frac{1}{2}s)
 +h(s+1)+1]ds.
\end{gather*}

\item[(H4)] There exist $u_1>R_0$ and $[\alpha, \beta]\subset  (a,1)$ such that
 $$
 \alpha(1-\beta-a)h_0(\frac{1}{2}u_1)\int^{\beta+a}_{\alpha+a}\phi_0(s)ds>u_1.
 $$
\end{itemize}

\begin{remark} \label{rmk1}\rm
The nonlinear term $f$ of the form \eqref{e1.3} in the case
$\phi_0(t)=p(t)=0$ for all $t\in [0,1]$ has been studied by many
authors  \cite{a1,o2,e1,x1}.
\end{remark}

 Let $Q=\{x\in P:x(t)\geq\|x\|t(1-t)$ for $t\in [0,1]\}$.
 It is easy to see that $Q$ is a cone of $C[-a,1]$. For each
$x\in  P$, let
 $$
[x(t-a)]^*=\max\{x(t-a)+x_0(t-a)-w(t-a),
 \widetilde{x}_0(t-a)\},\quad \forall t\in [0,1],
$$
 where
\begin{gather*}
 x_0(t)= \begin{cases}
 \mu(t), &t\in [-a,0],\\
 0,&t\in (0,1],
 \end{cases}
\\
\widetilde{x}_0(t)= \begin{cases}
0, &t\in [-a,0],\\
 \frac{1}{2}R_0t(1-t),&t\in (0,1],
 \end{cases}
\\
w(t)= \begin{cases}
 0, &t\in [-a,0],\\
 \int^1_0G(t,s)p(s)ds,&t\in (0,1],
 \end{cases}
 \\
 G(t,s)= \begin{cases}
 s(1-t), &s\leq t,\\
 t(1-s),&s>t.
 \end{cases}
\end{gather*}
For each positive integer $n$, let us define an operator
$T_n:P\to P$ by
\begin{equation} \label{e2.1}
(T_nx)(t)=\begin{cases}
0,&t\in [-a,0];\\
\int^1_0 G(t,s)[f(s,[x(s-a)]^*+n^{-1})+p(s)]ds,&t\in (0,1].\\
\end{cases}
\end{equation}


\begin{lemma}[\cite{g1}] \label{lem1}
 Let $X$ be a retract of the real Banach space $E$ and $X_1$ be
a bounded convex retract of $X$. Let $U$ be a nonempty open set
of $X$ and $U\subset X_1$. suppose that $A:X_1\to X$ is completely
continuous, $A(X_1)\subset X_1$ and
$A$ has no fixed points on $X_1\backslash U$. Then
$i(A, U, X)=1$.
\end{lemma}

\begin{lemma} \label{lem2}
Suppose that (H1) and (H2) hold.
Then $T_n:P\to Q$ is a completely continuous operator for each
positive integer $n$.
\end{lemma}

\begin{proof} Let $n$ be a fixed positive integer, and
$y=T_nx$ for some $x\in P$. Suppose that $\|y\|_{[0,1]}=y(t_0)$
for some $t_0\in (0,1)$, where
$\|y\|_{[0,1]}=\max_{t\in [0,1]}|y(t)|$. Since
$$
y(t)\geq y(s)=0,\quad  t\in [-a,1],\; s\in [-a,0],
$$
it follows that $\|y\|=\|y\|_{[0,1]}$. It is easy to see that
$$
y''(t)=-f(t, [x(t-a)]^*+n^{-1})-p(t)\leq 0,\ \forall t\in (0, 1].
$$
Therefore, the graph of $y(t)$ is concave down on $(0,1)$.
For any $0\leq t\leq t_0$, we have
$$
y(t)=y((1-\frac{t}{t_0})\cdot 0+\frac{t}{t_0}t_0)
 \geq \|y\|t(1-t).
$$
Similarly,
$$
y(t)=y(\frac{1-t}{1-t_0}t_0+(1-\frac{1-t}{1-t_0})\cdot 1)
\geq \|y\|t(1-t),\quad \forall  t_0\leq t\leq 1.
$$
Hence, $T_n:P\to Q$.

Now, we show that $T_n$ is a completely continuous operator for
every positive integer $n$. It is easy to see that $T_n$ is a continuous
and bounded operator for every positive integer $n$.
 Let $B\subset P$ be a bounded set
such that $\|x\|\leq L$ for all $x\in B$ and some $L>0$.
Then, we can easily see that
\begin{equation}
x_0(t-a)+\widetilde{x}_0(t-a)\leq [x(t-a)]^*\leq \|x\|+\|w\|
+\|\mu\|+R_0, \quad \forall x\in B, \; t\in [0,1]. \label{e2.2}
\end{equation}
Thus, for any $t_1,\ t_2\in [0,1]$, we
have
\begin{equation}
\begin{aligned}
&|(T_nx)(t_1)-(T_nx)(t_2)|\\
&\leq \int^1_0|G(t_1,s)-G(t_2,s)|[\phi(s)(g(x_0(s-a)
+\widetilde{x}_0(s-a))\\
&\quad +h(L+\|w\|+\|\mu\|+R_0+1))+p(s)]ds.
\end{aligned} \label{e2.3}
\end{equation}
Then the uniform continuity of $G(t,s)$ on
$[0,1]\times [0,1]$, \eqref{e2.3} and the assumption (H2) imply that
$T_n(B)$ is an equicontinuous set on [0,1]. Obviously, $T_n(B)$ is
equicontinuous on $[-a,0]$. Thus, $T_n:P\to Q$ is a completely
continuous operator. The proof is complete.
\end{proof}

\begin{lemma} \label{lem3}
Let $\Omega_0=\{x\in Q:\|x\|<R_0\}$. Suppose that (H1)-(H3) hold.
Then for every positive integer $n$,
$$
i(T_n, \Omega_0,Q)=1\,.
$$
\end{lemma}

\begin{proof} We claim that
\begin{equation}
z\not=\lambda T_n z, \quad  \lambda\in [0,1],\; z\in \partial
\Omega_0. \label{e2.4}
\end{equation}
 where $\partial\Omega_0$ denotes the boundary of $\Omega_0$ in $Q$.
In fact, if \eqref{e2.4} is not true, then
there exist $\lambda_0\in [0,1]$, $z_0\in \partial \Omega_0$,
and positive integer $n_0$ such that
$z_0=\lambda_0 T_{n_0}z_0$. From $z_0\in Q$, we have
\begin{equation}
z_0(t)\geq \|z_0\|t(1-t)=R_0t(1-t), \ t\in [0,1].\label{e2.5}
\end{equation}
On the other hand, using the fact that $G(t,s)\leq t(1-t)$ for
$(t,s)\in [0,1]\times [0,1]$, we have
\begin{equation}
w(t)=\int^1_0G(t,s)p(s)ds\leq
\Big(\int^1_0p(s)ds\Big)t(1-t),\quad  \forall t\in [0,1].\label{e2.6}
\end{equation}
 It follows from \eqref{e2.5} and \eqref{e2.6} that
\begin{equation}
z_0(t)-w(t)\geq \frac{1}{2}z_0(t)\geq \frac{1}{2}R_0t(1-t),
\forall t\in [0,1].\label{e2.7}
\end{equation}
From $z_0=\lambda_0T_{n_0}z_0$, by direct computation, we have
\begin{equation}
\begin{gathered}
z_0''(t)+\lambda_0[ f(t,[z_0(t-a)]^*+n^{-1})+p(t)]=0,\quad t\in (0,1],\\
z_0(t)=0, \quad t\in [-a,0),\\
z_0(1)=0.
\end{gathered} \label{e2.8}
\end{equation}
By \eqref{e2.7} and \eqref{e2.8}, we get that
\begin{equation}
[z_0(t-a)]^*=\begin{cases}
\mu(t-a),& t\in [0,a],\\
z_0(t-a)-w(t-a),& t\in (a,1].
\end{cases} \label{e2.9}
\end{equation}
 It follows from \eqref{e2.8} that
$z_0''(t)\leq 0$ for $t\in (0,1]$. Thus, the graph of
$z_0(t)$ is concave down on $(0,1)$, and so, there exists
$t_0\in (0,1)$ such that
\begin{gather*}
\|z_0\|=z_0(t_0),\quad z'(t_0)=0,\quad z_0'(t)\geq 0\quad
 \mbox{on }(0, t_0),\\
and z'(t)\leq 0\quad  \mbox{on }  (t_0,1).
\end{gather*}
 Therefore,  we have the following two cases:


\noindent\textbf{Case (a): $t_0\leq a$.}
By \eqref{e2.8} and \eqref{e2.9}, we have
 $$
-z_0''(t)\leq  \phi(t)(g(\mu(t-a))+h(\mu(t-a)+1))+p(t),\quad
\forall t\in (0,t_0).
$$
 Integrating from $t(t\in(0,t_0))$ to $t_0$, we have
 $$
 z_0'(t)\leq
  \int^{t_0}_0[\phi(s)(g(\mu(s-a)+h(\mu(s-a)+1))+p(s)]ds,\quad
  t\in   (0, t_0].
$$
 Then integrating from 0 to $t_0$, we have
\begin{equation}
z_0(t_0)\leq  \int^{t_0}_0s[\phi(s)(g(\mu(s-a))+h(\mu(s-a)+1))+p(s)]
ds\leq A. \label{e2.10}
\end{equation}
\textbf{Case (b): $t_0>a$.}
 By \eqref{e2.7}, \eqref{e2.8} and \eqref{e2.9}, we have
\begin{align*}
-z_0''(t)&\leq \phi(t)(g(z_0(t-a)-w(t-a))+h(z_0(t-a)+1))+p(t)\\
 &\leq \phi(t)(g(\frac{1}{2}z_0(t-a))+h(z_0(t-a)+1))+p(t),\quad
  \forall t\in [a,t_0].
\end{align*}
Since $z_0'(t-a)\geq z_0'(t)$ for $t\in [a,t_0]$, we
have
$$
-z_0''(t)z_0'(t)\leq
[\phi(t)(g(\frac{1}{2}z_0(t-a))+h(z_0(t-a)+1))+p(t)]z_0'(t-a),
$$
for all $t\in [a,t_0]$. Integrating from $t(t\in [a, t_0])$ to $t_0$, we
have
\begin{align*}
&[z_0'(t)]^2\\
&\leq 2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int^{t_0}_t[g(\frac{1}{2}z_0(s-a))
+h(z_0(s-a)+1)+1] z_0'(s-a)ds\\
&\leq 2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int^{z_0(t_0-a)}_{z_0(t-a)}
[g(\frac{1}{2}s)+h(s+1)+1]ds\\
&\leq 2\sup_{t\in [a,1]}[\phi(t)+p(t)]\int_0^{z_0(t_0)}
[g(\frac{1}{2}s)+h(s+1)+1]ds\\
&=B(R_0)\\
\end{align*}
and so
\begin{equation}
z_0'(t)\leq \sqrt{ B(R_0)},\quad  \forall t\in [a,t_0].\label{e2.11}
\end{equation}
Then integrating from $a$ to $t_0$, we
have
\begin{equation}
z_0(t_0)\leq z_0(a)+\sqrt{B(R_0)}.\label{e2.12}
\end{equation}
On the other hand, by \eqref{e2.8} and \eqref{e2.9}, we have
$$
-z_0''(t)\leq \phi(t)[g(\mu(t-a))+h(\mu(t-a)+1)]+p(t), \quad
 \forall t\in (0,a].
$$
Integrating from $t(t\in (0,a))$ to $a$, by \eqref{e2.11}, we have
 $$
 z'_0(t)\leq
 z_0'(a)+\int^a_0[\phi(s)(g(\mu(s-a))+h(\mu(s-a)+1)+p(s)]ds\\
 \leq  \sqrt{B(R_0)}+A.
 $$
Then integrating from 0 to $a$, we have
\begin{equation}
 z_0(a)\leq \sqrt{B(R_0)}+A.\label{e2.13}
\end{equation}
It follows from \eqref{e2.12} and \eqref{e2.13} that
\begin{equation}
z_0(t_0)\leq 2\sqrt{B(R_0)}+A\label{e2.14}
\end{equation}
Since $z_0(t_0)=R_0$, from \eqref{e2.10} and \eqref{e2.14}, we have
$$
R_0\leq 2\sqrt{B(R_0)}+A,
$$
 which is a contradiction to (H3). Thus \eqref{e2.4} holds. By the
 properties of fixed point index, we have
 $$
 i(T_n,\Omega_0, Q)=i(\theta, \Omega_0, Q)=1.
$$
 The proof is completed.
\end{proof}

\begin{remark} \label{rmk2} \rm
The inequality \eqref{e*} played an important
  role in the proof of Lemma \ref{lem3}. This type of inequality has been
employed extensively in the literature \cite{a1,e1,o1}. The main
idea of our proof of Lemma \ref{lem3} is derived from \cite{a1}.
\end{remark}

 \section{Main Results}

\begin{theorem} \label{thm1}
Suppose that (H1)-(H4) hold. Assume that
\begin{equation}
\lim_{x\to +\infty}\frac{h(x)}{x}=0.\label{e3.1}
\end{equation}
Then \eqref{e1.1} has at least two positive solutions.
\end{theorem}

\begin{proof} For each positive integer $n$, let us define
 the operator $T_n$ by \eqref{e2.1}. It follows from Lemma \ref{lem2} that
 $T_n:Q\to Q$ is a completely continuous operator for every
 $n$. Let $\delta$ be a positive number such that
$$
0<\delta<\min\big\{1,(\int^1_0\phi(s)ds)^{-1}\big\}.
$$
 It follows from \eqref{e3.1} that there exists $R>u_1$ such that
 $h(x)\leq \delta x$ for all
 $x\geq R$. Since $h:R^+\to R^+$ is increasing,
 then
 $$
h(x)\leq \delta x+h(R), \quad \forall x\in R^+.
$$
 By (H4), there exists $\widetilde{u}_1>u_1$ such that
\begin{equation}
\alpha(1-\beta-a)h(\frac{1}{2}\widetilde{u}_1)
 \int^{\beta+a}_{\alpha+a}\phi_0(s)ds>\widetilde{u}_1.
 \label{e3.2}
\end{equation}
Put
\begin{equation}
R_1=\max\big\{2\widetilde{u}_1,\frac{2[A+\|w\|+
(\|w\|+\|\mu\|+R_0+1+h(R))\int^1_0\phi(s)ds]}{1-\delta\int^1_0\phi(s)ds}
\big\},\label{e3.3}
\end{equation}
\begin{gather*}
\Omega_0=\{x\in Q:\|x\|<R_0\},\\
\Omega_1=\{x\in Q:\|x\|<R_1\},\\
\Omega_{01}=\{x\in Q:\|x\|<R_1, \quad
\inf_{t\in [\alpha,\beta]}x(t)>u_1\},\\
U_{01}=\{x\in Q:\|x\|<R_1 \quad
\inf_{t\in [\alpha,\beta]}x(t)>\widetilde{u}_1\}.
\end{gather*}
It is easy to see that $\Omega_0$, $\Omega_1$, $\Omega_{01}$ and
$U_{01}$ are bounded open convex sets of $Q$, and
that
$$
\Omega_0\subset \Omega_1,\quad
\Omega_{01}\subset \Omega_1,\quad
U_{01}\subset \Omega_1,\quad
 \Omega_0\cap \Omega_{01}=\emptyset, \quad
U_{01}\subset \Omega_{01}.
$$
For each positive integer $n$ and $x\in \overline{\Omega}_1$,
by \eqref{e2.2} and \eqref{e3.3}, we have
\begin{equation}
\begin{aligned}
&(T_nx)(t)\\
&\leq\int^1_0G(t,s)[\phi(s)(g([x(s-a)]^*+n^{-1})+h([x(s-a)]^*+n^{-1}))+p(s)]ds\\
&\leq \int^a_0\phi(s)(g(\mu(s-a))ds
 +\int_a^1\phi(s)(g(\frac{1}{2}R_0a(s-a))ds\\
&+h(\|x\|+\|w\|+\|\mu\|+R_0+1))\int^1_0\phi(s)ds+\|w\|\\
&\leq A+\|w\|+[\delta(\|x\|+\|w\|+\|\mu\|+R_0+1)+h(R)]\int^1_0\phi(s)ds\\
&\leq A+\|w\|+[\delta R_1+\|w\|+\|\mu\|+R_0+1+h(R)]\int^1_0\phi(s)ds\\
&<R_1,\quad  \forall t\in [0,1].
\end{aligned} \label{e3.4}
\end{equation}
Since $(T_nx)(t)=0$ for $t\in [-a,0]$, it follows that $\|T_nx\|<R_1$
for all $x\in \overline{\Omega}_1$. Hence,
$T_n(\overline{\Omega}_1)\subset \Omega_1$ for all
positive integer $n$. By Lemma \ref{lem1}, we have for each positive
integer $n$
\begin{equation}
i(T_n, \Omega_1, Q)=1.\label{e3.5}
\end{equation}
For any $x\in \overline{\Omega}_{01}$, by \eqref{e3.4}, we have
$\|T_nx\|<R_1$. It is easy to see that for
$x\in\overline{\Omega}_{01}$
$$
[x(t-a)]^*=x(t-a)-w(t-a)\geq\frac{1}{2}x(t-a)\geq\frac{1}{2}u_1,
\quad  t\in [\alpha+a, \beta+a].
$$
Then the assumption (H4) implies
\begin{align*}
(T_nx)(t)&\geq\int^{\beta+a}_{\alpha+a}G(t,s)\phi_0(s)h_0([x(s-a)]^*)ds\\
&\geq \alpha(1-\beta-a)h_0(\frac{1}{2}u_1)\int^{\beta+a}_{\alpha+a}\phi_0(s)ds\\
&>u_1,\quad \forall t\in [\alpha, \beta],
 \end{align*}
 and so,
$T_n(\overline{\Omega}_{01})\subset\Omega_{01}$ for every positive
integer $n$. Also by Lemma \ref{lem1}, we have
\begin{equation}
i(T_n, \Omega_{01}, Q)=1\label{e3.6}
\end{equation}
for every positive integer $n$. Similarly, by \eqref{e3.2} we can show that
\begin{equation}
i(T_n, U_{01},Q)=1\label{e3.7}
\end{equation}
for every positive integer $n$. It follows from \eqref{e3.7}  that for
every positive integer $n$, $T_n$ has at least one fixed point
$\widetilde{x}_n\in \overline{U}_{01}$.
It is easy to see that
\begin{align*}
[\widetilde{x}_n(t-a)]^*
&=\begin{cases} \mu(t-a),&t\in [0,a] \\
\widetilde{x}_n(t-a)-w(t-a),&t\in (a,1]
\end{cases}
\\
&\geq\begin{cases}
\mu(t-a),& t\in (0,a]\\
\frac{1}{2}\widetilde{x}_n(t-a),& t\in (a,1]
\end{cases}
\\
& \geq x_0(t-a)+\widetilde{x}_0(t-a),\quad t\in [0,1],
\end{align*}
and so
\begin{equation}
g([\widetilde{x}_n(t-a)]^*+n^{-1})\leq g(x_0(t-a)+\widetilde{x}_0(t-a)),
t\in (0,1]\backslash\{a\}.\label{e3.8}
\end{equation}
Using essentially the same argument as in  Lemma \ref{lem3}, we see that there
 exists $t_n\in(0,1)$ such that $\widetilde{x}'_n(t_n)=0$, and
$$
-\widetilde{x}''_n(t)\leq \phi(t)
[g(x_0(t-a)+\widetilde{x}_0(t-a)+n^{-1})+h(R_1+1)]+p(t),
$$
for all $t\in (0,1]$. Integration yields
$$
|\widetilde{x}_n'(t)|\leq \int^1_0[\phi(s)(g(x_0(s-a)+
\widetilde{x}_0(s-a))+h(R_1+1))+p(s)]ds,
$$
for all $t\in(0,1]$. This means that $\{ \widetilde{x}_n\}$ is
equicontinuous on $[0,1]$.
Since $ \widetilde{x}_n(t)=0$ for $t\in [-a,0]$, $\{\widetilde{x}_n\}$
is equicontinuous on $[-a, 0]$. Therefore, the
Arzela-Ascoli Theorem guarantees the existence of a subsequence
$\{ \widetilde{x}_{n_i}\}$ of $\{ \widetilde{x}_n\}$  and a
function $x_{01}\in  \overline{U}_{01}$ with
$\widetilde{x}_{n_i}$ converging uniformly on $[-a,1]$ to $x_{01}$
as $i\to \infty$.
   From $ \widetilde{x}_n =T_n \widetilde{x}_n$, by \eqref{e3.8} and using the Lebesgue dominated
convergence Theorem, we have
\begin{align*}
x_{01}(t)&=\begin{cases}
0,&t\in [-a,0];\\
\int^1_0G(t,s)[f(s,[x_{01}(s-a)]^*)+p(s)]ds,&t\in
(0,1].
\end{cases}
\\
&=\begin{cases}
0,&t\in [-a,0];\\
\int^1_0G(t,s)[f(s,x_{01}(s-a)+x_0(s-a)\\
-w(s-a))+p(s)]ds,& t\in (0,1].
\end{cases}
\end{align*}
Let $y_{01}(t)=x_{01}(t)+x_0(t)-w(t)$ for $t\in [-a,1]$. Then, we have
$$
y_{01}(t)=\begin{cases}
\mu(t), &t\in [-a,0];\\
\int^1_0G(t,s)f(s, y_{01}(s-a))ds,&t\in (0,1].
\end{cases}
$$
It is easily verfied that $y_{01}$  is a positive solution of \eqref{e1.1}.
It follows from \eqref{e3.5},
 \eqref{e3.6} and Lemma \ref{lem3} that for every positive integer $n$
$$
 i(T_n,\Omega_1\backslash(
\bar{\Omega}_{01}\cup\bar{\Omega}_0), Q)=i(T_n,
\Omega_1,Q)-i(T_n,\Omega_{01}, Q)-i(T_n, \Omega_0, Q)=-1.
$$
Therefore,  $T_n$ has at least one fixed point
$\bar{x}_n\in \Omega_1\backslash(\bar{\Omega}_{01}\cup \bar{\Omega}_0)$
for every positive integer $n$.  For every positive integer $n$, there
is at least one point $t_n\in [\alpha,\beta]$ such that
$\bar{x}_n(t_n)\leq u_1$. In a similar way as above, we can show
that there exist a subsequence $\{\bar{x}_{n_i}\}$ of
$\{\bar{x}_n\}$,
  $x_1\in \overline{\Omega_1\backslash( \bar{\Omega}_{01}\cup \bar{\Omega}_0)}$
and a point $t_0\in [\alpha,\beta]$
such that $\bar{x}_{n_i}$ convergence uniformly
 on [-a,1] to $x_1$ as $i\to \infty$, and  $x_1(t_0)\leq u_1$.
 Let $y_1(t)=x_1(t)+x_0(t)-w(t)$ for $t\in [-a,1]$.
Then $y_1$ is a positive solution of
 \eqref{e1.1}. Since
 $$ x_1(t_0)\leq u_1<\widetilde{u}_1\leq x_{01}(t_0),
$$
 $y_{01}$ and $y_1$ are two distinct positive solutions of \eqref{e1.1}.
 \end{proof}

\begin{theorem} \label{thm2}
Suppose that (H1)-(H4) hold, and that there exists $R_1>u_1$ such that
\begin{gather*}
A+\|w\|+h(R_1+\|w\|+\|\mu\|+1)\int^1_0\phi(s)ds<R_1,\label{e3.9}\\
\lim_{x\to+\infty}\frac{h_0(x)}{x}=+\infty.\label{e3.10}
\end{gather*}
 Then \eqref{e1.1} has at least three positive solutions.
\end{theorem}

\begin{proof}
For every positive integer $n$, let us define an operator $T_n$ by
\eqref{e2.1}. By \eqref{e3.9}, there exists a
 positive number $\bar{R}_1>R_1$ such that
\begin{equation}
A+\|w\|+h(\bar{R}_1+\|w\|+\|\mu\|+1)\int^1_0\phi(s)ds<\bar{R}_1.
\label{e3.11}
\end{equation}
 Let us define the open sets $\Omega_0$, $\Omega_{01}$, $\Omega_1$ and
$U_{01}$ as in Theorem \ref{thm1}. Let $ U_1 =\{x\in Q:\|x\|<\bar{R}_1\}$.
It is easy to see that for any $x\in \bar{\Omega}_1$ and $t\in [0,1]$
 $$
R_1+\|w\|+\|\mu\|\geq[x(t-a)]^*\geq
 \begin{cases}
 \mu(t-a),&t\in [0, a];\\
 \frac{1}{2}R_0a(t-a),&t\in[a,1].
 \end{cases}
$$
  Since $h:R^+\to R^+$ is increasing, we have
  $$
h([x(t-a)]^*+n^{-1})\leq h(R_1+\|w\|+\|\mu\|+1),\forall t\in [0,1].
$$
Therefore, by \eqref{e3.9}, we have
\begin{align*}
|(T_nx)(t)|
&\leq A+\|w\|+h(R_1+\|w\|+\|\mu\|+1)\int^1_0\phi(s)ds\\
&<R_1,\quad \forall x\in \bar{\Omega}_1,\; t\in [0,1]
\end{align*}
for any positive integer $n$. This means that
$T_n(\bar{\Omega}_1)\subset \Omega_1$.
Similarly, by \eqref{e3.11}, we can show $T_n(\bar{U}_1)\subset U_1$
for every positive integer $n$.
By Lemma \ref{lem1}, we have
\begin{equation}
i(T_n, U_1, Q)=1.\label{e3.12}
\end{equation}
In a similar way as Theorem \ref{thm1}, we can show that \eqref{e1.1} has
at least two positive solutions $y_1$ and $y_{01}$ such that
$$
y_1(t)=x_1(t)+x_0(t)-w(t),\quad
 y_{01}(t)=x_{01}(t)+x_0(t)-w(t),\quad  \forall t\in [-a,1],
$$
where $x_1\in \overline{\Omega_1\backslash(\bar\Omega_{01}\cup
\bar{\Omega}_0)}$ and $ x_{01}\in \bar{U}_{01}$.

Now, we shall show the existence of the third positive solution of
\eqref{e1.1}. Let
\begin{equation}
M> 2(\alpha(1-\beta)\sup_{t\in [0,1]}\int^{\beta+a}_{\alpha+a}G(t,s)
\phi_0(s)ds)^{-1}. \label{e3.13}
\end{equation}
By \eqref{e3.10}, there exists $\widetilde{R}>\bar{R}_1$ such that
$h_0(y)\geq My$ for any $y\geq \widetilde{R}$. Set
$R_2=2\widetilde{R }\alpha^{-1}(1-\beta)^{-1}$, $\Omega_2=\{x\in
Q:\|x\|<R_2\}$. Let  $\psi_0\in Q\backslash \{\theta\}$. We claim
that for every positive integer $n$
\begin{equation}
y\neq T_n y+\lambda \psi_0,\quad \lambda\geq 0, \quad y\in \partial \Omega_2.
\label{e3.14}
\end{equation}
If not, then there exist $n_0\in \textbf{N}$,
$y_0\in\partial \Omega_2$ and $\lambda_0\geq 0$ such that
$$
y_0=T_{n_0}y_0+\lambda_0\psi_0
$$
It is easy to see that
\begin{align*}
[y_0(t-a)]^*&=y_0(t-a)-w(t-a)\\
&\geq\frac{1}{2}\|y_0\|(t-a)(1-t+a)\\
&\geq\frac{1}{2} R_2\alpha(1-\beta)
>\widetilde{R},\quad  t\in [\alpha+a,\beta+a].
\end{align*}
Therefore,
\begin{align*}
R_2\geq y_0(t)&\geq \int^1_0G(t,s)\phi_0(s)h_0([y_0(s-a)]^*)ds\\
&\geq \int^{\beta+a}_{\alpha+a}G(t,s)\phi_0(s)h_0(y_0(s-a)-w(s-a))ds\\
&\geq \frac{1}{2}MR_2\alpha(1-\beta)\int^{\beta+a}_{\alpha+a}G(t,s)\phi_0(s)ds,
\quad t\in [0,1].
\end{align*}
Hence
$$
M\leq 2(\alpha(1-\beta)\sup_{t\in [0,1]}\int^{\beta+a}_{\alpha+a}G(t,s)
\phi_0(s)ds)^{-1},
$$
which is a contradiction to \eqref{e3.13}. Hence, \eqref{e3.14} holds.
 From the properties of the fixed point index, we have
$$
i(T_n, \Omega_2, Q)=0.\label{e3.15}
$$
It follows from \eqref{e3.12} and \eqref{e3.15} that
 $$
i(T_n, \Omega_2\backslash\bar{U}_1, Q)=i(T_n, \Omega_2, Q)-i(T_n, U_1, Q)=-1
$$
 for every positive integer $n$. Hence, $T_n$ has at least one fixed point
$\widetilde{x}_n\in \Omega_2\backslash\bar{U}_1$.
 Using essentially the same argument as in Theorem \ref{thm1}, we can show
 that there exist a
subsequence $\{ \widetilde{x}_{n_i}\}$ of $\{\widetilde{x}_n\}$, and
$x_3\in \overline{\Omega_2\backslash\bar{U}_1}$ such that
$\widetilde{x}_{n_i}\to x_3(i\to +\infty)$, and
$y_3=x_3+x_0-w$ is a positive solution of \eqref{e1.1}. The
proof is completed.
\end{proof}


\begin{corollary} \label{coro1}
Suppose that (H1)-(H3) hold. Moreover, there exist $R_i,\ u_i(i=1,2,\dots, n)$
with $R_0<u_1<R_1<u_2<R_2<\dots<u_n<R_n$ such that
\begin{gather*}
A+\|w\|+h(R_i+\|w\|+\|\mu\|+1)\int^1_0\phi(s)ds<R_i,\quad i=1,2,\dots, n,
\\
\alpha(1-\beta-a)h_0(\frac{1}{2}u_i)\int^{\beta+a}_{\alpha+a}\phi_0(s)ds>u_i,
\quad i=1,2,\dots, n,
\\
\lim_{x\to+\infty}\frac{h_0(x)}{x}=+\infty.
\end{gather*}
Then \eqref{e1.1} has at least $2n+1$ positive solutions.
\end{corollary}

\begin{corollary} \label{coro2}
Suppose that (H1)-(H3) hold, and
 $\lim_{x\to +\infty}\frac{h_0(x)}{x}=+\infty$.
 Then \eqref{e1.1} has at least one positive solution.
\end{corollary}

We remark that the multiplicity  results for positive solutions
of singular semi-positone delay  differential equations are new.
 Obviously,  we can use the ideas of this paper to establish
multiplicity results for positive solutions  of the more general
delay equation.


\subsection*{Example} Consider the  delay differential boundary-value problem
\begin{equation}
\begin{gathered}
y''+40\big[\frac{1}{\sqrt{y(t-\frac{1}{4})}}+h(y(t-\frac{1}{4}))\big]
=\frac{1}{t^{1/4}}, \quad t\in (0,1]\backslash\{\frac{1}{4}\},\\
y(t)=-t, \quad t\in [-\frac{1}{4}, 0],\\
y(1)=0,
\end{gathered} \label{e3.16}
\end{equation}
where $h:R^+\to R^+$ is increasing,  $h(y)=y^{1/4}$
for $y\in [0, 2\times 10^4]$, and
\[
\lim_{x\to+\infty}\frac{h(x)}{x}=0\,.
\]

\noindent\textbf{Claim:} If there exits $u_1>2\times 10^4$ such that $h(\frac{1}{2}u_1)>2u_1$,
then  the boundary-value problem \eqref{e3.16} has at least two
positive solutions.

\begin{proof} Let $\phi(t)=\phi_0(t)=40$ for $t\in [0,1]$,
$a=1/4$, $g(y)=1/\sqrt{y},$ $p(t)=1/t^{1/4}$,
$\mu(t)=-t$ for $t\in [-\frac{1}{4}, 0]$. It is
easy to see (H1) and (H2) hold. Set $R_0=10^4$. Then, we
have
\begin{align*}
A&\leq \int^{1/4}_0\big[40(\frac{1}{\sqrt{\frac{1}{4}-s}}+h(\frac{1}{4}-s+1))
+\frac{1}{s^{1/4}}\big]ds+\int^1_{1/4}
\frac{40}{\sqrt{\frac{R_0}{8}(s-\frac{1}{4})}}\\
&\leq 80\sqrt{\frac{1}{4}-s}\Big|^0_{1/4}+10h(\frac{5}{4})
+\frac{4}{3}+\frac{320}{\sqrt{R_0}}\\
&=40+10h(\frac{5}{4})+\frac{4}{3}+\frac{320}{\sqrt{R_0}}
\leq 80,
\end{align*}
\begin{align*}
B(R_0)&=2\times (40+\sqrt{2})\int^{R_0}_0
\big(\frac{1}{\sqrt{s/2}}+(s+1)^{\frac{1}{4}}+1\big)ds\\
&< 84(2\sqrt{2R_0}+\frac{4}{5}(R_0+1)^{5/4}+R_0)\\
&<340\sqrt{R_0}+80(R_0+1)^{5/4}+84R_0<8875040,
\end{align*}
and
$$
R_0-2\sqrt{B(R_0)}>R_0-2\sqrt{8875040}>80,
$$
which implies  (H3).
 Put $[\alpha,\beta]=[\frac{1}{3}, \frac{1}{2}]$,
$h(y)=h_0(y)$ for $y\in R^+$. It is easy to check that (H4)
holds. Thus, by Theorem \ref{thm1}, the conclusion holds. The proof is
completed.
\end{proof}


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\end{document}