\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 151, pp. 1--9.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/151\hfil Existence of solutions]
{Existence of solutions to first-order singular and nonsingular
initial value problems}

\author[P. S. Kelevedjiev\hfil EJDE-2008/151\hfilneg]
{Petio S. Kelevedjiev}

\address{Petio S. Kelevedjiev \newline
Department of Mathematics, Technical University of Sliven,
Sliven, Bulgaria}
\email{keleved@lycos.com}

\thanks{Submitted July 20, 2008. Published November 1, 2008.}
\subjclass[2000]{34B15, 34B16}
\keywords{Initial value problem; first order differential equation;
 singularity; \hfill\break\indent sign conditions}

\begin{abstract}
 Under barrier strip type arguments we investigate the  existence
 of global solutions to the initial value problem
 $x'=f(t,x,x')$, $x(0)=A$,
 where the scalar function $f(t,x,p)$ may be singular at $t=0$.
\end{abstract}

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

\section{Introduction}

Results presented in Kelevedjiev O'Regan \cite{k1} show the solvability
of the singular initial-value problem (IVP)
\begin{equation}
x'=f(t,x,x'),\quad x(0)=A,\label{e1.1}
\end{equation}
where the function $f$ may be unbounded when $t\to 0^{-}$.
In this paper we give existence results for problem \eqref{e1.1}
under less restrictive assumptions which alow $f$ to be unbounded
when $t\to 0$; i.e., here  $f$ may be unbounded for $t$ tending to $0$
from both sides. In fact, we consider the nonsingular problem \eqref{e1.1}
with $f:D_t\times D_x\times D_p\to \mathbb{R}$ continuous on a suitable
subset of $D_t\times D_x\times D_p$ containing $(0,A)$ and the singular
problem \eqref{e1.1} with $f(t,x,p)$ discontinuous for $(t,x,p)\in S$ and
defined at least for $(t,x,p)\in(D_t\times D_x\times D_p)\setminus S$, where
$D_t, D_x, D_p\subseteq \mathbb{R}$ may be bounded, and
$S=\{0\}\times{\sf X}\times{\sf P}$ for some sets ${\sf X}\subseteq D_x$
 and ${\sf P}\subseteq D_p$.

Singular and nonsingular IVPs for the equation $x'=f(t,x)$ have
been discussed extensively in the literature; see, for example,
\cite{a2,b1,b2,c1,c2,c3,f1,f2,h1,s1}.
Singular IVPs of the form \eqref{e1.1} have been received very
little attention; we mention only \cite{a1,k1}.

This paper is divided into three main sections.
For the sake of completeness, in {Section 2} we state the
Topological transversality theorem \cite{g1}. In Section 3 we discus the
nonsingular problem \eqref{e1.1}. Obtain a new existence result applying
the approach \cite{g1}. Moreover, we again use the barrier strips technique
initiated in \cite{k2}. In Section 4 we use the obtained existence result
for the nonsingular problem \eqref{e1.1} to study the solvability of the
singular problem \eqref{e1.1}.


\section{Topological preliminaries}

 Let $X$ be a metric space, and $Y$ be a convex subset of a Banach space
$E$. We say that the homotopy $\{H_\lambda:X\to Y\}$,
$0\le\lambda\le1$, is compact if the map
$H(x,\lambda):X\times[0,1]\to Y$ given by
$H(x,\lambda)\equiv H_\lambda(x)$ for $(x,\lambda)\in X\times[0,1]$
is compact.

Let $U\subset Y$ be open in $Y$, $\partial U$ be the boundary of
$U$ in $Y$, and $\overline {U}=\partial U\cup U$. The compact map
$F:\overline {U}\to Y$ is called admissible if it is fixed point free on
$\partial U$. We denote the set of all such maps by
$\mathbf{L}_{\partial U}(\overline {U},Y)$.


\begin{definition}[{\cite[Chapter I, Def. 2.1]{g1}}] \label{def2.1} \rm
The map $F$ in $\mathbf{L}_{\partial U}(\overline {U},Y)$ is inessential
if there is a fixed point free compact map $G:\overline {U}\to Y$ such
that $G|\partial U=F|\partial U$. The map
$F$ in $\mathbf{L}_{\partial U}(\overline {U},Y)$ which is not
inessential is called essential.
\end{definition}

\begin{theorem}[{\cite[Chapter I, Theorem 2.2]{g1}}] \label{thm2.2}
 Let $p\in U$ be arbitrary and $F\in \mathbf{L}_{\partial U}(\overline {U},Y)$
be the constant map $F(x)=p$ for $x\in\overline {U}$. Then $F$ is essential.
\end{theorem}

\begin{proof} Let $G:\overline {U}\to Y$ be a compact map such that
$G|\partial U=F|\partial U$. Define the map $H:Y\to Y$ by
$$
H(x)=\begin{cases} p &\text{for }x\in Y\backslash\overline {U},\\
G(x) &\text{for } x\in\overline {U}.
\end{cases}
$$
Clearly $H:Y\to Y$ is a compact map. By Shauder fixed point theorem,
$H$ has a fixed point $x_0\in Y$; i. e., $H(x_0)=x_0$.
By definition of $H$ we have $x_0\in U$. Thus, $G(x_0)=x_0$ since $H$
equals $G$ on $U$. So every compact map from $\overline {U}$ into $Y$ which
agrees with $F$ on $\partial U$ has a fixed point. That is, $F$
is essential.
\end{proof}

\begin{definition}[{\cite[Chapter I, Def. 2.3]{g1}}] \label{def2.3} \rm
 The maps $F,G\in \mathbf{L}_{\partial U}(\overline {U},Y)$ are called
homotopic $(F\sim G)$ if there is a compact homotopy
$ H_\lambda:\overline {U}\to Y$, such that $ H_\lambda$ is
admissible for each $\lambda\in[0,1]$ and $G=H_0,\;F=H_1$.
\end{definition}

\begin{lemma}[{\cite[Chapter I, Theorem 2.4]{g1}}] \label{lem2.4}
The map $F\in \mathbf{L}_{\partial U}(\overline {U},Y)$ is inessential
if and only if it is homotopic to a fixed point free map.
\end{lemma}

\begin{proof} Let $F$ be inessential and $G:\overline {U}\to Y$ be a
compact fixed point free map such that $G|\partial U=F|\partial U$. Then
the homotopy $ H_\lambda:\overline {U}\to Y$, defined by
$$
H_\lambda(x)=\lambda F(x)+(1-\lambda)G(x),\quad\lambda\in[0,1],
$$
is compact, admissible and such that $G=H_0$, $F=H_1$.

Now let $H_0:\overline {U}\to Y$ be a compact fixed point free map, and
$H_\lambda:\overline {U}\to Y$ be an admissible homotopy joining
$H_0$ and $F$. To show that $H_\lambda,\lambda\in[0,1]$, is an
inessential map consider the map $H:\overline {U}\times[0,1]\to Y$ such
that $H(x,\lambda)\equiv H_\lambda(x)$ for each
$x\in\overline {U}$ and $\lambda\in[0,1]$ and define the set
$B\subset\overline {U}$ by
$$
B=\{x\in\overline {U}:H_\lambda(x)\equiv H(x,\lambda)=x
\text{ for some }\lambda\in[0,1]\}.
$$
If $B$ is empty, then $H_1=F$ has no fixed point which means that
$F$ is inessential. So we may assume that $B$ is non-empty. In
addition $B$ is closed and such that $B\cap\partial U=\emptyset$
since $H_\lambda,\lambda\in[0,1]$, is an admissible map. Now consider
the Urysohn function $\theta:\overline {U}\to[0,1]$ with
$$
\theta(x)=1\text{ for } x\in\partial U\quad\text{and}\quad
\theta(x)=0\text{ for }x\in B
$$
and define the homotopy $H^{*}_\lambda:\overline {U}\to Y,\lambda\in[0,1]$,
by
$$
H^{*}_\lambda=H(x,\theta(x)\lambda)\quad\text{for }
(x,\lambda)\in\overline {U}\times[0,1].
$$
It easy to see that $H^{*}_\lambda:\overline {U}\to Y$ is inessential.
In particular $H_1=F$ is inessential, too. The proof is complete.
\end{proof}

Lemma \ref{lem2.4} leads to the Topological transversality theorem:

\begin{theorem}[{\cite[Chapter I, Theorem 2.6]{g1}}] \label{thm2.5}
 Let $Y$ be a convex subset of a Banach space $E$, and $U\subset Y$ be
open. Suppose that
\begin{itemize}
\item[(i)] $F,G:\overline {U}\to Y$ are compact maps.
\item[(ii)] $G\in \mathbf{L}_{\partial U}(\overline {U},Y)$ is
essential.
\item[(iii)] $H_\lambda(x),\lambda\in[0,1]$, is a compact homotopy
joining $F$ and $G$; i.e., $H_0(x)=G(x)$, $H_1(x)=F(x)$.
\item[(iv)] $H_\lambda(x),\lambda\in[0,1]$, is fixed point free on
$\partial U$.
\end{itemize}
Then $H_\lambda,\lambda\in[0,1]$, has a least one fixed point
$x_0\in U$, and in particular there is a $x_0\in U$ such that
$x_0=F(x_0)$.
\end{theorem}

\section{Nonsingular problem}
Consider the problem
\begin{equation}
x'=f(t,x,x'),\quad x(a)=A,\label{e3.1}
\end{equation}
where $f:D_t\times D_x\times D_p\to R$, and the sets
$D_t, D_x, D_p\subseteq \mathbb{R}$ may be bounded.
Assume that:
\begin{itemize}
\item[(R1)]  There are constants $T>a$, $Q>0$, $L_i,F_i$, $i=1,2$,
and a sufficiently small $\tau>0$ such that $[a,T]\subseteq D_t$,
$L_2-\tau\geq L_1\geq\max\{0,A\}$,
$F_2+\tau\leq F_1\leq\min\{0,A\}$,
$[F_2, L_2]\subseteq D_x$, $[h-\tau,H+\tau]\subseteq D_p$
for $h=-Q-L_1$ and $H=Q-F_1$,
\begin{equation}
\begin{gathered}
f(t,x,p)\leq0\quad\text{for }(t,x,p)\in[a,T]\times[L_1,L_2]\times D_p^+
\quad \text{where }D_p^+=D_p\cap(0,\infty),\\
f(t,x,p)\geq0\quad \text{for }(t,x,p)\in[a,T]\times[F_2,F_1]
 \times D_p^-\quad \text{where } D_p^-=D_p\cap(-\infty,0),\\
pf(t,x,p)\leq0\quad \text{for }(t,x,p)\in[a,T]\times[F_1-\tau,L_1+\tau]
\times(D_Q^-\cup D_Q^+),
\end{gathered}\label{e3.2}
\end{equation}
where $D_Q^-=\{p\in D_p:p<-Q\}$ and $D_Q^+=\{p\in D_p:p>Q\}$.
\end{itemize}

\subsection*{Remark}
The sets $D_p^-,D_p^+,D_Q^-$ and $D_Q^+$ are not empty because
$h-\tau<h=-Q-L_1<-Q<0$, $H+\tau>H=Q-F_1>Q>0$ and
$[h-\tau,H+\tau]\subseteq D_p$.

\begin{itemize}
\item[(R2)]   $f(t,x,p)$ and $f_p(t,x,p)$ are continuous for
 $(t,x,p)\in\Omega_\tau=[a,T]\times[F_1-\tau,L_1+\tau]\times[h-\tau,H+\tau]$
and for some $\varepsilon>0$
$$
f_p(t,x,p)\leq1-\varepsilon\quad \text{for }(t,x,p)\in\Omega_\tau,
$$
where $T,F_1,L_1,h,H$ and $\tau$ are as in (R1).
\end{itemize}

Now for $\lambda\in[0,1]$ construct the family of IVPs
\begin{equation}
x'+(1-\lambda)x=\lambda f(t,x,x'+(1-\lambda)x),\quad
x(a)=A.\label{e3.3lambda}
\end{equation}
Note that  \eqref{e3.3lambda} with $\lambda=1$ is  problem \eqref{e1.1},
and that when $\lambda=0$, this problem has a unique solution
 $x(t)=Ae^{a-t}$, $t\in \mathbb{R}$.

For the proof of the main result of this section we need the following
auxiliary result.

\begin{lemma}[{\cite[Lemma 3.1]{k1}}] \label{lem3.1}
 Let {\rm (R1)} hold and $x(t)\in C^1[a,T]$ be a solution to
\eqref{e3.3lambda} with $\lambda\in[0,1]$. Then
$$
F_1\leq x(t)\leq L_1\quad\text{and}\quad
-Q-L_1\leq x'(t)\leq Q-F_1\quad \text{for }t\in[a,T].
$$
\end{lemma}

We will omit the proof of the above lemma. Note only
that \eqref{e3.2} yields
\begin{equation}
-Q\leq x'(t)+(1-\lambda)x(t)\leq Q\quad\text{for }\lambda\in[0,1]
\text{ and }t\in[a,T],\label{e3.4}
\end{equation}
which together with the obtained bounds for $x(t)$ gives the bounds
for $x'(t)$.

\begin{lemma} \label{lem3.2}
Let {\rm (R1)} and {\rm (R2)} hold. Then there exists a function
$\Phi(\lambda,t,x)$ continuous for
$(\lambda,t,x)\in[0,1]\times[a,T]\times[F_1-\tau,L_1+\tau]$ and such that:
\begin{itemize}
\item[(i)] The family
$$
x'+(1-\lambda)x=\Phi(\lambda,t,x),\quad x(a)=A,
$$
 and  family \eqref{e3.3lambda}  are equivalent.

\item[(ii)] $\Phi(0,t,x)=0$ for $(t,x)\in[a,T]\times[F_1-\tau,L_1+\tau]$.
\end{itemize}
\end{lemma}

\begin{proof} (i) Consider the function
$$
G(\lambda,t,x,p)=\lambda f(t,x,p)-p\quad\text{for }
(\lambda,t,x,p)\in[0,1]\times\Omega_\tau.
$$
Since $h-\tau<-Q$ and $H+\tau>Q$, \eqref{e3.2} implies
$$
f(t,x,h-\tau)\geq0,\quad
f(t,x,H+\tau)\leq0\quad\text{for }(t,x)\in[a,T]\times[F_1-\tau,L_1+\tau],
$$
which together with the definition of the function $G$ yields
\begin{equation}
G(\lambda,t,x,h-\tau)\,G(\lambda,t,x,H+\tau)<0,\quad
(\lambda,t,x)\in[0,1]\times[a,T]\times[F_1-\tau,L_1+\tau].\label{e3.5}
\end{equation}
In addition, $G(\lambda,t,x,p)$ and
\begin{equation}
G_p(\lambda,t,x,p)=\lambda f_p(t,x,p)-1 \label{e3.6}
\end{equation}
are continuous for $(\lambda,t,x,p)\in[0,1]\times\Omega_\tau$
because $f(t,x,p)$ and $f_p(t,x,p)$ are continuous for
$(t,x,p)\in\Omega_\tau$. Besides, from
${f_p(t,x,p)\leq1-\varepsilon}$ for $(t,x,p)\in\Omega_\tau$ we have
\begin{equation}
G_p(\lambda,t,x,p)\leq\lambda(1-\varepsilon)-1\leq\max\{-\varepsilon,-1\}
\quad \text{for }(\lambda,t,x,p)\in[0,1]\times\Omega_\tau.
\label{e3.7}
\end{equation}
Using \eqref{e3.5}, \eqref{e3.6} and \eqref{e3.7} we conclude that
the equation
$$
G(\lambda,t,x,p)=0,\quad (\lambda,t,x,p)\in[0,1]\times\Omega_\tau
$$
defines a unique function $\Phi(\lambda,t,x)$ continuous for
$(\lambda,t,x)\in[0,1]\times[a,T]\times[F_1-\tau,L_1+\tau]$ and such that
$$
G(\lambda,t,x,\Phi(\lambda,t,x))=0\quad\text{for }
(\lambda,t,x)\in[0,1]\times[a,T]\times[F_1-\tau,L_1+\tau];
$$
i.e., $p=\Phi(\lambda,t,x)$ for
$(\lambda,t,x)\in[0,1]\times[a,T]\times[F_1-\tau,L_1+\tau]$.

Now write the differential equation \eqref{e3.3lambda} as
$$
\lambda f(t,x,x'+(1-\lambda)x)-(x'+(1-\lambda)x)=0
$$
and use that for $\lambda\in[0,1]$ and $t\in[a,T]$,
$$
x(t)\in[F_1,L_1]\subset[F_1-\tau,L_1+\tau],
$$
 by lemma \ref{lem3.1}, and
$$
x'(t)+(1-\lambda)x(t)\in[-Q,Q]\subset[h-\tau,H+\tau],
$$
according to \eqref{e3.4},
to conclude that the first part of the assertion is true.

\noindent(ii) It follows immediately from $G(0,t,x,0)=0$ for
$(t,x)\in\times[a,T]\times[F_1-\tau,L_1+\tau]$.
\end{proof}

We will only sketch the proof of the following result since
it is similar to the proof of  \cite[Theorem 2.3]{k1}.


\begin{theorem} \label{thm3.3}
Let {\rm (R1)} and {\rm (R2)} hold. Then the nonsingular
IVP \eqref{e1.1} has at least one solution in $C^1[a,T]$.
\end{theorem}

\begin{proof} Consider the family of IVPs
\begin{equation}
x'+(1-\lambda)x=\Phi(\lambda,t,x),\;x(a)=A,\label{e3.8lambda}
\end{equation}
where $\Phi$ is the function from Lemma \ref{lem3.2},
define the maps
\begin{gather*}
j:C^{1}_{I}[a,T]\to C[a,T]\quad\text{by}\quad jx=x,\\
V_\lambda:C^{1}_{I}[a,T]\to C[a,T]\quad \text{by}\quad
 V_\lambda x=x'+(1-\lambda)x,\lambda\in[0,1],\\
\Phi_\lambda: C[a,T]\to C[a,T]\quad \text{by}\quad
(\Phi_\lambda x)(t)=\Phi(\lambda,t,x(t)),\;t\in[a,T],\;\lambda\in[0,1],
\end{gather*}
where $C^{1}_{I}[a,T]=\{x(t)\in C^{1}[a,T]:x(a)=A\}$, and introduce the set
$$
U=\big\{x\in C^{1}_{I}[a,T]:F_1-\tau<x<L_1+\tau,\,h-\tau<x'<H+\tau\big\}.
$$
Next, define the compact homotopy
$$
H:\overline{U}\times[0,1]\to C^{1}_{I}[a,T]\quad\text{by}\quad
H(x,\lambda )\equiv H_{\lambda }(x)\equiv V^{-1}_\lambda\Phi_\lambda j(x).
$$
By Lemma \ref{lem3.1}, the $C^{1}[a,T]$-solutions to the family \eqref{e3.3lambda}
do not belong to $\partial U$. This means, according to (i) of
 Lemma \ref{lem3.2},
that the family \eqref{e3.8lambda} has no solutions in $\partial U$.
Consequently, the homotopy is admissible because its fixed points are
solutions to \eqref{e3.8lambda}. Besides, from (ii) of Lemma \ref{lem3.2}
it follows $(\Phi_0 x)(t)=0$ for each $x\in U$. Then for each $x\in U$ we have
$$
H_{0}(x)=V^{-1}_0\Phi_0 j(x)=V^{-1}_0(0)=Ae^{a-t}
$$
where $Ae^{a-t}$ is the unique solution to the problem
$$
x'+x=0,\quad x(a)=A.
$$
According to Theorem \ref{thm2.2} the constant map $H_{0}=Ae^{a-t}$ is essential.
Then, by Theorem \ref{thm2.5},
$H_1$ has a fixed point in $U$. This means that problem \eqref{e3.8lambda}
with $\lambda=1$ has at least one solution $x(t)\in C^{1}[a,T]$.
Finally, use Lemma \ref{lem3.2} to see that $x(t)$ is also a solution to
problem \eqref{e3.3lambda} with $\lambda=1$
which coincides with problem \eqref{e1.1}.
\end{proof}


The following  result is known, but we state it for completeness.
We will need it in Section 4.

\begin{lemma} \label{lem3.4}
 Suppose that there are constants $m_i,M_i$, $i=0,1$, such that:
\begin{itemize}
\item[(i)] $\;f(t,x,p)$ is continuously differentiable for
   $(t,x,p)\in[a,T]\times[m_0,M_0]\times[m_1,M_1]$.
\item[(ii)] $1-f_p(t,x,p)\ne0$ for
   $(t,x,p)\in[a,T]\times[m_0,M_0]\times[m_1,M_1]$.
\item[(iii)] $x(t)\in C^1[a,T]$ is a solution to the IVP \eqref{e1.1}
   satisfying the bounds
$$
m_0\leq x(t)\leq M_0,\quad m_1\leq x'(t)\leq M_1\quad
\text{for }t\in[a,T].
$$
\end{itemize}
Then $x''(t)$ exists and is continuous on $[a,T]$ and
$$
x''(t)=\frac{f_t(t,x(t),x'(t))+x'(t)f_x(t,x(t),x'(t))}{1-f_p(t,x(t),x'(t))}
$$
for $t\in[a,T]$.
\end{lemma}

\begin{proof} In view of (i) and (iii) for $t,\,t+h\in[a,T]$ we can work
out the identity
\begin{align*}
&f(t,x,x')-f(t,x,x')+f(t_h,x,x')-f(t_h,x,x')+f(t_h,x_h,x')\\
&-f(t_h,x_h,x')+f(t_h,x_h,x_h')-f(t_h,x_h,x_h')+x'-x'+x'_h-x'_h=0,
\end{align*}
where $t_h=t+h$, $x_h=x(t+h)$ and $x'_h=x'(t+h)$. Using that $x(t)$
is a solution to \eqref{e1.1} we obtain
\begin{align*}
&f(t_h,x,x')-f(t,x,x')+f(t_h,x_h,x')-f(t_h,x,x')\\
&+f(t_h,x_h,x_h')-f(t_h,x_h,x')+x'-x'_h=0
\end{align*}
and apply the mean value theorem to get
\begin{align*}
&\bigl(1-f_p(t_h,x_h,x'+\theta_p(x'_h-x'))\bigr)(x'_h-x')\\
&=f_t(t+\theta_th,x,x')h+f_x(t_h,x+\theta_x(x_h-x),x')(x_h-x),
\end{align*}
for some $\theta_t,\theta_x,\theta_p\in(0,1)$. Dividing by
$\bigl(1-f_p(t_h,x_h,x'+\theta_p(x'_h-x'))\bigr)h$, (ii) allows us to obtain
\begin{align*}
&\lim_{h\to0}\frac{x'(t+h)-x'(t)}{h}\\
&=\lim_{h\to0}\Big(f_t(t+\theta_th,x(t),x'(t))\\
&\quad +f_x\big(t+h,x(t) +\theta_x(x(t+h)-x(t)),x'(t)\big)
 \frac{x(t+h)-x(t)}{h}\Big)\\
&\quad \div\Big( {1-f_p(t+h,x(t+h),x'+\theta_p(x'(t+h)-x'(t)))}\Big),
\end{align*}
from where the lemma follows.
\end{proof}


\section{Singular problem}

 Consider problem \eqref{e1.1} for
\begin{equation}
\parbox{9cm}{
$f(t,x,p)$ is discontinuous for $(t,x,p)\in S$ and is defined
at least for $(t,x,p)\in(D_t\times D_x\times D_p)\setminus S$,
where $D_t,D_x,D_p\subseteq R$, $S=\{0\}\times{\sf X}\times{\sf P}$,
${\sf X}\subseteq D_x$ and ${\sf P}\subseteq D_p$.}
\label{e4.1}
\end{equation}
which allows $f$ to be unbounded at $t=0$.

In this section we assume the following:
\begin{itemize}
\item[(S1)]  There exist constants $T,\,Q>0$, $L_i,F_i$, $i=1,2$, and
a sufficiently small $\tau>0$ such that $(0,T]\subseteq D_t$,
$L_2-\tau\geq L_1\geq\max\{0,A\}$, $F_2+\tau\leq F_1\leq\min\{0,A\}$,
$[F_2, L_2]\subseteq D_x$, $[h-\tau,H+\tau]\subseteq D_p$ for $h=-Q-L_1$
and $H=Q-F_1$,
\begin{gather*}
f(t,x,p)\leq0\quad\text{for }(t,x,p)\in(0,T]\times[L_1,L_2]\times D_p^+,\\
f(t,x,p)\geq0\quad\text{for }(t,x,p)\in(0,T]\times[F_2,F_1]\times D_p^-,\\
pf(t,x,p)\leq0\quad\text{for }(t,x,p)\in(0,T]\times[F_1-\tau,L_1+\tau]
 \times(D_Q^-\cup D_Q^+),
\end{gather*}
where the sets $D_p^-,D_p^+,D_Q^-,D_Q^+$ are as in (R1).

\item[(S2)]  $f(t,x,p)$ and $f_p(t,x,p)$ are continuous for
$(t,x,p)$ in $(0,T]\times[F_1-\tau,L_1+\tau]\times [h-\tau,H+\tau]$,
and for some $\varepsilon>0$,
\begin{equation}
f_p(t,x,p)\leq1-\varepsilon\quad\text{for }(t,x,p)\in(0,T]
\times[F_1-\tau,L_1+\tau]\times[h-\tau,H+\tau],\label{e4.2}
\end{equation}
where the constants $T,F_1,L_1,h,H, \tau$ are as in (S1).

\item[(S3)]  $f_t(t,x,p)$ and $f_x(t,x,p)$ are continuous for
$(t,x,p)\in(0,T]\times[F_1,L_1]\times[h,H]$, where
$T,F_1,L_1,h,H,\tau$ are as in (S1).

\end{itemize}
Note, in [12] the condition \eqref{e4.2} has the form
$$
f_p(t,x,p)\leq-K_p<0\quad\text{for }(t,x,p)\in(0,T]\times[F_1-\tau,L_1
+\tau]\times[h-\tau,H+\tau]
$$
where $K_p$ is a positive constant. Besides, in contrast to \cite{k1},
here we do not need the assumption
$$
\Big|\frac{f_t(t,x,p)+pf_x(t,x,p)}{1-f_p(t,x,p)}\Big|\leq M,\quad
(t,x,p)\in(0,T]\times[F_1,L_1]\times[h,H],
$$
for some constant $M$.


Now we are ready to prove the main result of this paper.
It guarantees solutions to the problem \eqref{e1.1} in the case \eqref{e4.1}.

\begin{theorem} \label{thm4.1}
Let {\rm (S1), (S2), (S3)}  hold. Then the singular
initial-value problem \eqref{e1.1} has at least one solution
in $C[0,T]\cap C^1(0,T]$.
\end{theorem}

\begin{proof}
For $n\in N_T=\{n\in\mathbb{N}:n^{-1}<T\}$ consider the family of IVP's
\begin{equation}
x'=f(t,x,x'),\quad x(n^{-1})=A.  \label{e4.3n}
\end{equation}
It satisfies (R1) and (R2) with $a=n^{-1}$ for each $n\in N_T$.
By Theorem \ref{thm3.3}, \eqref{e4.3n} has a solution $x_n(t)\in C^1[n^{-1},T]$
for each $n\in N_T$; i.e., the sequence $\{x_n\}$, $n\in N_T$, of
$C^1[n^{-1},T]$-solutions to \eqref{e4.3n} exists.

Now, we take a sequence $\{\theta_n\}$, $n\in\mathbb{N}$, such that
$\theta_n\in(0,T)$, $\theta_{n+1}<\theta_n$ for $n\in\mathbb{N}$ and
$\lim_{n\to\infty}\theta_n=0$.

It is clear, $\{x_n\}\subset C^1[\theta_1,T]$ for
$n\in N_1=\{n\in N_T:n^{-1}<\theta_1\}$. In addition, by Lemma \ref{lem3.1},
we have the bounds
$$
F_1\leq x_n(t)\leq L_1,\quad h\leq x'_n(t)\leq H\quad\text{for }
t\in[\theta_1,T],
$$
independent of $n$. On the other hand, $f(t,x,p)$ is continuously
differentiable for $(t,x,p)\in[\theta_1,T]\times[F_1,L_1]\times[h,H]$ and
$$
1-f_p(t,x,p)\geq\varepsilon>0\quad\text{for }(t,x,p)\in[\theta_1,T]
\times[F_1,L_1]\times[h,H].
$$
The hypotheses of Lemma \ref{lem3.4} are satisfied. Consequently, $x_n''(t)$
exists for each $n\in N_1$ and is continuous on $[\theta_1,T]$ and
$$
x_n''(t)=\frac{f_t(t,x_n(t),x_n'(t))
+x_n'(t)f_x(t,x_n(t),x_n'(t))}{1-f_p(t,x_n(t),x_n'(t))}\quad\text{for }
t\in[\theta_1,T],\; n\in N_1.
$$
The a priori bounds for $x_n(t)$ and $x'_n(t)$ on $[\theta_1,T]$ alow us
to conclude that there is a constant $C_1$, independent of $n$, such that
$$
|x''_n(t)|\leq C_1,\quad \in[\theta_1,T],\;n\in N_1.
$$
Applying the Arzela-Ascoli theorem we extract a subsequence
$\{x_{n_1}\}$, $n_1\in N_1$, such that the sequences
$\{x_{n_1}^{(i)}\}$, $i=0,1$, are uniformly convergent on $[\theta_1,T]$
and if
$$
\lim_{n_1\to\infty}x_{n_1}(t)=x_{\theta_1}(t),\quad\text{then }
x_{\theta_1}(t)\in C^1[\theta_1,T]\quad\text{and}\quad
\lim_{n_1\to\infty}x'_{n_1}(t)=x'_{\theta_1}(t).
$$
It is clear that $x_{\theta_1}(t)$ is a solution to the differential
equation $x'=f(t,x,x')$ on $t\in[\theta_1,T]$. Besides, integrating
from  $n_1^{-1}$ to $t,\,t\in(n_1^{-1},T]$, the inequalities
$h\leq x'_{n_1}(t)\leq H$ we get
$$
ht-hn_1^{-1}+A\leq x_{n_1}(t)\leq Ht-Hn_1^{-1}+A\quad\text{for }
t\in[n_1^{-1},T],\; n_1\in N_1,
$$
which yields
$$
ht+A\leq x_{\theta_1}(t)\leq Ht+A\quad\text{for }t\in[\theta_1,T].
$$
Now we consider the sequence $\{x_{n_1}\}$ for
$n_1\in N_2=\{n\in N_T:n^{-1}<\theta_2\}$.
In a similar way we extract a subsequence  $\{x_{n_2}\},\,n_2\in N_2$,
converges uniformly on $[\theta_2,T]$ to a function $x_{\theta_2}(t)$
which is a $C^1[\theta_2,T]$-solution to the differential
equation $x'=f(t,x,x')$ on $[\theta_2,T]$,
$$
ht+A\leq x_{\theta_2}(t)\leq Ht+A\quad\text{for }t\in[\theta_2,T]
$$
and $x_{\theta_2}(t)=x_{\theta_1}(t)$ for $t\in[\theta_1,T]$.

Continuing this process, for $\theta_i\to 0$, we establish a
function $x(t)\in C^1(0,T]$ which is a solution to the differential
equation $x'=f(t,x,x')$ on $(0,T]$,
\begin{equation}
ht+A\leq x(t)\leq Ht+A\quad\text{for }t\in(0,T]\label{e4.4}
\end{equation}
and $x(t)\equiv x_{\theta_i}(t)$ for $t\in[\theta_i,T],\,i\in\mathbb{N}$.
 Also \eqref{e4.4} gives $x(0)=A$ and $x(t)\in C[0,T]$.
 Consequently, $x(t)$ is a $C[0,T]\cap C^1(0,T]$-solution to the
singular IVP \eqref{e1.1}.
\end{proof}


\subsection*{Example}
Consider the initial-value problem
$$
(0.5-x-\sqrt[3]{x'})e^{1/t}-2x'=0,\quad x(0)=1.
$$
Write this equation as
$$
x'=(0.5-x-\sqrt[3]{x'})e^{1/t}-x'
$$
and fix  $T>0$. Then
\begin{gather*}
f(t,x,p)=(0.5-x-\sqrt[3]{p})e^{1/t}-p<0\quad\text{for }
(0,T]\times[2,4]\times(0,\infty),\\
f(t,x,p)=(0.5-x-\sqrt[3]{p})e^{1/t}-p>0\quad\text{for }
(0,T]\times[-3,-1]\times(-\infty,0).
\end{gather*}
In addition, we have
\begin{gather*}
f(t,x,p)=(0.5-x-\sqrt[3]{p})e^{1/t}-p>0\quad\text{for }
(0,T]\times[-1.5,2.5]\times(-\infty,-10), \\
f(t,x,p)=(0.5-x-\sqrt[3]{p})e^{1/t}-p<0\quad\text{for }
(0,T]\times[-1.5,2.5]\times(10,\infty).
\end{gather*}
Consequently, (S1) holds for $Q=10$, $F_2=-3$, $F_1=-1$,
$L_1=2$, $L_2=4$ and $\tau=0.5$. Moreover, $h=-Q-L_1=-12$ and
$H=Q-F_1=11$. Condition  (S2) also holds because
$$
f(t,x,p)\quad\text{and}\quad f_p(t,x,p)=-\frac{e^{1/t}}{3\sqrt[3]{p^2}}-1
$$
are continuous for $(t,x,p)\in(0,T]\times[-1.5,2.5]\times[-12.5,11.5]$
and
$$
f_p(t,x,p)\leq-1\quad\text{for }(t,x,p)\in(0,T]\times[-1.5,2.5]
\times[-12.5,11.5].
$$
Finally, $f_t(t,x,p)=-t^{-2}(0.5-x-\sqrt[3]{p})e^{1/t}$ and
$f_x(t,x,p)=-e^{1/t}$ are continuous for
$(t,x,p)\in(0,T]\times[-1,2]\times[-12,11]$ which means
 (S3) holds.

According to Theorem \ref{thm4.1}, the problem under consideration
has at least one solution in $C[0,T]\cap C^1(0,T]$.


\subsection*{Acknowledgements}
The author is grateful to the anonymous referee for his/her
useful suggestions.
This research was partially supported by grant VU-MI-02/2005
from the Bulgarian NSC.

\begin{thebibliography}{00}

\bibitem{a1} {R. P. Agarwal and P. Kelevedjiev};
\emph{Existence of solutions to a singular initial value problem,}
Acta Mathematica Sinica, English Series, 23 (2007) 1797-1806.

\bibitem{a2} {J. Andres and L. J\"{u}tter};
\emph{Periodic solutions of discontinuous differential systems,}
 Nonlinear Analysis Forum 6, (2001) 391-407.

\bibitem{b1} {L. E. Bobisud and D. O'Regan};
\emph{Existence of solutions to some singular initial value problems,}
J. Math. Anal. Appl. 133, (1988) 215-230.

\bibitem{b2} {A. Bressan and W. Shen};
\emph{On discontinuous differential equations,}
Differential Inclusions and Optimal Control 2, (1988) 73-87.

\bibitem{c1} {A. Cabada and R.L. Pouso};
\emph{On first order discontinuous scalar differential equations,}
Nonlinear Studies 6, (1999) 161-170.

\bibitem{c2} {S. Carl, S. Heikkil\"{a} and M. Kumpulainen};
\emph{On solvability of first order discontinuous scalar differential
equations,} Nonlinear Times and Digest 2, (1995) 11-24.

\bibitem{c3} {S. Carl and S. Heikkil\"{a}};
\emph{Nonlinear Differential Equations in Ordered Spaces,}
Chapman and Hall, Boca Raton, (2000).

\bibitem{f1} {A. F. Filippov};
\emph{Differential Equations with Discontinuous Right-Hand Sides,}
Kluwer, Dordrecht, (1988).

\bibitem{f2} {M. Frigon and D. O'Regan};
 \emph{Existence results for some initial and boundary value
problems without growth restriction,}
Proc. Amer. Math. Soc. 123, (1995) 207-216.

\bibitem{g1} {A. Granas, R.B. Guenther, J.W. Lee};
 \emph{Nonlinear boundary value problems for ordinary differential
equations,} Dissnes. Math. 244 (1985) 1-128.

\bibitem{h1} {S. Heikkil\"{a} and V. Lakshmikantham};
 \emph{A unified theory for first order discontinuous scalar differential
equations,} Nonl. Anal. 26 (1996) 785-797.

\bibitem{k1} {P. Kelevedjiev and D. O'Regan};
 \emph{On the solvability of a singular initial value problem
$x'=f(t,x,x')$ using barrier strips}, Nonl. Anal. 64, (2006) 726-738.

\bibitem{k2} {P. Kelevedjiev};
 \emph{Existence of solutions for two-point boundary value problems},
 Nonl. Anal. 22, (1994) 217-224.

\bibitem{s1} {M. \v{S}enky\v{r}ik and R. B. Guenther};
 \emph{Boundary value problems with discontinuites in the spatial
variable,} J. Math. Anal. Appl. 193, (1995) 296-305.

\end{thebibliography}

\end{document}