\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) $$x'=f(t,x,x'),\quad x(0)=A,\label{e1.1}$$ 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 $$x'=f(t,x,x'),\quad x(a)=A,\label{e3.1}$$ 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 $H=Q-F_1$, $$\begin{gathered} f(t,x,p)\leq0\quad\text{for }(t,x,p)\in[a,T]\times[L_1,L_2]\times D_p^+ \quad \text{where }D_p^+=D_p\cap(0,\infty),\\ f(t,x,p)\geq0\quad \text{for }(t,x,p)\in[a,T]\times[F_2,F_1] \times D_p^-\quad \text{where } D_p^-=D_p\cap(-\infty,0),\\ pf(t,x,p)\leq0\quad \text{for }(t,x,p)\in[a,T]\times[F_1-\tau,L_1+\tau] \times(D_Q^-\cup D_Q^+), \end{gathered}\label{e3.2}$$ 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 $h-\tauH=Q-F_1>Q>0$ 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 $$x'+(1-\lambda)x=\lambda f(t,x,x'+(1-\lambda)x),\quad x(a)=A.\label{e3.3lambda}$$ 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 $$-Q\leq x'(t)+(1-\lambda)x(t)\leq Q\quad\text{for }\lambda\in[0,1] \text{ and }t\in[a,T],\label{e3.4}$$ 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 $$G(\lambda,t,x,h-\tau)\,G(\lambda,t,x,H+\tau)<0,\quad (\lambda,t,x)\in[0,1]\times[a,T]\times[F_1-\tau,L_1+\tau].\label{e3.5}$$ In addition, $G(\lambda,t,x,p)$ and $$G_p(\lambda,t,x,p)=\lambda f_p(t,x,p)-1 \label{e3.6}$$ 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 $$G_p(\lambda,t,x,p)\leq\lambda(1-\varepsilon)-1\leq\max\{-\varepsilon,-1\} \quad \text{for }(\lambda,t,x,p)\in[0,1]\times\Omega_\tau. \label{e3.7}$$ 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 $$x'+(1-\lambda)x=\Phi(\lambda,t,x),\;x(a)=A,\label{e3.8lambda}$$ 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 $$U=\big\{x\in C^{1}_{I}[a,T]:F_1-\tau0, L_i,F_i, i=1,2, and a sufficiently small \tau>0 such that (0,T]\subseteq D_t, L_2-\tau\geq L_1\geq\max\{0,A\}, F_2+\tau\leq F_1\leq\min\{0,A\}, [F_2, L_2]\subseteq D_x, [h-\tau,H+\tau]\subseteq D_p for h=-Q-L_1 and H=Q-F_1, \begin{gather*} f(t,x,p)\leq0\quad\text{for }(t,x,p)\in(0,T]\times[L_1,L_2]\times D_p^+,\\ f(t,x,p)\geq0\quad\text{for }(t,x,p)\in(0,T]\times[F_2,F_1]\times D_p^-,\\ pf(t,x,p)\leq0\quad\text{for }(t,x,p)\in(0,T]\times[F_1-\tau,L_1+\tau] \times(D_Q^-\cup D_Q^+), \end{gather*} where the sets D_p^-,D_p^+,D_Q^-,D_Q^+ are as in (R1). \item[(S2)] f(t,x,p) and f_p(t,x,p) are continuous for (t,x,p) in (0,T]\times[F_1-\tau,L_1+\tau]\times [h-\tau,H+\tau], and for some \varepsilon>0, $$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}$$ where the constants T,F_1,L_1,h,H, \tau are as in (S1). \item[(S3)] f_t(t,x,p) and f_x(t,x,p) are continuous for (t,x,p)\in(0,T]\times[F_1,L_1]\times[h,H], where T,F_1,L_1,h,H,\tau are as in (S1). \end{itemize} Note, in [12] the condition \eqref{e4.2} has the form$$ 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], $$for some constant M. Now we are ready to prove the main result of this paper. It guarantees solutions to the problem \eqref{e1.1} in the case \eqref{e4.1}. \begin{theorem} \label{thm4.1} Let {\rm (S1), (S2), (S3)} hold. Then the singular initial-value problem \eqref{e1.1} has at least one solution in C[0,T]\cap C^1(0,T]. \end{theorem} \begin{proof} For n\in N_T=\{n\in\mathbb{N}:n^{-1}0\quad\text{for }(t,x,p)\in[\theta_1,T] \times[F_1,L_1]\times[h,H].$$ 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 $(0,T]$, $$ht+A\leq x(t)\leq Ht+A\quad\text{for }t\in(0,T]\label{e4.4}$$ 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. 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