\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small Sixth Mississippi State Conference on Differential Equations and Computational Simulations, {\em Electronic Journal of Differential Equations}, Conference 15 (2007), pp. 29--39.\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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \setcounter{page}{29} \title[\hfilneg EJDE-2006/Conf/15\hfil noncontinuable solutions] {Existence of noncontinuable solutions} \author[M. Bartu\v sek\hfil EJDE/Conf/15 \hfilneg] {Miroslav Bartu\v sek} \address{Miroslav Bartu\v sek \newline Department of Mathematics, Masaryk University, Jan\'a\v ckovo n\'am. 2a, 602 00 Brno, Czech Republic} \email{bartusek@math.muni.cz} \thanks{Published February 28, 2007.} \subjclass[2000]{34C11} \keywords{Noncontinuable solutions; singular solutions of the second kind; \hfill\break\indent black-hole solutions} \begin{abstract} This paper presents necessary and sufficient conditions for an $n$-th order differential equation to have a non-continuable solution with finite limits of its derivatives up to the orders $n-2$ at the right-hand end point of the definition interval. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem*{remark}{Remark} \newtheorem*{example}{Example} \section{Introduction} Consider the $n$-th order differential equation $$\label{e1} y^{(n)} = f \big(t, y, y', \dots , y^{(n-2)}\big) g \big(y^{(n-1)}\big)$$ where $n\geq 2$, $f\in C^0 (R_+ \times R^{n-1})$, $g\in C^0 (R), R_+ = [0, \infty)$, $R= (-\infty , \infty)$ and $M>0$ exists such that $$\label{e2} g(x) >0 \quad \text{for } |x| \geq M.$$ This inequality will be assumed throughout the paper. So we study equations for which $g$ is nonzero in neighbourhoods of $\infty$ and $-\infty$; this case can be easily transformed into \eqref{e1} and \eqref{e2}. A solution $y$ of \eqref{e1} defined on $[T, \tau) \subset R_+$ is called noncontinuable if $\tau < \infty$ and $y$ cannot be defined at $t=\tau$. Sometimes such solutions are called singular of the second kind \cite{Ba,Cha,KiCha}. A noncontinuable solution $y$ is called nonoscillatory if $y\ne 0$ in a left neighbourhood of $\tau$. Sufficient conditions for the existence of noncontinuable solutions for the Cauchy problem can be found in \cite{KiCha}. For $f (t, x_1, \dots , x_{n-1}) \equiv r(t) |x_1|^\lambda \times \mathop{\rm sgn} x_1, r\ne 0$ in \cite{Cha}. For $n=2$ in \cite{BaCe,CoUl,Hei}. Sufficient conditions for the nonexistence of noncontinuable solutions of \eqref{e1} and of its special cases be found in \cite{CoWo,Gra1,Gra2,KiCha}. Jaro\v s and Kusano \cite{JaKu} investigated the differential equation $$\label{e3} y'' = r(t) |y|^\sigma |y'|^\lambda \mathop{\rm sgn} y$$ with $\sigma >0, r<0$ on $R_+$. They proved that there exists a noncontinuable solution $y$ of \eqref{e3} fulfilling $\lim_{t\to\tau_-} y(t) \in [0, \infty ), \lim_{t\to\tau_-} y'(t) = - \infty$ if, and only if $\lambda > 2$; they call it a black hole solution. In \cite{Ba}, a problem is formulated for \eqref{e1}: To find conditions under which \eqref{e1} has a noncontinuable solution $y$ fulfilling the conditions $$\label{e4} \begin{gathered} \tau \in (0, \infty ), \quad c_i \in R, \quad \lim_{t\to \tau_-} y(t) = c_i \quad i=0, 1, 2, \dots , n-2, \\ \lim_{t\to\tau_-} |y^{(n-1)} (t) | = \infty \end{gathered}$$ and $y$ is defined in a left neighbourhood of $\tau$. Note that \eqref{e4} is a boundary-value problem and a solution $y$ fulfilling \eqref{e4} is nonoscillatory. The obtained results are summed up in the following theorem. \begin{theorem}[\cite{Ba}] \label{thmA} Let $\tau \in (0, \infty )$, $f(t, x_1, \dots , x_{n-1}) x_1 \ne 0$ for $x_1 \ne 0$ and $g(x) \geq 0$ for $x\in R$. \begin{itemize} \item[(i)] If $M_1 \in (0, \infty )$ is such that $g(x) \leq x^2$ for $|x| \geq M_1$, then \eqref{e1} has no solution $y$ fulfilling \eqref{e4}. \item[(ii)] Let $\tau \in (0, \infty )$, $c_0 \ne 0$, $\lambda >2$, $M_1 \in (0, \infty )$ and $g (x) \geq |x|^\lambda$ for $|x| \geq M_1$, then \eqref{e1} has a solution $y$ fulfilling \eqref{e4} that is defined in a left neighbourhood of $\tau$. \end{itemize} \end{theorem} In the present paper, we generalize Theorem \ref{thmA} and the necessary and sufficient condition for the existence of a noncontinuable solution $y$ fulfilling \eqref{e4} will be stated if $f(\tau , c_0, \dots , c_{n-2}) \ne 0$. Sufficient conditions for the existence of a noncontinuable solution $y$ fulfilling \eqref{e4} are given in case $f (\tau , c_0, \dots , c_{n-2}) = 0$. \subsection*{Notation} Let $\int^\infty_M \frac{d\sigma }{g(\sigma )}<\infty$. Then we put $$F(z) = \int^\infty_z \frac{d\sigma}{g(\sigma )}, z\geq M$$ and $F^{-1}: (0, F(M)]\to [M, \infty )$ denotes the inverse function to $F$. Similarly, if $\int^{-M}_{-\infty} \frac{d\sigma}{g(\sigma )} < \infty$, put $$G(z) = \int^z_{-\infty} \frac{d\sigma }{g(\sigma )}, z\leq -M$$ and $G^{-1}: (0, G(-M)] \to (- \infty ,-M]$ is the inverse function to $G$. The next lemma follows from the definitions of $F$ and $G$. \begin{lemma}\label{L1} \begin{itemize} \item[(i)] Let $\int^\infty_M \frac{d\sigma}{g(\sigma )} < \infty$. Then functions $F$ and $F^{-1}$ are decreasing, $\lim_{z\to\infty} F(z) =0$ and $\lim_{z\to 0_+} F^{-1} (z) = \infty$. \item[(ii)] Let $\int^{-M}_{- \infty} \frac{d\sigma }{g(\sigma )} < \infty$. Then functions $G$ and $G^{-1}$ are increasing, $G>0, G^{-1}< 0, \lim_{z\to -\infty} G(z) = 0$ and $\lim_{z\to 0_+} G^{-1} (z) = -\infty$. \end{itemize} \end{lemma} Denote by $[[a]]$ the entire part of the number $a$. \section{Main results} The next theorem gives conditions for the nonexistence of a solution $y$ fulfilling \eqref{e4}. \begin{theorem}\label{T1} Let the following two assumptions hold. \begin{itemize} \item[(i)] Let either $$\label{e5} \int^\infty_M \frac{d\sigma}{g(\sigma )} = \infty$$ or $$\label{e6} \int^\infty_M \frac{d\sigma}{g(\sigma )} < \infty \quad \text{and} \quad \int^{F(M)}_0 F^{-1} (\sigma ) d\sigma = \infty ;$$ \item[(ii)] Let either $$\label{e7} \int^{-M}_{-\infty} \frac{d\sigma}{g(\sigma)} = \infty$$ or $$\label{e8} \int^{-M}_{-\infty} \frac{d\sigma}{g(\sigma )} < \infty \quad \text{and} \quad \int^{G(-M)}_0 |G^{-1} (\sigma )| d\sigma = \infty .$$ \end{itemize} Then \eqref{e1} has no noncontinuable solution $y$ fulfilling \eqref{e4} that is defined in a left neighbourhood of $\tau$. \end{theorem} \begin{proof} Suppose, contrarily, that $y$ is a noncontinuable solution of \eqref{e1} fulfilling \eqref{e4} that is defined on $[T, \tau) \subset R_+$. Furthermore, suppose that $\lim_{t\to\tau_-} y^{(n-1)}(t) = \infty$; the opposite case, if $\lim_{t\to\tau_-} y^{(n-1)} (t) = - \infty$, can be studied similarly using \eqref{e7} and \eqref{e8}. From this, $T_1 \in [T, \tau)$ and $M_1 >0$ exist such that $$f \big(t, y(t), \dots , y^{(n-2)} (t)\big) \leq M_1, \quad y^{(n-1)} (t) \geq M \quad \text{for } t\in [T_1, \tau ).$$ Hence, the integration of \eqref{e1} and \eqref{e2} yields \label{e9} \begin{aligned} \int^\infty_{y^{(n-1)}(t)} \frac{d\sigma}{g(\sigma)} &= \int^\tau_t \frac{y^{(n)}(\sigma)d\sigma}{g(y^{(n-1)}(\sigma))} \\ &= \int^\tau_t f \big(\sigma, y(\sigma ), \dots , y^{(n-2)} (\sigma )\big) d\sigma \\ &\leq M_1 (\tau -t) \leq M_1\tau, \quad t\in [T_1, \tau ). \end{aligned} It follows from this that \eqref{e5} is not valid and hence \eqref{e6} holds. Let $T_2 \in [T_1, \tau)$ be such that $\tau -T \leq F(M) M^{-1}_1$. From this and from \eqref{e9} $$F \big(y^{(n-1)}(t)\big) \leq M_1 (\tau -t) \in (0, F(M)] \quad \text{for } t\in [T_2, \tau );$$ hence, Lemma \ref{L1} yields $$y^{(n-1)} (t) \geq F^{-1} \left(M_1 (\tau -t)\right), \quad t\in [T_2, \tau)$$ and an integration on $[T_2, \tau)$ and \eqref{e6} yield \begin{align*} \infty &> c_{n-2} - y^{(n-2)} (T_2) = y^{(n-2)} (\tau) - y^{(n-2)}(T_2) = \int^\tau_{T_2} y^{(n-1)} (\sigma) d\sigma \\ &\geq \int^\tau_{T_2} F^{-1} (M_1 (\tau -s))\, ds = \frac{1}{M_1} \int^{M_1(\tau -T_2)}_0 F^{-1} (\sigma ) d\sigma = \infty . \end{align*} This contradiction proves that a noncontinuable solution $y$ fulfilling \eqref{e4} does not exist. \end{proof} The following theorem formulates necessary and sufficient conditions for the existence of a solution $y$ fulfilling \eqref{e4} in case $f(\tau , c_0, \dots , c_{n-2}) \ne 0$. \begin {theorem}\label{T2} Let $\tau >0$ and $c_i \in R$, $i=0,1, \dots , n-2$ be such that $$\label{e10} f(\tau , c_0, c_1, \dots , c_{n-2}) \ne 0.$$ Then \eqref{e1} has a noncontinuable solution $y$ fulfilling \eqref{e4} if and only if one of the following two conditions holds: \begin{gather}\label{e11} \int^\infty_M \frac{d\sigma}{g(\sigma)} < \infty \quad \text{and}\quad \int^{F(M)}_0 F^{-1}(\sigma ) \, d\sigma < \infty ; \\ \label{e12} \int^{-M}_{-\infty} \frac{d\sigma}{g(\sigma )} < \infty \quad \text{and} \quad \int^{G(-M)}_0 |G^{-1} (\sigma )| \, d\sigma < \infty . \end{gather} In this case $y$ is defined in a left neighbourhood of $\tau$. Moreover, let $f(0,c_0, c_1, \dots , c_{n-2})\ne 0$ and either \eqref{e11} or \eqref{e12} holds. Then there exists $\tau_0 >0$ such that for every $0<\tau \leq \tau_0$, a noncontinuable solution $y$ fulfilling \eqref{e4} exists and is defined on $[0,\tau)$. \end{theorem} \begin{proof} Necessity: This follows from Theorem \ref{T1}. Sufficiency: We prove the statement in case $f(\tau, c_0, \dots , c_{n-2})>0$; the opposite case can be studied similarly. There exist $N>0$ and $\bar{\tau} \in [0, \tau)$ such that $$\label{e13} f(t,x_1, \dots , x_{n-1})>0 \quad \text{for } t\in[\bar{\tau}, \tau], |x_i -c_{i-1}|\leq N, i=1,2,\dots , n-1.$$ From this, positive constants $M_1$ and $M_2$ exist such that \label{e14} \begin{gathered} 0 M_2 = \max \{f(t, x_1, \dots , x_{n-1}): t\in [\bar{\tau}, \tau], \; |x_i - c_{i-1}| \leq N, i= 1,2, \dots , n-1\}. \end{gathered} Consider the auxiliary problem $$\label{e15} \begin{gathered} y^{(n)} = f\left(t, \chi_0 (y), \chi_1 (y'), \dots, \chi_{n-2}(y^{(n-2)})\right) g(y^{(n-1)}),\\ y^{(i)} (\tau) = c_i,i=0,1,\dots,n-2, \: y^{(n-1)} (\tau) =k, \end{gathered}$$ where $k\in \{k_0,k_0+1, \dots\}, k_0 \geq [[2M]]$, $$\label{e16} \chi_i(s) = \begin{cases} s & \text{for } |s-c_i| \leq N\\ c_i+N & \text{for } s > c_i + N\\ c_i-N & \text{for } s < c_i - N, \end{cases}$$ where $i= 0,1,\dots , n-2$. Furthermore, let $J=[T,\tau)\subset[\bar{\tau},\tau)$ be such that $0<\tau -T\leq 1$, $$\label{e17} (\tau -T)\sum^{n-2}_{j=1} |c_j|+\frac{1}{M_1}\int^{M_1(\tau-T)}_0 F^{-1} (z) \, dz \leq N,$$ and $$\label{e18} M_2 (\tau-T) < \int^{2M}_M \frac{ds}{g(s)};$$ this choice is possible due to the second inequality in \eqref{e11}, \eqref{e2} and Lemma \ref{L1}; it does not depend on $k$. Denote by $y_k$ a solution of \eqref{e15} and by $J_k$ the intersection of its maximal definition interval and $[T, \tau)$. We prove that $$\label{e19} y^{(n-1)}_k (t) >M \quad \text{for} \quad t\in J_k.$$ As $k\geq k_0 \geq [[2M]]$, \eqref{e15} yields \eqref{e19} is valid in a left neighbourhood of $\tau$. Suppose, contrarily, that $T_1 \in J_k$ exists such that $y^{(n-1)}_k (T_1) =M$ and $y^{(n-1)}_k (t) > M$ on $(T_1,\tau]$. Then \eqref{e2}, \eqref{e14}, \eqref{e15} and \eqref{e16} yield $y^{(n)}_k (t)>0$ on $[T_1, \tau]$ and $$y^{(n)}_k (t) \leq M_2 \, g(y^{(n-1)}_k (t)), \, g(y^{(n-1)}_k (t)) >0, \, t\in [T_1, \tau].$$ From this, and an integration on $[T_1, \tau]$, we obtain $$\int^{2M}_M \frac{ds}{g(s)} \leq \int^k_M\frac{ds}{g(s)} \leq M_2 (\tau -T_1) \leq M_2 (\tau -T).$$ The contradiction with \eqref{e18} proves that \eqref{e19} holds. According to \eqref{e14}, \eqref{e15} and \eqref{e16}, we have $$y^{(n)}_k (t) \geq M_1 \, g(y^{(n-1)}_k (t)), \quad t\in J_k$$ and an integration on $[t, \tau]$, \eqref{e2}, and \eqref{e19} yield $$F(y^{(n-1)}_k (t)) \geq \int^k_{y^{(n-1)}_k (t)} \frac{ds}{g(s)} \geq M_1(\tau -t), \quad t\in J_k.$$ As, according to \eqref{e18}, $M_1 (\tau -T) \leq M_2 (\tau -T) < F(M)$, we have $M_1(\tau -t) \in(0, f(M)]$ and $$\label{e20} y^{(n-1)}_k (t) \leq F^{-1} (M_1 (\tau -t)),\quad t\in J_k.$$ From this and from Lemma \ref{L1}, $y^{(n-1)}_k$ is bounded on $J_k$ and hence $J_k = [T, \tau]$; moreover, $J_k$ is defined on the same interval $[T, \tau]$ for every $k=k_0, k_0 +1, \dots$. We estimate the functions $y^{(i)}_k$. Taylor's formula, $T-\tau \leq 1$, \eqref{e17}, and \eqref{e20} yield \label{e21} \begin{aligned} \big| y^{(i)}_k (t) -c_i\big| &\leq \sum^{n-2}_{j=i+1} \frac{|c_j|}{(j-i)!} (\tau -t)^{j-i}+ \Big|\int^t_\tau \frac{|(t-s)^{n-i-2}|}{(n-i-2)!} F^{-1}(M_1 (\tau -s))\, ds \Big|\\ &\leq (\tau -T) \sum^{n-2}_{j=1} |c_j| + \Big|\int^t_\tau F^{-1}(M_1(\tau -s))\, ds\Big|\\ &\leq (\tau -T) \sum^{n-2}_{j=1} |c_j| + \frac{1}{M_1} \int^{M_1(\tau -T)}_0 F^{-1} (z) \, dz \\ &\leq N,\quad t\in [T, \tau],\; i=0,1,\dots,n-2. \end{aligned} From this and from \eqref{e16}, $y_k$ is a solution of \eqref{e1}, as well. As the estimations \eqref{e19}, \eqref{e20} and \eqref{e21} and the definition interval of $y_k$ do not depend on $k$, then according to the Arzel-Ascoli Theorem (see \cite[Lemma 10.2]{Cha}) there exists a subsequence of $\{y_k\}^\infty_{k=k_0}$ that converges locally uniformly to a solution $y$ of \eqref{e1} on $[T, \tau)$ together with all derivatives up to the order $n-1$. Evidently conditions \eqref{e4} hold for $i= 0,1, \dots , n-2$ and we prove $\lim_{t\to\tau_-} y^{(n-1)} (t) = \infty$. As $y^{(n-1)}$ is increasing on $[T, \tau)$, there exists a limit as $t\to \tau_-$. Suppose, contrarily, that $\lim_{t\to\tau_-} y^{(n-1)} (t) = Q < \infty$. Then Lemma \ref{L1} yields the existence of $T_2 \in [T, \tau)$ such that $$\label{e21a}% rovnice c. 22 Q < F^{-1} (M_2 (\tau - T_2)), \quad M (\tau -T_2) \leq F(M).$$ Moreover, there exists a subsequence of $\{y_k^{(n-1)}\}^\infty_{k=k_0}$, we denote it $\{y_k^{(n-1)}\}^\infty_{k=k_0}$ for simplicity, that converges to $y^{(n-1)}$ on $[T, T_2]$. From this, $\bar{k}$ exists such that $$y^{(n-1)}_k (T_2) \leq 2Q \quad \text{for } k=\bar{k}, \bar{k} +1, \dots .$$ According to \eqref{e1} and \eqref{e21}, $y_k^{(n)} (t) \leq M_2 g(y_k^{(n-1)} (t) )$, we obtain, by integration on $[T_2, \tau)$, $$M_2 (\tau -T_2) \geq \int^k_{y_k^{(n-1)} (T_2)} \frac{ds}{g(s)} \geq \int^k_{2Q} \frac{ds}{g(s)}, \ k\geq \bar{k}.$$ Thus, $$\label{e21b}%rovnice c. 23 M_2 (\tau -T_2) \geq \int^\infty_{2Q} \frac{ds}{g(s)} = F(2Q),$$ so $2 Q \geq F^{-1} (M_2 (\tau - T_2))$, which contradicts \eqref{e21a}. Hence, $\lim_{t\to\tau_-} y(t) = \infty$. Let $f(0,c_0,c_1,\dots,c_{n-2})>0$. Then there exist $N>0$ and $\bar{\tau}_0\leq 1$ such that $$f(t,x_1,\dots, x_{n-1})>0 \quad \text{for } t\in [0, \bar{\tau}_0],\; [x_i-c_{i-1}| \leq N, \; i=1, \dots , n-1.$$ Define \begin{gather*} M_1 = \min \{f(t, x_1, \dots , x_{n-1}): t\in [0, \bar{\tau}_0], |x_i- c_{i-1}| \leq N, i=1,2,\dots,n-1\},\\ M_2 = \max \{f(t, x_1, \dots , x_{n-1}): t\in [0, \bar{\tau}_0], |x_i- c_{i-1}| \leq N, i=1,2,\dots,n-1\}. \end{gather*} Constants $N, M_1$ and $M_2$ are given by \eqref{e13} and \eqref{e14}, but for $[0, \bar{\tau}_0]$ instead of $[\bar{\tau}, \tau]$. Let $0<\tau_0 \leq \bar{\tau}_0$ be a number such that \eqref{e17} and \eqref{e18} hold with $T=0$ and $\tau =\tau_0$. It is clear that \eqref{e17} and \eqref{e18} are valid for $\tau\leq\tau_0$ and $T=0$ and a noncontinuable solution $y$ fulfilling \eqref{e4} exists according to the first part of the proof, and it is defined on $[0, \tau)$. \end{proof} Next, we prove a comparison theorem. \begin{theorem}\label{T3} Let $\tau >0$ and $c_i \in R_+$, be such that $f(\tau, c_0,\dots, c_{n-2}) \ne 0$. Let $f_1 \in C^0 (R_+ \times R^{n-1})$, $f_1(\tau, c_0,\dots,c_{n-1})\ne 0$ and let $\bar{g} \in C^0(R)$ exist such that $$\label{e22} \bar{g} (x) \geq g(x) >0 \quad \text{for } |x|\geq M.$$ \begin{itemize} \item[(i)] If \eqref{e1} has a solution fulfilling \eqref{e4}, then the equation $$\label{e23} y^{(n)} = f_1\big(t, y,\dots, y^{(n-2)}\big) \bar{g} \big(y^{(n-1)}\big)$$ has the same property. \item[(ii)] If \eqref{e23} has no solution fulfilling \eqref{e4}, then \eqref{e1} has the same property. \end{itemize} \end{theorem} \begin{proof} (i) According to Theorem \ref{T2} either \eqref{e11} or \eqref{e12} holds. Suppose that \eqref{e11} is valid; if \eqref{e12} holds the proof is similar. Then \eqref{e22} yields $$\label{e24} \int^z_M \frac{d\sigma}{\bar{g}(\sigma)} \leq \int^z_M \frac{d\sigma}{g(\sigma)}, \ z\leq M.$$ According to \eqref{e11} and \eqref{e24}, $\int^\infty_M \frac{d\sigma}{\bar{g}(\sigma)} < \infty$. Denote $F_1(z) =\int^\infty_z \frac{d\sigma}{\bar{g}(\sigma)}$, $z\geq M$ and let $F^{-1}_1$ be the inverse function to $F_1$. As $F_1(z) \leq F(z), z\geq M$, and as $F$ and $F_1$ are nonincreasing, then $F_1^{-1} (z) \leq F^{-1} (z), z\geq F_1(M)$ and, hence, \eqref{e11} yields $$\label{e25} \int^{F_1(M)}_0 F^{-1}_1 (\sigma ) d\sigma \leq \int^{F_1(M)}_0 F^{-1} (\sigma )\, ds <\infty .$$ Hence, Theorem \ref{T2} applied to \eqref{e23} proves that it has a noncontinuable solution $y$ fulfilling \eqref{e4}. \noindent (ii) Suppose, contrarily, that \eqref{e1} has a solution $y$ fulfilling \eqref{e4}. Then Theorem \ref{T2} yields either \eqref{e11} or \eqref{e12} holds. Suppose that \eqref{e11} holds. \eqref{e23} has no solution fulfilling \eqref{e4}; hence according to Theorem \ref{T2} (i) (applied to \eqref{e23} ) either $$\label{e26} \int^\infty_M \frac{d\sigma}{\bar{g}(\sigma)} = \infty$$ or $$\label{e27} \int^\infty_M \frac{d\sigma}{\bar{g}(\sigma)} = \infty \quad \text{and} \quad \int^{F_1(M)}_0 F^{-1}_1 (\sigma)\, d\sigma = \infty .$$ As \eqref{e11} and \eqref{e22} yield \eqref{e24}, \eqref{e26} is in a contradiction with \eqref{e11} and \eqref{e24}. As \eqref{e11} yields \eqref{e25}, the inequality \eqref{e25} contradicts \eqref{e27}. If \eqref{e12} holds, the proof is similar. \end{proof} \begin{example} \rm Consider problem \eqref{e1}, \eqref{e2} with $g(x)=|x|^\lambda$ for $|x|\geq M$, $\lambda \in R$. Let $\tau >0, c_i, i=0, \dots , n-2$ be such that $f(\tau, c_0, \dots, c_{n-2})\ne0$. Then, according to Theorem \ref{T2}, \eqref{e1} has a noncontinuable solution $y$ fulfilling \eqref{e4} if and only if $\lambda >2$. \end{example} \begin{remark}\rm Theorem \ref{thmA} (ii) follows from Theorem \ref{T3} and the Example. \end{remark} Let us turn our attention to the case when \eqref{e10} does not hold. \begin{theorem}\label{T4} Let $\beta \in \{-1, 1\}$, $\delta>0$, $\varepsilon >0$, $\tau \in (0, \infty)$, $\alpha \in \{-1, 1\}$, $s\in \{0, 1, \dots , n-2\}$ and $c_i \in R, i=0,1,\dots, n-2$ be such that $\tau >\varepsilon$, \begin{gather} \label{e28} \lambda > \delta (n-s-2)+2,\\ \label{e29} c_s =0, (-1)^{i-s}\, \beta \, c_i\geq 0 \quad \text{for } i=s+1, \dots , n-2,\\ \label{e30} n-s+\frac{1-\alpha}{2} \quad \text{be odd},\\ \label{e31} g(x) \geq |x|^\lambda \quad \text{for } \beta x\geq M. \end{gather} Let, moreover, a positive function $r$ exist such that $$\label{e32} \begin{gathered} \alpha f(t, x_1, \dots , x_{n-1}) \mathop{\rm sgn} x_{s+1} \geq r(t) |x_{s+1}|^\delta\\ \text{for } t\in [\tau - \varepsilon, \tau]\cap R_+, \; |x_i-c_{i-1}|\leq \varepsilon,\; i=1,2,\dots , n-1. \end{gathered}$$ Then there exists a solution $y$ of \eqref{e1} fulfilling \eqref{e4} that is defined in a left neighbourhood of $\tau$. \end{theorem} \begin{proof} Let $\alpha =1$ and $\beta =1$; thus $n-s$ is odd. For the other cases the proof is similar. Note that \eqref{e32} and $c_s =0$ yield $f(\tau, c_0,\dots, c_{n-2})=0$. Consider problem \eqref{e15} and \eqref{e16} with $N = \varepsilon$ and $\bar{\tau} = \max (0, \tau - \varepsilon)$. Put \begin{gather*} M_1 = \left( (n-s-1)!\right)^{-\delta} \min_{t\in [\bar{\tau}, \tau]} \, r(t) >0, \\ \delta_1 = \frac{\delta (n-s-1)+1}{\lambda +\delta -1},\\ M_2 = \max \{|f(t,x_1,\dots,x_{n-1})|: t\in [\bar{\tau}, \tau], \, |x_i-c_{i-1}|\leq \varepsilon, i=1,2,\dots, n-1\}, \\ M_3 = \Big(\frac{M_1(\lambda + \delta-1)}{\delta (n-s-1)+1} \Big) ^{-1/(\lambda +\delta -1)},\\ M_4 = (\lambda -1) \varepsilon^\delta \min_{t\in [0, \tau ]} r(t), \end{gather*} and $M_5 = M_4^{-1/(\lambda -1)}$. Note that due to \eqref{e28}, $\delta_1 \in(0,1)$. Furthermore, let $J =[T,\tau)\subset [\bar{\tau}, \tau)$ be such that $0 < \tau -T\leq 1$, $$\label{e2*} (\tau -T) \sum^{n-2}_{j=0} |c_j| + \frac{M_3}{1-\delta_1} (\tau -T)^{1-\delta_1} + \frac{\lambda -1}{\lambda -2} M_5 (\tau -T)^{\frac{\lambda -2}{\lambda -1}} \leq \varepsilon ,$$ and $$M_2 (\tau -T) < \int^{2M}_M \frac{ds}{g(s)}.$$ Denote by $y_k$ a solution of \eqref{e15} and by $J_k$ the intersection of its maximal definition interval and $[T, \tau]$. We prove, similarly as in the proof of Theorem \ref{T2}, (see \eqref{e19}) that $$\label{e33} y_k^{(n-1)} (t) > M \quad \text{for } t\in J_k;$$ hence \eqref{e29}, \eqref{e32} and \eqref{e33} yield $$\label{e34} c_{s+1} \leq 0, \quad c_{s+2} \geq 0, \dots , c_{n-2} \leq 0,$$ \begin{gather*} (-1)^{j-s} y^{(j)}_k (t) >0 \quad \text{for } j=s+1, s+2, \dots, n-2,\\ \mathop{\rm sgn} y^{(s)}_k (t) =1, \quad t\in J_k - \{\tau\}. \end{gather*} From this, \eqref{e15}, \eqref{e16} and \eqref{e32}, $$\label{e36} y_k^{(n)} (t) \geq 0 \quad \text{and}\quad y^{(n-1)}_k \text{ is nondecreasing on } J_k.$$ The Taylor formula at $t=\tau$, \eqref{e34}, \eqref{e36}, and $n-s$ being odd yield \begin{align*} y^{(s)}_k (t) &= \sum^{n-2}_{j=s} c_j \frac{(t-\tau)^{j-s}}{(j-s)!} + \int^t_\tau \frac{(t-\sigma)^{n-s-2}}{(n-s-2)!} \, y^{(n-1)}_k (\sigma ) \, d\sigma \\ &\geq \int^t_\tau \frac{(t-\sigma)^{n-s-2}}{(n-s-2)!} \, y_k^{(n-1)} (\sigma) \, d\sigma \\ &\geq \frac{(\tau-t)^{n-s-1}} {(n-s-1)!} \, y_k^{(n-1)} (t), \quad t\in J_k. \end{align*} Let $T^* \in [T, \tau )$ be a number such that $$0\leq y_k^{(s)} (T) \leq \varepsilon \quad \text{for } t\in [T^*, \tau ),$$ and, if $T^* >T$, $$y^{(s)}_k (T) >\varepsilon \quad \text{for } t\in [T, T^*);$$ this choice is possible due to \eqref{e36}. Let $T^* >T$ and $t\in [T, T^*)$. Then \eqref{e15}, \eqref{e16}, \eqref{e31}, \eqref{e32} and \eqref{e33} yield $$y^{(n)}_k (t) \geq r(t) \ \varepsilon^\delta \big(y_k^{(n-1)} (t)\big) ^\lambda ,$$ and since $\lambda >1$, an integration on $[t, T^*]$ shows $$\big(y^{(n-1)}_k (t)\big)^{1-\lambda} \geq \big(y^{(n-1)}_k (t)\big)^{1-\lambda} - \big(y^{(n-1)}_k (T^*)\big)^{1-\lambda} \geq M_4 (T^* -t)$$ and $$\label{e1*} y_k^{(n-1)} (t) \leq M_5 (T^* -t)^{-\frac{1}{\lambda -1}}, \ t\in [T, T^*).$$ Similarly, for $t\in [T^*, \tau )$, we have $$\label{e3*} y^{(n)}_k (t) \geq r(t)(y^{(s)}_k (t))^\delta (y_k^{(n-1)} (t))^\lambda \geq M_1 (\tau -t)^{\delta (n-s-1)} (y_k^{(n-1)} (t))^{\lambda +\delta}.$$ Hence, as $\lambda +\delta >1$, an integration on $[t, \tau]$ yields $$\label{e37} y_k^{(n-1)} (t) \leq M_3 (\tau -t)^{-\delta_1}, \quad t\in J_k, k=k_0, k_{0+1}, \dots$$ From this and from \eqref{e33} and \eqref{e1*} we have $J_k = [T, \tau]$. Moreover, as $\tau -T\leq 1$ and $\delta_1 <1$, Taylor's theorem, \eqref{e1*}, \eqref{e2*} and \eqref{e37} yield \begin{align*} \big|y_k^{(i)} (T) - c_i\big| & \leq \sum^{n-2}_{j=i+1} \frac{|c_j|}{(j-i)!} (\tau -T)^{j-i} + \big|\int^{T^*}_\tau \frac{(T-\sigma)^{n-i-2}}{(n-i-2)!}\, y_k^{(n-1)} (\sigma ) \, d\sigma \big|\\ &\quad + \big|\int^T_{T^*} \frac{(T-\sigma)^{n-i-2}}{(n-i-2)!} y_k^{(n-1)} (\sigma )\, d\sigma \big| \\ &\leq (\tau -T) \sum_{j=0}^{n-2} |c_j| + \frac{M_3}{1-\delta_1} (\tau -T)^{1-\delta_1} + \frac{\lambda -1}{\lambda -2} M_5 (\tau -T)^{\frac{\lambda -2}{\lambda -1}}\\ &\leq \varepsilon,\quad i=0, 1, \dots , n-2. \end{align*} From this and from \eqref{e16} and \eqref{e36}, $\chi_i(y^{(i)} (t)) = y^{(i)}(t), t\in [T, \tau ]$ and $y_k$ is the solution of \eqref{e1} fulfilling $y^{(i)}_k (\tau ) = c_i, i=0,1, \dots , n-2$ and $y_k^{(n-1)} = k$. Moreover, as $\chi_s (y^{(s)}(t)) = y^{(s)} (t)$, the estimations \eqref{e3*} and \eqref{e37} holds on $[T, \tau)$. The statement of the theorem follows from this and from the Arz\`el-Ascoli Theorem similarly as in the proof of Theorem \ref{T2}; when prooving $\lim_{t\to\tau_-} y(t) = \infty$, $T_2$ has to be defined such that $M_2 (\tau -T_2)<\int^\infty_{2Q} \frac{ds}{g(s)}$ (this is possible due to \eqref{e2}) and the inequality in \eqref{e21b} is in contradiction with the choice of $T_2$. \end{proof} The following Corollary shows that conditions \eqref{e28} and \eqref{e30} cannot be weakened. \begin{corollary}\label{C1} Let $c_i =0$, $i=0,1,\dots, n-2$, $\delta >0$, $s\in \{0,1,\dots, n-2\}$, $\alpha \in \{-1, 1\}$, $\tau\in (0, \infty)$, $r\in C^0(R_+)$ and $r>0$ on $[0, \tau]$. Then the equation $$\label{e38} y^{(n)} = \alpha r\, (t)|y^{(s)}|^\delta |y^{(n-1)}|^\lambda \, \mathop{\rm sgn} y^{(s)}$$ has a noncontinuable solution $y$ fulfilling \eqref{e4} if and only if $$\label{e39} \lambda > \delta (n-s-2)+2 \quad \text{and}\quad n-s+\frac{1-\alpha}{2} \quad \text{is odd}.$$ \end{corollary} \begin{proof} If \eqref{e39} holds the statement follows from Theorem \ref{T4}. Let \eqref{e39} be not valid. Let, contrarily, \eqref{e38} have a solution $y$ fulfilling \eqref{e4} defined on $[\bar{\tau}, \tau) \subset R_+$. Suppose, for simplicity, that $\alpha =1$ and $\lim_{t\to\tau_-} y^{(n-1)} (t) = \infty$. In the other cases the proof is similar. As $c_i =0$ for $i=0,1, \dots, n-2$, there exists $t_0 \in [\bar{\tau}, \tau)$ such that $$\label{e40} \begin{gathered} (-1)^{i-s+\beta} y^{(i)} (t) >0, \quad i=s,s+1, \dots, n-2, \\ y^{(n-1)} (t) \geq 1, \quad y^{(n)} (t)>0 \quad \text{on } J = [t_0, \tau), \end{gathered}$$ where $\beta =0\ \ (\beta =1)$ if $n-s$ is odd (is even). Let $n-s$ be even. Then \eqref{e40} yields $y^{(s)} (t) < 0$ on $J$ and according to \eqref{e38} $y^{(n)} (t) < 0$ on $J$ which contradicts \eqref{e40}. Let $n-s$ be odd and $\lambda \leq \delta (n-s-2)+2$. From this, from \eqref{e4}, \eqref{e40} and Taylor's theorem, we get \label{e41} 0