\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 33, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/33\hfil Blow-up solutions]
{Integrability of blow-up solutions to some non-linear differential equations}

\author[Michael Karls \& Ahmed Mohammed\hfil EJDE-2004/33\hfilneg]
{Michael Karls \& Ahmed Mohammed} % in alphabetical order

 \address{Michael Karls \hfill\break
Department of Mathematical Sciences\\
Ball State University\\
Muncie, IN 47306, USA}
\email{mkarls@bsu.edu}

 \address{Ahmed Mohammed \hfill\break
 Department of Mathematical Sciences\\
 Ball State University\\
 Muncie, IN 47306, USA}
 \email{amohammed@bsu.edu}

\date{}
\thanks{Submitted December 25, 2003. Published March 8, 2004.}
\subjclass{34C11, 34B15, 35J65}
\keywords{Blow-up solution, Keller-Osserman condition, integrability}

\begin{abstract}
  We investigate the integrability of solutions to the boundary
  blow-up problem
  $$
  r^{-\lambda}\bigl(r^{\lambda}(u')^{p-1}\bigr)'=H(r,u),\quad
   u'(0)\geq 0,\quad  u(R)=\infty
  $$
  under some appropriate conditions on the non-linearity $H$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{rem}[thm]{Remark}
\newtheorem{example}[thm]{Example}

\section{Introduction}

Let $\lambda\ge 0$, $p>1$, $R>0$. For $0<r<R$ we consider solutions
$u\in C^1([0,R))$ of the  problem
\begin{equation} \label{meqnre}
\begin{gathered}
r^{-\lambda}(r^{\lambda}|u'|^{p-2}u')'=H(r,u),\\
u(0)\geq 0,\quad u'(0)\geq 0 , \quad \lim_{r\to R}u(r)=\infty.
\end{gathered}
\end{equation}
Here $H$ satisfies the conditions
\begin{itemize}
\item[(H1)] $H:[0,R)\times[0,\infty)\to [0,\infty)$ is continuous,
\item[(H2)] $H(\cdot,s)$ is non-decreasing,
\item[(H3)] $H(0,s)>0$ for all $s>0$.
\end{itemize}
Further assumptions on $H$ will be given as needed.  In the
literature, solutions of (\ref{meqnre}) are known as blow-up
solutions, explosive solutions or large solutions.

These type of equations arise as radial solutions of the
$p$-Laplace equation, as well as the Monge Amp\'{e}re equation on
balls. Radial solutions $u$ of the $p$-Laplace equation $$
\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)=J(|x|,u), $$ in the
ball $B:=B(0,R)\subseteq \mathbb R^N$ satisfy the first equation
of (\ref{meqnre}) with $\lambda=N-1$, $H(r,u)=J(r,u)$. Likewise
radial solutions of the Monge Amp\'{e}re equation $$ \det
(D^2u)=J(|x|,u), $$ in the ball $B$ also satisfy the first
equation of (\ref{meqnre}) with $\lambda=0$, $p=N+1$ and
$H(r,u)=Nr^{N-1}J(r,u)$.

Noting that $u'$ is non-negative for any solution $u$ of
(\ref{meqnre}), we will find it convenient to rewrite equation
(\ref{meqnre}) as
\begin{equation} \label{meqn}
\begin{gathered}
\bigl((u')^{p-1}\bigr)'+\frac{\lambda}{r}(u')^{p-1}=H(r,u),\\
u(0)\geq0,\quad u'(0)\geq 0,\quad u(R)=\infty.
\end{gathered}
\end{equation}
A necessary and sufficient condition for the existence of a
solution to problem (\ref{meqnre}) with $u'(0)=0$, $H(r,u)=f(u)$,
and $f(0)=0$, is the (generalized) Keller-Osserman condition (see
\cite{KEL,OSS,MAT}).
\begin{equation} \label{ok}
\int_1^\infty\frac{ds}{F(s)^{1/p}}<\infty,\quad
F(s)=\int_0^sf(t)dt.
\end{equation}
If a nonnegative, non-decreasing continuous function $F$ defined
on $[0,\infty)$ satisfies the Keller-Osserman condition (\ref{ok})
for some $p>1$, we will indicate this by writing $F\in KO(p)$.

When $H(r,s)=f(s)$, and $\lambda=N-1$, problem (\ref{meqnre}) has
been studied extensively by several authors, (see
\cite{BAM, BAM3,KEL, LAI, LMK, MAT, OSS} and the references therein). The
questions of existence, uniqueness and asymptotic boundary
estimates have received particular attention. The case when $p=2$
and $H(r,s)=g(r)f(s)$ with $g\in C([0,R])$, possibly vanishing on
a set of positive measure, has been considered in \cite{LAI}. In
all these cases, the Keller-Osserman condition on $f$ remains the
key condition for the existence of solutions. However, if $g$ is
allowed to be unbounded the situation is completely different and
existence and boundary behavior of a blow-up solution depends on
how fast $g$ is allowed to grow near $R$. For such cases we refer
the reader to \cite{MOH} or \cite{MPP}. For a discussion on
solutions of (\ref{meqnre}) for general non-linearity $H$, we
refer the reader to the paper \cite{RAW}.

In this paper we are interested in studying the integrability
property of blow-up solutions to (\ref{meqnre}) for $F\in KO(p)$.
A blow-up solution may not have any integrability property at all,
as the following example, taken from \cite{MOP}, shows.

\begin{example} \label{ex1.1} \rm
Let $u(r)=-1+e^{(1-r)^{-1}}$. Then
\begin{gather*}
u''(r)=f(u),\quad 0<r<1,\\
u'(0)\geq 0,\quad u(1)=\infty,
\end{gather*}
 where
$f(s)=(s+1)[\log^4(s+1)+2\log^3(s+1)],\;s\geq 0$. Notice that
$u\notin L^{\gamma}(0,1)$ for any $\gamma>0$. The antiderivative
$F$ of $f$ that vanishes at zero is given by
$F(s)=((s+1)^2\log^4(s+1))/2$, and observe that $F\in KO(2)$, but
$F\notin KO(\alpha)$ for any $\alpha>2.$
\end{example}

On the other extreme any positive power of a blow-up solution
could be integrable. This can be seen from the following example.

\begin{example} \label{ex1.2} \rm
We fix $0<R<1/2$ and let
$$
f(s)=e^s-1,\quad s\in[0,\infty),\quad \mbox{and}\quad
g(r)=\frac{1}{(r-R+1)(R-r)},\quad r\in[0,R)
$$
Then $u(r)=-\log(R-r)$ is a solution of
\begin{gather*}
u''(r)=g(r)f(u),\quad 0<r<R,\\
u'(0)\geq 0,\quad u(r)\to\infty
\quad\text{as }r\to R.
\end{gather*}
 Note that $u\in L^\gamma(0,R)$
for all $\gamma>0$. In this example the primitive $F$ of $f$ with
$F(0)=0$ satisfies $F\in KO(\alpha)$ for all $\alpha>0$.
\end{example}



The outline of the paper is as follows.  In Section 2 we compare
solutions $u$ of (\ref{meqn}) with solutions of
\begin{equation} \label{seqn}
\begin{gathered}
((w')^{p-1})'+\frac{\lambda}{r}(w')^{p-1}=H(0,w),\\
 w(0)\geq 0,\quad w'(0)=0,\quad w(R)=\infty,
 \end{gathered}
\end{equation}
for $0<r<R$.

The main result of Section 2, Theorem \ref{scomp}, is used in
Section 3 to prove the following integrability result for
solutions of (\ref{meqn}).

\begin{thm}\label{uint}
Suppose in addition to (H1)--(H3), $H(r,\cdot)$ is non-decreasing
on $[0,R)$ and for $f(s)=H(0,s)$, $f(0)=0$ and $F\in KO(\alpha)$ for
some $\alpha>p$. Then $u\in L^{(\alpha-p)/p}(0,R)$ for any solution $u$
of (\ref{meqn}).
\end{thm}

In Section 3, we also show that for $H(r,s)=g(r)f(s)$, the
following result holds.

\begin{thm} \label{FKO}
 Let $H(r,s)=g(r)f(s)$ satisfy (H1)--(H3),
with $g(0)>0$ and $g$ positive, non-decreasing near $R$. Suppose
(\ref{meqn}) has a blow-up solution $u$ such that $u\in
L^{(\alpha-p)/p}(0,R)$ for some $\alpha>p$. If $g\in L^{1/\sigma}(0,R)$ with
$0<\sigma<p(\alpha-p)/\alpha$, then $F\in KO(\gamma)$ for some $p<\gamma<\alpha$.
\end{thm}

\begin{rem} \label{rmk1.5} \rm
When $H(r,s)=g(r)f(s)$, (H3) and the requirement that $g(0)>0$
imply that $f(s)>0$ for $s>0$.  Since $f(s)>0$, it follows from
(H1) that $g$ is non-negative on $[0,R)$.
\end{rem}

Finally, we give some corollaries to Theorem \ref{FKO}.

\section{A Comparison Result}

We will need the following comparison lemma (see \cite{RAW} for a
proof). For notational convenience in stating the lemma and in
this section, we let $L$ denote the differential operator on the
left hand side of equation (\ref{meqnre}) above.  In this lemma,
we use the following notation: $u(a+)<w(a+)$ means there exists
$\epsilon>0$ such that $u<w$ in $(a, a+\epsilon)$.


\begin{lem}\label{comp}
Let $0\leq a<b$, and suppose $u,w\in C^1([a,b])$ with
$(u')^{p-1}, (w')^{p-1}\in C^1((a,b])$ satisfy
\begin{gather*}
Lu-G(r,u)\leq Lw-G(r,w) \quad \text{in }(a,b]\\
u(a+)<w(a+),\quad u'(a)\leq w'(a)
\end{gather*}
for some function $G(r,s)$ which is non-decreasing in the second
variable $s$.
Then $u'\leq w'$ in $[a,b]$, which implies $u<w$ in $(a,b]$.
\end{lem}


Another result we will need is the following, which is a
consequence of Lemma 2.1 in \cite{GLP} via L'H\^{o}pital's Rule.

\begin{lem}\label{AF}
If $F\in KO(\alpha)$ for some $\alpha>1$, then
$$
\lim_{t\to\infty}\frac{t^\alpha}{F(t)}=0.
$$
\end{lem}


We need the following lemma, which shows that solutions
of (\ref{meqn}) with initial slope zero have non-decreasing slope
for $r \in [0,R)$.

\begin{lem}\label{prel}
Suppose in addition to (H1)--(H3), $H(r,\cdot)$ is
non-decreasing on $[0,R)$.  If for $0<r<R$, $w$ is a solution of
\begin{equation} \label{seqnn}
((w')^{p-1})'+\frac{\lambda}{r}(w')^{p-1}=H(r,w),\;\;w(0)\geq 0,\;\;
w'(0)=0,\;\;w(R)=\infty,
\end{equation}
then $w'$ is non-decreasing on $[0,R)$.
\end{lem}

\begin{proof}
Let $w$ be a solution of (\ref{seqnn}). Integrating the equation
$(r^\lambda(w')^{p-1})'=r^\lambda H(r,w)$ over the interval $(0,r)$ for
any $r\in (0,R)$ and recalling that $w'$ is non-negative, we
obtain
\begin{align*}
(w')^{p-1}&=r^{-\lambda}\int_0^r s^\lambda H(s,w(s))\,ds\\
&\leq r^{-\lambda}H(r,w(r))\int_0^rs^\lambda\,ds\\
&=\frac{r}{\lambda+1}H(r,w)
\end{align*}
Using this inequality back in the equation (\ref{seqnn}) we obtain
\begin{align*}
H(r,w)&=((w')^{p-1})'+\frac{\lambda}{r}(w')^{p-1}\\
&\leq ((w')^{p-1})'+\frac{\lambda}{r}\cdot\frac{r}{\lambda+1}H(r,w)
\end{align*}
so that
\begin{equation} \label{wine}
((w')^{p-1})'\geq \frac{1}{\lambda +1}H(r,w),\quad 0<r<R.
\end{equation}
The fact that $w'$ is non-decreasing on $(0,R)$ is a consequence
of (\ref{wine}) as follows. Let $0<r_1<r_2<R$. Integrating
(\ref{wine}) on $(r_1,r_2)$ leads to
$$
(w'(r_2))^{p-1}-(w'(r_1))^{p-1}\geq \frac{1}{\lambda+1}
\int_{r_1}^{r_2}H(s,w(s))\,ds \geq 0.
$$
\end{proof}

We are now ready to state and prove the main result of this
section.

\begin{thm}\label{scomp}
Suppose in addition to (H1)--(H3), $H(r,\cdot)$ is
non-decreasing on $[0,R)$ and for $f(s)=H(0,s)$, $f(0)=0$ and
$F\in KO(p)$. Then there is a solution $w$ of (\ref{seqn}) such
that for any solution $u$ of (\ref{meqn}),
$$
u(r)\leq w(r),\quad 0\leq r<R \,.
$$
\end{thm}

\begin{proof}
For each positive integer $k$, with $1/k<R$, let $w_k$ be a
solution, in $(0,R-1/k)$, of the problem
\begin{equation} \label{w}
\begin{gathered}
((w')^{p-1})'+\frac{\lambda}{r}(w')^{p-1}=H(0,w),\\
w(0)\geq 0,\quad w'(0)= 0,\quad w(R-1/k)=\infty.
\end{gathered}
\end{equation}
This is possible, since $f(s)=H(0,s)$ satisfies the Keller-Osserman
condition.

Since $H(0,u)\leq H(r,u)$ for all $0\leq r<R$, we first note that
$$
Lw_k-H(0,w_k)\leq Lu-H(0,u)\quad \text{on }(0,R-1/k).
$$
Suppose that $w_k(0)<u(0)$.  Then, since $0=w'_k(0)\leq u'(0)$, by Lemma
\ref{comp} we conclude that $w_k<u$ on $(0,R-1/k)$. But this is
obviously not possible since $w_k$ blows up at $R-1/k$ and $u$
does not. Thus we must have $u(0)\leq w_k(0)$.
Actually, we claim that
$$
u(r)\leq w_k(r),\quad \mbox{for all $r$
with}\;\;0\leq r<R-\frac{1}{k}.
$$
 Suppose to the contrary that
$u(r)>w_k(r)$ for some $0<r<R-1/k$.  Since $u(0)\leq w_k(0)$ the
function $u-w_k$ takes on a positive maximum inside $[0,r_1]$
where $r_1$ is taken sufficiently close to $R-1/k$. If $r^*$ is
such a maximum point, then we have
$$
w_k(r^*)<u(r^*),\quad \mbox{and}\quad w'_k(r^*)=u'(r^*)\,.
$$
 By the comparison Lemma \ref{comp} we conclude that $w_k<u$ on
$(r^*,R-1/k)$, which is impossible. Thus we must have
$u(r)\leq w_k(r)$, $r\in(0,R-1/k)$, as claimed.

By a similar argument as above, and using $w_{k+1}$ instead of
$u$, we also conclude that
$$
w_{k+1}(r)\leq w_k(r),\quad 0\leq r<R-\frac{1}{k}\,.
$$
 Using this and the
fact that $w_k$ and $w_{k+1}$ satisfy equation (\ref{w}) we obtain
\begin{align*}
(w'_{k+1}(r))^{p-1}&=r^{-\lambda}\int_0^r s^\lambda
H(0,w_{k+1}(s))\,ds\\ &\leq r^{-\lambda}\int_0^r s^\lambda
H(0,w_k(s))\,ds\\ &=(w'_k(r))^{p-1},\quad 0<r<R-1/k.
\end{align*}
This shows that
$ w'_{k+1}(r)\leq w'_k(r)$, $0\leq r<R-1/k$.
Therefore, we have
 \begin{equation} \label{wp} w'_n(r)\leq w'_m(r),\quad 0\leq r<R-1/m,
\end{equation}
whenever $n \geq m > 1/R$.

For $t,r\in (0,R-1/k)$, and $n>k$ we have
\[
|w_n(r)-w_n(t)|=\big|\int_t^rw'_n(s)\,ds\big| \leq
w'_n(\zeta)|r-t| \leq w'_{k+1}(R-1/k)|r-t|,
\]
where $\zeta=\max\{r,t\}$. The fact that $w'_{k+1}$ is
non-decreasing, by Lemma \ref{prel}, has been exploited in the
last inequality.

Thus $\{w_n\}_{n=k+1}^\infty$ is a bounded equicontinuous family
in $C([0,R-1/k])$, and hence has a uniformly convergent
subsequence. Let $w$ be the limit. For $r\in[0,R-1/k]$ and $n>k$
the solution $w_n$ satisfies the integral equation
$$
w_n(r)=w_n(0)+\int_0^r\Big(\int_0^t\left(\frac{s}{t}\right)^\lambda
H(0,w_n(s))\,ds\Big)^{1/(p-1)}\,dt\,.
$$
 Letting $n\to\infty$
we see that $w$ satisfies the same integral equation. Since $k$ is
arbitrary we conclude that $w$ satisfies equation (\ref{seqn}).
Since $u\leq w_n$ on $(0,R-1/k)$ for each $n\geq k$ we conclude
that $u\leq w$ on $(0,R)$.
\end{proof}

\section{Proofs of Main Results and Some Corollaries}

\begin{proof}[Proof of Theorem \ref{uint}]
By Theorem \ref{scomp} we take a solution $w$ of
(\ref{seqn}) such that $u(r)\leq w(r)$ for $0\leq r<R$. Using
$f(w):=H(0,w)$ in place of $H(r,w)$ in inequality (\ref{wine}), we
note that $w$ satisfies
$$
((w')^{p-1})'>\frac{1}{\lambda+1}f(w),\quad 0<r<R.
$$
Multiplying both sides of the above inequality by $w'$ and
integrating on $(0,r)$, we find that for $r$ close to $R$,
\begin{align*}
\frac{p-1}{p}(w'(r))^p&\geq\frac{1}{\lambda+1}[F(w(r))-F(w(0))]\\
&= \frac{1}{\lambda+1}F(w(r))\big[1-\frac{F(w(0))}{F(w(r))}\big].
\end{align*}
Thus, for some positive constants $C$ and $\tau$, which may change
in each line below, but depend only on the constants $\lambda$ and
the primitive $F$, we see that
$$
\frac{p-1}{p}(w'(r))^p\geq C F(w(r)),\quad \tau<r<R,
$$
 or \begin{equation} \label{F} F(w(r))^{1/p}\leq C w'(r),\quad \tau<r<R.
\end{equation}
From Lemma \ref{AF}, it follows that
\begin{equation} \label{op}
w(r)\leq F(w(r))^{\frac{1}{\alpha}},\quad \tau<r<R.
\end{equation}
Using (\ref{F}) and (\ref{op}),  we obtain
\begin{align*}
\int_\tau^{R} w(r)^{\frac{\alpha-p}{p}}\, dr
&\leq\int_\tau^{R} F(w(r))^{\frac1p-\frac{1}{\alpha}}\,dr\\
&\leq C\int_\tau^{R} \frac{w'(r)}{F(w(r))^{\frac{1}{\alpha}}}\,dr\\
&=C\int_{w(\tau)}^\infty\frac{1}{F(t)^{\frac{1}{\alpha}}}\,dt<\infty
\end{align*}
Thus, recalling that $u\leq w$ on $(0,R)$  we get
$$
\int_\tau^{R} u(r)^{\frac{\alpha-p}{p}}\, dr\leq C\int_{w(\tau)}^
\infty\frac{1}{F(t)^{\frac{1}{\alpha}}}\,dt<\infty,
$$
 giving the desired result.
\end{proof}


\begin{proof}[Proof of Theorem \ref{FKO}]
Suppose that $g$ is positive and non-decreasing on
$(r_*,R)$ for some $0<r_*<R$. Observe that from (\ref{meqn}) we
obtain
$$
((u')^{p-1})'\leq g(r)f(u),
$$
 and multiplying both sides of this
by $u'$ and integrating on $(r_*,r)$ we find that
$$
\frac{u'}{F(u)^{1/p}}\leq (qg(r))^{1/p}\big[1+\frac{u'(r_*)^p}
{qg(r)F(u(r))}\big]^{1/p},\quad r_*<r<R,
$$
 where $q$ is the
H\"{o}lder conjugate exponent of $p$.  From this we conclude that,
for some positive constants $C$ and $r_0$,
\begin{equation} \label{F1}
\frac{u'}{F(u)^{1/p}}\leq Cg(r)^{1/p},\quad r_0<r<R.
\end{equation}

Let $\gamma=\alpha(1-\sigma/p)$. The hypothesis
$0<\sigma<p(\alpha-p)/\alpha$ implies
that $p<\gamma<\alpha$. Using (\ref{F1}) and H\"{o}lder's inequality, we
obtain
\begin{align*}
&\int_{u(r_0)}^\infty\frac{1}{F(t)^{1/\gamma}}\,dt\\
&=\int_{u(r_0)}^\infty\frac{t^{(\alpha-p)/\alpha
}}{F(t)^{1/\gamma}}\cdot\frac{1}{t^{(\alpha-p)/\alpha}}\,dt\\
&\leq \Big(\int_{u(r_0)}^\infty\frac{t^{[(\alpha-p)/\alpha]\cdot
[\gamma/p]}} {F(t)^{1/p}}\,dt\Big)^{p/\gamma}
\Big(\int_{u(r_0)}^\infty\frac{1}
{t^{[(\alpha-p)/\alpha]\cdot[\gamma/(\gamma-p)]}}\,dt\Big)^{(\gamma-p)/\gamma}\\
&\leq C
\Big(\frac{\alpha}{p}\cdot\frac{\gamma-p}{\alpha-\gamma}\Big)^{(\gamma-p)/\gamma}
\Big(\int_{r_0}^R\frac{u'(r)u(r)^{[(\alpha-p)/p]
\cdot[\gamma/\alpha]}}{F(u(r))^{1/p}}\,dr\Big)^{p/\gamma}\\ &\leq
C\big(\frac{\alpha}{p}\cdot\frac{\gamma-p}{\alpha-\gamma}\big)
^{(\gamma-p)/\gamma} \Big(\int_0^R g(r)^{1/p}u(r)^{[(\alpha-p)/p]
\cdot[\gamma/\alpha]}\,dr\Big)^{p/\gamma}\\ &\leq
C\big(\frac{\alpha}{p}\cdot\frac{\gamma-p}{\alpha-\gamma}\big)^{(\gamma-p)/\gamma}
\Big(\int_0^R
g(r)^{1/p\cdot\alpha/(\alpha-\gamma)}\,dr\Big)^{\frac{p}{\gamma}
\cdot\frac{\alpha-\gamma}{\alpha}}\Big(\int_0^R
u(r)^{(\alpha-p)/p} \,dr\Big)^{p/\alpha}.
\end{align*}
Recalling that $1/p\cdot\alpha/(\alpha-\gamma)=1/\sigma$, by
hypothesis the right hand side of the last inequality is finite
and this proves the claim.
\end{proof}


Note that if $g$ is bounded on $[0,R)$, but not necessarily
non-decreasing near $R$, the right hand side of (\ref{F1}) can be
replaced by a constant.  The proof of Theorem \ref{FKO} shows that
$F \in KO(\gamma)$ for any $0<\gamma<\alpha$.  We record this as
follows.


\begin{cor}\label{coro3.1}
Let $H(r,s)=g(r)f(s)$ satisfy (H1)--(H3), with $g(0)>0$. Suppose
(\ref{meqn}) has a blow-up solution that belongs to
$L^{(\alpha-p)/p}(0,R)$ for some $\alpha>p$. If $g$ is bounded, then
$F\in KO(\gamma)$ for any $0<\gamma<\alpha$.
\end{cor}



\begin{rem} \label{rmk3.2}\rm
The conclusion of Corollary \ref{coro3.1} is false when $g$ is
unbounded near $R$ as the following example shows.

The function $u(r)=(1-r)^{-1}$ is a solution of
\begin{gather*}
u''(r)=g(r)f(u),\\
u(0)\geq 0,\quad u'(0)\geq 0,\quad u(1)=\infty,
\end{gather*}
 where
$$
g(r):=2/(1-r),\quad \text{and}\quad  f(s):=s^2.
$$
 Observe that
$u\in L^{(\alpha-2)/2}(0,1)$ for $2<\alpha<4$. However note that
$F\notin KO(3)$.
\end{rem}

\begin{cor} \label{coro3.2}
Suppose $H(r,s)=g(r)f(s)$ satisfies (H1)--(H3), with $g(0)>0$, $g$
non-decreasing on $[0,R)$, and $f(0)=0$. Further, let $g$ be
bounded on $[0,R)$, and let $F\in KO(p)$. Then a blow up solution
$u$ of (\ref{meqnre}) belongs to $L^q(0,R)$ for some $q>0$ if and
only if $F\in KO(\gamma)$ for some $\gamma>p$.
\end{cor}

\begin{proof}
Suppose $F\in KO(\gamma)$ for some $\gamma>p$. Then by Theorem
\ref{uint}, we see that $u\in L^q(0,R)$ for $q=(\gamma-p)/p$. For
the converse, suppose that $u\in L^q(0,R)$ for some $q>0$. Then
for $\alpha=p(q+1)$ we see that $q=(\alpha-p)/p$ so that by the
above corollary, $F\in KO(\gamma)$ for some $p<\gamma<p(q+1)$.
\end{proof}

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