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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 44, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/44\hfil Positive solutions for a nonpositone problem]
{Radial positive solutions for a nonpositone problem in a ball}

\author[S. Hakimi, A. Zertiti\hfil EJDE-2009/44\hfilneg]
{Said Hakimi, Abderrahim Zertiti}  % in alphabetical order

\address{Said Hakimi \newline
Universit\'e Abdelmalek Essaadi\\
Facult\'e des sciences \\
D\'epartement de Math\'ematiques \\
BP 2121, T\'etouan, Morocco}
\email{h\_saidhakimi@yahoo.fr}

\address{Abderrahim Zertiti \newline
Universit\'e Abdelmalek Essaadi\\
Facult\'e des sciences \\
D\'epartement de Math\'ematiques \\
BP 2121, T\'etouan, Morocco}
\email{zertitia@hotmail.com}

\thanks{Submitted November 10, 2008. Published March 24, 2009.}
\subjclass[2000]{35J25, 34B18}
\keywords{Nonpositone problem; radial positive solutions; shooting method}

\begin{abstract}
 In this paper, we study the existence of radial positive solutions for a
 nonpositone problem when the nonlinearity is superlinear and may
 have more than one zero.
\end{abstract}

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

\section{Introduction}

We  study the existence of radial positive solutions for the
boundary-value problem
\begin{equation}
\begin{gathered}
-\Delta u(x)=\lambda f(u(x))\quad x\in \Omega, \\
u(x)=0\quad  x\in \partial \Omega,
\end{gathered}  \label{eq1}
\end{equation}
where $\Omega $\ is the unit ball of\ $\mathbb{R}^N$, $\lambda >0$
and $N\geq2$.

The existence of radial positive solutions of
\eqref{eq1} is equivalent to the existence of positive solutions of
the problem
\begin{equation}
\begin{gathered}
-u''(r)-\frac{N-1}ru'(r)=\lambda f(u(r)) \quad r\in ( 0,1), \\
u'(0)=0, \quad u(1)=0.
\end{gathered} \label{eq2}
\end{equation}

When $f$ is a monotone nondecreasing nonlinearity and
has only one zero, problem \eqref{eq2} has been studied by  Castro and
Shivaji \cite{c1} in the ball,  and by Arcoya and Zertiti \cite{a1} in the
annulus.

Our main objective in this article is to prove that the result of
the existence of radial positive solutions of problem \eqref{eq1}
remains valid  when $f$ has more than one zero and is not strictly
increasing entirely on $[ 0,+\infty)$; see \cite[Theorem 1.1]{c1}.
 More precisely, we assume that the map $f$ : $[ 0,+\infty)
\to \mathbb{R}$ satisfies the following hypotheses
\begin{itemize}
\item[(F1)]  $f \in C^1( [ 0,+\infty),\mathbb{R})$ such that
$f'\geq 0$ on $[ \beta ,+\infty)$, where $\beta$ is the greatest
 zero of $f$.

\item[(F2)] $f(0)<0$,

\item[(F3)] $\lim_{u\to +\infty } \frac{f(u)}u=+\infty$,

\item[(F4)] For some $k\in (0,1) $, $\lim_{d \to +\infty }
\big(\frac {d}{f(d)}\big)^{N/2}\big(F(kd)-\frac{N-2}{2N}\,df(d)\big)
=+\infty$ where,
$F(x)=\int_0^xf(r)dr$.
\end{itemize}

\textbf{Remark.} We note that in hypothesis (F1) there is no
restriction on the function $f(u)$ for $0<u<\beta$.

\textbf{Remark.} It is well known (see \cite{g1}) that all positive
solutions u of problem \eqref{eq1} are radial with $\frac{\partial
u}{\partial r}<0$ in $\Omega $. This fact permits the authors in
\cite{a1,c1} to apply the shooting method when $f$ satisfies some
suitable hypotheses.

In this paper, we follow the work of  Castro and  Shivaji \cite{c1}
and we used similar ideas (adapted to our case) in the work of
Iaia and  Pudipeddi \cite{i1}.
So we apply the shooting method. For this, we consider the initial
auxiliary boundary-value problem
\begin{gather*}
-u''(r)-\frac{N-1}{r}u'(r)=\lambda
f(u(r))\quad r\in (0,1),\\
u'(0)=0, \quad u(0)=d,
\end{gather*}
where $d$ is the parameter of the shooting method.

It has been shown \cite{c1} that for each real number $d$, the above
auxiliary problem has a unique solution $u(t,d,\lambda )$
satisfying the following property:
if $(d_n,\lambda _n)\to ( d,\lambda)$ as $n\to +\infty$,
then $\{u(.,d_n,\lambda _n)\}$ converges to $u(.,d,\lambda )$
on $[0,1]$.

\section{The main Result}

In this section, we shall give the main result of this work. More
precisely, we have the following theorem.

\begin{theorem} \label{thm2.1}
Assume that the hypotheses {(F1)--(F4)} are satisfied. Then
there exists a positive real number $\lambda _0$ such that if
$\lambda \in ] 0,\lambda _0[ $, problem
\eqref{eq1} has at least one radial positive solution which is decreasing
on $[0,1]$.
\end{theorem}

The proof of this theorem is based on the three next technical
Lemmas in this section. We note that the proofs of the first two
Lemmas are analogous with those of \cite[Lemmas 3.1 and 3.2]{c1}.
On the opposite, the proof of the last
Lemma, in our work, is different from that of \cite[Lemma 3.3]{c1}. This is
due to that in our case $f$ may have many zeros and isn't increasing
entirely on $[0,+\infty)$.

Following \cite{c1}, we introduce the notation and the following
preliminaries. Denote by $\theta$ the greatest zero of $F$. From
(F4), we can choose $\gamma \geq \sup \{ \frac {\beta} {k},\frac
{\theta} {k}\} $ such that
\begin{equation} \label{e2.1}
2NF(kd)-(N-2)df(d) \geq 1,\quad \forall d\geq \gamma.
\end{equation}
Moreover, for $d\geq \gamma $ there exists $t_0\in (0,1]$ such that
\begin{equation} \label{e2.2}
u(t_0,d,\lambda )=kd \quad\text{and}\quad
kd\leq u(t,d,\lambda)\leq
d,\quad \forall t\in [0,t_0].
\end{equation}
Taking into account that $f$ is nondecreasing on
$[kd,d]\subset (\beta ,+\infty ) $ and that
$$
u'(t,d,\lambda )=-\lambda t^{-(N-1)}\int_0^tr^{N-1}f(u(r))dr,
$$
 we obtain from this and \eqref{e2.2}
\[
\frac {\lambda f(k d)}{N} t\leq -u'\leq \frac {\lambda f(d)}{N} t
\]
and integrating on $[0,t_0]$
\begin{equation} \label{e2.3}
c_1\Big(\frac d{\lambda f(kd)}\Big) ^{1/2}\geq t_0\geq
c_1\Big(\frac d{\lambda f(d)}\Big) ^{1/2},
\end{equation}
where $c_1=(2N(1-k)) ^{1/2}$.
Next given $d\in \mathbb{R}$ and $\lambda \in \mathbb{R}$, we define
\begin{gather*}
E(t,d,\lambda)=\frac{u'(t,d,\lambda )^2}{2}+\lambda
F(u(t,d,\lambda )). \\
H(t,d,\lambda)= t E(t,d,\lambda )+\frac{N-2}{2}u(t,d,\lambda
)u'(t,d,\lambda).
\end{gather*}
Then, we show the following identity of Pohozaev type:
\begin{align*}
t^{N-1}H(t,d,\lambda)
&= \widehat{t}^{N-1}H(\widehat{t},d,\lambda)\\
&\quad + \lambda \int_{\widehat{t}}^{t}r^{N-1}\{NF(u(r,d,\lambda))
-\frac{N-2}{2}f(u( r,d,\lambda))u(r,d,\lambda)\}dr.
\end{align*}
Taking $\widehat{t}=0$ and $t=t_0$ in the previous identity and
using \eqref{e2.2} and \eqref{e2.3}, we show as in \cite{c1}:
\begin{equation} \label{e2.4}
t_0^{N-1}H( t_0,d,\lambda) \geq c_2\lambda ^{1-\frac N2}
\{ F(kd) -\frac{N-2}{2N}f(d)d\}\{\frac d{f(d)}\} ^{N/2},
\end{equation}
where $c_2=c_1^N$.


\begin{lemma} \label{lem2.2}
There exists $\lambda _1>0$ such that $u(t,\gamma ,\lambda) \geq \beta $
for all $\lambda \in ( 0,\lambda_1)$ and for all $t\in [ 0,1] $.
\end{lemma}

\begin{proof}
Let $t_1=\sup\{t\leq 1 : u(r,\gamma,\lambda ) \geq \beta,
\forall r\in(0,t)\}$.
Since $f\geq 0$ on $[\beta ,+\infty ) $ and
$u'( t,\gamma ,\lambda) =-\lambda t^{-(N-1)}\int_0^t
r^{N-1}f(u(r,\gamma ,\lambda))dr$, $u $ is decreasing on
$[0,t_1]$. Again, for all $t\in [0,t_1]$,
we have
$$
 | u'(t,\gamma ,\lambda)| \leq
\frac{\lambda \;f\;(\gamma )}{N}<\gamma -\beta,
$$
by assuming that $\lambda <\lambda _1=\frac{N ( \gamma -\beta )}{f(\gamma)}$.
After that, by using the mean value theorem, we obtain
for all $\lambda \in (0,\lambda _1)$:
$$
u(t_1,\gamma ,\lambda) -u(0,\gamma ,\lambda)=u(t_1,\gamma ,\lambda)
-\gamma \geq -(\gamma -\beta ) t_1.
$$
If $t_1<1$, then $u(t_1,\gamma ,\lambda) >\beta$, which contradicts
the definition of $t_1$. Thus $t_1=1$ and the lemma is
proved.
\end{proof}

\begin{lemma} \label{lem2.3}
There exists $\lambda_2>0$ such that if $\lambda \in (0,\lambda_2)$, then
$$
{u(t,d,\lambda)}^2+{u'(t,d,\lambda)}^2>0,\quad
 \forall t\in [ 0,1],\; \forall d\geq \gamma.
$$
\end{lemma}

\begin{proof}
For $t\geq t_0$, the identity of Pohozaev type gives
$$
t^{N-1}H(t)=t_0^{N-1}H(t_0)+\lambda \int_{t_0}^tr^{N-1}
\{NF(u) -\frac{N-2}2f(u)u\}dr.
$$
Extending $f$ by $f(x) =f(0) <0$,
for all $x\in (-\infty ,0]$, there exists $B<0$ such that
$$
NF(s)-\frac{N-2}2f(s)s\geq B,\quad \forall s\in \mathbb{R}.
$$
By (F4), we can take $\gamma $ sufficiently large such that
$$
\{F(kd)-\frac{N-2}{2N}df(d)\} \{ \frac d{f(d)
}\} ^{N/2}\geq 1\quad \forall d\geq \gamma,
$$
and using inequality \eqref{e2.4}, we obtain
\begin{equation} \label{e2.5}
t^{N-1}H(t)\geq c_2\lambda ^{1-\frac {N}{2}}\{ F(kd)
-\frac{N-2}{2N}df(d)\} \{ \frac {d}{f(d)}\} ^{N/2}
+\lambda B\frac{t^N-t_0^N}{N},
\end{equation}
then
\[
t^{N-1}H(t)\geq \lambda (c_2\lambda ^{-\frac {N}{2}}+\frac {B}{N}),
\quad \forall t\in [t_0,1].
\]
Hence, there exists $\lambda _2$ such that for all $\lambda \in
(0,\lambda _2) $, $H(t)>0$ for all $t\in [0,1]$ and for all $d\geq \gamma $.
This implies that $u^2(t) +u'(t)^2>0$, for all $t\in [0,1]$, for all
 $d\geq \gamma $.
\end{proof}

\begin{lemma} \label{lem2.4}
 Given any $\lambda >0$, there exists $d\geq \gamma $
such that $u(t,d,\lambda ) <0$ for some $t\in [0,1]$.
\end{lemma}

\begin{proof}
Let $d\geq \gamma $. We put $\overline{t}=\sup \{ t\in (0,1) :
u(.,d,\lambda )\text{ is decreasing on }(0,t) \}$. Let $\omega $ be
such  that
\begin{gather*}
\omega''+\frac{N-1}r\omega'+\varrho \omega =0,\\
\omega(0) =1, \quad \omega'( 0) =0,
\end{gather*}
where $\varrho $ is chosen such that the first zero of $\omega $ is
$\frac{\overline{t}}4$.

We argue by contradiction. Suppose that $u(t,d,\lambda) \geq 0$
for all $t\in [0,1]$  and all $d\geq \gamma $.
By (F3), there exists $d_0\geq \gamma $ such that
\begin{equation} \label{e2.6}
\frac{f(x)}x\geq \frac \varrho \lambda ,\quad \forall x\geq
d_0 .
\end{equation}
On the other hand, since
$(d\omega)''+\frac{N-1}r(d\omega )'+ \varrho(d\omega ) =0$ and
$u''+\frac{N-1}ru'+\lambda f(u)=0$, we obtain
$$
t^{N-1}[u(t)v(t)'-v(t)u(t)']
=\int_0^ts^{N-1}[\lambda \frac{f(u(s))}{u(s}-\varrho]u(s)v(s)ds,
$$
where $v=d\omega $. Therefore, if $u(t,d,\lambda ) \geq d_0$, for
all $t\in [0,\frac{\overline{t}}4]$, we obtain from \eqref{e2.6},
\[
\int_0^ts^{N-1}[\lambda \frac{f(u(s))}{u(s)}-\varrho ]
u(s)v(s)ds\geq 0,
\]
so that
\begin{equation} \label{e2.7}
u(t)v'(t)-v(t)u'(t)>0,\quad \forall t\in (0,\frac{\overline{t}}{4}].
\end{equation}
Thus, taking into account that $v(\frac{\overline{t}}4) =0$,
 $v'(\frac{\overline{t}}4) <0$, we have
$$
u\big(\frac{\overline{t}}4\big) v'\big(\frac{\overline{t}}4\big)
-v\big(\frac{\overline{t}}4\big) u'\big(\frac{\overline{t}}
4\big) <0.
$$
This is a contradiction with \eqref{e2.7}.
Hence, there exists $t^{*}$ in $(0,\overline{t}/4)$
such that  $u(t^{*},d,\lambda ) =d_0$ and since
$d_0\geq \gamma >\beta $ there exists
$\widehat{t}\in (t^{*},\overline{t}) $ such that
\begin{equation} \label{e2.8}
\beta \leq u(t,d,\lambda )\leq d_0,\quad \forall t\in (t^{*},
\widehat{t}).
\end{equation}
Now, we consider the point $t_0$ defined in \eqref{e2.2}. It is clear
that $t_0<\overline{t}$.

On $[0,t_0]$, since $F$ is nondecreasing on $[ \beta,+\infty[$ and
$u(t,d,\lambda) \geq kd\geq \beta $, for all $t\in(0,t_0]$, we
have
\begin{equation} \label{e2.9}
E(t,d,\lambda) =\frac{u'(t,d,\lambda)^2}2+\lambda F(u( t,d,\lambda))
\geq \lambda F(kd).
\end{equation}
On the other hand, since $u(t,d,\lambda)u'(t,d,\lambda)\leq 0$
for all $t\in (t_0,\overline{t}]$, we have
\begin{align*}
t^NE(t,d,\lambda)
&= t^{N-1}H(t,d,\lambda) -\frac{ N-2}2t^{N-1}u(
t,d,\lambda)u'(t,d,\lambda)  \\
&\geq t^{N-1}H(t,d,\lambda),
\end{align*}
hence, by \eqref{e2.5}, we obtain
\begin{equation} \label{e2.10}
t^NE(t,d,\lambda)\geq c_2\lambda ^{1-\frac N2}\{ F(kd)
-\frac{N-2}{2N} df(d)\} \{ \frac d{f(d)}\} ^{N/2}
+\lambda B\frac{t^N-t_0^N}N.
\end{equation}
Now from (F4), \eqref{e2.9} and \eqref{e2.10} we obtain
$\lim_{d\to+\infty }E(t,d,\lambda) =+\infty$ uniformly
with respect to $t\in [0,\overline{t}]$.
Hence there exists $d_1\geq d_0$
such that for all $d\geq d_1$,
$$
E(t,d,\lambda) \geq \lambda F(d_0)+\frac 2{( \widehat{t}-t^{*})
^2}d_0^2,\quad  \forall t\in [0,\overline{t}].
$$
Taking into account \eqref{e2.8}, this implies
\begin{align*}
u'( t,d,\lambda)^2
&\geq \frac 4{( \widehat{t}-t^{*}) ^2}d_0^2+2\lambda (F(d_0) -F(
u(t,d,\lambda)) ) \\
&\geq \frac 4{( \widehat{t}-t^{*})^2}d_0^2\,,\quad \forall t\in (
t^{*},\widehat{t}),
\end{align*}
which implies
$u'(t,d,\lambda )\leq -\frac {2}{\widehat{t}-t^{*}}d_0$, for all
$t\in (t^{*},\widehat{t})$.

 The mean value theorem gives us
$c\in (t^{*},\frac{t^{*}+\widehat{t}}2)$ such that
\[
u(\frac{\widehat{t}+t^{*}}{2})-u(t^{*})
=u'(c)\frac{\widehat{t}-t^{*}}{2}
\leq -\frac{2d_0}{\widehat{t}-t^{*}}.\frac{\widehat{t}-t^{*}}{2}
=-{d_0}.
\]
Hence  $u(\frac{t^{*}+\widehat{t}}{2})\leq 0$ and since
$u'(\frac{t^{*}+\widehat{t}}2)\leq -\frac {2}{\widehat{t}-t^{*}}d_0<0$,
there exists $T\in (0,1)$ such that $u(T,d,\lambda )<0$ which is a
contradiction. So the proof is finished.
\end{proof}

\begin{remark} \label{rmk2.5}\rm
 In \cite{c1}, to prove the last lemma, the
authors use the fact that if $u(t,d,\lambda)\leq\beta$, where $\beta
$ is the only zero of $f$, then $ F(u(t,d,\lambda ))\leq 0$, which
simplifies the proof.

In our case, even if $u(t,d,\lambda )\leq \beta$ we do not have
$F(u(t,d,\lambda ))\leq 0$,  and we have overcome this problem.
\end{remark}

\begin{proof}[Proof of theorem \ref{thm2.1}]
 As in \cite{c1}, we take $\lambda \in (0,\lambda _0)$  where
$\lambda _0=\min\{\lambda _1,\lambda _2\}$.
 Let $\widehat{d}=\sup \{ d\geq \gamma :
u(t,d,\lambda)\geq 0\;,\;\forall t\in (0,1]\}$.

Lemma \ref{lem2.2} leads to the fact that the set
$\{ d\geq \gamma : u(t,d,\lambda )\geq 0,\, \forall t\in [0,1]\}$
is nonempty.
By Lemma \ref{lem2.4}, we have $\widehat{d}<+\infty $, and we claim
that $u(.,\widehat{d},\lambda )$ is the desired solution, which
satisfies the following properties
\begin{itemize}
\item[(i)] $u(t,\widehat{d},\lambda )>0$, for all $t\in [0,1)$,

\item[(ii)] $u(1,\widehat{d},\lambda )=0$,

\item[(iii)] $u'(1,\widehat{d},\lambda )<0$,

\item[(iv)] $u$ is decreasing in $[0,1]$.

\end{itemize}

To prove (i), we assume that there exists  $T_1<1$ such that
$u(T_1,\widehat{d},\lambda )=0$. By Lemma \ref{lem2.3},
$u'(T_1,\widehat{d},\lambda )\neq 0$. We can assume (see \cite{c1}) that
$u'(T_1,\widehat{d},\lambda )<0$, then there exists
$T_2\in (T_1,1)$ such that $ u(T_2,\widehat{d},\lambda )<0$, which is a
contradiction with the definition of $\widehat{d}$.

 For(ii), we assume $u(1,\widehat{d},\lambda )>0$, then there exists
$\eta >0$ such that $u(t,\widehat{d},\lambda )\geq \eta$ for all
$t\in(0,1]$. Again there exists $\delta >0$ such that
$u(t,\widehat{d}+\delta ,\lambda )\geq \frac \eta 2$ for all
$t\in (0,1]$, which is a contradiction with the definition of
$\widehat{d}$.

Statement (iii) is a consequence of Lemma \ref{lem2.3};
and (iv) is a result of \cite{g1}.
\end{proof}


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 Existence and non-existence
of radially symmetric non-negative solutions for a class of
semi-positone problems in annulus, Rendiconti di Matematica, serie
VII, Volume. 14, Roma (1994), 625-646.

\bibitem{c1} A. Castro and R. Shivaji; Non-negative
solutions for a class of radially symmetric nonpositone problems,
Proc. Amer. Math. Soc., 106, n.3 (1989), 735-740.

\bibitem{g1} B. Gidas, W.M. NI, L. Nirenberg;
symmetry and related properties via the maximum principle, commun.
Math Phys., 68 (1979), 209-243.

\bibitem{i1} J. Iaia, S. Pudipeddi; Radial Solutions to a
Superlinear Dirichlet Problem Using Bessel Functions, Electronic
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\end{thebibliography}

\end{document}