\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 $$\begin{gathered} -\Delta u(x)=\lambda f(u(x))\quad x\in \Omega, \\ u(x)=0\quad x\in \partial \Omega, \end{gathered} \label{eq1}$$ 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 $$\begin{gathered} -u''(r)-\frac{N-1}ru'(r)=\lambda f(u(r)) \quad r\in ( 0,1), \\ u'(0)=0, \quad u(1)=0. \end{gathered} \label{eq2}$$ 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 $00$ 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 $$\label{e2.5} t^{N-1}H(t)\geq c_2\lambda ^{1-\frac {N}{2}}\{ F(kd) -\frac{N-2}{2N}df(d)\} \{ \frac {d}{f(d)}\} ^{N/2} +\lambda B\frac{t^N-t_0^N}{N},$$ 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 $$\label{e2.6} \frac{f(x)}x\geq \frac \varrho \lambda ,\quad \forall x\geq d_0 .$$ 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 $$\label{e2.7} u(t)v'(t)-v(t)u'(t)>0,\quad \forall t\in (0,\frac{\overline{t}}{4}].$$ 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 $$\label{e2.8} \beta \leq u(t,d,\lambda )\leq d_0,\quad \forall t\in (t^{*}, \widehat{t}).$$ 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 $$\label{e2.9} E(t,d,\lambda) =\frac{u'(t,d,\lambda)^2}2+\lambda F(u( t,d,\lambda)) \geq \lambda F(kd).$$ 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 $$\label{e2.10} t^NE(t,d,\lambda)\geq c_2\lambda ^{1-\frac N2}\{ F(kd) -\frac{N-2}{2N} df(d)\} \{ \frac d{f(d)}\} ^{N/2} +\lambda B\frac{t^N-t_0^N}N.$$ 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} \begin{thebibliography}{0} \bibitem{a1} D. Arcoya and A. Zertiti; 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 Journal of Qualitative Theory of Differential Equations, 2008, no. 38, pp. 1-13. \end{thebibliography} \end{document}