\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 26, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/26\hfil Nonexistence of radial positive solutions] {Nonexistence of radial positive solutions for a nonpositone problem} \author[S. Hakimi, A. Zertiti\hfil EJDE-2011/26\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 June 14, 2010. Published February 10, 2011.} \subjclass[2000]{35J25, 34B18} \keywords{Nonpositone problem; radial positive solutions} \begin{abstract} In this article we study the nonexistence of radial positive solutions for a nonpositone problem when the nonliearity is superlinear and has more than one zero. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} We study the nonexistence 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 $\lambda >0$, $f: [ 0,+\infty ) \to \mathbb{R}$ is a continuous nonlinear function that has more than one zero, and $\Omega \subset \mathbb{R}^N$; is the annulus: $\Omega =C(0,R,\widehat{R}) =\{x\in \mathbb{R}^N: R<|x| <\widehat{R}\}$ ($N>2$, $00$. Our main objective in this article is to prove that the result of nonexistence of radial positive solutions of \eqref{eq1} remains valid when $f$ has more than one zero and is not increasing entirely on $[0,+\infty )$; see \cite[Theorem 3.1]{a1}. More precisely we assume that the map $f: [0,+\infty ) \to \mathbb{R}$ satisfies the following hypotheses \begin{itemize} \item[(H1)] $f\in C^1([ 0,+\infty ),\mathbb{R})$ such that $f$ has three zeros $\beta _1<\beta_2<\beta_3$ with $f'(\beta _i) \neq 0$ for all $i\in \{1,2,3\}$. Moreover, $f'\geq 0$ on $[ \beta _3,+\infty )$. \item[(H2)] $f(0)<0$. \item[(H3)] $\lim_{u\to +\infty } f(u)/u=+\infty$, \end{itemize} \section{The main result} In this section, we give the main result in this work. More precisely we shall prove the following theorem. \begin{theorem} \label{thm2.1} Assume that the hypotheses {\rm (H1)--(H3)} are satisfied. Then there exists a positive real number $\lambda _0$ such that if $\lambda >\lambda _0$, problem \eqref{eq1} has no radial positive solution. \end{theorem} \noindent\textbf{Remark.} We do not know for what radius $r\in (R,\widehat{R})$ the solution $u$ attains its maximum. In addition, $f$ changes sign in $(\beta _1,+\infty )$. These two facts make our study more difficult, and change the proof of nonexistence in \cite{a1}. To prove Theorem \ref{thm2.1}, we need the next three technical lemmas. We note that the proofs of the first and the last lemma are analogous to those of \cite[Lemma 3.2, Lemma 3.4]{a1}. On the other hand, the proof of the second lemma is different from that of \cite[Lemma 3.3]{a1}. This is so because our $f$ has no constant sign in $(\beta _1,+\infty )$. Denote by $u_\lambda (r)$ a positive solution of \eqref{eq1} (if it exists) and let $R_0=(R+\widehat{R})/2$. Following the work \cite{a1}, we introduce the following notation: $\beta =\beta _1$, $\theta =\min \{\beta _2\text{,}\min \theta _i\}$ where $\theta _i$ are the zeros of $F$ ($F(x)=\int_0^xf(t)dt$). \medskip \noindent\textbf{Remark.} In \cite[Theorem B iii]{g1}, $F$ has at most one zero. On the opposite, in our case $F$ may have more than one zero because $f$ has a finite number of zeros. In this paper we assume, with out loss of generality, that $f$ has three zeros. In fact, the number of zeros of $F$ depends on $f$, but $F$ has at most three zeros. \begin{lemma} \label{lem2.2} Let $f\in C^1([ 0,+\infty))$ satisfying {\rm (H3)} and consider $\lambda>2$. If $u_\lambda$ is a positive solution of \eqref{eq2}, then for every $r\in(R_0,\widehat{R}]$ there exists a positive number $M=M(r)>0$ (independent of $\lambda$) such that $u_\lambda (r)\leq M$. \end{lemma} \begin{proof} Let $\varphi _1$ be a positive eigenfunction associated to the first eigenvalue $\mu _1>0$ of the eigenvalue problem \begin{gather*} -(r^{N-1}v')'=\mu r^{N-1}v,\quad R\mu _1/2$,$c>0$such that $f(\zeta)\geq \mu \zeta -c,\quad \forall \zeta \geq 0$ (because$fis superlinear), we deduce \begin{align*} \mu _1\int_R^{\widehat{R}}r^{N-1}\varphi _1(r)u_\lambda(r)dr &= \lambda \int_R^{\widehat{R}}r^{N-1}f(u_\lambda (r))\varphi_1(r)dr\\ &\geq \lambda \mu\int_R^{\widehat{R}}r^{N-1}\varphi _1(r)u_\lambda (r)dr-\lambda c\int_R^{\widehat{R}}r^{N-1}\varphi _1(r)dr, \end{align*} from which $\int_R^{\widehat{R}}r^{N-1}\varphi _1(r)u_\lambda (r)dr \leq \frac{\lambda k}{\lambda \mu -\mu _1} \leq \frac k{\mu -\frac{\mu _1}2}:= A, \quad \forall \lambda >2,$ withk=c\int_R^{\widehat{R}}r^{N-1}\varphi _1(r)dr>0$and$A>0$is independent of$\lambda$. Now, let$r\in(R_0,\widehat{R}]$and choosing$\delta>0$such that$R_02\,. \end{align*} The proof is complete. \end{proof} \begin{lemma} \label{lem2.3} Assume {\rm (H1)--(H3)} and let $R_1\in (R_0,\widehat{R})$, $c\in (\beta ,\theta )$. Then there exists $\lambda _1>0$ such that for all positive solutions $u_\lambda$ of \eqref{eq2} with $\lambda \geq \lambda _1$, there exists $t_1=t_1(\lambda )\in (R_0,R_1)$ satisfying $u_\lambda (t_1)0$. Therefore, $$\lim_{n\to +\infty } u_{\lambda _n}'(r)=-\infty,\; \text{uniformly on compact subsets of} \left( R_0,R_1\right) . \label{eq5}$$ Now, let $r_1, r_2\in (R_0,R_1)$ such that $R_00$ (see Lemma \ref{lem2.2}) and $\inf_n \left(\underline{b}(n)-\overline{a}(n)\right)>0$ and $\lim_{n\to +\infty } u_{\lambda_n}'(r*)=-\infty$ (by \eqref{eq6}), it follows that $\lim_{n\to +\infty }u_{\lambda _n}(r_2)=-\infty$, which contradicts $u_{\lambda _n}\geq0$ for all $n\in \mathbb{N}$. \textbf{Case 2:} $u_{\lambda _n}(\overline{t}_n)\geq\beta _3$. Let $r_0\in \left[\underline{a}(n),\overline{b}(n)\right]$, then $$-r_0^{N-1}u_{\lambda _n}'(r_0)=\lambda _n\int_{\overline{t} _n}^{r_0}s^{N-1}f(u_{\lambda _n}(s))ds.$$ Consider $t_{\beta _2}$ such that $t_{\beta _2} =\max \{r_n\in (R,\widehat{R}] : u_{\lambda_n} (r_n)=\beta_2\}$. Then \begin{align*} -r_0^{N-1}u_{\lambda _n}'(r_0) &= \lambda _n\int_{\overline{t} _n}^{r_0}s^{N-1}f(u_{\lambda _n}(s))ds\\ &= \lambda_n\Big[\int_{\overline{t} _n}^{t_{\beta _2}}s^{N-1}f(u_{\lambda _n}(s))ds+\int_{t_{\beta _2}}^{r_0}s^{N-1}f(u_{\lambda _n}(s))ds\Big] \\ &\geq \lambda _n\int_{t_{\beta _2}}^{r_0}s^{N-1}f(u_{\lambda _n}(s))ds, \end{align*} because $\int_{\overline{t}_n}^{t_{\beta _2}}s^{N-1}f(u_{\lambda _n}(s))ds\geq0$. Then as in Case 1, we obtain a contradiction with the positivity of $u_{\lambda _n}$. \end{proof} \begin{lemma}\label{lem2.4} Assume {\rm (H2)}. Let $R_2\in (R_0,\widehat{R})$ and $\overline{c} >1$. Then there exists $\lambda _2>0$ such that every positive solution $u_\lambda$ of \eqref{eq2} satisfies $\frac \beta {\overline{c}}\in u_\lambda([ R_2,\widehat{R}] )$, for all $\lambda \geq \lambda _2$. Where $b_\lambda =\max \{r\in (R,\widehat{R}) :u_\lambda (r)=\frac \beta {\overline{c}}\}$. \end{lemma} \begin{proof} This lemma will be proved if we show that $$\lim_{\lambda \rightarrow +\infty } b_\lambda =\widehat{R} \label{eq7}$$ To do this, we multiply the equation in \eqref{eq2} by $r^{N-1}$, integrate it from $b_\lambda$ to $\widehat{R}$ and use that $u_\lambda(r)<\frac \beta {\overline{c}}$, for all $r\in (b_\lambda , \widehat{R}]$, to deduce that $$\int_{b_\lambda }^{\widehat{R}}(r^{N-1}u_\lambda '(r)) 'dr \geq \int_{b_\lambda }^{\widehat{R}}\lambda r^{N-1}Kdr$$ where $K=-\max \{f(\zeta ):\zeta \in [ 0,\frac \beta {\overline{c}}]\}>0$. Hence $$\widehat{R}^{N-1}u_\lambda '(\widehat{R})-b_\lambda ^{N-1}u_\lambda '(b_\lambda )\geq \frac \lambda NK(\widehat{R}^N-b_\lambda ^N) >0\,. \label{eq8}$$ On the other hand, multiplying the same equation by $r^{2(N-1)}u_\lambda '(r)$ and integrating from $b_\lambda$ to $\widehat{R}$, we have $$-\int_{b_\lambda}^{\widehat{R}}[ r^{N-1}u_\lambda '(r)] 'u_\lambda '(r)r^{N-1}dr=\lambda \int_{b_\lambda }^{\widehat{R}}[ F(u_\lambda (r))] 'r^{2(N-1)}dr$$ Computing the two integrals by parts, we obtain \begin{align*} &\frac 12[ b_\lambda ^{2(N-1)}u_\lambda '(b_\lambda )^2- \widehat{R}^{2(N-1)}u_\lambda '(\widehat{R})^2]\\ & =-\lambda b_\lambda ^{2(N-1)}F(\frac \beta {\overline{c}}) -2(N-1)\lambda \int_{b_\lambda }^{\widehat{R}}F(u_\lambda (r)) r^{2N-3}dr \end{align*} Since $u_\lambda (r)<\frac \beta {\overline{c}}$, for all $r\in (b_\lambda ,\widehat{R}]$ and $F$ is decreasing in $(0,\beta )$ by (H2), we deduce that \begin{align*} &\frac 12[ b_\lambda ^{2(N-1)}u_\lambda '(b_\lambda )^2- \widehat{R}^{2(N-1)}u_\lambda '(\widehat{R})^2]\\ &\leq -\lambda b_\lambda ^{2(N-1)}F(\frac \beta {\overline{c}})-2(N-1)F(\frac \beta {\overline{c}})\lambda \int_{b_\lambda }^{\widehat{R}}r^{2N-3}dr \\ &= -\lambda \widehat{R}^{2(N-1)}F(\frac \beta {\overline{c}}) \end{align*} By \eqref{eq8}, the left hand of the precedent inequality is positive (because $u_\lambda'(b_\lambda )\leq 0$ by definition of $b_\lambda$ and $u_\lambda '(\widehat{R})\leq 0$ by \cite{g2}). Consequently we can take square roots and using that $A-B\leq \sqrt{A^2-B^2}$ for all $A\geq B\geq 0$, we obtain (by \eqref{eq8} again) $\frac 1{N\sqrt{2}}K\frac 1{ \sqrt{-F(\frac \beta {\overline{c}})}}\sqrt{\lambda }(\widehat{R} ^N-b_\lambda ^N) \leq \widehat{R}^{N-1}$ and as a consequence \eqref{eq7} is satisfied. So the proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.1}] Let $c\in (\beta ,\theta )$, $\overline{c}>1$ and $R_1$, $R_2\in (R_0,\widehat{R})$ such that \$R_1