\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 155, 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 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/155\hfil Infinite semipositone problems] {Infinite semipositone problems with indefinite weight and asymptotically linear growth forcing-terms} \author[G. A. Afrouzi, S. Shakeri \hfil EJDE-2013/155\hfilneg] {Ghasem A. Afrouzi, Saleh Shakeri} % in alphabetical order \address{Ghasem Alizadeh Afrouzi \newline Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran} \email{afrouzi@umz.ac.ir} \address{Saleh Shakeri \newline Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran} \email{s.shakeri@umz.ac.ir} \thanks{Submitted September 17, 2012. Published July 8, 2013.} \subjclass[2000]{35J55, 35J25} \keywords{Infinite semipositone problem; indefinite weight; forcing term; \hfill\break\indent asymptotically linear growth; sub-supersolution method} \begin{abstract} In this work, we study the existence of positive solutions to the singular problem \begin{gather*} -\Delta_{p}u = \lambda m(x)f(u)-u^{-\alpha} \quad \text{in }\Omega,\\ u = 0 \quad \text{on }\partial \Omega, \end{gather*} where $\lambda$ is positive parameter, $\Omega$ is a bounded domain with smooth boundary, $0 <\alpha<1$, and $f:[0,\infty] \to\mathbb{R}$ is a continuous function which is asymptotically p-linear at $\infty$. The weight function is continuous satisfies $m(x)>m_0>0$, $\|m\|_{\infty}<\infty$. We prove the existence of a positive solution for a certain range of $\lambda$ using the method of sub-supersolutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} In this article, we consider the positive solution to the boundary-value problem $$\label{e1.1} \begin{gathered} -\Delta_{p}u = \lambda m(x)f(u)-\frac{1}{u^{\alpha}} \quad \text{in }\Omega,\\ u = 0 \quad \text{on }\partial \Omega, \end{gathered}$$ where $\lambda$ is positive parameter, $\Delta_{p}u=\operatorname{div}(|\nabla u|^{p-2}\nabla u)$, $p>1$, $\Omega$ is a bounded domain with smooth boundary $\partial\Omega$, $0 <\alpha<1$, and $f:[0,\infty] \to\mathbb{R}$ is a continuous function which is asymptotically p-linear at $\infty$. The weight function $m(x)$ satisfies $m(x) \in C(\Omega)$ and $m(x)>m_0>0$ for $x\in \Omega$ and also $\|m\|_{\infty}= l<\infty$. We prove the existence of a positive solution for a certain range of $\lambda$. We consider problem \eqref{e1.1} under the following assumptions. \begin{itemize} \item[(H1)] There exist $\sigma_1> 0$, $k > 0$ and $s_0 >1$ such that $f(s)\geq \sigma_1s^{p-1}-k$ for all $s\in[0,s_0]$; \item[(H2)] $\lim_{s \to+\infty} \frac{f(s)}{s^{p-1}}= 0$ for some $\sigma > 0$. \end{itemize} Let $F(u):=\lambda m(x)f(u)-\frac{1}{u^{\alpha}}$. The case when $F(0)< 0$ (and finite) is referred to in the literature as a semipositone problem. Finding a positive solution for a semipositone problem is well known to be challenging (see \cite{BeCaNi,Lion}). Here we consider the more challenging case when $\lim_{u\to 0^{+}} F(u)=-\infty$. which has received attention very recently and is referred to as an infinite semipositone problem. However, most of these studies have concentrated on the case when the nonlinear function satisfies a sublinear condition at $\infty$ (see \cite{LeShiYe,LeShiYe2,RamShiYe}). The only paper to our knowledge dealing with an infinite semipositone problem with an asymptotically linear nonlinearity is \cite{Hai}, where the author is restricted to the case $p = 2$. Also here the existence of a positive solution is focused near $\lambda_1/\sigma$ , where $\lambda_1$ is the first eigenvalue of$-\Delta$. See also \cite{AmArBu,Zhang}, where asymptotically linear nonlinearities have been discussed in the case of a nonsingular semipositone problem and an infinite positone problem. Recently, in the case when $m(x)=1$ problem \eqref{e1.1} has been studied by Shivaji et al \cite{HaSanShi}. The purpose of this paper is to improve the result of \cite{HaSanShi} with weight $m$. We shall establish our existence results via the method of sub and super-solutions. \begin{definition} \rm We say that $(\psi)$ (resp. $Z$) in $W^{1,p}(\Omega) \cap C(\overline{\Omega})$ are called a sub-solution (resp. super-solution) of \eqref{e1.1}, if $\psi$ satisfies $$\label{2} \begin{gathered} \int_{\Omega}|\nabla \psi(x)|^{p-2}\nabla \psi(x).\nabla w(x) dx \leq \int_{\Omega}(\lambda m(x)f(\psi )-\frac{1}{\psi^{\alpha}}) w(x)dx \quad\forall w \in W,\\ \psi >0 \quad \text{in }\Omega,\\ \psi=0 \quad \text{on }\partial \Omega, \end{gathered}$$ and $$\label{3} \begin{gathered} \int_{\Omega}|\nabla Z|^{p-2}\nabla Z .\nabla w(x) dx \geq \int_{\Omega} (\lambda m(x)f(Z )-\frac{1}{Z^{\alpha}}) w(x)dx \quad \forall w \in W,\\ Z>0 \quad \text{in }\Omega,\\ Z=0 \quad \text{on }\partial \Omega, \end{gathered}$$ where $W = \{\xi \in C_0^\infty(\Omega): \xi \geq 0 \text{in } \Omega \}$ \end{definition} The following lemma was established by Cui \cite{SCui}. \begin{lemma}[\cite{SCui}] \label{lem1.2} If there exist sub-supersolutions $\psi$ and $Z$, respectively, such that $\psi\leq Z$ on $\Omega$, Then \eqref{e1.1} has a positive solution $u$ such that $\psi\leq u\leq Z$ in $\Omega$. \end{lemma} In next Section, we will state and prove the existence of a positive solution for a certain range of $\lambda$. \section{Main result} Our main result for problem \eqref{e1.1} reads as follows. \begin{theorem}\label{the2.1} Assume {\rm (H1)--(H2)}. If there exist constants $s_0^*(\sigma,\Omega)$, $J(\Omega)$, $\underline{\lambda}$, and $\hat{\lambda}>\underline{\lambda}$ such that if $s_0\geq s_0^*$ and $m_0\sigma_1/(l\sigma)\geq J$, then \eqref{e1.1} has a positive solution for $\lambda \in [\underline{\lambda},\hat{\lambda}]$. Here $\mu_1$ is the principal eigenvalue of operator $-\Delta_{p}$ with zero Dirichlet boundary condition. \end{theorem} \begin{proof} By Anti-maximum principle \cite{Peter}, there exists $\xi=\xi(\Omega)>0$ such that the solution $z_{\mu}$ of \begin{gather*} -\Delta_{p} z-\mu |z|^{p-2}z=-1 \quad \text{in } \Omega,\\ z=0 \quad \text{on } \partial \Omega, \end{gather*}\ for $\mu\in(\mu_1,\mu_1+\xi)$, is positive in $\Omega$ and is such that $\frac{\partial z_{\mu}}{\partial \nu}<0$ on $\partial\Omega$, where $\nu$ is outward normal vector at $\partial \Omega$. Since $z_{\mu}>0$ in $\Omega$ and $\frac{\partial z_{\mu}}{\partial \nu}< 0$ there exist $m>0$, $A > 0$, and $\delta >0$ such that $|\nabla z_{\mu}|\geq m$ in $\overline{\Omega}_{ \delta}$ and $z_{\mu}\geq A$ in $\Omega \setminus \overline{ \Omega}_{\delta}$, where $W = \{x \in \Omega: d(x,\partial \Omega)\leq \delta\}$. We prove the existence of a solution by the comparison method \cite{DraKerTak}. It is easy to see that any sub-solution of $$\label{e2.1} \begin{gathered} -\Delta_{p}u = \lambda m_0f(u)-\frac{1}{u^\alpha} \quad \text{in } \Omega,\\ u=0 \quad \text{on }\partial \Omega, \end{gathered}$$ is a sub-solution of \eqref{e1.1}. Also any super-solution of $$\label{e2.2} \begin{gathered} -\Delta_{p}u =\lambda lf(u)-\frac{1}{u^\alpha} \quad \text{in }\Omega,\\ u=0 \quad \text{on }\partial \Omega, \end{gathered}$$ is a super-solution of \eqref{e1.1}, where $l$ is as defined above. We first construct a supersolution for\eqref{e1.1}.Let $Z=M_{\lambda}e_{p}$, where $M_{\lambda} \gg 1$ and $e_{p}$ is the unique positive solution of \begin{gather*} -\Delta_{p}e_{p} = 1 \quad \text{in }\Omega,\\ e_{p} = 0 \quad \text{on }\partial \Omega, \end{gather*} Let $\tilde{f}(s)=\max_{t\in [0,s]}f(t)$. Then $f(s)\leq \tilde{f}(s)$, $\tilde{f}(s)$ is increasing, and $\lim_{u \to+\infty} \frac{\tilde{f}(u)}{u^{p - 1}} = 0\,.$ Hence, we can choose $M_{\lambda} \gg 1$ such that $$2\sigma \geq \frac{\tilde{f}(M_{\lambda}\|e_{p}\|_{\infty})}{(M_{\lambda}\|e_{p}\|_{\infty})^{p-1}}$$ Now let $\hat{\lambda}=1/\big(2l\sigma\|e_{p}\|_{\infty}^{p-1}\big)$. For $\lambda \leq \hat{\lambda}$, $$-\Delta_{p}Z = M_{\lambda}^{p-1} \geq \frac{\tilde{f}(M_{\lambda}\|e_{p}\|_{\infty})}{2\sigma \|e_{p}\|_{\infty}^{p-1}} \geq \lambda l \tilde{f}(M_{\lambda}e_{p}) \geq\lambda l f(M_{\lambda }e_{p})\geq \lambda l f(Z)-\frac{1}{Z^{\alpha}}.$$ Thus, $Z$ is a supersolution of \eqref{e2.2}; therefore $Z$ is a supersolution of \eqref{e1.1} Define $$\psi:=k_0z_{\mu}^\frac{p}{p-1+\alpha},$$ where $k_0> 0$ is such that \begin{gather*} \frac{1}{k_0^{p-1+\alpha}}\Big(1+\frac{k k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}} {2l\sigma\|e_{p}\|_{\infty}^{p-1} }\Big) \leq \min\Big\{\Big(\frac{m^{p}(1-\alpha)(p-1)p^{p-1}}{(p-1+\alpha)^{p}}\Big), \Big(\frac{p}{p-1+\alpha}\Big)^{p-1}A \Big\}. \end{gather*} Then $\nabla \psi=k_0\Big(\frac{p}{p-1+\alpha}\Big)z_{\mu}^\frac{ 1-\alpha}{p-1+\alpha}\nabla z_{\mu},$ and $$\label{e2.3} \begin{split} -\Delta_{p} \psi &= -\operatorname{div}(|\nabla \psi|^{p-2}\nabla \psi)\\ &=-k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1}\operatorname{div}( z_{\mu}^\frac{(1-\alpha)(p-1)}{p-1+\alpha}|\nabla z_{\mu}|^{p-2}\nabla z_{\mu}) \\ &= -k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1}\Big\{ (\nabla z_{\mu}^\frac{(1-\alpha)(p-1)}{p-1+\alpha}).|\nabla z_{\mu}|^{p-2}\nabla z_{\mu} \\ &\quad + z_{\mu}^\frac{(1-\alpha)(p-1)}{p-1+\alpha}\Delta_{p}z_{\mu} \Big\} \\ &= -k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1}\Big\{\frac{(1-\alpha)(p-1)}{p-1+\alpha} z_{\mu}^\frac{-\alpha p}{p-1+\alpha}|\nabla z_{\mu}|^{p}\\ &\quad + z_{\mu}^\frac{(1-\alpha)(p-1)}{p-1+\alpha}(1-\mu z_{\mu}^{p-1}) \Big\} \\ &= k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1} \mu z_{\mu}^\frac{p(p-1)}{p-1+\alpha}-k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1} z_{\mu}^\frac{(1-\alpha)(p-1)}{p-1+\alpha}\\ &\quad - \frac{k_0^{p-1}p^{p-1}(1-\alpha)(p-1) |\nabla z_{\mu}|^{p}}{(p-1+\alpha)^{p}z_{\mu}^\frac{\alpha p}{p-1+\alpha}}. \end{split}$$ Now we let $s_0^*(\sigma,\Omega)= k_0\|z_{\mu}^\frac{p}{p-1+\alpha}\|_{\infty}$. If we can prove that $$\label{e2.4} -\Delta_{p}\psi \leq \lambda m_0 \sigma_1k_0^{p-1}z_{\mu}^\frac{p(p-1)}{p-1+\alpha}- \lambda k-\frac{1}{ k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}},$$ then (H1) implies that $-\Delta_{p}\psi \leq \lambda m_0 f(\psi)-\frac{1}{\psi^{\alpha}}$, and $\psi$ will be a subsolution of \eqref{e1.1}. We will now prove \eqref{e2.4} by comparing terms in \eqref{e2.3} and \eqref{e2.4}. Let $\underline{\lambda}=\frac{\mu (\frac{p}{p-1+\alpha})^{p-1} }{m_0\sigma_1}$. For $\lambda \geq\underline{\lambda}$, $$\label{e2.5} k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1} \mu z_{\mu}^\frac{p(p-1)}{p-1+\alpha} \leq \lambda m_0 \sigma_1k_0^{p-1}z_{\mu}^\frac{p(p-1)}{p-1+\alpha}$$ Also since $\lambda \leq \hat{\lambda}=\frac{1}{2l\sigma\|e_{p}\|_{\infty}^{p-1} }$, $$\label{e2.6} \begin{split} \lambda k+\frac{1}{ k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}} &\leq \frac{1}{ k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}}+\frac{k}{2l\sigma\|e_{p}\|_{\infty}^{p-1} }\\ &=\frac{k_0^{p-1}}{z_{\mu}^\frac{\alpha p}{p-1+\alpha}}\Big[\frac{1}{k_0^{p-1+\alpha}}(1+\frac{k k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}}{2l\sigma\|e_{p}\|_{\infty}^{p-1} })\Big] \end{split}$$ Now in ${\Omega}_{\delta}$, we have $|\nabla z_{\mu}|\geq m$, and by \eqref{e2.2}, $$\frac{1}{k_0^{p-1+\alpha}}(1+\frac{k k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}}{2l\sigma\|e_{p}\|_{\infty}^{p-1} })]\leq \frac{m^{p}(1-\alpha)(p-1)p^{p-1}}{(p-1+\alpha)^{p}}.$$ Hence $$\label{e2.7} \lambda k+\frac{1}{ k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}} \leq \frac{k_0^{p-1}p^{p-1}(1-\alpha)(p-1)|\nabla z_{\mu}|^{p}}{(p-1+\alpha)^{p}z_{\mu}^\frac{\alpha p}{p-1+\alpha}}\quad\text{ in }{\Omega}_{\delta}$$ From \eqref{e2.5}, \eqref{e2.7} it can be seen that \eqref{e2.4} holds in ${\Omega}_{\delta}$. We will now prove \eqref{e2.4} holds also in $\Omega \setminus { \Omega}_{\delta}.$ Since $z_{\mu}\geq A$ in $\Omega \setminus {\Omega}_{\delta}$ and by \eqref{e2.2} and \eqref{e2.6} we obtain $$\label{e2.8} \lambda k+\frac{1}{ k_0^{\alpha}z_{\mu}^\frac{\alpha p}{p-1+\alpha}} \leq \frac{k_0^{p-1}}{ z_{\mu}^\frac{\alpha p}{p-1+\alpha}}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1}z_{\mu} \leq k_0^{p-1}\Big(\frac{p}{p-1+\alpha}\Big)^{p-1} z_{\mu}^\frac{(1-\alpha)(p-1)}{p-1+\alpha}$$ in $\Omega \setminus {\Omega}_{\delta}$. From \eqref{e2.5} and \eqref{e2.8}, \eqref{e2.4} holds also in $\Omega \setminus \overline{ \Omega}_{\delta}$. Thus $\psi$ is a positive subsolution of\eqref{e1.1} if $\lambda \in [\underline{\lambda},\hat{\lambda}]$. We can now choose $M_{\lambda} \gg 1$ such that $\psi \leq Z$. 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