\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 19, pp. 1--10.\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-2004/19\hfil Existence of solutions] {Existence of solutions to nonlocal and singular elliptic problems via Galerkin method} \author[F. J. S. A. Corr\^ea \& S. D. B. Menezes\hfil EJDE-2004/19\hfilneg] {Francisco Julio S. A. Corr\^ea \& Silvano D. B. Menezes} % in alphabetical order \address{Francisco Julio S. A. Corr\^ea \hfill\break Departamento de Matem\'atica-CCEN \\ Universidade Federal do Par\'a \\ 66.075-110 Bel\'em Par\'a Brazil} \email{fjulio@ufpa.br} \address{Silvano D. B. Menezes\hfill\break Departamento de Matem\'atica-CCEN \\ Universidade Federal do Par\'a \\ 66.075-110 Bel\'em Par\'a Brazil} \email{silvano@ufpa.br} \date{} \thanks{Submitted December 15, 2003. Published February 11, 2004.} \subjclass[2000]{35J60, 35J25} \keywords{Nonlocal elliptic problems, Galerkin Method} \begin{abstract} We study the existence of solutions to the nonlocal elliptic equation $$ -M(\|u\|^2)\Delta u = f(x,u) $$ with zero Dirichlet boundary conditions on a bounded and smooth domain of $\mathbb{R}^n$. We consider the $M$-linear case with $f\in H^{-1}(\Omega )$, and the sub-linear case $f(u)=u^{\alpha}$, $0<\alpha <1$. Our main tool is the Galerkin method for both cases when $M$ continuous and when $M$ is discontinuous. \end{abstract} \date{} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{prop}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \section{Introduction} In this paper we study some questions related to the existence of solutions for the nonlocal elliptic problem \begin{equation}\label{eq:1.1} \begin{gathered} -M(\|u\|^2)\Delta u = f \quad \mbox{in }\Omega ,\\ u =0 \quad \mbox{on } \partial\Omega , \end{gathered} \end{equation} where $\Omega\subset \mathbb{R}^{N}$ is a bounded smooth domain, $f\in H^{-1}(\Omega )$ and $M:\mathbb{R}\to\mathbb{R}$ is a function whose behavior will be stated later, and the norm in $H_{0}^{1}(\Omega )$ is $\|u\|^2=\int_{\Omega}|\nabla u|^2$. The main purpose of this work is establishing properties on $M$ under which problem \eqref{eq:1.1}, and its nonlinear counterpart, possesses a solution. This equation has called our attention because the operator $$ Lu:=M(\|u\|^2)\Delta u $$ contains the nonlocal term $M(\|u\|^2)$ which poses some interesting mathematical questions. Also the operator $L$ appears in the Kirchhoff equation, which arises in nonlinear vibrations, namely \begin{gather*} u_{tt}-M\big(\int_{\Omega}|\nabla u|^2dx\big)\Delta u = f(x,u) \quad \mbox{in } \Omega \times (0,T),\\ u=0 \quad\mbox{on }\partial\Omega \times (0,T),\\ u(x,0)=u_{0}(x), \quad u_{t}(x,0)=u_{1}(x). \end{gather*} For more details on physical motivation of this problem the interested reader is invited to consult Eisley, Limaco-Medeiros \cite{eisley,limaco} and the references therein. In a previous paper Alves-Corr\^ea \cite{alves} focused their attention on problem \eqref{eq:1.1} in case $M(t)\geq m_{0}>0$, for all $t\geq 0$, where $m_{0}$ is a constant. Among other things they studied the above $M$-linear problem \eqref{eq:1.1} where $M$, besides the strict positivity mentioned before, satisfies the following assumption: \begin{quote} The function $H:\mathbb{R}\to \mathbb{R}$ with $$H(t)=M(t^2)t$$ is monotone and $H(\mathbb{R})=\mathbb{R}$. \end{quote} The above authors also studied the sublinear problem \begin{equation}\label{eq:1.2}% equation 1.2 \begin{gathered} -M(\|u\|^2)\Delta u = u^{\alpha}\quad \mbox{in }\Omega ,\\ u=0 \quad \mbox{on } \partial\Omega ,\\ u>0 \quad \mbox{in } \Omega \,, \end{gathered} \end{equation} where $0<\alpha <1$, $M$ is a non-increasing continuous function, $H$ is increasing, $H(\mathbb{R})=\mathbb{R}$ and $$ G(t)=t[M(t^2)]^{2/(1-\alpha)} $$ is injective. Under these assumptions it is proved that \eqref{eq:1.2} possesses a unique solution. A straightforward computation shows that the function $M(t)=\exp (-t)+C$, with $C$ a positive constant, satisfies the above assumptions. In the present paper we prove similar results by allowing $M$ to attain negative values and $M(t)\geq m_{0}>0$ only for $t$ large enough. This is possible thanks to a device explored by Alves-de Figueiredo \cite{alves-fig}, who use Galerkin method to attack a non-variational elliptic system. The technique can be conveniently adapted to problems such as \eqref{eq:1.1} and \eqref{eq:1.2}. In this way we improve substantially the existence result on the above problems mainly because our assumptions on $M$ are weakened. Indeed, we may also consider the case in which $M$ possesses a singularity. The method we use rests heavily on the following result whose proof may be found in Lions \cite[p.53]{lions}, and it is a well known variant of Brouwer's Fixed Point Theorem. \begin{prop} \label{prop1.1} Suppose that $F:\mathbb{R}^m \to \mathbb{R}^m $ is a continuous function such that $\langle F(\xi ),\xi \rangle \geq 0$ on $|\xi|=r$, where $\langle \cdot , \cdot \rangle$ is the usual inner product in $\mathbb{R}^m $ and $| \cdot |$ its related norm. Then, there exists $z_{0}\in \overline{B_{r}}(0)$ such that $F(z_{0})=0$. \end{prop} We recall that by a solution of \eqref{eq:1.1} we mean a weak solution, that is, a function $u\in H_{0}^{1}(\Omega )$ such that $$ M(\|u\|^2)\int_{\Omega}\nabla u\cdot\nabla\varphi =\int_{\Omega}f(x,u)\varphi ,\quad \mbox{for all } \varphi\in H_{0}^{1}(\Omega ). $$ We point out that, depending on the regularity of $f(\cdot ,u)$, a bootstrap argument may be used to show that a weak solution is a classical solution, i.e., a function in $C_{0}^2(\overline{\Omega})$. This happens, for instance, with the solution obtained in Theorem \ref{thm4.1}. This paper is organized as follows: Section 2 is devoted to the study of the $M$-linear problem in the continuous case. In Section 3 the $M$-linear is studied in case $M$ possesses a discontinuity. In Section 4 we focus our attention on the sublinear problem. In Section 5 we analyze another type of nonlocal problem. \section{The $M$-linear Problem: Continuous Case}% section 2 In this section we are concerned with the $M$-linear problem \eqref{eq:1.1} where $f\in H^{-1}(\Omega )$ and $M:\mathbb{R}\to \mathbb{R}$ is a continuous function satisfying \begin{itemize} \item[(M1)] There are positive numbers $t_{\infty}$ and $m_{0}$ such that $M(t)\geq m_{0}$, for all $t\geq t_{\infty}$. \end{itemize} \begin{theorem} \label{thm2.1} Under assumption (M1), for each $0\neq f\in H^{-1}(\Omega )$ problem \eqref{eq:1.1} possesses a weak solution. \end{theorem} \begin{proof} Inspired by Alves-de Figueiredo \cite{alves-fig} we use the Galerkin Method. First let us take $M^{+}=\max\{M(t),0\}$, the positive part of $M$, and consider the auxiliary problem \begin{equation}\label{eq:2.1}% equation 2.1 \begin{gathered} -M^{+}(\|u\|^2)\Delta u=f \quad \mbox{in }\Omega ,\\ u=0 \quad \mbox{on }\partial\Omega . \end{gathered} \end{equation} We will prove that problem \eqref{eq:2.1} possesses solution and such a solution also solves problem \eqref{eq:1.1}. We point out that $M^{+}$ also satisfies assumption (M1). We are ready to apply the Galerkin Method by using Proposition \ref{prop1.1}. Let $\sum =\{e_{1},\ldots ,e_{m},\ldots \}$ be an orthonormal basis of the Hilbert space $H_{0}^{1}(\Omega )$. For each $m\in \mathbb{N}$ consider the finite dimensional Hilbert space $$ \mathbb{V}_{m}=\mathop{\rm span}\{e_{1},\ldots ,e_{m}\}. $$ Since $(\mathbb{V}_{m},\|\cdot\|)$ and $(\mathbb{R}^m ,|\cdot |)$ are isometric and isomorphic, where $\|\cdot\|$ is the usual norm in $H^{1}_{0}(\Omega )$ and $|\cdot |$ is the Euclidian norm in $\mathbb{R}^m $, $\langle \cdot,\cdot\rangle$ its corresponding inner product, we make, with no additional comment, the identification $$ u=\sum_{j=1}^m \xi_{j}e_{j}\longleftrightarrow \xi =(\xi_{1},\ldots ,\xi_{m}),\quad \|u\|=|\xi |. $$ We will show that for each $m$ there is $u_{m}\in \mathbb{V}_{m}$, an approximate solution of \eqref{eq:2.1}, satisfying $$ M^{+}(\|u_{m}\|^2)\int_{\Omega}\nabla u_{m}\cdot\nabla e_{i}=\langle\langle f,e_{i}\rangle\rangle,\quad i=1,\ldots, m. $$ where $\langle\langle \; , \;\rangle\rangle$ is the duality pairing between $H^{-1}(\Omega )$ and $H^{1}_{0}(\Omega )$. First we consider the function $F:\mathbb{R}^m \to \mathbb{R}^m $ given by \begin{gather*} F(\xi )=(F_{1}(\xi ),\ldots ,F_{m}(\xi )), \\ F_{i}(\xi )=M^{+}(\|u\|)\int_{\Omega}\nabla u\cdot \nabla e_{i}-\langle\langle f,e_{i}\rangle\rangle, \end{gather*} where $i=1,\ldots,m$ and $u=\sum_{j=1}^m \xi_{j}e_{j}$. So that $$ F_{i}(\xi )=M^{+}(\|u\|^2)\xi_{i}-\langle\langle f,e_{i}\rangle\rangle. $$ With the above identifications one has $$ \langle F(\xi ),\xi \rangle=M^{+}(\|u\|^2)\|u\|^2-\langle\langle f,u\rangle\rangle. $$ Using (M1), H$\ddot{o}$lder and Poincar\'e inequalities we get $$ \langle F(\xi ),\xi \rangle \geq m_{0}\|u\|^2-C\|f\|_{H^{-1}}\|u\|\geq 0, $$ if $\|u\|=r$, for $r$ large enough, where $\|f\|_{H^{-1}}$ is the norm of the linear form $f$. Thus, because of Proposition \ref{prop1.1}, there is $u_{m}\in \mathbb{V}_{m}$, $\|u_{m}\|\leq r$, $r$ does not depend on $m$, such that $$ M^{+}(\|u_{m}\|^2)\int_{\Omega}\nabla u_{m}\cdot\nabla e_{i}=\langle\langle f,e_{i}\rangle\rangle,\quad i=1,\ldots ,m, $$ which implies that \begin{equation}\label{eq:2.2} M^{+}(\|u_{m}\|^2)\int_{\Omega}\nabla u_{m}\cdot \nabla\omega =\langle\langle f,\omega\rangle\rangle , \quad \text{for all }\omega\in\mathbb{V}_{m}. \end{equation} Because $(\|u_{m}\|^2)$ is a bounded real sequence and $M^{+}$ is continuous one has $$ \|u_{m}\|^2\to \tilde{t}_{0}, $$ for some $\tilde{t}_{0}\geq 0$, and $$ u_{m}\rightharpoonup u \text{ in } H_{0}^{1}(\Omega),\quad u_{m}\to u\text{ in }L^2(\Omega),\quad M^{+}(\|u_{m}\|^2)\to M^{+}(\tilde{t}_{0}), $$ perhaps for a subsequence. Take $k\leq m$, $\mathbb{V}_{k}\subset \mathbb{V}_{m}$. Fix $k$ and let $m\to \infty$ in equation \eqref{eq:2.2} to obtain $$ M^{+}(\tilde{t}_{0})\int_{\Omega}\nabla u\cdot \nabla \omega =\langle\langle f,\omega \rangle\rangle,\quad \text{for all } \omega\in\mathbb{V}_{k}. $$ Since $k$ is arbitrary we will have that the last equality remains true for all $\omega \in H_{0}^{1}(\Omega )$. If $M^{+}(\tilde{t}_{0})=0$ we would have $\langle\langle f,\omega \rangle\rangle=0$ for all $\omega\in H_{0}^{1}$ and so $f=0$ in $H^{-1}(\Omega )$ which is a contradiction. Consequently $M^{+}(\tilde{t}_{0})>0$ and so $M(\tilde{t}_{0})=M^{+}(\tilde{t}_{0})$. We now take $\omega =u_{m}$ in \eqref{eq:2.2} to obtain $$ M(\|u_{m}\|^2)\|u_{m}\|^2=\langle\langle f,u_{m}\rangle\rangle $$ and so $M(\tilde{t}_{0})\tilde{t}_{0}=\langle\langle f,u\rangle\rangle$. From this equality and $$ M(\tilde{t}_{0})\|u\|^2=\langle\langle f,u\rangle\rangle $$ we have $\|u\|^2=\tilde{t}_{0}$ which shows that the function $u$ is a weak solution of problem \eqref{eq:1.1} and the proof of Theorem \ref{thm2.1} is complete. \end{proof} \begin{remark} \label{rmk2.1} \rm It follows from the proof of Theorem \ref{thm2.1} that the solution $u$ obtained there satisfies $M(\|u\|^2)>0$ (of course, if we had used another device in order to obtain a solution of \eqref{eq:1.1} such a property might not be true). \end{remark} We claim that there is only one solution to \eqref{eq:1.1} satisfying this property. This may be proved as follows. Let $u$ and $v$ be solutions of \eqref{eq:1.1} obtained as before. Since $u$ and $v$ are weak solutions of \eqref{eq:1.1} one has $$ M(\|u\|^2)\int_{\Omega}\nabla u\cdot \nabla \omega =M(\|v\|^2)\int_{\Omega}\nabla v\cdot \nabla\omega ,\quad \text{for all } \omega\in H_{0}^{1}(\Omega )\,. $$ Hence $M(\|u\|^2)u$ and $M(\|v\|^2)v$ are both solutions of the problem \begin{gather*} -\Delta U = f \quad \mbox{in }\Omega ,\\ U=0 \quad \mbox{on } \partial\Omega . \end{gather*} By the uniqueness one has $M(\|u\|^2)u=M(\|v\|^2)v$ in $\Omega$ and so $M(\|u\|^2)\|u\|=M(\|v\|^2)\|v\|$. Supposing that the function $t\to M(t^2)t$ is increasing for $t>0$ one obtains that $\|u\|=\|v\|$. Consequently \begin{gather*} -\Delta u = -\Delta v\quad \mbox{in }\Omega ,\\ u=v \quad \mbox{on } \partial\Omega . \end{gather*} and then $u=v$ in $\Omega$. Hence, we have proved that problem \eqref{eq:1.1} possesses only one solution $u$ if $t\to M(t^2)t$ is increasing for $t>0$. \begin{remark} \label{rmk2.2} \rm If $M(t_{0})=0$ for some $t_{0}>0$ and $f=0$ in $H^{-1}(\Omega )$ then we lose uniqueness. In fact, let $u\neq 0$ be a function in $C_{0}^2(\overline{\Omega})$ and set $v=\sqrt{t_{0}}\,u/\|u\|$. In this case $\|v\|^2=t_{0}$ and so \begin{equation}\label{eq:2.3}% equation 2.3 \begin{gathered} -M(\|v\|^2)\Delta v=0 \quad \mbox{in }\Omega ,\\ v=0 \quad \mbox{on }\partial\Omega , \end{gathered} \end{equation} that is, for each nonzero function $u\in C_{0}^2(\overline{\Omega})$ the function $v$ defined above is a nontrivial solution of (\ref{eq:2.3}). \end{remark} \begin{remark}[A Dual Problem] \label{rmk2.3} \rm Suppose that $M:\mathbb{R}\to \mathbb{R}$ is a continuous function satisfying \begin{itemize} \item[(\~M1)] There are positive numbers $\tilde{t}_{\infty}$ and $ \tilde{m}_{0}$ such that $M(t)\leq -\widetilde{m}_{0}$ for all $t\geq\tilde{t}_{\infty}$. \end{itemize} In this case \eqref{eq:1.1} possesses a solution. Indeed, suppose $f\neq 0$ in $H^{-1}(\Omega )$ and consider the problem \begin{equation}\label{eq:2.4} \begin{gathered} -\widetilde{M}(\|u\|^2)\Delta u = f \quad \mbox{in }\Omega ,\\ u=0 \quad\mbox{on } \partial\Omega , \end{gathered} \end{equation} where $\widetilde{M}(t)=-M(t)$. Clearly $\widetilde{M}$ satisfies (M1) and so problem (\ref{eq:2.4}) possesses a solution $v\in H_{0}^{1}(\Omega )$ with $\widetilde{M}(\|v\|^2)>0$. Hence $u=-v$ is a solution of \eqref{eq:1.1} with $M(\|u\|^2)<0$. \end{remark} \section{The $M$-linear Problem: A Discontinuous Case}%section 3 In this section we concentrate our interest on problem \eqref{eq:1.1} when $M$ possesses a discontinuity. More precisely, we study problem \eqref{eq:1.1} with $M:\mathbb{R}/\{\theta \}\to \mathbb{R}$ continuous such that \begin{itemize} \item[(M2)] $\lim_{t\to \theta^{+}}M(t)=\lim_{t\to \theta^{-}}M(t)=+\infty$ \item[(M3)] $\limsup_{t\to +\infty}M(t^2)t=+\infty$ and (M1) is satisfied for some $t_{\infty}>\theta$. \end{itemize} \begin{theorem} \label{thm3.1} If $M$ satisfies (M1)--(M3) problem \eqref{eq:1.1} possesses a solution $u\in H_{0}^{1}(\Omega )$, for each $0\neq f\in H^{-1}(\Omega )$. \end{theorem} \begin{proof} We first consider the sequence of functions $M_{n}:\mathbb{R}\to \mathbb{R}$ given by $$ M_{n}(t)=\begin{cases} n, & \theta -\epsilon'_{n}\leq t\leq \theta +\epsilon''_{n},\\ M(t), & t\leq \theta -\epsilon'_{n} \mbox{ or } t\geq \theta +\epsilon''_{n}, \end{cases} $$ for $n>m_{0}$, where $\theta -\epsilon'_{n}$ and $\theta +\epsilon''_{n}$, $\epsilon'_{n}$, $\epsilon''_{n}>0$, are, respectively, the points closest to $\theta$, at left and at right, so that $$ M(\theta -\epsilon'_{n})=M(\theta +\epsilon''_{n})=n. $$ We point out that, in this case, $\epsilon'_{n},\epsilon''_{n}\to 0$ as $n\to \infty$. Take $n>m_{0}$ and observe that the horizontal lines $y=n$ cross the graph of $M$. Hence $M_{n}$ is continuous and satisfies (M1), for each $n>m_{0}$. In view of this, for each $n$ like above, there is $u_{n}\in H_{0}^{1}(\Omega )$ satisfying $$ M_{n}(\|u_{n}\|^2)\int\nabla u_{n}\cdot \nabla \omega =\langle\langle f, \omega \rangle\rangle , \quad \text{for all } \omega \in H_{0}^{1}(\Omega ). $$ Taking $\omega =u_{n}$ in the above equation one has $$ M_{n}(\|u_{n}\|^2)\|u_{n}\|^2=\langle\langle f,u_{n}\rangle\rangle , $$ and so $$ M_{n}(\|u_{n}\|^2)\|u_{n}\|\leq \|f\|_{H^{-1}} $$ Because of (M3) the sequence $(\|u_{n}\|)$ must be bounded. Hence \begin{gather*} u_{n}\rightharpoonup u \quad \text{in } H_{0}^{1}(\Omega ),\\ u_{n}\to u \quad \text{in } L^2(\Omega ), \\ \|u_{n}\|^2\to \theta_{0},\quad \text{for some } \theta_{0}, \end{gather*} perhaps for subsequences. If $M_{n}(\|u_{n}\|^2)\to 0$, then $\langle\langle f, \omega \rangle\rangle =0$, for all $\omega \in H_{0}^{1}(\Omega )$, which is impossible because $0\neq f\in H^{-1}(\Omega )$. Thus if $(M_{n}(\|u_{n}\|^2)$ converges its limit is different of zero. Suppose that $\|u_{n}\|^2\to \theta$. If $\|u_{n}\|^2>\theta +\epsilon''_{n}$ or $\|u_{n}\|^2<\theta -\epsilon'_{n}$, for infinitely many $n$, we would get $M_{n}(\|u_{n}\|^2)=M(\|u_{n}\|^2)$, for such $n$, and so $$ M(\|u_{n}\|^2)\|u_{n}\|^2=\langle\langle f, u_{n}\rangle\rangle \;\Rightarrow\; +\infty =\langle\langle f,u \rangle\rangle $$ which is a contradiction. On the other hand, if there are infinitely many $n$ so that $\theta -\epsilon'_{n}\leq \|u_{n}\|^2\leq \theta +\epsilon''_{n}\; \Rightarrow \;M_{n}(\|u_{n}\|^2)=n$ and so $n\|u_{n}\|^2=\langle\langle f,u_{n}\rangle\rangle \; \Rightarrow \; \infty =\langle\langle f, u \rangle\rangle$ and we arrive again in a contradiction. Consequently $\|u_{n}\|^2\to \theta_{0}\neq\theta $ which implies that for $n$ large enough $$ \|u_{n}\|^2<\theta -\epsilon'_{n} \quad \text{or} \quad \|u_{n}\|^2>\theta +\epsilon''_{n} $$ and so $M_{n}(\|u_{n}\|^2)=M(\|u_{n}\|^2)$ which yields $$ M(\|u_{n}\|^2)\int_{\Omega}\nabla u_{n}\cdot\nabla \omega =\langle\langle f, \omega \rangle\rangle , \quad \forall \omega \in H_{0}^{1}(\Omega ). $$ Consequently $ M(\theta_{0} )\int_{\Omega}\nabla u\cdot \nabla \omega =\langle\langle f, \omega \rangle\rangle, \; \text{for all} \; \omega \in H_{0}^{1}(\Omega )$ which implies $M(\|u_{n}\|^2)\|u_{n}\|^2=\langle\langle f,u_{n}\rangle\rangle $ and taking limits $$ M(\theta_{0} )\theta_{0} =\langle\langle f, u\rangle\rangle $$ Hence $M(\theta_{0} )\|u\|^2=M(\theta_{0} )\theta_{0}$. Reasoning as before we conclude that $M(\theta_{0} )\neq 0$ and so $\|u\|^2= \theta_{0}$ and the proof of the theorem is complete. \end{proof} \section{A Sublinear Problem}% section 4 In this section we focus our attention on problem \eqref{eq:1.2}. More precisely, we have the following result: \begin{theorem} \label{thm4.1} If $M$ satisfies assumption (M1), $M(t)\leq m_{\infty}$, for some positive constant $m_{\infty}$ and all $t\geq 0$, and $\lim_{t\to\infty} M(t^2)t^{1-\alpha}=+\infty$, then problem \eqref{eq:1.2} possesses a solution. \end{theorem} Since the proof of this theorem is quite similar to the one in Alves-de Figueiredo \cite{alves-fig} we omit it and make only some remarks giving some directions on how to proceed. First of all we have to consider the problem \begin{equation}\label{eq:4.1} \begin{gathered} -M(\|u\|^2)\Delta u = (u^{+})^{\alpha}+\lambda\phi (x) \quad\mbox{in } \Omega ,\\ u=0 \quad \mbox{on }\partial\Omega ,\\ u>0 \quad\mbox{in } \Omega , \end{gathered} \end{equation} where $\lambda >0$ is a parameter, $\phi >0$ is a function in $H_{0}^{1}(\Omega )$, and $u^{+}=\max \{u,0\}$ is the positive part of $u$. Proceeding as in the proof of Theorem \ref{thm2.1} we found, for each $\lambda \in (0,\tilde{\lambda })$, a solution $u_{\lambda}$ of equation (\ref{eq:4.1}) and, in view of $M(\|u_{\lambda }\|^2)>0$- we can prove that $u_{\lambda}\geq 0$-, using the maximum principle to conclude that $u_{\lambda}>0$. Hence \begin{equation}\label{eq:4.2} % equation 4.2 \begin{gathered} -M(\|u_{\lambda }\|^2)\Delta u_{\lambda } = (u_{\lambda})^{\alpha}+\lambda\phi (x)\geq u_{\lambda }^{\alpha} \quad\mbox{in } \Omega ,\\ u_{\lambda }=0 \quad \mbox{on }\partial\Omega ,\\ u_{\lambda }>0 \quad \mbox{in } \Omega , \end{gathered} \end{equation} which implies \begin{gather*} -\Delta u_{\lambda }\geq m_{\infty}^{-1}u_{\lambda}^{\alpha}\quad \mbox{in } \Omega ,\\ u_{\lambda }=0\quad \mbox{on }\partial\Omega . \end{gather*} Thanks to a result by Ambrosetti-Br\'ezis-Cerami \cite{abc}, one has $$ u_{\lambda }\geq m_{\infty}^{-1}\omega_{1} , $$ where $\omega_{1}>0$ in $\Omega$ is the only positive solution of \begin{gather*} -\Delta \omega_{1}=\omega_{1}^{\alpha}\quad \mbox{in }\Omega ,\\ \omega_{1}=0 \quad \mbox{on }\partial\Omega . \end{gather*} As in the proof of Theorem \ref{thm2.1} one has that $\|u_{\lambda }\|\leq r_{\lambda}$ where $r_{\lambda}$ is a positive constant that depends on $\lambda$. Let us consider $\lambda\in (0,\overline{\lambda })$ and make $\lambda\to 0^{+}$. For we have to guarantee that $(\|u_{\lambda}\|)$ is bounded for all $\lambda\in (0,\overline{\lambda })$. First observe that $$ M(\|u_{\lambda}\|^2)\|u_{\lambda}\|^2=\int_{\Omega}u_{\lambda}^{\alpha +1}+\lambda \int_{\Omega}\phi u_{\lambda} $$ Because $0<\alpha <1$ and using some standard arguments we have $$ M(\|u_{\lambda})\|^2\|u_{\lambda}\|^{1-\alpha}\leq C+\frac{C}{\|u_{\lambda}\|} $$ Since $M(t^2)t^{1-\alpha}\to +\infty$ as $t\to \infty$ we have that $(\|u_{\lambda}\|)$ is bounded for all $\lambda\in (0,\overline{\lambda})$. Finally, we may take $\lambda\to 0$ to obtain a solution $u$ of problem \eqref{eq:1.2}. \section{Another Nonlocal Problem}% section 5 Next, we make some remarks on a nonlocal problem which is a slight generalization of one studied by Chipot-Lovat \cite{chilo} and Chipot-Rodrigues \cite{chiro}. More precisely, the above authors studied the problem \begin{equation}\label{eq:5.1}% equation 5.1 \begin{gathered} -a\big(\int_{\Omega}u\big)\Delta u = f \quad \mbox{in }\Omega ,\\ u=0 \quad \mbox{on } \partial\Omega , \end{gathered} \end{equation} where $\Omega\subset \mathbb{R}^{N}$ is a bounded domain, $N\geq 1$, and $a:\mathbb{R}\to (0,+\infty )$ is a given function. Equation (\ref{eq:5.1}) is the stationary version of the parabolic problem \begin{gather*} u_{t}-a\big(\int_{\Omega} u(x,t)dx\big)\Delta u = f \quad \mbox{in } \Omega \times (0,T),\\ u=0 \quad \mbox{on } \partial\Omega \times (0,T),\\ u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x). \end{gather*} Here $T$ is some arbitrary time and $u$ represents, for instance, the density of a population subject to spreading. See \cite{chilo,chiro} for more details. In particular, \cite{chilo} studies problem (\ref{eq:5.1}), with $f\in H^{-1}(\Omega )$, and proves the following result. \begin{prop} \label{prop5.1} Let $a:\mathbb{R}\to (0,+\infty )$ be a positive function, $f\in H^{-1}(\Omega )$. Then problem (\ref{eq:5.1}) has as many solutions $\mu$ as the equation $$ a(\mu )\mu =\langle\langle f,\varphi \rangle\rangle , $$ where $\varphi$ is the function(unique) satisfying \begin{gather*} -\Delta \varphi = 1\quad \mbox{in }\Omega ,\\ \varphi =0 \quad \mbox{on } \partial\Omega , \end{gather*} \end{prop} Now, we study the nonlocal problem \begin{equation}\label{eq:5.2} \begin{gathered} -a\big(\int_{\Omega}|u|^{q}\big)\Delta u = f\quad \mbox{in }\Omega ,\\ u=0 \quad\mbox{on } \partial\Omega , \end{gathered} \end{equation} where $\Omega$ and $f$ are as before and $1=a(\|u\|_{q}^{q})\|u\|^2-\langle\langle f,u\rangle\rangle $$ We have to show that there is $r>0$ so that $\langle F(\xi ), \xi \rangle\geq 0$, for all $|\xi |=r$ in $\mathbb{V}_{m}$. Suppose, on the contrary, that for each $r>0$ there is $u_{r}\in \mathbb{V}_{m}$ such that $\|u_{r}\|=r$ and $$ \langle F(\xi_{r}),\xi_{r}\rangle <0, \quad \xi_{r}\leftrightarrow u_{r}. $$ Taking $r=n\in \mathbb{N}$ we obtain a sequence $(u_{n})$, $\|u_{n}\|=n$, $u_{n}\in \mathbb{V}_{m}$ and $$ \langle F(u_{n}),u_{n}\rangle =a(\|u_{n}\|_{q}^{q})\|u_{n}\|^2-\langle\langle f,u_{n}\rangle\rangle <0 $$ and so $$ a(\|u_{n}\|_{q}^{q})\|u_{n}\|0$ such that $\langle F(\xi ), \xi \rangle \geq 0,$ for all $|\xi |=r_{m}$. In view of Proposition \ref{prop1.1} there is $u_{m}\in \mathbb{V}_{m}$, $\|u_{m}\|\leq r_{m}$ such that $F_{i}(u_{m})=0, i=1,\ldots ,m$, that is, \begin{equation}\label{eq:5.3} a\left(\|u_{m}\|_{q}^{q}\right)\int_{\Omega}\nabla u_{m}\cdot\nabla\omega =\langle\langle f,\omega\rangle\rangle ,\quad \forall\omega\in\mathbb{V}_{m}. \end{equation} Reasoning as before, by using the facts that $t\to a(t)$ is decreasing for $t\geq 0$ and $\lim_{t\to +\infty}a(t^{q})t=+\infty$, we conclude that $\|u_{m}\|\leq C,\forall m=1,2\ldots $ for some constant $C$ that does not depend on $m$. Hence, $u_{m}\rightharpoonup u$ in $H_{0}^{1}(\Omega )$, $u_{m}\to u$ in $L^{q}(\Omega ), \; 10 \quad\mbox{in } \Omega. \end{gathered} \end{equation} where $0<\alpha <1$, and $a$ satisfies the assumptions in Theorem \ref{thm5.1} possesses a solution. \end{remark} \begin{thebibliography}{99} \bibitem{alves} C. O. Alves \& F. J. S. A. 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