\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 144, pp. 1--8.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/144\hfil Nonlinear boundary conditions] {Nonlinear boundary conditions for elliptic equations} \author[P. Amster, M. C. Mariani, O. Mendez\hfil EJDE-2005/144\hfilneg] {Pablo Amster, Mar\'\i a Cristina Mariani, Osvaldo M\'endez} % in alphabetical order \address{Pablo Amster \hfill\break Departamento de Matem\'atica, Facultad de Ciencias Exactas y Naturales \\ Universidad de Buenos Aires \\ Ciudad Universitaria, Pabell\'on I\\ (1428) Buenos Aires, Argentina\\ Consejo Nacional de Investigaciones Cient\'\i ficas y T\'ecnicas (CONICET)} \email{pamster@dm.uba.ar} \address{Mar\'\i a Cristina Mariani \hfill\break Department of Mathematical Sciences\\ New Mexico State University\\ Las Cruces, NM 88003-8001, USA} \email{mmariani@nmsu.edu} \address{Osvaldo M\'endez \hfill\break Department of Mathematics\\ 500W University Ave., 124 Bell Hall\\ The University of Texas at El Paso\\ El Paso, TX 79968, USA} \email{mendez@math.utep.edu} \date{} \thanks{Submitted November 15, 2004. Published December 8, 2005.} \subjclass[2000]{35J25, 35J60} \keywords{Nonlinear elliptic problems; nonlinear boundary conditions; \hfill\break\indent topological methods} \begin{abstract} This work is devoted to the study of the elliptic equation $\Delta u = f(x,u)$ in a bounded domain $\Omega\subset \mathbb{R}^n$ with a nonlinear boundary condition. We obtain various existence results applying coincidence degree theory and the method of upper and lower solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} %%Delete [theorem] to re-start numbering \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} In this paper we study the problem \begin{equation}\label{1} \Delta u=f(x,u) \quad \mbox{in }\Omega \end{equation} subject to the nonlinear boundary condition \begin{equation}\label{nbc2} \frac {\partial u}{\partial \nu} = g(x,u) \quad \mbox{on } \partial\Omega, \end{equation} where $f,g:\overline\Omega\times \mathbb{R}^n\to \mathbb{R}$ are continuous functions and $\Omega\subset \mathbb{R}^n$ is a bounded $C^{1,1}$ domain. There is a vast body of literature concerning nonlinear problems with nonlinear boundary conditions, see e.g. \cite{LM} for a survey. In section \ref{uls} we obtain solutions of \eqref{1}--\eqref{nbc2} by the method of upper and lower solutions. The proof relies in the associated maximum principle and the unique solvability of the linear Robin problem \begin{equation} \label{robin} \begin{gathered} \Delta u - \lambda u= \varphi \quad \mbox{in }\Omega, \\ \frac{\partial u}{\partial \nu} +\mu u = \xi \quad \mbox{on }\partial\Omega \end{gathered} \end{equation} for $\lambda,\mu >0$ (see \cite{LaMe} and its references). The method of super and subsolutions has been extensively used in nonlinear analysis, both for ODE's and PDE's problems. In particular, for elliptic problems with nonlinear boundary conditions, this method has been applied to obtain more general existence results for example in \cite{I}, \cite{MS1}-\cite{MS2}. However, the method presented here allows to relax the assumptions: firstly, less regularity is required for the domain $\Omega$; secondly, only continuity is assumed for the nonlinearities $f$ and $g$. More precisely, we prove Theorem \ref{upper-lower} below. \begin{theorem}\label{upper-lower} Assume there exist $\alpha\le \beta$ such that $$ \Delta \alpha \ge f(x,\alpha)\quad \mbox{in }\Omega,\quad \frac {\partial \alpha}{\partial \nu} \le g(x,\alpha) \quad \mbox{on }\partial\Omega$$ and $$\Delta \beta \le f(x,\beta) \quad \mbox{in }\Omega,\quad \frac {\partial \beta}{\partial \nu} \ge g(x,\beta) \quad \mbox{on }\partial\Omega.$$ Then \eqref{1}--\eqref{nbc2} admits at least one solution $u$, with $\alpha \le u \le \beta$. \end{theorem} In section \ref{ll} we study problem \eqref{1}--\eqref{nbc2} for bounded $f$ and $g$ applying coincidence degree theory. We obtain an existence result under Landesman-Lazer type conditions (\cite{ll}, \cite{LandLazer}. For further applications to problems of resonant type see \cite{ma2}). More precisely, we have \begin{theorem}\label{llteo} Assume that $f$ and $g$ are bounded, and define \begin{gather*} \limsup_{t\to \pm \infty} f(x,t) := f_{s}^\pm(x),\quad \liminf_{t\to \pm \infty} f(x,t) := f_{i}^\pm(x) \\ \limsup_{t\to \pm \infty} g(x,t) := g_{s}^\pm(x), \quad \liminf_{t\to \pm \infty} g(x,t) := g_{i}^\pm(x). \end{gather*} Then \eqref{1}--\eqref{nbc2} admits at least one solution, provided that one of the following assumptions holds: \begin{gather}\label{ll1} \int_{\partial\Omega} g_{i}^+ > \int_\Omega f_{s}^+ \quad\mbox{and} \quad \int_{\partial\Omega} g_{s}^- < \int_\Omega f_{i}^- \\ \label{ll2} \int_{\partial\Omega} g_{i}^- > \quad \int_\Omega f_{s}^- \quad\mbox{and}\quad \int_{\partial\Omega} g_{s}^+ < \int_\Omega f_{i}^+. \end{gather} \end{theorem} \section{Upper and lower solutions} \label{uls} In this section we present a proof of Theorem \ref{upper-lower}. First we recall the following classical result. \begin{lemma}\label{rob} Let $\lambda,\mu >0$ and $\varphi\in C(\overline\Omega)$, $\xi\in C(\partial\Omega)$. Then the Robin problem (\ref{robin}) admits a unique solution $u$. Furthermore, the operator $T:C(\overline\Omega)\times C(\partial\Omega) \to C(\overline\Omega)$ given by $T(\varphi,\xi) = u$ is compact. \end{lemma} \begin{remark} \label{cones} \rm If $\lambda =0$ it is possible to extend Lemma \ref{rob} (except the last statement) to Lipschitz domains, in particular domains with corners or edges, considering $\varphi$ in a suitable Sobolev space and $\xi$ in the corresponding trace space. In this case it is also possible to replace $\mu$ with a function $g\in L^{n-1}(\Omega)$, see \cite{LaMe}. \end{remark} Moreover, we shall use a maximum principle associated for the problem \begin{lemma} \label{max} Let $\lambda>0$, $\mu\ge 0$, and assume that $w$ satisfies \begin{gather*} \Delta w -\lambda w \ge 0\quad \mbox{in } U,\\ \frac{\partial w}{\partial \nu} + \mu w\le 0, \quad \mbox{on } \Gamma_1,\\ w\le 0 \quad \mbox{on } \Gamma_2. \end{gather*} where $U\subset \mathbb{R}^n$ is a bounded domain with boundary $\partial U = \Gamma_1 \cup \Gamma_2$. Then $w\le 0$ in $U$. \end{lemma} \begin{proof} Let $w^+= \max\{ w,0\}$, and $U^+= \{x\in U: w(x)>0\}$. From our assumptions we have $$ 0\le \int_U (\Delta w-\lambda w) w^+ = -\int_{U^+} |\nabla w|^2 - \lambda \int_{U^+} w^2 + \int_{\partial U}w^+\frac{\partial w}{\partial \nu}$$ Moreover, since $\frac{\partial w}{\partial \nu} \le -\mu w$ on $\Gamma_1$, we deduce that $$ \int_{\partial U}w^+\frac{\partial w}{\partial \nu} = \int_{\Gamma_1}w^+\frac{\partial w}{\partial \nu} \le -\mu \int_{\Gamma_1}w^+ w \le 0. $$ Hence $|U^+|=0$, and the proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{upper-lower}] Set two positive constants $\lambda,\mu$, and define the function $P:\overline\Omega\times \mathbb{R}\to \mathbb{R}$ by $$ P(x,u) = \begin{cases} \alpha(x) & \hbox{if } u< \alpha(x) \\ u & \hbox{if } \alpha(x)\le u\le\beta(x) \\ \beta(x) & \hbox{if } u > \beta(x). \end{cases} $$ Next, consider the compact fixed point operator $T:C(\overline\Omega)\to C(\overline\Omega)$ given by $Tv = u$, where $u$ is the unique solution of the Robin problem \begin{gather*} \Delta u - \lambda u =f(x,P(x,v)) - \lambda P(x,v) \quad \mbox{in }\Omega \\ \frac{\partial u}{\partial \nu} +\mu u = g(x,P(x,v)) +\mu P(x,v) \quad \mbox{on }\partial\Omega. \end{gather*} Using Schauder Theorem, it is straightforward to prove that $T$ has a fixed point $u$. We claim that $\alpha\le u\le \beta$, and hence $u$ is a solution the problem. Indeed, let $U=\{x\in \Omega: u(x)>\beta(x)\}$. For $x\in U$, we have $$ \Delta u(x) - \lambda u(x) = f(x,\beta(x)) - \lambda \beta(x)\ge \Delta \beta(x) - \lambda \beta(x). $$ Moreover, if $x\in\partial U\cap \partial\Omega:=\Gamma_1$, then $$\frac{\partial u}{\partial \nu}(x) +\mu u(x) = g(x,\beta(x)) +\mu \beta(x) \le \frac{\partial \beta}{\partial \nu}(x) +\mu \beta(x).$$ Thus, if $w=u-\beta$ we deduce from Lemma \ref{max} that $w\le 0$ in $U$, and hence $U$ is empty. In the same way we show that $u\ge \alpha$, and the proof is complete. \end{proof} \begin{example} \label{ex2.4} \rm In particular, when $\alpha\equiv R^-$ and $\beta \equiv R^+$ for some constants $R^-0$ the problem \begin{gather*} \Delta u = f(x,u) \quad \mbox{in }\Omega \\ \frac {\partial u}{\partial \nu} + \varphi(x)|u|^r = p(x) \quad \mbox{on }\partial\Omega \end{gather*} admits at least one solution $R^-\le u\le R^+$, provided that $$ f(x,R^+)\ge 0\ge f(x,R^-)$$ and $\varphi(x)|R^-|^r\le p(x)\le \varphi(x)|R^+|^r$. \end{example} \section{Landesman-Lazer type conditions}\label{ll} For the sake of completeness, we summarize in this section the main aspects of coincidence degree theory. This technique has been applied to many problems, see e.g. \cite{fang} and \cite{i}. For further details see \cite{gm,Mawhin}. Let $ X$ and $Y$ be real normed spaces, $L: D \subset X \to Y $ a linear Fredholm mapping of index $0$, and $ N:X\to Y $ continuous. Next, set two continuous projectors $ P:X\to X $ and $Q:Y\to Y $ such that $ \mathop{\rm Ran}(P)=\ker (L) $ and $ \ker (Q)=\mathop{\rm Ran}(L) $ and an isomorphism $J:\mathop{\rm Ran}(Q)\to \ker (L) $. It is readily seen that $$ L_{P}:=L|_{D\cap \ker (P)}:D\cap \ker (P)\to \mathop{\rm Ran}(L)$$ is one-to-one; denote its inverse by $ K_{P} $. If $ \mathcal{B} $ is a bounded open subset of $ X $, $N$ is called $ L$-compact on $ \mathcal{B}$ if $ QN(\mathcal{B} ) $ is bounded and $ K_{P}(I-Q)N:\mathcal{B} \to X $ is compact. The following continuation theorem was proved in Mawhin \cite{Mawhin}. \begin{theorem} \label{cont} Let $ L $ be a Fredholm mapping of index zero and $ N $ be $ L$-compact on a bounded domain $ \mathcal{B}\subset X $. Suppose: \begin{enumerate} \item $ Lx\neq \lambda Nx $ for each $ \lambda \in (0,1] $ and each $ x\in \partial\mathcal{B} $. \item $ QNx\neq 0 $ for each $ x\in \ker (L)\cap \partial \mathcal{B} $. \item $ d (JQN,\mathcal{B} \cap \ker (L),0)\neq 0 $, where $d $ denotes the Brouwer degree. \end{enumerate} Then the equation $ Lx= Nx $ has at least one solution in $ D \cap \mathcal{B} $. \end{theorem} In this context, we may consider $X = H^{1}(\Omega)$, $D = H^{2}(\Omega)$, $Y = L^{2}(\Omega)\times H^{-1/2}(\partial\Omega)$ and $L$, $N$ the operators given by \begin{gather*} Lu = \Big( \Delta u, \frac{\partial u}{\partial \nu} \big|_{\partial\Omega} \Big),\\ Nu = \Big( f(\cdot , u), g(\cdot , u|_{\partial\Omega})\Big). \end{gather*} We recall that the operator $N$ is defined in $X = H^1$. Hence, the trace of a function $u$ is well defined, and we can see that $g(\cdot,u|_{\partial\Omega})\in H^{-1/2}(\partial\Omega)$. Since $f$ and $g$ are bounded, it is immediate to prove that $N$ is well defined and continuous. Moreover, \begin{gather*} \ker (L) = \mathbb{R},\\ \mathop{\rm Ran}(L) = \{ (\varphi, \xi) \in Y : \int_\Omega \varphi = \int_{\partial\Omega}\xi \}. \end{gather*} Thus $L$ is a Fredholm mapping of index zero, and we may consider the projectors $P:X\to X$ and $Q:Y\to Y$ given by \begin{gather*} Pu = \overline u := \frac 1{|\Omega|} \int_\Omega u ,\\ Q(\varphi,\xi) = \frac 1{|\Omega|+|\partial\Omega|} \Big( \int_\Omega \varphi - \int_{\partial\Omega}\xi, \int_{\partial\Omega}\xi - \int_\Omega \varphi \Big). \end{gather*} Also define $J:\mathop{\rm Ran}(Q)\to \ker (L) $ as $$ J(c,-c) = c. $$ Hence, for $(\varphi,\xi) \in \mathop{\rm Ran}(L)$ it follows that $K_P(\varphi,\xi)$ is the unique solution $u\in H^2(\Omega)$ of the problem \begin{gather*} \Delta u = \varphi \\ \frac{\partial u}{\partial \nu}\big|_{\partial \Omega}= \xi \\ \overline u = 0. \end{gather*} We shall use the following estimate. \begin{lemma} \label{lem3.2} There exists a constant $c$ such that \begin{equation} \label{e6} \| u\|_{H^2} \le c\| \Delta u\|_{L^2} \end{equation} for every $u\in H^2(\Omega)$ such that $$ \frac{\partial u}{\partial \nu}\big|_{\partial \Omega} = 0, \quad \overline u =0. $$ \end{lemma} \begin{proof} >From the Neumann condition for $u$ we have that $\int_{\partial\Omega}u \frac{\partial u}{\partial \nu}=0$, and using Green's identity, $-\int_\Omega u\Delta u = \int_\Omega |\nabla u|^2$. It follows that $$ \|\nabla u\|^2_{L^2} \le \|u\|_{L^2} \|\Delta u\|_{L^2}. $$ On the other hand, as $\lambda_0=0$ is the first eigenvalue of the problem $$ -\Delta u = \lambda u, \quad \frac{\partial u}{\partial \nu}\big|_{\partial \Omega} = 0, $$ with eigenfunction $u\equiv 1$, it follows for the second eigenvalue $\lambda_1$ that $$ \lambda_1 = \inf_{\overline u =0,u\neq 0} \frac {\int_\Omega |\nabla u|^2}{\int_\Omega u^2}. $$ Hence $\|u\|_{L^2}\le \frac 1{\sqrt{\lambda_1}}\|\nabla u\|_{L^2}$, and it follows that $$ \|\nabla u\|_{L^2}\le \frac 1{\sqrt{\lambda_1}}\|\Delta u\|_{L^2}, \quad \|u\|_{L^2}\le \frac 1{{\lambda_1}}\|\Delta u\|_{L^2}. $$ Let us recall the following result (see e.g. \cite{gri}): There exists a constant $c$ such that if $u$ is a weak solution of the problem $\Delta u - u = f$, with homogeneous Neumann condition, then $\|u\|_{H^2}\le c\|f\|_{L^2}$. Since we already know that $\|u\|_{H^1}\le c\|\Delta u\|_{L^2}$, we may define $f = \Delta u - u$ and hence $$ \|u\|_{H^2}\le c\|f\|_{L^2} \le c(\|\Delta u\|_{L^2} + \|u\|_{L^2}) \le C\|\Delta u\|_{L^2}\,. $$ Thus, the proof is complete. \end{proof} \begin{lemma} \label{lem3.3} Let $L$ and $N$ be as before and assume that $f$ and $g$ are bounded. Then $N$ is $L$-compact on $\mathcal{B}$ for any bounded domain $\mathcal{B} \subset H^1(\Omega)$. \end{lemma} \begin{proof} If $\| w\|_{H^1}\le R$, and $(\varphi,\xi) = (I-Q)N(w)$, it follows that $\|\varphi\|_{L^2} +\|\xi\|_{L^2(\partial\Omega)} \le C$ for some constant $C$ depending only on $R$. Let $u = K_P(\varphi,\xi)$, and define $$c_\xi = \frac 1{|\Omega|}\int_{\partial\Omega}\xi,$$ and $v_\xi$ the unique solution of the problem \begin{gather*} \Delta v_\xi = c_\xi \\ \frac{\partial v_\xi}{\partial \nu}\big|_{\partial \Omega}= \xi \\ \overline v_\xi = 0. \end{gather*} Then $$ \|u-v_\xi\|_{H^2} \le c \|\varphi-c_\xi\|_{L^2} \le C $$ for some constant $C$ depending only on $R$. Moreover, $\|v_\xi\|_{L^2} \le c\|\nabla v_\xi\|_{L^2}$, and by Cauchy-Schwarz inequality and the continuity of the trace function $Tr:H^1(\Omega)\to L^2(\partial\Omega)$ we have $$ \|\Delta v_\xi\|_{L^2}.\|v_\xi\|_{L^2} \ge \|\nabla v_\xi\|^2_{L^2} - C \|v_\xi\|_{H^1} $$ for some constant $C$. Since $\|\Delta v_\xi\|_{L^2} \le C$ for some constant $C$ depending only on $R$, it follows that $\|v_\xi\|_{H^2} \le C$ for some constant $C$ depending only on $R$. Hence, the norm of $u$ is bounded by a constant depending only on $R$, and the result follows from the compactness of the imbedding $H^2(\Omega)\hookrightarrow H^1(\Omega)$. \end{proof} \begin{proof}[Proof of Theorem \ref{llteo}] We shall prove that if $R$ is large enough, then the assumptions of Theorem \ref{cont} are fulfilled for $\mathcal{B} = B_R(0)\subset H^1(\Omega)$. \noindent {\bf Step 1:} There exists a constant $R$ such that if $Lu = \lambda Nu$ for some $\lambda\in (0,1]$, then $\| u\|_{H^1} < R$. Indeed this is so, otherwise there exists a sequence $\{u_n\}_{n\in \mathbb{N}}$ such that $$ Lu_n = \lambda_n Nu_n,\quad \| u_n\|_{H^1}\to +\infty $$ with $\lambda_n\in (0,1]$. Let $v_n = u_n - \overline {u_n}$. Then $v_n$ satisfies \begin{equation} \label{eq} \begin{gathered} \Delta v_n = \lambda_n f(x,u_n) \\ \frac{\partial v_n}{\partial \nu}\big|_{\partial \Omega} = \lambda_n g(x,u_n|_{\partial\Omega}) \\ \overline {v_n} = 0. \end{gathered} \end{equation} As in the proof of the previous lemma, it follows that $\| v_n\|_{H^2}\le C$ for some constant $C$, and hence $|\overline {u_n}| \to +\infty$. Taking a subsequence, assume for example that $\overline {u_n}\to +\infty$. Integrating (\ref{eq}), as $\lambda_n\neq 0$ we obtain: $$\int_\Omega f(x,u_n) = \int_{\partial\Omega} g(x,u_n).$$ If (\ref{ll1}) holds, taking a subsequence if necessary we get, by Fatou's Lemma, \begin{align*} \limsup_{n\to\infty} \int_\Omega f(x,u_n) &\le \int_\Omega \limsup_{n\to\infty} f(x,u_n)\\ &= \int_\Omega f_s^+\\ &< \int_{\partial\Omega} g_i^+ \\ &\le \liminf_{n\to\infty} \int_{\partial\Omega} g(x,u_n), \end{align*} a contradiction. The proof for $\overline {u_n}\to -\infty$ is analogous. In the same way, we obtain a contradiction if (\ref{ll2}) holds. \noindent{\bf Step 2:} For $R$ is large enough, if $u \in \partial\mathcal{B}_R \cap \ker (L)$ then $QN(u) \neq 0$ and $$ d (JQN,\mathcal{B} \cap \ker (L),0)\neq 0. $$ Indeed, for $u \in \mathcal{B}_R \cap \ker (L) = [-R,R]$, by definition we have that $$ JQN(u) = \frac 1{|\Omega|+|\partial\Omega|} \Big(\int_\Omega f(x,u) - \int_{\partial\Omega} g(x,u)\Big). $$ In the same way as before, if (\ref{ll1}) holds, we obtain $$ \limsup_{n\to\infty} \int_\Omega f(x,u_n) - \liminf_{n\to\infty} \int_{\partial\Omega} g(x,u_n) < 0 $$ for $u_n\to +\infty$ and $$ \liminf_{n\to\infty} \int_\Omega f(x,u_n) - \limsup_{n\to\infty} \int_{\partial\Omega} g(x,u_n) > 0 $$ for $u_n\to -\infty$. Thus, for $R$ large enough $$ \int_\Omega f(x,R) - \int_{\partial\Omega} g(x,R) < 0 < \int_\Omega f(x,-R) - \int_{\partial\Omega} g(x,-R), $$ and the result holds. 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