\input amstex \documentstyle{amsppt} \loadmsbm \magnification=\magstephalf \hcorrection{1cm} \vcorrection{6mm} \nologo \TagsOnRight \NoBlackBoxes \headline={\ifnum\pageno=1 \hfill\else% {\tenrm\ifodd\pageno\rightheadline \else \leftheadline\fi}\fi} \def\rightheadline{EJDE--1995/14\hfil Picone's Identity \hfil\folio} \def\leftheadline{\folio\hfil W. Allegretto \& D. Siegel \hfil EJDE--1995/14} \def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt % Electronic Journal of Differential Equations, Vol. {\eightbf 1995}(1995), No. 14, pp. 1-13.\hfil\break ISSN: 1072-6691. URL: http://ejde.math.swt.edu (147.26.103.110) \hfill\break telnet (login: ejde), ftp, and gopher acess: ejde.math.swt.edu or ejde.math.unt.edu\bigskip} } \topmatter \title PICONE'S IDENTITY AND THE MOVING PLANE PROCEDURE \endtitle \thanks \noindent {\it Subject Classification:} 35B05, 35J60.\hfil\break {\it Key words and phrases:} symmetry, positive solutions, nonlinear elliptic, moving plane, Spectral Theory, Picone's Identity. \hfil\break \copyright 1995 Southwest Texas State University and University of North Texas.\hfil\break Submitted April 15, 1995. Published October 6 ,1995. \hfil\break Supported in part by NSERC (Canada) \endthanks \author Walter Allegretto \\ and \\ David Siegel \endauthor \address Department of Mathematics \newline\indent University of Alberta \newline\indent Edmonton, Alberta \newline\indent Canada\quad T6G 2G1 \endaddress \email retl\@retl.math.ualberta.ca \endemail \address Department of Applied Mathematics \newline\indent University of Waterloo \newline\indent Waterloo, Ontario\newline\indent Canada\quad N2L 3G1 \endaddress \email dsiegel\@math.uwaterloo.ca \endemail \abstract Positive solutions of a class of nonlinear elliptic partial differential equations are shown to be symmetric by means of the moving plane argument coupled with Spectral Theory results and Picone's Identity. The method adapts easily to situations where the moving plane procedure gives rise to variational problems with positive eigenfunctions. \endabstract \endtopmatter \document \def\Qbar{\text{\sl Q\kern-.45em{\vrule height.63em width.05em depth-.033em}}~} \def\qbar{{{\scriptstyle Q}\kern-.45em{\vrule height.41em width.035em depth-.03em}}~} \def\Cbar{\text{\sl C\kern-.35em{\vrule height.63em width.05em depth-.033em}}~} \def\cbar{{{\scriptstyle C}\kern-.41em{\vrule height.42em width.035em depth-.03em}}~} \def\ibid{\hbox to .5truein{\hrulefill}} \def\IH{\text{{\rm I}\kern-.13em{\rm H}}} \def\Z{\Bbb Z} \def\IR{\text{{\rm I}\kern-.13em{\rm R}}} \def\pd#1#2{\frac{\partial#1}{\partial#2}} \def\hra{\hookrightarrow} \def\pr{\prime} \def\bs{\backslash} \def\adb{\allowdisplaybreaks} \def\la{\langle} \def\ra{\rangle} \def\wt{\widetilde} \def\wh{\widehat} \def\ol{\overline} \def\os{\overset} \def\us{\underset} \def\hdotfill{\leaders\hbox to 1em{\hss .\hss}\hfill} \heading 0. Introduction \endheading Consider the problem: $$ \alignat2 -\Delta u &= \lambda p(x)g(u) &\qquad\text{in} &\quad \Omega \tag 1\\ u &= 0 &\qquad\text{on} &\quad \partial\Omega \endalignat $$ where $\Omega $ is a cylinder in $R^n: \;\Omega =(-1,1)\times \Omega ^\pr$ with $\Omega ^\pr$ a domain \quad (=\;bounded, open, connected set) of $R^{n-1}$. Are all $C^2(\ol\Omega )$ positive solutions symmetric in $x_1?$ If the boundary of $\Omega ^\pr$ is reasonably smooth, then under suitable conditions on $p,g$ the classic moving plane argument of Serrin, [21], as extended by Gidas, Ni, Nirenberg in [14], [15], Berestycki and Nirenberg, [5], Amick and Fraenkel, [3], and elsewhere, [8], [18], [25], applies and the answer is positive. More recently, [6], Berestycki and Nirenberg showed that the result is still true for general nonlinear equations if, for example, $u\in W^{2,n}_{\ell\text{oc}}(\Omega )\cap C(\ol\Omega )$ with no regularity assumed on $\partial\Omega ,$ while Dancer, [11], dealt with the case of $u\in H^{1,2}_0(\Omega )\cap L^\infty (\Omega )$. It is the purpose of this paper to discuss the symmetry of positive solutions under somewhat weaker conditions than those that to the best of our knowledge have been applied earlier. We avoid direct use of pointwise considerations near $\partial \Omega $ by employing related arguments from Spectral Theory and Picone's Identity. In this way, we are able in particular to bypass various ``corner Lemma'' and Maximum Principle procedures. Most of the paper is devoted to the problem $$ \alignat2 -\Delta u &= f(u) &\qquad\text{in} &\quad \Omega\tag 2 \\ u &= 0 &\qquad\text{on} &\quad \partial\Omega \endalignat $$ with $\Omega $ bounded, $\Omega \subset R^n $ (we consider explictly the case $n\ge 3),$ or $\Omega =R^n$ itself since these cases illustrate all the ideas. While our approaches are different, our results are closest to those obtained by Dancer, [11]. We feel that one possible advantage of our approach is that it adapts easily to variational problems with positive eigenfunctions. This is illustrated explicitly for the case of $\Omega =R^n,$ and at the end we indicate some extensions to more general problems. We were unable to extend our approach to the more general nonlinear equations considered in [6]. Our results in particular show: \proclaim {Theorem 0} Let $0n/2,$ symmetric in $x_1$. Assume $p$ is nonincreasing in $x_1$ for $x_1\ge 0$ and: $g\in C^{1+\theta }_{\ell\text{oc}},\;g(\xi )>0$ for $\xi \ge 0$. \item{\rm (a)} For $\lambda $ small enough, Problem~1 has a bounded (i.e. $L^\infty )$ positive solution. \item{\rm (b)} All bounded positive solutions to {\rm (1)} are symmetric with respect to $x_1,$ and if moreover $p\in L^{n+\varepsilon }(\Omega )$ then $\pd {u}{x_1} <0$ for $0n/2$. We then have: \proclaim {Theorem 1} \item {\rm (a)} For each $\lambda \in [a,b],$ the map $\ell _\lambda :\;H^{1,2}_0(\Omega _\lambda )\to L^2(\Omega _\lambda )$ formally given by $\ell_\lambda =-\Delta -p_\lambda I$ has a least eigenvalue $\mu _0(\lambda )$ to which corresponds a nonnegative eigenfunction $u_0(\lambda ),$ positive in at least one of the components of $\Omega _\lambda $. \item{\rm (b)} For any $K\Subset \Omega _\lambda $ there exists $\delta >0$ such that $u_0(\lambda )\in C^\delta (K)$. \item{\rm (c)} There exists a function $\omega \in \big(H^{1,2}(\wt\Omega _\lambda ) - H^{1,2}_0(\wt \Omega _\lambda )\big) \cap C(\wt\Omega _\lambda )$ with $\omega \ge 0$ and $\ell_\lambda (\omega )\ge 0$ {\rm a.e.} $\wt \Omega _\lambda $ for any component $\wt\Omega _\lambda $ of $\Omega _\lambda $ iff $\mu _0(\lambda )>0$. \item{\rm (d)} $\mu _0(\lambda ) =0$ iff there exists at least one $\omega \in H^{1,2}(\Omega _\lambda )\cap C(\Omega _\lambda ),$ with $\omega \ge 0$ and $\ell_\lambda (\omega )\ge 0$ {\rm a.e.,} nontrivial in each component of $\Omega _\lambda ,$ and, for any such $\omega $ there exists in each component of $\Omega _\lambda $ a constant $c\ge 0$ with $c\omega =u_0(\lambda )$. \item{\rm (e)} If $\Vert p_\lambda \Vert _{L^\alpha }$ is bounded, there exists a constant $C_0$ such that if \newline $\text{\rm meas}\;(\Omega _\lambda )0$. \endproclaim \demo {Proof} (a) If need be we add a positive constant to $p_\lambda $ and observe that $\ell^{-1}_\lambda $ defines a compact self-adjoint map $L^2(\Omega _\lambda )\to L^2(\Omega _\lambda )$ by Sobolev's Theorem since $\alpha >n/2,$ [16]. The existence of the least eigenvalue $\mu _0(\lambda )$ then follows. That the associated eigenfunction $u_0(\lambda )$ has the desired positivity properties is immediate, since by the Courant min.-max. characterization of $\mu _0(\lambda )$ we conclude that $u_0(\lambda )\ge 0,$ [16], and indeed it follows that $u_0(\lambda )>0$ in at least one component by the weak Harnack Inequality, [16]. (b) This is given in [16]. (c) It is here that we employ Picone's Identity. Once again by the weak Harnack Inequality, $\omega >0$ in $ \Omega _\lambda $ since $\omega $ is assumed nontrivial in each component. Let $\varphi \in C^\infty _0(\Omega _\lambda ),\quad \varepsilon >0$. We recall Picone's Identity, see eg. [2]: $$ \align \int_{\Omega_\lambda } (\omega +\varepsilon )^2\big[\nabla [\frac{\varphi }{\omega +\varepsilon }]\big]^2 &= \int_{\Omega_\lambda } \big[\vert \nabla \varphi \vert ^2 - p_\lambda \varphi ^2\big]\\ &\quad -\int_{\Omega_\lambda } \big[\nabla (\frac{\varphi ^2}{\omega +\varepsilon })\nabla (\omega )-\frac{p_\lambda \varphi ^2(\omega +\varepsilon )}{\omega +\varepsilon }\big]\\ &\le \int_{\Omega_\lambda } \big[\vert \nabla \varphi \vert ^2 - p_\lambda \varphi ^2\big] +\int_{\Omega_\lambda } \;\frac{p_\lambda \varepsilon \varphi ^2}{\omega +\varepsilon }\;. \endalign $$ Since $\vert \varepsilon /(\omega +\varepsilon )\vert \le 1,$ and $\varepsilon /(\omega +\varepsilon )\to 0$ pointwise, we let $\varepsilon \to 0,$ apply Lebesgue's Dominated Convergence Theorem, followed by letting $\varphi \to u_0(\lambda )$ in $H^{1,2}_0$ to conclude that either in each component $\wt\Omega _\lambda $ we have $\wt c\omega =u_0(\lambda )$ for some constant $\wt c\ge 0$ or else $\mu _0(\lambda )>0$. The first case is impossible since $\omega \notin H^{1,2}_0(\wt\Omega _\lambda )$ for any $\wt\Omega _\lambda ,$ and we conclude $\mu _o(\lambda )>0$. On the other hand, if $\mu _0(\lambda )>0$ then $\ell_\lambda \eta =p_\lambda $ has a solution $\eta \in H^{1,2}_0(\Omega _\lambda )$. Set $\omega =\eta +1$ then $\ell_\lambda (\omega ) =0 $ and Courant's min.-max. principle shows that if $\omega $ is nontrivial in a component then $\omega ^- =0$ a.e., whence $\omega >0$ (again by Harnack's inequality). Finally $\omega \in H^{1,2}(\wt\Omega _\lambda ) - H^{1,2}_0(\wt \Omega _\lambda )$ for otherwise $1=\omega -\eta \in H^{1,2}_0(\wt \Omega _\lambda )$. Since an equivalent norm on $H^{1,2}_0(\wt \Omega _\lambda )$ is $(\int_{\wt\Omega _\lambda } \vert \nabla u\vert ^2)^{1/2}$ then $C\;\text{meas}\; (\wt \Omega _\lambda ) \le \Vert 1\Vert ^2_{H^{1,2}(\wt \Omega _\lambda )} = 0$ and the result follows. (d) If $\mu _0(\lambda ) =0$ then for any such $\omega $ we conclude $c\omega =u_0(\lambda )$ by part~(c). Observe that we may always construct at least one such $\omega $ by using the eigenfunction itself in some components and solving $\ell_\lambda (\omega )=1$ in others. Conversely, since such a $\omega $ exists then $\mu _0(\lambda )\ge 0$ by part~(c). On the other hand, if $\mu _0(\lambda )>0$ then part~(c) shows there exists a $\omega ,$ as desired, with $\omega \in H^{1,2}(\wt\Omega _\lambda )-H^{1,2}_0(\wt\Omega _\lambda )$ whence $c\omega \ne u_0(\lambda )$ in at least one component. (e) We need only apply Sobolev's Estimate, [16] to obtain: $$ \int_{\Omega _\lambda } \vert \nabla\varphi \vert ^2 - p_\lambda (x)\varphi ^2\ge \int_{\Omega _\lambda }\vert \nabla\varphi \vert ^2[ 1-K\Vert p_\lambda \Vert _{L^\alpha }\;(\text{meas}\;(\Omega _\lambda )^{\beta })] $$ where $\beta =\frac{2\alpha -n}{\alpha n }\;,$ and $\varphi \in C^\infty _0(\Omega _\lambda ),$ and observe that $\Vert p_\lambda \Vert _{L^\alpha }$ is bounded. \enddemo As a consequence we have: \proclaim {Corollary 2} If $\mu _0\in C(a,b)$ with $ \mu _0(\lambda )>0$ for all $b-\lambda $ small enough, and for each $\lambda $ there exists a $\omega \ge 0,$ dependent on $\lambda ,$ in each $H^{1,2}(\wt\Omega _\lambda ) - H^{1,2}_0(\wt\Omega _\lambda )$ such that $\ell_\lambda (\omega )\ge 0,$ then $\mu _0(\lambda )>0$ for all $\lambda \in (a,b)$. \endproclaim Otherwise there would exist $\lambda _0$ with $\mu _0(\lambda _0) = 0$. The existence of such a $\omega $ then contradicts Theorem~1(d). The continuity of $\mu _0$ --~indeed of the entire spectrum~-- is a classical problem, discussed by Courant and Hilbert, [10], and studied more recently in a variety of papers: [4,13,20,26]. Based on these results we have: \proclaim {Theorem 3} Let $\Omega ^*_\lambda =\Omega \cap \{x_1>\lambda \}\ne\emptyset$ for $\lambda \in [a,b]$ and let $\ell^*_\lambda =-\Delta u - P_\lambda u$ defined on $H^{1,2}_0(\Omega ^*_\lambda )$ where $P\in C([a,b];\; L^\alpha (\Omega ^*_\lambda ))$ for some $\alpha >n/2,$ and $\Omega $ is a fixed bounded domain. Then $\mu ^*_0(\lambda ),$ the least eigenvalue of $\ell^*_\lambda $ in $\Omega ^*_\lambda ,$ is continuous. If $\text{\rm meas}\;(\Omega ^*_\lambda )$ is small enough, $\mu ^*_0(\lambda )>0$. \endproclaim By $P\in C\big([a,b];\;L^\alpha (\Omega ^*_\lambda )\big)$ we mean $P_\lambda \in L^\alpha (R^n)$ and $\chi _{\lambda ^*}P_\lambda \in C\big([a,b];\; L^\alpha (R^n)\big)$ where $\chi _{\lambda ^*}$ denotes the characteristic function of $\Omega _{\lambda ^*}$. \demo {Proof} Observe that these are nested domains. Choose $\lambda _0$ in the interval of interest, and without loss of generality pass to a subsequence and suppose first $\lambda _m\downarrow \lambda _0,$ with $\delta =\lim\;\big(\mu^*_0(\lambda _m)\big)$. Note that $\mu ^*_0(\lambda _m)$ are bounded since $\alpha >n/2$. Let $\{\omega _m\}$ be associated, normalized in $L^2,$ eigenfunctions and observe that $\omega _m \ge 0\quad (\omega _m>0$ if $\Omega ^*_\lambda $ is connected). By the trivial extension, $\Vert \omega _m\Vert _{L^2(\Omega ^*_{\lambda _0})} = 1$ and $\int_{\Omega ^*_{\lambda _0}} P_{\lambda_m}\omega ^2_m$ can be estimated. To see this, note that: $$ \vert \int_{\Omega ^*_{\lambda _0}}P_{\lambda_m}\omega ^2_m\vert =\vert \int_{\Omega ^*_{\lambda _0}} P_{\lambda_m}\omega ^\varepsilon _m\omega ^{2-\varepsilon }_m\vert \le K\Vert P_{\lambda_m}\Vert _{L^\alpha }\Vert \omega _m\Vert ^\varepsilon _{L^2}\Vert \nabla\omega _m\Vert ^{2-\varepsilon }_{L^2} $$ where $\varepsilon = 2-(n/\alpha )$. Whence: $$ \Vert \nabla \omega _m\Vert ^2_{L^2}\le K_1+K_2\Vert \nabla\omega _m\Vert ^{2-\varepsilon }_{L^2}. $$ We conclude that $\Vert \omega _m\Vert _{H^{1,2}_0(\Omega ^*_{\lambda _0})} \sim \Vert \nabla\omega _m\Vert _{L^2(\Omega ^*_{\lambda _0})}$ is bounded. Passing to a subsequence, also denoted by $\{\omega _m\},$ we may conclude convergence (weakly) in $H^{1,2}_0(\Omega ^*_{\lambda _0})$ and (strongly) in $L^q(\Omega ^*_{\lambda _0}),$ for $q<(2n)/(n-2),$ to some $\omega \in H^{1,2}_0(\Omega ^*_{\lambda _0})$. Obviously, $\omega \ge 0,$ nontrivial, and $\ell_{\lambda _0}(\omega ) = \delta \omega $ in $H^{1,2}_0(\Omega ^*_{\lambda _0})$. We claim that $\delta =\mu ^*_0(\lambda _0)$. If $\Omega ^*_{\lambda_0} $ is connected, this is immediate by the positivity of $\omega $ -- and again employing Picone's Identity. If $\delta \ne \mu ^*_0(\lambda _0),$ then $\mu ^*_0(\lambda _0)<\delta $ by the min.-max. principle. We conclude that there exists some $\varphi \in C^\infty _0(\Omega ^*_{\lambda _0})$ such that $(\ell^*_{\lambda _0}\varphi ,\varphi ) <(\delta-2\varepsilon ) (\varphi ,\varphi )$. But since $\varphi $ has compact support, then $(\ell^*_{\lambda _m}\varphi ,\varphi )<(\delta -\varepsilon )(\varphi ,\varphi )$ for all large $m$ and some $\varepsilon >0,$ i.e. $\mu ^*_0(\lambda _m)<\delta -\varepsilon $ contradicting the definition of $\delta $. Suppose next that $\lambda _m\uparrow \lambda _0$ and again set $\delta =\lim\;\big(\mu ^*_0(\lambda _m)\big)$. The same procedure as above shows the existence of a subsequence $\omega _m$ and function $\omega $. We need to show $\omega \in H^{1,2}_0(\Omega ^*_{\lambda _0})$. To see this, let $g$ be a cut off function: $g\in C^\infty (R,R),$ with $g(\xi ) =0$ if $\xi <2,\quad g(\xi )=1$ if $\xi >3$ and set $z_m(x) = g\big((x_1-\lambda _0)/(\lambda _0-\lambda _m)\big)$. We then have: $$ \Vert z_m\omega_m\Vert ^2_{H^{1,2}(\Omega ^*_{\lambda _0})} \le C\big[\Vert \omega _m\Vert ^2_{H^{1,2}(\Omega ^*_{\lambda _0})} +\Vert\; \vert \nabla z_m\vert \omega _m\Vert ^2_{L^2(\Omega ^*_{\lambda _0})}\big]. $$ The first term is clearly bounded. For the second, note: $$ \Vert \;\vert \nabla z_m \vert \omega _m\Vert ^2_{L^2(\Omega ^*_{\lambda _0})}\le C\;\frac{1}{(\lambda _0-\lambda_m )^2}\;\int_{\Omega ^*_{\lambda _m}\cap\{x_1<\lambda _0+3(\lambda _0-\lambda _m)\}}\omega ^2_m $$ Poincare's Inequality, [16], then shows this term is bounded as well. Observe that $z_m\omega _m$ is obviously in $H^{1,2}_0(\Omega ^*_{\lambda _0})$ and thus, without loss of generality, weakly convergent in this space. Since $z_m\to 1$ pointwise in $\Omega ^*_{\lambda _0}$ we conclude that $\omega \in H^{1,2}_0(\Omega ^*_{\lambda _0})$. Clearly $\delta $ is an eigenvalue as $\omega $ is nontrivial. By the min.-max. principle, it is the least. Finally the positivity of $\mu ^*_0(\lambda )$ for $\text{meas}\;(\Omega ^*_\lambda )$ small, follows from Theorem~1(e) with $p$ replaced by $P$. \enddemo \proclaim {Theorem 4} Let $T^{-1}_\lambda :R^n\to R^n$ by $y=T^{-1}_\lambda (x)$ where $y=(2\lambda -x_1,\ol x),$ with $\ol x = (x_2,\dots,x_n)$. Let $\Omega _\lambda =T^{-1}_\lambda (\Omega ^*_\lambda ),\quad p_\lambda (x) = P_\lambda \big(T_\lambda (x)\big)$ where $P_\lambda ,\Omega ^*_\lambda $ are as before. Then $\mu _0(\lambda ),$ the least eigenvalue of $\ell_\lambda = -\Delta -p_\lambda $ in $\Omega _\lambda ,$ is also continuous. \endproclaim We merely map the quadratic form associated with $\ell_\lambda $ in $\Omega _\lambda $ to that for $\ell^*_\lambda $ in $\Omega _{\lambda ^*}$. Notice that this leaves $-\Delta$ unchanged, and that the Jacobian is $-1$. \heading 2. The Nonlinear Problem\endheading Consider now the nonlinear problem (2). We recall $\Omega ^*_\lambda =\Omega \cap \{x_1>\lambda \}$ and $\Omega _\lambda =\{x\vert x^\lambda \in \Omega ^*_\lambda \}$ where $x^\lambda =(2\lambda -x_1,x_2,\dots,x_n),$ and $00,$ with $\Vert P_\lambda \Vert _{L^{\frac{n+\varepsilon }{2}}(\Omega ^*_\lambda )}$ bounded for $\lambda \in [\lambda _0,\lambda _1)$. Note that $\mu _0(\lambda )$ -- the least eigenvalue of $\ell_\lambda $ -- is then continuous in $\lambda ,$ and $p_\lambda (y) \equiv P_\lambda (y)$. The results in Section~1 then yield: \proclaim {Theorem 5} Let {\rm (I), (II)} hold. Then: \item {\rm (a)} $\mu _0(\lambda )>0$ for $\lambda _0<\lambda <\lambda _1,$ and $u>v_\lambda $ in $\Omega _\lambda $. \item {\rm (b)} If $\Omega _{\lambda _0} = \Omega \cap \{x_1<\lambda _0\}$ then $u\equiv v_{\lambda _0}$ on $\Omega _{\lambda _0}$. \item {\rm (c)} If $x_0\in \Omega \cap \{x_1=\lambda \}, \quad\pd {u}{x_1}\;(x_0)$ exists and $P_\lambda \in L^{n+\varepsilon }(\Omega ^*_\lambda ),$ then $\pd{u}{x_1}\;(x_0) <0$. \endproclaim \demo {Proof} We apply the earlier results using $\omega =u-v_\lambda $ in Theorem~1. We first show that Theorem~1-d can be used to conclude that $\mu _0(\lambda )>0$. Specifically, assume otherwise i.e. $\mu _0(\lambda ^\pr)=0, \; \mu _0(\lambda )>0$ for $\lambda >\lambda ^\pr,$ for some $\lambda ^\pr\in(\lambda _0,\lambda _1),$ and note that $\omega ^-\in H^{1,2}_0(\Omega _{\lambda ^\pr})$ and hence the min.-max. principle shows $\omega ^-=0,$ since, by continuity, if $\omega ^-\ne 0$ then $(u-v_\lambda )^-\ne 0$ for some $\lambda >\lambda ^\pr$ and thus $\mu _0(\lambda )\le 0$. Assume now that $\omega \in H^{1,2}_0(\wt\Omega _{\lambda^\pr})$. Choose a small ball $B\subset \wt\Omega _{\lambda ^\pr}$ and look at the cylinder $Z=(-a,\wt x_1)\times S$ where: $(\wt x_1,\ol {y^*})$ is the center of $B$ and $S=\{\ol y\vert (\wt x_1,\ol y)\in B\}$. For notational convenience denote $\lambda ^\pr$ by $\lambda $ henceforth. Since $v_\lambda $ can be approximated in $H^{1,2}$ by functions which vanish near $Z\cap \partial\wt\Omega _\lambda$ and $\omega \in H^{1,2}_0,$ then this is also true of $u. $ We may assume $u$ admits in $Z-\wt \Omega _\lambda $ a trivial extension (also denoted by $u)$ and thus if $a$ is large enough $u\equiv 0$ on $(-a)\times S$. Now by a fundamental result employing Fubini's Theorem, [24], there exists a $\ol y_1\in S$ such that $u(x_1,\ol y_1) = \int^{x_1}_{-a} \;\pd{u(\xi ,\ol y_1)}{x_1}\; d\xi $ for almost all $x_1$. Since $u$ is continuous in $Z\cap \Omega _\lambda $ and in $Z-Z\cap \ol \Omega _\lambda $ and clearly so is the integral, we conclude that equality must actually hold in these regions. We thus have that $\int^{x_1}_{-a} \;\pd{u}{x_1}\;(\xi ,\ol y_1)d\xi =0$ if $(x_1,\ol y_1) \in Z-Z\cap \ol\Omega _\lambda $ and $\int^{x_1}_{-a}\;\pd{u}{x_1}\;(\xi ,\ol y_1)d\xi =u(x_1,\ol y_1)\ge \delta >0$ if $(x_1,\ol y_1)\in Z\cap K, $ if $K\Subset \Omega ,$ by the positivity of the solution $u$ in the compacta of $\Omega $. Now let $\alpha $ be the least number such that we have $C=\{(x_1,\ol y_1)\vert \alpha 0$ small enough, and we conclude that $(\alpha ,\ol y_1)\in \Omega $. I.e. $\ol C\subset \Omega $ and thus $u>\delta >0$ in $\ol C$. This contradicts the absolute continuity of the integral and it follows that $\omega \notin H^{1,2}_0(\wt\Omega _{\lambda ^\pr})$ and thus $\mu _0(\lambda ^\pr)>0$ by Theorem~1-d. Since $\mu _0(\lambda )>0$ and $(u-v_\lambda )^- \in H^{1,2}_0(\Omega _\lambda ),$ then the min.-max. principle again implies $(u-v_\lambda )^-=0,$ whence $u\ge v_\lambda $. The earlier arguments show that $u\not\equiv v_\lambda $ in $\Omega _\lambda ,$ and then $u>v_\lambda $. This shows part (a). As for part (b), we have $u(x)\ge u(x^{\lambda _0})$ in $\Omega _{\lambda _0}$ by continuity, since $u>v_\lambda $ on $\Omega _\lambda ,\;\lambda >\lambda _0$. However $\Omega $ is symmetric in this case about $x_1=\lambda _0$. If we first perform a reflection about $x_1=\lambda _0,$ and then repeat the above procedure we would conclude $u(x^{\lambda _0})\ge u\big((x^{\lambda _0})^{\lambda _0}\big) = u(x)$ and the result. Finally, for part (c), note that $x_0\in \Omega $ and $P_\lambda \in L^{n+\varepsilon },$ imply that $\omega $ is, without loss of generality, in $C^{1+\theta} (\ol B)$ for some ball $B\subset \Omega _\lambda $ with $x_0\in \partial B,$ [16]. Choose a function $z\in C^{1+\theta }(\ol B)$ such that $-\Delta z-P_\lambda z=0$ in $B,\; z>0$ in $\ol B$. Then by considering the equation $\omega /z$ satisfies we conclude, again by [16], that $\omega /z\in C^2(B)\cap C^{1+\theta }(\ol B)$ and $\pd{}{x_1}\;(\omega /z)(x_0)<0,$ i.e. $\pd{\omega }{x_1}\;(x_0) = 2\;\pd{u}{x_1}\;(x_0) <0$. Next assume that $f$ depends on $x$ as well, i.e. $f\equiv f(x,u)$. As was the case in the previous references, this situation can also be dealt with if we assume $f$ is monotone in $x_1: \;f(x,\xi )\ge f(x^\lambda ,\xi )$ for $\lambda >0$. There is no significant change in the proofs. Note that we could thus deal with some cases where $f$ had singularities with respect to $x$ on $\{x_1=\lambda _0\}\cap \Omega ,$ or with singularities along planes $\{x_m =c\}\cap \Omega ,\quad m\ne 1,$ as examples with $f(x,u) = p(x)g(u)$ easily show, as long as the resulting $P_\lambda $ was in $L^\alpha (\Omega _\lambda )$. One limitation in the applicability of Theorem~5 is given by condition~(II). Observe that since $\Vert P_\lambda \Vert _{L^{\frac{n+\varepsilon }{2}}}$ is assumed bounded, then it suffices that $\chi _\lambda P_\lambda (x)$ be pointwise continuous, a.e. in $\lambda ,$ which will be immediately the case if $f$ is smooth in $u$. This follows from the observation that if $\{g_n\}$ is a sequence of functions bounded in $L^\alpha $ and $g_n\to g\in L^\alpha $ pointwise then $g_n\to g$ in $L^{\alpha -\varepsilon }$ for any $\varepsilon >0,$ and this is because we can find a constant $c$ --independent of $n\text{--}$ such that $g_n =\bar g^c_n +g^\pr_n$ with $\bar g^c_n = g_n$ if $\vert g_n\vert \le c,\;\vert \bar g^c_n\vert =c$ otherwise, and $g^\pr_n$ of small $L^{\alpha-\varepsilon } $ norm. We thus only need to check the boundedness of $\chi _\lambda P_\lambda $. Recall that we may also express $P_\lambda $ as $P_\lambda =\int^1_0 f^\pr\big(tu+(1-t)v_\lambda \big)dt$ and thus if, for example, $u\in L^\infty (\Omega )$ and $f\in C^{1+\theta }_{\ell\text{oc}},$ then Theorem~5 applies. Finally, observe that the results apply if $f$ is Lipschitz (locally Lipschitz if $u$ is in $L^\infty )$. This follows by setting $p_\lambda =[f(u)-f(v_\lambda )]/[u-v_\lambda ]$ if $u(x)\ne v_\lambda(x) ;\;p_\lambda =0$ otherwise, and this example also shows that the continuity of $\mu _0$ is really only needed from the left: If we let $\lambda ^\pr$ be given by $\mu _0(\lambda )>0$ if $\lambda >\lambda ^\pr,$ as before, then we repeat the above arguments, in particular Theorem~1-c, and conclude that both $\mu _0(\lambda ^\pr)>0$ and $u>v_{\lambda ^\pr}$. We next observe that $\chi _\mu P_\mu \to \chi _{\lambda ^\pr}P_{\lambda ^\pr}$ pointwise, and that $\vert P_\mu \vert 0$ for $\vert \mu -\lambda ^\pr\vert $ small. \enddemo \heading 3. $\pmb{\Omega =R^n}$\endheading The above approach also works for the case of $\Omega =R^n$ and we now consider the modifications needed to deal with this case. Detailed references of other results for this case may be found in [17]. Specifically, assume $-\Delta u =f(x,u)$ weakly in $R^n,$ with $00$ and $\xi >0,$ furthermore given any $\lambda >0,\;\varepsilon >0$ there exists $x\in \Omega _\lambda $ such that $f(x,\xi )>f(x^\lambda ,\xi )$ for some $0<\xi <\varepsilon $. Observe that in the special case: $f(x,\xi ) =p(x)\xi ^\delta ,$ then condition (IV) holds if, for example, $\pd {p}{x_1}\le 0,\quad \pd{p}{x_1}\;(\varepsilon ,\bar x)<0$ for $0<\varepsilon $ small enough. If $u\in C^1,$ say, decays fast enough at $\infty $ then $u\in E(R^n)$. We give explicit conditions for this to be the case, as well as other convenient results in Theorem~6. \proclaim {Theorem 6} \item{\rm (a)} If $u\in C^1$ and: $u\in L^{\frac{2n}{n-2}}, \quad uf(x,u)\in L^1$ then $u\in E(R^n)$. \item{\rm (b)} Let $\Omega _\lambda = \{x\vert x_1<\lambda \},$ then $\omega = u-v_\lambda \in E(\Omega _\lambda )$ if $u\in E(R^n)$. \endproclaim We recall that it is often possible to show that a solution $00$ follows from the next Lemma. In any case observe that $f(x,u)$ must be positive somewhere, since $-\Delta u = f(x,u)$ and $u>0$. Since we also have $p_\lambda \to p_\infty =f(x,u)/u$ pointwise, then $p_\lambda $ cannot be nonpositive for $\lambda $ large. I.e. for such $\lambda $ at least, $\xi _\lambda >0$. \proclaim {Lemma 8} \item{\rm (a)} Let $\omega =u-v_\lambda $ and $\xi _\lambda ,\;\eta _\lambda $ exist then \newline $(\xi _\lambda -1)(p_\lambda \eta _\lambda ,\omega )_{\Omega _\lambda }\ge 0$. \item{\rm (b)} $\xi _\lambda $ is continuous in $\lambda ,\quad \xi _\lambda \ge 1$ and $w>0$ for all $\lambda >0$. \endproclaim \demo {Proof} (a) Observe that $\ell_\lambda \omega =-\Delta \omega -p_\lambda \omega =f(x,v_\lambda ) -f(x^\lambda ,v_\lambda )=r(x)\ge 0$ whence $\xi _\lambda (p_\lambda \eta _\lambda ,\omega )_{\Omega _\lambda } = (\eta _\lambda ,\ell_\lambda \omega )_{\Omega _\lambda } + (p_\lambda \eta _\lambda ,\omega )_{\Omega _\lambda }$ and the result since $\eta _\lambda >0$. (b) We show first that $\xi _\lambda \ge 1$. We claim that this is true for $\lambda $ large enough since otherwise there exists a sequence $\lambda _m\to\infty $ for which this is not the case. But as $\lambda _m\to\infty , \quad p_{\lambda _m}\to p_\infty $ in $L^\alpha ,$ and thus $\xi _{\lambda _m}>0$ exists. Furthermore: as $\lambda _m\to\infty ,\quad\omega \to u$ pointwise and thus $\omega \to u$ in $L^\tau (B)$ for any large $\tau $ and fixed ball $B\subset R^n$ by the uniform boundedness of $\omega ,u$. Similarily, we note that $\xi _{\lambda _m}$ is bounded by the min.-max. principle and setting $\xi _\infty $ equal to the limit of $\xi _{\lambda _m},$ we may assume that $\Vert \eta _{\lambda _m}\Vert _E =1$ and thus $\eta _{\lambda _m}\to \eta $ weakly in $E(R^n)$ and strongly in $L^{\frac{2n-\varepsilon }{n-2}}(B)$ where $\eta $ denotes a (positive) eigenfunction of $-\Delta \eta =\xi _\infty p_\infty \eta $ in $E(R^n), $ since $\eta \not\equiv 0,$ as a consequence of $\xi _\infty (p_\infty \eta ,\eta )=1$. We thus have the existence of two positive eigenfunctions: $\eta ,u$ corresponding to the eigenvalues $\xi _\infty ,1$ respectively. However, Picone's Identity also shows the simplicity of the eigenvalue associated with a positive eigenfunction $\eta $ such that $(p_\infty \eta ,\eta )>0,$ and thus $\xi _\infty =1$ and $\eta \equiv u$. We conclude by part~(a) that $(p_\infty u,u) = \Vert u\Vert ^2_{E(R^n)}\le 0$ if $\xi _{\lambda _m}<1,$ which is a contradiction. Let $\lambda _0$ denote the least $\lambda $ such that $\xi _\lambda \ge 1$ and $\omega $ is nontrivial (and thus positive) for $\lambda >\lambda _0$. The next arguments also show that if $\lambda _0 =0$ we are done, hence suppose $\lambda _0>0$. Assume first $\omega $ is nontrivial in $\Omega _{\lambda _0}$. By the presumed continuity, $\xi _{\lambda _0} = 1$ since $\omega $ is also nontrivial for some $\lambda <\lambda _0,$ and $-\Delta\omega -p_{\lambda _0}\omega \ge 0$. We conclude $(p_{\lambda _0}\eta _{\lambda _0},\omega )_{\Omega _{\lambda_0} }=(p_{\lambda _0}\eta _{\lambda _0},\omega)_{\Omega _{\lambda _0}} + (r,\eta _{\lambda _0})_{\Omega _{\lambda _0}}, $ i.e. $r\equiv 0$ and since $\omega \in E(\Omega _{\lambda_0} ),$ it must be an eigenfunction corresponding to $\xi _{\lambda _0}$. Indeed, again by Picone's Identity, if $\omega \not\equiv 0$ then $\omega =c_{\lambda_0} \eta _{\lambda_0} $ in $\Omega _{\lambda_0} $ for some constant $c_{\lambda_0} $. Note that this result applies to all $\lambda $ for which $\xi _\lambda =1$ and $\omega \not\equiv 0$ in $\Omega _\lambda ,$ and, identically, if $\xi _\lambda >1$ then $(-\Delta \omega ,\omega^- )\ge (p_\lambda\omega ,\omega^- )$ whence $(-\Delta \omega ^-,\omega ^-)\le (p_\lambda \omega ^-,\omega ^-),$ i.e. $$ 1\le \frac{(p_\lambda \omega ^-,\omega ^-)}{\Vert \omega ^-\Vert ^2_{E(\Omega _\lambda )}} $$ and we have a contradiction unless $\omega ^- =0,$ i.e. once again, if $\omega\not\equiv 0$ then $\omega \ge 0$ and $\omega >0$ by the weak Harnack Inequality. It follows that since for any $\lambda >\lambda _0$ we have $\xi _\lambda \ge 1$ and $\omega $ is nontrivial then $\omega >0$ in $\Omega _\lambda ,\quad \omega =0$ on $x_1=\lambda $ and thus $\pd {u}{x_1} <0$ if $x_1> \lambda_0 $. Now suppose $\omega $ is trivial in $\Omega _{\lambda _0}$ and then observe $f\big(x,u(x)\big) \equiv f\big(x^{\lambda _0},u(x)\big)$ for $x\in \Omega _{\lambda _0},$ contradicting assumption (IV) on $f$. Hence if $\lambda _0>0,$ then $\xi _{\lambda _0}=1$ and $\omega >0$ is its associated eigenfunction. But this is impossible since then $r\equiv 0$ in $\Omega _{\lambda _0}$ and again this violates (IV). Finally, the continuity of $\xi _\lambda $ follows from the properties of $p_\lambda $ and the \; min.-max.\; definition of $\xi _{\lambda }$. Specifically, observe first that if $\xi _\mu >0$ then $p_\mu $ is somewhere positive and thus so is $p_\lambda $ for $\vert \lambda -\mu \vert $ small $(\lambda $ sufficiently large if $\mu =\infty )$. We conclude $\xi _\lambda $ is bounded above and below, and set $\xi ^*$ equal to any limit point of $\xi _\lambda $. The earlier arguments in this proof show that a subsequence of the normalized (in $E)$ positive eigenfunctions $\eta _\lambda $ converges to $\eta $ weakly in $E(R^n)$ and strongly in $L^{\frac{2n-\varepsilon }{n-2}}(B)$. We again note that $\xi ^*(p_\mu \eta ,\eta )_{\Omega _\mu } = 1,\;-\Delta \eta =\xi ^*p_\mu \eta $ in $\Omega _\mu ,$ and $\eta \ge 0,$ nontrivial. If $\lambda \uparrow \mu $ then $\eta \in E(\Omega _\mu )$ and again by Picone's Identity, $\xi ^*=\xi _\mu $ and $\eta =\eta _\mu $. If $\lambda \downarrow \mu $ then we need only show $\eta \in E(\Omega _\mu ),$ since the rest is the same. But this is immediate here by the smoothness of the (plane) boundary. The uniqueness of $\xi _\mu ,\eta _\mu $ then shows the continuity. \enddemo We then have under the above assumptions $\text{(II}^\pr),$ (III), (IV) on $f:$ \proclaim {Theorem 9} If $f(x,\xi )$ is symmetric (in $x_1)$ about $x_1=0,$ then $u$ is symmetric in $x_1,$ and $ x_1\;\pd{u}{x_1}<0$ if $x_1\ne 0$. \endproclaim \demo {Proof} We have by Lemma~8 that $\xi _\lambda \ge 1$ and $w>0,$ i.e. $u>v$ and $\pd{u}{x_1}<0$ for all $\lambda >0$ and thus $u\ge v_\lambda $ for $\lambda =0$. Repeating the procedure for $\lambda $ negative we obtain the result. \enddemo \heading 4. Extensions\endheading We now briefly and heuristically comment on some extensions where the eigenvalue arguments and Picone's Identity still work. Observe that the same procedure can yield some modest results even for problems not involving purely Dirichlet conditions. Consider for example the cylinder $(-1,1)\times \Omega ^\pr$ with Dirichlet conditions on $\{-1\}\times \Omega ^\pr,\quad \{+1\}\times \Omega ^\pr$ but Neumann elsewhere. The procedure in such a case is identical. Note that a key step involves the fact that the part of the boundary with Neumann conditions does not reflect inside the regions $\Omega _\lambda $. Assume now that $f$ is not smooth. Suppose first as in [3], [14] that $f=f_1+f_2$ with $f_1$ smooth and $f_2$ monotone increasing and continuous. Let $f_2(\xi )\equiv 0$ if $\xi 0,$ and suppose $\Omega $ is bounded. We now also require that $u$ be in $C(\ol\Omega )\cap C^1(\Omega )$. This is similar to the requirements of [6]. If $f_1(u),f_2(u)$ are in $L^p$ for some $p>n$ then it suffices to assume that $\Omega $ satisfies an exterior cone condition at every point of $\partial\Omega ,$ [16]. A more detailed study of the requirements on $\partial\Omega $ can be found in [23]. We now set $p_\lambda(x) =[f_1(u)-f_1(v_\lambda )] /(u-v_\lambda )$ in (II) and repeat the earlier procedures. Observe that $\big((u-v_\lambda )^-,\ell_\lambda ((u-v_\lambda ))\big) = \big(f_2(u)-f_2(v_\lambda ),(u-v_\lambda )^-\big)$. If $u \ngeq v_\lambda $ we have an immediate contradiction to $\mu _0(\lambda )\ge 0$ unless there is a point $x^0_\lambda $ such that $\big(f_2(u)-f_2(v_\lambda )\big)(u-v_\lambda )^-\vert (x^0_\lambda )<0$. On the other hand, $f_2(u) \equiv 0$ in a neighborhood of $\partial\Omega $ by the continuity of $u$ and thus we may keep our arguments away from $\partial \Omega $. We conclude in such eventuality that $\text{dist}\; (x^0_\lambda ,(\partial\Omega_\lambda -\{x_1=\lambda \} ))\ge \varepsilon _0$ for some $\varepsilon _0>0$. To apply these observations, suppose that for some value of $\lambda =\lambda ^*$ we have $u\ge v_{\lambda^*} $ in $\Omega _{\lambda ^*}$ and $\mu _0(\lambda ^*)>0,$ then by the assumed continuity, $\mu _0(\lambda )>0$ for $\lambda $ near $\lambda ^*$. If $u\ngeq v_\lambda $ for such $\lambda $ then we construct a sequence of points $\{x^0_\lambda \}$ with: $x^0_\lambda \to x_0,\quad \text{dist}\; (x_0,\partial\Omega ) \ge \varepsilon _0, \quad u(x^0_\lambda ) 0,$ since the maximum of $u$ is at $x=0$. We note that we can begin our eigenvalue procedures for any $\Omega _{\lambda _0}$ for which we can conclude that $\mu (\lambda _0)>0$. If some information is known about the norm of $u,$ then we need not start by considering a very thin domain $\Omega _\lambda $ to ensure $\mu (\lambda _0)>0$. We can bypass in this way the requirement that $\Omega _\lambda \subset \Omega $ for all relevant $\lambda ,$ and still obtain the monotonicity result $u>v_\lambda ,$ for $\lambda >\lambda _0$ say. We note that, as a consequence, we immediately have the observation that if $f$ is Lipschitz, with small constant, and $\Omega $ is symmetric in $x_1$ then all positive solutions must be symmetric, regardless of whether $\Omega $ is convex or not in $x_1$. This result is essentially known, [12]. This approach can also be applied if some extra conditions are imposed on $f,$ as the following arguments in particular indicate. Consider now the situation where $f(x,\xi)$ is not smooth at $\xi =0$. In general we could not obtain results for this case. In special situations, however, we could actually obtain better results than those obtained earlier. Specifically, assume now: $f\in \;\text{Lip}_{\ell\text{oc}} \big(\ol\Omega \times (0,\infty )\big)$ and $0\le f(x,\xi )/\xi $ is nonincreasing in $\xi $ for $\xi >0$. The prototype $f$ we have in mind is $f(x,\xi ) = p(x)\xi ^\theta +q(x)$ with $p,q\ge 0,$ smooth and $\theta \le 1$. Our result for these $f$ is as follows (see also [9]): \proclaim {Theorem 10} Suppose $\Omega $ is a bounded domain and that $u$ is a positive solution of $-\Delta u = f(x,u)$. Then $u$ is unique up to a constant multiple. If $f(x,c\xi ) \not\equiv cf(x,\xi )$ for any $c>0,\;\xi >0\quad (c\ne 1),$ and $x\in \Omega $ then $u$ is unique. If both $\Omega $ and $f(x,\cdot )$ are symmetric about $x_1=0$ then so is $u$. \endproclaim We observe that we do not require that $\Omega $ be convex in $x_1$ nor assume conditions on $\vec\nabla_xf$. \demo {Proof} Let $\varphi \in C^\infty _0(\Omega )$. We apply once again Picone's Identity and conclude: $$ \int_\Omega \vert \nabla\varphi \vert ^2 = \int_\Omega \;\frac{f(x,u)}{u}\;\varphi ^2 +\int_\Omega u^2\vert \nabla\;(\frac \varphi u)\vert ^2. $$ Now let $v$ denote another positive solution and without loss of generality, $(u-v)^-\not\equiv 0$. By our assumptions on $f(x,\xi )/\xi $ we have: $$ \aligned J(\varphi ) &\equiv \int_\Omega \vert \nabla \varphi \vert ^2 -\big[\frac{f(x,u)-f(x,v)}{u-v}\big]\varphi ^2\\ &= \int_\Omega u^2\vert \nabla\;(\frac \varphi u)\vert ^2 +\int_\Omega [\frac{f(x,u)}{u}\;-\;\frac{f(x,u)-f(x,v)}{u-v}]\varphi ^2\ge 0 \endaligned \tag 3 $$ since $\frac{f(x,u)-f(x,v)}{u-v} \le \frac{f(x,u)}{u}\;$. Note that here we set $[\frac{f(x,u)-f(x,v)}{u-v}]=0$ if $u=v$. Now let $w=(u-v)^-$ and observe that we may construct a sequence of functions $0\le \varphi _m\le (u-v)^-$ with compact support such that $\varphi _m\to (u-v)^-$ in $H^{1,2}$. We conclude that $$ \align \big\vert \big(\frac{f(x,u)-f(x,v)}{u-v}\big)\big\vert \varphi ^2_m &\le \big\vert \big[\frac{f(x,u)-f(x,v)}{u-v}\big]\big\vert [(u-v)^-]^2\\ &=\vert f(x,u)-f(x,v)\vert (u-v)^-, \endalign $$ and by integrating and recalling the equation $(u-v)$ satisfies, that $J(w) =0$ since: $$ \int_\Omega \vert \nabla \varphi _m\vert ^2\to \int_\Omega \vert \nabla w\vert ^2=-\int_\Omega \nabla (u-v)\cdot \nabla w. $$ In the same way, we obtain $0\le J(w+\varepsilon \varphi )$ for given $\varepsilon $ and any $\varphi \in C^\infty _0$. We conclude that $- \Delta w-[\frac{f(x,u)-f(x,v)}{u-v}]w=0$ in $\Omega $. But $w\ge 0$ in $\Omega $ and since $\frac{f(x,u)-f(x,v)}{u-v}$ is locally in $L^\infty (\Omega )$ by the fact that $u,v\in C^\delta (K),$ Harnack's Inequality, [16], implies that $w>0$ or $w\equiv 0$. By assumption, we have $w\not\equiv 0$. But then (3) yields $w=cu$ for some $c>0,$ i.e. $v=(1+c)u$. It also follows that $f\big(x,(1+c)u\big) = (1+c)f(x,u)$ for $x\in \Omega $ from the equations that $u,v$ satisfy, and if this is impossible, we conclude $u\equiv v$. Finally, suppose both $\Omega $ and $f(x,\cdot )$ are symmetric in $x_1,$ and now choose $v$ by $v(x,\bar x) = u(-x_1,\bar x)$. Clearly $00$ or $w\equiv 0$. Now $w>0$ is impossible since $w=0$ on $x_1=0,$ and it follows that $w\equiv 0,$ i.e. $u\ge v$. In the same way we obtain $v\ge u$ and the result. \enddemo The proof of Theorem~10 also yields monotonicity results: suppose, for example, $f=f(u)$ and $\Omega _\lambda $ is properly contained in $\Omega $ for some $\lambda >0$ then $u>v_\lambda $ follows by choosing $v=v_\lambda $ in the proof, and we thus have $\pd{u}{x_1} <0$ on $\Omega \cap \{x_1=\lambda \}$. As a final remark, we recall that the classic moving plane argument can be extended to systems: $-\Delta \vec u = \vec f(\vec u),$ [22], if we merely assume $\pd{f_i}{u_j}\ge 0$. We were not able to obtain a similar extension under our conditions on $\partial \Omega ,\; \vec u,\;\vec f$. \heading 5. Examples\endheading We conclude with some simple examples. We begin with: \demo {Proof of Theorem 0} (a) Since $p\in L^\alpha ,$ the linear problem: $-\Delta \wt u = p\ge 0,\quad \wt u\in H^{1,2}_0(\Omega )$ has a positive solution $\wt u$ in $L^\infty (\Omega )\cap C(\Omega ),$ [16]. Observe that we thus have $-\Delta \wt u\ge \lambda p(x)g(\wt u)$ for $\lambda $ such that $\lambda g(\wt u)\le 1$. Since $\us\sim\to u = 0$ is a subsolution we have the existence of a solution $0