\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 104, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/104\hfil Optimization of the energy integral] {Placement of a source or a well for optimizing the energy integral} \author[K. Feyissa, A. Tadesse, G. Porru \hfil EJDE-2018/104\hfilneg] {Kebede Feyissa, Abdi Tadesse, Giovanni Porru} \address{Kebede Feyissa \newline Adama Science and Technology University, Department of Applied Mathematics, Adama, Ethiopia} \email{feyissake11@gmail.com} \address{Abdi Tadesse \newline Department of Mathematics, College of Natural and Computational Sciences, Addis Ababa University, Addis Ababa, Ethiopia} \email{tadesse.abdi@aau.edu.et} \address{Giovanni Porru \newline Department of Mathematics and Informatics, University of Cagliari, Cagliari, Italy} \email{porru@unica.it} \dedicatory{Communicated by Jesus Ildefonso Diaz} \thanks{Submitted December 4, 2017. Published May 7, 2018.} \subjclass[2010]{35J25, 49K20, 49K30} \keywords{Energy integral; maximization; minimization} \begin{abstract} This article concerns the maximization and minimization of the energy integral associated with solutions to partial differential equations with coefficients depending on a suitable source or well. Under suitable geometrical conditions on the domain we find the optimal configurations. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction} Let $\Omega$ be a bounded domain in $\mathbb R^2$, and let $f=f(x,y)$ and $g=g(x,y)$ be non-negative bounded functions. We assume $f$ to be positive in a subset with a positive measure. Consider the boundary value problem \begin{equation*} -\Delta u+gu=f \text{ in } \Omega, \quad u=0 \text{ on } \partial\Omega. \end{equation*} This Dirichlet problem has a unique solution $u\in H^1_0(\Omega)$. By standard regularity results, $u\in C^1(\Omega)\cap C^0(\overline\Omega)$ and is positive on $\Omega$. The corresponding energy integral is the following $$ I=\int_\Omega (|\nabla u|^2+gu^2)\,dx\,dy =\int_\Omega f u\,dx\,dy. $$ Let $\mathcal F$ be the class of rearrangements of a given function $f_0$. A typical problem is the investigation of the maximum or the minimum of $I$ for $f\in\mathcal F$. Again, let $\mathcal G$ be the class of rearrangements of a given function $g_0$. One can investigate the maximum or the minimum of $I$ for $g\in\mathcal G$. These problems have been discussed in many papers, we refer to \cite{ATL, Bu1, Bu2} and references therein. In this article we consider subclasses of $\mathcal F$ and $\mathcal G$. More precisely, let $f_0=\chi_{D_0}$, where $D_0$ is a given subset of $\Omega$. We shall investigate the maximum and the minimum of $I(D)$ for $g$ fixed and $f=\chi_D$ where $D\subset\Omega$ is any translation or rotation of $D_0$. Furthermore, let $g_0=\chi_{D_0}$, where $D_0$ is a given subset of $\Omega$. We shall investigate the maximum and the minimum of $I(D)$ for $f$ fixed and $g=\chi_D$. These problems are inspired by the paper \cite{HKK}, where the case of eigenvalues is discussed. However the situation here is different for the simultaneous presence of $f$ and $g$. We shall consider only the case $D$ is a disc. In this case it is easy to prove existence for a maximizer or a minimizer. Our main effort will be the localization of a maximizer or a minimizer. This will be possible under suitable symmetry assumptions on $\Omega$. The equation $-\Delta u+gu=f$ models the temperature $u$ in $\Omega$ in case of a steady state situation. The term $gu$ corresponds to the density of heat absorbed, while $f$ corresponds to the density of the heat produced. The energy integral $\int_\Omega f u\,dx\,dy$ is related with the average temperature in $\Omega$. One may be interested in the maximization or minimization of the average temperature acting either on the data $g$ or the data $f$. At the end of Section 2 and Section 3, some precise situation is described. The paper is organized as follows. In Section 2 we study the case $g$ fixed. In Section 3 we discuss the case $f$ fixed. In Section 4 we assume $\Omega$ to be a ball in $\mathbb R^N$ and discuss the optimization of the energy integral in general classes of rearrangements. \section{First problem} Let $\Omega$ be a bounded plane domain, and let $D$ be a disc contained in $\Omega$. Consider the Dirichlet problem \begin{equation}\label{1} -\Delta u+gu=\chi_D \text{ in } \Omega, \quad u=0 \text{ on } \partial\Omega, \end{equation} where $g=g(x,y)$ is a non negative bounded function. The corresponding energy integral is $$ I(D)=\int_\Omega (|\nabla u|^2+gu^2)\,dx\,dy =\int_\Omega\chi_D u\,dx\,dy. $$ When the location of $D$ (that we call source) changes through $\Omega$, $I(D)$ may change. We are interested in finding the locations of $D$ which realize the maximum or the minimum value of $I(D)$. We first discuss the effect of a translation of the source. If $D_\epsilon$ is the domain $D$ shifted at a distance $\epsilon$ in the $x$ direction (assuming this is allowed), the new problem reads as \begin{equation}\label{2} -\Delta u_\epsilon+gu_\epsilon=\chi_{D_\epsilon}\text{ in } \Omega, \quad u_\epsilon=0 \text{ on } \partial\Omega, \end{equation} and the corresponding energy integral is $$ I(D_\epsilon)=\int_\Omega (|\nabla u_{\epsilon}|^2+g u_\epsilon^2)\,dx\,dy =\int_{\Omega}\chi_{D_\epsilon}u_\epsilon\,dx\,dy. $$ So, we are interested in finding the sign of $$ I(D_\epsilon)-I(D)= \int_\Omega \bigl[\chi_{D_\epsilon}u_\epsilon- \chi_Du\bigr]\,dx\,dy $$ for a small $\epsilon>0$. From the equations in \eqref{1} and \eqref{2} we find \begin{equation}\label{3} -\Delta(u_\epsilon-u)+g(u_\epsilon-u)=\chi_{D_\epsilon}-\chi_D. \end{equation} The Alexandrov-Bakelman-Pucci maximum principle (see \cite[Theorem 9.1]{GT}) applied to \eqref{3} yields \begin{equation}\label{4} \sup_{\Omega}| u_\epsilon-u| \le C\| \chi_{D_\epsilon}-\chi_D\|_{L^2(\Omega)}, \end{equation} where the constant $C$ depends on $\Omega$, but it is independent of $\epsilon$ and $g$ (see \cite{St}). Note that \[ | \chi_{D_\epsilon}-\chi_D|^2 =\begin{cases} 1 &x\in (D\setminus D_\epsilon)\cup (D_\epsilon\setminus D) \\ 0 & \text{elsewhere in } \Omega. \end{cases} \] Moreover, if $r$ is the radius of $D$ we have $|D\setminus D_\epsilon|=|D_\epsilon\setminus D|<\pi r\epsilon$. Therefore, we can find a constant $C$ such that \begin{equation}\label{5} \| \chi_{D_\epsilon}-\chi_D\|_{L^2(\Omega)}\le C\epsilon^\frac{1}{2}. \end{equation} Here and in what follows we denote by $C$ constants (sometimes different from line to line) independent of $\epsilon$. From \eqref{4} and \eqref{5} we find \begin{equation}\label{6} \sup_{\Omega}| u_\epsilon-u| \le C\epsilon^\frac{1}{2}. \end{equation} Note that \eqref{5} is equivalent to $$ \int_\Omega|\chi_{D_\epsilon}-\chi_D|\,dx\,dy\le C^2\epsilon. $$ From the latter estimate and \eqref{6} we find \begin{equation}\label{7} \int_\Omega (\chi_{D_\epsilon}-\chi_D)(u_\epsilon-u)\,dx\,dy =o(\epsilon). \end{equation} Moreover, multiplying \eqref{1} by $u_\epsilon$, \eqref{2} by $u$ and integrating over $\Omega$ we find \begin{equation}\label{8} \int_\Omega \chi_Du_\epsilon\,dx\,dy =\int_\Omega \chi_{D_\epsilon}u\,dx\,dy. \end{equation} Hence, \begin{align*} \frac{1}{\epsilon}\bigl[I(D_\epsilon)-I(D)\bigr] =&\frac{1}{\epsilon}\int_\Omega \bigl[\chi_{D_\epsilon}u_\epsilon -\chi_{D}u\bigr]\,dx\,dy\quad \text{by using \eqref{8}}\\ =&\frac{1}{\epsilon}\int_\Omega \bigl[\chi_{D_\epsilon}u_\epsilon-\chi_{D}u_\epsilon+\chi_{D_\epsilon}u -\chi_Du\bigr]\,dx\,dy\quad \text{by using \eqref{7}}\\ =&\frac{2}{\epsilon}\int_\Omega\bigl[\chi_{D_\epsilon}u-\chi_{D}u\bigr]\,dx\,dy +\frac{o(\epsilon)}{\epsilon}\\ =&2\int_D\frac{u(x+\epsilon,y)-u(x,y)}{\epsilon}\,dx\,dy +\frac{o(\epsilon)}{\epsilon}. \end{align*} As $\epsilon\to 0$, we find that \begin{equation*} \frac{dI_{D_\epsilon}}{d\epsilon}\Big|_{\epsilon=0}=2\int_D\frac{\partial u}{\partial x}(x,y)\,dx\,dy. \end{equation*} Recall the Green formula \begin{equation*} \int_D\frac{\partial u}{\partial x}(x,y)\,dx\,dy =\int_{\partial D}u(x,y)\cos(n,x)ds, \end{equation*} where $n$ is the exterior normal to $\partial D$. Hence, \begin{equation}\label{9} \frac{dI_{D_\epsilon}}{d\epsilon}\Big|_{\epsilon=0} =2\int_{\partial D}u(x,y)\cos(n,x)ds. \end{equation} To find the sign of the derivative in \eqref{9} we need some conditions on $\Omega$ and on the function $g(x,y)$. \begin{theorem}\label{thm1} Assume $\Omega$ to be Steiner symmetric with respect to the line $\{x=0\}$. The function $g(x,y)$ is assumed to satisfy $g(x,y)=g(-x,y)$ and to be non-increasing with respect to $x$ for $x<0$. Then, if $(x,y)$ is the center of $D$, the energy integral $I(D)$ associated with problem \eqref{1} increases as $(x,y)$ approaches the point $(0,y)$. \end{theorem} \begin{proof} Suppose the center of $D$ is located on $\{x<0\}$. The line $\{x=a\}$ passing through the center of $D$ divides $\Omega$ into $\Omega_1$ (the small part) and $\Omega_2$. Since $\Omega$ is Steiner symmetric, the reflection of $\Omega_1$ with respect to the line $\{x=a\}$ is strictly contained in $\Omega_2$. In $\Omega_1$, define $w=u(2a-x,y)-u(x,y)$. Of course, we have $w=0$ for $x=a$. Furthermore, recalling that $u(x,y)>0$ in $\Omega$ and $u(x,y)=0$ on $\partial\Omega$ we have $w(x,y)>0$ on the part of the boundary $\partial\Omega_1$ with $x 0$, using the assumption $g(x,y)=g(-x,y)$ we rewrite the inequality as $g(x-2a,y)\le g(-x,y)$ and apply the condition that $g(x,y)$ is non-decreasing with respect to $x$ for $x>0$. The claim follows. Hence, from \eqref{10} we find that \begin{equation}\label{11} -\Delta w+g(x,y)w(x,y)\ge 0\quad \text{in } \Omega_1. \end{equation} The strong maximum principle (see \cite[Theorem 8.19]{GT}), yields $w(x,y)>0$; that is, $u(2a-x,y)>u(x,y)$ in $\Omega_1$. Hence, $u(x,y)$ computed at the right hand side of $\partial D$ is larger than $u(x',y)$ computed at the left hand side of $\partial D$. If we denote $(\partial D)^r$ the part of $\partial D$ located at the right hand side with respect to the line $\{x=a\}$, and $(\partial D)^l$ the part of $\partial D$ located at the left hand side with respect to the same line we have \begin{equation*} \int_{\partial D}u(x,y)\cos(n,x)ds =\int_{(\partial D)^r}u(x,y)\cos(n,x)ds +\int_{(\partial D)^l}u(x,y)\cos(nx)ds. \end{equation*} Now, it is easy to note that $\cos(n,x)$ is positive at each point $(x,y)\in (\partial D)^r$. Moreover, if we take $(x,y)\in (\partial D)^r$ and $(x',y)\in (\partial D)^l$, the values of the corresponding $\cos(n,x)$ are opposite each other. This fact, coupled with the information that the value of $u(x,y)$ computed at $(x,y)\in (\partial D)^r$ is larger than the value of $u(x',y)$ computed at $(x',y)\in (\partial D)^l$ (for the same value of $y$) yields \begin{equation}\label{12} \frac{dI_{D_\epsilon}}{d\epsilon}\Big|_{\epsilon=0} =2\int_{\partial D}u(x,y)\cos(nx)ds>0. \end{equation} Therefore, the energy integral increases moving $D$ in the $x$ direction until the center of $D$ belongs to the $y$ axes.\\ By symmetry, if the center of $D$ is located on $\{x>0\}$, the energy integral increases moving $D$ in the opposite direction of $x$ until the center of $D$ belongs to the $y$ axes. The proof is complete. \end{proof} Theorem \ref{thm1} allows us to solve the optimization problem in some simple cases. \smallskip \noindent $\bullet$ Let $\Omega$ be a disc (larger than $D$) centered at the origin $(0,0)$. If $g(x,y)\ge 0$ is radially symmetric and non-decreasing with respect to the distance from $(x,y)$ to $(0,0)$, the maximum of $I(D)$ is attained when $D$ is concentric with $\Omega$, and the minimum is attained when $D$ is (internally) tangent to $\partial\Omega$. \smallskip \noindent $\bullet$ Let $\Omega$ be the rectangle $(-a_1,a_1)\times(-a_2,a_2)$. Let $g(x,y)\ge 0$ satisfy $g(x,y)=g(-x,y)=g(x,-y)$, and let $g(x,y)$ be non-increasing with respect to $x$ for $x<0$ and non-increasing with respect to $y$ for $y<0$. Then the maximum of $I(D)$ occurs when $D$ is centered at $(0,0)$, and the minimum occurs when $D$ is located at one corner of the rectangle (there are four locations for the minimum). \smallskip \noindent $\bullet$ Let $\Omega$ be a regular polygon of $n$ sides centered at the origin. If $g(x,y)\ge 0$ is radially symmetric and non-decreasing with respect to the distance from $(x,y)$ to $(0,0)$, the maximum of $I(D)$ is attained when $D$ is concentric with $\Omega$, and the minimum occurs when $D$ is located at one corner (there are $n$ locations for the minimum). \smallskip \noindent $\bullet$ By the proof of Theorem \ref{thm1} we can find indications for the maximum even if $\Omega$ is not Steiner symmetric as in the following example. Let $\Omega$ be the union of the half disc $\{x^2+y^2<1,\ x\le 0\}$ and the square $\{0< x<2,\ -1v(x)$, where $$ -v''=1,\quad v(1)=v(2)=0. $$ This fact follows from the familiar comparison principle in that, for $10,\quad u(2)=0. $$ Therefore, $$ J=\int_0^2 u(x)\,dx>\int_1^2 v(x)\,dx =\int_1^2\Bigl(-\frac{x^2}{2}+\frac{3}{2}x-1\Bigr)dx=\frac{1}{12}. $$ As a consequence, for $\lambda$ such that $\frac{2}{\lambda^2}\le \frac{1}{12}$ the energy integral $I$ is less than the energy integral $J$. Hence, the maximum of the energy integral does not correspond to the case $D$ is centered at the origin. Let us give a simple model described by problem \eqref{1} in case of $g(x,y)\equiv 0$. Let $\Omega$ be a plane thermal conductor subject to a source of heat with density $\chi_D(x,y)$. In other words, we have a stove which occupies $D$ and produces heat with density one (in $D$ only). Assume the temperature is equal to zero on the boundary $\partial\Omega$. Then, in a steady state situation, the solution $u(x,y)$ to problem \eqref{1} yields the temperature in $\Omega$. The energy integral $$ I(D)=\int_D u_D(x,y)\,dx\,dy $$ is related with the average of the temperature in $D$, and it depends on the location of our stove in $\Omega$. By our previous results, for particular domains we can find the exact location of $D$ which maximizes $I(D)$. This result agrees with the physical intuition in that, for maximizing the energy integral, $D$ has to be located as far as possible from the boundary $\partial\Omega$. \section{Second problem} Now we discuss a new problem which is, in some sense, complementary to the previous one. Using the previous notation, consider the boundary-value problem \begin{equation}\label{13} -\Delta u+ \chi_{D}u=f,\text{ in } \Omega, \quad u=0 \text{ on } \partial\Omega, \end{equation} where $f=f(x,y)\ge 0$ is a bounded function, positive in a subset of positive measure. Physically $f(x,y)$ represents the density of the heat produced (source) and $\chi_{D}u$ represents the density of the heat absorbed (well). The associated energy integral is $$ I(D)=\int_\Omega f(x,y)u_D\,dx\,dy. $$ Recall that $D$ is a disc with a fixed radius. We shall investigate the minimum and the maximum of $I(D)$ for $D\subset\Omega$. As in the previous case, we investigate the effect of a translation of $D$. If $D_\epsilon$ is the domain $D$ shifted at a distance $\epsilon$ in the $x$ direction (assuming this is allowed), the new problem reads as \begin{equation}\label{14} -\Delta u_\epsilon+\chi_{D_\epsilon}u_\epsilon=f \text{ in } \Omega, \quad u_\epsilon=0 \text{ on } \partial\Omega, \end{equation} and the corresponding energy integral is $$ I(D_\epsilon)=\int_\Omega f(x,y) u_{\epsilon}\,dx\,dy. $$ So, we are interested in finding the sign of $$ I({D_\epsilon})-I(D)= \int_\Omega f(x,y)(u_\epsilon-u)\,dx\,dy $$ for a small $\epsilon>0$. Here we have denoted with $u_\epsilon$ the solution to problem \eqref{14}, and with $u$ the solution to problem \eqref{13}. From equation \eqref{13} we find \begin{equation*} \int_\Omega\bigl( \nabla u\cdot\nabla u_\epsilon+\chi_D u u_\epsilon\bigr)\,dx\,dy =\int_\Omega f(x,y) u_\epsilon\,dx\,dy. \end{equation*} Similarly, from \eqref{14} we find \begin{equation*} \int_\Omega\bigl( \nabla u_\epsilon\cdot\nabla u+\chi_{D_\epsilon} u_\epsilon u\bigr)\,dx\,dy=\int_\Omega f(x,y) u\,dx\,dy. \end{equation*} It follows that \begin{equation}\label{15} \int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) u_\epsilon u\,dx\,dy =\int_\Omega f(x,y) (u_\epsilon-u)\,dx\,dy =I({D_\epsilon})-I(D). \end{equation} From equations \eqref{13} and \eqref{14} we also find $$ -\Delta(u_\epsilon-u)+\chi_D(u_\epsilon-u)=(\chi_D-\chi_{D_\epsilon}) u_\epsilon. $$ By Alexandrov-Bakelman-Pucci maximum principle applied to the latter equation we have \begin{equation}\label{16} \sup_{\Omega}| u_\epsilon-u| \le C\| (\chi_D-\chi_{D_\epsilon}) u_\epsilon\|_{L^2(\Omega)} \le C\| \chi_D-\chi_{D_\epsilon} \|_{L^2(\Omega)}\sup_\Omega u_\epsilon. \end{equation} The Alexandrov-Bakelman-Pucci maximum principle applied to \eqref{13} and \eqref{14} also yields \begin{equation}\label{17} \sup_{\Omega} u \le C\| f\|_{L^2(\Omega)}\quad \text{and}\quad \sup_{\Omega} u_\epsilon \le C\| f\|_{L^2(\Omega)}. \end{equation} Moreover, as already observed, it is easy to find a constant $C$ such that \begin{equation}\label{18} \int_\Omega|\chi_{D_\epsilon}-\chi_D|\,dx\,dy\le C\epsilon\quad \text{and}\quad \| \chi_D-\chi_{D_\epsilon} \|_{L^2(\Omega)} \le C\epsilon^\frac{1}{2}. \end{equation} Hence, \eqref{16} implies \begin{equation}\label{19} \sup_{\Omega}| u_\epsilon-u| \le C\epsilon^\frac{1}{2}. \end{equation} Since $$ \int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) u_\epsilon u\,dx\,dy =\int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) u^2\,dx\,dy +\int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) (u_\epsilon- u)u\,dx\,dy, $$ using \eqref{18} and \eqref{19} we find $$ \int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) u_\epsilon u\,dx\,dy =\int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) u^2\,dx\,dy+o(\epsilon). $$ Now, by \eqref{15} and the latter equation we have $$ I(D_\epsilon)-I(D)=\int_\Omega\bigl(\chi_D-\chi_{D_\epsilon}) u^2\,dx\,dy+o(\epsilon). $$ Finally, we get \begin{align*} \lim_{\epsilon\to 0}\frac{I(D_\epsilon)-I(D)}{\epsilon} &=-\lim_{\epsilon\to 0}\int_\Omega\frac{\chi_{D_\epsilon}-\chi_D}{\epsilon} u^2\,dx\,dy\\ &=-\lim_{\epsilon\to 0}\int_D\frac{u^2(x+\epsilon,y)-u^2(x,y)}{\epsilon} \,dx\,dy \\ &=-\int_D \frac{\partial u^2}{\partial x} \,dx\,dy. \end{align*} Hence, $$ \frac{dI_{D_\epsilon}}{d\epsilon}\Big|_{\epsilon=0} =-\int_D \frac{\partial u^2}{\partial x} \,dx\,dy. $$ By using Green's formula we find \begin{equation}\label{20} \frac{dI_{D_\epsilon}}{d\epsilon}\Big|_{\epsilon=0}=-\int_{\partial D}u^2\; \cos(n,x)ds. \end{equation} \begin{theorem}\label{thm2} Assume $\Omega$ to be Steiner symmetric with respect to the line $\{x=0\}$. The function $f(x,y)$ is assumed to satisfy $f(x,y)=f(-x,y)$ and to be non-decreasing with respect to $x$ for $x<0$. Then the energy integral $I(D)$ associated with problem \eqref{13} decreases as the center $(x,y)$ of the disc $D$ approaches the point $(0,y)$. \end{theorem} \begin{proof} To find the sign of the derivative \eqref{20}, we proceed as in the previous case. Suppose the center of $D$ is located on $\{x<0\}$. The line $\{x=a\}$ passing through the center of $D$ divides $\Omega$ into $\Omega_1$ and $\Omega_2$, and the reflection of $\Omega_1$ with respect to the line $\{x=a\}$ is strictly contained in $\Omega_2$. In $\Omega_1$, define $w=u(2a-x,y)-u(x,y)$. We have $w=0$ for $x=a$ and $u(x,y)=0$ on the part of the boundary $\partial\Omega_1$ with $xx$ in $\Omega_1$) the inequality holds since $f(x,y)$ is non-decreasing for $x<0$. If $2a-x> 0$, recalling that $f(-x,y)=f(x,y)$ we can rewrite the inequality as $$ f(x-2a,y)-f(-x,y)\ge 0, $$ and this holds since $x-2a<-x$ and $f(x,y)$ is non-increasing for $x>0$. The claim follows. Hence, \eqref{21} yields \begin{equation}\label{22} -\Delta w+\chi_Dw\ge 0\quad \text{in } \Omega_1. \end{equation} By the strong maximum principle we have $w(x,y)>0$, and, equivalently, $$ u(2a-x,y)>u(x,y)\quad \text{in } \Omega_1. $$ Hence, $u(x,y)$ computed at the right hand side of $\partial D$ is larger than $u(x',y)$ computed at the left hand side of $\partial D$ (for the same value of $y$). Arguing as in the previous case we find \begin{equation*} \frac{dI_{D_\epsilon}}{d\epsilon}\Big|_{\epsilon=0}=-\int_{\partial D}u^2(x,y)\; \cos(n,x)ds<0. \end{equation*} Therefore, the energy integral decreases moving $D$ in the $x$ direction until the center of $D$ belongs to the $y$ axes. By symmetry, if the center of $D$ is located on $\{x>0\}$, the energy integral decreases moving $D$ in the opposite direction of $x$ until the center of $D$ belongs to the $y$ axes. The proof is complete. \end{proof} We underline that in this second problem, $f(x,y)$ is non-decreasing with respect to $x$ for $x<0$, whereas, in the first problem, $g(x,y)$ was non-increasing. The conclusions concerning the location of the maximum and the minimum are reversed. As in the previous case, Theorem \ref{thm2} allows one to solve the optimization problems for special domains $\Omega$. Theorem \ref{thm2} can be easily extended to the case $\chi_D$ is replaced by $\lambda^2\chi_D$ for any $\lambda>0$. However, it fails to hold even in dimension $N=1$ without appropriate assumptions on $f(x)$. Indeed, let $$ -u''+\lambda^2\chi_{[-1,1]}u=\chi_{[0,2]},\quad u(-2)=u(2)=0. $$ As already noticed, the corresponding energy integral $J$ satisfies $$ J(\lambda)=\int_0^2 u(x)\,dx>\frac{1}{12}\quad \text{for any } \lambda>0. $$ On the other hand, let $$ -u''+\lambda^2\chi_{[0,2]}u=\chi_{[0,2]},\quad u(-2)=u(2)=0. $$ One finds \[ u(x)=\begin{cases} A(x+2) & \text{if } -2< x< 0\\ Be^{\lambda x}+Ce^{-\lambda x}+\frac{1}{\lambda^2} & \text{if } 0< x<2, \end{cases} \] with $$ 2A=B+C+\frac{1}{\lambda^2},\quad A=\lambda(B-C),\quad Be^{2\lambda }+Ce^{-2\lambda}+\frac{1}{\lambda^2}=0. $$ By easy computation one finds \begin{gather*} B=-\frac{(2\lambda+1)e^{2\lambda}-1}{\lambda^2 \bigl(2\lambda-1+e^{4\lambda}(2\lambda+1)\bigr)},\\ C=-\frac{(2\lambda-1)e^{2\lambda}+e^{4\lambda}}{\lambda^2 \bigl(2\lambda-1+e^{4\lambda}(2\lambda+1)\bigr)}. \end{gather*} Since $B$ and $C$ have a negative sign, the corresponding energy integral satisfies \begin{equation*} I(\lambda)=\int_0^2\Bigl( Be^{\lambda x}+Ce^{-\lambda x}+\frac{1}{\lambda^2}\Bigr) dx<\frac{2}{\lambda^2}. \end{equation*} As a consequence, for $\lambda$ such that $\frac{2}{\lambda^2}\le \frac{1}{12}$ we have $I(\lambda)