\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 31, 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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/31\hfil On a convex combination of solutions] {On a convex combination of solutions to elliptic variational inequalities} \author[M. Boukrouche, D. A. Tarzia\hfil EJDE-2007/31\hfilneg] {Mahdi Boukrouche, Domingo A. Tarzia} % in alphabetical order \address{Mahdi Boukrouche \newline Laboratory of Mathematics, University of Saint-Etienne, 23 rue du Dr Paul Michelon, Saint-Etienne, 42023, France \newline Phone: 33 4 77 48 15 35; fax: 33 4 77 25 60 71} \email{Mahdi.Boukrouche@univ-st-etienne.fr} \address{Domingo A. Tarzia \newline Departamento Matem\'atica-CONICET, FCE, Univ. Austral, Paraguay 1950, S2000FZF Rosario, Argentina} \email{Domingo.Tarzia@fce.austral.edu.ar} \thanks{Submitted September 2006. Published February 22, 2007.} \subjclass[2000]{35R35} \keywords{Elliptic variational inequalities; convex combination of its solutions.} \begin{abstract} Let $u_{g_i}$ the unique solutions of an elliptic variational inequality with second member $g_{i}$ ($i=1,2$). We establish necessary and sufficient conditions for the convex combination $t u_{g_{1}}+ (1-t) u_{g_{2}}$, to be equal to the unique solution of the same elliptic variational inequality with second member $t g_{1}+ (1-t) g_{2}$. We also give some examples where this property is valid. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} In the linear problems for partial differential equations this property leads to the well known superposition principle which is classically used for example in Fourier series, variational equalities, etc. In these cases the linear combination (and also a convex combination) of two solutions of a linear problem, associated to two data, is also solution of the same problem with linear (convex) combination of the two data. However, in general, this property is not true for the solutions of nonlinear problems, for example for the variational inequalities. The variational inequality theory is fundamental in order to solve free boundary problems for partial differential equations, e.g. the dam problem \cite{BC78}; the one-phase Stefan problem \cite{duv73}; the obstacle problem \cite{KS80,HB72,JFR87}; the mathematical foundation of the finite element method \cite{PG02} and its corresponding numerical analysis \cite{RG84}. The goal of this paper is to give necessary and sufficiently condition to obtain that this property in valid for a convex combination of the solutions of elliptic variational inequalities. Let $V$ be an Hilbert space, $V'$ its topological dual, $K$ be a closed convex non empty set in $V$, $g_{i}$ in $V'$ for $i=1$ and $2$, and a bilinear form $a : V\times V\to \mathbb{R}$, which is \begin{itemize} \item symmetric: $a(u , v) = a(v , u)$ for all $(v , u)\in V\times V$, \item continuous: there exists $M>0$ such that $|a(v , u)|\leq M \|v\|_{V}|u\|_{V}$ for all $(v , u)\in V\times V$, \item coercive: there exists $m>0$ such that $|a(v , v)|\geq m \|v\|^{2}_{V}$ for all $v\in V$. \end{itemize} It is known \cite{Stamp64,JLL1,Kind80} that for each $g_{i}\in V'$ there exists a unique solution $u_{i}\in K$, namely $$\label{eq1} a(u_{i} , v-u_{i}) \geq \langle g_{i} , v- u_{i}\rangle \quad \forall v\in K \quad i=1 , 2\,,$$ where $< u \, ,\, v >$ denotes the duality brackets between $u\in V'$ and $v\in V$. Then we can consider $g_{i} \mapsto u_{i}= u_{g_{i}}$ as function from $V'$ to $V$. We want to establish necessary and sufficiently conditions for the convex combination $u_{3}(t)= t u_{1}+ (1-t) u_{2}$, with $t\in [0 , 1]$, to be the unique solution of the elliptic variational inequality \eqref{eq1} with second member $g_{3}(t)= t g_{1}+ (1-t) g_{2}$, such that $$\label{u3} u_{t g_{1}+ (1-t) g_{2}} = t u_{g_{1}}+ (1-t) u_{g_{2}} \quad \forall t\in [0 , 1]$$ if and only if \begin{gather}\label{alp} a(u_{g_{1}} , u_{g_{2}} - u_{g_{1}}) -\langle g_{1} , u_{g_{2}} - u_{g_{1}}\rangle = 0,\\ \label{bet} a(u_{g_{2}} , u_{g_{1}} - u_{g_{2}}) -\langle g_{2} , u_{g_{1}} - u_{g_{2}}\rangle = 0. \end{gather} This means also that if $g_{1}$ and $g_{2}$ are two points in $V'$ and $u_{g_{1}}$ and $u_{g_{2}}$ are the corresponding closest points in the closed convex $K$ then the closest point $u_{t g_{1}+ (1-t) g_{2}}$ to $t g_{1}+ (1-t) g_{2}$ is equal to $t u_{g_{1}}+ (1-t) u_{g_{2}}$ for all $t\in [0 , 1]$ if and only if $u_{g_{1}}-u_{g_{2}}$ is orthogonal to both $u_{g_{1}}- g_{1}$ and $u_{g_{2}}-g_{2}$. This paper is organized as follows. In Section 2 we establish some preliminary results which allow us, in Section 3, to prove our main result that (\ref{u3}) is equivalent to (\ref{alp}) and (\ref{bet}). We also give in Section 4 some examples where this property is valid. \section{Preliminary results} For $t\in [0 , 1]$ and $v\in K$, we define the function $f : [0 , 1]\times K \to \mathbb{R}$, by $$\label{eqf} f(t , v) = a(u_{3}(t), v - u_{3}(t)) -\langle g_{3}(t), v - u_{3}(t) \rangle$$ with $u_{3}(t)= t u_{1}+ (1-t) u_{2}$ and $g_{3}(t)= t g_{1}+ (1-t) g_{2}$. \begin{lemma}\label{L1} For all $t\in [0 , 1]$ and all $v \in K$ there exist $A$, $B(v)$, and $C(v)$ such that $$\label{eq2} f(t , v) = A t^{2} + B(v) t + C(v)$$ where, for $u_{i}$ is the unique solution of (\ref{eq1}) with given data $g_{i}\in V'$ ($i=1 , 2$) \begin{gather}\label{eq3} A = a( u_{1} - u_{2} , u_{1} - u_{2}) -\langle g_{1} - g_{2} , g_{1} - g_{2}\rangle \geq 0, \\ \label{eq4} B(v)= a( u_{1} - u_{2} , v - 2u_{2}) -\langle g_{1} - g_{2} , v- u_{2}\rangle -\langle g_{2} , u_{1} - u_{2}\rangle ,\\ \label{eq5} C(v)= a(u_{2} , v- u_{2}) -\langle g_{2} , v- u_{2}\rangle \geq 0 \quad \forall v\in K \end{gather} Moreover, we have $$\label{eq7} A + B(v) + C(v) \geq 0 \quad \forall v\in K.$$ \end{lemma} \begin{proof} Since $v= tv + (1-t)v$ for all $t$ in $[0 , 1]$, we can write $f$ as \begin{align*} f(t , v)&= a(t u_{1}+ (1-t) u_{2} , t(v-u_{1})+ (1-t)(v- u_{2}))\\ &\quad -\langle t g_{1}+ (1-t) g_{2} , t(v- u_{1})+ (1-t)(v- u_{2})\rangle\\ &=t^{2} [ a(u_{1} , v-u_{1}) + a(u_{1} , v-u_{2}) -a(u_{2} , v-u_{1}) \\ &\quad + a(u_{2} , v-u_{2}) -\langle g_{1} , v-u_{1}\rangle + \langle g_{1} , v-u_{2}\rangle \\ & \quad +\langle g_{2} , v-u_{1}\rangle -\langle g_{2} , v-u_{2}\rangle ] \\ &\quad +t [(a(u_{1} , v-u_{2}) + a(u_{2} , v-u_{1}) -2a(u_{2} , v-u_{2}) \\ &\quad -\langle g_{1} , v-u_{2}\rangle -\langle g_{2} , v-u_{1}\rangle + 2\langle g_{2} , v-u_{2}\rangle] \\ &\quad + \left[ a(u_{2} , v-u_{2}) -\langle g_{2} , v-u_{2}\rangle \right] \\ &= At^{2} + B(v)t + C(v). \end{align*} So we have \begin{align*} A &= [a(u_{1} , v-u_{1}) -\langle g_{1} , v-u_{1}\rangle ] -a(u_{2} , v-u_{1})+ \langle g_{2} , v-u_{1}\rangle \\ &\quad +[a(u_{2} , v-u_{2}) -\langle g_{2} , v-u_{2}\rangle ] - a(u_{1} , v-u_{2}) + \langle g_{1} , v-u_{2}\rangle \\ &= a(u_{1} -u_{2}, v- u_{1}) -\langle g_{1}-g_{2} , v -u_{1}\rangle + a(u_{2}-u_{1} , v-u_{2}) \\ &\quad -\langle g_{2}- g_{1} , v-u_{2} \rangle \\ &= a(u_{1} -u_{2}, u_{2}-u_{1})-\langle g_{1}-g_{2} , u_{2} -u_{1}\rangle \end{align*} and we remark that $A= [a(u_{1} , u_{2}-u_{1})-\langle g_{1} , u_{2} -u_{1}\rangle] + [a(u_{2}, u_{1}-u_{2})-\langle g_{2} , u_{1} -u_{2}\rangle]$ as $u_{i}$ ($i=1 , 2$) is solution of \eqref{eq1} with $g_{i}$ then $A\geq 0$ and does not depend on $v$. So (\ref{eq3}) holds. \begin{align*} B(v) &= a(u_{1} , v-u_{2})+ a(u_{2} , v-u_{1}) - 2a(u_{2} , v-u_{2}) \\ & \quad -\langle g_{1} , v-u_{2}\rangle -\langle g_{2} , v-u_{1}\rangle + 2\langle g_{2} , v-u_{2}\rangle\\ &= a(u_{1}-u_{2}, v-u_{2})+ a(u_{2} , (v-u_{1})-(v-u_{2})) \\ & \quad -\langle g_{1} -g_{2} , v-u_{2}\rangle -\langle g_{2} , (v-u_{1})-(v-u_{2})\rangle \\ &= a(u_{1}-u_{2}, v-u_{2})+ a(u_{2} , u_{2}-u_{1}) \\ & \quad -\langle g_{1} -g_{2} , v-u_{2}\rangle -\langle g_{2} , u_{2}-u_{1}\rangle \\ &= a(u_{1}-u_{2}, v-2u_{2})-\langle g_{1} -g_{2} , v-u_{2}\rangle -\langle g_{2} , u_{2}-u_{1}\rangle \end{align*} So (\ref{eq4}) holds. Also $C(v) = a(u_{2} , v-u_{2}) -\langle g_{2} , v-u_{2}\rangle \geq 0,$ so as $u_{2}$ is the solution of the variational inequality \eqref{eq1} with second member $g_{2}$, then we have (\ref{eq5}). Moreover $A + B(v) + C(v) = f(1 , v) = a(u_{1} , v-u_{1}) -\langle g_{1} , v-u_{1}\rangle \geq 0 \quad \forall v\in K,$ then (\ref{eq7}) holds. \end{proof} Now we define \begin{gather}\label{alpha1} \alpha = a(u_{1} , u_{2}-u_{1}) -\langle g_{1} , u_{2}-u_{1}\rangle,\\ \label{beta1} \beta = a(u_{2} , u_{1} -u_{2}) -\langle g_{2} , u_{1} -u_{2}\rangle. \end{gather} \begin{lemma}\label{L2} Let $\alpha$ and $\beta$ be defined by \eqref{alpha1} and \eqref{beta1} respectively. Then $\alpha \geq 0$, $\beta \geq 0$, and for all $\lambda \in [0 , 1]$ we have \begin{gather}\label{eq8} C(\lambda u_{1} + (1- \lambda)u_{2}) = \lambda \beta \geq 0, \\ \label{eq9} A= \alpha + \beta\geq 0, \\ \label{eq10} B(\lambda u_{1} + (1- \lambda)u_{2}) =-\lambda \alpha -(1+\lambda) \beta\leq 0. \end{gather} \end{lemma} \begin{proof} As $u_{i}$ is a solution of \eqref{eq1} with $g_{i}$ ($i=1 , 2$) then $\alpha \geq 0$ and $\beta \geq 0$. Taking, in (\ref{eq5}), $v= \lambda u_{1} + (1- \lambda)u_{2}$ with $\lambda$ in $[0 , 1]$, we obtain \begin{align*} C(v) &= a(u_{2} , \lambda u_{1} + (1- \lambda)u_{2} - u_{2}) -\langle g_{2} ,\lambda u_{1} + (1- \lambda)u_{2} - u_{2}\rangle \\ &= \lambda [a(u_{2} , u_{1}- u_{2}) -\langle g_{2} , u_{1} - u_{2}\rangle] = \lambda \beta\geq 0. \end{align*} From (\ref{eq3}) we deduce that $A= \alpha + \beta \geq 0$. Taking, in (\ref{eq4}) $v= \lambda u_{1} + (1- \lambda)u_{2}$ with $\lambda$ in $[0 , 1]$, we have \begin{align*} B(v) &= a(u_{1} , \lambda u_{1} + (1- \lambda)u_{2}- u_{2})+ a(u_{2} , \lambda u_{1} + (1- \lambda)u_{2}- u_{1}) \\ &\quad -2a( u_{2} , \lambda u_{1} + (1- \lambda)u_{2}- u_{2}) -\langle g_{1} , \lambda u_{1} + (1- \lambda)u_{2}- u_{2}\rangle \\ &\quad -\langle g_{2} , \lambda u_{1} + (1- \lambda)u_{2}- u_{1}\rangle + 2\langle g_{2} , \lambda u_{1} + (1- \lambda)u_{2}- u_{2}\rangle \\ &= -\lambda [a(u_{1} , u_{2}- u_{1})-\langle g_{1}, u_{2}- u_{1}\rangle] \\ &\quad -(1+\lambda) [a(u_{2} , u_{1}- u_{2})-\langle g_{2}, u_{1}- u_{2}\rangle] \\ &= -\lambda \alpha -(1+\lambda)\beta \leq 0. \end{align*} \end{proof} \section{Main result} In this section we give a positive answer to our question when the equality (\ref{u3}) is valid. \begin{theorem}\label{th1} We have $$\label{eq2a} u_{t g_{1}+ (1-t) g_{2}} = t u_{g_{1}}+ (1-t) u_{g_{2}} \quad \forall t\in [0 , 1]$$ if and only if $$\label{alpha} \alpha= a(u_{g_{1}} , u_{g_{2}} - u_{g_{1}}) -\langle g_{1} , u_{g_{2}} - u_{g_{1}}\rangle = 0$$ and $$\label{beta} \beta= a(u_{g_{2}} , u_{g_{1}} - u_{g_{2}}) -\langle g_{2} , u_{g_{1}} - u_{g_{2}}\rangle = 0,$$ where $u_{4}(t) = u_{t g_{1}+ (1-t) g_{2}}$ is the unique solution of the inequality \eqref{eq1} with the second member $g_{3}(t)=t g_{1}+ (1-t) g_{2}$ and $u_{3}(t) = t u_{g_{1}}+ (1-t) u_{g_{2}}$ is the convex combination for $t\in [0 , 1]$, of $u_{g_{1}}$ and $u_{g_{2}}$ which are solutions of the variational inequality \eqref{eq1}, respectively with second members $g_{1}$ and $g_{2}$. \end{theorem} \begin{proof} Suppose that $\alpha = \beta = 0$. Therefore, $A=\alpha + \beta=0$, so $$f(t , v) = B(v)t + C(v),$$ and from (\ref{eq5}) and (\ref{eq7}) we have $$f(0 , v) = C(v) \geq 0 \quad \text{and} \quad f(1 , v) = B(v) + C(v)\geq 0 \quad \forall v\in K,$$ then we deduce that $f(t , v) \geq 0$ for all $(t, v)\in [0 , 1]\times K$. Therefore $u_{3}(t)=t u_{g_{1}}+ (1-t) u_{g_{2}}$ is the unique solution of \eqref{eq1} with the second member $g_{3}(t)= t g_{1}+ (1-t) g_{2}$, so we get (\ref{eq2a}), by the uniqueness of the variational inequality \eqref{eq1} for a given $g_{3}(t)$. Suppose now that $u_{3}(t)= u_{4}(t), \quad \forall t\in [0 , 1]$. Then from (\ref{eqf}) and (\ref{eq2}) we have $$f(t , v) = A t^{2} + B(v) t + C(v)\geq 0, \quad \forall t\in [0 , 1], \quad \forall v\in K.$$ Taking $v=u_{1}=u_{g_{1}}$ (i.e. $\lambda =1$ in Lemma \ref{L2}), we obtain $$A= \alpha +\beta\geq 0, \quad B(u_{1}) = -\alpha -2\beta \leq 0, \quad C(u_{1}) = \beta\geq 0\,.$$ Thus the discriminant $\Delta (u_{1}) = B(u_{1})^{2}-4 AC(u_{1})$ of the quadratic function $t\mapsto f(t , v)$ is equal to $\alpha^{2}$. Then there exist two roots $t_{1}\geq 0$, $t_{2} \geq 0$ with $$\min(t_{1} , t_{2}) \geq 1 \quad \text{or} \quad \max(t_{1} , t_{2}) \leq 0.$$ Then the two roots $t_{1}= 1$ and $t_{2}=\frac{\beta}{\alpha +\beta}$ must not be in $]0 , 1[$ so $t_{2}$ must be equal to $1$, which gives $\alpha=0$. Taking now $v=u_{2}=u_{g_{2}}$ (i.e. $\lambda =0$ in Lemma \ref{L2}) we obtain $$A= \alpha +\beta\geq 0, \quad B(u_{2}) = -\beta \leq 0, \quad C(u_{2}) = 0$$ thus the corresponding discriminant is $\Delta (u_{2}) = B(u_{1})^{2}-4 AC(u_{1})= \beta^{2}$. Then there exist two roots $t_{1}\geq 0$, $t_{2} \geq 0$ with $$\min(t_{1} , t_{2}) \geq 1 \quad \text{or} \quad \max(t_{1} , t_{2}) \leq 0.$$ Then the two roots $t_{1}= 0$ and $t_{2}=\frac{\beta}{\alpha +\beta}$ must not be in $]0 , 1[$ so $t_{2}=\frac{\beta}{\alpha +\beta}$ must be equal to $0$, which give $\beta=0$. \end{proof} \begin{corollary}\label{co1} For $K=V$ the variational inequality \eqref{eq1} becomes the following variational equality $$u\in V : \quad a(u , v) = \langle g, v\rangle \quad \forall v\in V$$ thus $\alpha=\beta=0$ so $$\label{E} u_{tg_{1}+(1-t)g_{2}}= tu_{g_{1}}+(1-t)u_{g_{2}} \quad \forall t\in [0 , 1].$$ \end{corollary} \begin{remark} \rm Property \eqref{E} has been used in \cite{GT} to prove the strict convexity of the cost functional for optimal control problems. \end{remark} \section{Applications} Let $\Omega$ be an open set in $\mathbb{R}^{n}$, $V=L^{2}(\Omega)$, so $V'=V$ and the duality brackets $<\cdot \, , \, \cdot>$ becomes the scalar product in $V$ denoted by $(\cdot \, , \, \cdot)_{V}$. We use the usual notation $G^{+}= \max( G , 0)$ and $G^{-}=(-G)^{+}$, and $$G\perp F \Longleftrightarrow ( G, F )_{V} =0.$$ We have a preliminary result. \begin{lemma}\label{L3} Let $G_{i}\in L^{2}(\Omega)$ for $i= 1 , 2$, then we have $$G_{1}^{-}\perp G_{2}^{+} \quad \text{and} \quad G_{2}^{-}\perp G_{1}^{+} \Longleftrightarrow G_{1}(x) G_{2}(x) \geq 0 \quad \text{a.e. in } \Omega$$ \end{lemma} \begin{proof} Suppose that $G_{1}(x) G_{2}(x) \geq 0$ a.e. in $\Omega$. If $G_{1}(x) >0$ which means $G_{1}^{-}(x)=0$ and $G_{1}^{+}(x)=G_{1}(x)$, then $G_{2}(x) \geq 0$ implies $$G_{2}^{-}(x)=0 \quad \text{and} \quad G_{2}^{+}(x)=G_{2}(x);$$ thus $$G_{1}^{-}(x) G_{2}^{+}(x)=0 \quad \text{and} \quad G_{2}^{-}(x) G_{1}^{+}(x)=0.$$ If $G_{1}(x) <0$ which means $G_{1}^{+}(x)=0$ and $G_{1}^{-}(x)=G_{1}(x)$, then $G_{2}(x) \leq 0$ implies $$G_{2}^{+}(x)=0 \quad \text{and} \quad G_{2}^{-}(x)=G_{2}(x);$$ thus we have also $$G_{1}^{-}(x) G_{2}^{+}(x)=0 \quad \text{and} \quad G_{2}^{-}(x) G_{1}^{+}(x)=0.$$ So we have \begin{align*} (G_{1}^{-} , G_{2}^{+})_{V} &= \int_{\Omega}G_{1}^{-}(x) G_{2}^{+}(x) dx \\ &= \int_{\Omega\cap \{G_{1} >0\}}G_{1}^{-}(x) G_{2}^{+}(x) dx +\int_{\Omega\cap \{G_{1} <0\}}G_{1}^{-}(x) G_{2}^{+}(x) dx \\ &\quad +\int_{\Omega\cap \{G_{1} =0\}}G_{1}^{-}(x) G_{2}^{+}(x) dx = 0 \quad \Rightarrow G_{1}^{-}\perp G_{2}^{+}, \end{align*} and \begin{align*} (G_{2}^{-} , G_{1}^{+})_{V} &= \int_{\Omega}G_{2}^{-}(x) G_{1}^{+}(x) dx \\ &= \int_{\Omega\cap \{G_{1} >0\}}G_{2}^{-}(x) G_{1}^{+}(x) dx +\int_{\Omega\cap \{G_{1} <0\}}G_{2}^{-}(x) G_{1}^{+}(x) dx \\ &\quad +\int_{\Omega\cap \{G_{1} =0\}}G_{2}^{-}(x) G_{1}^{+}(x) dx = 0 \quad \Rightarrow G_{2}^{-}\perp G_{1}^{+}. \end{align*} Conversely we have \begin{align*} 0&= (G_{1}^{-} , G_{2}^{+})_{V} = \int_{\Omega}G_{1}^{-}(x) G_{2}^{+}(x)dx =\int_{\Omega\cap \{G_{1} <0\}}(-G_{1})(x) G_{2}^{+}(x) dx \\ &=\int_{\Omega\cap \{G_{1} <0\}\cap\{G_{2} >0\} }(-G_{1})(x) G_{2}(x) dx \end{align*} then $$(G_{1}^{-} , G_{2}^{+})_{V}=0 \Longrightarrow \begin{cases} |\{G_{1} <0\}\cap\{G_{2} >0\}| = 0, \\ G_{1}<0 \Rightarrow G_{2}^{+}=0 \Rightarrow G_{2}\leq 0\Rightarrow G_{1}G_{2}\geq 0\\ G_{2}>0 \Rightarrow G_{1}^{-}=0 \Rightarrow G_{1}\geq 0\Rightarrow G_{1}G_{2}\geq 0\\ \end{cases}$$ where $|\omega|$ is the measure of the set $\omega$. We have also \begin{align*} 0&= (G_{2}^{-} , G_{1}^{+})_{V} = \int_{\Omega}G_{2}^{-}(x) G_{1}^{+}(x)dx =\int_{\Omega\cap \{G_{1} >0\}}G_{1}(x) G_{2}^{-}(x) dx \\ &=\int_{\Omega\cap \{G_{1} >0\}\cap\{G_{2} <0\} }G_{1}(x) (-G_{2})(x) dx \end{align*} then $$(G_{2}^{-} , G_{1}^{+})_{V}=0 \Longrightarrow \begin{cases} & |\{G_{2} <0\}\cap\{G_{1} >0\}| = 0, \\ & G_{2}<0 \Rightarrow G_{1}^{+}=0 \Rightarrow G_{1}\leq 0\Rightarrow G_{2}G_{1}\geq 0\\ & G_{1}>0 \Rightarrow G_{2}^{-}=0 \Rightarrow G_{2}\geq 0\Rightarrow G_{1}G_{2}\geq 0.\\ \end{cases}$$ This completes the proof. \end{proof} \begin{example}\label{ex1} \rm Let $\psi\in L^{2}(\Omega)$, $V=L^{2}(\Omega)$, and $$K=\{v \in L^{2}(\Omega) : v\geq \psi\}, \quad a(u \, ,\, v) = (u \, ,\, v)_{V}.$$ We have here easily the existence and uniqueness of $u\in K$ such that $[a(u , v-u) \geq (g \, ,\, v-u)_{V} \quad\forall v\in K]\Leftrightarrow [(u -g \, ,\, v-u)_{V} \geq 0 \quad\forall v\in K]$ which is also equivalent to $$\label{eq14} u=P_{K}(g) = \max(g , \psi)= g+(\psi -g)^{+}= \psi+(g-\psi)^{+}.$$ \end{example} \begin{proof} From (\ref{eq14}), following \cite{Kind80}, we have $$u-g= (\psi -g)^{+}\quad \text{and}\quad v-u= v-\psi -(g-\psi)^{+}$$ so \begin{align*} (u -g, v-u)_{V}&= ((\psi -g)^{+} , (v -\psi) -(g-\psi)^{+})_{V}\\ &=((\psi -g)^{+} , v-\psi)_{V}- ((\psi-g)^{+} , (g-\psi)^{+})_{V} \end{align*} as $$((\psi-g)^{+} , (g-\psi)^{+})_{V} =((g-\psi)^{-} , (g-\psi)^{+})_{V}=0$$ then $$(u -g, v-u)_{V}= ((\psi -g)^{+} , v-\psi)_{V} = \int_{\Omega}(\psi -g)^{+} (v-\psi) dx \geq 0 , \quad \forall v\in K.$$ \end{proof} \begin{theorem} Let $V=L^{2}(\Omega)$, $\psi\in V$, $K=\{v\in V : v\geq \psi\}$, $a(u , v) = (u , v)_{V}$. For a given $g_{i}\in V$, $i=1 , 2$, we associate $$u_{i}= u_{g_{i}}= g_{i} +(\psi-g_{i})^{+}= \max(\psi , g_{i}).$$ Then we have $$\label{eq13} u_{tg_{1} + (1-t)g_{2}} = tu_{g_{1}} + (1-t)u_{g_{2}} \, \forall t\in[0 , 1] \Leftrightarrow (g_{1}-\psi) (g_{2}-\psi)\geq 0 \quad \text{ a.e. in } \Omega.$$ \end{theorem} \begin{proof} >From the definition of $\alpha$ we have $$\alpha= a(u_{1} , u_{2}- u_{1}) - ( g_{1}, u_{2}- u_{1})_{V} = (u_{1}-g_{1}, u_{2}- u_{1})_{V}.$$ From $$u_{1}-g_{1} = (\psi -g_{1})^{+}, \quad u_{2}- u_{1}= (g_{2} -\psi)^{+} -(g_{1}-\psi)^{+},$$ we have \begin{align*} \alpha&= ((\psi -g_{1})^{+}, (g_{2} -\psi)^{+} -(g_{1}-\psi)^{+})_{V} \\ &= ((\psi -g_{1})^{+}, (g_{2} -\psi)^{+})_{V} - ((\psi -g_{1})^{+}, (g_{1}-\psi)^{+})_{V}\\ &= \int_{\Omega}(\psi -g_{1})^{+}(x)(g_{2} -\psi)^{+}(x)dx \\ &= \int_{\Omega}(g_{1}-\psi)^{-}(x) (g_{2} -\psi)^{+}(x)dx, \end{align*} then we deduce that $\alpha =0 \Leftrightarrow \int_{\Omega}(g_{1}-\psi)^{-}(x) (g_{2} -\psi)^{+}(x)dx =0 \Leftrightarrow (g_{1}-\psi)^{-}\perp (g_{2} -\psi)^{+}.$ We have also from the definition of $\beta$ that $$\beta= a(u_{2} , u_{1}- u_{2}) - ( g_{2}, u_{1}- u_{2})_{V} = (u_{2}-g_{2}, u_{1}- u_{2})_{V}$$ and from $$u_{2}-g_{2} = (\psi -g_{2})^{+} \quad u_{1}- u_{2}= (g_{1} -\psi)^{+} -(g_{2}-\psi)^{+}$$ we have \begin{align*} \beta&= ((\psi -g_{2})^{+}, (g_{1} -\psi)^{+} -(g_{2}-\psi)^{+})_{V} \\ &= ((\psi -g_{2})^{+}, (g_{1} -\psi)^{+})_{V} - ((\psi -g_{2})^{+}, (g_{2}-\psi)^{+})_{V}\\ &= \int_{\Omega}(\psi -g_{2})^{+}(x) (g_{1} -\psi)^{+}(x)dx\\ &= \int_{\Omega}(g_{2}-\psi)^{-}(x)\cdot (g_{1} -\psi)^{+}(x)dx, \end{align*} then we deduce that \begin{align*} \beta =0 \Leftrightarrow \int_{\Omega}(g_{2}-\psi)^{-}(x) (g_{1} -\psi)^{+}(x)dx =0 \Leftrightarrow (g_{2}-\psi)^{-}\perp (g_{1} -\psi)^{+}. \end{align*} Using now Lemma \ref{L3}, with $G_{i}=g_{i} -\psi$, we deduce that $$\left. \begin{array}{c} (g_{2}-\psi)^{-}\perp (g_{1} -\psi)^{+}\\ \text{and}\\ (g_{1}-\psi)^{-}\perp (g_{2} -\psi)^{+} \end{array} \right\} \Leftrightarrow (g_{1}-\psi) (g_{2}-\psi)\geq 0 \quad \text{a.e. in }\Omega.$$ Moreover $$(g_{2}-\psi)^{-}\perp (g_{1} -\psi)^{+}\quad \text{and}\quad (g_{1}-\psi)^{-}\perp (g_{2} -\psi)^{+} \Leftrightarrow \alpha=\beta =0,$$ and with Theorem \ref{th1} we have $$\alpha=\beta =0 \Leftrightarrow u_{tg_{1} + (1-t)g_{2}} = tu_{g_{1}} + (1-t)u_{g_{2}} \quad \forall t\in[0 , 1],$$ which gives us the equivalence (\ref{eq13}) and completes the proof. \end{proof} \begin{example}\label{ex2} \rm Let us consider the following free boundary problem of obstacle type \cite{Kind80}, $$\label{IV} \begin{gathered} V=H^{1}(\Omega), \quad V_{0}=H^{1}_{0}(\Omega), \quad H=L^{2}(\Omega), \quad K=\{v\in V_{0} : v\geq 0\}, \\ L(v)=(g \, ,\, v)_{H}, \quad \text{and} \quad a(u \, ,\, v) = \int_{\Omega}\nabla u\nabla v dx, \\ u\in K : \quad a(u \, ,\, v-u) \geq (g \, ,\, v-u)_{H} \quad \forall v\in K. \end{gathered}$$ \end{example} We recall here some usual notation. $a$ is a bilinear symmetric, coercive and continuous form, there exist $m,M$, such that $$m\|v\|^{2}_{V} \leq \||v\||^{2}_{V}= a(v \, ,\, v) \leq M \|v\|^{2}_{V}$$ and $(( \cdot \, ,\, \cdot )) : V\times V \to \mathbb{R}$ such that $$((u , v))= a(u , v) \quad \forall (u , v) \in V\times V$$ is the inner scalar product in $V$. $L$ is linear and continuous form on a Hilbert space $V$, and also on $V_{0}$. Then by Riesz theorem there exists a unique $g^{\star}\in V$ such that $$(g \, ,\, v)_{H}= ((g^{\star}, v))_{V} \quad \forall v\in V.$$ then (\ref{IV}) is equivalent to $$\label{IV1} u\in K : \quad ((u , v-u))_{V} \geq ((g^{\star} , v-u))_{V } \quad \forall v\in K.$$ Therefore, $u=\mathcal{P}_{K}(g^{\star})$ is the projection of $g^{\star}$ on $K$ with the norm $\||\cdot\||_{V}$. \noindent Remark that $V_{0} \hookrightarrow H\hookrightarrow V'$ so this exemple is a particular cas of (\ref{eq1}). \begin{lemma}\label{L4} With the above notation, we have $$g\geq 0 \quad\text{in} \quad \Omega \Longrightarrow g^{\star}\geq 0 \quad \text{in} \quad \Omega \Longrightarrow u=\mathcal{P}_{K}(g^{\star})= {g^{\star}}^{+}= g^{\star}.$$ \end{lemma} \begin{proof} As $g^{\star}= {g^{\star}}^{+}- {g^{\star}}^{-}$, then \begin{align*} (({g^{\star}}^{+} - g^{\star} , v- {g^{\star}}^{+}))_{V} &= (({g^{\star}}^{-} , v- {g^{\star}}^{+}))_{V} =(({g^{\star}}^{-} , v))_{V} -(({g^{\star}}^{-} , {g^{\star}}^{+}))_{V} \\ &= a( ({g^{\star}}^{-} , v)_{V}- a(({g^{\star}}^{-} , {g^{\star}}^{+})_{V} =\int_{\Omega}\nabla {g^{\star}}^{-} \nabla v dx. \end{align*} From $$\int_{\Omega}\nabla {g^{\star}} \nabla v dx = ((g^{\star}, v))_{V} =(g \, ,\, v)_{H}\quad \forall v\in V_{0} \quad \text{and} \quad g\in H(\Omega),$$ by Green formula we have the representation of $g^{\star}$ given by $$\label{eq16} -\Delta g^{\star}= g \quad \text{in } \Omega \quad \text{and} \quad g^{\star}_{|\partial\Omega}= 0.$$ Since $g\geq 0$ in $\Omega$, by the maximum principle we have $g^{\star}\geq 0$ thus ${g^{\star}}^{-}\equiv 0$, so $$(({g^{\star}}^{+} - g^{\star} , v- {g^{\star}}^{+}))_{V} = \int_{\Omega}\nabla {g^{\star}}^{-} \nabla v dx = 0$$ which gives $u={g^{\star}}^{+}$. \end{proof} \begin{theorem}\label{th3} Let $V=H^{1}(\Omega)$, $H=L^{2}(\Omega)$, $K=\{v\in V : v\geq 0\}$, $a(u , v) = (\nabla u , \nabla v)_{H^{n}}$. For a given $g_{i}\geq 0$ in $H$, $i=1 , 2$, we have $u_{i}= u_{g_{i}}= g^{\star}_{i}$ and $$\label{eq1A} u_{tg_{1} + (1-t)g_{2}} = tu_{g_{1}} + (1-t)u_{g_{2}} \quad \forall t\in \, [0 , 1].$$ \begin{proof} As $g_{i}\geq 0$ in $H$, then $u_{i}= u_{g_{i}}= g^{\star}_{i}$, for $i=1, 2$. Moreover \begin{align*} \alpha &= a(u_{1} , u_{2} - u_{1}) -(g_{1} , u_{2} - u_{1})_{H} =a(g^{\star}_{1} , g^{\star}_{2} - g^{\star}_{1}) -(g_{1} , g^{\star}_{2}- g^{\star}_{1})_{H} \\ &= ((g^{\star}_{1} , g^{\star}_{2} - g^{\star}_{1}))_{V} -(g_{1} , g^{\star}_{2}- g^{\star}_{1})_{H} \end{align*} and by the Riesz theorem $\alpha = ((g^{\star}_{1} , g^{\star}_{2} - g^{\star}_{1}))_{V} -(g^{\star}_{1} , g^{\star}_{2} - g^{\star}_{1})_{V}=0.$ We have \begin{align*} \beta &= a(u_{2} , u_{1} - u_{2}) -(g_{2} , u_{1} - u_{2})_{H} =a(g^{\star}_{2} , g^{\star}_{1} - g^{\star}_{2}) -(g_{2} , g^{\star}_{1}- g^{\star}_{2})_{H} \\ &= ((g^{\star}_{2} , g^{\star}_{1} - g^{\star}_{2}))_{V} -(g_{2} , g^{\star}_{1}- g^{\star}_{2})_{H} \end{align*} and also by the Riesz theorem $\beta = ((g^{\star}_{2} , g^{\star}_{1} - g^{\star}_{2}))_{V} -(g^{\star}_{2} , g^{\star}_{1} - g^{\star}_{2})_{V}=0.$ So by Theorem \ref{th1} we obtain \eqref{eq1A} and we finish the proof. \end{proof} \end{theorem} \begin{remark} \rm In the case $\Omega=]0 , 1[\subset \mathbb{R}$, (\ref{eq16}) becomes $$-{g^{\star}}''(x) = g(x) \quad in \quad ]0 , 1[, \quad g^{\star}(0)=g^{\star}(1)=0.$$ and we can obtain the explicit expression of $$g^{\star}(x)=x\int_{x}^{1} g(t)(1-t) dt + (1-x)\int_{x}^{1}tg(t)dt$$ so we have $$g(t) \geq 0 \quad in \quad [0 , 1] \Longrightarrow g^{\star}(x) \geq 0 \quad in \quad [0 , 1].$$ From Lemma \ref{L4} and Theorem \ref{th3} we deduce that $u_{g}=g^{\star}$. 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