Data: $\Omega \subset \mathbb{R}^{n}$ is an open connected (may be unbounded) set, and locally $\partial \Omega$ is aLipschitz graph. $S \subset \partial \Omega$ is measurabel and $H^{n-1}(S)>0.$ The Dirichlet data on $S$ are given by non-negative function $u^0 \in ^{1}_{Loc}(\Omega)$ with $\nabla u^0 \in L^{2}(\Omega)$. The given force function $Q$ is non-negative and measurable.

Consider the convex set \begin{equation} K:=\{ v \in L^{1}_{Loc}(\Omega): \nabla v \in L^{2}(\Omega) \quad \mbox{and} \quad v=u^0 \quad \mbox{on} S\}. \end{equation} We are looking for an absolute minimum of the functional \begin{equation} J(v):= \int_{\Omega}(|\nabla v|^{2} + \chi(\{v>0\})Q^2) \end{equation} in the class $K$.

Definition: We call $u \in K$ a local minimum if for some smal $\varepsilon>0$ we have $J(u)\le J(v)$ for every $v \in K$ with \begin{equation} \|\nabla (u-v)\|_{L^{2}(\Omega)} + \| \chi(\{v>0\}) -\chi(\{u>0\})\|_{L^{1}(\Omega)} \le \varepsilon. \end{equation}

Lemma: If $u$ is a minimum local, then $u$ is subharmonic, hence we can assume that \begin{equation} u(x) = \lim_{r\downarrow 0} \oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega, \end{equation} where $\oint $ denotes the mean value.

Proof: For non-negative functions $\xi \in C^{\infty}_{0}(\Omega)$ we have

\begin{equation} 0 \le \limsup_{\varepsilon\downarrow 0} \dfrac{1}{2\varepsilon} (J(u- \varepsilon \xi) - J(u)) \le - \int_{\Omega} \nabla \xi \nabla u, \end{equation} that is, $u$ is subharmonic. Then the limit in the assertuion exists for every $x \in \Omega$, and coincides with $u(x)$ for almost all $x$.

Lemma: If $u$ is a minimum local, then $0\le u\le\sup_{\Omega}u^0$.

Proof:For $|\varepsilon|\le 1$ use $u_\varepsilon:=u-\varepsilon \min (u,0)$ and $u_\varepsilon:=u+\varepsilon \min (\sup_{\Omega}u^0-u,0)$ as a first variation.

  1. The lemma suggests that $u$ is subharmonic if for all non-negative $\xi \in C^{\infty}_{0}(\Omega)$ we have \begin{equation} \int_{\Omega} \nabla \xi \nabla u \le 0 \end{equation} I'd like to know what the relation between this definition of subharmoninic and others. For example, by trudinger, if $u$ is subharmonic we have \begin{equation} u(x) \le \lim_{r\downarrow 0} \oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega. \end{equation} Please, correct me if I am wrong.
  2. In the last lemma, What is a first variariation?

3.\begin{eqnarray} \limsup_{\varepsilon \downarrow 0}\dfrac{1}{2\varepsilon} (J(u- \varepsilon \xi) - J(u))&=& \limsup_{\varepsilon \downarrow 0}\left \{\dfrac{1}{2\varepsilon}( \int_{\Omega} -2 \varepsilon \nabla \xi \nabla u + \varepsilon^2 \nabla \xi + \chi(\{u-\varepsilon \xi> 0\}) - \chi(\{u>0\}))\right \} \\ &\le & \limsup_{\varepsilon \downarrow 0}\left \{\dfrac{1}{2\varepsilon}( \int_{\Omega} -2 \varepsilon \nabla \xi \nabla u + \varepsilon^2 \nabla \xi \right \}\\ & & + \limsup_{\varepsilon \downarrow 0}\left \{ \chi(\{u-\varepsilon \xi>0\}) - \chi(\{u>0\}))\right \} \\ &\le& \int_{\Omega} -\nabla \xi \nabla u + \limsup_{\varepsilon \downarrow 0}\dfrac{1}{2\varepsilon}\left \{ \chi(\{u-\varepsilon \xi>0\}) - \chi(\{u>0\}))\right \} \end{eqnarray} Am I right here? Why $\limsup_{\varepsilon \downarrow 0}\dfrac{1}{2\varepsilon}\left \{ \chi(\{u-\varepsilon \xi>0\}) - \chi(\{u>0\}))\right \} \le 0$?

If you want the details can be found in the article Alt, H. M. and Caffarelli, L. A. Existence and regularity for a minimum problem with free boundary. J. Reine Angew. Math., 325, (1981), 105–144.. I thank any hint.

  • $\begingroup$ what is $\chi(\{u-\varepsilon\xi\})$? $\endgroup$ – Mercy King Aug 10 '12 at 23:28
  • $\begingroup$ I'm not sure, but I Think is characteristic function of $u-\varepsilon \xi$. $\endgroup$ – user29999 Aug 11 '12 at 0:52
  • $\begingroup$ $u-\varepsilon\xi$ isn't a set! $\endgroup$ – Mercy King Aug 11 '12 at 1:01
  • $\begingroup$ I was wrong, is $\{u -\varepsilon \xi > 0\}$. Haven't you noticed? $\endgroup$ – user29999 Aug 11 '12 at 13:25
  • $\begingroup$ You just changed it! Now it makes sense. $\endgroup$ – Mercy King Aug 11 '12 at 13:47

1). Note that this is just a weak form of subharmonicity: If $u$ was smooth, then $$ \int\nabla\xi\nabla u = -\int\xi\Delta u, $$ and so the definition coincides with the pointwise definition of subharmonicity (i.e., $\Delta u\geq0$).

Subharmonicity in Gilbarg-Trudinger is $$ u(x) \le \oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega, $$ and for all sufficiently small $r>0$.

2). Here a first variation means you plug $u_\varepsilon$ into $J(\cdot)$ and use that $J(u_\varepsilon)\geq J(u)$.

3). Your manipulation is correct, except there should be $\varepsilon^2|\nabla\xi|^2$ instead of $\varepsilon^2\nabla\xi$. For the last question, since $\varepsilon>0$ and $\xi\geq0$, we have $u-\varepsilon\xi\leq u$, so $$ \chi(\{u-\varepsilon\xi>0\})\subset\chi(\{u>0\}). $$

  • $\begingroup$ First, thank you for your answer I accept your anwer 3. However, what I meant in question 1 was by trudinger $u$ is subharmonic when $$ u(x) \le \oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega, $$ and for all sufficiently small $r>0$. Hence \begin{equation} u(x) \le \lim_{r \downarrow0} \oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega, \end{equation} if the limit exists, and the article says that if $u$ is subharmonic, then the limits exists and $$ u(x) \le \lim_{r\downarrow 0}\oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega, $$ $\endgroup$ – user29999 Aug 18 '12 at 1:33
  • $\begingroup$ I'd like to know exactly too, what is the definition of subharmonic function for them. Seens that is $u \in $(some space) is subharmonic if for all $\xi \in C^{\infty}_{0}(\Omega)$ we have $$ 0 \le -\int_{\Omega} \nabla \xi \nabla u$$ and compare with the definition in Gilbarg-trudinger. I accept your answer 2 too. But I do another ask. How to prove the last lemma. What I have tried was, in $\{u\ge0\}, u_\varepsilon=u$.Hence, $J(u_\varepsilon)\ge J(u)$ implies $$\int_{\{u\le 0\}}|\nabla u_\varepsilon|^2 -|\nabla u|^2 +Q^2(\chi_{\{u_\varepsilon>0\}}-\chi_{\{u>0\}} \ge 0$$ $\endgroup$ – user29999 Aug 18 '12 at 1:41
  • $\begingroup$ and as in $\{u\le0\}$ we have $u_\varepsilon = (1-\varepsilon)u\le0$ because |\varepsilon|\le 1 we have \begin{eqnarray} \int_{\{u\le 0\}}|\nabla u(1-\varepsilon)|^2 -|\nabla u|^2 &\ge& 0 \end{eqnarray} On the other hand we have \begin{eqnarray} \int_{\{u\le 0\}}|\nabla u(1-\varepsilon)|^2 -|\nabla u|^2 &\le& 0. \end{eqnarray} Then $|\nabla u|= 0 $ a.e and $u$ is constant a.e equals to 0 in $\{u<0\}$. This show that $u\ge0$. Is this right? $\endgroup$ – user29999 Aug 18 '12 at 1:41
  • $\begingroup$ @user29999: Yes it is correct. The intuition behind this is that if $u$ gets below zero, you lift the below-zero part slightly so that the gradient gets smaller in that part. $\endgroup$ – timur Aug 18 '12 at 2:08
  • $\begingroup$ I made a typo in my first comment. The correction is: and the article says that if $u$ is subharmonic $$ u(x) = \lim_{r\downarrow0} \oint_{B_r(x)}u \quad \mbox{for} \quad x \in \Omega, $$ $\endgroup$ – user29999 Aug 18 '12 at 17:37

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