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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.

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    You just changed it! Now it makes sense.2012-08-11

1 Answers 1

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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\}). $

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    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, $2012-08-18