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Thread: Maximum Principle

  1. #1
    Junior Member HAL9000's Avatar
    Aug 2010

    Question Maximum Principle

    The maximum principle for harmonic functions asserts:
    Let $\displaystyle u \in C^2(U) \cap C(\bar{U})$ be harmonic in $\displaystyle U \subset \mathbb{R}^n$.
    1) (Weak maximum principle)

    $\displaystyle \max_{\bar{U}}u=\max_{\partial U}u$;

    2) (Strong maximum principle)

    If $\displaystyle U$ is connected and there exists a point $\displaystyle x_0 \in U$ such that

    $\displaystyle u(x_0)=\max_{\bar{U}}u$,

    then $\displaystyle u=const$ in $\displaystyle U$.

    I wonder if I got this right. The weak principle means that the maximal value on $\displaystyle \bar{U}$ of a harmonic function is necessarily attained in a boundary point, i.e. there are no solely interior points which give maximal values for a harmonic function. Although it may be that there are points $\displaystyle x \in \partial U$ and $\displaystyle y \in U$ such that

    $\displaystyle u(y)=\max_{\bar{U}}u=u(x)$,

    in which case the shape of $\displaystyle U$ may cause the function to be constant (i.e. in case $\displaystyle U$ is connected and so 2) holds); but if $\displaystyle U$ is any open set nothing specific is known about $\displaystyle u$, i.e. it may rather be non-constant.

    I wonder why the conclusion of the strong version holds just for $\displaystyle U$ and not for $\displaystyle \bar{U}$. I mean 1) holds anyway and $\displaystyle u \in C(\bar{U})$. Doesn't this, together with 2), imply that $\displaystyle u$ is constant on the boundary? The proof of the strong version, however, works only with $\displaystyle U$, so that it's no big deal to accept it as it is.
    Last edited by HAL9000; Jan 16th 2013 at 08:48 AM.
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