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I've been told that the answer is no, but I'm having a hard time finding a reference. More precisely, I'm interested in the following. Let $\Omega \subset \mathbb{R}^n$ be a bonded open connected set, say with $C^{\infty}$ smooth boundary and

$L=-\sum_{i,j}\frac{\partial}{\partial_{x_j}}\left(a_{i,j}(x)\frac{\partial}{\partial_{x_i}}\right)+\sum_i a_i(x)\frac{\partial}{\partial_{x_i}}+a_0(x)$

a differential operator satisfiying $a_{i,j},~a_i \in C^{\infty}(\overline{\Omega})$ and the uniform ellipticity condition,

$\exists \lambda >0,~\forall x \in \Omega,~\forall \xi \in \mathbb{R}^n ,~\sum_{i,j}a_{i,j}(x)\xi_i \xi_j \ge \lambda |\xi|^2.$

Elliptic regularity estimates tell us that if $u$ is a solution to the Dirichlet problem $Lu=0$ in $\Omega$ and $u=0$ on $\partial \Omega$ then $u \in C^{\infty}(\overline{\Omega})$. What I would like to know is: assume that $x\in \Omega$ and that all derivatives of $u$ at $x$ are zero. Is $u$ zero on $\Omega$? And what if $x \in \partial \Omega$? My motivation is to understand the description of the nodal set of an eigenfuction of the Laplace operator.

Like I said, I'm quite sure this has been done, but I don't know where and I would be grateful if someone could point me in the right direction.

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To add to Piero's answer: the keyphrase you should search for in this case is "strong unique continuation". –  Willie Wong Jun 17 '11 at 13:52
    
...and maybe 'Aronszajn and Cordes' –  Piero D'Ancona Jun 17 '11 at 14:31

2 Answers 2

up vote 7 down vote accepted

Take a look at this paper by Protter; there are certainly newer references (which you can easily find starting from this one), but basically the problem was already solved at the time.

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This seems to be what I am looking for. Thank you very much for your help. –  mathapprentice Jun 17 '11 at 16:58

In your assumption, the solution itself must be zero. This is called the strong uniqueness of the solutions to ellpitic equations. You can find reference by searching "strong uniqueness" or "strong unique continuation property" in google.

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