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Suppose $u$ is harmonic in the interior of the unit square $0 \leq x \leq 1$, $0\leq y\leq1$. Suppose furthermore that $u$ and its first derivatives continuously extend to the bottom side $0\leq x \leq 1$, $y=0$ and that $u = \partial_y u = 0$ there. Is it true that $u \equiv 0$? If so, can this result be extended to the unit n-cube with 'side' above replaced with the n-1-cube obtained by fixing a coordinate to 0?

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It is true at least if we assume that $u$ extends harmonically across the bottom side.

Assume that it does. In that case, there is a harmonic conjugate $v$, such that $u+iv$ is holomorphic on a neighbourhood of the square with the bottom side added. By Cauchy-Riemann, $v'_y = u'_x = 0$ and $v'_x = -u'_y = 0$ on $A = \{ 0 \le x \le 1, y = 0 \}$, so $v$ is constant on $A$. Hence $f = u+iv$ is constant on $A$ and by the identity theorem for holomorphic functions, $f$ (and therefore $u$) is constant on the square.

For the general case, start by extending $u$ to a harmonic function on $0 \le x \le 1$, $-1 \le y \le 1$ using Schwarz' reflection principle, i.e. let $u(x,y) = -u(x,-y)$ for negative $y$.

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Any leads for the general case? –  user18063 Nov 6 '12 at 9:28
    
@user18063 How's that? –  mrf Nov 6 '12 at 9:51
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Well, reflection is the way to go. However, we don't need Schwarz' reflection principle per se, we can simply 0-extend and end up with a $C^1$ function. Then using the mean value property characterization of harmonic functions, we see that the extension is harmonic on the face of the cube where $u$ and $u_y$ vanish. –  user18063 Nov 11 '12 at 20:14

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