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Let $u(x)$ be a solution to the problem $\Delta u(x) = u(x)$ on a compact domain with smooth boundary. Furthermore demand that $u(x)=0$ on the boundary. Is there an easy argument why $u(x)$ has to be zero everywhere? I can prove it using the spectral theorem but it seems to be overkill and not so enlightening to me.

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up vote 1 down vote accepted

Assuming you want a solution that is twice differentiable and continuous up to the boundary (else you can just take any solution and "set" the boundary value to be zero...)

Multiply by (the complex conjugate) $\bar{u}$ (if $u$ is real valued, just take $\bar{u} = u$) and integrate (by parts)

$$ 0 \geq -\int_\Omega |\nabla u|^2 dx = \int_\Omega \bar{u} \triangle u dx = \int u^2 \geq 0$$

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Thank you! That's exactly what I was looking for. – user3620 Nov 25 '11 at 17:54

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