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Let $\Omega$ be a domain of $R^n,$ let $\omega$ be open subset of $\Omega$ and let $\theta \in W^{2,\infty}(\omega).$

I am wondering about the existence of a function $\tilde{\theta} \in W^{2,\infty}(\Omega)$ (eventually, under some conditions on the value of $\theta$ on $\partial \omega) $ such that :

1) $\tilde{\theta}=\theta $ on $\omega,$

2) $\Delta \tilde{\theta}=0$ on $\Omega-\omega.$

Thanks!

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    You need to say more about the relation between $\omega$ and $\Omega$. For example, if $\omega$ is dense in $\Omega$, then the extension is unique (if it exists at all), and there is nothing we can do about its Laplacian... Do the domains have smooth boundary? Is the closure of $\omega$ contained in $\Omega$? In what sense do you want the Laplacian to vanish? (on an open set Weyl's lemma removes any ambiguity, but $\Omega\setminus\omega$ is not open in general.2012-08-05
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    In my question we can assume that the closure of $\omega$ is contained in $\Omega,$ and take the equality in 1) in the closure of $\omega$ and the one of 2) in $\Omega-\overline{\omega}$, which is open.2012-08-06
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    $\omega $ has a smooth boundary and the Laplacian is in the classical sense.2012-08-06

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