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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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    $\omega $ has a smooth boundary and the Laplacian is in the classical sense.2012-08-06

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Counterexample in two dimensions, using complex notation: $\omega=\{z\in\mathbb C:|z|<1\}$, $\Omega=\{z\in\mathbb C:|z|<3\}$, $\theta(z)=\mathrm{Re}\,\frac{1}{z-2}$.

Indeed, suppose $\tilde \theta$ exists. Since it is in $W^{2,\infty}(\Omega)$, its second-order weak derivatives are functions that can be calculated by differentiating $\tilde\theta$ pointwise. In particular, $\Delta \tilde\theta=0$ holds in the weak sense because it holds a.e. By Weyl's lemma $\tilde \theta $ is harmonic in $\Omega$ in the classical sense. But then it must be different from $\mathrm{Re}\,\frac{1}{z-2}$ somewhere in $\Omega$, contradicting the uniqueness theorem for harmonic functions. QED

The obstruction lies in that $W^{2,\infty}$ requirement prescribes both the values and the normal derivative of $\tilde \theta$ on $\partial\omega$. This means you are trying to solve the Cauchy problem for an elliptic equation, famously shown to be ill-posed by Hadamard.