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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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    $\begingroup$ Which special cases have you already tried to do? What have you already tried doing? See mathoverflow.net/howtoask $\endgroup$
    – Yemon Choi
    Commented Aug 3, 2012 at 4:31
  • $\begingroup$ I sudied the case $n=1,\, \Omega = (a,b)$ and $ \omega=(c,d).$ Then I complete $\theta$ on $\Omega$ by the tangent equtions $\theta'(c)(x-c)+\theta(c), $ for $a<x<c $ and $ \theta'(d)(x-d)+\theta(d),$ for $d<x<b.$ $\endgroup$
    – hardy
    Commented Aug 3, 2012 at 14:24
  • $\begingroup$ $W^{2,\infty}$-estimates in dimension $n\ge 2$ are not always true without assuming further regularity on the other terms, so dimension $1$ might not generalize so nicely. As regards your problem, $\tilde\theta$ may only depends on the $\theta$ and $\frac{\partial\theta}{\partial n}$ at the boundary $\partial\omega$. The values inside $\omega$ don't matter so much as long as they match with the traces I mentioned. $\endgroup$
    – Francois Monard
    Commented Aug 3, 2012 at 16:16
  • $\begingroup$ If we try to extend the above idea to the case $n=2$ : for $x\in \Omega-\omega,$ we consider the normal to $\partial\omega$ at $\xi$ and passing from $x$. Then we set $ \tilde{\theta}(x)=\theta'(\xi)(x-\xi)+\theta(\xi).$ $\endgroup$
    – hardy
    Commented Aug 5, 2012 at 1:34
  • $\begingroup$ Cross-posted to MSE: math.stackexchange.com/questions/179109/… $\endgroup$
    – Yemon Choi
    Commented Aug 13, 2012 at 4:53

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