I have the following question: Suppose $\Omega$ is a bounded open subset of $\mathbb R^d$ with Lipschitz boundary. Say $g\in H^{1/2}(\partial\Omega)$ is known.

  1. Can I then write down a function $\tilde g\in H^1(\Omega)$ which satisfies $\tilde g\Big|_{\partial\Omega}=g$? Obviously such a $\tilde g$ will not be unique in general if it exists, but that's ok.

  2. If the previous question is answered in the affirmative, can you name some methods by which to construct such a $\tilde g$ from knowing $g$ and $\Omega$? I don't need a full answer to this question; I'm just curious to know how one can do it.


  • $\begingroup$ This question may be of some help: mathoverflow.net/questions/70740/image-of-the-trace-operator $\endgroup$ – Umberto P. Aug 5 '16 at 17:04
  • $\begingroup$ Thanks a lot. So it seems that the answer to question 1 is in the affirmative. That is great to know! Do you happen to know how one may go about "recovering" such an extension? $\endgroup$ – Lentes Aug 5 '16 at 17:18

As Umberto said, the the answer to the first question is "yes". In fact, $H^{1/2}(\partial \Omega)$ is the image of $H^1(\Omega)$ under the trace operator $T$.

For the second one, you can solve $$-\Delta u = 0 \quad \text{in } \Omega\,, \quad u = g \quad \text{on } \partial\Omega\,.$$

Otherwise, here is a way how to construct another approximate extension in the finite element framework. Let assume that $\Omega$ is a polygon/polyhedron/... . Then

  1. mesh it with simpleces (triangles, tetrahedra, ... , depending on $d$)
  2. create a piecewise linear approximation $g^h$ of $g$ on $\partial\Omega$ via interpolation
  3. extend $g^h$ to the volume by assigning its coefficients to linear Lagrangian basis functions that "live" in the volume

This way you can build a function $\tilde{g}^h\in H^1(\Omega)$ that satisfies $$\Vert T(\tilde{g}^h)-g \Vert_{H^{1/2}} = \mathcal{O}(h)\,.$$

If $\Omega$ has no flat facets, you can still use this procedure, but the first step (meshing the domain) introduces an additional error in the analysis.

  • $\begingroup$ Thank you for such a complete answer! $\endgroup$ – Lentes Aug 6 '16 at 15:08

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