Let $\Omega _1$ and $\Omega_2$ are smooth open sets and $A= \partial \Omega_1 \cap \partial \Omega_2.$ Let $\Omega ^\prime = \Omega_1 \cup \Omega_2 \cup A$ be an open set. Let $f$ is defined on $\Omega^\prime$ such that $ f |_{\Omega_1 }:=g_1 \in H^1 (\Omega_1 )$ and $ f |_{\Omega_2}:=g_2 \in H^1 (\Omega_2 )$. Let the traces of $g_1$ and $g_2$ agree on A. Then can we say that $f \in H^1(\Omega^\prime)$?


Then answer is yes. Here is a proof: For each $j=1,\ldots,n$ put $f_j:=\partial_jg_1 \chi_{\Omega_1}+\partial_j g_2 \chi_{\Omega_2}$. Clearly, $f_j\in L^2(\Omega')$. If we can show that the weak (distributional) derivative of $f$ coincides with $f_j$, we obtain $f\in H^1(\Omega')$. To this end, we simply verify for any $\phi\in C^\infty_0(\Omega')$ that: \begin{align*} \int_{\Omega'} f\,\partial_j\phi\,dx &= \int_{\Omega_1} f\,\partial_j\phi\,dx + \int_{\Omega_2} f\,\partial_j\phi\,dx\\ &= \int_{\Omega_1} g_1\,\partial_j\phi\,dx + \int_{\Omega_2} g_2\,\partial_j\phi\,dx\\ &= -\int_{\Omega_1} \partial_j g_1\, \phi\,dx + \int_{A} g_1\, n_j\, \phi\, d\sigma - \int_{\Omega_2} \partial_j g_2\,\phi\,dx + \int_{A} g_2\, (-n_j)\, \phi\, d\sigma\\ &= -\int_{\Omega_1} \partial_j g_1\, \phi\,dx - \int_{\Omega_2} \partial_j g_2\,\phi\,dx = -\int_{\Omega'} f_j\,\phi\,dx. \end{align*} Observe that the boundary integrals reduce to integrals over $A$ since $\phi=0$ on $\partial\Omega'\setminus A$. Observe also the sign in $-n_j$ in the second boundary integral due to $n$ denoting the outer normal on $\Omega_1$.

Edit: As pointed our @Apratim, the partial integration above is valid since $\Omega_1$ and $\Omega_2$ are assumed to be smooth. So no further regularity assumptions needed on $A$.

  • $\begingroup$ Thanks for the nice solution. Please note that we don't need any assumption on $A$, as you did integration by parts on $\Omega_1$ and $\Omega_2$, which are smooth. $\endgroup$ – Apratim Bhattacharya Jun 10 at 11:23
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    $\begingroup$ @Apratim: You are right, no assumptions are needed on $A$ since $\Omega_1$ and $\Omega_2$ are already assumed to be smooth. I edited the answer accordingly. $\endgroup$ – StarBug Jun 10 at 12:10

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