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I'm looking for an explanation of why the jump $[\nabla u]=[u] =0$ assuming $u \in H^2(\Omega)$.

We know that according to an embedding theorem, $H^1(\Omega)$ is a subspace of $C^0(\bar{\Omega})$ (space of continuous functions in $\bar{\Omega}$) so that $[u] = 0$ makes sense, but does $H^2(\Omega) \subset C^1(\bar{\Omega})$ ?

gbmreda
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    The embedding theorem depends on the dimension $n$, you have $H^s\subset C^{s-n/2}$. For $n=1$ and barely for $n=2$ this gives indeed your claims, but not in higher dimensions. // Maybe I do not completely understand this topic or maybe the sentence structure is too sparse, but please expand the question text a little more. What jump, does $[.]$ indicate the class of equivalent generalized functions? – Lutz Lehmann Jul 22 '22 at 13:33
  • I believe that it should be $[\nabla u\cdot \mathbf{n}]=0$ instead of $[\nabla u]=0$. The first one is a weaker condition. – Tulip Jul 25 '22 at 02:20

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Let $\Omega = \Omega_1 \cup \Omega_2$, where $\{ \Omega_i \}$ represents two elements that share an interface $\Gamma = \Omega_1 \cap \Omega_2$. Assume that $u \in H^2(\Omega)$. Multiplying $\Delta u$ by a smooth test function $v \in C^\infty$, we then can integrate by parts \begin{equation} \label{eq:omega} \tag{1} \int_\Omega (\Delta u) v \, dx = -\int_\Omega \nabla u \cdot \nabla v \, dx + \int_{\partial \Omega} v \nabla u \cdot \boldsymbol{n} \, ds \end{equation} where $\Delta u$ and $\nabla u$ are interpreted in the sense of weak derivatives.

The same also holds over each element $\Omega_i$, \begin{equation} \label{eq:omega-i} \tag{2} \int_{\Omega_i} (\Delta u) v \, dx = -\int_{\Omega_i} \nabla u \cdot \nabla v \, dx + \int_{\partial \Omega_i} v \nabla u \cdot \boldsymbol n_i \, ds, \end{equation} where $\boldsymbol n_i$ is the normal vector facing outwards from $\Omega_i$. Summing \eqref{eq:omega-i} over $i=1,2$ we obtain \begin{equation} \label{eq:sum} \tag{3} \int_{\Omega} (\Delta u) v \, dx = -\int_{\Omega} \nabla u \cdot \nabla v \, dx + \int_{\partial \Omega} v \nabla u \cdot \boldsymbol{n} \, ds + \int_\Gamma v [ \nabla u ] \, ds. \end{equation} We see that in order for both \eqref{eq:omega} and \eqref{eq:sum} to hold, we must have that $\int_\Gamma v [ \nabla u ] \, ds = 0$, and since this holds for all test functions $v \in C^\infty$, we have that $[ \nabla u ] = 0$.

In the above, I am using the (fairly standard) notation that for a vector-valued quantity $\boldsymbol \varphi$, the jump is defined by $[ \boldsymbol \varphi ] = \boldsymbol \varphi_1 \cdot \boldsymbol n_1 + \boldsymbol \varphi_2 \cdot \boldsymbol n_2$, where $\boldsymbol \varphi_i$ is the trace from within $\Omega_i$.

Will P.
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