let's consider a mixed boundary problem $-\Delta u = f$, $u=0$ on $ \Gamma_D$ and $\nabla u \cdot n = 0 $ on $\Gamma_N$. Then the weak formulation is to find $u \in H^1_D = \{u \in H^1 | u = 0 \text{ on } \Gamma_D\}$ such that $\int_\Omega \nabla u \cdot \nabla v = \int_\Omega f \ v $ in $\Omega$. That means the Neumann bc were imposed weakly. Why is it not possible to look for a solution in the space $H^1_{D,N} = \{u \in H^1 | u = 0 \text{ on } \Gamma_D\ ,\nabla u \cdot n = 0 \text{ on } \Gamma_N\}$?
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$\begingroup$ Just try to make sense out of $\nabla u \vert_{\Gamma_N}$. Does this exist? $\endgroup$– MeowdogMar 13, 2021 at 10:41
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$\begingroup$ $\nabla u$ is not well defined on the boundary as the trace of an L² function is not well defined. But is $\nabla u \cdot n$ well-defined? If not how can we then talk about the problem $\nabla u \cdot n = 0$ on $\Gamma_N$? $\endgroup$– taylor123123Mar 13, 2021 at 11:07
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$\begingroup$ I will write an answer later $\endgroup$– MeowdogMar 13, 2021 at 16:21
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1 Answer
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This is a rather techincal issue. A priori, we can not talk about $\nabla u \cdot n$ in an $H^1$, i.e. trace-sense. If our domian $\Omega$ has $C^2$-boundary, we can say that our weak solution $u \in H^1_0$ (that we are granted by Lax-Milgram-Lemma) even lives in $H^2(\Omega)$. This is Gilbarg and Trudinger Theorem 8.12. It is sometimes referred to as Friedrichs-shift-theorem.
Then, traces can be defined for $\nabla u$, and we can make sense of the Neumann BC.
So we only choose $H_0^1$ as test and solution space, because no further regularity is needed.
Since Neumann-BCs are integrated into the weak formulation (in our case, since the Neumann-values are $0$, they do not appear), they are often referred to as natural BCs or do-nothing BCs.
