# Why are Neumann Boundary conditions imposed weakly?

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 $$H^1_{D,N} = \{u \in H^1 | u = 0 \text{ on } \Gamma_D\ ,\nabla u \cdot n = 0 \text{ on } \Gamma_N\}$$?

• Just try to make sense out of $\nabla u \vert_{\Gamma_N}$. Does this exist? Mar 13, 2021 at 10:41
• $\nabla u$ 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 $\Gamma_N$? Mar 13, 2021 at 11:07
• I will write an answer later Mar 13, 2021 at 16:21

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.