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I've been studying Trace Theorem. From PDE Evans, we have

THEOREM 1 (Trace Theorem). Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $$T : W^{1,p}(U) \rightarrow L^p(\partial U)$$ such that

$\quad$(i) $Tu=u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\bar{U})$

and

$\quad$(ii) $$\|Tu\|_{L^p(\partial U)} \le C \| U \|_{W^{1,p} (U)},$$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.

DEFINITION. We call $Tu$ the trace of $u$ on $\partial U$.

Then, makes sense to find solutions in $W_{0}^{1,2}(\Omega) = H^{1}_{0}(\Omega)$ for the problem $-\Delta u = f$ when $u$ vanishes in $\partial\Omega$, because $Ker(T) = H^{1}_{0}(\Omega)$. So, analyzing from a naive perspective, I thought that "weak solutions" for the problem:

$$\left\{\begin{array}{c} −\Delta u = f(x) & ,\Omega \\ \dfrac{\partial u}{\partial \nu}=0 , \partial\Omega \end{array}\right.$$ were in the space $H^{2}_{0}(\Omega)$. But the solution space for the above problem is $H^{1}(\Omega)$, why?


My approach is, if $u \in H^{1}(\Omega) $ is a weak solution for the above problem, then

$$\displaystyle\int_{\Omega}\nabla u \nabla v = \displaystyle\int_{\Omega} f v, \forall v \in H^{1}(\Omega)$$

Futhermore if $u$ has $C^{2}(\overline{\Omega})$ regularity, by Green's formula we can conclude that

$$ \displaystyle\int_{\Omega} \dfrac{\partial u}{\partial \nu} v = 0, \forall v \in C^{\infty}_{0}(\Omega)$$

By the Du Bois Raymond Lemma, we can state that $\dfrac{\partial u}{\partial \nu} = 0$ in $\partial \Omega$.

Is my approach wrong?

Thanks in advance.

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  • $\begingroup$ In fact, the solution to your Neumann problem is unique up to a constant. $\endgroup$
    – Simon Pun
    Commented May 19, 2021 at 14:19

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Your explanation for the choice of the function space is right.

Note that $u \in H_0^2(\Omega)$ if and only if $u = 0$ and $\nabla u = 0$ on $\partial \Omega$. This, of course, is not the right function space for your Neumann problem!

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