= Solution
For this <Neumann Poisson problem>, interpret the forcing in $L^2(U)$, as is automatic if it is <smooth> up to the boundary. Literal interior smoothness alone does not ensure the integrals or bounded functionals required in this question: for example, $f(x)=x^{-2}$ on $(0,1)$ is interior <smooth> but even $\int f\cdot1$ diverges. Classical regularity in the converse is likewise understood up to the boundary.
Suppose the <weak solution> is <smooth> on $\overline U$. Testing against compactly supported <test functions> gives $-\Delta u=f$ in distributions and hence pointwise. Now the weak identity and <Green's first identity> imply
$$
0=\int_U\nabla u\cdot\nabla v-\int_Ufv=\int_{\partial U}(\partial_nu)v\,dS
$$
for every <smooth> $v$ on $\overline U$. Every <smooth> boundary function has such an extension, so $\partial_nu=0$ on $\partial U$. This proves both the interior equation and the boundary condition.
Conversely, for $u\in C^2(\overline U)$ satisfying the equation and zero <normal derivative>, <Green's first identity> gives the weak identity for all <smooth> $v$ on $\overline U$. The <density of smooth functions in a Sobolev space> and the <Cauchy-Schwarz inequality> extend it continuously to every $v\in H^1(U)$. Thus \b[the classical solution is a <weak solution>], and the smooth <weak solution> is classical.
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