# Proving that a weak solution of a BVP satisfies the boundary condition

I am given the smooth function $u$ which satisfies $\int_U (\nabla u \cdot \nabla v +uv)\,dx = \int_U fv\,dx$ for all functions $v$ in the Sobolev space $H^1(U)$, where $f\in C^{\infty}(\overline{U})$ and $U\in\mathbb{R}^n$ is an open, bounded and has smooth boundary. So basically $u$ is just the smooth weak solution of the boundary value problem $-\Delta u+u=f$. I need to prove that $u$ satisfies $(\nabla u\cdot n)|_{\partial U}=0$, with $n$ the outward pointing normal on the boundary.

Partial integration gives $\int_U (-\Delta u +u -f)v\, dx + \int_{\partial U} (\nabla u\cdot n)v\,dS=0$ for all $v \in H^1(U)$. Still I do not see why the Neumann boundary condition follows. My problems are:

• For arbitrary $v$, is there maybe some theorem that tells that both the volume integral and the surface integral must vanish? If not, how can we see that the volume integral vanishes if we only know that $u$ is a weak solution, but possibly not a strong solution? (We did not yet discuss regularity in class.)
• Again for arbitrary $v$, is there some theorem that then states that the integrand of the surface integral must vanish pointwise?

I think I must use the smoothness of $u$ for the first point, but I don't see how.

First, you consider functions $v$ that are smooth and compactly supported in $U$. Because $u$ is assumed to be smooth, it will imply that $-\Delta u + u - f=0$ in $U$.
Taking this into account, we have $\int_{\partial U} (n\cdot\nabla u)v=0$ for all $v\in C^\infty(\overline{U})$. This implies that $n\cdot \nabla u =0$ on the boundary. More precisely, suppose that $n\cdot\nabla u\neq0$ at some point $x\in\partial U$. Without loss of generality let us assume $n\cdot\nabla u>0$ at $x$. By continuity, we have $n\cdot\nabla u>0$ in a neighbourhood $G$ of $x$. Now we can produce a contradiction by picking a nonnegative function $v$ whose restriction to $\partial U$ vanishes outside $G$.
• Thanks, that's quite clear. I see that the surface integral vanishes for compactly supported $v$, but why does the smoothness of $u$ imply then that $-\Delta u + u =f$? Apr 8, 2013 at 8:28
• Ah wait, $-\Delta u + u - f = 0$ by the same reasoning you used for the surface integral: suppose $-\Delta u + u - f > 0$ at $x\in U$, then by smoothness it's nonzero in a neighborhood and we can pick a nonzero $v$ such that it vanishes outside that neighborhood, correct? Thanks. Apr 8, 2013 at 13:59
• @ScroogeMcDuck: Exactly! Note also that regularity is secretly taken into account through the assumption $u$ is smooth. Apr 8, 2013 at 15:36