# Can one obtain Neumann boundary conditions via Friedrichs extension?

According to the book "Applied Functional Analysis" vol. I by Zeidler, the Friedrichs extension of an operator $B\colon D(B)\subset X\to X$, where $X$ is a real Hilbert space, is obtained as follows. One needs:

1. that $B$ be symmetric, that is $(Bu, v)=(u, Bv)$ for all $u, v\in D(B)$;
2. that $B$ be strongly monotone (aka strictly positive), that is, $(Bu, u)\ge c\|u\|^2$ for all $u\in D(B)$ for a constant $c>0$.

If those assumptions are satisfies, one defines the energetic space $$X_E=\text{completion of }D(B)\text{ with norm }\|u\|_E^2=(Bu, u).$$ This gives rise to a triple (Hilbert triple as it is sometimes called) $$X_E\subset X\subset X_E^\star,$$ and one defines the energetic extension $$B_E\colon X_E\to X_E^\star,\qquad {}_{X_E^\star}\langle B_E u, v\rangle_{X_E}=(u, v)_{X_E},$$ that is, $B_E$ equals the Riesz isomorphism of the real Hilbert space $X_E$ with its dual. (It is here that one uses the fact that $X$ is real. In the complex case, this mapping is not linear but conjugate-linear. This is not a big deal nonetheless).

This said, the Friedrichs's extension is the maximal restriction of $B_E$ that does not leave $X$: $$Au=B_E u\qquad D(A)=\left\{ u\in X_E\ :\ B_E u\in X\right\}.$$

The standard example is obtained on $X=L^2(\Omega)$, where $\Omega$ is a bounded open set in $\mathbb{R}^n$, with $B=-\Delta$ and $D(B)=C^\infty_0(\Omega)$ In this case one has that $X_E=H^1_0(\Omega)$ and the relation $$Au=f$$ is the abstract reformulation of the boundary value problem $$\begin{cases} -\Delta u=0 & \Omega \\ u=0 & \text{on }\partial \Omega.\end{cases}$$ In particular, one has obtained Dirichlet boundary conditions.

Question. Can one abstractly obtain Neumann's boundary conditions by means of the same construction? Of course one can assume that $\Omega$ boundary is as smooth as one wishes. If it simplifies things one can even assume that $\Omega$ is the unit ball.

• One problem to address is the additional conditions on $\Omega$ so that the phrase "Neumann boundary conditions" makes sense. After that, the most obvious candidate to try would be $D(B)=$ the infinitely differentiable functions with vanishing normal derivative and with all derivatives bounded. Mar 7 '16 at 17:00
• @Justpassingby: That's surely a good start. I forgot to mention that the domain is as smooth as one wishes. The only obstacle is that one does not have $(Bu, u)\ge c(u,u)$, but I think that either considering the operator $B+I$ or quotienting constants out should do the trick. To be continued Mar 7 '16 at 19:11
• @Justpassingby: Your idea is the right one. The details are on Davies's book "Spectral theory and differential operators", section 7.2: "Neumann boundary conditions". Mar 8 '16 at 11:16

For smooth functions $u$ with vanishing normal derivative on $\partial\Omega$, \begin{align} (Bu,u) & = \int_{\Omega}(-\Delta u)udx \\ & =\int_{\Omega}\nabla\cdot((-\nabla u)u)+|\nabla u|^2dx \\ & = -\int_{\Omega}u\frac{\partial u}{\partial n}dS+\int_{\Omega}|\nabla u|^2dx \\ & = \int_{\Omega}|\nabla u|^2dx. \end{align} You cannot have $(Bu,u) \ge c\|u\|^2$ for some $c > 0$ because the constant functions $u$ are in the subspace you're considering and $Bu=0$ for a constant function. However, you do have $(Bu,u) \ge 0$. So $(B+\epsilon I)$ is positive definite.