# Proving an estimate for the Neumann problem on $\mathbb{R}^3 \setminus B_1$ in Weighted Sobolev spaces

Let $$M := \mathbb{R}^3 \setminus B_1$$ where $$B_1$$ is the unit ball.

It is known that for every $$g \in H^{\frac{1}{2}} (\partial M)$$, and for an appropriate $$\delta$$, there exists a unique solution $$u$$ in a weighted Sobolev space $$H^2_{\delta} (M)$$ satisfying

$$\begin{cases} \Delta u=0, & \text{on M}\\ \partial_r u = g, & \text{on \partial M} \end{cases}$$

Furthermore, the following estimate holds: $$\lVert u \rVert_{H^2_{\delta}} \leq C \lVert g \rVert_{H^{\frac{1}{2}}}$$

How is this estimate proven?

Here is my attempt:

I expanded $$u$$ into a series using spherical harmonics and reduced the PDE to the following ODEs: $$r^2 a_{ml}'' + 2ra_{ml}' - l(l+1)a_{ml} = 0$$

I define the norm $$\lVert a_{ml} \rVert^2 = \int_1^{\infty} r^4 a_{ml}''^2 + \int_1^{\infty} r^2 a_{ml}'^2 + [1+l(l+1)]^2\int_1^{\infty} a_{ml}^2$$

and so the $$H^2_{\delta}$$ norm for some $$\delta$$ can be defined as $$\lVert u \rVert_{H^2_{\delta}}^2 := \sum_{m,l} \lVert a_{ml} \rVert^2$$.

By multiplying the ODE by $$a_{ml}$$ and integrating by parts, I only managed to get $$\lVert a_{ml} \rVert^2 \leq C [1+l(l+1)] |b_{ml}|^2$$ where $$b_{ml}$$ is the coefficient of $$f$$, and so I get $$\lVert u \rVert_{H^2_{\delta}} \leq C \lVert g \rVert_{H^{1}}$$. This is weaker than what I wanted to get.

Any references are appreciated.