1
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

This paper concerns with your problem.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .