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.


This paper concerns with your problem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.