4
$\begingroup$

Let $B_1$ denote the unit ball in $\mathbb{R}^d$, let \begin{equation} \rho(x) = 1-|x| \quad \text{ for } x \in B_1, \end{equation} and let $\sigma >0$ be given. As per the comments, notice that $\rho(x)$ is in fact the distance of $x \in B_1$ to the boundary $\partial B_1$. For $k \geq 0$, consider the following norm \begin{equation} \| u\|_k^2 = \sum_{j=0\\ |\alpha|=j}^k \int_{B_1} \rho^{\sigma+j} |D^\alpha u|^2 dx, \end{equation} and we finally let $\mathcal{H}^k$ denote the completion of $C^\infty(\overline{B_1})$ with respect to the above norm, for $k \geq 0$. In particular, it can be shown that \begin{equation} \mathcal{H}^0 = L^2(B_1, \rho^\sigma dx) := L^2_\sigma, \end{equation} and also \begin{equation} \mathcal{H}^1 = L^2_\sigma \cap \dot{H}^1_{\sigma+1}, \end{equation} where $\dot{H}^1_{\sigma+1}$ is the homogeneous Sobolev space consisting of all $f \in L^1_{loc}(B_1)$ for which the seminorm \begin{equation} \int_{B_1} \rho^{\sigma+1} |\nabla u|^2 dx \end{equation} is finite, and similarly $\mathcal{H}^2 = L^2_\sigma \cap \dot{H}^1_{\sigma+1} \cap \dot{H}^2_{\sigma+2}$. An analogous characterization holds for the higher order $\mathcal{H}^k$ spaces.

Now, using standard integration by parts, it can be seen that

\begin{equation} \int_{B_1} \nabla u \cdot \nabla v \ \rho^{\sigma+1}dx = -\int_{B_1} \nabla \cdot (\rho^{\sigma+1} \nabla u) v dx \quad \forall u, v \in C^\infty(\overline{B_1}), \end{equation}

since the boundary term \begin{equation} \int_{B_1} \rho^{\sigma+1} \nabla u \cdot \nu \ v \ dS \end{equation} vanishes, due to the form of $\rho^{\sigma+1}$ given above.

My question is the following:

May one extend the above integration by parts formula to the case where, say, $u \in \mathcal{H}^2$ and $v \in \mathcal{H}^1$? Moreover, would this mean that $\rho^{\sigma+1} \nabla u \cdot \nu = 0$ on $\partial B_1$ in some "trace sense", a condition which clearly holds for any $u \in C^\infty(\overline{B_1})$?

Many thanks.

$\endgroup$
  • 1
    $\begingroup$ note that the weight is essentially the same as $ \delta(x)$ (the Euclidean distance from $x$ to the boundary. ) People look at various spaces with these weights...so googling should find something $\endgroup$ – Math604 Mar 2 at 5:00
  • 1
    $\begingroup$ If my (probably inappropriately superficial) estimates were not wrong, I would suppose you will need something like $|\nabla \rho|^2 \lesssim \rho$ to prove the identity by a density argument.. but you probably already tried that anway. $\endgroup$ – Hannes Mar 5 at 14:15
  • $\begingroup$ @Hannes This is a very good remark. So it might not be possible to argue by density for $u \in \mathcal{H}^2$, but I think considering the class of functions $u \in \mathcal{H}^1$ with $\nabla \cdot (\rho^{\sigma+1} \nabla u) \in L^2(B_1, \rho^{-\sigma}dx)$ might suffice. $\endgroup$ – bgsk Mar 5 at 16:23
  • $\begingroup$ Yes, that also would correspond to the way one would define the divergence-gradient operator in $L^2$ with the weight. It will however probably remain difficult to actually characterize that space, I guess. But maybe the abstract domain of definition as you suggested is enough for what you actually want to do. $\endgroup$ – Hannes Mar 6 at 12:07
  • $\begingroup$ @Hannes I think that for such $u$, the formula does indeed hold. But I don't see what this would mean for the term $\rho^{\sigma+1} \nabla u \cdot \nu$ on $\partial B_1$. $\endgroup$ – bgsk Mar 6 at 18:18

Your Answer

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

Browse other questions tagged or ask your own question.