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Assume $U$ is a bounded open subset of $\mathbb{R}^N$. Furthermore $\Delta: H^2(U) \subset L^2(U) \to L^2(U)$ is the Laplace operator. My question is:

What is the domain of $\vert \Delta\vert^{1/2}$?

I've seen this notation a lot recently and still don't know what to make of it. First of all the Laplace operator is not a positive operator. I guess, that is what the $\vert \cdot \vert$ is for.

Here is a definition of the root of $-\Delta$, which suggests that the domain could be $L^1 (U)$.

It would be perfect for my cause, if the domain could be taken as $H^2(U)$ again.

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1 Answer 1

up vote 4 down vote accepted

Strictly speaking, this is not a well-posed question: the definition of an operator, as well as of any function, must include the domain. If we don't know what the domain of $|\Delta|^{1/2}$ is, then we don't know what $|\Delta|^{1/2}$ means.

The fractional Laplacian is a nonlocal operator: we can't find the values of $(-\Delta)^{1/2}u$ on a domain just by looking at the values of $u$ on that domain. This is why the papers dealing with fractional Laplacian normally work on $\mathbb R^n$ rather than on $U$. (If the paper you are reading is an exception to this, please give a reference.) So, in order to find $(-\Delta)^{1/2}u$, we must decide how to extend the function $u$ outside of its original domain $U$.

You could extend by zero outside of $U$, but this may be a bad move. The boundary discontinuity can take $u$ out of all Sobolev spaces, and $(-\Delta)^{1/2} u$ will likely fail to be integrable on $U$. (It will be in $L^2$ locally, though.)

Harmonic extension outside of $U$ works better, and is easy to implement as follows: first take $\Delta u$ (local operator), then extend $\Delta u$ by zero outside of $U$, and finally apply $(-\Delta)^{-1/2}$, which is realized by convolution with integrable kernel $\|x\|^{-n+1}$. If $u\in H^2(U)$, this process yields $(-\Delta)^{1/2} u\in L^2(U)$ (in fact a little better), which apparently is what you wanted.

You may find the nonlocal PDE wiki helpful.

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