Ngô Quốc Anh

May 12, 2010

Half-Laplacian in R^n

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 1:13

In this topic, we are going to define the square root of the Laplacian (-\Delta )^\frac{1}{2} in \mathbb R^n. Our approach makes use of the harmonic extension.

Let u be a bounded continuous function in all of \mathbb R^n. There is a unique harmonic extension v of u in the half-space \mathbb R^{n+1}_+ = \mathbb R^n \times (0,\infty). That is,

\displaystyle\begin{cases}\Delta v = 0, & {\rm in}\; \mathbb R^{n+1}_+=\{ (x,y)\in \mathbb R^n \times (0,\infty)\},\\v=u, & {\rm on}\; \mathbb R^n=\partial \mathbb R^{n+1}_+.\end{cases}

Consider the operator T : u \mapsto -\frac{\partial }{\partial y}v(\cdot, 0). Since \partial_yv is still a harmonic function, if we apply the operator T twice, we obtain

\displaystyle (T \circ T)u = \frac{{{\partial ^2}}}{{\partial {y^2}}}v( \cdot ,0) = - \frac{\partial }{{\partial y}}{\Delta _x}v( \cdot ,0) = - \Delta u

in \mathbb R^n. Thus, we see that the operator T mapping the Dirichlet data u to the Neumann data -\frac{\partial }{\partial y}v(\cdot, 0) is actually a square root of the Laplacian, denoted by (-\Delta )^\frac{1}{2}. It is worth noticing that it is only left to check that T is indeed a positive operator, which follows by a simple integration by parts argument.

Example. In \mathbb R^3, function \frac{1}{|x|} satisfies

\displaystyle \Delta \left(\frac{1}{|x|}\right)=0.

However, it is no longer true for (-\Delta )^\frac{1}{2}. In fact, it should be

\displaystyle (-\Delta )^\frac{1}{2} \left(\frac{1}{|x|^2}\right)=0.

Indeed,

\displaystyle \frac{1}{|x|^2+|y|^2}

in the harmonic extension of \frac{1}{|x|^2} on \mathbb R_+^4. Therefore

\displaystyle (-\Delta )^\frac{1}{2}  \left(\frac{1}{|x|^2}\right)=\frac{\partial}{\partial y}\left(\frac{1}{|x|^2+|y|^2}\right) \Bigg|_{y=0}=0.

Following is the main reference of this entry [here]. In addition, the book entitled “Foundations of Modern Potential Theory” (Springer-Verlag, 1972) due to N.S. Landkof is also a good choice.

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