# 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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