# Ngô Quốc Anh

## May 20, 2010

### Fractional Laplacian in R^N

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 17:29

Half-Laplacian has been discussed in this entry using harmonic extension. Today, we derive a more general form, called fractional laplacian denoted by $(-\Delta)^s$, for a function $f : \mathbb R^n \to \mathbb R$ where the parameter $s$ is a real number between 0 and 1 and $n \geqslant 3$.

It is worth recalling the following fact known as Newtonian potential in theory of PDEs. The Newtonian potential $u$ of a compactly supported integrable function $f$, i.e. $f \in L_{loc}^1(\mathbb R^n)$, is defined as the convolution

$\displaystyle u(x) = \Gamma * f(x) = \int_{\mathbb{R}^n} \Gamma(x-y)f(y)dy$

where the Newtonian kernel $\Gamma$ in dimension $n$ is defined by

$\displaystyle\Gamma(x) = \begin{cases} \frac{1}{2}\left| x \right| & n=1 \\ \frac{1}{2\pi} \log{ | x | } & n=2 \\ \frac{1}{n(2-n)\omega_n} | x | ^{2-n} & n>2. \end{cases}$.

Here $\omega_d$ is the volume of the unit in $\mathbb R^n$. Coefficients in $\Gamma(x)$, denoted by $\frac{1}{C_n}$, are usually called normalization constants. The Newtonian potential $u$ of $f$ is a solution of the Poisson equation

$\Delta u = f$.

In case $n \geqslant 3$, $u$ is actually the Riesz potential

$\displaystyle u(x)=(I_2f)(x)=\frac{1}{C_{n,2}} \int_{\mathbb{R}^n}\frac{f(y)}{|x-y|^{n-2}}dy$

with $\alpha=2$ where the Riesz potential is defined by

$\displaystyle (I_{\alpha}f) (x)= \frac{1}{C_{n,\alpha}} \int_{{\mathbb{R}}^n} \frac{f(y)}{| x - y |^{n-\alpha}} dy$

where $C_{n,s}$ is some normalization constant given by

$\displaystyle C_{n,s} = \pi^{n/2}2^s\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{n-s}{2}\right)}$.

Observe that at least formally the Riezs potentials $I_\alpha$ verify the following rule

$\displaystyle \Delta I_{\alpha+2} = -I_\alpha, \quad 0 \leqslant \alpha \leqslant n$

which implies

$\displaystyle \Delta^2 I_{\alpha+2} = -\Delta I_\alpha=I_{\alpha-2}, \quad 2 \leqslant \alpha \leqslant n-2$.

Thus if we set $\alpha$ to be $0$ (in fact, this is impossible) we get

$\displaystyle \Delta f = \Delta^2 u=\Delta^2 (I_2f)=I_{-2}f$

which helps us to write down

$\displaystyle \Delta f=\frac{1}{C_{n,2}} \int_{\mathbb{R}^n}\frac{f(y)}{|x-y|^{n+2}}dy$.

Thus we wish to define the fractional Laplacian $(-\Delta)^s$ as follows

$\displaystyle (-\Delta )^s f(x) = \frac{1}{C_{n,s}} \int_{\mathbb{R}^n} \frac{-f(\xi)}{|x-\xi|^{n+2s}}d\xi$.

The above fractional Laplacian is also often called the Riesz fractional derivative [here]. In a paper entitled “An Extension Problem Related to the Fractional Laplacian” due to Luis Caffarelli et al. [here] published in Comm. Partial Differential Equations in 2007, the fractional Laplacian can also be defined using

$\displaystyle (-\Delta )^s f(x) = \frac{1}{C_{n,s}} \int_{\mathbb{R}^n} \frac{f(x)-f(\xi)}{|x-\xi|^{n+2s}}d\xi$.

Let us now study the terminology of weak solution to the following semilinear elliptic equation in the whole space

$\displaystyle (-\Delta)^\frac{\alpha}{2}u=u^\frac{n+\alpha}{n-\alpha}$.

By a weak solution we mean a function $u \in H^\frac{\alpha}{2}(\mathbb R^n)$ such that

$\displaystyle\int_{\mathbb R^n} (-\Delta)^\frac{\alpha}{4}u (-\Delta)^\frac{\alpha}{4} \varphi dx = \int_{\mathbb R^n} u^\frac{n+\alpha}{n-\alpha} \varphi dx$

for any positive test function $\varphi$ in the distribution sense. I will back to this stuff once I finish introducing the fractional Laplacian via pseudo-differential operators.

## 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.