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.

See also: Half-Laplacian in \mathbb R^n

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.

Create a free website or blog at WordPress.com.