Fractional Laplacian

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The fractional Laplacian $(-\Delta)^s$ is a classical operator which can be defined in several equivalent ways.

It is the most typical elliptic operator of order $2s$.



All the definitions below are equivalent.

As a pseudo-differential operator

The fractional Laplacian is the pseudo-differential operator with symbol $|\xi|^{2s}$. In other words, the following formula holds \[ \widehat{(-\Delta)^s f}(\xi) = |\xi|^{2s} \hat f(\xi).\] for any function (or tempered distribution) for which the right hand side makes sense.

This formula is the simplest to understand and it is useful for problems in the whole space. On the other hand, it is hard to obtain local estimates from it.

From functional calculus

Since the operator $-\Delta$ is a self-adjoint positive definite operator in a dense subset $D$ of $L^2(\R^n)$, one can define $F(-\Delta)$ for any continuous function $F:\R^+ \to \R$. In particular, this serves as a more or less abstract definition of $(-\Delta)^s$.

This definition is not as useful for practical applications, since it does not provide any explicit formula.

As a singular integral

If $f$ is regular enough and $s \in (0,1)$, $(-\Delta)^s f(x)$ can be computed by the formula \[ (-\Delta)^s f(x) = c_{n,s} \int_{\R^n} \frac{f(x) - f(y)} {|x-y|^{n+2s}} \mathrm d y .\]

Where $c_{n,s}$ is a constant depending on dimension and $s$.

This formula is the most useful to study local properties of equations involving the fractional Laplacian and regularity for critical semilinear problems.

As a generator of a Levy process

The operator can be defined as the generator of $\alpha$-stable Levy processes. More precisely, if $X_t$ is an $\alpha$-stable process starting at zero and $f$ is a smooth function, then \[ (-\Delta)^{\alpha/2} f(x) = \lim_{h \to 0^+} \frac 1 {h} \mathbb E [f(x) - f(x+X_h)]. \]

This definition is important for applications to probability.

Inverse operator

The inverse of the $s$ power of the Laplacian is the $-s$ power of the Laplacian $(-\Delta)^{-s}$. For $0<s<n/2$, there is an integral formula which says that $(-\Delta)^{-s}u$ is the convolution of the function $u$ with the Riesz potential: \[ (-\Delta)^{-s} u(x) = C_{n,s} \int_{\R^n} u(x-y) \frac{1}{|y|^{n-2s}} \mathrm d y,\] which holds as long as $u$ is regular enough for the right hand side to make sense.

Poisson kernel

Given a function $g : \R^n \setminus B_1 \to \R$, there exists a unique function $u$ which solves the Dirichlet problem \begin{align*} u(x) &= g(x) \qquad \text{if } x \notin B_1 \\ (-\Delta)^s u(x) &= 0 \qquad \text{if } x \in B_1. \end{align*}

The solution can be computed explicitly using the Poisson kernel \[ u(x) = \int_{\R^n \setminus B_1} g(y) P(y,x) \mathrm d y,\] where \[ P(y,x) = C_{n,s} \left( \frac{1-|x|^2}{|y|^2-1}\right)^s \frac 1 {|x-y|^n}.\]

The justification of this Poisson kernel can be found in the classical book of Landkof (1.6.11')[1].

Regularity issues

Any function $u$ which satisfies $(-\Delta)^s u=0$ in any open set $\Omega$, then $u \in C^\infty$ inside $\Omega$. This is a classical fact for pseudo-differential operators[citation needed].

Full space regularization of the Riesz potential

If $(-\Delta)^s u = f$ in $\R^n$, then of course $u = (-\Delta)^{-s}f$. It is simple to see that the operator $(-\Delta)^{-s}$ regularizes the functions up to $2s$ derivatives. In Fourier side, $\hat u(\xi) = |\xi|^{-2s} \hat f(\xi)$, thus $\hat u$ has a stronger decay than $\hat f$. More precisely, if $f \in C^\alpha$, then $u \in C^{2s+\alpha}$ as long as $2s+\alpha$ is not an integer (A proof of this using only the integral representation of $(-\Delta)^{-s}$ was given in the preliminaries section of [2], but the result is presumably very classical). More generally, if $f$ belongs to the Besov space $B_{p,q}^r$, then $u \in B_{p,q}^{r+2s}$.

Boundary regularity

From the Poisson formula, one can observe that if the boundary data $g$ of the Dirichlet problem in $B_1$ is bounded and smooth, then $u \in C^s(\overline B_1)$ and in general no better. The singularity of $u$ occurs only on $\partial B_1$, the solution $u$ would be $C^\infty$ in the interior of the unit ball (which is also a consequence of the explicit Poisson kernel).


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