Fractional Laplacian and Levy processes: Difference between pages

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The fractional Laplacian $(-\Delta)^s$ is a classical operator which can be defined in several equivalent ways.
A Lévy process is an important type of [[stochastic process]] (namely, a family of $\mathbb{R}^d$ valued random variables each indexed by a positive number $t\geq 0$).  In the context of parabolic integro-differential equations they play the same role that Brownian motion and more general diffusions play in the theory of second order parabolic equations.


It is the most typical elliptic operator of order $2s$.
Informally speaking, a Lévy process is a Brownian motion which may jump, the times, length and direction of the jumps being random variables. A prototypical example would be $X(t)=B(t)+N(t)$ where $B(t)$ is the standard [[Brownian motion]] and $N(t)$ is a [[Compound Poisson process]], the trajectory described by typical sample path of this process would look like the union of several disconnected Brownian motion paths.


== Definitions ==
== Definition ==


All the definitions below are equivalent.
A stochastic process $X=\{X(t)\}_{t \geq 0}$ with values in $\mathbb{R}^d$ is said to be a Lévy process if


=== As a pseudo-differential operator ===
1.For any sequence $0 \leq t_1 < t_2 <...<t_n$ the random variables $X(t_0),X(t_1)-X(t_0),...,X(t_n)-X(t_{n-1})$ are independent.
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.
2.For any positive times $s\leq t$ the random variables $X(t-s)$ and $X(t)-X(s)$ have the same probability law.


=== From functional calculus ===
3.Almost surely, the trajectory of $X(t)$ is continuous from the right, with limit from the left also known as "càdlàg" for its acronym in french.


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 $(-\Delta)^s$.
== Lévy-Khintchine Formula ==


This definition is not as useful for practical applications, since it does not provide any explicit formula.
It follows from the first two properties above that if $X$ is a Lévy process and we further assume $X(0)=0$ a.s. then for each fixed positive $t$ the random variable $X(t)$ is infinitely divisible, that is, it can be written as the sum of $n$ independent and identically distribued random variables, for all $n\in\mathbb{N}$. Indeed, let $h=\tfrac{t}{n}$, then


=== As a singular integral ===
\[X(t) = \left( X(h)-X(0)\right)+\left( X(2h)-X(h)\right)+...+\left( X(t)-X((n-1)h)\right)\]
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 $s$.
and by the above definition the differences $X(kh)-X((k-1)h)$ are independent and distributed the same as $X(h)$. From the infinite divisibility of $X(t)$ it follows by a theorem of Lévy and Khintchine that  for any $\xi \in \mathbb{R}^d$ we have


This formula is the most useful to study local properties of equations involving the fractional Laplacian and regularity for critical semilinear problems.
\[ \mathbb{E} \left [ e^{i\xi\cdot X_t}\right ] = e^{t\eta(\xi)}\]


=== As a generator of a [[Levy process]] ===
the function $\eta(\xi)$ given by
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.
\[\eta(\xi)=y\cdot b i  -\tfrac{1}{2}(A\xi,\xi)+\int_{\mathbb{R}^d} \left ( e^{\xi\cdot y}-1-i\xi\cdot y \chi_{B_1}(y) \right ) d\mu(y)  \]


== Inverse operator ==
where $b$ is a vector, $A$ is a positive matrix and $\mu$ is a Lévy measure, that is, a Borel measure in $\mathbb{R}^d$ such that
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 ==
\[ \int_{\mathbb{R}^d}\frac{|y|^2}{1+|y|^2}d\mu(y) <+\infty \]
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
== Connection to linear integro-differential operators ==
\[ u(x) = \int_{\R^n \setminus B_1} g(y) P(y,x) \mathrm d y,\]
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')<ref name="L"/>.
Any Lévy process $X(t)$ such that $X(0)=0$ almost surely defines a linear semigroup $\{U_t\}_{t\geq0}$ on the space  of continuous functions $f:\mathbb{R}^d\to\mathbb{R}^d$ as follows


== Regularity issues ==
\[(U_tf)(x)= \mathbb{E}\left [ f(x+X(t) \right ]\]
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 ===
Given the initial assumption on $X(0)$ it is clear that $U_0$ is the identity, and given that $X(t)-X(s)$ is distributed as $X(t-s)$ it follows that $U_t \circ U_s = U_{t+s}$.  
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 <ref name="S"/>, 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 ===
As a semigroup, $U_t$ has an infinitesimal generator which turns out to be a [[Linear integro-differential operator]]. More precisely, if we let $f(x,t):=(U_tf)(x)$, then, assuming that $f(x,t)$ has enough regularity it can be checked that
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).


== References ==
\[\partial_t f = Lf \;\;\;\mbox{ for all } (x,t)\in\mathbb{R}^d\times \mathbb{R}_+\]
{{reflist|refs=
 
<ref name="L">{{Citation | last1=Landkof | first1=N. S. | title=Foundations of modern potential theory | publisher=[[Springer-Verlag]] | location=Berlin, New York | year=1972}}</ref>
where for any smooth function $\phi$, we have
<ref name="S">{{Citation | last1=Silvestre | first1=Luis | title=Regularity of the obstacle problem for a fractional power of the Laplace operator | url=http://dx.doi.org/10.1002/cpa.20153 | doi=10.1002/cpa.20153 | year=2007 | journal=[[Communications on Pure and Applied Mathematics]] | issn=0010-3640 | volume=60 | issue=1 | pages=67–112}}</ref>
 
}}
\[ L\phi(x) = b \cdot \nabla \phi(x) +\mathrm{tr} \,( A\cdot D^2 \phi )+ \int_{\R^d} (\phi(x+y) - \phi(x) - y \cdot \nabla \phi(x) \chi_{B_1}(y)) \, \mathrm{d} \mu(y) \]
 
== Stochastic control and fully non-linear integro-differential operators ==
 
A similar connection holds via the [[Isaacs-Bellman]] equation arising in stochastic games/stochastic control problems. However, in this case the corresponding semigroup is not linear, however, using viscosity solutions instead of linear analysis one may construct a corresponding semigroup.

Revision as of 14:52, 22 January 2012

A Lévy process is an important type of stochastic process (namely, a family of $\mathbb{R}^d$ valued random variables each indexed by a positive number $t\geq 0$). In the context of parabolic integro-differential equations they play the same role that Brownian motion and more general diffusions play in the theory of second order parabolic equations.

Informally speaking, a Lévy process is a Brownian motion which may jump, the times, length and direction of the jumps being random variables. A prototypical example would be $X(t)=B(t)+N(t)$ where $B(t)$ is the standard Brownian motion and $N(t)$ is a Compound Poisson process, the trajectory described by typical sample path of this process would look like the union of several disconnected Brownian motion paths.

Definition

A stochastic process $X=\{X(t)\}_{t \geq 0}$ with values in $\mathbb{R}^d$ is said to be a Lévy process if

1.For any sequence $0 \leq t_1 < t_2 <...<t_n$ the random variables $X(t_0),X(t_1)-X(t_0),...,X(t_n)-X(t_{n-1})$ are independent.

2.For any positive times $s\leq t$ the random variables $X(t-s)$ and $X(t)-X(s)$ have the same probability law.

3.Almost surely, the trajectory of $X(t)$ is continuous from the right, with limit from the left also known as "càdlàg" for its acronym in french.

Lévy-Khintchine Formula

It follows from the first two properties above that if $X$ is a Lévy process and we further assume $X(0)=0$ a.s. then for each fixed positive $t$ the random variable $X(t)$ is infinitely divisible, that is, it can be written as the sum of $n$ independent and identically distribued random variables, for all $n\in\mathbb{N}$. Indeed, let $h=\tfrac{t}{n}$, then

\[X(t) = \left( X(h)-X(0)\right)+\left( X(2h)-X(h)\right)+...+\left( X(t)-X((n-1)h)\right)\]

and by the above definition the differences $X(kh)-X((k-1)h)$ are independent and distributed the same as $X(h)$. From the infinite divisibility of $X(t)$ it follows by a theorem of Lévy and Khintchine that for any $\xi \in \mathbb{R}^d$ we have

\[ \mathbb{E} \left [ e^{i\xi\cdot X_t}\right ] = e^{t\eta(\xi)}\]

the function $\eta(\xi)$ given by

\[\eta(\xi)=y\cdot b i -\tfrac{1}{2}(A\xi,\xi)+\int_{\mathbb{R}^d} \left ( e^{\xi\cdot y}-1-i\xi\cdot y \chi_{B_1}(y) \right ) d\mu(y) \]

where $b$ is a vector, $A$ is a positive matrix and $\mu$ is a Lévy measure, that is, a Borel measure in $\mathbb{R}^d$ such that

\[ \int_{\mathbb{R}^d}\frac{|y|^2}{1+|y|^2}d\mu(y) <+\infty \]

Connection to linear integro-differential operators

Any Lévy process $X(t)$ such that $X(0)=0$ almost surely defines a linear semigroup $\{U_t\}_{t\geq0}$ on the space of continuous functions $f:\mathbb{R}^d\to\mathbb{R}^d$ as follows

\[(U_tf)(x)= \mathbb{E}\left [ f(x+X(t) \right ]\]

Given the initial assumption on $X(0)$ it is clear that $U_0$ is the identity, and given that $X(t)-X(s)$ is distributed as $X(t-s)$ it follows that $U_t \circ U_s = U_{t+s}$.

As a semigroup, $U_t$ has an infinitesimal generator which turns out to be a Linear integro-differential operator. More precisely, if we let $f(x,t):=(U_tf)(x)$, then, assuming that $f(x,t)$ has enough regularity it can be checked that

\[\partial_t f = Lf \;\;\;\mbox{ for all } (x,t)\in\mathbb{R}^d\times \mathbb{R}_+\]

where for any smooth function $\phi$, we have

\[ L\phi(x) = b \cdot \nabla \phi(x) +\mathrm{tr} \,( A\cdot D^2 \phi )+ \int_{\R^d} (\phi(x+y) - \phi(x) - y \cdot \nabla \phi(x) \chi_{B_1}(y)) \, \mathrm{d} \mu(y) \]

Stochastic control and fully non-linear integro-differential operators

A similar connection holds via the Isaacs-Bellman equation arising in stochastic games/stochastic control problems. However, in this case the corresponding semigroup is not linear, however, using viscosity solutions instead of linear analysis one may construct a corresponding semigroup.

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