Levy processes and Obstacle problem for the fractional Laplacian: Difference between pages

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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.
The obstacle problem for the fractional Laplacian refers to the particular case of the [[obstacle problem]] when the elliptic operator $L$ is given by the [[fractional Laplacian]]: $L = -(-\Delta)^s$ for some $s \in (0,1)$. The equation reads
\begin{align}
u &\geq \varphi \qquad \text{everywhere}\\
(-\Delta)^s u &\geq 0 \qquad \text{everywhere}\\
(-\Delta)^s u &= 0 \qquad \text{wherever } u > \varphi.
\end{align}


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.
The equation is derived from an [[optimal stopping problem]] when considering $\alpha$-stable Levy processes. It serves as the simplest model for other optimal stopping problems with purely jump processes and therefore its understanding is relevant for applications to [[financial mathematics]].


== Definition ==
== Existence and uniqueness ==
The equation can be studied from either a variational or a non-variational point of view, and with or without boundary conditions.


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 variational inequality the equation emerges as the minimizer of the homoegeneous $\dot H^s$ norm from all functions $u$ such that $u \geq \varphi$. In the case when the domain is the full space $\mathbb R^d$, a decay at infinity $u(x) \to 0$ as $|x| \to \infty$ is usually assumed. Note that in low dimensions $\dot H^s$ is not embedded in $L^p$ for any $p<\infty$ and therefore the boundary condition at infinity cannot be assured. In low dimensions one can overcome this inconvenience by minimizing the full $H^s$ norm and therefore obtaining the equation with an extra term of zeroth order:
\begin{align}
u &\geq \varphi \qquad \text{everywhere}\\
(-\Delta)^s u + u &\geq 0 \qquad \text{everywhere}\\
(-\Delta)^s u + u &= 0 \qquad \text{wherever } u > \varphi.
\end{align}
This extra zeroth order term does not affect any regularity consideration for the solution.


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.
From a non variational point of view, the solution $u$ can be obtained as the smallest $s$-superharmonic function (i.e. $(-\Delta)^s u \geq 0$ such that $u \geq \varphi$. In low dimensions one cannot assure the boundary condition at infinity because of the impossibility of constructing barriers (this is related to the fact that the fundamental solutions $|x|^{-n+2s}$ fail to decay to zero at infinity if $2s \geq n$). This can be overcome with the addition of the zeroth order term or by the study of the problem in a bounded domain with Dirichlet boundary conditions in the complement.


2.For any positive times $s\leq t$ the random variables $X(t-s)$ and $X(t)-X(s)$ have the same probability law.
== Regularity considerations ==
=== Regularity of the solution ===
Assuming that the obstacle $\varphi$ is smooth, the optimal regularity of the solution is $C^{1,s}$.


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.  
The regularity $C^{1,s}$ coincides with $C^{1,1}$ when $s=1$, which is the optimal regularity in the classical case of the Laplacian. However, adapting the ideas of the classical proof to the fractional case suggests that the optimal regularity should be only $C^{2s}$. The optimal regularity in the case $s<1$ is better than the order of the equation and cannot be justified by any simple scaling argument.


== Lévy-Khintchine Formula ==
Below, we outline the steps leading to the optimal regularity with a sketch of the ideas used in the proofs.


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
==== Almost $C^{2s}$ regularity ====
This first step of the proof is the simplest and it is the only step which is an adaptation of the classical case $s=1$.  


\[X(t) = \left( X(h)-X(0)\right)+\left( X(2h)-X(h)\right)+...+\left( X(t)-X((n-1)h)\right)\]
From the statement of the equation we have $(-\Delta)^s u \geq 0$.


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
Since the average of two $s$-superharmonic function is also $s$-superharmonic, one can see that for any $h \in \mathbb R^d$, the function $v(x):=(u(x+h)+u(x-h))/2 + C|h|^2$ is $s$ superharmonic and $v \geq \varphi$ if $C = ||D^2 \varphi||_{L^\infty}$. By the comparison principle $v \geq u$. This means that $u$ is semiconvex: $D^2 u \geq -C I$.


\[ \mathbb{E} \left [ e^{i\xi\cdot X_t}\right ] = e^{t\eta(\xi)}\]
Interpolating the semiconvexity and $L^\infty$ boundedness of $u$, we obtain that $(-\Delta)^s u \leq C$ for some constant $C$.


the function $\eta(\xi)$ given by
The boundedness of $(-\Delta)^s u$ does not imply that $u \in C^{2s}$ but it does imply that $u \in C^\alpha$ for all $\alpha < s$.


\[\eta(\xi)=y\cdot b i  -\tfrac{1}{2}(A\xi,\xi)+\int_{\mathbb{R}^d} \left ( e^{\xi\cdot y}-1-i\xi\cdot y \chi_{B_1}(y) \right ) d\mu(y)  \]
==== $C^{2s+\alpha}$ regularity, for some small $\alpha>0$ ====
Let $w(x) = (-\Delta)^s u(x)$. A key observation is that the function $w$ satisfies the equation
\begin{align}
(-\Delta)^{1-s} w = -\Delta \varphi \qquad \text{in } \{u=\varphi\}, \\
w = 0 \varphi \qquad \text{outside } \{u=\varphi\}.
\end{align}


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
This is a Dirichlet problem for the conjugate fractional Laplacian. However there are two difficulties. First of all we need to prove that $w$ is continuous on the boundary $\partial \{u=\varphi\}$. Second, this boundary can be highly irregular a priori so we cannot expect to obtain any H\"older continuity of $w$ from the Dirichlet problem alone.


\[ \int_{\mathbb{R}^d}\frac{|y|^2}{1+|y|^2}d\mu(y) <+\infty \]
From the semiconvexity of $u$ we have $-\Delta u \leq C$, and therefore we derive the extra condition $(-\Delta)^{1-s} w \leq C$ in the full space $\R^d$ (in particular across the boundary $\partial \{u=\varphi\}$). Moreover, we also know that $w \geq 0$ everywhere.


== Semigroup and infinitesimal generator  ==
The $C^\alpha$ Holder continuity of $w$ on the boundary $\partial \{u=\varphi\}$ is obtained from an [[iterative improvement of oscillation]] procedure. Since $w \geq 0$ and $(-\Delta)^{1-s} u \leq C$, for any $x_0$ on $\partial \{u=\varphi\}$ we can show that $\max_{B_r(x_0)} w$ decays provided that $\{u > \varphi\} \cap B_r$ is sufficiently "thick" using the [[weak Harnack inequality]]. We cannot rule out the case in which $\{u > \varphi\} \cap B_r$ has a very small measure. However, in the case that $\{u > \varphi\} \cap B_r$ is too small in measure, we can prove that $u$ separates very slowly from $\varphi$. This slow separation is used to prove that $w$ must also improve its oscillation and this step is particularly tricky <ref name="S"/>.


Any Lévy process $X(t)$ such that $X(0)=0$ almost surely defines a semigroup $\{U_t\}_{t\geq0}$ on the space  of continuous functions $f:\mathbb{R}^d\to\mathbb{R}^d$ as follows
Once we know that $w(x) = (-\Delta)^s u(x)$ is $C^\alpha$, this implies that $u \in C^{2s+\alpha}$ by classical potential analysis theory.


\[(U_tf)(x)= \mathbb{E}\left [ f(x+X(t) \right ]\]
==== $C^{1,s}$ regularity ====
If the contact set $\{u=\varphi\}$ is convex or at least has an exterior ball condition, a fairly simple barrier function can be constructed to show that $w$ must be $C^{1-s}$ on the boundary $\partial \{u=\varphi\}$. This is the generic boundary regularity for solutions of fractional Laplace equations in smooth domains.


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 $U_t \circ U_s = U_{t+s}$.  
Without assuming anything on the contact set $\{u=\varphi\}$, one can still obtain that $w \in C^\alpha$ for every $\alpha < 1-s$ though an iterative use of barrier functions <ref name="S"/>. The sharp $w \in C^{1-s}$ regularity in full generality was obtained rewriting the equation as a [[thin obstacle problem]] using the [[extension technique]] and then applying blowup techniques, the Almgren monotonicity formula and classification of global solutions <ref name="CSS"/>.


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
=== Regularity of the free boundary ===


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


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


\[ L\phi(x) = b \cdot \nabla \phi(x) +\mathrm{tr} \,( A\cdot D^2 \phi )+ \int_{\R^d} (\phi(x+y) - \phi(x) - y \cdot \nabla \phi(x) \chi_{B_1}(y)) \, \mathrm{d} \mu(y) \]
 
== References ==
{{reflist|refs=
<ref name="S">{{Citation | last1=Silvestre | first1=Luis | title=Regularity of the obstacle problem for a fractional power of the Laplace operator | publisher=Wiley Online Library | year=2007 | journal=[[Communications on Pure and Applied Mathematics]] | issn=0010-3640 | volume=60 | issue=1 | pages=67–112}}</ref>
<ref name="CSS">{{Citation | last1=Caffarelli | first1=Luis A. | last2=Salsa | first2=Sandro | last3=Silvestre | first3=Luis | title=Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian | url=http://dx.doi.org/10.1007/s00222-007-0086-6 | doi=10.1007/s00222-007-0086-6 | year=2008 | journal=[[Inventiones Mathematicae]] | issn=0020-9910 | volume=171 | issue=2 | pages=425–461}}</ref>
}}

Revision as of 13:21, 28 January 2012

The obstacle problem for the fractional Laplacian refers to the particular case of the obstacle problem when the elliptic operator $L$ is given by the fractional Laplacian: $L = -(-\Delta)^s$ for some $s \in (0,1)$. The equation reads \begin{align} u &\geq \varphi \qquad \text{everywhere}\\ (-\Delta)^s u &\geq 0 \qquad \text{everywhere}\\ (-\Delta)^s u &= 0 \qquad \text{wherever } u > \varphi. \end{align}

The equation is derived from an optimal stopping problem when considering $\alpha$-stable Levy processes. It serves as the simplest model for other optimal stopping problems with purely jump processes and therefore its understanding is relevant for applications to financial mathematics.

Existence and uniqueness

The equation can be studied from either a variational or a non-variational point of view, and with or without boundary conditions.

As a variational inequality the equation emerges as the minimizer of the homoegeneous $\dot H^s$ norm from all functions $u$ such that $u \geq \varphi$. In the case when the domain is the full space $\mathbb R^d$, a decay at infinity $u(x) \to 0$ as $|x| \to \infty$ is usually assumed. Note that in low dimensions $\dot H^s$ is not embedded in $L^p$ for any $p<\infty$ and therefore the boundary condition at infinity cannot be assured. In low dimensions one can overcome this inconvenience by minimizing the full $H^s$ norm and therefore obtaining the equation with an extra term of zeroth order: \begin{align} u &\geq \varphi \qquad \text{everywhere}\\ (-\Delta)^s u + u &\geq 0 \qquad \text{everywhere}\\ (-\Delta)^s u + u &= 0 \qquad \text{wherever } u > \varphi. \end{align} This extra zeroth order term does not affect any regularity consideration for the solution.

From a non variational point of view, the solution $u$ can be obtained as the smallest $s$-superharmonic function (i.e. $(-\Delta)^s u \geq 0$ such that $u \geq \varphi$. In low dimensions one cannot assure the boundary condition at infinity because of the impossibility of constructing barriers (this is related to the fact that the fundamental solutions $|x|^{-n+2s}$ fail to decay to zero at infinity if $2s \geq n$). This can be overcome with the addition of the zeroth order term or by the study of the problem in a bounded domain with Dirichlet boundary conditions in the complement.

Regularity considerations

Regularity of the solution

Assuming that the obstacle $\varphi$ is smooth, the optimal regularity of the solution is $C^{1,s}$.

The regularity $C^{1,s}$ coincides with $C^{1,1}$ when $s=1$, which is the optimal regularity in the classical case of the Laplacian. However, adapting the ideas of the classical proof to the fractional case suggests that the optimal regularity should be only $C^{2s}$. The optimal regularity in the case $s<1$ is better than the order of the equation and cannot be justified by any simple scaling argument.

Below, we outline the steps leading to the optimal regularity with a sketch of the ideas used in the proofs.

Almost $C^{2s}$ regularity

This first step of the proof is the simplest and it is the only step which is an adaptation of the classical case $s=1$.

From the statement of the equation we have $(-\Delta)^s u \geq 0$.

Since the average of two $s$-superharmonic function is also $s$-superharmonic, one can see that for any $h \in \mathbb R^d$, the function $v(x):=(u(x+h)+u(x-h))/2 + C|h|^2$ is $s$ superharmonic and $v \geq \varphi$ if $C = ||D^2 \varphi||_{L^\infty}$. By the comparison principle $v \geq u$. This means that $u$ is semiconvex: $D^2 u \geq -C I$.

Interpolating the semiconvexity and $L^\infty$ boundedness of $u$, we obtain that $(-\Delta)^s u \leq C$ for some constant $C$.

The boundedness of $(-\Delta)^s u$ does not imply that $u \in C^{2s}$ but it does imply that $u \in C^\alpha$ for all $\alpha < s$.

$C^{2s+\alpha}$ regularity, for some small $\alpha>0$

Let $w(x) = (-\Delta)^s u(x)$. A key observation is that the function $w$ satisfies the equation \begin{align} (-\Delta)^{1-s} w = -\Delta \varphi \qquad \text{in } \{u=\varphi\}, \\ w = 0 \varphi \qquad \text{outside } \{u=\varphi\}. \end{align}

This is a Dirichlet problem for the conjugate fractional Laplacian. However there are two difficulties. First of all we need to prove that $w$ is continuous on the boundary $\partial \{u=\varphi\}$. Second, this boundary can be highly irregular a priori so we cannot expect to obtain any H\"older continuity of $w$ from the Dirichlet problem alone.

From the semiconvexity of $u$ we have $-\Delta u \leq C$, and therefore we derive the extra condition $(-\Delta)^{1-s} w \leq C$ in the full space $\R^d$ (in particular across the boundary $\partial \{u=\varphi\}$). Moreover, we also know that $w \geq 0$ everywhere.

The $C^\alpha$ Holder continuity of $w$ on the boundary $\partial \{u=\varphi\}$ is obtained from an iterative improvement of oscillation procedure. Since $w \geq 0$ and $(-\Delta)^{1-s} u \leq C$, for any $x_0$ on $\partial \{u=\varphi\}$ we can show that $\max_{B_r(x_0)} w$ decays provided that $\{u > \varphi\} \cap B_r$ is sufficiently "thick" using the weak Harnack inequality. We cannot rule out the case in which $\{u > \varphi\} \cap B_r$ has a very small measure. However, in the case that $\{u > \varphi\} \cap B_r$ is too small in measure, we can prove that $u$ separates very slowly from $\varphi$. This slow separation is used to prove that $w$ must also improve its oscillation and this step is particularly tricky [1].

Once we know that $w(x) = (-\Delta)^s u(x)$ is $C^\alpha$, this implies that $u \in C^{2s+\alpha}$ by classical potential analysis theory.

$C^{1,s}$ regularity

If the contact set $\{u=\varphi\}$ is convex or at least has an exterior ball condition, a fairly simple barrier function can be constructed to show that $w$ must be $C^{1-s}$ on the boundary $\partial \{u=\varphi\}$. This is the generic boundary regularity for solutions of fractional Laplace equations in smooth domains.

Without assuming anything on the contact set $\{u=\varphi\}$, one can still obtain that $w \in C^\alpha$ for every $\alpha < 1-s$ though an iterative use of barrier functions [1]. The sharp $w \in C^{1-s}$ regularity in full generality was obtained rewriting the equation as a thin obstacle problem using the extension technique and then applying blowup techniques, the Almgren monotonicity formula and classification of global solutions [2].

Regularity of the free boundary

The parabolic version

References

  1. ↑ 1.0 1.1 Silvestre, Luis (2007), "Regularity of the obstacle problem for a fractional power of the Laplace operator", Communications on Pure and Applied Mathematics (Wiley Online Library) 60 (1): 67–112, ISSN 0010-3640 
  2. ↑ Caffarelli, Luis A.; Salsa, Sandro; Silvestre, Luis (2008), "Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian", Inventiones Mathematicae 171 (2): 425–461, doi:10.1007/s00222-007-0086-6, ISSN 0020-9910, http://dx.doi.org/10.1007/s00222-007-0086-6 
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