Obstacle problem for the fractional Laplacian

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.

Almost $C^{2s}$ regularity

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 $C$.