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 $s \in (0,1)$. Given some smooth function $\varphi$, 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.

Outline of the proof.

The proof consists of the following steps that we sketch below.

• 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 \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 .

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 $\alpha < 1-s$ through an iterative use of barrier functions . 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 .

The $C^{1-s}$ regularity of $w$ implies that $u \in C^{1,s}$ by classical potential analysis.

Regularity of the free boundary

A regular point of the free boundary is where the solution $u$ is exactly $C^{1,s}$ an no better. This is classified explicitly in terms of the limits of the Almgren frequency formula . Around any regular point, the free boundary is a smooth $C^{1,\alpha}$ surface .

A singular point is defined as a point on the free boundary where the measure of the contact set has vanishing density. More precisely, if \[ \lim_{r \to 0} \frac{|\{u=\varphi\} \cap B_r|}{r^n} = 0.\]

In the case $s=1/2$, it was shown by Nicola Garofalo and Arshak Petrosyan that the singular points of the free boundary are contained inside a differentiable surface. The proof is done in the context of the thin obstacle problem and presumably can be extended to other powers of the Laplacian using the extension technique.

It is important to notice that the definitions of regular and singular points of the free boundary are mutually exclusive but they do not exhaust all possible free boundary points. It is an interesting open problem to understand what other type of free boundary points are possible if any.

The parabolic version

The parabolic version of the fractional obstacle problem was studied by Caffarelli and Figalli . They concluded that the solution $u$ has the following regularity estimates. \begin{align} u_t, (-\Delta)^s u \in LogLip_t C_x^{1-s}, \text{ if } s\leq 1/3,\\ u_t, (-\Delta)^s u \in C_{t,x}^{\frac{1-s}{2s},{1-s}}, \text{ if } s > 1/3. \end{align}

It turns out that it is crucial to consider solutions $u$ to be non decreasing in time (which is assured by taking the initial value coinciding with the obstacle). Otherwise the regularity of the solution is reduced to merely $C^{2s}$ in space.

The regularity of the free boundary has not been explored in the parabolic setting yet.