Numerical Approximation of the Fractional Laplacian via $hp$-finite Elements, with an Application to Image Denoising

Abstract

The fractional Laplacian operator $(-\varDelta )^s$ on a bounded domain $\varOmega$ can be realized as a Dirichlet-to-Neumann map for a degenerate elliptic equation posed in the semi-infinite cylinder $\varOmega \times (0,\infty )$. In fact, the Neumann trace on $\varOmega$ involves a Muckenhoupt weight that, according to the fractional exponent $s$, either vanishes $(s < 1/2)$ or blows up $(s > 1/2)$. On the other hand, the normal trace of the solution has the reverse behavior, thus making the Neumann trace analytically well-defined. Nevertheless, the solution develops an increasingly sharp boundary layer in the vicinity of $\varOmega$ as $s$ decreases. In this work, we extend the technology of automatic $hp$-adaptivity, originally developed for standard elliptic equations, to the energy setting of a Sobolev space with a Muckenhoupt weight, in order to accommodate for the problem of interest. The numerical evidence confirms that the method maintain exponential convergence. Finally, we discuss image denoising via the fractional Laplacian. In the image processing community, the standard way to apply the fractional Laplacian to a corrupted image is as a filter in Fourier space. This construction is inherently affected by the Gibbs phenomenon, which prevents the direct application to “spliced” images. Since our numerical approximation relies instead on the extension problem, it allows for processing different portions of a noisy image independently and combine them, without complications induced by the Gibbs phenomenon.

Appendix: Extension Problem: Separation of Variables

Our goal is to construct a solution to the extension problem (5) via separation of variables and show that the restriction of such a solution to $\varOmega$ coincides with the solution of the fractional Laplacian as defined in (3) through the eigenmodes of the Dirichlet Laplace operator. Let us recall 5:

$$\begin{aligned} - \mathrm{div }(y^\alpha \nabla u)&= 0 \qquad \text {in } \varOmega \times (0, \infty ) \end{aligned}$$

(9a)

$$\begin{aligned} u&= 0 \qquad \text {on } \partial \varOmega \times [0,\infty ) \,\, , \end{aligned}$$

(9b)

$$\begin{aligned} (y^\alpha \nabla u) \cdot \varvec{\nu }&= d_s f \quad \text {on } \varOmega \times \{0\} \end{aligned}$$

(9c)

where $d_s$ is a suitable constant to be determined, and (9b) (9b) are understood in the sense of the trace. We shall assume $f \in L^2(\varOmega )$. If we let $u(x,y)= X(x)Y(y)$, then Eq. (9a) can be separated as follows:

$$\begin{aligned} \frac{\alpha }{y} \frac{Y'}{Y} +\frac{Y''}{Y} = - \frac{\varDelta X}{X} = c^2\,\,. \end{aligned}$$

Using the boundary condition (9b), it is immediate that $X$ coincides with the eigenmodes $\{ \varphi _k\}$ of the Dirichlet Laplace problem on $\varOmega$, and the separation constant $c^2$ must be an eigenvalue $\mu _k$ of such a problem. Thus, $X= \varphi _k$, and $c^2 = \mu _k$. With the separation constant determined, we move to the ordinary differential equation for $Y$. In the case $\alpha = 0$, i.e., $s=1/2$, we obtain $Y = \exp (-\mu _k^{1/2} y)$. Namely, we discarded the solutions associated with $\exp (\mu _k^{1/2}y)$ because of the energy assumption $u \in H^1({\mathcal {C}},w)$. In the case $\alpha \ne 0$, function $Y$ satisfies the following equation:

$$\begin{aligned} y^2 \,Y'' + \alpha y \, Y' -(\mu _k^{1/2}y)^2 Y = 0\,\,. \end{aligned}$$

By seeking a solution $Y$ of the form $(\mu _k^{1/2}y)^s \, g(\mu _k^{1/2}y)$, after lengthy yet trivial computations, we arrive at a modified Bessel equation for $g$:

$$\begin{aligned} (\mu _k^{1/2}y)^2 g''(\mu _k^{1/2}y) + (\mu _k^{1/2}y) g'(\mu _k^{1/2}y) - \big ((\mu _k^{1/2}y)^2 + s^2) g(\mu _k^{1/2}y) = 0\,\,. \end{aligned}$$

The general solution is given by a linear combination of $I_s$ and $K_s$, i.e., the modified Bessel function of the first kind and second kind, respectively. As in the previous case, because of the energy assumption, we discard the solutions associated to $I_s$. Thus, if we define:

$$\begin{aligned} \psi _k = \exp (-\mu _k^{1/2}y) \quad \text {for }s = \tfrac{1}{2} ; \qquad \psi _k = (\mu _k^{1/2}y)^s \, K_s(\mu _k^{1/2}y) \quad \text {otherwise} \end{aligned}$$

we obtain the following expression:

$$\begin{aligned} u(x,y) = \sum \limits _k u_k \, \varphi _k(x) \psi _k(y) \end{aligned}$$

where the coefficients $u_k$ are to be determined through (9c). Since we are dealing with smooth functions, the left-hand side of (9c) reduces to the limit $\lim _{y \downarrow 0} -y^\alpha u_y$. In the case $s = 1/2$, we immediately obtain that:

$$\begin{aligned} u_y(x,0) = \sum \limits _k u_k \, \varphi _k(x) \psi _k'(0) = -\sum \limits _k u_k \mu _k^{1/2}\, \varphi _k(x) \end{aligned}$$

Since $\{\varphi _k\}$ is a complete orthogonal set for $L^2(\varOmega )$, we also have that $f = \sum \nolimits _k f_k\varphi _k$. Equation (9c) implies that $f_k = u_k\mu _k^{1/2}$ for all $k$’s and, thus, $(-\varDelta )^{1/2} u = f$ on $\varOmega$ as desired. In the case $s\ne 1/2$, in order to make use of Eq. (9c), we employ the following properties of function $K_s$:

• 1. $K_s(z) = K_{-s}(z)$;

• 2. $K_s'(z) = -\frac{1}{2}\big [ K_{s-1}(z) + K_{s+1}(z)\big ]$;

• 3. $K_s(z) \simeq \frac{1}{2} \varGamma (s)\big ( \frac{1}{2}z\big )^{-s}, \quad z\downarrow 0,\quad \quad s > 0$.

We have that:

$$\begin{aligned} \lim \limits _{y \downarrow 0}\, y^\alpha \psi _k'&= \lim \limits _{y \downarrow 0}\, y^\alpha \big [ s(\mu _k^{1/2} y)^{s-1} \mu _k^{1/2} K_s(\mu _k^{1/2} y) + (\mu _k^{1/2} y)^s \mu _k^{1/2} K_s'(\mu _k^{1/2} y) \big ] \\&= \lim \limits _{y \downarrow 0}\, y^\alpha \Big [ s\mu _k^{s/2} y^{s-1} K_s(\mu _k^{1/2} y) - \mu _k^{(s+1)/2} y^s \tfrac{1}{2}\big [ K_{s-1}(\mu _k^{1/2} y) + K_{s+1}(\mu _k^{1/2} y)\big ] \Big ] \\&= \lim \limits _{y \downarrow 0}\, y^\alpha \Big [ s\mu _k^{s/2} y^{s-1} K_s(\mu _k^{1/2} y) - \mu _k^{(s+1)/2} y^s \tfrac{1}{2}\big [ K_{1-s}(\mu _k^{1/2} y) + K_{s+1}(\mu _k^{1/2} y)\big ] \Big ] \\&= \lim \limits _{y \downarrow 0}\, y^\alpha \Big [ s\mu _k^{s/2} y^{s-1} \tfrac{1}{2} \varGamma (s)\big ( \tfrac{1}{2}\mu _k^{1/2} y\big )^{-s} + \\&\qquad - \mu _k^{(s+1)/2} y^s \tfrac{1}{2}\big [ \tfrac{1}{2} \varGamma (1-s)\big ( \tfrac{1}{2}\mu _k^{1/2} y\big )^{s-1} + \tfrac{1}{2} \varGamma (s+1)\big ( \tfrac{1}{2}\mu _k^{1/2} y\big )^{-s-1}\big ] \Big ] \\&= \lim \limits _{y \downarrow 0}\, y^\alpha \Big [\varGamma (s+1)\tfrac{1}{2}^{1-s}y^{-1} + \\&\qquad -\varGamma (1-s) \tfrac{1}{2}^{s+1} \mu _k^s y^{2s-1} - \varGamma (s+1) \tfrac{1}{2}^{1-s}y^{-1} \Big ] \\&= \lim \limits _{y \downarrow 0}\, y^{1-2s} \big [-\varGamma (1-s) \tfrac{1}{2}^{s+1} \mu _k^s y^{2s-1} \big ] \\&= - \varGamma (1-s) \tfrac{1}{2}^{s+1}\mu _k^s \\&= - d_s\mu _k^s \end{aligned}$$

where we defined $d_s := \varGamma (1-s) \tfrac{1}{2}^{s+1}$. Therefore Eq. (9c) can be equivalently written as

$$\begin{aligned} \lim \limits _{y \downarrow 0} y^\alpha \sum \limits _k u_k \varphi _k(x) \psi _k'(y) = d_s\sum \limits _k u_k \mu _k^s \,\varphi _k(x) = \sum \limits _k f_k \, \varphi _k(x) \end{aligned}$$

where, by virtue of the fact that $\{ \varphi _k\}$ is a complete set, we have that $f = \sum \nolimits _k f_k \varphi _k$. Thus, $f_k = d_s u_k \mu _k^s$ for all $k$’s and $(-\varDelta )^s u = d_s f$ on $\varOmega$.