Mellin definition of the fractional Laplacian

Abstract

It is known that at least ten equivalent definitions of the fractional Laplacian exist in an unbounded domain. Here we derive a further equivalent definition that is based on the Mellin transform and it can be used when the fractional Laplacian is applied to radial functions. The main finding is tested in the case of the space-fractional diffusion equation. The one-dimensional case is also considered, such that the Mellin transform of the Riesz (namely the symmetric Riesz–Feller) fractional derivative is established. This one-dimensional result corrects an existing formula in literature. Further results for the Riesz fractional derivative are obtained when it is applied to symmetric functions, in particular its relation with the Caputo and the Riemann–Liouville fractional derivatives.

Appendix: Derivation of the kernel function \(\mathcal {E}(\xi )\) (3.20)

By using the rule $\sin (\theta +\omega )=\sin \theta \cos \omega + \cos \theta \sin \omega$, formula (3.19) can be re-written as

$$\begin{aligned} \mathcal {M}\mathcal {E}(s)=\sin \left( \frac{\pi \alpha }{2}\right) \, \cot \left( \frac{\pi s}{2}\right) - \cos \left( \frac{\pi \alpha }{2}\right) \,, \end{aligned}$$

and then we want two functions $\varphi _1(r)$ and $\varphi _2(r)$ such that $\mathcal {M}\varphi _1 = \cot (\pi s/2)$ and $\mathcal {M}\varphi _2 = 1$.

By noting from [34, formula (2.24)] that

$$\begin{aligned} \int _0^\infty \frac{x^{s-1}}{1-x} \, dx = \pi \cot \pi s \,, \quad \Re (s) \in (0,1) \,, \end{aligned}$$

we have, after replacing $x \rightarrow r^2$, the first pair

$$\begin{aligned} \mathcal {M}\varphi _1 (s)= \frac{\pi }{2} \cot \frac{\pi s}{2} \,, \quad \text {with} \quad \varphi _1(r)=\frac{1}{1-r^2} \,. \end{aligned}$$

Now, let $\mathcal {M}^{-1}$ be the Mellin inverse transformation then for a given function $\psi (r)$ it holds $\psi =\mathcal {M}^{-1}\mathcal {M}\psi$. Therefore, after the exchange of integration order we have

$$\begin{aligned} \psi (r)=\int _0^\infty \psi (y) \left\{ \mathcal {M}^{-1} y^s \right\} \frac{dy}{y} \,. \end{aligned}$$

This integral can be understood as a Mellin convolution integral where by definition $\mathcal {M}^{-1} y^s = q(r/y)$ and, by using the properties of the delta-function, it holds $q(z)=z \delta (z-1)$ with $z=r/y$. Function q(r/y) is consistent with $\mathcal {M}q = y^s$. Thus, by setting $y=1$, we have also the second pair

$$\begin{aligned} \mathcal {M}\varphi _2(s)=1 \,, \quad \text {with} \quad \varphi _2(r)=r\delta (r-1) \,. \end{aligned}$$

Finally it results

$$\begin{aligned} \mathcal {M}\mathcal {E}(s)= \frac{2}{\pi }\sin \left( \frac{\pi \alpha }{2}\right) \mathcal {M}\varphi _1(s) - \cos \left( \frac{\pi \alpha }{2}\right) \mathcal {M}\varphi _2(s) \,, \quad \Re (s) \in (0,1) \,, \end{aligned}$$

from which formula (3.20) follows.

However, also setting $\mathcal {M}^{-1} y^s = q^*(r/y)= \delta (r/y-1)$ is appropriate for solving the integral through properties of delta-function and it is consistent with the Mellin convolution interpretation and with formula $\mathcal {M}q^*=y^s$. Thus, in our case, by setting $y=1$ another suitable choice for function $\varphi _2(r)$ is $\varphi _2(r)=\delta (r-1)$. Moreover, by disregarding the Mellin convolution interpretation, also setting $\mathcal {M}^{-1} y^s = q'(r,y)=y \delta (r-y)$ or $\mathcal {M}^{-1} y^s = q{''}(r,y)=r \delta (r-y)$ is appropriate and consistent with the formula $\mathcal {M}q'=\mathcal {M}q{''}=y^s$. Actually, this procedure does not solve the ambiguity emerging from different rearrangements of the delta-function like $a\delta (r-a)$ and $r\delta (r-a)$, with $a>0$, that provide the same Mellin transform. More extensive analysis on the Mellin transform and delta-function are available in literature [52, 53].