Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We want to know if there exists a fundamental theorem of fractional calculus for the Riesz Derivative (a type of fractional Laplacian), e.g. there exists an operator $L$ such that $-L_a^b((-\Delta)^\beta f)=f(b)-f(a)$ or something similar, where the fractional laplacian is defined via a Fourier space:

\begin{eqnarray} -(-\Delta)^\beta f = -{\mathcal{F}}^{-1}((|q|^{2\beta}\mathcal{F}(f)), \end{eqnarray} where $q$ is the variable in the Fourier domain.

Secondly, although the Riesz derivative is a left inverse for the Riesz potential, is there a way to get the right inverse? We need to solve some simple fractional differential equations, namely:

\begin{eqnarray} -(-\Delta)^\beta H&=&H,\;\lim_{|x|\to\infty}H=0, \end{eqnarray} and another equation that we try to solve is \begin{eqnarray} -(-\Delta)^\beta H&=&c,\; c\in\mathbb{R}. \end{eqnarray} Do we need to use other definitions of the fractional derivative for this to work nicely? One criterion that we need is that the fractional derivative should approach the classical derivative when the fractional exponent approaches an integer.

share|improve this question
add comment

3 Answers

The fractional Riez derivative can be written as a sum of Caputo or Reimann-Liouville fractional derivatives, so maybe this could help you?

share|improve this answer
add comment

The fractional Laplacean $(-\Delta)^{\alpha/2}$ in $\mathbb R^n$ is, up to some constant the Fourier multiplier $\vert\xi\vert^\alpha$. So its inverse is, at least formally, the Fourier multiplier $\vert\xi\vert^{-\alpha}$.

Let us assume that $0<\alpha < n$. Then $\vert\xi\vert^{-\alpha}$ is $L^1_{loc}$ and a tempered distribution homogeneous with degree $-\alpha$. Its Fourier transform is a tempered distribution homogeneous with degree $\alpha-n$ and in fact is also radial so is, up to a constant $\vert x\vert^{\alpha-n}$: a parametrix for $(-\Delta)^{\alpha/2}$ is the convolution by const. $\vert x\vert^{\alpha-n}$. Note that this is the case for $\alpha =2$, when $n\ge 3$. The constants are not uninteresting to compute.

When $\alpha =n$, $\vert \xi\vert^{-n}$ is not $L^1_{loc}$ and is not a tempered Fourier multiplier. The homogeneity is partly lost and you have to perform a direct calculation: for instance for $n=2$ you find $$ \frac{-1}{2\pi}\ln\vert x\vert $$ which is not homogeneous of degree 0 but whose derivatives are homogeneous of degree -1.

share|improve this answer
add comment

some time ago I defined derivatives that are similar to the Grunwald-Letnikov derivative, but that are two sided. One particular case has Fourier transform equal to |w|^beta.F(w). I showed that it is equivalent to the Riesz and has inverse if |beta| < 1. If you are interested, I can send you my papers: mdo@fct.unl.pt

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.