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A fractional derivative of distributions is usually introduced using definition of fractional integral as a convolution of two distributions. However there is another approach based on fractional integration by parts rule, suggested by Samko, Kilbas, Marichev. Let

$C_0^\infty(a,b)=\{ \phi \in C^\infty(a,b): \phi^{(k)}(a) = \phi^{(k)}(b) = 0, k=0, 1, 2,\ldots \}$

be a space of test functions. Linear functionals on $C_0^\infty(a,b)$ are distributions or generalized functions. Fractional derivative of a distribution is then defined in terms of adjoint fractional differentiation operators: $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle,\ 0 < \alpha < 1$.

However in this case Riemann--Liouville and regularized fractional derivatives are indistinguishable. Consider regularized (Caputo) derivative $D_{a+}^{(\alpha)} f(x) = D_{a+}^\alpha (f(x) - f(a))$. Obviously

$D_{a+}^{(\alpha)} \phi = D_{a+}^{\alpha} \phi\ \forall \phi \in C_0^\infty(a,b)$

and $\langle D_{a+}^\alpha f,\phi \rangle = \langle f, D_{b-}^\alpha \phi \rangle = \langle D_{a+}^{(\alpha)} f,\phi \rangle$.

On the other hand, we know that the regularized derivative of a constant is zero while for the Riemann--Liouville derivative we have $D^\alpha_{a+}(x-a)^{\alpha - 1}=0$.

Hence a question: which distribution has zero fractional derivative in the sense described above? Is this a constant or $(x-a)^{\alpha - 1}$ or both?

Thank you.

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3 Answers

Dear Ivan

Let me put things in a slightly different fashion. We work on R and so our definitions of fractional derivative must reflect this fact. If you incorporate a particular property of the function into the definition you must be creating problems that you cannot solve. this is the case of the usual RL and C derivatives that "start" at zero. This means that they do not distinguish the constant function from the Heaviside function. This originates several problems. See, for example, my paper "Ortigueira, M. D. and Coito, F.J., “On the Usefulness of Riemann-Liouville and Caputo Derivatives in Describing Fractional Shift-invariant Linear Systems”, Journal of Applied Nonlinear Dynamics, 1(2) (2012) 113–124, DOI:10.5890/JAND.2012.05.001.". I recomend also to read the paper "The infinite state approach: Origin and necessity" by J.C. Trigeassou, N. Maamri, A. Oustaloup, that is available at the site Computers and Mathematics with Applications. Returning back to the original question: the derivative of a constant is zero with the Grunwald-Letnikov derivative. I can send you my papers on the subject if you want mdo@fct.unl.pt

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Dear Prof. Ortigueira! Thank you for your answer. It really gives me a new avenue to investigate. If you don’t mind I will email you some details of my current research. I would also appreciate you sending me the papers you mention. –  Ivan Matan Feb 10 '13 at 15:44
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Consider a distribution defined by the function $\frac{(x-a)^{\alpha-1}}{\Gamma(\alpha)}$. We define its derivative of order $\alpha$ as a distribution that acts upon the functions from the test space $C_0^\infty(a,b)$ as follows:

$\int_a^b \frac{(x-a)^{\alpha-1}}{\Gamma(\alpha)} D_{b-}^\alpha \phi(x) dx$

$=\int_a^b \frac{(x-a)^{\alpha-1}}{\Gamma(\alpha)} D_{b-}^{(\alpha)} \phi(x) dx$

$=\int_a^b \frac{(x-a)^{\alpha-1}}{\Gamma(\alpha)} I_{b-}^{1-\alpha} \phi'(x) dx$

$=\int_a^b I_{a+}^{1-\alpha}\left(\frac{(x-a)^{\alpha-1}}{\Gamma(\alpha)}\right) \phi'(x) dx$

$=\int_a^b \phi'(x) dx = 0$.

The similar reasoning applied to a constant shows that distributional fractional derivative of a constant is not zero. Which is probably related to the fact that the integration-by-parts rule, generally speaking, doesn't hold for regularized derivatives.

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Yes. The problem is in the integration by parts, because the power function is irregular at x=a. Any way, for me, the problem is some how different. I think you should do a distinction between anti-derivative (derivative of negative order) and integral. They are not the same think. This may create some difficulties in looking for an integration constant that is not necessary to appear.

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