Letter to the editorA critical analysis of the Caputo–Fabrizio operator
Introduction
In recent years, several operators were proposed having the keyword “fractional”. One of the proposals consists of the Caputo-Fabrizio (CF) operator [1] that is defined by the expression where 0 ≤ α ≤ 1, and M(α) is a normalization function such that . We will assume that (1) is valid for functions having Laplace transform (LT).
In the following we analyse this operator and we verify that it implements an integer order highpass filter. We can prove that the CF operator is neither fractional, nor a derivative, by showing that it does not verify the Leibniz rule [11]. The proof is somehow involved and we shall adopt here a different perspective based on Laplace transform and Bode diagrams.
Section snippets
Is the CF operator fractional?
Let us consider expression (1) and just in order to simplify expressions. Recalling that the multiplication of 2 Laplace transforms corresponds to the convolution in the time domain, Eq. (1) can be written as where ε(t) stands for the Heaviside unit step and * denotes the convolution.
The bilateral Laplace transform (no initial conditions are needed [9]) of (2) gives where and . Eq. (3) can be
Conclusions
The analysis of the Caputo–Fabrizio operator revealed that it is neither a fractional nor a derivative operator [6]. Similarly, the generalization of the operator is not a fractional derivative.
In this short note was adopted a strategy based on Laplace transform and related properties. The results support the use in FC of mathematical tools well established in the area of applied sciences.
Conflict of Interest
The authors declare that they have no conflict of interest.
Acknowledgements
This work was partially funded by National Funds through the Foundation for Science and Technology of Portugal, under the projects PEst-UID/EEA/00066/2013.
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