Abstract

Fractional derivatives with three parameter generalized Mittag-Leffler kernels and their properties are studied. The corresponding integral operators are obtained with the help of Laplace transforms. The action of the presented fractional integrals on the Caputo and Riemann type derivatives with three parameter Mittag-Leffler kernels is analyzed. Integration by parts formulas in the sense of Riemann and Caputo are proved and then used to formulate the fractional Euler–Lagrange equations with an illustrative example. Certain nonconstant functions whose fractional derivatives are zero are determined as well.

Highlights

  • Fractional derivatives with three parameter generalized Mittag-Leffler kernels and their properties are studied

  • Keeping the above mentioned things in our minds, we discuss in our manuscript the fractional operators with three parameter generalized Mittag-Leffler kernels and we present the integration by parts formulas, and we obtain the related fractional Euler–Lagrange equations

  • 5 Conclusions The fractional derivatives studied in [15, 17] are of interest for real world problems since they contain nonsingular Mittag-Leffler kernels and their corresponding fractional integrals are expressed by mean of the classical Riemann fractional integrals

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Summary

ABR a

Remark 1 Note that the above generalized type fractional derivatives have singular kernels for 0 < μ < 1. The one parameter ML function kernel defined in [15] is nonsingular. The limiting process α, μ, γ → 1 gives the ordinary derivative. Eα–,γν (λ, x), ν →1–μ is a nonzero function such that its ABR and ABC derivatives are zero. By inspection we report that the function G(x) tends to 1 as μ → 1– and α → 1 with γ = 1. It is of interest to study the fractional polynomial function Gγ (x) with γ = 1, 2, 3, . We solve the equation (Aa BRDα,μ,γ f )(t) = u(t) with γ = 1 to find the fractional integral operator of two parameters.

From which it follows that
ABRDαb f
Proof From the relations
The discussion in the previous section shows that
If f
ABC a
Then b
Notice that μ
Conclusions
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