On conformable fractional calculus
By Thabet Abdeljawad. Journal of Computational and Applied Mathematics, 2015
1,954 citations recorded by the source corpus.
Subfields: Applied Mathematics, Modeling and Simulation, Numerical Analysis
View Paper: https://www.semanticscholar.org/paper/b0cdf3d62ae812722f1dda02c30b0681665d4a4b
How researchers see this paper
The researchers whose trusted networks rank it highest, strongest advocate first. Every researcher gets a different order.
In Applied Mathematics
- Y. Chu
- Thabet Abdeljawad
- Dumitru Băleanu
- Fahd Jarad
- M. Adil Khan
- Saima Rashid
- Pshtiwan Othman Mohammed
- Mehmet Yavuz
In Modeling and Simulation
- Dumitru Băleanu
- Thabet Abdeljawad
- Y. Chu
- Fahd Jarad
- Mehmet Yavuz
- Saima Rashid
- Pshtiwan Othman Mohammed
- M. Adil Khan
In Numerical Analysis
- Dumitru Băleanu
- Thabet Abdeljawad
- Mehmet Yavuz
- Fahd Jarad
- Pshtiwan Othman Mohammed
- Y. Chu
- Saima Rashid
- Delfim F. M. Torres
Citations in the corpus
- The flaw in the conformable calculus: It is conformable because it is not fractional (2018, background): “Many examples are given in [1] for functions differentiable on ]a, b[ such that T a αf(a) exists, but f ′ +(a) does not.”
- A Conformable Fractional Calculus on Arbitrary Time Scales (2015, background): “In particular, the new fractional calculus on time scales unifies and generalizes: the Hilger calculus (Bohner and Peterson, 2001; Hilger, 1990), obtained by choosing a 1⁄4 1; and the conformable fractional calculus (Abdeljawad, 2015; Khalil et al., 2014; Batarfi et al., 2015), obtained by choosing ”
- New Modified Conformable Fractional Integral Inequalities of Hermite–Hadamard Type with Applications (2020, background): “In [13], the following conformable derivative was defined:”
- Conformable Fractional Integrals Versions of Hermite-Hadamard Inequalities and Their Generalizations (2018, neutral): “(9) Theorem 8 (see [13] (integration by parts)).”
- Existence for fractional Dirichlet boundary value problem under barrier strip conditions (2017, background): “5, then, f(t, x,p) > 0 for (t, x,p) ∈ [0, 1]× [−1, 1]× ([4, 5]∪ [−2,−1]), f(t, x,p) 6 0 for (t, x,p) ∈ [0, 1]× [−1, 1]× ([1, 2]∪ [0, 0.”