Katugampola fractional operators

These operators have been defined on the following extended-Lebesgue space.

Let ${\displaystyle {\textit {X}}_{c}^{p}(a,b),\;c\in \mathbb {R} ,\,1\leq p\leq \infty }$ be the space of those Lebesgue measurable functions ${\displaystyle f}$ on ${\displaystyle [a,b]}$ for which ${\displaystyle \|f\|_{{\textit {X}}_{c}^{p}}<\infty }$, where the norm is defined by [1]

$${\displaystyle {\begin{aligned}\|f\|_{{\textit {X}}_{c}^{p}}=\left(\int _{a}^{b}|t^{c}f(t)|^{p}{\frac {dt}{t}}\right)^{1/p}<\infty ,\end{aligned}}}$$

for ${\displaystyle 1\leq p<\infty ,\,c\in \mathbb {R} }$ and for the case ${\displaystyle p=\infty }$

$${\displaystyle {\begin{aligned}\|f\|_{{\textit {X}}_{c}^{\infty }}={\text{ess sup}}_{a\leq t\leq b}[t^{c}|f(t)|],\quad (c\in \mathbb {R} ).\end{aligned}}}$$

It is defined via the following integrals [1][2][9][10][11]

${\displaystyle ({}^{\rho }{\mathcal {I}}_{a+}^{\alpha }f)(x)={\frac {\rho ^{1-\alpha }}{\Gamma (\alpha )}}\int _{a}^{x}{\frac {\tau ^{\rho -1}f(\tau )}{(x^{\rho }-\tau ^{\rho })^{1-\alpha }}}\,d\tau ,}$

(1)

  for ${\displaystyle x>a}$ and ${\displaystyle \operatorname {Re} (\alpha )>0.}$ This integral is called the left-sided fractional integral. Similarly, the right-sided fractional integral is defined by,  

${\displaystyle ({}^{\rho }{\mathcal {I}}_{b-}^{\alpha }f)(x)={\frac {\rho ^{1-\alpha }}{\Gamma ({\alpha })}}\int _{x}^{b}{\frac {\tau ^{\rho -1}f(\tau )}{(\tau ^{\rho }-x^{\rho })^{1-\alpha }}}\,d\tau .}$

(2)

  for ${\displaystyle \textstyle x<b}$ and ${\displaystyle \textstyle \operatorname {Re} (\alpha )>0}$.

These are the fractional generalizations of the ${\displaystyle n}$-fold left- and right-integrals of the form

${\displaystyle \int _{a}^{x}t_{1}^{\rho -1}\,dt_{1}\int _{a}^{t_{1}}t_{2}^{\rho -1}\,dt_{2}\cdots \int _{a}^{t_{n-1}}t_{n}^{\rho -1}f(t_{n})\,dt_{n}}$

and

${\displaystyle \int _{x}^{b}t_{1}^{\rho -1}\,dt_{1}\int _{t_{1}}^{b}t_{2}^{\rho -1}\,dt_{2}\cdots \int _{t_{n-1}}^{b}t_{n}^{\rho -1}f(t_{n})\,dt_{n}}$ for ${\displaystyle \textstyle n\in \mathbb {N} ,}$

respectively. Even though the integral operators in question are close resemblance of the famous Erdélyi–Kober operator, it is not possible to obtain the Hadamard fractional integrals as a direct consequence of the Erdélyi–Kober operators. Also, there is a corresponding fractional derivative, which generalizes the Riemann–Liouville and the Hadamard fractional derivatives. As with the case of fractional integrals, the same is not true for the Erdélyi–Kober operator.

As with the case of other fractional derivatives, it is defined via the Katugampola fractional integral.[3][9][10][11]

Let ${\displaystyle \alpha \in \mathbb {C} ,\ \operatorname {Re} (\alpha )\geq 0,n=[\operatorname {Re} (\alpha )]+1}$ and ${\displaystyle \rho >0.}$ The generalized fractional derivatives, corresponding to the generalized fractional integrals (1) and (2) are defined, respectively, for ${\displaystyle 0\leq a<x<b\leq \infty }$, by

The half-derivative of the function ${\displaystyle f(x)=x^{0.5}}$ for the Katugampola fractional derivative.

The half derivative of the function ${\displaystyle f(x)=x^{\nu }}$ for the Katugampola fractional derivative for ${\displaystyle \alpha =0.5}$ and ${\displaystyle \rho =2}$.

${\displaystyle {\begin{aligned}{\big (}{}^{\rho }{\mathcal {D}}_{a+}^{\alpha }f{\big )}(x)&={\bigg (}x^{1-\rho }\,{\frac {d}{dx}}{\bigg )}^{n}\,\,{\big (}{}^{\rho }{\mathcal {I}}_{a+}^{n-\alpha }f{\big )}(x)\\&={\frac {\rho ^{\alpha -n+1}}{\Gamma ({n-\alpha })}}\,{\bigg (}x^{1-\rho }\,{\frac {d}{dx}}{\bigg )}^{n}\int _{a}^{x}{\frac {\tau ^{\rho -1}f(\tau )}{(x^{\rho }-\tau ^{\rho })^{\alpha -n+1}}}\,d\tau ,\end{aligned}}}$

and

${\displaystyle {\begin{aligned}{\big (}{}^{\rho }{\mathcal {D}}_{b-}^{\alpha }f{\big )}(x)&={\bigg (}-x^{1-\rho }\,{\frac {d}{dx}}{\bigg )}^{n}\,\,{\big (}{}^{\rho }{\mathcal {I}}_{b-}^{n-\alpha }f{\big )}(x)\\&={\frac {\rho ^{\alpha -n+1}}{\Gamma ({n-\alpha })}}{\bigg (}-x^{1-\rho }{\frac {d}{dx}}{\bigg )}^{n}\int _{x}^{b}{\frac {\tau ^{\rho -1}f(\tau )}{(\tau ^{\rho }-x^{\rho })^{\alpha -n+1}}}\,d\tau ,\end{aligned}}}$

respectively, if the integrals exist.

These operators generalize the Riemann–Liouville and Hadamard fractional derivatives into a single form, while the Erdelyi–Kober fractional is a generalization of the Riemann–Liouville fractional derivative.[3] When, ${\displaystyle b=\infty }$, the fractional derivatives are referred to as Weyl-type derivatives.

As in the case of Laplace transforms, Mellin transforms will be used specially when solving differential equations. The Mellin transforms of the left-sided and right-sided versions of Katugampola Integral operators are given by [2][4]

Theorem

Let ${\displaystyle \alpha \in {\mathcal {C}},\ \operatorname {Re} (\alpha )>0,}$ and ${\displaystyle \rho >0.}$ Then,

$${\displaystyle {\begin{aligned}&{\mathcal {M}}{\bigg (}{}^{\rho }{\mathcal {I}}_{a+}^{\alpha }f{\bigg )}(s)={\frac {\Gamma {\big (}1-{\frac {s}{\rho }}-\alpha {\big )}}{\Gamma {\big (}1-{\frac {s}{\rho }}{\big )}\,\rho ^{\alpha }}}\,{\mathcal {M}}f(s+\alpha \rho ),\quad \operatorname {Re} (s/\rho +\alpha )<1,\,x>a,\\&{\mathcal {M}}{\bigg (}{}^{\rho }{\mathcal {I}}_{b-}^{\alpha }f{\bigg )}(s)={\frac {\Gamma {\big (}{\frac {s}{\rho }}{\big )}}{\Gamma {\big (}{\frac {s}{\rho }}+\alpha {\big )}\,\rho ^{\alpha }}}\,{\mathcal {M}}f(s+\alpha \rho ),\quad \operatorname {Re} (s/\rho )>0,\,x<b,\end{aligned}}}$$

for ${\displaystyle f\in {\textit {X}}_{s+\alpha \rho }^{1}(\mathbb {R} ^{+})}$, if ${\displaystyle {\mathcal {M}}f(s+\alpha \rho )}$ exists for ${\displaystyle s\in \mathbb {C} }$.

Katugampola operators satisfy the following Hermite-Hadamard type inequalities:[16]

Theorem

Let ${\displaystyle \alpha >0}$ and ${\displaystyle \rho >0}$. If ${\displaystyle f}$ is a convex function on ${\displaystyle [a,b]}$, then

$${\displaystyle f\left({\frac {a+b}{2}}\right)\leq {\frac {\rho ^{\alpha }\Gamma (\alpha +1)}{4(b^{\alpha }-a^{\alpha })^{\alpha }}}\left[{}^{\rho }{\mathcal {I}}_{a+}^{\alpha }F(b)+{}^{\rho }{\mathcal {I}}_{b-}^{\alpha }F(a)\right]\leq {\frac {f(a)+f(b)}{2}},}$$

where ${\displaystyle F(x)=f(x)+f(a+b-x),\;x\in [a,b]}$.

When ${\displaystyle \rho \rightarrow 0^{+}}$, in the above result, the following Hadamard type inequality holds:[16]

Corollary

Let ${\displaystyle \alpha >0}$. If ${\displaystyle f}$ is a convex function on ${\displaystyle [a,b]}$, then

$${\displaystyle f\left({\frac {a+b}{2}}\right)\leq {\frac {\Gamma (\alpha +1)}{4\left(\ln {\frac {b}{a}}\right)^{\alpha }}}\left[\mathbf {I} _{a+}^{\alpha }F(b)+\mathbf {I} _{b-}^{\alpha }F(a)\right]\leq {\frac {f(a)+f(b)}{2}},}$$

where ${\displaystyle \mathbf {I} _{a+}^{\alpha }}$ and ${\displaystyle \mathbf {I} _{b-}^{\alpha }}$ are left- and right-sided Hadamard fractional integrals.

These operators have been mentioned in the following works:

1. Fractional Calculus. An Introduction for Physicists, by Richard Herrmann [17]
2. Fractional Calculus of Variations in Terms of a Generalized Fractional Integral with Applications to Physics, Tatiana Odzijewicz, Agnieszka B. Malinowska and Delfim F. M. Torres, Abstract and Applied Analysis, Vol 2012 (2012), Article ID 871912, 24 pages [18]
3. Introduction to the Fractional Calculus of Variations, Agnieszka B Malinowska and Delfim F. M. Torres, Imperial College Press, 2015
4. Advanced Methods in the Fractional Calculus of Variations, Malinowska, Agnieszka B., Odzijewicz, Tatiana, Torres, Delfim F.M., Springer, 2015
5. Expansion formulas in terms of integer-order derivatives for the Hadamard fractional integral and derivative, Shakoor Pooseh, Ricardo Almeida, and Delfim F. M. Torres, Numerical Functional Analysis and Optimization, Vol 33, Issue 3, 2012, pp 301–319.[19]

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