[**On more general forms of proportional fractional operators**]{}

.20in Fahd Jarad$^{a}$, Manar A. Alqudah$^{b}$, Thabet Abdeljawad$^{c,d}$\

$^{a}$Department of Mathematics, Çankaya University, 06790 Ankara, Turkey\
email: fahd@cankaya.edu.tr\
$^{b}$ Department of Mathematical Sciences, Princess Nourah Bint Abdulrahman University\
P.O. Box 84428, Riyadh 11671, Saudi Arabia.\
email:maalqudah@pnu.edu.sa.\
$^{c}$Department of Mathematics and Physical Sciences, Prince Sultan University\
P. O. Box 66833, 11586 Riyadh, Saudi Arabia\
email:tabdeljawad@psu.edu.sa\
$^d$ Department of Medical Research, China Medical University, 40402, Taichung, Taiwan

.2in

Introduction
============

The fractional calculus, which is engaged in integral and differential operators of arbitrary orders, is as old as the conceptional calculus that deals with integrals and derivatives of non-negative integer orders. Since not all of the real phenomena can be modeled using the operators in the traditional calculus, researchers searched for generalizations of these operators. It turned out that the fractional operators are excellent tools to use in modeling long-memory processes and many phenomena that appear in physics, chemistry, electricity, mechanics and many other disciplines. Here, we invite the readers to read [@podlubny; @Samko; @f1; @f222; @f2; @f3] and the reference cited in these books. However, for the sake of better understanding and modeling real world problems, researchers were in need of other types of fractional operators that were confined to Riemann-Liouville fractional operators. In the literature, one can find many works that propose new fractional operators. We mention [@had; @Kat1; @Kat2; @fahd3; @fahd1; @fahd11]. Nonetheless, the fractional integrals and derivatives which were proposed in these works were just particular cases of what so called fractional integrals/derivatives of a function with respect to another function [@Samko; @f2; @fahd10]. There are other types of fractional operators which were suggested in the literature.

On the other hand, due to the singularities found in the traditional fractional operators which are thought to make some difficulties in the modeling process, some researches recently proposed new types of non-singular fractional operators. Some of these operators contain exponential kernels and some of them involve the Mittag-Leffler functions. For such types of fractional operators we refer to [@FCaputo; @Losada; @TD; @ROMP; @Abdon; @TD; @JNSA].

All the fractional operators considered in the references in the first and the second paragraphs are non-local. However, there are many local operators found in the literature that allow differentiation to a non-integer order and these are called local fractional operators. In [@kh], Khalil et al. introduced the so called conformable (fractional) derivative. The author in [@T11] presented other basic concepts of conformable derivatives. We would like to mention that the fractional operators proposed in [@Kat1; @Kat2] are the non-local fractional version of the local operators suggested in [@kh]. In addition, the non-local fractional version of the ones in [@T11] can be seen in [@fahd11].

It is customary that any derivative of order 0 when performed to a function should give the function itself. This essential property is dispossessed by the conformable derivatives. Notwithstanding, in [@Anderson1; @Anderson2], the authors introduced a newly defined local derivative that tend to the original function as the order tends to zero and hence improved the conformable derivatives. In addition to this, the non-local fractional operators that emerge from iterating the above-mentioned derivative were held forth in [@fahd12].

Motivated by the above mentioned background, we extend the work done in [@fahd12] introduce a new generalized fractional calculus based on the proportional derivatives of a function with respect to another function in paralel with the definition discussed in [@Anderson1]. The kernel obtained in the fractional operators which will be proposed contains an exponential function and is function dependent. The semi–group properties will be discussed.

The article is organized as follows: Section 2 presents some essential definitions for fractional derivatives and integrals. In Section 3 we present the general forms of the fractional proportional integrals and derivatives. In section 4, we present the general form of Caputo fractional proportional derivatives. In the end, we conclude our results.

Preliminaries
=============

In this section, we present some essential definitions of some fractional derivatives and integrals. We first present the traditional fractional operators and then the fractional proportional operators.

The conventional fractional operators and their general forms
-------------------------------------------------------------

For $\alpha \in \mathbb{C},~Re(\alpha)>0$, the left Riemann–Liouville fractional integral of order $\alpha $ has the f form $$\label{001}
(_{a}I^\alpha f)(x)=\frac{1}{\Gamma(\alpha)}\int_a^x (x-u)^{\alpha-1}f(u)du.$$

The right Riemann–Liouville fractional integral of order $\alpha >0$ is defined by $$\label{002}
(I_b^\alpha f)(x)=\frac{1}{\Gamma(\alpha)}\int_x^b (u-x)^{\alpha-1}f(u)du.$$ The left Riemann–Liouville fractional derivative of order $\alpha, Re(\alpha)\geq 0 $ is given as $$\label{003}
 (_{a}D^\alpha f)(x)=\Big(\frac{d}{dx}\Big)^n(_{a}I^{n-\alpha} f)(x),~~n=[\alpha]+1.$$ The right Riemann–Liouville fractional derivative of order $\alpha, Re(\alpha)\geq 0 $ reads $$\label{004}
(D_b^\alpha f)(t)=\Big(-\frac{d}{dt}\Big)^n(I_b^{n-\alpha} f)(t).$$

The left Caputo fractional derivative has the following form $$\label{005}
  (_{a}^{C}D^\alpha f)(x)=\big(_{a}I^{n-\alpha} f^{(n)}\big)(x),~~n=[\alpha]+1.$$

The right Caputo fractional derivative becomes $$\label{006}
(^CD_b^\alpha f)(x)=\big(I_b^{n-\alpha}(-1)^nf^{(n)}\big)(x).$$

The generalized left and right fractional integrals in the sense of Katugampola [@Kat1] are given respectively as $$\label{015}
(_{a}\textbf{I}^{\alpha,\rho} f)(x)=\frac{1}{\Gamma(\alpha)}\int_a^x(\frac{x^\rho-u^\rho}{\rho})^{\alpha-1} f(u)\frac{du}{u^{1-\rho}}$$ and $$\label{016}
(\textbf{I}_{b}^{\alpha,\rho}f)(x)=\frac{1}{\Gamma(\alpha)}\int_t^b (\frac{u^\rho- x^\rho}{\rho})^{\alpha-1} f(u)\frac{du}{u^{1-\rho}}.$$ The generalized left and right fractional derivatives in the sense of Katugampola [@Kat2] are defined respectively as $$\begin{aligned}
\label{017}\nonumber
(_{a}\textbf{D}^{\alpha,\rho} f)(x)&=&\gamma^n(_{a}\textbf{I}^{n-\alpha,\rho} f)(t)\\&=&\frac{\gamma^n}{\Gamma(n-\alpha)}\int_a^x(\frac{x^\rho-u^\rho}{\rho})^{n-\alpha-1} f(u)\frac{du}{u^{1-\rho}}\end{aligned}$$ and $$\begin{aligned}
\label{018}\nonumber
(\textbf{D}_{b}^{\alpha,\rho} f)(x)&=& (-\gamma)^n(\textbf{I}_b^{n-\alpha,\rho} f)(x)\\
&=&\frac{(-\gamma)^n}{\Gamma(n-\alpha)}\int_x^b(\frac{u^\rho-x^\rho}{\rho})^{n-\alpha-1} f(u)\frac{du}{u^{1-\rho}},
 \end{aligned}$$ where $\rho>0$ and $\gamma=x^{1-\rho}\frac{d}{dx}$. The Caputo modification of the left and right generalized fractional derivatives in the sense of Jarad et al. [@fahd3] are presented respectively as $$\begin{aligned}
\label{019}\nonumber
 (_{a}^C\textbf{D}^{\alpha,\rho} f)(x)&=&(_{a}\textbf{I}^{n-\alpha,\rho}\gamma^n f)(x)\\&=&\frac{1}{\Gamma(n-\alpha)}\int_a^x(\frac{x^\rho-u^\rho}{\rho})^{n-\alpha-1}\gamma^n f(u)\frac{du}{u^{1-\rho}},\end{aligned}$$ and $$\begin{aligned}
\label{020}\nonumber
(^C\textbf{D}_{b}^{\alpha,\rho} f)(x)&=& (_{a}\textbf{I}^{n-\alpha,\rho}(-\gamma)^n f)(x)\\
&=&\frac{1}{\Gamma(n-\alpha)}\int_x^b(\frac{u^\rho-x^\rho}{\rho})^{n-\alpha-1} (-\gamma)^nf(u)\frac{du}{u^{1-\rho}}.\end{aligned}$$ For $\alpha \in \mathbb{C},~Re(\alpha)>0$ the left Riemann-Liouville fractional integral of order $\alpha $ of $f$ with respect to a continuously differentiable and increasing function $g$ has the following form [@Samko; @f2] $$\label{3}  
~_{a}I^{\alpha,g} f(x)=\frac{1}{\Gamma(\alpha)}\int_{a}^x \Big(g(x)-g(u)\Big)^{\alpha-1}f(u)g'(u)du.$$ For $\alpha \in \mathbb{C},~Re(\alpha)>0$ the right Riemann-Liouville fractional integral of order $\alpha $ of $f$ with respect to a continuously differentiable and increasing function $g$ has the following form [@Samko; @f2] $$\label{333}  
I_b^{\alpha,g} f(x)=\frac{1}{\Gamma(\alpha)}\int_{x}^b \Big(g(u)-g(x)\Big)^{\alpha-1}f(u)g'(u)du.$$ For $\alpha \in \mathbb{C},~Re(\alpha)\geq 0$, the generalized left and right Riemann-Liouville fractional derivative of order $\alpha $ of $f$ with respect to a continuously differentiable and increasing function $g$ have respectively the form [@Samko; @f3] $$\begin{aligned}
\label{4}\nonumber
 ~_{a}D^{\alpha,g}  f(x)&=&\Big(\frac{1}{g'(x)}\frac{d}{dx}\Big)^n(~_{a}I^{n-\alpha,g} f)(x)\\&=&\frac{\Big(\frac{1}{g'(x)}\frac{d}{dx}\Big)^n}{\Gamma(n-\alpha)}\int_{a}^x \Big(g(x)-g(u)\Big)^{n-\alpha-1}f(u)g'(u)du
 \end{aligned}$$ and $$\begin{aligned}
\label{444}\nonumber
 D_b^{\alpha,g}  f(x)&=&\Big(-\frac{1}{g'(x)}\frac{d}{dx}\Big)^n(I_b^{n-\alpha,g} f)(x)\\&=&\frac{\Big(-\frac{1}{g'(x)}\frac{d}{dx}\Big)^n}{\Gamma(n-\alpha)}\int_{a}^x \Big(g(x)-g(u)\Big)^{n-\alpha-1}f(u)g'(u)du,
 \end{aligned}$$ where $n=[\alpha]+1$. It is easy to observe that if we choose $g(x)=x$, the integrals in (\[3\]) and (\[333\]) becomes the left and right Riemann-Liouville fractional integrals respectively and (\[4\]) and (\[444\]) becomes the left and right Riemann-Liouville fractional derivatives. When $g(x)=\ln x$, the Hadamard fractional operators are obtained [@Samko; @f2]. While if one considers $g(x)=\frac{x^\rho}{\rho}$, the fractional operators in the settings of Katugampola [@Kat1; @Kat2] are derived.

In left and right generalized Caputo derivatives of a function with respect to another function are presented respectively as [@fahd10] $$\label{555}
 ~_{a}^CD^{\alpha,g}  f(x)=\Big(~_aI^{n-\alpha,g} f^{[n]}\Big)(x)$$ and $$\label{666}
 ^CD_b^{\alpha,g}  f(x)=\Big(~_aI^{n-\alpha,g} (-1)^nf^{[n]}\Big)(x),$$ where $\displaystyle  f^{[n]}(x)=\Big(\frac{1}{g'(x)}\frac{d}{dx}\Big)^nf(x)$.

The proportional derivatives and their fractional integrals and derivatives
---------------------------------------------------------------------------

The conformable derivative was first introduced by Khalil et al. in [@kh] and then explored by the current author in [@T11]. In his distinctive paper [@Anderson1], Anderson et al. modified the conformable derivative by using the proportional derivative. Indeed, he gave the following definition.

\[D1\] *(Modified conformable derivatives)* For $\rho \in [0,1]$, let the functions $\kappa_0, \kappa_1:[0,1]\times \mathbb{R}\rightarrow [0,\infty)$ be continuous such that for all $t \in \mathbb{R}$ we have $$\lim_{\rho\rightarrow 0^+}\kappa_1(\rho,t)=1,~\lim_{\rho\rightarrow 0^+}\kappa_0(\rho,t)=0, \lim_{\rho\rightarrow 1^-}\kappa_1(\rho,t)=0,~\lim_{\rho\rightarrow 1^-}\kappa_0(\rho,t)=1,$$ and $\kappa_1(\rho,t)\neq 0,~~\rho \in [0,1),~~\kappa_0(\rho,t)\neq 0,~~\rho \in (0,1]$. Then, the modified conformable differential operator of order $\rho$ is defined by $$\label{anndy}
D^\rho f(t)=\kappa_1(\rho,t) f(t)+\kappa_0(\rho,t) f^\prime(t).$$

The derivative given in (\[anndy\]) is called a proportional derivative. For more details about the control theory of the proportional derivatives and its component functions $\kappa_0$ and $\kappa_1$, we refer the reader to [@Anderson1; @Anderson2].

Of special interest, we shall restrict ourselves to the case when $\kappa_1(\rho,t)=1-\rho$ and $\kappa_0(\rho,t)=\rho$. Therefore, (\[anndy\]) becomes $$\label{prop derivative}
  D^\rho f(t)=(1-\rho) f(t)+\rho f^\prime(t).$$ Notice that $\lim_{\rho \rightarrow 0^+}D^\rho f(t)= f(t)$ and $\lim_{\rho \rightarrow 1^-}D^\rho f(t)= f^\prime(t)$. It is clear that the derivative (\[prop derivative\]) is somehow more general than the conformable derivative which does not tend to the original function as $\rho$ tends to $0$. The associated fractional proportional integrals are defined as [@fahd12]

\[left and right integrals\]For $\rho>0$ and $\alpha \in \mathbb{C},~~Re(\alpha)>0$, the left fractional proportional integral of $f$ reads $$\label{LPI}
        (_{a}I^{\alpha,\rho} f)(x)=  \frac{1}{\rho^\alpha \Gamma(\alpha)}\int_a^x e^{\frac{\rho-1}{\rho}(x-\tau)} (x-\tau)^{\alpha-1} f(\tau)d\tau$$ and the right one reads $$\label{RPI}
          (I_b^{\alpha,\rho} f)(x)=  \frac{1}{\rho^\alpha \Gamma(\alpha)}\int_x^b  e^{\frac{\rho-1}{\rho}(\tau-x)} (\tau-x)^{\alpha-1} f(\tau)d\tau.$$

\[Prop fractional derivatives\][@fahd12] For $\rho>0$ and $\alpha \in \mathbb{C},~~Re(\alpha)\geq 0$, the left fractional proportional derivative is defined as

$$\begin{aligned}
\label{LPD}\nonumber
  (_{a}D^{\alpha,\rho}f)(x)&=& D^{n,\rho} ~_{a}I^{n-\alpha,\rho} f(x)\\&=&\frac{D_x^{n,\rho}}{\rho^{n-\alpha}\Gamma(n-\alpha)}
  \int_a^x e^{\frac{\rho-1}{\rho}(x-\tau)}(x-\tau)^{n-\alpha-1} f(\tau)d \tau.\end{aligned}$$

The right proportional fractional derivative is defined by [@fahd12] $$\begin{aligned}
\label{RPD}\nonumber
   (D_b^{\alpha,\rho}f)(x)&=& ~_{\ominus}D^{n,\rho} I_b^{n-\alpha,\rho} f(x)\\&=&\frac{~_{\ominus}D^{n,\rho}}{\rho^{n-\alpha}\Gamma(n-\alpha)}
   \int_x^b e^{\frac{\rho-1}{\rho}(\tau-x)}(\tau-x)^{n-\alpha-1} f(\tau)d \tau,
 \end{aligned}$$

where $n=[Re(\alpha)]+1$ and $\displaystyle (_{\ominus}D^\rho f)(t)=(1-\rho)f(t)-\rho f^\prime(t)$.

Lastly, the left and right fractional proportional derivatives in the Caputo settings respectively read [@fahd12] $$\begin{aligned}
\label{LPDC}\nonumber
  (_{a}^CD^{\alpha,\rho}f)(x)&=&  \Big(~_{a}I^{n-\alpha,\rho}D^{n,\rho}f\Big)(x)\\\nonumber &=&\frac{1}{\rho^{n-\alpha}\Gamma(n-\alpha)}
  \int_a^x e^{\frac{\rho-1}{\rho}(x-\tau)}(x-\tau)^{n-\alpha-1} (D^{n,\rho}f)(\tau)d \tau
  \\\end{aligned}$$ and $$\begin{aligned}
\label{RPDC}\nonumber
   (^CD_b^{\alpha,\rho}f)(x)&=& \Big( I_b^{n-\alpha,\rho}{~_\ominus}D^{n,\rho}f\Big)(x)\\\nonumber &=&\frac{1}{\rho^{n-\alpha}\Gamma(n-\alpha)}
   \int_x^b e^{\frac{\rho-1}{\rho}(\tau-x)}(\tau-x)^{n-\alpha-1} (~_{\ominus}D^{n,\rho}f)(\tau)d\tau.\\
 \end{aligned}$$

The fractional proportional derivative of a function with respect to another function
=====================================================================================

\[D2\] *(The proportional derivative of a function with respect to anothor function)*\
For $\rho \in [0,1]$, let the functions $\kappa_0, \kappa_1:[0,1]\times \mathbb{R}\rightarrow [0,\infty)$ be continuous such that for all $t \in \mathbb{R}$ we have $$\lim_{\rho\rightarrow 0^+}\kappa_1(\rho,t)=1,~\lim_{\rho\rightarrow 0^+}\kappa_0(\rho,t)=0, \lim_{\rho\rightarrow 1^-}\kappa_1(\rho,t)=0,~\lim_{\rho\rightarrow 1^-}\kappa_0(\rho,t)=1,$$ and $\kappa_1(\rho,t)\neq 0,~~\rho \in [0,1),~~\kappa_0(\rho,t)\neq 0,~~\rho \in (0,1]$. Let also $g(t)$ be a strictly increasing continuous function. Then, the proportional differential operator of order $\rho$ of $f$ with respect to $g$ is defined by $$\label{eq1}
D^{\rho,g} f(t)=\kappa_1(\rho,t) f(t)+\kappa_0(\rho,t)\frac{f^\prime(t)}{g'(t)}.$$

we shall restrict ourselves to the case when $\kappa_1(\rho,t)=1-\rho$ and $\kappa_0(\rho,t)=\rho$. Therefore, (\[eq1\]) becomes $$\label{eq2}
  D^{\rho,g} f(t)=(1-\rho) f(t)+\rho \frac{f^\prime(t)}{g'(t)}.$$ The corresponding integral of

$$\label{eq3}
_{a}I^{1,\rho,g}f(t)=\frac{1}{\rho}\int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(s))}f(s)g'(s)ds,$$

where we accept that $~_{a}I^{0,\rho}f(t)=f(t)$.

To produce a generalized type fractional integral depending on the proportional derivative, we proceed by induction through changing the order of integrals to show that $$\begin{aligned}
 \label{eq4}
 \nonumber
   (_{a}I^{n,\rho,g} f)(t) &=& \frac{1}{\rho}\int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(\tau_1))} g'(\tau_1)d \tau_1 \frac{1}{\rho} \int_a^{\tau_1}
   e^{\frac{\rho-1}{\rho}(g(\tau_1)-g(\tau_2))}g'(\tau_2)d \tau_2\cdot\cdot\cdot\\\nonumber&\cdot\cdot\cdot&\frac{1}{\rho} \int_a^{\tau_{n-1}} e^{\frac{\rho-1}{\rho}(g(\tau_{n-1})-g(\tau_n))}f(\tau_n)g'(\tau_n) d\tau_n\\
    &=&  \frac{1}{\rho^n \Gamma(n)}\int_a^t  e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))} (g(t)-g(\tau))^{n-1} f(\tau)g'(\tau)d\tau.
 \end{aligned}$$ Based on (\[eq4\]), we can present the following general proportional fractional integral.

\[general left and right integrals\]For $\rho \in (0,1]$, $\alpha \in \mathbb{C},~~Re(\alpha)>0$, we define the left fractional integral of $f$ with respect to $g$ by $$\label{eq5}
        (_{a}I^{\alpha,\rho,g} f)(t)=  \frac{1}{\rho^\alpha \Gamma(\alpha)}\int_a^t  e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))} (g(t)-g(\tau))^{\alpha-1} f(\tau)g'(\tau)d\tau.$$ The right fractional proportional integral ending at $b$ can be defined by $$\label{eq6}
          (I_b^{\alpha,\rho,g} f)(t)=  \frac{1}{\rho^\alpha \Gamma(\alpha)}\int_t^b  e^{\frac{\rho-1}{\rho}(g(\tau)-g(t))} (g(\tau)-g(t))^{\alpha-1} f(\tau)g'(\tau)d\tau.$$

To deal with the right proportional fractional case we shall use the notation $$(_{\ominus}D^{\rho,g} f)(t):=(1-\rho)f(t)-\rho \frac{f^\prime(t)}{g'(t)}.$$ We shall also write $$(_{\ominus}D^{n,\rho,g} f)(t)= (\underbrace{_{\ominus}D^{\rho,g}~ _{\ominus}D^{\rho,g}\ldots~_{\ominus}D^{\rho,g}}_{\texttt{n times}} f)(t).$$

\[general left and right derivatives\] For $\rho>0$, $\alpha \in \mathbb{C},~~Re(\alpha)\geq 0$ and $g\in C[a,b]$, where $g'(t)>0$, we define the general left fractional derivative of $f$ with respect to $g$ as

$$\begin{aligned}
\label{eq7}\nonumber
  (_{a}D^{\alpha,\rho,g}f)(t)&=& D^{n,\rho,g} ~_{a}I^{n-\alpha,\rho,g} f(t)\\\nonumber
  &=&\frac{D_t^{n,\rho,g}}{\rho^{n-\alpha}\Gamma(n-\alpha)}
  \int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))}(g(t)-g(\tau))^{n-\alpha-1} f(\tau)g'(\tau)d \tau\\   \end{aligned}$$

and the general right fractional derivative of $f$ with respect to $g$ as $$\begin{aligned}
\label{eq8}\nonumber
   (D_b^{\alpha,\rho,g}f)(t)&=& ~_{\ominus}D^{n,\rho,g} I_b^{n-\alpha,\rho,g} f(t)\\\nonumber
   &=&\frac{~_{\ominus}D_t^{n,\rho,g}}{\rho^{n-\alpha}\Gamma(n-\alpha)}
   \int_t^b e^{\frac{\rho-1}{\rho}(g(\tau)-g(t))}(g(\tau)-g(t))^{n-\alpha-1} f(\tau)g'(\tau)d \tau,\\
 \end{aligned}$$

where $n=[Re(\alpha)]+1$.

\[reduction\]

Clearly, if we let $\rho=1$ in Definition \[general left and right integrals\] and Definition \[general left and right derivatives\], we obtain the

-   the Riemann-Liouville fractional operators , , and if $g(t)=t$.

-   the fractional operators in the Katugampola setting, , and if $\displaystyle g(t)=\frac{t^{\mu}}{\mu}$.

-   The Hadamard fractional operators if $g(t)=\ln t$ [@Samko; @f2].

-   The fractional operators mentioned in [@fahd11] if $\displaystyle g(t)=\frac{(t-a)^{\mu}}{\mu}$.

\[2.4\] Let $\alpha, \beta \in \mathbb{C}$ be such that $Re(\alpha)\geq 0$ and $Re(\beta)>0$. Then, for any $\rho>0$ we have

-   \(a)  $\big(_{a}I^{\alpha,\rho,g} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a))^{\beta-1}\big)(t)=\frac{\Gamma(\beta)}{\Gamma(\beta+\alpha)\rho^\alpha}e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{\alpha+\beta-1},$\
    $~~~Re(\alpha)>0.$

-   (b)  $\big(I_b^{\alpha,\rho,g} e^{-\frac{\rho-1}{\rho}g(x)} (g(b)-g(x))^{\beta-1}\big)(t)=\frac{\Gamma(\beta)}{\Gamma(\beta+\alpha)\rho^\alpha}e^{-\frac{\rho-1}{\rho}g(t)}(g(b)-g(t))^{\alpha+\beta-1},$\
    $~~~Re(\alpha)>0.$

-   (c)  $\big(_{a}D^{\alpha,\rho} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a))^{\beta-1}\big)(t)=\frac{\rho^\alpha\Gamma(\beta)}{\Gamma(\beta-\alpha)}e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{\beta-1-\alpha},$\
    $~~~Re(\alpha)\geq 0.$

-   (d) $\big(D_b^{\alpha,\rho,g} e^{-\frac{\rho-1}{\rho}g(x)} (g(b)-g(x))^{\beta-1}\big)(t)=\frac{\rho^\alpha\Gamma(\beta)}{\Gamma(\beta-\alpha)}e^{-\frac{\rho-1}{\rho}g(t)}(g(b)-G
        g(t))^{\beta-1-\alpha},$\
    $~~~Re(\alpha)\geq 0.$

The proofs of relations (a) and (b) are very easy to handle. We will prove (c) while the proof of (d) is analogous.

By the definition of the left proportional fractional derivative and relation (a), we have $$\begin{aligned}
&\Big(_{a}D^{\alpha,\rho,g} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a))^{\beta-1}\Big)(t)\\ 
&=D^{n,\rho,g} \Big(_{a}I^{n-\alpha,\rho,g} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a))^{\beta-1}\Big)(t)\\
&=D^{n,\rho,g}\frac{\Gamma(\beta)}{\Gamma(\beta+n-\alpha)\rho^{n-\alpha}}e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{n-\alpha+\beta-1} \\
&=\frac{\rho^n\Gamma(\beta)(n-\alpha+\beta-1)(n-\alpha+\beta-1)\cdot \cdot \cdot
   (\beta-\alpha)}{\rho^{n-\alpha}\Gamma(n-\alpha+\beta)}
\times e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{\beta-1-\alpha}\\
& =\frac{\rho^\alpha\Gamma(\beta)}{\Gamma(\beta-\alpha)}e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{\beta-1-\alpha}.\end{aligned}$$ Here, we have used the fact that $\displaystyle D^{\rho,g} \Big(h(t)  e^{\frac{\rho-1}{\rho}g(t)} \Big)=\rho \frac{h'(t)}{g^\prime (t)}  e^{\frac{\rho-1}{\rho}g(t)} $.

Below we present the semi–group property for the general fractional proportional integrals of a function with respect to another function.

\[THM1\] Let $\rho\in (0,1],~Re(\alpha)>0$ and $Re(\beta)>0$. Then, if $f$ is continuous and defined for $t \geq a$ or $t\le b$, we have $$\label{Left Semi integrals}
  ~_aI^{\alpha,\rho,g} (_{a}I^{\beta,\rho,g} f)(t)= ~_aI^{\beta,\rho,g} (_{a}I^{\alpha,\rho} f)(t)=(~_{a}I^{\alpha+\beta,\rho,g} f)(t)$$ and $$\label{Right Semi integrals}
  I_b^{\alpha,\rho,g} (I_b^{\beta,\rho,g} f)(t)= ~I_b^{\beta,\rho,g} (I_b^{\alpha,\rho} f)(t)=(I_b^{\alpha+\beta,\rho,g} f)(t).$$

We will prove . is proved similarly. Using the definition, interchanging the order and making the change of variable $z=\frac{g(u)-g(\tau)}{g(t)-g(\tau)}$, we get

$$\begin{aligned}
 &~_aI^{\alpha,\rho,g} (_{a}I^{\beta,\rho,g} f)(t)\\
  &= \frac{1}{\rho^{\alpha+\beta}\Gamma(\alpha)\Gamma(\beta)}
  \int_a^t \int_a^ue^{\frac{\rho-1}{\rho}(g(t)-g(u))}e^{\frac{\rho-1}{\rho}(g(u)-g(\tau))}(g(t)-g(u))^{\alpha-1}\\
   & \times(g(u)-g(\tau))^{\beta-1}f(\tau)g'(\tau)d\tau g'(u)du\\
   &=\frac{1}{\rho^{\alpha+\beta}\Gamma(\alpha)\Gamma(\beta)}\int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))} f(\tau) \int_\tau^t
  (g(t)-g(u))^{\alpha-1} (g(u)-g(\tau))^{\beta-1}\\
  &\times g'(u)dug'(\tau) d\tau \\
   &= \frac{1}{\rho^{\alpha+\beta}\Gamma(\alpha)\Gamma(\beta)}
   \int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))}(g(t)-g(\tau))^{\alpha+\beta-1} f(\tau)g'(u)d\tau \\
   &\times \int_0^1 (1-z)^{\alpha-1} z^{\beta-1} dz\\
    &=\frac{1}{\rho^{\alpha+\beta}\Gamma(\alpha+\beta)}\int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))}(g(t)-g(\tau))^{\alpha+\beta-1} f(\tau)g'(\tau)d\tau\\
    &=(_{a}I^{\alpha+\beta,\rho} f)(t).\end{aligned}$$

\[THM2\]Let $0\leq  m< [Re(\alpha)]+1$. Then, we have $$\label{D on L}
  D^{m,\rho,g} (_{a}I^{\alpha,\rho,g}f)(t)=(_{a}I^{\alpha-m,\rho,g}f)(t)$$ and $$\label{LD on L}
  ~_{\ominus}D^{m,\rho,g} (I_b^{\alpha,\rho,g}f)(t)=(I_b^{\alpha-m,\rho,g}f)(t)$$

Here we prove , while one can prove likewise. Using the fact that $D_t^{\rho,g}e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))} =0$), we have $$\begin{aligned}
 & D^{m,\rho,g} (_{a}I^{\alpha,\rho,g}f)(t) D^{m-1,\rho,g} (D^{\rho,g}~_{a}I^{\alpha,\rho,g}f)(t) \\
   &= D^{m-1,\rho,g} \frac{1}{\rho^{\alpha-1}\Gamma(\alpha-1)}\int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(\tau))} (g(t)-g(\tau))^{\alpha-2}f(\tau)g'(\tau)d \tau.\end{aligned}$$ Proceeding $m-$times in the same manner we obtain (\[D on L\]).

\[D on I\] Let $0<Re(\beta) < Re(\alpha)$ and $m-1<Re(\beta)\leq m$. Then, we have $$\label{LD on LI}
_{a}D^{\beta, \rho,g} ~_{a}I^{\alpha,\rho,g} f(t)=~_{a}I^{\alpha-\beta,\rho,g} f(t)$$ and $$\label{RD on RI}
D_b^{\beta, \rho,g} I_b^{\alpha,\rho,g} f(t)=I_b^{\alpha-\beta,\rho,g} f(t).$$

By the help of Theorem \[THM1\] and Theorem \[THM2\], we have $$\begin{aligned}
  ~_{a}D^{\beta, \rho,g} ~_{a}I^{\alpha,\rho,g} f(t)&=& D^{m,\rho,g} _{a}I^{m-\beta,\rho,g} _{a}I^{\alpha,\rho,g}f(t)\\
   &=& D^{m,\rho,g}~_{a}I^{m-\beta+\alpha,\rho,g} f(t)=~_{a}I^{\alpha-\beta,\rho,g} f(t).\end{aligned}$$ This was the proof of . One can prove in a similar way.

\[THM4\] Let $f$ be integrable on $t\geq a$ or $t\le b$ and $Re[\alpha]>0, ~\rho \in (0,1],~~n=[Re(\alpha)]+1$. Then, we have $$\label{LD on LI sameorder}
~_{a}D^{\alpha, \rho,g} ~_{a}I^{\alpha, \rho,g} f(t)=f(t)$$ and $$\label{RD on RI sameorder}
D_b^{\alpha, \rho,g} I_b^{\alpha, \rho,g} f(t)=f(t).$$

By the definition and Theorem \[THM1\], we have $$~_{a}D^{\alpha, \rho,g} ~_{a}I^{\alpha, \rho,g} f(t)=D^{n,\rho,g}~_{a}I^{n-\alpha, \rho,g}  ~_{a}I^{\alpha, \rho,g} f(t)= D^{n,\rho,g}~_{a}I^{n, \rho,g} f(t)=f(t).$$

The Caputo fractional proportional derivative of a function with respect to another function 
=============================================================================================

For $\rho \in (0,1]$ and $\alpha \in \mathbb{C}$ with $Re(\alpha)\geq 0$ we define the left derivative of Caputo type as $$\begin{aligned}
\label{CFP}
 &(^{C}_{a}D^{\alpha,\rho,g} f)(t)=_{a}I^{n-\alpha,\rho,g} (D^{n,\rho,g}f)(t)\\\nonumber
 &=\frac{1}{\rho^{n-\alpha}\Gamma(n-\alpha)}\int_a^t e^{\frac{\rho-1}{\rho}(g(t)-g(s))}(g(t)-g(s))^{n-\alpha-1}(D^{n,\rho,g}f)(s)g'(s)ds.\end{aligned}$$ Similarly, the right derivative of Caputo type ending is defined by $$\begin{aligned}
\label{rCFP}
 &(^{C}D_b^{\alpha,\rho} f)(t)= I_b^{n-\alpha,\rho,g} (_{\ominus}D^{n,\rho,g}f)(t)\\\nonumber 
 &= \frac{1}{\rho^{n-\alpha}\Gamma(n-\alpha)}\int_t^b e^{\frac{\rho-1}{\rho}(g(s)-g(t))}(g(s)-g(t))^{n-\alpha-1}(~_{\ominus}D^{n,\rho,g}f)(s)g'(s)ds,\end{aligned}$$ where $n=[Re(\alpha)]+1$.

\[4.2\] Let $\alpha, \beta \in \mathbb{C}$ be such that $Re(\alpha)> 0$ and $Re(\beta)>0$. Then, for any $\rho \in (0,1]$ and $n=[Re(\alpha)]+1$ we have

1.  $\big(^{C}_{a}D^{\alpha,\rho,g} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a))^{\beta-1}\big)(t)=\frac{\rho^\alpha\Gamma(\beta)}{\Gamma(\beta-\alpha)}e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{\beta-1-\alpha},$\
    $~~~Re(\beta)> n.$

2.  $\big(^{C}D_b^{\alpha,\rho,g} e^{-\frac{\rho-1}{\rho}g(x)} (g(b)-g(x))^{\beta-1}\big)(t)=\frac{\rho^\alpha\Gamma(\beta)}{\Gamma(\beta-\alpha)}e^{-\frac{\rho-1}{\rho}g(t)}(g(b)-g(t))^{\beta-1-\alpha},$\
    $~~~Re(\beta)> n.$

For $k=0,1,\ldots,n-1$, we have $$\big(^{C}_{a}D^{\alpha,\rho,g} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a)^{k}\big)(t)=0\quad \mbox{and}\quad \big(^{C}D_b^{\alpha,\rho,g} e^{-\frac{\rho-1}{\rho}g(x)} (g(b)-g(x))^{k}\big)(t)=0.$$ In particular, $(~^{C}_{a}D^{\alpha,\rho} e^{\frac{\rho-1}{\rho}g(x})(t)=0$ and $(^{C}D_b^{\alpha,\rho} e^{-\frac{\rho-1}{\rho}g(x)})(t)=0$.

We only prove the first relation. The proof of the second relation is similar. We have $$\begin{aligned}
&(^{C}_{a}D^{\alpha,\rho,g} e^{\frac{\rho-1}{\rho}g(x)} (g(x)-g(a))^{\beta-1})(t)= ~_{a}I^{n-\alpha,\rho,g} D^{n,\rho,g} \left[e^{\frac{\rho-1}{\rho}g(t)} (g(t)-g(a))^{\beta-1} \right]\\
   &=~_{a}I^{n-\alpha,\rho,g} 
   \left[ \rho^n (\beta-1)(\beta-2)\ldots(\beta-1-n) (g(t)-g(a))^{\beta-n-1} e^{\frac{\rho-1}{\rho}g(t)}\right] \\
   &= \frac{\rho^n (\beta-1)(\beta-2)\ldots(\beta-1-n)\Gamma(\beta-n)} {\Gamma(\beta-\alpha)\rho^{n-\alpha}} (g(t)-g(a))^{\beta-\alpha-1} e^{\frac{\rho-1}{\rho}g(t)}\\
   &= \frac{\rho^\alpha\Gamma(\beta)}{\Gamma(\beta-\alpha)}e^{\frac{\rho-1}{\rho}g(t)}(g(t)-g(a))^{\beta-1-\alpha}.\end{aligned}$$

Conclusions
===========

We have used the proportional derivatives of a function with respect to another to obtain left and right generalized type of fractional integrals and derivatives involving two parameters $\alpha$ and $\rho$ and depending on a kernel function . The Riemann–Liouville and Caputo fractional derivatives in classical fractional calculus can obtained as $\rho$ tends to $1$ and by choosing $g(x)=1$. The integrals have the semi–group property and together with their corresponding derivatives have exponential functions as part of their kernels. It should be noted that other properties of these new operators can be obtained by using the Laplace transform proposed in [@fahd10]. Moreover, for a specific choice of $g$, the proportional fractional operators in the settings of Hadamard and Katugamplola can be extracted.

**Funding** This research was funded by the Deanship of Scientific Research at Princess Nourah bint Abdulrahman University through the Fast-track Research Funding Program.\
**Availability of data and materials** Not applicable.\
**Competing interests** The authors have no competing interest regarding this article.\
**Authors contributions** All authors have done equal contribution in this article. All authors read and approved the last version of the manuscript.

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