A conformable fractional calculus on arbitrary time scales

Definition 1

Let $f:\mathbb{T}\to \mathbb{R}\text{,}t\in {\mathbb{T}}^{\kappa }$ , and $\alpha \in ]0\text{,}1]$ . For $t>0$ , we define $T_{\alpha }(f)(t)$ to be the number (provided it exists) with the property that, given any $∊>0$ , there is a $\delta$ -neighborhood ${\mathcal{V}}_t\subset \mathbb{T}$ of $t\text{,}\delta >0$ , such that $\left|\left[f(\sigma (t))-f(s)\right]t^{1-\alpha }-T_{\alpha }(f)(t)\left[\sigma (t)-s\right]\right|⩽∊\left|\sigma (t)-s\right|$ for all $s\in {\mathcal{V}}_t$ . We call $T_{\alpha }(f)(t)$ the conformable fractional derivative of f of order $\alpha$ at t, and we define the conformable fractional derivative at 0 as $T_{\alpha }(f)(0)={\lim }_{t\to 0^{+}}{T}_{\alpha }(f)(t)$ .

Remark 2

If $\alpha =1$ , then we obtain from Definition 1 the delta derivative of time scales. The conformable fractional derivative of order zero is defined by the identity operator: $T_0(f)≔f$ .

Remark 3

Along the work, we also use the notation ${\left(f(t)\right)}^{(\alpha )}=T_{\alpha }(f)(t)$ .

Theorem 4

Let $\alpha \in ]0\text{,}1]$ and $\mathbb{T}$ be a time scale. Assume $f:\mathbb{T}\to \mathbb{R}$ and let $t\in {\mathbb{T}}^{\kappa }$ . The following properties hold.

1. If f is conformal fractional differentiable of order $\alpha$ at $t>0$ , then f is continuous at t.

2. If f is continuous at t and t is right-scattered, then f is conformable fractional differentiable of order $\alpha$ at t with

   (1)
   $$T_{\alpha }(f)(t)=\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }\text{.}$$

3. If t is right-dense, then f is conformable fractional differentiable of order $\alpha$ at t if, and only if, the limit ${\lim }_{s\to t}\frac{f(t)-f(s)}{(t-s)}{t}^{1-\alpha }$ exists as a finite number. In this case,

   (2)
   $$T_{\alpha }(f)(t)={\displaystyle \underset{s\to t}{\lim }}\frac{f(t)-f(s)}{t-s}{t}^{1-\alpha }\text{.}$$

4. If f is fractional differentiable of order $\alpha$ at t, then $f(\sigma (t))=f(t)+(\mu (t))t^{\alpha -1}{T}_{\alpha }(f)(t)$ .

Proof

1. Assume that f is conformable fractional differentiable at t. Then, there exists a neighborhood ${\mathcal{V}}_t$ of t such that $\left|\left[f(\sigma (t))-f(s)\right]t^{1-\alpha }-T_{\alpha }(f)(t)\left[\sigma (t)-s\right]\right|⩽∊\left|\sigma (t)-s\right|$ for $s\in {\mathcal{V}}_t$ . Therefore, $\left|f\left(t\right)-f\left(s\right)\right|⩽\left|\left[f(\sigma (t)-f(s)\right]-T_{\alpha }(f)(t)\left[\sigma (t)-s\right]t^{\alpha -1}\right|+\left|\left[f(\sigma (t))-f(t)\right]\right|+\left|f^{(\alpha )}(t)\right|\left|\left[\sigma (t)-s\right]\right|\left|t^{\alpha }-1\right|$ for all $s\in {\mathcal{V}}_t\cap \left]t-∊\text{,}t+∊\right[$ and, since t is a right-dense point, $$\left|f\left(t\right)-f\left(s\right)\right|⩽\left|\left[f^{\sigma }(t)-f(s)\right]-f^{(\alpha )}(t){\left[\sigma (t)-s\right]}^{\alpha }\right|+\left|f^{(\alpha )}(t){\left[t-s\right]}^{\alpha }\right|⩽∊\delta +\left|t^{\alpha -1}\right|\left|T_{\alpha }(f)(t)\right|\delta \text{.}$$ Since $\delta \to 0$ when $s\to t$ , and $t>0$ , it follows the continuity of f at t.

2. Assume that f is continuous at t and t is right-scattered. By continuity, $${\displaystyle \underset{s\to t}{\lim }}\frac{f(\sigma (t))-f(s)}{\sigma (t)-s}{t}^{1-\alpha }=\frac{f(\sigma (t))-f(t)}{\sigma (t)-t}{t}^{1-\alpha }=\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }\text{.}$$ Hence, given $∊>0$ and $\alpha \in ]0\text{,}1]$ , there is a neighborhood ${\mathcal{V}}_t$ of t such that $$\left|\frac{f(\sigma (t))-f(s)}{\sigma (t)-s}{t}^{1-\alpha }-\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }\right|⩽∊$$ for all $s\in {\mathcal{V}}_t$ . It follows that $$\left|\left[f(\sigma (t))-f(s)\right]t^{1-\alpha }-\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }(\sigma (t)-s)\right|⩽∊|\sigma (t)-s|$$ for all $s\in {\mathcal{V}}_t$ . The desired equality (1) follows from Definition 1.

3. Assume that f is conformable fractional differentiable of order $\alpha$ at t and t is right-dense. Let $∊>0$ be given. Since f is conformable fractional differentiable of order $\alpha$ at t, there is a neighborhood ${\mathcal{V}}_t$ of t such that $\left|[f(\sigma (t))-f(s)]t^{1-\alpha }-T_{\alpha }(f)(t)(\sigma (t)-s)\right|⩽∊|\sigma (t)-s|$ for all $s\in {\mathcal{V}}_t$ . Because $\sigma (t)=t$ , $$\left|\frac{f(t)-f(s)}{t-s}{t}^{1-\alpha }-T_{\alpha }(f)(t)\right|⩽∊$$ for all $s\in {\mathcal{V}}_t\text{,}s\ne t$ . Therefore, we get the desired result (2). Now, assume that the limit on the right-hand side of (2) exists and is equal to L, and t is right-dense. Then, there exists ${\mathcal{V}}_t$ such that $\left|(f(t)-f(s))t^{1-\alpha }-L(t-s)\right|⩽∊|t-s|$ for all $s\in {\mathcal{V}}_t$ . Because t is right-dense, $$\left|(f(\sigma (t))-f(s))t^{1-\alpha }-L(\sigma (t)-s)\right|⩽∊|\sigma (t)-s|\text{,}$$ which leads us to the conclusion that f is conformable fractional differentiable of order $\alpha$ at t and $T_{\alpha }(f)(t)=L$ .

4. If t is right-dense, i.e., $\sigma (t)=t$ , then $\mu (t)=0$ and $f(\sigma (t))=f(t)=f(t)+\mu (t)T_{\alpha }(f)(t)t^{1-\alpha }$ . On the other hand, if t is right-scattered, i.e., $\sigma (t)>t$ , then by (iii) $$f(\sigma (t))=f(t)+\mu (t)t^{\alpha -1}·\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }=f(t)+{(\mu (t))}^{\alpha -1}{T}_{\alpha }(f)(t)\text{,}$$

and the proof is complete.  □

Remark 5

In a time scale $\mathbb{T}$ , due to the inherited topology of the real numbers, a function f is always continuous at any isolated point $t\in \mathbb{T}$ .

Example 6

Let $h>0$ and $\mathbb{T}=h\mathbb{Z}≔\{\mathit{hk}:k\in \mathbb{Z}\}$ . Then $\sigma (t)=t+h$ and $\mu (t)=h$ for all $t\in \mathbb{T}$ . For function $f:t\in \mathbb{T}\mapsto t^2\in \mathbb{R}$ we have $T_{\alpha }(f)(t)={\left(t^2\right)}^{(\alpha )}=(2t+h)t^{1-\alpha }$ .

Example 7

Let $q>1$ and $\mathbb{T}=\overline{q^{\mathbb{Z}}}≔q^{\mathbb{Z}}\cup \{0\}$ with $q^{\mathbb{Z}}≔\{q^k:k\in \mathbb{Z}\}$ . In this time scale $$\sigma (t)=\left\{\begin{array}{cc}\mathit{qt}\hfill & \mathrm{if}t\ne 0\hfill \\ 0\hfill & \mathrm{if}t=0\hfill \end{array}\right.\mathrm{and}\mu (t)=\left\{\begin{array}{cc}(q-1)t\hfill & \mathrm{if}t\ne 0\hfill \\ 0\hfill & \mathrm{if}t=0\text{.}\hfill \end{array}\right.$$ Here 0 is a right-dense minimum and every other point in $\mathbb{T}$ is isolated. Now consider the square function f of Example 6. It follows that $$T_{\alpha }(f)(t)={\left(t^2\right)}^{(\alpha )}=\left\{\begin{array}{cc}(q+1)t^{2-\alpha }\hfill & \mathrm{if}t\ne 0\hfill \\ 0\hfill & \mathrm{if}t=0\text{.}\hfill \end{array}\right.$$

Example 8

Let $q>1$ and $\mathbb{T}=q^{{\mathbb{N}}_0}≔\{q^n:n\in {\mathbb{N}}_0\}$ . For all $t\in \mathbb{T}$ we have $\sigma (t)=\mathit{qt}$ and $\mu (t)=(q-1)t$ . Let $f:t\in \mathbb{T}\mapsto \log (t)\in \mathbb{R}$ . Then $T_{\alpha }(f)(t)={\left(\log (t)\right)}^{(\alpha )}=\frac{\log (q)}{(q-1)t^{\alpha }}$ for all $t\in \mathbb{T}$ .

Proposition 9

If $f:\mathbb{T}\to \mathbb{R}$ is defined by $f(t)=c$ for all $t\in \mathbb{T}\text{,}c\in \mathbb{R}$ , then $T_{\alpha }(f)(t)={(c)}^{(\alpha )}=0$ .

Proof

If t is right-scattered, then by Theorem 4 (ii) one has $T_{\alpha }(f)(t)=\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }=0$ . Otherwise, t is right-dense and, by Theorem 4 (iii), $T_{\alpha }(f)(t)={\lim }_{s\to t}\frac{c-c}{t-s}{t}^{1-\alpha }=0$ .  □

Proposition 10

If $f:\mathbb{T}\to \mathbb{R}$ is defined by $f(t)=t$ for all $t\in \mathbb{T}$ , then $$T_{\alpha }(f)(t)={(t)}^{(\alpha )}=\left\{\begin{array}{cc}{t}^{1-\alpha }\hfill & \mathrm{if}\alpha \ne 1\text{,}\hfill \\ 1\hfill & \mathrm{if}\alpha =1\text{.}\hfill \end{array}\right.$$

Proof

From Theorem 4 (iv), it follows that $\sigma (t)=t+\mu (t)t^{\alpha -1}{T}_{\alpha }(f)(t)\text{,}\mu (t)=\mu (t)t^{\alpha -1}{T}_{\alpha }(f)(t)$ . If $\mu (t)\ne 0$ , then $T_{\alpha }(f)(t)=t^{1-\alpha }$ and the desired relation is proved. Assume now that $\mu (t)=0$ , i.e., $\sigma (t)=t$ . In this case t is right-dense and, by Theorem 4 (iii), $T_{\alpha }(f)(t)={\lim }_{s\to t}\frac{t-s}{t-s}{t}^{1-\alpha }=t^{1-\alpha }$ . Therefore, if $\alpha =1$ , then $T_{\alpha }(f)(t)=1$ ; if $0<\alpha <1$ , then $T_{\alpha }(f)(t)=t^{1-\alpha }$ .  □

Corollary 11

Function $f:\mathbb{R}\to \mathbb{R}$ is conformable fractional differentiable of order $\alpha$ at point $t\in \mathbb{R}$ if, and only if, the limit ${\lim }_{s\to t}\frac{f(t)-f(s)}{t-s}{t}^{1-\alpha }$ exists as a finite number. In this case,

$$T_{\alpha }(f)(t)={\displaystyle \underset{s\to t}{\lim }}\frac{f(t)-f(s)}{t-s}{t}^{1-\alpha }\text{.}$$

Proof

Here $\mathbb{T}=\mathbb{R}$ , so all points are right-dense. The result follows from Theorem 4 (iii).  □

Remark 12

The identity (3) corresponds to the conformable derivative introduced in Khalil et al. (2014) and further studied in Abdeljawad (2015).

Corollary 13

Let $h>0$ . If $f:h\mathbb{Z}\to \mathbb{R}$ , then f is conformable fractional differentiable of order $\alpha$ at $t\in h\mathbb{Z}$ with $$T_{\alpha }(f)(t)=\frac{f(t+h)-f(t)}{h}{t}^{1-\alpha }\text{.}$$

Proof

Here $\mathbb{T}=h\mathbb{Z}$ and all points are right-scattered. The result follows from Theorem 4 (ii).  □

Example 14

Let $a\text{,}b>0$ and consider the time scale ${\mathbb{P}}_{a\text{,}b}={\bigcup }_{k=0}^{\infty }[k(a+b)\text{,}k(a+b)+a]$ . Then $$\sigma (t)=\left\{\begin{array}{cc}t\hfill & \mathrm{if}t\in {\displaystyle \bigcup _{k=0}^{\infty }}[k(a+b)\text{,}k(a+b)+a)\text{,}\hfill \\ t+b\hfill & \mathrm{if}t\in {\displaystyle \bigcup _{k=0}^{\infty }}\{k(a+b)+a\}\hfill \end{array}\right.$$ and $$\mu (t)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}t\in {\displaystyle \bigcup _{k=0}^{\infty }}[k(a+b)\text{,}k(a+b)+a)\text{,}\hfill \\ b\hfill & \mathrm{if}t\in {\displaystyle \bigcup _{k=0}^{\infty }}\{k(a+b)+a\}\text{.}\hfill \end{array}\right.$$ Let $f:{\mathbb{P}}_{a\text{,}b}\to \mathbb{R}$ be continuous and $\alpha \in ]0\text{,}1]$ . It follows from Theorem 4 that the conformable fractional derivative of order $\alpha$ of a function f defined on ${\mathbb{P}}_{a\text{,}b}$ is given by $$T_{\alpha }(f)(t)=\left\{\begin{array}{cc}{\displaystyle \underset{s\to t}{\lim }}\frac{f(t)-f(s)}{(t-s)}{t}^{1-\alpha }\hfill & \mathrm{if}t\in {\displaystyle \bigcup _{k=0}^{\infty }}[k(a+b)\text{,}k(a+b)+a)\text{,}\hfill \\ \frac{f(t+b)-f(t)}{b}{t}^{1-\alpha }\hfill & \mathrm{if}t\in {\displaystyle \bigcup _{k=0}^{\infty }}\{k(a+b)+a\}\text{.}\hfill \end{array}\right.$$

Theorem 15

Assume $f\text{,}g:\mathbb{T}\to \mathbb{R}$ are conformable fractional differentiable of order $\alpha$ . Then,

1. the sum $f+g:\mathbb{T}\to \mathbb{R}$ is conformable fractional differentiable with $T_{\alpha }(f+g)=T_{\alpha }(f)+T_{\alpha }(g)$ ;

2. for any $\lambda \in \mathbb{R}\text{,}\lambda f:\mathbb{T}\to \mathbb{R}$ is conformable fractional differentiable with $T_{\alpha }(\lambda f)=\lambda T_{\alpha }(f)$ ;

3. if f and g are continuous, then the product $\mathit{fg}:\mathbb{T}\to \mathbb{R}$ is conformable fractional differentiable with $T_{\alpha }(\mathit{fg})=T_{\alpha }(f)g+(f\circ \sigma )T_{\alpha }(g)=T_{\alpha }(f)(g\circ \sigma )+{\mathit{fT}}_{\alpha }(g)$ ;

4. if f is continuous, then $1/f$ is conformable fractional differentiable with $$T_{\alpha }\left(\frac{1}{f}\right)=-\frac{T_{\alpha }(f)}{f(f\circ \sigma )}\text{,}$$ valid at all points $t\in {\mathbb{T}}^{\kappa }$ for which $f(t)f(\sigma (t))\ne 0$ ;

5. if f and g are continuous, then $f/g$ is conformable fractional differentiable with $$T_{\alpha }\left(\frac{f}{g}\right)=\frac{T_{\alpha }(f)g-{\mathit{fT}}_{\alpha }(g)}{g(g\circ \sigma )}\text{,}$$ valid at all points $t\in {\mathbb{T}}^{\kappa }$ for which $g(t)g(\sigma (t))\ne 0$ .

Proof

Let us consider that $\alpha \in ]0\text{,}1]$ , and let us assume that f and g are conformable fractional differentiable at $t\in {\mathbb{T}}^{\kappa }$ .

1. Let $∊>0$ . Then there exist neighborhoods ${\mathcal{V}}_t$ and ${\mathcal{U}}_t$ of t for which $$\left|[f(\sigma (t))-f(s)]t^{1-\alpha }-T_{\alpha }(f)(t)\left(\sigma (t)-s\right)\right|⩽\frac{∊}{2}|\sigma (t)-s|\text{for all}s\in {\mathcal{V}}_t$$ and $$\left|[g(\sigma (t))-g(s)]t^{1-\alpha }-T_{\alpha }(g)(t)(\sigma (t)-s)\right|⩽\frac{∊}{2}|\sigma (t)-s|\text{for all}s\in {\mathcal{U}}_t\text{.}$$ Let ${\mathcal{W}}_t={\mathcal{V}}_t\cap {\mathcal{U}}_t$ . Then $\left|[(f+g)(\sigma (t))-(f+g)(s)]t^{1-\alpha }-\left[T_{\alpha }(f)(t)+T_{\alpha }(g)(t)\right](\sigma (t)-s)\right|⩽∊|\sigma (t)-s|$ for all $s\in \mathcal{W}$ . Thus, $f+g$ is conformable differentiable at t and $T_{\alpha }(f+g)(t)=T_{\alpha }(f)(t)+T_{\alpha }(g)(t)$ .

2. Let $∊>0$ . Then $\left|[f(\sigma (t))-f(s)]t^{1-\alpha }-T_{\alpha }(f)(t)(\sigma (t)-s)\right|⩽∊|\sigma (t)-s|$ for all s in a neighborhood ${\mathcal{V}}_t$ of t. It follows that $$\left|[(\lambda f)(\sigma (t))-(\lambda f)(s)]t^{1-\alpha }-\lambda T_{\alpha }(f)(t)(\sigma (t)-s)\right|⩽∊|\lambda ||\sigma (t)-s|\text{for all}s\in {\mathcal{V}}_t\text{.}$$ Therefore, $\lambda f$ is conformable fractional differentiable at t and $T_{\alpha }(\lambda f)=\lambda T_{\alpha }(f)$ holds at t.

3. If t is right-dense, then $$T_{\alpha }(\mathit{fg})(t)={\displaystyle \underset{s\to t}{\lim }}\left[\frac{f(t)-f(s)}{t-s}{t}^{1-\alpha }\right]g\left(t\right)+{\displaystyle \underset{s\to t}{\lim }}\left[\frac{g(t)-g(s)}{t-s}{t}^{1-\alpha }\right]f\left(s\right)=T_{\alpha }(f)(t)g(t)+T_{\alpha }(g)(t)f(t)=T_{\alpha }(f)(t)g(\sigma (t))+T_{\alpha }(g)(t)f(t)\text{.}$$ If t is right-scattered, then $$T_{\alpha }\left(\mathit{fg}\right)(t)=\left[\frac{f(\sigma (t))-f(t)}{\mu (t)}{t}^{1-\alpha }\right]g\left(\sigma (t)\right)+\left[\frac{g(\sigma (t))-g(t)}{\mu (t)}{t}^{1-\alpha }\right]f(t)=T_{\alpha }(f)(t)g(\sigma (t))+f(t)T_{\alpha }(g)(t)\text{.}$$ The other product rule formula follows by interchanging the role of functions f and g.

4. We use the conformable fractional derivative of a constant (Proposition 9) and property (iii) of Theorem 15 (just proved): from Proposition 9 we know that $T_{\alpha }\left(f·\frac{1}{f}\right)(t)={(1)}^{(\alpha )}=0$ . Therefore, by (iii) $$T_{\alpha }\left(\frac{1}{f}\right)(t)f(\sigma (t))+T_{\alpha }(f)(t)\frac{1}{f(t)}=0\text{.}$$

Since we are assuming $f(\sigma (t))\ne 0\text{,}{T}_{\alpha }\left(\frac{1}{f}\right)(t)=-\frac{T_{\alpha }(f)(t)}{f(t)f(\sigma (t))}$ .

(v) We use (ii) and (iv) to obtain $$T_{\alpha }\left(\frac{f}{g}\right)(t)=T_{\alpha }\left(f·\frac{1}{g}\right)(t)=f(t)T_{\alpha }\left(\frac{1}{g}\right)(t)+T_{\alpha }(f)(t)\frac{1}{g(\sigma (t))}=\frac{T_{\alpha }(f)(t)g(t)-f(t)T_{\alpha }(g)(t)}{g(t)g(\sigma (t))}\text{.}$$ This concludes the proof.  □

Theorem 16

Let c be a constant, $m\in \mathbb{N}$ , and $\alpha \in \left]0\text{,}1\right]$ .

1. If $f(t)={(t-c)}^m$ , then $T_{\alpha }(f)(t)=t^{1-\alpha }{\sum }_{p=0}^{m-1}{\left(\sigma (t)-c\right)}^{m-1-p}{(t-c)}^p$ .

2. If $g(t)=\frac{1}{{(t-c)}^m}$ and $(t-c)\left(\sigma (t)-c\right)\ne 0$ , then $T_{\alpha }(g)(t)=-t^{1-\alpha }{\sum }_{p=0}^{m-1}\frac{1}{{(\sigma (t)-c)}^{p+1}{(t-c)}^{m-p}}$ .

Proof

We prove the first formula by induction. If $m=1$ , then $f(t)=t-c$ and $T_{\alpha }(f)(t)=t^{1-\alpha }$ holds from Propositions 9 and 10 and Theorem 15 (i). Now assume that $$T_{\alpha }(f)(t)=t^{1-\alpha }{\displaystyle \sum _{p=0}^{m-1}}{(\sigma (t)-c)}^{m-1-p}{(t-c)}^p$$ holds for $f(t)={(t-c)}^m$ and let $F(t)={(t-c)}^{m+1}=(t-c)f(t)$ . We use Theorem 15 (iii) to obtain ${\left(F(t)\right)}^{(\alpha )}=T_{\alpha }(t-c)f(\sigma (t))+T_{\alpha }(f)(t)(t-c)=t^{1-\alpha }{\sum }_{p=0}^m{(\sigma (t)-c)}^{m-p}{(t-c)}^p$ . Hence, by mathematical induction, part (i) holds. (ii) Let $g(t)=\frac{1}{{(t-c)}^m}=\frac{1}{f(t)}$ . From (iv) of Theorem 15, $$g^{(\alpha )}(t)=-\frac{T_{\alpha }(f)(t)}{f(t)f(\sigma (t))}=-t^{1-\alpha }{\displaystyle \sum _{p=0}^{m-1}}\frac{1}{{(\sigma (t)-c)}^{p+1}{(t-c)}^{m-p}}\text{,}$$ provided $(t-c)\left(\sigma (t)-c\right)\ne 0$ .  □

Example 17

Let $\alpha \in \left]0\text{,}1\right]$ and $f(t)=t^m$ . Then $T_{\alpha }(f)(t)=t^{1-\alpha }{\sum }_{p=0}^{m-1}\sigma {(t)}^{m-1-p}{t}^p$ . Note that if t is right-dense, then $T_{\alpha }(f)(t)={\mathit{mt}}^{m-\alpha }$ . If we choose $\mathbb{T}=\mathbb{R}$ and $\alpha =1$ , then we obtain the usual derivative: $T_1(f)(t)={\mathit{mt}}^{m-1}=f^{\prime }(t)$ .

Example 18

Let $\alpha \in \left]0\text{,}1\right]$ and $f(t)=\frac{1}{t^m}$ . Then $T_{\alpha }(f)(t)=-t^{1-\alpha }{\sum }_{p=0}^{m-1}\frac{1}{t^{p-m}\sigma {(t)}^{p+1}}$ . If t is right-dense, then $T_{\alpha }(f)(t)=-\frac{m}{t^{m+\alpha }}$ . Moreover, if $\alpha =1$ , then we obtain $T_1(f)(t)=-\frac{m}{t^{m+1}}$ .

Example 19

If $f(t)={(t-1)}^2$ , then $T_{\alpha }(f)(t)=t^{1-\alpha }\left[{(\sigma (t)+1)}^2+(\sigma (t)+1)(t+1)+{(t+1)}^2\right]$ for all $\alpha \in \left]0\text{,}1\right]$ .

Example 20

Let $\alpha \in (0\text{,}1)\text{;}\mathbb{T}=\mathbb{N}=\{1\text{,}2\text{,}\dots \}$ , for which $\sigma (t)=t+1$ and $\mu (t)=1$ ; and $f\text{,}g:\mathbb{T}\to \mathbb{T}$ be given by $f(t)=g(t)=t$ . Then, $T_{\alpha }(f\circ g)(t)\ne T_{\alpha }(f)\left(g(t)\right)T_{\alpha }(g)(t):T_{\alpha }(f\circ g)(t)=t^{1-\alpha }$ , while $T_{\alpha }(f)\left(g(t)\right)T_{\alpha }(g)(t)=t^{2(1-\alpha )}$ .

Theorem 21

Theorem 21 Chain rule

Let $\alpha \in \left]0\text{,}1\right]$ . Assume $g:\mathbb{T}\to \mathbb{R}$ is continuous and conformable fractional differentiable of order $\alpha$ at $t\in {\mathbb{T}}^{\kappa }$ , and $f:\mathbb{R}\to \mathbb{R}$ is continuously differentiable. Then there exists c in the real interval $[t\text{,}\sigma (t)]$ with

$$T_{\alpha }(f\circ g)(t)=f^{\prime }(g(c))T_{\alpha }(g)(t)\text{.}$$

Proof

Let $t\in {\mathbb{T}}^{\kappa }$ . First we consider t to be right-scattered. In this case, $$T_{\alpha }(f\circ g)(t)=\frac{f(g(\sigma (t)))-f(g(t))}{\mu (t)}{t}^{1-\alpha }\text{.}$$ If $g(\sigma (t))=g(t)$ , then we get $T_{\alpha }(f\circ g)(t)=0$ and $T_{\alpha }(g)(t)=0$ . Therefore, (4) holds for any c in the real interval $[t\text{,}\sigma (t)]$ . Now assume that $g(\sigma (t))\ne g(t)$ . By the mean value theorem we have $$T_{\alpha }(f\circ g)(t)=\frac{f(g(\sigma (t)))-f(g(t))}{g(\sigma (t))-g(t)}·\frac{g(\sigma (t))-g(t)}{\mu (t)}{t}^{1-\alpha }=f^{\prime }(\xi )T_{\alpha }(g)(t)\text{,}$$ where $\xi \in ]g(t)\text{,}g(\sigma (t))[$ . Since $g:\mathbb{T}\to \mathbb{R}$ is continuous, there is a $c\in [t\text{,}\sigma (t)]$ such that $g(c)=\xi$ , which gives the desired result. Now let us consider the case when t is right-dense. In this case $$T_{\alpha }(f\circ g)(t)={\displaystyle \underset{s\to t}{\lim }}\frac{f(g(t))-f(g(s))}{g(t)-g(s)}·\frac{g(t)-g(s)}{t-s}{t}^{1-\alpha }\text{.}$$ By the mean value theorem, there exists ${\xi }_s\in ]g(t)\text{,}g(\sigma (t))[$ such that $$T_{\alpha }(f\circ g)(t)={\displaystyle \underset{s\to t}{\lim }}\left\{f^{\prime }({\xi }_s)·\frac{g(t)-g(s)}{t-s}{t}^{1-\alpha }\right\}\text{.}$$ By the continuity of g, we get that ${\lim }_{s\to t}{\xi }_s=g(t)$ . Then $T_{\alpha }(f\circ g)(t)=f^{\prime }(g(t))·T_{\alpha }(g)(t)$ . Since t is right-dense, we conclude that $c=t=\sigma (t)$ , which gives the desired result.  □

Example 22

Let $\mathbb{T}=2^{\mathbb{N}}$ , for which $\sigma (t)=2t$ and $\mu (t)=t$ . (i) Choose $f(t)=t^2$ and $g(t)=t$ . Theorem 21 guarantees that we can find a value c in the interval $[t\text{,}\sigma (t)]=[t\text{,}2t]$ , such that

$$T_{\alpha }(f\circ g)(t)=f^{\prime }(g(c))T_{\alpha }(g)(t)\text{.}$$ Indeed, from Theorem 4 it follows that $T_{\alpha }(f\circ g)(t)=3t^{1-\alpha }\text{,}{T}_{\alpha }(g)(t)=t^{1-\alpha }$ , and $f^{\prime }(g(c))=2c$ . Equality (5) leads to $3t^{1-\alpha }=2{\mathit{ct}}^{1-\alpha }$ and so $c=\frac{3}{2}t\in [t\text{,}2t]$ . (ii) Now let us take $f(t)=g(t)=t^2$ for all $t\in \mathbb{T}$ . We obtain $15t^{4-\alpha }=T_{\alpha }(f\circ g)(t)=f^{\prime }(g(c))T_{\alpha }(g)(t)=2c^{2}3t^{2-\alpha }$ . Therefore, $c=\sqrt{\frac{5}{2}}t\in [t\text{,}2t]$ .

Definition 23

Let $\mathbb{T}$ be a time scale, $\alpha \in (n\text{,}n+1]\text{,}n\in \mathbb{N}$ , and let f be n times delta differentiable at $t\in {\mathbb{T}}^{{\kappa }^n}$ . We define the conformable fractional derivative of f of order $\alpha$ as $T_{\alpha }(f)(t)≔T_{\alpha -n}\left(f^{{\mathrm{\Delta }}^n}\right)(t)$ . As before, we also use the notation ${(f(t))}^{(\alpha )}=T_{\alpha }(f)(t)$ .

Example 24

Let $\mathbb{T}=h\mathbb{Z}\text{,}h>0\text{,}f(t)=t^3$ , and $\alpha =2.1$ . Then, by Definition 23, we have $T_{2.1}(f)=T_{0.1}\left(f^{{\mathrm{\Delta }}^2}\right)$ . Since $\sigma (t)=t+h$ and $\mu (t)=h\text{,}{T}_{2.1}(f)(t)={\left(t^3\right)}^{(2.1)}={(6t+6h)}^{(0.1)}$ . By Proposition 9 and Theorem 15 (i) and (ii), we obtain that $T_{2.1}(f)(t)=6{(t)}^{(0.1)}$ . We conclude from Proposition 10 that $T_{2.1}(f)(t)=6t^{0.9}$ .

Theorem 25

Let $\alpha \in (n\text{,}n+1]\text{,}n\in \mathbb{N}$ . The following relation holds:

$$T_{\alpha }(f)(t)=t^{1+n-\alpha }{f}^{{\mathrm{\Delta }}^{1+n}}(t)\text{.}$$

Proof

Let f be a function n times delta-differentiable. For $\alpha \in (n\text{,}n+1]$ , there exist $\beta \in (0\text{,}1]$ such that $\alpha =n+\beta$ . Using Definition 23, $T_{\alpha }(f)=T_{\beta }\left(f^{{\mathrm{\Delta }}^n}\right)$ . From the definition of (higher-order) delta derivative and Theorem 4 (ii) and (iii), it follows that $T_{\alpha }(f)(t)=t^{1-\beta }{\left(f^{{\mathrm{\Delta }}^n}\right)}^{\mathrm{\Delta }}(t)$ .  □

Definition 26

Let $f:\mathbb{T}\to \mathbb{R}$ be a regulated function. Then the $\alpha$ -fractional integral of f, $0<\alpha ⩽1$ , is defined by $\int f(t){\mathrm{\Delta }}^{\alpha }t≔\int f(t)t^{\alpha -1}\mathrm{\Delta }t$ .

Remark 27

For $\mathbb{T}=\mathbb{R}$ Definition 26 reduces to the conformable fractional integral given in Khalil et al. (2014); for $\alpha =1$ Definition 26 reduces to the indefinite integral of time scales (Bohner and Peterson, 2001).

Definition 28

Suppose $f:\mathbb{T}\to \mathbb{R}$ is a regulated function. Denote the indefinite $\alpha$ -fractional integral of f of order $\alpha \text{,}\alpha \in (0\text{,}1]$ , as follows: $F_{\alpha }(t)=\int f(t){\mathrm{\Delta }}^{\alpha }t$ . Then, for all $a\text{,}b\in \mathbb{T}$ , we define the Cauchy $\alpha$ -fractional integral by ${\int }_a^{b}f(t){\mathrm{\Delta }}^{\alpha }t=F_{\alpha }(b)-F_{\alpha }(a)$ .

Example 29

Let $\mathbb{T}=\mathbb{R}\text{,}\alpha =\frac{1}{2}$ , and $f(t)=t$ . Then ${\int }_1^{{10}^{2/3}}f(t){\mathrm{\Delta }}^{\alpha }t=6$ .

Theorem 30

Let $\alpha \in (0\text{,}1]$ . Then, for any rd-continuous function $f:\mathbb{T}\to \mathbb{R}$ , there exists a function $F_{\alpha }:\mathbb{T}\to \mathbb{R}$ such that $T_{\alpha }\left(F_{\alpha }\right)(t)=f(t)$ for all $t\in {\mathbb{T}}^{\kappa }$ . Function $F_{\alpha }$ is said to be an $\alpha$ -antiderivative of f.

Proof

The case $\alpha =1$ is proved in Bohner and Peterson (2001). Let $\alpha \in (0\text{,}1)$ . Suppose f is rd-continuous. By Theorem 1.16 of Bohner and Peterson (2003), f is regulated. Then, $F_{\alpha }(t)=\int f(t){\mathrm{\Delta }}^{\alpha }t$ is conformable fractional differentiable on ${\mathbb{T}}^{\kappa }$ . Using (6) and Definition 26, we obtain that $T_{\alpha }\left(F_{\alpha }\right)(t)=t^{1-\alpha }{\left(F_{\alpha }(t)\right)}^{\mathrm{\Delta }}=f(t)\text{,}t\in {\mathbb{T}}^{\kappa }$ .  □