Reply to Kiskinov et al. Comment on “Vatsala et al. Analysis of Sequential Caputo Fractional Differential Equations versus Non-Sequential Caputo Fractional Differential Equations with Applications. Foundations 2022, 2, 1129–1142”

In our article [1], we are not trying to provide a counter-example to Theorems 1–3 of [2]. In fact, Theorem 1 is Theorem 2.2 of [3] (p. 93). In order to solve a Caputo fractional differential equation of order q, where $0<q<1$, we do not require the solution to be a $C^1$ function on the interval $[a,b]$. It is enough that the solution is a $C^q$ function, which essentially means the Caputo $q^{th}$ order derivative exists on the needed interval.

Here, we provide the following definitions of left and right Caputo fractional derivatives for clarity.

Definition 1.

The Caputo (left-sided) fractional derivative of $f\left(t\right)$ of order α, when $n-1<\alpha <n$, is given by equation

$${}^{c}D_{a^{+}}^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\int }_a^t{(t-s)}^{n-\alpha -1}{f}^{\left(n\right)}\left(s\right)ds,t>a,$$

(1)

and the (right-sided) fractional derivative is given by

$${}^{c}D_{b^{-}}^{\alpha }f\left(t\right)={\displaystyle \frac{{(-1)}^n}{\Gamma (n-\alpha )}}{\int }_t^b{(s-t)}^{n-\alpha -1}{f}^{\left(n\right)}\left(s\right)ds,t<b,$$

(2)

where $f^{\left(n\right)}\left(t\right)={\displaystyle \frac{d^n\left(f\right)}{d{t}^n}}$.

In this work, we use Caputo fractional derivatives of order q, where $0<q<1$.

Remark 1.

In order for Theorem 2.2 of [2], (p. 93) to hold true, the Caputo left fractional derivative can be computed when the function can be expressed as a function of $(t-a)$. Then, for the same function, the Caputo right fractional derivative has to be computed as a function of $(b-t)$. Furthermore, if $f\in C^1[a,b]$, then part (a) and (b) of Theorem 2.2 of [2] holds true.

However, it is to be noted that, from Theorem 2.2 of [3], both the Caputo left derivative and the Caputo right derivative have been defined to yield the integer derivative result $q\to n$ as a special case. In particular, when $0<q<1$, then $q\to 1$ yields the solution of the first order differential equations.

Remark 2.

Observe that the solution of the Caputo fractional differential equation of order α is not a $C^n$ solution on $[a,b]$, when $(n-1)<\alpha <n$. See Equation (4.1.62) of [3], which is the solution of an ${\alpha }^{th}$-order Caputo fractional differential equation when $n-1<\alpha <n$.

In short, when we are seeking the solution of a left Caputo fractional differential equation of order q, where $0<q<1$, the solution is not $C^1$ on $(a,b]$. Similarly, when we are seeking the solution of a right Caputo fractional differential equation of order q, where $0<q<1$, the solution is not $C^1$ on $[a,b)$.

The next comment is about the equivalence of the Caputo left derivative and Caputo right derivative of a function at any point on the interval $(a,b)$.

Consider the linear left Caputo fractional differential equation

$${}^{c}D_{a^{+}}^{q}u=\lambda u,a<t<b,u\left(a\right)=u_a.$$

Then, the solution of this left Caputo fractional differential equation is given by

$$u_l\left(t\right)=u_a{E}_{q,1}\left(\lambda {(t-a)}^q\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\lambda {(t-a)}^q\right)}^k}{\Gamma (qk+1)}}.$$

Now, let us consider the right Caputo fractional differential equation,

$${}^{c}D_{b^{-}}^{q}u=\lambda u,a<t<b,u\left(b\right)=u_a,$$

It is easy to check that the solution of this right Caputo fractional differential equation is

$$u_r\left(t\right)=u_a{E}_{q,1}\left(\lambda {(b-t)}^q\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\lambda {(b-t)}^q\right)}^k}{\Gamma (qk+1)}}.$$

Although the above solutions are solutions of Caputo fractional initial value problems, we want to show the equivalence of the left and right Caputo derivative of $u_l$ and $u_r$ at any point on $(a,b)$. Observe that the two solutions are equal at $t={\displaystyle \frac{a+b}{2}}$. Now, we claim that the two solutions are equal for any t in $(a,b)$, say $t=t_1$. Then, we need to substitute $t_1=a+\alpha$ in the left Caputo fractional derivative solution and $t_1=b-\alpha$ in the right Caputo fractional derivative solution. Then, it is easy to check that the two solutions are one and the same for any $t_1$ on $(a,b)$ when $\alpha \in (0,(b-a\left)\right]$. If $\alpha =0$, then $u_l\left(a\right)=u_a=u_r\left(b\right)$.

As an example, consider $a=1,b=2$, and $t_1=1.25$; then, in this case, $\alpha =0.25$. Then, the above left and right solution can be written as

$$u_l(1+0.25)=u_a{E}_{q,1}\left(\lambda {\left(0.25\right)}^q\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\lambda {\left(0.25\right)}^q\right)}^k}{\Gamma (qk+1)}},$$

$$u_r(2-0.25)=u_a{E}_{q,1}\left(\lambda {\left(0.25\right)}^q\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\lambda {\left(0.25\right)}^q\right)}^k}{\Gamma (qk+1)}}.$$

Next, we consider a few examples to demonstrates that Caputo fractional derivative of order q, where $0<q<1$ exists, even if it is not a $C^1$ function.

Let

$$f\left(t\right)={(t-a)}^{\omega }.$$

(3)

It is easy to see that the left Caputo fractional derivative of order q is

$${}^{c}D_{a^{+}}^q{(t-a)}^{\omega }={\displaystyle \frac{\Gamma (\omega +1){(t-a)}^{(\omega -q)}}{\Gamma (\omega -q+1)}}.$$

(4)

Similarly, the right Caputo fractional derivative of ${(b-t)}^{\omega }$ is

$${}^{c}D_{b^{-}}^q{(b-t)}^{\omega }={\displaystyle \frac{\Gamma (\omega +1){(b-t)}^{(\omega -q)}}{\Gamma (\omega -q+1)}}.$$

(5)

Observe that $f\left(t\right)={(t-a)}^{\omega }$ is not $C^1$ at $t=a$, and $f\left(t\right)={(b-t)}^{\omega }$, is not $C^1$ at $t=b$, when $\omega <1$. However, the Caputo left fractional derivative of ${(t-a)}^{\omega }$ exists on $[a,b]$ and zero at $t=a$. Similarly, the Caputo right fractional derivative of ${(b-t)}^{\omega }$ exists on $[a,b]$ and zero at $t=b$. This satisfies part (a) of Theorem 2.2 of [3]. If $\omega =1$, then part (a) of Theorem 2 of [3] holds true.

If $\omega =1$, then (4) and (5) become

$${}^{c}D_{a^{+}}^q(t-a)={\displaystyle \frac{{(t-a)}^{(1-q)}}{\Gamma (2-q)}},$$

(6)

and

$${}^{c}D_{b^{-}}^q(b-t)={\displaystyle \frac{{(b-t)}^{(1-q)}}{\Gamma (2-q)}}.$$

(7)

If $q=1$, then the left derivative equals $f^{\prime }\left(t\right)=1$ and the right derivative equals $-(-f^{\prime }\left(t\right))=1$. This validates Remark 1, as well as Theorem 2.2 of [3]

In particular, if $\omega =q$, then

$${}^{c}D_{a^{+}}^q{(t-a)}^q=\Gamma (q+1)={}^{c}D_{b^{-}}^q{(b-t)}^q.$$

It appears that the authors of [2] have computed the Caputo derivative of

$$f\left(x\right)={(x-a)}^q{(b-x)}^q,$$

at a specific point numerically by substituting the $x_1$ value in place of the s value in the integral. The integral of this function will not be in closed form.

In fact, for any function $f=f(t-a)$, where $f\in C^q$ on $(a,b]$, we have

$${}^{c}D_{a^{+}}^q\left(f(t-a)\right){{|}_{t=a+\alpha }=F(t-a)|}_{t=a+\alpha }=F\left(\alpha \right),$$

and

$${}^{c}D_{b^{-}}^q\left(f(b-t)\right){{|}_{t=b-\alpha }=F(b-t)|}_{t=b-\alpha }=F\left(\alpha \right).$$

This proves that

$${}^{c}D_{a^{+}}^q\left(f(t-a)\right){{|}_{t=a+\alpha }={}^{c}D_{b^{-}}^q\left(f(b-t)\right)|}_{t=b-\alpha }=F\left(\alpha \right).$$

This is our claim in [1].

The reference [4] is devoted to Caputo fractional differential equations.

Conclusions: We conjecture that this result can be extended to Caputo sequential fractional derivative of order $nq$, which is sequential of order q. Note that the integer derivative is sequential.