Stability and Logarithmic Decay of the Solution to Hadamard-Type Fractional Differential Equation

Abstract

In this paper, we study the stability and logarithmic decay of the solutions to fractional differential equations (FDEs). Both linear and nonlinear cases are included. And the fractional derivative is in the sense of Hadamard or Caputo–Hadamard with order $\alpha \,(0<\alpha <1)$. The solutions can be expressed by Mittag–Leffler functions through applying the modified Laplace transform. In view of the asymptotic expansions of Mittag–Leffler function, we discuss the stability and logarithmic decay of the solution to FDEs in great detail.

6 Appendix

In the Appendix section, we first introduce a modified Laplace transform and a modified Mellin transform. These integral transforms can be used to solve the left and right Riemann–Liouville fractional integral and Riemann–Liouville and Caputo derivatives with starting point a ($a\in {\mathbb {R}}$, not necessarily the origin) or final point $b\,(b\in {\mathbb {R}})$. Then, we discuss their properties. These integral transforms are displayed just for reference of the interested readers.

Definition 6.1

(Kilbas et al. 2006; Li and Cai 2019) The left and right Riemann–Liouville fractional integrals of order $\alpha \,(\alpha >0)$ are defined, respectively, by

$$\begin{aligned} _{RL}\mathrm{{D}}^{-\alpha }_{a,t}f(t)=\frac{1}{\Gamma (\alpha )}\int _{a}^{t} (t-\tau )^{-(1-\alpha )}f(\tau )\mathrm{{d}}\tau ,\,t>a,\,a\in {\mathbb {R}}, \end{aligned}$$

(6.1)

and

$$\begin{aligned} _{RL}\mathrm{{D}}^{-\alpha }_{t,b}f(t)=\frac{1}{\Gamma (\alpha )}\int _{t}^{b} (\tau -t)^{-(1-\alpha )}f(\tau )\mathrm{{d}}\tau ,\,t<b,\,b\in {\mathbb {R}}. \end{aligned}$$

(6.2)

Definition 6.2

(Kilbas et al. 2006; Li and Cai 2019) The left and right Riemann–Liouville fractional derivatives of order $\alpha \,(n-1<\alpha <n\in {\mathbb {Z}}^+)$ are defined, respectively, by

$$\begin{aligned} _{RL}\mathrm{{D}}_{a,t}^{\alpha }f(t)&=\frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}} \left( \,_{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f(t)\right) \nonumber \\&=\frac{1}{\Gamma (n-\alpha )}\frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}}\int _{a}^{t} (t-\tau )^{n-\alpha -1}f(\tau )\mathrm{{d}}\tau ,\, t>a,\,a\in {\mathbb {R}}, \end{aligned}$$

(6.3)

and

$$\begin{aligned} _{RL}\mathrm{{D}}_{t,b}^{\alpha }f(t)&=(-1)^{n}\frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}} \left( \,_{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f(t)\right) \nonumber \\&=\frac{(-1)^{n}}{\Gamma (n-\alpha )}\frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}}\int _{t}^{b} (\tau -t)^{n-\alpha -1}f(\tau )\mathrm{{d}}\tau ,\, t<b,\,b\in {\mathbb {R}}. \end{aligned}$$

(6.4)

Definition 6.3

(Kilbas et al. 2006; Li and Cai 2019) The left and right Caputo fractional derivatives of order $\alpha \,(n-1<\alpha <n\in {\mathbb {Z}}^+)$ are defined by

$$\begin{aligned} _{C}\mathrm{{D}}^{\alpha }_{a,t}f(t)&= \,_{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f^{(n)}(t)\nonumber \\&=\frac{1}{\Gamma (n-\alpha )}\int _{a}^{t} (t-\tau )^{n-\alpha -1}f^{(n)}(\tau ) \mathrm{{d}}\tau ,\, t>a,\,a\in {\mathbb {R}}, \end{aligned}$$

(6.5)

and

$$\begin{aligned} _{C}\mathrm{{D}}^{\alpha }_{t,b}f(t)&=(-1)^{n} \,_{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f^{(n)}(t)\nonumber \\&=\frac{(-1)^{n}}{\Gamma (n-\alpha )}\int _{t}^{b} (\tau -t)^{n-\alpha -1}f^{(n)}(\tau ) \mathrm{{d}}\tau ,\, t<b,\,b\in {\mathbb {R}}. \end{aligned}$$

(6.6)

1.1 6.1 The Modified Laplace Transform

Now we introduce the following left and right modified Laplace transforms.

Definition 6.4

Suppose that the given function f(t) is defined on $[a, +\infty )\, (a\in {\mathbb {R}})$. The left modified Laplace transform of f(t) is defined as

$$\begin{aligned} {\widetilde{f}}(s)={\mathscr {L}}_{lm}\{f(t)\} =\int _{a}^{\infty }e^{-s(t-a)}f(t) \mathrm{{d}}t,\,a\in {\mathbb {R}},\,s\in {\mathbb {C}}. \end{aligned}$$

(6.7)

The inverse left modified Laplace transform is given by

$$\begin{aligned} f(t)={\mathscr {L}}^{-1}_{lm}\{{\widetilde{f}}(s)\}=\frac{1}{2\pi i} \int _{c-i\infty }^{c+i\infty } e^{s(t-a)}{\widetilde{f}}(s)\mathrm{{d}}s,\, c>0,\, i^{2}=-1. \end{aligned}$$

(6.8)

The left modified Laplace transform defined by (6.7) is a special case of the generalized Laplace transform given by Definition 3.1 in Jarad and Abdeljawad (2020).

Definition 6.5

Suppose that the given function f(t) is defined on $(-\infty , b]\, (b\in {\mathbb {R}})$, the right modified Laplace transform of f(t) is defined as

$$\begin{aligned} {\widetilde{f}}(s)={\mathscr {L}}_{rm}\{f(t)\} =\int _{-\infty }^{b}e^{-s(b-t)}f(t) \mathrm{{d}}t,\,b\in {\mathbb {R}},\,s\in {\mathbb {C}}. \end{aligned}$$

(6.9)

The inverse right modified Laplace transform is given by

$$\begin{aligned} f(t)={\mathscr {L}}^{-1}_{rm}\{{\widetilde{f}}(s)\}=\frac{1}{2\pi i} \int _{c-i\infty }^{c+i\infty } e^{s(b-t)}{\widetilde{f}}(s)\mathrm{{d}}s,\, c>0,\, i^{2}=-1. \end{aligned}$$

(6.10)

In the following, we present the conditions for the existence of the modified Laplace transforms.

Theorem 6.1

Assume that the given function f(t) is defined on $[a, +\infty )\, (a\in {\mathbb {R}})$. If

(1) f(t) is continuous or piecewise continuous on every finite subinterval of $[a, +\infty )$,

(2) there exist positive constants $M>0$ and $\sigma >0$ such that for the given $T>a$,

$$\begin{aligned} |f(t)|\le M e^{\sigma (t-a)},\, \text{ when }\ t>T, \end{aligned}$$

then the left modified Laplace transform of f(t) exists with $\mathrm{{Re}}(s)>\sigma$.

Proof

In view of Definition 6.4 and the above conditions (1) and (2), one has

$$\begin{aligned} |{\widetilde{f}}(s)|&\le \int _{a}^{\infty } e^{-|s|(t-a)}|f(t)|\mathrm{{d}}t \le \int _{a}^{\infty }e^{-|s|(t-a)}M e^{\sigma (t-a)}\mathrm{{d}}t\\&= M\int _{a}^{\infty }e^{-(|s|-\sigma )(t-a)}\mathrm{{d}}t = \frac{M}{|s|-\sigma }. \end{aligned}$$

This proves the conclusion. $\square$

Theorem 6.2

Assume that the given function f(t) is defined on $(-\infty , b]\, (b\in {\mathbb {R}})$. If

(1) f(t) is continuous or piecewise continuous on every finite subinterval of $(-\infty , b]$,

(2) there exist positive constants $M>0$ and $\sigma >0$ such that for the given $T'<b$,

$$\begin{aligned} |f(t)|\le M e^{\sigma (b-t)},\, \text{ when }\ t<T', \end{aligned}$$

then the right modified Laplace transform of f(t) exists with $\mathrm{{Re}}(s)>\sigma$.

Proof

By means of Definition 6.5 and the above conditions (1) and (2), one gets

$$\begin{aligned} |{\widetilde{f}}(s)|&\le \int _{-\infty }^{b} e^{-|s|(b-t)}|f(t)|\mathrm{{d}}t \le \int _{-\infty }^{b}e^{-|s|(b-t)}M e^{\sigma (b-t)}\mathrm{{d}}t\\&= M\int _{-\infty }^{b}e^{-(|s|-\sigma )(b-t)}\mathrm{{d}}t = \frac{M}{|s|-\sigma }. \end{aligned}$$

The result is thus proved. $\square$

Definition 6.6

For the given functions f(t) and g(t) defined on $[a, +\infty )\, (a\in {\mathbb {R}})$, the integral $\int _{a}^{t}f(a+t-\tau )g(\tau )\mathrm{{d}}\tau$ is called the left convolution of f(t) and g(t), i.e.

$$\begin{aligned} f(t)*_{l}g(t)=(f*_{l}g)(t)=\int _{a}^{t}f(a+t-\tau )g(\tau )\mathrm{{d}}\tau . \end{aligned}$$

Theorem 6.3

(Left convolution theorem) If ${\mathscr {L}}_{lm}\{f(t)\}={\widetilde{f}}(s)$ and ${\mathscr {L}}_{lm}\{g(t)\}={\widetilde{g}}(s)$, then

$$\begin{aligned} {\mathscr {L}}_{lm}\{f(t)*_{l}g(t)\}={\mathscr {L}}_{lm}\{f(t)\}{\mathscr {L}}_{lm}\{g(t)\} ={\widetilde{f}}(s){\widetilde{g}}(s); \end{aligned}$$

Or equivalently,

$$\begin{aligned} {\mathscr {L}}_{lm}^{-1}\{{\widetilde{f}}(s){\widetilde{g}}(s)\}=f(t)*_{l}g(t). \end{aligned}$$

Proof

According to Definitions 6.4 and 6.6, and interchanging the order of integration, one gets

$$\begin{aligned} {\mathscr {L}}_{lm}\{f(t)*_{l}g(t)\}&=\int _{a}^{\infty }e^{-s(t-a)} \int _{a}^{t}f(a+t-\tau )g(\tau )\mathrm{{d}}\tau \mathrm{{d}}t\\&=\int _{a}^{\infty }\int _{\tau }^{\infty }e^{-s(t-a)} f(a+t-\tau )g(\tau )\mathrm{{d}}t\mathrm{{d}}\tau \\&=\int _{a}^{\infty }\int _{a}^{\infty }e^{-s(\tau +w-2a)} f(w)g(\tau )\mathrm{{d}}w\mathrm{{d}}\tau \\&={\mathscr {L}}_{lm}\{f(t)\}{\mathscr {L}}_{lm}\{g(t)\} ={\widetilde{f}}(s){\widetilde{g}}(s), \end{aligned}$$

in which the change of variable $a+t-\tau =w$ is used. $\square$

Definition 6.7

For the given functions f(t) and g(t) defined on $(-\infty , b]\, (b\in {\mathbb {R}})$, the integral $\int _{t}^{b}f(b+t-\tau )g(\tau )\mathrm{{d}}\tau$ is called the right convolution of f(t) and g(t), i.e.

$$\begin{aligned} f(t)*_{r}g(t)=(f*_{r}g)(t)=\int _{t}^{b}f(b+t-\tau )g(\tau )\mathrm{{d}}\tau . \end{aligned}$$

Theorem 6.4

(Right convolution theorem) If ${\mathscr {L}}_{rm}\{f(t)\}={\widetilde{f}}(s)$ and ${\mathscr {L}}_{rm}\{g(t)\}={\widetilde{g}}(s)$, then

$$\begin{aligned} {\mathscr {L}}_{rm}\{f(t)*_{r}g(t)\}={\mathscr {L}}_{rm}\{f(t)\}{\mathscr {L}}_{rm}\{g(t)\} ={\widetilde{f}}(s){\widetilde{g}}(s); \end{aligned}$$

Or equivalently,

$$\begin{aligned} {\mathscr {L}}_{rm}^{-1}\{{\widetilde{f}}(s){\widetilde{g}}(s)\}=f(t)*_{r}g(t). \end{aligned}$$

Proof

Taking into account Definitions 6.5 and 6.7, and interchanging the order of integration, one has

$$\begin{aligned} {\mathscr {L}}_{lm}\{f(t)*_{r}g(t)\}&=\int _{-\infty }^{b}e^{-s(b-t)} \int _{t}^{b}f(b+t-\tau )g(\tau )\mathrm{{d}}\tau \mathrm{{d}}t\\&=\int _{-\infty }^{b}\int _{-\infty }^{b}e^{-s(2b-\tau -w)} f(w)g(\tau )\mathrm{{d}}w\mathrm{{d}}\tau \\&={\mathscr {L}}_{rm}\{f(t)\}{\mathscr {L}}_{rm}\{g(t)\} ={\widetilde{f}}(s){\widetilde{g}}(s), \end{aligned}$$

where the change of variable $b+t-\tau =w$ is used. $\square$

We now present the differential property.

Lemma 6.1

If ${\mathscr {L}}_{lm}\{f(t)\}={\widetilde{f}}(s)$, then

$$\begin{aligned} {\mathscr {L}}_{lm}\{f^{(n)}(t)\} =s^{n}{\widetilde{f}}(s)-\sum _{k=0}^{n-1}s^{n-k-1}f^{(n)}(a),\,t>a,\, n \in {\mathbb {Z}}^{+}. \end{aligned}$$

(6.11)

If ${\mathscr {L}}_{rm}\{f(t)\}={\widetilde{f}}(s)$, then

$$\begin{aligned} {\mathscr {L}}_{rm}\{f^{(n)}(t)\} =\sum _{k=0}^{n-1}s^{n-k-1}f^{(k)}(b)-s^{n}{\widetilde{f}}(s),\,t<b,\, n \in {\mathbb {Z}}^{+}. \end{aligned}$$

(6.12)

Proof

The proof can be done by integration by parts. $\square$

Next, we can find the modified Laplace transform for Definitions 6.1–6.3.

Theorem 6.5

Let $n-1<\alpha <n\in {\mathbb {Z}}^+$ and $a\in {\mathbb {R}}$. Then, there hold:

$$\begin{aligned}&{\mathscr {L}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-\alpha }f(t)\} =s^{-\alpha }{\mathscr {L}}_{lm}\{f(t)\}, \end{aligned}$$

(6.13)

$$\begin{aligned}&{\mathscr {L}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{\alpha }f(t)\} =s^{\alpha }{\mathscr {L}}_{lm}\{f(t)\} -\sum _{k=0}^{n-1}s^{n-k-1}\, _{RL}\mathrm{{D}}_{a,t}^{\alpha +k-n}f(t)\Big |_{t=a}, \end{aligned}$$

(6.14)

$$\begin{aligned}&{\mathscr {L}}_{lm}\{_{C}\mathrm{{D}}_{a,t}^{\alpha }f(t)\} =s^{\alpha }{\mathscr {L}}_{lm}\{f(t)\} -\sum _{k=0}^{n-1}s^{\alpha -k-1}f^{(k)}(a). \end{aligned}$$

(6.15)

Proof

According to Definitions 6.1 and 6.6, Riemann–Liouville integral can be represented convolution form as follows:

$$\begin{aligned} _{RL}\mathrm{{D}}_{a,t}^{-\alpha }f(t)=\frac{1}{\Gamma (\alpha )}\int _{a}^{t} (t-\tau )^{-(1-\alpha )}f(\tau )\mathrm{{d}}\tau =\frac{(t-a)^{\alpha -1}}{\Gamma (\alpha )}*_{l}f(t). \end{aligned}$$

From Theorem 6.3, it holds that

$$\begin{aligned} {\mathscr {L}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-\alpha }f(t)\} ={\mathscr {L}}_{lm}\left\{ \frac{(t-a)^{\alpha -1}}{\Gamma (\alpha )}*_{l}f(t)\right\} =s^{-\alpha }{\mathscr {L}}_{lm}\{f(t)\}, \end{aligned}$$

which gives equality (6.13).

We prove (6.14). By using equality (6.3) and differential property (6.11), one gets

$$\begin{aligned} {\mathscr {L}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{\alpha }f(t)\}&={\mathscr {L}}_{lm}\left\{ \frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}} \left( _{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f(t)\right) \right\} \\&=s^{n}{\mathscr {L}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f(t)\} -\sum _{k=0}^{n-1}s^{n-k-1}\frac{\mathrm{{d}}^{k}}{\mathrm{{d}}t^{k}} \left( _{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f(t)\right) \Big |_{t=a}\\&=s^{\alpha }{\mathscr {L}}_{lm}\{f(t)\} -\sum _{k=0}^{n-1}s^{n-k-1}\, _{RL}\mathrm{{D}}_{a,t}^{\alpha +k-n}f(t)\Big |_{t=a}. \end{aligned}$$

To prove (6.15), using equalities (6.5), (6.13), and differential property (6.11) yields

$$\begin{aligned} {\mathscr {L}}_{lm}\{_{C}\mathrm{{D}}_{a,t}^{\alpha }f(t)\}&={\mathscr {L}}_{lm}\{_{C}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f^{(n)}(t)\} =s^{\alpha -n}{\mathscr {L}}_{lm}\{f^{(n)}(t)\}\\&=s^{\alpha -n}\left[ s^{n}{\mathscr {L}}_{lm}\{f(t)\} -\sum _{k=0}^{n-1}s^{n-k-1}f^{(k)}(a)\right] \\&=s^{\alpha }{\mathscr {L}}_{lm}\{f(t)\} -\sum _{k=0}^{n-1}s^{\alpha -k-1}f^{(k)}(a). \end{aligned}$$

The proof is completed. $\square$

Theorem 6.6

Let $n-1<\alpha <n\in {\mathbb {Z}}^+$ and $b\in {\mathbb {R}}$. Then, there hold:

$$\begin{aligned}&{\mathscr {L}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-\alpha }f(t)\} =s^{-\alpha }{\mathscr {L}}_{rm}\{f(t)\}, \end{aligned}$$

(6.16)

$$\begin{aligned}&{\mathscr {L}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{\alpha }f(t)\} =(-1)^{n}\left[ \sum _{k=0}^{n-1}s^{n-k-1}\, _{RL}\mathrm{{D}}_{t,b}^{\alpha +k-n}f(t)\Big |_{t=b} -s^{\alpha }{\mathscr {L}}_{rm}\{f(t)\}\right] , \end{aligned}$$

(6.17)

$$\begin{aligned}&{\mathscr {L}}_{rm}\{_{C}\mathrm{{D}}_{t,b}^{\alpha }f(t)\} =(-1)^{n}\left[ \sum _{k=0}^{n-1}s^{\alpha -k-1}f^{(k)}(b) -s^{\alpha }{\mathscr {L}}_{rm}\{f(t)\}\right] . \end{aligned}$$

(6.18)

Proof

In term of Definitions 6.1 and 6.7, Riemann–Liouville integral can be rewritten in the following convolution form:

$$\begin{aligned} _{RL}\mathrm{{D}}_{t,b}^{-\alpha }f(t)=\frac{1}{\Gamma (\alpha )}\int _{t}^{b} (\tau -t)^{-(1-\alpha )}f(\tau )\mathrm{{d}}\tau =\frac{(b-t)^{\alpha -1}}{\Gamma (\alpha )}*_{r}f(t). \end{aligned}$$

It follows from Theorem 6.4 that

$$\begin{aligned} {\mathscr {L}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-\alpha }f(t)\} ={\mathscr {L}}_{rm}\left\{ \frac{(b-t)^{\alpha -1}}{\Gamma (\alpha )}*_{r}f(t)\right\} =s^{-\alpha }{\mathscr {L}}_{rm}\{f(t)\}, \end{aligned}$$

which is (6.16).

We give the proof of (6.17). Recalling equality (6.4) and differential property (6.12), one has

$$\begin{aligned}&{\mathscr {L}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{\alpha }f(t)\}\\&\quad =(-1)^{n}\left[ {\mathscr {L}}_{rm}\left\{ \frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}} \left( _{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f(t)\right) \right\} \right] \\&\quad =(-1)^{n}\left[ \sum _{k=0}^{n-1}s^{n-k-1}\frac{\mathrm{{d}}^{k}}{\mathrm{{d}}t^{k}} \left( _{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f(t)\right) \Big |_{t=b} -s^{n}{\mathscr {L}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f(t)\}\right] \\&\quad =(-1)^{n}\left[ \sum _{k=0}^{n-1}s^{n-k-1}\, _{RL}\mathrm{{D}}_{t,b}^{\alpha +k-n}f(t)\Big |_{t=b} -s^{\alpha }{\mathscr {L}}_{rm}\{f(t)\}\right] . \end{aligned}$$

For (6.18), employing equalities (6.6), (6.16), and differential property (6.12) gives

$$\begin{aligned} {\mathscr {L}}_{rm}\{_{C}\mathrm{{D}}_{t,b}^{\alpha }f(t)\}&=(-1)^{n}{\mathscr {L}}_{rm}\{_{C}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f^{(n)}(t)\} =(-1)^{n}s^{\alpha -n}{\mathscr {L}}_{rm}\{f^{(n)}(t)\}\\&=(-1)^{n}s^{\alpha -n}\left[ \sum _{k=0}^{n-1}s^{n-k-1}f^{(k)}(b) -s^{n}{\mathscr {L}}_{rm}\{f(t)\}\right] \\&=(-1)^{n}\left[ \sum _{k=0}^{n-1}s^{\alpha -k-1}f^{(k)}(b) -s^{\alpha }{\mathscr {L}}_{rm}\{f(t)\}\right] . \end{aligned}$$

All this ends the proof. $\square$

Finally, we present the left and right modified Laplace transform of Mittag–Leffler function. Notice that Podlubny (1999)

$$\begin{aligned} \int _{0}^{\infty }e^{-st}t^{\alpha k+\beta -1} E_{\alpha ,\beta }^{(k)}(\pm \lambda t^{\alpha })\mathrm{{d}}t =\frac{k!s^{\alpha -\beta }}{(s^{\alpha }\mp \lambda )^{k+1}},\, \mathrm{{Re}}(s)>|\lambda |^{\frac{1}{\alpha }}. \end{aligned}$$

Thus, we can obtain

$$\begin{aligned} \int _{a}^{\infty }e^{-s(t-a)}(t-a)^{\alpha k+\beta -1} E_{\alpha ,\beta }^{(k)}(\pm \lambda (t-a)^{\alpha })\mathrm{{d}}t =\frac{k!s^{\alpha -\beta }}{(s^{\alpha }\mp \lambda )^{k+1}},\, \mathrm{{Re}}(s)>|\lambda |^{\frac{1}{\alpha }}, \end{aligned}$$

(6.19)

and

$$\begin{aligned} \int _{-\infty }^{b}e^{-s(b-t)}(b-t)^{\alpha k+\beta -1} E_{\alpha ,\beta }^{(k)}(\pm \lambda (b-t)^{\alpha })\mathrm{{d}}t =\frac{k!s^{\alpha -\beta }}{(s^{\alpha }\mp \lambda )^{k+1}},\, \mathrm{{Re}}(s)>|\lambda |^{\frac{1}{\alpha }}. \end{aligned}$$

(6.20)

1.2 The Modified Mellin Transform

In this subsection, we define a modified Mellin transform and investigate its properties.

Definition 6.8

The left modified Mellin transform of the known function f(t) with $t\in [a,\infty )\, (a\in {\mathbb {R}})$ is defined by

$$\begin{aligned} F(\xi )={\mathscr {M}}_{lm}\{f(t)\}=\int _{a}^{\infty } (t-a)^{\xi -1}f(t)\mathrm{{d}}t,\,a\in {\mathbb {R}},\, \gamma _{1}<\mathrm{{Re}}(\xi )<\gamma _{2}. \end{aligned}$$

(6.21)

The inverse left modified Mellin transform is given by

$$\begin{aligned} f(t)={\mathscr {M}}_{lm}^{-1}\{F(\xi )\}=\frac{1}{2\pi i} \int _{c-i\infty }^{c+i\infty }(t-a)^{-\xi }F(\xi )\mathrm{{d}}\xi ,\,t>a,\, c=\mathrm{{Re}}(\xi ). \end{aligned}$$

(6.22)

Definition 6.9

The right modified Mellin transform of the known function f(t) with $t\in [-\infty , b)\, (b\in {\mathbb {R}})$ is defined by

$$\begin{aligned} F(\xi )={\mathscr {M}}_{rm}\{f(t)\}=\int _{-\infty }^{b} (b-t)^{\xi -1}f(t)\mathrm{{d}}t,\,b\in {\mathbb {R}},\, \gamma _{3}<\mathrm{{Re}}(\xi )<\gamma _{4}. \end{aligned}$$

(6.23)

The inverse right modified Mellin transform is given by

$$\begin{aligned} f(t)={\mathscr {M}}_{rm}^{-1}\{F(\xi )\}=\frac{1}{2\pi i} \int _{c-i\infty }^{c+i\infty }(b-t)^{-\xi }F(\xi )\mathrm{{d}}\xi ,\,t<b,\, c=\mathrm{{Re}}(\xi ). \end{aligned}$$

(6.24)

In the sequel, we present the definition of convolution and convolution theorem in the sense of modified Mellin transform.

Definition 6.10

Suppose that functions f(t) and g(t) are defined on $[a, +\infty )\, (a\in {\mathbb {R}})$. The integral $\int _{a}^{\infty }f\left( a+\frac{t-a}{\tau -a}\right) g(\tau )\frac{\mathrm{{d}}\tau }{\tau -a}$ is called the left convolution of f(t) and g(t), i.e.

$$\begin{aligned} f(t)*_{l}g(t)=(f*_{l}g)(t) =\int _{a}^{\infty }f\left( a+\frac{t-a}{\tau -a}\right) g(\tau )\frac{\mathrm{{d}}\tau }{\tau -a}. \end{aligned}$$

Theorem 6.7

(Left convolution theorem) If ${\mathscr {M}}_{lm}\{f(t)\}=F(\xi )$ and ${\mathscr {M}}_{lm}\{g(t)\}=G(\xi )$, then

$$\begin{aligned} {\mathscr {M}}_{lm}\{f(t)*_{l}g(t)\}={\mathscr {M}}_{lm}\{f(t)\}{\mathscr {M}}_{lm}\{g(t)\} =F(\xi )G(\xi ); \end{aligned}$$

Or equivalently,

$$\begin{aligned} {\mathscr {M}}_{lm}^{-1}\{F(\xi )G(\xi )\}=f(t)*_{l}g(t). \end{aligned}$$

Proof

Noticing that Definitions 6.8 and 6.10, and changing the order of integration, one gets

$$\begin{aligned} {\mathscr {M}}_{lm}\{f(t)*_{l}g(t)\}&=\int _{a}^{\infty }(t-a)^{\xi -1} \int _{a}^{\infty }f\left( a+\frac{t-a}{\tau -a}\right) g(\tau )\frac{\mathrm{{d}}\tau }{\tau -a}\mathrm{{d}}t\\&=\int _{a}^{\infty }\int _{a}^{\infty }(t-a)^{\xi -1} f\left( a+\frac{t-a}{\tau -a}\right) g(\tau )\mathrm{{d}}t\frac{\mathrm{{d}}\tau }{\tau -a}\\&=\int _{a}^{\infty }\int _{a}^{\infty }(w-a)^{\xi -1}(\tau -a)^{\xi -1} f(w)g(\tau )\mathrm{{d}}w\mathrm{{d}}\tau \\&={\mathscr {M}}_{lm}\{f(t)\}{\mathscr {M}}_{lm}\{g(t)\} =F(\xi )G(\xi ), \end{aligned}$$

where the change of variable $a+\frac{t-a}{\tau -a}=w$ is utilized. $\square$

Definition 6.11

Suppose that functions f(t) and g(t) are defined on $(-\infty , b]\, (b\in {\mathbb {R}})$. The integral $\int _{-\infty }^{b}f\left( b-\frac{b-t}{b-\tau }\right) g(\tau )\frac{\mathrm{{d}}\tau }{b-\tau }$ is called the right convolution of f(t) and g(t), i.e.

$$\begin{aligned} f(t)*_{r}g(t)=(f*_{r}g)(t) =\int _{-\infty }^{b}f\left( b-\frac{b-t}{b-\tau }\right) g(\tau )\frac{\mathrm{{d}}\tau }{b-\tau }. \end{aligned}$$

Theorem 6.8

(Right convolution theorem) If ${\mathscr {M}}_{rm}\{f(t)\}=F(\xi )$ and ${\mathscr {M}}_{rm}\{g(t)\}=G(\xi )$, then

$$\begin{aligned} {\mathscr {M}}_{rm}\{f(t)*_{r}g(t)\}={\mathscr {M}}_{rm}\{f(t)\}{\mathscr {M}}_{rm}\{g(t)\} =F(\xi )G(\xi ); \end{aligned}$$

Or equivalently,

$$\begin{aligned} {\mathscr {M}}_{rm}^{-1}\{F(\xi )G(\xi )\}=f(t)*_{r}g(t). \end{aligned}$$

Proof

According to Definitions 6.9 and 6.11, and interchanging the order of integration, one has

$$\begin{aligned} {\mathscr {M}}_{rm}\{f(t)*_{r}g(t)\}&=\int _{-\infty }^{b}(b-t)^{\xi -1} \int _{-\infty }^{b}f\left( b-\frac{b-t}{b-\tau }\right) g(\tau )\frac{\mathrm{{d}}\tau }{b-\tau }\mathrm{{d}}t\\&={\mathscr {M}}_{rm}\{f(t)\}{\mathscr {M}}_{rm}\{g(t)\} =F(\xi )G(\xi ). \end{aligned}$$

$\square$

The following differential property is valid.

Lemma 6.2

If ${\mathscr {M}}_{lm}\{f(t)\}=F(\xi )$ and the limits

$$\begin{aligned}&\lim _{t\rightarrow a^{+}}\left[ (t-a)^{\xi -k-1}f^{(n-k-1)}(t)\right] ,\\&\quad \lim _{t\rightarrow \infty }\left[ (t-a)^{\xi -k-1}f^{(n-k-1)}(t)\right] \ (k=0,1,\ldots ,n-1) \end{aligned}$$

exist, then

$$\begin{aligned} {\mathscr {M}}_{lm}\{f^{(n)}(t)\}= & {} \sum _{k=0}^{n-1}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )}\left[ (t-a)^{\xi -k-1} f^{(n-k-1)}(t)\right] \Big |_{a}^{\infty }\nonumber \\&+\frac{\Gamma (1+n-\xi )}{\Gamma (1-\xi )}F(\xi -n),\,t>a,\, n \in {\mathbb {Z}}^{+}. \end{aligned}$$

(6.25)

If ${\mathscr {M}}_{rm}\{f(t)\}=F(\xi )$ and the limits

$$\begin{aligned}&\lim _{t\rightarrow b^{-}}\left[ (b-t)^{\xi -k-1}f^{(n-k-1)}(t)\right] ,\\&\quad \lim _{t\rightarrow -\infty }\left[ (b-t)^{\xi -k-1}f^{(n-k-1)}(t)\right] \ (k=0,1,\ldots ,n-1) \end{aligned}$$

exist, then

$$\begin{aligned} {\mathscr {M}}_{rm}\{f^{(n)}(t)\} =&\sum _{k=0}^{n-1}(-1)^{k}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )}\left[ (b-t)^{\xi -k-1} f^{(n-k-1)}(t)\right] \Big |_{-\infty }^{b}\nonumber \\&+(-1)^{n}\frac{\Gamma (1+n-\xi )}{\Gamma (1-\xi )}F(\xi -n),\,t<b,\, n \in {\mathbb {Z}}^{+}. \end{aligned}$$

(6.26)

Proof

Applying repeatedly the formula of integration by parts, we obtain

$$\begin{aligned}&{\mathscr {M}}_{lm}\{f^{(n)}(t)\} =\int _{a}^{\infty }(t-a)^{\xi -1}f^{(n)}(t)\mathrm{{d}}t\\&\quad =(t-a)^{\xi -1}f^{(n-1)}(t)|_{a}^{\infty } +(1-\xi )\int _{a}^{\infty }(t-a)^{\xi -2}f^{(n-1)}(t)\mathrm{{d}}t \\&\quad =(t-a)^{\xi -1}f^{(n-1)}(t)|_{a}^{\infty } +(1-\xi )(t-a)^{\xi -2}f^{(n-2)}(t)|_{a}^{\infty } \\&\qquad +(1-\xi )(2-\xi )\int _{a}^{\infty }(t-a)^{\xi -3}f^{(n-2)}(t)\mathrm{{d}}t\\&\quad =(-1)^{n}\frac{\Gamma (\xi )}{\Gamma (\xi -n)}F(\xi -n) +\sum _{k=0}^{n-1}(-1)^{k}\frac{\Gamma (\xi )}{\Gamma (\xi -k)}\left[ (t-a)^{\xi -k-1} f^{(n-k-1)}(t)\right] \Big |_{a}^{\infty }\\&\quad =\frac{\Gamma (1+n-\xi )}{\Gamma (1-\xi )}F(\xi -n) +\sum _{k=0}^{n-1}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )}\left[ (t-a)^{\xi -k-1} f^{(n-k-1)}(t)\right] \Big |_{a}^{\infty }, \end{aligned}$$

and

$$\begin{aligned}&{\mathscr {M}}_{rm}\{f^{(n)}(t)\} =\int _{-\infty }^{b}(b-t)^{\xi -1}f^{(n)}(t)\mathrm{{d}}t\\&\quad =(b-t)^{\xi -1}f^{(n-1)}(t)|_{-\infty }^{b} +(\xi -1)\int _{-\infty }^{b}(b-t)^{\xi -2}f^{(n-1)}(t)\mathrm{{d}}t\\&\quad =\frac{\Gamma (\xi )}{\Gamma (\xi -n)}F(\xi -n) +\sum _{k=0}^{n-1}\frac{\Gamma (\xi )}{\Gamma (\xi -k)}\left[ (b-t)^{\xi -k-1} f^{(n-k-1)}(t)\right] \Big |_{-\infty }^{b}\\&\quad =(-1)^{n}\frac{\Gamma (1+n-\xi )}{\Gamma (1-\xi )}F(\xi -n)\\&\qquad +\sum _{k=0}^{n-1}(-1)^{k}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )}\left[ (b-t)^{\xi -k-1} f^{(n-k-1)}(t)\right] \Big |_{-\infty }^{b}. \end{aligned}$$

The proof is thus complete. $\square$

Finally, we are ready to present the modified Mellin transform for Definitions 6.1–6.3.

Theorem 6.9

Let $n-1<\alpha <n\in {\mathbb {Z}}^+$ and $a\in {\mathbb {R}}$. If ${\mathscr {M}}_{lm}\{f(t)\}=F(\xi )$, $\alpha <\mathrm{{Re}}(1-\xi )$ and the limits, for $k=0,1,\ldots ,n-1$,

$$\begin{aligned}&\lim _{t\rightarrow a^{+}}\left[ (t-a)^{\xi -k-1}\,_{RL}\mathrm{{D}}_{a,t}^{\alpha -k-1}f(t)\right] ,\ \ \lim _{t\rightarrow \infty }\left[ (t-a)^{\xi -k-1}\,_{RL}\mathrm{{D}}_{a,t}^{\alpha -k-1}f(t)\right] ,\\&\lim _{t\rightarrow a^{+}}\left[ (t-a)^{\xi -\alpha +k}f^{(k)}(t)\right] ,\ \ \lim _{t\rightarrow \infty }\left[ (t-a)^{\xi -\alpha +k}f^{(k)}(t)\right] \end{aligned}$$

exist, then:

$$\begin{aligned}&{\mathscr {M}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-\alpha }f(t)\} =\frac{\Gamma (1-\xi -\alpha )}{\Gamma (1-\xi )}F(\xi +\alpha ), \end{aligned}$$

(6.27)

$$\begin{aligned}&{\mathscr {M}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{\alpha }f(t)\} =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha ) \nonumber \\&\ \ \ \ \ +\sum _{k=0}^{n-1}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )} \left[ (t-a)^{\xi -k-1}\, _{RL}\mathrm{{D}}_{a,t}^{\alpha -k-1}f(t)\right] \Big |_{a}^{\infty }, \end{aligned}$$

(6.28)

$$\begin{aligned}&{\mathscr {M}}_{lm}\{_{C}\mathrm{{D}}_{a,t}^{\alpha }f(t)\} =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha )\nonumber \\&\ \ \ \ \ +\sum _{k=0}^{n-1}\frac{\Gamma (\alpha -k-\xi )}{\Gamma (1-\xi )}\left[ (t-a)^{\xi -\alpha +k} f^{(k)}(t)\right] \Big |_{a}^{\infty }. \end{aligned}$$

(6.29)

Proof

We first prove (6.27). In term of Definitions 6.1 and 6.8, by interchanging order of integration and using the substitution $t-\tau =(\tau -a)w$, it yields

$$\begin{aligned} {\mathscr {M}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-\alpha }f(t)\} =&\;\int _{a}^{\infty }(t-a)^{\xi -1}\frac{1}{\Gamma (\alpha )}\int _{a}^{t} (t-\tau )^{\alpha -1}f(\tau )\mathrm{{d}}\tau \mathrm{{d}}t\\ =&\;\frac{1}{\Gamma (\alpha )}\int _{a}^{\infty }\int _{0}^{\infty } (1+w)^{\xi -1}w^{\alpha -1}\mathrm{{d}}w(\tau -a)^{\xi +\alpha -1}f(\tau )\mathrm{{d}}\tau \\ =&\;\frac{\Gamma (1-\xi -\alpha )}{\Gamma (1-\xi )}F(\xi +\alpha ), \end{aligned}$$

where we have used the integral equality $\int _{0}^{\infty } (1+w)^{\xi -1}w^{\alpha -1}\mathrm{{d}}w=B(\alpha , 1-\xi -\alpha )$. In fact, we have $\int _{0}^{\infty }(1+w)^{\xi -1}w^{\alpha -1}\mathrm{{d}}w =\int _{1}^{\infty }u^{\xi -1}(u-1)^{\alpha -1}\mathrm{{d}}w =\int _{0}^{1}(1-v)^{\alpha -1}v^{1-\xi -\alpha -1}\mathrm{{d}}w =B(\alpha , 1-\xi -\alpha )$ when $\alpha <\mathrm{{Re}}(1-\xi )$.

To prove (6.28), denoting $_{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f(t)=g_{1}(t)$ and exploiting differential property (6.25), one gets

$$\begin{aligned}&{\mathscr {M}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{\alpha }f(t)\} ={\mathscr {M}}_{lm}\{\frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}}\, _{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f(t)\} ={\mathscr {M}}_{lm}\{g_{1}^{(n)}(t)\}\\&\quad =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha ) +\sum _{k=0}^{n-1}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )} \left[ (t-a)^{\xi -k-1}\, _{RL}\mathrm{{D}}_{a,t}^{\alpha -k-1}f(t)\right] \Big |_{a}^{\infty }. \end{aligned}$$

We now show (6.29). By using (6.27) and differential property (6.25), and denoting $f^{(n)}(t)=g_{2}(t)$, there holds

$$\begin{aligned} {\mathscr {M}}_{lm}\{_{C}\mathrm{{D}}_{a,t}^{\alpha }f(t)\} =&{\mathscr {M}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}f^{(n)}(t)\} \\ =&{\mathscr {M}}_{lm}\{_{RL}\mathrm{{D}}_{a,t}^{-(n-\alpha )}g_{2}(t)\} =\frac{\Gamma (1-\xi -n+\alpha )}{\Gamma (1-\xi )}G_{2}(\xi +n-\alpha ) \\ =&\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha )\\&+\sum _{k=0}^{n-1}\frac{\Gamma (\alpha -k-\xi )}{\Gamma (1-\xi )}\left[ (t-a)^{\xi -\alpha +k} f^{(k)}(t)\right] \Big |_{a}^{\infty }. \end{aligned}$$

We thus complete the proof. $\square$

Theorem 6.10

Let $n-1<\alpha <n\in {\mathbb {Z}}^+$ and $b\in {\mathbb {R}}$. If ${\mathscr {M}}_{rm}\{f(t)\}=F(\xi )$, $\alpha <\mathrm{{Re}}(1-\xi )$ and the limits, for $k=0,1,\ldots ,n-1$,

$$\begin{aligned}&\lim _{t\rightarrow b^{-}}\left[ (b-t)^{\xi -k-1}\,_{RL}\mathrm{{D}}_{t,b}^{\alpha -k-1}f(t)\right] ,\ \ \lim _{t\rightarrow -\infty }\left[ (b-t)^{\xi -k-1}\,_{RL}\mathrm{{D}}_{t,b}^{\alpha -k-1}f(t)\right] ,\\&\lim _{t\rightarrow b^{-}}\left[ (b-t)^{\xi -\alpha +k}f^{(k)}(t)\right] ,\ \ \lim _{t\rightarrow -\infty }\left[ (b-t)^{\xi -\alpha +k}f^{(k)}(t)\right] \ \end{aligned}$$

exist, then:

$$\begin{aligned}&{\mathscr {M}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-\alpha }f(t)\} =\frac{\Gamma (1-\xi -\alpha )}{\Gamma (1-\xi )}F(\xi +\alpha ), \end{aligned}$$

(6.30)

$$\begin{aligned}&{\mathscr {M}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{\alpha }f(t)\} =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha ) \nonumber \\&\ \ \ \ \ +\sum _{k=0}^{n-1}(-1)^{k+n}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )} \left[ (b-t)^{\xi -k-1}\, _{RL}\mathrm{{D}}_{t,b}^{\alpha -k-1}f(t)\right] \Big |_{-\infty }^{b}, \end{aligned}$$

(6.31)

$$\begin{aligned}&{\mathscr {M}}_{rm}\{_{C}\mathrm{{D}}_{t,b}^{\alpha }f(t)\} =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha )\nonumber \\&\ \ \ \ \ +\sum _{k=0}^{n-1}(-1)^{k+n}\frac{\Gamma (\alpha -k-\xi )}{\Gamma (1-\xi )}\left[ (b-t)^{\xi -\alpha +k} f^{(k)}(t)\right] \Big |_{-\infty }^{b}. \end{aligned}$$

(6.32)

Proof

We start with the proof of (6.30). According to Definitions 6.1 and 6.9, by interchanging order of integration and using the substitution $\tau -t=(b-\tau )w$, one has

$$\begin{aligned} {\mathscr {M}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-\alpha }f(t)\}&=\int _{-\infty }^{b}(b-t)^{\xi -1}\frac{1}{\Gamma (\alpha )}\int _{t}^{b} (\tau -t)^{\alpha -1}f(\tau )\mathrm{{d}}\tau \mathrm{{d}}t\\&=\frac{1}{\Gamma (\alpha )}\int _{-\infty }^{b}\int _{0}^{\infty } (1+w)^{\xi -1}w^{\alpha -1}\mathrm{{d}}w(b-\tau )^{\xi +\alpha -1}f(\tau )\mathrm{{d}}\tau \\&=\frac{\Gamma (1-\xi -\alpha )}{\Gamma (1-\xi )}F(\xi +\alpha ). \end{aligned}$$

Now we deal with (6.31). Letting $_{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f(t)=h_{1}(t)$, it follows from differential property (6.26) that

$$\begin{aligned}&{\mathscr {M}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{\alpha }f(t)\} =(-1)^{n}{\mathscr {M}}_{rm}\{\frac{\mathrm{{d}}^{n}}{\mathrm{{d}}t^{n}}\, _{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f(t)\} =(-1)^{n}{\mathscr {M}}_{rm}\{h_{1}^{(n)}(t)\} \\&\quad =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha )\\&\qquad +\sum _{k=0}^{n-1}(-1)^{k+n}\frac{\Gamma (1+k-\xi )}{\Gamma (1-\xi )} \left[ (b-t)^{\xi -k-1}\, _{RL}\mathrm{{D}}_{t,b}^{\alpha -k-1}f(t)\right] \Big |_{-\infty }^{b}. \end{aligned}$$

Finally, we prove (6.32). Observing that (6.30) and differential property (6.26), and denoting $f^{(n)}(t)=h_{2}(t)$, it holds that

$$\begin{aligned}&{\mathscr {M}}_{rm}\{_{C}\mathrm{{D}}_{t,b}^{\alpha }f(t)\} =(-1)^{n}{\mathscr {M}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}f^{(n)}(t)\} \\&\quad =(-1)^{n}{\mathscr {M}}_{rm}\{_{RL}\mathrm{{D}}_{t,b}^{-(n-\alpha )}h_{2}(t)\} =(-1)^{n}\frac{\Gamma (1-\xi -n+\alpha )}{\Gamma (1-\xi )}H_{2}(\xi +n-\alpha ) \\&\quad =\frac{\Gamma (1-\xi +\alpha )}{\Gamma (1-\xi )}F(\xi -\alpha ) +\sum _{k=0}^{n-1}(-1)^{k+n}\frac{\Gamma (\alpha -k-\xi )}{\Gamma (1-\xi )}\left[ (b-t)^{\xi -\alpha +k} f^{(k)}(t)\right] \Big |_{-\infty }^{b}. \end{aligned}$$

Hence, we complete the proof of lemma.$\square$

Except Definitions 6.1–6.3, the other definitions and results seem to be novel. These definitions and conclusions are very useful in analyzing linear fractional differential equations with left (right) Riemann–Liouville or left (right) Caputo derivatives. We would like to list them here for reference of interested readers.