fractal and fractional

Article Uniqueness of Solutions of the Generalized Abel Integral Equations in Banach Spaces

Chenkuan Li 1,* and Hari M. Srivastava 2,3,4,5

1 Department of Mathematics and Computer Science, Brandon University, Brandon, MB R7A 6A9, Canada 2 Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada; [email protected] 3 Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan 4 Department of Mathematics and Informatics, University, 71 Jeyhun Hajibeyli Street, AZ1007, Azerbaijan 5 Section of Mathematics, International Telematic University Uninettuno, I-00186 Rome, Italy * Correspondence: [email protected]

Abstract: This paper studies the uniqueness of solutions for several generalized Abel’s integral equations and a related coupled system in Banach spaces. The results derived are new and based on Babenko’s approach, Banach’s contraction principle and the multivariate Mittag–Leffler function. We also present some examples for the illustration of our main theorems.

Keywords: Riemann–Liouville fractional integral; Banach’s fixed point theorem; Babenko’s approach; Wright’s generalized Bessel function; multivariate Mittag–Leffler function

 

Citation: Li, C.; Srivastava, H.M. 1. Introduction Uniqueness of Solutions of the Let T > 0. The space L[0, T] is given by Generalized Abel Integral Equations in Banach Spaces. Fractal Fract. 2021,  Z T  5, 105. https://doi.org/10.3390/ L[0, T] = u(x) : kuk = |u(x)|dx < ∞ . 0 fractalfract5030105 Clearly, L[0, T] is a Banach space. The product space L[0, T] × L[0, T] (which is also a Academic Editor: Paul Eloe Banach space) is defined as follows:

Received: 4 August 2021 L[0, T] × L[0, T] = u(x), v(x) : u(x), v(x) ∈ L[0, T] , Accepted: 28 August 2021 Published: 31 August 2021 with the norm given by k(u, v)k = kuk + kvk. Publisher’s Note: MDPI stays neutral α + with regard to jurisdictional claims in The Riemann–Liouville fractional integral I of order α ∈ R is defined for the published maps and institutional affil- function u(x) by (see [1,2]): iations. 1 Z x (Iαu)(x) = (x − t)α−1u(t)dt. Γ(α) 0

In particular, we have Copyright: © 2021 by the authors. (I0u)(x) = u(x). Licensee MDPI, Basel, Switzerland. < = ··· This article is an open access article Let 0 5 α0 αi for i 1, 2, , m and α0 5 α. In this paper, we begin to construct an distributed under the terms and explicit solution in L[0, T] to the following Abel’s integral equation by using Babenko’s conditions of the Creative Commons approach and the multivariate Mittag–Leffler function: Attribution (CC BY) license (https:// m creativecommons.org/licenses/by/ α0 αi α I u(x) + ∑ ai I u(x) = I f (x), (1) 4.0/). i=1

Fractal Fract. 2021, 5, 105. https://doi.org/10.3390/fractalfract5030105 https://www.mdpi.com/journal/fractalfract Fractal Fract. 2021, 5, 105 2 of 13

where each ai (i = 1, 2, ··· , m) is a constant and f (x) ∈ L[0, T]. We then further investigate the uniqueness of solutions in L[0, T] for the following nonlinear Abel’s integral equation by using Banach’s fixed point theorem:

m α0 αi α  I u(x) + ∑ ai I u(x) = I g x, u(x) , (2) i=1

where g is a mapping from [0, T] × R to R and satisfies certain conditions. Finally, the sufficient conditions are given for the uniqueness of solutions in the product space L[0, T] × L[0, T] to the associated system given by

 m α αi α  I 0 u(x) + ai I u(x) = I g1 x, u(x), v(x)  ∑  i=1 (3)  m  β0 βi β  I v(x) + ∑ bi I v(x) = I g2 x, u(x), v(x) , i=1

2 where g1 and g2 are mappings from [0, T] × R to R, 0 5 β0 5 β and β0 < βi for all i = 1, 2, ··· , m. Equations (1)–(3) are new and, to the best of our knowledge, have never been investigated earlier. The single-term (for m = 1) Equation (1) turns out to be

α1 u(x) + a1 I u(x) = f (x)(α = α0 = 0), (4)

which is the classical Abel’s integral equation of the second kind with the following solution given by Hille and Tamarkin (see, for details [3]; see also [4–6]):

Z x α1−1 α1 u(x) = f (x) − a1 (x − t) Eα1,α1 (−a1(x − t) ) f (t)dt, 0

where Eα,β(z) given by

∞ zj E (z) = (α, β > 0), α,β ∑ ( + ) j=0 Γ αj β

is the two-parameter Mittag–Leffler function. The above solution can also be easily deduced by the Laplace transform. Indeed,

α1 L(u(x) + a1 I u(x)) = L f (x) = f˜(s)

infers that a u˜(s) + 1 u˜(s) = f˜(s). sα1 Hence, we have  a  ( ) = − 1 ˜( ) u˜ s 1 α f s . a1 + s 1 Using the formula (1.80) from [7]

Z ∞ 1 −st α1−1 (− α1 ) = e t Eα1,α1 a1t dt α , 0 a1 + s 1 we arrive at

α1−1 α1 u(x) = f (x) − a1x Eα1,α1 (−a1x ) ∗ f (x) Z x α1−1 α1 = f (x) − a1 (x − t) Eα1,α1 (−a1(x − t) ) f (t)dt, 0 Fractal Fract. 2021, 5, 105 3 of 13

where φ ∗ ψ denotes the Laplace convolution given by

Z x (φ ∗ ψ)(x) = φ(x − t)ψ(t)dt. 0 On the other hand, Babenko’s approach is a potentially powerful tool for solving dif- ferential, integral and integro-differential equations by treating integral operators like vari- ables. The method itself is similar to the Laplace transform method while dealing with such equations with constant coefficients, but it can be used in other cases as well, such as han- dling integral equations with variable coefficients (see [8,9]). To demonstrate this method, we are going to solve Equation (4) with Babenko’s approach. Clearly, Equation (4) becomes

α1 (1 + a1 I )u(x) = f (x).

Therefore, we get

∞ α1 −1 k α1 k u(x) = (1 + a1 I ) f (x) = ∑ (−1) (a1 I ) f (x) k=0 ∞ k+1 α1 k+1 = f (x) + ∑ (−1) (a1 I ) f (x) k=0 ∞ 1 Z x = f (x) − a (−a )k (x − t)α1k+α1−1 f (t)dt 1 ∑ 1 ( + ) k=0 Γ α1k α1 0 Z x ∞ 1 = f (x) − a (x − t)α1−1 (−a )k (x − t)α1k f (t)dt 1 ∑ 1 ( + ) 0 k=0 Γ α1k α1 Z x α1−1 α1 = f (x) − a1 (x − t) Eα1,α1 (−a1(x − t) ) f (t)dt. 0

We now recall Wright’s generalized Bessel Function ϕ(β, δ; z) defined as follows:

∞ zj ϕ(β, δ; z) = (β, δ > 0). ∑ ( + ) j=0 j! Γ βj δ

We also define

µ Sm(x; z1, ··· , zm; β1, ··· , βm) = (h1 ∗ h2 ∗ · · · ∗ hm)(x),

where   µk−1 βk hk = hk(x) = x φ βk, µk; zkx (x, βk > 0; zk ∈ R),

m µ = ∑ µk, µk > 0. k=1 Let

µ Gm(x; γ1, ··· , γm; β1, ··· , βm) Z ∞ −t µ = e Sm(x; γ1t, ··· , ymt; β1, ··· , βm)dt, 0 and µ−α0 wµ(x) = Gm (x; −a1, ··· , −am; α1 − α0, ··· , αm − α0)(µ > α0). Fractal Fract. 2021, 5, 105 4 of 13

In 2013, Pskhu [10] constructed an explicit solution for the following Abel’s integral equation which is a special case of the Equation (1):

m α0 αi α0 I u(x) + ∑ ai I u(x) = I f (x)( f ∈ L[0, T]), i=1

as follows: µ u(x) = D0,x( f ∗ wµ)(x)(µ > α0), where the solution u(x) is independent of the parameter µ and

 1 Z x  φ(t)(x − t)−µ−1 dt = I−µφ(x)(µ < 0)  Γ(−µ)  0  µ  D0,xφ(x) := φ(x)(µ = 0)    n  d n−µ  {I φ(x)} (n − 1 < µ n; n ∈ N), dxn 5 N being the set of positive integers. We would also like to add that Gorenflo and Luchko [11] established an explicit solution to the following generalized Abel integral equation of the second kind, which was based on a modification of the Mikusi´nskioperational calculus and the Mittag–Leffler function of several variables (see, for details, [12]):

m αjµ u(x) − ∑ aj I u(x) = f (x)(αj > 0; m = 1; µ > 0; x > 0), j=1

which is also a special case of Equation (1). There are many analytic and numerical studies on Abel’s integral equation and its variants in distribution, as well as the existence and uniqueness of the corresponding solu- tions by using fixed point theorems [7,8,13–15]. For example, Brunner et al. [16] considered numerical solutions of Abel’s integral equation of the second kind:

Z x u(x) = f (x) + (x − t)−α κx, t, u(t)dt (x ∈ [0, T]), 0

where 0 < α < 1 and f ∈ C[0, T], and the kernel κ is continuous on S × R, with

S = {(t, s) : 0 5 s 5 t 5 T},

and satisfies the Lipschitz conditions in the third argument. The multivariate Mittag–Leffler function was studied by (among others) Hadid and Luchko [17] for solving linear fractional differential equations with constant coefficients by applying the operational calculus:

E (z ··· z ) (α1,··· ,αm),β 1, , m ∞   k1 km k z ··· zm = ∑ ∑ 1 , k1, ··· , km Γ(α1k1 + ··· + αmkm + β) k=0 k1+···+km=k

where αi > 0 (i = 1, 2, ··· , m) and β > 0.

2. A Set of Main Results In this section, we begin to establish an explicit solution to Equation (1) by using Babenko’s approach in [18]. Fractal Fract. 2021, 5, 105 5 of 13

Theorem 1. Assume that f ∈ L[0, T], 0 5 α0 < αi for i = 1, 2, ··· , m and α0 5 α. Then, Equation (1) has a unique solution in L[0, T] given by

∞  k  ( ) = (− )k k1 ··· km u x ∑ 1 ∑ a1 am k1, ··· , km k=0 k1+···+km=k · Ik1(α1−α0)+···+km(αm−α0)+α−α0 f (x), (5)

where each ai (i = 1, 2, ··· , m) is a constant.

Proof. Equation (1) becomes

m αi−α0 α−α0 u(x) + ∑ ai I u(x) = I f (x), i=1

α0 by applying the operator D0,x to both sides of Equation (1). This implies that, by Babenko’s method, we have

− m ! 1 αi−α0 α−α0 u(x) = 1 + ∑ ai I I f (x) i=1 k ∞ m ! k αi−α0 α−α0 = ∑ (−1) ∑ ai I I f (x) k=0 i=1 ∞  k  k α1−α0 k1 αm−α0 km α−α0 = ∑ (−1) ∑ a1 I ··· am I I f (x) k1, ··· , km k=0 k1+···+km=k ∞  k  = (− )k k1 ··· km ∑ 1 ∑ a1 am k1, ··· , km k=0 k1+···+km=k · Ik1(α1−α0)+···+km(αm−α0)+α−α0 f (x).

Let xγ Φ (x) = + , γ Γ(γ) where γ ∈ R and  xα (x > 0) α  x+ =  0 (otherwise). We then find from [19] that

Φγ1 ∗ Φγ2 = Φγ1+γ2 .

Clearly, we have

Ik1(α1−α0)+···+km(αm−α0)+α−α0 f (x) = ∗ f Φk1(α1−α0)+···+km(αm−α0)+α−α0 .

Moreover, it follows from [20] that

Φ ∗ f k1(α1−α0)+···+km(αm−α0)+α−α0

Φ k f k 5 k1(α1−α0)+···+km(αm−α0)+α−α0 Tk1(α1−α0)+···+km(αm−α0)+α−α0 5 k f k. Γ(k1(α1 − α0) + ··· + km(αm − α0) + α − α0 + 1) Fractal Fract. 2021, 5, 105 6 of 13

This implies that

∞  k  k1 km ku(x)k 5 ∑ ∑ |a1| · · · |am| k1, ··· , km k=0 k1+···+km=k Tk1(α1−α0)+···+km(αm−α0)+α−α0 · k f k Γ(k1(α1 − α0) + ··· + km(αm − α0) + α − α0 + 1) = Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0 k f k (α1−α0,··· ,αm−α0), α−α0+1 1 , , m < ∞.

Hence, we get u(x) ∈ L[0, T], and that the series on the right-hand side of Equation (5) is absolutely convergent in L[0, T]. The uniqueness of the solution follows immediately from the fact that the following fractional integral equation:

m α0 αi I u(x) + ∑ ai I u(x) = 0, i=1

has only the zero solution by Babenko’s method. It remains to show that the series on the right-hand side of Equation (5) is a solution of Equation (1). Indeed, we have

m α0 αi I u(x) + ∑ ai I u(x) i=1 ∞   k k = (− )k 1 ··· km · k1(α1−α0)+···+km(αm−α0)+α ( ) ∑ 1 ∑ a1 am I f x k1, ··· , km k=0 k1+···+km=k ∞   m k + + (− )k k1 ··· ki 1 ··· km ∑ 1 ∑ ∑ a1 ai am k1, ··· , km k=0 k1+···+km=k i=1 · Ik1(α1−α0)+···+(ki+1)(αi−α0)+···+km(αm−α0)+α f (x) ∞  k  = α ( ) + (− )k k1 ··· km I f x ∑ 1 ∑ a1 am k1, ··· , km k=1 k1+···+km=k · Ik1(α1−α0)+···+km(αm−α0)+α f (x) ∞   m k + + (− )k k1 ··· ki 1 ··· km ∑ 1 ∑ ∑ a1 ai am k1, ··· , km k=0 k1+···+km=k i=1 · Ik1(α1−α0)+···+(ki+1)(αi−α0)+···+km(αm−α0)+α f (x) = Iα f (x),

by noting that

∞   k k (− )k 1 ··· km · k1(α1−α0)+···+km(αm−α0)+α ( ) ∑ 1 ∑ a1 am I f x k1, ··· , km k=1 k1+···+km=k ∞   m k + + (− )k k1 ··· ki 1 ··· km ∑ 1 ∑ ∑ a1 ai am k1, ··· , km k=0 k1+···+km=k i=1 · Ik1(α1−α0)+···+(ki+1)(αi−α0)+···+km(αm−α0)+α f (x) = 0, Fractal Fract. 2021, 5, 105 7 of 13

after the sign changes and cancellations. Obviously, it is true that   1 k (− )1 1 ··· km · k1(α1−α0)+···+km(αm−α0)+α ( ) 1 ∑ a1 am I f x k1, ··· , km k1+···+km=1   m 0 + + (− )0 k1 ··· ki 1 ··· km 1 ∑ ∑ a1 ai am k1, ··· , km k1+···+km=0 i=1 · Ik1(α1−α0)+···+(ki+1)(αi−α0)+···+km(αm−α0)+α f (x) = 0.

This completes the proof of Theorem1.

As an example, we can deduce that the following integral equation:

1 I0.5u(x) + Iu(x) + I1.5u(x) = x2 2 has a unique solution given by

∞  k  x0.5k1+k2+1.5 u(x) = ∑ (−1)k ∑ , k1, k2 Γ(0.5k1 + k2 + 2.5) k=0 k1+k2=k

in the space L[0, T] by using the relation:

1 x2 = Φ (x) = Φ ∗ Φ = I21, 2 3 2 1 and Theorem1. Using Banach’s fixed point theorem, we are now ready to show the uniqueness of solutions to Equation (2) in the space L[0, T].

Theorem 2. Assume that 0 5 α0 < αi for i = 1, 2, ··· , m and α0 5 α. Suppose also that there exists a constant C = 0 such that

|g(x, y1) − g(x, y2)| 5 C|y1 − y2|,

for all x ∈ [0, T], y1, y2 ∈ R, g(x, 0) ∈ L[0, T] and

CTα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  < (α1−α0,··· ,αm−α0), α−α0+1 1 , , m 1.

Then the Equation (2) has a unique solution in the space L[0, T].

Proof. Let u ∈ L[0, T]. Then gx, u(x) ∈ L[0, T]. Indeed, we have    g x, u(x) = g x, u(x) − g(x, 0) + g(x, 0) 5 |g x, u(x) − g(x, 0)| + |g(x, 0)| 5 C|u(x)| + |g(x, 0)|,

which implies that

Z T  Z T Z T g x, u(x) dx 5 C |u(x)|dx + |g(x, 0)|dx < ∞. 0 0 0

We now define a nonlinear mapping S on L[0, T] as follows:

∞  k  ( )( ) = (− )k k1 ··· km Su x ∑ 1 ∑ a1 am k1, ··· , km k=0 k1+···+km=k k ( − )+···+k ( − )+ −  · I 1 α1 α0 m αm α0 α α0 g x, u(x) . Fractal Fract. 2021, 5, 105 8 of 13

It follows from the lines of the proof of Theorem1 that

kS(u)k Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  5 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m  · g x, u(x) < ∞,

which shows that S is a mapping from L[0, T] to itself. It now remains to show that the mapping S is contractive. In fact, we have

Z T   Z T g x, u(x) − g x, v(x) dx 5 C |u(x) − v(x)|dx 0 0 and

kS(u) − S(v)k Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  5 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m   · g x, u(x) − g x, v(x) CTα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0 ku − vk 5 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m = qku − vk,

where

q = CTα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  < (α1−α0,··· ,αm−α0), α−α0+1 1 , , m 1.

This completes the proof of Theorem2.

Finally, we present the sufficient conditions for the uniqueness of the solution of Equation (3) in the product space L[0, T] × L[0, T].

Theorem 3. Assume that 0 5 α0 < αi, 0 5 β0 < βi for i = 1, 2, ··· , m and α0 5 α, β0 5 β. Suppose also that there exist constants C1, C2, C3 and C4 such that

|g1(x, y1, y2) − g1(x, z1, z2)| 5 C1|y1 − z1| + C2|y2 − z2|

and

|g2(x, y1, y2) − g2(x, z1, z2)| 5 C3|y1 − z1| + C4|y2 − z2|,

for all x ∈ [0, T], y1, y2, z1, z2 ∈ R, g1(x, 0, 0), g2(x, 0, 0) ∈ L[0, T] and

q = {C C }Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  max 1, 2 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m + {C C }Tβ−β0 E max 3, 4 (β1−β0,··· ,βm−β0), β−β0+1   β1−β0 βm−β0 · |b1|T , ··· , |bm|T < 1.

Then, Equation (3) has a unique solution in the space L[0, T] × L[0, T].  Proof. Let u, v ∈ L[0, T]. Then g1 x, u(x), v(x) ∈ L[0, T]. Indeed, we have  g1 x, u(x), v(x)  = g1 x, u(x), v(x) − g1(x, 0, 0) + g1(x, 0, 0)  5 g1 x, u(x), v(x) − g1(x, 0, 0) + |g1(x, 0, 0)| 5 C1|u(x)| + C2|v(x)| + |g1(x, 0, 0)|. Fractal Fract. 2021, 5, 105 9 of 13

This implies that

Z T  g1 x, u(x), v(x) dx 0 Z T Z T Z T 5 C1 |u(x)|dx + C2 |v(x)|dx + |g(x, 0, 0)|dx 0 0 0 < ∞.  Similarly, we can see that g2 x, u(x), v(x) ∈ L[0, T]. Let us now define the mappings S1, S2 on L[0, T] × L[0, T] as follows:

∞  k  ( )( ) = (− )k k1 ··· km S1 u, v x ∑ 1 ∑ a1 am k1, ··· , km k=0 k1+···+km=k

k1(α1−α0)+···+km(αm−α0)+α−α0  · I g1 x, u(x), v(x)

and ∞  k  ( )( ) = (− )k k1 ··· km S2 u, v x ∑ 1 ∑ a1 am k1, ··· , km k=0 k1+···+km=k

k1(α1−α0)+···+km(αm−α0)+α−α0  · I g2 x, u(x), v(x) .

Furthermore, we define a mapping S on L[0, T] × L[0, T] as follows:  S(u, v) = S1(u, v), S2(u, v) ,

with kS(u, v)k = kS1(u, v)k + kS2(u, v)k. Thus, clearly, we see that

kS (u v)k Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  1 , 5 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m  · g1 x, u(x), v(x) < ∞

and   kS (u v)k Tβ−β0 E |b |Tβ1−β0 ··· |b |Tβm−β0 2 , 5 (β1−β0,··· ,βm−β0), β−β0+1 1 , , m  · g2 x, u(x), v(x) < ∞.

This implies that S is a mapping from L[0, T] × L[0, T] to itself. It remains to be shown that S is contractive. Indeed, we have

kS(u1, v1) − S(u2, v2)k

= kS1(u1, v1) − S1(u2, v2)k

+ kS2(u1, v1) − S2(u2, v2)k, Fractal Fract. 2021, 5, 105 10 of 13

and

kS1(u1, v1) − S1(u2, v2)k Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  5 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m

· kg1(x, u1, v1) − g1(x, u2, v2)k Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  5 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m  Z T Z T  · C1 |u1(x) − u2(x)|dx + C2 |v1(x) − v2(x)|dx 0 0 {C C }Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  5 max 1, 2 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m

· k(u1, v1) − (u2, v2)k.

Similarly, we obtain

kS2(u1, v1) − S2(u2, v2)k   {C C }Tβ−β0 E |b |Tβ1−β0 ··· |b |Tβm−β0 5 max 3, 4 (β1−β0,··· ,βm−β0), β−β0+1 1 , , m

· k(u1, v1) − (u2, v2)k.

We thus find that

kS(u1, v1) − S(u2, v2)k {C C }Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  5 max 1, 2 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m

· k(u1, v1) − (u2, v2)k   + {C C }Tβ−β0 E |b |Tβ1−β0 ··· |b |Tβm−β0 max 3, 4 (β1−β0,··· ,βm−β0), β−β0+1 1 , , m

· k(u1, v1) − (u2, v2)k

= qk(u1, v1) − (u2, v2)k,

where q is defined above and q < 1. This completes the proof of Theorem3.

3. An Illustrative Example In this section, we present the following example to illustrate the use of Theorem3.

Example 1. The integral system given by

 1 I0.5u(x) − I1.6u(x) + I1.5u(x) = I1.7 sin v(x) 8 1 2.3 I2.3v(x) + I3.3v(x) − I3.4v(x) − I3.5v(x) = I cos u(x), 53 has a unique solution in L[0, 1] × L[0, 1].

Demonstration of Example1 Clearly, we have

1 g x, u(x), v(x) = sin v(x) 1 8 and 1 g x, u(x), v(x) = cos u(x), 2 53

and C1 = 0, C2 = 1/8, C3 = 1/53 and C4 = 0, by noting that

| sin z1 − sin z2| 5 |z1 − z2| Fractal Fract. 2021, 5, 105 11 of 13

and

| cos z1 − cos z2| 5 |z1 − z2|,

for all z1, z2 ∈ R. Furthermore, we have

T = 1 and |a1| = |a2| = |b1| = |b2| = |b3| = 1.

Hence, we get

q = {C C }Tα−α0 E |a |Tα1−α0 ··· |a |Tαm−α0  max 1, 2 (α1−α0,··· ,αm−α0), α−α0+1 1 , , m   + {C C }Tβ−β0 E |b |Tβ1−β0 ··· |b |Tβm−β0 max 3, 4 (β1−β0,··· ,βm−β0), β−β0+1 1 , , m 1 1 = E (1, 1) + E (1, 1, 1). 8 (1.1,1),2.2 53 (1,1.1,1.2),1 It is now evident that

E(1.1,1),2.2(1, 1) ∞  k  1 = ∑ ∑ k1, k2 Γ(1.1k1 + k2 + 2.2) k=0 k1+k2=k ∞  k  1 ∞ 2k 5 ∑ ∑ = ∑ k1, k2 Γ(k + 2) (k + 1)! k=0 k1+k2=k k=0 2 2 · 2 2 · 2 · 2 2 · 2 · 2 · 2 2 · 2 · 2 · 2 · 2 = 1 + + + + + + ··· 2! 1 · 2 · 3 1 · 2 · 3 · 4 1 · 2 · 3 · 4 · 5 1 · 2 · 3 · 4 · 5 · 6 ! 2 2  2 2 1 + 1 + 1 + 1 + + + ··· 5 15 5 5 2 1 = 3 + 3.23. 15 2 5 1 − 5 On the other hand, we have

∞  k  1 E(1,1.1,1.2),1(1, 1, 1) = ∑ ∑ . k1, k2, k3 Γ(k1 + 1.1k2 + 1.2k3 + 1) k=0 k1+k2+k3=k

It follows from [21] that

1 1 5 Γ(k1 + 1.1k2 + 1.2k3 + 1) Γ(k + 1)

for all k = 0, 1, ··· . This implies that

E(1,1.1,1.2),1(1, 1, 1) ∞ 3k 5 ∑ k=0 k! 3 3 · 3 3 · 3 · 3 3 · 3 · 3 · 3 3 · 3 · 3 · 3 · 3 = 1 + + + + + + ··· 1 1 · 2 1 · 2 · 3 1 · 2 · 3 · 4 1 · 2 · 3 · 4 · 5  1  1 1 1 1 + 3 + 4.5 + 4.5 + 3.375 + 2.025 + + 1 + + + + ··· 5 80 2 22 23 = 20.4125.

Clearly, therefore, we get q < 1. This completes our demonstration of Example1. Fractal Fract. 2021, 5, 105 12 of 13

4. Conclusions By using Babenko’s approach, Banach’s contraction principle and the multivariate Mittag–Leffler function, we have studied several generalized forms of Abel’s integral equa- tions and a related coupled system with constant coefficients in Banach spaces. The results, which we have presented in this article, are new and provide interesting generalizations of the existing results in the literature. We have also included some examples, including one example that shows the application of our main theorem (Theorem3).

Author Contributions: Conceptualization, C.L. and H.M.S.; methodology, C.L.; software, C.L. and H.M.S.; validation, C.L. and H.M.S.; formal analysis, C.L.; investigation, C.L. and H.M.S.; resources, C.L. and H.M.S.; writing—original draft preparation, C.L.; writing—review and editing, C.L. and H.M.S.; visualization, C.L. and H.M.S. All authors have read and agreed to the published version of the manuscript. Funding: This work is supported by the Natural Sciences and Engineering Research Council of Canada (Grant No. 2019-03907). Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: The authors are grateful to the two reviewers for their careful reading of the paper with productive comments and suggestions. Conflicts of Interest: The authors declare no conflict of interest.

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