# Solution of Singular Integral Equations via Riemann–Liouville Fractional Integrals

Authors: Manar A. Alqudah; Pshtiwan Othman Mohammed; Thabet Abdeljawad
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this attempt, we introduce a new technique to solve main generalized Abel’s integral equations and generalized weakly singular Volterra integral equations analytically. This technique is based on the Adomian decomposition method, Laplace transform method, andΨ-Riemann–Liouville fractional integrals. Finally, some examples are proposed and they illustrate the rapidness of our new technical method.


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## Body

## 1. Introduction

Fractional dynamic varying systems with singular kernels either in the Riemann–Liouville sense or in the Caputo sense have been investigated in the literature [1–3]. To solve a fractional dynamic equation, we always apply a corresponding fractional integral operator. The action of this integral operator will transform the fractional dynamic equation into its corresponding integral equation whose singularity is reflected in the kernel. Motivated by this fact, in this article, we introduce a new technique to solve main generalized Abel’s integral equations and generalized weakly singular Volterra integral equations analytically.Letδ¯∈0,1 and 0<γ<b. Consider the main generalized Abel’s integral equation [4, 5]:(1)uγ=∫0γutdtgγ−gtδ¯,and the generalized weakly singular Volterra type integral equation of the second kind [4, 5]:(2)uγ=gγ+∫0γβ¯utdtgγ−gtδ¯,where g is a strictly monotonically increasing and differentiable function in the interval 0,b with gγ≠0 for every γ in 0,b and β¯ is a constant.Particularly, ifδ¯=1/2 and gγ=γ, then integral equation (1) reduces to the classical Abel’s integral equation in which Abel, in 1823, investigated the motion of a particle that slides down along a smooth unknown curve under the influence of the gravity in a vertical plane. The particle takes the time gγ to move from the highest point of vertical height γ to the lowest point 0 on the curve. This problem is derived to find the equation of that curve. Indeed, Abel’s integral equation is one of the most famous equations that frequently appear in many engineering problems and physical properties such as heat conduction, semiconductors, chemical reactions, and metallurgy (see, e.g., [6, 7]). Besides, over the past few years, many numerical methods for solving Abel’s integral equation have been developed, such as collocation methods [8], product integration methods [9, 10], fractional multistep methods [11–13], methods based on wavelets [14–16], backward Euler methods [9], Adomian decomposition method [17], and Tau approximation method [18].Recently, Wazwaz [4, 5] solved integral equations (1) and (2). Unlike [4, 5], in the present paper, we are going to introduce a new technique, which is based on the Adomian decomposition method, Laplace transform method, and Ψ-Riemann–Liouville fractional integrals, for solving main generalized Abel’s integral equations and generalized weakly singular Volterra integral equations. For related works on generalized fractional derivatives in the Riemann–Liouville and Caputo senses and their Laplace transforms, we refer to [19]. Moreover, there are several methods used in obtaining approximate solutions to linear and nonlinear fractional ordinary differential equations (FODEs) and fractional partial differential equations (FPDEs) and their real-world applications. For this reason, we advise the readers to visit [20–28].The paper is organized as follows. In Section2, we recall the definitions of Riemann–Liouville fractional integrals, Ψ-Riemann–Liouville fractional integrals, and some essential properties. Section 3 is devoted to deliver the main results for the generalized Abel’s integral equations and generalized weakly singular Volterra integral equations. In Section 4, several examples are considered to illustrate the applicability of our main results.


## 2. Preliminaries

Here, we give the definitions of Riemann–Liouville fractional integrals,Ψ-Riemann–Liouville fractional integrals, and some essential properties.Definition 1.
Letg:a,b⊆ℝ⟶ℝ and let a,b⊆−∞,∞ be a finite interval on the real-axis ℝ. The left and right-sided Riemann–Liouville fractional integrals of order δ¯>0 are, respectively, defined by [29](3)ℑa+δ¯gγ=1Γδ¯∫aγγ−tδ¯−1gtdt,γ>a,ℑb−δ¯gγ=1Γδ¯∫γbt−γδ¯−1gtdt,γ<b.Definition 2.
Leta,b⊆−∞,∞ be a finite or infinite interval of the real-axis ℝ and δ¯>0. Let ψγ be an increasing and positive function on the interval a,b with a continuous derivative ψ′γ on the interval a,b. Then, the left and right-sided ψ-Riemann–Liouville fractional integrals of a function f with respect to another function ψγ on a,b are defined by [29, 30](4)ℑa+δ¯:ψgγ=1Γδ¯∫aγψ′tψγ−ψtδ¯−1gtdt,ℑb−δ¯:ψgγ=1Γδ¯∫γbψ′tψt−ψγδ¯−1gtdt.
If we setψγ=γ in (4), then Definition 2 reduces to Definition 1.
The following lemmas hold in [29, 30].Lemma 1.
Letδ¯>0,μ¯>0, and gγ=γμ¯; then,(5)ℑa+δ¯gγ=Γμ¯+1Γδ¯+μ¯+1γδ¯+μ¯.Lemma 2.
Letδ¯>0,μ¯>0, and gγ=ψγ−ψaμ¯; then,(6)ℑa+δ¯:ψgγ=Γμ¯+1Γδ¯+μ¯+1ψγ−ψaδ¯+μ¯.Remark 1.
In this context,ℑδ¯gγ and ℑδ¯:ψgγ stand for ℑ0+δ¯gγ and ℑ0+δ¯:ψgγ, respectively.
In this paper, using the Adomian decomposition method and Laplace transform method combined with Lemma1, we produce a new powerful technique. By using this technique, we obtain exact solution for main generalized Abel’s integral equations and generalized weakly singular Volterra integral equations.


## 3. Main Results

Now, we give our main results.Lemma 3 (see [4, 5]).
Ifgγ is bounded on 0<γ<b, g is strictly monotonically increasing and differentiable function in some interval 0,b and gγ≠0 for every γ in 0,b. Then, Abel’s integral equation (1) has the following solution:(7)uγ=sinδ¯ππddγ∫0γg′tftgγ−gt1−δ¯dt,0<δ¯<1.Theorem 1.
Ifft is bounded on 0<γ<b, g is strictly monotonically increasing and differentiable function in some interval 0,b and gγ≠0 for every γ in 0,b. Abel’s integral equation (1) has the following solution:(8)uγ=Γδ¯sinδ¯ππddγℑδ¯:ψgγ.Proof.
From Lemma3 for δ¯∈0,1, we have(9)uγ=Γδ¯sinδ¯ππddγ1Γδ¯∫0γg′tgγ−gtδ¯−1ftdt.
From this and Definition2, we get (8). This completes the proof.Corollary 1.
Under the similar assumptions of Theorem1, if gγ=gγ−g0μ¯, then the solution of Abel’s integral equation (1) is(10)uγ=sinδ¯ππΓδ¯Γμ¯+1Γδ¯+μ¯+1ddγgγ−g0δ¯+μ¯.Now, the solution of (2) can be obtained in the following theorem.Theorem 2.
Ifgγ=θ¯gγ−g0μ¯, g is strictly monotonically increasing and differentiable function in some interval 0,b and gγ≠0 for every γ in 0,b. The generalized weakly singular Volterra type integral equation (2) has the following solution:(11)uγ=θ¯Γμ¯+1gγ−g0μ¯E1−δ¯,μ¯+1β¯Γ1−δ¯gγ−g01−δ¯,where Eδ¯,μ¯ is the 2-parameter Mittag–Leffler function which is defined by [31, 32](12)Eδ¯,μ¯γ=∑n=0∞γnΓδ¯n+μ¯.Proof.
From the Adomian decomposition method, we substitute the decomposition series(13)uγ=∑n=0∞unγ,into both sides of (2) to obtain(14)∑n=0∞unγ=gγ+∫0γβ¯gγ−gtδ¯∑n=0∞untdt.
The componentsu0γ,u1γ,u2γ,… are determined by using (3) in the Adomian recurrence relation:(15)u0γ=gγ=θ¯gγ−g0μ¯,u1γ=β¯∫0γgt−g0μ¯gγ−gtδ¯dt=θ¯β¯Γ1−δ¯Γμ¯+1Γ2−δ¯+μ¯gγ−g01−δ¯+μ¯,u2γ=θ¯β¯2∫0γu1tgγ−gtδ¯dt=θ¯β¯2Γ1−δ¯2Γμ¯+1Γ3−2δ¯+μ¯gγ−g021−δ¯+μ¯,u3γ=θ¯β¯3∫0γu2tgγ−gtδ¯dt=θ¯β¯3Γ1−δ¯3Γμ¯+1Γ4−3δ¯+μ¯gγ−g031−δ¯+μ¯.⋮
Thus, the exact solution is(16)uγ=θ¯gγ−g0μ¯+β¯Γ1−δ¯Γμ¯+1Γ1−δ¯+μ¯+1gγ−g01−δ¯+μ¯+β¯2Γ1−δ¯2Γμ¯+1Γ21−δ¯+μ¯+1gγ−g021−δ¯+μ¯+⋯=θ¯Γμ¯+1gγ−g0μ¯∑n=0∞β¯Γ1−δ¯gγ−g01−δ¯nΓ1−δ¯n+μ¯+1=θ¯Γμ¯+1gγ−g0μ¯E1−δ¯,μ¯+1β¯Γ1−δ¯gγ−g01−δ¯,which completes the proof.Corollary 2.
Under the similar assumptions of Theorem2, if gγ=γ, then the solution of the generalized weakly singular Volterra type integral equation (2) is(17)uγ=θ¯Γμ¯+1γμ¯E1−δ¯,μ¯+1β¯Γ1−δ¯γ1−δ¯.The noise terms may appear between components ofu0γ and u1γ of (15) with opposite signs. Hence, by canceling these noise terms between these components, we may give the exact solution that should be justified through substitution and thus minimize the size of the calculations. In this situation, we use the following corollary.Corollary 3.
Under the similar assumptions of Theorem2, the solution of the generalized weakly singular Volterra type integral equation (2) can be obtained as(18)uγ=θ¯Γμ¯+1β¯Γ1−δ¯γδ¯+μ¯−1E1−δ¯,δ¯+μ¯β¯Γ1−δ¯γ1−δ¯−1Γδ¯+μ¯.Proof.
To prove this, we use one of the Mittag–Leffler function’s properties, that is [31, 32],(19)Eδ¯,μ¯γ=1γEδ¯,μ¯−δ¯γ−1xΓμ¯−δ¯.
From this, we can write (17) as(20)uγ=θ¯Γμ¯+1γμ¯β¯Γ1−δ¯γ1−δ¯E1−δ¯,δ¯+μ¯β¯Γ1−δ¯γ1−δ¯−1Γδ¯+μ¯.
This completes the proof.Theorem 3.
Under the similar assumptions of Theorem2, if gγ=∑j=1∞θ¯jγμ¯j, then the exact of the generalized weakly singular Volterra type integral equation (2) is given by(21)uγ=∑j=1∞θ¯jΓμ¯j+1γμ¯jE1−δ¯,μ¯j+1β¯Γ1−δ¯γ1−δ¯,or equivalently(22)uγ=∑j=1∞θ¯jΓμ¯j+1β¯Γ1−δ¯γδ¯+μ¯j−1E1−δ¯,δ¯+μ¯jβ¯Γ1−δ¯γ1−δ¯−1Γδ¯+μ¯j.Proof.
By the same manner of Theorem2, we get(23)uγ=θ¯1Γμ¯1+1γμ¯1E1−δ¯,μ¯1+1β¯Γ1−δ¯γ1−δ¯+θ¯2Γμ¯2+1γμ¯2E1−δ¯,μ¯2+1β¯Γ1−δ¯γ1−δ¯+θ¯3Γμ¯3+1γμ¯3E1−δ¯,μ¯3+1β¯Γ1−δ¯γ1−δ¯+⋯.
This completes the proof of the first part of this theorem.
By using property (19), we easily obtain the proof of the second part of this theorem. Thus, the proof of Theorem 3 is completed.Remark 2.
Due to the occurrence of the noise terms betweenu0γ and u1γ, we can write uγ as(24)uγ=θ¯1Γμ¯1+1γμ¯1E1−δ¯,μ¯1+1β¯Γ1−δ¯γ1−δ¯+θ¯2Γμ¯2+1β¯Γ1−δ¯γδ¯+μ¯2−1E1−δ¯,δ¯+μ¯2β¯Γ1−δ¯γ1−δ¯−1Γδ¯+μ¯2+⋯,which is obtained from (21) and (22).
Now, we introduce some spaces of the continuous functions in order to obtain the boundness of the above solution.Definition 3.
Let0<μ¯≤1 and J=a,b be a finite interval on the half-axis ℝ+ with 0≤a<b<∞. Then,(i)
We denote byCa,b the space of continuous functions g on J with the norm(25)gC≔=maxgγ;γ∈a,b.(ii)
We define the weighted spaceCμ¯ψa,b of functions g with respect to an increasing function ψ on a,b by(26)Cμ¯ψa,b≔maxg:a,b⟶ℝ,ψγ−ψaμ¯gγ∈Ca,b,with the norm(27)gCμ¯ψ≔maxψγ−ψaμ¯gγ;γ∈a,b.
Note thatC0ψa,b=Ca,b.Remark 3.
Letδ¯,β¯,τ>0; then, we have(28)∫τtgt−gsδ¯gs−gτβ¯g′sds=gt−gτδ¯+β¯+1∫011−zδ¯zβ¯dz=Γδ¯+1Γβ¯+1Γδ¯+β¯+2gt−gτδ¯+β¯+1.Proof.
The proof follows directly from the substitutiongs=gτ+zgt−gτ and the definition of the beta function.Theorem 4.
Let0≤a<b<∞,δ¯∈0,1, and μ¯>0. Then,
Ifμ¯≤δ¯, the fractional operator ℑa+δ¯:ψ is bounded from Cμ¯ψa,b into Ca,b with(29)ℑa+δ¯:ψgC≤gCμ¯ψΓ1−μ¯Γδ¯−μ¯+1ψb−ψaδ¯−μ¯.(i)
Moreover, ifμ¯≤δ¯−1, the solution ux in Theorem 1 is bounded from Cμ¯ψa,b into Ca,b with(30)uC≤gCμ¯ψsinδ¯πΓδ¯Γ1−μ¯πΓδ¯−μ¯ψb−ψ0δ¯−μ¯−1ψ′b.(ii)
For anyg∈Ca,b, the fractional operator ℑa+δ¯:ψ is a mapping from Ca,b into Ca,b with(31)ℑa+δ¯:ψgC≤gC1Γδ¯+1ψb−ψaδ¯.
Moreover, for anyg∈Ca,b, the solution ux in Theorem 1 is bounded from Ca,b into Ca,b with(32)uC≤gCsinδ¯ππψbψ0δ¯−1ψ′b.Proof.
(i)
From Definition3, we get(33)ℑa+δ¯:ψgC=maxℑa+δ¯:ψgx;γ∈a,b=maxγ∈a,b1Γδ¯∫aγψ′sψγ−ψsδ¯−1gsds=maxγ∈a,b1Γδ¯∫aγψ′sψγ−ψsδ¯−1ψs−ψaμ¯ψs−ψa−μ¯gsds≤gCμ¯ψ1Γδ¯∫aγψ′sψγ−ψsδ¯−1ψs−ψa−μ¯ds.Then, by making use of Remark3, it follows that(34)ℑa+δ¯:ψgC≤gCμ¯ψΓ1−μ¯Γδ¯−μ¯+1ψγ−ψaδ¯−μ¯.Sinceψ is an increasing function, it follows that(35)ℑa+δ¯:ψgC≤gCμ¯ψΓ1−μ¯Γδ¯−μ¯+1ψb−ψaδ¯−μ¯.This completes the proof of the first part inequality in (i). By making use of Theorem1 and inequality (34) with a=0, we obtain(36)uC≤sinδ¯πΓδ¯πgCμ¯ψΓ1−μ¯Γδ¯−μ¯+1ddγψγ−ψ0δ¯−μ¯=gCμ¯ψsinδ¯πΓδ¯Γ1−μ¯πΓδ¯−μ¯ψγ−ψ0δ¯−μ¯−1ψ′x≤gCμ¯ψsinδ¯πΓδ¯Γ1−μ¯πΓδ¯−μ¯ψb−ψ0δ¯−μ¯−1ψ′b.Letg∈Ca,b; then, by using Definition 3 and Lemma 2 and since ψ is an increasing function, we have(37)ℑa+δ¯:ψgC=maxγ∈a,b1Γδ¯∫aγψ′sψγ−ψsδ¯−1gsds=gCℑa+δ¯:ψ1=gC1Γδ¯+1ψγ−ψaδ¯≤gC1Γδ¯+1ψb−ψaδ¯.
This completes the proof of the first inequality of (ii). By making use of Theorem1 and inequality (37) with a=0, we can deduce the second part inequality in (ii).Theorem 5.
For anyδ¯∈0,1,μ¯,a≥0, we have(38)ℑa+δ¯:ψψsψaμ¯γ=Γμ¯+1Γδ¯+μ¯+1ψγ−ψaδ¯+μ¯.Proof.
The proof is similar to Lemma2, so it is omitted.Theorem 6.
Let0≤a<b<∞,δ¯∈0,1,μ¯>0 and g∈Cμ¯ψa,b. If μ¯<δ¯, then we have(39)ℑa+δ¯:ψga=limx⟶aℑa+δ¯:ψgγ=0.Moreover, the solutionuγ in Theorem 1 vanishes at γ=0.Proof.
Letg∈Cμ¯ψa,b; then, ψγψaμ¯gγ∈Ca,b and there exists some M>0 such that(40)ψγ−ψaμ¯gγ≤M,γ∈a,b,and(41)ℑa+δ¯:ψgγ=1Γδ¯∫aγψ′tψγ−ψtδ¯−1gtdt=1Γδ¯∫aγψ′sψγ−ψsδ¯−1ψs−ψγ−μ¯ψs−ψγμ¯gsds.
Then, by making use of Remark3 and Theorem 5, it follows that(42)ℑa+δ¯:ψg≤MΓδ¯∫aγψ′sψγ−ψsδ¯−1ψs−ψγ−μ¯ds=Mℑa+δ¯:ψψs−ψγ−μ¯=Γ1−μ¯Γδ¯−μ¯+1ψγ−ψaδ¯−μ¯.
Taking the limit on both sides, it follows that(43)limγ⟶aℑa+δ¯:ψg=0,which completes the proof of the first part. By making use of Theorem 1 and formula (42) with a=0, we can deduce the second part of theorem.


## 4. Test Examples

In this section, we consider several test problems corresponding to the equations (1) and (2) to demonstrate the efficiency of our new mechanism.Example 1.
Consider the generalized Abel’s integral equation [4]:(44)43sinγ3/4=∫0γutdtsinγ−sint1/4,0<γ<π2,where δ¯=1/4,gγ=4/3sinγ3/4 and gγ=sinγ. It is clear that gγ is strictly monotonically increasing in 0<γ<π/2 and g′γ≠0 for each γ in 0<γ<π/2. Using Corollary 1 with μ¯=3/4, we get(45)uγ=12π433/4Γ1/4Γ3/4Γ2ddγsinγ=cosγ,which is the exact solution, where we have used the fact that Γ1/4Γ3/4=2π.Example 2.
Consider the generalized Abel’s integral equation [4]:(46)625γ25/6=∫0γutdtγ5−t51/6,0<γ<2.
Here, equation (7) takes the following form:(47)uγ=sinδ¯ππddγ∫0x5t4⋅6/25t25/6γ5−t51/6dt=12π625ddγ∫0x5t4t55/6γ5−t51/6dt.
From this, we haveδ¯=1/6,gγ=6/25γ25/6,gγ=γ5, and μ¯=5/6. Thus, Corollary 1 gives the exact solution:(48)uγ=12πddγ25πγ5=γ4,where we have used the fact that Γ1/6Γ5/6=2π.Example 3.
Consider the weakly singular second kind Volterra integral equation [4, 33, 34]:(49)uγ=2γ−∫0γutdtγ−t,x∈0,2.
Using Corollary3 with θ¯=2,μ¯=1/2,β¯=−1, and δ¯=1/2, we can easily obtain(50)uγ=1−E1/2,1−πγ=1−eπxerfcπγ,where erfc is the complementary error function which is defined as(51)erfcw=2π∫w∞e−z2dz.Example 4.
Consider the weakly singular second kind Volterra integral equation [4, 33, 34]:(52)uγ=γ2+1615γ52/−∫0γutdtγ−t,x∈0,1,where θ¯1=1,θ¯2=16/15,μ¯1=2,μ¯2=5/2,β¯=−1, and δ¯=1/2. Thus, by formula (24), we get(53)uγ=Γ2+1γ2E1/2,3−πγ−1615Γ5/2+1γ5/2Γ1/2γ1/2E1/2,3−πγ−1Γ3=2γ2E1/2,3−πγ−2γ2E1/2,3−πγ+γ2=γ2.
Hence, the exact solution isuγ=γ2.Example 5.
Consider the weakly singular second kind Volterra integral equation [9]:(54)uγ=γ−πγ+2∫0γutdtγ−t.
In this example,θ¯1=1,θ¯2=−π,μ¯1=1/2,μ¯2=1,β¯=2, and δ¯=1/2. Thus, formula (24) gives(55)uγ=12πγE1/2,3/22πγ−12πγE1/2,3/22πγ+γ=γ,which is the exact solution of integral equation (54).Example 6.
Consider the weakly singular second kind Volterra integral equation [9]:(56)uγ=12−γ+∫0γutdtγ−t.
From formula (24) with θ¯1=1/2,θ¯2=−1,μ¯1=0,μ¯2=1/2,β¯=1, and δ¯=1/2, we get(57)uγ=12E1/2,1πγ−12E1/2,1πγ+12=12,which is the exact solution of integral equation (56).Example 7.
Consider the Abel integral equation of the second kind [33]:(58)uγ=γ+43γ3/2−∫0γutdtγ−t.
From formula (24) with θ¯1=1,θ¯2=4/3,μ¯1=1,μ¯2=3/2,β¯=−1, and δ¯=1/2, we get(59)uγ=γE1/2,2−πγ−γE1/2,2−πγ+γ=γ,which is the exact solution of (58).


## 5. Conclusion

In the present article, a new technique involving Riemann–Liouville fractional integrals has been used to solve main generalized Abel’s integral equations and generalized weakly singular Volterra integral equations. Also, we have solved several examples with our proposed technique. We can observe that our developed technique is easy and straightforward to apply.



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