Int. J. Anal. Appl. (2022), 20:9 Solutions of Linear and Nonlinear Fractional Fredholm Integro-Differential Equations Abdelhalim Ebaid1,∗, Hind K. Al-Jeaid2 1Department of Mathematics, Faculty of Science, University of Tabuk, P.O. Box 741, Tabuk 71491, Saudi Arabia 2Department of Mathematical sciences, Umm Al-Qura University, Makkah, Saudi Arabia ∗Corresponding author: aebaid@ut.edu.sa, halimgamil@yahoo.com Abstract. The present paper analyzes a class of first-order fractional Fredholm integro differential equations in terms of Caputo fractional derivative. In the literature, such kind of fractional integro- differential equations have been solved using several numerical methods, while the exact solutions were not obtained. However, the exact solutions are obtained in this paper for various linear and nonlinear examples. It is shown that the exact solution of the linear problems is unique, while multiple exact solutions exist for the nonlinear ones. Moreover, the obtained results reduce to the classical ones in the relevant literature as the fractional order becomes unity. The obtained exact solutions can be further invested by other researchers to validate their numerical/approximation methods. 1. Introduction The fractional calculus (FC) has gained observable interest in recent years due to its applications several fields [1-14]. The FC has been also extended to integro-differential equations (FIDEs) as ob- served in the literature [15-28], where various numerical and analytical methods were applied to solve for approximate solutions. We are concerned here with fractional Fredholm integro-differential equa- tions (FFIDEs) of first-order. Although important results were reported [15-28] for FIDEs, obtaining the exact solution of FFIDEs is not an easy task, even for simple equations as will be shown later. So, Received: Nov. 3, 2021. 2010 Mathematics Subject Classification. 34K37. Key words and phrases. fractional calculus; analytic solution; Fredholm integro-differential equations. https://doi.org/10.28924/2291-8639-20-2022-9 ISSN: 2291-8639 © 2022 the author(s). https://doi.org/10.28924/2291-8639-20-2022-9 2 Int. J. Anal. Appl. (2022), 20:9 we consider in this paper the following class of FFIDEs: C 0 D α x u(x) = f (x) + λ ∫ b2 b1 K(x,τ,u(τ)) dτ, 0 < α ≤ 1, (1.1) u(0) = h, (1.2) where h, λ, b1 and b2 are given constants, f (x) is a given continuous function on [b1,b2]. The objective of this paper is to introduce a direct analytic approach for obtaining exact solutions for the class (1-2). It will be shown that the solution is unique when K(x,τ,u(τ)) is a linear function in the unknown function u(τ). In addition, it will be declared that multiple exact solutions exists when K(x,τ,u(τ)) is a nonlinear function in u(τ). The Caputo definition is chosen as a fractional derivative in Eq. (1) and the structure of the paper is as follows. In section 2, we give the main aspects of the FC. In addition, a basic Lemma will be provided for the formal exact solution of the class (1-2). Sections 3 investigates the application of the present approach on several linear and nonlinear problems. Besides, the way of obtaining exact dual solution for the nonlinear case will be demonstrated in section 3. Moreover, it will be shown that the present exact solutions reduce to the classical ones as α → 1. Finally, section 5 outlines the conclusions. 2. Main aspects of FC The Riemann-Liouville fractional integral of order α is defined as [1]: Jα0 u(x) = 1 Γ(α) ∫ x 0 (x −τ)α−1 u(τ)dτ α > 0. (2.1) The Caputo’s FD of order α of a function u(x) is defined by C 0 D α x u(x) = 1 Γ(n−α) ∫ x 0 (x −τ)n−α−1u(n)(τ)dτ, n− 1 < α ≤ n. (2.2) The Jα0 and C 0 D α x are related by: Jα0 ( C 0 D α x u(x) ) = u(x) − n−1∑ m=0 u(m)(0) m! xm(0), (2.3) which is useful when solving FDEs/FIEs. A basic property of the Jα0 is Jα0 (x r ) = Γ(r + 1) Γ(α + r + 1) xα+r, r > −1. (2.4) Int. J. Anal. Appl. (2022), 20:9 3 The Mittag-Leffler function (MLF) of one-parameter is defined as Eα(z) = ∞∑ m=0 zm Γ(αm + 1) , z ∈ C, (2.5) while the two-parameter MLF is given as Eα,β(z) = ∞∑ m=0 zm Γ(αm + β) , α > 0, β > 0. (2.6) The following properties are also hold: E1,2(z) = (e z − 1) / (z) , (2.7) E2,1(−z2) = cos(z), E2,2(−z2) = sin(z) z . (2.8) Lemma 1. The analytic solution of the first-order FFIDE (1-2) is given by u(x) = h + aλ ( xα Γ(α + 1) ) + Jα0 (f (x)) , (2.9) provided that the fractional integral of f (x), i.e., Jα0 (f (x)), exists and a is the constant given by a = ∫b2 b1 K(x,τ,u(τ)) dτ. Proof: The bounded integral involved in Eq. (1) can be assumed as a constant. Besides, we assume that such integral is given by the constant a as a = ∫ b2 b1 K(x,τ,u(τ)) dτ. (2.10) Operating with Jα0 on Eq. (1) and implementing (2), (5), and (14), it then follows u(x) −u(0) = Jα0 (aλ) + J α 0 (f (x)) , (2.11) or u(x) = h + aλJα0 (1) + J α 0 (f (x)) . (2.12) Calculating Jα0 (1) from Eq. (6) at r = 0, we have J α 0 (1) = xα Γ(α+1) . Substituting this last result into Eq. (14) we obtain Eq. (11) which completes the proofs. � 4 Int. J. Anal. Appl. (2022), 20:9 3. Examples Example 1: Consider the FFIDE [29] C 0 D α x u(x) = 2 ( 1 − ∫ 1 0 u(τ) dτ ) , u(0) = 0. (3.1) Let a1 = ∫ 1 0 u(τ) dτ, (3.2) where a1 is an unknown constant. Accordingly, Eq. (15) becomes C 0 D α x u(x) = 2 (1 −a1) . (3.3) Applying the integral operator Jα0 on Eq. (17) and making use of Eq. (5), we have u(x) = u(0) + 2 (1 −a1) Jα0 (1), (3.4) or u(x) = 2 (1 −a1) xα Γ(α + 1) . (3.5) The constant a1 is evaluated by inserting (19) into (16), this yields a1 = 2 (1 −a1) ∫ 1 0 τα Γ(α + 1) dτ = 2 (1 −a1) Γ(α + 2) . (3.6) Solving Eq. (20) for a1, we obtain a1 = 2 2 + Γ(α + 2) , (3.7) and hence, Eq. (19) becomes u(x) = ( 2Γ(α + 2) Γ(α + 1) (2 + Γ(α + 2)) ) xα. (3.8) As α → 1, Eq. (22) reduces to the exact solution u(x) = x for the classical form of Eq. (15), given by u′(x) = 2 ( 1 − ∫ 1 0 u(τ) dτ ) . Example 2: Consider the FFIDE [29] C 0 D α x u(x) = 3 + 6x + x ∫ 1 0 τu(τ) dτ, u(0) = 0. (3.9) Suppose that a2 = ∫ 1 0 τu(τ) dτ, (3.10) Int. J. Anal. Appl. (2022), 20:9 5 where a2 is a constant to be determined, then Eq. (23) becomes C 0 D α x u(x) = 3 + (6 + a2) x. (3.11) Operating with Jα0 on Eq. (25), it then follows u(x) = 3xα Γ(α + 1) + (6 + a2)x α+1 Γ(α + 2) . (3.12) From Eq. (24), we have a2 = ∫ 1 0 ( 3τα+1 Γ(α + 1) + (6 + a2)τ α+2 Γ(α + 2) ) dτ, = 3 (α + 2)Γ(α + 1) + (6 + a2) (α + 3)Γ(α + 2) , (3.13) which gives a2 = 3α2 + 18α + 21 (α + 2) [(α + 3)Γ(α + 2) − 1] . (3.14) Therefore, u(x) = 3xα Γ(α + 1) + ( 6 + 3α2 + 18α + 21 (α + 2) [(α + 3)Γ(α + 2) − 1] ) xα+1 Γ(α + 2) , (3.15) and reduces, as α → 1, to u(x) = 3x + 4x2 which is the same solution in Ref. [29] for the classical form: u′(x) = 3 + 6x + x ∫ 1 0 τu(τ) dτ. Example 3: This example considers the FFIDE [29] C 0 D α x u(x) = −1 + cos x + ∫ π/2 0 τu(τ) dτ, u(0) = 0, (3.16) which takes the form: C 0 D α x u(x) = (a3 − 1) + cos x, (3.17) where a3 is a constant defined by a3 = ∫ π/2 0 τu(τ) dτ. (3.18) Expressing cos x as Maclaurin series and then applying Jα0 on both sides of Eq. (31), gives u(x) = (a3 − 1) Jα0 (1) + J α 0 ( ∞∑ m=0 (−1)mx2m (2m)! ) , = (a3 − 1) xα Γ(α + 1) + ∞∑ m=0 (−1)m (2m)! × Γ(2m + 1)xα+2m Γ(α + 2m + 1) , (3.19) 6 Int. J. Anal. Appl. (2022), 20:9 which is simplified as u(x) = (a3 − 1) xα Γ(α + 1) + ∞∑ m=0 (−1)mxα+2m Γ(α + 2m + 1) , (3.20) or in terms of the MLF E2,α+1(−x2) as u(x) = (a3 − 1) xα Γ(α + 1) + xαE2,α+1(−x2). (3.21) From Eq. (32) and Eq. (35), we have a3 = ∫ π/2 0 ( (a3 − 1) τα+1 Γ(α + 1) + τα+1E2,α+1(−τ2) ) dτ, = (a3 − 1) (π/2)α+2 (α + 2)Γ(α + 1) + I, (3.22) where the integral I is defined by I = ∫ π/2 0 τα+1E2,α+1(−τ2) dτ. (3.23) Solving Eq. (36) for a3, we obtain a3 = − (π/2)α+2 (α + 2)Γ(α + 1) − (π/2)α+2 + (α + 2)Γ(α + 1) (α + 2)Γ(α + 1) − (π/2)α+2 I. (3.24) Therefore, u(x) is finally given by u(x) = ( − (π/2)α+2 (α + 2)Γ(α + 1) − (π/2)α+2 + (α + 2)Γ(α + 1) (α + 2)Γ(α + 1) − (π/2)α+2 I − 1 ) × xα Γ(α + 1) + xαE2,α+1(−x2), (3.25) and I is already defined by Eq. (37). The solution given by Eq. (39) reduces, as α → 1, to u(x) = ( − (π/2)3 3 − (π/2)3 + 3 3 − (π/2)3 ∫ π/2 0 τ2E2,2(−τ2) dτ − 1 ) x + xE2,2(−x2), = ( − (π/2)3 3 − (π/2)3 + 3 3 − (π/2)3 ∫ π/2 0 τ sin τ dτ − 1 ) x + x ( sin x x ) , = ( − (π/2)3 3 − (π/2)3 + (π/2)3 3 − (π/2)3 ) x + sin x, where ∫ π/2 0 τ sin τ dτ = 1, = sin x, (3.26) which is the corresponding solution for the classical form: u′(x) = −1 + cos x + ∫π/2 0 τu(τ) dτ. Int. J. Anal. Appl. (2022), 20:9 7 Example 4: In this example, we considers the FFIDE [29]: C 0 D α x u(x) = −10x + ∫ 1 −1 (x −τ) u(τ) dτ, u(0) = 1. (3.27) Assuming that a4 = ∫ 1 −1 u(τ) dτ, a5 = ∫ 1 −1 τu(τ) dτ, (3.28) then Eq. (41) becomes C 0 D α x u(x) = (a4 − 10) x −a5, (3.29) where a4 and a5 are constants. Following the same analysis of the previous examples, we obtain u(x) = 1 + (a4 − 10) xα+1 Γ(α + 2) − a5x α Γ(α + 1) . (3.30) Substituting Eq. (44) into Eqs. (42) and performing the associated integrals, we get the following system: ( 1 − ( 1 − (−1)α Γ(α + 3) )) a4 + ( 1 + (−1)α Γ(α + 2) ) a5 = 2 − 10 ( 1 − (−1)α Γ(α + 3) ) , (3.31)( 1 + (−1)α (α + 3)Γ(α + 2) ) a4 − ( 1 + 1 − (−1)α (α + 2)Γ(α + 1) ) a5 = 10 ( 1 + (−1)α (α + 3)Γ(α + 2) ) . (3.32) The solution of the system (45-46) can be obtained as a4 = ∆1 ∆ , a5 = ∆2 ∆ , (3.33) where ∆, ∆1, and ∆2 are given by the determinants: ∆ = ∣∣∣∣∣∣∣∣∣∣ 1 − 1 − (−1)α Γ(α + 3) 1 + (−1)α Γ(α + 2) 1 + (−1)α (α + 3)Γ(α + 2) −1 − 1 − (−1)α (α + 2)Γ(α + 1) ∣∣∣∣∣∣∣∣∣∣ , (3.34) and ∆1 = ∣∣∣∣∣∣∣∣∣∣ 2 − 10 ( 1−(−1)α Γ(α+3) ) 1 + (−1)α Γ(α + 2) 10 ( 1+(−1)α (α+3)Γ(α+2) ) −1 − 1 − (−1)α (α + 2)Γ(α + 1) ∣∣∣∣∣∣∣∣∣∣ , (3.35) ∆2 = ∣∣∣∣∣∣∣∣∣∣ 1 − 1 − (−1)α Γ(α + 3) 2 − 10 ( 1−(−1)α Γ(α+3) ) 1 + (−1)α (α + 3)Γ(α + 2) 10 ( 1+(−1)α (α+3)Γ(α+2) ) ∣∣∣∣∣∣∣∣∣∣ . (3.36) 8 Int. J. Anal. Appl. (2022), 20:9 Hence, the solution of Eq. (41) is finally given by u(x) = 1 + ( ∆1 ∆ − 10 ) xα+1 Γ(α + 2) − ( ∆2 ∆ ) xα Γ(α + 1) . (3.37) As α → 1, u(x) given by Eq. (44) implies that u(x) = 1 + 1 2 ([ ∆1 ∆ ] α→1 − 10 ) x2 − [ ∆2 ∆ ] α→1 x. (3.38) Calculating ∆, ∆1, and ∆2 when α → 1, we obtain ∆ = − 10 9 , ∆1 = 20 9 , ∆2 = 0. (3.39) Thus, Eq. (52) becomes u(x) = 1 − 6x2, (3.40) which is the corresponding solution of the classical form: u′(x) = −10x + ∫ 1 −1 (x −τ) u(τ) dτ. Example 5: In order to show how to apply the present direct approach to solving nonlinear FFIDEs, we consider here a simple example, given by the nonlinear FFIDE: C 0 D α x u(x) = ∫ 1 0 u2(τ) dτ, u(0) = 0, (3.41) which can be written as C 0 D α x u(x) = a6, (3.42) where a6 is a constant defined by a6 = ∫ 1 0 u2(τ) dτ. (3.43) On solving Eq. (56), we have u(x) = a6x α Γ(α + 1) . (3.44) Evaluating a6 from Eq. (57), we obtain a6 = a 2 6 ∫ 1 0 τ2α (Γ(α + 1)) 2 dτ, (3.45) which leads to a6 ( a6 (2α + 1) (Γ(α + 1)) 2 − 1 ) = 0. (3.46) Solving this equation for a6, we obtain a6 = 0, a6 = (2α + 1) (Γ(α + 1)) 2 . (3.47) Int. J. Anal. Appl. (2022), 20:9 9 The first value a6 = 0 leads u(x) = 0, which is a trivial solution. While the value a6 = (2α + 1) (Γ(α + 1)) 2 gives u(x) = (2α + 1)Γ(α + 1)xα, (3.48) as a second solution. As α → 1, we obtain the corresponding solution u(x) = 3x for the classical form u′(x) = ∫ 1 0 u2(τ) dτ. Example 6: We consider an additional nonlinear example: C 0 D α x u(x) = 10x − 5 + ∫ 1 0 u2(τ) dτ, u(0) = 0, (3.49) Following the above analysis, we can obtain u(x) = (a7 − 5)xα Γ(α + 1) + 10xα+1 Γ(α + 2) , (3.50) and a7 is given as a7 = ∫ 1 0 u2(τ) dτ. (3.51) Substituting Eq. (64) into Eq. (65), yields a7 = ∫ 1 0 [ (a7 − 5)2τ2α (Γ(α + 1))2 + 20(a7 − 5)τ2α+1 Γ(α + 1)Γ(α + 2) + 100τ2α+2 (Γ(α + 2))2 ] dτ. (3.52) Performing this integral, we find that a7 is governed by the equation: a7 = (a7 − 5)2 (2α + 1)(Γ(α + 1))2 + 10(a7 − 5) (Γ(α + 2))2 + 100 (2α + 3)(Γ(α + 2))2 . (3.53) Eq. (67) can be rewritten as A(a7 − 5)2 + B(a7 − 5) + C = 0, (3.54) where A = (a7 − 5)2 (2α + 1)(Γ(α + 1))2 , B = 10 (Γ(α + 2))2 − 1, C = 100 (2α + 3)(Γ(α + 2))2 − 5. (3.55) Solving Eq. (68) for the constant a7, we get a7 = 5 + 1 2A ( −B ± √ B2 − 4AC ) . (3.56) From Eq. (64) and Eq. (70), we obtain u(x) = 1 2A ( −B ± √ B2 − 4AC )( xα Γ(α + 1) ) + 10xα+1 Γ(α + 2) . (3.57) 10 Int. J. Anal. Appl. (2022), 20:9 It can be seen from Eq. (71) that there are two different solutions for the present nonlinear example, the first one is given by u1(x) = 1 2A ( −B + √ B2 − 4AC )( xα Γ(α + 1) ) + 10xα+1 Γ(α + 2) , (3.58) while the second solution is u2(x) = − 1 2A ( B + √ B2 − 4AC )( xα Γ(α + 1) ) + 10xα+1 Γ(α + 2) . (3.59) In order to check these two solutions, we evaluate them as α → 1. In this case, we have from Eqs. (69) that A = 1 3 , B = 3 2 , C = 0. (3.60) Hence, (u1(x))α→1 = 5x 2, (3.61) and (u2(x))α→1 = − 9 2 x + 5x2, (3.62) The solutions (75) and (76) are the same obtained one in Ref. [29] for the classical nonlinear version u′(x) = 10x − 5 + ∫ 1 0 u2(τ) dτ. 4. Conclusion A class of first-order FFIDEs was investigated in terms of Caputo definition in FC. The analytic solutions of several linear and nonlinear examples were obtained. 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Bushnaq, Numerical solutions to systems of fractional Voltera integro differential equations, using Chebyshev wavelet method, J. Taibah Univ. Sci. 12 (2018), 584–591. https://doi. org/10.1080/16583655.2018.1510149. [29] A.M. Wazwaz, Linear and Nonlinear Integral Equations, Methods and Applications, Higher Education Press, Beijing and Springer-Verlag, Berlin Heidelberg (2011). https://doi.org/10.1155/2014/103016 https://doi.org/10.1155/2014/103016 https://doi.org/10.1016/j.cam.2013.07.044 https://doi.org/10.1016/j.cam.2013.07.044 https://doi.org/10.1016/j.cnsns.2013.04.026 https://doi.org/10.1016/j.cnsns.2013.04.026 https://doi.org/10.1016/j.apm.2015.03.053 https://doi.org/10.1016/j.apm.2015.03.053 https://doi.org/10.15640/arms.v3n2a6 https://doi.org/10.1155/2017/3821870 https://doi.org/10.1155/2017/3821870 https://doi.org/10.1080/16583655.2018.1510149 https://doi.org/10.1080/16583655.2018.1510149 1. Introduction 2. Main aspects of FC 3. Examples 4. Conclusion References