EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 1, 2024, 546-568 ISSN 1307-5543 – ejpam.com Published by New York Business Global Utilization of the modified Adomian decomposition method on the Bagley-Torvik equation amidst Dirichlet boundary conditions Mariam Al-Mazmumy1, Mona Alsulami1,∗ 1 Department of Mathematics and Statistics, Faculty of Science, University of Jeddah, Jeddah 23218, Saudi Arabia Abstract. The Bagley-Torvik equation is an imperative differential equation that considerably arises in various branches of mathematical physics and mechanics. However, very few methods exist for the treatment of the model analytically; in fact, researchers frequently shop for semi-analytical and numerical methods in their studies. Therefore, the main goal of this research is to find the exact analytical solution for the fractional Bagley-Torvik equation fitted with Dirichlet boundary data, as well as a system of fractional Bagley-Torvik equations. Thus, this research aims to show that the modified Adomian decomposition method (MADM) via the proposed two algorithms is a very effective method for treating a class of Bagley-Torvik equations endowed with Dirichlet boundary data. Certainly, MADM is a very powerful approach for solving dissimilar functional equations without the need for either linearization, discretization, perturbation, or even unneces- sary restraining postulations. Additionally, the method reveals exact analytical solutions whenever obtainable or closed-form series solutions whenever exact solutions are not feasible. Lastly, some illustrative test problems of the governing model are examined to demonstrate the superiority of the proposed algorithms. 2020 Mathematics Subject Classifications: 26A33, 34A08, 34B15. Key Words and Phrases: Fractional calculus, Bagley-Torvik equation, Dirichlet boundary con- dition, Modified Adomian decomposition method, Boundary-value problem. 1. Introduction Fractional calculus is an aged area of research that recently reemerged more strongly with burning applications that cuts across all aspects of life. Indeed, the area started off by the ignition put forward by Leibniz (1695) and Euler (1730) [28, 30], and since then keeps propelling to date. Notably, various real-life models are discovered to be perfectly captured through the application of fractional differential equations (FDEs). These FDEs have, in recent times gained considerable relevance in modeling different ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i1.5050 Email addresses: mhalmazmumy@uj.edu.sa (M. Al-Mazmumy), mralsolami@uj.edu.sa (M. Alsulami) https://www.ejpam.com 546 © 2024 EJPAM All rights reserved. M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 547 emerging problems arising in, for instance, electrical networks, electromagnetic theory, fractals theory, viscoelasticity, control theory, material science, chemistry, potential theory, fluid flow, biology, and statistics to mention but a few [5, 6, 10, 11, 24, 26, 27, 33, 34, 39, 41]. In light of this, different researchers have in the past and recent times proposed a variety of methods, including analytical, semi-analytical, and computational to deal with FDEs. In this regard, the present paper shops for an elegant semi-analytical method that is founded on the utilization of the famous Adomian decomposition method (ADM) [1, 3, 4, 15– 17, 32, 38, 47, 49]. On the other hand, the boundary-value problems (BVPs) featuring fractional-order derivatives have - in recent years - fascinated or rather shaped the thoughts of various theoreticians and experimentalists in diverse stems of applied and pure sciences. In par- ticular, we make mention of the Bagley-Torvik equation, being an imperative differential equation that arises in various branches of mathematical physics and mechanics [46]. This equation is, however, used in modeling various processes, including viscoelasticity, the sub- mergence of solid structures in fluids, and the interaction of solid media with fluids among others; for more on the uniqueness and existence results of the model when prescribed with Dirichlet boundary data, an interested reader can consult [8, 25] and the references therewith. Moreover, as it is always thorny to tackle FDEs analytically, many mathemati- cians have introduced several efficient computational schemes based on various concepts to computationally treat the class of Bagley-Torvik equations. Here, we make mention of such approaches that are applied on the Bagley-Torvik equation in recent years to com- prise the quadratic spline solution [50], the cubic spline polynomials [51], the Chebyshev wavelet method [35], the shifted Legendre polynomials [45], the Taylor’s method [40], the exponential spline technique [7], the Chelyshkov-Tau approach [20], the Chebyshev collocation method [44], the quintic B-spline polynomial [23], the exponential spline ap- proximation [21], the Green’s function iterative approach [22] and the shifted Chebyshev operational matrix [29] to review but just a few; yet, read [14] for a mesmerizing study on the Bagley-Torvik equation with the aid of the differential transform approach. However, we, in the current paper aim to make use of the modified Adomian decom- position method (MADM) [2, 9, 12, 36, 37, 42, 48] by proposing two different algorithms to treat the Bagley-Torvik equation with Dirichlet boundary condition. MADM is a semi- analytical approach that was improved upon the classical ADM [3]-[38] to easily reveal exact analytical solutions whenever obtainable or closed-form series solutions whenever exact solutions are not feasible. In fact, the approach is very powerful in solving dis- similar functional equations without the need for either of linearization, discretization, perturbation, or even unnecessary restraining postulations. Besides, the method has been successfully used in the literature to solve various real-life models. In addition, we orga- nize the paper in the following pattern: Section 2 gives certain fundamental definitions of features with regard to fractional calculus, Section 3 gives the procedures of the devised MADM algorithms on BVP for the Bagley-Torvik equation, while Section 4 demonstrates the applicability of the devised algorithms, and Section 5 provides some concluding points. M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 548 2. Fractional calculus The current section introduces certain fundamental definitions and features for frac- tional calculus, consisting mainly of the fractional derivatives and fractional integrals that were put forward based on the Riemann-Liouville fractional (RLF) and the Caputo frac- tional (CF) integrals/derivatives. For more on some basics related to this work; an inter- ested reader can further read the famous book by Kilbas et al. [13]. 2.1. Preliminaries Here, we review some definitions of the fractional-order derivatives and integrals based on the definitions put forward by Riemann-Liouville and Caputo, respectively. Certainly, we will be considering the set χ = [a, b] ∈ R, such that a < b, a finite closed interval on R, the set of real numbers. Definition 1. (RLF integrals): The RLF right-sided Iαb−y and left-sided Iαa+y integrals of order α ∈ R are respectively defined as follows (Iαb−y)(x) := 1 Γ(α) ∫ b x y(t) (t− x)−α+1 dt, (x < b; α > 0), and (Iαa+y)(x) := 1 Γ(α) ∫ x a y(t) (x− t)−α+1 dt, (x > a; α > 0). Definition 2. (RLF derivatives): The RLF right-sided Dα b−y and left-sided Dα a+y derivatives of order α ∈ R are respectively defined as follows Dα b−y(x) := −dn dxn (In−α b− y)(x), := 1 Γ(n− α) −dn dxn ∫ b x y(t) (t− x)α+1−n dt, (x > b; α ≥ 0; n = [α] + 1), and Dα a+y(x) := dn dxn (In−α a+ y)(x), := 1 Γ(n− α) dn dxn ∫ x a y(t) (x− t)α+1−n dt, (x > a; α ≥ 0; n = [α] + 1), with [α] representing the integer part of the fractional-order α. Definition 3. (CF derivatives): The CF right-sided cDα b−y(x) and left-sided cDα a+y(x) derivatives of order α ∈ R+ on [a, b] are respectively defined via the RLF derivatives as follows cDα b−y(x) := ( Dα b− [ y(t)− n−1∑ k=0 y(k)(b) k! (b− t)k ]) (x), (1) M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 549 and cDα a+y(x) := ( Dα a+ [ y(t)− n−1∑ k=0 y(k)(a) k! (t− a)k ]) (x), (2) with n = [α] + 1, when α /∈ N, and n = α, for α ∈ N; with N denoting the set of positive whole numbers. Note that, as a peculiar case, when 0 < α < 1, the formulae given in (2) and (1) could be re-expressed as follows cDα b−y(x) = (Dα b− [y(t)− y(b)])(x), and cDα a+y(x) = (Dα a+ [y(t)− y(a)])(x). 2.2. Some useful properties This subsection recalls some useful properties of the aforementioned RLF integrals/derivatives and CF derivatives that will be greatly utilized in the course of the governing model. Lemma 1. Assume α > 0, and further assume n = [α] + 1, when α /∈ N, and n = α, when α ∈ N. Then, if y(x) ∈ Cn[a, b] or y(x) ∈ ACn[a, b], we accordingly have as follows (Iαb− cDα b−y)(x) = y(x)− n−1∑ k=0 (−)ky(k)(b) k! (b− x)k, (3) and (Iαa+ cDα a+y)(x) = y(x)− n−1∑ k=0 y(k)(a) k! (x− a)k. (4) Consequently, when 0 < α ≦ 1 and y(x) ∈ C[a, b] or y(x) ∈ AC[a, b], the above results respectively reduce to (Iαb− cDα b−y)(x) = y(x)− y(b), and (Iαa+ cDα a+y)(x) = y(x)− y(a). Lemma 2. The following results for the fractional integral/derivative hold [45] (i) Iα[c] = cxα Γ(α+ 1) , if c is constant, (5) (ii) Dαxn = Γ(n+ 1)xn−α Γ(n+ 1− α) . (6) (iii) Iα[xn] = Γ(n+ 1)xα+n Γ(α+ n+ 1) , (7) M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 550 3. Treatment of Bagley-Torvik BVPs via MADM The current section mainly makes use of MADM [2, 9, 12, 36, 37, 42, 48] to acquire the exact analytical solution of the Bagley-Torvik BVP where possible; moreover, when the acquisition of such an exact analytical solution is not feasible, a closed-form series solution of the governing model will be acquired. In fact, we will be making consideration to the BVP of the nonhomogeneous Bagley-Torvik equation endowed with Dirichlet boundary conditions as follows [51]-[29] D2y(x) +D3/2y(x) + y(x) = g(x), a < x < b, a, b ∈ R+, y(a) = λ1, y(b) = λ2, (8) with g(x) denoting the nonhomogeneous/source term, λ1 and λ2 are constants; while the function y(x) is the solution that we intend to determine. Certainly, the differential equation in (8) is a fractional-order differential equation defined in CF derivative sense with the highest integer-order of 2, and then followed by the CF-order of 3/2. In addition, this fractional-order model arises in many physical processes, and further has vast relevance in the study of fluid flows and viscoelastic materials [14], among others; more so, the highest-order of the equation remains 2, the integer-order, then follows by the fractional- order derivative of 3/2. Further, the Bagley-Torvik equation expressed above can equally be expressed in an operator notation upon making use of the differential operator notation D as follows Ly = D2y = g(x)−D3/2y(x)− y(x), (9) where L = D2 = d2 dx2 . Furthermore, to determine the explicit exact analytical solution of the Bagley-Torvik BVP expressed in (8) - using the version expressed in (9) through the differential operator - the MADM [36]-[42] will be utilized. In fact, two algorithms based on MADM will be proposed in this section for the governing model in what follows. 3.1. Algorithm 1 The first algorithm can be summed up in the following bulleted lists: • Applying the inverse operator L−1(.) = ∫ x a ∫ x a (.)dxdx, on (9), one gets y(x) = y(a) + xy′(a) + L−1[g(x)]− L−1[D3/2y(x)]− L−1[y(x)]. (10) Remarkably, we will further put y′(a) = c1; because we do not have the value of this condition at the present, thereafter, the prescribed boundary data will help in getting a hold of the real value of c1. M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 551 • Application of MADM requires that the term −pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] (11) should be added to (10), with p as a synthetic parameter, and ai, i ≥ 0 are obscure coefficients. Moreover, recall that the classical ADM [3]-[38] expresses the solution function y(x) as y(x) = ∑∞ n=0 yn(x). Consequently, one gets ∑∞ n=0 yn(x) = λ1 + c1x− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] + L−1[g(x)] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] . (12) • As a result, one may deduce the resulting recursive scheme from (12) as follows y0(x) = λ1 + c1x+ L−1 [ ∞∑ n=0 anx n ] , y1(x) = L−1[g(x)]− pL−1 [ ∞∑ n=0 anx n ] − L−1[D3/2y0(x)]− L−1[y0(x)], ... yn+1(x) = −L−1[D3/2yn(x)]− L−1[yn(x)], n ⩾ 0. (13) • Computation of the coefficients ai, for i ≥ 0 by taking y1 = 0 and subsequently fixing p = 1 reveals the solution of (8) in form y(x) = y0(x). • Lastly, substituting the values of ai into y0(x), and further upon utilizing the second boundary condition y(1) = λ2, the constant c1 is thus revealed. 3.2. Algorithm 2 Algorithm 2 is equally based on MADM, which possesses the following steps: • We begin by implementing the inverse operator L−1(.) that takes following repre- sentation [19, 31] L−1(.) = ∫ x a dx′ ∫ x′′ a (.)dx′′ − x− a b− a ∫ b a dx ∫ x′′ a (.)dx′′, (14) on (9), which reveals y(x)− y(a)− xy(b) + xy(a) = L−1[g(x)]− L−1[D3/2y(x)]− L−1[y(x)]. (15) M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 552 • MADM requires the addition of the expression given in (11) to (15) to yield the following equation ∑∞ n=0 yn(x) = λ1 + λ2x− λ1x− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] + L−1[g(x)] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] . (16) • Then, the formal recursive relation of the governing model is thus determined in this regard as follows y0(x) = λ1 + (λ2 − λ1)x+ L−1 [ ∞∑ n=0 anx n ] , y1(x) = L−1[g(x)]− pL−1 [ ∞∑ n=0 anx n ] − L−1[D3/2y0(x)]− L−1[y0(x)], ... yn+1(x) = −L−1[D3/2yn(x)]− L−1[yn(x)], n ⩾ 0. (17) Therefore, the values of the coefficients ai, for i ≥ 0 are then calculated by taking y1 = 0 and setting p = 1. Moreover, these values are then substituted into y0(x) to obtain y(x) = y0(x). 4. Applications This section applies the proposed MADM on several test models, featuring various forms of BVP of the Bagley-Torvik equation. Precisely, the proposed MADM through the devised algorithms 1 and 2 will be applied to the governing model and its variant, including the coupled system of the fractional differential equation, which emanates from the Bagley-Torvik equation. Moreover, the test models of interest will be considered from the open literature in order to draw a firm conclusion with well-known exact solutions. Example 1. Consider the BVP for the nonhomogeneous Bagley-Torvik equation as follows [43] D2y(x) +D3/2y(x) + y(x) = x3 + 5x+ 8√ π x3/2, y(0) = 0, y(1) = 0. (18) The exact analytical solution of the above BVP is thus found to be y(x) = x3 − x. Algorithm 1: Write (18) in an operator form as follows Ly = D2y = x3 + 5x+ 8√ π x3/2 −D3/2y(x)− y(x). (19) M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 553 Here, we define L−1(.) = ∫ x 0 ∫ x 0 (.)dxdx to be the inverse operator. Therefore, applying L−1 on both sides of (19) such that y′(0) = c1 gives y(x) = c1x− L−1[D3/2y(x)]− L−1[y(x)] + L−1[x3 + 5x+ 8√ π x3/2]. Then, on using MADM as explained in the procedure, we write ∞∑ n=0 yn(x) = c1x− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] + L−1[x3 + 5x+ 8√ π x3/2], thereby taking y0(x) = c1x+ L−1[a0 + a1x+ a2x 2 + ...], (20) and y1(x) = −pL−1[a0+a1x+a2x 2+...]+L−1[x3+5x+ 8√ π x3/2]−L−1[D3/2y0(x)]−L−1[y0(x)]. (21) Computing the values of y0(x) and y1(x) from (20) and (21), respectively by using L−1 and (6), one gets y0(x) = c1x+ 1 2 a0x 2 + 1 6 a1x 3 + 1 12 a2x 4 + ... y1(x) = −p[ 1 2 a0x 2 + 1 6 a1x 3 + 1 12 a2x 4 + ...] + 1 20 x5 + 5 6 x3 + 32 35 √ π x7/2 − 4c 3 √ π x3/2 − 8a0 15 √ π x5/2 − 16a1 105 √ π x7/2 − 64a2 945 √ π x9/2 − c 6 x3 − a0 24 x4 − a1 120 x5 − a2 360 x6 + .... Now, put y1(x) = 0 and collecting the like terms, we have y1(x) = −p 1 2 a0x 2 + [−p 1 6 a1 + 5 6 − c 6 ]x3 + [−p 1 12 a2 − a0 24 ]x4 + ... = 0. Then, on considering p = 1 and equating the coefficients of xn from the both sides of the above equation, one gains a0 = a2 = 0, and a1 = 5 − c1. Next, on substituting the values of a0, a1 and a2 in y0(x), where yn(x) = 0, for n ⩾ 1, we obtain y(x) = y0(x) = c1x+ 1 6 (5− c1)x 3 + ... Lastly, with the aid of the second boundary prescription, that is, y(1) = 0, we have c1 = −1, which finally yields y(x) = x3 − x, M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 554 that exactly matches the exact analytical solution for (18). Algorithm 2: Applying the inverse operator (14) on the left-hand side of (19), one gets L−1Ly(x) = ∫ x 0 dx′ ∫ x′ 0 d2y dx2 dx′′ − x ∫ 1 0 dx ∫ x′ 0 d2y dx2 dx′′, = y(x)− y(0)− xy(1) + xy(0), = y(x). So, y(x) = L−1[x3 + 5x+ 8√ π x3/2]− L−1[D3/2y(x)]− L−1[y(x)]. Therefore, upon deploying the MADM procedure, we have ∞∑ n=0 yn(x) = L−1[x3 + 5x+ 8√ π x3/2]− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] , which leads to the recurrent scheme as follows y0(x) = L−1 [ ∞∑ n=0 anx n ] , and y1(x) = L−1 [ x3 + 5x+ 8√ π x3/2 ] − pL−1 [ ∞∑ n=0 anx n ] − L−1[D3/2y0(x)]− L−1[y0(x)]. In fact, explicit expression for y0(x) is found to be y0(x) = a0 2 x2 + a1 6 x3 + a2 12 x4 − x (a0 2 + a1 6 + a2 12 ) . Also, on computing the component of y1(x) using the inverse operator (14), one explicitly gets y1(x) = 32 35 √ π x7/2 + 1 20 x5 + 5 6 x3 − (53 60 + 32 35 √ π ) x− p [a0 2 x2 + a1 6 x3 + a2 12 x4 − x (a0 2 + a1 6 + a2 12 )] − 1 945 √ π [ 64a2x 9/2 + 144a1x 7/2 + 504a0x 5/2 − ( 105a2 + 210a1 + 630a0 ) x3/2 ] − 1 210 √ π [82 9 a2 + 44 3 a1 + 28a0 ] x− a2 360 x6 − a1 120 x5 M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 555 − a0 24 x4 + (a2 72 + a1 36 + a0 12 ) x3 − (a2 90 + 7a1 360 + a0 24 ) x. More so, upon collecting the related terms of y1(x) from the above equation at p = 1, we get y1(x) = − 64a2 945 √ π x9/2 + ( 32 35 √ π − 16a1 105 √ π ) x7/2 − 8a0 15 √ π x5/2 + 1 945 √ π ( 105a2 + 210a1 + 630a0 ) x3/2 − a2 360 x6 + ( 1 20 − a1 120 ) x5 − (a2 12 + a0 24 ) x4 + (5 6 + a2 72 − 5a1 36 + a0 12 ) x3 − a0 2 x2 + [13a2 180 + 53a1 360 + 11a0 24 − 1 210 √ π (82a2 9 + 44a1 3 + 28a0 ) − 32 35 √ π + 53 60 ] x. Indeed, on taking y1(x) = 0, and thereafter equating the coefficients of xn, we find a0 = 0, a1 = 6, and a2 = 0. Hence, substituting these values into y0(x), gives y(x) = y0(x) = x3 − x, which matches the exact analytical solution of (18). Example 2. Consider the BVP for the nonhomogeneous Bagley-Torvik equation [29] D3/2y(x) + y(x) = 2x1/2 Γ(3/2) + x2 − x, y(0) = 0, y(1) = 0, (22) which admits the exact analytical solution as follows y(x) = x2 − x. To begin with, we write (22) in an operator form as follows D3/2y(x) = 2x1/2 Γ(3/2) + x2 − x− y(x). (23) Next, upon deploying the inverse operator I3/2 on both sides of (23) using the property (4) on the left-hand side with y′(0) = c1, one gets I3/2[D3/2y(x)] = I3/2 [ 2x1/2 Γ(3/2) + x2 − x ] − I3/2[y(x)], y(x)− 1∑ k=0 yk(0) xk Γ(k + 1) = I3/2 [ 2x1/2 Γ(3/2) + x2 − x ] − I3/2[y(x)], y(x) = c1x+ I3/2 [ 2x1/2 Γ(3/2) + x2 − x ] − I3/2[y(x)]. Using MADM and the properties mentioned in (5) and (7), we write ∞∑ n=0 yn(x) = c1x−pI3/2 [ ∞∑ n=0 anx n ] +I3/2 [ ∞∑ n=0 anx n ] +I3/2 [ 2x1/2 Γ(3/2) +x2−x ] −I3/2[yn(x)], M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 556 upon which we iteratively consider y0(x) = c1x+ I3/2 [ 2x1/2 Γ(3/2) + x2 − x ] + I3/2 [ ∞∑ n=0 anx n ] , = c1x+ x2 + 32 105 √ π x7/2 − 8 15 √ π x5/2 + 4a0 3 √ π x3/2 + 8a1 15 √ π x5/2 + 32a2 105 √ π x7/2 + 64a3 315 √ π x9/2 + ... and y1(x) = −pI3/2 [ ∞∑ n=0 anx n ] − I3/2[y0(x)], = −p [ 4a0 3 √ π x3/2 + 8a1 15 √ π x5/2 + 32a2 105 √ π x7/2 + 64a3 315 √ π x9/2 + ... ] − 8c1 15 √ π x5/2 − 32 105 √ π x7/2 − 1 60 x5 + 1 24 x4 − a0 6 x3 − a1 24 x4 − a2 60 x5 − 3a3 360 x6 + ... Now, setting y1(x) = 0 with p = 1 and collecting the related terms while equating the coefficients of xn in both sides of the above expression, one obtains y1(x) = −a0 6 x3 + ( 1 24 − a1 24 ) x4 − ( 1 60 + a2 60 ) x5 − 3a3 360 x6 + ... = 0. Hence, we acquire a0 = a3 = 0, a1 = 1, and a2 = −1, which when these values of a0, a1, a2 and a3 are substituted in y0(x), where yn(x) = 0, for n ⩾ 1 reveals y(x) = y0(x) = c1x+ x2 + 32 105 √ π x7/2 − 8 15 √ π x5/2 + 8 15 √ π x5/2 − 32 105 √ π x7/2 + ... = c1x+ x2 + ... Lastly, when cancelling the noise terms, alongside using the second boundary prescription of y(1) = 0, one obtains c1 = −1, so that y(x) = x2 − x, which matches the exact analytical solution for (22). Example 3. We make consideration to the BVP for the nonhomogeneous Bagley-Torvik [40] D2y(x) +D3/2y(x) + y(x) = x+ 1, y(0) = 1, y(1) = 2, (24) that admits y(x) = x+ 1 as its exact analytical solution. M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 557 Algorithm 1: We re-write (24) using operator notation as follows Ly(x) = x+ 1−D3/2y(x)− y(x). (25) Applying inverse operator L−1(.) = ∫ x 0 ∫ x 0 (.)dxdx on both sides of (25) such that y′(0) = c1 gives y(x) = 1 + c1x− L−1[D3/2y(x)]− L−1[y(x)] + L−1[x+ 1]. Further, using the MADM procedure along with (6), we gain ∞∑ n=0 yn(x) = 1 + c1x− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] + L−1[x+ 1], upon which the related iterates y0(x) and y1(x) are considered from the latter equation as follows y0(x) = 1 + c1x+ L−1 [ ∞∑ n=0 anx n ] , = 1 + c1x+ 1 2 a0x 2 + 1 6 a1x 3 + 1 12 a2x 4 + ... and y1(x) = L−1[1 + x]− pL−1 [ ∞∑ n=0 anx n ] − L−1 [ D3/2y0(x) ] − L−1[y0(x)], = 1 2 x2 + 1 6 x3 − p [1 2 a0x 2 + 1 6 a1x 3 + 1 12 a2x 4 + ... ] − 4c1 3 √ π x3/2 − 8a0 15 √ π x5/2, − 16a1 105 √ π x7/2 − 64a2 945 √ π x9/2 − 1 2 x2 − c1 6 x3 − a0 24 x4 − a1 120 x5 − a2 360 x6 + ... Next, collecting the like terms, and thereafter equating the coefficients of xn in both sides of the above expression, where y1(x) = 0 with p = 1 gives y1(x) = 1 6 x3 − 1 2 a0x 2 − 1 6 a1x 3 − 1 12 a2x 4 − 4c1 3 √ π x3/2 − 8a0 15 √ π x5/2 − 16a1 105 √ π x7/2 − 64a2 945 √ π x9/2 − c1 6 x3 − a0 24 x4 − a1 120 x5 − a2 360 x6 + ... = 0. M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 558 Therefore, we obtain a0 = 0, a1 = 1−c1, and a2 = 0. In addition, substituting these values in y0(x), where yn(x) = 0, for n ⩾ 1 yields y(x) = y0(x) = 1 + c1x+ 1 6 (1− c)x3. Moreover, on using the subsequent boundary condition y(1) = 2, one acquires c1 = 1, so that y(x) = x+ 1, which is indeed the reported exact analytical solution for BVP (24). Algorithm 2: Here, we make use of the equation represented in an operator notation (25), and further define L−1 as given in (14). Thus, implementing the inverse operator on equation (25) reveals y(x) = x+ 1 + L−1[x+ 1]− L−1[D3/2y(x)]− L−1[y(x)]. Using MADM and (6) ∞∑ n=0 yn(x) = x+ 1 + L−1[x+ 1]− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] . Now, let y0(x) = x+ 1 + L−1 [ ∞∑ n=0 anx n ] , and y1(x) = L−1[x+ 1]− pL−1 [ ∞∑ n=0 anx n ] − L−1[D3/2y0(x)]− L−1[y0(x)], then, we precisely compute the expressions for y0(x) and y1(x) via the inverse operator (14) as follows y0(x) = x+ 1 + a0 2 x2 + a1 6 x3 + a2 12 x4 − x (a0 2 + a1 6 + a2 12 ) , and y1(x) = 1 2 x2 + 1 6 x3 − 2 3 x− p [a0 2 x2 + a1 6 x3 + a2 12 x4 − x (a0 2 + a1 6 + a2 12 )] − 1 945 √ π [ 504a0x 5/2 + 144a1x 7/2 + 64a2x 9/2 − ( 630a0 + 210a1 + 105a2 − 1260 ) x3/2 ] − 1 210 √ π [ 28a0 + 44 3 a1 + 82 9 a2 − 280 ] x M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 559 − a0 24 x4 − a1 120 x5 − a2 360 x6 + (a0 12 + a1 36 + a2 72 − 1 6 ) x3 − 1 2 x2 − (a0 24 + 7a1 360 + a2 90 − 2 3 ) x. In fact, the explicit expression for y1(x) will be found after collecting the like terms from the equation at p = 1 as follows y1(x) = − 64a2 945 √ π x9/2 − 16a1 105 √ π x7/2 − 8a0 15 √ π x5/2 + 1 945 √ π ( 630a0 + 210a1 + 105a2 − 1260 ) x3/2 − a2 360 x6 − a1 120 x5 − (a2 12 + a0 24 ) x4 + (a2 72 − 5a1 36 + a0 12 ) x3 − a0 2 x2 + [13a2 180 + 53a1 360 + 11a0 24 − 1 210 √ π (82a2 9 + 44a1 3 + 28a0 − 280 )] x. Finally, setting y1(x) = 0, and thereafter equating the coefficients of xn, we obtain a0 = a1 = 0 = a2 = 0. We, therefore, obtain y(x) = y0(x) = x+1 as the aiming solution, which is indeed the reported exact analytical solution for (24). Example 4. We consider the BVP for the nonhomogeneous Bagley-Torvik equation as follows [35] D2y(x) +D3/2y(x) + y(x) = x2 + 2 + 4 √ x π , y(0) = 0, y(5) = 25, (26) that admits y(x) = x2 as its exact analytical solution. Algorithm 1: Illustrating (26) via operator notation, one writes Ly(x) = x2 + 2 + 4 √ x π −D3/2y(x)− y(x). (27) Next, on enforcing the inverse operator L−1(.) = ∫ x 0 ∫ x 0 (.)dxdx on both sides of (27) such that y′(0) = c1, we get y(x) = c1x− L−1[D3/2y(x)]− L−1[y(x)] + L−1 [ x2 + 2 + 4 √ x π ] . Using MADM with (6), we further get ∞∑ n=0 yn(x) = c1x− pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] + L−1 [ x2 + 2 + 4 √ x π ] − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] , M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 560 and thereafter take into account y0(x) and y1(x) as follows y0(x) = c1x+ L−1 [ ∞∑ n=0 anx n ] , = c1x+ 1 2 a0x 2 + 1 6 a1x 3 + 1 12 a2x 4 + ... and y1(x) = −pL−1 [ ∞∑ n=0 anx n ] + L−1 [ x2 + 2 + 4 √ x π ] − L−1 [ D3/2y0(x) ] − L−1[y0(x)], = −p [1 2 a0x 2 + 1 6 a1x 3 + 1 12 a2x 4 + ... ] + 1 12 x4 + x2 + 16 15 √ π x5/2 − 4c1 3 √ π x3/2 − 8a0 15 √ π x5/2 − 16a1 105 √ π x7/2 − 64a2 945 √ π x9/2 − c1 6 x3 − a0 24 x4 − a1 120 x5 − a2 360 x6 + ... More so, collecting the related terms, and subsequently the equating coefficients of xn in both sides of the above expression, where y1(x) = 0 with p = 1, one gets y1(x) = ( − 1 2 a0 + 1 ) x2 − (1 6 a1 + c1 6 ) x3 + ( 1 12 − 1 12 a2 − a0 24 ) x4 + 16 15 √ π x5/2 − 4c1 3 √ π x3/2 − 8a0 15 √ π x5/2 − 16a1 105 √ π x7/2 − 64a2 945 √ π x9/2 + ... = 0. Therefore, we get a0 = 2, a1 = −c1, and a2 = 0. More so, putting these values in y0(x), where yn(x) = 0, for n ⩾ 1, we get y(x) = y0(x) = c1x+ x2 − 1 6 c1x 3. Finally, deploying the second boundary condition y(5) = 25, we get c1 = 0, which leads to the acquisition of y(x) = x2, as the reported exact analytical solution. Algorithm 2: As represented in (27), we define the inverse operator L−1 in (14). Further, we deploy L−1(.) = ∫ x 0 dx′ ∫ x′′ 0 (.)dx′′ − x 5 ∫ 5 0 dx ∫ x′′ 0 (.)dx′′, on both sides of (27) to get y(x) = 5x+ L−1 [ x2 + 2 + 4 √ x π ] − L−1[D3/2y(x)]− L−1[y(x)] Using MADM and (6) ∞∑ n=0 yn(x) = 5x+ L−1 [ x2 + 2 + 4 √ x π ] − pL−1 [ ∞∑ n=0 anx n ] + L−1 [ ∞∑ n=0 anx n ] M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 561 − L−1 [ D3/2 ∞∑ n=0 yn(x) ] − L−1 [ ∞∑ n=0 yn(x) ] . In addition, the latter equation further results in y0(x) = 5x+ L−1 [ ∞∑ n=0 anx n ] , and y1(x) = L−1 [ x2 + 2 + 4 √ x π ] − pL−1 [ ∞∑ n=0 anx n ] − L−1[D3/2y0(x)]− L−1[y0(x)], then, evaluating y0(x) using the inverse operator (14) yields y0(x) = 5x+ a0 2 x2 + a1 6 x3 + a2 12 x4 + a3 20 x5 − x (5a0 2 + 25a1 6 + 125a2 12 + 125a3 4 ) , while y1(x) is found by collecting out the like terms by using the inverse operator (14) and setting p = 1 as follows y1(x) = − a3 840 x7 − a2 360 x6 − ( a1 120 + a3 20 ) x5 + ( 1 12 − a2 12 − a0 24 ) x4 + (125a3 24 + 125a2 72 − 19a1 36 + 5a0 12 − 25 6 ) x3 + ( 1− a0 2 ) x2 − [185 12 + 80 √ 5 15 √ π + 1125a3 14 + 875a2 36 + 575a1 72 + 65a0 24 − 652 6 + 1 3150 √ π ( 2100 √ 5a0 + 5500 √ 5a1 + 51250 √ 5 3 a2 + 643750 √ 5 11 a3 − 105000 √ 5 )] x− 128a3 3465 √ π x11/2 − 64a2 945 √ π x9/2 − 16a1 105 √ π x7/2 + ( 16 15 √ π − 8a0 15 √ π ) x5/2 + 1 10395 √ π ( 433125a3 + 144375a2 + 57750a1 + 34650a0 − 346500 ) x3/2. Moreover, setting y1(x) = 0 while equating the coefficients of xn, we acquire a0 = 2, and a1 = a2 = a3 = 0. Hence, y(x) = y0(x) = x2, which happens to be the reported exact analytical solution of (26). Example 5. Consider the BVP (24) after being transformed into a coupled system via [18] as follows { D1.5y1 = y2, y1(0) = 1, y1(1) = 2, D0.5y2 = −y2 − y1 + 1 + x, y2(0) = 0. (28) Here, we apply the inverse operator on both sides of the equations in (28) to get{ I1.5[D1.5y1] = I1.5[y2], I0.5[D0.5y2] = I0.5[−y2 − y1 + 1 + x]. (29) M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 562 More so, applying the property of the fractional derivative mentioned in (4) on the left- hand sides of the latter equations with y′1(0) = c1, we get{ y1(x) = 1 + c1x+ I1.5[y2], y2(x) = −I0.5[y2]− I0.5[y1] + I0.5[1 + x]. Next, upon using MADM, one gets ∞∑ n=0 y1,n(x) = 1 + c1x+ I1.5[ ∞∑ n=0 y2,n]− pI1.5[ ∞∑ n=0 anx n] + I1.5[ ∞∑ n=0 anx n], ∞∑ n=0 y2,n(x) = −I0.5[ ∞∑ n=0 y2,n]− I0.5[ ∞∑ n=0 y1,n] + I0.5[1 + x]− pI0.5[ ∞∑ n=0 anx n] + I0.5[ ∞∑ n=0 anx n], that expands to  y1,0(x) = 1 + c1x+ I1.5[ ∑∞ n=0 anx n], y1,1(x) = −pI1.5[ ∑∞ n=0 anx n] + I1.5[y2,0], y2,0(x) = I0.5[1 + x] + I0.5[ ∑∞ n=0 anx n], y2,1(x) = −pI0.5[ ∑∞ n=0 anx n]− I0.5[y2,0]− I0.5[y1]. (30) Additionally, upon deploying (7), one can iteratively solve the equations in (30) concur- rently as follows y1,0(x) = 1 + c1x+ 4a0 3 √ π x3/2 + 8a1 15 √ π x5/2 + 32a2 105 √ π x7/2 + 64a3 315 √ π x9/2 + ... y1,1(x) = −p [ 4a0 3 √ π x3/2 + 8a1 15 √ π x5/2 + 32a2 105 √ π x7/2 + 64a3 315 √ π x9/2 + ... ] + 1 2 x2 + 1 6 x3 + a0 2 x2 + a1 6 x3 + a2 12 x4 + a3 20 x5 + ... and y2,0(x) = 2√ π x1/2 + 4 3 √ π x3/2 + 2a0√ π x1/2 + 4a1 3 √ π x3/2 + 16a2 15 √ π x5/2 + 32a3 35 √ π x7/2 + ... y2,1(x) = −p [2a0√ π x1/2 + 4a1 3 √ π x3/2 + 16a2 15 √ π x5/2 + 32a3 35 √ π x7/2 + ... ] − x− 1 2 x2 − a0x− a1 2 x2 − a2 3 x3 − a3 12 x4 − I0.5[y1]. M. Al-Mazmumy, M. Alsulami / Eur. J. Pure Appl. Math, 17 (1) (2024), 546-568 563 Now, to compute the expression for the term I0.5[y1] in the last equation, we have to find y1(x) first. So, let y1,1(x) = 0 with p = 1, and equating coefficients of xn/2, we get a0 = a1 = a2 = a3 = 0. Then, y1(x) = y1,0(x) = 1 + c1x = 1 + x, (31) after using the second boundary condition y1(1) = 2. Further, we get I0.5[y1] = I0.5[1 + x] = 2√ π x1/2 + 4 3 √ π x3/2. Next, substituting I0.5[y1] in y2,1(x), and further letting y2,1(x) = 0 with p = 1, the re- sulting coefficients of xn/2 are equated to obtain a0 = a1 = −1, and a2 = a3 = 0. Hence, substituting these values in y2,0(x), one gets y2(x) = y2,0(x) = 2√ π x1/2 + 4 3 √ π x3/2 − 2√ π x1/2 − 4 3 √ π x3/2 + 0, = 0. (32) We, therefore, conclude from (31) and (32) that the coupled system expressed in (28) admits the following solution y1(x) = 1 + x, y2(x) = 0, (33) which is indeed the exact analytical solution for (24). 5. Conclusion In conclusion, we obtained the exact analytical solution of the fractional Bagley-Torvik equation, as well as a system of fractional Bagley-Torvik equations fitted with Dirichlet boundary data. we introduced two algorithms based on MADM to semi-analytically treat the class of fractional Bagley-Torvik equation endowed with Dirichlet boundary data. The algorithms were founded by the inverse linear operator theorems and infused in MADM to calculate only the first and second components y0 and y1; indeed, this is one of the advantages of MADM over the classical ADM, which computes several components most at times to arrive as the optimal solution. Further, to demonstrate the effectiveness and application of the new algorithms, Maple software 2023 has been used for the computa- tional simulation of different nonhomogeneous BVP for Bagley-Torvik equations, which are indeed more complex than their homogeneous counterparts. 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