EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 864-892 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Homotopy Analysis Method to Solve the Nonlinear System of Volterra Integral Equations and Applying the Genetic Algorithm to Enhance the Solutions Rasha F. Ahmed1, Waleed Al-Hayani1,∗, Abbas Y. Al-Bayati2 1 Department of Mathematics, College of Computer Science and Mathematics, University of Mosul, Mosul, Iraq 2 Department of Mathematics, College of Basic Education, University of Telafer, Mosul, Iraq Abstract. This paper presents the application of the Homotopy Analysis Method (HAM) for solving nonlinear system of Volterra integral equations used to obtain a reasonably approximate solution. The HAM contains the auxiliary parameter h which provides a simple way to adjust and control the convergence region of the solution series. The results show that the HAM is a very effective method as well. The results were compared with the solutions obtained by developing a homotopy analysis method using the genetic algorithm (HAM-GA), considering the residual error function as a fitness function of the genetic algorithm. 2020 Mathematics Subject Classifications: 45D05, 45G15, 68W50 Key Words and Phrases: System of Volterra Integral Equations (SVIEs), Homotopy Analysis Method (HAM), h-curves, Genetic Algorithm (GA). 1. Introduction The integral equations can be used to describe a wide variety of problems in science and engineering, such as the population dynamics, the spread of diseases, some Dirichlet problems in potential theory, electrostatics, mathematical modeling of radioactive equi- librium, particle transport problems of astrophysics and reactor theory, radiative energy and/or heat transfer problems, other general heat transfer problems, oscillation of strings and membranes, the problem of momentum representation in quantum mechanics, etc. However, various additional challenging problems in mechanics, astrophysics, chemistry, and mathematics can also be represented in terms of Volterra integral equations. In ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4693 Email addresses: rasha.20csp146@student.uomosul.edu.iq (Rasha F. Ahmed), waleedalhayani@uomosul.edu.iq (Waleed Al-Hayani), profabbasalbayati@yahoo.com (Abbas Y. Al-Bayati) https://www.ejpam.com 864 © 2023 EJPAM All rights reserved. Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 865 addition, there are several real-world problems where urges naturally develop. A dif- ferential equation, an integral equation, an integro-differential equation, or a system of these equations together can be used to describe problems that either arise naturally (as in population dynamics or many biological applications) or are brought about by a control system [9–11]. A numerical method is frequently required because systems of inte- gral and/or integro-differential equations, particularly nonlinear systems of Volterra inte- gral equations or with variable coefficients, are typically challenging to solve analytically. Therefore, there are many research papers to solve such systems using several analytical methods (see [4, 5, 7, 15, 29]), etc. Many researchers [3, 36, 37] have tried to solve the Volterra integral equation using different analytical and numerical methods. Recently, Younis and Al-Hayani [50] have been used Adomian decomposition method (ADM) for solving fuzzy system of Volterra integro-differential equations. The researchers [39, 40, 46– 49] have been applied different analytical methods such as optimal homotopy asymptotic method (OHAM), homotopy perturbation method (HPM) and Padé approximation tech- nique to solve the mathematical models as (porous rotating disk electrodes, ECE reactions at rotating disk electrodes, magnetohydrodynamic, etc.). So, in this paper, we consider the nonlinear system of Volterra Integral Equations (NLSVIEs) [2]: U (x) = F (x) + ∫ g(x) f(x) K (x, t)R [U (t)] dt, (1) where U (x) = [u1 (x) , u2 (x) , · · · , un (x)]T , F (x) = [f1 (x) , f2 (x) , · · · , fn (x)]T , (2) K (x, t) = [kij (x, t)] , i, j = 1, 2, · · · , n Consider the i− th equation of Eq 1 ui (x) = fi (x) + ∫ g(x) f(x) n∑ j=1 [Ki,j (x, t)R (ui (t))] dt, i = 1, 2, · · · , n (3) where the unknown functions ui (x) ∈ C [a, b] , x ∈ [a, b], the functions fi (x) ∈ C [a, b] , f, g ∈ C [a, b] , a ≤ f (x) ≤ g (x) ≤ b. The non-negative kernel functions Ki,j (x, t) ∈ C [a, b] × C [a, b] , i = 1, 2, · · · , n and R : C [a, b] → C [a, b], are nonlin- ear operator. Therefore, R, the operator, is assumed to satisfy the Lipschitz condition, ∥ R(v1) − R(v2) ∥≤ s ∥ v1 − v2 ∥ for every v1, v2 ∈ C [a, b] and some s > 0. As the norm of the function, we take the supremum norm ∥ v ∥= supx∈Ω |v(x)|, in particular ∥ K ∥= sup(x,t)∈[a,b]×[a,b] |K(x, t)| and ∥ F ∥= supx∈[a,b] |F (x)| Shijun Liao invented the HAM [24–27, 41] by which many types of equations can be resolved [24]. The method by Shijun Liao is mainly derived from topology, and can ap- proach to the exact solution. Some examples of using this method for solving integral equations well discussed [17–19]. However, the researcher was able to prove the conver- gence of HAM [31]. This method has been successfully applied to solve many types of Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 866 linear and nonlinear problems [1, 13, 14, 16, 30, 32, 33, 38, 45], In 2020, researchers were able to solve Two-Dimensional nonlinear Volterra-Fredholm fuzzy integral equations by HAM [15]. Also, others were able to solve Mixed Volterra-Fredholm integro-differential equations using same method [16]. In the last years, the researchers [35, 43] have been applied the HAM to solve the math- ematical models as (porous rotating disk electrodes and ECE reactions at rotating disk electrodes). Finally, Liao [28] published an explanation of how to avoid “small denomina- tor problems” by the HAM. The main objective of the present study is to improve the results obtained from HAM application to the system of Volterra integral equations using the genetic algorithm by considering the residual error function as a fitness function and finding the best values for the constants. In particular, this study demonstrates that for reasonable assumptions, the analyzed equations have both existent and unique solutions. Also, a comparison of the solutions with that obtained by developing a HAM using the genetic algorithm was per- formed considering the residual error function as a fitness function of the genetic algorithm, as shown in section 4 and listed in several examples. 2. Fundamental of the HAM The HAM is applied for solving the operator equations N [u (x)] = 0, x ∈ Ω (4) where N refers to general operator (liner or non-linear), while u (x) is an unknown function. First, we define the homotopy operator H as follows: H (ϕ, q) = (1 − q)L [ϕ (x; q) − u0 (x)] − qhN [ϕ (x; q)] , (5) where q ∈ [0, 1]is an embedding parameter, h is a non-zero auxiliary function (denotes the convergence control parameter [23, 26, 27, 44]), L is an auxiliary linear operator which has the property L (0) = 0, u0 (x) is an initial guess of u (x) and ϕ (x; q) is an unknown function. The zero-order deformation equation of the operator H (ϕ, q) = 0 is considered as follow: (1 − q)L [ϕ (x; q) − u0 (x)] = qhN [ϕ (x; q)] , (6) Obviously, when q=0 we have L [ϕ (x; q) − u0 (x)] = 0, that implies ϕ (x; 0) = u0 (x). While, when q = 1, we have N [ϕ (x; q)] = 0, which leads to ϕ (x; 1) = u (x), where u (x) is the solution of Equation 4. in this case, the variation of parameter q from zero to one coincides with the variation of problem from the initial guess u0 (x) to the non-trivial solution u (x). Using the Maclaurin series for expanding the function ϕ (x; q) with respect the parameter q, we get: ϕ (x; q) = ϕ (x; q) + ∞∑ m=1 1 m! ∂mϕ (x; q) ∂qm ∣∣∣∣ q=0 , (7) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 867 By considering um = 1 m! ∂mϕ (x; q) ∂qm ∣∣∣∣ q=0 , m = 1, 2, 3, · · · (8) Equation 7 can be formulated as follow: ϕ (x; q) = u0 (x) + ∞∑ m=1 um (x) qm, (9) If the series in Equation 9 converge at q = 1 then we obtain the solution u (x) = ∞∑ m=0 um (x). (10) we differentiate both sides of Equation 10 m times with respect to the embedding parameter q in order to determine functions um. Then, we divide the received finding by m!and then setting q = 0. In this way, we deduce the so-called mth order deformation equation (m > 0): L [um (x) −Xmum−1 (x)] = hRm (um−1) (11) where u⃗m−1 = {u0 (x) , u1 (x) , · · · , um−1 (x)} . and Xm = { 0, m ≤ 1 1, m > 1 , (12) and Rm (u⃗m−1 (x)) = 1 (m− 1)! ( ∂m−1 ∂qm−1 N ( ∞∑ i=0 ui (x) qi ))∣∣∣∣∣ q=0 , (13) In case of inability to find the sum of the series in Equation 10, the partial sum was determined of this series. ûn (x) = n∑ m=0 um (x) , (14) is the approximate solution of the considered equation. Choosing convenient amount of the parameter h has a great influence on the region of convergence of the series in Equation 10 and the convergence rate as well [27, 34, 42]. One of the methods for selecting the value of convergence control parameter is Genetic Algorithm (GA). Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 868 3. Description of the Method Consider the operators L and N are defined as follows L (ui) = ui, N (ui) = ui(x) − fi (x) − ∫ g(x) f(x) [∑n j=1Kij (x, t)R(ui (t)) ] dt, i = 1, 2, · · · , n, (15) Let ui,0 ∈ C [a, b], i = 1, 2, · · · , n. In this case, by considering the HAM, we obtain the following formula for function ui,m, i = 1, 2, · · · , n, m ≥ 1: ui,m = χmui,m−1 (x) + hRm (u⃗i,m−1(x)) (16) where χm and Rm are defined as in Equations 12 and 13 respectively. Using the definitions of the respective operators, we obtain Ri,m (u⃗i,m−1, x)) = 1 (m− 1)! ∂m−1 ∂qm−1N ( ∞∑ k=0 ui,k(x)qk )∣∣∣∣∣ q=0 , (17) = 1 (m− 1)! ∂m−1 ∂qm−1  ∞∑ k=1 ui,k(x)qk − fi (x) − ∫ g(x) f(x)  n∑ j=1 Kij (x, t)R ( ∞∑ k=1 uj,k(t)qk ) q=0 , = 1 (m− 1)! ∂m−1 ∂qm−1  ∞∑ k=1 ui,k(x)qk − fi (x) − ∞∑ k=1 ∫ g(x) f(x) n∑ j=1 Kij (x, t)R (ui,k(t)) qkdt  q=0 , = 1 (m− 1)! ( (m− 1)!ui,m−1 (x) − (1 − χm)fi (x) − ∫ g(x) f(x) n∑ j=1 [ Kij(x, t)(m− 1)!R (ui,m−1(t)) ] , = ui,m−1 (x) − 1 − χm (m− 1)! fi (x) − ∫ g(x) f(x)  n∑ j=1 Kij (x, t)R (ui,m−1 (t))  dt. Utilizing the above relation and Equation 3 we get the following formula: u1,i = h u0,i − fi (x) − ∫ g(x) f(x)  n∑ j=1 Ki,j (x, t)R(u0,i (t))  dt  . And form m ≥ 2: Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 869 ui,m (x) = (1 + h)ui,m−1 (x) − h (m− 1)! ∫ g(x) f(x)  n∑ j=1 Ki,j (x, t) ( ∂m−1 ∂qm−1R ( ∞∑ i=0 ui,m−1 (t) qi )) dt. (18) In literature, one can find the expression ∂m−1 ∂qm−1R (∑∞ i=0 ui,m−1 (t) qi ) , calculated for vari- ous nonlinear operators R. Most of these results were collected in monograph [27]. Now, we turn to prove under proper assumptions Equation 6 which has a unique solution. Theorem 1. Consider the system of VIEs as in Equation 3, such that ∥ Kij(x, t) ∥= Mi, and ∥ M ∥= max |Mi| , i, j = 1, 2, · · · , n. If the following condition is satisfied sn ∥ M ∥ (b− a) < 1. (19) Then Equation 3 possess at most one solution. Proof. Suppose that there exit two solutions ui,1 and ui,2, we have ∥ ui,1 − ui,2 ∥= ∥∥∥∥∥ ∫ g(x) f(x)  n∑ j=1 Kij(x, t)(R ( uj,1 (t) ) −R ( uj,2 (t) ) )  dt ∥∥∥∥∥ ≤∥ M ∥ ∫ g(x) f(x) n∑ j=1 ∥ R (ui,1) −R(ui,2) ∥ dt, ≤∥ M ∥ ∫ g(x) f(x) n∑ j=1 ∥ R (ui,1) −R(ui,2) ∥ dt ≤ sn ∥ M ∥ (b− a) < 1 ∥ ui,1 − ui,2 ∥ . Hence, we obtain ( 1 − sn ∥ M ∥ (b− a) ) ≤ 1. (20) So, if the condition 19 is satisfied, then the quality ui,1 = ui,2, i = 1, 2, · · · , n must be true. we now proceed to prove the theorem to ensure that the sum of the given series is the solution to the equation discussed. Theorem 2. Let the functionsui,m, i = 1, 2, · · · , nm ≥ 1, be defined in Equations 17 and 18. then, if (s < 1) and the series ui (x) = ∞∑ m=0 ui,m (x) , i = 1, 2, · · · , n, (21) is convergent, then the sum of this series is the solution of Equation 3. Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 870 Proof. Let the series in Equation 10 is convergent. Then, from the necessary condition for the convergent series, we conclude that for any x ∈ [a, b]: lim m→∞ ui,m (x) = 0, i = 1, 2, · · · , n. Now, let Ti,m has the form: Ti,m = 1 m! ( ∂m ∂qm R ( ∞∑ k=0 ui,k(x)qk ))∣∣∣∣∣ q=0 , i = 1, 2, · · · , n. If R is the contraction mapping and (s < 1), the series in Equation 21 converges to ui (x), then the series ∑∞ m=0 Ti,m converges to R(ui (x)) (see [12]). Utilizing the definition of the operator L we can write n∑ m=1 L (ui,m (x) − χmui,m−1 (x)) = n∑ m=1 (ui,m (x) − χmui,m−1 (x)) , = ui,1 (x) + (ui,2 (x) − ui,1 (x)) + (ui,3 (x) − ui,2 (x)) + · · · + (ui,n (x) − ui,n−1 (x)) = ui,n (x) . Hence ∞∑ m=1 L (ui,m (x) − χmui,m−1 (x)) = lim n→∞ ui,n (x) = 0 From Equation 11 we get h ∞∑ m=1 Ri,m (u⃗i,m−1, x) = ∞∑ m=1 L (ui,m (x) − χmui,m−1 (x)) , i = 1, 2, · · · , n. and since h ̸= 0 thus we have ∞∑ m=1 Ri,m (u⃗i,m−1, x) = 0, i = 1, 2, · · · , n. As a result of some transformations, we successively get 0 = ∞∑ m=1 Ri,m (u⃗i,m−1, x), = ∞∑ m=1 ( 1 (m− 1)! ∂m−1 ∂qm−1 ∞∑ k=1 ui,k (x) qk − fi(x) − ∫ g(x) f(x)  n∑ j=1 Ki,j (x, t)R ( ∞∑ k=1 ui,k(t)qk ) dt ∣∣∣∣∣ q=0 ) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 871 = ∞∑ m=1 ( ui,m−1 (x) − 1 − χm (m− 1)! fi (x) − ∫ g(x) f(x) n∑ j=1 Kij(x, t) [ 1 (m− 1)! ∂m−1 ∂qm−1R ( ∞∑ k=1 ui,k(t)qk )] ∣∣∣∣∣ q=0 dt ) = ∞∑ m=1 ui,m−1 (x) − 1 − χm (m− 1)! fi (x) − ∫ g(x) f(x)  n∑ j=1 Kij(x, t)Ti,m−1(t) ∣∣∣∣∣∣ q=0 dt  , = ∞∑ m=1 ui,m−1 (x) − fi (x) − ∫ g(x) f(x)  n∑ j=1 Kij(x, t) ∞∑ m=1 Ti,m−1(t)  dt, = ui (x) − fi (x) − ∫ g(x) f(x)  n∑ j=1 Kij (x, t)R (ui (t))  dt. 4. Genetic Algorithm (GA) The genetic algorithm is a random search technique used to optimize difficult problems and solve nonlinear systems of equations. Instead of employing deterministic principles, GA makes the use of probabilistic transition rules and manages a population of alternative solutions known as individuals or iteratively evolving chromosomes [22]. The iterations of algorithm are known as ”generations”. A fitness function and genetic operators like repro- duction, crossover, and mutation are used to mimic how solutions evolve [21]. The initial population of a genetic algorithm, as shown in Figure 1, is often random. A chromosome, which is a binary string or a real-valued number, is typically used to represent this popu- lation (mating pool). The objective function, which provides each person a corresponding number called its fitness, measures and evaluates the individual performance. The ob- jective function, which provides each person a corresponding number called its fitness, measures and evaluates the individual performance. Each chromosome’s fitness is evalu- ated, as well as a survival of the fittest approach is used. The fitness of each chromosome is evaluated in this work using the residual error value. A genetic algorithm primarily performs three operations: reproduction, crossover, and mutation. Figure 1 describes the GA operation sequences in detail. 4.1. Basic Steps of Genetic Algorithm [20] Step 1: Use a population of random solutions to initialize the parameter, including crossover rate, mutation rate, number of clusters, and generations. Discover the coding mode. Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 872 Figure 1: Flowchart of genetic algorithm. Step 2: Calculate and assess the fitness function’s value. Step 3: Continue with the crossover and mutation process to create the new cluster. Step 4: Repetition of Step 2 is necessary to get the best result. Here, we use the genetic algorithm to improve the results of using the HAM in the following ways: Algorithm 1 Genetic Algorithm for the best parameters h and λ1, λ2 in system of volterra integral equations Input: Set number of variables (var) , Set upper and lower limit for each variable (ub, lb), Set size of population (a), Set rate of crossover (rc), Set rate of mutation (rm), Set number of iterations (Maxiteration), Set fitness function name Output: solution λ1, λ2 and h Initialization 1: Generate individual feasible solutions randomly with limit boundary 2: Save them in the population p; 3: Find fitness value for each population F Loop until the terminal condition 4: for i = 1 to Max iteration do , Elitism based selection using Rank selection 5: fitness = Sort individual descending to find minimum fitness value. 6: Select the best rc solutions in pop1 and save them in popt according to fitness; 7: end for Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 873 Crossover 8: number of crossover nc = (α− ne)/2 9: for j = 1 to nc do 10: randomly select two solutions XA and XB from Popi; 11: generate XC and XD by Arithmetic crossover to XA and XB; 12: save XC and XD to Popt; 13: end for Mutation 14: for j = 1 to ne do 15: select a solution Xj from Popt; 16: mutate Random Resetting of XJ under the rate rm and generate a new solution X ′ j ; 17: if thenX ′ j is unfeasible 18: update X ′ j with a feasible solution by repairing X ′ j ; 19: end if 20: update Xj with X ′ j in Popt; 21: end for Updating 22: update popi + 1= Popi + popt; Returning the best solution 23: return the best solution X in Pop 5. Applications and numerical results In this section, the HAM applied to obtain an approximate-exact solution for NLSVIEs is displayed in the following three problems. To show the high accuracy of the solution results compared with the exact solution, the maximum absolute errors (MAE) are defined as: ∥ · ∥∞=∥ yExact (xi) − ϕn (xi) ∥∞ Moreover, giving the maximum residual error (MRE), the computations associated with the problems were performed using the Maple 18 package with a precision of 20 digits. Problem 1. Let us consider the following NLSVIEs [6] { u1 (x) = f1 (x) + λ1 ∫ x 0 ( u21 (t) + u32(t) ) dt, u2 (x) = f2 (x) + λ2 ∫ x 0 ( u31 (t) − u22(t) ) dt, (22) where { f1 (x) = ( − 1 10x 10 − 1 5x 5 ) λ1 + x2 f2 (x) = x3, (23) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 874 with the exact solutions uExact1 (x) = x2, uExact2 (x) = x3. (24) To solve 22 and 23 by means of the standard HAM, we choose the initial approximation u1,0 (x) = f1 (x) , u2,0 (x) = f2 (x) , (25) and the linear operator L [ϕ1 (x, q)] = ϕ1 (x, q) , L [ϕ2 (x, q)] = ϕ2 (x, q) . (26) Furthermore, the system 22 suggests that we define the non-linear operator as{ N1 [ϕ1 (x, q) , ϕ2 (x, q)] = ϕ1 (x, q) − f1 (x) − λ1 ∫ x 0 ( ϕ2 1 (t, q) + ϕ3 2 (t, q) ) dt, N2 [ϕ1 (x, q) , ϕ2 (x, q)] = ϕ2 (x, q) − f2 (x) − λ2 ∫ x 0 ( ϕ3 1 (t, q) − ϕ2 2 (t, q) ) dt, (27) Using the above definition, we construct the zeroth-order deformation equation as in 6 and 7 and the mth-order deformation equation for m ≥ 1 which is{ L [u1,m (x) − χmu1,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , L [u2,m (x) − χmu2,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , (28) where{ R1,m (u⃗1,m−1, u⃗2,m−1) = u1,m−1 (x) − f1 (x) − λ1 ∫ x 0 ( u21,m−1 (t) + u32,m−1 (t) ) dt, R2,m (u⃗1,m−1, u⃗2,m−1) = u2,m−1 (x) − f2 (x) − λ2 ∫ x 0 ( u31,m−1 (t) − u22,m−1 (t) ) dt, (29) Now, for m ≥ 1, the solutions of the mth-order deformation Equation 29 are{ u1,m (x) = χmu1,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , u2,m (x) = χmu2,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , (30) Now, we successively obtain u1,1 (x) = − 1 2100 hλ3 1x 21 − 1 400 hλ3 1x 16 + 1 65 hλ2 1x 13 − 1 275 hλ3 1x 11 + 1 20 hλ2 1x 8 − 1 5 hλ1x 5 − 1 10 hλ1x 10 u2,1 (x) = 1 31000 hλ3 1λ2x 31 + 3 13000 hλ3 1λ2x 26 − 3 2300 hλ2 1λ2x 23 + 1 1750 hλ3 1λ2x 21 − 1 150 hλ2 1λ2x 18 + 1 50 hλ1λ2x 15 + 1 2000 hλ3 1λ2x 16 − 3 325 hλ2 1λ2x 13 + 3 50 hλ1λ2x 10 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 875 ... Thus, the approximate solution in a series form is given by u1 (x) = u1,0 (x) + 6∑ i=1 u1,i(x), u2 (x) = u2,0 (x) + 6∑ i=1 u2,i(x), This series has the closed form as m → ∞ u1 (x) = x2, u2 (x) = x3. which are the exact solutions of the Problem 1. Tables 1 and 2 show a comparison of the numerical results with the errors applying the standard HAM (m = 6) and the numerical results applying the HAM developed by genetic algorithm HAM-GA with the exact solutions 24 within the interval 0 ≤ x ≤ 1 for various values of λ1, λ2and h. Tables 3 and 4, we list the MAE and the MRE by the HAM and HAM-GA on the interval [0, 1] for various values of λ1, λ2 and h. Table 5 gives the errors on the interval h-curves [−1.2,−0.8] when λ1 = λ2 = 1. Table 1: Numerical results when λ1 = λ2 = 1, for (Problem 1) x i uExacti (x) HAM h=-1 AE HAM-GA h=-0.98213 AE 0.1 1 0.010 0.0100000000 2.113E-24 0.0099999999 3.899E-15 2 0.001 0.0010000000 5.828E-26 0.0009999999 3.113E-18 0.3 1 0.090 0.0899999999 7.314E-15 0.0899999999 4.325E-12 2 0.027 0.0269999999 1.689E-15 0.0269999999 4.316E-13 0.5 1 0.250 0.2499999997 2.022E-10 0.2499999982 1.763E-09 2 0.125 0.1249999998 1.166E-10 0.1249999992 7.596E-10 0.7 1 0.490 0.4899997951 2.048E-07 0.4899995366 4.633E-07 2 0.343 0.3429998134 1.865E-07 0.3429996197 3.802E-07 0.9 1 0.810 0.8099446465 5.535E-05 0.8099235084 7.649E-05 2 0.729 0.7289423171 5.768E-05 0.7289244930 7.550E-05 Table 2: Numerical results when λ1 = 0.5, λ2 = 1.33838729 for (Problem 1) x i uExacti (x) HAM h=-1 AE HAM-GA h=-0.98213 AE 0.1 1 0.010 0.010000000 3.302E-26 0.009999999 1.884E-15 2 0.001 0.001000000 2.305E-27 0.000999999 2.043E-18 0.3 1 0.090 0.089999999 1.152E-16 0.089999999 1.040E-12 2 0.027 0.026999999 5.883E-17 0.026999999 1.718E-13 0.5 1 0.250 0.249999999 3.432E-12 0.249999999 1.340E-10 2 0.125 0.124999999 3.983E-12 0.124999999 9.760E-11 0.7 1 0.490 0.489999995 4.232E-09 0.489999979 2.034E-08 2 0.343 0.342999990 9.081E-09 0.342999975 2.470E-08 0.9 1 0.810 0.809999004 9.954E-07 0.809996725 3.274E-06 2 0.729 0.728994536 5.463E-06 0.728993970 6.029E-06 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 876 Table 3: MAE and MRE of HAM and HAM-GA for (Problem 1) HAM λ1 = λ2 = 1h = −1 HAM-GA λ1 = λ2 = 1, h = −0.98213 m i MAE MRE MAE MRE 2 1 5.877E-02 1.308E-02 6.308E-02 1.434E-02 2 6.413E-02 1.935E-02 6.298E-02 1.858E-02 3 1 2.574E-02 6.445E-03 2.698E-02 6.684E-03 2 1.335E-02 3.409E-03 1.513E-02 3.957E-03 4 1 5.897E-03 1.379E-03 6.974E-03 1.651E-03 2 6.426E-03 1.897E-03 6.839E-03 1.990E-03 5 1 2.390E-03 6.002E-04 2.672E-03 6.635E-04 2 1.947E-03 5.564E-04 2.273E-03 6.527E-04 6 1 6.738E-04 1.679E-04 8.337E-04 2.078E-04 2 7.095E-04 2.088E-04 8.304E-04 2.429E-04 Table 4: MAE and MRE of HAM-GA for (Problem 1) HAM-GA λ1 = 0.5, λ2 = 1.33838, h = −1 HAM-GA λ1 = 0.5, λ2 = 1.33838, h = −0.98213 m i MAE MRE MAE MRE 2 1 1.551E-02 3.520E-03 1.792E-02 4.297E-03 2 4.800E-02 1.655E-02 4.714E-02 1.596E-02 3 1 6.088E-03 1.764E-03 6.465E-03 1.833E-03 2 1.283E-04 3.189E-04 1.772E-03 3.028E-04 4 1 3.162E-04 5.617E-05 6.297E-04 1.475E-04 2 2.101E-03 6.938E-04 2.040E-03 6.561E-04 5 1 3.177E-04 9.251E-05 3.286E-04 9.307E-05 2 1.741E-06 1.322E-05 1.426E-04 3.460E-05 6 4.388E-06 2.567E-07 3.172E-05 7.968E-06 9.404E-05 3.052E-05 9.224E-05 2.888E-05 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 877 Table 5: Errors on the interval h-curves for (Problem 1) when λ1 = λ2 = 1 x i h=-1.2 h=-0.98213 h=-1 h=-0.8 -0.9 1 2.260E-03 1.646E-05 3.642E-05 5.252E-06 2 3.258E-03 5.517E-05 9.355E-05 2.752E-06 -0.7 1 9.179E-05 5.215E-08 2.007E-07 1.512E-06 2 7.466E-05 9.922E-08 3.115E-07 2.430E-06 -0.5 1 4.730E-06 2.689E-12 2.133E-10 1.005E-06 2 1.023E-06 3.624E-12 1.693E-10 2.430E-07 -0.3 1 1.895E-07 9.567E-14 7.451E-15 1.354E-07 2 3.866E-09 4.992E-14 2.062E-15 2.089E-09 -0.1 1 6.448E-10 3.398E-15 2.115E-24 6.368E-10 2 5.785E-14 2.886E-18 6.186E-26 3.829E-14 0.1 1 6.352E-10 3.899E-15 2.113E-24 6.432E-10 2 5.736E-14 3.113E-18 5.828E-26 3.849E-14 0.3 1 1.267E-10 4.325E-12 7.314E-15 1.778E-07 2 3.074E-09 4.316E-13 1.689E-15 2.412E-09 0.5 1 7.503E-07 1.763E-09 2.022E-10 3.644E-06 2 3.825E-07 7.596E-10 1.166E-10 4.909E-07 0.7 1 2.699E-06 4.633E-07 2.048E-07 5.064E-05 2 9.241E-06 3.802E-07 1.865E-07 2.237E-05 0.9 1 1.872E-04 7.649E-05 5.535E-05 8.640E-04 2 2.275E-04 7.550E-05 5.768E-05 6.502E-04 In Figures 2 and 3, an illustration shows how well the exact solution matches with the approximate solution using the genetic algorithm h=-0.98213 with λ1 = λ2 = 1. In Figures 4 and 5 present the h-curves for (Problem 1) when λ1 = λ2 = 1. We present the error residual ER (ui (x)) , i = 1, 2 of HAM-GA when λ1 = 0.5, λ2 = 1.33838, h = −0.98213 in Figures 6 and 7. The CPU time = 1.188 Seconds. Figure 2: Line : uExact1 (x) , o : HAM −GA1(x) Figure 3: Line : uExact2 (x) , o : HAM −GA2(x) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 878 Figure 4: h-curve for u′ 1 (0.4, h) Figure 5: h-curve for u′ 2 (0.4, h) Figure 6: ER (u1 (x)) of HAM-GA Figure 7: ER (u2 (x)) of HAM-GA Problem 2. Let us consider the following NLSVIEs [8] { u1 (x) = f1 (x) + λ1 ∫ x 0 ( u21 (t) + u22(t) ) dt, u2 (x) = f2 (x) + λ2 ∫ x 0 u1(t)u2(t)dt,, (31) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 879 where { f1 (x) = sin (x) − λ1x, f2 (x) = cos (x) − 1 2λ2sin 2(x), (32) with the exact solutions uExact1 (x) = sin (x), uExact2 (x) = cos (x), (33) To solve 31 and 32 by means of the standard HAM, we choose the initial approximation u1,0 (x) = f1 (x) , u2,0 (x) = f2 (x) , (34) and the linear operator L [ϕ1 (x, q)] = ϕ1 (x, q) , L [ϕ2 (x, q)] = ϕ2 (x, q) , (35) Furthermore, the system 31 suggests that we define the non-linear operator as{ N1 [ϕ1 (x, q) , ϕ2 (x, q)] = ϕ1 (x, q) − f1 (x) − λ1 ∫ x 0 ( ϕ2 1 (t, q) + ϕ2 2 (t, q) ) dt, N2 [ϕ1 (x, q) , ϕ2 (x, q)] = ϕ2 (x, q) − f2 (x) − λ2 ∫ x 0 (ϕ1 (t, q)ϕ2 (t, q)) dt, (36) Using the above definition, we construct the zeroth-order deformation equation as in 6 and 7 and the mth-order deformation equation for m ≥ 1 is{ L [u1,m (x) − χmu1,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , L [u2,m (x) − χmu2,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , (37) where{ R1,m (u⃗1,m−1, u⃗2,m−1 ) = u1,m−1 (x) − f1 (x) − λ1 ∫ x 0 ( u21,m−1 (t) + u22,m−1 (t) ) dt, R2,m (u⃗1,m−1, u⃗2,m−1) = u2,m−1 (x) − f2 (x) − λ2 ∫ x 0 (u1,m−1 (t)u2,m−1 (t)) dt,, (38) Now, for m ≥ 1, the solutions of the mth-order deformation Equation 38 are{ u1,m (x) = χmu1,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , u2,m (x) = χmu2,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1)] , (39) we now successively obtain u1,1 (x) = −1 3 hλ3 1x 3 − 2hλ2 1x cosx + 2hλ2 1 sinx− hλ1x + 1 16 hλ1λ 2 2 sin3 x cosx + 3 32 hλ1λ 2 2 sinx cosx− 3 32 hλ1λ 2 2x + 1 3 hλ1λ2 sin3 x u2,1 (x) = −hλ1λ2 + 1 3 hλ2 2 + 1 4 hλ1λ 2 2x sinx cosx− 1 8 hλ1λ 2 2x 2 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 880 −1 8 hλ1λ 2 2 sin2 x + hλ1λ2x sinx + hλ1λ2 cosx −1 6 hλ2 2 sin2 x cosx− 1 3 hλ2 2 cosx− 1 2 hλ2 sin2 x ... Thus, the approximate solution in a series form is given by u1 (x) = u1,0 (x) + 5∑ i=1 u1,i(x), u2 (x) = u2,0 (x) + 5∑ i=1 u2,i(x), This series has the closed form as m → ∞ u1 (x) = sinx, u2 (x) = cosx. which are the exact solutions of the Problem 2. Tables 6, 7 show a comparison of the numerical results with the errors applying the standard HAM (m = 5) and the numerical results applying the HAM developed by genetic algorithm HAM-GA with the exact solutions 33 within the interval 0 ≤ x ≤ 0.5 for various values of λ1, λ2 and h. Tables 8, 9, we list the MAE and the MRE by the HAM and HAM- GA on the interval [0, 0.5] for the various values of λ1, λ2 and h. Table 10 gives the errors on the interval h-curves [−1.2,−0.8] when λ1 = λ2 = 1. Table 6: Numerical results when λ1 = λ2 = 1, for (Problem 2) x i uExacti (x) HAM h = −1 AE HAM-GA h=-0.98401 AE 0.1 1 0.099833416 0.099829110 4.306E-06 0.099828113 5.303E-06 2 0.995004165 0.995003990 1.745E-07 0.995002111 2.054E-06 0.3 1 0.295520206 0.294524648 9.955E-04 0.294500710 1.019E-03 2 0.955336489 0.955219396 1.170E-04 0.955085971 2.505E-04 0.5 1 0.479425538 0.467847859 1.157E-02 0.467776042 1.164E-02 2 0.877582561 0.875456892 2.125E-03 0.874603732 2.978E-03 Table 7: Numerical results when λ1 = 0.5, λ2 = 0.5, for (Problem 2) x i uExacti (x) HAM h = −1 AE HAM-GA h=-0.98401 AE 0.1 1 0.099833416 0.099833280 1.356E-07 0.099833132 2.841E-07 2 0.995004165 0.995004155 9.445E-09 0.995004017 1.473E-07 0.3 1 0.295520206 0.295486811 3.339E-05 0.295479223 4.098E-05 2 0.955336489 0.955329938 6.550E-06 0.955320666 1.582E-05 0.5 1 0.479425538 0.478989428 4.361E-04 0.478926206 4.993E-04 2 0.877582561 0.877455542 1.270E-04 0.877391282 1.912E-04 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 881 Table 8: MAE and MRE of HAM and HAM-GA for (Problem 2) HAM λ1 = λ2 = 1 h = −1 HAM-GA λ1 = λ2 = 1, h = −0.98401 m i MAE MRE MAE MRE 2 1 7.494E-02 1.545E-02 8.174E-02 1.899E-02 2 1.170E-01 5.382E-02 1.170E-01 5.295E-02 3 1 7.716E-02 3.207E-02 7.720E-02 3.158E-02 2 1.534E-02 2.581E-03 1.857E-02 4.154E-03 4 1 1.114E-02 3.207E-03 1.425E-02 4.529E-03 2 1.577E-02 6.432E-03 1.583E-02 6.287E-03 5 1 1.157E-02 4.556E-03 1.164E-02 4.514E-03 2 2.125E-03 4.892E-04 2.978E-03 8.526E-04 6 1 1.642E-03 5.243E-04 2.413E-03 8.310E-04 2 2.204E-03 8.691E-04 2.232E-03 8.547E-04 Table 9: MAE and MRE of HAM-GA for (Problem 2) HAM-GA λ1 = 0.5, λ2 = 0.5, h = −1 HAM-GA λ1 = 0.5, λ2 = 0.5, h = −0.98401 m i MAE MRE MAE MRE 2 1 2.411E-02 8.656E-03 2.772E-02 1.058E-02 2 3.019E-02 1.452E-02 3.063E-02 1.451E-02 3 1 1.051E-02 4.590E-03 1.100E-02 4.749E-03 2 2.735E-03 9.328E-04 3.613E-03 1.351E-03 4 1 1.221E-03 4.665E-04 1.671E-03 6.618E-04 2 1.110E-03 4.649E-04 1.207E-03 4.968E-04 5 1 4.361E-04 1.779E-04 4.993E-04 2.020E-04 2 1.270E-04 4.611E-05 1.912E-04 7.297E-05 6 1 5.861E-05 2.268E-05 9.015E-05 3.552E-06 2 4.302E-05 1.721E-05 5.203E-05 2.049E-06 Table 10: Errors on the interval h-curves for (Problem 2) when λ1 = λ2 = 1 x i h=-1.2 h=-0.98213 h=-1 h=-0.8 -0.5 1 3.319E-02 1.164E-02 1.157E-02 2.387E-02 2 1.640E-02 2.978E-03 2.125E-03 1.287E-02 -0.3 1 6.576E-03 1.019E-03 9.955E-04 4.423E-03 2 3.711E-03 2.505E-04 1.170E-04 2.504E-03 -0.1 1 3.672E-04 5.303E-06 4.306E-06 2.826E-04 2 2.132E-04 2.054E-06 1.745E-07 1.520E-04 0.1 1 3.672E-04 5.303E-06 4.306E-06 2.826E-04 2 2.132E-04 2.054E-06 1.745E-07 1.520E-04 0.3 1 6.576E-03 1.019E-03 9.955E-04 4.423E-03 2 3.711E-03 2.505E-04 1.170E-04 2.504E-03 0.5 1 3.319E-02 1.164E-02 1.157E-02 2.387E-02 2 1.640E-02 2.978E-03 2.125E-03 1.287E-02 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 882 In Figures 8 and 9, an illustration shows how well the exact solution matches with the approximate solution using the genetic algorithm h=-0.98401 with λ1 = λ2 = 1. In Figures 10 and 11 the h-curves present for (Problem 2) when λ1 = λ2 = 1. We present the error residual ER (ui (x)) , i = 1, 2 of HAM-GA when λ1 = λ2 = 0.5, h = −0.98401 in Figures 12 and 13. The CPU time = 2.828 Seconds. Figure 8: Line : uExact1 (x) , o : HAM −GA1(x) Figure 9: Line : uExact2 (x) , o : HAM −GA2(x) Figure 10: h-curve for u′ 1 (0.1, h) Figure 11: h-curve for u′ 2 (0.4, h) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 883 Figure 12: ER (u1 (x)) of HAM-GA Figure 13: ER (u2 (x)) of HAM-GA Problem 3. Finally, let us consider the following NLSVIEs [8]  u1 (x) = f1 (x) + 4 ∫ x 0 u1(t)u2(t)dt, u2 (x) = f2 (x) + ∫ x 0 tu1(t)u 2 3(t)dt, u3 (x) = f3 (x) − 1 3 ∫ x 0 tu2(t)u3(t)dt, (40) where  f1 (x) = −2 ln (x)x2 + x2 + ln (x), f2 (x) = −1 6 ln (x)x6 + 1 36x 6 + x, f3 (x) = 1 15x 5 + x2, (41) with the exact solutions uExact1 (x) = ln (x) , uExact2 (x) = x, uExact3 (x) = x2. (42) To solve 40 and 41 by means of the standard HAM, we choose the initial approximation u1,0 (x) = f1 (x) , u2,0 (x) = f2 (x) , u3,0 (x) = f3 (x) , (43) and the linear operator L [ϕ1 (x, q)] = ϕ1 (x, q) , L [ϕ2 (x, q)] = ϕ2 (x, q) , L [ϕ3 (x, q)] = ϕ3 (x, q) . (44) Furthermore, the system 40 suggests that we define the non-linear operator as N1 [ϕ1 (x, q) , ϕ2 (x, q) , ϕ3 (x, q)] = ϕ1 (x, q) − f1 (x) − 4 ∫ x 0 ϕ1 (t, q)ϕ2 (t, q) dt, N2 [ϕ1 (x, q) , ϕ2 (x, q) , ϕ3 (x, q)] = ϕ2 (x, q) − f2 (x) − ∫ x 0 tϕ1 (t, q)ϕ2 3 (t, q) dt, N3 [ϕ1 (x, q) , ϕ2 (x, q) , ϕ3 (x, q)] = ϕ3 (x, q) − f3 (x) + 1 3 ∫ x 0 tϕ2 (t, q)ϕ3 (t, q) dt. (45) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 884 Using the above definition, we construct the zeroth-order deformation equation as in 6 and 7 and the mth-order deformation equation for m ≥ 1 is L [u1,m (x) − χmu1,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1) , R3,m (u⃗3,m−1)] , L [u2,m (x) − χmu2,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1) , R3,m (u⃗3,m−1)] , L [u3,m (x) − χmu3,m−1 (x)] = h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1) , R3,m (u⃗3,m−1)] . (46) where R1,m (u⃗1,m−1, u⃗2,m−1 , u⃗3,m−1 ) = u1,m−1 (x) − f1 (x) − 4 ∫ x 0 u1,m−1 (t)u2,m−1 (t) dt, R2,m (u⃗1,m−1, u⃗2,m−1, u⃗3,m−1 ) = u2,m−1 (x) − f2 (x) − ∫ x 0 tu1,m−1 (t)u23,m−1 (t) dt, R3,m (u⃗1,m−1, u⃗2,m−1, u⃗3,m−1 ) = u3,m−1 (x) − f3 (x) + 1 3 ∫ x 30 tu2,m−1 (t)u3,m−1 (t) dt. (47) Now, for m ≥ 1, the solutions of the mth-order deformation Equation 47 are u1,m (x) = χmu1,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1) , R3,m (u⃗2,m−1)] , u2,m (x) = χmu2,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1) , R3,m (u⃗2,m−1)] , u3,m (x) = χmu3,m−1 (x) + h [R1,m (u⃗1,m−1) , R2,m (u⃗2,m−1) , R3,m (u⃗2,m−1)] , (48) we now successively obtain u1,1 (x) = h ( −2ln (x)x2 + x2 − 4 27 ln2 (x)x9 + 32 243 ln(x)x9 − 59 2187 x9 + 2 ln (x)x4 − 3 2 x4 ) . u2,1 (x) = h ( −1 6 ln(x)x6 + 1 36 x6 + 1 1575 ln(x)x14 − 4 11025 x14 − 1 2700 ln(x)x12 + 1 32400 x12 + 4 165 ln(x)x11 − 26 1815 x11 − 2 135 ln(x)x9 + 2 1215 x9 + 1 4 ln(x)x8 − 5 32 x8 ) , u3,1 (x) = h ( 1 15 x5 − 1 3510 ln (x)x13 + 19 273780 x13 − 1 180 ln (x)x10 + 1 675 x10 + 1 360 x8 ) , ... Thus, the approximate solution in a series form is given by u1 (x) = u1,0 (x) + 6∑ i=1 u1,i(x), u2 (x) = u2,0 (x) + 6∑ i=1 u2,i(x), Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 885 u3 (x) = u3,0 (x) + 6∑ i=1 u3,i(x). This series has the closed form m → ∞ u1 (x) = ln(x), u2 (x) = x, u3 (x) = x2. which are the exact solutions of the Problem 3. Table 11 shows a comparison of the numerical results with the errors applying the standard HAM (m=6) and the numerical results applying the HAM developed by genetic algorithm (HAM-GA) with the exact solutions 42 within the interval 0 ≤ x ≤ 0.6. Table 12 the MAE is tabulated for different frequencies as well as the MRE. Table 13 gives the errors on the interval h-curves [−1.2,−0.8]. Table 11: Numerical results for (Problem 3) x i uExacti (x) HAM h = −1 AE HAM-GA h=-1.01064 AE 0.1 1 -2.30258509 -2.302585092 3.094E-13 -2.302585092 3.734E-13 2 0.10000000 0.1000000000 5.644E-18 0.099999999 1.234E-17 3 0.01000000 0.0099999999 2.932E-21 0.009999999 9.274E-17 0.3 1 -1.20397280 -1.203972712 9.189E-08 -1.203972808 4.117E-09 2 0.30000000 0.300000000 1.477E-10 0.300000000 5.030E-12 3 0.09000000 0.089999999 7.052E-13 0.089999999 9.033E-14 0.5 1 -0.69314718 -0.693128726 1.845E-05 -0.693144508 2.672E-06 2 0.50000000 0.500000343 3.436E-07 0.500000159 1.596E-07 3 0.25000000 0.249999995 4.934E-09 0.249999997 2.706E-09 Table 12: MAE and MRE of HAM and HAM-GA for (Problem 3) HAMh = −1 HAM-GA h = −1.01064 m i MAE MRE MAE MRE 2 1 3.266E-01 1.102E-01 3.223E-01 1.087E-01 2 4.773E-03 1.241E-03 4.768E-03 1.251E-03 3 7.305E-05 2.150E-05 1.289E-04 4.453E-05 3 1 8.783E-02 2.910E-02 8.276E-02 2.744E-02 2 1.582E-03 4.708E-04 1.514E-03 4.544E-04 3 1.783E-05 4.884E-06 1.606E-05 4.295E-06 4 1 1.626E-02 5.420E-03 1.404E-02 4.690E-03 2 3.342E-04 1.057E-04 2.950E-04 9.430E-05 3 4.897E-06 1.510E-06 4.505E-06 1.409E-06 5 1 2.020E-03 6.799E-04 1.452E-03 4.914E-04 2 5.048E-05 1.679E-05 3.905E-05 1.319E-05 3 8.725E-07 2.839E-07 7.070E-07 2.332E-07 6 1 8.969E-05 3.016E-05 2.672E-06 9.493E-07 2 5.256E-06 1.880E-06 3.111E-06 1.169E-06 3 1.138E-07 3.870E-08 7.714E-08 2.675E-08 Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 886 Table 13: Errors on the interval h-curves for (Problem 3) x i h=-1.2 h=-1.01064 h=-1 h=-0.8 -0.5 1 1.758E-04 1.337E-04 1.806E-04 9.697E-03 2 2.235E-06 8.834E-07 1.363E-06 8.807E-05 3 5.742E-07 9.568E-09 1.562E-08 1.219E-06 -0.3 1 4.996E-05 4.619E-08 2.375E-07 8.175E-04 2 6.683E-08 4.114E-11 3.324E-10 8.066E-07 3 5.017E-08 1.777E-13 1.546E-12 5.382E-08 -0.1 1 1.921E-05 8.053E-13 4.255E-13 3.286E-05 2 1.200E-10 1.517E-17 8.126E-18 2.866E-10 3 2.130E-10 8.919E-17 4.310E-21 2.135E-10 0.1 1 1.255E-05 3.734E-13 3.094E-13 2.206E-05 2 6.870E-11 1.234E-17 5.644E-18 1.815E-10 3 2.136E-10 9.274E-17 2.932E-21 2.131E-10 0.3 1 2.885E-05 4.117E-09 9.189E-08 4.264E-04 2 3.622E-08 5.030E-12 1.477E-10 3.910E-07 3 5.355E-08 9.033E-14 7.052E-13 4.991E-08 0.5 1 8.557E-05 2.672E-06 1.845E-05 4.074E-03 2 1.003E-06 1.596E-07 3.436E-07 3.433E-05 3 7.582E-07 2.706E-09 4.934E-09 3.348E-07 In Figures 14-16 an illustration of how well the exact solution matches with the ap- proximate solution using the genetic algorithm h=-1.01064. In Figures 17-19 present the h-curves for (Problem 3). We present the error residual ER (ui (x)) , i = 1, 2, 3 of HAM- GA when h = −1.01064 in Figures 20-22. The CPU time = 10.890 Seconds. Figure 14: Line : uExact1 (x) , o : HAM −GA1(x) Figure 15: Line : uExact2 (x) , o : HAM −GA2(x) Rasha F. Ahmed, W. Al-Hayani, A. Y. Al-Bayati / Eur. J. Pure Appl. Math, 16 (2) (2023), 864-892 887 Figure 16: Line : uExact3 (x) , o : HAM −GA3(x) Figure 17: h-curve for u′ 1 (0.15, h) Figure 18: h− curveforu′ 2 (0.15, h) Figure 19: h− curveforu′ 3 (0.15, h) Figure 20: ER (u1 (x)) of HAM-GA Figure 21: ER (u2 (x)) of HAM-GA REFERENCES 888 Figure 22: ER (u3 (x)) of HAM-GA 6. Conclusion The present study suggests a new algorithm (HAM-GA) merging both the HAM and the GA for solving the nonlinear system of Volterra integral equations. Based on four consecutive cases, the algorithm determines the solution. The algorithm deems the residual error function as a fitness function and based on which the optimum value of λ1, λ2 and h are selected. In the first case, the results were calculated according to the standard HAM, in the second case, the best value for h were chosen according to the genetic algorithm and based on which the results were calculated using HAM-GA. As for the third case, the best λ1and λ2 for the Volterra system of integral equations were chosen, then the results were calculated using HAM-GA. Finally, the results were calculated using HAM-GA depending on optimal λ1, λ2 and h. The results obtained using the best h were better than those obtained using the standard HAM. As for the results for the fourth case, they were optimal for the exact solution. The findings are in good agreement with the h-curves indicating that the proposed algorithm is successful in finding the solution. References [1] Hamdy R Abdl-Rahim, Mohra Zayed, and Gamal M Ismail. 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