# Optimal Homotopy Asymptotic and Multistage Optimal Homotopy Asymptotic Methods for Solving System of Volterra Integral Equations of the Second Kind

Authors: Jafar Biazar; Roya Montazeri
Journal: Journal of Applied Mathematics
 (2019)

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## Abstract

In this paper, optimal homotopy asymptotic method (OHAM) and its implementation on subinterval, called multistage optimal homotopy asymptotic method (MOHAM), are presented for solving linear and nonlinear systems of Volterra integral equations of the second kind. To illustrate these approaches two examples are presented. The results confirm the efficiency and ability of these methods for such equations. The results will be compared to find out which method is more accurate. Advantages of applying MOHAM are also illustrated.


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## Body

## 1. Introduction

Integral equations, differential equations, integrodifferential equations, and system of such equations, linear and nonlinear, usually appeared in mathematical modeling of different phenomena in physics, biology, and engineering [1–4].There are various numerical methods for finding approximate solutions of a system of integral equations and a system of integrodifferential equations such as differential transform method [5], Adomian decomposition method [6], modified homotopy perturbation method [7–9], homotopy perturbation method [10], block by block method [11], Galerkin method [12], rationalized Haar functions method [2, 3], Runge-Kutta method [13], homotopy analysis method [14], Tau method [15], and variational iteration method [16].Moreover, Sezar et al. applied Chebyshev polynomial method and Taylor collocation method for systems of linear differential equations and integrodifferential equations [1, 17]. Yusufoglu employed homotopy pertubation method to solve a system of Fredholm–Volterra type integral equations [18].One of the powerful and efficient methods for solving integral equations is OHAM. In this paper, we apply OHAM to solve systems of integral equations of the second kind. We will also consider a modified version of OHAM, that is called multistage optimal homotopy asymptotic method (MOHAM). This approach was introduced for the first time by Anakira et al. to approximate the solutions of differential equations with initial-values [19].The organization of this research is as follows: in Section2, OHAM and MOHAM are introduced. In Sections 3 and 4, applications of OHAM and MOHAM to system of Volterra integral equations of the second kind are explained, respectively. Section 5 is devoted to proving the convergence of OHAM. In section six, illustrative examples are presented, and conclusion appeared in the last section.


## 2. OHAM and MOHAM

The OHAM approach is usually applied to solve boundary value problems; say(1)Liuit+fit+Niuit=0,Bui,duidt=0,i=1,2,…,n,where Li and Ni are linear and nonlinear functional operators, respectively. fit is a known function, uit is an unknown function, and B is a boundary operator [20–24].According to OHAM we construct a homotopyφiuit,P;P:R×0,1→R for (1) as follows: (2)1-PLiuit,P+fit=HPLiuit,P+fit+Niuit,P,Buit,P,duit,Pdt=0,where P∈0,1 is an embedding parameter; HP, for P≠0, is a nonzero auxiliary function such that H0=0. For P=0 and P=1 we have uit,0=ui0t, and uit,1=uit, respectively. Thus, as P increases from 0 to 1, the solution ut,P varies from u0t to the solution ut, where u0t is an initial guess, for the solution, that satisfies the linear operator which is obtained from (2), for P=0;(3)Liui0t+fit=0,Bui0,dui0dt=0.The auxiliary function HP is considered as the following power series in P:(4)HP=C1P+C2P2+⋯,where C1, C2,… are constants that will be determined later.Let us consider the approximate solution,uit;P,C1,C2,…, as a power series about P(5)uit;P,C1,C2,…=ui0t+∑k≥1uikt,C1,C2,…,CkPk,i=1,2,…,n.Substitution in (2) from (5) and equating the coefficients of the terms with identical powers of P lead to governing equations of ui0t, ui1t,…,uikt, for i=1,2,…,n, which starts from (3), followed by(6)Liui1t=C1Ni0u0t,Bui1,dui1dt=0,(7)Liuikt-uik-1t=CkNi0ui0t+∑l=1k-1CiLiuik-lt+Nik-lui0t,ui1t,…,uik-lt,Buik,duikdt=0,k=2,3,…where Nmu0t,u1t,…,umt is the coefficient of Pm in the expansion of Nut;P,C1,…,Cm about the embedding parameter P(8)Niuit;P=Niui0t+∑k≥1Nikui0t,ui1t,…,uiktPk,i=1,2,…,nwhere uit;P is given by (5).Studying the rate of convergence of the series (5) depends upon the auxiliary constants Cq, q=1,2,3,…. If the series (5) converges for P=1, one has (9)uit,C1,C2,…=ui0t+∑k≥1uikt,C1,C2,…,Ck,i=1,2,…,n.Then the mth order approximation is as follows: (10)u~it,C1,C2,…,Cm=ui0t+∑k=1muikt,C1,C2,…,Ck,i=1,2,…,n.Substitution of (10) into (1) results in the following expression for residual: (11)Rit,C1,C2,…,Cm=Lu~it,C1,C2,…,Cm+fit+Niu~it,C1,C2,…,Cm,i=1,2,…,n.If Rit,C1,C2,…,Cm=0, then u~it,C1,C2,…,Cm will be an exact solution and this, in general, does not happen especially in nonlinear problems. In order to find the optimal values of Cq, q=1,2,3,…m, we apply least square minimization approach (12)∂Ji∂C1=∂Ji∂C2=⋯=∂Ji∂Cm=0,where(13)JiC1,C2,…,Cm=∫abRi2t,C1,C2,…,Cmdt,i=1,2,…,n,where a and b are two values, depending on the given problem. Knowing Cq, q=1,2,3,…m, from (12), the approximate solution of order m will be determined easily.If the interval of changes of the time variable is long, then OHAM fails to reach accurate solutions.MOHAM overcomes this shortcoming by partitioning the time interval,t0,T, into N subintervals t0,t1,…,tγ-1,tγ, where tγ=T and OHAM will be applied over each subintervals. The solution at the last point, in each subinterval, denotes an initial approximation to the solution, over the next interval. The process will continue until we achieve the preassigned time, TImplementation of MOHAM is almost the same as OHAM, with some minor changes.Equations (4), (10), (11), (12), and (13) change to (16), (17), (18), (20), and (19), respectively. Also, initial approximation in [tγ,tγ+1), γ=0,1,…,N-1, will be considered as (14)ui0jtj=αij,i=1,2,…,n,j=1,…,N,In addition, deformation equation in each subinterval for i=1,2,…,n, j=1,…,N, will change to the following [19]:(15)1-PLiuijt,P-ui0jt=HPLiuijt,P+fit+Niuijt,P.Moreover, auxiliary function HjP,t will be generalized as follows:(16)HjP,t=Ci1j+Ci2jt+Ci3jt2+⋯P.(17)u~ijt,Ciqj=ui0jt+∑k=1muk,jt,Ciqj,q=1,2,3,…,m,(18)Rijt,Ciqj=Lu~it,Ciqj+fit+Niu~it,Ciqj,q=1,2,3,…,m,(19)JijCiqj=∫tγtγ+hRij2s,Ciqjds,γ=0,1,…,N-1,q=1,2,3,…,m.The length of the subinterval tγ,tγ+1 is apparently h, and the number of subintervals is N=T/h. Now, we consider derivatives of (19), with respect to Ciqj, (i=1,2,…,n,j=1,…,N,q=1,2,3,…,m) to zero. In fact we define αij=u~ijtj, in each subinterval tγ,tγ+1. Therefore, the convergence control parameters can be determined from the solution of the following system of equations:(20)∂Jj∂Ci1j=∂Jj∂Ci2j=⋯=∂Jj∂Cimj=0.Approximate analytic solutions, on each subinterval, are as follows:(21)u~t=u~1t,t0≤t<t1,u~2t,t1≤t<t2,⋮u~Nt,tN-1≤t≤T.


## 3. Application of OHAM to Systems of Volterra Integral Equations

In this section, we apply OHAM on the following system of Volterra integral equations:(22)Ut=Ft+∫0tKt,sUsds,t∈0,1,where t∈R and(23)Kt,s=ki,jt,s,i,j=1,2,…,n,Ut=u1t,u2t,…,untt,Ft=f1t,f2t,…,fntt.In (22), K is the kernel that is known, F is given vector, and U is an unknown vector function. Consider i th equation of (22)(24)uit=fit+∫0t∑j=1nki,jt,sujsds,i=1,2,⋯,n.Using aforementioned OHAM procedure results in the following sequential equations, for i=1,2,…,n.(25)P0:ui0t=fit,P1:ui1t=-Ci1∫0t∑j=1nki,jt,suj0sds,P2:ui2t=1+Ci1ui1t-Ci1∫0t∑j=1nki,jt,suj1sds-Ci2∫0t∑j=1nki,jt,suj0sds,⋮Pk:uikt=1+Ci1uik-lt+∑l=2k-1Ciluik-lt-∑z=1kCiz∫0t∑j=1nki,jt,sujk-zsds,k=3,4,….Using (11)–(13), we find Cβ, β=1,…,m.Knowing these parameters, an approximate solution, of orderm, will be determined.


## 4. Application of MOHAM to Systems of Volterra Integral Equations

In this section, we apply MOHAM to (22). This procedure leads to the following sequence of equations, for i=1,2,…,n.(26)P0:ui0jt=αij,P1:ui1jt=Ci1j+Ci2jt+⋯ui0jt-fit-∫0t∑j=1nki,juj0jtds,P2:ui2jt=ui1jt+Ci1j+Ci2jt+⋯ui1jt-∫0t∑j=1nki,jt,suj1jtds,⋮Pk:uj1jt=uik-1jt+Cji1+Cji2t+⋯uik-1jt-∫0t∑j=1nki,jt,sujk-1jtds,k=3,4,….In addition, by using (18)–(20), we find Ciqj, i=1,2,3,…n, j=1,…,N, q=1,2,3,…,m.Knowing that the values are parameters, the approximate solution of orderm will be determined.


## 5. Convergence of the OHAM

There are two proofs presented in [19, 20] that have some oversight. The following proof covers these shortcoming.Theorem 1.
If series (9) convergences to ui(t) where uikt,C1,…,Ck∈L2c,d is produced by (3), (6), and (7), then ui(t) is the exact solution of (1).Proof.
Since the series(27)∑k=0∞uikt,C1,…,Ckconverges; it holds(28)limk→∞⁡uikt,C1,…,Ck=0.The left hand-side of (7) satisfies(29)ui1t,C1+∑k=2nuikt,C1,…,Ck-∑k=2nuik-1t,C1,…,Ck-1=uint,C1,…,Cn.According to (29), we have (30)ui1t,C1+∑k=2∞uikt,C1,…,Ck-∑k=2∞uik-1t,C1,…,Ck-1=limn→∞⁡uint,C1,…,Cn=0.Since the operator Li is linear, from (30), we have (31)Liui1t,C1+∑k=2∞Liuikt,C1,…,Ck-∑k=2∞Liuik-1t,C1,…,Ck-1=Liui1t,C1+Li∑k=2∞uikt,C1,…,Ck-Li∑k=2∞uik-1t,C1,…,Ck-1=0.Then(32)Liui1t,C1+Li∑k=2∞uikt,C1,…,Ck-Li∑k=2∞uik-1t,C1,…,Ck-1=Liui1t,C1+∑k=2∞CkNi0ui0t+∑m=1k-1CmLiuik-mt+Nik-mui0t,ui1t,…,uik-mt=0,Equation (32) can be written as follows:(33)Liui0t+fit+C1Ni0ui0t+∑k=2∞CkNi0ui0t+∑m=1k-1CmLiuik-mt+Nik-mui0t,ui1t,…,uik-mt=∑k=1∞∑m=1kCk-mLum-1t,C1,…,Cm-1+Nim-1ui0t,ui1t,…,uik-mt+fit=0.If the Cn, n=0,1,…,m, chosen properly, then (33) leads to (1).


## 6. Numerical Examples

In this section, two systems of Volterra integral equations of the second kind, a linear and a nonlinear, will be solved to show the efficiency of both OHAM and MOHAM. The results of applying OHAM and MOHAM will be compared. Matlab package is used to carry out computations, with double precision.Example 1.
let us consider the following linear system of Volterra integral equations of the second kind, with exact solutionsu1t=et and u2t=e-t.(34)u1t=1-t22+∫0tu1s+sesu2sds,u2t=1+t22-∫0tse-su1s+u2sds.The following aforementioned OHAM procedure results in(35)
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4
+
1.333333333
C
21
t
3
-
2
C
21
t
2
+
C
21
-
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C
21
t
e
-
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0.1666666667
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0.01666666667
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21
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t
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-
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21
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0.04166666667
e
-
t
+
0.075
t
5
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21
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C
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0.4583333333
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24
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1.25
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25.25
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0.1666666667
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1.333333333
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12
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5
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45
t
4
+
160
t
3
-
240
t
2
-
240
t
+
720
+
20
t
4
+
80
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3
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120
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2
-
480
t
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720
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0.1666666667
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1
.Using the same technique as presented by (11)–(13), we find Ci1, Ci2, and Ci3, i=1,2, as follows:(39)C11=-1.1515281477,C12=0.0312808541,C13=-0.0040455435,C21=-0.0602863302,C22=0.7021897011,C23=-1.1196100026.Substituting the values of C11, C12, C13, and C21, C22, and C23, into (37) and (38), one obtains(40)u1t=-0.1526946et-0.04241517t6+1.488772et-0.2163173t5-9.47979et+0.1627504t4+42.16258et-1.526946e-t+1.162051t3+-132.3468et-7.63473e-t+3.832068t2+277.0864et-16.79641e-t+22.74883t+1.017005t-1-298.8182+0.5085023t3-3t2+6t-6et-16.79641e-t+320.6826,u2t=0.000003651786t6+-0.000009129464e-t+0.003148389t5+0.005375128e-t-0.01171288t4+0.0216831e-t-0.1448423t3+0.00715933e-2t+0.03383929e-t+0.43758t2+0.01442821e-2t-0.1490952e-t-1.197225t+-1.179469t+1+0.0003527703et12t5-45t4+160t3-240t3-240t+720+20t4+80t3+120t2-480t-720-0.559805t3+3t2+6t+6-0.2269294e-t+0.00732366e-2t+5.757905.To derive a solution to (34), for 0≤t≤1, by MOHAM, we consider the following initial approximation:(41)ui0jtj=αij,i=1,2,j=1,…,5.Now, consider the auxiliary function HiP,t, as the following:(42)HiP,t=Ci1j+Ci2jt+Ci3jt2P.where Ciqj, i=1,2, j=1,…,5, q=1,2,3, are still unknown.
Regarding (17), the first-order approximate solution, for m=1, is as follows: (43)u~ijt=ui0jt+ui1jt,i=1,2,j=1,…,5,where(44)u11jt=C11j+C12jt+C13jt2u10jt-1+t22-∫0tu10js+sesu20jsds,(45)u21jt=C21j+C22jt+C23jt2u20jt-1-t22+∫0tsesu10js+u20jsds,Substitution of (44) and (45) into (43), regarding (41) and using (18)–(20), determines the residual and the functional J1jC1qj, and J2jC2qj for q=1,2,3, respectively(46)R1jt,C1qj=u~1jt-1+t22-∫0tu1js+sesu2jsds,R2jt,C2qj=u~2jt-1-t22+∫0tsesu1js+u2jsds,J1jC1qj=∫tγtγ+hR1j2s,C1qjds,J2jC2qj=∫tγtγ+hR2j2s,C2qjds.Minimization condition required(47)∂J1j∂C11j=∂J1j∂C12j=∂J1j∂C13j=0,∂J2j∂C21j=∂J2j∂C22j=∂J2j∂C23j=0.Parameters that control the convergence, Ciqj, i=1,2, j=1,…,5, q=1,2,3, are presented in Table 1, by considering α11=0, α21=0, and h=0.2, for t0=0 up to t5=T=1.
By substituting the values of the control parametersCiqj’s into (44) and (45), one obtains(48)u~1t=-0.5t2-11.5241t2+0.95441t+1.0008,0≤t<0.2,0.083435t2+0.28081t+0.956130.14906t5et+1.259t4et+5.8342t3et-18.53t2et-38.061et+1.0008t+0.5t2-11.5241t2+0.95441t+1.0008-0.15241t5-0.1193t4+0.34122t3-0.022794t2+38.061tet+39.061-0.5t2-11.5241t2+0.95441t+1.0008,0.2≤t<0.4,⋮u~2t=0.5t2+10.29812t2-1.0273t+1.0006,0≤t<0.2,--0.39731t2+0.061232t+0.915640.5t2+10.29812t2-1.0273t+1.0006-0.5t2+1.8504e-t+18et+161.6111t5-6.94t4-5.0717et+19+4.1183t5e17+e182.317t4+8.715t3+et+171.4383t3-2.776t2+5.4076t+e192.5629t2+5.0717t+5.0717-1+0.5t2+10.29812t2-1.0273t+1.0006,0.2≤t<0.4,⋮Expressions of the solution of the other interval are too long, that is why they are not presented here. Interested readers can calculate the solutions by the program appeared in Appendix A.
First-order MOHAM approximate solutions and three-order OHAM approximate solutions can be compared with exact solutions in Tables2 and 3, and plots are presented in Figures 1 and 2.
Absolute errors for OHAM and MOHAM are plotted in Figures3 and 4.
Exactui, OHAM ui, and MOHAM ui stand for exact solution of ui, the solution of ui by OHAM, and solution of by MOHAM, respectively.Table 1
Values of the control parametersCij.
j C 11 j C 12 j C 13 j C 21 j C 22 j C 23 j 1 -1.00082 -0.95441 -1.52407 -1.00061 1.02732 -0.29811 2 -0.95613 -0.28080 -0.08343 -0.91564 -0.06123 0.39730 3 -0.77599 -0.63069 -0.00753 0.00000 0.00000 0.00000 4 0.00000 0.00000 0.00000 -0.36062 -0.91215 0.19472 5 0.00000 0.00000 0.00000 -0.89126 -0.01338 0.00040Table 2
The results of applying OHAM, MOHAM, and the exact solution, for Example1.
t j Exactu1 Exactu2 OHAMu1 OHAMu2 MOHAMu1 MOHAMu2 0.0 1.000000 1.000000 1.000000 1.000000 1.000822 1.000615 0.1 1.105170 0.904837 1.106625 0.898382 1.105964 0.905368 0.2 1.221402 0.818730 1.225525 0.815433 1.227614 0.823217 0.3 1.349858 0.740818 1.360566 0.748515 1.350942 0.740830 0.4 1.491824 0.670320 1.515929 0.695399 1.494814 0.670041 0.5 1.648721 0.606530 1.696140 0.654166 1.648851 0.606287 0.6 1.822118 0.548811 1.906111 0.623127 1.821875 0.549821 0.7 2.013752 0.496585 2.151164 0.600764 2.012654 0.496810 0.8 2.225540 0.449328 2.437011 0.585686 2.223236 0.448796 0.9 2.459603 0.406569 2.769628 0.576594 2.456573 0.406549 1.0 2.718281 0.367879 3.154950 0.572265 2.717034 0.367906Table 3
Absolute errors of applying OHAM and MOHAM for Example1.
t j Ab. Error(OHAMu1) Ab. Error(OHAMu2) Ab. Error(MOHAMu1) Ab. Error(MOHAMu2) 0.0 0.000000 0.000000 0.000822 0.000615 0.1 0.001455 0.006455 0.000775 0.000531 0.2 0.004123 0.003296 0.006211 0.004487 0.3 0.010708 0.007697 0.001083 0.004486 0.4 0.024105 0.025079 0.002990 1.2469e-5 0.5 0.047419 0.047635 0.000130 0.000242 0.6 0.083993 0.074315 0.000242 0.001010 0.7 0.137412 0.104179 0.001098 0.000225 0.8 0.211471 0.136357 0.002304 0.000532 0.9 0.310025 0.170024 0.003030 2.0620e-5 1.0 0.436669 0.204386 0.001247 2.7127e-5Figure 1
The results of OHAM, MOHAM, and exact solution for u1 in Example1.Figure 2
The results of OHAM, MOHAM, and exact solution for u2 in Example1.Figure 3
Absolute errors of OHAM and MOHAM for u1 in Example1.Figure 4
Absolute errors of OHAM and MOHAM for u2 in Example1.Example 2.
let us consider the following nonlinear system of Volterra integral equations of the second kind, with the exact solutionsu1t=t and u2t=t. (49)u1t=t-t2-∫0tu1s+u2sds,u2t=t-t22-t33-∫0tu12s+u2sds.Following the procedure in Example 1, we have(50)C11=0,C12=-0.0000000008,C13=-1.9560359039,C21=0.0000000018,C22=-0.0000000033,C23=-1.4774464040,u1t=8.436182e-35t8+8.3237e-34t7+8.756976e-22t6-e-213.50279t5+2.189244t4-4.533431e-22t3-2t3+8t2+5t-30+1.313546e-20t3-0.163003t2t2+6t-12-t2+t,u2t=3.777777e-29t9-3.134551e-28t8-6.437708e-20t7-4.656609e-19t6+e-181.93775t5-1.464579t4+1.922551e-22t360t4+434t3-1806t2+1365t+420+0.02462411t212t3-35t2+10t+30-0.3333333t3-0.5t2+t,Parameters that control the convergence, Ciqj, i=1,2, j=1,…,5, q=1,2,3 are presented in Table 4, by considering α11=0, α21=0, h=0.2, for t0=0 up to t5=T=1.
Therefore, the approximate solutions can be written in the following form:(51)u~1t=t-t21.83126t2+0.916398t1.00294,0≤t<0.2,t-t21.83126t2+0.916398t1.00294,0.2≤t<0.4,-0.335879t2+0.39467t+0.8552311.54198e-19t2e183.03659t3-1.67635t2-6.49617+e172.50648t4+2.17358t+t-t2.1.83126t2+0.916398t+1.00294+t2-t+t-t21.83126t2+0.916398t+1.00294,0.4≤t<0.6,⋮u~2t=-0.486219t+0.695688t2+1.000450.5t2-t+0.333333t3,0≤t<0.2,-0.486219t+0.695688t2+1.000450.5t2-t+0.333333t3,0.2≤t<0.4,1.88661t2-0.994006t+1.14645t+2.7391e-33t21.36035e32t7-1.52911e32t6+6.01834e31t5-2.47255e32t4+9.73078e31t3-4.97417e30t2+1.20707e32t+1.82623e32-0.333333t3-0.5t2+0.695688t2+0.486219t+1.00045.0.333333t3+0.5t2-t-0.695688t2+0.486219t+1.00045.0.333333t3+0.5t2-t,0.4≤t<0.6,⋮Due to the long expressions of the solutions on other intervals, the results are not reported, but related programs will be appeared in Appendix B.
First-order MOHAM approximate solution and three-order OHAM approximate solution can be compared with an exact solution in Tables5 and 6, and plots are presented in Figures 5 and 6.
Absolute errors for OHAM and MOHAM are plotted in Figures7 and 8.Table 4
Values of the control parametersCij.
j C 11 j C 12 j C 13 j C 21 j C 22 j C 23 j 1 -1.002941 -0.916397 -1.831260 -1.000448 -0.486218 -0.695688 2 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 3 -0.855231 -0.394669 -0.335878 -1.146447 0.994005 -1.886613 4 -0.771655 0.098153 -0.628421 0.000000 0.000000 0.000000 5 -0.085517 -0.993931 0.069185 -0.666758 0.328380 -0.669028Table 5
The results of applying OHAM, MOHAM, and the exact solution for Example2.
t j Exactu1 Exactu2 OHAMu1 OHAMu2 MOHAMu1 MOHAMu2 0.00.10.20.30.40.50.60.70.80.91.0 0.00.10.20.30.40.50.60.70.80.91.0 0.00.10.20.30.40.50.60.70.80.91.0 0.0000000.1085660.2301560.3583160.4861990.6065690.7117950.7938600.8443510.8544670.815014 0.0000000.1022160.2075670.3128700.4135840.5041630.5784100.6298350.6520060.6389030.585276 0.0000000.1001600.2015150.3029610.3990000.4995310.6009590.7002960.7996660.8997601.000794 0.0000000.0999700.1995920.2973950.3901320.4999590.5999660.7002290.8009620.9004600.999225Table 6
Absolute errors of applying OHAM and MOHAM for Example2.
tj Ab. Error(OHAMu1) Ab. Error(OHAMu2) Ab. Error(MOHAMu1) Ab. Error(MOHAMu2) 0.0 0.000000 0.000000 0.000000 0.000000 0.1 0.008566 0.002216 0.000160 3.0000e-5 0.2 0.030156 0.007567 0.001515 0.000408 0.3 0.058316 0.012870 0.002961 0.002605 0.4 0.086199 0.013584 0.001000 0.009868 0.5 0.106569 0.004163 0.000469 4.1000e-5 0.6 0.111795 0.021589 0.000959 3.4000e-5 0.7 0.093860 0.070165 0.000296 0.000229 0.8 0.044351 0.147994 0.000334 0.000962 0.9 0.045533 0.261097 0.000240 0.000460 1.0 0.184986 0.414724 0.000794 0.000775Figure 5
The results of OHAM, MOHAM, and exact solution for u1 in Example2.Figure 6
The results of OHAM, MOHAM, and exact solution for u2 in Example2.Figure 7
Absolute errors of OHAM and MOHAM for u1 in Example2.Figure 8
Absolute errors OHAM and MOHAM for u2 in Example2.


## 7. Conclusion and Discussion

In this paper, two well known approaches, OHAM, and MOHAM, have been applied for solving linear and nonlinear systems of Volterra integral equations of the second kind. The results of applying these two approaches are presented in Tables and are plotted in Figures. Tables3 and 6 show the absolute errors of applying OHAM and MOHAM at some selected points. Comparing numerical results, reveal that MOHAM is more accurate than OHAM, especially for the points farther from the initial point. Moreover MOHAM is very efficient and convenient to use for finding approximation solution for system of Volterra integral equations of the second kind.



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*Source: 3037273-2019-01-03.xml*