# Application of Optimal Homotopy Asymptotic Method to Some Well-Known Linear and Nonlinear Two-Point Boundary Value Problems

Authors: Muhammad Asim Khan; Shafiq Ullah; Norhashidah Hj. Mohd Ali
Journal: International Journal of Differential Equations
 (2018)

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## Abstract

The objective of this paper is to obtain an approximate solution for some well-known linear and nonlinear two-point boundary value problems. For this purpose, a semianalytical method known as optimal homotopy asymptotic method (OHAM) is used. Moreover, optimal homotopy asymptotic method does not involve any discretization, linearization, or small perturbations and that is why it reduces the computations a lot. OHAM results show the effectiveness and reliability of OHAM for application to two-point boundary value problems. The obtained results are compared to the exact solutions and homotopy perturbation method (HPM).


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## Body

## 1. Introduction

Two-point boundary value problems (TPBVP) have many applications in the field of science and engineering [1, 2]. These problems arise in many physical situations like modeling of chemical reactions, heat transfer, viscous fluids, diffusions, deflection of beams, the solution of optimal control problems, etc. Due to the wide applications and importance of boundary value problems (BVP) in science and engineering we need solutions to these problems.There are many techniques available for the solution of-of BVP like Adomian Decomposition Method (ADM) [3–7], Extended Adomian Decomposition Method (EADM)[8], Differential Transformation Method (DTM) [9], Variational Iteration Method (VIM) [10], Perturbation methods(PMs) [1, 11–13], and so on. Perturbation methods are easy to solve but they require small parameters which are sometimes not an easy task. Recently V. Marinca et al. presented optimal homotopy asymptotic method (OHAM) [14] for the solution of BVP, which did not require small parameters. The method can also be applied to solve the stationary solution of some partial differential equations, e.g., gKdv equation, nonlinear parabolic problems, and so on [15–20]. In OHAM, the concept of homotopy is used together with the perturbation techniques. Here, OHAM is applied to TPBVP to check the applicability of OHAM for TPBVP.


## 2. Basics of OHAM

Let us take the BVP whose general form is the following:(1)Lwξ+Nwξ+Ϝξ=0,Bw,dwdξwhere L is a linear operator, ξis independent variable, N is the nonlinear operator, Ϝ(ξ) is a known function, and B is a boundary operator.Homotopy on OHAM can be constructed as(2)1-þLφξ,þ+Ϝξ=HþLφξ,þ+Ϝξ+Nφξ,þ,Bφξ,þ,∂φξ,þ∂ξwhere þ∈[0,1] is an embedding parameter, φ(ξ,þ) is an unknown function, H(þ) is a nonzero auxiliary function for þ≠0, and H(þ) is of the form(3)Hþ=þC1+þ2C2+þ3C3+⋯Clearly when þ=0 then H(0)=0. And obviously, when þ=0 then φ(ξ,0)=w0(ξ). When þ=1 then φ(ξ,1)=w(ξ). So as þ increases from 0 to 1, the solution φ(ξ,þ) varies from w0(ξ) to the exact solution w(ξ), where w0(ξ) is obtained from (2) for þ=0(4)Lw0ξ+Ϝξ=0The proposed solution of (1) will be of the form(5)φξ,þ,Ci=w0ξ+∑k≥1wkξ,Ciþk,i=1,2,3,…Substituting this value of φ(ξ,þ,Ci) into (1), after some calculations, we can obtain the governing equations of w0(ξ) by using (4) and wk(ξ), that is,(6)Lw1ξ=C1N0w0ξ,Bw1,dw1dξ=0(7)Lwkξ-wk-1ξ=CkN0w0ξ+∑i=1k-1CiLwk-1ξ+Nk-1w0ξ,w1ξ,w2ξ,…,wk-1ξ,k=2,3,4,…,Bwk,dwkdξ=0,where Nm(w0(ξ),w1(ξ),w2(ξ),…,wm(ξ)) is the coefficient of þm in the series expansion of N(φ(ξ,þ,Ci)) with respect to the embedding parameter þ. And(8)Nφξ,þ,Ci=N0w0ξ+∑m≥1Nmw0,w1,w2,…,wmþm,i=1,2,3,…,mwhere φ(ξ,þ,Ci) is given by (5). The convergence of series (5) depends on the convergence of the constants Ci′s, if these constants are convergent at þ=1, then the solution becomes(9)wξ,Ci=w0ξ+∑k≥1wkξ,Ci.Generally, the mth order solution of the problem can be obtained in the form(10)wmξ,Ci=w0ξ+∑k=1mwkξ,Ci,i=1,2,3,…,mPutting this solution in (1) we get the following residual:(11)Rξ,Ci=Lwmξ,Ci+Ϝξ+Nwmξ,Ci,i=1,2,3,…,mIf R(ξ,Ci)=0, then the solution is going to be exact, but generally, such a situation does not arise in nonlinear problems but the functional defined below can be minimized(12)Jξ,Ci=∫x0x1R2ξ,Cidξ,where x0 and x1 are two constants depending on the given problem. The values of Ci′s can be optimally found by the condition(13)∂J∂C1=∂J∂C2=⋯=∂J∂Cm=0After knowing these constants, the solution (10) is well determined.


## 3. Examples

To check the applicability of OHAM for TPBVP, in this section four examples of TPBVP are presented in which one example is linear and the remaining are nonlinear.


### 3.1. Example 1

Let us consider the linear problem [1] of second order(14)w′′ξ=w′ξ-eξ-1-1,0<ξ<1,w0=0,w1=1The exact solution of problem (14) is ξ(1-eξ-1). Now according to OHAM L(w0(ξ))=w′′(ξ)-w′(ξ), the nonlinear part N(w(ξ))=0 and Ϝ(ξ)=e(ξ-1)+1.The zeroth-order problem is(15)w0′′ξ-w0′ξ=1+eξ-1,w00=0,w01=1The solution of (15) is(16)w0ξ=e-eξξξThe first-order problem is(17)w1′′ξ-w1′ξ=-1-eξ-1-C1-C1eξ-1+w0′ξ+C1w0′ξ-w0′′ξ-C1w0′′ξ,w10=0,w11=0.The solution of (17) is(18)w1ξ=0The second-order problem is(19)w2′′ξ=-C2-expξ-1C2+C2w0′ξ+w1′ξ+C1w1′ξ-w2′ξ-C2w0′′ξ-w1′′ξ-C1w1′′ξ,w20=0,w21=0The solution of (19) is (20)w2ξ=0And the third-order approximate solution of the bvp (14) is as follows:(21)w2ξ=w0ξ+w1ξ+w2ξ(22)w2ξ=e-eξξξTable 1 shows the comparison between the exact solution and the approximate solution obtained by OHAM. Figure 1 of the solution also shows well agreement with the exact solution.Table 1
Comparison of the third-order OHAM solution with the exact solution and HPM.
ξ OHAM Solution ( w ( 2 )) E x a c t HPM [1] | w ( 2 ) - E x a c t | 0.1 0.059343 0.059343 0.05934820 1.38778 × 1 0 - 17  0.3 0.151024 0.151024 0.15103441 2.77556 × 1 0 - 17  0.5 0.196735 0.196735 0.19673826 2.77556 × 1 0 - 17  0.7 0.181427 0.181427 0.18142196 5.55112 × 1 0 - 17  0.9 0.0856463 0.0856463 0.08564186 5.55112 × 1 0 - 17Figure 1
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 1.


### 3.2. Example 2

Consider the nonlinear two-point boundary value problem [1] of the type(23)w′′ξ=w3ξ-wξw′ξ,ξ∈1,2,w1=12,w2=13According to OHAM L(w(x))=w′′(ξ) and N(w(ξ))=u(ξ)u′(ξ)-w3(ξ), while Ϝ(ξ)=0. The exact solution of (23) is 1/ξ+1. Now proceeding with the same lines as above we have the following zeroth-order problem:(24)w0′′ξ=0,w01=12,w02=13The solution of (24) is(25)w0ξ=4-ξ6Now the first-order problem is(26)w1′′ξ=C1w03ξ-C1w0ξw0′ξ-w0′′ξ-C1w0′′ξ(27)w11=0,w12=0The solution of (26) is(28)w1ξ=-930C1+1649ξC1-880ξ12C+180ξ13C-20ξ14C+ξ5C14320The second-order problem is(29)w2′′ξ=C2w03ξ+3C1w02ξw1ξ-C2w0ξw0′ξ-C1w1ξw0′ξ-C1w0ξw1′ξ-C2w0′′ξ-w1′′ξ-C1w1′′ξ,w21=0,w22=0.The solution of (29) is(30)w2ξ=1130636800ξ-2ξ-130240-465+ξ127+ξ-17ξC1-8985375+ξ4253423+ξ-3664113+ξ1100519+5ξ-36795+ξ3709+7-33+ξξC12+30240-465+ξ127+-17+ξξC2The third-order problem is (31)C3w03ξ+3C2w02ξw1ξ+3C1w0ξw12ξ+3C1w02ξw2ξ-3C3w0ξw0′ξ-C2w1ξw0′ξ-C1w2ξw0′ξ-C2w0ξw1′ξ-C1w1ξw1′ξ-C1w0ξw2′ξ-C3w0′′ξ-C2w1′′ξ-w2′′ξ-C1w2′′ξ+w3′′ξ=0The solution of the third-order problem results a large output, therefore not included here.Now the third-order approximate solution is(32)w3ξ=w0ξ+w1ξ+w2ξ+w3ξCi′s has the following values and then substituting in the above solution we will get the approximate solution. w(3)(ξ) is given in Appendix (A.1).(33)C1=-0.9637924142971654,C2=-0.0002296939939480446,C3=-0.000014314891134337846,The solution at the points given in Table 2 and the graph of the solution is shown in Figure 2. Here it is third-order OHAM solution while the HPM [1] gives the accuracy up to 9 decimal places in 7th order.Table 2
Comparison of second-order OHAM solution with the exact solution for example 2.
ξ OHAM Solution(w(3)) E x a c t | w ( 3 ) - E x a c t | 1.1 0.47619 0.47619 2.0597 × 1 0 - 7  1.3 0.434783 0.434783 5.38284 × 1 0 - 7  1.5 0.400001 0.4 1.39261 × 1 0 - 6  1.7 0.370371 0.37037 6.7426 × 1 0 - 7  1.9 0.344828 0.344828 8.10196 × 1 0 - 8Figure 2
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 2.


### 3.3. Example 3

Now we consider higher order TPBVP of order four. The problem is(34)d4wξdξ4=w2ξ+Ϝξ,0≤ξ≤1with the boundary conditions w(0)=0, w′(0)=0, w(1)=1, and w′(1)=1.WhereL(w(ξ))=d4w(ξ)/dx4, N(w(ξ))=w2(ξ), and Ϝ(ξ)=-ξ10+4ξ9-4ξ8-4ξ7+8ξ6-4ξ4+120ξ-48, the exact solution of problem (34) is wexact=ξ5-2ξ4+2ξ2. After solving this by the method described in Section 2, we have the following zeroth-order problem:(35)48-120ξ+4ξ4-8ξ6+4ξ7+4ξ8-4ξ9+ξ10+d4w0ξdξ4=0(36)w00=0,w0′0=0,w01=1,w0′1=1The solution to (35) is(37)w0ξ=110810802155683ξ2+8038ξ3-2162160ξ4+1081080ξ5-2574ξ8+1716ξ10-546ξ11-364ξ12+252ξ13-45ξ14The first-order problem is (38)-48+120ξ-4ξ4+8ξ6-4ξ7-4ξ8+4ξ9-ξ10-48C1+120ξC1-4ξ14C+8ξ6C1-4ξ7C1-4ξ8C1+4ξ9C1-ξ10C1+C1w02ξ-1+C1d4w0ξdξ4+d4w1ξdξ4=0(39)w10=0,w1′0=0,w11=0,w1′1=0.The second-order problem is (40)-48C2+120ξC2-4ξ4C2+8ξ6C2-4ξ7C2-4ξ8C2+4ξ9C2-ξ10C2+C2w02ξ+2C1w0ξw1ξ-C2w04ξ-w14ξ-C1w14ξ+w24ξ=0(41)w20=0,w2′0=0,w21=0,w2′1=0.The solutions of problem (38) and (40) are very large; therefore we did not write it here. The constants C1 and C2 have the values -1.0011320722175725 and -1.079468959963785×10-6, respectively. Table 3 and Figure 3 show a good agreement with the exact values. The approximate solution w2(ξ) is given in Appendix (A.2).Table 3
Comparison of second-order OHAM solution with the exact solution for example 3.
ξ OHAM Solution(w(2)) E x a c t | w ( 2 ) - E x a c t | 0.2 0.077119 0.0771200 6.152306 × 1 0 - 11 0.4 0.279039 0.2790400 1.314346 × 1 0 - 10 0.6 0.538559 0.5385600 1.001054 × 1 0 - 10 0.8 0.788479 0.7884800 2.415356 × 1 0 - 11Figure 3
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 3.


### 3.4. Example 4

At last, consider the second-order nonlinear TPBVP[1](42)w′′ξ=w2ξ+2π2cos2πξ-sin22πξ,0≤ξ≤1(43)w0=0,w1=0.The exact solution of (42) is sin2(πξ). Solving (42) by the method depicted in Section 2, we have the following zeroth order problem:(44)-2π2cos2πξ+sin4πξ+w0′′ξ=0,w00=0,w01=0The solution of (44) is given by(45)w0ξ=1128π215-64π2+24π2ξ-24π2ξ2-16cos2πξ-64π2cos2πξ+cos4πξ.The first-, second-, and third-order problems are given in (46), (47), and (48) respectively. (46)2π2cos2πξ1+C1-sin4πξ1+C1+C1w02ξ-w0′′ξ-C1w0′′ξ+w1′′ξ=0,w10=0,w11=0(47)2π2cos2πξ-sin4πξC2+C2w02ξ+2C1w0ξw1ξ-C2w0′′ξ-w1′′ξ-C1w1′′ξ+w2′′ξ=0,w20=0,w21=0(48)2π2cos2πξ-sin4πξC3+C3w02ξ+2C2w0ξw1ξ+C1w12ξ+2C1w0ξw2ξ+C3w0′′ξ-C2w1′′ξ-w2′′ξ-C1w2′′ξ+w3′′ξ=0(49)w30=0,w31=0The solutions of problem (46), (47), and (48) are very large and therefore cannot be written here but the table of values and the graph are shown in Table 4 and Figure 4, respectively. The approximate solution w(3)(ξ) is written in Appendix (A.3). The values of the constants Ci′s can be found by (13) which are given as follows:(50)C1=-0.9030981665320986,C2=-0.00569345107292796,C3=0.00021560218552318884Table 4
Comparison of third-order OHAM solution with the exact solution.
ξ OHAM Solution(w(3)) E x a c t | w ( 3 ) - E x a c t | 0.1 0.0954915 0.0954915 5.59262 × 1 0 - 9  0.3 0.654508 0.654508 1.23779 × 1 0 - 8  0.5 0.999999 1 . 1.51565 × 1 0 - 8  0.7 0.654508 0.654508 1.23779 × 1 0 - 8  0.9 0.0954915 0.0954915 5.59262 × 1 0 - 9Figure 4
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 4.


## 3.1. Example 1

Let us consider the linear problem [1] of second order(14)w′′ξ=w′ξ-eξ-1-1,0<ξ<1,w0=0,w1=1The exact solution of problem (14) is ξ(1-eξ-1). Now according to OHAM L(w0(ξ))=w′′(ξ)-w′(ξ), the nonlinear part N(w(ξ))=0 and Ϝ(ξ)=e(ξ-1)+1.The zeroth-order problem is(15)w0′′ξ-w0′ξ=1+eξ-1,w00=0,w01=1The solution of (15) is(16)w0ξ=e-eξξξThe first-order problem is(17)w1′′ξ-w1′ξ=-1-eξ-1-C1-C1eξ-1+w0′ξ+C1w0′ξ-w0′′ξ-C1w0′′ξ,w10=0,w11=0.The solution of (17) is(18)w1ξ=0The second-order problem is(19)w2′′ξ=-C2-expξ-1C2+C2w0′ξ+w1′ξ+C1w1′ξ-w2′ξ-C2w0′′ξ-w1′′ξ-C1w1′′ξ,w20=0,w21=0The solution of (19) is (20)w2ξ=0And the third-order approximate solution of the bvp (14) is as follows:(21)w2ξ=w0ξ+w1ξ+w2ξ(22)w2ξ=e-eξξξTable 1 shows the comparison between the exact solution and the approximate solution obtained by OHAM. Figure 1 of the solution also shows well agreement with the exact solution.Table 1
Comparison of the third-order OHAM solution with the exact solution and HPM.
ξ OHAM Solution ( w ( 2 )) E x a c t HPM [1] | w ( 2 ) - E x a c t | 0.1 0.059343 0.059343 0.05934820 1.38778 × 1 0 - 17  0.3 0.151024 0.151024 0.15103441 2.77556 × 1 0 - 17  0.5 0.196735 0.196735 0.19673826 2.77556 × 1 0 - 17  0.7 0.181427 0.181427 0.18142196 5.55112 × 1 0 - 17  0.9 0.0856463 0.0856463 0.08564186 5.55112 × 1 0 - 17Figure 1
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 1.


## 3.2. Example 2

Consider the nonlinear two-point boundary value problem [1] of the type(23)w′′ξ=w3ξ-wξw′ξ,ξ∈1,2,w1=12,w2=13According to OHAM L(w(x))=w′′(ξ) and N(w(ξ))=u(ξ)u′(ξ)-w3(ξ), while Ϝ(ξ)=0. The exact solution of (23) is 1/ξ+1. Now proceeding with the same lines as above we have the following zeroth-order problem:(24)w0′′ξ=0,w01=12,w02=13The solution of (24) is(25)w0ξ=4-ξ6Now the first-order problem is(26)w1′′ξ=C1w03ξ-C1w0ξw0′ξ-w0′′ξ-C1w0′′ξ(27)w11=0,w12=0The solution of (26) is(28)w1ξ=-930C1+1649ξC1-880ξ12C+180ξ13C-20ξ14C+ξ5C14320The second-order problem is(29)w2′′ξ=C2w03ξ+3C1w02ξw1ξ-C2w0ξw0′ξ-C1w1ξw0′ξ-C1w0ξw1′ξ-C2w0′′ξ-w1′′ξ-C1w1′′ξ,w21=0,w22=0.The solution of (29) is(30)w2ξ=1130636800ξ-2ξ-130240-465+ξ127+ξ-17ξC1-8985375+ξ4253423+ξ-3664113+ξ1100519+5ξ-36795+ξ3709+7-33+ξξC12+30240-465+ξ127+-17+ξξC2The third-order problem is (31)C3w03ξ+3C2w02ξw1ξ+3C1w0ξw12ξ+3C1w02ξw2ξ-3C3w0ξw0′ξ-C2w1ξw0′ξ-C1w2ξw0′ξ-C2w0ξw1′ξ-C1w1ξw1′ξ-C1w0ξw2′ξ-C3w0′′ξ-C2w1′′ξ-w2′′ξ-C1w2′′ξ+w3′′ξ=0The solution of the third-order problem results a large output, therefore not included here.Now the third-order approximate solution is(32)w3ξ=w0ξ+w1ξ+w2ξ+w3ξCi′s has the following values and then substituting in the above solution we will get the approximate solution. w(3)(ξ) is given in Appendix (A.1).(33)C1=-0.9637924142971654,C2=-0.0002296939939480446,C3=-0.000014314891134337846,The solution at the points given in Table 2 and the graph of the solution is shown in Figure 2. Here it is third-order OHAM solution while the HPM [1] gives the accuracy up to 9 decimal places in 7th order.Table 2
Comparison of second-order OHAM solution with the exact solution for example 2.
ξ OHAM Solution(w(3)) E x a c t | w ( 3 ) - E x a c t | 1.1 0.47619 0.47619 2.0597 × 1 0 - 7  1.3 0.434783 0.434783 5.38284 × 1 0 - 7  1.5 0.400001 0.4 1.39261 × 1 0 - 6  1.7 0.370371 0.37037 6.7426 × 1 0 - 7  1.9 0.344828 0.344828 8.10196 × 1 0 - 8Figure 2
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 2.


## 3.3. Example 3

Now we consider higher order TPBVP of order four. The problem is(34)d4wξdξ4=w2ξ+Ϝξ,0≤ξ≤1with the boundary conditions w(0)=0, w′(0)=0, w(1)=1, and w′(1)=1.WhereL(w(ξ))=d4w(ξ)/dx4, N(w(ξ))=w2(ξ), and Ϝ(ξ)=-ξ10+4ξ9-4ξ8-4ξ7+8ξ6-4ξ4+120ξ-48, the exact solution of problem (34) is wexact=ξ5-2ξ4+2ξ2. After solving this by the method described in Section 2, we have the following zeroth-order problem:(35)48-120ξ+4ξ4-8ξ6+4ξ7+4ξ8-4ξ9+ξ10+d4w0ξdξ4=0(36)w00=0,w0′0=0,w01=1,w0′1=1The solution to (35) is(37)w0ξ=110810802155683ξ2+8038ξ3-2162160ξ4+1081080ξ5-2574ξ8+1716ξ10-546ξ11-364ξ12+252ξ13-45ξ14The first-order problem is (38)-48+120ξ-4ξ4+8ξ6-4ξ7-4ξ8+4ξ9-ξ10-48C1+120ξC1-4ξ14C+8ξ6C1-4ξ7C1-4ξ8C1+4ξ9C1-ξ10C1+C1w02ξ-1+C1d4w0ξdξ4+d4w1ξdξ4=0(39)w10=0,w1′0=0,w11=0,w1′1=0.The second-order problem is (40)-48C2+120ξC2-4ξ4C2+8ξ6C2-4ξ7C2-4ξ8C2+4ξ9C2-ξ10C2+C2w02ξ+2C1w0ξw1ξ-C2w04ξ-w14ξ-C1w14ξ+w24ξ=0(41)w20=0,w2′0=0,w21=0,w2′1=0.The solutions of problem (38) and (40) are very large; therefore we did not write it here. The constants C1 and C2 have the values -1.0011320722175725 and -1.079468959963785×10-6, respectively. Table 3 and Figure 3 show a good agreement with the exact values. The approximate solution w2(ξ) is given in Appendix (A.2).Table 3
Comparison of second-order OHAM solution with the exact solution for example 3.
ξ OHAM Solution(w(2)) E x a c t | w ( 2 ) - E x a c t | 0.2 0.077119 0.0771200 6.152306 × 1 0 - 11 0.4 0.279039 0.2790400 1.314346 × 1 0 - 10 0.6 0.538559 0.5385600 1.001054 × 1 0 - 10 0.8 0.788479 0.7884800 2.415356 × 1 0 - 11Figure 3
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 3.


## 3.4. Example 4

At last, consider the second-order nonlinear TPBVP[1](42)w′′ξ=w2ξ+2π2cos2πξ-sin22πξ,0≤ξ≤1(43)w0=0,w1=0.The exact solution of (42) is sin2(πξ). Solving (42) by the method depicted in Section 2, we have the following zeroth order problem:(44)-2π2cos2πξ+sin4πξ+w0′′ξ=0,w00=0,w01=0The solution of (44) is given by(45)w0ξ=1128π215-64π2+24π2ξ-24π2ξ2-16cos2πξ-64π2cos2πξ+cos4πξ.The first-, second-, and third-order problems are given in (46), (47), and (48) respectively. (46)2π2cos2πξ1+C1-sin4πξ1+C1+C1w02ξ-w0′′ξ-C1w0′′ξ+w1′′ξ=0,w10=0,w11=0(47)2π2cos2πξ-sin4πξC2+C2w02ξ+2C1w0ξw1ξ-C2w0′′ξ-w1′′ξ-C1w1′′ξ+w2′′ξ=0,w20=0,w21=0(48)2π2cos2πξ-sin4πξC3+C3w02ξ+2C2w0ξw1ξ+C1w12ξ+2C1w0ξw2ξ+C3w0′′ξ-C2w1′′ξ-w2′′ξ-C1w2′′ξ+w3′′ξ=0(49)w30=0,w31=0The solutions of problem (46), (47), and (48) are very large and therefore cannot be written here but the table of values and the graph are shown in Table 4 and Figure 4, respectively. The approximate solution w(3)(ξ) is written in Appendix (A.3). The values of the constants Ci′s can be found by (13) which are given as follows:(50)C1=-0.9030981665320986,C2=-0.00569345107292796,C3=0.00021560218552318884Table 4
Comparison of third-order OHAM solution with the exact solution.
ξ OHAM Solution(w(3)) E x a c t | w ( 3 ) - E x a c t | 0.1 0.0954915 0.0954915 5.59262 × 1 0 - 9  0.3 0.654508 0.654508 1.23779 × 1 0 - 8  0.5 0.999999 1 . 1.51565 × 1 0 - 8  0.7 0.654508 0.654508 1.23779 × 1 0 - 8  0.9 0.0954915 0.0954915 5.59262 × 1 0 - 9Figure 4
Comparison between exact solution (dashed line) and approximate solution (dotted line) for example 4.


## 4. Conclusion

This paper reveals that OHAM is a very strong method for solving TPBVP and gives us a more accurate solution as compared to other methods. In these examples only second- and third-order solution gives us the accuracy up to 8 or 10 decimal places; therefore it is concluded that this method converges very fast to the exact solution and in some problems like example 1 it gives us the exact solution. The plots and tables show well agreement with the exact solution.



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