# Least Square Homotopy Perturbation Method for Ordinary Differential Equations

Authors: Mubashir Qayyum; Imbsat Oscar
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this study, a new modification of the homotopy perturbation method (HPM) is introduced for various order boundary value problems (BVPs). In this modification, HPM is hybrid with least square optimizer and named as the least square homotopy perturbation method (LSHPM). The proposed scheme is tested against various linear and nonlinear BVPs (second to seventh order DEs). Validity of the obtained solutions is confirmed by finding absolute errors. To analyze the efficiency of the proposed scheme, tested problems have also been solved through HPM and results are compared with LSHPM. Furthermore, obtained results are also compared with other numerical schemes available in literature. Analysis reveals that LSHPM is a consistent and effective scheme which can be used for more complex BVPs in science and engineering.


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## Body

## 1. Introduction

Most of the phenomena in mathematical physics, biological mathematics, and applied mathematics are modeled in the form of differential equations. Solutions to such differential equations are needed to analyze and predict changes in a physical system. In most cases, it is impossible to calculate an exact solution. Therefore, to solve these problems, various numerical and seminumerical methods have been modeled.Liao presented one of the first analytical methods for nonlinear problems which do not require a small parameter [1]. This method has been applied to many situations in various fields of science and technology [2–4]. In late 90s, Prof. He introduced the homotopy perturbation method (HPM) for highly nonlinear equations [5]. This technique matures into a full fledged theory for nonlinear problems with the efforts of many researchers, notably Ji-Huan He and his students. This method combines homotopy concepts with perturbation theory and does not rely on small or large parameter like other traditional perturbation techniques [6]. Wide range of problems has been solved using HPM [7–10].Various modifications of HPM have also been proposed by different researchers to tackle more complex problems. Darvishi et al. combine HPM with adomian polynomials to sine-Gorden type equations in [11]. Biazar and Eslami proposed a modification of HPM based on Taylor series expansion of the kernel and source term [12]. HPM with rank upgrading technique for the superior nonlinear oscillation is proposed in [13]. Qayyum et al. used coupling of HPM with Laplace transform for squeezing flows in [14]. Bota and Caruntu applied HPM with the least square method to fluid flow problems in [15]. For enhanced results, Le-He extension of HPM is proposed in [16]. Ji et al. further applied Le-He extension of HPM to nonlinear packaging system in [17]. Ain et al. introduced Li-He modified extension of HPM for micro-electro-mechanical systems in [18].In this article, HPM is combined with LS optimizer along with some refine inital guesses to obtain fast convergent semianalytical solutions of linear and nonlinear BVPs. This scheme takes few iterations (cycles) to achieve accurate solution, and hence, it has less computational cost with improved accuracy.


## 2. Basic Idea of LSHPM

Let us consider the following differential equations along with boundary conditions.(1)Lφ+Nφ−gr=0,r∈Ω,(2)Bφ,dnφdrn=0,r∈γ,where L and N represent the linear and nonlinear parts, B represents the boundary operator, γ is the boundary of the domain, φ represents an unknown value, and gx is a known function. We construct a homotopy such that ηr,p:φ×0,1⟶ℝ which satisfies(3)ψη,p=1−pLη−Lφ0+pLη+Nη−gr=0,r∈Ω,where p∈0,1 is the embedding parameter and Lφ0 is the initial guess of equation (1) that satisfies the boundary conditions:(4)ψη,0=Lη−Lφ0=0,ψη,1=Lη+Nη−gr=0.Therefore, asp changes from 0 to 1, the solution ηr,p approaches from φ0 to φ˜r. We expand ηr,p into a Taylor series expansion about p to obtain an approximate solution:(5)ηr,p=η0+∑k=1∞ηkpk.Settingp=1, the approximate HPM solution of (1) would be(6)φ˜=limp⟶1ηr,p=∑k=1∞ηk.We reassign the convergence controlling parametersci′s as coefficients of series (6) and substitute supposed solution φ˜ back in (1) to get the residual function:(7)ℝx,ci=Rx,φ˜.Now, we compute the sum of square of residuals:(8)Jci=∫Iℝ2x,cidx.After computingJci, we find the optimal values of ci′s from system of equations obtained from ∂J/∂ci=0. Then, putting these optimal values back into φ˜ will provide refined solution.


## 3. Application of LSHPM

Problem 1.
Second-order linear ODE:(9)G″x+2G′x+Gx=0,0<x<1,subject to boundary conditions,(10)G0=1,G1=3.The exact solution of this problem ise−x+3e−1xe−x. First, we construct a homotopy as(11)1−pG″x+pG″x+2G′x+Gx=0.Zeroth-order problem is(12)G0″x=0,G00=1,G01=3.The solution to (12) is(13)G0x=1+2x.First-order problem is(14)G0x+2G0′x+G1″x=0,G10=0,G11=0.The solution to (14) is(15)G1x=1617x−15x2−2x3.Second-order problem is(16)G1x+2G1′x+G2″x=0,G20=0,G21=0.The solution to (16) is(17)G2x=1360449x−1020x2+430x3+135x4+6x5.Third-order problem is(18)G2x+2G2′x+G3″x=0,G30=0,G31=0.The solution to (18) is(19)G3x=1151202351x−18858x2+25417x3−5460x4−3171x5−273x6−6x7.By combining (13), (15), (17), and (19) will give approximate solution of (9) as follows:(20)G˜x=1+2x+1617x−15x2−2x3+1360449x−1020x2+430x3+135x4+6x5+1151202351x−18858x2+25417x3−5460x4−3171x5−273x6−6x7.It follows that (20) consist of x0,x,x2,x4,x5,x6,x7, and hence, the required solution is(21)G˜=c0+c1x+c2x2+c3x4+c4x5+c5x6+c6x7.By applying the boundary conditions given in (27), we have(22)c0=1,c1=2−c2−c3−c4−c5−c6−c7.By puttingc0,c1 in G˜ gives(23)G˜x=1+2−c2−c3−c4−c5−c6−c7x+c2x2+c3x4+c4x5+c5x6+c6x7.Now, replacingGx with G˜x in (27)(24)Rx,c2,c3,c4,c5,c6,c7=1+2c2+6c3x+2−c2−c3−c4−c5−c6−c7x+c2x2+12c4x2+c3x3+20c5x3+c4x4+30c6x4+c5x5+42c7x5+c6x6+c7x7+22−c2−c3−c4−c5−c6−c7+2c2x+3c3x2+4c4x3+5c5x4+6c6x5+7c7x6.Next, we minimizeJc2,c3,c4,c5,c6=∫01R2c2,c3,c4,c5,c6dx to get optimal values of ci′s. Here,(25)c2=−6.65482,c3=3.4104,c4=−1.14897,c5=0.285225,c6=−0.0522252,c7=0.00554595.The third-order approximation solution is(26)G˜x=1+6.15485x−6.65482x2+3.4104x3−1.14897x4+0.285225x5−0.0522252x6+0.00554595x7.The results are presented in Table1.Table 1
Comparison of third-order absolute error of LSHPM with HPM in Problem1.
xExactLSHPMHPMSolutionSolutionErrorSolutionError011.0100.11.552231.552233.76×10−81.560348.10×10−30.21.990311.990314.82×10−82.004281.39×10−20.32.330952.330956.86×10−82.346831.58×10−20.42.588732.588733.12×10−82.602531.37×10−20.52.776352.776359.38×10−82.78528.85×10−30.62.904812.904812.57×10−82.907662.85×10−30.72.983682.983686.94×10−82.981442.23×10−30.83.021233.021234.41×10−83.016464.76×10−30.93.024623.024623.67×10−83.020733.88×10−3133030Problem 2.
Let us consider the following second-order nonhomogeneous nonlinear ODE [20]:(27)G″x+Gx2−x4−2=0,0<x<1,subject to boundary conditions,(28)G0=0,G1=1.This problem was studied by Momani [20] by applying the differential transform method, and later, Mohamad [19] by applying four different numerical methods for this problem. The exact solution to this problem is x2.The first step is to construct a homotopy:(29)1−pG″x+pG″x+Gx2+x4−2=0.Then, we linearize the problem as follows.Zeroth-order problem is(30)G0″x=0,G00=0,G01=1.The solution to (30) is(31)G0x=x.First-order problem is(32)−2+x4+G0x2+G1″x=0,G10=0,G11=0.The solution to (32) is(33)G1x=160−53x+60x2−5x4−2x6.Second-order problem is(34)2G0xG1x+G2″x=0,G20=0,G21=0.The solution to (34) is(35)G2x=17560−394x+1113x4−756x5+30x7+7x9.By combining (31), (33), and (35) will give an approximate series solution of (27) as(36)G˜x=x+160−53x+60x2−5x4−2x6+17560−394x+1113x4−756x5+30x7+7x9.(36) consists of x,x2,x4,x5,x6,x7,x9; hence, the required solution will take the form(37)G˜x=c1x+c2x2+c3x4+c4x5+c5x6+c6x7+c7x9.By applying boundary conditions from (27) gives(38)c1=1−1c2+c3+c4+c5+c6+c7.By puttingc1 in G˜x gives(39)G˜x=1−1c2+c3+c4+c5+c6+c7x+c2x2+c3x4+c4x5+c5x6+c6x7+c7x9.Now, replacingGx with G˜x in (27), we get the following residual function:(40)Rx,c2,c3,c4,c5,c6,c7=−2+2c2+12c3x2+20c4x3+x4+30c5x4+42c6x5+72c7x7+1−c2−c3−c4−c5−c6−c7x+c2x2+c3x4+c4x5+c5x6+c6x7+c7x92.Next, we minimizeJc2,c3,c4,c5,c6,c7=∫01R2c2,c3,c4,c5,c6,c7dx to get optimal values of ci′s.Here,(41)c1=0.0725613,c2=0.999994,c3=−0.0000429997,c4=−0.00942955,c5=−0.0618513,c6=−0.00444548,c7=0.00321356.Hence, second-order approximate solution is(42)G˜x=0.0725613x+0.999994x2−0.0000429997x4−0.00942955x5−0.0618513x6−0.00444548x7+0.00321356x9.The results are presented in Table2.Table 2
Comparison of second-order absolute error of LSHPM with HPM and RK4 in Problem2.
xExactLSHPMHPMRK4SolutionSolutionErrorSolutionErrorError [19]0.10.010.014.11×10−160.008868971.13×10−32.14×10−80.20.040.041.04×10−150.03781512.18×10−38.56×10−80.30.090.091.04×10−150.0869773.02×10−31.86×10−70.40.160.165.55×10−160.1564893.51×10−33.42×10−70.50.250.251.38×10−160.2464253.57×10−35.06×10−70.60.360.361.66×10−160.3567763.22×10−37.45×10−70.70.490.493.88×10−160.4874492.55×10−31.07×10−60.80.640.643.33×10−160.6382981.70×10−31.43×10−60.90.810.810.0.8091748.25×10−41.90×10−61.1.1.0.1.0.2.38×10−6Problem 3.
Third-order linear ODE [21]:(43)G‴x−xGx−x3−2x2−5x−3ex,0<x<1,subject to boundary conditions(44)G0=0,G′0=1,G1=0.The exact solution to this problem isx1−xex. After applying LSHPM proceedure, second-order approximate solution is(45)G˜x=−26.4063+26.4063ex−14.2359x−11.1703exx−2.77751x2+0.744709exx2−0.0605874exx3+0.118979x4−0.00260775exx4+0.029123x5+0.00352613x6−0.0000251847x9−3.92104×10−7x10.The results are presented in Table3.Table 3
Comparison of second-order absolute error of LSHPM with HPM, Q-spline, and B-spline in Problem3.
xExactLSHPMHPMQ-splineB-splineSolutionSolutionErrorSolutionErrorError [22]Error [23]00−1.77×10−151.77×10−1500000.10.09946540.09946543.36×10−100.09946513.05×10−71×10−700.20.1954240.1954245.43×10−100.1954231.22×10−62×10−72×10−30.30.283470.283472.01×10−100.2834682.72×10−63×10−75×10−30.40.3580380.3580387.95×10−100.3580334.73×10−65×10−77×10−30.50.412180.412184.16×10−100.4121737.04×10−66×10−78×10−30.60.4373090.4373093.07×10−100.4372999.25×10−67×10−78×10−30.70.4228880.4228884.05×10−100.4228771.06×10−57×10−77×10−30.80.3560870.3560871.68×10−110.3560761.03×10−57×10−76×10−30.90.2213640.2213641.17×10−100.2213577.15×10−64×10−74×10−310−7.52×10−157.52×10−15−1.59×10−131.59×10−1301.4×10−2Problem 4.
Third-order nonlinear ODE [24]:(46)G‴x−G2x+Gx+x2x2−1,0<x<1,subject to boundary conditions(47)G1=0,G′1=−2,G′0=0.The exact solution to this problem is1−x2. The first-order approximate solution using LSHPM is(48)G˜x=1−x2+1.45333×10−18x3−2.76238×10−17x5+3.64818×10−16x7−7.51562×10−16x8+1.13156×10−15x10−2.20449×10−15x12+1.83775×10−15x13−4.11003×10−16x15+5.91342×10−17x17.The results are presented in Table4.Table 4
Comparison of second-order absolute error of LSHPM with HPM and MADM in Problem4.
xExactLSHPMHPMMADM [24]SolutionSolutionErrorSolutionErrorError0.10.990.9900.9903683.67×10−42.78×10−30.20.960.9600.9603243.23×10−43.44×10−30.30.910.9100.9102652.65×10−46.41×10−30.40.840.8400.8402022.02×10−47.05×10−30.50.750.7500.7501421.42×10−46.14×10−30.60.640.6400.6400919.11×10−54.37×10−30.70.510.5100.5100515.08×10−52.31×10−30.80.360.361.11×10−160.36002222×10−53.98×10−40.90.190.1900.1900065.57×10−61.04×10−31000001.79×10−3Problem 5.
Fourth-order linear ODE [25]:(49)Givx−G″x−Gx−exx−3=0,0<x<1,subject to boundary conditions(50)G0=1,G′0=0,G1=0,G′1=−e.The exact solution to this problem isex−xex. The first-order approximate solution using LSHPM is(51)G˜x=ex−xex.The results are presented in Table5.Table 5
Comparison of first-order absolute error of LSHPM with HPM and VIM in Problem5.
xExactLSHPMHPMVIM [25]SolutionSolutionErrorSolutionErrorError01101000.10.9946540.99465400.994623.37×10−52.00×10−100.20.9771220.97712200.9771.21×10−47.00×10−100.30.9449010.94490100.9446722.29×10−41.35×10−90.40.8950950.89509500.8947773.17×10−42.00×10−90.50.8243610.82436100.8240063.54×10−42.50×10−90.60.7288480.72884800.7285213.26×10−42.72×10−90.70.6041260.60412600.6038832.42×10−42.21×10−90.80.4451080.44510800.4449761.32×10−41.80×10−90.90.245960.2459600.2459233.74×10−57.25×10−1010005.55112×10−165.55×10−169.94×10−14Problem 6.
Fourth-order nonlinear ODE [26]:(52)Givx+Rx−GxG‴x+3G″x=0,0<x<1,subject to boundary conditions(53)G1=1,G′1=0,G0=0,G″0=0,
whereR is the Reynolds number, and we have fixed it to R=0.1. The second-order approximate solution of this problem using LSHPM is(54)G˜x=1.50671x−0.513264x3+0.00639854x5+0.00016573x7−7.3875×10−6x9−1.43995×10−7x11.The results are presented in Table6.Table 6
Comparison of second-order absolute error of LSHPM with HPM in Problem6.
xLSHPMHPMSolutionErrorSolutionError000000.10.1501585.14×10−70.1501572.54×10−50.20.2972375.99×10−70.2972375.10×10−50.30.438171.56×10−70.4381697.55×10−50.40.56994.59×10−70.5698999.58×10−50.50.6893976.55×10−70.6893961.06×10−40.60.7936611.16×10−70.7936619.88×10−50.70.8797346.64×10−70.8797346.54×10−50.80.9447054.64×10−70.9447052.03×10−60.90.9857229.06×10−70.9857228.25×10−5112.46×10−611.51×10−4Problem 7.
Fifth-order linear ODE [26]:(55)Gvx−Gx+15ex+10xex=0,0<x<1,subject to boundary conditions(56)G0=0,G′0=1,G″0=0,G1=0,G′1=−e.The exact solution of this problem isx−x2ex. The second-order approximate solution using LSHPM is(57)G˜x=−36.8878+36.8878ex−24.1329x−11.7548exx−6.68904x2−0.770542x3+0.0888158x4+0.0573868x5+0.0133906x6+0.00206275x7+0.000226806x8+0.0000165216x9−1.3145×10−7x11−2.34824×10−8x13+2.1814×10−9x14.The results are presented in Table7.Table 7
Comparison of second-order absolute error of LSHPM with HPM, OHAM, and B-spline in Problem7.
xExactLSHPMHPMOHAMB-splineSolutionSolutionErrorSolutionErrorError [27]Error [28]000000000.10.09946540.09946541.57×10−140.09946544.85×10−129×10−118×10−30.20.1954240.1954243.63×10−130.1954243.26×10−114×10−101×10−30.30.283470.283471.24×10−120.283478.99×10−115×10−105×10−30.40.3580380.3580381.75×10−120.3580381.66×10−102×10−113×10−30.50.412180.412188.38×10−130.412182.38×10−101×10−98×10−30.60.4373090.4373097.74×10−130.4373092.76×10−102×10−96×10−30.70.4228880.4228881.35×10−120.4228882.56×10−102×10−900.80.3560870.3560876.56×10−130.3560871.74×10−101×10−99×10−30.90.2213640.2213647.23×10−140.2213646.34×10−114×10−109×10−3108.56×10−158.56×10−15−6.57×10−146.57×10−1400Problem 8.
Fifth-order nonlinear ODE [29]:(58)Gvx−G′x25+x3−235+x5=0,0<x<1,along with boundary conditions(59)G0=ln5,G′0=15,G″0=−125,G1=ln6,G′1=16.The exact solution to this problem islnx+5. The zeroth-order approximate solution using LSHPM is(60)G˜x=0.0644833+0.00801314x−0.000801314x2+0.00010495x3−0.0000119326x4−2.38377×10−9x6+0.959934ln5+x.The results are presented in Table8.Table 8
Comparison of zeroth-order absolute error of LSHPM with HPM in Problem8.
xExactLSHPMHPMSolutionSolutionErrorSolutionError01.609441.6094401.6094400.11.629241.629241.50×10−91.629242.36×10−60.21.648661.648669.38×10−91.648671.56×10−50.31.667711.667712.38×10−81.667754.22×10−50.41.68641.68644.09×10−81.686487.67×10−50.51.704751.704755.46×10−81.704861.08×10−40.61.722771.722775.95×10−81.722891.24×10−40.71.740471.740475.24×10−81.740581.15×10−40.81.757861.757863.42×10−81.757947.96×10−50.91.774951.774951.20×10−81.774982.93×10−511.791761.7917601.791760Problem 9.
Sixth-order linear ODE [14]:(61)Gvix−Gx+6ex=0,0<x<1,subject to boundary conditions(62)G0=1,G″0=−1,Giv0=−3,G1=0,G″1=−2e,Giv1=−4e.The exact solution to this problem isGx=1−xex. The first-order approximate solution using LSHPM is(63)G˜x=12.4982−11.4982ex+11.4982x+5.24912x2+1.58304x3+0.354093x4+0.0624853x5+0.00902533x6+0.00109092x7+0.000111561x8+9.64481×10−6x9+6.83273×10−7x10+4.08283×10−8x11.The results are presented in Table9.Table 9
Comparison of first-order absolute error of LSHPM with second-order HPM, HPLM, and ADM in Problem9.
xExactLSHPMHPMHPLMADMSolutionSolutionErrorSolutionErrorError [14]Error [30]011010000.10.9946540.9946541.88×10−150.9946583.98×10−61.1×10−134.0×10−40.20.9771220.9771227.77×10−160.977137.58×10−62.2×10−137.7×10−40.30.9449010.9449011.99×10−150.9449121.04×10−53.0×10−131.0×10−30.40.8950950.8950958.21×10−150.8951071.22×10−53.6×10−131.2×10−30.50.8243610.8243611.33×10−140.8243741.28×10−53.9×10−131.3×10−30.60.7288480.7288481.60×10−140.728861.22×10−53.7×10−131.2×10−30.70.6041260.6041268.99×10−150.6041361.04×10−53.2×10−131.0×10−30.80.4451080.4451081.11×10−150.4451167.58×10−62.4×10−134.0×10−40.90.245960.245962.41×10−150.2459643.98×10−61.3×10−137.7×10−410−3.19×10−163.19×10−161.59×10−161.59×10−161.9×10−150Problem 10.
Sixth-order nonlinear ODE [14]:(64)Gvix−e−xG2x=0,0<x<1,subject to boundary conditions(65)G0=1,G″0=1,Giv0=1,G1=e,G″1=e,Giv1=e.The exact solution to this problem isex. The zeroth-order approximate solution using LSHPM is(66)G˜x=1+0.999876x+0.5x2+0.166857x3+0.0416667x4+0.00833376x5+0.00121254x6+0.000335379x7.The results are presented in Table10.Table 10
Comparison of zeroth-order absolute error of LSHPM with second-order HPM, VIM, and ADM in Problem10.
xExactLSHPMHPMVIMADMSolutionSolutionErrorSolutionErrorError [31]Error [30]011010000.11.105171.105161.22×10−51.105614.43×10−41.2×10−41.2×10−40.21.22141.221382.32×10−51.222258.44×10−42.3×10−42.3×10−40.31.349861.349833.21×10−51.351021.16×10−33.2×10−43.2×10−40.41.491821.491793.78×10−51.493191.36×10−33.8×10−43.8×10−40.51.648721.648683.99×10−51.650161.44×10−34.0×10−44.0×10−40.61.822121.822083.79×10−51.823491.37×10−33.9×10−43.9×10−40.72.013752.013723.22×10−52.014921.16×10−33.3×10−43.3×10−40.82.225542.225522.34×10−52.226398.51×10−42.4×10−42.4×10−40.92.45962.459591.22×10−52.460054.48×10−41.2×10−41.2×10−412.718282.7182802.718284.44×10−162.0×10−92.0×10−9Problem 11.
Seventh-order linear ODE [32]:(67)Gviix−Gx+7ex=0,0<x<1,subject to boundary conditions(68)G0=1,G′0=0,G″0=−1,G‴0=−2,G1=0,G′1=−e,G″1=−2e.The exact solution to this problem is1−xex. The first-order approximate solution using LSHPM is(69)G˜x=7.99997−6.99997ex+6.99997x+2.99998x2+0.833328x3+0.166665x4+0.0249997x5+0.00277773x6+0.000198406x7−2.75823×10−6x9−5.4721×10−7x10−7.9003×10−8x11−6.14598×10−9x12−1.47065×10−9x13.The results are presented in Table11.Table 11
Comparison of first-order absolute error of LSHPM with second-order HPM and VIM in Problem11.
xExactLSHPMHPMVIMSolutionSolutionErrorSolutionErrorError [32]01101000.10.9946540.9946545.55×10−160.9946541.32×10−131.22×10−150.20.9771220.9771221.33×10−150.9771221.55×10−124.44×10−160.30.9449010.9449012.66×10−150.9449015.52×10−129.99×10−160.40.8950950.8950954.99×10−150.8950951.14×10−114.55×10−150.50.8243610.8243618.65×10−150.8243611.67×10−117.32×10−150.60.7288480.7288484.88×10−150.7288481.83×10−111.02×10−140.70.6041260.6041263.10×10−150.6041261.46×10−111.22×10−140.80.4451080.4451082.10×10−150.4451087.56×10−121.50×10−140.90.245960.245961.30×10−150.245961.52×10−121.06×10−1410−2.02×10−162.02×10−16001.17×10−14Problem 12.
Seventh-order nonlinear ODE [32]:(70)Gviix−e−xG2x=0,0<x<1,subject to boundary conditions,(71)G0=1,G′0=1,G″0=1,G‴0=1,G1=e,G′1=e,G″1=e.The exact solution to this problem isex. The zeroth-order approximate solution using LSHPM is(72)G˜x=1+x+0.5x2+0.166667x3+0.041736x4+0.00808598x5+0.00171879x6+0.0000744048x8.The results are presented in Table12.Table 12
Comparison of zeroth-order absolute error of LSHPM with HPM in Problem12.
xExactLSHPMHPMSolutionSolutionErrorSolutionError0110100.11.105171.105174.76×10−91.105174.11×10−80.21.22141.22145.04×10−81.22145.42×10−70.31.349861.349861.60×10−71.349862.28×10−60.41.491821.491822.99×10−71.491836.12×10−60.51.648721.648723.95×10−71.648731.30×10−50.61.822121.822123.90×10−71.822142.46×10−50.72.013752.013752.83×10−72.01384.33×10−50.82.225542.225541.32×10−72.225617.28×10−50.92.45962.45962.44×10−82.459721.18×10−412.718282.7182802.718471.84×10−4


## 4. Results and Discussion

In this article, a new modification of HPM has been introduced by hybriding HPM with LS optimizer. The proposed scheme has been tested agaist various order linear and nonlinear BVPs. The validity of LSHPM has been checked by comparing exact and approximate solutions. For testing the efficiency LSHPM, problems are also solved with HPM and results are compared with LSHPM. This can easily be observed in Tables1–12. These tables signify the efficiency of LSHPM in terms of high accuracy with less computational cost. The convergence of LSHPM can also be observed in Figures 1–12. These figures show that LSHPM is more consistent as compared other mentioned schemes.Figure 1
Graphical representation of solutions and errors in Problem1. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 2
Graphical representation of solutions and errors in Problem2. (a) Comparison of Exact, HPM, and LSHPM Solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 3
Graphical representation of solutions and errors in Problem3. (a) Comparison of Exact, HPM, and LSHPM Solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 4
Graphical representation of solutions and errors in Problem4. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 5
Graphical representation of solutions and errors in Problem5. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 6
Graphical representation of solutions and errors in Problem6. (a) Comparison of HPM and LSHPM solutions. (b) Comparison of absolute residual errors of HPM and LSHPM.
(a)(b)Figure 7
Graphical representation of solutions and errors in Problem7. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 8
Graphical representation of solutions and errors in Problem8. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 9
Graphical representation of solutions and errors in Problem9. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 10
Graphical representation of solutions and errors in Problem10. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 11
Graphical representation of solutions and errors in Problem11. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)Figure 12
Graphical representation of solutions and errors in Problem12. (a) Comparison of Exact, HPM, and LSHPM solutions. (b) Comparison of absolute errors of HPM and LSHPM.
(a)(b)


## 5. Conclusion

In present article, an efficient and reliable modification of HPM is introduced by mixing HPM with the LS optimizer. The proposed scheme is tested against different order linear and nonlinear BVPs. The obtained solutions are compared with HPM and other numerical schemes available in the literature. Analysis of results shows that LSHPM is more consistent in terms of accuracy with less computational cost and can be used in different areas of science and technology.



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*Source: 7059194-2021-10-05.xml*