# Solutions of Stiff Systems of Ordinary Differential Equations Using Residual Power Series Method

Authors: Mubashir Qayyum; Qursam Fatima
Journal: Journal of Mathematics
 (2022)

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## Abstract

The stiff differential equations occur in almost every field of science. These systems encounter in mathematical biology, chemical reactions and diffusion process, electrical circuits, meteorology, mechanics, and vibrations. Analyzing and predicting such systems with conventional numerical techniques require more time and memory; still accurate solution is completely uneconomical and uncertain. Most of the numerical techniques have stability issues while dealing with stiff systems. To overcome these limitations, residual power series method (RPSM) is proposed for stiff systems of differential equations (DEs). RPSM is applied to various linear and nonlinear stiff systems, and closed-form solutions are achieved. This indicates the effectiveness of proposed scheme for stiff family of DEs. Since this method leads to better results with less computational cost, it can be extended for more complex systems which arise in different areas of engineering and sciences.


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## Body

## 1. Introduction

Stiff system of ordinary differential equations (ODEs) arises in modeling of various real-world physical problems. Solutions of such systems are of great interest for the scientific community. Several numerical, analytical, and seminumerical schemes have been proposed for the solution of such systems. Spijker provided comprehensive review of different stiffness in [1]. He also shows theoretical analysis of numerical methods in stiff situation. Darvishi et al. performed comparative analysis for stiff systems using variation iteration method (VIM) and Adomian decomposition method (ADM) [2]. Abelman and Patidar attempted combustion equation which falls in stiff family of differential equations using finite difference methods [3].Classical numerical techniques, such as finite difference and finite element methods, are computationally expensive and usually affected by round-off errors, which may lead to inaccurate results. Analytical methods are widely used for solving functional equations which are very restrictive for stiff systems and can be used only in special cases [4, 5]. In 2013, Arqub et al. proposed a method based on Taylor series for the solution of functional equations and systems [6]. This method overcomes the limitation of pure numerical schemes to some extent as it does not require linearization, perturbation, or discretization. Also, it has less computational cost and does not effected by round-off errors. In this method, the solution is considered as the sum of an infinite series, which converges rapidly to the exact solution. This technique is used by many authors for solving linear and nonlinear DEs and systems. Arqub used RPSM to Lane Emden equation (LEE) in [7]. Al-Smadi used this method to different linear and nonlinear IVPs [8]. Mahmood and Yousaf apply the RPSM to Boussinesq–Burgers equations [9]. Kumar and Kumar applied this method to fractional burger type equations in [10]. Kumar et al. also extended this method to fractonal diffusion equation (16). Alquran analyzed fractional form of drainange equation using RPSM [11]. Arqub also altered RPSM to fuzzy DEs [12]. Yao and Kumar applied RPSM to fractional Fokker–Planck equations (21). Kumar and Baleanu apply this scheme to fractional coupled Boussinesq–Burgers equations [13]. El-Ajou et al. used RPSM to nonlinear fractional KdV-Burgers equation [14]. Arqub et al. also apply this method to solve the higher order IVPs [6]. Komashynska et al. used this method to solve the system of multi-panto-graph differential equations in [15]. Koashynska et al. also analyzed Fredhlom integral equation with the same method [16].Analysis of stiff systems with traditional numerical techniques requires more time and effort; still accurate solution is uncertain in most of the cases due to stability issues. To overcome these issues, RPSM is proposed for stiff systems in this manuscript. PRSM is tested against various linear and nonlinear stiff systems to check the efficiency and effectiveness of proposed scheme. In the rest of the manuscript, the proposed approach is presented in Section 2, which is then applied to various linear and nonlinear stiff systems in Section 3, while the conclusion is in Section 4.


## 2. Residual Power Series Method for Stiff Systems

In this section, RPSM is explained for thenth-order system of IVPs. Consider the following system:(1)ϕ1′r=f1r,ϕ1r,ϕ2r,…,ϕnr,ϕ2′r=f2r,ϕ1r,ϕ2r,…,ϕnr,⋮ϕn′r=fnr,ϕ1r,ϕ2r,…,ϕnr,with initial conditions(2)ϕ1r0=α1,ϕ2r0=α2,…ϕnr0=αn.where r∈r0,r0+a and fi:r0,r0+a×Rn⟶R are nonlinear continuous functions of r,ϕ1,ϕ2,…,ϕn and ϕi′s are unknown functions to be determined. r0 and a are real finite constants with a≥0. It is assumed that fi and ϕ are analytical functions on the given interval.Let us assume solution of the form(3)ϕir=∑j=0∞ϕi,jrj,i=1,2,…,n.where ϕi,j are terms of approximations and given as ϕi,jr  =  Ai,jrj. When j=0, ϕi,jr satisfy the initial conditions (2), where ϕi,jr are the initial guesses of ϕir, we have Ai,0=ϕi,0r0=ϕir0, i=1,2,…,n.If we chooseϕi,0r0=ϕir0 as initial guesses approximations of ϕir, then we can calculate ϕi,jr for j=1,2,⋯ and approximate the solutions ϕir of system of IVPs (1) by the kth truncated series(4)ϕikr=∑j=0kAi,jr−r0j,i=0,1,2,…,n.Rewrite the system in the following form(5)ϕi′r−fir,ϕ1r,ϕ2r,…,ϕnr=0,i=1,2,…,n.Using (4) in (5), the kth residual function is(6)Resikr=∑j=1kjAi,jr−r0j−1−fr,∑j=0kA1,jr−r0j,∑j=0kA2,jr−r0j,…,∑j=0kAn,jr−r0j,and the ∞th residual function is(7)Resi∞r=limk⟶∞Resikr,i=0,1,2,…,n.This show thatResi∞r is infinitely many times differentiable at r=r0.(8)dk−ndrk−nResikr0=0,k=1,2,⋯,which is iterative process of obtaining higher order solution.


## 3. Application of RPSM to Stiff Systems

### 3.1. Test Problem 1

Consider the following linear stiff system [17]:(9)ϕ1′r=−2ϕ1r+ϕ2r+2Sinr,(10)ϕ2′r=−ϵ−1+2ϕ1r+ϵ−1+1ϕ2r−Cosr+Sinr,with initial conditions(11)ϕ10=2,ϕ20=3.Exact solutions of the problem is(12)ϕ1r=2e−r+Sinr,ϕ2r=2e−r+Cosr.Let we supposeϵ=1. Also, the approximate solution for the given problem is(13)ϕ1r=∑x=0kAxrx,k=0,1,2,…n−1,(14)ϕ2r=∑x=0kBxrx,k=0,1,2,…n−1.Using initial condition withk=0 in (13) and (14) gives ϕ10=A0=2,ϕ20=B0=3.Rewrite (9) as(15)ϕ1′r+2ϕ1r−ϕ2r−2Sinr=0,(16)ϕ2′r+3ϕ1r−2ϕ2r−Cosr+Sinr=0.Now, plugging thekth truncated series ((13)–(14)) in (15) and (16), we obtain(17)Res1kr=∑x=1kxAxrx−1+2∑x=0kAxrx−∑x=0kBxrx−2Sinrand(18)Res2kr=∑x=1kxBxrx−1+3∑x=0kAxrx−2∑x=0kBxrx−Cosr+Sinr.To get1st approximate solution, use (17) and (18) along with (8) with k=1. This leads to A1=−1, B1=−2. Proceeding in a similar pattern, RPS solution for the given problem is(19)ϕ1r=2−r+r2−r32+r412−r5120+r6360−r71680+r820160−r9362880+r101814400−⋯,(20)ϕ2r=3−2r+r22−r33+r48−r560+r6720−r72520+r813440−r9181440+r103628800−⋯,(21)ϕ1r=2e−r+Sinr,ϕ2r=2e−r+Cosr,which is closed-form solution.


### 3.2. Test Problem 2

Consider the following nonlinear stiff system [17]:(22)ϕ1′r=−ϵ−1+2ϕ1r+ϵ−1ϕ2r2,(23)ϕ2′r=ϕ1r−ϕ2r1+ϕ2r,with initial conditions(24)ϕ10=1,ϕ20=1.Exact solutions of the given system are(25)ϕ1r=e−2r,ϕ2r=e−r.Let us supposeϵ=1. Also, suppose that approximate solution for the given system is(26)ϕ1r=∑x=0kAxrx,k=0,1,2,…,(27)ϕ2r=∑x=0kBxrx,k=0,1,2,….The values ofA0 and B0 can be computed from (26) and (27) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1Rewrite (22) and (23) as follows:(28)ϕ1′r+3ϕ1r−ϕ2r2=0,(29)ϕ2′r−ϕ1r+ϕ2r1+ϕ2r=0.Plugging thekth truncated series (26), (27) in (28) and (29), we get(30)Res1kr=∑x=1kxAxrx−1+3∑x=0kAxrx−∑x=0kBxrx2,(31)Res2kr=∑x=1kxBxrx−1−∑x=0kAxrx+∑x=0kBxrx1+∑x=0kBxrx.To obtain1st approximate solution put k=1 in (30) and (31), then use these residual functions in (8) will give A1=−2 and B1=−1. Proceeding in this way, the higher order coefficients can be obtained, and the RPS solution is(32)ϕ1r=1−2r+2r2−4r33+2r43−4r515+4r645−8r7315+2r8315−4r92835+4r1014175−⋯,(33)ϕ2r=1−r+r22−r36+r424−r5120+r6720−r75040+r840320−r9362880+r103628800−⋯,(34)ϕ1r=e−2r,ϕ2r=e−r.which is closed-form (exact) solution.


### 3.3. Test Problem 3

Consider the following linear stiff system [18]:(35)ϕ1′r=−ϕ1r+95ϕ2r,(36)ϕ2′r=−ϕ1r−97ϕ2r,with initial conditions(37)ϕ10=1,ϕ20=1.Exact solutions of the problem are(38)ϕ1r=95e−2r−48e−96r47,ϕ2r=48e−96r−e−2r47.Approximate solution for the given problem is(39)ϕ1r=∑x=0kAxrx,k=0,1,2,⋯,(40)ϕ2r=∑x=0kBxrx,k=0,1,2,⋯.The values ofA0 and B0 can be computed from (39) and (40) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1Rewriting the given (35) and (36) in the following form:(41)ϕ1′r=−ϕ1r+95ϕ2r,(42)ϕ2′r=−ϕ1r−97ϕ2r,and plugging the kth truncated series (39) and (40) in (41) and (42) to get the following residual functions(43)Res1kr=∑x=1kxAxrx−1+∑x=0kAxrx−95∑x=0kBxrxand(44)Res2kr=∑x=1kxBxrx−1+∑x=0kAxrx+97∑x=0kBxrx.To obtain1st approximate solution put k=1 in (43) and (44), then use these residual functions in (8) will give A1=94 and B1=−98 . Proceeding in this way, we can obtain higher order coefficients, and RPS solution is(45)ϕ1r=1+94r−4702r2+4517723r3−108427183r4+104090130815r5−4996326316445r6+4796473264504315r7−57557679174238315r8+55255372007272282835r9−26522578563490732414175r10+⋯⋯,(46)ϕ2r=1−98r+4706r2−4517803r3+108427223r4−104090131615r5+4996326317245r6−1370420932729r7+57557679174242315r8−789362457246748405r9+26522578563490733214175r10+⋯⋯(47)ϕ1r=95e−2r−48e−96r47,ϕ2r=48e−96r−e−2r47.which is closed-form solution.


### 3.4. Test Problem 4

Consider the following nonlinear stiff system [18]:(48)ϕ1′r=−1002ϕ1r+1000ϕ2r2,(49)ϕ2′r=ϕ1r−ϕ2rϕ2r2,with initial conditions(50)ϕ10=1,ϕ20=1.Exact solutions of the given problem is(51)ϕ1r=e−2r,ϕ2r=e−r.Approximate solution for the system is(52)ϕ1r=∑x=0kAxrx,k=0,1,2,⋯,(53)ϕ2r=∑x=0kBxrx,k=0,1,2,⋯.The values ofA0 and B0 can be computed from (52) and (53) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1Rewriting the given system in the following form:(54)ϕ1′r=−1002ϕ1r+1000ϕ2r2=0,(55)ϕ2′r=ϕ1r−ϕ2rϕ2r2=0,and plugging the kth truncated series (52) and (53) in (54) and (55) to obtain the following residual functions(56)Res1kr=∑x=1kxAxrx−1+1002∑x=0kAxrx−1000∑x=0kBxrx2and(57)Res2kr=∑x=1kxBxrx−1−∑x=0kAxrx+∑x=0kBxrx+∑x=0kBxrx2.To obtain1st approximate solution put k=1 in (56) and (57), we use these residual functions in (8) leading to A1=−2 and B1=−1 . Proceeding in this way, higher order coefficients can easily be found, and the RPS solution is(58)ϕ1r=1−2r+2r2−4r33+2r43−4r515+4r645−8r7315+2r8315−4r92835+4r1014175−⋯,(59)ϕ2r=1−r+r22−r36+r424−r5120+r6720−r75040+r840320−r9362880+r103628800−⋯,(60)ϕ1r=e−2r,ϕ2r=e−r.which is closed-form solution.


### 3.5. Test Problem 5

Consider the following system [19]:(61)ϕ1′r=ϕ3r−Cosr,(62)ϕ2′r=ϕ3r−er,(63)ϕ3′r=ϕ1r−ϕ2r,with initial conditions(64)ϕ10=1,ϕ20=0,ϕ30=2.Exact solutions of the problem are(65)ϕ1r=er,ϕ2r=Sinr,ϕ3r=er+Cosr.Approximate solution for the given problem is(66)ϕ1r=∑x=0kAxrx,ϕ2r=∑x=0kBxrx,ϕ3r=∑x=0kCxrx,k=0,1,2,⋯The values ofA0, B0, and Ci can be computed from (66) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=0,ϕ30=C0=2Rewriting the given IVP in the following form:(67)ϕ1′r−ϕ3r+Cosr=0,(68)ϕ2′r−ϕ3r+er=0,(69)ϕ3′r−ϕ1r+ϕ2r=0.Plugging thekth truncated series (66) in (67), (68), and (69) to obtain the following residual functions(70)Res1kr=∑x=1kxAxrx−1−∑x=0kCxrx+Cosr,(71)Res2kr=∑x=1kxBxrx−1−∑x=0kCxrx+er,and(72)Res3kr=∑x=1kxCxrx−1−∑x=0kAxrx+∑x=0kBxrx.For1st approximate solution, put k=1 in (70)–(72), we use these functions in (8) to get the following coefficients:(73)A1=1,B1=1,C1=1.Repeat the recursive process to obtain higher order coefficients. Hence, RPS solution is(74)ϕ1r=1+r+r22+r36+r424+r5120+r6720+r75040+r840320+r9362880+r103628800+⋯,(75)ϕ2r=r−r36+r5120−r75040+r9362880−⋯,(76)ϕ3r=2+r+r36+r412+r5120+r75040+r820160+r9362880+⋯(77)ϕ1r=er,ϕ2r=Sinr,ϕ3r=er+Cosr.which is closed-form solution.


### 3.6. Test Problem 6

Consider the following nonlinear system [19]:(78)ϕ1′r=2ϕ2r2,(79)ϕ2′r=e−rϕ1r,(80)ϕ3′r=ϕ2r+ϕ3r,with initial conditions(81)ϕ10=1,ϕ20=0,ϕ30=2.Exact solutions of the given problem is(82)ϕ1r=e2r,ϕ2r=er,ϕ3r=rer.Approximate solution for the given problem is(83)ϕ1r=∑x=0kAxrx,ϕ2r=∑x=0kBxrx,ϕ3r=∑x=0kCxrx,k=0,1,2,⋯The values ofA0, B0, and Ci can be computed from (83) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1,ϕ30=C0=0.Rewriting the given system in the following form:(84)ϕ1′r−2ϕ2r2=0,(85)ϕ2′r−e−rϕ1r=0,(86)ϕ3′r−ϕ2r−ϕ3r=0.Plugging thekth truncated series (83) in (84), (85), and (86) to obtain the following residual functions:(87)Res1kr=∑x=1kxAxrx−1−2∑x=0kBxrx2,(88)Res2kr=∑x=1kxBxrx−1−e−r∑x=0kAxrx,and(89)Res3kr=∑x=1kxCxrx−1−∑x=0kBxrx−∑x=0kCxrx.For1st approximate solution, put k=1 in (87)–(89), we use these functions in (8) to get the following coefficients:(90)A1=2,B1=1,C1=1.Repeat the process to get higher order coefficients of the series solution. Hence, RPS solution is(91)ϕ1r=1+2r+2r2+4r33+2r43+4r515+4r645+⋯,(92)ϕ2r=1+r+r22+r36+r424+r5120+r6720+⋯,(93)ϕ3r=r+r2+r32+r46+r524+r6120+⋯,(94)ϕ1r=e2r,ϕ2r=er,ϕ3r=rer.which is closed-form solution.


## 3.1. Test Problem 1

Consider the following linear stiff system [17]:(9)ϕ1′r=−2ϕ1r+ϕ2r+2Sinr,(10)ϕ2′r=−ϵ−1+2ϕ1r+ϵ−1+1ϕ2r−Cosr+Sinr,with initial conditions(11)ϕ10=2,ϕ20=3.Exact solutions of the problem is(12)ϕ1r=2e−r+Sinr,ϕ2r=2e−r+Cosr.Let we supposeϵ=1. Also, the approximate solution for the given problem is(13)ϕ1r=∑x=0kAxrx,k=0,1,2,…n−1,(14)ϕ2r=∑x=0kBxrx,k=0,1,2,…n−1.Using initial condition withk=0 in (13) and (14) gives ϕ10=A0=2,ϕ20=B0=3.Rewrite (9) as(15)ϕ1′r+2ϕ1r−ϕ2r−2Sinr=0,(16)ϕ2′r+3ϕ1r−2ϕ2r−Cosr+Sinr=0.Now, plugging thekth truncated series ((13)–(14)) in (15) and (16), we obtain(17)Res1kr=∑x=1kxAxrx−1+2∑x=0kAxrx−∑x=0kBxrx−2Sinrand(18)Res2kr=∑x=1kxBxrx−1+3∑x=0kAxrx−2∑x=0kBxrx−Cosr+Sinr.To get1st approximate solution, use (17) and (18) along with (8) with k=1. This leads to A1=−1, B1=−2. Proceeding in a similar pattern, RPS solution for the given problem is(19)ϕ1r=2−r+r2−r32+r412−r5120+r6360−r71680+r820160−r9362880+r101814400−⋯,(20)ϕ2r=3−2r+r22−r33+r48−r560+r6720−r72520+r813440−r9181440+r103628800−⋯,(21)ϕ1r=2e−r+Sinr,ϕ2r=2e−r+Cosr,which is closed-form solution.


## 3.2. Test Problem 2

Consider the following nonlinear stiff system [17]:(22)ϕ1′r=−ϵ−1+2ϕ1r+ϵ−1ϕ2r2,(23)ϕ2′r=ϕ1r−ϕ2r1+ϕ2r,with initial conditions(24)ϕ10=1,ϕ20=1.Exact solutions of the given system are(25)ϕ1r=e−2r,ϕ2r=e−r.Let us supposeϵ=1. Also, suppose that approximate solution for the given system is(26)ϕ1r=∑x=0kAxrx,k=0,1,2,…,(27)ϕ2r=∑x=0kBxrx,k=0,1,2,….The values ofA0 and B0 can be computed from (26) and (27) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1Rewrite (22) and (23) as follows:(28)ϕ1′r+3ϕ1r−ϕ2r2=0,(29)ϕ2′r−ϕ1r+ϕ2r1+ϕ2r=0.Plugging thekth truncated series (26), (27) in (28) and (29), we get(30)Res1kr=∑x=1kxAxrx−1+3∑x=0kAxrx−∑x=0kBxrx2,(31)Res2kr=∑x=1kxBxrx−1−∑x=0kAxrx+∑x=0kBxrx1+∑x=0kBxrx.To obtain1st approximate solution put k=1 in (30) and (31), then use these residual functions in (8) will give A1=−2 and B1=−1. Proceeding in this way, the higher order coefficients can be obtained, and the RPS solution is(32)ϕ1r=1−2r+2r2−4r33+2r43−4r515+4r645−8r7315+2r8315−4r92835+4r1014175−⋯,(33)ϕ2r=1−r+r22−r36+r424−r5120+r6720−r75040+r840320−r9362880+r103628800−⋯,(34)ϕ1r=e−2r,ϕ2r=e−r.which is closed-form (exact) solution.


## 3.3. Test Problem 3

Consider the following linear stiff system [18]:(35)ϕ1′r=−ϕ1r+95ϕ2r,(36)ϕ2′r=−ϕ1r−97ϕ2r,with initial conditions(37)ϕ10=1,ϕ20=1.Exact solutions of the problem are(38)ϕ1r=95e−2r−48e−96r47,ϕ2r=48e−96r−e−2r47.Approximate solution for the given problem is(39)ϕ1r=∑x=0kAxrx,k=0,1,2,⋯,(40)ϕ2r=∑x=0kBxrx,k=0,1,2,⋯.The values ofA0 and B0 can be computed from (39) and (40) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1Rewriting the given (35) and (36) in the following form:(41)ϕ1′r=−ϕ1r+95ϕ2r,(42)ϕ2′r=−ϕ1r−97ϕ2r,and plugging the kth truncated series (39) and (40) in (41) and (42) to get the following residual functions(43)Res1kr=∑x=1kxAxrx−1+∑x=0kAxrx−95∑x=0kBxrxand(44)Res2kr=∑x=1kxBxrx−1+∑x=0kAxrx+97∑x=0kBxrx.To obtain1st approximate solution put k=1 in (43) and (44), then use these residual functions in (8) will give A1=94 and B1=−98 . Proceeding in this way, we can obtain higher order coefficients, and RPS solution is(45)ϕ1r=1+94r−4702r2+4517723r3−108427183r4+104090130815r5−4996326316445r6+4796473264504315r7−57557679174238315r8+55255372007272282835r9−26522578563490732414175r10+⋯⋯,(46)ϕ2r=1−98r+4706r2−4517803r3+108427223r4−104090131615r5+4996326317245r6−1370420932729r7+57557679174242315r8−789362457246748405r9+26522578563490733214175r10+⋯⋯(47)ϕ1r=95e−2r−48e−96r47,ϕ2r=48e−96r−e−2r47.which is closed-form solution.


## 3.4. Test Problem 4

Consider the following nonlinear stiff system [18]:(48)ϕ1′r=−1002ϕ1r+1000ϕ2r2,(49)ϕ2′r=ϕ1r−ϕ2rϕ2r2,with initial conditions(50)ϕ10=1,ϕ20=1.Exact solutions of the given problem is(51)ϕ1r=e−2r,ϕ2r=e−r.Approximate solution for the system is(52)ϕ1r=∑x=0kAxrx,k=0,1,2,⋯,(53)ϕ2r=∑x=0kBxrx,k=0,1,2,⋯.The values ofA0 and B0 can be computed from (52) and (53) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1Rewriting the given system in the following form:(54)ϕ1′r=−1002ϕ1r+1000ϕ2r2=0,(55)ϕ2′r=ϕ1r−ϕ2rϕ2r2=0,and plugging the kth truncated series (52) and (53) in (54) and (55) to obtain the following residual functions(56)Res1kr=∑x=1kxAxrx−1+1002∑x=0kAxrx−1000∑x=0kBxrx2and(57)Res2kr=∑x=1kxBxrx−1−∑x=0kAxrx+∑x=0kBxrx+∑x=0kBxrx2.To obtain1st approximate solution put k=1 in (56) and (57), we use these residual functions in (8) leading to A1=−2 and B1=−1 . Proceeding in this way, higher order coefficients can easily be found, and the RPS solution is(58)ϕ1r=1−2r+2r2−4r33+2r43−4r515+4r645−8r7315+2r8315−4r92835+4r1014175−⋯,(59)ϕ2r=1−r+r22−r36+r424−r5120+r6720−r75040+r840320−r9362880+r103628800−⋯,(60)ϕ1r=e−2r,ϕ2r=e−r.which is closed-form solution.


## 3.5. Test Problem 5

Consider the following system [19]:(61)ϕ1′r=ϕ3r−Cosr,(62)ϕ2′r=ϕ3r−er,(63)ϕ3′r=ϕ1r−ϕ2r,with initial conditions(64)ϕ10=1,ϕ20=0,ϕ30=2.Exact solutions of the problem are(65)ϕ1r=er,ϕ2r=Sinr,ϕ3r=er+Cosr.Approximate solution for the given problem is(66)ϕ1r=∑x=0kAxrx,ϕ2r=∑x=0kBxrx,ϕ3r=∑x=0kCxrx,k=0,1,2,⋯The values ofA0, B0, and Ci can be computed from (66) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=0,ϕ30=C0=2Rewriting the given IVP in the following form:(67)ϕ1′r−ϕ3r+Cosr=0,(68)ϕ2′r−ϕ3r+er=0,(69)ϕ3′r−ϕ1r+ϕ2r=0.Plugging thekth truncated series (66) in (67), (68), and (69) to obtain the following residual functions(70)Res1kr=∑x=1kxAxrx−1−∑x=0kCxrx+Cosr,(71)Res2kr=∑x=1kxBxrx−1−∑x=0kCxrx+er,and(72)Res3kr=∑x=1kxCxrx−1−∑x=0kAxrx+∑x=0kBxrx.For1st approximate solution, put k=1 in (70)–(72), we use these functions in (8) to get the following coefficients:(73)A1=1,B1=1,C1=1.Repeat the recursive process to obtain higher order coefficients. Hence, RPS solution is(74)ϕ1r=1+r+r22+r36+r424+r5120+r6720+r75040+r840320+r9362880+r103628800+⋯,(75)ϕ2r=r−r36+r5120−r75040+r9362880−⋯,(76)ϕ3r=2+r+r36+r412+r5120+r75040+r820160+r9362880+⋯(77)ϕ1r=er,ϕ2r=Sinr,ϕ3r=er+Cosr.which is closed-form solution.


## 3.6. Test Problem 6

Consider the following nonlinear system [19]:(78)ϕ1′r=2ϕ2r2,(79)ϕ2′r=e−rϕ1r,(80)ϕ3′r=ϕ2r+ϕ3r,with initial conditions(81)ϕ10=1,ϕ20=0,ϕ30=2.Exact solutions of the given problem is(82)ϕ1r=e2r,ϕ2r=er,ϕ3r=rer.Approximate solution for the given problem is(83)ϕ1r=∑x=0kAxrx,ϕ2r=∑x=0kBxrx,ϕ3r=∑x=0kCxrx,k=0,1,2,⋯The values ofA0, B0, and Ci can be computed from (83) by taking k=0 along with initial conditions. ϕ10=A0=1,ϕ20=B0=1,ϕ30=C0=0.Rewriting the given system in the following form:(84)ϕ1′r−2ϕ2r2=0,(85)ϕ2′r−e−rϕ1r=0,(86)ϕ3′r−ϕ2r−ϕ3r=0.Plugging thekth truncated series (83) in (84), (85), and (86) to obtain the following residual functions:(87)Res1kr=∑x=1kxAxrx−1−2∑x=0kBxrx2,(88)Res2kr=∑x=1kxBxrx−1−e−r∑x=0kAxrx,and(89)Res3kr=∑x=1kxCxrx−1−∑x=0kBxrx−∑x=0kCxrx.For1st approximate solution, put k=1 in (87)–(89), we use these functions in (8) to get the following coefficients:(90)A1=2,B1=1,C1=1.Repeat the process to get higher order coefficients of the series solution. Hence, RPS solution is(91)ϕ1r=1+2r+2r2+4r33+2r43+4r515+4r645+⋯,(92)ϕ2r=1+r+r22+r36+r424+r5120+r6720+⋯,(93)ϕ3r=r+r2+r32+r46+r524+r6120+⋯,(94)ϕ1r=e2r,ϕ2r=er,ϕ3r=rer.which is closed-form solution.


## 4. Conclusion

In this manuscript, the residual power series method is applied to various linear and nonlinear stiff systems of ODEs and closed-form solutions are obtained. These solutions shows that RPSM is simple and easy to use technique for stiff systems and can effectively be used for wide variety of problems in science and engineering.



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*Source: 7887136-2022-03-04.xml*