# Modified Adomian Decomposition Method to Solve Generalized Emden–Fowler Systems for Singular IVP

Authors: Aisha Abdullah Alderremy; Tarig M. Elzaki; Mourad Chamekh
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

An adeptness modified Adomian decomposition method (MADM) is proposed to solve a generalized system of Emden–Fowler type. By a few examples, it is shown that this method can overcome a singular initial value problem.


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## Body

## 1. Introduction

Various methods can be used to approach the analytical solution for nonlinear differential systems, such as the Adomian method [1], the variation iteration method [2], differential transform [3], and novel iterative method [4]. The Adomian decomposition method (ADM) and MADM have made it possible to solve a few ODE problems in the recent years. For this reason, we propose MADM for solving the following system:(1a)u′′t+α1tu′t+ftFu,v=h1t,(1b)v′′t+α2tv′t+gtGu,v=h2t,(1c)u0=u0,v0=v0,u′0=u1,v′0=v1,whereα1,α2,u0,v0,u1,v1,∈R.This is a generalized system containing the equations which are called Emden–Fowler equations and that would be found in Wong [5]. An equation of this type has been of interest for many searchers in various areas of physics and mathematics because of its deep structure. Examples are as follows: in chemical field, Zhao et al. mention a problem of an autocatalytic chemical reaction [6], Zhai et al propose an application on the elasticity [7], and Li et al. present in [8] a developing model for an enterprise.As regards the Emden–Fowler systems (EFS), Bidaut-Veron et al. in [9] study an EFS of radial type, dealt specifically with the p-laplacian operator. Böhmer et al. in [10] present an EFS modeling an application for a polytropic perfect fluid.Here, motivated by the works of Wazwaz [11] where the ADM recommend for getting an analytic solution and Biazard et al. [12] in which the MADM is used, we propose an efficient method to calculate an exact and/or an approximate solution for a larger class of generalized EFS.This paper is organized as follows. Section2 contains the proposed MADM for resolving the EFS. A few examples are given in Section 3 followed by a comparison between the exact solutions obtained by MADM and many numerical results are treated in Section 4.


## 2. Modified Adomian Decomposition Method

The MADM is an excellent technique, which is helping us to take a step forward in the calculus of analytic approximate solutions for various physical problems. We need to solve serious mathematical problems which have not yet has an exact solution. The basic idea of MADM is to consider the nonlinear equation in the formal form:(2)LUt+NUt=Ht,with U(t)=(u(t),v(t))T and H=(h1(t),h2(t))T, where L is a linear operator and N is a nonlinear operator.The MADM admits the decomposition into an infinite series of components. We assume(3)U=∑i=0∞uit,∑i=0∞vit,and(4)Fu,v=∑n=0∞Anu0,⋯,un,v0,⋯,vn,Gu,v=∑n=0∞Bnu0,⋯,un,v0,⋯,vn,where An,Bn are called the Adomian polynomials for all n∈N.In the case of problem (1a), (1b), and (1c), the inverse operator L-1=∫0x∫0x(·)dxdx. Applying the inverse operator L-1, on both sides of (1a)-(1b) and considering (4), we get(5a)∑n=0∞unt=u0+tu′0+L-1∑n=0∞antn-pL-1∑n=0∞antn+L-1h1t-L-1R∑n=0∞unt+Anu0,⋯,un,v0,⋯,vn,(5b)∑n=0∞vnt=v0+tv′0+L-1∑n=0∞bntn-pL-1∑n=0∞bntn+L-1h2t-L-1Q∑n=0∞vnt+Bnu0,⋯,un,v0,⋯,vn,where R(u)=α1/tu′ and Q(v)=α2/tv′. We note that we have a similar expression obtained from u and v. We also note that we have intercalating the expression L-1(∑n=0∞antn)-pL-1(∑n=0∞antn) in (5a) similar to (5b), where p is an artificial parameter and for all i∈N, ai and bi are unknown coefficients.The calculus of the componentsun and vn, i∈N, could be summarized as follows using (5a) and (5b). We can see Adomian et al. in [1] for more details.We findU0=(u0,v0) as follow:(6a)u0t=u0+tu′0+L-1∑n=0∞antn,(6b)v0t=v0+tv′0+L-1∑n=0∞bntn.And U1=(u1,v1) is given by(7a)u1t=-pL-1∑n=0∞antn+L-1h1t-L-1Ru0+A0u0,v0,(7b)v1t=-pL-1∑n=0∞bntn+L-1h2t-L-1Qv0+B0u0,v0.For Un=(un,vn),n∈N∗ we have(8a)unt=-L-1Run-1+An-1u0,⋯,un-1,v0,⋯,vn-1,(8b)vnt=-L-1Qun-1+Bn-1u0,⋯,un-1,v0,⋯,vn-1.To find the coefficients an and bn, we put u1=v1=0; we immediately verify un=vn=0 for all n≥1, and in the last, set p=1, to find the solution of problem (1a), (1b), and (1c) in the form(9a)ut=u0+tu1+L-1∑n=0∞antn(9b)vt=v0+tv1+L-1∑n=0∞bntn


## 3. Numerical Results

Thanks to precise examples of a singular initial value linear system, nonlinear systems, two unknowns and three unknowns, we can illustrate the suggested method. As well, the system of Emden–Fowler equations is solved to demonstrate the applicability and suppleness of the method for solving even more problems. For simplicity reasons and without limiting the generality of the foregoing, we have considered the initial conditionsu(0)=v(0)=±1,u′(0)=v′(0)=0 in all the examples of this paper.Example 1.
First, we aim to illustrate the MADM method by using the following system of linear equations of Emden–Fowler type withα1=2 and α2=3.(10a)u′′t+2tu′t-6+4t2v=6,(10b)v′′t+3tv′t-8+4t2u=-4t4-8t2,(10c)u0=v0=1,u′0=v′0=0.
To solve this problem by the new method, let us write (10a), (10b), and (10c) using (5a), (5b) as follows:(11)∑n=0∞un=u0+tu′0+L-1∑n=0∞antn-pL-1∑n=0∞antn+L-16-L-1Run,vn+L-1∑n=0∞Anu0,⋯,un,v0,⋯,vn,∑n=0∞vnt=v0+tv′0+L-1∑n=0∞bntn-pL-1∑n=0∞bntn+L-1-4t4-8t2-L-1R∑n=0∞vnt+L-1Bnu0,⋯,un,v0,⋯,vn.
Using (6a) and (6b) we get(12)u0t=u0+tu′0+L-1∑n=0∞antn=1+a02t2+a16t3+a212t4+a320t5+⋯(13)v0t=v0+tv′0+L-1∑n=0∞bntn=1+b02t2+b16t3+b212t4+b320t5+⋯
The principle of the MADM is to compute the coefficientsan and bn, n≥0, by putting u1(t)=v1(t)=0 and using the expressions (7a) and (7b)(14)u1t=-pL-1∑n=0∞antn+L-16-L-1Ru0,v0,v1t=-pL-1∑n=0∞bntn+L-1-4t4-8t2-L-1Qu0,v0.Thus,(15)u1t=3t2-L-12a0+a1t+a23t2+a32t3+2a45t4+⋯+L-16+4t21+b02t2+b16t3+b212t4+⋯-pL-1a0+a1t+a2t2+a3t3+a4t4+⋯=0.
And(16)v1t=-215t6-23t4-L-13b0+3b12t+b2t2+3b34t3+3b45t4+⋯+L-18+4t21+a02t2+1a16t3+a212t4+⋯-pL-1b0+b1t+b2t2+b3t3+a4t4+⋯=0.
By equating the coefficients oftn and setting p=1, we can easily determine an and bn by (i)
To finda0 and b0(17)3-a0+3-a02=0⇒a0=4-3b02+4-b02=0⇒b0=2.(ii)
We havea1=b1=0.(iii)
To finda2 and b2(18)-a218+b04+13-a212=0⇒a2=6-23+-b212+a03+13-b212=0⇒b2=6.Also we can show that a3=b3=0 and a4=b4=5
From (9a) and (9b) we get the solution of system (10a), (10b), and (10c)(19)ut=1+L-14+6t2+5t4+⋯=1+2t2+t42+t66+⋯vt=1+L-12+6t2+5t4+⋯=1+t2+t42+t66+⋯.
This gives the exact solution of (10a), (10b), and (10c)(20)ut,vt=t2+expt2,expt2.Example 2.
Consider the cylindrical and spherical equation which is the system of nonlinear Emden–Fowler equations(21a)u′′t+1tu′t+vtu2t-4t2+5ut=0,(21b)v′′t+2tv′t+utv2t-4t2-5ut=0,(21c)u0=1,v0=1,u′0=v′0=0.In the same way as the above example, we substitute the expressions of R(u,v)=1/tu′-(4t2+5)u, Q(u,v)=2/tv′-(4t2-5)v, An=vnun2, and Bn=unvn2 in (5a) and (5b). And then we use (6a) and (6b); we obtain(22)u0t=1+a02t2+a16t3+a212t4+a320t5+⋯v0t=-1+b02t2+b16t3+b212t4+b320t5+⋯To Find an and bn, n≥0, we put u1(t)=v1(t)=0 and p=1 using (7a) and (7b). By equating the coefficients of t2,t4,⋯, we get(23)-a02-a02+52-12=0⇒a0=2and(24)-b02-b0-52-12=0⇒b0=-2,also a1=b1=0, a2=b2=6, a3=b3=0 and a4=5,b4=-5.
Then the analytic solution of system (21a), (21b), and (21c) is(25)ut,vt=1+t2+t42+t66+⋯,-1-t2+t42-t66+⋯=expt2,exp-t2.Example 3.
We can also extend the examples to a system of three unknowns containing linear or nonlinear Emden–Fowler equations. We offer the following system:(26a)u′′t+1tu′t+vt=t3+5,(26b)v′′t+2tv′t+wt=t4+12t+1,(26c)w′′t+3twt+ut=25t2+1,(26d)u0=v0=w0=1,u′0=v′0=w′0=0.Using the same way, we get(27)u0t=1+a02t2+a16t3+a212t4+a320t5+⋯,v0t=1+b02t2+b16t3+b212t4+b320t5+⋯,w0t=1+c02t2+c16t3+c212t4+c320t5+⋯ Putting u1(t)=v1(t)=0 and p=1 and using (7a) and (7b), we get(28)52-a02-a02-12=0⇒a0=2,12-b0-12-b02=0⇒b0=0,12-3c02-12-c02=0⇒c0=0, also a1=0,b1=6 and c1=0, a2=b2=0 and c2=12, a3=b3=0 and a4=5,b4=-5.We verifyai=0 if i≠0, bi=0 if i≠1, and ci=0 if i≠2. This gives the analytic solution of system (26a), (26b), (26c), and (26d) which is given by(29)ut,vt,wt=1+t2,1+t3,1+t4.This example was also studied by Wazwaz in [13], using the variational method but the solution contained noise terms. The noise terms are a phenomenon which often appears in the system of inhomogeneous equations of Emden–Fowler.As a result, this proposed method is relatively easy to calculate the coefficientsan,bn,cn,⋯ of a system with only two or three unknowns. We can extend this method to a system containing more unknowns, but we need help from symbolic computation Software.Example 4.
Next we study the nonlinear system of Emden–Fowler equations(30a)u′′t+2tu′t+v2t-u2t-4vt=0,(30b)v′′t+2tv′t+u2t-v2t+4ut=0,(30c)u0=1,v0=-1,u′0=v′0=0.
This system consists of two spherical equations becauseα1=α2=2. To solve this system by the new method, we substitute the expressions of h1(t)=0, h2(t)=0, R(u,v)=2/tu′-4v, Q(u,v)=2/tv′+4u, An=v2-u2, and Bn=u2-v2 in (5a) and (5b). And then using (6a) and (6b), we obtain(31)u0t=1+a02t2+a16t3+a212t4+a320t5+⋯,v0t=-1+b02t2+b16t3+b212t4+b320t5+⋯.In addition, we also calculate (u1,v1). To Find an and bn, n≥0, we put u1(t)=v1(t)=0 and p=1 using (7a) and (7b). By equating the coefficients of t2,t4,⋯, we have(32)3-a0+2-a02+12-12=0⇒a0=-43,-b0-2-b02+12-12=0⇒b0=+-43anda1=b1=0,-a218+b06-a212+a012+b012=0⇒a2=165,-b218-a06-b212-b012-a012=0⇒b2=-165.Also we can verify that a3=b3=0 and a4=b4=-80/49. We get the solution of system (30a), (30b), and (30c)(33)ut,vt=1-2t23+4t415-8t6147+⋯,-1-2t23-4t415-8t6147+⋯.For this system, the exact solution is no longer obvious. To validate the decomposition method, a comparison was made between the MADM method and Adams-Bashforth-Moulton (ABM) method, Table 1. Despite the high performance of ABM method, it cannot solve the EFS on an interval containing zero due to the singularity of the initial conditions. This requires changing the study domain of [0,T] to [ε,T](ε=1/1000). This comparison is made in the neighborhood of zero (see Figure 1) because, usually, the decomposition method gives the solutions in the Taylor series form, at zero.Table 1
Comparison between MADM and ABM method.
point ABM method MADM error 0,0010 1,0000 0,9999 6 × 1 0 - 7 0,0148 0,9998 0,9998 1.8 × 1 0 - 6 0,0315 0,9993 0,9993 1.4 × 1 0 - 6 0,0494 0,9983 0,9983 1.2 × 1 0 - 6 0,07503 0,9962 0,9962 1.4 × 1 0 - 5 0,16719 0,9811 0,9815 4.1 × 1 0 - 4 0,24911 0,9576 0,9596 0.0020 0,33103 0,9237 0,9300 0.0062 0,46285 0,8452 0,8688 0.0236 0,50000 0,8171 0,8491 0.0320Figure 1
The solution (u, v) using the MADM and ABM method.


## 4. Conclusion

In this paper, a new modification of Adomian method was introduced to solve systems of singular initial value problems of Emden–Fowler type of equations withα1,α2=1,2, or α1,α2>2. We demonstrate that this method can be well suited to attain analytical solutions by only requiring the calculation of the first Adomian polynomial. The obtained results demonstrate reliability and effectiveness of the employed method.



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