The Adomian decomposition method together with some properties of nested integrals is used to provide a solution to a class of nonlinear ordinary differential equations and a coupled system.

1. Introduction

Most scientific problems and phenomena such as heat transfer occur nonlinearly. We know that only a limited number of these problems have a precise analytical solution [15]. In the 1980t’s, George Adomian (1923–1996) introduced a powerful method for solving nonlinear functional equations. His method is known as the Adomian decomposition method (ADM) [6]. This technique is based on the representation of a solution to a functional equation as series of functions. Each term of the series is obtained from a polynomial generated by a power series expansion of an analytic function. Although the abstract formulation of the Adomian method is very simple, the calculations of the polynomials and the verification of convergence of the function series in specific situations are usually a difficult task [7, 8].

We will see that if the nested integral properties are used properly in the Adomian decomposition method, the analytical solution to the initial value problem is easily obtained.

Nested integrals integrals which are evaluated several times on the same variable. In contrast, multiple integrals consist of a number of integrals evaluated with respect to different variables. Concretely, if 𝑓 is a continuous function defined on a (open) interval 𝕀 and 𝑥0𝕀, 𝑥𝑥0𝑥1𝑥0𝑥2𝑥0𝑥𝑘1𝑥0𝑓𝑥1𝑓𝑥2𝑓𝑥2𝑥𝑓𝑘𝑑𝑥𝑘𝑑𝑥𝑘1𝑑𝑥2𝑑𝑥1=1𝑘!𝑥𝑥0𝑓(𝑥1)𝑑𝑥1𝑘.(1.1) Also, (see [9]),𝑥𝑥0𝑥𝑛𝑥0𝑥2𝑥0𝑥1𝑥0𝑓𝑥1𝑑𝑥1𝑑𝑥2𝑑𝑥𝑛1𝑑𝑥𝑛=1Γ(𝑛+1)𝑥𝑥0(𝑥𝑢)𝑛𝑓(𝑢)𝑑𝑢.(1.2)

2. Solution Method

Consider the IVP 𝑑𝑦𝑑𝑥=𝑓(𝑥)+𝑒𝑦(𝑥)𝑥;𝑦0=0,(2.1) where 𝑓 is a continuous function defined on an (open) interval 𝕀, and 𝑥0𝕀.

In operator form, (2.1) becomes 𝐿𝑦=𝑓(𝑥)+𝑒𝑦(𝑥),(2.2) where 𝐿()=(𝑑/𝑑𝑥)(). Then inverse of 𝐿 is, therefore, 𝐿1()=𝑥𝑥0()𝑑𝑠. Applying 𝑙1 to both sides of 4 we find that 𝑦(𝑥)=𝐹(𝑥)+𝑥𝑥0𝑒𝑦(𝑠)𝑑𝑠,(2.3) where 𝐹(𝑥)=𝑥𝑥0𝑓(𝑠)𝑑𝑠.

Adomiant’s technique consists in writing the solution of (1.2) as an infinite series 𝑦(𝑥)=𝑛=0𝑦𝑛(𝑥)(2.4) and decomposing the nonlinear term 𝑁(𝑦)=𝑒𝑦 as 𝑒𝑦=𝑛=0𝐴𝑛,(2.5) where each 𝐴𝑛 is an Adomian polynomial depending on 𝑦0,𝑦1,,𝑦𝑛, which is given by 𝐴𝑛𝑦0,𝑦1,,𝑦𝑛=1𝑑𝑛!𝑛𝑑𝜆𝑛𝑁𝑛𝑖=0𝑦𝑡𝜆𝑡𝜆=0.(2.6) (see [6, 7]).

Substituting (2.4) and (2.5) into (2.3), we obtain 𝑛=0𝑦𝑛(𝑥)=𝐹(𝑥)+𝑥𝑥0𝑛=0𝐴𝑛(𝑠)𝑑𝑠.(2.7) This leads to the following recurrence scheme 𝑦0(𝑥)=𝐹(𝑥)𝑦𝑘+1(𝑥)=𝑥𝑥0𝐴𝑘(𝑠)𝑑𝑠𝑘=0,1,2,.(2.8) We then define the solution 𝑦(𝑥) as 𝑦(𝑥)=lim𝑛𝑛𝑘=1𝑦𝑘(𝑥).(2.9) The following algorithm will be used in order to calculate the Adomian polynomials (see [10])𝐴0𝑦(𝑥)=𝑁0𝐴1𝑑(𝑥)=𝑁𝑦𝑑𝜆0+𝜆𝑦1|||𝜆=0=𝑑𝑒𝑑𝜆(𝑦0+𝜆𝑦1)|||𝜆=0=𝑦1(𝑥)𝑁𝑦0,𝐴21(𝑥)=𝑑2!2𝑑𝜆2𝑁𝑦0+𝜆𝑦1+𝜆2𝑦2||||𝜆=0=𝑦2(𝑥)𝑁𝑦0+12𝑦21𝑁𝑦0,𝐴31(𝑥)=𝑑3!3𝑑𝜆3𝑁3𝑛=0𝜆𝑛𝑦𝑛|||||(𝑥)𝜆=0=𝑦3(𝑥)𝑁𝑦0+𝑦1(𝑥)𝑦31(𝑥)+𝑦3!31𝑁𝑦0,𝐴41(𝑥)=𝑑4!4𝑑𝜆4𝑁4𝑛=0𝜆𝑛𝑦𝑛|||||(𝑥)𝜆=0=𝑦4(𝑥)𝑁𝑦0+𝑦1(𝑥)𝑦31(𝑥)+2𝑦22𝑁(𝑥)𝑦0+12𝑦21(𝑥)𝑦2(𝑥)𝑁𝑦0+1𝑦4!41(𝑥)𝑁(4)𝑦0.(2.10)

Combining this with (2.8), one obtains 𝑦0𝑦(𝑥)=𝐹(𝑥),1(𝑥)=𝑥𝑥0𝐴0(𝑠)𝑑𝑠=𝑥𝑥0𝑒𝑦0(𝑠)𝑑𝑠=𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠,(2.11) where 𝐴0(𝑥)=𝑁(𝑦0)=𝑒𝑦0=𝑒𝐹(𝑥).

By using 𝐴1𝑑(𝑥)=𝑁𝑦𝑑𝜆0+𝜆𝑦1|||𝜆=0=𝑑𝑒𝑑𝜆(𝑦0+𝜆𝑦1)|||𝜆=0=𝑒𝑦0(𝑥)𝑦1(𝑥),(2.12) we find that 𝑦2(𝑥)=𝑥𝑥0𝐴1(𝑠)𝑑𝑠=𝑥𝑥0𝑒𝐹(𝑠)𝑦1(𝑠)𝑑𝑠=𝑥𝑥0𝑒𝐹(𝑠)𝑠𝑥0𝑒𝐹(𝑢)𝑑𝑢𝑑𝑠.(2.13) Now, using property (1.1) in (2.13) yields 𝑥𝑥0𝑒𝐹(𝑠)𝑠𝑥0𝑒𝐹(𝑢)𝑑𝑢𝑑𝑠=𝑥𝑥0𝑠𝑥0𝑒𝐹(𝑠)𝑒𝐹(𝑢)1𝑑𝑢𝑑𝑠=2!𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠2.(2.14) Since 𝐴2(𝑥)=𝑦2(𝑥)𝑁𝑦0+12𝑦21𝑁𝑦0=12𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠𝑒𝐹(𝑥)+12𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠2𝑒𝐹(𝑥).(2.15) One finally obtains 𝐴2(𝑥)=𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠2𝑒𝐹(𝑥).(2.16) In order to obtain 𝑦3, we again use (1.1) and (2.16) 𝑦3(𝑥)=𝑥𝑥0𝐴2(𝑠)𝑑𝑠=𝑥𝑥0𝑒𝐹(𝑠)𝑥𝑥0𝑒𝐹(𝑢)𝑑𝑢21𝑑𝑠=3!𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠3;(2.17) continuing in this fashion, we obtain 𝑦𝑘(𝑥)=(1)𝑘+11𝑘!𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠𝑘,𝑘=1,2,3,.(2.18) The solution is given by 𝑦(𝑥)=𝑦0(𝑥)+𝑦1(𝑥)+𝑦2(𝑥)++𝑦𝑛(𝑥)+.(2.19) By replacing (2.18) into (2.19), one obtains 𝑦(𝑥)=𝐹(𝑥)+𝑥𝑥0𝑒𝐹(𝑠)1𝑑𝑠2!𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠2++(1)𝑘+11𝑘!𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠𝑘+.(2.20) Or, in a more compact form, 𝑦(𝑥)=𝐹(𝑥)+𝑛=1(1)𝑘+11𝑘!𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠𝑘.(2.21) The latter equation can be written as 𝑦(𝑥)=𝐹(𝑥)+ln1+𝑥𝑥0𝑒𝐹(𝑠)𝑑𝑠.(2.22) Observe that in this case, Adomiant’s method yields an exact analytical solution. The analytical solution to this probleme can be obtained by performing the substitution 𝑢=exp(𝑦(𝑥)), which leads to a Bernoulli differential equation whose solution is a straightforward exercise.

3. Examples

3.1. Example 1

Consider the nonlinear initial value problem 𝑑𝑦=1𝑑𝑥𝑥+𝑒𝑦,𝑥>0,𝑦(1)=0.(3.1) In this case, 𝑓(𝑥)=1/𝑥,𝑥>0 and consequently 𝐹(𝑥)=𝑥1(1/𝑠)𝑑𝑠=ln𝑥.

Thus, the analytical solution is given by 𝑦(𝑥)=ln𝑥+ln1+𝑥0𝑒ln𝑠𝑑𝑠=ln𝑥+ln1+𝑥11𝑠𝑑𝑠=ln𝑥+ln(1+ln𝑥),𝑥>𝑒10.36787944.(3.2)

3.2. Example 2

Consider the nonlinear initial value problem 𝑑𝑦𝑑𝑥=𝑥+𝑒𝑦,𝑦(0)=0.(3.3) In this case, 𝑓(𝑥)=𝑥, and therefore 𝐹(𝑥)=𝑥2/2. The analytical solution is 𝑥𝑦(𝑥)=22+ln1+𝜋2𝑥erf2,𝑥>1.2755,(3.4) where erf(𝑥)=(2/𝜋)𝑥0𝑒𝑡2𝑑𝑡.

4. Application of the Method to Coupled Systems of ODE’s

Consider the coupled system 𝑑𝑥𝑑𝑡𝑎𝑦(𝑡)=𝑓(𝑡),𝑑𝑦𝑑𝑡+𝑎𝑥(𝑡)=0(𝑎,𝑎0).(4.1) Together with the initial conditions 𝑥(0)=𝛼,𝑦(0)=𝛽,(4.2) we shall obtain its solution by using the Dirichlet's integral formula (1.2) and the Adomian decomposition method.

Equation (4.1) in operator form takes the form 𝐿𝑥(𝑡)=𝑓(𝑡)+𝑎𝑦(𝑡),𝐿𝑦(𝑡)=𝑎𝑥(𝑡),(4.3) where 𝐿()=(𝑑/𝑑𝑥)(). The inverse of 𝐿 is 𝐿1()=𝑡0()𝑑𝑠. Applying 𝐿1 to both sides of (4.3) and using (4.2), we find that 𝑥(𝑡)=𝛼+𝑡0𝑓(𝑠)𝑑𝑠+𝑎𝑡0𝑦(𝑠)𝑑𝑠,(4.4)𝑦(𝑡)=𝛽𝑎𝑡0𝑥(𝑠)𝑑𝑠.(4.5) In order to obtain 𝑥(𝑡), we apply the Adomian iterative scheme 𝑥0(𝑡)=𝛼+𝑡0𝑓(𝑠)𝑑𝑠,𝑥𝑘+1(𝑡)=𝑎𝑡0𝑦𝑘(𝑠)𝑑𝑠,𝑘=0,1,2,(4.6) in (4.4).

Similarly, 𝑦(𝑡) is obtained by applying the scheme 𝑦0(𝑡)=𝛽,𝑦𝑘+1(𝑡)=𝑎𝑡0𝑥𝑘(𝑠)𝑑𝑠,𝑘=0,1,2,(4.7) to (4.5).

Replacing 𝑦0(𝑡)=𝛽 in (4.6), we find 𝑥1(𝑡). In fact, 𝑥1(𝑡)=𝑎𝑡0𝑦0(𝑠)𝑑𝑠=𝑎𝑡0𝛽𝑑𝑠=𝑎𝛽𝑡.(4.8) Also, by replacing 𝑥0(𝑡)=𝛼+𝑡0𝑓(𝑠) and (1.2) in (4.5), we obtain 𝑦1(𝑡)=𝑎𝑡0𝑥0(𝑠)𝑑𝑠=𝑎𝑡0𝛼+𝑠0𝑓(𝑢)𝑑𝑢𝛽𝑑𝑠=𝑎𝛼𝑡𝑎𝑡0𝑠01𝑓(𝑢)𝑑𝑢𝑑𝑠=𝑎𝛼𝑡𝑎Γ(2)𝑡0(𝑡𝑢)𝑓(𝑢)𝑑𝑢.(4.9) Applying (1.2) to the right hand side of the last equation, one finds that 𝑥2(𝑡)=𝑎𝑡0𝑦1(𝑠)𝑑𝑠=𝑎𝑡01𝑎𝛼𝑠𝑎Γ(2)𝑠0(𝑠𝑢)𝑓(𝑢)𝑑𝑢𝑑𝑠=𝑎2𝛼𝑡22!𝑎21Γ(2)𝑡0𝑠0(𝑠𝑢)𝑓(𝑢)𝑑𝑢𝑑𝑠=𝑎2𝛼𝑡22!𝑎21Γ(3)𝑡0(𝑡𝑢)2𝑓(𝑢)𝑑𝑢.(4.10) To obtain 𝑦2(𝑡), (4.8) into (4.7), and (1.2), we have 𝑦2(𝑡)=𝑎𝑡0𝑥1(𝑠)𝑑𝑠=𝑎𝑡0𝑎2𝛼𝑡22!𝑎21Γ(3)𝑠0(𝑠𝑢)2𝑓(𝑢)𝑑𝑢𝑑𝑠=𝑎3𝛼𝑡33!+𝑎31Γ(3)𝑡0𝑠0(𝑠𝑢)2𝑓(𝑢)𝑑𝑢𝑑𝑠=𝑎3𝛼𝑡33!+𝑎21Γ(4)𝑡0(𝑡𝑢)3𝑓(𝑢)𝑑𝑢.(4.11) Continuing in this fashion, one arrives at the formula 𝑥𝑛(𝑡)=(1)𝑛𝑎𝑛𝛼𝑡𝑛𝑛!+𝑎𝑛1Γ(𝑛+1)𝑡0(𝑡𝑢)𝑛,𝑦𝑓(𝑢)𝑑𝑢𝑛(𝑡)=(1)𝑛+1𝑎𝑛𝛼𝑡𝑛𝑛!+𝑎𝑛1Γ(𝑛+1)𝑡0(𝑡𝑢)𝑛.𝑓(𝑢)𝑑𝑢(4.12) The solution 𝑥(𝑡) is the given by 𝑥(𝑡)=𝑥1(𝑡)+𝑥2(𝑡)+𝑥3(𝑡)++𝑥𝑛(𝑡)+=𝛼+𝑡0𝑓(𝑢)𝑑𝑢+𝑎𝛽𝑡+𝑎2𝛼𝑡22!𝑎21Γ(3)𝑡0(𝑡𝑢)2𝑓(𝑢)𝑑𝑢𝛼𝑎3𝑡3+𝑎3!4𝛼𝑡44!𝑎41Γ(5)𝑡0(𝑡𝑢)4𝑓(𝑢)𝑑𝑢+.(4.13) Rearranging terms and writing as a single integral, we have 𝑥(𝑡)=𝛼1(𝑎𝑡)2+2!(𝑎𝑡)44!+(1)𝑛(𝑎𝑡)2𝑛(2𝑛)!++𝛽(𝑎𝑡)(𝑎𝑡)2+3!(𝑎𝑡)45!+(1)𝑛(𝑎𝑡)2𝑛+1+(2𝑛+1)!+𝑡0(𝑎(𝑡𝑢))0Γ(1)(𝑎(𝑡𝑢))2+Γ(3)(𝑎(𝑡𝑢))4Γ(5)+(1)𝑛(𝑎(𝑡𝑢))2𝑛Γ(2𝑛+1)+𝑓(𝑢)𝑑𝑢.(4.14) This is easily recognized as 𝑥(𝑡)=𝛼cos(𝑎𝑡)+𝛽sin(𝑎𝑡)+𝑡0cos(𝑎(𝑡𝑢))𝑓(𝑢)𝑑𝑢=𝛼cos(𝑎𝑡)+𝛽sin(𝑎𝑡)+𝑓(𝑡)cos(𝑎𝑡),(4.15) where denotes convolution.

The analogous process gives 𝑦(𝑡)=𝑦1(𝑡)+𝑦2(𝑡)+𝑦3(𝑡)++𝑦𝑛1(𝑡)+=𝛽𝛼𝑎𝑡𝑎Γ(2)𝑡0(𝑡𝑢)𝑓(𝑢)𝑑𝑢𝑎2𝛽𝑡22!+𝛼𝑎3𝑡3+13!Γ(4)𝑡0(𝑡𝑢)3𝑓(𝑢)𝑑𝑢++(1)𝑛𝑎𝑛𝛼𝑡𝑛𝑛!+𝑎𝑛1Γ(2𝑛+2)𝑡0(𝑡𝑢)2𝑛+1𝑓(𝑢)𝑑𝑢+.(4.16) Rearranging, we obtain 𝑦(𝑡)=𝛽1(𝑎𝑡)2+2!(𝑎𝑡)44!+(1)𝑛(𝑎𝑡)2𝑛(2𝑛)!+𝛼(𝑎𝑡)(𝑎𝑡)2+3!(𝑎𝑡)45!+(1)𝑛(𝑎𝑡)2𝑛+1+(2𝑛+1)!+𝑡0𝑎𝑓(𝑢)𝑑𝑢2Γ(3)𝑡0(𝑡𝑢)2𝑎𝑓(𝑢)𝑑𝑢+4Γ(5)(𝑡𝑢)4𝑓(𝑢)𝑑𝑢++(1)𝑛𝑎2𝑛Γ(2𝑛+2)𝑡0(𝑡𝑢)2𝑛𝑓(𝑢)𝑑𝑢+.(4.17) Writting this as a single integral, we have 𝑦(𝑡)=𝛽cos(𝑎𝑡)𝛼sin(𝑎𝑡)+𝑡0sin(𝑎(𝑡𝑠))𝑓(𝑠)𝑑𝑠.(4.18) And then, 𝑦(𝑡)=𝛽cos(𝑎𝑡)𝛼sin(𝑎𝑡)+𝑓(𝑡)sin(𝑎𝑡).(4.19) It is important to observe that the analytical solution of the IVP given by (4.1) and (4.2) is precisely 𝑦𝑥(𝑡)=𝛼cos(𝑎𝑡)+𝛽sin(𝑎𝑡)+𝑓(𝑡)cos(𝑎𝑡),(𝑡)=𝛽cos(𝑎𝑡)𝛼sin(𝑎𝑡)+𝑓(𝑡)sin(𝑎𝑡).(4.20) In particular, let us consider the forced undamped system given by 𝑑2𝑦𝑑𝑡2+𝑎2𝑦(𝑡)=cos(𝑡),𝑦(0)=𝛼,𝑦(0)=𝛽.(4.21) (Note that (4.21) is equivalent to the system formed by (4.1) and (4.2) with 𝑓(𝑡)=cos(𝑡)).

There are two cases.

Case 1. 𝑎1. In this case, we obtain the solutions 1𝑥(𝑡)=𝛼cos(𝑎𝑡)+𝛽sin(𝑎𝑡)+1𝑎2𝑎sin(𝑡),𝑦(𝑡)=𝛽cos(𝑎𝑡)𝛼sin(𝑎𝑡)+1+𝑎2cos(𝑡).(4.22) Let us observe that the solutions are bounded in this case

Case 2. 𝑎=1. In this case, one obtain 1𝑥(𝑡)=𝛼cos(𝑡)+𝛽sin(𝑡)+21𝑡cos(𝑡),𝑦(𝑡)=𝛼21cos(𝑡)+𝛽sin(𝑡)2𝑡sin(𝑡).(4.23) Observe that the solutions are unbounded in this case, and we have resonance.

5. Conclusion

The results obtained in this paper show that the Adomian decomposition method is a powerful technique for finding the theoretical solutions of nonlinear initial value problems and coupled systems if properties of nested integrals are used properly. If a solution in closed form is not found, the method always provides a convergent series which solve the problem, see [11].