Comment on “Adomian Decomposition Method for a Class of Nonlinear Problems”

Abstract

Sánchez Cano in his paper “Adomian Decomposition Method for a Class of Nonlinear Problems” in application part pages 8, 9, and 10 had made some mistakes in context; in this paper we correct them.

1. Introduction

Adomian [1, 2] proposed a powerful method for solving nonlinear functional equation. The technique uses a decomposition of the nonlinear operator as a series of functions; each term of this series is a generalized polynomial called Adomian polynomial.

We will see that using the Adomian decomposition method together with some properties of the nested integral [3, 4] the solution of nonlinear ordinary differential equations system is obtained.

2. Correct Equations

In page 8, he showed that the solution 𝑥_𝑛(𝑡)is given by 𝑥(𝑡)=𝑥₀(𝑡)+𝑥₁(𝑡)+𝑥₂(𝑡)+⋯+𝑥_𝑛(𝑡)+⋯=∝+^𝑡₀𝑓(𝑢)𝑑𝑢+𝑎𝛽𝑡+−𝑎²∝𝑡²2!−𝑎²1Γ(3)×^𝑡₀(𝑡−𝑢)²𝑓(𝑢)𝑑𝑢−𝑎³𝛽𝑡³+𝑎3!⁴∝𝑡⁴4!+𝑎⁴1Γ(5)×^𝑡₀(𝑡−𝑢)⁴𝑓(𝑢)𝑑𝑢+⋯. (1) By rearranging he obtained 𝑥(𝑡)()=∝1−𝑎𝑡²+()2!𝑎𝑡⁴()4!+⋯−1^𝑛(𝑎𝑡)^2𝑛)−()2𝑛!+⋯+𝛽(𝑎𝑡𝑎𝑡³+()3!𝑎𝑡⁵()5!+⋯−1^𝑛(𝑎𝑡)^2𝑛+1()!+2𝑛+1+⋯_𝑡0(𝑎(𝑡−𝑢))⁰Γ(1)−(𝑎(𝑡−𝑢))²Γ(3)+⋯(−1)^𝑛(𝑎(𝑡−𝑢))^2𝑛Γ()𝑓(𝑢)2𝑛+1+⋯𝑑𝑢. (2) And similarly for 𝑦_𝑛(𝑡) we will have 𝑦(𝑡)=𝑦₀(𝑡)+𝑦₁(𝑡)+𝑦₂(𝑡)+⋯+𝑦_𝑛(𝑡)1+⋯=𝛽−∝𝑎𝑡−𝑎Γ(2)×^𝑡₀(𝑡−𝑢)𝑓(𝑢)𝑑𝑢−𝑎²𝛽𝑡²2!+𝑎³∝𝑡³3!+𝑎³1Γ(4)×^𝑡₀(𝑡−𝑢)³𝑓(𝑢)𝑑𝑢+𝑎⁴𝛽𝑡⁴4!+⋯. (3) By rearranging he obtained 𝑦(𝑡)()=𝛽1−𝑎𝑡²+()2!𝑎𝑡⁴()4!+⋯−1^𝑛(𝑎𝑡)^2𝑛)−()2𝑛!+⋯−∝(𝑎𝑡𝑎𝑡³+()3!𝑎𝑡⁵()5!+⋯−1^𝑛(𝑎𝑡)^2𝑛+1()!−2𝑛+1+⋯_𝑡0(𝑎(𝑡−𝑢))¹Γ(2)−(𝑎(𝑡−𝑢))³Γ(4)+⋯(−1)^𝑛(𝑎(𝑡−𝑢))^2𝑛+1Γ()𝑓(𝑢)2𝑛+⋯𝑑𝑢. (4) Continuing in this fashion, he concluded the following formulas: 𝑥_𝑛(𝑡)=(−1)^𝑛𝑎^𝑛𝑎𝑡^𝑛𝑛!+𝑎^𝑛1Γ(𝑛+1)^𝑡₀(𝑡−𝑢)^𝑛𝑓(𝑢),𝑦𝑑𝑢_𝑛(𝑡)=(−1)^𝑛+1𝑎^𝑛𝑎𝑡^𝑛𝑛!+𝑎^𝑛1Γ(𝑛+1)^𝑡₀(𝑡−𝑢)^𝑛𝑓(𝑢)𝑑𝑢. (5) But the correct formulas are given by 𝑥_2𝑛−1(𝑡)=(−1)^𝑛+1𝑎^2𝑛−1𝛽𝑡^2𝑛−1()!,𝑥2𝑛−1_2𝑛(𝑡)=(−1)^𝑛𝑎^2𝑛∝𝑡^2𝑛(2𝑛)!+𝑎^𝑛1Γ()×2𝑛+1_𝑡0(𝑡−𝑢)^2𝑛𝑓(𝑢),𝑦𝑑𝑢_2𝑛−1(𝑡)=(−1)^𝑛𝑎^2𝑛−1∝𝑡^2𝑛−1(2𝑛−1)!+𝑎^𝑛1Γ()×2𝑛_𝑡0(𝑡−𝑢)^2𝑛−1𝑓(𝑢),𝑦𝑑𝑢_2𝑛(𝑡)=(−1)^𝑛𝑎^2𝑛𝛽𝑡^2𝑛()!.2𝑛 (6) Writing (2) and (4) as a single integral, he had 𝑥(𝑡)=∝cos(𝑎𝑡)+𝛽sin(𝑎𝑡)+^𝑡₀(𝑎(𝑓(𝑢)()()(𝑡)().cos𝑡−𝑢))𝑑𝑢=∝cos𝑎𝑡+𝛽sin𝑎𝑡+𝑓∗cos𝑎𝑡 (7) Similarly, 𝑦(𝑡)=𝛽cos(𝑎𝑡)−∝sin(𝑎𝑡)−^𝑡₀(𝑎(𝑓(𝑢)()()(𝑡)().sin𝑡−𝑢))𝑑𝑢=𝛽cos𝑎𝑡−∝sin𝑎𝑡+𝑓∗sin𝑎𝑡 (8) In page 9, he uses 𝑑𝑥(𝑡)(𝑡),𝑑𝑡−𝑎𝑦=𝑓𝑑𝑦(𝑡)().𝑎𝑡+𝑎𝑥=0𝑎∈𝑅,𝑎≠0 (9) With 𝑥(0)=𝛼 and 𝑦(0)=𝛽, he obtained 𝑑²𝑦𝑑𝑡²+𝑎²𝑦(𝑡)(𝑡),𝑦(0)𝑦́(0)=𝑓=𝛼,=𝛽. (10) In fact, the correct solution is given by 𝑑𝑥(𝑡)(𝑡)⟶𝑑𝑡−𝑎𝑦=𝑓𝑑𝑥(𝑡)(𝑡),𝑑𝑡=𝑎𝑦+𝑓𝑑𝑦(𝑡)𝑑𝑑𝑡+𝑎𝑥=0⟶²𝑦𝑑𝑡²+𝑎𝑑𝑥⇓𝑑𝑑𝑡=0,₂𝑦𝑑𝑡²+𝑎²𝑦(𝑡)(𝑡),𝑦(0)𝑦́(0)=−𝑎𝑓=𝛽,=𝛼. (11) In page 10, he showed two cases for the solutions 𝑥(𝑡) and 𝑦(𝑡) using (2), (4), and 𝑓(𝑡)=cos(𝑡).

Case 1 (𝑎≠1). In this case, he obtained the solutions 𝑥(𝑡)()()+1=𝛼cos𝑎𝑡+𝛽sin𝑎𝑡1−𝑎²(𝑡),𝑦(𝑡)()()+𝑎sin=𝛽cos𝑎𝑡−𝛼sin𝑎𝑡1+𝑎²cos(𝑡). (12)But the correct method is the following: 𝑥(𝑡)=∝cos(𝑎𝑡)+𝛽sin(𝑎𝑡)+^𝑡₀cos(𝑎(𝑡−𝑢))𝑓(𝑢)𝑑𝑢=⟶^𝑡₀(𝑎(𝑓(𝑢)=1cos𝑡−𝑢))𝑑𝑢2^𝑡₀()()=1cos𝑎𝑡−𝑎𝑢+𝑢cos𝑎𝑡−𝑎𝑢−𝑢𝑑𝑢2()()+11−𝑎sin𝑎𝑡−𝑎𝑢+𝑢2()(−1−𝑎sin𝑎𝑡−𝑎𝑢−𝑢)^𝑡₀=11−𝑎²𝑎sin𝑡−1−𝑎²(𝑡)()()+1sin𝑎𝑡⟶𝑥=∝cos𝑎𝑡+𝛽sin𝑎𝑡1−𝑎²−𝑎sin𝑡1−𝑎²𝑥(𝑡)𝑎sin𝑎𝑡,=𝛼cos𝑎𝑡+𝛽−1−𝑎²1sin𝑎𝑡+1−𝑎²sin𝑡. (13) Similarly, 𝑦(𝑡)=𝑎𝛽−1−𝑎²𝑎cos𝑎𝑡−𝛼sin𝑎𝑡+1−𝑎²cos𝑡. (14)

Case 2 (𝑎=1). He arrived at the formulas 𝑥(𝑡) and 𝑦(𝑡) as follows: 𝑥(𝑡)(𝑡)(𝑡)+1=∝cos+𝛽sin2𝑦(𝑡)=1𝑡cos𝑡,∝−21cos𝑡+𝛽sin𝑡−2𝑡sin𝑡. (15) But the correct formulas are given by 𝑥(𝑡)=∝cos(𝑡)+𝛽sin(𝑡)+^𝑡₀cos((𝑡−𝑢))𝑓(𝑢)𝑑𝑢=⟶^𝑡₀=1cos((𝑡−𝑢))cos𝑢𝑑𝑢2^𝑡₀()=1cos𝑡+cos𝑡−2𝑢𝑑𝑢21𝑡cos𝑡−2sin(𝑡−2𝑢)^𝑡₀=121𝑡cos𝑡+2(𝑡)1sin𝑡⟹𝑥=𝛼cos𝑡+𝛽sin𝑡+21𝑡cos𝑡+2(𝑡)1sin𝑡⟹𝑥=𝛼cos𝑡+𝛽+21sin𝑡+2𝑡cos𝑡. (16)

Similarly, 𝑦(𝑡)1=𝛽cos𝑡−𝛼sin𝑡−2𝑡sin𝑡. (17)