# Resolution of nonlinear and non-autonomous ODEs by the ADM using a new practical Adomian polynomials

Idriss Noureddine Zaouagui, Toufik Badredine

► To cite this version:

Idriss Noureddine Zaouagui, Toufik Badredine. Resolution of nonlinear and non-autonomous ODEs by the ADM using a new practical Adomian polynomials. Bull. Comput. Appl. Math., 2020, 7 (1), pp.83-106. hal-03580878

HAL Id: hal-03580878

https://hal.science/hal-03580878

Submitted on 18 Feb 2022

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution - ShareAlike 4.0 International License
# Resolution of nonlinear and non-autonomous ODEs by the ADM using a new practical Adomian polynomials

Idriss Noureddine Zaouagui,<fnref n="1" />,<fnref n="2" /> Toufik Badredine <fnref n="3" />

CompAMa Vol.7, pp.83-106, 2019 - Received December 6, 2019 - Accepted February 9, 2020

## Abstract

In this paper, a new practical formulas of Adomian polynomials has been adapted to resolve nonlinear and non-autonomous ordinary differential equations by the Adomian decomposition method, a simple computational for this new polynomials has been suggested, Therefore new conditions of convergence have been generalized.

**Keywords:** Decomposition Method, Adomian Polynomials, nonlinear and non-autonomous ODEs, Cauchy's problems, initial value problems.

## Resumen

En este artículo, se ha adaptado una nueva fórmula práctica de polinomios de Adomian para resolver ecuaciones diferenciales ordinarias no lineales y no autónomas mediante el método de descomposición de Adomian, se ha sugerido un cálculo simple para estos nuevos polinomios. Por lo tanto, se han generalizado nuevas condiciones de convergencia.

**Palabras claves:** Métodos de descomposición, Polinomios de Adomian, ecua-

<footnote n="1">¹Department of Mathematics, Faculty of Sciences, University Ferhat Abbas - Setif 1, Algeria. (idriss.zaouagui@univ-batna.dz).</footnote>

<footnote n="2">²Corresponding author.</footnote>

<footnote n="3">³Department of Mathematics, Faculty of Sciences, University Ferhat Abbas - Setif 1, Algeria. (toufik.badredine@gmail.com).</footnote>
ciones diferenciales ordinarias no lineales y no autónomas, problema de Cauchy, problemas de valores iniciales.

# 1 Introduction

Since the eighties of the last century, G. Adomian invented a semi-analytic method it's called Adomian decomposition method, which was used to solve a large various kind of nonlinear equations (algebraic, differential, partial differential, integral, intgo-differential, etc.). This method has been developing continuously since its invention to this day by a number of scientists such as G.Adomian, R. Rach, Y. Cherruault, P. Grimalt, K. Abbaoui, S. Guellal, V. Seng, A-M. Wazwaz, J-S. Duan, J. Biazar, E. Babolian, A. R. Vahidi and others. The principle of this method consists in decomposing the nonlinear term into a series of functions (where each term is called Adomian Polynomial) and then to calculate the solution as a series of function where each term can be easily determined with Adomian's algorithm, and that converges quickly towards the exact solution, this method does not change the nature of the problem and it doesn't perform any linearization.

Let us recall the basic principles of the ADM. Consider a functional equation in the form (see [1-4])

$$Lu + Ru + Nu = g \quad (1)$$

where L and R are a linear operators, N is a nonlinear analytical operator and g is given function. For the initial value problem, we assume that $L^{-1}Lu = u - \phi$, where $L^{-1}$ is a reciprocal operator of L, and thus $\phi$ is determined by the initial values. Applying the operator $L^{-1}$ on the both sides of (1) gives:

$$u = \phi + L^{-1}g - L^{-1}Ru - L^{-1}Nu \quad (2)$$

The ADM consists in calculating the solution, in a series form:

$$u = \sum_{n \ge 0} u_n \quad (3)$$

and decomposes the nonlinear term Nu into a series:

$$Nu = \sum_{n \ge 0} A_n \quad (4)$$
where $A_n$ depending on $u_0, u_1, \dots, u_n$ are the Adomian polynomials, they are obtained by the formula:

$$A_n = \frac{1}{n!} \frac{d^n}{d\lambda^n} N \left( \sum_{n \ge 0} u_n \lambda^n \right) \bigg|_{\lambda=0} = \sum_{\sum_{i=1}^n ik_i = n} N^{\left[ \sum_{i=1}^n k_i \right]} (u_0) \prod_{i=1}^n \frac{u_i}{k_i!} \quad (5)$$

Each term of the series $\sum_{n \ge 0} u_n$ can be identified by the formula:

$$\left\{ \begin{array}{ll} u_0 = \phi + L^{-1}g & \\ u_{n+1} = -L^{-1}Ru_n - L^{-1}A_n & n \in \mathbb{N} \end{array} \right. \quad (6)$$

The first five Adomian polynomials are:

$$\begin{aligned} A_0 &= N(u_0) \\ A_1 &= N'(u_0)u_1 \\ A_2 &= N'(u_0)u_2 + \frac{1}{2}N''(u_0)u_1^2 \\ A_3 &= N'(u_0)u_3 + N''(u_0)u_1u_2 + \frac{1}{6}N'''(u_0)u_1^3 \\ A_4 &= N'(u_0)u_4 + N''(u_0)(u_1u_3 + \frac{1}{2}u_1^2) + \frac{1}{2}N'''(u_0)u_1^2u_2 + \frac{1}{24}N^{[4]}(u_0)u_1^4 \\ A_5 &= N'(u_0)u_5 + N'(u_0)(u_1u_4 + u_2u_3) + \frac{1}{2}N'''(u_0)(u_1^2u_3 + u_1u_2^2) + \frac{1}{6}N^{[4]}(u_0)u_1^3u_2 + \frac{1}{120}N^{[5]}(u_0)u_1^5 \end{aligned} \quad (7)$$

Some previous works concerning this method deal only with the case of non-linear but autonomous differential equation (see [1, 2, 5–9]), and some others treated the nonlinear and non-autonomous differential equation with the classical formulas (5) by determine the reciprocal operator $L^{-1}$ (see [3, 4, 10]), our objective in this work is to resolve the nonlinear and non-autonomous ODEs by a simple integral (without determining the reciprocal operator $L^{-1}$) and we dispense the operator $R$ and the function $g$.

In this paper, we determine new expressions of Adomian's polynomials $\tilde{A}_n$ permits to apply this technique on the non-linear and non-autonomous ODEs, by using the generalization of Faà di Bruno's theorem (see [11]), and we give a new proof of convergence more general than the proof in [8].

## 2 Application of the decomposition method to non-autonomous and non-linear ODEs

Let us consider the following non-linear and non-autonomous differential equation:

$$\frac{du}{dt} = f(t, u) \quad (8)$$
with the initial value

$$u(t_0) = c \qquad (9)$$

where f is the non-linear term.

By integrating both sides the equation (8) we obtain:

$$u = \int_{t_0}^{t} f(s, u) ds + c \qquad (10)$$

The ADM consists in setting the solution in a series form:

$$u = \sum_{n \ge 0} u_n \qquad (11)$$

and the non-linear term $f(t, u)$ is written by the form:

$$f(t, u) = \sum_{n \ge 0} \tilde{A}_n \qquad (12)$$

where the $\tilde{A}_n$'s are terms dependent on $t, u_0, u_1, \dots, u_n$ and are called Adomian's polynomials

Inserting equations (11) and (12) into equation (10) leads to:

$$\sum_{n \ge 0} u_n = c + \sum_{n \ge 0} \int_{t_0}^{t} \tilde{A}_n ds \qquad (13)$$

which implies (according to Adomian algorithm) that:

$$u_0 = c \quad \text{and} \quad u_n = \int_{t_0}^{t} \tilde{A}_{n-1} ds \quad \text{for } n \ge 1 \qquad (14)$$

**Remark 1.** In practice it's difficult to calculate all the terms of the series (11) that's why an approximation of the solution is used by truncating the series: $\phi_n = \sum_{i=0}^{n-1} u_i$

## 3 Notations

Let $n \in \mathbb{N}$, $k = (k_1, k_2, \dots, k_n) \in \mathbb{N}^n$, $t \in \mathbb{R}$ and $f$ is a two variables function, we note:

$$ \left\{ \begin{array}{l} |k| = k_1 + k_2 + \dots + k_n \\ |nk_n| = k_1 + 2k_2 + \dots + nk_n \\ \partial_{t^n u^m} f = \frac{\partial^n}{\partial t^n} \frac{\partial^m}{\partial u^m} f \end{array} \right. \qquad (15) $$
# 4 New Adomian's Polynomials

**Theorem 1.** The new terms $\tilde{A}_n$ are given by the formulas:

$$ egin{cases} 
\tilde{A}_0 = f(t_0, u_0) \\
\tilde{A}_n = \frac{1}{n!} \frac{d^n}{d\lambda^n} f \left( t_0 + \lambda(t-t_0), \sum_{m \ge 0} \lambda^m u_m \right) \Big|_{\lambda=0} & \text{for } n \ge 1 
\end{cases} 
\quad (16) $$

*Proof.* For the convenience of writing we define two operators $F$ and $L$ as follows:

$$ egin{cases} F(U) = (1, f(t, u)) \\ L(U) = \frac{d}{dt}U = (1, u') \end{cases} \quad \text{where} \quad U = (t, u) $$

From equation (8) we have: $L(U) = F(U)$, so $L^{-1}(L(U)) = L^{-1}(F(U))$ and it follows that:

$$ \left( \int_{t_0}^{t} ds, \int_{t_0}^{t} u'(s) ds \right) = L^{-1}(F(U)) $$

We obtain an equation of the form:

$$ U = N(U) + C \quad \text{where} \quad N(U) = L^{-1}F(U) \quad \text{and} \quad C = (t_0, c) $$

According to (5), the Adomian's polynomials associated to this equation are given by:

$$ egin{cases} A_0 = N(U_0) \\ A_n = \frac{1}{n!} \frac{d^n}{d\lambda^n} N \left( \sum_{m \ge 0} U_m \lambda^m \right) \Big|_{\lambda=0} & \text{for } n \ge 1 \end{cases} $$

where again,

$$ egin{cases} U_0 = C \\ U_1 = N(U_0) \\ U_n = \frac{1}{(n-1)!} \frac{d^{n-1}}{d\lambda^{n-1}} N \left( \sum_{m \ge 0} U_m \lambda^m \right) \Big|_{\lambda=0} & \text{for } n \ge 2 \end{cases} $$
This translates into:

$$ egin{cases} U_0 = (t_0, c) \\ U_1 = L^{-1}(F(U_0)) = L^{-1}(1, f(t_0, c)) \\ U_n = \frac{1}{(n-1)!} \frac{d^{n-1}}{d\lambda^{n-1}} L^{-1} \left( F \left( \sum_{m \ge 0} U_m \lambda^m \right) \right) \Big|_{\lambda=0} \\ = \frac{1}{(n-1)!} \frac{d^{n-1}}{d\lambda^{n-1}} L^{-1} \left( 1, f \left( \sum_{m \ge 0} U_m \lambda^m \right) \right) \Big|_{\lambda=0} \quad \text{for } n \ge 2 \end{cases} $$

As a result of this:

$$ egin{cases} U_0 = (t_0, c) \\ U_1 = \left( t - t_0, \int_{t_0}^t f(t_0, c) \, ds \right) \\ U_n = \frac{1}{(n-1)!} \left. \left( \frac{d^{n-1}}{d\lambda^{n-1}} (t-t_0) \right) \Big|_{\lambda=0}, \int_{t_0}^t \frac{d^{n-1}}{d\lambda^{n-1}} f \left( \sum_{m \ge 0} U_m \lambda^m \right) \Big|_{\lambda=0} \, ds \right) \\ = \left. \left( 0, \int_{t_0}^t \frac{1}{(n-1)!} \frac{d^{n-1}}{d\lambda^{n-1}} f \left( U_0 + U_1 \lambda + \sum_{m \ge 2} U_m \lambda^m \right) \right) \Big|_{\lambda=0} \, ds \right) \\ \text{for } n \ge 2 \end{cases} $$

Considering the fact that the sequence $(U_n)_{n \in \mathbb{N}}$ is defined by: $U_0 = (t_0, u_0)$, $U_1 = (t - t_0, u_1)$ and $U_n = (0, u_n)$ for $n \ge 2$, we obtain:

$$ egin{cases} u_0 = c \\ u_1 = \int_{t_0}^{t} f(t_0, c) \, ds \\ u_n = \int_{t_0}^{t} \frac{1}{(n-1)!} \frac{d^{n-1}}{d\lambda^{n-1}} f \left( (t_0, u_0) + (s-t_0, u_1) \lambda + \sum_{m \ge 2} (0, u_m) \lambda^m \right) \Big|_{\lambda=0} \, ds \\ = \int_{t_0}^{t} \frac{1}{(n-1)!} \frac{d^{n-1}}{d\lambda^{n-1}} f \left( t_0 + (s-t_0) \lambda, \sum_{m \ge 0} u_m \lambda^m \right) \Big|_{\lambda=0} \, ds \\ \text{for } n \ge 2 \end{cases} $$

which ends the proof by a simple identification with equation (14) □

**Lemma 1.** (see [11,12]) If $x, y$ are functions $C^\infty$ at the neighborhood of a and $f$ is two variables function and it's $C^\infty$ at the neighborhood of $(x(a), y(a))$,
then:

$$ \frac{d^n}{d\lambda^n} f(x(\lambda), y(\lambda))|_{\lambda=a} = \tag{17} \\ = \sum_{|nm_n|+|nk_n|=n} \frac{n!}{\prod_{i=1}^n m_i! k_i!} \partial_{x^{|m|}y^{|k|}} f(x(a), y(a)) \prod_{i=1}^n \left( \frac{x^{|i|}(a)}{i!} \right)^{m_i} \left( \frac{y^{|i|}(a)}{j!} \right)^{k_i} $$

*Proof.* Since $x$ and $y$ are analytic, then they are develop-able in the neighborhood of $a$, and for each $t$ belongs to the neighborhood of $a$, we have:

$$ x(\lambda) = \sum_{i \ge 0} \frac{x^{|i|}(a)}{i!} (\lambda - a)^i \quad y(\lambda) = \sum_{j \ge 0} \frac{y^{|j|}(a)}{j!} (\lambda - a)^j $$

Since $f$ is also analytic, then it is develop-able in the neighborhood of $(x(a), y(a))$, and for each $(x, y)$ belongs to the neighborhood of $(x(a), y(a))$, we have:

$$ f(x, y) = \sum_{p+q \ge 0} \frac{\partial_{x^p y^q} f(x(a), y(a))}{p!q!} (x-x(a))^p (y-y(a))^q $$

If we replace $x$ by $x(\lambda)$ and $y$ by $y(\lambda)$ on $f(x, y)$ we have:

$$ \begin{align*} f(x(\lambda), y(\lambda)) &= \sum_{p+q \ge 0} \frac{\partial_{x^p y^q} f(x(a), y(a))}{p!q!} (x(\lambda) - x(a))^p (y(\lambda) - y(a))^q \\ &= f(x(a), y(a)) + \sum_{p+q \ge 1} \frac{\partial_{x^p y^q} f(x(a), y(a))}{p!q!} \prod_{r=1}^{p} \left( \sum_{i_r \ge 1} \frac{x^{|i_r|}}{i_r!} (\lambda - a)^{i_r} \right) \prod_{s=1}^{q} \left( \sum_{j_s \ge 1} \frac{y^{|j_s|}}{j_s!} (\lambda - a)^{j_s} \right) \\ &= f(x(a), y(a)) + \sum_{p+q \ge 1} \frac{1}{p!q!} \sum_{i_1 \ge p+q} \cdots \sum_{i_p \ge 1} \sum_{j_1 \ge 1} \cdots \sum_{j_q \ge 1} \partial_{x^p y^q} f(x(a), y(a)) \\ & \qquad \prod_{r=1}^{p} \frac{x^{|i_r|}(a)}{i_r!} \prod_{s=1}^{q} \frac{y^{|j_s|}(a)}{j_s!} \cdot (\lambda - a)^{i_1 + \cdots + i_p + j_1 + \cdots + j_q} \\ &= f(x(a), y(a)) + \sum_{p+q \ge 1} \frac{1}{p!q!} \sum_{n \ge p+q} \sum_{\substack{|i|+|j|=n \\ i_r, j_s \ge 1}} \partial_{x^p y^q} f(x(a), y(a)) \prod_{r=1}^{p} \frac{x^{|i_r|}(a)}{i_r!} \prod_{s=1}^{q} \frac{x^{|j_s|}(a)}{j_s!} \cdot (\lambda - a)^n \\ &= f(x(a), y(a)) + \sum_{n \ge 1} \sum_{p+q=1}^{n} \frac{1}{p!q!} \sum_{\substack{|i|+|j|=n \\ i_r, j_s \ge 1}} \partial_{x^p y^q} f(x(a), y(a)) \prod_{r=1}^{p} \frac{x^{|i_r|}(a)}{i_r!} \prod_{s=1}^{q} \frac{y^{|j_s|}(a)}{j_s!} \cdot (\lambda - a)^n \end{align*} $$

The possible values of {$i_1, \dots, i_p, j_1, \dots, j_q$} are {1, 2, \dots, $n-p-q+1$}, since $i_r, j_s \ge 1$ for: $r = 1, \dots, p$ and $s = 1, \dots, q$.
Let $(m_1, \dots, m_{n-p-q+1}, k_1, \dots, k_{n-p-q+1})$ are the solutions of Diophantine equations:

$$m_1+2m_2+\dots+(n-p-q+1)m_{n-p-q+1}+k_1+2k_2+\dots+(n-p-q+1)k_{n-p-q+1}=n$$

for: $p+q=1, \dots, n$ that it checks: $m_1 + \dots + m_{n-p-q+1} = p, k_1 + \dots + k_{n-p-q+1} = q$, then:

$$\begin{aligned} f(x(\lambda), y(\lambda)) &= f(x(a), y(a)) + \sum_{n \ge 1} \sum_{p+q=1}^{n} \frac{1}{p!q!} \sum_{|(n-p-q+1)m_{n-p-q+1}| + |(n-p-q+1)k_{n-p-q+1}| = n} \\ & \qquad \partial_{x^{m_1+\dots+m_{n-p-q+1}} y^{k_1+\dots+k_{n-p-q+1}}} f(x(a), y(a)). \\ & \qquad \prod_{i=1}^{n-p-q+1} \left( \frac{x^{[i]}(a)}{i!} \right)^{m_i} \left( \frac{y^{[i]}(a)}{i!} \right)^{k_i} \cdot (\lambda - a)^n \\ &= f(x(a), y(a)) + \sum_{n \ge 1} \sum_{|nm_n| + |nk_n| = n} \frac{1}{p!q! m_1! \dots m_n! k_1! \dots k_n!} \partial_{x^{|m|} y^{|k|}} f(x(a), y(a)). \\ & \qquad \prod_{i=1}^{n} \left( \frac{x^{[i]}(a)}{i!} \right)^{m_i} \left( \frac{y^{[i]}(a)}{i!} \right)^{k_i} \cdot (\lambda - a)^n \\ &= f(x(a), y(a)) + \sum_{n \ge 1} \frac{1}{n!} \sum_{|nm_n| + |nk_n| = n} \frac{n!}{\prod_{i=1}^{n} m_i! k_i!} \partial_{x^{|m|} y^{|k|}} f(x(a), y(a)). \\ & \qquad \prod_{i=1}^{n} \left( \frac{x^{[i]}(a)}{i!} \right)^{m_i} \left( \frac{y^{[i]}(a)}{i!} \right)^{k_i} \cdot (\lambda - a)^n \end{aligned}$$

by identification with:

$$f(x(\lambda), y(\lambda)) = f(x(a), y(a)) + \sum_{n \ge 1} \frac{1}{n!} \frac{d^n}{d\lambda^n} f(x(\lambda), y(\lambda)) \Big|_{\lambda=a} \cdot (\lambda - a)^n$$

we obtain (17)

<figure><img src="image_2.png" /></figure>

**Theorem 2.** The $\tilde{A}_n$ are given by:

$$\left\{ \begin{array}{l} \tilde{A}_0 = f(t_0, u_0) \\ \tilde{A}_n = \sum_{m+|nk_n|=n} \partial_{t^m u^{|k|}} f(t_0, u_0) \left( \prod_{i=1}^{n} \frac{u_i^{k_i}}{k_i!} \right) \frac{(t-t_0)^m}{m!} \quad \text{for } n \ge 1 \end{array} \right. \quad (18)$$
*Proof.* By replacing in (17) the variables $x, y$ by $t, u$, we have:

$$ egin{cases} x(\lambda) = t_0 + (t - t_0)\lambda \implies \begin{cases} x'(\lambda) = t - t_0 \\ x^{[i]}(\lambda) = 0 \quad \text{for } i = 2, \dots, n \end{cases} \\ y(\lambda) = \sum_{m \ge 0} \lambda^m u_m \implies y^{[i]}(\lambda) = \sum_{m \ge i} \frac{m!}{(m-i)!} \lambda^{m-i} u_m \\ a = 0 \end{cases} $$

where:

$$ egin{cases} x_0(0) = t_0 \\ x'_0(0) = t - t_0 \\ x_0^{[i]}(0) = 0 \quad \text{for } i = 2, \dots, n \\ y(0) = u_0 \\ y^{[i]}(0) = i!u_i \quad \text{for } i = 1, \dots, n \end{cases} $$

from where:

$$ 
\begin{aligned}
\tilde{A}_n &= \frac{1}{n!} \left. \frac{d^n}{d\lambda^n} f \left( t_0 + \lambda (t-t_0), \sum_{m \ge 0} \lambda^m u_m \right) \right|_{\lambda=0} = \frac{1}{n!} \left. \frac{d^n}{d\lambda^n} f(x(\lambda), y(\lambda)) \right|_{\lambda=0} \\
&= \frac{1}{n!} \sum_{|nm_n|+|nk_n|=n} n! \partial_{x^{|m|}y^{|k|}} f(x(0), y(0)) \\
&= \left( \prod_{i=1}^{n} \frac{1}{m_i!} \left( \frac{x^{[i]}(0)}{i!} \right)^{m_i} \frac{1}{k_i!} \left( \frac{y^{[i]}(0)}{i!} \right)^{k_i} \right) \\
&= \sum_{\substack{|nm_n|+|nk_n|=n \\ m_1+|nk_n| \neq n}} \partial_{t^{|m|}u^{|k|}} f(t_0, u_0) \\
&\quad \left( \frac{1}{m_1!} \left( \frac{t-t_0}{1!} \right)^{m_1} \prod_{i=2}^{n} \frac{1}{m_i!} \left( \frac{0}{i!} \right)^{m_i} \prod_{i=1}^{n} \frac{1}{k_i!} \left( \frac{i!u_i}{i!} \right)^{k_i} \right) \\
&\quad + \sum_{m_1+|nk_n|=n} \partial_{t^{|m_1|}u^{|k|}} f(t_0, u_0) \left( \frac{1}{m_1!} \left( \frac{t-t_0}{1!} \right)^{m_1} \prod_{i=1}^{n} \frac{1}{k_i!} \left( \frac{i!u_i}{i!} \right)^{k_i} \right) \\
&= \sum_{m_1+|nk_n|=n} \partial_{t^{|m_1|}u^{|k|}} f(t_0, u_0) \left( \prod_{i=1}^{n} \frac{u_i^{k_i}}{k_i!} \right) \frac{(t-t_0)^{m_1}}{m_1!}
\end{aligned}
$$

□
**Corollary 1.** If $\alpha \in \mathbb{N}$ and $(\beta_i)_{i=1,...,n} \subset \mathbb{N}$ is a decreasing sequence of non-negative integers then:

$$ egin{cases} 
\tilde{A}_0 = f(t_0, u_0) \\ 
\tilde{A}_n = \sum_{\alpha+|\beta|=n} \partial_{t^\alpha u^{\beta_1}} \frac{u_1^{\beta_1-\beta_2}}{(\beta_1-\beta_2)!} \cdots \frac{u_{n-1}^{\beta_{n-1}-\beta_n}}{(\beta_{n-1}-\beta_n)!} \frac{u_n^{\beta_n}}{\beta_n!} \frac{(t-t_0)^\alpha}{\alpha!} & \text{for } n \ge 1 
\end{cases} $$

*Proof.* It's sufficient to take: $m = \alpha$, $k_i = \beta_i - \beta_{i+1}$ for $i = 1, ..., n-1$ and $k_n = \beta_n$. $\square$

By using the decreasing sequences: $(\beta_i)_{i=1,...,n}$ we are going to show that Adomian's polynomials are easily computed. We consider the system of equations:

$$ \alpha + \sum_{i=1}^{n} \beta_{i} = n \qquad (19) $$

with:

$$ \beta_1 \ge \beta_2 \ge \cdots \ge \beta_n \qquad (20) $$

If $n = 5$, we obtain the table:

<table><thead><tr><th>α</th><th>β₁</th><th>β₂</th><th>β₃</th><th>β₄</th><th>β₅</th></tr></thead><tbody><tr><td>5</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>4</td><td>1</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>3</td><td>2</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>3</td><td>1</td><td>1</td><td>0</td><td>0</td><td>0</td></tr><tr><td>2</td><td>3</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>2</td><td>2</td><td>1</td><td>0</td><td>0</td><td>0</td></tr><tr><td>2</td><td>1</td><td>1</td><td>1</td><td>0</td><td>0</td></tr><tr><td>1</td><td>4</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>1</td><td>3</td><td>1</td><td>0</td><td>0</td><td>0</td></tr><tr><td>1</td><td>2</td><td>2</td><td>0</td><td>0</td><td>0</td></tr><tr><td>1</td><td>2</td><td>1</td><td>1</td><td>0</td><td>0</td></tr><tr><td>1</td><td>1</td><td>1</td><td>1</td><td>0</td><td>0</td></tr><tr><td>0</td><td>5</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>4</td><td>1</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>3</td><td>2</td><td>0</td><td>0</td><td>0</td></tr><tr><td>0</td><td>3</td><td>1</td><td>1</td><td>0</td><td>0</td></tr><tr><td>0</td><td>2</td><td>2</td><td>1</td><td>0</td><td>0</td></tr><tr><td>0</td><td>2</td><td>1</td><td>1</td><td>1</td><td>0</td></tr><tr><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr></tbody></table>
We can theirs list the Adomian's polynomials for: n = 0, ..., 5

$$ \tilde{A}_0 = f(t_0, u_0) $$

$$ \tilde{A}_1 = \partial_t f(t_0, u_0)(t - t_0) + \partial_u f(t_0, u_0)u_1 $$

$$ \begin{aligned} \tilde{A}_2 = & \frac{1}{2} \partial_{t^2} f(t_0, u_0)(t - t_0)^2 + \partial_{tu} f(t_0, u_0)u_1(t - t_0) \\ & + \frac{1}{2} \partial_{u^2} f(t_0, u_0)u_1^2 + \partial_u f(t_0, u_0)u_2 \end{aligned} $$

$$ \begin{aligned} \tilde{A}_3 = & \frac{1}{6} \partial_{t^3} f(t_0, u_0)(t - t_0)^3 + \frac{1}{2} \partial_{t^2 u} f(t_0, u_0)u_1(t - t_0)^2 \\ & + \left( \frac{1}{2} \partial_{tu^2} f(t_0, u_0)u_1^2 + \partial_{tu} f(t_0, u_0)u_2 \right) (t - t_0) \\ & + \frac{1}{6} \partial_{u^3} f(t_0, u_0)u_1^3 + \partial_{u^2} f(t_0, u_0)u_1u_2 + \partial_u f(t_0, u_0)u_3 \end{aligned} $$

$$ \begin{aligned} \tilde{A}_4 = & \frac{1}{24} \partial_{t^4} f(t_0, u_0)(t - t_0)^4 + \frac{1}{6} \partial_{t^3 u} f(t_0, u_0)u_1(t - t_0)^3 \\ & + \left( \frac{1}{4} \partial_{t^2 u^2} f(t_0, u_0)u_1^2 + \frac{1}{2} \partial_{t^2 u} f(t_0, u_0)u_2 \right) (t - t_0)^2 \\ & + \left( \frac{1}{6} \partial_{tu^3} f(t_0, u_0)u_1^3 + \partial_{tu^2} f(t_0, u_0)u_1u_2 + \partial_{tu} f(t_0, u_0)u_3 \right) (t - t_0) \\ & + \frac{1}{24} \partial_{u^4} f(t_0, u_0)u_1^4 + \frac{1}{2} \partial_{u^3} f(t_0, u_0)u_1^2u_2 \\ & + \partial_{u^2} f(t_0, u_0) \left( \frac{1}{2}u_2^2 + u_1u_3 \right) + \partial_u f(t_0, u_0)u_4 \end{aligned} $$

$$ \begin{aligned} \tilde{A}_5 = & \frac{1}{120} \partial_{t^5} f(t_0, u_0)(t - t_0)^5 + \frac{1}{24} \partial_{t^4 u} f(t_0, u_0)u_1(t - t_0)^4 \\ & + \left( \frac{1}{12} \partial_{t^3 u^2} f(t_0, u_0)u_1^2 + \frac{1}{6} \partial_{t^3 u} f(t_0, u_0)u_2 \right) (t - t_0)^3 \\ & + \left( \frac{1}{12} \partial_{t^2 u^2} f(t_0, u_0)u_1^3 + \frac{1}{2} \partial_{t^2 u} f(t_0, u_0)u_1u_2 + \frac{1}{2} \partial_{t^2 u} f(t_0, u_0)u_3 \right) (t - t_0)^2 \\ & + \left( \frac{1}{24} \partial_{tu^4} f(t_0, u_0)u_1^4 + \frac{1}{2} \partial_{tu^3} f(t_0, u_0)u_1^2u_2 + \partial_{tu^2} f(t_0, u_0) \left( \frac{1}{2}u_2^2 + u_1u_3 \right) \right) (t - t_0) \\ & + \partial_{tu} f(t_0, u_0)u_4 (t - t_0) + \frac{1}{120} \partial_{u^5} f(t_0, u_0)u_1^5 + \frac{1}{6} \partial_{u^4} f(t_0, u_0)u_1^3u_2 \\ & + \frac{1}{2} \partial_{u^3} f(t_0, u_0) (u_1u_2^2 + u_1^2u_3) + \partial_{u^2} f(t_0, u_0) (u_2u_3 + u_1u_4) + \partial_u f(t_0, u_0)u_5 \end{aligned} $$
## 5 Convergence of the technique

**Proposition 1.** For all $n \in \mathbb{N}$ we have:

$$ \frac{d^n}{dt^n} f(t, u(t)) = \sum_{m+|nk_n|=n} \frac{n!}{m! \prod_{i=1}^n k_i!} \partial_{t^{m} u^{k}|} f(t, u(t)) \prod_{i=1}^n \left( \frac{u^{[i]}(t)}{i!} \right)^{k_i} \quad (21) $$

*Proof.* Inserting the variables $x, y$ by $t, u$ into (17), we have:

$$ \left\{ \begin{array}{l} x(t) = t \implies \begin{cases} x'(t) = 1 \\ x^{[i]}(t) = 0 \end{cases} \quad \text{for } i = 2, \dots, n \\ y(t) = u(t) \implies \begin{cases} y^{[i]}(t) = u^{[i]}(t) \end{cases} \quad \text{for } i = 1, \dots, n \end{array} \right. $$

so:

$$ \begin{aligned} \frac{d^n}{dt^n} f(t, u(t)) &= \left. \frac{d^n}{d\lambda^n} f(x(\lambda), y(\lambda)) \right|_{\lambda=t} \\ &= \sum_{|nm_n|+|nk_n|=n} \frac{n!}{\prod_{i=1}^n m_i! k_i!} \partial_{x^{|m|} y^{|k|}} f(x(t), y(t)) \\ & \quad \left( \prod_{i=1}^n \left( \frac{x^{[i]}(t)}{i!} \right)^{m_i} \left( \frac{y^{[i]}(t)}{i!} \right)^{k_i} \right) \\ &= \sum_{\substack{|nm_n|+|nk_n|=n \\ m_1+|nk_n| \neq n}} \frac{n!}{\prod_{i=1}^n m_i! k_i!} \partial_{t^{|m|} u^{|k|}} f(t, u(t)) \\ & \quad \left( \left( \frac{1}{1!} \right)^{m_1} \prod_{i=2}^n \left( \frac{0}{i!} \right)^{m_i} \prod_{i=1}^n \left( \frac{u^{[i]}(t)}{i!} \right)^{k_i} \right) \\ & \quad + \sum_{m_1+|nk_n|=n} \frac{n!}{m_1! \prod_{i=1}^n k_i!} \partial_{t^{m_1} u^{k}|} f(t, u(t)) \left( \left( \frac{1}{1!} \right)^{m_1} \prod_{i=1}^n \left( \frac{u^{[i]}(t)}{i!} \right)^{k_i} \right) \\ &= \sum_{m_1+|nk_n|=n} \frac{n!}{m_1! \prod_{i=1}^n k_i!} \partial_{t^{m_1} u^{k}|} f(t, u(t)) \prod_{i=1}^n \left( \frac{u^{[i]}(t)}{i!} \right)^{k_i} \end{aligned} $$

□

**Proposition 2.** For all $n \in \mathbb{N}$ we have:

$$ \tilde{A}_n = \frac{1}{n!} u^{[n+1]}(t_0) (t - t_0)^n \quad (22) $$
Proof. For $n = 0$ in (16), the relationship (22) is checked.
According the relationship (14), we have for $i = 1, \dots, n$:

$$u_i = \int_{t_0}^{t} \tilde{A}_{i-1} ds = \int_{t_0}^{t} \frac{1}{(i-1)!} u^{[i]}(t_0) (s-t_0)^{i-1} ds = \frac{1}{i!} u^{[i]}(t_0) (t-t_0)^i$$

According to the relationship (18), we have:

$$\begin{aligned}
\tilde{A}_n &= \sum_{m+|nk_n|=n} \partial_{t^m u^{[k]}} f_j(t_0, u_0) \left( \prod_{i=1}^{n} \frac{u_i^{k_i}}{k_i!} \right) \frac{(t-t_0)^m}{m!} \\
&= \sum_{m+|nk_n|=n} \frac{1}{\prod_{i=1}^{n} k_{i,j}!} \partial_{t^m u^{[k]}} f(t_0, u_0) \left\{ \prod_{i=1}^{n} \left[ \frac{1}{i!} u^{[i]}(t_0) (t-t_0)^i \right]^{k_i} \right\} \frac{(t-t_0)^m}{m!} \\
&= \frac{1}{n!} \sum_{m+|nk_n|=n} \frac{n!}{m! \prod_{i=1}^{n} k_i!} \partial_{t^m u^{[k]}} f(t_0, u_0) \left\{ \prod_{i=1}^{n} \left[ \frac{u^{[i]}(t_0)}{i!} \right]^{k_i} \right\} (t-t_0)^{m+|nk_n|} \\
&= \frac{1}{n!} \left[ \frac{d^n}{dt^n} f(t, u(t)) \bigg|_{t=t_0} \right] (t-t_0)^n \quad \text{according to (21)}
\end{aligned}$$

while $\frac{d}{dt}u = f(t, u)$ according to (8) so:

$$\frac{d^n}{dt^n} f(t, u) = \frac{d^{n+1}}{dt^{n+1}} u(t) = u^{[n+1]}(t)$$

where: (22).

<figure><img src="image_3.png" /></figure>

**Proposition 3.**

$$u = \sum_{n \ge 0} \frac{1}{n!} u^{[n]}(t_0) (t - t_0)^n \quad (23)$$

Proof. According to (14) and (22) we have:

$$u_n = \int_{t_0}^{t} \tilde{A}_{n-1} ds = \int_{t_0}^{t} \frac{1}{(n-1)!} u^{[n]}(t_0) (s-t_0)^{n-1} ds = \frac{1}{n!} u_j^{[n]}(t_0) (t-t_0)^n$$

using (11) we obtain:

$$u = \sum_{n \ge 0} u_n = \sum_{n \ge 0} \frac{1}{n!} u^{[n]}(t_0) (t - t_0)^n$$

<figure><img src="image_4.png" /></figure>
**Proposition 4.** For all $n \ge 2$

$$ \sum_{\substack{m+|nk_n|=n \\ m \ne 1}} \frac{1}{m! \prod_{i=1}^n k_i! i^{k_i}} + 2 \sum_{1+|nk_n|=n} \frac{1}{\prod_{i=1}^n k_i! i^{k_i}} = 1 \quad (24) $$

*Proof.* We take in (21), $f(t, u(t)) = t + \exp(u(t))$ with: $u(t) = \sum_{m \ge 1} \frac{1}{m} t^m$ and $t = 0$:

$$ u(t) = \sum_{m \ge 1} \frac{1}{m} t^m \implies u^{[i]}(t) = \sum_{m \ge i} \frac{(m-1)!}{(m-i)!} t^{m-i} \quad \text{for } i = 1, \dots, n $$

it becomes:

$$ \left\{ \begin{array}{l} u(0) = 0 \\ u^{[i]}(0) = (i-1)! \quad \text{for } i = 1, \dots, n \end{array} \right. $$

where:

$$ \begin{aligned} & \left. \frac{d^n}{dt^n} \left[ t + \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \right|_{t=0} = \left. \frac{d^n}{dt^n} f(t, u(t)) \right|_{t=0} \\ &= \sum_{m+|nk_n|=n} \frac{n!}{m! \prod_{i=1}^n k_i!} \partial_{t^m u^{|k|}} f(0, u(0)) \prod_{i=1}^n \left[ \frac{u^{[i]}(0)}{i!} \right]^{k_i} \\ &= \sum_{m+|nk_n|=n} \frac{n!}{m! \prod_{i=1}^n k_i!} \partial_{t^m u^{|k|}} \exp(0) \prod_{i=1}^n \left[ \frac{(i-1)!}{i!} \right]^{k_i} \\ & \quad + \sum_{|nk_n|=n} \frac{n!}{0! \prod_{i=1}^n k_i!} \partial_{t u^{|k|}} [0 + \exp(0)] \prod_{i=1}^n \left[ \frac{(i-1)!}{i!} \right]^{k_i} \\ & \quad + \sum_{1+|nk_n|=n} \frac{n!}{1! \prod_{i=1}^n k_i!} \partial_{t u^{|k|}} [1 + \exp(0)] \prod_{i=1}^n \left[ \frac{(i-1)!}{i!} \right]^{k_i} \\ &= n! \left( \sum_{\substack{m+|nk_{n,j}|=n \\ m \ne 1}} \frac{1}{m! \prod_{i=1}^n k_i! i^{k_i}} + 2 \sum_{1+|nk_{n,j}|=n} \frac{1}{\prod_{i=1}^n k_i! i^{k_i}} \right) \end{aligned} $$
we pose:

$$ \gamma_n = \frac{1}{n!} \left. \frac{d^n}{dt^n} \left[ t + \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \right|_{t=0} $$

For $n = 2$, we have:

$$ \begin{aligned} \gamma_2 &= \frac{1}{2!} \left. \frac{d^2}{dt^2} \left[ t + \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \right|_{t=0} \\ &= \frac{1}{2} \left. \frac{d}{dt} \left[ 1 + \left( \sum_{m \ge 1} \lambda^{m-1} \right) \exp \left( \sum_{m \ge 1} \frac{1}{m} \lambda^m \right) \right] \right|_{t=0} \\ &= \frac{1}{2} \left\{ \left[ \sum_{m \ge 2} (m-1) t^{m-2} \right] \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right. \\ & \qquad \left. + \left( \sum_{m \ge 1} t^{m-1} \right)^2 \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right\} \Big|_{t=0} = 1 \end{aligned} $$

We suppose that: $\gamma_i = 1$ for $i = 2, \dots, n-1$, then:

$$ \begin{aligned} \gamma_n &= \frac{1}{n!} \left. \frac{d^n}{dt^n} \left[ t + \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \right|_{t=0} \\ &= \frac{1}{n!} \left. \frac{d^{n-1}}{dt^{n-1}} \left\{ \frac{d}{dt} \left[ t + \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \right\} \right|_{t=0} \\ &= \frac{1}{n!} \left\{ \frac{d^{n-1}}{dt^{n-1}} \left[ 1 + \left( \sum_{m \ge 1} t^{m-1} \right) \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \right\} \Big|_{t=0} \\ &= \frac{1}{n!} \sum_{i=0}^{n-1} \left\{ \binom{i}{n-1} \frac{d^{n-i-1}}{dt^{n-i-1}} \left( \sum_{m \ge 1} t^{m-1} \right) \Bigg|_{t=0} \frac{d^i}{dt^i} \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \Bigg|_{t=0} \right\} \end{aligned} $$
$$ \gamma_n = \frac{1}{n!} \left\{ \left. \frac{d^{n-1}}{dt^{n-1}} \left( \sum_{m \ge 1} t^{m-1} \right) \right|_{t=0} \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \bigg|_{t=0} + (n-1) \left. \frac{d^{n-2}}{dt^{n-2}} \left( \sum_{m \ge 1} t^{m-1} \right) \right|_{t=0} \frac{d}{dt} \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \bigg|_{t=0} + \sum_{i=2}^{n-1} \left\{ \begin{pmatrix} i \\ n-1 \end{pmatrix} \frac{d^{n-i-1}}{dt^{n-i-1}} \left( \sum_{m \ge 1} t^{m-1} \right) \bigg|_{t=0} \frac{d^i}{dt^i} \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \bigg|_{t=0} \right\} \right\} $$

we could remark that:

$$ \begin{aligned} \left. \frac{d^k}{dt^k} \left( \sum_{m \ge 1} t^{m-1} \right) \right|_{t=0} &= k! \\ \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \bigg|_{t=0} &= \exp(0) = 1 \\ \left. \frac{d}{dt} \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right|_{t=0} &= \left( \sum_{m \ge 1} t^m \right) \exp \left( \frac{1}{m} t^m \right) \bigg|_{t=0} = \exp(0) = 1 \\ \left. \frac{d^i}{dt^i} \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right|_{t=0} &= \frac{d^i}{dt^i} \left[ t + \exp \left( \sum_{m \ge 1} \frac{1}{m} t^m \right) \right] \bigg|_{t=0} = i! \gamma_i \\ &= i! \quad \forall i = 2, \dots, n-1 \end{aligned} $$

then:

$$ \gamma_n = \frac{1}{n!} \left[ (n-1)! + (n-1)(n-2)! + \sum_{i=2}^{n-1} \frac{(n-1)!}{i!(n-i-1)!} (n-i-1)!i! \right] = 1 $$

<figure><img src="image_5.png" /></figure>

**Theorem 3.** If $|\partial_{tuk} f(t_0, u_0)| \le 2M^2$ and $|\partial_{t^m u^k} f(t_0, u_0)| \le M^{m+1}$ for $m \ne 1$ then: $|u_n| \le \frac{(|t-t_0|M)^n}{n}$ for $n \ge 1$. The series solution of the equation (8) is absolutely convergent within the interval $]t_0 - \frac{1}{M}, t_0 + \frac{1}{M}]$

*Proof.* According to (23), we have: $u = \sum_{n \ge 0} \frac{1}{n!} u^{[n]}(t_0) (t - t_0)^n$ so: $u_n = \frac{1}{n!} u^{[n]}(t_0) (t - t_0)^n$ where:
$|u_n| = \frac{1}{n!} |u^{[n]}(t_0)| |t - t_0|^n$. It is sufficient to show that: $\frac{1}{n!} |u^{[n]}(t_0)| \le \frac{M^n}{n}$
If $n = 1$ we have:

$$
\frac{1}{1!} |u'(t_0)| = |f(t_0, u_0)| \le M^{0+1} = \frac{M^1}{1}
$$

Suppose that: $\frac{1}{i!} |u^{[i]}(t_0)| \le \frac{M^i}{i}$ for $i = 1, \dots, n-1$ so:

$$
\begin{align*}
\frac{1}{n!} |u^{[n]}(t_0)| &= \frac{1}{n!} \left| \frac{d^{n-1}}{dt^{n-1}} f(t_0, u) \right| \\
&= \frac{1}{n!} \left| \sum_{m+|(n-1)k_{n-1}|=n-1} \frac{(n-1)!}{m! \prod_{i=1}^{n-1} k_i!} \partial_{t^m u^{[k]}} f(t_0, u_0) \prod_{i=1}^{n-1} \left( \frac{u^{[i]}(t)}{i!} \right)^{k_i} \right| \\
&\le \frac{1}{n} \sum_{m+|(n-1)k_{n-1}|=n-1} \frac{1}{m! \prod_{i=1}^{n-1} k_i!} |\partial_{t^m u^{[k]}} f(t_0, u_0)| \prod_{i=1}^{n-1} \left( \frac{1}{i!} |u^{[i]}(t)| \right)^{k_i} \\
&\le \frac{1}{n} \left( \sum_{m+|(n-1)k_{n-1}|=n-1} \frac{1}{m! \prod_{i=1}^{n-1} k_i!} |\partial_{t^m u^{[k]}} f(t_0, u_0)| \prod_{i=1}^{n-1} \left( \frac{1}{i!} |u^{[i]}(t)| \right)^{k_i} \right. \\
&\qquad \left. + \sum_{1+|(n-1)k_{n-1}|=n-1} \frac{1}{\prod_{i=1}^{n-1} k_i!} |\partial_{t u^{[k]}} f(t_0, u_0)| \prod_{i=1}^{n-1} \left( \frac{1}{i!} |u^{[i]}(t)| \right)^{k_i} \right) \\
&\le \frac{1}{n} \left( \sum_{m+|(n-1)k_{n-1}|=n-1} \frac{1}{m! \prod_{i=1}^{n-1} k_i!} M^{m+1} \prod_{i=1}^{n-1} \frac{M^{ik_i}}{i^{k_i}} \right. \\
&\qquad \left. + \sum_{1+|(n-1)k_{n-1}|=n-1} \frac{1}{\prod_{i=1}^{n-1} k_i!} 2M^2 \prod_{i=1}^{n-1} \frac{M^{ik_i}}{i^{k_i}} \right)
\end{align*}
$$
$$ \le \frac{1}{n} \left( \sum_{\substack{m+(n-1)k_{n-1}=n-1 \\ m \ne 1}} \frac{1}{m! \prod_{i=1}^{n-1} k_i! i^{k_i}} M^{m+1+(n-1)k_{n-1}}} + 2 \sum_{1+(n-1)k_{n-1}=n-1} \frac{1}{\prod_{i=1}^{n-1} k_i! i^{k_i}} M^{2+(n-1)k_{n-1}}} \right) \\ \le \frac{M^n}{n} \left( \sum_{\substack{m+(n-1)k_{n-1}=n-1 \\ m \ne 1}} \frac{1}{m! \prod_{i=1}^{n-1} k_i! i^{k_i}} + 2 \sum_{1+(n-1)k_{n-1}=n-1} \frac{1}{\prod_{i=1}^{n-1} k_i! i^{k_i}} \right) \\ = \frac{M^n}{n} $$

the absolute convergence follows the D'Alembert criterion. □

# 6 Numerical examples

## 6.1 Example

Let us solve the following equation:

$$ \begin{cases} u' = (t-1) \exp(u-t) \\ u(0) = 1 \end{cases} \qquad (25) $$

The exact solution of this equation is the function:

$$ u(t) = t - \ln t $$

By the Adomian decomposition method, we obtain table 1, from where, it follows that:
$$
\begin{align*}
\phi_5(t) &= 1 + \frac{1}{2}(t-1)^2 - \frac{1}{3}(t-1)^3 + \frac{1}{4}(t-1)^4 \\
\phi_6(t) &= 1 + \frac{1}{2}(t-1)^2 - \frac{1}{3}(t-1)^3 + \frac{1}{4}(t-1)^4 - \frac{1}{5}(t-1)^5 \\
\phi_7(t) &= 1 + \frac{1}{2}(t-1)^2 - \frac{1}{3}(t-1)^3 + \frac{1}{4}(t-1)^4 - \frac{1}{5}(t-1)^5 + \frac{1}{6}(t-1)^6
\end{align*}
$$

The graphs of the exact solution and the ADM approximatives solutions are
in Fig.1.

<table><caption>Tab. 1: ADM Solution of the equation (25)</caption>
  <thead>
    <tr>
      <th>n</th>
      <th>0</th>
      <th>1</th>
      <th>2</th>
      <th>3</th>
      <th>4</th>
      <th>5</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <th>Δ&#x0302;<sub>n</sub></th>
      <td>0</td>
      <td>t - 1</td>
      <td>-(t - 1)<sup>2</sup></td>
      <td>(t - 1)<sup>3</sup></td>
      <td>-(t - 1)<sup>4</sup></td>
      <td>(t - 1)<sup>5</sup></td>
    </tr>
    <tr>
      <th>u<sub>n+1</sub></th>
      <td>0</td>
      <td><sup>1</sup>&frasl;<sub>2</sub>(t - 1)<sup>2</sup></td>
      <td>-<sup>1</sup>&frasl;<sub>3</sub>(t - 1)<sup>3</sup></td>
      <td><sup>1</sup>&frasl;<sub>4</sub>(t - 1)<sup>4</sup></td>
      <td>-<sup>1</sup>&frasl;<sub>5</sub>(t - 1)<sup>5</sup></td>
      <td><sup>1</sup>&frasl;<sub>6</sub>(t - 1)<sup>6</sup></td>
    </tr>
  </tbody>
</table>

**6.2 Example**

Let us solve the following equation:

$$
\left\{
\begin{array}{l}
u' = \exp(-u + \sin t) + \sin t \exp(u - \sin t) \\
u(0) = 0
\end{array}
\right.
\qquad (26)
$$

The exact solution of this equation is the function:

$$
u(t) = \sin t - \ln(\cos t)
$$

By the Adomian decomposition method, we obtain table 2, from where, it
follows that:

$$
\begin{align*}
\phi_5(t) &= t + \frac{1}{2}t^2 - \frac{1}{6}t^3 + \frac{1}{12}t^4 \\
\phi_6(t) &= t + \frac{1}{2}t^2 - \frac{1}{6}t^3 + \frac{1}{12}t^4 + \frac{1}{120}t^5 \\
\phi_7(t) &= t + \frac{1}{2}t^2 - \frac{1}{6}t^3 + \frac{1}{12}t^4 + \frac{1}{120}t^5 + \frac{1}{45}t^6
\end{align*}
$$

The graphs of the exact solution and the ADM approximatives solutions are
in Fig.2.
<table><caption>Tab. 2: ADM Solution of the equation (26)</caption><thead><tr><th>n</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr></thead><tbody><tr><td>A&#x0302;<sub>n</sub></td><td>1</td><td>t</td><td>-1/2 t<sup>2</sup></td><td>1/3 t<sup>3</sup></td><td>1/24 t<sup>4</sup></td><td>2/15 t<sup>5</sup></td></tr><tr><td>u<sub>n+1</sub></td><td>t</td><td>1/2 t<sup>2</sup></td><td>-1/6 t<sup>3</sup></td><td>1/12 t<sup>4</sup></td><td>1/120 t<sup>5</sup></td><td>1/45 t<sup>6</sup></td></tr></tbody></table>

## 6.3 Example

Let us solve the following equation:

$$ \left\{ \begin{array}{l} u' = u \{1 - [\exp(t-2) + 1]u\} \\ u(2) = -1 \end{array} \right. \qquad (27) $$

The exact solution of this equation is the function:

$$ u(t) = \frac{1}{1 - t \exp(t - 2)} $$

By the Adomian decomposition method, we obtain table 3, from where, it follows that:

$$ \phi_5(t) = -1 + 3(t-2) - 7(t-2)^2 - \frac{95}{6}(t-2)^3 + \frac{143}{4}(t-2)^4 $$

$$ \phi_6(t) = -1 + 3(t-2) - 7(t-2)^2 - \frac{95}{6}(t-2)^3 + \frac{143}{4}(t-2)^4 \\ - \frac{3229}{40}(t-1)^5 $$

$$ \phi_7(t) = -1 + 3(t-2) - 7(t-2)^2 - \frac{95}{6}(t-2)^3 + \frac{143}{4}(t-2)^4 \\ - \frac{3229}{40}(t-1)^5 + \frac{10937}{60}(t-2)^7 $$

The graphs of the exact solution and the ADM approximatives solutions are in Fig.3.

<table><caption>Tab. 3: ADM Solution of the equation (27)</caption><thead><tr><th>n</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr></thead><tbody><tr><td>A&#x0302;<sub>n</sub></td><td>3</td><td>-14(t-2)</td><td>95/2(t-2)<sup>2</sup></td><td>-143(t-2)<sup>3</sup></td><td>3229/8(t-2)<sup>4</sup></td><td>-10937/10(t-2)<sup>5</sup></td></tr><tr><td>u<sub>n+1</sub></td><td>3(t-2)</td><td>-7(t-2)<sup>2</sup></td><td>95/6(t-2)<sup>3</sup></td><td>-143(t-2)<sup>4</sup></td><td>3229/40(t-2)<sup>5</sup></td><td>-10937/60(t-2)<sup>6</sup></td></tr></tbody></table>
<figure><img src="image_6.png" /><figcaption>Fig. 1: Graphs $u, \phi_7, \phi_6$ and $\phi_5$ solutions of the equation (25)</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Fig. 2: Graphs $u, \phi_7, \phi_6$ and $\phi_5$ solutions of the equation (26)</figcaption></figure>
<figure><img src="image_8.png" /><figcaption>Fig. 3: Graphs $u, \phi_7, \phi_6$ and $\phi_5$ solutions of the equation (27)</figcaption></figure>

# 7 Conclusion

Finally we have got a new formula of ADM (16) and (18) using to solve nonlinear and non-autonomous ODEs more general than the classical formula (5), we have proved that the series solution coincides with that the Taylor series. Moreover We have obtained also a new conditions of convergence more general than the conditions of K. Abbaoui and Y. Cherruault in [8]

# Acknowledgment

The authors of this paper are very thankful to the editor who suggested to accepting our article for publication in Bulletin of Computational Applied Mathematics - Bull CompAMa, without forgetting all other editors, especially the editor-in-chief of this journal, thank you very much.
References

[1] G. Adomian. *Solving Frontier Problems of Physics: The Decomposition Method*. Klumer, Boston, 1994. doi: 10.1007/978-94-015-8289-6, ISBN 978-0-7923-2644-1.

[2] K. Abbaoui. *Les fondements mathématiques de la méthode décompositionnelle d'Adomian et application à la résolution de problémes isus de la biologie et de la médecine*. PhD thesis, Paris 6, 1995. www.theses.fr/1995PA066495.

[3] J-S. Duan. New ideas for decomposing nonlinearities in differential equations. *Applied Mathematics and Computation*, 218(5):1774–1784, 2011. doi: 10.1016/j.amc.2011.06.061.

[4] J-S. Duan. New higher-order numerical one-step methods based on the Adomian and the modified decomposition methods. *Applied Mathematics and Computation*, 218(6):2810–2828, 2011. doi: 10.1016/j.amc.2011.08.024.

[5] G. Adomian and R. Rach. A nonlinear differential delay equation. *Journal of Mathematical Analysis and Applications*, 91(2):301–304, 1983. doi: 10.1016/0022-247X(83)90152-X.

[6] G. Adomian and R. Rach. Polynomial nonlinearities in differential equations. *Journal of Mathematical Analysis and Applications*, 109(1):90–95, 1985. doi: 10.1016/0022-247X(85)90178-7.

[7] G. Adomian and R. Rach. Nonlinear differential equations with negative power nonlinearities. *Journal of Mathematical Analysis and Applications*, 112(2):497–501, 1985. doi: 10.1016/0022-247X(85)90258-6.

[8] K. Abbaoui and Y. Cherruault. Convergence of Adomian's method applied to differential equations. *Computers and Mathematics with Applications*, 28(5):103–109, 1994. doi: 10.1016/0898-1221(94)00144-8.

[9] K. Abbaoui and Y. Cherruault. The decomposition method applied to the Cauchy problem. *Kybernetes*, 28(1):68–74, 1999. doi: 10.1108/03684929910253261.
[10] A.M. Wazwaz. A new algorithm for solving differential equations of Lane-Emden type. *Applied Mathematics and Computation*, 118(2):287-310, 2001. doi: 10.1016/S0096-3003(99)00223-4.

[11] R. L. Mishkov. Generalization of the formula of Faà di Bruno for a composite function with a vector argument, article id 498526. *International Journal of Mathematics and Mathematical Sciences*, 24(7):481-491, 2000. doi: 10.1155/S0161171200002970.

[12] W.P. Johnson. The curious history of Faà di Bruno's formula. *The American Mathematical Monthly*, 109(3):217-234, 2002. doi: 10.2307/2695352.