A note on the convergence of the Adomian decomposition method

TL;DR

This paper presents a new convergence result for the Adomian decomposition method, enhancing understanding of its mathematical properties and potential applications.

Contribution

It introduces a novel convergence theorem for the Adomian decomposition method, expanding theoretical insights into its behavior.

Findings

New convergence theorem established

Improved understanding of the method's reliability

Potential for broader application in nonlinear problems

Abstract

In this note we obtain a new convergence result for the Adomian decomposition method.

Equations

$$||f|{|}^{[n]}:=\underset{(x_1,\dots ,x_n)\in (X-\{0\}{)}^n}{\sup }\left(\frac{||f(x_1,\dots ,x_n)||}{||x_1||\dots ||x_n||}\right)$$

$$\mathfrak{R}\left(F,a\right):=\frac{1}{\underset{n\to +\infty }{\mathrm{limsup}}{\left(\frac{1}{n!}{\Vert {\mathfrak{A}}_n(a)\Vert }^{\left[n\right]}\right)}^{\frac{1}{n}}}$$

$$F^{\left(n\right)}\left(a\right)=\frac{1}{n!}{\mathfrak{A}}_n(a)$$

$$\mathfrak{u}=\mathfrak{N}(\mathfrak{u})+\mathfrak{f}$$

$$A_n:=\frac{1}{n!}\frac{d^n}{d{\epsilon }^n}{\left[\mathfrak{N}\left(\underset{j=0}{\sum ^{+\infty }}{\epsilon }^j{\mathfrak{u}}_j\right)\right]}_{\epsilon =0}$$

$$\underset{n=0}{\sum ^{+\infty }}{\mathfrak{u}}_n=\mathfrak{f}+\underset{n=0}{\sum ^{+\infty }}{A}_n$$

$$\left\{\begin{array}[]{c}\mathfrak{u}_{0}=\mathfrak{f}\\ \mathfrak{u}_{n+1}=A_{n}\end{array}\right.$$

$$A_n:=\frac{1}{n!}\frac{d^n}{d{\epsilon }^n}{\left[\mathfrak{N}\left(\underset{j=0}{\sum ^n}{\epsilon }^j{\mathfrak{u}}_j\right)\right]}_{\epsilon =0}$$

$$A_n=\sum _{j_1+2j_2+...+n{j}_n=n}\frac{1}{j_1!...j_n!}{\mathfrak{N}}^{\left(j_1+...+j_n\right)}(\mathfrak{f}).\left({\mathfrak{u}}_1^{\left[j_1\right]}\cdot \cdot \cdot {\mathfrak{u}}_n^{\left[j_n\right]}\right)$$

$$\mathfrak{N}\left(\underset{j=0}{\sum ^n}{\epsilon }^j{\mathfrak{u}}_j\right)=\mathfrak{N}(u_0)+\underset{p=0}{\sum ^{+\infty }}\frac{1}{p!}{\mathfrak{N}}^{\left(p\right)}({\mathfrak{u}}_0).{\left(\underset{j=0}{\sum ^n}{\epsilon }^j{\mathfrak{u}}_j\right)}^{\left[p\right]}$$

$$A_n=\sum _{j_1+2j_2+...+n{j}_n=n}\frac{1}{j_1!...j_n!}{\mathfrak{N}}^{\left(j_0+j_1+...+j_n\right)}(\mathfrak{f}).\left({\mathfrak{u}}_1^{\left[j_1\right]}\cdot \cdot \cdot {\mathfrak{u}}_n^{\left[j_n\right]}\right)$$

$$||{\mathfrak{N}}^{\left(n\right)}(\mathfrak{f})|{|}^{\left[n\right]}\le M{a}^{n}n!,n\in \mathbb{N}$$

$$Ma\le \frac{1}{4}$$

$$\left\{\begin{array}[]{c}\left\|\mathfrak{u}\right\|\leq\left\|\mathfrak{f}\right\|+\left(1+\frac{8}{\sqrt{3\pi}}\right)M\\ \left\|\mathfrak{u-}\underset{j=0}{\overset{n}{\sum}}\mathfrak{u}_{j}\right\|\leq\frac{8M}{\sqrt{3\pi}}\frac{1}{\sqrt{n}},\mathit{\ }n\in\mathbb{N}^{\ast}\end{array}\right.$$

$$\left\{\begin{array}[]{c}\left\|\mathfrak{u}\right\|\leq\left\|\mathfrak{f}\right\|+\left(1+\frac{32M^{3}a^{3}}{\sqrt{6\pi}\left(1-4Ma\right)}\right)M\\ \left\|\mathfrak{u-}\underset{j=0}{\overset{n}{\sum}}\mathfrak{u}_{j}\right\|\leq\frac{16M^{2}a}{\sqrt{3\pi}\left(1-4Ma\right)}\frac{\left(4Ma\right)^{n}}{\left(n+1\right)\sqrt{n+1}},\text{ }n\in\mathbb{N}^{\ast}\end{array}\right.$$

$$||{\mathfrak{u}}_n||\le w_n{M}^n{a}^{n-1},n\in {\mathbb{N}}^{\ast }$$

$$\left\{\begin{array}[]{c}w_{1}=1\\ w_{n+1}=\underset{j_{1}+2j_{2}+...+nj_{n}=n}{\sum}\frac{\left(j_{1}+...+j_{n}\right)!}{j_{1}!...j_{n}!}\left(w_{1}\right)^{j_{1}}...\left(w_{n}\right)^{j_{n}},\text{ }n\in\mathbb{N}^{\ast}\end{array}\right.$$

$$\sum _{n\ge 1}{w}_{n+1}{T}^{n+1}=T\sum _{n\ge 1}\left(\sum _{j_1+2j_2+...+n{j}_n=n}\frac{\left(j_1+...+j_n\right)!}{j_1!...j_n!}{\left(w_{1}T\right)}^{{}^{j_1}}...{\left(w_n{T}^n\right)}^{j_n}\right)$$

$$S-T=T\left(\frac{S}{1-S}\right)$$

$$S=T\left(\frac{1}{1-S}\right)$$

$$\left[T^n\right]S=\frac{1}{n!}{\left(\frac{d^{n-1}}{d{t}^{n-1}}\left[{\left(\frac{1}{1-t}\right)}^n\right]\right)}_{t=0}$$

$$w_n=\frac{\left(2n-1\right)!}{{\left(n!\right)}^2}$$

$$||{\mathfrak{u}}_n||\le \frac{\left(2n-1\right)!}{{\left(n!\right)}^2}{M}^n{a}^{n-1},n\in {\mathbb{N}}^{\ast }$$

$$\frac{\left(2n-1\right)!}{{\left(n!\right)}^2}{M}^n{a}^{n-1}\underset{n\to +\infty }{\sim }\frac{1}{2\sqrt{\pi }a}\frac{1}{n\sqrt{n}}{\left(4Ma\right)}^n$$

$$\mathfrak{U}(\epsilon ):=\stackrel{+\infty }{\sum _{n=0}}{\epsilon }^n{\mathfrak{u}}_n,\mathfrak{u}:=\stackrel{+\infty }{\sum _{n=0}}{\mathfrak{u}}_n=\mathfrak{U}(1)$$

$$\mathfrak{N}(\mathfrak{U}(\epsilon ))=\stackrel{+\infty }{\sum _{n=0}}{\epsilon }^n{A}_n$$

$$\mathfrak{N}(\mathfrak{u})$$

$$\Vert \mathfrak{u}-\underset{j=0}{\sum ^n}{\mathfrak{u}}_j\Vert$$

$$\sqrt{2\pi }{j}^{j+\frac{1}{2}}{e}^{-j}\le j!\le \sqrt{\frac{j}{j-1}}\sqrt{2\pi }{j}^{j+\frac{1}{2}}{e}^{-j}$$

$$\frac{\left(2j\right)!\left(2j+1\right)}{{\left(j!\right)}^2{\left(j+1\right)}^2}{\rho }^j\le \frac{4}{\sqrt{3\pi }j\sqrt{j}}{\left(4Ma\right)}^j$$

Full text

1. Introduction

The Adomian decomposition method (ADM) was developped in the $1980$’s by the American physicist G. Adomian $(1923-1996)$ ([1]-[4]) as a powerful method for solving functional equations. The main idea of this method lies in the decomposition of the solution $u$ of a vector nonlinear equation $u=\mathfrak{f+}N(u)$ where $N$ is an analytic operator and $\mathfrak{f}$ a given vector, into a series $u=\underset{n=0}{\overset{+\infty}{\sum}}u_{n}$ such that the term $u_{n+1}$ is determined from the terms $u_{0},u_{1},...,u_{n}$ by a reccurence relation involving a polynomial $A_{n}$ generated by the Taylor expansion of the operator $N.$ Many papers on the applications of the ADM to the problems arising from different areas of pure and applied sciences have been published ([5]-[7], [9], [15], [16], [23]). Many works have been also devoted to the convergence of the ADM ([8], [10]-[14]). However let us pointwise that some recent works on the application of the ADM to nonlinear problems avoid the theoretical treatment of the question of the convergence of the method. On the other hand we observe that the current convergence criteria of the ADM give rise to small regions of convergence (that is to small set of vectors $\mathfrak{f}$ for which the vector series $\sum u_{n}$ is convergent according to the criteria of convergence) which limits the application of this method. Our purpose in this note is to obtain a new convergence result for the ADM, which we believe will contribute to strenghting the method by enlarging the field of its applications.

2. Preliminary notes

2.1. Main definitions

Along this paper $\left(X,\text{ }\left\|.\right\|\right)$ is a given real or complex Banach space.

The following definitions were introduced in ([22]).

Definition 2.1**.**

$\mathfrak{L}_{n}(X)$ denotes for each $n\in\mathbb{N}^{\ast}$ a set of continuous symetric $n$-multilinear mappings from $X^{n}$ to $X$. It is well known that $\mathfrak{L}_{n}(X)$ is a Banach space for the norm $||.||^{[n]}$ defined by the relation $:$

[TABLE]

Let $x\in X$ and $F\in\mathfrak{L}_{n}(X)$. We will denote $F(x,\dots,x)$ by $F.x^{[n]}$. Given $x_{1},...,x_{p}\in X,$ $n_{1},...,n_{p}\in\mathbb{N}$ such that $n_{1}+...+n_{p}=n,$ we will write $F.\left(x_{1}^{[n_{1}]}\cdot\cdot\cdot x_{p}^{[n_{p}]}\right)$ for $F\left(\underset{n_{1}\text{times}}{\underbrace{x_{1},...,x_{1}}},\dots,\underset{n_{p}\text{ times}}{\underbrace{x_{p},...,x_{p}}}\right).$ For convenience we set $\mathfrak{L}_{0}(X):=X$ and $||.||^{[0]}:=||.||.$

Definition 2.2**.**

A power series in $x$ with values in $X$ ([22]) is a series of the form $\sum_{n=0}^{\infty}F_{n}.x^{[n]}$, where $F_{0}\in X$ and $F_{n}\in\mathfrak{L}_{n}(X)$ for $n\geq 1$.

Definition 2.3**.**

A mapping $F:X\rightarrow X$ is called an analytic operator on $X$ ([22]) if there exists for each vector $a\in X$ a sequence $A_{n}(a)\in\mathfrak{L}_{n}(X)~{}~{}(n\geq 1)$ and a open neighborhood $U_{a}$ such that the real series $\sum\frac{1}{n!}\|A_{n}(a)\|^{[n]}\|x-a\|^{n}$ is convergent when $x\in U_{a}$ and the power series $\sum\frac{1}{n!}A_{n}(a)(x-a)^{[n]}$ is convergent to $F(x)-F(a)$ for all $x\in U_{a}$.

Definition 2.4**.**

The radius of the power series $\sum\frac{1}{n!}A_{n}(a)(x-a)^{[n]}$ is the radius of convergence of the complex variable power series* $\sum\frac{1}{n!}\left\|\mathfrak{A}_{n}(a)\right\|^{\left[n\right]}t^{n}$*, that is the number :

[TABLE]

with the conventions that $\frac{1}{0}=+\infty$.

Definition 2.5**.**

Let $F:X\rightarrow X$ be an analytic operator on the Banach space $X$ and $a\in X.$ We say that $F$ has an infinite radius of convergence at the point $a$ if $\mathfrak{R}\left(F,a\right)=+\infty$.

Remark 2.6**.**

If a mapping $F:X\rightarrow X$ is an analytic operator on $X$ then the operator $F$ is of class $C^{\infty}$ on $X$ and the following relations hold for every $a\in X$ and $n\in\mathbb{N}:$

[TABLE]

Remark 2.7**.**

If an analytic operator $F$ has an infinite radius of convergence at some point $a\in X$ on the Banach space $X$, then thanks to $($[22]$)$ $F$ will have the same property at every point of $X$. Thence we will say, without loss of precision, that $F$ has an infinite radius of convergence.

3. Abstract presentation of the ADM

Let $\mathfrak{f}\in X$ and $\mathfrak{N}:X\rightarrow X$  an analytic operator on $X$ with infinite radius of convergence. We consider the vector equation :

[TABLE]

The ADM for solving the vector equation (3.1) consists in writing the unknown vector $\mathfrak{u}$ in the form of an absolutely convergent vector series $\mathfrak{u}=\sum_{n=0}^{+\infty}\mathfrak{u}_{n}$ and in splitting the nonlinear term $\mathfrak{N}(\mathfrak{u})$ into an absolutely convergent vector series $N(\mathfrak{u})=\sum_{n=0}^{+\infty}A_{n}$ where the term $A_{n}$ is obtained for all n\in\mathbb{N}\by the formula :

[TABLE]

Thence the equation (3.1) becomes :

[TABLE]

Then we set by formal identification :

[TABLE]

and the following relation holds for every $n\in\mathbb{N}$ :

[TABLE]

We denote by $\mathcal{A}dm\left(\mathfrak{N};\mathfrak{f}\right)$ the vector series $\sum\mathfrak{u}_{n}$ if it is well-defined. In the sequel we will continue to denote by $\mathfrak{u}_{n}$ the general term of the vector series $\mathcal{A}dm\left(\mathfrak{N};\mathfrak{f}\right).$

Theorem 1**.**

([ABBA])

Let $\mathfrak{f}\in X,$ $\mathfrak{N:}$ $X\rightarrow X$ be an analytic mapping on $X$ which has an infinite radius of convergence. Then the following formula holds for every $n\in\mathbb{N}:$

[TABLE]

Proof.

Since $\mathfrak{N}$ is analytic, we can write for all $n\in\mathbb{N}$ and $\varepsilon>0:$

[TABLE]

It follows then from the relation (3.2) that the following relation holds for each $n\in\mathbb{N}$ :

[TABLE]

4. Statement of the main result

Our main result in this paper is the following.

Theorem 2**.**

Let $\mathfrak{f}\in X,$ $\mathfrak{N:}$ $X\rightarrow X$ be an analytic mapping on $X$ on some open neighborhood of the vector $\mathfrak{f}$ and such that the following estimates hold $:$

[TABLE]

where the constants $M,$ $a>0$* satisfy the condition *:

[TABLE]

$1.$Then the vector series $\mathcal{A}dm\left(\mathfrak{N};\mathfrak{f}\right)$ is absolutely convergent in the  Banach space $X$ to a vector $\mathfrak{u}$ *which is a solution of the equation (3.1). *

$2.$ If $Ma=\frac{1}{4},$then the vector $\mathfrak{u}$ fullfiles the following estimates $:$

[TABLE]

$3.$ If $Ma<\frac{1}{4}$ then the vector $\mathfrak{u}$ fullfiles the following estimates $:$

[TABLE]

5. Proof of the main result

Relying on the formula (3.3) we can easily prove, under the assumption (4.1) of the main result, that :

[TABLE]

where $\left(w_{n}\right)_{n\geq 1}$is the sequence of positive coefficient defined by the relations :

[TABLE]

It follows that :

[TABLE]

Let us denote by $S$ the formal series $\underset{n\geq 1}{\sum}w_{n}T^{n}\in\mathbb{C}\left[\left[T\right]\right].$ Then we can write :

[TABLE]

It follows that :

[TABLE]

It follows from the well known Lagrange inversion formula for formal series ([17]) that the following equalities hold for all $n\geq 1:$

[TABLE]

Thence we have for all $n\geq 1:$

[TABLE]

The inequalities (5.1) become :

[TABLE]

But we have by virtue of the well known Stirling formula :

[TABLE]

It follows then from the assumption (4.2) that the vector series $\mathcal{A}dm\left(\mathfrak{N};\mathfrak{f}\right)$ is absolutely convergent in the Banach space $X.$ Let us then set for all $\varepsilon\in[0,1]:$

[TABLE]

Since $\mathfrak{N}$ is analytic on $E,$ it follows that we have for all $\varepsilon\in[0,1]:$

[TABLE]

If we choose $\varepsilon=1,$ we will then obtain the relations :

[TABLE]

It follows that $\mathfrak{u}$ is a solution of the equation (3.1). On the other hand we have for each $n\in\mathbb{N}^{\ast}:$

[TABLE]

But, according to ([SAND]) the double inequality holds for all integers $j\geq 2$ :

[TABLE]

It follows then, by easy computations, that we have for each $j\in\mathbb{N}^{\ast}\backslash\left\{1\right\}$ :

[TABLE]

• •

First case : $4Ma=1$

In this case the following estimate holds for all $n\in\mathbb{N}^{\ast}:$

[TABLE]

It follows that :

[TABLE]

• •

Second case : $4Ma<1$

In this case the following estimate holds for all $n\in\mathbb{N}^{\ast}:$

[TABLE]

It follows that :

[TABLE]

Thence the proof of the main result is complete.

$\square$