EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 3, 2021, 1044-1056
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Comparison of the Adomian decomposition method and
regular perturbation method on non linear equations

second kind of Volterra

Rasmane Yaro1, Bakari Abbo2, Bassono Francis3,Youssouf Paré3,∗,
1 Université de Dédougou, Dédougou, Burkina-Faso
2 Université de Ndjaména, Ndjaména, Tchad
3Université Joseph Ki-Zerbo, Ouagadougou, Burkina-Faso

Abstract. In this paper, we study convergence of Adomian decomposition method applied to
second kind Volterra general integral and show that this method and regular perturbation method
converges to the same solution.

2020 Mathematics Subject Classifications: 44Axx, 40C10,45I05

Key Words and Phrases: Adomian Decomposition Method (ADM), Regular Pertubation Method(RPM),
Volterra integral equation second kind

1. Introduction

In the literature, there are few analytical methods or digital successful for solving non
linear integral equations. This is due to the strong non linearity of integral equations and
the difficulty to find their exact solutions. In this paper, we examine second kind Volterra
general integral :

ϕ(x) = f(x) + ε

∫ x

0
K(x, t)ϕp(t)dt; p ≥ 2; 0 < ε� 1, a ≤ t ≤ x ≤ T ≺ +∞ (1)

Where

ϕ is the unknown function,

f is continuous function on
∑

t = [a;T ],

and K ∈ L2(Ω) is continuous on Ω = [a;T ]× [a;T ] .

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.4014

Email addresses: yarorasmane@yahoo.fr (R. Yaro), bakariabbo@yahoo.fr (B. Abbo),
sonobi2002@yahoo.fr ( F. Bassono) , pareyoussouf@gmail.com (Y. Paré)

http://www.ejpam.com 1044 c© 2021 EJPAM All rights reserved.



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1045

We used Adomian decomposition method and regular perturbation method permitted us
to find the solution of the problem (1).

The structure of the present study consists of an introductory section, Sections 2 and
3, and final conclusions. In Section 2, we prove the convergence of both methods. Section
3 contains numerical examples.

2. Convergence

In this section we show the theorem of convergence.

Theorem 1. Let us consider the non linear equations second kind of Volterra defined in
(1). Then the problem (P ) converges to a unique solution ϕ ∈ C([a;T ]) and the Adomian

algorithm [1–4, 6–10, 12]

(PADM )

{
ϕ0(x) = f(x)

ϕn(x) = ε
∫ x
0 K(x, t)An−1(t)dt; n ≥ 1

(2)

converges to this solution. In addition, Adomian and regular perturbation algorithms are
equivalents.

Proof. The Adomian polynomial are obtained by the following formula :

[
∂i(λkAk)

∂λi
]λ=0 = [

∂i(λkϕk)
p

∂λi
]λ=0 (3)

By unfolding, we get:

A0= ϕp0
A1= pϕp−10 ϕ1

A2=
1

2
p[2ϕp−10 ϕ2+(p− 1)ϕp−20 ϕ2

1

A3=
1

6
p[6ϕp−10 ϕ3−6pϕp−20 ϕ1ϕ2+(p2−3p+ 2)ϕp−30 ϕ3

1]

...

An=
1

n!
[
∂n(λkϕk)

p

∂λn
]λ=0; ∀n ≥ 0

(4)

We have following Adomian algorithm:

(PADM )


ϕ0(x) = f(x)

ϕ1(x) = ε
∫ x
0 K(x, t)A0(t)dt

ϕ2(x) = ε
∫ x
0 K(x, t)A1(t)dt
...

ϕn(x) = ε
∫ x
0 K(x, t)An−1(t)dt; n ≥ 1

(5)



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1046

Let us show that the algoritm (PADM ) converges

• f is assumed continuous on
∑

t = [a;T ], there existm > 0, such that ∀x ∈
∑

t,
∀(x, t) ∈ Ω = [0;T ]× [a;T ], |f(x)| ≤ m
• K being continuous on Ω = [0;T ]× [a;T ], there exist M > 0, such that ∀(x, t) ∈ Ω,

|K(x, t)| ≤M .

For ϕn(x), we have successively ∀n ≥ 1



|ϕ0(x)| ≤ m0

|ϕ1(x)| ≤ mp
0

[εM(x− a)]

1!
⇒ |ϕ1(x)| ≤ m1

[εM(x− a)]

1!

|ϕ2(x)| ≤ pmp
0m1

[λM(x− a)]2

2!
⇒ |ϕ2(x)| ≤ m2

[εM(x− a)]2

2!

|ϕk3(x)| ≤ 1
2p[2m

p−1
0 m2+(p− 1)mp−2

0 m2
1

[εM(x− a)]3

3!
⇒ |ϕk3(x)| ≤ m3

[εM(x− a)]3

3!
...

|ϕn(x)| ≤ mn
[ιM(x− a)]n

n!
(6)

Let’s put
m′ = Sup(m0,m1, ...,mn)

it follows that :

|ϕn(x)| ≤ m′ [εM(x− a)]n

n!

Let’s put

φn(x) = m′
[εM(x− a)]n

n!

The series: ∑
n≥0

φn(x) = m′
∑
n≥0

[εM(x− a)]n

n!

Converging geometrically toward the function:

g(x) = m
′
e[εM(x−a)] on [0;T ]

Therefore, the series
∑

n≥0 ϕ
k
n(x) converges normally and thus absolutely to ϕ(x) on

[a;T which is the solution to problem (2)

-Let us suppose that the problem (P ) admit two distinct solutions ϕ and ψ
For the function ψ ,we have following Adomian algorithm:



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1047

{
ψ0(x) = f(x)

ψn(x) = ε
∫ x
0 K(x, t)Bn−1(t)dt; n ≥ 1

(7)

Where



B0= ψp0
B1= pψp−10 ψ1

B2=
1
2p[2ψ

p−1
0 ϕ2+(p− 1)ψp−21 ψ2

1

B3=
1
6p[6ψ

p−1
0 ψ3−6pψp−20 ψ1ψ2+(p2−3p+ 2)ψp−30 ψ3

1]
...

Bn= 1
n! [

∂n(λkψk)
p

∂λn ]λ=0

(8)

are Adomian’s polynomial.
Let’s put ω(x) = ϕ(x)− ψ(x), w(x) checks the following Adomian algorithm:

{
ω0(x) = ϕ0(x)− ψ0(x)

ωn(x) = ε
∫ x
0 K(x, t)[An−1(t)−Bn−1(t)]dt; n ≥ 1

(9)

By unfolding the algorithm (9) for k ≥ 1,we get:


ω0(x) = f(x)− f(x) = 0
ω1(x) = ε

∫ x
0 K(x, t)[ϕp0(t)− ψ

p
0(t)]dt = 0

ω2(x) = ε
∫ x
0 K(x, t)[A1(t)−B1(t)]dt = ε

∫ x
0 K(x, t)[pϕp0(t)ϕ1(t)− pψp0(t)ψ1(t)]dt = 0

...
ωn(x) = ε

∫ x
0 K(x, t)[An−1(t)−Bn−1(t)]dt = 0;∀n ≥ 0

we have
ω(x) =

∑
n≥0

ωn(x) = 0

since
∀n ≥ 0, ωn(x) = 0

And thus
ω(x) = ϕ(x)− ψ(x) = 0⇔ ϕ(x) = ψ(x)

and
ϕ = ψ∀x ∈ Σt = [a;T ].

This proves the uniqueness of the solution of the equation (1) and the convergence of
the Adomian algorithm.



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1048

Let us show that the Adomian algorithm and the regular perturbation
converges to the same solution

Let us consider the problem (P )
• Applying the Adomian decomposition algorithm to (1) , we get:

(PADM )

{
ϕ0(x) = f(x)

ϕn(x) = ε
∫ x
0 K(x, t)An−1(t)dt; n ≥ 1

(10)

• Regular perturbation method [5, 11]

Let us suppose that ψ is another solution of the problem (P )
This method consist to search the approximate solution by an asymptotic expression :

ψ(x) =
∑
n≥0

εnψn(x) (11)

where ε is a small parameter of the problem.

Let’s introduce (11) to (1), we get:∑
n≥0

εnψn(x) = f(x) + ε

∫ x

0
K(x, t)(

∑
n≥0

εnψn(t))p(t)dt; (12)

Using Binomial Newton formula, we get

[ψ0 + (εψ1 + ε2ψ2)]
p = ψp0 + pψp−10 (εψ1 + ε2ψ2) +

p(p− 1)

2
ψp−20 (ε2ψ2

1 + 2ε3ψ1ψ2 + ε4ψ2
2) + ...+ (εψ1 + ε2ψ2)

p

By identification according to the growth power of ε, we get:



ε0 : ψ0(x) = f(x)
ε1 : ψ1(x) =

∫ x
0 K(x, t)ψp0(t)dt

ε2 : ψ2(x) =
∫ x
0 K(x, t)pψp−10 (t)ψ1(t)dt

ε3 : ψ3(x) =
∫ x
0 K(x, t)12p[2ψ

p−1
0 ψ2(t)+(p− 1)ψp−20 (t)ψ2

1(t)]dt
...
εn : ψn(x) = ...



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1049

• Comparision of the solution of the both methods

Let’s put φn(x) = ϕn(x)− εnψn(x)
By unfolding, we get:

ii



φ0(x) = ϕ0(x)− ψ0(x)f(x) = f(x)− f(x) = 0
φ1(x) = ε

∫ x
0 K(x, t)ϕp0(t)dt− ε

∫ x
0 K(x, t)ψp0(t)dt = ε

∫ x
0 K(x, t)[ϕp0(t)− ψ

p
0(t)]dt = 0

φ2(x) = ε
∫ x
0 K(x, t)pϕp−10 (t)ϕ1(t)dt− ε2

∫ x
0 K(x, t)pψp−10 (t)ψ1(t)]dt

= ε
∫ x
0 K(x, t)[pfp−10 (t)[ϕ1(t)− εψ1(t)]dt = 0

...
φn(x) = ϕn(x)− εnψn(x) = 0,∀n ≥ 0

we have ∑
n≥0

φn(x) = 0

⇐⇒
∑
n≥0

[ϕn(x)− εnψn(x)] = 0⇐⇒
∑
n≥0

[ϕn(x) =
∑
n≥0

εnψn(x)]⇐⇒ ϕ(x) = ψ(x)

and
ϕ = ψ∀x ∈ Σt = [a;T ].

And thus
ω(x) = ϕ(x)− ψ(x) = 0⇔ ϕ(x) = ψ(x)

and
ϕ = ψ,∀x ∈ Σt = [a;T ].

This proves that t of the Adomian decomposition algorithm and regular perturbation
method converges to the same solution .

3. Numerical examples

3.1. Example 1

Let’s consider the following non linear integral equation of second kind of Volterra :

ϕ(x) =
√
x− 16

15
εx2
√
x+ ε

∫ x

0

ϕ4(t)√
x− t

dt; 0 < ε� 1 (13)

• Solving by the Adomian decomposition method
- Applying the Adomian algorithm, it follows that (13)



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1050



ϕ0(x) =
√
x

ϕ1(x) = −16
15εx

2√x+ ε
∫ x
0

A0(t)√
x− t

dt

...

ϕn(x) = ε
∫ x
0

An−1(t)√
x− t

dt, n ≥ 2

(14)

where 

A0= ϕ4
0

A1= 4ϕ3
0 ϕ1

A2=4ϕ3
0ϕ2+6ϕ2

0ϕ
2
1

...

An=
1

n!
[
∂n(λkϕk)

p

∂λn
]λ=0

(15)

- Computation of

ε

∫ x

0

A0(t)√
x− t

dt

Let’s take T = x− t and integrating, we get:

ε

∫ x

0

A0(t)√
x− t

dt = ε

∫ x

0

ϕ4
0(t)√
x− t

dt = ε

∫ x

0

t2√
x− t

dt

= ε

∫ x

0
(x− T )2T

−1
2 dT =

16

15
εx2
√
x

By induction on n , we get:

ϕ0(x) =
√
x

ϕ1(x) = −16

15
εx2
√
x+ ε

∫ x
0

A0(t)√
x− t

dt = −16

15
εx2
√
x+

16

15
εx2
√
x = 0

ϕ2(x) = ε
∫ x
0

A1(t)√
x− t

dt = ε
∫ x
0

4ϕ3
0 (t)ϕ1(t)√
x− t

dt = 0

ϕ3(x) = ε
∫ x
0

A2(t)√
x− t

dt = ε
∫ x
0

[4ϕ3
0(t)ϕ2(t)+6ϕ2

0(t)ϕ
2
1(t)]√

x− t
dt = 0

...
ϕp(x) = 0, ∀ n ≥ 1

(16)

Let’s put

ϕ(x) =
∑
n≥0

ϕn(x)



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1051

= ϕ0(x)

Then the exact solution of the problem (13) is:

ϕ(x) =
√
x, (17)

• Solving by the regular perturbation method
Let us search the solution of (13) by an asymptotic expression :

ϕ

ϕ(x) =
∑
n≥0

εnϕn(x) (18)

where ε is a small parameter of the problem.

Let’s introduce (18) to (13), we get:

∑
n≥0

εnϕn(x) =
√
x− 16

15
εx2
√
x+ ε

∫ x

0

(
∑

n≥0 ε
nϕn(t))4

√
x− t

dt; (19)

Using Binomial Newton formula, we get

[ϕ0+(εϕ1+ε
2ϕ2)]

4 = ϕ4+4ϕ3
0(εϕ1+ε

2ϕ2)+6ϕ2
0(ε

2ϕ1+2ε3ϕ1ϕ2+ε
4ϕ2

2)+4ψ0(εϕ1+ε
2ϕ2)

3+(εϕ1+ε
2ϕ2)

4..

By identification and by unfolging according to the growth power of ε, we get:



ε0 : ϕ0(x) =
√
x

ε1 : ϕ1(x) = −16

15
εx2
√
x+

∫ x
0

ϕ4
0(t)√
x− t

dt = −16

15
x2
√
x+

16

15
x2
√
x = 0

ε2 : ϕ2(x) =
∫ x
0

4ϕ3
0(t)ϕ1(t)√
x− t

dt = 0

ε3 : ϕ3(x) =
∫ x
0

[4ϕ3
0(t)ϕ2(t)+6ϕ2

0(t)ϕ
2
1(t)]√

x− t
dt = 0

...
εn : ϕn(x) = 0;∀ n ≥ 1

Let’s put

ϕ(x) =
∑
n≥0

εnϕn(x)



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1052

= ϕ0(x)

Then the exact solution of the problem (13) is:

ϕ(x) =
√
x, (20)

3.2. Example 2

Let’s consider the following a system of non linear integral equations of second kind of
Volterra :


u(x) = ex +

ε(e−mx − ex)

m+ 1
+ ε

∫ x
0 e

x−tvm(t)dt; 0 < ε� 1;m ≥ 2

v(x) = e−x − ε(e2mx − ex)

2m− 1
+ ε

∫ x
0 e

x−twm(t)dt; 0 < ε� 1;m ≥ 2

w(x) = e2x − ε(emx − ex)

m− 1
+ ε

∫ x
0 e

x−tum(t)dt; 0 < ε� 1;m ≥ 2

(21)

• Solving by the Adomian decomposition method
- Applying the Adomian algorithm, it follows that (13) :


u0(x) = ex

u1(x) =
ε(e−mx − ex)

m+ 1
+ ε

∫ x
0 e

x−tA0(t)dt

...
un(x) = ε

∫ x
0 e

x−tAn−1(t)dt n ≥ 2


v0(x) = e−x

v1(x) = −ε(e
2mx − ex)

2m− 1
+ ε

∫ x
0 e

x−tB0(t)dt

...
vn(x) = ε

∫ x
0 e

x−tBn−1(t)dt n ≥ 2
(22)


w0(x) = e2x

w1(x) = −ε(e
mx − ex)

m− 1
+ ε

∫ x
0 e

x−tC0(t)dt

...
wn(x) = ε

∫ x
0 e

x−tCn−1(t)dt n ≥ 2

(23)

where the Adomian polynomial’s are given by:

A0= vm0
A1= mvm−10 v1
A2=

1
2m[2vm−10 v2+(m− 1)vm−21 v21

...

An= 1
n! [

∂n(λkvk)
m

∂λn ]λ=0



B0= wm0
B1= mwm−10 w1

B2=
1
2m[2wm−10 w2+(m− 1)wm−21 w2

1

...

Bn= 1
n! [

∂n(λkwk)
m

∂λn ]λ=0

(24)



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1053



C0= um0
C1= mum−10 u1

C2=
1

2
m[2um−10 u2+(m− 1)um−21 u21

...

Cn=
1

n!
[
∂n(λkuk)

m

∂λn
]λ=0

(25)

By unfoldind on n , we get:


u0(x)=ex

u1(x)= 0
u2(x)=0
...
un(x)=0;∀ n ≥ 1


v0=e

−x

v1(x)= 0
v2(x)= 0
...
vn(x)=0;∀ n ≥ 1


w0=e

2x

w1(x)= 0
w2(x)=0
...
wn(x)=0;∀ n ≥ 1

(26)

Let’s put

(u(x), v(x), w(x)) = (
∑
n≥0

un(x),
∑
n≥0

vn(x),
∑
n≥0

wn(x))

= (u0(x), v0(x), w0(x))

Then the exact solution of the problem (21) is:

(u(x), v(x), w(x)) = ( ex, e−x, e2x) (27)

• Solving by the regular perturbation method
Let us search the solution of (13) by an asymptotic expression :

(u(x), v(x), w(x)) = (
∑
n≥0

εnun(x),
∑
n≥0

εnvn(x),
∑
n≥0

εnwn(x)) (28)

where ε is a small parameter of the problem.

Let’s introduce (18) to (13) , we get:

∑
n≥0 ε

nun(x)=ex +
ε(e−mx − ex)

m+ 1
+ ε

∫ x
0 e

x−t(
∑

n≥0 ε
nvn(t))mdt∑

n≥0 ε
nvn(x)=e−x − ε(emx − ex)

2m− 1
+ ε

∫ x
0 e

x−t(
∑

n≥0 ε
nwn(t))mdt∑

n≥0 ε
nwn(x)=e2x − ε(e−mx − ex)

m− 1
+ ε

∫ x
0 e

x−t(
∑

n≥0 ε
nun(t))mdt

(29)

Using Binomial Newton formula, we get



Y. Paré et al. / Eur. J. Pure Appl. Math, 14 (3) (2021), 1044-1056 1054

[u0 + (εu1 + ε2u2)]
m = um0 + pum−10 (εu1 + ε2u2) +

m(m− 1)

2
um−20 (ε2u21 + 2ε3u1u2 + ε4u22) + ...

[v0 + (εu1 + ε2v2)]
m = vm0 + pvm−10 (εv1 + ε2v2) +

m(m− 1)

2
vm−20 (ε2v21 + 2ε3v1v2 + ε4v22) + ...

[w0 + (εw1 + ε2w2)]
m = wm0 + pwm−10 (εw1 + ε2w2) +

m(m− 1)

2
wm−20 (ε2w2

1 + 2ε3w1w2 + ε4w2
2) + ...

By identification according to the growth power of ε, we get:





ε0 : u0(x) = ex

ε1 : u1(x) =
(e−mx − ex)

m+ 1
+
∫ x
0 e

x−tvm0 (t)dt

ε2 : u2(x) =
∫ x
0 e

x−tmvm−10 (t)v1(t)dt

ε2 : u3(x) =
∫ x
0 e

x−t 1

2
m[2vm−10 (t)v2(t)+(m− 1)vm−21 (t)v21(t)]dt

...
εn:2un(x) = ...

ε0 : v0(x) = e−x

ε1 : v1(x) = −(emx − ex)

2m− 1
+
∫ x
0 e

x−twm0 (t)dt

ε2 : v2(x) =
∫ x
0 e

x−tmwm−10 (t)w1(t)dt

ε2 : v3(x) =
∫ x
0 e

x−t 1

2
m[2wm−10 (t)w2(t)+(m− 1)wm−21 (t)w2

1(t)]dt

...
εn:2vn(x) = ...

ε0 : w0(x) = e2x

ε1 : w1(x) = (e−mx−ex)
m+1 +

∫ x
0 e

x−tum0 (t)dt

ε2 : w2(x) =
∫ x
0 e

x−tmum−10 (t)u1(t)dt

ε2 : w3(x) =
∫ x
0 e

x−t 1
2m[2um−10 (t)u2(t)+(m− 1)um−21 (t)u21(t)]dt

...
εn:2wn(x) = ...

(30)

By unfoldind on n , we get :


u0(x)=ex

u1(x)= 0
u2(x)=0
...
un(x)=0;∀ n ≥ 1


v0=e

−x

v1(x)= 0
v2(x)= 0
...
vn(x)=0;∀ n ≥ 1


w0=e

2x

w1(x)= 0
w2(x)=0
...
wn(x)=0;∀ n ≥ 1

(31)

Let’s put



REFERENCES 1055

(u(x), v(x), w(x)) = (
∑
n≥0

εnun(x),
∑
n≥0

εnvn(x),
∑
n≥0

εnwn(x))

= (u0(x), v0(x), w0(x))

Then the exact solution of the problem (21) is:

(u(x), v(x), w(x)) = ( ex, e−x, e2x) (32)

4. Conclusion

In this paper, we first showed that the Adomian decomposition method converges when
applied to Volterra general integral equations of second kind.

Then we showed that the Adomian decomposition method and regular perturbation
method converges to the same solution when applied to Volterra general integral equations
of second kind.

Lastly, we used these both method to solve a non linear integral equation of second
kind and a system of non linear integral equations of second kind of Volterra .

We showed that using the both method, we get the same solution. There are then the
very powerful numerical tools for the resolution of non linear equations and systems of
nonlinear equations of Volterra .

References

[1] K. Abbaoui and Y. Cherruault. Decomposition method applied to the cauchy prob-
lem. Kybernetes, 28(1):68–74, 1999.

[2] S. Akhuri. Laplace decomposition algorithm applied to clas of nonlinear differential
equations. J.Math.Appl, 1(4):141–155, 2001.

[3] B.Abbo, N.Ngarhasta, B.Mampassi, B.Some, and L. Some. A new approach of the
adomian algorithm for solving nonlinear ordinary or partial differential equations. Far
East J.Appl. Math, 23(3):299–312, 2006.

[4] Pierre Baki-Tangou and Gabriel Bissanga. Application of adomian decomposition
method to solving the duffing-van der pol equation. Communications in Mathematical
Analysis, 4(0):30–40, 2008.

[5] Francis Bassono, Pare Youssouf, Gabriel Bissanga, and Blaise Some. Application of
the adomian decomposition method and the perturbation method to solving a sytem
of perturbation equations. Far East Journal of Applied Mathematics, 72(2):91–99,
2012.

[6] M. Hussain and Majid Khan. Modified laplace decomposition method. Applied Math-
ematical sciences, 4(36):1769–1783., 2010.



REFERENCES 1056

[7] K.Abbaoui and Y. Cherruault. Convergence of adomian method applied to differential
equations. Math. Comput. Modelling, 28(5):103–109, 1994.

[8] S. Khelifa and Yves Cherruault. The decomposition method for solving first order
partial differential equations. Kybernetes, 31(6):844–871, 2002.

[9] N.Ngarhasta., B.Some, K.Abbaoui, and Y. Cherruault. New numerical study of ado-
mian method applied to a diffusion model. Kybernetes, 31(1):61–75, 2002.

[10] Blaise Some. Méthode SBA de résolution des modèles mathématiques en environ-
nement. Editions Universitaires Européennes, 2019.

[11] Pare Youssouf. Résolution de quelques équations fonctionnelles par la méthode SBA
(Somé Blaise-Abbo). PhD thesis, Université de Ouagadougou, 2010.

[12] Pare Youssouf, Francis Bassono, and Blaise Some. A new technique for numerical res-
olution of few nonlinear integral equations of fredholm by sba method. International
Journal of Applied Mathematical Research, 70(1):21–33, 2012.