.


27 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

ReseaRch aRticle

Cooperating Newton’s Method with Series Solution Method 
for Solving System of Linear Mixed Volterra-Fredholm Integral 
Equation of the Second Kind
Talaat I. Hasan*

Department of Mathematics, College of Basic Education, Salahaddin University, Erbil, Iraq

ABSTRACT

In this paper, for the 1st time, we use Newton’s method with series solution method (SSM) for solving system of linear mixed Volterra-
Fredholm integral equations of the second kind (SLMVFIE-2). In this work, we use a new technique for studying the numerical solutions 
for SLMVFIE-2 which is Newton’s method with SSM, first solving this system using SSM and after that cooperation Newton’s method 
with SSM, suggesting an algorithm for the technique. The new results are achieved and some new theorems have proved for convergence 
of the method, several numerical examples are tested for illustrating the usefulness of the technique; the numerical results are obtained 
and compared with the exact solution, computing the least square error, and running times which are criterion of discussion. For 
programming the technique, we use general Matlab program.

Keywords: Newton’s method, series solution method, system of linear mixed Volterra-Fredholm integral equations of the second kind, 
Taylor series

INTRODUCTION

Integral equations have been one of the significant and principal 
instruments in various areas of sciences such as applied 
mathematics physics, biology, and engineering. On the other 
hand, it has many applications in different areas of science, 
involving potential theory, electricity, and quantum mechanics.[1,2]

In the past 20 years ago, many kinds of problems in 
integral equation making from various phenomena in applied 
mathematical physics. The mixed Volterra-Fredholm integral 
equations can be constructed form ordinary differential 
equations with changed argument,[3-5] which arising from the 
theory of parabolic boundary value problems. In addition, this 
type of problems appears in various physical and biological 
problems.[6,7]

Recently, many kinds of integral equations constructed and 
reformulated; the approximating methods take an important 
role for finding the numerical solution for these classes of 
problems in integral equations. It has many advantages 
witnessed by the increasing frequency of the integral equations 
in literature and in many areas, because some problems have 
their mathematical representation appear directly.[8,9]

Then, for the reasons, many scientists applied different 
techniques for finding numerical solution of integral equations, 
Maleknejad, in 2006, solved system of Volterra-Fredholm 
integral equations using computational method; Chen, in 
2013, studied the approximate solution for mixed linear 

Volterra-Fredholm integral equation. Wazwaz, in 2011, solved 
Volterra integral equation of the second kind using series 
solution method (SSM). Young, in 2015, used Newton’s 
Raphson for finding the approximate solution of non-linear 
equations. Hassan, in 2016, found numerical solution of linear 
system Volterra-Fredholm integral equation. Saleh, in 2017, 
studied system of Volterra-Fredholm integral equation of the 
second type, for extending this work, the system of linear 
mixed Volterra-Fredholm integral equations of the second kind 
(SLMVFIE-2) is studying and finding the numerical result of it 
by cooperating the NM with SSM and using power functions 
u x xi
m

i
m( ) ,=  for m=0, 1, 2, 3,…q.

SOME DEFINITIONS AND THEOREMS

In this section, discussing some definitions and theorem, 
which relate to the study.

Cihan University-Erbil Scientific Journal (CUESJ)

Corresponding Author: 
Talaat I. Hasan, Department of Mathematics, College 
of Basic Education, Salahaddin University, Erbil, Iraq. 
E-mail: talhat.hassan@su.edu.krd

Received: Nov 13, 2018 
Accepted: Dec 24, 2018 
Published: May 13, 2019

DOI: 10.24086/cuesj.v3n1y2019.pp27-33

Copyright © 2019 Talaat I. Hasan. This is an open-access article distributed 
under the Creative Commons Attribution License.

https://creativecommons.org/licenses/by-nc-nd/4.0/


Hasan: Cooperating Newton’s Method with Series Solution Method

28 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

Definition (1)

We assume SLMVFIE-2 as follows:

u x f x k x t u t dtdxi i
j

n

ij

a

x

ij j

a

b

( ) ( ) ( , ) ( )= +
=
∑ ∫ ∫

1

  (1)

For i=1, 2, 3,…r. Where f
i
(x) and k

ij
(x,t) are analytic 

functions on domain H= {(x,t), a≤t<x≤b} such that u
i
(x) is 

the unknown functions.

Definition (2)[4,5]

A real value function u(x) is called analytic if it has derivatives 
of all order such that the Taylor series at a point b in its domain.

u x
u b
k

x b
k

k

k

( )
( )
!
( ) ,= −

=

∞

∑
0

 converges to u(x) in the 

neighborhood of b.

Definition (3)[10]

Let f(x)=0 be the non-linear function and x
0
 be initial solution 

to the exact solution and then the Newton’s method is defined 

as the form x x
f x
f xn n

n

n

= −
′−

−

−
1

1

1

( )
( )

, n=1, 2,…. which generating 

the sequence {x
n
}.

USING SSM FOR SOLVING SLMVFIE-2

Suppose that the numerical solution of SLMVFIE-2 be analytic 
function and can be written of the form

  u x u xi im i
m

m

q

( ) ( )=
=
∑

0

 for i=1, 2, 3,…r (2)

Substituting it in equation (1), we get

α λ αip
m

q

i
m

i ij
j

n

ij im
m

q

i
mu x T f x k x t u t dtdx

= = =
∑ ∑ ∑= +

0 1 0

( ) ( ( )) ( , ) ( )
aa

b

a

x

i i i i i i iq i
q

i

u x u x u x u x

T f x

∫∫
+ + + +

=

α α α α0
0

1
1

2
2( ) ( ) ( ) ... ( )

( ( ))) ( , )[ ( )

( ) ( ) ..

+ +

+ +

∫∫∑
=

λ α

α α

ij ij i i
a

b

a

x

j

n

i i i i

k x t u t

u t u t

0
0

1

1
1

2
2 .. ( )]+ αiq i

qu t dtdx

 

 (3)

From equation (3), in the right-hand side calculating the 
integrals for variables t and x. Getting the system of (r×(q+1) 
equations. Collecting and equating the terms of the same 
powers of u xi

m( )  in both sides. Using Gauss elimination 
method, solving the resulting in equation (3) to obtain the 
values of a

im
 such that the determinate of algebraic system 

does not equal to zero. By putting the values of a
im

 in 
equation (2), we get the approximation solution of equation 
(1). The calculation more terms we use will enhance the level 
of accuracy of the approximate solutions. In this work, we use 
power functions u x xi

m
i
m( ) =  for m=0, 1, 2, 3,…q.

New Theorem (1)

Let V: B(A)→B(A) be a continuous linear integral operator on 
the Banach space B(A) f

i
(x) be analytic function on [a,b] and 

k
ij
(x,t) be continuous on H={(x,t), a≤t<x≤b} where |k

ij
(x,t)

u
1
(t)−k

ij
(x,t)u

2
(t)−|≤d

i
|u

1−
u

2
| such that 0<d

i
<1, then Vn is 

contractive mapping.

Proof: Suppose that

[ ]
0 0

( ( )) ( ) ( , ) ( ) , 0, ,
p x

i i ij i i iV u x f x k x t u t dtdx x p p 1.= + <∫∫

To prove that Vn is contraction mapping for any u
1
(s), 

u
2
(s)   B (A) for sufficient large n.

V u x V u x f x k x t u t dtdx

f x k

i ij

xp

i ij

( ( )) ( ( )) ( ) ( , ) ( )

( ( )

1 2
0

1
0

− = +

+

∫∫ 

00
2

0

0
1 2

0

xp

ij

xp

x t u t dtdx

k x t u t u t dtdx

∫∫

∫∫= −

( , ) ( ) )

( , )( ( ) ( )

V u x V u x
k x t u t

k x t u t dtdx

d

ij

ij

xp

i

( ( )) ( ( ))
( , ) ( )

( , ) ( )1 2
1

200

− =
−

≤

∫∫

[[ ]
p

u ui
2

1 22
−

By suppose |k
ij
(x,t)u

1
(t)−k

ij
(x,t)u

2
(t)|≤d

i
|u

1
−u

2
|, then 

we get

V u x V u x d u u dtdx

d u u dtdx d
p

i

xp

i

xp

i

( ( )) ( ( ))

[

1 2 1 2
00

1 2
00

− ≤ −

= − =

∫∫

∫∫ ii u u
2

1 22
] −

Now,

V u x V u x V f x

k x t V u t dtdx

V

i

ij

xp

2
1

2
2

0
1

0

( ( )) ( ( )) ( ( ))

( , ) ( ( ))

[

− = +

∫∫

 (( ( )) ( , )

( ( )) ]

f x k x t

V u t dtdx

i ij

xp

− ∫∫
00

2

= −∫∫ k x t V u t V u t dtdxij
xp

0
1 2

0

( , )( ( ( )) ( ( )))

V u x V u x d d
p

u u dtdx

d
p

u

i i

x
i

p

i
i

2

1
2

2
0

2

1 2
0

2
2

1

2

2

( ( )) ( ( )) [ ]

[ ]

− ≤ − ≤

−

∫∫

uu dtdx d
p

u u
xp

i
i

2
00

2
4

1 24∫∫ ≤ −[ ]
Using mathematical induction, we obtain 

V u x V u x d
p

u u
n n

i
n i

n

n
( ( )) ( ( )) [ ] .1 2

2

1 22
− ≤ −

Since by suppose 0<d
i
<1 and p

i
<1, then we obtain 

0<d
i
p

i
<1 and 0 12< <d pi

n
i
n  therefore 0

2
1

2

< <d
p

i
n i

n

n
[ ]

Then, by definition of contraction Vn is contraction on.



Hasan: Cooperating Newton’s Method with Series Solution Method

29 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

New Theorem (2)

Suppose that the hypotheses of theorem (1) are holds; then, 
system of mixed Volterra-Fredholm integral equation of the 
second kind has a unique solution.

Proof: Suppose that u
1
(x) and u

2
(x) be any two solution 

of SLVFIE-2.

u x f x k x t u t dtdxi ij
xp

1
0

1
0

( ) ( ) ( , ) ( )= + ∫∫

And

u x f x k x t u t dtdxi ij
xp

2
0

2
0

( ) ( ) ( , ) ( )= + ∫∫
Using theorem (1), we get 

V u x V u x d
p

u u
n n

i
n i

n

n
( ( )) ( ( )) [ ]1 2

2

1 22
− ≤ −

Since lim [ ] ,
n i

n i
n

n
d

p
→∞

=
2

2
0  then we obtain Vn(u

1
(x))=Vn(u

2
(x)), 

therefore u
1
(x)=u

2
(x)

Hence, SLVFIE-2 has a unique solution.

Algorithm for SSM

•	 Step	1:	Let	the	numerical	solution	of	equation	(1)	as	the	

form u x u xi im i
m

m

q

( ) ( )=
=
∑

0

, for i=1, 2,…,r.

•	 Step	2:	By	substituting	equation	(2)	in	equation	(1)	and	
calculating the integrals of variables.

•	 Step	3:	From	both	sides	collect	and	equate	the	terms	of	
the same power for m=1, 2, 3,…,q.

•	 Step	4:	By	solving	the	system	of	algebraic	equations	with	
unknown coefficients a

im
.

•	 Step	5:	Substitute	the	values	of	a
im

, in equation (2) for 
getting the numerical solution of SLMVFIE-2.

NUMERICAL EXAMPLES AND RESULTS

In this section, illustrating the technique from numerical 
examples.

Example (1): Solve LSMVFIE-2

u x e x e xt u t u t dtdxx
x

1
2 1

1 2
0

1

0

2 1( ) ( ) ( )( ( ) ( ))= − + − + −∫∫

u x xe
x e x

x u t u t dtdxx
x

2

3 1 3
2

1 2
0

1

0

2
2
3

8
3

( ) ( )( ( ) ( ))= − + − + −∫∫

Using SSM, where the exact solutions u
1
(x)=−2ex and 

u
2
(x)=−2xex?

Figure 2: (a) Graphs of the exact u
1
(x)=−2sin(x) and the approximate solutions. (b) Graphs of the exact u

2
(x)=−2cos(x) and the approximate 

solutions. (Source: Created by researcher)
AQ1

ba

Figure 1: (a) Graphs of the exact u
1
(x)=−2ex and the approximate 

solutions. (b) Graphs of the exact u
2
(x)=−2xex and the approximate 

solutions. (Source: Created by researcher)

AQ1
b

a



Hasan: Cooperating Newton’s Method with Series Solution Method

30 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

Solution: Let u x u xm
m

m

q

1 1 1
0

( ) ( )=
=
∑

Then, using SSM, obtaining the values of coefficients

a
10
=−2,	a

11
=−2,	a

12
=−1,	a

13
=−0.3333,	a

14
=−0.083333	 

and a
15
=−0.03333,

For u x u xm
m

m

q

2 2 2
0

( ) ( )=
=
∑

Then, using SSM, obtaining the values of coefficients

a
20

=0, a
21
=−2,	 a

22
=−2,	 a

23
=−1,	 a

24
=−0.33333	and	

a
25
=−0.08333

Example (2): Solve LSMVFIE-2

�� ( )
( )

sin( )
( )(sin( ) cos( ))

u x
x

x x
x t t
u dtdx

x

1
2

2

0 0

2

2 2= − − +
−

∫ ∫


�� ( )
( )

cos( )
( )(sin( ) cos( ))

u x
x

x x
t t t
u dtdx

x

2
10 0

2

2= − − +
−

∫ ∫


Using SSM, where the exact solutions u
1
(x)=−2sin(x) 

and u
2
(x)=−2cos(x)?

Table 1: Comparison between the exact and numerical results

x m Exact solution 
of u

1
(x)=−2ex

Approximate 
value of 

u
1
(x)

Absolute error 

m+1 m+1
i i ie = u (x) - u (x)

Exact 
solution of 

u
2
(x)=−2xex

Approximate 
value of 

u
2
(x)

Absolute error 

m+1 m+1
i i ie = u (x) - u (x)

0.2 1

3

5

7

9

11

13

15

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.40000000

−2.44266666

−2.44280533

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

4.2805×10−2

10.3884×10−4

1.8298×10−7

1.2986×10−9

1.1653×10−12

6.7659×10−14

5.3362×10−16

3.6521×10−18

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.45344832

−0.48898815

−0.48856177

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

3.5611×10−2

4.2735×10−4

6.7527×10−7

2.5998×10−9

5.8432×10−11

3.4562×10−13

4.6783×10−15

2.3351×10−17

Table 2: Comparison between the exact and numerical results

x m Exact 
solution of 

u
1
(x)=−2sin (x)

Approximate 
value of 

u
1
(x)

Absolute error 

m+1 m+1
i i ie = u (x) - u (x)

Exact 
solution of 

u
2
(x)=−2cos (x)

Approximate 
value of 

u
2
(x)

Absolute error 

m+1 m+1
i i ie = u (x) - u (x)

8
π 1

3

5

7

9

11

13

15

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.78539816

−0.76211785

−0.76536743

−0.76536686

−0.76536686

−0.76536686

−0.76536686

−0.76536686

2.0031×10−2

1.5507×10−4

5.7028×10−6

2.6223×10−8

1.7146×10−10

2.2083×10−12

5.0679×10−14

3.6582×10−16

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−2.00000000

−1.84578731

−1.84776922

−1.84775903

−1.84775906

−1.84775906

−1.84775906

−1.84775906

1.5224×10−1

1.9716×10−3

1.0159×10−5

2.8005×10−7

4.8013×10−9

5.6301×10−11

3.9014×10−13

4.3382×10−15

Table 3: Comparison between the exact and numerical results

x q Exact 
solution of 
u

1
(x)=−2ex

Approximate 
value of u

1
(x)

Absolute error

p
p 1 1e = u (x) - u (x)

Exact solution of

u
2
(x)=−2xex

Approximate 
valueof u

2
(x)

Absolute error

p
p 2 2e = y (x) - y (x)

0.2 1

3

5

7

9

11

13

15

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.44280551

−2.8698700

−2.4779392

−2.4424024

−2.44280875

−2.44280551

−2.44280551

−2.44280551

−2.44280551

4.2702×10−1

3.5134×10−2

1.4041×10−4

3.2468×10−6

5.8432×10−8

3.4562×10−9

4.6783×10−11

2.3351×10−13

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.48856110

−0.79987234

−0.4964455

−0.48553062

−0.48858445

−0.48856110

−0.48856110

−0.48856110

−0.48856110

3.1131×10−1

1.2214×10−2

3.0304×10−3

2.3342×10−5

6.3489×10−7

5.7741×10−9

4.5560×10−10

3.2981×10−12



Hasan: Cooperating Newton’s Method with Series Solution Method

31 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

Solution: Let u x u xm
m

m

q

1 1 1
0

( ) ( )=
=
∑  then using SSM, 

obtaining the values of coefficients

a
10

=0, a
11
=−2,	a

12
=0, a

13
=0.3333, a

14
=0, a

15
=−0.01666,	

and a
16

=0, for u x u xm
m

m

q

2 2 2
0

( ) ( )=
=
∑ then using SSM, obtaining 

the values of coefficients a
20
=−2,	 a

21
=0, a

22
=1, a

23
=0, 

a
24
=−0.08333,	a

25
=0, and a

26
=0.002777.

NEWTON’S METHOD COOPERATE WITH 
SSM FOR SOLVING SLMVFIE-2

Since the numerical solution u x u xi im i
m

m

q

( ) ( )=
=
∑

0

is analytic 

function, then using Taylor series expansion at a point x
0
 it can 

be written of the form

u x
u b
k

x b u b u b x b
u b

x b

i
i
k

k

k
i i

i( )
( )
!
( ) ( ) ( )( )

( )
!

(

= − = + − +

−

=

∞

∑
0

0 1 1
2

2

))
( )
!
( ) ...2

3
3

3
+ − +
u b

x bi

Which is similar to

u x
u b
k

x b u b u b x b
u b

x b

i
i
k

k

k
i i

i( )
( )
!
( ) ( ) ( )( )

( )
!

(

= − = + ′ − +
′′

−

=

∞

∑
0

1

2

))
( )
!

( ) ...2 3
3

+
′′′

− +
u b

x bi

Table 4: Comparison between the exact and numerical results

x q Exact solution of 
u

1
(x)=−2sin (x)

Approximate 
value of u

1
(x)

Absolute error

p
p 1 1e = u (x) - u (x)  

Exact solution of 
u

2
(x)=−2cos (x)

Approximate 
value of u

2
(x)

Absolute error 

p
p 2 2e = y (x) - y (x)

8
π 1

3

5

7

9

11

13

15

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.76366864

−0.65484390

−0.78998921

−0.76887972

−0.76539865

−0.76536922

−0.76536686

−0.76536686

−0.76536686

1.1122×10−1

2.6323×10−2

5.2214×10−3

3.2213×10−5

1.1426×10−6

5.7892×10−8

3.4753×10−9

5.2253×10−11

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.84775906

−1.52331434

−1.84992540

−1.84790077

−1.84779887

−1.84775968

−1.84775906

−1.84775906

−1.84775906

3.2444×10−1

4.2234×10−2

2.5971×10−4

4.1812×10−5

6.2342×10−7

4.6638×10−8

5.6754×10−10

3.8932×10−11

Table 5: Comparison between the results which depending on LSE and RT

At appoint Number of terms Criterion of SSM for u
1
(x)=−2ex Criterion of SSM for u

2
(x)=−2xex

x q LSE RT LSE RT

0.2 3 2.5480×10−4 0:0:1.852 3.2361×10−4 0:0:1.852

5 3.6531×10−8 0:0:2.224 4.3452×10−8 0:0:2.224

7 2.8874×10−10 0:0:3.036 2.6674×10−10 0:0:3.036

9 6.4456×10−13 0:0:4.368 9.8678×10−14 0:0:4.368

LSE: Least square error, RT: Running times, SSM: Series solution method

If we truncating it in the terms of order two, we get

u x u b u b x bi i i( ) ( ) ( )( )= + ′ −

And put h=x−b such that u
i
(x)=0, then we get

 

h
u b

u b
i

i

= −
′
( )

( )

Then, we obtain this iteration formula:

u x u x
u x
u xi

m
i

i
m

i
m

+ = −
′

1( ) ( )
( )

( ( ))
 (4)

For m=0, 1, 2, 3,… where m is the number of iteration.

COOPERATING NEWTON’S METHOD WITH 
SSM FOR SOLVING SLMVFIE-2

The main aim of this section is reformulating and cooperating 
NM with SSM for getting the approximate solutions of the 
problem. These proses are accelerating the convergence 
to exact solution for the problem. First solving SLMVFIE-2 
using SSM. Then, cooperating NM with SSM to obtain the 
new sequence u xi

m

m
( ){ }

=

∞

1
 of approximate solutions for the 

problem using the relationship in equation (4).

u x u x
u x
u x

mi
m

i
i
m

i
m

+ = −
′

=1 0 1 2 3( ) ( )
( )

( ( ))
, , , ,...for

NM on SSM Algorithm

Input: a, b, x, m, E

For i=1 to q

For m=0 to max



Hasan: Cooperating Newton’s Method with Series Solution Method

32 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

Step 1: Using SSM to obtain the numerical solutions 

u x u xi im i
m

m

q

( ) ( )=
=
∑

0

End for

End for

For i=1 to q

For m=0 to max

Step 2: Getting the new sequence using the relationship 
in equation (4).

Step 3: Compute the absolute error e u x u xi
m

i
m

i
+ += −1 1( ) ( ).

If e Ei
m+ <1

Go to output

End if

End for

End for

Output: The numerical results of the technique u xi
m+1( ) 

and ei
m+1 .

Table 7: Comparison between the results for cooperating MN with SSM which depending on LSE and RT

At appoint Number of terms Criterion of SSM for u
1
(x)=−2ex Criterion of SSM for u

2
(x)=−2xex

x q LSE RT LSE RT

0.2 3 3.5573×10−6 0:0:1.852 5.3428×10−6 0:0:1.852

5 4.2582×10−10 0:0:2.224 3.6680×10−12 0:0:2.224

7 2.4006×10−12 0:0:3.036 6.2314×10−13 0:0:3.036

9 5.8062×10−15 0:0:4.368 7.7451×10−16 0:0:4.368

LSE: Least square error, RT: Running times, SSM: Series solution method

Tables 8: Comparison between the results for cooperating MN with SSM which depending on LSE and RT

At a point Number of terms Criterion of SSM for u
1
(x)=−2sin (x) Criterion of SSM for u

2
(x)=−2cos (x)

x q LSE of SSM RT LSE RT

8
π 3 4.4116×10

−5 0:0:1.7734 3.5411×10−4 0:0:1.7734

5 3.2984×10−8 0:0:2.1453 4.6905×10−7 0:0:2.1453

7 5.5205×10−12 0:0:2.9784 4.8834×10−11 0:0:2.9784

9 6.8752×10−16 0:0:3.5621 5.9672×10−15 0:0:3.5621

LSE: Least square error, RT: Running times, SSM: Series solution method

Table 6: Comparison between the results which depending on LSE and RT

At a point Number of terms Criterion of SSM for u
1
(x)=−2sin (x) Criterion of SSM for u

2
(x)=−2cos (x)

x q LSE of SSM RT LSE RT

8
π 3 6.3489×10

−3 0:0:1.7734 2.0036×10−2 0:0:1.7734

5 5.7823×10−6 0:0:2.1453 3.4437×10−6 0:0:2.1453

7 3.5539×10−9 0:0:2.9784 5.7284×10−9 0:0:2.9784

9 5.6127×10−11 0:0:3.5621 4.3582×10−12 0:0:3.5621

LSE: Least square error, RT: Running times, SSM: Series solution method

NUMERICAL EXAMPLES AND RESULTS

In this section, illustrating the technique from numerical 
examples.

Test Example: Solving example (1) using new technique.

Solution: By cooperating NM with SSM, we get the 
following results in Table 1.

Test Example: Solving example (2) using new technique.

Solution: By cooperating NM with SSM, we get the 
following results in Table 2.

CONCLUSION

In this paper, the approximate solutions of SLMVFIE-2 by 
cooperating Newton’s method with SSM for solving SLMVFIE-2 
are discussed. Several examples were solved for illustrating the 
technique and numerical results were achieved. After testing 
the numerical technique and its algorithm on numerical 
examples, the results are written in Tables 1-4 for SSM and 
cooperating Newton’s method with SSM. When the number of 
terms m increasing, the absolute errors are decreasing and the 
convergence is satisfactory. In addition, the least square error 
and running times are written in Tables 5-8 which are criterion 



Hasan: Cooperating Newton’s Method with Series Solution Method

33 http://journals.cihanuniversity.edu.iq/index.php/cuesj CUESJ 2019, 3 (1): 27-33

of discussion for the technique which depending on the number 
of terms m. Then, we conclude that the cooperating Newton’s 
method with SSM is faster and better than SSM for getting the 
approximate solutions. Therefore, this technique is successful 
for getting the numerical solutions for this class of problems.

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Author Queries???
AQ1:Kindly cite figures 1 and 2 in the text part