# Laguerre Collocation Method for Solving Fredholm Integro-Differential Equations with Functional Arguments

Authors: Burcu Gürbüz; Mehmet Sezer; Coşkun Güler
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

Laguerre collocation method is applied for solving a class of the Fredholm integro-differential equations with functional arguments. This method transforms the considered problem to a matrix equation which corresponds to a system of linear algebraic equations. The reliability and efficiency of the proposed scheme are demonstrated by some numerical experiments. Also, the approximate solutions are corrected by using the residual correction method.


---

## Body

## 1. Introduction

Orthogonal polynomials occur often as solutions of mathematical and physical problems. They play an important role in the study of wave mechanics, heat conduction, electromagnetic theory, quantum mechanics, medicine and mathematical statistics, and so forth [1–11]. They provide a natural way to solve, expand, and interpret solutions to many types of important equations. Representation of a smooth function in terms of a series expansion using orthogonal polynomials is a fundamental concept in approximation theory and forms the basis of spectral methods of solution of Fredholm integro-differential equations with functional arguments. Laguerre polynomials L
n
(
x
) constitute complete orthogonal sets of functions on the semi-infinite interval [
0
,
∞
). The past couple of decades have seen a dramatic increase in the application of these models to problems in biology, physics, and engineering. In the field of Fredholm integro-differential equation with functional arguments the computation of its solution has been a great challenge and has been of great importance due to the versatility of such equations in the mathematical modeling of processes in various application fields. Mainly, we deal with the following equation:
(1)
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
y
(
k
)
(
α
h
k
x
+
β
h
k
)
=
g
(
x
)
+
λ
∫
a
b
‍
K
f
(
x
,
t
)
y
(
t
)
d
t
0
≤
a
≤
x
,
t
≤
b
<
∞
,
under the mixed conditions
(2)
∑
k
=
0
m
-
1
‍
a
j
k
y
(
k
)
(
0
)
+
b
j
k
y
(
k
)
(
b
)
=
λ
j
,
j
=
0,1
,
2
,
…
,
m
-
1
,
where y
(
x
) is an unknown function, the known functions P
h
k
(
x
),  g
(
x
), and K
f
(
x
,
t
) are defined on an interval [
0
,
b
], and also a
j
k, b
j
k, and λ
j are appropriate constants. Our aim is to find an approximate solution expressed in the form
(3)
y
N
(
x
)
=
∑
n
=
0
N
‍
a
n
L
n
(
x
)
,
where a
n are unknown coefficients, L
n
(
x
), n
=
0,1
,
…, are Laguerre polynomials, and N is chosen positive integer. The relation between the powers of x and the Laguerre polynomials L
n
(
x
) is
(4)
L
n
(
x
)
=
∑
r
=
0
n
‍
(
-
1
)
r
r
!
(
n
r
)
x
r
.
To obtain a solution in the form (3) of the problem (1) under conditions (2), we can use the collocation points defined by [12–15]
(5)
x
i
=
b
N
i
,
i
=
0,1
,
2
,
…
,
N
.
The remainder of the paper is organized as follows. Higher order linear mixed functional integro-differential equation with variable coefficients and fundamental relations is presented in Section 2. The method of finding approximate solution and the algorithm for the calculation are described in Section 3. Residual error analysis is described in Section 4. To support our finding, we present result of numerical experiments using Maple 12 in Section 5. Section 6 concludes this paper with a brief summary.


## 2. Fundamental Matrix Relations

Let us write (1) in the form
(6)
D
(
x
)
=
g
(
x
)
+
I
(
x
)
,
where the functional differential part is
(7)
D
(
x
)
=
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
y
(
k
)
(
α
h
k
x
+
β
h
k
)
.
Fredholm integral part is
(8)
I
(
x
)
=
∫
a
b
‍
K
f
(
x
,
t
)
y
(
t
)
d
t
.
We convert these parts and the mixed conditions equation (2) into the matrix form. Let us consider (1) and find the matrix forms of each term of the equation. We first consider the solution y
(
x
) and its derivative y
(
k
)
(
x
) defined by a truncated Laguerre series. Then, we can put series in the matrix form
(9)
y
(
x
)
=
L
(
x
)
A
,
y
(
k
)
(
x
)
=
L
(
k
)
(
x
)
A
,
where
(10)
L
(
x
)
=
[
L
0
(
x
)
L
1
(
x
)
⋯
L
N
(
x
)
]
,
L
(
k
)
(
x
)
=
[
L
0
(
k
)
(
x
)
L
1
(
k
)
(
x
)
⋯
L
N
(
k
)
(
x
)
]
,
A
=
[
a
0
a
1
⋯
a
N
]
T
.
By using (4) and taking n
=
0,1
,
…
,
N, we obtain the matrix relation as
(11)
L
T
(
x
)
=
H
X
T
(
x
)
,
L
(
x
)
=
X
(
x
)
H
T
,
where
(12)
X
(
x
)
=
[
1
x
1
⋯
x
N
]
,
H
=
[
(
-
1
)
0
0
!
(
0
0
)
0
0
…
0
(
-
1
)
0
0
!
(
1
0
)
(
-
1
)
1
1
!
(
1
1
)
0
…
0
(
-
1
)
0
0
!
(
2
0
)
(
-
1
)
1
1
!
(
2
1
)
(
-
1
)
2
2
!
(
2
2
)
…
0
⋮
⋮
⋮
⋱
⋮
(
-
1
)
0
0
!
(
N
0
)
(
-
1
)
1
1
!
(
N
1
)
(
-
1
)
2
2
!
(
N
2
)
…
(
-
1
)
N
N
!
(
N
N
)

]
.
Then, by taking into account (11), we have
(13)
L
(
k
)
(
x
)
=
X
(
k
)
(
x
)
H
T
,
k
=
0,1
,
2
,
…
.
Moreover, it is clear that the relation between the matrix X
(
x
) and its derivative X
(
k
)
(
x
) is
(14)
X
(
k
)
(
x
)
=
X
(
x
)
B
k
,
where
(15)
B
=
[
0
1
0
⋯
0
0
0
2
⋯
0
⋮
⋮
⋮
⋱
⋮
0
0
0
⋯
N
0
0
0
⋯
0
]
.
Hence, by means of (13) and (14) we have the matrix relation in the form
(16)
[
y
(
k
)
(
x
)
]
=
L
(
k
)
(
x
)
A
=
X
(
x
)
B
k
H
T
A
.


### 2.1. Matrix Representation for Differential-Difference Parts

The derivative of the matrixL
(
x
) that is defined in (13), by using relation (16), can be expressed as
(17)
L
(
k
)
(
x
)
=
X
(
x
)
B
k
H
T
.
By substituting x
=
α
h
k
x
+
β
h
k into (16) [16, 17], we obtain
(18)
[
y
(
k
)
(
α
h
k
x
+
β
h
k
)
]
=
L
(
k
)
(
α
h
k
x
+
β
h
k
)
A
=
X
(
α
h
k
x
+
β
h
k
)
B
k
H
T
A
,
where we have from (13)
(19)
L
(
k
)
(
α
h
k
x
+
β
h
k
)
=
X
(
k
)
(
α
h
k
x
+
β
h
k
)
H
T
,
X
(
α
h
k
x
+
β
h
k
)
=
[
1
(
α
h
k
x
+
β
h
k
)
1
⋯
(
α
h
k
x
+
β
h
k
)
N
]
.
Then, we consider (14) and we have
(20)
X
(
k
)
(
α
h
k
x
+
β
h
k
)
=
X
(
x
)
B
(
α
h
k
,
β
h
k
)
B
k
,
where
(21)
B
(
α
h
k
,
β
h
k
)
=
[
(
0
0
)
(
α
h
k
)
0
(
β
h
k
)
0
(
1
0
)
⋯
(
N
0
)
(
α
h
k
)
0
(
β
h
k
)
N
0
(
1
1
)
⋯
(
N
1
)
(
α
h
k
)
1
(
β
h
k
)
N
-
1
0
0
⋯
(
N
2
)
(
α
h
k
)
2
(
β
h
k
)
N
-
2
⋮
⋮
⋱
⋮
0
0
⋯
(
N
N
)
(
α
h
k
)
N
(
β
h
k
)
0
]
.
Therefore, by using relation (16), the matrix representation of differential-difference part can be given by
(22)
D
(
x
)
=
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
X
B
(
α
h
k
,
β
h
k
)
B
k
H
T
A
.


### 2.2. Matrix Representation for Fredholm Integral Part

K
f
(
x
,
t
) can be expanded by Laguerre series with respect to t as follows:
(23)
K
f
(
x
,
t
)
=
∑
i
,
j
=
0
N
‍
k
i
j
f
(
x
)
L
j
(
t
)
,
i
,
j
=
0,1
,
…
,
N
.
Then, the matrix representation of the kernel function K
f
(
x
,
t
) becomes
(24)
K
f
(
x
,
t
)
=
K
f
(
x
)
L
T
(
t
)
,
where
(25)
K
f
(
x
,
t
)
=
[
k
i
j
(
x
)
]
i
,
j
=
0,1
,
…
,
N
.
Substituting relations (9) and (24) in the Fredholm part, we obtain
(26)
I
(
x
)
=
∫
a
b
‍
X
(
x
)
K
f
L
T
L
A
d
t
=
L
K
f
Q
A
,
where
(27)
Q
=
∫
a
b
‍
L
T
L
d
t
=
∫
a
b
‍
H
X
T
(
t
)
X
(
t
)
H
T
d
t
=
H
[
∫
a
b
‍
X
T
(
t
)
X
(
t
)
d
t
]
H
T
.
We define
(28)
M
f
=
[
h
i
j
f
]
,
(29)
[
h
i
j
f
]
=
∫
a
b
‍
X
(
t
)
T
X
(
t
)
d
t
=
b
i
+
j
+
1
-
a
i
+
j
+
1
i
+
j
+
1
i
,
j
=
0,1
,
…
,
N
.
We also have the matrix relation Q
f
=
H
M
f
H
T. Hence, the matrix representation of Fredholm integral part can be given by
(30)
I
(
x
)
=
X
(
x
)
H
T
K
f
Q
f
A
.


### 2.3. Matrix Representation of the Conditions

Let us define the matrix form of the conditions given by (2) that can be written as
(31)
∑
k
=
0
m
-
1
‍
a
j
k
y
(
k
)
(
0
)
+
b
j
k
y
(
k
)
(
b
)
=
λ
j
,
j
=
0,1
,
2
,
…
,
m
-
1
.
Then, we have
(32)
[
y
(
k
)
(
0
)
]
=
X
(
0
)
(
B
T
)
k
H
T
A
,
[
y
(
k
)
(
b
)
]
=
X
(
b
)
(
B
T
)
k
H
T
A
.
Briefly, we demonstrate the conditions in matrix form as
(33)
U
j
=
∑
k
=
0
m
-
1
‍
a
j
k
y
(
k
)
(
0
)
+
b
j
k
y
(
k
)
(
b
)
(
B
T
)
k
H
T
A
=
[
λ
j
]
,
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
j
,
k
=
0,1
,
2
,
…
,
m
-
1
,
where
(34)
U
j
A
=
[
u
j
0
u
j
1
⋯
u
j
N
]
.


## 2.1. Matrix Representation for Differential-Difference Parts

The derivative of the matrixL
(
x
) that is defined in (13), by using relation (16), can be expressed as
(17)
L
(
k
)
(
x
)
=
X
(
x
)
B
k
H
T
.
By substituting x
=
α
h
k
x
+
β
h
k into (16) [16, 17], we obtain
(18)
[
y
(
k
)
(
α
h
k
x
+
β
h
k
)
]
=
L
(
k
)
(
α
h
k
x
+
β
h
k
)
A
=
X
(
α
h
k
x
+
β
h
k
)
B
k
H
T
A
,
where we have from (13)
(19)
L
(
k
)
(
α
h
k
x
+
β
h
k
)
=
X
(
k
)
(
α
h
k
x
+
β
h
k
)
H
T
,
X
(
α
h
k
x
+
β
h
k
)
=
[
1
(
α
h
k
x
+
β
h
k
)
1
⋯
(
α
h
k
x
+
β
h
k
)
N
]
.
Then, we consider (14) and we have
(20)
X
(
k
)
(
α
h
k
x
+
β
h
k
)
=
X
(
x
)
B
(
α
h
k
,
β
h
k
)
B
k
,
where
(21)
B
(
α
h
k
,
β
h
k
)
=
[
(
0
0
)
(
α
h
k
)
0
(
β
h
k
)
0
(
1
0
)
⋯
(
N
0
)
(
α
h
k
)
0
(
β
h
k
)
N
0
(
1
1
)
⋯
(
N
1
)
(
α
h
k
)
1
(
β
h
k
)
N
-
1
0
0
⋯
(
N
2
)
(
α
h
k
)
2
(
β
h
k
)
N
-
2
⋮
⋮
⋱
⋮
0
0
⋯
(
N
N
)
(
α
h
k
)
N
(
β
h
k
)
0
]
.
Therefore, by using relation (16), the matrix representation of differential-difference part can be given by
(22)
D
(
x
)
=
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
X
B
(
α
h
k
,
β
h
k
)
B
k
H
T
A
.


## 2.2. Matrix Representation for Fredholm Integral Part

K
f
(
x
,
t
) can be expanded by Laguerre series with respect to t as follows:
(23)
K
f
(
x
,
t
)
=
∑
i
,
j
=
0
N
‍
k
i
j
f
(
x
)
L
j
(
t
)
,
i
,
j
=
0,1
,
…
,
N
.
Then, the matrix representation of the kernel function K
f
(
x
,
t
) becomes
(24)
K
f
(
x
,
t
)
=
K
f
(
x
)
L
T
(
t
)
,
where
(25)
K
f
(
x
,
t
)
=
[
k
i
j
(
x
)
]
i
,
j
=
0,1
,
…
,
N
.
Substituting relations (9) and (24) in the Fredholm part, we obtain
(26)
I
(
x
)
=
∫
a
b
‍
X
(
x
)
K
f
L
T
L
A
d
t
=
L
K
f
Q
A
,
where
(27)
Q
=
∫
a
b
‍
L
T
L
d
t
=
∫
a
b
‍
H
X
T
(
t
)
X
(
t
)
H
T
d
t
=
H
[
∫
a
b
‍
X
T
(
t
)
X
(
t
)
d
t
]
H
T
.
We define
(28)
M
f
=
[
h
i
j
f
]
,
(29)
[
h
i
j
f
]
=
∫
a
b
‍
X
(
t
)
T
X
(
t
)
d
t
=
b
i
+
j
+
1
-
a
i
+
j
+
1
i
+
j
+
1
i
,
j
=
0,1
,
…
,
N
.
We also have the matrix relation Q
f
=
H
M
f
H
T. Hence, the matrix representation of Fredholm integral part can be given by
(30)
I
(
x
)
=
X
(
x
)
H
T
K
f
Q
f
A
.


## 2.3. Matrix Representation of the Conditions

Let us define the matrix form of the conditions given by (2) that can be written as
(31)
∑
k
=
0
m
-
1
‍
a
j
k
y
(
k
)
(
0
)
+
b
j
k
y
(
k
)
(
b
)
=
λ
j
,
j
=
0,1
,
2
,
…
,
m
-
1
.
Then, we have
(32)
[
y
(
k
)
(
0
)
]
=
X
(
0
)
(
B
T
)
k
H
T
A
,
[
y
(
k
)
(
b
)
]
=
X
(
b
)
(
B
T
)
k
H
T
A
.
Briefly, we demonstrate the conditions in matrix form as
(33)
U
j
=
∑
k
=
0
m
-
1
‍
a
j
k
y
(
k
)
(
0
)
+
b
j
k
y
(
k
)
(
b
)
(
B
T
)
k
H
T
A
=
[
λ
j
]
,
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
j
,
k
=
0,1
,
2
,
…
,
m
-
1
,
where
(34)
U
j
A
=
[
u
j
0
u
j
1
⋯
u
j
N
]
.


## 3. Method of Solution

We are ready to construct the fundamental matrix equation corresponding to (1). For this purpose, by substituting the matrix relations (22) and (30) into (1), we obtain
(35)
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
X
(
x
)
B
(
α
h
k
,
β
h
k
)
B
k
H
T
A
=
G
(
x
)
+
λ
X
(
x
)
H
T
K
f
Q
f
A
.
Then, we put equally spaced collocation points in (5) and relation (35) obtains
(36)
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
i
)
X
(
x
i
)
B
(
α
h
k
,
β
h
k
)
B
k
H
T
A
=
G
(
x
i
)
+
λ
X
(
x
i
)
H
T
K
f
Q
f
A
.
Briefly, the fundamental matrix equation is gained as follows:
(37)
(
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
X
B
(
α
h
k
,
β
h
k
)
B
k
H
T
-
λ
X
H
T
K
f
Q
f
)
A
=
G
,
where
(38)
P
h
k
=
[
P
h
k
(
x
0
)
0
0
…
0
0
P
h
k
(
x
1
)
0
…
0
0
0
P
h
k
(
x
2
)
…
0
⋮
⋮
⋮
⋱
⋮
0
0
0
…
P
h
k
(
x
N
)
]
,
X
=
[
1
x
0
x
0
2
…
x
0
N
1
x
1
x
1
2
…
x
1
N
1
x
2
x
2
2
…
x
2
N
⋮
⋮
⋮
⋱
⋮
1
x
N
x
N
2
…
x
N
N
]
,
K
f
=
[
K
f
(
x
0
)
K
f
(
x
1
)
K
f
(
x
2
)
⋮
K
f
(
x
N
)
]
,
G
=
[
g
(
x
0
)
g
(
x
1
)
g
(
x
2
)
⋮
g
(
x
N
)
]
.
The fundamental matrix equation (35) for (1) corresponds to a system of (
N
+
1
) algebraic equation for the (
N
+
1
) unknown coefficients a
0
,
a
1
,
…
,
a
N [18]. Briefly we can write (37) as
(39)
W
A
=
G
or
[
W
;
G
]
so that, for i
,
j
=
0,1
,
2
,
…
,
N,
(40)
W
=
[
w
i
j
]
=
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
X
B
(
α
h
k
,
β
h
k
)
B
k
H
T
-
λ
X
(
x
)
H
T
K
f
Q
f
.
Then, the matrix form of conditions (2) is
(41)
U
j
A
=
[
λ
j
]
or
[
U
j
;
λ
j
]
,
j
=
0,1
,
…
,
m
-
1
,
where
(42)
[
U
~
j
]
=
[
U
j
;
λ
j
]
.
To obtain the solution of (1) under conditions (2), by replacing the rows in matrix (39) by the last m rows of matrix (41), we have the required augmented matrix
(43)
W
~
A
=
G
~
.
If rank
⁡
(
W
~
)
=
rank
⁡
[
W
~
;
G
~
]
=
N
+
1, then we can write
(44)
A
=
(
W
~
)
-
1
G
~
.
Thus, the coefficients a
n, i
=
0,1
,
…
,
N, are uniquely determined by (43) [19].


## 4. Residual Correction and Error Estimation

In this section, we give an error estimation for the Laguerre polynomial solution (3) with the residual error function, and it supports the idea of corrected Laguerre polynomial solution with respect to the residual error function. Let us define (1) with L operator as follows:
(45)
L
[
y
(
x
)
]
=
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
y
(
k
)
(
α
h
k
x
+
β
h
k
)
-
λ
∫
a
b
‍
K
f
(
x
,
t
)
y
(
t
)
d
t
=
g
(
x
)
FFffFFFFF
0
≤
a
≤
x
,
t
≤
b
<
∞
.
Then, we get the residual function of the Laguerre collocation method as
(46)
R
N
(
x
)
=
L
[
y
N
(
x
)
]
-
g
(
x
)
,
where y
N
(
x
) is the approximate solution which is solved via Laguerre polynomials given by (3) of problem (1) with conditions (2). Thus, y
N
(
x
) satisfies the problem
(47)
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
y
N
(
k
)
(
α
h
k
x
+
β
h
k
)
=
g
(
x
)
+
λ
∫
a
b
‍
K
f
(
x
,
t
)
y
N
(
t
)
d
t
FFFFFFFFFI
0
≤
a
≤
x
,
t
≤
b
<
∞
,
(48)
∑
k
=
0
m
-
1
‍
a
j
k
y
N
(
k
)
(
0
)
+
b
j
k
y
N
(
k
)
(
b
)
=
λ
j
,
j
=
0,1
,
2
,
…
,
m
-
1
.
Also, the error function E
N
(
x
) can be defined as
(49)
E
N
(
x
)
=
y
(
x
)
-
y
N
(
x
)
,
where y
(
x
) is the exact solution of problem (1) with (2) conditions. Then, by using (1), (2), (45), and (47), we obtain the error differential equation
(50)
L
[
E
N
(
x
)
]
=
L
[
y
(
x
)
]
-
L
[
y
N
(
x
)
]
=
-
R
N
(
x
)
with the homogeneous conditions
(51)
∑
k
=
0
m
-
1
‍
a
j
k
E
N
(
k
)
(
0
)
+
b
j
k
E
N
(
k
)
(
b
)
=
0
,
j
=
0,1
,
2
,
…
,
m
-
1
.
Briefly, the error problem is
(52)
∑
h
=
0
J
‍
∑
k
=
0
m
‍
P
h
k
(
x
)
E
N
(
k
)
(
α
h
k
x
+
β
h
k
)
=
g
(
x
)
+
λ
∫
a
b
‍
K
f
(
x
,
t
)
E
N
(
t
)
d
t
FFFFFffFFFI
0
≤
a
≤
x
,
t
≤
b
<
∞
,
(53)
∑
k
=
0
m
-
1
‍
a
j
k
E
N
(
k
)
(
0
)
+
b
j
k
E
N
(
k
)
(
b
)
=
0
,
j
=
0,1
,
2
,
…
,
m
-
1
.
By solving problem (52) with the technique introduced in Section 3, we get the approximation
(54)
E
N
,
M
(
x
)
=
∑
n
=
0
M
‍
a
n
*
L
n
(
x
)
to E
N
(
x
). Consequently, by means of the polynomials y
N
(
x
) and E
N
,
M
(
x
), (M
>
N), we obtain the corrected Laguerre polynomial solution y
N
,
M
(
x
)
=
y
N
(
x
)
+
E
N
,
M
(
x
). Also, we construct the corrected error function
(55)
E
N
,
M
*
(
x
)
=
E
N
(
x
)
-
E
N
,
M
(
x
)
=
y
(
x
)
-
y
N
,
M
(
x
)
and the estimated error function E
N
,
M
(
x
) [20–22].


### 4.1. Algorithm

In this section, we will consider each step of the method which is solved by Maple 12 and we distinguish the algorithms in three steps. Algorithm1 demonstrates the calculation of approximate solution. Algorithm 2 shows each step of residual error function and its support to corrected error and corrected Laguerre polynomial solution. We plot the graphics of all functions in Algorithm 3.Algorithm 1.
Consider the following.(1)
Input initial data: the coefficients areP
h
k, kernel function is K
f
(
x
,
t
), and the coefficients of the conditions are a
j
k, b
j
k. The problem is defined on [
0,1
].
(2)
Develop the algorithm by matrix inputs which are constructed by collocation points in (4).
(3)
Use the conditions to construct augmented matrix and the system is done.
(4)
Output: the system is solved andA unknown matrix is found. Then, approximate solution is found with respect to coefficients of unknown matrix in truncated Laguerre series.
(5)
End of Algorithm1.Algorithm 2.
Consider the following.(1)
Input data: the approximate solution isy
N
(
x
), the error function E
N
(
x
), and truncated Laguerre series (3).
(2)
We check fory
N
(
x
) that equals truncated Laguerre series. Develop the algorithm to demonstrate R
N
(
x
), the residual function of the Laguerre collocation method, via absolute error function |
E
N
(
x
)
|.
(3)
The conditions are constructed by absolute error function. Then, obtain the estimated error functionE
N
,
M.
(4)
Output: the corrected error functionE
N
,
M
* is found.
(5)
End of Algorithm2.Algorithm 3.
Consider the following.(
1
)
Input data: the comparison between approximate and exact solutions is demonstrated by the graphics.
(
2
)
Output:
with(plots):
plot1:=plot(([y(x)], x=0..1), style = point, color = magenta, legend =[typeset(“Exact Solution”)]);
plot2:=plot(([y(N,M)(x), x = 0..1]), style = line, color = blue, legend =[typeset(“Approximate Solution”)]);
display(plot(plot1,plot2)).
(
3
)
Input: the comparison between error and residual error solutions is demonstrated by the graphics.
(
4
)
Output:
with(plots):
plot3:=plot(([E
N(x)], x = 0..1), style =point, color = magenta, legend =[typeset(“Error Function”)]);
plot4:=plot(([E(N,M)(x), x=0..1]), style=line, color = blue, legend =[typeset(“Residual Error Function”)]);
display(plot(plot3,plot4)).
(
5
)
End of Algorithm3.


## 4.1. Algorithm

In this section, we will consider each step of the method which is solved by Maple 12 and we distinguish the algorithms in three steps. Algorithm1 demonstrates the calculation of approximate solution. Algorithm 2 shows each step of residual error function and its support to corrected error and corrected Laguerre polynomial solution. We plot the graphics of all functions in Algorithm 3.Algorithm 1.
Consider the following.(1)
Input initial data: the coefficients areP
h
k, kernel function is K
f
(
x
,
t
), and the coefficients of the conditions are a
j
k, b
j
k. The problem is defined on [
0,1
].
(2)
Develop the algorithm by matrix inputs which are constructed by collocation points in (4).
(3)
Use the conditions to construct augmented matrix and the system is done.
(4)
Output: the system is solved andA unknown matrix is found. Then, approximate solution is found with respect to coefficients of unknown matrix in truncated Laguerre series.
(5)
End of Algorithm1.Algorithm 2.
Consider the following.(1)
Input data: the approximate solution isy
N
(
x
), the error function E
N
(
x
), and truncated Laguerre series (3).
(2)
We check fory
N
(
x
) that equals truncated Laguerre series. Develop the algorithm to demonstrate R
N
(
x
), the residual function of the Laguerre collocation method, via absolute error function |
E
N
(
x
)
|.
(3)
The conditions are constructed by absolute error function. Then, obtain the estimated error functionE
N
,
M.
(4)
Output: the corrected error functionE
N
,
M
* is found.
(5)
End of Algorithm2.Algorithm 3.
Consider the following.(
1
)
Input data: the comparison between approximate and exact solutions is demonstrated by the graphics.
(
2
)
Output:
with(plots):
plot1:=plot(([y(x)], x=0..1), style = point, color = magenta, legend =[typeset(“Exact Solution”)]);
plot2:=plot(([y(N,M)(x), x = 0..1]), style = line, color = blue, legend =[typeset(“Approximate Solution”)]);
display(plot(plot1,plot2)).
(
3
)
Input: the comparison between error and residual error solutions is demonstrated by the graphics.
(
4
)
Output:
with(plots):
plot3:=plot(([E
N(x)], x = 0..1), style =point, color = magenta, legend =[typeset(“Error Function”)]);
plot4:=plot(([E(N,M)(x), x=0..1]), style=line, color = blue, legend =[typeset(“Residual Error Function”)]);
display(plot(plot3,plot4)).
(
5
)
End of Algorithm3.


## 5. Illustrative Examples

In this section, several numerical examples are given to illustrate the accuracy and the effectiveness of the method. All of them are performed on the computer by using Maple 12.Example 1 (see [23]).
Let us first consider the functional Fredholm integro-differential equations as follows:(56)
y
(
x
+
1
)
+
x
y
(
x
)
-
x
y
(
x
-
1
)
=
(
2
+
3
x
2
+
(
1
-
8
3
e
x
)
x
)
+
∫
0
1
‍
(
2
x
e
x
)
y
(
t
)
d
t
,
ffffffffffffffffffffffffffffffffffffffffffffffffffffff
0
≤
x
≤
1
,
under the mixed conditions
(57)
y
(
0
)
=
1
,
y
(
1
2
)
=
5
4
,
and we seek the approximate solution y
N
(
x
) as a truncated Laguerre series:
(58)
y
N
(
x
)
=
∑
r
=
0
N
‍
a
r
L
r
(
x
)
,
0
≤
x
<
∞
.
Here,
(59)
P
00
(
x
)
=
x
,
P
10
(
x
)
=
-
x
,
P
20
(
x
)
=
1
,
g
(
x
)
=
e
x
-
e
+
e
-
1
,
K
f
(
x
,
t
)
=
2
x
e
x
,
λ
=
1
.
Then, for N
=
3, the collocation points are x
0
=
0, x
1
=
1
/
2, x
2
=
2
/
3, and x
3
=
1 and the problem is defined by
(60)
W
=
P
00
X
H
T
+
P
10
X
B
α
1
,
β
1
+
P
20
X
B
α
2
,
β
2
-
λ
X
H
T
K
f
Q
f
,
and matrices for conditions are
(61)
[
U
0
;
λ
0
U
1
;
λ
1
]
=
[
1
1
1
1
;
1
1
1
2
1
8
-
7
48
;
5
4
]
,
and the matrices are computed as
(62)
[
W
;
G
]
=
[
1
0
-
1
2
-
2
3
;
2
2
27
-
61
24
-
134
81
-
15
8
;
800
561
-
41
27
-
70
27
-
419
162
-
119
54
;
336
625
-
4
-
9
2
-
10
3
-
43
24
;
-
213
170
]
.
Then, we have the augmented matrix
(63)
[
W
~
;
G
~
]
=
[
1
0
-
1
2
-
2
3
;
2
2
27
-
61
24
-
134
81
-
15
8
;
800
561
1
1
1
1
;
1
1
1
2
7
8
-
7
48
;
5
4
]
.
Subsequently, we have
(64)
A
=
[
423
142
-
545
140
360
190
583
122
]
,
and we obtain the approximate solution of the problem for N
=
3 as
(65)
y
3
(
x
)
=
0.999999999999996
-
0.366738338031.1
0
-
3
x
+
1.01100215014091
x
2
-
0.733476676060432.1
0
-
3
x
3
.
We follow the same steps for N
=
10 and N
=
15:
(66)
y
10
(
x
)
=
-
0.12404861.1
0
-
6
x
+
1.000000772576
x
2
+
0.999999999999973
-
0.896642.1
0
-
6
x
3
-
0.12907822.1
0
-
5
x
4
+
0.195475243.1
0
-
5
x
5
+
0.709505509.1
0
-
6
x
6
-
0.14545410555.1
0
-
5
x
7
+
0.39178123286.1
0
-
7
x
8
+
0.3995801024017.1
0
-
6
x
9
-
0.109576123674046.1
0
-
6
x
10
,
and for N
=
15
(67)
y
15
(
x
)
=
0.99999999945109
x
2
+
0.33996.1
0
-
9
x
+
1.00000000000001
-
0.135090.1
0
-
8
x
3
+
0.2063949.1
0
-
8
x
4
+
0.14857110.1
0
-
8
x
5
-
0.26538783.1
0
-
8
x
6
-
0.386318369.1
0
-
9
x
7
+
0.153862741670.1
0
-
8
x
8
-
0.311133726104.1
0
-
9
x
9
-
0.32406068899445.1
0
-
9
x
10
+
0.1760808732067.1
0
-
9
x
11
-
0.30984313933219.1
0
-
10
x
12
+
0.187639980825480.1
0
-
11
x
13
-
0.458623284005214.1
0
-
13
x
14
+
0.399275908003723.1
0
-
15
x
15
.
The exact solution of this problem is y
(
x
)
=
x
2
+
1.Table1 shows the comparison between exact and approximate solutions for N
=
3, 10, 15 values. The results of residual error function support the evaluation of the corrected new Laguerre polynomial solutions for different N and M values. Figure 1 shows us the comparison between exact and approximate solutions for various N values. Figure 2 shows the absolute error and absolute residual error functions. Mainly, this comparison supports the idea of corrected Laguerre polynomial solution for different N and M values.Table 1
Numerical solution of Example1 for different N values.
Present method and corrected Laguerre polynomial solution
x
Exact solution
y
3
y
3,7
y
10
y
10,12
y
15
y
15,20
0.0
1.000000
0.999999
0.99999
0.999999
1.00000
1.000000
1.000000
0.1
1.010000
1.009736
1.01054
1.009999
1.00999
1.010000
1.010000
0.2
1.040000
1.039648
1.04075
1.039999
1.03999
1.040000
1.040000
0.3
1.090000
1.089692
1.09068
1.090000
1.08999
1.090000
1.090000
0.4
1.160000
1.159824
1.16040
1.160000
1.15999
1.160000
1.160000
0.5
1.250000
1.249999
1.25000
1.250000
1.24999
1.250000
1.250000
0.6
1.360000
1.360176
1.35957
1.359999
1.36000
1.359999
1.360000
0.7
1.490000
1.490308
1.48923
1.489999
1.49000
1.489999
1.490000
0.8
1.640000
1.640352
1.63909
1.639999
1.64000
1.639999
1.639999
0.9
1.810000
1.810264
1.80931
1.809999
1.80999
1.809999
1.809999
1.0
2.000000
2.000000
2.00202
2.000000
1.99999
1.999999
1.999999Figure 1
Numerical and exact solution of Example1 for N
=
3
,
10
,
15.Figure 2
Comparison of error and residual error functions fory
3,7 values for Example 1.Table2 shows the corrected absolute errors by our method for N
=
3, 10, 15 and M
=
7, 12, 20. The results support the idea that when N and M values are chosen large enough, the absolute error and residual error are decreased.Table 2
Comparison of the actual and estimated absolute errors forN
=
3,10,15 and M
=
7,12,20 of Example 1.
The actual absolute and estimated absolute errors
x
E
3
E
3,7
E
10
E
10,12
E
15
E
15,20
0.0
0.40000
E
-
14
0.48000
E
-
08
0.27000
E
-
13
0.13300
E
-
09
0.1000
E
-
13
0.34760
E
-
13
0.1
0.26405
E
-
03
0.54125
E
-
03
0.56847
E
-
08
0.11292
E
-
07
0.2738
E
-
10
0.10459
E
-
09
0.2
0.35206
E
-
03
0.74827
E
-
04
0.24926
E
-
08
0.16453
E
-
07
0.3885
E
-
10
0.15989
E
-
09
0.3
0.30806
E
-
03
0.67809
E
-
03
0.26114
E
-
08
0.14766
E
-
07
0.3452
E
-
10
0.16136
E
-
09
0.4
0.17603
E
-
03
0.40076
E
-
03
0.42220
E
-
08
0.82377
E
-
08
0.1916
E
-
10
0.10735
E
-
09
0.5
0.10000
E
-
13
0.28387
E
-
09
0.21000
E
-
12
0.50360
E
-
09
0.00000000
0.26070
E
-
13
0.6
0.17603
E
-
03
0.42664
E
-
03
0.88534
E
-
08
0.47667
E
-
08
0.1571
E
-
10
0.15464
E
-
09
0.7
0.30806
E
-
03
0.76824
E
-
03
0.18229
E
-
07
0.54107
E
-
08
0.2306
E
-
10
0.34641
E
-
09
0.8
0.35207
E
-
03
0.90156
E
-
03
0.22651
E
-
07
0.18877
E
-
08
0.2078
E
-
10
0.56197
E
-
09
0.9
0.26405
E
-
03
0.69278
E
-
03
0.17309
E
-
07
0.23159
E
-
08
0.1137
E
-
10
0.78552
E
-
09
1.0
0.000000000
0.12445
E
-
07
0.15600
E
-
11
0.11339
E
-
08
0.1000
E
-
13
0.10001
E
-
08Example 2 (see [24]).
Let us find the Laguerre series solution of the following second order linear functional Fredholm integro-differential equation with variable coefficients:(68)
x
2
y
′′
(
x
)
+
x
y
′
(
x
)
+
y
(
x
)
+
y
(
x
-
1
)
=
∫
0
1
‍
(
12
7
x
2
-
2
)
y
(
t
)
d
t
+
2
x
2
-
28
3
x
+
13
,
fffffffffffffffffffffffffffffffffffffffffffffffffff
0
≤
x
≤
1
,
with initial conditions y
(
0
)
=
4, y
′
(
0
)
=
-
4 and its exact solution is y
(
x
)
=
(
x
-
2
)
2. We obtained the approximate solution of the problem for N
=
8, 10, 18 and the matrices are computed. Hence linear algebraic system is gained. This system is approximately solved using Maple 12.Table3 shows us the comparison between exact solution, approximate solutions for different N values, and corrected Laguerre polynomial solutions with respect to the residual error analysis. Figure 3 shows us the comparison between exact and approximate solutions for different N values. Figure 4 shows the comparison of absolute error function of Example 2 for the same N values.Table 3
Numerical solution of Example2 for different N values.
Present method and corrected Laguerre polynomial solution
x
Exact solution
y
8
y
8,10
y
10
y
10,12
y
18
y
18,22
0.0
4.000000
4.000000
4.000000
3.999999
3.999999
4.000000
4.000000
0.1
3.610000
3.607781
3.608978
3.607685
3.610898
3.610363
3.608798
0.2
3.240000
3.230410
3.235483
3.230156
3.243601
3.240185
3.235706
0.3
2.890000
2.867005
2.878938
2.866678
2.898090
2.887697
2.881383
0.4
2.560000
2.516891
2.538843
2.516663
2.574315
2.551308
2.546356
0.5
2.250000
2.179554
2.214765
2.179633
2.272214
2.229593
2.231063
0.6
1.960000
1.854596
1.906326
1.855192
1.991717
1.921284
1.935850
0.7
1.690000
1.541700
1.613200
1.543010
1.732756
1.625252
1.661011
0.8
1.440000
1.240609
1.335103
1.242812
1.495265
1.340495
1.406788
0.9
1.210000
0.951112
1.071789
0.954366
1.279181
1.066119
1.173387
1.0
1.000000
0.673025
0.823042
0.677475
1.084448
0.801327
0.960984Figure 3
Numerical and exact solution of Example2 for N
=
8, 10, 18.Figure 4
Error function of Example2 for various N values.Figure5 shows us the comparison between absolute error and absolute residual error graphics.Figure 5
Comparison of error and residual error functions fory
8,10 values.Table4 shows the corrected absolute errors by our method for N
=
8, 10, 18 and M
=
10, 12, 22.Table 4
Comparison of the actual and estimated absolute errors forN
=
8,10,18 and M
=
10,12,22 of Example 2.
The actual absolute and estimated absolute errors
x
E
8
E
8,10
E
10
E
10,12
E
18
E
18,22
0.0
0.67000
E
-
15
0.34000
E
-
13
0.12300
E
-
12
0.82600
E
-
11
0.33307
E
-
08
0.10950
E
-
09
0.1
0.22193
E
-
02
0.10221
E
-
02
0.39880
E
-
04
0.89756
E
-
03
0.36289
E
-
03
0.12021
E
-
02
0.2
0.95897
E
-
02
0.45169
E
-
02
0.26746
E
-
04
0.36009
E
-
02
0.18489
E
-
03
0.42936
E
-
02
0.3
0.22995
E
-
01
0.11062
E
-
01
0.22276
E
-
03
0.80900
E
-
02
0.23027
E
-
02
0.86169
E
-
02
0.4
0.43109
E
-
01
0.21157
E
-
01
0.87128
E
-
03
0.14315
E
-
01
0.86922
E
-
02
0.13641
E
-
01
0.5
0.70446
E
-
01
0.35235
E
-
01
0.20625
E
-
02
0.22214
E
-
01
0.20407
E
-
01
0.18937
E
-
01
0.6
0.10540
E
-
01
0.53674
E
-
01
0.39232
E
-
02
0.31717
E
-
01
0.38716
E
-
01
0.24149
E
-
01
0.7
0.14829
E
-
01
0.76799
E
-
01
0.65657
E
-
02
0.42756
E
-
01
0.64748
E
-
01
0.28989
E
-
01
0.8
0.19939
E
-
01
0.10489
E
-
01
0.10089
E
-
01
0.55265
E
-
01
0.99505
E
-
01
0.33212
E
-
01
0.9
0.25889
E
-
01
0.13821
E
-
01
0.14583
E
-
01
0.69182
E
-
01
0.14389
E
-
01
0.36613
E
-
01
1.0
0.32697
E
-
01
0.17696
E
-
01
0.20124
E
-
01
0.84448
E
-
01
0.19867
E
-
01
0.39016
E
-
01Example 3 (see [25]).
Let us find the Laguerre series solution of the following second order linear functional Fredholm integro-differential equation with variable coefficients:(69)
(
x
+
4
)
2
y
′′
(
x
)
-
(
x
+
4
)
y
′
(
x
)
+
y
(
x
-
1
)
+
y
′
(
x
-
1
)
=
ln
⁡
(
x
+
3
)
+
1
x
+
3
+
3
ln
⁡
(
3
)
-
5
ln
⁡
(
5
)
+
∫
-
1
1
‍
y
(
t
)
d
t
,
0
≤
x
≤
1
.
with initial conditions y
(
0
)
=
ln
⁡
(
4
) and y
′
(
0
)
=
1
/
4 and its exact solution isy
(
x
)
=
ln
⁡
(
x
+
4
). We obtained the approximate solution of the problem forN
=
8, N
=
10, and N
=
20 which are tabulated and graphed.Table5 shows the exact solution comparison with approximate solutions for N
=
8, 10, 20 with present method. Absolute error comparisons are also given in the same table. Figure 6 shows the comparison between approximate solutions for different N values. We observe the absolute error differences for the same N values in Figure 7.Table 5
Numerical solution and absolute error of Example3 for different N values.
Present method
x
Exact solution
y
8
E
8
y
10
E
10
y
20
E
20
0.0
1.386294
1.386294
0.00000000
1.386294
0.00000000
1.386294
0.1200
E
-
17
0.1
1.410987
1.410833
0.15424
E
-
4
1.410833
0.15424
E
-
4
1.410833
0.15424
E
-
4
0.2
1.435085
1.434461
0.62350
E
-
4
1.434461
0.62350
E
-
4
1.434461
0.62350
E
-
4
0.3
1.458615
1.457198
0.14110
E
-
3
1.457198
0.14165
E
-
3
1.457198
0.14165
E
-
3
0.4
1.481605
1.479064
0.25409
E
-
3
1.479064
0.25409
E
-
3
1.479064
0.25409
E
-
3
0.5
1.504077
1.500007
0.40029
E
-
3
1.500007
0.40029
E
-
3
1.500074
0.40029
E
-
3
0.6
1.526056
1.520248
0.58081
E
-
3
1.520248
0.58081
E
-
3
1.526056
0.58081
E
-
3
0.7
1.547563
1.539602
0.79609
E
-
3
1.539602
0.79609
E
-
3
1.547563
0.79609
E
-
3
0.8
1.568616
1.558151
0.10465
E
-
2
1.558151
0.10465
E
-
2
1.568616
0.10465
E
-
2
0.9
1.589235
1.575911
0.13324
E
-
2
1.575911
0.13324
E
-
2
1.589235
0.13324
E
-
2
1.0
1.609438
1.592899
0.16540
E
-
2
1.592899
0.16539
E
-
2
1.609438
0.16539
E
-
2Figure 6
Numerical and exact solution of Example3 for N
=
8, 10, 20.Figure 7
Error function of Example3 for various N values, N
e
=
8,10,20.Example 4 (see [24]).
Consider the linear functional Fredholm integro-differential equation(70)
y
′′
(
x
)
+
x
y
′
(
x
)
+
x
y
(
x
)
-
y
′
(
x
-
1
)
+
y
(
x
-
1
)
=
x
(
sin
(
x
)
+
cos
⁡
(
x
)
)
-
cos
⁡
(
x
)
+
sin
⁡
(
x
-
1
)
+
cos
⁡
⁡
(
x
-
1
)
+
4
x
sin
(
1
)
∫
-
1
1
‍
(
3
t
-
2
x
)
y
(
t
)
d
t
,
0
≤
x
≤
1
,
with conditions y
(
0
)
=
1, y
′
(
0
)
=
0 and exact solution y
(
x
)
=
cos
⁡
(
x
).Table6 shows the comparison between exact and approximate solutions and corrected Laguerre polynomial solutions for the different N and M values. Figure 8 shows the comparison of absolute error functions for the same N values.Table 6
Numerical solution of Example4 for different N values.
Present method and corrected Laguerre polynomial solution
x
Exact solution
y
12
y
12,15
y
22
y
22,28
y
25
y
25,28
0.0
1.000000
0.999999
1.000000
0.999999
1.000000
1.000000
1.000000
0.1
0.995004
0.996593
0.996600
0.996593
0.993964
0.996593
0.993963
0.2
0.980067
0.986641
0.986365
0.986641
0.975909
0.986641
0.975909
0.3
0.955336
0.969544
0.969444
0.969544
0.946004
0.969544
0.946003
0.4
0.921061
0.946139
0.946036
0.946139
0.904531
0.946139
0.904529
0.5
0.877583
0.916382
0.916381
0.916382
0.851886
0.916382
0.851884
0.6
0.825336
0.880504
0.880768
0.880504
0.788573
0.880504
0.788570
0.7
0.764842
0.838776
0.839522
0.838776
0.715199
0.838776
0.715195
0.8
0.696707
0.791509
0.793010
0.791509
0.632466
0.791509
0.632461
0.9
0.621610
0.739046
0.741629
0.739046
0.541167
0.739046
0.541161
1.0
0.540302
0.681764
0.685810
0.681764
0.442174
0.681764
0.442167Figure 8
Numerical and exact solution of Example4 for N
=
12, 22, 25.Table7 shows the comparison between exact and approximate solutions and corrected Laguerre polynomial solutions for the different N and M values.Table 7
Comparison of the actual and estimated absolute errors forN
=
12,22,25 and M
=
15,28,28 of Example 4.
The actual absolute and estimated absolute errors
x
E
12
E
12,15
E
22
E
22,28
E
25
E
25,28
0.0
0.69500
E
-
17
7.000000
E
-
13
0.10000
E
-
18
0.0000000000
0.000000000
0.0000000000
0.1
0.15886
E
-
03
3.681538
E
-
06
0.15886
E
-
03
1.040134
E
-
03
0.15886
E
-
03
1.040221
E
-
03
0.2
0.63428
E
-
03
2.207214
E
-
05
0.63428
E
-
03
4.157153
E
-
03
0.63428
E
-
03
4.157504
E
-
03
0.3
0.14208
E
-
02
4.966673
E
-
05
0.14208
E
-
02
9.332459
E
-
03
0.14208
E
-
02
9.333242
E
-
03
0.4
0.25078
E
-
02
5.150761
E
-
05
0.25078
E
-
02
1.652982
E
-
02
0.25078
E
-
02
1.653119
E
-
02
0.5
0.38799
E
-
02
2.543790
E
-
11
0.38799
E
-
02
2.569605
E
-
02
0.38799
E
-
02
2.569816
E
-
02
0.6
0.55168
E
-
02
1.324508
E
-
04
0.55168
E
-
02
3.676180
E
-
02
0.55168
E
-
02
3.676477
E
-
02
0.7
0.73934
E
-
02
3.734395
E
-
04
0.73934
E
-
02
4.964259
E
-
02
0.73934
E
-
02
4.964652
E
-
02
0.8
0.94802
E
-
02
7.505606
E
-
04
0.94802
E
-
02
6.423988
E
-
02
0.94802
E
-
02
6.424488
E
-
02
0.9
0.11744
E
-
01
1.291408
E
-
03
0.11744
E
-
01
8.044241
E
-
02
0.11744
E
-
01
8.044855
E
-
02
1.0
0.14146
E
-
01
2.023575
E
-
03
0.14146
E
-
01
9.812752
E
-
02
0.14146
E
-
01
9.813488
E
-
02


## 6. Conclusion

In recent years, the studies of linear functional Fredholm integro-differential equation have attracted the attention of many mathematicians and physicists. [26–28]. Laguerre collocation methods are used to solve the linear integro-differential equation with functional arguments numerically. It is observed that the method has the best advantage when the known functions in equation can be expanded to Laguerre series. Another advantage of the method is that the Laguerre polynomial coefficients of the solution are found very easily by using computer programs. Shorter computation time and lower operation count result in reduction of cumulative truncation errors and improvement of overall accuracy. To get the best approximating solution of the equation, we have improved the Laguerre collocation method with the aid of the residual error function for solving the linear Fredholm integro-differential equation with piecewise intervals. For this purpose, the power of the employed method is confirmed. As a result, considerable solutions are shown in tables and figures. We assured the correctness of the obtained solutions by putting them back into the original equation with the aid of Maple 12. As a future work, the method can also be extended to the system of linear functional Fredholm integro-differential equations with variable coefficients and their residual error analysis, but some modifications are required.



---

*Source: 682398-2014-06-18.xml*