# Numerical Techniques for Solving Linear Volterra Fractional Integral Equation

Authors: Safa’ Hamdan; Naji Qatanani; Adnan Daraghmeh
Journal: Journal of Applied Mathematics
 (2019)

---

## Abstract

Two numerical techniques, namely, Haar Wavelet and the product integration methods, have been employed to give an approximate solution of the fractional Volterra integral equation of the second kind. To test the applicability and efficiency of the numerical method, two illustrative examples with known exact solution are presented. Numerical results show clearly that the accuracy of these methods are in a good agreement with the exact solution. A comparison between these methods shows that the product integration method provides more accurate results than its counterpart.


---

## Body

## 1. Introduction

Fractional integral equations of the second kind have attracted the attention of many scientists and researchers in recent years. These equations appear frequently in heat transformations and heat radiation, population growth model, biological species living together, porous media, rheology, control, and electro-chemistry (see [1–5] for more details). The concept of fractional calculus is now considered as a partial technique in many branches of science including physics (Oldham and Spanier [6]). Recently Srivastava et al., [7] gave the model of under actuated mechanical system with fractional order derivative and Sharma [8] studied advanced generalized fractional kinetic equation in Astrophysics. Caputo refermulated the more “classic” definition of the Riemann–Liouville fractional derivatives in order to use integer order initial conditions to solve his fractional order differential Equation [9]. Kowankar and Gangal reformulated the Riemann–Liouville fractional derivative in order to differentiate nowhere differentiable fractal functions [10]. Abdou et al., [11] have implemented the Toeplitz matrix and product Nystrom methods for solving singular integral equation with logarithmic kernel and Hilbert kernel. They concluded that these methods give a very accurate approximation for these types of kernels. In [12] Hamaydi and Qatanani have solved Linear Fuzzy Volterra Integral Equation. In addition, Hamdan [13] has employed several numerical methods for solving Volterra fractional integral equations. In recent years, numerous methods have been proposed for solving fractional Volterra integral Equations [2, 14, 15]. Lepik [16] has solved fractional integral equations by the Haar wavelet method. Fractional linear multistep methods have been employed by Lubich [17] to solve Abel’s Volterra integral equation of the second kind. Bruner [18] has solved Volterra integral and related functional equations by using the collocation method. Moreover, collocation method based on the orthogonal polynomials is presented to solve fractional integral Equations [19]. Other numerical methods on fractional integral equations are hybrid collocation [20], smoothing technique [15], piecewise constant orthogonal functions approximation [14], the Galerkin method [2], Berstein’s approximation [19, 21], the Simpson 3
/
8 rule method [22], mechanical quadrature [23], Legender Pseudo spectral [24], and iterative numerical method [25]. The fractional Volterra integral equation of the second kind under investigation is defined as follows:(1)
u
x
=
f
x
+
1
Γ
a
∫
a
x
x
−
t
α
−
1
k
x
,
t
u
t
d
t
,
0
<
α
<
1
,wherek
x
,
t is called the kernel or the nucleus of the integral equation. The function u
x to be determined appears under the integral sign. The kernel k
x
,
t and the function f
x are given.In this article, two numerical techniques, namely the Harr wavelet and the product integration method are implemented to solve the Volterra fractional integral Equation (1). A comparison between these methods is carried out by solving some test examples.The paper is organized as follows: In Section2 the Haar wavelet method is introduced. The product integration method used to approximate solution of the fractional integral equation is addressed in Section 3. The proposed methods are implemented using numerical examples with known analytical solution by applying MAPLE software in Section 4. Conclusions are given in Section 5.


## 2. The Haar Wavelet Method

The Haar functions are an orthogonal family of switched rectangular waveforms where amplitudes can differ from one function to another. They are defined in the interval0
,
1. However, we are interseted in defining these functions for the more general interval A
,
B. First, we define the integer M
=
2
I (I is the maximal level of resolution). We divide the interval A
,
B into 2
M subintervals of equal length; each subinterval has the length Δ
z
=
B
−
A
/
2
M. Next, two parameters are introduced: the dilation parameter i for which i
=
0
,
1
,
…
,
I and the translation parameter c
=
0
,
1
,
…
,
m
−
1 (here the notation m
=
2
i is introduced). The wavelet number j is identified as j
=
m
+
c
+
1, (m
>
c and m
≠
c).Thej
t
h Haar wavelet is defined as:(2)
h
j
x
=
1
,
τ
1
≤
x
≤
τ
2
,
−
1
,
τ
2
≤
x
≤
τ
3
,
0
,
otherwise,where(3)
τ
1
=
A
+
2
c
μ
Δ
z
,
τ
2
=
A
+
2
c
+
1
μ
Δ
z
,
τ
3
=
A
+
2
c
+
1
μ
Δ
z
,
μ
=
M
m
.Forj
=
1 and x
∈
0
,
1, the function h
1
x is the scaling function or the father wavelet for the family of the Haar wavelets which is defined as(4)
h
1
x
=
1
,
0
≤
x
<
1
,
0
,
otherwise
.Forj
=
2 and x
∈
0
,
1, the function h
2
x is the mother wavelet for the family of the Haar wavelet which is defined as(5)
h
2
x
=
1
,
0
≤
x
<
1
2
,
−
1
,
1
2
≤
x
<
1
,
0
,
otherwise
.Forj
=
3 and x
∈
0
,
1, the function h
3
x is defined as(6)
h
3
x
=
1
,
0
≤
x
<
1
4
,
−
1
,
1
4
≤
x
<
2
4
,
0
,
otherwise
.Forj
=
4 and x
∈
0
,
1, the function h
4
x is defined as(7)
h
4
x
=
1
,
2
4
≤
x
<
3
4
,
−
1
,
3
4
≤
x
<
4
4
,
0
,
otherwise
.In fact Table1 shows the index computations for Harr basis functions. The Haar wavelet series expansion of a given function g
x is given as [24]Table 1
Index computations for Haar basis function.
i
0
1
1
2
2
2
2
⋯
⋯
c
0
0
1
0
1
2
3
⋯
⋯
j
=
2
i
+
c
+
1
2
3
4
5
6
7
8
⋯
⋯(8)
g
x
=
∑
j
=
1
2
M
b
j
h
j
x
,wherex
∈
A
,
B and b
j
′
s are the wavelet coefficients. There are many methods for computing b
j
′
s, one of these is the collocation method.The collocation points are defined as(9)
x
k
=
A
+
k
−
0.5
Δ
z
,
k
=
1
,
2
,
…
,
2
M
.Substituting Equation (9) into Equation (8) gives the discrete version, i.e.,(10)
g
x
k
=
∑
j
=
1
2
M
b
j
h
j
x
k
.It is convenient to put Equation (10) into the matrix form as(11)
g
=
b
H
,whereg and b are 2
M-dimensional row vectors and the Haar coefficients H
j
,
k
=
h
j
x
k is the element of a 2
M
×
2
M matrix.Now, in virtue of Equation (9) and upon substituting Equation (10) into Equation (1) we get(12)
g
x
k
=
f
x
k
+
1
Γ
α
∫
0
x
k
x
k
−
t
α
−
1
k
x
k
,
t
g
t
d
tor(13)
∑
j
=
1
2
M
b
j
h
j
x
k
=
f
x
k
+
1
Γ
α
∫
0
x
k
x
k
−
t
α
−
1
k
x
k
,
t
∑
j
=
1
2
M
b
j
h
j
t
d
t
,
(14)
f
x
k
=
∑
j
=
1
2
M
b
j
h
j
x
k
−
1
Γ
α
∫
0
x
k
x
k
−
t
α
−
1
k
x
k
,
t
∑
j
=
1
2
M
b
j
h
j
t
d
t
,
(15)
f
x
k
=
∑
j
=
1
2
M
b
j
h
j
x
k
−
1
Γ
α
∫
0
x
k
x
k
−
t
α
−
1
k
x
k
,
t
h
j
t
d
t
.Assume that(16)
u
j
x
k
=
1
Γ
α
∫
0
x
k
x
k
−
t
α
−
1
k
x
k
,
t
h
j
t
d
tthen Equation (15) becomes(17)
f
x
k
=
∑
j
=
1
2
M
b
j
h
j
x
k
−
u
j
x
k
,
k
=
1
,
2
,
…
,
2
M
.The matrix form of Equation (17) is(18)
F
=
b
H
−
U
,whereU
j
,
k
=
u
j
x
k
,
F
k
=
f
x
k.The solution of Equation (18) is(19)
b
=
F
H
−
U
−
1
.Upon using Equation (2), then Equation (16) can be rewritten in the following form(20)
U
j
,
k
=
0
,
x
k
<
τ
1
,
U
j
,
k
=
1
Γ
α
∫
τ
1
x
k
x
k
−
t
α
−
1
k
x
k
,
t
d
t
,
τ
1
≤
x
≤
τ
2
,
U
j
,
k
=
1
Γ
α
∫
τ
1
τ
2
x
k
−
t
α
−
1
k
x
k
,
t
d
t
−
∫
τ
2
x
k
x
k
−
t
α
−
1
k
x
k
,
t
d
t
,
τ
2
≤
x
≤
τ
3
,
U
j
,
k
=
1
Γ
α
∫
τ
1
τ
2
x
k
−
t
α
−
1
k
x
k
,
t
d
t
−
∫
τ
2
τ
2
x
k
−
t
α
−
1
k
x
k
,
t
d
t
,
τ
3
≤
x
k
.


## 3. The Product Integration Method

We introduce the product integration method that can be used to solve linear Volterra fractional integral equation of the second kind.To use the product integration method as a numerical technique [3, 10], we consider the linear Volterra integral equation of the second kind which has the general form:(21)
g
x
=
f
x
+
λ
∫
a
x
k
x
,
t
g
t
d
t
,
a
≤
x
≤
b
,wherek
x
,
t is the kernel (and is known ), λ is a constant parameter and g
x is unknown function to be determined. This method is based on factoring of the singularity of the kernel k
x
,
t as follows:(22)
k
x
,
t
=
p
x
,
t
G
x
,
t
,whereG
x
,
t is well-behaved function and p
x
,
t is badly behaved function.Set Equation (22) into Equation (21) yields(23)
g
x
=
f
x
+
λ
∫
a
x
p
x
,
t
G
x
,
t
g
t
d
t
,
a
≤
x
≤
b
.First, we decompose the intervala
,
b into N subintervals h
i where:(24)
h
i
=
x
i
+
1
−
x
i
,
i
=
0
,
1
,
2
,
…
,
N
−
1and(25)
a
=
x
0
<
x
1
<
⋯
<
x
N
=
b
.If we use theN-points quadrature rule and collocate Equation (23) at the nodes {
x
i
}
i
=
1
N
∪
{
0
}, we get(26)
g
x
i
=
f
x
i
+
λ
∫
a
x
i
p
x
i
,
t
G
x
i
,
t
g
t
d
t
,
i
=
0
,
1
,
2
,
…
,
N
.Using Lagrange interpolation polynomial, Equation (26) can be written as:(27)
g
x
i
=
f
x
i
+
λ
∑
j
=
0
i
w
i
j
G
x
i
,
t
j
g
t
j
,wheret
i
=
x
i
=
a
+
i
h for i
=
0
,
1
,
…
,
N, with h
=
b
−
a
/
N and w
i
j are the weights which can be determined directly. Approximating the integral terms by a product integration using composite trapezoidal rule, where x
=
x
i, we get:(28)
∫
a
x
i
p
x
i
,
t
G
x
i
,
t
g
t
d
t
≈
∑
j
=
0
i
w
i
j
G
x
i
,
t
j
g
t
j
,
i
=
0
,
1
,
2
,
⋯
,
N
.It follows that(29)
∫
a
x
i
p
x
i
,
t
G
x
i
,
t
g
t
d
t
=
∑
j
=
0
i
−
1
∫
t
j
t
j
+
1
p
x
i
,
t
G
x
i
,
t
g
t
d
t
≈
∑
j
=
0
i
−
1
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
G
x
i
,
t
j
g
t
j
+
t
−
t
j
+
1
h
j
G
x
i
,
t
j
+
1
g
t
j
d
t
=
∑
j
=
0
i
w
i
j
G
x
i
,
t
j
g
t
j
,where(30)
w
i
0
=
∫
t
0
t
1
p
x
i
,
t
t
1
−
t
h
0
d
t
,
f
o
r
j
=
0
(31)
w
i
j
=
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
d
t
+
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
f
o
r
j
=
1
,
…
,
i
−
1
,
(32)
w
i
i
=
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
f
o
r
j
=
i
,The approximate solution to Equation (26) is determined recursively using:(33)
g
̂
x
i
=
f
x
i
+
λ
∑
j
=
0
i
w
i
j
G
x
i
,
t
j
g
t
j
,
i
=
1
,
2
,
…
,
Nwith(34)
g
̂
n
x
0
=
f
a
.Hence, Equation (26) yields a system of linear algebraic equations:(35)
B
G
̂
n
=
F
.Solving Equation (35) gives(36)
G
̂
n
=
B
−
1
F
,where the matrixB
=
b
i
j
,
0
≤
i
≤
N
,
0
≤
j
≤
i
−
1 with:(37)
b
00
=
1
b
i
j
=
0
,
∀
j
≤
i
+
1
b
0
j
=
0
,
∀
j
≥
1
b
i
j
=
0
,
∀
j
>
i
b
i
0
=
λ
w
i
0
G
x
i
,
x
0
,
j
=
0
b
i
i
=
1
−
λ
w
i
i
G
x
i
,
x
i
,
j
=
i
b
i
j
=
λ
w
i
j
G
x
i
,
x
j
,
j
=
0
,
1
,
…
,
i
−
1
(38)
B
=
1
0
0
0
…
0
b
10
b
11
0
0
…
0
b
20
b
21
b
22
0
…
0
⋮
⋮
⋮
⋱
⋮
⋮
b
N
0
b
N
1
b
N
2
0
…
b
N
N
,
(39)
F
=
f
x
0
=
f
a
,
f
x
1
,
…
,
f
x
N
=
f
b
T
,
(40)
G
̂
n
=
g
̂
x
0
=
g
̂
a
,
g
̂
x
1
,
…
,
g
̂
x
N
=
g
̂
b
T
.HereB is a lower triangular matrix N
+
1
∗
N
+
1 and 1
−
λ
w
i
i
G
x
i
,
x
i
≠
0.


### 3.1. Uniqueness of a Solution

To prove the uniqueness of the solution, we can apply the Cauchy–Minkoviski inequality (see [11] for more details) and get λ
≤
1 which represents a sufficient condition to have a unique solution, and the value of g
i satisfies the inequality:(41)
g
i
≤
f
i
1
−
λ
.


## 3.1. Uniqueness of a Solution

To prove the uniqueness of the solution, we can apply the Cauchy–Minkoviski inequality (see [11] for more details) and get λ
≤
1 which represents a sufficient condition to have a unique solution, and the value of g
i satisfies the inequality:(41)
g
i
≤
f
i
1
−
λ
.


## 4. Numerical Examples and Results

In this section, in order to examine the accuracy of the proposed methods, we solve two numerical examples of fractional Volterra integral equations. Moreover, the numerical results will be compared with exact solution.Example 1.
Consider the linear Volterra fractional integral equation of the second kind:(42)
g
x
=
x
1
/
2
−
3
4
π
x
2
+
8
3
π
x
3
/
2
−
1
+
1
π
∫
0
x
x
−
t
−
1
2
2
t
g
t
d
t
.
The analytical solution of Equation (42) is:(43)
g
x
=
x
1
/
2
−
1
.We implement Algorithm1 to solve Equation (42) using the Haar wavelet method. Table 2 displays the exact and the numerical results using the Haar wavelet method for Equation (42) with Δ
z
=
0.0156 and N
=
64. The maximum error with N
=
64 is 8.8745
e
−
04. Figure 1(a) compares both the exact and numerical solutions for the Volterra fractional integral Equation (42). Moreover, Figure 1(b) shows the absolute error between exact and numerical solutions. Now, we implement Algorithm 2 to solve Equation (42) using the Product Integration Method. We obtain the following results:Table 2
The exact and numerical solutions using the Haar wavelet method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0.0078
−0.9116116523
−0.9116009499
1.0702
e
−
05
0.1172
−0.6576734015
−0.6576324164
4.0985
e
−
05
0.2266
−0.5240141808
−0.5239489931
6.5188
e
−
05
0.3359
−0.4203988440
−0.4203067536
9.2090
e
−
05
0.4453
−0.3326826092
−0.3325562333
1.2638
e
−
04
0.5547
−0.2552265445
−0.2550525941
1.7395
e
−
04
0.6641
−0.1850996993
−0.1848551078
2.4459
e
−
04
0.7734
−0.1205470450
−0.1201906958
3.5635
e
−
04
0.8828
−0.0604189763
−0.0598738640
5.4511
e
−
04
0.9922
−0.0039139093
−0.0030264584
8.8745
e
−
04Algorithm 1: Numerical realization using the Haar wavelet method.
1: Input:•A
,
B: A
,
B is the interval for the solution function.
•I: The maximal level of resolution.
•i: The dilation parameter.
•c: The translation parameter.
•f
x: The function of the integral equation.
•k
x
,
t: The kernel function.
•h
1
x: The scaling function.
•λ: Is a constant parameter.
2: Calculate:• The quantityM
=
2
I.
•N
=
2
M
• The length of each subintervalsΔ
z
=
B
−
A
N.
•m
=
2
i
• The wavelet numberj
=
m
+
c
+
1.
3: Calculate: The collocation pointsx
k
=
A
+
k
−
0.5
Δ
z.
4: Calculate:h
j
x
k
,
u
j
x
k and f
x
k.
5: Calculate:• The matrixH
−
U.
• Calculateb
=
F
H
−
U
−
1.
6: Determine the solution of the linear systemg
=
b
H.Algorithm 2:Numerical realization using the product integration method.
1: Input:•a
,
b: a
,
b is the interval for the solution function.
•N: The number of subdivisions of a
,
b.
•f
x: The function of the integral equation.
•G
x
,
t
,
p
x
,
t: The kernel functions.
•λ: Is a constant parameter.
• Setx
0
=
a and x
N
=
b.
2: Calculate:•h
=
b
−
a
N
•t
i
=
x
i
=
a
+
i
h
•h
i
=
x
i
+
1
−
x
i.
3: Calculate:w
i
0
,
w
i
j
,
w
i
i
w
i
0
=
∫
t
0
t
1
p
x
i
,
t
t
1
−
t
h
0
d
t
,
f
o
r
j
=
0
w
i
j
=
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
d
t
+
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
f
o
r
j
=
1
,
⋯
,
i
−
1
w
i
i
=
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
f
o
r
j
=
i
4: Solve the recurrence relation:
g
̂
n
x
0
=
f
x
0
=
f
a
g
̂
x
i
=
f
x
i
+
λ
∑
j
=
0
i
w
i
j
G
x
i
,
t
j
g
t
j
,
i
=
1
,
2
,
⋯
,
N
or we can also solve the linear system of algebraic equations
G
̂
n
=
B
−
1
F


### 4.1. W Function

Ifp
x
i
,
t
=
x
i
−
t
−
1
/
2, then we have(44)
w
i
0
=
1
h
0
∫
t
0
t
1
p
x
i
,
t
t
1
−
t
d
t
,
f
o
r
j
=
0
,
(45)
w
i
0
=
1
h
∫
t
0
t
1
x
i
−
t
−
1
/
2
h
−
t
d
t
,
(46)
w
i
0
=
1
h
∫
t
0
t
1
i
h
−
t
−
1
/
2
h
−
t
d
t
,
(47)
w
i
0
=
2
3
h
1
/
2
3
i
1
/
2
+
2
i
−
1
3
/
2
−
i
3
/
2
,
(48)
w
i
j
=
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
d
t
+
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
j
=
1
,
…
,
i
−
1
,
(49)
w
i
j
=
∫
j
h
j
+
1
h
x
i
−
t
−
1
/
2
j
+
1
h
−
t
h
d
t
+
∫
j
−
1
h
j
h
x
i
−
t
−
1
/
2
t
−
j
−
1
h
h
d
t
,
(50)
w
i
j
=
4
3
h
1
/
2
i
−
j
−
1
3
/
2
+
i
−
j
+
1
3
/
2
−
2
i
−
j
3
/
2
,
(51)
w
i
i
=
1
h
i
−
1
∫
t
i
−
1
t
i
p
x
i
,
t
t
−
t
i
−
1
d
t
,
j
=
i
,
(52)
w
i
i
=
1
h
∫
i
−
1
h
i
h
x
i
−
t
−
1
/
2
t
−
i
−
1
h
d
t
,
(53)
w
i
i
=
1
h
∫
i
−
1
h
i
h
i
h
−
t
−
1
/
2
t
−
i
−
1
h
d
t
,
(54)
w
i
i
=
4
3
h
1
/
2
.From above equations we obtain the linear system of algebraic equations:(55)
G
n
=
B
−
1
F
.Table3 displays the exact and the numerical results using the product integration method for Equation (42) with h
=
0.0156. The maximum error with N
=
64 is 4.7435
e
−
04. Figure 2(a) compares both the analytical and numerical solutions for Equation (42) with N
=
64. Moreover, Figure 2(b) shows the absolute error between exact and numerical solutions.Table 3
The exact and numerical solutions using the product integration method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0
−1
−1
0
0.125
−0.6464466094
−0.6463939905
5.2619
e
−
05
0.250
−0.5000000000
−0.4999385093
6.1491
e
−
05
0.375
−0.3876275643
−0.3875532017
7.4363
e
−
05
0.500
−0.2928932188
−0.2927990146
9.4204
e
−
05
0.625
−0.2094305849
−0.2093044528
1.2613
e
−
04
0.750
−0.1339745962
−0.1337942129
1.8038
e
−
04
0.875
−0.0645856533
−0.0643065527
2.7910
e
−
04
1
0
4.7435
e
−
04
4.7435
e
−
04Figure 1
A comparison and absolute error between exact and numerical solutions in Example1 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)Figure 2
A comparison and absolute error between exact and numerical solutions in Example1 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)Example 2.
Consider the Abel’s integral equation of the second kind:(56)
g
x
=
1
3
x
2
−
x
+
0.2399
x
7
/
3
+
0.8399
x
4
/
3
+
0.3733
∫
0
x
x
−
t
−
2
/
3
g
t
d
t
.
The exact solution of Equation (56) is:(57)
g
x
=
1
3
x
2
−
x
.Applying Algorithms1 to Equation (56), we obtain the following results. Table 4 displays the exact and the numerical results using the Haar wavelet method for Equation (56) with Δ
z
=
0.0156 and N
=
64. The maximum error with N
=
64 is 1.1113
e
−
03. Figure 3(a) compares both the exact and numerical solutions for the Abel’s integral equation (56). Moreover, Figure 3(b) shows that the absolute error between exact and numerical solutions. Applying Algorithm 2 to Equation (56) using the product integration method, we getTable 4
The exact and numerical solutions using the Haar wavelet method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0.0078
−0.0077921549
−0.0083481389
5.5598
e
−
04
0.1172
−0.1126098632
−0.1131662257
5.5636
e
−
04
0.2266
−0.2094523111
−0.2100916889
6.3938
e
−
04
0.3359
−0.2983194986
−0.2990367942
7.1730
e
−
04
0.4453
−0.3792114257
−0.3800030241
7.9160
e
−
04
0.5547
−0.4521280924
−0.4529904969
8.6240
e
−
04
0.6641
−0.5170694986
−0.5179996508
9.3015
e
−
04
0.7734
−0.5740356445
−0.5750303556
9.9471
e
−
04
0.8828
−0.6230265299
−0.6240817729
1.0552
e
−
03
0.9922
−0.6640421549
−0.6651534675
1.1113
e
−
03Figure 3
A comparison and absolute error between exact and numerical solutions in Example2 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)


### 4.2. W Function

Ifp
x
i
,
t
=
x
i
−
t
−
2
/
3, then we have(58)
w
i
0
=
1
h
0
∫
t
0
t
1
p
x
i
,
t
t
1
−
t
d
t
,
f
o
r
j
=
0
,
(59)
w
i
0
=
1
h
∫
t
0
t
1
x
i
−
t
−
2
/
3
h
−
t
d
t
,
(60)
w
i
0
=
1
h
∫
t
0
t
1
i
h
−
t
−
2
/
3
h
−
t
d
t
,
(61)
w
i
0
=
3
4
h
1
/
3
4
i
1
/
3
+
3
i
−
1
4
/
3
−
i
4
/
3
,
(62)
w
i
j
=
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
d
t
+
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
j
=
1
,
…
,
i
−
1
,
(63)
w
i
j
=
∫
j
h
j
+
1
h
x
i
−
t
−
2
/
3
j
+
1
h
−
t
h
d
t
+
∫
j
−
1
h
j
h
x
i
−
t
−
2
/
3
t
−
j
−
1
h
h
d
t
,
(64)
w
i
j
=
∫
j
h
j
+
1
h
i
h
−
t
−
2
/
3
j
+
1
h
−
t
h
d
t
+
∫
j
−
1
h
j
h
i
h
−
t
−
2
/
3
t
−
j
−
1
h
h
d
t
,
(65)
w
i
j
=
9
4
h
1
/
3
i
−
j
−
1
4
/
3
+
i
−
j
+
1
4
/
3
−
2
i
−
j
4
/
3
,
(66)
w
i
i
=
1
h
i
−
1
∫
t
i
−
1
t
i
p
x
i
,
t
t
−
t
i
−
1
d
t
,
j
=
i
,
(67)
w
i
i
=
1
h
∫
i
−
1
h
i
h
x
i
−
t
−
2
/
3
t
−
i
−
1
h
d
t
,
(68)
w
i
i
=
1
h
∫
i
−
1
h
i
h
i
h
−
t
−
2
/
3
t
−
i
−
1
h
d
t
,
(69)
w
i
i
=
9
4
h
1
/
3
.Table5 displays the exact and the numerical results using the product integration method for Equation (56) with h
=
0.0156. The maximum error with N
=
64 is 2.1385
e
−
04. Figure 4(a) compares both the analytical and numerical solutions for Equation (56) with N
=
64. Moreover, Figure 4(b) shows the absolute error between exact and numerical solutions.Table 5
The exact and numerical solutions using the product integration method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0
0
0
0
0.125
−0.1197916666
−0.1197818969
9.7697
e
−
05
0.250
−0.2291666666
−0.2291501145
1.6552
e
−
05
0.375
−0.3281250000
−0.3280982056
2.6794
e
−
05
0.500
−0.4166666666
−0.4166237294
4.2937
e
−
05
0.625
−0.4947916666
−0.4947244207
6.7246
e
−
04
0.750
−0.5625000000
−0.5623979179
1.0208
e
−
04
0.875
−0.6197916666
−0.6196416593
1.5001
e
−
04
1
−0.6666666666
−0.6664528164
2.1385
e
−
04Figure 4
A comparison and absolute error between exact and numerical solutions in Example2 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)


## 4.1. W Function

Ifp
x
i
,
t
=
x
i
−
t
−
1
/
2, then we have(44)
w
i
0
=
1
h
0
∫
t
0
t
1
p
x
i
,
t
t
1
−
t
d
t
,
f
o
r
j
=
0
,
(45)
w
i
0
=
1
h
∫
t
0
t
1
x
i
−
t
−
1
/
2
h
−
t
d
t
,
(46)
w
i
0
=
1
h
∫
t
0
t
1
i
h
−
t
−
1
/
2
h
−
t
d
t
,
(47)
w
i
0
=
2
3
h
1
/
2
3
i
1
/
2
+
2
i
−
1
3
/
2
−
i
3
/
2
,
(48)
w
i
j
=
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
d
t
+
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
j
=
1
,
…
,
i
−
1
,
(49)
w
i
j
=
∫
j
h
j
+
1
h
x
i
−
t
−
1
/
2
j
+
1
h
−
t
h
d
t
+
∫
j
−
1
h
j
h
x
i
−
t
−
1
/
2
t
−
j
−
1
h
h
d
t
,
(50)
w
i
j
=
4
3
h
1
/
2
i
−
j
−
1
3
/
2
+
i
−
j
+
1
3
/
2
−
2
i
−
j
3
/
2
,
(51)
w
i
i
=
1
h
i
−
1
∫
t
i
−
1
t
i
p
x
i
,
t
t
−
t
i
−
1
d
t
,
j
=
i
,
(52)
w
i
i
=
1
h
∫
i
−
1
h
i
h
x
i
−
t
−
1
/
2
t
−
i
−
1
h
d
t
,
(53)
w
i
i
=
1
h
∫
i
−
1
h
i
h
i
h
−
t
−
1
/
2
t
−
i
−
1
h
d
t
,
(54)
w
i
i
=
4
3
h
1
/
2
.From above equations we obtain the linear system of algebraic equations:(55)
G
n
=
B
−
1
F
.Table3 displays the exact and the numerical results using the product integration method for Equation (42) with h
=
0.0156. The maximum error with N
=
64 is 4.7435
e
−
04. Figure 2(a) compares both the analytical and numerical solutions for Equation (42) with N
=
64. Moreover, Figure 2(b) shows the absolute error between exact and numerical solutions.Table 3
The exact and numerical solutions using the product integration method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0
−1
−1
0
0.125
−0.6464466094
−0.6463939905
5.2619
e
−
05
0.250
−0.5000000000
−0.4999385093
6.1491
e
−
05
0.375
−0.3876275643
−0.3875532017
7.4363
e
−
05
0.500
−0.2928932188
−0.2927990146
9.4204
e
−
05
0.625
−0.2094305849
−0.2093044528
1.2613
e
−
04
0.750
−0.1339745962
−0.1337942129
1.8038
e
−
04
0.875
−0.0645856533
−0.0643065527
2.7910
e
−
04
1
0
4.7435
e
−
04
4.7435
e
−
04Figure 1
A comparison and absolute error between exact and numerical solutions in Example1 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)Figure 2
A comparison and absolute error between exact and numerical solutions in Example1 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)Example 2.
Consider the Abel’s integral equation of the second kind:(56)
g
x
=
1
3
x
2
−
x
+
0.2399
x
7
/
3
+
0.8399
x
4
/
3
+
0.3733
∫
0
x
x
−
t
−
2
/
3
g
t
d
t
.
The exact solution of Equation (56) is:(57)
g
x
=
1
3
x
2
−
x
.Applying Algorithms1 to Equation (56), we obtain the following results. Table 4 displays the exact and the numerical results using the Haar wavelet method for Equation (56) with Δ
z
=
0.0156 and N
=
64. The maximum error with N
=
64 is 1.1113
e
−
03. Figure 3(a) compares both the exact and numerical solutions for the Abel’s integral equation (56). Moreover, Figure 3(b) shows that the absolute error between exact and numerical solutions. Applying Algorithm 2 to Equation (56) using the product integration method, we getTable 4
The exact and numerical solutions using the Haar wavelet method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0.0078
−0.0077921549
−0.0083481389
5.5598
e
−
04
0.1172
−0.1126098632
−0.1131662257
5.5636
e
−
04
0.2266
−0.2094523111
−0.2100916889
6.3938
e
−
04
0.3359
−0.2983194986
−0.2990367942
7.1730
e
−
04
0.4453
−0.3792114257
−0.3800030241
7.9160
e
−
04
0.5547
−0.4521280924
−0.4529904969
8.6240
e
−
04
0.6641
−0.5170694986
−0.5179996508
9.3015
e
−
04
0.7734
−0.5740356445
−0.5750303556
9.9471
e
−
04
0.8828
−0.6230265299
−0.6240817729
1.0552
e
−
03
0.9922
−0.6640421549
−0.6651534675
1.1113
e
−
03Figure 3
A comparison and absolute error between exact and numerical solutions in Example2 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)


## 4.2. W Function

Ifp
x
i
,
t
=
x
i
−
t
−
2
/
3, then we have(58)
w
i
0
=
1
h
0
∫
t
0
t
1
p
x
i
,
t
t
1
−
t
d
t
,
f
o
r
j
=
0
,
(59)
w
i
0
=
1
h
∫
t
0
t
1
x
i
−
t
−
2
/
3
h
−
t
d
t
,
(60)
w
i
0
=
1
h
∫
t
0
t
1
i
h
−
t
−
2
/
3
h
−
t
d
t
,
(61)
w
i
0
=
3
4
h
1
/
3
4
i
1
/
3
+
3
i
−
1
4
/
3
−
i
4
/
3
,
(62)
w
i
j
=
∫
t
j
t
j
+
1
p
x
i
,
t
t
j
+
1
−
t
h
j
d
t
+
∫
t
j
−
1
t
j
p
x
i
,
t
t
−
t
j
−
1
h
j
−
1
d
t
,
j
=
1
,
…
,
i
−
1
,
(63)
w
i
j
=
∫
j
h
j
+
1
h
x
i
−
t
−
2
/
3
j
+
1
h
−
t
h
d
t
+
∫
j
−
1
h
j
h
x
i
−
t
−
2
/
3
t
−
j
−
1
h
h
d
t
,
(64)
w
i
j
=
∫
j
h
j
+
1
h
i
h
−
t
−
2
/
3
j
+
1
h
−
t
h
d
t
+
∫
j
−
1
h
j
h
i
h
−
t
−
2
/
3
t
−
j
−
1
h
h
d
t
,
(65)
w
i
j
=
9
4
h
1
/
3
i
−
j
−
1
4
/
3
+
i
−
j
+
1
4
/
3
−
2
i
−
j
4
/
3
,
(66)
w
i
i
=
1
h
i
−
1
∫
t
i
−
1
t
i
p
x
i
,
t
t
−
t
i
−
1
d
t
,
j
=
i
,
(67)
w
i
i
=
1
h
∫
i
−
1
h
i
h
x
i
−
t
−
2
/
3
t
−
i
−
1
h
d
t
,
(68)
w
i
i
=
1
h
∫
i
−
1
h
i
h
i
h
−
t
−
2
/
3
t
−
i
−
1
h
d
t
,
(69)
w
i
i
=
9
4
h
1
/
3
.Table5 displays the exact and the numerical results using the product integration method for Equation (56) with h
=
0.0156. The maximum error with N
=
64 is 2.1385
e
−
04. Figure 4(a) compares both the analytical and numerical solutions for Equation (56) with N
=
64. Moreover, Figure 4(b) shows the absolute error between exact and numerical solutions.Table 5
The exact and numerical solutions using the product integration method withN
=
64.
x
k
Exact solutiong
x
=
x
1
/
2
−
1
Numerical solutiong
h
x
E
r
r
o
r
:
|
g
−
g
h
|
0
0
0
0
0.125
−0.1197916666
−0.1197818969
9.7697
e
−
05
0.250
−0.2291666666
−0.2291501145
1.6552
e
−
05
0.375
−0.3281250000
−0.3280982056
2.6794
e
−
05
0.500
−0.4166666666
−0.4166237294
4.2937
e
−
05
0.625
−0.4947916666
−0.4947244207
6.7246
e
−
04
0.750
−0.5625000000
−0.5623979179
1.0208
e
−
04
0.875
−0.6197916666
−0.6196416593
1.5001
e
−
04
1
−0.6666666666
−0.6664528164
2.1385
e
−
04Figure 4
A comparison and absolute error between exact and numerical solutions in Example2 with N
=
64. (a) A comparison. (b) Absolute error.
(a)
(b)


## 5. Conclusions

In this article, two numerical schemes namely, the Harr wavelet and the product integration methods have been employed to solve Volterra fractional integral equation of the second kind. The advantages and flexibility of these techniques are demonstrated with a variety of numerical examples. The results show that both techniques are in a good agreement with the analytical solution. According to comparison of numerical results, mentioned in tables and figures, we conclude that the product integration method provides more accurate results than its counterpart and therefore is more advantageous. Moreover, in comparison of our numerical results with those presented in the literature in particular Abdo et al., [11], we observe that the Toeplitz matrix and product Nystrom methods are more effective for integral equations with logarithmic and Hilbert kernels. However, the Haar wavelet and the product integration methods are more suitable for weakly singular kernels.



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