# Approximate Solution of Volterra-Stieltjes Linear Integral Equations of the Second Kind with the Generalized Trapezoid Rule

Authors: Avyt Asanov; Elman Hazar; Mustafa Eroz; Kalyskan Matanova; Elmira Abdyldaeva
Journal: Advances in Mathematical Physics
 (2016)

---

## Abstract

The numerical solution of linear Volterra-Stieltjes integral equations of the second kind by using the generalized trapezoid rule is established and investigated. Also, the conditions on estimation of the error are determined and proved. A selected example is solved employing the proposed method.


---

## Body

## 1. Introduction

Various issues concerning Volterra and Volterra-Stieltjes integral equations were studied in [1–13]. Some practical and theoretical investigations were made in paper [1] for nonclassical Volterra integral equations of the first kind. Also, the approximate solution for the integral equation considered is obtained. In paper [2], various inverse problems including Volterra operator equations were studied. Some properties for Volterra-Stieltjes integral operators were given in [3]. In the studies [6, 7], existence and uniqueness of the solutions were given for Volterra integral and Volterra operator equations of the first and the second kinds. In papers [4, 6], quadratic integral equations of Urysohn-Stieltjes type and their applications were investigated. Various numerical solution methods for integral equations were presented in the studies [8–13]. The notion of derivative of a function by means of a strictly increasing function was given by Asanov in [14]. In the study [15], the generalized trapezoid rule was proposed to evaluate the Stieltjes integral approximately by employing the notion of derivative of a function by means of a strictly increasing function.In this study, we investigate the numerical solution of linear Volterra-Stieltjes integral equations of the second kind by using the generalized trapezoid rule. Therefore, we need the concept of the derivative defined in the works [14, 15] and theorems connected with it.


## 2. Approximating Volterra-Stieltjes Integral Equations

Consider the linear integral equation of the second kind(1)
u
x
=
∫
a
x
K
x
,
s
u
s
d
φ
s
+
f
x
,
x
∈
a
,
b
,where K
x
,
s is a given continuous function on G
=
x
,
s
:
a
≤
s
≤
x
≤
b
,
f
x are given continuous functions on a
,
b
,
φ
s is a given strictly increasing continuous function on a
,
b, and u
x is the sought function on a
,
b.Definition 1.
The derivative of a functionf
x with respect to φ
x is the function f
φ
′
x, whose value at x
∈
a
,
b is the number(2)
f
φ
′
x
=
lim
Δ
→
0
⁡
f
x
+
Δ
-
f
x
φ
x
+
Δ
-
φ
x
,where φ
x is a given strictly increasing continuous function in a
,
b.If the limit in (2) exists, we say that f
x has a derivative (is differentiable) with respect to φ
x
. The first derivative f
φ
′
x may also be a differentiable function with respect to φ
x at every point x
∈
a
,
b. Then, its derivative(3)
f
φ
′
′
x
=
f
φ
′
x
φ
′is called the second derivative of f
x with respect to φ
x. Consequently, the nth derivative of f
x with respect to φ
x is defined by(4)
f
φ
n
x
=
f
φ
n
-
1
x
φ
′
.We need the following theorem which is given in [15].Theorem 2.
Letφ
x and ψ
x be two strictly increasing continuous functions on a
,
b and f
φ
′
′
x
,
f
ψ
′
′
x
∈
C
a
,
b. Then,(5)
I
-
A
n
=
M
0
12
φ
b
-
φ
a
ω
φ
h
2
+
M
0
′
12
ψ
b
-
ψ
a
ω
ψ
h
2
,where(6)
I
=
∫
a
b
f
x
d
φ
x
-
∫
a
b
f
x
d
ψ
x
,
M
0
=
f
φ
′
′
x
C
=
sup
x
∈
a
,
b
⁡
f
φ
′
′
x
,
M
0
′
=
f
ψ
′
′
x
C
=
sup
x
∈
a
,
b
⁡
f
ψ
′
′
x
,
ω
φ
h
=
sup
x
-
y
≤
h
⁡
φ
x
-
φ
y
,
ω
ψ
h
=
sup
x
-
y
≤
h
⁡
ψ
x
-
ψ
y
,
A
n
=
1
2
∑
i
=
1
n
f
x
i
+
f
x
i
-
1
φ
x
i
-
φ
x
i
-
1
-
1
2
∑
i
=
1
n
f
x
i
+
f
x
i
-
1
ψ
x
i
-
ψ
x
i
-
1
,and x
i
=
a
+
i
h
,
i
=
0,1
,
…
,
n
,
h
=
b
-
a
/
n
,
n
∈
N (N denotes the set of natural numbers).Corollary 3.
Letφ
x be a strictly increasing continuous function on a
,
b, ψ
x
=
0 for all x
∈
a
,
b and f
φ
′
′
x
∈
C
a
,
b. Then,(7)
I
-
A
n
≤
M
0
12
φ
b
-
φ
a
ω
φ
h
2
,
I
i
-
M
i
≤
M
0
12
φ
x
i
-
φ
x
i
-
1
3
,
i
=
1,2
,
…
,
n
,where(8)
I
i
=
∫
x
i
-
1
x
i
f
x
d
φ
x
,
M
i
=
1
2
f
x
i
+
f
x
i
-
1
φ
x
i
-
φ
x
i
-
1
.Theorem 4.
Letφ
x be a strictly increasing continuous function on a
,
b
,
K
x
,
s
∈
C
G, and f
x
∈
C
a
,
b. Then, the integral equation (1) has a unique solution u
x
∈
C
a
,
b and(9)
u
x
C
≤
c
1
f
x
C
,where c
1
=
exp
⁡
K
0
φ
b
-
φ
a and K
0
=
K
x
,
s
C
=
sup
x
,
s
∈
G
⁡
K
x
,
s.Then, we will need the following theorem which is given in [16].Theorem 5.
LetF
x
,
s
,
F
φ
x
′
x
,
s
∈
C
G
,
φ
x be strictly increasing continuous functions on a
,
b, and P
x
=
∫
a
x
F
x
,
s
d
φ
s
,
x
∈
a
,
b. Then,(10)
P
φ
x
′
x
=
F
x
,
x
+
∫
a
x
F
φ
x
′
x
,
s
d
φ
s
,
x
∈
a
,
b
,where(11)
F
φ
x
′
x
,
s
=
l
i
m
Δ
x
→
0
F
x
+
Δ
x
,
s
-
F
x
,
s
φ
x
+
Δ
x
-
φ
x
,
x
,
s
∈
x
,
s
:
a
<
s
<
x
<
b
,
P
φ
x
′
a
=
lim
Δ
x
→
0
+
⁡
P
a
+
Δ
x
-
P
a
φ
a
+
Δ
x
-
φ
a
,
P
φ
x
′
b
=
lim
Δ
x
→
0
-
⁡
P
b
+
Δ
x
-
P
b
φ
b
+
Δ
x
-
φ
b
.Corollary 6.
Letu
x
∈
C
a
,
b be a solution of the integral equation (1), K
φ
x
′
x
,
s
∈
C
G, and f
φ
x
′
x
∈
C
a
,
b. Then, u
φ
x
′
x
∈
C
a
,
b and(12)
u
φ
x
′
x
=
K
x
,
x
u
x
+
∫
a
x
K
φ
x
′
x
,
s
u
s
d
φ
s
+
f
φ
x
′
x
,where x
∈
a
,
b.Corollary 7.
Letu
x
∈
C
a
,
b be a solution of the integral equation (1), K
φ
x
′
′
x
,
s
∈
C
G
,
K
φ
x
′
x
,
x
∈
C
a
,
b, and f
φ
x
′
′
x
∈
C
a
,
b. Then, u
φ
x
′
′
x
∈
C
a
,
b and(13)
u
φ
x
′
′
x
=
K
x
,
x
u
φ
x
′
x
+
K
x
,
x
φ
x
′
+
K
φ
x
′
x
,
s
s
=
x
u
x
+
∫
a
x
K
φ
x
′
′
x
,
s
u
s
d
φ
s
+
f
φ
x
′
′
x
,where x
∈
a
,
b.In this paper, we assume thatK
x
,
x
φ
x
′
∈
C
a
,
b
,
K
φ
x
′
′
x
,
s
,
K
φ
s
′
′
x
,
s
∈
C
G, and f
φ
x
′
′
x
∈
C
a
,
b. Then, using Theorems 4 and 5 (and Corollaries 6 and 7), we show that the number M defined as(14)
M
=
sup
x
,
s
∈
G
⁡
K
x
,
s
u
s
φ
s
′
′can be determined in terms of quantities f
x
C, f
φ
′
x
C
,
f
φ
′
′
x
C
,
K
x
,
x
φ
x
′
C
,
K
φ
x
′
x
,
s
C, K
φ
s
′
x
,
s
C
,
K
φ
x
′
′
x
,
s
C, and K
φ
s
′
′
x
,
s
C.Under these circumstances, using Theorem2, the integral(15)
∫
a
x
K
x
,
s
u
s
d
scan be evaluated numerically by employing the generalized trapezoid rule.


## 3. Numerical Solution

In order to obtain the approximate solution of (1), we employ the generalized trapezoid rule given in [15] to the integral in (1). Let n
∈
N,(16)
h
=
b
-
a
n
,
x
k
=
a
+
k
h
,where k
=
0,1
,
…
,
n. Let us substitute x
=
x
k in the integral equation (1) and examine the following system of equations:(17)
u
x
0
=
f
x
0
,
x
0
=
a
,
u
x
k
=
∫
a
x
k
K
x
k
,
s
u
s
d
s
+
f
x
k
,
k
=
1,2
,
…
,
n
.To evaluate the integral term in (17), we employ the generalized trapezoid rule given in [15] at the nodes x
0
,
x
1
,
…
,
x
k. So, we get(18)
∫
a
x
k
K
x
k
,
s
u
s
d
s
=
∑
j
=
1
k
1
2
K
x
k
,
x
j
-
1
u
x
j
-
1
+
K
x
k
,
x
j
u
x
j
φ
x
j
-
φ
x
j
-
1
+
∑
j
=
1
k
R
j
n
u
,where(19)
R
j
n
u
≤
M
12
φ
x
j
-
φ
x
j
-
1
3
,
(20)
M
=
sup
x
,
s
∈
G
⁡
K
x
,
s
u
s
φ
s
′
′
=
sup
x
,
s
∈
G
⁡
K
x
,
s
u
φ
s
′
′
s
+
2
K
φ
s
′
x
,
s
u
φ
s
′
s
+
K
φ
s
′
′
x
,
s
u
s
.Substituting (18) in (17), we get(21)
u
x
0
=
f
x
0
,
x
0
=
a
,
u
x
k
=
∑
j
=
1
k
1
2
K
x
k
,
x
j
-
1
u
x
j
-
1
+
K
x
k
,
x
j
u
x
j
φ
x
j
-
φ
x
j
-
1
+
∑
j
=
1
k
R
j
n
u
+
f
x
k
,where k
=
1,2
,
…
,
n.Omitting the terms∑
j
=
1
k
R
j
(
n
)
u appearing in each equation of system (21) and u
k
≈
u
(
x
k
), we obtain(22)
u
0
=
f
x
0
,
x
0
=
a
,
u
k
=
∑
j
=
1
k
1
2
K
x
k
,
x
j
-
1
u
x
j
-
1
+
K
x
k
,
x
j
u
x
j
φ
x
j
-
φ
x
j
-
1
+
f
x
k
,where k
=
1,2
,
…
,
n.Let us assume that(23)
α
=
1
2
sup
x
∈
a
,
b
⁡
K
x
,
x
ω
φ
h
<
1
,where ω
φ
h denotes the modulus of continuity of the function φ; that is,(24)
ω
φ
h
=
sup
x
-
y
≤
h
⁡
φ
x
-
φ
y
.Under condition (23), the system of (22) has a unique solution which is given by the formulas(25)
u
0
=
f
a
,
u
1
=
1
-
1
2
K
x
1
,
x
1
φ
x
1
-
φ
x
0
-
1
·
1
2
K
x
1
,
x
0
φ
x
1
-
φ
x
0
u
0
+
f
x
1
,
u
k
=
1
-
1
2
K
x
k
,
x
k
φ
x
k
-
φ
x
k
-
1
-
1
·
1
2
∑
j
=
1
k
-
1
K
x
k
,
x
j
φ
x
j
+
1
-
φ
x
j
-
1
u
j
+
1
2
K
x
k
,
x
0
φ
x
1
-
φ
x
0
u
0
+
f
x
kfor k
=
2,3
,
…
,
n.We give a concrete example below.Example 8.
Let us take the integral equation (1) for a
=
0 and b
=
2 with(26)
φ
x
=
x
,
for
0
≤
x
≤
1
,
x
,
for
1
<
x
≤
2
K
x
,
s
=
a
1
x
b
1
s
,
a
1
x
=
2
x
+
1
,
for
0
≤
x
≤
1
,
2
x
+
1
,
for
1
<
x
≤
2
,
b
1
s
=
s
,
for
0
≤
s
≤
1
,
s
2
,
for
1
<
s
≤
2
,and using(27)
f
x
=
1
-
2
3
x
2
-
1
3
x
x
,
for
0
≤
x
≤
1
,
1
-
x
3
3
2
x
+
1
,
for
1
<
x
≤
2
.It is easily seen that u
x
≡
1
,
x
∈
0,2, is the unique solution of the integral equation (1) and the conditions f
φ
x
′
′
x
∈
C
0,2
,
K
x
,
x
φ
x
′
∈
C
0,2
,
K
φ
x
′
′
x
,
s, and K
φ
s
′
′
x
,
s
∈
C
G hold, where G
=
x
,
s
:
0
≤
s
≤
x
≤
2.Using the proposed method of this study, we get the following results. Here, 20 nodes are selected; that is,n
=
20. In Table 1, we give the values of the approximate solution obtained by the proposed method of this study and the error in absolute values at the given nodes.Table 1
The values of approximate solution, analytical solution, and the error at the nodes.
The nodes
x
k
Real value atx
k
u
(
x
k
)
Approx. value atx
k
u
k
The error atx
k
u
(
x
k
)
-
u
k
0
1
1
0
0.1
1
1.00271872
0.00271872
0.2
1
1.00331826
0.00331826
0.3
1
1.00376994
0.00376994
0.4
1
1.00417187
0.00417187
0.5
1
1.00455757
0.00455757
0.6
1
1.00494416
0.00494416
0.7
1
1.00534264
0.00534264
0.8
1
1.00576140
0.00576140
0.9
1
1.00620761
0.00620761
1.0
1
1.00668805
0.00668805
1.1
1
1.00720950
0.00720950
1.2
1
1.00777907
0.00777907
1.3
1
1.00840442
0.00840442
1.4
1
1.00909399
0.00909399
1.5
1
1.00985722
0.00985722
1.6
1
1.01070480
0.01070480
1.7
1
1.01164880
0.01164880
1.8
1
1.01270300
0.01270300
1.9
1
1.01388350
0.01388350
2.0
1
1.01520860
0.01520860


## 4. Estimation of the Error

In this section, we investigate the problem of convergence of the approximate solutionu
k to the solution of integral (1) at the nodes as n
→
∞.Theorem 9.
Letφ
x be a strictly increasing continuous function on a
,
b and for all x
,
y
∈
a
,
b the following inequality holds:(28)
φ
x
-
φ
y
≤
L
x
-
y
,where L
>
0 and L is independent of the variables x and y. Then, the inequality(29)
u
x
k
-
u
k
≤
M
L
2
φ
b
-
φ
a
12
1
-
α
e
x
p
K
0
L
b
-
a
1
-
α
h
2
,
k
=
1,2
,
…
,
n
,holds in which K
0
=
K
x
,
s
C
,
α
=
1
/
2
K
x
,
x
C
L
h
<
1, and the number M is determined by (20).Proof.
Let the error be denoted byv
k
=
u
x
k
-
u
k for k
=
0,1
,
…
,
n. Taking into account (21) and (22), we have the following system of equations:(30)
v
0
=
0
,
v
k
=
∑
j
=
1
k
1
2
K
x
k
,
x
j
-
1
v
j
-
1
+
K
x
k
,
x
j
v
j
φ
x
j
-
φ
x
j
-
1
+
∑
j
=
1
k
R
j
n
u
,where k
=
1,2
,
…
,
n.
Rearranging the above system of equations, we get(31)
1
-
1
2
K
x
1
,
x
1
φ
x
1
-
φ
x
0
v
1
=
R
1
n
u
,
1
-
1
2
K
x
k
,
x
k
φ
x
k
-
φ
x
k
-
1
v
k
=
1
2
∑
j
=
1
k
-
1
K
x
k
,
x
j
φ
x
j
+
1
-
φ
x
j
-
1
v
j
+
∑
j
=
1
k
R
j
n
u
,where k
=
1,2
,
…
,
n.
Along with the inequalityω
φ
h
≤
L
h, using conditions (19) and (23), we get the following inequality for v
k from (31):(32)
v
1
≤
1
1
-
α
R
h
,
v
k
≤
1
1
-
α
R
h
+
K
0
L
h
∑
j
=
1
k
-
1
v
j
,where k
=
1,2
,
…
,
n
,
R
h
=
M
/
12
L
2
h
2
φ
b
-
φ
a.
Let the termε
k for k
=
1,2
,
…
,
n be determined by(33)
ε
k
=
1
1
-
α
R
h
+
K
0
L
h
∑
j
=
1
k
-
1
ε
j
,
k
=
2,3
,
…
,
n
,and ε
1
=
R
h
/
1
-
α as an initial condition.
It is easily seen thatv
k
≤
ε
k for k
=
1,2
,
…
,
n. This can be verified by mathematical induction as follows: for k
=
1, it is trivial. Let v
j
≤
ε
j for j
=
1
,
…
,
k
-
1. Then, using inequality (32), we get(34)
v
k
≤
1
1
-
α
R
h
+
K
0
L
h
∑
j
=
1
k
-
1
ε
j
=
ε
k
.Let us show that(35)
ε
j
=
R
h
1
-
α
1
+
K
0
L
h
1
-
α
j
-
1
,
j
=
1,2
,
…
,
n
,are the solution of the system of (33). Taking (35) into account, we get(36)
1
1
-
α
R
h
+
K
0
L
h
∑
j
=
1
k
-
1
ε
j
=
R
h
1
-
α
1
+
K
0
L
h
1
-
α
∑
j
=
1
k
-
1
1
+
K
0
L
h
1
-
α
j
-
1
=
R
h
1
-
α
1
+
1
+
K
0
L
h
1
-
α
k
-
1
-
1
=
ε
k
,
k
≥
2
.Here, we use the equality(37)
1
+
γ
k
-
1
-
1
=
γ
∑
j
=
1
k
-
1
1
+
γ
j
-
1
,
k
≥
2
,where γ
=
K
0
L
h
/
1
-
α. Consequently, we get the following estimate for the error v
k for all values k
=
1
,
…
,
n:(38)
v
k
≤
R
h
1
-
α
1
+
K
0
L
h
1
-
α
k
-
1
.Using the fact that 1
+
t
1
/
t is increasing and approaches the number e as t
→
0
+, we get the following chain of inequalities:(39)
1
+
K
0
L
h
1
-
α
k
-
1
≤
1
+
K
0
L
h
1
-
α
b
-
a
/
h
=
1
+
K
0
L
1
-
α
h
1
-
α
/
K
0
L
h
K
0
L
b
-
a
/
1
-
α
≤
e
K
0
L
b
-
a
/
1
-
αfor k
≤
n
=
b
-
a
/
h. Hence, the proof is obtained.Remark 10.
The function(40)
φ
x
=
x
,
for
0
≤
x
≤
1
,
2
x
-
1
,
for
1
<
x
≤
2
,
3
x
-
3
,
for
2
≤
x
≤
3is a strictly increasing continuous function on 0,3
,
φ
′
x
∉
C
0,3. But, for all x
,
y
∈
0,3, the following inequality holds:(41)
φ
x
-
φ
y
≤
4
x
-
y
.Theorem 11.
Letφ
x be a strictly increasing continuous function on a
,
b and(42)
β
=
K
0
φ
b
-
φ
a
<
1
.Then, the inequality(43)
u
x
k
-
u
k
≤
M
12
1
-
β
ω
φ
h
2
φ
b
-
φ
a
,
k
=
1,2
,
…
,
n
,holds in which K
0
=
K
x
,
s
C.Proof.
Let the error be denoted byv
k
=
u
x
k
-
u
k and set up the system of equations(44)
v
1
=
1
2
K
x
1
,
x
1
φ
x
1
-
φ
x
0
v
1
+
R
1
n
u
,
v
k
=
∑
j
=
1
k
-
1
1
2
K
x
k
,
x
j
φ
x
j
+
1
-
φ
x
j
-
1
v
j
+
1
2
K
x
k
,
x
k
φ
x
k
-
φ
x
k
-
1
v
k
+
∑
j
=
1
k
R
j
n
ufor k
=
2,3
,
…
,
n. From this system of equations, we get(45)
v
k
≤
1
2
K
0
sup
j
⁡
v
j
∑
j
=
1
k
-
1
φ
x
j
+
1
-
φ
x
j
+
φ
x
j
-
φ
x
j
-
1
+
φ
x
k
-
φ
x
k
-
1
+
M
12
ω
φ
h
2
φ
b
-
φ
a
=
1
2
K
0
sup
j
⁡
v
j
φ
x
k
-
φ
x
1
+
φ
x
k
-
1
-
φ
x
0
+
φ
x
k
-
φ
x
k
-
1
+
M
12
ω
φ
h
2
φ
b
-
φ
a
≤
K
0
sup
j
⁡
v
j
φ
b
-
φ
a
+
M
12
ω
φ
h
2
φ
b
-
φ
afor k
=
2,3
,
…
,
n. Using condition (42), we get inequality (43). Therefore, Theorem 11 is proved.



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*Source: 1798050-2016-09-27.xml*