Research ArticleApproximate Solution of Volterra-Stieltjes LinearIntegral Equations of the Second Kind with the GeneralizedTrapezoid Rule

Avyt Asanov1 Elman Hazar23 Mustafa Eroz2 Kalyskan Matanova1 and Elmira Abdyldaeva1

1Department of Mathematics Kyrgyz-Turkish Manas University Bishkek Kyrgyzstan2Department of Mathematics Sakarya University Sakarya Turkey3Department of Applied Mathematics and Informatics Kyrgyz-Turkish Manas University Bishkek Kyrgyzstan

Correspondence should be addressed to Elman Hazar ealiyevsakaryaedutr

Received 21 June 2016 Revised 15 August 2016 Accepted 24 August 2016

Academic Editor Soheil Salahshour

Copyright copy 2016 Avit Asanov et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

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

1 Introduction

Various issues concerning Volterra and Volterra-Stieltjesintegral equations were studied in [1ndash13] Some practicaland theoretical investigations were made in paper [1] fornonclassical Volterra integral equations of the first kindAlso the approximate solution for the integral equationconsidered is obtained In paper [2] various inverse problemsincluding Volterra operator equations were studied Someproperties for Volterra-Stieltjes integral operators were givenin [3] In the studies [6 7] existence and uniqueness ofthe solutions were given for Volterra integral and Volterraoperator equations of the first and the second kinds In papers[4 6] quadratic integral equations of Urysohn-Stieltjes typeand their applications were investigated Various numericalsolutionmethods for integral equations were presented in thestudies [8ndash13]Thenotion of derivative of a function bymeansof a strictly increasing function was given by Asanov in [14]In the study [15] the generalized trapezoid rule was proposedto evaluate the Stieltjes integral approximately by employingthe notion of derivative of a function by means of a strictlyincreasing function

In this study we investigate the numerical solution oflinear Volterra-Stieltjes integral equations of the second kind

by using the generalized trapezoid rule Therefore we needthe concept of the derivative defined in the works [14 15] andtheorems connected with it

2 Approximating Volterra-StieltjesIntegral Equations

Consider the linear integral equation of the second kind

119906 (119909) = int

119909

119886

119870 (119909 119904) 119906 (119904) 119889120593 (119904) + 119891 (119909) 119909 isin [119886 119887] (1)

where 119870(119909 119904) is a given continuous function on 119866 = (119909 119904)

119886 le 119904 le 119909 le 119887 119891(119909) are given continuous functions on[119886 119887] 120593(119904) is a given strictly increasing continuous functionon [119886 119887] and 119906(119909) is the sought function on [119886 119887]

Definition 1 The derivative of a function 119891(119909) with respectto 120593(119909) is the function 119891

1015840

120593(119909) whose value at 119909 isin (119886 119887) is the

number

1198911015840

120593(119909) = lim

Δrarr0

119891 (119909 + Δ) minus 119891 (119909)

120593 (119909 + Δ) minus 120593 (119909)

(2)

Hindawi Publishing CorporationAdvances in Mathematical PhysicsVolume 2016 Article ID 1798050 6 pageshttpdxdoiorg10115520161798050

2 Advances in Mathematical Physics

where 120593(119909) is a given strictly increasing continuous functionin (119886 119887)

If the limit in (2) exists we say that 119891(119909) has a derivative(is differentiable) with respect to 120593(119909) The first derivative1198911015840

120593(119909) may also be a differentiable function with respect to

120593(119909) at every point 119909 isin (119886 119887) Then its derivative

11989110158401015840

120593(119909) = (119891

1015840

120593(119909))

1015840

120593

(3)

is called the second derivative of 119891(119909) with respect to 120593(119909)Consequently the 119899th derivative of 119891(119909) with respect to 120593(119909)

is defined by

119891(119899)

120593(119909) = (119891

(119899minus1)

120593(119909))

1015840

120593

(4)

We need the following theorem which is given in [15]

Theorem 2 Let 120593(119909) and 120595(119909) be two strictly increasingcontinuous functions on [119886 119887] and 119891

10158401015840

120593(119909) 119891

10158401015840

120595(119909) isin 119862[119886 119887]

Then1003816100381610038161003816119868 minus 119860

119899

1003816100381610038161003816=

1198720

12

(120593 (119887) minus 120593 (119886)) (120596120593(ℎ))

2

+

1198721015840

0

12

(120595 (119887) minus 120595 (119886)) (120596120595(ℎ))

2

(5)

where

119868 = int

119887

119886

119891 (119909) 119889120593 (119909) minus int

119887

119886

119891 (119909) 119889120595 (119909)

1198720=

1003817100381710038171003817100381711989110158401015840

120593(119909)

10038171003817100381710038171003817119862

= sup119909isin[119886119887]

1003816100381610038161003816100381611989110158401015840

120593(119909)

10038161003816100381610038161003816

1198721015840

0=

1003817100381710038171003817100381711989110158401015840

120595(119909)

10038171003817100381710038171003817119862

= sup119909isin[119886119887]

1003816100381610038161003816100381611989110158401015840

120595(119909)

10038161003816100381610038161003816

120596120593(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816

120596120595(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120595 (119909) minus 120595 (119910)

1003816100381610038161003816

119860119899

=

1

2

119899

sum

119894=1

[119891 (119909119894) + 119891 (119909

119894minus1)] [120593 (119909

119894) minus 120593 (119909

119894minus1)]

minus

1

2

119899

sum

119894=1

[119891 (119909119894) + 119891 (119909

119894minus1)] [120595 (119909

119894) minus 120595 (119909

119894minus1)]

(6)

and 119909119894= 119886 + 119894ℎ 119894 = 0 1 119899 ℎ = (119887 minus 119886)119899 119899 isin 119873 (119873

denotes the set of natural numbers)

Corollary 3 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] 120595(119909) = 0 for all 119909 isin [119886 119887] and 119891

10158401015840

120593(119909) isin

119862[119886 119887] Then1003816100381610038161003816119868 minus 119860

119899

1003816100381610038161003816le

1198720

12

(120593 (119887) minus 120593 (119886)) (120596120593(ℎ))

2

1003816100381610038161003816119868119894minus 119872119894

1003816100381610038161003816le

1198720

12

(120593 (119909119894) minus 120593 (119909

119894minus1))3

119894 = 1 2 119899

(7)

where

119868119894= int

119909119894

119909119894minus1

119891 (119909) 119889120593 (119909)

119872119894=

1

2

[119891 (119909119894) + 119891 (119909

119894minus1)] [120593 (119909

119894) minus 120593 (119909

119894minus1)]

(8)

Theorem 4 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] 119870(119909 119904) isin 119862(119866) and 119891(119909) isin 119862[119886 119887] Thenthe integral equation (1) has a unique solution 119906(119909) isin 119862[119886 119887]

and119906 (119909)

119862le 1198881

1003817100381710038171003817119891 (119909)

1003817100381710038171003817119862

(9)

where 1198881

= exp1198700(120593(119887) minus 120593(119886)) and 119870

0= 119870(119909 119904)

119862=

sup(119909119904)isin119866

|119870(119909 119904)|

Then we will need the following theorem which is givenin [16]

Theorem 5 Let 119865(119909 119904) 1198651015840

120593(119909)(119909 119904) isin 119862(119866) 120593(119909) be strictly

increasing continuous functions on [119886 119887] and 119875(119909) =

int

119909

119886

119865(119909 119904)119889120593(119904) 119909 isin [119886 119887] Then

1198751015840

120593(119909)(119909) = 119865 (119909 119909) + int

119909

119886

1198651015840

120593(119909)(119909 119904) 119889120593 (119904)

119909 isin [119886 119887]

(10)

where

1198651015840

120593(119909)(119909 119904) = lim

Δ119909rarr0

119865 (119909 + Δ119909 119904) minus 119865 (119909 119904)

120593 (119909 + Δ119909) minus 120593 (119909)

(119909 119904) isin (119909 119904) 119886 lt 119904 lt 119909 lt 119887

1198751015840

120593(119909)(119886) = lim

Δ119909rarr0+

119875 (119886 + Δ119909) minus 119875 (119886)

120593 (119886 + Δ119909) minus 120593 (119886)

1198751015840

120593(119909)(119887) = lim

Δ119909rarr0minus

119875 (119887 + Δ119909) minus 119875 (119887)

120593 (119887 + Δ119909) minus 120593 (119887)

(11)

Corollary 6 Let 119906(119909) isin 119862[119886 119887] be a solution of the integralequation (1) 1198701015840

120593(119909)(119909 119904) isin 119862(119866) and 119891

1015840

120593(119909)(119909) isin 119862[119886 119887] Then

1199061015840

120593(119909)(119909) isin 119862[119886 119887] and

1199061015840

120593(119909)(119909) = 119870 (119909 119909) 119906 (119909) + int

119909

119886

1198701015840

120593(119909)(119909 119904) 119906 (119904) 119889120593 (119904)

+ 1198911015840

120593(119909)(119909)

(12)

where 119909 isin [119886 119887]

Corollary 7 Let 119906(119909) isin 119862[119886 119887] be a solution of the integralequation (1) 11987010158401015840

120593(119909)(119909 119904) isin 119862(119866) 119870

1015840

120593(119909)(119909 119909) isin 119862[119886 119887] and

11989110158401015840

120593(119909)(119909) isin 119862[119886 119887] Then 11990610158401015840

120593(119909)(119909) isin 119862[119886 119887] and

11990610158401015840

120593(119909)(119909) = 119870 (119909 119909) 119906

1015840

120593(119909)(119909)

+ [(119870 (119909 119909))1015840

120593(119909)+ 1198701015840

120593(119909)(119909 119904)

10038161003816100381610038161003816119904=119909

] 119906 (119909)

+ int

119909

119886

11987010158401015840

120593(119909)(119909 119904) 119906 (119904) 119889120593 (119904) + 119891

10158401015840

120593(119909)(119909)

(13)

where 119909 isin [119886 119887]

Advances in Mathematical Physics 3

In this paper we assume that (119870(119909 119909))1015840

120593(119909)isin 119862[119886 119887]

11987010158401015840

120593(119909)(119909 119904) 119870

10158401015840

120593(119904)(119909 119904) isin 119862(119866) and 119891

10158401015840

120593(119909)(119909) isin 119862[119886 119887] Then

using Theorems 4 and 5 (and Corollaries 6 and 7) we showthat the number 119872 defined as

119872 = sup(119909119904)isin119866

10038161003816100381610038161003816[119870 (119909 119904) 119906 (119904)]

10158401015840

120593(119904)

10038161003816100381610038161003816 (14)

can be determined in terms of quantities 119891(119909)119862

1198911015840

120593(119909)119862 11989110158401015840

120593(119909)119862 (119870(119909 119909))

1015840

120593(119909)119862 1198701015840

120593(119909)(119909 119904)

119862

1198701015840

120593(119904)(119909 119904)

119862 11987010158401015840

120593(119909)(119909 119904)

119862 and 119870

10158401015840

120593(119904)(119909 119904)

119862

Under these circumstances usingTheorem 2 the integral

int

119909

119886

119870 (119909 119904) 119906 (119904) 119889119904 (15)

can be evaluated numerically by employing the generalizedtrapezoid rule

3 Numerical Solution

In order to obtain the approximate solution of (1) we employthe generalized trapezoid rule given in [15] to the integral in(1) Let 119899 isin 119873

ℎ =

119887 minus 119886

119899

119909119896= 119886 + 119896ℎ

(16)

where 119896 = 0 1 119899 Let us substitute 119909 = 119909119896in the integral

equation (1) and examine the following system of equations

119906 (1199090) = 119891 (119909

0) 119909

0= 119886

119906 (119909119896) = int

119909119896

119886

119870(119909119896 119904) 119906 (119904) 119889119904 + 119891 (119909

119896)

119896 = 1 2 119899

(17)

To evaluate the integral term in (17) we employ the general-ized trapezoid rule given in [15] at the nodes 119909

0 1199091 119909

119896

So we get

int

119909119896

119886

119870(119909119896 119904) 119906 (119904) 119889119904

=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(18)

where10038161003816100381610038161003816119877(119899)

119895(119906)

10038161003816100381610038161003816le

119872

12

[120593 (119909119895) minus 120593 (119909

119895minus1)]

3

(19)

119872 = sup(119909119904)isin119866

10038161003816100381610038161003816[119870 (119909 119904) 119906 (119904)]

10158401015840

120593(119904)

10038161003816100381610038161003816

= sup(119909119904)isin119866

10038161003816100381610038161003816119870 (119909 119904) 119906

10158401015840

120593(119904)(119904) + 2119870

1015840

120593(119904)(119909 119904) 119906

1015840

120593(119904)(119904)

+ 11987010158401015840

120593(119904)(119909 119904) 119906 (119904)

10038161003816100381610038161003816

(20)

Substituting (18) in (17) we get

119906 (1199090) = 119891 (119909

0) 119909

0= 119886

119906 (119909119896)

=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906) + 119891 (119909

119896)

(21)

where 119896 = 1 2 119899Omitting the terms sum

119896

119895=1119877(119899)

119895(119906) appearing in each equa-

tion of system (21) and 119906119896asymp 119906(119909

119896) we obtain

1199060= 119891 (119909

0) 119909

0= 119886

119906119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] + 119891 (119909

119896)

(22)

where 119896 = 1 2 119899Let us assume that

120572 =

1

2

sup119909isin[119886119887]

|119870 (119909 119909)| 120596120593(ℎ) lt 1 (23)

where 120596120593(ℎ) denotes the modulus of continuity of the

function 120593 that is

120596120593(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816 (24)

Under condition (23) the systemof (22) has a unique solutionwhich is given by the formulas

1199060= 119891 (119886)

1199061= [1 minus

1

2

119870 (1199091 1199091) (120593 (119909

1) minus 120593 (119909

0))]

minus1

sdot [

1

2

119870 (1199091 1199090) (120593 (119909

1) minus 120593 (119909

0)) 1199060+ 119891 (119909

1)]

119906119896= [1 minus

1

2

119870 (119909119896 119909119896) (120593 (119909

119896) minus 120593 (119909

119896minus1))]

minus1

sdot[

[

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) (120593 (119909

119895+1) minus 120593 (119909

119895minus1)) 119906119895

+

1

2

119870 (119909119896 1199090) (120593 (119909

1) minus 120593 (119909

0)) 1199060+ 119891 (119909

119896)]

]

(25)

for 119896 = 2 3 119899We give a concrete example below

Example 8 Let us take the integral equation (1) for 119886 = 0 and119887 = 2 with

4 Advances in Mathematical Physics

120593 (119909) =

radic119909 for 0 le 119909 le 1

119909 for 1 lt 119909 le 2

119870 (119909 119904) = 1198861(119909) 1198871(119904) 119886

1(119909) =

2radic119909 + 1 for 0 le 119909 le 1

2119909 + 1 for 1 lt 119909 le 2

1198871(119904) =

119904 for 0 le 119904 le 1

1199042

for 1 lt 119904 le 2

(26)

and using

119891 (119909) =

1 minus

2

3

1199092

minus

1

3

119909radic119909 for 0 le 119909 le 1

1 minus

1199093

3

(2119909 + 1) for 1 lt 119909 le 2

(27)

It is easily seen that 119906(119909) equiv 1 119909 isin [0 2] is the unique solu-tion of the integral equation (1) and the conditions 11989110158401015840

120593(119909)(119909) isin

119862[0 2] (119870(119909 119909))1015840

120593(119909)isin 119862[0 2] 119870

10158401015840

120593(119909)(119909 119904) and 119870

10158401015840

120593(119904)(119909 119904) isin

119862(119866) hold where 119866 = (119909 119904) 0 le 119904 le 119909 le 2

Using the proposed method of this study we get thefollowing results Here 20 nodes are selected that is 119899 = 20In Table 1 we give the values of the approximate solutionobtained by the proposed method of this study and the errorin absolute values at the given nodes

4 Estimation of the Error

In this section we investigate the problem of convergence ofthe approximate solution 119906

119896to the solution of integral (1) at

the nodes as 119899 rarr infin

Theorem 9 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and for all 119909 119910 isin [119886 119887] the followinginequality holds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 119871

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (28)

where 119871 gt 0 and 119871 is independent of the variables 119909 and 119910Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816

le

1198721198712

[120593 (119887) minus 120593 (119886)]

12 (1 minus 120572)

exp

1198700119871 (119887 minus 119886)

1 minus 120572

ℎ2

119896 = 1 2 119899

(29)

holds in which 1198700= 119870(119909 119904)

119862 120572 = (12)119870(119909 119909)

119862119871ℎ lt 1

and the number 119872 is determined by (20)

Proof Let the error be denoted by V119896

= 119906(119909119896) minus 119906

119896for

119896 = 0 1 119899 Taking into account (21) and (22) we havethe following system of equations

V0= 0

V119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) V119895minus1

+ 119870(119909119896 119909119895) V119895]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(30)

where 119896 = 1 2 119899Rearranging the above system of equations we get

(1 minus

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)]) V1= 119877(119899)

1(119906)

(1 minus

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)]) V119896

=

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

119896

sum

119895=1

119877(119899)

119895(119906)

(31)

where 119896 = 1 2 119899Along with the inequality 120596

120593(ℎ) le 119871ℎ using conditions

(19) and (23) we get the following inequality for V119896from (31)

1003816100381610038161003816V1

1003816100381610038161003816le

1

1 minus 120572

119877 (ℎ)

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

]

]

(32)

where 119896 = 1 2 119899 119877(ℎ) = (11987212)1198712

ℎ2

[120593(119887) minus 120593(119886)]Let the term 120576

119896for 119896 = 1 2 119899 be determined by

120576119896=

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

119896 = 2 3 119899 (33)

and 1205761= 119877(ℎ)(1 minus 120572) as an initial condition

It is easily seen that |V119896| le 120576119896for 119896 = 1 2 119899This can be

verified by mathematical induction as follows for 119896 = 1 it is

Advances in Mathematical Physics 5

Table 1 The values of approximate solution analytical solutionand the error at the nodes

Thenodes119909119896

Real value at 119909119896

119906(119909119896)

Approx value at 119909119896

119906119896

The error at 119909119896

|119906(119909119896) minus 119906119896|

0 1 1 001 1 100271872 00027187202 1 100331826 00033182603 1 100376994 00037699404 1 100417187 00041718705 1 100455757 00045575706 1 100494416 00049441607 1 100534264 00053426408 1 100576140 00057614009 1 100620761 00062076110 1 100668805 00066880511 1 100720950 00072095012 1 100777907 00077790713 1 100840442 00084044214 1 100909399 00090939915 1 100985722 00098572216 1 101070480 00107048017 1 101164880 00116488018 1 101270300 00127030019 1 101388350 00138835020 1 101520860 001520860

trivial Let |V119895| le 120576119895for 119895 = 1 119896minus1Then using inequality

(32) we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

= 120576119896 (34)

Let us show that

120576119895=

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

119895 = 1 2 119899 (35)

are the solution of the system of (33) Taking (35) intoaccount we get

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

=

119877 (ℎ)

1 minus 120572

1 +

1198700119871ℎ

1 minus 120572

119896minus1

sum

119895=1

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

=

119877 (ℎ)

1 minus 120572

1 + [(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

minus 1] = 120576119896

119896 ge 2

(36)

Here we use the equality

(1 + 120574)119896minus1

minus 1 = 120574

119896minus1

sum

119895=1

(1 + 120574)119895minus1

119896 ge 2 (37)

where 120574 = 1198700119871ℎ(1 minus 120572) Consequently we get the following

estimate for the error V119896for all values 119896 = 1 119899

1003816100381610038161003816V119896

1003816100381610038161003816le

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

(38)

Using the fact that (1 + 119905)1119905 is increasing and approaches

the number 119890 as 119905 rarr 0+ we get the following chain ofinequalities

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

le (1 +

1198700119871ℎ

1 minus 120572

)

(119887minus119886)ℎ

= [(1 +

1198700119871

1 minus 120572

ℎ)

(1minus120572)1198700119871ℎ

]

1198700119871(119887minus119886)(1minus120572)

le 1198901198700119871(119887minus119886)(1minus120572)

(39)

for 119896 le 119899 = (119887 minus 119886)ℎ Hence the proof is obtained

Remark 10 The function

120593 (119909) =

119909 for 0 le 119909 le 1

2119909 minus 1 for 1 lt 119909 le 2

3119909 minus 3 for 2 le 119909 le 3

(40)

is a strictly increasing continuous function on [0 3] 1205931015840

(119909) notin

119862[0 3] But for all 119909 119910 isin [0 3] the following inequalityholds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 4

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (41)

Theorem 11 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and

120573 = 1198700[120593 (119887) minus 120593 (119886)] lt 1 (42)

Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816le

119872

12 (1 minus 120573)

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

119896 = 1 2 119899

(43)

holds in which 1198700= 119870(119909 119904)

119862

6 Advances in Mathematical Physics

Proof Let the error be denoted by V119896= 119906(119909

119896) minus 119906119896and set up

the system of equations

V1=

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)] V1+ 119877(119899)

1(119906)

V119896=

119896minus1

sum

119895=1

1

2

119870 (119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)] V119896

+

119896

sum

119895=1

119877(119899)

119895(119906)

(44)

for 119896 = 2 3 119899 From this system of equations we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

sdot

119896minus1

sum

119895=1

[120593 (119909119895+1

) minus 120593 (119909119895) + 120593 (119909

119895) minus 120593 (119909

119895minus1)]

+ [120593 (119909119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887)

minus 120593 (119886)] =

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119909119896) minus 120593 (119909

1)

+ 120593 (119909119896minus1

) minus 120593 (1199090) + 120593 (119909

119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)] le 1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119887)

minus 120593 (119886)] +

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

(45)

for 119896 = 2 3 119899 Using condition (42) we get inequality(43) Therefore Theorem 11 is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] A S Apartsyn Nonclassical Linear Volterra Equations of theFirst Kind VSP TB Utrecht The Netherlands 2003

[2] A L Bukhgeim Volterra Equations and Inverse Problems VSPUtrecht The Netherlands 1999

[3] J Banas and D OrsquoRegan ldquoVolterra-Stieltjes integral operatorsrdquoMathematical and ComputerModelling vol 41 no 2-3 pp 335ndash344 2005

[4] J Banas J R Rodriguez and K Sadarangani ldquoOn a class ofUrysohn-Stieltjes quadratic integral equations and their appli-cationsrdquo Journal of Computational and Applied Mathematicsvol 113 no 1-2 pp 35ndash50 2000

[5] L M Delves and J Walsh Numerical Solution of IntegralEquations Oxford University Press New York NY USA 1974

[6] M Federson and R Bianconi ldquoLinear Volterrra-Stieltjes inte-gral equations in the sense of the Kurzweil-Henstock integralrdquoArchivum Mathematicum vol 37 no 4 pp 307ndash328 2001

[7] M Federson R Bianconi and L Barbanti ldquoLinear Volterraintegral equationsrdquo Acta Mathematicae Applicatae Sinica vol18 no 4 pp 553ndash560 2002

[8] A D Polyanin and A V Manzhirov Handbook of IntegralEquations Chapman amp HallCRC Boca Raton Fla USA 2ndedition 2008

[9] K Maleknejad and N Aghazadeh ldquoNumerical solution ofVolterra integral equations of the second kind with convolutionkernel by using Taylor-series expansionmethodrdquoAppliedMath-ematics and Computation vol 161 no 3 pp 915ndash922 2005

[10] M A Darwish and J Henderson ldquoNondecreasing solutions ofa quadratic integral equation of Urysohn-Stieltjes typerdquo RockyMountain Journal of Mathematics vol 42 no 2 pp 545ndash5662012

[11] S A Isaacson and R M Kirby ldquoNumerical solution of linearVolterra integral equations of the second kind with sharpgradientsrdquo Journal of Computational and Applied Mathematicsvol 235 no 14 pp 4283ndash4301 2011

[12] T Diogo N J Ford P Lima and S Valtchev ldquoNumericalmethods for a Volterra integral equation with non-smoothsolutionsrdquo Journal of Computational and Applied Mathematicsvol 189 no 1-2 pp 412ndash423 2006

[13] S Zhang Y Lin and M Rao ldquoNumerical solutions for second-kind Volterra integral equations by Galerkin methodsrdquo Appli-cations of Mathematics vol 45 no 1 pp 19ndash39 2000

[14] A Asanov ldquoThe derivative of a function by means of anincreasing functionrdquo Manas Journal of Engineering no 1 pp18ndash64 2001 (Russian)

[15] A Asanov M H Chelik and A Chalish ldquoApproximating theStieltjes integral by using the generalized trapezoid rulerdquo LeMatematiche vol 66 no 2 pp 13ndash21 2011

[16] A Asanov ldquoVolterra-stielties integral equations of the secondkind and the first kindrdquoManas Journal of Engineering no 2 pp79ndash95 2002 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Advances in Mathematical Physics

where 120593(119909) is a given strictly increasing continuous functionin (119886 119887)

If the limit in (2) exists we say that 119891(119909) has a derivative(is differentiable) with respect to 120593(119909) The first derivative1198911015840

120593(119909) may also be a differentiable function with respect to

120593(119909) at every point 119909 isin (119886 119887) Then its derivative

11989110158401015840

120593(119909) = (119891

1015840

120593(119909))

1015840

120593

(3)

is called the second derivative of 119891(119909) with respect to 120593(119909)Consequently the 119899th derivative of 119891(119909) with respect to 120593(119909)

is defined by

119891(119899)

120593(119909) = (119891

(119899minus1)

120593(119909))

1015840

120593

(4)

We need the following theorem which is given in [15]

Theorem 2 Let 120593(119909) and 120595(119909) be two strictly increasingcontinuous functions on [119886 119887] and 119891

10158401015840

120593(119909) 119891

10158401015840

120595(119909) isin 119862[119886 119887]

Then1003816100381610038161003816119868 minus 119860

119899

1003816100381610038161003816=

1198720

12

(120593 (119887) minus 120593 (119886)) (120596120593(ℎ))

2

+

1198721015840

0

12

(120595 (119887) minus 120595 (119886)) (120596120595(ℎ))

2

(5)

where

119868 = int

119887

119886

119891 (119909) 119889120593 (119909) minus int

119887

119886

119891 (119909) 119889120595 (119909)

1198720=

1003817100381710038171003817100381711989110158401015840

120593(119909)

10038171003817100381710038171003817119862

= sup119909isin[119886119887]

1003816100381610038161003816100381611989110158401015840

120593(119909)

10038161003816100381610038161003816

1198721015840

0=

1003817100381710038171003817100381711989110158401015840

120595(119909)

10038171003817100381710038171003817119862

= sup119909isin[119886119887]

1003816100381610038161003816100381611989110158401015840

120595(119909)

10038161003816100381610038161003816

120596120593(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816

120596120595(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120595 (119909) minus 120595 (119910)

1003816100381610038161003816

119860119899

=

1

2

119899

sum

119894=1

[119891 (119909119894) + 119891 (119909

119894minus1)] [120593 (119909

119894) minus 120593 (119909

119894minus1)]

minus

1

2

119899

sum

119894=1

[119891 (119909119894) + 119891 (119909

119894minus1)] [120595 (119909

119894) minus 120595 (119909

119894minus1)]

(6)

and 119909119894= 119886 + 119894ℎ 119894 = 0 1 119899 ℎ = (119887 minus 119886)119899 119899 isin 119873 (119873

denotes the set of natural numbers)

Corollary 3 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] 120595(119909) = 0 for all 119909 isin [119886 119887] and 119891

10158401015840

120593(119909) isin

119862[119886 119887] Then1003816100381610038161003816119868 minus 119860

119899

1003816100381610038161003816le

1198720

12

(120593 (119887) minus 120593 (119886)) (120596120593(ℎ))

2

1003816100381610038161003816119868119894minus 119872119894

1003816100381610038161003816le

1198720

12

(120593 (119909119894) minus 120593 (119909

119894minus1))3

119894 = 1 2 119899

(7)

where

119868119894= int

119909119894

119909119894minus1

119891 (119909) 119889120593 (119909)

119872119894=

1

2

[119891 (119909119894) + 119891 (119909

119894minus1)] [120593 (119909

119894) minus 120593 (119909

119894minus1)]

(8)

Theorem 4 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] 119870(119909 119904) isin 119862(119866) and 119891(119909) isin 119862[119886 119887] Thenthe integral equation (1) has a unique solution 119906(119909) isin 119862[119886 119887]

and119906 (119909)

119862le 1198881

1003817100381710038171003817119891 (119909)

1003817100381710038171003817119862

(9)

where 1198881

= exp1198700(120593(119887) minus 120593(119886)) and 119870

0= 119870(119909 119904)

119862=

sup(119909119904)isin119866

|119870(119909 119904)|

Then we will need the following theorem which is givenin [16]

Theorem 5 Let 119865(119909 119904) 1198651015840

120593(119909)(119909 119904) isin 119862(119866) 120593(119909) be strictly

increasing continuous functions on [119886 119887] and 119875(119909) =

int

119909

119886

119865(119909 119904)119889120593(119904) 119909 isin [119886 119887] Then

1198751015840

120593(119909)(119909) = 119865 (119909 119909) + int

119909

119886

1198651015840

120593(119909)(119909 119904) 119889120593 (119904)

119909 isin [119886 119887]

(10)

where

1198651015840

120593(119909)(119909 119904) = lim

Δ119909rarr0

119865 (119909 + Δ119909 119904) minus 119865 (119909 119904)

120593 (119909 + Δ119909) minus 120593 (119909)

(119909 119904) isin (119909 119904) 119886 lt 119904 lt 119909 lt 119887

1198751015840

120593(119909)(119886) = lim

Δ119909rarr0+

119875 (119886 + Δ119909) minus 119875 (119886)

120593 (119886 + Δ119909) minus 120593 (119886)

1198751015840

120593(119909)(119887) = lim

Δ119909rarr0minus

119875 (119887 + Δ119909) minus 119875 (119887)

120593 (119887 + Δ119909) minus 120593 (119887)

(11)

Corollary 6 Let 119906(119909) isin 119862[119886 119887] be a solution of the integralequation (1) 1198701015840

120593(119909)(119909 119904) isin 119862(119866) and 119891

1015840

120593(119909)(119909) isin 119862[119886 119887] Then

1199061015840

120593(119909)(119909) isin 119862[119886 119887] and

1199061015840

120593(119909)(119909) = 119870 (119909 119909) 119906 (119909) + int

119909

119886

1198701015840

120593(119909)(119909 119904) 119906 (119904) 119889120593 (119904)

+ 1198911015840

120593(119909)(119909)

(12)

where 119909 isin [119886 119887]

Corollary 7 Let 119906(119909) isin 119862[119886 119887] be a solution of the integralequation (1) 11987010158401015840

120593(119909)(119909 119904) isin 119862(119866) 119870

1015840

120593(119909)(119909 119909) isin 119862[119886 119887] and

11989110158401015840

120593(119909)(119909) isin 119862[119886 119887] Then 11990610158401015840

120593(119909)(119909) isin 119862[119886 119887] and

11990610158401015840

120593(119909)(119909) = 119870 (119909 119909) 119906

1015840

120593(119909)(119909)

+ [(119870 (119909 119909))1015840

120593(119909)+ 1198701015840

120593(119909)(119909 119904)

10038161003816100381610038161003816119904=119909

] 119906 (119909)

+ int

119909

119886

11987010158401015840

120593(119909)(119909 119904) 119906 (119904) 119889120593 (119904) + 119891

10158401015840

120593(119909)(119909)

(13)

where 119909 isin [119886 119887]

Advances in Mathematical Physics 3

In this paper we assume that (119870(119909 119909))1015840

120593(119909)isin 119862[119886 119887]

11987010158401015840

120593(119909)(119909 119904) 119870

10158401015840

120593(119904)(119909 119904) isin 119862(119866) and 119891

10158401015840

120593(119909)(119909) isin 119862[119886 119887] Then

using Theorems 4 and 5 (and Corollaries 6 and 7) we showthat the number 119872 defined as

119872 = sup(119909119904)isin119866

10038161003816100381610038161003816[119870 (119909 119904) 119906 (119904)]

10158401015840

120593(119904)

10038161003816100381610038161003816 (14)

can be determined in terms of quantities 119891(119909)119862

1198911015840

120593(119909)119862 11989110158401015840

120593(119909)119862 (119870(119909 119909))

1015840

120593(119909)119862 1198701015840

120593(119909)(119909 119904)

119862

1198701015840

120593(119904)(119909 119904)

119862 11987010158401015840

120593(119909)(119909 119904)

119862 and 119870

10158401015840

120593(119904)(119909 119904)

119862

Under these circumstances usingTheorem 2 the integral

int

119909

119886

119870 (119909 119904) 119906 (119904) 119889119904 (15)

can be evaluated numerically by employing the generalizedtrapezoid rule

3 Numerical Solution

In order to obtain the approximate solution of (1) we employthe generalized trapezoid rule given in [15] to the integral in(1) Let 119899 isin 119873

ℎ =

119887 minus 119886

119899

119909119896= 119886 + 119896ℎ

(16)

where 119896 = 0 1 119899 Let us substitute 119909 = 119909119896in the integral

equation (1) and examine the following system of equations

119906 (1199090) = 119891 (119909

0) 119909

0= 119886

119906 (119909119896) = int

119909119896

119886

119870(119909119896 119904) 119906 (119904) 119889119904 + 119891 (119909

119896)

119896 = 1 2 119899

(17)

To evaluate the integral term in (17) we employ the general-ized trapezoid rule given in [15] at the nodes 119909

0 1199091 119909

119896

So we get

int

119909119896

119886

119870(119909119896 119904) 119906 (119904) 119889119904

=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(18)

where10038161003816100381610038161003816119877(119899)

119895(119906)

10038161003816100381610038161003816le

119872

12

[120593 (119909119895) minus 120593 (119909

119895minus1)]

3

(19)

119872 = sup(119909119904)isin119866

10038161003816100381610038161003816[119870 (119909 119904) 119906 (119904)]

10158401015840

120593(119904)

10038161003816100381610038161003816

= sup(119909119904)isin119866

10038161003816100381610038161003816119870 (119909 119904) 119906

10158401015840

120593(119904)(119904) + 2119870

1015840

120593(119904)(119909 119904) 119906

1015840

120593(119904)(119904)

+ 11987010158401015840

120593(119904)(119909 119904) 119906 (119904)

10038161003816100381610038161003816

(20)

Substituting (18) in (17) we get

119906 (1199090) = 119891 (119909

0) 119909

0= 119886

119906 (119909119896)

=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906) + 119891 (119909

119896)

(21)

where 119896 = 1 2 119899Omitting the terms sum

119896

119895=1119877(119899)

119895(119906) appearing in each equa-

tion of system (21) and 119906119896asymp 119906(119909

119896) we obtain

1199060= 119891 (119909

0) 119909

0= 119886

119906119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] + 119891 (119909

119896)

(22)

where 119896 = 1 2 119899Let us assume that

120572 =

1

2

sup119909isin[119886119887]

|119870 (119909 119909)| 120596120593(ℎ) lt 1 (23)

where 120596120593(ℎ) denotes the modulus of continuity of the

function 120593 that is

120596120593(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816 (24)

Under condition (23) the systemof (22) has a unique solutionwhich is given by the formulas

1199060= 119891 (119886)

1199061= [1 minus

1

2

119870 (1199091 1199091) (120593 (119909

1) minus 120593 (119909

0))]

minus1

sdot [

1

2

119870 (1199091 1199090) (120593 (119909

1) minus 120593 (119909

0)) 1199060+ 119891 (119909

1)]

119906119896= [1 minus

1

2

119870 (119909119896 119909119896) (120593 (119909

119896) minus 120593 (119909

119896minus1))]

minus1

sdot[

[

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) (120593 (119909

119895+1) minus 120593 (119909

119895minus1)) 119906119895

+

1

2

119870 (119909119896 1199090) (120593 (119909

1) minus 120593 (119909

0)) 1199060+ 119891 (119909

119896)]

]

(25)

for 119896 = 2 3 119899We give a concrete example below

Example 8 Let us take the integral equation (1) for 119886 = 0 and119887 = 2 with

4 Advances in Mathematical Physics

120593 (119909) =

radic119909 for 0 le 119909 le 1

119909 for 1 lt 119909 le 2

119870 (119909 119904) = 1198861(119909) 1198871(119904) 119886

1(119909) =

2radic119909 + 1 for 0 le 119909 le 1

2119909 + 1 for 1 lt 119909 le 2

1198871(119904) =

119904 for 0 le 119904 le 1

1199042

for 1 lt 119904 le 2

(26)

and using

119891 (119909) =

1 minus

2

3

1199092

minus

1

3

119909radic119909 for 0 le 119909 le 1

1 minus

1199093

3

(2119909 + 1) for 1 lt 119909 le 2

(27)

It is easily seen that 119906(119909) equiv 1 119909 isin [0 2] is the unique solu-tion of the integral equation (1) and the conditions 11989110158401015840

120593(119909)(119909) isin

119862[0 2] (119870(119909 119909))1015840

120593(119909)isin 119862[0 2] 119870

10158401015840

120593(119909)(119909 119904) and 119870

10158401015840

120593(119904)(119909 119904) isin

119862(119866) hold where 119866 = (119909 119904) 0 le 119904 le 119909 le 2

Using the proposed method of this study we get thefollowing results Here 20 nodes are selected that is 119899 = 20In Table 1 we give the values of the approximate solutionobtained by the proposed method of this study and the errorin absolute values at the given nodes

4 Estimation of the Error

In this section we investigate the problem of convergence ofthe approximate solution 119906

119896to the solution of integral (1) at

the nodes as 119899 rarr infin

Theorem 9 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and for all 119909 119910 isin [119886 119887] the followinginequality holds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 119871

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (28)

where 119871 gt 0 and 119871 is independent of the variables 119909 and 119910Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816

le

1198721198712

[120593 (119887) minus 120593 (119886)]

12 (1 minus 120572)

exp

1198700119871 (119887 minus 119886)

1 minus 120572

ℎ2

119896 = 1 2 119899

(29)

holds in which 1198700= 119870(119909 119904)

119862 120572 = (12)119870(119909 119909)

119862119871ℎ lt 1

and the number 119872 is determined by (20)

Proof Let the error be denoted by V119896

= 119906(119909119896) minus 119906

119896for

119896 = 0 1 119899 Taking into account (21) and (22) we havethe following system of equations

V0= 0

V119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) V119895minus1

+ 119870(119909119896 119909119895) V119895]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(30)

where 119896 = 1 2 119899Rearranging the above system of equations we get

(1 minus

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)]) V1= 119877(119899)

1(119906)

(1 minus

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)]) V119896

=

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

119896

sum

119895=1

119877(119899)

119895(119906)

(31)

where 119896 = 1 2 119899Along with the inequality 120596

120593(ℎ) le 119871ℎ using conditions

(19) and (23) we get the following inequality for V119896from (31)

1003816100381610038161003816V1

1003816100381610038161003816le

1

1 minus 120572

119877 (ℎ)

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

]

]

(32)

where 119896 = 1 2 119899 119877(ℎ) = (11987212)1198712

ℎ2

[120593(119887) minus 120593(119886)]Let the term 120576

119896for 119896 = 1 2 119899 be determined by

120576119896=

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

119896 = 2 3 119899 (33)

and 1205761= 119877(ℎ)(1 minus 120572) as an initial condition

It is easily seen that |V119896| le 120576119896for 119896 = 1 2 119899This can be

verified by mathematical induction as follows for 119896 = 1 it is

Advances in Mathematical Physics 5

Table 1 The values of approximate solution analytical solutionand the error at the nodes

Thenodes119909119896

Real value at 119909119896

119906(119909119896)

Approx value at 119909119896

119906119896

The error at 119909119896

|119906(119909119896) minus 119906119896|

0 1 1 001 1 100271872 00027187202 1 100331826 00033182603 1 100376994 00037699404 1 100417187 00041718705 1 100455757 00045575706 1 100494416 00049441607 1 100534264 00053426408 1 100576140 00057614009 1 100620761 00062076110 1 100668805 00066880511 1 100720950 00072095012 1 100777907 00077790713 1 100840442 00084044214 1 100909399 00090939915 1 100985722 00098572216 1 101070480 00107048017 1 101164880 00116488018 1 101270300 00127030019 1 101388350 00138835020 1 101520860 001520860

trivial Let |V119895| le 120576119895for 119895 = 1 119896minus1Then using inequality

(32) we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

= 120576119896 (34)

Let us show that

120576119895=

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

119895 = 1 2 119899 (35)

are the solution of the system of (33) Taking (35) intoaccount we get

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

=

119877 (ℎ)

1 minus 120572

1 +

1198700119871ℎ

1 minus 120572

119896minus1

sum

119895=1

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

=

119877 (ℎ)

1 minus 120572

1 + [(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

minus 1] = 120576119896

119896 ge 2

(36)

Here we use the equality

(1 + 120574)119896minus1

minus 1 = 120574

119896minus1

sum

119895=1

(1 + 120574)119895minus1

119896 ge 2 (37)

where 120574 = 1198700119871ℎ(1 minus 120572) Consequently we get the following

estimate for the error V119896for all values 119896 = 1 119899

1003816100381610038161003816V119896

1003816100381610038161003816le

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

(38)

Using the fact that (1 + 119905)1119905 is increasing and approaches

the number 119890 as 119905 rarr 0+ we get the following chain ofinequalities

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

le (1 +

1198700119871ℎ

1 minus 120572

)

(119887minus119886)ℎ

= [(1 +

1198700119871

1 minus 120572

ℎ)

(1minus120572)1198700119871ℎ

]

1198700119871(119887minus119886)(1minus120572)

le 1198901198700119871(119887minus119886)(1minus120572)

(39)

for 119896 le 119899 = (119887 minus 119886)ℎ Hence the proof is obtained

Remark 10 The function

120593 (119909) =

119909 for 0 le 119909 le 1

2119909 minus 1 for 1 lt 119909 le 2

3119909 minus 3 for 2 le 119909 le 3

(40)

is a strictly increasing continuous function on [0 3] 1205931015840

(119909) notin

119862[0 3] But for all 119909 119910 isin [0 3] the following inequalityholds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 4

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (41)

Theorem 11 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and

120573 = 1198700[120593 (119887) minus 120593 (119886)] lt 1 (42)

Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816le

119872

12 (1 minus 120573)

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

119896 = 1 2 119899

(43)

holds in which 1198700= 119870(119909 119904)

119862

6 Advances in Mathematical Physics

Proof Let the error be denoted by V119896= 119906(119909

119896) minus 119906119896and set up

the system of equations

V1=

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)] V1+ 119877(119899)

1(119906)

V119896=

119896minus1

sum

119895=1

1

2

119870 (119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)] V119896

+

119896

sum

119895=1

119877(119899)

119895(119906)

(44)

for 119896 = 2 3 119899 From this system of equations we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

sdot

119896minus1

sum

119895=1

[120593 (119909119895+1

) minus 120593 (119909119895) + 120593 (119909

119895) minus 120593 (119909

119895minus1)]

+ [120593 (119909119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887)

minus 120593 (119886)] =

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119909119896) minus 120593 (119909

1)

+ 120593 (119909119896minus1

) minus 120593 (1199090) + 120593 (119909

119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)] le 1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119887)

minus 120593 (119886)] +

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

(45)

for 119896 = 2 3 119899 Using condition (42) we get inequality(43) Therefore Theorem 11 is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] A S Apartsyn Nonclassical Linear Volterra Equations of theFirst Kind VSP TB Utrecht The Netherlands 2003

[2] A L Bukhgeim Volterra Equations and Inverse Problems VSPUtrecht The Netherlands 1999

[3] J Banas and D OrsquoRegan ldquoVolterra-Stieltjes integral operatorsrdquoMathematical and ComputerModelling vol 41 no 2-3 pp 335ndash344 2005

[4] J Banas J R Rodriguez and K Sadarangani ldquoOn a class ofUrysohn-Stieltjes quadratic integral equations and their appli-cationsrdquo Journal of Computational and Applied Mathematicsvol 113 no 1-2 pp 35ndash50 2000

[5] L M Delves and J Walsh Numerical Solution of IntegralEquations Oxford University Press New York NY USA 1974

[6] M Federson and R Bianconi ldquoLinear Volterrra-Stieltjes inte-gral equations in the sense of the Kurzweil-Henstock integralrdquoArchivum Mathematicum vol 37 no 4 pp 307ndash328 2001

[7] M Federson R Bianconi and L Barbanti ldquoLinear Volterraintegral equationsrdquo Acta Mathematicae Applicatae Sinica vol18 no 4 pp 553ndash560 2002

[8] A D Polyanin and A V Manzhirov Handbook of IntegralEquations Chapman amp HallCRC Boca Raton Fla USA 2ndedition 2008

[9] K Maleknejad and N Aghazadeh ldquoNumerical solution ofVolterra integral equations of the second kind with convolutionkernel by using Taylor-series expansionmethodrdquoAppliedMath-ematics and Computation vol 161 no 3 pp 915ndash922 2005

[10] M A Darwish and J Henderson ldquoNondecreasing solutions ofa quadratic integral equation of Urysohn-Stieltjes typerdquo RockyMountain Journal of Mathematics vol 42 no 2 pp 545ndash5662012

[11] S A Isaacson and R M Kirby ldquoNumerical solution of linearVolterra integral equations of the second kind with sharpgradientsrdquo Journal of Computational and Applied Mathematicsvol 235 no 14 pp 4283ndash4301 2011

[12] T Diogo N J Ford P Lima and S Valtchev ldquoNumericalmethods for a Volterra integral equation with non-smoothsolutionsrdquo Journal of Computational and Applied Mathematicsvol 189 no 1-2 pp 412ndash423 2006

[13] S Zhang Y Lin and M Rao ldquoNumerical solutions for second-kind Volterra integral equations by Galerkin methodsrdquo Appli-cations of Mathematics vol 45 no 1 pp 19ndash39 2000

[14] A Asanov ldquoThe derivative of a function by means of anincreasing functionrdquo Manas Journal of Engineering no 1 pp18ndash64 2001 (Russian)

[15] A Asanov M H Chelik and A Chalish ldquoApproximating theStieltjes integral by using the generalized trapezoid rulerdquo LeMatematiche vol 66 no 2 pp 13ndash21 2011

[16] A Asanov ldquoVolterra-stielties integral equations of the secondkind and the first kindrdquoManas Journal of Engineering no 2 pp79ndash95 2002 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Advances in Mathematical Physics 3

In this paper we assume that (119870(119909 119909))1015840

120593(119909)isin 119862[119886 119887]

11987010158401015840

120593(119909)(119909 119904) 119870

10158401015840

120593(119904)(119909 119904) isin 119862(119866) and 119891

10158401015840

120593(119909)(119909) isin 119862[119886 119887] Then

using Theorems 4 and 5 (and Corollaries 6 and 7) we showthat the number 119872 defined as

119872 = sup(119909119904)isin119866

10038161003816100381610038161003816[119870 (119909 119904) 119906 (119904)]

10158401015840

120593(119904)

10038161003816100381610038161003816 (14)

can be determined in terms of quantities 119891(119909)119862

1198911015840

120593(119909)119862 11989110158401015840

120593(119909)119862 (119870(119909 119909))

1015840

120593(119909)119862 1198701015840

120593(119909)(119909 119904)

119862

1198701015840

120593(119904)(119909 119904)

119862 11987010158401015840

120593(119909)(119909 119904)

119862 and 119870

10158401015840

120593(119904)(119909 119904)

119862

Under these circumstances usingTheorem 2 the integral

int

119909

119886

119870 (119909 119904) 119906 (119904) 119889119904 (15)

can be evaluated numerically by employing the generalizedtrapezoid rule

3 Numerical Solution

In order to obtain the approximate solution of (1) we employthe generalized trapezoid rule given in [15] to the integral in(1) Let 119899 isin 119873

ℎ =

119887 minus 119886

119899

119909119896= 119886 + 119896ℎ

(16)

where 119896 = 0 1 119899 Let us substitute 119909 = 119909119896in the integral

equation (1) and examine the following system of equations

119906 (1199090) = 119891 (119909

0) 119909

0= 119886

119906 (119909119896) = int

119909119896

119886

119870(119909119896 119904) 119906 (119904) 119889119904 + 119891 (119909

119896)

119896 = 1 2 119899

(17)

To evaluate the integral term in (17) we employ the general-ized trapezoid rule given in [15] at the nodes 119909

0 1199091 119909

119896

So we get

int

119909119896

119886

119870(119909119896 119904) 119906 (119904) 119889119904

=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(18)

where10038161003816100381610038161003816119877(119899)

119895(119906)

10038161003816100381610038161003816le

119872

12

[120593 (119909119895) minus 120593 (119909

119895minus1)]

3

(19)

119872 = sup(119909119904)isin119866

10038161003816100381610038161003816[119870 (119909 119904) 119906 (119904)]

10158401015840

120593(119904)

10038161003816100381610038161003816

= sup(119909119904)isin119866

10038161003816100381610038161003816119870 (119909 119904) 119906

10158401015840

120593(119904)(119904) + 2119870

1015840

120593(119904)(119909 119904) 119906

1015840

120593(119904)(119904)

+ 11987010158401015840

120593(119904)(119909 119904) 119906 (119904)

10038161003816100381610038161003816

(20)

Substituting (18) in (17) we get

119906 (1199090) = 119891 (119909

0) 119909

0= 119886

119906 (119909119896)

=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906) + 119891 (119909

119896)

(21)

where 119896 = 1 2 119899Omitting the terms sum

119896

119895=1119877(119899)

119895(119906) appearing in each equa-

tion of system (21) and 119906119896asymp 119906(119909

119896) we obtain

1199060= 119891 (119909

0) 119909

0= 119886

119906119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) 119906 (119909119895minus1

) + 119870 (119909119896 119909119895) 119906 (119909

119895)]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] + 119891 (119909

119896)

(22)

where 119896 = 1 2 119899Let us assume that

120572 =

1

2

sup119909isin[119886119887]

|119870 (119909 119909)| 120596120593(ℎ) lt 1 (23)

where 120596120593(ℎ) denotes the modulus of continuity of the

function 120593 that is

120596120593(ℎ) = sup

|119909minus119910|leℎ

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816 (24)

Under condition (23) the systemof (22) has a unique solutionwhich is given by the formulas

1199060= 119891 (119886)

1199061= [1 minus

1

2

119870 (1199091 1199091) (120593 (119909

1) minus 120593 (119909

0))]

minus1

sdot [

1

2

119870 (1199091 1199090) (120593 (119909

1) minus 120593 (119909

0)) 1199060+ 119891 (119909

1)]

119906119896= [1 minus

1

2

119870 (119909119896 119909119896) (120593 (119909

119896) minus 120593 (119909

119896minus1))]

minus1

sdot[

[

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) (120593 (119909

119895+1) minus 120593 (119909

119895minus1)) 119906119895

+

1

2

119870 (119909119896 1199090) (120593 (119909

1) minus 120593 (119909

0)) 1199060+ 119891 (119909

119896)]

]

(25)

for 119896 = 2 3 119899We give a concrete example below

Example 8 Let us take the integral equation (1) for 119886 = 0 and119887 = 2 with

4 Advances in Mathematical Physics

120593 (119909) =

radic119909 for 0 le 119909 le 1

119909 for 1 lt 119909 le 2

119870 (119909 119904) = 1198861(119909) 1198871(119904) 119886

1(119909) =

2radic119909 + 1 for 0 le 119909 le 1

2119909 + 1 for 1 lt 119909 le 2

1198871(119904) =

119904 for 0 le 119904 le 1

1199042

for 1 lt 119904 le 2

(26)

and using

119891 (119909) =

1 minus

2

3

1199092

minus

1

3

119909radic119909 for 0 le 119909 le 1

1 minus

1199093

3

(2119909 + 1) for 1 lt 119909 le 2

(27)

It is easily seen that 119906(119909) equiv 1 119909 isin [0 2] is the unique solu-tion of the integral equation (1) and the conditions 11989110158401015840

120593(119909)(119909) isin

119862[0 2] (119870(119909 119909))1015840

120593(119909)isin 119862[0 2] 119870

10158401015840

120593(119909)(119909 119904) and 119870

10158401015840

120593(119904)(119909 119904) isin

119862(119866) hold where 119866 = (119909 119904) 0 le 119904 le 119909 le 2

Using the proposed method of this study we get thefollowing results Here 20 nodes are selected that is 119899 = 20In Table 1 we give the values of the approximate solutionobtained by the proposed method of this study and the errorin absolute values at the given nodes

4 Estimation of the Error

In this section we investigate the problem of convergence ofthe approximate solution 119906

119896to the solution of integral (1) at

the nodes as 119899 rarr infin

Theorem 9 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and for all 119909 119910 isin [119886 119887] the followinginequality holds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 119871

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (28)

where 119871 gt 0 and 119871 is independent of the variables 119909 and 119910Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816

le

1198721198712

[120593 (119887) minus 120593 (119886)]

12 (1 minus 120572)

exp

1198700119871 (119887 minus 119886)

1 minus 120572

ℎ2

119896 = 1 2 119899

(29)

holds in which 1198700= 119870(119909 119904)

119862 120572 = (12)119870(119909 119909)

119862119871ℎ lt 1

and the number 119872 is determined by (20)

Proof Let the error be denoted by V119896

= 119906(119909119896) minus 119906

119896for

119896 = 0 1 119899 Taking into account (21) and (22) we havethe following system of equations

V0= 0

V119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) V119895minus1

+ 119870(119909119896 119909119895) V119895]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(30)

where 119896 = 1 2 119899Rearranging the above system of equations we get

(1 minus

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)]) V1= 119877(119899)

1(119906)

(1 minus

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)]) V119896

=

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

119896

sum

119895=1

119877(119899)

119895(119906)

(31)

where 119896 = 1 2 119899Along with the inequality 120596

120593(ℎ) le 119871ℎ using conditions

(19) and (23) we get the following inequality for V119896from (31)

1003816100381610038161003816V1

1003816100381610038161003816le

1

1 minus 120572

119877 (ℎ)

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

]

]

(32)

where 119896 = 1 2 119899 119877(ℎ) = (11987212)1198712

ℎ2

[120593(119887) minus 120593(119886)]Let the term 120576

119896for 119896 = 1 2 119899 be determined by

120576119896=

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

119896 = 2 3 119899 (33)

and 1205761= 119877(ℎ)(1 minus 120572) as an initial condition

It is easily seen that |V119896| le 120576119896for 119896 = 1 2 119899This can be

verified by mathematical induction as follows for 119896 = 1 it is

Advances in Mathematical Physics 5

Table 1 The values of approximate solution analytical solutionand the error at the nodes

Thenodes119909119896

Real value at 119909119896

119906(119909119896)

Approx value at 119909119896

119906119896

The error at 119909119896

|119906(119909119896) minus 119906119896|

0 1 1 001 1 100271872 00027187202 1 100331826 00033182603 1 100376994 00037699404 1 100417187 00041718705 1 100455757 00045575706 1 100494416 00049441607 1 100534264 00053426408 1 100576140 00057614009 1 100620761 00062076110 1 100668805 00066880511 1 100720950 00072095012 1 100777907 00077790713 1 100840442 00084044214 1 100909399 00090939915 1 100985722 00098572216 1 101070480 00107048017 1 101164880 00116488018 1 101270300 00127030019 1 101388350 00138835020 1 101520860 001520860

trivial Let |V119895| le 120576119895for 119895 = 1 119896minus1Then using inequality

(32) we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

= 120576119896 (34)

Let us show that

120576119895=

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

119895 = 1 2 119899 (35)

are the solution of the system of (33) Taking (35) intoaccount we get

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

=

119877 (ℎ)

1 minus 120572

1 +

1198700119871ℎ

1 minus 120572

119896minus1

sum

119895=1

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

=

119877 (ℎ)

1 minus 120572

1 + [(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

minus 1] = 120576119896

119896 ge 2

(36)

Here we use the equality

(1 + 120574)119896minus1

minus 1 = 120574

119896minus1

sum

119895=1

(1 + 120574)119895minus1

119896 ge 2 (37)

where 120574 = 1198700119871ℎ(1 minus 120572) Consequently we get the following

estimate for the error V119896for all values 119896 = 1 119899

1003816100381610038161003816V119896

1003816100381610038161003816le

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

(38)

Using the fact that (1 + 119905)1119905 is increasing and approaches

the number 119890 as 119905 rarr 0+ we get the following chain ofinequalities

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

le (1 +

1198700119871ℎ

1 minus 120572

)

(119887minus119886)ℎ

= [(1 +

1198700119871

1 minus 120572

ℎ)

(1minus120572)1198700119871ℎ

]

1198700119871(119887minus119886)(1minus120572)

le 1198901198700119871(119887minus119886)(1minus120572)

(39)

for 119896 le 119899 = (119887 minus 119886)ℎ Hence the proof is obtained

Remark 10 The function

120593 (119909) =

119909 for 0 le 119909 le 1

2119909 minus 1 for 1 lt 119909 le 2

3119909 minus 3 for 2 le 119909 le 3

(40)

is a strictly increasing continuous function on [0 3] 1205931015840

(119909) notin

119862[0 3] But for all 119909 119910 isin [0 3] the following inequalityholds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 4

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (41)

Theorem 11 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and

120573 = 1198700[120593 (119887) minus 120593 (119886)] lt 1 (42)

Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816le

119872

12 (1 minus 120573)

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

119896 = 1 2 119899

(43)

holds in which 1198700= 119870(119909 119904)

119862

6 Advances in Mathematical Physics

Proof Let the error be denoted by V119896= 119906(119909

119896) minus 119906119896and set up

the system of equations

V1=

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)] V1+ 119877(119899)

1(119906)

V119896=

119896minus1

sum

119895=1

1

2

119870 (119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)] V119896

+

119896

sum

119895=1

119877(119899)

119895(119906)

(44)

for 119896 = 2 3 119899 From this system of equations we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

sdot

119896minus1

sum

119895=1

[120593 (119909119895+1

) minus 120593 (119909119895) + 120593 (119909

119895) minus 120593 (119909

119895minus1)]

+ [120593 (119909119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887)

minus 120593 (119886)] =

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119909119896) minus 120593 (119909

1)

+ 120593 (119909119896minus1

) minus 120593 (1199090) + 120593 (119909

119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)] le 1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119887)

minus 120593 (119886)] +

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

(45)

for 119896 = 2 3 119899 Using condition (42) we get inequality(43) Therefore Theorem 11 is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] A S Apartsyn Nonclassical Linear Volterra Equations of theFirst Kind VSP TB Utrecht The Netherlands 2003

[2] A L Bukhgeim Volterra Equations and Inverse Problems VSPUtrecht The Netherlands 1999

[3] J Banas and D OrsquoRegan ldquoVolterra-Stieltjes integral operatorsrdquoMathematical and ComputerModelling vol 41 no 2-3 pp 335ndash344 2005

[4] J Banas J R Rodriguez and K Sadarangani ldquoOn a class ofUrysohn-Stieltjes quadratic integral equations and their appli-cationsrdquo Journal of Computational and Applied Mathematicsvol 113 no 1-2 pp 35ndash50 2000

[5] L M Delves and J Walsh Numerical Solution of IntegralEquations Oxford University Press New York NY USA 1974

[6] M Federson and R Bianconi ldquoLinear Volterrra-Stieltjes inte-gral equations in the sense of the Kurzweil-Henstock integralrdquoArchivum Mathematicum vol 37 no 4 pp 307ndash328 2001

[7] M Federson R Bianconi and L Barbanti ldquoLinear Volterraintegral equationsrdquo Acta Mathematicae Applicatae Sinica vol18 no 4 pp 553ndash560 2002

[8] A D Polyanin and A V Manzhirov Handbook of IntegralEquations Chapman amp HallCRC Boca Raton Fla USA 2ndedition 2008

[9] K Maleknejad and N Aghazadeh ldquoNumerical solution ofVolterra integral equations of the second kind with convolutionkernel by using Taylor-series expansionmethodrdquoAppliedMath-ematics and Computation vol 161 no 3 pp 915ndash922 2005

[10] M A Darwish and J Henderson ldquoNondecreasing solutions ofa quadratic integral equation of Urysohn-Stieltjes typerdquo RockyMountain Journal of Mathematics vol 42 no 2 pp 545ndash5662012

[11] S A Isaacson and R M Kirby ldquoNumerical solution of linearVolterra integral equations of the second kind with sharpgradientsrdquo Journal of Computational and Applied Mathematicsvol 235 no 14 pp 4283ndash4301 2011

[12] T Diogo N J Ford P Lima and S Valtchev ldquoNumericalmethods for a Volterra integral equation with non-smoothsolutionsrdquo Journal of Computational and Applied Mathematicsvol 189 no 1-2 pp 412ndash423 2006

[13] S Zhang Y Lin and M Rao ldquoNumerical solutions for second-kind Volterra integral equations by Galerkin methodsrdquo Appli-cations of Mathematics vol 45 no 1 pp 19ndash39 2000

[14] A Asanov ldquoThe derivative of a function by means of anincreasing functionrdquo Manas Journal of Engineering no 1 pp18ndash64 2001 (Russian)

[15] A Asanov M H Chelik and A Chalish ldquoApproximating theStieltjes integral by using the generalized trapezoid rulerdquo LeMatematiche vol 66 no 2 pp 13ndash21 2011

[16] A Asanov ldquoVolterra-stielties integral equations of the secondkind and the first kindrdquoManas Journal of Engineering no 2 pp79ndash95 2002 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Advances in Mathematical Physics

120593 (119909) =

radic119909 for 0 le 119909 le 1

119909 for 1 lt 119909 le 2

119870 (119909 119904) = 1198861(119909) 1198871(119904) 119886

1(119909) =

2radic119909 + 1 for 0 le 119909 le 1

2119909 + 1 for 1 lt 119909 le 2

1198871(119904) =

119904 for 0 le 119904 le 1

1199042

for 1 lt 119904 le 2

(26)

and using

119891 (119909) =

1 minus

2

3

1199092

minus

1

3

119909radic119909 for 0 le 119909 le 1

1 minus

1199093

3

(2119909 + 1) for 1 lt 119909 le 2

(27)

It is easily seen that 119906(119909) equiv 1 119909 isin [0 2] is the unique solu-tion of the integral equation (1) and the conditions 11989110158401015840

120593(119909)(119909) isin

119862[0 2] (119870(119909 119909))1015840

120593(119909)isin 119862[0 2] 119870

10158401015840

120593(119909)(119909 119904) and 119870

10158401015840

120593(119904)(119909 119904) isin

119862(119866) hold where 119866 = (119909 119904) 0 le 119904 le 119909 le 2

Using the proposed method of this study we get thefollowing results Here 20 nodes are selected that is 119899 = 20In Table 1 we give the values of the approximate solutionobtained by the proposed method of this study and the errorin absolute values at the given nodes

4 Estimation of the Error

In this section we investigate the problem of convergence ofthe approximate solution 119906

119896to the solution of integral (1) at

the nodes as 119899 rarr infin

Theorem 9 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and for all 119909 119910 isin [119886 119887] the followinginequality holds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 119871

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (28)

where 119871 gt 0 and 119871 is independent of the variables 119909 and 119910Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816

le

1198721198712

[120593 (119887) minus 120593 (119886)]

12 (1 minus 120572)

exp

1198700119871 (119887 minus 119886)

1 minus 120572

ℎ2

119896 = 1 2 119899

(29)

holds in which 1198700= 119870(119909 119904)

119862 120572 = (12)119870(119909 119909)

119862119871ℎ lt 1

and the number 119872 is determined by (20)

Proof Let the error be denoted by V119896

= 119906(119909119896) minus 119906

119896for

119896 = 0 1 119899 Taking into account (21) and (22) we havethe following system of equations

V0= 0

V119896=

119896

sum

119895=1

1

2

[119870 (119909119896 119909119895minus1

) V119895minus1

+ 119870(119909119896 119909119895) V119895]

sdot [120593 (119909119895) minus 120593 (119909

119895minus1)] +

119896

sum

119895=1

119877(119899)

119895(119906)

(30)

where 119896 = 1 2 119899Rearranging the above system of equations we get

(1 minus

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)]) V1= 119877(119899)

1(119906)

(1 minus

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)]) V119896

=

1

2

119896minus1

sum

119895=1

119870(119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

119896

sum

119895=1

119877(119899)

119895(119906)

(31)

where 119896 = 1 2 119899Along with the inequality 120596

120593(ℎ) le 119871ℎ using conditions

(19) and (23) we get the following inequality for V119896from (31)

1003816100381610038161003816V1

1003816100381610038161003816le

1

1 minus 120572

119877 (ℎ)

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

]

]

(32)

where 119896 = 1 2 119899 119877(ℎ) = (11987212)1198712

ℎ2

[120593(119887) minus 120593(119886)]Let the term 120576

119896for 119896 = 1 2 119899 be determined by

120576119896=

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

119896 = 2 3 119899 (33)

and 1205761= 119877(ℎ)(1 minus 120572) as an initial condition

It is easily seen that |V119896| le 120576119896for 119896 = 1 2 119899This can be

verified by mathematical induction as follows for 119896 = 1 it is

Advances in Mathematical Physics 5

Table 1 The values of approximate solution analytical solutionand the error at the nodes

Thenodes119909119896

Real value at 119909119896

119906(119909119896)

Approx value at 119909119896

119906119896

The error at 119909119896

|119906(119909119896) minus 119906119896|

0 1 1 001 1 100271872 00027187202 1 100331826 00033182603 1 100376994 00037699404 1 100417187 00041718705 1 100455757 00045575706 1 100494416 00049441607 1 100534264 00053426408 1 100576140 00057614009 1 100620761 00062076110 1 100668805 00066880511 1 100720950 00072095012 1 100777907 00077790713 1 100840442 00084044214 1 100909399 00090939915 1 100985722 00098572216 1 101070480 00107048017 1 101164880 00116488018 1 101270300 00127030019 1 101388350 00138835020 1 101520860 001520860

trivial Let |V119895| le 120576119895for 119895 = 1 119896minus1Then using inequality

(32) we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

= 120576119896 (34)

Let us show that

120576119895=

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

119895 = 1 2 119899 (35)

are the solution of the system of (33) Taking (35) intoaccount we get

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

=

119877 (ℎ)

1 minus 120572

1 +

1198700119871ℎ

1 minus 120572

119896minus1

sum

119895=1

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

=

119877 (ℎ)

1 minus 120572

1 + [(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

minus 1] = 120576119896

119896 ge 2

(36)

Here we use the equality

(1 + 120574)119896minus1

minus 1 = 120574

119896minus1

sum

119895=1

(1 + 120574)119895minus1

119896 ge 2 (37)

where 120574 = 1198700119871ℎ(1 minus 120572) Consequently we get the following

estimate for the error V119896for all values 119896 = 1 119899

1003816100381610038161003816V119896

1003816100381610038161003816le

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

(38)

Using the fact that (1 + 119905)1119905 is increasing and approaches

the number 119890 as 119905 rarr 0+ we get the following chain ofinequalities

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

le (1 +

1198700119871ℎ

1 minus 120572

)

(119887minus119886)ℎ

= [(1 +

1198700119871

1 minus 120572

ℎ)

(1minus120572)1198700119871ℎ

]

1198700119871(119887minus119886)(1minus120572)

le 1198901198700119871(119887minus119886)(1minus120572)

(39)

for 119896 le 119899 = (119887 minus 119886)ℎ Hence the proof is obtained

Remark 10 The function

120593 (119909) =

119909 for 0 le 119909 le 1

2119909 minus 1 for 1 lt 119909 le 2

3119909 minus 3 for 2 le 119909 le 3

(40)

is a strictly increasing continuous function on [0 3] 1205931015840

(119909) notin

119862[0 3] But for all 119909 119910 isin [0 3] the following inequalityholds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 4

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (41)

Theorem 11 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and

120573 = 1198700[120593 (119887) minus 120593 (119886)] lt 1 (42)

Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816le

119872

12 (1 minus 120573)

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

119896 = 1 2 119899

(43)

holds in which 1198700= 119870(119909 119904)

119862

6 Advances in Mathematical Physics

Proof Let the error be denoted by V119896= 119906(119909

119896) minus 119906119896and set up

the system of equations

V1=

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)] V1+ 119877(119899)

1(119906)

V119896=

119896minus1

sum

119895=1

1

2

119870 (119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)] V119896

+

119896

sum

119895=1

119877(119899)

119895(119906)

(44)

for 119896 = 2 3 119899 From this system of equations we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

sdot

119896minus1

sum

119895=1

[120593 (119909119895+1

) minus 120593 (119909119895) + 120593 (119909

119895) minus 120593 (119909

119895minus1)]

+ [120593 (119909119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887)

minus 120593 (119886)] =

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119909119896) minus 120593 (119909

1)

+ 120593 (119909119896minus1

) minus 120593 (1199090) + 120593 (119909

119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)] le 1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119887)

minus 120593 (119886)] +

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

(45)

for 119896 = 2 3 119899 Using condition (42) we get inequality(43) Therefore Theorem 11 is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] A S Apartsyn Nonclassical Linear Volterra Equations of theFirst Kind VSP TB Utrecht The Netherlands 2003

[2] A L Bukhgeim Volterra Equations and Inverse Problems VSPUtrecht The Netherlands 1999

[3] J Banas and D OrsquoRegan ldquoVolterra-Stieltjes integral operatorsrdquoMathematical and ComputerModelling vol 41 no 2-3 pp 335ndash344 2005

[4] J Banas J R Rodriguez and K Sadarangani ldquoOn a class ofUrysohn-Stieltjes quadratic integral equations and their appli-cationsrdquo Journal of Computational and Applied Mathematicsvol 113 no 1-2 pp 35ndash50 2000

[5] L M Delves and J Walsh Numerical Solution of IntegralEquations Oxford University Press New York NY USA 1974

[6] M Federson and R Bianconi ldquoLinear Volterrra-Stieltjes inte-gral equations in the sense of the Kurzweil-Henstock integralrdquoArchivum Mathematicum vol 37 no 4 pp 307ndash328 2001

[7] M Federson R Bianconi and L Barbanti ldquoLinear Volterraintegral equationsrdquo Acta Mathematicae Applicatae Sinica vol18 no 4 pp 553ndash560 2002

[8] A D Polyanin and A V Manzhirov Handbook of IntegralEquations Chapman amp HallCRC Boca Raton Fla USA 2ndedition 2008

[9] K Maleknejad and N Aghazadeh ldquoNumerical solution ofVolterra integral equations of the second kind with convolutionkernel by using Taylor-series expansionmethodrdquoAppliedMath-ematics and Computation vol 161 no 3 pp 915ndash922 2005

[10] M A Darwish and J Henderson ldquoNondecreasing solutions ofa quadratic integral equation of Urysohn-Stieltjes typerdquo RockyMountain Journal of Mathematics vol 42 no 2 pp 545ndash5662012

[11] S A Isaacson and R M Kirby ldquoNumerical solution of linearVolterra integral equations of the second kind with sharpgradientsrdquo Journal of Computational and Applied Mathematicsvol 235 no 14 pp 4283ndash4301 2011

[12] T Diogo N J Ford P Lima and S Valtchev ldquoNumericalmethods for a Volterra integral equation with non-smoothsolutionsrdquo Journal of Computational and Applied Mathematicsvol 189 no 1-2 pp 412ndash423 2006

[13] S Zhang Y Lin and M Rao ldquoNumerical solutions for second-kind Volterra integral equations by Galerkin methodsrdquo Appli-cations of Mathematics vol 45 no 1 pp 19ndash39 2000

[14] A Asanov ldquoThe derivative of a function by means of anincreasing functionrdquo Manas Journal of Engineering no 1 pp18ndash64 2001 (Russian)

[15] A Asanov M H Chelik and A Chalish ldquoApproximating theStieltjes integral by using the generalized trapezoid rulerdquo LeMatematiche vol 66 no 2 pp 13ndash21 2011

[16] A Asanov ldquoVolterra-stielties integral equations of the secondkind and the first kindrdquoManas Journal of Engineering no 2 pp79ndash95 2002 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Advances in Mathematical Physics 5

Table 1 The values of approximate solution analytical solutionand the error at the nodes

Thenodes119909119896

Real value at 119909119896

119906(119909119896)

Approx value at 119909119896

119906119896

The error at 119909119896

|119906(119909119896) minus 119906119896|

0 1 1 001 1 100271872 00027187202 1 100331826 00033182603 1 100376994 00037699404 1 100417187 00041718705 1 100455757 00045575706 1 100494416 00049441607 1 100534264 00053426408 1 100576140 00057614009 1 100620761 00062076110 1 100668805 00066880511 1 100720950 00072095012 1 100777907 00077790713 1 100840442 00084044214 1 100909399 00090939915 1 100985722 00098572216 1 101070480 00107048017 1 101164880 00116488018 1 101270300 00127030019 1 101388350 00138835020 1 101520860 001520860

trivial Let |V119895| le 120576119895for 119895 = 1 119896minus1Then using inequality

(32) we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

= 120576119896 (34)

Let us show that

120576119895=

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

119895 = 1 2 119899 (35)

are the solution of the system of (33) Taking (35) intoaccount we get

1

1 minus 120572

[

[

119877 (ℎ) + 1198700119871ℎ

119896minus1

sum

119895=1

120576119895

]

]

=

119877 (ℎ)

1 minus 120572

1 +

1198700119871ℎ

1 minus 120572

119896minus1

sum

119895=1

(1 +

1198700119871ℎ

1 minus 120572

)

119895minus1

=

119877 (ℎ)

1 minus 120572

1 + [(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

minus 1] = 120576119896

119896 ge 2

(36)

Here we use the equality

(1 + 120574)119896minus1

minus 1 = 120574

119896minus1

sum

119895=1

(1 + 120574)119895minus1

119896 ge 2 (37)

where 120574 = 1198700119871ℎ(1 minus 120572) Consequently we get the following

estimate for the error V119896for all values 119896 = 1 119899

1003816100381610038161003816V119896

1003816100381610038161003816le

119877 (ℎ)

1 minus 120572

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

(38)

Using the fact that (1 + 119905)1119905 is increasing and approaches

the number 119890 as 119905 rarr 0+ we get the following chain ofinequalities

(1 +

1198700119871ℎ

1 minus 120572

)

119896minus1

le (1 +

1198700119871ℎ

1 minus 120572

)

(119887minus119886)ℎ

= [(1 +

1198700119871

1 minus 120572

ℎ)

(1minus120572)1198700119871ℎ

]

1198700119871(119887minus119886)(1minus120572)

le 1198901198700119871(119887minus119886)(1minus120572)

(39)

for 119896 le 119899 = (119887 minus 119886)ℎ Hence the proof is obtained

Remark 10 The function

120593 (119909) =

119909 for 0 le 119909 le 1

2119909 minus 1 for 1 lt 119909 le 2

3119909 minus 3 for 2 le 119909 le 3

(40)

is a strictly increasing continuous function on [0 3] 1205931015840

(119909) notin

119862[0 3] But for all 119909 119910 isin [0 3] the following inequalityholds

1003816100381610038161003816120593 (119909) minus 120593 (119910)

1003816100381610038161003816le 4

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (41)

Theorem 11 Let 120593(119909) be a strictly increasing continuousfunction on [119886 119887] and

120573 = 1198700[120593 (119887) minus 120593 (119886)] lt 1 (42)

Then the inequality

1003816100381610038161003816119906 (119909119896) minus 119906119896

1003816100381610038161003816le

119872

12 (1 minus 120573)

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

119896 = 1 2 119899

(43)

holds in which 1198700= 119870(119909 119904)

119862

6 Advances in Mathematical Physics

Proof Let the error be denoted by V119896= 119906(119909

119896) minus 119906119896and set up

the system of equations

V1=

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)] V1+ 119877(119899)

1(119906)

V119896=

119896minus1

sum

119895=1

1

2

119870 (119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)] V119896

+

119896

sum

119895=1

119877(119899)

119895(119906)

(44)

for 119896 = 2 3 119899 From this system of equations we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

sdot

119896minus1

sum

119895=1

[120593 (119909119895+1

) minus 120593 (119909119895) + 120593 (119909

119895) minus 120593 (119909

119895minus1)]

+ [120593 (119909119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887)

minus 120593 (119886)] =

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119909119896) minus 120593 (119909

1)

+ 120593 (119909119896minus1

) minus 120593 (1199090) + 120593 (119909

119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)] le 1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119887)

minus 120593 (119886)] +

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

(45)

for 119896 = 2 3 119899 Using condition (42) we get inequality(43) Therefore Theorem 11 is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] A S Apartsyn Nonclassical Linear Volterra Equations of theFirst Kind VSP TB Utrecht The Netherlands 2003

[2] A L Bukhgeim Volterra Equations and Inverse Problems VSPUtrecht The Netherlands 1999

[3] J Banas and D OrsquoRegan ldquoVolterra-Stieltjes integral operatorsrdquoMathematical and ComputerModelling vol 41 no 2-3 pp 335ndash344 2005

[4] J Banas J R Rodriguez and K Sadarangani ldquoOn a class ofUrysohn-Stieltjes quadratic integral equations and their appli-cationsrdquo Journal of Computational and Applied Mathematicsvol 113 no 1-2 pp 35ndash50 2000

[5] L M Delves and J Walsh Numerical Solution of IntegralEquations Oxford University Press New York NY USA 1974

[6] M Federson and R Bianconi ldquoLinear Volterrra-Stieltjes inte-gral equations in the sense of the Kurzweil-Henstock integralrdquoArchivum Mathematicum vol 37 no 4 pp 307ndash328 2001

[7] M Federson R Bianconi and L Barbanti ldquoLinear Volterraintegral equationsrdquo Acta Mathematicae Applicatae Sinica vol18 no 4 pp 553ndash560 2002

[8] A D Polyanin and A V Manzhirov Handbook of IntegralEquations Chapman amp HallCRC Boca Raton Fla USA 2ndedition 2008

[9] K Maleknejad and N Aghazadeh ldquoNumerical solution ofVolterra integral equations of the second kind with convolutionkernel by using Taylor-series expansionmethodrdquoAppliedMath-ematics and Computation vol 161 no 3 pp 915ndash922 2005

[10] M A Darwish and J Henderson ldquoNondecreasing solutions ofa quadratic integral equation of Urysohn-Stieltjes typerdquo RockyMountain Journal of Mathematics vol 42 no 2 pp 545ndash5662012

[11] S A Isaacson and R M Kirby ldquoNumerical solution of linearVolterra integral equations of the second kind with sharpgradientsrdquo Journal of Computational and Applied Mathematicsvol 235 no 14 pp 4283ndash4301 2011

[12] T Diogo N J Ford P Lima and S Valtchev ldquoNumericalmethods for a Volterra integral equation with non-smoothsolutionsrdquo Journal of Computational and Applied Mathematicsvol 189 no 1-2 pp 412ndash423 2006

[13] S Zhang Y Lin and M Rao ldquoNumerical solutions for second-kind Volterra integral equations by Galerkin methodsrdquo Appli-cations of Mathematics vol 45 no 1 pp 19ndash39 2000

[14] A Asanov ldquoThe derivative of a function by means of anincreasing functionrdquo Manas Journal of Engineering no 1 pp18ndash64 2001 (Russian)

[15] A Asanov M H Chelik and A Chalish ldquoApproximating theStieltjes integral by using the generalized trapezoid rulerdquo LeMatematiche vol 66 no 2 pp 13ndash21 2011

[16] A Asanov ldquoVolterra-stielties integral equations of the secondkind and the first kindrdquoManas Journal of Engineering no 2 pp79ndash95 2002 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Advances in Mathematical Physics

Proof Let the error be denoted by V119896= 119906(119909

119896) minus 119906119896and set up

the system of equations

V1=

1

2

119870 (1199091 1199091) [120593 (119909

1) minus 120593 (119909

0)] V1+ 119877(119899)

1(119906)

V119896=

119896minus1

sum

119895=1

1

2

119870 (119909119896 119909119895) [120593 (119909

119895+1) minus 120593 (119909

119895minus1)] V119895

+

1

2

119870 (119909119896 119909119896) [120593 (119909

119896) minus 120593 (119909

119896minus1)] V119896

+

119896

sum

119895=1

119877(119899)

119895(119906)

(44)

for 119896 = 2 3 119899 From this system of equations we get

1003816100381610038161003816V119896

1003816100381610038161003816le

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816

sdot

119896minus1

sum

119895=1

[120593 (119909119895+1

) minus 120593 (119909119895) + 120593 (119909

119895) minus 120593 (119909

119895minus1)]

+ [120593 (119909119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887)

minus 120593 (119886)] =

1

2

1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119909119896) minus 120593 (119909

1)

+ 120593 (119909119896minus1

) minus 120593 (1199090) + 120593 (119909

119896) minus 120593 (119909

119896minus1)]

+

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)] le 1198700sup119895

10038161003816100381610038161003816V119895

10038161003816100381610038161003816[120593 (119887)

minus 120593 (119886)] +

119872

12

(120596120593(ℎ))

2

[120593 (119887) minus 120593 (119886)]

(45)

for 119896 = 2 3 119899 Using condition (42) we get inequality(43) Therefore Theorem 11 is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] A S Apartsyn Nonclassical Linear Volterra Equations of theFirst Kind VSP TB Utrecht The Netherlands 2003

[2] A L Bukhgeim Volterra Equations and Inverse Problems VSPUtrecht The Netherlands 1999

[3] J Banas and D OrsquoRegan ldquoVolterra-Stieltjes integral operatorsrdquoMathematical and ComputerModelling vol 41 no 2-3 pp 335ndash344 2005

[4] J Banas J R Rodriguez and K Sadarangani ldquoOn a class ofUrysohn-Stieltjes quadratic integral equations and their appli-cationsrdquo Journal of Computational and Applied Mathematicsvol 113 no 1-2 pp 35ndash50 2000

[5] L M Delves and J Walsh Numerical Solution of IntegralEquations Oxford University Press New York NY USA 1974

[6] M Federson and R Bianconi ldquoLinear Volterrra-Stieltjes inte-gral equations in the sense of the Kurzweil-Henstock integralrdquoArchivum Mathematicum vol 37 no 4 pp 307ndash328 2001

[7] M Federson R Bianconi and L Barbanti ldquoLinear Volterraintegral equationsrdquo Acta Mathematicae Applicatae Sinica vol18 no 4 pp 553ndash560 2002

[8] A D Polyanin and A V Manzhirov Handbook of IntegralEquations Chapman amp HallCRC Boca Raton Fla USA 2ndedition 2008

[9] K Maleknejad and N Aghazadeh ldquoNumerical solution ofVolterra integral equations of the second kind with convolutionkernel by using Taylor-series expansionmethodrdquoAppliedMath-ematics and Computation vol 161 no 3 pp 915ndash922 2005

[10] M A Darwish and J Henderson ldquoNondecreasing solutions ofa quadratic integral equation of Urysohn-Stieltjes typerdquo RockyMountain Journal of Mathematics vol 42 no 2 pp 545ndash5662012

[11] S A Isaacson and R M Kirby ldquoNumerical solution of linearVolterra integral equations of the second kind with sharpgradientsrdquo Journal of Computational and Applied Mathematicsvol 235 no 14 pp 4283ndash4301 2011

[12] T Diogo N J Ford P Lima and S Valtchev ldquoNumericalmethods for a Volterra integral equation with non-smoothsolutionsrdquo Journal of Computational and Applied Mathematicsvol 189 no 1-2 pp 412ndash423 2006

[13] S Zhang Y Lin and M Rao ldquoNumerical solutions for second-kind Volterra integral equations by Galerkin methodsrdquo Appli-cations of Mathematics vol 45 no 1 pp 19ndash39 2000

[14] A Asanov ldquoThe derivative of a function by means of anincreasing functionrdquo Manas Journal of Engineering no 1 pp18ndash64 2001 (Russian)

[15] A Asanov M H Chelik and A Chalish ldquoApproximating theStieltjes integral by using the generalized trapezoid rulerdquo LeMatematiche vol 66 no 2 pp 13ndash21 2011

[16] A Asanov ldquoVolterra-stielties integral equations of the secondkind and the first kindrdquoManas Journal of Engineering no 2 pp79ndash95 2002 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of